Senior Secondary School Mathematics for Class 12 R S Aggarwal not Agarwal Bharati Bhawan

Table of contents :
Cover......Page 1
Title......Page 2
Preface......Page 4
Contents......Page 10
Relations and Functions......Page 12
Algebra......Page 163
Differential Calculus......Page 334
Integral Calculus......Page 599
Differential Equations......Page 904
Vector Algebra......Page 1005
3D Geometry......Page 1099
Probability......Page 1273
Linear Programming......Page 1365

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SSS Mathematics for Class 12

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12

[In accordance with the latest CBSE syllabus]

R S Aggarwal, MSc, PhD

(i)

SSS Mathematics for Class 12 (iii)

Preface It gives me great pleasure in presenting the new edition of this book. In this edition, the modifications have been dictated by the changes in the CBSE syllabus. The structure and the methods used in the previous editions, which have been appreciated by teachers using the book in classroom conditions, remain unchanged. The main consideration in writing the book was to present the considerable requirements of the syllabus in as simple a manner as possible. Special attention has been paid to the gradation of problems. This will help students gain confidence in problem-solving. One problem faced by students is the lack of a comprehensive and carefully selected set of solved problems in textbooks of this kind. I have given due weightage to this aspect. Each set of solved examples is followed by a comprehensive exercise section in which students will get enough questions for practice. Hints have been given to the more difficult questions. Students should take their help as a last resort. I have received many suggestions and letters of appreciation from teachers all over the country. I thank them all for contributing in the improvement of the book and for their encouragement. I hope they will like this edition as well. And as always, I would like to hear their views on the book.

R S Aggarwal

( iii )

SSS Mathematics for Class 12 (v)

Mathematics Syllabus For Class 12

UNIT I. Relations and Functions 1. Relations and Functions 15 Periods Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions, composite functions, inverse of a function. Binary operations. 2. Inverse Trigonometric Functions 15 Periods Definition, range, domain, principal value branches. Graphs of inverse trigonometric functions. Elementary properties of inverse trigonometric functions.

UNIT II. Algebra 1. Matrices 25 Periods Concept, notation, order, equality, types of matrices, zero matrix, transpose of a matrix, symmetric and skew-symmetric matrices. Addition, multiplication and scalar multiplication of matrices, simple properties of addition, multiplication and scalar multiplication. Non-commutativity of multiplication of matrices and existence of nonzero matrices whose product is the zero matrix (restrict to square matrices of order 2). Concept of elementary row and column operations. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). 2. Determinants 25 Periods Determinant of a square matrix (up to 3 ´ 3 matrices), properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

UNIT III. Calculus 1. Continuity and Differentiability 20 Periods Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inverse trigonometric functions, derivative of implicit functions. Concept of exponential and logarithmic functions. Derivatives of logarithmic and exponential functions. Logarithmic (v)

SSS Mathematics for Class 12 (vi)

differentiation, derivative of functions expressed in parametric forms. Second-order derivatives. Rolle’s and Lagrange’s Mean Value Theorems (without proof) and their geometric interpretation. 2. Applications of Derivatives 10 Periods Applications of derivatives: rate of change of bodies, increasing/ decreasing functions, tangents and normals, use of derivatives in approximation, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations). 3. Integrals 20 Periods Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, simple integrals of the following type to be evaluated: dx dx dx dx dx ò x 2 ± a 2 , ò x 2 ± a 2 , ò a 2 - x 2 , ò ax 2 + bx + c , ò ax 2 + bx + c , ( px + q) dx , 2 + bx + c

ò ax ò

ò

( px + q) ax 2 + bx + c

ax 2 + bx + c dx , ò ( px + q)

dx ,

ò

a 2 ± x 2 dx,

ò

x 2 - a 2 dx,

ax 2 + bx + c dx.

Definite integrals as a limit of a sum, Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals. 4. Applications of the Integrals 15 Periods Applications in finding the area under simple curves, especially lines, areas of circles/parabolas/ellipses (in standard form only). Area between the two above-said curves (the region should be clearly identifiable). 5. Differential Equations 15 Periods Definition, order and degree, general and particular solutions of a differential equation. Formation of a differential equation whose general solution is given. Solution of differential equations by method of separation of variables, homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type: dy + py = q , where p and q are functions of x or constants. dx dx + px = q , where p and q are functions of y or constants. dy

(vi)

SSS Mathematics for Class 12 (vii)

UNIT IV. Vectors and Three-dimensional Geometry 1. Vectors 15 Periods Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) product of vectors, projection of a vector on a line. Vector (cross) product of vectors. Scalar triple product of vectors. 2. Three-dimensional Geometry 15 Periods Direction cosines and direction ratios of a line joining two points. Cartesian and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane.

UNIT V. Linear Programming 1. Linear Programming 20 Periods Introduction, related terminology such as constraints, objective function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions, feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

UNIT VI. Probability 1. Probability 30 Periods Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’s theorem, Random variable and its probability distribution, mean and variance of random variable. Repeated independent (Bernoulli) trials and Binomial distribution.

(vii)

SSS Mathematics for Class 12 (viii)

Weightage Topic

1. 2. 3. 4. 5. 6.

Marks

Relations and Functions Algebra Calculus Vectors and 3-Dimensional Geometry Linear Programming Probability

10 13 44 17 06 10 100

Total

Type

No. of Questions

Marks of Each

Total Marks

Very-Short-Answer

4

1

4

Short-Answer

8

2

16

Long-Answer I

11

4

44

Long-Answer II

6

6

36

29

Total

Sl. Division of Types No. of Questions

1. 2. 3. 4. 5.

Remembering Understanding Application HOTS

Evaluation Total

100

Marks of Each VSA

SA

LA–I

LA–II

(1 Mark) 2 1 1 – – 4

(2 Marks) 2 3 – 3 – 8

(4 Marks) 2 4 3 1 1* 11

(6 Marks) 1 2 2 – 1 6

* VBQ

(viii)

SSS Mathematics for Class 12 (ix)

Contents Relations and Functions 1. Relations

1

2. Functions

25

3. Binary Operations

67

4. Inverse Trigonometric Functions

82

Algebra 5. Matrices

152

6. Determinants

215

7. Adjoint and Inverse of a Matrix

276

8. System of Linear Equations

298

Differential Calculus 9. Continuity and Differentiability

323

10. Differentiation

365

11. Applications of Derivatives

452

Integral Calculus 12. Indefinite Integral

588

13. Methods of Integration

609

14. Some Special Integrals

698

15. Integration Using Partial Fractions

747

16. Definite Integrals

793

17. Area of Bounded Regions

860

Differential Equations 18. Differential Equations and Their Formation

893

19. Differential Equations with Variable Separable

914

20. Homogeneous Differential Equations

941

21. Linear Differential Equations

966 (ix)

SSS Mathematics for Class 12 (x)

Vector Algebra 22. Vectors and Their Properties

994

23. Scalar, or Dot, Product of Vectors

1011

24. Cross, or Vector, Product of Vectors

1038

25. Product of Three Vectors

1062

3D Geometry 26. Fundamental Concepts of 3-Dimensional Geometry

1088

27. Straight Line in Space

1107

28. The Plane

1161

Probability 29. Probability

1262

30. Bayes’s Theorem and Its Applications

1290

31. Probability Distribution

1307

32. Binomial Distribution

1325

Linear Programming 33. Linear Programming

1354 1390

Sample Question Paper

(i) to (iv)

Log Table

(x)

SSS Mathematics for Class 12 1

1. RELATIONS In class 11 we discussed about the Cartesian product of sets. Now, we extend our ideas to relation in a set and then in next chapter we shall be taking up functions. RELATION IN A SET

A relation R in a set A is a subset of A ´ A. Thus, R is a relation in a set A Û R Í A ´ A. If ( a , b) Î R then we say that a is related to b and write, a R b. If ( a , b) Ï R then we say that a is not related to b and write, a R b. Example

Let A = {1, 2, 3, 4, 5, 6} and let R be a relation in A, given by R = {( a , b) : a - b = 2}. Then, R = {( 3 , 1), ( 4, 2), (5 , 3), ( 6, 4)}. Clearly, 3 R 1, 4 R 2, 5 R 3 and 6 R 4. But, 1 R 3 , 2 R 4, 5 R 6, etc.

DOMAIN AND RANGE OF A RELATION Let R be a relation in a set A. Then, the set of all first coordinates of elements of R is called the domain of R, written as dom (R) and the set of all second coordinates of R is called the range of R, written as range (R). \ dom ( R) = {a : ( a , b) Î R} and range ( R) = {b : ( a , b) Î R}. Example

Let A = {1, 2, 3, 4, …, 15, 16} and let R be a relation in A, given by R = {( a , b) : b = a 2}. Then, R = {(1, 1), (2, 4), (3, 9), (4, 16)}. \ dom ( R) = {1, 2, 3, 4} and range ( R) = {1, 4, 9, 16}.

Some Particular Types of Relations EMPTY RELATION (Or VOID RELATION) A relation R in a set A is called an empty relation, if no element of A is related to any element of A and we denote such a relation by f.

Thus, R = f Í A ´ A. Example

Let A = {1, 2, 3, 4, 5} and let R be a relation in A, given by R = {( a , b) : a - b = 6}. 1

SSS Mathematics for Class 12 2

2

Senior Secondary School Mathematics for Class 12

Clearly, no element ( a , b) Î A ´ A satisfies the property a - b = 6. \ R is an empty relation in A. UNIVERSAL RELATION A relation R in a set A is called a universal relation, if each element of A is related to every element of A.

Thus, R = ( A ´ A) Í ( A ´ A) is the universal relation on A. Example

Let A = {1, 2, 3}. Then, R = ( A ´ A) = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} is the universal relation in A.

IDENTITY RELATION

The relation I A = {( a , a) : a Î A} is called the identity relation

on A. Example

Let A = {1, 2, 3}. Then, I A = {(1, 1), (2, 2), (3, 3)} is the identity relation on A.

VARIOUS TYPES OF RELATIONS

Let A be a nonempty set. Then, a relation R on A is said to be (i) reflexive

if ( a , a) Î R for each a Î A, i.e., if a R a for each a Î A.

(ii) symmetric if ( a , b) Î R Þ ( b , a) Î R for all a , b Î A, i.e., if a R b Þ b R a for all a , b Î A. (iii) transitive if ( a , b) Î R , ( b , c) Î R Þ ( a , c) Î R for all a , b , c Î A, i.e., if a R b and b R c Þ a R c. EQUIVALENCE RELATION A relation R in a set A is said to be an equivalence relation if it is reflexive, symmetric and transitive.

SOLVED EXAMPLES EXAMPLE 1

Let A be the set of all triangles in a plane and let R be a relation in A, defined by R = {(C 1 , C 2) : C 1 @ C 2 }. Show that R is an equivalence relation in A.

SOLUTION

The given relation satisfies the following properties: (i) Reflexivity Let a be an arbitrary triangle in A. Then, C @ C Þ (C , C) Î R for all values of C in A. \ R is reflexive. (ii) Symmetry Let C 1 , C 2 Î A such that (C 1 , C 2) Î R. Then,

SSS Mathematics for Class 12 3

Relations

3

(C 1 , C 2) Î R Þ C 1 @ C 2 Þ C 2 @ C1 Þ (C 2 , C 1) Î R. \ R is symmetric. (iii) Transitivity Let C1 , C 2 , C 3 Î A such that (C 1 , C 2) Î R and (C2 , C 3) Î R. Then, (C1 , C2) Î R and (C2 , C 3) Î R Þ C1 @ C2 and C2 @ C 3 Þ C1 @ C2 Þ (C1 , C 3) Î R. \ R is transitive. Thus, R is reflexive, symmetric and transitive. Hence, R is an equivalence relation. EXAMPLE 2

Let A be the set of all lines in xy-plane and let R be a relation in A, defined by R = {( L1 , L2) : L1 | | L2}. Show that R is an equivalence relation in A. Find the set of all lines related to the line y = 3 x + 5.

SOLUTION

The given relation satisfies the following properties: (i) Reflexivity Let L be an arbitrary line in A. Then, L | | L Þ ( L, L) Î R " L Î A. Thus, R is reflexive. (ii) Symmetry Let L1 , L2 Î A such that ( L1 , L2) Î R. Then, ( L1 , L2) Î R Þ L1 | | L2 Þ L2 | | L1 Þ ( L2 , L1) Î R. \ R is symmetric. (iii) Transitivity Let L1 , L2 , L 3 Î A such that ( L1 , L2) Î R and ( L2 , L 3) Î R. Then, ( L1 , L2) Î R and ( L2 , L 3) Î R Þ L1 | | L2 and L2 | | L 3

SSS Mathematics for Class 12 4

4

Senior Secondary School Mathematics for Class 12

Þ L1 | | L 3 Þ ( L1 , L 3) Î R. \ R is transitive. Thus R is reflexive, symmetric and transitive. Hence, R is an equivalence relation. The family of lines parallel to the line y = 3 x + 5 is given by y = 3 x + k, where k is real. EXAMPLE 3

Let Z be the set of all integers and let R be a relation in Z, defined by R = {( a , b) : ( a - b) is even}. Show that R is an equivalence relation in Z.

SOLUTION

Here, R satisfies the following properties: (i) Reflexivity Let a be an arbitrary element of Z. Then, ( a - a) = 0, which is even. \ ( a , a) Î R " a Î Z. So, R is reflexive. (ii) Symmetry Let a , b Î Z such that ( a , b) Î R. Then, ( a , b) Î R Þ ( a - b) is even Þ - ( a - b) is even Þ ( b - a) is even Þ ( b , a) Î R. \ R is symmetric. (iii) Transitivity Let a , b , c Î Z such that ( a , b) Î R and ( b , c) Î R. Then, ( a , b) Î R and ( b , c) Î R Þ ( a - b) is even and ( b - c) is even Þ {( a - b) + ( b - c)} is even Þ ( a - c) is even Þ ( a , c) Î R. \ R is transitive. Thus, R is reflexive, symmetric and transitive. Hence, R is an equivalence relation in Z.

SSS Mathematics for Class 12 5

Relations

5

EXAMPLE 4

Let A be the set of all lines in a plane and let R be a relation in A defined by R = {( L1 , L2) : L1 ^ L2}. Show that R is symmetric but neither reflexive nor transitive.

SOLUTION

Clearly, any line L cannot be perpendicular to itself. \ ( L, L) Ï R for any L Î A. So, R is not reflexive. Again, let ( L1 , L2) Î R. Then, ( L1 , L2) Î R Þ L1 ^ L2 Þ L2 ^ L1 Þ ( L2 , L1) Î R. \ R is symmetric. Now, let L1 , L2 , L 3 Î A such that L1 ^ L2 and L2 ^ L 3 . Then, clearly L1 is not perpendicular to L 3 . Thus, ( L1 , L2) Î R and ( L2 , L 3) Î R, but ( L1 , L 3) Ï R. \ R is not transitive. Hence, R is symmetric but neither reflexive nor transitive.

EXAMPLE 5

Let S be the set of all real numbers and let R be a relation in S defined by R = {( a , b) : (1 + ab) > 0}. Show that R is reflexive and symmetric but not transitive.

SOLUTION

Let a be any real number. Then, (1 + aa) = (1 + a 2) > 0 shows that ( a , a) Î R " a Î S. \ R is reflexive. Also, ( a , b) Î R Þ (1 + ab) > 0 Þ (1 + ba) > 0 [Q ab = ba] Þ ( b , a) Î R. \ R is symmetric. In order to show that R is not transitive, consider ( -1, 0) and ( 0, 2). Clearly, ( -1, 0) Î R, since [1 + ( -1) ´ 0] > 0. And, ( 0, 2) Î R , since [1 + 0 ´ 2] > 0. But, ( -1, 2) Ï R , since [1 + ( -1) ´ 2] is not greater than 0. Hence, R is reflexive and symmetric but not transitive.

EXAMPLE 6

Let S be the set of all real numbers and let R be a relation in S, defined by R = {( a , b) : a £ b}. Show that R is reflexive and transitive but not symmetric.

SOLUTION

Here, R satisfies the following properties: (i) Reflexivity Let a be an arbitrary real number.

SSS Mathematics for Class 12 6

6

Senior Secondary School Mathematics for Class 12

Then, a £ a Þ ( a , a) Î R. Thus, ( a , a) Î R " a Î S. \ R is reflexive. (ii) Transitivity Let a , b , c be real numbers such that ( a , b) Î R and ( b , c) Î R. Then, ( a , b) Î R and ( b , c) Î R Þ a £ b and b £ c Þ a£c Þ ( a , c) Î R. \ R is transitive. (iii) Nonsymmetry Clearly, ( 4, 5) Î R since 4 £ 5. But, (5 , 4) Ï R since 5 £ 4 is not true. \ R is not symmetric. EXAMPLE 7

SOLUTION

Let S be the set of all real numbers and let R be a relation in S, defined by R = {( a , b) : a £ b 2}. Show that R satisfies none of reflexivity, symmetry and transitivity. (i) Nonreflexivity 2

Clearly,

1 1 æ1ö is a real number and £ ç ÷ is not true. 2 è 2ø 2

æ1 1ö \ ç , ÷ Ï R. è 2 2ø Hence, R is not reflexive. (ii) Nonsymmetry Consider the real numbers Clearly,

1 and 1. 2

1 æ1 ö £ 12 Þ ç , 1÷ Î R. 2 è2 ø 2

æ1ö æ 1ö But, 1 £ ç ÷ is not true and so ç1, ÷ Ï R. 2ø è 2ø è æ æ1 ö Thus, ç , 1÷ Î R but ç1, 2 è è ø

1ö ÷ Î R. 2ø

Hence, R is not symmetric.

SSS Mathematics for Class 12 7

Relations

7

(iii) Nontransitivity Consider the real numbers 2, –2 and 1. Clearly, 2 £ ( -2) 2 and -2 £ (1) 2 but 2 £ 12 is not true. Thus, ( 2, - 2) Î R and ( -2, 1) Î R, but ( 2, 1) Ï R. Hence, R is not transitive. EXAMPLE 8

SOLUTION

Let S be the set of all real numbers and let R be a relation in S, defined by R = {( a , b) : a £ b 3 }. Show that R satisfies none of reflexivity, symmetry and transitivity. (i) Nonreflexivity 3

Clearly, \

1 1 æ1ö is a real number and £ ç ÷ is not true. 2 è 2ø 2

æ1 1ö ç , ÷ Ï R. è 2 2ø

Hence, R is not reflexive. (ii) Nonsymmetry Take the real numbers Clearly,

1 and 1. 2

1 æ1 ö £ 1 3 is true and therefore, ç , 1÷ Î R. 2 è2 ø 3

æ1ö æ 1ö But, 1 £ ç ÷ is not true and so ç1, ÷ Ï R. 2 è ø è 2ø Hence, R is not symmetric. (iii) Nontransitivity Consider the real numbers 3 , 3

3 4 and × 2 3 3

3

3 æ 4ö æ 3ö æ 4ö Clearly, 3 £ ç ÷ and £ ç ÷ but 3 £ ç ÷ is not true. 2 è 3ø è 2ø è 3ø 3ö 4ö æ æ æ 3 4ö Thus, ç 3 , ÷ Î R and ç , ÷ Î R, but ç 3 , ÷ Ï R. 2ø 3ø è è è 2 3ø Hence, R is not transitive. Thus, R satisfies none of reflexivity, symmetry and transitivity. EXAMPLE 9

Let N be the set of all natural numbers and let R be a relation in N, defined by R = {(a, b)} : a is a factor of b }. Then, show that R is reflexive and transitive but not symmetric.

SSS Mathematics for Class 12 8

8 SOLUTION

Senior Secondary School Mathematics for Class 12

Here, R satisfies the following properties: (i) Reflexivity Let a be an arbitrary element of N. Then, clearly, a is a factor of a. \ ( a , a) Î R " a Î N . So, R is reflexive. (ii) Transitivity Let a , b , c Î N such that ( a , b) Î R and ( b , c) Î R. Now, ( a , b) Î R and ( b , c) Î R Þ

(a is a factor of b) and (b is a factor of c)

Þ

b = ad and c = be for some d , e Î N

Þ

c = ( ad) e = a( de)

Þ

a is a factor of c

[by associative law]

Þ ( a , c) Î R. \ ( a , b) Î R and ( b , c) Î R Þ ( a , c) Î R. Hence, R is transitive. (iii) Nonsymmetry Clearly, 2 and 6 are natural numbers and 2 is a factor of 6. \ ( 2, 6) Î R. But, 6 is not a factor of 2. \ ( 6, 2) Ï R. Thus, ( 2, 6) Î R and ( 6, 2) Ï R. Hence, R is not symmetric. EXAMPLE 10

Let N be the set of all natural numbers and let R be a relation in N, defined by R = {( a , b) : a is a multiple of b }. Show that R is reflexive and transitive but not symmetric.

SOLUTION

Here R satisfies the following properties: (i) Reflexivity Let a be an arbitrary element of N. Then, a = ( a ´ 1) shows that a is a multiple of a. \ ( a , a) Î R " a Î N . So, R is reflexive.

SSS Mathematics for Class 12 9

Relations

9

(ii) Transitivity Let a , b , c Î N such that ( a , b) Î R and ( b , c) Î R. Now, ( a , b) Î R and ( b , c) Î R Þ

(a is a multiple of b) and (b is a multiple of c)

Þ

a = bd and b = ce for some d Î N and e Î N

Þ

a = ( ce) d

Þ

a = c( ed)

Þ

a is a multiple of c

Þ ( a , c) Î R. \ ( a , b) Î R and ( b , c) Î R Þ ( a , c) Î R. Hence, R is transitive. (iii) Nonsymmetry Clearly, 6 and 2 are natural numbers and 6 is a multiple of 2. \ ( 6, 2) Î R. But, 2 is not a multiple of 6. \ ( 2, 6) Ï R. Thus, ( 6, 2) Î R and ( 2, 6) Ï R. Hence, R is not symmetric. EXAMPLE 11

Let X be a nonempty set and let S be the collection of all subsets of X. Let R be a relation in S, defined by R = {( A , B) : A Ì B}. Show that R is transitive but neither reflexive nor symmetric.

SOLUTION

Clearly, R satisfies the following properties: (i) Transitivity Let A , B , C Î S such that ( A , B) Î R and ( B , C) Î R. Now, ( A , B) Î R and ( B , C) Î R Þ A Ì B and B Ì C Þ AÌC Þ ( A , C) Î R. \ R is transitive. (ii) Nonreflexivity Let A be any set in S.

SSS Mathematics for Class 12 10

10

Senior Secondary School Mathematics for Class 12

Then, A Ë A shows that ( A , A) Ï R. \ R is not reflexive. (iii) Nonsymmetry Now ( A , B) Î R Þ A Ì B Þ BË A Þ ( B , A) Ï R. \ R is not symmetric. Hence, R is transitive but neither reflexive nor symmetric. EXAMPLE 12

SOLUTION

Give an example of a relation which is (i) reflexive and transitive but not symmetric; (ii) symmetric and transitive but not reflexive; (iii) reflexive and symmetric but not transitive; (iv) symmetric but neither reflexive nor transitive; (v) transitive but neither reflexive nor symmetric. Let A = {1, 2, 3}. Then, it is easy to verify that the relation (i) R1 = {(1, 1), (2, 2), (3, 3), (1, 2)} is reflexive and transitive. R1 is not symmetric, since (1, 2) Î R and ( 2, 1) Ï R. (ii) R 2 = {(1, 1), (2, 2), (1, 2), (2, 1)} is symmetric and transitive. But, R 2 is not reflexive, since ( 3 , 3) Ï R 2. (iii) R 3 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} is reflexive and symmetric. But, R 3 is not transitive, since (1, 2) Î R 3 , ( 2, 3) Î R 3 but (1, 3) Ï R 3. (iv) R 4 = {(2, 2), (3, 3), (1, 2), (2, 1)} is symmetric. But, R 4 is not reflexive since (1, 1) Ï R 4. Also, R 4 is not transitive, as (1, 2) Î R 4 and ( 2, 1) Î R 4 but (1, 1) Î R 4.

SSS Mathematics for Class 12 11

Relations

11

(v) R5 = {(2, 2), (3, 3), (1, 2)} is transitive. But, R5 is not reflexive, since (1, 1) Ï R. And, R5 is not symmetric as (1, 2) Î R5 but ( 2, 1) Ï R5. EXAMPLE 13

Let N be the set of all natural numbers and let R be a relation on N ´ N , defined by ( a , b) R ( c, d) Û ad = bc. Show that R is an equivalence relation.

SOLUTION

Here R satisfies the following properties: (i) Reflexivity Let ( a , b) Þ R. Then, ( a , b) R ( a , b), since ab = ba [by commutative law of multiplication on N]. Thus, ( a , b) R ( a , b) " ( a , b) Î R. \ R is reflexive. (ii) Symmetry Let ( a , b) R ( c, d). Then, ( a , b) R ( c, d) Þ ad = bc Þ bc = ad Þ cb = da [by commutativity of multiplication on N] Þ ( c, d) R ( a , b). \ R is symmetric. (iii) Transitivity Let ( a , b) R ( c, d) and ( c, d) R ( e, f ). Then, ad = bc and cf = de Þ adcf = bcde Þ ( af )( cd) = ( be)( cd) Þ af = be

[by cancellation law]

Þ ( a , b) R ( e, f ). \ ( a , b) R ( c, d) and ( c, d) R ( e, f ) Þ ( a , b) R ( e, f ). \ R is transitive. Thus, R is reflexive, symmetric and transitive. Hence, R is an equivalence relation.

SSS Mathematics for Class 12 12

12 EXAMPLE 14

SOLUTION

Senior Secondary School Mathematics for Class 12

If R1 and R 2 be two equivalence relations on a set A, prove that R1 Ç R 2 is also an equivalence relation on A. Let R1 and R 2 be two equivalence relations on a set A. Then, R1 Í A ´ A , R 2 Í A ´ A Þ ( R1 Ç R 2) Í A ´ A. So, ( R1 Ç R 2) is a relation on A. This relation on A satisfies the following properties. (i) Reflexivity R1 is reflexive and R 2 is reflexive Þ ( a , a) Î R1 and ( a , a) Î R 2 for all a Î A Þ ( a , a) Î R1 Ç R 2 for all a Î A Þ R1 Ç R 2 is reflexive. (ii) Symmetry Let ( a , b) be an arbitrary element of R1 Ç R 2 . Then, ( a , b) Î R1 Ç R 2 Þ ( a , b) Î R1 and ( a , b) Î R 2 Þ ( b , a) Î R1 and ( b , a) Î R 2 [Q R1 is symmetric and R 2 is symmetric] Þ ( b , a) Î R1 Ç R 2. This shows that R1 Ç R 2 is symmetric. (iii) Transitivity ( a , b) Î R1 Ç R 2 and ( b , c) Î R1 Ç R 2 Þ ( a , b) Î R1 , ( a , b) Î R 2 , and ( b , c) Î R1 , ( b , c) Î R 2 Þ {( a , b) Î R1 , ( b , c) Î R1 }, and {( a , b) Î R 2 , ( b , c) Î R 2} Þ ( a , c) Î R1 and ( a , c) Î R 2 [Q R1 is transitive and R 2 is transitive] Þ ( a , c) Î R1 Ç R 2 . This shows that ( R1 Ç R 2) is transitive. Thus, R1 Ç R 2 is reflexive, symmetric and transitive. Hence, R1 Ç R 2 is an equivalence relation.

EXAMPLE 15

Give an example to show that the union of two equivalence relations on a set A need not be an equivalence relation on A.

SOLUTION

Let R1 and R 2 be two relations on a set A = {1, 2, 3}, given by R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} and R 2 = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}. Then, it is easy to verify that each one of R1 and R 2 is an equivalence relation.

SSS Mathematics for Class 12 13

Relations

13

But, R1 È R 2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)} is not transitive, as ( 3 , 1) Î R1 È R 2 and (1, 2) Î R1 È R 2 but ( 3 , 2) Ï R1 È R 2. Hence, ( R1 È R 2) is not an equivalence relation. EQUIVALENCE CLASSES Let R be an equivalence relation in a set A and let a Î A . Then, the set of all those elements of A which are related to a, is called the equivalence class determined by a and it is denoted by [a].

Thus, [a] = {b Î A : ( a , b) Î R}. Two equivalence classes are either disjoint or identical. An Important Result An equivalence relation R on a set A partitions the set into mutually disjoint equivalence classes. EXAMPLE 16

On the set Z of all integers, consider the relation R = {( a , b) : ( a - b) is divisible by 3}. Show that R is an equivalence relation on Z. Also find the partitioning of Z into mutually disjoint equivalence classes.

SOLUTION

The relation R on Z satisfies the following properties: (i) Reflexivity Let a Î Z. Then, ( a - a) = 0, which is divisible by 3. \ a R a " a Î Z. So, R is reflexive. (ii) Symmetry Let a , b Î Z such that a R b. Then, a R b Þ ( a - b) is divisible by 3 Þ - ( a - b) is divisible by 3 Þ ( b - a) is divisible by 3 Þ b R a. \

a R b Þ b R a " a , b Î Z.

So, R is symmetric. (iii) Transitivity Let a , b , c Î Z such that a R b and b R c. Then, a R b , b R c Þ ( a - b) is divisible by 3 and ( b - c) is divisible by 3 Þ [( a - b) + ( b - c)] is divisible by 3 Þ ( a - c) is divisible by 3. Thus, a R b , b R c Þ a R c " a , b , c Î Z.

SSS Mathematics for Class 12 14

14

Senior Secondary School Mathematics for Class 12

\ R is an equivalence relation on Z. Now, let us consider [0], [1] and [2]. We have: [0] = {x Î Z : x R 0 } = {x Î Z : ( x - 0) is divisible by 3} = {... , - 6, - 3 , 0, 3 , 6, 9, ... }. \

[0] = {¼, - 6, - 3 , 0, 3 , 6, 9, ¼}.

Similarly, [1] = {x Î Z : x R 1 } = {x Î Z : ( x - 1) is divisible by 3} = {... , - 5 , - 2, 1, 4, 7 , 10, ¼}. \

[1] = {... , - 5 , - 2, 1, 4, 7 , 10, ... }.

And, [2] = {x Î Z : x R 2 } = {x Î Z : ( x - 2) is divisible by 3} = {... , - 4, - 1, 2, 5 , 8, 11, ... }. \

[2] = {... , - 4, - 1, 2, 5 , 8, 11, ... }.

Clearly, [0], [1] and [2] are mutually disjoint and Z = [0] È [1] È [2]. EXAMPLE 17

SOLUTION

Let A = {x Î Z : 0 £ x £ 12}. Show that R = {( a , b) : |a - b|is a multiple of 4} is (i) reflexive, (ii) symmetric and (iii) transitive. Find the set of elements related to 1. Clearly, A = {0, 1, 2, 3 , 4, ... , 10, 11, 12}. Here, R satisfies the following properties. (i) Reflexivity Let a be an arbitrary element of A. Then, a - a = 0, which is a multiple of 4. \

a R a for all a Î A.

So, R is reflexive. (ii) Symmetry Let a R b. Then, a R b Þ | a - b |is a multiple of 4 Þ | - ( a - b)|is a multiple of 4 Þ | b - a |is a multiple of 4 Þ b R a. \

R is symmetric.

[CBSE 2010]

SSS Mathematics for Class 12 15

Relations

15

(iii) Transitivity Let a R b and b R c. Then, a R b, b R c Þ |a - b|is a multiple of 4 and|b - c|is a multiple of 4. Let|a - b| = 4k1 and|b - c| = 4k2. Then, | a - c| = |( a - b) - ( b - c)| = | 4k1 - 4k2 | = | 4( k1 - k2)|= 4| k1 - k2 |, which is a multiple of 4. \

a R b , b R c Þ a R c. So, R is transitive.

Thus, R is reflexive, symmetric and transitive. Hence, R is an equivalence relation. Now, [1] = {x Î A : x R 1} = {x Î A : | x - 1|is a multiple of 4} = {1, 5 , 9 }. Hence, the required set is {1, 5, 9}. EXAMPLE 18

SOLUTION

Let A = {1, 2, 3 , 4, 5 , 6, 7 , 8, 9} and R be a relation in A ´ A , defined by ( a , b) R ( c, d) Û a + d = b + c for all ( a , b) and ( c, d) Î A ´ A. Prove that R is an equivalence relation. Also obtain the equivalence class determined by ( 2, 5). [CBSE 2014] (i) Reflexivity Let ( a , b) Î A ´ A. Then, ( a , b) Î A ´ A Þ a , b Î A Þ a+ b = b+ a Þ ( a , b) R ( a , b). \

R is reflexive.

(ii) Symmetry Let ( a , b) R ( c, d). Then, ( a , b) R ( c, d) Þ a + d = b + c Þ c+ b = d + a Þ ( c, d) R ( a , b). \

R is symmetric.

(iii) Transitivity Let ( a , b) R ( c, d) and ( c, d) R ( e, f ). Then, ( a , b) R ( c, d) and ( c, d) R ( e, f ) Þ

a + d = b + c and c + f = d + e

Þ

a+ d + c+ f = b+ c+ d + e

Þ

a+ f = b+ e

Þ

( a , b) R ( e, f ).

SSS Mathematics for Class 12 16

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Senior Secondary School Mathematics for Class 12

\

R is transitive.

Thus, R is reflexive, symmetric and transitive. Hence, R is an equivalence relation. [( 2, 5)] = {( a , b) : ( 2, 5) R ( a , b)} = {( a , b) : 2 + b = 5 + a} = {( a , b) : b - a = 3} = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

EXERCISE 1A Very-Short-Answer Questions 1. Find the domain and range of the relation R = {(–1, 1), (1, 1), (–2, 4), (2, 4)}. 2. Let R = {( a , a 3) : a is a prime number less than 5}. Find the range of R.

[CBSE 2014]

3. Let R = {( a , a 3) : a is a prime number less than 10}. Find (i) R (ii) dom ( R) (iii) range ( R). 4. Let R = {x , y) : x + 2y = 8} be a relation on N. Write the range of R.

[CBSE 2014]

5. Let R = {( a , b) : a , b Î N and a + 3 b = 12 }. Find the domain and range of R. 6. Let R = {( a , b) : b =|a - 1|, a Î Z and |a|< 3}. Find the domain and range of R. ìæ 1 ö ü 7. Let R = íç a , ÷ : a Î N and 1 < a < 5 ý. îè a ø þ Find the domain and range of R. 8. Let R = {( a , b) : a , b Î N and b = a + 5 , a < 4 }. Find the domain and range of R. 9. Let S be the set of all sets and let R = {( A , B) : A Ì B)}, i.e., A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric. 10. Let A be the set of all points in a plane and let O be the origin. Show that the relation R = {( P , Q) : P , Q Î A and OP = OQ} is an equivalence relation. 11. On the set S of all real numbers, define a relation R = {( a , b) : a £ b }. Show that R is (i) reflexive (ii) transitive (iii) not symmetric . 12. Let A = {1, 2, 3, 4, 5, 6} and let R = {( a , b) : a , b Î A and b = a + 1}. Show that R is (i) not reflexive, (ii) not symmetric and (iii) not transitive.

SSS Mathematics for Class 12 17

Relations ANSWERS (EXERCISE 1A)

1. dom ( R) = {-1, 1, - 2, 2 } and range ( R) = {1, 4} 2. range (R) = {8, 27} 3. (i) R = {(2, 8), (3, 27), (5, 125), (7, 343)} (ii) dom ( R) = {2, 3, 5, 7} (iii) range ( R) = {8, 27, 125, 343} 4. {3, 2, 1} 5. dom ( R) = { 3 , 6, 9 } and range ( R) = { 3 , 2, 1} 6. dom ( R) = {–2, –1, 0, 1, 2} and range ( R) = {3, 2, 1, 0} ì1 1 1 ü 7. dom ( R) = {2, 3, 4} and range ( R) = í , , ý î2 3 4þ 8. dom ( R) = {1, 2, 3} and range ( R) = {6, 7, 8}

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 1A) 2. R = {( 2 , 8 ), ( 3 , 27 )} 4. R = {( 2 , 3 ), ( 4 , 2 ), ( 6 , 1)} 5. R = {( 3 , 3 ), ( 6 , 2 ), ( 9 , 1)} 6. Clearly, a is an integer such that -3 < a < 3. \

R = {( -2 , 3 ), ( -1, 2 ), ( 0 , 1),( 1, 0 )( 2 , 1)}.

ìæ 1 ö æ 1 ö æ 1 ö ü 7. R = í ç 2 , ÷ , ç 3 , ÷ , ç 4 , ÷ ý îè 2 ø è 3 ø è 4 ø þ 8. R = {( 1, 6 ), ( 2 , 7 ), ( 3 , 8 )}. 9.

(i) ( A Ì B and B Ì C ) Þ ( A Ì C ). So, R is transitive. (ii) Clearly, A Ì C is not true. So, R is not reflexive. (iii) If A Ì B then B Ì C is not true. So, R is not symmetric.

10. Let O be the origin and let P , Q , X be any three points in a plane. Then, (i) OP = OP is always true. So, R is reflextive. (ii) OP = OQ Þ OQ = OP. So, R is symmetric. (iii) (OP = OQ and OQ = OX) Þ (OP = OX). So, R transitive. Hence, R is an equivalence relation. 11.

(i) For all a Î R , a £ a is alway true. So, R is reflexive. (ii) For all a, b , c Î R , we have ( a £ b , b £ c ) Þ ( a £ c ). \ R is transitive. (iii) But R is not symmetrtic, as 3 £ 4 is true, while 4 £ 3 is not true.

12.

(i) ( 1, 1) Ï R as 1 = 1 + 1 is not true. (ii) ( 2 = 1 + 1) Þ 1 R 2. But 1 = 2 + 1 is not true. So, 2 is not related to 1. \ R is not symmetric. (iii) 1 R 2 and 2 R 3. But, 1 is not related to 3.

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SSS Mathematics for Class 12 18

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Senior Secondary School Mathematics for Class 12

EXERCISE 1B 1. Define a relation on a set. What do you mean by the domain and range of a relation. Give an example. 2. Let A be the set of all triangles in a plane. Show that the relation R = {(C1 , C 2) : C1 ~ C 2} is an equivalence relation on A. 3. Let R = {( a , b) : a , b Î Z and ( a + b) is even}. Show that R is an equivalence relation on Z. 4. Let R = {( a , b) : a , b Î Z and ( a - b) is divisible by 5}. Show that R is an equivalence relation on Z. 5. Show that the relation R defined on the set A = {1, 2, 3 , 4, 5} , given by [CBSE 2009] R = {( a , b) :|a - b|is even} is an equivalence relation. 6. Show that the relation R on N ´ N , defined by ( a , b) R ( c, d) Û a + d = b + c is an equivalent relation.

[CBSE 2010]

7. Let S be the set of all real numbers and let R = {( a , b) : a , b Î S and a = ± b}. Show that R is an equivalence relation on S. 8. Let S be the set of all points in a plane and let R be a relation in S defined by R = {( A , B) : d( A , B) < 2 units}, where d( A , B) is the distance between the points A and B. Show that R is reflexive and symmetric but not transitive. 9. Let S be the set of all real numbers. Show that the relation R = {( a , b) : a 2 + b 2 = 1} is symmetric but neither reflextive nor transitive. [CBSE 2008]

10. Let R = {( a , b) : a = b 2} for all a , b Î N . Show that R satisfies none of reflexivity, symmetry and transitivity. 11. Show that the relation R = {( a , b) : a > b } on N is transitive but neither reflexive nor symmetric. 12. Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Show that R is reflexive but neither symmetric nor transitive. 13. Let A = {1, 2, 3, 4} and R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (3, 2)}. Show that R is reflexive and transitive but not symmetric. HINTS TO SOME SELECTED QUESTIONS (EXERCISE 1B) 3.

(i) Let a Î Z. Then, a + a = 2 a, which is even. So, a R a. (ii) Let a R b. Then, a R b Þ a + b is even Þ b + a is even Þ b R a.

SSS Mathematics for Class 12 19

Relations (iii) Let a R b and b R c. Then, a R b and b R c Þ ( a + b ) is even and ( b + c ) is even. \

( a + c ) = {( a + c ) + 2 b} - 2 b

= {( a + b ) + ( b + c ) - 2 b }, which is even Þ a R c [Q each of ( a + b ), ( b + c ) and 2b is even]. 4. ( a - c ) = ( a - b ) + ( b - c ). 5. (i) | a - a|= 0 , which is even. So, a R a. (ii) a R b Þ | a - b|is even Þ | - ( a - b )|is even Þ | b - a|is even Þ b R a. (iii) a R b , b R c Þ | a - b|is even and| b - c|is even Þ ( a - b ) is even and ( b - c ) is even Þ {( a - b ) + ( b - c )} is even Þ ( a - c ) is even Þ | a - c|is even Þ a R c. 8. (i) d( A , A ) = 0 < 2. So, A R A. (ii) A R B Þ d( A , B) < 2 Þ d( B, A ) < 2 [Q d( B, A ) = d( A , B)] Þ B R A. (iii) Consider the points A( 0 , 0 ), B( 1. 5 , 0 ) and C(3, 0) on the x-axis. Then, d( A , B)1. 5 , d( B, C ) = 1. 5 and d( A , C ) = 3. \ A R B and B R C. But, A is not related to C. So, R is not transitive. 9.

(i) R is symmetric, since a R b Þ a2 + b 2 = 1 Þ b 2 + a2 = 1 Þ b R a. (ii) R is not reflexive, since 1 is not related to 1, as 12 + 12 = 1 is not true. (iii) Clearly,

1 R 2

3 3 1 and R × 2 2 2 2

But,

2

1 1 æ 1ö æ 1ö is not related to as ç ÷ + ç ÷ ¹ 1. 2 2 è2ø è2ø

\ R is not transitive. 10.

(i) 2 ¹ 2 2 Þ 2 is not related to 2. (ii) 4 = 2 2 Þ 4 R 2. But 2 ¹ 4 2 . So, 2 R 4. (iii) 16 R 4 , 4 R 2. But 16 is not related to 2, as 16 ¹ 2 2 .

19

SSS Mathematics for Class 12 20

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Senior Secondary School Mathematics for Class 12

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 1. Let A = {1, 2, 3} and let R = {(1, 1), ( 2, 2), ( 3 , 3), (1, 3), ( 3 , 2), (1, 2)}. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation 2. Let A = {a , b , c } and let R = {( a , a), ( a , b), ( b , a)}. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation 3. Let A = {1, 2, 3} and let R = {(1, 1), ( 2, 2), ( 3 , 3), (1, 2), ( 2, 1), ( 2, 3), ( 3 , 2)}. Then, R is (a) reflexive and symmetric but not transitive (b) symmetric and transitive but not reflexive (c) reflexive and transitive but not symmetric (d) an equivalence relation 4. Let S be the set of all straight lines in a plane. Let R be a relation on S defined by a R b Û a ^ b. Then, R is (a) reflexive but neither symmetric nor transitive (b) symmetric but neither reflexive nor transitive (c) transitive but neither reflexive nor symmetric (d) an equivalence relation 5. Let S be the set of all straight lines in a plane. Let R be a relation on S defined by a R b Û a | | b. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation 6. Let Z be the set of all integers and let R be a relation on Z defined by a R b Û ( a - b) is divisible by 3. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation

SSS Mathematics for Class 12 21

Relations

21

7. Let R be a relation on the set N of all natural numbers, defined by a R b Û a is a factor of b. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation 8. Let Z be the set of all integers and let R be a relation on Z defined by a R b Û a ³ b. Then, R is (a) symmetric and transitive but not reflexive (b) reflexive and symmetric but not transitive (c) reflexive and transitive but not symmetric (d) an equivalence relation 9. Let S be the set of all real numbers and let R be a relation on S defined by a R b Û |a| £ b. Then, R is (a) reflexive but neither symmetric nor transitive (b) symmetric but neither reflexive nor transitive (c) transitive but neither reflexive nor symmetric (d) none of these 10. Let S be the set of all real numbers and let R be a relation on S, defined by a R b Û |a - b| £ 1. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation 11. Let S be the set of all real numbers and let R be a relation on S, defined by a R b Û (1 + ab) > 0. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) none of these 12. Let S be the set of all triangles in a plane and let R be a relation on S defined by D 1 S D 2 Û D 1 º D 2. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation

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Senior Secondary School Mathematics for Class 12

13. Let S be the set of all real numbers and let R be a relation on S defined by a R b Û a 2 + b 2 = 1. Then, R is (a) symmetric but neither reflexive nor transitive (b) reflexive but neither symmetric nor transitive (c) transitive but neither reflexive nor symmetric (d) none of these 14. Let R be a relation on N ´ N , defined by ( a , b) R ( c, d) Û a + d = b + c. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation 15. Let A be the set of all points in a plane and let O be the origin. Let R = {( P , Q) : OP = OQ }. Then, R is (a) reflexive and symmetric but not transitive (b) reflexive and transitive but not symmetric (c) symmetric and transitive but not reflexive (d) an equivalence relation 16. Let Q be the set of all rational numbers and * be the binary operation, defined by a * b = a + 2b , then (a) * is commutative but not associative (b) * is associative but not commutative (c) * is neither commutative nor associative (d) * is both commutative and associative 17. Let a * b = a + ab for all a , b Î Q. Then, (a) * is not a binary composition (b) * is not commutative (c) * is commutative but not associative (d) * is both commutative and associative 18. Let Q + be the set of all positive rationals. Then, the operation * on Q + ab for all a , b Î Q + is defined by a * b = 2 (a) commutative but not associative (b) associative but not commutative (c) neither commutative nor associative (d) both commutative and associative

SSS Mathematics for Class 12 23

Relations

23

19. Let Z be the set of all integers and let a * b = a - b + ab. Then, * is (a) commutative but not associative (b) associative but not commutative (c) neither commutative nor associative (d) both commutative and associative 20. Let Z be the set of all integers. Then, the operation * on Z defined by a * b = a + b - ab is (a) commutative but not associative (b) associative but not commutative (c) neither commutative nor associative (d) both commutative and associative 21. Let Z + be the set of all positive integers. Then, the operation * on Z + defined by a * b = a b is (a) commutative but not associative (b) associative but not commutative (c) neither commutative nor associative (d) both commutative and associative 22. Define * on Q - {-1} by a * a + b + ab. Then, * on Q - {-1} is (a) commutative but not associative (b) associative but not commutative (c) neither commutative nor associative (d) both commutative and associative ANSWERS (OBJECTIVE QUESTIONS)

1. (b)

2. (c)

3. (a)

4. (b)

5. (d)

6. (d)

7. (b)

8 (c)

9. (c) 10. (a)

11. (a) 12. (d) 13. (a) 14. (d) 15. (d) 16. (c) 17. (b) 18. (d) 19. (c) 20. (d) 21. (c) 22. (d) HINTS TO SOME SELECTED OBJECTIVE QUESTIONS 1. R is reflexive and transitive but not symmetric. 2. R is symmetric and transitive but not reflexive. 3. ( 1, 2 ) Î R and ( 2 , 3 ) Î R. But, ( 1, 3 ) Ï R. So, R is not transitive. 4. a ^ a is not true. So, R is not reflexive. a ^ b and b ^ c do not imply a ^ c. So, R is not transitive. But, a ^ b Þ b ^ a is always true. 9. (i)| - 3| £ - 3 is not true. So, R is not reflexive.

SSS Mathematics for Class 12 24

24

Senior Secondary School Mathematics for Class 12 (ii)| -1| £ 1 Þ ( -1) R 1. But| 1| £ - 1 is not true. \ R is not symmetric. (iii) a R b , b R c Þ | a| £ b and| b| £ c Þ | a| £ c. \ R is transitive.

10. (i)| a - a| = 0 £ 1 is always true. (ii) a R b Þ | a - b| £ 1 Þ | - ( a - b )| £ 1 Þ | b - a| £ 1 Þ b R a. 1 (iii) 2 R 1 and 1 R × 2 1 But, 2 is not related to × So, R is not transitive. 2 11. (i) a R a, since ( 1 + a2 ) > 0. (ii) a R b Þ ( 1 + ab ) > 0 Þ ( 1 + ba) > 0 Þ b R a. -1 1 (iii) Let a = , b = and c = R. 2 2 Then, a R b and b R c. But, a is not related to c. 13. (i) ( 12 + 12 ) ¹ 1. So, 1 R 1 is not true. (ii) a R b Þ a2 + b 2 = 1 Þ b 2 + a2 = 1 Þ b R a. (iii) 1 R 0 and 0 R 1. But, 1 is not related to 1.

SSS Mathematics for Class 12 25

2. FUNCTIONS FUNCTION Let A and B be two nonempty sets. Then, a rule f which associates to each element x Î A, a unique element, denoted by f ( x) of B, is called a function from A to B and we write, f : A Î B. f ( x) is called the image of x, while x is called the pre-image of f ( x).

Domain, Codomain and Range of a Function Let f : A ® B. Then, A is called the domain of f and B is called the codomain of f . And, f ( A) = { f ( x) : x Î A} is called the range of f . Example 1

Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16, 25}.

Consider the rule f : A ® B : f ( x) = x 2 " x Î A. Then, each element in A has its unique image in B. So, f is a function from A to B. f (1) = 12 = 1, f ( 2) = 22 = 4, f ( 3) = 3 2 = 9, f ( 4) = 42 = 16. Dom ( f ) = {1, 2, 3, 4} = A, codomain ( f ) = {1, 4, 9, 16, 25} = B and range ( f ) = {1, 4, 9, 16}. Clearly, 25 Î B does not have its pre-image in A. Example 2

Let N be the set of all natural numbers. Let f : N ® N : f ( x) = 2x " x Î N . Then, every element in N has its unique image in N. So, f is a function from N to N. Clearly, f (1) = 2, f ( 2) = 4, f ( 3) = 6 ..., and so on. Dom ( f ) = N , codomain ( f ) = N , range ( f ) = {2, 4, 6, 8, 10, ...}. 25

SSS Mathematics for Class 12 26

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Senior Secondary School Mathematics for Class 12

Various Types of Functions

A function f : A ® B is said to be many-one if two or more than two elements in A have the same image in B.

MANY-ONE FUNCTION

Example

Let A = {–1, 1, 2, 3} and B = {1, 4, 9}. Let f : A ® B : f ( x) = x 2 " x Î A. Then, each element in A has a unique image under f in B. \ f is a function from A to B such that f ( -1) = ( -1) 2 = 1; f (1) = 12 = 1; f ( 2) = 22 = 4 and f ( 3) = 3 2 = 9. Clearly, two different elements, namely –1 and 1, have the same image 1 Î B. Hence, f is many-one.

One-One or Injective Function A function f : A ® B is said to be one-one if distinct elements in A have distinct images in B. f is one-one when f ( x1) = f ( x 2) Þ x1 = x 2. Example

Let N be the set of all natural numbers. Let f : N ® N : f ( x) = 2x " x Î N . Then, f ( x1) = f ( x 2) Þ 2x1 = 2x 2 Þ x1 = x 2. \ f is one-one.

Onto or Surjective Function A function f : A ® B is said to be onto if every element in B has at least one pre-image in A. Thus, if f is onto then for each y Î B $ at least one element x Î A such that y = f ( x). Also, f is onto Û range ( f ) = B. Example

Let N be the set of all natural numbers and let E be the set of all even natural numbers. Let f : N ® E : f ( x) = 2x " x Î N . 1 Then, y = 2x Þ x = y. 2

SSS Mathematics for Class 12 27

Functions

Thus, for each y Î E there exists

27

1 y Î N such that 2

1 ö æ1 ö æ f ç y ÷ = ç 2 ´ y ÷ = y. 2 ø è2 ø è \

f is onto.

INTO FUNCTION A function f : A ® B is said to be into if there exists even a single element in B having no pre-image in A.

Clearly, f is into Û range ( f ) Ì B. Example

Let A = {2, 3, 5, 7} and B = {0, 1, 3, 5, 7}. Let f : A ® B : f ( x) = ( x - 2). Then, f ( 2) = ( 2 - 2) = 0, f ( 3) = ( 3 - 2) = 1, f (5) = (5 - 2) = 3 and f (7) = (7 - 2) = 5. Thus, every element in A has a unique image in B. Now, $ 7 Î B having no pre-image in A. \ f is into. Note that range ( f ) = {0, 1, 3 , 5} Ì B.

Bijective Function A one-one onto function is said to be bijective or a one-to-one correspondence. A function f : A ® B is called a constant function if every element of A has the same image in B.

CONSTANT FUNCTION

Example

Let A = {1, 2, 3} and B = {5, 7, 9}. Let f : A ® B : f ( x) = 5 for all x Î A. Clearly, every element in A has the same image. So, f is a constant function.

REMARK

The range of a constant function is a singleton set.

IDENTITY FUNCTION The function I A : A ® A : I A ( x) = x " x Î A is called an identity function on A.

Domain ( I A ) = A and range ( I A ) = A. EQUAL FUNCTIONS Two functions f and g having the same domain D are said to be equal if f ( x) = g( x) " x Î D.

SOLVED EXAMPLES EXAMPLE 1

Let f : N ® N : f ( x) = 2x for all x Î N . Show that f is one-one and into.

SOLUTION

We have f ( x1) = f ( x 2) Þ 2x1 = 2x 2 Þ x1 = x 2.

SSS Mathematics for Class 12 28

28

Senior Secondary School Mathematics for Class 12

\ f is one-one. Let y = 2x. Then, x =

y × 2

3 Ï N. 2 Thus, 3 Î N has no pre-image in N. \ f is into. Hence, f is one-one and into. If we put y = 3 then x =

EXAMPLE 2

SOLUTION

Show that the function f : R ® R : f ( x) = x 2 is neither one-one nor onto. We have f ( -1) = ( -1) 2 = 1 and f (1) = 12 = 1. Thus, two different elements in R have the same image. \ f is not one-one. If we consider –1 in the codomain R, then it is clear that there is no element in R whose image is -1. \ f is not onto. Hence, f is neither one-one nor onto.

EXAMPLE 3

Show that the function f : R ® R : f ( x) = |x|is neither one-one nor onto.

SOLUTION

We have f ( -1) = |- 1| = 1 and f (1) = |1| = 1. Thus, two different elements in R have the same image. \ f is not one-one. If we consider –1 in the codomain R, then it is clear that there is no real number x whose modulus is –1. Thus, -1 Î R has no pre-image in R. \ f is not onto. Hence, f is neither one-one nor onto.

EXAMPLE 4

For any real number x, we define [x] = greatest integer less than or equal to x. Prove that the greatest integer function f : R ® R : f ( x) = [x] is neither one-one nor onto.

SOLUTION

Clearly, [1.2] = 1 and [1. 3] = 1. \ f (1.2) = 1 and f (1.3) = 1. Thus, two different real numbers have the same image. \ f is not one-one. Clearly, there is no real number x such that f ( x) = [x] = 11. . So, f is not onto. Hence, f is neither one-one nor onto.

SSS Mathematics for Class 12 29

Functions

29

EXAMPLE 5

Let R 0 be the set of all nonzero real numbers. 1 Show that f : R 0 ® R 0 : f ( x) = is a one-one onto function. x

SOLUTION

We have f ( x1) = f ( x 2) Þ

1 1 = Þ x1 = x 2. x1 x 2

\ f is one-one. 1 1 Again, y = Þ x= × x y æ1ö Now, if y is a nonzero real number, then x = çç ÷÷ is a nonzero real èyø æ1ö number such that f çç ÷÷ = y. èyø Thus, each y in R 0 has its pre-image in R 0 . So, f is onto. Hence, f is one-one onto. EXAMPLE 6

Show that the function f : R ® R : f ( x) = x 3 is one-one and onto.

SOLUTION

We have f ( x1) = f ( x 2) Þ x13 = x 23 Þ ( x13 - x 23) = 0 Þ ( x1 - x 2) ( x12 + x1 x 2 + x 22) = 0

\ f is one-one.

2 éæ x ö 3 ù Þ ( x1 - x 2) ê ç x1 + 2 ÷ + x 22 ú = 0 2ø 4 úû êë è 2 é æ ù x ö 3 2 Þ ( x1 - x 2) = 0 êQ ç x1 + 2 ÷ + x 2 ¹ 0ú 2ø 4 êë è úû Þ x1 = x 2.

Let y Î R and let y = x 3 . Then, x = y

1

3

Î R.

Thus, for each y in the codomain R there exists y f (y

1

3)

= (y

1

3) 3

1

3

in R such that

= y.

\ f is onto. Hence, f is one-one onto. EXAMPLE 7

Show that the function f : R ® R : f ( x) = 3 - 4x is one-one onto and hence bijective.

SOLUTION

We have f ( x1) = f ( x 2) Þ 3 - 4x1 = 3 - 4x 2

SSS Mathematics for Class 12 30

30

Senior Secondary School Mathematics for Class 12

Þ - 4x1 = - 4x 2 Þ x1 = x 2. f is one-one. ( 3 - y) Now, let y = 3 - 4x. Then, x = × 4

\

Thus, for each y Î R (codomain of f ), there exists x =

( 3 - y) ÎR 4

( 3 - y) ü æ 3 - yö ì such that f ( x) = f ç ÷ = í3 - 4 × ý = 3 - ( 3 - y) = y. 4 þ è 4 ø î

This shows that every element in codomain of f has its pre-image in dom ( f ). \ f is onto. Hence, the given function is bijective. EXAMPLE 8

Show that the function f : N ® N , defined by ìx + 1, if x is odd f ( x) = í îx - 1, if x is even is one-one and onto.

SOLUTION

[CBSE 2012]

Suppose f ( x1) = f ( x 2). When x1 is odd and x 2 is even In this case, f ( x1) = f ( x 2) Þ x1 + 1 = x 2 - 1 Þ x 2 - x1 = 2. This is a contradiction, since the difference between an odd integer and an even integer can never be 2. Thus, in this case, f ( x1) ¹ f ( x 2). Similarly, when x1 is even and x 2 is odd, then f ( x1) ¹ f ( x 2). Case 2 When x1 and x 2 are both odd In this case, f ( x1) = f ( x 2) Þ x1 + 1 = x 2 + 1 Þ x1 = x 2. \ f is one-one. Case 3 When x1 and x 2 are both even In this case, f ( x1) = f ( x 2) Þ x1 - 1 = x 2 - 1 Þ x1 = x 2. \ f is one-one. In order to show that f is onto, let y Î N (the codomain). Case 1 When y is odd In this case, ( y + 1) is even. \ f ( y + 1) = ( y + 1) - 1 = y. Case 2 When y is even In this case, ( y - 1) is odd. \ f ( y - 1) = y - 1 + 1 = y. Case 1

SSS Mathematics for Class 12 31

Functions

31

Thus, each y Î N (codomain of f ) has its pre-image in dom ( f ). \ f is onto. Hence, f is one-one onto. EXAMPLE 9

SOLUTION

Show that f : N ® N , defined by ìn + 1 , if n is odd ïï f (n) = í 2 ïn , if n is even ïî 2 is a many-one onto function.

[CBSE 2012C]

We have f (1) =

(1 + 1) 2 = = 1 and 2 2

f ( 2) =

2 = 1. 2

Thus, f (1) = f ( 2) while 1 ¹ 2. \ f is many-one. In order to show that f is onto, consider an arbitrary element n Î N. If n is odd then ( 2n - 1) is odd, and ( 2n - 1 + 1) 2n f ( 2n - 1) = = = n. 2 2 If n is even then 2n is even and 2n f ( 2n) = = n. 2 Thus, for each n Î N (whether even or odd) there exists its pre-image in N. \ f is onto. Hence, f is many-one onto. EXAMPLE 10

Show that the signum function f : R ® R, defined by ì 1, if x > 0 ï f ( x) = í 0, if x = 0 ï-1, if x < 0 î is neither one-one nor onto.

SOLUTION

Clearly, f ( 2) = 1 and f ( 3) = 1. Thus, f ( 2) = f ( 3) while 2 ¹ 3. \ f is not one-one. Range ( f ) = {1, 0, - 1} Ì R. So, f is into. Hence, f is neither one-one nor onto.

SSS Mathematics for Class 12 32

32 EXAMPLE 11

Senior Secondary School Mathematics for Class 12

Let A = R - { 3} and B = R - {1}. x-2 for all values of x Î A. x-3 Show that f is one-one and onto.

Let f : A ® B : f ( x) =

SOLUTION

[CBSE 2012]

f is one-one, since x - 2 x2 - 2 f ( x1) = f ( x 2) Þ 1 = x1 - 3 x 2 - 3 Þ ( x1 - 2)( x 2 - 3) = ( x1 - 3)( x 2 - 2) Þ x1 x 2 - 3 x1 - 2x 2 + 6 = x1 x 2 - 2x1 - 3 x 2 + 6 Þ x1 = x 2. x-2 Let y Î B such that y = × x-3 Then, ( x - 3) y = ( x - 2) Þ x =

( 3 y - 2) × ( y - 1)

Clearly, x is defined when y ¹ 1. Also, x = 3 will give us 1 = 0, which is false. \ x ¹ 3. æ 3y - 2 ö çç - 2÷÷ 1 y ø = y. And, f ( x) = è æ 3y - 2 ö çç - 3 ÷÷ 1 y è ø Thus, for each y Î B, there exists x Î A such that f ( x) = y. \ f is onto. Hence, f is one-one onto. EXAMPLE 12

SOLUTION

Let A and B be two nonempty sets. Show that the function f : ( A ´ B) ® ( B ´ A) : f ( a , b) = ( b , a) is a bijective function. f is one-one, since f ( a1 , b1) = f ( a2 , b2) Þ ( b1 , a1) = ( b2 , a2) Þ a1 = a2 and b1 = b2 Þ ( a1 , b1) = ( a2 , b2). In order to show that f is onto, let ( b , a) be an arbitrary element of ( B ´ A). Then, ( b , a) Î ( B ´ A) Þ b Î B and a Î A Þ ( a , b) Î ( A ´ B). Thus, for each ( b , a) Î ( B ´ A), there exists ( a , b) Î A ´ B such that f ( a , b) = ( b , a). \ f is onto. Thus, f is one-one onto and hence bijective.

SSS Mathematics for Class 12 33

Functions EXAMPLE 13

33

Find the domain and range of the real function f ( x) = 9 - x 2 .

SOLUTION

It is clear that f ( x) = 9 - x 2 is not defined when ( 9 - x 2) < 0, i.e., when x 2 > 9, i.e., when x > 3 or x < –3. \ dom ( f ) = {x Î R : - 3 £ x £ 3}. Also, y = 9 - x 2 Þ y 2 = ( 9 - x 2) Þ x = 9 - y2 . Clearly, x is not defined when ( 9 - y 2) < 0. But, ( 9 - y 2) < 0 Þ y 2 > 9 Þ y > 3 or y < - 3 \ range ( f ) = {y Î R : - 3 £ y £ 3}.

EXAMPLE 14

SOLUTION

Find the domain and range of the real function, defined by 1 f ( x) = × (1 - x 2) Clearly,

1 (1 - x 2)

is not defined when 1 - x 2 = 0,

i.e., when x = ± 1. \ dom ( f ) = R - {-1, 1}. Also, y =

1 2

(1 - x )

Þ (1 - x 2) =

1 1 Þ x = 1- × y y

æ 1 1ö Clearly, x is not defined when çç1 - ÷÷ < 0 or 1 < or y < 1. y y è ø \ range ( f ) = R - {y Î R : y < 1} = {y Î R : y ³ 1}. EXAMPLE 15

Consider a function f : X ® Y and define a relation R in X by R = {( a , b) : f ( a) = f ( b)}. Show that R is an equivalence relation.

SOLUTION

Here, R satisfies the following properties: (i) Reflexivity Let a Î X. Then, f ( a) = f ( a) Þ ( a , a) Î R. \ R is reflexive. (ii) Symmetry Let ( a , b) Î R. Then, ( a , b) Î R Þ f ( a) = f ( b) Þ f ( b) = f ( a) Þ ( b , a) Î R. \ R is symmetric.

SSS Mathematics for Class 12 34

34

Senior Secondary School Mathematics for Class 12

(iii) Transitivity Let ( a , b) Î R and ( b , c) Î R. Then, ( a , b) Î R , ( b , c) Î R Þ f ( a) = f ( b) and f ( b) = f ( c) Þ f ( a) = f ( c) Þ ( a , c) Î R. \ R is transitive. Hence, R is an equivalence relation.

EXERCISE 2A 1. Define a function. What do you mean by the domain and range of a function? Give examples. 2. Define each of the following: (i) injective function (ii) surjective function (iii) bijective function (iv) many-one function (v) into function Give an example of each type of functions. 3. Give an example of a function which is (i) one-one but not onto (ii) one-one and onto (iii) neither one-one nor onto (iv) onto but not one-one. 4. Let f : R ® R be defined by ì2x + 3 , when x < - 2 ï f ( x) = íx 2 - 2, when - 2 £ x £ 3 ï 3 x - 1, when x > 3. î Find (i) f ( 2) (ii) f ( 4) (iii) f ( -1) (iv) f ( -3). 5. Show that the function f : R ® R : f ( x) = 1 + x 2 is many-one into. 6. Show that the function f : R ® R : f ( x) = x 4 is many-one and into. 7. Show that the function f : R ® R : f ( x) = x5 is one-one and onto. pù pù é é 8. Let f : ê 0, ú ® R : f ( x) = sin x and g : ê 0, ú ® R : g( x) = cos x. Show 2û 2û ë ë that each one of f and g is one-one but ( f + g) is not one-one. 9. Show that the function (i) f : N ® N : f ( x) = x 2 is one-one into. (ii) f : Z ® Z : f ( x) = x 2 is many-one into.

SSS Mathematics for Class 12 35

Functions

35

10. Show that the function (i) f : N ® N : f ( x) = x 3 is one-one into (ii) f : Z ® Z : f ( x) = x 3 is one-one into 11. Show that the function f : R ® R : f ( x) = sin x is neither one-one nor onto. 12. Prove that the function f : N ® N : f (n) = (n2 + n + 1) is one-one but not onto. 13. Show that the function f : N ® Z, defined by ì1 ïï 2 (n - 1), when n is odd f (n) = í ï- 1 n, when n is even ïî 2 is both one-one and onto. 14. Find the domain and range of the function f : R ® R : f ( x) = x 2 + 1. 15. Which of the following relations are functions? Give reasons. In case of a function, find its domain and range. (i) f = {(–1, 2), (1, 8), (2, 11), (3, 14)} (ii) g = {(1, 1), (1, –1), (4, 2), (9, 3), (16, 4)} (iii) h = {( a , b), ( b , c), ( c, b), ( d , c)} 16. Find the domain and range of the real function, defined by f ( x) =

x2 (1 + x 2)

Show that f is many-one. 17. Show that the function ì 1, if x is rational f : R ® R : f ( x) = í î-1, if x is irrational is many-one into. æ1ö Find (i) f ç ÷ (ii) f ( 2) (iii) f ( p) (iv) f ( 2 + è 2ø

3 ).

ANSWERS (EXERCISE 2A)

4. (i) 2

(ii) 11

(iii) –1

(iv) –3

14. dom ( f ) = R and range ( f ) = {y Î R : y ³ 1} 15.

(i) f is a function, dom ( f ) = {-1, 1, 2, 3} and range ( f ) = {2, 8, 11, 14} (ii) g is not a function (iii) h is a function, dom ( h) = {a , b , c, d} and range ( h) = {b , c}

17. (i) 1

(ii) –1

(iii) –1

(iv) –1

×

SSS Mathematics for Class 12 36

36

Senior Secondary School Mathematics for Class 12 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 2A)

2.

(i) f : N ® N : f ( x ) = 2 x is an injective function, as f ( x1 ) = f ( x2 ) Þ 2 x1 = 2 x2 Þ x1 = x2 . (ii) Let A = {1, –1, 2, 3} and B = {1, 4, 9}. Then, f : A ® B : f ( x ) = x 2 is surjective, since each element of B has at least one pre-image in A. (iii) Let E be the set of all even natural numbers. Then, f : N ® E : f ( x ) = 2 x is one-one and onto. f is one-one, since f ( x1 ) = f ( x2 ) Þ 2 x1 = 2 x2 Þ x1 = x2 . 1 æ1 ö f is onto, since for each y Î E, there exists y Î N such that f ç y ÷ = y. 2 è2 ø (iv) Example given in (ii) is many-one. (v) Let A = {1, 2, 3} and B = {1, 4, 9, 16}. Then, f : A ® B : f ( x ) = x 2 is an into function, since range ( f ) = {1, 4 , 9} Ì B.

5. f ( -1) = 2 = f ( 1). So, f is many-one. -1 Î R has no pre-image in R. So, f is into. 6. f ( -1) = f ( 1) = 1. So, f is many-one. -1 Î R has no pre-image in R. So, f is into. 7. f ( x1 ) = f ( x2 ) Þ x15 = x52 Þ x1 = x2 . So, f is one-one for each y Î R $ y

1

5

Î R s.t. f ( y

1

5

) = y.

é pù 8. If x1 ¹ x2 and x1 , x2 Î ê 0 , ú then sin x1 ¹ sin x2 and cos x1 ¹ cos x2 . ë 2û \ f is one-one and g is one-one. But ( f + g )( x ) = f ( x ) + g ( x ) = sin x + cos x. p pö æpö æ \ ( f + g )( 0 ) + (sin 0 + cos 0 ) = 1 and ( f + g ) ç ÷ = ç sin + cos ÷ = 1. 2 2ø è2ø è 9.

(i) f ( a) = f ( b ) Þ a2 = b 2 Þ a = b

[Q a, b Î N ]

\ f is one-one. Clearly, 2 Î N [codomain ( f )] has no pre-image in N. \ f is into. (ii) f ( -1) = f ( 1) = 1. So, f is many-one. 2 Î N [codomain ( f )] has no pre-image in N. \ f is into. 10.

(i) f ( x1 ) = f ( x2 ) Þ x13 = x23 Þ x1 = x2

[Q x1 , x2 Î N ].

\ f is one-one. 2 Î N. But, 2 (ii) f ( x1 ) = f ( x2 ) Þ 2 Î Z. But, 2

1

3

1

3

Ï N. So, f is into.

x13

= x23 Þ x1 = x2

Ï Z. So, f is into.

[Q x1 , x2 Î Z]

SSS Mathematics for Class 12 37

Functions

37

11. We know that sin ( 0 ) = 0 and sin ( p ) = 0. Thus, 0 and p have the same image. So, f is many-one. Range ( f ) = [-1, 1] Ì R. Hence, f is into. So, f is neither one-one nor onto. 12. f (n1 ) = f (n2 ) Þ n12 + n1 + 1 = n22 + n2 + 1 Þ (n12 - n22 ) + (n1 - n2 ) = 0 Þ (n1 - n2 )(n1 + n2 + 1) = 0 Þ n1 - n2 = 0 Þ n1 = n2 . \ f is one-one. But, f (n) = 1 Þ n2 + n + 1 = 1 Þ n2 + n = 0 Þ n(n + 1) = 0 Þ n = 0 or n = - 1. And, none of 0 and –1 is a natural number. Thus, 1Î N has no pre-image in N. \ f is into. 14. Dom ( f ) = R. Also, y = x 2 + 1 Þ x = y - 1. x is defined when y - 1 ³ 0, i.e., y ³ 1. \ range ( f ) = {y Î R : y ³ 1}. 15. g is not a function, since 1 has two images under g. 16. When x is real, 1 + x 2 ¹ 0. So, dom ( f ) = R. y=

x2 ( 1 + x2 )

For x to be real,

Þ x 2 ( 1 - y) = y Þ x =

y × 1- y

y ³ 0 and ( 1 - y ) ¹ 0. ( 1 - y)

\ range ( f ) = {y Î R : 0 £ y < 1}. æ 1ö Also, 1 and –1 have the same image ç ÷ × è2ø

Composition of Functions Let f : A ® B and g : B ® C be two given functions. Then, the composition of f and g, denoted by g o f is the function, defined by ( g o f ) : A ® C : ( g o f ) ( x) = g{ f ( x)} " x Î A. Clearly, dom ( g o f ) = dom ( f ). Also, g o f is defined only when range ( f ) Í dom ( g). REMARK

( f o g) is defined only when range ( g) Í dom ( f ). And, dom ( f o g) = dom ( g).

SSS Mathematics for Class 12 38

38

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Let f : {1, 3 , 4 } ® {1, 2, 5} and g : {1, 2, 5} ® {1, 3} be defined as f = {(1, 2), ( 3 , 5), ( 4, 1)} and g = {(1, 3), ( 2, 3), (5 , 1)}. Find ( g o f ) and ( f o g).

SOLUTION

Here range ( f ) = {1, 2, 5} and dom ( g) = {1, 2, 5}. Clearly, range ( f ) Í dom ( g). \ ( g o f ) is defined and dom ( g o f ) = dom ( f ) = {1, 3, 4}. Now, ( g o f ) (1) = g{ f (1)} = g( 2) = 3 ; ( g o f ) ( 3) = g{ f ( 3)} = g(5) = 1; ( g o f ) ( 4) = g{ f ( 4)} = g(1) = 3. Hence, ( g o f ) = {(1, 3), (3, 1), (4, 3)}. Again, range ( g) = {1, 3} and dom ( f ) = {1, 3, 4}. Clearly, range ( g) Í dom ( f ). \ ( f o g) is defined and dom ( f o g) = dom ( g) = {1, 2, 5}. Now, ( f o g) (1) = f {g(1)} = f ( 3) = 5 ; ( f o g) ( 2) = f {g( 2)} = f ( 3) = 5 ; ( f o g) (5) = f {g(5)} = f (1) = 2. Hence, ( f o g) = {(1, 5), (2, 5), (5, 2)}.

EXAMPLE 2

Let R be the set of all real numbers. Let f : R ® R : f ( x) = cos x and let g : R ® R : g( x) = 3 x 2 . Show that ( g o f ) ¹ ( f o g).

SOLUTION

Let x be an arbitrary real number. Then, ( g o f ) ( x) = g{ f ( x)} = g(cos x) = 3(cos x) 2 = 3 cos2 x. ( f o g ) ( x) = f {g( x)} = f ( 3 x 2) = cos ( 3 x 2). Taking x = 0, we have ( g o f ) ( 0) = 3 cos2 0 = ( 3 ´ 1) = 3. ( f o g ) ( 0) = cos ( 3 ´ 0) = cos 0 = 1. \ ( g o f ) ( 0) ¹ ( f o g) ( 0). Hence, g o f ¹ f o g.

EXAMPLE 3

Let R be the set of all real numbers. Let f : R ® R : f ( x) = sin x and g : R ® R : g( x) = x 2 . Prove that g o f ¹ f o g.

SOLUTION

Let x be an arbitrary real number. Then, ( g o f ) ( x) = g{ f ( x)} = g(sin x) = (sin x) 2. ( f o g) ( x) = f {g( x)} = f ( x 2) = sin x 2. Clearly, (sin x) 2 ¹ sin x 2. Hence, g o f ¹ f o g.

SSS Mathematics for Class 12 39

Functions

39 1

EXAMPLE 4

SOLUTION

Let f : R ® R : f ( x) = 8x 3 and g : R ® R : g( x) = x 3 . Find ( g o f ) and ( f o g) and show that g o f ¹ f o g. Let x Î R. Then, we have ( g o f ) ( x) = g{ f ( x)} = g( 8x 3) = ( 8x 3) 1

\ EXAMPLE 5 SOLUTION

( f o g) ( x) = f {g( x)} = f ( x 3 ) = 8( x g o f ¹ f o g.

1

1

= 2x.

3

3) 3

= 8x.

Let f : R ® R : f ( x) = ( x 2 - 3 x + 2), find ( f o f ) ( x). We have ( f o f )( x) = f { f ( x)} = f ( x 2 - 3 x + 2) = f ( y), where y = ( x 2 - 3 x + 2) = ( y 2 - 3 y + 2) = ( x 2 - 3 x + 2) 2 - 3( x 2 - 3 x + 2) + 2 = ( x 4 - 6x 3 + 10x 2 - 3 x). 1

EXAMPLE 6

If f : R ® R : f ( x) = ( 3 - x 3) 3 , show that ( f o f ) ( x) = x.

SOLUTION

We have ( f o f ) ( x) = f { f ( x)} = f ( 3 - x 3)

1

3

= f ( y), where y = ( 3 - x 3) =(3 -

1 y 3) 3

1

3

1

= [ 3 - ( 3 - x 3)]

3

[Q y 3 = ( 3 - x 3)]

1 ( x 3) 3

= = x. Hence, ( f o f ) ( x) = x. EXAMPLE 7

Let f : A ® B, and let I A and I B be identity functions on A and B respectively. Prove that ( f o I A ) = f and ( I B o f ) = f .

SOLUTION

Let x Î A and let f ( x) = y. Then, ( f o I A ) ( x) = f {I A ( x)} = f ( x) \ ( f o IA ) = f .

[Q I A ( x) = x].

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And, ( I B o f ) ( x) = I B{ f ( x)} = I B( y)

\

[Q f ( x) = y]

=y

[Q I B( y) = y]

= f ( x)

[Q y = f ( x)].

( IB o f ) = f .

Hence, ( f o I A ) = f and ( I B o f ) = f . EXAMPLE 8

(Associativity) Let f : A ® B , g : B ® C and h : C ® D. Then, prove that ( h o g) o f = h o ( g o f ).

SOLUTION

Let x Î A. Then, {( h o g) o f } ( x) = ( h o g) { f ( x)} = h [ g{ f ( x)}] = h [( g o f ) ( x)] = {h o ( g o f )} ( x). \

EXAMPLE 9

SOLUTION

( h o g) o f = h o ( g o f ).

Let f : Z ® Z : ìn ï , g(n) = í 3 ïî 0, Show that g o f

f (n) = 3n and let g : Z ® Z, defined by if n is a multiple of 3 if n is not a multiple of 3. = I Z and f o g ¹ I Z.

Let n be an arbitrary element of Z. Then, ( g o f ) (n) = g{ f (n)} 3n = g( 3n) = =n 3 = I Z(n). \ ( g o f ) = I Z. Also, we have ( f o g) (1) = f { g(1)} = f ( 0)

[Q g(1) = 0]

= ( 3 ´ 0) = 0

[Q f (n) = 3n].

I Z(1) = 1 [Q I Z( x) = x " x Î Z]. \ f o g ¹ I Z. EXAMPLE 10

ì3ü ì7 ü Let A = R - í ý and B = R - í ý × î5 þ î5 þ 7x + 4 3y + 4 Let f : A ® B : f ( x) = and g : B ® A : g( y) = × 5x - 3 5y - 7 Show that ( g o f ) = I A and ( f o g) = I B.

SSS Mathematics for Class 12 41

Functions SOLUTION

Let x Î A. Then, ( g o f ) ( x) = g[ f ( x)] æ 7x + = g çç è 5x -

4ö ÷ 3 ÷ø

= g( y), where y =

41

7x + 4 5x - 3

… (i)

æ 7x + 4 ö ÷+4 3 çç 5 x - 3 ÷ø 3y + 4 [using (i)] = è = 5y - 7 æ 7x + 4 ö ÷÷ - 7 5 çç è 5x - 3 ø ( 21x + 12 + 20x - 12) (5 x - 3) = ´ (5 x - 3) ( 35 x + 20 - 35 x + 21) 41x = = x = I A ( x). 41 \ ( g o f ) = IA . Again, let y Î B. Then, ( f o g) ( y) = f [ g( y)] æ 3y + 4ö ÷÷ = f çç è 5y - 7 ø 3y + 4 … (ii) = f (z), where z = 5y - 7 æ 3y + 4ö ÷+4 7 çç 5 y - 7 ÷ø 7z + 4 = = è æ 3y + 4ö 5z - 3 ÷÷ - 3 5 çç è 5y - 7 ø ( 21y + 28 + 20y - 28) (5 y - 7) = ´ (5 y - 7) (15 y + 20 - 15 y + 21) 41y = = y = I B( y). 41 \ ( f o g) = I B. Hence, ( g o f ) = I A and ( f o g) = I B. EXAMPLE 11

Let f : A ® B and g : B ® A such that ( g o f ) = I A . Show that f is one-one and g is onto.

SOLUTION

We have f ( x1) = f ( x 2) Þ Þ Þ Þ

g{ f ( x1)} = g{ f ( x 2)} ( g o f )( x1) = ( g o f )( x 2) I A ( x1) = I A ( x 2) x1 = x 2.

SSS Mathematics for Class 12 42

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\ f is one-one. In order to show that g is onto, let a Î A and let f ( a) = b Î B. Then, g( b) = g[ f ( a)] = ( g o f ) ( a) [Q g o f = I A ]. = I A ( a) Thus, for each a Î A, there exists b Î B such that g( b) = a. \ g is onto.

EXERCISE 2B 1. Let A = {1, 2, 3, 4}. Let f : A ® A and g : A ® A, defined by f = {(1, 4), (2, 1), (3, 3), (4, 2)} and g = {(1, 3), (2, 1), (3, 2), (4, 4)}. Find (i) g o f (ii) f o g (iii) f o f . 2. Let f : {3, 9, 12} ® {1, 3, 4} and g : {1, 3, 4, 5} ® {3, 9} be defined as f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3), (4, 9), (5, 9)}. Find (i) ( g o f ) (ii) ( f o g). 3. Let f : R ® R : f ( x) = x 2 and g : R ® R : g( x) = ( x + 1). Show that ( g o f ) ¹ ( f o g). 4. Let f : R ® R : f ( x) = ( 2x + 1) and g : R ® R : g( x) = ( x 2 - 2). Write down the formulae for (i) ( g o f ) (ii) ( f o g) (iii) ( f o f )

(iv) ( g o g).

2

5. Let f : R ® R : f ( x) = ( x + 3 x + 1) and g : R ® R : g( x) = ( 2x - 3). Write down the formulae for (i) g o f (ii) f o g (iii) g o g. 6. Let f : R ® R : f ( x) = |x|, prove that f o f = f . 7. Let f : R ® R : f ( x) = x 2 , g : R ® R : g( x) = tan x and h : R ® R : h( x) = log x. Find a formula for h o ( g o f ). p Show that [h o ( g o f )] = 0. 4 8. Let f : R ® R : f ( x) = ( 2x - 3) and g : R ® R : g( x) = Show that ( f o g) = I R = ( g o f ). 9. Let f : Z ® Z : f ( x) = 2x. Find g : Z ® Z : g o f = I Z. 10. Let f : N ® N : f ( x) = 2x , g : N ® N : g( y) = 3 y + 4 and h : N ® N : h(z) = sin z. Show that h o ( g o f ) = ( h o g) o f .

1 ( x + 3). 2

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Functions

43

11. If f be a greatest integer function and g be an absolute value function, find æ 4ö æ -3 ö the value of( f o g) ç ÷ + ( g o f ) ç ÷ × [CBSE 2007] è 3ø è 2ø 12. Let f : R ® R : f ( x) = x 2 + 2 and g : R ® R : g( x) = and g o f and hence find ( f o g)( 2) and ( g o f )( -3).

x , x ¹ 1. Find f o g x -1 [CBSE 2014]

ANSWERS (EXERCISE 2B)

1.

(i) ( g o f ) = {(1, 4), (2, 3), (3, 2), (4, 1)} (ii) ( f o g) = {(1, 3), (2, 4), (3, 1), (4, 2)} (iii) ( f o f ) = {(1, 2), (2, 4), (3, 3), (4, 1)}

2.

(i) ( g o f ) = {(3, 3), (9, 3), (12, 9)} (ii) ( f o g) = {(1, 1), (3, 1), (4, 3), (5, 3)}

4.

(i) ( g o f ) ( x) = ( 4x 2 + 4x - 1)

(ii) ( f o g) ( x) = ( 2x 2 - 3)

(iii) ( f o f ) ( x) = ( 4x + 3)

(iv) ( g o g) ( x) = ( x 4 - 4x 2 + 2)

(i) ( g o f ) ( x) = ( 2x 2 + 6x - 1) (iii) ( g o g) ( x) = ( 4x - 9)

(ii) ( f o g) ( x) = ( 4x 2 - 6x + 1)

5.

7. [h o ( g o f )] ( x) = log (tan x 2) 12. ( f o g)( x) =

x2 ( x - 1)

2

11. 2

+ 2, ( g o f )( x) =

( f o g)( 2) = 6, ( g o f )( -3) =

x2 + 2 x2 + 1

;

11 × 10

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 2B) 9. Let x Î Z. Then, ( g o f ) = I Z Þ ( g o f ) ( x ) = I Z( x ) Þ g [ f ( x )] = x Þ g( 2x) = x 1 Þ g ( y ) = y [where 2x = y]. 2 1 Thus, g : Z ® Z : g ( y ) = y. 2 ì æ -3 ö ü æ -3 ö 11. ( f o g )ç ÷ = f íg ç ÷ý = f è 2 ø î è 2 øþ

æ½-3½ö æ 3 ö é 3ù ÷ = f ç ÷ = ê ú = 1. ç è 2 ø ë2û è½ 2 ½ø

ì æ 4 öü æ 4ö é 4ù ( g o f ) ç ÷ = gí f ç ÷ ý = g ê ú = g(1) = | 1 | = 1. 3 3 è ø ë 3û î è øþ Required sum = (1 + 1) = 2.

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Senior Secondary School Mathematics for Class 12

x2 æ x ö 12. ( f o g )( x ) = f {g ( x )} = f ç + 2. ÷= x 1 è ø ( x - 1) 2

( g o f )( x) = g{ f ( x)} = g( x 2 + 2) =

x2 + 2 x2 + 2 - 1

=

x2 + 2 x2 + 1

×

Invertible Function Let f : A ® B. If there exists a function g : B ® A such that g o f = I A and f o g = I B then f is called an invertible function and g is called the inverse of f. We write, f -1 = g. Clearly, f

REMARK

o f = I A and f o f -1 = I B .

Let f : R ® R : f ( x) = 2x + 3. Let y = f ( x). Then, y = f ( x) Þ y = 2x + 3 1 Þ x = ( y - 3) 2 1 -1 Þ f ( y) = ( y - 3) [Q y = f ( x) Þ x = f -1( y)]. 2 Thus, we define: 1 f -1 : R ® R : f -1( y) = ( y - 3). 2

Example

If f : A ® B is one-one onto then prove that f is an invertible function. Let y Î B. Then, f being one-one onto, there exists a unique x Î A such that f ( x) = y. We define g : B ® A : g( y) = x. Then, ( g o f ) ( x) = g[ f ( x)] = g( y) [Q f ( x) = y] =x [Q g( y) = x] = I A ( x). \ ( g o f ) = IA . ( f o g) ( y) = f [ g( y)] = f ( x) [Q g( y) = x] =y [Q f ( x) = y] = I B( y). \ ( f o g) = I B . Hence, f is invertible and f -1 = g.

THEOREM 1 PROOF

-1

SSS Mathematics for Class 12 45

Functions

45

If f : A ® B is an invertible function, then prove that f is one-one onto. Let f : A ® B be an invertible function. Then, there exists a function g : B ® A such that g o f = I A and f o g = I B. Now, f ( x1) = f ( x 2) Þ g{ f ( x1)} = g{ f ( x 2)} Þ ( g o f ) ( x1) = ( g o f ) ( x 2) Þ I A ( x1) = I A ( x 2) [Q g o f = I A ] Þ x1 = x 2. \ f is one-one. Let y Î B. Then, g( y) Î A. Let g( y) = x. Then, g( y) = x Þ f { g( y)} = f ( x) Þ ( f o g) ( y) = f ( x) Þ I B( y) = f ( x) [Q f o g = I B] Þ y = f ( x). Thus, for each y Î B there exists x Î A such that y = f ( x). \ f is onto. Hence, f is one-one onto.

THEOREM 2 PROOF

REMARK

f is invertible Û f is one-one onto. SOLVED EXAMPLES

EXAMPLE 1

Let f : R ® R : f ( x) = 4x + 3 for all x Î R. Show that f is invertible and find f -1 .

SOLUTION

We have f ( x1) = f ( x 2) Þ 4x1 + 3 = 4x 2 + 3 Þ 4x1 = 4x 2 Þ x1 = x 2. \ f is one-one. ( y - 3) Again, y = 4x + 3 Þ x = × 4 Now, if y Î R (codomain of f ) then there exists x = ü æ y - 3 ö ì ( y - 3) such that f ( x) = f ç + 3 ý = y. ÷ = í4 × 4 þ è 4 ø î \ f is onto. Thus, f is one-one onto and therefore invertible. Now, y = f ( x) Þ y = 4x + 3 ( y - 3) Þ x= 4

( y - 3) ÎR 4

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Senior Secondary School Mathematics for Class 12

Þ

f -1( y) =

( y - 3) 4

[Q f ( x) = y Þ x = f -1( y)].

Thus, we define: f -1 : R ® R : f -1( y) =

( y - 3) for all y Î R. 4

EXAMPLE 2

Let R + be the set of all positive real numbers. Let f : R + ® [4, ¥ [ : f ( x) = x 2 + 4. Show that f is invertible and find f -1.

SOLUTION

We have f ( x1) = f ( x 2) Þ x12 + 4 = x 22 + 4 Þ x12 = x 22 Þ x12 - x 22 = 0 Þ ( x1 - x 2)( x1 + x 2) = 0 Þ x1 - x 2 = 0 [Q ( x1 + x 2) ¹ 0] Þ x1 = x 2. \ f is one-one. Now, y = x 2 + 4 Þ x = y - 4. For each y Î [4, ¥ [ there exists x = y - 4 in R + such that f ( x) = f ( y - 4) = ( y - 4) 2 + 4 = ( y - 4) + 4 = y. \ f is onto. Thus, f is one-one onto and therefore invertible. Now, y = f ( x) Þ y = x 2 + 4 Þ x= y-4 Þ f -1( y) = y - 4. \ f -1 : [4, ¥ [ ® R + : f -1( y) = y - 4.

EXAMPLE 3

Let R + be the set of all positive real numbers. Let f : R + ® R + : f ( x) = ex for all x Î R + . Show that f is invertible and hence find f -1 .

SOLUTION

f is one-one, since f ( x1) = f ( x 2) Þ ex1 = ex 2 Þ x1 = x 2. Now, for each y Î R + , there exists a positive real number, namely log y such that f (log y) = e

log y

= y.

\ f is onto. Thus, f is one-one onto and hence invertible.

SSS Mathematics for Class 12 47

Functions

47

We define: f -1 : R + ® R + : f -1( y) = log y for all y Î R + . EXAMPLE 4

pü -p ì Let A = íx : x Î R , £ x £ ý and B = {y : y Î R , - 1 £ y £ 1}. Show 2 2þ î that the function f : A ® B : f ( x) = sin x is invertible and hence find f -1 .

SOLUTION

é -p p ù Here, A = ê , ú and B = [-1, 1]. 2û ë 2 Also, f : A ® B : f ( x) = sin x. f is one-one, since f ( x1) = f ( x 2) Þ sin x1 = sin x 2 Þ x1 = x 2 \

ì é -p p ù ü , úý× íQ x1 , x 2 Î ê 2ûþ ë 2 î

f is one-one.

Also, range ( f ) = [-1, 1] = B. So, f is onto. Thus, f is one-one onto and hence invertible. Now, y = f ( x) Þ y = sin x Þ x = sin -1 y Þ f -1( y) = sin -1 y. Thus, we define: é - p p ù -1 f -1 : [-1, 1] ® ê , ú : f ( y) = sin -1 y. 2û ë 2 EXAMPLE 5

Let f : N ® Y : f ( x) = x 2 , where Y = range ( f ). Show that f is invertible and find f

SOLUTION

-1

.

We have f ( x1) = f ( x 2) Þ x12 = x 22 Þ x12 - x 22 = 0 Þ ( x1 - x 2)( x1 + x 2) = 0 Þ x1 - x 2 = 0 [Q x1 + x 2 ¹ 0] Þ x1 = x 2. \ f is one-one. Since range ( f ) = Y, so f is onto. Thus, f is one-one onto and therefore invertible. Let y Î Y. Then, there exists x Î N such that f ( x) = y.

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Senior Secondary School Mathematics for Class 12

Now, y = f ( x) Þ y = x 2 Þ x= y Þ f -1( y) = y [Q f ( x) = y Þ x = f -1( y)]. Thus, we define f -1 : Y ® N : f -1( y) = y . EXAMPLE 6

Let f : [-1, 1] ® Y : f ( x) =

x , x ¹ - 2 and Y = range ( f ). Show ( x + 2)

that f is invertible and find f -1 . SOLUTION

We have f ( x1) = f ( x 2) Þ Þ Þ Þ Þ

x1 x2 = x1 + 2 x 2 + 2 x1 x 2 + 2x1 = x1 x 2 + 2x 2 2( x1 - x 2) = 0 x1 - x 2 = 0 x1 = x 2.

\ f is one-one. Since range ( f ) = Y, so f is onto. Thus, f is one-one onto and therefore invertible. Let y Î Y. Then, there exists x Î [-1, 1] such that f ( x) = y. x Now, y = f ( x) Þ y = ( x + 2) 2y Þ x= (1 - y) 2y Þ f -1( y) = × (1 - y) Thus, we define: 2y f -1 : [-1, 1] ® Y : f -1( y) = , y ¹ 1. (1 - y) EXAMPLE 7

SOLUTION

Let f : N ® Y : f ( x) = 4x 2 + 12x + 15 and Y = range ( f ). Show that f is invertible and find f -1 . [CBSE 2013C] f is one-one, since f ( x1) = f ( x 2) Þ 4x12 + 12x1 + 15 = 4x 22 + 12x 2 + 15 Þ 4( x12 - x 22) + 12( x1 - x 2) = 0 Þ Þ Þ Þ

( x12 - x 22) + 3( x1 - x 2) = 0 ( x1 - x 2)( x1 + x 2 + 3) = 0 [Q x1 + x 2 + 3 ¹ 0] x1 - x 2 = 0 x1 = x 2.

SSS Mathematics for Class 12 49

Functions

49

Also, range ( f ) = Y. So, f is onto. Thus, f is one-one onto and therefore invertible. Let y Î Y. Then, f being onto, there exists x such that y = f ( x). Now, y = f ( x) Þ y = 4x 2 + 12x + 15 Þ y = ( 2x + 3) 2 + 6 Þ ( 2x + 3) = y - 6 1 ( y - 6 - 3) 2 1 Þ f -1( y) = ( y - 6 - 3). 2 Thus, we define: 1 f -1 : Y ® N : f -1( y) = ( y - 6 - 3). 2 Þ x=

EXAMPLE 8

Let f : R ® R : f ( x) = 10x + 7. Find the function g : R ® R such that g o f = f o g = IR . [CBSE 2011]

SOLUTION

Clearly, g = f

-1

… (i)

Now, f ( x1) = f ( x 2) Þ 10x1 + 7 = 10x 2 + 7 Þ 10x1 = 10x 2 Þ x1 = x 2. \ f is one-one. Now, y = f ( x) Þ y = 10x + 7 ( y - 7) Þ x= × 10 Clearly, for each y Î R (codomain of f ) there exists x Î R such that ü æy -7ö ì æy -7ö f ( x) = f ç ÷ = í10 × ç ÷ + 7 ý = y. 10 10 è ø î è ø þ \

f is onto.

Thus, f is one-one onto and therefore, f

-1

exists.

y -7 We define: f -1 : R ® R : f -1( y) = × 10 y -7 [using (i)]. Hence, g : R ® R : g( y) = 10 EXAMPLE 9

SOLUTION

ì(n - 1), when n is odd Let f : W ® W : f (n) = í î(n + 1), when n is even. Show that f is invertible. Find f -1 . Let f (n1) = f (n2).

[CBSE 2012]

SSS Mathematics for Class 12 50

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Senior Secondary School Mathematics for Class 12

When n1 is odd and n2 is even In this case, f (n1) = f (n2) Þ n1 - 1 = n2 + 1 Þ n1 - n2 = 2. If n1 is odd and n2 is even, then (n1 - n2) ¹ 2. Thus, we arrive at a contradiction. So, in this case, f (n1) ¹ f (n2). Similarly, when n1 is even and n2 is odd, then f (n1) ¹ f (n2). Case 2 When n1 and n2 are both odd In this case, f (n1) = f (n2) Þ n1 - 1 = n2 - 1 Þ n1 = n2. Case 3 When n1 and n2 are both even In this case, f (n1) = f (n2) Þ n1 + 1 = n2 + 1 Þ n1 = n2. Thus, from all the cases, we get f (n1) = f (n2) Þ n1 = n2. \ f is one-one. Now, we show that f is onto. Let n Î W. Case 1 When n is odd In this case, (n - 1) is even and f (n - 1) = (n - 1) + 1 = n. … (i) Case 2 When n is even In this case, (n + 1) is odd and f (n + 1) = (n + 1) - 1 = n. … (ii) Thus, each n Î W has its pre-image in W. \ f is onto. Thus, f is one-one onto and hence invertible. Clearly, we have ì(n - 1), when n is odd [using (i) and (ii)]. f -1(n) = í î(n + 1), when n is even Case 1

EXAMPLE 10

Let A = {1, 2, 3} and let f : A ® A, defined by f = {(1, 2), ( 2, 3), ( 3 , 1)}. Find f -1 , if it exists.

SOLUTION

We have f (1) = 2, f ( 2) = 3 and f ( 3) = 1. Dom ( f ) = {1, 2, 3} = A and range ( f ) = { 1, 2, 3} = A. Clearly, different elements in A have different images. \ f is one-one. Range ( f ) = A Þ f is onto. Thus, f is one-one onto and therefore invertible.

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Functions

Now, f (1) = 2, f ( 2) = 3 and f ( 3) = 1 Þ f -1( 2) = 1, f -1( 3) = 2 and f -1(1) = 3. Hence, f -1 = {(2, 1), (3, 2), (1, 3)}.

Some Results on Invertible Functions Prove that an invertible function has a unique inverse. Let f : A ® B, which is one-one onto and therefore, invertible. If possible, let it have two inverses, say g and h. Then, ( f o g) = I B and ( f o h) = I B [each = I B( y)] Þ ( f o g) ( y) = ( f o h) ( y) Þ f { g( y)} = f {h( y)} for all y Î B [Q f is one-one]. Þ g( y) = h( y) for all y Î B \ g = h. Hence, f has a unique inverse.

THEOREM 1 PROOF

THEOREM 2

PROOF

Let f be an invertible function. Then, prove that ( f -1 ) -1 = f .

Let f : A ® B, which is invertible. In order to prove that ( f -1) -1 = f , it is sufficient to show that f -1 o f = I A and f o f -1 = I B. Clearly f : A ® B is one-one onto. \ f -1 : B ® A is one-one onto. Let x Î A and let f ( x) = y. Then, f -1( y) = x.

\ ( f -1 o f ) ( x) = f -1{ f ( x)} = f -1( y) [Q f ( x) = y] =x = I A ( x). \ f

-1

o f = IA .

Again, let y Î B. Then, f being onto, there exists x Î A such that f ( x) = y and therefore, f -1( y) = x.

51

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Senior Secondary School Mathematics for Class 12

\ ( f o f -1) ( y) = f { f -1( y)} = f ( x) [Q f -1( y) = x] =y = I B( y). \ f o f -1 = I B . Thus, f -1 o f = I A and f o f -1 = I B. Hence, ( f -1) -1 = f . THEOREM 3

PROOF

Let f : A ® B and g : B ® C be one-one onto functions. Prove that ( g o f ) : A ® C which is one-one onto and ( g o f ) -1 = f -1 o g -1.

Let f : A ® B be one-one onto and g : B ® C be one-one onto.

We first show that g o f is one-one onto. ( g o f ) is one-one, since ( g o f ) ( x1) = ( g o f ) ( x 2) Þ g{ f ( x1)} = g{ f ( x 2)} [Q g is one-one] Þ f ( x1) = f ( x 2) [Q f is one-one]. Þ x1 = x 2 Let z Î C. Then, g being onto, there exists y Î B such that g( y) = z. Now, f being onto, there exists x Î A such that f ( x) = y. \ z = g( y) = g{ f ( x)} [Q y = f ( x)] = ( g o f ) ( x). Thus, for each z Î C, there exists x Î A such that ( g o f ) ( x) = z. \ ( g o f ) is onto. Thus, ( g o f ) is one-one onto. Now, f ( x) = y Þ f -1( y) = x. And, g( y) = z Þ g -1(z) = y. Also, ( g o f ) ( x) = z Þ ( g o f ) -1(z) = x. \ ( f -1 o g -1) (z) = f -1{ g -1(z)} = f -1( y)

[Q g -1(z) = y]

=x

[Q f -1( y) = x]

= ( g o f ) -1(z). Hence, ( g o f )

-1

= ( f - 1 o g - 1 ).

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Functions

53

EXERCISE 2C Very-Short-Answer Questions 1. Prove that the function f : R ® R : f ( x) = 2x is one-one and onto. 2. Prove that the function f : N ® N : f ( x) = 3 x is one-one and into. 3. Show that the function f : R ® R : f ( x) = x 2 is neither one-one nor onto. 4. Show that the function f : N ® N : f ( x) = x 2 is one-one and into. 5. Show that the function f : R ® R : f ( x) = x 4 is neither one-one nor onto. 6. Show that the function f : Z ® Z : f ( x) = x 3 is one-one and into. 7. Let R 0 be the set of all nonzero real numbers. Then, show that the function 1 f : R 0 ® R 0 : f ( x) = is one-one and onto. x 8. Show that the function f : R ® R : f ( x) = 1 + x 2 is many-one into. 9. Let f : R ® R : f ( x) =

2x - 7 be an invertible function. Find f -1. 4 [CBSE 2008C]

10. Let f : R ® R : f ( x) = 10x + 3. Find f

-1

.

ì 1, if x is rational 11. f : R ® R : f ( x) = í î-1, if x is rational. Show that f is many-one and into. 12. Let f ( x) = x + 7 and g( x) = x - 7 , x Î R. Find ( f o g)(7).

[CBSE 2008]

13. Let f : R ® R and g : R ® R defined by f ( x) = x 2 and g( x) = ( x + 1). Show that g o f ¹ f o g. 14. Let f : R ® R : f ( x) = ( 3 - x 3)1 / 3. Find f o f . 15. Let f : R ® R : f ( x) = 3 x + 2, find f { f ( x)}.

[CBSE 2010] [CBSE 2010C]

16. Let f = {(1, 2), ( 3 , 5), ( 4, 1)} and g = {(1, 3), ( 2, 3), (5 , 1)}. Write down g o f . [CBSE 2014C]

17. Let A = {1, 2, 3, 4} and f = {(1, 4), (2,1) (3, 3), (4, 2)}. Write down ( f o f ). 18. Let f ( x) = 8x 3 and g( x) = x1 / 3. Find g o f and f o g. 19. Let f : R ® R : f ( x) = 10x + 7. Find the function g : R ® R : g o f = f o g = I g . [CBSE 2011]

20. Let A = {1, 2, 3}, B = {4, 5 , 6, 7 } and let f = {(1, 4), ( 2, 5),( 3 , 6)} be a function [CBSE 2011] from A to B. State whether f is one-one.

SSS Mathematics for Class 12 54

54

Senior Secondary School Mathematics for Class 12 ANSWERS (EXERCISE 2C)

1 9. f -1( y) = ( 4y + 7) for all y Î R 2 14. ( f o f )( x) = x

12. 7

16. ( g o f ) = {(1, 3), ( 3 , 1), ( 4, 3)}

10. f -1( y) =

1 ( y - 3) for all y Î R 10

15. f { f ( x)} = ( 9x + 8) 17. f o f = {(1, 2), (2, 4), (3, 3), (4, 1)}

18. ( g o f )( x) = 2x and ( f o g)( x) = 8x 19. g( x) =

1 ( x - 7) for all x Î R 10

20. Yes

HINTS TO THE GIVEN QUESTIONS (EXERCISE 2C) 1. (i) f ( x1 ) = f ( x2 ) Þ 2 x1 = 2 x2 Þ x1 = x2 . So, f is one-one. 1 (ii) Let y = 2 x. Then, x = y. 2 1 Thus for each y in codomain R, there exists y such that 2 1 ö æ1 ö æ f ç y ÷ = ç 2 ´ y ÷ = y. 2 ø è2 ø è \ f is onto. 2. (i) f ( x1 ) = f ( x2 ) Þ 3 x1 = 3 x2 Þ x1 = x2 . So, f is one-one. (ii) If we consider 2 in codomain N , there is no natural number whose image is 2. So, f is into. 3. (i) Clearly, f( 1) = 12 = 1 and f( -1) = ( -1) 2 = 1. So, f is many-one. (ii) If we consider –1 in the codomain R then there is no element in R whose square is –1. \ -1 Î R has no pre-image in R. So, f is many-one into. 4.

(i) f ( x1 ) = f ( x2 ) Þ x12 = x22 Þ x1 = x2 [Q x1 , x2 Î N ]. \ f is one-one. (ii) If we consider 2 in the codomain N , then 2 Î N and f( 2 ) = ( 2 ) 2 = 2 . So, f is into.

5. (i) f( 1) = 14 = 1 and f( -1) = ( -1) 4 = 1. So, f is many-one. (ii) If we consider –1 in the codomain R then there exists no x Î R such that f ( x ) = x 4 = - 1. So, f is into. 6.

(i) Let x1 , x2 Î Z and x1 ¹ x2 . Then, x1 ¹ x2 Þ x13 ¹ x23 Þ f ( x1 ) ¹ f ( x2 ). (ii) Let 2 Î Z. Then, there exists no x Î Z such that x 3 = 2 . Thus, 2 Î Z has no pre-image in Z. So, f is into.

SSS Mathematics for Class 12 55

Functions

7.

(i) f ( x1 ) = f ( x2 ) Þ (ii) y =

55

1 1 = Þ x1 = x2 . So, f is one-one. x1 x2

1 1 Þ x= × x y

Thus, for each y in codomain R 0 , there exists

1 in domain R 0 such that y

æ 1ö 1 = y. So, f is onto. f çç ÷÷ = y æ è ø ç 1 ö÷ çy÷ è ø 8. (i) f( -1) = 1 + ( -1) 2 = 2 and f( 1) = ( 1 + 12 ) = 2 . So, f is many-one. (ii) y = ( 1 + x 2 ) Þ x = y - 1. So, when y < 1, then y - 1 is imaginary. In particular, 0 Î R has no pre-image in R. \ f is into. 2x - 7 1 9. y = Þ x = ( 4y + 7 ) 4 2 4y + 7 for all y Î R. Þ f -1 ( y ) = 2 y- 3 10. y = 10 x + 3 Þ x = 10 ( y - 3) -1 Þ f ( y) = × 10 11.

(i) Since all rationals have the same image, namely 1, so f is many-one. (ii) Range ( f ) = {-1 1} Ì R. So, f is into.

12. ( f o g )(7 ) = f {g (7 )} = f (7 - 7 ) = f ( 0 ) = ( 0 + 7 ) = 7 . 13. ( g o f )( x ) = g {f ( x )} = g ( x 2 ) = ( x 2 + 1). ( f o g )( x ) = f {g ( x )} = f ( x + 1) = ( x + 1) 2 . Hence, g o f ¹ f o g . 14. ( f o f )( x ) = f {f ( x )} = f {( 3 - x 3 )1/3 } = f ( y ), where y = ( 3 - x 3 )1/3 = ( 3 - y 3 )1/3 = { 3 - ( 3 - x 3 )}1/3 = ( x 3 )1/3 = x. \ ( f o f )( x ) = x. 15. f {f ( x )} = f ( 3 x + 2 ) = 3( 3 x + 2 ) + 2 = ( 9 x + 8 ). 16. Dom ( g o f ) = Dom ( f ) = {1, 3 , 4}. ( g o f )( 1) = g {f ( 1)} = g ( 2 ) = 3. ( g o f )( 3 ) = g {f ( 3 )} = g (5 ) = 1. ( g o f )( 4 ) = g {f ( 4 )} = g ( 1) = 3. \

g o f = {( 1, 3 ), ( 3 , 1), ( 4 , 3 )}.

17. ( f o f )( 1) = f {f ( 1)} = f ( 4 ) = 2. ( f o f )( 2 ) = f {f ( 2 )} = f ( 1) = 4.

SSS Mathematics for Class 12 56

56

Senior Secondary School Mathematics for Class 12 ( f o f )( 3 ) = f {f ( 3 )} = f ( 3 ) = 3. ( f o f )( 4 ) = f {f ( 4 )} = f ( 2 ) = 1. \

f o f = {( 1, 2 ), ( 2 , 4 ), ( 3 , 3 ), ( 4 , 1)}.

18. ( g o f )( x ) = g [f ( x )] = g ( 8 x 3 ) = g ( y ), where y = 8 x 3 =y

1

3

= ( 8x 3 )

1

( f o g )( x ) = f [g ( x )] = f ( x = 8 y 3 = 8( x

1

= 2 x.

3 1

3)

3 )3

= f ( y ), where y = x

(

= 8x

1

3

) = 8x.

1 ´3 3

19. g o f = I R Þ ( g o f )( x ) = I R ( x ) = x Þ g {f ( x )} = x Þ g {10 x + 7 } = x. (y - 7) Put 10 x + 7 = y. Then, x = × 10 (y - 7) × \ g( y) = 10 1 Hence, g : R ® R : g ( x ) = ( x - 7 ) for all x Î R. 10 20. f ( 1) = 4 , f ( 2 ) = 5 and f( 3 ) = 6. Thus, different elements in A have different images in B. Hence, f is one-one.

EXERCISE 2D 1. Let A = {2, 3, 4, 5} and B = {7, 9, 11, 13}, and let f = {(2, 7), (3, 9), (4, 11), (5, 13)}. Show that f is invertible and find f -1 . 2. Show that the function f : R ® R : f ( x) = 2x + 3 is invertible and find f -1 . 3. Let f : Q ® Q : f ( x) = 3 x - 4. Show that f is invertible and find f -1 . 1 4. Let f : R ® R : f ( x) = ( 3 x + 1). Show that f is invertible and find f -1 . 2 ( 4x + 3) 2 2 5. If f ( x) = , x ¹ , show that ( f o f )( x) = x for all x ¹ × ( 6x - 4) 3 3 Hence, find f -1 .

4x + 3 ì2ü is 6. Show that the function f on A = R - í ý, defined as f ( x) = 6x - 4 î3þ one-one and onto. Hence, find f -1. [CBSE 2013] ì -4 ü 7. Show that the function f on A = R - í ý into itself, defined by î3þ 4x is one-one and onto. Hence, find f -1 . f ( x) = ( 3 x + 4)

SSS Mathematics for Class 12 57

Functions

57

8. Let R + be the set of all positive real numbers. Show that the function f : R + ® [-5 , ¥ [ : f ( x) = ( 9x 2 + 6x - 5) is invertible. Find f -1 . 9. Let f : N ® R : f ( x) = 4x 2 + 12x + 15. Show that f : N ® range ( f ) is invertible. Find f -1.

[CBSE 2010, ’13C]

10. Let A = R - {2} and B = R - {1}. If f : A ® B : f ( x) = one-one and onto. Hence, find f -1.

x -1 , show that f is x-2 [CBSE 2013]

11. Let f and g be two functions from R into R, defined by f ( x) =| x |+ x and [CBSE 2014C] g( x) =| x |- x for all x Î R. Find f o g and g o f .

ANSWERS (EXERCISE 2D)

1. f -1 = {(7, 2), (9, 3), (11, 4), (13, 5)} 1 ( y + 4) 3 ( 4y + 3) 5. f -1( y) = ( 6y - 4) 3. f -1( y) =

7. f -1( y) = 9. f -1( y) =

( 4y) ( 4 - 3 y) y-6- 3 2

1 ( y - 3) 2 ( 2y - 1) 4. f -1( y) = 3 y + 3) ( 4 6. f -1( y) = 6y - 4 2. f -1( y) =

8. f -1( y) = 10. f -1( y) =

y + 6 -1 3 2y - 1 y -1

11. ( f o g)( x) =| x | - x and ( g o f )( x) = | x | - x HINTS TO SOME SELECTED QUESTIONS (EXERCISE 2D) 1.

Clearly f( 2 ) = 7 , f( 3 ) = 9 , f( 4 ) = 11 and f(5 ) = 13. Thus, different elements in A have different images in B. So, f is one-one. Range ( f ) = {7, 9, 11, 13} = B. So, f is onto. f -1 = {(7, 2), (9, 3), (11, 4), (13, 5)}.

2.

f ( x1 ) = f ( x2 ) Þ 2 x1 + 3 = 2 x2 + 3 Þ 2 x1 = 2 x2 Þ x1 = x2 . \

f is one-one. y- 3 Î R such that 2 æ y - 3 ö ì ( y - 3) ü f ( x) = f ç + 3 ý = y. ÷ = í2 × 2 è 2 ø î þ

If y Î R then there exists x =

\

f is onto.

SSS Mathematics for Class 12 58

58

Senior Secondary School Mathematics for Class 12 y = f ( x) Þ y = 2x + 3 1 1 Þ x = ( y - 3 ) Þ f -1 ( y ) = ( y - 3 ). 2 2

ì 4x + 3 ü 5. ( f o f )( x ) = f {f ( x )} = f í ý= î 6x - 4 þ = \ 6.

æ 4x + 3 ö 4ç ÷+ 3 è 6x - 4 ø æ 4x + 3 ö 6ç ÷-4 è 6x - 4 ø

( 16 x + 12 + 18 x - 12 ) 34 x = = x = I( x ). ( 24 x + 18 - 24 x + 16 ) 34

f o f = I Þ f -1 = f .

f ( x1 ) = f ( x2 ) Þ

4 x1 + 3 4 x2 + 3 = 6 x1 - 4 6 x2 - 4

Þ ( 4 x1 + 3 )( 6 x2 - 4 ) = ( 6 x1 - 4 )( 4 x2 + 3 ) Þ 24 x1 x2 - 16 x1 + 18 x2 - 12 = 24 x1 x2 + 18 x1 - 16 x2 - 12 Þ 34 x1 = 34 x2 Þ x1 = x2 . \ f is one-one. Let y be an arbitrary element of A. Then, 4x + 3 f ( x) = y Þ =y 6x - 4 Þ ( 6 x - 4 )y = 4 x + 3 Þ 6 xy - 4 y = 4 x + 3 Þ 6 xy - 4 x = 4 y + 3 Þ x( 6 y - 4 ) = ( 4 y + 3 ) 4y + 3 Þ x= × 6y - 4 Thus, for each y Î A , there exists x =

æ 4y + 3 ö ÷÷ = f ( x ) = f çç è 6y - 4 ø = \

4y + 3 such that 6y - 4

æ 4y + 3 ö ÷÷ + 3 4 × çç è 6y - 4 ø æ 4y + 3 ö ÷÷ - 4 6 × çç è 6y - 4 ø

16 y + 12 + 18 y - 12 34 y = = y. 24 y + 18 - 24 y + 16 34

f is onto.

Now, f ( x ) = y Þ x = f -1 ( y ). \ 7.

f -1 ( y ) =

4y + 3 , for all y Î A. 6y - 4

f ( x1 ) = f ( x2 ) Þ

x1 x2 4 x1 4 x2 = Þ = 3 x1 + 4 3 x2 + 4 3 x1 + 4 3 x2 + 4

Þ x1 ( 3 x2 + 4 ) = x2 ( 3 x1 + 4 ) Þ 3 x1 x2 + 4 x1 = 3 x1 x2 + 4 x2 \

Þ 4 x1 = 4 x2 Þ x1 = x2 . f is one-one.

SSS Mathematics for Class 12 59

Functions

59

Let y be an arbitrary element of A. Then, 4x f ( x) = y Þ =y 3x + 4 Þ 3 xy + 4 y = 4 x Þ 4 x - 3 xy = 4 y 4y Þ x( 4 - 3 y ) = 4 y Þ x = × ( 4 - 3y) Thus, for each y Î A , there exists an x Î A such that æ 4y ö ÷÷ = f ( x ) = f çç è 4 - 3y ø = \

æ 4y ö ÷÷ 4 × çç è 4 - 3y ø æ 4y ö ÷÷ + 4 3 × çç è 4 - 3y ø

16 y 16 y = = y. 12 y + 16 - 12 y 16

f is onto.

Now, f ( x ) = y Þ x = f -1 ( y ). \

f -1 ( y ) =

4y × 4 - 3y y+ 6 -1

Þ f -1 ( y ) =

y+ 6 -1

8.

y = ( 3 x + 1) 2 - 6 Þ x =

9.

f ( x1 ) = f ( x2 ) Þ 4 x12 + 12 x1 + 15 = 4 x22 + 12 x2 + 15

3

3

×

Þ 4( x12 - x22 ) + 12( x1 - x2 ) = 0 Þ ( x12 - x22 ) + 3( x1 - x2 ) = 0 Þ ( x1 - x2 )( x1 + x2 + 3 ) = 0 Þ x1 - x2 = 0 Þ x1 = x2 . Since f : N ® range ( f ), so f is onto. Now 4 x 2 + 12 x + 15 = y Þ ( 2 x + 3 ) 2 + 6 = y Þ ( 2x + 3) = y - 6 Þ x = \

f -1 ( y ) =

y-6 - 3 2

y-6 - 3 2

×

×

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 1. f : N ® N : f ( x) = 2x is (a) one-one and onto (c) many-one and onto

(b) one-one and into (d) many-one and into

2. f : N ® N : f ( x) = x 2 + x + 1 is (a) one-one and onto (c) many-one and onto

(b) one-one and into (d) many-one and into

SSS Mathematics for Class 12 60

60

Senior Secondary School Mathematics for Class 12

3. f : R ® R : f ( x) = x 2 is (a) one-one and onto

(b) one-one and into

(c) many-one and onto

(d) many-one and into

3

4. f : R ® R : f ( x) = x is (a) one-one and onto

(b) one-one and into

(c) many-one and onto +

+

(d) many-one and into

x

5. f : R ® R : f ( x) = e is (a) many-one and into

(b) many-one and onto

(c) one-one and into

(d) one-one and onto

é -p p ù 6. f : ê , ® [-1, 1] : f ( x) = sin x is ë 2 2 úû (a) one-one and into

(b) one-one and onto

(c) many-one and into

(d) many-one and onto

7. f : R ® R : f ( x) = cos x is (a) one-one and into (c) many-one and into 8. f : C ® R : f (z) = || z is (a) one-one and into

(b) one-one and onto (d) many-one and onto (b) one-one and onto

(c) many-one and into

(d) many-one and onto ( x - 2) is 9. Let A = R - { 3} and B = R - {1}. Then, f : A ® B : f ( x) = ( x - 3) (a) one-one and into

(b) one-one and onto

(c) many-one and into

(d) many-one and onto

ì1 ïï (n + 1), when n is odd 10. Let f : N ® N : f (n) = í 2 ïn , when n is even. ïî 2 Then, f is (a) one-one and into (b) one-one and onto (c) many-one and into

(d) many-one and onto

11. Let A and B be two non-empty sets and let f : ( A ´ B) ® ( B ´ A) : f ( a , b) = ( b , a). Then, f is (a) one-one and onto (b) one-one and into (c) many-one and onto

(d) many-one and into

12. Let f : Q ® Q : f ( x) = ( 2x + 3). Then, f -1( y) = ? (a) ( 2y - 3)

(b)

1 ( 2y - 3)

(c)

1 ( y - 3) 2

(d) none of these

SSS Mathematics for Class 12 61

Functions

61

4x ì4ü ì -4 ü 13. Let f : R - í ý ® R - í ý : f ( x) = × Then, f -1( y) = ? ( 3 x + 4) î3þ î3þ 4y 4y 4y (b) (c) (d) none of these (a) ( 4 - 3 y) ( 4y + 3) ( 3 y - 4) 14. Let f : N ® X : f ( x) = 4x 2 + 12x + 15. Then, f -1( y) = ? 1 1 (b) ( y - 6 - 3) ( y - 4 + 3) 2 2 1 (d) none of these (c) ( y - 4 + 5) 2 ( 4x + 3) 2 15. If f ( x) = , x ¹ then ( f o f )( x) = ? ( 6x - 4) 3 (a)

(b) ( 2x - 3)

(a) x

(c)

4x - 6 3x + 4

(d) none of these

16. If f ( x) = ( x 2 - 1) and g( x) = ( 2x + 3) then ( g o f )( x) = ? (a) ( 2x 2 + 3)

(b) ( 3 x 2 + 2)

(c) ( 2x 2 + 1)

(d) none of these

(c) ( x 2 - 2)

(d) none of these

1ö 1 æ 17. If f ç x + ÷ = x 2 + 2 then f ( x) = ? x ø è x (a) x 2

(b) ( x 2 - 1)

18. If f ( x) = (a)

1 then ( f o f o f )( x) = ? (1 - x)

1 (1 - 3 x)

19. If f ( x) = (a) x

1

3

(b)

x (1 + 3 x)

(c) x

(d) none of these

3 - x 3 then ( f o f )( x) = ?

3

1

(c) (1 - x 3 )

(b) x

(d) none of these

2

20. If f ( x) = x - 3 x + 2 then ( f o f )( x) = ? (a) x 4

(b) x 4 - 6x 3

21. If f ( x) = 8x 3 and g( x) = x (a) x

(b) 2x

1

3

(c) x 4 - 6x 3 + 10x 2 (d) none of these

then ( g o f )( x) = ? (c)

x 2

(d) 3 x 2

æ p 22. If f ( x) = x 2 , g( x) = tan x and h ( x) = log x then {h o ( g o f )} çç è 4 1 1 (d) log (a) 0 (b) 1 (c) x 2

ö ÷ =? ÷ ø p 4

23. If f = {(1, 2), ( 3 , 5), ( 4, 1)} and g = {( 2, 3), (5 , 1), (1, 3)} then ( g o f ) = ? (a) {(3, 1), (1, 3), (3, 4)} (b) {(1, 3), (3, 1), (4, 3)} (c) {(3, 4), (4, 3), (1, 3)} (d) {(2, 5), (5, 2), (1, 5)}

SSS Mathematics for Class 12 62

62

Senior Secondary School Mathematics for Class 12

24. Let f ( x) = 9 - x 2 . Then, dom ( f ) = ? (a) [-3 , 3] (c) [ 3 , ¥ ) 25. Let f ( x)

(b) ( -¥ , - 3] (d) ( -¥ , - 3] È ( 4, ¥ ) x -1 × Then, dom ( f ) = ? x-4

(a) [1, 4) (c) ( -¥ , 4]

(b) [1, 4] (d) ( -¥ , 1] È ( 4, ¥ )

26. Let f ( x) = e

x2 - 1

× log ( x - 1). Then, dom ( f ) = ?

(a) ( -¥ , 1]

(b) [-1, ¥ )

(c) (1, ¥ ) 27. Let f ( x) =

(d) ( -¥ , - 1] È (1, ¥ ) x ( x 2 - 1)

× Then, dom ( f ) = ? (b) R - {1}

(a) R

(c) R - {-1}

(d) R - {-1, 1}

sin -1 x 28. Let f ( x) = × Then, dom ( f ) = ? x (a) ( -1, 1)

(b) [-1, 1] -1

29. Let f ( x) = cos

(d) none of these

2x. Then, dom ( f ) = ? é -1 1 ù (b) ê , ú ë 2 2û

(a) [-1, 1]

(c) [-1, 1] - {0} é -p p ù (c) ê , ë 2 2 úû

é -p p ù (d) ê ú ë 4, 4 û

30. Let f ( x) = cos-1( 3 x - 1). Then, dom ( f ) = ? æ 2ö (a) ç 0, ÷ è 3ø

é 2ù (b) ê 0, ú ë 3û

é -2 2 ù (c) ê , ú ë 3 3û

(d) none of these

31. Let f ( x) = cos x . Then, dom ( f ) = ? é pù (a) ê 0, ú ë 2û é p ù é 3p ù (c) ê 0, ú È ê , 2p ú ë 2û ë 2 û

é 3p ù (b) ê , 2p ú ë 2 û (d) none of these

32. Let f ( x) = log ( 2x - x 2). Then, dom ( f ) = ? (a) (0, 2)

(b) [1, 2]

(c) ( -¥ , 1]

(d) none of these

33. Let f ( x) = x 2. Then, dom ( f ) and range ( f ) are respectively (b) R + and R +

(a) R and R

(c) R and R +

(d) R and R - {0}

34. Let f ( x) = x 3. Then, dom ( f ) and range ( f ) are respectively (b) R + and R +

(a) R and R (c) R and R

+

(d) R + and R

SSS Mathematics for Class 12 63

Functions

35. Let f ( x) = log (1 - x) + (a) (1, ¥ )

x 2 - 1. Then, dom ( f ) = ?

(b) ( -¥ , - 1]

36. Let f ( x) =

1 (1 - x 2)

(a) ( -¥ , 1] 37. Let f ( x) =

(1 + x 2)

(c) [-1, 1)

(d) (0, 1)

× Then, range ( f ) = ? (b) [1, ¥ )

x2

63

(c) [-1, 1]

(d) none of these

× Then, range ( f ) = ?

(a) [1, ¥ )

(b) [0, 1) 1 38. The range of f ( x) = x + is x (a) [-2, 2] (b) [2, ¥ )

(c) [–1, 1]

(d) (0, 1]

(c) ( -¥ , - 2]

(d) none of these

x

39. The range of f ( x) = a , where a > 0 is (a) ] - ¥ , 0]

(b) ] - ¥ , 0)

(c) [0, ¥ )

(d) ( 0, ¥ )

ANSWERS (OBJECTIVE QUESTIONS)

1. (b)

2. (b)

3. (d)

4. (a)

5. (d)

6. (b)

7. (c)

8 (c)

9. (b) 10. (d)

11. (a) 12. (c) 13. (a) 14. (b) 15. (a) 16. (c) 17. (c) 18. (c) 19. (b) 20. (d) 21. (b) 22. (a) 23. (b) 24. (a) 25. (d) 26. (c) 27. (d) 28. (c) 29. (b) 30. (b) 31. (c) 32. (d) 33. (c) 34. (a) 35. (b) 36. (b) 37. (b) 38. (d) 39. (d) HINTS TO SOME SELECTED OBJECTIVE QUESTIONS 1. 2 x = 3 Þ x =

3 Ï N . So, f is into. 2

2. f ( x1 ) = f ( x2 ) Þ x12 + x1 + 1 = x22 + x2 + 1 Þ ( x12 - x22 ) + ( x1 - x2 ) = 0 Þ ( x1 - x2 )( x1 + x2 + 1) = 0 Þ x1 - x2 = 0 Þ x1 = x2 . \ f is one-one. f ( x ) = 1 Þ x 2 + x + 1 = 1 Þ x( x + 1) = 0 Þ x = 0 or x = - 1. And, none of 0 and –1 is in N. So, f is into. 5. f ( x1 ) = f ( x2 ) Þ e x1 = e x2 Þ x1 = x2 . So, f is one-one. For each x Î R + $ log x Î R + s.t. f (log x ) = x. So, f is onto. 7. cos ( 2p - q) = cos q Þ f is many-one. Range ( f ) = [-1, 1] Ì R Þ f is into. 8. i ¹ - i. But f (i ) = f ( -i ) = 1. So, f is many-one. -1 Î R having no pre-image in C. So, f is into.

SSS Mathematics for Class 12 64

64

Senior Secondary School Mathematics for Class 12

9. f ( x1 ) = f ( x2 ) Þ

( x1 - 2 ) ( x2 - 2 ) = Þ x1 = x2 . So, f is one-one. ( x1 - 3 ) ( x3 - 3 )

Let

x-2 3y - 2 = y. Then, x = × Clearly, y ¹ 1 and x ¹ 3. x- 3 y-1

\

f ( x ) = y and so f is onto.

10. f ( 1) = f ( 2 ) shows that f is many-one. If n is odd then ( 2n - 1) is odd and f ( 2n - 1) = n. If n is even then 2n is even and f ( 2n) = n. f is onto. 1 1 12. y = 2 x + 3 Þ x = ( y - 3 ) Þ f -1 ( y ) = ( y - 3 ). 2 2 \

13. y =

4y 4y 4x Þ x= Þ f -1 ( y ) = × 3x + 4 ( 4 - 3y) ( 4 - 3y)

14. y = 4 x 2 + 12 x + 15 = ( 2 x + 3 ) 2 + 6 Þ x = \

f

15. f ( x ) =

-1

( y) =

1 ( y - 6 - 3) 2

1 ( y - 6 - 3 ). 2

4x + 3 = y (say). 6x - 4

4y + 3 Then f ( y ) = = 6y - 4

æ 4x + 3 ö ÷÷ + 3 4 çç 34 x è 6x - 4 ø = = x. 34 æ 4x + 3 ö ÷÷ - 4 6 çç è 6x - 4 ø

Þ f [f ( x )] = x Þ ( f o f )( x ) = x. 16. ( g o f )( x ) = g [f ( x )] = g ( x 2 - 1) = 2( x 2 - 1) + 3 = ( 2 x 2 + 1). 17. Let x +

1 = z. Then, x 2

1ö æ 1 ö æ 1ö æ f ( z) = f ç x + ÷ = ç x 2 + 2 ÷ = ç x + ÷ - 2 = ( z 2 - 2 ). xø è xø è x ø è Þ f ( x ) = ( x 2 - 2 ). æ 1 ö 1- x x - 1 1 ÷÷ = 18. ( f o f )( x ) = f {f ( x )} = f çç = = -x x 1 ö è 1 - x ø æç ÷ 1 ç 1 - x ÷ø è 1 æ x - 1ö = x. Þ {f o ( f o f )} ( x ) = f {( f o f )( x )] = f ç ÷= è x ø 1- x - 1 x 1

19. ( f o f )( x ) = f {f ( x )} - {( 3 - x 3 ) 3 } = f ( y ), where y = ( 3 - x 3 ) = ( 3 - y3 )

1

3

= [ 3 - { 3 - x 3 }]

1

3

= ( x3 )

1

3

= x.

1

3

SSS Mathematics for Class 12 65

Functions 20. ( f o f )( x ) = f {f ( x )} = f ( x 2 - 3 x + 2 ) = f ( y ), where y = ( x 2 - 3 x + 2 ) = y 2 - 3 y + 2 = ( x 2 - 3 x + 2 ) 2 - 3( x 2 - 3 x + 2 ) + 2 = ( x 4 - 6 x 3 + 10 x 2 - 3 x ). 21. ( g o f )( x ) = g [f ( x )] = g ( 8 x 3 ) = ( 8 x 3 )

1

3

= 2 x.

22. {h o ( g o f )}( x ) = ( h o g ){f ( x )} = ( h o g )( x 2 ) = h{g ( x 2 )} = h(tan x 2 ) = log (tan x 2 ). \ {h o ( g o f )}

p pö æ = log ç tan ÷ = log 1 = 0. 4 4ø è

23. Dom ( g o f ) = dom ( f ) = {1, 3 , 4}. ( g o f )( 1) = g {f ( 1)} = g ( 2 ) = 3 , ( g o f )( 3 ) = g {f ( 3 )} = g (5 ) = 1 ( g o f )( 4 ) = g {f ( 4 )} = g ( 1) = 3 \

g o f = {( 1, 3 ), ( 3 , 1), ( 4 , 3 )}.

24. f ( x ) is defined only when 9 - x 2 ³ 0 Þ x 2 £ 9 Þ - 3 £ x £ 3. \

dom ( f ) = [-3 , 3].

25. f ( x ) is defined when x - 4 ¹ 0 and

x-1 ³0 x-4

Þ x ¹ 4 and ( x ³ 4 or x £ 1) Þ ( x > 4 or x £ 1) Þ dom ( f ) = ( -¥ , 1] È ( 4 , ¥ ). 26. f ( x ) is defined only when ( x 2 - 1) ³ 0 and ( x - 1) > 0 Þ ( x - 1)( x + 1) ³ 0 and ( x - 1) > 0 Þ x + 1 ³ 0 and x - 1 > 0 Þ x > 1 \

dom ( f ) = ( 1, ¥ ).

27. f ( x ) is not defined when ( x 2 - 1) = 0 , i.e., when ( x - 1)( x + 1) = 0 , i.e., when x = 1 or x = - 1. \

dom ( f ) = R - {1, - 1}.

sin -1 x is defined only when x ¹ 0 and x Î [-1, 1]. 28. x \

dom ( f ) = [-1, 1] - {0}.

é -1 1 ù 29. sin -1 2x is defined only when 2 x Î [-1, 1] Þ x Î ê , ú × ë 2 2û 30. cos -1 ( 3 x - 1) is defined only when ( 3 x - 1) Î [-1, 1] é 2ù é 2ù Þ 3 x Î [0 , 2] Þ x Î ê 0 , ú Þ dom ( f ) = ê 0 , ú × ë 3û ë 3û 31. f ( x ) is defined only when cos x ³ 0 Þ x lies in 1st or 4th quadrant é p ù é 3p ù , 2p ú × Þ dom ( f ) = ê 0 , ú È ê ë 2û ë 2 û

65

SSS Mathematics for Class 12 66

66

Senior Secondary School Mathematics for Class 12

32. f ( x ) is defined only when log ( 2 x - x 2 ) ³ 0 Þ ( 2 x - x 2 ) ³ 1 Þ ( 1 + x 2 - 2 x ) £ 0 Þ ( 1 - x ) 2 £ 0 Þ ( 1 - x ) = 0 Þ x = 1. \

dom ( f ) = {1}.

33. f ( x ) = x 2 is clearly defined for each x Î R. So, dom ( f ) = R. y = x2 Þ x = ±

y.

When y < 0, there is no real value of x. So, y ³ 0. \

range ( f ) = R + .

34. f ( x ) = x 3 is defined for each x Î R. So, dom ( f ) = R. For each y Î R , y \

1

3

Î R and so x = y

1

3

is real.

range ( f ) = R.

35. Let f ( x ) = g ( x ) + h( x ), where g ( x ) = log ( 1 - x ) and h( x ) = x 2 - 1. g ( x ) is defined only when 1 - x > 0 Þ x < 1. So, dom ( g ) = ( -¥ , 1). h( x ) is defined only when x 2 - 1 ³ 0 Þ x ³ 1 or x £ - 1. \ \ 36. y =

dom ( h ) = ( -¥ , - 1] È [1, ¥ ). dom ( f ) = dom ( g ) Ç dom ( h ) = ( -¥ , - 1]. 1 ( 1 - x2 )

Þ x = 1-

1 × y

Clearly, x is not defined when y = 0 or 1 \ 37. y =

1 < 0 , i.e., y = 0 or y < 1. y

range ( f ) = [1, ¥ ). x2 2

(1 + x )

Þ x=

y × 1- y

Clearly, x is defined only when

y ³ 0 , and ( 1 - y ) ¹ 0 , i.e., when 0 £ y < 1. ( 1 - y)

So, range ( f ) = [0 , 1). 38. y =

y ± y2 - 4 x2 + 1 Þ x 2 - xy + 1 = 0 Þ x = × 2 x

x is defined when ( y 2 - 4 ) ³ 0 Þ y 2 ³ 4 Þ y ³ 2 or y £ -2. \

range ( f ) = ( -¥ , - 2] È [2 , ¥ ).

39. Clearly, ax > 0 whatever may be the value of x. \

range ( f ) = ( 0 , ¥ ).

SSS Mathematics for Class 12 67

3. BINARY OPERATIONS CLOSURE PROPERTY

closure property, if

An operation * on a nonempty set S is said to satisfy the

a Î S, b Î S Þ a * b Î S for all a , b Î S. Also, in this case, we say that S is closed for *. An operation * on a nonempty set S, satisfying the closure property is known as a binary operation. EXAMPLE 1

(i) Addition on the set N of all natural numbers is a binary operation, since a Î N , b Î N Þ a + b Î N for all a , b Î N . (ii) Multiplication on N is a binary operation, since a Î N , b Î N Þ a ´ b Î N for all a , b Î N . Similarly, addition as well as multiplication is a binary operation on each one of the sets Z , Q , R and C of integers, rationals, reals and complex numbers respectively.

EXAMPLE 2

Let S be a nonempty set and P( S) be its power set. Then, the union operation on P( S) is a binary operation, since A Î P( S), B Î P( S) Þ A È B Î P( S) for all A , B Î P( S). Similarly, intersection on P( S) is a binary operation, as A Î P( S), B Î P( S) Þ A È B Î P( S) for all A , B Î P( S).

EXAMPLE 3

Subtraction on N is not a binary operation, since 3 Î N , 5 Î N but ( 3 - 5) = - 2 Ï N . Subtraction on the set Z of all integers is a binary operation, since a Î Z , b Î Z Þ a - b Î Z for all a , b Î Z.

EXAMPLE 4

Addition on the set S of all irrationals is not a binary operation, since 2 + 3 Î S, 2 - 3 Î S but ( 2 + 3 ) + ( 2 - 3 ) = 4 Ï S. Multiplication on the set S of all irrationals is not a binary operation, since 2 Î S, - 2 Î S but ( 2)( - 2) = - 2 Ï S.

EXAMPLE 5

Let N be the set of all natural numbers. Then, the exponential operation ( a , b) Þ a b is a binary operation on N, since a Î N , b Î N Þ a b Î N for all a , b Î N . 67

SSS Mathematics for Class 12 68

68

Senior Secondary School Mathematics for Class 12

Let Z be the set of all integers. The exponential operation ( a , b) ® a b is not a binary operation on Z, since 2 Î Z , - 3 Î Z but 1 1 2-3 = 3 = Ï Z. 8 2 Properties of a binary operation (i) Associative law A binary operation * on a nonempty set S is said to be associative, if ( a * b) * c = a * ( b * c) for all a , b , c Î S. (ii) Commutative law A binary operation * on a nonempty set S is said to be commutative, if a * b = b * a for all a , b Î S. (iii) Distributive law Let * and º be two binary operations on a nonempty set S. We say that * is distributive over º, if a * ( b º c) = ( a * b) º ( a * c) for all a , b , c Î S. EXAMPLE 1

Let R be the set of all real numbers. Then, (i) addition on R satisfies the closure property, the associative law and the commutative law, (ii) multiplication on R satisfies the closure property, the associative law and the commutative law, (iii) multiplication distributes addition on R, since a × ( b + c) = a × b + a × c for all a , b , c Î R.

EXAMPLE 2

Let Z be the set of all integers. Then, subtraction on Z is clearly a binary operation. But, it is neither commutative nor associative, as ( 3 - 5) ¹ (5 - 3) and ( 3 - 4) - 5 ¹ 3 - ( 4 - 5).

IDENTITY ELEMENT Let * be a binary operation on a nonempty set S. An element e Î S, if it exists such that

a * e = e * a = a for all a Î S, is called an identity element of S, with respect to *. EXAMPLE 1

(i) For addition on R, zero is the identity element in R, since a + 0 = 0 + a = a for all a Î R. (ii) For multiplication on R, 1 is the identity element in R, since a ´ 1 = 1 ´ a = a for all a Î R.

EXAMPLE 2

Let P( S) be the power set of a nonempty set S. Then, f is the identity element for union on P( S) as A È f = f È A = A for all A Î P( S). Also, S is the identity element for intersection on P( S), since A Ç S = S Ç A = A for all A Î P( S).

SSS Mathematics for Class 12 69

Binary Operations

69

INVERSE OF AN ELEMENT Let * be a binary operation on a nonempty set S and let e be the identity element.

Let a Î S. We say that a is invertible if there exists an element b Î S such that a * b = b * a = e. Also, in this case, b is called the inverse of a, and we write, a -1 = b. EXAMPLE 1

Consider addition on Z. Clearly, the additive identity is 0, since a + 0 = 0 + a = a for all a Î Z. Also, corresponding to each a Î Z, there exists - a Î Z such that a + ( - a) = ( - a) + a = 0. Thus, the additive inverse of a is - a.

EXAMPLE 2

Consider multiplication on Z. Clearly, 1 is the multiplicative identity on Z, since a ´ 1 = 1 ´ a = a for all a Î Z. Since 1 ´ 1 = 1, so the multiplicative inverse of 1 is 1. Since ( -1) ´ ( -1) = 1, so the multiplicative inverse of ( -1) is ( -1). No integer other than 1 and -1 has its multiplicative inverse in Z. SOLVED EXAMPLES

EXAMPLE 1

Let Z be the set of all integers. Then, addition on Z satisfies the following properties: (i) Closure Property We know that the sum of two integers is always an integer, i.e., a Î Z , b Î Z Þ a + b Î Z for all a , b Î Z. (ii) Associative Law For all a , b , c Î Z, we have ( a + b) + c = a + ( b + c). (iii) Commutative Law For all a , b Î Z, we have a + b = b + a. (iv) Existence of Additive Identity Clearly, 0 Î Z is the additive identity, since 0 + a = a + 0 = a for all a Î Z. (v) Existence of Additive Inverse For each a Î Z, there exists - a Î Z such that a + ( - a) = ( - a) + a = 0. So, - a is the additive inverse of a.

SSS Mathematics for Class 12 70

70 EXAMPLE 2

Senior Secondary School Mathematics for Class 12

Let R 0 be the set of all nonzero real numbers. Then, multiplication on R 0 satisfies the following properties: (i) Closure Property We know that the product of two nonzero real numbers is a nonzero real number. Thus, a Î R 0 , b Î R 0 Þ ab Î R 0 for all a , b Î R 0 . (ii) Associative Law For all a , b , c Î R 0 , we have ( ab) c = a( bc). (iii) Commutative Law For all a , b Î R 0 , we have ab = ba. (iv) Existence of Multiplicative Identity Clearly, 1 Î R 0 is the multiplicative identity, since 1 ´ a = a ´ 1 for all a Î R 0 . (v) Existence of Multiplicative Inverse 1 For each a Î R 0 there exists Î R 0 such that a 1 1 a ´ = ´ a = 1. a a 1 Thus, the multiplicative inverse of a is × a

EXAMPLE 3

SOLUTION

Show that the operation * on Z, defined by a * b = a + b + 1 for all a , b Î Z satisfies (i) the closure property, (ii) the associative law and (iii) the commutative law. (iv) Find the identity element in Z. (v) What is the inverse of an element a Î Z ? (i) Closure Property Let a Î Z , b Î Z. Then, a * b = a + b + 1. Now, a Î Z , b Î Z Þ a + b Î Z Þ a + b + 1 Î Z. \ * on Z satisfies the closure property. (ii) Associative Law For all a , b , c Î Z, we have: ( a * b) * c = ( a + b + 1) * c = ( a + b + 1) + c + 1 = a + b + c + 2. a * ( b * c) = a * ( b + c + 1) = a + ( b + c + 1) + 1 = a + b + c + 2. \ ( a * b) * c = a * ( b * c).

SSS Mathematics for Class 12 71

Binary Operations

71

(iii) Commutative Law For all a , b Î Z, we have a*b=a+ b+1 =b+a+1 = b * a.

[Q a + b = b + a]

(iv) Existence of Identity Element Let e be the identity element in Z. Then, a * e = a Þ a + e + 1 = a Þ e = - 1. Thus, -1 Î Z is the identity element for *. (v) Existence of Inverse Let a Î Z and let its inverse be b. Then, a * b = -1 Þ a + b + 1 = -1 Þ b = - ( 2 + a). Clearly, 2 Î Z , a Î Z Þ - ( 2 + a) Î Z. Thus, each a Î Z has - ( 2 + a) Î Z as its inverse. EXAMPLE 4

SOLUTION

Show that the operation * on Q - {1}, defined by a * b = a + b - ab for all a , b Î Q - {1} satisfies (i) the closure property, (ii) the associative law, (iii) the commutative law. (iv) What is the identity element? (v) For each a Î Q - {1}, find the inverse of a. (i) Closure Property Let a Î Q - {1} and b Î Q - {1}. We know that Q is closed for addition, subtraction and multiplication. \ a + b - ab Î Q. But, a * b = 1 Þ a + b - ab = 1 Þ a(1 - b) = (1 - b) Þ a = 1, which is a contradiction since 1 Ï Q - {1}. \ a * b ¹ 1. Thus, a Î Q - {1}, b Î Q - {1} Þ a * b Î Q - {1}. \ * is a binary operation on Q - {1}. (ii) Associative Law Let a , b , c Î Q - {1}. Then, ( a * b) * c = ( a + b - ab) * c = ( a + b - ab) + c - ( a + b - ab) c = ( a + b + c) - ( ab + bc + ac) + abc. And, a * ( b * c) = a * ( b + c - bc) = a + ( b + c - bc) - a( b + c - bc) = ( a + b + c) - ( ab + bc + ac) + abc.

SSS Mathematics for Class 12 72

72

Senior Secondary School Mathematics for Class 12

\ ( a * b) * c = a * ( b * c). Hence, * is associative. (iii) Commutative Law Let a , b Î Q - {1}. Then, a * b = a + b - ab = b + a - ba [Q + and × are commutative on Q - {1}] = b * a. Hence, * is commutative. (iv) Existence of Identity Element Let e be the identity element. Then, for all a Î Q - {1}, we have a * e = a Þ a + e - ae = a Þ e(1 - a) = 0 Þ e = 0 Î Q - {1}. Now, a * 0 = a + 0 - a ´ 0 = a. And, 0 * a = 0 + a - 0 ´ a = a. Thus, 0 is the identity element in Q - {1}. (v) Existence of Inverse Let a Î Q - {1} and let a -1 = b. Then, a * b = 0 Þ a + b - ab = 0 Þ a = ab - b = ( a - 1) b a Þ b= Î Q - {1}. ( a - 1) a Î Q - {1}. \ a -1 = ( a - 1) Thus, each a Î Q - {1} has its inverse in Q - {1}. EXAMPLE 5

SOLUTION

On the set N of all natural numbers, define the operation * on N by m * n = gcd (m , n) for all m, n Î N . Show that * is commutative as well as associative. (i) Commutativity For all m , n Î N , we have gcd (m , n) = gcd (n, m). \ m * n = n * m " m, n Î N. Hence, * is commutative on N. (ii) Associativity Let m , n, p Î N . Then, (m * n) * p = [gcd (m , n)] * p = gcd [gcd {(m , n), p}] = gcd [{m, gcd (n, p)}] [Q gcd of three numbers = gcd {(gcd of any two, third)}] = gcd (m , n * p) = m * (n * p). Hence, * is associative on N.

SSS Mathematics for Class 12 73

Binary Operations EXAMPLE 6

73

Consider the set A = {-1, 1} with multiplication operation. We may prepare its composition table as shown below. ´

1

–1

1

1

–1

–1

–1

1

Multiplication on A satisfies the following properties: (i) Closure Property Since all the entries of the composition table are in A, so A is closed for multiplication. (ii) Associative Law Since multiplication of integers is associative, in particular, multiplication on A is associative. (iii) Commutative Law Since every row of the table coincides with the corresponding column, i.e., 1st row coincides with 1st column, 2nd row coincides with 2nd column. So, multiplication is commutative on A. (iv) Existence of Identity Clearly, 1 is the identity element in A, since 1 ´ 1 = 1 and ( -1) ´ 1 = 1 ´ ( -1) = - 1. (v) Existence of Inverse It is clear from the table that 1 ´ 1 = 1 Þ inverse of 1 is 1, ( -1) ´ ( -1) = 1 Þ inverse of (–1) is (–1). NEW OPERATIONS EXAMPLE 7

Let A = {1, 2, 3 , 4, 5}. Define an operation Ù by a Ù b = min {a , b}. Then, we may prepare its composition table as given below. Ú

1

2

3

4

5

1

1

1

1

1

1

2

1

2

2

2

2

3

1

2

3

3

3

4

1

2

3

4

4

5

1

2

3

4

5

SSS Mathematics for Class 12 74

74

Senior Secondary School Mathematics for Class 12

Closure Property Since all the entries of the composition table are from given set A, so A is closed for the operation Ù. Commutative Law In the given table every row coincides with the corresponding column, i.e., 1st row coincides with 1st column, 2nd row coincides with 2nd column, and so on. \ Ù on A is commutative. EXAMPLE 8

Let A = {1, 2, 3 , 4, 5}. Define an operation Ú by a Ú b = max {a , b}. Prepare its composition table. Show that A is closed for the given operation and that the given operation is commutative.

SOLUTION

We prepare the table as given below. Ú 1 2 3 4 5

1 1 2 3 4 5

2 2 2 3 4 5

3 3 3 3 4 5

4 4 4 4 4 5

5 5 5 5 5 5

Closure Property Since all the entries of the composition table are from the given set, so closure property is satisfied. Commutative Law Clearly, every row coincides with the corresponding column. So, commutative law is satisfied.

EXERCISE 3A Very-Short-Answer Questions 1. Let * be a binary operation on the set I of all integers, defined by [CBSE 2011C] a * b = 3 a + 4b - 2. Find the value of 4 * 5. 2. The binary operation * on R is defined by a * b = 2a + b. Find ( 2 * 3) * 4. [CBSE 2012]

3. Let * be a binary operation on the set of all nonzero real numbers, defined ab [CBSE 2014] by a * b = × Find the value of x given that 2 * ( x * 5) = 10. 5 4. Let * : R ´ R ® R be a binary operation given by a * b = a + 4b 2. Then, compute ( -5) * ( 2 * 0). [CBSE 2014C]

SSS Mathematics for Class 12 75

Binary Operations

75

5. Let * be a binary operation on the set Q of all rational numbers given as a * b = ( 2a - b) 2 for all a , b Î Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3? [CBSE 2008C]

6. Let * be a binary operation on N given by a * b = lcm of a and b. Find the value of 20 * 16. [CBSE 2008C] Is * (i) commutative, (ii) associative? 7. If * be the binary operation on the set Z of all integers defined by a * b = ( a + 3 b 2), find 2 * 4. [CBSE 2009] 8. Show that * on Z + defined by a * b = | a - b |is not a binary operation. 9. Let * be a binary operation on N, defined by a * b = a b for all a , b Î N . Show that * is neither commutative nor associative. 10. Let a * b = lcm ( a , b) for all values of a , b Î N . (i) Find (12 * 16). (ii) Show that * is commutative on N. (iii) Find the identity element in N. (iv) Find all invertible elements in N. 11. Let Q + be the set of all positive rational numbers.

1 (i) Show that the operation * on Q + defined by a * b = ( a + b) is a binary 2 operation. (ii) Show that * is commutative. [CBSE 2008] (iii) Show that * is not associative. 12. Show that the set A = {-1, 0, 1} is not closed for addition. a 13. Show that * on R - {-1}, defined by ( a * b) = is neither commutative ( b + 1) [CBSE 2007] nor associative. 14. For all a , b Î R , we define a * b = | a - b |. Show that * is commutative but not associative. 15. For all a , b Î N , we define a * b = a 3 + b 3 . Show that * is commutative but not associative.

16. Let X be a nonempty set and * be a binary operation on P(X), the power set of X, defined by A * B = A Ç B for all A , B Î P(X). (i) Find the identity element in P(X). (ii) Show that X is the only invertible element in P(X). 17. A binary operation * on the set {0, 1, 2, 3, 4, 5} is defined as if a + b < 6 ì a + b; a*b =í a + b ; 6 if a + b ³ 6. î Show that 0 is the identity for this operation and each element a has an [CBSE 2011C] inverse ( 6 - a).

SSS Mathematics for Class 12 76

76

Senior Secondary School Mathematics for Class 12 ANSWERS (EXERCISE 3A)

1. 30 5. 1, 49; No 10. (i) 48 (iii) 1

2. 18 3. x = 25 6. 20 * 16 = 80; (i) Yes (ii) Yes (iv) 1 16. (i) X

4. 11 7. 50

HINTS TO THE GIVEN QUESTIONS (EXERCISE 3A) 1. 4 * 5 = 3 ´ 4 + 4 ´ 5 - 2 = 12 + 20 - 2 = 30. 2. 2 * 3 = 2 ´ 2 + 3 = 4 + 3 = 7 . \ ( 2 * 3 ) * 4 = 7 * 4 = 2 ´ 7 + 4 = 14 + 4 = 18. æ x ´5 ö 3. 2 * ( x * 5 ) = 10 Þ 2 * ç ÷ = 10 è 5 ø Þ 2 * x = 10 Þ

2´x = 10 5

Þ 2 x = 50 Þ x = 25. 4. ( 2 * 0 ) = 2 + 4 ´ 0 2 = 2 + 4 ´ 0 = 2 + 0 = 2. \ ( -5 ) * ( 2 * 0 ) = ( -5 ) * 2 = ( -5 ) + 4 ´ 2 2 = ( -5 ) + 16 = 11. 5. ( 3 * 5 ) = ( 2 ´ 3 - 5 ) 2 = ( 6 - 5 ) 2 = 12 = 1. (5 * 3 ) = ( 2 ´ 5 - 3 ) 2 = ( 10 - 3 ) 2 = 7 2 = 49. \ ( 3 * 5 ) ¹ (5 * 3 ). 6. ( 20 * 16 ) = lcm (20, 16) = 80.

(i) For all a , b Î N , we know that lcm {a , b} = {b , a}. So, a * b = b * a. \ * is commutative on N. (ii) For all a , b , c Î N , we know that lcm {a , b and c} = lcm {( a , b) and c}. \ * is associative on N. 7. ( 2 * 4 ) = ( 2 + 3 ´ 4 2 ) = ( 2 + 3 ´ 16 ) = ( 2 + 48 ) = 50. 8. Let a be an arbitrary positive integer. Then, a * a =| a - a|= 0 , which is not a positive integer. \ * on Z + is not a binary operation. 9.

(i) 2 * 5 = 25 = 32 and 5 * 2 = 5 2 = 25. Hence, 2 * 5 ¹ 5 * 2. (ii) ( 2 * 1) * 4 = 21 * 4 = ( 2 * 4 ) = 2 4 = 16. 2 * ( 1 * 4 ) = 2 * 14 = 2 * 1 = 21 = 2.

10.

\ ( 2 * 1) * 4 ¹ 2 * ( 1 * 4 ). (i) 12 * 16 = lcm (12, 16) = ( 4 ´ 3 ´ 4 ) = 48. (ii) a * b = lcm ( a, b ) = lcm ( b , a) = b * a. (iii) Let e be the identity element in N.

4

12, 16 3, 4

SSS Mathematics for Class 12 77

Binary Operations Then, for all a Î N , we have a * e = a Þ lcm ( a, e ) = a Þ e = 1. \ 1 is the identity element in N. (iv) Let a Î N be an arbitrary element. Then, a * b = 1 Þ lcm ( a, b ) = 1 Þ a = b = 1. \ 1 is the only element in N, which is invertible. 11.

(i) a Î Q + , b Î Q + Þ a + b Î Q + 1 ( a + b) Î Q + 2 Þ a * b Î Q+. 1 1 (ii) a * b = ( a + b ) = ( b + a) = b * a. 2 2 (iii) We have 1 ö 1 æ7 æ7 ö 17 ( 3 * 4) * 5 = ( 3 + 4) * 5 = ç * 5 ÷ = ç + 5 ÷ = × 2 è2 ø 2 è2 ø 4 9 ö 15 9ö 1æ æ 3 * ( 4 * 5) = ç 3 * ÷ = ç 3 + ÷ = × 2ø 2è 2ø 4 è Þ

\ ( 3 * 4 ) * 5 ¹ 3 * ( 4 * 5 ). 12. -1 Î A and -1 Î A. But ( -1) + ( -1) = -2 Ï A. 2 2 1 3 3 13. (i) 2 * 3 = = = and 3 * 2 = = = 1. ( 3 + 1) 4 2 ( 2 + 1) 3 \ 2 * 3 ¹ 3 * 2. So, * is not commutative. 1 1 2 = 1× (ii) ( 2 * 3 ) * 1 = * 1 = 2 ( 1 + 1) 4 3 3 ( 3 * 3) = = × ( 1 + 1) 2 \ 2 * ( 3 * 1) = 2 *

14.

3 = 2

2

( 32 + 1)

2ö 4 æ = ç2 ´ ÷ = × 5ø 5 è

Thus, ( 2 * 3 ) * 1 ¹ 2 * ( 3 * 1). (i) For all a, b Î R we have a * b =| a - b| =| - ( a - b )| =| b - a| = ( b * a). \ * is commutative. (ii) We have ( 2 * 3 ) * 4 =| 2 - 3|* 4 =| - 1|* 4 = 1 * 4 =| 1 - 4| =| - 3| = 3. 2 * ( 3 * 4 ) = 2 *| 3 - 4| = 2 *| - 1| = 2 * 1 =| 2 - 1| =| 1| = 1. \ ( 2 * 3 ) * 4 ¹ 2 * ( 3 * 4 ).

15.

Hence, * is not associative. (i) For all a, b Î N we have a * b = a3 + b 3 = b 3 + a3 = b * a. \ * is commutative. (ii) ( 1 * 2 ) * 3 = ( 13 + 2 3 ) * 3 = ( 9 * 3 ) = ( 9 3 + 3 3 ) = (729 + 27 ) = 756.

77

SSS Mathematics for Class 12 78

78

Senior Secondary School Mathematics for Class 12 1 * ( 2 * 3 ) = 1 * ( 2 3 + 3 3 ) = 1 * ( 8 + 27 ) = 1 * 35 = {13 + ( 35 ) 3 }.

16.

\ ( 1 * 2 ) * 3 ¹ 1 * ( 2 * 3 ). (i) Since A Ç X = A for all A in P(X). \ X is the identity element. (ii) Let A be invertible in P(X) and let B be its inverse. Then, A Ç B = X. This is possible only when A = B = X. \ X is the only invertible element in P(X) and its inverse is X.

17. a * 0 = a + 0 = a [Q 0 £ a £ 5]. So, 0 is the identity. a * ( 6 - a) = a + ( 6 - a) - 6 = 0 [Q a + ( 6 - a) ³ 6]. \ The inverse of a is ( 6 - a).

EXERCISE 3B 1. Define * on N by m * n = lcm (m , n). Show that * is a binary operation which is commutative as well as associative. 2. Define * on Z by a * b = a - b + ab. Show that * is a binary operation on Z which is neither commutative nor associative. 3. Define * on Z by a * b = a + b - ab. Show that * is a binary operation on Z which is commutative as well as associative. 4. Consider a binary operation on Q - {1}, defined by a * b = a + b - ab. (i) Find the identity element in Q - {1}. (ii) Show that each a Î Q - {1} has its inverse. 5. Let Q0 be the set of all nonzero rational numbers. Let * be a binary ab operation on Q0 , defined by a * b = for all a , b Î Q0. 4 (i) Show that * is commutative and associative. (ii) Find the identity element in Q0 . (iii) Find the inverse of an element a in Q0 . 6. On the set Q + of all positive rational numbers, define an operation * on Q + ab by a * b = for all a , b Î Q + . 2 Show that (i) * is a binary operation on Q + ,

SSS Mathematics for Class 12 79

Binary Operations

79

(ii) * is commutative, (iii) * is associative. Find the identity element in Q + for *. What is the inverse of a Î Q + ? 7. Let Q + be the set of all positive rational numbers. 1 (i) Show that the operation * on Q + defined by a * b = ( a + b) is a binary 2 operation. (ii) Show that * is commutative. (iii) Show that * is not associative. 8. Let Q be the set of all rational numbers. Define an operation * on Q - {-1} by a * b = a + b + ab. Show that (i) * is a binary operation on Q - {-1}, (ii) * is commutative, (iii) * is associative, (iv) zero is the identity element in Q - {-1} for *, æ -a ö ÷÷ , where a Î Q - {-1}. (v) a -1 = çç è1 + aø 9. Let A = N ´ N . Define * on A by ( a , b) * ( c, d) = ( a + c, b + d). Show that (i) A is closed for *, (ii) * is commutative, (iii) * is associative, (iv) identity element does not exist in A. 10. Let A = {1, - 1, i , - i} be the set of four 4th roots of unity. Prepare the composition table for multiplication on A and show that (i) A is closed for multiplication, (ii) multiplication is associative on A, (iii) multiplication is commutative on A, (iv) 1 is the multiplicative identity, (v) every element in A has its multiplicative inverse. ANSWERS (EXERCISE 3B)

4. (i) 0

(ii) a -1 =

a 4 16 5. (ii) 4 (iii) a -1 = 6. e = 2, a -1 = ( a - 1) a a

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 3B) 1. We know that the lcm of two natural numbers is a natural number. So, N is closed for *.

SSS Mathematics for Class 12 80

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Senior Secondary School Mathematics for Class 12

We know that lcm (m , n) = lcm (n, m ). So, m * n = n * m. lcm (m , n, p ) = lcm {lcm (m, n), p} = lcm { m, lcm (n, p )}. 2. a Î Z , b Î Z Þ ( a - b ) Î Z and ab Î Z Þ {( a - b ) + ab} Î Z Þ a - b + ab Î Z. \ Z is closed for *. Show that 3 * 2 ¹ 2 * 3 and ( 4 * 3 ) * 2 ¹ 4 * ( 3 * 2 ). 4.

(i) Let e be the identity element. Then, for all a Î Q - {1}, we have a * e = a Þ a + e - ae = a Þ e( 1 - a) = 0 Þ e = 0 [Q a ¹ 1]. \ 0 is the identity element. (ii) Let a Î Q - {1} be an arbitrary element and let b be its inverse. Then, a * b = 0 Þ a + b - ab = 0 Þ ab - b = a a Þ b( a - 1) = a Þ b = × ( a - 1) Thus, each a Î Q - {1} has

5.

a as its inverse. ( a - 1)

(i) For all a, b , c Î Q 0, we have ab ba a*b = = = b * a. 4 4 ab × c ( ab )c ab 4 And, ( a * b ) * c = c = = × * 4 4 16 æ bc ö aç ÷ bc è 4 ø a( bc ) Also, a * ( b * c ) = a * = = × 4 4 16 But, ( ab )c = a( bc ). Hence, ( a * b ) * c = a * ( b * c ). (ii) Let e be the identity element and let a Î Q 0. Then, ae a*e= a Þ = a Þ e = 4. 4

\ 4 is the identity element in Q 0. (iii) Let a Î Q 0 and let its inverse be b. Then, ab 16 a*b = e Þ =4 Þ b= Î Q 0. 4 a 16 as its inverse. Thus, each a Î Q 0 has a ae 6. a * e = a Þ = a Þ e = 2. 2 ab 4 4 a*b = 2 Þ =2 Þ b= Þ a -1 = × 2 a a 9.

(i) Let ( a, b ) Î A and ( c , d ) Î A. Then, a, b , c , d Î N . ( a, b ) * ( c , d ) = ( a + c , b + d ) Î A [Q a + c Î N , b + d Î N ]. \ A is closed for *.

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Binary Operations

81

(ii) ( a, b ) * ( c , d ) = ( a + c , b + d ) = ( c + a, d + b ) [Q a + c = c + a and b + d = d + b] = ( c , d ) * ( a, b ). (iii) [( a, b ) * ( c , d )] * ( e , f ) = ( a + c , b + d ) * ( e , f ) = [( a + c ) + e , ( b + d ) + f ] = [a + ( c + e ), b + ( d + f )] = ( a, b ) * ( c + e , d + f ) = ( a, b ) * [( c , d ) * ( e , f )]. (iv) ( a, b ) * ( 0 , 0 ) = ( a + 0 , b + 0 ) = ( a, b ). But, ( 0 , 0 ) Ï A , since 0 Ï N. So, identity element does not belong to A. 10. We may prepare the composition table for multiplication on A as given below: ´

i

-i

–1

i

-i

1

-i

i

1

–1

1

1

–1

–1

i

i

-i

–1

1

-i

-i

i

1

–1

Clearly, 1 is the identity element. ( 1) -1 = 1, ( -1) -1 = - 1, (i ) -1 = - i and ( -i ) -1 = i.

SSS Mathematics for Class 12 82

4. INVERSE TRIGONOMETRIC FUNCTIONS A one-one onto function is called an invertible function. Let f : X ® Y be a one-one onto function. Then, for each y Î Y, there exists a unique element x Î X such that f ( x) = y. INVERTIBLE FUNCTIONS

INVERSE OF A FUNCTION

So, we define a new function, denoted by f -1 , called the inverse of f , as f -1 : Y ® X : f -1( y) = x Û f ( x) = y. Clearly, domain ( f -1) = range ( f ) and range ( f -1) = domain ( f ). Trigonometric functions are, in general, not one-one onto. Therefore, their inverses do not exist. However, if we restrict their domains, we can make them one-one onto, enabling us to have their inverses. EXAMPLE

The sine function restricted to any of the intervals [- p 2 , p 2], [ 3p 2 , 5p 2], etc., is one-one onto and in each case, the range is [–1, 1].

Therefore, we can define the inverse of the sine function, denoted by sin -1 x, in each of these intervals. Corresponding to each such interval, we get a branch of sin -1 x. The branch [- p 2 , p 2] is called the principal-value branch and the value belonging to it is called the principal value. We denote the inverse of the sine function as sin -1 x, called sine inverse x. é p pù \ sin -1 : [-1, 1] ® ê - , ú and sin -1 x = q Û x = sin q. ë 2 2û é p pù Thus, sin -1 x is a function whose domain is [–1, 1] and range is ê - , ú × ë 2 2û Note that sin -1 x does not mean (sin x) -1 . Similarly, we can define the principal-value branches of the remaining five trigonometrical functions. The following table shows the inverse trigonometric functions and their principal-value branches. 82

SSS Mathematics for Class 12 83

Inverse Trigonometric Functions

83

Domain

Range (Principal value)

1. y = sin -1 x

[–1, 1]

é -p p ù êë 2 , 2 úû

2. y = cos-1 x

[–1, 1]

[0, p]

3. y = tan -1 x

R

Function

æ -p p ö , ÷ ç 2ø è 2 é -p p ù êë 2 , 2 úû - {0}

4. y = cosec-1 x

R - ( -1, 1)

5. y = sec-1 x

R - [-1, 1]

ìp ü [0, p] - í ý î2þ

6. y = cot -1 x

R

( 0, p)

SOLVED EXAMPLES EXAMPLE 1

Find the principal value of each of the following: æ 3ö æ 1 ö ÷ (ii) cos-1 çç (i) sin -1 ç ÷ ÷ è 2ø è 2 ø (iii) tan -1( 3 )

SOLUTION

(iv) cosec-1( 2)

(i) We know that the range of the principal-value branch of é -p p ù , ú× sin -1 is ê 2û ë 2 æ 1 ö Let sin -1 ç ÷ = q. Then, è 2ø 1 p p é -p p ù sin q = = sin Þ q = Îê , ú× 4 4 ë 2 2û 2 æ 1 ö p Hence, the principal value of sin -1 ç ÷ is × è 2ø 4 (ii) We know that the range of the principal-value branch of cos-1 is [0, p ]. æ 3ö ÷ = q. Then, Let cos-1 çç ÷ è 2 ø 3 p p cos q = = cos Þ q = Î [0, p]. 2 6 6 æ 3ö p ÷ is × Hence, the principal value of cos-1 çç ÷ è 2 ø 6

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Senior Secondary School Mathematics for Class 12

(iii) We know that the range of the principal-value branch of æ -p p ö , ÷× tan -1 is ç 2ø è 2 Let tan -1( 3 ) = q. Then, p p æ -p p ö tan q = 3 = tan Þ q = Îç , ÷× 3 3 è 2 2ø p -1 Hence, the principal value of tan ( 3 ) is × 3 (iv) We know that the range of the principal-value branch of é -p p ù cosec-1 is ê , ú - {0}. 2û ë 2 Let cosec-1( 2) = q. Then, cosec q = 2 = cosec

p p é -p p ù Þ q = Îê , ú - {0}. 6 6 ë 2 2û

Hence, the principal value of cosec-1( 2) is EXAMPLE 2

SOLUTION

p × 6

Find the principal value of each of the following: æ- 3ö æ -1 ö æ -1 ö ÷ (iii) tan -1 ç (i) sin -1 ç ÷ (ii) sin -1 çç ÷ ÷ 2 2 è ø è 3ø è ø (iv) tan -1( -1) (v) cosec-1( -2) (vi) tan -1( - 3 ) é -p p ù (i) We know that the principal-value branch of sin -1 is ê , × ë 2 2 úû æ -1 ö Let sin -1 ç ÷ = q. Then, è 2ø sin q = -

1 -p é -p p ù æ -p ö , ú× = sin ç ÷ , where Î 2 6 6 êë 2 2û è ø

æ -1 ö - p Hence, the principal value of sin -1 ç ÷ is × 6 è 2ø (ii) We know that the principal-value branch of sin -1 is é -p p ù êë 2 , 2 úû × æ- 3ö ÷ = q. Then, Let sin -1 çç ÷ è 2 ø -p é -p p ù - 3 æ -p ö , ú× Î sin q = = sin ç ÷ , where 2 3 êë 2 2û è 3 ø

SSS Mathematics for Class 12 85

Inverse Trigonometric Functions

85

æ - 3 ö -p ÷ is Hence, the principal value of sin -1 çç × ÷ 3 è 2 ø (iii) We know that the principal-value branch of tan -1 is æ -p p ö , ÷× ç 2ø è 2 æ -1 ö Let tan -1 ç ÷ = q. Then, è 3ø tan q =

-1 -p æ -p p ö æ -p ö , ÷× = tan ç ÷ , where Îç 6 è 2 2ø 3 è 6 ø

æ -1 ö - p Hence, the principal value of tan -1 ç × ÷ is 6 è 3ø (iv) We know that the principal-value branch of tan -1 is æ -p p ö , ÷× ç 2ø è 2 Let tan -1( -1) = q. Then, -p æ - p p ö æ -p ö , ÷× tan q = - 1 = tan ç ÷ , where Îç 4 è 2 2ø è 4 ø Hence, the principal value of tan -1( -1) is

-p × 4

(v) We know that the principal-value branch of cosec-1 is é - p pù êë 2 , 2 úû - {0}. Let cosec-1( -2) = q. Then, -p é -p p ù æ -p ö , ú - {0}. cosec q = - 2 = cosec ç ÷ , where Î 6 êë 2 2û è 6 ø Hence, the principal value of cosec-1( -2) is

-p × 6

(vi) We know that the principal-value branch of tan -1 is æ -p p ö , ÷× ç 2ø è 2 Let tan -1( - 3 ) = q. Then, -p æ -p p ö æ -p ö , ÷× tan q = - 3 = tan ç ÷ , where Îç 3 è 2 2ø è 3 ø Hence, the principal value of tan -1( - 3 ) is

-p × 3

SSS Mathematics for Class 12 86

86 EXAMPLE 3

SOLUTION

Senior Secondary School Mathematics for Class 12

Find the principal value of each of the following: æ -1 ö æ -1 ö æ -1 ö (ii) cos-1 ç ÷ (i) cos-1 ç (iii) cot -1 ç ÷ ÷ è 2ø è 2ø è 3ø (i) We know that the range of the principal value of cos-1 is [0, p]. æ -1 ö Let cos-1 ç ÷ = q. Then, è 2ø cos q = \ q=

3p -1 p pö æ × = - cos = cos ç p - ÷ = cos 4 4 4 2 ø è

3p Î [0, p]. 4

æ -1 ö 3 p Hence, the principal value of cos-1 ç × ÷ is 4 è 2ø (ii) We know that the range of the principal value of cos-1 is [0, p]. æ -1 ö Let cos-1 ç ÷ = q. Then, è 2ø 2p -1 p pö æ × cos q = = - cos = cos ç p - ÷ = cos 2 3 3 3 ø è 2p \ q= Î [0, p]. 3 æ -1 ö 2p Hence, the principal value of cos-1 ç ÷ is × 3 è 2ø (iii) We know that the range of the principal value of cot -1 is ( 0, p). æ -1 ö Let cot -1 ç ÷ = q. Then, è 3ø 2p -1 p pö æ × cot q = = - cot = cot ç p - ÷ = cot 3 3ø 3 3 è 2p \ q= Î ( 0, p). 3 æ -1 ö 2p Hence, the principal value of cot -1 ç × ÷ is 3 è 3ø EXAMPLE 4

SOLUTION

Find the principal value of each of the following: (i) cot -1( - 3 ) (ii) sec-1( - 2) (iii) cosec-1( -1) (i) We know that the range of the principal value of cot -1 is ( 0, p). Let cot -1( - 3 ) = q. Then,

SSS Mathematics for Class 12 87

Inverse Trigonometric Functions

cot q = - 3 = - cot \

q=

5p Î ( 0, p). 6

87

5p p pö æ × = cot ç p - ÷ = cot 6 6ø 6 è

Hence, the principal value of cot -1( - 3 ) is

5p × 6

(ii) We know that the range of principal value of sec-1 is ìp ü [0, p] - í ý × î2þ Let sec-1( - 2) = q. Then, p pö 3p æ sec q = - 2 = - sec = sec ç p - ÷ = sec × 4 4 4 è ø 3p ìp ü Î [0, p] - í ý × \ q= 4 î2þ 3p Hence, the principal value of sec-1( - 2) is × 4 (iii) We know that the range of principal value of cosec-1 is é -p p ù êë 2 , 2 úû - {0}. Let cosec-1( -1) = q. Then, cosec q = - 1. p æ - pö cosec q = - 1 = - cosec = cosec ç ÷× 2 è 2 ø -p é -p p ù \ q= Î , ú - {0}. 2 êë 2 2û -p Hence, the principal value of cosec-1( -1) is × 2 EXAMPLE 5

SOLUTION

æ1ö Find the value of cos-1 ç ÷ + 2 sin -1 è 2ø

æ1ö ç ÷× è 2ø

[CBSE 2012C]

We know that the ranges of principal values of cos-1 and sin -1 are é -p p ù , ú respectively. [0, p] and ê 2û ë 2 æ1ö Let cos-1 ç ÷ = q1. Then, è 2ø p p 1 cos q1 = = cos Þ q1 = Î [0, p]. 2 3 3 1 æ ö Let sin -1 ç ÷ = q2. Then, è 2ø sin q2 =

p p é -p p ù 1 , ú× = sin Þ q2 = Î ê 2 6 6 ë 2 2û

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Senior Secondary School Mathematics for Class 12

æ1ö \ cos-1 ç ÷ + 2 sin -1 è 2ø EXAMPLE 6

SOLUTION

p ö æ p p ö 2p æ1ö p æ × ç ÷ = + ç2 ´ ÷ = ç + ÷ = 3 6ø è 3 3 ø è 2ø 3 è

æ -1 ö æ -1 ö Find the value of tan -1(1) + cos-1 ç ÷ + sin -1 ç ÷ × 2 è 2ø è ø We know that the ranges of principal values of tan -1 , cos-1 and æ -p p ö é -p p ù , ÷ , [0, p] and ê sin -1 are ç , ú respectively. 2ø 2û è 2 ë 2 Let tan -1(1) = q 1. Then, tan q1 = 1 = tan

p p Þ q1 = Î [0, p]. 4 4

æ -1 ö Let cos-1 ç ÷ = q2. Then, è 2ø -1 p pö 2p æ × = - cos = cos ç p - ÷ = cos cos q2 = 2 3 3ø 3 è 2p \ q2 = Î[0, p]. 3 æ -1 ö Let sin -1 ç ÷ = q 3 . Then, è 2ø -p é -p p ù -1 p æ -p ö Î = - sin = sin ç ÷ Þ q 3 = sin q 3 = , ú× 2 6 6 6 êë 2 2û è ø æ -1 ö æ p 2p p ö 3 p æ -1 ö - ÷= \ tan -1(1) + cos-1 ç ÷ + sin -1 ç ÷ = ç + × 4 3 6ø è 2 ø è4 è 2ø EXAMPLE 7 SOLUTION

Find the value of tan -1 3 - sec-1( -2). We know that the ranges of principal values of tan -1 and sec-1 are ìp ü æ - p pö , ÷ and [0, p] - í ý respectively. ç 2ø î2þ è 2 Let tan -1 3 = q1. Then, p p æ -p p ö Þ q1 = Î ç , ÷× 3 3 è 2 2ø Let sec-1( -2) = q2 . Then, p pö 2p æ sec q2 = - 2 = - sec = sec ç p - ÷ = sec × 3 3ø 3 è 2p ìp ü Î [0, p] - í ý × \ q2 = 3 î2þ æ p 2p ö - p Hence, tan -1 3 - sec-1( -2) = ç × ÷= 3ø 3 è3 tan q1 =

3 = tan

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Inverse Trigonometric Functions

89

EXERCISE 4A 1. Find the principal value of: æ 3ö ÷ (i) sin -1 çç ÷ è 2 ø

æ1ö (ii) sin -1 ç ÷ è 2ø

æ 1 ö (v) tan -1 ç ÷ è 3ø

æ 2 ö -1 (vi) sec-1 ç ÷ (vii) cosec ( 2) è 3ø

æ1ö (iii) cos-1 ç ÷ è 2ø

(iv) tan -1(1)

2. Find the principal value of: æ -1 ö (i) sin -1 ç ÷ è 2ø (iv) sec-1( -2) ìï cos ícos-1 ïî

3. Evaluate

HINT: cos

æ- 3ö ÷ (ii) cos-1 çç ÷ è 2 ø

(iii) tan -1( - 3 )

(v) cosec-1( - 2)

æ -1 ö (vi) cot -1 ç ÷ è 3ø

æ - 3 ö p üï ç ÷ ç 2 ÷ + 6 ýï × è ø þ

- 3 5p 5p - 3ö p pö æ ÷ ç 2 ÷ = q Þ cos q = 2 = - cos 6 = cos çè p - 6 ÷ø = cos 6 Þ q = 6 × ø è

-1 æç

æ 5p p ö \ given expression = cos ç + ÷ = cos p = - 1. 6ø è 6

ìï p sin í - sin -1 îï 2

4. Evaluate

æ - 3 ö üï ç ÷ ç 2 ÷ ýï × è øþ

p 1 5p pö ì p æ - p öü æ -ç = sin ç p - ÷ = sin = × ÷ ý = sin 6 2 6 6ø è î 2 è 3 øþ

HINT: Given exp. = sin í

ANSWERS (EXERCISE 4A)

1. (i)

p 3

2. (i)

-p 4

3. –1

(ii)

p 6

(ii)

5p 6

(iii)

p 3

(iii)

p 4

(v)

(iv)

2p 3

(iv) -p 3 4.

1 2

p 6

(vi)

(v)

-p 4

p 6

(vii) (vi)

2p 3

p 4

SSS Mathematics for Class 12 90

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Senior Secondary School Mathematics for Class 12

PROPERTIES OF INVERSE FUNCTIONS THEOREM 1

Prove that: é -p p ù (i) sin -1(sin x) = x , x Î ê , ú 2û ë 2

(ii) cos-1(cos x) = x , x Î [0, p] æ -p p ö (iii) tan -1(tan x) = x , x Î ç , ÷ 2ø è 2 p pù é (iv) cosec-1( cosec x) = x , x Î ê , ú - {0 } 2û ë 2 ìp ü -1 (v) sec ( sec x) = x , x Î [0, p] - í ý î2þ (vi) cot -1(cot x) = x , x Î ( 0, p)

PROOF

é -p p ù (i) Let sin x = y, where x Î ê , ú× 2û ë 2 Then, sin -1 y = x Þ sin -1(sin x) = x

[Q y = sin x].

-1

\ sin (sin x) = x. (ii) Let cos x = y, where x Î[0, p]. Then, cos-1 y = x Þ cos-1(cos x) = x

[Q y = cos x].

-1

\ cos (cos x) = x. Similarly, the other results may be proved. THEOREM 2

Prove that: (i) sin (sin -1 x) = x , x Î [-1, 1] (ii) cos (cos-1 x) = x , x Î [-1, 1] (iii) tan (tan -1 x) = x , x Î R (iv) cosec ( cosec-1 x) = x , x Î R - [-1, 1] (v) sec ( sec-1 x) = x , x Î R - [-1, 1] (vi) cot (cot -1 x) = x , x Î R

PROOF

(i) Let sin -1 x = q, where x Î [-1, 1]. Then, sin q = x Þ sin (sin -1 x) = x

[Q q = sin -1 x].

\ sin (sin -1 x) = x. (ii) Let cos-1 x = q, where x Î [-1, 1].

Then, cos q = x Þ cos (cos-1 x) = x

[Q q = cos-1 x].

\ cos (cos-1 x) = x. Similarly, the other results may be proved.

SSS Mathematics for Class 12 91

Inverse Trigonometric Functions THEOREM 3

Prove that: 1 = cosec-1 x , ( x ³ 1 or x £ - 1) x 1 (ii) cos-1 = sec-1 x , ( x ³ 1 or x £ - 1) x -1 1 (iii) tan = cot -1 x , ( x > 0) x (i) sin -1

PROOF

1 1 = q Þ sin q = x x Þ cosec q = x Þ q = cosec-1 x 1 Þ sin -1 = cosec-1 x. x -1 1 Hence, sin = cosec-1 x. x 1 1 (ii) cos-1 = q Þ cos q = x x Þ sec q = x Þ q = sec-1 x 1 Þ cos-1 = sec-1 x. x -1 1 -1 Hence, cos = sec x. x 1 1 (iii) tan -1 = q Þ tan q = x x Þ cot q = x Þ q = cot -1 x 1 Þ tan -1 = cot -1 x. x -1 1 -1 Hence, tan = cot x. x (i) sin -1

THEOREM 4

Prove that: (i) sin -1( - x) = - sin -1 x , x Î [-1, 1] (ii) tan -1( - x) = - tan -1 x , x Î R (iii) cosec-1( - x) = - cosec-1 x , | x | ³ 1

PROOF

(i) Let sin -1( - x) = q. Then, sin -1( - x) = q Þ - x = sin q Þ x = - sin q = sin ( - q) Þ - q = sin -1 x Þ q = - sin -1 x Þ sin -1( - x) = - sin -1 x.

91

SSS Mathematics for Class 12 92

92

Senior Secondary School Mathematics for Class 12

\ sin -1( - x) = - sin -1 x. Similarly, the other results may be proved. THEOREM 5

Prove that: (i) cos-1( - x) = p - cos-1 x , x Î [-1, 1] (ii) sec-1( - x) = p - sec-1 x , |x| ³ 1 (iii) cot -1( - x) = p - cot -1 x , x Î R

PROOF

(i) Let cos-1( - x) = q. Then, cos-1( - x) = q Þ - x = cos q Þ x = - cos q = cos ( p - q) Þ cos-1 x = ( p - q) = p - cos-1( - x) Þ cos-1( - x) = p - cos-1 x. \ cos-1( - x) = p - cos-1 x. Similarly, the other results may be proved.

THEOREM 6

Prove that:

p , x Î [-1, 1] 2 p (ii) tan -1 x + cot -1 x = , x Î R 2 p -1 -1 (iii) cosec x + sec x = , |x| ³ 1 2 -1 (i) Let sin x = q. Then, æp ö sin -1 x = q Þ x = sin q = cos ç - q÷ è2 ø p -1 Þ cos x = - q 2 p -1 Þ cos x = - sin -1 x 2 p -1 Þ sin x + cos-1 x = × 2 p -1 -1 \ sin x + cos x = × 2 Similarly, the other results may be proved. (i) sin -1 x + cos-1 x =

PROOF

THEOREM 7

Prove that: æx+yö ÷÷ , if xy < 1 (i) tan -1 x + tan -1 y = tan -1 çç è 1 - xy ø æ x-yö ÷÷ , if xy > - 1 (ii) tan -1 x - tan -1 y = tan -1 çç è 1 + xy ø æ 2x ö ÷ if| (iii) 2 tan -1 x = tan -1 ç ç 1 - x 2 ÷ , x| < 1 è ø

SSS Mathematics for Class 12 93

Inverse Trigonometric Functions PROOF

(i) Let xy < 1, tan -1 x = q and tan -1 y = f. Then, tan q = x and tan f = y tan q + tan f x+y Þ tan ( q + f) = = 1 - tan q tan f 1 - xy æx+yö ÷÷ Þ q + f = tan -1 çç è 1 - xy ø æx+yö ÷÷ × Þ tan -1 x + tan -1 y = tan -1 çç è 1 - xy ø æx+yö ÷÷ × \ tan -1 x + tan -1 y = tan -1 çç è 1 - xy ø (ii) Let xy > - 1. On replacing y by -y in (i), we get æ x-yö ÷÷ tan -1 x + tan -1( - y) = tan -1 çç è 1 + xy ø æ x-yö ÷÷ × Þ tan -1 x - tan -1 y = tan -1 çç è 1 + xy ø (iii) Let|x| < 1. Replacing y by x in (i), we get æ 2x ö ÷× 2 tan -1 x = tan -1 çç 2÷ è1 - x ø

THEOREM 8

Prove that:

æ 2x ö ÷ (i) 2 tan -1 x = sin -1 ç ç 1 + x 2 ÷ , |x| < 1 è ø 2ö 1 x -1 -1 æ ÷ , |x| ³ 0 (ii) 2 tan x = cos ç ç 1 + x2 ÷ è ø æ ö 2 x ÷ (iii) 2 tan -1 x = tan -1 ç ç 1 - x 2 ÷ , |x| < 1 è ø

PROOF

Let tan -1 x = q. Then, x = tan q. æ 2x ö æ ö ÷ sin -1 ç 2 tan q ÷ (i) sin -1 ç ç 1 + x2 ÷ = ç 1 + tan 2 q ÷ è ø è ø = sin -1(sin 2q) = 2q = 2 tan -1 x. æ 2x ö ÷× \ 2 tan -1 x = sin -1 çç 2÷ è1 + x ø

[Q x = tan q]

93

SSS Mathematics for Class 12 94

94

Senior Secondary School Mathematics for Class 12

æ 1 - x2 ö ÷ = cos-1 (ii) cos-1 ç ç 1 + x2 ÷ è ø

æ 1 - tan 2 q ö ç ÷ ç 1 + tan 2 q ÷ è ø

[Q x = tan q]

= cos-1(cos 2q) = 2q = 2 tan -1 x. æ 1 - x2 ö ÷× \ 2 tan -1 x = cos-1 ç ç 1 + x2 ÷ è ø æ 2x ö ÷ tan -1 (iii) tan -1 ç ç 1 - x2 ÷ = è ø

æ 2 tan q ö ç ÷ ç 1 - tan 2 q ÷ è ø

[Q x = tan q]

= tan -1(tan 2q) = 2q = 2 tan -1 x. æ 2x ö ÷ \ 2 tan -1 x = tan -1 ç ç 1 - x2 ÷ × è ø THEOREM 9 PROOF

Prove that

2 sin -1 x = sin -1[2x 1 - x 2 ],

-1 1 £x£ × 2 2

Let sin -1 x = q. Then, x = sin q. \ sin 2q = 2 sin q cos q = 2 sin q × (1 - sin 2 q) = 2x 1 - x 2 Þ 2q = sin -1 {2x 1 - x 2} Þ 2 sin -1 x = sin -1 {2x 1 - x 2}. Hence, 2 sin -1 x = sin -1{2x 1 - x 2}.

THEOREM 10 PROOF

Prove that

2 cos-1 x = cos-1( 2x 2 - 1),

Let cos-1 x = q. Then, x = cos q.

1 £ x £ 1. 2

\ cos 2q = ( 2 cos2 q - 1) = ( 2x 2 - 1) Þ 2q = cos-1( 2x 2 - 1) Þ 2 cos-1 x = cos-1( 2x 2 - 1)

[Q q = cos-1 x].

Hence, 2 cos-1 x = cos-1( 2x 2 - 1). THEOREM 11 PROOF

Prove that

é 1 1ù 3 sin -1 x = sin -1( 3 x - 4x 3), x Î ê - , ú × ë 2 2û

Put sin -1 x = q. Then, x = sin q.

Now, sin 3 q = ( 3 sin q - 4 sin 3 q) = ( 3 x - 4x 3) Þ 3 q = sin -1( 3 x - 4x 3) Þ 3 sin -1 x = sin -1( 3 x - 4x 3) [Q q = sin -1 x]. Hence, 3 sin -1 x = sin -1( 3 x - 4x 3).

SSS Mathematics for Class 12 95

Inverse Trigonometric Functions

é1 3 cos-1 x = cos-1( 4x 3 - 3 x), x Î ê , ë2 Put cos-1 x = q. Then, x = cos q. Now, cos 3 q = ( 4 cos 3 q - 3 cos q) = ( 4x 3 - 3 x) Þ 3 q = cos-1( 4x 3 - 3 x) Þ 3 cos-1 x = cos-1( 4x 3 - 3 x) Hence, 3 cos-1 x = cos-1( 4x 3 - 3 x).

THEOREM 12 PROOF

THEOREM 13

Prove that

95

ù 1ú × û

[Q q = cos-1 x].

Prove that: (i) sin -1 x + sin -1 y = sin -1{x 1 - y 2 + y 1 - x 2} (ii) sin -1 x - sin -1 y = sin -1{x 1 - y 2 - y 1 - x 2} (iii) cos-1 x + cos-1 y = cos-1{xy - (1 - x 2)(1 - y 2)} (iv) cos-1 x - cos-1 y = cos-1{xy + (1 - x 2)(1 - y 2)}

PROOF

(i) Let sin -1 x = q1 , and sin -1 y = q2 . Then, sin q1 = x and sin q2 = y Þ sin ( q1 + q2) = sin q1 cos q2 + cos q1 sin q2 = {sin q1 × 1 - sin 2 q2} + { 1 - sin 2 q1 × sin q2} = {x 1 - y 2 + y 1 - x 2} Þ ( q1 + q2) = sin -1{x 1 - y 2 + y 1 - x 2} Þ sin -1 x + sin -1 y = sin -1{x 1 - y 2 + y 1 - x 2}. Hence, sin -1 x + sin -1 y = sin -1{x 1 - y 2 + y 1 - x 2}. The other results can be proved similarly.

THEOREM 14

Prove that:

æ ö x ÷ = cosec-1 æ 1 ö (i) sin -1 x = cos-1 1 - x 2 = tan -1 ç ç ÷ ç 2÷ èxø 1 x è ø 2ö æ ö æ 1 ÷ = cot -1 ç 1 - x ÷ = sec-1 ç ç ÷ ç ÷ 2 x è 1-x ø è ø æ 1 - x2 ö ÷ (ii) cos-1 x = sin -1 1 - x 2 = tan -1 ç ç ÷ x è ø æ ö æ ö 1 1 x ÷ = sec-1 æ ö = cot -1 ç ÷ = cosec-1 ç ÷ ç ç ç 2÷ 2÷ èxø 1 x 1 x è ø è ø ö æ ö æ x 1 ÷ ÷ = cos-1 ç (iii) tan -1 x = sin -1 ç ç ç 2÷ 2÷ è 1+ x ø è 1+ x ø æ 1 + x2 ö ÷ = sec-1( 1 + x 2 ) = cot -1 æç 1 ö÷ = cosec-1 ç ç x ÷ èxø è ø

SSS Mathematics for Class 12 96

96

PROOF

Senior Secondary School Mathematics for Class 12

(i) Let sin -1 x = q. Then, sin q = x. \ cos q = 1 - x 2 , tan q =

sec q =

1 1 - x2

x 1-x

and cot q =

2

, cosec q =

1 , x

1 - x2 × x

æ ö x ÷ , q = cosec-1 1 , \ q = cos-1 1 - x 2 , q = tan -1 ç ç 2÷ x è 1-x ø ö æ 1 ÷ and q = cot -1 q = sec-1 ç ç 2÷ 1 x ø è

æ 1 - x2 ö ÷× ç ÷ ç x ø è

æ ö x ÷ Hence, sin -1 x = cos-1 1 - x 2 = tan -1 ç ç 2÷ è 1-x ø ö æ 1 1 ÷ = cosec-1 = sec-1 ç ç 2÷ x è 1-x ø æ 1 - x2 ö ÷× = cot -1 ç ç ÷ x è ø Similarly, the other results can be proved.

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Find the value of: pö æ (i) sin -1 ç sin ÷ 3ø è

2p ö æ (ii) cos-1 ç cos ÷ 3ø è

pö æ (iii) tan -1 ç tan ÷ 4ø è

We have p é -p p ù pö p æ (i) sin -1 ç sin ÷ = , since Î ê , ú× 3 ë 2 2û 3ø 3 è 2 p 2 p 2 p ö æ (ii) cos-1 ç cos ÷ = , since Î [0, p]. 3ø 3 3 è pö p p æ -p p ö æ (iii) tan -1 ç tan ÷ = , since Î ç , ÷× 4 4 4 è 2 2ø è ø

EXAMPLE 2

Find the value of: 2p ö æ (i) sin -1 ç sin ÷ 3ø è

7p ö æ (ii) cos-1 ç cos ÷ 6ø è

3p ö æ (iii) tan -1 ç tan ÷ 4 ø è

SSS Mathematics for Class 12 97

Inverse Trigonometric Functions SOLUTION

97

(i) We know that the principal-value branch of sin -1 is é -p p ù êë 2 , 2 úû × 2p ö 2p æ × \ sin -1 ç sin ÷¹ 3ø 3 è 2p ö p öü æ æ -1 ì Now, sin -1 ç sin ÷ = sin ísin ç p - ÷ ý 3 3 øþ è ø è î

Hence, sin -1

pö æ = sin -1 ç sin ÷ [Q sin ( p - q) = sin q] 3ø è p ì p é -p p ù ü = Î , úý× íQ 3 î 3 êë 2 2ûþ 2p ö p æ ÷= × ç sin 3ø 3 è

(ii) We know that the principal-value branch of cos-1 is [0, p ]. 7p ö 7p æ × \ cos-1 ç cos ÷ ¹ 6ø 6 è 7p ö 5p ö ü ì æ æ Now, cos-1 ç cos ÷ = cos-1 ícos ç 2p ÷ý 6ø 6 øþ è è î

Hence, cos-1

ì 5p ü = cos-1 ícos ý {Q cos ( 2p - q) = cos q} 6þ î 5p = × 6 7p ö 5p æ × ç cos ÷ = 6ø 6 è

æ -p p ö (iii) We know that the principal-value branch of tan -1 is ç , ÷× è 2 2ø 3p ö 3p æ \ tan -1 ç tan × ÷¹ 4 ø 4 è 3p ö pöü æ æ -1 ì Now, tan -1 ç tan ÷ = tan ítan ç p - ÷ ý 4 ø 4øþ è è î pü é pö pù ì æ = tan -1 í- tan ý êQ tan ç p - ÷ = - tan ú 4þ ë 4ø 4û î è p ì ü æ ö = tan -1 ítan ç ÷ ý [Q - tan q = tan ( - q)] è 4 øþ î -p = × 4 3 p ö -p æ Hence, tan -1 ç tan × ÷= 4 ø 4 è

SSS Mathematics for Class 12 98

98 EXAMPLE 3

SOLUTION

Senior Secondary School Mathematics for Class 12

Find the value of: 3p ö æ (i) sin -1 ç sin ÷ 5 ø è

13 p ö æ (ii) cos-1 ç cos ÷ 6 ø è

7p ö æ (iii) tan -1 ç tan ÷ 6ø è

é -p p ù (i) We know that the principal-value branch of sin -1 is ê , × ë 2 2 úû 3p ö 3p æ \ sin -1 ç sin × ÷¹ 5 ø 5 è 3p ö 2p ö ü æ æ -1 ì Now, sin -1 ç sin ÷ = sin ísin ç p ÷ý 5 ø 5 øþ è è î 2p ö æ = sin -1 ç sin ÷ [Q sin ( p - q) = sin q] 5 ø è =

2p 5

2p é - p p ù ü ì Î ê , úý× íQ 5 2ûþ ë 2 î

3 p ö 2p æ Hence, sin -1 ç sin × ÷= 5 ø 5 è (ii) We know that the principal-value branch of cos-1 is [0, p]. 13 p ö 13 p æ × \ cos-1 ç cos ÷¹ 6 ø 6 è 13 p ö pöü æ æ -1 ì Now, cos-1 ç cos ÷ = cos ícos ç 2p + ÷ ý 6 ø 6øþ è è î pü ì = cos-1 ícos ý [Q cos ( 2p + q) = cos q] 6þ î =

p × 6

13 p ö p æ Hence, cos-1 ç cos ÷= × 6 ø 6 è

æ -p p ö (iii) We know that the principal-value branch of tan -1 is ç , ÷× è 2 2ø 7p ö 7p æ \ tan -1 ç tan × ÷¹ 6ø 6 è 7p ö pöü æ æ -1 ì Now, tan -1 ç tan ÷ = tan ítan ç p + ÷ ý 6 6øþ è ø è î pö æ = tan -1 ç tan ÷ [Q tan( p + q) = tan q] 6ø è p = × 6

SSS Mathematics for Class 12 99

Inverse Trigonometric Functions

99

7p ö p æ Hence, tan -1 ç tan ÷= × 6ø 6 è EXAMPLE 4

SOLUTION

3p ü 3p ì and find its value. (i) Show that sin -1 ísin ý¹ 4 þ 4 î ì æ -p ö ü -p (ii) Show that cos-1 ícos ç ÷ ý ¹ and find its value. è 3 øþ 3 î 5p ü 5p ì and find its value. (iii) Show that tan -1 ítan ý¹ 6þ 6 î é -p p ù (i) We know that the principal-value branch of sin -1 is ê , × ë 2 2 úû 3p ü 3p ì × \ sin -1 ísin ý¹ 4 þ 4 î 3p ü pöü æ ì -1 ì Now, sin -1 ísin ý = sin ísin ç p - ÷ ý 4 þ 4øþ è î î pö pù pü é ì æ = sin -1ísin ý êQ sin ç p - ÷ = sin ú 4 4 4û þë î è ø p ì p é -p p ù ü = , úý× íQ Î 4 î 4 êë 2 2ûþ 3p ü p -1 ì \ sin ísin ý= × 4 þ 4 î (ii) We know that the principal-value branch of cos-1 is [0, p]. ì æ -p ö ü -p \ cos-1 ícos ç ÷ ý ¹ × è 3 øþ 3 î pü ì æ -p ö ü ì Now, cos-1ícos ç ÷ ý = cos-1 ícos ý [Q cos ( - q) = cos q] 3 3þ î è ø î þ p é p ù = Q Î [0, p]ú × 3 êë 3 û p p ì ü æ ö \ cos-1 ícos ç ÷ ý = × è 3 øþ 3 î æ -p p ö (iii) We know that the principal-value branch of tan -1 is ç , ÷× è 2 2ø 5p ü 5p ì × \ tan -1 ítan ý¹ 6þ 6 î 5p ü pöü ì æ ì Now, tan -1ítan ý = tan -1ítan ç p - ÷ ý 6þ 6øþ è î î ì æ -p ö ü = tan -1ítan ç ÷ ý î è 6 øþ [Q tan( p - q) = tan( - q)]

SSS Mathematics for Class 12 100

100

Senior Secondary School Mathematics for Class 12

-p é Q 6 êë 5p ü ì \ tan -1 ítan ý= 6þ î =

EXAMPLE 5

SOLUTION

EXAMPLE 6

Evaluate: 3ö æ (i) sin ç cos-1 ÷ 5ø è

-p æ -p p ö ù , ÷ú × Î ç 6 2øû è 2 -p × 6

3ö æ (ii) cos ç tan -1 ÷ 4ø è

3 = q, where q Î [0, p]. 5 3 Then, cos q = × 5 Since q Î [0, p ], we have sin q > 0. 9 16 4 \ sin q = 1 - cos2 q = 1 = = × 25 25 5 3ö 4 æ Hence, sin ç cos-1 ÷ = sin q = × 5ø 5 è æ -p p ö -1 3 , ÷× = q, where q Î ç (ii) Let tan 2ø 4 è 2 3 Then, tan q = × 4 æ -p p ö Since q Î ç , ÷ , so cos q > 0. 2ø è 2 1 1 1 16 4 \ cos q = = = = = × 2 sec q 25 5 9 1 + tan q 1+ 16 4 æ -1 3 ö Hence, cos ç tan ÷ = cos q = × 4ø 5 è (i) Let cos-1

Evaluate: ìp æ -1 ö ü (i) sin í - sin -1 ç ÷ ý [CBSE 2008] è 2 øþ î3 (iii) sin (cot -1 x)

SOLUTION

4ö æ1 (ii) sin ç cos-1 ÷ 5ø è2 1æ 5ö ÷ (iv) tan çç cos-1 2è 3 ÷ø

(i) We know that sin -1( - q) = - sin -1 q. 1ü ìp æ -1 ö ü ìp \ sin í - sin -1 ç ÷ ý = sin í + sin -1 ý 3 2 3 2þ î è øþ î 1 pù æ p pö é = sin ç + ÷ êQ sin -1 = ú 2 6û è 3 6ø ë = sin

p = 1. 2

SSS Mathematics for Class 12 101

Inverse Trigonometric Functions

101

4 = q, where q Î [0, p]. 5 4 Then, cos q = × 5

-1 (ii) Let cos

Since q Î [0, p] Þ

1 pù 1 é q Î ê 0, ú Þ sin q > 0. 2 2û 2 ë

4ö 1 1 - cos q æ1 \ sin ç cos-1 ÷ = sin q = = 2 5 2 2 è ø

æ 4ö 1-ç ÷ è5 ø = 1 × 2 10

(iii) Let cot -1 x = q. Then, q Î [0, p]. \ sin (cot -1 x) = sin q > 0. Now, sin q =

1 1 1 = = × 2 cosec q 1 + cot q 1 + x2

\ sin (cot -1 x) = sin q =

1 1 + x2

×

5 5 = q. Then, cos q = , where q Î [0, p]. 3 3 (1 - 5 3) 1æ 5ö 1 1 - cos q ÷ = tan q = = \ tan çç cos-1 2è 3 ÷ø 2 1 + cos q (1 + 5 3)

(iv) Let cos-1

( 3 - 5) ( 3 - 5) ( 3 - 5) ´ = × 2 ( 3 + 5) ( 3 - 5)

= é æ -3 ö ù sin ê 2 cos-1 ç ÷ ú × è 5 øû ë

EXAMPLE 7

Evaluate

SOLUTION

æ -3 ö Let cos-1 ç ÷ = q, where q Î [0, p]. è 5 ø Then, cos q =

-3 × 5

Since q Î [0, p ], we have sin q > 0. \ sin q = 1 - cos2 q = 1 -

9 = 25

16 4 = × 25 5

é æ -3 ö ù \ sin ê 2 cos-1 ç ÷ ú = sin 2q è 5 øû ë 4 æ -3 ö ü -24 ì = 2 sin q cos q = í2 ´ ´ ç ÷ ý = × î 5 è 5 ø þ 25

SSS Mathematics for Class 12 102

102

Senior Secondary School Mathematics for Class 12

EXAMPLE 8

Evaluate

SOLUTION

Let sin -1

3 5 ö æ cos ç sin -1 + sin -1 ÷× 5 13 ø è 3 5 = A and sin -1 = B. Then, 5 13

é -p p ù A, B Î ê , ú Þ cos A > 0 and cos B > 0. 2û ë 2 3 5 \ sin A = and sin B = 5 13 Þ cos A = 1 - sin 2 A = 1 -

9 = 25

16 4 = 25 5

and cos B = 1 - sin 2 B = 1 -

25 144 12 = = × 169 169 13

3 5 ö æ \ cos ç sin -1 + sin -1 ÷ = cos ( A + B) 5 13 ø è = cos A cos B - sin A sin B æ 4 12 ö æ 3 5 ö =ç ´ ÷-ç ´ ÷ è 5 13 ø è 5 13 ø æ 48 15 ö 33 × =ç ÷= è 65 65 ø 65 EXAMPLE 9

1öü ì æ Find the value of tan -1 í2 cos ç 2 sin -1 ÷ ý × 2ø þ è î

SOLUTION

We have 1öü pöü ì ì æ æ tan -1 í2 cos ç 2 sin -1 ÷ ý = tan -1í2 cos ç 2 ´ ÷ ý 2ø þ 6øþ è è î î

pù é -1 1 êëQ sin 2 = 6 úû

pü ì = tan -1 í2 cos ý 3þ î 1ö p æ = tan -1 ç 2 ´ ÷ = tan -1 1 = × 2ø 4 è EXAMPLE 10

SOLUTION

If tan -1 tan -1

4 = q, find the value of cos q. 3

4 æ -p p ö = q, where q Î ç , ÷× 2ø 3 è 2

\ tan q =

4 × 3

æ -p p ö We know that cos q > 0, when q Î ç , ÷× 2ø è 2

SSS Mathematics for Class 12 103

Inverse Trigonometric Functions

3 1 1 1 = = = × 2 5 sec q 16 1 + tan q 1+ 9

\ cos q =

EXAMPLE 11

SOLUTION

103

æ 1ö If cot -1 ç - ÷ = q, find the value of sin q. è 5ø æ 1ö cot -1 ç - ÷ = q, where q Î ( 0, p). è 5ø

Given:

-1 × 5

\ cot q =

sin q > 0 in ( 0, p). sin q =

EXAMPLE 12

SOLUTION

1 1 1 5 = = = × 2 cosec q 1 26 1 + cot q 1+ 25

Prove that

tan -1

1 1 2 + tan -1 = tan -1 × 7 13 9

æx+yö ÷÷ , xy < 1. We know that tan -1 x + tan -1 y = tan -1 çç è 1 - xy ø Let x =

1 1 1 and y = × Then, xy = < 1. 7 13 91

ìæ1 1 ö ü ïï ç 7 + 13 ÷ ïï 1 1 ø = tan -1 æ 20 ö = tan -1 2 × = tan -1 í è \ tan -1 + tan -1 ç ÷ 1 1 ý 7 13 9 è 90 ø ï1 - ´ ï ïî 7 13 ïþ EXAMPLE 13

SOLUTION

Prove that

tan -1

3 8 p 3 + tan -1 - tan -1 = × 4 5 19 4

We have ì î

LHS = ítan

= tan

-1

-1

3 3ü 8 + tan -1 ý - tan -1 4 5þ 19

ì æ 3 3ö ü ïï ç 4 + 5 ÷ ïï è ø -1 8 í ý - tan 3 3 æ ö 19 ï ç1 - ´ ÷ ï ïî è 4 5 ø ïþ

8 æ 27 ö = tan -1 ç ÷ - tan -1 19 è 11 ø

[CBSE 2009, ’12C]

SSS Mathematics for Class 12 104

104

Senior Secondary School Mathematics for Class 12

= tan -1

8ö æ 27 - ÷ ç è 11 19 ø = tan -1 27 8 ö æ ´ ÷ ç1 + 11 19 ø è

p æ 425 ö -1 ÷ = tan 1 = = RHS. ç 4 è 425 ø

\ LHS = RHS. EXAMPLE 14

1 1 1 1 p + tan -1 + tan -1 + tan -1 = × 3 5 7 8 4

Prove that tan -1

[CBSE 2008, ‘09C] SOLUTION

We have æ è

LHS = ç tan -1

= tan

-1

= tan

-1

= tan -1

= tan

-1

1 1ö æ 1 1ö + tan -1 ÷ + ç tan -1 + tan -1 ÷ 3 5ø è 7 8ø

æ 1 1ö ç + ÷ è 3 5 ø + tan -1 1 1ö æ ç1 - ´ ÷ 3 5ø è æ 8ö ç ÷ è 15 ø + tan -1 æ 14 ö ç ÷ è 15 ø

æ1 1ö ç + ÷ è 7 8ø 1 1ö æ ç1 - ´ ÷ 7 8ø è

æ 15 ö ç ÷ è 56 ø æ 55 ö ç ÷ è 56 ø

8 15 4 3 + tan -1 = tan -1 + tan -1 14 55 7 11 3ö æ4 æ 65 ö ç + ÷ ç ÷ è 7 11 ø = tan -1 è 77 ø = tan -1 1 = p = RHS. 4 3ö æ æ 65 ö 4 ç1 - ´ ÷ ç ÷ 7 11 ø è è 77 ø

\ LHS = RHS. EXAMPLE 15 SOLUTION

Prove that

2 tan -1

1 1 31 + tan -1 = tan -1 × 2 7 17

We have LHS = 2 tan

-1

= tan -1

= tan -1

1 1 + tan -1 2 7 1ö æ ç2 ´ ÷ 2ø è

2ü

ïì æ1ö ï í1 - ç ÷ ý è 2 ø þï îï

+ tan -1

1 7

é -1 -1 æ 2x ö ù ÷ú êQ 2 tan x = tan ç è 1 - x2 ø û ë

1 1 4 1 + tan -1 = tan -1 + tan -1 æ 3ö 7 3 7 ç ÷ è 4ø

SSS Mathematics for Class 12 105

Inverse Trigonometric Functions

= tan -1

105

æ 31 ö æ 4 1ö ç ÷ ç + ÷ è 3 7 ø = tan -1 è 21 ø = tan -1 31 = RHS. 4 1ö æ æ 17 ö 17 ç1 - ´ ÷ ç ÷ 3 7ø è è 21 ø

\ LHS = RHS. EXAMPLE 16

SOLUTION

Prove that

1 2ö 4 æ 2 ç tan -1 + tan -1 ÷ = tan -1 × 4 9ø 3 è

We have LHS = 2 tan -1

= tan

-1

1 2 + 2 tan -1 4 9 1ö æ ç2 ´ ÷ 4ø è

2 ìï æ 1 ö üï í1 - ç ÷ ý è 4 ø ïþ ïî

+ tan

æ1ö ç ÷ 2 = tan -1 è ø + tan -1 æ 15 ö ç ÷ è 16 ø

-1

2ö æ ç2 ´ ÷ 9ø è

2 ìï æ 2 ö üï í1 - ç ÷ ý è 9 ø ïþ ïî

æ 4ö ç ÷ è 9ø æ 77 ö ç ÷ è 81 ø

æ 1 16 ö æ 4 81 ö = tan -1 ç ´ ÷ + tan -1 ç ´ ÷ è 2 15 ø è 9 77 ø = tan -1

= tan

-1

8 36 + tan -1 15 77 æ 8 36 ö ÷ ç + è 15 77 ø = tan -1 ( 616 + 540) 8 36 ö æ (1155 - 288) ´ ÷ ç1 è 15 77 ø

æ 1156 ö -1 4 = tan -1 ç = RHS. ÷ = tan 3 è 867 ø \ LHS = RHS. EXAMPLE 17 SOLUTION

Prove that cot -1 7 + cot -1 8 + cot -1 18 = cot -1 3. [CBSE 2008C, ’14] We have LHS = cot

-1

7 + cot -1 8 + cot -1 18

1 1ö 1 æ = ç tan -1 + tan -1 ÷ + tan -1 7 8ø 18 è

SSS Mathematics for Class 12 106

106

Senior Secondary School Mathematics for Class 12

æ1 1ö æ 15 ö ç + ÷ ç ÷ 1 1 7 8ø 56 + tan -1 = tan -1 è = tan -1 è ø + tan -1 1 1 55 ö æ æ ö 18 18 ç1 - ´ ÷ ç ÷ 7 8ø è è 56 ø 1ö æ 3 ç + ÷ 18 ø æ -1 3 -1 1 ö -1 è 11 = ç tan + tan ÷ = tan 3 1ö æ 11 18 ø è ´ ÷ ç1 11 18 ø è æ 65 ö ÷ ç 1 198 ø = tan -1 = cot -1 3 = RHS. = tan -1 è 195 ö æ 3 ÷ ç è 198 ø \ LHS = RHS. EXAMPLE 18

SOLUTION

Prove that

sin -1

3 8 84 - sin -1 = cos-1 × 5 17 85

[CBSE 2012]

3 8 = x and sin -1 = y. Then, 5 17 3 8 sin x = and sin y = × 5 17 9 16 4 = = \ cos x = 1 - sin 2 x = 1 25 25 5 64 225 15 and cos y = 1 - sin 2 y = 1 = = × 289 289 17 Let sin -1

\ cos ( x - y) = cos x cos y + sin x sin y æ 4 15 ö æ 3 8 ö æ 12 24 ö 84 + =ç ´ ÷+ç ´ ÷=ç ÷= è 5 17 ø è 5 17 ø è 17 85 ø 85 Þ

EXAMPLE 19

SOLUTION

æ 84 ö x - y = cos-1 ç ÷ Þ è 85 ø

Prove that

tan -1

sin -1

3 8 84 - sin -1 = cos-1 × 5 17 85

1 2 1 3 + tan -1 = cos-1 × 4 9 2 5

1 2 æ 1 2ö 1 0). Solve tan -1 çç è1 + x ø 2

[CBSE 2008C, ’11, ’14C]

We have æ1 - x ö 1 ÷÷ = tan -1 x , ( x > 0) tan -1 çç è1 + x ø 2 1 Þ tan -1 1 - tan -1 x = tan -1 x 2 p 3 -1 -1 Þ tan x = tan 1 = 2 4 æp 2ö p -1 Þ x = tan Þ tan x = ç ´ ÷ = è 4 3ø 6 1 Hence, x = × 3

EXAMPLE 25 SOLUTION

1 æ pö × ç ÷= 6 3 è ø

Solve 2 tan -1 (cos x) = tan -1 ( 2cosec x). We have 2 tan -1 (cos x) = tan -1 ( 2cosec x) æ 2 cos x ö ÷ tan -1 ( 2cosec x) Þ tan -1 ç ç 1 - cos2 x ÷ = è ø é -1 æ 2 cos x ö ù Þ tan ê tan ç ÷ ú = 2cosec x è sin 2 x ø û ë 2 cos x = 2cosec x Þ cos x = sin x Þ sin 2 x p Þ tan x = 1 Þ x = × 4

PROBLEMS BASED ON TRIGONOMETRIC FORMULAE EXAMPLE 26

SOLUTION

Prove that

æ 1 - cos x ö x ÷ = , x < p. tan -1 çç ÷ è 1 + cos x ø 2

We have LHS = tan

-1

æ 1 - cos x ö -1 ç ÷ ç 1 + cos x ÷ = tan è ø

xö x æ = tan -1 ç tan ÷ = = RHS. 2ø 2 è æ cos 1 x ö÷ x \ tan -1 çç ÷= × è 1 + cos x ø 2

æxö 2 sin 2 ç ÷ è 2ø 2æ x ö 2 cos ç ÷ è 2ø

SSS Mathematics for Class 12 110

110

EXAMPLE 27 SOLUTION

Senior Secondary School Mathematics for Class 12

Prove that

æ cos x - sin x ö æ p ÷÷ = ç - x ö÷ , x < p. tan -1 çç cos sin + x x ø ø è4 è

We have æ cos x - sin x ö ÷÷ è cos x + sin x ø

LHS = tan -1 çç

æ 1 - tan x ö ÷÷ [dividing num. and denom. by cos x] = tan -1 çç è 1 + tan x ø ì æp öü æ p ö = tan -1 ítan ç - x ÷ ý = ç - x ÷ = RHS. è4 øþ è 4 ø î ö æ x x p cos sin ÷÷ = æç - x ö÷ × \ tan -1 çç ø è cos x + sin x ø è 4 EXAMPLE 28

SOLUTION

Prove that

æ cos x ö æ p x ö ÷÷ = ç - ÷ × tan -1 çç è 1 + sin x ø è 4 2 ø

[CBSE 2012]

We have æ cos x ö ÷÷ è 1 + sin x ø

LHS = tan -1 çç

ì æp ö ü ïï sin ç 2 - x ÷ ïï è ø = tan -1 í ý p æ ï1 + cos ç - x ö÷ ï ïî ø ïþ è2 ì ïï 2 sin = tan -1 í ï ïî

æp xöü æp xö ç - ÷ cos ç - ÷ ï 4 2 è ø è 4 2ø ï ý p x ö æ ï 2 cos2 ç - ÷ ïþ è 4 2ø

ì æ p x öü æ p x ö = tan -1 ítan ç - ÷ ý = ç - ÷ = RHS. è 4 2øþ è 4 2ø î æ cos x ö æ p x ö ÷÷ = ç - ÷ × \ tan -1 çç è 1 + sin x ø è 4 2 ø EXAMPLE 29

SOLUTION

Prove that

æ cos x ö æ p x ö -p p ö ÷÷ = ç + ÷ , x Î æç tan -1 çç , ÷ × [CBSE 2012] è 2 2ø è 1 - sin x ø è 4 2 ø

We have LHS = tan

-1

æ cos x ö çç ÷÷ è 1 - sin x ø

SSS Mathematics for Class 12 111

Inverse Trigonometric Functions

111

ì æp ö ü ïï sin ç 2 - x ÷ ïï è ø = tan í ý p æ ï1 - cos ç - x ö÷ ï ïî ø ïþ è2 p x ì æ ö æp xöü ïï 2 sin ç 4 - 2 ÷ cos ç 4 - 2 ÷ ïï è ø è ø = tan -1 í ý xö 2 æp ï ï 2 sin ç - ÷ ïî ïþ è 4 2ø é ì p æ p x ö üù ì æ p x öü = tan -1 ícot ç - ÷ ý = tan -1 ê tan í - ç - ÷ ýú 4 2 øþ è î î 2 è 4 2 ø þû ë ì æ p x öü = tan -1 ítan ç + ÷ ý è 4 2øþ î æp xö = ç + ÷ = RHS. è 4 2ø æ cos x ö æ p x ö ÷÷ = ç + ÷ × Hence, tan -1 çç è 1 - sin x ø è 4 2 ø -1

EXAMPLE 30

SOLUTION

Prove that

æ 3 a 2x - x 3 ö ÷ = 3 tan -1 x × tan -1 ç 3 ç a - 3 ax 2 ÷ a è ø

Putting x = a tan q, we get æ 3 a 2x - x 3 ö ÷ = tan -1 tan -1 ç 3 ç a - 3 ax 2 ÷ è ø

æ 3 a 3 tan q - a 3 tan 3 q ö ç ÷ ç a 3 - 3 a 3 tan 2 q ÷ è ø

æ 3 tan q - tan 3 q ö ÷ = tan -1(tan 3 q) = tan -1 ç ç 1 - 3 tan 2 q ÷ è ø = 3 q = 3 tan -1

x × a

æ 3 a 2x - x 3 ö ÷ = 3 tan -1 x × Hence, tan -1 ç 3 ç a - 3 ax 2 ÷ a è ø EXAMPLE 31

SOLUTION

Prove that

æ a cos x - b sin tan -1 çç è b cos x - a sin

xö a ÷ = tan -1 æç ö÷ - x. x ÷ø è bø

We have LHS = tan

-1

ì a cos x - b sin x ü xö b cos x ïï ïï ÷÷ = tan -1 í ý cos sin b x a x xø ï ï ïþ ïî b cos x [on dividing num. and denom. by b cos x]

æ a cos x - b sin çç è b cos x - a sin

SSS Mathematics for Class 12 112

112

Senior Secondary School Mathematics for Class 12

ì a - tan x ü ï ï a -1 æ p - q ö ÷÷ , where = p and tan x = q = tan í b ý = tan çç a pq 1 b è ø ï 1 - tan x ï b î þ a = tan -1 p - tan -1 q = tan -1 - tan -1 (tan x ) b a = æç tan -1 - x ö÷ = RHS. b è ø -1

æ a cos x - b sin x ö æ a ÷÷ = ç tan -1 - x ÷ö × Hence, tan -1 çç b x a x b cos sin è ø è ø EXAMPLE 32

ì 1 + sin x + 1 - sin x ü x Prove that cot -1 í ý= , xÎ î 1 + sin x - 1 - sin x þ 2

æ 0, p ö × ÷ ç è 4ø [CBSE 2011, ’14]

SOLUTION

We have ì ( 1 + sin x + 1 - sin x )

LHS = cot -1 í

î ( 1 + sin x - 1 - sin x )

´

( 1 + sin x + 1 - sin x ) ü ý ( 1 + sin x + 1 - sin x ) þ

2 ü ì ï (1 + sin x ) + (1 - sin x ) + 2 1 - sin x ï = cot -1 í ý (1 + sin x ) - (1 - sin x ) ïî ïþ

ì 2 (1 + cos x ) ü -1 = cot -1 í ý = cot 2 x sin î þ = cot

-1

æ 1 + cos x ö çç ÷÷ è sin x ø

x ì ü 2 cos2 æç ö÷ ïï ïï xö x -1 æ è2ø ý = cot ç cot ÷ = = RHS. í x x 2ø 2 æ ö æ ö è ï 2 sin ç ÷ cos ç ÷ ï ïî è2ø è 2 ø ïþ

Hence, LHS = RHS. EXAMPLE 33

SOLUTION

ì 1+x - 1-x ü p 1 -1 Prove that tan -1 í ý = - cos x . 4 2 1 + x + 1 x î þ

[CBSE 2011, ’14]

Putting x = cos q , we get LHS = tan

-1

= tan

-1

ì 1 + cos q - 1 - cos q ü í ý î 1 + cos q + 1 - cos q þ ì 2 æq ö 2æ q ö ü ï 2 cos ç ÷ - 2 sin ç ÷ ï ï è2ø è 2 ø ï = tan -1 í ý ï 2 cos2 æç q ö÷ + 2 sin 2 æç q ö÷ ï ïî è 2 ø ïþ è2ø

ì cos q - sin q ü ï 2 2ï í q qý ï cos + sin ï 2 2þ î

SSS Mathematics for Class 12 113

Inverse Trigonometric Functions

= tan

-1

ì ï 1 - tan í ï1 + tan î

qü 2ï qý ï 2þ

113

[dividing num. and denom. by cos

q ] 2

ì æ p qö ü æ p qö = tan -1 ítan ç - ÷ ý = ç - ÷ è 4 2ø þ è 4 2ø î ö æp 1 = ç - cos-1 x ÷ = RHS. ø è4 2 Hence, LHS = RHS. EXAMPLE 34

SOLUTION

Prove that

æ 1 + x2 - 1ö 1 ÷ = tan -1 x. tan -1 ç ÷ 2 ç x ø è

Putting x = tan q, we get æ 1 + x2 - 1ö ÷ = tan -1 tan -1 ç ÷ ç x ø è

æ 1 + tan 2 q - 1 ö ÷ ç ÷ ç tan q ø è

æ sec q - 1 ö ÷÷ = tan -1 = tan -1 çç è tan q ø = tan

=

-1

ì ü æ qö 2 sin 2 ç ÷ ïï ïï qö è 2ø -1 æ í ý = tan ç tan ÷ q q ö æ ö 2ø æ è ï 2 sin ç ÷ cos ç ÷ ï ïî ï 2 2 è øþ è ø

q 1 = tan -1 x. 2 2

æ 1 + x2 - 1ö 1 ÷ = tan -1 x. \ tan -1 ç ç ÷ 2 x è ø EXAMPLE 35

SOLUTION

Prove that

æ ö x ÷ = sin -1 x × tan -1 ç ç 2 2÷ a è a -x ø

Putting x = a sin q, we get LHS = tan

-1

æ 1 - cos q ö çç ÷÷ è sin q ø

ö æ x ÷ ç ç 2 2÷ è a -x ø

ö æ a sin q ÷ = tan -1 ç ç 2 2 2 ÷ è a - a sin q ø

SSS Mathematics for Class 12 114

114

Senior Secondary School Mathematics for Class 12

æ a sin q ö ÷÷ = tan -1(tan q) = tan -1 çç è a cos q ø x = q = sin -1 = RHS. a æ ö x ÷ = sin -1 x × \ tan -1 ç ç 2 2÷ a è a -x ø EXAMPLE 36

SOLUTION

Prove that

tan -1

x =

1 cos-1 2

æ1 - x ö çç ÷÷ , x Î [0, 1] × è1 + x ø

[CBSE 2010]

Putting x = tan 2 q , we get RHS =

1 cos-1 2

æ1 - x ö çç ÷÷ è1 + x ø

=

1 cos-1 2

æ 1 - tan 2 q ö ç ÷ ç 1 + tan 2 q ÷ è ø

=

1 cos-1(cos 2q) 2

æ1 ö = ç ´ 2q÷ = q = tan -1 x = LHS è2 ø [Q x = tan 2 q Þ tan q = x Þ q = tan -1 x ]. Hence, tan -1

EXAMPLE 37

SOLUTION

Prove that

x =

1 cos-1 2

æ1 - x ö çç ÷÷ × è1 + x ø

æ ö 1 ÷ = æ p - sec-1 x ö × tan -1 ç ÷ ç 2 ÷ çè 2 ø è x -1ø

Putting x = sec q, we get æ ö 1 ÷ = tan -1 tan -1 ç ç 2 ÷ è x -1ø

æ ö 1 ç ÷ ç ÷ 2 è sec q - 1 ø æ 1 ö ÷÷ = tan -1(cot q) = tan -1 çç tan q è ø æp öü æ p ö æp ö -1 ì = tan ítan ç - q÷ ý = ç - q÷ = ç - sec-1 x ÷ × è2 øþ è 2 ø è2 ø î

æ ö 1 ÷ = æ p - sec-1 x ö × \ tan -1 ç ÷ ç 2 ÷ çè 2 ø è x -1ø

SSS Mathematics for Class 12 115

Inverse Trigonometric Functions

EXAMPLE 38

SOLUTION

Prove that

115

æ 1 + x2 + 1 - x2 ö p 1 ÷ = + cos-1 x 2. tan -1 ç ç 2 2 ÷ 4 2 è 1+ x - 1-x ø

Putting x 2 = cos 2q, we get æ 1 + x2 + 1 - x2 ö ÷ = tan -1 æç 1 + cos 2q + 1 - cos 2q ö÷ tan -1 ç ç 1 + cos 2q - 1 - cos 2q ÷ ç 1 + x2 - 1 - x2 ÷ è ø è ø æ 2 cos2 q + 2 sin 2 q ö ÷ = tan -1 ç ç 2 cos2 q - 2 sin 2 q ÷ ø è æ cos q + sin q ö -1 æ 1 + tan q ö = tan -1 ç ÷ ÷ = tan ç è cos q - sin q ø è 1 - tan q ø [dividing num. and denom. by cos q] ì æp öü æ p ö = tan -1 ítan ç + q÷ ý = ç + q÷ è4 øþ è 4 ø î =

p 1 + cos-1 x 2. 4 2

æ 1 + x2 + 1 - x2 ö p 1 ÷ = + cos-1 x 2. \ tan -1 ç ç 2 2 ÷ 4 2 è 1+ x - 1-x ø EXAMPLE 39

SOLUTION

Prove that Let tan -1

2 tan -1

1 = sin -1 x

æ 2x ö ç ÷ ç x2 + 1÷ × è ø

1 1 = q. Then, = tan q Þ x = cot q. x x

\ LHS = 2 tan -1 RHS = sin

-1

1 = 2q. x

æ 2 cot q ö -1 ç ÷ ç cot 2 q + 1 ÷ = sin è ø

æ 2 tan q ö ÷ ç ç 1 + tan 2 q ÷ è ø

= sin -1(sin 2q) = 2q. \ LHS = RHS. Hence, 2 tan -1

EXAMPLE 40

1 = sin -1 x

ìï 1 tan í sin -1 îï 2

æ 2x ö ç ÷ ç x2 + 1÷ × è ø

æ 2x ö 1 -1 ç ÷ ç 1 - x 2 ÷ + 2 cos è ø where| x | < 1, y > 0 and xy < 1.

Prove that

æ 1 - y 2 ö ïü æ x + y ö ÷ =ç ç ÷, ç 1 + y 2 ÷ ýï ç 1 - xy ÷ ø øþ è è [CBSE 2013]

SSS Mathematics for Class 12 116

116 SOLUTION

Senior Secondary School Mathematics for Class 12

Putting x = tan q and y = tan f, we get ìï 1 sin -1 2 îï

LHS = tan í

ìï 1 = tan í sin -1 ïî 2

æ 2x ö 1 ç ÷ + cos-1 ç 1 - x2 ÷ 2 è ø

æ 1 - y 2 ö üï ÷ ç ç 1 + y 2 ÷ ýï øþ è

æ 2 tan q ö 1 -1 ç ÷ ç 1 - tan 2 q ÷ + 2 cos è ø

æ 1 - tan 2 f ö üï ç ÷ ç 1 + tan 2 f ÷ ýï è øþ

1 ü ì1 = tan í sin -1 (sin 2q) + cos-1(cos 2f) ý 2 þ î2 ìæ 1 ö æ1 öü = tan íç ´ 2q÷ + ç ´ 2f÷ ý ø è2 øþ îè 2 = tan ( q + f) = ìï 1 \ tan í sin -1 ïî 2 EXAMPLE 41

SOLUTION

Prove that

tan q + tan f ( x + y) = = RHS. 1 - tan q tan f (1 - xy)

æ 2x ö 1 -1 ç ÷ ç 1 - x 2 ÷ + 2 cos è ø

æ ab + 1 ö ÷÷ + cot -1 cot -1 çç a b è ø

æ 1 - y 2 ö üï x + y ç ÷ = × ç 1 + y 2 ÷ ýï 1 - xy è øþ æ bc + 1 ö çç ÷÷ + cot -1 b c è ø

æ ca + 1 ö çç ÷÷ = 0. è c-a ø

We have æ a-bö ÷÷ + tan -1 + 1 ab è ø

LHS = tan -1 çç

æ b-cö çç ÷÷ + tan -1 + 1 bc è ø

æ c-a ö çç ÷÷ è 1 + ca ø

= (tan -1 a - tan -1 b) + (tan -1 b - tan -1 c) + (tan -1 c - tan -1 a) = 0 = RHS. \ LHS = RHS. SOME MORE EXAMPLES EXAMPLE 1

Prove that

SOLUTION

We have

cos-1

-1

LHS = cos

4 12 33 + cos-1 = cos-1 × 5 13 65

4 12 + cos-1 5 13

2 2ü ì 4 12 ïæ ö æ 4ö æ 12 ö ï = cos-1íç ´ ÷ - 1 - ç ÷ × 1 - ç ÷ ý è 13 ø ï è5 ø ïîè 5 13 ø þ

144 ïü ïì 48 æ 16 ö = cos-1í - 1 - ç ÷ × 1 ý 65 25 169 ïþ è ø îï

[CBSE 2010, ’12]

SSS Mathematics for Class 12 117

Inverse Trigonometric Functions

117

ì 48 9 25 ü = cos-1í × ý 65 25 169 î þ æ 48 15 ö ì 48 3 5 ü - ÷ = cos-1í - ´ ý = cos-1 ç è 65 65 ø î 65 5 13 þ æ 33 ö = cos-1 ç ÷ = RHS. è 65 ø Hence, cos-1

EXAMPLE 2

Prove that

SOLUTION

LHS

4 12 33 + cos-1 = cos-1 × 5 13 65

sin -1

8 3 36 + sin -1 = cos-1 × 17 5 85

= sin -1

8 3 + sin -1 17 5

= cos-1

15 4 + cos-1 17 5

[CBSE 2014C]

2 2ü ì 15 4 ïæ ö æ 15 ö æ 4ö ï = cos-1íç ´ ÷ - 1 - ç ÷ × 1 - ç ÷ ý è 17 ø è5 ø ï ïîè 17 5 ø þ

16 ïü ïì 12 æ 225 ö = cos-1í - 1 - ç ÷ × 1ý 25 ïþ 289 17 è ø îï ì 12 64 9 ü 8 3ü -1 ì 12 = cos-1í × ´ ý ý = cos í 17 289 25 17 17 5þ î î þ æ 36 ö ì 12 24 ü = cos-1í - ý = cos-1 ç ÷ = RHS. è 85 ø î17 85 þ Hence, sin -1

EXAMPLE 3

Prove that

SOLUTION

We have

8 3 36 + sin -1 = cos-1 × 17 5 85

2 tan -1

LHS = tan

-1

3 17 p - tan -1 = × 4 31 4

3 17 - tan 4 31

ì 3 ü ï ï 2´ ï 4 ï - tan -1 17 = tan -1í 2ý 31 ï1 - æ 3 ö ï ïî çè 4 ÷ø ïþ

[CBSE 2011C]

SSS Mathematics for Class 12 118

118

Senior Secondary School Mathematics for Class 12

= tan

-1

ì æ 24 17 ö ü ï ç - 31 ÷ ïï 24 ø -1 17 -1 ï è 7 - tan = tan í ý 24 17 7 31 æ ï1 + ç ´ ö÷ ï îï è 7 31 ø ïþ

p ì 625 217 ü -1 = tan -1í ´ ý = tan 1 = = RHS. 4 î 217 625 þ Hence, 2 tan -1

EXAMPLE 4

Prove that

SOLUTION

Let sec-1

3 17 p - tan -1 = × 4 31 4

2 tan -1

1 5 2 1 p + sec-1 + 2 tan -1 = × 5 7 8 4

[CBSE 2014]

7 5 2 5 2 × So, cos q = × = q. Then, sec q = 7 7 5 2

Draw a right triangle with base = 7 units and hypotenuse = 5 2 units. \ perpendicular = (5 2) 2 - 7 2 = 50 - 49 = 1 unit. \ tan q =

1 1 Þ q = tan -1 × 7 7

1 1ö 1 æ \ LHS = 2 ç tan -1 + tan -1 ÷ + tan -1 5 8 7 è ø ì æ1 1ö ü ïï ç 5 + 8 ÷ ïï ø + tan -1 1 = 2 × tan -1í è ý 7 ï æç1 - 1 ´ 1 ö÷ ï ïî è 5 8 ø ïþ 1 1 1 æ 13 40 ö = 2 tan -1 ç ´ ÷ + tan -1 = 2 tan -1 + tan -1 7 3 7 è 40 30 ø ì 1 ü ï 2´ ï ï 3 ï + tan -1 1 = tan -1 3 + tan -1 1 = tan -1í 2ý 7 4 7 ï1 - æ 1 ö ï ç ÷ ïî è 3 ø ïþ = tan

=

-1

æ 3 1ö ç + ÷ è 4 7 ø = tan -1 25 = tan -1 1 3 1ö æ 25 ç1 - ´ ÷ 4 7ø è

p = RHS. 4

Hence, LHS = RHS.

SSS Mathematics for Class 12 119

Inverse Trigonometric Functions

119

ö p æx 1 cos-1 x + cos-1 ç + 3 - 3x2 ÷ = × ø 3 è2 2

EXAMPLE 5

Prove that

SOLUTION

Let cos-1 x = q. Then, x = cosq and 1 - x 2 = sin q.

[CBSE 2014C]

æx ö 3 1 - x 2 ÷÷ \ LHS = cos-1 x + cos-1 çç + 2 è2 ø p p æ = q + cos-1 ç cos cos q + sin sin 3 3 è

ö q÷ ø

ì æp öü æp ö p = q + cos-1ícos ç - q÷ ý = q + ç - q÷ = = RHS. 3 3 øþ è ø 3 î è Hence, LHS = RHS. x2 + 1

cos [tan -1{sin(cot -1 x)}] =

EXAMPLE 6

Prove that

SOLUTION

Let cot -1 x = q. Then, x = cot q.

x2 + 2

×

\ cosec q = 1 + cot 2 q = 1 + x 2 1 Þ sin q = 1 + x2 1 [Q q = cot -1 x] Þ sin (cot -1 x) = 1 + x2 1 = f (say) Þ tan -1{sin (cot -1 x)} = tan -1 1 + x2 Þ cos [tan -1{sin (cot -1 x}] = cos f. 1 1 Now, tan -1 = f Þ tan f = 2 1 + x2 1+ x Þ sec f = 1 + tan 2 f = 1 + Þ cos f =

1 2

(1 + x )

=

1 + x2

× 2 + x2 From (i) and (ii), we get

EXAMPLE 7

SOLUTION

Prove that

... (ii)

x2 + 1 x2 + 2

×

3 3ö 6 æ × cos ç sin -1 + cot -1 ÷ = 5 2 ø 5 13 è

3 3 = q. Then, sin q = × 5 5 Draw a right triangle in which Let sin -1

... (i)

2 + x2

1 + x2

cos [tan -1{sin (cot -1 x)}] =

[CBSE 2010]

[CBSE 2012]

SSS Mathematics for Class 12 120

120

Senior Secondary School Mathematics for Class 12

perp. = 3 units, hyp. = 5 units and base = (5) 2 - ( 3) 2 = 16 = 4 units. 3 3 Þ q = tan -1 × \ tan q = 4 4 æ æ -1 3 -1 3 ö -1 2 ö \ cos ç sin + cot ÷ ÷ = cos ç q + tan 5 2 3ø è ø è é ì 3 2 üù + ê æ -1 3 -1 2 ö -1 ï 4 3 ïú = cos ç tan + tan ÷ = cos ê tan í 3 2 ýú 4 3ø è ï1 - ´ ïú ê î 4 3 þû ë 17 6 6 ì ü æ ö = cos ç tan -1 ÷ = cosícos-1 ý= 6ø 325 þ 5 13 è î 17 6 6 6 ù é -1 êQ tan f = 6 Þ cos f = 325 = 5 13 Þ f = cos 5 13 ú ú× ê 6 ê ú Þ cos f = 5 13 û ë EXAMPLE 8

SOLUTION

Evaluate: ì1 5ü (i) tan í cos-1 ý 2 3 î þ (i) Let cos-1

1 pü ì (ii) tan í2 tan -1 - ý 5 4þ î

5 5 = q. Then, cosq = × 3 3

ì1 q 5ü \ tan í cos-1 ý = tan 2 3 2 î þ 1 - cos q = = 1 + cos q

= =

æ ç1 ç è æ ç1 + ç è

5ö ÷ 3 ÷ø

5ö ÷ 3 ÷ø

=

( 3 - 5) ( 3 + 5)

( 3 - 5) ( 3 - 5) ( 3 - 5) ´ = ( 3 + 5) ( 3 + 5) 9 -5 3 - 5 ( 3 - 5) = × 2 4

1 pü ì (ii) tan í2 tan -1 - ý 5 4þ î 1 ö ü ì æ ï ï -1 ç 2 ´ 5 ÷ ÷ - tan -1 1ý = tan ítan ç 1 ÷÷ çç 1 ï ï è 25 ø þ î

SSS Mathematics for Class 12 121

Inverse Trigonometric Functions

121

ì ü æ 2 25 ö = tan ítan -1 ç ´ ÷ - tan -1 1ý è 5 24 ø î þ 5 ü ì = tan ítan -1 - tan -1 1ý 12 þ î ì öü æ 5 ï -1 ç 12 - 1 ÷ ï -7 ü -7 ì ÷ ý = tan ítan -1 æç ö÷ ý = × = tan ítan ç 5 è 17 ø þ 17 çç 1 + ÷÷ ï î ï è 12 ø þ î EXAMPLE 9

Prove that

SOLUTION

Let cos-1

aü a ü 2b ìp 1 ìp 1 × tan í + cos-1 ý + tan í - cos-1 ý = bþ bþ a î4 2 î4 2 a a = q. Then, = cos q. b b

ìp 1 ü ìp 1 ü \ LHS = tan í + qý = tan í - qý î4 2 þ î4 2 þ æ qö æ qö 1 + tan ç ÷ 1 - tan ç ÷ è 2ø è 2ø + = æ qö æ qö 1 - tan ç ÷ 1 + tan ç ÷ è 2ø è 2ø 2

2

ì æ qö ü ì æ qö ü í1 + tan ç ÷ ý + í1 - tan ç ÷ ý è 2ø þ î è 2ø þ î = q ì ü ö æ 2 í1 - tan ç ÷ ý è 2ø þ î

ì 2 æ qö ü ïï1 + tan ç 2 ÷ ïï è ø = 2 = 2b = RHS. = 2í ý æ 2 ï 1 - tan ç q ö÷ ï cos q a ïî è 2 ø ïþ Hence, LHS = RHS. EXAMPLE 10

SOLUTION

Prove that

9p 9 1 9 2 2 - sin -1 = sin -1 × 8 4 3 4 3

We have 9p 9 1 - sin -1 8 4 3 9æp -1 1 ö = ç - sin ÷ 4è 2 3ø

LHS =

=

9 1 9 1 cos-1 = × sin -1 1 9 4 3 4

[CBSE 2011]

SSS Mathematics for Class 12 122

122

Senior Secondary School Mathematics for Class 12

=

æ2 2ö 9 ÷ = RHS. sin -1 çç ÷ 4 è 3 ø

Hence, LHS = RHS. EXAMPLE 11

SOLUTION

æxö æx-yö ÷÷ × Find the value of tan -1 çç ÷÷ - tan -1 çç èyø èx + yø

[CBSE 2011]

We have æxö æx-yö ÷÷ tan -1 çç ÷÷ - tan -1 çç y è ø èx + yø ü ìx ï - 1 ïï -1 æ x ö -1 ï y = tan çç ÷÷ - tan í ý x èyø ï + 1ï ïþ îï y ü æxö ì æxö = tan -1 çç ÷÷ - ítan -1 çç ÷÷ - tan -1(1) ý èyø î èyø þ p -1 = tan (1) = × 4

EXAMPLE 12 SOLUTION

Prove that tan -1

1 2 1 3 1 4 + tan -1 = cos-1 = sin -1 × [CBSE 2010C] 4 9 2 5 2 5

We have

tan

-1

ì 1 2 ü + ï ïï 1 1 -1 2 -1 ï + tan = tan í 4 9 ý = tan -1 × 1 2 ö æ 4 9 2 ï1 - ç ´ ÷ ï îï è 4 9 ø ïþ

2ö æ æ 1 - x2 ö ÷ Þ tan -1 x = 1 cos-1 ç 1 - x ÷ Now, 2 tan -1 x = cos-1 ç ç 1 + x2 ÷ ç 1 + x2 ÷ 2 è è ø ø

Þ tan

-1

1ö æ ç1- ÷ 1 1 1 -1 ç 4 ÷ = cos-1 3 × = cos 2 2 5 çç 1 + 1 ÷÷ 2 è 4ø

1 æ 2x ö æ 2x ö And, 2 tan -1 x = sin -1 ç Þ tan -1 x = sin -1 ç ÷ 2÷ 2 è1 + x ø è 1 + x2 ø 1ö æ ç 2´ ÷ 1 1 1 -1 ç 2 ÷ = sin -1 4 × Þ tan = sin 2 2 5 çç 1 + 1 ÷÷ 2 è 4ø 1 2 1 3 1 4 Hence, tan -1 + tan -1 = cos-1 = sin -1 × 4 9 2 5 2 5 -1

SSS Mathematics for Class 12 123

Inverse Trigonometric Functions

123

Problems on Trigonometric Equations EXAMPLE 13 SOLUTION

Solve for x:

2 tan -1(cos x) = tan -1( 2cosec x).

[CBSE 2009, ’10C]

We have 2 tan -1(cos x) = tan -1( 2cosec x) æ 2 cos x ö -1 Þ tan -1 ç ÷ = tan ( 2cosec x) è 1 - cos2 x ø Þ tan -1( 2cosec x cot x) = tan -1( 2cosec x) Þ 2cosec x cot x = 2cosec x Þ cot x = 1 Þ x = Hence, x =

EXAMPLE 14 SOLUTION

p × 4

p × 4

Solve for x: 2 tan -1(sin x) = tan -1( 2cosec x), x ¹

p × 2

[CBSE 2012]

We have 2 tan -1(sin x) = tan -1( 2sec x) æ 2 sin x ö Þ tan -1 ç ÷ = tan ( 2sec x) è 1 - sin 2 x ø Þ tan -1( 2sec x tan x) = tan -1( 2sec x) Þ 2sec x tan x = 2sec x Þ tan x = 1 Þ x = Hence, x =

EXAMPLE 15

SOLUTION

p × 4

p × 4

1 ö æ If sin ç sin -1 + cos-1 x ÷ = 1, find the value of x. 5 ø è We have 1 æ ö sin ç sin -1 + cos-1 x ÷ = 1 5 è ø Þ sin -1

1 p + cos-1 x = sin -1 1 = 5 2

1ö æp Þ cos-1 x = ç - sin -1 ÷ 5ø è2 Þ cos-1 x = cos-1 Hence, x =

1 × 5

1 1 Þ x= × 5 5

[CBSE 2014]

SSS Mathematics for Class 12 124

124

EXAMPLE 16 SOLUTION

Senior Secondary School Mathematics for Class 12

If sin -1

x 5 p + cosec-1 = then find the value of x. 5 4 2

The given equation may be written as x 4 p sin -1 + sin -1 = 5 5 2 x æp 4ö Þ sin -1 = ç - sin ÷ 5 è2 5ø x 4 æ 4ö = cos-1 = sin -1 1 - ç ÷ 5 5 è5 ø 16 -1 -1 3 = sin 1= sin 25 5 x 3 Þ = Þ x = 3. 5 5 Hence, x = 3. Þ sin -1

EXAMPLE 17

SOLUTION

2

3ö æ Solve for x: cos (tan -1 x) = sin ç cot -1 ÷ . 4ø è Let cot -1

[CBSE 2014]

3 3 = q. Then, cot q = × 4 4

Draw a right triangle in which base = 3 units and perpendicular = 4 units. Then, hypotenuse =

3 2 + 42 = 25 = 5 units.

4 4 Þ q = sin -1 × 5 5 3ö 4ö 4 æ æ So, sin ç cot -1 ÷ = sin ç sin -1 ÷ = × 4 5 ø 5 è ø è x Let tan -1 x = f. Then, tan f = × 1

\ sin q =

Draw a right triangle in which perpendicular = x and base = 1. Then, hypotenuse = 1 + x 2 . 1 × \ cos f = 1 + x2 So, cos (tan -1 x) = cos f =

1

× 1 + x2 1 4 1 16 Thus, = Þ = Þ 16x 2 = 9 (1 + x 2) 25 1 + x2 5 9 3 Þ x2 = Þ x=± × 16 4 3 Hence, x = ± × 4

SSS Mathematics for Class 12 125

Inverse Trigonometric Functions

EXAMPLE 18 SOLUTION

125

p æ x - 2ö -1 æ x + 2 ö If tan -1 ç ÷ = , find the value of x. [CBSE 2014] ÷ + tan ç è x + 4ø 4 è x - 4ø The given equation may be written as æ x - 2ö p -1 æ x + 2 ö tan -1 ç ÷ ÷ = - tan ç è x + 4ø è x - 4ø 4 æ x + 2ö = tan -1 1 - tan -1 ç ÷ è x + 4ø

p é -1 ù êëQ 4 = tan 1úû

x+ 2ü x + 4 ï = tan -1 æç 2 ö÷ = tan í x + 2ý è 2x + 6 ø ï ï1 + î x + 4þ ì

-1 ï

1-

æ 1 = tan -1 ç èx+ \

ö ÷. 3ø

x-2 1 = Þ ( x - 2)( x + 3) = ( x - 4) x-4 x+ 3

Þ x 2 + x - 6 = x - 4 Þ x 2 = 2 Þ x = ± 2. Hence, x = ± 2. EXAMPLE 19

SOLUTION

Solve for x:

æ 1 - x2 ö p æ 2x ö ÷ = , - 1 < x < 1. tan -1 ç + cot -1 ç 2÷ ç 2x ÷ 3 è1 - x ø è ø [CBSE 2011C]

The given equation may be written as p æ 2x ö -1 æ 2x ö tan -1 ç ÷ + tan ç ÷= è 1 - x2 ø è 1 - x2 ø 3 æ 2x ö p Þ 2 tan -1 ç ÷= è 1 - x2 ø 3 æ 2x ö p Þ tan -1 ç ÷= è 1 - x2 ø 6 1 2x p Þ = tan = 2 6 3 1-x Þ (1 - x 2) = 2 3 x Þ x2 + 2 3x - 1 = 0 Þ x=

-2 3 ± 12 + 4 -2 3 ± 4 = 2 2

= ( - 3 + 2) or ( - 3 - 2). But, -1 < x < 1. So, x ¹ ( - 3 - 2). Hence, x = ( 2 - 3 ).

SSS Mathematics for Class 12 126

126

Senior Secondary School Mathematics for Class 12

EXERCISE 4B Very-Short-Answer Questions Find the principal value of each of the following: æ -1 ö 1. sin -1 ç ÷ è 2ø

æ -1 ö 2. cos-1 ç ÷ è 2ø

3. tan -1 ( -1)

4. sec-1 ( -2)

5. cosec-1 ( - 2)

6. cot -1 ( -1)

7. tan -1 ( - 3 )

æ -2 ö 8. sec-1 ç ÷ è 3ø

9. cosec-1 ( 2)

Find the principal value of each of the following: 2p ö æ 10. sin -1 ç sin ÷ 3ø è 13 p ö æ 13. cos-1 ç cos ÷ 6 ø è

æ 11. tan -1 ç tan è æ 14. tan -1 ç tan è

3p ö ÷ 4 ø 7p ö ÷ 6ø

7p ö æ 12. cos-1 ç cos ÷ 6ø è 15. tan -1

3 - cot -1 ( - 3 )

ìp æ -1 ö ü 16. sin í - sin -1 ç ÷ ý è 2 øþ î3

17. cot (tan -1 x + cot -1 x)

18. cosec (sin -1 x + cos-1 x)

19. sin ( sec-1 x + cosec-1 x)

20. cos-1

æ 21. tan -1 1 + cos-1 ç è

1 1 + 2 sin -1 2 2

1ö -1 æ 1 ö ÷ + sin ç - ÷ 2ø è 2ø

3p ü ì 22. sin -1 ísin ý [CBSE 2009] 5 þ î ANSWERS (EXERCISE 4B)

-p 6 p 9. 6 1.

17. 0

2p 3 p 10. 3

-p 4 -p 11. 4

18. 1

19. 1

2.

3.

2p 3 5p 12. 6 2p 20. 3 4.

5p 4 p 13. 6 3p 21. 4 5.

3p 4 p 14. 6 2p 22. 5 6.

-p 3 5p 15. 6 7.

HINTS TO THE GIVEN QUESTIONS (EXERCISE 4B) é -p p ù 1. Range of sin -1 is ê , × ë 2 2 úû 1 æ 1ö æ -p ö \ sin -1 ç - ÷ = q Þ sin q = - = sin ç ÷ 2 è 2ø è 6 ø -p Þ q= × 6

8.

5p 6

16. 1

SSS Mathematics for Class 12 127

Inverse Trigonometric Functions 2. Range of cos -1 is [0 , p]. -1 p pö 2p æ -1 ö æ \ cos -1 ç ÷ = q Þ cos q = = - cos = cos ç p - ÷ = cos 2 3 3ø 3 è 2 ø è 2p Þ q= × 3 æ -p p ö 3. Range of tan -1 is ç , ÷× è 2 2ø \ tan -1 ( -1) = q Þ tan q = - 1 = - tan Þ q=

p æ -p ö = tan ç ÷ 4 è 4 ø

-p × 4

ìpü 4. Range of sec-1 is [0 , p] - í ý × î2þ sec-1 ( -2 ) = q Þ sec q = - 2 = - sec

p pö 2p æ = sec ç p - ÷ = sec 3 3ø 3 è

2p × 3 é -p p ù is ê , - {0}. ë 2 2 úû

Þ q= 5. Range of cosec -1

\ cosec-1 ( - 2 ) = q Þ cosec q = - 2 = - cosec Þ q=

p pö 5p æ = cosec ç p + ÷ = cosec 4 4ø 4 è

5p × 4

6. Range of cot -1 is ( 0 , p ). \ cot -1 ( -1) = q Þ cot q = - 1 = - cot

p pö 3p 3p æ = cot ç p - ÷ = cot × Þ q= 4 4ø 4 4 è

æ -p p ö 7. Range of tan -1 is ç , ÷× è 2 2ø \ tan -1 ( - 3 ) = q Þ tan q = -

-p æ -p ö 3 = tan ç × ÷ Þ q= 3 è 3 ø

ìpü 8. Range of sec-1 is [0 , p] - í ý × î2þ -2 p pö 5p æ -2 ö æ = - sec = sec ç p - ÷ = sec sec-1 ç ÷ = q Þ sec q = 6 6ø 6 3 è è 3ø 5p Þ q= × 6 é -p p ù 9. Range of cosec-1 is ê , - {0}. ë 2 2 úû \ cosec-1 ( 2 ) = q Þ cosec q = 2 = cosec Þ q=

p × 6

p 6

127

SSS Mathematics for Class 12 128

128

Senior Secondary School Mathematics for Class 12

é -p p ù 10. Range of sin -1 is ê , × ë 2 2 úû 2p ö p öü æ æ -1 ì \ sin -1 ç sin ÷ = sin í sin ç p - ÷ ý 3 ø 3 øþ è è î pü p ì = sin -1 í sin ý = × 3þ 3 î æ -p p ö 11. Range of tan -1 is ç , ÷× è 2 2ø 3p ö p öü æ æ -1 ì \ tan -1 ç tan ÷ = tan í tan ç p - ÷ ý 4 ø 4 øþ è è î p ü ì = tan -1 í - tan ý 4þ î ì æ - p öü - p = tan -1 í tan ç × ÷ý = è 4 øþ 4 î 12. Range of cos -1 is [0 , p]. 7p ö 5p ö ü æ æ -1 ì \ cos -1 ç cos ÷ = cos í cos ç 2 p ÷ý 6 ø 6 øþ è è î 5p ü 5p ì × = cos -1 í cos ý = 6þ 6 î 13. Range of cos -1 is [0 , p]. 13 p ö p öü æ æ -1 ì \ cos -1 ç cos ÷ = cos í cos ç 2 p + ÷ ý 6 ø 6 øþ è è î pü p ì = cos -1 í cos ý = × 6þ 6 î æ -p p ö 14. Range of tan -1 is ç , ÷× è 2 2ø 7p ö p öü æ æ -1 ì \ tan -1 ç tan ÷ = tan í tan ç p + ÷ ý 6 ø 6 øþ è è î pü p ì = tan -1 í tan ý = × 6þ 6 î 15. tan -1

3 = q1 Þ tan q1 =

3 = tan

cot -1 ( - 3 ) = q 2 Þ cot q 2 = -

p p Þ q1 = × 3 3

3 = - cot

5p × 6 æ p 5p ö 3 - cot -1 ( - 3 ) = ç ÷ 6 ø è3 Þ q2 =

\ tan -1

=

-3 p - p = × 6 2

p pö 5p æ = cot ç p - ÷ = cot 6 6ø 6 è

SSS Mathematics for Class 12 129

Inverse Trigonometric Functions

129

1 -p æ 1ö æ pö 16. Let sin -1 ç - ÷ = q. Then, sin q = - = sin ç - ÷ Þ q = × 2 6 è 2ø è 6ø é p æ - p öù 3p p æp pö ÷ ú = sin ç + ÷ = sin given exp. = sin ê - ç \ = sin = 1. 6 2 è3 6ø ë 3 è 6 øû p p Þ cot (tan -1 a + cot -1 a) = cot = 0. 2 2 p p -1 -1 -1 -1 18. (sin a + cos a) = Þ cosec (sin a + cos a) = cosec = 1. 2 2 p p -1 -1 -1 -1 19. ( sec a + cosec a) = Þ sin ( sec a + cosec a) = sin = 1. 2 2 p ö æ p p ö 2p -1 1 -1 1 æ p 20. cos + 2 sin =ç + 2´ ÷=ç + ÷= × 2 2 è3 6ø è 3 3ø 3 17. (tan -1 a + cot -1 a) =

æ -1 ö æ -1 ö ì p 2 p æ - p ö ü æ p 2 p p ö ÷ý = ç + 21. tan -1 1 + cos -1 ç ÷ + sin -1 ç ÷ = í + +ç - ÷ 3 3 6ø è 2 ø è 2 ø î4 è 6 øþ è 4 =

( 3p + 8p - 2p ) 9p 3p = = × 12 12 4

22. See hint for Q. 10.

EXERCISE 4C 1. Prove that: æ1 + x ö p (i) tan -1 ç ÷ = + tan -1 x , x < 1 è1- x ø 4 (ii) tan -1 x + cot -1( x + 1) = tan -1( x 2 + x + 1) 2. Prove that sin -1( 2x 1 - x 2 ) = 2 sin -1 x ,|x|£

1 × 2

[CBSE 2008]

3. Prove that: (i) sin -1( 3 x - 4x 3) = 3 sin -1 x ,|x|£ (ii) cos-1( 4x 3 - 3 x) = 3 cos-1 x ,

1 2

1 £ x £1 2

æ 3x - x 3 ö ÷ = 3 tan -1 x , |x|< 1 (iii) tan -1 ç ç 1 - 3x2 ÷ 3 è ø æ 3x - x 3 ö æ 2x ö ÷ (iv) tan -1 x + tan -1 ç ÷ = tan -1 çç 2 2 ÷ è1 - x ø è 1 - 3x ø 4. Prove that: (i) cos-1(1 - 2x 2) = 2 sin -1 x (ii) cos-1( 2x 2 - 1) = 2 cos-1 x

[CBSE 2010]

SSS Mathematics for Class 12 130

130

Senior Secondary School Mathematics for Class 12

æ 1 ö (iii) sec-1 ç 2 ÷ = 2 cos-1 x è 2x - 1 ø (iv) cot -1( 1 + x 2 - x) =

p 1 - cot -1 x 2 2

[CBSE 2007]

5. Prove that: æ x + yö ÷ = tan -1 x + tan -1 y (i) tan -1 çç ÷ 1 xy ø è æ x+ x ö ÷ = tan -1 x + tan -1 x (ii) tan -1 çç 3/2 ÷ è1 - x ø æ sin x ö x (iii) tan -1 ç ÷= è 1 + cos x ø 2 6. Prove that: (i) tan -1 (ii) tan -1

1 2 3 + tan -1 = tan -1 2 11 4

2 7 1 + tan -1 = tan -1 11 24 2

(iii) tan -1 1 + tan -1 (iv) 2 tan -1

1 1 p + tan -1 = 2 3 2

1 1 p + tan -1 = 3 7 4

(v) tan -1 2 - tan -1 1 = tan -1

1 3

(vi) tan -1 1 + tan -1 2 + tan -1 3 = p (vii) tan -1

1 1 1 p + tan -1 + tan -1 = 2 5 8 4

(viii) tan -1

1 2 1 4 + tan -1 = tan -1 4 9 2 3

[CBSE 2011, ’12C]

7. Prove that: (i) cos-1

4 12 33 + cos-1 = cos-1 5 13 65

(ii) sin -1

(iii) cos-1

3 12 56 + sin -1 = sin -1 5 13 65

(iv) cos-1

(v) tan -1

1 5 p + sec-1 = 3 2 4

(vi) sin -1

(vii) 2 sin -1

3 17 p - tan -1 = 5 31 4

1 2 p + sin -1 = 5 5 2 4 3 27 + sin -1 = sin -1 5 5 11 1 9 1 + cos-1 = tan -1 2 17 85

SSS Mathematics for Class 12 131

Inverse Trigonometric Functions

131

8. Solve for x: 8 31 2 (ii) tan -1( 2 + x) + tan -1( 2 - x) = tan -1 3 1 (iv) cos ( 2 sin -1 x) = 9 (i) tan -1( x + 1) + tan -1( x - 1) = tan -1

[CBSE 2008, ’08C]

1 9 8 15 p (v) sin -1 + sin -1 = x x 2

(iii) cos (sin -1 x) =

9. Solve for x: (i) cos (sin -1 x) =

1 2

(ii) tan -1 x = sin -1

(iii) sin -1 x - cos-1 x =

1 2

p 6

ANSWERS (EXERCISE 4C)

1 4 5 (ii) x = 3 or x = -3 (iii) x = ± 4 9 3 3 (ii) x = 1 (iii) x = 9. (i) x = ± 2 2 8. (i) x =

(iv) x = ±

2 3

(v) x = 17

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 4C) 1. (i) Put x = tan q (ii) LHS = tan -1 x +

1 tan -1( x + 1)

2. Put x = sin q 3. 4.

(i) Put x = sin q (ii) Put x = cos q (iii) Put x = tan q (iv) Use tan -1 A + tan -1 B (i) Put x = sin q (ii) Put x = cos q (iii) Put x = cos q (iv) Putting x = cot q, we get: cot -1( 1 + x 2 - x ) = cot -1( cosec q - cot q)

= cot

-1 æç 1 - cos q ö÷

ç è

sin q ø

q 2

ö ÷ ÷ q q÷ ç 2 sin cos ÷ è 2 2ø æ ç

-1 ç ÷ = cot ç

2 sin 2

é qö æ æ p q öù = cot -1 ç tan ÷ = cot -1 ê cot ç - ÷ ú 2ø è è 2 2 øû ë = 5.

p q æp 1 ö - = ç - cot -1 x ÷ × 2 2 è2 2 ø

(i) Put x = tan q and y = tan f on each side. (ii) Put x = tan q and x = tan f on each side.

SSS Mathematics for Class 12 132

132

Senior Secondary School Mathematics for Class 12

(iii) LHS = tan

ì æ x öü æxö ï 2 sin ç 2 ÷ cos ç 2 ÷ ïï xö x è ø è ø -1 æ ý = tan ç tan ÷ = = RHS. í x æ ö 2ø 2 è 2 ï 2 cos ç ÷ ï ïî è 2 ø ïþ

-1 ï

ì

6. (iv) 2 tan

-1 1

3

= tan

1ü 3 ï. í ý ï1- 1 ï î 9þ

-1 ï

2´

æ x-y ö ÷. (v) Here 2 ´ 1 = 2 > -1. Now, use tan -1 x - tan -1 y = tan -1 çç ÷ è 1 + xy ø (vi) Here 2 ´ 3 = 6 > 1. So, we use the formula ì 2+ 3 ü -1 -1 tan -1 2 + tan -1 3 = p + tan -1í ý = p + tan ( -1) = p - tan 1. î 1 - ( 2 ´ 3) þ \ tan -1 1 + tan -1 2 + tan -1 3 = p. æ 1 2 ö ç ÷ + æ -1 1 -1 2 ö -1 ç 4 9 ÷ = 2 tan -1 1 ÷ = 2 tan (viii) 2 ç tan + tan 1 2÷ 4 9ø 2 ç è ç 1- ´ ÷ è 4 9ø = tan

1ö æ ç2´ ÷ 2 ÷ = tan -1 4 × 1÷ 3 ç ç 1- ÷ è 4ø

-1 ç

3 = q. 5 Then, base = 3, hypotenuse = 5.

7. (iii) Let cos -1

So, perpendicular = (5 ) 2 - 3 2 = 16 = 4 . 4 4 Þ q = sin -1 × 5 5 4 (iv) Let cos -1 = q. 5 \ sin q =

Then, base = 4, hypotenuse = 5. \ perpendicular = 5 2 - 4 2 = 9 = 3. \ tan q = (v) Let sec -1

4 3 Þ q = tan -1 × 3 4 5 5 = q. Then, sec q = × 2 2

So, hypotenuse = 5 , base = 2. So, perpendicular = ( 5 ) 2 - 2 2 = 1. \ tan q =

1 1 Þ q = tan -1 × 2 2

SSS Mathematics for Class 12 133

Inverse Trigonometric Functions (vi) Let sin -1

1 = q. Then, sin q = 4

1 × 17

\ perp. = 1 and hyp. = 17 . \ base = ( 17 ) 2 - 12 = 16 = 4. 1 1 Þ q = tan -1 × 4 4 9 9 Again, let cos -1 = f. Then, cos f = . 85 85 \ base = 9, hyp. = 85.

\ tan q =

So, perp. = ( 85 ) 2 - 9 2 = 4 = 2. 2 2 Þ f = tan -1 × 9 9 3 3 (vii) sin -1 = tan -1 × 5 4 3 ö æ ç 2´ ÷ -1 3 -1 3 -1 ç 4 ÷ 2 sin = 2 tan = tan 9 ÷ 5 4 ç ç 1÷ è 16 ø 24 = tan -1 × 7 1 8. (iii) sin -1 x = cos -1 9 \ tan f =

4 5 × 9 p öù 1 p 5p é æ (i) cos (sin -1 x ) = = cos or cos i. e. , cos ç 2 p - ÷ ú 3 øû 2 3 3 êë è p 5p -1 -1 or sin x = Þ sin x = 3 3 p 5p 3 - 3 or x = sin or Þ x = sin Þ x= × 3 3 2 2 = sin -1

9.

GRAPHS OF INVERSE TRIGONOMETRIC FUNCTIONS 1. Graph of sin-1 x é -p p ù Let f : [-1, 1] ® ê , ú : f ( x) = sin -1 x. 2û ë 2 é -p p ù Here ê , ú is called the principal-value branch of sin -1 x. 2û ë 2 The other branches of sin -1 x are é p 3p ù é 3p 5p ù é -3 p - p ù é -5 p -3 p ù êë 2 , 2 úû , êë 2 , 2 úû , ¼and êë 2 , 2 úû , êë 2 , 2 úû , etc.

133

SSS Mathematics for Class 12 134

134

Senior Secondary School Mathematics for Class 12

Table for sin -1 x

x

0

1 = 0.5 2

sin -1 x

0

p 6

1 2 ´ 2 2 2 1. 41 = = = 0.70 2 2 p 4

3 1.73 = = 0. 86 2 2

1

p 3

p 2

Also, sin -1( - x) = - sin -1 x. -1 -p ö æ \ çx = Þ sin -1 x = ÷, 2 6 ø è -p ö æ ç x = - 0.7 Þ sin -1 x = ÷, 4 ø è -p ö æ ç x = - 0. 86 Þ sin -1 x = ÷, 3 ø è -p ö æ ç x = - 1 Þ sin -1 x = ÷× 2 ø è On a graph paper, we plot the points pö æ 1 pö æ O( 0, 0), A ç , ÷ , B ç 0.7 , ÷ , 4ø è 2 6ø è pö pö æ æ æ -1 - p ö C ç 0. 86, ÷ , D ç1, ÷ , E ç , ÷, 3ø 2ø 6 ø è è è 2 -p ö -p ö æ æ F ç -0.7 , ÷ , G ç -0. 86, ÷ and 4 ø 3 ø è è -p ö æ H ç -1, ÷× 2 ø è Join the points OA , AB, BC, CD, and OE, EF, FG, GH successively with a freehand to get the required graph, as shown in the given figure. Moreover, we have pö p 2p æ = sin ç p - ÷ = sin = 0.86. sin 3 3ø 3 è 5p pö p æ sin = sin ç p - ÷ = sin = 0.5 , sin p = 0. 6 6ø 6 è 2p æ -2p ö sin ç = - 0.86. ÷ = - sin 3 è 3 ø 5p æ -5 p ö sin ç = - 0.5. ÷ = - sin 6 è 6 ø Now, we may extend the graph as shown in the figure.

SSS Mathematics for Class 12 135

Inverse Trigonometric Functions

135

2. Graph of cos-1x Let f : [-1, 1] ® [0, p] : f ( x) = cos-1 x. Here, [0, p] is called the principal-value branch of cos-1 x. The other branches of cos-1 x are [p , 2p], [2p , 3 p], … , [-p , 0], [-2p , - p],etc. Table for cos-1 x

x

1

cos -1 x 0

1 2 ´ 2 2 3 1. 732 1 = = 0 .5 2 1. 41 2 2 = = 2 = 0.87 2 2 = 0 .7 p 6

p 4

p 3

0

p 2

-1 2 ´ 2 2 - 2 -1 = - 0 .5 = 2 2 -1. 41 = 2 = - 0.70 2p 3p 3 4

- 3 2 -1. 73 = 2 = - 0 . 86

–1

5p 6

p

On a graph paper, we plot the points pö pö p ö æ pö 2p ö æ æ æ æ A (1, 0), B ç 0. 87 , ÷ , C ç 0.7 , ÷ , D ç 0.5 , ÷ , E ç 0, ÷ , F ç - 0.5 , ÷, 6ø 4ø 3 ø è 2ø 3ø è è è è 3p ö 5p ö æ æ G ç - 0.7 , ÷ , H ç - 0. 86, ÷ and K( -1, p). 4 ø 6ø è è Join AB , BC , CD , DE, EF, FG, GH and HK successively freehand to obtain the graph of cos-1 x , as shown in the given figure.

SSS Mathematics for Class 12 136

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Senior Secondary School Mathematics for Class 12

3. Graph of tan-1x æ -p p ö Let f : R ® ç , ÷ : f ( x) = tan -1 x. 2ø è 2 æ -p p ö Here ç , ÷ is the principal-value branch of tan -1 x. 2ø è 2 æ p 3p ö æ 3p 5p ö æ -3 p - p ö The other branches of tan -1 x are ç , , , ÷, ç ÷ , ... , ç ÷ , etc. 2 ø è2 2 ø è 2 2 ø è 2 We take the positive values and then use tan -1( - x) = - tan -1 x. Table for tan -1 x x

0

tan -1 x

0

1 ´ 3

3 3 1.73 = = = 0.58 3 3 3 p 6

1 p 4

3 = 1.73 p 3

Also, tan -1( - x) = - tan -1 x. p p -p , tan -1( -1) = - , tan -1( -173 . )= × 6 4 3 pö æ æ pö On a graph paper, we plot the points O( 0, 0), A ç 0.58, ÷ , B ç1, ÷ , 6ø è è 4ø pö -p ö -p ö -p ö æ æ æ æ C ç1.73 , ÷ , and D ç - 0.58, ÷ , E ç -1, ÷ , F ç -1.73 , ÷× 3ø 6 ø 4 ø 3 ø è è è è

\ tan -1( - 0.58) = -

Join OA , AB , BC , and OD , DE, EF successively freehand to get the graph. p Now, when x ® ¥ , then tan -1 x ® × 2 -p -1 Also, when x ® - ¥ , then tan x ® × 2

SSS Mathematics for Class 12 137

Inverse Trigonometric Functions

137

4. Graph of cot -1x Let f : R ® ( 0, p) : f ( x) = cot -1 x. Here, ( 0, p) is the principal-value branch of cot -1 x. The other branches of cot -1 x are ( p , 2p), ( 2p , 3 p), ¼ , ( - p , 0), etc. Table for cot -1 x x cot -1 x

3 = 1.73 p 6

1

1 ´ 3

3 3 1.732 = = = 0.58 3 3 3

p 4

p 3

0 p 2

Using cot( p - x) = - cot x , we have pö p 3p pö 2p p æ æ = cot ç p - ÷ = - cot = -0.58, cot = cot ç p - ÷ = - cot = -1, cot 4 4ø 3 3ø 3 4 è è cot

pö p 5p æ = cot ç p - ÷ = - cot = - 1.73. 6 6ø 6 è

pö pö æ æ pö æ On a graph paper, we plot the points A ç1.73 , ÷ , B ç1, ÷ , C ç 0.58, ÷ , 6ø 3ø è è 4ø è 2p ö 3p ö 5p ö æ pö æ æ æ D ç 0, ÷ , E ç - 0.58, ÷ , F ç -1, ÷ , and G ç -1.73 , ÷× 3ø 4 ø 6ø è 2ø è è è We join the points AB , BC , CD , DE, EF, FG successively to get the graph. As x ® ¥ , then cot -1 x ® 0. And, as x ® - ¥ , then cot -1 x ® p.

SSS Mathematics for Class 12 138

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Senior Secondary School Mathematics for Class 12

5. Graph of sec-1x ìp ü Let f : R - ( -1, 1) ® [0, p] - í ý : f ( x) = sec-1x. î2þ ì 3p ü ì -p ü The other branches of sec-1x are [p , 2p] - í ý , ¼ , [- p , 0] - í ý , etc. î 2 þ î 2 þ Table for sec-1 x x

1

sec-1x

0

2 ´ 3

3 2 3 2 ´ 1.73 3 . 46 = = = = 1.15 3 3 3 3 p 6

2 = 1. 41 p 4

2 p 3

3p 2p ü ì 3p ü ì 2p -1 = - 2 = -1.41 Þ sec-1( -1.41) = ý, ísec = -2 Þ sec ( -2) = ý, ísec 4 3þ î 3 4 þ î ì 5 p -2 3 -2 3 -2 ´ 1.73 -3.46 5p ü = = = = -1.15 Þ sec-1( -1.15) = ´ ísec = ý, 6 3 3 3 6þ 3 3 î sec p = -1 Þ sec-1( -1) = p. On a graph paper, we plot the points pö pö æ æ æ pö A(1, 0), B ç1.15 , ÷ , C ç1. 41, ÷ , D ç 2, ÷ and 6ø 4ø è è è 3ø 5p ö 2p ö 3p ö æ æ æ E ç -2, ÷ , F ç -1.41, ÷ , G ç -1.15 , ÷ , H( -1, p). 3ø 4 ø 6ø è è è We join the points AB , BC , CD , DE, and HG, GF, FE.

SSS Mathematics for Class 12 139

Inverse Trigonometric Functions

139

6. Graph of cosec-1x é -p p ù Let f : R - ( -1, 1) ® ê , ú - {0 } : f ( x) = cosec-1x. 2û ë 2 é p 3p ù The other branches of cosec-1x are ê , - {p}, ..., etc. 2 úû ë2 Table for cosec-1 x x

2

cosec-1x

p 6

2 = 1.41 p 4

2 ´ 3

3 2 3 2 ´ 1.73 = = = 1.15 3 3 3 p 3

1 p 2

Since cosec-1( - x) = - cosec-1x , we have æ -p ö æ -p ö æ -p ö and . cosec-1 ç ÷ = - 2, cosec-1 ç ÷ = - 1.41, cosec-1 ç ÷ = - 115 è 6 ø è 4 ø è 3 ø æ -p ö cosec-1 ç ÷ = - 1. è 2 ø pö pö æ pö æ æ On a graph paper, we plot the points A ç 2, ÷ , B ç1.41, ÷ , C ç115 . , ÷, 4ø 3ø è 6ø è è -p ö -p ö -p ö -p ö æ pö æ æ æ æ D ç1, ÷ and E ç -2, . , ÷ , F ç -1.41, ÷ , G ç -115 ÷ , H ç -1, ÷× 6 ø 4 ø 3 ø 2 ø è 2ø è è è è Join the points as shown in the given figure, to get the graph. Also as x ® 0 from +ve values, then cosec -1 x ® ¥ . As x ® 0 from –ve values, then cosec -1 x ® - ¥ .

SSS Mathematics for Class 12 140

140

Senior Secondary School Mathematics for Class 12

EXERCISE 4D Write down the interval for the principal-value branch of each of the following functions and draw its graph: 1. sin -1 4. cot

-1

2. cos-1 x x

5. sec

-1

3. tan -1 x 6. cosec -1 x

x

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: æ 3ö ÷ is 1. The principal value of cos-1 çç ÷ è 2 ø p 5p 7p (a) (b) (c) 6 6 6

(d) none of these

2. The principal value of cosec-1( 2) is (a)

p 3

(b)

p 6

(c)

2p 3

æ -1 ö 3. The principal value of cos-1 ç ÷ is è 2ø -p p 3p (a) (b) (c) 4 4 4 æ -1 ö 4. The principal value of sin -1 ç ÷ is è 2ø -p 5p 7p (a) (b) (c) 6 6 6 1 æ ö 5. The principal value of cos-1 ç ÷ is è 2ø -p 2p 4p (a) (b) (c) 3 3 3

(d)

5p 6

(d)

5p 4

(d) none of these

(d)

p 3

6. The principal value of tan -1( - 3 ) is (a)

2p 3

(b)

4p 3

(c)

-p 3

(d) none of these

(c)

5p 4

(d)

3p 4

(d)

7p 6

7. The principal value of cot -1( -1) is (a)

-p 4

(b)

p 4

æ -2 ö 8. The principal value of sec-1 ç ÷ is è 3ø p -p 5p (a) (b) (c) 6 6 6

SSS Mathematics for Class 12 141

Inverse Trigonometric Functions

141

9. The principal value of cosec-1( - 2) is (a)

-p 4

(b)

3p 4

(c)

5p 4

(d) none of these

(c)

7p 6

(d)

(c)

p 3

(d) none of these

(c)

5p 6

(d)

(c)

p 6

(d) none of these

(c)

3p 4

(d) none of these

(c)

8p 5

(d) none of these

(c)

2p 3

(d) none of these

(c)

-p 4

(d) none of these

(c)

p 2

(d) p

10. The principal value of cot -1( - 3 ) is (a)

-p 6

(b)

p 6

2p ö æ 11. The value of sin -1 ç sin ÷ is 3ø è 2p 5p (a) (b) 3 3 13 p ö æ 12. The value of cos-1 ç cos ÷ is 6 ø è 13 p 7p (a) (b) 6 6 7p ö æ 13. The value of tan -1 ç tan ÷ is 6ø è 7p 5p (a) (b) 6 6 5p ö æ 14. The value of cot -1 ç cot ÷ is 4ø è p -p (a) (b) 4 4 8p ö æ 15. The value of sec-1 ç sec ÷ is 5 ø è 2p 3p (b) (a) 5 5 4p ö æ 16. The value of cosec-1 ç cosec ÷ is 3ø è p -p (a) (b) 3 3 3p ö æ 17. The value of tan -1 ç tan ÷ is 4 ø è 3p p (b) (a) 4 4 18.

5p 6

p 6

p æ -1 ö - sin -1 ç ÷ = ? 3 è 2ø (a) 0

(b)

2p 3

SSS Mathematics for Class 12 142

142

Senior Secondary School Mathematics for Class 12

1 1ö æ 19. The value of sin ç sin -1 + cos-1 ÷ = ? 2 2ø è (a) 0

(b) 1

20. If x ¹ 0 then cos (tan

-1

x + cot

(c) –1 -1

(a) –1 (c) 0

(d) none of these

x) = ? (b) 1 (d) none of these

3ö æ 21. The value of sin ç cos-1 ÷ is 5ø è 2 4 -2 (a) (b) (c) 5 5 5 2p ö 2p ö -1 æ -1 æ 22. cos ç cos ÷ + sin ç sin ÷ =? 3ø 3ø è è 4p p 3p (a) (b) (c) 3 2 4

(d) none of these

(d) p

23. tan -1( 3 ) - sec-1( -2) = ? (a)

p 3

(b)

-p 3

(c)

5p 3

1 1 + 2 sin -1 = ? 2 2 2p 3p (b) (c) 2p (a) 3 2 æ -1 ö æ -1 ö 25. tan -1 1 + cos-1 ç ÷ + sin -1 ç ÷ = ? 2 è ø è 2ø 2p 3p (a) p (b) (c) 3 4 pù é -1 1 26. tan ê 2 tan =? 5 4 úû ë 7 -7 7 (a) (b) (c) 17 17 12 1 æç -1 5 ö÷ 27. tan ç cos =? 2è 3 ÷ø

(d) none of these

24. cos-1

( 3 - 5) ( 3 + 5) (b) 2 2 æ -1 3 ö 28. sin ç cos ÷ =? 5ø è 3 4 (b) (a) 4 5 (a)

(d) none of these

(d)

p 2

(d)

-7 12

(5 + 3 ) 2

(c)

(5 - 3 ) 2

(d)

(c)

3 5

(d) none of these

SSS Mathematics for Class 12 143

Inverse Trigonometric Functions

3ö æ 29. cos ç tan -1 ÷ = ? 4ø è 3 4 (a) (b) 5 5 ìp æ -1 ö ü 30. sin í - sin -1 ç ÷ ý = ? è 2 øþ î3 (a) 1

(b) 0

4ö æ1 31. sin ç cos-1 ÷ = ? 5ø è2 1 2 (a) (b) 5 5

(c)

4 9

(d) none of these

(c)

-1 2

(d) none of these

1 10

(c)

ì 1öü æ 32. tan -1 í2 cos ç 2 sin -1 ÷ ý = ? 2ø þ è î p p (a) (b) 3 4 -1 æ -1 ö 33. If cot ç ÷ = x then sin x = ? è5 ø 1 5 (a) (b) 26 26

143

(c)

3p 4

1 24

(c)

(d)

(d)

2 10

2p 3

(d) none of these

æ- 3ö æ -1 ö ÷ =? 34. sin -1 ç ÷ + 2 cos-1 çç ÷ è 2ø è 2 ø (a)

p 2

(b) p

æ -1 ö 35. tan -1( -1) + cos-1 ç ÷ =? è 2ø p (a) (b) p 2

(c)

3p 2

(d) none of these

(c)

3p 2

(d)

2p 3

36. cot (tan -1 x + cot -1 x) = ? (a) 1

(b)

37. tan -1 1 + tan -1 (a) tan -1

4 3

(c) 0

(d) none of these

(c) tan -1 2

(d) tan -1 3

1 =? 3 (b) tan -1

38. tan -1

1 1 + tan -1 = ? 2 3

p 3

(b)

(a)

1 2

p 4

2 3

(c)

p 2

(d)

2p 3

SSS Mathematics for Class 12 144

144

Senior Secondary School Mathematics for Class 12

1 =? 3 3 (a) tan -1 2

39. 2 tan -1

(b) tan -1

3 4

(c) tan -1

1ö æ 40. cos ç 2 tan -1 ÷ = ? 2ø è 3 4 (b) (a) 5 5 5ù é 41. sin ê 2 tan -1 ú 8û ë 25 (a) 64 4ù é 42. sin ê 2 sin -1 ú 5û ë 12 (a) 25 43. If tan -1 x = (a)

1 2

(d) none of these

(c)

7 8

(d) none of these

(b)

80 89

(c)

75 128

(d) none of these

(b)

16 25

(c)

24 25

(d) none of these

(c)

1 6

(d) none of these

p 1 - tan -1 then x = ? 4 3 1 (b) 4

44. If tan -1(1 + x) + tan -1(1 - x) = (a) 1

4 3

p then x = ? 2

(b) –1

(c) 0

2p then (cos-1 x + cos-1 y) = ? 3 p (b) (c) p 3

(d)

1 2

(d)

2p 3

45. If sin -1 x + sin -1 y = (a)

p 6

46. (tan -1 2 + tan -1 3) = ? (a)

-p 4

(b)

p 4

(c)

3p 4

(d) p

47. If tan -1 x + tan -1 3 = tan -1 8 then x = ? (a)

1 3

(b)

1 5

48. If tan -1 3 x + tan -1 2x = (a)

1 or –2 2

(b)

(c) 3

(d) 5

p then x = ? 4

1 or –3 3

(c)

1 or –2 4

(d)

1 or –1 6

SSS Mathematics for Class 12 145

Inverse Trigonometric Functions

4 2ü ì 49. tan ícos-1 + tan -1 ý = ? 5 3þ î 13 17 (a) (b) 6 6 41 =? 4 p (b) 4

50. cot -1 9 + cosec-1 (a)

p 6

145

(c)

19 6

(d)

23 6

(c)

p 3

(d)

3p 4

51. Range of sin -1 x is é pù (a) ê 0, ú ë 2û

(b) [0, p]

é -p p ù (c) ê , ë 2 2 úû

(d) none of these

é pù (b) ê 0, ú ë 2û

é -p p ù (c) ê , ë 2 2 úû

(d) none of these

æ -p p ö (b) ç , ÷ è 2 2ø

é -p p ù (c) ê , ë 2 2 úû

(d) none of these

(b) [0, p]

ìp ü (c) [0, p] - í ý î2þ

(d) none of these

52. Range of cos-1 x is (a) [0, p] 53. Range of tan -1 x is æ pö (a) ç 0, ÷ è 2ø 54. Range of sec-1x is é pù (a) ê 0, ú ë 2û

55. Range of cosec-1x is æ -p p ö (a) ç , ÷ è 2 2ø

é -p p ù (b) ê , ë 2 2 úû

é -p p ù (c) ê , - {0 } ë 2 2 úû

(d) none of these

56. Domain of cos-1 x is (a) [0, 1]

(b) [–1, 1]

(c) [–1, 0]

(d) none of these

(c) R - [-1, 1]

(d) R - {-1, 1}

57. Domain of sec-1x is (b) R - {0 }

(a) [–1, 1]

ANSWERS (OBJECTIVE QUESTIONS)

1. (a) 11. (c) 21. (b) 31. (c) 41. (b) 51. (c)

2. (b) 12. (d) 22. (d) 32. (b) 42. (c) 52. (a)

3. (c) 13. (c) 23. (b) 33. (b) 43. (a) 53. (b)

4. (a) 14. (a) 24. (a) 34. (c) 44. (c) 54. (c)

5. (b) 15. (a) 25. (c) 35. (a) 45. (b) 55. (c)

6. (c) 16. (b) 26. (b) 36. (c) 46. (c) 56. (b)

7. (d) 17. (c) 27. (a) 37. (c) 47. (b) 57. (c)

8 (c) 18. (c) 28. (b) 38. (b) 48. (d)

9. (a) 19. (b) 29. (b) 39. (b) 49. (b)

10. (d) 20. (c) 30. (a) 40. (a) 50. (b)

SSS Mathematics for Class 12 146

146

Senior Secondary School Mathematics for Class 12 HINTS TO SOME SELECTED OBJECTIVE QUESTIONS

æ 3ö ÷ = x , where x Î [0 , p]. 1. Let cos -1 çç ÷ è 2 ø Then, cos x =

3 p p = cos Þ x = × 2 6 6

é -p p ù 2. Let cosec-1( 2 ) = x , where x Î ê , - {0 }. ë 2 2 úû Then, cosec x = 2 cosec

p p Þ x= × 6 6

æ -1 ö ÷ = x , where x Î [0 , p]. 3. Let cos -1 ç è 2ø Then, cos x =

-1 p pö 3p 3p æ = - cos = cos ç p - ÷ = cos Þ x= × 4 4ø 4 4 2 è

é -p p ù æ -1 ö 4. Let sin -1 ç ÷ = x , where x Î ê , × ë 2 2 úû è 2 ø Then, sin x =

p -p -1 æ -p ö ÷ Þ x= × = - sin = sin ç 2 6 6 è 6 ø

æ -1 ö 5. Let cos -1 ç ÷ = x , where x Î [0 , p]. è 2 ø Then, cos x =

-1 p pö 2p 2p æ = - cos = cos ç p - ÷ = cos Þ x= × 2 3 3ø 3 3 è

æ -p p ö 6. Let tan -1( - 3 ) = x , where x Î ç , ÷× è 2 2ø Then, tan x = -

3 = - tan

p -p æ -p ö ÷ Þ x= = tan ç × 3 3 è 3 ø

7. Let cot -1( -1) = x , where x Î [0 , p]. Then, cot x = -1 = - cot

p pö 3p 3p æ = cot ç p - ÷ = cot Þ x= × 4 4ø 4 4 è

æ -2 ö ìpü ÷ = x, where x Î [0 , p] - í ý × 8. Let sec-1 ç î2þ è 3ø Then, sec x =

-2 p pö 5p 5p æ = - sec = sec ç p - ÷ = sec Þ x= × 6 6ø 6 6 3 è

é -p p ù 9. Let cosec-1( - 2 ) = x , where x Î ê , - {0 }. ë 2 2 úû Then, cosec x = - 2 = - cosec

p -p æ -p ö ÷ Þ x= × = cosec ç 4 4 è 4 ø

10. Let cot -1( - 3 ) = x , where x Î [0 , p]. Then, cot x = - 3 = - cot

p pö 5p 5p æ = cot ç p - ÷ = cot Þ x= × 6 6ø 6 6 è

SSS Mathematics for Class 12 147

Inverse Trigonometric Functions 2p ö é -p p ù æ ÷ = x , where x Î ê 11. Let sin -1 ç sin , × 3 ø ë 2 2 úû è Then, sin x = sin

2p pö p p æ = sin ç p - ÷ = sin Þ x = × 3 3ø 3 3 è

13 p ö æ ÷ = x , where x Î [0 , p]. 12. Let cos -1 ç cos 6 ø è 13 p pö p p æ = cos ç 2 p + ÷ = cos Þ x = × 6 6ø 6 6 è

Then, cos x = cos

7p ö æ -p p ö æ ÷ = x , where x Î ç 13. Let tan -1 ç tan , ÷× 6 ø è 2 2ø è 7p pö p p æ = tan ç p + ÷ = tan Þ x = × 6 6ø 6 6 è

Then, tan x = tan

5p ö æ ÷ = x , where x Î [0 , p]. 14. Let cot -1 ç cot 4 ø è Then, cot x = cot

5p pö p p æ = cot ç p + ÷ = cot Þ x= × 4 4ø 4 4 è

8p ö æ ìpü ÷ = x, where x Î [0 , p] - í ý × 15. Let sec-1 ç sec 5 ø è î2þ Then, sec x = sec

8p 2p ö 2p 2p æ ÷ = sec = sec ç 2 p Þ x= × 5 5 ø 5 5 è

4p ö é -p p ù æ ÷ = x, where x Î ê 16. Let cosec-1 ç cosec , - {0 }. 3 ø ë 2 2 úû è Then, cosec x = cosec \

x=

4p pö p æ -p ö æ ÷× = cosec ç p + ÷ = - cosec = cosec ç 3 3ø 3 è è 3 ø

-p × 3

3p ö æ -p p ö æ ÷ = x , where x Î ç 17. Let tan -1 ç tan , ÷× 4 ø è 2 2ø è Then, tan x = tan

3p pö -p æ -p ö æ ÷ Þ x= × = tan ç p - ÷ = tan ç 4 4ø 4 è 4 ø è

é -p p ù æ -1 ö 18. Let sin -1 ç ÷ = x , where x Î ê , × ë 2 2 úû è 2 ø Then, sin x = \ 19. sin -1

1 p -p æ -p ö ÷ Þ x= = - sin = sin ç × 2 6 6 è 6 ø

given exp. =

p æ - p ö æ p p ö 3p p ÷=ç + ÷= = × -ç 3 è 6 ø è3 6ø 6 2

1 1 p 1 1ö p æ + cos -1 = Þ sin ç sin -1 + cos -1 ÷ = sin = 1. 2 2 2 2 2ø 2 è

20. cos (tan -1 x + cot -1 x ) = cos

p = 0. 2

147

SSS Mathematics for Class 12 148

148

Senior Secondary School Mathematics for Class 12 9 4 3 = sin -1 1 = sin -1 × 5 25 5 3ö 4ö 4 æ æ sin ç cos -1 ÷ = sin ç sin -1 ÷ = × 5ø 5ø 5 è è

21. cos -1 x = sin -1 1 - x 2 Þ cos -1 \

2p ö 2p ö æ æ ÷ = sin -1 ç sin ÷ 22. cos -1 ç cos 3 ø 3 ø è è ì ì p öü p öü æ æ = cos -1 í cos ç p - ÷ ý + sin -1 í sin ç p - ÷ ý 3 3 øþ è ø è î þ î pö pü æ ì æ -1 ö p = cos -1 ç - cos ÷ + sin -1 í sin ý = cos -1 ç ÷ + 3ø 3þ è î è 2 ø 3 æ 2p p ö =ç + ÷ = p. 3ø è 3 23. tan -1 3 - sec-1( -2 ) = tan -1 3 - ( p - sec-1 2 ) p p -p -p+ = × 3 3 3 1 1 p æ p ö æ p p ö 2p 24. cos -1 + 2 sin -1 = + ç 2 ´ ÷ = ç + ÷ = × 2 2 3 è 6ø è 3 3ø 3 =

æ -1 ö 25. tan -1 1 + cos -1 ç ÷ + sin -1 è 2 ø

1ö 1 æ -1 ö p æ ç ÷ = + ç p - cos -1 ÷ - sin -1 2ø 2 è 2 ø 4 è

p p ö 3p æp =ç + p- - ÷= × 3 6ø 4 è4 1 1 1 1 1 26. Let 2 tan -1 = q. Then, tan -1 = q Þ tan q = × 5 5 2 2 5 \

1ö æ 1 ç2 ´ ÷ q 5 ø æ 2 25 ö 5 è 2 = tan q = =ç ´ ÷= × 1 1 ö è 5 24 ø 12 æ 1 - tan 2 q ç 1 ÷ 2 25 ø è

\

p tan q - tan pö é æ -1 1 p ù 4 tan ê 2 tan = tan ç q - ÷ = p 5 4 úû 4ø ë è 1 + tan q × tan 4

2 tan

æ 5 ö æ -7 ö ç ç ÷ - 1÷ è 12 ø è 12 ø -7 = × = = 5 æ ö æ 17 ö 17 ç1+ ÷ ´ 1÷ ç 12 è ø è 12 ø 27. Let cos -1

5 5 = q. Then, cos q = × 3 3

æqö sin ç ÷ 1 æç 1 è2ø -1 5 ö÷ = tan q = tan ç cos æqö 2è 3 ÷ø 2 cos ç ÷ è2ø

SSS Mathematics for Class 12 149

Inverse Trigonometric Functions 1

ìæ 5 ö÷ ü ï çç 1 ï 3 ÷ø ï 1 - cos q ï è =í ý ö 1 + cos q ï æ ç1+ 5 ÷ï ïç 3 ÷ø ïþ îè

=

æ 3- 5 ö ÷ = çç ÷ è 3+ 5ø

1

2

149

2

1

ì 3- 5 3- 5ü =í ´ ý î3+ 5 3- 5þ

\

3 3 = x , where x Î [0 , p]. Then, cos x = × 5 5 since x Î [0 , p], sin x > 0.

\

sin x = 1 -

2

=

( 3 - 5) . 2

28. Let cos -1

29. Let tan -1

9 = 25

16 4 3ö 4 æ = Þ sin ç cos -1 ÷ = × 25 5 5ø 5 è

3 æ -p p ö , ÷× = x , where x Î ç 4 è 2 2ø

\

tan x =

3 æ -p p ö and since x Î ç , ÷ , we have cos x > 0. 4 è 2 2ø

\

cos x =

1 = sec x

\

3ö 4 æ cos ç tan -1 ÷ = cos x = × 4ø 5 è

1 2

1 + tan x

1

=

9 1+ 16

=

4 × 5

ìp 1ü p æ -1 ö ü æp pö ìp 30. sin í - sin -3 ç ÷ ý = sin í + sin -1 ý = sin ç + ÷ = sin = 1. 2þ 2 è 2 øþ è3 6ø î3 î3 4 4 = x , where x Î [0 , p]. Then, cos x = × 5 5 1 1 é pù Since x Î [0 , p] ® x Î ê 0 , ú Þ sin x > 0. 2 2 ë 2û

31. Let cos -1

4ö 1 æ1 sin ç cos -1 ÷ = sin x = 5ø 2 è2

4ö æ ç1- ÷ 5ø è = 2

1 - cos x = 2

1 × 10

ì ì 1 öü p öü æ æ 32. tan -1 í 2 cos ç 2 sin -1 ÷ ý = tan -1 í 2 cos ç 2 ´ ÷ ý 2 øþ 6 øþ è è î î pü 1ü p ì ì = tan -1 í 2 cos ý = tan -1 í 2 ´ ý = tan -1 1 = × 3þ 2þ 4 î î -1 æ -1 ö 33. cot -1 ç ÷ = x Þ cot x = , where x Î ( 0 , p ). 5 è 5 ø sin x > 0 in ( 0 , p ). 1 1 sin x = = = cosec x 1 + cot 2 x

1 1 1+ 25

=

5 × 26

SSS Mathematics for Class 12 150

150

Senior Secondary School Mathematics for Class 12

34. Range of cos -1 is [0 , p]. æ- 3ö ÷ = x Þ cos x = - 3 = - cos p = cos æç p - p ö÷ = cos 5 p Þ x = 5 p × cos -1 çç ÷ 6 6 2 2 6 6ø è è ø æ -1 ö sin -1 ç ÷ + 2 cos -1 è 2 ø

æ- 3ö 5p p 5p 9p 3p -1 1 ç ÷ ç 2 ÷ = - sin 2 + 2 ´ 6 = - 6 + 3 = 6 = 2 × è ø

æ -p p ö 35. Range of tan -1 is ç , ÷× è 2 2ø tan -1( -1) = x Þ tan x = - 1 = - tan

-p p æ -p ö ÷ Þ x= = tan ç 4 4 è 4 ø

Range of cos -1 is [0 , p]. -1 p pö 3p 3p æ -1 ö æ ÷ = y Þ cos y = = - cos = cos ç p - ÷ = cos Þ y= × cos -1 ç 4 4ø 4 4 2 è è 2ø p 3p 2p p given exp. = - + = = × \ 4 4 4 2 p 36. Given exp. = cot = 0. 2 1ü ì ï1+ 3 ï 1 -1 37. tan -1 1 + tan -1 = tan -1 í ý = tan 2 . 3 ï1- 1 ï î 3þ ì1

38. tan

-1 1

2

+ tan

-1 1

3

= tan

1ü 3 ï = tan -1 1 = p × í ý 1 4 ï 1- ï î 6 þ

-1 ï 2

+

æ 2x ö ÷× 39. Use 2 tan -1 x = tan -1 ç ç 1 - x2 ÷ è ø æ 1 - x2 ö ÷× 40. Use 2 tan -1 x = cos -1 ç ç 1 + x2 ÷ è ø æ 2x ö ÷× 41. Use 2 tan -1 x = sin -1 ç ç 1 + x2 ÷ è ø 42. Use 2 sin -1 x = sin -1[2 x 1 - x 2 ]. ìæ ïï ç 1 p 1 1 è 43. - tan -1 = tan -1 1 - tan1 = tan -1 í æ 4 3 3 ïç1+ ïî è 44. We know that tan -1 x + cot -1 x =

1öü ÷ 1 3 ø ïï -1 1 Þ x= × ý = tan 1 öï 2 2 ÷ 3 ø þï

p 1 p Þ tan -1 x + tan -1 = × 2 x 2

SSS Mathematics for Class 12 151

Inverse Trigonometric Functions \

tan -1( 1 + x ) + tan -1( 1 - x ) =

p 1 Þ ( 1 - x) = 2 ( 1 + x) Þ ( 1 - x 2 ) = 1 Þ x = 0.

45. sin -1 x + sin -1 y = \

2p æp ö æp ö 2p Þ ç - cos -1 x ÷ + ç - cos -1 y ÷ = × 3 3 è2 ø è2 ø

2p ö p æ ÷= × cos -1 x + cos -1 y = ç p 3 ø 3 è

46. ( x = 2 , y = 3 ) Þ xy > 1. \

æ 2+ 3 ö ÷ = p + tan -1( -1) tan -1 2 + tan -1 3 = p + tan -1 çç ÷ è 1- 2 ´ 3 ø p ö 3p æ = p - tan ( 1) = ç p - ÷ = × 4ø 4 è

æ 3+ x ö ÷ = tan -1 8. 47. tan -1 x + tan -1 3 = tan -1 8 Þ tan -1 çç ÷ è 1 - 3x ø 3+ x 1 = 8 Þ 3 + x = 8 - 24 x Þ x = × \ 1 - 3x 5 48. tan -1 3 x + tan -1 2 x = 5x

\

1 - 6x

2

= tan

p Þ tan -1 4

æ 3x + 2x ö p ç ÷= ç 1 - 6x 2 ÷ 4 è ø

p = 1 Þ 6x 2 + 5x - 1 = 0 4 Þ ( x + 1)( 6 x - 1) = 0 1 Þ x = - 1 or x = × 6

49. cos

-1

x = tan

-1 4

-1

1 - x2 x

-1 2

Þ cos

-1 4

5

-1 3

= tan

\

cos

\

17 ü 17 ì given exp. = tan í tan -1 × ý= 6 þ 6 î

5

+ tan

3

= tan

4

+ tan

-1 2

3

16 3 25 = tan -1 æ4ö 4 ç ÷ è5 ø

1-

= tan

41 = cot -1 4

-1

æ3 2ö ç + ÷ 17 è 4 3ø = tan -1 3 2ö æ 6 ç1- ´ ÷ 4 3ø è

41 5 - 1 = cot -1 16 4 æ 1 4 ö ç ÷ + 5 1 4 p cot -1 9 + cot -1 = tan -1 + tan -1 = tan -1 ç 9 5 ÷ = tan -1 1 = × 1 4÷ 4 9 5 4 ç ç 1- ´ ÷ è 9 5ø

50. cosec-1 x = cot -1 x 2 - 1 Þ cosec-1

\

-1

151

SSS Mathematics for Class 12 152

5. MATRICES A rectangular array of mn numbers in the form of m horizontal lines (called rows) and n vertical lines (called columns) is called a matrix of order m by n, written as an m ´ n matrix. Such an array is enclosed by [ ] or ( ). Each of the mn numbers constituting the matrix is called an element or an entry of the matrix. Usually, we denote a matrix by a capital letter. The plural of matrix is matrices.

MATRIX

Example

é 3 5 -4 ù (i) A = ê is a matrix, having 2 rows and 3 columns. 9 úû ë 0 1 Its order is 2 ´ 3 and it has 6 elements. é 9 4 2 -1 ù ú ê (ii) B = ê 1 8 -3 2 ú is a matrix, having 3 rows and 4 ê 6 0 5 7 úû ë columns. Its order is 3 ´ 4 and it has 12 elements.

How to Describe a Matrix In order to locate the position of a particular element of a matrix, we have to specify the number of the row and that of the column in which the element occurs. An element occurring in the ith row and jth column of a matrix A will be called the (i , j)th element of A , to be denoted by aij . In general, an m ´ n matrix A may be written as a12 a13 ¼ a1n ù é a11 ê a a22 a23 ¼ a2n ú ú ê 21 ¼ ú ¼ ¼ ¼ ê ¼ = [aij ]m ´ n . A=ê a ai 2 ai 3 ¼ ain ú ú ê i1 ¼ ú ¼ ¼ ¼ ê ¼ ú ê a a a ¼ a ë m1 mn û m2 m3 EXAMPLE 1

é 3 -2 5 ù Consider the matrix A = ê . 9 1 úû ë 6 Clearly, the element in the 1st row and 2nd column is –2. So, we write a12 = - 2. Similarly, a11 = 3 ; a12 = - 2; a13 = 5 ; a21 = 6; a22 = 9 and a23 = 1. 152

SSS Mathematics for Class 12 153

Matrices EXAMPLE 2 SOLUTION

153

Construct a 3 ´ 2 matrix whose elements are given by aij = (i + 2j). A 3 ´ 2 matrix has 3 rows and 2 columns. In general, a 3 ´ 2 matrix is given by é a11 a12 ù A = ê a21 a22 ú ú ê êë a 31 a 32 úû 3 ´ 2. Thus aij = (i + 2j) for i = 1, 2, 3 and j = 1, 2. \

a11 = (1 + 2 ´ 1) = 3 ; a12 = (1 + 2 ´ 2) = 5 ; a21 = ( 2 + 2 ´ 1) = 4; a22 = ( 2 + 2 ´ 2) = 6; a 31 = ( 3 + 2 ´ 1) = 5 ; a 32 = ( 3 + 2 ´ 2) = 7.

é 3 5 Hence, A = ê 4 6 ê êë 5 7 EXAMPLE 3 SOLUTION

ù ú ú úû

3 ´ 2.

1 Construct a 2 ´ 3 matrix whose elements are given by aij = |5i - 3 j |. 2 A 2 ´ 3 matrix has 2 rows and 3 columns. In general, a 2 ´ 3 matrix is given by é a11 a12 a13 ù A=ê ú ë a21 a22 a23 û 3 ´ 2. 1 Thus, aij = |5i - 3 j |where i = 1, 2 and j = 1, 2, 3. 2 1 1 1 \ a11 = |5 ´ 1 - 3 ´ 1| = ×| 2| = ´ 2 = 1; 2 2 2 1 1 1 1 1 a12 = |5 ´ 1 - 3 ´ 2| = ×|5 - 6| = ×| - 1| = ´ 1 = ; 2 2 2 2 2 1 1 1 1 a13 = |5 ´ 1 - 3 ´ 3 | = ×|5 - 9| = ×| - 4| = ´ 4 = 2; 2 2 2 2 1 1 1 1 7 a21 = |5 ´ 2 - 3 ´ 1| = ×|10 - 3 | = ×|7 | = ´ 7 = ; 2 2 2 2 2 1 1 1 1 a22 = |5 ´ 2 - 3 ´ 2| = ×|10 - 6| = ×| 4| = ´ 4 = 2; 2 2 2 2 1 1 1 1 1 a23 = |5 ´ 2 - 3 ´ 3 | = ×|10 - 9| = ×|1| = ´ 1 = × 2 2 2 2 2 é ê1 Hence, A = ê 7 ê ë2

1 2 2

ù 2ú × 1ú ú 2û

SSS Mathematics for Class 12 154

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Senior Secondary School Mathematics for Class 12

EXAMPLE 4

If a matrix has 12 elements, what are the possible orders it can have?

SOLUTION

We know that a matrix of order m ´ n has mn elements. Hence, all possible orders of a matrix having 12 elements are (12 ´ 1), (1 ´ 12), ( 6 ´ 2), ( 2 ´ 6), ( 4 ´ 3) and ( 3 ´ 4).

EXERCISE 5A é ê 5 ê 1. If A = ê 7 ê ê 2 êë

-2

6

0

8

3 5

4

ù 1 ú ú -3 ú then write ú 3ú úû

(i) the number of rows in A,

(ii) the number of columns in A,

(iii) the order of the matrix A,

(iv) the number of all entries in A,

(v) the elements a23 , a 31 , a14 , a 33 , a22 of A. 2. Write the order of each of the following matrices: é 6 -5 ù é ù 5 4 -2 ú ê 1 ê 3 3ú ú (i) A = ê (ii) B = ê ú 4 ê 2 4ú 3 -1 ê 0 ú êë -2 -1úû ë 9 û (iii) C = [7 é -2ù (v) E = ê 3 ú ê ú êë 0úû

- 2

5

0]

(iv) D = [8

- 3]

(vi) F = [6]

3. If a matrix has 18 elements, what are the possible orders it can have? 4. Find all possible orders of matrices having 7 elements. 5. Construct a 3 ´ 2 matrix whose elements are given by aij = ( 2i - j). i 6. Construct a 4 ´ 3 matrix whose elements are given by aij = × j 7. Construct a 2 ´ 2 matrix whose elements are aij =

(i + 2j) 2 × [CBSE 2002, ’07] 2

8. Construct a 2 ´ 3 matrix whose elements are aij =

(i - 2j) 2 × 2

[CBSE 2002]

1 9. Construct a 3 ´ 4 matrix whose elements are given by aij = | - 3i + j |. 2

SSS Mathematics for Class 12 155

Matrices

155

ANSWERS (EXERCISE 5A)

1. (i) 3 (ii) 4 (iii) 3 ´ 4 (iv) 12 (v) a23 = 8, a 31 = 2 , a14 = 1, a 33 = 4, a22 = 0 2. (i) ( 2 ´ 4)

(ii) ( 3 ´ 2)

(iii) (1 ´ 4)

(iv) (1 ´ 2)

3. (18 ´ 1), (1 ´ 18), ( 9 ´ 2), ( 2 ´ 9), ( 6 ´ 3), ( 3 ´ 6) 1 1ù é ê 1 2 3ú ê 2ú é 1 9ù ú ê2 1 3ú 6. ê 5. A = ê 3 2 ú ú ê ê 3 3 1ú êë 5 4 úû ú ê 2 4ú ê êë 4 2 3 úû 1 1ù é ê 1 2 0 2ú é 1 9 25 ù ú ê5 3 8. ê 2 2 9. ê 1ú 2 2ú ê ú 2 ú ê2 8û ë0 2 ê4 7 3 5ú êë 2 úû 2

(v) ( 3 ´ 1)

(vi) (1 ´ 1)

4. (7 ´ 1), (1 ´ 7)

é9 7. ê 2 ê ë8

25 ù 2ú ú 18 û

Various Types of Matrices ROW MATRIX

Examples

A matrix having only one row is known as a row matrix or a row vector. (i) A = [5 18] is a row matrix of order 1 ´ 2. (ii) B = [2 5 - 9 0] is a row matrix of order 1 ´ 4.

COLUMN MATRIX

A matrix having only one column is known as a column matrix or a

column vector.

Examples

é 2ù (i) A = ê 7 ú is a column matrix of order 3 ´ 1. ú ê êë -3 úû é 6ù (ii) B = ê ú is a column matrix of order 2 ´ 1. ë 4û

ZERO OR NULL MATRIX

A matrix each of whose elements is zero is called a zero

matrix or a null matrix. Examples

é 0 0ù é 0 0 0ù The matrices [0], [0 0], ê ú and ê 0 0 0 ú are null matrices 0 0 ë û ë û of order (1 ´ 1), (1 ´ 2), ( 2 ´ 2) and ( 2 ´ 3) respectively.

SSS Mathematics for Class 12 156

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Senior Secondary School Mathematics for Class 12

SQUARE MATRIX

A matrix having the same number of rows and columns is called a

square matrix. A matrix of order (n ´ n) is called a square matrix of order n or an n-rowed square matrix. A matrix of order m ´ n, where m ¹ n, is called a rectangular matrix. Examples

é 3 (i) The matrix ê ë 6 é ê 5 ê (ii) The matrix ê 7 ê ê -9 êë

2ù is a 2-rowed square matrix. -5 úû ù 3 6ú ú 2 -4 ú is a 3-rowed square matrix. ú 1 0ú úû 3

DIAGONAL ELEMENTS OF A MATRIX Let A = [a ij ]m ´ n be an m ´ n matrix. Then, the elements aij for which i = j , are called the diagonal elements of A.

Thus, the diagonal elements of A = [aij ]m ´ n are a11 , a22 , a 33 , a44 , etc. The line along which the diagonal elements lie is called the diagonal of the matrix.

Example

ù é 2 -1 ú ê 3 ú ê 5 Let A = ê 5 7 ú 8 ú ê ê 6 -4 2ú úû êë Then, the diagonal elements of A are: 5 a11 = 3 , a22 = , a 33 = 2. 8

DIAGONAL MATRIX A square matrix in which every nondiagonal element is zero is called a diagonal matrix.

as

If A = [aij ]n ´ n be a diagonal matrix then aij = 0 when i ¹ j and we write it A = diag [a11 , a22 , a 33 , ¼, ann].

Example

é 6 Let A = ê 0 ê êë 0

0 4 0

0 ù 0 ú . Then, A is a diagonal matrix. ú -2 úû

We may write it as, A = diag [6, 4, - 2]. SCALAR MATRIX A square matrix in which every nondiagonal element is zero and all diagonal elements are equal is known as a scalar matrix.

Examples

é5 (i) A = ê ë 0

0 5

ù ú is a scalar matrix of order 2. û

SSS Mathematics for Class 12 157

Matrices

é -3 (ii) B = ê 0 ê êë 0

0 -3 0

157

0 ù 0 ú is a scalar matrix of order 3. ú -3 úû

UNIT MATRIX A square matrix in which every nondiagonal element is 0 and every diagonal element is 1 is called a unit matrix or an identity matrix.

Thus, a square matrix [aij ]n ´ n is a unit matrix if ì 0 when i ¹ j , aij = í î 1 when i = j. A unit matrix of order n will be denoted by I n or simply by I. Examples

é 1 (i) I 2 = ê ë 0 é 1 (ii) I 3 = ê 0 ê êë 0

0ù is a unit matrix of order 2. 1 úû 0 0ù 1 0 ú is a unit matrix of order 3. ú 0 1 úû

COMPARABLE MATRICES Two matrices A and B are said to be comparable if they are of the same order, i.e., they have the same number of rows and the same number of columns.

Example

é 2 -5 1 ù é 3 7 0ù A=ê ú and B = ê 1 4 -9ú are comparable matrices, each ë 0 3 6û ë û being of order ( 2 ´ 3).

EQUAL MATRICES Two matrices A and B are said to be equal, written as A = B , if they are of the same order and their corresponding elements are equal. EXAMPLE 1

SOLUTION

EXAMPLE 2

SOLUTION

é5 3ù éy zù Find x , y , z when ê ú = ê 1 7ú . x 7 ë û ë û Since the corresponding elements of equal matrices are equal, we have é 5 3 ù é y z ù ê x 7 ú = ê 1 7 ú Û x = 1, y = 5 and z = 3. ë û ë û éx - y Find x, y, z, w when ê ë 2x - y

2x + z ù é -1 5 ù = × 3z + wúû êë 0 13 úû

[CBSE 2013]

We know that in equal matrices, the corresponding elements are equal. 2x + z ù é -1 5ù é x-y \ ê = 3z + wúû êë 0 13 úû ë 2x - y Û x - y = - 1, 2x - y = 0, 2x + z = 5 and 3z + w = 13.

SSS Mathematics for Class 12 158

158

Senior Secondary School Mathematics for Class 12

Solving the first two equations, we get x = 1 and y = 2. Putting x = 1 in 2x + z = 5 , we get z = 3. Putting z = 3 in 3z + w = 13 , we get w = 4. \ x = 1, y = 2, z = 3 and w = 4. é 0 0ù ê 0 ¹ ê 0 úû êë 0

0ù 0 ú . Why? ú 0 úû

EXAMPLE 3

é 0 ê 0 ë

SOLUTION

Since the given null matrices are not comparable, they are not equal.

0 0

Operations on Matrices ADDITION OF MATRICES Let A and B be two comparable matrices, each of order (m ´ n). Then, their sum ( A + B) is a matrix of order (m ´ n), obtained by adding the corresponding elements of A and B.

Thus, if

A = [aij ]m ´ n and B = [bij ]m ´ n then A + B = [aij + bij ]m ´ n.

REMARK

Example 1

Example 2

For two matrices A and B, the sum ( A + B) exists only when A and B are comparable. é 2 1ù é 3 4 5 ù If A = ê and B = ê ú ú then A and B are matrices ë 0 4û ë 1 2 3 û of order 2 ´ 2 and 2 ´ 3 respectively. So, A and B are not comparable. Hence, A + B is not defined. é 5 Let A = ê ë 3

0 2

-2 ù é 4 and B = ê -7 úû ë -1

-3 0

-6 ù . 4 úû

Clearly, each one of A and B is a 2 ´ 3 matrix. So, A and B are comparable matrices. \ A + B is defined. 0 + ( -3) -2 + ( -6) ù é 5+4 Now, A + B = ê ú 3 + ( 1 ) 2 + 0 -7 + 4 ë û é 9 -3 -8 ù =ê . 2 -3 úû ë 2 Some Results on Addition of Matrices THEOREM 1

Matrix addition is commutative, i.e., A + B = B + A for all comparable matrices A and B.

SSS Mathematics for Class 12 159

Matrices PROOF

159

Let A = [aij ]m ´ n and B = [bij ]m ´ n. Then, A + B = [aij ]m ´ n + [bij ]m ´ n = [aij + bij ]m ´ n [by the definition of addition of matrices] = [bij + aij ]m ´ n [Q addition of numbers is commutative] = [bij ]m ´ n + [aij ]m ´ n = B + A. Hence, A + B = B + A. Matrix addition is commutative, i.e, ( A + B) + C = A + ( B + C) for all comparable matrices A, B, C. Let A = [aij ]m ´ n, B = [bij ]m ´ n and C = [cij ]m ´ n. Then, ( A + B) + C = ([aij ]m ´ n + [bij ]m ´ n) + [cij ]m ´ n = [aij + bij ]m ´ n + [cij ]m ´ n = [( aij + bij ) + cij ]m ´ n = [aij + ( bij + cij )]m ´ n [Q addition of numbers is associative] = [aij ]m ´ n + [bij + cij ]m ´ n = [aij ]m ´ n + ([bij ] + [cij ]m ´ n) = A + ( B + C).

THEOREM 2 PROOF

Hence, ( A + B) + C = A + ( B + C). If A is an m ´ n matrix and O is an m ´ n null matrix then A + O = O + A = A. Let A = [aij ]m ´ n and O = [bij ]m ´ n, where bij = 0 for all suffixes i and j. Then, A + O = [aij ]m ´ n + [bij ]m ´ n = [aij + bij ]m ´ n = [aij + 0]m ´ n [Q bij = 0] = [aij ]m ´ n = A. \ A + O = A. Similarly, O + A = A. Hence, A + O = O + A = A.

THEOREM 3 PROOF

REMARK

EXAMPLE 3

SOLUTION

The null matrix O of order m ´ n is the additive identity in the set of all m ´ n matrices. 4ù é 3 5 é 0 Let A = ê and O = ê ú ë 1 2 -3 û ë 0 A + O = O + A = A.

0 0

0ù , then verify that 0 úû

Clearly, each one of A and O is a matrix of order ( 2 ´ 3). So, ( A + O) and (O + A) are both defined. 4ù é 0 0 0ù é 3 5 Now, A + O = ê ú+ê ú ë 1 2 -3 û ë 0 0 0 û é3+ 0 =ê ë 1+ 0

5+0 2+ 0

4 + 0ù é 3 + -3 + 0úû êë 1

5 2

4ù = A. -3 úû

SSS Mathematics for Class 12 160

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Senior Secondary School Mathematics for Class 12

0 é0 And, O + A = ê 0 ë0 é0 + 3 =ê ë 0+1

0ù é 3 5 +ê ú 0û ë 1 2 0+5 0+ 4 0+ 2

4ù -3 úû

ù é3 = 0 + ( -3) úû êë 1

5 2

4ù = A. -3 úû

Hence, A + O = O + A = A. NEGATIVE OF A MATRIX Let A = [aij ]m ´ n. Then, the negative of A is the matrix ( - A) = [- aij ]m ´ n , obtained by replacing each element of A with its corresponding additive inverse. ( -A) is called the additive inverse of A. EXAMPLE 4

SOLUTION

0 ù é 3 -2 If A = ê ú , find ( -A) and verify that 5 7 2 û ë A + ( - A) = ( - A) + A = 0. Clearly, we have: é -3 2 ( -A) = ê ë 5 -7

0ù ú - 2 û 0 ù é -3 0ù é 3 -2 2 A + ( -A) = ê ú+ê ú 5 7 5 7 - 2 û 2 û ë ë 0 + 0 ù é0 0 é 3 + ( -3) -2 + 2 =ê ú=ê + + 5 5 7 ( 7 ) + ( - 2) û ë 0 0 2 ë 0 ù é 3 -2 0ù é -3 2 And, ( - A) + A = ê ú+ê ú 7 2 û ë 5 -7 - 2 û ë -5 0 + 0 ù é0 é -3 + 3 2 + ( -2) =ê ú=ê -7 + 7 - 2 + 2 û ë0 ë 5 + ( -5) Hence, A + ( - A) = ( - A) + A = O. SUMMARY

Laws of Addition on Matrices (i) A + B = B + A [Commutative law] (ii) ( A + B) + C = A + ( B + C) [Associative law] (iii) A + O = O + A = A (iv) A + ( - A) = ( - A) + A = O

SOLVED EXAMPLES EXAMPLE 1

é 2 3 5ù é 4 -2 3 ù Let A = ê ú and B = ê 2 6 -1ú × 1 0 4 ë û ë û Verify that A + B = B + A.

0ù = O× 0úû

0 0ù 0 0úû

SSS Mathematics for Class 12 161

Matrices SOLUTION

161

Here, A is a 2 ´ 3 matrix and B is a 2 ´ 3 matrix. So, A and B are comparable. Therefore, ( A + B) and ( B + A) both exist and each is a 2 ´ 3 matrix. é 2 3 5 ù é 4 -2 3 ù Now, A + B = ê ú+ê ú ë -1 0 4û ë 2 6 -1û é 2+ 4 =ê ë -1 + 2

3 + ( -2) 0+ 6

-2 6

é4 And, B + A = ê ë2

5+ 3 ù é6 1 = ê ú 4 + ( -1) û ë1 6

8ù × 3 úû

3ù é 2 3 5ù + -1úû êë -1 0 4úû

é4 + 2 =ê ë 2 + ( -1)

-2 + 3

3 + 5ù é 6 1 8ù = ê ú× ( -1) + 4úû ë 1 6 3û

6+ 0

Hence, A + B = B + A.

EXAMPLE 2

é1 Let A = ê5 ê êë 3

-2ù é 3 4ú , B = ê 0 ê ú 0úû êë -3

1ù é 4 2ú and C = ê -2 ê ú 5 úû êë 1

3ù 2ú × ú 6úû

Verify that ( A + B) + C = A + ( B + C). SOLUTION

Clearly, each one of the matrices A , B , C is a ( 3 ´ 2) matrix. So, ( A + B) + C and A + ( B + C) are both defined and each one is a 3 ´ 2 matrix. 1ù -2ù é 3 é1 Now, ( A + B) = ê 5 4ú + ê 0 2ú ú ú ê ê 0úû êë -3 5 úû êë 3 é1 + 3 = ê5 + 0 ê êë 3 + ( -3) \

é4 ( A + B) + C = ê5 ê êë 0

-1ù é 4 6ú + ê -2 ú ê 5 úû êë 1

é4 + 4 = ê5 + ( -2) ê êë 0 + 1 é 3 Also, ( B + C) = ê 0 ê êë -3

-2 + 1ù é 4 4 + 2ú + ê5 ú ê 0 + 5 úû êë 0

-1ù 6ú × ú 5 úû

3ù 2ú ú 6úû

-1 + 3 ù é 8 2ù 6 + 2ú = ê 3 8ú × ú ú ê 5 + 6úû êë 1 11úû

1ù é 4 2ú + ê -2 ú ê 5 úû êë 1

3ù 2ú ú 6úû

SSS Mathematics for Class 12 162

162

Senior Secondary School Mathematics for Class 12

\

é 3+4 = ê 0 + ( -2) ê êë -3 + 1 -2ù é1 4ú + A + ( B + C) = ê5 ú ê 0úû êë 3 é1 + 7 = ê5 + ( -2) ê êë 3 + ( -2)

1 + 3ù é 7 2 + 2ú = ê -2 ú ê 5 + 6úû êë -2 4ù é 7 ê -2 4ú ú ê êë -2 11úû -2 + 4 ù é 8 4+ 4 ú = ê3 ú ê 0 + 11 úû êë 1

4ù 4ú ú 11ûú

2ù 8ú × ú 11úû

Hence, ( A + B) + C = A + ( B + C). EXAMPLE 3

SOLUTION

é 2 Find the additive inverse of the matrix A = ê ë 4

SOLUTION

é 2 If A = ê ë 0

If A and B are two comparable matrices then we define ( A - B) = A + ( - B). -3 7

1ù é 1 2 -3 ù and B = ê ú ú , find ( A - B). -9 û ë 4 8 -4û

é -1 We have, ( - B) = ê ë -4 \ ( A - B) = A + ( - B) é 2 -3 =ê 7 ë 0 é 2 + ( -1) =ê ë 0 + ( -4) é 1 Hence, ( A - B) = ê ë -4

EXAMPLE 5

SOLUTION

3

0ù . -1 úû

Clearly, the additive inverse of the given matrix A is the matrix -A, given by 0 ù 5 0ù é -2 -( -5) é -2 -A = ê = ê ú ú× -3 - ( -1) û ë -4 ë -4 -3 1 û

SUBTRACTION OF MATRICES

EXAMPLE 4

-5

é a+4 If ê ë 8

-2 -8

3 ù . 4 úû

1ù é -1 -2 3 ù + ê ú -9 úû ë -4 -8 4 û -3 + ( -2) 1 + 3 ù é 1 -5 4ù = . 7 + ( -8) -9 + 4úû êë -4 -1 -5 úû 4ù -5 × -1 -5 úû

3 b ù é 2a + 2 b + 2 = -6 úû êë 8 a - 8b

ù ú , then find the value of ( a - 2b). û [CBSE 2014]

Comparing the corresponding elements of given equal matrices, we have a + 4 = 2a + 2, 3 b = b + 2 and a - 8b = -6. From these equations, we get a = 2 and b = 1. \ ( a - 2b) = ( 2 - 2 ´ 1) = ( 2 - 2) = 0.

SSS Mathematics for Class 12 163

Matrices

EXAMPLE 6

SOLUTION

é 9 -1 If ê ë -2 1

163

4ù é 1 2 -1 ù =A+ê ú ú , then find the matrix A. 3û ë 0 4 9û [CBSE 2013]

Clearly, we have é 9 -1 4 ù é A=ê ú -ê ë -2 1 3 û ë é 9 - 1 -1 - 2 =ê ë -2 - 0 1 - 4

1 2 -1 ù 0 4 9 úû 4 + 1 ù é 8 -3 5 ù = . 3 - 9 úû êë -2 -3 -6 úû

SCALAR MULTIPLICATION Let A be a given matrix and k be a number. Then, the matrix obtained by multiplying each element of A by k is called the scalar multiple of A by k, to be denoted by kA. If A is an (m ´ n) matrix then kA is also an (m ´ n) matrix. If A = [aij ]m ´ n then kA = [k × aij ]m ´ n. EXAMPLE 7

SOLUTION

If

é5 A=ê ë 6

4 -1

-2 7

ù ú , find (i) 3A û

(ii)

1 A 2

(iii) -2A.

We have: é 3 ×5 (i) 3 A = ê ë 3×6 é 1 ×5 ê 1 (ii) A = ê 2 2 ê 1×6 êë 2

3×4 3 × ( -1)

1 ×4 2 1 × ( -1) 2 é ( -2) × 5 (iii) -2A = ( -2) × A = ê ë ( -2) × 6 é -10 =ê ë -12

3 × ( -2) ù é 15 12 -6 ù ú = ê 18 -3 21 ú . 3 ×7 û ë û 1 ù é 5 ù × ( -2) ú 2 -1 ú ê 2 2 = . ú ê 1 7 ú 1 ú ê 3 ú ×7 úû ë 2 2 û 2 ( -2) × 4 ( -2) × ( -1)

-8 2

4 -14

( -2) × ( -2) ù ( -2) × 7 úû

ù ú× û

Some Properties of Scalar Multiplication If A and B are two matrices of the same order and k is a scalar then prove that k( A + B) = kA + kB. Let A = [aij ]m ´ n and B = [bij ]m ´ n. Then, k( A + B) = k × ([aij ]m ´ n + [bij ]m ´ n) = k × [aij + bij ]m ´ n [by definition of addition of matrices] = [k( aij + bij )]m ´ n [by definition of scalar multiplication] = [k × aij + k × bij ]m ´ n [by distributive law] = [k × aij ]m ´ n + [k × bij ]m ´ n = kA + kB. Hence, k( A + B) = kA + kB.

THEOREM 1 PROOF

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If A is any matrix and k1 , k2 are any scalars then prove that (ii) k1( k2 A) = ( k1 k2) A. (i) ( k1 + k2) A = k1 A + k2 A Let A = [aij ]m ´ n. Then,

THEOREM 2

PROOF

(i) ( k1 + k2) A = ( k1 + k2) × [aij ]m ´ n = [( k1 + k2) × aij ]m ´ n [by definition of scalar multiplication] = [k1 × aij + k2 × aij ]m ´ n [by distributive law] = [k1 × aij ]m ´ n + [k2 × aij ]m ´ n [by definition of addition of matrices] = k1 A + k2 A. Hence, ( k1 + k2) A = k1 A + k2 A. (ii) k1( k2 A) = k1[k2 × aij ]m ´ n = [k1( k2 × aij )]m ´ n = [( k1 k2) × aij ]m ´ n [by associativity of multiplication in numbers] = ( k1 k2) × [aij ]m ´ n = ( k1 k2) A. Hence, k1( k2 A) = ( k1 k2) A. SUMMARY

I. If A and B are comparable matrices and k is a scalar then k( A + B) = ( kA + kB). II. If k1 , k2 are scalars and A is any matrix then (i) ( k1 + k2) A = ( k1 A + k2 A) (ii) k1( k2 A) = ( k1 k2) A

SOLVED EXAMPLES

EXAMPLE 1

SOLUTION

é 3 If A = ê ë 7

5 ù é 6 and B = ê -9 úû ë 2

-4 ù , find ( 4A - 3 B). 3 úû

( 4A - 3 B) = 4A + ( -3 B). 4×5 é 4× 3 ù é 12 Now, 4A = ê = ê ú 4 × ( -9) û ë 4×7 ë 28 é ( -3) × 6 ( -3) × ( -4) And, -3 B = ( -3) × B = ê ë ( -3) × 2 ( -3) × 3 \

4A - 3 B = 4A + ( -3 B) 20 ù é 12 é -18 =ê + ê ú -36 û ë 28 ë -6 12 + ( 18 ) 20 + 12 é =ê -36 + ( -9) ë 28 + ( -6)

ù ú× û ù é -18 12 ú = ê -6 -9 û ë 20 -36

ù ú. û

12 ù -9 úû ù é -6 ú = ê 22 û ë

32 -45

ù ú. û

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é -6 Hence, ( 4A - 3 B) = ê ë 22

165

32 ù × -45 úû

EXAMPLE 2

Let A = diag [ 3 , - 5 , 7] and B = diag [-1, 2, 4] × Find (i) ( A + B) (ii) ( A - B) (iii) -5A (iv) ( 2A + 3 B).

SOLUTION

We have é 3 A=ê 0 ê êë 0

0 -5 0

0ù é -1 0 ú and B = ê 0 ê ú 7 úû êë 0

0 2 0

0ù 0ú× ú 4 úû

0 0 é 3 0 0ù é -1 0 0ù é 2 \ (i) A + B = ê 0 -5 0ú + ê 0 2 0ú = ê 0 -3 0 ú ê ú ê ê 0 11 êë 0 0 7 úû êë 0 0 4úû êë 0 (ii) ( A - B) = A + ( - B) 0 0ù é3 ê = 0 -5 0ú + ú ê 0 7 úû êë 0

0 é 1 0 0ù é 4 ê 0 -2 0ú = ê 0 -7 ú ê ê 0 êë 0 0 -4úû êë 0

ù ú× ú úû 0ù 0ú × ú 3 úû

0 0ù é -15 0 0ù é3 (iii) -5 A = ( -5) × A = ( -5) × ê 0 -5 0ú = ê 0 25 0ú × ú ú ê ê 0 7 úû êë 0 0 -35 úû êë 0 0 0ù é -3 0 0ù é 3 é6 (iv) 2A + 3 B = ê 0 -10 0ú + ê 0 6 0ú = ê 0 ú ê ú ê ê 0 14úû êë 0 0 12úû êë 0 êë 0 EXAMPLE 3

Simplify

é cos q cos q × ê ë - sin q

sin q ù é sin q + sin q × ê cos q úû ë cos q

0 0ù -4 0ú . ú 0 26úû - cos q ù . sin q úû [CBSE 2012]

SOLUTION

We have é cos q cos q × ê ë - sin q

sin qù é sin q - cos qù + sin q × ê cos qúû sin qúû ë - cos q

- sin q cos qù é sin q cos qù é sin 2 q cos2 q =ê ú ú+ê coss2q û ë sin q cos q sin 2 q û ë - sin q cos q sin q cos q + ( - sin q cos q) ù é cos2 q + sin 2 q =ê ú cos2 q + sin 2 ë - sin q cos q + sin q cos q û é 1 =ê ë 0

0 1

ù ú. û

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EXAMPLE 4

SOLUTION

Senior Secondary School Mathematics for Class 12

é 3 If 2 ê ë 5

4ù é 1 y ù é 7 0ù + = , find ( x - y). x úû êë 0 1 úû êë 10 5 úû

We have é 3 4ù é 1 2ê ú+ê ë 5 x û ë 0 é 6 8 ù é 1 Þ ê ú+ê ë 10 2x û ë 0 é 6+1 8+ y ù Þ ê ú ë 10 + 0 2x + 1 û Þ Þ Þ

y ù é 7 = 1 úû êë 10 y ù é 7 = 1 úû êë 10

0ù 5 úû 0ù 5 úû

é 7 0ù =ê ú ë 10 5 û

2x + 1 = 5 and 8 + y = 0 [comparing the corresponding elements] x = 2 and y = -8. ( x - y) = 2 - ( -8) = 10.

EXAMPLE 5

é 4 Find a matrix X, if X + ê ë -3

SOLUTION

é 4 Let A = ê ë -3

6ù é 3 = 7 úû êë 5

6ù é 3 and B = ê ú 7 û ë 5

-6 -8

-6 -8

ù ú. û

ù ú. û

Then, the given matrix equation is X + A = B. \ X + A = B Þ X = B + ( -A) é 3 -6 ù é -4 -6 ù =ê ú+ê ú ë 5 -8 û ë 3 -7 û -6 + ( -6) ù é -1 é 3 + ( -4) =ê = -8 + ( -7) úû êë 8 ë5+ 3 é -1 -12 ù Hence, X = ê ú. ë 8 -15 û EXAMPLE 6

SOLUTION

[CBSE 2014]

-12 -15

ù ú. û

é -1 2ù Find a matrix X such that 2A + B + X = O , where A = ê ú and ë 3 4û é 3 -2ù [CBSE 2002C] B=ê ú× ë 1 5û We have 2A + B + X = O Þ X = - ( 2A + B). é -1 2ù é 3 -2ù Now, ( 2A + B) = 2 × ê ú+ê 5 úû ë 3 4û ë 1 é -2 4ù é 3 -2ù =ê ú+ê 5 úû ë 6 8û ë 1

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4 + ( -2) ù é 1 é -2 + 3 =ê = 8 + 5 úû êë7 ë 6+1 é -1 -2ù \ X = - ( 2A + B) = ê ú× ë -7 -13 û EXAMPLE 7

SOLUTION

é5 Find matrices X and Y, if X + Y = ê ë 0

2ù × 13 úû

2ù é 3 and X - Y = ê ú 9û ë 0

Adding the given matrices, we get é5 2ù é 3 (X + Y) + (X - Y) = ê + ê ú ë 0 9û ë 0 é 5 + 3 2+ 6 ù Þ 2X = ê ú 0 + 0 9 + ( 1 ) ë û é 8 8ù 1 é 8 8ù é Þ 2X = ê ú Þ X = 2×ê 0 8 ú = ê 0 8 ë û ë û ë

6ù . -1 úû

6ù -1 úû

4 0

4ù . 4 úû

On subtracting the given matrices, we get 6ù é5 2ù é 3 (X + Y) - (X - Y) = ê ú -ê ú ë 0 9 û ë 0 -1 û é 5 - 3 2-6 ù é 2 -4 ù Þ 2Y = ê ú = ê 0 10 ú 0 0 9 ( 1 ) ë û ë û 1 é 2 -4 ù é 1 -2 ù = . 5 úû 2 êë 0 10 úû êë 0 é 4 4ù é 1 Hence, X = ê and Y = ê ú ë 0 4û ë 0

Þ

Y=

-2 5

ù ú. û

EXERCISE 5B 2 -2 ù é 2 -3 5 ù é 3 1. If A = ê and B = ê , verify that ( A + B) = ( B + A). ú 1 úû 0 3û ë -1 ë 4 -3 2ù 5 ù é 0 é -1 -3 ù é 3 ê ú ê ú ê 2. If A = -2 0 ,B = 4 2 and C = 3 -4 ú , ú ê ú ê ú ê 6 úû êë 1 êë -2 3 úû êë 6 -1 úû verify that ( A + B) + C = A + ( B + C). 2ù 0 4ù é 3 1 é -2 3. If A = ê ú and B = ê 5 -3 2 ú , find ( 2A - B). 1 2 3 ë û ë û é 2 4ù é 1 3 ù é -2 5 ù 4. Let A = ê ú , B = ê - 2 5 ú and C = ê 3 4ú . Find: 3 2 ë û ë û ë û (i) A + 2B (ii) B - 4C (iii) A - 2B + 3C

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1 -2 ù é 0 é 1 -3 5. Let A = ê , B=ê ú ë 5 -1 -4 û ë 0 -2 Compute 5 A - 3 B + 4C. é 5 10 -15 ù 6. If 5 A = ê 2 3 4 ú , find A. ú ê 0 -5 úû êë 1 7. Find matrices A é A+B=ê ê êë

-1 ù é 2 and C = ê ú 5 û ë -4

and B , if 1 0 2ù é -5 ú 5 4 -6 and A - B = ê 11 ú ê 7 3 8 úû êë -1

8. Find matrices A and B, if é 6 -6 2A - B = ê 2 ë -4 9. Find matrix X, if 3 é -2 ê 10. If A = 4 5 ê êë 1 -6 A + B - C = O.

é 3 ê -1 ë

5 4

-4 2 7

0ù é 3 and 2B + A = ê 1 úû ë -2 -9 ù é 6 +X = ê ú -7 û ë 4

é 5 ù ú and B = ê -7 ê ú êë 6 úû

2 8

-5 0

1ù × 6 úû

8ù 0 ú. ú 4 úû 2 1

5 ù . -7 úû

3 ù . 6 úû

2 ù 3 ú , find a matrix C such that ú 4 úû

11. Find the matrix X such that 2A - B + X = O , é 3 1ù é -2 1 ù where A = ê and B = ê ú ú× ë 0 2û ë 0 3û é 1 -3 2ù é 2 -1 -1ù 12. If A = ê and B = ê , find a matrix C such that ú 0 2û 0 -1úû ë2 ë1 ( A + B + C) is a zero matrix. 13. If A = diag [2, - 5 , 9], B = diag [-3 , 7 , 14] and C = diag [4, - 6, 3], find: (i) A + 2B (ii) B + C - A (iii) 2A + B - 5C. 14. Find the values of x and y, when é x+y ù é 8ù (i) ê ú = ê 4ú x y ë û ë û 7 é 2x + 5 ù (ii) ê = 0 3 y - 7 úû ë 5 ù é x é 3 (iii) 2 ê + ê ú y-3 û ë 7 ë 1

é x-3 ê 0 ë -4 ù = 2 úû

7 ù -5 úû 6ù é 7 ê 15 14 ú ë û

[CBSE 2012]

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15. Find the value of ( x é 1 3 ù 2ê ú + ë 0 x û

169

+ y) from the following equation: é y 0ù é5 6ù ê 1 2ú=ê 1 8ú ë û ë û

[CBSE 2012, ’13C]

2y ù é 1 4ù é x-y 16. If ê ú=ê ú then write the value of ( x + y + z). [CBSE 2013C] ë 2y + z x + y û ë 9 5 û ANSWERS (EXERCISE 5B)

é 8 3. ê ë -3

é5 5. ê ë 9

0ù -8 úû

2 7

-6 1

é 4 4. (i) ê ë -1

10 ù 12 úû

é ê ê 6. ê ê ê êë

-3 ù -11 úû

1

2

2 5 1 5

3 5 0

é -2 7. A = ê 8 ê êë 3

-2 3 5

5 ù 2 -3 ù é 3 1 -3 ú -3 ú and B = ê -3 ê ú ú 6 úû 2úû êë 4 -2

é 3 8. A = ê ë -2

-2 1

1ù é 0 and B = ê -1 úû ë 0

é 3 9. ê ë 5 é -8 11. ê ë 0

-3 4 -1 ù -1 úû

13. (i) diag [-4, 11, 37] 14. (i) x = 6, y = 2 15. x + y = 6

é -6 (iii) ê ë 16

13 ù 4 úû

ù -3 ú 4ú ú 5 ú -1 ú úû

4 4ù 0 -6 úû

é 3 10. ê -3 ê êë 7 é -3 12. ê ë -3

12 ù 13 úû

-17 ù -11 úû

9 é (ii) ê 14 ë

5 ù 8ú ú -2 úû 4 0

(ii) diag [-1, 6, 8] 2 (ii) x = - 8, y = 3

-1 ù -1 úû (iii) diag [-19, 27 , 17] (iii) x = 2, y = 4

16. 10

Multiplication of Matrices For two given matrices A and B, we say that the product AB exists only when the number of rows in A equals the number of columns in B. When AB exists, we say that A is conformable to B for multiplication.

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PRODUCT OF MATRICES

Let A = [aij ]m ´ n and B = [bjk ]n ´ p be two matrices such that the number of columns in A equals the number of rows in B. Then, AB exists and it is an (m ´ p) matrix, given by n

AB = [cik ]m ´ p , where cik = ( ai 1 b1 k + ai 2 b2 k + ¼ ainbnk ) = S aij bjk . j =1

\ (i , k)th element of AB = sum of the products of corresponding elements of ith row of A and kth column of B. SUMMARY

(i) If A is an (m ´ n) matrix and B is an (n ´ p) matrix then AB exists and it is an (m ´ p) matrix. (ii) (i , k)th element of AB = sum of the products of corresponding elements of ith row of A and kth column of B. REMARKS

For two given matrices A and B: (i) AB may exist and BA may not exist; (ii) BA may exist and AB may not exist; (iii) AB and BA both may not exist; (iv) AB and BA both may exist. SOLVED EXAMPLES

EXAMPLE 1

SOLUTION

é b11 a13 ù ê b and B = ê 21 a22 a23 úû êë b 31 AB and BA both exist. Find AB and BA. é a If A = ê 11 ë a21

a12

b12 b22 b 32

Here, A is a 2 ´ 3 matrix and B is a 3 ´ 2 matrix. \ AB exists and it is a 2 ´ 2 matrix. c12 ù é c Let AB = ê 11 ú . Then, c ë 21 c22 û c11 = (1st row of A) ´ (1st column of B) = a11 b11 + a12 b21 + a13 b 31 ; c12 = (1st row of A) ´ ( 2nd column of B) = a11 b12 + a12 b22 + a13 b 32 ; c21 = ( 2nd row of A) ´ (1st column of B) = a21 b11 + a22 b21 + a23 b 31 ;

ù ú then show that ú úû

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171

and c22 = ( 2nd row of A) ´ ( 2nd column of B) = a21 b12 + a22 b22 + a23 b 32. \

é a11 AB = ê ë a21

a12 a22

b12 ù b22 ú ú b 32 úû

é b11 a13 ù ê b21 a23 úû ê êë b 31

é a b + a12 b21 + a13 b 31 = ê 11 11 ë a21 b11 + a22 b21 + a23 b 31

a11 b12 + a12 b22 + a13 b 32 ù × a21 b12 + a22 b22 + a23 b 32 úû

Again, B is a 3 ´ 2 matrix and A is a 2 ´ 3 matrix. So, BA exists and it is a 3 ´ 3 matrix. Proceeding as above, we get b12 ù é b11 é a11 a12 ê BA = b21 b22 ú ê ú ë a21 a22 ê êë b 31 b 32 úû é b11 a11 + b12 a21 = ê b21 a11 + b22 a21 ê êë b 31 a11 + b 32 a21 -1 4 5

a13 ù a23 úû

b11 a12 + b12 a22 b21 a12 + b22 a22 b 31 a12 + b 32 a22

ù ú and B = é -1 ê 2 ú ë úû

EXAMPLE 2

é 2 If A = ê 3 ê êë 1

SOLUTION

Here A is a 3 ´ 2 matrix and B is a 2 ´ 2 matrix.

3 ù , 1 úû

b11 a13 + b12 a23 ù b21 a13 + b22 a23 ú . ú b 31 a13 + b 32 a23 úû

find AB. Does BA exist?

Clearly, the number of columns in A equals the number of rows in B. \ AB exists and it is a 3 ´ 2 matrix. -1 ù é 2 é -1 3 ù Now, AB = ê 3 4úê ú ë 2 1 úû ê 5 úû êë 1 é 2 × ( -1) + ( -1) × 2 =ê 3 × ( -1) + 4 × 2 ê 1 × ( -1) + 5 × 2 êë é -4 =ê 5 ê êë 9

5 13 8

2 × 3 + ( -1) × 1 3 × 3 + 4×1 1× 3 + 5 ×1

ù ú ú úû

ù ú. ú úû

Further, B is a 2 ´ 2 matrix and A is a 3 ´ 2 matrix. So, the number of columns in B is not equal to the number of rows in A. So, BA does not exist.

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EXAMPLE 3

Senior Secondary School Mathematics for Class 12

é 1 Let A = ê ë -4

-2 2

é 2 3 ù and B = ê 4 ê 5 úû êë -2

3 ù 5 ú . Find AB and BA, and ú 1 úû

show that AB ¹ BA. SOLUTION

Here A is a 2 ´ 3 matrix and B is a 3 ´ 2 matrix. So, AB exists and it is a 2 ´ 2 matrix. é 1 Now, AB = ê ë -4

é 2 3 ù ê 4 5 úû ê êë -2

-2 2

3 ù ú ú úû

5 1

é 1 × 2 + ( -2) × 4 + 3 × ( -2) =ê ë ( -4) × 2 + 2 × 4 + 5 × ( -2) é -12 =ê ë -10

1 × 3 + ( -2) × 5 + 3 × 1 ù (-4) × 3 + 2 × 5 + 5 × 1 úû

-4 ù . 3 úû

Again, B is a 3 ´ 2 matrix and A is a 2 ´ 3 matrix. So, BA exists and it is a é 2 3 ù Now, BA = ê 4 5 ú ú ê êë -2 1 úû

3 ´ 3 matrix. é 1 ê -4 ë

é 2 × 1 + 3 × ( -4) =ê 4 × 1 + 5 × ( -4) ê êë ( -2) × 1 + 1 × ( -4) é -10 = ê -16 ê êë -6 Hence, AB ¹ BA. EXAMPLE 4

SOLUTION

2 2 6

-2 2

3 ù 5 úû

2 × ( -2) + 3 × 2 4 × ( -2) + 5 × 2 ( -2) × ( -2) + 1 × 2

2× 3 + 3 ×5 4× 3 + 5 ×5 ( -2) × 3 + 1 × 5

ù ú ú úû

21 ù 37 ú . ú -1 úû

é x ù If [2x 4] ê ú = O , find the positive value of x. ë -8 û

[CBSE 2014C]

The given matrix equation is AB = O , where A is a (1 ´ 2) matrix and B is a ( 2 ´ 1) matrix. So, AB is a (1 ´ 1) matrix. So, O is a (1 ´ 1) matrix. é x ù \ [2x 4] ê ú = [0] ë -8 û Þ

2x 2 - 32 = 0 Þ 2x 2 = 32

Þ

x 2 = 16 Þ x = 16 = 4.

Hence, x = 4.

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173

Properties of Matrix Multiplication 1. Commutativity Matrix multiplication is not commutative in general. PROOF

Let A and B be two given matrices. If AB exists then it is quite possible that BA may not exist. For example, if A is a 3 ´ 2 matrix and B is a 2 ´ 2 matrix then clearly, AB exists but BA does not exist. Similarly, if BA exists then AB may not exist. For example, if A is a 2 ´ 3 matrix and B is a 2 ´ 2 matrix then clearly, BA exists but AB does not exist. Further, if AB and BA both exist, they may not be comparable. For example, if A is a 2 ´ 3 matrix and B is a 3 ´ 2 matrix then clearly, AB as well as BA exists. But, AB is a 2 ´ 2 matrix while BA is a 3 ´ 3 matrix. Again, if AB and BA both exist and they are comparable, even then they may not be equal. é1 1ù é 1 2 ù For example, if A = ê ú and B = ê 0 3 ú then AB and BA are ë1 2û ë û both defined and each one is a 2 ´ 2 matrix. é1 But, AB = ê ë1

5 ù é 3 and BA = ê 8 úû ë 3

5 ù . 6 úû

This shows that AB ¹ BA. Hence, in general, AB ¹ BA. REMARKS

(i) When AB = BA , we say that A and B commute. (ii) When AB = - BA , we say that A and B anticommute.

2. Associative law For any matrices A, B, C for which ( AB)C and A( BC) both exist, we have ( AB)C = A( BC). 3. Distributive laws of multiplication over addition We have: (i) A × ( B + C) = ( AB + AC) (ii) ( A + B) × C = ( AC + BC) 4. The product of two nonzero matrices can be a zero matrix. Example

é 0 1ù é 1 A=ê ú and B = ê 0 0 2 ë û ë Then, A ¹ O and B ¹ O. Let

2 0

ù ú. û

é 0 1 ù é 1 2ù é 0 × 1 + 1 × 0 0 × 2 + 1 × 0 ù And, AB = ê ú×ê ú=ê ú ë 0 2û ë 0 0û ë 0 × 1 + 2 × 0 0 × 2 + 2 × 0û

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é 0 = ê ë 0 Left zero divisor Right zero divisor

0 0

ù ú× û

If AB = O and A ¹ O then A is called a left zero divisor of AB. If AB = O and B ¹ O then B is called a right zero divisor of AB.

5. If A is a given square matrix and I is an identity matrix of the same order as A then we have A × I = I × A = A. 6. If A is a given square matrix and O is the null matrix of the same order as A then O × A = A × O = O. Positive Integral Powers of a Square Matrix Let A be a square matrix of order n. Then, we define:

\ THEOREM 1

A2 = A × A; A 3 = A × A × A = A2 × A; A 4 = A × A × A × A = A 3 × A, and so on. A n = ( A × A × A × ... × n times). If A and B are square matrices of the same order then ( A + B) 2 = A 2 + AB + BA + B 2. Also, when AB = BA then ( A + B) 2 = A 2 + 2AB + B 2.

PROOF

Let A and B be n-rowed square matrices. Then, clearly, ( A + B) is a square matrix of order n. So, ( A + B) 2 is defined. Now, ( A + B) 2 = ( A + B) × ( A + B) = A × ( A + B) + B × ( A + B) = AA + AB + BA + BB = A 2 + AB + BA + B 2.

[by distributive law] [by distributive law]

Hence, ( A + B) 2 = ( A 2 + AB + BA + B 2). Particular case When AB = BA In this case, we have ( A + B) 2 = ( A 2 + AB + AB + B 2) = ( A 2 + 2AB + B 2) [Q BA = AB ]. THEOREM 2

If A and B are square matrices of the same order then ( A + B) ( A - B) = A 2 - AB + BA - B 2. Also, when AB = BA then ( A + B)( A - B) = A 2 - B 2.

PROOF

We have: ( A + B) × ( A - B) = A( A - B) + B( A - B) = AA - AB + BA - BB = A 2 - AB + BA - B 2. Hence,

[by distributive law] [Q A( B - C) = AB - AC ]

( A + B) ( A - B) = A 2 - AB + BA - B 2.

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175

Particular case When AB = BA In this case, ( A + B) ( A - B) = ( A 2 - B 2) MATRIX POLYNOMIAL

m

Let f ( x) = a0 x + a1 x

m-1

[Q BA = AB ].

+ a2 x m - 2 + ¼ + am - 1 x + am be

a polynomial of degree m and let A be a square matrix of order n. Then, we define f ( A) = a0 A m + a1 A m - 1 + a2 A m - 2 + ¼ + am - 1 A + amI , where I is a unit matrix of order n.

SOLVED EXAMPLES

EXAMPLE 1

SOLUTION

é5 4 ù é3 If A = ê ú and B = ê 6 2 3 ë û ë

5 1ù , find AB and BA whichever exists. 8 4úû

Here, A is a 2 ´ 2 matrix and B is a 2 ´ 3 matrix. Clearly, the number of columns in A = number of rows in B. \ AB exists and it is a 2 ´ 3 matrix. 4ù é 3 5 1ù é5 AB = ê úê 6 2 3 8 4 úû ë ûë 5 ×5 + 4× 8 é 5 × 3 + 4× 6 =ê × + × 2 3 3 6 2×5 + 3 × 8 ë 25 + 32 5 + 16 é 15 + 24 =ê 10 + 24 2 + 12 ë 6 + 18

5 ×1 + 4× 4 ù 2 × 1 + 3 × 4 úû 57 ù é 39 ú = ê 24 34 û ë

21 14

ù ú× û

Again, B is a 2 ´ 3 matrix and A is a 2 ´ 2 matrix. \ number of columns in B ¹ number of rows in A. So, BA does not exist.

EXAMPLE 2

é 2 If A = ê ë -4

-1 5

é 2 3 ù ê 4 and B = ê 1 úû êë 1

3ù -2ú then find AB and BA. ú 5 úû

Show that AB ¹ BA. SOLUTION

Here, A is a 2 ´ 3 matrix and B is a 3 ´ 2 matrix So, number of columns in A = number of rows in B. \ AB exists and it is a 2 ´ 2 matrix. 3ù é 2 3 ù ê -1 é 2 4 2ú AB = ê ú 5 1 úû ê ë -4 5 úû êë 1 é 2 × 2 + ( -1) × 4 + 3 × 1 =ê ë -4 × 2 + 5 × 4 + 1 × 1

2 × 3 + ( -1) × ( -2) + 3 × 5 ù ú -4 × 3 + 5 × ( -2) + 1 × 5 û

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é 4-4+ 3 =ê ë -8 + 20 + 1

6 + 2 + 15 ù é 3 = ê ú -12 - 10 + 5 û ë 13

23 ù × -17 úû

Again, B is a 3 ´ 2 matrix and A is a 2 ´ 3 matrix. So, number of columns in B = number of rows in A. \ BA exists and it is a 3 3 ù é 2 é 2 BA = ê 4 -2 ú ê ú ë -4 ê 5 úû êë 1 é 2 × 2 + 3 × ( -4) = ê 4 × 2 + ( -2) × ( -4) ê êë 1 × 2 + 5 × ( -4) -2 + 15 é 4 - 12 ê = 8+ 8 -4 - 10 ê -1 + 25 êë 2 - 20

´ 3 matrix. -1

3 ù 1 úû

5

2 × ( -1) + 3 × 5 4 × ( -1) + ( -2) × 5

2× 3 + 3 ×1 4 × 3 + ( -2) × 1 1× 3 + 5 ×1

1 × ( -1) + 5 × 5 6+ 3 ù 12 - 2 ú = ú 3 + 5 úû

é -8 ê 16 ê êë -18

ù ú ú ûú

9ù 10 ú × ú 8 úû

13 -14 24

Clearly, AB ¹ BA.

EXAMPLE 3

SOLUTION

é 3 é 1 -1 2 ù If A = ê 3 2 0ú , B = ê 0 ê ú ê 0 1 úû êë -2 êë -2 verify that ( AB)C = A( BC).

1ù é 2 2 ú and C = ê ú ë3 5 úû

1 0

-3 ù then -1 úû

We have -1 2 0

é 1 AB = ê 3 ê êë -2

2 0 1

é 3 -0-4 =ê 9+ 0-0 ê êë - 6 + 0 - 2 Þ

é -1 ( AB)C = ê 9 ê êë -8

é 3 BC = ê 0 ê êë -2

é 3 ê 0 ê êë -2

1 2 5

ù ú ú úû

1 - 2 + 10 ù é -1 ú 3 + 4+ 0 = ê 9 ê ú -2 + 0 + 5 úû êë -8

9ù 7 ú ú 3 úû

9ù é 2 1 -3 ù 7 ú ê ú ë 3 0 -1úû 3 úû

é -2 + 27 = ê 18 + 21 ê êë -16 + 9 Also,

ù ú ú úû

1 2 5

-1 + 0 9+ 0 -8 + 0 ù ú é 2 ú êë 3 úû

3 - 9ù é 25 -27 - 7 ú = ê 39 ú ê 24 - 3 úû êë -7 1 0

-3 ù -1 úû

-1 9 -8

-6ù -34ú × ú 21úû

SSS Mathematics for Class 12 177

Matrices

6+ 3 é ê 0+ 6 = ê êë - 4 + 15 Þ

3+0 0+ 0

2ù é 9 0ú ê 6 ú ê 1 úû êë 11

0

9 - 6 + 22 é = ê 27 + 12 + 0 ê êë -18 + 0 + 11 -1 9

é 25 = ê 39 ê êë -7

-9 - 1 ù é 9 0-2 ú =ê 6 ú ê 6 - 5 úû êë 11

-2 + 0

-1 2

é 1 A( BC) = ê 3 ê êë -2

177

-2

9+ 0-0 -6+ 0-2

0 -2

-10 -2 1

3 0

3 -0-4

3

-10 ù -2 ú ú 1 úû

ù ú ú úû

-10 + 2 + 2 ù -30 - 4 + 0 ú ú 20 - 0 + 1 úû

-6 ù -34 ú × ú 21 úû

-8

Hence, ( AB)C = A( BC). EXAMPLE 4

SOLUTION

é 3 2ù é 1 -2 5 ù é 8 If A = ê ú,B=ê 0 ú and C = ê 2 1 0 7 3 ë û ë û ë verify that A( B + C) = ( AB + AC).

-6 ù , 0 úû

1 -5

We have é 3 A( B + C) = ê ë 1 é 3 =ê ë 1

é 3 Now, AB = ê ë 1

-1 2

2ù é 9 0 úû êë 2

é 3 × 9 + 2× 2 =ê ë 1× 9 + 0× 2 é 31 =ê ë 9

-2 7

2 ù ìé 1 × í 0 úû îêë 0

-2

é 3 =ê ë 1

8 -2

21 5

é 3 And, AC = ê ë 1

2ù 0 úû

é 8 ê 2 ë

7

3 × ( -1) + 2 × 3 ù 1 × ( -1) + 0 × 3 úû

5 ù 3 úû

3 × ( -2) + 2 × 7 1 × ( -2) + 0 × 7 ù ú. û 1 -5

-6 ù ü ý 0 úû þ

-1 ù 3 úû

3 × ( -1) + 2 × 2 1 × ( -1) + 0 × 2

é 1 ê 0 ë

é 3 ×1 + 2× 0 =ê ë 1×1 + 0× 0

1 -5

3 ù . -1 úû

1 -1

2ù 0 úû

5 ù é 8 + 3 úû êë 2

-6 ù 0 úû

3 ×5 + 2× 3 ù 1 × 5 + 0 × 3 úû

SSS Mathematics for Class 12 178

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Senior Secondary School Mathematics for Class 12

\

3 × 1 + 2 × ( -5) é 3 × 8 + 2× 2 =ê 1 × 1 + 0 × ( -5) ë 1× 8 + 0× 2 é 28 -7 -18 ù =ê . 1 -6 úû ë 8 8 21 ù é 3 é 28 ( AB + AC) = ê ú + ê 8 1 2 5 ë û ë é 31 =ê ë 9

Hence, EXAMPLE 5

SOLUTION

1 -1

3 -1

3 × ( -6) + 2 × 0 ù 1 × ( -6) + 0 × 0 úû

-7 1

-18 ù -6 úû

ù ú. û

A( B + C) = AB + AC.

Give an example of two matrices A and B such that A ¹ O , B ¹ O and AB = BA = O. é1 1ù é 1 -1 ù Let A = ê × Then, ú and B = ê -1 1 1 1 úû ë û ë é 1 1 ù é 1 -1 ù AB = ê ú ê 1 úû ë 1 1 û ë -1 é 1 - 1 -1 + 1 ù é 0 0ù =ê = ê ú ú ë 1 - 1 -1 + 1 û ë 0 0û é 1 -1 ù é 1 1 ù and BA = ê 1 úû êë 1 1 úû ë -1 é 1-1 =ê ë -1 + 1

1-1 ù é 0 = ê -1 + 1 úû ë 0

0ù × 0 úû

Thus, A ¹ O , B ¹ O. But, AB = BA = O. EXAMPLE 6

SOLUTION

cos q sin qù cos f sin fù é cos2 q é cos2 f If A = ê ú and B = ê ú, 2 sin q û sin 2 f û ë cos q sin q ë cos f sin f p show that AB is a zero matrix if q and f differ by an odd multiple of . 2 We have é cos2 q AB = ê ë cos q sin q

cos q sin q ù é cos2 f ê ú sin 2 q û ë cos f sin f

cos f sin f ù ú sin 2 f û

é cos2 q cos2 f + cos q sin q cos f sin f =ê êë cos q sin q cos2 f + sin 2 q cos f sin f cos2 q cos f sin f + cos q sin q sin 2 fù ú cos q sin q cos f sin f + sin 2 q sin 2 fúû

SSS Mathematics for Class 12 179

Matrices

é =ê ë é =ê ë

179

cos q cos f× cos ( q - f) sin q cos f× cos ( q - f) 0 0

EXAMPLE 7

é 3 If A = ê ë -4

SOLUTION

We have

cos q sin f× cos ( q - f) ù sin q sin f× cos ( q - f) úû 0 ù é Q ( q - f) being an odd multiple of (p 2), ù ú× 0 úû êë we have cos ( q - f) = 0 û -5 ù , show that A 2 - 5 A - 14I = O. 2 úû

é 3 -5 ù é 3 A2 = ê 2 úû êë -4 ë -4 é 3 × 3 + ( -5) ( -4) =ê ë -4 × 3 + 2 × ( -4)

[CBSE 2004]

-5 ù 2 úû

3 × ( -5) + ( -5) × 2 ù é 29 -25 = ê ú -4 × ( -5) + 2 × 2 û 24 ë -20 25 ù é 3 -5 ù é -15 -5 A = ( -5) ê = ê ú; 2 úû ë -4 ë 20 -10 û 0ù é 1 0ù é -14 -14I = ( -14) ê ú. ú = ê 0 1 0 14 û ë û ë \ A 2 - 5 A - 14I = A 2 + ( -5) A + ( -14I) 0ù é 29 -25 ù é -15 25 ù é -14 =ê +ê +ê ú ú 24û ë 20 -10û ë 0 -14úû ë -20 -25 + 25 + 0 ù é 29 + ( -15) + ( -14) =ê -20 + 20 + 0 24 + ( -10) + ( -14) ûú ë

ù ú; û

é 0 0ù =ê ú. ë 0 0û Hence, A 2 - 5 A - 14I = O. EXAMPLE 8

é 1 If A = ê ë -1

SOLUTION

We have

0ù , find k so that A 2 = 8A + kI. 7 úû

0 ù é 1 0 ù é 1-0 = 7 úû êë -1 7 úû êë -1 - 7 é 1 0ù é 1 0ù ( 8A + kI) = 8 × ê + k×ê ú ú ë -1 7 û ë 0 1û 0ù é k 0ù é é 8 =ê ú+ê ú=ê ë -8 56 û ë 0 k û ë é 1 A2 = ê ë -1

[CBSE 2005C]

0+ 0 ù é 1 = 0 + 49 úû êë -8

8+ k -8

ù 56 + k úû 0 ù é 1 0 ù é 8+ k \ A 2 = 8A + kI Þ ê ú = ê -8 8 49 56 + k úû ë û ë Þ 8 + k = 1 and 56 + k = 49 Þ k = -7. Hence, k = - 7. 0

0 49

ù ú; û

SSS Mathematics for Class 12 180

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Senior Secondary School Mathematics for Class 12

EXAMPLE 9

é 3 1ù If f ( x) = x 2 - 5 x + 7 and A = ê ú , find f ( A). [CBSE 2004C, ’07C] ë -1 2û

SOLUTION

We have

f ( A) = A 2 - 5 A + 7 I.

é 3 1ù é 3 1ù Now, A 2 = ê ú ê ú ë -1 2 û ë -1 2 û 3+2ù é 9-1 é 8 5 ù =ê ú = ê -5 3 ú ; ë -3 - 2 -1 + 4 û ë û ( -5) × 1 ù -5 ù é ( -5) × 3 é -15 -5 A = ê ; ú = ê ( 5 ) × ( 1 ) ( 5 ) × 2 5 10 úû ë û ë é7 0ù é 1 0ù 7I = 7 × ê ú = ê 0 7 ú. 0 1 û ë û ë f ( A) = A 2 - 5 A + 7 I

\

-5 ù é 8 5 ù é -15 =ê +ê + ú 5 -10 úû ë -5 3 û ë é 8 + ( -15) + 7 5 + ( -5) + 0 =ê 3 + ( -10) + 7 ë -5 + 5 + 0 é 0 Hence, f ( A) = ê ë 0

EXAMPLE 10

SOLUTION

0 0

é7 ê 0 ë

0ù 7 úû

ù é 0 ú = ê 0 û ë

0ù . 0 úû

ù ú. û

é 2 Find the matrix A such that ê 1 ê êë -3

-1 0 4

é -1 ù ú×A = ê 1 ê ú êë 9 úû

-8 -2 22

-10 ù -5 ú . ú 15 úû

Clearly, the product is a 3 ´ 3 matrix and the prefactor is a 3 ´ 2 matrix. So, A must be a 2 ´ 3 matrix. é x y z ù Let A = ê ú. ë u v wû Then, the given equation becomes é -1 é 2 -1 ù é x y z ù ê ê 1 = 0úê 1 ú ë u v w úû ê ê 4 úû êë 9 êë -3 or

é ê ê êë

2x - u x -3 x + 4u

2y - v y -3 y + 4v

-8 -2 22

2z - w z -3z + 4w

ù ú ú úû

-10 ù -5 ú ú 15 úû é -1 -8 = ê 1 -2 ê êë 9 22

-10 -5 15

ù ú. ú úû

So, by the definition of equal matrices, we have 2x - u = - 1; 2y - v = - 8; 2z - w = - 10; x = 1; y = - 2; z = - 5. \ x = 1; y = - 2; z = - 5 ; u = 3 ; v = 4 and w = 0.

SSS Mathematics for Class 12 181

Matrices

é 1 Hence, A = ê ë 3 EXAMPLE 11

SOLUTION

-2 4

181

-5 ù . 0 úû

Find the value of x, if 3 é 1 [1 x 1] ê 2 5 ê 3 êë 15

2ù 1ú ú 2 úû

é 1 ù ê 2 ú = O. ê ú êë x úû

[CBSE 2006C]

We have [1 x 1]1 ´ 3

é 1 ê 2 ê êë 15

3 5 3

2ù 1ú ú 2 úû

3´3

é 1 ù ê 2ú ê ú êë x úû

é 1 3 + 5 x + 3 2 + x + 2] ê 2 ê êë x

=O 3 ´1

ù ú =O ú úû

Þ

[1 + 2x + 15

Þ

é 1 ù [16 + 2x 6 + 5 x 4 + x] ê 2 ú = O ê ú êë x úû

Þ

[(16 + 2x) × 1 + ( 6 + 5 x) × 2 + ( 4 + x) × x] = O

Þ

(16 + 2x) + (12 + 10x) + ( 4x + x 2) = 0

Þ

x 2 + 16x + 28 = 0

Þ ( x + 14) ( x + 2) = 0 Þ x + 14 = 0 or x + 2 = 0 Þ x = - 14 or x = - 2. Hence, x = - 14 or x = - 2. EXAMPLE 12

SOLUTION

é x y ù é 1ù é 3 ù Solve for x and y, given that ê úê ú=ê ú× ë 3 y x û ë 2û ë 5 û

[CBSE 2003C]

We have é x ê 3y ë Þ Þ

y ù é 1ù é 3 ù = ê ú x úû êë 2 úû ë 5 û é x + 2y ù é 3 ù ê 3 y + 2x ú = ê 5 ú ë û ë û ì x + 2y = 3 í î 2x + 3 y = 5

Multiplying (i) by 2 and subtracting (ii) from it, we get y = 1. Putting y = 1 in (i), we get x = 1. Hence, x = 1 and y = 1.

... (i) ... (ii)

SSS Mathematics for Class 12 182

182

EXAMPLE 13

SOLUTION

Senior Secondary School Mathematics for Class 12

a ù 2 ú and is the identity matrix of order 2. I ú ú 0 úû é cos a - sin a ù Show that ( I + A) = ( I - A) × ê . cos a úû ë sin a é ê 0 Let A = ê ê tan a êë 2

Let tan

a = t. 2

Then, cos a = and

- tan

sin a =

1 - tan 2( a /2) 2

1 + tan ( a /2) 2 tan( a /2) 2

1 + tan ( a /2)

=

1 - t2 1 + t2

=

é 1 ( I + A) = ê ë 0 é 1 And, ( I - A) = ê ë 0

0ù é 0 + ê ú 1û ë t 0ù é 0 - ê 1 úû ë t

é cos a \ ( I - A) × ê ë sin a

- sin a ù cos a úû

\

é 1 =ê ë -t

é 1 - t2 ê t ù ê 1 + t2 1 úû ê 2t ê 2 êë 1 + t

2t 1 + t2

.

-t ù é 1 -t ù = ê . ú 0û 1 úû ë t -t ù é 1 t ù = ê ú. 0 úû ë -t 1 û

- 2t ù 1 + t2 ú ú 1 - t2 ú ú 1 + t 2 ûú

é 1 - t2 t(1 - t 2) 2t 2 - 2t + + ê 1 + t2 1 + t2 1 + t2 1 + t2 =ê ê t(1 t 2) 2t 2t 2 1 - t2 ê - + + 2 2 2 êë 1 + t 1+t 1+t 1 + t2 é 1 -t ù =ê = ( I + A). 1 úû ë t é cos a - sin a ù Hence, ( I + A) = ( I - A) ê . cos a úû ë sin a EXAMPLE 14

SOLUTION

é cos q If A = ê ë - sin q cos nq é An = ê ë sin nq

sin q ù then prove that cos q úû sin nq ù , n Î N. cos nq úû

ù ú ú ú ú úû

[CBSE 2004, ’05, ’06]

We shall prove the result by using the principle of mathematical induction.

SSS Mathematics for Class 12 183

Matrices

When n = 1, we have é cos 1 × q A1 = ê ë - sin 1 × q

183

sin 1 × q ù é cos q = cos 1 × q úû êë - sin q

sin q ù × cos q úû

Thus, the result is true for n = 1. Let the result be true for n = k. sin kq é cos kq Then, A k = ê sin q cos kq k ë \

ù ú× û

Ak + 1

é cos q = A × Ak = ê ë - sin q

sin q ù é cos kq cos q úû êë - sin kq

é cos q cos kq - sin q sin kq =ê ë - sin q cos kq - cos q sin kq é cos ( q + kq) =ê ë - sin ( q + kq)

sin ( q + kq) ù cos ( q + kq) úû

é cos ( k + 1) q =ê ë - sin ( k + 1) q

sin ( k + 1) qù cos ( k + 1) qúû

sin kq ù cos kq úû

cos q sin kq + sin q cos kq ù - sin q sin kq + cos q cos kq úû

Thus, the result is true for n = ( k + 1), whenever it is true for n = k. é cos nq sin nq ù Hence, A n = ê ú for all values of n Î N . ë - sin nq cos nq û EXAMPLE 15

SOLUTION

é 3 If A = ê ë 1 of n Î N .

-4 ù é1 + 2n -4n ù then prove that A n = ê for all values -1 úû 1 - 2núû ë n

We shall prove the result by using the principle of mathematical induction. When n = 1, we have -4 × 1 ù -4 ù é 1 + 2×1 é 1+ 2 é 3 -4 ù A1 = ê ú = ê 1 ú = ê 1 -1 ú × 1 1 2 × 1 1 2 ë û ë û ë û Thus, the result is true for n = 1. é 1 + 2k Let the result be true for n = k. Then, A k = ê ë k \

-4k ù × 1 - 2k úû

Ak + 1 = A × Ak é 3 =ê ë 1

-4 ù é 1 + 2k -1 úû êë k

é 3 + 6k - 4k =ê ë 1 + 2k - k

-4k ù 1 - 2k úû

-12k - 4 + 8k ù é 3 + 2k = -4k - 1 + 2k úû êë k + 1

-4k - 4ù 1 - 2k úû

SSS Mathematics for Class 12 184

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Senior Secondary School Mathematics for Class 12

é 1 + 2( k + 1) =ê k+1 ë

-4( k + 1) ù × 1 - 2( k + 1) úû

Thus, the result is true for n = ( k + 1), whenever it is true for n = k. So, the result is true for all n Î N . -4n ù é 1 + 2n for all values of n Î N . Hence, A n = ê 1 - 2n úû ë n

EXAMPLE 16

SOLUTION

é 3n - 1 1ù ê 1 1 ú , prove that A n = ê 3 n - 1 ú ê n-1 1 1 úû êë 3 all values of n Î N . é1 If A = ê 1 ê êë 1

1

We shall prove the result by using the induction. When n = 1, we have é 31 - 1 31 - 1 31 - 1 ù é 3 0 ê ú ê A1 = ê 31 - 1 31 - 1 31 - 1 ú = ê 3 0 ê 1-1 ú ê 31 - 1 31 - 1 ú ê 3 0 êë 3 û ë

3n - 1 ù ú 3 n - 1 ú for ú 3 n - 1 úû

3n - 1 3n - 1 3n - 1

principle of mathematical

30 30 30

30 ù é 1 ú 30 ú = ê 1 ú ê 3 0 ú êë 1 û

1 1 1

1ù 1ú× ú 1 ûú

Thus, the result is true for n = 1. é 3k - 1 ê Let it be true for n = k. Then, A k = ê 3 k - 1 ê k -1 êë 3 \

3k - 1 3k - 1 3k - 1

3k - 1 ù ú 3k - 1 ú × ú 3k - 1 ú û

Ak + 1 = A × Ak é1 =ê1 ê êë 1

1 1 1

k -1 1ù é 3 ê 1 ú ê 3k - 1 ú 1 úû ê 3 k - 1 êë

é 3( 3 k - 1) ê = ê 3( 3 k - 1) ê k -1 ) êë 3( 3

3( 3 k - 1) 3( 3

k -1

)

3( 3 k - 1)

3k - 1 3k - 1 3k - 1

3k - 1 ù ú 3k - 1 ú ú 3k - 1 ú û

3( 3 k - 1) ù é 3 k ú ê 3( 3 k - 1) ú = ê 3 k ú ê 3( 3 k - 1) ú ê 3 k û ë

3k 3

k

3k

3k ù ú 3k ú × ú 3k ú û

Thus, the result is true for n = ( k + 1), whenever it is true for n = k. So, the result is true for all n Î N . é 3n - 1 3n - 1 3n - 1 ù ê ú Hence, A n = ê 3 n - 1 3 n - 1 3 n - 1 ú for all values of n Î N . ê n-1 ú 3n - 1 3n - 1 ú êë 3 û

SSS Mathematics for Class 12 185

Matrices

EXAMPLE 17

SOLUTION

185

é 0 1ù If A = ê ú , prove that for all n Î N , ë 0 0û ( aI + bA) n = a nI + na n - 1 bA , where I is the identity matrix of order 2. We shall prove the result by mathematical induction. When n = 1, we have: LHS = ( aI + bA)1 = ( aI + bA) = ( a1 I + 1a 0 bA) = RHS.

So, the result is true for n = 1. Let it be true for n = m, so that ( aI + bA) m = a mI + ma m - 1 bA \

( aI + bA)

m+1

… (i)

= ( aI + bA) × ( aI + bA)

m

= ( aI + bA) × ( a mI + ma m - 1 bA) m

= aI( a I + ma =a

m+1

m-1

m

[using (i)]

bA) + bA( a I + ma

m

m

I + ma bA + a bA + ma

m-1

m-1 2

b A

bA)

2

[Q II = I , IA = A = AI] = a m + 1 + (m + 1) a mbA

[Q A 2 = 0].

This shows that the result is true for n = (m + 1), whenever it is true for n = m. Hence, by the principle of mathematical induction, the result is true for all n Î N . EXAMPLE 18 SOLUTION

If A = diag [a , b , c], show that A n = diag [a n , b n , cn] for all n Î N . We shall prove the result by mathematical induction. When n = 1, we have A1 = diag [a1 , b1 , c1] = diag [a , b , c] = A. So, the result is true for n = 1. Let it be true for n = m , so that A m = diag [a m , b m , cm] \

A

m+1

= A×A

… (i)

m

= diag [a , b , c] × diag [a m , b m , cm] [using (i)] éa = ê0 ê êë 0

0 b 0

é am + 1 ê =ê0 ê0 ë

0ù é a m 0 ê bm 0ú × ê 0 ú c úû ê 0 0 ë 0 ù 0 ú m+1 0 b ú m+1 ú 0 c û

0 ù ú 0 ú cm úû = diag [a m + 1 , b m + 1 , cm + 1 ].

SSS Mathematics for Class 12 186

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Senior Secondary School Mathematics for Class 12

This shows that the result is true for n = (m + 1), whenever it is true for n = m. Hence, by the principle of mathematical induction, the result is true for all n Î N . EXAMPLE 19

If A is an m ´ n matrix and B is a matrix given in such a way that AB and BA are both defined, show that B is an n ´ m matrix.

SOLUTION

Since AB is defined, we have number of rows in B = number of columns in A = n. Again, since BA is defined, we have number of columns in B = number of rows in A = m. Hence, the order of B is n ´ m.

EXAMPLE 20

If A and B are two matrices given in such a way that AB and A + B are both defined, show that A and B are square matrices of the same order.

SOLUTION

Since ( A + B) is defined, it follows that both A and B are of the same order, say (m ´ n). Thus, order of A is m ´ n and order of B is m ´ n. But, AB is defined. So, number of columns in A = number of rows in B. Consequently, n = m. \ A and B are square matrices of the same order.

VALUE BASED QUESTION The cooperative store of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are ` 65 .70, ` 43.20 and ` 76.50 respectively. Find by matrix method the total amount received by the store from selling all these items.

EXAMPLE 21

SOLUTION

We have Number of books é physics [Total cost] = ê ë 120

chemistry 96

é ` 65 .70 ù mathematics ù ê ú ú ê ` 43 . 20ú 60 û êë ` 76.50 úû

= [ ` (120 ´ 65 .70 + 96 ´ 43 . 20 + 60 ´ 76.50) ] = [ ` 16621. 20 ]. \ the total amount realised = ` 16621. 20.

EXERCISE 5C 1. Compute AB and BA , whichever exists when -1 ù é 2 3 ù é -2 (i) A = ê 3 0 ú and B = ê ú ê 0 4 úû ë 4 úû êë -1

Rates

SSS Mathematics for Class 12 187

Matrices

é -1 (ii) A = ê -2 ê êë -3

1 ù 2 ú ú 3 úû

é 0 (iii) A = ê ë 2

4

1ù 2ú ú 1 úû

4

1ù 2ú ú -5 úû 3 ù 0 ú ú 5 úû

é 1 and B = ê -1 ê êë 0

é ê (iv) A = [1 2 3 4] and B = ê ê ê ë é 2 (v) A = ê 3 ê êë -1

-2 1

é 3 and B = ê 0 ê êë -3 -5 ù 0 úû

1

187

1 ù 2 ú ú 3 ú 4 úû

é 1 and B = ê ë -1

0 2

1ù 1 úû

2. Show that AB ¹ BA in each of the following cases: é 5 -1 ù é 2 1ù (i) A = ê and B = ê ú ú 7 û ë 6 ë 3 4û 1 0ù é -1 é 1 2 3 ù ê ú ê (ii) A = 0 1 0 and B = 0 -1 1 ú ú ê ú ê 3 4 úû êë 2 êë 1 1 0 úû 3. Show that AB = BA in each of the following cases: é cos q - sin q ù é cos f - sin f ù (i) A = ê and B = ê cos q úû cos f úû ë sin q ë sin f é 10 -4 -1 ù é 1 2 1ù (ii) A = ê 3 4 2 ú and B = ê -11 5 0ú ú ê ú ê 9 -5 1 ûú êë êë 1 3 2 úû é 1 (iii) A = ê 2 ê êë 3

3 2 0

-1 ù -1 ú ú -1 úû

é 2 -3 -5 4. If A = ê -1 4 5 ê êë 1 -3 -4 AB = A and BA = B. é 0 5. If A = ê - c ê êë b

c 0 -a

ù ú ú úû

é -2 and B = ê -1 ê êë -6

3 2 9

-1 ù -1 ú ú -4 úû

é 2 and B = ê -1 ê êë 1

-2 3 -2

-4 4 -3

é a2 -bù ê a ú and B = ê ab ú ê ac 0úû ë

ab 2

b bc

ù ú , show that ú úû

ac ù ú bc ú , show that AB is a zero matrix. c2 úû

SSS Mathematics for Class 12 188

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Senior Secondary School Mathematics for Class 12

6. For the following matrices, verify that A( BC) = ( AB)C: 3 0ù é 1ù é 2 é 1 2 5 ù ú and C = ê 4 ú ê 1 (i) A = ê , 0 4 B = ú ê ú ú ê ë 0 1 3 û êë 5 úû êë 1 - 1 2 úû é 1ù é 2 3 -1 ù (ii) A = ê , B = ê 1 ú and C = [1 - 2] ê ú 2 úû ë 3 0 êë 2 úû 7. Verify that A( B + C) = ( AB + AC), when 0ù é 1 2ù é 2 é 1 (i) A = ê , B=ê and C = ê ú ú ë 3 4û ë 1 -3 û ë 0 2 3 ù é é 5 -3 ù é -1 (ii) A = ê -1 4 ú , B = ê and C = ê ú ú ê 1û ë2 ë 3 êë 0 1 úû

-1 1

ù ú. û

2ù . 4 úû

é 0 5 -4ù é 1 0 -2ù é 1 5 2ù ú ê ú ê 8. If A = 3 -1 0 , B = -2 1 3 and C = ê -1 1 0ú ; ú ê ú ê ú ê êë -1 0 2úû êë -2 1 1úû êë 0 -1 1 úû verify that A( B - C) = ( AB - AC). é ab b2 ù 2 9. If A = ê 2 ú , show that A = O. ë - a - ab û -2 3

é 2 10. If A = ê -1 ê êë 1 é 4 11. If A = ê 3 ê êë 3

-2 -1 0 -1

-4 ù 4 ú , show that A 2 = A. ú -3 úû -4 -4 -3

ù ú , show that A 2 = I. ú úû

é 2 12. If A = ê ë 3

-1 ù é 0 and B = ê ú 2û ë -1

é 2 13. If A = ê ë -3

-2 ù then find ( - A 2 + 6A). 4 úû

é 3 14. If A = ê ë -1

1ù , show that ( A 2 - 5 A + 7 I) = O. 2 úû

é2 15. Show that the matrix A = ê ë1 é 3 16. If A = ê ë 4

-2 -2

4 7

ù 2 ú , find ( 3 A - 2B + I). û

[CBSE 2005]

[CBSE 2003C]

3ù satisfies the equation A 3 - 4A 2 + A = O. 2 úû [CBSE 2005]

ù 2 ú , find k so that A = kA - 2I. û

[CBSE 2003]

SSS Mathematics for Class 12 189

Matrices

189

é -1 17. If A = ê ë 3

2ù , find f ( A), where f ( x) = x 2 - 2x + 3. 1 úû

é 1 18. If A = ê ë 4

2ù and f ( x) = 2x 3 + 4x + 5 , find f ( A). -3 úû

[CBSE 2004C]

19. Find the values of x and y, when é 2 -3 ù é x ù é 1 ù = × ê 1 1 úû êë y úû êë 3 úû ë

[CBSE 2003C]

20. Solve for x and y, when é 3 ê 1 ë é 3 21. If A = ê ë 7 é 3 22. If A = ê ë 1

-4 ù é x ù é 3 ù = ê ú× 2 úû êë y úû ë 11 û 1 5

[CBSE 2005C]

ù 2 ú , find x and y such that A + xI = yA. û

[CBSE 2005]

2ù , find the values of a and b such that A 2 + aA + bI = O. 1 úû [CBSE 2006]

-7 ù é -16 ×A =ê 3 úû ë 7

é 5 23. Find the matrix A such that ê ë -2 é 2 24. Find the matrix A such that A × ê ë 4 é 1 -1 ù é a -1 A=ê , B=ê ú ë 2 -1 û ë b -1 values of a and b. é cos x - sin x 0 ù 26. If F( x) = ê sin x cos x 0 ú , ú ê 0 1 úû êë 0 25. If

3 ù é 0 = 5 úû êë 10

-4 3

show that F( x) × F( y) = F( x + y).

sin a ù é cos 2a , show that A 2 = ê cos a úû ë - sin 2a

é 1 28. If [1 x 1] ê 4 ê êë 3

2 5 2

3 6 5

é 2 4 1] ê 1 ê êë 0

1 0 2

2 2 -4

29. If [x

ù ú ú úû

ù ú× û

ù 2 2 2 ú and ( A + B) = ( A + B ) then find the û

é cos a 27. If A = ê ë - sin a

ù ú ú úû

-6 ù . 2 úû

é 1ù ê -2 ú = O , find x. ú ê êë 3 úû é x ù ê 4 ú = O , find x. ú ê êë -1 úû

sin 2a ù × cos 2a úû [CBSE 2005]

SSS Mathematics for Class 12 190

190

Senior Secondary School Mathematics for Class 12

30. Find the values of a and b for which bù é 2ù é a é5 ù ê - a 2b ú ê -1 ú = ê 4 ú × ë û ë û ë û é 3 31. If A = ê ë -4 é 1 32. If A = ê ë 0

[CBSE 2003C]

4ù , find f ( A), where f ( x) = x 2 - 5 x + 7. -3 úû 1ù é 1 , prove that A n = ê 1 úû ë 0

[CBSE 2004C]

nù for all n Î N . 1 úû

33. Give an example of two matrices A and B such that A ¹ O , B ¹ O , AB = O and BA ¹ O. 34. Give an example of three matrices A , B , C such that AB = AC but B ¹ C. é 1 35. If A = ê ë -1 é 2 36. If ê ë5

0 7

ù é 0 ú and B = ê -1 û ë

3 ù é 1 7 úû êë -2

-3 ù é -4 = 4 úû êë -9

4 7

ù 2 ú , find ( 3 A - 2B + I). û

[CBSE 2005]

6ù , find the value of x. x úû

[CBSE 2012]

ANSWERS (EXERCISE 5C)

1.

é -4 (i) AB = ê -6 ê êë 2 é -2 (ii) BA = ê -8 ê êë 10

2ù 9 ú and BA does not exist. ú 13 úû 2ù 8ú ú -10 úû

and AB does not exist.

é 6 13 -25 ù and BA = ê 0 -1 ê 6 úû êë 10 20 4ù é 1 2 3 ê 2 4 6 8ú ú (iv) AB = [ 30] and BA = ê ê 3 6 9 12 ú ê 4 8 12 16 ú û ë é 1 2 3 ù é 1 2ù (v) AB = ê 1 4 5 ú and BA = ê ú ú ê ë 3 4û êë -2 2 0 úû é 4 -20 ù é 2 0ù 12. ê 13. ê 16. k = 1 ú ú ë 38 -10 û ë 0 2û é -1 (iii) AB = ê ë -2

-5 ù 5 ú ú 0 úû

é 12 17. ê ë -6

-4 ù 8 úû

SSS Mathematics for Class 12 191

Matrices

é -5 18. f ( A) = ê ë 136

68 ù -141 úû

191

19. x = 2, y = 1

é1 22. a = - 4, b = 1 23. A = ê ë3 -5 29. x = - 2 25. a = 1, b = 4. 28. x = 3 é -15 -20 ù 33. 30. a = 1, b = - 3 31. f ( A) = ê 15 úû ë 20 21. x = 8, y = 8

é 1 0ù é0 34. A = ê ú , B = ê1 0 0 ë û ë

0ù é0 , C=ê 0úû ë0

0ù 1 úû

20. x = 5 , y = 3

-4ù é -8 4 ù 24. A = ê ú ú -2û ë -19 12û or x = - 1

é 1 0ù é 0 0ù A=ê ú, B=ê 1 0ú 0 0 ë û ë û é 4 -8 ù 36. x = 13 35. ê ú ë -22 134û

Transpose of Matrices TRANSPOSE OF A MATRIX Let A be an (m ´ n) matrix. Then, the matrix obtained by interchanging the rows and columns of A is called the transpose of A, denoted by A ¢ or AT .

Thus, if A = [aij ]m ´ n then A ¢ = [aji ]n ´ m. REMARKS

(i) If A is an (m ´ n) matrix then A ¢ is an (n ´ m) matrix. (ii) (i , j)th element of A = ( j , i)th element of A ¢.

EXAMPLE

é ê 2 If A = ê ê 3 ë

2 -2

é ê 2 ù 0ú ê , then A ¢ = ê 2 2ú ê ú 5 û ê 0 êë

ù 3ú ú -2 ú × ú 2ú 5 úû

Some Results on Transpose of Matrices For any matrix A, prove that ( A ¢) ¢ = A. Let A = [aij ]m ´ n. Then, A is an m ´ n matrix Þ A ¢ is an n ´ m matrix Þ ( A ¢) ¢ is an m ´ n matrix. \ A and ( A ¢) ¢ are matrices of the same order. Also, (i , j)th element of A = ( j , i)th element of A ¢ = (i , j)th element of ( A ¢) ¢. \ (i , j)th element of A = (i , j)th element of ( A ¢) ¢. Thus, A and ( A ¢) ¢ are comparable matrices having their corresponding elements equal. Hence, ( A ¢) ¢ = A.

THEOREM 1 PROOF

SSS Mathematics for Class 12 192

192

Senior Secondary School Mathematics for Class 12

If A is any matrix and k is a scalar, prove that ( kA) ¢ = kA ¢. Let A = [aij ]m ´ n be an m ´ n matrix and k be a scalar. Then, A is an m ´ n matrix Þ kA is an m ´ n matrix

THEOREM 2 PROOF

Þ ( kA) ¢ is an n ´ m matrix. Again, A is an m ´ n matrix Þ A ¢ is an n ´ m matrix Þ kA ¢ is an n ´ m matrix. Thus, ( kA) ¢ and kA ¢ are matrices of the same order. Now, ( j , i)th element of ( kA) ¢ = (i , j)th element of kA = k times (i , j)th element of A = k times ( j , i)th element of A ¢ = ( j , i)th element of kA ¢. \ ( j , i)th element of ( kA) ¢ = ( j , i)th element of kA ¢. Thus, ( kA) ¢ and kA ¢ are comparable matrices having their corresponding elements equal. Hence, ( kA) ¢ = kA ¢. If A and B are two matrices of the same order then prove that ( A + B) ¢ = A ¢ + B ¢. Let A = [aij ]m ´ n and B = [bij ]m ´ n. Then, A is an m ´ n matrix and B is an m ´ n matrix

THEOREM 3 PROOF

Þ ( A + B) is an m ´ n matrix Þ ( A + B) ¢ is an n ´ m matrix. Also, A is an m ´ n matrix and B is an m ´ n matrix Þ A ¢ is an n ´ m matrix and B¢ is an n ´ m matrix Þ ( A ¢ + B ¢) is an n ´ m matrix. \ ( A + B) ¢ and ( A ¢ + B ¢) are comparable matrices. Also, ( j , i)th element of ( A + B ¢) = (i , j)th element of ( A + B) = (i , j)th element of A + (i , j)th element of B = ( j , i)th element of A ¢ + ( j , i)th element of B¢ = ( j , i)th element of ( A ¢ + B ¢). \ ( j , i)th element of ( A + B ¢) = ( j , i)th element of ( A ¢ + B ¢). Thus, ( A + B) ¢ and ( A ¢ + B ¢) are comparable and their corresponding elements are equal. Hence, ( A + B) ¢ = A ¢ + B ¢. THEOREM 4 REMARK

If A and B are square matrices of the same order then ( AB) ¢ = B ¢ A ¢. The proof of this theorem is beyond the scope of this book.

SSS Mathematics for Class 12 193

Matrices

193

SUMMARY

(i) ( A ¢) ¢= A

(ii) ( kA) ¢ = kA ¢

(iii) ( A + B) ¢ = A ¢ + B ¢

(iv) ( AB) ¢ = B ¢ A ¢

SOLVED EXAMPLES EXAMPLE 1

é 2 3 -1ù Let A = ê ú × Verify that ( A ¢) ¢ = A. ë 0 -5 7 û

SOLUTION

We have 0ù é 2 -1ù ¢ ê ú 3 5 = ú ê 7 úû êë -1 7 úû é 2 0ù ¢ é 2 3 -1ù ( A ¢) ¢ = ê 3 - 5 ú = ê ú = A. ú ê ë 0 -5 7 û 1 7 úû êë é2 3 A¢ = ê ë 0 -5

Þ

Hence, ( A ¢) ¢ = A. EXAMPLE 2

é 2 3 -5 ù If A = ê , verify that ( 3 A) ¢ = 3 A ¢. 4úû ë 0 -1

SOLUTION

We have é3 ´2 3 ´ 3 3A = ê ë 3 ´ 0 3 ´ ( -1) \

0ù é 6 9 -15 ù ¢ ê é6 = ( 3 A) ¢ = ê 9 -3 ú × ú ê 12úû ë 0 -3 êë -15 12úû

é2 Also, A ¢ = ê ë0

\

3 ´ ( -5) ù é 6 9 -15 ù = × 3 ´ 4 úû êë 0 -3 12úû

3 -1

é 2 3A ¢ = 3 × ê 3 ê êë -5

é 2 -5 ù ¢ ê = 3 ê 4úû êë -5 0ù é 6 -1ú = ê 9 ú ê 4úû êë -15

0ù -1ú × ú 4úû 0ù -3 ú × ú 12úû

Hence, ( 3 A) ¢ = 3 A ¢.

EXAMPLE 3

4ù é 2 -3 ù é 3 Let A = ê -2 6ú × Verify that ( A + B) ¢ = A ¢ + B ¢. 0ú and B = ê 5 ú ê ú ê 8úû êë -1 êë 7 -5 úû

SSS Mathematics for Class 12 194

194 SOLUTION

Senior Secondary School Mathematics for Class 12

We have 4ù é 2 -3 ù é 3 0ú + ê 5 6ú A + B = ê -2 ú ú ê ê 8úû êë 7 -5 úû êë -1 é 3 + 2 4 + ( -3) ù é 5 1ù 0 + 6ú = ê 3 6ú × = ê -2 + 5 ú ú ê ê -5 + 8úû êë 6 3 úû êë7 + ( -1) \

é5 ( A + B) ¢ = ê 3 ê êë 6

1ù ¢ é5 6ú = ê ú ë1 3 úû

3 6

6ù × 3 úû

4ù ¢ é 3 é 3 -2 7 ù Also, A ¢ = ê -2 0ú = ê ú× ú ê ë 4 0 -5 û êë 7 -5 úû é 2 -3 ù ¢ é 2 5 -1ù And, B¢ = ê 5 6ú = ê ú× ú ê ë -3 6 8û 8úû êë -1 \

é 3 -2 7 ù é 2 5 -1ù ( A ¢+ B ¢) = ê ú+ê ú ë 4 0 -5 û ë -3 6 8û é3 + 2 =ê ë 4 + ( -3)

-2 + 5 0+ 6

7 + ( -1) ù é5 ú = ê1 -5 + 8 û ë

3 6

6ù × 3 úû

Hence, ( A + B) ¢ = A ¢ + B ¢.

EXAMPLE 4

é -3 ù If A = ê 5 ú and B = [ 1 6 - 4 ] then verify that ( AB) ¢ = B ¢ A ¢. ú ê êë 2 úû [CBSE 2002]

SOLUTION

é -3 ù We have A = ê 5 ú and B = [ 1 6 - 4 ]. ú ê êë 2 úû é -3 ù \ AB = ê 5 ú [ 1 6 - 4 ] ú ê êë 2 úû é -3 =ê 5 ê êë 2

-18 30 12

12 ù ú× ú úû

-20 -8

SSS Mathematics for Class 12 195

Matrices

é -3 So, ( AB) ¢ = ê -18 ê êë 12

195

2ù 30 12 ú × ú -20 -8 úû é 1 Also, A ¢ = [ - 3 5 2 ] and B¢ = ê 6 ê êë -4 \

5

ù ú× ú úû

é 1ù B ¢ A ¢ = ê 6 ú [ - 3 5 2] ú ê êë -4 úû 5 2ù é -3 30 12 ú × = ê -18 ú ê êë 12 -20 -8 úû

Hence, ( AB) ¢ = B ¢ A ¢. SYMMETRIC AND SKEW-SYMMETRIC MATRICES A square matrix A is said to be symmetric if A ¢ = A. A is symmetric Û aji = aij .

SYMMETRIC MATRIX

EXAMPLE 1

EXAMPLE 2

é 4 2 ù ú Consider the matrix A = ê 1 ú× ê 2 êë 3 úû ¢ é é 4 2 ù 2 ù 4 ú ú =ê Then, A ¢ = ê 1 ú 1 ú = A. ê 2 ê 2 êë êë 3 úû 3 úû Hence, A is symmetric. é 6 ê Consider the matrix B = ê -7 ê 4 ë é 6 -7 ê Then, B ¢ = ê -7 3 ê 4 0 ë Hence, B is symmetric.

SKEW-SYMMETRIC MATRIX

A ¢ = - A. REMARK

-7 3 0

¢ é 6 4ù ú ê 0 ú = ê -7 ê 4 5 úû ë

ù ú ú× 5 úû 4 0

-7 3 0

4ù ú 0 ú = B. 5 úû

A square matrix A is said to be skew-symmetric if

Û A ¢ = - A Û aji = - aij Û aii = - aii Û 2aii = 0 Û aii = 0. Thus, every diagonal element of a skew-symmetric matrix is zero. A is skew-symmetric

SSS Mathematics for Class 12 196

196

Senior Secondary School Mathematics for Class 12

EXAMPLE 1

é 0 Let A = ê ë -5 é 0 A¢ = ê ë5

5 ù × Then, 0 úû -5 ù = - A. 0 úû

Hence, A is skew-symmetric.

EXAMPLE 2

é 0 Let A = ê - h ê êë - g é 0 A¢ = ê h ê êë g

h 0 -f -h 0 f

g f 0

ù ú × Then, ú úû

-g -f 0

ù ú = - A. ú úû

Hence, A is skew-symmetric. SOME RESULTS ON SYMMETRIC AND SKEW-SYMMETRIC MATRICES Prove that the sum of two symmetric matrices is a symmetric matrix. Let A and B be symmetric matrices of the same order. Then, A ¢ = A and B ¢ = B. [Q A ¢ = A and B ¢ = B]. \ ( A + B) ¢ = ( A ¢ + B ¢) = ( A + B) Hence, ( A + B) is symmetric.

THEOREM 1 PROOF

If A is a symmetric matrix then prove that kA is symmetric. Since A is symmetric, we have A ¢ = A. \ ( kA) ¢ = k × A ¢ = kA [Q A ¢ = A]. This shows that kA is symmetric.

THEOREM 2 PROOF

Prove that the sum of two skew-symmetric matrices is a skew-symmetric matrix. Let A and B be two skew-symmetric matrices. Then, A ¢ = - A and B ¢ = - B. \ ( A + B) ¢ = ( A ¢ + B ¢) = ( - A) + ( - B) = - ( A + B). Hence, ( A + B) is skew-symmetric.

THEOREM 3 PROOF

If A is a skew-symmetric matrix then prove that kA is skew-symmetric. Since A is skew-symmetric, we have A ¢ = - A. [Q A ¢ = - A] \ ( kA) ¢ = k × A ¢ = k × ( - A) = - ( k × A ). Thus, ( kA) ¢ = - ( kA). Hence, ( kA) is skew-symmetric.

THEOREM 4 PROOF

THEOREM 5

For any square matrix A with real number entries, prove that (i) ( A + A ¢) is symmetric and (ii) ( A - A ¢) is skew-symmetric.

SSS Mathematics for Class 12 197

Matrices PROOF

Let A be a square matrix with real number entries. Then, we have: (i) ( A + A ¢) ¢ = A ¢ + ( A ¢) ¢ [Q ( A + B) ¢ = A ¢ + B ¢] = A¢ + A [Q ( A ¢) ¢ = A] = ( A + A ¢) [Q ( A + B) = ( B + A)]. Thus, ( A + A ¢) ¢ = ( A + A ¢). Hence, ( A + A ¢) is symmetric. (ii) ( A - A ¢) ¢= A ¢ - ( A ¢) ¢ [Q ( A - B) ¢ = A ¢ - B ¢] = A ¢- A [Q ( A ¢) ¢ = A] = - ( A - A ¢ ). Thus, ( A - A ¢) ¢= - ( A - A ¢) Hence, ( A - A ¢) is skew-symmetric. Prove that every square matrix is expressible as the sum of a symmetric and a skew-symmetric matrix. Let A be any square matrix. Then, we can write 1 1 A = ( A + A ¢) + ( A - A ¢) = P + Q (say), 2 2 1 1 where P = ( A + A ¢) and Q = ( A - A ¢). 2 2 ü¢ ì1 Now, P ¢ = í ( A + A ¢) ý þ î2 1 = ( A + A ¢) ¢ [Q ( kA) ¢ = kA ¢] 2 1 = {A ¢ + ( A ¢) ¢} [Q ( A + B) ¢ = ( A ¢ + B ¢)] 2 1 = ( A ¢ + A) [Q ( A ¢) ¢ = A] 2 1 = ( A + A ¢) [Q ( A ¢ + A) = ( A + A ¢)] 2 = P. Thus, P ¢ = P and therefore, P is symmetric. ü¢ ì1 And, Q ¢ = í ( A - A ¢) ý þ î2 1 = ( A - A ¢) [Q ( kA) ¢ = kA ¢] 2 1 = {A ¢ - ( A ¢) ¢} [Q ( A - B) ¢ = A ¢ - B ¢] 2 1 = ( A ¢ - A) [Q ( A ¢) ¢ = A] 2 1 = - ( A - A ¢) = - Q. 2

THEOREM 6 PROOF

197

SSS Mathematics for Class 12 198

198

Senior Secondary School Mathematics for Class 12

Thus, Q ¢ = - Q and therefore, Q is skew-symmetric. \ A = P + Q, where P is symmetric and Q is skew-symmetric. SOLVED EXAMPLES EXAMPLE 1

SOLUTION

é 3 Express the matrix A = ê ë 1 a skew-symmetric matrix. We know that A =

-4 ù as the sum of a symmetric matrix and -1 úû

1 1 1 ( A + A ¢) + ( A - A ¢), where ( A + A ¢) is 2 2 2

1 ( A - A ¢) is skew-symmetric. 2 1ù é 3 -4 ù é 3 Here, A = ê Þ A¢ = ê ú ú× ë 1 -1 û ë -4 -1 û 1ù é 3 -4 ù é 3 \ ( A + A ¢) = ê = ê ú ú ë 1 -1 û ë -4 -1 û symmetric and

é 3+ 3 =ê ë 1 + ( -4)

-4 + 1 ù é 6 -3 ù = ê ú× -1 + ( -1) úû ë -3 -2 û -3 ù é ê 3 1 1 é 6 -3 ù 2 ú× Let P = ( A + A ¢) = × ê = ê ú -3 2 2 ë -3 -2 úû -1 ú ê ë 2 û 1ù é 3 -4ù é 3 And, ( A - A ¢) = ê ú -ê ú ë 1 -1û ë -4 -1û é 3-3 =ê ë1 - ( -4) Let Q =

1 1 é ( A - A ¢) = × ê 2 2 ë

-3 é ê 3 2 Then, P ¢ = ê -3 -1 ê ë 2 \ P is symmetric.

-4 - 1 ù é 0 -5 ù é 0 -5 ù =ê = × ú -1 - ( -1) û ë1 + 4 -1 + 1úû êë5 0úû -5 ù é 0 -5 ù ê 0 2 ú× = ê ú 5 5 0 úû 0ú ê ë 2 û

ù¢ é ú ê 3 ú = ê -3 ú ê û ë 2

-5 ù ¢ é é 0 ê 0 2 ú =ê And, Q ¢ = ê ú ê 5 -5 0ú ê ê û ë 2 ë 2 \ Q is skew-symmetric.

-3 ù 2 ú = P. ú -1 ú û 5 ù 2 ú = - Q. ú 0ú û

SSS Mathematics for Class 12 199

Matrices

-3 ù é 0 2 ú+ê ú ê 5 -1 ú ê û ë 2

é ê 3 Now, P + Q = ê -3 ê ë 2

199

-5 ù 2 ú = é 3 ú ê ë 1 0 ú û

-4 ù = A. -1 úû

Hence, A = P + Q, where P is symmetric and Q is skew-symmetric.

EXAMPLE 2

SOLUTION

é 1 Express the matrix A = ê -6 ê êë -4 a skew-symmetric matrix.

3 8 6

5 ù 3 ú as the sum of a symmetric and ú 5 úû [CBSE 2006]

We have é 1 A = ê -6 ê êë -4 \

3 8 6

5 ù é 1 3 ú Þ A¢ = ê 3 ê ú 5 úû êë 5

3

-4 ù 6ú× ú 5 úû

5 ù é 1 -6 -4ù 3ú + ê 3 8 6ú ú ú ê 5 úû êë 5 3 5 úû

é 1 3 ( A + A ¢) = ê - 6 8 ê êë -4 6 é 1+1 = ê -6 + 3 ê êë -4 + 5

3 + ( -6) 5 + ( -4) ù é 2 -3 1ù 8+ 8 3 + 6 ú = ê -3 16 9ú × ú ê ú 6+ 3 5 + 5 úû êë 1 9 10úû -3 16 9

é 1 1 ( A + A ¢) = × ê - 3 2 2 ê êë 1 2

Let P =

-6 8

é ê 1 1ù ê -3 9ú = ê ú ê 2 10 úû ê 1 ëê 2

-3 2 8 9 2

ù ú ú ú× ú 5 ú úû 1 2 9 2

é 1 3 5 ù é 1 -6 -4ù And, ( A - A ¢) = ê -6 8 3 ú - ê 3 8 6ú ú ú ê ê êë -4 6 5 úû êë 5 3 5 úû é 1-1 = ê -6 - 3 ê êë -4 - 5 é 0 = ê -9 ê êë -9

9 0 3

3 - ( -6) 5 - ( -4) ù é 0 8-8 3 - 6 ú = ê -9 ú ê 6- 3 5 - 5 úû êë -9 9ù ú× ú úû

-3 0

3 + 6 5 + 4ù 0 -3 ú ú 3 0 úû

SSS Mathematics for Class 12 200

200

Senior Secondary School Mathematics for Class 12

Let Q =

é 0 1 1 ( A - A ¢) = × ê - 9 2 2 ê êë -9

é ê ê P¢ = ê ê ê êë

-3 2 1 2

0 3

ù¢ é ê 1 ú ê -3 ú ú =ê ê 2 ú ê 1 ú 5 êë 2 úû

-3 2

1

9

1 2 9 2

8 9 2

é ê 9ù ê ú -3 = ê ú ê 0 úû ê êë -3 2

0 -9 2 -9 2

9 2 0 3 2

9ù 2ú -3 ú ú× 2ú 0ú úû

ù ú ú ú = P. ú 5 ú úû 1 2 9 2

8 9 2

\ P is symmetric. é ê ê And, Q¢ = ê ê ê êë

0 -9 2 -9 2

9 2 0 3 2

ù¢ é ê 0 ú ê 9 ú ú =ê ê 2 ú ê 9 ú 0 êë 2 úû

9 2 -3 2

-9 2 0 -3 2

-9 2 3 2

ù ú ú ú = -Q. ú 0 ú úû

\ Q is skew-symmetric. é ê ê Now, ( P + Q) = ê ê ê êë

1 -3 2 1 2

é 1 = ê -6 ê êë -4

-3 2 8 9 2 3 8 6

ù é ú ê ú ê ú+ê ú ê 5 ú ê úû êë 1 2 9 2

0 -9 2 -9 2

9 2 0 3 2

ù ú ú ú ú 0 ú úû

9 2 -3 2

5 ù 3 ú = A. ú 5 úû

Hence, A = P + Q, where P is symmetric and Q is skew-symmetric. EXAMPLE 3

SOLUTION

If A and B are symmetric matrices of the same order then show that AB is symmetric if and only if AB = BA. Let A and B be symmetric matrices. Then, A ¢ = A and B ¢ = B. Let AB be symmetric. Then, AB = ( AB) ¢ [by definition] = B ¢ A ¢ = BA \

AB = BA.

[Q B ¢ = B and A ¢ = A].

SSS Mathematics for Class 12 201

Matrices

201

Conversely, let AB = BA. Then, ( AB) ¢ = B ¢ A ¢ [Q B ¢ = B and A ¢ = A] = BA [given]. = AB Thus, ( AB) ¢ = AB and hence, AB is symmetric. EXAMPLE 4

If A and B are symmetric matrices, prove that ( AB - BA) is skew-symmetric.

SOLUTION

Let A and B be symmetric matrices. Then, A ¢ = A and B ¢ = B. Now, ( AB - BA) ¢ = ( AB) ¢ - ( BA) ¢ = ( B ¢ A ¢) - ( A ¢ B ¢) [Q A ¢ = A and B ¢ = B] = ( BA - AB) = - ( AB - BA). Thus, ( AB - BA) ¢ = - ( AB - BA). Hence, ( AB - BA) is skew-symmetric.

EXAMPLE 5 SOLUTION

If A is symmetric, show that B ¢ AB is symmetric. Let A be symmetric. Then, A ¢ = A. \ ( B ¢ AB) ¢ = B ¢ A ¢( B ¢) ¢ = B ¢ A ¢ B [Q ( B ¢) ¢ = B] = B ¢ AB [Q A ¢ = A]. Thus, ( B ¢ AB) ¢ = B ¢ AB. Hence, ( B ¢ AB) is symmetric.

EXERCISE 5D é 2 1. If A = ê ë 0 é 3 2. If A = ê -2 ê êë 4

-3 7

5 ù , verify that ( A ¢) ¢ = A. -4 úû

5 0 -6

ù ú , verify that ( 2A) ¢ = 2A ¢. ú úû

é 3 2 -1ù é -4 -5 -2ù 3. If A = ê and B = ê , verify that ( A + B) ¢ = ( A ¢+ B ¢). ú 1 8úû ë -5 0 -6û ë 3 4ù é 7 ú and Q = ê -4 ú ê úû êë 2

-5 0 6

é 3 4. If P = ê 2 ê êë 0

-1 5

é4 5. If A = ê ë5

1ù , show that ( A + A ¢) is symmetric. 8úû

ù ú , verify that ( P + Q) ¢ = ( P ¢ + Q ¢). ú úû [CBSE 2001]

SSS Mathematics for Class 12 202

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Senior Secondary School Mathematics for Class 12

é3 6. If A = ê ë1

-4ù , show that ( A - A ¢) is skew-symmetric. -1úû

é 0 7. Show that the matrix A = ê - a ê êë - b HINT:

a 0 -c

b c 0

[CBSE 2001C]

ù ú is skew-symmetric. ú úû

Show that A ¢ = - A.

é 2 3ù 8. Express the matrix A = ê ú as the sum of a symmetric matrix and a ë -1 4 û skew-symmetric matrix. é 3 -4ù 9. Express the matrix A = ê ú as the sum of a symmetric matrix and a ë 1 -1û skew-symmetric matrix. é -1 5 1 ù 10. Express the matrix A = ê 2 3 4 ú as the sum of a symmetric and a ú ê 0 9 úû êë 7 skew-symmetric matrix. 11. Express the matrix A as the sum of a symmetric and a skew-symmetric matrix, where é 3 -1 0 ù [CBSE 2005C] 0 3 ú× A=ê 2 ú ê êë 1 -1 2 úû é 3 2 5 ù 12. Express the matrix A = ê 4 1 3 ú as sum of two matrices such that one ú ê êë 0 6 7 úû is symmetric and the other is skew-symmetric. [CBSE 2008] 13. For each of the following pairs of matrices A and B, verify that ( AB) ¢ = ( B ¢ A ¢): é 1 3 ù é 1 4ù (i) A = ê and B = ê ú ú ë 2 4û ë 2 5 û é 3 -1 ù é 1 -3 ù (ii) A = ê and B = ê ú ú ë 2 -2 û ë 2 -1 û é -1 ù (iii) A = ê 2 ú and B = [ - 2 - 1 - 4 ] ú ê êë 3 úû é 3 2 -3 ù é -1 ê 2 (iv) A = ê and B = ú ê 4 5 6 ë û êë -1

[CBSE 2002]

-4 1 0

ù ú ú úû

SSS Mathematics for Class 12 203

Matrices

é cos a 14. If A = ê ë - sin a

203

sin a ù , show that A ¢ A = I. cos a úû

15. If matrix A = [1 2

3], write AA ¢.

[CBSE 2009]

ANSWERS (EXERCISE 5D)

é 2 8. A = ê ë 1

1ù é 0 + ê ú 4û ë -2

é ê -1 ê7 10. A = ê ê2 ê4 êë é ê3 ê 12. A = ê 3 ê ê5 êë 2

ù é 4ú ê 0 ú ê -3 2ú+ê ú ê 2 9ú ê 3 úû êë

7 2 3 2

ù ú ú 1 ú+ ú 9 7 ú úû 2 3

5 2 9 2

é ê ê ê ê ê êë

-3 ù é 0 2 ú+ê ú ê 5 -1 ú ê û ë 2

é ê 3 9. A = ê -3 ê ë 2

2ù 0 úû 3 2 0 -2

0

-1

1

0

-5 2

3 2

ù -3 ú ú 2ú ú 0ú úû

é ê3 ê1 11. A = ê ê2 ê1 êë 2

ù ú ú ú ú 0 ú úû

15. [14]

5 2 -3 2

1 2 0 1

1ù é 2ú ê ú ê 1ú + ê ú ê 2ú ê úû ëê

0 3 2 1 2

-5 ù 2 ú ú 0 ú û -3 2 0 -2

-1 ù 2ú ú 2ú ú 0ú úû

Elementary Operations on Matrices Given below are three row operations and three column operations on a matrix, which are called elementary operations or transformations. EQUIVALENT MATRICES Two matrices A and B are said to be equivalent if one is obtained from the other by one or more elementary operations and we write, A ~ B.

THREE ELEMENTARY ROW OPERATIONS (i) Interchange of any two rows

by R i « R j .

EXAMPLE

é ê Let A = ê ê ë

3 2 5

2 4 -3

The interchange of ith and jth rows is denoted -1 6 7

ù ú ú× ú û

SSS Mathematics for Class 12 204

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Senior Secondary School Mathematics for Class 12

é ê Applying R 2 « R 3 , we get A ~ ê ê ë

3

2

5

-3

2

4

-1 ù ú 7 ú× 6 úû

(ii) Multiplication of the elements of a row by a nonzero number Suppose each element of ith row of a given matrix is multiplied by a nonzero number k.

Then, we denote it by Ri ® k Ri .

EXAMPLE

é ê Let A = ê ê ë

3

2

3

-5 8

1

ù ú ú× 4 úû

-1 6

3 é ê Applying R 2 ® 4R 2 , we get A ~ ê 4 3 ê 1 ë

2 -20 8

-1 ù ú 24 ú × 4 úû

(iii) Multiplying each element of a row by a nonzero number and then adding them to the corresponding elements of another row Suppose each element of jth row

of a matrix A is multiplied by a nonzero number k and then added to the corresponding elements of ith row. We denote it by Ri ® Ri + kR j .

EXAMPLE

é 2 ê Let A = ê -3 ê 7 ë

-1 4 6

5 ù ú 2 ú× 3 úû

é 16 ê Applying R1 ® R1 + 2R 3 , we get A ~ ê -3 ê 7 ë

11 4 6

11 ù ú 2 ú× 3 úû

THREE ELEMENTARY COLUMN OPERATIONS (i) Interchange of any two columns

denoted by Ci « C j .

EXAMPLE

é ê 2 ê Let A = ê -1 ê ê 6 êë

1 5 3

The interchange of ith and jth columns is

ù -3 ú ú 4 ú× ú 1 ú 2 úû

SSS Mathematics for Class 12 205

Matrices

é ê ê Applying C1 « C 2 , we get A ~ ê ê ê êë

205

ù 2 -3 ú ú -1 4ú× ú 1 ú 6 2 úû

1 5 3

(ii) Multiplying each element of a column by a nonzero number Suppose each element of ith column of matrix A is multiplied by a nonzero number k. Then, we write, Ci ® kCi .

EXAMPLE

é ê Let A = ê ê ë

3

1

2

-2 2

6

ù ú ú× 8 úû

-5 4

é ê Applying C 3 ® 2C 3 , we get A ~ ê ê ë

3

1

2

-2

6

2

-10 ù ú 8ú× 16 úû

(iii) Multiplying each element of a column of a given matrix A by a nonzero number and then adding to the corresponding elements of another column

Suppose each element of jth column of a given matrix A is multiplied by a nonzero number k and then added to the corresponding elements of ith column. Then, we write, Ci ® Ci + kC j . EXAMPLE

é 2 Let A = ê -1 ê êë 5

0 3 -2

4ù 1ú× ú 6 úû

é 2 Applying C 3 ® C 3 + 2C1 , we get A ~ ê -1 ê êë 5

0 3 -2

8 -1 16

ù ú× ú úû

INVERTIBLE MATRICES A square matrix A of order n is said to be invertible if there exists a square matrix B of order n such that AB = BA = I.

Also, then B is called the inverse of A and we write, A-1 = B. EXAMPLE

é 3 5 ù é 2 -5 ù Let A = ê and B = ê × Then, ú 3 úû ë 1 2û ë -1 é 3 5 ù é 2 -5 ù é 6 - 5 -15 + 15 ù é 1 0ù AB = ê = = I, úê ú=ê -5 + 6úû êë 0 1úû ë 1 2û ë -1 3 û ë 2 - 2 é 2 -5 ù é 3 5 ù é 6 - 5 10 - 10ù é 1 0ù BA = ê ú=ê ú = I. úê ú=ê ë -1 3 û ë 1 2û ë -3 + 3 -5 + 6û ë 0 1û

SSS Mathematics for Class 12 206

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Senior Secondary School Mathematics for Class 12

\

AB = BA = I.

Hence, A -1 = B. (Uniqueness of Inverse) Every invertible square matrix has a unique inverse. Let A be an invertible square matrix of order n. If possible, let B as well as C be the inverse of A. Then, AB = BA = I and AC = CA = I. Now, AC = I Þ B( AC) = B × I = B, BA = I Þ ( BA)C = I × C = C. But, B( AC) = ( BA)C [by associative law of multiplication] \ B = C. Hence, an invertible matrix has a unique inverse.

THEOREM 1 PROOF

INVERSE OF A MATRIX BY ELEMENTARY ROW OPERATIONS Let A be a square matrix of order n. We can write, A = I × A.

… (i)

Now, let a sequence of elementary row operations reduce A on LHS of (i) to I and I on RHS of (i) to a matrix B. Then, I = BA Þ I × A -1 = ( BA) A -1 = B( AA -1) = BI Þ A -1 = B. We can summarise the above method as given below. METHOD

Write A = I × A. By using elementary row operations on A, transform it into a unit matrix. Step 3. In the same order we apply elementary operations on I to convert it into a matrix B. Step 1.

Step 2.

Step 4. REMARK

Then, A -1 = B.

If on applying one or more elementary row operations on A, we obtain all zeros in one or more rows, then we say that A -1 does not exist. SOLVED EXAMPLES

EXAMPLE 1

By using elementary row operations, find the inverse of the matrix é 1 -2 ù A=ê ú× ë 2 -6 û

SOLUTION

We have é 1 -2 ù é 1 ê 2 -6 ú = ê 0 ë û ë

0 1

ù ú×A û

SSS Mathematics for Class 12 207

Matrices

é 1 ê 0 ë

-2 ù é 1 = ê ú -2 û ë -2

0 1

é 1 Þ ê ë 0

é -2 ù ê 1 = ê 1 úû ê 1 ë

ù 0 ú ×A -1 ú ú 2 û

é 0ù ê 3 = ê ú 1û ê 1 ë

ù -1 ú ×A 1ú - ú 2û

Þ

Þ

é 1 ê 0 ë

-1 1 2

é 3 Hence, A -1 = ê ê 1 ë EXAMPLE 2

ù ú × A [R 2 ® R 2 - 2R1] û é ê R2 ® ë

ù æ -1 ö ç ÷ R2 ú è 2ø û

[R1 ® R1 + 2R 2].

ù ú× ú û

By using elementary row operations, find the inverse of the matrix -1 2

é 3 A=ê ë -4 SOLUTION

207

ù ú× û

We have -1 ù é 1 = ê 2 úû ë 0

é 3 A=ê ë -4

0ù ×A 1 úû

Þ

é -1 ê -4 ë

1ù é 1 = ê ú 2û ë 0

Þ

é 1 ê -4 ë

-1 ù é -1 = ê 2 úû ë 0

Þ

é 1 ê 0 ë

-1 ù é -1 = ê ú -2 û ë -4

Þ

é 1 ê 0 ë

é ù -1 ù ê -1 -1 ú ×A = ê 3ú 1 úû ê 2 ú ë 2û

é ù æ -1 ö ê R2 ® ç 2 ÷ R2 ú è ø ë û

Þ

é 1 ê 0 ë

é 0ù ê 1 = ê 1 úû ê 2 ë

1 2 3 2

[R1 ® R1 + R 2].

é ê 1 =ê ê 2 ë

ù ú ú× ú û

Hence, A

-1

1 2 3 2

1 1

ù ú × A [R1 ® R1 + R 2] û -1 1

-1 -3

ù ú × A [R1 ® ( -1) × R1] û

ù ú × A [R 2 ® R 2 + 4R1] û

ù ú ú×A ú û

SSS Mathematics for Class 12 208

208

EXAMPLE 3

SOLUTION

Senior Secondary School Mathematics for Class 12

-3 ù , show that A -1 does not exist. 1 úû

é 6 If A = ê ë -2 We have

-3 ù é 1 = ê 1 úû ë 0

é 6 ê -2 ë Þ

é ê 1 ê ê -2 ë

Þ

é ê 1 ê ê 0 ë

0 1

ù ú×A û

1 ù é 1 2 ú = ê 6 ú ê 1 ú ê 0 û ë

ù 0ú ú×A 1ú û

1 ù é êë R1 ® 6 R1 úû

1ù é 1 2ú = ê 6 ú ê 1 0 ú ê û ë 3

ù 0 ú ú×A 1 ú û

[R 2 ® R 2 + 2R1].

-

-

Thus, we have all zeros in second row of the left-hand side matrix. Hence, A -1 does not exist. EXAMPLE 4

SOLUTION

By using elementary row operations, find the inverse of the matrix é 1 3 -2 ù A = ê -3 0 -5 ú × ú ê 0 úû êë 2 5 We have é 1 ê -3 ê êë 2

3

3

Þ

é 1 ê 0 ê êë 0

Þ

é 1 ê 0 ê êë 0

3 -1

Þ

é 1 ê 0 ê êë 0

0 -1 0

Þ

é 1 ê 0 ê êë 0

0 1 0

0 5 9 -1

9

-2 ù é 1 ú -5 = ê 0 ê ú 0 úû êë 0

0ù 0 ú×A ú 1 úû

0 1 0

-2 ù é 1 ú -11 = ê 3 ê ú 4 úû êë -2

0

-2 ù 4ú = ú -11 úû

é 1 ê -2 ê êë 3

0 0

é -5 ù ú = ê -2 ê ú êë -15 úû

10 4 25 10 -4 25

é -5 ù ú = ê 2 ê ú êë -15 úû

0ù 0 ú×A ú 1 úû

é R 2 ® R 2 + 3 R1 ù ê R ® R - 2R ú ë 3 3 1 û

[R 2 « R 3]

1

0ù 1 ú×A ú 0 úû

0 0 1

3 1 9

ù ú×A ú úû

é R1 ® R1 + 3 R 2 ù ê R ® R + 9R ú ë 3 3 2 û

3 -1 9

ù ú×A ú úû

[R 2 ® ( -1) × R 2]

1 0

0 0 1

SSS Mathematics for Class 12 209

Matrices

Þ

Þ

é 1 ê 0 ê êë 0

é 1 ê 0 ê êë 0

é ê -5 10 ù ê -4 ú = ê 2 ú ê -3 1 úû ê ë 5

0 1 0

é ê 0ù ê ú 0 = ê ú ê 1 úû ê êë

0 1 0

Hence, A -1

é ê ê =ê ê ê êë

1 -2 5 -3 5

0

-2 5 -3 5

-3 5 11 25 9 25

ù ú ú ú×A 9 ú ú 25 û

1 é ù êë R 3 ® 25 R 3 úû

ù ú ú ú×A ú ú úû

é R1 ® R1 - 10R 3 ù ê R ® R + 4R ú × ë 2 2 3 û

3 -1

0 1 25 -2 5 4 25 1 25

1

-2 5 4 25 1 25

209

-3 5 11 25 9 25

ù ú ú ú× ú ú úû

EXAMPLE 5

By using elementary row operations, find the inverse of the matrix é 3 -1 -2 ù 0 -1 ú × A=ê 2 ú ê 0 úû êë 3 -5

SOLUTION

We have é 3 ê 2 ê êë 3 é 1 Þ ê 2 ê êë 3 Þ

Þ

é 1 ê 0 ê êë 0

-1 0 -5 -1 0 -5 -1 2 -2

-2 ù é 1 -1 ú = ê 0 ê ú 0 úû êë 0

0

-1 ù é 1 ú -1 = ê 0 ê ú 0 úû êë 0

-1

-1 ù 1ú = ú 3 úû

é 1 ê -2 ê êë -3

ù é é ê 1 -1 -1ú ê 1 -1 ê 1ú ê 3 ú = ê -1 ê 0 1 2 2 ú ê ê ê 0 -2 3 ú ê -3 3 úû êë êë

0ù 0 ú×A ú 1 úû

1 0

0ù 0 ú×A ú 1 úû

1 0 -1 3 3

0ù 0 ú×A ú 1 úû

ù 0ú ú 0 ú×A ú 1ú úû

[R1 ® R1 - R 2]

é R 2 ® R 2 - 2R1 ù ê R ® R - 3R ú ë 3 3 1 û

1 ü ì íR 2 ® R 2 ý 2 þ î

SSS Mathematics for Class 12 210

210

Senior Secondary School Mathematics for Class 12

Þ

Þ

Þ

é ê 1 ê ê 0 ê ê 0 êë

ù é ú ê 0 ú ê ú = ê -1 ú ê 4 ú ê -5 úû êë

1 0

é ê 1 ê ê 0 ê ê 0 êë

é 1 ê 0 ê êë 0

-1 2 1 2

0

1 0

0 1 0

Hence, A -1

6

-1 ù é ê 0 2 ú ê ú 1 ú = ê -1 2 ú ê ê -5 1 ú êë 4 úû

0

é ê 0ù ê 0ú = ê ú ê 1 úû ê ê êë é ê ê =ê ê ê ê ë

-5 8 -3 8 -5 4

-5 8 -3 8 -5 4 5 4 3 4 3 2

ù 0 ú ú ì R ® R1 + R 2 ü 0 ú×A í 1 ý î R 3 ® R 3 + 2R 2 þ ú 1 ú úû

1 2 3 2

ù 0ú ú 1 ü ì 0 ú × A íR 3 ® R 3 ý 4 þ î ú 1 ú 4 úû

1 2 3 2 3 2 5 4 3 4 3 2

1 8 -1 8 1 4

1 8 -1 8 1 4

ù ú ú ú × A. ú ú ú úû

1 ì ïï R1 ® R1 + 2 R 3 í ï R2 ® R2 - 1 R 3 ïî 2

ü ïï ý ï ïþ

ù ú ú ú× ú ú ú û

EXAMPLE 6

By using elementary row transformations, find the inverse of the matrix é 2 0 -1 ù 0ú× A=ê5 1 ú ê 3 úû êë 0 1

SOLUTION

We have é 2 0 ê5 1 ê êë 0 1

-1 ù 0ú = ú 3 úû

é 1 ê 0 ê êë 0

0ù 0 ú×A ú 1 úû

0 1 0

Þ

é 2 ê 1 ê êë 0

0 1 1

-1 2 3

é 1 ù ú = ê -2 ê ú êë 0 úû

0 1 0

0 0 1

ù ú×A ú úû

[R 2 ® R 2 - 2R1]

Þ

é 1 ê 2 ê êë 0

1 0 1

2 -1 3

é -2 ù ú = ê 1 ê ú êë 0 úû

1 0 0

0 0 1

ù ú×A ú úû

[R1 « R 2]

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Matrices

211

2ù 1 0ù é -2 ê ú Þ 0 -2 -5 = 5 -2 0 ú × A ú ê ú 0 1 3 úû 0 1 úû êë 0 1 1 2ù 1 0ù é -2 Þ 0 1 3 ú = ê 0 0 1 ú×A ú ê ú 0 -2 -5 úû êë 5 -2 0 úû 1 0 -1 ù 1 -1 ù é -2 Þ 0 1 3 ú = ê 0 0 1 ú×A ú ê ú 0 0 1 úû 2 úû êë 5 -2 1 0 0ù 3 -1 1ù é Þ 0 1 0 ú = ê -15 6 -5 ú × A ú ê ú 0 0 1 úû 5 -2 2 úû êë 3 -1 1ù é Hence, A -1 = ê -15 6 -5 ú × ú ê 5 -2 2 úû êë é ê ê êë é ê ê êë é ê ê êë é ê ê êë

EXAMPLE 7

SOLUTION

1

1

é 1 If A = ê 2 ê êë -1 We have é 1 ê 2 ê êë -1

-1 1 -2

[R 2 ® R 2 - 2R1]

[R 2 « R 3] é R1 ® R1 - R 2 ù ê R ® R + 2R ú ë 3 3 2 û é R1 ® R1 + R 3 ù ê R ® R - 3R ú × ë 2 2 3 û

1ù ú , show that A -1 does not exist. ú úû

-1 2

-1

1ù é 1 ú 1 -1 = ê 0 ê ú -2 2 úû êë 0 -1 1ù é 1 ú 3 -3 = ê -2 ê ú -3 3 úû êë 1 -1 1ù é 1 3 -3 ú = ê -2 ê ú 0 0 úû êë -1

0 1 0

0ù 0 ú×A ú 1 úû

0 0ù é 1 é R ® R 2 - 2R1 ù ê Þ 0 1 0 ú×A ê 2 ú ú ê ë R 3 ® R 3 + R1 û 0 1 úû êë 0 0 0ù é 1 Þ ê 0 1 0 ú × A [R 3 ® R 3 + R 2]. ú ê 1 1 úû êë 0 Thus, we have all zeros in 3rd row of the left-hand side matrix. Hence, A -1 does not exist.

EXERCISE 5E Using elementary row transformations, find the inverse of each of the following matrices: é 1 1. ê ë 3

2ù 7 úû

é 1

[CBSE 2007] 2. ê ë 2

2ù -1 úû

é 2 3. ê ë -3

5 ù 1 úû

SSS Mathematics for Class 12 212

212

Senior Secondary School Mathematics for Class 12

é 2 4. ê ë 4

-3 ù 5 úû

é 0 7. ê 1 ê êë 3

1 2

é 1 10. ê 2 ê ëê 3 é 1 ê 13. ê 2 ê -2 ë

2ù 3 ú ú 1 úû

1

-3 ù 2ú ú -4 úû

2 3 -3

3 ù ú 7 ú -5 ú û

2 5 -4

é 4 5. ê ë 2

0ù 5 úû

é 2 8. ê 2 ê êë 3

-3

é 3 11. ê 2 ê êë 3

-1

é 3 14. ê 2 ê êë 0

[CBSE 2008]

3 ù 3 ú ú 2 úû

2 -2

-2 ù -1 ú ú 0 úû

0 -5

-1 ù 0ú ú 1 úû

0 3 4

é 6 6. ê ë 8

7 ù 9 úû

é 3 9. ê 1 ê êë 6

5

2ù 9ú ú 7 úû

0 4

é 1 3 -2ù 12. ê -3 0 -1ú [CBSE 2011] ú ê 0úû êë 2 1 é -1 15. ê 1 ê êë 3

[CBSE 2009]

2ù 3 ú ú 1 úû

1 2 1

[CBSE 2012]

ANSWERS (EXERCISE 5E)

é 7 1. ê ë -3

-2 ù 1 úû

é ê 4. ê ê êë

3 22 1 11

7.

10.

5 22 -2 11

é 1 1 ê × -8 2 ê êë 5

é ê 2. ê ê êë

ù ú ú ú úû -1 6 -3

é 1 ê 5. ê 4 -1 ê ë 10 1 -2 1

ù ú ú úû

é -6 17 13 ù 1 ê × 14 5 -8ú ú 67 ê êë -15 9 -1úû

é 3 -2 -1ù 13. ê - 4 1 -1ú ú ê êë 2 0 1úû

1 5 2 5

2 5 -1 5

ù ú ú ú úû

ù 0ú 1 ú ú 5 û

8.

é 2 0 -3 ù -1 ê 0ú × 1 -1 ú 5 ê 2úû êë -2 -1

11.

é 5 -10 -1ù -1 ê × 3 -6 1ú ú 8 ê êë10 -12 -2úû

é 3 - 4 3ù ê ú 14. ê -2 3 -2ú ê 8 -12 9ú ë û

3.

1 17

é 1 ×ê ë 3

é -9 ê 6. ê 2 ê 4 ë

9.

-5 ù 2 úû

7 ù 3 ú ú -3 ú û

8 -10ù é -1 -1 ê 9 -25 ú × 47 ú 55 ê êë -26 -12 15 úû

é 1 -2 12. ê -2 4 ê êë -3 5

-3 ù 7 ú ú 9 úû

é 1 -1 1ù ê ú 15. ê -8 7 -5 ú ê 5 -4 3 ú ë û

SSS Mathematics for Class 12 213

Matrices

213

EXERCISE 5F Very-Short-Answer Questions 1. Construct a 3 ´ 2 matrix whose elements are given by 1 aij = (i - 2j) 2. 2 2. Construct a 2 ´ 3 matrix whose elements are given by 1 aij = |- 3i + j|. 2 é x + 2y - y ù é - 4 3 ù 3. If ê = , find the values of x and y. 4 úû êë 6 4 úû ë 3x 4. Find the values of x and y, if é 1 3 ù é y 0ù é5 2ê ú=ê ú+ê ë 0 x û ë 1 2û ë 1

[CBSE 2008C]

6ù × 8 úû

[CBSE 2008]

é 2ù é -1 ù é 10 ù 5. If x × ê ú + y × ê ú=ê ú , find the values of x and y. 3 ë û ë 1û ë 5 û é x 6. If ê ë 2x + z

3x - y ù é 3 = 3 y - w úû êë 4

2ù , find the values of x , y , z , w. 7 úû

6ù é 4 x + yù é x éx yù 7. If ê ú + êz + w ú = 3 ê z wú , find the values of x , y , z , w. 2 3 1 w ë û ë û ë û 8. If A = diag ( 3 - 2 5) and B = diag (1 3 - 4), find ( A + B). é cos q sin q ù é sin q - cos q ù 9. Show that cos q × ê = I. ú + sin q × ê cos q sin q úû ë - sin q cos q û ë 1ù é 3 é 1 -5 ù 10. If A = ê -3 2 ú and B = ê 2 -1 ú , find the matrix C such that ú ê ú ê 3 úû êë -2 êë 4 -2 úû A + B + C is a zero matrix. é cos a - sin a ù 11. If A = ê then find the least value of a for which A + A ¢ = I. cos a úû ë sin a 12. Find the values of x and y for which é 2 -3 ù é x ù é 1 ù = × ê 1 1 úû êë y úû êë 3 úû ë 13. Find the values of x and y for which é x y ù é 1ù é 3 ù ê 3y x ú ê 2 ú = ê 5 ú × ë û ë û ë û é 4 14. If A = ê ë 1

5 ù , show that ( A + A ¢) is symmetric. 8 úû

[CBSE 2003]

[CBSE 2003C]

[CBSE 2001]

SSS Mathematics for Class 12 214

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Senior Secondary School Mathematics for Class 12

é 2 15. If A = ê ë 4

3 ù , show that ( A - A ¢) is skew-symmetric. 5 úû

é 2 16. If A = ê ë 4

-3 ù é -1 2 ù and B = ê ú ú , find a matrix X such that 5 û ë 0 3 û A + 2B + X = O.

é 4 17. If A = ê ë 1

2ù é -2 and B = ê ú 3 û ë 3

1ù , find a matrix X such that 2 úû

3 A - 2B + X = O. é cos a 18. If A = ê ë - sin a

[CBSE 2001]

[CBSE 2000]

sin a ù , show that A ¢ A = I. cos a úû

19. If A and B are symmetric matrices of the same order, show that ( AB - BA) is a skew-symmetric matrix. é 2 3 ù 2 20. If A = ê ú and f ( x) = x - 4x + 1, find f ( A). 1 2 ë û 21. If the matrix A is both symmetric and skew-symmetric, show that A is a zero matrix.

ANSWERS (EXERCISE 5F)

é ê ê 1. ê ê ê êë

1 2 0 1 2

9 ù 2 ú ú 2 ú ú 1 ú 2 úû

é ê 1 2. ê 5 ê ë 2

1 2 2

ù 0 ú 3 ú ú 2 û

3. x = 2, y = - 3

4. x = 3 , y = 3

5. x = 3 , y = - 4

6. x = 3 , y = 7 , z = - 2, w = 14

7. x = 2, y = 4, z = 1, w = 3

8. diag ( 4 1 1)

é -4 10. C = ê 1 ê êë -2

4ù -1 ú ú -1 úû

13. x = 1, y = 1 é -16 17. X = ê ë 3

-4 ù -5 úû

11. a =

p 3

12. x = 2, y = 1

é 0 16. X = ê ë -4

-1 ù -11 úû

é 0 20. f ( A) = ê ë 0

0ù 0 úû

SSS Mathematics for Class 12 215

6. DETERMINANTS Determinant of a Square Matrix Corresponding to each square matrix é a11 a12 a13 êa 21 a 22 a 23 A=ê ê¼ ¼ ¼ ê ë an1 an2 an 3

¼ a1n ù ¼ a2n ú ú, ¼ ¼ ú ¼ ann úû

there is associated an expression, called the determinant of A , denoted by det A , or|A|, written as

det A = |A| =

a13 a23

¼ a1n ¼ a2n

a11 a21

a12 a22

¼ an1

¼ ¼ ¼ ¼ an2 an 3 ¼ ann

×

A matrix is an arrangement of numbers and so it has no fixed value, while each determinant has a fixed value. A determinant having n rows and n columns is known as a determinant of order n. The determinants of nonsquare matrices are not defined. VALUE OF A DETERMINANT OF ORDER 1

[a] is defined as|a| = a. VALUE OF A DETERMINANT OF ORDER 2

a11 a21 EXAMPLE 1

SOLUTION

The value of a determinant of a (1 ´ 1) matrix We define

a12 = ( a11 a22 - a21 a12). a22 Evaluate: 6 -3 (i) 7 -2

(ii)

x2 - x + 1 x - 1 x+1 x+1

(i)

6 -3 = 6( -2) - 7( -3) = -12 + 21 = 9. 7 -2

(ii)

x2 - x + 1 x - 1 = ( x 2 - x + 1)( x + 1) - ( x + 1)( x - 1) x+1 x+1 = ( x 3 + 1) - ( x 2 - 1) = ( x 3 - x 2 + 2). 215

SSS Mathematics for Class 12 216

216

Senior Secondary School Mathematics for Class 12

3x 7

EXAMPLE 2

If

SOLUTION

We have

-2 4

=

8 7 6 4

, find the value of x.

3x 7 -2 4

=

[CBSE 2014]

8 7 6 4

Û ( 3 x ´ 4) - ( -2) ´ 7 = ( 8 ´ 4) - ( 6 ´ 7) Û 12x + 14 = 32 - 42 Û 12x + 14 = -10 Û 12x = -24 Û

EXAMPLE 3

SOLUTION

x = -2.

Hence, x = -2. x + 1 x -1 4 -1 If = , find the value of x. x-3 x+2 1 3 We have

x+1

x -1

x-3

x+2

=

[CBSE 2013]

4 -1 1

3

Û ( x + 1)( x + 2) - ( x - 3)( x - 1) = ( 4 ´ 3) - 1 ´ ( -1) Û ( x 2 + 3 x + 2) - ( x 2 - 4x + 3) = 12 + 1 Û 7 x - 1 = 13

Û 7 x = 14 Û

x = 2.

Hence, x = 2. EXAMPLE 4

Show that

sin 10° - cos 10° = 1. sin 80° cos 80°

SOLUTION

We have

sin 10° - cos 10° sin 80° cos 80°

= (sin 10°)(cos 80°) - (sin 80°)( - cos 10°) = (sin 10° cos 80° + cos 10° sin 80°) = sin (10° + 80°)

[Q sin A cos B + cos A sin B = sin( A + B)]

= sin 90° = 1. For finding the value of a determinant of order 3 or more, we need the following definitions. VALUE OF A DETERMINANT OF ORDER 3 OR MORE

MINOR OF aij IN | A | The minor of an element a ij in|A| is defined as the value of the determinant obtained by deleting the ith row and jth column of|A|, and it is denoted by Mij . COFACTOR OF a ij IN | A |

C ij = ( -1) i + j × Mij .

The cofactor C ij of an element aij in |A| is defined as

SSS Mathematics for Class 12 217

Determinants EXAMPLE 1

SOLUTION

217

Find the minors and cofactors of the elements of the determinant a11 a12 a13 D = a21

a22

a23 ×

a 31

a 32

a 33

Let Mij denote the minor of aij in D. Now, a11 occurs in the 1st row and 1st column. So, in order to find the minor of a11 , we delete the 1st row and 1st column of D. The a a23 minor M11 of a11 is given by M11 = 22 = ( a22 a 33 - a 32 a23). a 32 a 33 Similarly, M12

we have a21 a23 = = ( a21 a 33 - a 31 a23); a 31 a 33

M13 =

a21 a 31

a22 = ( a21 a 32 - a 31 a22); a 32

M21 =

a12 a 32

a13 = ( a12 a 33 - a 32 a13). a 32

Similarly, we may obtain the minor of each of the remaining elements. Now, if we denote the cofactor of aij by C ij then C11 = ( -1)1 + 1 × M11 = M11 = ( a22 a 33 - a 32 a23);

C12 = ( -1)1 + 2 × M12 = - M12 = ( a 31 a23 - a21 a 33); C13 = ( -1)1 + 3 × M13 = M13 = ( a21 a 32 - a 31 a22); C 21 = ( -1) 2 + 1 × M21 = - M21 = ( a 32 a13 - a12 a 33).

EXAMPLE 2

SOLUTION

Similarly, the cofactor of each of the remaining elements of D can be determined. 1 -3 2 Find the minor and cofactor of each element of D = 4 -1 2 × 3 5 2 The minors of the elements of D are given by -1 2 4 2 4 -1 = -12; M12 = = 2; M13 = = 23 ; M11 = 5 2 3 2 3 5 M21 =

-3 5

2 1 2 1 -3 = -16; M22 = = - 4; M23 = = 14; 2 3 2 3 5

M 31 =

-3 2 1 2 1 -3 = - 4; M 32 = = -6; M 33 = = 11. -1 2 4 2 4 -1

So, the cofactors of the corresponding elements of D are C11 = ( -1)1 +1 × M11 = M11 = -12; C12 = ( -1)1 + 2 × M12 = - M12 = -2;

SSS Mathematics for Class 12 218

218

Senior Secondary School Mathematics for Class 12

C13 = ( -1)1 + 3 × M13 = M13 = 23 ; C 21 = ( -1) 2 +1 × M21 = - M21 = 16; C 22 = ( -1) 2 + 2 × M22 = M22 = - 4; C 23 = ( -1) 2 + 3 × M23 = - M23 = -14; C 31 = ( -1) 3 +1 × M 31 = M 31 = - 4; C 32 = ( -1) 3 + 2 × M 32 = - M 32 = 6; C 33 = ( -1) 3 + 3 × M 33 = M 33 = 11. Value of a Determinant The value of a determinant is the sum of the products of elements of a row (or a column) with their corresponding cofactors. We may expand a determinant by any arbitrarily chosen row or column. Expansion of a Determinant Expanding the given determinant by 1st row, we have a11 a12 a13 a21 a22 a23 = a11 × (its cofactor) + a12 × (its cofactor) + a13 × (its cofactor) a 31

a 32

a 33 = a11C11 + a12C12 + a13C13 = a11 M11 - a12 M12 + a13 M13 [Q C12 = - M12 ] a22 a23 a21 a23 a21 a22 = a11 × - a12 × + a13 × a 32 a 33 a 31 a 33 a 31 a 32

= a11 × ( a22 a 33 - a 32 a23) - a12 × ( a21 a 33 - a 31 a23) + a13 × ( a21 a 32 - a 31 a22). We may expand it by any row or column. REMARK 1

If we expand a determinant by any row or column using minors, we keep in view the following symbols for a determinant of order three: + - + - + +

REMARK 2

EXAMPLE 1

SOLUTION

- +

If a row or a column of a determinant consists of all zeros, the value of the determinant is zero. 3 4 5 Evaluate D = -6 2 -3 × 8 1 7 Expanding the given determinant by 1st row, we get 2 -3 -6 -3 -6 2 D = 3× -4× +5× 1 7 8 7 8 1 = 3 × (14 + 3) - 4 × ( -42 + 24) + 5 × ( -6 - 16) = 13.

SSS Mathematics for Class 12 219

Determinants

EXAMPLE 2

a h g Expand the determinant D = h b f × g

SOLUTION

219

f

c

Expanding by 1st column, we get b f h g h g D = a× - h× + g× f c f c b f = a( bc - f 2) - h × ( ch - fg) + g × ( fh - bg) = abc - af 2 - ch2 + fgh + fgh - bg 2 = ( abc + 2 fgh - af 2 - bg 2 - ch2).

Properties of Determinants The properties of a determinant serve the purpose of a useful tool for finding its value. We will mention these properties and verify them for a third-order determinant. THEOREM 1

PROOF

The value of a determinant remains unchanged if its rows and columns are interchanged.

a1 Let D = a2 a3

b1 b2 b3

c1 c2 and let D¢ be the determinant obtained by c3

interchanging the rows and columns of D. a1 a2 a 3 Then, D¢ = b1 b2 b 3 c1 c2 c 3 = a1 ×

b2

b3

c2

c3

- b1 ×

a2

a3

c2

c3

+ c1 ×

a2

a3

b2

b3

[expanded by 1st column] = a1( b2 c 3 - b 3 c2) - b1( a2 c 3 - c2 a 3) + c1( a2 b 3 - a 3 b2) =D COROLLARY THEOREM 2

PROOF

[expanded by 1st row].

If A is a square matrix then|A ¢| = |A|. If two rows or columns of a determinant are interchanged then the determinant retains its absolute value but its sign is changed.

a1 Let D = a2 a3

b1 b2 b3

c1 c2 and let D¢ be the determinant obtained by c3

interchanging any two rows, say 1st and 3rd rows, of D. Then,

SSS Mathematics for Class 12 220

220

Senior Secondary School Mathematics for Class 12

a3

b3

c3

D¢ = a2

b2

c2

a1

b1

c1

= a3 ×

b2 b1

c2 b - a2 × 3 c1 b1

c3 b + a1 × 3 c1 b2

c3 c2

[expanded by 1st column] = a 3( b2 c1 - b1 c2) - a2( b 3 c1 - b1 c 3) + a1( b 3 c2 - b2 c 3) = - a1( b2 c 3 - b 3 c2) + a2( b1 c 3 - b 3 c1) - a 3( b1 c2 - b2 c1) = - [a1( b2 c 3 - b 3 c2) - a2( b1 c 3 - b 3 c1) + a 3( b1 c2 - b2 c1)] = -D

If any two rows or columns of a determinant are identical then its value is zero.

THEOREM 3

PROOF

If we interchange the identical rows of the given determinant D then clearly there is no change in D. But, interchanging any two rows of a determinant changes its sign. \

D = -D Û

2D = 0, i.e., D = 0.

If each element of a row or a column of a determinant is multiplied by a constant k then the value of the new determinant is k times the value of the original determinant.

THEOREM 4

PROOF

[expanded by 1st column].

a1 Let D = a2 a3

b1 b2 b3

c1 c2 c3

and let D¢ be the determinant obtained by

multiplying each element of a row, say the second row of D, by k. Then, a1 b1 c1 D¢ = ka2 kb2 kc2 a 3 b3 c3 = a1 ×

kb2

kc2

b3

c3

- ka2 ×

b1

c1

b3

c3

+ a3 ×

b1

c1

kb2

kc2

= a1( kb2 c 3 - kb 3 c2) - ka2( b1 c 3 - b 3 c1) + a 3( kb1 c2 - kb2 c1) = k [a1( b2 c 3 - b 3 c2) - a2( b1 c 3 - b 3 c1) + a 3( b1 c2 - b2 c1)] = k D. COROLLARY THEOREM 5

PROOF

For a square matrix A of order n,|k A| = k n ×|A|. If any two rows or columns of a determinant are proportional then its value is zero.

Let

a1 D = a2 ka1

b1 b2 kb1

c1 c2 kc1

SSS Mathematics for Class 12 221

Determinants

a1

b1

c1

= k a2

b2

c2

a1

b1

c1

= k ´0 = 0 THEOREM 6

[by Theorem 4] [Q 1st and 3rd rows are identical].

If each element of a row (or column) of a determinant is expressed as a sum of two or more terms then the determinant can be expressed as the sum of two or more determinants. a1 + a 1

PROOF

221

D=

Let

b1 + b1

c1 + g1

a2

b2

c2

a3

b3

c3

×

Then, on expanding D by first row, we get D = ( a1 + a 1)( b2 c 3 - b 3 c2) - ( b1 + b1)( a2 c 3 - a 3 c2) + ( c1 + g1)( a2 b 3 - a 3 b2) = [a1( b2 c 3 - b 3 c2) - b1( a2 c 3 - a 3 c2) + c1( a2 b 3 - a 3 b2)] + [a 1( b2 c 3 - b 3 c2) - b1( a2 c 3 - a 3 c2) + g1( a2 b 3 - a 3 b2)] a1 = a2 a3 THEOREM 7

PROOF

b1 b2 b3

a1 c1 c2 + a2 c3 a3

b1

g1

b2 b3

c2 × c3

If to any row or column of a determinant, a multiple of another row or column is added, the value of the determinant remains the same.

Let

a1 D = a2 a3

b1 b2 b3

Let

D¢ =

Then,

a1 D¢ = a2 a3

b1 b2 b3

c1 ka 3 c2 + a2 c3 a3

a1 = a2 a3

b1 b2 b3

c1 c2 c3

a1 + ka 3 a2 a3

c1 c2 × c3 b1 + kb 3 b2 b3

c1 + kc 3 c2 × c3 kb 3 b2 b3

kc 3 c2 c3

[Q 2nd det. is 0, its 1st and 3rd rows being proportional]

= D. THEOREM 8

The sum of the products of the elements of any row (or column) of a determinant with cofactors of the corresponding elements of any other row (or column) is zero.

SSS Mathematics for Class 12 222

222

Senior Secondary School Mathematics for Class 12

a11 PROOF

Let

a12

a13

D = a21

a22

a23 ×

a 31

a 32

a 33

Let us take the sum of the products of elements of first row with the cofactors of the corresponding elements of second row. a11 A 21 + a12 A 22 + a13 A 23 = a11 × ( -1) 2 +1 ×

a12 a 32

a13 a + a12 × ( -1) 2 + 2 × 11 a 33 a 31

a13 a 33

+ a13 × ( -1) 2 + 3 ×

a11 a 31

a12 a 32

= - a11( a12 a 33 - a 32 a13) + a12( a11 a 33 - a 31 a13 ) - a13 ( a11 a 32 - a 31 a12) = 0. To Find the Value of a Determinant The main theme behind the simplification of a determinant lies in obtaining the maximum possible number of zeros in a row (or a column) by using the above properties and then to expand the determinant by that row (or column). We denote the 1st, 2nd, 3rd rows of a determinant by R1 , R 2 , R 3 respectively and the 1st, 2nd, 3rd columns by C1 , C 2 , C 3 respectively. We shall also express the (i) interchange of the ith and jth rows by R i « R j ; (ii) addition of k times the elements of the jth row with the corresponding elements of the ith row by R i ® R i + kR j . We use similar notations for operations on columns, replacing R by C. EXAMPLES

Operation

(i) Interchange of 2nd and 3rd rows.

Notation

(i) R 2 « R 3

(ii) Interchange of 1st and 3rd columns

(ii) C1 « C 3

(iii) Multiplying each element of 2nd row by (–5)

(iii) R 2 ® ( -5) R 2

(iv) Multiplying each element of 1st column 1 by × 3

(iv) C1 ®

(v) Multiplying each element of 3rd row by 6 and adding it to the corresponding element of 2nd row.

(v) R 2 ® R 2 + 6R 3

(vi) Multiplying each element of 2nd column by (–3) and adding it to the corresponding element of 1st column.

(vi) C1 ® C1 + ( -3)C 2

1 C1 3

SSS Mathematics for Class 12 223

Determinants

223

SOLVED EXAMPLES (Short-Answer Questions)

2 7 65 8 75 × 9 86

EXAMPLE 1

Evaluate

SOLUTION

Given determinant

3 5

2 7 = 3 5

65

8 75 9

2 7

86 2

= 3

8

3

5

9

5

[applying C 3 ® C 3 - 9C 2 ] [Q C1 and C 3 are identical].

=0

EXAMPLE 2

Prove that

a-b b-c c-a b - c c - a a - b = 0. c-a

SOLUTION

[CBSE 2014C]

[CBSE 2009]

a-b b-c

Given determinant a-b b-c c-a = b-c c-a a-b c-a a-b b-c

EXAMPLE 3

Prove that

0 b-c c-a = 0 c-a a-b 0 a-b b-c

[applying C1 ® C1 + C 2 + C 3 ]

=0

[Q C1 consists of all zeros].

1 a b+ c 1 b c + a = 0. 1 c a+b

SOLUTION

The given determinant 1 a b+ c = 1 b c+ a 1 c a+b 1 a a+ b+ c = 1 b a+ b+ c 1 c a+ b+ c

[applying C 3 ® C 3 + C 2 ]

SSS Mathematics for Class 12 224

224

Senior Secondary School Mathematics for Class 12

1 a 1 = ( a + b + c) × 1 b 1 [taking ( a + b + c) common from C 3 ] 1 c 1 = ( a + b + c) ´ 0 = 0

[Q C1 and C 3 are identical].

1 bc a ( b + c) 1 ca b ( c + a) = 0. 1 ab c( a + b)

EXAMPLE 4

Prove that

SOLUTION

The given determinant 1 bc a( b + c) = 1 ca b ( c + a) 1 ab c( a + b) 1 bc ab + bc + ca = 1 ca ab + bc + ca 1 ab ab + bc + ca

[applying C 3 ® C 3 + C 2 ]

1 bc 1 = ( ab + bc + ca) × 1 ca 1 [taking ( ab + bc + ca) common from C 3 ] 1 ab 1 = ( ab + bc + ca) ´ 0 = 0

EXAMPLE 5

Without expanding prove that

[Q C1 and C 3 are identical]. x+ y y+z z+ x z x y = 0. 1

SOLUTION

1

1

The given determinant =

x+y z 1

y+z z+ x x y 1 1

x+ y+z x+ y+z x+ y+z = z x y 1 1 1 1 1 1 = ( x + y + z) × z x y 1 1 1 = ( x + y + z) ´ 0 = 0

[R1 ® R1 + R 2]

[taking ( x + y + z) common from R1 ] [Q R1 and R 3 are identical].

SSS Mathematics for Class 12 225

Determinants EXAMPLE 6

If w is a complex cube root of unity, prove that w2

w

1

2

w w2 SOLUTION

225

w 1

1 = 0. w

The given determinant 1

w

= w w2

w2 1

w2 1 w

1 + w + w2 1 + w + w2 1 + w + w2 =

w2 1

w w2 0

0

0

= w

w2

1

1

w

2

w

EXAMPLE 7

SOLUTION

[R1 ® R1 + R 2 + R 3]

[Q 1 + w + w2 = 0] [Q R1 consists of all zeros].

=0 Show that

1 w

2 5

3 6

4 8

6x

9x 12x

= 0.

[CBSE 2009]

The given determinant 2 3 4 = 5 6 8 6x 9x 12x 2 = ( 3 x) 5 2

3 4 6 8 3 4

= ( 3 x) ´ 0 = 0

EXAMPLE 8

Prove that

1 1 1 1+ x 1

SOLUTION

1

The given determinant 1 1 1 = 1 1+ x 1 1 1 1+ y

[taking 3x common from R 3 ] [Q R1 and R 3 are identical]. 1 1 1+ y

= xy.

SSS Mathematics for Class 12 226

226

Senior Secondary School Mathematics for Class 12

1 0

0

= 1 x

0

[applying C 2 ® C 2 - C1 and C 3 ® C 3 - C1 ]

1 0 y = 1×

x 0 = xy [expanding by R1 ]. 0 y sin q cos q

x - sin q cos q

-x 1

is independent of q.

EXAMPLE 9

Prove that

SOLUTION

Let the given determinant be D. Then, expanding by first row, we get: x sin q cos q D = - sin q cos q

-x

1

1

x

1 x

= x( - x 2 - 1) - sin q( - x sin q - cos q) + cos q( - sin q + x cos q) = - x 3 - x + x(sin 2 q + cos2 q) = - x 3 - x + x = - x 3 , which is independent of q. EXAMPLE 10

SOLUTION

Prove that

0 - sin a

sin a 0

cos a

- sin b

- cos a sin b = 0. 0

Let the given determinant be D. Then, expanding by first row, we get: - sin a sin b - sin a 0 D = ( - sin a ) × + ( - cos a ) × cos a 0 cos a - sin b = ( - sin a ) × {0 - cos a sin b} + ( - cos a ) × (sin a sin b - 0) = ( - sin a )( - cos a sin b) + ( - cos a )(sin a sin b) = (sin a cos a sin b - sin a (cos a sin b) = 0. Hence, D = 0.

EXAMPLE 11

Prove that

a b c a + 2x b + 2y c + 2z = 0. x

SOLUTION

y

z

Let the given determinant be D. Then, applying R 2 ® R 2 - R1 , we get: a b c a b c D = 2x 2y 2z = 2 x y z = 0 [Q R 2 and R 3 are identical]. x y z x y z

SSS Mathematics for Class 12 227

Determinants EXAMPLE 12

SOLUTION

227

Without expanding the determinant, prove that a + b 2a + b 3 a + b 2a + b 3 a + b 4a + b = 0. 4a + b 5 a + b 6a + b Let the given determinant be D. Then, applying C 3 ® C 3 - C 2 , we get: a + b 2a + b a D = 2a + b

3a + b a

4a + b 5 a + b a+b a a

a

= 2a + b a a

[applying C 2 ® C 2 - C1 ]

4a + b a a [Q C 2 and C 3 are identical].

=0 Hence, D = 0.

ILLUSTRATIVE EXAMPLES [Long-Answer Questions]

EXAMPLE 1

Evaluate

265 240 219 240 225 198 × 219 198 181

SOLUTION

Let the given determinant be D. Then, 265 240 219 D = 240 225 198 219 198 181 46 21 219 = 42 27 198 38 17 181

[C1 ® (C1 - C 3) and C 2 ® (C 2 - C 3)]

4 21 9 = -12 27 -72 4 17 11

[C1 ® (C1 - 2C 2) and C 3 ® (C 3 - 10C 2)]

0 4 -2 = 0 78 -39 4 17 11

[R1 ® ( R1 - R 3) and R 2 ® ( R 2 + 3 R 3)]

0 2 -1 = 2( 39) 0 2 -1 4 17 11 [taking 2 common from R1 and 39 common from R 2 ] = (78 ´ 0) = 0 [Q R1 and R 2 are identical].

SSS Mathematics for Class 12 228

228 EXAMPLE 2

SOLUTION

Senior Secondary School Mathematics for Class 12

Without expanding, prove that sin a

cos a

cos ( a + d)

sin b sin g

cos b cos g

cos (b + d) = 0. cos ( g + d)

[CBSE 2007]

Let the given determinant be D. Then, sin a

cos a

cos a cos d - sin a sin d

D = sin b

cos b

cos b cos d - sin b sin d

sin g

cos g

cos g cos d - sin g sin d

sin a

cos a

0

= sin b

cos b

0

sin g

cos g

0

=0

[C 3 ® C 3 + (sin d)C1 - (cos d)C 2]

[expanded by C 3 ].

Hence, D = 0.

EXAMPLE 3

a 2a

Prove that

a+b 3 a + 2b

a+ b+ c 4a + 3 b + 2c = a 3 .

[CBSE 2012]

3 a 6a + 3 b 10a + 6b + 3 c SOLUTION

Let the given determinant be D. Then, a a+b a+ b+ c D = 2a 3 a + 2b 4a + 3 b + 2c 3 a 6a + 3 b 10a + 6b + 3 c a a+ b a+ b+ c = 0 a 2a + b 0 3a 7a + 3b

[R 2 ® R 2 - 2R1 and R 3 ® R 3 - 3 R1 ]

a a+ b a+ b+ c = 0 a 2a + b 0 0 a

[R 3 ® R 3 - 3 R 2 ]

= a×

a 2a + b 0

a

[expanding along C1 ]

= a × ( a 2 - 0) = a 3. Hence, D = a 3.

EXAMPLE 4

Prove that

-a 2 ab ac ab - b 2 bc = 4a 2 b 2 c2 . ac bc - c2

[CBSE 2011, ’11C]

SSS Mathematics for Class 12 229

Determinants SOLUTION

Let the given determinant be D. Then, -a 2 D=

ab

ac

ab - b 2

bc

bc - c2

ac

-a = ( abc) ×

a

a

b -b

b

c -c

c -a =

0

b

0 [C 2 ® (C 2 + C1) and C 3 ® (C 3 + C1)]

0 2b

c 2c = ( abc) × ( - a)

é taking out a , b , c common from ù êC , C , C respectively ú ë 1 2 3 û

0 0 2b = ( abc)( - a)( -4bc) 2c 0

= 4a 2 b 2 c2. EXAMPLE 5

Using properties of determinants, prove that 0 ab 2 a 2b 0 a 2 c cb 2

SOLUTION

ac2 bc2 = 2a 3 b 3 c 3 . 0

Let the given determinant be D. Then, ab 2 D= a b 0 0 2

a 2 c cb 2

ac2 bc2 0

0 a a = (a b c ) × b 0 b c c 0 2 2 2

0 1 1 = ( a 3b 3c 3) × 1 0 1 1 1 0

é taking a 2 , b 2 , c2 common from ù ê ú ëC1 , C 2 and C 3 respectively û é taking a , b , c common from ù ê R , R and R respectively ú ë 1 2 û 3

0 1 = ( a 3b 3c 3) × 1 0

1 1 0 1 -1

= ( a 3 b 3 c 3) × ( -1) ×

1 1 1 -1

[R 3 ® R 3 - R 2]

[expanded by C1 ]

= - ( a 3 b 3 c 3)( -1 - 1) = 2a 3 b 3 c 3. Hence, D = 2a 3 b 3 c 3.

229

SSS Mathematics for Class 12 230

230 EXAMPLE 6

Senior Secondary School Mathematics for Class 12

Using properties of determinants, prove that x+y x x 5 x + 4y 4x 2x = x 3 . 10x + 8y 8x

SOLUTION

[CBSE 2009, ‘14]

3x

Let the given determinant be D. Then, x+y x x D = 5 x + 4y

4x

2x

10x + 8y 8x 3x x+y x x = 3 x + 2y

2x

0

7x + 5y 5x 0 3 x + 2y 2x = x× 7x + 5y 5x 3 x + 2y 2 = x2 × 7x + 5y 5

[R 2 ® ( R 2 - 2R1) and R 3 ® ( R 3 - 3 R1)]

[expanded by C 3 ] [taking x common from C 2 ]

= x 2 × [5( 3 x + 2y) - 2(7 x + 5 y)] = ( x 2 × x) = x 3 . Hence, D = x 3 . EXAMPLE 7

Prove that

1 a a2 1 b b 2 = ( a - b)( b - c)( c - a). 1 c

SOLUTION

c2

Let the given determinant be D. Then, 1 a a2 D = 1 b b2 1 c c2 1 a a2 = 0 b - a b 2 - a 2 [applying R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1)] 0 c - a c2 - a 2 1 a a2 = ( b - a)( c - a) × 0 1 b + a 0 1 c+ a [taking ( b - a) common from R 2 and ( c - a) common from R 3 ] 1 b+ a [expanded by C1 ] = ( b - a)( c - a) ´ 1 × 1 c+ a = ( b - a)( c - a){( c + a) - ( b + a)} = ( b - a)( c - a)( c - b) = ( a - b)( b - c)( c - a). Hence, D = ( a - b)( b - c)( c - a).

SSS Mathematics for Class 12 231

Determinants

EXAMPLE 8

Prove that

231

1 a

1 b

1 c = ( a - b)( b - c)( c - a)( a + b + c).

a3

b3

c3 [CBSE 2009C, ’11, ‘12, ‘13C]

SOLUTION

Let the value of the given determinant be D. Then, 1 1 1 D= a

b

c

a3

b3

c3

0 =

0

1

a-c

b-c

c

a 3 - c3

b3 - c3

c3

[applying C1 ® (C1 - C 3) and C 2 ® (C 2 - C 3)] 0 0 1 = ( a - c)( b - c) ×

1

1

c

a 2 + ac + c2

b 2 + bc + c2

c3

[taking out ( a - c) and ( b - c) common from C1 and C 2 ] 1 1 [expanded by R1 ] = ( a - c)( b - c) × 1 × 2 2 2 a + ac + c b + bc + c2 = ( a - c)( b - c) × [( b 2 + bc + c2) - ( a 2 + ac + c2)] = ( a - c)( b - c)[( b 2 - a 2) + ( bc - ac)] = ( a - c)( b - c)[( b 2 - a 2) + ( b - a) c] = ( a - c)( b - c)( b - a)( b + a + c) = ( a - b)( b - c)( c - a)( a + b + c). Hence, D = ( a - b)( b - c)( c - a)( a + b + c). EXAMPLE 9

Prove that a b 2 a b2 b+ g g+ a

g g 2 = ( a - b)(b - g)(g - a)(a + b + g). a +b [CBSE 2007C, ’08, ’10C, ’11C, ’12C]

SOLUTION

Let the value of the given determinant be D. Then, a b g 2 2 D= a b g2 b+ g g+ a a +b =

a a2 a +b+ g

b b2 a +b+ g

g [applying R 3 ® ( R 3 + R1)] g2 a +b+ g

SSS Mathematics for Class 12 232

232

Senior Secondary School Mathematics for Class 12

a 2

b 2

= (a + b + g) × a b 1 1 a -g

g g 2 [taking ( a + b + g) common from R 3 ] 1 b-g g

= (a + b + g) × a 2 - g 2 b 2 - g 2 0 0

g2 1

[applying C1 ® (C1 - C 3) and C 2 ® (C 2 - C 3)] 1 1 g = ( a + b + g)( a - g)(b - g) × a + g b + g 0

g2

0

1

[taking ( a - g) common from C1 and (b - g) common from C 2 ] 1 1 [expanded by R 3 ] = ( a + b + g)( a - g)(b - g) × 1 × a + g b+ g

EXAMPLE 10

SOLUTION

= ( a + b + g)( a - g)(b - g)[(b + g) - (a + g)] = ( a + b + g)( a - g)(b - g)(b - a) = ( a - b)(b - g)( g - a )( a + b + g). Hence, D = ( a - b)(b - g)( g - a )( a + b + g). Using properties of determinants, show that a+x y z x a+y z = a 2( a + x + y + z).

[CBSE 2003, ’14]

x y a+z Let the value of the given determinant be D. Then, a+x y z D= x a+y z x y a+z a+ x+ y+z y z = a+ x+ y+z a+ y z [C1 ® C1 + C 2 + C 3] a+ x+ y+z y a+z 1 y z = ( a + x + y + z) × 1 a + y z 1 y a+z [taking ( a + x + y + z) common from C1 ] 1 y z = ( a + x + y + z) × 0 a 0 [R 2 ® R 2 - R1 and R 3 ® R 3 - R1] 0 0 a a 0 [expanded by C1 ] = ( a + x + y + z) × 1 × 0 a = a 2( a + x + y + z). Hence, D = a 2( a + x + y + z).

SSS Mathematics for Class 12 233

Determinants EXAMPLE 11

Using properties of determinants, prove that 1 x x

SOLUTION

233

x2

x 2

= (1 - x 3) 2 .

1 x x2 1

[CBSE 2008C, ’13]

Let the value of the given determinant be D. Then, 1 D = x2 x

x2

x 1

x x2 1

1 + x + x2 = 1 + x + x2 1 + x + x2

x2

x 1

[applying C1 ® (C1 + C 2 + C 3)]

x x2 1

1 x 2

= (1 + x + x ) 1 1

x2 x

[taking (1 + x + x 2) common from C1 ]

1 x2 1 1 x x2 = (1 + x + x ) 0 1 - x x(1 - x) 0 x( x - 1) (1 - x)(1 + x) 2

[applying R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1)] 1 x x2 = (1 + x + x 2)(1 - x) 2 × 0 1 x 0 - x (1 + x) [taking (1 - x) common from each of R 2 and R 3 ] 1 x [expanded by C1 ] = (1 + x + x 2)(1 - x) 2 × 1 × - x (1 + x) = (1 + x + x 2)(1 - x) 2 × (1 + x + x 2) = {(1 - x)(1 + x + x 2)} 2 = (1 - x 3) 2 . Hence, D = (1 - x 3) 2 . EXAMPLE 12

SOLUTION

Using the properties of determinants, prove that 2y y -z - x 2y 2z 2z z - x - y = ( x + y + z) 3 . x - y -z 2x 2x Let the value of the given determinant be D. Then, 2y y -z - x 2y D= 2z 2z z-x-y x - y -z 2x 2x

[CBSE 2014]

SSS Mathematics for Class 12 234

234

Senior Secondary School Mathematics for Class 12

=

2y

y -z - x

2z

2z

2y z - x - y [applying R 3 ® ( R1 + R 2 + R 3)]

x+ y+z x+ y+z x+ y+z 2y y - z - x = ( x + y + z) × 2z

2z

1

1 0

= ( x + y + z) × ( x + y + z) 0

2y z-x-y 1

[taking ( x + y + z) common from R 3 ] - ( x + y + z) 2y ( x + y + z)

z-x-y

0

1

[applying C1 ® (C1 - C 3) and C 2 ® (C 2 - C 3)] 0 - ( x + y + z) [expanded by R 3 ] = ( x + y + z) × 1 × ( x + y + z) ( x + y + z) = ( x + y + z){0 + ( x + y + z) 2} = ( x + y + z)( x + y + z) 2 = ( x + y + z) 3 . Hence, D = ( x + y + z) 3 . EXAMPLE 13

Using properties of determinants, prove that 3 a -a + b -a + c a-b 3b c - b = 3( a + b + c)( ab + bc + ca). [CBSE 2006, ’13] a-c

SOLUTION

b-c

3c

Let the value of the given determinant be D. Then, 3 a -a + b -a + c D = a-b 3b c-b a-c b-c 3c a + b + c -a + b -a + c = a+ b+ c 3b c - b [C1 ® (C1 + C 2 + C 3)] a+ b+ c b-c 3c 1 -a + b -a + c = ( a + b + c) × 1 3b c-b 1 b-c 3c [taking ( a + b + c) common from C1 ] 1 -a + b -a + c = ( a + b + c) × 0 2b + a a - b 0 a - c 2c + a [R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1)] 2b + a a - b [expanded by C1 ] = ( a + b + c) × 1 × a - c 2c + a = ( a + b + c)[( 2b + a)( 2c + a) - ( a - c)( a - b)]

SSS Mathematics for Class 12 235

Determinants

235

= ( a + b + c)[( 4bc + 2ab + 2ac + a 2) - ( a 2 - ab - ac + bc)] = 3( a + b + c)( ab + bc + ca). Hence, D = 3( a + b + c)( ab + bc + ca). EXAMPLE 14

Prove that b+ c

c+ a

a+b

c + a a + b b + c = 2( a + b + c)( ab + bc + ca - a 2 - b 2 - c2). a+ b b+ c c+ a SOLUTION

[CBSE 2004, ‘09C]

Let the given determinant be D. Then, b+ c D = c+ a

c+ a

a+b

a+ b b+ c

a+ b b+ c c+ a 2( a + b + c) 2( a + b + c) =

2( a + b + c)

c+ a

a+b

b+ c

a+b

b+ c

c+ a

[R1 ® ( R1 + R 2 + R 3)]

1 1 1 = 2( a + b + c) × c + a a + b b + c a+ b b+ c c+ a [taking ( a + b + c) common from R1] 1 0 0 éC 2 ® (C 2 - C1) and ù = 2( a + b + c) × c + a b - c b - a êC ® (C - C ) ú ë 3 û 3 1 a+ b c-a c-b b-c b-a [expanded by R1 ] = 2( a + b + c) × 1 × c-a c-b = 2( a + b + c) × [( b - c)( c - b) - ( c - a)( b - a)] = 2( a + b + c)( ab + bc + ca - a 2 - b 2 - c2). Hence, D = 2( a + b + c)( ab + bc + ca - a 2 - b 2 - c2). EXAMPLE 15

Using properties of determinants, prove that a-b-c 2a 2b b-c-a 2c

SOLUTION

2c

2a 2b

= ( a + b + c) 3 . [CBSE 2001, ‘04, ‘06C, ‘07]

c-a-b

Let the value of the given determinant be D. Then, a-b-c 2a 2a D= 2b b-c-a 2b 2c 2c c-a-b =

a+ b+ c a+ b+ c a+ b+ c 2b b-c-a 2b 2c 2c c-a-b

[R1 ® ( R1 + R 2 + R 3)]

SSS Mathematics for Class 12 236

236

Senior Secondary School Mathematics for Class 12

1

1

1

= ( a + b + c) × 2b b - c - a 2c

2b c-a-b

2c

1

0

[taking ( a + b + c) common from R1 ] 0

= ( a + b + c) × 2b - ( a + b + c)

0

2c

- ( a + b + c)

0

[C 2 ® (C 2 - C1) and C 3 ® (C 3 - C1)] - ( a + b + c) 0 [expanded by R1] = ( a + b + c) × 1 × 0 - ( a + b + c) = ( a + b + c)( a + b + c) 2 = ( a + b + c) 3. Hence, D = ( a + b + c) 3. EXAMPLE 16

Prove that a + b + 2c a c b + c + 2a c

b b

= 2( a + b + c) 3 .

c + a + 2b

a

[CBSE 2008, ‘12C, ’14C] SOLUTION

Let the value of the given determinant be D. Then, a + b + 2c a b D= c b + c + 2a b c a c + a + 2b =

a + b + c - ( a + b + c) 0 c b + c + 2a b 0 - ( a + b + c) ( a + b + c)

ìby R1 ® ( R1 - R 2) ü í ý îand R 3 ® ( R 3 - R 2) þ

-1 1 0 = ( a + b + c) 2 × c b + c + 2a b -1 0 1 [taking ( a + b + c) common from each one of R1 and R 3] -1 1 0 = ( a + b + c) 2 × 0 b + 2c + 2a b [R 2 ® R 2 - cR1] -1 0 1 = ( a + b + c) 2 × 1 ×

b + 2c + 2a b -1 1

[expanded by C1 ]

= ( a + b + c) 2 × ( b + 2c + 2a + b) = 2( a + b + c) 3 . Hence, D = 2( a + b + c) 3 .

SSS Mathematics for Class 12 237

Determinants EXAMPLE 17

Using properties of determinants, prove that a b c a 2 b 2 c2 = ( a - b)( b - c)( c - a)( ab + bc + ca). [CBSE 2011C, ’13C] bc

SOLUTION

237

ca

ab

Let the value of the given determinant be D. Then, a b c D = a2

b2

c2

bc

ca

ab

a-c = a 2 - c2 b( c - a)

b-c

c

b 2 - c2 a( c - b)

c2

[C1 ® (C1 - C 3), C 2 ® (C 2 - C 3)]

ab

1

1

c

= ( a - c)( b - c) × a + c b + c c2 -b

-a

ab

[taking ( a - c) common from C1 and ( b - c) common from C 2 ] 1 0 0 = ( a - c)( b - c) × a + c b - a - ca - b b - a ( a + c) b = ( a - c)( b - c) × 1 ×

b-a

- ca

[C 2 ® (C 2 - C1), C 3 ® (C 3 - cC1)]

b - a ( a + c) b

= ( a - c)( b - c) × ( b - a)

1

{expanded by R1 }

- ca

1 ( a + c) b

= ( a - b)( b - c)( c - a)( ab + bc + ca). Hence, D = ( a - b)( b - c)( c - a)( ab + bc + ca). EXAMPLE 18

Using properties of determinants, prove that a b a-b b-c

c c - a = ( a 3 + b 3 + c 3 - 3 abc).

[CBSE 2006, ‘09, ‘12C]

b+ c c+ a a+ b SOLUTION

Let the value of the given determinant be D. Then, a b c D = a-b b-c c-a b+ c c+ a a+ b =

a+ b+ c 0 2( a + b + c)

b c b-c c-a c+ a a+ b

[C1 ® (C1 + C 2 + C 3)]

SSS Mathematics for Class 12 238

238

Senior Secondary School Mathematics for Class 12

1

b

c

= ( a + b + c) × 0 b - c

c-a

2 c+ a a+ b 1

[taking ( a + b + c) common from C1 ] c

b

= ( a + b + c) × 0

b-c

c-a

[R 3 ® R 3 - 2R1]

0 c + a - 2b a + b - 2c = ( a + b + c) × 1 × = ( a + b + c) ×

b-c c-a c + a - 2b a + b - 2c

b-c c-a a-b b-c

[expanded by C1 ]

[R 2 ® R 2 + R1]

= ( a + b + c) × [( b - c) 2 - ( a - b)( c - a)] = ( a + b + c)( a 2 + b 2 + c2 - ab - bc - ca) = ( a 3 + b 3 + c 3 - 3 abc). Hence, D = ( a 3 + b 3 + c 3 - 3 abc).

EXAMPLE 19

Prove that

y+z z z z+x y

SOLUTION

x

Given determinant =

y x

= 4xyz.

x+y

y+z z y z z+x x y x x+y

0 -2x -2x = z z+x x y x x+y

[R1 ® R1 - ( R 2 + R 3)]

0 1 1 = ( -2x) × z z + x x y x x+y [taking ( -2x) common from R1] 0 0 1 = ( -2x) × z z x y -y x + y = ( -2x) × 1 ×

z z y -y

[C 2 ® (C 2 - C 3)]

[expanded by R1 ]

= ( -2x) × 1 × ( - yz - yz) = ( -2x)( -2yz) = 4xyz.

SSS Mathematics for Class 12 239

Determinants

EXAMPLE 20

Prove that

a2 SOLUTION

a2 bc c2 + ac 2 = 4a 2 b 2 c2 . a + ab b ac 2 2 ab b + bc c 2

bc

c2 + ac

2

ac c2

2

a + ab b 2 ab b + bc

c

c+ a

b

a

b+ c

c

a = ( abc) × a + b b

239

[CBSE 2014]

[taking a , b , c common from C1 , C 2 and C 3 respectively] c

c+ a

-2c

-2c

a = ( abc) × 0

b b+ c

a -a c + a = ( abc) × 0 0 -2c b b c = ( abc)( 2c) ×

a -a b

[R 2 ® R 2 - ( R1 + R 3)]

c

b

[C 2 ® C 2 - C 3]

[expanded by R 2]

= 2 abc2 × ( ab + ab) = 2abc2( 2ab) = 4a 2 b 2 c2. EXAMPLE 21

If a , b , c are positive and unequal, show that the value of the a b c determinant b c a is negative. c a b

SOLUTION

The given determinant a b c = b c a c a b a+ b+ c b c = a+ b+ c c a a+ b+ c a b 1 b c = ( a + b + c) × 1 c a 1 a b

[C1 ® (C1 + C 2 + C 3)]

SSS Mathematics for Class 12 240

240

Senior Secondary School Mathematics for Class 12

1

b

c

= ( a + b + c) × 0 c - b a - c [R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1 )] 0 a-b b-c = ( a + b + c) × [( c - b)( b - c) - ( a - b)( a - c)] 2

2

[expanded by C1]

2

= ( a + b + c) × ( -a - b - c + ab + bc + ca) 1 = - ( a + b + c)( 2a 2 + 2b 2 + 2c2 - 2ab - 2bc - 2ca) 2 1 = - ( a + b + c)[( a - b) 2 + ( b - c) 2 + ( c - a) 2], which is negative 2 [Q ( a + b + c) > 0, ( a - b) 2 > 0, ( b - c) 2 > 0 and ( c - a) 2 > 0]. m

C1

m

C2

m

n

C1 p C1

n

C2 p C2

n

C3 C3 × p C3

EXAMPLE 22

Evaluate

SOLUTION

Let the given determinant be D. Then, 1 1 m m(m - 1) × m(m - 1)(m - 2) 6 2 1 1 D= n n(n - 1) × n(n - 1)(n - 2) 2 6 1 1 p( p - 1) × p( p - 1)( p - 2) p 2 6 1 (m - 1) (m - 1)(m - 2) æ1 1 ö = ç ´ ´ mnp ÷ × 1 (n - 1) (n - 1)(n - 2) è2 6 ø 1 ( p - 1) ( p - 1)( p - 2) 1 m - 1 (m - 1)(m - 2) 1 = × mnp × 0 n - m (n - m)(n + m - 3) 12 0 p - m ( p - m)( p + m - 3) [R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1)] 1 m - 1 (m - 1)(m - 2) 1 = × (mnp)(n - m)( p - m) × 0 1 (n + m - 3) 12 0 1 ( p + m - 3) [taking (n - m) common from R 2 and ( p - m) common from R 3 ] 1 (n + m - 3) 1 = × (mnp)(n - m)( p - m) × 1 × 1 ( p + m - 3) 12 1 × (mnp)(n - m)( p - m)[( p + m - 3) - (n + m - 3)] 12 1 = × (mnp)(n - m)( p - m)( p - n) 12 =

SSS Mathematics for Class 12 241

Determinants

241

1 × (mnp)(m - n)(n - p)( p - m). 12 1 Hence, D = × (mnp)(m - n)(n - p)( p - m). 12 =

( b + c) 2

a2

2

( c + a) c2

a2 2

b2 = 2 abc( a + b + c) 3 . ( a + b) 2

EXAMPLE 23

Prove that

SOLUTION

Let the given determinant be D. Then,

b c2

( b + c) 2 D=

b c

2

a2 ( c + a)

2

c

( b + c) 2 - a 2 =

0 2 c - ( a + b) 2

2

a2 2

b2 ( a + b) 2

0 ( c + a) 2 - b 2

a2 b2

c2 - ( a + b) 2 ( a + b) 2

( a + b + c)( b + c - a) =

[CBSE 2010, ’10C]

[C1 ® C1 - C 3 and C 2 ® C 2 - C 3] 0 a2

0 ( a + b + c)( c + a - b) b2 ( a + b + c)( c - a - b) ( a + b + c)( c - a - b) ( a + b) 2 ( b + c - a) 2

= ( a + b + c) ×

0 c-a-b

a2 0 c+ a-b b2 c - a - b ( a + b) 2

[taking ( a + b + c) common from C1 and C 2 both] ( b + c - a) a2 0 2 = ( a + b + c) × c + a - b b2 0 [R 3 ® R 3 - ( R1 + R 2)] -2b -2a 2ab = ( a + b + c) 2[( b + c - a){( c + a - b) × 2ab + 2ab 2} + a 2{0 + 2b( c + a - b)}] = ( a + b + c) 2[( b + c - a) × 2ab{( c + a - b + b)} + 2a 2 b( c + a - b)] = 2ab ( a + b + c) 2{( b + c - a)( c + a) + a( c + a - b)} = 2ab ( a + b + c) 2 × {bc + ab + c2 + ac - ac - a 2 + ac + a 2 - ab} = 2ab ( a + b + c) 2{bc + c2 + ac} = 2abc( a + b + c) 3. Hence, D = 2abc( a + b + c) 3. TYPE: EXPRESSING A GIVEN DET. AS SUM OF TWO DETERMINANTS

EXAMPLE 24

Prove that

b+ c a b c + a c a = ( a + b + c)( a - c) 2 . a+b b c

[CBSE 2005C, ’07]

SSS Mathematics for Class 12 242

242 SOLUTION

Senior Secondary School Mathematics for Class 12

Let the given determinant be D. Then, b+ c a b D = c+ a

c a

a+b b c b a b = c

c a b

c a + a

a b c

c a

b b c

a+ b+ c a+ b+ c a+ b+ c =

c

c

a

a

b

c

a+ b+ c a+ b+ c a+ b+ c +

a

c

a

b

b

c

[R1 ® ( R1 + R 2 + R 3) in each determinant] 1 1 1 1 1 1 = ( a + b + c) × c

c a + ( a + b + c) × a

a b c

c a

b b c

[taking out ( a + b + c) common from R1 in each determinant] 1 0 0 1 0 0 = ( a + b + c) × c 0 a - c + ( a + b + c) × a c - a 0 a b-a c-a b 0 c-b [C 2 ® (C 2 - C1) and C 3 ® (C 3 - C1) in each] a-c c-a 0 = ( a + b + c) × 1 × + ( a + b + c) × 1 × b-a c-a 0 c-b 0

[each det. expanded by R1 ] = ( a + b + c) × [0 - ( b - a)( a - c)] + ( a + b + c)( c - a)( c - b) = ( a + b + c)( a - b)( a - c) - ( a + b + c)( a - c)( c - b) = ( a + b + c)( a - c){( a - b) - ( c - b)} = ( a + b + c)( a - c) 2. Hence, D = ( a + b + c)( a - c) 2. EXAMPLE 25

Prove that

1 a a 2 - bc 1 b b 2 - ca = 0. 1 c c2 - ab

SOLUTION

Let the given determinant be D. Then, 1 a a 2 - bc D = 1 b b 2 - ca 1 c c2 - ab 1 a a2 1 a - bc 2 = 1 b b + 1 b - ca 1 c c2 1 c - ab

[CBSE 2005C]

SSS Mathematics for Class 12 243

Determinants

243

1 - bc a a2 a 2 = 0 b - a b - a 2 + 0 b - a c( b - a) 0 c - a c2 - a 2 0 c - a b( c - a) 1

[R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1) in each determinant] 1 a

a2

1 a - bc

= ( b - a)( c - a) × 0 1 b + a + ( b - a)( c - a) × 0 1 0 1 c+ a

0 1

c b

= ( b - a)( c - a) × 1 × {( c + a) - ( b + a)} + ( b - a)( c - a) × 1 × ( b - c) = ( b - a)( c - a)( c - b) + ( b - a)( c - a)( b - c) = ( a - b)( b - c)( c - a) - ( a - b)( b - c)( c - a) = 0. Hence, D = 0. x x2 1 + x 3 EXAMPLE 26

If x ¹ y ¹ z and y y 2 1 + y 3 = 0 then prove that xyz = -1. z z2 1 + z 3 [CBSE 2008, ‘09C, ’11]

SOLUTION

Let the given determinant be D. Then, x D= y z

x2 1 + x 3 x x2 1 x x2 2 2 3 y 1 + y = y y 1 + y y2 z2 1 + z 3 z z2 1 z z2

x = y z

x2 1 1 x x2 2 y 1 + ( xyz) × 1 y y 2 z2 1 1 z z2

x = y z

x2 1 x x2 1 2 2 y 1 + ( xyz)( -1) × y y 2 1 z2 1 z z2 1

x3 y3 z3

[interchanging the columns of the 2nd det. twice] x x2 1 = (1 + xyz) y y 2 1 z z2 1 x x2 1 2 2 = (1 + xyz) ( y - x) ( y - x ) 0 [R 2 ® ( R 2 - R1 ), R 3 ® ( R 3 - R1 )] (z - x) (z 2 - x 2) 0 x x2 1 = (1 + xyz)( y - x)(z - x) 1 y + z 0 1 z+x 0

SSS Mathematics for Class 12 244

244

Senior Secondary School Mathematics for Class 12

1 y+x [expanding by C 3] 1 z+x = (1 + xyz) ( y - x) (z - x) (z - y). \ D = 0 Þ (1 + xyz)( y - x)(z - x)(z - y) = 0 Þ (1 + xyz) = 0 [Q ( y - x) ¹ 0, (z - x) ¹ 0, (z - y) ¹ 0] Þ xyz = -1. Hence, xyz = -1. For any scalar p, prove that = (1 + xyz)( y - x)(z - x) × 1 ×

EXAMPLE 27

SOLUTION

x

x 2 1 + px 3

y z

y 2 1 + py 3 = (1 + pxyz)( x - y)( y - z)(z - x). z 2 1 + pz 3

[CBSE 2010]

Let the given determinant be D. Then, x D= y

x 2 1 + px 3 y 2 1 + py 3

z

z2

x

x2 1 x x2 y2 1 + y y2 z2 1 z z2

= y z

1 + pz 3 px 3 py 3 pz 3

x 1 x2 x x2 = ( -1) × y 1 y 2 + p × y y 2 z 1 z2 z z2 [by C 2 1 x = ( -1)( -1) × 1 y 1 z

[say D1 + D 2 ] x3 y3 z3

« C 3 in D1 ] [taking p common from C 3 in D 2 ] 1 x x2 x2 2 y + ( pxyz) × 1 y y 2 1 z z2 z2

[by C1 « C 2 ] [taking x, y, z common from R1 , R 2 , R 3 resp.] 1 x x2 = (1 + pxyz) 1 y y 2 1 z z2 1 x x2 = (1 + pxyz) 0 y - x y 2 - x 2 0 z - x z2 - x 2 [by R 2 ® R 2 - R1 and R 3 ® R 3 - R1] 1 x x2 = (1 + pxyz)( y - x)(z - x) × 0 1 y + x 0 1 z+x [taking ( y - x) common from R 2 and (z - x) common from R 3 ]

SSS Mathematics for Class 12 245

Determinants

= (1 + pxyz)( y - x)(z - x) × 1 ×

245

1 y+x 1 z+x

= (1 + pxyz) ( y - x) (z - x) × {(z + y) - ( y + x)} = (1 + pxyz) ( y - x)(z - x)(z - y) = (1 + pxyz) ( x - y)( y - z)(z - x) = RHS. LHS = RHS. \ TYPE: SOME MORE PROBLEMS EXAMPLE 28

Prove that 1+ a 1 1

1

1

æ1 1 1 ö 1+ b 1 = ( abc) ç + + + 1÷ = ( bc + ca + ab + abc). ø èa b c 1 1+ c [CBSE 2004, ‘08, ‘09, ‘12, ’14]

SOLUTION

The given determinant 1+ a 1 1 =

1+ b 1 1 1+ c 1 1 1 +1 a a a 1 1 1 = ( abc) × +1 b b b 1 1 1 +1 c c c [taking a , b , c common from R1 , R 2 and R 3 respectively] 1 1

1 æ1 1 1 ö 1 = ( abc) ç + + + 1÷ × ø b èa b c 1 c

1

1

1 1 +1 b b 1 1 +1 c c

æ1 1 1 ö [applying R1 ® R1 + R 2 + R 3 and taking out ç + + + 1÷ ø èa b c common from R1 ] 0

0

1

1 æ1 1 1 ö = ( abc) ç + + + 1÷ × 0 1 a b c b è ø 1 -1 -1 +1 c [applying C1 ® (C1 - C 3) and C 2 ® (C 2 - C 3)] 0 1 æ1 1 1 ö = ( abc) ç + + + 1÷ × (1) × -1 -1 ø èa b c

[expanding by 1st row]

SSS Mathematics for Class 12 246

246

Senior Secondary School Mathematics for Class 12

1 1 ö + + 1÷ × 1 b c ø

æ1 = ( abc) ç + èa æ1 = ( abc) ç + èa

1 1 ö + + 1÷ = ( bc + ca + ab + abc). b c ø

Hence, the result follows.

EXAMPLE 29

SOLUTION

Prove that

1 a 1 b 1 c

a2

bc

b2

ca = 0.

c2

ab

Let the given determinant be D. Then, 1 a 2 bc a 1 b 2 ca D= b 1 c2 ab c 1 1 = × 1 abc 1

a3

abc

3

cba abc

b c3

1 1 = × abc × 1 abc 1 = (1 ´ 0) = 0 Hence, D = 0. EXAMPLE 30

a3 b3 c3

[multiplying R1 , R 2 , R 3 by a , b , c respectively and dividing D by abc] 1 1 1

[Q C1 and C 3 are identical].

Prove that a2 + 1 ab ab b2 + 1 ca

cb

ac bc

= (1 + a 2 + b 2 + c2).

2

c +1 [CBSE 2007, ‘08, ‘11C, ’14]

SOLUTION

Let the given determinant be D. Then, 1ö æ aç a + ÷ ab ac aø è 1ö æ ab bc D= çb + ÷b bø è 1ö æ ca cb çc + ÷c cø è

SSS Mathematics for Class 12 247

Determinants

a+ = ( abc) ×

1 a

1 b

b

1 c [taking a , b , c common from C1 , C 2 , C 3 respectively] c+

c

a2 + 1 b

a

b+

b c

( abc) = × ( abc)

a

247

2

a2

a2

2

b +1

b2

c2

c2 + 1

c2

[applying R1 ® aR1 , R 2 ® bR 2 , R 3 ® cR 3 and dividing the whole det. by abc] 1 + a 2 + b 2 + c2 1 + a 2 + b 2 + c2 1 + a 2 + b 2 + c2 b2 b2 + 1 b2 = 2 2 2 c c c +1 [by R1 ® R1 + R 2 + R 3] 1 b2 c2

= (1 + a 2 + b 2 + c2) ×

1 1 b2 + 1 b2 c2 c2 + 1

[taking out (1 + a 2 + b 2 + c2) common from R1 ] 1 = (1 + a + b + c ) × b 2 c2 2

2

2

= (1 + a 2 + b 2 + c2) × 1 ×

0 1 0

1 0 0 1

0 0 [by C 2 ® C 2 - C1 and C 3 ® C 3 - C1 ] 1 [expanded by R1 ]

= (1 + a 2 + b 2 + c2) × 1 × (1 - 0) = (1 + a 2 + b 2 + c2). Hence, D = (1 + a 2 + b 2 + c2).

EXAMPLE 31

Prove that

- bc b 2 + bc a 2 + ac - ac a 2 + ab b 2 + ab

SOLUTION

c2 + bc c2 + ac = ( ab + bc + ac) 3 . [CBSE 2007] - ab

We have - bc b 2 + bc a 2 + ac - ac a 2 + ab b 2 + ab

c2 + bc - abc ab 2 + abc ac2 + abc 1 c2 + ac = a 2 b + abc - abc c2 b + abc abc 2 2 - ab a c + abc b c + abc - abc

[R1 ® aR1 , R 2 ® bR 2 and R 3 ® cR 3 and dividing the det. by abc]

SSS Mathematics for Class 12 248

248

Senior Secondary School Mathematics for Class 12

- bc ab + ac ac + ab æ abc ö - ac bc + ab =ç ÷ ab + bc è abc ø ac + bc bc + ac - ab [taking out a , b , c common from C1 , C 2 , C 3 respectively] ab + bc + ac ab + bc + ac ab + bc + ac [R1 ® ( R1 + R 2 + R 3)] = ab + bc - ac bc + ab ac + bc

bc + ac 1

= ( ab + bc + ac) × ab + bc

- ab 1

1

- ac

bc + ab

ac + bc bc + ac 0 = ( ab + bc + ac) ×

0 ( ab + bc + ac)

- ab 0

1

- ( ab + bc + ac) ( ab + bc + ac)

bc + ab - ab

[C1 ® (C1 - C 3) and C 2 ® (C 2 - C 3)] = ( ab + bc + ac) 3 EXAMPLE 32

[expanding by R1].

Prove that a2 b2

a 2 - ( b - c) 2 b 2 - ( c - a) 2

bc ca = ( a 2 + b 2 + c2)( a - b)( b - c)( c - a)( a + b + c).

c2

c2 - ( a - b) 2

ab [CBSE 2012C]

SOLUTION

Let the value of the given determinant be D. Then, a2 D = b2 c2 a2 = b2 c2

a 2 - ( b - c) 2 b 2 - ( c - a) 2 c2 - ( a - b) 2 - ( b - c) 2 - ( c - a) 2 - ( a - b) 2

bc ca ab

a 2 ( b - c) 2 = ( -1) × b 2 ( c - a) 2 c2 ( a - b) 2 2

2

bc ca ab [applying C 2 ® (C 2 - C1)] bc ca ab

2

a = ( -1) × b 2 c2

b + c - 2bc bc c2 + a 2 - 2ca ca a 2 + b 2 - 2ab ab

a2 = ( -1) × b 2 c2

a 2 + b 2 + c2 a 2 + b 2 + c2 a 2 + b 2 + c2

bc ca ab

[C 2 ® C 2 + C1 + 2C 3]

SSS Mathematics for Class 12 249

Determinants

249

a 2 1 bc 2

2

2

= ( -1)( a + b + c ) × b 2 1 ca c2 1 ab a2 2

2

2

2

0 ca - bc 0 ab - bc

a 2

2

bc

c2 - a 2

= ( -1)( a + b + c ) × b - a

2

1 2

2

2

1

bc

2

0 ( a - b) c

c2 - a 2

0 ( a - c) b

= ( -1)( a + b + c ) × b - a

a2 2

2

1

bc

2

= ( -1)( a + b + c )( b - a)( c - a) × b + a 0 - c c+ a

0 -b

[taking out ( b - a) common from R 2 and ( c - a) common from R 3 ] b + a -c = ( -1)( a 2 + b 2 + c2)( b - a)( c - a)( -1) × c + a -b = ( a 2 + b 2 + c2)( b - a)( c - a) × ( -1) ×

b+ a

c

c+ a

b

= ( a 2 + b 2 + c2)( a - b)( c - a) × {b ( b + a) - c( c + a)} = ( a 2 + b 2 + c2)( a - b)( c - a) × {( b 2 - c2) + ( ab - ac)} = ( a 2 + b 2 + c2)( a - b)( c - a) × {( b 2 - c2) + a( b - c)} = ( a 2 + b 2 + c2)( a - b)( b - c)( c - a)( a + b + c). Hence, D = ( a 2 + b 2 + c2)( a - b)( b - c)( c - a)( a + b + c). TYPE: DETERMINANTS ON TRIGONOMETRY cos a cos b cos a sin b - sin a = 1. - sin b cos b 0 sin a cos b sin a sin b cos a

EXAMPLE 33

Prove that

SOLUTION

Let the given determinant be D. Then, cos a cos b cos a sin b - sin a D=

- sin b

cos b

sin a cos b sin a sin b 1 = × sin a cos a

0 cos a

sin a cos a cos b sin a cos a sin b - sin 2 a cos b - sin b 0 sin a cos a cos b sin a cos a sin b cos2 a

[R1 ® (sin a ) R1 , R 3 ® (cos a ) R 3 and dividing D by (sin a cos a )]

SSS Mathematics for Class 12 250

250

Senior Secondary School Mathematics for Class 12

1 = × sin a cos a

0

0

-1

- sin b cos b 0 sin a cos a cos b sin a cos a sin b cos2 a

[R1 ® ( R1 - R 3)] - sin b cos b 1 = × ( -1) × sin a cos a cos b sin a cos a sin b sin a cos a =

- sin b cos b (sin a cos a ) × ( -1) × cos b sin b (sin a cos a )

[taking sin a cos a common from R 2 ] = ( -1) × [- sin b - cos b] = (sin 2 b + cos2 b) = 1. 2

2

Hence, D = 1. EXAMPLE 34

If A + B + C = p, prove that sin ( A + B + C) sin ( A + C) - sin B cos ( A + B)

SOLUTION

0 tan ( B + C)

cos C tan C = 0. 0

Let the given determinant be D. Then, sin ( A + B + C) sin ( A + C) cos C D= - sin B 0 tan C 0 cos ( A + B) tan ( B + C) sin p = - sin B cos ( p - C)

sin ( p - B) 0 tan ( p - A)

0 sin B = - sin B 0 - cos C - tan A

cos C tan C 0

cos C tan A 0

[Q sin p = 0, sin( p - B) = sin B ; cos( p - C) = - cos C and tan( p - A) = - tan A] sin B cos C sin B cos C = sin B × - cos C × - tan A 0 0 tan A {expanded by C1 } = (sin B × tan A cos C) - (sin B tan A × cos C) = 0. Hence, D = 0. EXAMPLE 35

If A + B + C = p, show that sin 2 A sin 2 B sin 2 C

sin A cos A sin B cos B sin C cos C

cos2 A cos2 B cos2 C

= - sin( A - B) sin( B - C) sin(C - A).

SSS Mathematics for Class 12 251

Determinants SOLUTION

251

Let the given determinant be D. Then, 1 (1/2) sin 2A (1/2)(1 + cos 2A) D = 1 (1/2) sin 2B

(1/2)(1 + cos 2B)

[C1 ® C1 + C 3]

1 (1/2) sin 2C (1/2)(1 + cos 2C) 1 sin 2A 1 + cos 2A 1 = × 1 sin 2B 1 + cos 2B 4 1 sin 2C 1 + cos 2C 1 sin 2A 1 = × 0 sin 2B - sin 2A 4 0 sin 2C - sin 2A

1 + cos 2A cos 2B - cos 2A cos 2C - cos 2A

1 sin 2A 1 = × 0 2 cos ( A + B) sin ( B - A) 4 0 2 cos ( A + C) sin (C - A)

ìR 2 ® R 2 - R1 ü í ý îR 3 ® R 3 - R1 þ

1 + cos 2A 2 sin ( A + B) sin ( A - B) 2 sin ( A + C) sin ( A - C)

1 sin 2A = sin ( A - B) sin ( A - C) × 0 - cos ( A + B) 0 - cos ( A + C)

1 + cos 2A sin ( A + B) sin ( A + C)

[taking 2sin ( A - B) common from R 2 and 2sin ( A - C) common from R 3 ] 1 sin 2A 1 + cos 2A = sin ( A - B) sin ( A - C) × 0 - cos ( p - C) 0 - cos ( p - B)

sin( p - C) sin( p - B)

[Q A + B + C = p ] 1 sin 2A 1 + cos 2A = sin ( A - B) sin ( A - C) × 0 cos C sin C 0 cos B sin B = sin ( A - B) sin ( A - C) [sin B cos C - cos B sin C] = sin ( A - B) sin ( A - C) sin ( B - C) = - sin ( A - B) sin ( B - C) sin (C - A). Hence, D = - sin ( A - B) sin ( B - C) sin (C - A). sin a sin b sin g

cos a cos b cos g

cos ( a + d) cos (b + d) = 0. cos ( g + d)

EXAMPLE 36

Prove that

SOLUTION

Let the given determinant be D. Then, applying C 3 ® C 3 + (sin d) C1 - (cos d) C 2 , we get

SSS Mathematics for Class 12 252

252

Senior Secondary School Mathematics for Class 12

sin a

cos a

0

D = sin b

cos b

0 = 0 [Q each element in C 3 is 0].

sin g

cos g

0

TYPE: EQUALITY OF TWO DETERMINANTS EXAMPLE 37

Without expanding prove that a + bx c + dx p + qx

SOLUTION

We have a + bx

c + dx

p + qx

LHS = ax + b

cx + d

px + q

u

v

a - ax

2

= ax + b u

a

p

w 2

p - px 2

cx + d v

px + q w

c - cx

a(1 - x 2) = ax + b u

c

px + q = (1 - x 2) b d q × w u v w

ax + b cx + d u v

c(1 - x 2) cx + d v

[applying R1 ® R1 - xR 2 ]

p(1 - x 2) px + q w

a c = (1 - x ) ax + b cx + d u v

p px + q w

2

[taking out (1 - x 2) common from R1 ] a c p = (1 - x 2) b d q u v w

[applying R 2 ® R 2 - xR1 ]

= RHS. Hence, LHS = RHS. EXAMPLE 38

Without expanding the determinant, prove that a a2 b b2 c

SOLUTION

c2

bc 1 a2 ca = 1 b 2 ab

1 c2

We have a a2 2 LHS = b b 2 c c

bc ca ab

a3 b3 × c3

[CBSE 2001C]

SSS Mathematics for Class 12 253

Determinants

a2 1 = × b2 abc 2 c

a2

abc

b2

abc

2

abc

c

253

[R1 ® aR1 , R 2 ® bR 2 , R 3 ® cR 3 and dividing the whole det. by abc] a2 1 = × abc × b 2 abc c2 1 a2 = 1 b2 1 c2

a3 1 b3 1

[taking abc common from C 3 ]

c3 1

a3 b3 c3

[C 2 « C 3 and C1 « C 2]

= RHS. Hence, LHS = RHS. EXAMPLE 39

Using the properties of determinants, prove that a+ b b+ c c+ a a b c b+ c c+ a a+ b = 2 b c a × c+ a

SOLUTION

a+ b b+ c

[CBSE 2007, ’10C]

c a b

We have a+ b b+ c c+ a c+ a a+ b c+ a a+ b b+ c

LHS = b + c

-2c b + c c + a = -2a c + a a + b -2b a + b b + c c b+ c c+ a = ( -2) a c + a a + b b a+ b b+ c c b a = ( -2) × a c b b a c a b c =2b c a c a b = RHS. Hence, LHS = RHS.

[applying C1 ® C1 - (C 2 + C 3)]

[taking out (–2) common from C1 ]

[applying C 2 ® C 2 - C1 and C 3 ® C 3 - C1 ]

[applying C1 « C 3 ]

SSS Mathematics for Class 12 254

254 EXAMPLE 40

Senior Secondary School Mathematics for Class 12

Show that b+ c

c+ a

a+b

a

b c

q+r r + p p+ q = 2 p q r × y+z z+ x x+ y x y z SOLUTION

[CBSE 2008, ’10C, ’12C, ’14]

We have b+ c

c+ a

a+b

LHS = q + r

r+p

p+q

y+z z+ x x+ y c+ a

a+b

= 2 p+ q+r r + p

a+ b+ c

p+q

x+ y+z z+ x x+ y [applying C1 ® (C1 + C 2 + C 3) and taking out 2 common from C1 ] a+ b+ c

-b -c

= 2 p+ q+r

- q -r

x + y + z -y

[C 2 ® (C 2 - C1), C 3 ® (C 3 - C1)]

-z

a+ b+ c b c = 2( -1)( -1) × p + q + r q r x+ y+z y z [taking out (–1) common from each one of C 2 and C 3 ] a =2p x

b c q r = RHS y z

[applying C1 ® C1 - (C 2 + C 3)]

Hence, LHS = RHS. EXAMPLE 41

Show that D = D1 , where Ax x 2 1 A B C D = By y 2 1 and D1 = x y z × Cz z 2 1 zy zx xy

SOLUTION

[CBSE 2014C]

We have Ax x 2 1 D = By y 2 1 Cz z 2 1 Ax By Cz = x 2 y 2 z2 1 1 1

[interchanging the rows and columns]

SSS Mathematics for Class 12 255

Determinants

A = ( xyz) x 1 x

255

B C y 1 y

z 1 z 1 1 1 C1 , C 2 ® C 2 , C 3 ® C 3 and x y z multiplying the whole determinant by xyz]

[applying C1 ®

=

A B ( xyz) × x y ( xyz) yz zx

C z xy

[applying R 3 ® ( xyz) R 3 and dividing the whole determinant by xyz] A

B

C

= x

y

z = D1.

zy zx xy Hence, D = D1.

EXAMPLE 42

a If D1 = x p

b c q y z and D 2 = - p q r r

-b y a - x then without expanding D1 -c z

and D 2 , prove that D1 + D 2 = 0. SOLUTION

We have q D 2 = -p r

-b y a -x -c z

q = ( -1) × p r

-b -a -c

q = ( -1)( -1) × p r q = p r

y x z

[taking (–1) common from R 2 ]

b y a x c z

[taking (–1) common from C 2 ]

b y q p r a x = b a c c z y x z

p q r = ( -1) a b c x y z

[interchanging rows and columns]

[applying C1 « C 2 ]

SSS Mathematics for Class 12 256

256

Senior Secondary School Mathematics for Class 12

a = ( -1)( -1) × p

b

c [applying R1 « R 2 ]

q r

x y z a = p

b

c

q r

x

y z a b c = ( -1) × x y z = - D1 p q r

[applying R 2 « R 3].

Thus, D 2 = - D1 and hence D1 + D 2 = 0. TYPE: EQUATIONS OF DETERMINANTS EXAMPLE 43

Solve for x: a+ x a-x a-x a-x

SOLUTION

a-x

a + x a - x = 0. a-x a+ x

[CBSE 2004, ’05, ’11]

Let the given determinant be D. Then, a+ x a-x a-x D = a-x a+ x a-x a-x a-x a+ x 3a - x a - x a - x = 3 a - x a + x a - x [C1 ® (C1 + C 2 + C 3)] 3a - x a - x a + x 1 a-x a-x = ( 3 a - x) × 1 a + x a - x [taking ( 3a - x) common from C1 ] 1 a-x a+ x 1 a-x a-x = ( 3 a - x) × 0 2x 0 [R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1)] 0 0 2x 2x 0 [expanding by C1 ] = ( 3 a - x) × 1 × 0 2x = 4( 3 a - x) x 2. \ D = 0 Û 4( 3 a - x) x 2 = 0 Û x = 0 or x = 3 a. Hence, solution set = {0, 3 a}.

EXAMPLE 44

Solve

x - 2 2x - 3 x - 4 2x - 9 x - 8 2x - 27

3x - 4 3 x - 16 = 0. 3 x - 64

[CBSE 2011]

SSS Mathematics for Class 12 257

Determinants SOLUTION

257

Let the given determinant be D. Then, x - 2 2x - 3 3 x - 4 D = x - 4 2x - 9

3 x - 16

x - 8 2x - 27

3 x - 64

x-2

1

2

= x-4

-1

-4

[C 2 ® (C 2 - 2C1), C 3 ® (C 3 - 3C1)]

x - 8 -11 -40 x-2 = -2 -6

1

2

-2

-6

[R 2 ® ( R 2 - R1), R 3 ® ( R 3 - R1)]

-12 -42 x-2 1

= ( -2) × ( -6) × x-3 1 = 12 ×

0 -1

2

1

1

3

1

2 7 2

1 3 2 7

[C1 ® (C1 - C 2)]

= 12 × [( x - 3) (7 - 6) - 1 × ( 3 - 2)] = 12 × ( x - 4). \ D = 0 Û 12( x - 4) = 0 Û x = 4. Hence, solution set = {4}. EXAMPLE 45

a-x If a + b + c = 0 and c b

c b-x

b a

a

c-x

= 0 then show that

x = 0 or x = ( 3 2)( a 2 + b 2 + c2). SOLUTION

Let the given determinant be D. Then, a+ b+ c-x c b D = a+ b+ c-x b-x a [C1 ® (C1 + C 2 + C 3)] a+ b+ c-x a c-x 1 c = ( a + b + c - x) 1 b - x 1 a

b a c-x

1 c b = ( a + b + c - x) 0 b - x - c a-b 0 a-c c-x -b [R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1)]

SSS Mathematics for Class 12 258

258

Senior Secondary School Mathematics for Class 12

= ( a + b + c - x)[( b - x - c)( c - x - b) - ( a - c)( a - b)] = ( a + b + c - x)[x 2 - ( a 2 + b 2 + c2 - ab - bc - ca)]. Now, D = 0 Û ( a + b + c - x)[x 2 - ( a 2 + b 2 + c2 - ab - bc - ca)] = 0 Û x = a + b + c or x = ± ( a 2 + b 2 + c2 - ab - bc - ca) Û x = 0 or x = ± ( 3/2)( a 2 + b 2 + c2) [Q a + b + c = 0 Û ( a 2 + b 2 + c2) = -2( ab + bc + ca) Û ( ab + bc + ca) = - (1/2)( a 2 + b 2 + c2)].

EXERCISE 6A Very-Short-Answer Questions 1. If A is a 2 ´ 2 matrix such that|A| ¹ 0 and|A| = 5 , write the value of|4A|. 2. If A is a 3 ´ 3 matrix such that|A| ¹ 0 and|3A| = k|A|then write the value of k. [CBSE 2014] 3. Let A be a square matrix of order 3, write the value of|2A|, where|A|= 4. [CBSE 2012]

2 -3 4. If A ij is the cofactor of the element aij of 6 0 1

5

( a 32 A 32).

5 4 then write the value of -7 [CBSE 2013]

5. Evaluate

x2 - x + 1 x - 1 × x+1 x+1

[CBSE 2008]

6. Evaluate

a + ib - c + id

[CBSE 2008]

c + id × a - ib

7. If

3x 7 8 7 = , find the value of x. -2 4 6 4

[CBSE 2014]

8. If

2x 5 6 -2 = , write the value of x. 8 x 7 3

[CBSE 2014]

9. If

2x 2( x + 1)

x+ 3 1 = x+1 3

5 , find the value of x. 3

é3 10. If A = ê ë1

4ù , find the value of 3|A|. 2úû

11. Evaluate

2

7 -2 × -10 5

[CBSE 2013C]

[CBSE 2011C]

[CBSE 2009C]

SSS Mathematics for Class 12 259

Determinants

12. Evaluate

13. Evaluate

14. Evaluate 15. Evaluate 16. Evaluate 17. Evaluate

18. Evaluate

6

5

20

24

×

[CBSE 2009C]

2 cos q -2 sin q sin q

cos q

cos a

- sin a

sin a

cos a

sin 60°

×

[CBSE 2008]

×

cos 60°

- sin 30° cos 30° cos 65 ° sin 65 ° sin 25 ° cos 25 ° cos 15 ° sin 15 ° sin 75 ° cos 75 ° 0 2

2 3

4 5

259

×

× ×

[CBSE 2011]

0 4× 6

41 1 5 19. Without expanding the determinant, prove that 79 7 9 = 0 × 29 5 SINGULAR MATRIX

3

A square matrix A is said to be singular if|A| = 0.

Also, A is called nonsingular if|A| ¹ 0. é 3 - 2x x + 1ù 20. For what value of x, the given matrix A = ê is a singular 4 úû ë 2 matrix? [CBSE 2013C] 21. Evaluate

14 9 × -8 -7

22. Evaluate

3 - 5

5 × 3 3

ANSWERS (EXERCISE 6A)

1. 80

2. k = 27 3. 32

4. 110

7. x = -2 8. x = ±6 9. x = 1 10. 6 15. 1 16. 0 17. 0 18. 8

5. x 3 - x 2 + 2

6. ( a 2 + b 2 + c2 + d 2)

11. 30 12. 2 13. 2 20. x = 1 21. –26 22. 14

14. 1

SSS Mathematics for Class 12 260

260

Senior Secondary School Mathematics for Class 12 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 6A)

1. Since A is a 2 ´ 2 matrix, so|4 A| = ( 4 ´ 4 ) ×|A|. 2. Since A is a 3 ´ 3 matrix, so|3 A| = ( 3 ´ 3 ´ 3 ) ×|A|. 2 5 4. M 32 = = ( 8 - 30 ) = -22. 6 4 So, A 32 = ( -1) 3 + 2 × M 32 = ( -1)( -22 ) = 22. \ ( a32 A 32 ) = (5 ´ 22 ) = 110. 5. D = ( x 2 - x + 1)( x + 1) - ( x + 1)( x - 1) = ( x 3 + 1) - ( x 2 - 1) = ( x 3 - x 2 + 2 ). 6. D = ( a + ib )( a - ib ) - ( c + id )( - c + id ) = {a2 - (ib ) 2 } - {(id ) 2 - c 2 } = ( a2 + b 2 ) - ( - d 2 - c 2 ) = ( a2 + b 2 + c 2 + d 2 ). 7. ( 3 x ´ 4 ) - 7 ´ ( -2 ) = 32 - 42 Þ 12 x + 14 = -10 Þ 12 x = -24 Þ x = -2. 8. 2 x 2 - 40 = 18 + 14 Þ 2 x 2 = 32 + 40 = 72 Þ x 2 = 36 Þ x = ±6. 9. 2 x( x + 1) - 2( x + 1)( x + 3 ) = 3 - 15 Þ 2( x + 1){x - ( x + 3 )} = -12 Þ - 6( x + 1) = -12 Þ x + 1 = 2 Þ x = 1. 3 4 10. 3|A| = 3 × = 3 × ( 6 - 4 ) = 3 ´ 2 = 6. 1 2 11. 2 D = 2( 35 - 20 ) = ( 2 ´ 15 ) = 30. 12. D = ( 6 ´ 24 ) - ( 20 ´ 5 ) = { 6 ´ 24 - 20 ´ 5 } = { 144 - 100 } = ( 12 - 10 ) = 2. 13. D = ( 2 cos 2 q + 2 sin 2 q) = 2(cos 2 q + sin 2 q) = ( 2 ´ 1) = 2. 14. D = (cos 2 a + sin 2 a ) = 1. 15. D = (sin 60 ° cos 30 ° + cos 60 ° sin 30 ° ) = sin ( 60 ° + 30 ° ) = sin 90 ° = 1. 16. D = {cos 65 ° cos 25 ° - sin 65 ° sin 25 °} = cos ( 65 ° + 25 ° ) = cos 90 ° = 0. 2 4 18. D = ( -2 ) × = ( -2 )( 12 - 16 ) = ( -2 ) ´ ( -4 ) = 8. 4 6 19. Applying C 1 ® C 1 - 8C 3 , we get 1 1 5 D = 7 7 9 = 0 [Q C 1 and C 2 are identical]. 5 5 3 20. A is singular Û |A| = 0 3 - 2x x + 1 Û =0 2 4 Û 4( 3 - 2 x ) - 2( x + 1) = 0 Û 10 x = 10 Û x = 1.

EXERCISE 6B Evaluate: 67 19 21 1. 39 13 14 81 24 26

29 26 22 2. 25 31 27 63 54 46

SSS Mathematics for Class 12 261

Determinants

102 18 3.

1 17

36

3 3

4 6

261

12

22

32

2

2

42 52

4. 2 32

3 42

Using properties of determinants prove that: 1 1 1 5. a b c = ( a - b)( b - c)( c - a). bc ca ab

[CBSE 2006]

b 2 + c2

1 b+ c

6. 1 c + a c2 + a 2 = ( a - b)( b - c)( c - a). 1 a + b a 2 + b2 1+ p

1 7. 2 3 8.

1+ p + q

3 + 2p 1 + 3 p + 2q = 1. 6 + 3 p 1 + 6p + 3 q

a+x y x a+y x

z z

= a 2( a + x + y + z).

a x

a a = ( x + 2a)( x - a) 2.

a

a

x

x + 4 2x 2x x + 4 2x

11.

2x 2x

= (5 x + 4)( x - 4) 2.

[CBSE 2009, ’11]

x+4

2x

x + l 2x 2x x+l 2x

[CBSE 2014]

a+z

y

x 9. a

10.

[CBSE 2009]

2x 2x

= (5 x + l)( l - x) 2.

[CBSE 2014]

x+l

2x

a 2 + 2a 2a + 1 1 12. 2a + 1 a + 2 1 = ( a - 1) 3. 3

3

x 13. x + 2y

x+y x

x+y

x + 2y

3 x -x + y 14. x - y 3y x -z y -z

1 x + 2y x + y = 9y 2( x + y).

[CBSE 2013]

x -x + z z - y = 3( x + y + z)( xy + yz + zx). 3z

[CBSE 2013]

SSS Mathematics for Class 12 262

262

Senior Secondary School Mathematics for Class 12

x

y

2

2

z

y z 2 = xyz( x - y)( y - z)(z - x). y3 z3

15. x x3

b+ c a-b a 16. c + a b - c b = 3 abc - a 3 - b 3 - c 3. a+ b c-a c b+ c 17.

a

[CBSE 2006, ’12C]

a

c+ a b = 4abc. c a+b

b c

[CBSE 2010C, ’11]

[CBSE 2006, ’12, ’14C]

a a + 2b a + 2b + 3 c 18. 3 a 4a + 6b 5 a + 7 b + 9c = - a 3. 6a 9a + 12b 11a + 15 b + 18c a+ b+ c 19.

20.

-c -b

-c

-b

a+ b+ c -a = 2( a + b)( b + c)( c + a). -a a+ b+ c ax + by bx + cy = ( b 2 - ac)( ax 2 + 2bxy + cy 2).

a b

b c

ax + by

bx + cy

0

a2 b2 c2 2 2 21. ( a + 1) ( b + 1) ( c + 1) 2 = -4( a - b)( b - c)( c - a). ( a - 1) 2 ( b - 1) 2 ( c - 1) 2 ( x - 2) 2 ( x - 1) 2 22. ( x - 1) 2 x2 x2

x2 ( x + 1) 2 = -8.

( x + 1) 2 ( x + 2) 2

(m + n) 2 l 2 mn 23. (n + l) 2 m 2 ln = ( l 2 + m 2 + n2)( l - m)(m - n)(n - l). ( l + m) 2

n2

lm

( b + c) 2 24. ( c + a) 2 ( a + b) 2

a2 b2 c2

b 2 + c2 25. b2 c2

a2 c2 + a 2 c2

bc ca = ( a 2 + b 2 + c2)( a - b)( b - c)( c - a)( a + b + c). ab a2 = 4a 2 b 2 c2. b2 2 2 a +b

SSS Mathematics for Class 12 263

Determinants

1 + a 2 - b2

a

-2b

2ab

1 - a 2 + b2 2a = (1 + a 2 + b 2) 3. -2a 1 - a 2 - b2

2ab 2b

26.

263

b-c

[CBSE 2008, ’09, ’10C]

c+ b

27. a + c b c - a = ( a + b + c)( a 2 + b 2 + c2). a-b a+ b c b 2 c2 28. c2 a 2 a 2b2

bc b + c ca c + a = 0. ab a + b

( b + c) 2 29.

ab ac

ab

ca

( a + c) 2 bc = 2abc( a + b + c) 3. bc ( a + b) 2

[CBSE 2010]

b 2 - ab b - c bc - ac 30. ab - a 2 a - b b 2 - ab = 0. bc - ac c - a ab - a 2 - a( b 2 + c2 - a 2) 31. 2a 3 2a 3 x- 3 x-4 32. x - 2 x - 3 x -1

x-2

( a + 1)( a + 2) 33. ( a + 2)( a + 3) ( a + 3)( a + 4)

2b 3 - b( c2 + a 2 - b 2) 2b 3

2c 3 2c 3 2

2

= ( abc)( a 2 + b 2 + c2) 3. 2

- c( a + b + c )

x-a x - b = 0, where a , b , g are in AP. x-g a+2 1 a + 3 1 = -2. a+4 1

x 34. If x ¹ y ¹ z and y

x3 y3

x4 - 1 y 4 - 1 = 0, prove that

z

z3

z4 - 1

xyz( xy + yz + zx) = ( x + y + z). 35. Prove that 1 a 2 + bc a 3 1 b 2 + ca b 3 = - ( a - b)( b - c)( c - a)( a 2 + b 2 + c2). 1 c2 + ab c 3

[CBSE 2007]

SSS Mathematics for Class 12 264

264

Senior Secondary School Mathematics for Class 12

Without expanding the determinant, prove that: 1 a

bc

1 a a2

1 a a2

2

2

36. 1 b ca = 1 b b 1 c ab 1 c c2

1 bc b + c

37. 1 b b = 1 ca c + a 1 c c2 1 ab a + b -6

x 38. Show that x = 2 is a root of the equation 2 -3

-1

-3 x x - 3 = 0. 2x 2 + x

[CBSE 2007]

Solve the following equations: x3

1 x

3x - 8

3

x 43. 2 7

x+1

3

3 7 x 2 =0

44.

x 2 -3

c

3

5

x+2 5 =0 3 x+4

2 2

42.

6 x

45. Prove that a b-c a+c b

b

x+b c =0 b x+c

a b

40.

3x - 8 3 =0 3 3x - 8

3 3

41.

x+a

b3 = 0 c3

39. 1 b 1 c

-6 -1 -3 x x - 3 = 0 2x

x+2

c+ b c - a = ( a + b + c)( a 2 + b 2 + c2).

a-b b+ a

[CBSE 2012C]

c ANSWERS (EXERCISE 6B)

1. -43 2. 132 3. 0

4. –8 39. x = b or x = c or x = - ( b + c) 2 11 40. x = 0 or x = - ( a + b + c) 41. x = or x = 42. x = 1 or x = -9 3 3 43. x = -9 or x = 2 or x = 7 44. x = 1 or x = 2 or x = -3 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 6B) 1. Apply C 1 ® C 1 - 3C 2 , C 3 ® C 3 - C 2 . Now, apply C 2 ® C 2 - 13C 3 . Expand by R 2 . 2. Apply C 1 ® C 1 - C 2 and C 3 ® C 3 - C 2 . Take 3 common from C 1 and -4 common from C 3 . Apply C 2 ® C 2 - 26C 3 and C 3 ® C 3 - C 1 . 3. Apply R1 ® R1 - 6 R 3 . 4. Apply C 3 ® C 3 - C 2 and C 2 ® C 2 - C 1 . Now, apply C 3 ® C 3 - C 2 . 5. Apply C 1 ® C 1 - C 3 and C 2 ® C 2 - C 3 .

SSS Mathematics for Class 12 265

Determinants 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

265

Apply R 2 ® R 2 - R1 and R 3 ® R 3 - 3 R1 . Apply R 2 ® R 2 - 2 R1 and R 3 ® R 3 - 3 R1 . Apply C 1 ® C 1 + C 2 + C 3 and take ( a + x + y + z) common from C 1 . Apply C 1 ® C 1 + C 2 + C 3 and take ( x + 2 a) common from C 1 . Apply R1 ® R1 + R 2 + R 3 and take (5 x + 4 ) common from R1 . Apply R1 ® R1 + R 2 + R 3 and take (5x + l) common from R1 . Apply R1 ® R1 - R 3 , R 2 ® R 2 - R 3 and then expand along C 3 . Apply R1 ® R1 + R 2 + R 3 and take 3( x + y ) common from R1 . Apply C 1 ® C 1 + C 2 + C 3 and take ( x + y + z) common from C 1 . Take x, y and z common from C 1 , C 2 , C 3 respectively. Apply C 1 ® C 1 + C 3 and C 2 ® C 2 - C 3 . Apply R1 ® R1 - ( R 2 + R 3 ). Take ( -2 ) common from R1 . Now, apply R 2 ® R 2 - R1 and R 3 ® R 3 - R1 . Apply R 2 ® R 2 - 3 R1 and R 3 ® R 3 - 6 R1 . Apply R 2 ® R 2 + R1 and R 3 ® R 3 + R1 . Apply R 3 ® ( R 3 - xR1 - yR 2 ). Apply R 2 ® ( R 2 - R 3 ) and take 4 common from R 2 . Now, apply R 3 ® R 3 - R1 + 2 R 2 . Apply R1 ® R1 - R 3 and R 2 ® R 2 - R 3 . Then, apply R 2 ® R 2 + 2 R1 . Now, C 2 ® C 2 - C 1 and C 3 ® C 3 - C 1 .

23. Apply C 1 ® (C 1 + C 2 - 2C 3 ) and take ( l 2 + m 2 + n2 ) common from C 1 . 24. Apply C 1 ® (C 1 + C 2 - 2C 3 ). 25. Apply R1 ® R1 - ( R 2 + R 3 ) and expand by C 1 . 26. Apply C 1 ® C 1 - bC 3 and C 2 ® C 2 + aC 3 . Now, take ( 1 + a2 + b 2 ) common from each of C 1 and C 2 . 27. Apply C 1 ® aC 1 , C 2 ® bC 2 , C 3 ® cC 3 and divide the whole determinant by abc. Now, apply C 1 ® (C 1 + C 2 + C 3 ). 28. Taking bc , ca and ab common from R1 , R 2 and R 3 respectively, we get 1 1 bc 1 + b c 1 1 D = ( a2 b 2 c 2 ) × ca 1 + × c a 1 1 ab 1 + a b Now, R 2 ® R 2 - R1 and R 3 ® R 3 - R1 . 29. Apply R1 ® aR1 , R 2 ® bR 2 , R 3 ® cR 3 and divide D by abc. Take a, b, c common from C 1 , C 2 and C 3 respectively. b( b - a) b - c c ( b - a) 30. D = a( b - a) a - b b( b - a) c ( b - a) c - a a ( b - a) b b-c c = ( b - a) 2 × a a - b b c c-a a

[taking ( b - a) common from each of C 1 and C 3 ].

Now, apply C 2 ® (C 2 - C 1 + C 3 ).

SSS Mathematics for Class 12 266

266

Senior Secondary School Mathematics for Class 12

31. Take a, b, c common from C 1 , C 2 and C 3 respectively. Now, apply C 1 ® (C 1 + C 2 + C 3 ). 32. Apply R1 ® ( R1 + R 2 + R 3 ) and use a + g = 2b. Then, R1 and R 2 are proportional and hence D = 0. a2 + 3 a + 2 a + 2 1 33. D = a2 + 5 a + 6 a + 3 1 × a2 + 7 a + 12 a + 4 1 Now, apply R 2 ® R 2 - R1 and R 3 ® R 3 - R1 . 1 a2 35. D = 1 b 2 1 c2

a3 1 bc a3 b 3 + 1 ca b 3 × c3 1 ab c 3

36. Applying R1 ® aR1 , R 2 ® bR 2 and R 3 a a2 abc a a2 1 1 2 LHS = × b b abc = b b 2 1 = abc 2 c c abc c c2 1 37. Applying C 3 1 bc RHS = 1 ca 1 ab

® cR 3 , we get 1 a a2 1 b b2 × 1 c c2

® C 3 - ( a + b + c )C 1 , we get -a 1 a bc - b = 1 b ca -c 1 c ab

a a2 1 = × b b2 abc c c2

abc a a2 abc = b b 2 abc c c2

1 1 a a2 1 = 1 b b 2 = LHS. 1 1 c c2

38. Apply R1 ® ( R1 - R 2 ) and take ( x - 2 ) common from R1 . 39. Apply R1 ® R1 - R 2 and R 3 ® R 3 - R 2 . Take ( x - b ) common from R1 . 40. Apply C 1 ® C 1 + C 2 + C 3 and take ( x + a + b + c ) common from C 1 . 41. Apply C 1 ® C 1 + C 2 + C 3 and take ( 3 x - 2 ) common from C 1 . 42. Apply C 1 ® C 1 + C 2 + C 3 and take ( x + 9 ) common from C 1 . 43. Apply R1 ® R1 + R 2 + R 3 and take ( x + 9 ) common from R1 . 44. Apply R1 ® R1 - R 2 and take ( x - 2 ) common from R1 . 45. Apply C 1 ® aC 1 and divide D by a. a2 b-c c+ b 1 2 \ D = × a + ac b c-a a 2 a - ab b + a c a2 + b 2 + c 2 1 2 = a + b 2 + c2 a 2 a + b 2 + c2

b-c c+ b b c-a b+ a c

[applying C 1 ® C 1 + bC 2 + cC 3 ]

SSS Mathematics for Class 12 267

Determinants

267

Applications of Determinants AREA OF A TRIANGLE IN DETERMINANT FORM We know that the area of a triangle whose vertices are ( x1 , y1), ( x 2 , y 2) and ( x 3 , y 3) is given by

1 C = [x1( y 2 - y 3) - x 2( y1 - y 3) + x 3( y1 - y 2)] 2 x1 1 = x2 2 x3

y1

1

y2 1 y3 1

(in determinant form).

REMARK 1.

Since area is a positive quantity, we always take the absolute value of the above determinant for the area.

REMARK 2.

If three points A , B , C are collinear then ar(C ABC) = 0.

CONDITION FOR COLLINEARITY OF THREE POINTS

Let A( x1 , y1), B( x 2 , y 2) and C( x 3 , y 3) be three given points. Then, A , B , C are collinear Û ar(C ABC) = 0 Û

x1 1 x2 2 x3

y1 1 x1 y2 1 = 0 Û x2 y3 1 x3

y1 1 y2 1 = 0 y3 1

Û x1( y 2 - y 3) + x 2( y 3 - y1) + x 3( y1 - y 2) = 0. \ A , B , C are collinear Û x1( y 2 - y 3) + x 2( y 3 - y1) + x 3( y1 - y 2) = 0. SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Find the area of a triangle whose vertices are A( -2, - 3), B( 3 , 2) and C( -1, -8). Here, ( x1 = -2, y1 = -3), ( x 2 = 3 , y 2 = 2) and ( x 3 = -1, y 3 = -8). \

x1 1 x2 2 x3

y1 1 -2 1 y2 1 = × 3 2 y3 1 -1 -2 1 = × 5 2 1

-3 2 -8 -3 5 -5

1 1 1 1 0 0

[R 2 ® ( R 2 - R1) and R 3 ® ( R 3 - R1)] =

5 5 1 1 ×1× = ´ ( -25 - 5) = -15. 1 -5 2 2

Hence, ar(C ABC) = |-15| = 15 square units.

SSS Mathematics for Class 12 268

268

Senior Secondary School Mathematics for Class 12

EXAMPLE 2

Show that the points A( a , b + c), B( b , c + a) and C( c, a + b) are collinear.

SOLUTION

Points A , B , C are collinear Û ar(C ABC) = 0. a b+ c 1 1 Now, ar(C ABC) = × b c + a 1 2 c a+b 1 =

a a+ b+ c 1 1 × b a+ b+ c 1 2 c a+ b+ c 1

[C 2 ® (C 2 + C1)]

a 1 1 1 = ( a + b + c) × b 1 1 2 c 1 1 1 = ( a + b + c) ´ 0 = 0 [B C 2 and C 3 are identical]. 2 Hence, the given points are collinear. EXAMPLE 3

If the points ( a , b), ( a ¢ , b ¢) and ( a - a ¢ , b - b ¢) are collinear, show that ab ¢ = a ¢ b.

SOLUTION

The given points are collinear a b 1 Û a¢ b¢ 1 =0 a - a ¢ b - b¢ 1 a b 1 Û a ¢ - a b ¢ - b 0 = 0 [R 2 ® R 2 - R1 and R 3 ® R 3 - R1] -a ¢ -b¢ 0 Û 1 × [( a ¢ - a)( - b ¢) + a ¢( b ¢ - b)] = 0 [expanding by C 3] Û ab ¢ - a ¢ b = 0 Û ab ¢ = a ¢ b.

EXAMPLE 4

Find the value of k in order that the points (5 , 5), ( k , 1) and (10, 7) are collinear.

SOLUTION

The given points are collinear 5 5 1 Û k 1 1 =0 10 7 1 5 Û k -5 5

5 1 -4 0 = 0 [R 2 ® R 2 - R1 and R 3 ® R 3 - R1] 2 0

SSS Mathematics for Class 12 269

Determinants

269

Û 1 × [2( k - 5) + 20] = 0 Û 2k + 10 = 0 Û k = -5. Hence, k = -5. EXAMPLE 5

SOLUTION

Let A(1, 3), B( 0, 0) and C( k , 0) be three ar(C ABC) = 3 sq units. Find the value of k. We have ar(C ABC) = 3 sq units 1 3 1 1 3 1 1 Û × 0 0 1 = ±3 Û 0 0 1 = ±6 2 k 0 1 k 0 1 Û ( -1) ×

EXAMPLE 6

SOLUTION

1 k

points

such

that

3 = ±6 Û 3 k = ±6 Û k = ±2. 0

Hence, k = ±2. Find the equation of the line joining the points A(1, 2) and B( 3 , 6), using determinants. Let P( x , y) be a point on AB. Then, the points A , P and B are collinear. \ ar(C APB) = 0 1 2 1 1 2 1 1 Þ × x y 1 =0 Þ x y 1 =0 2 3 6 1 3 6 1 1 2 1 Þ x y 1 = 0 [R 3 ® R 3 - 3 R1] 0 0 -2 Þ ( -2) ×

1

2

x

y

= 0 Þ ( y - 2x) = 0 Þ y = 2x.

Hence, the required equation is y = 2x.

EXERCISE 6C 1. Find the area of the triangle whose vertices are: (i) A( 3 , 8), B( -4, 2) and C(5 , -1) (ii) A( -2, 4), B( 2, -6) and C(5 , 4) (iii) A( -8, -2), B( -4, -6) and C( -1, 5) (iv) P( 0, 0), Q( 6, 0) and R( 4, 3) (v) P(1, 1), Q( 2, 7) and R(10, 8)

[CBSE 2007]

2. Use determinants to show that the following points are collinear. (i) A( 2, 3), B( -1, -2) and C(5 , 8)

SSS Mathematics for Class 12 270

270

Senior Secondary School Mathematics for Class 12

(ii) A( 3 , 8), B( -4, 2) and C(10, 14) (iii) P( -2, 5), Q( -6, - 7) and R( -5 , -4) 3. Find the value of k for which the points A( 3 , -2), B( k , 2) and C(8, 8) are collinear. 4. Find the value of k for which the points P(5 , 5), Q( k , 1) and R(11, 7) are collinear. 5. Find the value of k for which the points A(1, -1), B( 2, k) and C(4, 5) are collinear. 6. Find the value of k for which the area of a ABC having vertices A( 2, - 6), B(5 , 4) and C( k , 4) is 35 sq units. 7. If A( -2, 0), B( 0, 4) and C( 0, k) be three points such that area of a ABC is 4 sq units, find the value of k. 1 1 8. If the points A( a , 0), B( 0, b) and C(1, 1) are collinear, prove that + = 1. a b ANSWERS (EXERCISE 6C)

1. (i) 37.5 sq units

(ii) 35 sq units

(iii) 28 sq units

(iv) 9 sq units

(v) 23.5 sq units 3. k = 5 4. k = -7 5. k = 1 6. k = -2 or k = 12 7. k = 0 or k = 8

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: cos 70° sin 20° 1. =? sin 70° cos 20° (a) 1 (b) 0 cos 15 ° sin 15 ° 2. =? sin 15 ° cos 15 ° (a) 1 3.

(b)

1 2

(c) cos 50°

(c)

3 2

(d) sin 50°

(d) none of these

sin 23 ° - sin 7 ° =? cos 23 ° cos 7 °

1 3 (b) 2 2 a + ib c + id 4. =? - c + id a - ib (a)

(c) sin 16°

(d) cos 16°

(a) ( a 2 + b 2 - c2 - d 2)

(b) ( a 2 - b 2 + c2 - d 2)

(c) ( a 2 + b 2 + c2 + d 2)

(d) none of these

SSS Mathematics for Class 12 271

Determinants

w

1 5. If w is a complex root of unity then w w2 (a) 1

(b) –1

271

2

w 1

(c) 0

w2 1 =? w (d) none of these 1 w 1+ w

6. If w is a complex cube root of unity then the value of 1 + w 1 w 1+ w (a) 2 2

2

1 7. 22 32

(c) 0

(d) –3

(b) –8

(c) 16

(d) 142

(c) 24

(d) 120

is

2

2

3

32 42

42 = ? 52

(a) 8 1! 2!

(b) 4

w 1

3!

8. 2 ! 3 ! 4 ! = ? 3 ! 4! 5 ! (a) 2 (b) 6 a-b b-c c-a 9. b - c c - a a - b = ? c-a

(a) ( a + b + c) (b) 3( a + b + c) 1 1+ p 1+ p + q 10. 2 3 + 2p 1 + 3 p + 2q = ? 3

[CBSE 2009]

a-b b-c (c) 3abc

(d) 0 [CBSE 2009]

6 + 3 p 1 + 6p + 3 q

(a) 0 1 1 11. a b a3 b3

(b) 1

(d) none of these

1 c =? c3

[CBSE 2009]

(a) ( a - b)( b - c)( c - a) (c) ( a - b)( b - c)( c - a)( a + b + c) sin a cos a sin ( a + d) 12. sin b cos b sin (b + d) = ? sin g

(c) –1

cos g

(b) - ( a - b)( b - c)( c - a) (d) abc( a - b)( b - c)( c - a)

sin ( g + d)

(a) 0 (b) 1 (c) sin ( a + d) + sin (b + d) + sin ( g + d) (d) none of these

SSS Mathematics for Class 12 272

272

Senior Secondary School Mathematics for Class 12

a b c 13. If a, b, c be distinct positive real numbers then the value of b c a is c a b (a) positive

(b) negative

(c) a perfect square x+y x x

(d) 0

14. 5 x + 4y 10x + 8y

4x 8x

2x = ? 3x (b) x 3

(a) 0

(c) y 3

(d) none of these

(c) ( a - 1) 3

(d) none of these

(c) 0

(d) none of these

2

a + 2a 2a + 1 1 15.

2a + 1 3

a+2 3

1 =? 1 (b) ( a - 1) 2

(a) ( a - 1) a 16. 3 a 6a

a + 2b 4a + 6b

a + 2b + 3 c 5 a + 7 b + 9c

=?

9a + 12b 11a + 15 b + 18c 3

(b) -a 3

(a) a b+ c a b 17. c + a c a = ? a+b b c (a) ( a + b + c)( a - c) (c) ( a + b + c)( a - c) 1 1 18. 1 1 + x 1

1

1 1

(b) ( a + b + c)( b - c) 2

(d) ( a + b + c)( b - c) 2

=?

1+ y

(a) ( x + y) bc b + c 1 19. ca c + a 1 = ?

(b) ( x - y)

(c) xy

(d) none of these

ab a + b 1 (a) ( a - b)( b - c)( c - a) (c) ( a + b)( b + c)( c + a) b+ c a a 20. b c+ a b =? c c a+b (a) 4abc

(b) 2( a + b + c)

(b) - ( a - b)( b - c)( c - a) (d) none of these

(c) ( ab + bc + ca)

(d) none of these

SSS Mathematics for Class 12 273

Determinants

273

a 1 b+ c 21. b 1 c + a = ? c 1 a+b (a) a + b + c x+1

x+2

(b) 2( a + b + c)

(c) 4abc

(d) a 2 b 2 c2

(c) x 2 - 2

(d) x 2 + 2

(c) –6

(d) 9

x+4

22. x + 3 x + 5 x + 8 = ? x + 7 x + 10 x + 14 (a) -2 5 23. If -7 9

(b) 2 3 -1 x 2 = 0 then x = ? 6 -2

(a) 0

(b) 6

x 24. The solution set of the equation 2 7 (a) {2, - 3 , 7}

3 7 x 2 = 0 is 6 x

(b) {2, 7, –9}

(c) [–2, 3, –7]

x-2 25. The solution set of the equation x - 4

2x - 3 2x - 9

3x - 4 3 x - 16 = 0 is

x - 8 2x - 27 (a) {4}

(b) {2, 4}

(d) none of these

3 x - 64

(c) {2, 8}

(d) {4, 8}

a+x 26. The solution set of the equation a - x

a-x a+x

a-x a - x = 0 is

a-x

a-x

a+x

(a) {a , 0}

(b) { 3 a , 0}

(c) {a , 3 a}

3x - 8 27. The solution set of the equation 3 3 ì2 8ü (a) í , ý î3 3þ

ì 2 11 ü (b) í , ý î3 3 þ

(d) none of these

3 3x - 8

3 3

3

3x - 8

ì3 8ü (c) í , ý î2 3þ

= 0 is

(d) none of these

28. The vertices of a a ABC are A( -2, 4), B( 2, -6) andC(5 , 4). The area of a ABC is (a) 17.5 sq units

(b) 35 sq units

(c) 32 sq units

(d) 28 sq units

29. If the points A( 3 , -2), B( k , 2) and C( 8, 8) are collinear then the value of k is (a) 2

(b) –3

(c) 5

(d) –4

SSS Mathematics for Class 12 274

274

Senior Secondary School Mathematics for Class 12 ANSWERS (OBJECTIVE QUESTIONS)

1. (b) 2. (c) 10. (b) 11. (c) 19. (a) 20. (a) 28. (b) 29. (c)

3. (b) 4. (c) 5. (c) 6. (b) 7. (b) 8. (c) 9. (d) 12. (a) 13. (b) 14. (b) 15. (c) 16. (b) 17. (c) 18. (c) 21. (c) 22. (a) 23. (c) 24. (b) 25. (a) 26. (b) 27. (b)

HINTS TO SOME SELECTED OBJECTIVE QUESTIONS 2. D = (cos 2 15 ° - sin 2 15 ° ) = cos ( 2 ´ 15 ° ) = cos 30 ° =

3 × 2

4. D = ( a + ib )( a - ib ) + ( c - id )( c + id ) = ( a2 + b 2 + c 2 + d 2 ). 5. Apply R1 ® ( R1 + R 2 + R 3 ) and use the result ( 1 + w + w 2 ) = 0. 6. 1 + w + w 2 = 0 Þ ( 1 + w ) = - w 2 . Put ( 1 + w ) = - w 2 and expand. 9. R1 ® ( R1 + R 2 + R 3 ) gives all zeros in R1 and hence D = 0. 10. Applying R 2 ® ( R 2 - 2 R1 ) and R 3 ® ( R 3 - 3 R1 ) and expanding by C 1 , we get D = 1. 12. C 3 ® C 3 + (sin d )C 1 - (cos d )C 2 makes all zeros in C 3 and hence D = 0. 13. C 1 ® (C 1 + C 2 + C 3 ) and take ( a + b + c ) common from C 1 . Simplify to get D = - ( a + b + c )( a2 + b 2 + c 2 - ab - bc - ca) 1 = - ( a + b + c ){( a - b ) 2 + ( b - c ) 2 + ( c - a) 2 }, which is negative. 2 14. Take x common from each of C 2 and C 3 . Apply R 3 ® R 3 - 2 R 2 . 15. Apply R1 ® ( R1 - R 2 ) and R 3 ® ( R 3 - R 2 ). Expand by C 3 . 16. Apply R 2 ® R 2 - 3 R1 and R 3 ® R 3 - 6 R1 . b a b c a b 17. D = c c a + a c a a b c b b c Apply R1 ® R1 + R 2 + R 3 in each and simplify. bc b 1 bc c 1 19. D = ca c 1 + ca a 1 × ab a 1 ab b 1 20. Apply R1 ® R1 - ( R 2 + R 3 ). Take ( -2 ) common from R1 . Apply R 2 ® ( R 2 - R1 ) and R 3 ® ( R 3 - R1 ). 21. R 2 ® R 2 - R1 and R 3 ® R 3 - R1 . Take ( a - b ) common from R 2 and ( c - a) common from R 3 . 22. Apply C 2 ® (C 2 - C 1 ) and C 3 ® (C 3 - C 1 ). Now, apply R 2 ® R 2 - R1 and R 3 ® R 3 - R1 . 23. Apply C 1 ® (C 1 - C 3 ) and take ( x - 7 ) common from C 1 . 25. Apply C 2 ® (C 2 - 2C 1 ) and C 3 ® (C 3 - 3C 1 ). Now, apply R 2 ® ( R 2 - R1 ) and R 3 ® ( R 3 - R1 ). 26. Apply C 1 ® (C 1 + C 2 + C 3 ) and take ( 3a - x ) common from C 1 .

SSS Mathematics for Class 12 275

Determinants 27. Apply C 1 ® (C 1 + C 2 + C 3 ) and take ( 3 x - 2 ) common from C 1 . -2 4 1 -2 4 1 1 1 28. D = 2 -6 1 = 4 -10 0 = 35 sq units. 2 2 5 4 1 7 0 0 3 -2 1 29. If D = k 2 1 then we must have D = 0. 8 8 1

275

SSS Mathematics for Class 12 276

7. ADJOINT AND INVERSE OF A MATRIX ADJOINT OF A MATRIX Let A = [aij ] be a square matrix of order n and let A ij denote the cofactor of aij in|A|. Then, the adjoint of A , denoted by adj A , is defined as

adj A = [A ji ]n ´ n. Thus adj A is the transpose of the matrix of the corresponding cofactors of elements of|A|. If

é a11 A = ê a21 ê êë a 31

é A11 adj A = ê A 21 ê êë A 31

a12 a22 a 32 A12 A 22 A 32

a13 ù a23 ú , then ú a 33 úû A13 ù ¢ é A11 A 23 ú = ê A12 ú ê A 33 úû êë A13

A 21 A 22 A 23

A 31 A 32 A 33

ù ú, ú úû

where A ij denotes the cofactor of aij in|A|. é 2 A=ê ë 1

5 ù , find adj A. 3 úû

EXAMPLE 1

If

SOLUTION

½2 Clearly, |A| = ½ ½1

5½ ½. 3½

The cofactors of the elements of|A|are given by A11 = 3 , A12 = - 1; A 21 = - 5 , A 22 = 2. \

-1ù ¢ é 3 =ê 2úû ë -1

é 3 adj A = ê ë -5 é 1 A=ê 0 ê êë -4

-2 2 5

-5 2

4 ù 1 ú , find adj A. ú 3 úû

EXAMPLE 2

If

SOLUTION

½ 1 We have, |A| = ½ 0 ½ ½ -4

-2 2 5 276

4½ 1 ½. ½ 3½

ù ú. û

SSS Mathematics for Class 12 277

Adjoint and Inverse of a Matrix

277

The cofactors of the elements of|A|are given by ½2 A11 = ½ ½5

½ -2 A 21 = -½ ½ 5 ½ -2 A 31 = ½ ½ 2

2½ ½ = 8; 5½

4½ 1 4½ 1 -2 ½ ½ ½ = 3; ½ ½ = 19; A 23 = -½ ½ = 26; A 22 = ½ 3½ ½- 4 5½ ½- 4 3½ 4½ 1 ½ ½ = - 10; A 32 = -½ 1½ ½0

1 é \ adj A = ê 26 ê êë -10

-4 19 -1

8 ù¢ 3 ú = ú 2 úû

4½ 1 ½ = - 1; A 33 = ½ ½ 1½ ½0

é 1 ê -4 ê ëê 8

26 19 3

-2 ½ ½ = 2. 2½

-10 ù -1 ú . ú 2 úû

If A is a square matrix of order n then prove that

THEOREM 1

PROOF

1½ 0 ½ ½ = - 4; A13 = ½ 3½ ½- 4

1½ 0 ½ ½ = 1; A12 = -½ 3½ ½- 4

Let

é a11 êa ê 21 ê¼ ê¼ A=ê ê ai 1 ê¼ ê ê¼ êa ë n1

A × ( adj A) = ( adj A) × A = |A|× I. a12 ¼ a1n ù a22 ¼ a2n ú ú ¼ ¼ ¼ú ¼ ¼ ¼ú ú . Then, ai 2 ¼ ain ú ¼ ¼ ¼ú ú ¼ ¼ ¼ú an2 ¼ ann úû

é A11 êA ê 21 ê ¼ adj A = ê ¼ ê ê A k1 ê ¼ ê êë A n1

A12 A 22 ¼ ¼ Ak2 ¼ A n2

¢

¼ A1n ù ¼ A 2n ú ú é A11 A 21 ¼ ¼ú êA 12 A 22 ¼ ¼ ú =ê ú ¼ ¼ ê ¼ A kn ú êA ë 1n A 2n ¼ ¼ú ú ¼ A nn úû

¼ A k1 ¼ A n1 ù ¼ A k 2 ¼ A n2 ú ú. ¼ ¼ ¼ ¼ú ¼ A kn ¼ A nn úû

Now, the (i , k)th element of A × ( adj A) = ai 1 A k1 + ai 2 A k 2 + ¼ + ainA kn ì|A|, when i = k =í when i ¹ k. î 0, This shows that each diagonal element of A ×( adj A) is |A| and each one of its nondiagonal elements is 0.

SSS Mathematics for Class 12 278

278

Senior Secondary School Mathematics for Class 12

\

¼

0 ù 0 ú ¼ ú ¼ ¼ ú ¼ |A| úû

0 0 é |A| ê 0 |A| 0 A( adj A) = ê ¼ ¼ ê ¼ ê 0 0 0 ë é ê = |A| ê ê ê ë

1

0

0

¼

0

1

0

¼

¼ 0

¼ 0

¼ 0

¼ ¼

0 ù 0 ú ú = |A|× I. ¼ú 1 úû

Thus, A( adj A) = |A|× I. Similarly, ( adj A) A = |A|× I. Hence, A( adj A) = ( adj A) A = |A|× I. SUMMARY

For every square matrix A, we have A × ( adj A) = ( adj A) × A = |A|× I. EXAMPLE 3

If

é 3 1ù A=ê ú , verify that ë 7 5 û A × ( adj A) = ( adj A) × A = |A|× I.

SOLUTION

We have é 3 1ù A=ê ú = (15 - 7) = 8 ¹ 0. ë 7 5 û The cofactors of the elements of|A|are given by A11 = 5 , A12 = - 7; A 21 = - 1, A 22 = 3. é 5 -7 ù ¢ é 5 -1ù \ (adj A) = ê × ú = ê -7 3 úû ë -1 3 û ë é 3 1 ù é 5 -1 ù \ A × ( adj A) = ê ú ê ú ë 7 5 û ë -7 3 û -3 + 3 ù é 8 0 ù é 15 - 7 =ê ú=ê ú ë 35 - 35 -7 + 15 û ë 0 8 û é 1 0ù = 8× ê [Q |A| = 8]. ú = 8I = |A|× I ë 0 1û é 5 -1 ù é 3 1 ù And, ( adj A) × A = ê ú ê ú ë -7 3 û ë 7 5 û 15 - 7 5 -5 ù é 8 0ù é =ê ú=ê ú ë -21 + 21 -7 + 15 û ë 0 8 û

SSS Mathematics for Class 12 279

Adjoint and Inverse of a Matrix

é 1 0ù = 8× ê ú = 8I = |A|× I ë 0 1û

279

[Q |A| = 8].

Hence, A × ( adj A) = ( adj A) × A = |A|× I. EXAMPLE 4

é 1 If A = ê 3 ê êë 0

0 4 -6

-1 ù 5 ú , verify that ú -7 úû

A × ( adj A) = ( adj A) × A = |A|× I. SOLUTION

We have ½1 |A| = ½ 3 ½ ½0

0 4 -6

-1 ½ 5 ½ = [1 × ( -28 + 30) - 1 × ( -18 - 0)] = 20. ½ -7 ½

Now, the cofactors of the elements of|A|are given by 3 5½ 3 4½ ½ 4 5½ ½ ½ = 21, A13 = ½ ½ ½ = - 18; ½ = 2, A12 = -½ A11 = ½ ½ 0 - 6½ ½ 0 -7 ½ ½ - 6 -7 ½ 1 -1 ½ 1 0½ ½ 0 -1 ½ ½ ½ = 6; ½ ½ = - 7 , A 23 = -½ ½ = 6, A 22 = ½ A 21 = - ½ ½ 0 - 6½ ½ 0 -7 ½ ½ - 6 -7 ½ 1 -1 ½ 1 0½ ½ 0 -1 ½ ½ ½ = 4. ½ ½ = - 8, A 33 = ½ ½ = 4, A 32 = - ½ A 31 = ½ ½ 3 4½ ½ 3 5½ ½4 5½ 6 4ù é 2 é 2 21 -18ù ¢ 6ú = ê 21 -7 -8ú . \ adj A = ê 6 -7 ú ú ê ê 4úû 6 4úû êë -18 êë 4 -8 0 -1 ù é 2 6 4ù é 1 So, A × ( adj A) = ê 3 4 5 ú ê 21 -7 -8 ú ú ê ê ú 6 4 úû êë 0 - 6 -7 úû êë -18 0 0ù é 1 0 0½ é 20 0 ú = 20 × ê 0 1 0 ½ = |A|× I. = ê 0 20 ê ½ ú ê 0 20 úû êë 0 0 1 ½ êë 0 Thus, A × ( adj A) = |A|× I. 2 6 4ù é 1 0 -1 ù é ê ú ê Further, ( adj A) × A = 21 -7 -8 × 3 4 5 ú ú ê ê ú 6 4 úû êë 0 - 6 -7 úû êë -18 é 1 0 0½ é 20 0 0 ù = ê 0 20 0 ú = 20 × ê 0 1 0 ½ = |A|× I. ê ½ ú ê êë 0 0 1 ½ êë 0 0 20 úû Thus, ( adj A) × A = |A|× I. Hence, A × ( adj A) = ( adj A) × A = |A|× I.

SSS Mathematics for Class 12 280

280

Senior Secondary School Mathematics for Class 12

SINGULAR AND NONSINGULAR MATRICES

(i) singular if|A| = 0, Examples

A square matrix A is said to be (ii) nonsingular if|A| ¹ 0.

é 1 2ù A=ê ú × Then, ë 4 8û ½1 2½ ½ = ( 8 - 8) = 0 and therefore, A is singular. |A| = ½ ½ 4 8½ é 1 2ù (ii) Let B = ê ú × Then, ë 3 4û ½1 2½ ½ = ( 4 - 6) = - 2 ¹ 0, and |B| = ½ ½ 3 4½ therefore, B is nonsingular. (i) Let

INVERTIBLE MATRIX A nonzero square matrix A of order n is said to be invertible if there exists a square matrix B of order n such that AB = BA = I.

We say that the inverse of A is B and we write, A -1 = B. RESULTS ON INVERTIBLE MATRICES THEOREM 2 PROOF

THEOREM 3

PROOF

An invertible matrix possesses a unique inverse.

Let A be an invertible square of order n. If possible, let B as well as C be the inverse of A. Then, AB = BA = I and AC = CA = I. Now, AC = I Þ B( AC) = B × I = B. And, BA = I Þ ( BA)C = I × C = C. But B( AC ) = ( BA)C [by associative law of multiplication]. \ B = C. Hence, an invertible has a unique inverse. A square matrix A is invertible if and only if A is nonsingular, i.e., A is invertible Û |A| ¹ 0.

Let A be an invertible square matrix of order n. Then, there exists a square matrix B of order n such that AB = BA = I. Now, AB = I Þ |AB| = |I| Þ |A|| × B| = 1 [Q |AB| = |A|| × B| and |I| = 1 ] Þ |A| ¹ 0 [Q ab = 1 Þ a ¹ 0 and b ¹ 0 ]. This shows that A is nonsingular. Conversely, Let A be nonsingular. Then,|A| ¹ 0. \ A × ( adj A) = ( adj A) × A = |A|× I and |A| ¹ 0 æ 1 ö æ 1 ö Þ A × çç × adj A ÷÷ = çç × adj A ÷÷ × A = I è|A| ø è|A| ø

SSS Mathematics for Class 12 281

Adjoint and Inverse of a Matrix

Þ A -1 =

281

1 × adj A. |A|

This shows that A is invertible. Hence, A is invertible Û A is nonsingular. FORMULA FOR FINDING A -1

Let A be a square matrix such that|A| ¹ 0. 1 Then, A -1 = × ( adj A). |A| EXAMPLE 5

é 2 Find the inverse of the matrix, A = ê ë -4

SOLUTION

We have ½ 2 |A| = ½ ½- 4

-3 ù . 7 úû

-3 ½ ½ = (14 - 12) = 2 ¹ 0. 7½

So, A -1 exists. The cofactors of the elements of|A|are given by A11 = 7 , A12 = - ( - 4) = 4; A 21 = - ( -3) = 3 , A 22 = 2. ¢ é 7 4ù é7 3 ù \ ( adj A) = ê ú =ê 4 2ú× 3 2 ë û ë û 1 -1 Hence, A = × ( adj A) |A| =

EXAMPLE 6

SOLUTION

1 é7 × 2 êë 4

é 7 3 ù ê = 2 ú 2û ê ë 2

é 3 Find the inverse of the matrix ê -2 ê êë 2 é 3 Let ê -2 ê êë 2

-10 8 -4

½3 |A| = ½-2 ½ ½2

-1 ù 2 ú . Then, ú -2 úû -10 -1 ½ 8 2½ = ½ - 4 -2 ½

½ 0 ½ 4 ½ ½- 4

3 2 1

ù ú× ú û

-10 8 -4

0 -12 16

-1 2 -2

ù ú. ú úû

-1 ½ 2½ ½ -2 ½

[C1 ® C1 + 3C 3 and C 2 ® C 2 - 10C 3] = ( -1) × ( 64 - 48) = - 16 ¹ 0.

SSS Mathematics for Class 12 282

282

Senior Secondary School Mathematics for Class 12

Thus,|A| ¹ 0 and therefore, A -1 exists. Now, the cofactors of the elements of|A|are given by -2 -2 2 ½ 8½ ½ 8 2½ ½ ½ = - 8; ½ ½ = 0, A13 = ½ ½ = - 8, A12 = -½ A11 = ½ ½ 2 - 4½ ½ 2 -2 ½ ½ - 4 -2 ½ 3 -10½ 3 -1½ ½-10 -1 ½ ½ ½ = - 8; ½ ½ = - 4, A 23 = -½ ½ = - 16, A 22 = ½ A 21 = -½ 4 2 2 2 ½ 2 - 4½ ½ ½ ½ ½ 3 -1½ 3 -10½ ½-10 -1½ ½ ½ = 4. ½ ½ = - 4, A 33 = ½ ½ = - 12, A 32 = - ½ A 31 = ½ 8½ 8 2 2 2 ½ 2 ½ ½ ½ ½ ¢ 0 -8 ù é -8 -16 -12 ù é -8 ú ê -4 -4 ú. \ ( adj A) = -16 - 4 -8 = ê 0 ú ú ê ê 4 úû -8 4 úû êë -8 êë -12 - 4 1 Hence, A -1 = × ( adj A) |A| 3 ù é 1 1 ê 2 4 ú é -8 -16 -12 ù ú ê 1 ê 1 1 ú . 0 = × -4 -4 ú = ê 0 ú ê -16 ê 4 4 ú 8 8 4 úû êë ê 1 1 ú 1 - ú ê 4 û 2 ë 2

Some Results on Invertible Matrices Let A, B, C be three square matrices, each of order n such that AB = AC. If A is nonsingular then B = C. Let A , B , C be square matrices, each of order n such that AB = AC and A is nonsingular. Then,|A| ¹ 0 and therefore, A -1 exists. \ AB = AC Þ A -1( AB) = A -1( AC) Þ ( A -1 A) B = ( A -1 A)C Þ IB = IC Þ B = C. Thus, AB = AC and A is nonsingular Þ B = C.

THEOREM 1 (Cancellation law) PROOF

REMARK

If AB = AC and|A| = 0 then B and C are not necessarily equal.

Example

Let

é 1 A=ê ë 3

½1 Then,|A| = ½ ½3

2 6

ù é 2 ú, B=ê 0 û ë

0ù é 0 and C = ê ú 0û ë 1

2½ ½ = ( 6 - 6) = 0. 6½

0 0

ù ú. û

SSS Mathematics for Class 12 283

Adjoint and Inverse of a Matrix

283

é 1 Here, AB = ê ë 3

2ù é 2 6 úû êë 0

0ù é 2 = ê ú 0û ë 6

0ù 0 úû

é 1 AC = ê ë 3

2ù é 0 6 úû êë 1

0ù é 2 = ê 0 úû ë 6

0ù . 0 úû

and

Thus, AB = AC but B ¹ C. THEOREM 2

PROOF

(Reversal law) If A and B are invertible square matrices of the same order then AB is also invertible and ( AB) -1 = B -1 A -1.

Let A and B be two invertible square matrices, each of order n. Then,|A| ¹ 0 and |B| ¹ 0. \ |AB| = |A|| × B| ¹ 0. Thus, AB is invertible. Now,

( AB) ( B -1 A -1) = A( BB -1) A -1

And, ( B

-1

= ( AI ) A -1

[Q BB -1 = I ]

= AA - 1 = I

[Q AI = A ].

-1

A ) ( AB) = B

-1

=B

-1

(A

-1

A) B

( IB)

= B -1 B = I \ ( AB) ( B

-1

Hence, ( AB) THEOREM 3

PROOF

-1

A ) = (B

-1

=B

-1

[by associativity]

-1

[by associativity] [Q A - 1 A = I ] [Q IB = B ].

-1

A ) ( AB) = I.

-1

A .

If A is an invertible square matrix then prove that A ¢ is also invertible and ( A ¢) -1 = ( A -1) ¢.

Let A be an invertible square matrix of order n. Then, |A| ¹ 0. \ |A ¢| = |A| ¹ 0. This shows that A ¢ is invertible. Now,

AA -1 = A -1 A = I

Þ ( AA -1) ¢ = ( A -1 A) ¢ = I ¢ Þ ( A -1 ) ¢ × A ¢ = A ¢ × ( A -1 ) ¢ = I

[Q ( AB) ¢ = B ¢ A ¢ and I ¢ = I]

Þ ( A ¢) - 1 = ( A - 1 ) ¢

[Q AB = BA = I Þ B -1 = A].

Hence, ( A ¢) -1 = ( A -1) ¢. THEOREM 4 PROOF

If A is an invertible symmetric then prove that A -1 is also symmetric.

Let A be an invertible symmetric matrix. Then, A ¢ = A. We know that ( A -1) ¢ = ( A ¢) -1 .

SSS Mathematics for Class 12 284

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Senior Secondary School Mathematics for Class 12

\ ( A -1 ) ¢ = A -1 Hence, A THEOREM 5

PROOF

-1

[Q A ¢ = A].

is symmetric.

If A and B are nonsingular matrices of the same order then prove that ( adj AB) = ( adj B) × ( adj A).

We have | A | ¹ 0 and | B | ¹ 0. \ | AB | = |A|| × B| ¹ 0. So, ( AB) -1 exists. Now, A -1 = \

adj A adj B adj ( AB) and ( AB) -1 = × ; B -1 = |AB| |A| |B|

adj ( AB) = |AB|×

adj ( AB) = |A|| × B|× ( AB) -1 |AB| [Q |AB| = |A|| × B|and

= |A|| × B|× ( B -1 A -1) = |A|| × B|×

adj ( AB) = ( AB) -1 ] |AB|

[Q ( AB) -1 = B -1 A -1 ]

adj B adj A × = ( adj B) ( adj A). |B| |A|

Hence, adj ( AB) = ( adj B) ( adj A). For any square matrix A, prove that ( adj A) ¢ = adj A ¢. Let A be a square matrix of order n. Then, each one of ( adj A) ¢ and ( adj A ¢) is a square matrix of order n. Also, (i , j)th element of ( adj A) ¢ = ( j , i)th element of (adj A) = cofactor of (i , j)th element of A = cofactor of ( j , i)th element of A ¢ = (i , j)th element of ( adj A ¢). Hence, ( adj A) ¢ = ( adj A ¢).

THEOREM 6 PROOF

THEOREM 7

PROOF

If A is a nonsingular square matrix of order n then prove that |adj A| = |A|n - 1 .

We have A × ( adj A) = |A|× I Þ | A × ( adj A)| = | | A |× I| n

[ Q A = B Þ |A| = |B|]

Þ |A||adj A| = |A| |I|

[ Q |kI| = k n|I|]

Þ |A|| × adj A| = |A|n

[ Q |I| = 1 ]

n-1

Þ |adj A| = |A|

Hence,|adj A| = |A|n - 1 .

SSS Mathematics for Class 12 285

Adjoint and Inverse of a Matrix THEOREM 8

PROOF

285

If A is a nonsingular square matrix of order n then prove that adj ( adj A) = |A|n - 2 A.

For any nonsingular B of order n, we have B( adj B) = |B|× I. Taking B = adj A , we get ( adj A) [adj ( adj A)] = |adj A| I Þ

( adj A) [adj ( adj A)] = |A|n - 1 I

[Q |adj A| = |A|n - 1 ]

Þ

A( adj A) [adj ( adj A)] = |A|n - 1 A

[Q A I n = A ]

Þ Þ

n-1

|A| I n [adj ( adj A)] = |A| n- 2

[adj ( adj A)] = |A|

[Q A( adj A) = |A| I ]

A

A.

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

é 3 If A = ê ë 7 We have

2 5

ù é 6 ú and B = ê 8 û ë

½3 |A| = ½ ½7

7 9

ù -1 -1 -1 ú , verify that ( AB) = B A . û

2½ ½ = (15 - 14) = 1 ¹ 0. 5½

Cofactors of the elements of|A|are A11 = 5 , A12 = - 7 ; A 21 = - 2, A 22 = 3. ¢ é 5 -7 ù é 5 -2 ù \ adj A = ê =ê . ú 3 û 3 úû ë -2 ë -7 é 5 -2 ù 1 Hence, A -1 = × adj A = ê [Q |A| = 1 ]. 3 úû |A| ë -7 ½6 Further,|B| = ½ ½8

7½ ½ = (54 - 56) = - 2 ¹ 0. 9½

Cofactors of the elements of|B|are B11 = 9, B12 = - 8; B21 = - 7 , B22 = 6. é 9 -8 ù ¢ é 9 \ adj B = ê =ê 6 úû ë -7 ë -8 1 1é 9 Hence, B -1 = × adj B = - ê |B| 2 ë -8

-7 ù . 6 úû -7 ù . 6 úû

Now, |AB| = |A|| × B| = 1 ´ ( -2) = - 2 ¹ 0, and adj AB = ( adj B) × ( adj A)

SSS Mathematics for Class 12 286

286

Senior Secondary School Mathematics for Class 12

-7 6

é 9 =ê ë -8 \

( AB) -1 =

ù é 5 ú × ê -7 û ë

-2 ù é 94 = ê ú 3 û ë -82 -39 34

1 1 é 94 × ( adj AB) = × |AB| -2 êë -82

Also, B -1 A -1 = -

-39 34

ù ú. û

ù ú. û

1 é 94 -39 ù 1 é 9 -7 ù é 5 -2ù . ×ê ×ê = - ×ê ú ú 8 6 7 3 2 ë -82 34 úû 2 ë û ë û

Hence, ( AB) -1 = B -1 A -1.

EXAMPLE 2

SOLUTION

é 1 2 2ù Show that the matrix A = ê 2 1 2 ú satisfies the equation ú ê êë 2 2 1 úû [CBSE 2008] A 2 - 4A - 5 I = O, and hence find A -1. We leave it to the reader to show that A 2 - 4A - 5 I = O. Now,

A 2 - 4A - 5 I = O

Þ

AA - 4A = 5 I

Þ

( AA) × A -1 - 4A × A -1 = 5 I × A -1

Þ

A( AA -1) - 4I = 5 A -1

Þ

AI - 4I = 5 A -1

Þ

A - 4I = 5 A -1

Þ

A -1 =

1 ( A - 4I). 5

ìé 1 1 ïê ×í 2 5 ïê îêë 2 ìé 1 1 ï = × íê 2 5 ïê îëê 2 é -3 1 ê 2 = × 5 ê êë 2

\ A -1 =

é 1 0 0 ùü ù ú - 4 × ê 0 1 0 úï úý ê ú êë 0 0 1 úû ïþ úû 2 2ù é 4 0 0ù ü ï 1 2ú - ê 0 4 0ú ý ú ê ú 2 1 úû êë 0 0 4 úû ïþ 2 2ù 2 ú. -3 ú 2 -3 úû

2 1 2

2 2 1

EXAMPLE 3

2ù é 4 é3 Find a matrix X such that X × ê = 1 1 úû êë 2 ë

SOLUTION

Let

é 3 A=ê ë 1

2ù é 4 and B = ê ú -1 û ë 2

1 3

ù ú. û

1ù . 3 úû

SSS Mathematics for Class 12 287

Adjoint and Inverse of a Matrix

Here,

½3 |A| = ½ ½1

287

2½ ½ = ( -3 - 2) = - 5 ¹ 0. -1 ½

So, A is nonsingular and therefore, invertible. The given equation is XA = B. XA = B Þ (XA) × A -1 = BA -1

Now,

Þ X( AA -1) = BA -1 Þ XI = BA -1 Þ X = BA -1. Now, cofactors of elements of|A|are A11 = - 1, A12 = - 1; A 21 = - 2, A 22 = 3. ¢ é -1 -1 ù é -1 -2 ù \ adj A = ê =ê . ú 2 3 3 úû ë û ë -1 1 1 é -1 -2 ù . × adj A = - × ê \ A -1 = 3 úû |A| 5 ë -1 Hence, X = BA -1 é 4 =ê ë 2

1 ù æ 1 ö é -1 × ç- ÷ × 3 úû è 5 ø êë -1

1 ù é -1 3 úû êë -1 -5 -5 ù é1 = ê ú -5 5 û ë1

æ 1ö é 4 = ç- ÷ × ê è 5ø ë 2 æ 1ö é = ç- ÷ × ê è 5ø ë EXAMPLE 4

SOLUTION

-2 ù 3 úû -2 ù 3 úû 1 -1

ù ú. û

Find the matrix A satisfying the equation 2ù é 2 1ù é -3 é 1 ê 3 2 ú × A × ê 5 -3 ú = ê 0 ë û ë û ë 1ù 2ù é -3 and C = ê ú ú× 2û ë 5 -3 û ½ 2 1½ ½ = ( 4 - 3) = 1 ¹ 0. Clearly,|B| = ½ ½ 3 2½ 2½ ½ -3 ½ = ( 9 - 10) = - 1 ¹ 0. And, |C| = ½ ½ 5 -3 ½ é 2 Let B = ê ë 3

This shows that B as well as C is invertible. The given matrix equation is BAC = I. Now,

BAC = I Þ B -1 BA CC -1 = B -1 I C -1 Þ I A I = B -1 C -1 Þ A = B -1 C -1 .

0ù . 1 úû

SSS Mathematics for Class 12 288

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Senior Secondary School Mathematics for Class 12

Now, the cofactors of the elements of|B|are B11 = 2, B12 = - 3 ; B21 = - 1, B22 = 2. é 2 \ adj B = ê ë -1 So, B -1 = Again,

-3 2

ù¢ é 2 ú = ê -3 ë û

-1 ù . 2 úû

-1 ù [Q |B|= 1]. 2 úû

é 2 1 × adj B = ê |B| ë -3

the cofactors of the elements of|C|are C11 = - 3 , C12 = - 5 ; C 21 = - 2, C 22 = - 3.

é -3 \ adj C = ê ë -2

-5 -3

ù ¢ é -3 ú = ê -5 ë û

1 1 × ( adj C) = |C| ( -1)

Þ

C -1 =

Þ

é 2 A = ( B -1 C -1 ) = ê ë -3

EXAMPLE 5

é 2 If A = ê ë 4

SOLUTION

We have,

-2 ù -3 úû

é -3 ×ê ë -5

-2 ù é 3 = -3 úû êë 5

-1 ù é 3 2 úû êë 5

2 ù 3 úû

2ù é1 = ê ú 3 û ë1

1 0

-3 ù , verify that ( adj A) -1 = ( adj A -1). 6 úû ½2 |A| = ½ ½4

-3 ½ ½ = (12 + 12) = 24 ¹ 0. 6½

Cofactors of the elements of|A|are A11 = 6, A12 = - 4; A 21 = 3 , A 22 = 2. \

é 6 adj A = ê ë 3

So, A

Now,

-1

-4 2

ù¢ é 6 ú = ê- 4 ë û

1 1 é 6 = × adj A = × |A | 24 êë - 4

½ 1 ½ |A | = ½ 4 ½- 1 ½ 6 -1

3 2

ù ú× û é 1 3 ù ê 4 =ê 2 úû ê 1 êë 6

1 ½ 8 ½= æ 1 + 1 ö = 1 . ç ÷ 1 ½ ½ è 48 48 ø 24 12 ½

Cofactors of the elements of|A -1 |are C11 =

1 1 1 1 , C12 = ; C 21 = - , C 22 = . 12 6 8 4

1 8 1 12

ù ú ú. ú úû

ù ú. û

SSS Mathematics for Class 12 289

Adjoint and Inverse of a Matrix

é 1 ê 12 =ê ê -1 êë 8

1 6 1 4

ù¢ é 1 ú ê 12 ú = ê ú ê 1 úû êë 6

\

adj A -1

\

é 6 ( adj A) ( adj A ) = ê ë -4

é ê And, ( adj A ) ( adj A) = ê ê êë

ù ú ú× ú ûú -1 8 1 4

é 1 3 ù ê 12 ê 2 úû ê 1 êë 6

-1

-1

-1 8 1 4

289

1 12 1 6

-1 8 1 4

ù úé 6 úê ú ë -4 úû

ù ú é 1 ú=ê ú ë 0 úû

3 ù é 1 = 2 úû êë 0

0ù = I. 1 úû

0ù = I. 1 úû

Thus, ( adj A) ( adj A -1) = ( adj A -1) ( adj A) = I. Hence, ( adj A) -1 = ( adj A -1).

EXAMPLE 6

é 1 If A = ê -2 ê êë 1

SOLUTION

We have

-2 3 1

é 1 |A| = ê -2 ê êë 1

1 1 5

ù ú , verify that ( adj A) -1 = ( adj A -1). ú úû

-2 3 1

1 1 5

é 1 ù ú = ê 0 ê ú êë 0 úû

-2 -1 3

1 3 4

ù ú ú úû

é R 2 ® R 2 + 2R1 ù êR ® R - R ú ë 3 3 1 û

= 1 × ( - 4 - 9) = - 13 ¹ 0. So, A

-1

exists.

The cofactors of the elements of|A|are -2 3 ½ -2 1 ½ ½ 3 1½ ½ = - 5; ½ ½ ½ = 11, A13 = ½ ½ = 14, A12 = - ½ A11 = ½ 1 5 1 5 ½ 1 1½ ½ ½ ½ ½ 1 1½ 1 -2 ½ ½ -2 1 ½ ½ = - 3; ½ ½ ½ = 4, A 23 = - ½ ½ = 11, A 22 = ½ A 21 = - ½ 1 5 1 5 ½1 1 ½ ½ ½ ½ ½ 1 1½ 1 -2 ½ ½ -2 1 ½ ½ ½ = - 1. ½ ½ = - 3 , A 33 = ½ ½ = - 5 , A 32 = - ½ A 31 = ½ ½ -2 3 ½ ½ -2 1 ½ ½ 3 1½ é 14 \ ( adj A) = ê 11 ê êë -5

11 4 -3

-5 -3 -1

ù ¢ é 14 ú = ê 11 ú ê úû êë -5

11 4 -3

-5 -3 -1

ù ú ú úû

SSS Mathematics for Class 12 290

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Senior Secondary School Mathematics for Class 12

ÞA

-1

é 14 11 -5 ù 1 1 ê 4 -3 ú = × ( adj A) = × 11 ú |A| -13 ê êë -5 -3 -1 úû

é ê ê =ê ê ê ê ë

-14 13 -11 13 5 13

-11 5 ù 13 13 ú ú -4 3 ú × 13 13 ú 1 ú 3 ú 13 13 û

The cofactors of the elements of|A -1 |are

\

C11

½ -4 ½ = ½ 13 ½ 3 ½ 13

3 ½ ½ -11 13 ½ = -1 , C = - ½ 13 12 ½ 5 1 ½ ½ ½ 13 ½ 13 13 ½

C13

½ -11 ½ 13 =½ ½ 5 ½ 13

C 21

½ -11 ½ = - ½ 13 ½ 3 ½ 13

5 ½ 13 ½ = 2 , C 22 1 ½ ½ 13 13 ½

C 23

½ -14 ½ = - ½ 13 ½ 5 ½ 13

-11 ½ 13 ½ = -1 ; 3 ½ ½ 13 13 ½

3 ½ 13 ½ = 2 , 1 ½ ½ 13 13 ½

-4 ½ 13 ½ = -1 ; 3 ½ ½ 13 13 ½ ½ -14 ½ = ½ 13 ½ 5 ½ 13

5 ½ 13 ½ -3 = , 1 ½ ½ 13 13 ½

C 31

½ -11 ½ = ½ 13 ½ -4 ½ 13

5 ½ ½ -14 ½ 13 13 ½ = -1 , C 32 = - ½ ½ 3 ½ 13 ½ -11 ½ 13 13 ½

5 ½ 13 ½ = -1 , 3 ½ ½ 13 13 ½

C 33

½ -14 ½ = ½ 13 ½ -11 ½ 13

-11 ½ 13 ½ -5 = × -4 ½ ½ 13 13 ½

é ê ê ( adj A -1) = ê ê ê ê ë

-1 13 2 13 -1 13

2 13 -3 13 -1 13

-1 13 -1 13 -5 13

é -1 ù¢ ê 13 ú ê ú ú = ê 2 ú ê 13 ú ê -1 ú ê ë 13 û

2 13 -3 13 -1 13

-1 13 -1 13 -5 13

ù ú ú ú× ú ú ú û

SSS Mathematics for Class 12 291

Adjoint and Inverse of a Matrix

\

é 14 -1 ( adj A)( adj A ) = ê 11 ê êë -5 é 1 =ê 0 ê êë 0

-5 -3 -1

11 4 -3 0 1 0

ù ú ú úû

291

é ê ê ê ê ê ê ë

-1 13 2 13 -1 13

-1 13 -1 13 -5 13

2 13 -3 13 -1 13

ù ú ú ú ú ú ú û

0ù 0ú× ú 1 úû

-1

Thus, ( adj A) ( adj A ) = I. Similarly, ( adj A -1) ( adj A) = I. Hence, ( adj A) -1 = ( adj A -1). EXAMPLE 7

é 2 If A = ê ë 1

SOLUTION

Given: \

3 ù , verify that ( A ¢) -1 = ( A -1) ¢. 5 úû

é 2 A=ê ë 1

½2 |A| = ½ ½1

3 ù é 2 , and therefore A ¢ = ê 5 úû ë 3

1ù × 5 úû

3½ ½ = (10 - 3) = 7 ¹ 0. 5½

So, A -1 exists. The cofactors of the elements of|A|are

\

Þ

A11 = 5 , A12 = - 1; A 21 = - 3 , A 22 = 2. ¢ é 5 -1 ù é 5 -3 ù ( adj A) = ê =ê ú 3 2 2 úû ë û ë -1 A

-1

1 1 é 5 = × ( adj A) = × ê |A| 7 ë -1

é ê Þ (A )¢ = ê ê ë -1

5 7 -3 7

é 5 -3 ù ê 7 = 2 úû ê -1 ê ë 7

-1 ù 7 ú. 2 ú ú 7 û

Also,|A ¢| = |A| = 7 ¹ 0. So, ( A ¢) -1 exists. The cofactors of elements of|A ¢|are

\

C11 = 5 , C12 = - 3 ; C 21 = - 1, C 22 = 2. é 5 -3 ù ¢ é 5 -1 ù ( adj A ¢) = ê =ê 2 úû 2 úû ë -1 ë -3

-3 7 2 7

ù ú ú ú û … (i)

SSS Mathematics for Class 12 292

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Senior Secondary School Mathematics for Class 12

Þ

Þ

( A ¢) -1 =

( A ¢) -1

-1 ù 2 úû

1 1 é 5 × ( adj A ¢) = × ê |A ¢| 7 ë -3

é 5 ê = ê 7 ê -3 êë 7

-1 7 2 7

ù ú ú. ú úû

… (ii)

Hence, from (i) and (ii), we get ( A ¢) -1 = ( A -1) ¢. EXAMPLE 8

é 1 If A = ê ë - tan x

SOLUTION

We have

tan x ù é cos 2x - sin 2x ù , show that A ¢ A -1 = ê × ú cos 2x úû 1 û ë sin 2x

1 é |A| = ê ë - tan x

tan x ù = (1 + tan 2 x) = sec2 x ¹ 0. 1 úû

So, A is invertible. The cofactors of the elements of|A|are A11 = 1, A12 = tan x ; A 21 = - tan x , A 22 = 1. \

1 é ( adj A) = ê tan x ë

Þ

A -1 =

¢ tan x ù é 1 = ê 1 úû ë tan x

é 1 1 1 ×ê ( adj A) = 2 |A| sec x ë tan x

- tan x ù ú 1 û

- tan x ù ú 1 û

cos2 x - tan x cos2 x ù é 1 - tan x ù é = cos2 x × ê =ê ú ú 1 ë tan x û ëê tan x cos2 x cos2 x úû - sin x cos x ù ú cos2 x û

é cos2 x =ê ë sin x cos x Þ

é 1 A ¢ A -1 = ê ë tan x

- tan x ù é cos2 x ê ú 1 û ë sin x cos x

é cos2 x - sin 2 x =ê ë 2 sin x cos x é cos 2x =ê ë sin 2x

-2 sin x cos x

ù ú - sin x + cos x û 2

- sin 2x ù × cos 2x úû

é cos 2x Hence, A ¢ A -1 = ê ë sin 2x

- sin x cos x ù ú cos2 x û

- sin 2x ù × cos 2x úû

2

SSS Mathematics for Class 12 293

Adjoint and Inverse of a Matrix

293

é cos b é cos a - sin a 0 ù ú ê Let F( a ) = sin a 0 and G(b) = ê 0 cos a ê ú ê 0 1 úû êë - sin b êë 0 -1 Show that [F( a ) × G(b)] = G( -b) × F( - a ).

EXAMPLE 9

0 1 0

sin b ù 0 ú× ú cos b úû

We have

SOLUTION

é cos a F( a ) × F( - a ) = ê sin a ê êë 0 é cos a = ê sin a ê êë 0

0 ù é cos ( - a ) - sin ( - a ) 0 ú ê sin ( - a ) cos ( - a ) úê 0 0 1 úû êë sin a - sin a 0 ù é cos a ê ú cos a 0 - sin a cos a ú ê 0 1 úû êë 0 0

- sin a cos a 0

0ù 0ú ú 1 úû 0ù 0ú ú 1 úû

0 é cos2 a + sin 2 a 0ù é 1 0 0ù ê ú 2 2 =ê 0 sin a + cos a 0 ú = ê 0 1 0 ú = I. ú ê ê 0 0 1 ú êë 0 0 1 úû ë û Thus, F( a ) × F( - a ) = I Þ {F( a )} -1 = F( - a ). Similarly, G(b) × G( -b) = I Þ {G(b)} -1 = G( -b). {F( a ) × G(b)} -1 = {G(b)} -1 × {F( a )} -1

\

[by reversal law]

= G( -b) × F( - a ). Hence, {F( a ) × G(b)} -1 = G( -b) × F( - a ).

EXERCISE 7 Find the adjoint of the given matrix and verify in each case that A × (adj A) = (adj A) × A =| A| × I. é 2 1. ê ë5

3 ù 9 úû

é cos a 3. ê ë sin a

é 9 7. ê 5 ê êë 6

sin a ù cos a úû -1 6 -2

3 é 5. ê -15 ê 5 êë 7 -1 8

-5 ù 2 úû

é 3 2. ê ë -1

1 -5 2 3 4 2

ù ú ú úû

ù ú ú úû

é 1 4. ê 3 ê êë 1

-1 1

é 0 6. ê 1 ê êë 3

1 2 1

2 3 1

ù ú ú úû

é 4 8. ê 1 ê êë 2

5 0 7

3 6 9

ù ú ú úû

2ù -2 ú ú 3 úû

0

SSS Mathematics for Class 12 294

294

Senior Secondary School Mathematics for Class 12

é cos a 9. ê sin a ê êë 0

- sin a cos a

0ù 0ú ú 1 úû

0

é -4 10. If A = ê 1 ê êë 4

-3 0 4

-3 1 3

é -1 11. If A = ê 2 ê êë 2

-2 1

-2 -2 1

-2

ù ú , show that adj A = A. ú úû

ù ú , show that adj A = 3 A ¢. ú úû

Find the inverse of each of the matrices given below: é 3 -5 ù é 4 1 ù é 2 12. ê 13. ê 14. ê ú ú 2û ë -1 ë 2 3 û ë 4 é a 15. ê ë c é 1 16. ê 1 ê êë 2

-3 ù 6 úû

b ù , when ( ad - bc) ¹ 0 d úû -1 3

é 0 19. ê 3 ê êë -2

0 4 -4

é 2 22. If A = ê ë5

-1 0 6

1ù -1 ú ú 0 úû

-1 ù 4ù é 2 -1 20. ê -3 5 ú 0 1ú ú ú ê -7 úû 1 2 úû êë -1 3 ù 1 , show that A -1 = A. -2 úû 19

é 1 23. If A = ê 2 ê êë 1

-1 -1

é 3 24. If A = ê 2 ê êë 0

-3 -3 -1

25. If A =

é 2 17. ê 3 ê êë 2

5 ù -1 ú ú -1 úû

2

0

é -8 1ê 4 9ê êë 1

é 2 -3 3 18. ê 2 2 3 ê êë 3 -2 2 é 8 - 4 1ù 21. ê10 0 6ú ú ê 1 6úû êë 8

ù ú ú úû

1ù 0 ú , show that A -1 = A 2. ú 0 úû 4 4 1 1 4 -8

ù ú , prove that A -1 = A 3. ú úû 4ù ú , show that A -1 = A ¢. ú úû

7 4

26. Let D = diag [d1 , d 2 , d 3],

where none of d1 , d 2 , d 3 is 0; prove that

D -1 = diag [d1-1 , d 2-1 , d 3-1].

é 3 2ù é 6 7 ù -1 -1 -1 27. If A = ê and B = ê ú ú , verify that ( AB) = B A . ë 7 5 û ë 8 9û

SSS Mathematics for Class 12 295

Adjoint and Inverse of a Matrix

é 9 28. If A = ê ë 6

-1 ù é -4 and B = ê ú -2 û ë 5

29. Compute ( AB)

-1

3 ù , -4 úû

é 1 when A = ê 0 ê êë 3

1 2 -2

verify that ( AB) -1 = B -1 A -1. 2ù é 1 -1 ú -3 and B = ê 0 ú ê 4 úû êë 1

é 1 30. Obtain the inverses of the matrices ê 0 ê êë 0

p 1 0

0ù p ú and ú 1 úû

é 1 + pq And, hence find the inverse of the matrix ê q ê êë 0 é 3 31. If A = ê ë 2

é 3 34. If A = ê ë 7

é 1 ê q ê êë 0 p

1 + pq q

2 3 0 0 1 q

0ù -1 ú × ú 2 úû 0ù 0 ú. ú 1 úû

0ù p ú. ú 1 úû

2ù , verify that A 2 - 4A - I = O , and hence find A -1. 1 úû

é -8 32. Show that the matrix A = ê ë 2 -1 and hence find A . é -1 33. If A = ê ë 2

295

-1 -2

5 ù satisfies the equation x 2 + 4x - 42 = 0 4 úû

ù 2 -1 ú , show that A + 3 A + 4I 2 = O and hence find A . û

1ù , find x and y such that A 2 + xI = yA. Hence, find A -1. 5 úû [CBSE 2005]

é 3 -2ù 2 -1 35. If A = ê ú , find the value of l so that A = lA - 2I. Hence, find A . ë 4 -2û [CBSE 2007]

é 1 36. Show that the A = ê -2 ê êë 3

0 -1 4

-2 2 1

ù ú satisfies the equation ú úû

A 3 - A 2 - 3 A - I = O , and hence find A -1.

37. Prove that:

(i) adj I = I

(ii) adj O = O

(iii) I -1 = I.

SSS Mathematics for Class 12 296

296

Senior Secondary School Mathematics for Class 12 ANSWERS (EXERCISE 7)

é 9 1. ê ë -5

-3 ù 2 úû

é 3 4. ê -11 ê êë -1

3 1

é 2 2. ê ë 1 0ù 8ú ú 4 úû

-1

31 é -34 10 ê 7. 14 0 -21 ê êë 46 -30 - 44

ù ú ú úû

é 3 ê 13. ê 10 ê-1 êë 5

é2 5ù 12. ê ú ë1 3û

16.

é 4 1 ê × -1 27 ê êë 5

19.

é -8 1 ê × 11 4 ê êë - 4

21.

é -6 1 ê × -12 10 ê êë 10

4 -2 0

4 -3 0

25 40 - 40

é -1 1 -1 ù 6. ê 8 -6 2 ú ú ê êë -5 3 -1 úû

é - 42 -24 30 ù 8. ê 3 30 -21 ú ú ê 7 -18 -5 úû êë

é cos a 9. ê - sin a ê 0 êë

-1 ù 10 ú ú 2 ú 5 úû 17.

é ê 14. ê ê êë

1 4 -1 6

1 ù 8 ú ú 1 ú 12 úû

é 6 1 ê × -2 32 ê êë 8

ù ú ú úû -24 -38 40

ù ú ú úû

-6 -8 1

1 14 3

ù ú ú úû

29.

é 16 1 ê × 21 19 ê êë 10

12 11 -2

1 -7 3

ù ú ú úû

1 é -4 32. × 42 êë 2

é -9 36. A -1 = ê 8 ê êë -5

-8 7 -4

-2 ù 2ú ú -1 úû

-1 ù 3 úû

5 ù 8 úû

0ù cos a 0 ú ú 0 1 úû

é -2 1 ê × -1 5 ê êë 2

é 1 1 ê × -5 19 ê êë 3

é -1 31. ê ë 2

1 é 5 × 8 êë -7

18.

20.

é 1 0 0ù ê ú 1 0 ú and ê - q ê 2 -q 1 ú û êë q

34. x = 8, y = 8 and A -1 =

1 ù 5 ú ú 3 úû

6

sin a

é d - bù 1 × a úû ( ad - bc) êë - c

15.

-2 -14

é 1 -p p 2 ù é 1 ê ú ê 30. ê 0 1 -p ú , ê - q ê0 0 1 ú ê q2 ë û ë 2ù -3 úû

- sin a ù cos a úû

é 2 0 -1 ù 5. ê 5 1 0ú ú ê êë 0 1 3 úû

3 ù 6ú ú -3 úû

17 -11 1

é cos a 3. ê ë - sin a

5 ù 3 úû

3 ù 0ú ú -2 úû

ù ú - qp - p ú ú p 2 q 2 + pq + 1 ú û p2

-p pq + 1

2

- pq 2 - q

33. A

0 1 1

-1

é 1 ê =ê 2 ê -1 êë 2

35. l = 1, A -1 =

1 4 1 4

ù ú ú ú úû

1 é -2 2 ù × 2 êë - 4 3 úû

SSS Mathematics for Class 12 297

Adjoint and Inverse of a Matrix

297

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 7) 29. Find A -1 . Then, ( AB) -1 = B -1 A -1 . 30. Let the first two matrices be A and B. Then, the third matrix is AB. Now, ( AB) -1 = ( B -1 × A -1 ). ì 1, when i = j 37. (i)|I| = 1. Cofactors of aij in| I |are A ij = í î 0 , when i ¹ j \ ( adj I ) = [A ij ]¢ = I ¢ = I. (ii)|O| = 0 and cofactor of each element of|O|is 0. \ adj O = O . (iii) I × I = I Þ I -1 = I.

SSS Mathematics for Class 12 298

8. SYSTEM OF LINEAR EQUATIONS

Solving a System of Linear Equations by Matrix Method CONSISTENT SYSTEM OF EQUATIONS A given system of equations is said to be consistent if it has one or more solutions. INCONSISTENT SYSTEM OF EQUATIONS

A given system of equations is said to be

inconsistent if it has no solution. Consider the system of equations a1 x + b1 y + c1z = d1 , a2 x + b2 y + c2z = d 2 , a 3 x + b 3 y + c 3z = d 3 . b1 c1 ù é x ù é a1 é d1 ù Let A = ê a2 b2 c2 ú , X = ê y ú and B = ê d 2 ú × ê ú ú ú ê ê êë z úû êë a 3 b 3 c 3 úû êë d 3 úû Then, the given system can be written as b1 c1 ù é x ù é a1 ú ê ú ê a b c 2 ú ê y ú = 2 ê 2 êë a 3 b 3 c 3 úû êë z úû

é d1 ù ê d ú× ê 2 ú êë d 3 úû

\ AX = B. When|A| ¹ 0 In this case, A -1 exists.

CASE 1

\ AX = B Þ A -1( AX) = A -1 B Þ ( A -1 A ) X = A -1 B

[by associative law]

Þ I × X = A -1 B

[Q A -1 A = I]

Þ X = A -1 B. -1

CASE 2

CASE 3

Since A is unique, the given system has a unique solution. Thus, when |A| ¹ 0, then the given system is consistent and it has a unique solution. When|A| = 0 and ( adj A) B ¹ O In this case, the given system has no solution and hence it is inconsistent. When|A| = 0 and ( adj A) B = O In this case, the given system has infinitely many solutions. 298

SSS Mathematics for Class 12 299

System of Linear Equations

299

SUMMARY

Let AX = B be the given system of equations. (i) If|A| ¹ 0, the system has a unique solution. (ii) If|A| = 0 and ( adj A) B ¹ O then the given system has no solution. (iii) If|A| = 0 and ( adj A) B = O then the system has infinitely many solutions.

SOLVED EXAMPLES EXAMPLE 1

Use matrix method to show that the system of equations 2x + 5 y = 7 , 6x + 15 y = 13 is inconsistent.

SOLUTION

The given equations are 2x + 5 y = 7, 6x + 15 y = 13. 5 ù é 2 é x ù é 7 ù Let A = ê ú , X = ê y ú and B = ê 13 ú . 6 15 ë û ë û ë û

… (i) … (ii)

Then, the given system in matrix form is AX = B. 5½ ½2 ½ = 0. Now, |A| = ½ ½ 6 15 ½ The system will be inconsistent if (adj A) B ¹ O. The minors of the elements of|A|are M11 = 15 , M12 = 6; M21 = 5 , M22 = 2. So, the cofactors of the elements of|A|are A11 = 15 , A12 = - 6; A 21 = - 5 , A 22 = 2. ¢ é 15 - 6 ù é 15 -5 ù \ adj A = ê =ê ú 2û 2 úû ë -5 ë -6 é 15 -5 ù é 7 ù é 105 - 65 ù é 40 ù Þ (adj A) B = ê = = ¹ O. 2úû êë13 úû êë - 42 + 26úû êë -16 úû ë- 6 Thus,|A| = 0 and ( adj A) B ¹ O. Hence, the given system of equations is inconsistent. EXAMPLE 2

Use matrix method to show that the following system of equations is inconsistent: 3 x - y + 2z = 3 , 2x + y + 3z = 5 , x - 2y - z = 1.

SSS Mathematics for Class 12 300

300 SOLUTION

Senior Secondary School Mathematics for Class 12

Let us take é 3 A=ê 2 ê êë 1

-1 1 -2

2ù é 3 ù é x ù ú , X = ê y ú and B = ê 5 ú × ê ú ê ú ú êë 1 ûú êë z úû úû

3 -1

The given system in matrix form is AX = B. 2½ ½ 3 -1 Now,|A| = ½ 2 1 3½ ½ ½ ½ 1 -2 -1 ½ = 3( -1 + 6) + 1 × ( -2 - 3) + 2 × ( - 4 - 1) = (15 - 5 - 10) = 0. So, the system will be inconsistent if ( adj A) × B ¹ O. The minors of the elements of|A|are M11 = 5 , M12 = - 5 , M13 = - 5; M21 = 5 , M22 = - 5 , M23 = - 5; M 31 = - 5 , M 32 = 5 , M 33 = 5. So, the cofactors of the elements of|A|are A11 = 5 , A12 = 5 , A13 = - 5; A 21 = - 5 , A 22 = - 5 , A 23 = 5;

\

Þ

A 31 = - 5 , A 32 = - 5 , A 33 = 5. 5 -5 ù ¢ é 5 é 5 ê 5 ú =ê 5 ( adj A) = -5 -5 ú ê ê 5 úû êë -5 êë -5 -5 é 5 -5 -5 ù é 3 ù ( adj A) B = ê 5 -5 -5 ú ê 5 ú ú ê ú ê 5 5 úû êë 1 úû êë -5 é 5 × 3 + ( -5) × 5 + ( -5) × 1 = ê 5 × 3 + ( -5) × 5 + ( -5) × 1 ê êë ( -5) × 3 + 5 × 5 + 5 × 1

-5 -5 5

-5 ù -5 ú ú 5 úû

ù é 15 - 25 - 5 ú = ê 15 - 25 - 5 ú ê ûú êë -15 + 25 + 5

ù ú ú úû

é -15 ù = ê -15 ú ¹ O. ú ê êë 15 úû Thus,|A| = 0 and ( adj A) B ¹ O. Hence, the given system of equations is inconsistent. EXAMPLE 3

Show that the following system of equations is consistent and solve it: 2x + 5 y = 1, 3 x + 2y = 7.

SSS Mathematics for Class 12 301

System of Linear Equations SOLUTION

301

The given system of equations is 2x + 5 y = 1,

… (i)

3 x + 2y = 7. é 2 5 ù é x ù é 1ù Let A = ê ú , X = ê y ú and B = ê 7 ú × 3 2 ë û ë û ë û

… (ii)

Then, the given system is AX = B. ½ 2 5½ ½ = ( 4 - 15) = - 11 ¹ 0. Now,|A| = ½ ½ 3 2½ Hence, the given system has a unique solution. The minors of the elements of|A|are M11 = 2, M12 = 3; M21 = 5 , M22 = 2. So, the cofactors of the elements of|A|are A11 = 2, A12 = - 3; A 21 = - 5 , A 22 = 2. \

é 2 ( adj A) = ê ë -5

ù¢ é 2 ú = ê -3 û ë

-3 2

-1 é 2 1 = × ( adj A) = 11 êë -3 |A|

Þ

A -1

Þ

X = A -1 B

Þ

é -2 é x ù ê 11 ê y ú=ê 3 ë û ê êë 11

Þ

x = 3 and y = - 1.

5 11 -2 11

-5 ù 2 úû é -5 ù ê =ê 2 úû ê ëê

ù é -2 35 ú é 1 ù ê 11 + 11 ú ê ú=ê ú ë 7 û ê 3 - 14 úû êë 11 11

-2 11 3 11

5 11 -2 11

ù ú ú ú úû

ù ú é 3 ù ú=ê ú ú ë -1 û úû

Hence, x = 3 and y = - 1.

EXAMPLE 4

SOLUTION

é 1 2 -3 ù If A = ê 2 3 2 ú , find A -1 and hence solve the system of linear ú ê êë 3 -3 - 4 úû equations: x + 2y - 3 z = - 4; 2x + 3 y + 2z = 2; 3 x - 3 y - 4z = 11. [CBSE 2012C] The given equations are

x + 2y - 3z = - 4.

… (i)

2x + 3 y + 2z = 2,

… (ii)

3 x - 3 y - 4z = 11.

… (iii)

SSS Mathematics for Class 12 302

302

Senior Secondary School Mathematics for Class 12

2 -3 ù é -4 ù é x ù é 1 ú ê ú ê Let A = 2 3 2 , X = y and B = ê 2 ú . ú ê ê ú ú ê êë 11 úû êë z úû êë 3 -3 -4 úû So, the given system in matrix form is AX = B. 2 -3½ ½ 1 2 -3 ½ ½1 Now, |A| = ½ 2 3 2½ = ½ 0 -1 8½ ½ ½ ½ ½ 5½ ½ 3 -3 - 4½ ½ 0 -9 [R 2 ® R 2 - 2R1 and R 3 ® R 3 - 3 R1] = 1 × ( -5 + 72) = 67 ¹ 0. Thus, A is invertible. So, the system has a unique solution, X = A -1 B. Now,

\

So, \ or

\ EXAMPLE 5

SOLUTION

the cofactors of the elements of|A|are A11 = - 6, A12 = 14, A13 = - 15 ; A 23 = 9; A 21 = 17 , A 22 = 5 , A 31 = 13 , A 32 = - 8, A 33 = - 1. é - 6 14 -15 ù ¢ é - 6 17 13 ù 5 9 ú = ê 14 5 -8 ú . adj A = ê 17 ú ú ê ê 9 -1 úû -1 úû êë -15 êë 13 -8 é - 6 17 13 ù 1 1 ê A -1 = × adj A = × 14 5 -8 ú . ú |A| 67 ê êë -15 9 -1 úû X = A -1 B

é - 6 17 é x ù ê y ú = 1 × ê 14 5 ê ú 67 ê 9 êë z úû ëê -15 é 201 ù é 1 ê = × -134 ú = ê ú ê 67 ê êë 67 úû êë x = 3 , y = - 2 and z = 1.

ù é -4 ù ú×ê 2 ú ú ú ê úû êë 11 úû 3 ù -2 ú × ú 1 úû

13 -8 -1

Using matrices, solve the following system of linear equations: 3 x + 4y + 2z = 8, 2y - 3 z = 3 , x - 2y + 6z = -2. [CBSE 2006C] The given equations are 3 x + 4y + 2z = 8, 2y - 3 z = 3, x - 2y + 6z = -2.

… (i) … (ii) … (iii)

SSS Mathematics for Class 12 303

System of Linear Equations

Let

é 3 A=ê 0 ê êë 1

4 2 -2

303

2ù é 8ù é x ù ú ê ú -3 , X = y and B = ê 3 ú × ú ê ê ú ú 6 úû êë -2 úû êë z úû

So, the given system in matrix form is AX = B. 4 2 ½ ½ 0 10 -16 ½ ½3 Now, |A| = ½ 0 2 -3 ½ = ½ 0 2 -3 ½ [R1 ® R1 - 3 R 3] ½ ½ ½ ½ 6 ½ ½ 1 -2 6½ ½ 1 -2 = 1 × ( -30 + 32) = 2 ¹ 0. So, A is invertible. Therefore, the given system has a unique solution, X = A -1 B. Now, the minors of the elements of|A|are M11 = 6, M12 = 3 , M13 = - 2; M21 = 28, M22 = 16, M23 = - 10; M 31 = - 16, M 32 = - 9, M 33 = 6. The cofactors of the elements of|A|are A11 = 6, A12 = - 3 , A13 = -2; A 21 = -28, A 22 = 16, A 23 = 10; A 31 = - 16, A 32 = 9, A 33 = 6. ¢ 6 -3 -2 ù é 6 -28 -16 ù é ú ê 16 9ú \ ( adj A) = -28 16 10 = ê -3 ú ú ê ê 9 6 úû 10 6 úû êë -2 êë -16 1 Þ A -1 = × ( adj A) |A| ù é 3 -14 -8ú é 6 -28 -16 ù ê ê -3 1 9ú 16 9ú=ê 8 = × ê -3 ú× ú 2 ê 2 2ú ê 10 6 ûú ëê -2 ê -1 5 3ú úû êë ù é 3 -14 -8ú é x ù ê é 8ù ê -3 9ú ê \ X = A -1 B Þ ê y ú = ê 3 ú 8 ú ú ê ú 2 2ú ê ê êë z úû êë -2 úû ê -1 5 3ú êë ûú é x ù é 24 - 42 + 16 ù é -2 ù Þ ê y ú = ê -12 + 24 - 9 ú = ê 3 ú ú ú ê ê ú ê êë z úû êë - 8 + 15 - 6 úû êë 1 úû Þ x = -2, y = 3 and z = 1. Hence, x = -2, y = 3 and z = 1.

SSS Mathematics for Class 12 304

304 EXAMPLE 6

SOLUTION

Senior Secondary School Mathematics for Class 12

Using matrices, solve the following system of equations: 2 3 10 + + = 4; x y z 4 6 5 - + = 1; x y z 6 9 20 + = 2. ( x , y , z ¹ 0) x y z Putting

[CBSE 2011]

1 1 1 = u, = v and = w, the given equations become: x y z

2u + 3 v + 10w = 4,

… (i)

4u - 6v + 5 w = 1,

… (ii)

6u + 9v - 20w = 2. 3 10 é 2 ê Let A = 4 - 6 5 ê 9 -20 êë 6

… (iii) é 4ù é u ù ù ú , Y = ê v ú and B = ê 1 ú × ê ú ê ú ú êë 2 úû êë w úû úû

Then, the given system in matrix form is AY = B. 3 10 ½ ½2 Now,|A| = ½ 4 - 6 5½ ½ ½ 9 -20 ½ ½6 ½2 =½ 0 ½ ½0

3 -12 0

10 ½ -15 ½ ½ -50 ½

[R 2 ® R 2 - 2R1 ; R 3 ® R 3 - 3 R1]

= ( -50) × ( -24 - 0) = 1200 ¹ 0. Thus, A is invertible. So, the given system has a unique solution, Y = A -1 B. The minors of the elements of|A|are M11 = 75 , M12 = - 110, M13 = 72; M21 = -150, M22 = - 100, M23 = 0; M 31 = 75 , M 32 = - 30, M 33 = - 24. So, the cofactors of the elements of|A|are A11 = 75 , A12 = 110, A13 = 72; A 21 = 150, A 22 = - 100, A 23 = 0;

\

A 31 = 75 , A 32 = 30, A 33 = - 24. 110 72 ù ¢ é 75 é 75 ê 0 ú = ê 110 ( adj A) = 150 -100 ú ê ê 30 -24 úû êë 72 êë 75

150 -100 0

75 30 -24

ù ú× ú úû

SSS Mathematics for Class 12 305

System of Linear Equations

\ \ Þ

Þ Þ

A

-1

é 75 1 1 ê = × ( adj A) = × 110 |A| 1200 ê êë 72

305

150 -100 0

75 ù 30 ú × ú -24 úû

-1

Y=A B 150 75 ù é 4 ù é 75 é u ù ê v ú = 1 × ê 110 -100 30 ú ê 1 ú ú ê ú ê ú 1200 ê 0 -24 ûú êë 2 úû êë w úû ëê 72 é 600 ù é 300 + 150 + 150 ù 1 ê 1 ê × 400 ú = × 440 - 100 + 60 ú = ú ú 1200 ê 1200 ê úû êë 240 úû êë 288 + 0 - 48 é 600 ù é 1 ù ê 1200 ú ê 2 ú ú ú ê ê 400 ú ê 1 ú = =ê ê 1200 ú ê 3 ú ê 240 ú ê 1 ú ú ú ê ê ë 1200 û ë 5 û 1 1 1 u= , v= , w= 2 3 5 1 1 1 1 1 1 = , = and = x 2 y 3 z 5

Þ x = 2, y = 3 and z = 5. Hence x = 2, y = 3 and z = 5.

EXAMPLE 7

é 1 Use the product ê 0 ê êë 3

-1 2 -2

2 -3 4

ù ú ú úû

é -2 ê 9 ê êë 6

0 2 1

1 -3 -2

ù ú to solve the ú úû

following system of equations: x - y + 2z = 1; 2y - 3 z = 1; 3 x - 2y + 4z = 2. SOLUTION

The given equations are x - y + 2z = 1, 2y - 3z = 1, 3 x - 2y + 4z = 2. 2ù é 1 é x ù é 1 -1 Let A = ê 0 2 -3 ú , X = ê y ú and B = ê 1 ê ê ú ú ê 4 úû êë 2 êë z úû êë 3 -2 Then, the given system in matrix form is AX = B.

… (i) … (ii) … (iii) ù ú× ú úû

SSS Mathematics for Class 12 306

306

Senior Secondary School Mathematics for Class 12

2 ù é -2 0 -3 ú ê 9 2 ú ê -2 4 úû êë 6 1 é -2 - 9 + 12 0 - 2 + 2 = ê 0 + 18 - 18 0 + 4 - 3 ê êë -6 - 18 + 24 0 - 4 + 4

é 1 Now, ê 0 ê êë 3

-1 2

Þ

é -2 A×ê 9 ê êë 6

Þ

é -2 A -1 = ê 9 ê êë 6

0 2 1

ù ú ú úû 1+ 3 -4 ù é 1 0-6+ 6 ú = ê 0 ê ú 3 + 6 - 8 úû êë 0

1 -3 -2

0 1 0

0ù 0ú ú 1 úû

1ù ú=I ú úû 0 1ù 2 -3 ú × ú 1 -2 úû -3 -2

Now, AX = B Þ Þ

X = A -1 B 1ù é 1ù é x ù é -2 0 ê y ú = ê 9 2 -3 ú ê 1 ú ú ê ú ê ú ê êë z úû êë 6 1 -2 úû êë 2 úû é -2 + 0 + 2 ù é 0 ù =ê 9+ 2-6 ú =ê 5 ú ú ê ú ê êë 6 + 1 - 4 úû êë 3 úû

Þ x = 0, y = 5 and z = 3. Hence, x = 0, y = 5 and z = 3. EXAMPLE 8

SOLUTION

4 4ù 1ù é -4 é 1 -1 Given A = ê 1 -2 -2 ú and B = ê -7 1 3 ú , find AB and ú ê ú ê 1 3 úû êë 5 -3 -1 úû êë 2 use this result in solving the following system of equations: x - y + z = 4; x - 2y - 2z = 9; 2x + y + 3z = 1. [CBSE 2006C, ‘10C, ‘12C] The given equations are x - y + z = 4, x - 2y - 2z = 9, 2x + y + 3 z = 1. 1ù é 4 é x ù é 1 -1 ú ê ú ê Let A = 1 -2 -2 , X = y and C = ê 9 ê ê ú ú ê 1 3 úû êë 1 êë z úû êë 2 Then, the given system of equations is AX = C.

… (i) … (ii) … (iii) ù ú× ú úû

SSS Mathematics for Class 12 307

System of Linear Equations

-1 -2 1

é 1 Now, AB = ê 1 ê êë 2

1ù -2 ú ú 3 úû

é -4 ê -7 ê êë 5

4 1 -3

307

4ù 3 ú ú -1 úû

4 -1 - 3 4 - 3 -1 ù é 8 0 é -4 + 7 + 5 ê = -4 + 14 - 10 4 - 2 + 6 4 - 6 + 2 ú = ê 0 8 ú ê ê êë -8 - 7 + 15 8 + 1 - 9 8 + 3 - 3 úû êë 0 0

0ù 0ú ú 8 úû

= 8I Þ

æ1 ö A × ç B÷ = I è8 ø

Þ

A -1 =

é -4 1 1 B = × ê -7 8 8 ê êë 5

4ù ú× ú úû

4 1

3 -1

-3

Now, AX = C Þ Þ

X = A -1C é -4 é x ù ê y ú = 1 × ê -7 ê ú 8 ê êë z úû ëê 5

4 1 -3

4 3 -1

é -16 + 36 + 4 1 ê = × -28 + 9 + 3 8 ê êë 20 - 27 - 1 Þ

ù ú ú úû

é 4ù ê 9ú ê ú êë 1 úû

é 24 ù é 3 ù ù ú = 1 × ê -16 ú = ê -2 ú ú ú ê ú 8 ê úû ëê -8 úû êë -1 úû

x = 3 , y = -2 and z = - 1.

Hence, x = 3 , y = -2 and z = - 1. EXAMPLE 9

The sum of three numbers is 6. Twice the third number when added to the first number gives 7. On adding the sum of the second and third numbers to thrice the first number, we get 12 . Find the numbers, using method.

SOLUTION

Let the first, second and third numbers be x , y , z respectively. Then, x + y + z = 6,

… (i)

x + 2z = 7,

… (ii)

3 x + y + z = 12. é 1 Let A = ê 1 ê êë 3

1 0 1

1 2 1

… (iii) é 6 é x ù ù ú , X = ê y ú and B = ê 7 ê ê ú ú êë z úû úû ëê 12

Then, the given system in matrix form is AX = B.

ù ú× ú úû

SSS Mathematics for Class 12 308

308

Senior Secondary School Mathematics for Class 12

½1 Now,|A| = ½ 1 ½ ½3

1 0 1

1½ ½ 1 2 ½ =½ 0 ½ ½ 1½ ½ 0

1½ 1½ ½ -2 ½

1 -1 -2

é R 2 ® R 2 - R1 ; ù ê R ® R - 3R ú ë 3 3 1û

= 1 × ( 2 + 2) = 4 ¹ 0. \ A is invertible. So, the given system has a unique solution, X = A -1 B. The minors of the elements of|A|are M11 = - 2, M12 = - 5 , M13 = 1; M21 = 0, M22 = - 2, M23 = - 2; M 31 = 2, M 32 = 1, M 33 = - 1. The cofactors of the elements of|A|are A11 = - 2, A12 = 5 , A13 = 1; A 21 = 0, A 22 = - 2, A 23 = 2;

\

Þ Þ Þ

Þ

A 31 = 2, A 32 = - 1, A 33 = - 1. 5 1 ù ¢ é -2 é -2 ê 0 -2 2ú =ê 5 ( adj A) = ú ê ê êë 1 êë 2 -1 -1 úû 0 é -2 1 1 A -1 = × ( adj A) = × ê 5 -2 |A| 4 ê 2 êë 1

2ù -1 ú ú -1 úû

0 -2 2 2 -1 -1

ù ú ú úû

X = A -1 B 0 2ù é -2 é x ù ê y ú = 1 × ê 5 -2 -1 ú ú ê ú 4 ê 2 -1 úû êë 1 êë z úû é -12 + 0 + 24 ù 1 = × ê 30 - 14 - 12 ú ú 4 ê êë 6 + 14 - 12 úû

é 6ù ê 7 ú ú ê êë 12 úû =

é 12 ù é 3 1 ê × 4ú=ê 1 ú ê 4 ê êë 8 úû êë 2

ù ú ú úû

x = 3 , y = 1, z = 2.

Hence, the required numbers are 3, 1, 2. EXAMPLE 10

A school wants to award its students for the value of Honesty, Regularity and Hardwork with a total cash award of ` 6000. Three times the award money for Hardwork added to that given for Honesty amounts to `11000. The award money given for Honesty and Hardwork together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from the given three values, suggest one more value which the school must include for awards. [CBSE 2013]

SSS Mathematics for Class 12 309

System of Linear Equations SOLUTION

309

Let the amount of award for Honesty, Regularity and Hardwork be ` x, ` y and `z respectively. Then, x + y + z = 6000,

… (i)

x + 0y + 3z = 11000

… (ii)

and x + z = 2y Þ x - 2y + z = 0. é1 Let A = ê 1 ê êë 1

… (iii)

1ù é 6000 ù é x ù 3 ú , X = ê y ú and B = ê 11000 ú × ú ê ê ú ú 1 úû êë 0 úû êë z úû

1 0 -2

Then, the matrix equation is AX = B. \

X = A -1 B.

½1 Now,|A| = ½ 1 ½ ½1

1 0 -2

1 ½ ½1 3 ½ =½ 1 ½ ½ 1 ½ ½1

0½ 2½ ½ 0½

0 -1 -3

{C 2 ® (C 2 - C1) and C 3 ® (C 3 - C1)} ½ -1 2 ½ ½ = ( 0 + 6) = 6 ¹ 0. =1× ½ ½ -3 0 ½ \ A -1 exists. Now, the cofactors of the elements of|A|are given by ½ 0 C11 = ½ ½ -2

3½ 1 3½ ½ = 6, C12 = - ½ ½ ½ = 2, C13 1½ ½1 1 ½

½ 1 1½ ½ = -3 , C 22 C 21 = - ½ ½ -2 1 ½

½1 0 ½ ½ = -2; =½ ½ 1 -2 ½

1 1½ ½1 1 ½ ½ = 0, C 23 = - ½ ½ ½ = 3; =½ 1 1 ½ ½ ½ 1 -2 ½

1 1½ ½ 1 1½ ½ = 3 , C 32 = - ½ ½ ½ = -2, C 33 C 31 = ½ 0 3 ½ ½ ½1 3 ½ \

é 6 ( adj A) = ê -3 ê êë 3

2 0 -2

-2 3 -1

t

é 6 ù ú =ê 2 ê ú êë -2 úû

é 6 1 1 × ( adj A) = × ê 2 |A| 6 ê êë -2

Þ

A -1 =

Þ

X = A -1 B =

é 6 1 ê × 2 6 ê êë -2

-3 0 3

3 -2 -1

-3 0 3 ù ú ú úû

-3 0 3 3 -2 -1

ù ú ú úû

é 6000 ù ê 11000 ú ê ú êë 0 úû

½1 1 ½ ½ = -1. =½ ½1 0 ½

3 ù -2 ú ú -1 úû

SSS Mathematics for Class 12 310

310

Senior Secondary School Mathematics for Class 12

Þ

é 36000 - 33000 + 0 1 ê X = × 12000 + 0 + 0 6 ê êë -12000 + 33000 - 0

Þ

é x ù é 500 ê y ú = ê 2000 ê ú ê êë z úû êë 3500

Þ

x = 500, y = 2000 and z = 3500.

ù é 3000 ù ú = 1 ê 12000 ú ú ú 6ê êë 21000 úû úû

ù ú ú úû

Hence, the award money for Honesty, Regularity and Hardwork is ` 500, ` 2000 and ` 3500 respectively. Apart from honesty, regularity and hardwork, the school must include an award for a student to be well-behaved.

EXERCISE 8A Show that each one of the following systems of equations is inconsistent. 1. x + 2y = 9; 2x + 4y = 7.

2. 2x + 3 y = 5 ; 6x + 9y = 10.

3. 4x - 2y = 3 ; 6x - 3 y = 5.

4. 6x + 4y = 5 ; 9x + 6y = 8.

5. x + y - 2z = 5; x - 2y + z = - 2; -2x + y + z = 4.

6. 2x - y + 3z = 1; 3 x - 2y + 5z = - 4; 5 x - 4y + 9z = 14.

7. x + 2y + 4z = 12; y + 2z = - 1; 3 x + 2y + 4z = 4.

8. 3 x - y - 2z = 2; 2y - z = - 1; 3 x - 5 y = 3.

Solve each of the following systems of equations using matrix method. 9. 5 x + 2y = 4; 7 x + 3 y = 5.

10. 3 x + 4y - 5 = 0; x - y + 3 = 0.

11. x + 2y = 1; 3 x + y = 4.

12. 5 x + 7 y + 2 = 0; 4x + 6y + 3 = 0.

13. 2x - 3 y + 1 = 0; x + 4y + 3 = 0.

14. 4x - 3 y = 3 ; 3 x - 5 y = 7.

15. 2x + 8y + 5z = 5 ; x + y + z = - 2; x + 2y - z = 2. [CBSE 2009C]

16.

x - y + z = 1; 2x + y - z = 2; x - 2y - z = 4.

[CBSE 2006C]

SSS Mathematics for Class 12 311

System of Linear Equations

311

17. 3 x + 4y + 7z = 4; 2x - y + 3z = - 3 ; x + 2y - 3z = 8. [CBSE 2012]

18. x + 2y + z = 7; x + 3z = 11; 2x - 3 y = 1.

19. 2x - 3 y + 5z = 16; 3 x + 2y - 4z = - 4; x + y - 2z = - 3. [CBSE 2005C]

20. x + y + z = 4; 2x - y + z = - 1; 2x + y - 3z = - 9.

21. 2x - 3 y + 5z = 11; 3 x + 2y - 4z = - 5 ; x + y - 2z = - 3. [CBSE 2009]

22. x + y + z = 1; x - 2y + 3z = 2; 5 x - 3 y + z = 3.

[CBSE 2004, ‘09C]

23. x + y + z = 6; x + 2z = 7; 3 x + y + z = 12. [CBSE 2009]

24. 2x + 3 y + 3z = 5 ; x - 2y + z = - 4; 3 x - y - 2z = 3.

[CBSE 2008C, ‘12]

[CBSE 2005]

x - y + 2z = 7; 3 x + 4y - 5 z = - 5 ; 2x - y + 3 z = 12.

[CBSE 2012]

27. 6x - 9y - 20z = - 4; 4x - 15 y + 10z = - 1; 2x - 3 y - 5z = - 1.

28. 3 x - 4y + 2z = - 1; 2x + 3 y + 5 z = 7 ; x + z = 2.

[CBSE 2011C]

29. x + y - z = 1; 3 x + y - 2z = 3 ; x - y - z = - 1. [CBSE 2004]

30. 2x + y - z = 1; x - y + z = 2; 3 x + y - 2z = - 1.

[CBSE 2004C]

25.

4x - 5 y - 11z = 12; x - 3 y + z = 1; 2x + 3 y - 7z = 2. [CBSE 2007]

[CBSE 2005, ‘08, ‘11]

31. x + 2y + z = 4; - x + y + z = 0; x - 3 y + z = 4. 33. 5 x - y = - 7 ; 2x + 3 z = 1; 3 y - z = 5.

26.

32. x - y - 2z = 3; x + y = 1; [CBSE 2012C] x + z = - 6. 34.

x - 2y + z = 0; y - z = 2; 2x - 3 z = 10.

35. x - y = 3; 36. 4x + 3 y + 2z = 60; 2x + 3 y + 4z = 17 ; x + 2y + 3 z = 45; [CBSE 2003C, ‘07C] y + 2z = 7. 6x + 2y + 3 z = 70. 5 ù é 2 -3 37. If A = ê 3 2 - 4 ú , find A -1 . [CBSE 2007C, ‘08C] ú ê 1 -2 úû êë 1 Using A -1 , solve the following system of equations: 2x - 3 y + 5 z = 11; 3 x + 2y - 4z = - 5 ; x + y - 2z = - 3. 1 1ù é 2 38. If A = ê 1 -2 -1 ú , find A -1 . ú ê 3 -5 úû êë 0

[CBSE 2011]

SSS Mathematics for Class 12 312

312

Senior Secondary School Mathematics for Class 12

Using A -1 , solve the following system of linear equations: 2x + y + z = 1; 3 x - 2y - z = ; 2 3 y - 5z = 9. é2 ê HINT: Here A = ê 1 êë 0

1 -2 3

é 1 ù éxù 1ù ê ú ú ê ú -1 ú , X = ê y ú and B = ê 3 2 ú × êë 9 úû êë z úû -5 úû

é 7 2 é 1 -2 0 ù 39. If A = ê 2 1 3 ú and B = ê -2 1 ê ú ê êë -4 2 êë 0 -2 1 úû Hence, solve the system of equations:

-6 ù -3 ú , find AB. ú 5 ûú

x - 2y = 10, 2x + y + 3z = 8 and -2y + z = 7. HINT:

æ 1 ö æ 1ö AB = ( 11) I Þ A ç B ÷ = I Þ A -1 = ç ÷ B. è 11 ø è 11 ø

Using matrices, solve the following system of equations. 2 3 3 1 1 1 3 1 2 40. - + = 10, + + = 10, - + = 13 x y z x y z x y z 41.

[CBSE 2011]

[CBSE 2007C]

1 1 1 2 1 3 1 1 1 - + = 4; + - = 0; + + = 2. ( x , y , z ¹ 0) x y z x y z x y z

VALUE BASED QUESTIONS 42. The sum of three numbers is 2. If twice the second number is added to the sum of first and third, we get 1. On adding the sum of second and third numbers to five times the first, we get 6. Find the three numbers by using matrices. 43. The cost of 4 kg potato, 3 kg wheat and 2 kg rice is ` 60. The cost of 1 kg potato, 2 kg wheat and 3 kg rice is ` 45. The cost of 6 kg potato, 2 kg wheat and 3 kg rice is ` 70. Find the cost of each item per kg by matrix method. 44. An amount of ` 5000 is put into three investments at 6%, 7% and 8% per annum respectively. The total annual income from these investments is ` 358. If the total annual income from first two investments is ` 70 more than the income from the third, find the amount of each investment by the matrix method. HINT:

Let these investments be ` x, ` y and ` z respectively. Then, x + y + z = 5000, 7y 6x 8z + + = 358 Þ 6 x + 7 y + 8z = 35800 100 100 100 7y 8z 6x and, + = + 70 Þ 6 x + 7 y - 8 z = 7000. 100 100 100

… (i) … (ii) … (iii)

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45. Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ` x each, ` y each and ` z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ` 1,600. School B wants to spend ` 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ` 900, using matrices, find the award money for each value. Apart from these three values, suggest [CBSE 2014] one more value which should be considered for award. HINT:

By the given data, we have 3 x + 2 y + z = 1600 4 x + y + 3z = 2300 x + y + z = 900

ü ï ý ï þ

ANSWERS (EXERCISE 8A)

9. x = 2, y = - 3 12. x =

9 -7 , y= 2 2

10. x = - 1, y = 2 13. x =

-13 -5 , y= 11 11

7 -1 , y= 5 5 -6 -19 14. x = , y= 11 11 11. x =

15. x = -3 , y = 2, z = -1 16. x = 1, y = -1, z = -1 17. x = 1, y = 2, z = -1 18. x = 2, y = 1, z = 3 21. x = 1, y = 2, z = 3

19. x = 2, y = 1, z = 3 1 1 22. x = , y = 0, z = 2 2

20. x = 2, y = 1, z = 3 23. x = 3 , y = 1, z = 2

24. x = 1, y = 2, z = -1 1 1 1 27. x = , y = , z = 2 3 5

25. x = -1, y = -1, z = -1

26. x = 1, y = 2, z = 5

30. x = 1, y = 2, z = 3

31. x = 2, y = 0, z = 2

32. x = - 2, y = 3 , z = -4

33. x = -1, y = 2, z = 1

34. x = 2, y = 0, z = -2

36. x = 5 , y = 8, z = 8

37. x = 1, y = 2, z = 3

39. x = 4, y = -3 , z = 1

40. x =

35. x = 2, y = -1, z = 4 1 -3 38. x = 1, y = , z = 2 2 1 41. x = , y = - 1, z = 1 2

42. 1, –1, 2

43. ` 5, ` 8, ` 8

28. x = 3 , y = 2, z = - 1 29. x = 2, y = 1, z = 2

1 1 1 , y = ,z= 2 3 5

44. ` 1000, ` 2200, ` 1800

45. ` 200 for sincerity, ` 300 for truthfulness and ` 400 for helpfulness. One more value may be like honesty, kindness, etc.

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OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 1. If A and B are 2-rowed square matrices such that é 4 -3 ù é -2 -1ù then AB = ? ( A + B) = ê and( A - B) = ê ú 6û 2úû ë1 ë 5 5ù é -7 (a) ê 1 5 úû ë

é 7 -5 ù (b) ê ú ë 1 5û

é7 (c) ê ë5

-1ù -5 úû

é 7 -1ù (d) ê ú ë -5 5 û

é 3 -2ù é 5 6ù 2. If ê ú + 2A = ê -7 10ú then A = ? 5 6 ë û ë û é 1 (a) ê ë -5

3ù 4úû

é -1 5 ù (b) ê ú ë -3 4û

é 1 4ù (c) ê ú ë -6 2û

(d) none of these

é 2 0ù é 4 -3 ù 3. If A = ê are such that 4A + 3X = 5 B then X = ? ú and B = ê -6 3 1 2úû ë û ë é 4 -5 ù (a) ê 2úû ë -6

é 4 5ù (b) ê ú ë -6 -2û

é -4 5 ù (c) ê ú ë 6 -2û

(d) none of these

2ù é 1 -2ù é -2 4. If ( A - 2B) = ê ú and ( 2A - 3 B) = ê 3 -3 ú then B = ? 3 0 ë û ë û é 6 -4ù (a) ê ú ë -3 3 û

6ù é -4 (b) ê ú ë -3 -3 û

é 4 -6ù (c) ê ú ë 3 -3 û

(d) none of these

é 6 -6 0ù é 3 2 5ù 5. If ( 2A - B) = ê and ( 2B + A) = ê ú ú then A = ? ë -4 2 1û ë -2 1 -7 û é -3 2 1ù (a) ê ú ë 2 1 -1û

2 -1ù é3 (b) ê 2 1 1úû ë

é 3 -2 1ù (c) ê ú ë -2 1 -1û

(d) none of these

é3 6. If 2 ê ë5

4ù é 1 y ù é 7 0ù then + = x úû êë 0 1úû êë10 5 úû

(a) ( x = -2, y = 8)

(b) ( x = 2, y = -8)

(c) ( x = 3 , y = -6) (d) ( x = -3 , y = 6) é x - y 2x - y ù é -1 0ù 7. If ê ú=ê ú then ë 2x + z 3z + wû ë 5 13 û (a) z = 3 , w = 4

(b) z = 4, w = 3

(c) z = 1, w = 2

(d) z = 2, w = - 1

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System of Linear Equations

é x 8. If ê ë 3y

yù x úû

315

é 1ù é 3 ù ê 2ú = ê 5 ú then ë û ë û

(a) x = 1, y = 2

(b) x = 2, y = 1 (c) x = 1, y = 1 (d) none of these é 3 - 2x x + 1ù 9. If the matrix A = ê is singular then x = ? 4 úû ë 2 (a) 0 é cos a 10. If A a = ê ë - sin a

(b) 1 (c) -1 sin a ù then ( A a ) 2 = ? cos a úû

é cos2 a sin 2 a ù (a) ê ú êë - sin 2 a cos2 a úû é 2 cos a 2 sin a ù (c) ê ú ë - sin a 2 cos a û é cos a 11. If A = ê ë - sin a

é cos 2a (b) ê ë - sin 2a

(d) -2

sin 2a ù cos 2a úû

(d) none of these

sin a ù be such that A + A ¢ = I , then a = ? cos a úû

(a) p

(b)

p 3

(c) p

3ù é1 k ê 12. If A = 3 k -2ú is singular then k = ? ú ê êë 2 3 -4úû 16 34 33 (a) (b) (c) 3 5 2 a b é ù 13. If A = ê ú then adj A = ? ë c dû bù é -d é d -bù é d - cù (b) ê (c) ê (a) ê ú ú c a a úû b a ë û ë -c ë û é 2x 0 ù é 1 0ù -1 14. If A = ê ú and A = ê -1 2ú then x = ? x x ë û ë û 1 (a) 1 (b) 2 (c) 2

(d)

2p 3

(d) none of these

é -d -bù (d) ê a úû ë c

(d) -2

15. If A and B are square matrices of the same order then ( A + B)( A - B) = ? (a) ( A 2 - B 2)

(b) A 2 + AB - BA - B 2

(c) A 2 - AB + BA - B 2

(d) none of these

16. If A and B are square matrices of the same order then ( A + B) 2 = ? (a) A 2 + 2AB + B 2

(b) A 2 + AB + BA + B 2

(c) A 2 + 2BA + B 2

(d) none of these

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17. If A and B are square matrices of the same order then ( A - B) 2 = ? (a) A 2 - 2AB + B 2

(b) A 2 - AB - BA + B 2

(c) A 2 - 2BA + B 2

(d) none of these

18. If A and B are symmetric matrices of the same order then ( AB - BA) is always (a) a symmetric matrix (b) a skew-symmetric matrix (c) a zero matrix (d) an identity matrix 19. Matrices A and B are inverses of each other only when (a) AB = BA (b) AB = BA = O (c) AB = O , BA = I (d) AB = BA = I 20. For square matrices A and B of the same order, we have adj ( AB) = ? (a) (adj A)(adj B) (b) (adj B)(adj A) (c)|AB| (d) none of these 21. If A is a 3-rowed square matrix and|A| = 4 then adj (adj A) = ? (a) 4A (b) 16A (c) 64A (d) none of these 22. If A is a 3-rowed square matrix and|A| = 5 then|adj A| = ? (a) 5 (b) 25 (c) 125 (d) none of these 23. For any two matrices Aand B, (a) AB = BA is always true (b) AB = BA is never true (c) sometimes AB = BA and sometimes AB ¹ BA (d) whenever AB exists, then BA exists 2ù é 4 1ù é3 24. If A × ê ú=ê ú then A = ? ë 1 -1û ë 2 3 û é1 -1ù (a) ê ú ë1 1û

é 1 1ù (b) ê ú ë -1 1û

é1 1ù (c) ê ú ë1 -1û

(d) none of these

25. If A is an invertible square matrix then|A -1| = ? (a)|A|

(b)

1 |A|

(c) 1

(d) 0

26. If A and B are invertible matrices of the same order then ( AB) -1 = ? (a) ( A -1 ´ B -1)

(b) ( A ´ B -1)

(c) ( A -1 ´ B)

(d) ( B -1 ´ A -1)

27. If A and B are two nonzero square matrices of the same order such that AB = 0 then (a)|A| = 0 or|B| = 0 (b)|A| = 0 and|B| = 0 (c)|A| ¹ 0 and|B| ¹ 0 (d) none of these 28. If A is a square matrix such that|A| ¹ 0 and A 2 - A + 2I = O then A -1 = ? (a) ( I - A)

(b) ( I + A)

(c)

1 ( I - A) 2

(d)

1 ( I + A) 2

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é 1 l 2ù 29. If A = ê 1 2 5 ú is not invertible then l = ? ú ê êë 2 1 1 úû (b) 1 (c) -1 cos q sin q é ù 30. If A = ê then A -1 = ? cos qúû ë sin q

(d) 0

(b) -A é ab b2 ù 31. The matrix A = ê 2 ú is ë - a - ab û

(d) -adj A

(a) 2

(a) A

(c) adj A

(a) idempotent (c) nilpotent

(b) orthogonal (d) none of these

é 2 -2 -4ù 32. The matrix A = ê -1 3 4ú is ú ê êë 1 -2 -3 úû (a) nonsingular (c) nilpotent

(b) idempotent (d) orthogonal

33. If A is singular then A( adj A) = ? (a) a unit matrix (c) a symmetric matrix

(b) a null matrix (d) none of these é 8 0ù 34. For any 2-rowed square matrix A , if A × ( adj A) = ê ú then the value of ë 0 8û |A|is (a) 0 (b) 8 (c) 64 (d) 4 é -2 3 ù -1 35. If A = ê ú then|A | = ? ë 1 1û (a) -5

(b)

-1 5

(c)

1 25

(d) 25

é 3 1ù 2 36. If A = ê ú and A + xI = yA then the values of x and y are 7 5 ë û (a) x = 6, y = 6

(b) x = 8, y = 8

(c) x = 5 , y = 8

(d) x = 6, y = 8

37. If matrices A and B anticommute then (a) AB = BA

(b) AB = - BA

é2 5 ù 38. If A = ê ú then adj A = ? ë1 3û é 3 -5 ù é 3 -1ù (a) ê (b) ê ú 2û 2úû ë -1 ë -5

(c) ( AB) = ( BA) -1 (d) none of these

2ù é -1 (c) ê ú ë 3 -5 û

(d) none of these

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é 3 -4ù 39. If A = ê ú and B is a square matrix of order 2 such that AB = I then ë -1 2û B =? 1ù é 1 2ù é ê ú ê 1 2ú é 1 2ù (d) none of these (c) (b) (a) ê ê 1 3ú ê ú ú 3 2 3 ë û ê 2 2ú ê2 ú ë û ë 2û 40. If A and B are invertible square matrices of the same order then ( AB) -1 = ? (a) AB -1 (b) A -1 B é 2 -1ù -1 41. If A = ê ú , then A = ? ë 1 3û é3 ê (a) ê 7 ê1 êë 7

-1 ù 7 ú ú 2ú 7 úû

é 3 1ù ê ú (b) ê 7 7 ú ê -1 2 ú êë 7 7 úû

é 3 42. If|A| = 3 and A -1 = ê -5 ê ë 3

(c) A -1 B -1

é1 ê (c) ê 3 ê1 êë 7

1ù 7ú ú 2ú 7 úû

(d) B -1 A -1

(d) none of these

-1ù 2 ú then adj A = ? ú 3û é 9 -3 ù (b) ê 2úû ë -5 é 9 -3 ù (d) ê ú ë5 -2û

é 9 3ù (a) ê ú ë -5 -2û é -9 3 ù (c) ê ú ë 5 -2û

é 3 4ù 43. If A is an invertible matrix and A -1 = ê ú then A = ? ë 5 6û é 6 -4ù (a) ê ú ë -5 3 û

é1 ê (b) ê 3 ê1 êë 5

1ù 4ú ú 1ú 6 úû

é -3 (c) ê 5 ê êë 2

2ù -3 ú ú 2 úû

(d) none of these

2ù é1 2 44. If A = ê ú and f ( x) = 2x - 4x + 5 then f ( A) = ? 4 3 ë û é 19 -32ù (a) ê ú ë -16 51û

é 19 -16ù (b) ê ú ë -32 51û

é 19 -11ù (c) ê 51úû ë -27

(d) none of these

é 1 4ù 2 45. If A = ê ú then A - 4A = ? ë 2 3û (a) I

(b) 5I

(c) 3I

(d) 0

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46. If A is a 2-rowed square matrix and|A| = 6 then A × adj A = ? é 6 0ù é 3 0ù (a) ê (b) ê ú ú 0 6 ë û ë 0 3û é 1 0ù (c) ê 6 1 ú ê ú ë 0 6û

(d) none of these

47. If A is an invertible square matrix and k is a non-negative real number then ( kA) -1 = ? (a) k × A -1

(b)

1 × A -1 k

(c) - k × A -1

(d) none of these

4 1ù é 3 48. If A = ê 1 0 -2ú then A -1 = ? ú ê êë -2 -1 2úû é 2 9 -8ù (a) ê -2 8 7 ú ú ê êë -1 5 -4úû

9 -8ù é -2 ê (b) 2 8 7ú ú ê 4úû êë -1 -5

é -2 -9 -8ù (c) ê 2 8 7ú ú ê êë -1 -5 -4úû

(d) none of these

49. If A is a square matrix then ( A + A ¢) is (a) a null matrix

(b) an identity matrix

(c) a symmetric matrix

(d) a skew-symmetric matrix

50. If A is a square matrix then ( A - A ¢) is (a) a null matrix

(b) an identity matrix

(c) a symmetric matrix

(d) a skew-symmetric matrix

51. If A is a 3-rowed square matrix and|3A| = k|A|then k = ? (a) 3

(b) 9

(c) 27

(d) 1

52. Which one of the following is a scalar matrix? é 6 0ù é1 1ù (a) ê (b) ê ú ú 1 1 ë 0 3û ë û é -8 0ù (c) ê ú ë 0 -8û

(d) none of these

1ù é 1 -1ù éa 2 2 2 53. If A = ê ú and B = ê b -1ú and ( A + B) = ( A + B ) then 2 1 ë û ë û (a) a = 2, b = - 3

(b) a = - 2, b = 3

(c) a = 1, b = 4

(d) none of these

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1. (b)

2. (c)

3. (a)

4. (b)

5. (c)

6. (b)

7. (a)

8 (c)

9. (b) 10. (b)

11. (b) 12. (c) 13. (c) 14. (c) 15. (c) 16. (b) 17. (b) 18. (b) 19. (d) 20. (b) 21. (a) 22. (b) 23. (c) 24. (c) 25. (b) 26. (d) 27. (b) 28. (c) 29. (b) 30. (c) 31. (c) 32. (b) 33. (b) 34. (b) 35. (b) 36. (b) 37. (b) 38. (a) 39. (c) 40. (d) 41. (b) 42. (b) 43. (c) 44. (b) 45. (b) 46. (a) 47. (b) 48. (c) 49. (c) 50. (d) 51. (c) 52. (c) 53. (c)

HINTS TO SOME SELECTED OBJECTIVE QUESTIONS 1. 2A = ( A + B) + ( A - B) and 2B = ( A + B) - ( A - B). é 5 6 ù é 3 -2 ù é 2 8 ù 2. 2 A = ê ú× ú=ê ú-ê ë -7 10 û ë 5 6 û ë -12 4 û 1 3. 4 A + 3X = 5 B Þ 3X = (5 B - 4 A ) Þ X = (5 B - 4 A ). 3 4. B = ( 2 A - 3 B) - 2( A - 2 B). 5. 5 A = 2( 2 A - B) + ( 2 B + A ). Then, A =

1 (5 A ). 5

6. [2 x + 1 = 5 Þ x = 2] and [8 + y = 0 Þ y = - 8]. 7. ( x - y = -1 and 2 x - y = 0 ) Þ ( x = 1, y = 2 ) ( 2 x + z = 5 Þ z = 3 ) and ( 3z + w = 13 Þ w = 4 ). 8. Solve x + 2 y = 3 and 3 y + 2 x = 5. 9. A is singular Û |A| = 0. 12. A is singular Û |A| = 0. 14. Use AA -1 = I. 15. Using distributive law, we have ( A + B) × ( A - B) = A( A - B) + B( A - B) = ( A 2 - AB + BA - B 2 ). 16. ( A + B) 2 = ( A + B) × ( A + B) = A( A + B) + B( A + B) = ( A 2 + AB + BA + B 2 ). 17. ( A - B) 2 = ( A - B) × ( A - B) = A( A - B) - B( A - B) = ( A 2 - AB - BA + B 2 ). 18. Given A ¢ = A and B¢ = B. \ \

( AB - BA )¢ = ( AB)¢ - ( BA )¢ = ( B¢ A ¢ - A ¢ B¢ ) = ( BA - AB ) = - ( AB - BA ) ( AB - BA ) is skew symmetric.

19. A and B are inverses of each other only when AB = BA = I. 20. adj ( AB) = ( adj B)( adj A ). 21. adj ( adj A ) =|A|(n - 2 ) =|A|(3 - 2 ) × A =|A|× A = 4 A. 22. |adj A| =|A|(n - 1 ) =|A|2 = 5 2 = 25. é a b ù é 3 2ù é 4 1ù 24. Let ê ú × Find a, b , c and d. ú=ê úê ë c d û ë 1 -1û ë 2 3 û

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321

25. AA -1 = I Þ |AA -1| =|I| = 1 Þ |A|×|A -1| = 1 Þ |A -1| =

1 × |A|

26. ( AB) -1 = B -1 A -1 . 27. [AB = 0 and A ¹ 0 , B ¹ 0] Þ |A| = 0 and| B| = 0. 28. 2 I = ( A - A 2 ) Þ 2 A -1 = A -1 A - A -1 AA = I - IA = ( I - A ) \

Þ A -1 =

1 ( I - A ). 2

29. A is not invertible Û |A| = 0. 1 30. |A| = 1 Þ A -1 = adj A = ( adj A ). |A| 31. A 2 = O Þ A is nilpotent. 32. A 2 = A Þ A is idempotent. 33. Given|A| = 0. So, A( adj A ) =|A|× I = 0 × I = 0. \

A( adj A ) is a null matrix. é 1 0ù 34. A × adj A =|A|× I = 8 ´ ê ú = 8 I ® |A| = 8. ë 0 1û 35. AA -1 = I Þ |AA -1| =|I| Þ |A|×|A -1| = 1 Þ |A -1| =

1 × |A|

-1 ½ -2 3½ ½ = ( -2 - 3 ) = ( -5 ) Þ |A -1| = |A | =½ × 5 ½ 1 1½ é 3 1ù é 1 0ù é 3 1ù é 3 1ù 36. ê ú ê7 5 ú + x ê 0 1ú = y ê7 5 ú 7 5 û ë û ë û ûë ë Þ

8 ù é 3y yù é 16 + x é 16 8 ù é x 0 ù é 3 y y ù . = ê 56 32 ú + ê 0 x ú = ê 7 y 5 y ú Þ ê 56 32 + x úû êë 7 y 5 y úû û ë û ë û ë ë

\

y = 8 and ( 16 + x = 3 y = 3 ´ 8 = 24 ) Þ x = 8.

37. A and B anticommute Û AB = - BA. 39. AB = I Þ B = A -1 . 40. ( AB) -1 = B -1 A -1 . ½ 2 -1½ ½ = ( 6 + 1) = 7 ¹ 0 41. |A| =½ ½ 1 3½ \ Þ

Þ

M11 = 3 , M12 = 1, M 21 = - 1 and M 22 = 2 C 11 = 3 , C 12 = - 1, C 21 = 1 and C 22 = 2 ¢ é 3 -1ù é 3 1ù Adj A = ê =ê ú ú ë 1 2 û ë -1 2 û é 3 1ù 1 1 é 3 1ù ê 7 7 ú -1 A = ( adj A ) = ê ú× ú=ê |A| 7 ë -1 2 û ê 1 2 ú êë 7 7 úû

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42. A -1 =

é 9 -3 ù 1 × adj A ® adj A =|A|× A -1 = 3 A -1 = ê ú× |A| ë -5 2 û

43. A = ( A -1 ) -1 . So, find the inverse of A -1 . 44. f ( A ) = 2 A 2 - 4 A + 5 I. é 1 4 ù é 1 4 ù é 4 16 ù é 9 16 ù é 4 16 ù é5 0 ù 45. A 2 - 4 A = ê ú = 5I. ú=ê ú-ê ú=ê ú-ê úê ë 2 3 û ë 2 3 û ë 8 12 û ë 8 17 û ë 8 12 û ë 0 5 û é 1 0ù é 6 0ù 46. A × ( adj A ) =|A|× I = 6 × ê ú× ú=ê ë 0 1û ë 0 6 û 47. ( kA ) -1 = 48. A -1 =

1 × A -1 is true. k

1 × adj A. |A|

49. A is a square matrix Þ ( A + A ¢ ) is symmetric. 50. A is a square matrix Þ ( A - A ¢ ) is skew-symmetric. 51. | 3 A| = ( 3 ´ 3 ´ 3 )|A| = 27 ×|A|. 52. A scalar matrix is a square matrix each of whose non-diagonal elements is 0 and all diagonal elements are equal. 53. ( A + B) 2 = ( A 2 + B 2 ) Û A 2 + B 2 + AB + BA = ( A 2 + B 2 ) Û AB = - BA. é a - b 2 ù é - a - 2 a + 1ù AB = - BA Û ê ú ú=ê ë 2 a - b 3 û ë - b + 2 b - 1û Now, ( a + 1 = 2 and b - 1 = 3 ) Þ ( a = 1 and b = 4).

Senior Secondary School Mathematics for Class 12 Pg-323

9. CONTINUITY AND DIFFERENTIABILITY FUNCTIONS Real Functions Let R be the set of all real numbers, and let X and Y be any two nonempty subsets of R. Then, a rule f which associates to each x Î X, a unique real number f ( x) Î Y is called a real function from X to Y and we write, f : X ® Y. f ( x) is called the image of x or the value of the function at x. The sets X and Y are respectively known as the domain and the codomain of f . Also, the set { f ( x) : x Î X} is called the range of f . Clearly, range ( f ) Í Y. However, if range ( f ) = Y , we say that f is an onto function; otherwise f is said to be an into function. If two or more than two elements in X have the same image in Y then f is said to be a many-one function. On the other hand, if different elements in X have different images in Y, we say that f is one-one. Clearly, f is one-one Û [ f ( x1) = f ( x 2) Þ x1 = x 2]. A one-one onto function is called a one-to-one correspondence. REMARK 1

Sometimes a function is described only by a formula and the domain of the function is not explicitly stated. In such cases, the domain of the function is the set of all those real numbers for which the formula is meaningful.

REMARK 2

Usually the domain of a real function is an interval. For any two real numbers a and b, where a < b , we define (i) closed interval [a , b] = {x Î R : a £ x £ b} (ii) open interval ]a , b[ = {x Î R : a < x < b} (iii) right-half open interval [a , b[ = {x Î R : a £ x < b} (iv) left-half open interval ]a , b] = {x Î R : a < x £ b} (v) ]a , ¥ [ = {x Î R : x > a} (vi) [a , ¥ [ = {x Î R : x ³ a} (vii) ] -¥ , a[ = {x Î R : x < a} (viii) ] -¥ , a] = {x Î R : x £ a} Sometimes, we write, R = ] -¥ , ¥ [.

Some Important Functions Let c be a fixed real number. Then, the function defined by f ( x) = c for all x Î R is called a constant function c.

1. CONSTANT FUNCTION

323

Senior Secondary School Mathematics for Class 12 Pg-324

324

Senior Secondary School Mathematics for Class 12

Clearly, dom ( f ) = R and range ( f ) = {c}. The function defined by f ( x) = x for all x Î R is called the identity function. Clearly, its domain is R and its range is R.

2. IDENTITY FUNCTION

The function defined by ì x , when x ³ 0 f ( x) = |x| = í î- x , when x < 0 is called the modulus function. Since the modulus of every real number is a unique non-negative real number, so dom f ( x) = R. Since|x|is either 0 or a positive real number, we have range ( f ) = {|x|: x Î R} = set of non-negative real numbers.

3. MODULUS FUNCTION

4. RECIPROCAL FUNCTION

The function defined by f ( x) =

1 is called the reciprocal x

function. 1 is not defined when x = 0. x \ dom ( f ) = R - {0}. 1 1 Also, y = Û x= . x y Clearly,

Clearly, x is defined for all real numbers y except when y = 0. \ range ( f ) = R - {0}. 5. SIGNUM FUNCTION

This function is defined by

ì|x| ï , when x ¹ 0 f ( x) = í x ïî 0, when x = 0.

ì 1, when x > 0 ï f ( x) = í 0, when x = 0 ï-1, when x < 0. î Clearly, its domain is R and range = {-1, 0, 1}. Thus, we have

6. SQUARE-ROOT FUNCTION

Let f ( x) = + x .

We know that the negative real numbers do not have real square roots. So, f ( x) is not defined when x is a negative real number. dom ( f ) = set of all non-negative real numbers = [0, ¥ [. \ Clearly, range ( f ) = {+ x : x Î [0, ¥ [} = [0, ¥ [. 7. STEP FUNCTION OR THE GREATEST INTEGER FUNCTION

If x Î R then [x] is defined as the greatest integer not exceeding x. For example, we have [2.01] = 2; [2.9] = 2; [-1.3] = - 2; [ 3] = 3 and [-1] = -1, etc.

Senior Secondary School Mathematics for Class 12 Pg-325

Continuity and Differentiability

325

Now, if we consider f ( x) = [x] then clearly for each x Î R , [x] is defined. So dom ( f ) = R. By definition, [x] is an integer. So, range ( f ) = {[x] : x Î R} = set of all integers. EXAMPLE SOLUTION

Find a set of all real numbers x such that [x] = 2. Clearly, for all x such that 2 £ x < 3 , we have f ( x) = [x] = 2. \ required set = {x Î R : 2 £ x < 3} = [2, 3[.

8. SMALLEST INTEGER FUNCTION (OR CEILING FUNCTION) For any real number x, we define éx ù as the smallest integer greater than or equal to x.

For example, é6.3 ù = 7 , é7.01ù = 8, é-6.1ù = - 6, é-2.9ù = -2, é-3 ù = -3 , é5 ù = 5. The function f : R Î R : f ( x) = éx ù , x Î R is called the smallest integer function or ceiling function. Clearly, domain ( f ) = R and range ( f ) = I. 9. POLYNOMIAL FUNCTION

A function of the form

p( x) = a0 x n + a1 x n-1 + a2 x n- 2 + ¼ + an-1 x + an , where a0 , a1 , a2 , ¼, an - 1 , an are real numbers, a0 ¹ 0 and n is a non-negative integer, is called a polynomial function of degree n. Polynomials of degree 1, 2, 3 and 4 are respectively called linear, quadratic, cubic and biquadratic polynomials. Thus, (i) f ( x) = ax + b , a ¹ 0, is a linear polynomial. (ii) f ( x) = ax 2 + bx + c, a ¹ 0, is a quadratic polynomial. (iii) f ( x) = ax 3 + bx 2 + cx + d , a ¹ 0, is a cubic polynomial. 10. RATIONAL FUNCTION

A function of the form f ( x) =

p( x) , where p( x) and q( x) are q( x)

polynomials and q( x) ¹ 0, is called a rational function. æ x2 + 1 ö ÷ is a rational function, where x 3 - 2x + 5 ¹ 0. Thus, f ( x) = ç 3 ç x - 2x + 5 ÷ è ø

Graphs Graph of a Function For a given function f ( x), the aggregate of the points {x , f ( x)} is called the graph or the curve representing the function. In practice, we plot some of the points and join them freehand to obtain the graph.

Senior Secondary School Mathematics for Class 12 Pg-326

326 EXAMPLE 1

SOLUTION

Senior Secondary School Mathematics for Class 12

Draw the graphs of the constant functions (i) f ( x) = 2 (ii) f ( x) = 0 (iii) f ( x) = -2 (i) When f ( x) = 2 for all x Î R , some of the points on the graph may be taken as ( 0, 2), ( -1, 2), (1, 2), ( -2, 2), ( 2, 2), etc. Joining these points, we obtain a line y = 2, drawn parallel to the x-axis at a distance of 2 units from it, as the required graph.

Y f(x) = 2 X

Y

(ii) The graph of the function f ( x) = 0 is the line y = 0, i.e., the x-axis. (iii) The graph of the function f ( x) = -2 is the line y = -2, drawn parallel to the x-axis at a distance of 2 units below the x-axis.

X

O

Y

X

X

O f(x) = –2 Y

EXAMPLE 2

SOLUTION

Draw the graphs of the linear functions (i) f ( x) = 1 - x (ii) f ( x) = 2x + 1 (i) When f ( x) = 1 - x , some of the points on the graph are ( 0, 1), (1, 0), ( 2, - 1), ( -1, 2), etc. Joining these points, we get a line as the graph of the function.

Y (0, 1)

X

Y

(ii) Let f ( x) = 2x + 1.

Y

Some of the points on the graph are ( 0, 1), (1, 3), ( -1, - 1), ( 2, 5), etc. X (–1, –1)

2x+ 1

(0, 1) X

O

f(x )=

Joining these points, we obtain a line as the graph.

(1, 0) X f(x )= 1– x

O

Y

SOLUTION

The graph of a linear function is a straight line. Draw the graph of the identity function f ( x) = x. f ( x) = x is clearly a linear function whose graph must be a line. Plotting the points ( 0, 0), (1, 1), ( -1, -1), etc., and joining them, we get the required graph.

Y (1, 1)

X

O

(–1, –1)

x

EXAMPLE 3

f(x )=

REMARK

Y

X

Senior Secondary School Mathematics for Class 12 Pg-327

Continuity and Differentiability EXAMPLE 4

327

Draw the graphs of the polynomial functions (ii) f ( x) = 1 - x 2 (i) f ( x) = x 2

Y

(iii) f ( x) = x 3 - x SOLUTION

(–2, 4) 2

(i) The function f ( x) = x is a quadratic function. Some of the points on the graph are (0, 0), (1, 1), ( -1, 1), ( -2, 4), ( 2, 4), (-3, 9), (3, 9), etc. Joining these points, we get a parabola as the graph.

(–1, 1) X

(1, 1) X

(0, 0) O

Y

(ii) f ( x) = 1 - x 2 is also a quadratic function. Some of the points on the graph are ( 0, 1), (1, 0), ( -1, 0), ( 2, - 3), ( -2, - 3), ( -3 , - 8), ( 3 , - 8), etc.

Y (0, 1)

X

(–1, 0)

(1, 0) O

Joining these points, we obtain a parabola as its graph.

X

Y

(iii) Let f ( x) = x 3 - x.

Y

This is a cubic function. Some of the points on the graph are ( 0, 0), ( -1, 0), (1, 0), ( -0.5 , 0.375), ( 0.5 , - 0.375), ( 2, 6), ( -2, - 6), etc. Joining these points, we obtain the required graph. REMARK

(2, 4)

X

X

O

Y

It may be observed here that whenever ( x , y) is a point on the graph then ( - x , - y) is also a point on the graph. So, the graph is symmetrical about the origin. This is an important property possessed by the graph of an odd function.

EXAMPLE 5

Draw the graph of the modulus function f ( x) = |x|.

SOLUTION

ì x when x ³ 0 f ( x) = |x| = í î- x when x < 0

(–2, 2)

Some of the points on the graph are ( 0, 0), ( -1, 1), ( -2, 2), (1, 1), ( 2, 2), etc.

X

Joining these points, we get the required graph.

Y (2, 2)

(–1, 1)

(1, 1) O (0, 0)

Y

X

Senior Secondary School Mathematics for Class 12 Pg-328

328

Senior Secondary School Mathematics for Class 12

It may be observed here that whenever ( x , y) is a point on the graph, then ( -x , y) is also a point of it. Thus, the graph is symmetrical about the y-axis. This is an important property possessed by the graphs of even functions. 1 EXAMPLE 6 Draw the graph of the reciprocal function f ( x) = . x 1 SOLUTION Clearly, f ( x) = is not defined at x = 0. Some points on the graph x Y are (1, 1), ( -1, -1), (1/2, 2), ( 2, 1/2), ( -1/2, - 2), ( -2, - 1/2), (1/ 3 , 3), ( -1/ 3 , - 3), ( 3 , 1/ 3), etc. Joining these points, we get the X X O 1 required graph. Since f ( x) = is an x odd function, it is symmetrical about REMARK

Y

the origin. EXAMPLE 7

Draw the graph of the square-root function f ( x) = x .

SOLUTION

Let f be a real-valued function which associates to each non-negative real number x, its non-negative square root. Y Then, f : R 0+ ® R 0+ : f ( x) = x , is called the square-root function. y=

Domain f ( x) = R 0+ , and range ( f ) = R 0+ .

Some of the points on the graph are (0, 0), (1, 1), (2, 1.4), (3, 1.7), (4, 2), (5, 2.2), etc. Joining these points, we get the required graph. EXAMPLE 8 SOLUTION

X

Draw the graph of the rational function f ( x) =

O

x

X

Y 2

x -1 × x -1

x2 - 1 × x -1 0 Now, f (1) = , which is meaningless. 0 So, the function is not defined at x = 1. Also, when x ¹ 1, we have æ x2 - 1ö ÷ = ( x + 1). f ( x) = ç ç x -1 ÷ è ø This being a linear function, its graph is a straight line. Some of the points on the graph are ( 0, 1), ( -1, 0), ( 2, 3), ( 3 , 4), ( -2, -1), ( -3 , - 2), etc. Let f ( x) =

Senior Secondary School Mathematics for Class 12 Pg-329

Continuity and Differentiability

329

Joining these points, we obtain the required graph. Clearly, the point (1, 2) does not lie on the graph. So, it is a broken graph, and we shall say that the given function is discontinuous at x = 1. EXAMPLE 9 SOLUTION

EXAMPLE 10 SOLUTION

Draw the graph of the step function f ( x) = [x]. Y As the definition of the function indicates, (3, 2) (2, 2) for all x such that -2 £ x < - 1, (2, 1) (1, 1) we have f ( x) = - 2; O X X for all x such that -1 £ x < 0, (1, 0) (–1, –1) (0, –1) we have f ( x) = -1; (–2, –2) for all x such that 0 £ x < 1, (–1, –2) Y we have f ( x) = 0; for all x such that 1 £ x < 2, we have f ( x) = 1, and so on, é -2 when x Î [-2, - 1[ ê -1 when x Î [-1, 0[ ê i.e., f ( x) = ê 0 when x Î [0, 1[ ê 1 when x Î [1, 2[ ê êë and so on. Clearly, the function jumps at the points ( -1, - 2), ( 0, -1), (1, 0), ( 2, 1), etc. In other words, the given function is discontinuous at each integral value of x. Draw the graph of the smallest integer function f ( x) = éx ù . As the definition of the function suggests, for all x such that -3 < x £ - 2, we have f ( x) = - 2; for all x such that -2 < x £ - 1, we have f ( x) = - 1; for all x such that -1 < x £ 0, we have f ( x) = 0; for all x such that 0 < x £ 1, we have f ( x) = 1; and so on. é -2 when x Î ] - 3 , - 2] ê -1 when x Î ] - 2, - 1] ê ê 0 when x Î ] - 1, 0] i.e., f ( x) = ê 1 when x Î ]0, 1] ê ê 2 when x Î ]1, 2] ê 3 when x Î ]2, 3] ê êë and so on. Plotting these points, we can get the required graph. The function jumps at the points (–2, –1), (–1, 0), (0, 1), (1, 2), etc., or is discontinuous at each integral value of x.

Senior Secondary School Mathematics for Class 12 Pg-330

330

EXAMPLE 11

SOLUTION

EXAMPLE 12

SOLUTION

Senior Secondary School Mathematics for Class 12

ì|x| when x ¹ 0 ï Draw the graph of the signum function f ( x) = í x ïî 0 when x = 0. Clearly, ( 0, 0) is a point on the graph. Now, when x > 0, we have |x| = x , and so in this case, we have, f ( x) = 1, i.e., f ( x) = 1 for all values of x > 0. Y And, when x < 0, we have|x| = - x and therefore, y=1 f ( x) = -1 for all values of x < 0. X X Hence the graph may be drawn, as O y = –1 shown in the adjoining figure. Clearly, the function is broken (i.e., it is discontinuous) at each of the points Y x = -1, 0 and 1.

y=

ì x 2 , when x < 0 ï Draw the graph of the function f ( x) = í x , when 0 £ x £ 1 ï 1/x , when 1 £ x < ¥ . î Y Here, the graph consists of three parts. Some of the points of the graph are ( -3 , 9), ( -2, 4), ( -1, 1), æ 1ö æ1 1ö æ 3 3 ö ( 0, 0), ç , ÷ , ç , ÷ , (1, 1), ç 2, ÷ , y = 1/x è 2ø è 2 2ø è 4 4 ø X O æ 1ö ç 3 , ÷ , etc. 3ø è Y And, the graph may now be drawn, as shown in the adjoining figure. x

y=

x

2

EXAMPLE 13

Draw the graph of the function f ( x) = |x| + |x - 1|.

SOLUTION

Let us consider the following cases. Case I When x < 0 In this case, ( x - 1) < 0. \ |x| = - x and |x - 1| = - ( x - 1) = 1 - x. Consequently, |x| + |x - 1| = - x + 1 - x = 1 - 2x. Case II

When 0 £ x £ 1 In this case, |x| = x and |x - 1| = - ( x - 1) = 1 - x. \

Case III

|x| + |x - 1| = x + 1 - x = 1.

When x > 1 In this case, |x| = x and |x - 1| = x - 1. \

|x| + |x - 1| = x + ( x - 1) = ( 2x - 1).

Thus, we may define the above function as

X

Senior Secondary School Mathematics for Class 12 Pg-331

Continuity and Differentiability

331

Y ì 1 - 2x , when x < 0 (–2, 5) ï f ( x) = í 1, when 0 £ x £ 1 (–1, 3) (2, 3) ï 2x - 1, when x > 1. î (0, 1) So, we have (1, 1) (i) a linear function 1 - 2x when X X O x < 0; Y (ii) a constant function 1 when 0 £ x £ 1; (iii) a linear function 2x - 1 when x > 1. The corresponding points on these parts of the graph are ( -1, 3), ( -2, 5), ( 0, 1), (1, 1), (2, 3), (3, 5), etc. Joining these points, we obtain the graph as shown. EXAMPLE 14

Draw the graph of the exponential function: æ1ö (ii) f ( x) = ç ÷ è 3ø

(i) f ( x) = 2x

x

(i) Let f ( x) = 2x . Some of the points on the graph are 1ö æ 1ö æ 1ö æ (0, 1), (1, 2), (2, 4), (3, 8), ç -1, ÷ , ç -2, ÷ , ç -3 , ÷ , etc. 2ø è 4ø è 8ø è

SOLUTION

y=

2 x

And so the graph takes the form, shown in the adjoining figure. It may be observed here that the given (2, 4) function is strictly increasing. Also, as the value of x decreases, the corresponding (1, 2) (–2, 1/4) (–1, 1/2) (0, 1) value of the function decreases, and X X O therefore, on the left-hand side of the y-axis, the curve comes closer and closer to the Y x-axis. x REMARK This is the case of the exponential function a , where a > 1. Y

(3, 8)

x

1 æ1ö (ii) Let f ( x) = ç ÷ = x × 3 è ø 3 Y y=

(–2, 9)

1/3

X

x

(–1, 3)

(1, 1/3)

(0, 1) O Y

REMARK

Some of the points on the graph are ( -2, 9), æ 1 ö æ 1ö ( -1, 3), ( 0, 1), ç1, ÷ , ç 2, ÷ , etc. è 3 ø è 9ø Joining these points, we obtain the graph as shown. It follows from the graph that the given (2, 1/9) function is strictly decreasing. X On the RHS of the y-axis, the curve comes closer and closer to the x-axis. This is the case of the exponential function a x , where 0 < a < 1.

Senior Secondary School Mathematics for Class 12 Pg-332

332 EXAMPLE 15

Senior Secondary School Mathematics for Class 12

Draw the graphs of the logarithmic functions (i) log a x , when a > 1

SOLUTION

(ii) log a x , when 0 < a < 1.

(i) We know that when Y y = ax a > 1, the function a x is y=x strictly increasing, i.e., y = loga x different values of x X X give different values of (1, 0) O a x . Also, the range of this function is R. So, the function Y f ( x) = a x is one-one onto and therefore invertible. Its graph is of the form shown in the adjoining figure. The graph of the function g( x) = log a x is the reflection of the graph of f ( x) = a x in the line y = x. It may be noted that the graph passes through (1, 0). Y (ii) We know that when y=x 0 < a < 1, the function a x y = ax is strictly decreasing, (0 < a < 1) i.e., different values of x X X O (1, 0) give different values of x a . So, the function is one-one. Also, its range y = loga x Y is R. So, it is onto. Thus, the function a x is invertible. Its inverse is the log function reflected by the line y = x , as shown. The graph clearly passes through (1, 0).

EXAMPLE 16

On the same scale draw the graphs of ex and log e x.

SOLUTION

Since e lies between 2 and 3, it follows that e > 1. So, it is a particular case of a x , where a > 1. The function ex is strictly increasing. Also, its range is R. So, the function is one-one onto and therefore invertible. Y The graph passes through the point ( 0, 1) and comes closer and closer to the x-axis and the values of x decrease. Thus, the graph of ex may be drawn as shown in the figure. Its reflection in the line y = x gives the graph of log e x.

y=x

y=e X

x

(0, 1) X

O y = loge x Y

Senior Secondary School Mathematics for Class 12 Pg-333

Continuity and Differentiability

333

Graphs of Trigonometric Functions 1. GRAPH OF A SINE FUNCTION We know that a sine function is periodic with period 2p. So, we have to sketch the graph in the interval [0, 2p] and then we may complete the graph by repeating it over intervals, each of length 2p. We first draw it in the interval [0, p/2]. We know that it is strictly increasing in this interval. Also, we may use the table, given below.

x sin x

0°

30°

0

1 2

45° 1 = 0.707 2

60°

90°

3 = 0.87 2

1

é pù Thus, we may draw the graph in the interval ê 0, ú × ë 2û Y (–3 /2, 1) X

( /2, 1) (2 , 0)

(– , 0) (–2 , 0)

0

( , 0)

X

(3 /2, –1)

(– /2, –1) Y

Now, we know that sin ( p - x) = sin x. So, we may get some other values of the function in the interval [p/2, p] Moreover, the function is strictly decreasing in this interval. Thus, we may draw it in the interval [p/2, p]. Finally, we draw it in ( p , 2p), using the fact that sin ( p + x) = - sin x. The graph may be completed now by making repetitions over each interval of length 2p. We know that a cosine function is periodic with period 2p. By making use of the table given below, we may first draw it in the interval [0, p/2], keeping in view that it is strictly decreasing in this interval. 2. GRAPH OF A COSINE FUNCTION

x

0°

p 6

p 4

cos x

1

3 = 0.87 2

1 = 0.707 2

p 3 1 2

p 2 0

Now, cos ( p - x) = - cos x , so, we may draw the graph in the interval [p/2, p]. It is strictly decreasing in this interval also. Further, making use of cos ( p + x) = cos x , we may draw the graph from p to 2p, as shown in the figure.

Senior Secondary School Mathematics for Class 12 Pg-334

334

Senior Secondary School Mathematics for Class 12

The graph may now be completed by making repetitions over each interval of length 2p. Y (0, 1)

(–2 , 1) X

(2 , 1) ( /2, 0)

(– /2, 0) O

(–3 /2, 0) (– , –1)

(3 /2, 0)

X

( , –1) Y

3. GRAPH OF A TANGENT FUNCTION We know that a tangent function is a periodic function with period p. Therefore, it is enough to draw the graph over an interval of length p. The complete graph then consists of infinitely many repetitions of the same to the left as well as to the right.

Since tan ( - x) = - tan x , therefore, if ( x , tan x) is any point on the graph then ( - x , - tan x) is also a point on the graph. Thus, we can say that the graph of y = tan x is symmetrical in opposite quadrants. Some of the values of the function are given below: x

0°

p 6

p 4

tan x

0

3 = 0.58 3

1

x

2p 3

3p 4

tan x

- 3 = -1.73

-1

p 3 3 = 1.73

p 2 undefined

5p 6 -

3 = - 0.58 3

The graph may thus be drawn as shown below. Y

/2 O

3 /2 2

X

Similarly, the graphs of cosec x , sec x and cot x may be drawn.

p 0

Senior Secondary School Mathematics for Class 12 Pg-335

Continuity and Differentiability

335

Continuity CONTINUITY AT A POINT A real function f ( x) is said to be continuous at a point a of its domain if lim f ( x) exists and equals f ( a). x ®a

Thus, f ( x) is continuous at x = a if lim f ( x) = lim f ( x) = f ( a).

x ® a+

x ® a-

If f ( x) is not continuous at a point, it is said to be discontinuous at that point. f ( x) is discontinuous at x = a in each of the following cases:

REMARK

(i) f ( a) is not defined (ii) lim f ( x) does not exist x ®a

(iii) lim f ( x) ¹ f ( a)

{removable discontinuity}

x ®a

SOLVED EXAMPLES EXAMPLE 1

Show that f ( x) = x 3 is continuous at x = 2.

SOLUTION

We have

f ( 2) = 2 3 = 8;

lim f ( x) = lim ( 2 + h) 3 = lim ( 8 + h 3 + 12h + 6h2) = 8;

x ® 2+

h® 0

h® 0

3

lim f ( x) = lim ( 2 - h) = lim ( 8 - h 3 - 12h + 6h2 ) = 8.

x ® 2-

\

h® 0

h® 0

lim f ( x) = lim f ( x) = f ( 2).

x ® 2+

x ® 2-

Hence, f ( x) is continuous at x = 2. EXAMPLE 2

Show that f ( x) = [x] is not continuous at x = n, where n is an integer.

SOLUTION

We have f (n) = [n] = n; lim f ( x) = lim f (n + h) = lim [n + h] = n {Q [n + h] = n} ; x ® n+

h® 0

h® 0

lim f ( x) = lim f (n - h) = lim [n - h] = (n - 1) {Q [n - h] = (n - 1)}.

x ® n-

h® 0

h® 0

Thus, lim f ( x) ¹ lim f ( x) and therefore, lim f ( x) does not exist. x ® n+

x ® n-

x ®n

Hence, f ( x) is discontinuous at x = n. ì x , if x is an integer ; f ( x) = í î 0, if x is not an integer is discontinuous at each integral value of x.

EXAMPLE 3

Show that the function

SOLUTION

Let x = n, where n is an integer. Then, f (n) = n;

Senior Secondary School Mathematics for Class 12 Pg-336

336

Senior Secondary School Mathematics for Class 12

lim f ( x) = lim f (n + h) = 0

x ® n+

h® 0

[Q (n + h) is not an integer Þ f (n + h) = 0];

lim f ( x) = lim f (n - h) = 0

and

x ® n-

h® 0

[Q (n - h) is not an integer Þ f (n - h) = 0].

lim f ( x) = lim f ( x) = 0.

\

x ® n+

x ® n-

lim f ( x) = 0 ¹ f (n).

So,

x ®n

Hence, f ( x) is discontinuous at x = n. EXAMPLE 4

SOLUTION

Discuss the continuity of the function f ( x) at x = 0, if ì 2x - 1, x < 0 f ( x) = í î 2x + 1, x ³ 0.

[CBSE 2002]

Clearly, f ( 0) = ( 2 ´ 0 + 1) = 1. lim f ( x) = lim f ( 0 + h) x ® 0+

h® 0

= lim [2( 0 + h) + 1] = lim ( 2h + 1) = 1. h® 0

h® 0

lim f ( x) = lim f ( 0 - h)

x ® 0-

h® 0

= lim [2( 0 - h) - 1)] = lim ( -2h - 1) = -1. h® 0

h® 0

Thus, lim f ( x) ¹ lim f ( x) and therefore, lim f ( x) does not exist. x ® 0+

x ® 0-

x ®0

Hence, f ( x) is discontinuous at x = 0. EXAMPLE 5

Show that the function

ì 3 x - 2, when x £ 0 f ( x) = í î x + 1, when x > 0

is discontinuous at x = 0. SOLUTION

We have, f ( 0) = ( 3 ´ 0 - 2) = -2. lim f ( x) = lim f ( 0 + h)

x ® 0+

h® 0

= lim ( h + 1) = 1. h® 0

lim f ( x) = lim f ( 0 - h)

x ® 0-

h® 0

= lim [ 3( - h) - 2] = lim ( -3 h - 2) = -2. h® 0

\

h® 0

lim f ( x) ¹ lim f ( x) and therefore, lim f ( x) does not exist.

x ® 0+

x ® 0-

Hence, f ( x) is discontinuous at x = 0.

x ®0

Senior Secondary School Mathematics for Class 12 Pg-337

Continuity and Differentiability

EXAMPLE 6

337

ì x , when x ¹ 0 ï f ( x) = í |x| ï 1, when x = 0 î

Show that the function is discontinuous at x = 0.

SOLUTION

It is being given that f ( 0) = 1.

h

lim f ( x) = lim f ( 0 + h) = lim

x ® 0+

h® 0

h ® 0 |h|

lim f ( x) = lim f ( 0 - h) = lim

x ® 0-

\

h® 0

-h

h ® 0 |h|

= lim

h = 1. h

= lim

-h = -1. h

h® 0

h® 0

lim f ( x) ¹ lim f ( x).

x ® 0+

x ® 0-

So, lim f ( x) does not exist. x ®0

Hence, f ( x) is discontinuous at x = 0. EXAMPLE 7

Examine the continuity of the function ì |sin x| , ï f ( x) = í x ïî 1,

SOLUTION

x¹0

[CBSE 2004]

x = 0 at x = 0.

We have f ( 0) = 1. lim f ( x) = lim f ( 0 + h) x ® 0+

h® 0

|sin ( 0 + h)| |sin h| sin h = lim = lim = 1. h® 0 h® 0 h® 0 ( 0 + h) h h

= lim

lim f ( x) = lim f ( 0 - h)

x ® 0-

h ®0

= lim

h® 0

\

|sin ( - h)| |- sin h| sin h = lim = lim = -1. h® 0 h® 0 -h -h -h

lim f ( x) ¹ lim f ( x). So, lim f ( x) does not exist.

x ® 0+

x ® 0-

x ®0

Hence, f ( x) is discontinuous at x = 0. EXAMPLE 8

Show that the function f ( x) = 2x - |x|is continuous at x = 0. [CBSE 2002]

SOLUTION

We have, f ( 0) = ( 2 ´ 0) - |0| = 0. lim f ( x) = lim f ( 0 + h)

x ® 0+

h® 0

= lim ( 2h - |h|) = lim ( 2h - h) = lim h = 0. h® 0

lim f ( x) = lim f ( 0 - h)

x ® 0-

h® 0

h® 0

h® 0

Senior Secondary School Mathematics for Class 12 Pg-338

338

Senior Secondary School Mathematics for Class 12

= lim {2( - h) - |- h|} = lim ( -2h - h) = lim ( -3 h) = 0. h® 0

h® 0

h® 0

Thus, lim f ( x) = lim f ( x) = 0 and therefore, lim f ( x) = 0. x ® 0+

\

x ® 0-

x ®0

lim f ( x) = f ( 0) = 0.

x ®0

Hence, f ( x) is continuous at x = 0.

EXAMPLE 9

SOLUTION

ì|x| + 3 , if x £ - 3 ï Let f ( x) = í - 2x , if - 3 < x < 3 ï6x + 2, if x ³ 3. î Show that f ( x) is continuous at x = -3 but discontinuous at x = 3. We have f ( -3) = |- 3| + 3 = ( 3 + 3) = 6. lim

x ®( -3 )+

f ( x) = lim f ( -3 + h) h® 0

= lim {-2( -3 + h)} = lim ( 6 - 2h) = 6. h® 0

lim

x ®( -3 )-

h® 0

f ( x) = lim f ( -3 - h) h® 0

= lim (|- 3 - h| + 3) = lim {|-( 3 + h)| + 3} h® 0

h® 0

= lim {( 3 + h) + 3} = 6. h® 0

\

lim

x ®( -3 )+

Thus,

lim

f ( x) =

x ®( -3 )

lim

x ®( -3 )-

f ( x) = 6. So,

lim

x ®( -3 )

f ( x) = 6.

f ( x) = f ( -3) = 6.

Hence, f ( x) is continuous at x = - 3. Now, f ( 3) = ( 6 ´ 3 + 2) = 20. lim f ( x) = lim f ( 3 - h)

x ® 3-

h® 0

= lim {-2( 3 - h)} = lim ( -6 + 2h) = -6. h® 0

h® 0

Thus, f ( 3) ¹ lim f ( x). x ® 3-

Hence, f ( x) is discontinuous at x = 3. EXAMPLE 10

Find the value of k for which ìkx + 5 , when x £ 2 f ( x) = í î x - 1, when x > 2 is continuous at x = 2.

SOLUTION

We have, f ( 2) = ( k ´ 2 + 5) = ( 2k + 5). lim f ( x) = lim f ( 2 + h)

x ® 2+

h® 0

[CBSE 2002]

Senior Secondary School Mathematics for Class 12 Pg-339

Continuity and Differentiability

339

= lim {( 2 + h) - 1} = lim (1 + h) = 1. h® 0

h® 0

lim f ( x) = lim f ( 2 - h)

x ® 2-

h® 0

= lim {k( 2 - h) + 5} = lim {( 2k + 5) - kh} = ( 2k + 5). h® 0

h® 0

Now, lim f ( x) exists only when 2k + 5 = 1, i.e., when k = -2. x ®2

When k = -2, we have lim f ( x) = f ( 2) = 1. x ®2

Hence, f ( x) is continuous at x = 2 when k = - 2. EXAMPLE 11

If the following function f ( x) is continuous at x = 0, find the value of k: ì1 - cos 2x , ï f ( x) = í 2x 2 ïî k,

SOLUTION

x¹0

[CBSE 2008]

x = 0.

We have, f ( 0) = k. lim f ( x) = lim f ( 0 + h)

x ® 0+

h® 0

= lim

(1 - cos 2h) 2h2

h® 0

= lim

h® 0

2 sin 2 h 2h2

hö ÷ ø

2

æ sin h ö = lim ç ÷ h ®0 è h ø

2

æ sin = lim ç h® 0 è h

2

ì æ sin h ö ü 2 = í lim ç ÷ ý = (1) = 1. h ® 0 è h øþ î lim f ( x) = lim f ( 0 - h)

x ® 0-

h® 0

ì1 - cos 2( - h) ü {1 - cos ( -2h)} = lim í ý = lim 2 h® 0 î 2( - h) 2h2 þ h® 0 = lim

(1 - cos 2h) 2h2

h® 0

= lim

h® 0

2 sin 2 h 2h2

2

sin h ü ì 2 = í lim ý = 1 = 1. h ® 0 h þ î Since f ( x) is continuous at x = 0, we must have f ( 0) = lim f ( x) = lim f ( x) Þ x ® 0+

EXAMPLE 12

x ® 0-

For what value of k is the following function continuous at x = 0? ì1 - cos 4x , ï f ( x) = í 8x 2 ïî k,

SOLUTION

k = 1.

We have, f ( 0) = k.

x¹0 x=0

[CBSE 2014C]

Senior Secondary School Mathematics for Class 12 Pg-340

340

Senior Secondary School Mathematics for Class 12

lim f ( x) = lim f ( 0 + h)

x ® 0+

h® 0

= lim

(1 - cos 4h) 8h2

h® 0

= lim

2 sin 2 2h

h® 0

8h2

æ sin 2h ö = lim ç ÷ h ® 0 è 2h ø

2

= 12 = 1. lim f ( x) = lim f ( 0 - h)

x ® 0-

h® 0

= lim

{1 - cos 4( - h)} 8( - h) 2

h® 0

= lim

{1 - cos ( -4h)} 8h2

h® 0

= lim

(1 - cos 4h)

h® 0

8h2

[Q cos ( - q) = cos q] = lim

2

2 sin 2h

h® 0

8h2

æ sin 2h ö = lim ç ÷ h ® 0 è 2h ø

2

= 12 = 1. \

lim f ( x) = 1.

x ®0

For continuity at x = 0, we must have é ù f ( 0) = lim f ( x) Þ k = 1 êQ f ( 0) = k and lim f ( x) = 1ú × x ®0 x ® 0 ë û EXAMPLE 13

Given that ì ï (1 - cos 4x) , if x < 0 ïï x2 a , if x = 0 f ( x) = í ï x , if x > 0. ï îï 16 + x - 4 If f ( x) is continuous at x = 0, find the value of a. [CBSE 2006C, ‘12C, ’13C]

SOLUTION

We have, f ( 0) = a. lim f ( x) = lim f ( 0 + h)

x ® 0+

h® 0

ìï üï h = lim í ý h ® 0 ï 16 + h - 4 ï î þ ìï 16 + h + 4 üï h = lim í ´ ý h ® 0 ï 16 + h - 4 16 + h + 4 þï î = lim

h® 0

h × { 16 + h + 4} = lim h® 0 {(16 + h) - 16}

= lim { 16 + h + 4} = ( 4 + 4) = 8. h® 0

h × { 16 + h + 4} h

Senior Secondary School Mathematics for Class 12 Pg-341

Continuity and Differentiability

341

lim f ( x) = lim f ( 0 - h) = lim f ( - h)

x ® 0-

h® 0

= lim

h® 0

{1 - cos 4( - h)} ( - h)

h® 0

= lim

2

(1 - cos 4h) h2

h® 0

= ( 2 ´ 4) × lim

= lim

h2

h® 0

= lim

h® 0

(sin 2 2h)

h® 0

{1 - cos ( -4h)}

( 2h)

2

2 sin 2 2h h2

æ sin 2h ö = 8 × lim ç ÷ h ® 0 è 2h ø

2

= ( 8 ´ 12) = 8. \

lim f ( x) = lim f ( x) = 8 Þ lim f ( x) = 8.

x ® 0+

x ® 0-

x ®0

But, by continuity of f at x = 0, we have lim f ( x) = f ( 0) Þ a = 8

x ®0

[Q f ( 0) = a and lim f ( x) = 8]. x ®0

Hence, a = 8. EXAMPLE 14

If the following function f ( x) is continuous at x = 0, find the values of a, b and c. ì ï sin ( a + 1) x + sin x , if x < 0 ï x ï c, if x = 0 f ( x) = í ï x + bx 2 - x ï , if x > 0. 3 ïî bx 2 [CBSE 2008]

SOLUTION

We have, f ( 0) = c. lim f ( x) = lim f ( 0 - h)

x ® 0-

h® 0

= lim

h® 0

sin ( a + 1)( - h) + sin ( - h) - {sin ( a + 1) h + sin h} = lim h ® 0 ( - h) -h

sin ( a + 1) h + sin h = lim = lim h® 0 h® 0 h

ah æa ö 2 sin ç + 1÷ h × cos 2 è2 ø h

ì ü æa ö ïï sin ç 2 + 1÷ h æ a ah ïï ö è ø × ç + 1÷ × cos ý = 2 × lim í h® 0 2ï ï æç a + 1ö÷ h è 2 ø ïþ îï è 2 ø

Senior Secondary School Mathematics for Class 12 Pg-342

342

Senior Secondary School Mathematics for Class 12

ö æa = 2 ç + 1÷ × lim è 2 ø h® 0

ö æa sin ç + 1÷ h è 2 ø × lim cos ah h® 0 ö æa 2 ç + 1÷ h è2 ø

= ( a + 2) ´ 1 ´ 1 = ( a + 2). lim f ( x) = lim f ( 0 + h)

x ® 0+

h® 0

ìï[ h + bh2 - h] [ h + bh2 + h] üï = lim í ´ ý 3 h® 0 ï [ h + bh2 + h] ïþ bh 2 î = lim

h® 0

= lim

h® 0

( h + bh2 - h) 3

bh 2 ( h + bh2 + h)

= lim

h® 0

bh2 2

bh ( 1 + bh + 1)

1 1 = × ( 1 + bh + 1) 2

Since f ( x) is continuous at x = 0, we have f ( 0) = lim f ( x) = lim f ( x) Þ c = x ® 0-

\

EXAMPLE 15

c=

x ® 0+

1 1 and a + 2 = × 2 2

1 -3 and a = × 2 2

If the function

ì 3 ax + b , for x > 1 ï f ( x) = í 11, for x = 1 ï5 ax - 2b , for x < 1 î

is continuous at x = 1, find the values of a and b. SOLUTION

[CBSE 2012C]

We have, f (1) = 11. lim f ( x) = lim f (1 + h) x ®1 +

h® 0

= lim { 3 a(1 + h) + b} = lim {( 3 a + b) + 3 ah} h® 0

h ®0

= ( 3a + b). lim f ( x) = lim f (1 - h)

x ®1 -

h® 0

= lim {5 a(1 - h) - 2b} = lim {(5 a - 2b) - 5 ah} h® 0

h® 0

= (5 a - 2b). Since f ( x) is continuous at x = 1, we have lim f ( x) = lim f ( x) = f (1).

x ®1 +

x ®1 -

\ 3 a + b = 5 a - 2b = 11. On solving ( 3 a + b = 11) and (5 a - 2b = 11), we get a = 3 , b = 2. Hence, a = 3 , b = 2.

Senior Secondary School Mathematics for Class 12 Pg-343

Continuity and Differentiability EXAMPLE 16

343

For what value of k is the function ìk( x 2 - 2x), if x £ 0 f ( x) = í 4x + 1, if x > 0 î (i) continuous at x = 0?

(ii) continuous at x = 1?

(iii) continuous at x = - 1? SOLUTION

(i) We have f ( 0) = k( 0 - 2 ´ 0 ) = 0. lim f ( x) = lim f ( 0 + h)

x ® 0+

h® 0

= lim {4( 0 + h) + 1} = lim ( 4h + 1) = 1. h® 0

h® 0

lim f ( x) = lim f ( 0 - h)

x ® 0-

h® 0

= lim k{( - h) 2 - 2( - h)} = lim k( h2 + 2h) = 0. h® 0

h® 0

Thus, lim f ( x) ¹ lim f ( x), and thus lim f ( x) does not exist. x ® 0+

x ® 0-

x ®0

So, f ( x) is not continuous at x = 0 for any value of k. (ii) f (1) = ( 4 ´ 1 + 1) = 5. lim f ( x) = lim f (1 + h)

x ®1 +

h® 0

= lim {4(1 + h) + 1} = lim (5 + 4h) = 5. h® 0

h® 0

lim f ( x) = lim f (1 - h)

x ®1 -

h® 0

= lim {4(1 - h) + 1} = lim (5 - 4h) = 5. h® 0

h® 0

lim f ( x) = lim f ( x) = lim f ( x) = 5.

Thus,

x ®1 +

x ®1 -

x ®1

Hence, f ( x) is continuous at x = 1 for every real value of k. (iii) f ( -1) = k {( -1) 2 - 2 ´ ( -1)} = 3 k. lim f ( x) = lim f ( -1 + h)

x ®( -1 )+

h® 0

= lim k {( -1 + h) 2 - 2( -1 + h)} h® 0

= lim k {1 + h2 - 2h + 2 - 2h} h® 0

= lim k ( 3 + h2 - 4h) = 3 k. h® 0

lim f ( x) = lim f ( -1 - h)

x ®( -1 )-

h® 0

= lim k {( -1 - h) 2 - 2( -1 - h)} = lim k{1 + h2 + 2h + 2 + 2h} h® 0

= lim k ( 3 + 4h + h2) = 3 k. h® 0

h® 0

Senior Secondary School Mathematics for Class 12 Pg-344

344

Senior Secondary School Mathematics for Class 12

Thus,

lim f ( x) = lim f ( x) = f ( -1) = 3 k. x ®( -1 )+

x ®( -1 )-

Hence, f ( x) is continuous at x = -1 for each real value of k.

EXAMPLE 17

ì sin 2 ax , when x ¹ 0 ï f ( x) = í x 2 ï 1, when x = 0 î

Show that the function is discontinuous at x = 0.

Redefine the function in such a way that it becomes continuous at x = a. SOLUTION

Clearly, f ( 0) = 1. Also, Since

lim f ( x) = lim

x ®0

x ®0

sin 2 ax x2

2

æ sin ax ö 2 = a 2 × lim ç ÷ =a . ax ® 0 è ax ø

lim f ( x) ¹ f ( 0), f ( x) is discontinuous at x = 0.

x ®0

However, it becomes continuous if f ( 0) = a 2. So, the desired function is

EXAMPLE 18

Is the function f ( x) =

ì sin 2 ax , when x ¹ 0 ï f ( x) = í x 2 ï a 2 , when x = 0. î

( 3 x + 4 tan x) continuous at x = 0? If not, how x

may the function be defined to make it continuous at this point? SOLUTION

Since f ( x) is not defined at x = 0, it cannot be continuous at x = 0. é sin x 1 ù æ 3 x + 4 tan x ö However, lim f ( x) = lim ç × ÷ = lim ê 3 + 4 × x ®0 x ®0 è x x cos x úû ø x ®0 ë 1 ü ì sin x ü ì = 3 + 4 × lim í ý = 7. ý × í lim x ® 0 î x þ îx ® 0 cos x þ So, in order to make f ( x) continuous at x = 0, we define it as ì ( 3 x + 4 tan x) , ï f ( x) = í x ïî 7,

EXAMPLE 19

Show that the function is discontinuous at x = 0.

SOLUTION

Clearly, f ( 0) = 0.

when x ¹ 0 when x = 0.

ìæ e1/x - 1 ö ÷ , when x ¹ 0 ïç f ( x) = íç e1/x + 1 ÷ ø è ï 0, when x = 0 î

Senior Secondary School Mathematics for Class 12 Pg-345

Continuity and Differentiability

345

æ e1/h - 1 ö ÷ Now, lim f ( x) = lim f ( 0 + h) = lim f ( h) = lim ç 1/h x ®0 h® 0 h® 0 h® 0 ç e + 1 ÷ø è 1 ö 1 ö æ æ e1/h ç1 - 1/h ÷ ç1 - 1/h ÷ è e ø = lim è e ø = 1. = lim 1 ö h® 0 æ 1 ö h ® 0 1/h æ e ç1 + 1/h ÷ ç1 + 1/h ÷ è è e ø e ø æ e-1/h - 1 ö ÷ lim f ( x) = lim f ( 0 - h) = lim f ( - h) = lim ç -1/h x ®0 h® 0 h® 0 h® 0 ç e + 1 ÷ø è æ 1 ö ç 1/h - 1÷ è ø = -1. e = lim h® 0 æ 1 ö ç 1/h + 1÷ èe ø Thus, lim f ( x) ¹ lim f ( x), and therefore, lim f ( x) does not exist. And,

x ® 0+

x ® 0-

Hence, f ( x) is discontinuous at x = 0.

x ®0

EXERCISE 9A 1. Show that f ( x) = x 2 is continuous at x = 2. 2. Show that f ( x) = ( x 2 + 3 x + 4) is continuous at x = 1. Prove that ìx2 - x - 6 ï , 3. f ( x) = í x - 3 ï 5, î

when x ¹ 3 ;

is continuous at x = 3.

when x = 3

ì x 2 - 25 ï , when x ¹ 5 ; is continuous at x = 5. 4. f ( x) = í x - 5 ï = x when 5 , 10 î ì sin 3 x , when x ¹ 0; ï 5. f ( x) = í is discontinuous at x = 0. x ïî 1, when x = 0 ì 1 - cos x , when x ¹ 0; ï is discontinuous at x = 0. 6. f ( x) = í x2 ïî 1, when x = 0 ì 2 - x, 7. f ( x) = í î 2 + x, ì 3 - x, 8. f ( x) = í 2 î x ,

when x < 2; is discontinuous at x = 2. when x ³ 2 when x £ 0; is discontinuous at x = 0. when x > 0

[CBSE 2000]

Senior Secondary School Mathematics for Class 12 Pg-346

346

Senior Secondary School Mathematics for Class 12

ì 5 x - 4, when 0 < x £ 1; is continuous at x = 1. 9. f ( x) = í 2 when 1 < x < 2 4 x 3 x , î ì x - 1, when 1 £ x < 2; 10. f ( x) = í is continuous at x = 2. î 2x - 3 , when 2 £ x £ 3 ì cos x , 11. f ( x) = í î - cos x , ì |x - a| , ï 12. f ( x) = í x - a ïî 1,

[CBSE 2001C]

when x ³ 0; is discontinuous at x = 0. when x < 0 when x ¹ a ;

is discontinuous at x = a.

when x = a

ì1 ï ( x - |x|), when x ¹ 0; 13. f ( x) = í 2 is discontinuous at x = 0. when x = 0 ïî 2, 1 ì ï sin , when x ¹ 0; 14. f ( x) = í is discontinuous at x = 0. x ïî 0, when x = 0 ì ï ï 15. f ( x) = í ï ïî ì ï 16. f ( x) = í ï î

2x , when x < 2; 2, when x = 2; is discontinuous at x = 2. x 2 , when x > 2 - x , when x < 0; 1, when x = 0; is discontinuous at x = 0. x , when x > 0

17. Find the value of k for which ì sin 2x , when x ¹ 0; ï is continuous at x = 0. f ( x) = í 5 x ïî k , when x = 0 18. Find the value of l for which ì x 2 - 2x - 3 ï , when x ¹ - 1; is continuous at x = - 1. f ( x) = í x + 1 ï when 1 x = l, î 19. For what value of k is the following function continuous at x = 2? ì 2x + 1, x < 2 ï [CBSE 2008] f ( x) = í k, x = 2 ï 3 x - 1, x > 2 î 20. For what value of k is the function ì x2 - 9 ï , when x ¹ 3 ; is continuous at x = 3? f ( x) = í x - 3 ï k , when x = 3 î

Senior Secondary School Mathematics for Class 12 Pg-347

Continuity and Differentiability

347

21. Find the value of k for which the function p ì ïï k cos x , if x ¹ 2 ; p is continuous at x = × [CBSE 2012C] f ( x) = í p - 2x p 2 ï 3 , if x = ïî 2 1 ì 2 ïx sin , if x ¹ 0; 22. Show that the function: f ( x) = í is continuous at x = 0. x ïî 0, if x = 0 ì x + 1, if x ³ 1; is continuous at x = 1. [CBSE 2007] 23. Show that: f ( x) = í 2 îx + 1, if x < 1 ìï x 3 - 3 , if x £ 2; is continuous at x = 2. [CBSE 2007] 24. Show that: f ( x) = í ïî x 2 + 1, if x > 2 25. Find the values of a and b such that the following function is continuous: ì 5 , when x £ ï [CBSE 2011] f ( x) = í ax + b , when 2 < x < 10 ï 21, when x ³ 10. î 26. Find the values of a for which the function f , defined as p ì ïï a sin 2 ( x + 1), x £ 0 is continuous at x = 0. f ( x) = í ï tan x - sin x , x > 0 îï x3

[CBSE 2011]

27. Prove that the function f given by f ( x) = |x - 3|, x Î R is continuous but not differentiable at x = 3. [CBSE 2012C] ANSWERS (EXERCISE 9A)

17. k =

2 1 18. l = -4 19. k = 5 20. k = 6 21. k = 6 25. a = 2, b = 1 26. a = 5 2 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 9A)

|a + h - a| |h| h = lim = lim = 1. ( a + h - a) h ® 0 h h®0 h |a - h - a| | - h| h lim f ( x ) = lim = lim = lim = -1. x ® ah ® 0 ( a - h - a) h ® 0 - h h®0 -h

12. lim f ( x ) = lim x ® a+

h®0

1 ( h -|h|) = 0 2 1 and lim f ( x ) = lim ( - h -|- h|) = - lim h = 0. x ® 0h®0 2 h®0

13. lim f ( x ) = lim x ® 0+

h®0

1 does not exist, since as x ® 0, the value of sin ( 1/ x ) oscillates between -1 x and 1, and it does not approach a definite number.

14. lim sin x®0

Senior Secondary School Mathematics for Class 12 Pg-348

348

Senior Secondary School Mathematics for Class 12

æp ö k cos ç - h ÷ k sin h k sin h æ k ö k è2 ø = lim = ç ´ 1÷ = × = lim æp ö h ® 0 2h 2 h®0 h è2 ø 2 p - 2ç - h ÷ ø è2 æp ö k cos ç + h ÷ - k sin h k sin h æ k æp ö ö k è2 ø f ç + 0 ÷ = lim = lim = ç ´ 1÷ = × = lim æp ö h ® 0 -2 h 2 h®0 h è2 ø h®0 è2 ø 2 p - 2ç + h÷ è2 ø k \ = 3 Þ k = 6. 2 1 æ -1 ö 22. f ( 0 - 0 ) = lim f ( 0 - h ) = lim ( - h ) 2 sin ç ÷ = lim -h 2 sin = ( 0 ´ a finite quantity) = 0. h h®0 h®0 è h ø h®0 1 f ( 0 + 0 ) = lim f ( 0 + h ) = lim h 2 sin = ( 0 ´ a finite quantity) = 0. h h®0 h®0 \ f ( 0 - 0 ) = f ( 0 + 0 ) = f ( 0 ). Hence, f ( x ) is continuous at x = 0.

æp ö 21. f ç - 0 ÷ = lim è2 ø h®0

Continuous Functions CONTINUITY IN AN INTERVAL A function f ( x) is said to be continuous in an open interval ]a , b[ if it is continuous at each point of ]a , b[.

If f ( x) is defined on a closed interval [a , b], we say that (i) f is continuous at a if lim f ( x) = f ( a); x ® a+

(ii) f is continuous at b if lim f ( x) = f ( b); x ®b-

(iii) f is continuous in [a , b] if it is continuous at a, at b and at each point of ]a , b[. CONTINUOUS FUNCTIONS

A function f ( x) is said to be continuous if it is continuous

at each point of its domain. Algebra of Continuous Functions THEOREM 1 PROOF

Every constant function is continuous.

Let f ( x) = c, where c is constant. Clearly, the domain of a constant function is R. Let a be an arbitrary real number. Then, \

f ( a) = c and lim f ( x) = lim c = c

lim f ( x) = f ( a).

x ®a

x ®a

[Q f ( x) = c].

x ®a

Thus,

f ( x) is continuous at x = a for all a Î R.

Hence, f ( x) is continuous.

Senior Secondary School Mathematics for Class 12 Pg-349

Continuity and Differentiability THEOREM 2 PROOF

349

Show that the identity function is continuous.

Let f ( x) = x for all x Î R. Let a be an arbitrary real number. Then, f ( a) = a. And,

lim f ( x) = lim x = a

x ®a

[Q f ( x) = x].

x ®a

lim f ( x) = f ( a) = a.

\

x ®a

This shows that f ( x) is continuous at x = a for all a Î R. Hence, the identity function is continuous. THEOREM 3

If f and g be continuous functions then (i) f + g is continuous (ii) f - g is continuous (iii) cf is continuous (iv) fg is continuous æfö (v) çç ÷÷ is continuous at those points where g( x) ¹ 0 ègø

PROOF

(i) Let dom( f ) = D1 and dom( g) = D2. Then, dom( f + g) = D1 Ç D2. Let a Î D1 Ç D2. Then, a Î D1 and a Î D2. By continuity of f and g, we have lim f ( x) = f ( a) and lim g( x) = g( a).

x ®a

\

x ®a

lim ( f + g)( x) = lim [ f ( x) + g( x)] = lim f ( x) + lim g( x)

x ®a

x ®a

x ®a

x ®a

= f ( a) + g( a) = ( f + g)( a). This shows that ( f + g) is continuous at a for all a Î D1 Ç D2. Hence, ( f + g) is continuous. Similarly, (ii) may be proved. (iii) Let dom( f ) = D1. Then, dom( cf ) = D1. Let a Î D1. Then, by the continuity of f , we have lim f ( x) = f ( a). x ®a

\

lim ( cf )( x) = lim c × f ( x) = c × lim f ( x) = c × f ( a) = ( cf )( a).

x ®a

x ®a

x ®a

This shows that ( cf ) is continuous at a for all a Î D1. Hence, cf is continuous. (iv) Let dom( f ) = D1 and dom( g) = D2. Then, dom( fg) = D1 Ç D2. Let a Î D1 Ç D2. Then, a Î D1 and a Î D2. By continuity of f and g, we have lim f ( x) = f ( a) and lim g( x) = g( a).

x ®a

\

x ®a

lim ( fg)( x) = lim [ f ( x) × g( x)]

x ®a

x ®a

= lim f ( x) × lim g( x) = f ( a) × g( a) = ( fg)( a). x ®a

x ®a

This shows that fg is continuous.

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(v) Let dom( f ) = D1 and dom( g) = D2. æfö Then, dom çç ÷÷ = [( D1 Ç D2) - {x : g( x) = 0}] = D (say). ègø Let a Î D. Then a Î D1 , a Î D2 and g( a) ¹ 0. By continuity of f and g, we have lim f ( x) = f ( a) and lim g( x) = g( a). x ®a

x ®a

lim f ( x) æfö f ( x) x ® a f ( a) æ f ö = = ç ÷( a), since g( a) ¹ 0. \ lim çç ÷÷( x) = lim = x ®a è g ø x ® a g( x) lim g( x) g( a) çè g ÷ø x ®a

æfö æfö Thus, çç ÷÷ is continuous at a for all a Î D. Hence, çç ÷÷ is continuous. ègø ègø THEOREM 4 PROOF

Every polynomial function is continuous.

Let f ( x) = a0 + a1 x + a2 x 2 + ¼ + anx n be a polynomial. We shall prove the theorem by induction on n. When n = 0 then f ( x) = a0 , which being a constant function is continuous. When n = 1 then f ( x) = a0 + a1 x. Clearly, f ( x) is the sum of a constant function and a multiple of the identity function. It, being the sum of two continuous functions, is continuous. Let every polynomial of degree at most n be continuous. Consider a general polynomial of degree (n + 1), namely, a0 + a1 x + a2 x 2 + ¼ + anx n + an+1 x n+1. This can be written as a0 + x( a1 + a2 x + ¼ + anx n-1 + an+1 x n). This is the sum of a constant function a0 (which is continuous) and the product of the identity function x (which is continuous) and the polynomial function a1 + a2 x + ¼ + an+1 x n of degree at most n (which we assumed to be continuous). Therefore, it is continuous. Thus, the continuity of a polynomial of degree n implies the continuity of a polynomial of degree (n + 1). Hence, by the principle of induction, every polynomial function is continuous.

THEOREM 5 PROOF

Every rational function is continuous.

Let f ( x) =

p( x) , where p( x) and q( x) are polynomials. q( x)

dom[p( x)] = R and dom[q( x)] = R. ì p( x) ü \ dom [ f ( x)] = dom í ý = R Ç R - {x : q ( x) = 0} = R - {x : q ( x) = 0}. î q( x) þ Then,

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351

Let a be an arbitrary element of dom f ( x). Then, a Î R and q( a) ¹ 0. But, every polynomial function being continuous, we have lim p( x) = p( a) and lim q( x) = q( a).

x ®a

x ®a

lim p( x)

\

lim f ( x) = lim =

x ®a

x ®a

x ®a

lim q( x)

=

x ®a

p( a) , where q( a) ¹ 0. q( a)

This shows that f ( x) is continuous at x = a for all a Î dom f ( x). Hence, every rational function is continuous. THEOREM 6

PROOF

Let f and g be real functions such that f o g is defined. If g is continuous at a and f is continuous at g( a), show that f o g is continuous at a.

Since f o g is defined, we have range ( g) Í dom ( f ). lim g( x) = g( a).

Since g is continuous at a, we have

x ®a

Also, f being continuous at g( a), we have \ lim ( f o g)( x) = lim f {g( x)} = x ®a

x ®a

lim

g ( x ) ® g ( a)

… (i)

lim f ( y) = f {g( a)}. … (ii)

y ® g ( a)

f {g( x)}

[Q x ® a Þ g( x) ® g( a) from (i)] = lim f ( y) = f {g( a)} y ® g ( a)

[using (ii)]

= ( f o g)( a). Thus, ( f o g) is continuous at a. THEOREM 7 PROOF

The composite of two continuous functions is continuous.

Let f and g be continuous functions such that g o f is defined. Then, range ( f ) Í dom( g). Let a Î dom( f ). Then,

a Î dom ( f ) Þ f ( a) Î range ( f ) Þ f ( a) Î dom ( g)

[Q range ( f ) Í dom( g)].

Thus, f is continuous at a , and g is continuous at f ( a). Consequently, And, \

lim f ( x) = f ( a).

… (i)

lim g( y) = g[ f ( a)].

… (ii)

x ®a

y ® f ( a)

lim ( g o f )( x) = lim g[ f ( x)]

x ®a

=

x ®a

lim

f ( x )® f ( a)

g[ f ( x)] [Q x ® a Þ f ( x) ® f ( a) from (i)]

= lim g( y) = g[ f ( a)] y ® f ( a)

= ( g o f )( a).

[using (ii)]

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This shows that g o f is continuous at a for all a Î dom( f ). Hence g o f is continuous. An important result

For continuity of f at a, it is sufficient to show that lim f ( a + h) = f ( a), h® 0 since f is continuous Û lim f ( x) = f ( a) Û

x ®a

lim

( a+ h )® a

f ( a + h) = f ( a)

[putting x = ( a + h)]

Û lim f ( a + h) = f ( a). h® 0

THEOREM 8 PROOF

The exponential function is continuous.

Let f ( x) = ex . Then, clearly, dom( f ) = R. Let a be an arbitrary real number. Then, lim ex = lim ea+ h = lim ea × eh = ea × lim eh = ea ´ 1 = ea. x ®a

h® 0

h® 0

h® 0

Thus, f ( x) = ex is continuous at x = a for all a Î R. Hence, the exponential function is continuous. THEOREM 9

PROOF

(i) The sine function is continuous. (ii) The cosine function is continuous. (iii) The tangent function is continuous.

(i) Let f ( x) = sin x. Then, clearly dom( f ) = R. Let a be an arbitrary real number. Then, lim f ( x) = lim sin x = lim sin ( a + h) = lim [sin a cos h + cos a sin h] x ®a

x ®a

h® 0

h® 0

Thus,

h® 0

= sin a × lim cos h + cos a × lim sin h h® 0

= ( sin a ´ 1 + cos a ´ 0) = sin a = f ( a). lim f ( x) = f ( a) for all a Î R.

x ®a

\ f ( x) = sin x is continuous at a for all a Î R. Hence, sin x is continuous. (ii) Let f ( x) = cos x. Clearly, dom( f ) = R. Let a be an arbitrary real number. Then, lim f ( x) = lim cos x = lim cos ( a + h) = lim [cos a cos h - sin a sin h] x ®a

x ®a

h® 0

h® 0

Thus,

h® 0

= cos a × lim cos h - sin a × lim sin h = (cos a ´ 1 - sin a ´ 0) h® 0

= cos a = f ( a). lim f ( x) = f ( a) for all a Î R.

x ®a

\ f ( x) = cos x is continuous at a for all a Î R. Hence, cos x is continuous. sin x p (iii) We have tan x = and dom (tan x) = R - {( 2n + 1) : n Î I}. cos x 2 p Let a Î R - {( 2n + 1) : n Î I}. Then, 2

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Continuity and Differentiability

lim tan x = lim

x ®a

x ®a

353

lim sin ( a + h) sin x h ® 0 = cos x lim cos ( a + h) h® 0

lim (sin a cos h + cos a sin h)

=

h® 0

lim (cos a cos h - sin a sin h)

h ®0

æ sin a ´ cos 0 + cos a ´ sin 0 ö æ sin a ´ 1 + cos a ´ 0 ö ÷÷ = çç ÷÷ = çç è cos a ´ cos 0 - sin a ´ sin 0 ø è cos a ´ 1 - sin a ´ 0 ø = tan a. p ü ì Thus, tan x is continuous at a for all a Î R - í( 2n + 1) : n Î I ý× 2 þ î Hence, tan x is continuous. THEOREM 10 PROOF

The logarithmic function is continuous.

Let f ( x) = log x. Then, dom( f ) = set of positive real numbers. Let a be any positive real number. Then, lim f ( x) = lim log x x ®a

x ®a

xù é æ x öù é = lim ê log ç a × ÷ ú = lim ê log a + log ú x ® a a aû è øû ë ë = log a + lim log x ®a

x x é ù = log a = f ( a) êQ lim log = log 1 = 0ú × x ® 0 a a ë û

Thus, f ( x) = log x is continuous at a Î R + . Hence, f ( x) = log x is continuous. THEOREM 11 PROOF

sin|x|is continuous.

Let f ( x) = |x| and g( x) = sin x. Then, ( g o f )( x) = g{ f ( x)} = g(|x|) = sin|x|. Now, f and g being continuous, it follows that their composite ( g o f ) is continuous. Hence, sin|x|is continuous. SOLVED EXAMPLES

EXAMPLE 1

SOLUTION

ì x if x ³ 1 Let f ( x) = í 2 Is f a continuous function? Why? î x if x < 1. Let a be any real number. Then, three possibilities arise. Case I When a > 1 In this case, f ( a) = a. Also, lim f ( x) = lim f ( a + h) = lim ( a + h) = a. x ® a+

h® 0

h® 0

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Senior Secondary School Mathematics for Class 12

lim f ( x) = lim f ( a - h) = lim ( a - h) = a

And, \

x ® a-

h® 0

h® 0

[Q a > 1 and h is very small Þ ( a - h) > 1]. lim f ( x) = lim f ( x) = a = f ( a).

x ® a+

a ® a-

So, f ( x) is continuous at each a > 1. Case II When a < 1 In this case, f ( a) = a 2. lim f ( x) = lim f ( a + h) = lim ( a + h) 2 = a 2

x ® a+

h® 0

h® 0

[Q a < 1 and h is very small Þ ( a + h) < 1]. And, lim f ( x) = lim f ( a - h) = lim ( a - h) 2 = a 2 x ® a-

h® 0

h® 0

[Q a < 1 and h is very small Þ ( a - h) < 1]. \

lim f ( x) = lim f ( x) = a 2 = f ( a).

x ® a+

x ® a-

Case III When a = 1

In this case, f (1) = 1. lim f ( x) = lim f (1 + h) = lim (1 + h) = 1 [Q (1 + h) > 1]. x ®1 +

h® 0

h® 0

lim f ( x) = lim f (1 - h) = lim (1 - h) 2 = 1 [Q (1 - h) < 1].

x ®1 -

\

h® 0

h® 0

lim f ( x) = lim f ( x) = 1.

x ®1 +

x ®1 -

So, lim f ( x) = 1 = f (1). x ®1

So, f is continuous at a = 1. Thus, from all the above three cases, it follows that f ( x) is continuous at x = a for all a Î R. Hence, f ( x) is continuous. EXAMPLE 2 SOLUTION

Prove that f ( x) = |x|is a continuous function. ì x , if x ³ 0 Let f ( x) = |x| = í î- x , if x < 0. Clearly, dom( f ) = R. Let a be any real number. Then, the following cases arise. Case I When a < 0 In this case, f ( a) = - a. lim f ( x) = lim f ( a + h) = lim {- ( a + h)} = - a x ® a+

h® 0

h® 0

[Q a < 0 and h is very small and positive Þ a + h < 0]. And, lim f ( x) = lim f ( a - h) = lim {- ( a - h)} = - a. x ® a-

h® 0

h® 0

[Q a < 0 and h is very small and positive Þ a - h < 0]. So, lim f ( x) = lim f ( x) = - a = f ( a). x ® a+

x ® a-

Thus, f ( x) is continuous at each a < 0.

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355

Case II When a > 0

In this case, f ( a) = a. lim f ( x) = lim f ( a + h) = lim ( a + h) = a [Q ( a + h) > 0]. x ® a+

h® 0

h® 0

x ® a-

h® 0

h® 0

lim f ( x) = lim f ( a - h) = lim ( a - h) = a

[Q ( a - h) > 0].

\ lim f ( x) = lim f ( x) = f ( a). x ® a+

x ® a-

So, f ( x) is continuous at each a > 0. Case III When a = 0 Clearly, f ( 0) = 0. lim f ( x) = lim f ( 0 + h) = lim f ( h) = lim h = 0. x ® 0+

h® 0

h® 0

h® 0

lim f ( x) = lim f ( 0 - h) = lim f ( - h) = lim {- ( - h)}

x ® 0-

h® 0

= lim h = 0.

h® 0

h® 0

h® 0

\ lim f ( x) = lim f ( x) = 0 = f ( 0). x ® 0+

EXAMPLE 3

x ® 0-

So, f ( x) is continuous at x = 0. Thus, from all the above cases, it follows that f ( x) = |x|is continuous at a for all a Î R. Hence, f ( x) = |x|is continuous. ì 2x - 1, if x < 0 Discuss the continuity of the function f ( x) = í î 2x + 1, if x ³ 0. [CBSE 2002]

SOLUTION

When x < 0, we have f ( x) = 2x - 1, which being a polynomial function, is continuous at each point where x < 0. Also, when x > 0, we have f ( x) = 2x + 1, which being a polynomial function, is continuous at each point where x > 0. Let us consider the point x = 0. f ( 0) = ( 2 ´ 0 + 1) = 1. lim f ( x) = lim f ( 0 + h) = lim f ( h) = lim ( 2h + 1) = 1. x ® 0+

h® 0

h® 0

h® 0

And, lim f ( x) = lim f ( 0 - h) = lim f ( - h) = lim [( 2( - h) - 1)] = -1. x ® 0-

h® 0

lim f ( x) ¹ lim f ( x).

\

x ® 0+

h® 0

h® 0

x ® 0-

So, lim f ( x) does not exist, and therefore f ( x) is not continuous at x ®0

x = 0.

EXAMPLE 4

SOLUTION

Thus, the given function is continuous at each point except at x = 0, where it is discontinuous. ì sin x , if x < 0 ï Discuss the continuity of the function f ( x) = í x ïî ( x + 1), if x ³ 0. We know that sin x as well as the identity function x is continuous.

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sin x is continuous at each x < 0. x Also, when x > 0, we have f ( x) = ( x + 1), which being a polynomial function, is continuous. Let us consider the point x = 0. Clearly, f ( 0) = ( 0 + 1) = 1. We have f ( 0) = ( 0 + 1) = 1; lim f ( x) = lim f ( 0 + h) = lim ( h + 1) = 1

So, the quotient function,

x ® 0+

and,

\ So, Thus, Hence, EXAMPLE 5

SOLUTION

h® 0

h® 0

lim f ( x) = lim f ( 0 - h) = lim f ( - h)

x ® 0-

h® 0

h® 0

sin ( - h) sin h æ - sin h ö = lim ç = 1. = lim ÷ = lim h® 0 h® 0 è -h ø h® 0 -h h

lim f ( x) = lim f ( x) = 1.

x ® 0+

x ® 0-

lim f ( x) = 1 = f ( 0).

x ®0

f ( x) is continuous at x = 0 also. f ( x) is continuous at all points.

ì x , if x ¹ 0 ï Discuss the continuity of the function f ( x) = í |x| ï 0, if x = 0. î We know that the identity function x is continuous and the modulus function|x|is continuous. x is continuous at each x ¹ 0. So, the quotient function |x| It has already been proved that f ( x) is discontinuous at x = 0. Hence, the given function is continuous at each point, except at x = 0.

EXAMPLE 6

SOLUTION

ì x 4 - 16 ï , if x ¹ 2 Locate the point of discontinuity of the function f ( x) = í x - 2 ï 16, if x = 2. î 4 x - 16 being a rational function, is continuous The function f ( x) = x-2 at all points of its domain, i.e., for all real numbers except 2. Now, consider the given function at x = 2. We have f ( 2) = 16. ì( 2 + h) 4 - 16 ü ì( 2 + h) 4 - 24 ü lim f ( x) = lim f ( 2 + h) = lim í ý = lim í ý x ®2 h® 0 h® 0 î 2 + h - 2 þ 2 + h ®2 î ( 2 + h) - 2 þ = 4 ´ 24 -1 = 32

ù é æ x n - an ö ÷ = na n-1 ú × êQ lim çç ÷ úû êë x ® a è x - a ø

Senior Secondary School Mathematics for Class 12 Pg-357

Continuity and Differentiability

357

Thus, lim f ( x) ¹ f ( 2), and therefore lim f ( x) ¹ f ( 2). x ® 2+

Hence, EXAMPLE 7

SOLUTION

x ®2

f ( x) is discontinuous at x = 2.

Determine the value of k so that the function ìkx 2 , if x £ 2; is continuous. f ( x) = í î 3 , if x > 2

[CBSE 1990C, ‘91]

Since a polynomial function is continuous and a constant function is continuous, the given function is continuous for all x < 2 and for all x > 2. So, consider the point x = 2. We have f ( 2) = 4k. lim f ( x) = lim f ( 2 + h) = lim 3 = 3. x ® 2+

h® 0

h® 0

And, lim f ( x) = lim f ( 2 - h) = lim k( 2 - h) 2 = 4k. x ® 2-

EXAMPLE 8

SOLUTION

h® 0

h® 0

\

for continuity, we must have 4k = 3

Let

1, if x £ 3 ì ï f ( x) = íax + b , if 3 < x < 5 ï 7 , if 5 £ x. î

or k =

3 . 4

Find the values of a and b so that f ( x) is continuous. We know that a constant function is continuous, and a polynomial function is continuous. So, the given function is continuous for all x < 3; for all x > 5 and for all x lying in ] 3 , 5[. Now, consider the point x = 3. We have f ( 3) = 1. lim f ( x) = lim f ( 3 + h) = lim [a( 3 + h) + b] = ( 3 a + b). And,

x ® 3+

h® 0

x ® 3-

h® 0

h® 0

lim f ( x) = lim ( 3 - h) = lim 1 = 1. h® 0

Since f ( x) is given to be continuous, it must be continuous at x = 3 also. So, we must have f ( 3) = lim f ( x) = lim f ( x), i.e., 3 a + b = 1. … (i) x® 3-

x® 3+

Again, consider the point x = 5. We have f (5) = 7. lim f ( x) = lim f (5 + h) = lim 7 = 7. And,

x ®5 +

h® 0

h® 0

x ®5 -

h® 0

h® 0

lim f ( x) = lim f (5 - h) = lim [a(5 - h) + b] = (5 a + b).

Now, by continuity of f ( x) at x = 5, we have f (5) = lim f ( x) = lim f ( x) or 5 a + b = 7. x® 5+

x® 5-

… (ii)

Solving (i) and (ii), we get a = 3 and b = - 8. EXAMPLE 9

Show that the function f ( x) = x 4 + 3 is continuous at each point.

SOLUTION

Let g( x) = x 4 + 3 and h( y) = y.

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Since every polynomial function is continuous at each point where x = a , it follows that g( x) is continuous at x = a. Clearly, g( a) is positive. Moreover, h[g( a)] = g( a) = lim y. y ®g ( a)

\ h( y) is continuous at g( a). Since the composite of continuous functions is continuous, it follows that ( h o g) ( x) is continuous. But,

( h o g)( x) = h[g( x)] = h( x 4 + 3) = x 4 + 3 = f ( x).

Hence, f ( x) is continuous. EXAMPLE 10

Show that the function f ( x) = |sin x + cos x|is continuous at x = p.

SOLUTION

Let g( x) = sin x + cos x and k( y) = |y|. We shall first show that g is continuous at x = p and k is continuous at y = g( p). Now, lim g( x) = lim (sin x + cos x) = sin p + cos p = -1. x ®p

Also,

x ®p

g( p) = (sin p + cos p) = -1.

\

lim g( x) = g( p).

x ®p

So, g is continuous at x = p. Now, g( p) = -1. \ k [g( p)] = k ( -1) = |-1| = 1. Now, lim k ( y) = lim |y| = lim |-1 + h| = 1. y ®( -1 )+

And, \

lim

y ®( -1 )-

y ®( -1 )+

k ( y) =

h® 0

lim |y| = lim |-1 - h| = 1.

y ®( -1 )-

h® 0

k [g( p)] = lim k ( y). This shows that k is continuous at g( p). y ®g (p )

Consequently, ( k o g) is continuous at x = p. But, ( k o g)( x) = k [g( x)] = k (sin x + cos x) = |sin x + cos x| = f ( x). Hence, f ( x) is continuous at x = p.

EXERCISE 9B ì (7 x + 5), when x ³ 0; 1. Show that the function f ( x) = í is a continuous î (5 - 3 x), when x < 0 function. ì sin x , if x < 0; 2. Show that the function f ( x) = í is continuous. x , if x ³ 0 î ì xn - 1 ï , when x ¹ 1; is continuous. 3. Show that the function f ( x) = í x - 1 ï when x = 1 , n î 4. Show that sec x is a continuous function. 5. Show that cos|x|is a continuous function.

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359

ì sin x , when x ¹ 0; ï f ( x) = í x ïî 2, when x = 0 is continuous at each point except 0.

6. Show that the function

7. Discuss the continuity of f ( x) = [x]. ì ( 2x - 1), if x < 2; ï 8. Show that f ( x) = í is continuous. 3x , if x ³ 2 ïî 2 9. Show that

ì x , if x ¹ 0; is continuous at each point except 0. f ( x) = í î 1, if x = 0

10. Locate the point of discontinuity of the function ì ( x 3 - x 2 + 2x - 2), if x ¹ 1; f ( x) = í 4, if x = 1. î 11. Discuss the continuity of the function f ( x) = |x| + |x - 1| in the interval [-1, 2]. ANSWERS (EXERCISE 9B)

7. discontinuous at x = n, where n is an integer 10. discontinuous at x = 1

11. continuous

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 9B) 1. Since a polynomial function is continuous, the given function is continuous at each x > 0 as well as at each x < 0. Test its continuity at x = 0. 2. Let a Î R. Case I When a > 0 then, lim f ( x ) = lim f ( x ) = f ( a) = a.

7. 8. 10. 11.

x ® a+

x ® a-

x ® a+

x ® a-

Case II

When a < 0 then, lim f ( x ) = lim f ( x ) = f ( a) = sin a.

Case III

When a = 0 then, lim f ( x ) = lim f ( x ) = f ( 0 ) = 0. x ® 0+

x ® 0-

f ( x ) is discontinuous at each integral value of x. Take three cases: (i) a > 2 (ii) a < 2 (iii) a = 2. Test at x = 1. The given function may be expressed as ì ( 1 - 2 x ), when - 1 £ x < 0 ; ï f ( x) = í 1, when 0 £ x £ 1; ï ( 2 x - 1), when 1 < x £ 2. î Since a polynomial function as well as a constant function is continuous, it follows that f ( x ) is continuous when -1 < x < 0 ; 0 < x < 1 and 1 < x < 2 . So, test the continuity at -1, 0 , 1 and 2. At x = - 1, lim f ( x ) = f ( -1), so it is continuous at x = -1. x ® (-1 ) +

Senior Secondary School Mathematics for Class 12 Pg-360

360

Senior Secondary School Mathematics for Class 12 lim f ( x ) = lim f ( x ) = f ( 0 ) = 1, so it is continuous at x = 0.

At x = 0,

x ® 0+

x ® 0-

lim f ( x ) = lim f ( x ) = f ( 1) = 1.

At x = 1,

x ®1 +

At x = 2,

x ®1 -

lim f ( x ) = f ( 2 ) = 3 , so it is continuous at x = 2.

x®2-

Differentiability Let f ( x) be a real function and a be any real number. Then, we define f ( a + h) - f ( a) (i) Right-hand derivative lim , if it exists, is called the righth® 0 h hand derivative of f ( x) at x = a , and it is denoted by Rf ¢( a). f ( a - h) - f ( a) (ii) Left-hand derivative lim , if it exists, is called the lefth ®0 -h hand derivative of f ( x) at x = a , and it is denoted by Lf ¢( a). A function f ( x) is said to be differentiable at x = a , if Rf ¢( a) = Lf ¢( a). The common value of Rf ¢( a) and Lf ¢( a) is denoted by f ¢( a) and it is known as the derivative of f ( x) at x = a. If, however, Rf ¢( a) ¹ Lf ¢( a), we say that f ( x) is not differentiable at x = a.

DIFFERENTIABILITY

REMARK

In each case, h is taken as positive and very small. SOLVED EXAMPLES

EXAMPLE 1

Show that f ( x) = x 2 is differentiable at x = 1 and find f ¢(1).

SOLUTION

Rf ¢(1) = lim

h® 0

f (1 + h) - f (1) (1 + h) 2 - (1) 2 = lim h® 0 h h

æ 1 + h2 + 2h - 1 ö ÷ = lim ( h + 2) = 2. = lim ç ÷ h® 0 h® 0 ç h è ø f (1 - h) - f (1) (1 - h) 2 - (1) 2 And, Lf ¢(1) = lim = lim h® 0 h® 0 -h -h æ 1 + h2 - 2h - 1 ö ÷ = lim ( - h + 2) = 2. = lim ç ÷ h® 0 h® 0 ç -h è ø \ Rf ¢(1) = Lf ¢(1) = 2. This shows that f ( x) is differentiable at x = 1 and f ¢(1) = 2. EXAMPLE 2

Show that f ( x) = [x] is not differentiable at x = 1.

SOLUTION

We have Rf ¢(1) = lim

h® 0

f (1 + h) - f (1) [1 + h] - [1] = lim =0 h® 0 h h {Q [1 + h] = 1 and [1] = 1} ,

Senior Secondary School Mathematics for Class 12 Pg-361

Continuity and Differentiability

361

f (1 - h) - f (1) [1 - h] - [1] = lim =¥ h® 0 -h -h {Q [1 - h] = 0 and [1] = 1}. Thus, Rf ¢(1) ¹ Lf ¢(1). Hence, f ( x) = [x] is not differentiable at x = 1. and

EXAMPLE 3

Lf ¢(1) = lim

h® 0

(i) Show that f ( x) = x 4/ 3 is differentiable at x = 0, and hence find f ¢( 0). (ii) Show that g( x) = x 3/2 is not differentiable at x = 0.

SOLUTION

(i) We have

Rf ¢( 0) = lim

h® 0

f ( 0 + h) - f ( 0) f ( h) - f ( 0) = lim h® 0 h h

h4/ 3 - 0 h4/ 3 = lim = lim h1/ 3 = 0. h® 0 h® 0 h h® 0 h f ( 0 - h) - f ( 0) f ( - h) - 0 And, Lf ¢( 0) = lim = lim h® 0 h® 0 -h -h = lim

( - h) 4/ 3 - 0 ( - h) 4/ 3 = lim = lim ( - h)1/ 3 = 0. h® 0 h ® 0 ( - h) h® 0 ( - h)

= lim

Thus, Rf ¢( 0) = Lf ¢( 0) = 0. Hence, f ( x) = x 4/ 3 is differentiable at x = 0 and f ¢( 0) = 0. (ii) Consider g( x) = x 3/2. Now,

Rg ¢( 0) = lim

h® 0

g( 0 + h) - g( 0) g( h) - g( 0) = lim h® 0 h h

h 3/2 - 0 h 3/2 = lim = lim h1/2 = 0. h® 0 h® 0 h h® 0 h g( 0 - h) - g( 0) g( - h) - g( 0) Lg ¢( 0) = lim = lim h® 0 h® 0 -h -h = lim

And,

( - h) 3/2 - 0 ( - h) 3/2 = lim h® 0 h ® 0 ( - h) ( - h)

= lim

= lim ( - h)1/2 , which is imaginary. h® 0

Thus, Lg¢( 0) does not exist. Hence, g( x) = x 3/2 is not differentiable at x = 0. EXAMPLE 4

SOLUTION

ì 1 + x , if x £ 2; Show that the function f ( x) = í is not differentiable at î 5 - x , if x > 2 x = 2. f ( 2 + h) - f ( 2) é 5 - ( 2 + h) - 3 ù = lim ê Rf ¢( 2) = lim úû h® 0 h® 0 ë h h -h = lim = lim ( -1) = -1. h® 0 h h® 0

Senior Secondary School Mathematics for Class 12 Pg-362

362

Senior Secondary School Mathematics for Class 12

f ( 2 - h) - f ( 2) -h -h é 1 + ( 2 - h) - 3 ù = lim ê = lim 1 = 1. úû = hlim h® 0 ë ® 0 -h h® 0 -h Thus, Rf ¢( 2) ¹ Lf ¢( 2). Hence, f ( x) is not differentiable at x = 2. p ì ï (1 + sin x), when 0 £ x < Let f ( x) = í 2 ïî 1, when x < 0. Show that f ¢( 0) does not exist. f ( 0 + h) - f ( 0) f ( h) - f ( 0) We have Rf ¢( 0) = lim = lim h® 0 h® 0 h h (1 + sin h) - 1 sin h = lim = lim = 1. h® 0 h® 0 h h f ( 0 - h) - f ( 0) f ( - h) - f ( 0) (1 - 1) And, Lf ¢( 0) = lim = lim = lim = 0. h® 0 h® 0 h® 0 -h -h -h Thus, Rf ¢( 0) ¹ Lf ¢( 0). Hence, f ¢( 0) does not exist. And,

EXAMPLE 5

SOLUTION

EXAMPLE 6 SOLUTION

Lf ¢( 2) = lim

h® 0

Let f ( x) = mx + c and f ( 0) = f ¢( 0) = 1. Find f ( 2). Clearly, f ( 0) = (m ´ 0 + c) = c = 1 (given). f ( 0 + h) - f ( 0) f ( h) - f ( 0) Also, f ¢( 0) = lim = lim h® 0 h® 0 h h (mh + c) - 1 mh + 1 - 1 = lim = lim [Q c = 1] h® 0 h® 0 h h mh = lim = lim m = m. h® 0 h h® 0 Thus, f ¢( 0) = 1 Þ m = 1. So, f ( x) = 1 × x + 1, i.e., f ( x) = ( x + 1). Hence, f ( 2) = ( 2 + 1) = 3.

Relation between Continuity and Differentiability THEOREM

PROOF

Every differentiable function is continuous. But, every continuous function need not be differentiable.

Let f ( x) be a differentiable function and let a be any real number in its f ( a + h) - f ( a) … (i) domain. Then, lim = f ¢( a). h® 0 h é f ( a + h) - f ( a) ù Now, lim [ f ( a + h) - f ( a)] = lim ê ´ hú h® 0 h® 0 ë h û f ( a + h) - f ( a) = lim ´ lim h h® 0 h® 0 h [using (i)]. = f ¢( a) ´ 0 = 0 Thus, lim [ f ( a + h) - f ( a)] = 0 or lim f ( a + h) = f ( a). h® 0

h® 0

This shows that f ( x) is continuous at a for all a.

Senior Secondary School Mathematics for Class 12 Pg-363

Continuity and Differentiability

363

Hence, every differentiable function is continuous. In order to show that a continuous function need not be differentiable, it is sufficient to give an example of a function which is continuous but not differentiable. Consider f ( x) = |x|at x = 0. Clearly,

f ( 0) = 0. lim f ( x) = lim f ( 0 + h) = lim f ( h) = lim |h| = lim h = 0.

x ® 0+

h® 0

h® 0

h® 0

h® 0

lim f ( x) = lim f ( 0 - h) = lim f ( - h) = lim |- h| = lim h = 0.

And,

x ® 0-

h® 0

x ® 0+

x ® 0-

h® 0

lim f ( x) = lim f ( x) = f ( 0).

Thus,

h® 0

h® 0

f ( x) = |x|is continuous at x = 0. f ( 0 + h) - f ( 0) f ( h) - f ( 0) Rf ¢( 0) = lim = lim h® 0 h® 0 h h |h| - 0 h = lim = lim = 1. h® 0 h® 0 h h f ( 0 - h) - f ( 0) f ( - h) - f ( 0) Lf ¢( 0) = lim = lim h® 0 h® 0 -h -h |- h| - 0 h = lim = lim = -1. h® 0 h® 0 -h -h

So, But,

And,

Thus, Rf ¢( 0) ¹ Lf ¢( 0). This shows that f ( x) = |x| is not differentiable at x = 0. Thus, f ( x) = |x| is continuous but not differentiable at x = 0. Hence, a continuous function need not be differentiable. EXAMPLE 7

SOLUTION

ì x sin (1/x), when x ¹ 0; f ( x) = í 0, when x = 0 î is continuous but not differentiable at x = 0. We have already discussed the above function for continuity at x = 0. f ( 0 + h) - f ( 0) h sin (1/h) - 0 Now, Rf ¢( 0) = lim = lim h® 0 h® 0 h h Show that the function

= lim sin (1/h), which does not exist. h® 0

Hence, f ( x) is not differentiable at x = 0. f ( x) = |x - 2|is continuous but not differentiable at x = 2.

EXAMPLE 8

Show that

SOLUTION

We have f ( 2) = |2 - 2| = 0. lim f ( x) = lim f ( 2 + h) = lim |2 + h - 2| = lim |h| = lim h = 0. x ® 2+

h® 0

h® 0

h® 0

h® 0

lim f ( x) = lim f ( 2 - h) = lim |2 - h - 2| = lim |- h| = lim h = 0.

x ® 2-

\ So,

h® 0

h® 0

lim f ( x) = lim f ( x) = f ( 2) = 0.

x ® 2+

x ® 2-

f ( x) is continuous at x = 2.

h® 0

h® 0

Senior Secondary School Mathematics for Class 12 Pg-364

364

Senior Secondary School Mathematics for Class 12

f ( 2 + h) - f ( 2) |2 + h - 2| - 0 = lim h® 0 h h |h| h = lim = lim = 1. h® 0 h h® 0 h f ( 2 - h) - f ( 2) |2 - h - 2| - 0 And, Lf ¢( 2) = lim = lim h® 0 h® 0 -h -h |- h| h = lim = lim = -1. h® 0 -h h® 0 -h Thus, Rf ¢( 2) ¹ Lf ¢( 2). This shows that f ( x) is not differentiable at x = 2. But,

Rf ¢( 2) = lim

h® 0

EXERCISE 9C 3

1. Show that f ( x) = x is continuous as well as differentiable at x = 3. 2. Show that f ( x) = ( x - 1)1/ 3 is not differentiable at x = 1. 3. Show that a constant function is always differentiable. 4. Show that f ( x) = |x - 5|is continuous but not differentiable at x = 5. ì ( 2 - x), when x ³ 1 5. Let f ( x) = í x , when 0 £ x £ 1. î Show that f ( x) is continuous but not differentiable at x = 1. 6. Show that f ( x) = [x] is neither continuous nor derivable at x = 2. ì (1 - x), when x < 1; 7. Show that the function f ( x) = í 2 î ( x - 1), when x ³ 1 is continuous but not differentiable at x = 1. ì ( 2 + x), if x ³ 0 8. Let f ( x) = í Show that f ( x) is not derivable at x = 0. î ( 2 - x), if x < 0. 9. If f ( x) = |x|, show that f ¢( 2) = 1. 10. Find the values of a and b so that the function ì ( x 2 + 3 x + a), when x £ 1; is differentiable at each x Î R. f ( x) = í ( bx + 2), when x > 1 î ANSWERS (EXERCISE 9C)

10. a = 3, b = 5 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 9C) 9. Show that Rf ¢( 2 ) = Lf ¢( 2 ) = 1. 10. Rf ¢( 1) = Lf ¢( 1) = 5.

Senior Secondary School Mathematics for Class 12 Pg-365

10. DIFFERENTIATION

1. Derivatives of Some Functions In class 11, we have studied about the derivatives of algebraic and trigonometric functions. We derived the following results. d n ( x ) = nxn-1 dx

(ii)

d (sin x) = cos x dx

(iii)

d (cos x) = - sin x dx

(iv)

d (tan x) = sec2 x dx

(v)

d (cot x) = - cosec2 x dx

(vi)

d (sec x) = sec x tan x dx

(i)

(vii)

d (cosec x) = - cosec x cot x dx

In algebra of derivatives, we have established the following rules. Some Rules of Differentiation (i)

d æ du dv ö (u + v) = ç + ÷ dx è dx dx ø

(ii)

d æ du dv ö (u - v) = ç ÷ dx è dx dx ø

(iii)

d du ö æ dv (uv) = çu × + v× ÷ dx dx ø è dx

dv ö æ du -u× ÷ çv × d æ u ö è dx dx ø (iv) ç ÷= dx è v ø v2 DERIVATIVES OF COMPOSITE FUNCTIONS (CHAIN RULE)

dy æ dy dt ö =ç ´ ÷× dx è dt dx ø dy æ dy dt du ö (ii) Let y = f (t), t = g(u) and u = h( x). Then, =ç ´ ´ ÷× dx è dt du dx ø This rule may be extended further on more variables. (i) Let y = f (t) and t = g( x). Then,

365

Senior Secondary School Mathematics for Class 12 Pg-366

366

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Differentiate each of the following w.r.t. x: (ii) cos 3 x (i) sin x 3

SOLUTION

3

(iii) tan

x

(i) Let y = sin x . Putting x 3 = t, we get y = sin t and t = x 3 dy dt = cos t and = 3x2 dt dx dy æ dy dt ö Þ =ç ´ ÷ dx è dt dx ø = (cos t × 3 x 2) = 3 x 2 cos t = 3 x 2 cos x 3. d Hence, (sin x 3) = 3 x 2 cos x 3. dx Þ

(ii) Let y = cos 3 x = (cos x) 3. Putting cos x = t, we get y = t 3 and t = cos x dy dt = 3t 2 and = - sin x dt dx dy æ dy dt ö Þ =ç ´ ÷ dx è dt dx ø = ( -3t 2 sin x) = ( -3 sin x)t 2 = ( -3 sin x cos2 x). d Hence, (cos 3 x) = -3 sin x cos2 x. dx Þ

(iii) Let y = tan

x.

Putting x = t, we get y = tan t and t = x dy dt 1 -1 2 1 Þ = sec2 t and = x = dt dx 2 2 x dy æ dy dt ö Þ =ç ´ ÷ dx è dt dx ø 1 ö sec2 x æ = ç sec2 t × [Q t = x ]. ÷= 2 xø 2 x è d sec2 x Hence, (tan x ) = × dx 2 x EXAMPLE 2

Differentiate each of the following w.r.t. x: (i) ( ax + b) m

(ii) ( 2x + 3)5

(iii) ax 2 + 2bx + c

Senior Secondary School Mathematics for Class 12 Pg-367

Differentiation SOLUTION

367

(i) Let y = ( ax + b) m. Putting ( ax + b) = t, we get y = t m and t = ( ax + b) dy dt = mt m-1 and =a dt dx dy æ dy dt ö Þ =ç ´ ÷ = (mt m-1 ´ a) = mat m-1 = ma( ax + b) m-1. dx è dt dx ø d ( ax + b) m = ma( ax + b) m-1. \ dx Þ

(ii) Let y = ( 2x + 3)5. Putting ( 2x + 3) = t, we get y = t5 and t = 2x + 3 dy dt Þ = 5t 4 and =2 dt dx dy æ dy dt ö Þ =ç ´ ÷ = 10t 4 = 10( 2x + 3) 4. dx è dt dx ø (iii) Let y = ax 2 + 2bx + c. Putting ( ax 2 + 2bx + c) = t, we get y = t and t = ( ax 2 + 2bx + c) dy 1 -1 2 1 dt and = t = = ( 2 ax + 2b) = 2( ax + b) dt 2 dx 2 t 1 dy æ dy dt ö Þ =ç ´ ÷= ´ 2( ax + b) dx è dt dx ø 2 t ( ax + b) ( ax + b) = = × t ax 2 + 2bx + c Þ

EXAMPLE 3 SOLUTION

EXAMPLE 4 SOLUTION

Differentiate

sin 3 x cos 5 x w.r.t. x. 1 Let y = sin 3 x cos 5 x = {2 cos 5 x sin 3 x} 2 1 = {sin (5 x + 3 x) - sin (5 x - 3 x)} 2 1 1 1 = × (sin 8x - sin 2x) = sin 8x - sin 2x. 2 2 2 dy 1 d 1 d = × (sin 8x) - × (sin 2x) \ 2 dx dx 2 dx 1 æ1 ö = ç × 8 cos 8x - ´ 2 cos 2x ÷ = ( 4 cos 8x - cos 2x). 2 è2 ø Differentiate

sin 2x sin 4x w.r.t. x. 1 Let y = sin 2x sin 4x = ( 2 sin 4x sin 2x) 2

Senior Secondary School Mathematics for Class 12 Pg-368

368

Senior Secondary School Mathematics for Class 12

1 = {cos ( 4x - 2x) - cos ( 4x + 2x)} 2 1 = [cos 2x - cos 6x]. 2 dy 1 d 1 d = × (cos 2x) - × (cos 6x) \ dx 2 dx 2 dx 1 1 = × ( -2 sin 2x) - × ( -6 sin 6x) 2 2 = ( 3 sin 6x - sin 2x). EXAMPLE 5

Differentiate

SOLUTION

Let y =

1 - tan x w.r.t. x. 1 + tan x

1 - tan x × 1 + tan x

(1 - tan x) = t, we get (1 + tan x) (1 - tan x) y = t and t = × (1 + tan x) dy 1 -1/2 1 = t = × \ dt 2 2 t d d (1 + tan x) × (1 - tan x) - (1 - tan x) × (1 + tan x) dt dx dx And, = dx (1 + tan x) 2 Putting

= \

(1 + tan x)( - sec2 x) - (1 - tan x)( sec2 x) (1 + tan x) 2

-2sec2 x (1 + tan x) 2

1 -2 sec2 x dy æ dy dt ö ´ =ç ´ ÷= dx è dt dx ø 2 t (1 + tan x) 2 = = y=

- sec2 x (1 + tan x)

2

´

1 + tan x 1 - tan x

- sec2 x

(1 + tan x) 3/2(1 - tan x)1/2 1 dy , find × 2 2 dx a -x

×

EXAMPLE 6

If

SOLUTION

Putting ( a 2 - x 2) = t, we get 1 -1 y= = t 2 and t = ( a 2 - x 2) t -1 dy 1 -3 dt Þ = - t 2 = 3 and = - 2x dt 2 dx 2 2t Þ

=

dy æ dy dt ö -1 x x =ç ´ ÷= ´ ( -2x) = 3 = × 3 2 dx è dt dx ø 2t 3 2 2 ( a - x 2) 2 t

.

Senior Secondary School Mathematics for Class 12 Pg-369

Differentiation

dy × dx y = (cos x 2) 2.

EXAMPLE 7

If y = cos2 x 2 , find

SOLUTION

Given:

Putting x 2 = t and cos t = u, we get y = u2 , u = cos t and t = x 2 dy du dt = 2u, = - sin t and = 2x du dt dx dy æ dy du dt ö Þ =ç ´ ´ ÷ dx è du dt dx ø Þ

= ( 2u) ´ ( - sin t) ´ 2x = - 4xu sin t = - 4x cos t sin t = - 4x cos x 2 sin x 2. dy = - 4x cos x 2 sin x 2. dx dy If y = sin (cos x 2), find × dx Putting x 2 = t and cos x 2 = cos t = u, we get

\ EXAMPLE 8 SOLUTION

y = sin u, u = cos t and t = x 2 dy du dt = cos u, = - sin t and = 2x du dt dx dy æ dy du dt ö Þ =ç ´ ´ ÷ dx è du dt dx ø Þ

= {cos u ´ ( - sin t) ´ 2 x} = -2 x sin t cos u = -2 x sin t cos (cos t) = -2 x sin x 2 cos (cos x 2). dy = -2 x sin x 2 cos (cos x 2). dx dy If y = sin ( sin x + cos x ), find × dx Putting (sin x + cos x) = t and t = u, we get

\ EXAMPLE 9 SOLUTION

y = sin u, u = t and t = (sin x + cos x) dy du 1 - 1 2 1 dt and Þ = cos u, = t = = (cos x - sin x) du dt 2 dx 2 t dy æ dy du dt ö Þ =ç ´ ´ ÷ dx è du dt dx ø 1 ì ü cos t = ícos u ´ ´ (cos x - sin x) ý = × (cos x - sin x) 2 t 2 t î þ =

cos ( sin x + cos x ) 2 sin x + cos x

× (cos x - sin x).

369

Senior Secondary School Mathematics for Class 12 Pg-370

370

EXAMPLE 10 SOLUTION

Senior Secondary School Mathematics for Class 12

dy . dx Putting x = t , sin x = sin t = u and sin x = u = v , we get y = sin v , v = u , u = sin t and t = x . If y = sin [ sin

dy dv 1 -1/2 1 du dt 1 ; = cos v ; = u = = cos t and = × dv du 2 dx 2 x 2 u dt

\

dy æ dy dv du dt ö é 1 1 ù =ç ´ ´ ´ ÷ = cos v × × cos t × dx è dv du dt dx ø êë 2 u 2 x úû 1 1 ù é = ê cos u × [Q v = u] cos t × 2 u 2 x úû ë 1 1 1 = cos ( sin t ) × × cos x × [Q u = sin t] 4 sin t x 1 1 1 = cos ( sin x ) × × cos x × [Q t = x ] 4 x sin x

So,

=

EXAMPLE 11 SOLUTION

x ], find

If y =

4 x

5x 3

x)

cos ( sin

1-x

2

x

sin

× cos x .

+ sin 2( 2x + 3), find

dy × dx

[CBSE 2000C]

We have -13

y = 5 x(1 - x 2) Þ

+ sin 2( 2x + 3)

dy d d -1 = {5 x(1 - x 2) 3} + {sin 2( 2x + 3)} dx dx dx ì ü -4 -1 æ -1 ö = í5 x × ç ÷(1 - x 2) 3 × ( -2x) + (1 - x 2) 3 - 5 ý è 3ø î þ + {2 sin ( 2x + 3) cos ( 2x + 3) × 2} = = =

10x 2 3(1 - x 2)

4

+ 3

5 (1 - x 2)

10x 2 + 15(1 - x 2) 3(1 - x 2) (15 - 5 x 2) 3(1 - x 2)

4

4

1

+ 2 sin ( 4x + 6) 3

+ 2 sin ( 4x + 6)

3

+ 2 sin ( 4x + 6).

3

EXERCISE 10A Differentiate each of the following w.r.t. x: 1. sin 4x 2. cos 5x 5. cot 2 x 4. cos x 3

3. tan 3x 6. tan 3 x

Senior Secondary School Mathematics for Class 12 Pg-371

Differentiation

x

7. cot

8.

10. ( 3 - 4x)5 13.

1 ( x - 3 x + 5)

14.

3

2 3

22.

3

2

2

2

2

a -x a +x

12. ( ax 2 + bx + c) 6 15.

17. sec 3( x 2 + 1)

16. cos x 19.

9. (5 + 7 x) 6

tan x

11. ( 2x 2 - 3 x + 4)5

2

sin 2x

20.

1 + cot x

sin x 3

23.

x sin x

3 2

371

18.

cos 3x 1 21. cosec 3 2 x

ax + b )

25. cot x

26. cos (sin

28. sin 5 x cos 3 x

29. sin 2x sin x

1 + sin x 1 - sin x

x

24.

cot

27.

cosec ( x 3 + 1)

30. cos 4x cos 2x

dy Find , when: dx æ 1 + x2 ö (sin x + x 2) ÷ 32. y = 31. y = sin ç ç 1 - x2 ÷ cot 2x è ø (cos x - sin x) dy 33. If y = , prove that + y 2 + 1 = 0. (cos x + sin x) dx pö (cos x + sin x) dy æ 34. If y = , prove that = sec2 ç x + ÷ × 4ø (cos x - sin x) dx è ANSWERS (EXERCISE 10A)

1. 4 cos 4x

2. -5 sin 5 x

3. 3 sec2 3 x

4. -3 x 2 sin x 3

5. -2 cot x cosec2 x

6. 3 tan 2 x sec2 x

7.

2

-1 cosec2 x 2 x

8.

10. -20( 3 - 4x) 4 12. 6( ax 2 + bx + c)( 2ax + b) 14.

-2a 2 x 3

( a 2 + x 2) 2 ( a 2 - x 2)

1

sec x 2 tan x

9. 42(5 + 7 x)5 11. 5( 2x 2 - 3 x + 4) 4( 4x - 3) -3( 2x - 3) 13. 2 ( x - 3 x + 5) 4 15. sec x( sec x + tan x)

2

16. -3 x 2 sin ( 2x 3) -3 sin 3 x 18. × 2 cos 3 x

17. 6x sec 3( x 2 + 1) tan ( x 2 + 1) cos 2x 2 19. × 3 (sin 2x) 2 3

1 cosec2 x 20. - × 2 1 + cot x

21.

3 2 cos x 3 x × 2 sin x 3

23.

22.

6 x

3

cosec 3

1 x

2

cot

( x cos x + sin x) 2 x sin x

1 x2

Senior Secondary School Mathematics for Class 12 Pg-372

372

24.

Senior Secondary School Mathematics for Class 12

-cosec2 x 4 x

cot

25. -6x cot 2 x 2 cosec2 x 2

x

- a cos ax + b 26. × sin {sin 2 ax + b

3 2 x cosec ( x 3 + 1) × cot ( x 3 + 1) 2 1 ö æ3 28. ( 4 cos 8x + cos 2x) 29. ç sin 3 x - sin x ÷ 2 ø è2 æ 4x 1 + x 2 ö÷ 30. - ( 3 sin 6x + sin 2x) 31. × cos ç 2 2 2÷ ç (1 - x ) è1- x ø 32. 2(sin x + x 2) sec2 2x + (cos x + 2x) tan 2x ax + b }

27. -

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 10A) 1 + sin x

1 + sin x ( 1 + sin x ) ´ = = ( sec x + tan x) cos x 1 - sin x 1 + sin x dy Þ = ( sec x tan x + sec2 x ) = sec x( sec x + tan x ). dx 32. y = (sin x + x 2 ) tan 2 x dy d d Þ = (sin x + x 2 ) × (tan 2 x ) + tan 2 x × (sin x + x 2 ). dx dx dx æ 1 - tan x ö ÷ [on dividing num. and denom. by cos x] 33. y = çç ÷ è 1 + tan x ø p æ ö Þ y = tan ç - x ÷ è4 ø dy ö 2æ p Þ = - sec ç - x ÷ dx è4 ø dy ö æp ö æp 2 Þ + y + 1 = - sec2 ç - x ÷ + tan 2 ç - x ÷ + 1 = 0. dx è4 ø è4 ø æ 1 + tan x ö ç ÷ 34. y = ç ÷ è 1 - tan x ø æp ö Þ y = tan ç + x ÷ è4 ø dy ö 2æ p Þ = sec ç + x ÷ × dx è4 ø 15. y =

2. Derivatives of Exponential and Logarithmic Functions EXPONENTIAL FUNCTION

Let a be a real number such that a > 1. Then, f ( x) = a x is called an exponential function. Its domain = R and range = R + .

Senior Secondary School Mathematics for Class 12 Pg-373

Differentiation

373

When a = e, we have the exponential function, f ( x) = ex . This is called natural exponential function. LOGARITHMIC FUNCTION

Let a be a real number such that a > 1. If a x = b, we define, log a x = b. We say that log of x to the base a is equal to b. log10 x is called common logarithm of x. log e x is called natural logarithm of x and we denote it simply by log x. It is easy to verify the following results: æxö (ii) log çç ÷÷ = (log x) - (log y) èyø

(i) log ( xy) = (log x) + (log y) (iii) log ( x n) = n log x

(iv) log a x =

1 log x a

(v) log x x = 1 and log 1 = 0 Derivatives of Exponential and Logarithmic Functions We have

(i)

d x ( e ) = ex dx

(ii)

d 1 (log x) = × dx x

Derivative of a x Let y = a x . Then, log y = x log a.

...

(i) On differentiating both sides of (i) w.r.t. x, we get 1 dy dy × = log a Þ = y(log a) y dx dx Þ Hence,

dy = a x (log a). dx

d x ( a ) = a x (log a). dx

SUMMARY

d x ( e ) = ex dx d 1 (iii) (log x) = dx x (i)

d x ( a ) = a x (log a) dx 1 d (iv) (log a x) = dx x log a (ii)

Senior Secondary School Mathematics for Class 12 Pg-374

374

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Differentiate each of the following w.r.t. x: (i) ex

2

(ii) e-3x

(iii) ecos x

2

SOLUTION

(i) Let y = ex . Putting x 2 = t, we get y = et and t = x 2 dy dt Þ = et and = 2x dt dx 2 2 dy æ dy dt ö Þ =ç ´ ÷ = ( et ´ 2 x) = ( ex ´ 2 x) = 2 x ex . dx è dt dx ø 2 d x2 Hence, ( e ) = 2 x ex . dx (ii) Let y = e-3x . Putting -3x = t, we get y = et and t = -3 x dy dt = et and =-3 dt dx dy æ dy dt ö Þ =ç ´ ÷ = - 3 e t = - 3 e -3 x . dx è dt dx ø d -3 x Hence, ( e ) = - 3 e -3 x . dx Þ

(iii) Let y = ecos x . Putting cos x = t, we get y = et and t = cos x dy dt = et and = - sin x dt dx dy æ dy dt ö Þ =ç ´ ÷ = ( - et sin x) = ( - ecos x sin x). dx è dt dx ø d cos x Hence, (e ) = - ecos x sin x. dx Þ

EXAMPLE 2

SOLUTION

Differentiate each of the following w.r.t. x: (i) sin (log x), x > 0 (ii) log (log x), x > 1 (i) Let y = sin (log x). Putting log x = t, we get y = sin t and t = log x dy dt 1 Þ = cos t and = dt dx x

Senior Secondary School Mathematics for Class 12 Pg-375

Differentiation

375

dy æ dy dt ö æ 1ö 1 cos (log x) =ç ´ ÷ = ç cos t ´ ÷ = cos (log x) ´ = × dx è dt dx ø è xø x x d cos (log x) Hence, {sin (log x)} = × dx x (ii) Let y = log (log x). Putting log x = t, we get y = log t and t = log x dy 1 dt 1 Þ = and = dt t dx x dy æ dy dt ö æ 1 1 ö æ 1 1ö 1 Þ =ç ´ ÷ = ç ´ ÷ = çç ´ ÷÷ = × dx è dt dx ø è t x ø è log x x ø ( x log x) d 1 \ {log (log x)} = × dx ( x log x) Þ

cot x

dy × dx

EXAMPLE 3

If y = e

SOLUTION

Given: y = e . Putting cot x = t and cot x = t = u, we get

, find

cot x

y = e u , u = t and t = cot x 1 dy du 1 - 1 2 dt and = eu, = t = = -cosec2 x du dt 2 dx 2 t dy æ dy du dt ö Þ =ç ´ ´ ÷ dx è du dt dx ø 1 1 ì ü cot x = íe u × × ( - cosec2 x) ý = e × × ( - cosec2 x) 2 t 2 cot x î þ Þ

= EXAMPLE 4

SOLUTION

( - cosec2 x) e 2 cot x

cot x

×

x dy , find × 2 dx x Given: y = log tan × 2 x x Putting = t and tan = tan t = u, we get 2 2 x y = log u, u = tan t and t = 2 dy 1 du dt 1 2 Þ = , = sec t and = du u dt dx 2 dy æ dy du dt ö Þ =ç ´ ´ ÷ dx è du dt dx ø If y = log tan

Senior Secondary School Mathematics for Class 12 Pg-376

376

EXAMPLE 5 SOLUTION

Senior Secondary School Mathematics for Class 12

ö 1ö æ 1 1 1 æ1 = = cosec x. = ç ´ sec2t ´ ÷ = çç ´ sec2t ÷÷ = 2 ø è 2 tan t èu ø sin 2t sin x dy Hence, = cosec x. dx 1 dy If y = , find × log cos x dx y = (log cos x) -1.

Given:

Putting cos x = t and log cos x = log t = u, we get 1 y = u-1 = , u = log t and t = cos x u dy -1 du 1 dt Þ = , = and = - sin x du u2 dt t dx dy æ dy du dt ö Þ =ç ´ ´ ÷ dx è du dt dx ø 1 1 ì -1 1 ü ì = í 2 ´ ´ ( - sin x) ý = í × × sin t îu þ î(log cos x) 2 cos x tan x = × (log cos x) 2 EXAMPLE 6 SOLUTION

ü xý þ

dy × dx Putting x = t , e x = et = u, we get If y = e

x

, find

y = u , u = et and t = x dy 1 - 1 2 1 du dt 1 - 1 2 1 = u = , = et and = x = du 2 dx 2 2 u dt 2 x dy æ dy du dt ö Þ =ç ´ ´ ÷ dx è du dt dx ø Þ

1

t

u e2 1 ö ì 1 1 ü æ 1 =ç ´ et ´ ´u ´ = ÷ =í ý= 2 x ø î2 u 2 xþ 4 x 4 x è2 u 1

x

e2 = × 4 x dy × dx

EXAMPLE 7

If y = log log log x 3 , find

SOLUTION

Let x 3 = t , log x 3 = log t = u and log log x 3 = log u = v. Then, y = log v , v = log u, u = log t and t = x 3 Þ

dy æ dy dv du dt ö =ç ´ ´ ´ ÷ dx è dv du dt dx ø

Senior Secondary School Mathematics for Class 12 Pg-377

Differentiation

377

2 æ1 1 1 ö 3x = ç ´ ´ ´ 3x2 ÷ = èv u t ø tuv 3x2 3 = 3 = x (log t)(log u) x(log t)(log log t) 3 1 = = × 3 3 x(log x )(log log x ) x(log x)(log log x 3)

EXAMPLE 8

SOLUTION

EXAMPLE 9 SOLUTION

ìï æ x2 ö üï dy If y = log ísin ç × - 1÷ ý, find ç ÷ dx ïî è 3 ø ïþ æ x2 ö æ x2 ö Put ç - 1÷ = t , sin ç - 1÷ = sin t = u ç 3 ÷ ç 3 ÷ è ø è ø ìï æ x2 ö üï and log ísin ç - 1÷ ý = log u = v. ç 3 ÷ï ïî è øþ

æ x2 ö Then, y = v , where v = log u, u = sin t and t = ç - 1÷ × ç 3 ÷ è ø dy 1 -1/2 1 dv 1 du dt 2x \ ; = v = = ; = cos t and = × dv 2 dx 3 2 v du u dt dy æ dy dv du dt ö So, =ç ´ ´ ´ ÷ dx è dv du dt dx ø 2x ö æ 1 1 =ç × × cos t × ÷ 3ø è2 v u x cos t x cos t = × = × 3 u × log u 3 sin t log sin t [Q v = log u and u = sin t] æ x2 ö - 1÷ x cot ç ç 3 ÷ é öù æ x2 x cot t è ø = = - 1÷ ú × êQ t = çç ÷ 3 log sin t æ x2 ö êë ø úû è 3 - 1÷ 3 × log sin ç ÷ ç 3 ø è dy x If y = e log (sin 2x), find × dx We have dy d x = {e log (sin 2x)} dx dx d d x = ex × {log (sin 2x)} + log (sin 2x) × (e ) dx dx ü ì 1 = ex × í × cos 2x × 2ý + log (sin 2x) × ex þ î sin 2x = 2ex cot 2x + ex log (sin 2x) = ex {2 cot 2x + log (sin 2x)}.

Senior Secondary School Mathematics for Class 12 Pg-378

378

EXAMPLE 10 SOLUTION

Senior Secondary School Mathematics for Class 12

If y = eax cos ( bx + c), find

dy × dx

We have dy d d ax = eax × {cos ( bx + c)} + cos ( bx + c) × (e ) dx dx dx = eax × {- b sin ( bx + c)} + cos ( bx + c) × aeax = eax × {a cos ( bx + c) - b sin ( bx + c)}.

EXAMPLE 11

SOLUTION

Differentiate

log

1 + cos2 x (1 - e2x )

w.r.t. x.

[CBSE 2003C] 1/2

æ 1 + cos2 x ö æ 1 + cos2 x ö 1 ç ç ÷ ÷× Let y = log = og log = l ç 1 - e2x ÷ ç 1 - e2x ÷ 2 (1 - e2x ) è è ø ø 1 1 \ y = log (1 + cos2 x) - log (1 - e2x ) 2 2 dy 1 1 1 1 Þ = × × ( -2e2x ) ( 2 cos x)( - sin x) - × 2 dx 2 (1 + cos x) 2 (1 - e2x ) 1 + cos2 x

ì - sin x cos x e2x ü =í + ý× 2 (1 - e2x ) þ î (1 + cos x) Hence,

d dx

EXAMPLE 12

If y = log

SOLUTION

We have

ìï ílog îï

1 + cos2 x üï ì - sin x cos x e2x ü = + ý× 2x ý í 2 (1 - e ) þï î (1 + cos x) (1 - e2x ) þ

1 + sin 2 x dy , find × 1 - sin x dx

[CBSE 2003C]

1 y = {log (1 + sin 2 x) - log (1 - sin x)}. 2 On differentiating (i) w.r.t. x, we get dy 1 é d d ù {log (1 + sin 2 x)} - {log (1 - sin x)} ú = × dx 2 êë dx dx û =

1 ì 2 sin x cos x ( - cos x) ü ×í ý 2 î (1 + sin 2 x) (1 - sin x) þ

=

1 ì sin 2x cos x ü ×í + ý× 2 2 î(1 + sin x) (1 - sin x) þ

EXAMPLE 13

If y = log sin ( ex + 5 x + 8), find

SOLUTION

Given:

... (i)

dy × dx

y = log sin ( ex + 5 x + 8).

... (i)

Senior Secondary School Mathematics for Class 12 Pg-379

Differentiation

379

On differentiating both sides of (i) w.r.t. x, we get 1 dy d x = × cos ( ex + 5 x + 8) × ( e + 5 x + 8) x dx sin ( e + 5 x + 8) dx = {cot ( ex + 5 x + 8)} ( ex + 5) = ( ex + 5) × cot ( ex + 5 x + 8). EXAMPLE 14 SOLUTION

ì1 1 ü dy y = x 2 + 1 - log í + 1 + 2 ý , find × x dx x þ î We have ìï1 + x 2 + 1 üï y = x 2 + 1 - log í ý x ïî ïþ If

Þ Þ

[CBSE 2008]

y = x 2 + 1 - log {1 + x 2 + 1} + log x -1 -1 dy 1 2 1 ü 1 ì1 = ( x + 1) 2 × 2x × í ( x 2 + 1) 2 × 2x ý + 2 dx 2 2 x î þ {1 + x + 1} x

= = = =

2

x +1

-

1 2

{1 + x + 1} 2

x {1 + x + 1} - x ( x 2 + 1) {1 + x 2 + 1} x {1 + x 2 + 1}

+

x

×

2

x +1

+

1 x

1 1 x x2 + 1 = + x ( x 2 + 1){1 + x 2 + 1} x

+

1 ( x 2 + 1) + x 2 + 1 = x x{1 + x 2 + 1}

( x 2 + 1){( x 2 + 1) + 1} x {1 + x 2 + 1}

x2 + 1 × x

=

EXERCISE 10B Differentiate each of the following w.r.t. x: 1. (i) e4x

(ii) e-5x

3. (i) ecot x

(iii) ex

(ii) e- sin 2x

4. (i) tan (log x)

3

(iii) e

2. (i) e

x

9. e

2x

x

(ii) log sec x (iii) log sin

(iii) e-2

x

(iii) log ( x + 1 + x 2 )

[CBSE 2003]

x2 + 1

[CBSE 2003]

8. log sin

sin 3 x

3x

13. log ( cosec x - cot x)

x

x 2

log x

11. e-5x cot 4x

(ii) e

sin x

5. (i) log 3 x (ii) 2-x (iii) 3 x + 2 1ö æ 6. (i) log ç x + ÷ (ii) log sin 3x xø è 7. e

2

10. e

cos 2x

12. ex log (sin 2x) x xö æ 14. log ç sec + tan ÷ 2 2ø è

Senior Secondary School Mathematics for Class 12 Pg-380

380

Senior Secondary School Mathematics for Class 12

1 + ex

15.

16.

1 - ex sin x

17. xe 19. e

1- x 2

ex + e- x ex - e- x

18. esin x sin ( ex )

tan x

20.

21. x 3 ex cos x

ex (1 + cos x)

22. ex cos x ANSWERS (EXERCISE 10B) 3

1. (i) 4e4x (ii) -5 e-5x (iii) ( 3 x 2) ex 2 1 - 1 -2 2. (i) - 2 × e2/x (ii) e e x (iii) x 2 x x 3. (i) -ecot x ( cosec2 x)

x

(ii) ( -2 cos 2x) e- sin 2x

(iii)

cos x ×e 2 sin x

sin x

sec2(log x) 1 x (ii) tan x (iii) cot x 2 2 1 (ii) -2- x log 2 (iii) ( 9 ´ 3 x log 3) 5. (i) x(log 3)

4. (i)

6. (i) 7.

e

( x 2 - 1) 2

x( x + 1) x

(ii) 3 cot 3 x (iii)

( 2 + x log x) 2x

8.

1 + x2

x 2

x +1

10. e 3x ( 3 cos 2x - 2 sin 2x) 12. ex { 2 cot 2x + log sin 2x} 15.

1

ex

(1 - ex ) 1 - e2x ö x cos x sin x æ 17. e ×ç + 1÷ ç 2 sin x ÷ è ø ì 2 x tan x üï 1- x 2 ï 19. e ý ísec x ïî 1 - x 2 ïþ 21. ex x 2( x cos x - x sin x + 3 cos x)

cot

x2 + 1

9. e2x ( 3 cos 3 x + 2 sin 3 x)

11. - e-5x (5 cot 4x + 4cosec2 4x) 1 x 13. cosec x 14. sec 2 2 -4 16. x ( e - e- x ) 2 18. esin x ( ex cos ex + cos x sin ex ) 20.

ex (1 + cos x + sin x) (1 + cos x) 2

22. ex cos x (cos x - x sin x)

3. Derivatives of Inverse Trigonometric Functions In the table given below, we mention the domain and range of various inverse trigonometric functions.

Senior Secondary School Mathematics for Class 12 Pg-381

Differentiation

Function

Domain

381

Range é p pù êë - 2 , 2 úû

(i) sin -1 x

[-1, 1]

(ii) cos-1 x

[-1, 1]

[ 0, p ]

(iii) tan -1 x

R

é p pù êë - 2 , 2 úû

(iv) cot -1 x

R

] 0, p [

(v) sec-1 x

R - ] -1, 1[

ìp ü [0, p ] - í ý î2þ

(vi) cosec-1 x

R - ]-1, 1[

é p pù êë - 2 , 2 úû - {0}

Derivatives of Inverse Trigonometric Functions EXAMPLE 1

SOLUTION

d 1 (sin -1 x) = , where x Î ] - 1, 1[. dx 1 - x2 ù p pé Let y = sin -1 x, where x Î ] -1, 1[ and y Î ú - , ê × Then, û 2 2ë Prove that

y = sin -1 x Þ x = sin y dx ù p pé Þ = cos y ³ 0 since y Î ú - , ê dy û 2 2ë dx Þ = 1 - sin 2 y = 1 - x 2 dy dy 1 Þ = × dx 1 - x2 d 1 Hence, × (sin -1 x) = dx 1 - x2 EXAMPLE 2

SOLUTION

-1 d (cos-1 x) = , where x Î ] -1, 1[. dx 1 - x2 ù pé Let y = cos-1 x, where x Î ] -1, 1[ and y Î ú 0, ê × Then, û 2ë y = cos-1 x Þ x = cos y dx ù pé Þ = - sin y, where sin y > 0, since y Î ú 0, ê dy û 2ë dx 2 2 Þ = - 1 - cos y = - 1 - x dy Prove that

Þ

dy -1 = × dx 1 - x2

Senior Secondary School Mathematics for Class 12 Pg-382

382

Senior Secondary School Mathematics for Class 12

Hence,

-1 d × (cos-1 x) = dx 1 - x2 d 1 (tan -1 x) = , where x Î R. dx (1 + x 2)

EXAMPLE 3

Prove that

SOLUTION

ù p pé Let y = tan -1 x, where x Î R and y Î ú - , ê × Then, û 2 2ë x = tan y dx Þ = sec2 y = (1 + tan 2 y) = (1 + x 2) dy dy 1 Þ = × dx (1 + x 2) d 1 Hence, × (tan -1 x) = dx (1 + x 2) -1 d (cot -1 x) = , where x Î R. dx (1 + x 2)

EXAMPLE 4

Prove that

SOLUTION

Let y = cot -1 x, where x Î R and y Î] 0, p [. Then, x = cot y dx Þ = - cosec2 y = - (1 + cot 2 y) = - (1 + x 2) dy Þ

dy -1 = × dx (1 + x 2)

Hence,

-1 d × (cot -1 x) = dx (1 + x 2) d 1 ( sec-1 x) = , where x Î R - [-1, 1]. dx |x| x 2 - 1

EXAMPLE 5

Prove that

SOLUTION

ìp ü Let y = sec-1 x, where x Î R - [-1, 1] and y Î ] 0, p [ - í ý× Then, î2þ x = sec y dx Þ = sec y tan y > 0 dy Þ

dy 1 1 = = dx sec y tan y sec y × sec2 y - 1

Þ

dy 1 = × dx |x| x 2 - 1

Hence,

d 1 × ( sec-1 x) = dx |x| x 2 - 1

Senior Secondary School Mathematics for Class 12 Pg-383

Differentiation

383

-1 d ( cosec-1 x) = , where x Î R - [-1, 1]. dx |x| x 2 - 1

EXAMPLE 6

Prove that

SOLUTION

ù p pé Let y = cosec-1 x, where x Î R - [-1, 1] and y Î ú - , ê - {0}. Then, û 2 2ë x = cosec y dx Þ = - cosec y cot y , where cosec y cot y > 0 dy Þ

dy -1 -1 -1 = = = × 2 dx cosec y cot y ( cosec y) cosec y - 1 |x| x 2 - 1

Hence,

-1 d × ( cosec-1 x) = dx |x| x 2 - 1

SUMMARY

d 1 (sin -1 x) = dx 1 - x2 d 1 (iii) (tan -1 x) = dx (1 + x2) (i)

(v)

d 1 (sec-1 x) = dx |x| x2 - 1

d -1 (cos-1 x) = dx 1 - x2 d -1 (iv) (cot -1 x) = dx (1 + x2) (ii)

(vi)

d -1 (cosec-1 x) = dx |x| × x2 - 1

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Differentiate the following w.r.t. x: (ii) tan -1 x (i) sin -1 2x

(iii) cos-1(cot x)

(i) Let y = sin -1 2x. Putting 2x = t, we get y = sin -1t and t = 2x. dy 1 Now, y = sin -1t Þ = × dt 1 - t2 dt And, t = 2x Þ = 2. dx 2 2 dy æ dy dt ö = =ç ´ ÷= \ 2 dx è dt dx ø 1 -t 1 - 4x 2 2 d Hence, × (sin -1 2x) = dx 1 - 4x 2 (ii) Let y = tan -1 x . Putting x = t, we get y = tan -1t and t = x .

[Q t = 2x].

Senior Secondary School Mathematics for Class 12 Pg-384

384

Senior Secondary School Mathematics for Class 12

dy 1 = × dt (1 + t 2) dt 1 - 1 2 1 And, t = x Þ = x = × dx 2 2 x 1 1 1 dy æ dy dt ö =ç ´ ÷= × = \ dx è dt dx ø (1 + t 2) 2 x 2 x (1 + x) d 1 Hence, × (tan -1 x ) = dx 2 x (1 + x) Now, y = tan -1t Þ

[Q t = x ].

(iii) Let y = cos-1(cot x). Putting cot x = t, we get y = cos-1t and t = cot x. dy -1 Now, y = cos-1t Þ = × dt 1 - t2 dt And, t = cot x Þ = -cosec2 x. dx dy æ dy dt ö cosec2 x cosec2 x \ [Q t = cot x]. =ç ´ ÷= = dx è dt dx ø 1 - t2 1 - cot 2 x Hence, EXAMPLE 2

SOLUTION

d cosec2 x × {cos-1(cot x)} = dx 1 - cot 2 x

Differentiate the following w.r.t. x: (i) sec (tan -1 x) (ii) sin (tan -1 x)

(iii) cot (cos-1 x)

(i) Let y = sec (tan -1 x). Putting tan -1 x = t, we get y = sec t and t = tan -1 x. dy Now, y = sec t Þ = sec t tan t. dt dt 1 And, t = tan -1 x Þ = × dx (1 + x 2) \

2 dy æ dy dt ö sec t tan t ( 1 + tan t )(tan t) = =ç ´ ÷= dx è dt dx ø (1 + x 2) (1 + x 2)

=

( 1 + x2 )x (1 + x 2)

=

x 1+ x

2

[Q t = tan -1 x Þ tan t = x]

. Hence,

d x × {sec (tan -1 x)} = dx 1 + x2

(ii) Let y = sin (tan -1 x). Putting tan -1 x = t, we get y = sin t and t = tan -1 x. dy Now, y = sin t Þ = cos t. dt

Senior Secondary School Mathematics for Class 12 Pg-385

Differentiation

385

dt 1 = × dx (1 + x 2) 1 1 dy æ dy dt ö = \ =ç ´ ÷ = cos t × 3 dx è dt dx ø (1 + x 2) (1 + x 2) 2 And, t = tan -1 x Þ

é 1 ù êQ tan t = x Þ cos t = ú× êë 1 + x 2 úû d 1 Hence, × {sin (tan -1 x)} = 3 dx (1 + x 2) 2

(iii) Let y = cot (cos-1 x).

Putting cos-1 x = t, we get y = cot t and t = cos-1 x. dy Now, y = cot t Þ = - cosec2t. dt dt -1 And, t = cos-1 x Þ = × dx 1 - x2

1 dy æ dy dt ö cosec2t =ç ´ ÷= = 3 2 dx è dt dx ø 1-x (1 - x 2) 2 é 1 ù 2 ú× êQ cos t = x Þ cosec t = (1 - x 2) û ë d 1 Hence, × {cot (cos-1 x)} = 3 dx (1 - x 2) 2

\

dy 2 = × dx (1 + 4x 2) 3 2

EXAMPLE 3

If y = sin (tan -1 2x), prove that

SOLUTION

Putting tan -1 2x = t, we get y = sin t and t = tan -1 2x. dy = cos t. dt ü dt ì 1 2 And, t = tan -1 2x Þ =í ´ 2ý = × 2 dx î(1 + 4x ) þ (1 + 4x 2) ü 2 dy æ dy dt ö ì × =ç ´ ÷ = ícos t ´ \ 2 ý dx è dt dx ø î (1 + 4x ) þ Now, t = tan -1 2x Þ tan t = 2x Now, y = sin t Þ

Þ sec t = 1 + tan 2t = 1 + 4x 2 1 1 Þ cos t = = × sec t 1 + 4x 2 Putting the value of cos t from (ii) in (i), we get üï 1 2 2 dy ìï × =í ´ ý= 2 dx ïî 1 + 4x 2 (1 + 4x ) ïþ (1 + 4x 2) 3 2

... (i)

... (ii)

Senior Secondary School Mathematics for Class 12 Pg-386

386

Senior Secondary School Mathematics for Class 12

EXAMPLE 4

Differentiate cot -1 x w.r.t. x.

SOLUTION

Let y = cot -1 x . Putting x = t and cot -1 x = cot -1t = u, we get y = u, where u = cot -1t and t = x . dy 1 - 1 2 1 = u = ; du 2 2 u du -1 u = cot -1t Þ = × dt (1 + t 2)

Now, y = u Þ

dt 1 - 1 2 1 = x = × dx 2 2 x

And, t = x Þ \

dy æ dy du dt ö -1 =ç ´ ´ ÷= dx è du dt dx ø 4 u(1 + t 2) x =

-1 -1

4( cot t )(1 + t ) x -1

= 4( cot EXAMPLE 5 SOLUTION

etan

Differentiate Let y = e

tan

-1

x

-1

x

2

-1

x )(1 + x) x

w.r.t. x.

.

Putting x = t and tan -1 x = tan -1t = u, we get y = e u, where u = tan -1t and t = x . dy = eu; du du 1 u = tan -1t Þ = × dt (1 + t 2)

Now, y = e u Þ

And, t = x Þ \

dt 1 - 1 2 1 = x = × dx 2 2 x

1 1 dy æ dy du dt ö =ç ´ ´ ÷ = eu × × 2 dx è du dt dx ø (1 + t ) 2 x = etan

-1

t

×

1 1 × (1 + t 2) 2 x

[Q u = tan -1 t]

-1

=

etan x 2 x (1 + x) -1

Hence,

[Q t = x ].

dy etan x = × dx 2 x (1 + x)

[Q u = cot -1t] [Q t = x ].

Senior Secondary School Mathematics for Class 12 Pg-387

Differentiation

EXAMPLE 6

SOLUTION

x sin -1 x

If y = y=

1-x

2

, find

387

dy × dx

x sin -1 x

... (i)

1 - x2

Þ y 1 - x 2 = x sin -1 x.

... (ii)

On differentiating both sides of (ii) w.r.t. x, we get y×

d dy d d ( 1 - x2 ) + ( 1 - x2 ) = x× (sin -1 x) + sin -1 x × ( x) dx dx dx dx 1

1 dy 1 Þ y × (1 - x 2) 2 × ( -2x) + ( 1 - x 2 ) = x× + sin -1 x × 1 2 dx 1 - x2

-x y

Þ Þ

1-x

+ ( 1 - x2 )

2

- x 2 sin -1 x (1 - x 2)

dy x = + sin -1 x 2 dx 1-x

+ ( 1 - x2 )

Þ - x 2 sin -1 x + (1 - x 2) Þ (1 - x 2) Þ

3

2

3

2

[using (i)]

dy = x( 1 - x 2 ) + (1 - x 2) sin -1 x dx

dy = x( 1 - x 2 ) + sin -1 x dx

dy x( 1 - x 2 ) + sin -1 x = × 3 dx (1 - x 2) 2 d [ 1 - x 2 sin -1 x - x]. dx

EXAMPLE 7

Find

SOLUTION

We have d [( 1 - x 2 ) sin -1 x - x] dx =

dy x = + sin -1 x 2 dx 1-x

d d [( 1 - x 2 sin -1 x] - ( x) dx dx

= ( 1 - x2 ) × = ( 1 - x2 ) ×

d d (sin -1 x) + (sin -1 x) × ( 1 - x 2 ) - 1 dx dx 1

1 -1 + (sin -1 x) × (1 - x 2) 2 × ( -2x) - 1 2 ( 1-x ) 2

-1 -1 ïì x sin x üï - x sin x × = í1 - 1ý = ïþ 1 - x2 1 - x2 îï

Senior Secondary School Mathematics for Class 12 Pg-388

388

Senior Secondary School Mathematics for Class 12

d éx 2 a2 xù sin -1 ú = a 2 - x 2 . a - x2 + ê dx ë 2 2 aû

EXAMPLE 8

Show that

SOLUTION

We have d éx a2 xù 2 2 × sin -1 ú ê × a -x + dx ë 2 2 aû 2 d éx ù a d é -1 x ù = × a2 - x 2 ú + × sin dx êë 2 a úû û 2 dx êë x d d æ x ö a2 × = × ( a2 - x 2 ) + ( a2 - x 2 ) × ç ÷ + 2 dx dx è 2 ø 2 =

[CBSE 2004C]

1 1-

x

2

a2

2

x 1 2 1 a -1 × ( a - x 2) 2 × ( -2x) + ( a 2 - x 2 ) × + 2 2 2 2 a2 - x 2 -x 2

a2 - x 2 a2 + 2 2 a2 - x 2 2 a2 - x 2 2 2 2 2 ( a 2 - x 2) -x + ( a - x ) + a = = = ( a 2 - x 2). 2 a2 - x 2 ( a 2 - x 2) =

Hence,

+

d éx a2 xù 2 2 × sin -1 ú = a 2 - x 2 . ê × a -x + dx ë 2 2 aû

EXERCISE 10C Differentiate each of the following w.r.t. x: 1. cos-1 2x

2. tan -1 x 2

3. sec-1 x

5. tan -1(log x)

6. cot -1( ex )

7. log (tan -1 x)

9. sin -1(cos x)

10. (1 + x 2) tan -1 x 11. tan -1(cot x)

12. log (sin -1 x 4)

13. (cot -1 x 2) 3

15. tan (sin -1 x)

16. etan

-1

14. tan -1(cos x )

x

17.

18. If y = sin -1(cos x) + cos-1(sin x), prove that 19. Prove that

x a 8. cot -1 x 3 4. sin -1

sin -1 x 2

dy = -2. dx

d { 2x tan -1 x - log (1 + x 2 )} = 2 tan -1 x. dx ANSWERS (EXERCISE 10C)

1.

-2 1 - 4x

2

2.

2x 4

(1 + x )

3.

1 2x x - 1

4.

1 2

a - x2

×

1 a

Senior Secondary School Mathematics for Class 12 Pg-389

Differentiation

5.

1 2

x{1 + (log x) }

-ex

7.

2x

(1 + e )

10. (1 + 2x tan -1 x)

9. –1 13.

6.

-6x(cot -1 x 2) 2

14.

(1 + x 4) etan x 2 x (1 + x)

17.

1 -1

2

(1 + x ) tan x

11. –1

- sin x ( 2 x )(1 + cos2 x )

-1

16.

389

8. 12.

15.

-3 x 2 (1 + x 6) 4x 3 (sin -1 x 4) 1 - x 8

1 (1 - x 2)

3

2

x ( 1 - x ) sin -1 x 2 4

4. Differentiation by Trigonometrical Transformations SOME USEFUL RESULTS

æ xö (i) (1 - cos x) = 2 sin 2 ç ÷ è 2ø

æ xö (ii) (1 + cos x) = 2 cos2 ç ÷ è 2ø

(iii) sin 3 x = (3 sin x - 4 sin 3 x) (v) sin x =

2 tan ( x/2) 2

1 + tan ( x/2)

(iv) cos 3 x = (4cos3 x - 3cos x) (vi) cos x =

1 - tan 2( x /2) 1 + tan 2( x /2)

ì x -y ü -1 -1 -1 ì x + y ü (vii) tan -1 x - tan -1 y = tan -1 í ý (viii) tan x + tan y = tan í ý + 1 xy þ î î1 - xy þ SOME USEFUL SUBSTITUTIONS -1

Suppose we are given sin

f ( x), cos-1 f ( x), tan -1 f ( x), etc.

Rule 1.

If f ( x) = a 2 - x 2 , put x = a sin q or x = a cos q.

Rule 2.

If f ( x) = a 2 + x 2 , put x = a tan q or x = a cot q.

Rule 3.

If f ( x) = x 2 - a 2 , put x = a sec q or x = a cosec q.

Rule 4.

If f ( x) = a - x , put x = a cos 2q.

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Differentiate each of the following w.r.t. x: æ 1 - cos x ö æ cos x - sin x ö ÷÷ ÷÷ (ii) tan -1 çç (i) tan -1 çç sin x è ø è cos x + sin x ø ì ü æ 1 - cos x ö 2 sin 2( x/2) ÷÷ = tan -1í (i) Let y = tan -1 çç ý è sin x ø î 2 sin ( x/2) cos ( x/2) þ xü x ì = tan -1ítan ý = × 2þ 2 î

Senior Secondary School Mathematics for Class 12 Pg-390

390

Senior Secondary School Mathematics for Class 12

x × 2 dy d æxö 1 Hence, = ç ÷= × dx dx è 2 ø 2

\ y=

æ cos x - sin x ö æ 1 - tan x ö ÷÷ = tan -1 çç ÷÷ (ii) Let y = tan -1 çç cos x + sin x è ø è 1 + tan x ø [on dividing num. and denom. by cos x] æp öü æ p ö -1 ì = tan ítan ç - x ÷ ý = ç - x ÷ × è4 øþ è 4 ø î ö æp \ y = ç - x÷ × ø è4 d æ pö d dy d æp ö Hence, = ç ÷ - ( x) = ( 0 - 1) = -1. ç - x÷ = dx dx è 4 ø dx è 4 ø dx EXAMPLE 2

SOLUTION

Differentiate w.r.t. x: æ cos x ö ÷÷ (i) tan -1 çç è 1 + sin x ø

(ii) tan -1( sec x + tan x)

ì æp ö ü ï sin ç 2 - x ÷ ïï cos x ö è ø -1 ï ÷÷ = tan í (i) Let y = tan çç ý p + sin x æ 1 è ø ï1 + cos ç - x ö÷ ï ïî ø ïþ è2 ì æp xö æp xöü ïï 2 sin ç 4 - 2 ÷ cos ç 4 - 2 ÷ ïï è ø è ø = tan -1í ý p x æ ö ï ï 2 cos2 ç - ÷ ïî ïþ è 4 2ø ì æ p x öü æ p x ö = tan -1ítan ç - ÷ ý = ç - ÷ × è 4 2øþ è 4 2ø î æp xö \ y = ç - ÷× è 4 2ø 1 ö -1 dy d æp xö d æ pö d æ x ö æ Hence, × = ç - ÷= ç ÷ç ÷ = ç0 - ÷ = 2 dx dx è 4 2 ø dx è 4 ø dx è 2 ø è 2ø -1 æ

(ii) Let y = tan -1( sec x + tan x) æ 1 æ 1 + sin x ö sin x ö ÷÷ = tan -1 çç ÷÷ = tan -1 çç + è cos x cos x ø è cos x ø ì æp öü ïï1 - cos ç 2 + x ÷ ïï è ø = tan -1í ý ï sin æç p + x ö÷ ï ïî ø ïþ è2 ì ü æp ö æp ö íQ cos ç + x ÷ = - sin x ; sin ç + x ÷ = cos x ý è2 ø è2 ø î þ

Senior Secondary School Mathematics for Class 12 Pg-391

Differentiation

391

ì ü æp xö 2 sin 2 ç + ÷ ï ïï 4 2ø è = tan í ý ï 2 sin æç p + x ö÷ cos æç p + x ö÷ ï ïî è 4 2 ø ïþ è 4 2ø p x p xö ì ü æ ö æ = tan -1ítan ç + ÷ ý = ç + ÷ × 4 2 4 2ø è øþ è î p x ö æ \ y = ç + ÷× è 4 2ø dy d æp xö d æ pö d æ x ö æ 1ö 1 Hence, = ç + ÷= ç ÷+ ç ÷ = ç0 + ÷ = × dx dx è 4 2 ø dx è 4 ø dx è 2 ø è 2ø 2 -1 ï

EXAMPLE 3

Differentiate w.r.t. x: ì 1 + cos x ü -1 ì 1 + sin x ü (i) tan -1í ý (ii) tan í ý cos 1 x î þ î 1 - sin x þ

SOLUTION

[CBSE 2003C, ‘04C]

2 ì ü ì 1 + cos x ü -1 ï 2 cos ( x/2) ï (i) Let y = tan -1í ý = tan í ý 2 ïî 2 sin ( x/2) ïþ î 1 - cos x þ

xö ì æ æ p x öü æ p x ö = tan -1 ç cot ÷ = tan -1ítan ç - ÷ ý = ç - ÷ × 2ø è è 2 2øþ è 2 2ø î æp xö \ y = ç - ÷× è 2 2ø 1 ö -1 dy d æp xö d æ pö d æ x ö æ Hence, × = ç - ÷= ç ÷ç ÷ = ç0 - ÷ = 2ø 2 dx dx è 2 2 ø dx è 2 ø dx è 2 ø è 1

ì æp ö ü2 1 - cos ç + x ÷ ï ï ì ü 1 + sin x è2 øï -1 ï (ii) Let y = tan -1í ý = tan í ý p æ î 1 - sin x þ ï1 + cos ç + x ö÷ ï ïî è2 ø ïþ 1

x ö ü2 ì 2æ p ïï 2 sin ç 4 + 2 ÷ ïï è ø = tan -1ìtan æ p + x ö ü = æ p + x ö × = tan -1í ÷ ç ÷ý ç í ý p x æ è 4 2øþ è 4 2ø î ï 2 cos2 ç + ö÷ ï ïî è 4 2 ø ïþ æp xö \ y = ç + ÷× è 4 2ø 1ö 1 dy d æp xö d æ pö d æ x ö æ Hence, = ç ÷ = ç0 + ÷ = × ç ÷+ ç + ÷= 2ø 2 dx dx è 4 2 ø dx è 4 ø dx è 2 ø è EXAMPLE 4

ì 1 + cos x ü Differentiate cos-1í ý w.r.t. x. 2 î þ

Senior Secondary School Mathematics for Class 12 Pg-392

392

SOLUTION

Senior Secondary School Mathematics for Class 12 2 ì ü ì 1 + cos x ü -1 ï 2 cos ( x/2) ï Let y = cos-1í ý = cos í ý 2 2 î þ îï þï x -1 = cos {cos ( x/2)} = × 2 x \ y= × 2 dy d æxö 1 Hence, = ç ÷= × dx dx è 2 ø 2

EXAMPLE 5

If y = cot -1

SOLUTION

We have y = cot -1

= cot

\

-1

dy 1 - sin x , find × 1 + sin x dx

1 - sin x = cot -1 1 + sin x æp 2 cos2 ç + è4 2æ p 2 sin ç + è4

[CBSE 2004C]

æp ö 1 + cos ç + x ÷ 2 è ø æp ö 1 - cos ç + x ÷ è2 ø

xö ÷ ì 2ø æ p x öü æ p x ö = cot -1ícot ç + ÷ ý = ç + ÷ × xö è 4 2øþ è 4 2ø î ÷ 2ø

dy d æp xö d æ pö d æ x ö æ = ç + ÷= ç ÷+ ç ÷ = ç0 + dx dx è 4 2 ø dx è 4 ø dx è 2 ø è

EXAMPLE 6

If y = cot -1

SOLUTION

We have

1 + sin x + 1 - sin x 1 + sin x - 1 - sin x

and 0 < x
0, x = çt + ÷ and y = a è t ø , find × dx è tø We have a ( a-1 ) ( a-1 ) 1ö dx d æ 1ö æ æ 1ö æ 1ö æ 1ö x = çt + ÷ Þ = a çt + ÷ × çt + ÷ = a çt + ÷ × ç1 - 2 ÷ × dt dt è t ø è tø è tø è tø è t ø

Senior Secondary School Mathematics for Class 12 Pg-442

442

Senior Secondary School Mathematics for Class 12 æ 1ö çt + ÷ tø

And, y = a è

æ

Þ

1ö

çt + ÷ d æ 1ö dy = a è t ø log a × çt + ÷ dt è t ø dt æ 1ö çt + ÷ tø log

= aè

dy ( dy/dt) = = \ dx ( dx/dt)

1ö æ a × ç1 - 2 ÷ × è t ø

æ 1ö çt + ÷ a è t ø log

æ 1 ö 1ö æ ç t + -1 ÷ a × ç1 - 2 ÷ t è ø × log a a è t ø = × ( a-1 ) ( a-1 ) 1ö æ æ 1ö æ 1ö a çt + ÷ 1 t + ç ÷ ÷ ç è tø è tø è t2 ø

EXERCISE 10I Find

dy , when dx

1. x = at 2 , y = 2at 2

2. x = a cos q, y = b sin q 2

3. x = a cos q, y = b sin q 5. x = a (1 - cos q), y = a ( q + sin q) 7. x = (log t + cos t), y = ( et + sin t)

4. x = a cos 3 q, y = a sin 3 q 6. x = a log t , y = b sin t

8. x = cos q + cos 2q, y = sin q + sin 2q

9. x = sin 2q , y = cos 2q

10. x = eq(sin q + cos q), y = eq(sin q - cos q) 11. x = a (cos q + q sin q), y = a (sin q - q cos q) [CBSE 2003C] 3 at 3 at 2 1 - t2 2t 12. x = 13. x = , y= , y= 1 + t2 1 + t2 (1 + t 2) (1 + t 2) t 1 14. x = cos-1 , y = sin -1 2 1+t 1 + t2 dy 3 15. If x = 3 cos t - 2 cos t , y = 3 sin t - 2 sin 3t , show that = cot t. dx 1 + log t 3 + 2log t dy and y = 16. If x = , show that = t. 2 t dx t p dy [CBSE 2003C] at q = × 17. If x = a ( q - sin q), y = a (1 - cos q), find 2 dx dy 3q 18. If x = 2 cos q - cos 2q and y = 2 sin q - sin 2q, show that = tan × dx 2

[CBSE 2013C]

sin 3t cos 3t dy 19. If x = , y= , find × dx cos 2t cos 2t

æ d 2y ö 20. If x = ( 2 cos q - cos 2q) and y = ( 2 sin q - sin 2q), find ç 2 ÷ × [CBSE 2005C] ç dx ÷ p è øq = 21. If x = a ( q - sin q), y = a (1 + cos q), find

d 2y dx 2

2

×

[CBSE 20011]

Senior Secondary School Mathematics for Class 12 Pg-443

Differentiation

443

ANSWERS (EXERCISE 10I)

1.

-b cot q a bt cos t 6. a

1 t

2.

5. cot

q 2

9. -(tan 2q) 3/2 13.

(t 2 - 1) 2t

10. tan q 14. 1

17. 1

-b 4. - tan q a æ cos q + 2 cos 2q ö t( et + cos t) ÷÷ 7. 8. - çç (1 - t sin t) è sin q + 2 sin 2q ø 2t 11. tan q 12. (1 - t 2)

3.

19. - cot 3t

20.

-3 2

21.

q 1 cosec4 4a 2

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 10I) 14. Put t = tan q. 1 1 log cos 2 t and log y = 3 log cos t - log cos 2t 2 2 dy 1 dx 3 1 1 Þ × = × cos t - × = y × ( -3 tan t + tan 2t ) ( -2 sin 2t ), x dt sin t 2 cos 2t dt dy dx Þ = x × [ 3 cot t + tan 2 t], = y × [-3 tan t + tan 2 t] dt dt æ 2 tan t ö÷ cot 3t × ç -3 tan t + ç - tan 2t ÷ø 1 dy ( dy/ dt ) y ( -3 tan t + tan 2 t ) è Þ = = × = æ ö dx ( dx/ dt ) x ( 3 cot t + tan 2 t ) ç 3 + 2 tan t ÷ ç tan t 1 - tan 2t ÷ è ø

19. log x = 3 log sin t -

= 20.

- ( 1 - 3 tan 2t ) ( 3 tan t - tan 3t )

=

-1 = - cot 3 t. tan 3 t

dy ( dy/ dq) 3q = = tan dx ( dx/ dq) 2 d2 y

3 = sec2 dx 2 2 æ d2 y ö Þ ç 2÷ = ç dx ÷ p è øq = Þ

2

3 q dq 3 3q 1 × = sec2 × 2 dx 2 2 2(sin 2 q - sin q) 1 -3 3 3p 1 3 × = × sec2 = ´2´ pö 4 ( 0 - 1) 2 4 4 æ ç sin p - sin ÷ 2ø è

10. Second-Order Derivatives Let y = f ( x) be a differentiable function of x whose second-order derivative d 2y exists. We denote the second-order derivative of y w.r.t. x by 2 or y 2 . dx

Senior Secondary School Mathematics for Class 12 Pg-444

444

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

SOLUTION

Find the second-order derivative of: (ii) log x (iii) tan -1 x (i) x10 (i) Let y = x10 . Then, dy = 10x 9 . dx d 2y \ = (10 ´ 9) x 8 = 90x 8. dx 2 d 2y Hence, 2 ( x10) = 90x 8. dx (ii) Let y = log x. Then, dy 1 = = x -1 . dx x -1 d 2y d ( -1 - 1 ) \ = ( x -1) = ( -1) x = - x -2 = 2 × dx 2 dx x d2 -1 Hence, 2 (log x) = 2 × dx x (iii) Let y = tan -1 x. Then, dy 1 = = (1 + x 2) -1 . dx (1 + x 2) d 2y

\

dx

2

Hence,

=

-2x d (1 + x 2) -1 = ( -1)(1 + x 2) -2 × ( 2x) = × dx (1 + x 2) 2

d2 dx

2

{tan -1 x} =

-2x (1 + x 2) 2 d 2y

× cos x

EXAMPLE 2

If y = (tan x + sec x), prove that

SOLUTION

y = (tan x + sec x). æ 1 sin x ö÷ dy 1 = sec2 x + sec x tan x = ç + × \ ç cos2 x cos x cos x ÷ dx è ø æ 1 + sin x ö æ 1 + sin x ö 1 ÷ ç ÷ × =ç ç cos2 x ÷ = ç 1 - sin 2 x ÷ = (1 - sin x) è ø è ø

dx

2

=

(1 - sin x) 2

Given that

ü d d ì 1 (1 - sin x) -1 ý= í ( sin ) dx 1 x dx þ dx î cos x = ( -1)(1 - sin x) -2( - cos x) = × (1 - sin x) 2 2 d y cos x Hence, = × 2 dx (1 - sin x) 2

\

d 2y 2

=

×

Senior Secondary School Mathematics for Class 12 Pg-445

Differentiation

d 2y

EXAMPLE 3

If y = e4x sin 3 x , find

SOLUTION

Let y = e4x sin 3 x. Then,

dx 2

×

dy = 3 e4x cos 3 x + 4 e4x sin 3 x = e4x ( 3 cos 3 x + 4 sin 3 x). dx \

d 2y dx

2

=

d æ dy ö d 4x {e ( 3 cos 3 x + 4 sin 3 x)} ç ÷= dx è dx ø dx

= e4x ( - 9 sin 3 x + 12 cos 3 x) + 4 e4x ( 3 cos 3 x + 4 sin 3 x) = e4x (7 sin 3 x + 24 cos 3 x). EXAMPLE 4

If y = ( x 4 + cot x), find

SOLUTION

We have

d 2y dx 2

×

y = ( x 4 + cot x) Þ Þ

dy = 4x 3 - cosec2 x dx d 2y

d ( 4x 3 - cosec 2 x) dx dx d d = 4× ( x 3) ( cosec 2 x) dx dx 2

=

= ( 4 ´ 3 x 2) - 2 cosec x ( - cosec x cot x) = (12x 2 + 2 cosec 2 x cot x). EXAMPLE 5 SOLUTION

Find the second derivative of log (log x) w.r.t. x. Let y = log (log x). Then, dy 1 1 1 = × = = ( x log x) -1 dx (log x) x ( x log x) Þ

d 2y dx

2

=

d ( x log x) -1 dx

d ( x log x) dx -1 ö - (1 + log x) æ 1 × = × ç x × + log x × 1÷ = ø ( x log x) 2 è x ( x log x) 2 = ( -1)( x log x) -2 ×

d 2y

EXAMPLE 6

If y = sin (log x), find

SOLUTION

We have y = sin (log x) dy d 1 cos (log x) Þ = {sin (log x)} = cos (log x) × = dx dx x x

dx 2

×

445

Senior Secondary School Mathematics for Class 12 Pg-446

446

Senior Secondary School Mathematics for Class 12

Þ

d 2y dx 2

d ì cos (log x) ü ý í dx î x þ d d x× {cos (log x)} - cos (log x) × ( x) dx dx = x2 1ü ì x í- sin (log x) × ý - cos (log x) × 1 xþ = î x2 - {sin (log x) + cos (log x)} = × x2 =

EXAMPLE 7

If ey( x + 1) = 1, prove that

SOLUTION

We have

d 2y

2

æ dy ö =ç ÷ × 2 è dx ø dx

ey( x + 1) = 1 Þ ey =

1 ( x + 1)

... (i)

ì 1 ü Þ y = log í ý = log 1 - log ( x + 1) î( x + 1) þ Þ y = - log ( x + 1). dy -1 \ = dx ( x + 1) Þ

d 2y dx

2

Hence, EXAMPLE 8 SOLUTION

=

2

æ dy ö =ç ÷ × è dx ø ( x + 1) 1

2

d 2y

2

æ dy ö =ç ÷ × è dx ø dx 2

If y = A cos nx + B sin nx, prove that

d 2y

dx We have y = A cos nx + B sin nx dy d d Þ = ( A cos nx) + ( B sin nx) dx dx dx = - An sin nx + Bn cos nx = n( B cos nx - A sin nx) Þ

Þ

d 2y dx 2

2

d y dx 2

... (ii)

2

+ n2 y = 0.

d ( B cos nx - A sin nx) dx d ü ì d (cos nx) - A × (sin nx) ý = n × íB × dx þ î dx = n × {- Bn sin nx - An cos nx}

= n×

= - n2( A cos nx + B sin nx) = - n2 y + n2 y = 0.

Senior Secondary School Mathematics for Class 12 Pg-447

Differentiation

EXAMPLE 9

If y = ex (sin x + cos x), prove that

SOLUTION

We have

447

d 2y dx

2

-2

dy + 2y = 0. [CBSE 2002, ‘09] dx

y = ex (sin x + cos x) Þ

dy d d x = ex × (sin x + cos x) + (sin x + cos x) × (e ) dx dx dx = ex (cos x - sin x) + (sin x + cos x) × ex = 2ex cos x

Þ

d 2y dx

= 2×

2

d x ( e cos x) dx

d x ü d ì = 2 × íex × (cos x) + cos x × (e )ý dx dx þ î = 2 × {ex ( - sin x) + (cos x) ex} = 2ex (cos x - sin x). æ d 2y ö dy \ ç 2 - 2 + 2y ÷ ç dx ÷ dx è ø x = 2e (cos x - sin x) - 4 ex cos x + 2ex (sin x + cos x) = 0. Hence,

d 2y dx

2

-2

dy + 2y = 0. dx

EXAMPLE 10

If y = 3 e2x + 2e 3x , prove that

SOLUTION

We have

d 2y dx 2

-5

dy + 6y = 0. dx

... (i) y = 3 e2x + 2e 3x dy d 2x d 3x Þ = 3× (e ) + 2× ( e ) [on differentiating (i) w.r.t. x] dx dx dx = ( 3 ´ 2e2x ) + ( 2 ´ 3 e 3x ) = ( 6e2x + 6e 3x ) dy ... (ii) Þ = 6( e2x + e 3x ). dx On differentiating (ii) w.r.t. x, we get d 2y

d 3x ü ìd (e )ý = 6 × í ( e2x ) + dx þ î dx dx = 6 × ( 2 e2x + 3 e 3x ). æ d 2y ö dy \ ç 2 -5 + 6y ÷ ç dx ÷ dx è ø = 6( 2e2x + 3 e 3x ) - 30( e2x + e 3x ) + (18 e2x + 12 e 3x ) 2

= (12 - 30 + 18) e2x + (18 - 30 + 12) e 3x = 0. æ d 2y ö dy Hence, ç 2 - 5 + 6y ÷ = 0. ç dx ÷ dx è ø

Senior Secondary School Mathematics for Class 12 Pg-448

448

Senior Secondary School Mathematics for Class 12

EXAMPLE 11

If y = sin -1 x, prove that (1 - x 2)

SOLUTION

Given:

d 2y dx

2

-x

dy = 0. dx

[CBSE 2012]

y = sin -1 x.

... (i)

On differentiating both sides of (i) w.r.t. x, we get 1 y1 = 1 - x2 1 Þ y12 = (1 - x 2) Þ (1 - x 2) y12 = 1. On differentiating both sides of (ii) w.r.t. x, we get (1 - x 2) × 2y1 y 2 + y12( -2x) = 0

... (ii)

Þ (1 - x 2) y 2 - xy1 = 0. Hence, (1 - x 2) -1

d 2y dx

2

-x

dy = 0. dx

EXAMPLE 12

2

If y = (tan x) , prove that (1 + x 2) 2 y 2 + 2x(1 + x 2) y1 = 2. [CBSE 2012]

SOLUTION

Given:

y = (tan -1 x) 2 .

... (i)

On differentiating both sides of (i) w.r.t. x, we get 1 y1 = 2 tan -1 x × (1 + x 2) Þ (1 + x 2) y1 = 2 tan -1 x Þ (1 + x 2) 2 y12 = 4(tan -1 x) 2

[on squaring both sides]

x 2) 2 y12

Þ (1 + - 4y = 0. On differentiating both sides of (ii) w.r.t. x, we get (1 + x 2) 2 × 2y1 y 2 + y12 × 2(1 + x 2) × 2x - 4y1 = 0

... (ii)

Þ (1 + x 2) 2 y 2 + 2x(1 + x 2) y1 - 2 = 0. Hence, (1 + x 2) 2 y 2 + 2x(1 + x 2) y1 = 2. EXAMPLE 13

If x = a ( q + sin q) and y = a (1 - cos q), find

d 2y dx 2

at q =

p × 2 [CBSE 2005, ‘09]

SOLUTION

We have x = a ( q + sin q) and y = a (1 - cos q) dx dy Þ = a (1 + cos q) and = a sin q dq dq dy ( dy/dq) Þ = dx ( dx/dq) a sin q sin q 2 sin ( q/2) cos ( q/2) q = = = = tan 2 2 a (1 + cos q) (1 + cos q) 2 cos ( q/2)

Senior Secondary School Mathematics for Class 12 Pg-449

Differentiation

Þ

d 2y dx 2

=

449

d æ 1 qö 1 qö æ1 2 q dq = ç sec2 ÷ ´ ç tan ÷ = sec × dx è 2ø 2 2 dx è 2 2 ø a(1 + cos q)

æ d 2y ö 1 p 1 Þ ç 2÷ = sec2 × ç dx ÷ p 2 4 æ è øq = a ç1 + cos 2 è

pö ÷ 2ø

=

1 × a d 2y

EXAMPLE 14

If x = ( 2 cos q - cos 2q) and y = ( 2 sin q - sin 2q), find

SOLUTION

We have x = ( 2 cos q - cos 2q) and y = ( 2 sin q - sin 2q) dx dy Þ = ( -2 sin q + 2 sin 2q) and = ( 2 cos q - 2 cos 2q) dq dq dy ( dy/dq) ( 2 cos q - 2 cos 2q) (cos q - cos 2q) Þ = = = dx ( dx/dq) ( -2 sin q + 2 sin 2q) (sin 2q - sin q)

dx 2

at q =

p × 2

[CBSE 2005C]

æ 3 qö 2 sin ç ÷ sin è 2 ø = æ 3 qö 2 cos ç ÷ sin è 2 ø Þ

d 2y dx 2

=

q 3q 2 = tan q 2 2

3 3q 1 3 qö 3 d æ 2 3 q dq = sec 2 × × ç tan ÷ = sec 2 2(sin 2q - sin q) 2 ø 2 2 dx 2 dx è

æ d 2y ö 3 1 æ 3p ö Þ ç 2÷ = × sec 2 ç ÷× ç dx ÷ p 2 è 4 ø æ 2 sin p - sin p ö è øq = ÷ ç 2 2ø è -3 p -3 -3 = sec 2 = ´ ( 2) 2 = 4 4 4 2 3p pö pù é æ êQ sec 4 = sec ç p - 4 ÷ = - sec 4 ú × è ø ë û

EXERCISE 10J 1. Find the second derivative of: (i) x11 (ii) 5 x

(iii) tan x

2. Find the second derivative of: (i) x sin x (ii) e2x cos 3 x

(iii) x 3 log x

3. If y = x + tan x, show that cos2 x ×

d 2y dx 2

4. If y = 2 sin x + 3 cos x, show that y +

- 2y + 2x = 0. d 2y dx 2

= 0.

(iv) cos-1 x

Senior Secondary School Mathematics for Class 12 Pg-450

450

Senior Secondary School Mathematics for Class 12

5. If y = 3 cos (log x) + 4 sin (log x), prove that x 2 y 2 + xy1 + y = 0. 6. If y = e

-x

cos x, show that

d 2y dx 2

[CBSE 2009, ’12]

= 2e

-x

7. If y = sec x - tan x, show that (cos x)

sin x. d 2y dx 2

= y 2. d 2y

- y 2 = 0. dx 2 d 2y dy 9. If y = tan -1 x, show that (1 + x 2) 2 + 2x = 0. dx dx d 2y dy 10. If y = sin (sin x), prove that 2 + (tan x) + y cos2 x = 0. dx dx 8. If y = ( cosec x + cot x), prove that (sin x)

11. If y = a cos (log x) + b sin (log x), prove that x 2 y 2 + xy1 + y = 0. 12. Find the second derivative of e

3x

[CBSE 2009]

sin 4x.

13. Find the second derivative of sin 3 x cos 5 x. HINT:

2 sin A cos B = sin ( A + B) + sin ( A - B).

14. If y = etan x , prove that (cos2 x)

d 2y 2

- (1 + sin 2x) ×

dx ( 2 log x - 3)

dy = 0. dx

[CBSE 2009C]

log x d 2y , show that 2 = × x dx x3 d 2y dy 16. If y = eax cos bx, show that 2 - 2a + ( a 2 + b 2) y = 0. dx dx -1 d 2y dy 17. If y = ea cos x , - 1 £ x £ 1, show that (1 - x 2) 2 - x - a 2 y = 0. dx dx d 2y 2 18. If x = at and y = 2at, find 2 at t = 2. dx d 2y 19. If x = a ( q - sin q) and y = a(1 - cos q), find 2 at q = p. dx 2 dy 2d y 20. If y = sin (log x), prove that x [CBSE 2007] +x + y = 0. dx dx 2 15. If y =

21. If y =

sin -1 x 1-x

2

, show that (1 - x 2)

22. If y = ex sin x , prove that

d 2y dx

2

-2

d 2y dx

2

- 3x

dy - y = 0. dx

dy + 2y = 0. dx

[CBSE 2009] [CBSE 2009C]

qö d 2y æ 23. If x = a ç cos q + log tan ÷ and y = a sin q, show that the value of at 2ø è dx 2 p 4 [CBSE 2009, ’13] q = is × 4 a

Senior Secondary School Mathematics for Class 12 Pg-451

Differentiation

24. If x = cos t + log tan t=

p × 4

451

t d 2y d 2y and at , y = sin t then find the values of 2 2 dt dx 2 [CBSE 2012C]

25. If y = x x , prove that

d 2y dx

2

2

-

1 æ dy ö y ç ÷ - = 0. y è dx ø x

26. If y = (cot -1 x) 2 , then show that ( x 2 + 1) 2

d 2y dx 2

[CBSE 2014]

+ 2x( x 2 + 1)

dy = 2. dx [CBSE 2010C]

d 2y

dy 27. If y = {x + x 2 + 1} m , then show that ( x 2 + 1) 2 + x - m 2 y = 0. dx dx [CBSE 2013C]

d 2y

dy 28. If y = log [x + x + a ], then prove that ( x + a ) 2 + x = 0. dx dx 2

2

2

2

[CBSE 2013C]

29. If x = a (cos q + q sin q) and y = a (sin q - q cos q), show that

d 2y dx 2

1 æç sec 3 q ö÷ × a çè q ÷ø

=

[CBSE 2011C, 12C]

30. If x = a cos q + b sin q and y = a sin q - b cos q, show that d 2y dy y2 2 - x + y = 0. dx dx

[CBSE 2014]

ANSWERS (EXERCISE 10J)

1. (i) 110x 9

(ii) 5 x (log 5) 2

2. (i) - x sin x + 2 cos x 3x

12. e ( 24 cos 4x - 7 sin 4x) -1 18. 16 a

(iii) 2sec2 x tan x

(iv)

(ii) - e2x (5 cos 3 x + 12 sin 3 x)

-x (1 - x 2)

3

2

(iii) (5 x + 6x log x)

13. ( 2 sin 2x - 32 sin 8x) -1 -1 19. 24. ,2 2 4a 2

Senior Secondary School Mathematics for Class 12 Pg-452

11. APPLICATIONS OF DERIVATIVES

1. Derivative as a Rate Measure Rate of Change of Quantities dy denotes the rate of change of y w.r.t. x and its value at dx é dy ù × x = a is denoted by ê ú ë dx û x = a

Let y = f ( x). Then,

If x = f (t), y = g(t) then by chain rule, we have dy ( dy/dt) æ dy dt ö = = ç × ÷× dx ( dx/dt) è dt dx ø SOLVED EXAMPLES EXAMPLE 1

Find the rate of change of the area of a circle with respect to its radius r when r = 6 cm.

SOLUTION

Let A be the area of a circle of radius r. Then, dA d A = pr 2 Þ = ( pr 2) = 2pr dr dr é dA ù Þ ê = ( 2p ´ 6) cm 2 /cm = (12p) cm 2 /cm . ë dr úû r = 6 cm Hence, the area is changing at the rate of (12p) cm 2 /cm .

EXAMPLE 2

SOLUTION

A stone is dropped into a quiet lake and the waves move in circles. If the radius of a circular wave increases at the rate of 4 cm/sec, find the rate of increase in its area at the instant when its radius is 10 cm. At any instant t, let the radius of the circle be r cm and its area be A cm 2 . Then, dr = 4 cm/sec dt Now, A = pr 2 Þ

(given)

... (i)

dA æ dA dr ö =ç × ÷ dt è dr dt ø d dr é ù = ( pr 2) × 4 êQ A = pr 2 and = 4ú dr dt ë û = ( 2pr ´ 4) cm 2 /sec = ( 8pr) cm 2 /sec 452

Senior Secondary School Mathematics for Class 12 Pg-453

Applications of Derivatives

Þ

453

é dA ù = ( 8p ´ 10) cm 2 /sec = ( 80p) cm 2 /sec. êë dt úû r = 10

Hence, the area of the circle is increasing at the rate of ( 80p) cm 2 /sec at the instant when r = 10 cm. EXAMPLE 3

A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area increasing when its radius is 5 cm? (Take p = 3.14.)

SOLUTION

A soap bubble is in the form of a sphere. At an instant t, let its radius be r and surface area S. Then, dr ... (i) = 0.02 cm/sec (given) dt Now, S = 4pr 2 dS dr Þ = 8pr × = ( 8 ´ 3.14 ´ r ´ 0.02) cm 2 /sec dt dt dS é ù Þ ê ú = ( 8 ´ 3.14 ´ 5 ´ 0.02)cm 2 /sec = 2.512 cm 2 /sec. ë dt û r = 5 Hence, the surface area of the bubble is increasing at the rate of 2.512 cm 2 /sec at the instance when its radius is 5 cm.

EXAMPLE 4

SOLUTION

The volume of a spherical balloon is increasing at the rate of 20 cm 3 /sec. Find the rate of change of its surface area at the instant when its radius is 8 cm. At any instant t, let r be the radius, V the volume and S the surface area of the balloon. Then, dV ... (i) = 20 cm 3 /sec (given) dt 4 dV dV dr Now, V = pr 3 Þ = × 3 dt dr dt d æ 4 3 ö dr Þ 20 = ç pr ÷ × dr è 3 ø dt 4 dr dr Þ 20 = p ´ 3r 2 ´ = 4pr 2 × 3 dt dt dr 5 ... (ii) Þ = dt pr 2 dS dS dr \ S = 4pr 2 Þ = × dt dr dt d 5 = ( 4pr 2) × 2 dr pr 5 ö 40 æ = ç 8pr ´ 2 ÷ = è pr ø r

Senior Secondary School Mathematics for Class 12 Pg-454

454

Senior Secondary School Mathematics for Class 12

é dS ù æ 40 ö Þ ê ú = ç ÷ cm 2 /sec = 5 cm 2 /sec. ë dt û r = 8 cm è 8 ø Hence, the rate of change of surface area at the instant when r = 8 cm is 5 cm 2 /sec. EXAMPLE 5

SOLUTION

EXAMPLE 6

SOLUTION

The surface area of a spherical balloon is increasing at 2 cm 2 /sec. At what rate is the volume of the bubble increasing when the radius of the bubble is 6 cm? [CBSE 2005] At any instant t, let r be the radius, V the volume and S the surface area of the balloon. Then, dS ... (i) = 2 cm 2 /sec (given) dt dS dS dr Now, S = 4pr 2 Þ = × dt dr dt d dr Þ 2 = ( 4pr 2) × dr dt dr Þ 8pr × =2 dt dr 1 ... (ii) Þ = dt 4pr 4 dV dV dr \ V = pr 3 Þ = × 3 dt dr dt dr d æ 4 3 ö dr = 4pr 2 × = ç pr ÷ × dr è 3 dt ø dt 1 ö æ 2 = ç 4pr × ÷ = r [using (ii)] 4pr ø è é dV ù Þ ê = 6 cm 3 /sec. ë dt úû r = 6 cm The volume of a cube is increasing at the rate of 7 cm 3 /sec. How fast is its surface area increasing at the instant when the length of an edge of the cube is 12 cm? [CBSE 2006C] At any instant t, let the length of each edge of the cube be x , V be its volume and S be its surface area. Then, dV ... (i) = 7 cm 3 /sec (given) dt dV dV dx Now, V = x 3 Þ = × dt dx dt d 3 dx Þ 7= (x ) × [Q V = x 3] dx dt dx Þ 3x2 × =7 dt

Senior Secondary School Mathematics for Class 12 Pg-455

Applications of Derivatives

455

dx 7 = dt 3 x 2 dS dS dx Þ = × dt dx dt d 7 = ( 6x 2) × 2 dx 3x 7 ö 28 æ = ç12x ´ 2 ÷ = è 3x ø x 1 dS é ù æ 28 ö Þ ê ú = ç ÷ cm 2 /sec = 2 cm 2 /sec. 3 ë dt û x =12 è 12 ø Þ

\ S = 6 x2

... (ii)

Hence, the surface area of the cube is increasing at the rate of 1 2 cm 2 /sec at the instant when its edge is 12 cm. 3 EXAMPLE 7

SOLUTION

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of (i) the perimeter, and (ii) the area of the rectangle. dx dy Given that = -5 cm/minute and = 4 cm/minute. dt dt (i) Let P be the perimeter of the rectangle at any instant. Then, dP æ dx dy ö = 2ç + P = 2( x + y) Þ ÷ = 2( -5 + 4) cm/minute dt è dt dt ø = - 2 cm/minute. Hence, the perimeter of the rectangle is decreasing at the rate of 2 cm/minute. (ii) Let A be the area of the rectangle at any instant. Then, dA dx dy A = x×y Þ = ×y + x× dt dt dt = [( -5) ´ 6 + 8 ´ 4] cm 2 /minute = 2 cm 2 /minute. Hence, the area of the rectangle is increasing at the rate of 2 cm 2 /minute.

EXAMPLE 8

Water is leaking from a conical funnel at the rate of 5 cm 3 /sec. If the radius of the base of the funnel is 5 cm and its altitude is 10 cm, find the rate at which the water level is dropping when it is 2.5 cm from the top. [CBSE 2004]

SOLUTION

At any instant t, let r be the radius of the water level, h the height of the water level and V the volume of the water in the conical funnel. Then,

Senior Secondary School Mathematics for Class 12 Pg-456

456

Senior Secondary School Mathematics for Class 12

dV ... (i) = -5 (given) dt From similar as OAB and OCD, we have AB CD r 5 1 1 = Þ = = Þ r = h. OA OC h 10 2 2 2

Now, V = \

1 2 1 1 æ1 ö pr h = p ´ ç h÷ ´ h = ph 3. 3 3 2 12 ø è

ph2 dh dV æ dV dh ö d æ 1 3 ö dh = × =ç ´ ÷= ç ph ÷ × 4 dt dt è dh dt ø dh è 12 ø dt

Þ -5 =

ph2 dh × 4 dt -20

dh = dt ph2 -20 æ dh ö Þ ç ÷ = è dt ø h = 7.5 cm p ´ (7.5) 2 Þ

=

... (i) [Q h = (10 - 2.5) cm = 7.5 cm]

-16 cm/sec. 45p

Hence, the rate of change of water level at h = 7.5 cm is EXAMPLE 9

SOLUTION

-16 cm/sec. 45p

Sand is pouring from a pipe at the rate of 12 cm 3 /sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing, when the height is 4 cm? [CBSE 2011] At any instant t, let r be the radius, h the height and V the volume of the cone. r Then, h = Þ r = 6 h. 6 1 2 1 \ V = pr h = p( 6h) 2 h = 12ph 3. 3 3 dV dV dh 3 Now, V = 12ph Þ = ´ dt dh dt d dh Þ 12 = (12ph 3) ´ dh dt dh Þ 12 = 36ph2 ´ dt dh 1 Þ = dt 3 p h2 1 æ dh ö Þ ç ÷ = cm/sec è dt ø h = 4 cm ( 3 ´ p ´ 4 ´ 4) =

1 cm/sec. 48p

Senior Secondary School Mathematics for Class 12 Pg-457

Applications of Derivatives

457

Hence, the rate of increase of the height of the sand cone at the 1 instant when h = 4 cm is cm/sec. 48p EXAMPLE 10

A 5-m long ladder is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall at the rate of 2 m/sec. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

SOLUTION

Let OC be the wall. At a certain instant t, let AB be the position of the ladder such that OA = x and OB = y. Length of the ladder AB = 5 m. Given that

dx = 2 m/sec. dt

From right a AOB, we have x 2 + y 2 = 25 Þ 2x ×

dx dy + 2y × =0 dt dt

Þ 2x × 2 + 2y × Þ

dy =0 dt

dx é ù êëQ dt = 2úû

dy -2x = dt y

... (i)

Now, x = 4 Þ y = 5 2 - 42 = 3. Putting x = 4, y = 3 in (i) we get

dy -8 = × dt 3

Hence, the required rate of decrease in the height of the ladder on the wall is (8/3) m/sec. EXAMPLE 11

The two equal sides of an isosceles triangle with fixed base b cm are decreasing at the rate of 3 cm/sec. How fast is the area decreasing when each of the equal sides is equal to the base? [CBSE 2006C]

SOLUTION

At a certain instant t, let a ABC be an isosceles triangle in which AB = AC = x cm and BC = b cm. dx Then, ... (i) = 3 cm/sec (given) dt b Draw AD ^ BC. Then, BD = DC = × 2 \ AD = x 2 -

b2 = 4

4x 2 - b 2 2

Senior Secondary School Mathematics for Class 12 Pg-458

458

Senior Secondary School Mathematics for Class 12

1 4x 2 - b 2 b 4x 2 - b 2 b× = 2 2 4 dA dA dx Þ = × dt dx dt dx é ù ö æ dA Q = 3 cm/secú =ç ´ 3÷ ê dt ë û ø è dx 2 2ü ì d ï b 4x - b ï = í ý´ 3 dx ï 4 ïþ î 3b 1 3 bx -1 = × ( 4x 2 - b 2) 2 × ( 8x) = × 4 2 4x 2 - b 2

\ A=

é dA ù \ ê = ë dt úû x = b EXAMPLE 12

SOLUTION

3 b2 4b 2 - b 2

=

3 b2 = 3b

3 b.

A point source of light along a straight road is at a height of ‘a’ metres. A boy ‘b’ metres in height is walking along the road. How fast is his shadow increasing if he is walking away from the light at the rate of ‘c’ metres per minute? [CBSE 2006C] Let AB be the lamp post, the lamp being at B. Then, AB = a metres. At any instant t, let CD be the position of the boy and CE be his shadow. Then, CD = b metres. Let AC = x metres and CE = y metres. dx Given that = c metres/min. dt Clearly, a BAE ~ a DCE. AB AE \ = CD CE a x+y Þ = b y Þ ( a - b) y = bx dy dx Þ ( a - b) = b× = bc dt dt dy bc m/min. Þ = dt ( a - b)

dx é ù êëQ dt = cúû

Hence, the shadow is increasing at the rate of EXAMPLE 13

SOLUTION

bc m/min. ( a - b)

A man 160 cm tall, walks away from a source of light situated at the top of a pole 6 m high, at the rate of 1.1 m/s. How fast is the length of his shadow increasing when he is 1 m away from the pole? Let AB be the lamp post, the lamp being at B.

Senior Secondary School Mathematics for Class 12 Pg-459

Applications of Derivatives

Then,

459

AB = 6 m.

At any instance t, let MN be the position of the man and MS be his shadow. Then, MN = 1.6 m. Let AM = x metres and MS = s metres. dx Given that = 1.1 m/s. dt Clearly, a SAB ~ a SMN . AS AB 6 15 \ = = = MS MN 1.6 4 x + s 15 11 Þ = Þ x= s, where MS = s. s 4 4 dx 11 ds 11 ds Þ = × Þ 1.1 = × dt 4 dt 4 dt ds æ 1.1 ´ 4 ö Þ =ç ÷ = 0.4 m/s. dt è 11 ø Hence, the length of the shadow is increasing at the rate of 0.4 m/s. EXAMPLE 14

SOLUTION

A particle moves along the curve 6y = x 3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate. [CBSE 2002] Let the required point be ( x , y). dy dx Given: =8 dt dt Given curve is 6y = x 3 + 2

... (i) ... (ii)

On differentiating both sides of (ii) w.r.t. t, we get dy dx 6 = 3x2 dt dt dx æ dx ö [using (i)] Þ 6ç 8 ÷ = 3x2 dt è dt ø Þ 3 x 2 = 48 Þ x 2 = 16 Þ x = ±4. Putting x = 4 in (ii), we get y = 11. - 62 -31 Putting x = - 4 in (ii), we get y = = × 6 3 -31 ö æ Hence, the required points are (4, 11) and ç - 4, ÷× 3 ø è EXAMPLE 15

Find the points on the curve y 2 = 8x for which the abscissa and ordinate change at the same rate.

SOLUTION

Let the required point be ( x , y).

Senior Secondary School Mathematics for Class 12 Pg-460

460

Senior Secondary School Mathematics for Class 12

Given:

dy dx = dt dt

... (i)

Given curve is y 2 = 8x

... (ii)

On differentiating both sides of (ii) w.r.t. t, we get dy dx 2y =8 dt dt dx dx [using (i)] Þ 2y × =8 dt dt Þ 2y = 8 Þ y = 4. Putting y = 4 in (ii), we get 8 x = 16 and therefore, x = 2. Hence, the required point is (2, 4). EXAMPLE 16

At what points of the ellipse 16x 2 + 9y 2 = 400 does the ordinate decrease at the same rate at which the abscissa increases?

SOLUTION

Let the required point be ( x , y). Then, dy dx =dt dt Given curve is 16x 2 + 9y 2 = 400

... (i) ... (ii)

On differentiating both sides of (ii) w.r.t. t, we get dx dy 32x × + 18y × =0 dt dt dx æ dx ö [using (i)] Þ 32x × + 18y × ç - ÷ = 0 dt è dt ø 16 Þ ( 32 x - 18y) = 0 Þ y = x. 9 16 Putting y = x in (ii), we get 9 2

æ 16x ö 2 16x 2 + 9 ´ ç ÷ = 400 Þ x = 9 Þ x = ±3. è 9 ø ö 16 æ 16 Now, x = 3 Þ y = ç ´ 3 ÷ = , ø 3 è 9 é 16 ù -16 x = -3 Þ y = ê ´ ( -3) ú = × 3 ë 9 û -16 ö æ 16 ö æ Hence, the required points are ç 3 , ÷ and ç -3 , ÷× 3 3 ø è ø è MARGINAL COST AND MARGINAL REVENUE

Marginal Cost:

Let C be the total cost of producing and marketing x units of a dC product. Then, the marginal cost ( MC) is defined as, MC = × dx

Senior Secondary School Mathematics for Class 12 Pg-461

Applications of Derivatives

461

Marginal Revenue: The rate of change of total revenue with respect to the dR quantity sold is called the marginal revenue ( MR) and therefore, MR = × dx EXAMPLE 17

The total cost C( x) of producing x items in a firm is given by C( x) = 0.005 x 3 - 0.02x 2 + 30x + 6000. Find the marginal cost when 4 units are produced.

SOLUTION

Given: C( x) = 0.005 x 3 - 0.02x 2 + 30x + 6000 dC dx d = ( 0.005 x 3 - 0.02x 2 + 30x + 6000) dx

Þ MC =

= {( 0.005 ´ 3 x 2) - ( 0.02 ´ 2x) + 30} Þ [MC]x = 4 = {( 0.005 ´ 3 ´ 42) - ( 0.02 ´ 2 ´ 4) + 30} = (0.24 - 0.16 + 30) = 30.08. Hence, the required marginal cost is ` 30.08. EXAMPLE 18

The total revenue received from the sale of x units of a product is given by R( x) = 3 x 2 + 40x + 10. Find the marginal revenue when x = 5.

SOLUTION

Given:

R( x) = 3 x 2 + 40x + 10

dR dx d = ( 3 x 2 + 40x + 10) = 6x + 40 dx Þ [MR]x = 5 = ( 6 ´ 5 + 40) = 70. Hence, the required marginal revenue is ` 70. Þ

MR =

EXERCISE 11A 1. The side of a square is increasing at the rate of 0.2 cm/s. Find the rate of increase of the perimeter of the square. 2. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? 3. The radius of a circle is increasing uniformly at the rate of 0.3 centimetre per second. At what rate is the area increasing when the radius is 10 cm? (Take p = 3.14.) 4. The side of a square sheet of metal is increasing at 3 centimetres per minute. At what rate is the area increasing when the side is 10 cm long? 5. The radius of a circular soap bubble is increasing at the rate of 0.2 cm/s. Find the rate of increase of its surface area when the radius is 7 cm.

Senior Secondary School Mathematics for Class 12 Pg-462

462

Senior Secondary School Mathematics for Class 12

6. The radius of an air bubble is increasing at the rate of 0.5 centimetre per second. At what rate is the volume of the bubble increasing when the radius is 1 centimetre? 7. The volume of a spherical balloon is increasing at the rate of 25 cubic centimetres per second. Find the rate of change of its surface at the instant when its radius is 5 cm. 8. A balloon which always remains spherical is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm. 9. The bottom of a rectangular swimming tank is 25 m by 40 m. Water is pumped into the tank at the rate of 500 cubic metres per minute. Find the rate at which the level of water in the tank is rising. 10. A stone is dropped into a quiet lake and waves move in circles at a speed of 3.5 cm per second. At the instant when the radius of the circular wave is 7.5 cm, how fast is the enclosed area increasing? (Take p = 22/7.) 11. A 2-m tall man walks at a uniform speed of 5 km per hour away from a 6-metre-high lamp post. Find the rate at which the length of his shadow increases. 12. An inverted cone has a depth of 40 cm and a base of radius 5 cm. Water is poured into it at a rate of 1.5 cubic centimetres per minute. Find the rate at which the level of water in the cone is rising when the depth is 4 cm. 13. Sand is pouring from a pipe at the rate of 18 cm 3 /s. The falling sand forms a cone on the ground in such a way that the height of the cone is one-sixth of the radius of the base. How fast is the height of the sand cone increasing when its height is 3 cm? 14. Water is dripping through a tiny hole at the vertex in the bottom of a conical funnel at a uniform rate of 4 cm 3 /s. When the slant height of the water is 3 cm, find the rate of decrease of the slant height of the water, given that the vertical angle of the funnel is 120°. 15. Oil is leaking at the rate of 16 mL/s from a vertically kept cylindrical drum containing oil. If the radius of the drum is 7 cm and its height is 60 cm, find the rate at which the level of the oil is changing when the oil level is 18 cm. 16. A 13-m long ladder is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 m/s. How fast is its height on the wall decreasing when the foot of the ladder is 5 m away from the wall? 17. A man is moving away from a 40-m high tower at a speed of 2 m/s. Find the rate at which the angle of elevation of the top of the tower is changing when he is at a distance of 30 metres from the foot of the tower. Assume that the eye level of the man is 1.6 m from the ground.

Senior Secondary School Mathematics for Class 12 Pg-463

Applications of Derivatives

463

18. Find an angle x which increases twice as fast as its sine. 19. The radius of a balloon is increasing at the rate of 10 cm/s. At what rate is the surface area of the balloon increasing when the radius is 15 cm? 20. An edge of a variable cube is increasing at the rate of 5 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long? 21. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which the area is increasing when the side is 10 cm. [CBSE 2014C] ANSWERS (EXERCISE 11A)

dP = 0.8 cm/s dt dA 4. = 60 cm 2 /min dt dS 7. = 10 cm 2 /s dt 1.

10. 165 cm 2 /s 1 cm/s 18p 5 16. m/s 6 19. 1200p cm 2 /s

13.

dC = 4.4 cm/s dt dS 5. = 35.2 cm 2 /s dt dr 8. = 0.32 cm/s dt 2.

11. 14. 17. 20.

dA = 18.84 cm 2 /s dt dV 6. = 6.28 cm 3 /s dt dh 9. = 0.5 m/min dt 1 2.5 km/h 12. cm/s 10p 32 16 cm/s 15. cm/s 27p 49p p 0.032 radian/second 18. 3 21. 10 3 cm 2 /s 1500 cm 3 /s 3.

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11A) dS dr = ( 8 pr ) × = ( 8 pr )( 0.2 ) dt dt dS é ù = ( 1.6 p ´ 7 ) cm 2 /s. Þ ê ú ë dt û r = 7

5. S = 4 pr 2 Þ

4 3 dV dr pr Þ = ( 4 pr 2 ) × = ( 4 pr 2 ) ´ 0.5. 3 dt dt 4 dV dr dr 25 7. V = pr 3 Þ = ( 4 pr 2 ) × Þ = × 3 dt dt dt 4 pr 2 dS dr æ 25 ö 50 S = 4 pr 2 Þ = 8 pr × = ç 8 pr × ÷= dt dt è 4 pr 2 ø r é dS ù æ 50 ö Þ ê ú = ç ÷ cm 2 /s. ë dt û r = 5 è 5 ø 6. V =

8. V =

4 3 dV dr dr 900 225 pr Þ = 4 pr 2 × Þ = = 3 dt dt dt 4 pr 2 pr 2 225 1 é dr ù = = cm/s. Þ ê ú ë dt û r = 15 p ´ ( 15 ) 2 p

Senior Secondary School Mathematics for Class 12 Pg-464

464

Senior Secondary School Mathematics for Class 12 dV æ dh ö = ç 1000 ´ ÷ dt è dt ø

9. V = ( 25 ´ 40 ´ h ) Þ

10. A = pr 2

dh ö dh æ Þ 500 = ç 1000 ´ ÷ Þ = 0.5 m/min. dt ø dt è dr dA Þ = 2 pr × = ( 2 pr ´ 3.5 ) = 7 pr dt dt é dA ù Þ ê = (7 p ´ 7.5 ) cm 2 /s = 165 cm 2 /s. ë dt úû r = 7.5

r 1 1 Þ r = 6 h. So, V = pr 2 h = p( 6 h ) 2 h = 12 ph 3 . 6 3 3 dh 18 1 dV 2 dh 2 dh \ Þ 18 = 36 ph × Þ = = = 36 ph × dt dt 36 ph 2 2 ph 2 dt dt 1 é dh ù cm/s. = Þ ê ú ë dt û h = 3 ( 2 p ´ 9 )

13. h =

1 p p ( l sin 60 ° ) 2 ( l cos 60 ° ) = l 3 . 3 8 dV 3 pl 2 dl 3 pl 2 dl dl -32 -32 é dl ù cm/s. = × Þ × = -4 Þ = Þ ê ú = \ dt 8 dt 8 dt dt 3 pl 2 ë dt û l = 3 27 p

14. V =

dV dh dh dh 16 cm/s. = 49 p × Þ 49 p × = 16 Þ = dt dt dt dt 49 p 17. Let at any time t, the man be at a distance of x metres from tower AB and let q be the angle of elevation at that time. dx dq Then, x = 40 cot q Þ = -40 cosec2 q × × dt dt 2 ( 40 ) 2 1 æ AB ö 1 dq -2 sin 2 q = =×ç × \ × ÷ =40 20 è BE ø 20 ( 40 ) 2 + x 2 dt

15. V = ( p ´ 7 ´ 7 ´ h ) Þ

Find its value when x = 30 m. 1 p dx d dx 18. = 2 × (sin x ) = 2 cos x × × So, cos x = or x = × 2 3 dt dt dt dS dr 2 2 19. S = 4 pr Þ = 8 pr × = ( 8 p ´ 15 ´ 10 ) cm /s. dt dt dV dx 20. V = x 3 Þ = 3x 2 × = [ 3 ´ ( 10 ) 2 ´ 5] cm 3 /s. dt dt 21. A =

3 2 3 dA dt 3 dA a Þ = aÞ × = a 4 da 2 dt da 2 dA 1 3 dA é dA ù Þ ´ = = 3a Þ ê = aÞ dt 2 2 dt ë dt úû a = 10

3 ´ 10 cm 2 .

2. Errors and Approximation Let y = f ( x). Then, lim

dx ® 0

\

f ( x + dx) - f ( x) = f ¢( x). dx

f ( x + dx) - f ( x) = f ¢( x) + Î, where ή 0 when dx ® 0 dx

Senior Secondary School Mathematics for Class 12 Pg-465

Applications of Derivatives

465

Þ f ( x + dx) - f ( x) = f ¢( x) × dx + Î× dx Þ f ( x + dx) - f ( x) = f ¢( x) × dx Þ dy = f ¢( x) × dx

(approximately)

[Q f ( x + dx) - f ( x) = dy].

Thus, if dx is an error in x then the corresponding error in y is dy. These small values dx and dy are called differentials. dx is called an absolute error in x. dx is called the relative error. (ii) Relative Error: x æ dx ö (iii) Percentage Error: ç ´ 100÷ is called the percentage error. è x ø (i) Absolute Error:

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Using differentials, find the approximate value of ( 82) places of decimal. 1 1 Let f ( x) = x 4 . Then, f ¢( x) = 3 × 4x 4 Now, { f ( x + dx) - f ( x)} = f ¢( x) × dx 1 Þ { f ( x + dx) - f ( x)} = 3 × dx 4x 4 We may write, 82 = ( 81 + 1). Putting x = 81 and dx = 1 in (i), we get 1 f ( 81 + 1) - f ( 81) = ×1 3 4 ´ ( 81) 4 1 1 Þ f ( 82) - f ( 81) = = ( 4 ´ 3 3) 108

1

4

up to three [CBSE 2005]

... (i)

1 1 ü ì 1 ü ì Þ f ( 82) = í f ( 81) + ý = 3 + 0.009 = 3.009. ý = í( 81) 4 + 108 þ î 108 þ î

EXAMPLE 2 SOLUTION

Find the approximate value of the cube root of 127. 1 1 Let f ( x) = x 3 . Then, f ¢( x) = × 2 3x 3 Now, { f ( x + dx) - f ( x)} = f ¢( x) × dx 1 Þ { f ( x + dx) - f ( x)} = × dx 2 3x 3 We may write, 127 = (125 + 2). Putting x = 125 and dx = 2 in (i), we get 1 f (125 + 2) - f (125) = ´2 2 3 ´ (125) 3

... (i)

Senior Secondary School Mathematics for Class 12 Pg-466

466

Senior Secondary School Mathematics for Class 12

Þ f (127) - f (125) =

2 75

Þ f (127) = f (125) +

1 2 ö 377 2ü æ 2 ì = í(125) 3 + ý = ç5 + ÷ = 75 ø 75 75 þ è 75 î

Þ EXAMPLE 3

3

127 =

377 = 5.026. 75

Using differentials find the approximate value of the square root of 26. [CBSE 2000]

SOLUTION

Let f ( x) = x . Then, f ¢( x) =

1 × 2 x

Now, { f ( x + dx) - f ( x)} = f ¢( x) × dx 1 Þ { f ( x + dx) - f ( x)} = × dx 2 x

... (i)

We may write, 26 = ( 25 + 1). Putting x = 25 and dx = 1 in (i), we get 1 f ( 26) - f ( 25) = ´1 2 25 Þ f ( 26) = f ( 25) + Þ

1 æ 1ö æ 1ö = ç 25 + ÷ = ç5 + ÷ = (5 + 0.1) = 5.1 10 ø è 10 ø 10 è

26 = 5.1.

Hence, 26 = 5.1. EXAMPLE 4

Using differentials find the approximate value of 0.037 . [CBSE 2005C]

SOLUTION

Let f ( x) = x . Then, f ¢( x) =

1 × 2 x

Now, { f ( x + dx) - f ( x)} = f ¢( x) × dx 1 Þ { f ( x + dx) - f ( x)} = × dx 2 x We may write, 0.037 = (0.04 - 0.003). Putting x = 0.04 and dx = - 0.003 in (i), we get 1 f ( 0.04 - 0.003) - f (0.04) = ´ ( - 0.003) 2 0.04 0.003 Þ f ( 0.037) = f (0.04) 0.4 3 ö æ 3 ö 77 æ Þ 0.037 = ç 0.04 ÷= ÷ = ç 0.2 400 400 ø 400 ø è è Þ

0.037 = 0.1925.

... (i)

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467

EXAMPLE 5

Find the approximate value of tan 46°, it is being given that 1° = 0.01745 radian.

SOLUTION

Let f ( x) = tan x. Then, f ¢( x) = sec2 x. Now, f ( x + dx) - f ( x) = f ¢( x) × dx Þ f ( x + dx) - f ( x) = sec2 x × dx

... (i)

Putting x = 45 ° , dx = 1° = 0.01745 in (i), we get f ( 46°) - f ( 45 °) = ( sec2 45 °) ´ 0.01745 Þ tan 46° - tan 45 ° = ( sec2 45 °) ´ 0.01745 Þ tan 46° = tan 45 ° + ( sec2 45 °) ´ 0.01745 = (1 + 2 ´ 0.01745) = 1.03490. Hence, tan 46° = 1.03490. EXAMPLE 6

SOLUTION

Find the approximate value of log10 10.1, it being given that log10 e = 0.4343. 1 Let f ( x) = log10 x. Then, f ¢( x) = (log10 e). x Now, f ( x + dx) - f ( x) = f ¢( x) × dx 1 Þ f ( x + dx) - f ( x) = (log10 e) × dx x 0.4343 ... (i) Þ f ( x + dx) - f ( x) = × dx x Putting x = 10 and dx = 0.1 in (i), we get 0.4343 f (10.1) - f (10) = ´ 0.1 10 Þ log10 (10.1) - log10 10 = 0.004343 Þ log10 (10.1) = (log10 10) + 0.004343 = (1 + 0.004343) = 1.004343. Hence, log10 (10.1) = 1.004343.

EXAMPLE 7

If f ( x) = 3 x 2 + 15 x + 5, then find the approximate value of f (3.02), using differentials.

SOLUTION

[CBSE 2008C, ’14]

2

f ( x) = 3 x + 15 x + 5 Þ f ¢( x) = 6x + 15 Now, f ( x + dx) - f ( x) = f ¢( x) × dx Þ f ( x + dx) - f ( x) = ( 6x + 15) × dx Putting x = 3 and dx = 0.02 in (i), we get f ( 3.02) - f ( 3) = ( 6 ´ 3 + 15) ´ (0.02) [Q f ( 3) = 3 ´ 3 2 + 15 ´ 3 + 5 = 77]

Þ

f (3.02) - 77 = 0.66

Þ

f (3.02) = 77 + 0.66 = 77.66.

... (i)

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Senior Secondary School Mathematics for Class 12

Use of the Formula:

dy =

dy × dx dx

EXAMPLE 8

If the radius of a circle increases from 5 cm to 5.1 cm, find the increase in area.

SOLUTION

Area of a circle of radius r is given by A = pr 2 . dA Now, A = pr 2 Þ = 2pr dr é dA ù Þ ê = ( 2p ´ 5) cm 2 = 10p cm 2. ë dr úû r = 5 Also, dr = (5.1 - 5) cm = 0.1 cm . dA × dr = (10p ´ 0.1) cm 2 = p cm 2. \ dA = dr Hence, the increase in area is p cm 2 .

EXAMPLE 9

SOLUTION

If y = ( x 4 - 12) and if x changes from 2 to 1.99, what is the approximate change in y? [CBSE 2002] dy 4 3 y = ( x - 12) Þ = 4x dx é dy ù Þ ê ú = ( 4 ´ 2 3) = 32. ë dx û x = 2 Let dx = (1.99 - 2) = - 0.01. dy × dx = 32 ´ ( - 0.01) = - 0.32. \ dy = dx

EXAMPLE 10

The time T of oscillation of a simple pendulum of length l is given by l × Find the percentage error in T, corresponding to an error of T = 2p g 2% in the value of l.

SOLUTION

T = 2p

l 1 1 Þ log T = log 2 + log p + log l - log g g 2 2 1 dT 1 1 dT 1 × = Þ × × dl = dl T dl 2l T dl 2l 1 1 dT Þ × dT = × dl [Q dT = × dl ] T 2l dl æ dT ö 1 æ dl ö 1 Þ ç ´ 100÷ = × ç ´ 100÷ = ´ 2 = 1. è T ø 2 è l ø 2 Þ

\ percentage error in T = 1%. EXAMPLE 11

If the error committed in measuring the radius of a circle be 0.01%, find the corresponding error in calculating the area.

Senior Secondary School Mathematics for Class 12 Pg-469

Applications of Derivatives

SOLUTION

469

dA dA = 2pr Þ × dr = 2pr × dr dr dr dr dA Þ dA = 2pr 2 × [Q × dr = dA ] r dr dA dr Þ = 2× [Q A = pr 2] A r ö æ dr ö æ dA Þ ç ´ 100÷ = 2 × ç ´ 100÷ A r ø è ø è

A = pr 2 Þ

Þ percentage error in area = (2 ´ 0.01) = 0.02%. Hence, the percentage error in area = 0.02%. EXAMPLE 12

SOLUTION

If in a triangle ABC, the side c and the angle C remain constant while the da db remaining elements are changed slightly, show that + = 0. cos A cos B c As given, we have = k (constant). sin C a b = = k Þ a = k sin A and b = k sin B. \ sin A sin B \

da = k cos A × dA and db = k cos B × dB da db or + = k( dA + dB) = kd( A + B) = kd( p - C) = 0. cos A cos B Hence,

EXAMPLE 13

SOLUTION

da db + = 0. cos A cos B

The area S of a triangle is calculated by measuring the sides b and c, and ÐA. If there be an error dA in the measurement of ÐA, show that the relative error in area is given by dS = cot A × dA. S 1 dS 1 dS 1 S = bc sin A Þ = bc cos A Þ × dA = bc cos A × dA 2 dA 2 dA 2 1 1 Þ dS = bc cos A × dA Þ dS = bc sin A × ( cot A) × dA 2 2 dS Þ dS = S × ( cot A) × dA Þ = ( cot A) × dA. S dS Hence, = (cot A) × dA. S

EXERCISE 11B Using differentials, find the approximate values of: 1.

37

[CBSE 2000]

2.

3

29

Senior Secondary School Mathematics for Class 12 Pg-470

470

Senior Secondary School Mathematics for Class 12

3. 3 27 5.

49.5

4. 6.

[CBSE 2012]

0.24 [CBSE 2001] 4

15

7.

1 (2.002) 2

8. log e 10.02, given that log e 10 = 2.3026 9. log10 (4.04) , it being given that log10 4 = 0.6021 and log10 e = 0.4343 10. cos 61° , it being given that sin 60° = 0.86603 and 1° = 0.01745 radian 11. If y = sin x and x changes from

p 22 to , what is the approximate change 2 14

in y? 12. A circular metal plate expands under heating so that its radius increases by 2%. Find the approximate increase in the area of the plate, if the radius of the plate before heating is 10 cm. 13. If the length of a simple pendulum is decreased by 2%, find the percentage l decrease in its period T, where T = 2p × g 14. The pressure p and the volume V of a gas are connected by the relation, pV1/4 = k , where k is a constant. Find the percentage increase in the pressure, corresponding to a diminution of 0.5% in the volume. 15. The radius of a sphere shrinks from 10 cm to 9.8 cm. Find approximately the decrease in (i) volume, and (ii) surface area. 16. If there is an error of 01 . % in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere. 17. Show that the relative error in the volume of a sphere, due to an error in measuring the diameter, is three times the relative error in the diameter.

ANSWERS (EXERCISE 11B)

1. 6.08 5. 7.0357 9. 0.606443 13. 1% 16. 0.3%

2. 3.074 63 6. 32

3. 5.027

4. 0.49

7. 0.2495

8. 2.3046 12.

2p

cm 2

10. 0.4849

11. No change

14. 0.125%

15. (i) 80p cm 3 (ii) 16p cm 2

5

Senior Secondary School Mathematics for Class 12 Pg-471

Applications of Derivatives

471

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11B) 1 -2 × Then f ( x + dx ) - f ( x ) = 3 × dx. x2 x Take x = 2 and dx = 0.002.

7. Let f ( x ) =

dA = 2 pr = 20 p , when r = 10. dr 2 ö 2p dA æ \ dA = × dr = ç 20 p ´ × ÷= dr 100 ø 5 è

12. A = pr

2

Þ

æ 2p ö 1 ÷ + log l Þ 1 × dT = 1 × 13. log T = log ç ç g÷ 2 T dl 2 l è ø ü T ö ì 1 æ -2 ö æ dT ö æ 1 dl ´ 100 ÷ = ç × ´ 100 ÷ = í ´ ç \ dT = × dl Þ ç ÷ ´ 100 ý = -1. 2l ø î 2 è 100 ø è T ø è2 l þ 1 log V = log k 4 1 dp 1 1 dp 1 Þ × + =0 Þ × × dV = × dV p dV 4V p dV 4V æ dp ö 1 1 Þ × dp = × dV Þ çç ´ 100 ÷÷ p 4V è p ø æ dp ö 1 æ dV 1öæ 1ö 1 ö æ Þ çç ´ 100 ÷÷ = - ç ´ 100 ÷ = ç - ÷ ç - ÷ = × p 4 V 4 è ø è øè 2ø 8 è ø

14. pV1 / 4 = k Þ log p +

3. Rolle’s and Lagrange’s Theorems

Rolle’s Theorem Let f be a real-valued function, defined in the closed interval [a , b] such that (i) f is continuous on [a , b]; (ii) f is differentiable on ]a , b[; (iii) f ( a) = f ( b). Then, there exists a real number c in the open interval ]a , b[ such that f ¢( c) = 0. PROOF

If f ( x) = c, where c is a constant then f ¢( x) = 0 for all x Î]a , b[. So, in this case, the theorem follows. Now, consider the case, when f is not a constant function. Since f ( a) = f ( b), the function should either increase or decrease, when x takes values slightly greater than a.

Senior Secondary School Mathematics for Class 12 Pg-472

Senior Secondary School Mathematics for Class 12 C

Y

O

a

B

b

X

O

a

c

f(b)

f(c)

f(c–h)

C f(a)

f(c)

c

f(c+h)

B f(c–h)

f(a)

A

A

f(c+h)

Y

f(b)

472

b

X

Let f ( x) be increasing at x > a. Since f ( b) = f ( a), f ( x) must cease to increase and begin to decrease at some point x = c Î]a , b[. Clearly, at such a point, the function has a maximum value. Thus, for a very small and positive value of h, we have \ So, But, if

f ( c + h) - f ( c) < 0 and f ( c - h) - f ( c) < 0. f ( c + h) - f ( c) f ( c - h) - f ( c) < 0 and > 0. h -h f ( c + h) - f ( c) f ( c - h) - f ( c) lim £ 0 and lim ³ 0. h® 0 h® 0 h -h lim

h® 0

f ( c + h) - f ( c) f ( c - h) - f ( c) then ¹ lim h® 0 h -h

Rf ¢( c) ¹ Lf ¢( c) and this would imply that f ( x) is not differentiable at x = c, contradicting the given hypothesis. f ( c + h) - f ( c) f ( c - h) - f ( c) \ lim = lim = 0. h® 0 h® 0 h -h Hence, f ¢( c) = 0, where a < c < b. Similarly, the theorem can be proved for the case when f ( x) is initially decreasing for values of x > a , and it ceases to decrease at x = c. Hence, there exists c Î]a , b[ such that f ¢( c) = 0. REMARK

Rolle’s theorem is applicable to a function only when all the three properties above listed are satisfied by it. If even a single property is violated, we say that the theorem is not applicable on such a function.

GEOMETRICAL SIGNIFICANCE OF ROLLE’S THEOREM Let f be a real function defined on [a , b] and let Rolle’s theorem be applicable on it. Then, f being continuous on [a , b], it follows that we can draw a graph of f ( x) between the values x = a and x = b.

Also, f ( x) being differentiable in ]a , b[, it follows that the graph of f ( x) has a tangent at each point of ]a , b[. Now, the existence of a real number c Î]a , b[ such that f ¢( c) = 0 shows that the tangent to the curve at x = c has a slope 0, i.e., it is parallel to the x-axis.

Senior Secondary School Mathematics for Class 12 Pg-473

Applications of Derivatives

473

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Verify Rolle’s theorem for the function f ( x) = x 3 - 6x 2 + 11x - 6 in the interval [1, 3]. Here, we observe that (i) f ( x) being a polynomial function of x, is continuous on the interval [1, 3]. (ii) f ¢( x) = 3 x 2 - 12x + 11, which clearly exists for all values of x Î[1, 3]. So, f ( x) is differentiable on the open interval ]1, 3[. (iii) f (1) = (1 3 - 6 ´ 12 + 11 ´ 1 - 6) = 0 and f ( 3) = ( 3 3 - 6 ´ 3 2 + 11 ´ 3 - 6) = 0. \ f (1) = f ( 3). Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist some c Î]1, 3[ such that f ¢( c) = 0. Now,

f ¢( c) = 0 Þ 3 c2 - 12 c + 11 = 0 Þ c=

12 ± 144 - 132 1 ö æ Þ c = ç2 ± ÷× 6 3ø è

Clearly, both the values of c lie in the interval ]1, 3[. Hence, Rolle’s theorem is verified. EXAMPLE 2

Verify Rolle’s theorem for the function f ( x) = x( x - 1) 2 in the interval [0, 1].

SOLUTION

We have

f ( x) = x 3 - 2x 2 + x.

We observe here that (i) f ( x) being a polynomial function, is continuous on [0, 1]. (ii) f ¢( x) = ( 3 x 2 - 4x + 1), which clearly exists for all values of x Î[0, 1]. So, f ( x) is differentiable on the interval ]0, 1[. (iii) f ( 0) = 0 and f (1) = 0. \ f ( 0) = f (1). Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist a real number c Î]0, 1[ such that f ¢( c) = 0. Now, f ¢( c) = 0 Þ 3 c2 - 4c + 1 = 0 Þ ( c - 1) ( 3 c - 1) = 0 1 Þ c = 1 or c = × 3 1 Out of these two values, clearly Î]0, 1[. 3 1 Thus, c = Î ]0, 1[ such that f ¢( c) = 0. 3 Hence, Rolle’s theorem is satisfied.

Senior Secondary School Mathematics for Class 12 Pg-474

474

Senior Secondary School Mathematics for Class 12

EXAMPLE 3

Verify Rolle’s theorem for the function f ( x) = ( x - a) m( x - b) n in the

SOLUTION

interval [a , b], where m and n are positive integers. Let f ( x) = ( x - a) m ( x - b) n , where a £ x £ b. On expanding ( x - a) m and ( x - b) n by the binomial theorem and then taking the product, we find that f ( x) is a polynomial of degree (m + n). But, a polynomial function being continuous everywhere, it follows that f ( x) is continuous in [a , b]. Also, f ¢( x) = m( x - a) m - 1( x - b) n + n( x - a) m ( x - b) n - 1 = ( x - a) m - 1 ( x - b) n - 1 × [m( x - b) + n( x - a)], which clearly exists for all x Î]a , b[. \ f ( x) is differentiable on ]a , b[. Also, f ( a) = f ( b) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. Consequently, there must exist some c Î]a , b[ such that f ¢( c) = 0. Now, f ¢( c) = 0 Û ( c - a) m - 1 ( c - b) n - 1 [m( c - b) + n( c - a)] = 0 Û ( c - a) = 0 or ( c - b) = 0 or m( c - b) + n( c - a) = 0 æ mb + na ö Û c = a or c = b or c = ç ÷× è m+n ø æ mb + na ö ÷÷ Î]a , b[ such that f ¢( c) = 0 Clearly, c = çç è m+n ø ü ì æ mb + na ö ÷ divides ]a , b[ in the ratio m : n ý× íQ ç î è m+n ø þ Hence, Rolle’s theorem is verified.

EXAMPLE 4

Verify Rolle’s theorem for each of the following functions: é pù (i) f ( x) = sin 2x in ê 0, ú ë 2û é pù (ii) f ( x) = (sin x + cos x) in ê 0, ú ë 2û pö é pù æ (iii) f ( x) = cos 2 ç x - ÷ in ê 0, ú 4ø ë 2û è

[CBSE 2006]

(iv) f ( x) = (sin x - sin 2x) in [0, p] SOLUTION

é pù (i) Consider f ( x) = sin 2x in ê 0, ú × ë 2û Since the sine function is continuous at each x Î R , it follows é pù that f ( x) = sin 2x is continuous on ê 0, ú × ë 2û é pù Also, f ¢( x) = 2 cos 2x , which clearly exists for all x Î ê 0, ú × ë 2û

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Applications of Derivatives

475

ù pé So, f ( x) is differentiable on ú 0, ê × û 2ë Also,

æ pö f ( 0) = f ç ÷ = 0. è 2ø

Thus, all the conditions of Rolle’s theorem are satisfied. ù pé So, there must exist a real number c Î ú 0, ê such that f ¢( c) = 0. û 2ë Now,

f ¢( c) = 0 Û 2 cos 2c = 0 Û cos 2c = 0 p p Û 2c = , i.e., c = × 2 4

Thus, c =

p ù pé such that f ¢( c) = 0. Î 0, 4 úû 2 êë

Hence, Rolle’s theorem is verified. é pù (ii) Consider f ( x) = (sin x + cos x) in ê 0, ú × ë 2û By the continuity of the sine function, the cosine function and the sum of continuous functions, it follows that f ( x) is é pù continuous on ê 0, ú × ë 2û Also, f ¢( x) = (cos x - sin x), which clearly exists for all values é pù of x Î ê 0, ú × ë 2û p é ù æ pö So, f ( x) is differentiable on ú 0, ê × Also, f ( 0) = f ç ÷ = 1. 2 ë û è 2ø Thus, all the conditions of Rolle’s theorem are satisfied. p é ù So, there must exist some c Î ú 0, ê such that f ¢( c) = 0. 2 ë û Now, f ¢( c) = 0 Û cos c - sin c = 0 Û cos c = sin c Û c = Thus, c =

p × 4

p ù p é such that f ¢( c) = 0. Î 0, 4 úû 2 êë

Hence, Rolle’s theorem is verified. pö pù é æ (iii) Consider f ( x) = cos 2 ç x - ÷ in ê 0, ú × 4ø 2û ë è Since the cosine function is continuous everywhere, it pö æ é pù follows that f ( x) = cos 2 ç x - ÷ is continuous on ê 0, ú × 4ø è ë 2û

Senior Secondary School Mathematics for Class 12 Pg-476

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Senior Secondary School Mathematics for Class 12

pö æ Also, f ¢( x) = -2 sin ç 2x - ÷ = 2 cos 2x, which clearly exists 2ø è p é ù for all x Î ú 0, ê × 2 ë û ù pé \ f ( x) is differentiable on ú 0, ê × û 2ë p p æ ö Further, f ( 0) = cos 2 ç - ÷ = cos = 0. 2 è 4ø p æ p pö æ pö And, f ç ÷ = cos 2 ç - ÷ = cos = 0. 2 è 2 4ø è 2ø p æ ö \ f ( 0) = f ç ÷ = 0. è 2ø Thus, all the conditions of Rolle’s theorem are satisfied. So, ù pé there must exist c Î ú 0, ê such that f ¢( c) = 0. û 2ë p p Now, f ¢( c) = 0 Û 2 cos 2c = 0 Û 2c = Þ c= × 2 4 p ù pé Thus, c = Î ú 0, ê such that f ¢( c) = 0. 4 û 2ë Hence, Rolle’s theorem is verified. (iv) Consider f ( x) = (sin x - sin 2x) in [0, p]. Since the sine function is continuous, it follows that g( x) = sin x and h( x) = sin 2x are both continuous and so their difference is also continuous. Consequently, f ( x) = g( x) - h( x) is differentiable on [0, p]. Also, f ¢( x) = (cos x - 2 cos 2x), which clearly exists for all x Î[0, p]. \ f ( x) is differentiable on ]0, p[. And, f ( 0) = f ( p) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist a real number c Î]0, p[ such that f ¢( c) = 0. Now, f ¢( c) = 0 Û cos c - 2 cos 2c = 0 Û cos c - 2( 2 cos2 c - 1) = 0 Û 4 cos2 c - cos c - 2 = 0 1 ± 33 Û cos c = = 0.8431 or - 0.5931 8 Û cos c = 0.8431 or cos(180° - c) = 0.5931. Û c = 32°× 32¢ or c = 126°23 ¢. Thus, c Î]0, p[ such that f ¢( c) = 0. Hence, Rolle’s theorem is satisfied.

Senior Secondary School Mathematics for Class 12 Pg-477

Applications of Derivatives EXAMPLE 5

SOLUTION

477

Verify Rolle’s theorem for each of the following functions: (i) f ( x) = sin 2 x in 0 £ x £ p p p (ii) f ( x) = ex cos x in - £ x £ 2 2 sin x (iii) f ( x) = x in 0 £ x £ p e (i) Consider f ( x) = sin 2 x in [0, p]. Let g( x) = sin x and h( x) = x 2. Then, the sine function being continuous for all x Î R and every polynomial function being continuous for all x Î R , it follows that g and h are both continuous for all x Î R. But the composite of continuous functions is continuous. \ ( h o g) ( x) = f ( x) = sin 2 x is continuous on [0, p]. Also, f ¢( x) = 2 sin x cos x = sin 2x , which clearly exists for all x Î]0, p[. So, f ( x) is differentiable on ]0, p[. Also, f ( 0) = f ( p) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist c Î]0, p[ such that f ¢( c) = 0. p Now, f ¢( c) = 0 Û sin 2c = 0 Û 2c = p Û c = × 2 p Thus, c = Î ]0, p[ such that f ¢( c) = 0. 2 Hence, Rolle’s theorem is verified. é p pù (ii) Consider f ( x) = ex cos x in ê - , ú × ë 2 2û

Let g( x) = ex and h( x) = cos x. Then, the exponential function as well as the cosine function being continuous, it follows that g( x) × h( x) = f ( x) is é p pù continuous on ê - , ú × ë 2 2û Also, which f ¢( x) = - ex sin x + ex cos x = ex (cos x - sin x), ù p pé clearly exists for all x Î ú - , ê × û 2 2ë ù p pé So, f ( x) is differentiable on ú - , ê × û 2 2ë p æ pö æ pö æ pö -p /2 Also, f ç - ÷ = e cos ç - ÷ = 0. And, f ç ÷ = ep /2 cos = 0. 2 è 2ø è 2ø è 2ø æ pö æ pö \ f ç - ÷ = f ç ÷ = 0. è 2ø è 2ø Thus, all the conditions of Rolle’s theorem are satisfied. ù p pé So, there must exist some c Î ú - , ê such that f ¢( c) = 0. û 2 2ë

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Senior Secondary School Mathematics for Class 12

Now,

f ¢( c) = 0 Û ec(cos c - sin c) = 0 Û cos c - sin c = 0 Û sin c = cos c Û c =

Thus, c =

p × 4

p ù p pé such that f ¢( c) = 0. Î - , 4 úû 2 2 êë

Hence, Rolle’s theorem is verified. sin x (iii) Consider f ( x) = x in [0, p]. e Since ex ¹ 0 for any x Î[0, p] and f ( x) is the quotient of two continuous functions, it follows that f ( x) is continuous on [0, p]. ex cos x - ex sin x (cos x - sin x) Also, f ¢( x) = = , which clearly e2x ex exists for all x Î]0, p[. So, f ( x) is differentiable on ]0, p[. Also, f ( 0) = f ( p) = 0. So, all the conditions of Rolle’s theorem are satisfied. \ there must exist c Î]0, p[ such that f ¢( c) = 0. (cos c - sin c) Now, f ¢( c) = 0 Û =0 ec p Û cos c - sin c = 0 Û sin c = cos c Û c = × 4 p Thus, c = Î ]0, p[ such that f ¢( c) = 0. 4 Hence, Rolle’s theorem is verified. -( x /2 )

in [-3 , 0].

EXAMPLE 6

Verify Rolle’s theorem for the function f ( x) = x( x + 3) e

SOLUTION

Since a polynomial function as well as an exponential function is continuous and the product of two continuous functions is continuous, it follows that f ( x) is continuous on the given interval [–3, 0]. -( x /2 ) 1 -( x /2 ) 2 Now, f ¢( x) = ( 2x + 3) e - e ( x + 3 x) 2 2ö -( x /2 ) æ ç x + 6 - x ÷, =e ç ÷ 2 è ø which is clearly finite for all values of x in ] -3 , 0[. So, f ( x) is differentiable on ] -3 , 0[. Also, f ( -3) = f ( 0) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist c Î ] - 3 , 0[ such that f ¢( c) = 0. But,

f ¢( c) = 0 Û e

-( c/2 ) æ çc+

ç è

6 - c2 ö÷ = 0 Û c + 6 - c2 = 0 ÷ 2 ø

Û ( 3 - c)( c + 2) = 0 Û c = 3 or c = -2.

Senior Secondary School Mathematics for Class 12 Pg-479

Applications of Derivatives

479

Thus, c = -2 Î ] - 3 , 0[ such that f ¢( c) = 0. Hence, Rolle’s theorem is verified. EXAMPLE 7

Verify Rolle’s theorem for the function f ( x) = {log ( x 2 + 2) - log 3} on [-1, 1].

SOLUTION

Clearly, f ( x) = {log ( x 2 + 2) - log 3} has a unique and definite value for each x Î [-1, 1]. So, at each point of [-1, 1], the limit of the function is equal to the value of the function. So, f ( x) is continuous on [-1, 1]. 2x Also, f ¢( x) = 2 , which is clearly finite for each x Î [-1, 1]. ( x + 2) So, f ( x) is differentiable on ] -1, 1[. Also, f ( -1) = f (1) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist c Î ] -1, 1[ such that f ¢( c) = 0. 2c Now, f ¢( c) = 0 Û = 0 Û 2c = 0, i.e., c = 0. 2 ( c + 2) Thus, c = 0 Î ] -1, 1[ such that f ¢( c) = 0. Hence, Rolle’s theorem is verified.

EXAMPLE 8

Verify Rolle’s theorem for the following functions: (i) f ( x) = 4 - x 2 in [-2, 2] æ x 2 + ab ù (ii) f ( x) = log ç in [a , b], where 0 < a < b ç ( a + b) x ú è û

SOLUTION

(i) Consider f ( x) = 4 - x 2 in [-2, 2]. Clearly, f ( x) = 4 - x 2 has a unique and definite value for each x Î [-2, 2]. So, at each point of [-2, 2], the limit is equal to the value of the function. \ f ( x) is continuous on [-2, 2]. -x Also, f ¢( x) = , which clearly exists for each 4 - x2 x Î [-2, 2]. So, f ( x) is differentiable on ] -2, 2[. Also, f ( -2) = f ( 2) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist c Î ] -2, 2[ such that f ¢( c) = 0. -c Now, f ¢( c) = 0 Û = 0 Û c = 0. 4 - c2 Thus, c = 0 Û ] - 2, 2[ such that f ¢( c) = 0. Hence, Rolle’s theorem is verified.

Senior Secondary School Mathematics for Class 12 Pg-480

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Senior Secondary School Mathematics for Class 12

(ii) Consider

é x 2 + ab ù 2 f ( x) = log ê ú = log ( x + ab) - log ( a + b) - log x. ( a + b ) x ë û Clearly, f ( x) has a unique and definite value for each x Î[a , b], where 0 < a < x < b. So, the value of the function is equal to the limit of the function at each point of [a , b]. \ f ( x) is continuous on [a , b]. é 2x 1ù Also, f ¢( x) = ê 2 - ú , which is clearly definite and ë ( x + ab) x û finite for all values of x in ]a , b[. So, f ( x) is differentiable on ]a , b[. Moreover, f ( a) = f ( b) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist c Î]a , b[ such that f ¢( c) = 0. 2c 1 But, f ¢( c) = 0 Û - =0 2 ( c + ab) c ( c2 - ab)

= 0 Û c2 - ab = 0 Û c = ab . c ( c2 + ab) Now, c being the geometric mean of a and b, it follows that a < c < b. Thus, c Î]a , b[ such that f ¢( c) = 0. Hence, Rolle’s theorem is verified. Û

EXAMPLE 9

Verify Rolle’s theorem for the function f ( x) = 2x 3 + x 2 - 4x - 2.

SOLUTION

Since a polynomial function is everywhere continuous and differentiable, the given function is continuous as well as differentiable on every interval. To identify the interval, we solve the equation f ( x) = 0. Now,

f ( x) = 0 Û 2x 3 + x 2 - 4x - 2 = 0 Û ( x 2 - 2)( 2x + 1) = 0 Û x 2 = 2 or x = -

1 Û x = 2 or x = - 2 or x = - × 2 So, we consider the given function in [- 2 , 2]. Clearly, f ( - 2) = f ( 2) = 0. Thus, all the conditions of Rolle’s theorem are satisfied. So, there must exist c Î ] - 2 , 2[ such that f ¢( c) = 0. But, f ¢( x) = ( 6x 2 + 2x - 4). \ f ¢( c) = 0 Û 6c2 + 2c - 4 = 0 Û 2( 3 c - 2)( c + 1) = 0 Û c = 2/ 3 or c = -1. Clearly, both these points lie in [- 2 , 2]. Hence, Rolle’s theorem is verified.

1 2

Senior Secondary School Mathematics for Class 12 Pg-481

Applications of Derivatives EXAMPLE 10

SOLUTION

481

Discuss the applicability of Rolle’s theorem to the functions: [CBSE 1999] (i) f ( x) = x 2 in [1, 2] (ii) f ( x) = x 2/ 3 in [-1, 1] (i) Consider f ( x) = x 2 in [1, 2]. Since a polynomial function is continuous and differentiable everywhere, the first two conditions of Rolle’s theorem are satisfied. But, f (1) = 12 = 1 and f ( 2) = 22 = 4. So, f (1) ¹ f ( 2). And, therefore, the condition f ( a) = f ( b) is violated. Hence, Rolle’s theorem is not applicable for f ( x) = x 2 in [1, 2]. (ii) Consider f ( x) = x 2/ 3 in [-1, 1]. We have f ( 0 + h) - f ( 0) h2/ 3 - 0 1 Rf ¢( 0) = lim = lim = lim 1/ 3 = ¥. h ®0 h ®0 h ®0 h h h f ( 0 - h) - f ( 0) ( - h) 2/ 3 - 0 -1 = lim = lim = -¥ . h ®0 h ®0 h ®0 ( h)1/ 3 -h -h

Lf ¢( 0) = lim

\ Rf ¢( 0) ¹ Lf ¢( 0), i.e., f ¢( 0) does not exist. Thus, f ( x) is not differentiable at x = 0 Î ] -1, 1[. So, the condition of differentiability at each point of the given interval is not satisfied. Hence, Rolle’s theorem is not applicable to f ( x) = x 2/ 3 in [–1, 1]. EXAMPLE 11

SOLUTION

Discuss the applicability of Rolle’s theorem on: (i) f ( x) = |x|in [-1, 1] (ii) f ( x) = tan x in [0, p] (i) Consider f ( x) = |x|in [-1, 1]. ì- x when - 1 £ x < 0 We may express it as f ( x) = í î x when 0 £ x £ 1 Clearly, f ( -1) = f (1) = 1. f ( 0 + h) - f ( 0) |h| h But, Rf ¢( 0) = lim = lim = lim = 1. h ®0 h ®0 h h® 0 h h And,

f ( 0 - h) - f ( 0) |- h| h = lim = lim = -1. h ®0 - h h ®0 - h -h \ Rf ¢( 0) ¹ Lf ¢( 0). This shows that f ( x) is not differentiable at x = 0. Thus, the condition of differentiability at each point of the given interval is not satisfied. (ii) Consider f ( x) = tan x in [0, p]. p Clearly, f ( 0) = f ( p) = 0. But, tan does not have a definite 2 value. Lf ¢( 0) = lim

h ®0

Senior Secondary School Mathematics for Class 12 Pg-482

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Senior Secondary School Mathematics for Class 12

p × 2 Hence, the condition of continuity at each point of the given interval is violated. So, f ( x) = tan x is not continuous at x =

EXAMPLE 12

Discuss the applicability of Rolle’s theorem on the function ì( x 2 + 1), when 0 £ x £ 1 f ( x) = í î( 3 - x), when 1 < x £ 2.

SOLUTION

The given function has been defined in [0, 2]. And, f ( 0) = f ( 2) = 1. Now, we consider the differentiability of f ( x) at x = 1. f (1 + h) - f (1) [ 3 - (1 + h)] - 2 -h Rf ¢(1) = lim = lim = lim = -1. h® 0 h ® 0 h ® 0 h h h Lf ¢(1) = lim

h® 0

f (1 - h) - f (1) (1 - h) 2 + 1 - 2 = lim = lim ( 2 - h) = 2. h® 0 h® 0 -h -h

Thus, Rf ¢(1) ¹ Lf ¢(1). So, f ( x) is not differentiable at x = 1 Î ]0, 2[. Thus, the condition of differentiability at each point of the given interval is not satisfied. EXAMPLE 13

If Rolle’s theorem holds for the function f ( x) = x 3 + bx 2 + ax + 5 on 1 ö æ [1, 3] with c = ç 2 + ÷ , find the values of a and b. 3ø è

SOLUTION

We have, \

f ¢( x) = 3 x 2 + 2bx + a.

f ¢( c) = 0 Û 3 c2 + 2bc + a = 0 Û c=

-2b ± 4b 2 - 12a - b ± b 2 - 3 a = × 6 3

-b + b2 - 3 a æ 1 ö 1 ö æ Now, c = ç 2 + = ç2 + ÷ Û ÷ 3 3ø 3ø è è Û

-b = 2 and 3

b2 - 3 a 1 = 3 3

Û b = -6 and b 2 - 3 a = 3 Û b = -6 and a = 11. Hence, a = 11 and b = -6. EXAMPLE 14

SOLUTION

At what points on the curve y = (cos x - 1) in [0, 2p], is the tangent parallel to the x-axis? Consider f ( x) = (cos x - 1) in [0, 2p]. Now, the cosine function being continuous and the constant function being continuous, it follows that their difference (cos x - 1) is continuous. So, f ( x) is continuous on [0, 2p]. f ¢( x) = - sin x , which exists for all x Î]0, 2p [.

Senior Secondary School Mathematics for Class 12 Pg-483

Applications of Derivatives

\

483

f ( x) is differentiable on ]0, 2p[.

Also, f ( 0) = f ( 2p) = 0. So, all the conditions of Rolle’s theorem are satisfied. So, there must exist some c Î]0, 2p[ such that f ¢( c) = 0. Now, f ¢( c) = 0 Û - sin c = 0 Û sin c = 0 Û c = 0 or c = p or c = 2p. Thus, the tangent to the curve is parallel to the x-axis at each of the points x = 0, x = p and x = 2p. Now, x = 0 and y = cos x - 1 Þ y = 0; x = p and y = cos x - 1 Þ y = -2; x = 2p and y = cos x - 1 Þ y = 0. \ the required points are ( 0, 0), ( p , - 2) and ( 2p , 0).

EXERCISE 11C Verify Rolle’s theorem for each of the following functions: 1. f ( x) = x 2 on [–1, 1] 3. f ( x) = x 2 - 5 x + 6 in [2, 3]

2. f ( x) = x 2 - x - 12 in [–3, 4][CBSE 2001] [CBSE 2002C]

2

4. f ( x) = x - 3 x - 18 in [–3, 6]

5. f ( x) = x 2 - 4x + 3 in [1, 3] [CBSE 2007]

6. f ( x) = x( x - 4) 2 in [0, 4]

7. f ( x) = x 3 - 7 x 2 + 16x - 12 in [2, 3]

3

2

8. f ( x) = x + 3 x - 24x - 80 in [– 4, 5] 9. f ( x) = ( x - 1)( x - 2)( x - 3) in [1, 3] 10. f ( x) = ( x - 1)( x - 2) 2 in [1, 2]

[CBSE 2006]

11. f ( x) = ( x - 2) 4( x - 3) 3 in [2, 3]

12. f ( x) = 1 - x 2 in [–1, 1]

é p pù 13. f ( x) = cos x in ê - , ú ë 2 2û

14. f ( x) = cos 2x in [0, p]

15. f ( x) = sin 3 x in [0, p] 17. f ( x) = e- x sin x in [0, p] 19. f ( x) = sin x - sin 2x in [0, 2p]

é pù 16. f ( x) = sin x + cos x in ê 0, ú ë 2û é p 5p ù 18. f ( x) = e- x (sin x - cos x) in ê , ë 4 4 úû x 20. f ( x) = x( x + 2) e in [–2, 0]

21. Show that f ( x) = x( x - 5) 2 satisfies Rolle’s theorem on [0, 5] and that the value of c is (5/3). Discuss the applicability of Rolle’s theorem, when: 22. f ( x) = ( x - 1)( 2x - 3), where 1 £ x £ 3 23. f ( x) = x

1

2

on [–1, 1]

24. f ( x) = 2 + ( x - 1)

2

3

on [0, 2]

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Senior Secondary School Mathematics for Class 12

25. f ( x) = cos

1 on [–1, 1] x

26. f ( x) = [x] on [–1, 1], where [x] denotes the greatest integer not exceeding x 27. Using Rolle’s theorem, find the point on the curve y = x( x - 4), x Î[0, 4], where the tangent is parallel to the x-axis. [CBSE 2000]

ANSWERS (EXERCISE 11C)

22. Not applicable, since f (1) ¹ f ( 3) 23. Not applicable, since f ¢( 0) does not exist 24. Not applicable, since f ¢(1) does not exist 25. Not applicable, since f ( x) is discontinuous at x = 0 26. Not applicable, since f ( x) is discontinuous at x = 0 27. ( 2, - 4) HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11C) 3

9. f ( x ) = ( x - 6 x 2 + 11x - 6 ). 22. f ( 1) ¹ f ( 3 ). 1 23. f ¢( x ) = Þ f ¢( 0 ) does not exist, where 0 Î [-1, 1]. 2 x 24. f ¢( 1) does not exist. 25. f ( x ) is discontinuous at x = 0. 26. f ( x ) is not continuous at x = 0.

Lagrange’s Mean-value Theorem Let f be a real function such that (i) f ( x) is continuous on [a , b],

(ii) f ( x) is differentiable on ]a , b[. f ( b) - f ( a) Then, there exists a real number c Î]a , b[ such that f ¢( c) = × ( b - a)

PROOF

Let F( x) = f ( x) + Ax , where A is a constant to be chosen in such a way that F( a) = F( b). f ( b) - f ( a) Now, F( a) = F( b) Û f ( a) + Aa = f ( b) + Ab Û A = × ( a - b) f ( b) - f ( a) … (i) \ F( x) = f ( x) + ×x ( a - b) ì f ( b) - f ( a) ü Now, f ( x) is continuous on [a , b], and í × x ý being a þ î ( a - b) polynomial function, is continuous on [a , b].

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Applications of Derivatives

485

And, the sum of two continuous functions being continuous, it follows from (i) that F( x) is continuous. Also, f ( x) is differentiable on ]a , b[, and since the polynomial function f ( b) - f ( a) x is differentiable on ]a , b[, it follows that F( x) is ( a - b) differentiable on ]a , b[. Also, F( a) = F( b). Thus, all the conditions of Rolle’s theorem are satisfied by F( x). So, there must exist some c Î]a , b[ such that F¢( c) = 0. f ( b) - f ( a) f ( b) - f ( a) Now, F( x) = f ( x) + × × x Þ F¢( x) = f ¢( x) + ( a - b) ( a - b) \

F¢( c) = 0 Û f ¢( c) +

f ( b) - f ( a) f ( b) - f ( a) × = 0 Û f ¢( c) = ( a - b) ( b - a)

GEOMETRICAL SIGNIFICANCE OF THE MEAN-VALUE THEOREM Let y = f ( x) be a given function defined on [a , b], which is continuous on [a , b] and differentiable on ]a , b[.

Then, by Lagrange’s mean-value theorem, there exists some c Î]a , b[ f ( b) - f ( a) … (i) f ¢( c) = ( b - a)

such that

Now, if we draw the curve y = f ( x) and take the points A[a , f ( a)] and B[b , f ( b)] on the curve then f ( b) - f ( a) slope of chord AB = ( b - a)

Y

[a, A

… (ii)

f(a)

]

[b,

] f(b) B

C

Thus, from (i) and (ii), we have f ¢( c) = slope of chord AB. This shows that the tangent to the curve y = f ( x) at the point x = c is parallel to the chord AB.

O

x=a

x=c x=b

X

PHYSICAL SIGNIFICANCE OF THE MEAN-VALUE THEOREM Let a particle be moving in a straight line and let f ( a) and f ( b) be its positions from the starting point, at the times a and b respectively.

f ( b) - f ( a) × ( b - a) We also know that f ¢( c) denotes the instantaneous speed of the particle at time c. Then, average speed of the particle =

f ( b) - f ( a) × ( b - a) So, the mean-value theorem says that at some time c between a and b, the instantaneous speed of the particle would be equal to the average speed. Now, by the mean-value theorem, f ¢( c) =

Senior Secondary School Mathematics for Class 12 Pg-486

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Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Verify Lagrange’s mean-value theorem for the following functions: (i) f ( x) = x( 2 - x) in [0, 1] (ii) f ( x) = x( x + 4) 2 in [0, 4] 1 (iii) f ( x) = x + in [1, 3] x (iv) f ( x) = ( x - 1)( x - 2)( x - 3) in [0, 4] (v) f ( x) = x 2 - 4

SOLUTION

[CBSE 2002]

(i) Consider f ( x) = x( 2 - x) in [0, 1]. The given function is f ( x) = 2x - x 2. It, being a polynomial function, is continuous on [0, 1]. Also, f ¢( x) = 2 - 2x , which exists for all x in [0, 1]. So, f ( x) is differentiable on ]0, 1[. Thus, both the conditions of Lagrange’s mean-value theorem are satisfied. So, there must exist some c Î]0, 1[ such that f (1) - f ( 0) f ¢( c) = = 1. (1 - 0) 1 Now, f ¢( c) = 1 Û 2 - 2c = 1 Û c = Î ]0, 1[. 2 1 f (1) - f ( 0) Thus, c = Î ]0, 1[ such that f ¢( c) = × 2 1-0 Hence, Lagrange’s mean-value theorem is verified. (ii) Consider f ( x) = x( x + 4) 2 in [0, 4]. The given function is f ( x) = x 3 + 8x 2 + 16x. It, being a polynomial function, is continuous on [0, 4]. Also, f ¢( x) = ( 3 x 2 + 16x + 16), which exists for all x Î]0, 4[. So, f ( x) is differentiable on ]0, 4[. Thus, both the conditions of Lagrange’s mean-value theorem are satisfied. So, there must exist some c Î]0, 4[ such that f ( 4) - f ( 0) æ 256 - 0 ö =ç f ¢( c) = ÷ = 64. ( 4 - 0) è 4 ø Now,

f ¢( c) = 64 Û 3 c2 + 16c + 16 = 64

-8 ± 4 13 Û 3 c2 + 16c - 48 = 0 Û c = × 3 -8 + 4 13 Clearly, c= = 2.13 Î ]0, 4[. 3 Hence, Lagrange’s mean-value theorem is verified.

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Applications of Derivatives

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1 in [1, 3]. The given function is x x2 + 1 f ( x) = × x Since it is a rational function such that x ¹ 0, it is continuous in [1, 3]. 1 ö æ x2 - 1ö æ Also, f ¢( x) = ç1 - 2 ÷ = ç 2 ÷ , which clearly exists for all è x ø çè x ÷ø values of x in ]1, 3[. So, f ( x) is differentiable on ]1, 3[. Thus, both the conditions of Lagrange’s mean-value theorem are satisfied. So, there must exist some c Î]1, 3[ such that æ 10 ö ç - 2÷ f ( 3) - f (1) è 3 ø = 2× = f ¢( c) = ( 3 - 1) 2 3

(iii) Consider f ( x) = x +

2 c2 - 1 2 Û = Û c = 3 Î ]1, 3[. 3 3 c2 f ( 3) - f (1) Thus, c = 3 Î ]1, 3[ such that f ¢( c) = × ( 3 - 1) Hence, Lagrange’s mean-value theorem is verified. (iv) Consider f ( x) = ( x - 1)( x - 2)( x - 3) in [0, 4]. The given function is f ( x) = x 3 - 6x 2 + 11x - 6. Being a polynomial function, it is continuous on [0, 4]. Also, f ¢( x) = ( 3 x 2 - 12x + 11), which exists for all x Î]0, 4[. So, f ( x) is differentiable on ]0, 4[. Thus, both the conditions of Lagrange’s mean-value theorem are satisfied. So, there exists some c Î]0, 4[ such that f ( 4) - f ( 0) 6 - ( -6) f ¢( c) = = = 3. ( 4 - 0) 4 Now,

f ¢( c) =

Now, f ¢( c) = 3 Û 3 c2 - 12c + 11 = 3 Û 3 c2 - 12c + 8 = 0 12 ± 144 - 96 6 ± 2 3 Û c= = × 6 3 Û c = 3.155 or c = 0.845 [Q 3 = 1.732]. Clearly, both these values of c lie in ]0, 4[. Hence, Lagrange’s mean-value theorem is verified. (v) Consider f ( x) = x 2 - 4 in [2, 4]. Clearly, f ( x) has a definite and unique value for each x Î[2, 4]. So, at every point [2, 4], the value of f ( x) is equal to the limit of f ( x).

Senior Secondary School Mathematics for Class 12 Pg-488

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Senior Secondary School Mathematics for Class 12

So, f ( x) is continuous on [2, 4]. x Also, f ¢( x) = , which exists for all x Î]2, 4[. x2 - 4 So, f ( x) is differentiable on ]2, 4[. So, there must exist some c Î]2, 4[ such that f ¢( c) =

f ( 4) - f ( 2) 12 - 0 = = ( 4 - 2) 2

3.

Now, f ¢( c) =

3 Û

c 2

c -4

=

3 Û c2 = 3( c2 - 4) Û c = ± 6.

Clearly, c = 6 Î ]2, 4[ such that f ¢( c) =

f ( 4) - f ( 2) × ( 4 - 2)

Hence, Lagrange’s mean-value theorem is verified. EXAMPLE 2

Verify the hypothesis and conclusion of Lagrange’s mean-value theorem 1 for the function f ( x) = , 1 £ x £ 4. ( 4x - 1)

SOLUTION

Clearly, for each x Î[1, 4], f ( x) has a definite and unique value. So, f ( x) is continuous on [1, 4]. -4 Also, f ¢( x) = , which exists for all x Î]1, 4[. ( 4x - 1) 2 So, f ( x) is differentiable on ]1, 4[. Thus, both the conditions of Lagrange’s mean-value theorem are satisfied. So, there must exist some c Î]1, 4[ such that f ( 4) - f (1) 1 æ 1 1 ö -4 × f ¢( c) = = ç - ÷= ( 4 - 1) 3 è 15 3 ø 45 Now,

f ¢( c) =

-4 -4 -4 Û = Û ( 4c - 1) 2 = 45 2 45 45 ( 4c - 1) æ1 ± 3 5 ö ÷× Û ( 4c - 1) = ±3 5 Û c = çç 4 ÷ø è

Clearly,

c=

1 + 3 5 æ 1 + 3 ´ 2.23 ö =ç ÷ = 1.92 Î ]1, 4[. 4 4 ø è

Hence, Lagrange’s mean-value theorem is verified. EXAMPLE 3

Find ‘c’ of the mean-value theorem for the functions: (i) f ( x) = 2x 2 - 10x + 29 in [2, 7] é 1ù (ii) f ( x) = x( x - 1)( x - 2) in ê 0, ú ë 2û

Senior Secondary School Mathematics for Class 12 Pg-489

Applications of Derivatives SOLUTION

489

(i) f ( x) = 2x 2 - 10x + 29 Û f ¢( x) = 4x - 10. f (7) - f ( 2) (57 - 17) Now, f ¢( c) = Û ( 4c - 10) = (7 - 2) 5 Û 4c = 18 Û c = 4.5 Î ]2, 7[. \ c = 4.5. (ii) The given function is f ( x) = x 3 - 3 x 2 + 2x. \ f ¢( x) = 3 x 2 - 6x + 2. Now, ö æ3 æ1ö f ç ÷ - f ( 0) ç - 0÷ 8 2 ø Û 3 c2 - 6c + 2 = è f ¢( c) = è ø æ1 ö æ1ö ç - 0÷ ç ÷ è2 ø è 2ø 2 Û 12c - 24c + 5 = 0 24 ± 576 - 240 6 ± 21 = × 24 6 6 - 21 6 - 4.58 é 1ù Clearly, c = = = 0.24 Î ê 0, ú × 6 6 ë 2û Hence, c = 0.24. Û c=

EXAMPLE 4

Using Lagrange’s mean-value theorem, find a point on the curve y = x - 2, defined in the interval [2, 3], where the tangent is parallel to the chord joining the end points of the curve.

SOLUTION

Let f ( x) = x - 2. Clearly, for each x Î[2, 3], the function f ( x) has a definite and unique value. So, f ( x) is continuous on [2, 3]. 1 Also, f ¢( x) = , which exists for all x in ]2, 3[. 2 x-2 So, f ( x) is differentiable on ]2, 3[. Thus, both the conditions of Lagrange’s mean-value theorem are satisfied. f ( 3) - f ( 2) So, there must exist some c Î]2, 3[ such that f ¢( c) = = 1. ( 3 - 2) Now,

f ¢( c) = 1 Û

Now,

x=

1 1 9 = 1 Û c - 2 = , i.e., c = Î ]2, 3[. 4 4 2 c-2

9 æ9 ö 1 and y = x - 2 Û y = ç - 2÷ = × 4 è4 ø 2

æ9 \ the tangent to the given curve at the point ç , è4 the chord joining the end points of the curve.

1ö ÷ is parallel to 2ø

Senior Secondary School Mathematics for Class 12 Pg-490

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Senior Secondary School Mathematics for Class 12

EXAMPLE 5

Find a point on the parabola y = ( x - 3) 2 , where the tangent is parallel to

SOLUTION

the chord joining ( 3 , 0) and ( 4, 1). [CBSE 2000C] Let us apply Lagrange’s mean-value theorem for the function f ( x) = ( x - 3) 2 in the interval [ 3 , 4]. Now, f ( x) being a polynomial function, it is continuous on [ 3 , 4]. Also, f ¢( x) = 2( x - 3), which exists for all x Î] 3 , 4[. So, f ( x) is differentiable on ] 3 , 4[. Thus, both the conditions of Lagrange’s mean-value theorem are satisfied. So, there must exist a point c Î] 3 , 4[ such that f ( 4) - f ( 3) f ¢( c) = = 1. ( 4 - 3) Now, f ¢( c) = 1 Û 2( c - 3) = 1 Û c = Now, x =

7 Î ] 3 , 4[. 2

7 1 and y = ( x - 3) 2 Û y = × 2 4

æ7 1ö Thus, at the point ç , ÷ on the given curve the tangent is parallel è 2 4ø to the chord joining (3, 0) and (4, 1).

EXERCISE 11D Verify Lagrange’s mean-value theorem for each of the following functions: 1. f ( x) = x 2 + 2x + 3 on [4, 6]

[CBSE 2006]

2

2. f ( x) = x + x - 1 on [0, 4] 2

3. f ( x) = 2x - 3 x + 1 on [1, 3]

[CBSE 2002] 3

2

4. f ( x) = x + x - 6x on [–1, 4]

5. f ( x) = ( x - 4)( x - 6)( x - 8) on [4, 10] 6. f ( x) = ex on [0, 1] 7. f ( x) = x

2

3

on [0, 1]

9. f ( x) = tan -1 x on [0, 1]

8. f ( x) = log x on [1, e] é p 5p ù 10. f ( x) = sin x on ê , ë 2 2 úû

é pù 11. f ( x) = (sin x + cos x) on ê 0, ú ë 2û 12. Show that Lagrange’s mean-value theorem is not applicable to f ( x) = |x| on [–1, 1]. 1 13. Show that Lagrange’s mean-value theorem is not applicable to f ( x) = on x [–1, 1].

Senior Secondary School Mathematics for Class 12 Pg-491

Applications of Derivatives

491

14. Find ’c‘ of Lagrange’s mean-value theorem for é 1ù (i) f ( x) = ( x 3 - 3 x 2 + 2x) on ê 0, ú ë 2û (ii) f ( x) = 25 - x 2 on [1, 5] (iii) f ( x) = x + 2 on [4, 6] 15. Using Lagrange’s mean-value theorem, find a point on the curve y = x 2 , where the tangent is parallel to the line joining the points (1, 1) and (2, 4). [CBSE 2000C] 3

16. Find a point on the curve y = x , where the tangent to the curve is parallel to the chord joining the points (1, 1) and (3, 27). 17. Find the points on the curve y = x 3 - 3 x , where the tangent to the curve is parallel to the chord joining (1, –2) and (2, 2). 18. If f ( x) = x(1 - log x), where x > 0, show that ( a - b) log c = b(1 - log b) - a(1 - log a), where 0 < a < c < b. ANSWERS (EXERCISE 11D)

14. (i) 1 -

7 12

(ii) 15

æ 39 13 39 ö ÷ 16. çç , 9 ÷ø è 3

(iii) 4.964 æ 17. çç è

æ 3 9ö 15. ç , ÷ è 2 4ø 7 -2 7 ö ÷, , 3 3 3 ÷ø

æ 7 2 ç, ç 3 3 è

7 3

ö ÷ ÷ ø

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11D) 6. c = log( e - 1). 8. 2 < e < 3. 12. f ¢( 0 ) does not exist. 13. f ( x ) is discontinuous at x = 0. 18. f ( x ) is continuous for all x > 0 and f ¢( x ) = - log x , which exists for all x > 0. f ( b ) - f ( a) b( 1 - log b ) - a( 1 - log a) \ f ¢( c ) = Þ - log c = × ( b - a) ( b - a)

4. Maxima and Minima ABSOLUTE MAXIMUM VALUE OF A FUNCTION A function f ( x) is said to have the greatest value or absolute maximum value at a point ‘a’ in its domain if f ( x) £ f ( a) for all x in the domain of f ( x).

In this case, ‘a’ is called the point of maximum. ABSOLUTE MINIMUM VALUE OF A FUNCTION A function f ( x) is said to have the smallest value or absolute minimum value at a point ‘a’ in its domain if f ( a) £ f ( x) for all x in the domain of f ( x).

Senior Secondary School Mathematics for Class 12 Pg-492

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Senior Secondary School Mathematics for Class 12

The point a in this case, is called the point of minimum. Examples (i) Consider f ( x) = sin x on the interval [0, p].

Clearly, the greatest value of the function, i.e., the absolute maximum, is f ( p/2) = 1. Also, the smallest value of the function, i.e., the absolute minimum, is f ( 0) = f ( p) = 0. (ii) Consider f ( x) = x 2 for all x Î R.

It is clear that the smallest value of the function, i.e., the absolute minimum, is 0. But, we can say nothing about the greatest value of the function. In other words, we can say that the absolute maximum does not exist. (iii) Consider f ( x) = x 3 for all x Î R.

Here, we observe that as the value of x increases, the value of the function increases, and as x decreases, the value of f ( x) decreases. But, the function has neither a greatest value nor a least value. Thus, the function has neither an absolute maximum nor an absolute minimum. 2

(iv) Consider f ( x) = - ( x - 1) + 5 for all x Î R.

For this function, we obtain the absolute maximum when its negative part is 0, i.e, when ( x - 1) 2 is 0. So, the greatest value of the function is 5. Clearly, it does not have a least value. EXAMPLE 1

Without using the derivative, find the maximum or minimum values, if any, of the function f ( x) = 4x 2 - 4x + 7 for all x Î R.

SOLUTION

We have f ( x) = 4x 2 - 4x + 7 = ( 2x - 1) 2 + 6. Clearly, ( 2x - 1) 2 is non-negative for all x Î R. The least value of ( 2x - 1) 2 is 0. So, the least value of the function is 6. Clearly, this happens when 2x - 1 = 0, i.e., x = (1/2). Thus, x = (1/2) is a point of absolute minimum. However, f ( x) does not have an absolute maximum.

EXAMPLE 2

SOLUTION

Find the maximum or minimum values, if any, of the following functions, without using the derivatives: (i) f ( x) = |x + 2| (ii) f ( x) = -|x + 1| + 3 (iii) f ( x) = |sin ( 4x + 3)| (iv) f ( x) = sin (sin x) (i) Clearly,|x + 2|is non-negative for all x Î R. The least value of|x + 2|is 0.

Senior Secondary School Mathematics for Class 12 Pg-493

Applications of Derivatives

493

So, the minimum value of the function is 0. Clearly, f ( x) = |x + 2|does not have a maximum value. (ii) The value of f ( x) = -|x + 1| + 3 is maximum when its negative part has a value 0. Thus, the maximum value of f ( x) is 3. Clearly, there is no minimum value of f ( x). (iii) The maximum value of|sin q|is 1, so the maximum value of |sin ( 4x + 3)| is 1. Also, the value of |sin ( 4x + 3)| is non-negative. So, the minimum value of|sin ( 4x + 3)|is 0. (iv) -1 £ sin x £ 1 Þ sin ( -1) £ sin (sin x) £ sin 1 {Q sin x is increasing on [-1, 1]} Þ - sin 1 £ sin (sin x) £ sin 1. This shows that the maximum value of f ( x) is sin 1 and the minimum value is - sin 1. Local Maxima and Local Minima Here, our main interest lies in finding the maximum and minimum values of a function in a small interval rather than considering it for the whole of the domain. So, we shall deal with the topics of local maxima and local minima. LOCAL MAXIMUM VALUE OF A FUNCTION We say that c is a point of local maximum of a function f ( x) if there is an open interval I containing c such that f ( x) < f ( c) for all x Î I.

Also, in this case, f ( c) is called a local maximum value of f ( x). LOCAL MINIMUM VALUE OF A FUNCTION We say that c is a point of local minimum of a function f ( x) if there is an open interval I containing c such that f ( c) < f ( x) for all x Î I. BEHAVIOUR OF f ¢(x ) AT LOCAL MAXIMA AND LOCAL MINIMA

Case I

When c is a point of local maxima In this case, f ( x) is increasing for values of x slightly less than c; it ceases to increase at x = c and then decreases.

Y

A f(c)

\ for small positive values of h, we have x=c f ¢( x) > 0 on [c - h, c] O X and, f ¢( x) < 0 on [c, c + h]. Thus, f ¢( x) changes sign from positive to negative as x increases through c.

Senior Secondary School Mathematics for Class 12 Pg-494

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Senior Secondary School Mathematics for Class 12

Case II When c is a point of local minima

Y

In this case f ( x) is decreasing for values of x slightly less than c; it ceases to decrease at x = c B and then increases for values of x slightly greater than c. f(c) X O x=c \ for small positive values of h, we have f ¢( x) < 0 on [c - h, c] and, f ¢( x) > 0 on [c, c + h]. Thus, f ¢( x) changes sign from negative to positive as x increases through c. EXTREMUM VALUES The maximum or minimum value of a function is called an extreme or extremum value of the function.

If f ¢( x) exists on an interval [a , b] and f ( x) has a maximum or minimum value at x = c Î]a , b[ then f ¢( c) = 0. The proof is beyond the scope of this book.

THEOREM

PROOF

STATIONARY POINTS

The points at which f ¢( c) = 0 are known as stationary points or

turning points. (i) A function can have at the most one absolute maximum and at the most one absolute minimum. C Y (ii) A function may have more than one local maximum and more D A than one local minimum. (iii) A local minimum may be greater B than a local maximum. (iv) Local maximums and local X O minimums occur alternately. In the given figure, A and C are the points of local maxima, while B and D are the points of local minima. Clearly, the local minimum at D is greater than the local maximum at A. REMARKS

Tests for Finding Local Extremum Values We must remember that (i) c is a point of local maximum if f ¢( c) = 0 and f ¢( x) changes sign from positive to negative as x increases through c (ii) c is a point of local minimum if f ¢( c) = 0 and f ¢( x) changes sign from negative to positive as x increases through c. FIRST-DERIVATIVE TEST

Working Rule for Finding Extremum Values In order to find the extremum values of a given function f ( x) by using the first-derivative test, we proceed according to the following steps.

Senior Secondary School Mathematics for Class 12 Pg-495

Applications of Derivatives

495

1. Find f ¢( x). 2. Solve f ¢( x) = 0. Let its roots be a, b, c, etc. Then, these are the candidates for maxima or minima. 3. We test the function at each one of these values. Let us take x = c. 4. Determine the sign of f ¢( x) for values of x slightly less than c and for values of x slightly greater than c. (i) If f ¢( x) changes sign from positive to negative as x increases through c then x = c is a point of maximum.

CONCLUSION

(ii) If f ¢( x) changes sign from negative to positive as x increases through c then x = c is a point of minimum. (iii) If f ¢( x) does not change sign as x increases through c, we say that x is neither a point of maximum nor a point of minimum. In this case, x = c is called a point of inflexion. Similarly, we may deal with the other roots of f ¢( x) = 0 and examine them for maxima or minima. SOLVED EXAMPLES EXAMPLE 1

Find the local maxima or local minima, if any, of 1 (ii) f ( x) = ( x 3 - 3 x) (i) f ( x) = 2 ( x + 2) In each case, find the local maximum or the local minimum values, as the case may be.

SOLUTION

(i) f ( x) =

1

Þ f ¢( x) =

-2x

× ( x + 2) ( x + 2) 2 For a local maxima or minima, we have f ¢( x) = 0. -2x \ f ¢( x) = 0 Þ = 0 Þ x = 0. ( x 2 + 2) 2 Now, when x is slightly less than 0, i.e., when x is negative, -2x is positive. then f ¢( x) = 2 ( x + 2) 2 -2x When x is slightly greater than 0 then f ¢( x) = 2 < 0. ( x + 2) 2 Thus, f ¢( x) changes sign from positive to negative. So, x = 0 is a point of local maximum. 1 1 And, local maximum value is f ( 0) = 2 = × ( 0 + 2) 2 2

2

(ii) f ( x) = ( x 3 - 3 x) Þ f ¢( x) = ( 3 x 2 - 3). Now, for a maximum or a minimum, we have f ¢( x) = 0. Now, f ¢( x) = 0 Þ 3 x 2 - 3 = 0 Þ 3( x - 1)( x + 1) = 0 Þ x = 1 or x = - 1.

Senior Secondary School Mathematics for Class 12 Pg-496

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Senior Secondary School Mathematics for Class 12

Thus, x = 1 and x = -1 are the candidates for local maxima or local minima. Consider x = 1. We know that f ¢( x) = 3( x 2 - 1). When x is slightly less than 1 then x 2 is slightly less than 1 and therefore, 3( x 2 - 1) is negative. Again, when x is slightly more than 1 then x 2 is slightly more than 1 and therefore, 3( x 2 - 1) is positive. Thus, f ¢( x) changes values from negative to positive as x increases through 1. So, x = 1 is a point of local minimum. Local minimum value = f (1) = (1 3 - 3 ´ 1) = -2. Consider x = - 1. When x is slightly less than -1 then x 2 is slightly more than 1. So, 3( x 2 - 1) is positive. Again, when x is slightly more than -1 then x 2 is slightly less than 1. So, 3( x 2 - 1) is negative. Thus, f ¢( x) changes sign from positive to negative as x increases through -1. So, x = - 1 is a point of local maximum. Local maximum value = f ( -1) = ( -1) 3 - 3 ´ ( -1) = 2. EXAMPLE 2

Find the local maxima or local minima of f ( x) = x 3 - 6x 2 + 9x + 15. Also, find the local maximum or local minimum values as the case may be.

SOLUTION

Here, f ( x) = x 3 - 6x 2 + 9x + 15 Þ f ¢( x) = 3 x 2 - 12x + 9. For a local maxima or minima, we must have f ¢( x) = 0. Now, f ¢( x) = 0 Þ 3( x 2 - 4x + 3) = 0 Þ 3( x - 3)( x - 1) = 0 Þ x = 3 or x = 1. Case I When x = 3 In this case, when x is slightly less than 3 then f ¢( x) = 3( x - 3)( x - 1) is negative and when x is slightly more than 3 then f ¢( x) is positive. Thus, f ¢( x) changes from negative to positive as x increases through 3. So, x = 3 is a point of local minimum. Local minimum value = f ( 3) = 15. Case II When x = 1 In this case, when x is slightly less than 1 then f ¢( x) = 3( x - 3)( x - 1) is positive and when x is slightly more than 1 then f ¢( x) is negative.

Senior Secondary School Mathematics for Class 12 Pg-497

Applications of Derivatives

497

Thus, f ¢( x) changes sign from positive to negative as x increases through 1. So, x = 1 is a point of local maximum. Local maximum value = f (1) = 19. EXAMPLE 3

Show that ( x5 - 5 x 4 + 5 x 3 - 1) has a local maximum when x = 1, a local minimum when x = 3, and neither when x = 0.

SOLUTION

Let f ( x) = x5 - 5 x 4 + 5 x 3 - 1. Then, f ¢( x) = 5 x 4 - 20x 3 + 15 x 2 = 5 x 2( x 2 - 4x + 3) = 5 x 2( x - 3)( x - 1). For a local maxima or minima, we have f ¢( x) = 0. Now, f ¢( x) = 0 Þ 5 x 2( x - 3)( x - 1) = 0 Þ x = 0 or x = 3 or x = 1. Case I

When x = 1 In this case, when x is slightly less than 1 then f ¢( x) = 5 x 2( x - 3)( x - 1) is positive and when x is slightly more than 1 then f ¢( x) is negative. Thus, f ¢( x) changes sign from positive to negative. So, x = 1 is a point of local maximum.

Case II When x = 3

Clearly,

when

x

is

slightly

less

than

3

then

2

f ¢( x) = 5 x ( x - 3)( x - 1) is negative and when x is slightly greater than 3 then f ¢( x) is positive. Thus, f ¢( x) changes sign from negative to positive. So, x = 3 is a point of local minimum. Case III When x = 0

Clearly,

when

x

is

slightly

less

than

0

then

2

f ¢( x) = 5 x ( x - 3)( x - 1) is positive and also when x is slightly more than 0 then f ¢( x) is positive. Thus, f ¢( x) does not change sign as it passes through 0. So, it is neither a point of local maximum nor a point of local minimum, i.e., x = 0 is a point of inflexion. EXAMPLE 4

Find the points of local maxima and local minima as well as the corresponding local maximum and local minimum values for the function f ( x) = ( x - 1) 3( x + 1) 2.

Senior Secondary School Mathematics for Class 12 Pg-498

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Senior Secondary School Mathematics for Class 12

SOLUTION

f ( x) = ( x - 1) 3( x + 1) 2 Þ f ¢( x) = 2( x - 1) 3( x + 1) + 3( x - 1) 2( x + 1) 2 = ( x - 1) 2( x + 1)(5 x + 1). \

f ¢( x) = 0 Þ ( x - 1) 2( x + 1)(5 x + 1) = 0 1 Þ x = 1 or x = -1 or x = - × 5

Consider x = 1. When x is slightly less than 1 then clearly f ¢( x) is +ve. When x is slightly greater than 1, then clearly f ¢( x) is +ve. Thus, f ¢( x) does not change sign as x passes through 1. So, x = 1 is neither a point of maximum nor a point of minimum. Consider x = - 1. When x is slightly less than -1 then f ¢( x) = ( + )( -)( -) = + ve. When x is slightly greater than -1 then f ¢( x) = ( + )( + )( -) = - ve. Thus, f ¢( x) changes sign from +ve to -ve as x passes through -1. So, x = - 1 is a point of local maximum. Local maximum value = f ( -1) = ( -2) 3( -1 + 1) 2 = 0. 1 Consider x = - × 5 1 then f ¢( x) = ( + )( + )( -) = - ve. 5 1 When x is slightly greater than - then f ¢( x) = ( + )( + )( + ) = + ve. 5 1 Thus, f ¢( x) changes sign from -ve to +ve as x passes through - × 5 1 So, x = - is a point of local minimum. 5 æ 1 ö -3456 Local minimum value = f ç - ÷ = × 3125 è 5ø When x is slightly less than -

SECOND-DERIVATIVE TEST

Working Rule for Finding Extremum Values 1. Find f ¢( x). 2. Solve f ¢( x) = 0. Each value of x so obtained is a candidate for maximum or minimum. Let x = c be one of its points. 3. Find f ¢¢( x) and put x = c to get f ¢¢( c). Now, if f ¢¢( c) < 0 then x = c is a point of local maximum; if f ¢¢( c) > 0 then x = c is a point of local minimum; if f ¢¢( c) = 0 then use the first-derivative test.

Senior Secondary School Mathematics for Class 12 Pg-499

Applications of Derivatives EXAMPLE 5

SOLUTION

499

Find all the points of local maxima and minima and the corresponding maximum and minimum values of the function 3 45 2 f ( x) = - x 4 - 8x 3 x + 105. 4 2 3 45 2 f ( x) = - x 4 - 8x 3 x + 105. 4 2 \ f ¢( x) = -3 x 3 - 24x 2 - 45 x = -3 x( x 2 + 8x + 15). Now, f ¢( x) = 0 Þ - 3 x( x 2 + 8x + 15) = 0 Þ - 3 x( x + 5)( x + 3) = 0 Þ x = 0 or x = -5 or x = -3. Thus, x = 0, x = -5 and x = -3 are the candidates for local maxima or minima. Moreover, f ¢¢( x) = ( -9x 2 - 48x - 45). Case I

When x = 0 We have f ¢¢( 0) = -45 < 0. So, x = 0 is a point of local maximum. And, local maximum value at x = 0 is f ( 0) = 105.

Case II When x = - 5

We have f ¢¢( -5) = -30 < 0. So, x = - 5 is a point of local maximum. And, local maximum value at x = -5 is f ( -5) =

295 × 4

Case III When x = -3

We have f ¢¢( -3) = 18 > 0. So, x = - 3 is a point of local minimum. Local minimum value at x = -3 is f ( -3) =

231 × 4

EXAMPLE 6

Find the local maxima and local minima of the functions: (i) f ( x) = sin 2x , where 0 < x < p p p (ii) f ( x) = (sin 2x - x), where - £ x £ 2 2

SOLUTION

(i) f ( x) = sin 2x Þ f ¢( x) = 2 cos 2x. \ f ¢( x) = 0 Þ 2 cos 2x = 0, i.e., cos 2x = 0 p 3p p 3p or 2x = Þ x= or x = × 2 2 4 4 p 3p are the candidates for local maxima Thus, x = and x = 4 4 or minima. Þ 2x =

Moreover, f ¢( x) = 2 cos 2x Þ f ¢¢( x) = - 4 sin 2x.

Senior Secondary School Mathematics for Class 12 Pg-500

500

Senior Secondary School Mathematics for Class 12 Case I

EXAMPLE 7

When x = ( p/4)

p æ pö We have f ¢¢ ç ÷ = - 4 sin = - 4 < 0. 2 è 4ø p \ x = is a point of local maximum. 4 æ pö Local maximum value = f ç ÷ = 1. è 4ø Case II When x = ( 3 p/4) 3p We have f ¢¢( 3 p/4) = - 4 sin = 4 > 0. 2 3p is a point of local minimum. \ x= 4 æ 3p ö Local minimum value = f ç ÷ = -1. è 4 ø (ii) f ( x) = (sin 2x - x) Þ f ¢( x) = ( 2 cos 2x - 1) and f ¢¢( x) = - 4 sin 2x. 1 \ f ¢( x) = 0 Þ ( 2 cos 2x - 1) = 0 Þ cos 2x = 2 p p -p p Þ 2x = or 2x = Þ x= or x = × 3 3 6 6 -p p Thus, x = and x = are the candidates for local maxima 6 6 or local minima. Case I When x = - ( p/6) p æ pö We have f ¢¢ ç - ÷ = 4 sin = 2 3 > 0. 6 3 ø è -p So, x = is a point of local minimum. 6 The local minimum value æ 3 pö p pö æ pö æ = f ç - ÷ = ç - sin - ÷ = - çç + ÷÷ × 6 3 6 6ø è ø è ø è 2 p Case II When x = 6 æ pö æ pö We have f ¢¢ ç ÷ = - 4 sin ç ÷ = -2 3 < 0. 6 è ø è 3ø p \ x = is a point of local maximum. 6 And, the local maximum value p pö æ 3 pö æ pö æ - ÷÷ × = f ç ÷ = ç sin - ÷ = çç 3 6ø è 2 6ø è 6ø è Find the local maxima and local minima of the functions: p (i) f ( x) = (sin x - cos x), where 0 < x < 2 (ii) f ( x) = ( 2 cos x + x), where 0 < x < p

Senior Secondary School Mathematics for Class 12 Pg-501

Applications of Derivatives SOLUTION

501

(i) f ( x) = (sin x - cos x). \ f ¢( x) = (cos x + sin x) and f ¢¢( x) = ( - sin x + cos x). Now,

f ¢( x) = 0 Þ cos x + sin x = 0 Þ tan x = -1 Þ x =

3p 7p or × 4 4

3p 7p and x = are the candidates for local maxima 4 4 or local minima. Thus, x =

Case I

Case II

When x = ( 3 p/4) 3p 3p ö æ 1 1 ö -2 æ 3p ö æ f ¢¢ ç + cos ÷= ÷ = ç - sin ÷ = ç4 4 ø è 2 2ø 2 è 4 ø è = - 2 < 0. So, x = ( 3 p/4) is a point of local maximum. æ 3p ö Local maximum value = f ç ÷ = 2. è 4 ø When x = (7 p/4) 7p 7p ö æ p pö æ 7p ö æ f ¢¢ ç ÷ = f ¢¢ ç - sin + cos ÷ = ç sin + cos ÷ 4 4 4 4 4ø è ø è ø è

= 2 > 0. So, x = (7 p/4) is a point of local minimum. 7p 7p æ 7p ö Local minimum value = f ç ÷ = sin - cos 4 4 4 è ø p p ö æ = ç - sin - cos ÷ = - 2. 4 4ø è (ii) f ( x) = ( 2 cos x + x). \ f ¢( x) = ( -2 sin x + 1) and f ¢¢( x) = -2 cos x. Now, f ¢( x) = 0 Þ - 2 sin x + 1 = 0 1 p 5p in ]0, p [. Þ sin x = Þ x = or x = 2 6 6 Thus, x = ( p/6) and x = (5 p/6) are the candidates for local maxima or local minima. Case I

When x = ( p/6) p æ 3ö æ pö ÷ = - 3 < 0. We have f ¢¢ ç ÷ = -2 cos = çç -2 ´ 6 è 2 ÷ø è 6ø \ x = ( p/6) is a point of local maximum. pö æ pö æ Local maximum value = f ç ÷ = ç 3 + ÷ × 6ø è 6ø è

Case II

When x = (5 p/6) p 3 5p æ 5p ö We have f ¢¢ ç ÷ = -2 cos = 2 cos = 2 ´ = 2 6 6 è 6ø

3 > 0.

Senior Secondary School Mathematics for Class 12 Pg-502

502

EXAMPLE 8

SOLUTION

Senior Secondary School Mathematics for Class 12

So, x = (5 p/6) is a point of local minimum. ö æ 5p ö æ 5p Local minimum value = f ç ÷ = ç - 3÷× 6 6 ø è ø è Find the points of local maxima or local minima of the function p f ( x) = (sin 4 x + cos4 x) in 0 < x < × 2 f ( x) = sin 4 x + cos4 x Þ f ¢( x) = 4 sin 3 x cos x - 4 cos 3 x sin x = -4 sin x cos x(cos2 x - sin 2 x) = -2 sin 2x cos 2x = - sin 4x. And, f ¢¢( x) = - 4 cos 4x. p Now, f ¢( x) = 0 Þ - sin 4x = 0 Þ 4x = p , i.e., x = × 4 \ x = ( p/4) is a point of local maximum or local minimum. æ pö Now, f ¢¢ ç ÷ = - 4 cos p = 4 > 0. è 4ø \ x = ( p/4) is a point of local minimum. æ pö 1 Local minimum value = f ç ÷ = × è 4ø 2

EXAMPLE 9

Find the local maxima and local minima, and the corresponding local maximum and local minimum values of the following functions: (i) f ( x) = x 1 - x , where x > 0 x (ii) f ( x) = , where 1 < x < 4 ( x - 1)( x - 4)

SOLUTION

(i) f ( x) = x 1 - x . ( 2 - 3 x) ( 3 x - 4) and f ¢¢( x) = × \ f ¢( x) = 2 1-x 4(1 - x) 3/2 2 Now, f ¢( x) = 0 Þ ( 2 - 3 x) = 0 Þ x = ; 3 2 ö æ ç 3 ´ - 4÷ 3/2 3 æ 2ö ø = -3 and, f ¢¢ ç ÷ = è < 0. 3/2 2 è 3ø 2ö æ 4 ç1 - ÷ 3ø è 2 \ x = is a point of local maximum. 3 2 æ 2ö Local maximum value = f ç ÷ = × 3 3 3 è ø x x (ii) f ( x) = = × ( x - 1)( x - 4) ( x 2 - 5 x + 4) ( 4 - x 2) \ f ¢( x) = 2 × ( x - 5 x + 4) 2 And, ( x 2 - 5 x + 4) 2( -2x) - ( 4 - x 2) 2( x 2 - 5 x + 4)( 2x - 5) f ¢¢( x) = × ( x 2 - 5x + 4) 4

Senior Secondary School Mathematics for Class 12 Pg-503

Applications of Derivatives

503

f ¢( x) = 0 Þ ( 4 - x 2) = 0 Þ x = 2 -16 f ¢¢( 2) = = -1 < 0. 16 \ x = 2 is a point of local maximum. Local maximum value = f ( 2) = - 1.

[Q 1 < x < 4].

x

EXAMPLE 10

SOLUTION

æ1ö Prove that the maximum value of ç ÷ is e1/e . èxø x æ1ö Let y = ç ÷ = x - x . Then, log y = - x log x. èxø 1 dy dy \ × = - (1 + log x) or = - y(1 + log x). y dx dx d 2y

1 dy æ1ö - (1 + log x) × = -ç ÷ x dx èxø

x +1

dy - (1 + log x) × × dx dx 2 dy Now, = 0 Þ 1 + log x = 0 dx Þ log x = -1 = - log e = log (1/e) Þ x = (1/e). And,

= -y ×

d 2y

dy 1 +(1/e ) at x = (1/e) is - e 0. So, y = 2 is a point of minimum. y2 4 Now, y = 2 Þ x = = = 2. So, the required point is ( 2, 2). 2 2 Show that none of the following functions has a maxima or minima. (ii) log x (iii) x 3 + x 2 + x + 1 (i) ex x x (i) f ( x) = e Þ f ¢( x) = e . For a maxima or a minima, we must have f ¢( x) = 0. But, f ¢( x) = 0 Þ ex = 0, which is not possible. So, f ( x) = ex does not have a maxima or a minima. 2

EXAMPLE 12 SOLUTION

2

2

Senior Secondary School Mathematics for Class 12 Pg-504

504

Senior Secondary School Mathematics for Class 12

1 × x For a maxima or a minima, we must have f ¢( x) = 0. 1 But, f ¢( x) = 0 Þ = 0, which is not possible for any value of x. x \ f ( x) = log x does not have a maxima or a minima. (iii) f ( x) = x 3 + x 2 + x + 1 Þ f ¢( x) = 3 x 2 + 2x + 1. For a maxima or a minima, we must have f ¢( x) = 0. Now, f ¢( x) = 0 Þ 3 x 2 + 2x + 1 = 0 -2 ± 4 - 12 , which are imaginary. Þ x= 6 Thus, f ¢( x) ¹ 0 for any real value of x. Hence, f ( x) does not have a maxima or a minima. (ii) f ( x) = log x Þ f ¢( x) =

EXAMPLE 13

Show that sin p q cosq q attains a maximum when q = tan -1

SOLUTION

Let y = sin p q cosq q.

p × q

dy = p sin p -1 q cosq +1 q - q cosq -1 q sin p +1 q dq = (sin p -1 q cosq -1 q)( p cos2 q - q sin 2 q). dy Now, for maxima or minima, we have = 0. dq dy But = 0 Þ sin p -1 q = 0 or cosq -1 q = 0 or p cos2 q - q sin 2 q = 0 dq p p Þ q = 0 or q = or q = tan -1 × 2 q Then,

Moreover, we may write dy y( p cos2 q - q sin 2 q) = = y( p cot q - q tan q). dq sin q cos q \

d 2y dq2

= y( - p cosec2 q - q sec2 q) + ( p cot q - q tan q) ×

Thus, at q = tan -1

dy × dq

dy p = 0 and therefore, , we have dq q

é d 2y -1 p ù p q 2 2 ê 2 at q = tan ú = sin q cos q( - p cosec q - q sec q) < 0. q q d ë û p Hence, y is maximum when q = tan -1 × q MAXIMA AND MINIMA ON A CLOSED INTERVAL Let f ( x) be a given function defined on [a , b]. Let the local minimum value of f ( x) be m, and let the local maximum value of f ( x) be M.

Senior Secondary School Mathematics for Class 12 Pg-505

Applications of Derivatives

505

Then, minimum value of f ( x) on [a , b] is the smallest of m , f ( a) and f ( b). The maximum value of f ( x) on [a , b] is the greatest of M , f ( a) and f ( b). EXAMPLE 14

Find the maximum and minimum values of ( 3 x 4 - 8x 3 + 12x 2 - 48x + 25) on [0, 3].

SOLUTION

Let

f ( x) = 3 x 4 - 8x 3 + 12x 2 - 48x + 25.

Then, f ¢( x) = 12x 3 - 24x 2 + 24x - 48 and

f ¢¢( x) = 36x 2 - 48x + 24.

Now, f ¢( x) = 0 Þ 12( x 3 - 2x 2 + 2x - 4) = 0 Þ 12( x - 2)( x 2 + 2) = 0 Þ x = 2 [neglecting the imaginary values]. And, f ¢¢( 2) = 72 > 0. So, x = 2 is a point of local minima. Now, f ( 2) = -39; f ( 0) = 25 and f ( 3) = 16. \ minimum value of f ( x) on [0, 3] is -39 at x = 2. Maximum value of f ( x) on [0, 3] is 25 at x = 0. EXAMPLE 15 SOLUTION

Find the maximum and minimum values of ( x + sin 2x) on [0, 2p]. Let f ( x) = ( x + sin 2x). Then, f ¢( x) = (1 + 2 cos 2x) and f ¢¢( x) = -4 sin 2x. 1 Now, f ¢( x) = 0 Þ cos 2x = - , where x Î[0, 2p] 2 2p 4p Þ 2x = or 2x = Þ x = ( p/ 3) or x = ( 2p/ 3). 3 3 p æ 2p ö æ pö f ¢¢ ç ÷ = - 4 sin ç ÷ = - 4 sin = -2 3 < 0. 3 è 3ø è 3ø \ x = ( p/ 3) is a point of local maxima. æ p ö 2p + 3 3 fç ÷ = , f ( 0) = 0 and f ( 2p) = 2p . 6 è 3ø \ maximum value of f ( x) is 2p at x = 2p.

Now,

Again, we consider the point x = ( 2p/ 3). 2p ö 4p p æ 2p ö æ f ¢¢ ç ÷ = - 4 sin ç 2 ´ ÷ = - 4 sin = 4 sin = 2 3 > 0. 3ø 3 3 è 3ø è 2p is a point of local minima. So, x= 3 4p 2p p 4p - 3 3 æ 2p ö 2p Now, f ç ÷ = + sin = - sin = ; 3 3 3 3 6 è 3ø f ( 0) = 0 and f ( 2p) = 2p + sin 4p = 2p. \ the minimum value of f ( x) is 0 at x = 0. EXAMPLE 16

Show that sin x(1 + cos x), x Î [0, p] is maximum at x = ( p/ 3).

SOLUTION

Let f ( x) = sin x(1 + cos x).

Senior Secondary School Mathematics for Class 12 Pg-506

506

Senior Secondary School Mathematics for Class 12

Then, f ¢( x) = - sin 2 x + cos x(1 + cos x) = 2 cos2 x + cos x - 1 = ( 2 cos x - 1)(cos x + 1). And, f ¢¢( x) = ( -4 cos x sin x - sin x) = - sin x(1 + 4 cos x). Now, f ¢( x) = 0 Þ ( 2 cos x - 1)(cos x + 1) = 0 1 p Þ cos x = or cos x = -1 Þ x = or x = p. 2 3 pæ p ö -3 3 æ pö But, < 0. f ¢¢ ç ÷ = - sin ç1 + 4 cos ÷ = 3è 3ø 2 è 3ø \ x = ( p/ 3) is a point of local maximum. æ pö 3 3 Now, f ( 0) = 0, f ( p) = 0 and f ç ÷ = × 4 è 3ø p 3 3 \ the maximum value of f ( x) is at x = × 4 3

EXERCISE 11E Find the maximum or minimum values, if any, without using derivatives, of the functions: 1. (5 x - 1) 2 + 4

2. - ( x - 3) 2 + 9

4. sin 2x + 5

5. |sin 4x + 3|

3. -|x + 4| + 6

Find the points of local maxima or local minima and the corresponding local maximum and minimum values of each of the following functions: 6. f ( x) = ( x - 3) 4

7. f ( x) = x 2

8. f ( x) = 2x 3 - 21x 2 + 36x - 20 9. f ( x) = x 3 - 6x 2 + 9x + 15 11. f ( x) = - x 3 + 12x 2 - 5 13. f ( x) = - ( x - 1) 3( x + 1) 2

10. f ( x) = x 4 - 62x 2 + 120x + 9 12. f ( x) = ( x - 1)( x + 2) 2 x 2 14. f ( x) = + , x > 0 2 x

15. Find the maximum and minimum values of 2x 3 - 24x + 107 on the interval [-3 , 3]. 16. Find the maximum and minimum values of 3 x 4 - 8x 3 + 12x 2 - 48x + 1 on the interval [1, 4]. 17. Find the maximum and minimum values of 1 p ö æ f ( x) = ç sin x + cos x ÷ in 0 £ x £ × 2 2 ø è 1/x 1/e is e . 18. Show that the maximum value of x 1ö æ 19. Show that ç x + ÷ has a maximum and minimum, but the maximum value xø è is less than the minimum value.

Senior Secondary School Mathematics for Class 12 Pg-507

Applications of Derivatives

507

20. Find the maximum profit that a company can make, if the profit function is given by p( x) = 41 + 24x - 18x 2. 21. An enemy jet is flying along the curve y = ( x 2 + 2). A soldier is placed at the point ( 3 , 2). Find the nearest point between the soldier and the jet. 22. Find the maximum and minimum values of f ( x) = ( - x + 2 sin x) on [0, 2p].

ANSWERS (EXERCISE 11E)

1. 4. 6. 8. 9.

min. value = 4 2. max. value = 9 3. max. value = 6 max. value = 4, min. value = 6 5. max. value = 4, min. value = 2 local max. value is 0 at x = 3 7. local min. value is 0 at x = 0 local max. value is -3 at x = 1 and local min. value is - 128 at x = 6 local max. value is 19 at x = 1 and local min. value is 15 at x = 3

10. local max. value is 68 at x = 1 and local min. values are -1647 at x = -6 and -316 at x = 5 11. local max. value is 251 at x = 8 and local min. value is -5 at x = 0 12. local max. value is 0 at x = -2 and local min. value is -4 at x = 0 13. local max. value is 0 at each of the points x = 1 and x = -1 and local min. 1 -3456 at x = value is 5 3125 14. local min. value is 2 at x = 2 15. max. value is 139 at x = -2 and min. value is 89 at x = 3 16. max. value is 257 at x = 4 and min. value is -63 at x = 2 17. max. value is

p p 3 1 at x = and min. value is at x = 6 2 4 2

20. 49

21. (1, 3)

æ p 22. max. value is ç - + è 3

p ö æ 5p 3 ÷ at x = and min. value is ç + 3 ø è 3

5p ö 3 ÷ at x = 3 ø

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11E) 4. -1 £ sin 2 x £ 1. 5. -1 £ sin 4 x £ 1 Þ ( -1 + 3 ) £ (sin 4 x + 3 ) £ ( 1 + 3 ). 6. Use the first-derivative test. 21. Find a point ( x , x 2 + 2 ) nearest to (3, 2).

Senior Secondary School Mathematics for Class 12 Pg-508

508

Senior Secondary School Mathematics for Class 12 SOME MORE SOLVED EXAMPLES

EXAMPLE 1

SOLUTION

Amongst all pairs of positive numbers with sum 24, find those whose product is maximum. Let the numbers be x and ( 24 - x). Let

P = x( 24 - x) = ( 24x - x 2).

Then, Now, Thus,

dP d 2P = ( 24 - 2 x) and 2 = -2. dx dx dP = 0 Þ ( 24 - 2x) = 0 Þ x = 12. dx ì d 2P ü = -2 < 0. í 2ý î dx þx = 12

\ x = 12 is a point of maximum. Hence, the required numbers are 12 and 12. EXAMPLE 2

SOLUTION

Amongst all pairs of positive numbers with product 256, find those whose sum is the least. 256 Let the required numbers be x and × x 256 ö d 2 S 512 dS æ 256 ö æ Let S = ç x + = ç1 - 2 ÷ and 2 = 3 × ÷ × Then, dx è x ø è dx x x ø dS For a maxima or minima, we have = 0. dx dS 256 ö æ Now, = 0 Þ ç1 - 2 ÷ = 0 Þ x 2 = 256 or x = 16. dx è x ø Also,

é d 2S ù 512 ö 1 æ =ç ÷ = > 0. ê 2ú 16 16 16 ´ ´ ø 8 è dx ë û x = 16

So, x = 16 is a point of minimum. Hence, the required numbers are 16 and 16. EXAMPLE 3

Find two positive numbers x and y such that ( x + y) = 60 and xy 3 is

SOLUTION

maximum. Let ( x + y) = 60 and let P = xy 3. Then,

P = ( 60 - y) y 3

\

dP = 3 y 2( 60 - y) + y 3( -1) = (180y 2 - 4y 3) = 4y 2( 45 - y) dy

and, Now,

d 2P dy 2

[Q x = ( 60 - y)].

= ( 360y - 12y 2) = 12y( 30 - y).

dP = 0 Þ 4y 2( 45 - y) = 0 Þ y = 0 or y = 45. dy

Senior Secondary School Mathematics for Class 12 Pg-509

Applications of Derivatives

509

Neglecting y = 0, we are left with y = 45. é d 2P ù Now, ê 2 ú = (12 ´ 45)( 30 - 45) = -8100 < 0. ë dy û y = 45 So, y = 45 is a point of maximum. Hence, the numbers are 45 and 15. EXAMPLE 4

SOLUTION

Find two numbers whose sum is 16 and the sum of whose cubes is minimum. Let the numbers be x and (16 - x). Let S = x 3 + (16 - x) 3. dS d 2S = 3 x 2 - 3(16 - x) 2 and = 6x + 6(16 - x) = 96. dx dx 2 dS Now, = 0 Þ 3 x 2 - 3(16 - x) 2 = 0 Þ x = 8. dx é d 2S ù And, ê 2 ú = 96 > 0. ë dx û x = 8 Then,

\ x = 8 is a point of minimum. Hence, the required numbers are 8 and 8. EXAMPLE 5

SOLUTION

Show that of all the rectangles with a given perimeter, the square has the largest area. Let a be the fixed perimeter. Consider a rectangle with sides x and y and perimeter a. Let A be the area of the rectangle. Then, 2x + 2y = a … (i) æ a - 2x ö [using (i)]. And, A = xy = x ç ÷ è 2 ø \ Now, Also,

dA æ 1 d 2A ö 1 = -2. = ç a - 2x ÷ = ( a - 4x) and dx è 2 ø 2 dx 2 dA 1 a = 0 Þ ( a - 4x) = 0 Þ x = × dx 2 4 é d 2A ù = -2 < 0. ê 2ú ë dx û x = ( a/4)

\

x = ( a/4) is a point of maximum. 1 1æ aö 1 a Also, from (i), we have y = ( a - 2x) = ç a - ÷ = a , when x = × 2 2è 2ø 4 4 Thus, x = y [each = ( a/4)]. Hence, the rectangle is a square.

Senior Secondary School Mathematics for Class 12 Pg-510

510 EXAMPLE 6

SOLUTION

Senior Secondary School Mathematics for Class 12

Show that of all the rectangles of a given area, the square has the smallest [CBSE 2011] perimeter. Let A be the fixed area. Consider a rectangle with sides x and y, and area A. Let P be the perimeter of this rectangle. A … (i) Then, A = xy Þ y = x 2A And, [using (i)]. P = 2x + 2y = 2x + x 2A ö dP æ 2A ö d 2 P 4A æ = ç 2 - 2 ÷ and P = ç 2x + = × ÷ Þ x ø dx è è x ø dx 2 x 3 dP 2A = 0 Þ 2- 2 = 0 Þ x = A dx x é d 2P ù 4A 4 = 3/2 = > 0. ê 2ú A dx A ë ûx = A

Now, Now, and,

So, x = A is a point of minimum. A A Moreover, x = A Þ y = = = A = x. x A So, when the perimeter is smallest, the rectangle is a square. EXAMPLE 7

SOLUTION

Prove that the area of a right-angled triangle of a given hypotenuse is maximum when the triangle is isosceles. [CBSE 2012C] Let h be the hypotenuse of the triangle and x be its altitude. Then, the base of the triangle = h2 - x 2 . 1 Now, A = x h2 - x 2 2 dA 1 ì 1 2 ü \ = íx × ( h - x 2) -1/2( -2x) + h2 - x 2 ý dx 2 î 2 þ =

and,

d 2A dx 2

1 æç h2 - 2x 2 ö÷ 2 çè h2 - x 2 ÷ø 1 ( h2 - x 2 )( - 4x) - ( h2 - 2x 2) × ( h2 - x 2) -1/2( -2x) 2 = 2( h2 - x 2) =

( 2x 3 - 3 xh2) 2( h2 - x 2) 3/2

×

dA h = 0 Þ ( h2 - 2x 2) = 0 Þ x = × dx 2 é d 2A ù \ ê 2ú = -2 < 0. ë dx û x = h Now,

2

Senior Secondary School Mathematics for Class 12 Pg-511

Applications of Derivatives

511

Thus, A is maximum at x = ( h/ 2). æ h2 ö h = altitude. \ base = h2 - x 2 = ç h2 - ÷ = ç ÷ 2 2 è ø Hence, the triangle is isosceles. EXAMPLE 8

SOLUTION

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is ( p/ 3). [CBSE 2009] Let us consider a right-angled triangle with base = x and hypotenuse = y. Let x + y = k , where k is a constant. Let q be the angle between the base and the hypotenuse. Let A be the area of the triangle. Then, A 1 1 2 2 A = ´ BC ´ AC = x y - x . y 2 2 2 2 2 2 2 2 ( ) [( ) ] x y x x k x x = [Q y = ( k - x)] \ A2 = 4 4 x C B k 2 x 2 - 2kx 3 2 or A = … (i) 4 dA 2k 2 x - 6kx 2 … (ii) On differentiating (i), we get 2A × = dx 4 dA k 2 x - 3 kx 2 or = × dx 4A dA k [neglecting x = 0]. Now, = 0 Þ ( k 2 x - 3 kx 2) = 0 Þ x = dx 3 Now, differentiating (ii), we get 2

EXAMPLE 9

SOLUTION

d 2 A 2k 2 - 12kx æ dA ö … (iii) 2ç ÷ + 2A × 2 = 4 è dx ø dx k dA d 2A -k 2 Putting = < 0. = 0 and x = in (iii), we get 4A 3 dx dx 2 Thus, A is maximum when x = (k/3). k k ö 2k æ Now, x = × Þ y = çk - ÷ = 3 3ø 3 è x ( k/ 3) 1 p \ = cos q Þ cos q = = Þ q= × y ( 2k/ 3) 2 3 p Hence, the area is maximum when q = × 3 Two sides of a triangle are given. Find the angle between them such that the area is maximum. Let a and b be the lengths of the given sides and let q be the angle between them. Let A be the area of the triangle. 1 Then, A = ab sin q. 2

Senior Secondary School Mathematics for Class 12 Pg-512

512

Senior Secondary School Mathematics for Class 12

dA 1 d 2A 1 = ab cos q and = - ab sin q. dq 2 2 dq2 dA 1 p Now, = 0 Þ ab cos q = 0 Þ cos q = 0 Þ q = × dq 2 2 A é d 2A ù 1 p 1 And, ê 2 ú = - ab sin = - ab < 0. b 2 2 2 d q ë û q = (p /2)

\

p a is a point of maximum. B C 2 Hence, the area of the triangle is maximum when the angle between the given sides is ( p/2).

\

EXAMPLE 10

SOLUTION

q=

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area. [CBSE 2008, ’13C] Let ABCD be a rectangle inscribed in a given circle with centre O and radius r. Let ÐCAB = q. Then, AC = 2r , AB = 2r cos q and BC = 2r sin q. Let A be the area of rectangle ABCD. Then, A = AB ´ BC = 4r 2 sin q cos q = 2r 2 sin 2q. Thus, A = 2r 2 sin 2q, where r is constant. dA d 2A = 4r 2 cos 2q and = -8r 2 sin 2q. dq dq2 p dA p Now, = 0 Þ 4r 2 cos 2q = 0 Þ cos 2q = 0 Þ 2q = , i.e., q = × 4 dq 2 é d 2A ù p And, ê 2 ú = -8r 2 sin = -8r 2 < 0. 2 q d ë û q = (p /4)

\

\ q = ( p/4) is a point of maximum. Thus, area is maximum when q = ( p/4). p In this case, AB = 2r cos = r 2 4 p and, BC = 2r sin = r 2. 4 Thus, AB = BC and therefore, ABCD is a square. EXAMPLE 11

SOLUTION

Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle. Let ABC be a triangle inscribed in a given circle with centre O and radius r. For maximum area, the vertex A should be at a maximum distance from the base BC.

Senior Secondary School Mathematics for Class 12 Pg-513

Applications of Derivatives

513

Therefore, A must lie on the diameter, perpendicular to BC. Thus, AD ^ BC. So, triangle ABC must be isosceles. Let ÐCAD = q. Now, BC = 2CD = 2OC sin 2q = 2r sin 2q and, AD = (OA + OD) = (r + r cos 2q). Let A be the area of the triangle. 1 Then, A = BC ´ AD = r 2 sin 2q (1 + cos 2q). 2 dA 2 \ = r [sin 2q ( -2 sin 2q) + (1 + cos 2q) × 2 cos 2q] dq = r 2[2(cos2 2q - sin 2 2q) + 2 cos 2q] = 2r 2 [cos 4q + cos 2q]. d 2A

= 2r 2[- 4 sin 4q - 2 sin 2q] = - 4r 2( 2 sin 4q + sin 2q). dq2 dA Now, = 0 Þ cos 4q + cos 2q = 0 dq Þ cos 4q = - cos 2q = cos( p - 2q) p Þ 4q = p - 2q Þ q = × 6 é d 2A ù pö 2p æ And, = - 4r 2 ç 2 sin + sin ÷ = - 6r 2 3 < 0. ê 2ú 3 3ø è q d ë û q = (p /6) And,

p is a point of maximum. 6 So, in this case, each angle of the triangle is ( p/ 3).

\ q=

Hence, ABC is an equilateral triangle. EXAMPLE 12

SOLUTION

The combined resistance R of two resistors R1 and R 2 where R1 , R 2 > 0 1 1 1 is given by = + × R R1 R 2 If R1 + R 2 = C (constant), show that the maximum resistance R is obtained by choosing R1 = R 2. R1 R 2 and R1 + R 2 = C. We have R = ( R1 + R 2) R1(C - R1) C

\

R=

So,

dR C - 2R1 d 2R 2 and = =- × 2 dR1 C C dR1

[Q R 2 = (C - R1)].

For a maxima or minima, we have

dR = 0. dR1

Senior Secondary School Mathematics for Class 12 Pg-514

514

Senior Secondary School Mathematics for Class 12

Now,

dR C - 2R1 C =0 Þ = 0 or R1 = × 2 dR1 C

And,

é d 2R ù -2 = < 0. ê 2ú dR ë 1 û R1 = ( C/ 2) C

C is a point of maximum. 2 C Cö C æ When, R1 = , we have R 2 = çC - ÷ = × Hence, R1 = R 2. 2ø 2 2 è

Thus, R1 =

EXAMPLE 13

SOLUTION

A beam of length l is supported at one end. If W is the uniform load per unit length, the bending moment M at a distance x from the end is given æ lx Wx 2 ö ÷ × Find the point on the beam at which the bending by M = ç ç2 2 ÷ø è moment has the maximum value. 2 æ lx Wx 2 ö ÷ Þ dM = æç l - Wx ö÷ and d M = -W. M=ç ç2 dx è 2 2 ÷ø ø dx 2 è dM For a maxima or minima, we have = 0. dx dM l l Now, =0 Þ - Wx = 0 Þ x = × dx 2 2W

Also,

d 2M dx 2

= -W < 0 for all values of x.

l is a point of maxima. 2W So, the required point is at a distance of ( l/2W) from the supporting end.

\

EXAMPLE 14

SOLUTION

x=

A wire of length 25 m is to be cut into two pieces. One of the wires is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum? [CBSE 1996] Let x metres and ( 25 - x) metres be the required lengths. Let a be the side of the square formed and r be the radius of the circle formed. Then, x 25 - x and r = × 4 a = x and 2pr = 25 - x or a = 4 2p æ x2 ö \ area of the square = ç ÷ sq metres. ç 16 ÷ è ø 2

2

( 25 - x) æ 25 - x ö Area of the circle = p ç × ÷ = 4p è 2p ø

Senior Secondary School Mathematics for Class 12 Pg-515

Applications of Derivatives

Now, the combined area A =

515

x 2 ( 25 - x) 2 + × 16 4p

dA x ( 25 - x) ( p + 4) x - 100 d 2 A ( p + 4) = = , and = × dx 8 2p 8p 8p dx 2 dA For a maxima or minima, we have = 0. dx dA 100 Now, = 0 Þ ( p + 4) x - 100 = 0 Þ x = × dx ( p + 4)

\

And, \

d 2A

( p + 4) > 0 for all values of x. 8p 100 is a point of minimum. x= ( p + 4) dx 2

=

æ 100 ö æ 25 p ö \ lengths of the pieces are ç ÷ metres and ç ÷ metres. p + 4 ø è è p + 4ø Some Examples Related to Volume and Area of Solids EXAMPLE 15

SOLUTION

Show that a cylinder of a given volume which is open at the top has minimum total surface area, provided its height is equal to the radius of its base. [CBSE 2009C, ’14] Let r be the radius and h the height of the cylinder of given volume V. V Then, V = pr 2 h Þ h = 2 … (i) pr Let the total surface area be S. Then, V ö æ [using (i)]. S = pr 2 + 2prh = ç pr 2 + 2pr × 2 ÷ è pr ø 2V ö æ S = ç pr 2 + \ ÷× r ø è dS æ d 2S æ 2V ö 4V ö = ç 2pr - 2 ÷ and = ç 2p + 3 ÷ × dr è r ø dr 2 è r ø dS For a maxima or minima, we have = 0. dr dS 2V Now, = 0 Þ 2pr - 2 = 0 Þ 2pr 3 = 2V = 2pr 2 h Þ r = h. dr r æ d 2S ö 4V ö æ ç ÷ = ç 2p + 3 ÷ > 0. ç dr 2 ÷ h ø è ø r= h è So,

Thus, r = h is a point of minimum. This shows that the total surface area of the cylinder is minimum when h = r , i.e., when the height of the cylinder is equal to the radius of the base.

Senior Secondary School Mathematics for Class 12 Pg-516

516 EXAMPLE 16

Senior Secondary School Mathematics for Class 12

Show that the height of a cylinder which is open at the top, having a given surface and the greatest volume, is equal to the radius of its base. [CBSE 2004, 2010]

SOLUTION

Let r be the radius, h the height and S the given surface area of an æ S - pr 2 ö ÷ … (i) open cylinder. Then, S = ( pr 2 + 2prh) or h = ç ç 2pr ÷ è ø Let V be the volume of the cylinder. æ S - pr 2 ö æ rS - pr 3 ö ÷=ç ÷ [using (i)]. Then, V = pr 2 h = pr 2 × ç ç 2pr ÷ ç ÷ 2 è ø è ø dV æç S - 3 pr 2 ö÷ d 2V \ = and = -3 pr. ç ÷ dr è 2 dr 2 ø dV Now, = 0 Þ S - 3 pr 2 = 0 Þ pr 2 + 2prh - 3 pr 2 = 0 Þ h = r. dr æ d 2V ö And, ç 2 ÷ = -3 ph < 0. Thus, r = h is a point of maximum. ç dr ÷ è ør= h So, V is maximum when r = h, i.e., when the height of the cylinder is equal to the radius of its base.

EXAMPLE 17

SOLUTION

Show that the semivertical angle of a cone of maximum volume and of 1 given slant height is cos-1 × [CBSE 2008C, ’14] 3 O Let a be the semivertical angle of a cone of given slant height l. Then, height of the cone = l cos a , radius of the base = l sin a . Let V be the volume of the cone. Then, 1 1 V = pr 2 h = p( l 2 sin 2 a )( l cos a ) 3 3 C B A 1 or V = pl 3 sin 2 a cos a . 3 dV 1 3 1 \ = pl ( - sin 3 a + 2 sin a cos2 a ) = pl 3 sin a ( 2 cos2 a - sin 2 a). da 3 3 d 2V 1 3 2 2 And, = pl cos a ( 2 cos a - sin a ) da 2 3 1 + pl 3 sin a ( - 4 sin a cos a - 2 sin a cos a ) 3 1 = pl 3 cos 3 a ( 2 - 7 tan 2 a ). 3 dV Now, for a maxima or minima, we have = 0. da dV 1 3 2 2 Now, =0 Þ pl sin a ( 2 cos a - sin a ) = 0 da 3 Þ sin a = 0 or 2 cos2 a - sin 2 a = 0

Senior Secondary School Mathematics for Class 12 Pg-517

Applications of Derivatives

517

Þ a = 0 or a = tan -1 2. Neglecting a = 0, we have a = tan -1 2 , i.e., tan a = 2. Now,

tan a = 2 Þ sec2 a = 3 Þ cos a =

\

é d 2V ù ê 2ú ë da û a = tan -1

1 1 Þ a = cos-1 3 3

3

= 2

1 3 æ 1 ö 4pl 3 < 0. pl × ç ÷ ( 2 - 14) = 3 3 3 è 3ø

Thus, V is maximum when a = cos-1 semivertical angle of the cone is cos-1

1 × 3

1 , i.e., when the 3

EXAMPLE 18

Show that the semivertical angle of a right circular cone of given surface area and maximum volume is sin -1(1/3).

SOLUTION

Let r be the radius, l the slant height and h the height of the cone. Let S denote the surface area and V the volume of the cone. Then, \ Now, \

S = ( pr 2 + prl ) = constant. æ S ö l = ç -r÷ è pr ø V= V2 =

… (i)

1 2 1 pr h = pr 2 l 2 - r 2 . 3 3 2 1 2 4 2 2 1 ïìæ S ïü ö p r ( l - r ) = p 2r 4 × íç - r ÷ - r 2 ý [using (i)] r 9 9 p ø îïè þï

1 S( Sr 2 - 2pr 4). 9 2p 4 ö æ1 Thus, V 2 = ç S2r 2 Sr ÷ 9 è9 ø dV æ 2 2 8pS 3 ö 2rS … (ii) =ç Srr ÷= ( S - 4pr 2) \ 2V × dr è 9 9 9 ø dV S (neglecting r = 0). \ = 0 Þ r = 0 or ( S - 4pr 2) = 0 Þ r 2 = dr 4p =

2

d 2V 1 æ dV ö 2 On differentiating (ii), we get 2 ç ÷ + 2V × 2 = S( 2S - 24pr ). 9 è dr ø dr dV S Putting , we get = 0 and r 2 = dr 4p 2V ×

d 2V 2

=

1 4 S( 2S - 6S) = - S2 < 0. 9 9

dr \ when the volume is maximum, we have S ( pr 2 + prl) = Þ l = 3r. r2 = 4p 4p

Senior Secondary School Mathematics for Class 12 Pg-518

518

Senior Secondary School Mathematics for Class 12

Now, if a is the semivertical angle of the cone then 1 r r æ1ö = sin a Þ = sin a Þ sin a = Þ a = sin -1 ç ÷ × 3r 3 l è 3ø Hence, the semivertical angle of a right cone of a given surface and maximum volume is sin -1(1/ 3). EXAMPLE 19

SOLUTION

EXAMPLE 20

SOLUTION

Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube. [CBSE 2005] Let V be the fixed volume of a closed cuboid with length a, breadth a and height h. Let S be its surface area. V … (i) Then, V = ( a ´ a ´ h) or h = 2 a 2V ö æ [using (i)] Now, S = 2( a 2 + ah + ah) = 2( a 2 + 2ah) = 2 ç a 2 + ÷ a ø è 2V ö dS d 2S æ 2V ö 8V ö æ æ i.e., S = 2 ç a2 + = 2 ç 2a - 2 ÷ and 2 = ç 4 + 3 ÷ × ÷× \ a ø da è è è a ø da a ø dS 3 3 Now, = 0 Þ V = a Þ a ´ a ´ h = a Þ h = a. da Now, when h = a , we have V = a 3. é d 2S ù æ 8a 3 ö \ = ç 4 + 3 ÷ = 12 > 0. ê 2ú ç a ÷ø ë da û h = a è So, S is minimum when length = a, breadth = a and height = a, i.e., when it is a cube. Show that the height of a closed cylinder of given surface and maximum volume is equal to the diameter of its base. Let S be the fixed surface area of a closed cylinder of radius r and height h. Let V be its volume. æ S - 2pr 2 ö ÷ Then, … (i) S = ( 2pr 2 + 2prh) Þ h = ç ç 2pr ÷ è ø æ S - 2pr 2 ö æ rS - 2pr 3 ö ÷=ç ÷ [using (i)]. And, V = pr 2 h = pr 2 ç ç 2pr ÷ ç ÷ 2 è ø è ø dV æç S - 6pr 2 ö÷ d 2V Þ = and = -6pr. ç ÷ dr è 2 dr 2 ø dV Now, = 0 Þ ( S - 6pr 2) = 0 Þ S = 6pr 2 dr 1 Þ ( 2pr 2 + 2prh) = 6pr 2 Þ h = 2r , i.e., r = h. 2 é d 2V ù 1 And, ê 2 ú = -6p ´ h = -3 ph < 0. 2 ë dr û r = ( h/ 2 )

So, V is maximum when h = 2r.

Senior Secondary School Mathematics for Class 12 Pg-519

Applications of Derivatives

519

EXAMPLE 21

A closed right circular cylinder has a volume of 2156 cubic units. What should be the radius of its base so that its total surface area may be maximum?

SOLUTION

Let r be the radius and h the height of the cylinder. æ 2156 ö Then, pr 2 h = 2156 Þ h = ç 2 ÷ × Let S be its total surface area. è pr ø 2156 ö æ 4312 ö æ Then, S = ( 2pr 2 + 2prh) = ç 2pr 2 + 2pr ´ 2 ÷ = ç 2pr 2 + ÷× r ø è pr ø è dS æ d 2S æ 4312 ö 8624 ö = ç 4pr - 2 ÷ and = ç 4p + 3 ÷ × dr è r ø dr 2 è r ø dS 4312 ö æ Now, = 0 Þ ç 4pr - 2 ÷ = 0 dr è r ø

\

æ 4312 ö Þ 4pr 3 = 4312 Þ r = ç ÷ è 4p ø And,

SOLUTION

æ 4312 ´ 7 ö =ç ÷ è 4 ´ 22 ø

1/ 3

= 7.

é d 2S ù 8624 ö æ = ç 4p + ÷ > 0. \ S is maximum when r = 7. ê 2ú 343 ø è dr ë ûr= 7

Show that the height of the cylinder of maximum volume that can be 2R inscribed in a sphere of radius R is × Find the volume of the largest 3 cylinder inscribed in a sphere of radius R. [CBSE 2006, ‘09C, ‘12C, ’13C] Let r be the radius and h the height of the inscribed cylinder ABCD. Let V be its volume. Then, V = pr 2 h Clearly, AC = 2R. Also,

… (i)

AC 2 = AB 2 + BC 2

C

D

h

2R

EXAMPLE 22

1/ 3

Þ ( 2R) 2 = ( 2r) 2 + h2 1 2r … (ii) Þ r 2 = ( 4R 2 - h2) A B 4 Using (ii) in (i), we get dV æ 2 3 2 ö d 2V ph 3 V = ( 4R 2 - h2) Þ = ç pR - ph ÷ and = - ph. 2 dh è 4 4 2 ø dh For a maxima or minima, we have ( dV/dh) = 0. dV 3 2R Now, = 0 Þ pR 2 - ph2 = 0 Þ h = × dh 4 3 \

é d 2V ù ê 2ú ë dh û h = ( 2R /

=3)

So, V is maximum when h =

3 2R p´ = - pR 3 < 0. 2 3 2R × 3

Senior Secondary School Mathematics for Class 12 Pg-520

520

Senior Secondary School Mathematics for Class 12

2R × 3 1é 4R 2 ù 2R 4pR 3 Largest volume of the cylinder = p ´ ê 4R 2 = × ú´ 4ë 3 û 3 3 3 Show that the cone of greatest volume which can be inscribed in a given sphere is such that three times its altitude is twice the diameter of the sphere. Find the volume of the largest cone inscribed in a sphere of radius R. [CBSE 2010C] Let R be the radius of the given sphere with centre O, and let V be the volume of the inscribed cone, h be its height and r be the radius of the base. In the given figure, we have OD = AD - AO = ( h - R). Hence, the height of the cylinder of maximum volume is

EXAMPLE 23

SOLUTION

R 2 = ( h - R) 2 + r 2 or r 2 = h( 2R - h) 1 1 Now, V = pr 2 h = ph2( 2R - h) [using (i)]. 3 3 dV 1 d 2V æ 4 ö \ = ph( 4R - 3 h), and = ç pR - 2ph÷ × dh 3 ø dh2 è 3 dV For a maxima or minima, we have = 0. dh B dV 1 Now, =0 Þ ph( 4R - 3 h) = 0 dh 3 4 Þ h = 0 or ( 4R - 3 h) = 0 Þ h = R 3 é d 2V ù 4pR And, ê 2 ú =< 0. 3 ë dh û \

… (i) A

O R D C

[Q h ¹ 0].

h = ( 4/ 3 )R

4 R , i.e., when 3 h = 2( 2R) 3 i.e., 3 times the height = 2 times the diameter. 1 16R 2 æ 4R ö 32pR 3 Volume of the largest cone = p ´ × ´ ç 2R ÷= 3 9 3 ø 81 è So, V is maximum when h =

EXAMPLE 24

SOLUTION

Prove that the perimeter of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles. Let h be the hypotenuse and q be the angle between the hypotenuse and the base. Then, base = h cos q and altitude = h sin q. Let P be the perimeter. Then, P = h + h cos q + h sin q. dP \ = - h sin q + h cos q = h(cos q - sin q). dq

Senior Secondary School Mathematics for Class 12 Pg-521

Applications of Derivatives

521

d 2P

= h( - sin q - cos q) = - h(sin q + cos q). dq2 dP Now, = 0 Þ h(cos q - sin q) = 0 dq Þ cos q = sin q, i.e., tan q = 1 Þ q = ( p/4). é d 2P ù p pù é Also, ê 2 ú = - h ê sin + cos ú = - h 2 < 0. 4 4û ë ë dq û q = (p /4)

And,

\ P is maximum when q = ( p/4). p h p h In this case, base = h cos = and altitude = h sin = × 4 4 2 2 \ base = altitude, and hence the triangle is isosceles. EXAMPLE 25

An open box is to be made out of a piece of cardboard measuring (24 cm ´ 24 cm) by cutting off equal squares from the corners and turning up the sides. Find the height of the box when it has maximum volume. [CBSE 2004]

SOLUTION

Let the length of the side of each square cut off from the corners be x cm. x x x x Then, height of the box = x cm. \ V = ( 24 - 2x) 2 ´ x = 4x 3 - 96x 2 + 576 x Þ

dV = 12( x 2 - 16x + 48) dx d 2V

x = 24( x - 8). x 24 – 2x dx 2 dV 2 Now, = 0 Þ x - 16x + 48 = 0, i.e., ( x - 12)( x - 4) = 0 dx

and,

Þ

x

x

x = 4 [Q x ¹ 12]

d 2V ù = -96 < 0. ú dx 2 û x = 4

\ V is maximum at x = 4.

Hence, the volume of the box is maximum when its height is 4 cm.

EXERCISE 11F 1. Find two positive numbers whose product is 49 and the sum is minimum. 2. Find two positive numbers whose sum is 16 and the sum of whose squares is minimum. 3. Divide 15 into two parts such that the square of one number multiplied with the cube of the other number is maximum.

Senior Secondary School Mathematics for Class 12 Pg-522

522

Senior Secondary School Mathematics for Class 12

4. Divide 8 into two positive parts such that the sum of the square of one and the cube of the other is minimum. 5. Divide a into two parts such that the product of the pth power of one part and the qth power of the second part may be maximum. 6. The rate of working of an engine is given by 6000 R = 15v + , where 0 < v < 30 v and v is the speed of the engine. Show that R is the least when v = 20. 7. Find the dimensions of the rectangle of area 96 cm 2 whose perimeter is the least. Also, find the perimeter of the rectangle. 8. Prove that the largest rectangle with a given perimeter is a square. 9. Given the perimeter of a rectangle, show that its diagonal is minimum when it is a square. 10. Show that a rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square of side 2 a. 11. The sum of the perimeters of a square and a circle is given. Show that the sum of their areas is least when the side of the square is equal to the diameter of the circle. 12. Show that the right triangle of maximum area that can be inscribed in a circle is an isosceles triangle. 13. Prove that the perimeter of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles. 14. The perimeter of a triangle is 8 cm. If one of the sides of the triangle be 3 cm, what will be the other two sides for maximum area of the triangle? 15. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 metres. Find the dimensions of the window to admit maximum light through it. [CBSE 2000, ‘02, ‘06]

16. A square piece of tin of side 12 cm is to be made into a box without a lid by cutting a square from each corner and folding up the flaps to form the sides. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume. 17. An open box with a square base is to be made out of a given cardboard of c3 area c2 (square) units. Show that the maximum volume of the box is 6 3 (cubic) units. [CBSE 2001] 18. A cylindrical can is to be made to hold 1 litre of oil. Find the dimensions which will minimize the cost of the metal to make the can. 19. Show that the right circular cone of least curved surface and given volume [CBSE 2007] has an altitude equal to 2 times the radius of the base.

Senior Secondary School Mathematics for Class 12 Pg-523

Applications of Derivatives

523

20. Find the radius of a closed right circular cylinder of volume 100 cm 3 which has the minimum total surface area. 21. Show that the height of a closed cylinder of given volume and the least surface area is equal to its diameter. 22. Prove that the volume of the largest cone that can be inscribed in a sphere 8 of the volume of the sphere. is [CBSE 2004, ‘04C, ‘05C] 27 23. Which fraction exceeds its pth power by the greatest number possible? 24. Find the point on the curve y 2 = 4x which is nearest to the point ( 2, - 8). 25. A right circular cylinder is inscribed in a cone. Show that the curved surface area of the cylinder is maximum when the diameter of the cylinder is equal to the radius of the base of the cone. [CBSE 2006C] 26. Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube. 27. A rectangle is inscribed in a semicircle of radius r with one of its sides on the diameter of the semicircle. Find the dimensions of the rectangle so that its area is maximum. Find also this area. [CBSE 2004] 28. Two sides of a triangle have lengths a and b and the angle between them is q. What value of q will maximize the area of the triangle? [CBSE 2002C] 29. Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius 5 3 cm is (500p) cm 3 . [CBSE 2004, 05C] 30. A square tank of capacity 250 cubic metres has to be dug out. The cost of the land is ` 50 per square metre. The cost of digging increases with the depth and for the whole tank, it is ` ( 400 ´ h2), where h metres is the depth of the tank. What should be the dimensions of the tank so that the cost is minimum? [CBSE 2003] 31. A square piece of tin of side 18 cm is to be made into a box without the top, by cutting a square piece from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find the maximum volume of the box. [CBSE 2003] 32. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width. [CBSE 2003, ‘07C] 33. A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum. [CBSE 2005]

Senior Secondary School Mathematics for Class 12 Pg-524

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Senior Secondary School Mathematics for Class 12

34. Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm. [CBSE 2006C]

ANSWERS (EXERCISE 11F)

1. 7, 7

2. 8, 8

3. 6, 9

4. 6, 2

5.

ap aq , p+q p+q

7. length = 4 6 cm, breadth = 4 6 cm, perimeter = 16 6 cm 20 10 14. b = c = 2.5 cm 15. breadth = m , height = m ( p + 4) ( p + 4) æ 500 ö 16. 2 cm, 128 cm 3 18. r = ç ÷ è p ø æ1ö 23. çç ÷÷ èpø

1/ 3

cm, h =

1000 1/ 3

p

(500)

2/ 3

æ 50 ö 20. ç ÷ èpø

cm

1/ 3

cm

1/( p -1 )

24. ( 4, - 4)

27. length = r 2, breadth =

r 2 , area = r 2 sq units 2

30. Side = 10 cm, depth = 2.5 m 144 3 324 33. cm , cm 4 3+9 4 3+9

28.

p 2

31. 3 cm, 432 cm 3 25 34. sq cm 4

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11F) 7. Let P be the fixed perimeter and x , y be the sides. 96 Then, 96 = xy Þ y = × x 96 ö æ P = 2( x + y ) Þ P = 2 ç x + ÷× x ø è 1 1 P = 2( x + y ) Þ y = ( P - 2 x ). \ A = xy Þ A = x( P - 2 x ). 2 2 1 9. P = 2( x + y ) Þ y = ( P - 2 x ). 2 1 \ z = D 2 = x 2 + y 2 Þ z = x 2 + ( P - 2x)2 . 4 Y Now, D is minimum when z is minimum. 10. Let ABCD be the rectangle inscribed in a circle of radius a and with centre O. Join OC. Let ÐCOX = q. Then, the coordinates of C are ( a cos q, a sin q). \ OM = a cos q and MC = a sin q. BC = 2 MC = 2 a and CD = 2OM = 2 a cos q. \ So,

D X

P = 2 a (cos q + sin q). dP d2 P = 2 a( - sin q + cos q) and 2 = -2 a(cos q + sin q). dq dq

C X

M

O A

B

Y

Senior Secondary School Mathematics for Class 12 Pg-525

Applications of Derivatives

525

d2 P dP p and at this value of q, < 0. =0 Þ q= dq dq 4 So, P is maximum at q = ( p/ 4 ). Then, BC = 2 a = CD. ( S - 4x) × 2p ( S - 4x)2 Let A = x 2 + py 2 . Then, A = x 2 + × 4p

11. Let S = 4 x + 2 py. Then, y =

13. Let ABC be the right triangle with given hypotenuse h. Let, base BC = x and altitude BA = a.

A

Then, h 2 = x 2 + a2 Þ a = h 2 - x 2 . \

P = a + x + h Þ P = ( h 2 - x 2 ) + x + h.

So,

dP = dx

-x h2 - x2

+ 1 and

d2 P dx

2

=

- h2 ( h - x 2 ) 3/2 2

h

a B

×

x

C

dP h =0 Þ x= × dx 2 2 h d P -2 3/2 And for x = , = < 0. h 2 dx 2 h h \ P is maximum when x = and a = , i.e., when x = a. 2 2

Now,

15. Let the length and breadth of the rectangle be x and y metres respectively. Then, radius of the semicircle = ( x/ 2 ) metres. So, the perimeter of the window is given by px æ2+ pö 10 = x + 2 y + Þ y=5- ç ÷ x. 2 è 4 ø 2 ( 2 + p ) ù px 2 1 æxö é y × xú + \ A = xy + p ç ÷ Þ A = x ê5 2 è2ø 4 8 ë û dA æ px ö d2 A pö æ = ç5 - x = - ç 1 + ÷ < 0. ÷ and dx è 4 ø 4ø è dx 2 dA 20 Now, =0 Þ x= × dx ( p + 4) 20 10 and y = × \ A is maximum when x = ( p + 4) ( p + 4)

x

\

16. Let the length cut off from each corner be x cm. Then, length of the box = ( 12 - 2x ) cm = breadth and height = x cm. \ V = ( 12 - 2 x ) 2 ´ x Þ V = 4 x 3 - 48 x 2 + 144 x.

x

x

dV d 2V = 12( x - 6 )( x - 2 ) and = 24( x - 4 ). dx dx 2 x dV Then, = 0 Þ x = 2 [Q x = 6 is not possible]. x dx 2 d V Also at x = 2 , < 0. dx 2 \ V is maximum when x = 2 and maximum V = [( 12 - 4 ) 2 ´ 2] cm 3 .

x

x

\

x x

Senior Secondary School Mathematics for Class 12 Pg-526

526

Senior Secondary School Mathematics for Class 12

17. Let each side of the base be a and height be h. Then, ( c 2 - a2 ) c 2 = ( a2 + 4 ah ) Þ h = × 4a ( c 2 a - a3 ) × 4 2 2 2 d V dV ( c - 3 a ) 3 = and = - a < 0. \ da 4 2 da2 c2 c Thus, V is maximum when a2 = and h = × 3 2 3 1000 [Q V = 1 litre = 1000 cm 3 ]. 18. pr 2 h = 1000 Þ h = pr 2 1000 ö æ \ S = ( 2 pr 2 + 2 prh ) Þ S = 2 ç pr 2 + ÷× r ø è 22. Let the radius of the given sphere be a. Let V be the volume of the inscribed cone, h be its height and r be its radius. 1 Then, V = pr 2 h. A 3 Now, OD = ( AD - OA ) = ( h - a). So, V = a2 ´ h Þ V =

OC 2 = OD 2 + DC 2 Þ a2 = ( h - a) 2 + r 2 Þ r 2 = h ( 2 a - h ).

O

é 2 pah 2 1 3 ù 1 - ph ú × \ V = ph 2 ( 2 a - h ) = ê 3 3 úû êë 3

B

dV æ 4 pah d 2V æ 4 pa ö ö =ç - ph 2 ÷ and =ç - 2 ph ÷ × \ dh è 3 ø ø dh 2 è 3 é d 2V 4a 4a ù 4 pa dV So, < 0. and ê 2 at h = ú = =0 Þ h= 3 3 úû 3 dh êë dh 4a 8 a2 and r 2 = × \ V is maximum when h = 3 9 2 1 8a 4a 8 æ4 ö Maximum volume = p ´ ´ = ´ ç p a3 ÷ × 3 9 3 27 è 3 ø 23. Let the required fraction be x. Let y = ( x - x p ). Then, 1/(p -1 )

æ 1ö dy = 0 Þ x = çç ÷÷ dx èpø

d2 y

×

dx 2

1/(p -1 )

æ 1ö \ y is maximum when x = çç ÷÷ èpø

Þ

y4 + 16 y + 68 16

.

é y2 ù ú êQ x = 4 úû êë

ö d2 D 3 2 dD æç 4 y 3 = + 16 ÷ and = y . ç ÷ dy è 16 4 dy 2 ø

æ p -2 ö ç ÷

æ 1 ö çè p -1 ÷ø at this point = - p ( p - 1) çç ÷÷ < 0. èpø

24. Let the required point be ( x , y ). Then, D = ( x - 2 ) 2 + ( y + 8 ) 2 Þ D = x 2 + y 2 - 4 x + 16 y + 68 Þ D=

C

D

Senior Secondary School Mathematics for Class 12 Pg-527

Applications of Derivatives

Now,

527

dD = 0 Þ y 3 = - 64 Þ y = - 4 . dy

é d2 D ù 3 = ´ 16 = 12 > 0. ê 2ú 4 dy úû y = - 4 êë \ D is minimum when y = - 4. At this value, we have x = 4. Hence, the required point is ( 4 , - 4 ). 25. Let r1 be the radius of the cone and h1 be its height and let r be the radius and h the height of the inscribed cylinder. Clearly, a OAD and a OBC are similar. O r h -h r … (i) \ = 1 Þ r = 1 ( h1 - h ) r1 h1 h1 Let S be the curved surface area of the cylinder. D A 2 pr1 Then, S = 2 prh Þ S = ( h1 - h )h h1 Þ

dS 2 pr1 d 2 S - 4 pr1 ( h1 - 2 h ) and = = < 0. h1 dh h1 dh 2

B

C

dS d2 S = 0 Û h1 = 2 h. Also, 2 < 0. dh dh So, S is maximum when h1 = 2 h. 2h - h 1 r = = Û r1 = 2 r = diameter of the cylinder. \ 2h 2 r1

Now,

26. Let V be the volume, x the side of the square base and h the height of the cuboid. V Then, V = x 2 h Þ h = 2 … (i) x 2 2 \ S = 2( x + xh + xh ) Þ S = 2 x + 4 xh. 4V [using (i)] \ S = 2x 2 + x 8V dS 4V d2 S Þ = 4 x - 2 and 2 = 4 + 3 × dx x dx x dS Now, = 0 Þ x 3 = V. dx 8V d2 S For this value of x, 2 = 4 + = 4 + 8 = 12 > 0. V dx \ surface area is minimum when x 3 = V. Also, h =

V

=

x3

= x. x x2 \ S is minimum when the cuboid is a cube. 27. Let ABCD be the rectangle of length 2x and breadth y, inscribed in a semicircle of radius r and centre O. Let ÐBOC = q. y x Then, = sin q and = cos q. D C r r r \ area of the rectangle is given by y A = 2 xy = r 2 sin 2 q 2 dA d A O x B A = - 4r 2 sin 2 q. Þ = 2r 2 cos 2 q and dq dq 2 2

Senior Secondary School Mathematics for Class 12 Pg-528

528

Senior Secondary School Mathematics for Class 12

dA p p = 0 Þ cos 2 q = 0 Þ 2 q = Þ q = × dq 2 4 d2 A 2 For this value of q, we have = - 4r < 0. dq 2 p p r 2 × \ area is maximum when 2 x = 2r cos = r 2 and y = r sin = 4 4 2 r 2 ´r 2 Maximum area = 2 xy = = r 2 sq units. 2 Now,

28. Let A be the area of a PQR. 1 Then, A = ab sin q 2 dA 1 Þ = ab cos q dq 2 Þ

P b

d2 A

1 = - ab sin q < 0. 2 dq 2

For maxima or minima, we have dA 1 p = 0 Û ab cos q = 0 Û cos q = 0 Û q = × dq 2 2 p \ A is maximum, when q = × 2

Q

a

R

30. Let the side of the square tank be x metres. Then, total cost is given by C = 50 x 2 + 400 h 2 250 Also, x 2 h = 250 Þ h = 2 x 400 ´ 62500 2 \ C = 50 x + x4 Þ

dC = ( 100 x - 10 8 × x -5 ) and dx

Now,

… (i) … (ii)

æ 10 8 ´ 5 ö÷ = ç 100 + × ç dx x 6 ÷ø è

d 2C 2

é d 2C ù dC = 600 > 0. = 0 Û x = 10 and ê 2 ú dx êë dx úû (x = 10)

æ 250 ö \ cost is minimum when x = 10 and h = ç ÷ m = 2.5 m. è 100 ø Hence, for minimum cost, the tank must have a square base of side 10 m and depth 2.5 m. 31. Let the side of the square piece cut from each corner of the given square plate of side 18 cm be x cm. Then, the dimensions of the open box are ( 18 - 2 x ) cm, ( 18 - 2 x ) cm and x cm. \ V = ( 18 - 2 x ) 2 ´ x Þ Þ

V = 4 x 3 - 72 x 2 + 324 x dV = 12( x 2 - 12 x + 27 ) and dx

… (i) 2

d V dx 2

= 12( 2 x - 12 ).

Senior Secondary School Mathematics for Class 12 Pg-529

Applications of Derivatives

529

dV = 0 Û x 2 - 12 x + 27 = 0 Û ( x - 3 )( x - 9 ) = 0 Û x = 3 [Q x ¹ 9] dx é d 2V ù and ê 2 ú = 12( 2 ´ 3 - 12 ) = - 72 < 0. êë dx úû (x = 3 ) Now,

\

V is maximum at x = 3 cm, and maximum volume = [4 ´ 3 3 - 72 ´ 3 2 + 324 ´ 3] cm 3 = 432 cm 3 .

32. Let x be the side of the square base and h be the depth. Then, V = x 2 h. Let the cost per square metre be ` p. Then, V ö 4V ö æ æ C = ( x 2 + 4 xh ) p Þ C = ç x 2 + 4 x ´ 2 ÷ p Þ C = ç x 2 + ÷ p. x ø è è x ø \

dC æ 4V ö = ç 2 x - 2 ÷ p and dx è x ø

Now,

d 2C dx 2

8V ö æ = ç 2 + 3 ÷ p. è x ø

dC 4V ö æ = 0 Û ç 2 x - 2 ÷ = 0 Û x = ( 2V )1/3 dx è x ø

é d 2C ù 8V ö æ And, ê 2 ú = pç2 + ÷ = 6 p > 0. 2V ø è êë dx úû x = (2V )1/3 \ C is minimum, when x = ( 2V )1/3 . Then, h =

V x2

=

1 1 ( 2V )1/3 = x. 2 2

Hence, for minimum cost, the depth of the tank should be equal to half of the side of its square base. 33. Let the perimeter of the square be x cm. Then the perimeter of the triangle is ( 36 - x ) cm. x \ side of the square = cm. 4 1 And, side of the triangle = ( 36 - x ) cm. 3 2 ö x2 xö x2 x2 3æ 3 æç \ A= + 144 + + - 8 x ÷÷ ç 12 - ÷ = 16 4 è 3ø 16 4 çè 9 ø æ 3 ö 2 1 ÷ x - 2 3 x + 36 3 + Þ A = çç ÷ è 36 16 ø dA ( 4 3 + 9 ) d2 A 4 3 + 9 = >0 = ´ 2 x - 2 3 and Þ 72 dx 144 dx 2 dA 144 3 cm. \ =0 Û x= dx ( 4 3 + 9) 34. Consider a right-angled a ABC in which hyp. AC = 5 cm and let ÐBAC = q. Then, AB = 5 cos q and BC = 5 sin q. 1 1 \ area, A = AB ´ BC = ´ 5 cos q ´ 5 sin q 2 2 25 Þ A= sin 2 q 4

Senior Secondary School Mathematics for Class 12 Pg-530

530

Senior Secondary School Mathematics for Class 12 dA 25 d2 A = cos 2 q and = - 25 sin 2 q. dq 2 dq 2 dA p p Now, = 0 Þ cos 2 q = 0 Þ 2 q = Þ q= × dq 2 4 é d2 A ù æpö = - 25 sin ç ÷ = - 25 < 0. ê 2ú p è2ø êë dq úû

Þ

q=

4

p \ q = is a point of maxima. 4 25 p ö 25 æ Maximum area = sq cm. ´ sin ç 2 ´ ÷ = 4 4ø 4 è

5. Increasing and Decreasing Functions INCREASING FUNCTION

or

A function f(x) defined on ]a , b[ is said to be increasing if

x1 < x 2 Þ

f ( x1) £ f ( x 2) for all x1 , x 2 Î ]a , b[

x1 > x 2 Þ

f ( x1) ³ f ( x 2) for all x1 , x 2 Î ]a , b[.

STRICTLY INCREASING FUNCTION

A function f(x) defined on ]a, b[ is said to be strictly

increasing if

or

x1 < x 2 Þ

f ( x1) < f ( x 2) for all x1 , x 2 Î ]a , b[

x1 > x 2 Þ

f ( x1) > f ( x 2) for all x1 , x 2 Î ]a , b[.

DECREASING FUNCTION

or

A function f(x) defined on ]a, b[ is said to be decreasing if

x1 < x 2 Þ

f ( x1) ³ f ( x 2) for all x1 , x 2 Î ]a , b[

x1 > x 2 Þ

f ( x1) £ f ( x 2) for all x1 , x 2 Î ]a , b[.

STRICTLY DECREASING FUNCTION

A function f(x) defined on ]a, b[ is said to be

strictly decreasing if

or

x1 < x 2 Þ

f ( x1) > f ( x 2) for all x1 , x 2 Î ]a , b[

x1 > x 2 Þ

f ( x1) < f ( x 2) for all x1 , x 2 Î ]a , b[.

EXAMPLE 1 SOLUTION

Show that f ( x) = 3 x + 5 is a strictly increasing function on R. Let x1 , x 2 Î R such that x1 < x 2 . Then, x1 < x 2 Þ 3 x1 < 3 x 2 Þ 3 x1 + 5 < 3 x 2 + 5 Þ f ( x1) < f ( x 2).

Senior Secondary School Mathematics for Class 12 Pg-531

Applications of Derivatives

Thus, x1 < x 2 Þ

531

f ( x1) < f ( x 2) for all x1 , x 2 Î R .

Hence, f ( x) is strictly increasing on R. EXAMPLE 2

Show that the function f ( x) = ex is strictly increasing on R.

SOLUTION

Let x1 , x 2 Î R such that x1 > x 2. Then, x1 > x 2 Þ Þ Thus, x1 > x 2 Þ

e x1 > e x 2

[Q e > 1 and x1 > x 2 Þ ex1 > ex 2 ]

f ( x1) > f ( x 2). f ( x1) > f ( x 2) for all x1 , x 2 Î R.

Hence, f ( x) is strictly increasing on R. EXAMPLE 3

Show that f ( x) = e- x is a strictly decreasing function on R.

SOLUTION

Let x1 , x 2 Î R such that x1 > x 2 . Then, x1 > x 2 Þ Þ Þ Thus, x1 > x 2 Þ

e x1 > e x 2 1 1 < x x1 e e 2 - x1 e < e- x 2

[Q e > 1 and x1 > x 2 Þ ex1 > ex 2 ]

e- x1 < e- x 2 for all x1 , x 2 Î R.

Hence, f ( x) = e- x is strictly decreasing on R. EXAMPLE 4

If a is a real number greater than 1, show that the function f ( x) = a x is strictly increasing on R.

SOLUTION

Let x1 , x 2 Î R such that x1 > x 2. Then, x1 > x 2 Þ Þ Thus, x1 > x 2 Þ

a x1 > a x 2

[Q a > 1 and x1 > x 2 Þ

a x1 > a x 2 ]

f ( x1) > f ( x 2). f ( x1) > f ( x 2) for all x1 , x 2 Î R.

Hence, f ( x) is strictly increasing on R. EXAMPLE 5

If a is a real number such that 0 < a < 1, show that the function f ( x) = a x is strictly decreasing on R.

SOLUTION

Let x1 , x 2 Î R such that x1 < x 2. Then, x1 < x 2 Þ Þ Thus, x1 < x 2 Þ

a x1 > a x 2 [Q 0 < a < 1 and x1 < x 2 Þ a x1 > a x 2 ] f ( x1) > f ( x 2). f ( x1) > f ( x 2) for all x1 , x 2 Î R.

Hence, f ( x) is strictly decreasing on R. THEOREM 1

Let f ( x) be a function continuous on [a , b] and differentiable on ]a , b[. Then, f ¢( x) > 0 for all x Î ]a , b[ Þ f ( x) is increasing on ]a , b[.

Senior Secondary School Mathematics for Class 12 Pg-532

532

Senior Secondary School Mathematics for Class 12

PROOF

Let f ¢( x) > 0 for all x Î]a , b[. Let x1 , x 2 Î ]a , b[ such that x1 < x 2. Then, clearly f ( x) is continuous on [x1 , x 2] and differentiable on ]x1 , x 2[. So, by the mean-value theorem, there exists a real number c Î]x1 , x 2[ f ( x 2) - f ( x1) such that f ¢( c) = × ( x 2 - x1) Since f ¢( x) > 0 for all x Î]a , b[, in particular, f ¢( c) > 0. f ( x 2) - f ( x1) Now, f ¢( c) > 0 Þ >0 ( x 2 - x1) Þ f ( x 2) - f ( x1) > 0 [Q x 2 - x1 > 0 when x1 < x 2] Þ f ( x 2) > f ( x1) Þ f ( x1) < f ( x 2). Thus, x1 < x 2 Þ f ( x1) < f ( x 2). Hence, f ( x) is an increasing function on ]a , b[.

THEOREM 2

PROOF

Let f ( x) be a function, continuous on [a , b] and differentiable on ]a , b[. Then, f ¢( x) < 0 for all x Î ]a , b[ Þ f ( x) is decreasing on ]a , b[.

Let it be given that f ¢( x) < 0 for all x Î]a , b[. Let x1 , x 2 Î ]a , b[ such that x1 < x 2. Then, f ( x) is continuous on [x1 , x 2] and differentiable on ]x1 , x 2[. So, by the mean-value theorem, there exists a real number c Î]x1 , x 2[ f ( x 2) - f ( x1) such that f ¢( c) = × ( x 2 - x1) Since, f ¢( x) < 0 for all x Î]a , b[, in particular, f ¢( c) < 0. f ( x 2) - f ( x1) Now, f ¢( c) < 0 Þ 0 when x1 < x 2] Þ f ( x 2) < f ( x1) Þ f ( x1) > f ( x 2). Thus, x1 < x 2 Þ f ( x1) > f ( x 2). This shows that f ( x) is decreasing on ]a , b[.

SUMMARY

(i) If f ¢( x) > 0 for all x Î] a, b [ then f ( x) is increasing on ] a, b [. (ii) If f ¢( x) < 0 for all x Î] a, b[ then f ( x) is decreasing on ] a, b[.

If f ¢( x) > 0 for all x Î]a , b[, except for a finite number of points where f ¢( x) = 0 then the function is increasing on ]a , b[. Let f ¢( x) > 0 for all x Î]a , b[ except at one point c, where f ¢( c) = 0. Then, f ¢( x) > 0 for all x Î]a , c[ and f ¢( x) > 0 for all x Î]c, b[. Thus, f ( x) is increasing on ]a , c[ as well as on ]c, b[.

COROLLARY 1

PROOF

Senior Secondary School Mathematics for Class 12 Pg-533

Applications of Derivatives

533

Hence, f ( x) is increasing on ]a , b[. The result may now be extended to the case when f ¢( x) = 0 at a finite number of points. If f ¢( x) < 0 for all x Î]a , b[, except for a finite number of points where f ¢( x) = 0, then f ( x) is decreasing on ]a , b[. It is similar to that of Corollary 1.

COROLLARY 2

PROOF

Two Important Results I. (i) If f ¢( x) ³ 0 for all x Î]a , b[ then f ( x) is increasing on [a , b]. (ii) If f ¢( x) > 0 for all x Î]a , b[ then f ( x) is strictly increasing on [a , b]. II. (i) If f ¢( x) £ 0 for all x Î]a , b[ then f ( x) is decreasing on [a , b]. (ii) If f ¢( x) < 0 for all x Î]a , b[ then f ( x) is strictly decreasing on [a , b].

SOLVED EXAMPLES EXAMPLE 1

Show that the function f ( x) = ( x 3 - 6x 2 + 12x - 18) is an increasing function on R.

[CBSE 2002C]

f ( x) = ( x 3 - 6x 2 + 12x - 18)

SOLUTION

Þ

f ¢( x) = 3 x 2 - 12x + 12 = 3( x 2 - 4x + 4) = 3( x - 2) 2 ³ 0 for all x Î R.

Thus, f ¢( x) ³ 0 for all x Î R. Hence, f ( x) is an increasing function on R. EXAMPLE 2

Show that the function f ( x) = ex is strictly increasing on R. f ( x) = ex Þ f ¢( x) = ex .

SOLUTION

Case I

When x > 0 æ ö x2 x 3 In this case, ex = ç1 + x + + + ¼÷ > 1 ç ÷ 2! 3 ! è ø Þ

Case II

ex > 0 for all x > 0

Þ f ¢( x) > 0 for all x > 0. When x = 0 In this case, ex = e0 = 1 > 0 Þ

Case III

f ¢( x) > 0 when x = 0.

When x < 0 Let x = - y , where y is positive.

Senior Secondary School Mathematics for Class 12 Pg-534

534

Senior Secondary School Mathematics for Class 12

Then,

ex = e- y =

1 y

=

1 > 0. (a +ve quantity)

e So, f ¢( x) > 0 when x < 0. Thus, from all the cases, we have f ¢( x) > 0 for all x Î R. Hence, f ( x) is strictly increasing for all x Î R. EXAMPLE 3

Show that f ( x) = e1/x is a strictly decreasing function for all x > 0.

SOLUTION

f ( x) = e1/x Þ f ¢( x) = -

1 x2

e1/x < 0 for all x > 0 [Q x 2 > 0 and e1/x > 0 when x > 0].

Thus, f ¢( x) < 0 for all x > 0. Hence, f ( x) is strictly decreasing for all x > 0. EXAMPLE 4

Show that f ( x) = ( x - 1) ex + 1 is a strictly increasing function for all x > 0.

SOLUTION

f ( x) = ( x - 1) ex + 1 Þ f ¢( x) = ( x - 1) ex + ex Þ f ¢( x) = x ex > 0 for all x > 0 [Q x > 0 and ex > 0 when x > 0]. Thus, f ¢( x) > 0 for all x > 0. Hence, f ( x) is a strictly increasing function for all x > 0.

EXAMPLE 5

Show that the function f ( x) = ( x - sin x) is increasing for all x Î R. Þ

f ( x) = ( x - sin x) f ¢( x) = (1 - cos x) ³ 0 for all x Î R [Q - 1 £ cos x £ 1]

Þ

f ¢( x) ³ 0 for all x Î R.

SOLUTION

Hence, f ( x) = ( x - sin x) is increasing for all x Î R. EXAMPLE 6

ù pé Show that the function f ( x) = cos2 x is strictly decreasing on ú 0, ê × û 2ë f ( x) = cos2 x

SOLUTION

ù pé f ¢( x) = - 2 cos x sin x = - (sin 2x) < 0 in ú 0, ê û 2ë [Q sin 2x > 0 in the 1st quadrant] ù pé Þ f ¢( x) < 0 for all x Î ú 0, ê × û 2ë ù pé 2 Hence, f ( x) = cos x is strictly decreasing on ú 0, ê × û 2ë Þ

EXAMPLE 7

ù pé Show that f ( x) = log sin x is (a) strictly increasing on ú 0, ê and û 2ë ùp é (b) strictly decreasing on ú , p ê × û 2 ë

Senior Secondary School Mathematics for Class 12 Pg-535

Applications of Derivatives SOLUTION

Þ

f ( x) = log sin x 1 f ¢( x) = × cos x Þ f ¢( x) = cot x sin x

535

¼(i)

ù pé (a) We know that for each x Î ú 0, ê , û 2ë f ¢( x) = cot x > 0. ù pé \ f ( x) is strictly increasing in ú 0, ê × û 2ë ùp é (b) We know that for each x Î ú , p ê , û 2 ë f ¢( x) = cot x < 0. ùp é \ f ( x) is strictly decreasing in ú , p ê × û 2 ë EXAMPLE 8

SOLUTION

Show that the function f ( x) = sin x is ù pé (a) strictly increasing on ú 0, ê û 2ë ùp é (b) strictly decreasing on ú , p ê 2 û ë (c) neither increasing nor decreasing on ]0, p [ f ( x) = sin x Þ f ¢( x) = cos x. ù pé (a) We know that for each x Î ú 0, ê , cos x > 0. û 2ë ù pé \ f ¢( x) > 0 for all x Î ú 0, ê × û 2ë ù pé Hence, f ( x) is increasing on ú 0, ê × û 2ë ùp é (b) We know that for each x Î ú , p ê , cos x < 0. û 2 ë ùp é \ f ¢( x) < 0 for all x Î ú , p ê × û 2 ë ùp é Hence, f ( x) is decreasing on ú , p ê × û 2 ë (c) It follows from the above two results that f ( x) = sin x is ù pé ùp é increasing on ú 0, ê and decreasing on ú , p ê × û 2ë û 2 ë So, it is neither increasing nor decreasing on ]0, p[.

EXAMPLE 9

Show that the function f ( x) = ( x 2 - x + 1) is neither increasing nor decreasing on ]0, 1[.

SOLUTION

f ( x) = ( x 2 - x + 1) Þ f ¢( x) = ( 2x - 1).

Senior Secondary School Mathematics for Class 12 Pg-536

536

Senior Secondary School Mathematics for Class 12

Now, f ¢( x) > 0 Û ( 2x - 1) > 0 Û

x>

ù1 é \ f ( x) is increasing when x Î ú , 1 ê × û2 ë And, f ¢( x) < 0 Û ( 2x - 1) < 0 Û

x
0 for all x Î R.

Hence, f ( x) = 10x is strictly increasing on R. EXAMPLE 11

SOLUTION

Show that f ( x) = tan -1(cos x + sin x) is a strictly increasing function æ pö on the interval ç 0, ÷ × [CBSE 2007] è 4ø f ( x) = tan -1(cos x + sin x) Þ

f ¢( x) = =

1 2

×

d (cos x + sin x) dx

1 + (cos x + sin x) ( - sin x + cos x)

(1 + cos2 x + sin 2 x + 2 sin x cos x) (cos x - sin x) = ( 2 + sin 2x) p Now, when 0 < x < , we have cos x > sin x and sin 2x > 0. 4 \ (cos x - sin x) > 0 and ( 2 + sin 2x) > 0. p \ f ¢( x) > 0 for all x when 0 < x < × 4 æ pö Hence, f ( x) is strictly increasing in ç 0, ÷ × è 4ø EXAMPLE 12

Find the intervals on which the function f ( x) = 10 - 6x - 2x 2 is (a) strictly increasing (b) strictly decreasing. f ( x) = 10 - 6x - 2x 2

SOLUTION

3ö æ f ¢( x) = - 6 - 4x = - 2 ( 3 + 2x) = - 4 ç x + ÷ 2ø è (a) f ( x) is strictly increasing Û f ¢( x) > 0 3ö æ [from (i)] Û - 4 çx + ÷ > 0 2ø è

Þ

¼(i)

Senior Secondary School Mathematics for Class 12 Pg-537

Applications of Derivatives

Û

3ö æ çx + ÷ < 0 Û 2ø è

x
0 2ø è -3 Û x> × 2

[from (i)]

ù -3 é , ¥ ê× \ f ( x) is strictly decreasing on the interval ú û 2 ë -3 é ù and Hence, f ( x) is strictly increasing on the interval ú - ¥ , 2 êë û ù -3 é strictly decreasing on ú , ¥ ê× û 2 ë EXAMPLE 13

Find the intervals on which the function f ( x) = -2x 3 - 9x 2 - 12x + 1

SOLUTION

is (a) strictly increasing (b) strictly decreasing. f ( x) = -2x 3 - 9x 2 - 12x + 1 Þ f ¢( x) = -6x 2 - 18x - 12 = - 6( x 2 + 3 x + 2) = - 6( x + 2)( x + 1) ¼(i) (a) f ( x) is strictly increasing Û f ¢( x) > 0 [from (i)] Û - 6( x + 2)( x + 1) > 0 Û ( x + 2)( x + 1) < 0 Û - 2 < x < -1 Û x Î ] - 2, -1[. \ f ( x) is strictly increasing on the interval ] - 2, -1[. (b) f ( x) is strictly decreasing Û f ¢( x) < 0 [from (i)] Û - 6( x + 2)( x + 1) < 0 Û ( x + 2)( x + 1) > 0 Û [( x + 2) > 0 and ( x + 1) > 0] or [( x + 2) < 0 and ( x + 1) < 0] Û (x > -2 and x > -1) or (x < -2 and x < -1) Û ( x > -1) or ( x < -2) Û x Î ] -1, ¥ [ or x Î ] -¥ , - 2[ Û x Î ] -¥ , - 2[ È ] -1, ¥ [. \ f ( x) is strictly decreasing on ] - ¥ , - 2[ È ] -1, ¥ [.

Senior Secondary School Mathematics for Class 12 Pg-538

538 EXAMPLE 14

Senior Secondary School Mathematics for Class 12

Find the intervals on which the function f ( x) = 2x 3 - 15 x 2 + 36x + 6 is (a) increasing (b) decreasing.

[CBSE 2004C, ‘05, ‘09C]

f ( x) = 2x 3 - 15 x 2 + 36x + 6

SOLUTION

Þ

f ¢( x) = 6x 2 - 30x + 36 = 6( x 2 - 5 x + 6) = 6( x - 3)( x - 2)

¼(i)

(a) f ( x) is increasing Û f ¢( x) ³ 0 [from (i)] Û 6( x - 3)( x - 2) ³ 0 Û ( x - 3)( x - 2) ³ 0 Û [( x - 3) ³ 0 and ( x - 2) ³ 0] or [( x - 3) £ 0 and ( x - 2) £ 0] Û [x ³ 3 and x ³ 2] or [x £ 3 and x £ 2] Û [x ³ 3] or [x £ 2] Û x Î [ 3 , ¥ [ or x Î ] - ¥ , 2] Û x Î ] - ¥ , 2] È [ 3 , ¥ ]. \ f ( x) is increasing on ] - ¥ , 2] È [ 3 , ¥ [.

EXAMPLE 15

(b) f ( x) is decreasing Û f ¢( x) £ 0 [from (i)] Û 6( x - 3)( x - 2) £ 0 Û ( x - 3)( x - 2) £ 0 Û 2£x £ 3 Û x Î [2, 3]. \ f ( x) is decreasing on [2, 3]. Hence, f ( x) is increasing on ] - ¥ , 2] È [ 3 , ¥ [ and decreasing on [2, 3]. Find the intervals on which the function f ( x) = x 3 + 2x 2 - 1 is (a) increasing

(b) decreasing.

3

f ( x) = x + 2x 2 - 1

SOLUTION

4ö æ f ¢( x) = 3 x 2 + 4x = 3 x ç x + ÷ ¼(i) 3ø è (a) f ( x) is increasing Û f ¢( x) ³ 0 4ö æ [from (i)] Û 3x ç x + ÷ ³ 0 3ø è 4ö æ Û x çx + ÷ ³ 0 3ø è 4ö 4ö æ æ Û [x ³ 0 and ç x + ÷ ³ 0] or [x £ 0 and ç x + ÷ £ 0] 3ø 3ø è è 4 4 Û [x ³ 0 and x ³ - ] or [x £ 0 and x £ - ] 3 3 4ö æ Û ( x ³ 0) or ç x ³ - ÷ 3ø è

Þ

Senior Secondary School Mathematics for Class 12 Pg-539

Applications of Derivatives

539

4 x Î [0, ¥ [ or x Î ] - ¥ , - ] 3 4ù ù Û x Î ú - ¥ , - ú È [0, ¥ [. 3û û 4ù ù \ f ( x) is increasing on ú - ¥ , - ú È [0, ¥ [. 3û û (b) f ( x) is decreasing Û f ¢( x) £ 0 4ö æ [from (i)] Û 3x ç x + ÷ £ 0 3ø è 4ö æ Û x çx + ÷ £ 0 3ø è 4 Û - £x£0 3 é 4 ù Û x Î ê - , 0ú × ë 3 û é 4 ù \ f ( x) is decreasing on ê - , 0ú × ë 3 û 4ù ù Hence, f ( x) is increasing on ú - ¥ , - ú È [0, ¥ [ and decreasing 3û û é 4 ù on ê - , 0ú × ë 3 û Find the intervals on which the function f ( x) = x 3 + 3 x 2 - 105 x + 25 is (a) increasing (b) decreasing. f ( x) = x 3 + 3 x 2 - 105 x + 25 Û

EXAMPLE 16

SOLUTION

Þ f ¢( x) = 3 x 2 + 6x - 105 = 3( x 2 + 2x - 35) = 3( x + 7)( x - 5) ¼(i) (a) f ( x) is increasing Û f ¢( x) ³ 0 [from (i)] Û 3( x + 7)( x - 5) ³ 0 Û ( x + 7)( x - 5) ³ 0 Û [( x + 7) ³ 0 and ( x - 5) ³ 0] or [( x + 7) £ 0 and ( x - 5) £ 0] Û [x ³ - 7 and x ³ 5] or [x £ - 7 and x £ 5] Û ( x ³ 5) or ( x £ - 7) Û x Î [5 , ¥ [ or x Î ] - ¥ , - 7] Û x Î ] - ¥ , - 7] È [5 , ¥[. \ f ( x) is increasing on ] - ¥ , - 7] È [5 , ¥[ . (b) f ( x) is decreasing Û f ¢( x) £ 0 Û 3( x + 7)( x - 5) £ 0 Û ( x + 7)( x - 5) £ 0 Û -7 £ x £ 5 Û x Î [ - 7 , 5].

[from (i)]

Senior Secondary School Mathematics for Class 12 Pg-540

540

Senior Secondary School Mathematics for Class 12

\ f ( x) is decreasing on [ - 7 , 5] × Hence, f ( x) is increasing on ] - ¥ , - 7] È [5 , ¥[ and decreasing on [ - 7 , 5]. EXAMPLE 17

SOLUTION

Find the intervals on which the function f ( x) = 5 + 36x + 3 x 2 - 2x 3 is (a) increasing (b) decreasing. f ( x) = 5 + 36x + 3 x 2 - 2x 3 Þ

f ¢( x) = 36 + 6x - 6x 2 = - 6( x 2 - x - 6) = - 6( x + 2)( x - 3)

¼(i)

(a) f ( x) is increasing Û f ¢( x) ³ 0 [from (i)] Û - 6( x + 2)( x - 3) ³ 0 Û ( x + 2)( x - 3) £ 0 Û -2£ x £ 3 Û x Î [ - 2, 3]. \ f ( x) is increasing on [ - 2, 3]. (b) f ( x) is decreasing Û f ¢( x) £ 0 [from (i)] Û - 6( x + 2)( x - 3) £ 0 Û ( x + 2)( x - 3) £ 0 Û [( x + 2) ³ 0 and ( x - 3) ³ 0] or [( x + 2) £ 0 and ( x - 3) £ 0] Û [x ³ - 2 and x ³ 3] or [x £ -2 and x £ 3] Û ( x ³ 3) or ( x £ -2) Û x Î [ 3 , ¥ [ or x Î ] - ¥ , - 2] Û x Î ] - ¥ , - 2] È [ 3 , ¥[. \ f ( x) is decreasing on ] - ¥ , - 2] È [ 3 , ¥[. Hence, f ( x) is increasing on [-2, 3] and decreasing on ] - ¥ , - 2] È [ 3 , ¥[. EXAMPLE 18

SOLUTION

Find the intervals on which the function f ( x) = ( x + 1) 3( x - 3) 3 is (a) increasing (b) decreasing. [CBSE 2001C] f ( x) = ( x + 1) 3( x - 3) 3 d d Þ f ¢( x) = ( x + 1) 3 × ( x - 3) 3 + ( x - 3) 3 × ( x + 1) 3 dx dx = ( x + 1) 3 × 3( x - 3) 2 + ( x - 3) 3 × 3( x + 1) 2 = 3( x + 1) 2( x - 3) 2[( x + 1) + ( x - 3)] = 6( x + 1) 2( x - 3) 2( x - 1) (a) f ( x) is increasing Û f ¢( x) ³ 0 Û 6( x + 1) 2( x - 3) 2( x - 1) ³ 0 Û ( x - 1) ³ 0 Û x ³1 Û x Î [1, ¥ [. \ f ( x) is increasing on [1, ¥ [.

¼(i)

[from (i)]

Senior Secondary School Mathematics for Class 12 Pg-541

Applications of Derivatives

541

(b) f ( x) is decreasing Û f ¢( x) £ 0 Û

EXAMPLE 19

SOLUTION

6( x + 1) 2( x - 3) 2( x - 1) £ 0

[from (i)]

Û ( x - 1) £ 0 Û x £ 1 Û x Î ] - ¥ , 1]. \ f ( x) is decreasing on ] - ¥ , 1]. Hence, f ( x) is increasing on [1, ¥ [ and decreasing on ] - ¥ , 1]. 4 x2 + 1 Find the intervals on which the function f ( x) = , ( x ¹ 0) is x [CBSE 2004] (a) increasing (b) decreasing. 4 x2 + 1 ,x¹0 x 1ö æ Þ f ( x) = ç 4x + ÷ , x ¹ 0 xø è 1 ö æ Þ f ¢( x) = ç 4 - 2 ÷ è x ø ( 4x 2 - 1) Þ f ¢( x) = x2 (a) f ( x) is increasing f ( x) =

Û Û Û Û Û or Û Û Û Û \

f ¢( x) ³ 0 Û

( 4 x 2 - 1)

[from (i)] ³0 x2 [Q x 2 > 0] ( 4x 2 - 1) ³ 0 1ö æ 1ö æ ( 2x - 1)( 2x + 1) ³ 0 Û 2 ç x - ÷ × 2 ç x + ÷ ³ 0 2ø è 2ø è 1öæ 1ö æ çx - ÷çx + ÷ ³ 0 2ø è 2ø è 1ö 1ö éæ ù æ ê ç x - 2 ÷ ³ 0 and ç x + 2 ÷ ³ 0ú ø è ø ëè û 1ö 1ö éæ ù æ ê ç x - 2 ÷ £ 0 and ç x + 2 ÷ £ 0ú ø è ø ëè û 1 1ù é 1 1ù é êë x ³ 2 and x ³ - 2 úû or êë x £ 2 and x £ - 2 úû 1ö æ 1ö æ ç x ³ ÷ or ç x £ - ÷ 2ø è 2ø è 1ù é1 é ù x Î ê , ¥ ê or x Î ú - ¥ , - ú 2û ë2 ë û 1ù é1 ù é x Îú - ¥, - ú È ê , ¥ ê × 2û ë 2 û ë 1ù é1 ù é f ( x) is increasing on ú - ¥ , - ú È ê , ¥ ê × 2û ë 2 û ë

¼(i)

Senior Secondary School Mathematics for Class 12 Pg-542

542

Senior Secondary School Mathematics for Class 12

(b) f ( x) is decreasing Û

EXAMPLE 20

( 4x 2 - 1)

[from (i)] £0 x2 [Q x 2 > 0] Û ( 4x 2 - 1) £ 0 1ö 1ö æ æ Û ( 2x - 1)( 2x + 1) £ 0 Û 2 ç x - ÷ × 2 ç x + ÷ £ 0 2ø 2ø è è 1ö æ 1ö æ Û çx - ÷ çx + ÷ £ 0 2ø è 2ø è 1 1 Û - £x£ 2 2 é 1 1ù Û x Îê- , ú × ë 2 2û é 1 1ù \ f ( x) is decreasing on ê - , ú × ë 2 2û 1ù é1 ù é Hence, f ( x) is increasing on ú - ¥ , - ú È ê , ¥ ê and decreasing 2û ë 2 û ë é 1 1ù on ê - , ú × ë 2 2û x Find the intervals on which the function f ( x) = 2 is ( x + 1) [CBSE 2003] (a) increasing (b) decreasing. f ( x) =

SOLUTION

f ¢( x) £ 0 Û

x ( x 2 + 1)

Þ

f ¢( x) =

Þ

f ¢( x) =

( x 2 + 1) ×

d d ( x) - x × ( x 2 + 1) ( x 2 + 1) × 1 - x × 2x dx dx = ( x 2 + 1) 2 ( x 2 + 1) 2

(1 - x 2)

¼(i)

( x 2 + 1) 2

(a) f ( x) is increasing Û f ¢( x) ³ 0 (1 - x 2) Û ³0 ( x 2 + 1) 2 Û (1 - x 2) ³ 0

[from (i)] [Q ( x 2 + 1) 2 > 0]

Û - (1 - x 2) £ 0 Û ( x 2 - 1) £ 0 Û ( x - 1)( x + 1) £ 0 Û -1 £ x £ 1 Û x Î [ - 1, 1]. \ f ( x) is increasing on [ - 1, 1]. (b) f ( x) is decreasing Û f ¢( x) £ 0

Senior Secondary School Mathematics for Class 12 Pg-543

Applications of Derivatives

Û

(1 - x 2) ( x 2 + 1) 2

£0

Û (1 - x 2) £ 0 Û

543

[from (i)] [Q ( x 2 + 1) 2 > 0]

- (1 - x 2) ³ 0

Û ( x 2 - 1) £ 0 Û ( x - 1)( x + 1) ³ 0

EXAMPLE 21

Û [( x - 1) ³ 0 and ( x + 1) ³ 0] or [( x - 1) £ 0 and ( x + 1) £ 0] Û [x ³ 1 and x ³ - 1] or [x £ 1 and x £ - 1] Û ( x ³ 1) or ( x £ - 1) Û x Î [1, ¥ [ or x Î ] - ¥ , - 1] Û x Î ] - ¥ , - 1] È [1, ¥ [. \ f ( x) is decreasing on ] - ¥ , - 1] È [1, ¥ [. Hence, f ( x) is increasing on [ - 1, 1] and decreasing on ] - ¥ , - 1] È [1, ¥ [. Find the intervals on which the function f ( x) = ( x + 2) e- x is

SOLUTION

(a) increasing (b) decreasing. f ( x) = ( x + 2) e- x

[CBSE 2000C]

Þ f ¢( x) = - ( x + 2) e- x + e- x × 1 Þ f ¢( x) = - ( x + 1) e- x (a) f ( x) is increasing Û f ¢( x) ³ 0 Û - ( x + 1) e- x ³ 0 Û ( x + 1) e

-x

¼(i)

[from (i)]

£0

Û ( x + 1) £ 0

[Q e- x > 0]

Û x £ -1 Û x Î ] -¥ , -1]. \ f ( x) is increasing on ] -¥ , -1]. (b) f ( x) is decreasing Û f ¢( x) £ 0 [from (i)] Û - ( x + 1) e- x £ 0 Û ( x + 1) e- x ³ 0

EXAMPLE 22

SOLUTION

Û ( x + 1) ³ 0 [Q e- x > 0] Û x ³ -1 Û x Î [ -1, ¥ [. \ f ( x) is decreasing on [ -1, ¥ [. Hence, f ( x) is increasing on ] -¥ , -1] and decreasing on [ -1, ¥ [. x is Find the intervals on which the function f ( x) = log (1 + x) (1 + x) (a) increasing (b) decreasing. [CBSE 2000C] x f ( x) = log (1 + x) (1 + x)

Senior Secondary School Mathematics for Class 12 Pg-544

544

Senior Secondary School Mathematics for Class 12

ì 1 (1 + x) × 1 - x × 1 ü Þ f ¢( x) = í ý ( 1 x ) + (1 + x) 2 î þ ì 1 1 ü (1 + x - 1) =í ý= 1 + ( x ) 1 + ( x) 2 þ (1 + x) 2 î x Þ f ¢( x) = (1 + x) 2 (a) f ( x) is increasing Û f ¢( x) ³ 0 x Û ³0 (1 + x) 2 Û Û

x³0 x Î [0, ¥ [.

¼(i)

[from (i)] [Q (1 + x) 2 ³ 0]

(b) f ( x) is decreasing Û f ¢( x) £ 0 x [from (i)] Û £0 (1 + x) 2 Û x£0 Û x Î ] -¥ , 0]. Hence, f ( x) is increasing on [0, ¥ [ and decreasing on ] -¥ , 0]. EXAMPLE 23

Find the intervals on which the function f ( x) = (sin x - cos x), 0 < x < 2p, is (a) increasing (b) decreasing. [CBSE 2000C, ‘09, ‘11]

SOLUTION

Þ

f ( x) = (sin x - cos x), 0 < x < 2p 1 æ 1 ö f ¢( x) = (cos x + sin x) = 2 ç cos x + sin x ÷ 2 2 è ø

p p æ ö f ¢( x) = 2 ç sin cos x + cos sin x ÷ 4 4 è ø p ö æ Þ f ¢( x) = 2 sin ç + x ÷ ø è4 p æp ö æp ö Now, 0 < x < 2p Þ < ç + x ÷ < ç + 2p ÷ 4 è4 ø è4 ø p æp ö 9p Þ < ç + x÷ < 4 è4 4 ø (a) f ( x) is increasing Û f ¢( x) ³ 0 æp ö [from (i)] Û 2 sin ç + x ÷ ³ 0 è4 ø æp ö Û sin ç + x ÷ ³ 0 è4 ø Þ

… (i)

¼(ii)

Senior Secondary School Mathematics for Class 12 Pg-545

Applications of Derivatives

545

ìp æ p ü ì ö æp ö 9p ü Û í < ç + x ÷ £ p ý or í2p £ ç + x ÷ < ý 4þ ø è4 ø î4 è 4 þ î 3p ü ü ì ì7 p Û í0 < x £ ý or í £ x < 2p ý 4 þ þ î î4 3p ù ù é 7p é or x Î ê , 2p ê Û x Î ú 0, 4 úû û ë 4 ë 3p ù é 7p ù é , 2p ê × Û x Î ú 0, È 4 úû êë 4 û ë ù 3p ù é 7p é È , 2p ê × \ f ( x) is increasing on ú 0, 4 úû êë 4 û ë (b) f ( x) is decreasing Û f ¢( x) £ 0 ö æp 2 sin ç + x ÷ £ 0 Û 4 ø è ö æp Û sin ç + x ÷ £ 0 ø è4 æp ö Û p £ ç + x ÷ £ 2p è4 ø 3p 7p Û £x£ 4 4 é 3p 7p ù Û x Îê , × ë 4 4 úû

[from (i)]

é 3p 7p ù , × \ f ( x) is decreasing on ê ë 4 4 úû

EXAMPLE 24

é pù Separate the interval ê 0, ú into subintervals in which ë 2û f ( x) = (sin 4 x + cos4 x) is (a) increasing

SOLUTION

4

(b) decreasing. [CBSE 2000C]

4

f ( x) = (sin x + cos x) Þ f ¢( x) = 4 sin 3 x cos x - 4 cos 3 x sin x = - 4 sin x cos x(cos2 x - sin 2 x) = -2 sin 2x cos 2x = - sin 4x p Also, 0 £ x £ Û 0 £ 4x £ 2p. 2 (a) f ( x) is increasing Û f ¢( x) ³ 0 [from (i)] Û - sin 4x ³ 0 Û sin 4x £ 0 Û p £ 4x £ 2p p p Û £x£ 4 2

¼(i)

Senior Secondary School Mathematics for Class 12 Pg-546

546

Senior Secondary School Mathematics for Class 12

é p pù Û x Îê , ú × ë 4 2û

EXAMPLE 25

SOLUTION

é p pù \ f ( x) is increasing on ê , ú × ë 4 2û (b) f ( x) is decreasing [from (i)] Û f ¢( x) £ 0 Û - sin 4x £ 0 Û sin 4x ³ 0 Û 0 £ 4x £ p p Û 0£x £ 4 é pù Û x Î ê 0, ú × ë 4û é pù \ f ( x) is decreasing on ê 0, ú × ë 4û p pù é pù é Hence, f ( x) is increasing on ê , ú and decreasing on ê 0, ú × ë 4û ë 4 2û é pù Separate ê 0, ú into subintervals in which f ( x) = sin 3 x is ë 2û (a) increasing (b) decreasing. f ( x) = sin 3 x Þ f ¢( x) = 3 cos 3 x p 3p Also, 0 £ x £ Û 0 £ 3x £ 2 2 (a) f ( x) is increasing Û f ¢( x) ³ 0 Û 3 cos 3 x ³ 0 Û cos 3 x ³ 0 p Û 0 £ 3x £ 2 p Û 0£x £ 6 é pù Û x Î ê 0, ú × ë 6û é pù \ f ( x) is increasing on ê 0, ú × ë 6û (b) f ( x) is decreasing Û f ¢( x) £ 0 Û 3 cos 3 x £ 0 Û cos 3 x ³ 0 p 3p Û £ 3x £ 2 2 p p Û £x£ 6 2 é p pù Û xÎê , ú× ë 6 2û

¼(i) ¼(ii)

[from (i)]

[from (i)]

Senior Secondary School Mathematics for Class 12 Pg-547

Applications of Derivatives

EXAMPLE 26 SOLUTION

EXAMPLE 27

547

é p pù \ f ( x) is decreasing on ê , ú × ë 6 2û é pù é p pù Hence, f ( x) is increasing on ê 0, ú and decreasing on ê , ú × ë 6û ë 6 2û é pù Prove that tan x > x for all x Î ê 0, ú × ë 2û p é ù Let c be an arbitrary real number such that c Î ú 0, ê × 2 ë û Let f ( x) = tan x - x for all x Î [0, c]. \ f ¢( x) = sec2 x - 1 = tan 2 x > 0 for all x Î ]0, c[. Thus, f ( x) is increasing on ]0, c[. Now, x > 0 Þ f ( x) > f ( 0) Þ f ( x) > 0 Þ tan x - x > 0 Þ tan x > x. Find the values of x for which the function f ( x) = x x , x > 0 is (a) increasing (b) decreasing. f ( x) = x x

SOLUTION

x log x

Þ

f ( x) = e

Þ

f ¢( x) = e

Þ

d ( x log x) dx f ¢( x) = x x (1 + log e x) x log x

×

(a) f ( x) is increasing Û f ¢( x) ³ 0 Û x x (1 + log e x) ³ 0 Û (1 + log e x) ³ 0

¼(i)

[from (i)] [Q x x > 0 when x > 0]

Û log e x ³ - 1

Û x ³ e -1 é1 é Û x Î ê , ¥ê × ëe ë

é1 é \ f ( x) is increasing on ê , ¥ ê × ëe ë (b) f ( x) is decreasing Û f ¢( x) £ 0 Û x x (1 + log e x) £ 0 Û (1 + log e x) £ 0 Û log e x £ - 1 Û x £ e -1

Û 0£x £

1 e

[Q x x > 0]

Senior Secondary School Mathematics for Class 12 Pg-548

548

Senior Secondary School Mathematics for Class 12

é Û x Î ê 0, ë

1ù × e úû

é 1ù \ f ( x) is decreasing on ê 0, ú × ë eû é1 ù é 1ù Hence, f ( x) is increasing on ê , ¥ ú and decreasing on ê 0, ú × ëe û ë eû EXAMPLE 28

Find the intervals on which the function f ( x) = ( x 4 - 2x 2) is increasing or decreasing.

SOLUTION

f ( x) = x 4 - 2x 2 Þ f ¢( x) = 4x 3 - 4x = 4x( x - 1)( x + 1). Now, f ¢( x) = 0 Þ x = -1 or x = 0 or x = 1. These points divide the whole real line into four disjoint open intervals, namely, ] - ¥ , - 1[, ] - 1, 0[, ]0, 1[ and ]1, ¥ [. Case I When x Î ] - ¥ , - 1[ In this case, x < - 1. So, f ¢( x) = ( -) ( -) ( -) = - ve. So, f ¢( x) < 0 for all x Î ] - ¥ , - 1[. \ f ( x) is decreasing for all x Î ] - ¥ , - 1[. Case II When x Î ] - 1, 0[ In this case, -1 < x < 0. So, f ¢( x) = ( -) ( -) ( + ) = + ve. \ f ¢( x) > 0 for all x Î ] - 1, 0[. \ f ( x) is increasing on ] - 1, 0[. When x Î]0, 1[ In this case, 0 < x < 1. So, f ¢( x) = ( + ) ( -) ( + ) = - ve. Thus, f ¢( x) < 0 for all x Î]0, 1[. \ f ( x) is decreasing on ]0, 1[. Case IV When x Î ]1, ¥ [ In this case, 1 < x < ¥ . So, f ¢( x) = ( + ) ( + ) ( + ) = + ve. Thus, f ¢( x) > 0 for all x Î ]1, ¥ [. \ f ( x) is increasing on ]1, ¥ [. From all the four cases, we conclude that f ( x) is increasing on ] - 1, 0[ È ]1, ¥ [ and, f ( x) is decreasing on ] - ¥ , - 1[ È ]0, 1[. x Prove that < log(1 + x) < x for x > 0. (1 + x) x Let f ( x) = log (1 + x) × (1 + x) é 1 1 ù x Then, f ¢( x) = ê > 0 for x > 0. ú= 2 2 ë 1 + x (1 + x) û (1 + x) Case III

EXAMPLE 29 SOLUTION

Senior Secondary School Mathematics for Class 12 Pg-549

Applications of Derivatives

549

Since f ¢( x) > 0 for all x > 0 and f ¢( 0) = 0, it follows that f ( x) is increasing on [0, ¥ [. Now, x > 0 Þ f ( x) > f ( 0) Þ f ( x) > 0 [Q f ( 0) = 0] é x ù x Þ ê log(1 + x) ú > 0 Þ log(1 + x) > (1 + x) … (i) + ( 1 x ) ë û Again, let g( x) = [x - log (1 + x)]. é x 1 ù Then, g ¢( x) = ê1 > 0 for x > 0. ú= ë (1 + x) û (1 + x) Now, g ¢( x) > 0 for all x > 0 and g¢( 0) = 0. \ g( x) is strictly increasing on [0, ¥ [. Also, g( 0) = 0. Now, x > 0 Þ g( x) > g( 0) Þ g( x) > 0 [Q g( 0) = 0] … (ii) Þ [x - log (1 + x)] > 0 Þ x > log (1 + x) x From (i) and (ii), we get < log (1 + x) < x for x > 0. (1 + x)

EXERCISE 11G 1. Show that the function f ( x) = 5 x - 2 is a strictly increasing function on R. 2. Show that the function f ( x) = -2x + 7 is a strictly decreasing function on R. 3. Prove that f ( x) = ax + b , where a and b are constants and a > 0, is a strictly increasing function on R. 4. Prove that the function f ( x) = e2x is strictly increasing on R. 5. Show that the function f ( x) = x 2 is (a) strictly increasing on [0, ¥ [ (b) strictly decreasing on ] - ¥ , 0[ (c) neither strictly increasing nor strictly decreasing on R 6. Show that the function f ( x) = |x|is (a) strictly increasing on ]0, ¥ [ (b) strictly decreasing on ] - ¥ , 0[ 7. Prove that the function f ( x) = log e x is strictly increasing on ]0, ¥ [. 8. Prove that the function f ( x) = log a x is strictly increasing on ]0, ¥ [ when a > 1 and strictly decreasing on ]0, ¥ [ when 0 < a < 1. 9. Prove that f ( x) = 3 x is strictly increasing on R. 10. Show that f ( x) = x 3 - 15 x 2 + 75 x - 50 is increasing on R. 1ö æ 11. Show that f ( x) = ç x - ÷ is increasing for all x Î R , where x ¹ 0. xø è æ3 ö 12. Show that f ( x) = ç + 5 ÷ is decreasing for all x Î R , where x ¹ 0. èx ø

Senior Secondary School Mathematics for Class 12 Pg-550

550

Senior Secondary School Mathematics for Class 12

13. Show that f ( x) =

1 (1 + x 2)

is increasing for all x £ 0.

1 ö æ [CBSE 2009] 14. Show that f ( x) = ç x 3 + 3 ÷ is decreasing on ] - 1, 1[. è x ø x ù pé is increasing on ú 0, ê × 15. Show that f ( x) = sin x û 2ë 2x 16. Prove that the function f ( x) = log (1 + x) is increasing for all ( x + 2) [CBSE 2000C] x > -1. 17. Let I be an interval disjoint from ] - 1, 1[. Prove that the function 1ö æ f ( x) = ç x + ÷ is strictly increasing on I. xø è ( x - 2) is increasing for all x Î R , except at x = -1. 18. Show that f ( x) = ( x + 1) 19. Find the intervals on which the function f ( x) = ( 2x 2 - 3 x) is (a) strictly increasing (b) strictly decreasing. 20. Find the intervals on which the function f ( x) = 2x 3 - 3 x 2 - 36x + 7 is (a) strictly increasing (b) strictly decreasing. [CBSE 2005C] 21. Find the intervals on which the function f ( x) = 6 - 9x - x 2 is (a) strictly increasing (b) strictly decreasing. Find the intervals on which each of the following functions is (a) increasing (b) decreasing. æ x 3 ö÷ 22. f ( x) = ç x 4 ç 3 ÷ø è 3 23. f ( x) = x - 12x 2 + 36x + 17 3

2

24. f ( x) = ( x - 6x + 9x + 10) 2

[CBSE 2006, ‘09C] [CBSE 2000, ‘04C]

3

25. f ( x) = ( 6 + 12x + 3 x - 2x ) 26. f ( x) = 2x 3 - 24x + 5 27. f ( x) = ( x - 1)( x - 2) 2 4

3

[CBSE 2009C] 2

28. f ( x) = ( x - 4x + 4x + 15) 29. f ( x) = 2x 3 + 9x 2 + 12x + 15

[CBSE 2008C]

30. f ( x) = x 4 - 8x 3 + 22x 2 - 24x + 21

[CBSE 2012C]

4

3

2

31. f ( x) = 3 x - 4x - 12x + 5 is (a) strictly increasing (b) strictly decreasing. [CBSE 2014]

3 4 36 32. f ( x) = x 4 - x 3 - 3 x 2 + x + 11 is (a) strictly increasing (b) strictly 10 5 5 decreasing. [CBSE 2014C]

Senior Secondary School Mathematics for Class 12 Pg-551

Applications of Derivatives ANSWERS (EXERCISE 11G)

ù 3 é 19. (a) f ( x) is strictly increasing on ú , ¥ ê × û 4 ë 3 ù é (b) f ( x) is strictly decreasing on ú -¥ , ê × 4ë û 20. (a) f ( x) is strictly increasing on ] -¥ , - 2[ È ] 3 , ¥ [. (b) f ( x) is strictly decreasing on ] -2, 3[. 9é ù 21. (a) f ( x) is strictly increasing on ú -¥ , - ê × 2ë û ù 9 é (b) f ( x) is strictly decreasing on ú - , ¥ ê × 2 û ë é1 é 22. (a) f ( x) is increasing on ê , ¥ ê × ë4 ë 1ù ù (b) f ( x) is decreasing on ú -¥ , ú × 4û û 23. (a) f ( x) is increasing on ] -¥ , - 2[ È ] 6, ¥ [. (b) f ( x) is decreasing on ]2, 6[. 24. (a) f ( x) is increasing on ] -¥ , 1] È [ 3 , ¥ [. (b) f ( x) is decreasing on [-1, 3]. 25. (a) f ( x) is increasing on [-1, 2]. (b) f ( x) is decreasing on ] -¥ , -1] È [2, ¥ [. 26. (a) f ( x) is increasing on ] - ¥ , - 2] È [1, ¥ [. (b) f ( x) is decreasing on [-2, 2]. 4ù ù 27. (a) f ( x) is increasing on ú -¥ , ú È [2, ¥ [. 3û û é4 (b) f ( x) is decreasing on ê , ë3

ù 2ú × û

28. (a) f ( x) is increasing on [0, 1] È [2, ¥ [. (b) f ( x) is decreasing on ] -¥ , 0] È [1, 2]. 29. (a) f ( x) is increasing on ] -¥ , - 2] È [-1, ¥ [.

(b) f ( x) is decreasing on [-2, - 1]. 30. (a) f ( x) is increasing on [1, 2] È [ 3 , ¥ [

(b) f ( x) is decreasing on ] -¥ , 1] È [2, 3] 31. (a) Strictly increasing in ] -1, 0[ È ]2, ¥ [. (b) Strictly decreasing in ] -¥ , -1[ È ]0, 2[. 32. (a) Strictly increasing in ] - 2, 1[ È ] 3 , ¥ [. (b) Strictly decreasing in ] -¥ , - 2[ È ]1, 3[.

551

Senior Secondary School Mathematics for Class 12 Pg-552

552

Senior Secondary School Mathematics for Class 12 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11G)

1. x1 > x2 Þ 5 x1 >5 x2 Þ (5 x1 - 2 ) >(5 x2 - 2 ) Þ f ( x1 ) > f ( x2 ). 2. x1 > x2 Þ - 2 x1 < - 2 x2 Þ - 2 x1 + 7 < - 2 x2 + 7 Þ f ( x1 ) < f ( x2 ). 5. (a) Let x1 , x2 Î [0 , ¥ [ such that x1 > x2 × Then, x1 > x2 Þ x12 > x1 x2 and x1 x2 > x22 Þ x12 > x22 Þ f ( x1 ) > f ( x2 ). (b) Let x1 , x2 Î ] - ¥ , 0] such that x1 > x2 . Then, x1 > x2 Þ x1 x2 < x22 Also, x1 > x2 Þ \

x12

x12

< x1 x2

[Q x2 < 0]. [Q x1 < 0].

< x1 x2 < x22 .

Thus, x1 > x2 Þ x12 < x22 Þ f ( x1 ) < f ( x2 ). (c) Since f ( x ) = x 2 is strictly increasing on [0 , ¥ [ and strictly decreasing on ] - ¥ , 0], it is neither strictly increasing nor strictly decreasing on the whole real line. 6. (a) Let x1 , x2 Î ]0 , ¥ [ such that x1 > x2 . Then, x1 > x2 Þ |x1| >|x2| Þ f ( x1 ) > f ( x2 ). \ f ( x ) =|x| is strictly increasing on ]0 , ¥[. (b) Let x1 , x2 Î ] - ¥ , 0[ such that x1 > x2 . Then, x1 > x2 Þ |x1| x2 Þ f ( x1 ) < f ( x2 ). \ f ( x ) =|x| is strictly decreasing on ] - ¥ , 0[. 7. For all x1 , x2 Î ]0 , ¥[, we have x1 ³ x2 Þ log e x1 ³ log e x2 Þ f ( x1 ) ³ f ( x2 ). log e x 8. f ( x ) = log a x = × log e a Case I

Let x1 , x2 Î ]0 , ¥[ and let a > 1. Then, x1 > x2 Þ log x1 > log x2 log x1 log x2 Þ > log a log a Þ log a x1 > log a x2 . \ f ( x ) is strictly increasing on [0 , ¥[ when a > 1.

Senior Secondary School Mathematics for Class 12 Pg-553

Applications of Derivatives Case II Let x1 , x 2 Î ] 0 , ¥[ and let 0 < a < 1.

Then, x1 > x2 Þ log x1 > log x2 Þ

ù é 1 < 0ú êQ 0 < a < 1 Þ log a < 0 Þ log a û ë

log x1 log x2 < log a log a

Þ (log

a x1 ) < (log a

x2 ).

\ f ( x ) is strictly decreasing on ]0 , ¥[ when 0 < a < 1. 9. f ( x ) = 3 x Þ f ¢( x ) = ( 3 x log 3 ) > 0 for all x Î R. 10. f ( x ) = x 3 - 15 x 2 + 75 x - 50 Þ f ¢( x ) = 3 x 2 - 30 x + 75 = 3( x 2 - 10 x + 25 ) = 3( x - 5 ) 2 ³ 0. 1ö 1 ö æ æ 11. f ( x ) = ç x - ÷ Þ f ¢( x ) = ç 1 + 2 ÷ > 0 for all x Î R , where x ¹ 0. xø è è x ø -3 æ3 ö 12. f ( x ) = ç + 5 ÷ Þ f ¢( x ) = 2 < 0 for all x Î R , where x ¹ 0. èx ø x 13. f ¢( x ) = - ( 1 + x 2 ) -2 × 2 x = 14. f ¢( x ) < 0 Û

3( x 6 - 1) x4

-2 x ( 1 + x 2 )2

³ 0 when x £ 0.

< 0 Û ( x 6 - 1) < 0

Û ( x - 1)( x + 1)( x 4 + x 2 + 1) < 0 Û ( x - 1) ( x + 1) < 0 Û - 1 < x < 1. sin x - x cos x 15. f ¢( x ) > 0 Û > 0 Û (sin x - x cos x ) > 0 sin 2 x ù p Û tan x > x , which is true on ú 0 , û 2

é êë ×

16. Clearly, log ( 1 + x ) is defined only when ( 1 + x ) > 0 , i.e., when x > - 1. 2x Now, f ( x ) = log ( 1 + x ) ( x + 2) ì 1 [( x + 2 ) × 2 - 2 x × 1] ü x2 Þ f ¢( x ) = í > 0 [Q ( 1 + x ) > 0] × ý= 2 2 ( x + 2) î( 1 + x) þ ( 1 + x )( x + 2 ) 17. f ( x ) = x +

1 1 ö ( x 2 - 1) æ Þ f ¢( x ) = ç 1 - 2 ÷ = × x è x ø x2

Now, f ¢( x ) > 0 Û ( x 2 - 1) > 0 Û ( x - 1) ( x + 1) > 0 Û [( x - 1) > 0 and ( x + 1) > 0] or [( x - 1) < 0 and ( x + 1) < 0] Û [x > 1 and x > - 1] or [x < 1 and x < - 1] Û ( x > 1) or ( x < - 1) Û x Î ] - ¥ , - 1[ or x Î ]1, ¥ [. Û x Î ] - ¥ , - 1[ È ]1, ¥ [.

553

Senior Secondary School Mathematics for Class 12 Pg-554

554

Senior Secondary School Mathematics for Class 12

6. Tangents and Normals Some Important Theorems THEOREM 1

Prove that the equation of a tangent to a curve y = f ( x) at a point y - y1 æ dy ö =ç ÷ , P( x1 , y1) is given by x - x1 è dx ø ( x 1, y1 ) dy æ dy ö denotes the value of at x = x1 and y = y1 , where ç ÷ dx è dx ø ( x 1, y1 ) and the equation of the normal at P( x1 , y1) is given by -1 y - y1 = × x - x1 æ dy ö ç ÷ è dx ø ( x 1, y1 )

PROOF

We know that the slope of the tangent to the curve y = f ( x) at a point æ dy ö P( x1 , y1) is ç ÷ × è dx ø ( x 1, y1 ) Thus, the tangent to the given curve at a point P is a line passing æ dy ö through ( x1 , y1) with a slope of ç ÷ × è dx ø ( x 1, y1 ) So, the equation of the tangent at P is

y - y1 æ dy ö × =ç ÷ x - x1 è dx ø ( x 1, y1 )

Again, the normal to the given curve at P is a line perpendicular to the tangent at P. -1 So, the slope of the normal is × æ dy ö ç ÷ è dx ø ( x 1, y1 ) Also, the normal at P passes through ( x1 , y1). y - y1 -1 So, the equation of the normal at P is = × x - x1 æ dy ö ç ÷ è dx ø ( x 1, y1 ) THEOREM 2 The tangent to a curve y = f ( x) at a point P( x1 , y1 ) is parallel to the æ dy ö x-axis if and only if ç ÷ = 0. è dx ø ( x 1, y1 ) The tangent is parallel to the x-axis Û its slope is 0 æ dy ö = 0. Û ç ÷ è dx ø ( x 1, y1 ) THEOREM 3 The tangent to a curve y = f ( x) at a point P( x1 , y1 ) is parallel to the æ dx ö y-axis if and only if çç ÷÷ = 0. è dy ø ( x , y ) PROOF

1

1

Senior Secondary School Mathematics for Class 12 Pg-555

Applications of Derivatives

555

The tangent is parallel to the y-axis æ dx ö æ dy ö Û its slope is 90° Û ç ÷ = tan 90° = ¥ Û çç ÷÷ = 0. è dx ø ( x 1, y1 ) è dy ø ( x 1, y1 ) REMARK If x = f (t) and y = g(t), where f ¢(t) ¹ 0 then y - g(t) g ¢(t) (i) equation of the tangent at ‘t‘ is = x - f (t) f ¢(t) y - g(t) f ¢(t) and, (ii) equation of the normal at ‘t‘ is =× x - f (t) g ¢(t) PROOF

SOLVED EXAMPLES EXAMPLE 1

Find the equations of the tangent and the normal to the curve y = x 4 - 6x 3 + 13 x 2 - 10x + 5 at the point (1, 3).

SOLUTION

The equation of the given curve is y = x 4 - 6x 3 + 13 x 2 - 10x + 5. dy = 4x 3 - 18x 2 + 26x - 10. dx æ dy ö So, ç ÷ = ( 4 ´ 1 3 - 18 ´ 12 + 26 ´ 1 - 10) = 2. è dx ø (1, 3)

\

\ the required equation of the tangent is y-3 = 2 or 2x - y + 1 = 0. x -1 And, the required equation of the normal is y - 3 -1 or x + 2y - 7 = 0. = x -1 2 EXAMPLE 2

Find the equations of the tangent and the normal to the curve y = x 2 + 4x + 1 at the point where x = 3.

SOLUTION

When x = 3, we have y = ( 3 2 + 4 ´ 3 + 1) = 22. So, the point of contact is ( 3 , 22). dy æ dy ö Now, y = x 2 + 4x + 1 Þ = 2x + 4 Þ ç ÷ = ( 2 ´ 3 + 4) = 10. dx è dx ø ( 3, 22) y - 22 = 10 Þ 10x - y - 278 = 0. x - 30 y - 22 -1 And, equation of the normal is = Þ x + 10y - 223 = 0. x - 3 10 Show that the equation of the tangent to the ellipse x2 y2 xx yy + 2 = 1 at ( x1 , y1) is 21 + 21 = 1. 2 a b a b

Equation of the tangent is

EXAMPLE 3

SOLUTION

x2 a

2

+

y2 b2

=1 Þ

2x a2

+

2y dy × = 0 [on differentiating w.r.t. x] b 2 dx

Senior Secondary School Mathematics for Class 12 Pg-556

556

Senior Secondary School Mathematics for Class 12

-b2x dy - b 2 x æ dy ö = 2 1× = 2 Þ ç ÷ dx è dx ø ( x 1, y1 ) a y a y1 So, the equation of the tangent at ( x1 , y1) is y - y1 - b 2 x1 = 2 x - x1 a y1

Þ

b 2 xx1 + a 2 yy1 = b 2 x12 + a 2 y12.

On dividing throughout by a 2 b 2 , we get i.e., EXAMPLE 4

xx1 a

2

+

yy1 b

2

xx1 a2

+

yy1

x12 a2

+

y12 b2

,

é ù x2 y2 x2 y2 = 1 êQ ( x1 , y1) lies on 2 + 2 = 1 Þ 12 + 12 = 1ú × a b a b ë û

Find the equation of the tangent to the curve y = 5 x - 3 - 2, which is parallel to the line 4x - 2y + 3 = 0.

SOLUTION

b2

=

[CBSE 2005, ’13C]

3 The given line is 4x - 2y + 3 = 0 or y = 2x + × 2 \ slope of the given line = 2. So, the slope of the tangent = 2. Let the point of contact be ( x1 , y1). dy 5 Now, y = 5 x - 3 - 2 Þ = dx 2 × 5 x - 3 5 æ dy ö = × Þ ç ÷ è dx ø ( x 1, y1 ) 2 × 5 x1 - 3 \

5 25 73 =2 Þ = 4, i.e., x1 = × 4(5 x1 - 3) 80 2 × 5 x1 - 3

73 -3 æ ö - 3÷ - 2 = \ y1 = 5 x1 - 3 - 2 = ç5 ´ × 80 4 è ø æ 73 -3 ö So, the point of contact is ç , ÷× è 80 4 ø

3ö æ çy + ÷ 4ø Hence, the required equation of the tangent is è = 2, 73 ö æ çx - ÷ i.e., 80x - 40y - 103 = 0. 80 ø è

EXAMPLE 5

SOLUTION

Find the equations of the normals to the curve 3 x 2 - y 2 = 8, parallel to the line x + 3 y = 4. 1 4 The given line is x + 3 y = 4 or y = - x + × 3 3 1 \ slope of the given line = 3 1 So, the slope of the required normal = 3 Let the point of contact be ( x1 , y1).

… (i)

Senior Secondary School Mathematics for Class 12 Pg-557

Applications of Derivatives

557

dy = 0 [on differentiating w.r.t. x] dx dy 3 x 3x æ dy ö = 1× Þ = Þ ç ÷ dx y y1 è dx ø ( x 1, y1 )

Now, 3 x 2 - y 2 = 8 Þ 6x - 2y ×

\ slope of the normal =

-1 -y = 1 æ dy ö 3 x1 ç ÷ è dx ø ( x 1, y1 )

Thus, from (i) and (ii) we get

… (ii)

1 - y1 = - Þ y1 = x1. 3 x1 3

Also, since ( x1 , y1) lies on the given curve, we have 3 x12 - y12 = 8 or 3 x12 - x12 = 8 or

x12

[Q

y1 = x1]

= 4 or x1 = ± 2.

\ y1 = ± 2

[Q

y1 = x1].

Thus, the points of contact are ( 2, 2) and ( -2, - 2). So, the equation of the required normal at ( 2, 2) is i.e., x + 3 y - 8 = 0. The equation of the required normal at ( - 2, - 2) is i.e., x + 3 y + 8 = 0. EXAMPLE 6

n

n

n

æxö æyö The equation of the curve is ç ÷ + ç ÷ = 2. a è ø è bø n-1 n-1 1 1 dy æyö æxö On differentiating w.r.t. x, we get n × ç ÷ × × =0 × + n× ç ÷ a b dx è bø è aø n x n - 1 n y n - 1 dy dy b nx n - 1 Þ + × =0 Þ = - n n-1 × n n dx dx a b a y - b na n - 1 b æ dy ö = n n-1 = - × \ ç ÷ a è dx ø ( a, b ) a b So, the equation of the tangent at ( a , b) is i.e., ay + bx = 2 ab. x y + = 2, which is independent of n. \ a b

EXAMPLE 7

y + 2 -1 = , x+2 3

æxö æyö Prove that the curve ç ÷ + ç ÷ = 2 touches the straight line è aø è bø x y + = 2 at the point ( a , b), whatever be the value of n. [CBSE 2007C] a b n

SOLUTION

y - 2 -1 = , x-2 3

y - b -b = , x-a a

At what points will the tangent to the curve y = 2x 3 - 15 x 2 + 36x - 21 be parallel to the x-axis? Also, find the equations of tangents to the curve at these points. [CBSE 2008]

Senior Secondary School Mathematics for Class 12 Pg-558

558 SOLUTION

Senior Secondary School Mathematics for Class 12

Let the required point be P( x1 , y1). é dy ù Then, ê ú = 0. ë dx û ( x 1, y 1 ) Now, y = 2x 3 - 15 x 2 + 36x - 21 dy Þ = 6x 2 - 30x + 36 = 6( x 2 - 5 x + 6) dx é dy ù Þ ê ú = 6( x12 - 5 x1 + 6). ë dx û ( x 1, y 1 ) é dy ù =0 Þ \ ê ú ë dx û ( x 1, y 1 ) Þ

EXAMPLE 8

SOLUTION

6( x12 - 5 x1 + 6) = 0 x12 - 5 x1 + 6 = 0

Þ ( x1 - 2)( x1 - 3) = 0 Þ x1 = 2 or x1 = 3. Since P( x1 , y1) lies on the given curve, we have y1 = 2x13 - 15 x12 + 36x1 - 21. \ x1 = 2 Þ y1 = ( 2 ´ 8 - 15 ´ 4 + 36 ´ 2 - 21) = 7. x1 = 3 Þ y1 = ( 2 ´ 27 - 15 ´ 9 + 36 ´ 3 - 21) = 6. So, the required points are P( 2, 7) and P¢( 3 , 6). The equations of tangents at these points are y = 7 and y = 6 respectively. xù é Prove that all points of the curve y 2 = 4 a ê x + a sin ú at which the aû ë tangent is parallel to the axis of x, lie on a parabola. Let the required point be ( x1 , y1). æ dy ö Then, we must have ç ÷ = 0. è dx ø ( x 1, y 1 ) xù dy xù é é Now, y 2 = 4 a ê x + a sin ú Þ 2y × = 4 a ê1 + cos ú aû dx a ë û ë xù x ù é é 2 a ê1 + cos ú 2 a ê1 + cos 1 ú dy dy æ ö aû a û Þ ë Þ = ë = × ç ÷ dx y y dx è ø ( x 1, y 1 ) 1 æ dy ö \ ç ÷ =0 Þ è dx ø ( x 1, y 1 )

x ù é 2 a ê1 + cos 1 ú x ù é aû ë = 0 Þ ê1 + cos 1 ú = 0 y1 aû ë

x1 x x = -1 Þ sin 1 = 1 - cos2 1 = 0. a a a But, ( x1 , y1) lies on the given curve. x ù x é é ù \ y12 = 4 a ê x1 + a sin 1 ú Þ y12 = 4 ax1 Q sin 1 = 0ú × ê aû a ë ë û This shows that ( x1 , y1) lies on the parabola y 2 = 4 ax. Þ cos

Senior Secondary School Mathematics for Class 12 Pg-559

Applications of Derivatives

559

EXAMPLE 9

Tangents are drawn from the origin to the curve y = sin x. Prove that their points of contact lie on the curve x 2 y 2 = ( x 2 - y 2).

SOLUTION

Let the point of contact be ( x1 , y1). dy æ dy ö Now, y = sin x Þ = cos x1. = cos x Þ ç ÷ dx è dx ø ( x 1, y 1 ) y - y1 = cos x1. x - x1 - y1 Since the tangent passes through the origin, we have = cos x1 , - x1 y … (i) i.e., 1 = cos x1 x1 But, ( x1 , y1) lies on the curve y = sin x. \ y1 = sin x1 … (ii) \ the equation of the tangent at ( x1 , y1) is

Squaring (i) and (ii) and adding, we get y12 + y12 = 1 Þ y12 + x12 y12 = x12 x12 Þ x12 y12 = ( x12 - y12). This shows that ( x1 , y1) lies on the curve x 2 y 2 = ( x 2 - y 2). EXAMPLE 10

SOLUTION

Determine the points on the curve 2y = ( 3 - x 2) at which the tangent is parallel to the line x + y = 0. Let the required point be ( x1 , y1). Then, slope of the tangent at ( x1 y1) = slope of the given line. æ dy ö = -1 \ ç ÷ [Q x + y = 0 Þ y = - x]. è dx ø ( x 1, y 1 ) Now, 2y = ( 3 - x 2) Þ 2 ×

dy dy æ dy ö = -2x Þ = -x Þ ç ÷ = - x1 . dx dx è dx ø ( x 1, y 1 )

\ - x1 = - 1 Þ x1 = 1. Since ( x1 , y1) lies on the curve 2y = ( 3 - x 2), we have 2y1 = ( 3 - x12). æ 3 - 12 ö ÷ = 1. y1 = ç ç 2 ÷ è ø Hence, the required point is (1, 1).

When x1 = 1, we have

EXAMPLE 11

Find the points on the curve 4x 2 + 9y 2 = 1, where the tangents are perpendicular to the line 2y + x = 0.

SOLUTION

1 The equation of the given line is y = - x. 2 1 \ slope of this line = 2 Let the required point be ( x1 , y1).

… (i)

Senior Secondary School Mathematics for Class 12 Pg-560

560

Senior Secondary School Mathematics for Class 12

dy =0 dx dy æ -4x ö dy - 4x1 ÷ Þ æç ö÷ Þ =ç = × dx çè 9y ÷ø 9y1 è dx ø ( x 1, y 1 )

Now, 4x 2 + 9y 2 = 1 Þ 8x + 18y ×

\ slope of the tangent at ( x1 , y1) = From (i) and (ii), we have

- 4x1 9y1

… (ii)

2 - 4x1 æ 1 ö ´ ç - ÷ = -1 or y1 = - x1 … (iii) 9y1 è 2 ø 9

Since ( x1 , y1) lies on the curve 4x 2 + 9y 2 = 1, we have 4 [using (iii)]. 4x12 + 9y12 = 1 Þ 4 x12 + 9 ´ x12 = 1 81 9 3 \ x12 = or x1 = ± × 40 2 10 2 3 1 So, y1 = ± ´ =± × 9 2 10 3 10 Hence, the required points are 1 ö -1 ö æ 3 æ -3 , , ç ÷ and ç ÷× è 2 10 3 10 ø è 2 10 3 10 ø EXAMPLE 12

SOLUTION

Find the coordinates of the points on the curve y = x 2 + 3 x + 4, the tangents at which pass through the origin. Let the required point be ( x1 , y1). dy æ dy ö Now, y = x 2 + 3 x + 4 Þ = ( 2x1 + 3). = 2x + 3 Þ ç ÷ dx è dx ø ( x 1, y 1 ) So, the equation of the tangent at ( x1 , y1) is

y - y1 = ( 2x1 + 3). x - x1

Since the tangent passes through the origin, we have - y1 = ( 2x1 + 3) - x1 Þ

y1 = ( 2x12 + 3 x1)

But, ( x1 , y1) lies on the given curve. So, y1 =

… (i) x12

+ 3 x1 + 4

… (ii)

From (i) and (ii), we get 2x12 + 3 x1 = x12 + 3 x1 + 4 or x12 = 4 or x1 = ± 2. Now, x1 = 2 Þ y1 = ( 22 + 3 ´ 2 + 4) = 14. x1 = -2 Þ y1 = [( -2) 2 + 3 ´ ( -2) + 4] = 2. Hence, the required points are ( 2, 14) and ( -2, 2).

And,

EXAMPLE 13

If the straight line x cos a + y sin a = p touches the ellipse prove that p 2 = a 2 cos2 a + b 2 sin 2 a .

x2 a2

+

y2 b2

= 1,

Senior Secondary School Mathematics for Class 12 Pg-561

Applications of Derivatives SOLUTION

561

Let the point of contact be ( x1 , y1). Then, the equation of the tangent at ( x1 , y1) is This is identical with x cos a + y sin a = p Comparing coefficients, we have Þ \

a2

+

yy1 b2

= 1 …(i) … (ii)

( x1 /a 2) ( y1 /b 2) 1 = = cos a sin a p ( x1 /a) ( y1 /b) 1 = = × a cos a b sin a p

x1 a cos a y b sin a = and 1 = × a p b p

Squaring and adding, we get Þ Þ EXAMPLE 14

xx1

x12 a

2

+

y12 b

2

=

a 2 cos2 a + b 2 sin 2 a p2

é x12 y12 ù = 1 êQ 2 + 2 = 1ú 2 p b ë a û a 2 cos2 a + b 2 sin 2 a = p 2. a 2 cos2 a + b 2 sin 2 a

If x cos a + y sin a = p touches the curve x my n = a m+ n , prove that p m+ n × m m × nn = (m + n) m+ n cosm a sin n a .

SOLUTION

Let the point of contact be ( x1 , y1). The equation of the given curve is x my n = a m+ n. This, on differentiation, gives dy mx m -1 y n + x m × ny n -1 × =0 Þ dx - my1 æ dy ö = × \ ç ÷ nx1 è dx ø ( x 1, y 1 )

dy - my = × dx nx

So, the equation of the tangent at ( x1 , y1) is

y - y1 - my1 , = x - x1 nx1

i.e., my1 x + nx1 y = (m + n) x1 y1 … (i) This is identical with x cos a + y sin a = p … (ii) (m + n) x1 y1 my1 nx1 Comparing coefficients, we get = = × cos a sin a p pm pn and y1 = × \ x1 = (m + n) cos a (m + n) sin a Since ( x1 , y1) lies on the given curve, x1my1n = a m+ n. m

\

n

ü ü ì ì pm pn m+ n ý =a ý ×í í + m n m n ( + ) cos a ( ) sin a þ þ î î

or p m+ n × m m × nn = (m + n) m+ n × a m+ n cosm a sin n a .

Senior Secondary School Mathematics for Class 12 Pg-562

562 EXAMPLE 15 SOLUTION

Senior Secondary School Mathematics for Class 12

Find the equation of the normal to the curve y = 2 sin 2 3 x at x =

p × 6

p æ 3p ö , we have y = 2 sin 2 ç ÷ = 2. 6 è 6 ø æp ö So, the point of contact is ç , 2÷ × è6 ø dy 2 Now, y = 2 sin 3 x Þ = ( 4 sin 3 x) ´ 3 ´ (cos 3 x) dx dy Þ = 12 sin 3 x cos 3 x = 6 sin 6x. dx pö pö æ æ dy ö æ \ ç ÷ at ç x = ÷ = 6 sin ç 6 ´ ÷ = 6 sin p = 0. 6ø 6ø è è dx ø è So, the tangent is parallel to the x-axis. Thus, the normal is parallel to the y-axis and passes through the æp ö point ç , 2÷ × è6 ø p So, the equation of the normal is x = × 6 When x =

EXAMPLE 16

Find the equations of the tangent and the normal to the curve y( x - 2)( x - 3) - x + 7 = 0 at the point where it cuts the x-axis.

SOLUTION

The curve cuts the x-axis, where y = 0. Putting y = 0 in the given equation, we get \ the point of contact is (7 , 0).

[CBSE 2010]

EXAMPLE 17

SOLUTION

x = 7.

Now, y( x - 2)( x - 3) - x + 7 = 0 Þ y( x 2 - 5 x + 6) - x + 7 = 0 dy dy 1 + 5 y - 2xy Þ ( x 2 - 5 x + 6) × + y( 2x - 5) - 1 = 0 Þ = 2 × dx dx x - 5x + 6 æ 1 + 5 ´ 0 - 2 ´7 ´ 0 ö 1 æ dy ö × =ç \ ç ÷ ÷= 49 - 35 + 6 ø 20 è dx ø ( 7, 0) è y-0 1 So, the equation of the tangent is or x - 20y - 7 = 0. = x - 7 20 y-0 Also, the equation of the normal is = -20 or 20x + y - 140 = 0. x -7 x y Show that + = 1 touches the curve y = be- x / a at the point where the a b [CBSE 2005C, ‘07C] curve crosses the axis of y. The equation of the curve is y = be- x / a. It crosses the y-axis at the point where x = 0. Putting x = 0 in the equation of the curve, we get y = b. So, the point of contact is ( 0, b). Now,

y = be- x / a Þ

dy - be- x / a = × dx a

Senior Secondary School Mathematics for Class 12 Pg-563

Applications of Derivatives

\

563

b æ dy ö =- × ç ÷ a è dx ø ( 0, b )

So, the equation of the tangent is x y y - b -b + = 1. = Þ bx + ay = ab Þ a b x-0 a x y Hence, + = 1 touches the curve y = be- x / a at ( 0, b). a b EXAMPLE 18

Find the equations of the tangent and the normal at the point ‘t’ on the [CBSE 2009C, ’14] curve x = a sin 3 t , y = b cos 3 t.

SOLUTION

We have \

dx = 3 a sin 2 t cos t and dt dy dy/dt -3 b cos2 t sin t = = = dx dx/dt 3 a sin 2 t cos t

dy = -3 b cos2 t sin t. dt - b cos t × a sin t

\ the equation of the tangent at the point ‘t‘ is y - b cos 3 t

- b cos t = a sin t x - a sin 3 t Þ ay sin t + bx cos t = ab sin t cos t x y Þ + = 1 [on dividing throughout by ab sin t cos t]. a sin t b cos t Also, the equation of the normal at the point ‘t‘ is y - b cos 3 t a sin t = x - a sin 3 t b cos t Þ ax sin t - by cos t = ( a 2 sin 4 t - b 2 cos4 t). EXAMPLE 19

SOLUTION

Find the equations of the tangent and normal to the curve p [CBSE 2008] x = a sin 3t , y = cos 2t at t = × 4 We have dx x = sin 3t Þ = 3 cos 3t dt dy and y = cos 2t Þ = -2 sin 2t. dt dy ( dy dt) -2 sin 2t = = \ 3 cos 3t dx ( dx dt) pö æ -2 sin ç 2 ´ ÷ é dy ù 4 ø -2 sin(p 2) 2 2 è = × Þ ê ú = = p p ö -3 cos(p 4) 3 æ ë dx û t = 3 3 ´ cos ç ÷ 4 4ø è p 3p pö p 1 æ When t = , we have x = sin = sin ç p - ÷ = sin = 4 4 4ø 4 2 è 2p p and y = cos = cos = 0. 4 2

Senior Secondary School Mathematics for Class 12 Pg-564

564

Senior Secondary School Mathematics for Class 12

æ 1 ö , 0÷ × \ point of contact is P ç è 2 ø Equation of the tangent at the point P is given by y-0 2 2 = 1 3 x2 2y 2 2 Þ = 3 2x - 1 Þ 3 2y = 4x - 2 2 Þ 2 2x - 3 y - 2 = 0. Equation of the normal at the point P is given by y-0 -3 -3 2y = Þ = 1 2 2 2x - 1 2 2 x2 Þ 4y = -3 2x + 3 Þ 3 2x + 4y - 3 = 0. EXAMPLE 20

For the curve y = 4x 3 - 2x5 , find all the points on the curve at which the

SOLUTION

tangent passes through the origin. The equation of the given curve is

[CBSE 2013C]

… (i) y = 4x 3 - 2x5. On differentiating (i) w.r.t. x, we get dy = 12x 2 - 10x 4. dx Let the required point be P( x , y). The equation of tangent to the given curve at P( x , y) is given by … (ii) Y - y = (12x 2 - 10x 4)(X - x). If it passes through (0, 0) then, we have y = (12x 2 - 10x 4) x Þ 4x 3 - 2x5 = 12x 3 - 10x5

[using (i)]

Þ 8x 3 - 8x5 = 0 Þ 8x 3(1 - x 2) = 0 Þ x = 0 or x = 1 or x = -1. Putting x = 0 in (i), we get y = 0. Putting x = 1 in (i), we get y = 2. Putting x = -1 in (i), we get y = -2. So, the required points are (0, 0), (1, 2) and (–1, –2).

EXERCISE 11H 1. Find the slope of the tangent to the curve (i) y = ( x 3 - x) at x = 2 (ii) y = ( 2x 2 + 3 sin x) at x = 0 p (iii) y = (sin 2x + cot x + 2) 2 at x = 2

Senior Secondary School Mathematics for Class 12 Pg-565

Applications of Derivatives

565

Find the equations of the tangent and the normal to the given curve at the indicated point: æ a 2a ö 2. y = x 3 - 2x + 7 at (1, 6) 3. y 2 = 4ax at ç 2 , ÷ èm m ø 4.

x2

+

y2

= 1 at ( a cos q, b sin q)

a 2 b2 6. y = x 3 at P(1, 1)

[CBSE 2006C]

x2

-

y2

= 1 at ( a sec q, b tan q) a 2 b2 7. y 2 = 4ax at ( at 2 , 2at) 5.

p 9. 16x 2 + 9y 2 = 144 at ( 2, y1), where y1 > 0 4 10. y = x 4 - 6x 3 + 13 x 2 - 10x + 5 at the point where x = 1 æ a2 a2 ö 11. Find the equation of the tangent to the curve x + y = a at ç , ÷ × ç 4 4÷ è ø 8. y = cot 2 x - 2 cot x + 2 at x =

12. Show that the equation of the tangent to the hyperbola ( x1 , y1) is

xx1 a2

-

yy1 b2

x2 a2

-

y2 b2

= 1 at

= 1.

13. Find the equation of the tangent to the curve y = ( sec4 x - tan 4 x) at x =

p × 3

14. Find the equation of the normal to the curve y = (sin 2x + cot x + 2) 2 at p x= × 2 15. Show that the tangents to the curve y = 2x 3 - 4 at the points x = 2 and [CBSE 2001C] x = -2 are parallel. 16. Find the equation of the tangent to the curve x 2 + 3 y = 3 , which is parallel [CBSE 2005] to the line y - 4x + 5 = 0. 17. At what points on the curve x 2 + y 2 - 2x - 4y + 1 = 0, is the tangent parallel [CBSE 2000, ‘01C, ‘02C] to the y-axis? 18. Find the points on the curve x 2 + y 2 - 2x - 3 = 0 where the tangent is [CBSE 2011] parallel to the x-axis. 19. Prove that the tangents to the curve y = x 2 - 5 x + 6 at the points (2, 0) and (3, 0) are at right angles. 20. Find the points on the curve y = x 2 + 3 x + 4 at which the tangent passes through the origin. 21. Find the point on the curve y = x 3 - 11x + 5 at which the equation of [CBSE 2012] tangent is y = x - 11. 22. Find the equations of the tangents to the curve 2x 2 + 3 y 2 = 14, parallel to the line x + 3 y = 4.

Senior Secondary School Mathematics for Class 12 Pg-566

566

Senior Secondary School Mathematics for Class 12

23. Find the equation of the tangent to the curve x 2 + 2y = 8, which is perpendicular to the line x - 2y + 1 = 0. 24. Find the point on the curve y = 2x 2 - 6x - 4 at which the tangent is parallel to the x-axis. 25. Find a point on the parabola y = ( x - 3) 2 , where the tangent is parallel to the chord joining the points (3, 0) and (4, 1). [CBSE 2005C] 26. Show that the curves x = y 2 and xy = k cut at right angles if 8k 2 = 1. [CBSE 2005, ‘07C]

27. Show that the curves xy = a 2 and x 2 + y 2 = 2a 2 touch each other. [CBSE 2002] 28. Show that the curves x 3 - 3 xy 2 + 2 = 0 and 3 x 2 y - y 3 - 2 = 0 cut orthogonally. 29. Find the equation of tangent to the curve x = ( q + sin q), y = (1 + cos q) at p [CBSE 2006C] q= × 4 p 30. Find the equation of the tangent at t = for the curve x = sin 3t , y = cos 2t. 4 ANSWERS (EXERCISE 11H)

1. (i) 11 (ii) 3 (iii) –12 2. x - y + 5 = 0, x + y - 7 = 0 3. m 2 x - my + a = 0, m 2 x + m 3 y - 2am 2 - a = 0 4. bx cos q + ay sin q = ab , ax sec q - by cosec q = ( a 2 - b 2) 5. bx sec q - ay tan q = ab , by cot q + ax cos q = ( a 2 + b 2) 6. y = 3 x - 2, x + 3 y = 4 7. x - ty + at 2 = 0, tx + y = at 3 + 2 at p 9. 8x + 3 5 y - 36 = 0, 9 5 x - 24y + 14 5 = 0 8. y = 1, x = 4 10. 2x - y + 1 = 0, x + 2y - 7 = 0 11. 2( x + y) = a 2 13. 3 y - 48 3 x + 16 3 p - 21 = 0 16. 4x - y + 13 = 0 18. (1, 2), (1, - 2) 21. (2, –9) 23. 2x + y - 6 = 0 29. y = (1 - 2) x +

14. 24y - 2x + p - 96 = 0

17. ( -1, 2) and (3, 2) 20. ( -2, 2), ( 2, 14) 22. x + 3 y = 7 , x + 3 y = -7 3 17 ö æ æ7 1ö 25. ç , ÷ 24. ç , ÷ è2 2 ø è 2 4ø ( 2 - 1) p +2 4

30. 4x - 3 2y - 2 2 = 0

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 11H) 15. x = 2 Þ y = ( 2 ´ 8 - 4 ) = 12. x = - 2 Þ y = 2 ´ ( -8 ) - 4 = - 20.

Senior Secondary School Mathematics for Class 12 Pg-567

Applications of Derivatives

567

dy æ dy ö æ dy ö = 24 Þ m1 = m 2 . = 6x 2 Þ ç ÷ = 24 and ç ÷ dx è dx ø (x = - 2 ) è dx ø (2 , 12 ) dy dy -2 x =0 Þ = × 3 dx dx y - 4 x + 5 = 0 Þ y = 4 x + 5 Þ its slope = 4. \ the slope of the tangent = 4. -2 x So, = 4 Þ x = - 6. 3 2 x + 3 y = 3 , x = - 6 Þ y = -11.

16. x 2 + 3 y = 3 Þ 2 x + 3

The equation of the tangent at ( -6 , -11) with slope = 4 is y + 11 = 4 ( x + 6 ). 17. x 2 + y 2 - 2 x - 4 y + 1 = 0 Þ 2 x Now,

dx dx æ y - 2 ö dx ÷× + 2y - 2 -4=0 Þ =ç dy dy dy è 1 - x ø

y-2 dx =0 Þ = 0 Þ y - 2 = 0 Þ y = 2. dy 1- x

x 2 + 4 - 2 x - 8 + 1 = 0 Þ x 2 - 2 x - 3 = 0 Þ ( x - 3 )( x + 1) = 0 Þ x = 3 or x = -1. \ the required points are (3, 2) and ( -1, 2 ). dy dy 18. 2 a2 = 0 Þ 3 x( x - 2 a) = 0 Þ x = 0 or x = 2 a. = 3 x 2 - 6 ax and so, dx dx dy = ( 2 x + 3 ). dx Let the required point be P( a, b ).

20. y = x 2 + 3 x + 4 Þ

é dy ù Then, ê ú = 2 a + 3. ë dx û (a, b) \ the tangent is

y-b -b = 2a + 3 Þ = 2 a + 3 [B the tangent passes through (0, 0)]. -a x-a

\ b = 2 a2 + 3 a

… (i)

Also, ( a, b ) lies on the given curve. \ b = a2 + 3 a + 4 2

2

… (ii)

2

\ 2 a + 3 a = a + 3 a + 4 Þ a = 4 Þ a = ± 2. Now, ( a = 2 Þ b = 14 ) and ( a = - 2 Þ b = 2 ). 21. The given curve is y = x 3 - 11x + 5. Let ( a, b ) be the point on the given curve, the tangent at which is y = x - 11. \ slope of the tangent = 1. dy æ dy ö Now, y = x 3 - 11x + 5 Þ = ( 3 x 2 - 11) Þ ç ÷ = ( 3 a2 - 11). dx è dx ø (a, b) 3 a2 - 11 = 1 Þ 3 a2 = 12 Þ a2 = 4 Þ a = ±2 . Since the point ( a, b ) lies on the curve y = x 3 - 11x + 5, we have b = a3 - 11a + 5. Now, a = 2 Þ b = ( 2 3 - 11 ´ 2 + 5 ) = ( 13 - 22 ) = - 9. And, a = -2 Þ b = [( -2 ) 3 - 11 ´ ( -2 ) + 5] = ( -8 + 27 ) = 19. Thus, the desired points can be (2, –9) and (–2, 19). But, (2, –9) satisfies (i) while (–2, 19) does not satisfy it. Hence, the required point is (2, –9).

… (i)

Senior Secondary School Mathematics for Class 12 Pg-568

568

Senior Secondary School Mathematics for Class 12

1 4 x+ × 3 3 -1 Slope of tangent = slope of given line = × 3 Let the point of contact be ( x1 , y1 ). dy dy -4 x -2 x -2 x æ dy ö Now, 2 x 2 + 3 y 2 = 14 Þ 4 x + 6 y = × =0 Þ = = Þ ç ÷ 6y 3y dx dx è dx ø (x1, y1 ) 3 y1 -2 x1 -1 \ = Þ 2 x1 = y1 3 y1 3 Also, ( x1 , y1 ) lies on the given curve.

22. The given line is x + 3 y = 4 Þ y = -

\ 2 x12 + 3 y12 = 14 Þ 2 x12 + 3 ´ 4 x12 = 14 Þ x12 = 1 Þ x1 = ±1. ( x1 = 1 Þ y1 = 2 ) and ( x1 = -1 Þ y1 = -2 ). So, the points of contact are (1, 2) and (–1, –2). The respective tangents are y - 2 -1 y + 2 -1 and = = x-1 3 x+ 1 3 Þ 3 y - 6 = - x + 1 and 3 y + 6 = - x - 1 Þ 3 y + x = 7 and 3 y + x = -7 . 25. Let the required point be P( x1 , y1 ). ( 1 - 0) Slope of the line joining A( 3 , 0 ) and B( 4 , 1) is = 1. Now, y = ( x - 3 ) 2 ( 4 - 3) dy 7 æ dy ö = 2( x - 3 ) Þ ç ÷ = 2( x1 - 3 ) Þ 2( x1 - 3 ) = 1 Þ x1 = × 2 dx è dx ø (x1, y1 ) 2 1 ö æ7 2 2 \ y = ( x - 3 ) Þ y1 = ( x1 - 3 ) = ç - 3 ÷ = × 4 è2 ø æ7 1ö Hence, the required point is ç , ÷ × è2 4ø 26. Let the point of intersection of the given curves be ( x1 , y1 ). Þ

dy 1 1 æ dy ö =1 Û ç ÷ = Û m1 = × dx 2 y1 è dx ø (x1, y1 ) 2 y1 dy dy - y -y æ dy ö xy = k Û x + y=0 Û = Û ç ÷ = m2 = 1 × dx dx x x1 è dx ø (x1, y1 ) - y1 1 1 m1m 2 = -1 Û ´ = - 1 Û x1 = × 2 y1 x1 2 1 2 2 2 ( x1 , y1 ) lies on y = x Û y1 = x1 Û y1 = × 2 ( x1 , y1 ) lies on xy = k Û x1 y1 = k Û x12 y12 = k 2 æ 1 1ö 1 Û k2 = ç ´ ÷ = × è4 2ø 8 Û 8 k 2 = 1.

Then, y 2 = x Û 2 y

\

27. Solving the given equations, we get their points of intersection as ( a, a) and ( - a, - a). xy = a2 Þ

dy - y æ dy ö = Þ ç ÷ = -1 = m1 . dx x è dx ø (a, a)

x 2 + y 2 = 2 a2 Þ

dy - x æ dy ö = Þ ç ÷ = -1 = m 2 . dx y è dx ø (a, a)

Senior Secondary School Mathematics for Class 12 Pg-569

Applications of Derivatives ½ m - m2 ½ ½ = 0 Þ f = 0. \ tan f =½ 1 ½1 + m1m 2½ So, the given curves touch each other. 28. m1m 2 = -1. dx 29. x = ( q + sin q) Þ = ( 1 + cos q). dq dy y = ( 1 + cos q) Þ = - sin q. dq dy dy ( dq) - sin q = = × \ dx ( dx dq) ( 1 + cos q) - sin ( p4 ) p -1 1 æ dy ö ç ÷ at q = = = × 4 ( 1 + cos p4 ) 2 (1 + 1 è dx ø

2

569

)

( 2 - 1) -1 ´ = ( 1 - 2 ). ( 2 + 1) ( 2 - 1) p pö æp 1 ö æp ÷ , and At q = , we have x = ç + sin ÷ = ç + 4 4ø è4 2ø è4 pö æ 1 ö æ ÷× y = ç 1 + cos ÷ = ç 1 + 4ø è 2ø è Required equation of tangent is 1 ö æ ÷ y - ç1+ ( 2 - 1)p 2ø è = ( 1 - 2 ) Þ y = ( 1 - 2 )x + + 2. 1 ö æp 4 ÷ x-ç + 2ø è4 =

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 1. If y = 2x then

dy =? dx

(a) x( 2x -1)

(b)

2. If y = log10 x then (a)

1 x

3. If y = e1/x then

2x (log 2)

dy =? dx 1 (b) (log 10) x

(a) x x log x

(d) none of these

1 x(log 10)

(d) none of these

(c) e1/x log x

(d) none of these

(c)

dy =? dx

1 (1/x -1) -e1/x (b) 2 ×e x x dy x 4. If y = x then =? dx (a)

(c) 2x (log 2)

(b) x x (1 + log x) (c) x(1 + log x)

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-570

570

Senior Secondary School Mathematics for Class 12

5. If y = x sin x then

dy =? dx

(sin x -1 )

(sin x - 1 )

(b) (sin x cos x) × x (a) (sin x) × x sin x ì sin x + x log x × cos x ü (d) none of these (c) x ý í x þ î dy 6. If y = x x then =? dx x x log x ì 2 + log x ü ( x - 1) (b) (c) x x í (a) x × x ý (d) none of these 2 x î 2 x þ dy 7. If y = esin x then =? dx esin x cos x esin x (d) none of these (c) (a) esin x × cos x (b) 2 x 2 x dy 8. If y = (tan x) cot x then =? dx (a) cot x × (tan x) cot x - 1 × sec2 x

(b) - (tan x) cot x × cosec2 x

(c) (tan x) cot x × cosec2 x(1 - log tan x) (d) none of these 9. If y = (sin x)

log x

then

dy =? dx

(log x - 1 )

(a) (log x) × (sin x) × cos x log x ì x log x + log sin x ü (b) (sin x) ×í ý x þ î log x ì( x log x) cot x + log sin (c) (sin x) ×í x î (d) none of these dy 10. If y = sin ( x x ) then =? dx

xü ý þ

(a) x x cos ( x x )

(b) x x cos x x (1 + log x)

(c) x x cos x x log x

(d) none of these

dy 11. If y = x sin x then =? dx ( x cos x + sin x) (a) 2 x sin x 1 (c) 2 x sin x 12. If ex + y = xy then (a)

x(1 - y) y( x - 1)

(b)

1 ( x cos x + sin x) × x sin x 2

(d) none of these

dy =? dx (b)

y(1 - x) x( y - 1)

(c)

( x - xy) ( xy - y)

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-571

Applications of Derivatives

13. If ( x + y) = sin ( x + y) then (a) –1

dy =? dx

(b) 1

- x y

(c)

1 - cos ( x + y) cos2( x + y)

(d) none of these

dy =? dx

14. If x + y = a then (a)

571

1 y (b) - × 2 x

(c)

- y

(d) none of these

x

dy =? dx ( y - x log y) y( y - x log y) y( y + x log y) (b) (c) (d) none of these (a) ( x - y log x) x( x - y log x) x( x + y log x)

15. If x y = y x then

16. If x p y q = ( x + y) (a)

( p+ q )

x y

then

(b)

dy =? dx

y x

(c)

1 dy then =? x dx 1 1 (a) x sin - cos x x 1 1 (c) - x sin + cos x x

xp -1

(d) none of these

yq -1

17. If y = x 2 sin

18. If y = cos2 x 3 then

(b) - cos

(d) none of these

dy =? dx

(a) -3 x 2 sin ( 2x 3) (b) -3 x 2 sin 2 x 3 x 2 + a 2 ) then

19. If y = log ( x + (a) (c)

2( x +

(c) -3 x 2 cos2( 2x 3) (d) none of these

dy =? dx

1 2

1 1 + 2x sin x x

(b)

2

x +a )

1

-1 2

x + a2

(d) none of these

2

x + a2

æ1 + x ö dy ÷ then 20. If y = log çç =? ÷ dx è1 - x ø (a)

1 x (1 - x)

(b)

-1 x(1 - x ) 2

(c)

- x 2(1 - x )

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-572

572

Senior Secondary School Mathematics for Class 12

æ 1 + x2 + x ö ÷ then dy = ? 21. If y = log ç ç 1 + x2 - x ÷ dx è ø 2

(a)

1+ x

2

(b)

2 1 + x2 x

2

-2

(c)

1 + x2

(d) none of these

dy 1 + sin x then =? 1 - sin x dx

22. If y =

1 æp xö sec2 ç - ÷ 2 è 4 2ø 1 æp xö æp xö (c) cosec ç - ÷ cot ç - ÷ 2 4 2 è ø è 4 2ø

(a)

23. If y =

(b)

1 æp xö cosec2 ç - ÷ 2 è 4 2ø

(d) none of these

dy sec x - 1 then =? dx sec x + 1

(a) sec2 x

(b)

x 1 sec2 2 2

(c)

x -1 cosec2 2 2

(d) none of these

1 + tan x dy then =? 1 - tan x dx

24. If y =

pö 1 æ (a) sec2 x × tan ç x + ÷ 2 4ø è æxö sec2 ç ÷ è 4ø

(c)

(b)

pö æ sec2 ç x + ÷ 4ø è pö æ 2 tan ç x + ÷ 4ø è

(d) none of these

pö æ tan ç x + ÷ 4ø è

æ 1 - cos x ö dy ÷÷ then 25. If y = tan -1 çç =? dx sin x è ø (a) 1

(b) –1

(c)

1 2

(d)

-1 2

(c)

1 2

(d)

-1 2

ì cos x + sin x ü dy 26. If y = tan -1í =? ý then dx cos x sin x þ î (a) 1

(b) –1

ì cos x ö dy ÷÷ then 27. If y = tan -1í =? dx î1 + sin x ø (a)

1 2

(b)

-1 2

(c) 1

(d) –1

Senior Secondary School Mathematics for Class 12 Pg-573

Applications of Derivatives

dy 1 - cos x then =? dx 1 + cos x

28. If y = tan -1 (a)

-1 2

(b)

1 2

(c)

1 (1 + x 2)

æ a cos x - b sin x ö dy ÷÷ then 29. If y = tan -1 çç =? dx b cos x + a sin x è ø a -b (a) (b) (c) 1 b a dy 30. If y = sin -1( 3 x - 4x 3) then =? dx 3 -4 3 (b) (c) (a) 2 2 1-x 1-x 1 + x2 dy 31. If y = cos-1( 4x 3 - 3 x) then =? dx 3 -3 4 (b) (c) (a) 2 2 1-x 1-x 1 - x2 æ a+ xö dy ÷ then 32. If y = tan -1 çç =? ÷ dx 1 ax ø è 1 1 (a) (b) (1 + x) x (1 + x) æ x2 - 1 ö ÷ then dy = ? 33. If y = cos-1 ç 2 ç x + 1÷ dx è ø 2 -2 (b) (a) (1 + x 2) (1 + x 2) æ 1 + x2 ö ÷ then dy = ? 34. If y = tan -1 ç ç 1 - x2 ÷ dx è ø 2x -2x (b) (a) 4 (1 + x ) (1 + x 4) 35. If y = cos-1 x 3 then (a)

-1 (1 + x)

-1 1-x

6

2 x (1 + x)

(c)

(c)

(c)

2x (1 + x 2)

x (1 + x 4)

(d) none of these

(d) –1

(d) none of these

(d)

(d)

-4 ( 3 x 2 - 1)

1 2 x (1 + x)

(d) none of these

(d) none of these

dy =? dx (b)

36. If y = cos-1 x 3 then (a)

573

2 (1 + x)

(c)

-1 2 x (1 + x)

(d) none of these

dy =? dx (b)

-3 x 2 1-x

6

-3

(c) x

2

1 - x6

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-574

574

Senior Secondary School Mathematics for Class 12

37. If y = tan -1( sec x + tan x) then 1 2

(a)

(b)

dy =? dx

-1 2

dy æ1- x ö 38. If y = cot -1 ç =? ÷ then dx è1 + x ø -1 1 (b) (a) 2 (1 + x ) (1 + x 2) 1+ x dy then =? 1-x dx 2

39. If y = (a) (c)

(1 - x)

2

(1 - x)

3

1 2

× (1 + x)

1

(c) 1

(c)

(b)

(d) none of these

1 (1 + x 2)

3

(d) none of these 2

x (1 - x)

3

2

(d) none of these 2

æ x2 + 1ö ÷ then dy = ? 40. If y = sec-1 ç 2 ç x -1÷ dx è ø -2 2 (b) (a) (1 + x 2) (1 + x 2) dy æ 1 ö 41. If y = sec-1 ç 2 =? ÷ then dx è 2x - 1 ø -2 -2 (a) (b) (1 + x 2) (1 - x 2) ìï 1 + x 2 - 1 üï dy 42. If y = tan -1í =? ý then x dx ïî ïþ 1 2 (b) (a) 2 (1 + x ) (1 + x 2)

(c)

(c)

(c)

-1 (1 - x 2)

-2 1 - x2

1 2(1 + x 2)

ì 1+ x + 1-x ü dy 43. If y = sin -1í =? ý then 2 dx î þ -1 1 1 (b) (c) (a) 2 2 2 1 ( + x 2) 2 1-x 2 1-x dy =? dx -1 (b) 2 t

(d) none of these

(d) none of these

(d) none of these

(d) none of these

44. If x = at 2 , y = 2at then (a)

1 t

(c)

-2 t

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-575

Applications of Derivatives

45. If x = asec q, y = b tan q then b sec q a

dy =? dx

b cosec q a dy 46. If x = a cos2 q, y = b sin 2 q then =? dx -a a (a) (b) cot q b b (a)

(b)

(c)

b cot q a

(d) none of these

(c)

-b a

(d) none of these

47. If x = a(cos q + q sin q) and y = a(sin q - q cos q) then (a) cot q 48. If y = x x (a)

x¼ ¥

(b) tan q dy then =? dx

y x(1 - log x)

(b)

1 ( 2y - 1)

(b)

1 2

( y - 1)

dy =? dx

(c) a cot q

y2 x(1 - log x)

49. If y = x + x + x + ¼ ¥ then (a)

575

(c)

y2 x(1 - y log x)

(d) a tan q

(d) none of these

dy =? dx (c)

2y 2

( y - 1)

dy =? dx cos x (c) ( 2y - 1)

(d) none of these

50. If y = sin x + sin x + sin x + ¼ ¥ then (a)

sin x ( 2y - 1)

(b)

cos x ( y - 1)

dy =? dx y (b) (1 - y)

(d) none of these

51. If y = ex + ex + ¼ ¥ then (a)

1 (1 - y)

(c)

y ( y - 1)

(d) none of these

ì sin 5 x , if x ¹ 0 ï 52. The value of k for which f ( x) = í 3 x is continuous at x = 0 is ïî k , if x = 0 (a)

1 3

(b) 0

(c)

3 5

(d)

5 3

1 ì ïx sin , if x ¹ 0 53. Let f ( x) = í x ïî 0 , when x = 0. Then, which of the following is the true statement? (a) f ( x) is not defined at x = 0

(b) lim f ( x) does not exist

(c) f ( x) is continuous at x = 0

(d) f ( x) is discontinuous at x = 0

x ®0

Senior Secondary School Mathematics for Class 12 Pg-576

576

Senior Secondary School Mathematics for Class 12

ì 3 x + 4 tan x , when x ¹ 0 ï 54. The value of k for which f ( x) = í x ïî k , when x = 0 is continuous at x = 0, is (a) 7 (b) 4 (c) 3 (d) none of these 3

55. Let f ( x) = x 2 . Then, f ¢( 0) = ? (a)

3 2

(b)

1 2

(c) does not exist (d) none of these

56. The function f ( x) = |x| V x Î R is (a) continuous but not differentiable at x = 0 (b) differentiable but not continuous at x = 0 (c) neither continuous nor differentiable at x = 0 (d) none of these ì1 + x , when x £ 2 57. The function f ( x) = í is î5 - x , when x > 2 (a) continuous as well as differentiable at x = 2 (b) continuous but not differentiable at x = 2 (c) differentiable but not continuous at x = 2 (d) none of these ìkx + 5 , when x £ 2 58. If the function f ( x) = í is continuous at x = 2 then k = ? î x - 1, when x > 2 (a) 2

(b) –2 (c) 3 (d) –3 ì1 - cos 4x ,x¹0 ï is continuous at x = 0 then k = ? 59. If the function f ( x) = í 8x 2 ïî k,x=0 1 -1 (a) 1 (b) 2 (c) (d) 2 2 ì sin 2 ax , when x ¹ 0 ï is continuous at x = 0 then k = ? 60. If the function f ( x) = í x 2 ï k , when x = 0 î (a) a

(b) a 2

(c) –2

p ì k cos x ïï( p - 2x) , when x ¹ 2 61. If the function f ( x) = í p ï 3 , when x = ïî 2 p be continuous at x = , then the value of k is 2 (a) 3 (b) –3 (c) –5

(d) – 4

(d) 6

Senior Secondary School Mathematics for Class 12 Pg-577

Applications of Derivatives

577

62. At x = 2, f ( x) = [x] is (a) continuous but not differentiable (b) differentiable but not continuous (c) continuous as well as differentiable (d) none of these ì x 2 - 2x - 3 , when x ¹ - 1 ï 63. Let f ( x) = í x + 12 ï k , when x = - 1. î If f ( x) is continuous at x = - 1 then k = ? (a) 4 (b) – 4 (c) –3 3

(d) 2

2

64. The function f ( x) = x - 6x + 15 x - 12 is (a) strictly decreasing on R (b) strictly increasing on R (c) increasing in ( -¥ , 2] and decreasing in ( 2, ¥ ) (d) none of these 65. The function f ( x) = 4 - 3 x + 3 x 2 - x 3 is (a) decreasing on R (c) strictly decreasing on R 66. The function f ( x) = 3 x + cos 3 x is (a) increasing on R (c) strictly increasing on R

(b) increasing on R (d) strictly increasing on R (b) decreasing on R (d) strictly decreasing on R

67. The function f ( x) = x 3 - 6x 2 + 9x + 3 is decreasing for (a) 1 < x < 3

(b) x > 1

(c) x < 1

(d) x < 1 or x > 3

3

68. The function f ( x) = x - 27 x + 8 is increasing when (a)|x| < 3 (b)|x| > 3 69. f ( x) = sin x is increasing in æp ö æ 3p ö (b) ç p , (a) ç , p ÷ ÷ 2 ø è2 ø è 2x is increasing in 70. f ( x) = log x

(c) -3 < x < 3

(d) none of these

(c) ( 0, p)

æ -p p ö (d) ç , ÷ è 2 2ø

(a) (0, 1) (b) (1, e) (c) ( e, ¥ ) 71. f ( x) = (sin x - cos x) is decreasing in æ 3p ö æ 3p 7p ö æ 7p ö (b) ç (c) ç , 2p ÷ (a) ç 0, , ÷ ÷ 4 ø è è 4 4ø è 4 ø x is 72. f ( x) = sin x (a) increasing in (0, 1) (b) decreasing in (0, 1)

(d) ( -¥ , e) (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-578

578

Senior Secondary School Mathematics for Class 12

æ (c) increasing in ç 0, è (d) none of these

1ö æ1 ö ÷ and decreasing in ç , 1÷ 2ø è2 ø

73. f ( x) = x x is decreasing in the interval æ 1ö (b) ç 0, ÷ è eø

(a) ( 0, e)

(c) (0, 1)

(d) none of these

(c) ( 2, ¥ )

(d) ( -¥ , ¥ )

74. f ( x) = x 2 e- x is increasing in (a) ( -2, 0)

(b) (0, 2)

75. f ( x) = sin x - kx is decreasing for all x Î R , when (a) k < 1

(b) k £ 1 3

(c) k > 1

(d) k ³ 1

(c) ( 3 , ¥ )

(d) (1, ¥ )

3

76. f ( x) = ( x + 1) ( x - 3) is increasing in (a) ( -¥ , 1)

(b) ( -1, 3) 2

77. f ( x) = [x( x - 3)] is increasing in (a) ( 0, ¥ )

(b) ( -¥ , 0) æ 3ö (d) ç 0, ÷ È ( 3 , ¥ ) è 2ø

(c) (1, 3)

78. If f ( x) = kx 3 - 9x 2 + 9x + 3 is increasing for every real number x, then (a) k > 3 (b) k ³ 3 x 79. f ( x) = 2 is increasing in ( x + 1) (a) ( -1, 1) (c) ( -¥ , - 1) È (1, ¥ )

(c) k < 3

(d) k £ 3

(b) ( -1, ¥ ) (d) none of these

80. The least value of k for which f ( x) = x 2 + kx + 1 is increasing on (1, 2), is (a) –2

(b) –1

(c) 1

(d) 2

81. f ( x) = |x|has (a) minimum at x = 0 (b) maximum at x = 0 (c) neither a maximum nor a minimum at x = 0 (d) none of these 82. When x is positive, the minimum value of x x is (a) ee

1

(b) e

e

æ log x ö 83. The maximum value of ç ÷ is è x ø 2 æ 1ö (b) (a) ç ÷ e è eø

(c) e

(c) e

-1

e

(d) ( 1 e )

(d) 1

Senior Secondary School Mathematics for Class 12 Pg-579

Applications of Derivatives

579

84. f ( x) = cosec x in ( -p , 0) has a maxima at -p -p (a) x = 0 (b) x = (c) x = 4 3

(d) x =

85. If x > 0 and xy = 1, the minimum value of ( x + y) is (a) –2 (b) 1 (c) 2 æ 2 250 ö 86. The minimum value of ç x + ÷ is x ø è (a) 0

(b) 25

(d) none of these

(c) 50 4

-p 2

(d) 75

3

87. The minimum value of f ( x) = 3 x - 8x - 48x + 25 on [0, 3] is (a) 16

(b) 25

(c) –39

(d) none of these

88. The maximum value of f ( x) = ( x - 2)( x - 3) 2 is (a)

7 3

(b) 3

(c)

4 27

(d) 0

89. The least value of f ( x) = ( ex + e- x ) is (a) –2

(b) 0

(c) 2

(d) none of these

ANSWERS (OBJECTIVE QUESTIONS)

1. (c)

2. (c)

3. (b)

4. (b)

5. (c)

6. (c)

7. (b)

8 (c)

9. (c) 10. (b)

11. (a) 12. (b) 13. (a) 14. (c) 15. (b) 16. (b) 17. (b) 18. (a) 19. (c) 20. (a) 21. (a) 22. (b) 23. (b) 24. (b) 25. (c) 26. (a) 27. (b) 28. (b) 29. (d) 30. (a) 31. (b) 32. (d) 33. (b) 34. (a) 35. (c) 36. (b) 37. (a) 38. (b) 39. (c) 40. (a) 41. (c) 42. (c) 43. (a) 44. (a) 45. (b) 46. (c) 47. (b) 48. (c) 49. (a) 50. (c) 51. (b) 52. (d) 53. (c) 54. (a) 55. (c) 56. (a) 57. (b) 58. (b) 59. (c) 60. (b) 61. (d) 62. (d) 63. (b) 64. (b) 65. (a) 66. (a) 67. (a) 68. (b) 69. (d) 70. (c) 71. (b) 72. (a) 73. (b) 74. (b) 75. (c) 76. (d) 77. (d) 78. (a) 79. (a) 80. (a) 81. (a) 82. (c) 83. (a) 84. (d) 85. (c) 86. (d) 87. (c) 88. (c) 89. (c)

HINTS TO THE GIVEN OBJECTIVE QUESTIONS 1.

dy = 2 x(log 2 ) dx

2. y =

log x log 10

3.

dy e1/x = dx - x 2

4. log y = x log x. Now differentiate w.r.t. x. 5. log y = sin x(log x ) 8. log y = cot x × log (tan x )

6. log y = x (log x )

7. log y = sin x

9. log y = (log x ) × (log sin x ) dy æ dy dz ö 10. Let x x = z. Then, y = sin z. Then, =ç ´ ÷× dx è dz dx ø

Senior Secondary School Mathematics for Class 12 Pg-580

580

Senior Secondary School Mathematics for Class 12

dy = ( x cos + sin x ). dx dy 1 1 dy 12. ( x + y ) = log x + log y Þ 1 + = + × × dx x y dx

11. y 2 = x sin x Þ 2 y ×

13. ( x + y ) = sin ( x + y ) Þ 1 + 14.

x+

y= a Þ

dy dy ù é = cos ( x + y ) × ê 1 + × dx dx úû ë

1 1 dy + × = 0. 2 x 2 y dx

15. x y = y x Þ y log x = x log y. Now, differentiate both sides w.r.t. x. 16. p log x + q log y = ( p + q) log ( x + y ). 17.

dy 1 ö æ -1 ö 1 æ = x 2 × ç cos ÷ ç 2 ÷ + 2 x sin × dx xøèx ø x è

18. y = (cos x 3 ) 2 Þ 19.

dy = dx ( x +

20. y = log ( 1 + Þ

dy = 2(cos x 3 )( - sin x 3 )( 3 x 2 ) = - 3 x 2 sin ( 2 x 3 ). dx

ì ü 1 ï ï × í1 + ´ 2xý = 2 2 2 2 2 x +a x + a ïî ïþ

1

1

2

x + a2

×

x ) - log ( 1 - x )

dy ì 1 =í dx î 2 x ( 1 +

x)

+

ü 1 1 2 × = ý= 2 x( 1 - x ) þ 2 x ( 1 - x )

1 × x( 1 - x )

21. y = log ( 1 + x 2 + x ) - log ( 1 + x 2 - x ). Now, differentiate. 1

ì æp ö ü2 ì 2 ïï 1 + cos ç 2 - x ÷ ïï ïï 2 cos è ø 22. y = í ý =í ï 1 - cos æç p - x ö÷ ï ï 2 sin 2 ïî ïî è2 ø ïþ

1

æ p x ö ü2 ç - ÷ï æp xö è 4 2 øï ý = cot ç - ÷ × æ p xöï è 4 2ø ç - ÷ è 4 2 ø ïþ

1

1

æ 1 - cos x ö 2 ïì 2 sin 2 ( x 2 ) üï 2 x ÷ =í 23. y = çç = tan × ÷ 2 x ý 1 2 cos + x ï ï 2 cos ( ) è ø î 2 þ 1

-

dy 1 ì ì p ö ü2 p öü æ æ 24. y = í tan ç x + ÷ ý Þ = í tan ç x + ÷ ý dx 2 î 4 øþ 4 øþ è è î

1 2

pö æ × sec2 ç x + ÷ × 4ø è

ì üï æ 1 - cos x ö 2 sin 2 ( x 2 ) ì ÷ = tan -1 ïí 25. y = tan -1 çç = tan -1 í tan ÷ x ) cos ( x ) ý 2 sin x sin ( î ï è ø 2 þ 2 îï

xü x ý= × 2þ 2

ì cos x + sin x ü æp öü æ p ö -1 ì 1 + tan x ü -1 ì 26. y = tan -1 í ý = tan í tan ç + x ÷ ý = ç + x ÷ × ý = tan í è4 øþ è 4 ø î î 1 - tan x þ î cos x - sin x þ ì æp ö ü ï sin ç 2 - x ÷ ïï æ p x öü æ p x ö è ø -1 ì 27. y = tan í ý = tan í tan ç - ÷ ý = ç - ÷ × p æ ö è 4 2 øþ è 4 2 ø î ï 1 + cos ç - x ÷ ï ïî è2 ø ïþ -1 ï

Senior Secondary School Mathematics for Class 12 Pg-581

Applications of Derivatives 1

2 x ü2 ì ( 2) ï xü x -1 ì 28. y = tan í ý = tan í tan ý = × 2þ 2 î ïî 2 cos 2 ( x 2 ) ïþ -1 ï 2 sin

29. y = tan

æ a ö ç - tan x ÷ æ tan q - tan x ö b ÷ = tan -1 ç ÷ ç 1 + tan q tan x ÷ a ç 1 + tan x ÷ è ø ç ÷ è ø b

-1 ç

a æ ö = tan -1 tan ( q - x ) = q - x = ç tan -1 - x ÷ × b è ø \

dy = -1. dx

30. Putting x = sin q, we get y = sin -1 (sin 3 q) = 3 q = 3 sin -1 x. 31. Putting x = cos q, we get y = cos -1 (cos 3 q) = 3 q = 3 cos -1 x. 32. Put a = tan q and x = tan f. Then, y = tan1 {tan ( q + f)} = q + f = tan -1 a + tan -1 x . 33. Putting x = cot q, we get: æ 1 - tan 2 q ö ÷ = cos -1 (cos 2 q) = 2 q = 2 cot -1 x. y = cos -1 ç ç 1 + tan 2 q ÷ è ø 34. Putting x 2 = tan q, we get: ì æp ö öü æ p ö æp y = tan -1 í tan ç + q ÷ ý = ç + q ÷ = ç + tan -1 x 2 ÷ × 4 4 è ø ø è ø è4 î þ 35. y = cot -1 x Þ

dy -1 -1 1 = × = × dx ( 1 + x ) 2 x 2 x ( 1 + x )

36. y = cos -1 x 3 Þ

dy -3 x 2 × = dx 1 - x6

x x é {cos ( x ) + sin ( x )} 2 ù æ 1 + sin x ö -1 ì cos ( 2 ) + sin ( 2 ) ü 2 2 ÷ = tan -1 ê 37. y = tan -1 çç ú = tan í ÷ 2 x 2 x x ) - sin ( x ) ý cos x cos ( êë cos ( 2 ) - sin ( 2 ) úû è ø î 2 2 þ

ì 1 + tan ( x 2 ) ü ì æ p x öü æ p x ö = tan -1 í = tan -1 í tan ç + ÷ ý = ç + ÷ × x ý è 4 2 øþ è 4 2 ø î î 1 - tan ( 2 ) þ é ìp æ p æp ö ö ö üù æ p 38. Put x = tan q. Then, y = cot -1 × tan ç - q ÷ = cot -1 ê cot í - ç - q ÷ ýú = ç + q ÷ × è4 ø ø ø þû è 4 î2 è 4 ë \ y= 39. log y =

p + tan -1 x. 4 1 1 dy 1 ì 1 1 ü {log ( 1 + x ) - log ( 1 - x )} Þ × = í + ý× 2 y dx 2 î ( 1 + x ) ( 1 - x ) þ

æ 1 + tan 2 q ö æ ö ÷ = sec-1 ç 1 ÷ = sec-1 ( sec 2 q) 40. Put x = cot q. Then, y = sec-1 ç ç cos 2 q ÷ ç 1 - tan 2 q ÷ è ø è ø = 2 q = 2 cot -1 x.

581

Senior Secondary School Mathematics for Class 12 Pg-582

582

Senior Secondary School Mathematics for Class 12

æ ö 1 ÷ = sec-1 ( sec 2 q) = 2 q = 2 cos -1 x. 41. Put x = cos q. Then, y = sec-1 ç ç 2 cos 2 q - 1 ÷ è ø 1 42. Put x = tan q. Then, y = tan -1 x. 2 p 1 43. Put x = cos q. Then, y = + cos -1 x. 4 2 44.

dy ( dy dt) 1 dy dx = = × = 2 at , = 2 a. So, dx ( dx dt) t dt dt

45.

dy dy ( dy dq) b dx = a sec q tan q, = b sec2 q × = = cosec q. dq dq dx ( dx dq) a

46.

dy dy ( dy dq) - b dx = - 2 a cos q sin q, = 2 b sin q cos q Þ = = × a dq dq dx ( dx dq)

47.

dy ( dy dq) = × dx ( dx dq)

48. y = x y Þ log y = y log x Þ

dy 1 dy y = + (log x ) × y dx x dx

dy dy = 1+ × dx dx dy dy 50. y = sin x + y Þ y 2 = sin x + y Þ 2 y = cos x + × dx dx dy 1 dy 51. y = e x + y Þ x + y = log y Þ 1 + = × × dx y dx

49. y = x + y Þ y 2 = x + y Þ 2 y

52. For continuity at x = 0, we must have lim f ( x ) = f ( 0 ). x®0

sin 5 x 5 5 sin 5 x æ 5 ö 5 lim f ( x ) = lim ´ = lim = ç ´ 1÷ = × 3 3 5 x ® 0 5x x®0 x ® 0 5x è3 ø 3 5 5 Û k= × \ we must have, f ( 0 ) = 3 3 53. f( 0 ) = 0. lim f ( x ) = lim x sin

x®0

x®0

1 = 0 ´ (a finite quantity) = 0. x

\ f ( x ) is continuous at x = 0. 54. f ( 0 ) = k. 3 x + 4 tan x 4 tan x ü ì lim f ( x ) = lim = lim í 3 + ý = ( 3 + 4) = 7. x x þ x®0 x®0 x®0î \ f ( x ) is continuous at x = 0 Û f ( 0 ) = 7 Û k = 7 . 3

1 f ( 0 - h) - f ( 0) ( h 2 - 0) = lim = lim ( - h) 2 , which does not exist, since -h -h h®0 h®0 h®0

55. f ¢( 0 ) = lim ( -h )

1

2

is imaginary.

56. f ( 0 + 0 ) = lim|0 + h| = lim|h| = 0 , h®0

h®0

f ( 0 - 0 ) = lim|0 - h| = lim|- h| = lim|h| = 0 and f( 0 ) = 0 h®0

h®0

h®0

Senior Secondary School Mathematics for Class 12 Pg-583

Applications of Derivatives \ f ( x ) is continuous at x = 0 f ( 0 + h) - f ( 0) f ( h) - 0 |h| h Rf ¢( 0 ) = lim = lim = lim = lim = 1 h h h®0 h®0 h®0 h h®0 h f ( 0 - h) - f ( 0) f ( - h) - 0 |- h| h Lf ¢( 0 ) = lim = lim = -1. = lim = lim -h -h h®0 h®0 h®0 -h h®0 -h \ Rf ¢( 0 ) ¹ Lf ¢( 0 ), which shows that f ( x ) is not differentiable at x = 0. 57. f ( 2 + 0 ) = lim f ( 2 + h ) = lim {5 - ( 2 + h )} = 3 ; h®0

h®0

f ( 2 - 0 ) = lim f ( 2 - h ) = lim {1 + ( 2 - h )} = 3 and f( 2 ) = 3. h®0

h®0

\ f ( x ) is continuous at x = 2 f ( 2 + h) - f ( 2) ( 3 - h) - 3 -h Rf ¢( 2 ) = lim = lim = lim = -1. h h h®0 h®0 h®0 h f ( 2 - h) - f ( 2) ( 3 - h) - 3 h Lf ¢( 2 ) = lim = lim = lim = 1. -h -h h®0 h®0 h®0 h \ f ( x ) is not differentiable at x = 2. 58. f ( 2 ) = lim f ( x ). x®2

lim f ( x ) = lim f ( 2 + h ) = lim ( 2 + h - 1) = 1

x®2+

h®0

h®0

lim f ( x ) = lim f ( 2 - h ) = lim {k( 2 - h ) + 5} = 2 k + 5

x®2-

h®0

h®0

\ 2 k + 5 = 1 Þ k = - 2. Also, f ( 2 ) = 2 k + 5 = 1. Hence, k = - 2. 1 - cos 4 h 2 sin 2 2 h 59. lim f ( x ) = lim f ( 0 + h ) = lim = lim 2 + h ® 0 h ® 0 h ® 0 8h 2 8h x®0 2

1 æ sin 2 h ö æ1 ö 1 ÷ = ç ´ 12 ÷ = × lim ç 2 h ® 0 è 2h ø è2 ø 2 1 - cos 4( - h ) ( 1 - cos 4 h ) 1 = × lim f ( x ) = lim f ( 0 - h ) = lim = lim 2 h®0 h®0 h®0 8h 2 8( - h ) 2 x ® 01 \ lim f ( x ) = 2 x ® 0+ =

1 × 2 2 2 sin ax æ sin ax ö ÷ = a2 ´ 12 = a2 . 60. lim f ( x ) = lim 2 2 ´ a2 = a2 × lim ç x®0 x®0 a x ax ® 0 è ax ø For continuity, we must have f ( 0 ) = a2 . æp ö k cos ç - h ÷ sin h æ k k sin h k æp ö ö k è2 ø 61. f ç - 0 ÷ = lim = lim = ç ´ 1÷ = × = lim æp ö h ® 0 2h 2 h®0 h è2 è2 ø h®0 ø 2 p - 2 ç - h÷ è2 ø æp ö k cos ç + h ÷ - k sin h k sin h k æp ö è2 ø f ç + 0 ÷ = lim = lim = = lim æp ö h ® 0 -2 h 2 h®0 h 2 è2 ø h®0 p - 2 ç + h÷ è2 ø k = 3 Þ k = 6. \ 2 For continuity, we must have f( 0 ) =

583

Senior Secondary School Mathematics for Class 12 Pg-584

584

Senior Secondary School Mathematics for Class 12

62. f( 2 ) = [2] = 2. f ( 2 + 0 ) = lim f ( 2 + h ) = lim [2 + h] = 2. h®0

h®0

f ( 2 - 0 ) = lim f ( 2 - h ) = lim [2 - h] = 1. h®0

h®0

\ f ( x ) is not continuous at x = 2. f ( 2 + h) - f ( 2) [2 + h] - [2] ( 2 - 2) = 0. Rf ¢( 2 ) = lim = lim = lim h h h h®0 h®0 h®0 Lf ¢( 2 ) = lim

h®0

( 1 - 2) f ( 2 - h) - f ( 2) [2 - h] - [2] 1 = lim = ¥ = lim = lim -h -h h®0 h®0 -h h®0 h

\ f ( x ) is not differentiable at x = 2. ( x - 3 )( x + 1) 63. lim f ( x ) = lim = lim ( x - 3 ) = -4. ( x + 1) x ® -1 x ® -1 x ® -1 For continuity, we must have, f( -1) = - 4. 64. f ¢( x ) = 3 x 2 - 12 x + 15 = 3 ( x 2 - 4 x + 5 ) = 3 [( x - 2 ) 2 + 1] > 0. Þ f ¢( x ) > 0 for all x Î R Þ f ( x ) is strictly increasing on R. 65. f ¢( x ) = -3 + 6 x - 3 x 2 = -3 ( x 2 - 2 x + 1) = -3 ( x - 1) 2 £ 0. Þ f ¢( x ) £ 0 for all x Î R Þ f ( x ) is decreasing on R. 66. f ¢( x ) = 3 - 3 sin 3 x = 3( 1 - sin 3 x ) ³ 0 since -1 £ sin 3 x £ 1. \ f ¢( x ) ³ 0 for all x Î R Þ f ( x ) is increasing on R. 67. f ¢( x ) = 3 x 2 - 12 + 9 = 3 ( x 2 - 4 x + 3 ) = 3 ( x - 1)( x - 3 ) f ¢( x ) = 0 Þ x = 1 or x = 3. There are two factors in f ¢( x ), so we start with +ve sign. \ f ( x ) is decreasing for 1 < x < 3. 68. f ¢( x ) = 3 x 2 - 27 = 3 ( x 2 - 9 ) = 3 ( x + 3 )( x - 3 ) \ f ¢( x ) = 0 Þ x = -3 or x = 3. There are two factors in f ¢( x ), so we start with +ve sign. \ f ( x ) is increasing when x < -3 or x > 3, i.e., when|x| > 3. æ -p p ö æ -p p ö 69. f ¢( x ) = cos x > 0 in ç , ÷ Þ f ( x ) is increasing in ç , ÷× è 2 2ø è 2 2ø 70. f ¢( x ) =

(log x ) × 2 - 2 x × (log x ) 2

1 x = 2(log x - 1) (log x ) 2

\ f ¢( x ) > 0 Û log x - 1 > 0 Û log x > 1 Û log x > log e Û x > e. \ f ( x ) is increasing in ( e , ¥ ). 1 æ 1 ö æp ö 71. f ¢( x ) = (cos x + sin x ) = 2 ç cos x + sin x ÷ = 2 sin ç + x ÷ 2 è4 ø è 2 ø p 3p 7p æp ö x and cos x > 0 Þ cos x(tan x - x ) > 0 Þ f ¢( x ) > 0 \ f ( x ) is increasing in (0, 1). 73. f ¢( x ) = x x( 1 + log x ) f ¢( x ) < 0 Û ( 1 + log x ) < 0 Þ log x < -1 = log æ 1ö \ f ( x ) is decreasing in ç 0 , ÷ × è eø

1 1 Þ x > 0 and x < × e e

74. f ¢( x ) = 2 xe - x - x 2 e - x = xe - x( 2 - x ) \ f ¢( x ) > 0 Û x > 0 and ( 2 - x ) > 0 Û 0 < x < 2. \ f ( x ) is increasing in (0, 2). 75. f ¢( x ) = (cos x - k ) and therefore, f ( x ) is decreasing Û f ¢( x ) < 0 Þ cos x - k < 0 Þ cos x < k Þ k > cos x Þ k > 1. 76. f ( x ) = ( x + 1) 3 ( x - 3 ) 3 \ f ¢( x ) = 3 ( x + 1) 3 ( x - 3 ) 2 + 3 ( x + 1) 2 ( x - 3 ) 3 = 3 ( x + 1) 2 ( x - 3 ) 2 [( x + 1) + ( x - 3 )] = 3 ( x + 1) 2 ( x - 3 ) 2 ( x - 1) Þ f ¢( x ) > 0 when ( x - 1) > 0 , i.e., when x > 1. \ f ( x ) is increasing in ( 1, ¥ ). 77. f ( x ) = [x( x - 3 )] 2 Þ f ¢( x ) = 2 x( x - 3 )( 2 x - 3 ). 3 or x = 3. 2 3 f ( x ) is increasing when 0 < x < or x > 3. 2 æ 3ö \ f ( x ) is increasing in ç 0 , ÷ È ( 3 , ¥ ). è 2ø 78. f ¢( x ) = 3 kx 2 - 18 x + 9 = 3 ( kx 2 - 6 x + 3 ). \ f ¢( x ) = 0 Þ x = 0 or x =

This is positive when k > 0 and ( 36 - 12 k ) < 0 Þ k > 3. 79. f ¢( x ) =

( x 2 + 1) × 1 - x × 2 x 2

( x + 1)

2

=

( 1 - x2 ) ( 1 + x 2 )2

×

f ¢( x ) > 0 Û ( 1 - x 2 ) > 0 Û x 2 < 1 Û - 1 < x < 1. \ f ( x ) is increasing in ( -1, 1). 80. f ¢( x ) = ( 2 x + k ). 1 < x < 2 Þ 2 < 2 x < 4 Þ 2 + k < 2 x + k < 4 + k Þ 2 + k < f ¢( x ) < 4 + k. f ( x ) is increasing Û ( 2 x + k ) ³ 0 Û 2 + k ³ 0 Û k ³ - 2. \ least value of k is -2. 81. f ( x ) =|x| ³ 0 for all x Î R. The least value of|x|is 0 at x = 0. \ f ( x ) =|x|has minima at x = 0.

585

Senior Secondary School Mathematics for Class 12 Pg-586

586

Senior Secondary School Mathematics for Class 12

é1 ù 82. f ( x ) = x 2 Þ f ¢( x ) = x x( 1 + log x ) and f ¢¢( x ) = x x ê + ( 1 + log x ) 2 ú × ëx û æ 1ö æ 1ö f ¢( x ) = 0 Þ 1 + log x = 0 Þ log x = - 1 = log ç ÷ Þ x = ç ÷ × èeø èeø [f ¢¢( x )] x = (1 \ x=

e)

æ 1ö =eç ÷ èeø

1

e

> 0.

1 is a point of minima. e 1

-1 æ 1ö e Minimum value of x x is ç ÷ = e e . èeø 1 x × - log x × 1 ( 1 - log x ) log x x 83. f ( x ) = Þ f ¢( x ) = = x x2 x2 1 x 2 × - ( 1 - log x ) × 2 x ( -3 + 2 log x ) x Þ f ¢¢( x ) = = x4 x3

\ f ¢( x ) = 0 Þ 1 - log x = 0 Þ log x = 1 = log e Þ x = e. æ -3 + 2 ö -1 ÷= f ¢¢( e ) = ç < 0. è e3 ø e3 \ x = e is a point of maxima. 1 1 Maximum value of f ( x ) is (log e ) = × e e 84. f ( x ) = cosec x Þ f ¢( x ) = - cosec x cot x Þ f ¢¢( x ) = cosec3 x + cosec x (cot 2 x ) = cosec x ( cosec2 x + cot 2 x ) = cosec x ( 2 cosec2 x - 1) -p f ¢( x ) = 0 Þ cot x = 0 Þ x = 2 ù æ -p ö é æ -p ö 2 æ -p ö f ¢¢ ç ÷ ê 2 cosec ç ÷ - 1ú = ( -1)( 2 - 1) = - 1 < 0. ÷ = cosec ç è 2 øë è 2 ø è 2 ø û -p is a point of maxima. \ x= 2 1 85. xy = 1 Þ y = × x Let S = x + y = x +

ds æ 1 ö ( x 2 - 1) 1 d 2s 2 and 2 = 3 × = ç1- 2 ÷ = × Then, 2 dx è x x ø x dx x

ds = 0 Þ x 2 - 1 = 0 Þ x = ± 1. dx d 2s ù d 2s ù = -2 < 0 and 2 ú = 2 >0 2ú dx úû (x = -1 ) dx úû (x = 1 ) \ S is minimum at x = 1 and minimum value of S = ( 1 + 1) = 2. 250 ö 500 ö 250 ö æ æ æ 86. Let f ( x ) = ç x 2 + ÷ × Then, f ¢( x ) = ç 2 x - 2 ÷ and f ¢¢( x ) = ç 2 + 3 ÷ x ø è è è x ø x ø

Senior Secondary School Mathematics for Class 12 Pg-587

Applications of Derivatives f ¢( x ) = 0 Þ 2 x 3 - 250 = 0 Þ x 3 = 125 Þ x = 5. 500 ö æ f ¢¢(5 ) = ç 2 + ÷ = 6 >0 125 ø è 250 ö æ \ f ( x ) is minimum at x = 5 and minimum value = ç 25 + ÷ = 75. 5 ø è 87. f ¢( x ) = 12 x 3 - 24 x 2 + 24 x - 48 = 12( x - 2 )( x 2 + 2 ) f ¢¢( x ) = 36 x 2 - 48 x + 24 = 12( 3 x 2 - 4 x + 2 ). f ¢( x ) = 0 Þ x = 2 and f ¢¢( 2 ) = 12( 3 ´ 4 - 4 ´ 2 + 2 ) = 72 > 0 \ x = 2 is a point of minima. Minimum value = min {f ( 0 ), f ( 2 ), f ( 3 )} = min {25 , - 39 , 16} = - 39. 88. f ( x ) = ( x - 2 )( x - 3 ) 2 Þ f ¢( x ) = ( x - 3 )( 3 x - 7 ) and f ¢¢( x ) = ( 6 x - 16 ). f ¢( x ) = 0 Þ x = 3 or x =

7 3

æ7 ö f ¢¢( 3 ) = 2 > 0 and f ¢¢ ç ÷ = -2 < 0. è 3ø 7 \ x = is a point of maxima. 3 2

4 öæ 7 ö æ7 Maximum value = ç - 2 ÷ ç - 3 ÷ = × 27 øè 3 ø è3 89. f ¢( x ) = e x - e - x and f ¢¢( x ) = e x + e - x. f ¢( x ) = 0 Þ e x - e - x = 0 Þ e x = e - x Þ e 2 x = e 0 Þ x = 0. f ¢¢( 0 ) = e 0 +

1 e0

= ( 1 + 1) = 2 > 0.

\ f ( x ) is minimum at x = 0 and minimum value of f ( x ) is 2.

587

Senior Secondary School Mathematics for Class 12 Pg-588

12. INDEFINITE INTEGRAL INTEGRATION

It is the inverse process of differentiation.

If the derivative of F( x) is f ( x) then we say that the antiderivative or integral of f ( x) is F( x) and we write,

ò f ( x) dx = F( x). d [F( x)] = f ( x) Þ dx

Thus, Example

Since

ò f ( x) dx = F( x).

d (sin x) = cos x , we have ò cos x dx = sin x. dx

Moreover, if C is any constant then

d (sin x + C) = cos x. dx

ò cos x dx = (sin x + C).

So, in general,

Clearly, different values of C will give different integrals. Thus, a given function may have an indefinite number of integrals. Because of this property, we call these integrals indefinite integrals. d [F( x)] = f ( x) Þ ò f ( x) dx = F( x) + C , where C is a constant, dx called the constant of integration. Any function to be integrated is known as an integrand. Thus,

The following two results are a direct consequence of the definition of an integral.

òx

RESULT 1

PROOF

n

dx =

We have,

( n+1 )

x + C , when n ¹ - 1. (n + 1)

d æç x n+1 ö÷ (n + 1) x n = = x n. (n + 1) dx çè n + 1 ÷ø

òx

\

n

dx =

( n+1 )

x + C. (n + 1)

Thus, we have (i)

òx

6

dx =

( 6 +1 )

x x7 +C = + C. ( 6 + 1) 7 588

Senior Secondary School Mathematics for Class 12 Pg-589

Indefinite Integral

589

æ2 ö ç +1 ÷

xè 3 ø 3 (ii) ò x 2/ 3 dx = + C = x5/ 3 + C. 2 ö 5 æ ç + 1÷ ø è3 æ 3 ö ç - +1 ÷

(iii)

RESULT 2 PROOF

1

òx

-3/4

xè 4 ø dx = = 4x1/4 + C. æ 3 ö ç - + 1÷ è 4 ø

ò x dx = log | x| + C ,

where x ¹ 0.

Clearly, either x > 0 or x < 0.

When x > 0 In this case, | x| = x. d d 1 \ [log | x|] = (log x) = . dx dx x 1 So, we have, ò dx = log | x| + C. x Case II When x < 0 In this case |x| = - x. d d 1 1 [log |x|] = [log ( - x)] = × ( -1) = × \ dx dx ( - x) x 1 So, we have ò dx = log |x| + C. x 1 Thus, from both the cases, we have ò dx = log |x| + C. x Case I

FORMULAE

On the basis of differentiation and the definition of integration, we have the following results. x n+1 d æç x n+1 ö÷ 1. = x n , n ¹ -1 Þ ò x ndx = +C dx çè n + 1 ÷ø (n + 1) d 1 1 2. (log |x|) = Þ ò dx = log |x| + C dx x x d x x x 3. ( e ) = e Þ ò e dx = ex + C dx ax d æç a x ö÷ 4. = a x Þ ò a x dx = +C dx çè log a ÷ø log a d (sin x) = cos x Þ ò cos x dx = sin x + C dx d 6. ( - cos x) = sin x Þ ò sin x dx = - cos x + C dx 5.

Senior Secondary School Mathematics for Class 12 Pg-590

590

7. 8. 9. 10. 11.

Senior Secondary School Mathematics for Class 12

d (tan x) = sec2 x Þ ò sec2 x dx = tan x + C dx d ( - cot x) = cosec2 x Þ ò cosec2 x dx = - cot x + C dx d ( sec x) = sec x tan x Þ ò sec x tan x dx = sec x + C dx d ( - cosec x) = cosec x cot x Þ ò cosec x cot x dx = - cosec x + C dx d 1 1 Þ ò (sin -1 x) = dx = sin -1 x + C 2 dx 1-x 1 - x2

d 1 1 Þ ò (tan -1 x) = dx = tan -1 x + C 2 dx (1 + x ) (1 + x 2) d 1 1 13. Þ ò ( sec-1 x) = dx = sec-1 x + C 2 2 dx x x -1 x x -1

12.

With the help of the above formulae, it is easy to evaluate the following integrals. EXAMPLE 1

Evaluate: (i) ò x 9 dx (iv)

SOLUTION

1

ò x 2 dx

(ii)

ò 3 x dx

(v)

ò x1/3 dx

1

(iii)

ò dx

(vi)

ò5

Using the standard formulae, we have (i)

òx

9

dx =

( 9 +1 )

x x10 +C = + C. ( 9 + 1) 10 æ1 ö ç +1 ÷

xè 3 ø 3 (ii) ò 3 x dx = ò x1/ 3 dx = + C = x 4/ 3 + C. 4 æ1 ö ç + 1÷ 3 è ø (iii)

0 ò dx = ò x dx =

(iv)

ò x 2 dx = ò x

1

1

-2

ò x1/3 dx = ò x

(vi)

ò5

dx =

dx =

( -2 + 1 )

1 x + C = - + C. ( -2 + 1) x æ 1 ö ç - +1 ÷

(v)

x

( 0 +1 )

x + C = x + C. ( 0 + 1)

-1/ 3

3 xè 3 ø dx = + C = x 2/ 3 + C. æ 1 ö 2 ç - + 1÷ è 3 ø

5x + C. log 5

x

dx

Senior Secondary School Mathematics for Class 12 Pg-591

Indefinite Integral

591

Some Standard Results on Integration d THEOREM 1 { f ( x) dx} = f ( x). dx ò PROOF

ò f ( x) dx = F( x).

Let Then, \

PROOF

d [by def. of integral ]. {F( x)} = f ( x) dx d [using (i)]. { f ( x) dx} = f ( x) dx ò

ò k × f ( x) dx = k × ò f ( x) dx ,

THEOREM 2

… (i)

where k is a constant.

ò f ( x) dx = F( x).

Let

… (i)

d Then, {F( x)} = f ( x). dx d d [using (ii)]. {k × F( x)} = k × {F( x)} = k × f ( x) \ dx dx So, by the definition of an integral, we have

ò {k × f ( x)} dx = k × F( x) = k × ò f ( x) dx Evaluate: (i) ò 3 x 2 dx

EXAMPLE 2

(ii) ò 2

dx

x + C = x 3 + C. 3 ( x + 3) 2x 2 ( x + 3) (ii) ò 2 dx = ò 2x × 2 3 dx = 8ò 2x dx = 8 × +C = + C. log 2 log 2 (i)

SOLUTION

THEOREM 3

PROOF

[using (i)]. ( x + 3)

3

… (ii)

(i)

ò 3x

2

dx = 3 ò x 2 dx = 3 ×

ò { f1( x) + f 2( x)} dx = ò f1( x) dx + ò f 2( x) dx (ii) ò { f1( x) - f 2( x)} dx = ò f1( x) dx - ò f 2( x) dx Let ò f1( x) dx = F1( x) and ò f 2( x) dx = F2( x). (i)

d d {F1( x)} = f1( x) and {F2( x)} = f 2( x). dx dx d d d Now, {F1( x) + F2( x)} = {F1( x)} + {F2( x)} dx dx dx = f1( x) + f 2( x) [using (ii)]. Then,

\

ò { f1( x) + f 2( x)} dx = F1( x) + F2( x) = ò f1( x) dx + ò f 2( x) dx

[using (i)].

Similarly, (ii) can be proved. REMARK

In general, we have ò {k1 × f1( x) ± k2 × f 2( x) ± ¼ ± kn × f n( x)} dx

= k1 × ò f1( x) dx ± k2 × ò f 2( x) dx ± ¼ ± kn × ò f n( x) dx.

… (i) … (ii)

Senior Secondary School Mathematics for Class 12 Pg-592

592

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Evaluate: 1 5ö æ (i) ò ç5 x 3 + 2x -5 - 7 x + + ÷ dx x xø è (ii) (iii)

SOLUTION

ò ( 3 sin x - 4 cos x + 5 sec x - 2cosec x) dx ò (1 - x)( 2 + 3 x)(5 - 4x) dx 2

2

æ 3 x 4 - 5 x 3 + 4x 2 - x + 2 ö ÷ dx ÷ x3 è ø

(iv)

ò çç

(i)

ò çè5 x

æ

3

+ 2x -5 - 7 x +

(v)

æ

ò çè x

2

3

+

1 ö ÷ dx x2 ø

1 5ö + ÷ dx x xø

= 5 ò x 3 dx + 2ò x -5 dx - 7 ò x dx + ò x -1/2 dx + 5 ò =5×

1 dx x

x4 x -4 x 2 x1/2 + 2× -7 × + + 5 log |x| + C 4 2 (1/2) ( - 4)

5x4 1 7x2 - 4+ 2 x + 5 log |x| + C. 4 2 2x (ii) ò ( 3 sin x - 4 cos x + 5 sec2 x - 2cosec2 x) dx =

= 3 ò sin x dx - 4ò cos x dx + 5 ò sec2 x dx - 2ò cosec2 x dx = 3( - cos x) - 4 sin x + 5 tan x - 2( - cot x) + C = ( -3 cos x - 4 sin x + 5 tan x + 2 cot x + C). (iii) ò (1 - x)( 2 + 3 x)(5 - 4x) dx = ò (10 - 3 x - 19x 2 + 12x 3) dx = 10ò dx - 3 ò x dx - 19ò x 2 dx + 12ò x 3 dx = 10x - 3 × = 10x (iv)

x2 x3 x4 - 19 × + 12 × +C 2 3 4

3 x 2 19x 3 + 3 x 4 + C. 2 3

æ 3 x 4 - 5 x 3 + 4x 2 - x + 2 ö ÷ dx = æç 3 x - 5 + 4 - 1 + 2 ö÷ dx òè 3 ÷ x x2 x 3 ø x è ø [dividing each term by x 3] 1 = 3 ò x dx - 5 ò dx + 4ò dx - ò x -2 dx + 2ò x -3 dx x 2 æ x -2 ö x æ 1ö ÷ +C = 3× - 5 x + 4 log |x| - ç - ÷ + 2 ç ç -2 ÷ 2 è xø è ø 3x2 1 1 = - 5 x + 4 log |x| + - 2 + C. 2 x x

ò çç

Senior Secondary School Mathematics for Class 12 Pg-593

Indefinite Integral

(v)

æ

ò çè x

2

593

3

+

1 ö 3ö æ 6 1 2 ÷ dx = ò ç x + 6 + 3 x + 2 ÷ dx è x2 ø x x ø = ò x 6 dx + ò x -6 dx + 3 ò x 2 dx + 3 ò =

1 x2

dx

x 7 x -5 x3 æ 1ö + + 3× + 3 × ç- ÷ + C 7 ( -5) 3 è xø

1 3 x7 + x 3 - + C. 7 5 x5 x æ x4 + 1ö ( x 3 + 4x 2 - 3 x - 2) ÷ dx [CBSE 2011] Evaluate: (i) ò dx (ii) ò ç 2 ç x + 1÷ ( x + 2) è ø =

EXAMPLE 2

SOLUTION

(i) On dividing ( x 3 + 4x 2 - 3 x - 2) by ( x + 2), we get

ò

( x 3 + 4x 2 - 3 x - 2) 12 ü ì dx = ò íx 2 + 2x - 7 + ý dx ( x + 2) x + 2þ î = ò x 2 dx + 2ò x dx - 7 ò dx + 12ò

1 dx x+2

=

x3 x2 + 2× - 7 x + 12 log |x + 2| + C 3 2

=

x3 + x 2 - 7 x + 12 log |x + 2| + C. 3

(ii) On dividing ( x 4 + 1) by ( x 2 + 1), we get æ x4 + 1ö

é

ù ú dx ( x + 1) û ë 1 = ò x 2 dx - ò dx + 2ò 2 dx x +1

ò çç x 2 + 1 ÷÷ dx = ò ê x è

ø

= EXAMPLE 3

SOLUTION

2

2

-1 +

2

x3 - x + 2tan -1 x + C. 3

Evaluate: (i)

ò tan

2

x dx

(i)

ò tan

2

x dx = ò ( sec2 x - 1) dx

(ii)

ò cot

2

x dx

(iii)

ò sin

= ò sec2 x dx - ò dx = tan x - x + C. (ii)

ò cot

2

x dx = ò ( cosec2 x - 1) dx = ò cosec2 x dx - ò dx = - cot x - x + C.

(iii) We know that 2 sin 2

x = (1 - cos x). 2

2

x dx 2

Senior Secondary School Mathematics for Class 12 Pg-594

594

Senior Secondary School Mathematics for Class 12

\

EXAMPLE 4

Evaluate

SOLUTION

ò

ò sin ò

2

x 1 dx = ò (1 - cos x) dx 2 2 1 1 1 = [ò dx - ò cos x dx ] = x - sin x + C. 2 2 2

1 - sin 2x dx.

[CBSE 2004]

1 - sin 2x dx = ò (cos2 x + sin 2 x - 2 sin x cos x)1/2 dx = ò (cos x - sin x) 2 dx = ò (cos x - sin x) dx = ò cos x dx - ò sin x dx = sin x - ( - cos x) + C = sin x + cos x + C.

EXAMPLE 5

SOLUTION

Evaluate: æ sin x ö dx ÷÷ dx [CBSE 2002C] (ii) ò çç (i) ò 1 + sin x è 1 + sin x ø dx 1 (1 - sin x) (i) ò =ò ´ dx (1 + sin x) (1 + sin x) (1 - sin x) =ò

(1 - sin x) (1 - sin 2 x)

dx = ò

(1 - sin x) cos2 x

dx

æ 1 sin x ö÷ = òç dx = ò (sec2 x - sec x tan x) dx ç cos2 x cos2 x ÷ è ø = ò sec2 x dx - ò sec x tan x dx = tan x - sec x + C. (ii)

æ sin x ö

ò ççè 1 + sin x ÷÷ø dx = ò

(1 + sin x) - 1 dx (1 + sin x)

æ ö 1 1 ÷÷ dx = ò dx - ò = ò çç1 dx 1 + sin x ( 1 + sin x) è ø 1 (1 - sin x) = ò dx - ò ´ dx (1 + sin x) (1 - sin x) = ò dx - ò

(1 - sin x) 2

cos x

æ 1 sin x ö÷ dx = ò dx - ò ç dx ç cos2 x cos2 x ÷ è ø

= ò dx - ò sec2 x dx + ò sec x tan x dx = x - tan x + sec x + C. sec x

ò ( sec x + tan x) dx.

EXAMPLE 6

Evaluate

SOLUTION

ò ( sec x + tan x) dx = ò ( sec x + tan x) ´ ( sec x - tan x) dx

sec x

sec x

=ò

( sec x - tan x)

( sec2 x - sec x tan x) ( sec2 x - tan 2 x)

dx

Senior Secondary School Mathematics for Class 12 Pg-595

Indefinite Integral

595

= ò ( sec2 x - sec x tan x) dx = ò sec2 x dx - ò sec x tan x dx = tan x - sec x + C. EXAMPLE 7

SOLUTION

Evaluate: æ 4 - 5 cos x ö ÷ dx (i) ò ç ç sin 2 x ÷ è ø 1 (iii) ò dx [CBSE 2014] sin 2 x cos2 x (i)

æ 1 - cos 2x ö

(ii)

ò ççè 1 + cos 2x ÷÷ø dx

(iv)

ò cos2x sin 2x dx

cos 2x

æ 4 - 5 cos x ö æ ö ÷ dx = ò ç 4 - 5 cos x ÷ dx 2 2 2 ÷ ÷ ç è sin x ø è sin x sin x ø

ò çç

= ò ( 4cosec2 x - 5 cosec x cot x) dx = 4ò cosec2 x dx - 5 ò cosec x cot x dx = 4( - cot x) - 5( - cosec x) + C = -4 cot x + 5 cosec x + C. (ii)

æ 1 - cos 2x ö

2 sin 2 x

ò ççè 1 + cos 2x ÷÷ø dx = ò 2 cos2x dx = ò tan

2

x dx

= ò ( sec2 x - 1) dx = ò sec2 x dx - ò dx = tan x - x + C. (iii)

æ sin 2 x + cos2 x ö ÷ dx 2 2 ÷ è sin x cos x ø

1

ò sin 2 x cos2 x dx = ò çç

æ 1 1 ö÷ = òç + dx ç cos2 x sin 2 x ÷ è ø = ò sec2 x dx + ò cosec2 x dx = tan x - cot x + C. (iv)

cos 2x

æ cos2 x - sin 2 x ö ÷ dx 2 2 ÷ è cos x sin x ø

ò cos2x sin 2x dx = ò çç

æ 1 1 ö÷ = òç dx ç sin 2 x cos2 x ÷ è ø = ò cosec2 x dx - ò sec2 x dx = - cot x - tan x + C. EXAMPLE 8

SOLUTION

Evaluate

æ cos 2x - cos 2a ö ÷ dx. cos x - cos a ÷ø

ò ççè

æ cos 2x - cos 2a ö ( 2 cos2 x - 1) - ( 2 cos2 a - 1) ÷÷ dx = ò dx cos x - cos a ø (cos x - cos a )

ò ççè

[CBSE 2013]

Senior Secondary School Mathematics for Class 12 Pg-596

596

Senior Secondary School Mathematics for Class 12

= 2ò

(cos2 x - cos2 a ) dx = 2ò (cos x + cos a ) dx (cos x - cos a )

= 2ò cos x dx + 2 cos a × ò dx = 2 sin x + 2x cos a + C. EXAMPLE 9

Evaluate

SOLUTION

ò tan

ò tan

-1 ì

1 - cos 2x ü í ý dx. î 1 + cos 2x þ

2 ì 1 - cos 2x ü -1 ï 2 sin x í ý dx = ò tan í 2 î 1 + cos 2x þ îï 2 cos x

-1 ì

[CBSE 2003]

üï ý dx ïþ

= ò tan -1(tan x) dx = ò x dx = EXAMPLE 10

Evaluate

SOLUTION

ò sin

-1

ò sin

-1

x2 + C. 2

(cos x) dx.

ì (cos x) dx = ò sin -1ísin î

æp öü ç - x ÷ ý dx è2 øþ

p px x 2 æp ö + C. = ò ç - x ÷ dx = × ò dx - ò x dx = 2 2 2 è2 ø EXAMPLE 11

Evaluate

SOLUTION

ò tan

-1

ò tan

-1

( sec x + tan x) dx.

æ 1 sin x ö ÷ dx (sec x + tan x) dx = ò tan -1 çç + cos cos x ÷ø x è ì æp öü ïï1 - cos ç 2 + x ÷ ïï æ ö + sin 1 x è ø dx ÷÷ dx = ò tan -1 í = ò tan -1 çç ý p æ ö cos x è ø ï sin ç + x ÷ ï ïî è2 ø ïþ ì ïï = ò tan -1í ï 2 sin ïî

ü æp xö 2 sin 2 ç + ÷ ïï è 4 2ø ý dx p x p x æ ö æ ö ç + ÷ cos ç + ÷ ï è 4 2ø è 4 2 ø ïþ

ì æ p x öü æp xö = ò tan -1ítan ç + ÷ ý dx = ò ç + ÷ dx è 4 2øþ è 4 2ø î p 1 px 1 2 = × ò dx + ò x dx = + x + C. 4 2 4 4 EXAMPLE 12

Evaluate

SOLUTION

I=ò

æ 1 + sin x ö

ò ççè 1 - sin x ÷÷ø dx.

(1 + sin x) (1 + sin x) ´ dx (1 - sin x) (1 + sin x)

[CBSE 2008C]

Senior Secondary School Mathematics for Class 12 Pg-597

Indefinite Integral

=ò

(1 + sin x) 2 2

(1 - sin x)

dx = ò

597

(1 + sin 2 x + 2 sin x) cos2 x

dx

æ 1 sin 2 x 2 sin x ö÷ = òç + + dx = ò (sec2 x + tan 2 x + 2 sec x tan x) dx ç cos2 x cos2 x cos2 x ÷ è ø = ò ( 2sec2 x - 1 + 2sec x tan x) dx = 2ò sec2 x dx - ò dx + 2ò sec x tan x dx = 2 tan x - x + 2sec x + C.

ò

(sin 6 x + cos6 x)

EXAMPLE 13

Evaluate

SOLUTION

Using the formula, ( a 3 + b 3) = ( a + b) 3 - 3 ab ( a + b), we get I=ò =ò

sin 2 x cos2 x

dx.

[CBSE 2014]

(sin 2 x + cos2 x) 3 - 3 sin 2 x cos2 x(sin 2 x + cos2 x) sin 2 x cos2 x (1 - 3 sin 2 x cos2 x) sin 2 x cos2 x

dx

ì ü 1 - 3 ý dx dx = ò í 2 2 sin x cos x î þ

ì(sin 2 x + cos2 x) ü ì 1 ü 1 = òí - 3 ý dx = ò í 2 + - 3 ý dx 2 2 2 x x x x sin cos cos sin î þ î þ = ò (sec2 x + cosec2 x - 3) dx = tan x - cot x - 3 x + C.

EXERCISE 12 Very-Short-Answer Questions Evaluate: 1.

ò x dx (iv) ò x5/ 3 dx (vii)

2.

3.

7

(i)

ò

3

(ii) (v)

x 2 dx

(viii)

-7

ò x dx -5/4 ò x dx ò4

1 x

3

dx

2 3 æ ö (i) ò ç 6x5 - 4 - 7 x + - 5 + 4ex + 7 x ÷ dx x è ø x 6 æ ö (ii) ò ç 8 - x + 2x 3 - 3 + 2x -5 + 5 x -1 ÷ dx è ø x (i) (iii)

ò ( 2 - 5 x)( 3 + 2x)(1 - x) dx æ

ò çç è

ö - 6ex + 1÷ dx ÷ x ø

7

3

x - x4 +

3

2

(iii) (vi) (ix)

-1

ò x dx x ò 2 dx 2

ò x 2 dx æx

a

(iii)

ò çè a + x + x

(ii)

ò

a

ö + a x + ax ÷ dx ø

x ( ax 2 + bx + c) dx

Senior Secondary School Mathematics for Class 12 Pg-598

598

Senior Secondary School Mathematics for Class 12

4.

(i)

æ

ò çè x

2

3

-

1 ö ÷ dx x2 ø

(ii)

2

5.

(iii)

1 ö æ ò çè x + x ÷ø dx

(v)

ò

(1 + x) 3 dx x

é

1

ò êê1 + (1 + x 2) ë

6.

(i)

2 1-x

2

+

(vi)

ò

(ii)

ø

æ

6

2

8.

ò çç sin x - tan

2

è

è

x-

x4

dx

2x 2 + x - 2 dx ( x - 2)

æ x6 - 1 ö

ò çç x 2 + 1 ÷÷ dx è

ø

æ x2 ö ÷ dx (iv) ò ç ç 1 + x2 ÷ è ø

ò çç 9 sin x - 7 cos x - cos2x + sin 2x + cot 2

(1 + 2x) 3

ù + a x ú dx úû x x -1

æ x2 - 1 ö

æ cot x

1 ö ÷ dx xø

2

7.

(ii)

ò

x-

5

æ x4 ö ÷ dx (iii) ò ç ç 1 + x2 ÷ è ø

9. (i)

(iv)

ò çç x 2 + 1 ÷÷ dx è

æ

ò çè

ö x ÷ dx ÷ ø

tan x 2 ö÷ + dx cos x cos2 x ÷ø

ò sec x( sec x + tan x) dx ò cosec x( cosec x - cot x) dx

10. (i)

ò (tan x + cot x)

11. (i)

ò (1 - cos x) dx

12. (i)

ò ( sec x + tan x) dx

13. (i)

ò 1 + cos x dx

14. (i)

ò

15. (i)

ò (1 + cos 2x) dx

2

dx

1

tan x

cos x

1 + cos 2x dx 1

16.

ò

18.

ò tan

1 + sin 2x dx -1 æ

sin 2x ö çç ÷ dx cos 2x ÷ø 1 + è

[CBSE 2011]

æ 1 + 2 sin x ö ÷ dx (iii) 2 ÷ è cos x ø 1 (ii) ò dx (1 - sin x) cosec x (ii) ò dx ( cosec x - cot x) sin x (ii) ò dx (1 - sin x) (ii)

ò çç

(ii)

ò

(ii)

ò (1 - cos 2x) dx

1 - cos 2x dx 1

æ sin 3 x + cos 3 x ö ò çç sin 2 x cos2 x ÷÷ dx è ø 2 æ 1 - tan x ö ÷ dx 19. ò cos-1 ç ç 1 + tan 2 x ÷ è ø 17.

æ 3 cos x + 4 ö ÷ dx 2 ÷ è sin x ø

ò çç

[CBSE 2002C]

Senior Secondary School Mathematics for Class 12 Pg-599

Indefinite Integral -1

20.

ò cos

22.

ò ( 3 cot x - 2 tan x)

24.

ò(

26.

599

1 - sin x dx [CBSE 2003, ‘06C] 1 + sin x

-1

21.

ò tan

23.

ò ( 3 sin x + 4cosec x)

25.

ò(

ò ççè 1 - cos x ÷÷ø dx

27.

ò (1 - tan x)

28.

ò sin ( x + b) dx

[CBSE 2006C] 29.

30.

ò (1 - x) x dx

[CBSE 2012]

32.

òí

(sin x) dx

dx x + 1 + x + 2)

2

dx [CBSE 2002]

æ 1 + cos x ö cos ( x + a)

ì 2 - 3 sin x ü ý dx 2 î cos x þ

31.

dx x + 3 - x + 2)

(1 + tan x)

2

dx [CBSE 2002]

dx

sin ( x - a )

ò sin ( x + a) dx sec2 x

ò cosec2x dx

[CBSE 2013]

[CBSE 2012C]

[CBSE 2011] ANSWERS (EXERCISE 12)

8

1.

2.

3.

x +C 8 3 (iv) x 8/ 3 + C 8 3 (vii) x5/ 3 + C 5 (i)

1 6x 6

+C

(iii) log |x| + C

(v) -4x -1/4 + C (viii) 4x1/4 + C

2x +C log 2 -2 (ix) +C x

(vi)

7x2 7x + 3 log |x| - 5 x + 4ex + +C 2 log 7 3x x2 x4 3 1 (ii) 8x + + 2 - 4 + 5 log |x| + C 2 2 x 2x ( a +1 ) x2 x ax ax 2 (iii) + a log |x| + + + +C 2a 2 ( a + 1) log a (i) x 6 +

2

3

-

17 x 2 x 3 5 x 4 (ii) + + +C 2 3 2 2 3/ 2 3 7/ 3 x - x + 21x1/ 3 - 6ex + x + C 3 7 1 3 x7 (ii) + 5 -x3 - +C 7 5x x 2 x (iv) + log |x| + 2x + C 2 2 6 2 x + x 7/ 2 + 2x 3/ 2 + x5/ 2 + C (vi) 7 5

(i) 6x (iii)

4.

(ii) -

(i) (iii) (v)

5. x + tan -1 x - 2 sin -1 x + 5sec-1 x +

ax +C log a

2ax 7/ 2 2bx5/ 2 2cx 3/ 2 + + +C 7 5 3

2 3/ 2 x - 2x1/ 2 + C 3 1 3 12 + 8 log |x| - 2 - + C 3 x 3x x x 2 + 5 x + 8 log |x - 2| + C

Senior Secondary School Mathematics for Class 12 Pg-600

600

6.

Senior Secondary School Mathematics for Class 12

(i) x - 2 tan -1 x + C (iii)

x3 - x + tan -1 x + C 3

(ii)

x5 x 3 + x - 2 tan -1 x + C 5 3

(iv) x - tan -1 x + C

7. -9 cos x - 7 sin x - 6 tan x - 3 cot x - x + C 8. - cosec x + tan x + x - sec x + C 9. (i) tan x + sec x + C 10. (i) tan x - cot x + C (iii) -3 cosec x - 4 cot x + C 11. (i) - cot x - cosec x + C 12. (i) sec x - tan x + x + C 13. (i) - cosec x + cot x + x + C 14. (i) 2 sin x + C 1 15. (i) tan x + C 2 16. sin x - cos x + C x2 18. +C 2 æ px x 2 ö 20. ç + C÷ ç 2 ÷ 2 è ø 22. 4 tan x - 9 cos x - 25 x + C

(ii) - cot x + cosec x + C (ii) tan x + 2sec x + C tan x + sec x + C - cot x - cosec x + C sec x + tan x - x + C - 2 cos x + C 1 (ii) - cot x + C 2 17. sec x - cosec x + C

(ii) (ii) (ii) (ii)

19. x 2 + C 21.

px x 2 +C 4 4

57 9 x - sin 2x - 16 cot x + C 2 4 2 2 25. ( x + 3) 3/2 + ( x + 2) 3/2 + C 3 3

23.

2 2 ( x + 2) 3/2 - ( x + 1) 3/2 + C 3 3 x 26. -2 cot - x + C 27. - log |cos x - sin x| + C 2 28. cos ( a - b) log |sin ( x + b)| - x sin ( a - b) + C 2 29. x cos 2a - sin 2a × log |sin ( x + a )| + C 30. x x (5 - 3 x) + C 15 31. tan x - x + C 32. 2 tan x - 3 sec x + C

24.

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 12)

21. I = ò tan

-1

= ò tan

-1

æp ö 1 - cos ç - x ÷ è2 ø dx æp ö 1 + cos ç - x ÷ è2 ø æp xö 2 sin 2 ç - ÷ ì æ p x öü è 4 2ø dx = ò tan -1 í tan ç - ÷ ý dx p x æ ö è 4 2 øþ î 2 cos 2 ç - ÷ è 4 2ø

Senior Secondary School Mathematics for Class 12 Pg-601

Indefinite Integral

601

px x 2 æp xö = ò ç - ÷ dx = + C. 4 4 è 4 2ø 22. I = ò ( 9 cot 2 x + 4 tan 2 x - 12 ) dx

= ò {9 ( cosec2 x - 1) + 4 ( sec2 x - 1) - 12} dx.

23. I = ò ( 9 sin 2 x + 16 cosec2 x + 24 ) dx

ü ì æ 1 - cos 2 x ö ÷ + 16 cosec2 x + 24 ý dx = ò í9ç 2 ø þ î è æ 57 9 2 ö =òç - cos 2 x + 16 cosec x ÷ dx. è 2 2 ø 1 ( x+ 2 +

24. I = ò

= ò x + 2 dx - ò 26. I = ò

´

( x + 2 - x + 1)

dx ( x + 2 - x + 1) 2 2 x + 1 dx = ( x + 2 ) 3/2 - ( x + 1) 3/2 + C . 3 3

x + 1)

( 1 + cos x ) 2 cos 2 ( x/ 2 ) dx = ò dx ( 1 - cos x ) 2 sin 2 ( x/ 2 )

x ö æ = ò cot 2 ( x/ 2 ) dx = ò ç cosec2 - 1 ÷ dx 2 è ø =

- cot ( x/ 2 ) x - x + C = -2 cot - x + C . ( 1/ 2 ) 2

27. I = ò

æ ç1+ ç è æ ç1ç è

= -ò

sin x ö ÷ cos x ÷ø

sin x ö ÷ cos x ÷ø

dx = ò

(cos x + sin x ) dx (cos x - sin x )

dt , where (cos x - sin x ) = t t

= - log|t| + C = - log|cos x - sin x| + C . cos ( x + b + a - b ) 28. I = ò dx sin ( x + b ) cos ( x + b ) cos ( a - b ) - sin ( x + b ) sin ( a - b ) dx =ò sin ( x + b ) = cos ( a - b ) ò cot ( x + b ) dx - sin ( a - b ) ò dx. sin ( x + a - 2 a ) dx sin ( x + a ) sin ( x + a ) cos 2 a - cos ( x + a ) sin 2 a cos ( x + a ) dx = cos 2 a ò dx - sin 2a × ò dx =ò sin ( x + a ) sin ( x + a )

29. I = ò

= x cos 2 a - sin 2 a × log|sin ( x + a )| + C . ì 2 3 sin x ü 32. I = ò í dx = ò ( 2 sec2 x - 3 sec x tan x ) dx. 2 2 ý î cos x cos x þ

Senior Secondary School Mathematics for Class 12 Pg-602

602

Senior Secondary School Mathematics for Class 12

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 1.

òx

6

dx = ?

(a) 7 x 7 + C 2.

òx

5

3 dx

(a) 3.

8 83 x +C 3

(c)

=?

3 23 x +C 5

-3 x2

+C

(b)

-1 2x 2

+C

3 34 x +C 4

4 34 x +C 3

(b)

(d)

5 83 x +C 3

(d)

x -2 +C 2

(d)

4 43 x +C 3

+C

(d)

2 32 x +C 3

(c)

5 35 x +C 3

(d)

3 35 x +C 5

(c)

3x +C log 3

(d)

(b)

x +C (log 2 + 1)

(c)

(c)

ò

3

3 23 x +C 2

3

(b)

2x

2

+C

3x2

+C

3 43 x +C 4

3x

2

3

2

x dx = ?

ò3 ò2

2

(c)

3

x

5 53 x +C 3

(b)

3 53 x +C 5

dx = ?

log x

log 3 3x

dx = ? log x +1

(a)

2 +C (log x + 1)

(c)

2 +C log 2

(log 2 +1 )

log x

9.

-1

1

(a) 3 x (log 3) + C (b) 3 x + C 8.

3 83 x +C 8

ò 3 x dx = ?

(a) 7.

(b)

(d) 6x 7 + C

ò 3 x dx = ?

(a) 6.

(c) 6x5 + C

1

(a) 5.

x7 +C 7

ò x 3 dx = ? (a)

4.

(b)

ò cosec x( cosec x + cot x) dx = ? (a) cot x - cosec x + C (c) cot x + cosec x + C

(d)

2

log x

2

+C

(b) - cot x + cosec x + C (d) - cot x - cosec x + C

+C

Senior Secondary School Mathematics for Class 12 Pg-603

Indefinite Integral

10.

11.

12.

13.

14.

15.

16.

sec x

ò ( sec x + tan x) dx = ? (a) tan x + sec x + C (c) - tan x + sec x + C (1 - cos 2x) ò (1 + cos 2x) dx = ? (a) tan x + x + C (c) - tan x + x + C 1 ò sin 2x cos2x dx = ? (a) tan x + cot x + C (c) tan x - cot x + C cos 2x ò cos2x sin 2x dx = ? (a) - cot x - tan x + C (c) cot x - tan x + C (cos 2x - cos 2a ) ò (cos x - cos a) dx = ? (a) 2 sin x + 2x cos a + C (c) -2 sin x + 2x cos a + C

ò ò

ò

(b) tan x - sec x + C (d) - tan x - sec x + C

(b) tan x - x + C (d) - tan x - x + C

(b) - tan x + cot x + C (d) none of these

(b) - cot x + tan x + C (d) cot x + tan x + C

(b) 2 sin x - 2x cos a + C (d) -2 sin x - 2x cos a + C

1 + cos 2x dx = ? (a) 2 cos x + C

(b) 2 sin x + C

(c) - 2 cos x + C (d) - 2 sin x + C

1 + sin 2x dx = ?

(a) sin x + cos x + C (c) sin x - cos x + C cos 2x 17. ò dx = ? sin 2 x cos2 x (a) cot x + tan x + C (c) cot x - tan x + C dx 18. ò =? (1 - cos 2x) 1 (a) cot x + C 2 1 (c) - cot x + C 2 19.

603

(b) - sin x + cos x + C (d) - sin x - cos x + C

(b) - cot x + tan x + C (d) - cot x - tan x + C

(b) 2cot x + C (d) -2cot x + C

sin 2x dx = ? sin x (a) 2sin x + C

1 (b) sin x + C 2

(c) 2cos x + C

1 (d) cos x + C 2

Senior Secondary School Mathematics for Class 12 Pg-604

604

20.

Senior Secondary School Mathematics for Class 12

ò

(1 - sin x) cos2 x

dx = ?

(a) tan x + sec x + C (c) - tan x + sec x + C 21.

ò cot

2

(b) tan x - sec x + C (d) - tan x - sec x + C

x dx = ?

(a) - cot x - x + C (c) - cot x + x + C 22.

(b) cot x - x + C (d) cot x + x + C

ò sec x( sec x + tan x) dx = ? (a) tan x - sec x + C (c) tan x + sec x + C

23.

(b) - tan x + sec x + C (d) - tan x - sec x + C

sec2 x

ò cosec2x dx = ? (a) tan x + x + C (c) - tan x + x + C

24.

(b) tan x - x + C (d) - tan x + x + C

sin 2 x

ò (1 + cos x) dx = ?

(a) x + sin x + C (c) sin x - x + C cot x 25. ò dx = ? ( cosec x - cot x)

(b) x - sin x + C (d) - sin x - x + C

(a) - cosec x - cot x - x + C (c) - cosec x + cot x - x + C sin x 26. ò dx = ? (1 + sin x) (a) sec x + tan x + x + C (c) - sec x + tan x + x + C (1 + sin x) 27. ò dx = ? (1 - sin x) (a) 2 tan x + 2sec x + x + C (c) tan x + sec x - x + C 1 28. ò dx = ? (1 + cos x) (a) - cot x + cosec x + C (c) cot x + cosec x + C

(b) cosec x - cot x - x + C (d) cosec x + cot x - x + C

29.

ò sin

-1

(b) sec x - tan x + x + C (d) none of these

(b) 2 tan x + 2sec x - x + C (d) none of these

(b) cot x - cosec x + C (d) none of these

(cos x) dx = ?

(a) cosec x + C

(b)

px x 2 + +C 2 2

(c)

px x 2 +C 2 2

(d)

x 2 px +C 2 2

Senior Secondary School Mathematics for Class 12 Pg-605

Indefinite Integral

30.

31.

32.

1 - cos 2x ü í ý dx = ? 1 î + cos 2x þ 1 1 (a) (b) +C +C 2 (1 + x ) 1 + x2

ò tan

-1 ì

sin 2x ö çç ÷ dx = ? cos 2x ÷ø 1 è -1 -1 (a) (b) +C +C (1 + x 2) (1 - x 2)

ò cot

ò sin

36.

(b) x 2 + C

(d)

x2 +C 2

(c)

x2 +C 2

(d) 2x 2 + C

(c)

x2 +C 2

(d) 2x 2 + C

-1 æ ç 1 - tan

2 ö x÷ dx = ? ç 1 + tan 2 x ÷ è ø

ò cos

ò tan

-1

(b) - x 2 + C

(c)

1 1 + x2

+C

(d)

1 1 - x2

( cosec x - cot x) dx = ?

x2 +C 4

(b)

-x 2 +C 4

(c)

x2 +C 2

x3 - x - 2 tan -1 x + C 3

(d)

-x 2 +C 2

æ ( x 4 + 1) ö ò çç ( x 2 + 1) ÷÷ dx = ? è ø (a)

x3 + x - tan -1 x + C 3

(b)

(c)

x3 + x - 2 tan -1 x + C 3

(d) none of these

( ax + b)

ò ( cx + d) dx = ? ax + log |cx + d| + C c ax ( bc - ad) (c) + log |cx + d| + C c c2

(a)

37.

1 - x2

+C

2 tan x ö ç ÷ dx = ? è 1 + tan 2 x ø

(a) 35.

1

-1 æ

(a) x 2 + C 34.

(c)

-1 æ

(a) - x 2 + C 33.

605

ò

(sin 3 x + cos 3 x) sin 2 x cos2 x

(b)

a + log |cx + d| + C c

(d) none of these

dx = ?

(a) sin x - cos x + C

(b) tan x - cos x + C

(c) sec x - cosec x + C

(d) none of these

+C

Senior Secondary School Mathematics for Class 12 Pg-606

606

38.

Senior Secondary School Mathematics for Class 12

sin x

ò sin( x - a) dx = ? (a) x cos a + (sin a ) log|sin ( x - a )| + C (b) x sin a + (sin a ) log|sin ( x - a )| + C (c) x cos a - (sin a ) log|sin( x - a )| + C (d) x sin a - (sin a ) log|sin( x - a )| + C

39.

ò sin 3 x sin 2x dx = ? 1 (a) - cos 5 x + C 5 1 1 (c) sin x - sin 5 x - C 2 10

40.

ò cos 3 x sin 2x dx = ? 1 1 (a) cos x - cos 5 x + C 2 10 1 1 (c) - cos x + cos 5 x + C 2 10

41.

1 1 (b) sin x + sin 5 x - C 2 10 1 1 (d) - cos 3 x - sin 2x + C 3 2 1 1 (b) - sin x + sin 5 x + C 2 10 (d) none of these

ò cos 4x cos x dx = ? 1 1 (a) sin 5 x + sin 3 x + C 5 3 1 1 (c) sin 5 x + sin 3 x + C 10 6

1 1 (b) cos 5 x - cos 3 x + C 5 3 (d) none of these

ANSWERS (OBJECTIVE QUESTIONS)

1. (b) 2. (c) 3. (b) 4. (c) 5. (a) 6. (b) 7. (c) 8. (b) 9. (d) 11. (b) 12. (c) 13. (a) 14. (a) 15. (b) 16. (c) 17. (d) 18. (c) 19. (a) 21. (b) 22. (c) 23. (b) 24. (b) 25. (a) 26. (b) 27. (b) 28. (a) 29. (c) 31. (c) 32. (b) 33. (a) 34. (a) 35. (b) 36. (c) 37. (c) 38. (a) 39. (c) 41. (c)

10. (b) 20. (b) 30. (d) 40. (a)

HINTS TO SOME SELECTED OBJECTIVE QUESTIONS 8. 2 log x = x log 2 . 12. Write 1 = (sin 2 x + cos 2 x ).

ì sec x ( sec x - tan x ) ü 10. I = ò í ´ ý dx. ( sec tan ) ( sec x - tan x ) þ x + x î 13. cos 2 x = (cos 2 x - sin 2 x ).

14. cos 2 x = ( 2 cos 2 x - 1) and cos 2 a = ( 2 cos 2 a - 1). 15.

1 + cos 2 x = 2 cos 2 x = 2 cos x.

16. ( 1 + sin 2 x ) = sin 2 x + cos 2 x + 2 sin x cos x = (sin x + cos x ) 2 \

1 + sin 2 x = (sin x + cos x ).

Senior Secondary School Mathematics for Class 12 Pg-607

Indefinite Integral

17. I = ò

(cos 2 x - sin 2 x ) (sin 2 x cos 2 x )

1 ö æ 1 dx = ò ç ÷ dx è sin 2 x cos 2 x ø

= ò ( cosec2 x - sec2 x ) dx. 18.

23.

1 1 1 = = cosec2 x. ( 1 - cos 2 x ) 2 sin 2 x 2 sec2 x cosec2 x

=

sin 2 x cos 2 x

21. cot 2 x = ( cosec2 x - 1).

= tan 2 x = ( sec2 x - 1).

ì cot x ( cosec x + cot x ) ü 25. I = ò í ´ ý dx x x ( cosec cot ) ( cosec x + cot x ) þ î = ò ( cosec x cot x + cot 2 x )dx = ò ( cosec x cot x + cosec2 x - 1)dx. ì sin x ( 1 - sin x ) ü sin x( 1 - sin x ) 26. I = ò í dx ´ ý dx = ò ( 1 sin ) ( 1 sin ) + x x ( 1 - sin 2 x ) î þ =ò

(sin x - sin 2 x ) 2

cos x

æ sin x ö dx = ò ç - tan 2 x ÷ dx ç cos 2 x ÷ è ø

= ò ( sec x tan x - sec2 x + 1) dx. ì ( 1 + sin x ) ( 1 - sin x ) ü ( 1 + sin x ) 2 27. I = ò í dx ´ ý dx = ò ( 1 - sin 2 x ) î ( 1 - sin x ) ( 1 + sin x ) þ =ò

( 1 + sin 2 x + 2 sin x ) cos 2 x

dx = ò ( sec2 x + tan 2 x + 2 sec x tan x ) dx

= ò ( 2 sec2 x - 1 + 2 sec x tan x ) dx. ì ( 1 - cos x ) ü ( 1 - cos x ) 1 28. I = ò í ´ dx ý dx = ò + x x 1 1 ( cos ) ( cos ) sin 2 x î þ ìï -1 cos x üï 2 =í ý dx = ò ( cosec x - cosec x cot x ) dx. 2 ïî sin x sin 2 x ïþ ì ö öü æ p æp 29. sin -1 (cos x ) = sin -1 í sin ç - x ÷ ý = ç - x ÷ × ø è2 øþ è 2 î æ 2 tan x ö ÷ = sin -1 (sin 2 x ) = 2 x. 32. sin -1 ç ç 1 + tan 2 x ÷ è ø 2 ö 2 2 ö æ æ 1 x tan ÷ = cos -1 ç cos x - sin x ÷ 33. cos -1 ç ç 1 + tan 2 x ÷ ç cos 2 x + sin 2 x ÷ è ø è ø = cos -1 (cos 2 x - sin 2 x ) = cos -1 (cos 2 x ) = 2 x. æ 1 - cos x ö æ 1 cos x ö ÷ ÷ = tan -1 ç 34. tan -1 ( cosec x - cot x ) = tan -1 çç ç sin x ÷ ÷ è ø è sin x sin x ø ì ü æxö 2 sin 2 ç ÷ ïï ï æxö x è2ø -1 = tan í ý = tan tan ç ÷ = × x x æ ö æ ö è2ø 2 ï 2 sin ç ÷ cos ç ÷ ï ïî è2ø è 2 ø ïþ -1 ï

607

Senior Secondary School Mathematics for Class 12 Pg-608

608

Senior Secondary School Mathematics for Class 12

35. On dividing ( x 4 + 1) by ( x 2 + 1), we get ( x 4 + 1) ( x 2 + 1)

= ( x 2 - 1) +

2 ( 1 + x2 )

×

36. On dividing ( ax + b ) by ( cx + d ), we get ì a ( bc - ad ) ü ( ax + b ) ò ( cx + d ) dx = ò í c + c ( cx + d ) ý dx î þ ( bc - ad ) a c = ò dx + ×ò dx c ( cx + d ) c2 ax ( bc - ad ) + log|cx + d| + C . c c2

= 37. I = ò

sin 3 x 2

2

sin x cos x

dx +

= ò sec x tan x dx +

cos 3 x

ò sin 2 x cos2 x dx

ò cosec x cot x dx

= sec x - cosec x + C. sin ( x - a + a ) sin ( x - a ) cos a + cos ( x - a ) sin a 38. I = dx = ò dx sin ( x - a ) sin ( x - a ) = (cos a ) ò dx + (sin a )ò cot ( x - a )dx = x cos a + (sin a ) log|sin( x - a )| + C . 1 39. I = ò 2 sin 3 x sin 2 x dx 2 1 1 1 = ò (cos x - cos 5 x ) dx = sin x sin 5 x + C . 2 2 10 1 40. I = 2 cos 3 x sin 2 x dx 2 -1 1 1 = ò (sin 5 x - sin x ) dx = cos 5 x + cos x + C . 2 10 2 1 1 41. I = 2 cos 4 x cos x dx = ò (cos 5 x + cos 3 x ) dx 2 2 1 1 = sin 5 x + sin 3 x + C . 10 6

Senior Secondary School Mathematics for Class 12 Pg-609

13. METHODS OF INTEGRATION Integration by Substitution If we have to evaluate an integral of the type ò f {f( x)} × f¢( x) dx then we put f( x) = t and f¢( x) dx = dt. With this substitution, the integrand becomes easily integrable. Case I

Case II

When the integrand is of the form f ( ax + b), we put ( ax + b) = t and 1 dx = dt. a When the integrand is of the form x n-1 × f ( x n), we put x n = t and nx n-1 dx = dt.

When the integrand is of the form { f ( x)} n × f ¢( x), we put f ( x) = t and f ¢( x) dx = dt. f ¢( x) Case IV When the integrand is of the form , we put f ( x) = t and f ( x) f ¢( x) dx = dt. Case III

THEOREM 1

PROOF

n

dx =

( ax + b) n+1 + C, a(n + 1)

Putting ax + b = t , we get \

THEOREM 2

PROOF

ò ( ax + b)

ò ( ax + b)

n

dx =

where n ¹ - 1.

a dx = dt or dx =

( ax + b) n+1 1 n 1 t n+1 t dt = × +C = + C. ò a a (n + 1) a(n + 1) 1

(i)

ò cos ( ax + b) dx = a sin ( ax + b) + C

(ii)

ò cosec ( ax + b) dx = - a cot ( ax + b) + C

1

2

(i) Put ( ax + b) = t so that dx = \

1 dt. a

1 dt. a

1

ò cos ( ax + b) dx = a ò cos t dt 1 sin t + C a 1 = sin ( ax + b) + C. a =

609

Senior Secondary School Mathematics for Class 12 Pg-610

610

Senior Secondary School Mathematics for Class 12

(ii) Put ( ax + b) = t so that dx = \

1 dt. a

1

ò cosec ( ax + b) dx = a ò cosec t dt 2

2

1 1 = - cot t + C = - cot ( ax + b) + C. a a

SOLVED EXAMPLES EXAMPLE 1

Evaluate:

(i) (iii)

SOLUTION

ò ( 3 x + 5)

7

(ii)

ò ( 4 - 9x)

(iv)

ò

dx

1

ò ( 2 - 3 x) 4 dx

7 ò ( 3 x + 5) dx =

dx

ax + b dx

(i) Put ( 3 x + 5) = t so that 3 dx = dt or dx = \

5

1 dt. 3

1 7 1 t8 ( 3 x + 5) 8 t dt = × + C = + C. 3ò 3 8 24

1 dt. 9 6 1 1 t - ( 4 - 9x) 6 \ ò ( 4 - 9x)5 dx = - ò t5 dt = - × + C = + C. 9 9 6 54

(ii) Put ( 4 - 9x) = t so that -9 dx = dt or dx = -

(iii) Put ( 2 - 3 x) = t so that -3 dx = dt or dx = \

1

1

1

1

1 dt. 3

1

1

ò ( 2 - 3 x) 4 dx = - 3 ò t 4 dt = - 3 × ( -3t 3) + C = 9( 2 - 3 x) 3 + C.

(iv) Put ( ax + b) = t so that a dx = dt. \ EXAMPLE 2

Evaluate:

ò

ax + b dt =

2 3/2 2( ax + b) 3/2 1 t C t dt = + = + C. aò 3a 3a

(i) ò cos 2x dx

(iii) ò sec2( 3 x + 5) dx SOLUTION

(5 x + 3 )

(ii) ò e

dx

(iv) ò sin 3 x dx

1 dt. 2 1 1 1 \ ò cos 2x dx = ò cos t dt = sin t + C = sin 2x + C. 2 2 2 1 (ii) Put (5 x + 3) = t so that 5 dx = dt or dx = dt. 5 1 1 1 (5x + 3) (5 x + 3 ) dx = ò et dt = × et + C = e + C. \ òe 5 5 5 (i) Put 2x = t so that 2dx = dt

or dx =

Senior Secondary School Mathematics for Class 12 Pg-611

Methods of Integration

611

(iii) Put ( 3 x + 5) = t so that 3 dx = dt or dx = \

1

1

1 dt. 3

ò sec ( 3 x + 5) dx = 3 ò sec t dt = 3 tan t + C 2

=

2

1 tan ( 3 x + 5) + C. 3

sin 3 x = 3 sin x - 4 sin 3 x.

(iv) We know that

1 sin 3 x = ( 3 sin x - sin 3 x). 4

\

ò sin

So,

3

1 æ3 ö x dx = ò ç sin x - sin 3 x ÷ dx 4 è4 ø 3 1 sin x dx - ò sin 3 x dx 4ò 4 3 1 ( - cos 3 x) = ( - cos x) - × +C 4 4 3 3 cos 3 x = - cos x + + C. 4 12 =

EXAMPLE 3

Evaluate:

(i)

(iii)

SOLUTION

ò

log x dx x etan

-1

(ii)

ò

(iv)

ò

x

ò (1 + x 2) dx

sec2(log x) dx x x

sin x

dx

1 dx = dt. x 1 1 log x \ ò dx = ò t dt = t 2 + C = (log x) 2 + C. 2 2 x

(i) Put log x = t so that

(ii) Put log x = t so that \

ò

1 dx = dt. x

sec2(log x) dx = ò sec2t dt = tan t + C = tan (log x) + C. x

(iii) Put tan -1 x = t so that \

etan

-1

x

(1 + x 2)

ò (1 + x 2) dx = ò e dt = e

(iv) Put x = t so that or

1

1 dx = 2 dt. x

t

t

dx = dt.

+ C = etan

1 - 1/2 x dx = dt 2

-1

x

+ C.

Senior Secondary School Mathematics for Class 12 Pg-612

612

Senior Secondary School Mathematics for Class 12

\

ò

x

sin x

dx = 2ò sin t dt = 2( - cos t) + C = -2 cos t + C = -2 cos x + C.

EXAMPLE 4

Evaluate:

ò cos

(i)

(ii)

x sin x dx 2

cosec x

ò (1 + cot x) dx

(iii) SOLUTION

3

(iv)

ò(

sin x ) cos x dx sin x

ò ( 3 + 4 cos x) 2 dx

(i) Put cos x = t so that sin x dx = - dt. t4 1 + C = - cos4 x + C. 4 4 (ii) Put sin x = t so that cos x dx = dt. 2 2 \ ò ( sin x ) cos x dx = ò t dt = t 3 / 2 + C = (sin x) 3 / 2 + C. 3 3 2 (iii) Put (1 + cot x) = t so that - cosec x dx = dt. \

ò cos

\

ò (1 + cot x) dx = -ò t

3

x sin x dx = - ò t 3 dt = -

cosec2 x

1

dt

= - log t + C = - log |(1 + cot x)| + C. (iv) Put ( 3 + 4 cos x) = t so that - 4 sin x dx = dt. 1 1 1 1 sin x dx = - ò 2 dt = \ ò +C = + C. 2 4 t 4t 4( 3 + 4 cos x) ( 3 + 4 cos x) EXAMPLE 5

SOLUTION

Evaluate:

(i)

2x

ò ( 2x + 1) 2 dx

(ii)

( 2 + 3 x)

ò ( 3 - 2x)

(i) Put ( 2x + 1) = t so that 2x = (t - 1) and dx = \

1 (t - 1)

2x

ò ( 2x + 1) 2 dx = 2 ò

t2

dx

1 dt. 2

dt

1 1 1 1 1 1 dt - ò 2 dt = log || t + +C ò 2 t 2 t 2 2t 1 1 = log |( 2x + 1)| + + C. 2 2( 2x + 1) =

1 æ 3 -t ö (ii) Put ( 3 - 2x) = t so that x = ç ÷ and dx = - dt. 2 2 ø è 9 3 t é ù æ ö 2+ ç ÷ú 1 (13 - 3t) ( 2 + 3 x) 1 êë è 2 øû dt = - ò dt = - ò dt \ ò 4 t ( 3 - 2x) 2 t -13 1 3 13 3 = dt + ò dt = - log || t + t+C 4 òt 4 4 4 13 3 = - log |( 3 - 2x)| + ( 3 - 2x) + C. 4 4

Senior Secondary School Mathematics for Class 12 Pg-613

Methods of Integration EXAMPLE 6

Evaluate: 3x2 (i) ò dx (1 + x 6)

(ii)

x3

ò ( x 2 + 1) 3

613

dx

(iii)

x8 (1 - x 3)1 / 3

dx

(i) Put x 3 = t so that 3 x 2 dx = dt. 3x2 dt dx = ò = tan -1t + C = tan -1 x 3 + C. \ ò 6 (1 + x ) (1 + t 2) 1 (ii) Put ( x 2 + 1) = t so that x 2 = (t - 1) and x dx = dt. 2 x3 x2 × x dx dx \ ò 2 = ò ( x 2 + 1) 3 ( x + 1) 3

SOLUTION

1 (t - 1) 1 1 1 1 dt = ò 2 dt - ò 3 dt 2ò t3 2 t 2 t -1 1 -1 1 = + +C = + +C 2t 4t 2 2( x 2 + 1) 4( x 2 + 1) 2 =

- (1 + 2x 2)

=

4( x 2 + 1) 2

+ C.

(iii) Put (1 - x 3) = t so that x 3 = (1 - t) and x 2 dx = \

x8

x6 × x2

1 dt. 3

ò (1 - x 3)1/3 dx = ò (1 - x 3)1/3 dx 1 (1 - t) 2 1 (1 + t 2 - 2t) dt = - ò dt ò / 1 3 3 t 3 t1/ 3 1 1 2 = - ò t -1/ 3 dt - ò t5/ 3 dt + ò t 2/ 3 dt 3 3 3 1 2/ 3 1 8/ 3 2 5/ 3 =- t - t + t +C 2 8 5 1 1 = - (1 - x 3) 2/ 3 - (1 - x 3) 8/ 3 2 8 2 + (1 - x 3)5/ 3 + C. 5 =-

ò

dx

×

EXAMPLE 7

Evaluate

SOLUTION

Put x 3 = t so that 3 x 2 dx = dt \

ò

x × x6 - 1

dx x × x6 - 1

=ò

x2 x 3 × x6 - 1

or x 2 dx =

1 dt. 3

dx

[multiplying numerator and denominator by x 2] 1 1 1 1 dt = sec-1t + C = sec-1 x 3 + C. = ò 2 3 t t -1 3 3

Senior Secondary School Mathematics for Class 12 Pg-614

614

Senior Secondary School Mathematics for Class 12

EXAMPLE 8

Evaluate

SOLUTION

ò(

ò(

1 dx. x + x)

[CBSE 2003]

1 1 dx = ò dx x + x) x (1 + x )

Now, put (1 + x ) = t so that \

ò(

1 dx = 2 dt. x

1 1 dx = ò dx x + x) x (1 + x ) 1 = 2ò dt = 2 log|t| + C = 2 log|(1 + x )| + C. t

EXAMPLE 9

Evaluate:

(i)

(iii) SOLUTION

ò

( x - 1) dx x+4

ò ( 4x + 2)

x 2 + x + 1 dx

(ii)

òx

x + 2 dx

(iv)

ò

( 4x + 3) 2x 2 + 3 x + 1

dx

(i) Put ( x + 4) = t 2 so that x = (t 2 - 4) and dx = 2t dt. \

ò

( x - 1) (t 2 - 5)t dx = 2ò dt t x+4 2t 3 - 10t + C 3 2 = ( x + 4) 3 / 2 - 10( x + 4)1 / 2 + C. 3 = 2ò t 2 dt - 10ò dt =

(ii) Put ( x + 2) = t 2 so that x = (t 2 - 2) and dx = 2t dt. \

òx

x + 2 dx = ò (t 2 - 2) 2t 2 dt = 2ò t 4 dt - 4ò t 2 dt =

2t5 4t 3 2( x + 2)5/2 4( x + 2) 3/2 +C = + C. 5 3 5 3

(iii) Put ( x 2 + x + 1) = t so that ( 2x + 1) dx = dt. \ ò ( 4x + 2)( x 2 + x + 1) dx = 2ò t dt =

4 3/2 4 t + C = ( x 2 + x + 1) 3/2 + C. 3 3

(iv) Put ( 2x 2 + 3 x + 1) = t so that ( 4x + 3) dx = dt. \

ò

( 4x + 3) 2

2x + 3 x + 1

dx = ò

dt = 2 t + C = 2 2x 2 + 3 x + 1 + C. t

Senior Secondary School Mathematics for Class 12 Pg-615

Methods of Integration EXAMPLE 10

615

Evaluate: ( 2x + 5) (i) ò 2 dx ( x + 5 x + 9) (cos x - sin x) (iii) ò dx (cos x + sin x)

( 6x - 7)

(ii)

ò ( 3 x 2 - 7 x + 5) 2 dx

(iv)

ò log ( sec x + tan x) dx

sec x

(i) Put ( x 2 + 5 x + 9) = t so that ( 2x + 5) dx = dt. ( 2x + 5) 1 dx = ò dt = log|t| + C \ ò 2 t ( x + 5 x + 9)

SOLUTION

= log|( x 2 + 5 x + 9)| + C. (ii) Put ( 3 x 2 - 7 x + 5) = t so that ( 6x - 7) dx = dt. ( 6x - 7) 1 1 -1 dx = ò 2 dt = - + C = + C. \ ò 2 2 2 t ( 3 x - 7 x + 5) t ( 3 x - 7 x + 5) (iii) Put (cos x + sin x) = t so that (cos x - sin x) dx = dt. (cos x - sin x) 1 \ ò dx = ò dt (cos x + sin x) t = log |t| + C = log |(cos x + sin x)| + C. (iv) Put log ( sec x + tan x) = t. Then, on differentiation, we get 1 × ( sec x tan x + sec2 x) dx = dt ( sec x + tan x) or sec x dx = dt. sec x 1 dx = ò dt \ ò log ( sec x + tan x) t = log || t + C = log |log ( sec x + tan x)| + C. sin 2x

ò ( a 2 sin 2 x + b2 cos2 x) dx.

EXAMPLE 11

Evaluate

SOLUTION

Put ( a 2 sin 2 x + b 2 cos2 x) = t so that

[CBSE 2005]

2( a 2 - b 2) sin x cos x dx = dt Û sin 2x dx = \

I=ò

sin 2x 2

2

2

2

( a sin x + b cos x) = =

EXAMPLE 12

Evaluate

ò

x 2 tan -1 x 3 (1 + x 6)

dx.

dx =

dt 2

( a - b2)

×

dt 2

( a - b 2)t

1 ( a 2 - b2) 1 ( a 2 - b2)

log |t| + C log ( a 2 sin 2 x + b 2 cos2 x) + C.

Senior Secondary School Mathematics for Class 12 Pg-616

616

SOLUTION

Senior Secondary School Mathematics for Class 12

Put tan -1 x 3 = t so that \

x 2 tan -1 x 3

ò

(1 + x 6)

3x2 6

dx = dt or

x2 6

dx =

1 dt 3

(1 + x ) (1 + x ) 1 1 2 1 dx = ò t dt = t + C = (tan -1 x 3) 2 + C. 3 6 6

tan x

ò sin x cos x dx.

EXAMPLE 13

Evaluate

SOLUTION

ò sin x cos x dx = ò (

tan x

=ò

tan x sec2 x dx dx = ò tan x ) × sin x cos x tan x

1 dt, where tan x = t and sec2 x dx = dt t

= 2 t + C = 2 tan x + C. EXAMPLE 14

Evaluate

SOLUTION

ò(

dx 2x + 3 + 2x - 3 ) =ò

1 ( 2x + 3 ) - 2x - 3 ) ´ dx ( 2x + 3 + 2x - 3 ) ( 2x + 3 ) - 2x - 3 )

=ò

1 1 ( 2x + 3 - 2x - 3 ) dx = ò ( 2x + 3)1 / 2 dx - ò ( 2x - 3)1 / 2 dx 6 6 [( 2x + 3) - ( 2x - 3)]

= EXAMPLE 15

dx × 2x + 3 + 2x - 3 )

ò(

1 1 ( 2x + 3) 3 / 2 - ( 2x - 3) 3 / 2 + C. 18 18

Evaluate: (i) ò

1 dx (1 + tan x)

æ 1 - tan x ö ÷÷ dx [CBSE 2000C] (iii) ò çç è 1 + tan x ø SOLUTION

(i)

(ii) ò

1 dx (1 + cot x)

(iv) ò

tan x dx ( sec x + cos x)

1

1 dx sin x ö çç1 + ÷ cos x ÷ø è cos x (cos x + sin x) + (cos x - sin x) dx =ò dx = ò (cos x + sin x) 2(cos x + sin x) 1 1 (cos x - sin x) = ò dx + ò dx 2 2 (cos x + sin x) 1 1 1 = ò dx + ò dt, where (cos x + sin x) = t and 2 2 t (cos x - sin x) dx = dt

ò (1 + tan x) dx = ò æ

=

1 1 1 1 x + log |t| + C = x + log |cos x + sin x| + C. 2 2 2 2

Senior Secondary School Mathematics for Class 12 Pg-617

Methods of Integration

617

sin x 1 dx = ò dx ö + cos x) x (sin cos x çç1 + ÷÷ sin x ø è (sin x + cos x) - (cos x - sin x) =ò dx 2(sin x + cos x) 1 1 (cos x - sin x) = ò dx - ò dx 2 2 (sin x + cos x) 1 1 1 = ò dx - ò dt, 2 2 t where sin x + cos x = t and (cos x - sin x) dx = dt 1 1 1 1 = x - log |t| + C = x - log |sin x + cos x | + C. 2 2 2 2 æ sin x ö çç1 ÷ æ 1 - tan x ö cos x ÷ø (cos x - sin x) è (iii) ò çç ÷÷ dx = ò dx dx = ò æ (cos x + sin x sin x ö è 1 + tan x ø çç1 + ÷÷ cos x ø è (ii)

1

ò (1 + cot x) dx = ò æ

1 = ò dt, where (cos x + sin x) = t and t (cos x - sin x) dx = dt = log |t| + C = log |(cos x + sin x)| + C. (iv)

tan x

sin x

ò ( sec x + cos x) dx = ò 1 + cos2x dx = -ò

1 (1 + t 2)

dt , where cos x = t and sin x dx = - dt

= - tan -1t + C = - tan -1(cos x) + C. EXAMPLE 16

SOLUTION

Evaluate:

(i) ò tan x dx

(ii) ò cot x dx

(iii) ò sec x dx (iv) ò cosec x dx sin x (i) ò tan x dx = ò dx cos x 1 = - ò dt, where cos x = t and sin x dx = - dt t = - log |t| + C = - log |cos x| + C. \ (ii)

ò tan x dx = - log |cos x | + C. cos x

ò cot x dx = ò sin x dx 1 = ò dt, where sin x = t and cos x dx = dt t = log |t| + C = log |sin x| + C. \

ò cot x dx = log |sin x| + C.

Senior Secondary School Mathematics for Class 12 Pg-618

618

Senior Secondary School Mathematics for Class 12

(iii)

ò sec x dx = ò

sec x( sec x + tan x) dx ( sec x + tan x)

[multiplying numerator and denominator by ( sec x + tan x)] 1 = ò dt, where (sec x + tan x) = t t and sec x( sec x + tan x) dx = dt = log |t| + C = log |(sec x + tan x)| + C. \

ò sec x dx = log |(sec x + tan x)| + C.

Alternative form

æ 1 sin x ö (1 + sin x) ÷= sec x + tan x = çç + cos x cos x ÷ø cos x è sin x =

Putting

\ sec x + tan x = \

2 tan ( x/2) 1 + tan 2( x/2)

and cos x =

1 + tan ( x/2) = tan 1 - tan ( x/2)

1 - tan 2( x/2) 1 + tan 2( x/2)

.

æp xö ç + ÷× è 4 2ø

ò sec x dx = log |sec x + tan x| + C æp xö = log tan ç + ÷ + C. è 4 2ø

(iv)

ò cosec x dx = ò

cosec x( cosec x - cot x) dx ( cosec x - cot x)

[multiplying numerator and denominator by (cosec x - cot x)] 1 = ò dt, where (cosec x - cot x) = t t and cosec x ( cosec x - cot x) dx = dt = log |t| + C = log |cosec x - cot x| + C. \

ò cosec x dx = log |cosec x - cot x| + C.

Alternative form

æ 1 cos x ö 1 - cos x ÷= cosec x - cot x = çç sin x sin x ÷ø sin x è = \

2 sin 2( x/2) x = tan × 2 sin ( x/2) cos ( x/2) 2

ò cosec x dx = log |cosec x - cot x| + C x = log |tan | + C. 2

Senior Secondary School Mathematics for Class 12 Pg-619

Methods of Integration

619

As a consequence of the above results, the integral of trigonometric functions may be listed as given below:

ò sin x dx = - cos x + C (iii) ò tan x dx = - log |cos x| + C (iv) ò cot x dx = log |sin x| + C

EXAMPLE 17

SOLUTION

ò cos x dx = sin x + C

(i)

(ii)

(v)

ò sec x dx = log |sec x + tan x| + C = log

(vi)

ò cosec x dx = log |cosec x - cot x| + C = log

Evaluate: (1 + cos x) (i) ò dx [CBSE 2000C] (1 - cos x) (i)

(1 + cos x)

ò (1 - cos x)

dx = ò

æp xö tan ç + ÷ + C è 4 2ø

(ii)

tan

x +C 2

(1 + sin x)

ò (1 + cos x) dx

2 cos2( x/2)

dx 2 sin 2( x/2) x æxö = ò cot 2 ç ÷ dx = ò (cosec2 - 1) dx 2 è 2ø = ò cosec2

x dx - ò dx 2

= 2ò cosec2t dt - ò dx, where

x = t and dx = 2dt 2

æxö = -2 cot t - x + C = -2 cot ç ÷ - x + C. è 2ø (ii)

æ 1 + sin x ö

sin x

1

ò ççè 1 + cos x ÷÷ø dx = ò (1 + cos x) dx + ò (1 + cos x) dx =ò =

1 2 cos2( x/2)

dx + ò

2 sin ( x/2) cos ( x/2) 2 cos2( x/2)

x 1 æxö sec2 ç ÷ dx + ò tan dx ò 2 2 è 2ø

= ò sec2t dt + 2ò tan t dt , where = tan t - 2 log |cos t| + C æxö æxö = tan ç ÷ - 2 log cos ç ÷ + C. è 2ø è 2ø

x =t 2

dx

Senior Secondary School Mathematics for Class 12 Pg-620

620 EXAMPLE 18

Senior Secondary School Mathematics for Class 12

Evaluate: (i)

dx

ò1+

(ii)

x

ò

x+ x+1 dx x+2

(i) Put x = t so that x = t 2 and dx = 2t dt.

SOLUTION

\

dx

ò1+

2t 2(1 + t) - 2 dt = ò dt (1 + t) (1 + t) dt = 2ò dt - 2ò = 2t - 2 log |1 + t| + C 1+t = 2 x - 2 log |1 + x | + C.

=ò

x+1 =t

(ii) Put \

x

so that x + 1 = t 2 and dx = 2t dt.

(t 2 - 1 + t)t x+ x+1 dx = 2 ò ( x + 2) ò (t 2 + 1) dt æ t 3 + t2 - t ö ÷ dt = 2ò ç 2 ç t +1 ÷ è ø 2t + 1 ö æ [by division] = 2ò çt + 1 - 2 ÷ dt è t + 1ø 2t 1 ö æ = 2ò çt + 1 - 2 ÷ dt è t + 1 t2 + 1ø 2t 1 = 2ò t dt + 2ò dt - 2ò 2 dt - 2ò 2 dt t +1 t +1 = t 2 + 2t - 2 log |t 2 + 1| - 2 tan -1t + C = ( x + 1) + 2 x + 1 - 2 log |x + 2| - 2 tan - 1 x + 1 + C.

EXAMPLE 19

Evaluate

SOLUTION

ò

ò

1+ x dx. 1-x

[CBSE 2006]

1+ x 1+ x 1+ x dx = ò ´ dx 1-x 1-x 1+ x 1+ x dx x =ò +ò dx = ò dx 2 2 1-x 1-x 1 - x2 =ò

dx 1 - x2

-

1 1 dt , where (1 - x 2) = t 2ò t

= sin -1 x - t + C = sin -1 x - 1 - x 2 + C. EXAMPLE 20

Evaluate

( 3 sin x - 2) cos x

ò (5 - cos2x - 4 sin x) dx

[CBSE 2013C]

Senior Secondary School Mathematics for Class 12 Pg-621

Methods of Integration

621

We have

SOLUTION

I=ò =ò

( 3 sin x - 2) cos x {5 - (1 - sin 2 x) - 4 sin x} ( 3 sin x - 2) cos x {4 + sin 2 x - 4 sin x}

dx

dx = ò

( 3 sin x - 2) cos x ( 2 - sin x) 2

dx

… (i)

Putting 2 - sin x = t , we get sin x = 2 - t and cos x dx = - dt. { 3( 2 - t) - 2} ( 4 - 3t) ( 3t - 4) dt = - ò dt = ò dt \ I = -ò t2 t2 t2 4 æ3 4ö = ò ç - 2 ÷ dt = 3 log |t| + + C t t è t ø 4 = 3 log |( 2 - sin x)| + + C. ( 2 - sin x)

EXERCISE 13A Evaluate the following integrals: Very-Short-Answer Questions 1.

ò ( 2x + 9)

dx

2.

ò (7 - 3 x)

3.

ò

3 x - 5 dx

4.

ò

5.

ò

1 dx 3 - 4x

6.

ò ( 2x - 3) 3/2 dx

7.

dx òe ( 2- 3x ) dx ò3

8.

dx òe ò sin 3x dx

9.

5

( 2 x -1 )

10.

4

dx

1 dx 4x + 3 1

(1 - 3 x )

Short-Answer Questions 11. 13. 15. 17.

ò cos (5 + 6x) dx 2 ò cosec ( 2x + 5) dx 3 ò sin x cos x dx ò

sin x 1 - x2

dx

cos (log x) ò x dx 1 21. ò dx x log x

ò

14. 16.

-1

19.

23.

12.

(log x) 2 dx x

18.

ò sin x 1 + cos 2x dx ò sin x cos x dx ò ( cos x ) sin x dx ò

sin ( 2 tan -1 x) (1 + x 2)

dx

cosec2(log x) dx ò x ( x + 1)( x + log x) 2 22. ò dx x 20.

24.

ò

cos x dx x

[CBSE 2006C]

[CBSE 2002]

[CBSE 2001] [CBSE 2002]

Senior Secondary School Mathematics for Class 12 Pg-622

622

25. 27.

Senior Secondary School Mathematics for Class 12

ò e sec x dx ò sin ( ax + b) cos( ax + b) dx tan x

1

2

-1/x

26. 28.

cos2 x

ò e sin 2x dx 3 ò cos x dx æ1ö

1

29.

ò x2 e

dx

30.

ò x 2 cos çè x ÷ø dx

31.

ò ( ex + e- x )

32.

ò ( e2x - 2) dx

33.

ò cot x log (sin x) dx

34.

ò log (sin x) dx

35.

ò 2x sin ( x

36.

ò sec x log ( sec x + tan x) dx

dx

2

+ 1) dx 2

x sec x

tan

37.

ò

39.

ò

41.

ò

43.

ò (1 + cos x) dx

x sin -1 x 2 1-x

4

x

dx

dx

( 2 + log x) dx x sin x

45. (i)

(1 + tan x)

ò ( x + log sec x) dx sin 2x

46.

ò ( a 2 + b2 sin 2x) dx

48.

ò ççè 3 cos x + 2 sin x ÷÷ø dx

50.

ò ( x 2 + 2x - 3) dx

52.

ò ( 3 x 3 - 2x 2 + 5 x + 1) dx

54.

ò ( x + sin x) 3 dx

56.

ò

æ 2 cos x - 3 sin x ö ( x + 1)

( 9x 2 - 4x + 5)

(1 + cos x) ( 2x + 3) 2

dx

x + 3x - 2 dx 58. ò ( x + a + x + b) 60.

x2

ò (1 + x 6)

dx

e2x

cot x

x tan -1 x 2

38.

ò

40.

ò

42.

ò (1 + tan x) dx

44.

ò ççè 1 - tan x ÷÷ø dx

(ii)

ò ( x + cos2 x) dx

47.

ò ( a 2cos2x + b2sin 2x) dx

49.

ò ( 2x 2 + 3) dx

51.

ò ( 2x 2 - 5 x + 1) dx

53.

ò

55.

ò (1 + cos x) 2 dx

(1 + x 4)

dx

1 ( 1 - x ) sin -1 x 2

dx

sec2 x

æ 1 + tan x ö

(1 - sin 2x)

sin 2x

4x

( 4x - 5)

sec x cosec x dx log (tan x) sin x

ò

( 2x - 1)

dx x2 - x - 1 dx 59. ò ( 1 - 3x - 5 - 3x) 57.

61.

x3

ò (1 + x 8) dx

[CBSE 2000C] [CBSE 2000] [CBSE 2005]

Senior Secondary School Mathematics for Class 12 Pg-623

Methods of Integration

x

62.

ò (1 + x 4) dx

64.

ò

66.

òx òx

68.

623

x5

63.

ò

65.

ò

x - 1 dx

67.

x 2 - 1 dx

69.

ò (1 - x) 1 + x dx ò x 3 x - 2 dx

x dx 1+ x

dx

1+ x3 1

dx

x x4 - 1

dx

70.

ò x cos2(1 + log x)

71.

òx

72.

ò ( 2x + 4)

73.

ò (sin x - cos x) dx

74.

ò (1 - tan x)

75.

ò (1 - cot x)

76.

ò (sin x + cos x) 2 dx

77.

ò

79.

ò x sin

81.

òe

( x 2 + 3) tan ( x 2 + 3) dx

83.

ò ( a + b cos x) 2 dx

ò

86.

òx

x 2 + 4x + 3 dx

dx

cos 2x

( x + 1)( x + log x) 2 dx x sec2 x 80. ò dx 1 - tan 2 x

78.

ò

82.

ò 2x sec

84.

ò ( 3 - 5 x)

3

dx

em tan

-1

85.

x

87.

ò (1 + x 2)

90.

ò

dx

88.

ex - 1 dx sec2( 2 tan -1 x)

1 + x dx ( x + 1) ex

ò cos2( x ex )

2

sin x 3 dx sin x

dx

(cos x - sin x) dx (1 + sin 2x)

-x

3 2

x cos x 2 dx

cosec2( 2e- x + 5) dx sin 2x

e

2 x3

e

cos( ex ) dx

3

x

x

cos( e ò x dx 91. ò (x - x)

dx 89.

)

dx

æ 1 + sin 2x ö ÷ 93. ç ç x + sin 2 x ÷ dx (1 + x ) è ø æ 1 - tan x ö (1 + cot x) ÷÷ dx [CBSE 2000] 95. ò 94. ò çç dx ( x + log sin x) è x + log cos x ø 92.

ò

2

tan x sec2 x

96.

ò (1 - tan 2x) dx

98.

ò ( x1/2 + x1/3)

100.

dx

ò

2x tan -1 x 2 (1 + x 4)

dx

dx

sin ( 2 tan -1 x)

97.

ò

99.

ò ( sin

[CBSE 2006] 101.

(1 + x 2) -1

[CBSE 2005]

x) 2 dx

( x 2 + 1)

ò ( x 4 + 1) dx

dx

[CBSE 2000]

[CBSE 2000]

[CBSE 2002] [CBSE 2004] [CBSE 2006C, ‘07]

Senior Secondary School Mathematics for Class 12 Pg-624

624

102.

Senior Secondary School Mathematics for Class 12

ò

(sin x + cos x) dx sin 2x

[CBSE 2009C, ‘10C]

ANSWERS (EXERCISE 13A)

( 2x + 9) 6 +C 12 1 4. 4x + 3 + C 2 1.

7. 10. 13. 16. 19. 22.

1 ( 2x - 1) e +C 2 - cos 3 x +C 3 1 - cot ( 2x + 5) + C 2 2 - (cos x) 3/2 + C 3 sin (log x) + C 1 ( x + log x) 3 + C 3

25. etan x + C 28.

- (7 - 3 x)5 +C 15 1 5. 3 - 4x + C 2

2.

8. 11. 14. 17. 20. 23.

1 (1 - 3x ) e +C 3

1 sin (5 + 6x) + C 6 1 sin 2 x + C 2 1 (sin - 1 x) 2 + C 2 - cot (log x) + C 1 (log x) 3 + C 3

26. - ecos

2

x

+C

sin 3 x 3 + sin x + C 29. e-1/x + C 12 4

31. tan -1( ex ) + C

32.

1 log |e2x - 2| + C 2

2 ( 3 x - 5) 3/2 + C 9 -1 6. +C 2x - 3 3.

( 2- 3x )

9. 12. 15. 18. 21.

-3 +C 3 log 3 1 sin 2 x + C 2 1 sin 4 x + C 4 1 - cos ( 2 tan -1 x) + C 2 log |log x| + C

24. 2sin 27.

x +C

sin 2( ax + b) +C 2a

æ1ö 30. - sin ç ÷ + C èxø 33.

1 [log (sin x)]2 + C 2

34. log |log (sin x)| + C 35. - cos ( x 2 + 1) + C 1 1 36. [log ( sec x + tan x)]2 + C 37. tan 2 x + C 38. (tan -1 x 2) 2 + C 2 4 1 2 40. log |sin -1 x| + C 41. ( 2 + log x) 3/2 + C 39. (sin -1 x 2) 2 + C 4 3 42. log |1 + tan x| + C 43. - log |1 + cos x| + C 44. - log|cos x - sin x| + C 45. (i) log |x + log ( sec x)| + C 1 46. 2 log|a 2 + b 2 sin 2 x| + C b

(ii) log|x + cos2 x| + C 1 47. 2 log|a 2 cos2 x + b 2 sin 2 x| + C ( b - a 2)

48. log| 3 cos x + 2 sin x| + C 1 50. log|x 2 + 2x - 3| + C 2

49. log|2x 2 + 3| + C 51. log|2x 2 - 5 x + 1| + C

Senior Secondary School Mathematics for Class 12 Pg-625

Methods of Integration

625

53. log|log tan x| + C 52. log| 3 x 3 - 2x 2 + 5 x + 1| + C -1 1 54. 56. 2 x 2 + 3 x - 2 + C 55. +C +C (1 + cos x) 2( x + sin x) 2 57. 2 x 2 - x - 1 + C 59. 61. 63. 65. 67. 69. 71. 73. 75. 77.

58.

2 [( x + a) 3/2 - ( x + b) 3/2] + C 3( a - b)

1 1 [(1 - 3 x) 3/2 + (5 - 3 x) 3/2] + C 60. tan -1 x 3 + C 18 3 1 1 62. tan -1 x 2 + C tan -1 x 4 + C 4 2 2 2 2 3 3/2 3 1/2 (1 + x ) - (1 + x ) + C 64. (1 + x) 3/2 - 2 1 + x + C 9 3 3 1 2 2 -1 2 66. ( x - 1)5/2 + ( x - 1) 3/2 + C sec x + C 2 5 3 4 2 1 2 3/2 5/2 3/2 68. ( x - 1) (1 + x) - (1 + x) +C +C 3 5 3 2 4 ( 3 x - 2)5/2 + ( 3 x - 2) 3/2 + C 70. tan (1 + log x) + C 45 27 1 2 72. ( x 2 + 4x + 3) 3/2 + C - cos x 3 + C 3 3 1 1 1 1 74. x - log |sin x - cos x| + C x + log |sin x - cos x| + C 2 2 2 2 1 1 76. log |sin x + cos x| + C x + log |sin x - cos x| + C 2 2 -1 1 1 79. sin 4 x 2 + C 78. ( x + log x) 3 + C +C (sin x + cos x) 3 8

1 cot( 2e- x + 5) + C 2 ù a -2 é 83. 2 ê log|a + b cos x| + +C ( a + b cos x) úû b ë

80. sin -1(tan x) + C

81.

2 (1 + x) 3/2 + C 3

86.

85.

3 1 sin ( ex ) + C 3

sec 3( x 2 + 3) +C 3 - log | 3 - 5 x| 84. +C 5

82.

87.

em tan m

-1

x

+C

89. 2sin ( e x ) + C 90. 2 ex - 1 - 2 tan -1 ex - 1 + C 1 92. tan ( 2 tan -1 x) + C 93. log |x + sin 2 x| + C 91. 2 log | x - 1| + C 2 94. log|x + log cos x| + C 95. log |x + log sin x| + C 1 1 2 97. - cos ( 2 tan -1 x) + C 96. - log (1 - tan x) + C 2 2 88. tan ( x ex ) + C

98. 2 x - 3 x1/ 3 + 6x1/6 - 6 log|1 + x1/6| + C 99. x(sin -1 x) 2 + 2 1 - x 2 sin -1 x - 2x + C

100.

1 (tan -1 x 2) 2 + C 2

Senior Secondary School Mathematics for Class 12 Pg-626

626

101.

Senior Secondary School Mathematics for Class 12

ì tan x - 1 ü 102. tan -1í ý î 2 tan x þ

1 1 öü ì 1 æ tan -1í çx - ÷ý+ C x 2 2 è øþ î

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 13A) 12. I = ò sin x 2 cos 2 x dx = 2 ò sin x cos x dx. Put sin x = t and cos x dx = dt.

1ö 1ö æ æ 22. I = ò ç 1 + ÷ ( x + log x ) 2 dx. Put ( x + log x ) = t and ç 1 + ÷ dx = dt xø xø è è 25. Put tan x = t. 26. Put cos 2 x = t. 24. Put x = t. 3 1 -1 28. cos 3 x = cos x + cos 3 x. 29. Put = t. 4 4 x

27. Put sin ( ax + b ) = t. 32. Put ( e 2 x - 2 ) = t.

33. Put log (sin x ) = t.

35. Put ( x 2 + 1) = t. 36. Put log ( sec x + tan x ) = t.

37. Put tan x = t.

38. Put tan -1 x 2 = t.

39. Put sin -1 x 2 = t.

40. Put sin -1 x = t.

41. Put ( 2 + log x ) = t.

44. I = ò

(cos x + sin x ) dx. Put (cos x - sin x ) = t. (cos x - sin x )

45. to 53. In each of these questions, put the denominator equal to t. 60. Put x 3 = t.

2 2 t(t 2 - 1) dt = ò (t 5/2 - t 1/2 )dt. 3 3ò t

63. Put ( 1 + x 3 ) = t 2 . Then, I =

64. Put ( 1 + x ) = t 2 . 65. Multiply num. and denom. by x and put x 2 = t. 73. I = ò

(sin x - cos x ) + (sin x + cos x ) dx. 2(sin x - cos x )

74. I = ò

(sin x - cos x ) - (sin x + cos x ) - cos x dx = ò dx. (sin x - cos x ) 2(sin x - cos x ) (cos 2 x - sin 2 x )

76. I = ò

(sin x + cos x ) 2

77. I = ò

(sin x + cos x ) 2

(cos x - sin x )

dx = ò

(cos x - sin x ) dx. (sin x + cos x )

dx. Put (sin x + cos x ) = t.

1ö æ 78. I = ò ç 1 + ÷ ( x + log x ) 2 dx. Put ( x + log x ) = t. xø è 80. Put tan x = t.

81. Put ( 2 e - x + 5 ) = t.

79. Put sin x 2 = t. 82. Put sec ( x 2 + 3 ) = t.

( t - a) 1 and - sin x dx = dt. b b 2 1 æt - aö 2 ì1 a ü \ I = - ×ò 2ç ÷ dt = - 2 × ò í - 2 ý dt. b t è b ø ît t þ b

83. Put ( a + b cos x ) = t. Then, cos x =

3

86. Put e x = t.

88. Put ( xe x ) = t.

89. Put e

x

= t.

Senior Secondary School Mathematics for Class 12 Pg-627

Methods of Integration

627

2t dt. Then, (t 2 + 1) æ 2t 2 1 ö÷ I=ò dt = 2 ò ç 1 dt. 2 ç 1 + t 2 ÷ø (1 + t ) è dx 1 91. I = ò × Put x - 1 = t and dx = 2 dt. x ( x - 1) x 90. Put ( e x - 1) = t 2 and dx =

92. Put 2 tan -1 x = t. 97. Put tan -1 x = t and 98. I = ò

dx 1/3

x

( 1 + x1/6 )

96. Put ( 1 - tan 2 x ) = t 1 ( 1 + x2 )

dx = dt.

× Now put x = t 6 and dx = 6 t 5 dt.

t3 1 ö æ dt = 6 × ò ç t 2 - t + 1 ÷ dt. 1+ t 1+ t ø è 1ö æ 101. Divide num. and denom. by x 2 . Put ç x - ÷ = t. xø è dt × \ I=ò 2 {t + ( 2 )2 } Then, I = 6 ò

OBJECTIVE QUESTIONS I Mark (3) against the correct answer in each of the following: 1.

ò ( 2x + 3) (a)

2.

5

dx = ?

( 2x + 3) 6 ( 2x + 3) 4 ( 2x + 3) 6 + C (b) + C (c) + C (d) none of these 6 8 12

ò ( 3 - 5 x)

7

dx = ?

(a) -5( 3 - 5 x) 6 + C -5( 3 - 5 x) 8 +C 8 1 3. ò dx = ? ( 2 - 3 x) 4 1 (a) +C 15( 2 - 3 x)5 1 (c) +C 9( 2 - 3 x) 3 (c)

4.

ò

(b)

( 3 - 5 x) 8 +C -40

(d) none of these

(b)

1 -12( 2 - 3 x) 3

+C

(d) none of these

ax + b dx = ? 2( ax + b) 3/2 +C 3a 1 (c) +C 2 ax + b

(a)

(b)

3( ax + b) 3/2 +C 2a

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-628

628

5.

Senior Secondary School Mathematics for Class 12

ò sec (7 - 4x) dx = ? 2

-1 tan (7 - 4x) + C 4 (d) -4 tan (7 - 4x) + C

1 (a) tan (7 - 4x) + C 4 (c) 4 tan (7 - 4x) + C 6.

ò cos 3x dx = ? (a) -

7.

1 1 sin 3 x + C (b) sin 3 x + C 3 3

(c) 3 sin 3 x + C

(d) -3 sin 3 x + C

(5 - 3 x )

òe

dx = ? (5 - 3 x )

(a) -3 e 8.

(b)

( 3x + 4)

ò2

+C

(b)

1 (5 - 3x ) e +C 3

(5 - 3 x )

(c)

e

-3

+C

(d) none of these

dx = ?

(a)

3 ( 3x + 4) ×2 +C (log 2)

(c)

2 +C 2(log 3)

( 3x + 4)

(b)

2 3(log 2) + C

( 3x + 4)

9.

10.

x dx = ? 2 x (a) tan - x + C 2 x (c) 2 tan + x + C 2

ò tan

ò

2

ò

x x -1 x + C (b) -2 2 cos + C (c) cos + C 2 2 2 2

(d)

-1 x cos + C 2 2

1 + sin x dx = ? æp xö (a) - 2 sin ç - ÷ + C è 4 2ø æp xö (c) -2 2 sin ç - ÷ + C è 4 2ø

12.

x + x+C 2 x (d) 2 tan - x + C 2 (b) tan

1 - cos x dx = ? (a) - 2 cos

11.

(d) none of these

ò sin

3

æp xö (b) 2 sin ç - ÷ + C è 4 2ø (d) none of these

x dx = ?

3 cos 3 x cos x + +C 4 12 3 cos 3 x (c) - cos x +C 4 12 (a) -

(b)

3 cos 3 x cos x + +C 4 12

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-629

Methods of Integration

log x dx = ? x 1 (a) (log x) 2 + C 2 2 (c) 2 + C x sec2(log x) 14. ò dx = ? x (a) log (tan x) + C (c) tan (tan x) + C 1 15. ò dx = ? x(log x) 13.

ò

(a) log |x| + C 16.

òe

e

x

x

(b)

-2 x2

etan

-1

x

+C

(d) log |log x| + C

(b)

1 x3 e +C 3

(c)

1 x3 e +C 6

(d) none of these

1 (b) e 2

x

x

+C

(c) 2e

+C

x

(c) ex tan -1 x + C (d) none of these

(d) none of these

x

ò (1 + x 2) dx = ? -1

etan x +C x sin x 19. ò dx = ? x

20.

(c) (log x) 2 + C

+C

dx = ?

(a) e 18.

(b) - log (tan x) + C (d) - tan (log x) + C

x dx = ? 3

ò

1 (b) - (log x) 2 + C 2 -2 (d) 2 + C x

x3 2

(a) ex + C 17.

629

ò(

-1

(a)

(b) etan

(a) 2cos x + C

(b) -2cos x + C (c) -

+C

sin x ) cos x dx = ?

3 2 (cos x) 2 + C 3 3 2 (c) (sin x) 2 + C 3 1 21. ò dx = ? (1 + x 2) tan -1 x 1 (a) log |tan -1 x| + C 2 1 (c) +C 2 tan -1 x

(a)

cos x +C 2

3 3 (cos x) 2 + C 2 3 3 (d) (sin x) 2 + C 2

(b)

(b) 2 tan -1 x + C (d) none of these

(d)

cos x +C 2

Senior Secondary School Mathematics for Class 12 Pg-630

630

22.

Senior Secondary School Mathematics for Class 12

cot x

ò log(sin x) dx = ? (a) log |cot x| + C (c) log |log sin x| + C

23.

1

ò x cos2(1 + log x) dx = ? (a) tan (1 + log x) + C (c) sec (1 + log x) + C

24.

ò

x 2 tan -1 x 3 (1 + x 6)

1 (tan -1 x 3) 2 + C 3 1 (c) (tan -1 x 3) 2 + C 6

ò sec

5

(c) 5 log |cos x| + C

ò cosec

3

28.

29.

ò

ò

(d) none of these

1 (b) tan5 x + C 5 (d) none of these

( 2x + 1) cot ( 2x + 1) dx = ?

1 (a) cosec4( 2x + 1) + C 4 1 (c) - cosec 3( 2x + 1) + C 6 27.

(b) log |tan -1 x 3| + C

x tan x dx = ?

(a) 5 tan5 x + C

26.

(b) cot (1 + log x) + C (d) none of these

dx = ?

(a)

25.

(b) log |cot x cosec x| + C (d) none of these

tan(sin -1 x) 1 - x2

(b) -

1 cosec 3( 2x + 1) + C 3

1 (d) cosec ( 2x + 1) cot ( 2x + 1) + C 2

dx = ?

(a) log|sec (sin -1 x)| + C

(b) log|cos(sin -1 x)| + C

(c) tan (sin1 x) + C

(d) none of these

tan (log x) dx = ? x (a) x tan (log x) + C

(b) log|tan x| + C

(c) log|cos (log x)| + C

(d) - log|cos (log x)| + C

òe

x

cot ( ex ) dx = ?

(a) cot ( ex ) + C

(b) log|sin ex | + C

(c) log|cosec ex | + C

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-631

Methods of Integration

30.

ò

ex 1 + ex

dx = ?

(a) 2 1 + ex + C 1

(c) 31.

ò

1 + ex

x 1 - x2

(d) none of these

(b) sin -1 x + C

(c) 1 - x 2 + C

(b) cot ( xex ) + C

(c) ex x tan x + C (d) none of these

35.

(d) - 1 - x 2 + C

ex (1 + x)

ò cos2( xex ) dx = ? dx

ò ( ex + e- x ) = ? (a) cot -1( ex ) + C (c) log|ex + 1| + C

34.

1 1 + ex + C 2

dx = ?

(a) tan ( xex ) + C 33.

(b)

+C

(a) sin -1 x + C 32.

631

(b) tan -1( ex ) + C (d) none of these

2x

ò 1 - 4x dx = ?

(a) sin -1( 2x ) + C

(b) (log e2) sin -1( 2x ) + C

(c) (log e2) cos-1( 2x ) + C

(d) (log 2 e) sin -1( 2x ) + C

dx

ò ( ex - 1) = ? (a) log|ex - 1| + C

(b) log|1 - e- x| + C

(c) log|ex - 1| + C

(d) none of these

1 36. ò dx = ? ( x + x) (a) log|1 + x| + C 1 (c) tan -1 x + C x dx 37. ò =? (1 + sin x) (a) tan x + sec x + C 1 x (c) tan + C 2 2 sin x 38. ò dx = ? (1 + sin x) (a) x + tan x - sec x + C (c) x - tan x + sec x + C

(b) 2 log|1 + x| + C (d) none of these

(b) tan x - sec x + C (d) none of these

(b) x - tan x - sec x + C (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-632

632

39.

Senior Secondary School Mathematics for Class 12

sin x

ò (1 - sin x) dx = ? (a) - x + sec x - tan x + C (c) - log|1 - sin x| + C

40.

dx

ò (1 + cos x) = ? 1 x (a) tan + C 2 2 x (c) tan + C 2

41.

(b) x + cos x - sin x + C (d) none of these

(b) - cot

x +C 2

(d) none of these

dx

ò (1 - cos x) = ? (a)

1 +C ( x - sin x)

(b) log |x - sin x| + C

x (c) log ½tan ½+ C 2½ ½

(d) - cot

x +C 2

ì æxö ü ïï 1 - tan ç 2 ÷ ïï è ø dx = ? 42. ò í ý ï1 + tan æç x ö÷ ï ïî è 2 ø ïþ

43.

ò

x (a) 2 log ½sec ½+ C 2½ ½

x (b) 2 log ½cos ½+ C 2½ ½

½ æ p x ö½ (c) 2 log ½sec ç - ÷½+ C ½ è 4 2 ø½

½ æ p x ö½ (d) 2 log ½cos ç - ÷½+ C ½ è 4 2 ø½

ex dx = ? (a) ex + C

44.

(b) 2 ex + C

1 x e +C 2

(d) none of these

cos x

ò (1 + cos x) dx = ? x +C 2 x (c) x - tan + C 2 (a) x + tan

45.

(c)

(b) - x + tan

x +C 2

(d) none of these

ò sec x cosec x dx = ? 2

2

(a) tan x - cot x + C (c) - tan x + cot x + C

(b) tan x + cot x + C (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-633

Methods of Integration

46.

(1 - cos 2x)

ò (1 + cos 2x) dx = ?

(a) tan x + x + C (1 + cos x) 47. ò dx = ? (1 - cos x) (a) -2 cot (c) 2 cot 48.

633

(b) tan x - x + C

x -x+C 2

(c) - tan x + x + C (d) none of these

(b) -2 cot

x + x+C 2

x + x+C 2

(d) none of these

1

ò sin 2x cos2x dx = ?

(a) tan x + cot x + C (c) - tan x + cot x + C cos 2x 49. ò dx = ? cos2 x sin 2 x

(b) tan x - cot x + C (d) none of these

(a) cot x + tan x + C (c) cot x - tan x + C (cos 2x - cos 2a ) 50. ò dx = ? (cos x - cos a )

(b) - cot x + tan x + C (d) - cot x - tan x + C

(a) sin x + x cos a + C (c) 2 sin x + 2x cos a + C ì 1 - cos 2x ü 51. ò tan -1í ý dx = ? î 1 + cos 2x þ

(b) 2sin x + x cos a + C (d) none of these

(a) 2x 2 + C 52.

ò tan

-1

(a) 53.

(b)

x2 +C 2

2 (1 + x 2)

+C

(d) none of these

+C

(d) none of these

( sec x + tan x) dx = ?

px x 2 + +C 4 4

(b)

px x 2 +C 4 4

(c)

1 (1 + x 2)

(1 + sin x)

ò (1 - sin x) dx = ? (a) 2 tan x + x - 2sec x + C (c) 2 tan x - x - 2sec x + C

54.

(c)

(b) 2 tan x - x + 2sec x + C (d) none of these

x4

ò (1 + x 2) dx = ? x3 + x + tan -1 x + C 3 x3 (c) - x + tan -1 x + C 3 (a)

(b)

-x 3 + x - tan -1 x + C 3

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-634

634

55.

Senior Secondary School Mathematics for Class 12

sin ( x - a )

ò sin ( x + a) dx = ?

(a) x cos 2a - sin 2a × log|sin( x + a )| + C (b) x cos 2a + sin 2a × log|sin( x + a )| + C (c) x cos 2a + sin a × log|sin( x + a )| + C (d) none of these 1 56. ò dx = ? ( x + 3 - x + 2) 3 3 3 3 2 2 2 2 (a) ( x + 3) 2 - ( x + 2) 2 + C (b) ( x + 3) 2 + ( x + 2) 2 + C 3 3 3 3 3 3 3 3 (d) none of these (c) ( x + 3) 2 - ( x + 2) 2 + C 2 2 (1 + tan x) 57. ò dx = ? (1 - tan x) (a) - log|cos x - sin x| + C (b) log|cos x - sin x| + C (c) log|cos x + sin x| + C (d) none of these 58.

3x2

ò (1 + x 6 ) dx = ? (a) sin -1 x 3 + C -1 3

(c) tan x + C 59.

60.

ò

(d) cot -1 x 3 + C

dx

=? x x6 - 1 1 (a) sec-1 x 3 + C 3 1 (c) cot -1 x 3 + C 3

ò {( 2x + 1)

(b)

1 cosec-1 x 3 + C 3

(d) none of these

x 2 + x + 1} dx = ?

3 3 2 ( x + x + 1) 2 + C 2 3 3 (c) ( 2x + 1) 2 + C 2 dx 61. ò =? { 2x + 3 + 2x - 3} 3 3 1 1 (a) ( 2x + 3) 2 + ( 2x - 3) 2 + C 18 18 3 3 1 1 2 (c) ( 2x + 3) - ( 2x - 3) 2 + C 12 12

(a)

62.

(b) cos-1 x 3 + C

ò tan x dx = ?

(a) log |cos x| + C (c) log |sin x| + C

(b)

3 2 2 ( x + x + 1) 2 + C 3

(d) none of these

(b)

3 3 1 1 ( 2x + 3) 2 - ( 2x - 3) 2 + C 18 18

(d) none of these

(b) - log |cos x| + C (d) - log |sin x| + C

Senior Secondary School Mathematics for Class 12 Pg-635

Methods of Integration

63.

ò sec x dx = ? (a) log |sec x - tan x| + C (c) log |sec x + tan x| + C

64.

(1 + sin x)

x x + 2 log ½cos ½ + C 2 2½ ½ x x (c) tan - 2 log ½cos ½ + C 2 2½ ½

67.

68.

x x + 2 log ½cos ½ + C 2 2½ ½

(d) none of these

tan x

ò

(a) tan -1(cos x) + C

(b) - tan -1(cos x) + C

(c) cot -1(cos x) + C

(d) none of these

1+ x dx = ? 1-x (a) sin -1 x + 1 - x 2 + C

(b) sin -1 x + (1 + x 2) + C

(c) sin -1 x - 1 - x 2 + C

(d) none of these

1

ò x2 e

-1/x

dx = ? (b) - e-1/x + C

(c)

e-1/x +C x

x3

ò (1 + x 8) dx = ? (a) tan -1 x 4 + C 1 (c) tan -1 x 4 + C 4

70.

(b) - tan

ò ( sec x + cos x) dx = ?

(a) e-1/x + C 69.

(b) - log |cosec x - cot x| + C (d) none of these

ò (1 + cos x) dx = ? (a) tan

66.

(b) - log |sec x + tan x| + C (d) none of these

ò cosec x dx = ? (a) log |cosec x - cot x| + C (c) log |cosec x + cot x| + C

65.

635

ò

(b) 4 tan -1 x 4 + C (d) none of these

( x + 1)( x + log x) 2 dx = ? x 1 ( x + log x) 3 + C 3 x 3 x2 (c) + + x+C 3 2 (a)

(b)

x2 +x+C 2

(d) none of these

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-636

636

71.

Senior Secondary School Mathematics for Class 12

ò

2x tan -1 x 2 (1 + x 4)

dx = ?

(a) (tan -1 x 2) 2 + C 1 (c) (tan -1 x 2) 2 + C 2 dx 72. ò =? ( 2 - 3 x)

(b) 2 tan -1 x 2 + C (d) none of these

1 log |2 - 3 x| + C 3 (d) none of these

(a) -3 log |2 - 3 x| + C

(b) -

(c) - log |2 - 3x| + C 73.

òx

x 2 - 1 dx = ?

3 2 2 ( x - 1) 2 + C 3 1 (c) +C 2 x -1

(a)

74.

(5 - 3 x )

ò3

(b)

(d) none of these

dx = ?

(5 - 3 x )

(a)

(5 - 3 x )

76.

òe

tan x

log 3 + C

(d) none of these

(b) etan x × tan x + C

(c) etan x + C

(d) none of these

òe

cos2 x

sin 2x dx = ? 2

x

(b) - ecos

+C

2

x

+C

(c) esin

2

x

+C

(d) none of these

ò x sin x cos x dx = ? 3 2

ò

x

2

1 (b) sin 4 x 2 + C 8

1 (c) sin 4 x 2 + C 2

cos ( e x

x

)

(a) sin ( e

x

)+C

(b)

x

(d) none of these

e

(c) 2sin ( e 79.

3 +C (log 3)

(a) etan x + tan x + C

1 (a) sin 4 x 2 + C 4 78.

(b)

sec2 x dx = ?

(a) ecos 77.

( 4- 3x )

-3 +C 3(log 3)

(c) -3 75.

3 1 2 ( x - 1) 2 + C 3

òx

2

(d) none of these

dx = ?

)+C

1 sin ( e 2

x

)+C

3

sin x dx = ?

(a) cos x 3 + C

(b) - cos x 3 + C

(c) -

1 cos x 3 + C (d) none of these 3

Senior Secondary School Mathematics for Class 12 Pg-637

Methods of Integration

80.

( x + 1) ex

ò cos2( xex ) dx = ? (a) tan ( xex ) + C (c) cot ( xex ) + C

81.

ò

1 x x4 - 1

òx

(b) - tan ( xex ) + C (d) none of these

dx = ?

(a) sec-1 x 2 + C 82.

1 (b) sec-1 x 2 + C (c) cosec-1 x 2 + C (d) none of these 2

x - 1 dx = ?

3 2 ( x - 1) 2 + C 3 5 3 2 3 (c) ( x - 1) 2 + ( x - 1) 2 + C 5 2

(a)

83.

òx

3 1 2 ( x - 1) 2 + C 3 1 (c) +C 2 x -1 dx 84. ò =? (1 + x )

ò

5 2 (b) ( x - 1) 2 + C 5

(d) none of these

x 2 - x dx = ?

(a)

85.

637

(b)

3 2 2 ( x - 1) 2 + C 3

(d) none of these

(a) x - log |1 + x| + C

(b) x + log |1 + x| + C

(c) 2 x - 2 log |1 + x| + C

(d) none of these

ex - 1 dx

3 3 x ( e - 1) 2 + C 2 3 2 x (c) ( e - 1) 2 + C 3 sin x 86. ò dx = ? (sin x - cos x)

(a)

1 1 (b) ( ex - 1) 2 + C 2

(d) none of these

1 1 (a) x - log |sin x - cos x| + C 2 2 (c) log |sin x - cos x| + C dx 87. ò =? (1 - tan x)

1 1 (b) x + log |sin x - cos x| + C 2 2 (d) none of these

1 (a) log |sin x - cos x| + C 2 1 1 (c) x - log |sin x - cos x| + C 2 2

1 1 (b) x + log |sin x - cos x| + C 2 2 (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-638

638

88.

Senior Secondary School Mathematics for Class 12

dx

ò (1 - cot x) = ? (a) log |sin x - cos x| + C 1 1 x - log |sin x - cos x| + C 2 2

(c) 89.

90.

91.

ò

sec2 x 1 - tan 2 x

dx = ?

(a) sin -1(tan x) + C

(b) cos-1(sin x) + C

(c) tan -1(cos x) + C

(d) tan -1(sin x) + C

( x 2 + 1)

ò ( x 4 + 1) dx = ? 1 1ö æ tan -1 ç x - ÷ + C xø 2 è

(b)

(c)

1 1 öü ì 1 æ tan -1í çx - ÷ý+ C x øþ 2 î 2 è

(d) none of these

sin 6 x

ò cos8 x dx = ? 1 (b) sec7 x + C 7 (d) none of these

ò sec x tan x dx = ? 5

1 (a) tan5 x + C 5 (c) 5 log |cos x| + C 93.

1 1 öü ìæ cot -1íç x - ÷ ý + C x øþ 2 è î

(a)

1 (a) tan 7 x + C 7 (c) log |cos6 x| + C 92.

1 (b) log |sin x - cos x| + C 2 1 1 (d) x + log |sin x - cos x| + C 2 2

1 (b) sec5 x + C 5 (d) none of these

ò tan x dx = ? 5

1 (a) tan 6 x + C 6 1 1 (b) tan 4 x + tan 2 x + log |sec x| + C 4 2 1 1 4 (c) tan x - tan 2 x + log |sec x| + C 4 2 (d) none of these

94.

ò sin

3

x cos 3 x dx = ?

1 1 (a) - cos4 x + cos6 x + C 4 6 1 1 4 (c) sin x + cos6 x + C 4 6

1 1 (b) sin 4 x - sin 6 x + C 4 6 (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-639

Methods of Integration

95.

639

ò sec x tan x dx = ? 4

1 1 1 1 (a) sec2 x + sec4 x + C (b) tan 2 x + tan 4 x + C 2 4 2 4 1 (c) sec x + log |sec x + tan x| + C (d) none of these 2 log tan x 96. ò dx = ? sin x cos x 1 (a) log {log (tan x)} + C (b) (log tan x) 2 + C 2 (c) log (sin x cos x) + C (d) none of these 97.

ò sin

3

( 2x + 1) dx = ?

1 (a) sin 4( 2x + 1) + C 8 1 1 (b) cos ( 2x + 1) + cos 3( 2x + 1) + C 2 3 1 1 (c) - cos ( 2x + 1) + cos 3( 2x + 1) + C 2 6 (d) none of these tan x 98. ò dx = ? sin x cos x (a) 2 tan x + C (b) 2 cot x + C (c) 2 sec x + C

(d) none of these

(cos + sin x) 99. ò dx = ? (1 - sin 2x) 1 +C (cos x - sin x) (d) none of these

(a) log |sin x - cos x| + C

(b)

(c) log |cos x + sin x| + C 100.

ò

ex - 1 dx = ? (a)

3 2 x ( e - 1) 2 + C 3

(c) 2 ex - 1 - 2 tan -1 ex - 1 + C dx 101. ò =? 3 sin x cos x (a) 2 tan x + C

(b) 2 cot x + C

1 ex (b) × +C 2 ex - 1 (d) none of these

(c) -2 tan x + C (d)

-2 +C tan x

ANSWERS (OBJECTIVE QUESTIONS I)

1. (c)

2. (b)

3. (c)

4. (a)

5. (b)

6. (b)

7. (c)

8 (b)

9. (d) 10. (b)

11. (c) 12. (a) 13. (a) 14. (c) 15. (d) 16. (b) 17. (c) 18. (b) 19. (b) 20. (c)

Senior Secondary School Mathematics for Class 12 Pg-640

640

Senior Secondary School Mathematics for Class 12

21. (b) 22. (c) 23. (a) 24. (c) 25. (b) 26. (c) 27. (a) 28. (d) 29. (b) 30. (a) 31. (d) 32. (a) 33. (b) 34. (d) 35. (b) 36. (b) 37. (b) 38. (c) 39. (a) 40. (c) 41. (d) 42. (d) 43. (b) 44. (c) 45. (a) 46. (b) 47. (a) 48. (b) 49. (d) 50. (c) 51. (b) 52. (a) 53. (b) 54. (c) 55. (a) 56. (b) 57. (a) 58. (c) 59. (a) 60. (b) 61. (b) 62. (b) 63. (c) 64. (a) 65. (c) 66. (b) 67. (c) 68. (a) 69. (c) 70. (a) 71. (c) 72. (b) 73. (b) 74. (a) 75. (c) 76. (b) 77. (b) 78. (c) 79. (c) 80. (a) 81. (b) 82. (c) 83. (a) 84. (c) 85. (d) 86. (b) 87. (c) 88. (d) 89. (a) 90. (c) 91. (a) 92. (b) 93. (c) 94. (b) 95. (b) 96. (b) 97. (c) 98. (a) 99. (b) 100. (c) 101. (d) HINTS TO THE GIVEN OBJECTIVE QUESTIONS I 1. Put ( 2 x + 3 ) = t and 2dx = dt. 2. Put ( 3 - 5 x ) = t and -5dx = dt. 3. Put ( 2 - 3 x ) = t and -3dx = dt. 4. Put ( ax + b ) = t and a dx = dt. 3

\

I=

1 1 t 2 t dt = × 3 + C . ò a a ( 2)

5. Put (7 - 4 x ) = t and -4dx = dt. 1 1 \ I = - ò sec2t dt = - tan t + C . 4 4 6. Put 3x = t and 3dx = dt. 7. Put (5 - 3 x ) = t and -3dx = dt. 8. Put ( 3 x + 4 ) = t and 3dx = dt. 1 1 2t I = ò 2 t dt = × + C. \ 3 3 log 2 x x æ ö 9. I = ò ç sec2 - 1 ÷ dx = ò sec2 dx - ò dx. 2 2 è ø x 2 = 2 ò sec t dt - x + C , where = t 2 x = 2 tan t - x + C = 2 tan + C . 2 10. I = ò 2 sin 2 ( x 2 ) dx = 2 ò sin ( x 2 ) dx. Put \

x = t and dx = 2 dt. 2 I = 2 2 ò sin t dt = -2 2 cos t + C .

æp ö æp xö 11. ( 1 + sin x ) = 1 + cos ç - x ÷ = 2 cos 2 ç - ÷ × è2 ø è 4 2ø \

-1 æp xö æp xö I = 2 ò cos ç - ÷ dx. Put ç - ÷ = t and dx = dt. 2 è 4 2ø è 4 2ø I = -2 2 ò cos t dt = - 2 2 sin t + C .

Senior Secondary School Mathematics for Class 12 Pg-641

Methods of Integration 12. sin 3 x = 3 sin x - 4 sin 3 x Þ sin 3 x = \

3

1 ( 3 sin x - sin 3 x ). 4

1

ò sin 3 x dx = 4 ò sin x dx - 4 ò sin 3 x dx =

cos 3 x 3 3 1 ( - cos 3 x ) ( - cos x ) +C= - cos x + C . 4 4 3 12 4

1 dx = dt. x 1 14. Put log x = t and dx = dt. x 1 15. Put log x = t and dx = dt. x 13. Put log x = t and

16. Put x 3 = t and 3 x 2 dx = dt. 17. Put x = t and

1 -1 2 x dx = dt Þ 2

18. Put tan -1 x = t and

1 ( 1 + x2 )

1 dx = 2 dt. x

dx = dt.

1 dx = 2 dt. x I = 2 ò sin dt = -2 cos t + C = -2 cos x + C .

19. Put x = t and \

20. Put sin x = t and cos x dx = dt. 3

\

I = ò t dt =

21. Put tan -1 x = t and \

I=ò

3 t 2 2 + C = (sin x ) 2 + C . ( 3 2) 3

1 ( 1 + x2 )

dx = dt.

1 dt = 2 t + C . t

22. Put log (sin x ) = t and cot x dx = dt. 1 23. Put ( 1 + log x ) = t and dx = dt. x 1 \ I=ò dt = ò sec2t dt = tan t + C . cos 2 t 24. Put tan -1 x 3 = t and \

3x 2

dx = dt. ( 1 + x6) 1 1 1 I = ò t dt = t 2 + C = (tan -1 x 3 ) 2 + C . 3 6 6

25. Put sec x = t and sec x tan x dx = dt. \

I = ò sec4 x ( sec x tan x ) dx = ò t 4 dt =

26. Put ( 2 x + 1) = t and dx = \

I=

1 dt. 2

1 cosec2t( cosec t cot t ) dt 2ò

t5 + C. 5

641

Senior Secondary School Mathematics for Class 12 Pg-642

642

Senior Secondary School Mathematics for Class 12

1 2 u du, where cosec t = u 2ò 1 1 -1 = - u3 + C = - cosec3t + C = cosec3 ( 2 x + 1) + C . 6 6 6 1 27. Put sin -1 x = t and dx = dt. 1 - x2 =-

\

I = ò tan t dt = log|sec t| + C . 1 dx = dt. x I = ò tan t dt = - log|cos t| + C

28. Put log x = t and \

= - log|cos (log x )| + C .

29. Put e x = t and e x dx = dt. I = ò cot t dt = log|sin t| + C .

\

30. Put ( 1 + e x ) = t and e x dx = dt. 1 \ I=ò dt = 2 t + C . t 31. Put ( 1 - x 2 ) = t and -2x dx = dt. \

I=-

t 1 1 1 -1 1 dt = - ò t 2 dt = - × 1 + C = - t + C . 2ò t 2 2 ( 2)

32. Put xe x = t and ( xe x + e x ) dx = dt. 1 \ I=ò dt = ò sec2t dt = tan t + C . cos 2 t dx

33. I = ò

=ò

ex

dx. 1ö æ x ( 1 + e2x) çe + x ÷ è e ø 1 =ò dt , where e x = t and e x dx = dt. (1 + t2 )

= tan -1 (t ) + C = tan -1 ( e x ) + C . 34. Put 2 x = t and 2 x(log 2 ) dx + dt. 1 1 1 \ I=ò × dt = sin -1 (t ) + C (log 2 ) 1 - t 2 (log e 2 ) = (log 2 e ) sin -1 ( 2 x ) + C . 35. I = ò

dx x

x

-x

e (e (1 - e )

=ò

e -x ( 1 - e -x )

dx.

Put ( 1 - e - x ) = t and e - xdx = dt \ 36. I = ò \

1 I = ò dt = log|t| + C + log|1 - e - x| + C . t dx 1 × Put ( 1 + x ) = t and dx = dt. x( 1 + x ) 2 x 1 I = 2 × ò dt = 2 log|t| + C . t

Senior Secondary School Mathematics for Class 12 Pg-643

Methods of Integration ì ( 1 - sin x ) ü ( 1 - sin x ) 1 37. I = ò í ´ dx. ý dx = ò + x x ( 1 sin ) ( 1 sin ) ( 1 - sin 2 x ) î þ =ò

( 1 - sin x ) 2

cos x

æ 1 sin x ö÷ dx = ò ç dx ç cos 2 x cos 2 x ÷ è ø

= ò ( sec2 x - sec x tan x ) dx = tan x - sec x + C . ì sin x ( 1 - sin x ) ü sin x( 1 - sin x ) 38. I = ò í ´ dx. ý dx = ò ( 1 sin ) ( 1 sin ) + x x cos 2 x î þ æ sin x sin 2 x ö ÷ dx = ( sec x tan x - tan 2 x ) dx =òç ò ç cos 2 x cos 2 x ÷ è ø = ò sec x tan x dx - ò ( sec2 x - 1) dx = sec x - tan x + x + C . æ sin x + sin 2 x ö ì sin x ( 1 + sin x ) ü ÷ dx. 39. I = ò í ´ ý dx = ò çç ÷ cos 2 x î ( 1 - sin x ) ( 1 + sin x ) þ è ø ìï sin x sin 2 x üï dx = ò ( sec x tan x + tan 2 x ) dx = òí + 2 2 ý îï cos x cos x þï = ò ( sec x tan x + sec2 x - 1) dx = sec x + tan x - x + C . 40. I = ò

dx 2 cos 2 ( x 2 )

=

1 æxö sec2 ç ÷ dx. 2ò è2ø

= ò sec2t dt , where

x = t and dx = 2 dt 2

x + C. 2 dx 1 æxö = ò cosec2 ç ÷ dx. è2ø 2 sin 2 ( x 2 ) 2

= tan t + C = tan 41. I = ò

x = t and dx = 2 dt 2 x = - cot t + C = - cot ( 2 ) + C . 1 p x æp xö 42. I = ò tan ç - ÷ dx. Put - = t and - dx = dt. 2 4 2 è 4 2ø = ò cosec2t dt , where

\

½ æ p x ö½ I = -2 ò tan t dt = 2 log|cos t| + C = 2 log½cos ç - ÷½ + C. ½ è 4 2 ø½

43. Put e x = t and e xdx = dt Þ dx =

1 dt. t 1

\

I=ò

-1 t t 2 1 dt = ò dt = ò t 2 dt = 1 + C = 2 t + C = 2 e x + C . t ( t 2)

ì ( 1 + cos x ) - 1 ü ì ü 1 44. I = ò í ý = ò í1 ý dx. ( cos ) ( cos ) x x 1 1 + + î þ î þ dx dx x 1 = ò dx - ò =x-ò = x - ò sec2 dx ( 1 + cos x ) 2 2 2 cos 2 ( x 2 )

643

Senior Secondary School Mathematics for Class 12 Pg-644

644

Senior Secondary School Mathematics for Class 12 1 x ´ 2 ´ ò sec2t dt , where = t 2 2 x = x - tan t + C = x - tan + C . 2 =x-

45. I = ò sec2 x ( 1 + cot 2 x ) dx = ò sec2 x dx + = tan x +

ò sec2 x cot 2 x dx.

cos 2 x

1

ò cos2 x × sin 2 x dx = tan x + ò cosec2 x dx

= tan x - cot x + C . 2 sin 2 x 46. I = ò dx = ò tan 2 x dx = ò ( sec2 x - 1) dx = tan x - x + C . 2 cos 2 x 2 cos 2 ( x 2 )

x x ö æ dx = ò cot 2 dx = ò ç cosec2 - 1 ÷ dx. 2 2 è ø 2 sin 2 ( x 2 ) 2 x 2 = ò cosec dx - ò dx = 2 ò cosec t dt - x + C 2 x = -2 cot t - x + C = -2 cot - x + C . 2

47. I =

48. I = ò =ò 49. I = ò =ò

(sin 2 x + cos 2 x ) sin x cos x

dx = ò

1

1

2

cos 2 x

2

dx +

dx = ò

2

cos x sin x dx sin 2 x

-ò

2

2

sin x cos x

dx +

cos 2 x

ò sin 2 x cos2 x dx.

ò sin 2 x dx = ò sec2 x dx + ò cosec2 x dx = tan x - cot x + C.

(cos 2 x - sin 2 x ) 2

sin 2 x

dx cos 2 x

cos 2 x 2

2

cos x sin x

dx - ò

sin 2 x cos 2 x sin 2 x

dx.

= ò cosec2 x dx - ò sec2 x dx

= - cot x - tan x + C . 50. I = ò

( 2 cos 2 x - 1) - ( 2 cos 2 a - 1) (cos 2 x - cos 2 a ) dx. dx = 2 ò (cos x - cos a ) (cos x - cos a )

= 2 ò (cos x + cos a ) dx = ò 2 cos x dx + 2 cos a × ò dx = 2 sin x + 2 x cos a + C .

51. I = ò tan -1 = ò x dx =

2 sin 2 x 2 cos 2 x

dx = ò tan -1 (tan x ) dx.

x2 + C. 2

ì 1 + sin x ü æ 1 sin x ö ÷ = tan -1 í 52. tan -1 ( sec x + tan x ) = tan -1 çç + ý× ÷ è cos x cos x ø î cos x þ ü ì ì æp æp xö öü 2 sin 2 ç + ÷ ï ïï ï 1 - cos ç 2 + x ÷ ïï è è 4 2ø ø -1 ï = tan í ý ý = tan í ï 2 sin æç p + x ö÷ cos æç p + x ö÷ ï ï sin æç p + x ö÷ ï ï ïî ï è2 è 4 2ø ø þ è 4 2 ø ïþ î -1 ï

ì æ p x öü æ p x ö = tan -1 í tan ç + ÷ ý = ç + ÷ è 4 2 øþ è 4 2 ø î

Senior Secondary School Mathematics for Class 12 Pg-645

Methods of Integration px x 2 æp xö I = ò ç + ÷ dx = + + C. 4 2 4 4 è ø

\ 53. I = ò =ò

( 1 + sin x ) ( 1 + sin x ) dx. ´ ( 1 - sin x ) ( 1 + sin x ) ( 1 + sin x ) 2 2

( 1 - sin x )

dx = ò

1 + sin 2 x + 2 sin x cos 2 x

dx

sin 2 x 2 sin x ïü ïì 1 = òí + + ý dx ïî cos 2 x cos 2 x cos 2 x ïþ = ò ( sec2 x + tan 2 x + 2 sec x tan x ) dx = ò ( 2 sec2 x - 1 + 2 sec x tan x ) dx = 2 tan x - x + 2 sec x + C . 54. On dividing x 4 by ( x 2 + 1), we get: 1 ïü x3 ïì - x + tan -1 x + C . I = ò í( x 2 - 1) + 2 ý dx = 3 ïî ( x + 1 ïþ 55.

sin ( x - a ) sin ( x + a - 2 a ) sin ( x + a ) cos 2 a - cos ( x + a ) sin 2 a = = × sin ( x + a ) sin ( x + a ) sin ( x + a ) = cos 2 a - sin 2 a cot ( x + a ) I = ò { cos 2 a - sin 2 a × cot ( x + a ) } dx

\

= (cos 2 a ) × ò dx - sin 2 a × ò cot ( x + a ) dx = x cos 2 a - sin 2 a × log|sin ( x + a )| + C . 56. I = ò

{ x+ 3 + 1 ´ { x + 3 - x + 2} { x + 3 +

x + 2} x + 2}

ìï x + 3 + x + 2 üï = òí ý dx = ò x + 3 dx + îï ( x + 3 ) - ( x + 2 ) þï =

( x + 3) ( 3 2)

57. I = ò

æ ç1+ ç è æ ç1ç è

= -ò

3

2

+

( x + 2) ( 3 2)

sin x ö ÷ cos x ÷ø

sin x ö ÷ cos x ÷ø

dx = ò

3

2

+C=

dx.

ò

x + 2 dx

3 3 2 2 ( x + 3) 2 + ( x + 2) 2 + C . 3 3

(cos x + sin x ) dx. (cos x - sin x )

dt , where (cos x - sin x ) = t t

= - log|t| + C = - log|cos x - sin x| + C . 58. Putting x 3 = t and 3 x 2 dx = dt , we get: dt I=ò = tan -1t + C = tan -1 x 3 + C . (1+ t2 )

645

Senior Secondary School Mathematics for Class 12 Pg-646

646

Senior Secondary School Mathematics for Class 12

59. I = ò

x2

dx. x x6 - 1 1 dt , where x 3 = t = ò 3 t t2 - 1 1 1 = sec-1t + C = sec-1 x 3 + C. 3 3 60. Put x 2 + x + 1 = t and ( 2 x + 1) dx = dt. 3

3

3 t 2 2 + C = ( x 2 + x + 1) 2 + C . ( 3 2) 3 { 2x + 3 - 2x - 3 } 1 61. I = ò ´ dx. { 2 x + 3 + 2 x - 3} { 2 x + 3 - 2 x - 3 }

I = ò t dt =

{ 2x + 3 - 2x - 3 } 1 dx = ò { 2 x + 3 - 2 x - 3 } dx {( 2x + 3) - ( 2x - 3)} 6 3 3 1 1 = ( 2x + 3) 2 ( 2x + 3) 2 + C . 18 18 sin x - dt 62. I = ò , where cos x = t. dx = ò cos x t =ò

= - log|t| + C = - log|cos x| + C . sec x ( sec x + tan x ) 1 63. I = ò dx = ò dt , where ( sec x + tan x ) = t. ( sec x + tan x ) t = log|t| + C = log|sec x + tan x| + C . cosec x ( cosec x - cot x ) 1 64. I = ò dx = ò dt , where ( cosec x - cot x ) = t. ( cosec x - cot x ) t = log|t| + C = log|cosec x - cot x| + C . sin x 1 65. I = ò dx + ò dx. ( 1 + cos x ) ( 1 + cos x ) x 2 sin ( 2 ) cos ( x 2 ) dx =ò +ò dx 2 x 2 cos ( 2 ) 2 cos 2 ( x 2 ) x x 1 = ò sec2 dx + ò tan dx 2 2 2 x = ò sec2t dt + 2 ò tan t dt , where = t 2 x x = tan t - 2 log|cos t| + C = tan - 2 log½cos ½ + C . 2 2½ ½ sin x 1 66. I = ò dx = - ò dt , where cos x = t. ( 1 + cos 2 x ) (1 + t2 ) = - tan -1t + C = - tan -1 (cos x ) + C . 1 + x ïü ( 1 + x) ïì 1 + x 67. I = ò í ´ dx ý dx = ò 1 + x ïþ ïî 1 - x 1 - x2 1 1 2x 1 1 dx = sin -1 x - ò dt , where ( 1 - x 2 ) = t dx - ò =ò 2 2 2 2 t 1- x 1- x = sin -1 x - t + C = sin -1 x - 1 - x 2 + C .

Senior Secondary School Mathematics for Class 12 Pg-647

Methods of Integration 68. Put \

1 1 = t and 2 dx = dt. x x I = ò e tdt = e t + C = e -1/x + C .

69. Put x 4 = t and 4 x 3 dx = dt. \

I=

dt 1 1 1 = tan -1t + C = tan -1 x 4 + C . 4 ò (1 + t2 ) 4 4

1ö æ 70. Put ( x + log x ) = t and ç 1 + ÷ dx = dt. xø è 1ö æ x + 1ö æ ÷ × ( x + log x ) 2 dx = ò ç 1 + ÷ ( x + log x ) 2 dx I =òç xø è è x ø

\

t3 1 + C = ( x + log x ) 3 + C . 3 3 2x -1 2 71. Put tan x = t and dx = dt. ( 1 + x4 ) = ò t 2 dt =

I = ò t dt =

\

t2 1 + C = (tan -1 x 2 ) 2 + C . 2 2

72. Put ( 2 - 3 x ) = t and -3dx = dt. 1 1 1 1 I = - ò dt = - log|t| + C = - log|2 - 3 x| + C . \ 3 t 3 3 73. Put ( x 2 - 1) = t and 2x dx = dt. 3

\

I=

3 1 1 t 2 1 3 1 t dt = × 3 + C = t 2 + C = ( x 2 - 1) 2 + C . ò 2 2 ( 2) 3 3

74. Put 5 - 3 x = t and -3dx = dt. 1 1 3t 1 3 (5 - 3 x) \ I = - ò 3 t dt = - ´ +C=- ´ + C. log 3 3 3 log 3 3 75. Put tan x = t and sec2 x dx = dt. I = ò e tdt = e t + C = e tan x + C .

\

76. Put cos 2 x = t and -2 cos x sin x dx = dt. 2x

I = - ò e tdt = - e t + C = - e cos

\

+ C.

77. Put sin x 2 = t and 2 x cos x 2 dx = dt. t4 1 3 1 t4 1 t dt = ´ + C = + C = sin 4 x 2 + C . ò 2 2 4 8 8 1 = t and e x × dx = dt. 2 x

I=

\

x

78. Put e

I = 2 ò cos t dt + C = 2 sin t + C = 2 sin ( e

\ 3

2

x

) + C.

79. Put x = t and 3 x dx = dt. 1 1 1 I = ò sin t dt = ( - cos t ) + C = - cos x 3 + C . 3 3 3

647

Senior Secondary School Mathematics for Class 12 Pg-648

648

Senior Secondary School Mathematics for Class 12

80. Put xe x = t and ( xe x + e x × 1) dx = dt , i.e., ( x + 1)e x dx = dt. 1 \ I=ò dt = ò sec2t dt = tan t + C = tan ( xe x ) + C . cos 2 t dt x 1 81. I = ò dx = ò , where x 2 = t. 2 4 2 t t2 - 1 x x -1 1 1 = sec-1t + C = sec-1 x 2 + C. 2 2 82. I = ò {( x - 1 + 1) x - 1} dx = ò {( x - 1) =

3

2

1

+ ( x - 1) 2 } dx.

5 3 3 2 ( x - 1) 2 + ( x - 1) 2 + C . 5 2

83. Put x 2 - 1 = t and 2x dx = dt. 3

I=

3 1 1 t 2 1 3 1 t dt = × 3 + C = t 2 + C = ( x 2 - 1) 2 + C . ò 2 2 3 3 2

84. Put x = t , so that x = t 2 and dx = 2t dt. {2( 1 + t ) - 2} dx 2t I=ò =ò dt = ò dt \ (1 + t) (1 + t) (1+ x) ì 2 ü = ò í2 ý dt = 2 t - 2 log|1 + t| + C ( 1 + t)þ î = 2 x - 2 log|1 +

x| + C .

2t × (t 2 + 1) æ {(t 2 + 1) - 1} 2t 2 1 ö÷ I=ò 2 dt = 2 ò dt = 2 ò ç 1 - 2 dt 2 ç t + 1 ø÷ (t + 1) (t + 1) è

85. Put ( e x - 1) = t 2 . Then e x dx = 2t dt. So, dx = \

= 2[t - tan -1t] + C = 2[ e x - 1 - tan -1 { e x - 1} + C . 86. I = ò =ò

(sin x - cos x ) + (cos x + sin x ) dx. 2(sin x - cos x ) 1 1 (cos x + sin x ) 1 1 1 dx + ò dx = x + ò dt , where t = (sin x - cos x ) 2 2 (sin x - cos x ) 2 2 t

1 1 1 1 x + log|t| + C = x + log|sin x - cos x| + C . 2 2 2 2 cos x - cos x 87. I = ò dx = ò dx. (cos x - sin x ) (sin x - cos x ) =

=ò

ì 1 1 (sin x + cos x ) ü (sin x - cos x ) - (sin x + cos x ) dx = ò í - × ý dx 2(sin x - cos x ) î 2 2 (sin x - cos x ) þ

1 1 x - log|sin x - cos x| + C . 2 2 sin x 88. I = ò dx , which is the same as Q. 86. (sin x - cos x ) =

Senior Secondary School Mathematics for Class 12 Pg-649

Methods of Integration 89. Put tan x = t and sec2 x dx = dt. dt = sin -1 (t ) + C = sin -1 (tan x ) + C . \ I=ò 1- t2 90. On dividing the numerator and denominator by x 2 , we get 1 ö 1 ö æ æ ç1+ 2 ÷ ç1+ 2 ÷ è è x ø x ø I=ò dx = ò dx 2 1 ö üï ìï æ æ 2 1ö çx + 2 ÷ ç x - ÷ + 2ý í è x ø xø þï îï è =ò =ò 91. I = ò

dt (t 2 + 2 )

1 ö 1ö æ æ , where ç x - ÷ = t and ç 1 + 2 ÷ dx = dt xø è è x ø

dt 2

2

[t + ( 2 ) ]

=

ì 1 æ t 1 1 1 öü +C= tan -1 tan -1 í ç x - ÷ ý + C. x 2 2 2 2 è øþ î

sin 6 x

1 × dx = ò tan 6 x sec2 x dx = ò t 6dt , where tan x = t. cos 6 x cos 2 x t7 1 = + C = tan 7 x + C . 7 7

92. I = sec4 x × sec x tan x dx = ò t 4 dt , where sec x = t. 1 1 = t 5 + C = sec5 x + C . 5 5 93. I = ò tan 3 x tan 2 x dx = ò tan 3 x ( sec2 x - 1) dx. = ò tan 3 x sec2 x dx - ò tan 3 x dx = ò t 3 dt - ò tan x ( sec2 x - 1) dx , where tan x = t t4 - tan x sec2 x dx + ò tan x dx 4 ò 1 = tan 4 x - ò u du + log|sec x| + C , where tan x = u 4 1 1 1 1 = tan 4 x - u2 + log|sec x| + C = tan 4 x - tan 2 x + log|sec x| + C . 4 2 4 2 =

94. I = ò sin 3 cos 2 x cos x dx = ò sin 3 x ( 1 - sin 2 x ) cos x dx. = ò t 3 ( 1 - t 2 )dt , where sin x = t

= ò (t 3 - t 5 ) dt =

t4 t6 1 1 - + C = sin 4 x - sin 6x + C . 4 6 4 6

95. I = ò sec2 x × sec2 x × tan x dx.

= ò sec2 x ( 1 + tan 2 x ) tan x dx = ò ( 1 + t 2 ) t dt , where tan x = t

t2 t4 1 1 + + C = tan 2 x + tan 4 x + C . 2 4 2 4 1 1 96. Put log tan x = t. Then, × sec2 x dx = dt , i.e., dx = dt. tan x sin x cos x 1 1 I = ò t dt = t 2 + C = (log tan x ) 2 + C . \ 2 2 = ò (t + t 3 ) dt =

649

Senior Secondary School Mathematics for Class 12 Pg-650

650

Senior Secondary School Mathematics for Class 12

97. Put ( 2 x + 1) = t and dx = \

I=

1 dt. 2

1 1 sin 3 t dt = ò ( 1 - cos 2 t ) sin t dt 2ò 2

ù 1 1 1 é u3 ( 1 - u2 )( - du) = ò (u2 - 1) du = ê - uú + C , where cos t = u ò 2 2 2 êë 3 úû 1 1 3 1 1 1 1 3 = u - u + C = cos t - cos t + C = cos 3 ( 2 x + 1) - cos ( 2 x + 1) + C . 6 2 6 2 6 2 =

98. Dividing Nr and Dr by cos 2 x , we get I=ò =ò

tan x tan x

× sec2 x dx = ò

sec2 x dx tan x

dt , where tan x = t t

= 2 t + C = 2 tan x + C . 99. I = ò =ò =

(cos x + sin x ) (cos 2 x + sin 2 x - 2 sin x cos x ) (cos x + sin x ) (sin x - cos x ) 2

dx = ò

1 t2

dx

dt , where (sin x - cos x ) = t

-1 -1 1 +C= +C= + C. t (sin x - cos x ) (cos x - sin x )

100. Putting ( e x - 1) = t 2 and e xdx = 2t dt. 2t \ dx = 2 dt. (t + 1) 2t 2 2 ïü ïì -1 I=ò dt = ò í 2 - 2 \ ý dt = 2t - 2 tan t + C 2 ïî (t + 1) (t + 1) ïþ = 2 e x - 1 - 2 tan -1 e x - 1 + C . 101. On dividing Nr and Dr by cos 2 x , we get I=ò

sec2 x 3

tan x

dx = ò

dt t3

, where tan x = t.

Integration Using Trigonometric Identities When the integrand consists of trigonometric functions, we use known identities to convert it into a form which can easily be integrated. Some of the identities useful for this purpose are given below: æxö æxö (ii) 2 cos2 ç ÷ = (1 + cos x) (i) 2 sin 2 ç ÷ = (1 - cos x) 2 è ø è 2ø (iii) 2sin a cos b = sin ( a + b) + sin ( a - b) (iv) 2cos a sin b = sin ( a + b) - sin ( a - b)

Senior Secondary School Mathematics for Class 12 Pg-651

Methods of Integration

651

(v) 2cos a cos b = cos ( a + b) + cos ( a - b) (vi) 2sin a sin b = cos ( a - b) - cos ( a + b) SOLVED EXAMPLES EXAMPLE 1

Evaluate: (i) (iii) (v)

SOLUTION

x

ò sin 2 dx 2 ò cos nx dx 7 ò sin x dx 2

(ii) (iv) (vi)

x

ò tan 2 dx 5 ò cos x dx 3 ò sin ( 2x + 1) dx 2

x 1 x dx = ò 2 sin 2 dx 2 2 2 1 1 1 = ò (1 - cos x) dx = ò dx - ò cos x dx 2 2 2 1 1 = x - sin x + C. 2 2 x x ö x æ (ii) ò tan 2 dx = ò ç sec2 - 1÷ dx = ò sec2 dx - ò dx 2 2 2 è ø x = 2ò sec2t dt - ò dx , where = t 2 x = 2 tan t - x + C = 2 tan - x + C. 2 1 2 2 (iii) ò cos nx dx = ò 2 cos nx dx 2 1 1 1 = ò (1 + cos 2nx) dx = ò dx + ò cos 2nx dx 2 2 2 x 1 = + sin 2nx + C. 2 4n (i)

(iv)

ò sin

2

ò cos x dx = ò cos x × cos x dx = ò (1 - sin 2 x) 2 × cos x dx = ò (1 - t 2) 2 dt , where sin x = t = ò (1 + t 4 - 2t 2) dt = ò dt + ò t 4 dt - 2ò t 2 dt 5

4

1 2 t 5 2t 3 + C = sin x + sin5 x - sin 3 x + C. 5 3 5 3 (v) ò sin 7 x dx = ò sin 6 x × sin x dx =t+

= ò (1 - cos2 x) 3 sin x dx

= - ò (1 - t 2) 3 dt , where cos x = t = ò (t 6 - 3t 4 + 3t 2 - 1) dt = =

t 7 3t5 + t 3 -t + C 7 5

1 3 cos7 x - cos5 x + cos 3 x - cos x + C. 7 5

Senior Secondary School Mathematics for Class 12 Pg-652

652

Senior Secondary School Mathematics for Class 12

(vi)

ò sin

3

( 2x + 1) dx = ò {1 - cos2( 2x + 1)} × sin ( 2x + 1) dx 1 (1 - t 2) dt , where cos ( 2x + 1) = t 2ò 1 1 1 1 = - ò dt + ò t 2 dt = - t + t 3 + C 2 2 2 6 1 1 3 = - cos ( 2x + 1) + cos ( 2x + 1) + C. 2 6 =-

EXAMPLE 2 SOLUTION

Evaluate

ò cos mx cos nx dx ,

when (i) m ¹ n (ii) m = n.

(i) When m ¹ n, we have 1 ò cos mx cos nx dx = 2 ò [cos (m + n) x + cos (m - n) x] dx 1 1 = ò cos (m + n) x dx + ò cos (m - n) x dx 2 2 sin (m + n) x sin (m - n) x = + + C. 2(m + n) 2(m - n) (ii) When m = n, we have

ò cos mx cos nx dx = ò cos nx dx 2

1 1 2 cos2 nx dx = ò (1 + cos 2nx) dx 2ò 2 1 1 x sin 2nx = ò dx + ò cos 2nx dx = + + C. 2 2 2 4n =

EXAMPLE 3

Evaluate: (i) ò sin 3 x sin 2x dx (iii)

SOLUTION

ò cos 4x cos x dx

(ii) [CBSE 2001]

(iv)

ò cos 3 x sin 2x dx 3 3 ò sin x cos x dx

(i) Using 2sin a sin b = cos ( a - b) - cos ( a + b), we have 1 ò sin 3 x sin 2x dx = 2 ò 2 sin 3 x sin 2x dx 1 = ò (cos x - cos 5 x) dx 2 1 1 = ò cos x dx - ò cos 5 x dx 2 2 1 sin 5 x = sin x + C. 2 10 (ii) Using 2cos a sin b = sin ( a + b) - sin ( a - b), we get 1 ò cos 3 x sin 2x dx = 2 ò 2 cos 3 x sin 2x dx 1 = ò (sin 5 x - sin x) dx 2

Senior Secondary School Mathematics for Class 12 Pg-653

Methods of Integration

653

1 1 sin 5 x dx - ò sin x dx 2ò 2 - cos 5 x cos x = + + C. 10 2 =

(iii) Using 2cos a cos b = cos ( a + b) + cos ( a - b), we get 1 ò cos 4x cos x dx = 2 ò 2 cos 4x cos x dx 1 = ò (cos 5 x + cos 3 x) dx 2 1 1 = ò cos 5 x dx + ò cos 3 x dx 2 2 sin 5 x sin 3 x = + + C. 10 6 (iv)

ò sin

3

x cos 3 x dx = ò sin 3 x cos2 x cos x dx = ò sin 3 x(1 - sin 2 x) cos x dx = ò t 3(1 - t 2) dt , where sin x = t = ò t 3 dt - ò t5 dt = =

t4 t6 - +C 4 6

1 1 sin 4 x - sin 6 x + C. 4 6

ò cos x cos 2x cos 3 x dx.

EXAMPLE 4

Evaluate

SOLUTION

ò cos x cos 2x cos 3 x dx 1 ( 2 cos x cos 2x) cos 3 x dx 2ò 1 1 = ò (cos 3 x + cos x) cos 3 x dx = ò (cos2 3 x + cos x cos 3 x) dx 2 2 1 1 = ò ( 2 cos2 3 x) dx + ò ( 2 cos x cos 3 x) dx 4 4 =

1 1 (1 + cos 6x) dx + ò (cos 4x + cos 2x) dx 4 4ò 1 1 1 1 = ò dx + ò cos 6x dx + ò cos 4x dx + ò cos 2x dx 4 4 4 4 1 1 sin 6x 1 sin 4x 1 sin 2x = x+ × + × + × +C 4 4 6 4 4 4 2 x sin 6x sin 4x sin 2x = + + + + C. 4 24 16 8 =

Senior Secondary School Mathematics for Class 12 Pg-654

654

EXAMPLE 5 SOLUTION

Senior Secondary School Mathematics for Class 12

ò sec x tan x dx. 4 2 2 ò sec x tan x dx = ò sec x × sec x tan x dx = ò (1 + tan 2 x) sec2 x tan x dx 4

Evaluate

= ò (1 + t 2)t dt , where tan x = t = ò t dt + ò t 3 dt = = EXAMPLE 6 SOLUTION

EXAMPLE 7 SOLUTION

Evaluate

ò sin

4

t2 t4 + +C 2 4

1 1 tan 2 x + tan 4 x + C. 2 4

x dx.

[CBSE 2004]

1 ( 2 sin 2 x) 2 dx 4ò 1 1 = ò (1 - cos 2x) 2 dx = ò (1 + cos2 2x - 2 cos 2x) dx 4 4 1 = ò ( 2 + 2 cos2 2x - 4 cos 2x) dx 8 1 = ò [2 + (1 + cos 4x) - 4 cos 2x] dx 8 3 1 1 = ò dx + ò cos 4x dx - ò cos 2x dx 8 8 2 3 sin 4x sin 2x = x+ + C. 8 32 4 sin x Evaluate ò dx. [CBSE 2003C, ‘04] sin ( x - a )

ò sin

4

x dx =

Put ( x - a ) = t so that x = (t + a ) and dx = dt. sin x sin (t + a ) \ ò dx = ò dt sin ( x - a ) sin t =ò

sin t cos a + cos t sin a dt sin t

= cos a × ò dt + sin a × ò cot t dt

EXAMPLE 8

SOLUTION

Evaluate: (i)

= cos a × t + sin a × log |sin t | + C = (cos a )( x - a ) + sin a × log |sin ( x - a )| + C. sin 4x sin 4x (i) ò (ii) ò dx dx cos 2x sin x

sin 4x

ò cos 2x dx = ò

2 sin 2x cos 2x dx cos 2x

= 2ò sin 2x dx = - cos 2x + C.

Senior Secondary School Mathematics for Class 12 Pg-655

Methods of Integration

(ii)

ò

655

sin 4x 2 sin 2x cos 2x dx = ò dx sin x sin x 4 sin x cos x cos 2x =ò dx = 2ò 2 cos x cos 2x dx sin x = 2ò (cos 3 x + cos x) dx = 2ò cos 3 x dx + 2ò cos x dx =

EXAMPLE 9

Evaluate

SOLUTION

ò

EXAMPLE 10

Evaluate

SOLUTION

ò

2 sin 3 x + 2 sin x + C. 3

1 + sin x dx.

x x x x 1 + sin x dx = ò sin 2 + cos2 + 2 sin cos dx 2 2 2 2 x xö x x æ = ò ç sin + cos ÷ dx = ò sin dx + ò cos dx 2 2ø 2 2 è x x = -2 cos + 2 sin + C. 2 2 sin 2 x

ò (1 + cos x) 2 dx. 2

2

é 2 sin ( x/2) cos( x/2) ù æ sin x ö ò (1 + cos x) 2 dx = ò ççè 1 + cos x ÷÷ø dx = ò ê 2 cos2( x/2) ú dx ë û x ö æ = ò tan 2( x/2) dx = ò ç sec2 - 1÷ dx 2 ø è sin 2 x

= ò sec 2( x/2) dx - ò dx = 2 tan ( x/2) - x + C. EXAMPLE 11

SOLUTION

Evaluate: (i)

(i) ò sin x 1 - cos 2x dx

ò sin x 1 - cos 2x dx = ò sin x × 2 sin 2 x dx =

(ii) ò

2 ò sin 2 x dx =

cos 2x dx 1 + cos 4x

1 2 sin 2 x dx 2ò

1 1 1 (1 - cos 2x) dx = dx cos 2x dx 2ò 2ò 2ò 1 sin 2x = x+ C. 2 2 2 cos 2x cos 2x 1 x (ii) ò dx = ò dx = ò dx = 2 + C. 2 1 + cos 4x 2 2 cos 2x =

EXAMPLE 12 SOLUTION

Evaluate:

(i) ò

dx a sin x + b cos x

(ii) ò

dx sin x + cos x

(i) Put a = r cos q and b = r sin q so that r 2 = ( a 2 + b 2) and q = tan -1( b/a).

Senior Secondary School Mathematics for Class 12 Pg-656

656

Senior Secondary School Mathematics for Class 12

\

dx

dx

ò a sin x + b cos x = ò r cos q sin x + r sin q cos x =

1 dx 1 = × cosec ( x + q) dx r ò sin ( x + q) r ò

1 ì = log ítan r î

æ q+ x öü ç ÷ý+ C è 2 øþ

é ì1 æ b ö x üù log ê tan í tan -1 ç ÷ + ýú + C. 2 è a ø 2 þû î ë a +b 1

=

2

2

(ii) We can write, dx 1 dx ò sin x + cos x = 2 ò 1 1 sin x + cos x 2 2 1 dx = p p ö 2òæ ç cos sin x + sin cos x ÷ 4 4 è ø

EXAMPLE 13

SOLUTION

Evaluate:

(i) ò

1 dx æp ö = cosec ç + x ÷ dx æp ö 4 2ò è ø sin ç + x ÷ è4 ø

=

1 2ò

=

1 æp xö log tan ç + ÷ + C. 2 è 8 2ø

dx 4 cos x + 3 sin x

(ii) ò

dx ( 2 sin x + 3 cos x) 2

(i) Put 4 = r sin q and 3 = r cos q so that æ 4ö r 2 = 25 and q = tan -1 ç ÷ × è 3ø dx dx = \ ò 4 cos x + 3 sin x ò r sin q cos x + r cos q sin x =

1 dx 1 = cosec ( q + x) dx r ò sin ( q + x) r ò

1 ì æ q+ x öü = log ítan ç ÷ý+ C r è 2 øþ î =

é 1 ì1 æ 4 ö x üù log ê tan í tan -1 ç ÷ + ýú + C. 5 è 3 ø 2 þû î2 ë

æ 3ö (ii) Put 2 = r cos q and 3 = r sin q so that r 2 = 13 and q = tan -1 ç ÷ × è 2ø dx dx =ò Q ò ( 2 sin x + 3 cos x) 2 (r cos q sin x + r sin q cos x) 2

Senior Secondary School Mathematics for Class 12 Pg-657

Methods of Integration

=

EXAMPLE 14

SOLUTION

òæ

Evaluate

òæ

1 r

1

1 cot ( q + x) + C 13

=-

1 ì ü æ 3ö cot ítan -1 ç ÷ + x ý + C. 13 2 è ø î þ

cos x

dx = ò

dx

ò sin 2( q + x) = 13 × ò cosec ( q + x) dx

=-

x xö ç cos + sin ÷ 2 2ø è

cos x

2

657

3

2

dx.

cos2( x/2) - sin 2( x/2)

dx {cos ( x/2) + sin ( x/2)} 3 x xö ç cos + sin ÷ 2 2ø è x x cos ( x/2) - sin ( x/2) 1 =ò dx = 2ò 2 dt , where t = cos + sin 2 2 2 t x xö æ ç cos + sin ÷ 2 2ø è -2 -2 = +C = + C. t cos ( x/2) + sin ( x/2) 3

EXAMPLE 15

Evaluate

SOLUTION

ò

dx × 1 - sin x

ò

dx dx =ò 1 /2 1 - sin x x xù é 2 2 êë sin ( x/2) + cos ( x/2) - 2 sin 2 cos 2 úû 1 dx dx =ò = ò x x 1 x x 1 ö æ ö 2 æ × sin - cos × ÷ ç ç sin - cos ÷ 2 2ø 2 2 2ø è è 2 =

=

1 dx ×ò x p x pö 2 æ ç sin cos - cos sin ÷ 2 4 2 4ø è 1 1 æ x pö cosec ç - ÷ dx = 2 × log 2ò 2 è 2 4ø

æ x pö = 2 log tan ç - ÷ + C. è 4 8ø EXAMPLE 16

Evaluate

SOLUTION

ò

ò

sin x dx. 1 + sin x

sin x (1 + sin x) - 1 dx = ò dx 1 + sin x 1 + sin x

é æ x pöù ê tan ç 4 - 8 ÷ ú + C è øû ë

Senior Secondary School Mathematics for Class 12 Pg-658

658

Senior Secondary School Mathematics for Class 12

dx 1 + sin x

= ò 1 + sin x dx - ò = ò cos2 -ò

x x x x + sin 2 + 2 sin cos dx 2 2 2 2 dx

cos2( x/2) + sin 2( x/2) + 2 sin ( x/2) cos ( x/2)

= ò [cos ( x/2) + sin ( x/2)]dx - ò

dx [cos ( x/2) + sin ( x/2)]

x xö 1 dx æ × = ç 2 sin - 2 cos ÷ 2 2ø 2 ò 1 cos x + 1 sin x è 2 2 2 2 x æ = ç 2 sin - 2 cos 2 è

xö 1 × ÷2ø 2 ò

x æ = ç 2 sin - 2 cos 2 è

xö 1 æ x pö cosec ç + ÷ dx ÷ò 2ø 2 è 2 4ø

x æ = 2 ç sin - cos 2 è

dx æ x pö sin ç + ÷ è 2 4ø

xö 1 ½ ´ 2 log ½tan ÷2ø 2 ½

æ x p ö½ ½ ç + ÷ +C è 4 8 ø½

x xö æ æ x p ö½ ½ = 2 ç sin - cos ÷ - 2 log ½tan ç + ÷½+ C. 2 2ø è è 4 8 ø½ ½

EXERCISE 13B Evaluate the following integrals: 1. (i) 2. (i) 3. (i) 4.

ò sin x dx 2 ò cos ( x/2) dx 2 ò sin nx dx

ò cos x dx (ii) ò cot 2( x/2) dx (ii) ò sin5 x dx 5. ò sin 7( 3 - 2x) dx

2

ò cos

3

(ii)

( 3 x + 5) dx

æ 1 - cos 2x ö

6. (i)

ò ççè 1 + cos 2x ÷÷ø dx

7. (i)

ò ççè 1 + cos x ÷÷ø dx

8. 10.

æ 1 - cos x ö

ò sin 3 x cos 4x dx ò sin 4x sin 8x dx

æ 1 + cos 2x ö

(ii)

ò ççè 1 - cos 2x ÷÷ø dx

(ii)

ò ççè 1 - cos x ÷÷ø dx

9. [CBSE 2007]

2

11.

æ 1 + cos x ö

ò cos 4x cos 3 x dx ò sin 6x cos x dx

[CBSE 2007]

Senior Secondary School Mathematics for Class 12 Pg-659

Methods of Integration

12. 14. 16. 18. 20.

ò sin x 1 + cos 2x dx ò cos 2x cos 4x cos 6x dx 4 ò sec x dx 4 3 ò cos x sin x dx 3/5 3 ò cos x sin x dx cos 2x dx cos x

13. 15. 17. 19. 21.

659

ò cos x dx 3 ò sin x cos x dx 3 4 ò cos x sin x dx 2/ 3 3 ò sin x cos x dx 4 ò cosec 2x dx 4

cos x

22.

ò

24.

ò cos

26.

ò cos 2x dx

28.

ò ( 3 cos x + 4 sin x)

30.

ò (cos x - sin x)

32.

ò sin x sin 2x sin 3 x dx [CBSE 2012] 33. ò ççè 1 + cot x ÷÷ø dx

34.

ò ( 2 sin x + cos x + 3)

3

x sin 2x dx

4

dx

dx

[CBSE 2000, ‘03]

23.

ò cos ( x + a) dx

25.

ò sin x dx

27.

ò (1 + cos x) 2 dx

29.

ò ( a cos x + b sin x) 2 , a > 0 and b > 0

31.

ò ( 2 tan x - 3 cot x)

cos9 x

sin 2 x

dx

2

dx

æ 1 - cot x ö

dx

[CBSE 2000C, ‘04]

ANSWERS (EXERCISE 13B)

1 1 x - sin 2x + C 2 4 1 1 2. (i) x + sin x + C 2 2 1. (i)

5. 6. 7. 8.

1 1 x + sin 2x + C 2 4

(ii) -2 cot ( x/2) - x + C

cos5 x 2 cos 3 x 1 sin 2nx (ii) - cos x + +C x+C 5 3 2 4n 1 1 sin ( 3 x + 5) - sin 3( 3 x + 5) + C 3 9 1 1 1 3 cos ( 3 - 2x) - cos7( 3 - 2x) - cos 3( 3 - 2x) + cos5( 3 - 2x) + C 2 14 2 10 (i) tan x - x + C (ii) - cot x - x + C x x (i) 2 tan - x + C (ii) -2 cot - x + C 2 2 - cos 7 x cos x sin 7 x 1 9. + +C + sin x + C 14 2 14 2

3. (i) 4.

(ii)

Senior Secondary School Mathematics for Class 12 Pg-660

660

Senior Secondary School Mathematics for Class 12

10.

sin 4x sin 12x +C 8 24

11.

- cos 7 x cos 5 x +C 14 10

12.

sin 2 x +C 2

13.

3 x sin 4x sin 2x + + +C 8 32 4

14. 16. 18. 20. 22. 24. 26. 28.

x sin 4x sin 8x sin 12x sin 4 x +C + + + + C 15. 4 4 16 32 48 1 1 1 17. sin5 x - sin 7 x + C tan x + tan 3 x + C 3 5 7 1 1 3 3 19. sin5/ 3 x - sin11/ 3 x + C cos7 x - cos5 x + C 7 5 5 11 5 5 1 1 21. - cot 2x - cot 3 2x + C - cos8/5 x + cos18/5 x + C 8 18 2 6 23. x cos a - sin a × log |cos ( x + a )| + C 2sin x - log |sec x + tan x| + C 2 3 2 1 25. log |sin x | - 2 sin 2 x + sin 4 x - sin 6 x + sin 8 x + C - cos5 x + C 5 2 3 8 x 3 x sin 4x sin 8x 27. 2 tan - x + C + + +C 2 8 8 64 x 1 3 1 bü ì é1 ù 29. 2 log tan ê tan -1 + ú + C tan íx - tan -1 ý + C 2 5 2 4 2 aþ î ë û (a + b )

-1 æ p x ö½ ½ 31. 4 tan x - 9 cot x - 25 x + C log ½tan ç - ÷½+ C 2 è 8 2 ø½ ½ 1 1 1 32. cos 6x - cos 4x - cos 2x + C 24 16 8 xö æ 33. - log |sin x + cos x | + C 34. tan -1 ç1 + tan ÷ + C 2ø è 30.

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 13B) 12. I = ò sin x × 2 cos 2 x dx = 2 ò sin x cos x dx. Put sin x = t. 1 1 ( 2 cos 2 x ) 2 dx = ò ( 1 + cos 2 x ) 2 dx 4ò 4 1 é 1 æ 1 + cos 4 x ) ö ù é3 1 ù ÷ ú dx = ò ê + cos 2 x + cos 4 x ú dx. = ò ê( 1 + 2 cos 2 x ) + ç 4 ë 8 2 ë8 2 û è øû

13. I =

16. I = ò ( 1 + tan 2 x ) sec2 x dx. Put tan x = t. 17. I = ò cos x ( 1 - sin 2 x ) sin 4 x dx. Put sin x = t. 19. I = ò sin 2/3 x ( 1 - sin 2 x ) cos x dx. Put sin x = t. 21. I = ò ( 1 + cot 2 2 x )cosec2 2 x dx. Put cot 2x = t. 22. I = ò

( 2 cos 2 x - 1) dx = 2 ò cos x dx - ò sec x dx. cos x

Senior Secondary School Mathematics for Class 12 Pg-661

Methods of Integration

661

24. I = 2 ò cos 4 x sin x dx. Put cos x = t. 25. I = ò 27. I = ò

cos 8 x cos x ( 1 - sin 2 x ) 4 cos x dx = ò dx. Put sin x = t. sin x sin x ( 1 - cos 2 x ) ( 1 + cos x ) 2

dx = ò

( 1 - cos x ) x x æ ö dx = ò tan 2 dx = ò ç sec2 - 1 ÷ dx. 2 ( 1 + cos x ) 2 è ø

28. Put 3 = r sin q and 4 = r cos q. 29. Put a = r sin q and b = r cos q. dx dx 1 1 1 æp ö 30. I = ×ò = ò æ p ö = 2 × ò cosec çè 4 - x ÷ø dx. 1 æ 1 ö 2 2 sin ç - x ÷ ç cos x sin x ÷ 2 è4 ø è 2 ø 1 32. I = ò (sin 2 x + sin 4 x - sin 6 x ) dx. 4 (cos x - sin x ) 33. I = - ò dx. Put (sin x + cos x ) = t. (sin x + cos x ) 34. tan

2t 1- t2 x 2 dt = t , dx = ; sin x = and cos x = × 2 2 2 1+ t2 (1+ t ) (1+ t )

Integration by Parts THEOREM

PROOF

If u and v are two functions of x then ü ì du ò (uv) dx = [u × ò v dx] - ò íî dx × ò v dx ýþ dx.

For any two functions f1( x) and f 2( x), we have d [ f1( x) × f 2( x)] = f1( x) × f 2 ¢( x) + f 2( x) × f1 ¢( x). dx \ or or Let \

ò { f1( x) × f 2 ¢( x) + f 2( x) × f1 ¢( x)} dx = f1( x) × f 2( x) ò f1( x) × f 2 ¢( x) dx + ò f 2( x) × f1 ¢( x) dx = f1( x) × f 2( x) ò f1( x) × f 2 ¢( x) dx = f1( x) × f 2( x) - ò f 2( x) × f1 ¢( x) dx. f1( x) = u and f 2 ¢( x) = v so that f 2( x) = ò v dx. ì du

ü

ò (uv) dx = u × ò v dx - ò íî dx × ò v dx ýþ dx.

We can express this result as given below: Integral of product of two functions

= (1st function ) ´ (integral of 2nd ) - ò {(derivative of 1st) ´ (integral of 2nd)} dx.

Senior Secondary School Mathematics for Class 12 Pg-662

662 REMARKS

REMARKS

Senior Secondary School Mathematics for Class 12

(i) If the integrand is of the form f ( x) x n , we consider x n as the first function and f ( x) as the second function. (ii) If the integrand contains a logarithmic or an inverse trigonometric function, we take it as the first function. In all such cases, if the second function is not given, we take it as 1.

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Evaluate: (i) ò x sec2 x dx

(ii)

ò x sin 2x dx

(i) Integrating by parts, taking x as the first function, we get ìd

ü

ò x sec x dx = x × ò sec x dx - ò íî dx ( x) × ò sec x dx ýþdx 2

2

2

= x tan x - ò 1 × tan x dx = x tan x + log |cos x | + C. (ii) Integrating by parts, taking x as the first function, we get ìd

ü

ò x sin 2x dx = x × ò sin 2x dx - ò íî dx ( x) × ò sin 2x dx ýþdx æ - cos 2x ö æ - cos 2x ö = x×ç ÷ dx ÷ - 1× ç 2 2 ø ò è ø è - x cos 2x 1 = + ò cos 2x dx 2 2 - x cos 2x 1 sin 2x = + × +C 2 2 2 - x cos 2x 1 = + sin 2x + C. 2 4

òx

n

EXAMPLE 2

Evaluate

SOLUTION

Integrating by parts, taking log x as the first function, we get

òx

n

log x dx.

ü ìd log x = (log x) × ò x ndx - ò í (log x) × ò x ndx ýdx þ î dx = (log x) ×

1 x n +1 x n +1 -ò × dx (n + 1) x (n + 1)

=

1 x n +1 log x x ndx (n + 1) (n + 1) ò

=

x n +1 log x x n +1 + C. (n + 1) (n + 1) 2

Senior Secondary School Mathematics for Class 12 Pg-663

Methods of Integration

òx

2

EXAMPLE 3

Evaluate

SOLUTION

Integrating by parts, taking x 2 as the first function, we get é d ù 2 2 2 ò x sin x dx = x ò sin x dx - ò ëê dx ( x ) × ò sin x dx ûú dx

663

sin x dx.

= x 2( - cos x) - ò 2x( - cos x) dx

= - x 2 cos x + 2ò x cos x dx

é ü ù ìd = - x 2 cos x + 2 ê x(sin x) - ò í ( x) × ò cos x dx ýdx ú þ û î dx ë [integrating x cos x by parts]

= - x 2 cos x + 2[x sin x - ò sin x dx] = - x 2 cos x + 2[x sin x + cos x] + C. EXAMPLE 4

Evaluate

SOLUTION

ò x cos

2

ò x cos

2

x dx.

1 1 æ 1 + cos 2x ö x dx = ò x ç ÷ dx = ò x dx + ò x cos 2x dx 2 2 2 ø è =

=

x2 1 é ü ù ìd + × ê x × ò cos 2x dx - ò í ( x) × ò cos 2x dx ý dx ú 4 2 ë þ û î dx [integrating x cos 2x by parts] sin 2x ù x 2 1 é x sin 2x + -ò dx ú 4 2 êë 2 2 û

x 2 x sin 2x 1 ( - cos 2x) + - × +C 4 4 4 2 x 2 x sin 2x cos 2x = + + + C. 4 4 8 =

ò log x dx.

EXAMPLE 5

Evaluate

SOLUTION

Integrating by parts, taking log x as the first function and 1 as the second function, we get

ò log x dx = ò (log x × 1) dx ü ìd = (log x) × ò 1 dx - ò í (log x) × ò 1 dx ý dx þ î dx 1 æ ö = (log x) × x - ò ç × x ÷ dx = x log x - ò dx èx ø = x log x - x + C = x(log x - 1) + C.

ò log (1 + x

2

EXAMPLE 6

Evaluate

SOLUTION

Integrating by parts, taking log (1 + x 2) as the first function and 1

) dx.

Senior Secondary School Mathematics for Class 12 Pg-664

664

Senior Secondary School Mathematics for Class 12

as the second function, we get

ò log (1 + x

2

) dx = ò {log (1 + x 2) × 1} dx

é d ù = log (1 + x 2) × ò 1 dx - ò ê {log (1 + x 2)} × ò 1 dx ú dx ë dx û 2x 2 = log (1 + x ) × x - ò × x dx (1 + x 2) x2 = x log (1 + x 2) - 2ò dx (1 + x 2) 1 ö æ = x log (1 + x 2) - 2ò ç1 ÷ dx è 1 + x2 ø dx = x log (1 + x 2) - 2ò dx + 2ò (1 + x 2) = x log (1 + x 2) - 2x + 2 tan - 1 x + C.

ò (log x)

2

EXAMPLE 7

Evaluate

SOLUTION

Integrating by parts, taking (log x) 2 as the first function and 1 as the second function, we get

ò (log x)

2

dx.

dx = ò {(log x) 2 × 1} dx

ü ìd = (log x) 2 × ò 1 dx - ò í (log x) 2 × ò 1 dx ý dx þ î dx æ 2 log x ö 2 = x (log x) - ò ç × x ÷ dx ø è x = x (log x) 2 - 2ò (log x × 1) dx

é ü ù ìd = x (log x) 2 - 2 ê(log x) ò dx - ò í (log x) × ò dx ý dx ú þ û î dx ë 1 é ù 2 = x (log x) - 2 ê x log x - ò × x dx ú x ë û = x (log x) 2 - 2x log x + 2x + C. EXAMPLE 8 SOLUTION

Evaluate

ò

log x

dx. x2 1 Integrating by parts, taking log x as the first function and 2 as the x second function, we get log x 1 ò x 2 dx = ò (log x) × x 2 dx 1 1 ìd ü = (log x) × ò 2 dx - ò í (log x) × ò 2 dx ý dx î dx þ x x 1 æ 1ö log x 1 æ 1ö = (log x) ç - ÷ - ò × ç - ÷ dx = + ò 2 dx x x x x è ø è ø x log x 1 - (log x + 1) =- +C = + C. x x x

Senior Secondary School Mathematics for Class 12 Pg-665

Methods of Integration

òe

2x

EXAMPLE 9

Evaluate

SOLUTION

Integrating by parts, we get

òe

665

sin x dx.

ü ìd sin x dx = ( e2x × ò sin x dx) - ò í ( e2x ) × ò sin x dx ý dx þ î dx = e2x × ( - cos x) - 2ò e2x ( - cos x) dx

2x

= - e2x cos x + 2 ò e2x cos x dx

é ü ù ìd = - e2x cos x + 2 ê( e2x × ò cos x dx) - ò í ( e2x ) × ò cos x dx ý dx ú dx þ û î ë [integrating e2x cos x by parts]

= - e2x cos x + 2e2x sin x - 4ò e2x sin x dx + C.

\

5 ò e2x sin x dx = e2x ( 2 sin x - cos x) + C

or

òe

1 2x e ( 2 sin x - cos x) + C. 5 æ x - sin x ö ò ççè 1 - cos x ÷÷ø dx.

2x

sin x dx =

EXAMPLE 10

Evaluate

SOLUTION

ò ççè 1 - cos x ÷÷ø dx = ò (1 - cos x) dx - ò (1 - cos x) dx

æ x - sin x ö

x

=ò

sin x

x 2 sin 2( x/2)

dx - ò

2 sin ( x/2) cos( x/2) 2 sin 2( x/2)

dx

1 x cosec2( x/2) dx - ò cot ( x/2) dx 2ò 1é ü ù ìd = ê x × ò cosec2( x/2) dx - ò í ( x) × ò cosec2( x/2) dx ýdx ú dx 2ë þ û î =

- ò cot ( x/2) dx

=

1é æ ê x × ç -2 cot 2ë è

= - x cot

ò x tan

-1

xö é æ ÷ - ò ê1 × ç -2 cot 2ø ë è

[integrating by parts]

x öù ÷ dx - ò cot ( x/2)dx + C 2 ø úû

x x x x + cot dx - ò cot dx + C = - x cot + C. 2 ò 2 2 2

EXAMPLE 11

Evaluate

SOLUTION

Integrating by parts, taking tan -1 x as the first function, we get

ò x tan

-1

x dx.

[CBSE 2000C, ‘04C]

ü ìd x dx = (tan -1 x) × ò x dx - ò í (tan -1 x) × ò x dx ýdx þ î dx = (tan -1 x) × =

x2 x2 1 dx -ò × 2 (1 + x 2) 2

1 ö x 2 tan -1 x 1 æ 2 2 - ò ç1 ÷ dx [on dividing x by 1 + x ] 2 2 è 1 + x2 ø

Senior Secondary School Mathematics for Class 12 Pg-666

666

Senior Secondary School Mathematics for Class 12

=

1 1 x 2 tan -1 x 1 - ò dx + ò dx 2 2 2 (1 + x 2)

x 2 tan -1 x x 1 - + tan -1 x + C 2 2 2 1 1 2 = (1 + x ) tan -1 x - x + C. 2 2 =

òx

2

sin -1 x dx.

EXAMPLE 12

Evaluate

SOLUTION

Integrating by parts, taking sin -1 x as the first function we get

òx

EXAMPLE 13

SOLUTION

2

sin -1 x dx = (sin -1 x) ×

Evaluate: (i) ò cos-1 x dx

1 x3 x3 -ò × dx 3 1 - x2 3

=

x 3 sin -1 x 1 x3 - ò dx 3 3 1 - x2

=

x 3 sin -1 x 1 x × x 2 - ò dx 3 3 1 - x2

=

x 3 sin -1 x 1 t(1 - t 2) + ò dt , where (1 - x 2) = t 2 3 3 t

=

1 x 3 sin -1 x 1 + ò dt - ò t 2 dt 3 3 3

=

1 x 3 sin -1 x 1 + t - t3 +C 3 3 9

=

1 x 3 sin -1 x 1 1 - x 2 - (1 - x 2) 3/2 + C. + 3 3 9 (ii)

ò tan

-1

(iii)

x dx

-1

ò sec

x dx

(i) Put cos-1 x = t so that x = cos t and dx = - sin t dt. \ ò cos-1 x dx = - ò t sin t dt

= - [t × ( - cos t) - ò 1 × ( - cos t) dt] [integrating by parts]

= t cos t - ò cos t dt = t cos t - sin t + C

= x cos-1 x - 1 - x 2 + C [Q cos t = x Þ sin t = 1 - x 2 ].

(ii) Put tan -1 x = t so that x = tan t and dx = sec2t dt. \

ò tan

-1

x dx = ò tsec2t dt = t × tan t - ò 1 × tan t dt

[integrating by parts]

= t × tan t + log |cos t| + C

Senior Secondary School Mathematics for Class 12 Pg-667

Methods of Integration

667

½ 1 ½ = (tan -1 x) × x + log ½ ½+ C ½ 1 + x 2½ é 1 ù êQ tan t = x Þ cos t = ú êë 1 + x 2 úû 1 = x(tan -1 x) - log |1 + x 2 | + C. 2 (iii) Put sec-1 x = t so that x = sec t and dx = sec t tan t dt. -1

ò sec

\

x dx = ò t( sec t tan t) dt

= t ( sec t) - ò 1 × sec t dt

[integrating by parts]

= t( sec t) - log |sec t + tan t| + C = t( sec t) - log |sec t + sec2t - 1| + C = x ( sec-1 x) - log |x + x 2 - 1| + C.

ò (sin

-1

x) 2 dx.

EXAMPLE 14

Evaluate

SOLUTION

Putting x = sin t and dx = cos t dt , we get

ò (sin

-1

[CBSE 2004]

x) 2 dx = ò t 2 cos t dt

= t 2 × (sin t) - ò 2t(sin t) dt

[integrating by parts]

= t sin t - 2[t( - cos t) - ò 1 × ( - cos t) dt] 2

[integrating t(sin t) by parts] 2

= t sin t + 2t cos t - 2 sin t + C = x(sin -1 x) 2 + 2(sin -1 x) 1 - x 2 - 2x + C. sin -1 x

ò (1 - x 2) 3/2 dx.

EXAMPLE 15

Evaluate

SOLUTION

Put x = sin t so that dx = cos t dt and t = sin -1 x. \

sin -1 x

ò (1 - x 2) 3/2 dx = ò

t cos t 2

(1 - sin t)

3/2

dt = ò

t cos t cos 3t

dt

= ò t sec2t dt

= t × (tan t) - ò 1 × tan t dt

[integrating by parts]

= t × (tan t) + log |cos t| + C x = (sin -1 x) × + log | 1 - x 2 | + C. 2 1-x é x ù 2 êQ cos t = 1 - x and tan t = ú êë 1 - x 2 úû x(sin -1 x) 1 = + log |(1 - x 2)| + C. 2 1 - x2

Senior Secondary School Mathematics for Class 12 Pg-668

668

Senior Secondary School Mathematics for Class 12

x tan -1 x

ò (1 + x 2) 3/2 dx.

EXAMPLE 16

Evaluate

SOLUTION

Put x = tan t so that dx = sec2t dt. \

x tan -1 x

(tan t) t

ò (1 + x 2) 3/2 dx = ò (1 + tan 2 t) 3/2 × sec t dt =ò

2

(tan t)t dt = ò t sin t dt sec t

= t( - cos t) - ò 1 × ( - cos t) dt [integrating by parts] = - t cos t + sin t + C =

- tan -1 x 1+ x

2

+

x 1 + x2

+C

é x 1 ù and cos t = êQ sin t = ú× 2 êë 1+ x 1 + x 2 úû EXAMPLE 17

Evaluate:

SOLUTION

2x ö ç ÷ dx è 1 + x2 ø 1-x x (iii) ò tan -1 (iv) ò sin -1 dx dx 1+ x a+x (i) Put x = sin t so that dx = cos t dt. (i)

ò sin

\

-1

( 3 x - 4x 3) dx

ò sin

-1

(ii)

ò sin

-1 æ

[CBSE 2002]

( 3 x - 4x 3) dx = ò sin -1( 3 sin t - 4 sin 3 t) cos t dt = ò sin -1(sin 3t) cos t dt = 3 ò t cos t dt

= 3[t(sin t) - ò 1 × sin t dt] [integrating by parts] = 3 t sin t + 3 cos t + C = 3 x (sin -1 x) + 3 1 - x 2 + C.

(ii) Put x = tan t so that dx = sec2t dt. æ 2x ö -1 æ 2 tan t ö 2 \ ò sin -1 ç ÷ dx = ò sin ç ÷ sec t dt è 1 + x2 ø è 1 + tan 2 t ø = ò sin -1(sin 2t) sec2t dt = 2ò t × sec2t dt

= 2[t × tan t - ò 1 × tan t dt]

= 2t × tan t + 2 log |cos t| + C 1 = 2x (tan -1 x) + 2 log +C 1 + x2 æ 1ö = 2x (tan -1 x) + 2 × ç - ÷ log |1 + x 2| + C è 2ø

= 2x (tan -1 x) - log |1 + x 2 | + C.

Senior Secondary School Mathematics for Class 12 Pg-669

Methods of Integration

669

(iii) Put x = cos t so that dx = - sin t dt. 1-x 1 - cos t ( - sin t) dt dx = ò tan -1 \ ò tan -1 1+ x 1 + cos t = ò tan -1

2 sin 2(t/2) 2 cos2(t/2)

( - sin t) dt

t öù 1 é æ = ò ê tan -1 ç tan ÷ ú ( - sin t) dt = - ò t (sin t) dt 2ø û 2 è ë 1 = - [t ( - cos t) - ò 1 × ( - cos t) dt] [integrating by parts] 2 1 1 1 1 = t × cos t - sin t + C = x (cos-1 x) 1 - x 2 + C. 2 2 2 2 (iv) Put x = a tan 2t so that dx = ( 2 a sec2t tan t) dt. ìï x a tan 2 t üï dx = ò sin -1í 2a sec2t tan t dt \ ò sin -1 2 ý a+x ïî a(1 + tan t) ïþ = 2a ò t ( sec2t × tan t) dt 1 é 1 ù = 2a êt × tan 2 t - ò 1 × tan 2 t dt ú 2 ë 2 û [integrating by parts and using ò sec2t tan t dt =

1 tan 2t] 2

= at (tan 2t) - a ò ( sec2t - 1) dt = at (tan 2t) - a ò sec2t dt + a ò dt

= at (tan 2t) - a tan t + at + C

æ x ö æxö x x ÷ ×ç ÷ - a× = a çç tan - 1 + a tan -1 +C ÷ a ø è aø a a è = x tan -1

ò x cos

3

x x - ax + a tan -1 + C. a a

EXAMPLE 18

Evaluate

SOLUTION

Take x as the first function and (cos 3 x sin x) as the second. 1 Putting cos x = t , we can evaluate ò cos 3 x sin x dx as - cos4 x. 4 So, integrating by parts, we get æ 1ö æ -1 3 4 ö 4 ò x cos x sin x dx = x × çè 4 cos x ÷ø - ò 1 × çè - 4 ÷ø cos x dx

x sin x dx.

2

x 1 æ 1 + cos 2x ö = - cos4 x + ò ç ÷ dx 4 4 è 2 ø

Senior Secondary School Mathematics for Class 12 Pg-670

670

Senior Secondary School Mathematics for Class 12

ö x cos4 x 1 æç 1 cos2 2x + ò + + cos 2x ÷ dx ÷ 4 4 çè 4 4 ø 4 1 1 x cos x 1 2 cos2 2x dx =+ ò dx + ò cos 2x dx + 4 16 4 32 ò =-

=-

x cos4 x x sin 2x 1 + + + (1 + cos 4x) dx + C 4 16 8 32 ò

=-

1 x cos4 x x sin 2x 1 + + + dx + cos 4x dx 4 16 8 32 ò 32 ò

=-

x cos4 x 3 x sin 2x sin 4x + + + + C. 4 32 8 128

ò sin (log x) dx.

EXAMPLE 19

Evaluate

SOLUTION

Put log x = t so that x = et and \

ò sin (log x) dx = ò e

t

1 dx = dt or dx = et dt. x

sin t dt

… (i)

Now, ò et sin t dt = et ( - cos t) - ò et × ( - cos t) dt [integrating by parts] = - et cos t + ò et cos t dt

= - et cos t + [et sin t - ò et sin t dt]

[integrating et cos t by parts]

= - et cos t + et sin t - ò et sin t dt. \

2ò et sin t dt = - et cos t + et sin t

1 sin t dt = ( - et cos t + et sin t) + C. 2 Putting this value in (i), we get

or

òe

t

ò sin (log x) dx = ò e

t

sin t dt

1 = ( - et cos t + et sin t) + C 2 1 = [- x cos (log x) + x sin (log x)] + C 2 1 1 = - x cos (log x) + x sin (log x) + C. 2 2

ò sin

EXAMPLE 20

Evaluate

SOLUTION

Put

x = t so that

\

ò sin

x dx. 1 dx = dt or dx = 2t dt. 2 x

x dx = 2ò t sin t dt = 2[t ( - cos t) - ò 1 × ( - cos t) dt] [integrating t sin t by parts]

Senior Secondary School Mathematics for Class 12 Pg-671

Methods of Integration

671

= -2t cos t + 2 sin t + C = -2 x cos x + 2 sin x + C. EXAMPLE 21 SOLUTION

ò sec x dx. 3 2 ò sec x dx = ò sec x × sec x dx = sec x × (tan x) - ò sec x tan x(tan x) dx 3

Evaluate

= sec x tan x - ò sec x( sec2 x - 1) dx

[CBSE 2003]

[integrating by parts]

= sec x tan x - ò sec 3 x dx + ò sec x dx æp xö \ 2ò sec 3 x dx = sec x tan x + log tan ç + ÷ + C è 4 2ø

ò sec

or EXAMPLE 22

Evaluate

SOLUTION

Put \

3

x dx =

ò tan

1 1 æp xö sec x tan x + log tan ç + ÷ + C ¢. 2 2 è 4 2ø

-1

x = t so that

ò tan

-1

x dx. 1 dx = dt or dx = 2t dt. 2 x

x dx = 2ò t(tan -1t) dt

é ì 1 t2 t2 ü ù = 2 ê(tan -1t) × - ò í × ýdt ú + C 2 2 êë î(1 + t ) 2 þ úû t2 = t 2(tan -1t) - ò dt + C (1 + t 2) = t 2(tan -1t) - ò

[(1 + t 2) - 1]

dt + C (1 + t 2) 1 = t 2(tan -1t) - ò dt + ò dt + C (1 + t 2) = t 2(tan -1t) - t + tan -1t + C = (t 2 + 1) tan -1t - t + C = ( x + 1) tan -1 x - x + C. tan -1 x

ò (1 + x) 2 dx.

EXAMPLE 23

Evaluate

SOLUTION

Integrating by parts, taking tan -1 x as the first function and as the second function, we get ( -1) 1 ( -1) I = tan -1 x × -ò × dx 2 (1 + x) (1 + x ) (1 + x) =

[CBSE 2003]

1 (1 + x) 2

- tan -1 x 1 ì 1 - tan -1 x dx (1 - x) ü + × òí +ò = + ý 2 + + (1 + x) ( ) ( 1 x 2 1 x) (1 + x)(1 + x ) (1 + x 2) þ î [by partial fractions]

Senior Secondary School Mathematics for Class 12 Pg-672

672

Senior Secondary School Mathematics for Class 12

=

1 1 - tan -1 x 1 + log |1 + x | + tan -1 x - log (1 + x 2) + C. 2 2 4 (1 + x) ì sin -1 x - cos-1 x ü ýdx. x + cos-1 x þ î

ò í sin -1

EXAMPLE 24

Evaluate

SOLUTION

We have ì -1 æp öü x - ç - sin -1 x ÷ ý ísin p è2 øþ î I=ò dx [Q sin -1 x + cos-1 x = ] 2 æ pö ç ÷ è 2ø pö 2 æ = ò ç 2 sin -1 x - ÷ dx p è 2ø 4 4 -1 … (i) x dx - ò dx = ò sin -1 x dx - x + C. = ò sin p p Putting x = sin 2 q and dx = 2 sin q cos q dq = sin 2q dq, we get

ò sin

-1

[CBSE 2014C]

x dx = ò q sin 2 q dq I

II

( - cos 2 q) æ - cos 2 q ö = qç dq ÷ - ò1× 2 2 ø è [integrating by parts] q 1 = - cos 2q + ò cos 2q dq 2 2 1 1 = - q cos 2q + sin 2q 2 4 1 1 2 = - q(1 - 2 sin q) + sin q 1 - sin 2 q 2 2 1 1 -1 … (ii) = - sin x (1 - 2x) + x × 1 - x. 2 2 From (i) and (ii), we get 4 ì 1 1 ü I = × í- sin -1 x (1 - 2x) + x 1 - x ý- x + C p î 2 2 þ -2 2 -1 \ I= sin x (1 - 2x) + x 1 - x - x + C. p p EXAMPLE 25

Evaluate

SOLUTION

We have I=ò =ò

ò

x 2 + 1{log ( x 2 + 1) - 2 log x} x4

æ x2 + 1ö × log ç 2 ÷ dx ç x ÷ x è ø 1 1 1 ö æ 1 + 2 × 3 log ç1 + 2 ÷ dx è x x x ø

dx.

[CBSE 2012, ’14C]

x2 + 1 4

… (i)

Senior Secondary School Mathematics for Class 12 Pg-673

Methods of Integration

673

1 ö -2 1 1 æ Putting ç1 + 2 ÷ = t and 3 dx = dt, i.e., 3 dx = - dt in (i), we get 2 è x x x ø 1 I = - ò (log t) t dt 2 II I 1ì 2 2 æ1 ö ü = - í (log t) t 3/2 - ò ç ´ t 3/2 ÷ dt ý 2î 3 3 èt ø þ 1 3/2 1 1/2 = - t (log t) + ò t dt 3 3 1 3/2 2 3/2 = - t (log t) + t +C 3 9 1 2ü ì = - t 3/2ílog t - ý + C 3 3þ î =-

EXAMPLE 26

Evaluate

SOLUTION

Putting

1æ 1 ö ç1 + 2 ÷ 3è x ø

3/2

1 - sin x

1 ö 2ü ì æ ílog ç1 + 2 ÷ - ý + C. è x ø 3þ î

ò (1 + cos x) e

- x /2

dx.

[CBSE 2013C]

-x = t , we get x = -2t and dx = -2dt. 2 1 - sin x - x /2 I=ò e dx (1 + cos x)

\

=ò

1 - sin ( -2t) t 1 + sin 2t t e ( -2dt) = -2ò e dt {1 + cos ( -2t)} (1 + cos 2t)

= -2ò

cos2t + sin 2t + 2 sin t cos t

et dt 2 cos2t (cos t + sin t) t = -2ò e dt = - ò (sec t + sec t tan t) et dt 2 cos2t = - ò et { f (t) + f ¢(t)} dt , where f (t) = sec t

x æ -x ö = - et f (t) + C = - e- x /2 sec ç ÷ + C = - e- x /2 sec + C. 2 è 2 ø Integrals of the form

òe

THEOREM 1 PROOF

òe

x

x

òe

x

[ f ( x) + f ¢( x)] dx

{ f ( x) + f ¢( x)} dx = ex × f ( x) + C.

{ f ( x) + f ¢( x)} dx = ò ex × f ( x) dx + ò ex × f ¢( x) dx = f ( x) × ò ex dx - ò { f ¢( x) × ò ex dx} dx + ò ex f ¢( x) dx + C [evaluating the first integral by parts]

Senior Secondary School Mathematics for Class 12 Pg-674

674

Senior Secondary School Mathematics for Class 12

= ex f ( x) - ò ex f ¢( x) dx + ò ex f ¢( x) dx + C = ex f ( x) + C. \ EXAMPLE 1

òe

x

{ f ( x) + f ¢( x)} dx = ex f ( x) + C.

Evaluate: 1 ö ç - 2 ÷ dx èx x ø 1 ïü ïì (iii) ò ex ísin -1 x + ý dx ïî 1 - x 2 ïþ (i)

SOLUTION

òe

xæ1

(ii)

òe

(iv)

òe

We have ì1 æ 1 ö ü (i) I = ò ex í + ç - 2 ÷ ý dx îx è x ø þ = ò ex { f ( x) + f ¢( x)} dx , where f ( x) = = ex × f ( x) + C = ex ×

xæ

1 2 ö ç 2 - 3 ÷ dx èx x ø

x

(tan x + log sec x) dx

1 -1 and f ¢( x) = 2 x x

1 ex +C = + C. x x

ì 1 æ -2 ö ü (ii) I = ò ex í 2 + ç 3 ÷ ý dx è x øþ îx 1 -2 = ò ex { f ( x) + f ¢( x)} dx , where f ( x) = 2 and f ¢( x) = 3 x x 1 ex = ex × f ( x) + C = ex × 2 + C = 2 + C. x x ì -1 1 üï xï (iii) I = ò e ísin x + ý dx ïî 1 - x 2 ïþ 1 = ò ex { f ( x) + f ¢( x)} dx , where f ( x) = sin -1 x and f ¢( x) = 1 - x2 x x -1 x -1 = e × f ( x) + C = e × sin x + C = e sin x + C. (iv) I = ò ex (tan x + log sec x) dx = ò ex { f ( x) + f ¢( x)} dx , where f ( x) = log ( sec x) and f ¢( x) =

1 × sec x tan x = tan x sec x

= ex f ( x) + C = ex log ( sec x) + C. x ex

ò (1 + x) 2 dx.

EXAMPLE 2

Evaluate

SOLUTION

We have ì(1 + x) - 1 ü ì x ü I = ò ex × í dx = ò ex × í dx 2ý 2 ý î (1 + x) þ î(1 + x) þ

Senior Secondary School Mathematics for Class 12 Pg-675

Methods of Integration

675

ì (1 + x) ì 1 1 ü 1 ü dx = ò ex × í = ò ex × í ý dx 2 2ý + x ( 1 ) + x + x + x) 2 þ ( 1 ) ( 1 ) ( 1 î þ î 1 -1 and f ¢( x) = = ò ex × { f ( x) + f ¢( x)} dx , where f ( x) = 1+ x (1 + x) 2 1 ex = ex × f ( x) + C = ex × +C = + C. (1 + x) (1 + x)

òe

x æ 1 - sin

xö çç ÷ dx. x ÷ø 1 cos è

EXAMPLE 3

Evaluate

SOLUTION

ì æ 1 - sin x ö 1 sin x ü ÷÷ dx = ò ex × í We have I = ò ex × çç ý dx î(1 - cos x) (1 - cos x) þ è 1 - cos x ø ì 1 2 sin ( x/2) cos ( x/2) ü = ò ex × í ý dx 2 x 2 2 2 sin 2( x/2) sin ( / ) î þ xü x ì1 2x = ò e × í cosec - cot ý dx 2 2þ î2 x 1 xü ì = ò ex × í- cot + cosec2 ý dx 2 2 2þ î = ò ex × { f ( x) + f ¢( x)} dx ,

x 1 x and f ¢( x) = cosec2 2 2 2 xö x x xæ x = e × f ( x) + C = e ç - cot ÷ + C = - e cot + C. 2ø 2 è where f ( x) = - cot

òe

x

ì 2 + sin 2x ü ×í ý dx. î 1 + cos 2x þ

EXAMPLE 4

Evaluate

SOLUTION

We have ì 2 + sin 2x ü I = ò ex × í ý dx î 1 + cos 2x þ ì 2 sin 2x ü = ò ex × í + ý dx ( 1 cos 2 ) ( 1 cos 2x) þ + x + î ì 2 2 sin x cos x ü x 2 = ò ex × í + ýdx = ò e × {sec x + tan x} dx 2 2 cos2 x þ î 2 cos x = ò ex × {tan x + sec2 x} dx

= ò ex × { f ( x) + f ¢( x)} dx , where f ( x) = tan x and f ¢( x) = sec2 x = ex × f ( x) + C = ex tan x + C.

EXAMPLE 5

Evaluate

ò

( x 2 + 1) ex ( x + 1) 2

dx.

[CBSE 2005C, ‘06]

Senior Secondary School Mathematics for Class 12 Pg-676

676

Senior Secondary School Mathematics for Class 12

SOLUTION

We have ì( x + 1) 2 - 2x ü dx = ò ex × í ý dx 2 ( x + 1) î ( x + 1) þ ì ü 2 x x = ò ex × í1 dx = ò ex dx - 2ò ex × dx 2ý ( x + 1) 2 î ( x + 1) þ

I = ò ex ×

( x 2 + 1) 2

= ex - 2 × ò ex ×

{( x + 1) - 1} ( x + 1)

2

ì 1 1 ü dx = ex - 2 × ò ex × í dx 2ý î( x + 1) ( x + 1) þ

= e - 2 × ò e × { f ( x) + f ¢( x)} dx , x

x

where f ( x) =

1 -1 and f ¢( x) = ( x + 1) ( x + 1) 2

1 +C ( x + 1) ì 2 ü x æ x -1ö = ex × í1 ÷ + C. ý+ C = e ç ( x 1 ) + è x + 1ø þ î æ sin 4x - 2 ö ÷÷ dx. Evaluate ò e2x çç è 1 - cos 4x ø = ex - 2ex × f ( x) + C = ex - 2ex ×

EXAMPLE 6

SOLUTION

1 dt, we get 2 æ 2 sin t cos t - 2 ö 1 t æ sin 2t - 2 ö 1 ÷ dt ÷÷ dt = ò et ç e çç ò ç ÷ 2 2 2 sin 2t è 1 - cos 2t ø è ø 1 t ì sin t cos t - 1 ü 1 t 2 eí ý dt = ò e (cot t - cosec t) dt 2ò î 2 sin 2t þ 1 t e { f (t) + f ¢(t)} dt , where f (t) = cot t 2ò 1 t 1 1 e × f (t) + C = et cot t + C = e2x cot 2x + C. 2 2 2

Putting 2x = t and dx = I= = = =

Integrals of the form THEOREM 2 PROOF

òe

[CBSE 2007]

òe kx

kx

òe

kx

× { k × f ( x) + f ¢( x)} dx

{ k × f ( x) + f ¢( x)} dx = ekx × f ( x) + C.

× { k × f ( x) + f ¢( x)} dx = k × ò ekx f ( x) dx + ò ekx f ¢( x) dx

é ekx ekx ù = k × ê f ( x) × - ò f ¢( x) × dx ú + ò ekx f ¢( x) dx + C k k ë û [evaluating the first integral by parts] = ekx × f ( x) - ò ekx × f ¢( x) dx + ò ekx f ¢( x) dx + C = ekx × f ( x) + C.

\

òe

kx

× {k × f ( x) + f ¢( x)} dx = ekx × f ( x) + C.

Senior Secondary School Mathematics for Class 12 Pg-677

Methods of Integration EXAMPLE 7

Evaluate

SOLUTION

We have

òe

2x

677

× ( - sin x + 2 cos x) dx.

I = ò e2x × { 2 cos x - sin x} dx = 2ò e2x cos x dx - ò e2x sin x dx

é e2x e2x ù = 2 × ê cos x × - ò ( - sin x) × dx ú - ò e2x sin x dx 2 2 ë û [integrating e2x cos x by parts]

= e2x cos x + ò e2x sin x dx - ò e2x sin x dx + C = e2x cos x + C. Integrals of the form e ax cos ( bx + c) and e ax sin ( bx + c)

òe

ax

cos ( bx + c) dx.

EXAMPLE 8

Evaluate

SOLUTION

Integrating by parts, taking eax as the second function, we get ì eax ü eax ax ò e cos ( bx + c) dx = cos ( bx + c) × a - ò í-b sin ( bx + c) × a ý dx î þ ax e b = cos ( bx + c) + ò eax sin ( bx + c) dx a a ì eax üù eax bé eax = × cos ( bx + c) + ê sin ( bx + c) × - ò íb cos ( bx + c) × ýú dx + C a þúû a a êë a î [integrating eax sin ( bx + c) by parts]

[CBSE 2002]

eax b b2 × cos ( bx + c) + 2 eax sin ( bx + c) - 2 ò eax cos ( bx + c) dx + C a a a 2ö ax æ b e b cos ( bx + c) + 2 eax sin ( bx + c) + C \ ç1 + 2 ÷ ò eax cos ( bx + c) dx = ç ÷ a a ø a è é a cos ( bx + c) + b sin ( bx + c) ù or ò eax cos ( bx + c) dx = eax ê ú + C ¢. ( a 2 + b2) ë û =

REMARK

Put a = r cos q and b = r sin q so that

\

æ bö r = a 2 + b 2 and q = tan -1 ç ÷ × è aø ax r e cos ( bx + c - q) ax ò e cos ( bx + c) dx = 2 ( a + b2) = eax ×

Similarly,

òe

ax

sin ( bx + c) dx = eax ×

cos [bx + c - tan -1( b/a)] a 2 + b2

sin [bx + c - tan -1( b/a)] a 2 + b2

× ×

Senior Secondary School Mathematics for Class 12 Pg-678

678

Senior Secondary School Mathematics for Class 12

EXERCISE 13C Evaluate the following integrals: 1. 3. 5. 7. 9. 11. 13. 15.

ò x e dx 2x ò x e dx ò x cos 2x dx 2 ò x cosec x dx 2 ò x sin x dx 2 x ò x e dx 2 3x ò x e dx 3 ò x log 2x dx x

log x

2. 4. 6. 8. 10. 12. 14. 16.

ò x cos x dx ò x sin 3 x dx ò x log 2x dx 2 ò x cos x dx 2 ò x tan x dx 2 3 ò x cos x dx 2 2 ò x sin x dx ò x × log ( x + 1) dx 3 x2

17.

ò

dx

18.

ò 2x

19.

ò x sin x dx 3 2 ò x cos x dx ò x sin x cos x dx 3 ò cosec x dx

20.

ò x cos x dx ò sin x log (cos x) dx ò cos x dx 3 ò x sin x cos x dx

27.

ò sin x log (cos x) dx

28.

29.

ò log ( 2 + x

31.

ò í log x - (log x) 2 ý dx

32.

òe

35.

ò

38.

ò x sin x dx [CBSE 2000C, ‘09] -1 ò cot x dx 2 -1 ò x cot x dx [CBSE 2006C] -1 ò cos x dx

21. 23. 25.

40. 42. 44. 46.

x

n 3

2

24. 26.

1

ü þ

-x

1-x

2

dx [CBSE 2012]

-1

-1 æ ç1- x

ò cos

ö ÷ dx ç 1 + x2 ÷ è ø

dx

3

ò

[CBSE 2005]

cos 2x cos 4x dx

x sin -1 x

[CBSE 2008C]

log (log x) dx x x 30. ò dx (1 + sin x)

) dx [CBSE 2001]

ì 1 î

22.

e

[CBSE 2007]

33.

òe

36.

ò

x

dx

x 2 tan -1 x (1 + x 2) 39. 41. 43. 45.

2

47.

dx

34.

òe

37.

ò

sin x

log ( x + 2) ( x + 2) 2

-1

ò x cos x dx -1 ò x cot x dx -1 ò sin x dx -1 3 ò cos ( 4x - 3 x) dx ò tan

-1 æ

sin 2x dx

2x ö ç ÷ dx è 1 - x2 ø

dx

Senior Secondary School Mathematics for Class 12 Pg-679

Methods of Integration -1 æ ç

3 x - x 3 ö÷ dx ç 1 - 3x2 ÷ è ø

48.

ò tan

50.

ò (1 - tan 2x)

52.

ò e sin x dx 2x ò e cos ( 3 x + 4) dx x ò e (sin x + cos x) dx

54. 56.

tan x sec2 x

dx

2x

sin -1 x

49.

ò

51.

òe

53.

ò e sin x cos x dx -x ò e cos x dx x 2 ò e (cot x - cosec x) dx

55. 57.

x2

òe

x

sec x(1 + tan x) dx [CBSE 2012] 59.

òe

60.

òe

x

(cot x + log sin x) dx

61.

òe

62.

òe

x

[sec x + log ( sec x + tan x)] dx 63.

òe

xæ ç sin

x cos x - 1 ö÷ dx ç ÷ sin 2 x è ø æ 2 - sin 2x ö ÷÷ dx 66. ò ex çç è 1 - cos 2x ø

òe

4x - 4 ö çç ÷÷ dx [CBSE 2010] è 1 - cos 4x ø æ 1 + x log x ö 70. ò ex ç ÷ dx x ø è

68.

òe

x æ sin

( x - 1)

72.

òe

x

74.

òe

x

76.

ò ( x + 2) 2 e

dx [CBSE 2006]

( x + 1) 3 ×

( x - 3) ( x - 1)

( x + 1)

3

x

dx

dx

- 1ö ç ÷ dx è 4x 2 ø log x 80. ò dx (1 + log x) 2 78.

82.

òe

2 x æ 2x

ì 1

1

ü

ò í log x - (log x) 2 ý dx [CBSE 2005]

î þ æ sin -1 x - cos-1 x ö ÷ dx 84. ò ç ç sin -1 x + cos-1 x ÷ è ø

3x

dx

[CBSE 2002C, ‘04]

sin 4x dx

2x

58.

64.

679

65.

òe

67.

òe

69.

ò

xæ

ç tan è

x

-1

x+

1 ö ÷ dx 1 + x2 ø

(tan x - log cos x) dx [CBSE 2000]

xæ ç1 +

sin x cos x ö÷ dx ÷ cos2 x ø

x æ cos

x + sin x ö ÷ dx ø cos2 x

ç è

ç è

x æ1 +

sin x ö çç ÷ dx 1 + cos x ÷ø è

ex [ 1 - x 2 sin -1 x + 1] 1 - x2

dx

x dx (1 + x) 2 ( 2 - x) 73. ò ex dx (1 - x) 2 71.

òe

75.

òe

77.

ò (1 + 2x) 2 dx

79.

òe

81.

ò {sin (log x) + cos (log x)} dx

83.

ò ílog (log x) + (log x) 2 ý dx

85.

x

×

3x - 1ö ç ÷ dx è 9x 2 ø

3x æ

x e2x

xæ

1 ö ç log x + 2 ÷ dx è x ø

ì

1

î

ü þ

x 55

ò5

5x

×5

× 5 x dx

Senior Secondary School Mathematics for Class 12 Pg-680

680

86.

Senior Secondary School Mathematics for Class 12

òe

sin 2x ö ÷÷ dx [CBSE 2010] 87. çç è 1 + cos 2x ø

2x æ 1 +

òe

2x ö ÷÷ dx çç è 1 - cos 2x ø

2 x æ 1 - sin

[CBSE 2013C]

ANSWERS (EXERCISE 13C)

1. ex ( x - 1) + C 1 1 3. x e2x - e2x + C 2 4 1 1 5. x sin 2x + cos 2x + C 2 4 7. - x cot x + log |sin x| + C 9.

2. x sin x + cos x + C - x cos 3 x sin 3 x 4. + +C 3 9 1 1 6. x 2 log 2x - x 2 + C 2 4 2 8. x sin x + 2x cos x - 2 sin x + C

1 x2 1 - x sin 2x - cos 2x + C 4 4 8

11. ex ( x 2 - 2x + 2) + C

12.

æ x 2 2x 2 ö÷ 13. e 3x ç + +C ç 3 9 27 ÷ø è 15. 17. 19. 20. 21. 23. 25. 26. 27. 29.

10. x tan x + log |cos x| -

14.

x2 +C 2

2x sin 3 x cos 3 x + ( 4 + 3 x 2) 9 36

1 3 1 2 1 1 x - x sin 2x - x cos 2x + sin 2x + C 6 4 4 8

x 4 log 2x x 4 1 1 1 16. ( x 2 - 1) log ( x + 1) - x 2 + x + C +C 4 16 2 4 2 1- n 1- n 2 x x 18. ex ( x 2 - 1) + C × log |x| +C (1 - n) (1 - n) 2 -3 x cos x 3 sin x x cos 3 x sin 3 x + + +C 4 4 12 36 1 1 3 3 x sin 3 x + cos 3 x + x sin x + cos x + C 12 36 4 4 1 2 1 2 2 22. - cos x log (cos x) + cos x + C x sin x + cos x + C 2 2 1 1 24. 2[ x sin x + cos x ] + C - x cos 2x + sin 2x + C 4 8 1 1 x - cosec x cot x + log tan + C 2 2 2 1 3 1 1 x sin 4 x x + sin 2x sin 4x + C 4 32 16 128 28. (log x)[log (log x) - 1] + C cos x(1 - log cos x) + C 2 -1 æ x ö x log ( x + 2) - 2x + 2 2 tan ç ÷ +C è 2ø

æp xö ½ æ p x ö½ 30. - x tan ç - ÷ + 2 log ½cos ç - ÷½ + C è 4 2ø ½ è 4 2 ø½ 31.

x +C (log x)

1 é1 ù 32. e- x ê ( 6 sin 6x - cos 6x) + ( 2 sin 2x - cos 2x) ú + C 10 ë 74 û

Senior Secondary School Mathematics for Class 12 Pg-681

Methods of Integration

33. 2e

x

( x - 1) + C

35. - 1 - x 2 sin -1 x + x + C

34. 2esin x (sin x - 1) + C

1 1 36. x tan -1 x - log (1 + x 2) - (tan -1 x) 2 + C 2 2 38. 39. 41. 42. 43. 44.

681

37.

-1 log ( x + 2) +C ( x + 2) ( x + 2)

1 2 1 1 x sin -1 x + x 1 - x 2 - sin -1 x + C 2 4 4 1 2 1 1 1 40. x cot -1 x + log (1 + x 2) + C x cos-1 x - x 1 - x 2 + sin -1 x + C 2 4 4 2 1 2 1 -1 -1 x (cot x) + ( x - tan x) + C 2 2 1 3 1 2 1 -1 x cot x + x - log|x 2 + 1| + C 3 6 6 1 1 -1 ( 2x - 1) sin x+ x(1 - x) + C 2 2 1 1 ( 2x - 1) cos-1 x x(1 - x) + C 45. 3 x cos-1 x - 3 1 - x 2 + C 2 2

46. 2x tan -1 x - log (1 + x 2) + C 48. 3 x tan -1 x -

47. 2x tan -1 x - log (1 + x 2) + C

3 log (1 + x 2) + C 2

49.

½1 - sin -1 x 1 - x 2½ + log ½ ½+ C x x x ½ ½

e 3x × ( 3 sin 4x - 4 cos 4x) + C 25 1 53. × e2x (sin 2x - cos 2x) + C 8

1 50. - log|1 - tan 2 x| + C 2 1 2x 52. × e ( 2 sin x - cos x) + C 5

51.

e2x × {2 cos ( 3 x + 4) + 3 sin ( 3 x + 4)} + C 13 1 56. ex sin x + C 55. e- x (sin x - cos x) + C 2

54.

57. ex cot x + C

58. ex sec x + C

59. ex tan -1 x + C

60. ex log (sin x) + C

61. ex log ( sec x) + C

62. ex log ( sec x + tan x) + C

63. ex tan x + C

65. ex sec x + C

66. - ex cot x + C

67. ex tan

69. ex sin -1 x + C

70. ex log x + C

71.

ex +C (1 + x)

72.

75.

e 3x +C 9x

76.

73.

ex +C (1 - x)

74.

77.

e2x +C 4(1 + 2x)

78.

ex ( x - 1) 2 e2x +C 4x

+C

x +C 2

64. ex cot x + C 68. ex cot 2x + C

1ö æ 79. ex ç log x - ÷ + C xø è

ex ( x + 1) 2

+C

ex +C ( x + 2)

Senior Secondary School Mathematics for Class 12 Pg-682

682

Senior Secondary School Mathematics for Class 12

x +C (1 + log x) x 82. +C (log x) 80.

81. e

log x

sin (log x) + C

ì 1 ü 83. x × ílog (log x) ý+ C log xþ î 5x

2 84. { x - x 2 - (1 - 2x) sin -1 x } - x + C p 86.

55

+C (log 5) 3 1 87. e2x tan x + C 2

85.

1 2x e tan x + C 2

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 13C) 9. Write sin 2 x =

1 ( 1 - cos 2 x ). 2

10. tan 2 x = ( sec2 x - 1).

15. Integrate by parts, taking (log 2x ) as the 2nd function. 16. Integrate by parts, taking x as the second function. 3 1 19. sin 3 x = sin x - sin 3 x. 18. Put x 2 = t. 4 4 1 3 22. Put cos x = t. 21. Put x 2 = t. 20. cos 3 x = cos 3 x + cos x. 4 4 1 24. Put x = t and dx = 2 t dt. 23. I = ò x sin 2 x dx. 2 25. I = ò cosec x ( cosec2 x ) dx. Now, integrate by parts. 26. Take (sin 3 x cos x ) as 2nd function. Its integral is

1 sin 4 x. 4

2

æ 1 - cos 2 x ö ÷ × Further, use sin 4 x = ç 2 è ø 29. Take log ( 2 + x 2 ) as 1st function.

27. Put cos x = t.

1 x æp xö dx = ò x sec2 ç - ÷ dx æp ö 2 è 4 2ø 1 + cos ç - x ÷ è2 ø 1 31. Integrate by parts, taking 1 as 2nd function. (log x )

30. I = ò

32. cos 2 x cos 4 x =

1 (cos 6 x + cos 2 x ). 2

33. Put

x = t and dx = 2t dt.

34. I = 2 ò e sin x sin x cos x dx = 2 ò t e tdt , where sin x = t. 35. Put x = sin t and dx = cos t dt. 36. Put x = tan t , dx = sec2t dt and then write tan 2t = ( sec2t - 1). 1 as the 2nd 37. Integrate by parts, taking log ( x + 2 ) as the 1st function and ( x + 2)2 function. 38. Integrate by parts, taking sin -1 x as the 1st function and x as 2nd.

Senior Secondary School Mathematics for Class 12 Pg-683

Methods of Integration

I = (sin -1 x ) ×

x2 x2 1 1 1 ( 1 - x2 ) - 1 dx × dx = x 2 (sin -1 x ) + × ò 2 ò 1 - x2 2 2 2 1 - x2

=

dx 1 2 1 1 x (sin -1 x ) + × ò 1 - x 2 dx - ò 2 2 2 1 - x2

=

2 ù é 1 2 1 x 1- x 1 1 x (sin -1 x ) + × ê + sin -1 x ú - sin -1 x + C. ú 2 2 2 ê 2 2 û ë

x = sin t . Then, x = sin 2t and dx = sin 2t dt. 1 1 I = ò t sin 2t dt = - t cos 2t + sin 2t + C 2 4 1 1 1 1 = t ( 2 sin 2t - 1) + sin t cos t = sin -1 x ( 2 x - 1) + x( 1 - x ) + C. 2 2 2 2

43. Put

45. Put x = cos t and dx = - sin t dt . Use ( 4 cos 3t - 3 cos t ) = cos 3t . æ 1 - tan 2t ö ÷ = cos 2t 46. Put x = tan t and dx = sec2t dt. Use ç ç 1 + tan 2t ÷ è ø 47. Put x = tan t and dx = sec2t dt. æ 3 tan t - tan 3t ö ÷ = tan 3t. 48. Put x = tan t and dx = sec2t dt. Use ç ç 1 - 3 tan 2t ÷ è ø 49. Put x = sin t and dx = cos t dt. Then, I = ò t( cosec t cot t ) dt = -t cosec t + log|cosec t - cot t| + C =

½1 - sin -1 x + log½ ½x x ½

1 - x 2½ ½+ C. x ½ ½

50. Put tan x = t and sec2 x dx = dt. 51. Integrating by parts, taking e 3 x as 2nd function, we get e3x e3x 1 4 I = (sin 4 x ) × - ò 4 cos 4 x × dx = e 3 x(sin 4 x ) - × ò e 3 x cos 4x dx 3 3 3 3 e3x e3x ù 1 3x 4 é = e (sin 4 x ) - × ê(cos 4 x ) × - ò ( - 4 sin 4 x ) dx ú 3 3 êë 3 3 úû 1 3x 4 16 e (sin 4 x ) - e 3 x(cos 4 x ) I 3 9 9 16 ö 1 4 æ Û ç1+ ÷ I = e 3 x(sin 4 x ) - e 3 x(cos 4 x ). 9 ø 3 9 è 1 53. I = × ò e 2 x sin 2 x dx. 2 =

61. I = ò e x(log sec x + tan x ) dx = ò e x{ f ( x ) + f ¢( x ) } dx , where f ( x ) = log sec x. 62. I = ò e x[log ( sec x + tan x ) + sec x] dx = ò e x{ f ( x ) + f ¢( x )} dx , where f ( x ) = log ( sec x + tan x ).

63. I = ò e x( sec2 x + tan x ) dx = ò e x(tan x + sec2 x ) dx = e x tan x + C . 64. I = ò e x(cot x - cosec2 x ) dx = e x cot x + C .

683

Senior Secondary School Mathematics for Class 12 Pg-684

684

Senior Secondary School Mathematics for Class 12

65. I = ò e x ( sec x + sec x tan x ) dx = e x sec x + C . 2 sin x cos x ïü ïì 2 x 2 x 66. I = ò e x í ý dx = ò e ( - cot x + cosec x ) dx = - e cot x + C . ïî 2 sin 2 x 2 sin 2 x ïþ 2 sin ( x/ 2 ) cos ( x/ 2 ) üï 1 ïì 67. I = ò e x í + ý dx ïî 2 cos 2 ( x/ 2 ) ïþ 2 cos 2 ( x/ 2 ) x x x xü x 1 1 ì ü ì = ò e x × í sec2 + tan ý dx = ò e x í tan + sec2 ý dx = e x tan + C . 2 2 2þ 2 2 2þ î2 î üï ìï 2 sin 2 x cos 2 x 4 x 2 x 68. I = ò e x í ý dx = ò e {cot 2 x - 2 cosec 2 x } dx = e cot 2 x + C . 2 ïî 2 sin 2 2 x 2 sin 2 x ïþ ïì 69 I = ò e x í sin -1 x + ïî

ïü x -1 ý dx = e sin x + C . ï 1- x þ 1

2

1ö æ1 ö æ 70. I = ò e x ç + log x ÷ = ò e x ç log x + ÷ dx = e x log x + C . xø èx ø è ù é 1 1 dx = ò e x ê ú dx 2 ( 1 + x) êë ( 1 + x ) ( 1 + x ) úû 1 = ò e x { f ( x ) + f ¢( x ) } dx, where f ( x ) = × ( 1 + x)

71. I = ò e x

72. I = ò

( 1 + x - 1) 2

e x( x + 1 - 2 ) ( x + 1) 3

2 ïì ( x + 1) ïü dx = ò e x í ý dx ïî ( x + 1) 3 ( x + 1) 3 ïþ

üï ìï 1 2 1 = ò exí dx = ò e x × { f ( x ) + f ¢( x ) } dx , where f ( x ) = × 2 3ý ( x + 1) þï ( x + 1) 2 îï ( x + 1) 73. I = ò e x

( 1 - x + 1) ( 1 - x)2

1 ïü ïì ( 1 - x ) dx = ò e x í + ý dx ïî ( 1 - x ) 2 ( 1 - x ) 2 ïþ

ìï 1 1 üï 1 = ò exí + dx = ò e x × { f ( x ) + f ¢( x ) } dx , where f ( x ) = × 2ý ( 1 - x) îï ( 1 - x ) ( 1 - x ) þï 74. I = ò

e x ( x - 1 - 2) ( x - 1) 3

üï 2 ïì ( x - 1) dx = ò e x í ý dx ïî ( x - 1) 3 ( x - 1) 3 ïþ

üï ìï 1 2 1 = ò exí dx = ò e x × { f ( x ) + f ¢( x ) } dx , where f ( x ) = × 2 3ý ïî ( x - 1) ( x - 1) ïþ ( x - 1) 2 1 dt , we get 3 e3x 1 1 1 1 æt - 1ö æ1 1 ö I = × ò e t ç 2 ÷ dt = × ò e t ç - 2 ÷ dt = e t × + C = + C. t 9x 3 3 3 è t ø èt t ø

75. Putting 3 x = t and dx =

ì 1 1 1 ïü ïì ( x + 2 ) - 1 ïü x ï x 76. I = ò e x × í + C. ý dx = ò e × í ý dx = e × ( x + 2) ïî ( x + 2 ) 2 ïþ ïî ( x + 2 ) ( x + 2 ) 2 ïþ

Senior Secondary School Mathematics for Class 12 Pg-685

Methods of Integration 1 dt , we get 2 2x × e 2 x 1 1 ( 1 + t - 1) t te t 1 I = ×ò dx = × ò dt = × ò e dt 2 4 4 ( 1 + t )2 2 ( 1 + 2x) ( 1 + t )2

77. Putting 2 x = t and dx =

=

1 t ìï 1 1 üï et e2x +C= × e ×í + C. ý dt = 2 4 4( 1 + t ) 4 ( 1 + 2x ) îï 1 + t ( 1 + t ) þï

78. Putting 2 x = t and dx = I=

1 dt , we get 2

1 1 1 1 1 e2x æt - 1ö ì1 1 ü × ò e t × ç 2 ÷ dt = × ò e t × í - 2 ý dt = e t × + C = × + C. t 2 2 2 4 x è t ø ît t þ

1 1 1ü 1ü 1 ö ì æ1 ì 79. I = ò e xí(log x ) + - + 2 ý dx = ò e xí log x + ý dx - ò e x ç - 2 ÷ dx x x x þ xþ î î èx x ø = e x(log x ) - e x ×

1 1ö æ + C = e x ç log x - ÷ + C . x xø è

80. Putting log x = t , x = e t and dx = e tdt , we get I=ò

t ( 1 + t )2

= et

× e tdt = ò e t ×

( 1 + t - 1) ( 1 + t )2

1 üï ïì 1 dt = ò e t × í ý dt ïî ( 1 + t ) ( 1 + t ) 2 ïþ

x 1 +C= + C. (1 + t) ( 1 + log x )

81. Putting log x = t , x = e t and dx = e tdt , we get I = ò e t(sin t + cos t ) dt = e t sin t + C = e log x × sin (log x ) + C . 82. Putting log x = t , x = e t and dx = e tdt , we get x 1 æ1 1 ö I = ò e t ç - 2 ÷ dt = e t × + C = + C. t log x èt t ø 83. Putting log x = t , x = e t and dx = e tdt , we get 1ü 1 1 1ü ì ì I = ò í log t + 2 ý e tdt = ò e tí log t + - + 2 ý dt t t t þ î î t þ 1ö 1 æ1 1 ö æ = ò e t ç log t + ÷ dt - ò e t ç - 2 ÷ dt = e t log t - e t × + C tø t è èt t ø ì 1 ü = x í log (log x ) ý + C. log x þ î 84. I = ò

æp ö sin -1 x - ç - sin -1 x ÷ 2 æ pö è2 ø dx = × ò ç 2 sin -1 x - ÷ dx ( p/ 2 ) p è 2ø

4 4 sin -1 x dx - ò dx = × ò sin -1 x dx - x + C pò p 4 = × [ò t sin 2t dt] - x + C , where sin -1 x = t. p 1 85. Putting 5 x = t Û 5 x log 5 dx = dt Û 5 x dx = dt. log 5 =

685

Senior Secondary School Mathematics for Class 12 Pg-686

686

Senior Secondary School Mathematics for Class 12 t

\ I = ò 55 5 t

1 1 dt = ò 5u × du, where 5 t = u Û 5 t log 5 dt = du log 5 (log 5 ) 2

1 dv , where 5u = v Û 5u log 5 du = dv (log 5 ) 3 t 5x v 5u 55 55 = +C= +C= +C= + C. (log 5 ) 3 (log 5 ) 3 (log 5 ) 3 (log 5 ) 3 =ò

ìï 1 + 2 sin x cos x üï 87. I = ò e 2 x í ý dx ïþ ïî 2 cos 2 x ì 2 sin x cos x üï ï 1 = ò e2xí + ý dx 2 2 cos 2 x þï îï 2 cos x 1 2x e (sec2 x ) dx + ò e 2 x tan x dx 2ò 1 = { e 2 x tan x - ò 2 e 2 x tan x dx } + ò e 2 x tan x dx + C 2 1 = e 2 x tan x + C . 2 =

OBJECTIVE QUESTIONS II Mark (3) against the correct answer in each of the following: 1.

òxe

x

dx = ?

(a) ex (1 - x) + C 2.

3.

òxe

2x

(b) ex ( x + 1) + C

dx = ? 1 1 (b) x e2x - e2x + C 2 4

(c) 2x e2x + 4e2x + C

(d) none of these

ò x cos 2x dx = ? 1 1 (b) x sin 2x - cos 2x + C 2 4 (d) none of these

ò x sec x dx = ? 2

(a) x tan x - log |cos x| + C (c) x tan x + log |sec x| + C 5.

(d) none of these

1 1 (a) x e2x + e2x + C 2 4

1 1 (a) x sin 2x + cos 2x + C 2 4 (c) 2x sin 2x + 4 cos 2x + C 4.

(c) ex ( x - 1) + C

(b) x tan x + log |cos x| + C (d) none of these

ò x sin 2x dx = ? 1 1 (a) x cos 2x + sin 2x + C 2 4 1 1 (c) - x cos 2x + sin 2x + C 2 4

1 1 (b) - x cos 2x - sin 2x + C 2 4 (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-687

Methods of Integration

6.

ò x log x dx = ? 1 2 x +C 2 1 1 (c) x 2 log x - x 2 + C 2 4 (a) x log x +

7.

2

2

x 2 x sin 2x cos 2x + +C 4 4 8

x 2 x sin 2x cos 2x + +C 4 4 8 log x 10. ò 2 dx = ? x 1 (a) - (log x + 1) + C x 1 (c) (log x + 1) + C x (c)

1 +C x (c) x(log x + 1) + C

x 2 x sin 2x cos 2x + + +C 4 4 8

(d) none of these

1 (b) (log x - 1) + C x (d) none of these

1 (b) (log x) 2 + C 2 (d) x(log x - 1) + C

ò log10 x dx = ? 1 log e 10 + C x (c) x(log x - 1) log e 10 + C (a)

13.

(b)

ò log x dx = ? (a)

12.

1 1 (b) x cos 2x - sin 2x + C 4 8 1 1 (d) - x cos 2x + sin 2x + C 4 8

ò x cos x dx = ? (a)

11.

(b) - x cot x + log |sin x| + C (d) none of these

ò x sin x cos x dx = ? 1 1 (a) - x sin 2x + cos 2x + C 4 8 1 1 (c) x sin 2x + cos 2x + C 2 4

9.

(d) none of these

ò x cosec x dx = ? (a) x cot x - log |sin x| + C (c) x tan x - log |sec x| + C

8.

1 1 (b) x 2 log x + x 2 + C 2 4

ò (log x) (a)

2

1 log10 e + C x (d) x(log x - 1) log10 e + C

(b)

dx = ?

2log x +C x

(c) x(log x) 2 - 2x log x + 2x + C

(b)

1 (log x) 3 + C 3

(d) x(log x) 2 + 2x log x - 2x + C

687

Senior Secondary School Mathematics for Class 12 Pg-688

688

14.

Senior Secondary School Mathematics for Class 12

òe

x

dx = ? x

(a) e

x

(c) 2e 15.

ò cos

( x - 1) + C

( x + 1) + C

(d) none of these

1 (b) ( x sin 2

x + cos x + C

(c) 2[ x sin

17.

x

x dx = ?

(a) sin

16.

1 (b) e 2

+ x +C

x + cos x ] + C

x - cos x ) + C

(d) none of these

ò cos (log x) dx = ? x (a) [cos (log x) - sin (log x)] + C 2

x (b) [cos (log x) + sin (log x)] + C 2

(c) 2x[cos (log x) + sin (log x)] + C

(d) 2x[cos (log x) - sin (log x)] + C

ò sec

3

x dx = ?

1 (a) {sec x tan x - log |sec x + tan x|} + C 2 1 (b) {sec x tan x + log |sec x + tan x|} + C 2 (c) 2{sec x tan x + log |sec x + tan x|} + C (d) none of these 18.

ì

1

1

ü

ò í(log x) - (log x) 2 ý dx = ? î

þ

(a) x log x + C 1 +C log x

(c) x + 19.

ò 2x

3 x2

e

(b)

x +C log x

(d) none of these

dx = ?

2

2

2

(a) ex ( x 2 - 1) + C (b) ex ( x 2 + 1) + C (c) ex ( x + 1) + C (d) none of these 20.

ò(x2

x

) dx = ?

(a)

2x ( x + log 2) + C (log 2)

(b)

2x

(c)

x × 2x 2x + +C (log 2) (log 2) 2

(d) none of these

(log 2) 2

( x log 2 - 1) + C

Senior Secondary School Mathematics for Class 12 Pg-689

Methods of Integration

21.

ò x cot

2

x dx = ?

x2 + log |sin x| + C 2 x2 (c) - x cot x + - log |sin x| + C 2 (a) - x cot x +

22.

ò sin

24.

òe

ò

sin x

x +C

(d) 2esin x (sin x - 1) + C

sin -1 x (1 - x 2)

3

dx = ? 2

sin -1 x

2

x tan -1 x (1 - x 2)

3

2

1+ x x tan -1 x

-

1 + x2 -1

ò tan

-1

x 2

+C

1+ x ½ x ½ 1 + log ½ ½+ C 2 ½ 1 + x 2½

-1

(d) none of these

(b)

- tan -1 x 1+ x

2

+

x 1 + x2

+C

(d) none of these

x dx = ? 1 1 (b) x 2 tan -1 x + x + C 2 2 (d) none of these

x dx = ?

(a) ( x - 1) tan -1 x + x + C 1 1 (c) x tan -1 x x +C 2 2

ò cos

1 log |1 - x 2| + C 2

dx = ?

tan -1 x

ò x tan

(b) x sin -1 x +

2

1 1 (a) tan -1 x + log (1 + x 2) - x + C 2 2 1 1 2 -1 (c) (1 + x ) tan x - x + C 2 2

28.

x -C

(c) 2esin x (sin x + 1) + C

(c)

27.

(b) - x cos x - 2 sin (d) none of these (b) ( 2 cos x) esin x + C

(a)

26.

(d) none of these

sin 2x dx = ?

1 - log |1 - x 2| + C 2 1-x -1 x sin x 1 (c) + log |1 - x 2| + C 2 1 - x2

ò

x2 + log |sin x| + C 2

(a) ( 2 sin x) esin x + C

(a)

25.

(b) - x cot x -

x dx = ?

(a) - x cos x + C (c) -2 x cos x + 2 sin

23.

689

(b) ( x + 1) tan -1 x - x + C (d) none of these

x dx = ?

(a) x cos-1 x - 1 - x 2 + C

(b) x cos-1 x + 1 - x 2 + C

(c) x sin -1 x - 1 - x 2 + C

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-690

690

29.

Senior Secondary School Mathematics for Class 12

ò tan

-1

x dx = ?

1 log |1 + x 2| + C 2 1 (c) - x tan -1 x + log |1 + x 2| + C 2 (a) x tan -1 x +

30.

31.

-1

ò sec

(a) x sec-1 x + log |x + x 2 - 1| + C

(b) x sec-1 x - log |x + x 2 - 1| + C

(c) x sec-1 x + log |x - x 2 - 1| + C

(d) none of these

ò sin

-1

(c)

33.

34.

35.

( 3 x - 4x 3) dx = ?

ò sin

3x2 +C 2

(b) 3[x sin -1 x - 1 - x 2 ] + C (d) none of these

-1 æ

2x ö ç ÷ dx = ? è 1 + x2 ø

(a) 2x tan -1 x + log |1 + x 2| + C

(b) 2x tan -1 x - log |1 + x 2| + C

(c) 2x sin -1 x + log |1 + x 2| + C

(d) none of these

1-x dx = ? 1+ x 1 1 (a) x (cos-1 x) + 1 - x2 + C 2 2 1 1 (c) x (cos-1 x) 1 - x2 + C 2 2

1 1 (b) x (sin -1 x) + 1 - x2 + C 2 2

ò tan

-1

òx

2

(d) none of these

-1 æ ç

3 x - x 3 ö÷ dx = ? ç 1 - 3x2 ÷ è ø 3 (a) 3 x tan -1 x + log (1 + x 2) + C 2 3 -1 (c) 3 x cos x 1 - x2 + C 2

ò tan

3 log (1 + x 2) + C 2 3 (d) 3 x sin -1 x + 1 - x2 + C 2 (b) 3 x tan -1 x -

cos x dx = ?

(a) x 2 sin x + 2x cos x - 2 sin x + C 2

(c) x sin x - 2x sin x + 2 sin x + C 36.

(d) none of these

x dx = ?

(a) 3[x sin -1 x + 1 - x 2 ] + C

32.

1 (b) x tan -1 x - log |1 + x 2| + C 2

(b) 2x cos x - x sin x + 2 sin x + C (d) none of these

ò sin x log (cos x) dx = ? (a) cos x log (cos x) - cos x + C (c) cos x log (cos x) + cos x + C

(b) - cos x log (cos x) + cos x + C (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-691

Methods of Integration

37.

ò x sin x cos x dx = ? 1 1 (a) - x cos 2x + sin 2x + C 4 8 1 1 (c) x cos 2x - sin 2x + C 4 8

38.

39.

40.

òx

3

(d) none of these

cos x 2 dx = ?

(a) x 2 sin x 2 + cos x 2 + C

1 1 (b) x 2 sin x 2 + cos x 2 + C 2 2

1 1 (c) - x 2 sin x 2 + cos x 2 + C 2 2

(d) none of these

-1 æ ç1- x

ö ÷ dx = ? ç 1 + x2 ÷ è ø

ò cos

2

(a) 2x tan -1 x + log (1 + x 2) + C

(b) -2x tan -1 x - 2 log (1 + x 2) + C

(c) 2x tan -1 x - log (1 + x 2) + C

(d) none of these

ò x tan

-1

x dx = ?

1 1 (a) ( x 2 + 1) tan -1 x - x + C 2 2 1 2 1 (c) ( x + 1) tan -1 x + x + C 2 2 41.

1 1 (b) x cos 2x + sin 2x + C 4 8

1 1 (b) ( x 2 - 1) tan -1 x - x + C 2 2 (d) none of these

ò sin (log x) dx = ? 1 1 (a) x sin log x + x cos (log x) + C 2 2 1 1 (b) x sin log x - x cos (log x) + C 2 2 1 1 (c) - x sin (log x) + x cos (log x) + C 2 2 (d) none of these

42.

ò (sin (a) (b)

-1

x) 2 dx = ?

2 sin -1 x 1 - x2

+C

1 1 (sin -1 x) 3 + +C 3 1 - x2

(c) x(sin -1 x) 2 + (sin -1 x) 1 - x 2 + 2x + C (d) x(sin -1 x) 2 + 2(sin -1 x) 1 - x 2 - 2x + C

691

Senior Secondary School Mathematics for Class 12 Pg-692

692

Senior Secondary School Mathematics for Class 12 x ì1

1 ü í - 2 ý dx = ? îx x þ 1ü ì (a) ex ílog x + ý + C xþ î x 1 (c) e × + C x 2 ö æ 1 44. ò ex ç 2 - 3 ÷ dx = ? èx x ø -ex (a) 2 + C x æ -1 1 ö (c) ex ç + ÷ +C è x x2 ø

43.

45.

46.

47.

48.

49.

òe

ì -1 1 üï ý dx = ? ísin x + ïî 1 - x 2 ïþ 1 (a) ex × +C 1 - x2 -ex (c) +C sin -1 x

òe

òe

(d) none of these

(b)

ex x2

+C

(d) none of these

xï

x

(b) ex sin -1 x + C (d) none of these

(tan x + log sec x) dx = ?

(a) ex log sec x + C

(b) ex tan x + C

(c) ex (log cos x) + C

(d) none of these

òe

x

(cot x + log sin x) dx = ?

(a) ex cot x + C

(b) ex log sin x + C

(c) ex sin x + C

(d) none of these

òe

x

[sec x + log ( sec x + tan x)] dx = ?

(a) ex log ( sec x + tan x) + C

(b) ex sec x + C

(c) ex log tan x + C

(d) none of these

1 ü -1 ítan x + ý dx = ? (1 + x 2) þ î 1 (a) ex × +C (1 + x 2)

òe

xì

(c) - ex cot -1 x + C 50.

(b) xex - ex + C

òe

x

(b) ex tan -1 x + C (d) none of these

(tan x - log cos x) dx = ?

(a) ex tan x + C

(b) ex log cos x + C

(c) ex log sec x + C

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-693

Methods of Integration

51.

òe

x

(cot x - cosec2 x) dx = ?

(a) - ex cosec2 x + C (b) ex cot x + C 52.

òe

x

òe

x

(b) ex cos x + C

(c) ex tan x + C

(d) none of these

(a) ex (1 + tan x) + C

(b) ex sec x + C

(c) ex tan x + C

(d) none of these

òe

x æ1 +

ç è

(a) ex × 56.

(d) none of these

sec x(1 + tan x) dx = ?

x log x ö ÷ dx = ? x ø x 1 (b) ex log x + C (a) e × + C x x 55. ò ex × dx = ? (1 + x) 2 54.

(c) - ex cot x + C

(sin x + cos x) dx = ?

(a) ex sin x + C 53.

693

1 1 + C (b) ex × + C (1 + x) x

(c) ex ×

sin x ö çç ÷÷ dx = ? è 1 + cos x ø x x (a) ex sin + C (b) ex cos + C 2 2

òe

(c) x ex log x + C (d) none of these

x + C (d) none of these (1 + x)

x æ1 +

(c) ex tan

x +C 2

(d) none of these

ANSWERS (OBJECTIVE QUESTIONS II)

1. (c)

2. (b)

3. (a)

4. (b)

5. (c)

6. (c)

7. (b)

8 (d)

9. (b) 10. (a)

11. (d) 12. (d) 13. (c) 14. (c) 15. (c) 16. (b) 17. (b) 18. (b) 19. (a) 20. (b) 21. (b) 22. (c) 23. (d) 24. (c) 25. (b) 26. (c) 27. (b) 28. (a) 29. (b) 30. (b) 31. (a) 32. (b) 33. (c) 34. (b) 35. (a) 36. (b) 37. (a) 38. (b) 39. (c) 40. (a) 41. (b) 42. (d) 43. (c) 44. (b) 45. (b) 46. (a) 47. (b) 48. (a) 49. (b) 50. (c) 51. (b) 52. (a) 53. (b) 54. (b) 55. (a) 56. (c) HINTS TO THE GIVEN OBJECTIVE QUESTIONS II 1.

ò xI eIIx dx

2.

ò xI eII2x dx

3.

ò xI cosII 2 x dx

4.

5.

ò xI sinII2 x dx

6.

ò IIx log x dx

7.

2 x dx ò xI cosec II

8. I =

I

1 1 1 9. I = ò x × ( 1 + cos 2 x ) dx = ò x dx + ò x cos 2 x dx 2 2 2 I II

ò xI secII2 x dx 1 x sin 2 x dx 2 ò I II

Senior Secondary School Mathematics for Class 12 Pg-694

694

10.

Senior Secondary School Mathematics for Class 12

ò

log x x2

12. I = ò

dx = ò (log x ) × I

1 x2

11. I = ò (log x ) × 1 dx

dx

I

II

II

ì ü log x 1 dx = × (log x ) × 1 dx = (log10 e ) × ò í(log x ) × 1 ý dx log 10 log e 10 ò I î I II II þ

ì ü 2 log x 13. I = í(log x ) 2 × 1 ý dx = (log x ) 2 × x - ò × x dx x î I II þ = x (log x ) 2 - 2 ò log x dx = x (log x ) 2 - 2 [x (log x - 1)] + C . 1 dx = dt , we get I = 2 ò t e t dt I II 2 x 1 15. Putting x = t and dx = 2 dt , we get dx = 2t dt x \ I = 2 ò t cos t dt.

14. Putting x = t and

I

II

16. Putting log x = t , x = e t and dx = e tdt , we get I = ò e t cos t dt = e t cos t + e t sin t - ò e t cos t dt I

II

t

2 I = ( e cos t + e t sin t ) Þ I =

\

1 t ( e cos t + e t sin t ) + C . 2

17. I = ò sec2 x × sec x dx = sec x tan x - ò sec x tan 2 x dx II

\

I

= sec x tan x - ò sec x ( sec2 x - 1) dx = sec x tan x - I +

ò sec x dx

2I = sec x tan x + log|sec x + tan x| + C . ì ü 1 ï 1 ï 18. I = ò í × 1 ý dx - ò dx (log x ) 2 ï (log x ) II ï î I þ 1 1 1 1 = × x - ò ( -1) × × × x dx - ò dx + C (log x ) (log x ) 2 x (log x ) 2 x = + C. (log x ) 2

19. I = ò ( 2 x )x 2 e x dx = ò t e t dt , where x 2 = t. I II

2x 2x 20. I = ò x × 2 dx = x × dx (log 2 ) ò (log 2 ) I II x

=

x × 2x 2x +C (log 2 ) (log 2 ) 2

21. I = x ( cosec2 x - 1) dx = ò x cosec2 x dx - ò x dx I

II

= x ( - cot x ) - ò ( - cot x ) dx 22. Put x = t and \

1 dx = dt or dx = 2t dt 2 x

I = 2 ò t sin t dt. I

x2 x2 + C = - x cos x + log|sin x| + C. 2 2

II

Senior Secondary School Mathematics for Class 12 Pg-695

Methods of Integration

695

23. I = 2 ò e sin x sin x cos x dx = 2 ò t e t dt , where sin x = t. I II

24. Put x = sin t so that dx = cos t dt and t = sin -1 x. I=ò

\

t cos t 2

( 1 - sin t )

3

2

dt = ò

t cos t

dt = ò t sec2t dt.

cos 3 t

I

II

25. Put x = tan t , so that dx = sec2t dt and t = tan -1 x. I=ò

\

t(tan t ) ( 1 + tan 2 t )

3

2

sec2t dt = ò t sin t dt. I

II

26. I = ò x (tan -1 x ) dx. II

I

27. Put x = t and \

1 dx = dt or dx = 2t dt. 2 x

I = 2 ò t (tan -1t ) dt. II

I

-1

28. Put cos x = t , so that x = cos t and dx = - sin t dt. \

I = - ò t sin t dt. I

II

-1

29. Put tan x = t , so that x = tan t and dx = sec2t dt. \

I = ò t sec2t dt. I

II

-1

30. Put sec x = t , so that x = sec t and dx = sec t tan t dt. \

I = ò t ( sec t tan t ) dt. I

II

31. Put x = sin t and dx = cos t dt. \

I = ò sin -1 (sin 3t ) cos t dt = 3 ò t cos t dt. I

II

2

32. Put x = tan t and dx = sec t dt. \

æ 2 tan t ö ÷ sec2t dt = sin -1 (sin 2t ) sec2t dt I = ò sin -1 ç ò ç 1 + tan 2t ÷ è ø = 2 ò t sec2t dt. I

II

33. Put x = cos t and dx = - sin t dt. Then, I = ò tan -1

1 - cos t × ( - sin t ) dt = ò tan -1 1 + cos t

2 sin 2 ( t 2 ) 2 cos 2 ( t 2 )

tö 1 æ = - ò tan -1 ç tan ÷ ( - sin t ) dt = ò t (sin t ) dt. 2ø 2 I II è 34. Put x = tan t and dx = sec2t dt. I = ò tan -1 (tan 3 t ) sec2t dt = 3 ò t sec2t dt. I

II

( - sin t ) dt

Senior Secondary School Mathematics for Class 12 Pg-696

696

Senior Secondary School Mathematics for Class 12

35. I = ò x 2 cos x dx = x 2 sin x - ò 2 x sin x dx I

II

I

II

= ( x sin x ) - 2 [x ( - cos x ) - ò ( - cos x ) dx] + C 2

= ( x 2 sin x ) + 2 x cos x - 2 sin x + C . 36. Put cos x = t and - sin x dx = dt. æ ö 1 ù é I = - ò log t dt = - ò çç log t × 1 ÷÷ dt = - ê(log t ) t - ò × t dt ú \ t II ë û è I ø = -t(log t ) + t + C = - cos t[log (cos t )] + cos x + C . 1 37. I = ò x sin 2 x dx. 2 I II 1 38. Put x 2 = t , so that x dx = dt. 2 1 1 \ I = ò t cos t dt = ò t cos t. 2 2 I II 39. Put x = tan t and dx = sec2t dt. æ 1 - tan 2t ö ÷ sec2t dt = cos -1 (cos 2t ) sec2t dt Then, I = ò cos -1 ç ò ç 1 + tan 2t ÷ è ø = 2 ò t sec2t dt. I

II

40. I = ò x tan -1 x dx = (tan -1 x ) × II

I

x2 x2 1 -ò × dx 2 2 (1 + x ) 2

1 2 1 æ 1 ö÷ 1 1 1 x (tan -1 x ) - ò ç 1 dx = x 2 tan -1 x - x + tan -1 x + C ç 2 2 è 2 2 2 1 + x 2 ÷ø 1 1 = ( x 2 + 1) tan -1 x - x + C . 2 2 ì ü cos (log x ) 41. I = ò í sin (log x ) × 1 ý dx = (sin log x )x - ò × x dx x II þ î I ì ü = x sin (log x ) - ò í cos (log x ) × 1 ý dx II î þ I - sin (log x ) = x sin (log x ) - cos (log x ) × x + ò × x dx x = x sin (log x ) - x cos (log x ) - I =

\

2I = x sin (log x ) - x cos (log x ) + C .

42. Putting x = sin t and dx = cos t dt , we get I = ò {sin -1 (sin t )} 2 cos t dt = ò t 2 cos t dt I

II

= t 2 sin t - 2 ò t sin t dt = t 2 sin t - 2 [ t ( - cos t ) - ò ( - cos t ) dt] I

II

= t sin t + 2 t cos t - 2 sin t + C = x (sin -1 x ) 2 + 2 (sin -1 x ) 1 - x 2 - 2 x + C. 2

43. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = = e xf ( x ) + C = e x ×

1 + C. x

1 x

Senior Secondary School Mathematics for Class 12 Pg-697

Methods of Integration 44. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = = e xf ( x ) + C = e x ×

1 x2

1 x2

+ C.

45. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = sin -1 x. 46. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = log sec x = e x f ( x ) + C = e x log sec x + C . 47. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = log sin x = e x f ( x ) + C = e x log sin x + C . 48. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = log ( sec x + tan x ) = e x f ( x ) + C = e x log ( sec x + tan x ) + C . 49. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = tan -1 x = e x f ( x ) + C = e x tan -1 x + C .

50. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = log sec x = e x f ( x ) + C = e x log sec x + C . 51. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = cot x = e x f ( x ) + C = e x cot x + C . 52. I = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = sin x = e x f ( x ) + C = e x sin x + C . 53. I = ò e x { f ( x ) + f ¢( x )} dx , where f ( x ) = sec x = e xsec x + C . ö æ1 54. I = ò e x ç + log x ÷ dx èx ø = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = log x = e x f ( x ) + C = e x log x + C . ìï 1 üï ìï ( 1 + x ) - 1 üï 1 55. I = ò e x í dx = ò e x í dx 2 ý 2ý + x ( 1 ) ( 1 + x ) þï îï ( 1 + x ) þï îï 1 = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = ( 1 + x) 1 = e x f ( x) + C = e x × + C. ( 1 + x) æ 1 + 2 sin ( x ) cos ( x ) ö x 1 xö æ 2 2 ÷ 56. I = ò e x ç dx = ò e x ç tan + sec2 ÷ dx 2 x ç ÷ 2 2 2 è ø 2 cos ( ) è ø 2 x = ò e x { f ( x ) + f ¢( x ) } dx , where f ( x ) = tan 2 x x x = e f ( x ) + C = e tan + C . 2

697

Senior Secondary School Mathematics for Class 12 Pg-698

14. SOME SPECIAL INTEGRALS Three Special Integrals THEOREM

PROOF

(i)

dx

1

½a + x½

dx

1

½x - a½

dx

1

(i)

ò ( a 2 - x 2) = 2a log ½½a - x½½+ C.

(ii)

ò ( x 2 - a 2) = 2a log ½½x + a½½+ C.

(iii)

ò ( x 2 + a 2) = a tan dx

-1

x + C. a

dx

ò ( a 2 - x 2) = ò ( a + x)( a - x) =ò

1 ì( a - x) + ( a + x) ü 1 é dx dx ù × + ×í ý dx = 2a î ( a + x)( a - x) þ 2a êë ò ( a + x) ò ( a - x) úû

1 × [log |a + x| - log |a - x|] + C 2a 1 a + x½ = × log ½ + C. 2a ½a - x½ dx 1 a + x½ \ ò 2 = × log ½ + C. 2 2 a a - x½ ½ (a - x ) =

(ii)

dx

dx

ò ( x 2 - a 2) = ò ( x - a)( x + a) 1 ì( x + a) - ( x - a) ü 1 é dx dx ù × ×í ý dx = 2a î ( x - a)( x + a) þ 2a êë ò ( x - a) ò ( x + a) úû 1 1 x - a½ = × [log |x - a|- log |x + a|] + C = × log ½ + C. 2a 2a ½x + a½ x-a dx 1 ò ( x 2 - a 2) = 2a × log½½x + a½½+ C. =ò

\ (iii)

dx

1

dx

1

2ö ç1 + x ÷ 2 ç a ÷ø è a dt

ò ( x 2 + a 2) = a 2 × ò æ =

2

×ò

2

[putting

x = t and dx = a dt] a

a (1 + t ) x 1 1 -1 = tan t + C = tan -1 + C. a a a 698

Senior Secondary School Mathematics for Class 12 Pg-699

Some Special Integrals

\ REMARK

dx

1

ò ( x 2 + a 2) = a tan

-1

699

x + C. a

If we have an integral of the form

dx

ò ( ax 2 + bx + c)

then we put the

denominator in the form [( x + a ) 2 ± b 2] and then integrate. SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Evaluate: dx (i) ò (1 - 4x 2)

(ii)

dx

ò ( 32 - 2x 2)

(iii)

x2

ò (1 - x 6) dx

We have (i)

dx

1

ò (1 - 4x 2) = 4 × ò æ 1

dx

2ö ç -x ÷ è4 ø 1 dx = ×ò ü 4 ìïæ 1 ö 2 2ï íç ÷ - x ý ïþ ïîè 2 ø ½1 + x½ 1 1 ½ ½ × log 2 +C = × ½ 1 ½ 4 æ 1ö ½ - x½ ç2 ´ ÷ ½2 ½ 2ø è é a + x½ ù dx 1 log ½ = + Cú êQ ò 2 2 a 2 ½a - x½ û (a - x ) ë

1 1 + 2x½ log ½ + C. 4 ½1 - 2x½ dx 1 dx (ii) ò = ×ò ( 32 - 2x 2) 2 (16 - x 2) 1 dx = ×ò 2 2 {( 4) - x 2} 1 1 4 + x½ = × log ½ +C 2 ( 2 ´ 4) ½4 - x½ é a + x½ ù dx 1 log ½ = + Cú êQ ò 2 2 ½a - x½ û ( a - x ) 2a ë 1 4 + x½ = log ½ + C. 16 ½4 - x½ =

(iii) Putting x 3 = t and 3 x 2 dx = dt, we get x2 1 1 ò (1 - x 6) dx = 3 × ò (1 - t 2) dt

Senior Secondary School Mathematics for Class 12 Pg-700

700

Senior Secondary School Mathematics for Class 12

1 1 1+t × × log ½ ½+ C 3 ( 2 ´ 1) ½1 - t½

=

é êQ ë =

EXAMPLE 2

SOLUTION

Evaluate:

(i)

ò

dx ( a 2 - x 2)

=

1 a + x½ ù log ½ + Cú 2a ½a - x½ û

½1 + x 3½ 1 ½+ C. log ½ 3 6 ½1 - x ½ sin x

ò (1 - 4 cos2x)

(ii)

dx

cosec2 x

ò (1 - cot 2x) dx

(i) Putting cos x = t and - sin x dx = dt, we get sin x dt ò (1 - 4 cos2x) dx = -ò (1 - 4t 2) 1 dt dt 1 = - ×ò = - ×ò 4 æ 1 2ö 4 ìïæ 1 ö 2 2 üï ç -t ÷ íç ÷ - t ý è4 ø ïþ ïîè 2 ø ½1 + t½ 1 1 ½ ½ log 2 +C =- ´ ½ 1 ½ 4 æ 1ö ½ ½ t 2 ´ ÷ ç ½2 ½ 2ø è 1 1 1 + 2 cos x½ ½1 + 2t½ ½+ C. ½ ½+ C = - log ½ = - log ½ t 4 1 2 4 ½1 - 2 cos x½ ½ ½ (ii) Putting cot x = t and - cosec2 x dx = dt, we get cosec2 x

dt

1

½1 + t½

ò (1 - cot 2x) dx = -ò (1 - t 2) = - ( 2 ´ 1) log ½½1 - t½½+ C 1 ½1 + cot x½ ½+ C. = - log ½ 2 ½1 - cot x½ EXAMPLE 3

SOLUTION

Evaluate: dx (i) ò ( 9x 2 - 1)

(ii)

x

ò ( x 4 - 9) dx

We have (i)

dx

1

ò ( 9x 2 - 1) = 9 × ò æ

dx

2 1ö çx - ÷ 9ø è 1 dx = ×ò 9 ìï 2 æ 1 ö 2 üï íx - ç ÷ ý è 3 ø ïþ ïî

(iii)

x2

ò ( x 2 - 9) dx

Senior Secondary School Mathematics for Class 12 Pg-701

Some Special Integrals

½x 1 1 ½ log = × ½ 1ö 9 æ ½x + ç2 ´ ÷ ½ 3ø è é êQ ë =

701

1½ 3 ½+ C 1½ ½ 3½ dx

1

½x - a½

ù

ò ( x 2 - a 2) = 2a log ½½x + a½½+ C ú û

1 3 x - 1½ log ½ + C. 6 ½3 x + 1½

(ii) Putting x 2 = t and 2x dx = dt, we get x dt dt 1 1 ò ( x 4 - 9) dx = 2 ò (t 2 - 9) = 2 × ò {t 2 - ( 3) 2}

(iii)

=

t - 3½ 1 1 × log ½ +C 2 ( 2 ´ 3) ½t + 3½

=

½x 2 - 3½ 1 ½+ C. log ½ 2 12 ½x + 3½

x2

ì

9

ü

ò ( x 2 - 9) dx = ò íî1 + x 2 - 9 ýþdx = ò dx + 9ò

dx {x 2 - ( 3) 2}

é 1 x - 3½ù = x + 9× ê log ½ +C ( 2 ´ 3 ) + 3½úû x ½ ë 3 x - 3½ = x + log ½ + C. 2 ½x + 3½ EXAMPLE 4

Evaluate

SOLUTION

We have dx

dx

ò ( 4 + 25 x 2) × 1

dx 4 ö + x2 ÷ ç è 25 ø dx

ò ( 4 + 25 x 2) = 25 × ò æ

EXAMPLE 5

Evaluate

=

1 × 25 ò {( 25 ) 2 + x 2}

=

x 1 1 tan -1 × +C ( 25 ) 25 ( 25 )

=

1 æ 5x ö tan -1 ç ÷ + C. 10 è 2ø 3x

ò (1 + 2x 4) dx.

é ù dx x 1 = tan -1 + C ú êQ ò 2 2 a a ( ) a x + ë û

Senior Secondary School Mathematics for Class 12 Pg-702

702 SOLUTION

Senior Secondary School Mathematics for Class 12

Putting x 2 = t and 2x dx = dt, we get 3x

3

dt

ò (1 + 2x 4) dx = 2 × ò (1 + 2t 2) 3 dt 3 dt × = × 4 ò æ 1 2 ö 4 ò ìïæ 1 ö 2 2 üï ç +t ÷ ÷ +t ý íç è2 ø îïè 2 ø þï t 3 1 -1 tan = × +C ( 1 2) 4 ( 1 2) =

3 3 tan -1( 2 t) + C = tan -1( 2x 2) + C. 2 2 2 2 dx ò (1 + 5 sin 2x) × =

EXAMPLE 6

Evaluate

SOLUTION

On dividing num. and denom. by cos2 x, we get dx

ò (1 + 5 sin 2x)

=ò

(

1

cos2 x

)

dx æ 1 sin 2 x ö÷ ç + 5 × ç cos2 x cos2 x ÷ø è sec2 x sec2 x =ò = dx dx ò {(1 + tan 2 x) + 5 tan 2 x} ( sec2 x + 5 tan 2 x) sec2 x dt =ò dx = ò , where tan x = t (1 + 6 tan 2 x) (1 + 6t 2) 1 dt 1 dt = ò = × 6 æ 1 2 ö 6 ò ìïæ 1 ö 2 2 üï ç +t ÷ ÷ +t ý íç è6 ø îïè 6 ø þï t 1 1 1 -1 tan tan -1( 6t) + C = × +C = ( 1 6) 6 ( 1 6) 6

1 tan -1( 6 tan x) + C. 6 dx Evaluate ò × ( 2 + sin 2 x) On dividing num. and denom. by cos2 x, we get =

EXAMPLE 7 SOLUTION

dx

ò ( 2 + sin 2x)

=ò =ò

sec2 x ( 2 sec2 x + tan 2 x) sec2 x

dx = ò

dx = ò

dt

sec2 x {2(1 + tan 2 x) + tan 2 x}

, where tan x = t 2 + 3 tan x ( 2 + 3t 2) 1 dt 1 dt = ×ò = ×ò 2 ö 3 æ 2 ö2 3 æ 2 2 çt + ÷ ÷ ç 3ø è ç 3÷ +t ø è 2

dx

Senior Secondary School Mathematics for Class 12 Pg-703

Some Special Integrals

703

t 1 1 1 × × tan -1 tan -1 +C = æ 2ö 3 2 6 ç ÷ ç 3÷ 3 è ø 1 3 tan x ö -1 æ ÷ + C. = tan çç 6 2 ÷ø è dx ò ( x 2 + 6x + 13) × =

EXAMPLE 8

Evaluate

SOLUTION

We have dx

3t +C 2

dx

ò ( x 2 + 6x + 13) = ò {( x 2 + 6x + 9) + 4} =ò

EXAMPLE 9

Evaluate

SOLUTION

We have dx

dx 2

2

=ò

dt

{( x + 3) + 2 } (t + 22) t 1 = tan -1 + C 2 2 1 -1 1 = tan ( x + 3) + C. 2 2 dx ò ( x 2 + 8x + 20) × dx

2

, where ( x + 3) = t

[CBSE 2002C]

dt

ò ( x 2 + 8x + 20) = ò {( x + 4) 2 + 22} = ò (t 2 + 22) , where ( x + 4) = t t 1 tan -1 + C 2 2 1 -1 1 = tan ( x + 4) + C. 2 2 dx ò ( 9x 2 - 12x + 8) × =

EXAMPLE 10

Evaluate

SOLUTION

We have 4 8ö æ ( 9x 2 - 12x + 8) = 9 ç x 2 - x + ÷ 3 9ø è

2 2 ìïæ 4 4ö 4 8ü 2ö ìæ æ 2 ö üï = 9 íç x 2 - x + ÷ - + ý = 9 íç x - ÷ + ç ÷ ý 3 9ø 9 9þ 3ø è 3 ø ïþ ïîè îè dx 1 dx \ ò = ×ò 2 2 ( 9x 2 - 12x + 8) 9 ìïæ 2ö æ 2 ö üï íç x - ÷ + ç ÷ ý 3ø è 3 ø ïþ ïîè 2öü ìæ ï ç x - 3 ÷ ïï 1 1 ø + C = 1 tan -1 æ 3 x - 2 ö + C. -1 ï è = × tan í ç ÷ ý 2 ö æ 6 9 æ 2ö è 2 ø ï ç ÷ ï ç ÷ è 3ø îï è 3 ø þï

Senior Secondary School Mathematics for Class 12 Pg-704

704

Senior Secondary School Mathematics for Class 12

x

ò ( x 4 - x 2 + 1) dx.

EXAMPLE 11

Evaluate

SOLUTION

Putting x 2 = t and 2x dx = dt, we get x

dt

1

[CBSE 2003, ‘07C]

1

ò ( x 4 - x 2 + 1) dx = 2 × ò (t 2 - t + 1) = 2 × ò ì

dt

2 æ 3 ö üï ïæ 1 ö ÷ ç íç t - ÷ + ç ÷ ý è 2 ø ïþ ïîè 2 ø æ 1ö çt - ÷ 1 1 2ø -1 è +C = × tan æ 3ö 2 æ 3ö ÷ ç ÷ ç ç 2 ÷ ç 2 ÷ ø è ø è æ 2x 2 - 1 ö t 1 2 1 1 ö æ ÷ + C. = × tan -1 ç tan -1 ç ÷ +C = ç 3 3 3 ÷ø è 3 ø è

EXAMPLE 12

Evaluate

SOLUTION

We have dx

2

dx

ò ( 2x 2 + x - 1) × 1

dx 1 1ö 2 çx + x - ÷ 2 2ø è dx

ò ( 2x 2 + x - 1) = 2 × ò æ

1 1 dx = ò ò 2 2 2 2 éìï 1 æ 1 ö üï 1 1 ù 2 é æ x + 1 ö - æ 3 ö ù êç ê íx 2 + x + ç ÷ ý - - ú ç ÷ ú ÷ 4ø 2 è 4 ø úû è 4 ø ïþ 16 2 úû êë è êëïî 1 ö 3½ ½æ x+ ÷½çè 1 1 1 ½ 2x - 1 ½ 4 ø 4½ log ½ = × ½+ C = 3 log ½2( x + 1)½+ C. 3 1 3 2 2× æ ö ½ ½ ½ç x + ÷ + ½ 4 4 ø 4½ ½è =

EXAMPLE 13

Evaluate

SOLUTION

We have

dx

ò ( 3 x 2 + 13 x - 10) ×

13 10 ö æ ( 3 x 2 + 13 x - 10) = 3 ç x 2 + x- ÷ 3 3ø è 2 2 ìïæ ìïæ 13 ö 169 10 üï 13 ö 289 üï = 3 íç x + ÷ - ý = 3 íç x + ÷ ý 6ø 36 3 ïþ 6ø 36 ïþ ïîè ïîè 2 2 ìïæ 13 ö æ 17 ö üï = 3 íç x + ÷ - ç ÷ ý × 6ø è 6 ø ïþ ïîè dx dx \ ò =ò 2 2 2 ì ( 3 x + 13 x - 10) 13 ö ïæ æ 17 ö üï 3 íç x + ÷ - ç ÷ ý 6ø è 6 ø þï îïè

Senior Secondary School Mathematics for Class 12 Pg-705

Some Special Integrals

705

13 ö 1 dt æ , where ç x + ÷ = t ò 2 6ø 3 ïì 2 æ 17 ö ïü è ít - ç ÷ ý 6 è ø þï îï ½t - 17 ½ 1 1 ½ 6 ½+ C log = × ½ 17½ 3 æ 17 ö ½ t+ ½ ç2 ´ ÷ ½ 6½ 6ø è é x - a½ù dx 1 log ½ = êQ ò 2 ú 2 a x 2 + a½û ½ (x - a ) ë =

=

1 ½6t - 17½ ½+ C log ½ 17 ½6t + 17½

13 ö ½ ½ æ 6 x + ÷ - 17 ½ 1 ½ çè 1 6x - 4 ½ 6ø log ½ = log ½ +C = ½ 13 17 ½6x + 30½ ½6 æç x + ö÷ + 17½ 17 6ø ½ ½ è 1 3 x 2 ( 3 x - 2)½ ½+ C = 1 log ½ ½+ C ½ = log ½ 17 17 ½3 x + 15½ ½3( x + 5)½ 1 ì 1 ½3 x - 2½ü + C ílog + log ý 17 î 3 ½ x + 5 ½þ 1 3 x - 2½ log ½ = + k, 17 ½x +5½ 1 1 where log + C = k = constant. 17 3 =

EXAMPLE 14

Evaluate

SOLUTION

We have dx

dx

ò (1 + x - x 2) × dx

ò (1 + x - x 2) = -ò ( x 2 - x - 1) = -ò

dx

= -ò

dx

2 ì æ 5 ö üï 1ö ïæ ÷ ý íç x - ÷ - çç ÷ 2ø è è 2 ø ïþ îï 1ö dx dx æ , where ç x - ÷ = u =ò =ò 2 2ü ì ü ìæ 5 ö 2 2ø è 1 ïæç 5 ö÷ ï ïç ÷ - æç x - ö÷ ïý - u2 ý íç íç ÷ ÷ 2ø ï è ïîè 2 ø ïþ ïîè 2 ø þ ½ 5 ½ ½ + u½ 1 × log ½ 2 = ½+ C æ 5 5ö ½ ÷ ç2 ´ - u½ ç ½2 ½ 2 ÷ø è

1ö 5 ü ìæ 2 íç x - x + ÷ - ý 4ø 4þ îè

2

Senior Secondary School Mathematics for Class 12 Pg-706

706

Senior Secondary School Mathematics for Class 12

EXAMPLE 15

Evaluate

SOLUTION

We have dx

1 ö½ æ ½ 5 + 2ç x - ÷ ½ ½ 5 + 2u½ 1 1 2 ø½ è ½+ C = log ½ log ½ = ½+ C 5 5 ½ 5 - 2u½ ½ 5 - 2 çæ x - 1 ö÷½ 2 ø½ è ½ ½ 5 + 2x - 1½ ½( 5 - 1) + 2x½ 1 1 ½+ C = ½+ C. = log ½ log ½ 5 5 ½ 5 - 2x + 1½ ½( 5 + 1) - 2x½ dx ò (5 - 8x - x 2) × dx

ò (5 - 8x - x 2) = -ò ( x 2 + 8x - 5) = -ò

EXAMPLE 16

Evaluate

SOLUTION

We have dx

dx

= -ò

2

dx

{( x + 8x + 16) - 21} {( x + 4) - ( 21) 2} dx dt =ò =ò , 2 2 {( 21) - ( x + 4) } {( 21) 2 - t 2} where ( x + 4) = t ½ 21 + t½ 1 ½+ C = × log ½ 2 21 ½ 21 - t½ ½ 21 + 4 + x½ 1 ½+ C. = × log ½ 2 21 ½ 21 - 4 - x½ dx ò (1 - 6x - 9x 2) × dx

1

2

dx 2 1ö 2 çx + x - ÷ 3 9ø è

ò (1 - 6x - 9x 2) = -ò ( 9x 2 + 6x - 1) = - 9 ò æ

1 dx = - ×ò 1 ö 2ü 9 ìæ 2 2 íç x + x + ÷ - ý 3 9ø 9þ îè 1 dx 1 dx = ×ò = - ×ò 2 2 2 2 9 ì æ 2 ö üï 9 ìïæ 2 ö 1 ü 1ö ïæ ÷ - æç x + ö÷ ïý ÷ ý ç í íç x + ÷ - çç ÷ ç 3 ÷ 3ø ï 3ø è è ø è 3 ø þï îïè þ îï 1ö 1 dx æ = ×ò , where ç x + ÷ = t 3ø 9 ìæ 2 ö 2 ü è ïç ï ÷ - t2 ý íç ÷ ïîè 3 ø ïþ ½ 2 ½ ½ + t½ ½ 2 + 3t½ 1 1 1 = × log ½ 3 ½+ C = 6 2 log ½ 2 - 3t½+ C 9 2 2 ½ ½ ½ - t½ 2× ½3 ½ 3

Senior Secondary School Mathematics for Class 12 Pg-707

Some Special Integrals

707

æ ½ 2 + 3ç x + ½ è log ½ = æ 6 2 ½ 2 - 3ç x + è ½ 1

=

1 ö½ ÷ 3 ø½ +C 1 ö½ ÷½ 3 ø½

½ 2 + 1 + 3 x½ 1 ½+ C. log ½ 6 2 ½ 2 - 1 - 3 x½

cos x

ò (sin 2x + 4 sin x + 5) dx.

EXAMPLE 17

Evaluate

SOLUTION

Putting sin x = t and cos x dx = dt, we get cos x dt dt ò (sin 2 x + 4 sin x + 5) dx = ò (t 2 + 4t + 5) = ò {(t 2 + 4t + 4) + 1} du , (u2 + 1) where u = (t + 2) = tan -1u + C = tan -1(t + 2) + C

=ò

dt

{(t + 2) 2 + 12}

=ò

= tan -1(sin x + 2) + C. ex

ò ( e2x + 6ex + 5) dx.

EXAMPLE 18

Evaluate

SOLUTION

Putting ex = t and ex dx = dt, we get ex

dt

dt

ò e2x + 6ex + 5 dx = ò (t 2 + 6t + 5) = ò {(t 2 + 6t + 9) - 4} =ò

Integrals of the form METHOD

dt 2

2

{(t + 3) - 2 }

=ò

du 2

(u - 22)

, where (t + 3) = u

=

t + 3 - 2½ u - 2½ 1 1 log ½ + C = log ½ +C ( 2 ´ 2) 4 ½u + 2½ ½t + 3 + 2½

=

½ex + 1½ 1 t + 1½ 1 ½+ C. log ½ + C = log ½ x 4 4 ½t + 5½ ½e + 5½ ( px + q)

ò (ax2 + bx + c) dx.

Let ( px + q) = A ×

d ( ax 2 + bx + c) + B dx

Find A and B. Now, the integrand so obtained can be integrated easily. (5 x - 2)

ò (1 + 2x + 3 x 2) dx.

EXAMPLE 19

Evaluate

SOLUTION

Let (5 x - 2) = A ×

d (1 + 2x + 3 x 2) + B. dx

[CBSE 2013]

Senior Secondary School Mathematics for Class 12 Pg-708

708

EXAMPLE 20

SOLUTION

Senior Secondary School Mathematics for Class 12

Then, (5 x - 2) = A( 6x + 2) + B. … (i) Comparing the coefficients of like powers of x on both sides, we get 6A = 5 and 2A + B = -2. 5 -11 This gives A = and B = × 6 3 11 ü ì5 í ( 6x + 2) - ý 6 3þ dx \ I = òî (1 + 2x + 3 x 2) 5 6x + 2 11 dx = ×ò dx - ò 6 (1 + 2x + 3 x 2) 3 ( 3 x 2 + 2x + 1) dx 5 11 1 = log |1 + 2x + 3 x 2| - × ò 1ö 6 3 3 æ 2 2 çx + x + ÷ 3 3ø è 5 11 dx = log |1 + 2x + 3 x 2| - × ò 2 6 9 ìïæ 1ö æ 1 1 ö üï íç x + ÷ + ç - ÷ ý 3ø è 3 9 ø þï ïîè dx 5 11 +C = log |1 + 2x + 3 x 2| - × ò 2 2 6 9 ì æ 2 ö üï 1ö ïæ ÷ ý íç x + ÷ + çç ÷ 3ø è 3 ø ïþ ïîè ì ü ïx + 1 ï 5 11 1 ï 3 ï+ C = log |1 + 2x + 3 x 2| - × tan -1í ý 6 9 æ 2ö æ öï 2 ïç ç ÷ ÷ ç 3 ÷ ï çè 3 ø÷ ï è ø î þ 5 11 2 -1 æ 3 x + 1 ö = log |1 + 2x + 3 x | tan ç ÷ + C. 6 3 2 2 ø è ( 3 x + 1) Evaluate ò [CBSE 2006] dx. ( 2x 2 - 2x + 3) d Let ( 3 x + 1) = A × ( 2x 2 - 2x + 3) + B. Then, dx … (i) ( 3 x + 1) = A( 4x - 2) + B Comparing the coefficients of like powers of x, we get 3 5ö æ (4A = 3 and B - 2A = 1) Þ ç A = and B = ÷ × 4 2ø è ( 3 x + 1) A × ( 4x - 2) + B \ ò dx = ò ( 2x 2 - 2x + 3) ( 2x 2 - 2x + 3) 3 5 × ( 4x - 2) + 5 dx ( 4x - 2) 2 dx = 3 × =ò 4 2 dx + ò 3ö 2 æ 2 4 ò ( 2x 2 - 2x + 3) ( 2x - 2x + 3) 2ç x - x + ÷ 2ø è

Senior Secondary School Mathematics for Class 12 Pg-709

Some Special Integrals

709

dx 3 5 log|2x 2 - 2x + 3| + × ò 1ö 5 ü 4 4 ìæ 2 íç x - x + ÷ + ý 4ø 4þ îè 3 5 dx 2 = log|2x - 2x + 3| + × ò 2 2 4 4 ì æ 5 ö üï 1ö ïæ ÷ ý íç x - ÷ + çç ÷ 2ø è 2 ø ïþ ïîè ìæ 1ö ü ïç x - ÷ ï 3 5 1 ï 2ø ï = log|2x 2 - 2x + 3| + × tan -1í è ý+ C 4 4 æ 5ö ï æç 5 ö÷ ï ç ÷ ç 2 ÷ ï èç 2 ÷ø ï è ø î þ =

= EXAMPLE 21

SOLUTION

3 5 æ 2x - 1 ö log|2x 2 - 2x + 3| + tan -1 ç ÷ + C. 4 2 è 5 ø

Evaluate

( 2x + 1)

ò ( 4 - 3 x - x 2) dx.

[CBSE 2004C]

d ( 4 - 3 x - x 2) + B. dx Then, ( 2x + 1) = A( -3 - 2x) + B … (i) Comparing the coefficients of like terms, we get ( -2A = 2 and - 3 A + B = 1) Þ ( A = -1, B = -2). ì( -1) × ( -3 - 2x) - 2 ü ( 2x + 1) \ ò dx = ò í ý dx 2 ( 4 - 3 x - x 2) î (4 - 3x - x ) þ ( -3 - 2x) dx = -ò dx - 2 ò ( 4 - 3 x - x 2) ( 4 - 3 x - x 2) dx = - log|4 - 3 x - x 2| + 2 ò 2 ( x + 3 x - 4) dx = - log|4 - 3 x - x 2| + 2 ò 2 3ö 9ö æ æ çx + ÷ - ç4 + ÷ 2ø 4ø è è dx 2 = - log|4 - 3 x - x | + 2 ò 2 2 ìïæ 3ö æ 5 ö üï íç x + ÷ - ç ÷ ý 2ø è 2 ø ïþ ïîè 3 ö 5½ ½æ çx + ÷ - ½ ½ 2 2ø 2 = - log|4 - 3 x - x 2| + log ½è ½+ C 3 æ 5ö æ ½ç x + ÷ö + 5½ ç2 ´ ÷ 2ø 2 ø 2½ è ½è 2 x - 1½ ½ 2 = - log|4 - 3 x - x | + log + C. 5 ½x + 4½ Let ( 2x + 1) = A ×

Senior Secondary School Mathematics for Class 12 Pg-710

710

Senior Secondary School Mathematics for Class 12

EXAMPLE 22

Evaluate

SOLUTION

We have

æ x2 + 5x + 3 ö

ò çç x 2 + 3 x + 2 ÷÷ dx. è

ø

( x 2 + 5 x + 3) ( x 2 + 3 x + 2) Þ

( 2x + 1) ü ì = í1 + 2 ý î x + 3x + 2þ

( x 2 + 5 x + 3)

( 2x + 1)

ò ( x 2 + 3 x + 2) dx = ò dx + ò ( x 2 + 3 x + 2) dx.

… (i)

d 2 ( x + 3 x + 2) + B. Then, dx ( 2x + 1) = A( 2x + 3) + B. Comparing the coefficients of like powers of x, we get ( 2A = 2 and 3 A + B = 1) Þ ( A = 1 and B = -2). \ ( 2x + 1) = ( 2x + 3) - 2. ( 2x + 1) {( 2x + 3) - 2} dx = x + ò 2 dx \ I =x+ò 2 ( x + 3 x + 2) ( x + 3 x + 2) ( 2x + 3) dx =x+ò 2 dx - 2 ò 2 ( x + 3 x + 2) ( x + 3 x + 2) dx = x + log|x 2 + 3 x + 2| - 2 ò 9ö æ 9ö ü ìæ 2 íç x + 3 x + ÷ + ç 2 - ÷ ý 4ø è 4ø þ îè dx 2 = x + log|x + 3 x + 2| - 2 ò 2 2 3ö ïìæ æ 1 ö ïü íç x + ÷ - ç ÷ ý 2ø è 2 ø ïþ ïîè ½x + 3 - 1½ 1 ½ 2 2 2½+ C = x + log|x + 3 x + 2| - 2 × log ½ 3 1½ æ 1ö ½ x+ + ½ ç2 ´ ÷ ½ 2 2½ 2ø è + 1 x ½+ C. = x + log|x 2 + 3 x + 2| - 2 log ½ ½x + 2½ Let ( 2x + 1) = A ×

Integrals of the form and METHOD

dx

… (ii)

dx

ò a + b cos2 x , ò a + b sin 2 x dx

ò a cos2 x + b sin x cos x + c sin 2 x × In each such an integral, we divide the numerator and denominator by cos2 x and put tan x = t, sec2 x dx = dt and then integrate. dx

ò a 2 sin 2x + b2 cos2x ×

EXAMPLE 23

Evaluate

SOLUTION

Dividing the numerator and the denominator of the given integrand by cos2 x , we get

[CBSE 2003C]

Senior Secondary School Mathematics for Class 12 Pg-711

Some Special Integrals

711

sec2 x

dx

ò ( a 2 sin 2x + b2 cos2x) = ò a 2 tan 2x + b2 dx =ò =

= EXAMPLE 24

[putting tan x = t ]

t 1 1 1 dt tan -1 = × +C ò æ bö a2 é 2 æ b ö 2 ù a2 æ b ö ç ÷ ç ÷ êt + ç ÷ ú è aø è aø è a ø úû êë 1 1 æ at ö æ a tan tan -1 ç ÷ + C = tan -1 ç ab ab èbø è b

xö ÷ + C. ø

Evaluate: (i)

SOLUTION

dt ( a 2t 2 + b 2)

dx

ò (1 + 3 sin 2x)

(ii)

dx

ò ( 3 + 2 cos2x)

(i) Dividing the numerator and denominator by cos2 x , we get sec2 x

dx

sec2 x

ò (1 + 3 sin 2x) = ò sec2x + 3 tan 2x dx = ò (1 + 4 tan 2x) dx =ò

dt (1 + 4t 2)

[putting tan x = t]

1 dt t 1 1 = × +C tan -1 4 ò é 2 æ 1 ö 2 ù 4 (1/2) (1/2) + t ê ç ÷ ú è 2 ø úû êë 1 1 = tan -1( 2t) + C = tan -1( 2 tan x) + C. 2 2 =

(ii) Dividing the numerator and denominator by cos2 x , we get sec2 x

dx

sec2 x

ò ( 3 + 2 cos2x) = ò ( 3 sec2x + 2) dx = ò 5 + 3 tan 2x dx dt [putting tan x = t] (5 + 3t 2) 1 1 dt dt = × = ò 2 3 é 2 æ 5 öù 3 ò é æ 5ö ù êt 2 + ç êt + ç 3 ÷ ú ÷ ú è øû ç 3÷ ú ë ê ø û è ë =ò

æ 3t ö t 1 1 1 ÷ +C × tan -1 tan -1 çç +C = ÷ 3 æ 5ö æ 5ö 15 è 5 ø ç ÷ ç ÷ ç 3÷ ç 3÷ è ø è ø ö æ tan x 1 3 ÷ + C. = tan -1 çç ÷ 15 5 ø è =

Senior Secondary School Mathematics for Class 12 Pg-712

712

Senior Secondary School Mathematics for Class 12

dx

ò ( 4 sin 2x + 5 cos2x) ×

EXAMPLE 25

Evaluate

SOLUTION

On dividing the numerator and denominator by cos2 x , we get sec2 x

dx

ò ( 4 sin 2x + 5 cos2x) = ò ( 4 tan 2x + 5) dx dt [putting tan x = t] 4t 2 + 5 1 dt dt 1 = ò = × 2 4 æ 2 5ö 4 ò é æ 5ö ù çt + ÷ êt 2 + ç ÷ ú 4ø è ç 2 ÷ ú ê è ø û ë t 1 1 = × tan -1 +C 4 æ 5ö æ 5ö ç ÷ ç ÷ ç 2 ÷ ç 2 ÷ è ø è ø 1 1 æ 2 tan x ö -1 æ 2t ö = tan ç tan -1 ç ÷ +C = ÷ + C. 2 5 2 5 5 ø è 5ø è dx ò (1 + 3 sin 2x + 8 cos2x) × =ò

EXAMPLE 26

Evaluate

SOLUTION

On dividing the numerator and denominator by cos2 x , we get sec2 x dx

dx

ò (1 + 3 sin 2x + 8 cos2x) = ò sec2x + 3 tan 2x + 8 =ò

sec2 x 9 + 4 tan 2 x

dx = ò

dt

9 + 4t 2 [putting tan x = t] 1 dt dt 1 = ò = × 4 æ 2 9ö 4 ò é 2 æ 3 ö 2 ù + t ç ÷ êt + ç ÷ ú 4ø è è 2 ø úû êë ì t ü 1 1 = × × tan -1í ý+ C 4 ( 3/2) î( 3/2) þ 1 1 æ 2 tan x ö æ 2t ö = tan -1 ç ÷ + C = tan -1 ç ÷ + C. 6 6 è 3ø è 3 ø sin x

ò sin 3 x dx.

EXAMPLE 27

Evaluate

SOLUTION

We have sin x sin x ò sin 3 x dx = ò ( 3 sin x - 4 sin 3x) dx 1 =ò dx [dividing num. and denom. by sin x] ( 3 - 4 sin 2 x)

Senior Secondary School Mathematics for Class 12 Pg-713

Some Special Integrals

sec2 x

=ò

2

3 sec x - 4 tan 2 x

[dividing num. and denom. by cos2 x]

dx

sec2 x

=ò

3(1 + tan x) - 4 tan x

=ò

( 3 - t 2)

2

dt

=

2

dx = ò

sec2 x ( 3 - tan 2 x)

dx

, where tan x = t and sec2 x dx = dt

dt

=ò

713

2

2

[( 3 ) - t ]

½ 3 + t½ ½+ C log ½ 2 3 ½ 3 - t½ 1

=

½ 3 + tan x½ ½ + C. log ½ 2 3 ½ 3 - tan x½ 1

cos x

ò cos 3 x dx.

EXAMPLE 28

Evaluate

SOLUTION

We have cos x cos x ò cos 3 x dx = ò ( 4 cos 3x - 3 cos x) dx =ò =ò

dx 4 cos2 x - 3

dx

=ò

4 cos2 x - 3(sin 2 x + cos2 x)

dx 2

=ò

2

cos x - 3 sin x

sec2 x 1 - 3 tan 2 x

dx

[on dividing the num. and denom. by cos2 x] =ò

dt 1 - 3t 2

, where tan x = t and sec2 x dx = dt

½ 1 + t½ 1 dt dt 1 1 1 ½ ½ = × = ×ò = × log ½ 3 ½ + C 1 3 æ 1 2 ö 3 ò é æ 1 ö 2 2 ù 3 2× 1 ½ - t½ ç -t ÷ êç ÷ -t ú 3 3 ½ ½ è3 ø úû êë è 3 ø =

½1 + 3 t½ ½1 + 3 tan x½ 1 1 ½+ C = ½ + C. log ½ log ½ 2 3 1 3 t 2 3 ½ ½ ½1 - 3 tan x½

EXAMPLE 29

Evaluate

SOLUTION

We have dx

dx

ò ( 2 + cos x) × dx

dx

ò ( 2 + cos x) = ò 1 + (1 + cos x) = ò 1 + 2 cos2( x/2) = ò

sec2( x/2) dx sec2( x/2) + 2

[dividing the num. and denom. by cos2( x/2)]

Senior Secondary School Mathematics for Class 12 Pg-714

714

Senior Secondary School Mathematics for Class 12

sec2( x/2)

dt dx = 2ò , where tan ( x/2) = t 3 + tan 2( x/2) 3 + t2 1 dt t = 2× ò = 2× +C tan -1 2 2 3 3 ( 3) + t 2 é tan ( x/2) ù = tan -1 ê + C. 3 3 úû ë =ò

Some More Special Integrals EXAMPLE 30

Evaluate

SOLUTION

We have

( x 2 + 1)

ò ( x 4 + 1) dx.

[CBSE 2006C, ‘07, ‘11C]

1 ö æ ç1 + 2 ÷ æ x2 + 1ö è ø x 2 ò çç x 4 + 1 ÷÷ dx = ò æ 2 1 ö dx [dividing num. and denom. by x ] è ø çx + 2 ÷ è x ø 1 ö æ ç1 + 2 ÷ è ø dx x =ò 2 1ö æ çx - ÷ + 2 xø è dt 1ö 1 ö æ æ =ò 2 , where ç x - ÷ = t and ç1 + 2 ÷ dx = dt xø è è x ø [t + ( 2) 2] 1ö æ çx - ÷ t 1 1 æ ö x÷ +C = tan -1 ç tan -1 ç ÷ +C = 2 2 è 2ø çç 2 ÷÷ ø è 2 æ ö x - 1÷ 1 = tan -1 ç + C. ç 2x ÷ 2 è ø EXAMPLE 31

Evaluate

SOLUTION

We have

( x 2 + 4)

ò ( x 4 + 16) dx.

[CBSE 2007C, ‘11C]

4ö æ ç1 + 2 ÷ æ x2 + 4 ö è ø x 2 ò çç x 4 + 16 ÷÷ dx = ò æ 2 16 ö dx [dividing num. and denom. by x ] è ø çx + 2 ÷ è x ø 4ö æ ç1 + 2 ÷ x ø dx =ò è 2 4ö æ çx - ÷ + 8 xø è

Senior Secondary School Mathematics for Class 12 Pg-715

Some Special Integrals

715

4ö 4ö æ æ [putting ç x - ÷ = t and ç1 + 2 ÷ dx = dt] xø è è x ø

dt

=ò

(t + 8)

=ò

t 2 + ( 8) 2

2

dt

=

1 æ t ö tan -1 ç ÷ +C 8 è 8ø

4ö æ çx - ÷ æ x2 - 4ö 1 1 xø -1 è ÷ + C. +C = = tan tan -1 ç ç 2 2x ÷ 8 8 2 2 è ø EXAMPLE 32

Evaluate

SOLUTION

We have

( x 2 - 1)

ò ( x 4 + x 2 + 1) dx.

1 ö æ ç1 - 2 ÷ è x ø ò ( x 4 + x 2 + 1) dx = ò æ 2 1 ö dx ç x + 2 + 1÷ è ø x ( x 2 - 1)

[dividing num. and denom. by x 2] =ò

1 ö æ ç1 - 2 ÷ è x ø

2 éæ ù 1ö ê ç x + ÷ - 1ú xø êë è úû

dx = ò

dt (t 2 - 1)

1ö 1 ö æ æ [putting ç x + ÷ = t and ç1 - 2 ÷ dx = dt] xø è è x ø ½x + 1 t - 1½ 1 ½ ½ + C = log = log ½ 2 t 1 2 + ½ ½ ½x + ½ =

EXAMPLE 33

Evaluate

SOLUTION

We have

½x 2 - x + 1½ 1 ½+ C. log ½ 2 2 ½x + x + 1½

dx

ò ( x 4 + 1) ×

dx

ò ( x 4 + 1) = ò =

( x 2 + 1) - ( x 2 - 1) 2( x 4 + 1)

dx

1 ( x 2 + 1) 1 ( x 2 - 1) dx - ò 4 dx ò 4 2 ( x + 1) 2 ( x + 1)

1 ½ -1 x ½+ C 1 ½ + 1½ x ½

Senior Secondary School Mathematics for Class 12 Pg-716

716

Senior Secondary School Mathematics for Class 12

ù 1 ö 1 ö é æ æ ç1 + 2 ÷ ç1 - 2 ÷ ú 1ê è ø è ø x x dx - ò = êò dx ú 2ê æ 2 1 ö æ 2 1 ö ú çx + 2 ÷ çx + 2 ÷ è x ø úû x ø ëê è [dividing num. and denom. of each integral by x 2] ù é 1 ö 1 ö æ æ ú ê ç1 + 2 ÷ ç1 - 2 ÷ ú 1ê è ø è ø x x = êò dx ú dx - ò 2 2 éæ ù ú ù 2 ê éæ 1ö 1ö x - ÷ + 2ú ê ç x + ÷ - 2ú ú ê ê çè x x ø ø êë è úû û úû ë êë =

ù 1é dt du -ò 2 ú êò 2 2 2 2 ë [t + ( 2) ] [u - ( 2) ] û

1ö 1ö æ æ [putting ç x - ÷ = t in the 1st integral, and ç x + ÷ = u in the 2nd] xø xø è è ½u - 2½ïü 1 ïì 1 1 æ t ö ½ý + C = í tan -1 ç log ½ ÷2 îï 2 2 2 2 ø è ½u + 2½ïþ 1ö æ ½x + 1 - 2½ çx - ÷ 1 1 ½ ½ -1 ç x x ÷ = tan log +C ½ 1 2 2 2 4 2 ½x + + 2½ ½ ÷÷ çç ½ ½ ø è x =

EXAMPLE 34

Evaluate

SOLUTION

We have I=

æ x2 - 1ö ½ 2 ½ ÷ - 1 log ½x + 1 - 2x½ + C. tan -1 ç 2 ç 2x ÷ 4 2 2 2 ½x + 1 + 2x½ è ø 1

x2

ò ( x 4 + x 2 + 1) dx.

[CBSE 2008C]

1 2x 2 dx 2 ò ( x 4 + x 2 + 1)

=

1 ( x 2 - 1) + ( x 2 + 1) × dx 2 ò ( x 4 + x 2 + 1)

=

1 ( x 2 - 1) 1 ( x 2 + 1) + dx dx 2 ò ( x 4 + x 2 + 1) 2 ò ( x 4 + x 2 + 1)

1 ö 1 ö æ æ ç1 - 2 ÷ ç1 + 2 ÷ 1 1 è x ø dx x ø dx + = ò è 1 ö 1 ö 2ò æ 2 2 æ 2 çx + 1 + 2 ÷ çx + 1 + 2 ÷ è è x ø x ø [on dividing num. and denom. of each by x 2 ]

Senior Secondary School Mathematics for Class 12 Pg-717

Some Special Integrals

717

1 ö 1 ö æ æ ç1 + 2 ÷ ç1 - 2 ÷ 1 1 è x ø dx + x ø = ò è dx 2 2 2 ò ïìæ ü 2 ïìæ 1ö 1ö ïü 2ï íç x - ÷ + ( 3 ) ý íç x + ÷ - 1ý xø xø ïþ ïþ ïîè ïîè 1 du dv 1 = ò 2 + 2 (u - 1) 2 ò {v 2 + ( 3 ) 2} 1ö 1 ö ì ü æ æ ïputting ç x + x ÷ = u and ç1 - 2 ÷ = du in I1 , ï ø è è ï ï x ø í ý 1 1 æ ö æ ö ï and ç x - ÷ = v and ç1 + 2 ÷ dx = dv in I 2 ï ïî ïþ xø è è x ø u - 1½ 1 1 v 1 = log ½ + × tan -1 +C 2 3 ½u + 1½ 2 3 ½x + 1 - 1½ 2 1 ½ x ½+ 1 tan -1 æç x - 1 ö÷ + C = log ç 3x ÷ ½ 2 ½x + 1 + 1½ ½ 2 3 ø è ½ x ½ =

æ x2 + 1ö ½x 2 - x + 1½ 1 1 ÷ + C. ½+ log ½ 2 tan -1 ç ç 3x ÷ 2 ½x + x + 1½ 2 3 è ø

ò

EXAMPLE 35

Evaluate

SOLUTION

Put cot x = t 2 so that - cosec2 x dx = 2t dt or dx = \

ò

cot x dx.

cot x dx = - ò = -ò

2t 2 4

(t + 1)

-2t (1 + t 4)

dt.

dt

[(t 2 + 1) + (t 2 - 1)] 4

(t + 1)

dt = - ò

(t 2 + 1) 4

(t + 1)

dt - ò

(t 2 - 1) (t 4 + 1)

dt

1ö 1ö æ æ ç1 + 2 ÷ ç1 - 2 ÷ è ø è ø dt t t dt - ò = -ò æ 2 1ö æ 2 1ö çt + 2 ÷ çt + 2 ÷ è è t ø t ø 1ö 1ö æ æ ç1 + 2 ÷ ç1 - 2 ÷ è ø è ø t t dt = -ò dt - ò é æ 1ö 2 ù é æ 1ö 2 ù ê çt + ÷ - 2ú ê çt - ÷ + 2ú êë è t ø úû êë è t ø úû =-ò

du [u2 + ( 2) 2]

-ò

dv [v 2 - ( 2) 2]

æ 1ö æ 1ö [putting çt - ÷ = u in the 1st integral, and çt + ÷ = v in the 2nd] è tø è tø

Senior Secondary School Mathematics for Class 12 Pg-718

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Senior Secondary School Mathematics for Class 12

½v - 2½ 1 1 æ u ö ½+ C tan -1 ç log ½ ÷2 è 2ø 2 2 ½v + 2½ æ 1ö ½t + 1 - 2½ çt - ÷ 1 1 ½ t ½ -1 ç t÷=tan log +C ½ ½ 1 2 2 2 2 ½ ÷÷ çç t + + 2½ ½ t ½ ø è 2 2 æ ö ½ t - 1÷ t - 2 t + 1½ 1 1 ½+ C =tan -1 ç log ½ 2 ç ÷ 2 ½t + 2 t + 1½ è 2t ø 2 2 æ cot x - 1 ö ½cot x - 2 cot x + 1½ 1 ÷ - 1 log ½ ½+ C. =tan -1 ç ç 2 cot x ÷ 2 2 2 ½cot x + 2 cot x + 1½ è ø =-

EXAMPLE 36

Evaluate

SOLUTION

We have

ò(

ò(

tan x + cot x ) dx.

[CBSE 2010C, ’13C]

tan x + cot x ) dx

æ 1 ö÷ (tan x + 1) = ò ç tan x + dx = ò dx ç ÷ tan x tan x è ø (t 2 + 1) 2t =ò × dt, where tan x = t 2 Þ x = tan -1t 2 t (1 + t 4) 2t Þ dx = dt (1 + t 4) 1ö æ ç1 + 2 ÷ (t 2 + 1) t ø dt [on dividing num. and denom. by t 2 ] = 2ò 4 dt = 2ò è æ 2 1ö (t + 1) çt + 2 ÷ è t ø 1ö æ ç1 + 2 ÷ t ø dt = 2ò è 2 æ 1ö çt - ÷ + 2 è tø 1ö du æ æ 1ö = 2ò 2 , where çt - ÷ = u and ç1 + 2 ÷ dt = du t è ø è (u + 2) t ø æ 1ö çt - ÷ 1 u é æ 1ö ù tø +C = 2× tan -1 + C = 2 tan -1 è êQ u = çt - t ÷ ú 2 2 2 è øû ë 2 æ ö (t - 1) tan x - 1 ÷ = 2 tan -1 + C = 2 tan -1 ç [Q t 2 = tan x]. +C ç 2 tan x ÷ ( 2 t) è ø EXAMPLE 37 SOLUTION

Evaluate

ò

tan q dq.

Putting tan q = t 2 , we get q = tan -1t 2 Þ dq =

[CBSE 2006]

2t (1 + t 4)

dt.

Senior Secondary School Mathematics for Class 12 Pg-719

Some Special Integrals

\ I = òt × =ò

2t 4

(1 + t )

dt = ò

2t 2 4

(t + 1)

(t 2 + 1) + (t 2 - 1) (t 4 + 1)

dt = ò

719

dt (t 2 + 1) (t 4 + 1)

dt + ò

(t 2 - 1) (t 4 + 1)

dt

1ö 1ö æ æ ç1 + 2 ÷ ç1 - 2 ÷ è ø è ø dt t t dt + ò =ò æ 2 1ö æ 2 1ö çt + 2 ÷ çt + 2 ÷ è è t ø t ø 1ö 1ö æ æ ç1 - 2 ÷ ç1 + 2 ÷ è ø dt è ø t t =ò dt + ò 2 üï ìïæ 1 ö 2 æ 1ö çt - ÷ + 2 íçt + ÷ - 2ý t ø è ïþ ïîè t ø du dv =ò 2 +ò 2 , (u + 2) (v - 2) æ 1ö æ 1ö where çt - ÷ = u and çt + ÷ = v in I1 and I 2 t è ø è tø =

½v - 2½ u 1 1 ½+ C tan -1 + × log ½ 2 2 2 2 ½v + 2½

æ 1ö ½t + 1 - 2½ çt - ÷ 1 1 ½ t ½ -1 ç t ÷+ × = tan +C 2 ½t + 1 + 2½ ½ çç 2 ÷÷ 2 2 ½ ½ t ½ ø è =

æ t2 - 1ö ½2 ½ 1 ÷ + 1 ×½t - 2 t + 1½+ C , where t = tan q. tan -1 ç 2 ç ÷ 2 è 2 t ø 2 2 ½t + 2 t + 1½

EXERCISE 14A Evaluate: dx 1. ò (1 - 9x 2)

2.

dx

4.

ò ( 4 + 9x 2)

8.

ò ( 9 + 4x 2) dx

12.

x2

3 x5

ò (1 + x12)

dx

dx

ò ( 25 - 4x 2) dx

5.

ò (50 + 2x 2)

9.

ò ( e2x + 1) dx

13.

ex

2x 3

ò ( 4 + x 8) dx

3. 6.

dx

ò ( x 2 + 16)

[CBSE 2011]

dx

ò (16x 2 - 25) sin x

10.

ò (1 + cos2x) dx

14.

ò ( ex + e- x )

dx

7.

( x 2 - 1)

ò ( x 2 + 4) dx cos x

11.

ò (1 + sin 2 x) dx

15.

ò (1 - x 4) dx

x

Senior Secondary School Mathematics for Class 12 Pg-720

720

Senior Secondary School Mathematics for Class 12

x2

dx

16.

ò ( a 6 - x 6) dx

19.

ò ( 2x 2 + x + 3)

22.

ò ( x 2 + 3 x + 2) dx

25.

ò ( x 2 + 6x - 3) dx

28.

ò ( 2 + x - x 2) dx

31.

ò ( a 2 cos2x + b2 sin 2x) 32. ò (cos2x - 3 sin 2x)

34.

ò (sin x cos x + 2 cos2x)

36.

ò ( 6 - cos2f - 4 sin f) df

38.

dx x

x2

2x

17.

ò ( x 2 + 4x + 8)

20.

ò ( 2x 2 - x - 1)

23.

ò ( x 2 + 2x - 4) dx

26.

ò ( 2x 2 + 2x + 1) dx

29.

ò (1 + cos2x)

dx

( x - 3)

( 2x - 1) dx

dx

dx

dx

( 2 sin 2f - cos f)

(1 - x 2)

ò (1 + x 4)

37.

dx [CBSE 2007]

39.

dx

18.

ò ( 4x 2 - 4x + 3)

21.

ò ( 3 - 2x - x 2)

24.

ò ( x 2 + 3 x - 18) dx

27.

ò ( 3 x 2 + 4x + 2) dx

30.

ò

33.

ò (sin 2x - 4 cos2x)

35.

ò (sin 4x + cos4x) dx

dx

( 2x - 3)

(1 - 3 x) dx

( 2 + sin 2 x) dx sin 2x

dx

ò (sin x - 2 cos x)( 2 sin x + cos x)

( x 2 + 1)

ò ( x 4 + x 2 + 1) dx

40.

dx

ò (sin 4x + cos4x)

ANSWERS (EXERCISE 14A)

1 1 + 3 x½ log ½ +C 6 ½1 - 3 x½ 1 æ 3x ö 4. tan -1 ç ÷ +C 6 è 2 ø x 5 7. x - tan -1 + C 2 2 1.

10. - tan -1(cos x) + C 13.

æ x4 ö 1 tan -1 ç ÷ + C ç 2÷ 4 è ø

1 5 + 2x½ log ½ +C 20 ½5 - 2x½ x 1 5. tan -1 + C 10 5 x 3 æ 2x ö 8. - tan -1 ç ÷ + C 4 8 è 3ø 2.

11. tan -1(sin x) + C 14. tan -1( ex ) + C

½a 3 + x 3½ 1 æ x + 2ö ½+ C 17. tan -1 ç log ½ 3 ÷ +C 3 2 è 2 ø 6a a x ½ ½ 1 2 æ 4x + 1 ö ½2( x - 1)½+ C 19. tan -1 ç ÷ + C 20. log 3 23 ½ 2x + 1 ½ è 23 ø

16.

22.

1

3

x + 1½ 1 3 log |x 2 + 3 x + 2| - log ½ +C 2 2 ½x + 2½

x 1 tan -1 + C 4 4 1 4 x - 5½ 6. log ½ +C 40 4 x ½ + 5½ 3.

9. tan -1( ex ) + C 1 tan -1( x 6) + C 2 ½1 + x 2½ 1 ½+ C 15. log ½ 2 4 ½1 - x ½ 12.

18.

1 æ 2x - 1 ö tan -1 ç ÷ +C 2 2 è 2 ø

21.

1 3 + x½ log ½ +C 4 ½1 - x ½

Senior Secondary School Mathematics for Class 12 Pg-721

Some Special Integrals

½x + 1 1 2 log |x 2 + 2x - 4| log ½ 2 5 ½x + 1 + 2 x - 3½ 24. log |x 2 + 3 x - 18| - log ½ +C 3 ½x + 6½ 23.

25. x - 3 log|x 2 + 6x - 3| +

721

5½ ½+ C 5½

½x + 3 - 2 3 ½ 7 3 ½+ C log ½ 4 ½x + 3 + 2 3½

1 log |2x 2 + 2x + 1| - 2 tan -1( 2x + 1) + C 2 1 3 æ 3x + 2ö 27. - log |3 x 2 + 4x + 2| + tan -1 ç ÷+C 2 2 2 ø è 1 1 + x½ 28. - log |2 + x - x 2| + log ½ +C 3 ½2 - x½ 26.

30. 32.

æ 3 tan 1 tan -1 çç 6 2 è

xö ÷ +C ÷ ø

½1 + 3 tan x½ 1 ½+ C log ½ 2 3 ½1 - 3 tan x½

29.

1 æ tan x ö tan -1 ç ÷ +C 2 è 2 ø

31.

1 ö æb tan -1 ç tan x ÷ + C ab ø èa

33.

1 ½tan x - 2½ ½+ C log ½ 4 ½tan x + 2½

35. tan -1(tan 2 x) + C

34. log |tan x + 2| + C

36. 2 log |sin 2 f - 4 sin f + 5| + 7 tan -1(sin f - 2) + C 37.

1 ½ tan x - 2 ½ ½+ C log ½ 5 ½2 tan x + 1½

39.

( x 2 - 1) 1 tan -1 +C 3 3x

½ 2x + x 2 + 1½ ½+ C log ½ 2 2 2 ½ 2x - x - 1½ 1

38.

40.

æ tan 2 x - 1 ö 1 ÷ +C tan -1 ç ç 2 tan x ÷ 2 è ø

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 14A) 1. On dividing, we get

üï ìï 5 = í1 - 2 ý× ( x + 4 ) îï ( x + 4 ) þï ( x 2 - 1) 2

8. On dividing x 2 by ( 4 x 2 + 9 ), we get

x2 2

( 9 + 4x )

=

(9 ) 1 - 24 × 4 4x + 9

dx 1 9 × × \ I = ò dx 4 19 ò ìï 2 æ 3 ö 2 ïü x + ç ÷ ý í è 2 ø ïþ ïî 9. Put e x = t and e xdx = dt . 10. Put cos x = t and - sin x dx = dt. 12. Put x 6 = t and 6 x5 dx = dt . 14. Multiply numerator and denominator by e x and put e x = t .

Senior Secondary School Mathematics for Class 12 Pg-722

722

Senior Secondary School Mathematics for Class 12

15. Put x 2 = t and 2x dx = dt. 16. Put x 3 = t and 3 x 2 dx = dt. d ( x 2 + 3 x + 2 ) + B. Then, x = A × ( 2 x + 3 ) + B. dx 1 æ æ -3 ö \ (2 A = 1 and 3 A + B = 0 ) Þ ç A = and B = ç ÷× 2 è è 2 ø

22. Let x = A ×

d 2 ( x + 2 x - 4 ) + B Þ ( x - 3 ) = A( 2 x + 2 ) + B. dx 1 æ ö \ (2 A = 1 and 2 A + B = - 3 ) Þ ç A = and B = - 4 ÷ × 2 è ø

23. Let ( x - 3 ) = A ×

24. Let ( 2 x - 3 ) = A ×

d 2 ( x + 3 x - 18 ) + B Þ ( 2 x - 3 ) = A ( 2 x + 3 ) + B. dx

25. On dividing x 2 by ( x 2 + 6 x - 3 ), we get ( 2 x - 1) 6x - 3 ü ì I = ò í1 - 2 dx. ý dx = x - 3 ò 2 î x + 6x - 3 þ ( x + 6x - 3) Let ( 2 x - 1) = A ×

d ( x 2 + 6 x - 3 ) + B Þ ( 2 x - 1) = A ( 2 x + 6 ) + B. dx

29. On dividing num. and denom. by cos 2 x, we get I=ò

sec2 x ( 2 + tan 2 x )

dx = ò

dt (2 + t2 )

, where tan x = t.

30. Divide num. and denom. by cos 2 x. 31. Divide num. and denom. by cos 2 x and put tan x = t. 32. Divide num. and denom. by cos 2 x and put tan x = t. 33. Divide num. and denom. by cos 2 x and put tan x = t. 34. Divide num. and denom. by cos 2 x and put tan x = t. 35. On dividing num. and denom. by cos 4 x and putting tan x = t , sec2 x dx = dt , we get I=ò

2t (t 4 + 1)

dt . Now put t 2 = u.

( 4 sin f - 1) cos f ( 4t - 1) df = ò 2 dt , where sin f = t. (sin 2 f - 4 sin f + 5 ) (t - 4 t + 5 ) d Let ( 4 t - 1) = A × (t 2 - 4 t + 5 ) + B. dt dx 37. I = ò × ( 2 sin 2 x - 3 sin x cos x - 2 cos 2 x )

36. I = ò

Now, divide num. and denom. by cos 2 x and put tan x = t. 1 ö 1 ö æ æ ç1- 2 ÷ ç1- 2 ÷ è ø è ø x x 38. I = - ò dx = - ò 2 1 ö æ 2 1 æ ö çx + 2 ÷ çx + ÷ - 2 è x ø xø è

Senior Secondary School Mathematics for Class 12 Pg-723

Some Special Integrals 1 ö 1ö æ æ , where ç x + ÷ = t and ç 1 - 2 ÷ dx = dt xø è è x ø

(t 2 - 2 )

=ò

( 2 )2 - t 2

=

39.

- dt

=ò

dt

1 2 2

log

½ ½ ½ ½ ½

½ 2 + t½ 1 ½+ C log½ 2 2 ½ 2 - t½ 1 2 + x+ ½ 2 x + x 2 + 1½ x½ + C = 1 log½ ½ ½+ C. 2 1½ 2 2 ½ 2 x - x - 1½ 2 -x- ½ x½ =

1 (sin 4 x + cos 4 x )

=

sec4 x (tan 4 x + 1)

=

( 1 + tan 2 x ) sec2 x (tan 4 x + 1)

×

Now, put tan x = t.

Three More Special Integrals THEOREM

PROOF

(i)

ò

(ii)

ò

(iii)

ò

dx 2

a -x dx

2

x 2 - a2 dx x 2 + a2

= sin -1

x + C. a

= log x + x 2 - a 2 + C. = log x +

x 2 + a 2 + C.

(i) Put x = a sin q so that dx = a cos q dq. dx a cos q x =ò dq = ò dq = q + C = sin -1 + C. \ ò 2 2 a cos q a a -x dx -1 x Hence, ò = sin + C. a a2 - x 2 (ii) Put x = a sec q so that dx = a sec q tan q dq. dx a sec q tan q =ò dq \ ò 2 2 a tan q x -a

= ò sec q dq = log |sec q + tan q| + c

= log sec q + sec2 q - 1 + c ½x ½x + x 2 - a2 ½ æ x2 ö½ = log ½ + ç 2 - 1÷ ½+ c = log ½ ½+ c ça ÷ a a è ø½ ½ ½ ½ = log x + x 2 - a 2 - log a + c = log x + x 2 - a 2 + C [taking - log a + c = C]. Hence, ò

dx 2

x - a2

= log x + x 2 - a 2 + C.

723

Senior Secondary School Mathematics for Class 12 Pg-724

724

Senior Secondary School Mathematics for Class 12

(iii) Put x = a tan q so that dx = a sec2 q dq.

ò

\

dx x 2 + a2

=ò

a sec2 q dq a sec q

= ò sec q dq = log |sec q + tan q| + c = log

1 + tan 2 q + tan q + c

½ x2 x½ = log ½ 1 + 2 + ½+ c a a ½ ½ = log x + x 2 + a 2 - log a + c = log x + x 2 + a 2 + C [taking - log a + c = C]. Hence,

ò

dx 2

x + a2

= log x + x 2 + a 2 + C.

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Evaluate: dx (i) ò 9 - 25 x 2 We have dx 1 (i) ò = ×ò 9 - 25 x 2 5

(ii)

dx

ò

(iii)

4x 2 - 9

ò

dx dx 1 = ×ò 2 5 9 2 æ 3ö -x - x2 ÷ ç 25 è5ø æ x ö 1 1 5x ÷÷ + C = sin -1 æç ö÷ + C. = sin -1 çç / 5 3 5 5 è 3ø è ø

(ii)

ò

dx 2

4x - 9

=

1 2

ò

dx 9 x2 4

=

1 2

ò

dx æ 3ö x2 - ç ÷ è 2ø

2

½ 1 9½ log½x + x 2 - ½+ C 2 4½ ½ 1 = log 2x + 4x 2 - 9 + C. 2 dx 1 dx dx 1 = ×ò = ×ò 2 2 4 4 25 16x + 25 æ5 ö x2 + x2 + ç ÷ 16 è 4ø =

(iii)

ò

½ 1 25 ½ × log½x + x 2 + ½+ C 4 16 ½ ½ 1 = log 4x + 16x 2 + 25 + C. 4 =

dx 16x 2 + 25

Senior Secondary School Mathematics for Class 12 Pg-725

Some Special Integrals

Evaluate

SOLUTION

We have

SOLUTION

EXAMPLE 4 SOLUTION

15 - 8x 2

dx

×

[CBSE 2002C]

1 dx ×ò 8 15 15 - 8x - x2 8 ì ü ï ï æ 8 ö 1 1 -1 ï x ï = sin í sin -1 çç x ÷÷ + C. ý+ C = 2 2 2 2 è 15 ø ï 15 ï îï 8 ïþ cos x Evaluate ò dx. 4 - sin 2 x Putting sin x = t and cos x dx = dt, we get cos x dt dt =ò dx = ò ò 2 2 2 4 - sin x 4 -t 2 - t2 t æ sin x ö = sin -1 + C = sin -1 ç ÷ + C. 2 è 2 ø dx Evaluate ò × 1 - e2x Multiplying the numerator and denominator by e- x , we get

ò

EXAMPLE 3

dx

ò

EXAMPLE 2

725

2

=

dx

ò

1 - e2x

=ò = -ò

e- x dx e-2x (1 - e2x ) dt 2

t -1

=ò

e- x e -2 x - 1

dx

[putting e- x = t ]

= - log t + t 2 - 1 + C = - log e- x + e-2x - 1 + C. EXAMPLE 5 SOLUTION

Evaluate

dx. 1 - 4x Putting 2x = t and ( 2x log 2) dx = dt, we get

ò

EXAMPLE 6 SOLUTION

2x

ò

2x x

1-4

Evaluate

ò

dt 1 × (log 2) ò 1 - t 2 1 1 = × sin -1 t + C = × sin -1( 2x ) + C. (log 2) (log 2)

dx =

x2

dx. x6 - 1 1 Putting x 3 = t and x 2 dx = dt , we get 3

Senior Secondary School Mathematics for Class 12 Pg-726

726

Senior Secondary School Mathematics for Class 12

ò

EXAMPLE 7

SOLUTION

x2 6

x -1

1 1 dt = log t + t 2 - 1 + C 3 ò t2 - 1 3 1 = log x 3 + x 6 - 1 + C. 3

dx =

Evaluate: sin x (i) ò dx 4 cos2 x - 1

(ii) ò

sec2 x tan 2 x - 4

dx

(i) Putting cos x = t and - sin x dx = dt, we get - dt sin x 1 dt =- ò dx = ò ò 2 2 2 2 4 cos x - 1 4t - 1 t - (1/2) 2 ½ 1 1½ = - × log ½t + t 2 - ½+ C 2 4½ ½ 1 log 2t + 4t 2 - 1 + C 2 1 = - log 2 cos x + 4 cos2 x - 1 + C. 2 =-

(ii) Putting tan x = t and sec2 x dx = dt , we get

ò

sec2 x 2

tan x - 4

dx = ò

dt 2

t -4

= log t + t 2 - 4 + C

= log tan x + tan 2 x - 4 + C.

ò

x2

EXAMPLE 8

Evaluate

SOLUTION

Putting x 3 = t and x 2 dx =

EXAMPLE 9 SOLUTION

x 6 + a6

dx.

1 dt , we get 3 2 x 1 dt 1 2 6 ò 6 6 dx = 3 ò 2 3 2 = 3 log t + t + a + C x +a t + (a ) 1 = log x 3 + x 6 + a 6 + C. 3 2 sec x Evaluate ò dx. 16 + tan 2 x Putting tan x = t and sec2 x dx = dt , we get sec2 x

ò

2

16 + tan x

dx = ò

dt 16 + t

2

= log t + t 2 + 16 + C

= log tan x + tan 2 x + 16 + C. EXAMPLE 10

Evaluate

ò

1-x dx. 1+ x

[CBSE 2006]

Senior Secondary School Mathematics for Class 12 Pg-727

Some Special Integrals SOLUTION

727

We have

ò

ì 1-x 1-x 1-x ü dx = ò í ´ ý dx 1+ x 1-x þ î 1+ x (1 - x) dx x =ò -ò dx = ò dx 1 - x2 1 - x2 1 - x2 ( -2x) 1 = sin -1 x + × ò dx 2 1 - x2 1 dt = sin -1 x + ò , where (1 - x 2) = t and ( -2x) dx = dt 2 t 1

1 -1 2 1 t 2 t dt = sin -1 x + × +C ò 2 2 ( 1 2)

= sin -1 x +

= sin -1 x + 1 - x 2 + C.

METHOD EXAMPLE 11 SOLUTION

dx

ò

Integrals of the form

( ax 2 + bx + c)

×

Put( ax 2 + bx + c) in the form a{( x + a ) 2 ± b 2} and then integrate. dx

ò

Evaluate

2

x - 3x + 2

×

[CBSE 2006]

We have

ò

dx 2

x - 3x + 2

=ò

dx 9ö 1 æ 2 ç x - 3x + ÷ 4ø 4 è

=ò

dx 2

3ö æ1ö æ çx - ÷ - ç ÷ 2ø è 2ø è 3 dz æ ö , where ç x - ÷ = z =ò 2 2 è ø 1 æ ö z2 - ç ÷ è 2ø ½ 1½ = log½z + z 2 - ½+ C 4½ ½ é êQ êë

ò

2

ù = log z + z 2 - a 2 + C ú úû z -a dz

2

2

3ö ½æ ½ = log½ç x - ÷ + x 2 - 3 x + 2½ + C. 2 ø è ½ ½ EXAMPLE 12

Evaluate

SOLUTION

We have

ò

ò

dx 5 x 2 - 2x

dx 5 x 2 - 2x

=

×

1 × 5 ò

dx 2 x - x 5 2

=

1 × 5 ò

dx 2

2 æ1ö æ1ö x2 - x + ç ÷ - ç ÷ 5 è5 ø è5 ø

2

Senior Secondary School Mathematics for Class 12 Pg-728

728

Senior Secondary School Mathematics for Class 12

1 dx dt 1 = ×ò , × 2 2 2 5 ò æ 5 1ö æ1ö 2 æ1ö t -ç ÷ çx - ÷ - ç ÷ 5ø è5 ø è5 ø è 1ö æ where ç x - ÷ = t 5ø è 2½ ½ 1 2 æ1ö = × log½ t + t ½ ç ÷ ½ ½+ C 5 è5 ø ½ ½ ½æ 1 1ö 2x ½ = log½ç x - ÷ + x 2 - ½+ C. 5ø 5 ½ 5 ½è cos x Evaluate ò [CBSE 2006C] dx. sin 2 x - 2 sin x - 3 Putting sin x = t and cos x dx = dt, we get cos x dt dt =ò dx = ò ò 2 2 sin x - 2 sin x - 3 t - 2t - 3 (t - 1) 2 - 22 =

EXAMPLE 13 SOLUTION

= log (t - 1) + (t - 1) 2 - 22 + C = log (sin x - 1) + sin 2 x - 2 sin x - 3 + C.

ò

ex

EXAMPLE 14

Evaluate

SOLUTION

Putting ex = t and ex dx = dt, we get

ò

5 - 4 ex - e2x

ex x

5 - 4e - e

2x

Evaluate

SOLUTION

We have

ò

ò

[CBSE 2005C, ‘09]

dt

dx = ò

5 - 4t - t dt

=ò

EXAMPLE 15

dx.

2

=ò

dt 2

5 - (t + 4t + 4) + 4 dt

=ò 9 - (t + 2) 2 3 2 - (t + 2) 2 dz =ò , where (t + 2) = z 2 3 - z2 z (t + 2) = sin -1 + C = sin -1 +C 3 3 x ( e + 2) = sin -1 + C. 3 dx × 2 - 4x + x 2

dx 2 - 4x + x

2

=ò

dx 2

x - 4x + 4 - 2

=ò

dx ( x - 2) 2 - ( 2) 2

= log ( x - 2) + ( x - 2) 2 - 2 + C = log x - 2 + x 2 - 4x + 2 + C.

Senior Secondary School Mathematics for Class 12 Pg-729

Some Special Integrals EXAMPLE 16

Evaluate

SOLUTION

We have

ò

dx

ò

2

3 x + 6x + 12 dx

729

×

[CBSE 2004C]

1 dx ×ò 2 3 3 x + 6x + 12 x + 2x + 4 1 dx = ×ò 3 ( x + 1) 2 + ( 3 ) 2 1 dt = × , where ( x + 1) = t 3 ò t2 + ( 3)2 =

2

1 log t + t 2 + 3 + C 3 1 = log ( x + 1) + x 2 + 2x + 4 + C. 3 =

EXAMPLE 17

Evaluate

SOLUTION

We have

ò

EXAMPLE 18 SOLUTION

EXAMPLE 19 SOLUTION

ò

dx 8 + 3x - x2

dx

=ò

× dx

=ò

dx

dx 1 = x 2ò - x2 2

dx

9ö 9 æ 8 + 3x - x 8 - ( x - 3 x) 8 - ç x2 - 3x + ÷ + 4ø 4 è ìæ 3öü ïç x - ÷ ï dx ï 2øï = sin -1í è =ò ý+ C 2 æ öï 2 41 ïç æ 41 ö 3ö ÷ æ ÷ ç ï çè 2 ÷ø ï ç 2 ÷ - çè x - 2 ÷ø î þ ø è æ 2x - 3 ö = sin -1 ç ÷ + C. è 41 ø dx Evaluate ò × 2x - x 2 We have dx dx =ò ò 2 2 2x - x 1 - ( x - 2x + 1) dx =ò = sin -1( x - 1) + C. 2 1 - ( x - 1) dx Evaluate ò × x(1 - 2x) We have dx dx ò x(1 - 2x) = ò x - 2x 2 2

=

1 × 2 ò

2

x 1ö 1 æ - ç x2 - + ÷ + 2 16 ø 16 è

Senior Secondary School Mathematics for Class 12 Pg-730

730

Senior Secondary School Mathematics for Class 12

dx 1 ×ò 2 2 2 1 ì 2 x 1ü 1ö æ æ1ö - íx - + ý - çx - ÷ ÷ ç 16 î 2 16 þ 4ø è è 4ø 1ö 1 dt æ = , where ç x - ÷ = t 4ø 2 ò æ 1ö2 è 2 ç ÷ -t è 4ø t 1 1 1 1ö æ = sin -1 +C = sin -1( 4t) + C = sin -1 4 ç x - ÷ + C 4ø ( 1 4) 2 2 2 è 1 -1 = sin ( 4x - 1) + C. 2 =

dx

1 2ò

Integrals of the form METHOD

EXAMPLE 20

SOLUTION

Evaluate

ò

=

( px + q) ( ax 2 + bx + c)

dx.

d ( ax 2 + bx + c) + B. dx Now, the value of the integral can be obtained easily. Let ( px + q) = A ×

ò

(5 x + 3)

[CBSE 2011, 12] dx. x 2 + 4x + 10 d Let (5 x + 3) = A × ( x 2 + 4x + 10) + B. dx Then, (5 x + 3) = A( 2x + 4) + B. Comparing the coefficients of like powers of x, we get 5 æ ö ( 2A = 5 and 4A + B = 3) Þ ç A = and B = -7 ÷ 2 è ø ì5 ü ï × ( 2x + 4) - 7 ï (5 x + 3) \ ò dx = ò í 2 ý dx 2 x 2 + 4x + 10 ï x + 4x + 10 ï î þ 5 ( 2x + 4) dx = ×ò dx - 7 ò 2 2 2 x + 4x + 10 x + 4x + 10

dx 5 1 dt - 7 ò , where t = x 2 + 4x + 10 × 2 ò t ( x + 2) 2 + ( 6) 2 æ5 ö = ç ´ 2 t ÷ - 7 log ( x + 2) + x 2 + 4x + 10 + C è2 ø =

= 5 x 2 + 4x + 10 - 7 log ( x + 2) + x 2 + 4x + 10 + C. EXAMPLE 21

Evaluate

SOLUTION

Let

ò

( x + 1)

dx. 4 + 5x - x2 d ( x + 1) = A × ( 4 + 5 x - x 2) + B. Then, dx ( x + 1) = A(5 - 2x) + B.

Senior Secondary School Mathematics for Class 12 Pg-731

Some Special Integrals

731

Comparing the coefficients of like powers of x, we get 1 7ö æ ( -2A = 1 and 5 A + B = 1) Þ ç A = - and B = ÷ × 2 2ø è 1 7 \ ( x + 1) = - (5 - 2x) + × 2 2 1 7 - (5 - 2x) + ( x + 1) 2 2 dx dx = ò \ ò 2 2 4 + 5x - x 4 + 5x - x 1 (5 - 2x) 7 1 dx + ò dx ò 2 2 2 4 + 5x - x 4 + 5x - x2 1 1 7 1 dt + × ò dx , where t = 4 + 5 x - x 2 =- ò 2 2 t 4 - ( x 2 - 5 x) =-

dx

1 7 = - ×2 t + ×ò 2 2 =- t+

25 ö 25 æ 4 - ç x2 - 5x + ÷+ 4ø 4 è dx

7 × 2 2 ò æ 2 41 ö 5 ÷ - æç x - ö÷ ç ç 2 ÷ 2ø è ø è

5ö æ çx - ÷ 7 2ø -1 è = - 4 + 5 x - x + × sin +C æ 41 ö 2 ç ÷ ç 2 ÷ è ø 7 æ 2x - 5 ö = - 4 + 5 x - x 2 + sin -1 ç ÷ + C. 2 è 41 ø 2

EXAMPLE 22

SOLUTION

Evaluate

ò

( x + 3)

dx. 5 - 4x - x 2 d Let ( x + 3) = A × (5 - 4x - x 2) + B. Then, dx ( x + 3) = A( -4 - 2x) + B. Comparing the coefficients of like powers of x, we get -1 æ ö ( -2A = 1 and - 4A + B = 3) Þ ç A = , B = 1÷ × 2 è ø ( x + 3) 1 ( -4 - 2x) dx \ ò dx = - × ò dx + ò 2 5 - 4x - x 2 5 - 4x - x 2 5 - 4x - x 2 =-

1 dt dx +ò , ò 2 2 t 5 - ( x + 4x + 4) + 4 where (5 - 4x - x 2) = t

Senior Secondary School Mathematics for Class 12 Pg-732

732

Senior Secondary School Mathematics for Class 12 1

1 t 2 =- × + 2 ( 1 2) ò

dx 2

3 - ( x + 2) 2

( x + 2) +C 3 ( x + 2) = - 5 - 4x - x 2 + sin -1 + C. 3 = - t + sin -1

EXERCISE 14B Evaluate: dx 1. ò 16 - x 2 dx 4. ò x2 - 4 dx 7. ò x2 + 9 10.

ò

13.

ò

16.

ò

19.

ò

21.

ò

23.

ò

25.

ò

27.

ò

29.

ò

31.

ò

x 9 - x4

dx

sin x

dx

4 + cos2 x 2ex

dx

2x

4-e dx 2

x + 6x + 5

2.

ò

5.

ò

8.

ò

11.

ò

14.

ò

17.

ò

dx 1 - 9x dx

3x2 9 - 16x 6

9 sin 2 x - 1

dx

dx 1-e

[CBSE 2001C]

( x - 3) - 1 dx 2

16 - 6x - x 2 dx x - x2 dx

2x 2 + 4x + 6

dx

cos x

2

x2 - 3x + 2 dx

9.

ò

12.

ò

15.

ò

18.

ò

1 + 4x 2

[CBSE 2004C]

x

20.

ò

22.

ò

24.

ò

26.

ò

28.

ò

30.

ò

32.

ò

dx

ò

6.

4x 2 - 1 dx

dx

2 + 2x - x dx

3.

2

15 - 8x 2 dx 9x 2 - 7 dx

dx ( 2 - x) 2 + 1

9 + 4x 2 sec2 x 16 + tan 2 x ex 4 + e2x

dx

dx

a-x dx a+x [CBSE 2001C]

dx 2

x - 6x + 10 dx 8 - 4x - 2x 2 dx 7 - 6x - x 2 dx 8 + 2x - x 2 dx 2x 2 + 3 x - 2 dx 1 + 2x - 3 x 2

[CBSE 2003C] [CBSE 2002] [CBSE 2002]

Senior Secondary School Mathematics for Class 12 Pg-733

Some Special Integrals

33.

ò

35.

ò

37.

ò

39.

ò

41.

ò

43.

ò

dx 5-x

34.

ò

dx

36.

ò

dx [CBSE 2012C] x 2 + 4x + 10 ( 3 - 2x) dx 2 + x - x2 ( 3 x + 1) dx 5 - 2x - x 2 1+ x dx x

38.

ò

40.

ò

42.

ò

44.

ò

x

x2 6

3

x + 2x + 3 (5 x + 3)

733

dx 3 + 4x - 2x 2 ( 2x + 3) x2 + x + 1 ( 4x + 3)

dx

2x 2 + 2x - 3 ( x + 2) dx x 2 + 2x - 1 ( 6x + 5) dx 6 + x - 2x 2 ( x + 2) dx x2 + 5x + 6

[CBSE 2001]

[CBSE 2014]

ANSWERS (EXERCISE 14B)

1. sin -1

x +C 4

2.

4. log x + x 2 - 4 + C

1 1 8 3. sin -1 3 x + C sin -1 x+C 15 3 2 2 1 5. log 2x + 4x 2 - 1 + C 2

1 log 3 x + 9x 2 - 7 + C 3 1 8. log 2x + 4x 2 + 1 + C 2 æ x2 ö 1 10. sin -1 ç ÷ + C ç 3÷ 2 è ø

1 log 2x + 9 + 4x 2 + C 2 æ 4x 3 ö 1 ÷ +C 11. sin -1 ç ç 3 ÷ 4 è ø

12. log tan x + 16 + tan 2 x + C

13. - log cos x + 4 + cos2 x + C

1 log 3 sin x + 9 sin 2 x - 1 + C 3 æ ex ö 16. 2 sin -1 ç ÷ + C ç 2÷ è ø x 18. a sin -1 + a 2 - x 2 + C a

15. log ex + 4 + e2x + C 17. -2 log e- x /2 + e- x - 1 + C

20. - log ( 2 - x) + x 2 - 4x + 5 + C

21. log ( x - 3) + x 2 - 6x + 8 + C

6.

14.

22. log ( x - 3) + x 2 - 6x + 10 + C 24.

1 æ x + 1ö sin -1 ç ÷ +C 2 è 5 ø

7. log x + x 2 + 9 + C 9.

19. log ( x + 3) + x 2 + 6x + 5 + C

æ x - 1ö 23. sin -1 ç ÷ +C è 3 ø æx + 3ö 25. sin -1 ç ÷ +C è 5 ø

Senior Secondary School Mathematics for Class 12 Pg-734

734

Senior Secondary School Mathematics for Class 12

æx + 3ö 26. sin -1 ç ÷ +C è 4 ø æ x - 1ö 28. sin -1 ç ÷ +C è 3 ø 30.

27. sin -1( 2x - 1) + C 3ö ½æ ½ 29. log ½ç x - ÷ + x 2 - 3 x + 2½+ C 2 ø è ½ ½

½ ½ 1 3 3 1 log ½x + + x 2 + x - 1½+ C 31. log ( x + 1) + x 2 + 2x + 3 + C 4 2 2 2 ½ ½

1 æ 3x - 1ö sin -1 ç ÷ +C 3 è 2 ø ì 2( x - 1) ü 1 34. sin -1í ý+ C 2 5 î þ 32.

æ 2x - 5 ö 33. sin -1 ç ÷ +C è 5 ø 35.

1 log ( x 3 + 1) + x 6 + 2x 3 + 3 + C 3

½x + 1 + x 2 + x + 1½ ½+ C 36. 2 x 2 + x + 1 + 2 log ½ 2 ½ ½ 37. 5 x 2 + 4x + 10 - 7 log ( x + 2) + x 2 + 4x + 10 + C 38. 2 2x 2 + 2x - 3 +

½ 1 1 log ½x + + x 2 + x 2 2 ½

3½ ½+ C 2½

æ 2x - 1 ö 39. 2 2 + x - x 2 + 2 sin -1 ç ÷ +C è 3 ø 40.

x 2 + 2x - 1 + log ( x + 1) + x 2 + 2x - 1 + C

æ x + 1ö 41. -3 5 - 2x - x 2 - 2 sin -1 ç ÷ +C è 6 ø 13 æ 4x - 1 ö 42. -3 6 + x - 2x 2 + sin -1 ç ÷ +C 2 2 è 7 ø 1 ½x + 1 + x + x 2 ½ ½+ C 43. x + x 2 + log ½ 2 2 ½ ½ 1 5ö æ 44. x 2 + 5 x + 6 - log ç x + ÷ + x 2 + 5 x + 6 + C 2 2ø è HINTS TO SOME SELECTED QUESTIONS (EXERCISE 14B) 1 dt. 2 3 2 11. Put x = t and 3 x dx = dt. 10. Put x 2 = t and x dx =

12. Put tan x = t and sec2 x dx = dt. 13. Put cos x = t and sin x dx = - dt. 14. Put sin x = t and cos x dx = dt. 15. Put e x = t and e xdx = dt. 16. Put e x = t and e xdx = dt.

Senior Secondary School Mathematics for Class 12 Pg-735

Some Special Integrals

735

17. Multiplying the numerator and denominator by e - x/2 , we get e - x/2

I=ò 18.

e -x - 1

æ -1 ö dx. Now, put e - x/2 = t and e - x/2 × ç ÷ dx = dt . è 2 ø

ìï a - x a- x a- x dx = ò í ´ a+ x a- x îï a + x

ò

= aò

dx a2 - x 2

= a sin -1

+

üï ( a - x) dx ý dx = ò a2 - x 2 þï

( -2 x ) x 1 1 1 dx = a sin-1 + ò dt , where ( a2 - x 2 ) = t × a 2 2 ò a2 - x 2 t

x x 1 + ´ 2 t + C = a sin -1 + a 2 a

a2 - x 2 + C .

20. Put ( 2 - x ) = t and dx = - dt. 25. ( 16 - 6 x - x 2 ) = {16 - ( x 2 + 6 x + 9 ) + 9} = [25 - ( x + 3 ) 2 ]. 27.

1ö 1 æ ( x - x2 ) = - ç x2 - x + ÷ + = 4ø 4 è

32. I =

=

33. I = ò

34. I =

1 3

ò

1 ×ò 3

dx 1 = ×ò 2 1 3 - x2 + x + 3 3 dx × 2 2 1ö æ æ2ö ç ÷ - çx - ÷ 3ø è è 3ø

dx 5x - x 2

=ò

2

2

1ö æ 1ö æ ç ÷ - çx - ÷ × 2ø è2ø è dx 1 æ 2 2 1ö 1 - çx - x + ÷ + 3 è 3 9ø 9

dx 25 ö 25 æ - ç x 2 - 5x + ÷+ 4 ø 4 è

=ò

dx 2

5ö æ æ5 ö ç ÷ - çx - ÷ 2ø è è2ø

2

×

dx dx dx 1 1 1 ×ò = ×ò = ×ò × 2 2 2 2 2 2 + + x x 3 2 1 1 { ( ) } - x + 2x + 3 2 - ( x + 1) 2

35. Put x 3 = t and 3 x 2 dx = dt. æ 1+ x ( 1 + x) ( x + 1) 1 + x ö÷ 43. I = ò çç dx. ´ ÷ dx = ò x( 1 + x ) dx = ò x + x 1 è ø x2 + x d 2 ( x + 5 x + 6 ) + B = A( 2 x + 5 ) + B dx 1 -1 \ ( 2 A = 1 and 5 A + B = 2 ) Þ A = and B = × 2 2 1ü ì1 í ( 2x + 5) - ý ( 2x + 5) dx 1 1 2þ î2 dx = ò dx - ò \ I=ò 2 2 2 2 2 2 x + 5x + 6 x + 5x + 6 5ö æ æ 1ö çx + ÷ - ç ÷ 2ø è è2ø 1 5ö æ 2 2 = x + 5 x + 6 - log ç x + ÷ + x + 5 x + 6 + C . 2 2ø è

44. Let ( x + 2 ) = A ×

Senior Secondary School Mathematics for Class 12 Pg-736

736

Senior Secondary School Mathematics for Class 12

Three More Special Integrals THEOREM

PROOF

(i)

ò

(ii)

ò

(iii)

ò

x 2 a2 x a - x2 + sin -1 + C. 2 2 a 2 x a x 2 - a 2 dx = x 2 - a 2 - log x + x 2 - a 2 + C. 2 2 x a2 2 2 2 2 x + a dx = x +a + log x + x 2 + a 2 + C. 2 2

a 2 - x 2 dx =

(i) Integrating by parts, taking 1 as the second function, we get I = ò a 2 - x 2 dx = ò ( a 2 - x 2 × 1) dx

1 = ( a 2 - x 2 ) × x - ò ( a 2 - x 2) -1/2( -2x) × x dx 2 x2 a 2 - ( a 2 - x 2) 2 2 = x( a - x ) + ò dx = x( a 2 - x 2 ) + ò dx a2 - x 2 a2 - x 2 dx = x( a 2 - x 2 ) + a 2 ò - ò a 2 - x 2 dx a2 - x 2 x = x( a 2 - x 2 ) + a 2 sin -1 - I + c. a 2 2 2 -1 x \ 2I = x( a - x ) + a sin +c a x a2 x c sin -1 + C [taking = C]. Þ I = ( a2 - x 2 ) + 2 2 a 2 x a2 x Hence, ò a 2 - x 2 dx = ( a 2 - x 2 ) + sin -1 + C. 2 2 a (ii) Integrating by parts, taking 1 as the second function, we get I = ò x 2 - a 2 dx = ò ( x 2 - a 2 × 1) dx

1 = ( x 2 - a 2 ) × x - ò ( x 2 - a 2) -1/2( 2x) × x dx 2 x2 ( x 2 - a 2) + a 2 2 2 = x( x - a ) - ò dx = x( x 2 - a 2 ) - ò dx x 2 - a2 x 2 - a2 a2 = x( x 2 - a 2 ) - ò x 2 - a 2 dx - ò dx x 2 - a2 = x( x 2 - a 2 ) - I - a 2 log x + x 2 - a 2 + c.

\ 2I = x( x 2 - a 2 ) - a 2 log x + x 2 - a 2 + c x a2 c ( x 2 - a 2 ) - log x + x 2 - a 2 + C [taking = C]. 2 2 2 x a2 Hence, ò x 2 - a 2 dx = ( x 2 - a 2 ) - log x + x 2 - a 2 + C. 2 2 Þ I=

Senior Secondary School Mathematics for Class 12 Pg-737

Some Special Integrals

737

(iii) Integrating by parts, taking 1 as the second function, we get I = ò x 2 + a 2 dx = ò ( x 2 + a 2 × 1) dx

1 = ( x 2 + a 2 ) x - ò ( x 2 + a 2) -1/2( 2x) × x dx 2 x2 ( x 2 + a 2) - a 2 = x( x 2 + a 2 ) - ò dx = x( x 2 + a 2 ) - ò dx x 2 + a2 x 2 + a2 dx = x( x 2 + a 2 ) - ò x 2 + a 2 dx + a 2 ò 2 x + a2 = x × ( x 2 + a 2 ) - I + a 2 log x + x 2 + a 2 + c.

\

2I = x( x 2 + a 2 ) + a 2 log x + x 2 + a 2 + c

x a2 c [taking = C]. ( x 2 + a2 ) + log x + x 2 + a 2 + C 2 2 2 x a2 Hence, ò x 2 + a 2 dx = ( x 2 + a 2 ) + log x + x 2 + a 2 + C. 2 2

or I =

SOLVED EXAMPLES EXAMPLE 1

Evaluate: (i)

SOLUTION

ò

9 - x 2 dx

(ii)

ò

We know that ò a 2 - x 2 dx = \ (i)

ò

(iii)

ò

(iii)

ò

x 2

16 - 9x 2 dx

x 2 a2 x a - x2 + sin -1 + C. 2 2 a

9 - x 2 dx = ò 3 2 - x 2 dx

9 x sin -1 + C. 2 3 2 ì ü ï æ1ö ï æ1 ö 1 - 4x 2 dx = 2ò ç - x 2 ÷ dx = 2ò í ç ÷ - x 2 ýdx 2 è4 ø è ø ïî ïþ éx 1 æ x öù 1 ÷ú + C = 2ê - x 2 + sin -1 çç 1 ÷ 2 4 8 êë è ( 2) ø úû x 1 = 1 - 4x 2 + sin -1( 2x) + C. 2 4 ì 4 2 ü ì ï æ 16 ï æ ö ï ö üï 16 - 9x 2 dx = 3 ò í ç - x 2 ÷ ý dx = 3 ò í ç ÷ - x 2 ý dx 3 ïî è 9 ø ïþ è ø îï þï =

(ii)

1 - 4x 2 dx

9 - x2 +

é x 16 8 x ù = 3ê - x 2 + sin -1 ú +C 4 9 ( 3) û ë2 9 x 8 æ 3x ö = 16 - 9x 2 + sin -1 ç ÷ + C. 2 3 è 4 ø

Senior Secondary School Mathematics for Class 12 Pg-738

738 EXAMPLE 2

Senior Secondary School Mathematics for Class 12

Evaluate: (i)

SOLUTION

ò

x 2 - 16 dx

(ii)

ò

4x 2 - 5 dx

We know that ò x 2 - a 2 dx = \ (i)

ò

(iii)

ò

17 x 2 - 11 dx

x a2 x 2 - a 2 - log x + x 2 - a 2 + C. 2 2

x 2 - 16 dx = ò x 2 - 42 dx x 16 × x 2 - 42 log x + x 2 - 16 + C 2 2 x = × x 2 - 16 - 8 log x + x 2 - 16 + C. 2 =

2

(ii)

ò

(iii)

4x 2 - 5 dx = 2 ò x 2 -

ò

æ 5ö 5 ÷ dx dx = 2 × ò x 2 - çç ÷ 4 è 2 ø

éx = 2× ê êë 2

x2 -

= x x2 -

½ 5 5 5½ - log½x + x 2 - ½+ C. 4 4 4½ ½ 11 dx 17 ½ 11 11 11 ½ïü x2 log½x + x 2 - ½ý + C 17 34 17 ½ ½þï

17 x 2 - 11 dx = 17 × ò x 2 ïì x = 17 × í îï 2 =

ò

½ x 11 17 11 ½ log ½x + x 2 - ½+ C. 17 x 2 - 11 2 34 17 ½ ½

16x 2 + 25 dx.

EXAMPLE 3

Evaluate

SOLUTION

We know that ò x 2 + a 2 dx = \

EXAMPLE 4 SOLUTION

ò

½ 5 5 5 ½ù - log½x + x 2 - ½ú + C 4 8 4 ½úû ½

a2 log x + x 2 + a 2 + C. 2 2ü ì 25 ü ï 2 æ5 ö ï x2 + ý dx = 4 ò í x + ç ÷ ý dx 16 þ è 4ø ï ïî þ ½ 25 25 25 ½üï 2 2 ½ý + C + x + log½x + x + 16 32 16 ½ïþ ½ x 2

x 2 + a2 +

ì 16x 2 + 25 dx = 4 ò í î ìï x = 4×í ïî 2 ½ x 25 25 ½ ½+ C. = × 16 x 2 + 25 + log ½x + x 2 + 2 8 16 ½ ½

16 + (log x) 2 dx. x 1 Putting log x = t and dx = dt, we get x Evaluate

ò

I = ò 16 + t 2 dt = ò 42 + t 2 dt

[CBSE 2005]

Senior Secondary School Mathematics for Class 12 Pg-739

Some Special Integrals

739

t 16 16 + t 2 + log t + 16 + t 2 + C 2 2 1 = log x × 16 + (log x) 2 + 8 log log x + 16 + (log x) 2 + C. 2 =

EXAMPLE 5 SOLUTION

Evaluate

òx

x 4 + 1 dx.

[CBSE 2003]

1 Putting x 2 = t and x dx = dt, we get 2 1 I = ò t 2 + 1 dt 2 1 ét 1 ù = × ê t 2 + 1 + log t + t 2 + 1 ú + C 2 ë2 2 û =

1 x2 x 4 + 1 + log x 2 + x 4 + 1 + C. 4 4

òe

x

e2x + 4 dx.

EXAMPLE 6

Evaluate

SOLUTION

Putting ex = t and ex dx = dt, we get I = ò t 2 + 4 dx t 4 × t 2 + 4 + log t + t 2 + 4 + C 2 2 1 x 2x = e e + 4 + 2 log ex + e2x + 4 + C. 2 =

ò cos x

4 - sin 2 x dx.

EXAMPLE 7

Evaluate

SOLUTION

Putting sin x = t and cos x dx = dt, we get I = ò 4 - t 2 dt t 4 t 4 - t 2 + sin -1 + C 2 2 2 1 æ1 ö = sin x 4 - sin 2 x + 2 sin -1 ç sin x ÷ + C. 2 è2 ø =

Integrals of the form METHOD

EXAMPLE 8

Evaluate

SOLUTION

We have

ò

( ax 2 + bx + c) dx

Express ( ax 2 + bx + c) as a[( x + a ) 2 ± b 2] and obtain an integral which can be evaluated easily.

ò

x 2 + 3 x dx.

2 2 2 2 ìï 3ö æ 3ö æ 3 ö üï æ æ 3ö ( x 2 + 3 x) = íx 2 + 3 x + ç ÷ - ç ÷ ý = ç x + ÷ - ç ÷ × 2ø è 2ø è 2 ø þï è è 2ø îï

Senior Secondary School Mathematics for Class 12 Pg-740

740

Senior Secondary School Mathematics for Class 12

\ I = ò x 2 + 3 x dx 2

2

2

3ö 3ö æ æ æ 3ö æ 3ö = ò ç x + ÷ - ç ÷ dx = ò t 2 - ç ÷ dt, where ç x + ÷ = t 2ø 2ø è è è 2ø è 2ø ½ 1 9 9 9½ = t t 2 - - log ½t + t 2 - ½+ C 2 4 8 4½ ½ x a2 ïì ïü 2 2 x 2 - a2 log x + x 2 - a 2 + C ý íusing ò x - a dx = 2 2 ïî ïþ 9 3ö 1æ 3ö ½æ ½ = ç x + ÷ x 2 + 3 x - log ½ç x + ÷ + x 2 + 3 x½+ C. 8 2ø 2è 2ø ½è ½ EXAMPLE 9 SOLUTION

Evaluate We have

ò

2x 2 + 3 x + 4 dx.

3 æ ö ( 2x 2 + 3 x + 4) = 2 ç x 2 + x + 2÷ 2 è ø

2 2 ì æ 23 ö üï 9ö æ 9 öü 3ö ìæ 2 3 ïæ ç ÷ = 2 × íç x + x + ÷ + ç 2 - ÷ ý = 2 × íç x + ÷ + ç ÷ ý× 2 16 ø è 16 ø þ 4ø îè è 4 ø ïþ ïîè

2 æ 23 ö 3ö æ ÷ 2x + 3 x + 4 = 2 × ç x + ÷ + çç ÷ 4 ø è è 4 ø

2

2

\

2

Þ

ò

2 æ 23 ö 3ö æ ÷ dx 2x 2 + 3 x + 4 dx = 2 × ò ç x + ÷ + çç ÷ 4ø è è 4 ø 2

æ 23 ö 3 ÷ dt, where æç x + ö÷ = t = 2 ò t 2 + çç ÷ 4 4ø è è ø ìï t ½ 23 23 23 ½ïü = 2 × í × t2 + + log ½t + t 2 + ½ý + C 16 32 16 ½þï ½ îï 2 é êQ êë

ò

t 2 + a 2 dt =

ù t 2 a2 t + a2 + log t + t 2 + a 2 + C ú 2 2 úû

2

=

2æ 3ö æ 3ö 23 çx + ÷ çx + ÷ + 2 è 4ø è 4ø 16 +

½æ ½ 23 2 3ö 3 log ½ç x + ÷ + x 2 + x + 2½+ C 32 4 2 è ø ½ ½

=

½( 4x + 3) ( 4x + 3) 3 23 2 × x2 + x + 2 + + log ½ 2 32 4 4 2 ½

=

½( 4x + 3) ( 4x + 3) 2x 2 + 3 x + 4 23 2 + + log ½ 8 32 4 ½

2x 2 + 3 x + 4½ ½+ C 2 ½ 2x 2 + 3 x + 4½ ½+ C 2 ½

Senior Secondary School Mathematics for Class 12 Pg-741

Some Special Integrals

EXAMPLE 10

Evaluate

SOLUTION

We have

ò

741

3 - 2x - 2x 2 dx.

ì3 ü ( 3 - 2x - 2x 2) = 2 × í - x - x 2 ý î2 þ é3 æ 2 = 2× ê - ç x + x + ë2 è

1ö 1ù ÷+ 4 ø 4 úû 2 ìæ 7 ö 2 é7 æ 1ö ù ï ÷ - æç x + = 2 × ê - ç x + ÷ ú = 2 × íçç ÷ 2 4 2 è ø úû è êë ø ïîè 2

æ 7ö ÷ - æç x + 3 - 2x - 2x 2 = 2 × çç ÷ 2 è è ø

\

1ö ÷ 2ø

1ö ÷ 2ø

2ü

ï ý× ïþ

2

2

Þ

ò

2 æ 7ö 1 ÷ - æç x + ö÷ dx 3 - 2x - 2x 2 dx = 2 × ò çç ÷ 2ø è è 2 ø 2

= =

=

= =

æ 7ö 1 ÷ - t 2 dt, where æç x + ö÷ = t and dx = dt 2 × ò çç ÷ 2 2ø è è ø ìï t 7 2 7 t üï 2 ×í × - t + sin -1 ý+ C 8 ( 7 2 ) ïþ ïî 2 4 é ù t a2 t 2 2 × sin -1 + C ú a2 - t 2 + êQ ò a - t dt = a 2 2 êë ûú 1 ü ì ö æ çx + ÷ ï 2 ïï 1 æ 1ö 7 æ 1ö 7 2ø ï -1 è - ç x + ÷ + sin 2 ×í ç x + ÷ ý+ C 2ø 4 è 2ø 8 ( 7 2) ï ï2 è ïþ ïî ì1 ü 3 7 2 x + 1 æ ö 2í ( 2x + 1) × - x - x 2 + sin -1 ç ÷ý+ C 2 8 è 7 øþ î4 1 7 æ 2x + 1 ö ( 2x + 1) 3 - 2x - 2x 2 + sin -1 ç ÷ + C. 4 4 2 è 7 ø

Integrals of the form METHOD

ò

( px + q) ( ax 2 + bx + c) dx

Let ( px + q) = A ×

d ( ax 2 + bx + c) + B. dx

Find A and B. Then, we get the integrand which is easily integrable.

ò ( x + 3)

EXAMPLE 11

Evaluate

SOLUTION

Let ( x + 3) = A ×

3 - 4x - x 2 dx.

[CBSE 2005C]

d ( 3 - 4x - x 2) + B Û ( x + 3) = A( - 4 - 2x) + B . dx

Senior Secondary School Mathematics for Class 12 Pg-742

742

Senior Secondary School Mathematics for Class 12

Comparing the coefficients of like powers of x, we get 1 -2A = 1 and - 4A + B = 3 Û A = and B = 1. 2 1 \ ( x + 3) = - ( - 4 - 2x) + 1 2 ì 1 ü Þ I = ò í- ( - 4 - 2x) + 1ý 3 - 4x - x 2 dx î 2 þ 1 = - ò ( - 4 - 2x) 3 - 4x - x 2 dx + ò 3 - 4x - x 2 dx 2 1 = - ò t dt + ò 7 - ( 4x + x 2 + 4) dx , where 3 - 4x - x 2 = t 2 æ 1 3/2 2 ö = ç - ´t ´ ÷ + ò ( 7 ) 2 - ( x + 2) 2 dx 3ø è 2 7 1 1 æ x + 2ö = - ( 3 - 4x - x 2) 3/2 + ( x + 2) 3 - 4x - x 2 + sin -1 ç ÷ +C 2 2 3 è 7 ø é êQ êë EXAMPLE 12 SOLUTION

Evaluate

ò ( 3 x - 2)

ò

a 2 - x 2 dx =

ù x 2 a2 x × sin -1 + C ú × a - x2 + a 2 2 úû

x 2 + x + 1 dx.

d 2 ( x + x + 1) + B. dx Then, ( 3 x - 2) = A( 2x + 1) + B. Comparing the coefficients of like powers of x, we get 3 -7 and B = × 2A = 3 and A + B = -2. So, A = 2 2 3 7 \ ( 3 x - 2) = ( 2x + 1) - × 2 2 Let ( 3 x - 2) = A ×

So,

ò ( 3 x - 2)

x 2 + x + 1 dx

7ü ì3 = ò í ( 2x + 1) - ý x 2 + x + 1 dx 2þ î2 3 7 2 = ò ( 2x + 1) x + x + 1 dx - ò x 2 + x + 1 dx 2 2 3 7 1ö 3 æ 2 = ò t dt - ò ç x + x + ÷ + dx, 2 2 è 4ø 4 where x 2 + x + 1 = t in the 1st integral = t 3/2 -

7 2ò

2 2 ì æ 3 ö üï 1ö ïæ ç ÷ x + + ÷ íç ç 2 ÷ ý dx 2ø è ø ïþ ïîè 2

æ 3ö 7 ÷ du, where u = æç x + = ( x 2 + x + 1) 3/2 - × ò u2 + çç ÷ 2 2 è è ø

1ö ÷ 2ø

Senior Secondary School Mathematics for Class 12 Pg-743

Some Special Integrals

743

½ 7 ïìu 3 3 3 ½ïü = ( x 2 + x + 1) 3/2 - í × u2 + + log ½u + u2 + ½ý + C 2 ïî 2 4 8 4 ½ïþ ½ ïì íQ ïî

ò

u2 + a 2 du =

= ( x 2 + x + 1) 3/2 -

u 2 a2 ïü u + a2 + log u + u2 + a 2 + C ý 2 2 ïþ

½ 7u 21 3½ 4u2 + 3 - log ½u + u2 + ½+ C 8 16 4½ ½

7æ = ( x 2 + x + 1) 3/2 - ç x + 8è

2

1ö æ 1ö ÷ 4ç x + ÷ + 3 2ø è 2ø ½æ 21 1ö - log ½ ½ç x + 2 ÷ + 16 è ø ½ 7( 2x + 1) = ( x 2 + x + 1) 3/2 x2 + x + 1 8 21 ½( 2x + 1) - log ½ 16 ½ 2 EXAMPLE 13 SOLUTION

òx

Evaluate

x + x 2 dx.

1ö 3½ æ +C çx + ÷ + ½ 2ø 4½ è ½ 2

½+ C. + x 2 + x + 1½ ½ [CBSE 2005C]

d ( x + x 2) + B. Then, dx x = A(1 + 2x) + B Comparing the coefficients of various powers of x, we get 1 -1 ö æ ( 2A = 1 and A + B = 0) Þ ç A = and B = ÷× 2 2ø è 1 1 \ x = (1 + 2x) 2 2 Let x = A ×

Þ

òx

… (i)

x + x 2 dx 1ü ì1 = ò í (1 + 2x) - ý x + x 2 dx 2þ î2 1 1 = ò (1 + 2x) x + x 2 dx - ò x + x 2 dx 2 2 1 1 ìæ 2 1ö 1ü = ò t dt - ò íç x + x + ÷ - ý dx, 2 2 îè 4ø 4þ where ( x + x 2) = t in the first integral 2 2 ìïæ 1ö æ 1 ö üï íç x + ÷ - ç ÷ ý dx 2ø è 2 ø ïþ ïîè

=

1 t 3/2 1 × 2 ( 3/2) 2

=

1ö 1 1 æ1ö æ ( x + x 2) 3 / 2 - × ò u2 - ç ÷ du, where ç x + ÷ = u 2ø 3 2 è 2ø è

ò

2

Senior Secondary School Mathematics for Class 12 Pg-744

744

Senior Secondary School Mathematics for Class 12

=

½ 1 1 ïìu -1 1 1 ½ïü ( x + x 2) 3/2 - í × u2 - log ½u + u2 - ½ý + C 3 2 ïî 2 4 8 4½ïþ ½ 2 ìï üï x a 2 2 x 2 - a2 log x + x 2 - a 2 + C ý íQ ò x - a dx = 2 2 ïþ îï 2

1ö 1 æ çx + ÷ 2ø 4 è ½æ 1 1ö æ + log½ ½ç x + 2 ÷ ç x + 16 è ø è ½ 1 1 1 ½( 2x + 1) × = ( x + x 2) 3 / 2 - ( 2x + 1) x + x 2 + log ½ 16 3 8 ½ 2 =

1 1æ 1ö ( x + x 2) 3/2 - ç x + ÷ 3 4è 2ø

2 1ö 1½ ½+ C ÷ - ½ 2ø 4 ½ ½ 2½ x + x + C. ½

EXERCISE 14C Evaluate the following integrals: 1. 3. 5. 7. 9. 11. 13. 15. 17. 18. 20. 22. 24. 26.

2. ò 4 - 9x dx ò 4 - x dx [CBSE 2003C, ‘06] 2 4. ò 2x 2 - 3 dx ò x - 2 dx 2 6. ò 4x 2 + 9 dx ò x + 5 dx 2 8. ò cos x 9 - sin 2 x dx ò 3 x + 4 dx 2 10. x 2 + 6x - 4 dx ò x - 4x + 2 dx 2 12. ò 1 - 4x - x 2 dx ò 2x - x dx 2 14. ò 2x 2 + 3 x + 4 dx ò 2ax - x dx 2 16. ò x 2 + x + 1 dx ò x + x dx 2 ò ( 2x - 5) x - 4x + 3 dx [CBSE 2005C] 2 19. ò ( x - 5) x 2 + x dx ò ( x + 2) x + x + 1 dx 2 21. ò ( x + 1) 2x 2 + 3 dx ò ( 4x + 1) x - x - 2 dx 2 23. ò ( 2x - 5) 2 + 3 x - x 2 dx ò x 1 + x - x dx 2 25. ò ( x + 1) 1 - x - x 2 dx ò ( 6x + 5) 6 + x - 2x dx 2 ò ( x - 3) x + 3 x - 18 dx [CBSE 2014] 2

2

ANSWERS (EXERCISE 14C)

1.

x x × 4 - x 2 + 2 sin -1 + C 2 2

2.

x æ 3x ö × 4 - 9x 2 + 2 sin -1 ç ÷ +C 2 è 2 ø

Senior Secondary School Mathematics for Class 12 Pg-745

Some Special Integrals

3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.

745

1 x x 2 - 2 - log x + x 2 - 2 + C 2 x 3 × 2x 2 - 3 log 2x + 2x 2 - 3 + C 2 2 2 x 5 × x 2 + 5 + log x + x 2 + 5 + C 2 2 x 9 2 × 4x + 9 + log 2x + 4x 2 + 9 + C 2 4 x 2 2 × 3x + 4 + log 3 x + 3 x 2 + 4 + C 2 3 sin x 9 æ sin x ö 9 - sin 2 x + sin -1 ç ÷ +C 2 2 è 3 ø ( x - 2) × x 2 - 4x + 2 - log ( x - 2) + x 2 - 4x + 2 + C 2 1 13 ( x + 3) x 2 + 6x - 4 log ( x + 3) + x 2 + 6x - 4 + C 2 2 1 1 ( x - 1) 2x - x 2 + sin -1( x - 1) + C 2 2 5 1 æ x + 2ö 2 ( x + 2) 1 - 4x - x + sin -1 ç ÷ +C 2 2 è 5 ø a2 æ x - aö sin -1 ç ÷ +C 2 è a ø 23 2 3ö 1 1 ½æ ½ × 2x 2 + 3 x + 4½+ C ( 4x + 3) 2x 2 + 3 x + 4 + log ½ç x + ÷ + 32 4ø 8 2 ½è ½ 1 1 1 ½x + + x 2 + x ½ ½+ C ( 2x + 1) x 2 + x - log½ 4 8 2 ½ ½ 1 3 1 ½ 2 2 ½+ C ( 2x + 1) x + x + 1 + log½x + + x + x + 1½ 4 8 2 ½ ½ 2 2 1 3/2 1 2 ( x - 4x + 3) - ( x - 2) x - 4x + 3 + log x - 2 + x 2 - 4x + 3 + C 3 2 2 1 2 3 9 1ö ½æ ½ ( x + x + 1) 3/2 + ( 2x + 1) x 2 + x + 1 + log ½ç x + ÷ + x 2 + x + 1½+ C 3 8 16 2 ø ½è ½ 1 2 11 1 ½ ½ 3/2 11 2 2 ½ ½ ( x + x) - ( 2x + 1) x + x + log x + + x + x + C 3 8 16 2 ½ ½ 27 1 3 4 2 ö ½ æ ½ ( x - x - 2) 3/2 + ( 2x - 1) x 2 - x - 2 log ½ç x - ÷ + x 2 - x - 2 ½+ C 8 2 4 3 ø ½è ½ 1 ( x - a) 2

2ax - x 2 +

x 1 3 2 ( 2x 2 + 3) 3/2 + log 2x 2 + 3 + 4 6 2

22. -

2x + 2x 2 + 3 + C

5 1 1 æ 2x - 1 ö (1 + x - x 2) 3/2 + ( 2x - 1) 1 + x - x 2 + sin -1 ç ÷ +C 16 8 3 è 5 ø

Senior Secondary School Mathematics for Class 12 Pg-746

746

Senior Secondary School Mathematics for Class 12

23. -

1 2 ( 2 + 3 x - x 2) 3/2 - ( 2x - 3) 2 3

24. - ( 6 + x - 2x 2) 3/2 + 25. 26.

13 ( 4x - 1) 16

æ 2x - 3 ö ç ÷ +C è 17 ø 637 æ 4x - 1 ö 6 + x - 2x 2 + sin -1 ç ÷ +C 32 2 è 7 ø 2 + 3x - x2 -

17 sin -1 4

5 1 1 æ 2x + 1 ö (1 - x - x 2) 3/2 + ( 2x + 1) 1 - x - x 2 + sin -1 ç ÷ +C 16 8 3 è 5 ø

1 2 9 ( x + 3 x - 18) 3/2 - {( 2x + 3) x 2 + 3 x - 18} 3 8 81 3 log x + + x 2 + 3 x - 18 + C 2 2

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 14C) 2. I = 3 × ò

4 - x 2 dx. 9

4. I = 2 × ò x 2 6. I = 2 × ò x 2 + 7. I =

3 dx. 2

9 dx. 4

3 × ò x2 +

4 dx. 3

8. Put sin x = t and cos x dx = dt. 9. I = ò ( x - 2 ) 2 - ( 2 ) 2 dx. 11. I = ò -( x 2 - 2 x + 1) + 1 dx = ò 1 - ( x - 1) 2 dx.

12. I = ò 1 - ( x 2 + 4x + 4) + 4 dx = ò 5 - ( x + 2) 2 dx. 13. I = ò -( x 2 - 2 ax + a2 ) + a2 dx = ò a2 - ( x - a) 2 dx. 3 æ ö 14. I = 2 × ò ç x 2 + x + 2 ÷ dx = 2 ò 2 è ø

2 ìï æ 3ö 23 üï ý dx. íç x + ÷ + 4ø 16 ïþ ïî è

d ( x 2 - 4 x + 3 ) + B = A ( 2 x - 4 ) + B. dx d 26. Let ( x - 3 ) = A × ( x 2 + 3 x - 18 ) + B. Then, dx 1 -9 × ( x - 3 ) = A( 2 x + 3 ) + B. This gives A = and B = 2 2 1 9 \ I = ò ( 2 x + 3 ) x 2 + 3 x - 18 dx - ò x 2 + 3 x - 18 dx 2 2 17. Let ( 2 x - 5 ) = A ×

2

=

2

1 9 3ö æ æ9ö t dt - ò ç x + ÷ - ç ÷ dx. 2ò 2 2ø è è2ø

Senior Secondary School Mathematics for Class 12 Pg-747

15. INTEGRATION USING PARTIAL FRACTIONS Partial Fractions If f ( x) and g( x) are polynomial functions such that g( x) ¹ 0 f ( x) then is called a rational function. g( x) f ( x) is called a proper rational function. If degree f ( x) < degree g( x) then g( x) RATIONAL FUNCTIONS

If degree f ( x) ³ degree g( x) then

f ( x) is called an improper rational function. g( x)

f ( x) is an improper rational function then by dividing f ( x) by g( x), we g( x) f ( x) as the sum of a polynomial and a proper rational function. can express g( x) If

p( x) can be expressed as the sum q( x) of rational functions, each having a simplest factor q( x). Each such fraction is known as a partial fraction and the process of obtaining them is called the decomposition or resolving of the given function into partial fractions. PARTIAL FRACTIONS

Any proper rational function

METHOD We first resolve the denominator of the given fraction into simplest factors. On the basis of these factors, we obtain the corresponding partial fraction as per rules given below:

Factor in the denominator

Corresponding partial fraction

(i) ( x - a)

A ( x - a)

(ii) ( x - b) 2

A B + ( x - b) ( x - b) 2

(iii) ( x - c) 3

A B C + + ( x - c) ( x - c) 2 ( x - c) 3 Ax + B

(iv) ( ax 2 + bx + c)

( ax 2 + bx + c) 747

Senior Secondary School Mathematics for Class 12 Pg-748

748

Senior Secondary School Mathematics for Class 12

The values of A, B, C, etc., can be obtained as shown below. SOLVED EXAMPLES ON PARTIAL FRACTIONS EXAMPLE 1

Resolve

SOLUTION

Let

2x + 3 into partial fractions. ( x - 3)( x + 1)

( 2x + 3) A B = + × Then, ( x - 3)( x + 1) ( x - 3) ( x + 1) 2x + 3 A( x + 1) + B( x - 3) = ( x - 3)( x + 1) ( x - 3)( x + 1)

or ( 2x + 3) º A( x + 1) + B( x - 3). Putting ( x - 3) = 0 or x = 3 in (i), we get Putting ( x + 1) = 0 or x = -1 in (i), we get ( 2x + 3) 9 1 \ = × ( x - 3)( x + 1) 4( x - 3) 4( x + 1) EXAMPLE 2 SOLUTION

Resolve

x 3 - 2x 2 - 13 x - 12

x 2 - 3 x - 10 On dividing, we get x 3 - 2x 2 - 13 x - 12

into partial fractions.

2 × ( x 2 - 3 x - 10) A B 2 = = + ( x 2 - 3 x - 10) ( x - 5)( x + 2) x - 5 x + 2

x 2 - 3 x - 10 2

Let Then,

= ( x + 1) -

Hence, EXAMPLE 3

Resolve

SOLUTION

Let

… (i)

2 A( x + 2) + B( x - 5) = ( x - 5)( x + 2) ( x - 5)( x + 2)

or 2 º A( x + 2) + B( x - 5). Putting ( x - 5) = 0 or x = 5 in (ii), we get Putting ( x + 2) = 0 or x = - 2 in (ii), we get 2 2 2 \ = × ( x 2 - 3 x - 10) 7( x - 5) 7( x + 2)

or

… (i) A = ( 9/4). B = ( -1/4).

x 3 - 2x 2 - 13 x - 12 2

x - 3 x - 10 16 ( x - 2)( x + 2) 2 16

( x - 2)( x + 2)

2

=

2

=

16 ( x - 2)( x + 2)

= ( x + 1) -

… (ii) A = ( 2/7). B = ( -2/7).

2 2 + × 7( x - 5) 7( x + 2)

into partial fractions.

A B C + + x - 2 x + 2 ( x + 2) 2 A( x + 2) 2 + B( x - 2)( x + 2) + C( x - 2) ( x - 2)( x + 2) 2

Senior Secondary School Mathematics for Class 12 Pg-749

Integration Using Partial Fractions

749

16 º A( x + 2) 2 + B( x - 2)( x + 2) + C( x - 2)

\

… (i)

2

16 º ( A + B) x + ( 4A + C) x + ( 4A - 4B - 2C).

or

… (ii)

Putting ( x - 2) = 0 or x = 2 in (i), we get A = 1. Putting ( x + 2) = 0 or x = -2 in (i), we get C = - 4.

EXAMPLE 4

SOLUTION

Comparing the coefficients of x 2 on both sides of (ii), we get A + B = 0 or B = - A = -1. Thus A = 1, B = -1 and C = - 4. é 1 16 1 4 ù \ =ê ú× 2 ( x ) ( x ) 2 2 + ( x - 2)( x + 2) ( x + 2) 2 û ë 2x + 1 into partial fractions. Resolve ( x - 1)( x 2 + 1) 2x + 1 A Bx + C Let = + 2 2 ( 1 ) x ( x - 1)( x + 1) ( x + 1) or

( 2x + 1) 2

( x - 1)( x + 1)

=

A( x 2 + 1) + ( Bx + C)( x - 1) ( x - 1)( x 2 + 1)

×

\

2x + 1 º A( x 2 + 1) + ( Bx + C)( x - 1)

or

2x + 1 º ( A + B) x 2 + (C - B) x + ( A - C).

… (i)

Equating the like powers of x on both sides of (i), we get A + B = 0, C - B = 2 and A - C = 1. On solving these equations, we get 3 -3 1 A= , B= and C = × 2 2 2 1ö æ -3 x+ ÷ ç (1 - 3 x) ù 2x + 1 3 2ø é 3 è 2 =ê \ = + + ú× 2 2 2 ( x - 1)( x + 1) 2( x - 1) x +1 ë 2( x - 1) 2( x + 1) û Integration Using Partial Fractions SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Evaluate Let

( x - 1)

ò ( x + 1)( x - 2) dx .

[CBSE 2001]

( x - 1) A B = + × ( x + 1)( x - 2) ( x + 1) ( x - 2)

Then, ( x - 1) º A( x - 2) + B( x + 1). 2 Putting x = -1 in (i), we get A = × 3 1 Putting x = 2 in (i), we get B = × 3

… (i)

Senior Secondary School Mathematics for Class 12 Pg-750

750

Senior Secondary School Mathematics for Class 12

( x - 1) 2 1 = + ( x + 1)( x - 2) 3( x + 1) 3( x - 2) ( x - 1) 2 dx 1 dx Þ ò + ò dx = ò ( x + 1)( x - 2) 3 ( x + 1) 3 ( x - 2)

\

=

2 1 log|x + 1| + log|x - 2| + C. 3 3

( x 2 + 1)

ò ( x 2 - 5 x + 6) dx.

EXAMPLE 2

Evaluate

SOLUTION

Here the integrand is not a proper rational function; on dividing ( x 2 + 1) by ( x 2 - 5 x + 6), we get ( x 2 + 1) 2

( x - 5 x + 6)

=1+

(5 x - 5) 2

( x - 5 x + 6)

=1+

(5 x - 5) × ( x - 2)( x - 3)

(5 x - 5) A B Now, let = + ( x - 2)( x - 3) ( x - 2) ( x - 3) Þ

(5 x - 5) A( x - 3) + B( x - 2) = ( x - 2)( x - 3) ( x - 2)( x - 3)

Þ (5 x - 5) º A( x - 3) + B( x - 2). Putting x = 2 on both sides of (i), we get A = - 5. Putting x = 3 on both sides of (i), we get B = 10. ( x 2 + 1) 5 10 =1+ \ 2 x x ( ) ( 2 - 3) ( x - 5 x + 6) Þ

EXAMPLE 3

SOLUTION

( x 2 + 1)

dx

… (i)

dx

ò ( x 2 - 5 x + 6) dx = ò dx - 5 ò ( x - 2) + 10ò ( x - 3)

= x - 5 log|x - 2| + 10 log|x - 3| + C. ( 3 x - 2) Evaluate ò [CBSE 2006C] dx. ( x + 1) 2( x + 3) ( 3 x - 2) A B C Let = + + ( x + 1) 2( x + 3) ( x + 1) ( x + 1) 2 ( x + 3) … (i) Þ ( 3 x - 2) º A( x + 1)( x + 3) + B( x + 3) + C( x + 1) 2 . -11 Putting x = -3 on both sides of (i), we get C = × 4 -5 Putting x = -1 on both sides of (i), we get B = × 2 Comparing the coefficients of x 2 on both sides of (i), we get 11 A + C = 0 Þ A = -C = × 4 11 5 11 ( 3 x - 2) = \ 2 2 4 ( x + 1 ) 4 ( x + 3) 2( x + 1) ( x + 1) ( x + 3)

Senior Secondary School Mathematics for Class 12 Pg-751

Integration Using Partial Fractions

Þ

( 3 x - 2)

ò ( x + 1) 2( x + 3) dx =

11 dx 5 1 11 dx × - × dx - × ò 4 ò ( x + 1) 2 ò ( x + 1) 2 4 ( x + 3)

=

11 5 11 × log |x + 1| + - × log|x + 3| + C 4 2( x + 1) 4

=

x + 1½ 11 5 × log ½ + + C. x x 4 + 3 2 ( + 1) ½ ½

dx

ò ( x 3 + x 2 + x + 1) ×

EXAMPLE 4

Evaluate

SOLUTION

We have Let

751

1 ( x 3 + x 2 + x + 1) 1

( x + 1)( x 2 + 1)

=

=

[CBSE 2006]

1 x 2( x + 1) + ( x + 1)

=

1 ( x + 1)( x 2 + 1)

×

A Bx + C + ( x + 1) ( x 2 + 1)

Þ 1 º A( x 2 + 1) + ( Bx + C)( x + 1).

… (i)

1 Putting x = -1 on both sides of (i), we get A = × 2 Comparing the coefficients of x 2 on both sides of (i), we get -1 A + B = 0 Þ B = -A = × 2 Comparing the coefficients of x on both sides of (i), we get 1 B + C = 0 Þ C = -B = × 2 -1 1 x+ 1 1 2 \ = + 2 2 ( x + 1)( x 2 + 1) 2( x + 1) x +1 \

EXAMPLE 5

SOLUTION

dx

dx

ò ( x 3 + x 2 + x + 1) = ò ( x + 1)( x 2 + 1) =

1 dx 1 x dx 1 × dx + ò 2 2 ò ( x + 1) 2 ò ( x 2 + 1) 2 ( x + 1)

=

1 dx 1 2x 1 dx × - × dx + ò 2 2 ò ( x + 1) 4 ò ( x 2 + 1) 2 ( x + 1)

=

1 1 1 log|x + 1| - log|x 2 + 1| + tan -1 x + C. 2 4 2

x4

ò ( x - 1)( x 2 + 1) dx.

Evaluate x4 2

( x - 1)( x + 1)

=

x4 3

2

( x - x + x - 1)

= ( x + 1) +

1 3

2

( x - x + x - 1)

Senior Secondary School Mathematics for Class 12 Pg-752

752

Senior Secondary School Mathematics for Class 12

x4

1 × ( x - 1)( x 2 + 1) A Bx + C Let = + × Then, ( x - 1)( x 2 + 1) ( x - 1) ( x 2 + 1)

Þ

( x - 1)( x 2 + 1) 1

= ( x + 1) +

… (i)

… (ii) 1 º A( x 2 + 1) + ( Bx + C)( x - 1). 1 Putting x = 1 in (ii), we get A = × 2 Comparing the coefficients of x 2 on both sides of (ii), we get 1 A + B = 0 Þ B = -A = - × 2 Comparing the constant terms on both sides of (ii), we get æ 1 ö -1 A - C = 1 Þ C = ( A - 1) = ç - 1÷ = × 2 è2 ø 1 1 - x1 1 2 2 \ = + ( x - 1)( x 2 + 1) 2( x - 1) ( x 2 + 1) \

x4 ( x - 1)( x 2 + 1)

Þ

= ( x + 1) +

1 1 ( x + 1) - × 2( x - 1) 2 ( x 2 + 1)

x4

dx

1

ò ( x - 1)( x 2 + 1) dx = ò ( x + 1) dx + 2 ò ( x - 1) 1 2x 1 dx - ×ò 2 dx - ò 2 4 ( x + 1) 2 ( x + 1) 1 1 1 x2 + x + log |x - 1| - log ( x 2 + 1) - tan -1 x + C. 2 2 4 2 ( 3 x + 5) [CBSE 2005C, ’13C] ò ( x 3 - x 2 - x + 1) dx. =

EXAMPLE 6

Evaluate

SOLUTION

( x 3 - x 2 - x + 1) = x 2( x - 1) - ( x - 1) = ( x - 1)( x 2 - 1) = ( x - 1) 2( x + 1). Let

3x + 5 ( x 3 - x 2 - x + 1)

=

3x + 5 ( x - 1) 2( x + 1)

=

A B C + + ( x - 1) ( x - 1) 2 ( x + 1)

Þ ( 3 x + 5) º A( x - 1)( x + 1) + B( x + 1) + C( x - 1) 2 . Putting x = 1 on both sides of (i), we get B = 4. 1 Putting x = - 1 on both sides of (i), we get C = × 2 Comparing the coefficient of x 2 on both sides of (i), we get -1 A + C = 0 Þ A = -C = × 2 -1 4 1 ( 3 x + 5) = + + \ 3 2 2 2 ( x 1 ) 2 ( x + 1) ( x - x - x + 1) ( x - 1)

… (i)

Senior Secondary School Mathematics for Class 12 Pg-753

Integration Using Partial Fractions

Þ

ò

( 3 x + 5) 3

2

( x - x - x + 1)

dx = -

753

1 1 dx dx dx + 4ò + ò 2 2 ò ( x - 1) 2 ( x + 1) ( x - 1)

1 4 1 = - log|x - 1| + log|x + 1| + C. 2 ( x - 1) 2 ( x 3 - 1)

ò ( x 3 + x) dx.

EXAMPLE 7

Evaluate

SOLUTION

We have ( x 3 - 1) ( x 3 + x)

=1-

( x + 1)

[on dividing]

( x 3 + x)

( x + 1) × x( x 2 + 1) ( x + 1) A Bx + C Let = + 2 × 2 x( x + 1) x ( x + 1) =1-

… (i)

Then, ( x + 1) º A( x 2 + 1) + ( Bx + C) x.

… (ii)

Putting x = 0 in (ii), we get A = 1. Comparing the coefficients of x in (ii), we get C = 1. Comparing the coefficients of x 2 in (ii), we get A + B = 0 Þ B = - A = -1. \ A = 1, B = -1 and C = 1. 1 (1 - x) ( x + 1) Thus, = + 2 2 x( x + 1) x ( x + 1) Þ

ò

( x 3 - 1) 3

( x + x)

dx = ò dx - ò

( x + 1) x( x 2 + 1)

dx

ì dx ü (1 - x) = x - íò +ò 2 dx ý ( x + 1) þ î x dx dx 1 2x = x-ò + dx x ò ( x 2 + 1) 2 ò ( x 2 + 1) 1 = x - log|x| - tan -1 x + log ( x 2 + 1) + C. 2 cos x [CBSE 2006C] ò (1 - sin x)( 2 - sin x) dx.

EXAMPLE 8

Evaluate

SOLUTION

Putting sin x = t and cos x dx = dt, we get dt cos x I= dx = ò × (1 - sin x)( 2 - sin x) (1 - t)( 2 - t) 1 A B Let = + (1 - t)( 2 - t) (1 - t) ( 2 - t) Þ 1 º A( 2 - t) + B(1 - t). Putting t = 1 in (i), we get A = 1.

… (i)

Senior Secondary School Mathematics for Class 12 Pg-754

754

Senior Secondary School Mathematics for Class 12

Putting t = 2 in (i), we get B = - 1. 1 1 1 = \ (1 - t)( 2 - t) (1 - t) ( 2 - t)

ò

Þ

cos x dt dx = ò (1 - sin x)( 2 - sin x) (1 - t)( 2 - t) ì 1 1 ü = òí ý dt - t) þ ( 1 t ) ( 2 î dt dt =ò (1 - t) ò ( 2 - t) = - log|1 - t| + log|2 - t| + C 2 -t ½2 - sin x½ ½+ C. = log ½ ½+ C = log ½ ½1 - t½ ½1 - sin x½ dx

ò x{6(log x) 2 + 7 log x + 2} ×

EXAMPLE 9

Evaluate

SOLUTION

Putting log x = t and I=ò Let

1 dx = dt, we get x

dx x{6(log x) 2 + 7 log x + 2}

=ò

dt ( 6t 2 + 7t + 2)

=ò

dt × ( 2t + 1)( 3t + 2)

1 A B = + × ( 2t + 1)( 3t + 2) ( 2t + 1) ( 3t + 2)

Then, 1 º A( 3t + 2) + B( 2t + 1). 1 Putting t = - in (i), we get A = 2. 2 -2 in (i), we get B = -3. Putting t = 3 1 2 3 \ = ( 2t + 1)( 3t + 2) ( 2t + 1) ( 3t + 2) Þ I=ò =ò

dt ( 2t + 1)( 3t + 2) 2dt 3 dt -ò ( 2t + 1) ( 3t + 2)

= log|2t + 1| - log|3t + 2| + C 2t + 1½ = log ½ +C ½3t + 2½ ½2 log x + 1½ ½+ C. = log ½ ½3 log x + 2½

… (i)

Senior Secondary School Mathematics for Class 12 Pg-755

Integration Using Partial Fractions

x2

ò (1 + x 3)( 2 + x 3) dx.

EXAMPLE 10

Evaluate

SOLUTION

Putting x 3 = t and x 2 dx = I=ò Let

x2 3

(1 + x )( 2 + x 3)

755

[CBSE 2003C]

1 dt, we get 3

dx =

dt 1 × × 3 ò (1 + t)( 2 + t)

1 A B = + × Then, (1 + t)( 2 + t) (1 + t) ( 2 + t)

1 º A( 2 + t) + B(1 + t). Putting t = - 1 in (i), we get A = 1. Putting t = - 2 in (i), we get B = -1. 1 1 1 \ = (1 + t)( 2 + t) (1 + t) ( 2 + t) dt dt dt Þ I=ò =ò -ò (1 + t)( 2 + t) (1 + t) ( 2 + t)

… (i)

= log|1 + t| - log|2 + t| + C 1 + t½ = log ½ +C ½2 + t½ ½1 + x 3½ ½+ C. = log ½ 3 ½2 + x ½ EXAMPLE 11

SOLUTION

Evaluate

dx

ò ( ex - 1) ×

[CBSE 2003]

1 Putting ex = t and ex dx = dt, i.e., dx = dt, we get t dx dt I=ò x =ò × t(t - 1) ( e - 1) Let

1 A B = + × t (t - 1) t (t - 1)

Then, 1 º A(t - 1) + Bt. Putting t = 0 in (i), we get A = -1. Putting t = 1 in (i), we get B = 1. 1 -1 1 = + × \ t(t - 1) t (t - 1) dx Hence, I = ò x ( e - 1) -1 dt 1 =ò = dt + ò dt t(t - 1) ò t (t - 1) = - log|t| + log|t - 1| + C

… (i)

Senior Secondary School Mathematics for Class 12 Pg-756

756

Senior Secondary School Mathematics for Class 12

t -1 = log ½ ½+ C ½ t ½ ½ex - 1½ = log ½ x ½+ C. ½ e ½ dx

ò x( x n + 1) ×

EXAMPLE 12

Evaluate

SOLUTION

Putting x n = t, we get nx n-1 dx = dt. 1 1 nx n dx = dt Þ dx = dt (note). x x nt 1 dx dt dt \ ò =ò = ×ò × n nt(t + 1) n t(t + 1) x( x + 1) \

Let

1 A B = + × t(t + 1) t (t + 1)

Then, 1 º A(t + 1) + Bt. Putting t = 0 in (i), we get A = 1. Putting t = - 1 in (i), we get B = - 1. ì1 1 1 ü =í \ ý t (t + 1) î t (t + 1) þ Þ

… (i)

dx

… (ii)

dt

1

ò x( x n + 1) = n ò t (t + 1) =

ù 1é 1 1 dt - ò dt ú ê ò në t (t + 1) û

1 × {log|| t - log|t + 1|} + C n 1 t = × log ½ ½+ C n t + ½ 1½ =

= EXAMPLE 13

Evaluate

SOLUTION

Let I = ò

½ xn ½ 1 log ½ n ½+ C. n ½x + 1½

dx

ò x( x5 + 3) × dx x( x5 + 3)

=ò

[CBSE 2013]

x4 x5( x5 + 3)

dx.

Putting x5 = t and 5 x 4 dx = dt in (i), we get dt 1 I= ò × 5 t (t + 3) 1 A B Let = + × t (t + 3) t t + 3

… (i)

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-757

Integration Using Partial Fractions

Then, 1 º A(t + 3) + Bt.

757

… (iii)

1 × 3 -1 Putting t = -3 on both sides of (iii), we get B = × 3 1 1 1 = \ t (t + 3) 3t 3(t + 3) ü 1 ì1 1 Þ I = òí ý dt 5 î 3t 3(t + 3) þ 1 1 1 1 = dt - ò dt ò 15 t 15 (t + 3) 1 1 = log |t| - log |t + 3| + C 15 15 1 1 = log |x5| - log |x5 + 3| + C. 15 15 Putting t = 0 on both sides of (iii), we get A =

EXAMPLE 14

Evaluate

SOLUTION

Let Let

x2

ò ( x 2 + 4)( x 2 + 9) dx. x2

2

( x + 4)( x 2 + 9)

=

[CBSE 2013]

y , where x 2 = y. ( y + 4)( y + 9)

y A B = + × ( y + 4)( y + 9) ( y + 4) ( y + 9)

Then, y º A( y + 9) + B( y + 4).

-4 Putting y = -4 on both sides of (i), we get A = × 5 9 Putting y = -9 on both sides of (i), we get B = × 5 y -4 9 \ = + ( y + 4)( y + 9) 5( y + 4) 5( y + 9) Þ Þ

x2 2

( x + 4)( x 2 + 9) x2

=

-4 5( x 2 + 4)

ò ( x 2 + 4)( x 2 + 9) dx =

EXAMPLE 15

Evaluate

SOLUTION

Let

+

… (i)

9 5( x 2 + 9)

dx dx 9 -4 + 5 ò ( x 2 + 4) 5 ò ( x 2 + 9)

x æ9 1ö x æ -4 1 ö =ç ´ ÷ tan -1 + ç ´ ÷ tan -1 + C 2 è5 3 ø 3 è 5 2ø -2 3 -1 x -1 x = tan + tan + C. 5 2 5 3 ( x - 1)( x - 2)( x - 3) ò ( x - 4)( x - 5)( x - 6) dx.

( x - 1)( x - 2)( x - 3) A B C × =1+ + + ( x - 4)( x - 5)( x - 6) ( x - 4) ( x - 5) ( x - 6)

Senior Secondary School Mathematics for Class 12 Pg-758

758

Senior Secondary School Mathematics for Class 12

Then, ( x - 1)( x - 2)( x - 3) º ( x - 4)( x - 5)( x - 6) + A( x - 5)( x - 6) + B( x - 4)( x - 6) + C( x - 4)( x - 5). Putting x = 4 on both sides of (i), we get A = 3. Putting x = 5 on both sides of (i), we get B = -24. Putting x = 6 on both sides of (i), we get C = 30. ( x - 1)( x - 2)( x - 3) dx \ I=ò ( x - 4)( x - 5)( x - 6) ì 3 24 30 ü = ò í1 + + ý dx ( x 4 ) ( x 5 ) ( x - 6) þ î = ò dx + 3 ò

… (i)

dx dx dx - 24 ò + 30ò ( x - 4) ( x - 5) ( x - 6)

= x + 3 log|x - 4| - 24 log|x - 5| + 30 log|x - 6| + C. ( x 2 + 1)( x 2 + 2)

ò ( x 2 + 3)( x 2 + 4) dx.

EXAMPLE 16

Evaluate

SOLUTION

We have ( x 2 + 1)( x 2 + 2) 2

2

( x + 3)( x + 4)

= =

Let

(t + 1)(t + 2) , where x 2 = t (t + 3)(t + 4) (t 2 + 3t + 2) (t 2 + 7t + 12)

=1-

( 4t + 10) × (t + 3)(t + 4)

( 4t + 10) A B = + (t + 3)(t + 4) (t + 3) (t + 4)

Þ ( 4t + 10) º A(t + 4) + B(t + 3). Putting t = -3 in (i), we get A = - 2. Putting t = -4 in (i), we get B = 6. ( 4t + 10) -2 6 = + × \ (t + 3)(t + 4) (t + 3) (t + 4) Thus,

( x 2 + 1)( x 2 + 2) 2

2

( x + 3)( x + 4)

= =

… (i)

… (ii)

(t + 1)(t + 2) , where x 2 = t (t + 3)(t + 4) (t 2 + 3t + 2) 2

=1-

( 4t + 10) (t + 3)(t + 4)

(t + 7t + 12) ì -2 6 ü = 1 -í + ý [from (ii)] î(t + 3) (t + 4) þ ì 2 6 ü = í1 + ý ( t 3 ) ( t 4) þ + + î ì 2 6 ü = í1 + 2 - 2 ý× î ( x + 3) ( x + 4) þ

Senior Secondary School Mathematics for Class 12 Pg-759

Integration Using Partial Fractions

\

ì

( x 2 + 1)( x 2 + 2)

2

759

6

ü

ò ( x 2 + 3)( x 2 + 4) dx = ò í1 + ( x 2 + 3) - ( x 2 + 4) ý dx î

= ò dx + 2ò

EXAMPLE 17

Evaluate

SOLUTION

We have I=ò =ò =ò

=x+

2 3

=x+

2 3

dx

þ dx

- 6ò 2 ( x 2 + 3) ( x + 4) x 6 æ ö -1 æ x ö tan -1 ç ÷ - tan ç ÷ + C 2 3 è 2ø è ø æ x ö -1 æ x ö tan -1 ç ÷ - 3 tan ç ÷ + C. è 2ø è 3ø

( 3 sin q - 2) cos q

ò (5 - cos2q - 4 sin q) dq. ( 3 sin q - 2) cos q {5 - (1 - sin 2 q) - 4 sin q} ( 3 sin q - 2) cos q ( 4 + sin 2 q - 4 sin q) ( 3 sin q - 2) cos q (sin q - 2)

( 3t - 2)

2

dq

dq

dq = ò

( 3t - 2) (t - 2) 2

dt , where sin q = t.

A B + × Then, (t - 2) (t - 2) 2 (t - 2) … (i) ( 3t - 2) º A(t - 2) + B. Putting t = 2 in (i), we get B = 4. Comparing the coefficients of t on both sides of (i), we get A = 3. Thus, A = 3 and B = 4. ( 3t - 2) 3 4 \ = + 2 ( t ) 2 (t - 2) (t - 2) 2 ( 3t - 2) 3 4 Þ I=ò dt = ò dt + ò dt (t - 2) (t - 2) 2 (t - 2) 2 4 = 3 log|t - 2| +C (t - 2) 4 = 3 log|sin q - 2| +C (sin q - 2) 4 = 3 log ( 2 - sin q) + + C [Q ( 2 - sin q) > 0]. ( 2 - sin q) Let

2

EXAMPLE 18

Evaluate

SOLUTION

We have

ò

=

(tan q + tan 3 q) (1 + tan 3 q)

(tan q + tan 3 q) (1 + tan 3 q)

=

dq.

tan q (1 + tan 2 q) (1 + tan 3 q)

[CBSE 2009C]

=

tan q sec2 q (1 + tan 3 q)

×

Senior Secondary School Mathematics for Class 12 Pg-760

760

Senior Secondary School Mathematics for Class 12

\ I=ò

Let

(tan q + tan 3 q) (1 + tan 3 q) tan q sec2 q

=ò

(1 + tan 3 q)

=ò

(1 + t 3)

t

dq

dq

dt = ò

t (1 + t)(1 - t + t 2)

t (1 + t)(1 - t + t 2)

=

dt, where tan q = t.

( Bt + C) A + × Then, (1 + t) (1 - t + t 2)

t º A(1 - t + t 2) + ( Bt + C)(1 + t).

… (i)

-1 Putting t = -1 on both sides of (i), we get A = × 3 Comparing the coefficients of t 2 on both sides of (i), we get 1 A + B = 0 Þ B = -A = × 3 Comparing the constant terms on both sides of (i), we get 1 A + C = 0 Þ C = -A = × 3 1ö æ1 ç t+ ÷ t -1 3 3ø è × + \ = (1 + t)(1 - t + t 2) 3(1 + t) (1 - t + t 2) Now, I = ò

t (1 + t)(1 - t + t 2)

dt

=-

1 dt 1 2t 1 dt + dt + ò 2 3 ò (1 + t) 6 ò (t 2 - t + 1) 3 (t - t + 1)

=-

1 dt 1 ( 2t - 1) + 1 1 dt + dt + ò 2 3 ò (1 + t) 6 ò (t 2 - t + 1) 3 (t - t + 1)

=-

1 dt 1 ( 2t - 1) 1 dt + dt + ò 2 3 ò (1 + t) 6 ò (t 2 - t + 1) 2 (t - t + 1)

=-

dt 1 1 1 log |1 + t| + log |t 2 - t + 1| + ò 1ö 3 3 6 2 æ 2 çt - t + ÷ + 4ø 4 è

=-

dt 1 1 1 log |1 + t| + log |t 2 - t + 1| + ò 3 6 2 (t - 1/2) 2 + ( 3 /2) 2

æ 1ö çt - ÷ 1 1 1 2 2ø 2 -1 è +C = - log |1 + t| + log |t - t + 1| + × tan 3 6 2 3 ( 3 /2)

Senior Secondary School Mathematics for Class 12 Pg-761

Integration Using Partial Fractions

==-

761

1 1 1 æ 2t - 1 ö log |1 + t| + log |t 2 - t + 1| + tan -1 ç ÷ +C 6 3 3 è 3 ø 1 1 log |1 + tan q| + log |tan 2 q - tan q + 1| 3 6 1 æ 2 tan q - 1 ö + tan -1 ç ÷ + C. 3 3 è ø dx

ò (sin x - sin 2x) ×

EXAMPLE 19

Evaluate

SOLUTION

ò (sin x - sin 2x) = ò (sin x - 2 sin x cos x)

dx

[CBSE 2010]

dx

=ò =ò

dx sin x = dx sin x(1 - 2 cos x) ò sin 2 x(1 - 2 cos x) sin x 2

(1 - cos x)(1 - 2 cos x)

= -ò

dt (1 - t 2)(1 - 2t)

dx

, where cos x = t

dt × (t - 1)(t + 1)(1 - 2t) 1 A B C = + + × (t - 1)(t + 1)(1 - 2t) (t - 1) (t + 1) (1 - 2t) =ò

Let

1 º A(t + 1)(1 - 2t) + B(t - 1)(1 - 2t) + C(t - 1)(t + 1). -1 Putting t = 1 in (ii), we get A = × 2 -1 Putting t = - 1 in (ii), we get B = × 6 1 -4 Putting t = in (ii), we get C = × 2 3 dt dt dt 1 1 4 \ I=- ò - ×ò - ×ò 2 (t - 1) 6 (t + 1) 3 (1 - 2t) Then,

… (i)

… (ii)

-2dt 1 1 2 = - log |t - 1| - log |t + 1| + × ò 2 6 3 (1 - 2t)

EXAMPLE 20

SOLUTION

1 1 2 = - log |t - 1| - log |t + 1| + log|1 - 2t| + C 2 6 3 1 1 2 = - log |cos x - 1| - log |cos x + 1| + log |1 - 2 cos x| + C. 2 6 3 (1 - cos x) Evaluate ò cos x(1 + cos x) dx. (1 - cos x) (1 - t) A B Let = = + , where t = cos x. cos x(1 + cos x) t(1 + t) t (1 + t)

Senior Secondary School Mathematics for Class 12 Pg-762

762

Senior Secondary School Mathematics for Class 12

Then,

(1 - t) º A(1 + t) + Bt.

Putting t = 0 in this identity, we get A = 1. Putting t = -1 in the identity, we get B = -2. (1 - t) 1 2 = \ t (1 + t) t 1 + t (1 - cos x) 1 2 or = cos x(1 + cos x) cos x (1 + cos x) (1 - cos x) dx dx \ ò dx = ò - 2ò cos x(1 + cos x) cos x (1 + cos x) x = ò sec x dx - ò sec2 dx 2 = log |sec x + tan x| - 2 tan ( x 2 + 1)

ò ( x + 1) 2 dx.

EXAMPLE 21

Evaluate

SOLUTION

On dividing ( x 2 + 1) by ( x 2 + 2x + 1), we get ( x 2 + 1) ì 2x ü × = í1 2 2ý ( x + 1) î ( x + 1) þ 2x A B Let = + 2 ( x + 1) ( x + 1) 2 ( x + 1)

x + C. 2 [CBSE 2006]

Þ 2x º A( x + 1) + B. On equating the coefficients of x, we get A = 2.

… (i)

On equating constant terms, we get A + B = 0 Þ B = - A = -2. 2x 2 2 = \ 2 ( x ) + 1 ( x + 1) ( x + 1) 2 ì 2x ü \ I = ò í1 2ý î ( x + 1) þ ì 2 2 ü = ò í1 + ý dx 1 ( x ) + ( x + 1) 2 þ î 2 = x - 2 log|x + 1| + C. ( x + 1)

EXERCISE 15A Evaluate: dx 1. ò x( x + 2)

2.

( 2x + 1)

ò ( x + 2)( x - 3) dx

[CBSE 2007]

Senior Secondary School Mathematics for Class 12 Pg-763

Integration Using Partial Fractions

x

3.

ò ( x + 2)( 3 - 2x) dx

5.

ò ( x - 1)( x + 2)( x - 3) dx

6.

ò ( x 2 - 1)( 2x + 3) dx

8.

( 2x - 1)

( 2x - 3)

( x 2 + 5 x + 3)

ò ( x 2 + 3 x + 2)

[CBSE 2002]

12.

ò ( x - 1)( x - 2) dx

x3

( 2x + 1) ( 4 - 3 x - x 2)

dx [CBSE 2004C]

cos x

16.

ò (1 + sin x)( 2 + sin x) dx

18.

ò (cos2x - cos x - 2) dx ex

ò ( e 3x - 3 e2x - ex + 3) dx

22.

ò (1 - cot 2x) dx

24.

ò (1 + sin x)( 2 + sin x) dx

25.

cosec2 x

sin 2x

ex x

e ( ex - 1)

dx

(1 - x 2)

27.

ò x(1 - 2x) dx

29.

ò ( x + 2)( x - 3) 2 dx ( x 2 + 1)

31.

ò ( x + 3)( x - 1)

33.

ò ( 2x + 1) 2 dx

35.

[CBSE 2010]

( 2x + 9)

dx 2

2x

(5 x + 8)

ò x 2( 3 x + 8)

dx

dx

ò x( x - 2)( x - 4) ( 2x + 5)

ò ( x 2 - x - 2) dx ( x 2 + 1)

ò ( x 2 - 1) dx ( 3 + 4x - x 2)

11.

ò ( x + 2)( x - 1) dx

13.

ò

15.

ò ( x 2 + 1)( x 2 + 3) dx

( x 3 - x - 2) (1 - x 2) 2x

17.

[CBSE 2003C, ‘07]

sin x cos x

20.

7. 9.

x3

ò ( x 2 - 4) dx

4.

[CBSE 2005]

dx

10.

14.

[CBSE 2003]

763

dx [CBSE 2011]

sec2 x

ò ( 2 + tan x)( 3 + tan x) dx

ex

19.

ò ( e2x + 5 ex + 6) dx

21.

ò x[2(log x) 2 - log x - 3] dx

23.

ò (tan 3x + 4 tan x) dx

2 log x

sec2 x

[CBSE 2004]

dx

26.

ò x( x 4 - 1)

28.

ò ( x + 2)( x + 1) 2 dx

30.

ò ( x - 1) 2( x + 3) dx

( x 2 + x + 1) ( x 2 + 1)

( x 2 + x + 1)

32.

ò ( x + 2)( x 2 + 1) dx

34.

ò ( x + 2)( x - 2) 2 dx

36.

3x + 1

(5 x 2 - 18x + 17)

ò ( x - 1) 2( 2x - 3) dx

[CBSE 2006C]

[CBSE 2012]

[CBSE 2009C] [CBSE 2007C]

Senior Secondary School Mathematics for Class 12 Pg-764

764

Senior Secondary School Mathematics for Class 12

8

37.

ò ( x + 2)( x 2 + 4) dx

39.

ò ( x 2 + 1)( x 2 + 3) dx

41.

ò ( x 3 - 1)

43.

ò ( x + 1) 2( x 2 + 1)

45.

ò ( x 2 + 2)( x 2 + 4)

48.

ò x( x5 + 1)

51.

ò cos x(5 - 4 sin x)

54.

ò (sin x + sin 2x)

56.

ò ( x 2 + 1)( x 2 + 9)( x 2 + 16) dx

58.

2x

dx

dx

[CBSE 2002]

dx

dx

dx

dx

x4

2

ò (1 - x)(1 + x 2)

( 3 x + 5)

38.

ò ( x 3 - x 2 + x - 1) dx

40.

ò ( x 4 - 1) dx

42.

ò ( x 3 + 1)

44.

ò ( 2x + 1)( x 2 + 4) dx

x2

dx

17

x2

46.

ò ( x 2 + 4)( x 2 + 25) dx

49.

ò x( x 6 + 1)

52.

ò sin x cos2x

55.

ò ( x 4 - x 2 - 12) dx

dx

dx

[CBSE 2013] 47.

dx

ò ( ex - 1) 2

dx

50.

ò sin x( 3 + 2 cos x)

53.

ò (1 - sin x) dx

tan x

x2

57.

dx [CBSE 2012]

sin 2x

ò (1 - cos 2x)( 2 - cos 2x) dx 59.

2x 2 + 1

ò x 2( x 2 + 4) dx

[CBSE 2007]

[CBSE 2013]

ANSWERS (EXERCISE 15A)

1. 3. 4. 5. 6. 7. 9. 11. 12.

1 x ½ 3 7 2. log |x + 2| + log |x - 3| + C log ½ +C 2 x + 2 5 5 ½ ½ -2 3 log |x + 2| - log |3 - 2x| + C 7 14 1 1 1 log |x| - log |x - 2| + log |x - 4| + C 8 4 8 1 1 1 - log |x - 1| - log |x + 2| + log |x - 3| + C 6 3 2 1 5 12 - log |x - 1| + log |x + 1| - log |2x + 3| + C 10 2 5 8. x + 3 log |x + 2| - log |x + 1| + C 3 log |x - 2| - log |x + 1| + C x2 10. x + log |x - 1| - log | x + 1| + C + 2 log |x 2 - 4| + C 2 - x + 3 log |x + 2| + 2 log |x - 1| + C 1 - x½ x2 x2 + 3 x - log|x - 1| + 8 log|x - 2| + C 13. + log ½ +C 2 2 ½1 + x½

Senior Secondary School Mathematics for Class 12 Pg-765

Integration Using Partial Fractions

14. -

3 7 log|1 - x| - log|4 + x| + C 5 5

15.

½1 + sin x½ ½+ C 16. log ½ ½2 + sin x½ 18. -

765

1 log 2

æ x2 + 1 ö ç ÷ +C ç x2 + 3 ÷ è ø

½2 + tan x½ ½+ C 17. log ½ ½3 + tan x½

1 2 log |cos x + 1| - log |cos x - 2| + C 3 3

½2 + ex ½ ½+ C 19. log ½ x ½3 + e ½

1 1 1 log | ex - 3| - log | ex - 1| + log | ex + 1| + C 8 4 8 2 3 21. log |1 + log x| + log |2 log x - 3| + C 5 5 1 1 + cot x½ ½ ½+ C 22. - log ½ 2 ½1 - cot x½ 1 1 23. log |tan x| - log |tan 2 x + 4| + C 4 8 20.

½ex - 1½ 25. log ½ x ½+ C ½ e ½ 1 3 27. x + log|x| - log|1 - 2x| + C 2 4

24. -2 log |1 + sin x| + 4 log |2 + sin x| + C 1 26. log |x| - log |x 4 + 1| + C 4

1 +C ( x + 1) 1 1 3 log |x + 2| - log |x - 3| +C 5 5 ( x - 3) 3 1 5 log |x - 1| + log |x + 3| + C 8 2( x - 1) 8 5 3 1 log |x - 3| - log |x - 1| + +C 2 2 ( x - 1) 3 1 1 log |x + 2| + log |x 2 + 1| + tan -1 x + C 5 5 5 1 1 log |2x + 1| + +C 2 2( 2x + 1) -5 5 7 log |x + 2| + log |x - 2| +C 16 16 4( x - 2) 1 1 1 4 5 log |x| - - log |3 x + 8| + C 36. + log |2x - 3| + C 4 x 4 ( x - 1) 2 1 x log |x + 2| - log |x 2 + 4| + tan -1 + C 2 2 2 -1 4 log |x - 1| - 2 log |x + 1| - tan x + C x - 1½ 1 1 1 1 40. log ½ log|x 2 + 1| - log|x 2 + 3| + C + tan -1 x + C 2 2 4 ½x + 1½ 2

28. 3 log|x + 2| - 2 log|x + 1| 29. 30. 31. 32. 33. 34. 35. 37. 38. 39.

Senior Secondary School Mathematics for Class 12 Pg-766

766

41. 42. 43. 44. 45. 47. 49. 50. 51. 52. 54. 55. 56. 57. 58.

Senior Secondary School Mathematics for Class 12

1 1 1 æ 2x + 1 ö log |x - 1| - log |x 2 + x + 1| tan -1 ç ÷ +C 6 3 3 è 3 ø 1 1 1 æ 2x - 1 ö log |x + 1| - log |x 2 - x + 1| + tan -1 ç ÷ +C 6 3 3 è 3 ø 1 1 1 log |x + 1| - log |x 2 + 1| + C 2 2( x + 1) 4 x 2 log |2x + 1| - log |x 2 + 4| + 2 tan -1 + C 2 -1 8 1 æxö æxö -1 æ x ö 1 -1 x 46. tan ç tan -1 ç ÷ + tan -1 ç ÷ + C +C ÷ - tan 14 2 2 2 è5 ø è 2 ø 35 è 2ø 4 1 1 x 5 48. log |x| - log |x + 1| + C x - log |e - 1| - x +C 5 ( e - 1) 1 log |x| - log |x 6 + 1| + C 6 1 1 2 - log |cos x + 1| + log |cos x - 1| + log |2 cos x + 3| + C 2 10 5 1 1 4 - log |1 - sin x| + log |1 + sin x| + log |5 - 4 sin x| + C 2 18 9 1 cos x + 1½ 1 1 ½ ½1 - sin x½ ½+ C ½+ 53. log ½ +C sec x - log ½ 2 cos 1 x x 4 1 + sin 2 ( 1 sin x) ½ ½ ½ ½ 1 1 2 log |cos x + 1| + log |cos x - 1| - log |2 cos x + 1| + C 2 6 3 x - 2½ 1 3 ½ -1 x log tan + +C 7 3 ½x + 2½ 7 x x 1 27 64 tan -1 x tan -1 + tan -1 + C 120 56 3 105 4 1 1 log |1 - cos 2x| - log |2 - cos 2x| + C 2 2 x 1 -1 7 59. - log (1 - x) + log (1 + x 2) + tan -1 x + C + tan -1 + C 2 4x 8 2 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 15A)

7. I = ò

( 2x + 5) dx. ( x - 2 )( x + 1)

ì ( 2 x + 1) ü 8. I = ò í 1 + ý dx. ( x 2 )( x + 1) þ + î

üï ì ü 2 2 ïì 9. I = ò í 1 + 2 ý dx = ò í 1 + ý dx. + x 1 x 1 ( )( ) ïî ( x - 1) ïþ î þ ìï ì ü 4 x üï 4x 10. I = ò í x + 2 ý dx = ò í x + ý dx. ( x - 2 )( x + 2 ) þ ( x - 4 ) þï î îï 11.

( 3 + 4x - x 2 ) ( - x 2 + 4x + 3) ì (5 x + 1) ü = = í -1 + ý× ( x + 2 )( x - 1) ( x + 2 )( x - 1) þ ( x 2 + x - 2) î

Senior Secondary School Mathematics for Class 12 Pg-767

Integration Using Partial Fractions

12.

13.

ü 7x - 6 ù ì x3 x3 7x - 6 é = = x+ 3+ 2 ú = í x + 3 + ( x - 2 )( x - 1) ý × ( x - 1)( x - 2 ) ( x 2 - 3 x + 2 ) êë x - 3x + 2 û î þ ( x 3 - x - 2)

=

2

(1 - x )

üï ì ìï ü 2 2 = í- x = -x ý× 2 ý í ( 1 - x )( 1 + x ) þ ( - x + 1) ( 1 - x ) ïþ î îï

( x 3 - x - 2) 2

15. Putting x 2 = t and 2x dx = dt , we get dt I=ò × ( 2 + t )( 3 + t ) 16. Putting sin x = t and cos x dx = dt , we get dt I=ò × ( 1 + t )( 2 + t ) 17. Put tan x = t and sec2 x dx = dt. 18. Put cos x = t and - sin x dx = dt. 19. Put e x = t and e xdx = dt . 20. Put e x = t and e xdx = dt . 21. Put log x = t and

1 dx = dt . x

22. Put cot x = t and - cosec2 x dx = dt. 23. Put tan x = t and sec2 x dx = dt. 24. Write sin 2 x = 2 sin x cos x. Put sin x = t and cos x dx = dt . 25. Put e x = t and e xdx = dt . 26. I =

x3 4

x ( x 4 + 1)

dx. Put x 4 = t and 4 x 3 dx = dt .

ì 1 öü æ ç1÷ ( 1 - x2 ) ( - x 2 + 1) ïï 1 è 2 x ø ïï 27. =í + = ý× 2 x( 1 - 2 x ) ( -2 x + x ) ï 2 x( 1 - 2 x) ï ïî ïþ 1 ö æ ç1÷ 1 ö B æ 2x ø A è Let = + × Then, ç 1 ÷ º A( 1 - 2 x ) + Bx. 2x ø x( 1 - 2 x ) x ( 1 - 2 x ) è 1 3ö æ ( x = 0 Þ A = 1) and ç x = Þ B = ÷ 2 2ø è ì1 1 ü 3 \ I = òí + + ý dx. î 2 x 2( 1 - 2 x ) þ 32. Let 33.

( x 2 + x + 1) x 2 ( x + 2) 2x

( 2 x + 1) 2

34. Let

=

A B C + + × x x 2 ( x + 2)

A B + × ( 2 x + 1) ( 2 x + 1) 2

( 3x + 5) ( x - 2)2

=

=

A B + × ( x - 2) ( x - 2)2

767

Senior Secondary School Mathematics for Class 12 Pg-768

768

Senior Secondary School Mathematics for Class 12

35. Let

(5 x + 8 ) x 2 ( 3x + 8)

=

A B C + + × x x 2 ( 3x + 8)

39. Put x 2 = t and 2x dx = dt. Then, dt I=ò × (t + 1)(t + 3 ) 1 A B Let = + × Then, (t + 1)(t + 3 ) (t + 1) (t + 3 ) 1 º A(t + 3 ) + B(t + 1).

… (i)

… (ii)

1 × 2 -1 Putting t = -3 in (ii), we get B = × 2

Putting t = -1 in (ii), we get A =

43. Let

1 ( x + 1) 2 ( x 2 + 1)

=

Cx + D A B + + × ( x + 1) ( x + 1) 2 ( x 2 + 1)

\ 1 º A( x + 1)( x 2 + 1) + B( x 2 + 1) + (Cx + D )( x + 1) 2 .

… (i)

1 Putting x = -1 in (i), we get B = × 2 Comparing coefficients of various powers of x, we get A + B + D = 0 , A + C + 2 D = 0 , A + B + 2C + D = 0 , A + C = 0 ( A + C = 0 , A + C + 2 D = 0 ) Þ D = 0. -1 ù 1ö æ é êë A + B + D = 0 Þ A + B = 0 Þ A = 2 úû and çè C = - A = 2 ÷ø × 45. Put x 2 = t . Then, 1 ( x 2 + 2 )( x 2 + 4 )

=

1 A B = + × (t + 2 )(t + 4 ) (t + 2 ) (t + 4 )

\ 1 º A(t + 4 ) + B(t + 2 ).

… (i)

1 Putting t = -2 in (i), we get A = × 2 -1 Putting t = - 4 in (i), we get B = × 2 dx dx 1 1 \ I= ò 2 × 2 ( x + 2) 2 ò ( x 2 + 4) x2 + 1

y+ 1 = , where x 2 = y. ( x + 4 )( x 2 + 25 ) ( y + 4 )( y + 25 ) y+ 1 A B Let = + × ( y + 4 )( y + 25 ) ( y + 4 ) ( y + 25 )

46. Let

2

Then, y + 1 º A( y + 25 ) + B( y + 4 ). -3 -1 = × 21 7 -24 8 Putting y = -25 in (i), we get B = = × -21 7 -1 dx dx 8 \ I= + × 7 ò ( x 2 + 4 ) 7 ò ( x 2 + 25 ) Putting y = -4 in (i), we get A =

… (i)

Senior Secondary School Mathematics for Class 12 Pg-769

Integration Using Partial Fractions

769

x æ 8 1ö x æ -1 1 ö =ç ´ ÷ tan -1 + ç ´ ÷ tan -1 + C 2 è7 5 ø 5 è 7 2ø -1 x x 8 = tan -1 + tan -1 + C . 14 2 35 5 dt 47. Put e x - 1 = t and e xdx = dt , i.e., dx = × (t + 1) dt 1 A B C × Let 2 \ I=ò 2 = + 2 + × t+1 t (t + 1) t (t + 1) t t Then, 1 º At(t + 1) + B(t + 1) + Ct 2 . 48. I = ò

x4 5

x ( x5 + 1) x5

dx. Put x5 = t and 5 x 4 dx = dt .

49. I = ò

x ( x 6 + 1)

50. I = ò

sin 2 x( 3 + 2 cos x )

6

… (i)

dx. Now, put x 6 = t and 6 x5 dx = dt.

sin x

dx = ò

sin x ( 1 - cos 2 x )( 3 + 2 cos x )

dx.

Now, put cos x = t and - sin x dx = dt. cos x cos x 51. I = ò dx = ò dx. 2 2 cos x(5 - 4 sin x ) ( 1 - sin x )(5 - 4 sin x ) Now, put sin x = t and cos x dx = dt. sin x sin x 52. I = ò dx = ò dx. 2 2 sin x cos x ( 1 - cos 2 x ) cos 2 x Put cos x = t and - sin x dx = dt. sin x sin x cos x 53. I = ò dx = ò dx cos x( 1 - sin x ) cos 2 x( 1 - sin x ) sin x cos x t =ò dt , where sin x = t dx = ( 1 - sin 2 x )( 1 - sin x ) ( 1 - t 2 )( 1 - t ) t =ò dt. ( 1 + t )( 1 - t ) 2 dx dx = (sin x + 2 sin x cos x ) ò sin x( 1 + 2 cos x ) sin x =ò dx. Now put cos x = t . 2 ( 1 - cos x )( 1 + 2 cos x )

54. I = ò

x2

t t = = × ( x - x 2 - 12 ) (t 2 - t - 12 ) (t - 4 )(t + 3 ) t A B Let = + × (t - 4 )(t + 3 ) (t - 4 ) (t + 3 )

55. Let x 2 = t . Then,

4

Then, t º A(t + 3 ) + B(t - 4 ). 3 Putting t = -3, we get B = × 7 4 Putting t = 4, we get A = × 7

Senior Secondary School Mathematics for Class 12 Pg-770

770

Senior Secondary School Mathematics for Class 12 x2

\

4

( x - x 2 - 12 )

=

4 1 3 1 × + × × 7 ( x 2 - 4) 7 ( x 2 + 3)

Now, integrate. 56. Let x 2 = t . Then,

x4 2

2

2

( x + 1)( x + 9 )( x + 16 )

=

t2 × (t + 1)(t + 9 )(t + 16 )

EXERCISE 15B Very-Short-Answer Questions Evaluate: -6

æ

1 ö ÷ dx xø

1.

òx

dx

2.

ò çè

3.

ò sin 3x dx

4.

ò (1 + x 3) dx

5.

ò 3 sin 2x dx

6.

ò (5 - cos2f - 4 sin f) df

7.

ò sin

8.

ò

9.

( x + 1)( x + log x) 2 dx ò x

2 cos x

2

[CBSE 2008]

x dx

(1 + tan x)

11.

ò (1 - tan x) dx

13.

ò ( x + log sin x) dx

15.

ò

17.

ò (1 - tan 2x) dx

19.

ò tan

21.

ò cos x cos 3 x dx

23.

ò ( x + 1) 2 dx

25.

ò (1 - sin x)

(1 + cot x)

sec2(log x) dx x tan x sec2 x -1

1 - sin x dx 1 + sin x

xex

dx

x+ x2

[CBSE 2008]

( 3 sin f - 2) cos f

(log x) 2 dx x sin x

10.

ò (1 + cos x) dx

12.

ò (1 + cot x) dx

14.

ò ( x + cos2x) dx

16.

ò

18.

ò ( x 2 + 1) dx

20.

ò log (1 + x

22.

ò sin 3x sin x dx

24.

òe

26.

ò x cos x

(1 - cot x)

(1 - sin 2x)

sin ( 2 tan -1 x) (1 + x 2)

dx

( x 4 + 1)

x

2

) dx

{tan x - log cos x} dx 2

dx

[CBSE 2007]

Senior Secondary School Mathematics for Class 12 Pg-771

Integration Using Partial Fractions

cot x dx sin x

27.

ò

29.

ò sin

31.

ò2

33. 35.

x

-1

28.

37.

ò

39.

ò

sec2 x

ò cosec2x dx

dx x + 2 + x + 1) (1 + tan x) 32. ò dx ( x + log sec x)

(cos x) dx

dx

sec2(log x) dx ò x dx

ò

771

2

9x + 16 dx 4x 2 - 25 9 + x 2 dx

30.

ò(

34.

ò ( 2x + 1)(

36.

ò

38.

ò

4 - x 2 dx

40.

ò

x 2 - 16 dx

x 2 + x + 1) dx

dx 4 - 9x 2

ANSWERS (EXERCISE 15B)

1. 3. 5. 7. 9. 11. 13.

-1 +C 5 x5 -1 cos 3 x + C 3 -2 cosec x + C 3 1 1 x - sin 2x + C 2 4 1 ( x + log x) 3 + C 3 - log |cos x + sin x| + C log |x + log sin x| + C

15. tan (log x) + C 1 17. - log |1 - tan 2 x| + C 2 px x 2 19. +C 4 4 sin 4x sin 2x 21. + +C 8 4 ex 23. +C ( x + 1) 25. tan x + sec x + C

2 3/2 x + 2 x +C 3 1 4. log |1 + x 3| + C 3 2.

6. 3 log |sin f - 2| 8.

4 +C (sin f - 2)

1 (log x) 3 + C 3

10. - log |1 + cos x| + C 12. - log |sin x + cos x| + C 14. log |x + cos2 x| + C 1 16. - cos ( 2 tan -1 x) + C 2 x3 18. - x + 2 tan -1 x + C 3 20. x log (1 + x 2) - 2x + 2 tan -1 x + C 22.

sin 2x sin 4x +C 4 8

24. - ex log cos x + C 26.

1 sin x 2 + C 2

Senior Secondary School Mathematics for Class 12 Pg-772

772

27.

Senior Secondary School Mathematics for Class 12

-2 +C sin x

28. tan x - x + C

px x 2 +C 2 2 x 2 31. +C log 2

29.

30.

2 2 ( x + 2) 3/2 - ( x + 1) 3/2 + C 3 3

32. log | x + log sec x| + C 2 2 ( x + x + 1) 3/2 + C 3 1 æ 3x ö 36. sin -1 ç ÷ +C 3 è 2 ø x x 38. × 4 - x 2 + 2 sin -1 + C 2 2

33. tan (log x) + C

34.

1 log 3 x + 9x 2 + 16 + C 3 1 37. log 2x + 4x 2 - 25 + C 2 x 9 39. 9 + x 2 + log x + 9 + x 2 + C 2 2 1 40. x x 2 - 16 - 8 log x + x 2 - 16 + C 2 35.

OBJECTIVE QUESTIONS I Mark (3) against the correct answer in each of the following: dx 1. ò =? ( 9 + x 2) x x x 1 (b) tan -1 + C (c) 3 tan -1 + C (d) none of these (a) tan -1 + C 3 3 3 3 dx 2. ò =? ( 4 + 16x 2) x 1 1 (b) tan -1 + C (a) tan -1 4x + C 32 16 2 1 1 -1 -1 x (c) tan 2x + C (d) tan +C 8 4 2 dx 3. ò dx = ? ( 9 + 4x 2) 1 2x 1 2x (a) tan -1 (b) tan -1 +C +C 2 3 6 3 1 3x (c) tan -1 (d) none of these +C 6 2 sin x 4. ò dx = ? (1 + cos2 x) (a) - tan -1(cos x) + C

(b) cot -1(cos x) + C

(c) - cot -1(cos x) + C

(d) tan -1(cos x) + C

Senior Secondary School Mathematics for Class 12 Pg-773

Integration Using Partial Fractions

5.

6.

773

cos x

ò (1 + sin 2x) dx = ? (a) - tan -1(sin x) + C

(b) tan -1(cos x) + C

(c) tan -1(sin x) + C

(d) - tan -1(cos x) + C

e

x

ò ( e2x + 1) dx = ? (a) cot -1( ex ) + C (b) tan -1( ex ) + C (c) 2 tan -1( ex ) + C (d) none of these

7.

3 x5

ò (1 + x12) dx = ? (a) tan -1 x 6 + C

8.

1 1 (b) tan -1 x 6 + C (c) tan -1 x 6 + C (d) none of these 4 2

2x 3

ò ( 4 + x 8) dx = ?

x4 1 (a) tan -1 +C 2 2 1 (c) tan -1 x 4 + C 2 dx 9. ò 2 =? ( x + 4x + 8) 1 æ x + 2ö (a) tan -1 ç ÷ +C 2 è 2 ø 1 (c) tan -1( x + 2) + C 2 dx 10. ò =? 2 ( 2x + x + 3) 1 æ 4x + 1 ö (a) tan -1 ç ÷ +C 23 è 23 ø 2 æ 4x + 1 ö (c) tan -1 ç ÷ +C 23 è 23 ø dx 11. ò x =? ( e + e- x ) (a) tan -1( ex ) + C (c) - tan -1( e- x ) + C 12.

x4 1 (b) tan -1 +C 4 2 (d) none of these

1 æ x + 2ö (b) tan -1 ç ÷ +C 2 è 2 ø æ x + 2ö (d) tan -1 ç ÷ +C è 2 ø

(b)

1 æ x + 1ö tan -1 ç ÷ +C 23 è 23 ø

(d) none of these

(b) tan -1( e- x ) + C (d) none of these

x2

ò ( 9 + 4x 2) = ? x 1 x - tan -1 + C 4 8 3 x 3 -1 2x (c) - tan +C 4 8 3 (a)

(b)

x 3 x - tan -1 + C 4 8 3

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-774

774

13.

Senior Secondary School Mathematics for Class 12

( x 2 - 1)

ò ( x 2 + 4) dx = ?

x +C 2 x 5 5 (c) x - tan -1 +C 2 2 dx 14. ò =? ( 4 + 9x 2) 2 3x (a) tan -1 +C 3 2 1 3x (c) tan -1 +C 6 2 dx 15. ò =? ( 4x 2 - 4x + 3) 1 æ 2x - 1 ö (a) tan -1 ç ÷ +C 2 è 2 ø 1 æ 2x - 1 ö (c) tan -1 ç ÷ +C 2 è 2 ø dx 16. ò =? (sin 4 x + cos4 x) æ tan 2 x - 1 ö 1 ÷ +C (a) tan -1 ç ç 2 tan x ÷ 2 è ø ö 1 1 -1 æ ÷ +C (c) tan çç ÷ 2 è 2 tan x ø (a) x - 5 tan -1

17.

x 5 (b) x - tan -1 + C 2 2 (d) none of these

1 (b) tan -1 3 x + C 6 (d) none of these

(b)

æ 2x - 1 ö tan -1 ç ÷ +C 2 2 è 2 ø 1

(d) none of these

(b)

æ tan 2 x - 1 ö 1 ÷ +C tan -1 ç ç tan x ÷ 2 è ø

(d) none of these

( x 2 + 1)

ò ( x 4 + x 2 + 1) dx = ?

( x 2 - 1) +C 3 2 ( x - 1) 1 (c) tan -1 +C 3 3x sin 2x 18. ò dx = ? 4 (sin x + cos4 x) (a) tan -1

(a) tan -1(tan 2 x) + C (c) - tan -1(tan 2 x) + C dx 19. ò =? (1 - 9x 2) 1 1 + 3 x½ (a) log ½ +C 3 ½1 - 3 x½ 1 1 + 3 x½ (c) log ½ +C 6 ½1 - 3 x½

(b)

( x 2 - 1) 1 tan -1 +C 3 3

(d) none of these

(b) x 2 + C (d) none of these

1 1 - 3 x½ log ½ +C 3 ½1 + 3 x½ 1 1 - 3 x½ (d) log ½ +C 6 ½1 + 3 x½

(b)

Senior Secondary School Mathematics for Class 12 Pg-775

Integration Using Partial Fractions

20.

dx

ò (16 - 4x 2) = ?

1 2 - x½ (a) log ½ +C 8 ½2 + x½ 1 2 + x½ (c) log ½ +C 8 ½2 - x½ x2 21. ò dx = ? (1 - x 6 ) ½1 + x 3½ 1 ½+ C (a) log ½ 3 6 ½1 - x ½ ½1 - x 3½ 1 ½+ C log ½ 3 3 ½1 + x ½ x 22. ò dx = ? (1 - x 4)

24.

25.

½1 - x 3½ 1 ½+ C (b) log ½ 3 6 ½1 + x ½

(c)

(d) none of these

½1 + x 2½ 1 ½+ C (a) log ½ 2 4 ½1 - x ½

½1 - x 2½ 1 ½+ C (b) log ½ 2 4 ½1 + x ½

(c) 23.

1 2 - x½ log ½ +C 16 ½2 + x½ 1 2 + x½ (d) log ½ +C 16 ½2 - x½ (b)

½1 + x 2½ 1 ½+ C log ½ 2 2 ½1 - x ½

(d) none of these

x2

ò ( a 6 - x 6) dx = ? (a)

½a 3 + x 3½ ½+ C log ½ 3 3 3a ½a - x ½

(b)

(c)

½a 3 - x 3½ ½ 3 ½+ C log 3 6a 3 ½a + x ½

(d) none of these

1

3

1

½a 3 + x 3½ ½+ C log ½ 3 3 6a ½a - x ½ 1

3

dx

ò ( 3 - 2x - x 2) = ? 1 3 + x½ (a) log ½ +C 4 ½3 - x½

1 1 + x½ (b) log ½ +C 4 ½1 - x½

1 3 + x½ (c) log ½ +C 4 ½1 - x ½

(d) none of these

dx

ò (cos2x - 3 sin 2x) = ? (a)

½ 3 + tan x½ 1 ½+ C log ½ 3 ½ 3 - tan x½

(c)

½1 + 3 tan x½ ½+ C log ½ 2 3 ½1 - 3 tan x½ 1

(b)

½1 - 3 tan x½ 1 ½+ C log ½ 3 ½1 + 3 tan x½

(d) none of these

775

Senior Secondary School Mathematics for Class 12 Pg-776

776

26.

27.

Senior Secondary School Mathematics for Class 12

cosec2 x

ò (1 - cot 2x) dx = ? 1 ½1 + cot x½ ½+ C (a) log ½ 2 ½1 - cot x½

1 ½1 + cot x½ ½+ C (b) - log ½ 2 ½1 - cot x½

1 ½1 - cot x½ ½+ C (c) log ½ 2 ½1 + cot x½

(d) none of these

dx

ò ( 4x 2 - 1) = ? 1 2x - 1½ (a) log ½ +C 2 ½2x + 1½ 1 2x - 1½ (c) log ½ +C 4 ½2x + 1½

28.

x

(c)

30.

(d) none of these

ò ( x 4 - 16) dx = ? ½x 2 + 4½ 1 (a) log ½ 2 ½+ C 4 ½x - 4½

29.

1 2x + 1½ (b) log ½ +C 2 ½2x - 1½

½x 2 - 4½ 1 log ½ 2 ½+ C 16 ½x + 4½

(b)

½x 2 + 4½ 1 log ½ 2 ½+ C 16 ½x - 4½

(d) none of these

dx

ò (sin 2x - 4 cos2x) = ? 1 ½tan x - 2½ ½+ C (a) log ½ 4 ½tan x + 2½

1 ½tan x + 2½ ½+ C (b) log ½ 4 ½tan x - 2½

1 ½1 - tan x½ ½+ C (c) log ½ 4 ½1 + tan x½

(d) none of these

dx

ò ( 4 sin 2x + 5 cos2x) = ?

1 æ tan x ö (a) tan -1 ç ÷ +C 2 è 5 ø 1 æ 2 tan x ö (c) tan -1 ç ÷ +C 2 5 5 ø è sin x 31. ò dx = ? sin 3 x ½ log ½ 2 3 ½ ½ 1 (c) log ½ 2 3 ½ (a)

1

(b)

1 æ tan x ö tan -1 ç ÷ +C 5 è 5 ø

(d) none of these

3 + sin x½ ½+ C 3 - sin x½

(b)

½ 3 + cos x½ ½+ C log ½ 2 3 ½ 3 - cos x½

3 + tan x½ ½+ C 3 - tan x½

(d) none of these

1

Senior Secondary School Mathematics for Class 12 Pg-777

Integration Using Partial Fractions

32.

777

( x 2 + 1)

ò ( x 4 + 1) dx = ? æ x2 + 1ö 1 ÷ +C (a) tan -1 ç ç 2x ÷ 2 è ø (c)

æ x2 - 1ö 1 ÷ +C (b) tan -1 ç ç 2x ÷ 2 è ø

æ x2 + 1ö ç ÷ +C ç x2 - 1 ÷ è ø

1 log 2

(d) none of these

ANSWERS (OBJECTIVE QUESTIONS I)

1. (b)

2. (c)

3. (b)

4. (a)

5. (c)

6. (b)

7. (c)

8 (b)

9. (a) 10. (c)

11. (a) 12. (c) 13. (b) 14. (c) 15. (b) 16. (a) 17. (c) 18. (a) 19. (c) 20. (d) 21. (a) 22. (a) 23. (b) 24. (c) 25. (c) 26. (b) 27. (c) 28. (c) 29. (a) 30. (c) 31. (c) 32. (b)

HINTS TO THE GIVEN OBJECTIVE QUESTIONS I 1. I = ò

dx ( 32 + x2 )

=

x 1 tan -1 + C . 3 3

2. I =

dx dx x 1 1 1 1 1 × = × 1 tan -1 1 + C = tan -1 2 x + C . = ×ò 2 ö ü ì ( ) ( ) 16 ò æ 1 16 16 8 2 2 ïæ 1 ö 2ï ç + x2 ÷ íç ÷ + x ý è4 ø ïþ îï è 2 ø

3. I =

dx dx 1 1 = ×ò ü 4ò æ9 4 ìï æ 3 ö 2 2ö 2ï ç +x ÷ íç ÷ + x ý è4 ø ïþ ïî è 2 ø

=

x 1 1 1 2x × + C. tan -1 3 + C = tan -1 ( 2) 4 ( 3 2) 6 3

4. Putting cos x = t and - sin x dx = dt , we get dt I=-ò = - tan -1t + C = - tan -1 (cos x ) + C . (1 + t2 ) 5. Putting sin x = t and cos x dx = dt , we get dt I=ò = tan -1t + C = tan -1 (sin x ) + C . (1+ t2 ) 6. Putting e x = t and e xdx = dt , we get I=ò

dt (t 2 + 1)

= tan -1t + C = tan -1 ( e x ) + C .

Senior Secondary School Mathematics for Class 12 Pg-778

778

Senior Secondary School Mathematics for Class 12

7. Putting x 6 = t and 6 x5 dx = dt , i.e., 3 x5 dx = I=

dt 1 1 1 = tan -1t + C = tan -1 x 6 + C . 2 ò (1 + t2 ) 2 2

8. Putting x 4 = t and 4 x 3 dx = dt , i.e., 2 x 3 dx = I= 9. I = ò = 10. I =

=

1 dt , we get 2

1 dt , we get 2

dt t x4 1 1 1 1 = ´ tan -1 + C = tan -1 + C. ò 2 2 2 (2 + t ) 2 2 2 4 2 dx

{( x + 2 ) 2 + 2 2 }

=ò

dt (t 2 + 2 2 )

, where x + 2 = t

t 1 1 æx+ 2ö ÷ + C. tan -1 + C = tan -1 ç 2 2 2 è 2 ø dx dx 1 1 = 2 ò æ 2 x 3 ö 2 ò ìï 2 x æ 1 ö 2 üï æ 3 1 ö çx + + ÷ ÷ íx + + ç ÷ ý + ç 2 2ø è 2 è 4 ø ïþ è 2 16 ø ïî 1 dx 1 dx 1ö æ = ò , where ç x + ÷ = t 2ü 2 2òì 2 ìï 2 æ 23 ö ïü 4ø è æ ö 1 23 ïæ ö ÷ý ÷ ïý ít + çç í ç x + ÷ + çç ÷ ÷ ïî 4ø è 4 ø ïþ è 4 ø ïþ ïî è

1ö æ 4ç x + ÷ 2 2 t 1 4 4ø -1 -1 4t -1 è = ´ tan +C= tan +C= tan +C æ 23 ö 2 23 23 23 23 23 ç ÷ ç 4 ÷ è ø ( x + ) 4 1 2 = tan -1 + C. 23 23 11. I = ò

dx 1ö æ x çe + x ÷ è e ø

=ò

e xdx (e

2x

+ 1)

=ò

dt (t 2 + 1)

, where e x = t

= tan -1t + C = tan -1 ( e x ) + C . 12. On dividing x 2 by ( 4 x 2 + 9 ), we get \

dx x 9 x 1 9 1 1 dx - × ò tan -1 3 + C = × ( 2) 4ò 4 4 æ 2 9 ö 4 16 ( 3 2 ) çx + ÷ 4ø è x 3 2 x = - tan -1 + C. 4 8 3

I=

13. On dividing, we get \

( 9 4) 1 × 4 ( 4x 2 + 9)

I = ò dx - 5 ò

üï ìï 5 = í1 - 2 ý× ( x + 4 ) îï ( x + 4 ) þï ( x 2 - 1) 2

dx ( x 2 + 4)

=x-

5 x tan -1 + C . 2 2

Senior Secondary School Mathematics for Class 12 Pg-779

Integration Using Partial Fractions 14. I =

=

dx dx 1 1 = ×ò ü 9ò æ4 9 ìï æ 1 ö 2 2ö 2ï ç +x ÷ íç ÷ + x ý è9 ø 3 è ø ïþ ïî x 1 1 1 3x × + C. tan -1 2 + C = tan -1 ( 3) 9 ( 2 3) 6 2

dx dx dx 1 1 1 = = 2 2 3ö 4ò æ 2 1ö 1 4 ò æ 4ò æ 2 1 æ 1 ö ö çx - x + ÷ çx - x + ÷ + ÷ çx - ÷ + ç 4ø 4ø 2 è è 2ø è è 2ø 1 1 dt 1 1 t -1 -1 = × = ò tan ( 2 t ) + C tan +C= 2 æ 1 ö 4 ïì 2 2 æ 1 ö ïü 4 æç 1 ö÷ 2 ç ÷ ÷ ý ít + ç è 2ø è 2ø è 2 ø þï îï

15. I =

= 16.

ì 2 ( 2 x - 1) ü 1 æ 2x - 1 ö ÷ + C. tan -1 í tan -1 ç ý+ C = 2 2 2 2 2 2 ø è î þ 1

1 4

4

(sin x + cos x )

=

sec4 x 4

(tan x + 1)

=

( 1 + tan 2 x ) sec2 x (tan 4 x + 1)

2

Put tan x = t and sec x dx = dt. \

1ö æ ç1+ 2 ÷ du 1ö æ è ø t I=ò dt = ò 2 , where ç t - ÷ = u dt = ò 2 2 t è ø u + ( 2 ) (1+ t4 ) 1ö æ ç t - ÷ + ( 2 )2 tø è 1ö æ çt - ÷ u 1 1 t è ø +C = tan -1 +C= tan -1 2 2 2 2 (1+ t2 )

=

(t 2 - 1) (tan 2 x - 1) 1 1 tan -1 +C= tan -1 + C. t 2 2 2 tan x

1 ö 1 ö æ æ ç1+ 2 ÷ ç1+ 2 ÷ ø è ø è x x 17. = = × 2 ( x 4 + x 2 + 1) æç x 2 + 1 + 1 ö÷ æ 1ö 2 ( ) x + 3 ç ÷ è x2 ø è xø ( x 2 + 1)

1 ö 1ö æ æ Put ç x - ÷ = t and ç 1 + 2 ÷ dx = dt. xø è è x ø \

1ö æ çx - ÷ 1 1 xø -1 t -1 è I=ò 2 = +C= tan tan +C 3 3 3 3 t + ( 3 )2 dt

=

( x 2 - 1) 1 tan -1 + C. 3 3x

18. On dividing Nr and Dr by cos 4 x , we get 2 sin x I=ò

2 cos 3 x dx = 2 tan x sec x dx ò 4 4 (tan x + 1) (tan x + 1)

779

Senior Secondary School Mathematics for Class 12 Pg-780

780

Senior Secondary School Mathematics for Class 12 t dt , where tan x = t (t 4 + 1) du = 2 = tan -1u + C , where t 2 = u (u + 1) = 2ò

= tan -1 t 2 + C = tan -1 (tan 2 x ) + C . dx dx 1 1 × = ×ò ü 9 òæ1 9 ìï æ 1 ö 2 2ö 2ï ç -x ÷ íç ÷ - x ý è9 ø ïþ ïî è 3 ø 1 ½ + x½ 1 1 1 ½1 + 3 x½ ½ ½ ½+ C. = × log 3 + C = log½ ½ ½ 1 1ö 9 æ 6 ½ ½ ½1 - 3x½ -x ç2 ´ ÷ ½3 ½ 3ø è

19. I =

20. I =

dx dx 1 1 1 1 1 ½2 + x½ ½2 + x½ ½+ C = ½+ C. = × = × log½ log½ 4 ò ( 4 - x 2 ) 4 ( 2 2 - x 2 ) 4 ( 2 ´ 2) 16 ½2 - x½ ½2 - x½

21. Put x 3 = t and 3 x 2 dx = dt , we get I=

½1 + x 3½ dt 1 1 1 1 ½1 + t½ ½ ½ + C = log½ = × log ½+ C. ò 3 3 (1 - t2 ) 3 2 6 ½1 - t½ ½1 - x ½

22. Put x 2 = t and 2x dx = dt. \

I=

½1 + x 2½ dt 1 1 1 1 ½1 + t½ ½ + C = log½ = × log½ ½+ C. ò 2 2 2 (1 - t ) 2 2 4 ½1 - t½ ½1 - x ½

23. Put x 3 = t and 3 x 2 dx = dt. \

I=

½a3 + x 3½ ½a3 + t½ dt 1 1 1 1 = ´ log½ 3 ½+ C = 3 log½ 3 ½+ C . 3 3 ò [( a3 ) 2 - t 2 ] 3 3 a3 6a ½a - x ½ ½a - t½

24. ( 3 - 2 x - x 2 ) = 4 - ( 1 + 2 x + x 2 ) = [2 2 - ( 1 + x ) 2 ]. dx dt \ I=ò 2 =ò , where ( 1 + x ) = t {2 - ( 1 + x ) 2 } (4 - t2 ) =ò =

dt ( 22 - t 2 )

=

1 1 ½2 + t½ ½2 + ( 1 + x )½ ½ + C = log½ ½+ C log½ 2´2 4 ½2 - t½ ½2 - ( 1 + x )½

1 ½3 + x½ ½+ C. log½ 4 ½ 1 - x½

25. On dividing Nr and Dr by cos 2 x , we get I=ò

sec2 x ( 1 - 3 tan 2 x )

dx = ò

dt ( 1 - 3t 2 )

, where tan x = t

½ 1 + t½ 1 dt 1 dt 1 1 ½ 3 ½ = ò = ×ò = × log½ ½+ C 1 1 üï 3 3 æ1 3 ìï æ 1 ö 2 2ö ½ ½ 2 ´ 2 t ç -t ÷ ç ÷ t ý í 3 ½ 3 ½ è3 ø ïþ ïî è 3 ø ½1 + 3 tan x½ 1 ½+ C . = log½ 2 3 ½1 - 3 tan x½

Senior Secondary School Mathematics for Class 12 Pg-781

Integration Using Partial Fractions

781

26. Put cot x = t and - cosec2 x dx = dt. -1 1 ½1 + t½ ½1 + cot x½ ½+ C = ½+ C. log½ log½ t ( 2 ´ 1) 1 2 ½ ½ ½1 - cot x½ ½x - 1½ 2 x - 1½ 1 1 dx 1 ½ 2½ + C = 1 log½ ½ ½+ C. 27. I = ò = × log ½ 1½ 1ö 4 ïì 2 æ 1 ö 2 üï 4 æ 4 2 x + 1½ ½ ½ ½ x + ´ 2 ç ÷ íx - ç ÷ ý ½ 2½ 2ø è è 2 ø ïþ ïî I = -ò

\

dt

=-

(1 - t2 )

28. Put x 2 = t and 2x dx = dt . I=

\

½x 2 - 4½ dt 1 1 1 ½t - 4½ 1 ½= = × log½ log½ 2 ½+ C. ò 2 2 2 (t - 4 ) 2 2 ´ 4 ½t + 4½ 16 ½x + 4½

29. On dividing Nr and Dr by cos 2 x , we get I=ò =ò

sec2 x

dx = ò

(tan 2 x - 4 ) dt {t 2 - 2 2 }

=

dt (t 2 - 4 )

, where tan x = t and sec2 x dx = dt

1 1 ½t - 2½ ½tan x - 2½ ½ + C = log½ ½+ C. log½ ( 2 ´ 2) 4 ½t + 2½ ½tan x + 2½

30. On dividing Nr and Dr by cos 2 x , we get sec2 x

dt dx = ò , where tan x = t ( 4 tan 2 x + 5 ) ( 4t 2 + 5 ) 1 dt 1 dt 1 1 = ò tan -1 = × = 2ü æ ö 4 æ 2 5ö 4òì 4 5 ï 2 æç 5 ö÷ ï ç ÷ çt + ÷ ç 2 ÷ ít + ç 4ø è ÷ ý è ø è 2 ø ïþ ïî 1 æ 2 tan x ö ÷ + C. = tan -1 ç 2 5 5 ø è

I=ò

31. I = ò =ò =ò

sin x ( 3 sin x - 4 sin 3 x ) sec2 x 2

3 sec x - 4 tan 2 x sec2 x ( 3 - tan 2 x )

dx =

1 ( 3 - 4 sin 2 x )

t +C æ 5ö ç ÷ ç 2 ÷ è ø

dx

dx [on dividing Nr and Dr by cos 2 x]

dx = ò

dt ( 3 - t2)

, where tan x = t

½ 3 + t½ ½ 3 + tan x½ 1 1 ½+ C = ½+ C . log½ log½ t 2 3 3 2 3 ½ ½ ½ 3 - tan x½ 32. On dividing Nr and Dr by x 2 , we get 1 ö 1 ö æ æ ç1+ 2 ÷ ç1+ 2 ÷ dt 1 è ø è ø x x I=ò dx = ò dx = 2 , where x - = t 2 1 ö ü ìï æ æ 2 x + t 2 1ö ï çx + 2 ÷ íç x - ÷ + 2ý è x ø xø þï îï è =ò

dt

{( 3 ) 2 - t 2 }

=ò

=

dt 2

t + ( 2)

2

=

( x 2 - 1) t 1 1 +C= + C. tan -1 tan -1 2 2 2 2x

Senior Secondary School Mathematics for Class 12 Pg-782

782

Senior Secondary School Mathematics for Class 12

OBJECTIVE QUESTIONS II Mark (3) against the correct answer in each of the following: 1.

ò

dx 4 - 9x 2

=?

x 1 sin -1 + C 3 3 1 -1 æ 3 x ö (c) sin ç ÷ +C 3 è 2 ø (a)

2.

ò

dx

ò

cos x 4 - sin 2 x (a) sin -1

x 1 (b) sin -1 + C 4 2

5.

6.

ò

ò

ò

2x 1 - 4x

(d) none of these

x 1 (c) sin -1 + C 2 4

(d) none of these

=?

x +C 2

(c) sin -1( 2 sin x) + C 4.

2 æ 2x ö sin -1 ç ÷ + C 3 è 3ø

=?

16 - 4x 2

x 1 (a) sin -1 + C 2 2 3.

(b)

ö æ1 (b) sin -1 ç cos x ÷ + C ø è2 1 ö æ (d) sin -1 ç sin x ÷ + C ø è2

dx = ? sin -1( 2x ) +C log 2

(a) sin -1( 2x ) log 2 + C

(b)

(c) sin -1( 2x ) + C

(d) none of these

dx 2x - x 2

=?

(a) sin -1( x + 1) + C

(b) sin -1( x - 2) + C

(c) sin -1( x - 1) + C

(d) none of these

dx =? x(1 - 2x) (a)

1 sin -1( 2x - 1) + C 2

(b)

1 sin -1( 2x + 1) + C 2

(c)

1 sin -1( 4x + 1) + C 2

(d)

1 sin -1( 4x - 1) + C 2

Senior Secondary School Mathematics for Class 12 Pg-783

Integration Using Partial Fractions

7.

ò

3x2 9 - 16x 6

dx = ?

æx3ö 1 (a) sin -1 ç ÷ + C ç 3 ÷ 4 è ø æx3ö (c) 4 sin -1 ç ÷ + C ç 4 ÷ è ø

8.

9.

10.

11.

12.

ò

ò

ò

ò

ò

dx 2 + 2x - x 2

æ 4x 3 ö 1 ÷ +C (b) sin -1 ç ç 3 ÷ 4 è ø (d) none of these

=?

æ x - 1ö (a) sin -1 ç ÷ +C è 3 ø

æ x - 1ö (b) sin -1 ç ÷ +C è 2 ø

(c) sin -1 3 ( x - 1) + C

(d) none of these

dx 16 - 6x - x 2

=?

æx - 3ö (a) sin -1 ç ÷ +C è 5 ø

æx + 3ö (b) sin -1 ç ÷ +C è 5 ø

1 (c) sin -1( x + 3) + C 5

(d) none of these

dx x - x2

=?

(a) sin -1( x - 1) + C

(b) sin -1( x + 1) + C

(c) sin -1( 2x - 1) + C

(d) none of these

dx 1 + 2x - 3 x 2

=?

(a)

1 æ 3x - 1ö sin -1 ç ÷ +C 3 è 2 ø

(b)

(c)

1 æ 2x - 1 ö sin -1 ç ÷ +C 3 è 3 ø

(d) none of these

dx 2

x - 16

1 æ 2x - 1 ö sin -1 ç ÷ +C 2 è 3 ø

=?

æxö (a) sin -1 ç ÷ + C è 4ø

(b) log x + x 2 - 16 + C

(c) log x - x 2 - 16 + C

(d) none of these

783

Senior Secondary School Mathematics for Class 12 Pg-784

784

13.

14.

15.

Senior Secondary School Mathematics for Class 12

ò

ò

ò

dx

=? 4x 2 - 9 1 (a) log 2x + 4x 2 - 9 + C 2

1 (b) log x + 4x 2 - 9 + C 4

(c) log 2x + 4x 2 - 9 + C

(d) none of these

x2

dx = ? x6 - 1 1 (a) log x 3 + x 6 - 1 + C 2 1 (c) log x 3 - x 6 - 1 + C 3 sin x =? 4 cos2 x - 1

(b)

1 log x 3 + x 6 - 1 + C 3

(d) none of these

1 (a) - log 2 cos x + 4 cos2 x - 1 + C 2 1 (b) - log 2 cos x + 4 cos2 x - 1 + C 3 1 (c) - log cos x + 2 cos2 x - 1 + C 6 (d) none of these 16.

ò

sec2 x tan 2 x - 4

dx = ?

(a) log tan x - tan 2 x - 4 + C 1 log tan x + tan 2 x - 4 + C 2 dx 17. ò =? (1 - e2x )

18.

ò

(b) log tan x + tan 2 x - 4 + C

(c)

(d) none of these

(a) log ex + e2x - 1 + C

(b) log e- x + e-2x - 1 + C

(c) - log e- x + e-2x - 1 + C

(d) none of these

dx 2

x - 3x + 2

=?

3ö ½æ ½ (a) log ½ç x - ÷ + x 2 - 3 x + 2½+ C (b) log x + x 2 - 3 x + 2 + C 2 ø ½è ½ (c) log x - x 2 - 3 x + 2 + C

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-785

Integration Using Partial Fractions

19.

ò

cos x 2

sin x - 2 sin x - 3

785

dx = ?

(a) log sin x + sin 2 x - 2 sin x - 3 + C (b) log (sin x - 1) + sin 2 x - 2 sin x - 3 + C (c) log (sin x - 1) - sin 2 x - 2 sin x - 3 + C

20.

21.

22.

23.

24.

ò

ò

ò

ò

ò

(d) none of these dx =? 2 - 4x + x 2 (a) log ( x - 2) + x 2 - 4x + 2 + C

(b) log x + x 2 - 4x + 2 + C

(c) log x - x 2 - 4x + 2 + C

(d) none of these

dx 2

x + 6x + 5

=?

(a) log x + x 2 + 6x + 5 + C

(b) log x - x 2 + 6x + 5 + C

(c) log ( x + 3) + x 2 + 6x + 5 + C

(d) none of these

dx ( x - 3) 2 - 1

=?

(a) log ( x - 3) + x 2 - 6x + 8 + C

(b) log x + x 2 - 6x + 8 + C

(c) log ( x - 3) - x 2 - 6x + 8 + C

(d) none of these

dx 2

x - 6x + 10

=?

(a) log x + x 2 - 6x + 10 + C

(b) log ( x - 3) + x 2 - 6x + 10 + C

(c) log x - x 2 - 6x + 10 + C

(d) none of these

x 2 dx x 6 + a6

dx = ?

1 log |x 6 + a 6| + C 3 1 (c) log x 3 + x 6 + a 6 + C 3

(a)

(b)

æx3ö 1 tan -1 ç 3 ÷ + C ça ÷ 3 è ø

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-786

786

25.

26.

Senior Secondary School Mathematics for Class 12

ò

ò

sec2 x 16 + tan 2 x

dx = ?

(a) log tan x + tan 2 x + 16 + C

(b) log x + tan 2 x + 16 + C

(c) log tan x - tan 2 x + 16 + C

(d) none of these

dx 2

3 x + 6x + 12

=?

(a) log ( x + 1) + x 2 + 2x + 4 + C

(b)

1 log ( x + 1) + x 2 + 2x + 4 + C 3

1 log ( x + 1) + x 2 + 2x + 4 + C (d) none of these 3 dx =? 2 2x + 4x + 6 1 (a) log ( x + 1) + x 2 + 2x + 3 + C 2 1 (b) log ( x + 1) + x 2 + 2x + 3 + C 2 1 (c) log x + x 2 + 2x + 3 + C 2 (d) none of these

(c) 27.

28.

ò

ò

x2

dx = ? x 6 + 2x 3 + 3 1 (a) log ( x 3 + 1) + x 6 + 2x 3 + 3 + C 3 (b) log x 3 + x 6 + 2x 3 + 3 + C 1 log ( x 3 + 1) - x 6 + 2x 3 + 3 + C 3 (d) none of these (c)

29.

ò

4 - x 2 dx = ? x x 4 - x 2 + 2 sin -1 + C 2 2 1 2 -1 x (c) x 4 - x - 2 sin +C 2 2 (a)

30.

ò

(b) x 4 - x 2 + sin -1

x +C 2

(d) none of these

1 - 9x 2 dx = ? x 1 1 - 9x 2 + sin -1 3 x + C 18 2 x 1 (c) 1 - 9x 2 + sin -1 3 x + C 2 6 (a)

(b)

3x 1 1 - 9x 2 + sin -1 3 x + C 6 2

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-787

Integration Using Partial Fractions

31.

ò

9 - 4x 2 dx = ? x 9 2x +C 9 - 4x 2 + sin -1 2 4 3 x 9 2x (c) +C 9 - 4x 2 - sin -1 2 4 3 (a)

32.

787

ò cos x

(b) x 9 - 4x 2 +

9 2x sin -1 +C 2 3

(d) none of these

9 - sin 2 x dx = ?

1 9 æ sin x ö (a) sin x 9 - sin 2 x + sin -1 ç ÷ +C 2 2 è 3 ø (b)

sin x 9 æ sin x ö 9 - sin 2 x + sin -1 ç ÷ +C 2 2 è 3 ø

9 1 æ sin x ö (c) cos x 9 - sin 2 x + sin -1 ç ÷ +C 2 2 è 3 ø (d) none of these 33.

ò

x 2 - 16 dx = ? (a) x x 2 - 16 - 4 log x + x 2 - 16 + C x x 2 - 16 - 8 log x + x 2 - 16 + C 2 x (c) x 2 - 16 + 8 log x + x 2 - 16 + C 2 (d) none of these (b)

34.

ò

x 2 - 4x + 2 dx = ? 1 (a) ( x - 2) x 2 - 4x + 2 + log ( x - 2) + x 2 - 4x + 2 + C 2 1 (b) ( x - 2) x 2 - 4x + 2 + log ( x - 2) + x 2 - 4x + 2 + C 2 1 2 (c) ( x - 2) x - 4x + 2 - log ( x - 2) + x 2 - 4x + 2 + C 2 (d) none of these

35.

ò

9x 2 + 16 dx = ? x 9x 2 + 16 + 2 x (b) 9x 2 + 16 2 (a)

8 log 3 x + 9x 2 + 16 + C 3 8 log 3 x + 9x 2 + 16 + C 3

(c) x 9x 2 + 16 + 24 log 3 x + 9x 2 + 16 + C (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-788

788

36.

Senior Secondary School Mathematics for Class 12

òe

x

e2x + 4 dx = ?

1 (a) ex e2x + 4 - 2 log ex + e2x + 4 + C 2 1 (b) ex e2x + 4 + 2 log ex + e2x + 4 + C 2 1 (c) ex e2x + 4 + log ex + e2x + 4 + C 2 (d) none of these

37.

16 + (log x) 2 dx = ? x 1 (a) log x × 16 + (log x) 2 + 8 log log x + 16 + (log x) 2 + C 2 1 (b) log x × 16 + (log x) 2 + 4 log log x + 16 + (log x) 2 + C 2

ò

(c) log x × 16 + (log x) 2 + 16 log log x + 16 + (log x) 2 + C (d) none of these ANSWERS (OBJECTIVE QUESTIONS II)

1. (c) 2. (a) 3. (d) 4. (b) 5. (c) 6. (d) 7. (b) 8 (a) 9. (b) 10. (c) 11. (a) 12. (b) 13. (a) 14. (b) 15. (a) 16. (b) 17. (c) 18. (a) 19. (b) 20. (a) 21. (c) 22. (a) 23. (b) 24. (c) 25. (a) 26. (c) 27. (b) 28. (a) 29. (a) 30. (c) 31. (a) 32. (a) 33. (b) 34. (c) 35. (a) 36. (b) 37. (a) HINTS TO THE GIVEN OBJECTIVE QUESTIONS II 1. I = ò

dx 4 - 9x 2

=

1 × 3 ò

dx dx 1 = ò 2 3 4 2 æ2ö -x - x2 ç ÷ 9 è 3ø

æ x ö 1 sin -1 çç 2 ÷÷ + C . 3 è 3ø dx dx dx x 1 1 1 2. I = ò = ò = ò = sin -1 + C . 2 2 2 2 2 2 2 2 16 - 4 x 4-x 2 -x =

3. Putting sin x = t and cos x dx = dt , we get dt dt t æ1 ö I=ò =ò = sin -1 + C = sin -1 ç sin x ÷ + C . 2 è2 ø 4 - t2 22 - t 2 4. Put 2 x = t and 2 x log 2 dx = dt . dt 1 1 1 \ I= = sin -1t + C = sin -1 ( 2 x ) + C . log 2 ò 1 - t 2 log 2 (log 2 )

Senior Secondary School Mathematics for Class 12 Pg-789

Integration Using Partial Fractions 5.

789

2 x - x 2 = 1 - ( 1 - 2 x + x 2 ) = 1 - ( x - 1) 2 dx dt =ò , where ( x - 1) = t I=ò \ 1 - ( x - 1) 2 1- t2 = sin -1t + C = sin -1 ( x - 1) + C .

2 éæ 1 ö2 æ é 1 æ 1ö ù x 1 öù æx ö 6. x( 1 - 2 x ) = ( x - 2 x 2 ) = 2 ç - x 2 ÷ = 2 ê - ç x 2 - + ÷ú = 2 êç ÷ - ç x - ÷ ú 4 ø úû 2 16 ø û è è2 ø ë 16 è êë è 4 ø 1ö dx dx 1 1 æ \ I= = ×ò , where ç x - ÷ = t ò 2 2 2 4 2 2 è ø 1ö æ 1ö æ æ 1ö ç ÷ - çx - ÷ ç ÷ - t2 4 4 4 è ø è ø è ø ì æ 1 1 1 1 öü -1 t -1 = sin +C= sin 4 t + C = sin -1 í 4 ç x - ÷ ý + C ( 1 4) 4 2 2 2 è øþ î 1 = sin -1 ( 4 x - 1) + C . 2

7. Put x 3 = t and 3 x 2 dx = dt . dt 1 \ I=ò = ò 2 4 9 - 16 t

dt dt 1 = ×ò 2 4 9 2 æ 3ö -t ç ÷ - t2 16 è4ø 3 ö æ t x 1 1 4 ÷ + C. = sin -1 3 + C = sin -1 çç ÷ ( 4) 4 4 è 3 ø

8. ( 2 + 2 x - x 2 ) = 3 - ( 1 + x 2 - 2 x ) = ( 3 ) 2 - ( x - 1) 2 \

I=ò

dx ( 3 ) 2 - ( x - 1) 2

= sin -1

t 3

dt

=ò

+ C = sin -1

( 3 )2 - t 2

( x - 1) 3

, where ( x - 1) = t and dx = dt

+ C.

9. ( 16 - 6 x - x 2 ) = [- ( x 2 + 6 x + 9 ) + 25] = {25 - ( x + 3 ) 2 } \

I=ò

10. ( x - x 2 ) = \

dx 5 2 - ( x + 3)2

=ò

dt 52 - t 2

= sin -1

2

1 æ 2 1ö æ 1ö 1ö æ - çx - x + ÷ = ç ÷ - çx - ÷ 4 è 4ø è2ø 2ø è

I=ò

dx 2

1ö æ 1ö æ ç ÷ - çx - ÷ 2ø è2ø è

2

=ò

2

dx 2

t æx+ 3ö ÷ + C. + C = sin -1 ç 5 è 5 ø

æ 1ö ç ÷ - t2 è2ø

= sin -1

t +C ( 1 2)

1ö æ = sin -1 2t + C = sin -1 2 ç x - ÷ + C = sin -1 ( 2 x - 1) + C . 2ø è 2 æ 11. ( 1 + 2 x - 3 x 2 ) = - 3 ç x 2 - x 3 è ìï æ 2 ö 2 æ = 3 íç ÷ - ç x è îï è 3 ø

ì4 æ 1ö 2 1 öü ÷ = 3 í - ç x2 - x + ÷ý 3ø 3 9 øþ î9 è 2 1 ö üï ÷ ý× 3 ø þï

Senior Secondary School Mathematics for Class 12 Pg-790

790

Senior Secondary School Mathematics for Class 12 æ t ö dx dt 1 1 1 ×ò = ×ò sin -1 çç 2 ÷÷ + C = 2 2 2 3 3 3 è 3ø 1ö æ2ö æ æ2ö ç ÷ - çx - ÷ ç ÷ - t2 3ø è 3ø è è 3ø ( 3 x - 1) 1 1 æ 3t ö = sin -1 ç ÷ + C = sin -1 + C. 2 3 3 è 2 ø

\

I=

12. I = ò 13. I =

dx

dx

1 2ò

x 2 - 16 + C .

= log x +

x2 - 42

æ 3ö x2 - ç ÷ è2ø

2

=

½ 1 log½x + 2 ½

9½ 1 x 2 - ½ + C = log 2 x + 4½ 2

4x 2 - 9 + C .

14. Put x 3 = t and 3 x 2 dx = dt. Then, dt 1 1 1 = log t + t 2 - 1 + C = log x 3 + I= ò 2 3 3 3 t -1

x6 - 1 + C.

15. Put cos x = t and sin x dx = - dt. Then, ½ dt dt 1 1 1½ I = -ò =- ò = - log½t + t 2 - ½ + C 2 2 4½ 1 2 ½ 4t 2 - 1 t 4 1 1 = - log 2 t + 4t 2 - 1 + C = - log 2 cos x + 4 cos 2 x - 1 + C . 2 2 16. Put tan x = t and sec2 x dx = dt. Then, dt = log t + t 2 - 4 + C = log tan x + tan 2 x - 4 + C . I=ò t 2 - 22 17. I = ò

e - xdx e

-2 x

2x

(1 - e )

=ò

e -x e

-2 x

-1

dx = - ò

dt 2

(t - 1)

, where e - x = t

= - log t + t 2 - 1 + C = - log e - x + 18. I = ò

dx 9ö 1 æ 2 ç x - 3x + ÷ 4ø 4 è

=ò

e -2 x - 1 + C .

dx 2

3ö æ æ 1ö çx - ÷ - ç ÷ 2ø è è2ø

2

½ 1½ 3ö æ = log½t + t 2 - ½ + C , where ç x - ÷ = t 4 2 è ø ½ ½ æ 1ö t2 - ç ÷ è2ø 3ö æ ½ ½ = log½ ç x - ÷ + x 2 - 3 x + 2 ½ + C . 2ø ½è ½ =ò

dt

2

19. Putting sin x = t and cos x dx = dt , we get dt dt dt =ò =ò I=ò 2 2 t - 2t - 3 (t - 2t + 1) - 4 (t - 1) 2 - 2 2 = log (t - 1) + (t - 1) 2 - 2 2 + C = log (t - 1) + t 2 - 2t - 3 + C = log (sin x - 1) +

sin 2 x - 2 sin x - 3 + C .

Senior Secondary School Mathematics for Class 12 Pg-791

Integration Using Partial Fractions 20. I = ò

dx ( x - 4x + 4) - 2

= log ( x - 2 ) + 21. I = ò

23. I = ò

( x - 2)2 - ( 2 )2

x 2 - 4x + 2 + C .

dx

dx

=ò

( x 2 + 6x + 9) - 4

22. I = log ( x - 3 ) +

dx

=ò

2

791

( x + 3)2 - 2 2

x 2 + 6x + 5 + C .

= log ( x + 3 ) +

x 2 - 6x + 8 + C .

dx

= log ( x - 3 ) +

( x - 3 ) 2 + 12

x 2 - 6 x + 10 + C .

24. Put x 3 = t and 3 x 2 dx = dt. Then, dt 1 1 1 = log t + t 2 + a6 + C = log x 3 + I= ò 3 3 3 t 2 + ( a3 ) 2 25. Put tan x = t and sec2 x dx = dt. Then, dt = log t + t 2 + 16 = log tan x + I=ò 2 t + 42 dx

26. I = ò

=

2

1 3

dx

ò

2

=

1 3

ò

=

1 2

ò

x + 2x + 4 3 ( x + 2x + 4) 1 dt 1 log t + t 2 + 3 + C = = ò 2 3 3 t + ( 3 )2 =

27. I = ò =

1 × log ( x + 1) + 3 dx 2( x 2 + 2 x + 3 ) 1 log ( x + 1) + 2

x 6 + a6 + C .

tan 2 x + 16 + C . dx

( x + 1) 2 + ( 3 ) 2

x 2 + 2x + 4 + C . =

1 2

ò

dx x 2 + 2x + 3

dx ( x + 1) 2 + ( 2 ) 2

x 2 + 2x + 3 + C .

28. Put x 3 = t and 3 x 2 dx = dt. Then, dt dt 1 1 = 3 ò t 2 + 2t + 3 3 ò (t + 1) 2 + ( 2 ) 2 1 = log (t + 1) + t 2 + 2t + 3 + C 3 1 = log ( x 3 + 1) + x 6 + 2 x 3 + 3 + C . 3

I=

x 2 a2 x a - x2 + sin -1 + C , a 2 2 x x 4 - x 2 dx = ò 2 2 - x 2 dx = 4 - x 2 + 2 sin -1 + C . 2 2

29. Using ò a2 - x 2 dx =

ò 30.

ò

1 - 9 x 2 dx = 3 ò =

éx 1 1 1 x ù - x 2 dx = 3 ê - x2 + sin -1 1 ú + C 9 18 ( 3) û ë2 9

3x 1 1 x 1 - x 2 = sin -1 3 x + C = 1 - 9 x 2 + sin -1 3 x + C . 2 9 6 2 6

Senior Secondary School Mathematics for Class 12 Pg-792

792

31.

Senior Secondary School Mathematics for Class 12

ò

9 - 4 x 2 dx = 2 ò

éx 9 9 9 x ù - x 2 dx = 2 ê - x 2 + sin -1 3 ú + C 4 2 4 8 ( ë 2) û

x 9 2x + C. 9 - 4 x 2 + sin -1 2 4 3 32. Put sin x = t and cos x dx = dt. Then, t 9 t I = ò 9 - t 2 dt = 9 - t 2 + sin -1 + C 2 2 3 sin x sin x 9 æ ö ÷ + C. 9 - sin 2 x + sin -1 ç = 2 2 è 3 ø =

33.

ò

x 2 - 16 dx = ò x 2 - 4 2 dx = =

x 2 16 x - 42 log x + 2 2

x 2 x - 16 - 8 × log x + 2

x 2 - 16 + C

x 2 - 16 + C .

34. I = ò ( x - 2 ) 2 - ( 2 ) 2 dx 1 2 ( x - 2 ) x 2 - 4 x + 2 - log ( x - 2 ) + x 2 - 4 x + 2 + C 2 2 1 2 = ( x - 2 ) x - 4 x + 2 - log ( x - 2 ) + x 2 - 4 x + 2 + C . 2 =

35.

ò \

x 2 + a2 dx =

x 2 a2 x + a2 + log x + 2 2

I = 3ò x 2 + ìï x = 3í îï 2

x 2 + a2 + C 2

16 æ4ö dx = 3 ò x 2 + ç ÷ dx 9 è 3ø

½ 16 8 16 ½üï ½ý + C + log½x + x 2 + 9 9 9 ½þï ½ x 8 = 9 x 2 + 16 + log 3 x + 9 x 2 + 16 + C . 2 3 x2 +

36. Put e x = t and e xdx = dt. Then, I = ò t 2 + 4 dt = ò t 2 + 2 2 dt =

t 2 4 t + 4 + log t + t 2 + 4 + C 2 2

1 x 2x e e + 4 + 2 log e x + e 2 x + 4 + C . 2 1 37. Put log x = t and dx = dt. Then, x t 2 16 I = ò t 2 + ( 4)2 = t + 16 + log t + t 2 + 16 + C 2 2 1 = log x (log x ) 2 + 16 + 8 log log x + (log x ) 2 + 16 + C . 2 =

Senior Secondary School Mathematics for Class 12 Pg-793

16. DEFINITE INTEGRALS Fundamental Theorem of Integral Calculus Let f ( x) be a continuous function defined on an interval [a , b] and let the antiderivative of f ( x) be F( x). Then, the definite integral of f ( x) over [a , b], denoted by b

ò f ( x) dx , is given by

b

a

b

ò f ( x) dx = [F( x)]a = F( b) - F( a). a

NOTE

Here, a and b are respectively known as the lower limit and the upper limit of the integral.

The value of a definite integral is unique, for, if ò f ( x) dx = F( x) + C then b

b

ò f ( x) dx = [F( x) + C]a = {F( b) + C} - {F( a) + C} = F( b) - F( a). a

SOLVED EXAMPLES [Integrals based on formulae and integration by parts] EXAMPLE 1

Evaluate: 4

(i)

1

(iv)

ò

0 p/ 2

(vii)

9

dx òx 2

(ii)

2

ò

(iii)

x dx

4 2

dx 5x + 3

ò cos x dx 2

(v)

ò 1

ò

6x + 4 dx

0 p

dx

(vi)

x( x 2 - 1)

ò sin 5 x dx 0

p/ 4

[CBSE 2002]

(viii)

0

ò tan

2

x dx

0

p/ 4

(ix)

ò sin 2x sin 3 x dx

[CBSE 2006]

0

4

SOLUTION

dx 4 = [log x]2 = (log 4 - log 2) = ( 2 log 2 - log 2) = log 2. x 2

(i)

ò

(ii)

ò

9 4

9

38 2 é2 ù x dx = ê x 3/ 2 ú = × [( 9) 3/ 2 - ( 4) 3/ 2] = × 3 ë3 û4 3 793

Senior Secondary School Mathematics for Class 12 Pg-794

794

Senior Secondary School Mathematics for Class 12 2

é 2 ( 6x + 4) 3 / 2 ù 1 56 3/ 2 6x + 4 dx = ê × - ( 4) 3/ 2] = × ú = × [(16) 3 6 9 9 ë û0

2

(iii)

ò

0

1

1 é (5 x + 3)1/ 2 ù 1 dx = ò (5 x + 3) -1/ 2 dx = ê 2 × ú 5 5x + 3 ë û0 0

1

(iv)

ò

0

2 = ( 8 - 3 ). 5 2

(v)

ò 1

æp ö p 2 = [sec-1 x]1 = [sec-1( 2) - sec-1(1)] = ç - 0÷ = × è4 ø 4 x( x - 1) dx 2

p

(vi)

0 p/ 2

(vii)

2 ò cos x dx =

0 p/4

(viii)

ò tan

2

x dx =

0 p/ 4

(ix)

p

2 1 é - cos 5 x ù úû = - 5 [cos 5 p - cos 0] = 5 × 5 0

ò sin 5 x dx = êë

ò

1 2

p/ 2

1é

ò (1 + cos 2x) dx = 2 êë x +

0 p/4

p/4

ò (sec x - 1) dx = [tan x - x]0 2

0

sin 2x sin 3 x dx =

0

1 2

Evaluate:

(i)

ò

ò

ò

=

1 ( cos x - cos 5 x) dx 2 ò0

=

1é sin 5 x ù sin x 2 êë 5 úû 0

=

p sin (5 p/4) ö ù 3 2 1 éæ × ÷ú = ç sin ê 2 ëè 4 5 10 øû

p/4

p/ 2

1 + cos 2x dx

ò

cos2 x + sin 2 x + 2 sin x cos x dx

0 p/4

p/4

ò (cos x + sin x) dx = [sin x - cos x]0

0 p/ 2

p/ 2 0

ò

p/4

1 + sin 2x dx = =

ò

(ii)

0

0

(ii)

pö æ = ç1 - ÷ × 4ø è

( 2 sin 2x sin 3 x) dx

1 + sin 2x dx

p/4

(i)

p × 4

0 p/4

0

SOLUTION

=

p/ 4

p/4

EXAMPLE 2

p /2

sin 2x ù 2 úû 0

1 + cos 2x dx =

ò

0

= 2

2 cos2 x dx p/ 2

ò cos x dx = 0

p/ 2

2[sin x]0

= 2.

= 1.

Senior Secondary School Mathematics for Class 12 Pg-795

Definite Integrals

795

p/ 2

EXAMPLE 3

Evaluate:

(i)

p/ 2

3 ò cos x dx

ò sin

(ii)

0

SOLUTION

(i)

p/ 2

p/ 2

0

0

3 ò cos x dx =

ò

4

x dx

0

æ 3 cos x + cos 3 x ö ç ÷ dx 4 è ø [Q cos 3 x = 4 cos 3 x - 3 cos x ]

= =

3 × 4

p/ 2

1

p/ 2

ò cos x dx + 4 × ò cos 3 x dx 0

0

p/ 2

3 1 é sin 3 x ù p/ 2 × [sin x]0 + × ê 4 4 ë 3 úû 0

2 æ3 1ö 8 =ç - ÷= = × 4 12 12 3 è ø p/ 2

(ii)

4 ò sin x dx = 0

1 4

p/ 2

ò ( 2 sin

1 = × 4 = = = = =

1 × 4 1 × 4 1 × 4

2

x) 2 dx

0 p/ 2

ò (1 - cos 2x)

2

dx

0

p/ 2

ò (1 - 2 cos 2x + cos 2x) dx 2

0

p/ 2

ò

0

p/ 2

3 × 8

ò

0

(1 + cos 4x) ù é êë1 - 2 cos 2x + úû dx 2 1 æ3 ö ç - 2 cos 2x + cos 4x ÷ dx 2 è2 ø

p/ 2

ò

p/ 2

dx -

0

1 2

ò

p/ 2

cos 2x dx +

0 p/ 2

3 p/ 2 1 é sin 2x ù × [x]0 - × ê 8 2 ë 2 úû 0

1 8

+

ò cos 4x dx 0

p/2

1 é sin 4x ù × 8 êë 4 úû 0

æ 3p ö 3p =ç - 0 + 0÷ = × è 16 ø 16 4

EXAMPLE 4

Evaluate:

(i)

ò

0

4

SOLUTION

(i)

ò

0

(ii)

2

x + 2x + 3

dx 2

1

dx

x + 2x + 3

4

=ò

0

0

dx ( x + 1) 2 + ( 2) 2

dx

ò (1 + x + x 2)

Senior Secondary School Mathematics for Class 12 Pg-796

796

Senior Secondary School Mathematics for Class 12 4

= é log ( x + 1) + x 2 + 2x + 3 ù êë úû 0 = {log |5 + 27| - log |1 + 3|}. 1

(ii)

1

dx

1

dx

ò (1 + x + x 2) = ò é æ

dx 2 2 é æ 3ö ù 0 æ 1 ê ç x + ö÷ + ç ÷ ú ç 2 ÷ ú 2ø êè ø û è ë

=ò

1ö 3 ù 0 ç x2 + x + ÷ + ê 4 ø 4 úû ëè

0

1

1ö ù é æ çx + ÷ ú ê 2 2ø =ê tan -1 è ú = / 3 3 2) ú ( ê úû 0 êë 2 é æ = tan -1( 3 ) - tan -1 ç 3 êë è 2 é p pù p = × = × 3 êë 3 6 úû 3 3 EXAMPLE 5

Evaluate: a dx (i) ò 2 0 ax - x a

SOLUTION

(i)

2

(ii)

ax - x

0

1 öù ÷ 3 ø úû

2 - x 2 dx

0

a

dx

ò

ò

1

2 é æ 2x + 1 ö ù tan -1 ç ÷ú 3 êë è 3 øû0

2

=ò

0

dx æ a2 ö a2 - ç x 2 - ax + ÷ + ç 4 ÷ø 4 è

a

=ò

0

dx 2

aö æ aö æ ç ÷ - çx - ÷ 2ø è 2ø è

2

a

aö ù é æ a çx - ÷ ú ê é æ 2x - a ö ù 2ø = ê sin -1 è ú = ê sin -1 ç ÷ú æ aö ú è a øû0 ë ê ç ÷ ú êë 2 è ø û0 = [sin -1(1) - sin -1( -1)] pö æ = 2 sin -1(1) = ç 2 ´ ÷ = p. 2ø è 2

(ii)

ò

0

2 - x 2 dx =

2

ò

( 2) 2 - x 2 dx

0

éx ( 2) 2 x ù 2 - x2 + =ê × sin -1 ú 2 2 2û ë 0 p -1 -1 = [0 + sin (1)] - [0 + sin 0] = × 2

2

Senior Secondary School Mathematics for Class 12 Pg-797

Definite Integrals EXAMPLE 6

Evaluate: p/ 2

(i)

p

ò x cos x dx

(ii)

0 2

(iii)

ò 1

SOLUTION

797

log x x2

ò cos 2x log sin x dx

0 p/6

(iv)

dx

ò(2 + 3x

2

) cos 3 x dx

0

(i) Integrating by parts, we get p/ 2

p/ 2

ò x cos x dx = [x sin x]0

p/ 2

ò 1 × sin x dx

-

0

0

p p/ 2 æp ö = + [cos x]0 = ç - 1÷ × 2 è2 ø (ii) Integrating by parts, taking log (sin x) as the first function, we get p

ò cos 2x log sin x dx 0

p

p

sin 2x ù sin 2x ö é æ = ê(log sin x) × - ò ç cot x × ÷ dx ú 2 2 ø ë û0 0 è p

p

cos x 2 sin x cos x × dx = - ò cos2 x dx 2 sin x 0 0

= 0-ò =-

p

p

1 1 2 cos2 x dx = - ò (1 + cos 2x) dx 20 2 ò0 p

1 é sin 2x ù p = - × êx + =- × 2 ë 2 úû 0 2 (iii) Integrating by parts, taking (log x) as the first function, we get 2

ò 1

log x x2

2

dx = ò (log x) × x -2 dx 1

2

2

1 æ 1ö é æ 1 öù = ê(log x) ç - ÷ ú - ò × ç - ÷ dx x è ø û1 1 x è x ø ë 2

dx é log 2 log 1 ù = ê+ + 2 1 úû 1ò x 2 ë 2

=

- log 2 é 1 ù - log 2 ì 1 ü æ 1 - log 2 ö -ê ú = - í - 1ý = ç ÷× 2 2 2 ø î2 þ è ë x û1

Senior Secondary School Mathematics for Class 12 Pg-798

798

Senior Secondary School Mathematics for Class 12 p/6

ò(2 + 3x

(iv)

0

2

) cos 3 x dx

p/6

p/6

= 2 ò cos 3 x dx + 3 ò x 2 cos 3 x dx 0

0

p/6

é sin 3 x ù = 2ê ë 3 úû 0

ìïé æ sin 3 x ö ù p/ 6 p/6 æ sin 3 x ö üï + 3 íê x 2 ç - ò 2x ç ÷ú ÷ dx ý è 3 ø ïþ ïîë è 3 ø û 0 0 [integrating by parts]

p/6

=

=

2 p2 + - 2 ò x sin 3 x dx 3 36 0 p/6 p/6 2 p2 ïìé æ - cos 3 x ö ù æ - cos 3 x ö ïü - ò1× ç + - 2 íê x ç ÷ú ÷ dx ý 3 36 3 3 øû0 ø ïþ ïîë è 0 è [integrating by parts] p/6

=

2 p2 2 2 é sin 3 x ù p/6 + + [x cos 3 x]0 - × ê 3 36 3 3 ë 3 úû 0

=

2 p 2 2 æç p 2 + - = + 3 36 9 çè 36 2

EXAMPLE 7

4 ö÷ 1 2 ( p + 16). = 9 ÷ø 36

dx

ò x(1 + x 2) ×

Evaluate

1

SOLUTION

Let

1 x(1 + x 2)

Then,

=

A Bx + C + × x (1 + x 2)

1 º A(1 + x 2) + ( Bx + C) x. Putting x = 0, we get A = 1.

Comparing the coefficients of x 2 , we get

A + B = 0 or B = - 1.

Comparing coefficients of x, we get C = 0. 1

\

x(1 + x 2) 2

So,

dx

ò x(1 + x 2)

1

x ù é1 =ê ë x 1 + x 2 úû 2

2

dx 1 2x - ò dx x 2 + 1 x2 1 1

=ò

2 1 2 = [log x]1 - [log (1 + x 2)]1 2 1 é3 ù = ê (log 2) - (log 5) ú × 2 ë2 û

Senior Secondary School Mathematics for Class 12 Pg-799

Definite Integrals

799

4

( x 2 + x) dx. 2x + 1 2

ò

EXAMPLE 8

Evaluate

SOLUTION

Integrating by parts, taking ( x 2 + x) as the first function and 1 as the second function, we get 2x + 1 4

4

( x 2 + x) 4 2 ò 2x + 1 dx = [( x + x) × 2x + 1]2 - ò ( 2x + 1) × 2x + 1 dx 2 2 4

= ( 60 - 6 5 ) - ò ( 2x + 1) 3/ 2 dx 2

1 4 = ( 60 - 6 5 ) - × [( 2x + 1)5 / 2]2 5 æ 243 ö = ( 60 - 6 5 ) - ç -5 5 ÷ è 5 ø 57 - 5 5 æ 57 ö = ç - 5÷ = × 5 5 è ø 1/ 2

EXAMPLE 9

Evaluate:

(i)

0

1/ 2

SOLUTION

(i)

ò

0

dx = 1-x

1/ 2

ò (1 - x)

(ii)

æ1- x ö

ò çè 1 + x ÷ø dx 0

-1/ 2

0

1

(ii)

1

dx 1-x

ò

1/ 2

é2 1-x ù dx = ê ú ë -1 û 0

= ( 2 - 2).

1

2 ö æ1- x ö æ ò çè 1 + x ÷ø dx = ò çè -1 + x + 1 ÷ø dx [on dividing ( -x + 1) by ( x + 1)] 0 0 1

= [- x + 2 log |x + 1|]0 = [( 2 log 2) - 1].

EXERCISE 16A Very-Short-Answer Questions Evaluate: 4

3

1.

4 ò x dx

2.

2

3.

-5 ò x dx

16

4.

1

dx x -4

ò

x dx

òx

3/4

0

-1

5.

ò 1

1

4

6.

ò 1

dx x

dx

Senior Secondary School Mathematics for Class 12 Pg-800

800

Senior Secondary School Mathematics for Class 12 1

7.

8.

ò 3 dx

10.

0 4

9.

2 ¥

11.

8

dx x

ò3

1 1

dx

ò (1 + x 2)

12.

ò sec x dx

14.

2

ò cot

2

x dx

16.

x dx

18.

ò tan x dx

20.

ò cos ò

p/4

x dx

22.

ò

(1 - 3 cos x) sin 2 x

dx

24.

dx (1 + cos 2x)

28.

ò sin 2x sin 3 x dx

30.

ò

0

ò sin 2x cos 3 x dx

32.

0 p/ 2

ò

0 2

35.

ò

x dx

1 + cos 2x dx

dx + ( 1 sin x) -p/4

ò

dx cos 1 2x p/4

ò

ò cos x cos 2x dx ò

dx

1 + sin x dx

0

2

1 + cos x dx

ò ( x + 1)( x + 2) 1

3

0 p/ 2

p

33.

ò sin

p/6

0

31.

ò cosec x dx

p/ 2

p/4

29.

2

p/4

26.

0

27.

ò cos x dx

0

1 - sin 2x dx [CBSE 2004]

p/ 4

x dx

0 p/4

p/ 4

25.

2

p/ 2 3

0

p/ 2

ò tan

p/6

p/ 3

23.

2

0 p/4

0

21.

ò cosec x dx

p/4 2

0 p/ 3

19.

2

0

p/ 2

ò sin

0 1-x p/4

p/4

p/4

17.

dx

ò

-p/4

0

p/ 2

15.

dx

ò (1 + x 2) 0 1

0 p/6

13.

dx

ò x 2/3

34.

0 2

36.

( x 4 + 1)

ò ( x 2 + 1) dx ( x + 3)

ò x( x + 2) dx 1

[CBSE 2008]

Senior Secondary School Mathematics for Class 12 Pg-801

Definite Integrals 4

37.

4

dx

ò ( x 2 - 4)

38.

3

2

39.

2

x + 2x + 3

1

40.

2

x + 4x + 3

1

dx

ò

0

dx

ò

801

dx

ò (1 + x + 2x 2) 0

Short-Answer Questions Evaluate: p/4

p/ 2

ò ( a cos x + b sin 2

41.

2

x) dx

42.

p/ 2

a

ò cos x dx 4

44.

1/ 2

3

47.

46.

x - x2

1/4

ò

x(1 - x) dx

0 2

dx

ò x 2( x + 1)

48.

1

p /2

x ò x e dx

òx

50.

0

òx

sin x dx

òx

52.

sin 3 x dx

54.

0

e

56.

1

1

0î

ü þ

x ex

ò (1 + x) 2 dx 0

2

61.

5x2

ò ( x 2 + 4x + 3) dx

1

2

cos2 x dx

log x

ò (1 + x) 2 dx

1

ò í(log x) - (log x) 2 ý dx

1

59.

ì

[CBSE 2006C]

0

1

57.

òx

3

ò log x dx 2

cos 2x dx

p/ 2 3

2

55.

2

0

p/ 2

òx

cos x dx

p/ 2 2

0

53.

2

0

p/4

51.

dx

ò x (1 + 2x) 2 1

1

49.

dx

ò ( ax + a 2 - x 2) 1

dx

ò

dx

0

0

45.

2

p/ 3

0

43.

ò ( tan x + cot x)

e

58.

òe

x æ1 +

1

ç è

x log x ö ÷ dx x ø

p/ 2

60.

ò 2 tan 0

3

x dx

[CBSE 2004]

Senior Secondary School Mathematics for Class 12 Pg-802

802

Senior Secondary School Mathematics for Class 12 ANSWERS (EXERCISE 16A)

1.

242 5

2.

5. - log 4 9. 6 1 3 p 17. 4

13.

14 3

6. 2 10.

p 4

14. –2 æp 18. ç + è8

1ö ÷ 4ø

24. 1 1 28. 2 32. 2

33. 2

3)

35. ( 2 log 3 - 3 log 2) 37.

4.

8. 3 p 2 pö æ 16. ç1 - ÷ 4ø è

12.

21.

2 3 26. 2 5 30. 12 ö æ2 34. ç + 2 tan -1 2÷ 3 ø è

22.

1 (log 2 + log 3) 2 æ5 + 3 3 ö ÷ 38. log çç ÷ è 1+ 3 ø 2 ì -1 5 1 ü 40. - tan -1 ítan ý 7î 7 7þ

36.

1 (log 5 - log 3) 4

39. log ( 4 + 15 ) - log ( 3 + 8) 41.

p ( a + b) 4

42.

-2 3

43.

45.

p 6

46.

p 8

47. log 2 - log 3 +

48. log 6 - log 5 -

512 7

19. log 2 3 3 8 25. ( 2 - 1) 3 29. 5 2

20. log ( 2 - 1) + log ( 2 + 23. ( 4 - 3 2) 1 27. 2 -4 31. 5

15 64 3 7. 2 p 11. 2 pö æ 15. ç1 - ÷ 4ø è 3.

2 15

æ ö p p2 51. ç 2 + - 2÷ ç ÷ 2 2 16 2 è ø æ p3 pö 55. ( 2 log 2 - 1) 54. ç - ÷ ç 48 8 ÷ è ø æe ö 59. ç - 1÷ 58. ee è2 ø 5æ 5 3ö 61. 5 - ç 9 log - log ÷ 2è 4 2ø

3p 16

44.

ì7 + 3 5 ü 1 log í ý 2 5a î þ

2 3

æ p2 ö 50. ç - 2÷ ç 4 ÷ è ø æ 2 p2 ö -p 53. ç 52. - ÷ ç 27 12 ÷ 4 è ø æ ö 3 e2 57. ç - e÷ 56. log 3 - log 2 ç ÷ 4 è 2 ø

49. 1

60. (1 - log 2)

Senior Secondary School Mathematics for Class 12 Pg-803

Definite Integrals

803

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 16A) 15. cot 2 x = ( cosec2 x - 1). 17. sin 2 x = 19.

16. tan 2 x = ( sec2 x - 1).

1 ( 1 - cos 2 x ). 2

18. cos 2 x =

ò tan x dx = log sec x.

20.

1 ( 1 + cos 2 x ). 2

ò cosec x dx = log|cosec x - cot x|.

1 21. cos 3 x = 4 cos 3 x - 3 cos x Þ cos 3 x = (cos 3 x + 3 cos x ). 4 1 22. sin 3 x = 3 sin x - 4 sin 3 x Þ sin 3 x = ( 3 sin x - sin 3 x ). 4 æ 1 3 cos x ö÷ 23. I = ò ç dx = ò ( cosec2 x - 3 cosec x cot x ) dx. ç sin 2 x sin 2 x ÷ è ø 25.

1 - sin 2 x = (cos 2 x + sin 2 x - 2 sin x cos x )1 /2 = (cos x - sin x ) ì é p ùü íQ cos x > sin x in ê 0 , ú ý × ë 4 ûþ î

26. I = ò

sin x ö ( 1 - sin x ) ( 1 - sin x ) 1 æ 1 ÷ dx ´ dx = ò dx = ò ç ( 1 + sin x ) ( 1 - sin x ) è cos 2 x cos 2 x ø cos 2 x

= ò ( sec2 x - sec x tan x ) dx. 1 /2

32. 34.

ì æxö æxö æxö æ x öü 1 + sin x = í cos 2 ç ÷ + sin 2 ç ÷ + 2 sin ç ÷ cos ç ÷ ý è2ø è2ø è2ø è 2 øþ î

x xö æ = ç cos + sin ÷ × 2 2ø è

üï ìï 2 = íx2 - 1 + 2 ý× ( x + 1) îï ( x + 1) þï ( x 4 + 1) 2

41. cos 2 x =

1 1 ( 1 + cos 2 x ) and sin 2 x = ( 1 - cos 2 x ). 2 2

42. I = ò (tan 2 x + cot 2 x + 2 ) dx = ò {( sec2 x - 1) + ( cosec2 x - 1) + 2} dx = ò ( sec2 x + cosec2 x ) dx. dx

45. I = ò

ì1 æ 2 1 öü í - ç x - x + ÷ý 4 øþ î4 è

=ò

dx 2

1ö æ 1ö æ ç ÷ - çx - ÷ 2ø è2ø è

2

= sin -1 ( 2 x - 1).

2 ìï æ 1 ö 2 æ 1 ö üï 46. I = ò í ç ÷ - ç x - ÷ ý dx. 2 ø ïþ è îï è 2 ø

47. Let

1 2

x ( x + 1)

=

A B C + + × Then Ax ( x + 1) + B ( x + 1) + Cx 2 º 1. x x2 x + 1

This, gives A = - 1, B = 1, C = 1. 54. I = ò log x × ( 1 + x ) -2 dx. I

II

dx dx 55. I = ò = I1 - I 2 . (log x ) ò (log x ) 2 I1 = ò {(log x ) -1 × 1} dx. Integrate by parts.

Senior Secondary School Mathematics for Class 12 Pg-804

804

Senior Secondary School Mathematics for Class 12

1ö æ 58. I = ò e x ç log x + ÷ dx. xø è üï ìï 1 1 1 x 59. I = ò e xí × ý dx = e × 2 ( 1 + x) ïî ( 1 + x ) ( 1 + x ) ïþ 60. I = 2 ò tan x tan 2 x dx = 2 ò tan x ( sec2 x - 1) dx. 61. Let

5x 2 2

( x + 4x + 3)

=5+

A B + × ( x + 1) ( x + 3 )

To Evaluate a Definite Integral by Substitution b

In

ò f ( x) dx , when the variable x is converted into a new variable t by some a

relation then we put x = a and x = b in that relation to obtain the corresponding values of t, giving the lower limit and the upper limit respectively of the new integrand in t.

SOLVED EXAMPLES EXAMPLE 1

Evaluate: 4

2

(i)

x/2 ò e dx 0 1

(iii)

-1

ò cos

(ii)

2 1

(iv)

x dx

( 2x + 3)

ò (5 x 2 + 1) dx 0

0

SOLUTION

x

ò ( x 2 + 1) dx

x (i) Put = t so that dx = 2 dt. 2 Also, ( x = 0 Þ t = 0) and ( x = 2 Þ t = 1). 2

òe

\

0

x /2

1

dx = 2ò et dt = 2[et ]0 = 2( e - 1). 1

0

1 dt. 2 Also, ( x = 2 Þ t = 5) and ( x = 4 Þ t = 17). 2

(ii) Put ( x + 1) = t so that x dx = 4

\

x

17

1 dt 1 1 17 = [log ||] t 5 = (log 17 - log 5). 2 2 t 5

ò ( x 2 + 1) dx = 2 ò 2

(iii) Put x = cos t so that dx = - sin t dt. pö æ Also, ç x = 0 Þ t = ÷ and ( x = 1 Þ t = 0). 2ø è

Senior Secondary School Mathematics for Class 12 Pg-805

Definite Integrals 1

\

-1

ò cos 0

805

0

x dx = - ò cos-1 (cos t) sin t dt =

p/ 2

ò t sin t dt

p/ 2

p/ 2 = [t ( - cos t)]0

0

p/ 2

-

ò 1 × ( - cos t) dt 0

[integrating by parts] = [sin

p/ 2 t]0

= 1.

d (5 x 2 + 1) + B. dx Then, ( 2x + 3) º (10x) A + B.

(iv) Let ( 2x + 3) º A ×

Comparing the coefficients of like powers of x, we get 1 10A = 2 or A = and B = 3. 5 1 \ ( 2x + 3) = (10x) + 3. 5 1 1 1 (10x) + 3 ( 2x + 3) So, ò (5 x 2 + 1) dx = ò 5(5 x 2 + 1) dx 0 0 1

1

=

1 10x dx dx + 3 ò 2 5 ò0 (5 x 2 + 1) + 1) ( 5 x 0 1

3 1 1 = [log |5 x 2 + 1|]0 + ò 50 5

dx æ 1 ö x2 + ç ÷ è 5ø

2

1

é x ù 1 3 log 6 + × 5 ê tan -1 ú 5 5 ( / 1 5) û 0 ë 1 3 = log 6 + (tan -1 5 ). 5 5 =

EXAMPLE 2

Evaluate:

p/ 2

3

(i)

cos (log x) dx x 1

ò

p/ 2

(iii)

ò

0 p/ 2

(v)

ò

0

SOLUTION

(ii)

0 p/ 2

cos x dx [CBSE 2004] (iv) (1 + sin x)( 2 + sin x)

ò

0

cos q sin 3 q dq dx (1 - 2 sin x)

p/ 2

dx ( 3 + 2 cos x)

(i) Put log x = t so that

ò

(vi)

ò

0

dx 2

( 4 sin x + 5 cos2 x)

1 dx = dt. x

Also, ( x = 1 Þ t = log 1 = 0) and ( x = 3 Þ t = log 3).

Senior Secondary School Mathematics for Class 12 Pg-806

806

Senior Secondary School Mathematics for Class 12 3

cos (log x) dx = ò x 1

\

log 3

log 3

ò cos t dt = [sin t]0

= sin (log 3).

0

(ii) Put cos q = t so that sin q dq = - dt. p ö æ Also, ( q = 0 Þ t = 1) and ç q = Þ t = 0÷ × 2 ø è p/ 2

ò

\

p/ 2

ò

cos q sin 3 q dq =

0

cos q × (1 - cos2 q) sin q dq

0 0

1

= - ò t (1 - t 2) dt = ò (t1 /2 - t5 /2) dt 1

1

0

2 8 é2 ù æ 2 2ö = ê t 3/2 - t 7/2 ú = ç - ÷ = × 7 ë3 û 0 è 3 7 ø 21 (iii) Put sin x = t so that cos x dx = dt. p æ ö Also, ( x = 0 Þ t = 0) and ç x = Þ t = 1÷ × 2 è ø p/ 2

\

ò

0

cos x dx (1 + sin x)( 2 + sin x) 1

dt ( 1 t )( 2 + t) + 0

=ò

1 é 1 1 ù dt = òê ( t ) ( + + t) úû 1 2 0ë 1

[by partial fractions]

1

dt dt -ò + + t) ( 1 t ) ( 2 0 0

=ò

1

1

= [log |1 + t|]0 - [log |2 + t|]0 = [(log 2 - log 1) - (log 3 - log 2)] = ( 2 log 2) - (log 3). p/ 2

(iv)

ò

0

dx = (1 - 2 sin x)

=

p/2

dx

ò

ì 2 tan ( x/2) ü 1-2í ý 2 î1 + tan ( x/2) þ p/ 2 sec2( x/2) 0

ò

0 1

[1 + tan 2( x/2) - 4 tan ( x/2)]

= 2ò

dt

2 0 (1 + t - 4t)

1

= 2ò

0 (t

, where tan

dt - 2) 2 - ( 3 ) 2

dx

x =t 2 éx = 0 Þ t = 0 ù ú ê p êx = Þ t = 1ú û ë 2 1

= 2×

1 é t -2- 3 ù ê log ú 2 3 êë t - 2 + 3 úû 0

Senior Secondary School Mathematics for Class 12 Pg-807

Definite Integrals

= (v)

p/ 2

ò

0

1 3

é ê log êë

dx = ( 3 + 2 cos x)

p/ 2

dx é 1 - tan 2 ( x/2) ù 3 + 2× ê ú 2 ë 1 + tan ( x/2) û

ò

0

sec2 ( x/2)

ò

tan 2 ( x/2) + 5

0 1

dt

= 2ò

2 2 0 t + ( 5)

= 2× p/ 2

(vi)

ò

0

( 4 sin x + 5 cos x)

x =t 2

,where tan

éx = 0 Þ t = 0 ù ú ê p êx = Þ t = 1ú û ë 2 1

p/ 2 2

dx

2 1 é -1 t ù tan -1 = × êë tan 5 úû 0 5 5

1 5

dx 2

3 + 2ù ú× 3 - 2 úû

æ 3 + 1ö ç ÷ ç 3 - 1 ÷ - log è ø

p/ 2

=

807

ò

=

sec2 x ( 4 tan 2 x + 5)

0

dx

[dividing num. and denom. by cos2 x] =

¥

dt

ò ( 4t 2 + 5) ,

where tan x = t

0

=

éx = 0 Þ t = 0 ù ê ú p êx = Þ t = ¥ú ë û 2

¥

1 4 ò0

dt æ 5ö ÷ t 2 + çç ÷ è 2 ø

2

=

¥

1 2 é 2t ù tan -1 × 4 5 êë 5 úû 0

1 [tan -1( ¥ ) - tan -1( 0)] 2 5 1 æp p ö × = ç - 0÷ = 2 5 è2 4 5 ø

=

EXAMPLE 3

Evaluate: 1

(i)

x tan -1 x

ò (1 + x 2) 3/2 dx 0

SOLUTION

(i) Put

1/ 2

(ii)

ò

0

sin -1 x (1 - x 2) 3/2

dx

[CBSE 1996, ‘97]

x = tan q so that dx = sec2 q dq.

Clearly, x = 0 Þ q = 0 and x = 1 Þ q = ( p/4).

Senior Secondary School Mathematics for Class 12 Pg-808

808

Senior Secondary School Mathematics for Class 12 p/4

x tan -1 x

1

ò (1 + x 2) 3/2 dx = ò

\

0

= [- q cos

p/4 q]0

0 p/4

q tan q sec 3 q

p/4

× sec2 q dq =

0

ò ( - cos q) dq

-

ò q sin q dq

[integrating by parts]

0

p p p = - cos + sin 4 4 4 1 ö ( 4 - p) 2( 4 - p) æ -p =ç + = × ÷= 8 2ø 4 2 è4 2 p/4

= [- q cos q]0

p/4

+ [sin q]0

x = sin q so that dx = cos q dq. pö 1 æ Clearly, ( x = 0 Þ q = 0) and ç x = Þ q= ÷× 4ø 2 è

(ii) Put

1/ 2

ò

\

0

p/4

sin -1 x (1 - x 2) 3/2

dx = p /4

= [q tan q]0

ò

q cos 3 q

p/4

× cos q dq =

2

0

0 p /4

-

ò qsec q dq

ò 1 × tan q dq

[integrating by parts]

0

=

p p pö p /4 æ + [log (cos q)]0 = + log ç cos ÷ 4 4 4ø è

=

p æ 1 ö æp 1 ö + log ç ÷ = ç - log 2÷ × 4 ø è 2ø è 4 2

p/ 2

EXAMPLE 4

Evaluate:

(i)

ò

0

p/ 2

SOLUTION

(i)

ò

0

p/ 2

cos x x xö æ ç cos + sin ÷ 2 2ø è

cos x x xö æ ç cos + sin ÷ 2 2ø è

3

dx =

3

dx (ii)

ò

0

æ 2 p/ 2 ç cos

ò

0

è

cos x dx (1 + cos x + sin x)

x xö - sin 2 ÷ 2 2ø

x xö æ ç cos + sin ÷ 2 2ø è

3

dx

x xö æ 2ç cos - sin ÷ 2 2 dt é -2 ù 2 2ø dx = ò è = 2 = × ò t 2 êë t úû1 = 2( 2 - 1). 2 0 æ cos x + sin x ö 1 ç ÷ 2 2ø è x x 1æ x xö [putting cos + sin = t and ç cos - sin ÷ dx = dt ; 2 2 2è 2 2ø p/ 2

also, p/ 2

(ii)

ò

0

x = 0 Þ t = 1 and x = ( p/2) Þ t = 2]

cos x dx = (1 + cos x + sin x)

p/ 2

ò

0

cos x dx (1 + cos x) + sin x

Senior Secondary School Mathematics for Class 12 Pg-809

Definite Integrals p/ 2

=

ò

0

=

1 2

809

cos2( x/2) - sin 2( x/2) [2 cos2( x/2) + 2 sin ( x/2) cos ( x/2)]

p/ 2

ò

0

dx

1 - tan 2( x/2) dx 1 + tan ( x/2) [dividing num. and denom. by cos2( x/2)]

=

1 2

p/ 2

ò

[1 - tan ( x/2)] dx =

0

1 2

p/ 2

ò

p/ 2

dx -

0

1 2

ò

0

sin ( x/2) dx cos ( x/2)

1 p/ 2 p/ 2 = × [x]0 + [log cos ( x/2)]0 2 =

a

EXAMPLE 5

Evaluate

ò

-a

SOLUTION

p p p æ 1 ö æp 1 ö + log cos = + log ç ÷ = ç - log 2÷ × 4 4 4 ø è 2ø è 4 2

a-x dx. a+x

[CBSE 2008, ‘11C]

Put x = a cos q so that dx = - a sin q dq. Also, ( x = - a Þ q = p) and ( x = a Þ q = 0). a

\

ò

-a

0

a-x 1 - cos q dx = ò × ( - a sin q) dq a+x 1 + cos q p p

= aò

0

2 sin 2( q/2) 2 cos2( q/2)

× 2 sin ( q/2) cos( q/2) dq

p

p

= a ò 2 sin 2( q/2) dq = a ò (1 - cos q) dq 0 p

p

0

0

0

= a ò dq - a ò cos q dq = 1

EXAMPLE 6

Evaluate

òx× 0

SOLUTION

p a × [q]0

1 - x2 1 + x2

p

- a[sin q]0 = ap.

dx.

1 dt. Then, 2 [x = 0 Þ t = 0] and [x = 1 Þ t = 1].

Put x 2 = t and x dx =

[CBSE 2007]

Senior Secondary School Mathematics for Class 12 Pg-810

810

Senior Secondary School Mathematics for Class 12 1

\

I=

1 1 -t × dt 2 ò0 1 + t

=

1 1 1 ì 1 -t 1 -t ü 1 (1 - t) × òí ´ dt ý dt = × ò 2 0î 1 + t 2 0 1 - t2 1 -t þ

=

1 1 dt t × - × dt 2 ò0 1 - t 2 2 ò0 1 - t 2

=

( -2t) 1 1 1 × [sin -1 t]0 + × ò dt 2 4 0 1 - t2

=

1 1 1 × [sin -11 - sin -1 0] + × ò du, where (1 - t 2) = u 4 1 u 2

=

1æp ö 1 du p 1 1 = - [2 u]0 ç - 0÷ - × ò 2è 2 4 4 4 u ø 0

=

p 1 æ p 1ö - [ 1 - 0] = ç - ÷ × 4 2 è 4 2ø

1

1

1

0

1

p/ 2

EXAMPLE 7

Evaluate

ò

0

SOLUTION

cos x dx. ( 3 cos x + sin x)

d ( 3 cos x + sin x) dx Then, cos x = A( 3 cos x + sin x) + B × ( -3 sin x + cos x) Let cos x = A( 3 cos x + sin x) + B ×

Comparing the coefficients of cos x , we get 3 A + B = 1. Comparing the coefficients of sin x , we get A - 3 B = 0. 3 1 Solving 3 A + B = 1 and A - 3 B = 0, we get A = and B = × 10 10 3 1 \ cos x = ( 3 cos x + sin x) + ( -3 sin x + cos x). 10 10 p/ 2

So,

ò

cos x dx ( 3 cos x + sin x)

=

3 10

0

=

3 10

p/ 2

ò

0

( 3 cos x + sin x) 1 dx + ( 3 cos x + sin x) 10

p/ 2

ò

0

p/ 2

dx +

1 10

ò

0

p/ 2

ò

0

( -3 sin x + cos x) dx ( 3 cos x + sin x)

( -3 sin x + cos x) dx ( 3 cos x + sin x)

3 1 p/ 2 p/ 2 æ 3 p log 3 ö = × [x]0 + × [log|3 cos x + sin x|]0 = ç ÷× 10 ø 10 10 è 20

Senior Secondary School Mathematics for Class 12 Pg-811

Definite Integrals

811

p/ 2

ò{

Evaluate

EXAMPLE 8

tan x + cot x} dx.

[CBSE 2003, ‘12]

0

We have

SOLUTION

p/ 2

I=

ò

0

ì sin x + í î cos x p/ 2

= 2×

ò

0

cos x ü ý dx = sin x þ

p/ 2

ò

0

(sin x + cos x) dx sin x cos x

(sin x + cos x) dx = 2 × 2 sin x cos x

p/ 2

ò

(sin x + cos x) 1 - (sin x - cos x) 2

0

dx

Put (sin x - cos x) = t and (cos x + sin x) dx = dt. p é ù Also, [x = 0 Þ t = - 1] and ê x = Þ t = 1ú × 2 ë û 1

\

I = 2×

ò

-1

dt 1 -t

= 2[sin -1t]-1 1

2

-1

= 2{sin (1) - sin -1( -1)} = 2{2 sin -1(1)} pö æ = ç 2 ´ 2 ´ ÷ = 2p. 2ø è

EXERCISE 16B Very-Short-Answer Questions Evaluate the following integrals: 1

1

1.

dx ò ( 2x - 3) 0 2

3.

e

4.

7.

òxe

x2

[CBSE 1997]

6.

p/6

ò

0

1

11.

2 1/x

dx

8.

e

ò

x2

1

cos x dx ( 3 + 4 sin x) dx

ò ( ex + e- x ) 0

2x

ò (1 + x 4) dx 0

0

9.

ò (1 + x 2) dx 0 1

x

0

1

[CBSE 2008]

tan -1 x

1

3x

ò (1 + e2x ) dx

2x

ò (1 + x 2) dx 0

ò ( 9x 2 - 1) dx 1 1

5.

2.

p/ 2

10.

ò

0

e

12.

ò

dx sin x

(1 + cos2 x) dx

1/ 3 1/e x(log x)

dx

[CBSE 1997]

Senior Secondary School Mathematics for Class 12 Pg-812

812

Senior Secondary School Mathematics for Class 12

Short-Answer Questions Evaluate the following integrals: tan -1 x

1

13.

ò (1 + x 2)

p/ 2

dx

14.

0

0

p/ 2

p/ 2

15.

ò

sin x × cos5 x dx

16.

a

ò

2

a 2 - x 2 dx

18.

0

ò

0

x 2

a -x

2

dx

20.

22.

24.

ò

0

2

2

2

2

( a cos x + b sin x)

[CBSE 2003C] 26.

( 4 + 9 cos2 x) dx

ò

0

p/4

33.

ò

0

28.

ò

30.

0

tan 3 x dx (1 + cos 2x)

34.

39.

4

(sin x + cos x)

ò sin 0

-1

dx

ò (5 + 4 cos x)

[CBSE 2005]

dx

ò ( 3 + 2 sin x + cos x)

p/ 2

ò

0

dx [CBSE 2003] 36.

sin x cos x (cos2 x + 3 cos x + 2)

ò

1 + cos x

p/ 3 (1 - cos 1

38.

ò x(tan

[CBSE 2004]

-1

x)5/2

dx

dx

x) 2 dx

0 a

x dx

dx (5 + 4 sin x)

0

0

1

(1 + cos2 x)

p/ 2

sin 2x

-1 2 ò (cos x) dx

dx

p

1

37.

[CBSE 2002]

0

32.

4

ò

0

dx (cos x + 2 sin x)

p/ 2

35.

2x ö ç ÷ dx è 1 + x2 ø

1 + sin x dx

p

ò ( 6 - cos x)

p/ 2

ò

0

0

31.

ò

p/ 2

dx

p

29.

-1 æ

p/ 2

dx

p/ 2

27.

ò sin

dx

0

p/ 2 0

a + x2

p/ 2

1 + cos x dx

0

ò

x 2

0

p/ 2

25.

dx

2 - x 2 dx

1

ò x 2 - x dx ò

ò

0

0

23.

(1 + sin 4 x)

a

4

2

21.

ò

sin x cos x

0

a

19.

ò

0

0

17.

sin x dx 1 + cos x

ò

40.

ò sin 0

-1

x dx a+x

[CBSE 2008]

Senior Secondary School Mathematics for Class 12 Pg-813

Definite Integrals 9

41.

43.

1

dx

ò (1 + 0

x)

1

2

òx

42.

3

813

1 + 3 x 4 dx

0

2

(1 - x )

ò (1 + x 2) 2 dx

dx

ò

44.

1 (x

0

+ 1) x 2 - 1

p/ 2

45.

ò(

tan x + cot x ) dx [CBSE 2002, ‘03, ‘05, ‘12]

0

3

46.

5x - 6 - x

2

48.

p/ 2

( 2 - x)

ò

(p/ 2 )1/ 3 2

òx

2

dx

47.

p/ 2

ò

49.

1 + cosec2 x

p/4 æ cos

ç è

q qö + sin ÷ 2 2ø

3

dx

ò x(1 + log x) 2

dq

[CBSE 2003]

1

cosec x cot x

p/6

cos q

2

sin x 3 dx

0

50.

ò

dx

ANSWERS (EXERCISE 16B)

1 1. - log 3 2 4.

p2 32

æ e - 1ö 7. ç ÷ è 2 ø p 10. 4 1 3/2 13. p 12 17.

pa 2 4

21.

16 2 15

25.

p 2ab

29.

p 35

2. log 2

3.

1 (log 35 - log 8) 6

pö æ 5. ç tan -1 e - ÷ 4ø è

6.

p 4

8. ( e - e)

9.

1 (log 5 - log 3) 4

pö æ 11. ç tan -1 e - ÷ 4ø è 14. 2( 2 - 1) 18.

p 2

ö æp 22. ç - log 2÷ 2 ø è p 26. 2 2 30.

p 3

12. 0 p 8

15.

64 231

16.

19.

3 pa 4 16

20. a( 2 - 1)

23. 2 27. 31.

p 4 13

24. 2 28.

½3 - 5½ 1 ½ log ½ 5 ½ 2 ½

2 æ1ö tan -1 ç ÷ 3 è 3ø

Senior Secondary School Mathematics for Class 12 Pg-814

814

Senior Secondary School Mathematics for Class 12

p 4 p 35. 2 p 39. 4 1 43. 2

1 8 3 36. 2

32.

34. (log 9 - log 8)

33.

37. ( p - 2)

æp ö 40. a ç - 1÷ è2 ø 1 44. 3

41. ( 6 - 4 log 2) 45. p 2

2 - 2 p pö æ ç cos + sin ÷ 8 8ø è -1 1 50. tan 3

47.

1 3

48.

pæp ö ç - 1÷ 4è4 ø 7 42. 18 p 46. 2 log 2 49. (1 + log 2) 38.

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 16B) 5. Put e x = t. 1/2

8. Putting

1 = t , we get I = x

1

ò e tdt = ò e tdt = [e t]1/2 . 1

1

1/2 5

9. Putting 3 + 4 sin x = t , we get I =

1 dt × × 4 ò3 t 1

11. I = ò

10. Put cos x = t.

0 (1+

p/2

15. I =

ò

ex e2x)

dx.

sin x × ( 1 - sin 2 x ) 2 cos x dx. Now, put sin x = t.

0

16. Put sin 2 x = t.

18. Put x = 2 sin t.

21. Put ( x + 2 ) = t 2 .

22. Put x = tan t and integrate by parts.

20. Put x = a tan q.

25. Divide num. and denom. by cos 2 x and put tan x = t. 32. Write sin x =

2 tan ( x/ 2 ) 2

1 + tan ( x/ 2 )

, cos x =

1 - tan 2 ( x/ 2 ) 1 + tan 2 ( x/2)

and put tan

p/4

33. Using cos 2 x = ( 2 cos 2 x - 1), we get I =

ò tan 3 x sec2 x dx. 0

Now, put tan x = t. 34. Put cos x = t. 35. Divide num. and denom. by cos 4 x and put tan 2 x = t. p/2

36. I =

ò

cos ( x/ 2 )

5 p/3 sin ( x/ 2 )

dx. Put sin

x = t. 2

37. Put x = cos t. 38. Integrate by parts with x as the second function.

x = t. 2

Senior Secondary School Mathematics for Class 12 Pg-815

Definite Integrals

815

39. Put x = sin 2t.

40. Put x = a tan 2 q.

41. Put x = t 2 .

42. Put ( 1 + 3 x 4 ) = t.

43. Put x = tan t.

44. Put ( x + 1) =

p/2

45. I =

ò

2 (sin x + cos x ) 2 sin x cos x

0

p/2

= 2×

ò

p/2 0

(sin x + cos x ) 1 - (sin x - cos x ) 2

0

ò

dx = 2 ×

(sin x + cos x ) 1 - ( 1 - sin 2x )

dx

dx. Put (sin x - cos x ) = t.

d (5 x - 6 - x 2 ) + B. dx q qù é p/2 ê cos - sin ú q q 2 2û ë 47. I = ò dx. Now, put cos + sin = t. 1/2 2 2 q q æ ö p/4 ç cos + sin ÷ 2 2ø è

46. Let ( 2 - x ) = A ×

Properties of Definite Integrals b

THEOREM 1

b

ò f ( x) dx = ò f (t) dt. a

PROOF

a

ò f ( x) dx = F( x). Then, ò f (t) dt = F(t).

Let

b

b

ò f ( x) dx = [F( x)]a = F( b) - F( a).

\

a b

And,

b

ò f (t) dt = [F(t)]a = F( b) - F( a). a b

b

Hence, ò f ( x) dx = ò f (t) dt. a

THEOREM 2

PROOF

a

b

a

a

b

ò f ( x) dx = -ò f ( x) dx.

Let

ò f ( x) dx = F( x). b

Then,

b

ò f ( x) dx = [F( x)]a = F( b) - F( a). a a

a

And, - ò f ( x) dx = - [F( x)]b = - [F( a) - F( b)] = F( b) - F( a). b

Hence,

b

a

a

b

ò f ( x) dx = -ò f ( x) dx.

1 × t

Senior Secondary School Mathematics for Class 12 Pg-816

816

Senior Secondary School Mathematics for Class 12

PROOF

b

c

b

a

a

c

ò f ( x) dx = ò f ( x) dx + ò f ( x) dx , where a < c < b.

THEOREM 3

b

ò f ( x) dx = F( x). Then,

Let

c

b

a

c

b

ò f ( x) dx = [F( x)]a = F( b) - F( a). a

c

b

ò f ( x) dx + ò f ( x) dx = [F( x)]a + [F( x)]c = {F( c) - F( a)} + {F( b) - F( c)} b

b

= F( b) - F( a) = [F( x)]a = ò f ( x) dx. c

b

b

a

c

a

ò f ( x) dx + ò f ( x) dx = ò f ( x) dx.

Hence,

If a < c1 < c2 < ¼ < cn < b then

REMARK

b

c1

c2

a

a

c1

ò f ( x) dx = THEOREM 4 PROOF

a

a

a

0

0

ò f ( x) dx +

ò f ( x) dx + ¼ +

ò f ( x) dx = ò f ( a - x) dx.

a

0

a

a

a

0

0

ò f ( a - x) dx = -ò f (t) dt = ò f (t) dt = ò f ( x) dx. 0

a

a

0

0

Hence, ò f ( x) dx = ò f ( a - x) dx. THEOREM 5

PROOF

b

b

a

a

ò f ( a + b - x) dx = ò f ( x) dx.

Putting a + b - x = t , we get dx = - dt. Now, x = a Þ t = b. And, x = b Þ t = a. b

\

a

b

b

a

a

ò f ( a + b - x) dx = -ò f (t) dt = ò f (t) dt = ò f ( x) dx. a

b

b

b

Hence, ò f ( a + b - x) dx = ò f ( x) dx. a

THEOREM 6

PROOF

a

b

b

b

a

a

a

ò { f ( x) + g( x)} dx = ò f ( x) dx + ò g( x) dx.

Let ò f ( x) dx = F( x) and

ò f ( x) dx.

cn

[CBSE 2002C, ‘05C, ‘06]

In the RHS integral, put ( a - x) = t so that dx = - dt. Now, when x = 0 we have t = a. And, when x = a we have t = 0. \

b

ò g( x) dx = G( x).

Senior Secondary School Mathematics for Class 12 Pg-817

Definite Integrals

Then, b

\

817

ò [ f ( x) + g( x)] dx = ò f ( x) dx + ò g( x) dx = F( x) + G( x). b

ò { f ( x) + g( x)} dx = [F( x) + G( x)]a a

= [F( b) + G( b)] - [F( a) + G( a)] = [F( b) - F( a)] + [G( b) - G( a)] b

b

= ò f ( x) dx + ò g( x) dx a

a

b

b

b

a

a

a

Hence, ò { f ( x) + g( x)} dx = ò f ( x) dx + ò g( x) dx.

THEOREM 7

ì0, when f ( x) is an odd function a ï a ò f ( x) dx = í2ò f ( x) dx , when f ( x) is an even function. ï -a î 0 a

PROOF

ò f ( x) dx =

We have

-a

0

a

-a

0

[CBSE 2005C]

ò f ( x) dx + ò f ( x) dx

… (i)

In the first integral on the RHS of (i), put x = -t so that dx = - dt. When x = - a , we have t = a. And, when x = 0, we have t = 0. 0

0

a

a

a

0

0

ò f ( x) dx = -ò f ( -t) dt = ò f ( -t) dt = ò f ( -x) dx.

\

-a 0

a

-a

0

ò f ( x) dx = ò f ( -x) dx

Thus,

… (ii)

Using (ii) in (i), we have a

a

a

a

0

0

0

ò f ( x) dx = ò f ( -x) dx + ò f ( x) dx = ò [ f ( -x) + f ( x)] dx

-a

ì0, when f ( x) is odd ï =í a ï2ò f ( x) dx , when f ( x) is even. î 0 éQ f ( x) is odd Þ f ( - x) = - f ( x) ù ê f ( x) is even Þ f ( - x) = f ( x) ú ë û

ì0, if f ( 2a - x) = - f ( x) ï a ò f ( x) dx = í2 ò f ( x) dx , if f ( 2a - x) = f ( x). ï 0 î 0

2a

THEOREM 8

PROOF

We have

2a

a

2a

0

0

a

ò f ( x) dx = ò f ( x) dx + ò f ( x) dx

… (i)

Senior Secondary School Mathematics for Class 12 Pg-818

818

Senior Secondary School Mathematics for Class 12

Now, in the second integral on the RHS of (i), put x = 2a - t so that dx = - dt. When x = a , we have t = a. When x = 2a , we have t = 0. 2a

0

a

a

0

ò f ( x) dx = -ò f ( 2a - t) dt = ò f ( 2a - t) dt

\

a

a

= ò f ( 2a - x) dx. 0

Thus,

2a

a

a

0

ò f ( x) dx = ò f ( 2a - x) dx

… (ii)

Using (ii) in (i), we get 2a

a

a

0

0 a

0

ò f ( x) dx = ò f ( x) dx + ò f ( 2a - x) dx = ò { f ( x) + f ( 2a - x)} dx 0

Hence,

ì0, when f ( 2a - x) = - f ( x) ï =í a ï2ò f ( x) dx , when f ( 2a - x) = f ( x). î 0 ì0, when f ( 2a - x) = - f ( x) 2a ï a ò f ( x) dx = í2ò f ( x) dx , when f ( 2a - x) = f ( x). ï 0 î 0

SOLVED EXAMPLES p/ 2

EXAMPLE 1

ò

Prove that

0

p/ 2

SOLUTION

Let I =

ò

0

sin x p dx = × 4 (sin x + cos x)

[CBSE 2002C, ‘07C]

sin x dx (sin x + cos x) a

a

0

0

… (i)

Using the result ò f ( x) dx = ò f ( a - x) dx in (i), we get p/ 2

I=

ò

sin [( p/2) - x] dx sin [( p/2) - x] + cos[( p/2) - x]

ò

cos x dx = (cos x + sin x)

0 p/ 2

or I =

0

p/ 2

ò

0

cos x dx (sin x + cos x)

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-819

Definite Integrals

819

Adding (i) and (ii), we get p/ 2

2I =

ò

sin x dx + (sin x + cos x)

ò

(sin x + cos x) dx = (sin x + cos x)

0 p/ 2

=

0

\ I=

p/ 2

p , i.e., 4

ò

0

ò

Prove that

0

ò

Let I =

0

0 p/ 2

cos x dx (sin x + cos x) p/ 2

ò dx = [x]0

=

0

cos x ( sin x + cos x )

p/ 2

SOLUTION

ò

p × 2

sin x p dx = × 4 (sin x + cos x)

p/ 2

EXAMPLE 2

p/ 2

cos x ( sin x + cos x ) a

a

0

0

dx =

p × 4

dx

… (i)

Using the result ò f ( x) dx = ò f ( a - x) dx in (i), we get p/ 2

ò

I=

0 p/ 2

ò

or I =

cos[( p/2) - x] p/ 2

sin x cos x + sin x

0

dx

sin [( p/2) - x] + cos[( p/2) - x] dx =

sin x

ò

sin x + cos x

0

dx

… (ii)

Adding (i) and (ii), we get p/ 2

2I =

ò

0 p/ 2

=

ò

0

\ I=

( sin x + cos x ) ( sin x + cos x ) p/ 2

p , i.e., 4

ò

0

Evaluate:

(i)

(i) We have

0 p/ 2

dx =

sin x ( sin x + cos x)

n

n

p/ 2

ò dx = [x]0

=

0

cos x

ò x(1 - x) 0

ò

( sin x + cos x )

ò x(1 - x)

0 1

SOLUTION

dx +

( sin x + cos x )

1

EXAMPLE 3

p/ 2

cos x

dx =

p × 2

p × 4 1

(ii)

dx

dx

ò x(1 - x)

3/2

dx

0

1

dx = ò (1 - x)[1 - (1 - x)]n dx 0

a

[Q

a

ò f ( x) dx = ò f ( a - x) dx] 0

0

1

1

1

0

0

0

= ò (1 - x) x ndx = ò x ndx - ò x n +1 dx

Senior Secondary School Mathematics for Class 12 Pg-820

820

Senior Secondary School Mathematics for Class 12 1

1

é x n+1 ù é x n+ 2 ù 1 ö æ 1 =ê ÷ ú -ê ú =ç ( 1 ) 2 1 2ø + + + + n n n n è ë û0 ë û0 = 1

ò x(1 - x)

(ii) We have

3/2

0

1 × (n + 1)(n + 2) 1

dx = ò (1 - x)[1 - (1 - x)] 3/2 dx 0

a

[Q

a

ò f ( x) dx = ò f ( a - x) dx] 0

1

0

1

1

= ò (1 - x) x 3/2 dx = ò x 3/2 dx - ò x5/2 dx 0

0

0

1

4 é2 ù é2 ù æ 2 2ö × = ê x5/2 ú - ê x 7/2 ú = ç - ÷ = ë5 û0 ë7 û 0 è 5 7 ø 35

p/ 2

EXAMPLE 4

ò log(tan x) dx = 0.

Show that

0

p/ 2

SOLUTION

Let

ò log(tan x) dx

I=

… (i)

0 p/ 2

Then, I =

or \

1

é æp öù ò log êë tan çè 2 - x ÷ø úû dx [Q 0

p/ 2

p/ 2

0

0

æ

1

a

a

ò f ( x) dx = ò f ( a - x) dx ] 0

0

p/2

ö

ò log(cot x) dx = ò log ççè tan x ÷÷ø dx = - ò log tan x dx = -I.

I=

0

I = - I or 2I = 0 or I = 0. p/ 2

ò log(tan x) dx = 0.

Hence,

0

p/4

EXAMPLE 5

Prove that

p

ò log(1 + tan x) dx = 8 (log 2).

[CBSE 2007, ‘11, ‘13C]

0 p/4

SOLUTION

Let

I=

ò log(1 + tan x) dx 0

Then,

I=

p/4

é

æp

… (i) öù

ò log êë1 + tan çè 4 - x ÷ø úû dx 0

p/4

or

I=

é

1 - tan x ù

ò log êë1 + 1 + tan x úû dx 0

[Q

a

a

0

0

ò f ( x) dx = ò f ( a - x) dx ]

Senior Secondary School Mathematics for Class 12 Pg-821

Definite Integrals p /4

æ

ö

2

ò log ççè 1 + tan x ÷÷ø dx

I=

or

821

0 p /4

ò [log 2 - log(1 + tan x)] dx

I=

or

0

I = (log 2) ×

or

p /4

p /4

0

0

ò dx -

ò log(1 + tan x) dx

... (ii)

Adding (i) and (ii), we get p /4

2I = (log 2)

p /4

ò dx = (log 2) × [x]0 0

\

I=

0

0 p/ 2

SOLUTION

I=

Let

p

ò log(1 + tan x) dx = 8 (log 2).

ò log (sin x) dx =

Prove that

p (log 2). 4

p /4

p (log 2), i.e., 8 p/ 2

EXAMPLE 6

=

p/ 2

p

ò log (cos x) dx = - 2 (log 2). 0

[CBSE 2004, ‘07C, ‘08]

ò log (sin x) dx

… (i)

0 p/ 2

Then, I =

é æp öù ò log êë sin çè 2 - x ÷ø úû dx 0

p/ 2

I=

or

a

a

0

0

ò f ( x) dx = ò f ( a - x) dx ]

[Q

ò log (cos x) dx

… (ii)

0

Adding (i) and (ii), we get p/ 2

2I =

ò [log (sin x) + log (cos x)] dx

0 p/ 2

= =

=

0 p/ 2

p/ 2

æ sin 2x ö ÷ dx 2 ø

0

0

0

1

p

p/ 2

0

0

ò log (sin 2x) dx - ò (log 2) dx = 2 ò log sin t dt - (log 2) × ò dx [putting 2x = t in the 1st integral]

p

1 p/ 2 log sin t dt - (log 2) × [x]0 2 ò0

æ1 ö = ç ´ 2÷ × è2 ø p/ 2

=

p/ 2

ò log (sin x cos x) dx = ò log çè

p/ 2

p

ò log sin t dt - 2 (log 2) 0

p

ò log (sin x) dx - 2 (log 2). 0

Senior Secondary School Mathematics for Class 12 Pg-822

822

Senior Secondary School Mathematics for Class 12

p p (log 2) or I = - (log 2). 2 2

\ 2I = I p/ 2

\

ò log (sin x) dx =

p/ 2

p

ò log (cos x) dx = - 2 (log 2).

0

0

p/ 2

EXAMPLE 7

Prove that:

ò sin 2x log (tan x) dx = 0

(a)

0 1

æ1

[CBSE 2003C, ‘06C]

ö

ò log çè x - 1÷ø dx = 0

(b)

0

p/ 2

SOLUTION

(a) Let I =

ò sin 2x log (tan x) dx 0

p/ 2

Then, I =

æp

… (i) æp

ö

ö

ò sin 2 çè 2 - x ÷ø log tan çè 2 - x ÷ø dx 0

p/ 2

or

a

a

0

0

ò f ( x) dx = ò f ( a - x) dx]

[Q

ò sin 2x log (cot x) dx

I=

… (ii)

0

Adding (i) and (ii), we get p/ 2

2I =

ò [sin 2x log (tan x) + sin 2x log (cot x)] dx

0 p/ 2

= =

ò sin 2x{log (tan x) + log (cot x)} dx

0 p/ 2

p/ 2

0

0

ò sin 2x × log (tan x × cot x) dx =

ò sin 2x × log (1) dx = 0 [Q log 1 = 0]

p/ 2

\ I=0 Þ

ò sin 2x log (tan x) dx = 0. 0

(b) Put x = cos2 t so that dx = - sin 2t dt. When x = 0, we have cos t = 0 or t =

p . 2

When x = 1, we have cos t = 1 or t = 0. 1

\

æ1

ö

0

ò log çè x - 1÷ø dx = - ò log (tan 0

2

t) × sin 2t dt

p/ 2 p/ 2

= 2 ò sin 2t × log (tan t) dt = 0 0

[from (a)].

Senior Secondary School Mathematics for Class 12 Pg-823

Definite Integrals p/ 2

EXAMPLE 8

ò

Prove that: (a)

0

p/ 2

(b)

ò

0 p/ 2

SOLUTION

(a) Let

I=

Then,

sin 2 x 1 dx = log ( 2 + 1) (sin x + cos x) 2 [CBSE 2012, ‘14C]

sin 2 x p dx = × (1 + sin x cos x) 3 3

sin 2 x dx (sin x + cos x) 0 ö æp sin 2 ç - x ÷ p/2 2 ø è I= ò dx p p é öù æ ö æ 0 sin ç - x ÷ + cos ç - x ÷ ê ú øû è2 ø è2 ë

ò

p/ 2

or

823

I=

ò

0

cos2 x dx = (cos x + sin x)

p/ 2

ò

0

cos2 x dx (sin x + cos x)

… (i)

… (ii)

Adding (i) and (ii), we get p/ 2

2I =

ò

0 p/ 2

=

ò

0

p/ 2

=

ò

0

(sin 2 x + cos2 x) dx = (sin x + cos x)

p/2

ò

0

dx (sin x + cos x)

dx é 2 tan( x/2) 1 - tan 2( x/2) ù + ê ú 2 2 ë 1 + tan ( x/2) 1 + tan ( x/2) û sec2( x/ 2) [1 - tan 2( x / 2) + 2 tan( x/2)]

dx

1

x dt , where tan = t 2 2 1 + 2 ( t t ) 0 p é ù êë x = 0 Þ t = 0 and x = 2 Þ t = 1úû 1 1 dt dt =2ò = 2ò 2 2 2 1 2 ( t ) ( ) (t - 1) 2 0 0 = 2ò

1

æ 2 -1öù 1 ìï ½ 2 + t - 1½üï 1 é ÷ú ½ý = ê 0 - log çç ílog ½ ÷ 2 2 ïî ½ 2 - t + 1½ïþ 2 êë è 2 + 1 ø úû 0 æ 2 + 1ö 1 1 ( 2 + 1) ( 2 + 1) ÷= = log çç log ´ ÷ 2 2 ( 2 - 1) ( 2 + 1) è 2 -1ø 1 1 = log ( 2 + 1) 2 = ´ 2 log ( 2 + 1) 2 2 = 2 log ( 2 + 1) 1 1 \ I= 2 log ( 2 + 1) = log ( 2 + 1). 2 2 = 2×

Senior Secondary School Mathematics for Class 12 Pg-824

824

Senior Secondary School Mathematics for Class 12 p/ 2

(b) Let

ò

I=

sin 2 x dx (1 + sin x cos x)

0 p/ 2

Then, I =

ò

sin 2[( p/2) - x] dx 1 + sin[( p/2) - x] cos[( p/2) - x]

ò

cos2 x dx (1 + sin x cos x)

0 p/ 2

I=

or

… (i)

0

… (ii)

Adding (i) and (ii), we get p/ 2

2I =

ò

0

(sin 2 x + cos2 x) dx = (1 + sin x cos x)

ò

0

ò

0

dx (1 + sin x cos x) é êQ êë

p/ 2

=

p/ 2

sec2 x ( sec2 x + tan x)

a ù f ( x ) dx = ò ò f ( a - x) dx úú û 0 0 a

dx [dividing num. and denom. by cos2 x]

p/ 2

=

ò

0

sec2 x 2

(1 + tan x + tan x)

dx =

¥

dt

ò (t 2 + t + 1) ,

where tan x = t

0

p é ù êë x = 0 Þ t = 0 and x = 2 Þ t = ¥ úû ¥

¥

é 2 æ 2t + 1 ö ù =ê tan -1 ç ÷ú ò 2 2 è 3 øû0 ë 3 æ 3ö 1ö 0 æ ÷ çt + ÷ + çç ÷ 2ø è è 2 ø 2 é 2 æ p pö 2p æ 1 öù ×ç - ÷ = × = tan -1( ¥ ) - tan -1 ç ÷ú = 3 êë 3 è 2 6ø 3 3 è 3 øû =

Hence, I =

p 3 3 p

EXAMPLE 9

dt

× x tan x

æp

ö

ò ( sec x + tan x) dx = pçè 2 - 1÷ø ×

Prove that

[CBSE 2008, ’10]

0

p

SOLUTION

Let

x tan x dx x + tan x) ( sec 0

I=ò

p

( p - x) tan ( p - x) dx [Q [sec ( p - x) + tan ( p - x)] 0

Then, I = ò

… (i) a

a

0

0

ò f ( x) dx = ò f ( a - x) dx]

p

or

( p - x) tan x I=ò dx sec x + tan x) ( 0

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-825

Definite Integrals

825

Adding (i) and (ii), we get p p tan x tan x( sec x - tan x) dx = p ò dx 2I = p ò 2 2 + x x ( sec tan ) 0 ( sec x - tan x) 0 p ép ù = p × ê ò sec x tan x dx - ò tan 2 x dx ú êë 0 úû 0 p ü ìï ï p = p × í[sec x]0 - ò (sec2 x - 1) dx ý ïþ ïî 0 p p = p × {-2 - [tan x]0 + [x]0} = p( p - 2).

\

æp ö I = p ç - 1÷ i.e. , è2 ø p

EXAMPLE 10

Evaluate

p

æp

x tan x

ö

ò sec x + tan x dx = p çè 2 - 1÷ø × 0

x

ò (1 + sin x) dx.

[CBSE 2010, ‘11, ‘12C]

0

p

SOLUTION

Let

x dx ( 1 sin + x) 0

I=ò

… (i)

p

p

( p - x) ( p - x) dx = ò dx 1 1 + x + sin x) sin ( p ) ( 0 0

Then, I = ò

…(ii)

Adding (i) and (ii), we get p p dx (1 - sin x) 1 = p× ò ´ dx 2I = p ò + x + x ( sin ) ( sin ) (1 - sin x) 1 1 0 0 or

p p ép ù æ 1 - sin x ö 2 2I = p ò ç ÷ dx = p × ê ò sec x dx - ò sec x tan x dx ú 2 êë 0 úû 0 è cos x ø 0 p

p

= p × {[tan x]0 - [sec x]0 } = 2p \ I = p , i.e.,

p

x

ò (1 + sin x) dx = p. 0

p

EXAMPLE 11

Evaluate

x sin x

ò (1 + cos2x) dx.

[CBSE 2009C, ‘11C, ‘12, ’14]

0

p

SOLUTION

Let

I=ò p

Then, I = ò

0 p

or

x sin x

2 0 (1 + cos x)

I=ò

0

dx

( p - x) sin ( p - x) 1 + cos2( p - x) ( p - x) sin x (1 + cos2 x)

dx

… (i) dx … (ii)

Senior Secondary School Mathematics for Class 12 Pg-826

826

Senior Secondary School Mathematics for Class 12

Adding (i) and (ii), we get p

0 (1 + 1

2

cos x)

dt

= pò

2 -1 (1 + t )

\

ò

Evaluate

0 p/ 2

SOLUTION

Let

dt

dx = - p ò

, where cos x = t (1 + t 2) [ x = 0 Þ t = 1 and x = p Þ t = - 1 ] 1

= p [tan -1t]-1 1

é p æ p ö ù p2 = p[tan -1 1 - tan -1( -1)] = p ê - ç - ÷ ú = × 2 ë 4 è 4øû p2 I= × 4 p/ 2

EXAMPLE 12

-1

sin x

2I = p ò

I=

ò

0

x dx. (sin x + cos x)

[CBSE 2003C, ‘08]

x dx (sin x + cos x)

… (i)

ö æp ç - x÷ 2 ø è Then, I = ò dx + 2 sin [( / ) ] cos[( p p/2) - x] x 0 ö æp ö æp p/ 2 p/ 2 ç - x÷ ç - x÷ 2 2 ø è ø è or dx I= ò dx = ò (cos x + sin x) (sin x + cos x) 0 0 p/ 2

… (ii)

Adding (i) and (ii), we get 2I = \

I=

=

p 2 p 4 p 4

p/ 2

ò

dx (sin x + cos x)

ò

dx p = (sin x + cos x) 4

0 p/ 2 0

p/ 2

ò

1

=

0

p/ 2

ò

0

dx é 2 tan ( x/2) 1 - tan 2( x/2) ù + ê ú 2 2 ë 1 + tan ( x/2) 1 + tan ( x/2) û

sec2( x/2) 2

1 - tan ( x/2) + 2 tan ( x/2)

dx

x p 2dt , where t = tan 4 ò0 (1 - t 2 + 2t) 2

p é ù êë x = 0 Þ t = 0 and x = 2 Þ t = 1úû

1

=

p dt dt 2 ò0 [( 2) 2 - (t - 1) 2]

=

½ 2 + (t - 1)½ ½ 2 + 1½ p 1 p ½ = ½× × log ½ log ½ 2 2 2 ½ 2 - (t - 1)½0 4 2 ½ 2 - 1½

1

Senior Secondary School Mathematics for Class 12 Pg-827

Definite Integrals

827

p/ 2

EXAMPLE 13

p

ò ( 2 log sin x - log sin 2x) dx = - 2 (log 2).

Prove that

[CBSE 2009]

0 p/ 2

SOLUTION

Let

ò ( 2 log sin x - log sin 2x) dx.

I=

0 p/ 2

ò

Then, I =

0 p/ 2

or

… (i)

é æp ö æp öù ê 2 log sin ç 2 - x ÷ - log sin 2 ç 2 - x ÷ ú dx è ø è øû ë

ò ( 2 log cos x - log sin 2x) dx

I=

… (ii)

0

Adding (i) and (ii), we get p/ 2

ò [2(log sin x + log cos x) - 2 log sin 2x] dx

2I =

0 p/ 2

or

ò [log (sin x cos x) - log sin 2x] dx

I=

0 p/ 2

ò

=

0 p/ 2

é ù æ sin 2x ö ê log ç 2 ÷ - log sin 2x ú dx è ø ë û

ò (log sin 2x - log 2 - log sin 2x) dx

=

0

p/ 2

p/ 2

ò dx = ( - log 2) × [x]0

= - log 2 ×

0

p

EXAMPLE 14

p = - (log 2). 2

x

ò ( a 2 cos2x + b2 sin 2x) dx.

Evaluate

[CBSE 2003C, ‘09]

0

p

SOLUTION

Let

I=ò

0 (a p

x 2

2

cos x + b 2 sin 2 x)

dx

… (i)

( p - x)

Then, I = ò

2 2 2 2 0 ( a cos ( p - x) + b sin ( p - x)]

p

or

I=ò

0 (a

( p - x) 2

2

cos x + b 2 sin 2 x)

dx

dx.

… (ii)

Adding (i) and (ii), we get p

2I = ò

2

2

ò

2

0 ( a cos p/ 2

= 2p

p

( x + p - x)

0

2

2

x + b sin x)

dx = p ò

2

2

ò

2

0 ( a cos p/2

dx 2

dx 2

2

( a cos x + b sin x)

= 2p

0

x + b 2 sin 2 x) sec2 x

( a + b 2 tan 2 x)

dx

[dividing num. and denom. by cos2 x]

Senior Secondary School Mathematics for Class 12 Pg-828

828

Senior Secondary School Mathematics for Class 12 ¥

dt

= 2p ò

2 2 2 0 (a + b t )

2p

¥

, where tan x = t ¥

é 2p b æ bt ö ù = ê 2 × tan -1 ç ÷ ú ò a ö è a øû0 b2 0 æ a 2 b ë ç + t2 ÷ ç b2 ÷ è ø 2 2p 2p æ p ö æ 2p p ö p ´ ÷= × = [tan -1( ¥ ) - tan -1( 0)] = ç - 0÷ = ç ab ab è 2 ø è ab 2 ø ab =

\

I=

dt

p

p2 Þ 2ab

ò ( a 2 cos2x + b2 sin 2x)

Prove that

dx =

0

p

EXAMPLE 15

x

ò x sin

3

x dx =

0

p2 × 2ab

2p × 3

p

SOLUTION

Let

I = ò x sin 3 x dx 0

Then,

… (i)

p

I = ò ( p - x) sin 3( p - x) dx 0 p

or

I = ò ( p - x) sin 3 x dx

… (ii)

0

Adding (i) and (ii), we get p

p

2I = ò p sin 3 x dx = p ò sin 2 x × sin x dx 0

0

p

= p ò (1 - cos2 x) sin x dx 0

-1

= - p ò (1 - t 2) dt , where cos x = t 1

[x = 0 Þ t = 1 and x = p Þ t = -1] 1

é t3ù 4p = p ò (1 - t ) dt = p êt - ú = × 3 3 ë û -1 -1 1

Hence, I = 1

EXAMPLE 16

Evaluate

2

2p Þ 3

ò cot

-1

p

ò x sin

3

x dx =

0

{1 - x + x 2} dx.

0

SOLUTION

We have 1

I = ò cot -1{1 - x + x 2} dx 0

2p × 3 [CBSE 2008]

Senior Secondary School Mathematics for Class 12 Pg-829

Definite Integrals 1

829

1

1 æ ö ì x + (1 - x) ü = ò tan -1 ç dx = ò tan -1í ý dx 2÷ è ø î1 - x + x2 þ + 1 x x 0 0 1

= ò {tan -1 x + tan -1(1 - x)} dx 0 1

1

= ò tan -1 x dx + ò tan -1(1 - x) dx 0 1

0

1

-1

= ò tan x dx + ò tan -1{1 - (1 - x)} dx 0

0

1

1

1

0

0

é êQ êë

a ù f x dx ( ) = ò ò f ( a - x) dx úú û 0 0 a

= ò tan -1 x dx + ò tan -1 x dx = 2ò tan -1 x dx 0

1

= 2ò (tan -1 x × 1) dx 0

1

1 é ù 1 = 2 ê(tan -1 x) x - ò × x dx ú 2 êë úû 0 0 (1 + x ) 1

= 2[(tan -1 x) × x]0 - 2ò 1

x

0 (1 +

x 2)

dx

= 2{(tan -1 1) × 1 - 0} - [log (1 + x 2)]0

EXAMPLE 17

pö æ æp ö = ç 2 ´ ÷ - (log 2 - log 1) = ç - log 2÷ × 4ø è è2 ø 3p/10 sin x Evaluate ò (sin x + cos x) dx. p/5 3p/10

SOLUTION

1

Let

ò

I=

p/5

Here, a = b

sin x dx (sin x + cos x)

[CBSE 2009]

… (i)

p 3p æ p 3p ö p and b = Þ ( a + b) = ç + ÷= × 5 10 è 5 10 ø 2 b

Using ò f ( x) dx = ò f ( a + b - x) dx , we have a

a

3p/10

I=

ò

p/5 3p/10

Þ

I=

ò

p/5

ö æp sin ç - x ÷ dx ø è2 dx æp ö æp ö sin ç - x ÷ + cos ç - x ÷ è2 ø è2 ø

cos x dx (sin x + cos x)

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-830

830

Senior Secondary School Mathematics for Class 12

Adding (i) and (ii), we get 3p/10

3p/10

ò dx = [x]p/5

2I =

p/5

Þ

æ 3p p ö p =ç - ÷= è 10 5 ø 10

p I= × 20 a

Integrals of the form

ò f ( x) dx , where f ( x) is even or odd

-a

We know that (i) f ( x) is odd, if f ( - x) = - f ( x); (ii) f ( x) is even, if f ( - x) = f ( x); a

ò f ( x) dx = 0, when f ( x) is odd;

(iii)

-a a

a

ò f ( x) dx = 2ò f ( x) dx , when f ( x) is even.

(iv)

-a p/ 2

EXAMPLE 18

0

ò sin

Show that

7

x dx = 0.

-p/ 2

SOLUTION

Let

f ( x) = sin 7 x. Then,

f ( - x) = [sin ( - x)]7 = - sin 7 x = - f ( x). \ f ( x) is an odd function of x. a

But,

ò f ( x) dx = 0, when f ( x) is odd.

-a p/ 2

\

ò sin

7

x dx = 0.

-p/ 2

p/ 2

EXAMPLE 19

Evaluate

ò sin

2

x dx.

-p/ 2

SOLUTION

Let

f ( x) = sin 2 x.

Then, f ( - x) = [sin ( - x)]2 = ( - sin x) 2 = sin 2 x = f ( x). \ f ( x) is an even function. p/ 2

So,

p/ 2

-p/ 2

0 p/ 2

= 1

EXAMPLE 20

p/ 2

2 2 ò sin x dx = 2 ò sin x dx = 2

Prove that

0

æ 1 - cos 2x ö ç ÷ dx 2 è ø p/ 2

sin 2x ù é ò (1 - cos 2x) dx = êë x - 2 úû 0 0

æ 2-xö

ò log çè 2 + x ÷ø dx = 0.

-1

ò

=

p × 2 [CBSE 2005C]

Senior Secondary School Mathematics for Class 12 Pg-831

Definite Integrals

SOLUTION

831

æ 2-xö f ( x) = log ç ÷× è2+ xø -1 æ 2-xö æ2+ xö æ 2-xö Then, f ( - x) = log ç ÷ = - log ç ÷ = log ç ÷ = - f ( x). è2+ xø è 2-xø è2+ xø \ f ( x) is an odd function of x. Let

a

But, we know that

ò f ( x) dx = 0, when f ( x) is an odd function of x.

-a 1

æ 2-xö \ ò log ç ÷ dx = 0. è2+ xø -1 4

EXAMPLE 21

1 2

4

SOLUTION

ì 4x + 3 , if 1 £ x £ 2 if 2 £ x £ 4.

ò f ( x) dx , where f ( x) = íî 3 x + 5 ,

Evaluate

4

ò f ( x) dx = ò f ( x) dx + ò f ( x) dx 1

1 2

2

4

= ò ( 4x + 3) dx + ò ( 3 x + 5) dx 1

2

4

2 ù 2 é 3x = [2x 2 + 3 x]1 + ê + 5 x ú = ( 9 + 28) = 37. ë 2 û2

INTEGRALS OF MODULUS FUNCTIONS EXAMPLE 22

Evaluate: 2

(i)

(ii) ò|5 x - 3| dx

-1

SOLUTION

p

1

ò| x| dx

(iii) ò|cos x| dx

0

0

ì- x when - 1 £ x £ 0 (i) Clearly, | x| = í î x when 0 £ x £ 2. 2

\

0

2

ò| x| dx = ò| x| dx + ò| x| dx

-1

-1

0

0

2

é -x 2 ù é x2 ù ö 5 æ1 = ò ( -x) dx + ò x dx = ê ú + ê ú = ç + 2÷ = × 2 2 2 ø 2 è ë û -1 ë û 0 -1 0 3 ì ïï- (5 x - 3) when 0 £ x £ 5 (ii) Clearly, |5 x - 3| = í ï (5 x - 3) when 3 £ x £ 1. ïî 5 0

\

2

1

3/5

1

0

0 3/5

3/5 1

ò|5 x - 3| dx = ò|5 x - 3| dx + ò|5 x - 3| dx =

ò -(5 x - 3) dx + ò (5 x - 3) dx 0

3/5

Senior Secondary School Mathematics for Class 12 Pg-832

832

Senior Secondary School Mathematics for Class 12 3/5

1

é é 5x2 ù 5x2 ù = ê 3x +ê - 3xú ú 2 û ë ë 2 û 3/5 0 æ 9 9 ö æ -1 9 ö 13 =ç - ÷+ç + ÷= × è 5 10 ø è 2 10 ø 10 p ì ïï cos x when 0 £ x £ 2 (iii) Clearly, |cos x| = í ï- cos x when p £ x £ p. ïî 2 p

p/ 2

p

0

0 p/ 2

p/ 2 p

ò|cos x|dx = ò|cos x|dx + ò|cos x|dx

\

=

ò cos x dx + ò - cos x dx p/ 2

0

p/ 2

= [sin x]0

p

- [sin x]p/ 2 = (1 + 1) = 2.

4

EXAMPLE 23

Evaluate

ò f ( x) dx , where f ( x) =|x - 1| + |x - 2| + |x - 3|.

[CBSE 2013]

1 4

SOLUTION

2

3

4

1

2

3

ò f ( x) dx = ò f ( x) dx + ò f ( x) dx + ò f ( x) dx 1

2

3

1

2 4

= ò {( x - 1) - ( x - 2) - ( x - 3)} dx + ò {( x - 1) + ( x - 2) - ( x - 3)} dx + ò {( x - 1) + ( x - 2) + ( x - 3)} dx 2

3

3

4

= ò ( - x + 4) dx + ò x dx + ò ( 3 x - 6) dx 1

EXAMPLE 24

2

2

3

p/ 2

(i)

ò|sin x|dx

-p/ 2

SOLUTION

3

4

é -x 2 ù é x2 ù é 3x2 ù æ5 5 =ê + 4x ú + ê ú + ê - 6x ú = ç + + ë 2 û1 ë 2 û 2 ë 2 û3 è2 2 Evaluate: 1

(ii)

òe

|x |

9 ö 19 ÷= × 2ø 2

1

(iii)

dx

ò|2x + 1|dx

-2

-1

(i) Clearly, |sin x|is an even function of x. p/ 2

\

p/ 2

ò|sin x| dx = 2 ò|sin x|dx

-p/ 2

0 p/ 2

pù é = 2 ò sin x dx êQ sin x ³ 0, when 0 £ x £ ú 2û ë 0 p/ 2

= [-2 cos x]0

= 2.

Senior Secondary School Mathematics for Class 12 Pg-833

Definite Integrals

833

(ii) Clearly, e|x | is an even function of x. 1

òe

\

|x |

-1

1

dx = 2ò e|x | dx 0 1

= 2ò ex dx

[Q |x| = x , when 0 £ x £ 1 ]

0

1

= [2ex ]0 = ( 2e - 2) = 2( e - 1). 1 é ù é 1 ù (iii) ê -2 £ x < - Þ 2x + 1 < 0ú and ê - £ x £ 1 Þ 2x + 1 ³ 0ú 2 ë û ë 2 û 1

-1/2

1

-2

-2 -1/2

-1/2 1

ò|2x + 1|dx =

\

=

ò|2x + 1|dx + ò|2x + 1|dx

ò -( 2x + 1) dx + ò ( 2x + 1) dx

-2

= [- x

2

-1/2 - x]-2

-1/2 1 + [x 2 + x]-1/2

é æ 1 1ö æ 1 1öù = ç - + ÷ - ( - 4 + 2) + ê 2 - ç - ÷ ú è 4 2ø è 4 2ø û ë 1 9 9 = + 2+ = × 4 4 2

2p

ò|sin x|dx.

EXAMPLE 25

Evaluate

SOLUTION

We know that sin x is positive, when 0 £ x £ p and sin x is negative when p £ x £ 2p. ì sin x , when 0 £ x £ p \ |sin x| = í î- sin x , when p £ x £ 2p.

0

\

2p

p

2p

0

0 p

p 2p

ò|sin x|dx = ò|sin x|dx + ò|sin x|dx = ò sin x dx + ò ( - sin x) dx 0

p

p

2p

= [- cos x]0 + [cos x]p = ( 2 + 2) = 4. p/ 2

EXAMPLE 26

Evaluate

ò f ( x) dx , where f ( x) = sin|x| + cos|x|.

-p/ 2

SOLUTION

f ( x) = sin|x| + cos|x| Þ f ( - x) = sin|- x| + cos|- x| = sin|x| + cos|x| = f ( x) Þ f ( x) is an even function. p/ 2

\ I=

ò f ( x) dx

-p/ 2

[CBSE 2000]

Senior Secondary School Mathematics for Class 12 Pg-834

834

Senior Secondary School Mathematics for Class 12 p/ 2

=2

p/ 2

ò f ( x) dx = 2 ò {sin|x| + cos|x|} dx

0 p /2

0

p [Q |x| = x in 0 < x < ] 2

= 2 ò (sin x + cos x) dx 0

p pö éæ ù = 2 × [- cos x + sin x]p0 /2 = 2 ê ç - cos + sin ÷ - ( - cos 0 + sin 0) ú 2 2 è ø ë û = 4.

EXERCISE 16C Prove that p/ 2

1.

ò

0

p/ 2

3. (i)

(ii)

4. (i)

(ii)

3

sin x

ò

0 (sin p/ 2

3

0 (sin p/ 2

3

0 (sin p/ 2

7

ò

ò

0 p/ 2

7.

[CBSE 2000C]

5

dx =

p 4

[CBSE 2000C]

cos4 x 4

(sin x + cos x) sin

3/2

dx =

x

p/ 2

p 4

6. p 4

ò

0 p/ 2

cos1/4 x (sin

1/4

ò

dx p = (1 + tan x) 4

ò

p = 3 (1 + tan x) 4

ò

dx p = [CBSE 2006C, ‘07C, ‘11] 16. (1 + tan x ) 4

0

3/2

x + cos

3/2

x)

tan x

dx

dx =

dx =

p 4

8.

ò

(sin x + cosnx)

ò

( tan x + cot x )

0 p/ 2

10.

x + cos1/4 x)

sin nx

( tan x + cot x )

0 p/ 2

15.

p 4

ò

0 p/ 2

13.

dx =

sin5 x

p 4

dx =

p 4

[CBSE 2004C]

7

x + cos x)

4

p 4

dx =

[CBSE 1998, 2000C, ‘04]

(sin

0 p/ 2

11.

sin 7 x

p 4

sin x ( sin x + cos x )

ò

0 p/ 2

9.

=

(sin x + cos x)

p/ 2

ò

3

x + cos x)

ò

0

dx =

x + cos x)

5

2.

3

cos 3 x dx

ò

0

5.

p/ 2

cos x p dx = 4 (sin x + cos x)

0

n

cot x p/ 2

[CBSE 2003C, ‘06C]

dx =

12.

ò

0 p/ 2

14.

dx =

p 4

dx p = (1 + cot x) 4

ò

dx

ò

cot x

0 (1 + p/ 2 0

p 4

3

cot x)

=

(1 + cot x )

p 4

dx =

p 4

Senior Secondary School Mathematics for Class 12 Pg-835

Definite Integrals p/ 2

17.

ò

0

tan x (1 + tan x )

1

dx =

p/ 2

p 4

18.

0 2

1 19. ò x(1 - x) dx = 42 0 5

p

21.

ò x cos x dx = 2

0 p

22.

23.

x tan x

ò

0

x tan x

ò ( sec x + cos x) dx =

0 p

[CBSE 2005C]

[CBSE 2008, ‘09C, ‘11C]

p

p2 4

25.

p x dx = 4 x + a - x)

ò(

0 p

ò sin

2

ò sin

2m

0 p

ö p

ò ( 2 log cos x - log sin 2x) dx = - 2 (log 2)

27.

0

a

x

a

æp

x sin x

ò (1 + sin x) dx = pçè 2 - 1÷ø 0 p/ 2

p 28. ò dx = 2 4 + + ( 1 x )( 1 x ) 0

32.

16 2 15

p2 4

x

¥

31.

2 - x dx =

0

p2 26. ò dx = 2 2 2 0 (1 + sin x)

30.

òx

cos2 x 1 dx = log ( 2 + 1) (sin x + cos x) 2

p

24.

20.

(sin x - cos x) dx = 0 (1 + sin x cos x)

p2 4

ò ( sec x cosec x) dx =

0 p/ 2

ò

835

29.

ò

0

dx 2

x+ a -x

2

=

p 4

[CBSE 2008C]

x cos 3 x dx = 0 x cos2m +1 x dx = 0, where m is a positive integer

0 p/ 2

33.

ò (sin x - cos x) log (sin x + cos x) dx = 0

0 p/ 2

34.

p

p log (sin 2x) dx = - (log 2) 2

35.

ò log (1 + cos x) dx = -p(log 2)

37.

ò

0

0 p/ 2

p

36.

0 3p /8

38.

ò

p /8 p /3

39.

ò x log (sin x) dx = -

ò log (tan x + cot x) dx = p(log 2) 0

cos x p dx = 8 (cos x + sin x)

1 p [CBSE 2005, ‘06C, ‘07, ’11, ’12C] dx = 12 1 ( + tan x ) p /6

ò

p2 (log 2) 2

Senior Secondary School Mathematics for Class 12 Pg-836

836

Senior Secondary School Mathematics for Class 12 3p /4

ò

40.

p /4 3 a/4

ò

42.

3p /4

dx =2 (1 + cos x)

a/4 p/2

p /4 4

a x dx = 4 ( a - x + x)

43.

p

ò x cot x dx = 2 (log 2)

44.

45.

0

1

46.

0 a

48.

òx

-a p

50.

3

òx

12

47.

x dx = p( 2 - 1) (1 + sin x)

x 3 dx = 2 5 - x + x)

p sin -1 x ö÷ dx = (log 2) ÷ x 2 ø 0è

ò çç ò

log (1 + x) (1 + x 2)

a 2 - x 2 dx = 0

49.

ò (sin

-p 1

9

sin x dx = 0

51.

dx =

p (log 2) 8 [CBSE 2011C]

p

òe

75

x + x125) dx = 0

|x |

dx = 2( e - 1)

-1 8

ò|x + 1|dx = 6

53. ò|x - 5|dx = 17

ò|cos x|dx = 4.

55.

-2 2p

54.

1 1æ

0

-p 2

52.

ò(

1

p dx = - (log 2) 2 1 - x2

log x

ò

ò

41.

0 p /4

ò|sin x|dx = ( 2 -

2)

-p /4

0

ì2x + 1, when 1 £ x £ 2 56. Let f ( x) = í 2 îx + 1, when 2 £ x £ 3. 3 34 Show that ò f ( x) dx = × 3 1 ì 3 x 2 + 4, when 0 £ x £ 2 57. Let f ( x) = í î9x - 2, when 2 £ x £ 4. 4

Show that ò f ( x) dx = 66. 0

4

58. Prove that

ò {|x| + |x - 2| + |x - 4| dx} = 20 0

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 16C) p/2

11. I =

ò

0

p/2

14. I =

ò

0

dx æ sin x ö ç1+ ÷ ç cos x ÷ø è dx 3 ö æ ç 1 + cos x ÷ 3 ÷ ç sin x ø è

p/2

=

ò

0

cos x dx. (sin x + cos x )

p/2

=

ò

0

sin 3 x (sin 3 x + cos 3 x )

dx.

[CBSE 2013]

Senior Secondary School Mathematics for Class 12 Pg-837

Definite Integrals sin x cos x

p/2

17. I =

ò

0

p/2

sin x ö÷ cos x ÷ø

æ ç1+ ç è

ò

dx =

0

sin x ( sin x +

1

1

0

0

cos x )

837

dx.

1

19. I = ò ( 1 - x )[1 - ( 1 - x )]5 dx = ò ( 1 - x )x5 dx = ò ( x5 - x 6 ) dx. 0

2

2

2

0

0

20. I = ò ( 2 - x )[2 - ( 2 - x )]1/2 dx = ò ( 2 - x ) x dx = ò ( 2 x - x 3/2 ) dx. 0

p

p

p

0 p

0

p

p

21. I = ò x cos 2 x dx and I = ò ( p - x ) cos 2 ( p - x ) dx = ò ( p - x ) cos 2 x dx. 0

p

p p2 × \ 2 I = p ò cos x dx = × ò ( 1 + cos 2 x ) dx Þ I = 2 4 0 0 2

p

22. I = ò x sin 2 x dx and I = ò ( p - x ) sin 2 ( p - x ) dx = ò ( p - x ) sin 2 x dx. 0

0

p

0

p

p \ 2 I = p × ò sin x dx = × ò ( 1 - cos 2 x ) dx. 2 0 0 p

24. I = ò

2

p

x sin x 2

0 ( 1 + cos x ) p

dx and I = ò

0 (1 +

( 1 + cos 2 x )

0

sin x

\ 2I = p × ò

( p - x ) sin x

cos 2 x )

dx.

dx. Now, put cos x = t.

p

p æ ö sin x 1 ÷ dx dx = p × ò 20 çç 1 ÷ 1 1 ( sin ) sin + x + x è ø 0 0

25. 2 I = p × ò

p

= p2 - p × ò

( 1 - sin x ) 2

0 ( 1 - sin x )

p

dx = p 2 - p × ò

0

p

( 1 - sin x ) cos 2 x

dx

= p 2 - p × ò ( sec2 x - sec x tan x ) dx. 0

p/2

26. 2 I = 2 p ×

ò

0

dx ( 1 + sin 2 x )

×

Divide num. and denom. by cos 2 x and put tan x = t. p/2

28. Putting x = tan q, we get I =

ò

0

sin q dq. (cos q + sin q)

29. Put x = a sin q and dx = a cos q dq. 30. Apply

a

a

0

0

ò f ( x ) dx = ò f ( a - x ) dx. a

a

0

0

In Q. 31 to Q. 33, apply ò f ( x ) dx = ò f ( a - x ) dx to get I = - I.

Senior Secondary School Mathematics for Class 12 Pg-838

838

Senior Secondary School Mathematics for Class 12 p

34. Putting 2x = t , we get I =

1 × log (sin t ) dt = 2 ò0

p

p/2

0

0

35. 2 I = p × ò log (sin x ) dx = 2 p ×

p/2

ò log (sin x ) dx.

[by Thm. 8]

p

p/2

0

0

0

p/2

æ sin x

ò log çç cos x +

cos x ö ÷ dx = sin x ÷ø

è é p/2 = - ê ò log (sin x ) dx + êë 0 0

p/2

æ

[by Thm. 8]

0

p

36. 2 I = ò log ( 1 - cos 2 x ) dx = 2 ò log (sin x ) dx = 4 37. I =

ò log (sin t ) dt.

ò log (sin x ) dx.

1

[Thm. 8]

ö

ò log ççè sin x cos x ÷÷ø dx 0

p/2

ù

ò log (cos x ) dx úú × û

0

b

b

a

a

In Q. 38 to Q. 43, apply, ò f ( x ) dx = ò f ( a + b - x ) dx. 44. Integrating by parts, we get p/2

I = [x log (sin x )] 0

-

p/2

p/2

0

0

ò log (sin x ) dx = -

ò log (sin x ) dx.

45. Put x = sin q and integrate by parts. 46. Put x = sin q.

47. Put x = tan q.

In Q.48 to Q.50, f ( - x ) = - f ( x ) Þ I = 0. 51. I =

0

1

-1

0

ò e -xdx + ò e xdx.

2

4

58. I = ò {| x| + |x - 2| + | x - 4|} dx + ò| x| + | x - 2| + | x - 4|} dx 0 2

2

4

= ò {x - ( x - 2 ) - ( x - 4 )} dx + ò {x + ( x - 2 ) - ( x - 4 )} dx 0 2

2

4

= ò ( x - x + 2 - x + 4 ) dx + ò ( x + x - 2 - x + 4 ) dx 0 2

4

2

0

2

= ò ( - x + 2 ) dx + ò ( x + 2 ) dx.

Definite Integral as the Limit of a Sum Let f ( x) be a continuous real-valued function, defined in the closed interval [a , b]. Then we define b

h[ f ( a) + f ( a + h) + f ( a + 2h) + ¼ + ò f ( x) dx = hlim ®0 a

f [a + (n - 1) h)], where nh = ( b - a).

Senior Secondary School Mathematics for Class 12 Pg-839

Definite Integrals

839

b

The method of evaluating ò f ( x) dx by using the above definition is called a

integration from first principles. Some Useful Results for Direct Applications 1 (i) 1 + 2 + 3 + ¼ + (n - 1) = n(n - 1). 2 1 (ii) 12 + 22 + 3 2 + ¼ + (n - 1) 2 = (n - 1) n( 2n - 1). 6 2

ìn(n - 1) ü (iii) 1 3 + 2 3 + 3 3 + ¼ + (n - 1) 3 = í ý × î 2 þ (iv) a + ar + ar 2 + ¼ + ar n-1 =

a(r n - 1) × (r - 1)

(v) sin a + sin ( a + h) + sin ( a + 2h) + ¼ + sin [a + (n - 1) h] ì æ nh ö æ n - 1ö ü sin ía + ç ÷ hý sin ç ÷ è 2 ø þ è 2ø î = × sin ( h/2) (vi) cos a + cos( a + h) + cos( a + 2h) + ¼ + cos[a + (n - 1) h] æ nh ö ì (n - 1) ü cos ía + hý sin ç ÷ 2 þ è 2ø× î = sin ( h/2)

SOLVED EXAMPLES EXAMPLE 1

Evaluate the following integrals as limit of sums: 5

3

(i) ò ( x + 1) dx

(ii) ò ( 2x + 3) dx

0

SOLUTION

[CBSE 2000C]

1

(i) Let f ( x) = ( x + 1); a = 0; b = 5 and nh = (5 - 0) = 5. Then, 5

h[ f ( 0) + f ( 0 + h) + f ( 0 + 2h) + ¼ + f {0 + (n - 1) h}] ò ( x + 1) dx = hlim ®0 0

= lim h[1 + ( h + 1) + ( 2h + 1) + ¼ + {(n - 1) h + 1}] h® 0

= lim h[n + {h + 2h + 3 h + ¼ + (n - 1) h}] h® 0

= lim h[n + {1 + 2 + 3 + ¼ + (n - 1)} h] h® 0

n(n - 1) ù nh(nh - h) ù é é = lim h ên + hú = lim ênh + úû h® 0 ë 2 2 û h® 0 ë

Senior Secondary School Mathematics for Class 12 Pg-840

840

Senior Secondary School Mathematics for Class 12

5(5 - h) ù é = lim ê5 + [Q nh = 5] h® 0 ë 2 úû 35 = × 2 (ii) Let f ( x) = ( 2x + 3); a = 1; b = 3 and nh = ( 3 - 1) = 2. 3

Then, ò ( 2x + 3) dx = lim h[ f (1) + f (1 + h) + ¼ + f {1 + (n - 1) h}] h® 0

1

= lim h[5 + (5 + 2h) + (5 + 4h) + ¼ + {5 + 2(n - 1) h}] h® 0

[Q f (1) = 5 , f (1 + h) = 5 + 2h, etc.] = lim h[5n + {2h + 4h + ¼ + 2(n - 1) h}] h® 0

= lim h[5n + 2{1 + 2 + 3 + ¼ + (n - 1)} h] h® 0

n(n - 1) ù é = lim h ê5n + 2 × hú h® 0 ë 2 û = lim [5nh + nh(nh - h)] h® 0

[Q nh = 2]

= lim [10 + 2( 2 - h)] h® 0

= 14. EXAMPLE 2

Evaluate the following integrals as limit of sums: 2

(i)

3

2 ò ( x + 1) dx

(ii) ò ( x 2 + x) dx

0

SOLUTION

[CBSE 2006C]

1

(i) Let f ( x) = ( x 2 + 1); a = 0; b = 2 and nh = ( 2 - 0) = 2. 2

\

ò(x 0

2

+ 1) dx = lim h[ f ( 0) + f ( 0 + h) + ¼ + f {0 + (n - 1) h}] h® 0

= lim h[1 + ( h2 + 1) + ( 4h2 + 1) + ( 9h2 + 1) + h® 0

¼ + {(n - 1) 2 h2 + 1}] [Q f ( 0) = 1, f ( 0 + h) = ( h2 + 1), f ( 0 + 2 h) = ( 4h2 + 1), etc.] = lim h[n + {12 + 22 + 3 2 + ¼ + (n - 1) 2} h2] h® 0

(n - 1)n ( 2n - 1) 2 ù é = lim h ên + ×h ú h® 0 ë 6 û (nh - h)nh ( 2nh - h) ù é = lim ênh + úû h® 0 ë 6 ( 2 - h) 2 ( 4 - h) ù é = lim ê 2 + úû h® 0 ë 6 14 = × 3

[Q nh = 2]

Senior Secondary School Mathematics for Class 12 Pg-841

Definite Integrals

841

(ii) Let f ( x) = ( x 2 + x); a = 1; b = 3 and nh = ( 3 - 1) = 2. 3

ò(x

\

2

+ x) dx = lim h[ f (1) + f (1 + h) + f (1 + 2h) + ¼ h® 0 + f {1 + (n - 1) h}]

1

= lim h[(12 + 1) + {(1 + h) 2 + (1 + h)} + {(1 + 2h) 2 h® 0

+ (1 + 2h)} + ¼ + {1 + (n - 1) h} 2 + {1 + (n - 1) h}] = lim h[2n + h2{12 + 22 + ¼ + (n - 1) 2} + 3 h{1 + 2 + ¼ + (n - 1)}] h® 0

(n - 1)n ( 2n - 1) n(n - 1) ù é = lim h ê 2n + h2 × + 3h× h® 0 ë 6 2 úû (nh - h)nh ( 2nh - h) 3 é ù = lim ê 2nh + + × nh(nh - h) ú h® 0 ë 6 2 û ( 2 - h) 2 ( 4 - h) 3 é ù = lim ê 4 + + × 2( 2 - h) ú [Q nh = 2] h® 0 ë 6 2 û 38 = × 3 1

EXAMPLE 3

Evaluate

ò( 3x

2

+ 2x + 1) dx as limit of sums.

[CBSE 2007C]

0

SOLUTION

Let f ( x) = ( 3 x 2 + 2x + 1); a = 0, b = 1 and nh = (1 - 0) = 1. 1

Then,

ò( 3x

2

+ 2x + 1) dx

0

= lim h[ f ( 0) + f ( 0 + h) + f ( 0 + 2h) + ¼ + f {0 + (n - 1) h}] h® 0

= lim h[1 + ( 3 h2 + 2h + 1) + ( 3 × 22 h2 + 2 × 2h + 1) + ¼ h® 0

+ { 3 × (n - 1) 2 h2 + 2 × (n - 1) h + 1}] = lim h[n + 3 h2{12 + 22 + 3 2 + ¼ + (n - 1) 2} h® 0

+ 2 h{1 + 2 + 3 + ¼ + (n - 1)}] ( n 1 ) n ( 2 n - 1) n(n - 1) ù é = lim h ên + 3 h2 × + 2h × h® 0 ë 6 2 úû 1 é ù = lim ênh + (nh - h)(nh)( 2nh - h) + (nh)(nh - h) ú h® 0 ë 2 û 1 é = lim ê1 + (1 - h) × 1 × ( 2 - h) + 1 × (1 - h)] [Q nh = 1] h® 0 ë 2 = 3. 1

EXAMPLE 4

Evaluate

òe

x

dx as limit of sums.

-1

SOLUTION

Let f ( x) = ex and nh = {1 - ( -1)} = 2. Then,

Senior Secondary School Mathematics for Class 12 Pg-842

842

Senior Secondary School Mathematics for Class 12 1

òe

-1

x

dx = lim h[ f ( -1) + f ( -1 + h) + ¼ + f {-1 + (n - 1) h}] h® 0

( -1 + h )

= lim h[e-1 + e h® 0

( -1 + 2 h )

{ -1 +( n-1 )h}

+e

+ ¼+ e

] ( -1 + h )

[Q f ( -1) = e-1 , f ( -1 + h) = e -1

h

= lim e h[1 + e + e h® 0

2h

( n-1 ) h

+ ¼+ e

, etc.]

]

é ( eh) n - 1 ù 1 h[enh - 1] 1 = × lim h ê h ú = × lim e h ® 0 ë ( e - 1) û e h ® 0 ( eh - 1) h( e2 - 1) 1 = × lim e h ® 0 ( eh - 1)

[Q nh = 2]

h( e2 - 1) 1 = × lim e h® 0 ì ü h2 h 3 + + ¼ý - 1 í1 + h + ë ë 2 3 î þ e2 - 1 1 1 æ 1ö = ( e2 - 1) = ç e - ÷ × = × lim 2 ö e e h® 0 æ eø è ç1 + h + h + ¼÷ ÷ ç ë ë 2 3 ø è b

EXAMPLE 5

Evaluate

ò sin x dx

from first principles.

a

SOLUTION

Let f ( x) = sin x and let nh = ( b - a). Then, b

h[ f ( a) + f ( a + h) + ¼ + f {a + (n - 1) h}] ò sin x dx = hlim ®0 a

= lim h[sin a + sin ( a + h) + ¼ + sin {a + (n - 1) h}] h® 0

ì æ nh ö æ n - 1ö ü sin ía + ç ÷ hý sin ç ÷ è 2 ø þ è 2ø î = lim h × h® 0 sin ( h/2) é 1 1 ü ( h/2) ù æ nh ö ì = lim ê 2 sin ía + nh - hý sin ç ÷ ´ ú h® 0 ë 2 2 þ è 2 ø sin ( h/2) û î

é 1 1 ü ( h/2) ù æ b - aö ì = lim ê 2 sin ía + ( b - a) - hý sin ç ÷´ h® 0 ë in ( h/2) úû 2 2 2 s þ è ø î [Q nh = ( b - a)] ( h/2) ìæ b + a ö 1 ü æ b - aö = lim 2 sin íç ÷ × lim ÷ - hý sin ç h® 0 h ® 0 2 2 2 sin ( h/2) ø è ø îè þ æ b - aö æ b + aö = 2 sin ç ÷ ÷ sin ç è 2 ø è 2 ø = (cos a - cos b) [Q 2 sin A sin B = cos( A - B) - cos( A + B)].

Senior Secondary School Mathematics for Class 12 Pg-843

Definite Integrals

843

EXERCISE 16D Evaluate each of the following integrals as the limit of sums: 2

1.

2

ò ( x + 4) dx

2.

0

3

4.

ò(x

2

3

ò(x

5.

2

7.

2 ò ( x + 5 x) dx [CBSE 2008C, ’10]

3

9.

- 5) dx

2

+ 2x) dx

2

+ 5 x) dx

[CBSE 2009C]

11.

[CBSE 2012C]

ò(x

2

- 3 x + 2) dx

2

2

3

2 ò ( x + x) dx [CBSE 2005]

13.

0

ò ( 2x

2

+ 3 x + 5) dx

[CBSE 2007]

0

1

2

14. ò|3 x - 1| dx

15.

0

-x

ò ( 2x

1

0

3

2

ò( 3x

4

3 ò x dx

òe

dx

1

2

16.

ò( 3x 4

+ 2x) dx

1

12.

2

2

3

10.

òx

1 5

+ 1) dx [CBSE 2001]

0

8.

3.

1

0

6.

3

ò ( 3 x - 2) dx

òe

x

dx

0 b

dx

17.

ò cos x dx a

1

ANSWERS (EXERCISE 16D)

1. 10 9. 16.

2.

112 3

5 2

10. 4

( e2 - 1)

26 3 14 11. 3 3.

4. 12 12.

14 3

5. 102 13.

93 2

6. 18 14. 1

7. 78 15. ( e2 - 1)

17. (sin b - sin a)

e3

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 16D) 1

1/3

1

0

0

1/3

14. ò|3 x - 1| =

ò ( 1 - 3 x ) dx + ò ( 3 x - 1) dx = I1 + I 2 .

8.

86 3

Senior Secondary School Mathematics for Class 12 Pg-844

844

Senior Secondary School Mathematics for Class 12

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 4

1.

òx

x dx = ?

1

(a) 12.8

(b) 12.4

(c) 7

(d) none of these

(b) 7

(c)

56 9

(d)

(c)

2 8 5

(d) none of these

(c)

p 4

(d) none of these

(c)

p 4

(d) none of these

2

2.

ò

6x + 4 dx = ?

0

(a) 1

3.

64 9

dx =? 5x + 3

ò

0

2 (a) ( 8 - 3 ) 5 1

4.

60 9

2 (b) ( 8 + 5

3)

1

ò (1 + x 2) dx = ? 0

(a) 2

5.

p 2

dx

ò

4 - x2

0

(b) =?

(a) 1 8

6.

òx

p 3

(b) sin -1

1 2

1 + x 2 dx = ?

3

(a) 1

7.

19 3

(b)

19 6

(c)

38 3

(d)

9 4

(b)

p 4

(c)

p 8

(d)

p 16

(b)

1 3 e 3

(c)

1 3 ( e - 1) 3

(d) none of these

x3

ò (1 + x 8) dx = ? 0

(a)

p 2

e

8.

(log x) 2 dx = ? x 1

ò

(a)

1 3

Senior Secondary School Mathematics for Class 12 Pg-845

Definite Integrals p

9.

2

ò cot x dx = ?

p

4

(a) log 2 p

10.

4

ò tan 0

p

11.

2

p

(c)

pö æ (b) ç1 + ÷ 4ø è

pö æ (c) ç1 - ÷ 2ø è

(b) p

(c)

1 (b) log 3 2

(c) - log 2

p

x dx = ?

p 2

(d) 1

3

1 (a) log 2 2 2

ò cos p

3

4

(d) none of these

x dx = ? (b)

3 4

(c)

2 3

(d) none of these

etan x

ò cos2x dx = ? 0

(a) ( e - 1) p

2

(b) ( e + 1)

æ1 ö (c) ç + 1÷ èe ø

æ1 ö (d) ç - 1÷ èe ø

cos x

ò (1 + sin 2x) dx = ? 0

(a) 2

p

ò

1

p 2

sin ( 1 x) x2

p

p

(b)

p 4

(c) p

(d) none of these

(b)

1 2

(c)

3 2

(d) none of these

dx = ?

(a) 1 17.

p 4

2

(a) 1

16.

pö æ (d) ç1 + ÷ 2ø è

2

0

15.

(d) none of these

ò cosec x dx = ?

p

14.

1 log 2 2

(b) 2 log 2

ò cos x dx = ? (a)

13.

2

pö æ (a) ç1 - ÷ 4ø è

0

12.

845

dx

ò (1 + sin x) = ? 0

(a)

1 2

(b) 1

(c) 2

(d) 0

Senior Secondary School Mathematics for Class 12 Pg-846

846

Senior Secondary School Mathematics for Class 12 p

18.

2

ò(

sin x cos x) 3 dx = ?

0

(a) 1

19.

2 9

(b)

2 15

(c)

8 45

(d)

5 2

xex

ò (1 + x) 2 dx = ? 0

æe ö (a) ç - 1÷ è2 ø p

20.

2

òe 0

p

4

ò

(c) e( e - 1)

(d) none of these

x æ1 +

sin x ö çç ÷÷ dx = ? è 1 + cos x ø

(a) 0 21.

(b) ( e - 1)

p 4

(c) e

(b) 1

(c) 2

(b)

p

2

(d) ( e

p

2

- 1)

1 + sin 2x dx = ?

0

(a) 0 p

22.

2

ò

(d) 2

1 + cos 2x dx = ?

0

(a) 2 1

23.

(b)

3 2

(c)

3

(d) 2

(1 - x)

ò (1 + x) dx = ? 0

1 (a) log 2 2 p

24.

2

ò sin 0

(a) p

25.

2

(b) ( 2 log 2 + 1)

(c) ( 2 log 2 - 1)

æ1 ö (d) ç log 2 - 1÷ è2 ø

x dx = ?

p 3

(b)

p 4

(c)

p 2

(d)

2p 3

(b)

5 12

(c)

1 3

(d)

7 12

(b)

3 4

(c)

5 6

(d)

3 5

6

ò cos x cos 2x dx = ? 0

(a) p

26.

1 4

2

ò sin x sin 2x dx = ? 0

(a)

2 3

Senior Secondary School Mathematics for Class 12 Pg-847

Definite Integrals

847

p

27.

ò (sin 2x cos 3 x) dx = ? 0

(a) 1

28.

4 5

(b) -

4 5

(c)

5 12

(d) -

12 5

dx

ò ( ex + e- x ) = ? 0

pö æ (a) ç1 - ÷ 4ø è 9

29.

dx

ò (1 +

x)

0

30.

(d) tan -1 e -

p 4

(c) tan -1 e +

(b) ( 3 + 2 log 2)

(c) ( 6 - 2 log 4)

(d) ( 6 + 2 log 4)

æp ö (b) ç - 1÷ è2 ø

æp ö (c) ç + 1÷ è2 ø

(d) none of these

p 3 3

(d) none of these

æp ö (b) ç - 1÷ è2 ø

ö æp (c) ç + 1÷ ø è2

(d) none of these

(b) (log 2 - 1)

(c) ( 2 log 2 - 1)

(d) ( 2 log 2 + 1)

(c) 2ap

(d) none of these

p 2

(d) none of these

=?

(a) ( 3 - 2 log 2) p

p 4

(b) tan -1 e

2

ò x cos x dx = ? 0

(a) 1

31.

dx

ò (1 + x + x 2) = ? 0

(a) 1

32.

p 3

(b)

p 3

(c)

1-x dx = ? 1+ x

ò

0

(a) 1

33.

p 2

p 2

(1 - x)

ò (1 + x) dx = ? 0

(a) (log 2 + 1) a

34.

a-x dx = ? a+x

ò

-a

(a) ap 2

35.

ò

(b)

ap 2

2 - x 2 dx = ?

0

(a) p

(b) 2p

(c)

Senior Secondary School Mathematics for Class 12 Pg-848

848

Senior Secondary School Mathematics for Class 12 2

36.

ò| x| dx = ?

-2

(a) 4

(b) 3.5

(c) 2

(d) 0

1

37. ò| 2x - 1|dx = ? 0

(a) 2

(b)

1 2

(c) 1

(d) 0

(b)

7 2

(c)

9 2

(d) 4

1

38.

ò| 2x + 1|dx = ?

-2

(a)

5 2

1

39.

|x| dx = ? x -2

ò

(a) 3

(b) 2.5

(c) 1.5

(d) none of these

(b) 2a

(c)

2a 3 3

(d) none of these

a

40.

ò x|x|dx = ?

-a

(a) 0 p

41. ò|cos x|dx = ? 0

(a) 2

3 2

(c) 1

(d) 0

(b) 4

(c) 1

(d) none of these

(c) 0

(d)

(c) p

(d) 0

(b)

2p

42.

ò|sin x|dx = ? 0

(a) 2 p

43.

2

sin x

ò (sin x + cos x) dx = ? 0

(a) p p

44.

2

ò( 0

cos x cos x + sin x )

(a) p

45.

2

(b)

p 2

ò (sin 4x + cos4x) dx = ? 0

p 4

dx = ? (b)

sin 4 x

p 2

p 4

Senior Secondary School Mathematics for Class 12 Pg-849

Definite Integrals

(a) p

46.

p 4

0

cos 4 x 1

1

(sin 4 x + cos 4 x)

p

(c) 1

(d) 0

dx = ?

(a) 0

47.

p 2

1

2

ò

(b)

849

(b) 1

(c)

p 4

(d) none of these

sin nx

2

ò (sin nx + cosnx) dx = ? 0

(a) p

48.

p 2

(b)

2

ò( 0

cot x cot x + tan x )

(a) 0 p

49.

3

(c) 1

(d) 0

dx = ? (b)

2

p 4

p 2

(c)

p 4

(d) none of these

tan x

ò ( 3 tan x + 3 cot x ) dx = ? 0

(a) 0 p

50.

2

(b)

p 4

(c)

p 2

(d) p

(b)

p 2

(c)

p 4

(d) p

(b)

p 4

(c)

p 2

(d) p

(b) 0

(c)

p 2

(d) none of these

(c)

p 4

(d) p

1

ò (1 + tan x) dx = ? 0

(a) 0 p

51.

2

ò (1 + 0

1 dx = ? cot x )

(a) 0 p

52.

2

1

ò (1 + tan 3x) dx = ? 0

(a) p

53.

p 4 sec5 x

2

ò ( sec5x + cosec5x) dx = ? 0

(a)

p 2

(b) 0

Senior Secondary School Mathematics for Class 12 Pg-850

850

Senior Secondary School Mathematics for Class 12 p

54.

2

ò (1 +

cot x cot x )

0

p 4

(a) p

55.

2

dx = ? (b)

p 2

(c) 0

(d) 1

p 4

(d) p

tan x

ò (1 + tan x) dx = ? 0

(a) 0 p

56.

òx

4

(b) 1

(c)

(b) p

(c) 0

(d) none of these

(c) 2p

(d) 0

sin x dx = ?

-p

(a) 2p p

57.

òx

3

cos 3 x dx = ?

-p

(a) p p

58.

(b)

p 4

ò sin x dx = ? 5

-p

(a) -2

59.

òx

3

3p 4

(b) 2p

(c)

5p 16

(d) 0

40 3

(c)

5 6

(d) 0

(c) 0

(d) 1

(1 - x 2) dx = ?

-1

(a) a

60.

40 3

(b)

æa-xö

ò log çè a + x ÷ø dx = ?

-a

(a) 2a p

61.

ò (sin

61

(b) a

x + x123) dx = ?

-p

(a) 2p

(b) 0

(c)

p 2

(d) 125p

p

62.

ò tan x dx = ?

-p

(a) 2

(b)

1 2

(c) -2

(d) 0

Senior Secondary School Mathematics for Class 12 Pg-851

Definite Integrals 1

63.

ò log ( x +

851

x 2 + 1) dx = ?

-1

(a) log p

1 2

(b) log 2

1 (c) log 2 2

(d) 0

(b) 2

(c) -1

(d) none of these

2

ò cos x dx = ?

64.

- p2

(a) 0 a

65.

ò( 0

x dx = ? x + a - x) (a)

p

66.

a 2

(b) 2a

(c)

2a 3

(d)

(c)

p log 2 8

(d) 0

a 2

4

ò log (1 + tan x) dx = ? 0

(a)

p 4

(b)

p log 2 4

a

67.

ò f ( x) dx = ?

-a

a

a

(a) 2ò { f ( x) + f ( - x)} dx

(b) 2ò { f ( x) - f ( - x)} dx

0

0

a

(c) ò { f ( x) + f ( - x)} dx

(d) none of these

0

68. Let [x] denote the greatest integer less than or equal to x. 1.5

Then, ò [x] dx = ? 0

1 (a) 2

(b)

3 2

(c) 2

(d) 3

69. Let [x] denote the greatest integer less than or equal to x. 1

Then, ò [x] dx = ? -1

(a) -1

(b) 0

(c)

1 2

(d) 2

1 6

(c)

1 3

(d)

2

70. ò|x 2 - 3 x + 2|dx = ? 1

(a)

-1 6

(b)

2 3

Senior Secondary School Mathematics for Class 12 Pg-852

852

Senior Secondary School Mathematics for Class 12 2p

71.

ò|sin x|dx = ?

p

(a) 0 1

ò

(b) 1

(c) 2

sin -1 x

2

dx = ? 3 (1 - x 2) 2 1 (a) ( p - log 2) 2 ö æp 1 (c) ç - log 2÷ ø è4 2

72.

(d) none of these

0

1

73.

ò sin

æp ö (b) ç - 2 log 2÷ 2 è ø (d) none of these

-1 æ

2x ö ç ÷ dx = ? è 1 + x2 ø

0

æp ö (b) ç - log 2÷ è2 ø

1 (a) ( p - log 2) 2

(c) ( p - 2 log 2)

(d) none of these

ANSWERS (OBJECTIVE QUESTIONS)

1. (b) 2. (c)

3. (a)

4. (c)

5. (d)

6. (a)

7. (d)

8 (a)

9. (c) 10. (a)

11. (c) 12. (b) 13. (c) 14. (a) 15. (b) 16. (a) 17. (c) 18. (c) 19. (a) 20. (c) 21. (b) 22. (a) 23. (c) 24. (b) 25. (b) 26. (a) 27. (b) 28. (d) 29. (c) 30. (b) 31. (c) 32. (b) 33. (c) 34. (a) 35. (c) 36. (a) 37. (b) 38. (c) 39. (d) 40. (a) 41. (a) 42. (b) 43. (d) 44. (b) 45. (a) 46. (c) 47. (b) 48. (c) 49. (b) 50. (c) 51. (b) 52. (a) 53. (c) 54. (a) 55. (c) 56. (c) 57. (d) 58. (d) 59. (d) 60. (c) 61. (b) 62. (d) 63. (d) 64. (b) 65. (a) 66. (c) 67. (c) 68. (a) 69. (a) 70. (b) 71. (c) 72. (c) 73. (b)

HINTS TO THE GIVEN OBJECTIVE QUESTIONS 4

1. I = ò x

3

4

2 dx

1

62 é2 5 ù = ê x 2ú = = 12.4. 5 ë5 û1 2

2

2. I = ò ( 6 x + 4 )

1

2 dx

0

1

1

3. I = ò (5 x + 3 ) 0

1

4. I = ò

3 ù é 2 ( 6x + 4) 2 ú 56 =ê × = × ê3 ú 6 9 ë û0

dx

0 (1 +

x2 )

-1

2 dx

1 ù é (5 x + 3 ) 2 ú 2 = ê2 × = ( 8ê ú 5 5 ë û0

= [tan -1 x] 0 = (tan -1 1 - tan -1 0 ) = 1

p × 4

3 ).

Senior Secondary School Mathematics for Class 12 Pg-853

Definite Integrals 2

853

2

xù p é = ê sin -1 ú = (sin -1 1 - sin -1 0 ) = × 2 û0 2 ë 4-x dx

5. I = ò

2

0

6. Put ( 1 + x 2 ) = t and 2x dx = dt. [x =

3 Þ t = 4] and [x = 8 Þ t = 9]

\

1 19 é1 2 3 ù × I = ò t dt = ê ´ t 2 ú = ( 27 - 8 ) = 2 3 3 3 û ë 4 4

9

9

7. Put x 4 = t and 4 x 3 dx = dt. [x = 0 Þ t = 0] and [x = 1 Þ t = 1] 1 dt 1 1 1 æ1 pö p 1 I= ò = [tan -1t] 0 = [tan -1 1 - tan -1 0] = ç ´ ÷ = \ × 4 0 (1 + t2 ) 4 4 è 4 4 ø 16 1 dx = dt. x [x = 1 Þ t = 0] and [x = e Þ t = log e = 1] 1 1 ét 3 ù 1 I = ò t 2 dt = ê ú = × \ 3 3 ê ú ë û 0

8. Put log x = t and

0

p

9. I =

p

p

10. I =

2

p

é

ò cot x dx = [log sin x] p24 = êë log 1 - log 4

2

p

æ

1 ù 1 = log 2. 2 úû 2

pö

ò ( sec2 x - 1) dx = [tan x - x]0 2 = çè 1 - 4 ÷ø × 0

p

11. I =

1 2 p

12. I =

p

14. I =

ò 2 cos2 x dx = 0

2

p

sin 2 x ù 2 p 1 2 1é ( 1 + cos 2 x ) = ê x + = × 2 ò0 2ë 2 úû 0 4

é

xù

p

ò cosec x dx = êë log tan 2 úû p

p

13. I =

p

2

3

p

2

2 3

1 ö 1 æ ÷ = log 3. = ç log 1 - log 3ø 2 è

2

1

0

0

ò cos2 x × cos x dx = ò ( 1 - sin 2 x ) cos x dx = ò ( 1 - t 2 ) dt , where sin x = t.

0 p 4

ò e tan xsec2 x dx = [e tdt]10 , where tan x = t. 0

15. Put sin x = t and cos x dx = dt. p é ù [x = 0 Þ t = 0] and ê x = Þ t = 1ú × 2 ë û 1

\

I=ò

dt

2 0 (1 + t )

= [tan -1t] 0 = 1

p × 4

1 1 = t and 2 dx = - dt. x x 1 2 pù é ù é êë x = p Þ t = p úû and êë x = p Þ t = 2 úû

16. Put

Senior Secondary School Mathematics for Class 12 Pg-854

854

Senior Secondary School Mathematics for Class 12 p

p

p

( 1 - sin x )

17. I = ò

p

2

I = - ò sin t dt =

\

2 0 ( 1 - sin x )

p

p

dx = ò

ò sin t dt = [- cos t]pp2 = 1. 2

( 1 - sin x )

dx

cos 2 x

0

p

p

= ò [sec2 x - sec x tan x] dx = [tan x - sec x] 0 = 2 . 18. I =

0 p 2

ò (sin x )

0 1

= ò (t

3

2

-t

1

3

2

cos 2 x cos x dx = ò t

3

2 ( 1 - t 2 ) dt , where

sin x = t

0

7

1

2 ) dt

0 1

2 9 ù æ2 2ö 8 é2 5 = ê t 2 - t 2ú = ç - ÷ = × 9 ë5 û 0 è 5 9 ø 45

1 ì üï 1 ïì ( 1 + x ) - 1 ïü ï 1 19. I = ò e xí dx dx = ò e xí 2 ý 2ý + ( 1 ) x ï ïþ ï ï + ( 1 ) ( 1 + x x ) þ 0 î 0 î 1

= ò e x{ f ( x ) + f ¢( x )} dx , where f ( x ) = 0

1 ( 1 + x)

1

é 1 ù ö æe 1 = [e xf ( x )] 0 = ê e x × ú = ç - 1÷ × ( 1 + x) û 0 è 2 ø ë p

20. I =

2

ì

0

î

p

=

2

ìï

0

î

þ

1

ò e xíï 2 cos2 ( x p

=

ü

sin x

1

ò e xí( 1 + cos x ) + ( 1 + cos x ) ý dx 2)

+

2 sin ( x 2 ) cos ( x 2 ) üï ý dx 2 cos 2 ( x 2 ) þï

2

x 1 xü ì ò e xíî tan 2 + 2 sec2 2 ýþ dx = 0 p

p

2

ò e x{ f ( x ) + f ¢( x )} dx 0

p xù é = ê e x tan ú = e 2 . 2 û0 ë p

21. I =

4

ò

2

p

cos 2 x + sin 2 x + 2 sin x cos x dx =

0

4

ò (cos x + sin x ) dx 0

p

= [sin x - cos x] 0 4 = 1. p

22. I =

2

ò

0

p

2

p

2 cos x dx = 2 ò cos x dx = 2 [sin x] 0 2 = 2 . 2

0

23. On dividing ( - x + 1) by ( x + 1) we get: 1 ì 2 ü 1 I = ò í- 1 + ý dx = [- x + 2 log ( x + 1)]0 = ( 2 log 2 - 1). 1 ( ) x + þ 0î

Senior Secondary School Mathematics for Class 12 Pg-855

Definite Integrals p

855

p

24. I =

sin 2 x ù 2 p 1 2 1é × ò ( 1 - cos 2 x ) dx = ê x = × 2 0 2ë 2 úû 0 4

25. I =

1 6 1 é sin 3 x ù 6 15 [cos 3 x + cos x] dx = ê + sin x ú = × ò 2 0 2ë 3 12 û0

26. I =

1 2 1 2 [cos ( 2 x - x ) - cos ( 2 x + x )] dx = ò (cos x - cos 3 x ) dx ò 2 0 2 0

p

p

p

p

p

=

sin 3 x ù 2 1 æ 1é 1ö 2 sin x = ç1+ ÷ = × ê 2ë 3 úû 0 2è 3ø 3

27. 2 cos A sin B = sin ( A + B) - sin ( A - B) p p 1 1 é - cos 5 x -4 ù I = ò (sin 5 x - sin x ) dx = × ê + cos x ú = \ × 20 2 ë 5 û0 5 1

28. I = ò

1

dx

=ò

e xdx

e

dt

=ò

2x 1ö æ 0 ç ex + ÷ 0 ( e + 1) è ex ø pö æ e = [tan -1t]1 = ç tan -1 e - ÷ × 4ø è

2 1 ( t + 1)

, where e x = t.

29. Put x = t 2 and dx = 2t dt. [x = 0 Þ t = 0] and [x = 9 Þ t 2 = 9 Þ t = 3] 3

3 ì 2t 2 ü 3 dt = ò í 2 ý dt = [2t - 2 log ( 1 + t )] 0 = ( 6 - 2 log 4 ). + t + t ( ) ( ) 1 1 î þ 0 0

I=ò

\

30. Integrating by parts, we get: p

I=

2

p

ò x cos x dx = [x sin x]0 0

p 2

-

2

p

dx dx =ò 2 ì ì ü 1 3 æ ö 0 íç x2 + x + ÷ + ý 0 ï æç x + 1 ö÷ + í 4ø 4þ îè 2ø è îï

31. I = ò

æ ç ç è

é ê ê 2 =ê tan -1 2ü 3 ö÷ ï ê 3 ý 2 ÷ø ï êë þ

1

0

dx 1 - x2

1

+

0

-2 x 1 1 1 1 dx = [sin -1 x] 0 + ò dt , where ( 1 - x 2 ) = t 2 ò0 1 - x 2 21 t 1

1 é 1 ù 1 1 æp æp ö 1 t 2 ö = [sin 1 - sin 0] - × ò dt = ç - 0 ÷ - × ê 1 ú = ç - 1 ÷ × ê ú 2 0 t 2 2 ( ) è è2 ø ø 2 ë û0 -1

-1

1 æ 2 ö ÷ dx = [- x + 2 log|x + 1|]1 = ( 2 log 2 - 1). 33. I = ò çç -1 + 0 x + 1 ÷ø 0è

1

1öù æ çx + ÷ú p 2 è øú = × æ 3ö ú 3 3 ç ÷ ú ç 2 ÷ ú è ø û0

1ì 1 1 - x ïü ( 1 - x) ï 1- x 32. I = ò í ´ dx. ý dx = ò 2 + x x 1 1 ï ï þ 0î 0 1- x

=ò

ö

0

1

1

æp

p

ò ( 1 × sin x ) dx = 2 + [cos x]0 2 - çè 2 - 1 ÷ø ×

Senior Secondary School Mathematics for Class 12 Pg-856

856

Senior Secondary School Mathematics for Class 12

34. Put x = a cos q and dx = - a sin q dq. [x = - a Þ q = p] and [x = a Þ q = 0]. 0

I=ò

\

p

p

= aò 2 sin 2 0

\

I=

2

ò

2

æqö æqö 2 sin ç ÷ cos ç ÷ dq è2ø è2ø

p

q dq = aò ( 1 - cos q) dq = ap. 2 0

35. Put x = 2 sin t and dx = 2 cos t dt. pù é [x = 0 Þ t = 0] and ê x = 2 Þ t = ú 2û ë p

1

pì 2 1 - cos q ï 2 sin ( q 2 ) üï × ( - a sin q) dq = aò í ý 1 + cos q ï 2 cos 2 ( q 2 ) þï 0î

2

2 - 2 sin t × 2 cos t dt = 2

0

p

2

p

0

0

p

2

ò cos t dt = 2 ò 2

1 ( 1 + cos 2 t ) dt 2

sin 2 t ù 2 p é = êt + = × 2 úû 0 2 ë 0

2

36. I = ò|x| dx + ò|x| dx = -2

0

0

0

2

-2

0

ò - x dx + ò x dx.

2

é x2 ù é x2 ù = ê - ú + ê ú = 0 - ( -2 ) + ( 2 - 0 ) = 4. 2 êë úû -2 ëê 2 úû 0 1 37. Put 2 x - 1 = t and dx = dt. 2 [x = 0 Þ t = - 1] and [x = 1 Þ t = 1] 1 0 1 ü 1 1 ìï ï I = ò|t| dt = í ò -t dt + ò t ý \ 2 -1 2 ïî -1 þ 0 ï 0 1 ìé 2 ù 2ù ü é 1 ï -t t 1ö æ 1 ï 1 ìæ öü 1 = íê ú + ê ú ý = íç 0 + ÷ + ç - 0 ÷ ý = × 2 ï êë 2 úû 2 úû ï 2 î è 2ø è2 øþ 2 ê ë -1 0þ î 1 38. Put 2 x + 1 = t and dx = dt 2 [x = -2 Þ t = -3] and [x = 1 Þ t = 3] 3 0 3 üï 1 1 ìï I = ò|t| dt = í ò ( -t ) dt + ò t dt ý \ 2 -3 2 îï -3 þï 0 =

0 3 ì é t 2 ù üï 1 éì 1 ï é -t 2 ù æ -9 ö ü æ 9 öù 9 ÷ý + ç - 0 ÷ú = × ú + ê ú ý = êí 0 - ç íê 2 2 ï êë 2 úû 2 2 2 è ø è øû 2 þ ëî ëê úû 0 ïþ -3 î

0

39. I =

-x dx + x -2

ò

0

x

1

-2

0

a

0

40. I =

1

ò x dx = ò - dx + ò dx = [- x]-2 + [x]0 = ( -2 + 1) = -1. 0

0

0

a

-a

0

ò x( - x ) dx + ò x × x dx = ò - x 2 dx + ò x 2 dx

-a

0

0

a

é - x3 ù é x3 ù a3 1 3 =ê = 0. ú + ê ú = + ( - a) + 3 3 3 3 êë úû - a ëê úû 0

1

Senior Secondary School Mathematics for Class 12 Pg-857

Definite Integrals p

41. I =

857

p

2

ò cos x dx + ò - cos x dx. p

0

2

p

= [sin x] 0 2 - [sin x] pp = 1 - ( 0 - 1) = 2. 2

p

2p

0

p

42. I = ò sin x dx +

ò

- sin x dx.

p

2p

= [- cos x] 0 + [cos x] p = ( 2 + 2 ) = 4. p

43. I =

2

0

p

I=

sin x

ò (sin x + cos x ) dx 2

ò

0

æp ö sin ç - x ÷ è2 ø dx = ì æp ö æp öü í sin ç - x ÷ + cos ç - x ÷ ý è2 ø è2 øþ î p

\

... (i)

2I =

2

p

p

ò dx = [x]0 2 = 2

Þ I=

0

p

2

cos x

ò (sin x + cos x ) dx

... (ii)

0

p × 4

Q. 44 to 49: Similarly, each of the questions from 44 to 49 may be attempted. p

50. I =

2

ò

0

p

52. I =

2

cos x dx. (cos x + sin x ) cos 3 x

ò (cos3 x + sin 3 x )

55. I =

2

2

ò( 0

p

dx.

0

p

p

51. I =

54. I =

sin x sin x +

2

ò( 0

cos x )

cos x sin x +

cos x )

dx.

dx.

sin x

ò (cos x + sin x ) dx. 0

56. f ( x ) = x 4 sin x. \

f ( - x ) = ( - x ) 4 sin ( - x ) = - ( x 4 sin x ) = - f ( x ).

\

f ( x ) is odd and hence I = 0.

57. f ( x ) = x 3 cos 3 x. \

f ( - x ) = ( - x ) 3 [cos ( - x )] 3 = - x 3 (cos x ) 3 = - ( x 3 cos 3 x ) = - f ( x )

\

f ( x ) is odd and hence I = 0.

58. f ( x ) = sin5 x. f ( - x ) = [sin ( - x )]5 = ( - sin x )5 = - (sin5 x ) = - f ( x ). \

f ( x ) is odd and hence I = 0.

59. f ( x ) = x 3 ( 1 - x 2 ) = ( x 3 - x5 ). f ( - x ) = ( - x ) 3 - ( - x )5 = - x 3 - ( - x5 ) = - x 3 + x5 = - x 3 ( 1 - x 2 ) = - f ( x ). \

f ( x ) is odd and hence I = 0.

Senior Secondary School Mathematics for Class 12 Pg-858

858

Senior Secondary School Mathematics for Class 12

æ a- xö ÷× 60. f ( x ) = log çç ÷ èa+ xø

ì æ a - x ö -1 ü æ a- xö æa+ xö ç ÷ ÷ = log ïí ç ÷ ï f ( - x ) = log çç ÷ ç a + x ÷ ý = - log ç a + x ÷ = - f ( x ) a x è ø è ø ø ïþ ïî è

Þ \

f ( x ) is odd and hence I = 0.

61. f ( x ) = sin 61 x + x123 f ( - x ) = [sin ( - x )] 61 + ( - x )123 = ( - sin x ) 61 - x123 = - sin 61 x - x123

Þ

= - [sin 61 x + x123 ] = - f ( x ). \

f ( x ) is odd and hence I = 0.

62. f ( x ) = tan x Þ f ( - x ) = tan ( - x ) = - tan x = - f ( x ). \

f ( x ) is odd and hence I = 0.

63. Let f ( x ) = log ( x +

x 2 + 1 ). Then,

f ( - x ) = log ( - x +

x 2 + 1)

ì ü ìï ( x 2 + 1 - x ) ( x 2 + 1 + x ) üï 1 ï ï = log í ý ý = log í 2 2 ïî ïþ ( x + 1 + x) ïî ( x + 1 + x ) ïþ = log ( x + \

x 2 + 1) -1 = - log ( x +

x 2 + 1) = - f ( x )

f ( x ) is odd and hence I = 0.

64. f ( x ) = cos x Þ f ( - x ) = cos ( - x ) = cos x = f ( x ). \

f ( x ) is an even function.

\

I=2

p

2

ò

p

cos x dx = 2 [sin x] 0 2 = ( 2 ´ 1) = 2.

0

a

x dx and I = x + a - x) ( 0

65. I = ò

a

( x+

a - x)

( x+ 0

a - x)

2I = ò

\ p

66. I =

-a

ò{

a- x +

0

a

a- x dx a - ( a - x )}

dx = ò dx = [x] 0 = a Þ I = a

0

a × 2

4

ò log ( 1 + tan x ) dx

... (i)

0

p

I=

4

ò

0

p

=

é æp öù log ê 1 + tan ç - x ÷ ú dx = è4 øû ë

4

ò

0 p

4

ò {log 2 - log ( 1 + tan x )} dx = 0

Þ

p

p

2 I = (log 2 )[x] 0

4

ì 1 - tan x ü log í 1 + ý dx 1 + tan x þ î 4

ò (log 2 ) dx - I 0

p p = (log 2 ) Þ I = (log 2 ). 4 8

Senior Secondary School Mathematics for Class 12 Pg-859

Definite Integrals

67. We have:

a

a

-a

0

ò f ( x ) dx = ò { f ( x ) + f ( - x )} dx.

68. {0 £ x < 1 Þ [x] = 0} and {1 £ x < 1.5} Þ [x] = 1 1

\

I = ò [x] dx + 0

=0+

[x]1.5 1

1.5

1

1.5

1

0

0

ò [x] dx = ò 0 dx + ò 1 × dx

1 = ( 1.5 - 1) = 0.5 = × 2

69. {-1 £ x < 0 Þ [x] = -1} and {0 £ x < 1 Þ [x] = 0} 0

\

I=

1

0

1

0

-1

0

ò [x] dx + ò [x] dx = ò ( -1) dx + ò 0 × dx

-1

0

= [- x] -1 = 0 - {- ( -1)} = - 1. 70. ( x 2 - 3 x + 2 ) = ( x - 1)( x - 2 ) 1 £ x < 2 Þ ( x 2 - 3x + 2) £ 0 Þ |x 2 - 3 x + 2| = - ( x 2 - 3 x + 2 ) 2

\

2

I = ò - ( x 2 - 3 x + 1) dx = ò ( - x 2 + 3 x - 2 ) dx 1

1

2

é - x3 ù 3x 2 æ -2 5 ö 1 =ê + - 2xú = ç + ÷= × 2 6ø 6 êë 3 úû1 è 3 71. p £ x < 2 p Þ sin x £ 0 Þ |sin x| = - sin x 2p

\

I=

2p

ò - sin x dx = [cos x]p

= 2.

p

72. Put x = sin t and dx = cos t dt

1 1 pù é Þ sin t = Þ t= ú [x = 0 Þ sin t = 0 Þ t = 0] and ê x = 4û 2 2 ë p

\

I=

4

ò

0

t cos t 3

cos t

p

dt =

4

p

ò t sec t = [t tan t]0 2

0

p

4

-

4

ò tan t dt 0

p 1 p p æp 1 ö = + [log cos t] 0 4 = + log = ç - log 2 ÷ 4 4 2 è4 2 ø

73. Put x = tan t and dx = sec2t dt. pù é [x = 0 Þ t = 0] and ê x = 1 Þ tan t = 1 Þ t = ú 4û ë p

\

I=

4

p

4

ò sin -1 (sin 2t ) sec2t dt = 2 ò t sec2t dt 0

0

p é ù 4 p p ü ìp = 2 ê[t tan t] 0 4 - ò tan t dt ú = 2 í + [log cos t] 0 4 ý ê ú 4 î þ 0 êë úû p 1 æp ö = + 2 log = ç - log 2 ÷ × 2 2 è2 ø

859

Senior Secondary School Mathematics for Class 12 Pg-860

17. AREA OF BOUNDED REGIONS 1. AREA OF A CURVE BETWEEN TWO ORDINATES

Let y = f ( x) be a continuous and

finite function in [a , b]. Case I When the curve y = f ( x) lies above the x-axis

Y

The area bounded by the curve y = f ( x), the x-axis, and the ordinates x = a and x = b is given by

y=

f(x

)

b

area = ò y dx. a

a

O

Y

Case II When the curve y = f ( x) lies below the x-axis

The area between the curve y = f ( x), the x-axis, and the ordinates x = a and x = b is given by

X

b

a A

O

X'

b B

X

b

area = ò ( - y) dx. a

C

y=

f( x )

D

Y

2. AREA OF A CURVE BETWEEN TWO ABSCISSAE

Case I When the curve x = f ( y) lies to the right of the y-axis

Y

The area bounded by the curve x = f ( y), the y-axis, B and the abscissae y = c and y = d is given by

y=d

D

d

area = ò x dy.

A

c

x = f(y) y=c

C X

O

Case II When the curve x = f ( y) lies to the left of the y-axis

The area bounded by the curve x = f ( y), the y-axis, and the abscissae y = c and y = d is given by

Y y=d x = f(y)

d

area = ò ( - x) dy.

X'

c

y=c O Y'

860

X

Senior Secondary School Mathematics for Class 12 Pg-861

Area of Bounded Regions

861 Y

3. AREA BETWEEN TWO CURVES The area bounded by two curves y = f ( x) and y = g( x), which are intersected by the ordinates x = a and x = b , is given by

y = f(x) y = g(x)

b

area = ò { f ( x) - g( x)} dx.

O

b

a

X

a

SOLVED EXAMPLES EXAMPLE 1

Using integration, find the area of the region bounded by the line [CBSE 2002C] 2y + x = 8, the x-axis and the lines x = 2 and x = 4.

SOLUTION

The given line AB is 1 2y + x = 8 Þ y = 4 - x. … (i) 2 Required area = area PLMQP 1 = area between the line y = 4 - x , 2 and the x-axis between x = 2 and x=4 4

Y

B P Q O

M L (2, 0) (4, 0)

A

X

4

1 ö æ = ò y AB dx = ò ç 4 - x ÷ dx 2 ø è 2 2 4

1 ù é = ê 4x - x 2 ú = (12 - 7) sq units 4 û2 ë = 5 sq units. Hence, the required area is 5 sq units. EXAMPLE 2

Using integration, find the area of a ABC whose vertices are A( 2, 3), [CBSE 2001] B( 4, 7) and C( 6, 2).

SOLUTION

The equation of AB is y-3 7-3 = Þ y = 2x - 1. x -2 4-2 The equation of BC is y -7 2 -7 -5 = Þy= x + 17. x -4 6-4 2 The equation of AC is y - 3 2- 3 -1 7 = Þy= x+ × x -2 6-2 4 2

Y

B(4, 7)

(2, 3) A C(6, 2) O

L

M

N

X

Senior Secondary School Mathematics for Class 12 Pg-862

862

Senior Secondary School Mathematics for Class 12

Draw AL ^ OX, BM ^ OX and CN ^ OX. Area of a ABC = ( area ALMBA + area BMNCB) - ( area ALNCA) 4

6

2 4

4

6

= ò y AB dx + ò y BC dx - ò y AC dx 2

6

6

7ö æ 5 ö æ 1 = ò ( 2x - 1) dx + ò ç - x + 17 ÷ dx - ò ç - x + ÷ dx 4 2ø 2 ø 2 4è 2è 6

6

7 ù ù é 1 4 é 5 = [x 2 - x]2 + ê - x 2 + 17 x ú - ê - x 2 + x ú 2 û2 ë 4 û4 ë 8 é æ 33 13 ö ù = ê(12 - 2) + (57 - 48) - ç - ÷ ú sq units 2 øû è 2 ë = 9 sq units. Hence, the required area is 9 sq units. EXAMPLE 3

SOLUTION

Calculate the area bounded by the parabola y 2 = 4ax and its latus rectum. Let S( a , 0) be the focus of the parabola y 2 = 4ax. Then, its latus rectum LSL¢ is the line parallel to the y-axis at a distance a from it. So, its equation is x = a. Since the equation of the parabola contains only even powers of y, it is symmetrical about the x-axis. \ required area = area LOL¢ L = ( area LOSL) + ( area SOL¢ S) = 2 ´ ( area LOSL) a

a

0

0

= 2 ò y dx = 2 × ò 2 ax dx 2 8 a = 4 a × [x 3/ 2]0 = a 2 sq units. 3 3 EXAMPLE 4

SOLUTION

Using integration, find the area of the region bounded by the parabola y 2 = 16x and the line x = 4. [CBSE 1996, ‘97] y 2 = 16x is a right-handed parabola with its vertex at the origin. And, x = 4 is the line parallel to the y-axis at a distance of 4 units from it.

Senior Secondary School Mathematics for Class 12 Pg-863

Area of Bounded Regions

863

Also, y 2 = 16x contains only even powers of y. So, it is symmetrical about the x-axis. \ required area = area AOCA + area BOCB = 2( area AOCA) 4

4

0

0

= 2 ò y dx = 2 ò 16x dx 4

= 8 ò x dx = 8 ´ 0

2 4 ´ [x 3/ 2]0 3

16 æ 16 ö 128 sq units. = ´ ( 4) 3/2 = ç ´ 8÷ = 3 3 ø è 3 Hence, the required area is

128 sq units. 3

EXAMPLE 5

Using integration, find the area enclosed by the parabola y 2 = 4 ax and

SOLUTION

the chord y = mx. The given equations are and

y 2 = 4 ax

… (i)

y = mx

… (ii)

Y

A B

4a , 4a m2 m

Clearly, y 2 = 4 ax is a rightX O M handed parabola, passing through the origin. And y = mx is a line passing through the origin. In order to find the points of intersection of the given parabola and the given line, we solve (i) and (ii) simultaneously. Putting y = mx from (ii) into (i), we get m 2 x 2 = 4 ax Þ x(m 2 x - 4 a) = 0 Þ x = 0 or x =

4a m2 4a

×

4a ö æ Now, ( x = 0 Þ y = 0) and ç x = 2 Þ y = ÷× mø è m So, the points of intersection of the given parabola and the chord are æ 4a 4a ö O(0, 0) and A ç 2 , ÷× èm m ø Draw AM ^ OX. Required area = ( area OBAMO) - ( area OAMO)

Senior Secondary School Mathematics for Class 12 Pg-864

864

Senior Secondary School Mathematics for Class 12 4 a/ m2

=

ò ( y for the parabola) dx -

4 a/ m2

0

4 a/ m2

=

ò2

ò ( y for the line) dx 0

4 a/ m2

ax dx -

0

ò mx dx 0

4 a/ m2

é mx 2 ù 2 4 a/m2 = 2 a × [x 3/ 2]0 -ê ú 3 ë 2 û0 é 4 a 8 3/ 2 m 16a 2 ù =ê × 3a - × 4 ú 2 m û ë 3 m

æ 32a 2 8a 2 ö æ 8a 2 ö ÷=ç ÷ sq units. =ç ç 3m 3 m 3 ÷ ç 3m 3 ÷ è ø è ø æ 8a 2 ö ÷ sq units. Hence, the required area is ç ç 3m 3 ÷ è ø EXAMPLE 6

Find the area of the region {( x , y) : x 2 £ y £ x}.

SOLUTION

Consider the equations y = x2

[CBSE 2007, ’13C]

… (i)

and y = x

… (ii)

Clearly, y = x 2 is an upward parabola and y = x is a line passing through (0, 0). Solving (i) and (ii) simultaneously, we get 2

2

x = x Þ x - x = 0 Þ x( x - 1) = 0

y=

x

A(1, 1)

C X

X

Þ x = 0 or x = 1. From (ii), ( x = 0 Þ y = 0) and ( x = 1 Þ y = 1). So, the points of intersection of (i) and (ii) are O( 0, 0) and A(1, 1). Draw AM ^ OX. Required area = shaded area = ( area OMAO) - ( area OMACO) 1

1

= ò ( y for line) dx - ò ( y for parabola) dx 0

0

1

1

é x2 ù éx3ù é1 1 ù 1 = ò x dx - ò x 2 dx = ê ú - ê ú = ê - ú = sq unit. ë 2 û0 ë 3 û0 ë 2 3 û 6 0 0 1

1

Hence, the required area is

1 sq unit. 6

Senior Secondary School Mathematics for Class 12 Pg-865

Area of Bounded Regions EXAMPLE 7

865

Find the area of the region bounded by the curve x 2 = 4y and the line x = 4y - 2.

SOLUTION

[CBSE 2010, ‘13]

The given curve is x 2 = 4y

… (i)

The given line is x = 4y - 2

… (ii)

B(2, 1)

H 1

(–1, 4 ) X

A L

O

M

X

Putting 4y = ( x + 2) from (ii) into (i), we get x 2 = ( x + 2) Û ( x 2 - x - 2) = 0 Û ( x - 2)( x + 1) = 0 Û x = 2 or x = -1. Putting x = 2 in (i), we get y = 1. 1 Putting x = -1 in (i), we get y = × 4 Thus, the points of intersection of the given curve (i) and the line 1ö æ (ii) are A ç -1, ÷ and B( 2, 1). 4ø è Draw AL and BM as perpendiculars on the x-axis. \ required area = area AOBA = ( area ALMBA) - ( area AOBMLA) 2

=

2

2

x æ x + 2ö ò çè 4 ÷ø dx - ò 4 dx -1 -1 2

ì( x + 2) x 2 ü 1 é x2 x3ù = òí - ý dx = ê + 2x ú 4 4þ 4ë 2 3 û -1 î -1 2

1 éæ 8ö æ1 1 öù 9 ç 2 + 4 - ÷ - ç - 2 + ÷ ú = sq units. 4 êë è 3ø è2 3 øû 8 9 Hence, the required area is sq units. 8 =

EXAMPLE 8

Find the area bounded by the circle x 2 + y 2 = 16 and the line y = x in the first quadrant.

SOLUTION

[CBSE 2004]

The given circle is x 2 + y 2 = 16

… (i)

The given line is y = x

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-866

866

Senior Secondary School Mathematics for Class 12

Putting y = x from (ii) into (i), we get 2x 2 = 16 Û x 2 = 8 Û x = 2 2 [Q x is + ve in the first quad.]. Thus, the point of intersection of (i) and (ii) in the first quadrant is A( 2 2 , 2 2). Draw AL perpendicular on the x-axis. \ required area = ( area OLA) + area( LBAL) =

2 2

4

0

2 2

ò x dx + ò 2 ù2 2

éx =ê ú ë 2 û0

16 - x 2 dx 4

é x 16 - x 2 16 xù sin -1 ú +ê + 2 2 4ú êë û2

2

1 1 ù é = [( 2 2) 2 - 0] + ê ( 0 + 8 sin -1 1) - ( 4 + 8 sin -1 2 2 úû ë pö pöù é æ æ = ê 4 + ç 8 ´ ÷ - 4 - ç 8 ´ ÷ ú = ( 2p) sq units. 2 4øû è ø è ë EXAMPLE 9

SOLUTION

Using integration, find the area of a ABC, whose vertices are A( 2, 0), [CBSE 2003C, ‘06C] B( 4, 5) and C( 6, 3). The equation of side AB is y - 0 (5 - 0) 5 … (i) = Þ y = ( x - 2) x - 2 ( 4 - 2) 2 B(4, 5)

C(6, 3)

A(2, 0) L

M

X

Senior Secondary School Mathematics for Class 12 Pg-867

Area of Bounded Regions

867

The equation of side BC is y - 5 ( 3 - 5) = Þ y = -x + 9 x - 4 ( 6 - 4)

… (ii)

The equation of side AC is y - 0 ( 3 - 0) 3 = Þ y = ( x - 2) x - 2 ( 6 - 2) 4

… (iii)

Draw perpendiculars BL and CM on the x-axis. \ area of a ABC = ar(a ALB) + ar(trap. BLMC) - ar(a AMC) 4

6

6

= ò y AB dx + ò y BC dx - ò y AC dx 2

4

2

4

=

6

6

3 5 ( x - 2) dx + ò ( 9 - x) dx - ò ( x - 2) dx 42 2 ò2 4 4

6

6

é x2 ù é ù x2 ù 3 é x2 - 2x ú + ê 9x - ú - × ê - 2x ú ê 2û 4 ë 2 ë 2 û2 ë û2 4 5 3 = [0 - ( -2)] + ( 36 - 28) - [6 - ( -2)] 2 4 = (5 + 8 - 6) sq units = 7 sq units. =

5 2

EXAMPLE 10

Find the area cut off from the parabola 4y = 3 x 2 by the straight line

SOLUTION

[CBSE 2007, ‘09] 3 x - 2y + 12 = 0. Clearly, 4y = 3 x 2 is an upward parabola with its vertex at (0, 0).

And, 3 x - 2y + 12 = 0 is a line. The given equations are 4y = 3 x 2

… (i)

and 3 x - 2y + 12 = 0 … (ii) The points of intersection of the ( 2, 3)A given parabola and the given line will be obtained by solving (i) and X L (ii) simultaneously. ( 2, 0) 3 Putting y = x 2 from (i) in (ii), 4 we get 3 3 x - x 2 + 12 = 0 Þ x 2 - 2x - 8 = 0 2 Þ x 2 - 4x + 2x - 8 = 0 Þ x( x - 4) + 2( x - 4) = 0 Þ ( x - 4)( x + 2) = 0 Þ x = -2 or x = 4.

B(4, 12)

X (4, 0)

Senior Secondary School Mathematics for Class 12 Pg-868

868

Senior Secondary School Mathematics for Class 12

Now, ( x = -2 Þ y = 3) and ( x = 4 Þ y = 12). So, the points of intersection of (i) and (ii) are A( -2, 3) and B( 4, 12). Draw AL ^ OX¢ and BM ^ OX. Required area = ( area ALMBA) - ( area ALOMBOA) 4

4

-2

-2

= ò ( y of the line) dx - ò ( y of the parabola) dx 4

4

4 é 3x2 ù éx3ù ( 3 x + 12) 3x2 ò 2 dx - ò 4 dx = ê 4 + 6x ú - ê 4 ú ë û -2 ë û -2 -2 -2 4

=

= ( 45 - 18) = 27 sq units. Hence, the required area is 27 sq units. EXAMPLE 11

Find the area bounded by the line y = x , the x-axis and the ordinates x = -1, x = 2.

SOLUTION

We know that y = x is the line passing through the origin and making an angle of 45° with the x-axis, as shown in the given figure. Now, we have to find the area of the shaded region. B Y

X

( 1, 0)C

O D(2, 0)

X

A Y

Required area = ( area ODBO) + ( area OACO) 2

0

0

-1

2

0

= ò y dx + ò ( - y) dx

[Q area OACO is below the x-axis]

= ò x dx + ò ( - x) dx -1

0

2

0

é x2 ù é -x 2 ù 1ù 5 é =ê ú +ê ú = ê 2 + ú = sq units. 2 2 2û 2 ë ë û0 ë û -1 Hence, the required area is

5 sq units. 2

Senior Secondary School Mathematics for Class 12 Pg-869

Area of Bounded Regions

869

EXAMPLE 12

Find by integration, the area of the region bounded by the curve y = 2x - x 2 and the x-axis.

SOLUTION

The given curve is y = 2x - x 2 . 2

… (i)

2

Now, y = 2x - x Þ ( x - 2x + 1) = ( - y + 1) Þ ( x - 1) 2 = -1( y - 1) Y

Þ X2 = -Y ,

A

where ( x - 1) = X and ( y - 1) = Y. Clearly, X2 = -Y is a downward parabola with its vertex at (X = 0, Y = 0).

B (2, 0)

O

X

Now, X = 0, Y = 0 Þ x - 1 = 0 and y - 1 = 0 Þ x = 1 and y = 1. Thus, the vertex of the parabola is A(1, 1). Also, y = 0 Þ 2x - x 2 = 0 Þ x( 2 - x) = 0 Þ x = 0 or x = 2. Thus, the curve cuts the x-axis at O( 0, 0) and B( 2, 0). The rough sketch of the curve can now be drawn, as shown in the given figure. 2

\ required area = ò y dx 0

2

2 é x3ù = ò ( 2x - x 2) dx = ê x 2 ú 3 û ë 0 0

8ö 4 æ = ç 4 - ÷ = sq units. 3ø 3 è Hence, the required area is

4 sq units. 3

EXAMPLE 13

Using integration, find the area of the region bounded by the ellipse x2 y2 + = 1. a 2 b2

SOLUTION

The given equation contains only even powers of y. So, the curve is symmetrical about the x-axis. Also, the given equation contains only even powers of x. So, the curve is symmetrical about the y-axis. A rough sketch of the ellipse can be drawn, as shown in the figure.

Senior Secondary School Mathematics for Class 12 Pg-870

870

Senior Secondary School Mathematics for Class 12

Now,

x2 a

2

+

y2 b2

=1Þ y =

b × a2 - x 2 . a

Y

Area of the given ellipse = 4 ´ ( area OBCO) a

X

a

b = 4 ´ ò y dx = 4 ´ ò × a 2 - x 2 dx a 0 0 =

4b × a

A ( a, 0)

O

B (a, 0)

X

Y

p/ 2

2 2 ò a cos q dq [putting x = a sin q so that dx = a cos q dq] 0

p/ 2

= ( 4ab) ×

ò

0

p/ 2

(1 + cos 2q) sin 2q ù é dq = ( 2ab) × ê q + 2 2 úû 0 ë

= ( pab) sq units.

EXAMPLE 14

SOLUTION

x2

x2 + y2 = r 2

2

+

y2

= 1 is ( pab) sq units. a b2 By using integration, prove that the area of a circle of radius r units is pr 2 square units. The equation of a circle of radius r units with its centre at the origin is Y Hence, the area of the ellipse

… (i)

This equation contains only even powers of y. So, the curve is symmetrical about the x-axis. Also, the above equation contains only even powers of x. So, the curve is symmetrical about the y-axis.

B

X

A

X

O

Y

Now, x 2 + y 2 = r 2 Þ y = r 2 - x 2 . Area of the circle = 4 ´ ( area OABO) r

r

0 p/ 2

0

= 4 ´ ò y dx = 4 ´ ò r 2 - x 2 dx = 4 ´ ò r 2 cos2 q dq [putting x = r sin q] 0 p/ 2

= 4r 2 ×

ò

0

p/ 2

(1 + cos 2q) sin 2q ù é dq = 2r 2 × ê q + 2 ûú 0 2 ë

= ( pr 2) sq units. Hence, the required area is ( pr 2) sq units.

Senior Secondary School Mathematics for Class 12 Pg-871

Area of Bounded Regions

EXAMPLE 15

SOLUTION

871

x2 y2 Find the area of the smaller region bounded by the ellipse 2 + 2 = 1 a b x y [CBSE 2004, ‘05C] and the straight line + = 1. a b 2 2 Y x y The ellipse 2 + 2 = 1 and the a b x y b line + = 1 can be drawn, as a b a O X X shown in the given figure. Then, we have to find the area of the shaded region. Required area = {area between

x

2

a

2

+

y

2

= 1 and the x-axis from x = 0 to x = a}

b2

– {area between the line a

Y

x y + = 1 and the x-axis from x = 0 to x = a} a b a

= ò ( y of the ellipse) dx - ò ( y of the line) dx 0 a

0

a

b( a - x) b dx = ò × a 2 - x 2 dx - ò a a 0 0 a

a x2 ù b é x a2 - x 2 a2 xù b é sin -1 ú - × ê ax - ú = ×ê + a ê aú a ë 2 2 2û 0 ë û0

=

ab ab ö æ pab ab ö æ (sin -1 1 - sin -1 0) - ç ab - ÷ = ç - ÷ sq units. 2 2ø è 4 2ø è

æ pab ab ö Hence, the required area is ç - ÷ sq units. 2ø è 4 EXAMPLE 16

Find the area of the region bounded by the parabolas x 2 = y and y 2 = x.

SOLUTION

Clearly, x 2 = y is an upward parabola with its vertex at ( 0, 0) and y 2 = x is a right-handed parabola with its vertex also at (0, 0). The shaded region shows the area bounded by these parabolas.

[CBSE 2012C] Y

2

x

= y

X

A(1, 1)

B C O

X

L

y2 = Y

x

Senior Secondary School Mathematics for Class 12 Pg-872

872

Senior Secondary School Mathematics for Class 12

The given equations are x2 = y and y 2 = x Using (i) in (ii), we get

… (i) … (ii)

x 4 = x Þ x( x 3 - 1) = 0 Þ x = 0 or x = 1. Also, ( x = 0 Þ y = 0) and ( x = 1 Þ y = 1). So, the given curves intersect at O( 0, 0) and A(1, 1). Draw AL ^ OX. Required area = ( area OLABO) - ( area OLACO) = {area bounded by y 2 = x from x = 0 to x = 1} – {area bounded by x 2 = y from x = 0 to x = 1} 1

1

0

0 1

= ò x dx - ò x 2 dx 1

éx3ù é2 ù æ 2 1ö 1 = ê x 3/ 2 ú - ê ú = ç - ÷ = sq unit. ë3 û0 ë 3 û0 è 3 3 ø 3 Hence, the required area is EXAMPLE 17

1 sq unit. 3

Find the area of the region included between the parabolas y 2 = 4 ax and x 2 = 4 ay , where a > 0.

[CBSE 2003, ‘04, ‘08, ‘09, ‘12C]

The given parabolas are … (i)

and x 2 = 4 ay

… (ii)

In order to find the points of intersection of the given curves, we solve (i) and (ii) simultaneously. Putting x = we get y4 16a 2

4ay

Y

y 2 = 4 ax

x 2=

SOLUTION

B X

y2 from (i) in (ii), 4a

O

2 4 y =

ax

A(4a, 4a) C D(4a,0)

X

Y

= 4 ay Þ y 4 - 64 a 3 y = 0 Þ y( y 3 - 64 a 3) = 0 Þ y = 0 or y = 4 a.

Now, ( y = 0 Þ x = 0) and ( y = 4 a Þ x =

16a 2 = 4 a). 4a

Thus, the points of intersection of the two parabolas are O( 0, 0) and A( 4 a , 4 a).

Senior Secondary School Mathematics for Class 12 Pg-873

Area of Bounded Regions

873

Draw AD ^ OX. Then, point D is ( 4 a , 0). Required area = area OCABO = ( area OBADO) - ( area OCADO) 4a

4a

= ò y dx for ( y 2 = 4 ax) - ò y dx for ( x 2 = 4 ay) 0 4a

0

4a 2

x = ò 2 ax dx - ò dx a 4 0 0 4a

4a 2 1 éx3ù é ù = ê 2 a × × x 3/ 2 ú ê ú 3 ë û 0 4a ë 3 û 0

é4 a ù 1 =ê × ( 4 a) 3/ 2 ´ 64 a 3 ú 3 12 a ë û æ 32 a 2 16 a 2 ö æ 16 a 2 ö ÷=ç ÷ sq units. =ç ç 3 3 ÷ø çè 3 ÷ø è æ 16 a 2 ö ÷ sq units. Hence, the required area is ç ç 3 ÷ è ø EXAMPLE 18

SOLUTION

Using integration, find the area of the region common to the circle x 2 + y 2 = 16 and the parabola y 2 = 6x. [CBSE 2012C] 2 2 x + y = 16 is a circle with its centre at (0, 0) and radius = 4 units. And, y 2 = 6x is a right-handed parabola with its vertex at (0, 0). Now, we have to find the area of the shaded region. The given equations are x 2 + y 2 = 16 and

2

y = 6x

… (i) … (ii)

Using (ii) in (i), we get x 2 + 6x - 16 = 0 Þ ( x + 8)( x - 2) = 0 Þ x = - 8 or x = 2. Now, x = -8 Þ y 2 = - 48 Þ y is imaginary. And, x = 2 Þ y 2 = ( 6 ´ 2) = 12 Þ y = ± 2 3. Thus, the points of intersection of the given curves are A( 2, 2 3 ) and B( 2, - 2 3 ).

Senior Secondary School Mathematics for Class 12 Pg-874

874

Senior Secondary School Mathematics for Class 12

Since each of the given equations contains only even powers of y, each one is symmetrical about the x-axis. \ required area = 2( area OCDAO) = 2( area OCAO + area CDAC) 4 é2 ù = 2 ê ò y dx for curve (ii) + ò y dx for curve (i) ú êë 0 úû 2 4 é2 ù = 2 ê ò 6x dx + ò 16 - x 2 dx ú êë 0 úû 2 4 ì 2 é ù üï 16 - x 2 16 ïé 2 6 3/2 ù -1 x ú ý = 2í ê x ú + êx + sin 2 2 4ú ï û 0 êë ïë 3 û2þ î

ìïæ 2 6 3/2 ö 1 ö üï æ = íçç ×2 - 0÷÷ + 8 sin -1 1 - ç 2 3 + 8 sin -1 ÷ ý 2 ø ïþ ïîè 3 è ø æ 2 3 8p ö æ8 3 4p ö ÷ + = 2 çç + 4p - 2 3 - ÷÷ = 2 çç 3 3 3 ÷ø ø è 3 è 4 = ( 3 + 4p) sq units. 3 4 Hence, the required area is ( 3 + 4p) sq units. 3 EXAMPLE 19

Using integration, find the area of the region enclosed between the circles x 2 + y 2 = 4 and ( x - 2) 2 + y 2 = 4. [CBSE 2009, ’13C]

SOLUTION

x 2 + y 2 = 4 is a circle with its centre at O(0, 0) and radius = 2 units. And, ( x - 2) 2 + y 2 = 4 is a circle with its centre at C(2, 0) and radius = 2 units. The given circles are x2 + y2 = 4

… (i)

and ( x - 2) 2 + y 2 = 4

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-875

Area of Bounded Regions

875

Eliminating y from (i) and (ii), we get 4 - x 2 = 4 - ( x - 2) 2 Þ 4x = 4 Þ x = 1. Putting x = 1 in (i), we get y 2 = 3 Þ y = ± 3 . Thus, the points of intersection of the two circles are A(1, B(1, - 3 ). Both the circles are symmetrical about the x-axis. Required area = 2( area AOCA) = 2( area AODA + area CADC) 1

3 ) and

2

= 2ò y dx for circle (ii) + 2ò y dx for circle (i) 0 1

1

2

= 2ò 4 - ( x - 2) dx + 2ò 4 - x 2 dx 2

0

1

1

é ( x - 2) 4 - ( x - 2) 2 4 ( x - 2) ù ú =ê + × sin -1 2 2 2 ú ê ë û0 2

é x 4 - x2 4 xù +ê + × sin -1 ú 2 2 2ú êë û1

éì - 3 æ -1 ö ü = 2 êí + 2 sin -1 ç ÷ ý - {0 + 2 sin -1( -1)} 2 è 2 øþ êëî ì 3 æ 1 ö üù + í2 sin -1(1) - 2 sin -1 ç ÷ ýú 2 è 2 ø þúû î ìï - 3 3 p ö üï æ pö æ pö æ p = 2í + 2 ç - ÷ - 2 ç - ÷ + çç 2 × - 2 ´ ÷÷ ý 2 6 ø ïþ ïî 2 è 6ø è 2ø è 2 ö æ 4p = 2ç - 3 ÷ sq units. ø è 3 ö æ 4p Hence, the required area is 2 ç - 3 ÷ sq units. ø è 3 EXAMPLE 20

Using integration, find the area of the region {( x , y) : y 2 £ 4 x , 4 x 2 + 4 y 2 £ 9}.

SOLUTION

Clearly, we have to find the area enclosed between the curves y 2 = 4 x and 4 x 2 + 4 y 2 = 9.

[CBSE 2008, ‘13]

The curve y 2 = 4 x is a right-handed parabola with its vertex at (0, 0). 2

æ 3ö 4 x 2 + 4 y 2 = 9 Þ x 2 + y 2 = ç ÷ , which represents a circle with its è 2ø centre at O(0, 0) and radius equal to ( 3/2) units.

Senior Secondary School Mathematics for Class 12 Pg-876

876

Senior Secondary School Mathematics for Class 12

The shaded region shows the required area. y2 = 4x

Now, and

2

4x + 4y2 = 9

… (i) … (ii)

Putting the value of y 2 from (i) into (ii), we get 4 x 2 + 16x - 9 = 0 Þ 4 x 2 + 18x - 2x - 9 = 0 Þ 2x( 2x + 9) - ( 2x + 9) = 0 Þ ( 2x + 9)( 2x - 1) = 0 9 1 Þ x = - or x = × 2 2 9 Putting x = - in (i), we get 2 y 2 = -18 Þ y is imaginary. Putting x =

1 in (i), we get y 2 = 2 Þ y = ± 2. 2

æ1 So, the given curves intersect at the points A ç , è2 æ1 ö B ç , - 2÷ × è2 ø

ö 2 ÷ and ø

Since the equation of each of the given curves contains only even powers of y, each curve is symmetrical about the x-axis. Required area = 2 ( area ODCAO) = 2 ( area ODAO + area DCAD) 3/ 2 ü ì1/ 2 ï ï = 2 í ò ( y of the parabola) dx + ò ( y of the circle) dx ý ïî 0 ïþ 1/ 2

Senior Secondary School Mathematics for Class 12 Pg-877

Area of Bounded Regions

877

3/ 2 ì1/ 2 ü 9 ï ï = 2 í ò 2 x dx + ò - x 2 dx ý 4 1/ 2 îï 0 þï

ì é ï 1/ 2 ê x ïé 4 3/ 2 ù = 2 íê x ú +ê û0 ê ïë 3 ê ï ë î =

9 - x2 9 4 + sin -1 2 8

3/ 2 ü ù ï æ x öú ï çç ÷÷ ú ý è 3/2 ø ú ï úû 1/ 2 ï þ

1ö æ9 ö æ 2 9 + sin -1 ÷÷ + ç sin -1 1÷ - çç 4 3ø 3 2 è4 ø è 2 4

æ2 2 2 ö 9 æ -1 1 ÷ + ç sin 1 - sin -1 ö÷ = çç ÷ 2 ø 4è 3ø è 3 ì 2 9æp 1 öü =í + ç - sin -1 ÷ ý sq units 4è 2 3 øþ î 6 ì 2 9 æ 1 öü =í + cos-1 ç ÷ ý sq units. 4 è 3 øþ î 6 ì 2 9 1ü Hence, the required area is í + cos-1 ý sq units. 4 3þ î 6 EXAMPLE 21

Find the area of the region {( x , y) : x 2 + y 2 £ 1 £ x + y}.

SOLUTION

Let R = {( x , y) : x 2 + y 2 £ 1 £ x + y} = {( x , y) : x 2 + y 2 £ 1} Ç {( x , y) : x + y ³ 1} = R1 Ç R 2. Clearly, R1 is the interior of the circle x 2 + y 2 = 1 with its centre at O(0, 0) and radius = 1 unit. And, R 2 is the region lying above the line x + y = 1. Y Consider the equations x2 + y2 = 1

… (i)

and … (ii) x+ y =1 Putting y = (1 - x) from (ii) in (i), we get x 2 + (1 - x) 2 = 1 Þ 2x 2 - 2x = 0

A(0, 1) C X

B(1, 0) O

X

Þ 2x( x - 1) = 0 Þ x = 0 or x = 1. Y Now, ( x = 0 Þ y = 1) and ( x = 1 Þ y = 0). Thus, the points of intersection of (i) and (ii) are A( 0, 1) and B(1, 0). So, the required area is the shaded region.

Senior Secondary School Mathematics for Class 12 Pg-878

878

Senior Secondary School Mathematics for Class 12

Required area = area BCAB = ( area AOBCA) - ( area OBAO) 1

1

0

0

= ò 1 - x 2 dx - ò (1 - x) dx 1

1 é x x2 ù é1 ù 1 - x2 ú - ê x - ú = ê sin -1 x + 2 2û ë2 û0 ë 0

æ1 ö 1 æ 1 pö 1 æ p 1ö = ç sin -1 1÷ - = ç ´ ÷ - = ç - ÷ sq units. 2 è ø 2 è 2 2 ø 2 è 4 2ø æ p 1ö Hence, the required area is ç - ÷ sq units. è 4 2ø EXAMPLE 22

Find the area of the region {( x , y) : x 2 + y 2 £ 2 ax , y 2 ³ ax , x ³ 0, y ³ 0}.

SOLUTION

Clearly, we have to find the area of the region lying in the first quadrant ( x ³ 0, y ³ 0), included between the circle x 2 + y 2 = 2 ax and the parabola y 2 = ax. Thus, the equations of the given curves are and

x 2 + y 2 = 2 ax

… (i)

2

… (ii)

y = ax 2

2

Now, clearly x + y = 2 ax is a circle with its centre B( a , 0) and radius = a units. And, y 2 = ax is a parabola with O(0, 0) as its vertex and the x-axis as its axis. We can draw the figure, as shown. Their points of intersection may be obtained by solving (i) and (ii) and keeping in view that x ³ 0 and y ³ 0. Using (ii) in (i), we get x 2 - ax = 0 Þ x( x - a) = 0 Þ x = 0 or x = a. Now, ( x = 0 Þ y = 0) and ( x = a Þ y = a). Thus, the two curves intersect at O( 0, 0) and A( a , a). a

a

0

0

\ required area = ò 2ax - x 2 dx - ò ax dx a

a

0

0

= ò a 2 - ( x - a) 2 dx - a × ò x dx

Senior Secondary School Mathematics for Class 12 Pg-879

Area of Bounded Regions

é ( x - a) =ê ê ë

879 a

ù a 2 - ( x - a) 2 a 2 æ x - aöú + sin -1 ç ÷ 2 2 è a øú û0 a

é2 ù - a ê x 3/2 ú ë3 û0 ì a2 2 ü a2 sin -1( -1) - a 2 ý = í sin -1( 0) 3 þ 2 2 î æ pa 2 2 2 ö =ç - a ÷ sq units. ç 4 3 ÷ø è æ pa 2 2 2 ö Hence, the required area is ç - a ÷ sq units. ç 4 3 ÷ø è EXAMPLE 23 SOLUTION

Find the area of the region {( x , y) : x 2 £ y £ |x|}. [CBSE 2009, ‘12C, ‘13] Consider the equations x2 = y and

… (i)

y =|x|

… (ii)

Clearly, x 2 = y represents an upward parabola with its vertex at O( 0, 0). All the points inside this parabola represent x 2 £ y. ì x , when x ³ 0 Also, y = |x| = í î- x , when x < 0. The lines OA and OB, each equally inclined to the axes, represent y =|x|. All the points below the lines OA and OB, and above the x-axis represent y £|x|. Thus, the shaded portion is the required region. In each of the given equations, the equation remains unchanged when x is replaced by - x. So, each of the given curves is symmetrical about the y-axis. \ required area = 2( area OEAO). In this region, we have and

x2 = y y=x

Using (iv) in (iii), we get

… (iii) … (iv) x 2 = x Þ x( x - 1) = 0 Þ x = 0 or x = 1.

Now, ( x = 0 Þ y = 0) and ( x = 1 Þ y = 1). Thus, the line y = x and the curve x 2 = y intersect at O( 0, 0) and A(1, 1). Draw AD ^ OX and BC ^ OX¢.

Senior Secondary School Mathematics for Class 12 Pg-880

880

Senior Secondary School Mathematics for Class 12

\ required area = 2( area OEAO) = 2[( area ODAO) - ( area ODAEO)] = 2[( area between y = x and the x-axis from x = 0 to x = 1) – (area between x 2 = y and the x-axis from x = 0 to x = 1)] 1 æ1 ö = 2 ç ò y dx for the line OA - ò y dx for the curve OEA ÷ ç ÷ è0 ø 0 1 ìé 2 ù 1 é 3 ù 1 ü é1 ù x ï x ï é1 1 ù = 2 ê ò x dx - ò x 2 dx ú = 2íê ú - ê ú ý = 2 ê - ú 2 3 ë2 3û êë 0 úû ïîë û 0 ë û 0 ïþ 0 æ 1ö 1 = ç 2 ´ ÷ = sq unit. 6ø 3 è 1 Hence, the required area is sq unit. 3

EXAMPLE 24

Find the area bounded by the line y = x and the curve y = x 3.

SOLUTION

The given equations are y=x y=x

and

… (i)

3

… (ii)

Using (i) in (ii), we get x - x 3 = 0 Þ x(1 - x 2) = 0 Þ x(1 - x)(1 + x) = 0 Þ x = 0 or x = 1 or x = -1. Also, ( x = 0 Þ y = 0), ( x = 1 Þ y = 1) and ( x = -1 Þ y = -1). So, the given curve and the line intersect at the points O( 0, 0), A(1, 1) and B( -1, - 1). Now, y = x is a line passing through the origin and making an angle of 45° with the x-axis. Thus, the line y = x can be drawn. For the curve y = x 3 some values for x and the corresponding values of y are given below: Y

x y

–1

1 2

–1

1 8

0

1 2

0

1 8

A(1, 1)

1 1

Plotting the points ( -1, - 1), 1ö æ 1 æ1 1ö ç - , - ÷ , ( 0, 0), ç , ÷ , and 8ø è 2 è 2 8ø

X

O

M D

C L

X

Y

(1, 1), and joining them, we get a rough sketch of y = x 3 , as shown in the given figure.

Senior Secondary School Mathematics for Class 12 Pg-881

Area of Bounded Regions

881

Required area = ( area ACOA) + ( area ODBO) = ( area OALO) - ( area OCALO) + ( area OBMO) - ( area ODBMO) 1

1

0

0

0

-1

= ò {y for (i)} dx - ò {y for (ii)} dx + ò {( - y) for (i)} dx 0

- ò {( - y) for (ii)} dx -1

1

1

0

0

= ò x dx - ò x dx + 3

1

0

0

-1

-1

ò -x dx - ò -x

1

3

dx 0

0

é x2 ù é x4 ù é -x 2 ù é x4 ù = ê ú -ê ú + ê ú +ê ú ë 2 û 0 ë 4 û 0 ë 2 û -1 ë 4 û -1 æ1 1 1 1ö 1 = ç - + - ÷ = sq unit. è 2 4 2 4ø 2 Hence, the required area is 0.5 sq unit. EXAMPLE 25

Find the area bounded by the curve y = sin x between x = 0 and x = 2p.

SOLUTION

The given curve is y = sin x. Some values of x and the corresponding values of y are given below: x

0

p 6

p 2

p

3p 2

2p

y

0

1 2

1

0

–1

0

Taking a fixed unit (distance) for p along the x-axis, we can plot æ p 1ö æ p ö æ 3p ö the points (0, 0), ç , ÷ , ç , 1÷ , ( p , 0), ç , - 1÷ and ( 2p , 0). è 6 2ø è 2 ø è 2 ø Join these points freehand to obtain a rough sketch of the given curve. Y A O

B

D C

Y

X

Senior Secondary School Mathematics for Class 12 Pg-882

882

Senior Secondary School Mathematics for Class 12

Required area = ( area OABO) + ( area BCDB) p

2p

0

p

= ò y dx + ò ( - y) dx [Q area BCDB is below the x-axis] p

2p

= ò sin x dx - ò sin x dx p

0

= [- cos

p x]0

2p

- [- cos x]p = ( 2 + 2) = 4 sq units.

Hence, the required area is 4 sq units. EXAMPLE 26

Find the area of the region bounded by the curve y = x 2 + 2, and the lines

SOLUTION

y = x , x = 0 and x = 3. y = x 2 + 2 Þ x 2 = ( y - 2). Clearly, x 2 = ( y - 2) represents an upward parabola with its vertex at A(0, 2).

Also, y = x represents the straight line, making an angle of 45° with the positive direction of the x-axis. And, x = 0 is the y-axis, while x = 3 represents a line parallel to the y-axis at a distance of 3 units from it. Thus, the shaded region in the given figure is the required area. \ required area = ( area ODCAO) - ( area ODBO) 3

3

= ò ( x 2 + 2) dx - ò x dx 0

0

3

3

éx ù é x2 ù 9 ö 21 æ sq units. =ê + 2x ú - ê ú = ç15 - ÷ = 3 2 2ø 2 ë û0 ë û0 è 3

Hence, the required area is EXAMPLE 27

21 sq units. 2

Find the area of the region {( x , y) : 0 £ y £ ( x 2 + 1), 0 £ y £ ( x + 1), 0 £ x £ 2}.

[CBSE 2009]

Senior Secondary School Mathematics for Class 12 Pg-883

Area of Bounded Regions

Let R = {( x , y) : 0 £ y £ ( x 2 + 1), 0 £ y £ ( x + 1), 0 £ x £ 2} = {( x , y) : 0 £ y £ ( x 2 + 1)} Ç {( x , y) : 0 £ y £ ( x + 1)} Ç {( x , y) : 0 £ x £ 2} = R1 Ç R 2 Ç R 3. Clearly, R1 is the region consisting of the right-hand side of the y-axis, lying below the parabola y = x 2 + 1.

x 2+

1

Also, R 2 is the region consisting of the right-hand side of the y-axis, lying below the line y = ( x + 1). And, R 3 is the region above the x-axis, lying between the ordinates x = 0 and x = 2. Thus, R1 Ç R 2 Ç R 3 is the shaded region.

0=x

y=

y=

B O

1

E

C

A

X

x+

x=2

SOLUTION

883

D (1, 0)

X

F (2, 0)

We have, y = x 2 + 1 and y = x + 1 Þ x 2 + 1 = x + 1 Þ x( x - 1) = 0 Þ x = 0 or x = 1. Now, ( x = 0 Þ y = 1) and ( x = 1 Þ y = 2). Thus, the parabola y = ( x 2 + 1) and the line y = x + 1 intersect at the points A( 0, 1) and C(1, 2). \ required area = area of the shaded region = ( area ODCBA ) + ( area CDFEC ) 1

2

= ò ( y of the parabola) dx + ò ( y of the line) dx 0 1

1

2

= ò ( x 2 + 1) dx + ò ( x + 1) dx 0

1

1

2

éx ù é x2 ù =ê + xú + ê + xú 3 2 ë û0 ë û1 3

3 ö 23 æ1 ö æ sq units. = ç + 1÷ + ç 4 - ÷ = 2ø 6 è3 ø è Hence, the required area is

23 sq units. 6

Senior Secondary School Mathematics for Class 12 Pg-884

884

Senior Secondary School Mathematics for Class 12

EXAMPLE 28

Find the area of the region bounded by the curve y 2 = 2y - x and the

SOLUTION

y-axis. y 2 = 2y - x Þ y 2 - 2y = - x

Y

B(0, 2) Þ y 2 - 2y + 1 = - x + 1 Þ ( y - 1) 2 = - ( x - 1) Þ Y 2 = -X, A(1, 1) where y - 1 = Y and ( x - 1) = X. This is a left-handed parabola with X X O vertex at (X = 0, Y = 0). X = 0, Y = 0 Þ - x + 1 = 0 and y - 1 = 0 Þ x = 1 and y = 1. Thus, the vertex of the given parabola is A(1, 1). Also, x = 0 Þ y 2 - 2y = 0 Þ y( y - 2) = 0 Þ y = 0 or y = 2. Thus, the curve meets the y-axis at O( 0, 0) and B( 0, 2). A rough sketch of the curve can be drawn, as shown in the figure.

2

2

\ required area = ò x dy = ò ( 2y - y 2) dy 0

0

3 ù2

é 8ö 4 y æ = ê y2 ú = ç 4 - ÷ = sq units. 3 3ø 3 ë û0 è 4 Hence, the required area is sq units. 3

EXERCISE 17 1. Find the area of the region bounded by the curve y = x 2 , the x-axis, and the lines x = 1 and x = 3. 2. Find the area of the region bounded by the parabola y 2 = 4x , the x-axis, and the lines x = 1 and x = 4. 3. Find the area under the curve y = 6x + 4 (above the x-axis) from x = 0 to x = 2. 4. Determine the area enclosed by the curve y = x 3 , and the lines y = 0, x = 2 and x = 4. 5. Determine the area under the curve y = a 2 - x 2 , included between the lines x = 0 and x = 4. 6. Using integration, find the area of the region bounded by the line 2y = 5 x + 7 , the x-axis, and the lines x = 2 and x = 8. 7. Find the area of the region bounded by the curve y 2 = 4x and the line x = 3.

Senior Secondary School Mathematics for Class 12 Pg-885

Area of Bounded Regions

8. Evaluate the area bounded by the ellipse

885

x2 y2 + = 1 above the x-axis. 4 9

9. Using integration, find the area of the region bounded by the lines y = 1 + |x + 1|, x = -2, x = 3 and y = 0. 10. Find the area bounded by the curve y = ( 4 - x 2), the y-axis and the lines y = 0, y = 3. 11. Using integration, find the area of the region bounded by the triangle whose vertices are A( -1, 2), B(1, 5) and C( 3 , 4).

[CBSE 2014]

12. Using integration, find the area of the a ABC, the equations of whose sides AB , BC and AC are given by y = 4x + 5 , x + y = 5 and 4y = x + 5 respectively. 13. Using integration, find the area of the region bounded between the line x = 2 and the parabola y 2 = 8x. 14. Using integration, find the area of the region bounded by the line y - 1 = x , the x-axis, and the ordinates x = -2 and x = 3. 15. Sketch the region lying in the first quadrant and bounded by y = 4x 2 , x = 0, y = 2 and y = 4. Find the area of the region using integration. 16. Sketch the region lying in the first quadrant and bounded by y = 9x 2 , x = 0, y = 1 and y = 4. Find the area of the region, using integration. 17. Find the area of the region enclosed between the circles x 2 + y 2 = 1 and ( x - 1) 2 + y 2 = 1.

[CBSE 2007, ’13C]

18. Sketch the region common to the circle x 2 + y 2 = 16 and the parabola x 2 = 6y. Also, find the area of the region, using integration. 19. Sketch the region common to the circle x 2 + y 2 = 25 and the parabola y 2 = 8x. Also, find the area of the region, using integration. 20. Draw a rough sketch of the region {( x , y) : y 2 £ 3 x , 3 x 2 + 3 y 2 £ 16} and find the area enclosed by the region, using the method of integration. 21. Draw a rough sketch and find the area of the region bounded by the parabolas y 2 = 4x and x 2 = 4y , using the method of integration. 22. Find by integration the area bounded by the curve y 2 = 4ax and the lines y = 2a and x = 0. 23. Find the area between the curve y = x = 0 and x = p.

x + 2 sin 2 x , the x-axis and the ordinates p

Senior Secondary School Mathematics for Class 12 Pg-886

886

Senior Secondary School Mathematics for Class 12

24. Find the area bounded by the curve y = cos x , the x-axis and the ordinates x = 0 and x = 2p. 25. Compare the areas under the curves y = cos2 x and y = sin 2 x between x = 0 and x = p. 26. Using integration, find the area of the triangle, the equations of whose sides are y = 2x + 1, y = 3 x + 1 and x = 4. 27. Find the area of the region {( x , y) : x 2 £ y £ x}. 28. Find the area of the region bounded by the curve y 2 = 2y - x and the y-axis. 29. Draw a rough sketch of the curves y = sin x and y = cos x , as x varies from p 0 to , and find the area of the region enclosed between them and the 2 x-axis. 30. Find the area of the region bounded by the parabola y 2 = 2x + 1 and the line x - y = 1. 31. Find the area of the region bounded by the curve y = 2x - x 2 and the straight line y = - x. 32. Find the area of the region bounded by the curve ( y - 1) 2 = 4( x + 1) and the line y = x - 1. 33. Find the area of the region bounded by the curve y = x and the line y = x. 34. Find the area of the region included between the parabola y 2 = 3 x and the circle x 2 + y 2 - 6x = 0, lying in the first quadrant. 35. Find the area bounded by the curve y = cos x between x = 0 to x = 2p. 36. Using integration, find the area of the region in the first quadrant, enclosed by the x-axis, the line y = x and the circle x 2 + y 2 = 32. [CBSE 2014] 37. Using integration, find the area of the triangle whose vertices are A( 2, 3), [CBSE 2012C] B( 4, 7) and C( 6, 2). 38. Using integration, find the area of the triangle whose vertices are A(1, 3), [CBSE 2009C] B( 2, 5) and C( 3 , 4). 39. Using integration, find the area of the triangular region bounded by the lines y = 2x + 1, y = 3 x + 1 and x = 4. [CBSE 2011] ANSWERS (EXERCISE 17)

1.

26 sq units 3

4. 60 sq units 7. 8 3 sq units 14 sq units 3 32 sq units 13. 3 10.

28 sq units 3 2 pa 5. sq units 4 3p 8. sq units 2 2.

11. 4 sq units 14. 8.5 sq units

3.

56 sq units 9

6. 96 sq units 27 sq units 2 15 12. sq units 2 1 15. ( 8 - 2 2) sq units 3 9.

Senior Secondary School Mathematics for Class 12 Pg-887

Area of Bounded Regions

887

æ 2p 3ö 16p + 4 3 ÷ sq units 18. sq units 17. çç ÷ 3 3 2 ø è ì 8 2 3 / 2 25 p aü 19. í ×a + - a 25 - a 2 - 25 sin -1 ý sq units, where a = 41 - 4. 2 5þ î 3 æ 3 a ö ïü 16 2 16 -9 + 273 ïì 4 3 / 2 8p ÷ ý sq units, where a = 20. í a + -a -a sin -1 çç ÷ï 3 3 3 4 6 ïî 3 è øþ 16.

14 sq units 9

21.

16 sq units 3

22.

24. 4 sq units

25.

1 sq unit 6 16 30. sq units 3 1 33. sq unit 6

28.

27.

31. 34.

36. 4p sq units

2a 2 sq units 3 p sq units each 2 4 sq units 3 9 sq units 2 3 ( 3 p - 8) sq units 4

37. 9 sq units

23.

3p sq units 2

26. 8 sq units 29. ( 2 - 2) sq units 32.

64 sq units 3

35. 4 sq units 38.

3 sq units 2

39. 8 sq units HINTS TO SOME SELECTED QUESTIONS (EXERCISE 17) 7. Since the equation y 2 = 4 x contains only even powers of y, the curve is symmetrical about the y-axis. 3

\ required area = 2 × ò 2 x dx. 0

8. The given ellipse meets the x-axis at A( -2 , 0 ) and B( 2 , 0 ). x2 y2 3 + = 1Þ y = × 4 - x2 . 4 9 2 2 2 3 \ required area = ò y dx = × ò 4 - x 2 dx 2 -2 -2 2

=2´

3 ´ 4 - x 2 dx 2 ò0

p/2

= 6×

ò cos2 q dq, where x = 2 sin q

0 p/2

= 3×

ò ( 1 + cos 2 q) dq. 0

ì 1 + ( x + 1), when x + 1 ³ 0 9. y = 1 + |x + 1| = í î 1 - ( x + 1), when x + 1 < 0

Senior Secondary School Mathematics for Class 12 Pg-888

888

Senior Secondary School Mathematics for Class 12 ì x + 2 , when x ³ -1 =í î - x , when x < -1 3

-1

-2 -1

-2

3

ò y dx = ò y dx + ò y dx

Required area = =

-1

3

ò ( - x ) dx + ò ( x + 2 ) dx.

-2

-1

3

3

0

0

X

X

3

10. Required area = ò x dy = ò 4 - y dy. 11. Equations of AB, BC and AC are 1 1 y = ( 3 x + 7 ), y = ( 11 - x ) 2 2 1 and y = ( x + 5 ) respectively. 2 1 3 3 1 1 1 \ D = ò ( 3 x + 7 ) dx + ò ( 11 - x ) dx - ò ( x + 5 ) dx. 2 -1 21 2 -1

B(0, 5)

=

3

-1

0

+ = y

(-1, 1)A

3

ò yAB dx + ò yBC dx - ò yCA dx

5

Draw AL , BM and CN perpendiculars on the x-axis. 0

y=

4x

points A( -1, 1), B( 0 , 5 ) and C( 3 , 2 ).

Required area =

x+

5

12. Solving the given equations in pairs, we get the

C(3, 2)

4y = x + 5

-1

0

3

-1

0

1

3

L

N

M

ò ( 4 x + 5 ) dx + ò ( - x + 5 ) dx - 4 × ò ( x + 5 ) dx. -1

Y

2

x=2

13. Required area = 2 × ò 8 x dx. 0

O

y 2=

14. Let AB be the given line, intersecting the x-axis at C( -1, 0 ). Required area = ( area CDAC + area CBEC ) = =

3

-1

-1 3

-2

8x

A

ò y dx + ò ( - y ) dx -1

ò ( x + 1) dx + ò - ( x + 1) dx.

-1

-2

X

E

D (3, 0)

B Y

X

Senior Secondary School Mathematics for Class 12 Pg-889

Area of Bounded Regions

x=0

15. Clearly, we have to find the area of the region bounded by the curve y = 4 x 2 , the y-axis, and the lines y = 2 and y = 4. 4

4

2

2

Required area = ò x dy = ò

889

1 y dy. 2

y=4 y=2

X

X

17. ( x 2 + y 2 ) = 1 is a circle with its centre at (0, 0) and radius = 1 unit. Y A

X

O

1 2

1 D 2

3 2

C(1, 0)

X

B Y

And, ( x - 1) 2 + y 2 = 1 is a circle with its centre at (1, 0) and radius = 1 unit. The given equations are ( x2 + y2 ) = 1

… (i)

and ( x - 1) 2 + y 2 = 1.

… (ii)

Using (i) in (ii), we get 1 - 2 x = 0 Þ x =

1 × 2

1 3 in (i), we get y = ± × 2 2 æ1 æ1 - 3ö 3 ö÷ ÷× So, the two circles intersect at A çç , and B çç , ÷ 2 ÷ø è2 2 ø è2

Putting x =

Required area = area AOBCA = 2 ( area AOCA ) = 2 ( area ODAO + area DCAD ) 1 ì1/2 ü ï ï = 2 í ò 1 - ( x - 1) 2 dx + ò 1 - x 2 ý 1/2 îï 0 þï

18. The given curves are x 2 + y 2 = 16 and

2

x = 6y

… (i) … (ii)

Senior Secondary School Mathematics for Class 12 Pg-890

890

Senior Secondary School Mathematics for Class 12 Y

On solving (i) and (ii), we get the points

C

A( -2 3 , 2 ) and B( 2 3 , 2 ). Required area

B(2 3, 2)

A

= 2( area OBCO ) = 2[area ODBCO - area ODBO ] ì2 3

ï = 2í ïî

ò

16 - x 2 dx -

0

X

ò

0

D

O

x 2 üï dx ý × 6 ïþ

2 3

X

Y

19. The given curves are x 2 + y 2 = 25

… (i)

y 2 = 8x

… (ii)

and

On solving (i) and (ii), we get x 2 + 8 x - 25 = 0 Þ x =

-8 ±

Þ x = -4 +

64 + 100 2 41

(rejecting –ve value).

Putting y = 0 in (i), we get x = ± 5. Thus, the circle (i) cuts the x-axis at C(5 , 0 ) and C ¢( -5 , 0 ). Required area = 2 [area ODCAO ] = 2 [( area ODAO ) + ( area ADCA )] æa = 2 ç ò 8 x dx + ç è0

5

ò

a

ö 25 - x 2 dx ÷ , where a = - 4 + ÷ ø

41.

Y

A

X

( 5, 0) C

O

D

C(5, 0)

B

Y

21. See Example 14, taking a = 1.

X

Senior Secondary School Mathematics for Class 12 Pg-891

Area of Bounded Regions

22. Required area =

2a

2a 2

0

0

891 Y

2 a2 sq units . 3

y

ò x dy = ò 4 a dy =

X

X

O

Y Y

28. y 2 = 2 y - x Þ y 2 - 2 y + 1 = - x + 1 2

Þ ( y - 1) = - ( x - 1).

y=2

So, the given equation represents a parabola that opens on the left, having vertex at (1, 1). Now, x = 0 Þ y 2 = 2 y Þ y 2 - 2 y = 0

(1, 1)

Þ y( y - 2 ) = 0 Þ y = 0 or y = 2. Thus, the curve meets the y-axis at (0, 0) and (0, 2)

X

X

O

2

Required area = ò ( 2 y - y 2 ) dy.

Y

0

Y

29. A rough sketch of y = sin x and y = cos x from p x = 0 to x = is shown herewith. 2 p/4

Required area =

ò

y=

p/2

sin x dx +

ò cos x dx.

cos

y=

x

sin

x

p/4

0

O

35. Tabular values of y = cos x : x

0

p 2

p

3p 2

2p

y

1

0

-1

0

1

A rough sketch of the curve is given below: Y

1 O 2

2

Y

Required area =

p/2

3 p/2

2p

0

p/2

3 p/2

ò cos x dx + ò - cos x dx + ò cos x dx

X 4

2

Senior Secondary School Mathematics for Class 12 Pg-892

892

Senior Secondary School Mathematics for Class 12

36. Analogous to Example 27. 4

4 2

0

4

Required area = ò x dx +

ò

32 - x 2 dx 4 2

4 é x 32 - x 2 é x2 ù x ù 32 ú + =ê ú + ê sin -1 2 2 2 4 2ú êë úû 0 ëê û4 = 4p sq units. æ -1 ö 39. y = 2 x + 1 is the line passing through A ç , 0 ÷ and B( 0 , 1). è 2 ø

æ -1 ö y = 3 x + 1 is the line passing through C ç , 0 ÷ and B( 0 , 1). è 3 ø x = 4 is the line parallel to y-axis and passing through D( 4 , 0 ). Required area = ( area ODFB) - ( area ODEB) 4

4

0

0

= ò ( 3 x + 1) dx - ò ( 2 x + 1) dx = ( 28 - 20 ) sq units = 8 sq units

Senior Secondary School Mathematics for Class 12 Pg-893

18. DIFFERENTIAL EQUATIONS AND THEIR FORMATION DIFFERENTIAL EQUATION An equation containing an independent variable, a dependent variable and the derivatives of the dependent variable is called a differential equation.

Examples

Each of the following equations is a differential equation: dy d 2y dy (i) (ii) + 5 y = ex -2 + 3 y = sin x 2 dx dx dx dy x3 -y3 (iv) x 2 dx + y 2 dy = 0 (iii) = 2 dx xy - x 2 y

ORDER OF A DIFFERENTIAL EQUATION The order of the highest-order derivative occurring in a differential equation is called the order of the differential equation. DEGREE OF A DIFFERENTIAL EQUATION The power of the highest-order derivative occurring in a differential equation, after it is made free from radicals and fractions, is called the degree of the differential equation. EXAMPLE 1

SOLUTION

Write the order and the degree of the differential equation d 2y dy +5 + 3 y = 0. 2 dx dx d 2y In the given equation, the highest-order derivative is and its dx 2 power is 1. \

EXAMPLE 2

its order = 2 and degree = 1.

Write the order and the degree of the differential equation 2

4 æ d 3y ö æ dy ö x ç 3 ÷ + ç ÷ + y 2 = 0. ç dx ÷ è dx ø è ø

SOLUTION

In the given equation, the highest-order derivative is power is 2. \

EXAMPLE 3

its order = 3 and degree = 2.

Write the order and the degree of the differential equation 2

y=x

dy æ dy ö + 1+ ç ÷ × dx è dx ø 893

d 3y dx 3

and its

Senior Secondary School Mathematics for Class 12 Pg-894

894

Senior Secondary School Mathematics for Class 12

SOLUTION

The given equation may be written as 2

dy ö æ dy ö æ 1 + ç ÷ = çy - x ÷ dx dx ø è ø è 2

dy ö æ dy ö æ Þ 1 + ç ÷ = çy - x ÷ dx ø è dx ø è

2

[on squaring both sides]

2

2

dy æ dy ö æ dy ö Þ 1 + ç ÷ = y 2 + x 2 ç ÷ - 2xy dx è dx ø è dx ø 2

dy æ dy ö Þ (1 - x 2) ç ÷ + 2xy + (1 - y 2) = 0. dx è dx ø Clearly, it is a differential equation of order = 1 and degree = 2. EXAMPLE 4

Write the order and degree of the differential equation 2 ïì æ dy ö ïü í1 + ç ÷ ý ïî è dx ø ïþ

SOLUTION

3/2

æ d 2y ö = kç 2 ÷ × ç dx ÷ è ø

On squaring both sides, we get 3

2

2 ìï æ dy ö 2 üï 2æ d y ö í1 + ç ÷ ý = k çç 2 ÷÷ × ïî è dx ø ïþ è dx ø

In this equation, the highest-order derivative is

d 2y dx 2

whose power

is 2. \ its order = 2 and degree = 2. EXAMPLE 5

SOLUTION

Find the order and the degree of the differential equation dy y = px + a 2 p 2 + b 2 , where p = × dx y = px + a 2 p 2 + b 2 Þ

y - px = a 2 p 2 + b 2

Þ ( y - px) 2 = a 2 p 2 + b 2 Þ

2

2 2

[on squaring both sides] 2 2

y + x p - 2xyp = a p + b 2

Þ ( x 2 - a 2) p 2 - 2xyp + ( y 2 - b 2) = 0 2

æ dy ö æ dy ö Þ ( x 2 - a 2) ç ÷ - 2xy × ç ÷ + ( y 2 - b 2) = 0. è dx ø è dx ø Clearly, it is a differential equation of order 1 and degree 2. In case of differential equations involving one or more terms of æ dy ö æ dy ö æ dy ö , log ç ÷ , sin ç ÷ , cos ç ÷ , etc., the degree is not defined. è dx ø è dx ø è dx ø

AN IMPORTANT NOTE ( dy/dx )

the form e

However, the degree of the differential equation containing terms like ex , ey , log x , log y , tan x , tan y , etc., is defined as usual.

Senior Secondary School Mathematics for Class 12 Pg-895

Differential Equations and Their Formation EXAMPLE 6

SOLUTION

895

Find the order and degree (if any) of each of the differential equations given below: 2 dy æ dy ö (ii) ç ÷ + y = ex (i) - tan x = 0 dx è dx ø 2 d y (iv) ( y ¢¢) 2 + cos y ¢ = 0 (iii) 2 = sin 3 x + cos 3 x dx æ d 3y ö d 4y (v) y ¢¢ + 2y ¢ + sin y = 0 (vi) 4 + sin ç 3 ÷ = 0 ç dx ÷ dx è ø 2 2 d y æ dy ö (viii) 3 2 + 5 ç ÷ = log x (vii) y ¢¢¢ + y 2 + ey¢ = 0 è dx ø dx (i) The given equation is

dy - tan x = 0. dx

In this equation, the highest-order derivative is

dy whose dx

power is 1. \ its order = 1 and degree = 1. 2

æ dy ö (ii) The given equation is ç ÷ + y = ex . è dx ø dy In this equation, the highest-order derivative is whose dx power is 2. \ its order = 1 and degree = 2. d 2y

= sin 3 x + cos 3 x. dx 2 d 2y In this equation, the highest-order derivative is and its dx 2 power is 1. \ its order = 2 and degree = 1.

(iii) The given equation is

2

æ d 2y ö æ dy ö (iv) The given equation is ç 2 ÷ + cos ç ÷ = 0. ç dx ÷ è dx ø è ø In this equation, the highest-order derivative is order is 2.

d 2y dx 2

, so its

æ dy ö It has a term cos ç ÷ , so its degree is not defined. è dx ø d 2y dy (v) The given equation is 2 + 2 + sin y = 0. dx dx d 2y and its In this equation, the highest-order derivative is dx 2 power is 1. \ its order = 2 and degree = 1.

Senior Secondary School Mathematics for Class 12 Pg-896

896

Senior Secondary School Mathematics for Class 12

æ d 3y ö + sin ç 3 ÷ = 0. ç dx ÷ dx è ø d 4y In this equation, the highest-order derivative is , so its dx 4 order is 4. æ d 3y ö It has a term sin ç 3 ÷ , so its degree is not defined. ç dx ÷ è ø

(vi) The given equation is

(vii) The given equation is

d 4y 4

d 3y dx 3

( dy/dx )

+ y2 + e

= 0.

In this equation, the highest-order derivative is

d 3y dx 3

, so its

order is 3. ( dy/dx ) It has a term e , so its degree is not defined. d 2y

2

æ dy ö + 5 ç ÷ = log x. 2 è dx ø dx d 2y In this equation, the highest-order derivative is and its dx 2 power is 1. \ its order = 2 and degree = 1.

(viii) The given equation is 3

EXERCISE 18A Write order and degree (if defined) of each of the following differential equations. 4 æ d 2y ö æ dy ö [CBSE 2008C, ’13C] 1. ç ÷ + 3 y ç 2 ÷ = 0 ç dx ÷ è dx ø è ø 2 4 æ d 2y ö æ dy ö 2. x 3 ç 2 ÷ + x ç ÷ = 0 [CBSE 2013] ç dx ÷ è dx ø è ø 2 3 æ d 2s ö æ ds ö 3. ç 2 ÷ + ç ÷ + 4 = 0 [CBSE 2013] ç dt ÷ è dt ø è ø 2 3 4 2 æ d 3y ö æ d 2y ö d 2 y æ dy ö æ dy ö 5. 4. ç 3 ÷ + ç 2 ÷ + ç ÷ + y5 = 0 + + 2y = 0 ç ÷ ç dx ÷ ç dx ÷ è dx ø dx 2 è dx ø è ø è ø dy d 2y ( dy/dx ) 7. 6. + y = ex + y2 + e =0 dx dx 2 æ d 3y ö dy d 4y æ dy ö 9. 8. + sin ç ÷ = 0 - cos ç 3 ÷ = 0 4 ç dx ÷ dx è dx ø dx è ø 2 d 2y æ dy ö 10. [CBSE 2010] + 5 x ç ÷ - 6y = log x è dx ø dx 2

Senior Secondary School Mathematics for Class 12 Pg-897

Differential Equations and Their Formation 3

2

æ dy ö æ dy ö 11. ç ÷ - 4 ç ÷ + 7 y = sin x è dx ø è dx ø 2 æ dy ö 13. x ç ÷ + + 9 = y2 è dx ø æ dy ö ç ÷ è dx ø 15.

12.

14.

1 - y 2 dx + 1 - x 2 dy = 0

d 3y dx

+2

3

d 2y dx 2

+

897

dy =0 dx

2 æ d 2y ö æ dy ö 1 - ç ÷ = ça 2 ÷ ç dx ÷ è dx ø è ø

1/ 3

16. ( y ¢¢) 3 + ( y ¢) 2 + sin y ¢ + 1 = 0

17. ( 3 x + 5 y) dy - 4x 2 dx = 0

18. y =

dy 5 + dx æ dy ö ç ÷ è dx ø

ANSWERS (EXERCISE 18A)

1. 2, 1

2. 2, 2

8. 1, not defined 14. 2, 2

15. 1, 1

3. 2, 2

4. 3, 2

5. 2, 1

6. 1, 1

9. 4, not defined

10. 2, 1

11. 1, 3

16. 2, not defined

17. 1, 1

18. 1, 2

7. 2, not defined 12. 3, 1

13. 1, 2

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 18A) 13. The given differential equation on simplification becomes 2 æ dy ö æ dy ö æ dy ö xç ÷ + 2 + 9ç ÷ = y 2 ç ÷ × è dx ø è dx ø è dx ø So, its order = 1 and degree = 2. 2/3

2 2 ìï æ dy ö üï æ d y ö 14. On squaring both sides, í 1 - ç ÷ ý = ç a 2 ÷ × ç è dx ø ïþ è dx ÷ø ïî On cubing both sides of (i), we get 2 2 3 ìï æ d2 y ö æ dy ö üï ç ÷ ç ÷ 1 = a × So, its order = 2 and degree = 2. í ý ç dx 2 ÷ è dx ø ïþ ïî è ø

15.

2 dy - 1 - y = × So, its order =1 and degree = 1. 2 dx 1- x

17.

dy 4x 2 = × So, its order = 1 and degree = 1. dx ( 3 x + 5 y )

… (i)

SOLUTION OF A DIFFERENTIAL EQUATION A function of the form y = f ( x) + C which satisfies a given differential equation is called its solution. GENERAL SOLUTION OF A DIFFERENTIAL EQUATION Suppose a differential equation of order n is being given. If its solution contains n arbitrary constants then it is called a general solution.

Senior Secondary School Mathematics for Class 12 Pg-898

898

Senior Secondary School Mathematics for Class 12

PARTICULAR SOLUTION OF A DIFFERENTIAL EQUATION Giving particular values to arbitrary constants in the general solution of a differential equation, we get its particular solutions.

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

EXAMPLE 2

Verify that y = A cos x - B sin x is a solution of the differential equation d 2y [CBSE 2005C, ‘06] + y = 0. dx 2 Given: y = A cos x - B sin x … (i) dy Þ = - A sin x - B cos x dx d 2y = - A cos x + B sin x Þ dx 2 = - ( A cos x - B sin x) = - y [from (i)] d 2y + y = 0. Þ dx 2 Hence, y = A cos x - B sin x is a solution of the differential d 2y equation 2 + y = 0. dx Verify that y = ae2x + be- x is a solution of the differential equation d 2y 2

SOLUTION

-

dx + be- x

dy - 2y = 0. dx

[CBSE 2003C]

Given: y = ae2x … (i) dy … (ii) Þ = 2ae2x - be- x dx 2 d y … (iii) = 4ae2x + be- x Þ dx 2 æ d 2 y dy ö \ ç 2- 2y ÷ = ( 4ae2x + be- x ) - ( 2ae2x - be- x ) - 2( ae2x + be- x ) ç dx ÷ dx è ø [using (i), (ii) and (iii)]. =0 2x -x Hence, y = ae + be is a solution of the differential equation d 2y

EXAMPLE 3

dy - 2y = 0. dx dx B Verify that y = Ax + is a solution of the differential equation x d 2y dy x2 2 + x - y = 0. dx dx

SOLUTION

Given:

2

y = Ax +

B x

-

… (i)

Senior Secondary School Mathematics for Class 12 Pg-899

Differential Equations and Their Formation

Þ Þ

dy B =A- 2 dx x d 2 y 2B = dx 2 x 3

899

… (ii) … (iii)

Substituting the values of y ,

dy d 2y and from (i), (ii) and (iii) dx dx 2

respectively, we get æ 2 d 2y ö Bö æ Bö dy 2B æ çx +x - y ÷ = x 2 × 3 + x ç A - 2 ÷ - ç Ax + ÷ 2 ç dx ÷ dx xø è x x è ø è ø B Bö æ 2B =ç + Ax - - Ax - ÷ = 0. x xø è x 2 B d y dy Thus, y = Ax + satisfies x 2 2 + x - y = 0. x dx dx Hence, y = Ax + EXAMPLE 4

SOLUTION

EXAMPLE 5

SOLUTION

B d 2y dy is a solution of x 2 2 + x - y = 0. x dx dx

Verify that y = a cos (log x) + b sin (log x) is a solution of the d 2y dy [CBSE 2007] differential equation x 2 + 2 + x + y = 0. dx dx Given: y = a cos (log x) + b sin (log x) ... (i) dy - a sin (log x) b cos (log x) [on differentiating (i) w.r.t. x] Þ = + dx x x dy … (ii) Þ x = - a sin (log x) + b cos (log x) dx d 2 y dy - a cos (log x) b sin (log x) Þ x 2+ = dx x x dx [on differentiating (ii) w.r.t. x] d 2y dy = - a cos (log x) - b sin (log x) Þ x2 2 + x dx dx d 2y dy + y = 0 [using (i)]. Þ x2 2 + x dx dx Hence, y = a cos (log x) + b sin (log x) is a solution of d 2y dy x2 2 + x + y = 0. dx dx -1

Verify that y = em sin x is a solution of the differential equation d 2y dy (1 - x 2) 2 - x - m 2 y = 0. dx dx Given: Þ

y = em sin m sin -1 x

-1

x

dy e = ×m dx 1 - x2

... (i) [on differentiating (i)]

Senior Secondary School Mathematics for Class 12 Pg-900

900

Senior Secondary School Mathematics for Class 12

Þ

æ dy ö 1 - x 2 ç ÷ = my è dx ø

... (ii)

[Q em sin

-1

x

= y]

2

æ dy ö … (iii) [on squaring both sides of (ii)] Þ (1 - x 2) ç ÷ = m 2 y 2 è dx ø 2 dy dy æ d 2 y ö æ dy ö Þ (1 - x 2) 2 × ç 2 ÷ - 2x ç ÷ = 2m 2 y ç ÷ dx dx è dx ø è dx ø [on differentiating both sides of (iii)] 2 ì üï d y dy æ dy ö ï Þ 2 ç ÷ × í(1 - x 2) 2 - x - m 2y ý = 0 dx ïþ è dx ø ïî dx 2 d y dy - m 2 y = 0. Þ (1 - x 2) 2 - x dx dx Hence, y = em sin

-1

x

is a solution of the differential equation

(1 - x 2) EXAMPLE 6

SOLUTION

d 2y dx 2

-x

dy - m 2 y = 0. dx

A + B is a solution of the differential equation r d 2v 2 dv + × = 0. dr 2 r dr Since the given relation contains two arbitrary constants, we differentiate it two times w.r.t. r , and eliminate A and B. A dv - A … (i) v= +B Þ = 2 r dr r d 2v 2A … (ii) Þ = dr 2 r 3 On dividing (ii) by (i), we get ( d 2v/dr 2) ìï 2A r 2 üï -2 =í 3 ´ ý= ( dv/dr) ( - A) þï r îï r Verify that v =

d 2v

-2 dv × r dr 2 dv Þ + × = 0. dr 2 r dr A Hence, v = + B is a solution of the differential equation r d 2v 2 dv + × = 0. dr 2 r dr

Þ

EXAMPLE 7

dr 2 d 2v

=

Prove that ( x 2 - y 2) = c( x 2 + y 2) 2 is the general solution of the differential equation ( x 3 - 3 xy 2) dx = ( y 3 - 3 x 2 y) dy , where c is a parameter.

Senior Secondary School Mathematics for Class 12 Pg-901

Differential Equations and Their Formation SOLUTION

901

We have … (i) ( x 2 - y 2 ) = c( x 2 + y 2 ) 2 . On differentiating (i) w.r.t. x, we get dy dy ö æ 2x - 2y = 2c( x 2 + y 2) ç 2x + 2y ÷ dx dx ø è dy ö dy ö æ 2 2 æ … (ii) Þ ç x - y ÷ = 2c( x + y ) ç x + y ÷ dx ø dx ø è è dy ö 2( x 2 - y 2) 2 dy ö æ æ ( x + y 2) ç x + y ÷ Þ çx - y ÷ = 2 dx ø ( x + y 2) 2 dx ø è è [putting the value of c from (i) in (ii)] dy ö dy ö æ æ Þ ( x 2 + y 2) ç x - y ÷ = 2( x 2 - y 2) ç x + y ÷ dx ø dx ø è è dy 2 2 2 2 2 2 Þ {x( x + y ) - 2x( x - y )} = {2y( x - y ) + y( x 2 + y 2)} dx dy Þ ( 3 xy 2 - x 3) = ( 3 x 2 y - y 3) dx Þ ( x 3 - 3 xy 2) dx = ( y 3 - 3 x 2 y) dy , which is the required differential equation.

EXAMPLE 8

Prove that xy = aex + be- x + x 2 is the general solution of the differential equation x

d 2y dx 2

+2

dy - xy + x 2 - 2 = 0. dx

SOLUTION

We have … (i) xy = aex + be- x + x 2 . On differentiating (i) w.r.t. x, we get dy … (ii) x + y = aex - be- x + 2x. dx On differentiating (ii) w.r.t. x, we get d 2y dy x 2+2 = aex + be- x + 2 dx dx d 2y dy Þ x 2+2 = xy - x 2 + 2 [Q from (i), ( aex + be- x ) = ( xy - x 2)] dx dx d 2y dy Þ x 2 + 2 - xy + x 2 - 2 = 0, dx dx which is the required differential equation.

EXAMPLE 9

Verify that y = Aeax cos bx + Beax sin bx , where A and B are arbitrary

SOLUTION

constants, is the general solution of the differential equation d 2y dy - 2a + ( a 2 + b 2) y = 0. 2 dx dx We have … (i) y = Aeax cos bx + Beax sin bx.

Senior Secondary School Mathematics for Class 12 Pg-902

902

Senior Secondary School Mathematics for Class 12

Differentiating (i) on both sides w.r.t. x, we get dy = A × {eax ( - b sin bx) + aeax cos bx} + B × {eax ( b cos bx) + aeax sin bx} dx dy = a{Aeax cos bx + Beax sin bx} + b{- Aeax sin bx + Beax cos bx} Þ dx dy … (ii) [using (i)] Þ = ay + beax {B cos bx - A sin bx} dx Differentiating (ii) on both sides w.r.t. x, we get d 2y dx Þ

Þ Þ

2

d 2y dx 2 d 2y dx 2 d 2y dx 2

dy + b[eax ( - bB sin bx - bA cos bx) dx + ( B cos bx - A sin bx) aeax ] dy =a - b 2{Aeax cos bx + Beax sin bx} dx + a{beax ( B cos bx - A sin bx)} dy æ dy ö =a - b2y + a ç - ay ÷ [using (i) and (ii)] dx è dx ø dy - 2a + ( a 2 + b 2) y = 0, dx which is the required differential equation. =a

EXERCISE 18B 1. Verify that x 2 = 2y 2 log y is a solution of the differential equation dy ( x 2 + y 2) - xy = 0. dx 2. Verify that y = ex cos bx is a solution of the differential equation d 2y dy - 2 + 2y = 0. [CBSE 2006] 2 dx dx -1

3. Verify that y = em cos x is a solution of the differential equation d 2y dy (1 - x 2) 2 - x - m 2 y = 0. dx dx 4. Verify that y = ( a + bx) e2x is the general solution of the differential d 2y dy equation 2 - 4 + 4y = 0. dx dx 5. Verify that y = ex ( A cos x + B sin x) is the general solution of the d 2y dy [CBSE 2009] differential equation 2 - 2 + 2y = 0. dx dx 6. Verify that y = A cos 2x - B sin 2x is the general solution of the differential d 2y equation 2 + 4y = 0. [CBSE 2007, ’09] dx

Senior Secondary School Mathematics for Class 12 Pg-903

Differential Equations and Their Formation

903

7. Verify that y = ae2x + be- x is the general solution of the differential equation

d 2y dx

2

-

dy - 2y = 0. dx

8. Show that y = ex ( A cos x + B sin x) is the solution of the differential equation

d 2y dx

2

-2

dy + 2y = 0. dx

9. Verify that y 2 = 4a( x + a) is a solution of the differential equation ìï æ dy ö 2 üï dy yí1 - ç ÷ ý = 2x × dx dx è ø îï þï 10. Verify that y = cetan (1 + x 2)

2

d y dx 2

+ ( 2x - 1)

-1

x

is a solution of the differential equation

dy = 0. dx

11. Verify that y = aebx is a solution of the differential equation 12. Verify that y = d 2y dx

2

+

d 2y dx 2

2

=

1 æ dy ö ç ÷ × y è dx ø

a + b is a solution of the differential equation x

2 æ dy ö ç ÷ = 0. x è dx ø

13. Verify that y = e- x + Ax + B is a solution of the differential equation æ d 2y ö ex ç 2 ÷ = 1. ç dx ÷ è ø 14. Verify that Ax 2 + By 2 = 1 is a solution of the differential equation ìï d 2 y æ dy ö 2 üï dy xíy 2 + ç ÷ ý = y × dx dx è ø dx îï þï 15. Verify that y = (1 + x 2)

c-x 1 + cx

is a solution of the differential equation

dy + (1 + y 2) = 0. dx

16. Verify that y = log ( x + x 2 + a 2 ) satisfies the differential equation d 2y dx 2

+x

17. Verify 2

d y

dy = 0. dx that

y = e-3x

dy + - 6y = 0. dx 2 dx

is

a

solution

of

the

differential

equation

Senior Secondary School Mathematics for Class 12 Pg-904

904

Senior Secondary School Mathematics for Class 12 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 18B)

7. Find

dy d2 y and 2 × dx dx

Put these values in

d2 y dx 2

-

dy - 2 y to get 0. dx

14. Let Ax 2 + By 2 = 1 be given. Then, dy dy 2 Ax + 2 By = 0 Þ Ax + By =0 dx dx dy Þ Ax = - By Þ dx On differentiating (i) w.r.t. x, we get é d 2 y æ dy ö 2 ù d2 y A + Bê y 2 + ç ÷ ú = 0 Þ y 2 + è dx ø úû dx êë dx

… (i) A - y dy = × × B x dx

… (ii)

2

-A æ dy ö ç ÷ = B è dx ø

ìï d 2 y æ dy ö 2 üï y dy Þ íy 2 + ç ÷ ý = × è dx ø þï x dx îï dx

[from (ii)].

15. Let c = tan A and x = tan B. Then, tan A - tan B y= = tan ( A - B) Þ tan -1 y = A - B 1 + tan A tan B Þ tan -1 y = tan -1 ( c ) - tan -1 x dy 1 1 Þ =0( 1 + y 2 ) dx ( 1 + x2 ) dy Þ ( 1 + x2 ) + ( 1 + y 2 ) = 0. dx 16. y = log ( x +

x 2 + a2 ) ìï 2x × í1 + 2 x 2 + a2 x + a ) îï

üï ý Þ y1 = þï

Þ

dy = dx ( x +

Þ

y12 ( x 2 + a2 ) = 1 Þ y12 ( 2 x ) + ( x 2 + a2 )2 y1 y2 = 0

1

2

2

1 x 2 + a2

Þ ( x 2 + a2 )y2 + xy1 = 0.

Formation of a Differential Equation whose General Solution is Given METHOD Suppose an equation of a family of curves contains n arbitrary constants (called parameters).

Then, we obtain its differential equation as given below. Step 1.

Differentiate the equation of the given family of curves n times to get n more equations.

Step 2.

Eliminate n constants, using these (n + 1) equations.

This gives us the required differential equation of order n.

Senior Secondary School Mathematics for Class 12 Pg-905

Differential Equations and Their Formation

905

SOLVED EXAMPLES

EXAMPLE 2

Write the differential equation representing the family of curves y = mx , where m is an arbitrary constant. [CBSE 2013] The equation of the given family of curves is … (i), where m is a constant. y = mx Since the given equation contains one arbitrary constant, we differentiate it once only. On differentiating (i) w.r.t. x, we get dy … (ii) = m. dx Putting this value of m from (ii) in (i), we get æ dy ö æ dy ö y = ç ÷ x Þ x ç ÷ - y = 0. è dx ø è dx ø æ dy ö Hence, x ç ÷ - y = 0 is the required differential equation. è dx ø Find the differential equation of the family of curves y = Aex + Be- x ,

SOLUTION

where A and B are arbitrary constants. The equation of the given family of curves is

EXAMPLE 1

SOLUTION

EXAMPLE 3

SOLUTION

… (i) y = Aex + Be- x . Since the given equation contains two arbitrary constants, we differentiate it two times w.r.t. x. dy Now, = Aex - Be- x dx d 2y Þ = Aex + Be- x dx 2 d 2y d 2y Þ = Þ - y = 0. y dx 2 dx 2 2 d y Hence, 2 - y = 0 is the required differential equation. dx Find the differential equation of the family of curves y = ex ( A cos x + B sin x), where A and B are arbitrary constants. The equation of the given family of curves is … (i) y = ex ( A cos x + B sin x). Since the given equation contains two arbitrary constants, we differentiate it two times w.r.t. x. dy Now, = ex ( - A sin x + B cos x) + ex ( A cos x + B sin x) dx dy … (ii) - y = ex ( - A sin x + B cos x) Þ dx d 2 y dy Þ = ex ( - A cos x - B sin x) + ex ( - A sin x + B cos x) dx 2 dx

Senior Secondary School Mathematics for Class 12 Pg-906

906

Senior Secondary School Mathematics for Class 12

d 2y

dy æ dy ö = -y + ç - y ÷ [using (i) and (ii)] dx è dx ø dy Þ - 2 + 2y = 0, which is the required differential equation dx dx 2 of the given family of curves.

Þ

EXAMPLE 4 SOLUTION

EXAMPLE 5

dx 2 d 2y

-

Find the differential equation of the family of all straight lines. The general equation of the family of all straight lines is given by y = mx + c, where m and c are parameters. dy Now, y = mx + c Þ =m dx d 2y Þ = 0. dx 2 So, the required differential equation is d 2y = 0. dx 2 Form the differential equation of the family of all circles of radius r. [CBSE 2010]

SOLUTION

The equation of the family of all circles of radius r is ( x - a) 2 + ( y - b) 2 = r 2 , where a and b are arbitrary constants. Differentiating (i) w.r.t. x, we get 2( x - a) + 2( y - b) y1 = 0 Þ ( x - a) + ( y - b) y1 = 0. On differentiating (ii) again w.r.t. x, we get

… (i)

... (ii)

1 + ( y - b) y 2 + ( y1) 2 = 0 {1 + ( y1) 2} × y2 Putting the value of ( y - b) from (iii) in (ii), we get

Þ ( y - b) = -

… (iii)

{1 + ( y1) 2} y1 {1 + ( y1) 2} y1 … (iv) = 0 Þ ( x - a) = × y2 y2 Putting the values of ( y - b) and ( x - a) from (iii) and (iv) in (i), we get ( x - a) -

{1 + ( y1) 2} 2 ´ ( y1) 2 ( y 2) Þ

2

+

{1 + ( y1) 2} 2

{1 + ( y1) 2} 2 × {( y1) 2 + 1} ( y 2) 2

( y 2) 2

= r2

= r2

Þ {1 + ( y1) 2} 3 = r 2( y 2) 2 , which is the required differential equation.

Senior Secondary School Mathematics for Class 12 Pg-907

Differential Equations and Their Formation EXAMPLE 6

SOLUTION

907

Form the differential equation of the family of all circles in first quadrant [CBSE 2010] and touching the coordinate axes. The general equation of a circle in first quadrant and touching the coordinate axes is given by … (i), where a is a parameter. ( x - a) 2 + ( y - a) 2 = a 2 Since this equation contains one parameter, so we will differentiate it only once to get the differential equation. On differentiating (i) w.r.t, x, we get dy 2( x - a) + 2( y - a) =0 dx dy =0 Þ ( x - a) + ( y - a) dx dy =p Þ ( x - a) + ( y - a) p = 0, where dx ( x + yp) … (ii) × Þ x + yp = a(1 + p) Þ a = (1 + p) Putting the value of a from (ii) in (i), we get 2

2

æ x + yp ö æ æ x + yp ö x + yp ö ÷÷ = çç çç x ÷÷ + çç y ÷÷ 1+ p ø 1+ p ø è 1+ p ø è è Þ ( xp - yp) 2 + ( y - x) 2 = ( x + yp) 2

2

Þ ( x - y) 2 p 2 + ( x - y) 2 = ( x + yp) 2 Þ ( x - y) 2( p 2 + 1) = ( x + yp) 2 2 ìïæ dy ö 2 üï æ dy ö Þ ( x - y) 2íç ÷ + 1ý = ç x + y ÷ , dx ø îïè dx ø þï è which is the required differential equation. EXAMPLE 7

SOLUTION

Form the differential equation of the family of all circles touching the x-axis at the origin. [CBSE 2005, ’08, ’10C] The general equation of a circle touching the x-axis at the origin, is given by ( x - 0) 2 + ( y - a) 2 = a 2 , where a is a parameter Þ x 2 + y 2 - 2ay = 0. Since this equation contains one parameter, so we shall differentiate it only once to get the differential equation. On differentiating (i) w.r.t. x, we get dy dy 2x + 2y - 2a =0 dx dx dy dy Þ x+y =a dx dx x + yp dy … (ii), where Þ x + yp = ap Þ a = = p. p dx

… (i)

Senior Secondary School Mathematics for Class 12 Pg-908

908

Senior Secondary School Mathematics for Class 12

Putting the value of a from (ii) in (i), we get 2( x + yp) × y x2 + y2 = p 2xy 2 2 Þ x +y = + 2y 2 p 2xy Þ x2 - y2 = p Þ ( x 2 - y 2) p = 2xy. dy Hence, ( x 2 - y 2) = 2xy is the required differential equation. dx EXAMPLE 8

SOLUTION

Form the differential equation of the family of all parabolas having vertex at the origin and axis along the positive direction of the x-axis. The general equation of a parabola having vertex at the origin and axis along the positive direction of the x-axis is given by … (i), where a is the parameter. y 2 = 4ax Since this equation contains one parameter, so we shall differentiate it only once to get the requisite differential equation. Differentiating (i) w.r.t. x, we get dy dy 2y = 4a Þ y = 2a dx dx 1 dy Þ a = y × … (ii) 2 dx Putting the value of a from (ii) in (i), we get 1 dy dy y2 = 4 ´ y ´ x Þ y 2 - 2xy = 0. 2 dx dx dy Hence, y 2 - 2xy = 0 is the required differential equation. dx

EXAMPLE 9

SOLUTION

Form the differential equation of the family of all ellipses having foci on the x-axis and centre at the origin. [CBSE 2009C] The general equation of an ellipse having foci on the x-axis and centre at the origin, is given by x2

+

y2

= 1 … (i), where a and b are the parameters. a b2 Since this equation contains two parameters, so we shall differentiate it twice to get the required differential equation. Differentiating (i) w.r.t. x, we get: 2x 2y dy y dy x + × =0 Þ 2× + =0 a 2 b 2 dx b dx a 2 2

Senior Secondary School Mathematics for Class 12 Pg-909

Differential Equations and Their Formation

Þ

yy1 2

=

-x 2

, where

b a yy1 - b 2 Þ = 2 × x a Differentiating (ii) w.r.t. x, we get d d x × ( yy1) - yy1 × ( x) dx dx =0 x2 Þ x[yy 2 + ( y1) 2] - yy1 = 0

909

dy = y1 dx … (ii)

Þ ( xy) y 2 + x( y1) 2 - yy1 = 0. Hence, ( xy)

2

d 2y

æ dy ö æ dy ö + x ç ÷ - y ç ÷ = 0 is the required differential è dx ø è dx ø dx 2

equation. EXAMPLE 10

Form the differential equation for the family of the curves ( y - b) 2 = 4( x - a), where a and b are parameters.

SOLUTION

The general equation of the given family of curves is … (i), where a and b are parameters. ( y - b) 2 = 4( x - a) Since the given equation contains two parameters a and b, so we shall differentiate it twice to get the required differential equation. Differentiating (i) w.r.t. x, we get dy dy …(ii), where 2( y - b) = 4 Þ ( y - b) y1 = 2 = y1 . dx dx Differentiating (ii) w.r.t. x, we get d 2y … (iii), where 2 = y 2. ( y - b) y 2 + ( y1) 2 = 0 dx 2 from (ii) in (iii), we get Putting ( y - b) = y1 2y 2 + ( y1) 2 = 0 Þ 2y 2 + ( y1) 3 = 0. y1 Hence, 2

d 2y

3

æ dy ö + ç ÷ = 0 is the required differential equation. dx 2 è dx ø

EXAMPLE 11

Form the differential equation for the family of the curves ay 2 = ( x - c) 3 ,

SOLUTION

where c is a parameter. The general equation of the given family of curves is … (i) ay 2 = ( x - c) 3 , where c is a parameter. Since the given equation has only one parameter, so we shall differentiate it only once to get the required differential equation. Differentiating (i) w.r.t. x, we get 2ayy1 = 3( x - c) 2.

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-910

910

Senior Secondary School Mathematics for Class 12

On dividing (i) and (ii) on each side, we get ( x - c) 3 ( x - c) 3y ay 2 y = Þ = Þ ( x - c) = × 2ayy1 3( x - c) 2 2y1 3 2y1

… (iii)

Putting the value of ( x - c) from (iii) in (i), we get 3 æ 3y ö ÷÷ Þ 8ay13 = 27 y. ay 2 = çç è 2y1 ø 3

æ dy ö Hence, 8a ç ÷ - 27 y = 0 is the required differential equation. è dx ø EXAMPLE 12

SOLUTION

Form the differential equation for the family of the curves y 2 = a( b 2 - x 2), where a and b are arbitrary constants. The general equation of the given family of curves is …(i), where a and b are the parameters. y 2 = a( b 2 - x 2) Since the given equation has two parameters, so we shall have to differentiate it two times to get the required differential equations. Differentiating (i) w.r.t. x we get … (ii) 2yy1 = -2ax Þ yy1 = - ax. On differentiating (ii) w.r.t. x, we get yy 2 - ( y1) 2 = - a yy Þ yy 2 - ( y1) 2 = 1 [using (ii)] x Þ ( xy) y 2 - x( y1) 2 - yy1 = 0. Hence, ( xy)

d 2y

2

æ dy ö æ dy ö - x ç ÷ - y ç ÷ = 0 is the required differential è dx ø è dx ø dx 2

equation.

EXERCISE 18C 1. Form the differential equation of the family of straight lines y = mx + c, where m and c are arbitrary constants. 2. Form the differential equation of the family of concentric circles x 2 + y 2 = a 2 , where a > 0 and a is a parameter. 3. Form the differential equation of the family of curves, y = a sin ( bx + c), where a and c are parameters. 4. Form the differential equation of the family of curves x = A cos nt + B sin nt , where A and B are arbitrary constants. [CBSE 2007] 5. Form the differential equation of the family of curves y = aebx , where a and b are arbitrary constants. 6. Form the differential equation of the family of curves y 2 = m( a 2 - x 2), where a and m are parameters.

Senior Secondary School Mathematics for Class 12 Pg-911

Differential Equations and Their Formation

911

7. Form the differential equation of the family of curves given by ( x - a) 2 + 2y 2 = a 2 , where a is an arbitrary constant. 8. Form the differential equation of the family of curves given by x 2 + y 2 - 2ay = a 2 , where a is an arbitrary constant.

[CBSE 2005]

9. Form the differential equation of the family of all circles touching the y-axis at the origin. [CBSE 2008C, ’10C] 10. Form the differential equation of the family of circles having centres on the y-axis and radius 2 units. 11. Form the differential equation of the family of circles in second quadrant and touching the coordinate axes. 12. Form the differential equation of the family of circles having centres on the x-axis and radius unity. 13. Form the differential equation of the family of circles passing through the fixed points ( a , 0) and ( -a , 0), where a is the parameter. 14. Form the differential equation of the family of parabolas having vertex at the origin and axis along positive y-axis. [CBSE 2010, ’11] 15. Form the differential equation of the family of ellipses having foci on the y-axis and centre at the origin. 16. Form the differential equation of the family of hyperbolas having foci on the x-axis and centre at the origin. ANSWERS (EXERCISE 18C)

1.

d 2y dx

2

=0

2. x + y

2

dy =0 dx

3.

d 2y dx

2

+ b2y = 0

4.

d 2x dt 2

+ n2x = 0

2

dy dy 2y 2 - x 2 æ dy ö 7. - xç ÷ - y =0 = dx dx 4xy è dx ø dx dy dy 9. y 2 - x 2 - 2xy 8. p 2( x 2 - 2y 2) - 4pxy - x 2 = 0, where p = =0 dx dx 2 ì æ dy ö 2 üï æ ü ì 1 dy ö 2ï 11. 10. x 2í1 + = = + 4 ( x + y ) 1 + x y ç ÷ ç ÷ ý í ý 2 dx ø ïî è dx ø ïþ è î ( dy/dx) þ d 2y æ dy ö 5. ç ÷ = y × 2 è dx ø dx

2

æ dy ö 12. y 2 ç ÷ + y 2 = 1 è dx ø

6. xy

d 2y 2

13. ( x 2 - y 2 - a 2)

2 æ d 2y ö dy æ dy ö 15. xy ç 2 ÷ + x ç ÷ - y =0 ç dx ÷ dx dx è ø è ø

16. xy

dy = 2xy dx d 2y

14. x 2

dy = 2y dx

dy æ dy ö + xç ÷ - y =0 dx è dx ø dx 2

Senior Secondary School Mathematics for Class 12 Pg-912

912

Senior Secondary School Mathematics for Class 12 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 18C)

6. See Example 12. 7. Given family of curves is x 2 + 2 y 2 - 2 ax = 0.

… (i)

Differentiating (i) w.r.t. x, we get dy æ dy ö 2x + 4y - 2a = 0 Þ a = ç 2y + x÷× dx è dx ø Put this value of a in (i). 8. Given family of curves is x 2 + y 2 - 2 ay = a2 .

… (i)

Differentiating (i) w.r.t. x, we get dy dy dy dy 2x + 2y - 2a =0 Þ x+ y =a dx dx dx dx x + yp dy Þ a= , where p = × p dx x + yp Putting a = in (i), we get: p 2 2( x + yp )y æ x + yp ö ÷ = çç x2 + y2 ÷ p è p ø 2 2 2 Þ ( x + y )p - 2 yp( x + yp ) = ( x + yp ) 2 Þ x 2 p 2 + y 2 p 2 - 2 xyp - 2 y 2 p 2 = x 2 + y 2 p 2 + 2 xyp Þ x 2 p 2 - 2 y 2 p 2 - 4 xyp - x 2 = 0 dy × Þ ( x 2 - 2 y 2 )p 2 - 4 xyp - x 2 = 0 , where p = dx 9. The equation of given family of circles is ( x - a) 2 + ( y - 0 ) 2 = a2 Þ x 2 + y 2 - 2 ax = 0. … (i) Differentiating (i) w.r.t. x, we get dy dy … (ii) 2x + 2y - 2a = 0 Þ x + y = a. dx dx Put the value of a from (ii) in (i), we get the desired differential equation. 10. The given family is x 2 + ( y - a) 2 = 4. 11. The equation of the given family of circles is ( x + a) 2 + ( y - a) 2 = a2 … (i), where a is a parameter. Differentiating (i) w.r.t. x, we get dy 2( x + a) + 2( y - a) = 0 Þ ( x + a) + ( y - a) p = 0 dx ( x + yp ) Þ a= × ( p - 1) Put this value of a in (i) to get the required equation. 12. The equation of the given family of circles with centre ( a, 0 ) and radius 1, is given by ( x - a) 2 + ( y - 0 ) 2 = 12 , where a is the parameter Þ ( x - a) 2 + y 2 = 1.

… (i)

Senior Secondary School Mathematics for Class 12 Pg-913

Differential Equations and Their Formation 13. The general equation of a circle is x 2 + y 2 + 2 gx + 2 fy + c = 0.

913

… (i)

If it passes through the points A( - a, 0 ) and B( a, 0 ), we have … (ii) and a2 + 2 ga + c = 0 a2 - 2 ga + c = 0

… (iii)

Adding (ii) and (iii), we get 2( a2 + c ) = 0 Þ a2 + c = 0 Þ c = - a2 . Putting c = - a2 in (iii), we get 2 ga = 0 Þ g = 0. Putting g = 0 and c = - a2 in (i), we get x 2 + y 2 + 2 fy - a2 = 0

… (iv), where f is the parameter.

Differentiating (iv) w.r.t. x, we get 2 x + 2 yy1 + 2 fy1 = 0 Þ fy1 = - ( x + yy1 ) Þ f =

- ( x + yy1 ) × y1

Putting this value of f in (iv), we get 2 y ( x + yy1 ) - a2 = 0 x2 + y2 y1 Þ

x 2 y1 + y 2 y1 - 2 xy - 2 y 2 y1 - a2 y1 = 0 Þ ( x 2 - y 2 - a2 )y1 = 2 xy.

14. The equation of the given family of parabolas is given by x 2 = 4 ay. 15. The equation of the family of ellipses having centre at the origin and foci on the y-axis, is given by x2 y2 … (i), where b > a and a, b are the parameters. + 2 =1 a2 b Differentiating (i), w.r.t. x, we get yy 2 x 2 y dy x … (ii) + = 0 Þ 2 + 21 = 0. a2 b 2 dx a b Differentiating (ii) w.r.t. x, we get yy y2 xyy2 xy12 1 x … (iii) + 22 + 12 = 0 Þ 2 + + 2 = 0. 2 a b b a b2 b Subtracting (ii) from (iii), we get 2 æ d2 y ö 1 æ dy ö æ dy ö {xyy2 + xy12 - yy1 } = 0 Þ xy ç 2 ÷ + x ç ÷ - y ç ÷ = 0. 2 ç ÷ è dx ø è dx ø b è dx ø 16. The equation of the family of hyperbolas having foci on the x-axis and centre at the origin is given by x2 y2 … (i), where a and b are the parameters. =1 a2 b 2 Differentiating (i), w.r.t. x, we get 2 x 2 yy1 x yy … (ii) - 2 = 0 Þ 2 - 21 = 0. a2 b a b Differentiating (ii), w.r.t. x, we get 1 yy2 y12 … (iii) - 2 - 2 = 0. a2 b b Multiplying (iii) by x and subtracting from (ii), we get 2 æ d2 y ö dy 1 æ dy ö = 0. {xyy2 + xy12 - yy1 } = 0 Þ xy ç 2 ÷ + x ç ÷ - y 2 ç ÷ dx dx ø è b dx è ø

Senior Secondary School Mathematics for Class 12 Pg-914

19. DIFFERENTIAL EQUATIONS WITH VARIABLE SEPARABLE Method of Solving Differential Equations with Variables Separable GENERAL SOLUTION

g( y) dy = f ( x) dx.

Let the given differential equation be expressed in the form

Then, the general solution of the above differential equation is given by

ò g(y )dy = ò f ( x)dx + C , where C is an arbitrary constant.

PARTICULAR SOLUTION Let x = a be given and let the corresponding value of y be given as b. Putting x = a and y = b in the general solution, we get the value of C. With this value of C, we get the particular solution of the given differential equation.

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Find the general solution of the differential equation dy ( x + 2) = x 2 + 5 x - 3( x ¹ -2). dx The given differential equation may be written as

Þ Þ Þ

dy x 2 + 5 x - 3 = dx x+2 æ x2 + 5x - 3 ö ÷ dx [separating the variables] dy = ç ç x + 2 ÷ø è æ x2 + 5x - 3 ö ò dy = ò çç x + 2 ÷÷ dx è ø ü ì 9 y = ò íx + 3 ý dx + C , where C is an arbitrary constant ( x + 2) þ î [on dividing ( x 2 + 5 x - 3) by ( x + 2)]

x2 + 3 x - 9 log |x + 2| + C. 2 x2 Hence, y = + 3 x - 9 log |x + 2| + C is the required general 2 solution.

Þ

EXAMPLE 2

y=

Find the general solution of the differential equation dy (1 + x 2) - x = 2 tan -1 x. dx 914

[CBSE 2007]

Senior Secondary School Mathematics for Class 12 Pg-915

Differential Equations with Variable Separable SOLUTION

EXAMPLE 3

915

The given differential equation may be written as dy 2 tan -1 x x = + dx (1 + x 2) (1 + x 2) ì 2 tan -1 x x ü Þ dy = í + dx [separating the variables] 2 2 ý î (1 + x ) (1 + x ) þ x 2 tan -1 x Þ ò dy = ò dx + ò dx + C , (1 + x 2) (1 + x 2) where C is an arbitrary constant 1 2x dx + C Þ y = 2ò t dt + ò 2 (1 + x 2) 1 [putting tan -1 x = t and dx = dt in 1st integral] (1 + x 2) 1 Þ y = t 2 + log |1 + x 2| + C 2 1 -1 2 Þ y = (tan x) + log |1 + x 2| + C. 2 1 Hence, y = (tan -1 x) 2 + log |1 + x 2| + C is the required solution. 2 dy Find the general solution of the differential equation = log ( x + 1). dx [CBSE 2006]

SOLUTION

We have dy = log ( x + 1) dx Þ dy = log ( x + 1) dx Þ Þ

ò dy = ò log ( x + 1) dx [integrating both sides] y = ò {log ( x + 1) × 1} dx + C , where C is an arbitrary constant I II = {log ( x + 1) × x} - ò

1 × x dx + C [integrating by parts] ( x + 1)

( x + 1) - 1 dx + C ( x + 1) ì 1 ü = x log ( x + 1) - ò í1 ý dx + C ( x + 1) þ î = x log ( x + 1) - ò

= x log ( x + 1) - x + log ( x + 1) + C = ( x + 1) log ( x + 1) - x + C. Hence, y = ( x + 1) log ( x + 1) - x + C is the required solution. EXAMPLE 4

Find the general solution of the differential equation dy = sin -1 x. dx

Senior Secondary School Mathematics for Class 12 Pg-916

916 SOLUTION

Senior Secondary School Mathematics for Class 12

We have dy = sin -1 x dx Þ dy = sin -1 x dx Þ Þ

[separating the variables]

-1

ò dy = ò sin x dx [integrating both sides] y = ò {(sin -1 x) × 1} dx + C , where C is an arbitrary constant I II = (sin -1 x) × x - ò

1

× x dx + C [integrating by parts] 1 - x2 -2x 1 = (sin -1 x) × x + ò dx + C 2 1 - x2 1 1 = (sin -1 x) × x + ò dt + C , where (1 - x 2) = t 2 t 1 = (sin -1 x) × x + ´ 2 t + C 2 = (sin -1 x) x + 1 - x 2 + C. Hence, y = (sin -1 x) x + 1 - x 2 + C is the required solution. Hence, y = x(sin -1 x) + 1 - x 2 + C is the required solution of the given differential equation. EXAMPLE 5

SOLUTION

Find the general solution of the differential equation dy = 4 - y 2 ( -2 < y < 2). dx

[CBSE 2002C]

dy = 4 - y2 dx dy [on separating the variables] Þ = dx 4 - y2 dy Þ ò = ò dx [integrating both sides] 2 2 - y2 æyö Þ sin -1 ç ÷ = x + C , where C is an arbitrary constant. è 2ø -1 æ y ö Hence, sin ç ÷ = x + C is the required solution. è 2ø We have

EXAMPLE 6

Find the general solution of the differential equation

SOLUTION

dy = 2x 2 + x. dx The given differential equation may be written as ( x 3 + x 2 + x + 1)

[CBSE 2010]

dy 2x 2 + x ( 2x 2 + x) ( 2x 2 + x) = 3 = = dx ( x + x 2 + x + 1) x 2( x + 1) + ( x + 1) ( x + 1)( x 2 + 1)

Senior Secondary School Mathematics for Class 12 Pg-917

Differential Equations with Variable Separable

Þ

ì ( 2x 2 + x) ü dy = í ý dx 2 î( x + 1)( x + 1) þ

Þ

ò dy = ò ( x + 1)( x 2 + 1) dx

Þ

y=ò

Let

( 2x 2 + x)

( 2x 2 + x) ( x + 1)( x 2 + 1)

2x 2 + x 2

( x + 1)( x + 1)

=

… (i)

917

[integrating both sides]

dx + C , where C is an arbitrary constant.

A Bx + C + × ( x + 1) ( x 2 + 1)

Then, 2x 2 + x º A( x 2 + 1) + ( Bx + C)( x + 1).

… (ii)

1 Putting x = -1 in (ii), we get 2A = 1 Þ A = × 2 Putting x = 0 in (ii), we get A + C = 0 Þ C = - A =

-1 × 2

Putting x = 1 in (ii), we get 2A + 2B + 2C = 3. 3 1 1 3 3 Þ +B- = Þ B= × \ A + B+C = 2 2 2 2 2 1 3 -1 and C = × \ A= ,B= 2 2 2 1ö æ3 ç x- ÷ dx 1 2 2ø + è dx + C \ y= ò 2 ( x + 1) ò ( x 2 + 1) Þ

y=

dx x dx 1 3 1 + dx - ò 2 +C 2 ò ( x + 1) 2 ò ( x 2 + 1) 2 ( x + 1)

Þ

y=

1 3 2x 1 log |x + 1| + ò 2 dx - tan -1 x + C 2 4 ( x + 1) 2

1 3 1 log |x + 1| + log |x 2 + 1| - tan -1 x + C. 2 4 2 1 3 1 2 Hence, y = log |x + 1| + log |x + 1| - tan -1 x + C is the 2 4 2 required solution.

Þ

EXAMPLE 7

y=

Find the general solution of the differential equation dy 1 + y 2 = × dx 1 + x 2

SOLUTION

dy 1 + y 2 = dx 1 + x 2 1 1 Þ dy = dx [separating the variables] 2 (1 + y ) (1 + x 2)

Senior Secondary School Mathematics for Class 12 Pg-918

918

Senior Secondary School Mathematics for Class 12

1

1

Þ

ò (1 + y 2) dy = ò (1 + x 2) dx

Þ

tan -1 y = tan -1 x + C1 , where C1 is an arbitrary constant

Þ

tan -1 y - tan -1 x = C1

æ y-x ö ÷÷ = C1 tan -1 çç è 1 + yx ø y-x Þ = tan C1 = C , where C = tan C1. 1 + yx y-x Hence, = C is the required solution. 1 + yx

Þ

EXAMPLE 8

Find the general solution of the differential equation æ dy ö log ç ÷ = ( ax + by). è dx ø

SOLUTION

æ dy ö log ç ÷ = ax + by è dx ø dy Þ = eax + by = eax × eby dx 1 Þ dy = eax dx [on separating the variables] eby Þ ò e- by dy = ò eax dx [integrating both sides] Þ Þ

e- by eax = +C -b a ae- by + beax = C ¢, where C ¢ = - abC.

Thus, ae- by + beax = C ¢ is the required solution. EXAMPLE 9

Find the general solution of the differential equation 1 + x 2 + y 2 + x 2 y 2 + xy

SOLUTION

dy = 0. dx

The given differential equation may be written as dy (1 + x 2) + y 2(1 + x 2) + xy =0 dx dy Þ (1 + x 2)(1 + y 2) + xy =0 dx dy Þ ( 1 + x 2 )( 1 + y 2 ) + xy =0 dx dy Þ xy = - ( 1 + x 2 )( 1 + y 2 ) dx Þ

y 1 + y2

dy =

- 1 + x2 dx x

[CBSE 2010]

Senior Secondary School Mathematics for Class 12 Pg-919

Differential Equations with Variable Separable

Þ

ò

y 1 + y2

dy = - ò

1 + x2 x2

919

× x dx.

… (i)

Putting 1 + x 2 = u2 and 1 + y 2 = v 2 , we get 2x dx = 2u du and 2y dy = 2v dv Þ x dx = u du and y dy = v dv. Substituting these values in (i), we get v dv u ´ u du ò v = -ò (u2 - 1) Þ

u2

ò dv = -ò (u2 - 1) du = -ò

(u2 - 1) + 1 (u2 - 1)

du

ì 1 ü Þ v = - ò í1 + 2 ý du + C , where C is an arbitrary constant î (u - 1) þ 1 u -1 +C Þ v = -u - log 2 u+1 Þ

1 1 + y 2 = - 1 + x 2 - log 2

1 + x2 - 1 1 + x2 + 1

+ C,

which is the required solution. EXAMPLE 10

Find the general solution of the differential equation ( x cos y) dy = ex ( x log x + 1) dx.

SOLUTION

The given differential equation may be written as æ ex ö cos y dy = ç ex log x + ÷ dx ç x ÷ø è ex Þ ò cos y dy = ò (log x) ex dx + ò dx + C , II x I

[CBSE 2007]

where C is an arbitrary constant

EXAMPLE 11

Þ

ex 1 sin y = (log x) ex - ò × ex dx + ò dx + C [integrating by parts] x x

Þ

sin y = (log x) ex + C , which is the required solution.

Find the general solution of the differential equation x 1 - y 2 dx + y 1 - x 2 dy = 0.

SOLUTION

We have

Þ

x 1 - y 2 dx + y 1 - x 2 dy = 0 x y dx + dy = 0 [on separating the variables] 2 1-x 1 - y2

Senior Secondary School Mathematics for Class 12 Pg-920

920

Senior Secondary School Mathematics for Class 12

Þ Þ Þ Þ Þ Þ Þ

x

ò

dx + ò

y

dy = C [integrating both sides] 1-x 1 - y2 1 ( -2x) 1 ( -2y) - ×ò dx - × ò dy = C 2 2 2 1-x 1 - y2 1 1 1 1 - ×ò dt - × ò ds = C, where (1 - x 2) = t and (1 - y 2) = s 2 2 t s 1 1 - ò t -1/2 dt - ò s-1/2 ds = C 2 2 - t - s =C t + s = k [where k = - C] 2

1 - x 2 + 1 - y 2 = k.

Hence, 1 - x 2 + 1 - y 2 = k is the required solution. EXAMPLE 12

Find the general solution of the differential equation y-x

SOLUTION

dy dy ö æ = aç y 2 + ÷× dx dx ø è

[CBSE 2008]

We have dy ö dy æ = aç y 2 + ÷ dx ø dx è dy Þ ( y - ay 2) = ( a + x) dx dy dx [on separating the variables] Þ = y(1 - ay) ( a + x) dy dx [integrating both sides] Þ ò =ò y(1 - ay) ( a + x) y-x

æ1

a

ö

dx

Þ

ò ççè y + 1 - ay ÷÷ø dy = ò ( a + x)

Þ

log |y| - log |1 - ay| = log |a + x| + log |C1|, where C1 is an arbitrary constant y = log |C1| log (1 - ay)( a + x) y = ± C1 = C (say). (1 - ay)( a + x)

Þ Þ

[by partial fractions]

Hence, y = C(1 - ay)( a + x) is the required solution. EXAMPLE 13

SOLUTION

Find the general solution of the differential equation dy ( a + x) + x = 0. dx We have dy -x = dx a+x

Senior Secondary School Mathematics for Class 12 Pg-921

Differential Equations with Variable Separable

Þ Þ Þ Þ Þ Þ EXAMPLE 14

SOLUTION

921

-x dx [on separating the variables] a+x -x ò dy = ò a + x dx [integrating both sides] [( a + x) - a] y = -ò dx a+x a ö æ y = -ò ç a + x ÷ dx a+xø è y = - ò a + x dx + a ò ( a + x) -1 / 2 dx dy =

2 y = - ( a + x) 3 / 2 + 2a a + x + C , which is the required solution. 3

Solve the differential equation x cos y dy = ( xex log x + ex ) dx. We have Þ Þ Þ

x cos y dy = ( xex log x + ex ) dx 1ö æ cos y dy = ex ç log x + ÷ dx [on separating the variables] xø è 1ö xæ ò cos y dy = ò e çè log x + x ÷ø dx [integrating both sides] sin y = ex log x + C [Q ò ex { f ( x) + f ¢( x)} dx = ex f ( x)].

Hence, sin y = ex (log x) + C is the required solution. EXAMPLE 15

Solve the differential equation dy ex (sin 2 x + sin 2x) = × dx y( 2 log y + 1)

SOLUTION

We have dy ex (sin 2 x + sin 2x) = dx y( 2 log y + 1) Þ

y( 2 log y + 1) dy = ex (sin 2 x + sin 2x) dx

Þ

2ò y log y dy + ò y dy = ò ex (sin 2 x + sin 2x) dx II

Þ

Þ Þ \

I

é y2 1 y2 ù 1 2 ê(log y) × - ò × dy ú + y 2 = ò ex (sin 2 x + sin 2x) dx 2 y 2 ë û 2 [integrating by parts] 1 2 2 x 2 y (log y) - ò y dy + y = ò e (sin x + sin 2x) dx 2 y 2(log y) = ex sin 2 x + C [Q ò ex { f ( x) + f ¢( x)} dx = ex f ( x) + C]. y 2(log y) = ex sin 2 x + C is the required solution.

Senior Secondary School Mathematics for Class 12 Pg-922

922

Senior Secondary School Mathematics for Class 12

EXAMPLE 16

Solve the differential equation (1 + x)(1 + y 2) dx + (1 + y)(1 + x 2) dy = 0.

SOLUTION

(1 + x)(1 + y 2) dx + (1 + y)(1 + x 2) dy = 0

EXAMPLE 17

SOLUTION

(1 + x) 2

dx +

(1 + y)

dy = 0 [on separating the variables] (1 + y 2) (1 + y) Þ ò dx + ò dy = C [integrating both sides] (1 + x 2) (1 + y 2) ì 1 ì 1 x ü y ü + Þ òí + dx + ò í dy = C 2 2 ý 2 2 ý î(1 + y ) (1 + y ) þ î(1 + x ) (1 + x ) þ 1 1 2x 1 1 2y dx + × ò dx + ò dy + × ò dy = C Þ ò 2 2 2 2 (1 + x ) 2 (1 + y 2) (1 + x ) (1 + y ) 1 1 Þ tan -1 x + log (1 + x 2) + tan -1 y + log (1 + y 2) = C 2 2 1 -1 -1 2 Þ tan x + tan y + {log (1 + x ) + log (1 + y 2)} = C , 2 which is the required solution. Solve the differential equation dy cosec x log y + x 2 y 2 = 0. [CBSE 2014] dx The given differential equation may be written as dy cosec x log y = -x 2y 2 dx log y Þ dy = - x 2 sin x dx [separating the variables] y2 æ 1 ö Þ ò (log y) ç 2 ÷ dy = - ò x 2 sin x dx [integrating both sides]. çy ÷ I II I è ø Þ

(1 + x ) (1 + x)

II

Integrating by parts on each side, we get: æ -1 ö 1 æ -1 ö (log y) çç ÷÷ - ò × çç ÷÷ dy = - [x 2( - cos x) - ò 2x( - cos x) dx] + C1 , y y è yø è ø where C1 is an arbitrary constant - log y 1 2 + ò 2 dy = x cos x - 2ò x cos x dx + C1 Þ I II y y - log y 1 2 Þ - = x cos x - 2x sin x + 2ò sin x dx + C1 y y [integrating by parts] - log y 1 2 Þ - = x cos x - 2x sin x - 2 cos x + C1 y y log y 1 Þ + = - x 2 cos x + 2x sin x + 2 cos x + C , where -C1 = C. y y This is the required solution of the given differential equation.

Senior Secondary School Mathematics for Class 12 Pg-923

Differential Equations with Variable Separable

923

EXAMPLE 18

Show that the general solution of the differential equation dy y 2 + y + 1 + = 0 is given by ( x + y + 1) = C(1 - x - y - 2xy), where C dx x 2 + x + 1 is an arbitrary constant.

SOLUTION

We have dy y 2 + y + 1 + =0 dx x 2 + x + 1 Þ Þ Þ Þ

Þ

Þ Þ

dy - ( y 2 + y + 1) = dx ( x 2 + x + 1) dy dx =- 2 ( y 2 + y + 1) ( x + x + 1) dy dx ò ( y 2 + y + 1) = -ò ( x 2 + x + 1) [on integrating both sides] dy dx = -ò òì 2ü 2 2 2 ì æ 3ö ï æ 3 ö üï 1ö 1ö ïæ ïæ ÷ ý ç ÷ ý + x + ç ÷ íç y + ÷ j + çç í ÷ ç 2 ÷ 2ø 2ø è è è 2 ø þï è ø þï îï îï 1 æ 3ö ÷ ç ç 2 ÷ ø è

tan

æ æ 1ö 1ö çy + ÷ çx+ ÷ 2 ÷ = -1 tan -1 ç 2÷ +C , 1 ç 3 ÷ æ 3ö ç 3 ÷ ÷ ç ç ÷ ç ÷ è 2 ø çè 2 ÷ø è 2 ø where C1 is an arbitrary constant

-1 ç

æ 2 ö -1 æ 2y + 1 ö æ 2 ö -1 æ 2x + 1 ö ç ÷ tan ç ÷+ç ÷ tan ç ÷ = C1 è 3ø è 3 ø è 3ø è 3 ø 3 æ 2y + 1 ö -1 æ 2x + 1 ö tan -1 ç C1 ÷ + tan ç ÷= 2 è 3 ø è 3 ø 3 æ 2y + 1 ö æ 2x + 1 ö C1 , where ç ÷ = a and ç ÷=b 2 è 3 ø è 3 ø

Þ

tan -1 a + tan -1 b =

Þ

3 æ a+bö C1 , tan -1 ç ÷= 1 ab 2 è ø where ( a + b) =

2( x + y + 1) ( 4xy + 2x + 2y + 1) and ab = 3 3

æ 3 ö ç ÷ ç 2 C1 ÷ è ø

Þ

æ a+bö ç ÷ = tan è 1 - ab ø

Þ

ì 2( x + y + 1) ü í ý 3 î þ = tan ì ( 4xy + 2x + 2y + 1) ü 1 í ý 3 î þ

æ 3 ö ç ÷ ç 2 C1 ÷ è ø

Senior Secondary School Mathematics for Class 12 Pg-924

924

Senior Secondary School Mathematics for Class 12

æ 3 ö 1 ( x + y + 1) = tan çç C1 ÷÷ = C (say) (1 - x - y - 2xy) 3 ø è 2 Þ ( x + y + 1) = C(1 - x - y - 2xy), Þ

which is the solution of the given DE. EXAMPLE 19

Solve the differential equation x(1 + y 2) dx - y(1 + x 2) dy = 0, given that y = 0 when x = 1.

SOLUTION

[CBSE 2006C, ’14]

The given differential equation is x(1 + y 2) dx - y(1 + x 2) dy = 0 Þ Þ

x(1 + y 2) dx = y(1 + x 2) dy y 2

(1 + y )

dy =

y

x (1 + x 2)

dx

x

Þ

ò (1 + y 2) dy = ò (1 + x 2) dx

Þ

1 2y 1 2x dy = ò dx 2 ò (1 + y 2) 2 (1 + x 2)

Þ

1 1 1 log (1 + y 2) = log (1 + x 2) + log |C1|, 2 2 2 where C1 is an arbitrary constant

Þ Þ

2

2

log (1 + y ) = log (1 + x ) + log |C1| (1 + y 2) (1 + x 2)

= ± C1 = C (say)

Þ (1 + y 2) = C(1 + x 2).

… (i)

1 Putting x = 1 and y = 0 in (i), we get C = × 2 1 2 2 2 2 \ (1 + y ) = (1 + x ) Þ ( x - 2y ) = 1. 2 Hence, ( x 2 - 2y 2) = 1 is the required solution. EXAMPLE 20

Find the particular solution of the differential dy xy = ( x + 2)( y + 2), it being given that y = -1 when x = 1. dx

equation

[CBSE 2012]

SOLUTION

The given differential equation is dy xy = ( x + 2)( y + 2) dx y ( x + 2) Þ dy = dx ( y + 2) x

Senior Secondary School Mathematics for Class 12 Pg-925

Differential Equations with Variable Separable

y

( x + 2) dx + C , where C is an arbitrary constant x

Þ

ò ( y + 2) dy = ò

Þ

ò íî1 - ( y + 2) ýþ dy = ò çè1 + x ÷ø dx + C

ì

2

925

ü

æ

2ö

Þ y - 2 log |y + 2| = x + 2 log |x| + C. Putting x = 1 and y = -1 in (i), we get C = -2. Hence, the required solution is y - 2 log |y + 2| = x + 2 log |x| - 2. EXAMPLE 21

SOLUTION

… (i)

Find the particular solution of the differential equation x( x 2 - 1) it being given that y = 0 when x = 2. We have dy x( x 2 - 1) =1 dx 1 dx Þ dy = 2 x( x - 1) dx

Þ

ò dy = ò x( x - 1)( x + 1) ×

Let

1 A B C = + + × x( x - 1)( x + 1) x ( x - 1) ( x + 1)

Then, A( x - 1)( x + 1) + Bx( x + 1) + Cx( x - 1) º 1. Putting x = 0 in (ii), we get A = -1. 1 Putting x = 1 in (ii), we get B = × 2 1 Putting x = -1 in (ii), we get C = × 2 1 -1 1 1 = + + × \ x( x - 1)( x + 1) x 2( x - 1) 2( x + 1)

dy = 1, dx

[CBSE 2012]

… (i)

… (ii)

Putting this value in (i), we get - dx 1 dx 1 dx ò dy = ò x + 2 ò ( x - 1) + 2 ò ( x + 1) + C1 , where C1 is an arbitrary constant 1 1 … (iii) Þ y = - log |x| + log |x - 1| + log |x + 1| + C1. 2 2 We are given that y = 0 when x = 2. Putting x = 2 and y = 0 in (iii), we get 1 - log 2 + log 3 + C1 = 0 2 1 1 1 1 æ 4ö Þ C1 = log 2 - log 3 = log 4 - log 3 = log ç ÷ × 2 2 2 2 è 3ø

Senior Secondary School Mathematics for Class 12 Pg-926

926

Senior Secondary School Mathematics for Class 12

1 æ 4ö log ç ÷ in (iii) we get 2 è 3ø 1 1 1 4 y = - log |x| + log |x - 1| + log |x + 1| + log 2 2 2 3 1 1 1 4 y = - log x 2 + log |( x - 1)( x + 1)| + log 2 2 2 3 1 4( x 2 - 1) y = log , which is the required solution. 2 3x2

Putting C1 =

Þ Þ EXAMPLE 22

SOLUTION

Find

the particular solution of the differential equation dy ( x + 1) = 2e- y - 1, it being given that y = 0 when x = 0. [CBSE 2012] dx We have dy ( x + 1) = 2e- y - 1 dx 1 1 Þ dy = dx ( x + 1) ( 2e- y - 1) Þ

ey ( 2 - ey) ey

dy =

1 dx ( x + 1) -1

Þ

ò ( ey - 2) dy = ò ( x + 1) dx

Þ

log |ey - 2| = - log |x + 1| + log |C1|, where C1 is an arbitrary constant

Þ

log |ey - 2| + log |x + 1| = log |C1|

Þ

log |( x + 1)( ey - 2)| = log |C1|

Þ ( x + 1)( ey - 2) = ± C1 = C. It is given that when x = 0, then y = 0. Putting x = 0 and y = 0 in (i), we get C = -1. Putting C = -1 in (i), we get -1 ( x + 1)( ey - 2) = -1 Þ ( ey - 2) = ( x + 1) ì 1 ü ( 2x + 1) × Þ e y = í2 ý= x ( + 1) þ ( x + 1) î 2x + 1 \ y = log , x ¹ -1 is the required solution. x+1

… (i)

EXAMPLE 23

Solve the differential equation (1 + y 2)(1 + log x) dx + x dy = 0, it being given that y = 1 when x = 1.

SOLUTION

We have

[CBSE 2000, ‘03, ‘11]

(1 + y 2)(1 + log x) dx + x dy = 0

Senior Secondary School Mathematics for Class 12 Pg-927

Differential Equations with Variable Separable

Þ Þ Þ

927

(1 + log x) 1 dx + dy = 0 [on separating the variables] x (1 + y 2) (1 + log x) 1 dx + ò dy = C [integrating both sides] ò x (1 + y 2)

ò t dt + tan

-1

y = C , where (1 + log x) = t

1 2 t + tan -1 y = C , where C is an arbitrary constant 2 1 … (i) (1 + log x) 2 + tan -1 y = C. Þ 2 Putting x = 1 and y = 1 in (i), we get 1 æ 1 pö … (ii) C = + tan -11 Þ C = ç + ÷ × 2 è 2 4ø 1 æ 1 pö (1 + log x) 2 + tan -1 y = ç + ÷ [using (ii) in (i)] \ 2 è 2 4ø 1 p 2 -1 (log x) + log x + tan y = , which is the required solution. Þ 2 4 Þ

EXAMPLE 24

SOLUTION

EXAMPLE 25

SOLUTION

Solve the differential equation (1 + e2x ) dy + ex (1 + y 2) dx = 0, it being given that y = 1 when x = 0. We have

[CBSE 2004, ‘08C, ’11]

(1 + e2x ) dy + ex (1 + y 2) dx = 0 1 ex Þ dy + dx = 0 [separating the variables] (1 + y 2) (1 + e2x ) 1 ex Þ ò dy + ò dx = C [integrating both sides] 2 (1 + y ) (1 + e2x ) dt = C, where ex = t Þ tan -1 y + ò (1 + t 2) Þ tan -1 y + tan -1t = C … (i) [Q t = ex ] Þ tan -1 y + tan -1 ex = C Putting x = 0 and y = 1 in (i), we get æ p pö p C = tan -11 + tan -1 e0 = (tan -11 + tan -11) = ç + ÷ = × è 4 4ø 2 p -1 -1 x \ tan y + tan e = is the required solution. 2 Find the equation of the curve that passes through the point (1, 2) and dy -2xy satisfies the differential equation = × dx ( x 2 + 1) We have dy -2xy = 2 dx ( x + 1)

Senior Secondary School Mathematics for Class 12 Pg-928

928

Senior Secondary School Mathematics for Class 12

Þ Þ

dy -2x = dx [on separating the variables] y ( x 2 + 1) -2x dy ò y = ò ( x 2 + 1) dx [integrating both sides]

log y = - log ( x 2 + 1) + log C , where log C is an arbitrary constant Þ log y + log ( x 2 + 1) = log C Þ log {y( x 2 + 1)} = log C … (i) Þ y( x 2 + 1) = C Now, it is given that the curve passes through (1, 2). So, putting x = 1 and y = 2 in (i), we get C = 4. \ y( x 2 + 1) = 4 is the required equation of the curve. Þ

EXAMPLE 26

SOLUTION

Find the equation of a curve which passes through the point ( -2, 3) and 2x the slope of whose tangent at any point ( x , y) is 2 × y dy We know that the slope of a curve at a point ( x , y) is × dx dy 2x … (i) \ = dx y 2 Þ

y 2 dy = 2x dx [separating the variables]

Þ

òy

2

dy = ò 2x dx

1 3 … (ii) Þ y = x2 + C 3 where C is a constant. Thus, (ii) is the equation of the curve whose differential equation is given by (i). Since the given curve passes through the point ( -2, 3), we have æ1 ö C = ç ´ 27 ÷ - ( -2) 2 = ( 9 - 4) = 5. 3 è ø Hence, the required equation of the curve is 1 3 y = x 2 + 5 Þ y 3 = 3 x 2 + 15. 3 EXAMPLE 27

SOLUTION

In a bank principal increases at the rate of 5% per annum. In how many years will ` 1000 double itself ? Let P be the principal at any time t. Then, dP æ 5 ö =ç ÷P dt è 100 ø dP P Þ = dt 20 dP 1 = dt Þ P 20

Senior Secondary School Mathematics for Class 12 Pg-929

Differential Equations with Variable Separable

929

1 dP = dt [on integrating both sides] P ò 20 1 Þ log P = t + log C , where C is an arbitrary constant 20 t æ Pö Þ log ç ÷ = è C ø 20 P Þ = et / 20. C When t = 0, we have P = 1000 (given). Putting P = 1000 and t = 0 in (i), we get C = 1000. \ P = (1000) et/20. Þ

ò

… (i)

… (ii)

Let ` 1000 double itself in n years. Thus, when t = n, then P = 2000. Putting these values in (ii), we get (1000) ´ en/20 = 2000 Þ en/20 = 2 n Þ = log e 2 Þ n = 20(log e 2). 20 Hence, the required time is {20(log e 2)} years.

EXERCISE 19A Very-Short-Answer Questions Find the general solution of each of the following differential equations: dy dy 1. 2. x 4 = -y 4 = (1 + x 2)(1 + y 2) dx dx dy dy 4. 3. = 1 + x + y + xy [CBSE 2012] = 1 - x + y - xy dx dx dy dy 6. 5. ( x - 1) = 2x 3 y = ex + y dx dx dy 7. ( ex + e- x ) dy - ( ex - e- x ) dx = 0 [CBSE 2006] 8. = ex - y + x 2e- y dx 9. e2x - 3y dx + e2y - 3x dy = 0 10. ex tan y dx + (1 - ex ) sec2 y dy = 0 2

[CBSE 2011]

2

11. sec x tan y dx + sec y tan x dy = 0 12. cos x(1 + cos y) dx - sin y(1 + sin x) dy = 0 For each of the following differential equations, find a particular solution satisfying the given condition: æ dy ö 13. cos ç ÷ = a , where a Î R and y = 2 when x = 0. è dx ø dy 14. = -4xy 2 , it being given that y = 1 when x = 0. dx

Senior Secondary School Mathematics for Class 12 Pg-930

930

Senior Secondary School Mathematics for Class 12

15. x dy = ( 2x 2 + 1) dx ( x ¹ 0), given that y = 1 when x = 1. dy 16. = y tan x , it being given that y = 1 when x = 0. dx ANSWERS (EXERCISE 19A)

1. tan -1 y = x +

x3 +C 3

4. log |1 + y| = x 6. ex + e- y = C

2.

x2 +C 2

1 x3

+

1

3. log |1 + y| = x +

=C

y3

2x 3 + x 2 + 2x + 2 log |x - 1| + C 3

5. log |y| =

7. y = log ( ex + e- x ) + C

9. e5x + e5y = C

8. ey = ex +

10. tan y = C(1 - ex )

12. (1 + sin x)(1 + cos y) = C

x2 +C 2

x3 +C 3

11. tan x tan y = C

1 æ y - 2ö 13. cos ç ÷ = a 14. y = 2 è x ø ( 2x + 1)

15. y = x 2 + log |x| 16. y = sec x HINTS TO SOME SELECTED QUESTIONS (EXERCISE 19A) 1. 2. 3.

1

ò ( 1 + y 2 ) dy = ò ( 1 + x 2 ) dx + C . -1 y4

1

dy =

x4

dx Þ

ò x -4 dx + ò y -4 dy = C1

Þ

dy = ( 1 + x ) + y ( 1 + x ) = ( 1 + x )( 1 + y ) Þ dx

1 x3

+

1 y3

= -3C 1 = C.

dy

ò ( 1 + y ) = ò ( 1 + x ) dx + C .

é 1 2x 3 1 ù 3 dy = = 2 êx2 + x + 1 + ú [on dividing x by ( x - 1)] ( x - 1) ( x - 1) û y ë dy 1 6. = e x × e y Þ y dy = e xdx Þ ò e - y dy = ò e xdx + C dx e 5.

7. dy 8.

( e x - e -x ) -x

x

(e + e )

dx = 0 Þ

e x - e -x

ò dy - ò e x + e -x dx + C

dy = e x × e -y + x 2 e -y = ( e x + x 2 )e -y Þ dx

9. e 2 x × e -3 y dx + e 2 y × e -3 xdy = 0 Þ 10. - ò

-ex x

(1- e )

\ log

dx +

tan y ( 1 - ex)

ò e y dy = ò ( e x + x 2 ) dx + C .

ò e5 xdx + ò e5 y dy = C .

sec2 y

ò tan y dy = log|C1| Þ = log|C 1| Þ

Þ y - log ( e x + e - x ) = C .

tan y ( 1 - ex)

log|tan y| - log|1 - e x| = log|C 1|.

= ± C1 = C .

Senior Secondary School Mathematics for Class 12 Pg-931

Differential Equations with Variable Separable

11.

sec2 x

sec2 y

ò tan x dx + ò tan y dy = log|C1| Þ

931

log|tan x| + log|tan y| = log|C 1|

\ log|tan x tan y| = log|C 1| Þ |tan x tan y| =|C 1|. \ tan x tan y = ± C 1 = C . cos x - sin y 12. ò dx + ò dy = log|C 1| ( 1 + sin x ) ( 1 + cos y ) Þ log|1 + sin x| + log|1 + cos y| = log|C 1| Þ log|( 1 + sin x )( 1 + cos y )| = log|C 1| Þ |( 1 + sin x )( 1 + cos y )| = C 1 Þ ( 1 + sin x )( 1 + cos y ) = ± C 1 = C . dy 13. = cos -1 a Þ ò dy = ò (cos -1 a) dx Þ y = (cos -1 a)x + C . dx Putting x = 0 and y = 2 in (i), we get C = 2. æy- 2ö æy- 2ö ÷ Þ cos ç ÷ = a. \ cos -1 a = ç è x ø è x ø 14.

dy

ò y 2 = ò -4 x dx Þ

-1 1 = -2 x 2 + C Þ y = × y ( 2x 2 - C )

… (i)

… (i)

Putting x = 0 and y = 1 in (i), we get C = -1. 1 Hence, y = × ( 2 x 2 + 1) ( 2 x 2 + 1) 1ö æ dx = ò ç 2 x + ÷ dx Þ y = x 2 + log|x| + C . x xø è Putting x = 1 and y = 1 in (i), we get C = 0. Hence, y = x 2 + log|x|.

15.

ò dy = ò

16.

ò y dy = ò tan x dx Þ

1

log|y| = log|sec x| + log|C 1|

\ ( y cos x ) = ± C 1 = C . When x = 0 and y = 1, we get C = 1. Hence, y = sec x.

EXERCISE 19B Find the general solution of each of the following differential equations: dy x - 1 = dx y + 2 dy 3. = (1 + x)(1 + y 2) dx 1.

dy + y = 1 ( y ¹ 1) dx dy 7. x + y = y2 dx dy 9. y(1 - x 2) = x(1 + y 2) dx

5.

dy x = dx ( x 2 + 1) dy 4. (1 + x 2) = xy dx 2.

6.

8. x 2( y + 1) dx + y 2( x - 1) dy = 0 10. y log y dx - x dy = 0

11. x( x 2 - x 2 y 2) dy + y( y 2 + x 2 y 2) dx = 0 12. (1 - x 2) dy + xy(1 - y) dx = 0

dy 1 - y2 + =0 dx 1 - x2

… (i)

Senior Secondary School Mathematics for Class 12 Pg-932

932

Senior Secondary School Mathematics for Class 12

13. (1 - x 2)(1 - y) dx = xy(1 + y) dy 14. ( y + xy) dx + ( x - xy 2) dy = 0

[CBSE 2002C]

15. ( x 2 - yx 2) dy + ( y 2 + xy 2) dx = 0

[CBSE 2004C]

2

2

2

2

16. ( x y - x ) dx + ( xy - y ) dy = 0 17. x 1 + y 2 dx + y 1 + x 2 dy = 0 18.

dy = ex + y + x 2ey dx

19.

dy 3 e2x + 3 e4x = dx ex + e- x

[CBSE 2006]

20. 3 ex tan y dx + (1 - ex ) sec2 y dy = 0 21. ey(1 + x 2) dy -

x dx = 0 y

dy = ex + y + ex - y dx dy xy + y 24. + =0 dx xy + x 22.

26. cosec x log y

dy + x 2y = 0 dx

1 dy × = tan -1 x x dx dy 1 - cos x 30. = dx 1 + cos x 28.

32.

dy (1 + cos 2y) + =0 dx (1 - cos 2x)

23. ( ey + 1) cos x dx + ey sin x dy = 0 25.

1 - x 4 dy = x dx

27. y dx + (1 + x 2) tan -1 x dy = 0 29. ex 1 - y 2 dx + 31. (cos x) 33.

y dy = 0 x

dy + cos 2x = cos 3 x dx

dy cos x sin y + =0 dx cos y

34. cos x(1 + cos y) dx - sin y(1 + sin x) dy = 0 35. sin 3 x dx - sin y dy = 0 dy + sin ( x + y) = sin ( x - y) dx 1 1 37. cos2 y dy + cos2 x dx = 0 x y

36.

38.

dy = sin 3 x cos2 x + xex dx

39. Find the particular solution of the differential equation given that y = 0 when x = 1.

dy = 1 + x + y + xy , dx [CBSE 2014]

40. Find the particular solution of the differential equation x(1 + y 2) dx - y(1 + x 2) dy = 0, given that y = 1 when x = 0. [CBSE 2014]

Senior Secondary School Mathematics for Class 12 Pg-933

Differential Equations with Variable Separable

41. Find the particular solution of the æ dy ö log ç ÷ = 3 x + 4y , given that y = 0 when x = 0. è dx ø

933

differential

equation [CBSE 2014]

42. Solve the differential equation ( x 2 - yx 2) dy + ( y 2 + x 2 y 2) dx = 0, given that y = 1 when x = 1. [CBSE 2014] 43. Find the particular solution of the differential equation y [CBSE 2014] ex 1 - y 2 dx + dy = 0, given that y = 1 when x = 0. x 44. Find the particular solution of the differential equation p dy x( 2 log x + 1) [CBSE 2014] = , given that y = when x = 1. 2 dx (sin y + y cos y) 45. Solve the differential equation

dy = y sin 2x , given that y( 0) = 1. [CBSE 2004] dx

46. Solve the differential equation ( x + 1)

dy = 2xy , given that y( 2) = 3. dx [CBSE 2004]

dy 47. Solve = x( 2 log x + 1), given that y = 0 when x = 2. dx dy 48. Solve ( x 3 + x 2 + x + 1) = 2x 2 + x , given that y = 1 when x = 0. dx dy 49. Solve = y tan x , given that y = 1 when x = 0. dx dy 50. Solve = y 2 tan 2x , given that y = 2 when x = 0. dx p dy 51. Solve = y cot 2x , given that y = 2 when x = × 4 dx p 2 2 52. Solve (1 + x ) sec y dy + 2x tan y dx = 0, given that y = when x = 1. 4 æ pö 53. Find the equation of the curve passing through the point ç 0, ÷ whose è 4ø differential equation is sin x cos y dx + cos x sin y dy = 0. 54. Find the equation of a curve which passes through the origin and whose dy differential equation is = ex sin x. dx 55. A curve passes through the point (0, –2) and at any point ( x , y) of the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point. Find the equation of the curve. 56. A curve passes through the point (–2, 1) and at any point ( x , y) of the curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point ( -4, - 3). Find the equation of the curve.

Senior Secondary School Mathematics for Class 12 Pg-934

934

Senior Secondary School Mathematics for Class 12

57. In a bank, principal increases at the rate of r% per annum. Find the value of r if ` 100 double itself in 10 years. (Given log e 2 = 0.6931) 58. In a bank, principal increases at the rate of 5% per annum. An amount of ` 1000 is deposited in the bank. How much will it worth after 10 years? (Given e0.5 = 1.648) 59. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds. 60. In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present? ANSWERS (EXERCISE 19B)

1. y 2 + 4y - x 2 + 2x = C

2. y =

x2 1 log ( x 2 + 1) + C 3. tan -1 y = x + +C 2 2

4. y = C1 1 + x 2

5. x + log |1 - y| = C

7. y = 1 + C1 xy

8.

9. (1 + y 2)(1 - x 2) = C 12. y = C(1 - y) 1 - x 2 14. log |xy| + x -

y2 =C 2

6. sin -1 y + sin -1 x = C

x2 y2 + + x - y + log |x - 1| + log |y + 1| = C 2 2 10. x = C log y 13. log |x(1 - y 2)| = 15. log

11. 2

-1 2y

2

-

1 2x

2

+ log

x =C y

2

x y - 2y + C 2 2

x 1 1 = + +C y x y

1 16. ( x 2 + y 2) + ( x + y) + log |( x - 1)( y - 1)| = C 2

17. 1 + y 2 + 1 + x 2 = C

x3 20. tan y = C(1 - e- x ) 3 19. y = e 3x + C =C 3 1 21. ( y - 1) ey = log (1 + x 2) + C 22. tan -1( ey) = ex + C 2 1 23. (1 + ey) sin x = C 24. x + y + log |xy| = C 25. y = sin -1( x 2) + C 2 1 26. (log y) 2 + ( 2 - x 2) cos x + 2x sin x = C 27. y tan -1 x = C 2 1 1 29. ( x - 1) ex - 1 - y 2 = C 28. y = ( x 2 + 1) tan -1 x - x + C 2 2 x 30. y = 2 tan - x + C 31. y = sin 2x - 2 sin x - x + log |sec x + tan x| + C 2 32. tan y = cot x + C 33. log |sin y| + sin x = C

18. ex + e- y +

Senior Secondary School Mathematics for Class 12 Pg-935

Differential Equations with Variable Separable

34. (1 + sin x)(1 + cos y) = C

935

35. 12 cos y = 9 cos x - cos 3 x + C

36. log |cosec y - cot y| + 2 sin x = C 37. 2( x 2 + y 2) + 2x sin 2x + 2y sin 2y + cos 2x + cos 2y = C 38. y = -

1 1 cos 3 x + cos5 x + xex - ex + C 3 5

39. log |1 + y| = x +

41. 4e 3x + 3 e-4y = 7

40. y = 2x 2 + 1

43. ex ( x - 1) = 1 - y 2 - 1

42. log |y| +

44. y sin y = x 2 log x +

( 2x - 4)

46. y( x + 1) 2 = 27 e

p 2

x2 3 2 2

1 1 + -x =1 y x 2

45. y = esin

x

47. y = x 2 log x - 4 log 2

1ì 3 ü 48. y = ílog |x + 1| + log ( x 2 + 1) - tan -1 x ý + 1 49. y cos x = 1 2î 2 þ 52. (1 + x 2) tan y = 2

50. y(1 + log |cos 2x|) = 2

51. y = 2 sin 2x

æ 1 ö 53. y = cos-1 ç sec x ÷ è 2 ø

54. 2y = ex (sin x - cos x) + 1 55. y 2 = x 2 + 4

56. y + 3 = ( x + 4) 2 57. r = 6.931 60.

59. r = ( 63t + 27)1/ 3

58. ` 1648

2 log 2 æ 11 ö log ç ÷ è 10 ø HINTS TO SOME SELECTED QUESTIONS (EXERCISE 19B)

4.

1

1

2x

ò y dy = 2 ò ( 1 + x 2 ) dx Þ

log|y| y

Þ log 5.

1 log|1 + x 2| = log|C| 2

1+ x

2

= log|C| Þ

y 1 + x2

= ± C = C1 .

- dy 1 = dx dy = dx Þ - ò ( 1 - y) ( 1 - y) ò - dy Þ ò dx + ò = C Þ x + log|1 - y| = C . ( 1 - y) 1

1

ì

1

1ü

Þ

ò çè x + 1 +

1

7.

ò y( y - 1) dy = ò x dx Þ ò í( y - 1) - y ý dy = ò x dx

8.

ò ( x - 1) dx + ò ( y + 1) dy = C

9.

ò ( 1 + y 2 ) dy = - ò ( 1 - x 2 ) dx Þ

x2

2y

[by partial fractions] î þ Þ log|y - 1| - log|y| = log|x| + log|C| Þ ( y - 1) = ± Cxy Þ ( y - 1) = C 1 xy Þ y = 1 + C 1 xy.

y2

-2 x

æ

1 ö ÷ dx + x - 1ø

æ

ò ççè y - 1 +

1 y+

log|1 + y 2| + log|1 - x 2| = log|C 1|

Þ ( 1 + y 2 )( 1 - x 2 ) = ± C 1 = C .

ö ÷ dy = C. 1 ÷ø

Senior Secondary School Mathematics for Class 12 Pg-936

936

10.

Senior Secondary School Mathematics for Class 12

dt , where log y = t. t \ log|x| = log|| t + log|C 1| Þ log|x| - log|log y| = log|C 1| x x Þ log = log C 1 Þ = ± C1 = C . log y log y 1

1

1

ò x dx = ò y log y dy Þ ò x dx = ò

11. x 3 ( 1 - y 2 )dy + y 3 ( 1 + x 2 )dx = 0 Þ \ 12.

1ö

æ

ò

( 1 - y2 ) y

3

dy +

ò

( 1 + x2 ) x3

dx = C .

y -2 x -2 1ö - log|y| + + log|x| = C . ÷ dx = C Þ -2 xø -2

æ

ò ççè y -3 - y ÷÷ø dy + ò çè x -3 + x

1

ò y( 1 - y ) dy + ò ( 1 - x 2 ) dx = log|C1| Þ

ì1

1

ü

1

î

þ

Þ log|y| - log|1 - y| Þ

-2 x

ò í y + ( 1 - y ) ý dy - 2 ò ( 1 - x 2 ) dx = log|C1| y ( 1 - y) 1 - x 2

y 1 log|1 - x 2| = log|C 1| Þ log = log|C 1| 2 ( 1 - y) 1 - x 2

= ± C1 = C .

( 1 - x2 ) ( y 2 + y) y ( 1 + y) æ1 ö dy dx = ò dy Þ ò ç - x ÷ dx = ò x ( 1 - y) ( - y + 1) èx ø æ 2 ö æ1 ö ÷ dy [on dividing ( y 2 + y ) by ( - y + 1)]. \ ò ç - x ÷ dx = ò çç - y - 2 + 1 - y ÷ø èx ø è dy 3 e 2 x( 1 + e 2 x ) 3 e 3 x( 1 + e 2 x ) 19. = = = 3 e 3x. 1ö æ x dx ( 1 + e2x) çe + x ÷ è e ø sec2 y 3e x 20. ò dx + ò dy = log|C 1| x tan y (1 - e ) 13.

ò

Þ -3 ò

-ex x

(1 - e )

dx +

sec2 y

ò tan y dy = log|C1| Þ

Þ log|( 1 - e - x ) -3 tan y| = log|C 1| Þ 22.

ey

- 3 log|1 - e - x| + log|tan y|= log|C 1|

tan y ( 1 - e -x ) 3

= ± C1 = C .

dt

ò ( e 2 y + 1) dy = ò e xdx Þ ò (t 2 + 1) = e x + C , where e y = t

Þ tan -1 (t ) = e x + C Þ tan -1 ( e y ) = e x + C . log y 26. ò dy + ò x 2 sin x dx = C y I II Þ

ò t dt + x 2 ( - cos x ) - ò 2 x( - cos x ) dx = C , where log y = t

t2 - x 2 cos x + 2 ò x cos x dx = C . 2 I II dx 1 27. ò + ò dy = log|C 1|. y ( 1 + x 2 ) tan -1 x dx -1 Put tan x = t and = dt. ( 1 + x2 ) Þ

Senior Secondary School Mathematics for Class 12 Pg-937

Differential Equations with Variable Separable

28.

30.

x2 x2 1 × dx + C 2 ò ( 1 + x2 ) 2 1 1 ïì ( 1 + x 2 ) - 1 ïü Þ y = x 2 (tan -1 x ) - ò í ý dx + C 2 2 ïî ( 1 + x 2 ) ïþ üï 1 1 ìï 1 Þ y = x 2 (tan -1 x ) - ò í 1 ý dx + C . 2 2 2 ïî ( 1 + x ) ïþ

ò dy = ò (tanI-1 x ) IIx dx Þ

y = (tan -1 x )

dy 1 - cos x 2 sin 2 ( x/ 2 ) x æ x ö = = = tan 2 = ç sec2 - 1 ÷ 2 dx 1 + cos x 2 cos 2 ( x/ 2 ) 2 è ø æ ö 2x \ ò dy = ò ç sec - 1 ÷ dx. 2 è ø

dy cos 3 x - cos 2 x ( 4 cos 3 x - 3 cos x ) - ( 2 cos 2 x - 1) = = dx cos x cos x dy 2 = 4 cos x - 2 cos x - 3 + sec x Þ dx dy 4( 1 + cos 2 x ) = - 2 cos x - 3 + sec x = 2 cos 2 x - 2 cos x - 1 + sec x. Þ dx 2 2 dy 2 cos y dy cosec2 x 32. 0 + = Þ + = 0 Þ ò (sec2 y )dy + ò cosec2 x dx = C . dx 2 sin 2 x dx sec2 y 31.

33.

ò cot y dy + ò cos x dx = C .

34.

ò ( 1 + sin x ) dx - ò ( 1 + cos y ) dy = log|C1|

cos x

sin y

Þ log|1 + sin x| + log|1 + cos y| = log|C 1| Þ log|( 1 + sin x )( 1 + cos y )| = log|C 1| Þ ( 1 + sin x )( 1 + cos y ) = ± C 1 = C. ( 3 sin x - sin 3 x ) dx [Q sin 3 x = 3 sin x - 4 sin 3 x] 4 3 1 \ ò sin y dy = ò sin x dx - ò sin 3 x dx + C . 4 4 ì( x - y) - ( x + y) ü ì( x - y) + ( x + y) ü 36. sin ( x - y ) - sin ( x + y ) = 2 cos í ý ý sin í 2 2 î î þ þ = 2 cos x sin ( - y ) = -2 cos x sin y. 1 dy = -2 ò cos x dx Þ ò cosec y dy = -2 sin x + C \ ò sin y 35. sin y dy = sin 3 x dx =

Þ log|cosec y - cot y| = -2 sin x + C . 2

2

37. y cos y dy + x cos x dx = 0 Þ y ( 2 cos 2 y )dy + x( 2 cos 2 x )dx = 0 Þ y ( 1 + cos 2 y )dy + x( 1 + cos 2 x )dx = 0

38.

Þ

ò y dy + ò y cos 2 y dy + ò x dx + ò x cos 2 x dx = C

Þ

x2 y2 + + 2 2

ò y cos 2 y dy + ò xI cosII 2 x dx = C . I

II

ò dy = ò cos x( 1 - cos2 x ) sin x dx + ò xe xdx + C Þ y = - ò t 2 ( 1 - t 2 )dt + ò xe xdx + C , where t = cos x Þ y = - ò t 2 dt + ò t 4 dt + ò x e x dx + C . I II 2

937

Senior Secondary School Mathematics for Class 12 Pg-938

938

39.

Senior Secondary School Mathematics for Class 12 dy = ( 1 + x )( 1 + y ) Þ dx

dy

ò ( 1 + y ) = ò ( 1 + x ) dx Þ

Putting x = 1 and y = 0 in (i), we get C =

log|1 + y| = x +

x2 + C. 2

… (i)

-3 × 2

x2 3 - × 2 2 2y 1 2x 1 40. ò dx - ò dy = log|C 1| 2 ( 1 + x2 ) 2 ( 1 + y2 ) 1 1 Þ log ( 1 + x 2 ) - log ( 1 + y 2 ) = log|C 1| 2 2 \ log|1 + y| = x +

Þ log ( 1 + x 2 ) - log ( 1 + y 2 ) = log|C 12| = log C Þ Putting x = 0 and y = 1 in (i), we get C =

( 1 + x2 ) ( 1 + y2 )

= C.

… (i)

1 × 2

\ ( 1 + y 2 ) = 2( 1 + x 2 ) Þ y = 2 x 2 + 1. 41.

dy = e 3 x+ 4 y = e 3 x × e 4 y Þ dx e 3 x e -4 y \ = + C. 3 -4

ò e 3 xdx = ò e -4 y dy

æ 1 1ö 7 Putting x = 0 and y = 0 in (i), we get C = ç + ÷ = × è 3 4 ø 12 3x -4 y 7 e e = + Þ 4 e 3 x + 3 e -4 y = 7 . \ 3 -4 12 æ 1 + x2 ( 1 - y) 42. x 2 ( 1 - y )dy + y 2 ( 1 + x 2 )dx = 0 Þ ò dy + ò çç 2 2 y è x 1 1 1 1 \ - - log|y| - + x = C 1 Þ log|y| + + - x = C . y x y x Putting x = 1 and y = 1 in (i), we get C = 1. -2 y 1 43. ò x e xdx = ò dy Þ e x( x - 1) = 1 - y 2 + C . 2 I II 1 - y2

… (i)

ö ÷ dx = C . 1 ÷ ø … (i)

… (i)

Putting x = 0 and y = 1 in (i), we get C = -1. 44.

ò (sin y + y cos y ) dy = ò ( 2 x log x + x ) dx + C Þ ò sin y dy + ò y cos y dy = ò 2 x log x dx + ò x dx + C

é x2 1 x2 ù x2 Þ ( - cos y ) + y sin y - ò 1 × sin y dy = 2 ê(log x ) × +C -ò × dx ú + 2 x 2 2 êë ûú Þ y sin y = x 2 log x + C . 45.

1

1 cos 2 x + C . 2 1 Putting x = 0 and y = 1 in (i), we get C = × 2 1 æ1 ö \ log|y|= ( 1 - cos 2 x ) = ç ´ 2 sin 2 x ÷ = sin 2 x. 2 è2 ø

ò y dy = ò sin 2 x dx + C

2

Hence, y = e sin x.

Þ log|y| = -

… (i) … (i)

Senior Secondary School Mathematics for Class 12 Pg-939

Differential Equations with Variable Separable

46.

939

ì ( x + 1) - 1 1 ü dx = 2 ò í 1 ý dx + 1) þ x ( x + 1) ( î Þ log|y| = 2 x - 2 log|x + 1| + log|C 1| Þ log|y| = log|e 2 x| - log|( x + 1) 2| + log|C 1| 2x

1

ò y dy = ò ( x + 1) dx = 2 ò

Þ log

y( x + 1) 2 e

2x

= log|C 1| Þ

y( x + 1) 2 e2x

= ± C 1 = C (say).

Then, y( x + 1) 2 = Ce 2 x.

… (i) -4

Putting x = 2 and y = 3 in (i), we get C = 27 e . \ y( x + 1) 2 = 27 e 2 x- 4 is the required solution. 1 ïì 1 3 x - 1 ïü 48. dy = í + ý dx [by partial fractions] 2 ïî ( x + 1) ( x 2 + 1) ïþ üï 1 ìï 1 3 2x 1 + × 2 - 2 Þ ò dy = ò í ý dx + C 2 îï ( x + 1) 2 ( x + 1) ( x + 1) þï

1ì 3 2 -1 ü í log|x + 1| + log|x + 1| - tan x ý + C . 2î 2 þ When x = 0 and y = 1, then C = 1. dy cos 2 x 1 51. ò dx Þ log|y| = log|sin 2 x| + log|C 1| = y ò sin 2 x 2 Þ y=

\

y sin 2 x

= C1 Þ

y sin 2 x

= ± C 1 = C (say).

… (i)

p and y = 2 in (i), we get C = 2. 4 Hence, y = 2 sin 2 x .

Putting x =

53. tan x dx + tan y dy = 0 Þ ò tan x dx + ò tan y dy = constant

Þ - log|cos x| - log|cos y| = log|C 1| 1 1 Þ cos x cos y = ± = C (say). Þ log|cos x cos y| = log C1 C1 p 1 Putting x = 0 and y = in (i), we get C = × 4 2 1 1 æ 1 ö \ cos x cos y = Þ cos y = sec x Þ y = cos -1 ç sec x ÷ × 2 2 è 2 ø ax e ( a sin bx b cos bx ) 54. Use the formula ò e ax sin bx dx = × ( a2 + b 2 )

Put a = 1 and b = 1. dy y2 x2 55. y = x Þ ò y dy = ò x dx Þ = + C. 2 2 dx Put x = 0 and y = -2 to get C = 2. dy 2( y + 3 ) 1 1 56. = Þ ò dy = 2 ò dx dx ( x + 4 ) ( y + 3) ( x + 4) ( y + 3) \ log|y + 3| = 2 log|x + 4| + log|C 1| Þ = ± C1 = C . ( x + 4)2 Putting x = -2 and y = 1, we get C = 1.

… (i)

Senior Secondary School Mathematics for Class 12 Pg-940

940

57.

Senior Secondary School Mathematics for Class 12

dP æ r ö dP rt æ r ö =ç = ç + C. ÷P Þ ò ÷ dt Þ log P = 100 dt è 100 ø P ò è 100 ø At t = 0, we have P = P0. So, log P0 = C . æP ö rt rt Þ log çç ÷÷ = × \ log P - log P0 = 100 è P0 ø 100

… (i)

… (ii)

Putting P0 = 100 , P = 2 P0 = 200 and t = 10, we get r = log 2 Þ r = ( 10 ´ 0.6931) = 6.931. 10 dP 5 P dP 1 t 58. = Þ ò = dt Þ log P = + log C . dt 100 P ò 20 20 At t = 0, we have P = 1000 and so log C = log 1000. t + log 1000. \ log P = 20 P æ P ö Putting t = 10, we get log ç = e 0.5 = 1.648. ÷ = 0.5 Þ 1000 è 1000 ø \ P = ( 1000 ´ 1.648 ) = 1648. 4 59. The volume of a spherical balloon of radius r is given by V = pr 3 . 3 dV Now, = - k , where k > 0 [note that V is decreasing] dt d æ4 3ö dr Þ = -k ç pr ÷ = - k Þ ( 4 pr 2 ) dt è 3 dt ø Þ

ò ( 4 pr 2 )dr = ò ( - k ) dt

4 3 … (i), where C is an arbitrary constant. pr = - kt + C 3 Putting t = 0 and r = 3 in (i), we get C = 36p. 4 3 \ pr = - at + 36 p. 3 It is being given that when t = 3, then r = 6. Putting t = 3 and r = 6 in (ii), we get k = -84p. Putting k = -84p in (ii), we get r 3 = ( 63t + 27 ) Þ r = ( 63t + 27 )1/3 . Þ

60. Let at any time t, the bacteria count be N. Then, dN dN 1 µN Þ = kN Þ ò dN = ò k dt Þ log N = kt + log (C ). dt dt N At t = 0, we have N = 100000. \ log C = log 100000 Þ log N = kt + log 100000. At t = 2, we have N = 110000. 1 11 Putting these values in (i), we get k = log × 2 10 1 11 æ ö \ log N = t log ç ÷ + log 100000. 2 è 10 ø When N = 200000, let t = T , then 2 log 2 T æ 11 ö log 200000 = log ç ÷ + log 100000 Þ T = × æ 11 ö 2 10 è ø log ç ÷ è 10 ø

… (ii)

… (i)

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-941

20. HOMOGENEOUS DIFFERENTIAL EQUATIONS HOMOGENEOUS FUNCTION A function f ( x , y) in x and y is said to be a homogeneous function of degree n, if the degree of each term is n.

Examples

(i) f ( x , y) = ( x 2 + y 2 - xy) is a homogeneous function of degree 2. (ii) g( x , y) = ( x 3 - 3 xy 2 + 3 x 2 y + y 3) is a homogeneous function of degree 3.

In general, a homogeneous function f ( x , y) of degree n is expressible as æyö f ( x , y) = x n f ç ÷ × èxø HOMOGENEOUS DIFFERENTIAL EQUATION

An equation of the form

dy f ( x , y) = , dx g( x , y)

where both f ( x , y) and g ( x , y) are homogeneous functions of degree n, is called a homogeneous differential equation of order 1 and degree 1. dy æyö Such a differential equation may be expressed in the form = fç ÷× dx èxø dy x 2 - y 2 f ( x , y) = = , where f ( x , y) and g( x , y) are dx xy g( x , y) homogeneous functions of degree 2 each. So, it is a homogeneous differential equation of order 1 and degree 1. ìï æ y ö 2 üï í1 - ç ÷ ý dy x 2 - y 2 îï è x ø þï [on dividing Nr and Dr by x 2 ] We may write, = = dx xy ìy ü í ý îx þ dy æyö Thus, = fç ÷× dx èxø For example,

Method of Solving a Homogeneous Differential Equation Let

dy f ( x , y) be a homogeneous differential equation of order 1 and degree 1. = dx g( x , y)

Putting y = vx and v+x

dv ö dy æ = çv + x ÷ in the given equation, we get dx ø dx è

dv = F(v) dx 941

Senior Secondary School Mathematics for Class 12 Pg-942

942

Senior Secondary School Mathematics for Class 12

Þ

dv dx = [F(v) - v] x

Þ

ò [F(v) - v)] = ò

Þ

ò [F(v) - v] = log |x| + C.

dv

dx x

dv

Now, replace v by ( y/x) to obtain the required solution. NOTE

æxö dx = f çç ÷÷ then we put x = vy dy èyø and proceed in a manner similar to as above.

If the differential equation is of the form

SOLVED EXAMPLES

that

the

differential

equation

2x 2

dy - 2xy + y 2 = 0 dx

EXAMPLE 1

Show

SOLUTION

homogeneous and solve it. The given differential equation may be written as

is

[CBSE 2012]

dy 2xy - y 2 = × dx 2x 2

… (i)

On dividing the Nr and Dr of RHS of (i) by 2x 2 , we get 2 dy ìï y 1 æ y ö üï =í - ç ÷ ý= dx îï x 2 è x ø þï

æyö fç ÷× èxø

So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv 2vx 2 - v 2 x 2 = dx 2x 2 1 ö dv æ = çv - v 2 ÷ Þ v+x 2 ø dx è dv 1 2 dv 1 dx Þ x =- v Þ 2 + = 0. dx 2 2 x v v+x

On integrating (ii), we get dv 1 1 ò v 2 + 2 ò x dx = C , where C is an arbitrary constant -1 1 + log |x| = C Þ v 2 -x 1 Þ + log |x| = C , which is the required solution. y 2

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-943

Homogeneous Differential Equations

EXAMPLE 2

Show that the differential equation

943

dy y - x is homogeneous and = dx y + x

solve it. SOLUTION

[CBSE 2004]

The given differential equation is dy y - x = × dx y + x

… (i)

On dividing the Nr and Dr of RHS of (i) by x, we get ü ìy -1ï dy ï x æyö =í ý= fç ÷× dx ï y + 1 ï èxø þ îx So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv vx - x v+x = dx vx + x dv v - 1 = Þ v+x dx v + 1 dv æ v - 1 ö Þ x =ç - v÷ dx è v + 1 ø Þ Þ

dv - (1 + v 2) = dx (1 + v)

(1 + v) 2

(1 + v )

dv =

-1 dx x

(1 + v)

dx x

Þ

ò (1 + v 2) dv = -ò

Þ

ò (1 + v 2) dv + 2 ò (1 + v 2) dv = -ò

Þ

tan -1 v +

Þ

tan -1 v + log x 1 + v 2 = C

Þ

tan -1

Þ EXAMPLE 3

x

1

1

2v

dx x

1 log |1 + v 2| = - log |x| + C 2

y y + log x 2 + y 2 = C [putting v = ] x x 1 -1 y 2 2 tan + log ( x + y ) = C , which is the required solution. x 2

Show that the differential equation x and solve it.

dy - y = x 2 + y 2 is homogeneous dx [CBSE 2005, ’07, ’11]

Senior Secondary School Mathematics for Class 12 Pg-944

944 SOLUTION

Senior Secondary School Mathematics for Class 12

The given differential equation may be written as 2 2 dy y + x + y = × dx x On dividing the Nr and Dr of RHS of (i) by x, we get 2ü dy ìï y æyö ï æyö =í + 1+ ç ÷ ý= f ç ÷× dx ï x èxø ï èxø î þ So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx

v+x Þ Þ Þ Þ

… (i)

dv vx + x 2 + v 2 x 2 = = v + 1 + v2 dx x

dv = 1 + v2 dx dv 1 = dx 2 x 1+v dv dx ò (1 + v 2) = ò x x

log v + 1 + v 2 = log |x| + log |C1|, where C1 is an arbitrary constant v + 1+v x

2

Þ

log

Þ

v + 1 + v2 = ± C1 = C (say) x

= log |C1|

Þ v + 1 + v 2 = Cx Þ

yù é y + x 2 + y 2 = Cx 2 , which is the required solution êQ v = ú × xû ë

EXAMPLE 4

Show that the differential equation ( x x 2 + y 2 - y 2) dx + xydy = 0 is

SOLUTION

homogeneous and solve it. The given differential equation may be written as 2 2 2 dy y - x x + y = × dx xy

On dividing the Nr and Dr of RHS of (i) by x 2 , we get 2 ì y 2 y ü ï æç ö÷ - 1 + æç ö÷ ï dy ï è x ø æyö èxø ï =í ý= fç ÷× y dx ï æ ö èxø ï ç ÷ ï ï èxø þ î

… (i)

Senior Secondary School Mathematics for Class 12 Pg-945

Homogeneous Differential Equations

945

So, the given differential equation is homogeneous. dy dv in (i), we get Putting y = vx and =v+ x dx dx v+x Þ

x

Þ

x

Þ

ò

dv v 2 x 2 - x x 2 + v 2 x 2 = dx vx 2

ö dv æç v 2 - 1 + v 2 = - v÷ ÷ dx ç v è ø dv - 1 + v 2 = dx v v dx dv = - ò 2 x 1+v

Þ

1 + v 2 = - log |x| + C

Þ

x 2 + y 2 + x log |x| = Cx , which is the required solution.

EXAMPLE 5

Show that the differential equation ( x 2 + xy) dy = ( x 2 + y 2) dx is

SOLUTION

homogeneous and solve it. The given differential equation is dy x 2 + y 2 = × dx x 2 + xy On dividing the Nr and Dr of RHS of (i) by x 2 , we get ìï æ y ö 2 üï í1 + ç ÷ ý dy îï è x ø þï æyö = = fç ÷× dx ì yü èxø í1 + ý î xþ So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx v+x Þ Þ Þ Þ

dv x 2 + v 2 x 2 1 + v 2 = 2 = dx 1+v x + vx 2

dv (1 + v 2) 1 + v 2 - v - v 2 (1 - v) = -v = = dx (1 + v) (1 + v) (1 + v) (1 + v) 1 dv = dx (1 - v) x (1 + v) 1 ò (1 - v) dv = ò x dx [on integrating both sides] {2 - (1 - v)} 1 ò (1 - v) dv = ò x dx x

[CBSE 2005]

… (i)

Senior Secondary School Mathematics for Class 12 Pg-946

946

Senior Secondary School Mathematics for Class 12

ì

ü

2

1

Þ

ò íî(1 - v) - 1ýþ dv = ò x dx

Þ

-2 log |1 - v| - v = log |x| + log |C|, where C is an arbitrary constant log |x| + log |C| + 2 log |1 - v| = -v log |Cx(1 - v) 2| = -v

Þ Þ

Þ |Cx(1 - v) 2| = e- v 2

Þ

yö æ Cx ç1 - ÷ = e- y/x xø è

yù é êëQ v = x úû

Þ |C|( x - y) 2 = |x|e- y/x , which is the required solution. EXAMPLE 6

SOLUTION

Show that the differential equation y 2 dx + ( x 2 - xy + y 2) dy = 0 is homogeneous and solve it. The given differential equation may be written as dy -y 2 = 2 × dx ( x - xy + y 2) On dividing the Nr and Dr of RHS of (i) by x 2 , we get æyö -ç ÷ èxø

2

dy æyö = fç ÷× = dx ìï y æ y ö 2 üï èxø í1 - + ç ÷ ý îï x è x ø þï Thus, the given differential equation is homogeneous. dy dv in (i), we get Putting y = vx and =v+ x dx dx dv -v 2 v+x = dx (1 - v + v 2) Þ Þ Þ Þ Þ Þ Þ

é ù dv v2 = -ê + vú 2 dx ë (1 - v + v ) û dv - (v + v 3) x = dx (1 - v + v 2) x

(1 - v + v 2)

1 dv = - dx x v(1 + v 2) 2 (1 + v ) - v dx ò v(1 + v 2) dv = -ò x dv dv dx ò v - ò (1 + v 2) + ò x = log C

log |v| - tan -1 v + log |x| = log C |vx| tan -1 v = log C

… (i)

Senior Secondary School Mathematics for Class 12 Pg-947

Homogeneous Differential Equations

y = log x

tan -1

Þ

|y| tan -1 ( y/x ) =e C

Þ |y| = Ce EXAMPLE 7

SOLUTION

æ|y|ö ç ÷ èCø

Þ

tan -1 ( y/x )

947

yù é êëQ v = x úû

, which is the required solution.

æ dy ö Show that the differential equation x 2 ç ÷ = ( x 2 - 2y 2 + xy) is è dx ø homogeneous and solve it. The given differential equation may be written as dy ( x 2 - 2y 2 + xy) = × dx x2 On dividing the Nr and Dr of RHS of (i) by x 2 , we get 2 dy ìï æ y ö üï æyö æyö = í1 - 2ç ÷ + ç ÷ ý = f ç ÷ × dx ïî èxø èxø è x ø ïþ So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv x 2 - 2v 2 x 2 + vx 2 = dx x2 dv 2 = 1 - 2v + v Þ v+x dx dv Þ x = 1 - 2v 2 dx dv dx Þ = x (1 - 2v 2) dv dx =ò Þ ò x (1 - 2v 2) 1 dv dx Þ =ò 2ò æ1 x 2ö ç -v ÷ è2 ø 1 dv dx Þ =ò 2 ò ìïæ 1 ö 2 x ü 2ï ÷ -v ý íç ïîè 2 ø ïþ 1 +v 1 1 Þ × log 2 = log |x| + C 1 1 ö 2 æ -v ç2 ´ ÷ 2 2ø è y 1 x + 2y log - log |x| = C [Q v = ]. Þ x 2 2 x - 2y This is the required solution. v+x

… (i)

Senior Secondary School Mathematics for Class 12 Pg-948

948 EXAMPLE 8

Senior Secondary School Mathematics for Class 12

Show that the differential equation ( x 3 + y 3) dy - x 2 ydx = 0 is homogeneous and solve it.

SOLUTION

[CBSE 2008]

The given differential equation may be written as dy x 2y = 3 × dx ( x + y 3)

… (i)

On dividing the Nr and Dr of RHS of (i) by x 3 , we get æyö ç ÷ dy æyö èxø = fç ÷× = dx ìï æ y ö 3 üï èxø í1 + ç ÷ ý ïî è x ø ïþ So, the given differential equation is homogeneous. dy dv in (i), we get Putting y = vx and =v+ x dx dx dv vx 3 v+x = 3 dx x + v 3 x 3 dv v Þ v+x = dx 1 + v 3 Þ Þ Þ Þ Þ Þ Þ

EXAMPLE 9

x

dv æ v -v 4 ö =ç - v÷ = 3 dx è 1 + v ø (1 + v 3)

(1 + v 3) 4

dv =

-1 dx x

v 1 æ 1 1ö ò çè v 4 + v ÷ø dv = -ò x dx 1 æ 1 1ö ò çè v 4 + v ÷ø dv + ò x dx = C -1 + log |v| + log |x| = C 3v 3 -1 + log |vx| = C 3v 3 -x 3 + log |y| = C , which is the required solution. 3y 3

Show that the differential equation ( y 2 - x 2) dy = 3 xy dx is homogeneous and solve it.

SOLUTION

[CBSE 2006, ’08]

The given differential equation may be written as dy 3 xy = × dx ( y 2 - x 2)

… (i)

Senior Secondary School Mathematics for Class 12 Pg-949

Homogeneous Differential Equations

949

On dividing the Nr and Dr of RHS of (i) by x 2 , we get æyö 3ç ÷ dy èxø = f æyö× = ç ÷ dx ìïæ y ö 2 üï èxø íç ÷ - 1ý ïþ ïîè x ø Thus, the given differential equation is homogeneous. dy dv in (i), we get Putting y = vx and =v+ x dx dx dv 3v v+x = dx (v 2 - 1) ü ( 4v - v 3) dv ì 3v Þ x =í 2 - vý = 2 dx î(v - 1) þ (v - 1) (v 2 - 1) 1 Þ dv = dx x ( 4v - v 3) Þ

(v 2 - 1)

1

ò v( 2 - v)( 2 + v) dv = ò x dx.

… (ii)

(v 2 - 1) A B C = + + × v( 2 - v)( 2 + v) v ( 2 - v) ( 2 + v) … (iii) Then, (v 2 - 1) º A( 2 - v)( 2 + v) + Bv( 2 + v) + Cv( 2 - v). -1 Putting v = 0 on each side of (iii), we get A = × 4 3 Putting v = 2 on each side of (iii), we get B = × 8 -3 Putting v = -2 on each side of (iii), we get C = × 8 2 (v - 1) -1 3 3 … (iv) = + × \ v( 2 - v)( 2 + v) 4v 8( 2 - v) 8( 2 + v) Putting these values from (iv) in (ii), we get 1 dv 3 dv 3 dv 1 - ò + ò = dx 4 v 8 ( 2 - v) 8 ò ( 2 + v) ò x dv 1 1 dv 3 - dv 3 + ò + = log |C1|, Þ ò dx + ò x 4 v 8 ( 2 - v) 8 ò ( 2 + v) where C1 is an arbitrary constant 1 3 3 Þ log |x| + log |v|+ log |2 - v| + log |2 + v| = log |C1| 4 8 8 Þ 8 log |x| + 2 log |v| + 3 log |2 - v| + 3 log |2 + v| = 8 log |C1| Þ log |x 8v 2( 2 - v) 3( 2 + v) 3| = log {(C1) 8} Let

Þ |x 8v 2( 2 - v) 3( 2 + v) 3| = C18 = C (say) Þ

x8

3

3

y2 æ yö æ yö ç2 - ÷ ç2 + ÷ = C xø è xø x2 è

Senior Secondary School Mathematics for Class 12 Pg-950

950

EXAMPLE 10

SOLUTION

Senior Secondary School Mathematics for Class 12

Þ

y 2( 2x - y) 3( 2x + y) 3 = C , where C is an arbitrary constant.

Þ

y 2( 4x 2 - y 2) 3 = C , which is the required solution.

Show that the differential equation ( x 3 - 3 xy 2) dx = ( y 3 - 3 x 2 y) dy is homogeneous and solve it. The given differential equation may be written as dy x 3 - 3 xy 2 = × dx y 3 - 3 x 2 y

… (i)

On dividing the Nr and Dr of RHS of (i) by x 3 , we get 2

æyö 1 - 3ç ÷ dy æyö èxø = fç ÷× = dx æ y ö 3 èxø æyö ç ÷ - 3ç ÷ èxø èxø Thus, the given differential equation is homogeneous. dy dv in (i), we get Putting y = vx and =v+ x dx dx dv x 3 - 3v 2 x 3 v+x = 3 3 dx v x - 3vx 3 dv 1 - 3v 2 Þ v+x = dx v 3 - 3v ö dv æç 1 - 3v 2 Þ x = - v÷ ÷ dx çè v 3 - 3v ø 4 dv (1 - v ) Þ x = dx (v 3 - 3v) Þ Þ Let

( 3v - v 3) 4

(v - 1)

ò

dv =

( 3v - v 3) 4

(v - 1)

( 3v - v 3) 4

(v - 1)

=

dx x

dv = ò

dx × x

… (ii)

A B Cv + D + + × Then, (v - 1) (v + 1) (v 2 + 1)

( 3v - v 3) º A(v + 1)(v 2 + 1) + B(v - 1)(v 2 + 1) + (Cv + D)(v - 1)(v + 1). … (iii) 1 Putting v = 1 on each side of (iii), we get A = × 2 1 Putting v = -1 on each side of (iii), we get B = × 2 Comparing coefficients of v 3 on both sides of (iii), we get 1 1 A + B + C = -1 Þ + + C = -1 Þ C = -2. 2 2

Senior Secondary School Mathematics for Class 12 Pg-951

Homogeneous Differential Equations

951

Comparing the independent terms on both sides of (iii), we get æ1 1ö A - B - D = 0 Þ D = ( A - B) = ç - ÷ = 0. è 2 2ø ( 3 v - v 3) 1 1 2v = + - 2 × \ 4 ( v ) ( v + ) 2 1 2 1 (v - 1) (v + 1) Putting this value in (ii), we get 1 dv 1 dv 2v dx + dv = ò 2 ò (v - 1) 2 ò (v + 1) ò (v 2 + 1) x 1 1 Þ log |v - 1| + log |v + 1| - log |v 2 + 1| = log |x| + log |C|, 2 2 where C is an arbitrary constant Þ

log |v - 1| + log |v + 1| - 2 log |v 2 + 1| = 2 log |x| + 2 log |C|

Þ

log

(v - 1)(v + 1) 2

(v + 1)

2

= log |C 2 x 2| Þ log

(v 2 - 1) (v 2 + 1) 2

= log |C 2 x 2|

yù é êëQ v = x úû × Hence, ( y 2 - x 2) = C 2( y 2 + x 2) 2 is the required solution.

Þ ( y 2 - x 2) = C 2( y 2 + x 2) 2

EXAMPLE 11

SOLUTION

Show that the differential equation ( x - y)

dy = ( x + 2y) is homogeneous dx

and solve it. The given differential equation can be expressed as dy x + 2y = × dx x-y

[CBSE 2013C]

… (i)

On dividing the Nr and Dr of (i) by x, we get yü ì í1 + 2 × ý dy î æyö xþ = fç ÷× = y ì ü dx èxø í1 - ý î xþ So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv 1 + 2v v+x = dx 1 -v 2 dv ì1 + 2v v2 + v + 1 ü (1 + 2v) - v + v = =í - vý = dx î 1 - v (1 - v) (1 - v) þ (v - 1) 1 (v - 1) 1 dv = - dx Þ ò 2 dv = - ò dx. Þ 2 x x (v + v + 1) (v + v + 1) d 2 Let (v - 1) = A × (v + v + 1) + B. dv

Þ

x

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-952

952

Senior Secondary School Mathematics for Class 12

Then, (v - 1) = A( 2v + 1) + B. Comparing coefficients of like powers of v, we get 1 -3 ö æ ( 2A = 1 and A + B = -1) Þ ç A = and B = ÷× 2 2ø è 1 3 Putting (v - 1) = ( 2v + 1) - in (ii), we get 2 2 1 3 ( 2v + 1) ò 2(v 2 + v + 1)2 dv = - log |x| + C Þ Þ

Þ

EXAMPLE 12

1 ( 2v + 1) 3 dv dv - ò 2 = - log |x| + C 2 ò (v 2 + v + 1) 2 (v + v + 1) 1 3 1 log |v 2 + v + 1| - ò dv = - log |x|+ C 2 2 2 2 æ æ 3ö 1ö ÷ çv + ÷ + çç ÷ 2ø è è 2 ø 3 2 1 æ 2v + 1 ö log |v 2 + v + 1| - × tan -1 ç ÷ = - log |x| + C 2 2 3 è 3 ø

Þ

1 1 log |v 2 + v + 1| + log x 2 = 2 2

æ 2v + 1 ö 3 tan -1 ç ÷ +C è 3 ø

Þ

y2 y 1 1 log 2 + + 1 + log x 2 = x 2 2 x

æ 2y + x ö 3 tan -1 ç ÷ +C è 3x ø

Þ

ì|( y 2 + xy + x 2)| 2 ü 1 log í ×x ý= 2 x2 î þ

æ 2y + x ö 3 tan -1 ç ÷ +C è 3x ø

Þ

æ x + 2y ö log |x 2 + xy + y 2| = 2 3 tan -1 ç ÷ + C, è 3x ø which is the required solution.

Show that the differential equation x

dy y = y - x tan is homogeneous dx x

and solve it. SOLUTION

… (iii)

[CBSE 2006C, ’09]

The given differential equation may be written as dy y y = - tan × dx x x dy æyö This is of the form, = fç ÷× dx èxø So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv v+x = v - tan v dx

… (i)

Senior Secondary School Mathematics for Class 12 Pg-953

Homogeneous Differential Equations

Þ Þ

953

dv = - tan v dx dv dx =tan v x

x

dx x

Þ

ò cot v dv = -ò

Þ

ò sin v dv + ò x dx = log |C1|, where C1 is an arbitrary constant

Þ Þ Þ

log |sin v| + log |x| = log |C1| log |x sin v| = log |C1| x sin v = ± C1 = C (say) y x sin = C , which is the required solution. x

Þ

cos v

1

EXAMPLE 13

Show that the differential equation yö yö æ æ ç x cos ÷( y dx + x dy) = ç y sin ÷( x dy - y dx) xø xø è è is homogeneous and solve it. [CBSE 2010, ’13C]

SOLUTION

The given differential equation may be written as yö y yö y æ æ 2 2 ç xy cos + y sin ÷ dx = ç xy sin - x cos ÷ dy xø x xø x è è yü ý xþ yü ý xþ

Þ

ì íxy cos dy î = dx ì íxy sin î

Þ

dy ( y/x) cos ( y/x) + ( y/x) 2 sin ( y/x) æyö = fç ÷ = dx ( y/x) sin ( y/x) - cos ( y/x) èxø

y + y 2 sin x y - x 2 cos x

… (i)

[dividing Nr and Dr by x 2] So, the given differential equation is homogeneous. dy dv in (i), we get Putting y = vx and =v+ x dx dx v+x Þ Þ Þ

dv (v cos v + v 2 sin v) = dx (v sin v - cos v)

ü dv ì(v cos v + v 2 sin v) =í - vý dx î (v sin v - cos v) þ dv 2v cos v x = dx (v sin v - cos v) x

ò

(v sin v - cos v) 2 dv = ò dx v cos v x

Senior Secondary School Mathematics for Class 12 Pg-954

954

Senior Secondary School Mathematics for Class 12

Þ Þ Þ Þ Þ Þ

dv 2 = ò dx v x - log |cos v| - log |v| - 2 log |x| = constant log |cos v| + log |v| + 2 log |x| = log |C1|, where C1 is an arbitrary constant log |x 2v cos v| = log |C1|

ò tan v dv - ò

x 2v cos v = ± C1 = C (say) y xy cos = C , which is the required solution x

yù é êëQ v = x úû ×

EXAMPLE 14

Show that the differential equation ì æ y öü æyö xy log ç ÷ dx + íy 2 - x 2 log ç ÷ ý dy = 0 is homogeneous and solve it. èxø è x øþ î

SOLUTION

The given differential equation may be written as æyö xy log ç ÷ dy èxø = × dx ì 2 æyö 2ü íx log ç ÷ - y ý èxø î þ On dividing the Nr and Dr of RHS of (i) by x 2 , we get æyö æyö ç ÷ log ç ÷ dy æyö xø è èxø = fç ÷× = 2ü dx ìï èxø æyö æyö ï ílog ç ÷ - ç ÷ ý è x ø è x ø ïþ ïî So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv vx 2 log v v log v v+x = 2 = dx x (log v - v 2) (log v - v 2)

[CBSE 2010C]

Þ Þ Þ Þ

Þ

… (i)

ü dv ì v log v v3 =í - vý = 2 2 dx î(log v - v ) þ (log v - v ) (log v - v 2) 1 dv = dx x v3 1 ì log v 1 ü ò íî v 3 - v ýþ dv = ò x dx [integrating both sides] 1 1 -3 ò (logI v) vII dv - ò v dv = ò x dx + C1 , where C1 is an arbitrary constant æ v -2 ö æ v -2 ö 1 ÷÷ dv - log |v| = log |x| + C1 (log v) × ç ×ç ç -2 ÷ ò v ç -2 ÷ è ø è ø x

[integrating by parts]

Senior Secondary School Mathematics for Class 12 Pg-955

Homogeneous Differential Equations

Þ Þ Þ

Þ

Þ Þ

- log v

955

1 -3 v dv - log |v| = log |x| + C1 2ò 2v - log v 1 - 2 - log |v| = log |x| + C1 2v 2 4v log v 1 + 2 + log |v| + log |x| = - C1 = C (say) 2 2v 4v y log y x2 x + 2 + log |y| = C [Q v = ] æ y 2 ö 4y x 2ç 2 ÷ çx ÷ è ø y 2x 2 log + x 2 + 4y 2 log |y| = 4Cy 2 x y ö æ x 2 ç 2 log + 1÷ + 4y 2 log |y| = 4Cy 2 , x ø è 2

+

which is the required solution. EXAMPLE 15

SOLUTION

æyö æyö ç ÷ + x - y sin ç ÷ = 0 is èxø èxø homogeneous. Find the particular solution of this differential equation, p [CBSE 2013] given that x = 1 when y = × 2 The given differential equation may be written as dy ( y/x) sin ( y/x) - 1 æyö … (i) = = fç ÷× sin ( y/x) dx èxø So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv v sin v - 1 v+x = dx sin v Show that the differential equation x

Þ Þ Þ Þ

dy sin dx

-1 dv ìv sin v - 1 ü =í - vý = sin dx î sin v v þ -1 sin v dv = dx x 1 ò sin v dv = -ò x dx [on integrating both sides] - cos v = - log |x| + C , where C is an arbitrary constant x

æyö log |x| - cos ç ÷ = C èxø p Putting x = 1 and y = in (ii), we get C = 0. 2

Þ

… (ii)

yù é êëQ v = x úû ×

Senior Secondary School Mathematics for Class 12 Pg-956

956

Senior Secondary School Mathematics for Class 12

æyö Hence, log |x| = cos ç ÷ is the required solution. èxø EXAMPLE 16

Find the particular solution of the differential ( 3 xy + y 2) dx + ( x 2 + xy) dy = 0 for x = 1 and y = 1.

SOLUTION

We may write the given differential equation as

equation

[CBSE 2004C, ’07, ’13C]

dy - ( 3 xy + y 2) = × dx ( x 2 + xy)

… (i)

On dividing the Nr and Dr of RHS of (i) by x 2 , we get 2 ïì æ y ö æ y ö ïü - í3ç ÷ + ç ÷ ý ï è x ø è x ø þï dy æyö = î = fç ÷× ì yü xø dx è í1 + ý î xþ So, the given differential equation is homogeneous. dy dv Putting y = vx and in (i), we get =v+ x dx dx dv - ( 3vx 2 + v 2 x 2) - ( 3v + v 2) v+x = = dx (v + 1) ( x 2 + vx 2) 2 2 ì ü -2(v + 2v) - ( 3v + v ) dv - vý = =í Þ x (v + 1) dx î (v + 1) þ (v + 1) -2 Þ dv = dx x (v 2 + 2v) 1 2(v + 1) 2 Þ dv + ò dx = log |C1| 2 ò (v 2 + 2v) x 1 Þ log |v 2 + 2v| + 2 log |x| = log |C1| 2 Þ log |x 2 v 2 + 2v| = log |C1| yù é Þ log |x y 2 + 2xy| = log |C1| êQ v = ú × xû ë Þ

x y 2 + 2xy = ± C1

Þ x 2( y 2 + 2xy) = C , where C = C12. Putting x = 1 and y = 1 in (ii), we get C = 3. \ x 2( y 2 + 2xy) = 3 is the required solution. EXAMPLE 17

SOLUTION

… (ii)

Find the particular solution of the differential equation dy y x - y + x cosec = 0, given that y = 0 when x = 1. [CBSE 2009, ’14C] dx x The given differential equation may be written as dy y y … (i) = - cosec × dx x x

Senior Secondary School Mathematics for Class 12 Pg-957

Homogeneous Differential Equations

957

dy æyö = f ç ÷ × So, it is homogeneous. dx èxø dy dv Putting y = vx and in (i), we get =v+ x dx dx dv v+x = v - cosec v dx dv Þ x = -cosec v dx 1 Þ - sin v dv = dx x 1 Þ ò ( - sin v) dv = ò dx [on integrating both sides] x This is of the form

Þ Þ

cos v = log |x| + C , where C is an arbitrary constant y yù é … (ii) êQ v = ú × cos = log |x| + C xû x ë

Putting x = 1 and y = 0 in (ii), we get C = 1. y Hence, cos = 1 + log |x|is the required solution. x EXAMPLE 18

Find the particular solution of the differential equation p é ù 2æ y ö ê x sin ç x ÷ - y ú dx + x dy = 0, given that y = 4 when x = 1. è ø ë û

SOLUTION

The given differential equation may be written as dy y æyö = - sin 2 ç ÷ × dx x èxø dy æyö This is of the form = f ç ÷ × So, it is homogeneous. dx èxø dy dv Putting y = vx and in (i), we get =v+ x dx dx dv v+x = v - sin 2 v dx dv Þ x = - sin 2 v dx 1 Þ - cosec2 v dv = dx x 1 2 Þ ò ( - cosec v) dv = ò dx [on integrating both sides] x Þ Þ

… (i)

cot v = log |x| + C , where C is an arbitrary constant y yù é … (ii) êQ v = ú × cot = log |x| + C xû x ë

Senior Secondary School Mathematics for Class 12 Pg-958

958

Senior Secondary School Mathematics for Class 12

Putting x = 1 and y = \

cot

p in (ii), we get C = 1. 4

y = log |x| + 1 is the desired solution. x

EXAMPLE 19

Find the equation of the family of curves for which the slope of tangent at x2 + y2 any point ( x , y) on it, is × 2xy

SOLUTION

We know that the slope of the tangent at any point ( x , y) of the dy curve is × dx \

dy x 2 + y 2 = × 2xy dx

… (i)

On dividing the Nr and Dr of RHS of (i) by x 2 , we get ìï æ y ö 2 üï í1 + ç ÷ ý dy ïî è x ø ïþ æyö = = fç ÷× y dx æ ö èxø 2ç ÷ èxø So, the given differential equation is homogeneous. dy dv in (i), we get Putting y = vx and =v+ x dx dx dv 1 + v 2 v+x = dx 2v ü (1 - v 2) dv ì(1 + v 2) =í - vý = Þ x 2v dx î 2v þ 2v 1 Þ dv = dx x (1 - v 2) 2v 1 Þ ò 2 dv = - ò dx [on integrating both sides] x (v - 1)

Þ

log |v 2 - 1| = - log |x| + log |C1|, where C1 is an arbitrary constant log |v 2 - 1| + log |x| = log |C1|

Þ

log |(v 2 - 1) x| = log |C1|

Þ

Þ (v 2 - 1) x = ± C1 æ y2 ö Þ ç 2 - 1÷ x = ± C1 çx ÷ è ø 2 2 Þ ( y - x ) = ± C1 x Þ ( x 2 - y 2) = Cx , where ± C1 = C. Hence, ( x 2 - y 2) = Cx is the equation of the required family of curves.

Senior Secondary School Mathematics for Class 12 Pg-959

Homogeneous Differential Equations

959

EXAMPLE 20

Show that the differential equation 2yex /y dx + ( y - 2xex /y) dy = 0 is

SOLUTION

homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1. [CBSE 2013] The given differential equation may be written as dx ( 2xex /y - y) = × dy 2yex /y

… (i)

On dividing the Nr and Dr of RHS of (i) by y, we get ì x x /y ü - 1ý í2 × e dx î y þ = f æç x ö÷ × = çy÷ x / y dy 2e è ø

… (ii)

So, the given differential equation is homogeneous. dx dv Putting x = vy and in (ii), we get =v+ y dy dy v+y Þ Þ Þ Þ

dv ( 2vev - 1) = dy 2ev

ü -1 dv ì( 2vev - 1) =í - vý = v v dy î 2e þ 2e -1 2ev dv = dy y 1 2ò ev dv = - ò dy [on integrating both sides] y y

2ev = - log |y| + C

é xù êQ v = y ú × ë û Putting y = 1 and x = 0 in (iii), we get C = 2. \ 2ex /y + log |y| = 2 is the required solution. Þ

EXAMPLE 21

SOLUTION

2ex /y + log |y| = C

… (iii)

æ xö Show that the differential equation (1 + ex /y) dx + ex /y çç1 - ÷÷ dy = 0 is yø è homogeneous and solve it. The given differential equation may be written as æx ö ex /y çç - 1÷÷ dx èy ø × … (i) = dy (1 + ex /y) æxö dx = f çç ÷÷ × So, it is homogeneous. dy èyø dx dv in (i), we get Putting x = vy and =v+ y dy dy

This is of the form

Senior Secondary School Mathematics for Class 12 Pg-960

960

Senior Secondary School Mathematics for Class 12

v+y Þ Þ

y

dv ev(v - 1) = dy (1 + ev)

ü - (v + ev) dv ì ev(v - 1) =í v ý= v dy î (1 + ev) þ (1 + e )

(1 + ev) (v + ev)

1 dv = - dy y

(1 + ev)

1

Þ

ò (v + ev) dv = -ò y dy

Þ

log |v + ev| = - log |y| + log |C1|,

[on integrating both sides]

where C1 is an arbitrary constant v

Þ

log |v + e | + log |y| = log |C1|

Þ

log |(v + ev) y| = log |C1|

Þ (v + ev) y = ± C1 = C (say) Þ Þ EXAMPLE 22

SOLUTION

æx ö çç + ex / y ÷÷ y = C èy ø x + yex / y = C , which is the required solution.

æyö Show that the differential equation y dx + x log ç ÷ dy - 2x dy = 0 is èxø homogeneous and solve it. The given differential equation may be written as dx æyö y = 2x - x log ç ÷ dy èxø ì 1 ü = 2x - x × log í ý î( x/y) þ é xù = 2x - x × ê log 1 - log ú yû ë x = 2x + x log × y æxö æxö æxö dx x \ = 2çç ÷÷ + çç ÷÷ log = f çç ÷÷ × dy y y y è ø è ø èyø So, the given differential equation is homogeneous. dx dv in (i), we get Putting x = vy and =v+ y dy dy dv v+y = 2v + v(log v) dy dv Þ y = v(1 + log v) dy

… (i)

Senior Secondary School Mathematics for Class 12 Pg-961

Homogeneous Differential Equations

1

1

1

961

1

Þ

ò v(1 + log v) dv = ò y dy

Þ

log || t = log |y| + log |C1| Þ log |1 + log v|- log |y| = log |C1|

Þ

log

\

1 + log

Þ

ò t dt = ò y dy , where (1 + log v) = t

1 + log v 1 + log v = log |C1| Þ = ± C1 = C y y x = Cy is the required solution. y

EXERCISE 20 In each of the following differential equations show that it is homogeneous and solve it. 2. ( x 2 - y 2) dx + 2xy dy = 0

1. x dy = ( x + y) dx

3. x 2 dy + y( x + y) dx = 0 [CBSE 2010] 4. ( x - y) dy - ( x + y) dx = 0 5. ( x + y) dy + ( y - 2x) dx = 0

6. ( x 2 + 3 xy + y 2) dx - x 2 dy = 0 [CBSE 2006]

7. 2xy dx + ( x 2 + 2y 2) dy = 0 8. 10.

[CBSE 2008] 2

dy x - 2y + =0 dx 2x - y

9.

dy x 2 + y 2 = 2xy dx

11.

dy = 2xy + y 2 dx dy 14. y 2 + ( x 2 - xy) =0 dx 12. x 2

2

dy x - y + =0 3 xy dx dy 2xy = dx ( x 2 - y 2)

dy = x 2 + xy + y 2 dx dy 15. x - y = 2 y2 - x2 dx dy 17. ( x - y) = x + 3y dx

13. x 2

16. y 2 dx + ( x 2 + xy + y 2) dy = 0

18. ( x 3 + 3 xy 2) dx + ( y 3 + 3 x 2 y) dy = 0 19. ( x - xy ) dy = y dx dy dy 21. x + y 2 = xy = y(log y - log x + 1) [CBSE 2007] dx dx dy y dy æyö 22. x - y + x sin = 0 [CBSE 2012] 23. x = y - x cos2 ç ÷ dx x dx èxø

20. x 2

æ 24. ç x cos è

y ö dy æ = ç y cos ÷ x ø dx è

yö ÷+x xø

25. Find the particular solution of the differential equation dy 2xy + y 2 - 2x 2 = 0, it being given that y = 2 when x = 1. dx

[CBSE 2012C]

Senior Secondary School Mathematics for Class 12 Pg-962

962

Senior Secondary School Mathematics for Class 12

26. Find the particular solution of the differential p ü ì 2y - y ý dx + x dy = 0, it being given that y = when x = 1. íx sin x 4 þ î

equation

[CBSE 2011, ’14C]

27. Find the particular solution of the differential equation

dy y( 2y - x) = , dx x( 2y + x)

given that y = 1 when x = 1. 28. Find the particular solution of the differential equation ( xey/x + y) dx = x dy , given that y(1) = 0. dy 29. Find the particular solution of the differential equation xey/x - y + x = 0, dx given that y( e) = 0. 30. The slope of the tangent to a curve at any point ( x , y) on it is given by y æ y öæ yö - ç cot ÷ ç cos ÷ , where x > 0 and y > 0. If the curve passes through the x è x øè xø æ pö point ç1, ÷ , find the equation of the curve. è 4ø ANSWERS (EXERCISE 20)

1. y = x log |x| + Cx

2. x 2 + y 2 = Cx

3. x 2 y = C 2( y + 2x)

y 1 5. y 2 + 2xy - 2x 2 = C = log ( x 2 + y 2) + C x 2 x 7. 3 x 2 y + 2y 3 = C 8. ( y - x) = C( y + x) 3 6. log |x| + =C ( y + x) 4. tan -1

9. ( x 2 + 2y 2) 3 = Cx 2 12. y = Cx( x + y)

13. tan -1

15. y + y 2 - x 2 = C|x|3 17. log |x + y| + 19. 2

2x =C ( x + y)

x + log |y| = C y

æ y ö 22. x tan ç ÷ = C è 2x ø æyö 24. sin ç ÷ = log |x| + C èxø 27.

y 1 + log |xy| = 1 x 2

10. ( x 2 - y 2) = Cx y = log |x| + C x 16. log

11. y = C( y 2 + x 2)

14. y = x{C + log |y|}

x y + log |x| + =C y+x ( y + x)

18. y 4 + 6x 2 y 2 + x 4 = C 20. Cx = ex /y

21. y = xeCx

23. tan

y + log |x| = C x

25. y =

2x (1 - log |x|)

28. log |x| + e- y/x = 1

26. cot

y = log |x| + 1 x

29. y = - x log (log |x|)

Senior Secondary School Mathematics for Class 12 Pg-963

Homogeneous Differential Equations

30. sec

963

y + log |x| = 2 x

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 20)

2.

3.

dy y 2 - x 2 = , which is homogeneous. dx 2 xy 2v -1 ò ( 1 + v 2 ) dv = ò x dx Þ log|1 + v 2| + log|x| = log|C|. dv

1 ì1 1 ü 1 í ý dv = - ò dx [by partial fractions] 2 ò î v (v + 2 ) þ x

Þ 4.

æ 1-v ö

1

dv

1

2v

1

ò çè 1 + v 2 ÷ø dv = ò x dx Þ ò ( 1 + v 2 ) - 2 ò ( 1 + v 2 ) dv = ò x dx \ tan -1 v -

5.

1

ò v (v + 2 ) = - ò x dx

y 1 1 log ( 1 + v 2 ) = log|x| + C Þ tan -1 = log ( x 2 + y 2 ) + C. 2 x 2

(v + 1)

ò (v 2 + 2v - 2 ) dv = - ò

( 2v + 2 ) dx dx 1 Þ ò 2 dv = - ò x x 2 (v + 2v - 2 )

1 log|v 2 + 2v - 2| + log|x| = log|C 1|. 2

\ 6. x

dv = ( 1 + 2v + v 2 ) = (v + 1) 2 Þ dx

7. x

dv - ( 3v + 2v 3 ) = Þ dx ( 1 + 2v 2 )

dv

1

ò (v + 1)2 = ò x dx Þ

( 1 + 2v 2 )

ò ( 3v + 2v 3 ) dv + ò

log|x| -

1 = C. (v + 1)

dx = log|C 1|. x

1 ( 3 + 6v 2 ) dv + log|x| = log|C 1| 3 ò ( 3v + 2v 3 ) 1 Þ log|3v + 2v 3| + log|x| = log|C 1| 3 Þ log|3v + 2v 3| + 3 log|x| = 3 log|C 1|

\

Þ log|3v + 2v 3| + log|x 3| = log|C 13|. 8.

( 2 - v)

1

ò (v 2 - 1) dv = ò x dx. ì 1 3 ü =í ý [by partial fractions]. (v - 1) î 2(v - 1) 2(v + 1) þ ( 2 - v) 2

3v 1 dv - ( 1 + 2v 2 ) = Þ ò dv = - ò dx 3v dx x ( 1 + 2v 2 ) 3 4v 1 3 Þ ò dv = - ò dx Þ log|1 + 2v 2| + log|x| = log|C 1|. 4 ( 1 + 2v 2 ) x 4

9. x

Senior Secondary School Mathematics for Class 12 Pg-964

964

Senior Secondary School Mathematics for Class 12

10. x

dv ( 1 - v 2 ) = Þ 2v dx

-2v

1

ò ( 1 - v 2 ) dv = - ò x dx

Þ log|1 - v 2| + log|x| = log|C 1|. 11. x

dv (v + v 3 ) = Þ dx ( 1 - v 2 ) ìï 1

( 1 - v2 )

üï

2v

1

ò íïv - ( 1 + v 2 ) ýï dv = ò x dx

Þ

þ

î

1

1

ò v ( 1 + v 2 ) dv = ò x dx [by partial fractions].

ì1

1

ü

1

1

12.

ò v( 1 + v ) dv = ò x dx Þ ò ív - ( 1 + v ) ý dv = ò x dx

13.

ò ( 1 + v 2 ) dv = ò x dx Þ

14.

ò v (v + 1) dv = ò x dx Þ ò íîv + 1 - v ýþ dv = ò x dx.

15.

ò

16.

ò v(v + 1)2 dv = - ò x dx.

î

1

1

(v - 1)

1

1 2

v -1

dv = ò

[by partial fractions].

tan -1 v = log|x| + C . ì 2

1ü

1

2 dx Þ log|v + v 2 - 1| = 2 log|x| + log|C 1|. x

1

Let

þ

1

1 v(v + 1)

2

=

A B C + + × v (v + 1) (v + 1) 2

This gives A = 1, B = -1 and C = -1. 17. x

dv ( 1 + v ) 2 = Þ dx ( 1 - v )

dx ïì (v + 1) - 2 ïü Þ dv = - ò 2 ý x ï ( v + 1 ) þ î

ò íï

Þ

18.

(v - 1)

ò (v + 1)2 dv = - ò

(v 3 + 3 v )

dx x 1

1

ò (v + 1) dv - 2 ò (v + 1)2 dv + ò

1

ò (v 4 + 6v 2 + 1) dv = - ò x dx 1 1 ( 4 v 3 + 12 v ) dv = - ò dx x 4 ò (v 4 + 6 v 2 + 1)

Þ

1 log|v 4 + 6 v 2 + 1| + log|x| = log|C 1| 4 Þ (v 4 + 6 v 2 + 1)x 4 = C 14 = C (say) Þ y 4 + 6 x 2 y 2 + x 4 = C . Þ

19.

ò

1- v

Þ

v

3/2

òv

Þ 2

dv = ò

-3 / 2

1 dx Þ x

ì 1

v ü

1

ò ív 3 / 2 - v 3 / 2 ý dv = ò x dx

î þ -2 1 dv - ò dv = log|x| - C Þ - log|v| = log|x| - C v v

x + log|y| = C . y

dx = C. x

Senior Secondary School Mathematics for Class 12 Pg-965

Homogeneous Differential Equations dv 1 1 1 = -v 2 Þ ò - 2 dv = ò dx Þ = log x + log C . dx x v v x \ log Cx = Þ Cx = e x/ y . y

20. x

21.

dy y ì y ü æyö = í log + 1ý = f ç ÷ × dx x î x þ èxø dy dv Putting y = vx and = v + x , we get dx dx 1 1 = Þ dv dx log (log v ) = log x + log C ò v log v òx \ log v = Cx Þ v = e Cx Þ y = xe Cx.

dy y y æyö = - sin = f ç ÷ × dx x x èxø -1 v cosec v dv = ò ò x dx Þ log tan 2 + log x = log C . 2 dy 2 xy + y 2 ìï æ y ö 1 æ y ö üï æ y ö ç ÷ ç ÷ 25. = = + ý= fç ÷× í dx 2x 2 îï è x ø 2 è x ø þï è x ø 1 1 -2 -2 x 2 ò 2 dv = ò dx Þ = log|x| + C Þ = log|x| + C . x v y v 22.

Putting x = 1 and y = 2, we get C = -1. dy ì y y yü æ y ö 30. = í - cot cos ý = f ç ÷ which is homogeneous. dx î x x xþ è x ø dy dv Putting y = vx and = v + x , we get dx dx dx sec v tan v dv = Þ sec v = - log|x| + C ò òx y \ sec + log|x| = C . x p Putting x = 1 and y = , we get C = 2. 4

965

Senior Secondary School Mathematics for Class 12 Pg-966

21. LINEAR DIFFERENTIAL EQUATIONS Linear Differential Equations dy + Py = Q , where P dx is a constant and Q is a constant or a function of x. The other common form of dx linear differential equations is + Px = Q , where P is a constant and Q is a dy constant or a function of y. The most general form of linear differential equations is

Solution of First, we find eò

P dx

dy + Py = Q dx

, which is known as the integrating factor, i.e., IF.

dy P dx dy P dx P dx Now, + Py = Q Þ eò + Py eò = Q × eò dx dx Þ eò

P dx

Þ y × eò

dy + Py × eò

P dx

P dx

= ò Q × eò

dx = Q × eò

P dx

P dx

dx Þ d æç y × eò è

P dx ö

÷ = Q × eò ø

P dx

dx … (i)

dx + C , which is the required solution [integrating both sides of (i)].

Working Rule for Solving (i) Find IF = eò

P dx

dy + Py = Q dx

.

(ii) The solution is given by y ´ IF = ò {Q ´ IF} dx + C. SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Solve the differential equation dy x - y = 2x 3 , x > 0. [CBSE 2004, ’07, ’11] dx The given differential equation may be written as dy æ -1 ö … (i) + ç ÷ y = 2x 2. dx è x ø 966

Senior Secondary School Mathematics for Class 12 Pg-967

Linear Differential Equations

967

-1 dy and Q = 2x 2. + Py = Q , where P = x dx Thus, the given differential equation is linear. This is of the form,

IF = e ò

1

P dx

=e

ò - x dx

=e

-log x

=e

æ1 ö log ç ÷ èx ø

=

1 × x

So, the required solution is given by y ´ IF = ò {Q ´ IF} dx + C , where C is an arbitrary constant, i.e., y ´ Þ

1 1ö æ = ç 2x 2 ´ ÷ dx + C = 2ò x dx + C = x 2 + C x òè xø

y = x 3 + Cx.

Hence, the required solution is y = x 3 + Cx. EXAMPLE 2

SOLUTION

Solve the differential equation dy 2 ( x 2 - 1) + 2xy = 2 × dx ( x - 1) The given differential equation may be written as 2 dy ì 2x ü × +í ýy = 2 dx î( x 2 - 1) þ ( x - 1) 2 dy This is of the form + Py = Q , dx 2x 2 where P = 2 and Q = 2 × ( x - 1) ( x - 1) 2 Thus, the given differential equation is linear. ò P dx

ò

2x ( x 2 - 1)

dx

EXAMPLE 3

… (i)

log ( x 2 - 1 )

=e =e = ( x 2 - 1). So, the required solution is given by y ´ IF = ò {Q ´ IF} dx + C IF = e

[CBSE 2010, ’14]

ì ü 2 i.e., y ´ ( x 2 - 1) = ò í 2 ´ ( x 2 - 1) ý dx + C 2 î( x - 1) þ dx = 2ò 2 +C ( x - 1) 1ì 1 1 ü = 2ò í ý dx + C [by partial fraction] 2 î( x - 1) ( x + 1) þ x -1 = log + C. x+1 x -1 Hence, y( x 2 - 1) = log + C is the required solution. x+1 -1 dy Solve (1 + x 2) [CBSE 2014] + y = etan x . dx

Senior Secondary School Mathematics for Class 12 Pg-968

968 SOLUTION

Senior Secondary School Mathematics for Class 12

The given differential equation may be written as -1

dy 1 etan x + × = × y dx (1 + x 2) (1 + x 2)

-1

1 etan x dy and Q = × + Py = Q , where P = 2 dx (1 + x ) (1 + x 2) Thus, the given equation is linear. This is of the form 1

dx ò -1 P dx (1+x 2 ) IF = e ò =e = etan x . So, the required solution is given by

y ´ IF = ò|(Q ´ ( IF)| dx + C , i.e., y ´ etan

-1

x

ì etan -1 x -1 ü ï ï = òí ´ etan x ý dx + C 2 + x ( 1 ) þï îï

=ò

e2 tan

-1

x

2

(1 + x )

dx + C

= ò e2t dt + C , where tan -1 x = t =

-1 1 2t 1 e + C = e2 tan x + C. 2 2

-1 1 tan -1 x e + Ce- tan x is the required solution. 2 dy 2 [CBSE 2009, ’10, ’14] ( x log x) + y = log x. dx x

Hence, y = EXAMPLE 4

Solve

SOLUTION

The given differential equation may be written as dy 1 2 + ×y = 2 × dx ( x log x) x 1 2 dy This is of the form and Q = 2 × + Py = Q , where P = ( x log x) dx x Thus, the given differential equation is linear. 1

IF = e ò

P dx

=e

ò x log x

dx

1

=e

òt

dt

, where log x = t

log t

=e = t = log x. So, the solution of the given differential equation is y ´ IF = ò {Q ´ ( IF)} dx + C ,

æ 2 i.e., y(log x) = ò ç 2 log èx

ö x ÷ dx + C ø 1 = 2ò (log x) × 2 dx + C x 1 æ 1ö ù é æ 1ö = 2 ê(log x) ç - ÷ - ò × ç - ÷ dx ú + C x è xø û è xø ë [integrating by parts]

Senior Secondary School Mathematics for Class 12 Pg-969

Linear Differential Equations

969

-2 log x 2 - + C. x x -2 Hence, y(log x) = (log x + 1) + C is the required solution. x dy Solve ( x 2 + 1) [CBSE 2008, ’10] + 2xy = x 2 + 4. dx =

EXAMPLE 5 SOLUTION

The given differential equation may be written as dy æ 2x ö x2 + 4 +ç 2 × ÷y = 2 dx è x + 1 ø ( x + 1) dy This is of the form + Py = Q , dx

… (i)

where P =

2x 2

( x + 1)

and Q =

x2 + 4 ( x 2 + 1)

×

Thus, the given differential equation is linear. P dx IF = e ò =e

ò

2x ( x 2 +1 )

dx

=e

log ( x 2 +1 )

= ( x 2 + 1).

So, the required solution is given by y ´ IF = ò {Q ´ IF} dx + C, i.e., y( x 2 + 1) = ò Þ

x2 + 4 ( x 2 + 1)

´ ( x 2 + 1) dx

y( x 2 + 1) = ò x 2 + 4 dx

1 1 x x 2 + 4 + ´ 22 ´ log |x + x 2 + 4| + C 2 2 1 2 = x x + 4 + 2 log |x + x 2 + 4| + C. 2 1 2 Hence, y( x + 1) = x x 2 + 4 + 2 log |x + x 2 + 4| + C 2 required solution. =

dy + y = tan -1 x. dx

EXAMPLE 6

Solve (1 + x 2)

SOLUTION

The given differential equation may be written as dy 1 tan -1 x + ×y = × 2 dx (1 + x ) (1 + x 2)

is

the

[CBSE 2008C, ’09]

… (i)

1 dy tan -1 x and Q = × + Py = Q , where P = 2 dx (1 + x 2) (1 + x ) Thus, the given equation is linear. This is of the form

IF = e ò

P dx

=e

ò

1 1+ x 2

dx

= etan

-1

x

.

Senior Secondary School Mathematics for Class 12 Pg-970

970

Senior Secondary School Mathematics for Class 12

\

the required solution is y ´ IF = ò {Q ´ IF} dx + C ,

i.e., y ´ etan

-1

x

ì tan -1 x tan -1 x ü = òí ×e ý dx + C 2 î(1 + x ) þ

= ò (t et ) dt + C , where tan -1 x = t

= t et - ò 1 × et dt + C [integrating by parts] = t et - et + C = et (t - 1) + C = etan Þ

-1

x

(tan -1 x - 1) + C

y = (tan -1 x - 1) + Ce- tan

-1

x

Hence, y = (tan -1 x - 1) + Ce- tan EXAMPLE 7

SOLUTION

-1

x

is the required solution.

Find the general solution of the differential equation dy [CBSE 2012C] - y = cos x. dx The given differential equation is dy … (i) - y = cos x. dx dy This is of the form + Py = Q , where P = -1 and Q = cos x. dx Thus, the given differential equation is linear. P dx - dx IF = e ò = eò = e- x . So, the required solution is y ´ IF = ò {Q ´ IF} dx + C ,

i.e., y ´ e- x = ò (cos x) e- x dx + C.

… (ii)

Let I = ò (cos x) e- x dx I

II -x

= (cos x)( - e

) - ò ( - sin x)( - e- x ) dx [integrating by parts]

= - (cos x) e- x - ò (sin x)( e- x ) dx I

II

= - (cos x) e- x - {(sin x)( - e- x )} - ò (cos x)( - e- x ) dx

= - (cos x) e- x + (sin x)( e- x ) - I. 1 \ 2I = (sin x - cos x) e- x Þ I = (sin x - cos x) e- x . 2 Putting this value in (ii), we get 1 1 y ´ e- x = (sin x - cos x) e- x + C Þ y = (sin x - cos x) + Cex . 2 2 1 x Hence, y = (sin x - cos x) + Ce is the required solution. 2

Senior Secondary School Mathematics for Class 12 Pg-971

Linear Differential Equations EXAMPLE 8

SOLUTION

Solve the differential equation dy x + y - x + xy cot x = 0, x ¹ 0. dx

971

[CBSE 2011, ’12]

The given differential equation may be written as dy x + (1 + x cot x) y = x dx dy æ 1 ö + ç + cot x ÷ y = 1. Þ dx è x ø This is of the form

dy æ1 + Py = Q , where P = ç + cot dx èx

… (i) ö x ÷ and Q = 1. ø

Thus, the given differential equation is linear. æ1

ö

ò ç + cot x ÷ø dx P dx log x +log sin x IF = e ò = e èx =e =e

log ( x sin x )

= x sin x.

So, the solution of the given differential equation is given by y ´ (IF) = ò (Q ´ IF) dx + C,

i.e., y( x sin x) = ò 1 ´ ( x sin x) dx + C = ò x sin x dx + C I

II

= x( - cos x) - ò 1 × ( - cos x) dx + C = - x cos x + ò cos x dx + C \ EXAMPLE 9

SOLUTION

= - x cos x + sin x + C. 1 C is the required solution. y = - cot x + x x sin x

Solve the differential equation ìï e-2 x y üï dx ý = 1 ( x ¹ 0). í x ïþ dy ïî x

[CBSE 2012]

The given differential equation may be written as e -2 x dy 1 =×y + dx x x Þ

e -2 x dy 1 + ×y = × dx x x

This is of the form IF = e ò

P dx

=e

ò

1 x

dx

… (i)

1 e-2 x dy and Q = × + Py = Q , where P = dx x x = e2

x

.

Senior Secondary School Mathematics for Class 12 Pg-972

972

Senior Secondary School Mathematics for Class 12

So, the solution of the given differential equation is given by y ´ IF = ò (Q ´ IF) dx + C , i.e., y ´ e2

x

e -2 x ´ e2 x dx + C x 1 =ò dx + C = 2 x + C. x =ò

Hence, ye2 x = 2 x + C is the required solution. 1 C is the required solution. \ y = - cot x + x x sin x dy + 2y = x cos x. dx

EXAMPLE 10

Solve x

SOLUTION

The given differential equation may be written as dy 2 + × y = cos x. dx x 2 dy This is of the form + Py = Q , where P = and Q = cos x. x dx Thus, the given equation is linear.

… (i)

2

ò dx P dx 2 log x log ( x 2 ) IF = e ò =e x =e =e = x 2. So, the required solution is y ´ IF = ò {Q ´ ( IF)} dx + C , i.e., yx 2 = ò x 2 cos x dx + C

= x 2 sin x - ò 2x sin x dx + C [integrating by parts] = x 2 sin x - 2 × [x( - cos x) - ò 1 × ( - cos x) dx] + C

[integrating by parts] = x 2 sin x + 2x cos x - 2 sin x + C. 2 2 C Hence, y = sin x + cos x - 2 sin x + 2 is the required solution. x x x dy + (sec x) y = tan x. dx

EXAMPLE 11

Solve

SOLUTION

The given equation is of the form dy + Py = Q , where P = sec x and Q = tan x. dx Thus, the given equation is linear. P dx sec x dx log (sec x + tan x ) IF = e ò = eò =e = (sec x + tan x). So, the required solution is y ´ IF = ò {Q ´ ( IF)} dx + C ,

[CBSE 2006, ’08C, ’12]

Senior Secondary School Mathematics for Class 12 Pg-973

Linear Differential Equations

973

i.e., y(sec x + tan x) = ò tan x(sec x + tan x) dx + C = ò sec x tan x dx + ò tan 2 x dx + C = sec x + ò (sec2 x - 1) dx + C = sec x + tan x - x + C. Hence, y(sec x + tan x) = sec x + tan x - x + C solution. EXAMPLE 12

SOLUTION

is

the

required

dy - 2y = cos 3 x. dx dy The given differential equation is of the form + Py = Q , where dx P = -2 and Q = cos 3 x. Thus, the given differential equation is linear. P dx -2 dx IF = e ò = eò = e -2 x . Find the general solution of the differential equation

So, the required solution is y ´ IF = ò {Q ´ IF } dx + C , i.e., y ´ e-2x = ò e-2x cos 3 x dx + C é -2 cos 3 x + 3 sin 3 x ù = e -2 x ê ú +C {( -2) 2 + 3 2} ë û ù é ax ax ì a cos bx + b sin bx ü êQ ò e cos bx dx = e í ýú × 2 2 (a + b ) êë î þ úû ( 3 sin 3 x - 2 cos 3 x) 2x \ y= + Ce , which is the required solution. 13 EXAMPLE 13

SOLUTION

Solve the differential equation pö dy æ (cos2 x) + y = tan x ç 0 £ x < ÷ × dx 2ø è

[CBSE 2008, ’11]

The given differential equation may be written as dy … (i) + (sec2 x) y = (sec2 x) tan x. dx dy This is of the form + Py = Q , where P = sec2 x and dx Q = sec2 x tan x. Thus, the given differential equation is linear. P dx sec2 x dx IF = e ò = eò = etan x .

So, its solution is given by y ´ IF = ò (Q ´ IF) dx + C ,

Senior Secondary School Mathematics for Class 12 Pg-974

974

Senior Secondary School Mathematics for Class 12

i.e., y ´ etan x = ò etan x (sec2 x) tan x dx + C = ò t et dt , where tan x = t and sec2 x dx = dt I II t

= te - ò 1 × et dt + C

= tet - et + C = et (t - 1) + C

= etan x (tan x - 1) + C. Hence, y = (tan x - 1) + C e- tan x is the required solution. EXAMPLE 14

SOLUTION

Solve the differential equation dy + y tan x = 2x + x 2 tan x. dx The given differential equation is of the form dy + Py = Q , where P = tan x and Q = 2x + x 2 tan x. dx Thus, the given differential equation is linear. P dx tan x dx log sec x IF = e ò = eò =e = sec x. So, its solution is given by y ´ IF = ò (Q ´ IF) dx + C , i.e., y ´ sec x = ò ( 2x + x 2 tan x) sec x dx + C = ò 2x sec x dx + ò x 2 sec x tan x dx + C I

II

= ò 2x sec x dx + {x 2 sec x - ò 2x sec x dx} + C

= x 2 sec x + C. \ EXAMPLE 15

y = x 2 + C cos x is the required solution.

Solve the differential equation x

dy æ pö + y = x cos x + sin x , given y ç ÷ = 1. dx è 2ø [CBSE 2014C]

SOLUTION

The given differential equation may be written as dy 1 sin x + × y = cos x + × dx x x 1 dy This is of the form + Py = Q , where P = and x dx sin x Q = cos x + × x Thus, the given differential equation is linear. 1

ò dx log x =e x =e = x. So, its solution is given by y ´ IF = ò (Q ´ IF) dx + C , IF = e ò

P dx

… (i)

Senior Secondary School Mathematics for Class 12 Pg-975

Linear Differential Equations

975

sin x ö æ i.e., y ´ x = ò ç cos x + ÷ x dx + C x ø è = ò x cos x dx + ò sin x dx + C I

II

= x sin x - ò sin x dx + ò sin x dx + C \

= x sin x + C. C y = sin x + × x

It is being given that when x =

… (ii) p , then y = 1. 2

p and y = 1 in (ii), we get C = 0. 2 Hence, y = sin x is the required solution. Putting x =

EXAMPLE 16

SOLUTION

Find

the particular solution of the differential equation dy cos x + y = sin x , given that y = 2 when x = 0. dx The given differential equation may be written as dy … (i) + (sec x) y = tan x. dx dy This is of the form + Py = Q , where P = sec x and Q = tan x. dx Thus, the given differential equation is linear. P dx sec x dx log (sec x + tan x ) IF = e ò = eò =e = (sec x + tan x). So, its solution is given by y ´ IF = ò (Q ´ IF) dx + C , i.e., y(sec x + tan x) = ò tan x(sec x + tan x) dx + C = ò sec x tan x dx + ò tan 2 x dx + C = sec x + ò (sec2 x - 1) dx + C = sec x + ò sec2 x dx - ò dx + C = sec x + tan x - x + C. Thus, y(sec x + tan x) = sec x + tan x - x + C. … (ii) It is given that when x = 0, then y = 2. \ putting x = 0 and y = 2 in (ii), we get C = 1. Hence, y(sec x + tan x) = sec x + tan x - x + 1 is the required solution.

EXAMPLE 17

Find the particular solution of the differential dy x - y = ( x + 1) e- x , given that y = 0 when x = 1. dx

equation

Senior Secondary School Mathematics for Class 12 Pg-976

976 SOLUTION

Senior Secondary School Mathematics for Class 12

The given differential equation may be written as dy 1 ( x + 1) e- x … (i) - ×y = × dx x x -1 dy ( x + 1) e- x This is of the form and Q = × + Py = Q , where P = x x dx Thus, the given differential equation is linear. 1

- ò dx 1 P dx - log x log (1 / x ) IF = e ò =e x =e =e = ×

x

So, its solution is given by y ´ IF = ò (Q ´ IF) dx + C ,

1 ( x + 1) e- x =ò dx + C x x2 1 1 1 y ´ = ò e- x dx + ò 2 e- x dx + C x x II x

i.e., y ´ Þ

I

1 1 1 1 Þ y ´ = × ( - e- x ) + ò 2 ( - e- x ) dx + ò 2 e- x dx + C x x x x y -e- x = +C Þ x x Þ y = - e- x + Cx. It is being given that when x = 1, then y = 0. Putting x = 1 and y = 0 in (ii), we get C = e-1. \ EXAMPLE 18

SOLUTION

… (ii)

y = - e- x + xe-1 is the required solution.

Find the particular solution of the differential equation p dy - 3 y cot x = sin 2x , it being given that y = 2 when x = × 2 dx The given differential equation is dy … (i) - 3 y cot x = sin 2x. dx dy This is of the form + Py = Q , where P = -3 cot x and Q = sin 2x. dx Thus, the given equation is linear. 1 P dx -3 cot x dx log (sin x )-3 IF = e ò = eò =e = (sin x) -3 = × sin 3 x So, the solution of (i) is given by y ´ IF = ò {Q ´ ( IF)} dx + C , i.e., y ´

1 sin 3 x

1 ö æ = ò ç sin 2x ´ ÷ dx + C è sin 3 x ø cos x = 2ò dx + C [Q sin 2x = 2 sin x cos x] sin 2 x

Senior Secondary School Mathematics for Class 12 Pg-977

Linear Differential Equations

= 2ò

977

1

dt + C, where sin x = t t2 -2 -2 = +C = + C. t sin x Thus, y = -2 sin 2 x + C sin 3 x. p It is being given that when x = , then y = 2. 2 p Putting x = and y = 2 in (ii), we get 2 p p -2 sin 2 + C sin 3 = 2 Þ C - 2 = 2 Þ C = 4. 2 2 Hence, y = -2 sin 2 x + 4 sin 3 x is the required solution. EXAMPLE 19

Find

the particular solution of the differential dy (1 - x ) - xy = x 2 , given that y = 2 when x = 0. dx The given differential equation may be written as

… (ii)

equation

2

SOLUTION

dy x x2 ×y = × 2 dx (1 - x ) (1 - x 2)

… (i)

-x x2 dy and Q = × + Py = Q , where P = 2 dx (1 - x ) (1 - x 2) Thus, the given differential equation is linear. This is of the form

P dx IF = e ò =e

ò

-x (1 - x 2 )

dx

=e

1 -2 x dx ò 2 (1 - x 2 )

1

= e2

log (1 - x 2 )

log 1 - x 2

=e = 1 - x2 So, its solution is given by y ´ IF = ò (Q ´ IF) dx + C , i.e., y ´ 1 - x 2 = ò =ò

\

x2

(1 - x 2) x2

´ 1 - x 2 dx + C

dx + C 1 - x2 {- (1 - x 2) + 1} =ò dx + C 1 - x2 1 = - ò 1 - x 2 dx + ò dx + C 1 - x2 ìï x 1 - x 2 1 üï = -í + sin -1 x ý + sin -1 x + C 2 2 ïî ïþ 2 -x 1 - x 1 = + sin -1 x + C. 2 2 -x C sin -1 x y= + + × 2 2 1 - x2 1 - x2

… (ii)

Senior Secondary School Mathematics for Class 12 Pg-978

978

Senior Secondary School Mathematics for Class 12

It is being given that when x = 0, then y = 2. Putting x = 0 and y = 2 in (ii), we get C = 2. 2 -x sin -1 x is the required solution. Hence, y = + + 2 2 1 - x2 1 - x2 EXAMPLE 20

SOLUTION

Find the equation of a curve passing through the point (0, 1), it being given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of the point. dy We know that the slope of the tangent to the curve is × dx dy dy … (i) = x + xy Þ - xy = x. \ dx dx dy This is of the form + Py = Q , where P = - x and Q = x. dx So, the given differential equation is linear. -x 2

P dx - x dx IF = e ò = eò =e 2 . Hence, the solution of the given differential equation is given by y ´ IF = ò (Q ´ IF) dx + C ,

i.e., y

-x 2 ´e 2

=ò

-x 2 xe 2 dx

+C

= ò e- t dt + C , where = -e \

y=

-t

+C =

-x 2 -e 2

x2 =t 2

+ C.

x2 -1 + Ce 2 .

…(ii)

We have to find a curve satisfying (ii) and passing through (0, 1). Putting x = 0 and y = 1 in (ii), we get C = 2. Hence, y = -1 + EXAMPLE 21

SOLUTION

x2 2e 2

is the equation of the required curve.

An equation relating to the stability of an aeroplane is given by dv = g cos a - kv , where v is the velocity and g , a , k are constants. dt Find an expression for the velocity if v = 0 at t = 0. The given differential equation is dv … (i) + kv = g cos a . dt dv This is of the form + Pv = Q , where P = k and Q = g cos a . dt

Senior Secondary School Mathematics for Class 12 Pg-979

Linear Differential Equations

979

Thus, the given equation is linear. P dt k dt IF = e ò = eò = ekt . So, the solution of the given differential equation is v ´ IF = ò {Q ´ IF } dt + C , i.e., vekt = ò ( g cos a ) ekt dt + C ( g cos a ) ekt + C. k Now, it is given that v = 0 when t = 0. =

… (ii)

Putting t = 0 and v = 0 in (ii), we get C = \

vekt =

- g cos a × k

( g cos a ) ekt g cos a k k

1 Þ v = ( g cos a )(1 - e- kt ), which is the required expression. k SOLUTION OF

Working Rule for Solving (i) Find IF = eò

P dy

dx + Px = Q dy

dx + Px = Q dy

.

(ii) The solution is given by x ´ IF = ò {Q ´ IF } dy + C. EXAMPLE 22

Find

the general solution of the differential dy ( x + 2y ) = y , y ¹ 0. dx The given differential equation may be written as dx y = x + 2y 3 dy

equation

3

SOLUTION

dx 1 - × x = 2y 2. dy y 1 dx This is of the form + Px = Q , where P = - and Q = 2y 2. y dy

Þ

Thus, the given equation is linear. 1

P dy IF = e ò =e

ò - y dy

=e

- log y

So, the required solution is x ´ IF = ò {Q ´ IF} dy + C ,

=e

log ( y -1 )

= y -1 =

1 × y

… (i)

Senior Secondary School Mathematics for Class 12 Pg-980

980

Senior Secondary School Mathematics for Class 12

i.e., x ´

æ 1 1ö = ç 2y 2 ´ ÷÷ dy + C y ò çè yø = ò 2y dy + C = y 2 + C.

Hence, x = y 3 + Cy is the required solution. EXAMPLE 23

Find the particular solution of the differential (tan -1 y - x) dy = (1 + y 2) dx , given that when x = 0, y = 0.

SOLUTION

The given differential equation may be written as

equation

[CBSE 2009, ’13]

dx (tan -1 y - x) = dy (1 + y 2) Þ

dx 1 tan -1 y + × x = × dy (1 + y 2) (1 + y 2)

… (i)

1 dx tan -1 y and Q = × + Px = Q , where P = 2 dy (1 + y 2) (1 + y ) Thus, the given differential equation is linear. This is of the form

1

dy ò -1 P dy (1 + y 2 ) IF = e ò =e = etan y. So, the solution of the given differential equation is given by x ´ IF = ò {Q ´ IF } dy + C , where C is an arbitrary constant,

i.e., x ´ etan

-1

y

ì tan -1 y -1 ü = òí ´ etan y ý dy + C 2 î(1 + y ) þ

= ò t et dt + C , where tan -1 y = t and I II

1 (1 + y 2)

dy = dt

= tet - ò 1 × et dt + C [integrating by parts]

= tet - et + C = (t - 1) et + C \

xe

tan

-1

= (tan -1 y - 1) etan y

-1

= (tan y - 1) e

tan

-1

-1

y

y

+ C.

+ C.

… (ii)

Putting x = 0 and y = 0 in (ii), we get C = 1. \

xetan

-1

y

= (tan -1 y - 1) etan -1

Hence, x = (tan y - 1) + e EXAMPLE 24

Find

the

particular

-1

- tan

y

-1

+ 1.

y

solution

is the required solution. of

the

differential

equation

( 1 - y 2 ) dx = (sin -1 y - x) dy , it being given that when y = 0, then x = 0. SOLUTION

The given differential equation may be written as dx sin -1 y - x = dy 1 - y2

Senior Secondary School Mathematics for Class 12 Pg-981

Linear Differential Equations

Þ

981

dx 1 sin -1 y + ×x = × dy 1 - y2 1 - y2

This is of the form

… (i)

1 dx sin -1 y and Q = × + Px = Q , where P = dy 1 - y2 1 - y2

Thus, the given equation is linear. 1

dy ò -1 P dy IF = e ò = e 1 - y2 = esin y. So, the solution of the given differential equation is given by x ´ IF = ò (Q ´ IF) dy + C ,

i.e., x ´ esin

-1

y

ì sin -1 y -1 ü ï ï = òí × esin y ý dy + C 2 ïþ ïî 1 - y = ò t et dt + C , where sin -1 y = t and I II

1 1 - y2

dy = dt

= tet - ò 1 × et dt + C = tet - ò et dt + C = (tet - et ) + C = (t - 1) et + C = (sin -1 y - 1) esin \

x = (sin -1 y - 1) + Ce- sin

-1

y

-1

y

+C

[Q t = sin -1 y]. … (ii)

.

Putting y = 0 and x = 0 in (ii), we get C = 1. Hence, x = (sin -1 y - 1) + e- sin EXAMPLE 25

-1

y

is the required solution.

Find the particular solution of the differential equation p dx + x cot y = 2y + y 2 cot y ( y ¹ 0), given that x = 0 when y = × 2 dy [CBSE 2013C]

SOLUTION

The given differential equation is of the form P = cot y and Q = ( 2y + y 2 cot y). So, the given differential equation is linear. P dy cot y dy log sin y IF = e ò = eò =e = sin y. Therefore, the solution is given by x ´ IF = ò (Q ´ IF) dy + C , i.e., x ´ sin y = ò ( 2y + y 2 cot y) sin y dy + C = ò 2y sin y dy + ò y 2 cos y dy + C I

II

dx + Px = Q , where dy

Senior Secondary School Mathematics for Class 12 Pg-982

982

Senior Secondary School Mathematics for Class 12

= ò 2y sin y dy + [y 2 sin y - ò 2y sin y dy] + C = y 2 sin y + C. \

x = y 2 + C cosec y.

… (i)

p -p2 Putting y = and x = 0 in (i), we get C = × 4 2 p2 Hence, the required solution is x = y 2 cosec y. 4

EXERCISE 21 Find the general solution for each of the following differential equations. dy dy 1 1. 2. x + 2y = x 2 + × y = x2 dx dx x dy dy 3. 2x 4. x + y = 6x 3 [CBSE 2003C] + y = 3 x 2 - 2, x > 0 dx dx dy dy 5. x [CBSE 2006] 6. x - y = 2x 3 -y = x+1 dx dx dy 1 dy 8. (1 - x 2) 7. (1 + x 2) + 2xy = + xy = x 1 - x 2 dx dx (1 + x 2) dy dy 9. (1 - x 2) + xy = ax [CBSE 2006C] 10. ( x 2 + 1) - 2xy = ( x 2 + 1)( x 2 + 2) dx dx dy dy [CBSE 2007] 12. 11. + 2y = 6ex + 3 y = e -2 x dx dx dy dy 13. 14. x + 8y = 5 e-3x [CBSE 2007] - y = ( x - 1) ex , x > 0 dx dx dy dy 15. 16. ( x log x) - y tan x = ex sec x + y = 2 log x [CBSE 2009] dx dx dy dy 17. x 18. x + y = x log x [CBSE 2008] + 2y = x 2 log x dx dx dy 19. (1 + x) - y = e 3x (1 + x) 2 [CBSE 2008] dx dy 4x 1 20. + y+ 2 =0 dx ( x 2 + 1) ( x + 1) 2 dx 22. x dy - ( y + 2x 2) dx = 0 [CBSE 2011] 21. ( y + 3 x 2) = x [CBSE 2011] dy dy 23. x dy + ( y - x 3) dx = 0 [CBSE 2011] 24. + 2y = sin x dx dy dy [CBSE 2009C] 25. - y = sin x + y = cos x - sin x [CBSE 2009] 26. sec x dx dx 2 27. (1 + x ) dy + 2xy dx = cot x [CBSE 2011C] dy dy 28. (sin x) + (cos x) y = cos x sin 2 x 29. + 2y cot x = 3 x 2 cosec2 x dx dx

Senior Secondary School Mathematics for Class 12 Pg-983

Linear Differential Equations

983

dy dy 31. - y = 2x 2 sec x = y tan x - 2 sin x dx dx dy dy [CBSE 2008] 32. + y cot x = sin 2x [CBSE 2008] 33. + 2y tan x = sin x dx dx dy 34. [CBSE 2005] + y cot x = x 2 cot x + 2x dx Find a particular solution satisfying the given condition for each of the following differential equations. dy 35. x + y = x 3 , given that y = 1 when x = 2. dx p dy [CBSE 2012] 36. + y cot x = 4x cosec x , given that y = 0 when x = × 2 dx dy 37. + 2xy = x , given that y = 1 when x = 0. dx dy 38. + 2y = e-2x sin x , given that y = 0 when x = 0. dx dy 39. (1 + x 2) + 2xy = 4x 2 , given that y = 0 when x = 0. dx dy 40. x - y = log x , given that y = 0 when x = 1. dx dy 41. + y tan x = 2x + x 2 tan x , given that y = 1 when x = 0. dx 42. A curve passes through the origin and the slope of the tangent to the curve at any point ( x , y) is equal to the sum of the coordinates of the point. Find the equation of the curve. 43. A curve passes through the point (0, 2) and the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5. Find the equation of the curve. 30. x

Find the general solution for each of the following differential equations. 45. ydx + ( x - y 2) dy = 0 44. ydx - ( x + 2y 2) dy = 0 dy dy 46. ( x - y 3) + y = 0 [CBSE 2011] 47. ( x + 3 y 2) = y , ( y > 0) dx dx dy dy 49. ( x + y + 1) 48. ( x + y) =1 =1 dx dx dy 50. Solve ( x + 1) = 2e- y - 1, given that x = 0 when y = 0. dx 51. Solve (1 + y 2) dx + ( x - e- tan

-1

y

) dy = 0, given that when y = 0, then x = 0.

ANSWERS (EXERCISE 21)

1. y =

x3 C + 4 x

2. y =

x2 C + 4 x2

3. y =

6 3 C x + 7 x

4. y = x 2 - 2 +

C x

Senior Secondary School Mathematics for Class 12 Pg-984

984

Senior Secondary School Mathematics for Class 12

5. y = x 3 + Cx 8. y =

-1 1 - x 2 log (1 - x 2) + C 1 - x 2 2

10. y = ( x 2 + 1)( x + tan -1 x + C) 13. y =

- 5 -3 x e + Ce-2x 4

x2 C ( 4 log x - 1) + 2 16 x -x C 20. y = 2 + 2 2 ( x + 1) ( x + 1) 2

19. y =

2

41. y = x 2 + cos x y2 C + 3 y

y

33. y = cos x + C cos2 x x3 2 4 x

1 36. y sin x = 2x 2 - p 2 2 4x 3

38. y = sin x

39. y =

42. x + y + 1 = ex

43. y = 4 - x - 2ex

y3 C + 4 y

48. ( x + y + 1) e- y = C -1

31. 2y cos x = cos 2x + C

35. y =

46. x =

25. y = cos x + Ce- x

29. y sin 2 x = x 3 + C

34. y = x 2 + C cosec x

51. xetan

22. y = 2x 2 + Cx

27. y(1 + x 2) = log |sin x| + C

30. y = Cx + 2x log |sec x + tan x| 2 32. y sin x = sin 3 x + C 3

45. x =

1 (1 + x) e 3x + C(1 + x) 3

1 24. y = ( 2 sin x - cos x) + Ce-2x 5

1 sin 3 x + C 3

37. 2y = 1 + e- x

12. y = e-2x + Ce-3x

15. y cos x = ex + C

21. y = 3 x 2 + Cx

26. y = Cesin x - (1 + sin x) 28. y sin x =

11. y = 2ex + Ce-2x

17. 4xy = 2x 2 log x - x 2 + C

18. y =

x3 C + 4 x

9. y = a + C 1 - x 2

14. y = ex + Cx

16. y(log x) = (log x) 2 + C

23. y =

7. y(1 + x 2) = tan -1 x + C

6. y = x log x - 1 + Cx

3(1 + x 2)

40. y = x - 1 - log x 44. x = 2y 2 + Cy

47. x = 3 y 2 + Cy 49. x = Cey - ( y + 2)

50. y = log

= tan -1 y

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 21) 8.

dy x + ×y= dx ( 1 - x 2 ) -

IF = e

-2 x 1 dx ò 2 (1 - x2 )

=e

x 1 - x2 -

×

1 log (1 - x2 ) 2

log

=e

1 1 - x2

=

1 1 - x2

2x + 1 x+1

Senior Secondary School Mathematics for Class 12 Pg-985

Linear Differential Equations \

its solution is -1 -2 x 1 1 x y´ =ò × dx = dx + C 2 ò ( 1 - x2 ) 1 - x2 1 - x2 1 - x2 1 = - log ( 1 - x 2 ) + C . 2 dy x ax 9. + ×y= × dx ( 1 - x 2 ) ( 1 - x2 ) -

IF = e

-2 x 1 dx ò 2 (1 - x2 )

=e

1 - log (1 - x2 ) 2

=

1 1 - x2

×

\ its solution is given by 1 1 ax y´ =ò ´ dx + C 2 2 ( 1 - x2 ) 1- x 1- x -2 x a a dt =- ò dx + C = - ò 3/2 + C , where ( 1 - x 2 ) = t 2 ( 1 - x 2 ) 3/2 2 t a a a = - ò t -3/2 dt + C = +C= + C. 2 t 1 - x2 10. Write ( x 2 + 2 ) = {( x 2 + 1) + 1}. 14.

dy 1 ( x - 1) x - ×y= e . dx x x IF = e

-ò

1 dx x

= e - log x = e log (1/x) =

1 × x

\ its solution is given by 1 1 ö æ x - 1ö æ1 y ´ = ò e x ç 2 ÷ dx + C = ò e x ç - 2 ÷ dx + C x è x ø èx x ø 1 = e x × + C . [Q ò e x{ f ( x ) + f ¢( x )} dx = e xf ( x )]. x dy 1 2 16. + ×y= × dx x log x x 1

IF = e

ò x log x

dx = e

ò

1 dt t

= e log t = t = log x.

21.

dy y + 3 x 2 dy 1 = Þ - × y = 3 x. dx x dx x

22.

dy y + 2 x 2 dy 1 = Þ - × y = 2 x. dx x dx x

23.

dy x 3 - y dy 1 = Þ + × y = x2. dx x dx x

2 dx 24. IF = e ò = e2x \ y ´ e 2 x = ò e 2 x sin x dx + C

Þ

ìï 2 sin x - cos x üï = e 2 xí ý+ C ïî ( 2 2 + 12 ) ïþ ( 2 sin x - cos x ) y= + Ce -2 x. 5

é ìï a sin bx + b cos bx üïù ê ò e ax sin bx dx = e ax í ýú ïþúû ïî ( a2 + b 2 ) êë

985

Senior Secondary School Mathematics for Class 12 Pg-986

986

Senior Secondary School Mathematics for Class 12

dx 25. IF = e ò = e x. y ´ e x = ò e x(cos x - sin x ) dx + C

= ò (cos x ) e x dx - ò (sin x )e xdx + C I

II

= (cos x )e x - ò ( - sin x )e xdx - ò (sin x )e xdx + C = (cos x )e x + C . 26.

dy - (cos x )y = sin x cos x. dx - cos x dx IF = e ò = e - sin x. \ y ´ e - sin x = ò (sin x cos x ) e - sin xdx + C

= ò t e - t dt + C , where sin x = t. I II

32. IF = e ò

cot x dx

= e log sin x = sin x.

\ y ´ sin x = ò sin x sin 2 x dx + C

= 2 ò sin 2 x cos x dx + C = 2 ò t 2 dt + C , where sin x = t.

33. IF = e ò

2 tan x dx

= e -2 log cos x = e log (cos x)

-2

= (cos x ) -2 =

1 cos 2 x

×

1 1 ü ì = ò í sin x ´ ý dx + C = ò sec x tan x dx + C = sec x + C î cos 2 x cos 2 x þ \ y = cos x + C cos 2 x. y´

34. IF = e ò

cot x dx

= e log sin x = sin x.

y ´ sin x = ò ( x 2 cot x + 2 x ) sin x dx + C = ò x 2 cos x dx +

ò 2 x sin x dx + C ò 2 x sin x dx + C = x 2 sin x + C .

= x 2 sin x - ò 2 x sin x dx + 37. IF = e ò

2 x dx

\ y´e

x2

2

= ex .

2

= ò xe x dx =

1 x2 e + C. 2 1 C \ y= + 2 × 2 ex

1 t e dt + C , where x 2 = t 2ò

=

Putting y = 1 and x = 0, we get C = 40. IF = e

-ò

1 dx x

= e - log x = e log (1 / x) =

æ ç 1 1 = ò ç log x × 2 x x ç è I II dy 43. ( x + y ) = 5. dx y´

ö ÷ ÷ dx + C . ÷ ø

44. y dx = ( x + 2 y 2 )dy Þ \

dx 1 - × x = 2 y. dy y

1 × 2

1 × x

dx x = + 2y dy y

Senior Secondary School Mathematics for Class 12 Pg-987

Linear Differential Equations

45. y dx = ( y 2 - x )dy Þ \

dx y 2 - x = dy y

dx 1 - × x = y. dy y

dy dx ( y 3 - x ) =y Þ = dx dy y dx 1 \ + × x = y2. dy y

46. ( y 3 - x )

dy dx x + 3 y 2 =y Þ = dx dy y dx 1 \ - × x = 3 y. dy y

47. ( x + 3 y 2 )

48. ( x + y ) \

dy dx =1Þ = x + y. dx dy

dx - x = y. dy

- dy IF = e ò = e -y .

\ x ´ e - y = ò y e - y dy + C . I II

dx 49. - x = ( y + 1). dy - dy IF = e ò = e -y .

x ´ e - y = ò ( y + 1)e - y dy + C = ò y e - y dy + I II -y

= - ye

ò e -y dy + C

+ 2 ò e - y dy + C

= - ye - y - 2 e - y + C . \ x = Ce y - ( y + 2 ). 50. ( x + 1) \

dy ( 2 - e y ) dx ( x + 1)e y = Þ = y dx dy ( 2 - e y ) e

dx ey ey ×x= × y dy ( 2 - e ) ( 2 - ey )

IF = e

ò

-e y dy (2 - e y )

= e log (2 - e

\ x ´ ( 2 - ey ) = ò

e

y)

y

( 2 - ey )

= ( 2 - e y ). ´ ( 2 - e y )dy + C = ò e y dy + C = e y + C .

Now, y = 0 and x = 0 Þ C = -1. \ x( 2 - e y ) = e y - 1 Þ ( 2 x + 1) = ( x + 1)e y ( 2 x + 1) æ 2x + 1 ö Þ y = log ç \ ey = ÷× ( x + 1) è x+ 1 ø

987

Senior Secondary School Mathematics for Class 12 Pg-988

988

Senior Secondary School Mathematics for Class 12

51. ( 1 + y 2 )dx = ( e - tan

-1 y

-1

- x )dy Þ

dx e - tan y - x = × dy ( 1 + y2 )

-1

\

dx 1 e - tan y + ×x= × 2 dy ( 1 + y ) ( 1 + y2 ) ò

1

2 IF = e (1 + y )

\ x ´ e tan Þ

dy

-1 y

x ´ e tan

-1

= e tan y . ü ì - tan -1 y -1 ï 1 ïe = òí ´ e tan y ý dy = C = ò dy + C 2 ( 1 + y2 ) ïþ ïî ( 1 + y )

-1 y

= tan -1 y + C .

… (i)

Putting x = 0 and y = 0 in (i), we get C = 0.

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: dy 1. The solution of the DE = ex + y is dx (b) ex - e- y = C (c) ex + e- y = C (a) ex + ey = C dy 2. The solution of the DE = 2x + y is dx (a) 2x + 2y = C (b) 2x + 2- y = C (c) 2x - 2- y = C

(d) none of these

(d) none of these

3. The solution of the DE ( ex + 1) y dy = ( y + 1) ex dx is (a) ey = C( ex + 1)( y + 1)

(b) ey = ex + y + 1

(c) y = ( ex + 1)( y + 1)

(d) none of these

4. The solution of the DE x dy + y dx = 0 is (a) x + y = C (b) xy = C (c) log ( x + y) = C (d) none of these dy 5. The solution of the DE x = cot y is dx (a) x cos y = C (b) x tan y = C (c) x sec y = C (d) none of these 6. The solution of the DE

dy (1 + y 2) is = dx (1 + x 2)

(a) ( y + x) = C(1 - yx) (c) y = (1 + x)C 7. The solution of the DE (a) log (1 + y) = x (c) ey = x -

x2 +C 2

(b) ( y - x) = C(1 + yx) (d) none of these dy = 1 - x + y - xy is dx

x2 +C 2

(1 + y )

(b) e

=x-

x2 +C 2

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-989

Linear Differential Equations

8. The solution of the DE

dy = ex + y + x 2 × ey is dx

x3 +C 3 x3 (c) ex - e- y = +C 3 (a) ex - y +

9. The solution of the DE

989

(b) ex + e- y +

x3 = C¢ 3

(d) none of these dy 1 - y2 + = 0 is dx 1- x 2

(a) y + sin -1 y = sin -1 x + C

(b) sin -1 y - sin -1 x = C

(c) sin -1 y + sin -1 x = C

(d) none of these

10. The solution of the DE (a) y = 2 tan

dy 1 - cos x is = dx 1 + cos x

x -x+C 2

(b) y = tan

(c) y = tan x - x + C

x - 2x + C 2

(d) none of these

dy -2xy 11. The solution of the DE is = dx ( x 2 + 1) (a) y 2( x + 1) = C

(b) y( x 2 + 1) = C (c) x 2( y + 1) = C

(d) none of these

12. The solution of the DE cos x(1 + cos y) dx - sin y(1 + sin x) dy = 0 is (a) 1 + sin x cos y = C

(b) (1 + sin x)(1 + cos y) = C

(c) sin x cos y + cos x = C

(d) none of these

13. The solution of the DE x cos y dy = ( xex log x + ex ) dx is (a) sin y = ex + log x + C

(b) sin y - ex + log x = C

(c) sin y = ex (log x) + C

(d) none of these

14. The solution of the DE

dy + y log y cot x = 0 is dx

(a) cos x log y = C

(b) sin x log y = C

(c) log y = C sin x

(d) none of these

15. The general solution of the DE (1 + x 2) dy - xy dx = 0 is (a) y = C(1 + x 2)

(b) y 2 = C(1 + x 2) (c) y 1 + x 2 = C

(d) none of these

16. The general solution of the DE x 1 + y 2 dx + y 1 + x 2 dy = 0 is (a) sin -1 x + sin -1 y = C

(b) 1 + x 2 + 1 + y 2 = C

(c) tan -1 x + tan -1 y = C

(d) none of these

Senior Secondary School Mathematics for Class 12 Pg-990

990

Senior Secondary School Mathematics for Class 12

æ dy ö 17. The general solution of the DE log ç ÷ = ( ax + by) is è dx ø - e- by eax = +C b a (c) beax + aeby = C

(b) eax - e- by = C

(a)

(d) none of these dy 18. The general solution of the DE = ( 1 - x 2 )( 1 - y 2 ) is dx (a) sin -1 y - sin -1 x = x 1 - x 2 + C (b) 2 sin -1 y - sin -1 x = x 1 - x 2 + C (c) 2 sin -1 y - sin -1 x = C 19. The general solution of the DE

(d) none of these dy y 2 - x 2 is = 2xy dx

(a) x 2 - y 2 = C1 x

(b) x 2 + y 2 = C1 y

(c) x 2 + y 2 = C1 x

(d) none of these

20. The general solution of the DE x 2 y = log x + C x y (c) tan -1 = log y + C x (a) tan -1

dy = x 2 + xy + y 2 is dx x (b) tan -1 = log x + C y (d) none of these

dy y = y + x tan is dx x æyö æyö (b) sin ç ÷ = Cx (c) sin ç ÷ = Cy èxø èxø

21. The general solution of the DE x æyö (a) sin ç ÷ = C èxø

(d) none of these

22. The general solution of the DE 2xy dy + ( x 2 - y 2) dx = 0 is (a) x 2 + y 2 = Cx

(b) x 2 + y 2 = Cy (c) x 2 + y 2 = C

(d) none of these

23. The general solution of the DE ( x - y) dy + ( x + y) dx is y tan -1 ( y/x ) (b) e (a) tan -1 = C x 2 + y 2 = C x2 + y2 x æyö (d) none of these (c) tan -1 ç ÷ = x 2 + y 2 + C èxø dy y y 24. The general solution of the DE = + sin is dx x x y y y (c) tan (d) none of these (a) tan = Cx (b) tan = Cx =C 2x x 2x dy 25. The general solution of the DE + y tan x = sec x is dx (a) y = sin x - C cos x (b) y = sin x + C cos x (c) y = cos x - C sin x (d) none of these

Senior Secondary School Mathematics for Class 12 Pg-991

Linear Differential Equations

991

dy + y cot x = 2 cos x is dx (a) ( y + sin x) sin x = C (b) ( y + cos x) sin x = C (c) ( y - sin x) sin x = C (d) none of these dy y 27. The general solution of the DE + = x 2 is dx x (b) 4xy = x 4 + C (c) 3 xy = x 3 + C (d) none of these (a) xy = x 4 + C 26. The general solution of the DE

ANSWERS (OBJECTIVE QUESTIONS)

1. (c) 2. (b) 3. (a) 4. (b) 5. (a) 6. (b) 7. (a) 8. (b) 9. (c) 10. (a) 11. (b) 12. (b) 13. (c) 14. (b) 15. (b) 16. (b) 17. (a) 18. (b) 19. (c) 20. (a) 21. (b) 22. (a) 23. (b) 24. (a) 25. (b) 26. (c) 27. (b) HINTS TO THE GIVEN OBJECTIVE QUESTIONS 1. The given DE is e xdx = e - y dy. \

ò e xdx = ò e -y dy Þ

e x = - e -y + C Þ e x + e -y = C .

2. We have 2 x dx = 2 - y dy Þ \ 3.

ò 2 xdx = ò 2 -y dy.

2 x log 2 = -2 - y log 2 = log C Þ 2 x + 2 - y = C , where

Þ

æ 1 ö ÷ dy log ( e x + 1) = ò çç 1 y + 1 ÷ø è y - log ( y + 1) = log ( e x + 1) + log C Þ y = log [C ( e x + 1)( y + 1)]

Þ

e y = C ( e x + 1)( y + 1).

ex

æ

log C = C. log 2

ö

y

ò ( e x + 1) dx = ò ççè y + 1 ÷÷ø dy Þ

4. x dy + y dx = 0 Þ

1

1

ò y dy = - ò x dx Þ

log y = - log x + log C .

\

log xy = log C Þ xy = C . 1 5. ò tan y dy = ò dx Þ log x = log sec y + log C . x Þ x = C sec y Þ x cos y = C . dy dx 6. ò =ò Þ tan -1 y = tan -1 x + C 1 . ( 1 + y2 ) ( 1 + x2 ) æ y-x ö y-x ÷ = C1 Þ Þ tan -1 y - tan -1 x = C 1 Þ tan -1 çç = tan C 1 = C . ÷ ( 1 + yx ) è 1 + yx ø 7. 8.

dy

ò ( 1 + y ) = ò ( 1 - x ) dx Þ

log ( 1 + y ) = x -

x2 + C. 2

dy = e y ( e x + x 2 ) Þ ò ( e x + x 2 ) dx = ò e - y dy. dx x3 x3 Þ - e -y = e x + + C Þ e x + e -y + = C ¢. 3 3

Senior Secondary School Mathematics for Class 12 Pg-992

992

9.

Senior Secondary School Mathematics for Class 12 1

ò

1- y

2

dy +

1

ò

1- x

2

dx = C Þ sin -1 y + sin -1 x = C .

dy 1 - cos x 2 sin 2 ( x 2 ) x æ x ö 10. = = = tan 2 = ç sec2 - 1 ÷ dx 1 + cos x 2 cos 2 ( x 2 ) 2 è 2 ø x x æ ö Þ ò dy = ò ç sec2 - 1 ÷ dx Þ y = 2 tan - x + C . 2 2 è ø -2 x 1 2 11. ò dy = ò 2 dx Þ log y + log ( x + 1) = log C . y ( x + 1) y( x 2 + 1) = C . cos x sin y 12. ò dx - ò dy = log C ( 1 + sin x ) ( 1 + cos y ) \

Þ

log ( 1 + sin x ) + log ( 1 + cos y ) = log C Þ ( 1 + sin x )( 1 + cos y ) = C . 1ö æ 13. ò cos y dy = ò e x ç log x + ÷ dx Þ sin y = e x log x + C . xø è dy 1 14. dy + cot x dx = 0 Þ + cot x dx = log C y log y y log y ò Þ log (log y ) + log sin x = log C Þ log (sin x + log y ) = log C Þ (sin x ) log y = C . 1 1 2x 1 15. ò dy = ò dx Þ log y = log ( 1 + x 2 ) + log C y 2 ( 1 + x2 ) 2 Þ 16.

ò

log y = log (C 1 + x 2 ) Þ y = C 1 + x 2 . 2x

2 × 1 + x2

dx +

ò

2y 2 × 1 + y2

=C Þ

1 dt 1 du + =C, 2ò t 2ò u where 1 + x 2 = t and 1 + y 2 = u

Þ

1 + x2 +

t + u =C Þ

1 + y2 = C.

dy e - by e ax = e ax + by Þ ò e - by dy = ò e axdx Þ = + C. -b dx a dy x 1 18. ò = ò 1 - x 2 dx Þ sin -1 y = 1 - x 2 + sin -1 x + C 2 2 1 - y2 17.

Þ

2 sin -1 y - sin -1 x = x 1 - x 2 + C .

19. The given DE is homogeneous. Put y = vx and \

2v

1

ò ( 1 + v 2 ) dv = - ò x dx Þ

dy dv =v+ x × dx dx

log ( 1 + v 2 ) + log x = log C .

\ x 2 + y 2 = C 1 x. 20.

dy x 2 + xy + y 2 =d , which is homogeneous. dx x2 dy dv dv Put y = vx and to get v + x =v+ x = ( 1 + v + v 2 ). dx dx dx y dv 1 = ò dx Þ tan -1 v = log x + C Þ tan -1 = log x + C . \ ò x x ( 1 + v2 )

Senior Secondary School Mathematics for Class 12 Pg-993

Linear Differential Equations

21.

22.

dy y y = + tan , which is homogeneous. dx x x dy dv 1 Put y = vx and to get ò cot v dv = ò dx =v+ x dx dx x y Þ log sin v = log x + log C Þ sin v = Cx Þ sin = Cx. x dy y 2 - x 2 , which is homogeneous. = 2 xy dx dy dv Put y = vx and = v + x × Then, dx dx 2v 1 dv = dx ò ( 1 + v2 ) ò x Þ log ( 1 + v 2 ) = - log x + log C Þ

23.

x( 1 + v 2 ) = C Þ x 2 + y 2 = Cx.

dy x + y = , which is clearly homogeneous. dx x - y dy ( 1 - v) dv 1 Put y = vx and =v+ x Þ ò dv = ò dx dx x dx ( 1 + v2 ) dv 1 2v 1 Þ ò dv = ò dx - ò x ( 1 + v2 ) 2 ( 1 + v2 ) 1 Þ tan -1v - log ( 1 + v 2 ) = log x + log C Þ tan -1v = log (Cx 1 + v 2 ) 2 -1 y Þ tan -1 = log (C x 2 + y 2 ) Þ e tan y/x = C x 2 + y 2 . x

24. Given DE is homogeneous. dy dv Put y = vx and =v+ x × dx dx dv 1 Then, v + x = v + sin v Þ ò cosec v dv = ò dx dx x y v v = Cx. Þ log tan = log x + log C Þ tan = Cx Þ tan 2 2 2x 25. Given DE is linear. tan x dx IF = e ò = e log (sec x) = sec x. \

its solution is y(sec x ) = ò sec2 x dx Þ y(sec x ) = tan x + C Þ y = sin x + C cos x.

26. Given DE is linear. cot x dx IF = e ò = e log sin x = sin x. \ \

y(sin x ) = ò 2 sin x cos x dx Þ y sin x = sin 2 x + C .

( y - sin x ) sin x = C .

27. Given DE is linear. IF = e

ò

1 dx x

= e log x = x.

\

y ´ x = ò ( x ´ x 2 ) dx = ò x 3 dx =

Þ

4 xy = x 4 + C .

x4 +C 4

993

SSS Mathematics for Class 12 994

22. VECTORS AND THEIR PROPERTIES In our daily life, we generally come across two types of quantities, namely scalars and vectors. SCALARS

A quantity that has magnitude only is known as a scalar. Each of the quantities mass, length, time, temperature, density, speed, etc., is a scalar.

Examples

VECTORS

A quantity that has magnitude as well as direction is called a vector. Each of the quantities force, velocity, acceleration and momentum is a vector.

Examples

However, we define a vector as given below. ‘A directed line segment is called a vector.’ A directed line segment with initial point A and the terminal point B, is ¾®

the vector denoted by AB. ¾®

¾®

The magnitude of AB is denoted by|AB|. REMARK

We usually denote a vector by a single letter with an arrow on it and its magnitude is denoted by this letter only. ¾®

®

¾®

®

Thus, AB = a and|AB|=|a|= a. UNIT VECTOR

®

®

^

A vector a is called a unit vector if| a | = 1 and it is denoted by a. ®

®

Two vectors a and b are said to be equal if they have the same magnitude and the same direction regardless of the positions of their initial points. EQUAL VECTORS

A vector having the same magnitude as that of a given

NEGATIVE OF A VECTOR ®

®

®

vector a and the direction opposite to that of a is called the negative of a , to be denoted ®

by – a . ¾® ¾® ® ® Thus, if AB = a then BA = - a . ZERO OR NULL VECTOR

A vector whose initial and terminal points coincide is called

®

a zero vector, denoted by 0 . Clearly, the magnitude of a zero vector is 0 but it cannot be assigned a definite direction. ¾®

®

Thus, AA = 0. 994

SSS Mathematics for Class 12 995

Vectors and their Properties COINITIAL VECTORS

995

Two or more vectors having the same initial point are called

coinitial vectors. ¾®

¾®

In the given figure, OA and OB are the two coinitial vectors having the same initial point O. COLLINEAR VECTORS Vectors having the same or parallel supports are known as collinear vectors. ¾®

¾®

¾®

In the given figure AB, BC and AC are collinear vectors. Collinear vectors having the same direction are called like vectors.

LIKE VECTORS

¾® ¾®

¾®

Thus, AB, BC and AC shown above are like vectors. UNLIKE VECTORS

Collinear vectors having opposite directions are known as unlike

vectors. In the given figure, PQ | | RS. ¾®

¾®

\ PQ and SR are unlike vectors. FREE VECTORS

If the initial point of a vector is not specified then it is said to be a

free vector. LOCALISED VECTORS A vector drawn parallel to a given vector through a specified point as the initial point is called a localised vector. COPLANAR VECTORS Three or more nonzero vectors lying in the same plane or parallel to the same plane are said to be coplanar, otherwise they are called noncoplanar. POSITION VECTOR OF A POINT ¾® ®

Let O be the origin and let A be a point such that ®

OA = a , then we say that the position vector of A is a .

SOLVED EXAMPLES EXAMPLE 1

SOLUTION

Classify the following measures as scalars and vectors: (i) 40 seconds (ii) 100 m 2 (iii) 30 gm/cm 3 (iv) 60 km/hr (v) 56 m/s towards south (i) 40 seconds represents time, which is scalar. (ii) 100 m 2 represents an area, which is scalar. (iii) 30 gm/cm 3 represents density, which is scalar. (iv) 60 km/hr represents speed, which is scalar. (v) 56 m/s towards south represents velocity, which is a vector.

SSS Mathematics for Class 12 996

996

Senior Secondary School Mathematics for Class 12

EXAMPLE 2

Represent graphically a displacement of 50 km, 60° west of north. ¾®

SOLUTION

The vector OA given below represents a displacement of 50 km, 60° west of north.

VECTOR ADDITION ® ®

¾®

Let a and b be any two vectors. Take any point O and draw segments OA and ¾®

®

¾®

®

¾®

¾®

AB such that OA = a and AB = b . Join OB. Then, OB is called the sum, or ®

®

resultant of a and b .

®

®

¾®

¾®

¾®

\ ( a + b ) = OA + AB = OB. TRIANGLE LAW OF ADDITION OF VECTORS ¾®

®

¾®

®

In a COAB, if OA and AB represent a and b ®

¾®

®

respectively, then OB represents ( a + b ). This is known as Triangle Law of Addition of Vectors. PARALLELOGRAM LAW OF ADDITION OF VECTORS ¾®

¾®

®

®

In a | |gm OABC, if OA and AB represent a and b ¾®

®

®

respectively, then OB represents ( a + b ). This is known as Parallelogram Law of Addition of Vectors.

SSS Mathematics for Class 12 997

Vectors and their Properties

997

Laws of Addition of Vectors Vector addition is commutative,

(Commutative Law)

THEOREM 1

®

®

®

®

i.e., a + b = b + a . PROOF

®

®

¾®

Then,

\

¾®

¾®

®

¾®

¾®

®

OC = AB = b

CB = OA = a .

and ®

®

¾®

¾®

¾®

®

®

¾®

¾®

¾®

a + b = OA + AB = OB

and b + a = OC + CB = OB . ®

®

®

®

Hence, a + b = b + a . THEOREM 2

Vector addition is associative,

(Associative Law) ®

®

®

®

®

®

i.e., ( a + b ) + c = a + ( b + c ). ® ¾®

¾®

PROOF

®

¾®

®

Let OA = a , AB = b and BC = c . Join OB , OC and AC. ®

®

®

¾®

¾®

¾®

¾®

¾®

( a + b ) + c = (OA + AB) + BC

¾®

¾®

¾®

= (OB + BC) [Q OA + AB = OB ] ¾®

= OC ®

®

®

¾®

¾®

¾®

a + ( b + c ) = OA + ( AB + BC) ¾®

¾®

¾®

¾®

¾®

= OA + AC [Q AB + BC = AC] ¾®

= OC ®

®

®

®

®

®

\ ( a + b ) + c = a + ( b + c ). THEOREM 3

®

For any vector a , prove that

(Existence of Additive Identity) ®

®

®

®

®

a + 0 = 0 + a = a.

¾®

PROOF

®

®

®

¾®

¾®

¾®

®

Let OA = a . Then, a + 0 = OA + AA = OA = a ®

®

¾®

®

®

®

¾®

¾®

®

and, 0 + a = OO + OA = OA = a. \ REMARK

¾®

Let a and b be the given vectors represented by OA and AB respectively. Complete the parallelogram OABC.

®

®

a + 0 = 0 + a = a. ®

The vector 0 is called the additive identity for vectors.

SSS Mathematics for Class 12 998

998

Senior Secondary School Mathematics for Class 12 ®

For any vector a , prove that

(Existence of Additive Inverse)

THEOREM 4

®

®

®

®

®

a + ( - a ) = ( - a ) + a = 0. ®

¾®

¾®

®

Let OA = a . Then, AO = - a .

PROOF

®

\

®

¾®

¾®

¾®

®

¾®

¾®

¾®

®

a + ( - a ) = OA + AO = OO = 0 ®

®

and, ( - a ) + a = AO + OA = AA = 0 . ®

®

®

®

®

Hence, a + ( - a ) = ( - a ) + a = 0 . ®

®

The vector - a is called the additive inverse of a .

REMARK

®

®

®

®

For any two vectors a and b , we define

DIFFERENCE OF TWO VECTORS ®

®

a - b = a + ( - b ). ®

¾®

¾®

®

¾®

®

Let OA = a and OB = b . Then, BO = - b . ®

®

®

®

\ ( a - b) = a + (- b ) ¾®

¾®

¾®

¾®

¾®

= OA + BO = BO + OA = BA ¾®

¾®

¾®

\ (OA - OB) = BA . In a similar way, we can prove that ¾®

¾®

¾®

(OB - OA) = AB . AN IMPORTANT REMARK: ¾®

(i) AB = (position vector of B) – (position vector of A) ¾®

(ii) BA = (position vector of A) – (position vector of B)

Scalar Multiplication of a Vector ®

®

The scalar multiple of a by a scalar k is the vector k a such that ®

®

(i) | k a | = | k || a |, ®

®

(ii) direction of k a is the same as that of a , when k > 0 and opposite to that ®

of a when k < 0. Examples

®

(i) 5 a is the vector whose magnitude is 5 times the magnitude of ®

®

a and whose direction is the same as that of a . ®

(ii) -2 a is the vector whose magnitude is 2 times the magnitude ®

®

of a and whose direction is opposite to that of a .

SSS Mathematics for Class 12 999

Vectors and their Properties

999

Components of a Vector ^ ^ ^

Let O be the origin and let P( x , y , z) be any point in space. Let i , j , k be unit vectors along the x-axis, y-axis and z-axis respectively. Let the position vector ®

of P be r . ®

^

^

^

Then, r = ( x i + y j + z k ). This form of a vector is called its component form. ®

^

^

^

Here x , y , z are called the scalar components of r and x i , y j , z k are called its vector components. ®

^

^

^

Also,| r | = | x i + y j + z k | = x 2 + y 2 + z 2 . Direction Ratios and Direction Cosines of a Vector ®

^

^

^

Consider a vector r = a i + b j + c k.

®

Then, the numbers a , b , c are called the direction ratios of r . ®

Direction cosines of r are given by

a 2

2

a +b +c

2

,

b 2

2

a +b +c

2

and

c 2

a + b 2 + c2

×

NOTE

If l , m , n are the direction cosines of a vector then we always have ( l 2 + m 2 + n2) = 1.

REMARK

If A( x1 , y1 , z1) and B( x 2 , y 2 , z2) be any two points in space then ¾®

direction ratios of AB are ( x 2 - x1), ( y 2 - y1), (z2 - z1) and direction ¾®

cosines of AB are ( x 2 - x1) ( y 2 - y1) (z2 - z1) , , , r r r where r = ( x 2 - x1) 2 + ( y 2 - y1) 2 + (z2 - z1) 2 . SOLVED EXAMPLES ®

EXAMPLE 1

® SOLUTION

^

^

^

®

^

^

^

^

^

®

®

Let a = a1 i + 3 j + a 3 k and b = 2 i + b2 j + k. If a = b , find the values of a1 , b2 and a 3. ®

^

^

^

^

a = b Û a1 i + 3 j + a 3 k = 2 i + b2 j + k Û a1 = 2, b2 = 3 , a 3 = 1.

SSS Mathematics for Class 12 1000

1000

Senior Secondary School Mathematics for Class 12 ®

^

®

^

^

®

^

®

®

®

EXAMPLE 2

Let a = 3 i + 2 j and b = 2 i + 3 j . Is| a | = | b |? Is a = b ?

SOLUTION

We have ®

®

3 2 + 22 = 13 and| b | = 22 + 3 2 = 13 .

| a |= ®

\

®

| a | = | b |. ^

^

^

®

^

®

But, 3 i + 2 j ¹ 2 i + 3 j and therefore, a ¹ b . ®

EXAMPLE 3

® SOLUTION

^

^

^

Find a unit vector in the direction of the vector a = i + 2 j + 3 k. ^

^

®

^

a = i + 2 j + 3 k Þ | a | = 12 + 22 + 3 2 = 14. ®

Unit vector in the direction of a is given by ®

^

^

^

( i + 2 j + 3 k) æ 1 ^ = çç a= ® = i + 14 è 14 |a|

^

EXAMPLE 4

a

2 ^ j + 14

3 ^ö k÷ × 14 ÷ø

Write a vector of magnitude 15 units in the direction of the vector ^

^

^

( i - 2 j + 2 k). ®

SOLUTION

^

[CBSE 2010]

^

^

Let a = ( i - 2 j + 2 k). Then, ®

a unit vector in the direction of a is given by ®

^

a=

a

®

^

^

^

( i - 2 j + 2k)

=

12 + ( -2) 2 + 22

| a|

=

^ 1 ^ ^ ( i - 2 j + 2 k ). 3 ®

\ a vector of magnitude 15 in the direction of a ^ ^ ^ ^ 1 ^ ^ = 15 ´ ( i - 2 j + 2 k ) = (5 i - 10 j + 10 k ). 3 EXAMPLE 5

^

^

^

^

®

^

^

( 2 i - 3 j + 5 k ). SOLUTION

^

Find a unit vector parallel to the sum of the vectors ( i + j + k ) and ^

[CBSE 2012] ®

^

^

^

^

Let a = ( i + j + k) and b = ( 2 i - 3 j + 5 k). Then, ®

®

^

^ ^

^

^

^

^

^

^

( a + b ) = ( i + j + k ) + ( 2 i - 3 j + 5 k ) = ( 3 i - 2 j + 6 k ). ®

®

Required unit vectors parallel to ( a + b ) are ^

±

^

^

( 3 i - 2 j + 6k)

^ ^ 1 ^ = ± ( 3 i - 2 j + 6 k ). 7 3 2 + ( -2) 2 + 62

SSS Mathematics for Class 12 1001

Vectors and their Properties ®

EXAMPLE 6

^

®

^

^

^

®

^

^

^

^

If a = ( i + j + k ), b = ( 4 i - 2 j + 3 k ) and c = ( i - 2 j + k ), find a vector of magnitude 6 units which is parallel to the vector ®

SOLUTION

^

1001

®

®

( 2 a - b + 3 c ). We have ® ®

[CBSE 2011C] ®

^

^

^

^

^

^

^

^

^

( 2 a - b + 3 c ) = 2( i + j + k ) - ( 4 i - 2 j + 3 k ) + 3( i - 2 j + k ) ^

^

^

= ( 2 - 4 + 3) i + ( 2 + 2 - 6) j + ( 2 - 3 + 3) k ^

^

^

= ( i - 2 j + 2 k ). ®

®

®

Unit vectors parallel to ( 2 a - b + 3 c ) are ^

±

EXAMPLE 7

SOLUTION

^

^

( i - 2 j + 2k) 12 + ( -2) 2 + 22

=±

^ ^ 1 ^ ( i - 2 j + 2 k ). 3

Required vectors of magnitude 6 units are ^ ü ^ ^ ^ ì 1 ^ ^ ± í6 ´ ( i - 2 j + 2 k ) ý = ± 2( i - 2 j + 2 k ). 3 î þ ¾® Find a unit vector in the direction of AB, where A(1, 2, 3) and B( 4, 5 , 6) are the given points. We have

^

^

^

^

^

^

^

^

p.v. of A = ( i + 2 j + 3 k ) and p.v. of B = ( 4 i + 5 j + 6 k ). \

¾®

AB = (p.v. of B) - (p.v. of A) ^

^

^

^

^

^

^

= ( 4 i + 5 j + 6 k ) - ( i + 2 j + 3 k ) = ( 3 i + 3 j + 3 k ), and ¾®

3 2 + 3 2 + 3 2 = 27 .

| AB | =

¾®

\ unit vector in the direction of AB ¾®

=

AB

¾®

^

=

| AB |

^

^

^

æ 1 ^ 1 ^ 1 ^ö i + j + k÷ × = çç 3 3 3 ÷ø è EXAMPLE 8

SOLUTION

^

^

^

^

^

( 3 i + 3 j + 3 k ) 3( i + j + k ) ( i + j + k ) = = 27 3 3 3

^

^

^

^

^

^

For what value of a, the vectors ( 2 i - 3 j + 4 k ) and ( a i + 6 j - 8 k ) collinear? [CBSE 2011] The given vectors are collinear only when ^

^

^

^

^

^

( a i + 6 j - 8 k ) = l( 2 i - 3 j + 4 k ) for some nonzero scalar l . ^

^

^

^

^

^

Now, a i + 6 j - 8 k = 2l i - 3 l j + 4l k

SSS Mathematics for Class 12 1002

1002

Senior Secondary School Mathematics for Class 12

Û 2l = a, -3l = 6 and 4l = -8 a a Û l = and l = -2 Û = -2 Û a = -4. 2 2 Hence, the given vectors are collinear when a = -4. ^

EXAMPLE 9

SOLUTION

^

^

^

^

^

^ ^

Show that the points A( -2 i + 3 j + 5 k ), B( i + 2 j + 3 k ) and C(7 i - k ) are collinear. [CBSE 2009] Clearly, we have ¾®

AB = (position vector of B) – (position vector of A) ^

^

^

^

^

^

^

^

^

= ( i + 2 j + 3 k ) - ( -2 i + 3 j + 5 k ) = ( 3 i - j - 2 k ). ¾®

BC = (position vector of C) – (position vector of B) ^

^

^

^

^

^

^

^

= (7 i - k ) - ( i + 2 j + 3 k ) = ( 6 i - 2 j - 4 k ). ¾®

¾®

¾®

¾®

\ AB = 2 BC , which shows that AB and BC are parallel vectors, having a common end point B. Hence, the points A, B and C are collinear. ^ ^

EXAMPLE 10

®

SOLUTION

^

Write the direction cosines of the vector ( -2 i + j - 5 k ). ^ ^

[CBSE 2011]

^

The given vector is a = ( -2 i + j - 5 k ). ®

Direction ratios of a are –2, 1, –5. ®

| a|= ( -2 )2 + 12 + ( -5 )2 = 30 . ®

Hence, the direction cosines of a are

-2 1 -5 , , × 30 30 30 ^ ^

^

EXAMPLE 11

What is the cosine of the angle which the vector ( 2 i + j + k ) makes with the y-axis? [CBSE 2010]

SOLUTION

The given vector is a = ( 2 i + j + k ).

®

^ ^

^

®

Direction ratios of a are 2 , 1, 1. ®

| a|= ( 2 )2 + 12 + 12 = 4 = 2 . ®

\ direction cosines of a are ®

2 1 1 , , × 2 2 2

Let a make angle b with the y-axis. 1 Then, clearly cos b = × 2

SSS Mathematics for Class 12 1003

Vectors and their Properties ^

EXAMPLE 12

^

^

p( i + j + k ) is a unit vector ^

^

^ 2

Û p( i + j + k )

= 1 Û ( p 2 + p 2 + p 2) = 1

Û 3p2 = 1 Û p2 = EXAMPLE 13

^

Find the value of p for which p( i + j + k ) is a unit vector. [CBSE 2009] ^

SOLUTION

^

1003

1 1 Û p=± × 3 3

If A(1, 2, - 3) and B( -1, - 2, 1) are the two given points in space then ¾®

¾®

find (i) the direction ratios of AB and (ii) the direction cosines of AB . ¾®

^ ^

^

Express AB in terms of i , j and k . ¾®

SOLUTION

(i) The direction ratios of AB are ( -1 - 1), ( -2 - 2), (1 + 3), i.e., –2, –4, 4. ¾®

(ii) |AB|= ( -2) 2 + ( -4) 2 + 42 =

36 = 6.

¾®

\ the direction cosines of AB are -2 -4 4 -1 -2 2 , , , i.e., , , × 6 6 6 3 3 3 ¾®

Clearly, AB = -

1^ 2^ 2^ i - j + k. 3 3 3

EXAMPLE 14

Find the scalar and vector components of the vector with initial point A( 3 , - 1, 2) and terminal point B( -5 , 4, 3).

SOLUTION

We have ¾®

AB = (position vector of B) – (position vector of A) ^

^

^

^

^

^

^

^

^

= ( -5 i + 4 j + 3 k ) - ( 3 i - j + 2 k ) = ( -8 i + 5 j + k ). ¾®

The scalar components of AB are –8, 5, 1. ¾®

^

^ ^

The vector components of AB are -8 i , 5 j , k . EXAMPLE 15

Write two different vectors having same magnitude. ®

SOLUTION

^

^

®

^

^

^

^

Consider the vectors a = 2 i + 3 j - 4 k and b = 4 i + 2 j + 3 k. ®

®

Clearly, a ¹ b . ®

®

But,|a|= 22 + 3 2 + ( -4) 2 = 29 and|b|= 42 + 22 + 3 2 = 29. ®

®

®

®

Thus, |a| = |b|and a ¹ b .

SSS Mathematics for Class 12 1004

1004

Senior Secondary School Mathematics for Class 12

EXAMPLE 16

Write two different vectors having same direction. ¾®

¾®

SOLUTION

Clearly, 3 AB and 5 AB are two different vectors having the same direction.

EXAMPLE 17

Show that the points with position vectors a = ( 3 i - 4 j - 4 k ),

®

®

^

^

®

^

^

^

^

^

^

^

b = ( 2 i - j + k ) and c = ( i - 3 j - 5 k ) respectively, form the vertices of a right-angled triangle. SOLUTION

We have ¾®

AB = (position vector of B) – (position vector of A) ®

®

^ ^

^

^

^

^

= ( b - a ) = ( 2 i - j + k ) - ( 3 i - 4 j - 4 k) ^

^

^

= ( - i + 3 j + 5 k ). ¾®

BC = (position vector of C) – (position vector of B) ®

®

^

^

^

^ ^

^

= ( c - b) = ( i - 3 j - 5 k ) - (2 i - j + k ) ^

^

^

= ( - i - 2 j - 6 k ). ¾®

CA = (position vector of A) – (position vector of C) ®

®

^

^

^

^

^

^

= ( a - c) = ( 3 i - 4 j - 4 k) - ( i - 3 j - 5 k) ^

^

^

= ( 2 i - j + k ). \

¾®

|AB|2 = {( -1) 2 + 3 2 + 5 2} = (1 + 9 + 25) = 35 , ¾®

|BC|2 = {( -1) 2 + ( -2) 2 + ( -6) 2} = (1 + 4 + 36) = 41, ¾®

|CA|2 = {22 + ( -1) 2 + 12} = ( 4 + 1 + 1) = 6. \

|AB|2 + |CA|2 = |BC|2 Þ AB 2 + CA 2 = BC 2.

Hence, CABC is right-angled at A.

Section Formulae THEOREM 1

(Section Formula for Internal Division) ®

®

Let A and B be two points with

position vectors a and b respectively and let P be a point dividing AB ¾®

®

internally in the ratio m : n. Let OP = r . Then, prove that ®

r =

¾®

PROOF

®

®

(m b + n a ) × (m + n) ®

¾®

®

Let O be the origin. Then, OA = a and OB = b .

SSS Mathematics for Class 12 1005

Vectors and their Properties

Let P be a point on AB such that

1005

AP m = × Then, PB n

AP m = PB n Þ n × AP = m × PB ¾®

¾®

Þ n( AP) = m( PB ) ¾®

¾®

¾®

¾®

Þ n(OP - OA) = m(OB - OP) ¾®

¾®

®

¾®

®

Þ (m + n) OP = m OB + n OA = m b + n a ¾®

®

Þ OP =

®

m b +n a (m + n)

Þ

®

r =

®

®

(m b + n a ) × (m + n)

The position vector of the midpoint of the join of two points with ® ® 1 ® ® position vectors a and b is ( a + b ). 2 ® ® PROOF Let A and B be two points with position vectors a and b respectively. Let P be the midpoint of AB. Then, P divides AB in the ratio 1 : 1. COROLLARY

®

®

(1 × b + 1 × a ) 1 ® ® = ( a + b ). \ OP = (1 + 1) 2 ¾®

THEOREM 2

Let A and B be two points with

(Section Formula for External Division) ®

®

position vectors a and b respectively and let P be a point dividing AB ¾®

®

externally in the ratio m : n. Let OP = r . Then, prove that ®

®

¾®

PROOF

®

(m b - n a ) × (m - n)

r =

®

¾®

®

Let O be the origin. Then, OA = a and OB = b . Let AB be produced to P such that AP : BP = m : n. AP m Now, = BP n ¾®

¾®

Þ

n × AP = m × BP Þ n × AP = m × BP

Þ

n(OP - OA) = m(OP - OB)

Þ

(m - n) OP = (m OB - n OA) = (m b - n a )

Þ

OP =

¾®

¾®

¾®

¾®

®

¾®

¾®

®

(m b - n a ) Þ (m - n)

¾®

¾®

®

r =

®

®

®

®

(m b - n a ) × (m - n)

SSS Mathematics for Class 12 1006

1006

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Find the position vector of a point R which divides the line joining the ^

^

^

^

^

^

points P( i + 2 j - k ) and Q( - i + j + k ) in the ratio 2 : 1, (i) internally and (ii) externally. ®

SOLUTION

^

^

®

^

^

^

^

Here a = ( i + 2 j - k ) and b = ( - i + j + k ). Also, m = 2, n = 1. (i) When R divides PQ internally in the ratio 2 : 1; then ®

®

(m b + n a ) position vector of R = (m + n) ^

^

^

^

^

^

2( - i + j + k ) + 1 × ( i + 2 j - k ) = ( 2 + 1) ^

^

®

®

^

( - i + 4 j + k) = × 3 (ii) When R divides PQ externally in the ratio 2 : 1; then position vector of R =

(m b - n a ) (m - n) ^

^

^

^

^

^

2( - i + j + k ) + 1 × ( i + 2 j - k ) = ( 2 - 1) ^

^

= ( -3 i + 3 k ) × ®

®

®

®

EXAMPLE 2

P and Q are two points with position vectors ( 3 a - 2 b ) and ( a + b ) respectively. Write the position vector of a point R which divides the line segment PQ in the ratio 2 : 1 externally. [CBSE 2013]

SOLUTION

The position vectors of the given points are ®

®

®

®

P( 3 a - 2 b ) and Q( a + b ). We have to divide PQ in the ratio 2 : 1 externally at the point R. The position vector of R is ®

®

®

®

® ® 2( a + b ) - 1 × ( 3 a - 2 b ) = ( - a + 4 b ). ( 2 - 1) ®

®

Hence, the position vector of R is ( - a + 4 b ). EXAMPLE 3

Find the position vector of a point R which divides the line segment joining the points A( 2, - 3 , 4) and B( 3 , 1, - 2) externally in the ratio 3 : 2.

SSS Mathematics for Class 12 1007

Vectors and their Properties ^

SOLUTION

^

1007

^

The position vector of A is ( 2 i - 3 j + 4 k ). ^

^

^

The position vector of B is ( 3 i + j - 2 k ). Let R divide AB externally in the ratio 3 : 2. Then, position vector of R ^ ^ ^ ^ ^ ^ æ ® ®ö ç 3 b - 2 a ÷ 3( 3 i + j - 2 k ) - 2( 2 i - 3 j + 4 k ) =ç ÷= 1 ç 3-2 ÷ è ø ^

^

^

= (5 i + 9 j - 14 k ).

^

^

^

Hence, the position vector of R is (5 i + 9 j - 14 k ). EXAMPLE 4

Find the position vector of the mid-point of the vector joining the points P( 2, 3 , 4) and Q( 4, 1, - 2). [CBSE 2011]

SOLUTION

The position vectors of the given points P and Q are ®

^

^

®

^

^

^

^

a = ( 2 i + 3 j + 4 k ) and b = ( 4 i + j - 2 k ) respectively.

EXAMPLE 5

Show that the three points A(1, - 2, - 8), B(5 , 0, - 2) and C(11, 3 , 7) are collinear and find the ratio in which B divides AC.

SOLUTION

The position vectors of A, B and C are ( i - 2 j - 8 k ), (5 i - 2 k ) and

^

^

^

^

^

^

^

^

(11 i + 3 j + 7 k ) respectively. \

¾®

AB = (position vector of B) – (position vector of A) ^

^

^

^

^

^

^

^

= (5 i - 2 k ) - ( i - 2 j + 8 k ) = ( 4 i + 2 j + 6 k ). ¾®

BC = (position vector of C) – (position vector of B) ^

^

^

^

^

^

^

^

= (11 i + 3 j + 7 k ) - (5 i - 2 k ) = ( 6 i + 3 j + 9 k ), and ¾®

AC = (position vector of C) – (position vector of A) ^

^

^

^

^

^

^

^

^

^

^

^

= (11 i + 3 j + 7 k ) - ( i - 2 j - 8 k ) = (10 i + 5 j + 15 k ). ¾®

^

^

^

Now, AB = ( 4 i + 2 j + 6 k ) = 2( 2 i + j + 3 k ) ^ ^ 2 ¾® 2 ^ = (10 i + 5 j + 15 k ) = AC. 5 5 ¾®

¾®

\ AB and AC are parallel vectors having same end point A. Hence, the points A , B and C are collinear. ¾®

^

^

^

^

^

^

Also, AB = ( 4 i + 2 j + 6 k ) = 2( 2 i + j + 3 k ) =

^ ^ 2 ^ 2 ¾® ( 6 i + 3 j + 9 k ) = BC . 3 3

SSS Mathematics for Class 12 1008

1008

Senior Secondary School Mathematics for Class 12 ¾®

|AB|

\

¾®

=

|BC|

2 × 3

Hence, B divides AC in the ratio 2 : 3.

EXERCISE 22 1. Write down the magnitude of each of the following vectors: ®

^

^

®

^

(i) a = i + 2 j + 5 k

^

^

^

(ii) b = 5 i - 4 j - 3 k

® æ 1 ^ 1 ^ 1 ^ö (iii) c = çç i j + k÷ 3 3 ÷ø è 3

®

^

^

(iv) d = ( 2 i +

^

3 j - 5 k)

2. Find a unit vector in the direction of the vector: ^

^

^

^

(i) ( 3 i + 4 j - 5 k ) ^

^

^

(iii) ( i + k ) ®

^

®

^

^

^

(ii) ( 3 i - 2 j + 6 k ) ^

[CBSE 2012]

^

(iv) ( 2 i + j + 2 k ) ^

[CBSE 2009] ®

^

3. If a = ( 2 i - 4 j + 5 k ) then find the value of l so that l a may be a unit vector. ^

®

^

^

^

^

4. If a = ( - i + j - k ) and b = ( 2 i - j + 2 k ) then find the unit vector in the ®

®

direction of ( a + b ). ®

^

^

®

^

^

^

^

^

^

5. If a = ( 3 i + j - 5 k ) and b = ( i + 2 j - k ) then find a unit vector in the ®

®

direction of ( a - b ). ®

^

^

®

^

^

6. If a = ( i + 2 j - 3 k ) and b = ( 2 i + 4 j + 9 k ) then find a unit vector parallel ®

®

to ( a + b ).

[CBSE 2008]

7. Find a vector of magnitude 9 units in the direction of the vector ^

^

^

( -2 i + j + 2 k ).

[CBSE 2010]

8. Find a vector of magnitude 8 units in the direction of the vector ^

^

^

(5 i - j + 2 k ). 9. Find a vector of magnitude 21 units in the direction of the vector ^

^

^

( 2 i - 3 j + 6 k ). ®

^

^

®

[CBSE 2014] ^

^

®

^

^

®

®

®

10. If a = ( i - 2 j ), b = ( 2 i - 3 j ) and c = ( 2 i + 3 k ), find ( a + b + c ). [CBSE 2012]

SSS Mathematics for Class 12 1009

Vectors and their Properties

1009

11. If A(–2, 1, 2) and B(2, –1, 6) are two given points, find a unit vector in the ¾®

direction of AB. 12. Find ®

the

direction

^

^

ratios

and

direction

cosines

of

the

vector

^

a = (5 i - 3 j + 4 k ).

13. Find the direction ratios and the direction cosines of the vector joining the points A(2, 1, –2) and B(3, 5, –4). ^

^

^

14. Show that the points A, B and C having position vectors ( i + 2 j + 7 k ), ^

^

^

^

^

^

( 2 i + 6 j + 3 k ) and ( 3 i + 10 j - 3 k ) respectively, are collinear. ^

^

^

^

^

^

15. The position vectors of the points A, B and C are ( 2 i + j - k ), ( 3 i - 2 j + k ) ^

^

^

and ( i + 4 j - 3 k ) respectively. Show that the points A, B and C are collinear. 16. If the position vectors of the vertices A, B and C of a A ABC be ^

^

^

^

^

^

^

^

^

( i + 2 j + 3 k ), ( 2 i + 3 j + k ) and ( 3 i + j + 2 k ) respectively, prove that A ABC is equilateral. ^

^

^

17. Show that the points A, B and C having position vectors ( 3 i - 4 j - 4 k ), ^

^

^

^

^

^

( 2 i - j + k ) and ( i - 3 j - 5 k ) respectively, form the vertices of a right-angled triangle. 18. Using vector method, show that the points A(1, –1, 0), B(4, –3, 1) and C( 2, - 4, 5) are the vertices of a right-angled triangle. 19. Find the position vector of the point which divides the join of the points ®

®

®

®

( 2 a - 3 b ) and ( 3 a - 2 b ) (i) internally and (ii) externally in the ratio 2 : 3. ®

®

®

®

20. The position vectors of two points A and B are ( 2 a + b ) and ( a - 3 b ) respectively. Find the position vector of a point C which divides AB externally in the ratio 1 : 2. Also, show that A is the mid-point of the line segment CB. [CBSE 2010] 21. Find the position vector of a point R which divides the line joining A( -2, 1, 3) and B( 3 , 5 , - 2) in the ratio 2 : 1 (i) internally (ii) externally. 22. Find the position vector of the mid-point of the vector joining the points ^

^

^

^

^

^

A( 3 i + 2 j + 6 k ) and B( i + 4 j - 2 k ). ¾®

^

^

^

23. If AB = ( 2 i + j - 3 k ) and A(1, 2, - 1) is the given point, find the coordinates of B. ¾®

24. Write a unit vector in the direction of PQ , where P and Q are the points (1, 3 , 0) and (4, 5, 6) respectively. [CBSE 2014]

SSS Mathematics for Class 12 1010

1010

Senior Secondary School Mathematics for Class 12 ANSWERS (EXERCISE 22)

1. (i) 30 (ii) 5 2 (iii) 1

(iv) 10

æ 3 ^ 4 ^ 5 ^ö 2. (i) çç i + j k÷ 5 2 5 2 ÷ø è5 2

æ 3 ^ 2^ 6^ö (ii) çç i - j + k ÷÷ 7 7 ø è7

æ 1 ^ 1 ^ö (iii) çç i + k÷ 2 2 ÷ø è 3. ± 5.

æ 2 ^ 1 ^ 2 ^ö (iv) çç i + j + k ÷÷ 3 3 ø è3

1 3 5 ^ ^ ^ 1 (2 i - j - 4 k) 21 ^

^

^

^

^

^

æ 1 13. (1, 4, - 2), ç , è 21

^

-3 4 ö æ 1 12. (5 , - 3 , 4), ç , , ÷ è 2 5 2 5 2ø -2 ö 4 , ÷ 21 21 ø

19. (i)

® 12 ® 13 ® a b (ii) -5 b 5 5

æ 4 ^ 11 ^ 1 ^ ö 21. (i) çç i + j - k ÷÷ 3 3 ø è3

®

20. ( 3 a + 5 b ) ^

^

10. (5 i - 5 j + 3 k )

æ 2 ^ 1 ^ 2 ^ö 11. çç i - j + k ÷÷ 3 3 ø è3

^

^ ^ ^ 8 (5 i - j + 2 k ) 30

8.

9. ( 6 i - 9 j + 18 k )

®

^ ^ 1 ^ ( i + 2 j + 2 k) 3

6. ±

7. ( - 6 i + 3 j + 6 k ) ^

1 ^ ^ ( i + k) 2

4.

^

22. P( 2 i + 3 j + 2 k )

^

^

^

(ii) ( 8 i + 9 j - 7 k )

^ ^ 1 ^ ( 3 i + 2 j + 6k) 7

23. (3, 3, –4) 24.

HINTS TO SOME SELECTED QUESTIONS ¾®

¾®

¾®

16. Show that| AB|=|BC |=|CA|. ¾®

¾®

¾®

®

¾®

¾®

¾®

17. Show that AB + BC + CA = 0 and| AB|2 + |CA|2 =|BC |2 . \ C ABC is right angled at A. ^

^

^

^

^

^

^

^

18. The position vectors of A, B, C are ( i - j ), ( 4 i - 3 j + k ) and ( 2 i - 4 j + 5 k ) ¾®

¾®

¾®

®

¾®

¾®

¾®

respectively. Show that AB + BC + CA = 0 and| AB|2 + |BC |2 =|CA|2 . \ C ABC is right angled at B. ¾®

^

^

^

23. Let O be the origin. Then, OA = ( i + 2 j - k ). ¾®

¾®

¾®

¾®

¾®

¾®

\ AB = ( OB - OA ) Þ OB = ( AB + OA ).

SSS Mathematics for Class 12 1011

23. SCALAR, OR DOT, PRODUCT OF VECTORS ¾®

¾®

Let PQ and RS be two given vectors. Take any point O, and draw OA | | PQ and OB | | RS. Then, ÐAOB = q is called the angle ANGLE BETWEEN TWO VECTORS ¾®

¾®

between PQ and RS , provided 0 £ q £ p.

¾®

¾®

If q = 0 or q = p then PQ | | RS. ¾® ¾® p then PQ and RS are called perpendicular, or orthogonal, vectors. 2

If q =

®

SCALAR PRODUCT, OR DOT PRODUCT, OF TWO VECTORS

®

Let a and b be two vectors,

®

and let q be the angle between them. Then, the scalar product, or dot product, of a ®

® ®

® ®

and b , denoted by a × b , is defined as

®

®

a × b = | a || b |cos q = a b cos q.

Clearly, the scalar product of two vectors is a scalar. ANGLE BETWEEN TWO VECTORS IN TERMS OF SCALAR PRODUCT ® ®

between two nonzero vectors a and b . Then, ® ®

®

®

a × b =| a || b |cos q. ® ®

\

cos q =

a×b

®

Þ

®

-1

q = cos

| a || b | ® LENGTH OF A VECTOR Let a be ® ® ® ®

®

| a |=

®

® ®

a × a.

®

REMARKS

any vector.

a × a = | a || a |cos 0° = | a |2 .

Then,

\

æ ® ® ö ç a×b ÷ ç ® ® ÷× ç| a || b |÷ è ø

®

®

®

® ®

(i) If a = 0 or b = 0 , we define a × b = 0. ®

®

(ii) If a and b are like vectors, we have q = 0. 1011

Let q be the angle

SSS Mathematics for Class 12 1012

1012

Senior Secondary School Mathematics for Class 12 ® ®

\ a × b = a b cos 0° = ab. ®

®

(iii) If a and b are unlike vectors, we have q = p. ® ®

\ a × b = a b cos p = - ab. ®

®

(iv) If a and b are orthogonal vectors, we have q = ® ®

\ a × b = a b cos

p = 0. 2

p × 2

SOLVED PROBLEMS ®

EXAMPLE 1

®

®

®

Let a and b be two given vectors such that | a| = 3, | b | = 4 and the ® ®

angle between them is 60°. Find a × b . SOLUTION

Clearly, we have ® ®

® ® 1ö æ a × b = | a || b |cos 60° = ç 3 ´ 4 ´ ÷ = 6. 2ø è

®

EXAMPLE 2

®

®

Let a and b be two given vectors such that | a | =

® ®

®

®

®

3 , | b | = 2 and

a × b = 6. Find the angle between a and b . ®

SOLUTION

®

Let q be the angle between a and b . Then, ® ®

®

®

a × b = 6 Þ | a || b |cos q = 6 Þ ( 3 )( 2) cos q = 6

6 1 æ 1 ö p Þ q = cos-1 ç = ÷= × 2 3 2 è 2ø 4 p Hence, the required angle is × 4 Þ cos q =

®

EXAMPLE 3

®

®

®

® ®

If a and b are two vectors such that| a | = | b |= 2 and a × b = - 1, find ®

®

the angle between a and b . ®

SOLUTION

®

® ®

Let q be the angle between a and b . Then, a × b = - 1 ®

®

Û | a |×| b |× cos q = - 1 Û -1 2p Û q= Û cos q = 2 3 ®

2 ´ 2 ´ cos q = - 1 [Q 0 £ q £ 2p]. ®

Hence, the angle between a and b is

2p × 3

SSS Mathematics for Class 12 1013

Scalar, or Dot, Product of Vectors

1013

PROJECTION ® ¾®

¾®

®

Let OA = a , OB = b and ÐBOA = q. Draw AM ^ OB. ®

®

Then, OM is the projection of a on b . ® ®

®

OM = (OA) cos q =|a|cos q =

a× b ®

×

|b| ®

® ®

®

a× b

\ projection of a on b =

®

×

|b|

PROPERTIES OF SCALAR PRODUCT (Commutative Law)

THEOREM 1 ®

PROOF

Prove that

®

® ®

® ®

a × b = b × a.

® ®

® ®

If a or b is a zero vector then a × b = 0 and b × a = 0. ® ®

® ®

So, in this case, a × b = b × a . ®

®

Now, let a and b be any two nonzero vectors, and let q be the angle between them. Then, ® ®

®

®

® ®

®

®

a × b = | a || b |cos q = a b cos q, and b × a = | b || a |cos ( - q) = b a cos q = a b cos q. ® ®

Hence,

® ®

® ®

a × b = b × a. ® ®

THEOREM 2 PROOF

® ®

a × b = b × a.

So, in this case also,

®

®

®

®

®

®

Prove that a × b = 0 Û a = 0 or b = 0 or a ^ b . ® ®

a × b = 0.

Let

® ®

Then, a × b = 0 Þ a b cos q = 0 Þ a = 0 or b = 0 or cos q = 0 p Þ a = 0 or b = 0 or q = 2 ®

®

®

®

®

®

®

®

®

®

®

®

Þ a = 0 or b = 0 or a ^ b . ® ®

Thus, a × b = 0 Þ a = 0 or b = 0 or a ^ b . ®

®

®

®

®

®

®

®

Conversely, let a = 0 or b = 0 or a ^ b . If

®

®

® ®

a = 0 or b = 0 then by definition, a × b = 0.

SSS Mathematics for Class 12 1014

1014

Senior Secondary School Mathematics for Class 12 ®

®

® ®

If a ^ b then a × b = a b cos \

®

®

®

®

®

p = 0. 2 ®

® ®

a = 0 or b = 0 or a ^ b Þ a × b = 0. ® ®

®

®

®

®

®

®

Hence, a × b = 0 Û a = 0 or b = 0 or a ^ b . ®

THEOREM 3

®

For any two vectors a and b , prove that: ®

®

® ®

®

®

(i) a × ( - b ) = - ( a × b ) = ( - a ) × b ®

®

® ®

(ii) ( - a ) × ( - b ) = a × b ¾®

PROOF

®

¾®

®

Let OA = a and OB = b . Produce AO and BO to A ¢ and B¢ respectively, such that OA ¢ = OA and OB ¢ = OB. ®

¾®

Let ÐAOB = q. Then, ÐAOB ¢ = p - q. (i)

®

¾®

OA ¢ = - a and OB ¢ = - b .

Then,

®

®

®

®

a × ( - b ) = | a || - b |cos( p - q) ®

®

= | a || b |× ( - cos q) ® ®

= - a b cos q = - ( a × b ). \

®

®

® ®

a × ( - b ) = - ( a × b ). ®

®

® ®

Similarly, ( - a ) × b = - ( a × b ). Hence,

®

®

®

®

® ®

a × ( - b ) = ( - a ) × b = - ( a × b ).

®

®

®

®

(ii) ( - a ) × ( - b ) = | - a || - b |cos ÐA ¢OB ¢ ®

®

= | a || b |cos q [Q Ð A ¢OB ¢ = q] ® ®

= a b cos q = a × b . ®

®

® ®

Hence, ( - a ) × ( - b ) = ( a × b ). THEOREM 4

Prove that

(Distributive Law)

¾®

PROOF

® ¾®

®

®

®

® ®

® ®

a × ( b + c ) = a × b + a × c. ®

¾®

®

Let O be the origin, OA = a , OB = b and BC = c . Then, ¾®

¾®

¾®

®

®

OC = ( OB + BC ) = ( b + c ). Draw BM ^ OA and CN ^ OA. Then, ®

®

®

® ¾®

®

a × ( b + c ) = a × OC = | a |{(OC) cos q } , where ÐAOC = q = a ´ ON [Q (OC) cos q = ON ]

SSS Mathematics for Class 12 1015

Scalar, or Dot, Product of Vectors

1015

= a(OM + MN ) = a(OM) + a( MN ) ®

®

= a( projection of b on a )

®

®

+ a( projection of c on a ) ® ®

® ®

= ( a × b + a × c ). ®

®

®

® ®

® ®

Hence, a × ( b + c ) = ( a × b + a × c ). THEOREM 5

®

PROOF

®

®

®

®

®

®

®

Prove that | a × b | £ | a || b |.

(Cauch Schwartz Inequality)

®

®

®

®

When a = 0 or b = 0 , then| a × b | = 0 = | a || b |. ®

®

®

®

So, let us assume that a ¹ 0 and b ¹ 0. Then, ® ®

®

®

® ®

®

®

( a × b ) = | a || b |cos q Þ | a × b | = | a || b ||cos q| ® ®

Þ

|a × b| ®

®

® ®

®

®

= |cos q| £ 1 Þ | a × b | £ | a || b |.

| a || b | ® ®

®

®

Hence,| a × b | £ | a || b |. ®

THEOREM 6

®

PROOF

®

®

®

®

Prove that | a + b | £ | a | + | b |.

(Triangle Inequality) ®

When a = 0, then| a | = 0. ®

®

®

®

®

®

®

®

®

\ | a + b | = | 0 + b | = | b |and| a | + | b | = 0 + | b | = | b |. ®

®

®

®

®

®

So, in this case,| a + b | = | a | + | b |. ®

®

®

®

Similarly, when b = 0, we have| a + b | = | a | + | b |. ®

®

®

®

Let us consider the case when a ¹ 0 and b ¹ 0. ®

®

®

®

Now,| a + b |2 = ( a + b ) 2 ®

®

®

®

=( a + b ) ×( a + b ) ® ®

® ®

® ®

® ®

= a×a + a×b + b×a + b×b ®

® ®

®

®

® ®

®

®

®

® ®

® ®

= | a |2 + 2( a × b ) + | b |2 [Q b × a = a × b ] £ | a |2 + 2| a × b | + | b |2 [Q a £ | a |" a Î R] ®

®

®

®

£ | a |2 + 2| a |×| b| + | b |2 = {| a | + | b |} 2. ® ®

®

®

× b |] [Q | a × b | £ | a || ®

®

®

®

\ | a + b | £ | a | + | b |.

SSS Mathematics for Class 12 1016

1016

Senior Secondary School Mathematics for Class 12 ORTHONORMAL VECTOR TRIAD

^

^

^

Let i , j , k be unit vectors along three mutually perpendicular coordinate axes namely, the x-axis, y-axis and z-axis respectively. Then, these vectors are said to form an orthonormal triad of vectors. Clearly, we have: ^ ^

^

^

(i) i × i = | i || i |cos 0° = 1. ^ ^ ^ ^

^ ^

k × k = 1.

^ ^

i × i = j × j = k × k = 1.

Thus, ^ ^

^ ^

j × j = 1 and

Similarly,

^

p = 0. 2

^

(ii) i × j = | i || j |cos ^ ^

^ ^

^ ^

^ ^

^ ^

i × k = 0, j × k = 0, j × i = 0, k × i = 0 and k × j = 0.

Similarly, ^ ^

^ ^

^ ^

^ ^

^ ^

^ ^

Hence, i × j = i × k = j × i = j × k = k × i = k × j = 0. SUMMARY

^ ^ ^

i , j , k are unit vectors such that: ^ ^

^ ^

^ ^

^ ^

^ ^

(i) i × i = j × j = k × k = 1. ^ ^

^ ^

^

^

^ ^

^ ^

(ii) i × j = j × i = 0, j × k = k × j = 0 and i × k = k × i = 0. ®

THEOREM 7

^

®

^

^

^

^

^

^

If a = a1 i + a2 j + a 3 k and b = b1 i + b2 j + b 3 k the prove that ® ®

a × b = ( a1b1 + a2b2 + a 3b 3).

PROOF

We have ® ®

^

^

^

a × b = ( a1 i + a2 j + a 3 k ) × ( b1 i + b2 j + b 3 k ) ^

^

^

^

^

^

^

^

^

^

= a1 i × ( b1 i + b2 j + b 3 k ) + a2 j × ( b1 i + b2 j + b 3 k ) ^

^

+ a 3 k × ( b1 i + b2 j + b 3 k ) ^ ^

^ ^

^ ^

= ( a1b1)( i × i ) + ( a1b2)( i × j ) + ( a1b 3)( i × k ) ^ ^

^ ^

^ ^

^ ^

^ ^

+ ( a2b1)( j × i ) + ( a2b2)( j × j ) + ( a2b 3)( j × k ) ^ ^

+ ( a 3b1)( k × i ) + ( a 3b2)( k × j ) + ( a 3b 3)( k × k ) = ( a1b1 + a2b2 + a 3b 3) ^ ^

^ ^

^ ^

^

^

^ ^

^ ^

^ ^

[Q i × i = j × j = k × k = 1, i × j = i × k = j × i = ... = 0]. ^

^

^

^

Hence, ( a1 i + a2 j + a 3 k ) × ( b1 i + b2 j + b 3 k ) = ( a1b1 + a2b2 + a 3b 3).

SSS Mathematics for Class 12 1017

Scalar, or Dot, Product of Vectors

1017

CONDITION OF PERPENDICULARITY ®

^

^

®

^

^

^

^

Let a = a1 i + a2 j + a 3 k and b = b1 i + b2 j + b 3 k . ®

®

®

®

® ®

Then, a ^ b Û a × b = 0 Û

a1b1 + a2b2 + a 3b 3 = 0.

Thus, a ^ b Û a1b1 + a2b2 + a 3b 3 = 0. ANGLE BETWEEN TWO VECTORS ®

^

^

®

^

^

^

^

Let a = a1 i + a2 j + a 3 k and b = b1 i + b2 j + b 3 k , and let q be the angle between them. Then, ® ®

®

®

® ®

a × b =| a || b |cos q Û a1b1 + a2b2 + a 3b 3 = |a||b|cos q

Û cos q =

a1b1 + a2b2 + a 3b 3 ®

®

| a || b | ì ï a1b1 + a2b2 + a 3b 3 Û q = cos- 1 í 2 æ ï ç a1 + a22 + a 23 ö÷ æç b12 + b22 + b 23 ö÷ ø øè îè

ü ï ý× ï þ

SOLVED EXAMPLES ^

EXAMPLE 1

^

^

Find the projection of the vector ( i + 3 j + 7 k ) on the vector ^

^

^

( 2 i - 3 j + 6 k). ®

SOLUTION

^

[CBSE 2014] ®

^

^

^

^

^

Let a = ( i + 3 j + 7 k ) and b = ( 2 i - 3 j + 6 k ). Then, ®

® ®

®

projection of a on b =

( a × b) ®

|b| ^

^

^

^

^

^

®

^

( i + 3 j + 7 k) ×(2 i - 3 j + 6 k ) = 4 + 9 + 36 ( 2 - 9 + 42) 35 = = = 5. 7 49 ®

EXAMPLE 2

®

®

®

^

^

^

®

^

®

^

^

^

b = ( i + 2 j - 2 k ) and c = ( 2 i - j + 4 k ).

SOLUTION

^

^

Write the projection of ( b + c ) on a , where a = ( 2 i - 2 j + k ),

We have ®

^

^

^

^

[CBSE 2013C]

^

( b + c ) = ( i + 2 j - 2 k ) + ( 2 i - j + 4 k) ^

^

^

^ ^

^

= (1 + 2) i + ( 2 - 1) j + ( -2 + 4) k = ( 3 i + j + 2 k ).

SSS Mathematics for Class 12 1018

1018

Senior Secondary School Mathematics for Class 12

\

®

®

®

®

projection of ( b + c ) on a =

®

®

(b + c)× a ®

| a| ^

=

^

^

^

( 6 - 2 + 2) 6 = = 2. 3 9 ^

^

®

^

^

^

^

Find l when the projection of a = l i + j + 4 k on b = ( 2 i + 6 j + 3 k ) is 4 units. ®

SOLUTION

^

22 + ( -2) 2 + 12

®

EXAMPLE 3

^

( 3 i + j + 2 k) ×(2 i - 2 j + k)

=

[CBSE 2012]

® ®

®

Projection of a on b =

( a × b) ®

|b| ^

=

^

^

^

^

^

(l i + j + 4 k) ×(2 i + 6 j + 3 k)

22 + 62 + 3 2 ( 2l + 6 + 12) ( 2l + 18) 2( l + 9) = = = × 7 4 + 36 + 9 49

2( l + 9) = 4 Þ 2( l + 9) = 28 Þ l + 9 = 14 Þ l = 5. 7 Hence, l = 5.

\

®

EXAMPLE 4

^

^

^

^

®

® ®

b = i - 2 j + 3 k are perpendicular to each other. ®

SOLUTION

^

^

Write the value of l so that the vectors a = 2 i + l j + k and

®

[CBSE 2013C]

a^b Û a × b = 0 ^

^

^

^

^

^

Û ( 2 i + l j + k) × ( i - 2 j + 3 k) = 0 Û ( 2 - 2l + 3) = 0 Û 2l = 5 Û l = Hence, l =

5 × 2 ^

EXAMPLE 5

5 × 2

^

^

The scalar product of the vector ( i + j + k ) with the unit vector along ^

^

^

^

^

^

the sum of the vectors ( 2 i + 4 j - 5 k ) and ( l i + 2 j + 3 k ) is equal to 1. Find the value of l. SOLUTION

Let

®

^

^

^

[CBSE 2009, ’14] ®

^

^

^

^

^

a = ( i + j + k ), b = ( 2 i + 4 j - 5 k ) and

®

^

^

^

c = ( l i + 2 j + 3 k ).

Then, ®

®

^

^

^

^

^

^

^

( b + c ) = ( 2 i + 4 j - 5 k ) + ( l i + 2 j + 3 k ) = ( 2 + l) i + 6 j - 2 k .

SSS Mathematics for Class 12 1019

Scalar, or Dot, Product of Vectors ®

®

Unit vector along ( b + c ) =

®

®

®

®

(b + c)

1019 ^

=

|b + c |

^

( 2 + l) 2 + 62 + ( -2) 2 ^

=

But,

®

®

®

®

(b + c)

^

( 2 + l) i + 6 j - 2 k ^

^

( 2 + l) i + 6 j - 2 k l2 + 4l + 44

×

®

× a = 1 (given).

|b + c | ^

\

^

^

( 2 + l) i + 6 j - 2 k 2

l + 4l + 44 ^

^

^

^

^

×( i + j + k) = 1

^

^

^

^

Þ

( 2 + l) i + 6 j - 2 k ) × ( i + j + k ) = l2 + 4l + 44

Þ

( 2 + l) + 6 - 2 = l2 + 4l + 44 Þ ( 6 + l) = l2 + 4l + 44

Þ l2 + 4l + 44 = ( 6 + l) 2 Þ l2 + 4l + 44 = 36 + l2 + 12l Þ 8l = 8 Þ l = 1. Hence, the required value of l is 1. ^

EXAMPLE 6

^

^

^

^

^

^

( i + j + k ) are respectively 4, 0 and 2. Find the vector. ^

SOLUTION

^

^

Dot products of a vector with the vectors ( i - j + k ), ( 2 i + j - 3 k ) and ^

[CBSE 2013C]

^

Let the required vector be ( x i + y j + z k ). Then, ^

^

^

^

^

^

^

^

^

^

^

^

( x i + y j + z k ) × ( i - j + k) = 4 Þ x - y + z = 4 ^

^

... (i)

^

( x i + y j + z k ) × ( 2 i + j - 3 k ) = 0 Þ 2x + y - 3z = 0. ^

^

... (ii)

^

( x i + y j + z k ) × ( i + j + k ) = 2 Þ x + y + z = 2. On subtracting (i) from (iii) we get 2y = -2 Þ y = -1. On adding (i) and (ii), we get 3 x - 2z = 4 On adding (i) and (iii), we get 2x + 2z = 6 On solving (iv) and (v), we get x = 2 and z = 1. \ x = 2, y = -1 and z = 1. ^ ^ ^ Hence, the required vector is ( 2 i - j + k ). ®

EXAMPLE 7

^

^ ®

^

^

^

^

®

^

... (iii) ... (iv) ... (v)

^

^

Let a = i + 4 j + 2 k , b = 3 i - 2 j + 7 k and c = 2 i - j + 4 k . Find a ®

®

®

® ®

vector p which is perpendicualr to both a and b and p × c = 18. ®

SOLUTION

^

^

[CBSE 2012]

^

Let p = ( x i + y j + z k ). Then, ®

®

®

®

® ®

p ^ a , p ^ b and p × c = 18

Þ

® ®

® ®

® ®

p × a = 0, p × b = 0 and p × c = 18

SSS Mathematics for Class 12 1020

1020

Senior Secondary School Mathematics for Class 12 ^

^

^

^

^

^

^

^

^

^

^

^

^

^

( x i + y j + z k ) × ( i + 4 j + 2 k ) = 0 Þ x + 4y + 2z = 0 Þ

^

... (i)

^

( x i + y j + z k ) × ( 3 i - 2 j + 7 k ) = 0 Þ 3 x - 2y + 7z = 0... (ii) ^

^

( x i + y j + z k ) × ( 2 i - j + 4 k ) = 18 Þ 2x - y + 4z = 18 ... (iii) On solving (i) and (ii) by cross multiplication, we get x y z = = = k (say) ( 28 + 4) ( 6 - 7) ( -2 - 12) Þ x = 32k , y = - k and z = -14k. Substituting these values in (iii), we get: 64k + k - 56k = 18 Þ 9k = 18 Þ k = 2. \ x = ( 32 ´ 2) = 64, y = -2 and z = ( -14) ´ 2 = -28. ^

^

^

Hence, the required vector is ( 64 i - 2 j - 28 k ). EXAMPLE 8

Find a vector whose magnitude is 3 units and which is perpendicular to ®

^

^

^

®

^

^

^

each of the vectors a = 3 i + j - 4 k and b = 6 i + 5 j - 2 k . [CBSE 2000C] ®

SOLUTION

^

^

^

Let the required vector be c = c1 i + c2 j + c 3 k. ®

c12 + c22 + c 32 = 3 Û c12 + c22 + c 32 = 9. … (i)

Then,| c | = 3 Û ®

®

® ®

Also, c ^ a Þ c × a = 0 ^

^

^

^

^

^

Þ ( c1 i + c2 j + c 3 k ) × ( 3 i + j - 4 k ) = 0 Þ 3 c1 + c2 - 4 c 3 = 0. ®

®

… (ii)

® ®

And, c ^ b Þ c × b = 0 ^

^

^

^

^

^

Þ ( c1 i + c2 j + c 3 k ) × ( 6 i + 5 j - 2 k ) = 0 Þ 6c1 + 5 c2 - 2c 3 = 0. From (ii) and (iii), by cross multiplication, we get c1 c2 c3 = = = k (say) ( -2 + 20) ( -24 + 6) (15 - 6) c1 c c c c c = 2 = 3 =k Þ 1 = 2 = 3 =k Þ 18 -18 9 2 -2 1 Þ c1 = 2 k , c2 = - 2 k and c 3 = k. Substituting these values in (i), we get 4k \

2

+ 4k

2

+k

2

=9 Þ k

2

= 1 Þ k = 1.

c1 = 2, c2 = - 2 and c 3 = 1. ®

^

^

^

Hence, c = ( 2 i - 2 j + k ) is the required vector.

… (iii)

SSS Mathematics for Class 12 1021

Scalar, or Dot, Product of Vectors ®

EXAMPLE 9

®

®

®

® ®

® ®

1021 ® ®

If a and b are vectors such that |a| = 2,|b| = 3 and a × b = 4, find ®

®

|a - b|. SOLUTION

We have ®

®

®

®

®

®

| a - b |2 = ( a - b ) × ( a - b ) ® ®

® ®

= a×a - a×b - b×a + b×b ®

® ®

(by distributive law)

®

® ®

= | a |2 - 2 a × b + | b |2

® ®

[Q a × b = b × a ]

= ( 22 - 2 ´ 4 + 3 2) = 5. ®

®

Hence,| a - b | = 5. ®

EXAMPLE 10

If a makes equal angles with the coordinate axes and has magnitude 3, ®

find the angle between a and each of the three coordinate axes. ®

SOLUTION

^

^

®

^

Let a = a1 i + a2 j + a 3 k and let a be the angle between a and each of the coordinate axes. ®

^ ^

^

Then, a is the angle between a and each one of i , j and k . ® ^

\

a×i

cos a =

® ^

=

|a||i |

® ^ ® ^ a1 Þ a1 = 3 cos a [Q a × i = a1 ,|a| = 3 ,|i | = 1]. 3

Similarly, a2 = 3 cosa and a 3 = 3 cosa. ®

®

Now, |a|= 3 Þ |a|2 = 9 Þ a12 + a22 + a 32 = 9 Þ 9 cos2 a + 9 cos2 a + 9 cos2 a = 9 1 Þ 27 cos2 a = 9 Þ cos2 a = 3 1 -1 æ 1 ö Þ cos a = Þ a = cos ç ÷ 3 è 3ø æ 1 ö Hence, the required angle is cos-1 ç ÷× è 3ø ®

EXAMPLE 11

^

^

If a unit vector a makes angles p with i , p with j and an acute angle 4 3 ^

q with k then find the value of q. Also, find the scalar and vector ®

components of a along the axes. ®

SOLUTION

^

^

^

[CBSE 2013] ®

Let a = ( a1 i + a2 j + a 3 k ). Then, a being a unit vector, we have ( a12 + a22 + a 32) = 1.

SSS Mathematics for Class 12 1022

1022

Senior Secondary School Mathematics for Class 12 ® ^

®

® ^ p 1 [Q |a|= 1,| i | = 1] = a1 Þ a1 = 4 2 ® ^ ® ^ ® ^ p 1 [Q |a|= 1,|j | = 1] a × j = a2 Þ |a||j |cos = a2 Þ a2 = 3 2

^

Now, a × i = a1 Þ |a||i |cos

® ^

® ^

®

^

a × k = a 3 Þ |a||k |cos q = a 3 Þ a 3 = cos q [Q |a|= 1,|k | = 1] ®

®

Now,|a|= 1 Þ |a|2 = 1 Þ a12 + a22 + a 32 = 1 1 1 1 1 Þ + + cos2 q = 1 [Q a1 = , a2 = , a 3 = cosq ] 2 4 2 2 1 1 p 2 Þ cos q = Þ cos q = Þ q = × 4 2 3 ® 1 1 1 Hence, the scalar componets of a are , and × 2 2 2 ®

And, the vector components of a are ®

EXAMPLE 12

®

1 ^ 1^ 1^ i , j and k× 2 2 2 ®

®

If a and b are two unit vectors such that a + b is also a unit vector, ®

®

then find the angle between a and b .

[CBSE 2014] ®

SOLUTION

®

Let q be the angle between the unit vectors a and b . ®

®

®

®

Since a and b are unit vectors, we have|a|= 1 and|b|= 1. ®

®

Again, since ( a + b ) is a unit vector, we have ®

®

®

®

|a + b |2 = 1 Þ

®

®

® ®

® ®

® ®

® ®

( a + b ) ×( a + b ) = 1 Þ a × a + a × b + b × a + b × b = 1 ®

® ®

®

® ®

Þ |a|2 + 2 ( a × b ) + |b|2 = 1

® ® -1 [Q |a|2 = 1,|b|2 = 1] 2 ® ® ® ® -1 -1 2p [Q |a|= 1,|b|= 1]. Þ cosq = Þ q= Þ |a||b|cos q = 2 2 3 ® ® 2p Hence, the angle between a and b is × 3

Þ

EXAMPLE 13

® ®

® ®

[Q b × a = a × b ]

® ®

2 ( a × b ) = -1 Þ a × b =

^

magnitude of their difference is ^

SOLUTION

^

If the sum of two unit vectors a and b is a unit vector, show that the ^

^

^

3.

Let a + b = c , where c is a unit vector. Then,

[CBSE 2012C]

SSS Mathematics for Class 12 1023

Scalar, or Dot, Product of Vectors ^

^

^

^

^

^

^

1023

^ ^

a + b = c Þ ( a + b) × ( a + b ) = c × c ^ ^

^ ^

^ ^

^ ^

^ ^

Þ a× a + a× b + b× a + b×b = c× c ^

^ ^

^

^

^ ^

Þ |a |2 + 2( a × b ) + |b|2 =|c |2

^ ^

[Q b × a = a × b]

^ ^

^

^

^

[Q | a |=|b |=| c |= 1]

Þ 1 + 2( a × b ) + 1 = 1 ^ ^

Þ 2( a × b ) = -1 ^

^ 2

^

^

... (i) ^

^

Now,|a - b | = ( a - b ) × ( a - b ) ^ ^

^ ^

^ ^

^ ^

= a ×a - a × b - b × a + b ×b ^

^ ^

^

^ ^

= |a |2 - 2( a × b) + |b |2 ^ ^

= 1 - 2( a × b ) + 1 = 1 + 1 + 1 = 3 ^

^

Hence,|a - b|= ® ®

EXAMPLE 14

[using (i)].

3.

®

®

®

®

If a , b and c are three vectors such that|a| = 5 , |b| = 12,| c| = 13 and

®

®

®

®

®

®

® ®

® ®

® ®

® ®

a + b + c = 0, find the value of ( a × b + b × c + c × a ).

SOLUTION

^ ^

[Q b × a = a × b]

®

®

[CBSE 2012]

®

a + b + c=0 Þ a + b =-c ®

®

®

®

®

® ®

® ®

®

Þ ( a + b ) × c = ( - c ) × c Þ a × c + b × c = -| c|2 ® ®

® ®

Þ c × a + b × c = -169

... (i) ® ®

® ®

®

[Q a × c = c × a and | c|= 13] ®

®

®

®

®

®

®

Again, a + b + c = 0 Þ b + c = - a ®

®

®

®

®

Þ ( b + c ) × a = (- a ) × a ® ®

® ®

® ®

® ®

®

Þ b × a + c × a = -|a|2 Þ a × b + c × a = -25

... (ii)

® ®

® ®

®

[Q b × a = a × b and |a|= 5] ®

®

®

®

®

®

®

®

®

Also, a + b + c = 0 Þ a + c = - b ®

®

®

Þ ( a + c ) × b = (- b ) × b ® ®

® ®

® ®

® ®

®

Þ a × b + c × b = -|b|2 Þ a × b + b × c = -144 ® ®

... (iii) ® ®

®

[Q c × b = b × c and |b|= 12]

SSS Mathematics for Class 12 1024

1024

Senior Secondary School Mathematics for Class 12

Adding the corresponding sides of (i), (ii) and (iii), we get ® ®

® ®

® ®

2( a × b + b × c + c × a ) = - (169 + 25 + 144) ® ®

® ®

® ®

(a × b + b × c + c × a) =

Þ

® ®

® ®

-338 = -169. 2

® ®

Hence, ( a × b + b × c + c × a ) = -169. ® ® ®

EXAMPLE 15

®

®

®

If a , b , c are three vectors such that |a| = 3 ,|b| = 4 and | c| = 5 and each one of them is perpendicular to the sum of the two then find ®

®

®

|a + b + c|.

[CBSE 2011C]

® ® ®

SOLUTION

Let a , b , c be the given vectors such that ®

®

®

{|a| = 3 ,|b|= 4,| c|= 5}, ®

®

... (i)

®

ü a × ( b + c) = 0 ï ® ® ® ï b × ( c + a) = 0 ý× ® ® ® ï c × ( a + b) = 0 ï þ \

®

®

®

... (ii)

®

®

®

®

®

®

|a + b + c|2 = ( a + b + c ) × ( a + b + c ) ® ®

®

®

®

®

®

®

® ®

= a × a + a × ( b + c) + b × ( c + a) + b × b ®

®

®

® ®

+ c × ( a + b) + c × c ®

®

®

= |a|2 + |b|2 + | c|2 2

2

[using (ii)]

2

= ( 3 + 4 + 5 ) = ( 9 + 16 + 25) = 50 ®

®

®

Hence,|a + b + c| = 50 = 5 2. ® ®

EXAMPLE 16

®

If a , b and c are three mutually perpendicular vectors of the same ®

®

®

®

magnitude, prove that ( a + b + c ) is equally inclined to the vectors a , ®

®

b and c . Also find this angle.

SOLUTION

[CBSE 2006C, ’11C, ’13C]

It is given that ®

®

®

|a| = |b| = | c| = a (say) ® ®

... (i)

®

Since a , b and c are mutually perpendicular vectors, we have ® ®

® ®

® ®

a × b = b × c = c × a = 0. ®

®

®

®

®

... (ii) ®

®

®

®

Now,| a + b + c |2 = ( a + b + c ) × ( a + b + c )

SSS Mathematics for Class 12 1025

Scalar, or Dot, Product of Vectors ® ®

® ®

® ®

® ®

1025

® ®

® ®

= a × a + b × b + c × c + 2( a × b + b × c + c × a ) ®

®

®

= |a|2 + |b|2 + | c|2 [using (ii)] = 3 a 2 [using (i)]. \

®

®

®

®

®

®

|a + b + c|=

3 a.

... (iii) ®

®

®

Let ( a + b + c ) make angles a , b and g with a , b and c respectively. ®

®

®

®

®

®

® ®

Then ( a + b + c ) × a = |a + b + c||a|cos a = ( 3 a ´ a ´ cos a ) = Þ

® ®

® ®

® ®

3 a 2 cos a

( a × a + b × a + c × a) = ®

Þ |a|2 =

3 a 2 cos a

3 a 2 cos a Þ a 2 =

3 a 2 cos a

1 æ 1 ö Þ a = cos-1 ç ÷× 3 è 3ø æ 1 ö -1 æ 1 ö Similarly, b = cos-1 ç ÷ and g = cos ç ÷× è 3ø è 3ø æ 1 ö \ a = b = g = cos-1 ç ÷. è 3ø Þ

cos a =

®

®

®

®

®

®

Hence, ( a + b + c ) is equally inclined to a , b and c and the æ 1 ö required angle is cos-1 ç ÷× è 3ø ® ® ®

EXAMPLE 17

®

®

®

®

If a , b , c are unit vectors such that a + b + c = 0 then find the value ® ®

® ®

® ®

of ( a × b + b × c + c × a ). ® ® ®

SOLUTION

Since a , b , c are unit vectors, we have ®

®

®

| a | = 1,| b | = 1 and| c | = 1.

®

®

®

®

Now, a + b + c = 0 Þ Þ

®

®

®

®

®

®

® ®

(a + b + c) ×( a + b + c) = 0 ® ®

® ®

® ®

® ®

[Q 0 × 0 = 0] ® ®

® ®

a × a + b × b + c × c + 2( a × b + b × c + c × a ) = 0 ®

®

®

® ®

® ®

® ®

® ®

® ®

® ®

Þ | a |2 + | b |2 + | c |2 + 2( a × b + b × c + c × a ) = 0 Þ

( a × b + b × c + c × a) = ® ®

® ®

® ®

3 2

Hence, ( a × b + b × c + c × a ) = -

®

®

®

[Q | a |2 = 1,| b |2 = 1, | c |2 = 1]. 3 × 2

SSS Mathematics for Class 12 1026

1026

Senior Secondary School Mathematics for Class 12 ® ®

EXAMPLE 18

® ®

® ® SOLUTION

®

®

®

®

®

®

®

®

®

If a × b = a × c , show that a = 0 or b = c or a ^ ( b - c ). ® ®

® ®

® ®

®

a × b = a × c Þ a × b - a × c = 0 Þ a ×( b - c ) = 0 ®

®

®

®

®

®

®

®

®

®

Þ a = 0 or ( b - c ) = 0 or a ^ ( b - c ) ®

®

®

®

®

Þ a = 0 or b = c or a ^ ( b - c ). ® ®

® ®

®

®

®

®

®

®

®

Hence, a × b = a × c Þ a = 0 or b = c or a ^ ( b - c ). ®

EXAMPLE 19

®

Let a and b be two nonzero vectors. Prove that ®

®

®

®

®

®

a ^ b Û | a + b | = | a - b |.

®

SOLUTION

®

® ®

Let a ^ b . Then, ( a × b ) = 0. ®

®

®

®

… (i)

®

®

Now,| a + b |2 = ( a + b ) × ( a + b ) ® ®

® ®

® ®

® ®

= a×a + a×b + b×a + b×b ®

®

® ®

® ®

® ®

= | a |2 + | b |2 {Q a × b = 0 and b × a = a × b = 0}. ®

®

®

®

®

®

Also, | a - b |2 = ( a - b ) × ( a - b ) ® ®

® ®

® ®

® ®

= a×a - a×b - b×a + b×b ®

®

® ®

= | a |2 + | b |2 ®

®

®

® ®

® ®

{Q a × b = 0 and b × a = a × b = 0}.

®

®

®

®

®

Thus,| a + b |2 = | a - b |2 , and therefore,| a + b | = | a - b |. \

®

®

®

®

®

®

a ^ b Þ | a + b | = | a - b |. ®

®

®

®

Conversely, suppose that| a + b | = | a - b |. Then, ®

®

®

®

®

®

®

®

®

®

| a + b | = | a - b | Þ | a + b |2 = | a - b |2 ®

®

®

®

®

®

Þ ( a + b ) × ( a + b ) = ( a - b) × ( a - b ) ® ®

® ®

® ®

® ®

Þ a×a + a×b + b×a + b×b ® ®

® ®

® ®

® ®

= a×a - a×b - b×a + b×b ® ®

® ®

Þ 2( a × b + b × a ) = 0 ® ®

Þ 4( a × b ) = 0 ® ®

® ®

®

®

Þ a × b = 0 Þ a ^ b. ®

®

®

®

®

®

Thus,| a + b | = | a - b | Þ a ^ b . ®

®

®

®

®

®

Hence,| a + b | = | a - b | Û a ^ b .

® ®

b×a = a×b]

[Q

SSS Mathematics for Class 12 1027

Scalar, or Dot, Product of Vectors ®

EXAMPLE 20

^

^

1027

^

Express the vector a = (5 i - 2 j + 5 k ) as sum of two vectors such that ®

^

^

one is parallel to the vector b = ( 3 i + k ) and the other is perpendicular ®

to b .

[CBSE 2005] ®

SOLUTION

®

Any vector parallel to b is of the form l b for some scalar l. ®

®

®

®

®

®

®

®

®

Let a = l b + c , where c ^ b . Then, c = ( a - l b ) ^ b ®

®

®

® ®

® ®

Û ( a - l b ) × b = 0 Û ( a × b ) - l ( b × b) = 0 ^

^

^

^

^

^

^

^

^

Û (5 i - 2 j + 5 k ) × ( 3 i + k) - l ( 3 i + k) × ( 3 i + k) = 0 Û (15 - 0 + 5) - l ( 9 + 1) = 0 Û 10l = 20 Û l = 2. \

®

®

^

^

l b = 2 b = ( 6 i + 2 k ). ®

®

®

^

^

^

^

^

^

^

^

And, c = ( a - 2 b ) = (5 i - 2 j + 5 k ) - 2( 3 i + k ) = ( - i - 2 j + 3 k ). ^

^

^

^

^

Hence, the required vectors are ( 6 i + 2 k ) and ( - i - 2 j + 3 k ). EXAMPLE 21

Find the values of l for which the angle between the vectors

®

^

^

^

®

^

^

^

a = 2l2 i + 4l j + k and b = 7 i - 2 j + l k is obtuse. ®

SOLUTION

[CBSE 2013C]

®

Let q be the angle between a and b . Then, ® ®

cos q =

a×b

× |a||b|

... (i)

® ®

® ®

Clearly, q is obtuse Û cosq < 0 Û a × b < 0 [from (i)] Û 14l2 - 8l + l < 0 Û 14l2 - 7 l < 0 Û 2l2 - l < 0 Û l( 2l - 1) < 0. Now, l( 2l - 1) < 0 Þ either {l < 0 and ( 2l - 1) > 0} or {l > 0 and ( 2l - 1) < 0} 1ü 1ü ì ì Þ íl < 0 and l > ý or íl > 0 and l < ý 2 2þ þ î î 1ü ü ì ì1 Þ í < l < 0ý or í0 < l < ý 2þ þ î î2 1 é 1 ù Þ 0 < l < êQ < l < 0 is not possibleú 2 ë 2 û 1 Þ l Î ] 0, [. 2 \

the required values of l are all real values in ] 0,

1 [. 2

SSS Mathematics for Class 12 1028

1028

Senior Secondary School Mathematics for Class 12

EXAMPLE 22

Let A( 0, 1, 1), B( 3 , 1, 5) and C( 0, 3 , 3) be the vertices of a CABC. Using vectors, show that C ABC is right angled at C.

SOLUTION

In order to show that CA ^ CB , we must show that CA × CB = 0.

¾®

¾®

¾®

¾®

Let O be the origin. Then, ¾®

^

^

¾®

^

^

^

¾®

^

^

^

^

OA = 0 i + j + k , OB = 3 i + j + 5 k and OC = 0 i + 3 j + 3 k. ¾®

¾®

¾®

CA = ( OA - OC )

\

^

^

^

^

^

^

= ( 0 i + j + k ) - (0 i + 3 j + 3 k ) ^

^

^

= ( 0 - 0) i + (1 - 3) j + (1 - 3) k ^

^

^

= ( 0 i - 2 j - 2 k ). ¾®

\

¾®

¾®

CB = (OB - OC ) ^

^

^

^

^

^

= ( 3 i + j + 5 k ) - (0 i + 3 j + 3 k ) ^

^

^

^

^

= ( 3 i - 2 j + 2 k ). ¾®

¾®

^

^

^

^

CA × CB = ( 0 i - 2 j - 2 k ) × ( 3 i - 2 j + 2 k )

\

= [( 0 ´ 3) + ( -2) ´ ( -2) + ( -2) ´ 2] = 0

¾®

¾®

CA ^ CB .

Þ

Hence, CABC is right angled at C. EXAMPLE 23

^

^

^

Show that the points A, B and C having position vectors ( 2 i - j + k ), ^

^

^

^

^

^

( i - 3 j - 5 k ) and ( 3 i - 4 j - 4 k ) respectively are the vertices of a rightangled triangle. Also, find the remaining angles of the triangle. SOLUTION

We have ¾®

AB = (p.v. of B) – (p.v. of A) ^

^

^

= ( - i - 2 j - 6 k), ¾®

BC = (p.v. of C) – (p.v. of B) ^

^

^

= ( 2 i - j + k) and ¾®

^

^

^

CA = (p.v. of A) – (p.v. of C) = ( - i + 3 j + 5 k ). ¾®

¾®

¾®

®

Clearly, we have AB + BC + CA = 0 . \

A, B, and C are the vertices of a triangle. ¾®

¾®

^

^

^

^

^

^

Now, BC × CA = ( 2 i - j + k ) × ( - i + 3 j + 5 k ) = ( -2 - 3 + 5) = 0. \

¾®

¾®

BC ^ CA and therefore, ÐC = 90°.

SSS Mathematics for Class 12 1029

Scalar, or Dot, Product of Vectors ¾®

1029

¾®

Now, ÐA is the angle between AB and AC. \

¾®

¾®

¾®

¾®

¾®

¾®

¾®

( AB × AC )

¾®

AB × AC =|AB ||AC |cos A Þ cos A =

|AB ||AC | ^

=

^

^

^

^

^

(- i - 2 j - 6 k ) ×( i - 3 j - 5 k )

{

( -1) 2 + ( -2) 2 + ( -6) 2 } × { 12 + ( -3) 2 + ( -5) 2 } ¾®

¾®

[Q AC = - CA ] = Þ

( -1 + 6 + 30) = 41 ´ 35

35 = 41 ´ 35

35 × 41

A = cos-1

¾®

35 = 41

35 41

¾®

Further, ÐB is the angle between BC and BA. ¾®

¾®

¾®

¾®

\

BC × BA =|BC ||BA |cos B

Þ

cos B =

¾®

¾®

¾®

¾®

( BC × BA ) |BC ||BA | ^ ^

= =

^

^

^

^

(2 i - j + k) × ( i + 2 j + 6 k)

{

22 + ( -1) 2 + 12} × { 12 + 22 + 62}

( 2 - 2 + 6) = 6 ´ 41

6 = 6 ´ 41

6 = 41

6 × 41

Þ

B = cos-1

6 × 41

\

A = cos-1

35 6 , and C = 90°. , B = cos-1 41 41

^

^

^

^

^

^

^

^

^

^

^

EXAMPLE 24

Let ( i + j + k ), ( 2 i + 5 j ), ( 3 i + 2 j - 3 k ) and ( i - 6 j - k ) be the position vectors of points A, B, C, D respectively. Find the angle between AB and CD. Hence, show that AB | | CD. [CBSE 2008]

SOLUTION

Let the angle between AB and CD be q.

¾®

¾®

¾®

Now, AB = (p.v. of B) – (p.v. of A) ^

^

^

^

^

^

^

^

= (2 i + 5 j ) - ( i + j + k ) = ( i + 4 j - k) ¾®

and, CD = (p.v. of D) – (p.v. of C) ^

^

^

^

^

^

^

^

^

= ( i - 6 j - k ) - ( 3 i + 2 j - 3 k ) = ( -2 i - 8 j + 2 k ).

SSS Mathematics for Class 12 1030

1030

Senior Secondary School Mathematics for Class 12 ¾®

| AB | = 12 + 42 + ( -1) 2 = 18 = 3 2

\

¾®

and, |CD | = ( -2) 2 + ( -8) 2 + 22 = 72 = 6 2. ¾®

¾®

^

^

^

^

^

^

Now, AB × CD = ( i + 4 j - k ) × ( -2 i - 8 j + 2 k ) = ( -2 - 32 - 2) = - 36. ¾®

¾®

AB × CD

\ cos q =

¾®

=

¾®

| AB ||CD | Hence, AB | | CD.

-36 -36 = = - 1 Þ q = p. 36 ( 3 2 ´ 6 2)

EXERCISE 23 1.

® ®

Find a × b when ®

^

^

®

^

^

®

^

®

^

^

^

^

(i) a = i - 2 j + k and b = 3 i - 4 j - 2 k ®

^

^

^

(ii) a = i + 2 j + 3 k and b = - 2 j + 4 k ^

®

^

^

^

(iii) a = i - j + 5 k and b = 3 i - 2 k 2.

®

®

Find the value of l for which a and b are perpendicular, where ®

^

^

^

®

^

®

^

®

^

^

^

(i) a = 2 i + l j + k and b = ( i - 2 j + 3 k ) ®

^

®

^

^

^

^

[CBSE 2012C] ^

(ii) a = 3 i - j + 4 k and b = - l i + 3 j + 3 k ^

^

^

^

(iii) a = 2 i + 4 j - k and b = 3 i - 2 j + l k ®

^

^

®

^

^

[CBSE 2003C]

^

(iv) a = 3 i + 2 j - 5 k and b = - 5 j + l k 3.

®

^

^

®

^

^

^

^

®

®

(i) If a = i + 2 j - 3 k and b = 3 i - j + 2 k , show that ( a + b ) is ®

®

perpendicular to ( a - b ). ®

^

^

[CBSE 2002] ®

^

^

^

®

^

®

(ii) If a = (5 i - j - 3 k ) and b = ( i + 3 j - 5 k ) then show that ( a + b ) ®

®

and ( a - b ) are orthogonal. ®

^

^

^

®

^

[CBSE 2004] ^

^

4. If a = ( i - j + 7 k ) and b = (5 i - j + l k ) then find the value of l so that ®

®

® ®

( a + b ) and ( a - b ) are orthogonal vectors.

[CBSE 2013]

5. Show that the vectors ^ ^ ^ ^ ^ ^ ^ 1 1 ^ 1 ^ ( 2 i + 3 j + 6 k ), ( 3 i - 6 j + 2 k ) and ( 6 i + 2 j - 3 k ) 7 7 7 are mutually perpendicular unit vectors.

SSS Mathematics for Class 12 1031

Scalar, or Dot, Product of Vectors ®

^

^ ®

^

^

^

®

^

1031

^

^

^

6. Let a = 4 i + 5 j - k , b = i - 4 j + 5 k and c = 3 i + j - k . ®

®

®

Find a vector d which is perpendicular to both a and b , and is such that ® ®

d × c = 21. ®

^

^

®

^

^

^

^

7. Let a = ( 2 i + 3 j + 2 k ) and b = ( i + 2 j + k ). ®

®

®

®

Find the projection of (i) a on b and (ii) b on a . ^

^

^

^

^

8. Find the projection of ( 8 i + j ) in the direction of ( i + 2 j - 2 k ). ^

^

^

®

® ®

^

9. Write the projection of vector ( i + j + k ) along the vector j . ®

®

^

[CBSE 2014]

^

^

10. (i) Find the projection of a on b if a × b = 8 and b = ( 2 i + 6 j + 3 k ). [CBSE 2009] ^

^

^

^

(ii) Write the projection of the vector ( i + j ) on the vector ( i - j ). ®

[CBSE 2011]

®

11. Find the angle between the vectors a and b , when ®

^

^

^

®

^

^

^

®

^

^

^

^

^

^

(i) a = i - 2 j + 3 k and b = 3 i - 2 j + k ®

^

^

(ii) a = 3 i + j + 2 k and b = 2 i - 2 j + 4 k ®

^

®

^

^

^

(iii) a = i - j and b = j + k . ®

^

^

®

^

^

12. If a = ( i + 2 j - 3 k ) and b = ( 3 i - j + 2 k ) then calculate the angle ®

®

®

®

between ( 2 a + b ) and ( a + 2 b ). ®

®

®

®

®

®

13. If a is a unit vector such that ( x - a ) × ( x + a ) = 8, find|x|. ®

^

^

^

14. Find the angles which the vector a = 3 i - 6 j + 2 k makes with the coordinate axes. ®

^

^

^

15. Show that the vector a = ( i + j + k ) is equally inclined to the coordinate axes. ® 16. Find a vector a of magnitude 5 2, making an angle p/4 with x-axis, p/2 with y-axis and an acute angle q with z-axis. [CBSE 2014] ®

®

® ®

®

^

^

^

17. Find the angle between ( a + b ) and ( a - b ), if a = ( 2 i - j + 3 k ) and ®

^

^

^

b = ( 3 i + j + 2 k ).

[CBSE 2006] ®

^

^

^

18. Express the vector a = ( 6 i - 3 j - 6 k ) as sum of two vectors such that one ®

^

^

^

®

is parallel to the vector b = ( i + j + k ) and the other is perpendicular to b .

SSS Mathematics for Class 12 1032

1032

Senior Secondary School Mathematics for Class 12 ®

®

®

®

®

®

®

®

®

®

19. Prove that ( a + b ) × ( a - b ) = |a|2 + |b|2 Û a ^ b , where a ¹ 0 and ®

®

b ¹ 0. ®

®

®

® ®

®

®

®

®

20. If a + b + c = 0 ,|a|= 3 ,|b|= 5 and| c|= 7 , find the angle between a and b . ®

[CBSE 2008]

®

21. Find the angle between a and b , when ®

®

® ®

(i) |a|= 2,|b|= 1 and a × b = ®

®

3

® ®

®

®

® ®

(ii) |a|= |b|= 2 and a × b = -1. ®

®

22. If|a| = 2,|b| = 3 and a × b = 4, find|a - b |. ®

®

®

®

®

®

®

®

23. If ( a + b ) × ( a - b ) = 8 and|a| = 8|b|, find|a|and|b|. ^

^

24. If a and b are unit vectors inclined at an angle q then prove that: (i) cos

q 1 ^ ^ = |a + b | 2 2

^

(ii) tan

^

q |a - b | = ^ ^ 2 |a + b | ^

^

^

^

^

^

25. The dot products of a vector with the vectors ( i + j - 3 k ), ( i + 3 j - 2 k ) ^

^

^

and ( 2 i + j + 4 k ) are 0, 5 and 8 respectively. Find the vector. [CBSE 2003] ¾®

^

^

^

26. If AB = ( 3 i - j + 2 k ) and the coordinates of A are (0, –2, –1), find the coordinates of B. ¾®

27. If A(2, 3, 4), B(5, 4, –1), C(3, 6, 2) and D(1, 2, 0) be four points, show that AB ¾®

is perpendicular to CD . ^

^

^

28. Find the value of l for which the vectors ( 2 i + l j + 3 k ) and ^

^

^

( 3 i + 2 j - 4 k ) are perpendicular to each other. 29. Show ®

that ^

the

^

vectors

®

^

^

^

a = ( 3 i - 2 j + k ),

[CBSE 2010] ®

^

^

^

b = ( i - 3 j + 5 k)

^

c = ( 2 i + j - 4 k ) form a right-angled triangle.

and

[CBSE 2005]

30. Three vertices of a triangle are A( 0, - 1, - 2), B( 3 , 1, 4) and C(5 , 7 , 1). Show that it is a right-angled triangle. Also, find its other two angles. 31. If the position vectors of the vertices A, B and C of a C ABC be (1, 2, 3), (–1, 0, 0) and (0, 1, 2) respectively then find ÐABC. 32. If

®

a and

®

®

®

®

®

b are two unit vectors such that |a + b|= ®

®

( 2 a - 5 b ) × ( 3 a + b ).

3 , find

SSS Mathematics for Class 12 1033

Scalar, or Dot, Product of Vectors ®

®

®

®

1033

®

33. If a and b are two vectors such that |a + b |=|a| then prove that vector ®

®

®

( 2 a + b ) is perpendicular to vector b . ®

^

®

^

^

[CBSE 2013]

^

^

®

®

®

34. If a = ( 3 i - j ) and b = ( 2 i + j - 3 k ) then express b in the form ®

®

®

®

®

b = ( b 1 + b 2 ), where b 1|| a and b 2 ^ a .

[CBSE 2013C]

ANSWERS (EXERCISE 23)

1. (i) 9

(ii) 8

^

4. l = ±5 8.

10 3

10. (i)

^

p 6

17.

^

^

p 2

2p 3

(ii)

8 (ii) 0 7

æ -6 ö æ 3ö æ 2ö 14. cos-1 ç ÷ , cos-1 ç ÷ , cos-1 ç ÷ è7 ø è7ø è7 ø

13. |x|= 3

16. 5( i + k ) 21. (i)

^

®

æ 31 ö 12. cos-1 ç ÷ è 50 ø ^

^

6. d = 7( i - j - k )

9. 1

22.

^

25. ( i + 2 j + k )

5 (ii) l = 3 (iii) l = -2 (iv) l = -2 2 5 6 10 17 7. (i) (ii) 3 17 æ 3ö 5 ö æ ÷ (iii) 120° 11. (i) cos-1 ç ÷ (ii) cos-1 çç ÷ 7 è ø è 7ø

2. (i) l =

(iii) –7

®

^

^

^

^

^

^

® 8 and |b|= 63

8 63

18. a = - ( i + j + k ) + (7 i - 2 j - 5 k ) 5

®

23. |a| = 8

26. B(3, –3, 1)

20.

28. l = 3

-11 æ 10 ö 32. 31. cos-1 ç ÷ 2 è 102 ø ® ® ® ® ® ^ö æ 3 ^ 1 ^ö æ1 ^ 3 ^ 34. b = ( b 1 + b 2), where b 1 = çç i - j ÷÷ and b2 = çç i + j - 3 k ÷÷ 2 ø 2 è2 è2 ø 30. ÐA = 45 ° , ÐB = 90° , ÐC = 45 °

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 23) ® ® ®

5. Let the given vectors be a , b , c respectively. ®

®

®

® ®

® ®

® ®

Then, show that| a | =| b | =| c | = 1, and a × b = b × c = c × a = 0. ®

^

^

^

6. Let d = d1 i + d2 j + d 3 k . Then, ® ®

® ®

® ®

d × a = 0 , d × b = 0 and d × c = 21

Þ Þ

ì 4 d1 + 5 d2 - d 3 = 0 ü ý and 3 d1 + d2 - d 3 = 21 í î d1 - 4 d2 + 5 d 3 = 0 þ d1 d2 d3 = = = k (say) and 3 d1 + d2 - d 3 = 21 ( 25 - 4 ) ( -1 - 20 ) ( -16 - 5 )

p 3

SSS Mathematics for Class 12 1034

1034

Senior Secondary School Mathematics for Class 12

Þ

( d1 = 21 k , d2 = - 21 k , d 3 = - 21 k ) and 3 d1 + d2 - d 3 = 21 1 63 k - 21 k + 21 k = 21 Þ k = Þ d1 = 7 , d2 = - 7 and d 3 = - 7 . 3

Þ

11. (iii) Let the required angle be q. Then, ® ®

a × b = ( 1 ´ 0 ) + ( -1) ´ 1 + ( 0 ´ 1) = - 1 ®

®

Þ | a || b |cos q = - 1 Þ ( 2 )( 2 ) cos q = - 1 Þ cos q = ®

®

®

®

®

®

®

1 Þ q = 120 °. 2

®

12. ( 2 a + b ) × ( a + 2 b ) =|2 a + b |×| a + 2 b |cos q . Find q. ®

®

®

®

® ®

® ®

® ®

13. ( x - a ) × ( x + a ) = 8 Þ x × x - a × a = 8 [Q ®

®

® ®

x× a = a×x ]

®

Þ | x |2 -| a |2 = 8 Þ | x |2 = ( 8 + 1) = 9 Þ | x| = 3. ® ^

®

^

3 × 7

14. a × i =| a|| i |cos a Þ 7 ´ 1 ´ cos a = 3 Þ cos a = ® ^

®

^

® ^

®

^

® ^

®

a × j =| a|| j |cos b Þ 7 ´ 1 ´ cos b = -6 Þ cos b =

a × k =| a|| k |cos g Þ 7 ´ 1 ´ cos g = 2 Þ cos g = ^

15. a × i =| a||i|cos a Þ

-6 × 7

2 × 7

3 ´ 1 ´ cos a = 1 Þ cos a =

1 × 3

1 1 and cos g = × 3 3

Similarly, cos b =

æ 1 ö a = b = g = cos -1 ç ÷. è 3ø

\

®

^

^

^

®

16. Let a = a1 i + a2 j + a 3 k . Then,| a |2 = 50 Þ ( a1 2 + a2 2 + a 3 2 ) = 50. ® ^

®

^

p 1 =5 2 ´ = 5. 4 2

® ^

®

^

p =0 2

® ^

®

^

a × i =| a||i|cos a Þ a1 = 50 ´ 1 ´ cos a × j =| a|| j|cos b Þ a2 = 50 ´ 1 ´ cos

a × k =| a||k|cos g Þ a 3 = 50 ´ 1 ´ cos q

( a1 2 + a2 2 + a 3 2 ) = 50 Þ 25 + 0 + 50 cos 2 q = 50 Þ cos 2 q = \ \

a 3 = 50 ´ cos ®

^

1 1 p Þ cos q = Þ q= × 2 2 2

p æ 1 ö = ç5 2 ´ ÷ = 5. 4 è 2ø

^

a = 5( i + k ).

®

®

®

^

^

®

®

®

^

^

^

17. Let u = ( a + b ) = (5 i + k ) and v = ( a - b ) = ( - i - 2 j + 5 k ).

SSS Mathematics for Class 12 1035

Scalar, or Dot, Product of Vectors ®

1035

®

Then,| u | = 5 2 + 12 = 26 and| v | = ( -1) 2 + ( -2 ) 2 + 5 2 = ® ®

( u×v )

cos q =

®

( -5 - 0 + 5 )

=

®

26 × 30

| u || v | ®

®

®

®

p × 2

=0 Þ q=

®

30 .

®

®

®

®

18. Let a = l b + c , where c ^ b . Then, c = ( a - l b ) ^ b . \

®

®

®

® ®

® ®

( a - l b ) × b = 0 Þ ( a × b ) - l( b × b ) = 0 Þ ( 6 - 3 - 6 ) - l( 1 + 1 + 1) = 0 Þ l =

\ ®

®

®

®

®

®

®

®

®

®

-3 = -1. 3

a = - b + c Þ c = ( a + b ).

®

®

®

®

®

20. a + b = - c Þ ( a + b ) × ( a + b ) = ( - c ) × ( - c ) ®

®

® ®

®

® ®

Þ | a |2 + | b |2 + 2 a × b =| c |2 Þ a × b = Þ ab cos q =

15 1 15 Þ ( 3 ´ 5 ) cos q = Þ cos q = × 2 2 2

® ®

a×b

21. cos q =

® ®

×

| a|| b | ®

®

®

®

®

®

22. | a - b|2 = ( a - b ) × ( a - b ) ® ®

® ®

® ®

® ®

= a× a - a×b - b× a + b×b ®

® ®

®

®

®

=| a|2 - 2( a × b ) +| b |2 = 5. ®

®

®

23. | a|2 - | b |2 = 8 Þ 64| b |2 -| b |2 = 8 Þ | b |2 = \ ^

®

8 × 63

® 8 8 and| a|= 8 × × 63 63

| b |= ^

^

^

^

^

^ ^

^ ^

^ ^

^ ^

24. |a + b |2 = ( a + b ) × ( a + b ) = a × a + a × b + b × a + b × b ^

^ ^

^

^

^

=|a|2 + 2( a × b ) + |b|2 = 1 + 2|a||b|cos q + 1 q q = 2( 1 + cos q) = 2 ´ 2 cos 2 = 4 cos 2 × 2 2 q 1 ^ ^ cos = |a + b |. \ 2 2 ^ ^ q Similarly, we have|a - b |2 = 4 sin 2 × 2 q 1 ^ ^ sin = |a - b |. \ 2 2 ^

Hence, tan

^

q |a - b | = × ^ ^ 2 |a + b |

15 2

SSS Mathematics for Class 12 1036

1036

Senior Secondary School Mathematics for Class 12 ® ®

®

25. Let the given vectors be a , b and c respectively and let the required vector be ®

®

^

^

d = d1 i + d2 j + d 3 k. ® ®

® ®

® ®

Then, d × a = 0 , d × b = 5 and d × c = 8 Þ d1 + d2 - 3 d 3 = 0 ... (i), d1 + 3 d2 - 2 d 3 = 5 ... (ii) and 2 d1 + d2 + 4 d 3 = 8. ... (iii) ... (iv) On subtracting (i) from (ii), we get 2 d2 + d 3 = 5. ... (v) On multiplying (i) by 2 and subtracting (iii) from it, we get d2 - 10 d 3 = -8. On solving (iv) and (v), we get d2 = 2 , d 3 = 1 and therefore, d1 = 1. ¾®

¾®

¾®

¾®

¾®

¾®

¾®

^

¾®

¾®

26. AB = ( OB - OA ) Þ OB = ( AB + OA ). ^

¾®

^

¾®

¾®

^

^

^

27. AB = ( OB - OA ) = ( 3 i + j - 5 k ), CD = ( OD - OC ) = ( -2 i + 4 j - 2 k ). ®

®

®

® ®

29. Clearly, b + c = a and a × c = 0. \

C ABC is right angled at B.

30. Let O be the origin. Then, ¾®

^

^

¾®

¾®

¾®

¾®

^

^

¾®

^

^

^

^

OA = ( - j - 2 k ), OB = ( 3 i + j + 4 k ) and OC = (5 i + 7 j + k ). \

^

^

¾®

^

¾®

¾®

^

¾®

^

^

^

^

^

CA = ( -5 i - 8 j - 3 k ). ¾® ¾®

^

^

^

^

Now, AB × BC = ( 3 i + 2 j + 6 k ) × ( 2 i + 6 j - 3 k ) = ( 6 + 12 - 18 ) = 0. \

¾®

¾®

AB ^ BC and therefore, ÐB = 90 °.

Hence, CABC is right angled at B. ¾®

¾®

^

^

^

Now, AC = - CA = (5 i + 8 j + 3 k ). ¾® ¾®

cos A =

Þ

AB × AC

( 3 ´ 5 + 2 ´ 8 + 6 ´ 3) ì 32 + 22 + 62 ü ì 52 + 82 + 32 ü ýí ý | AB||AC| îí þî þ 49 1 = = ( 49 )( 98 ) 2 ¾®

¾®

=

ÐA = 45 ° and therefore, ÐC = 180 °- ( 90 °+ 45 ° ) = 45 °.

31. Let ÐABC = q. ¾®

^

^

^

Now, BA = (p.v. of A) – (p.v.of B) = ( 2 i + 2 j + 3 k ) and

^

^

AB = ( OB - OA ) = ( 3 i + 2 j + 6 k ), BC = ( OC - OB ) = ( 2 i + 6 j - 3 k ) and

¾®

^

^

^

BC = (p.v. of C) – (p.v.of B) = ( i + j + 2 k ).

SSS Mathematics for Class 12 1037

Scalar, or Dot, Product of Vectors ¾® ¾®

cos q =

Þ

BA × BC ¾®

¾®

( 2 ´ 1 + 2 ´ 1 + 3 ´ 2)

=

ì 2 2 2 üì 2 2 2ü |BA||BC | íî 2 + 2 + 3 ýí 1 + 1 + 2 ý þî þ 10 10 = = 17 ´ 6 102

æ 10 ö q = cos -1 ç ÷. è 102 ø ®

®

®

®

32. Given:| a|= 1,| b |= 1 and| a + b | = \

1037

®

®

®

®

®

3.

®

| a + b |2 = ( a + b ) × ( a + b ) ® ®

® ®

® ®

= ( a × a ) + 2( a × b ) + ( b × b ) ®

® ®

®

® ®

=| a|2 + 2( a × b ) +| b |2 = 2 + 2( a × b ). \

® ®

® ®

2[1 + ( a × b )] = 3 Þ ( a × b ) = ®

®

®

1 × 2

®

Now, find ( 2 a - 5 b ) × ( 3 a + b ). ®

®

®

®

®

®

33. | a + b |=| a| Þ | a + b |2 =| a|2 ®

®

®

®

®

Þ ( a + b ) × ( a + b ) =| a|2 ® ®

® ®

® ®

®

Þ a × a + 2 ( a × b ) + b × b =| a|2 ® ®

® ®

®

®

® ®

® ®

® ®

[Q ( b × a ) = ( a × b )] ® ®

Þ 2( a × b ) + b × b = 0 Þ 2 a × b + b × b = 0 ®

Þ ( 2 a + b ) × b = 0. ®

®

®

®

®

®

®

®

^

^

34. Let b = b 1 + b 2 , where b 1|| a and b 2 ^ a . ®

^

^

^

Let b 1 = l( 3 i - j ) and let b 2 = x i + y j + z k . Then,

\

® ® ^ ^ ^ ^ ^ b 2 ^ a Þ ( x i + y j + z k ) × ( 3 i - j ) = 0 Þ 3 x - y = 0. ® ® ® ^ ^ ^ ^ ^ ^ ^ ^ b = ( b 1 + b 2 ) Þ ( 2 i + j - 3 k ) = l( 3 i - j ) + ( x i + y j + z k ) ^

^

^

^

^

^

Þ ( 2 i + j - 3 k ) = ( 3 l + x ) i + ( y - l) j + z k

Þ z = -3 , y - l = 1 and 3 l + x = 2 Þ z = -3 , l = ( y - 1) and 3( y - 1) + x = 2 Þ z = -3 , l = ( y - 1) and 3 y + x = 5. 1 3 1 On solving 3 x - y = 0 and x + 3 y = 5, we get x = and y = ,z = -3 and l = × 2 2 2 \

® ® ^ 1 ^ ^ 1^ 3^ b 1 = ( 3 i - j ) and b 2 = i + j - 3 k.

2

2

2

SSS Mathematics for Class 12 1038

24. CROSS, OR VECTOR, PRODUCT OF VECTORS ®

®

Let a and b be two nonzero, nonparallel vectors, and let q be the angle between them such that 0 < q < p. VECTOR PRODUCT OF TWO VECTORS

®

®

Then, the vector product of a and b is defined as ®

®

®

®

^

a ´ b = (| a || b |sin q) n , ®

^

®

where n is a unit vector perpendicular to both a and b , ® ® ^

such that a , b , n form a right-handed system.

®

®

^

When a right-handed screw is rotated from a to b and it advances along n then the system is said to be right handed. ®

REMARK 1

®

If a and b are parallel or collinear, i.e., when q = 0 or q = p then we ®

®

®

define, a ´ b = 0. ®

®

®

In particular, a ´ a = 0. ®

®

®

®

®

®

®

REMARK 2

If a = 0 or b = 0 , we define, a ´ b = 0 .

REMARK 3

For any vector a , we have a ´ a = (| a || a |sin 0) n = 0 n = 0 .

®

®

®

®

®

^

®

ANGLE BETWEEN TWO VECTORS ®

®

^

®

®

Let q be the angle between a and b . Then, ®

^

®

a ´ b = ( a b sin q) n , where| a | = a and| b | = b ®

®

Þ | a ´ b | = a b sin q ®

®

®

®

|a ´ b| |a ´ b| = ® ® ab | a || b |

Þ

sin q =

Þ

ì ® ® ü ï| a ´ b |ï q = sin -1 í × ® ® ý ïî| a || b |ïþ

UNIT VECTOR PERPENDICULAR TO TWO GIVEN VECTORS ® ®

perpendicular to each one of a and b is given by ^

n=

®

®

®

®

(a ´ b ) |a ´ b | 1038

×

^

A unit vector n

SSS Mathematics for Class 12 1039

Cross, or Vector, Product of Vectors

1039

Properties of Vector Product RESULT 1

Vector product is not commutative. ®

®

®

®

In fact, we have ( b ´ a ) = - ( a ´ b ). PROOF

® ® ^

Let a , b , n form a right-handed system. Then, ®

®

^

( a ´ b ) = ( a b sin q) n . ®

®

... (i)

^

Then, b , a , - n form a right-handed system. ®

®

^

\ ( b ´ a ) = ( b a sin q) ( - n ) ®

®

^

Þ - ( b ´ a ) = ( a b sin q) n .

... (ii)

®

®

®

®

From (i) and (ii), we get ( a ´ b ) = - ( b ´ a ). ®

RESULT 2

®

For any two vectors a and b , prove that ®

®

®

®

®

®

( - b ) ´ a = ( a ´ b ) = b ´ ( - a ). ®

PROOF

®

Let q be the angle between a and b . ®

®

Then, the angle between ( - b ) and a is ( p - q). ®

® ^

®

®

And, ( - b ), a , n form a right-handed system. \

®

®

^

( - b ) ´ a = | - b || a |sin ( p - q) n ®

®

^

= (| b || a |sin q) n ®

^

®

= ( a b sin q) n = ( a ´ b ). \

®

®

®

®

( - b ) ´ a = ( a ´ b ). ®

®

®

®

Similarly, b ´ ( - a ) = ( a ´ b ). ®

®

®

®

®

®

Hence, ( - b ) ´ a = ( a ´ b ) = b ´ ( - a ). ®

RESULT 3

®

®

®

®

®

For any scalar m, we have (m a ´ b ) = m( a ´ b ) = a ´ (m b ).

AN IMPORTANT NOTE At a later stage, we ® ® ® ® ® ® ® ® ®

shall prove that for any three vectors

a , b , c , we have a ´ b × c = a × b ´ c , i.e., the position of dot and cross can be interchanged. However, we shall use this fact in proving the following theorem. ®

RESULT 4

®

®

(Distributive law) For any vectors a , b , c , prove that ®

®

®

®

®

®

®

a ´ ( b + c ) = ( a ´ b ) + ( a ´ c ).

®

PROOF

®

®

®

®

®

®

®

®

Let p = [ a ´ ( b + c ) - ( a ´ b ) - ( a ´ c )], and let r be any arbitrary vector.

SSS Mathematics for Class 12 1040

1040

Senior Secondary School Mathematics for Class 12

Then, ® ®

®

®

®

®

®

®

®

®

r × p = r × [ a ´ ( b + c ) - ( a ´ b ) - ( a ´ c )] ® ®

®

®

®

®

®

®

®

®

= r × [ a ´ ( b + c )] - r × ( a ´ b ) - r × ( a ´ c ) ®

®

®

®

®

®

®

®

®

®

®

®

=(r ´ a) ×(b + c) -(r ´ a) × b -(r ´ a) × c ®

®

®

®

®

®

®

®

®

®

=(r ´ a ) × b + (r ´ a ) × c -(r ´ a ) × b -(r ´ a) × c = 0. ® ®

®

®

®

®

®

®

Now, r × p = 0 Þ r = 0 or p = 0 or ( r ^ p ). ®

Since r is an arbitrary vector, we may choose it in such a way that ®

®

®

®

r ¹ 0 and r is not perpendicular to p . ®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

p = 0

Then,

®

Þ a ´( b + c ) - ( a ´ b ) - ( a ´ c ) = 0 Þ a ´ ( b + c ) = ( a ´ b ) + ( a ´ c ). ®

®

®

®

®

®

®

Hence, a ´ ( b + c ) = ( a ´ b ) + ( a ´ c ). ® ® ®

RESULT 5

For any three vectors a , b , c , prove that ®

®

®

®

®

®

®

a ´ ( b - c ) = ( a ´ b ) - ( a ´ c ).

PROOF

We have ®

®

®

®

®

®

a ´ ( b - c ) = a ´ [ b + ( - c )] ®

®

®

®

®

®

®

®

= ( a ´ b ) + a ´ (- c ) ®

[by the distributive law]

®

= ( a ´ b ) + [- ( a ´ c )] [Q ®

®

®

®

®

®

®

®

®

®

®

®

®

Prove that two nonzero vectors a and b are parallel or collinear if and ®

®

®

only if ( a ´ b ) = 0. ®

PROOF

®

a ´ ( - c ) = - ( a ´ c )]

= ( a ´ b ) - ( a ´ c ).

Hence, a ´ ( b - c ) = ( a ´ b ) - ( a ´ c ). RESULT 6

®

®

®

®

Let a ¹ 0 and b ¹ 0 and let q be the angle between them. ®

®

Let a and b be parallel or collinear. Then, q = 0 or q = p Þ

sin q = 0 Þ ®

®

®

®

^

^

®

( a ´ b ) = ( ab sin q) n = 0 n = 0 . ®

®

®

Thus, when a and b are parallel or collinear, then ( a ´ b ) = 0 .

SSS Mathematics for Class 12 1041

Cross, or Vector, Product of Vectors ®

® ®

®

®

®

1041

®

Again, let a ¹ 0, b ¹ 0 and ( a ´ b ) = 0. Then, ®

®

®

®

®

( a ´ b ) = 0 Þ |a ´ b | = 0 Þ ab sin q = 0 Þ sin q = 0 [a ¹ 0 and b ¹ 0] Þ q = 0 or q = p ®

®

®

®

Þ ( a | | b ) or ( a and b are collinear). Prove that

RESULT 7

®

®

®

®

®

®

®

®

®

®

®

( a ´ b ) = 0 Þ a = 0 or b = 0 or a | | b or a and b are collinear. ®

®

Let q be the angle between a and b . Then,

PROOF

®

®

®

®

^

a ´ b = 0 Þ ( a b sin q) n = 0 Þ a b sin q = 0 Þ a = 0 or b = 0 or sin q = 0 ®

®

®

®

®

®

Þ a = 0 or b = 0 ®

®

or q = 0 or q = p

®

®

Þ a = 0 or b = 0 or a and b are parallel or collinear. ®

®

®

®

®

®

Show that a ´ b = a ´ c does not imply that b = c .

COROLLARY ®

®

®

®

®

®

®

®

®

a ´b = a ´c Þ a ´b - a ´c = 0

PROOF

®

®

®

®

Þ a ´( b - c ) = 0 ®

®

®

®

®

®

®

®

®

Þ a = 0 or ( b - c ) = 0 or a | | ( b - c ) ®

®

®

®

®

®

Þ a = 0 or b = c or a | | ( b - c ).

®

®

®

®

®

®

®

a ´ b = a ´ c does not always mean that b = c .

\ SUMMARY ®

®

(i) ( a ´ b ) = - ( b ´ a ) r ® ® ® ® ® (ii) ( - b ) ´ a = ( a ´ b ) = b ´ ( - a ) ®

®

®

®

®

®

(iii) m a ´ b = m ( a ´ b ) = a ´ (m b ) ®

®

®

®

®

®

®

®

®

®

®

®

®

®

(iv) a ´ ( b + c ) = ( a ´ b ) + ( a ´ c ) (v) a ´ ( b - c ) = ( a ´ b ) - ( a ´ c ) ®

®

®

®

®

®

®

®

®

®

®

(vi) a ´ b = 0 Û ( a | | b ) or ( a and b are collinear), when a ¹ 0 and b ¹ 0

SSS Mathematics for Class 12 1042

1042

Senior Secondary School Mathematics for Class 12

Area of a Parallelogram THEOREM 1

®

®

Two adjacent sides of a | |gm are represented by a and b respectively. ®

®

Prove that the area of the | |gm = | a ´ b |. PROOF

Let ABCD be a | |gm. Join BD. ¾®

® ¾®

®

Let AB = a , AD = b and ÐBAD = q. Draw DL ^ AB. Then, AB = a and DL = AD sin q = b sin q. 1 \ ar (C ABD) = ´ base ´ altitude 2 æ1 ö 1 = ç ´ AB ´ DL÷ = a b sin q è2 ø 2 1 ® ® = | a ´ b |. 2 ®

®

\ ar (| |gm ABCD) = 2 ´ ar(C ABD) = | a ´ b |. ®

®

( a ´ b ) is called the vector area of the | |gm.

REMARK

Area of a Triangle THEOREM 2

¾® ® ¾® ® 1 ® ® Prove that the area of C ABC, where AB = a and AC = b , is |a ´ b|. 2 ¾®

®

¾®

®

In C ABC, let AB = a and AC = b and ÐBAC = q. Draw CL ^ AB. Then, AB = a and CL = ( AC ) sin q = b sin q 1 and ar (C ABC) = ´ AB ´ CL 2 1 1 ® ® = a b sin q = | a ´ b |. 2 2 1 ® ® \ ar(C ABC ) = | a ´ b |. 2 1 ® ® REMARK ( a ´ b ) is called the vector area of C ABC. 2 PROOF

Area of a Quadrilateral Prove that the area of a quadrilateral ABCD with diagonals AC and BD 1 ¾® ¾® is | AC ´ BD |. 2 Vector area of quad. ABCD = (vector area of C ABC) + (vector area of C ACD) 1 ¾® ¾® 1 ¾® ¾® = ( AB ´ AC ) + ( AC ´ AD ) 2 2

THEOREM 3

PROOF

SSS Mathematics for Class 12 1043

Cross, or Vector, Product of Vectors

1043

1 ¾® ¾® 1 ¾® ¾® ( AC ´ AB ) + ( AC ´ AD ) 2 2 1 ¾® ¾® ¾® 1 ¾® ¾® = × { AC ´ ( AD - AB )} = ( AC ´ BD ). 2 2 =-

1 ¾® ¾® \ ar(quad. ABCD) = | AC ´ BD |. 2 SUMMARY ®

®

¾®

®

¾®

®

(i) ar (| |gm ABCD) = | a ´ b |, where AB = a and AD = b . ® ® 1 ® ® (ii) ar (| |gm ABCD) = | d 1 ´ d 2 |, where d 1 and d 2 are the diagonal vectors. 2 ¾® ® ¾® ® 1 ® ® (iii) ar(C ABC) = | a ´ b |, where AB = a and AC = b . 2 1 ¾® ¾® (iv) ar (quad. ABCD) = | AB ´ BD |, where AC and BD are its diagonals. 2 VECTOR PRODUCT OF AN ORTHONORMAL VECTOR TRIAD

^

^

^

For mutually perpendicular unit vectors i , j , k , we have ^

^

^

^

^

^

^

^

^

^

i ´ i = j ´ j = k ´ k = 0, ^

^

^ ^

^

^

^

^

^

^

^

^

i ´ j = k = - j ´ i , j ´ k = i = - k ´ j and k ´ i = j = - i ´ k .

Vector Product in Terms of Components ®

^

^

®

^

^

^

^

Let a = a1 i + a2 j + a 3 k and b = b1 i + b2 j + b 3 k . Then, ^

^

^

a ´ b = a1

a2

a3 ×

b1

b2

b3

i

®

PROOF

®

j

k

We have ®

®

^

^

^

^

^

^

a ´ b = ( a1 i + a2 j + a 3 k ) ´ ( b1 i + b2 j + b 3 k ) ^

^

^

^

^

^

= a1b1( i ´ i ) + a1b2( i ´ j ) + a1b 3( i ´ k )

SSS Mathematics for Class 12 1044

1044

Senior Secondary School Mathematics for Class 12 ^

^

^

^

^

^

^

^

+ a2b1( j ´ i ) + a2b2( j ´ j ) + a2b 3( j ´ k ) ^

^

^

^

+ a 3b1( k ´ i ) + a 3b2( k ´ j ) + a 3b 3( k ´ k ) ^

^

^

= ( a2b 3 - a 3b2) i + ( a 3b1 - a1b 3) j + ( a1b2 - a2b1) k é ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ù êQ i ´ i = j ´ j = k ´ k = 0 , i ´ j = k , j ´ k = i , k ´ i = j , ú ê ^ ^ ^ ^ ^ ^ ^ ^ ^ú and j ´ i = - k , k ´ j = - i and i ´ k = - j úû êë ^

^

^

= a1

a2

a3 ×

b1

b2

b3

i

®

j

^

k

SOLVED EXAMPLES ® ^ ^ ^

^

®

^

®

®

®

EXAMPLE 1

If a = ( 3 i + j - 4 k ) and b = ( 6 i + 5 j - 2 k ), find ( a ´ b ) and| a ´ b |.

SOLUTION

We have ^

^

^

(a ´ b) = 3

1

-4

6

5

-2

i

®

j

®

k

^

^

^

= ( -2 + 20) i - ( - 6 + 24) j + (15 - 6) k ^

^

^

= (18 i - 18 j + 9 k ). ®

®

|a ´ b |2 = {(18) 2 + ( -18) 2 + 92} = 729 Þ ®

EXAMPLE 2

®

®

|a ´ b | = 729 = 27. ^

^

®

^

^

^

®

^

®

If a = ( i - 2 j + 3 k ) and b = ( 2 i + 3 j - 5 k ) then find ( a ´ b ) and ®

®

®

®

verify that ( a ´ b ) is perpendicular to each one of a and b . SOLUTION

We have ^

^

^

(a ´ b) = 1

-2

3

2

3

-5

i

®

®

j

k

^

^

^

^

^

^

= (10 - 9) i - ( -5 - 6) j + ( 3 + 4) k = ( i + 11 j + 7 k ).

SSS Mathematics for Class 12 1045

Cross, or Vector, Product of Vectors ®

®

®

^

^

^

^

1045

^

^

Now, ( a ´ b ) × a = ( i + 11 j + 7 k ) × ( i - 2 j + 3 k ). ®

®

®

®

®

\

( a ´ b ) × a = (1 - 22 + 21) = 0.

\

( a ´ b ) ^ a. ®

®

®

®

®

^

®

^

^

^

And, ( a ´ b ) × b = ( i + 11 j + 7 k ) × ( 2 i + 3 j - 5 k ) ®

= ( 2 + 33 - 35) = 0.

®

( a ´ b ) ^ b.

\ ®

EXAMPLE 3

®

^

^

®

^

^

®

^

®

®

® ®

and a × c = 3. ®

SOLUTION

®

If a = i + j + k and b = j - k , find a vector c such that a ´ c = b

^

[CBSE 2013] ^

®

^

^

^

Here a = i + j + k and b = j - k . ®

^

^

^

Let c = c1 i + c2 j + c 3 k. Then, ® ®

®

®

®

a × c = 3 and a ´ c = b

Þ

^

^

^

^

^

^

^

^

^

( i + j + k) × ( c1 i + c2 j + c 3 k) = 3 ^

^

^

^

^

and ( i + j + k ) ´ ( c1 i + c2 j + c 3 k) = ( j - k ) ^

^

^

... (i) and 1

1

1 = ( j - k)

c1

c2

c3

i

Þ

c1 + c2 + c 3 = 3

^

j

k

^

^

^

^

^

... (ii)

^

Now, (ii) gives: ( c 3 - c2) i - ( c 3 - c1) j + ( c2 - c1) k = ( j - k) Þ

c 3 - c2 = 0, c1 - c 3 = 1 and c2 - c1 = -1

Þ

c 3 = c2 and c1 - c2 = 1.

Putting c 3 = c2 in (i), we get c1 + 2c2 = 3. On solving c1 + 2c2 = 3 and c1 - c2 = 1, we get c2 =

2 5 and c1 = × 3 3

5 2 2 , c2 = and c 3 = × 3 3 3 ® 5 ^ ® ^ ^ 2^ 2^ 1 ^ Hence, c = i + j + k Þ c = (5 i + 2 j + 2 k). 3 3 3 3

\

EXAMPLE 4

c1 =

Find a unit vector perpendicular to each one of the vectors ®

^

^

^

®

^

^

^

a = ( 4 i - j + 3 k ) and b = ( 2 i + 2 j - k ).

SSS Mathematics for Class 12 1046

1046

Senior Secondary School Mathematics for Class 12 ®

SOLUTION

®

®

We know that ( a ´ b ) is a vector perpendicular to each one of a ®

and b . So, the required vector is

®

®

®

®

(a ´ b)

×

|a ´ b| ^

^

^

Now, ( a ´ b ) = 4

-1

3

2

2

-1

i

®

®

j

k

^

^

^

= (1 - 6) i - ( - 4 - 6) j + ( 8 + 2) k ^

^

^

= ( - 5 i + 10 j + 10 k ). ®

®

| a ´ b | = ( -5) 2 + (10) 2 + (10) 2 = 225 = 15. ^

^

^

( -5 i + 10 j + 10 k ) Hence, the required unit vector = ^ 1 ^ 15^ = ( - i + 2 j + 2 k ). 3 EXAMPLE 5

Find a vector of magnitude 15, which is perpendicular to both the vectors ^

^

^

^

^

( 4 i - j + 8 k ) and ( - j + k ). ®

SOLUTION

^

^

®

^

^

^

Let a = ( 4 i - j + 8 k ) and b = ( - j + k ). ®

®

A unit vector perpendicular to both a and b =

®

®

®

®

(a ´ b)

×

|a ´ b | ^

^

^

Now, a ´ b = 4

-1

8

0

-1

i

®

j

®

k

1 ^

^

^

= ( -1 + 8) i - ( 4 - 0) j + ( - 4 - 0) k ^

^

^

= (7 i - 4 j - 4 k ). \

®

®

| a ´ b | = 7 2 + ( - 4) 2 + ( - 4) 2 = 81 = 9. ®

®

So, a unit vector perpendicular to both a and b ®

=

®

(a ´ b ) ®

®

|a ´ b|

^

=

^

The required vector =

^

^

7 i -4 j -4k × 9 ^

^

^ ^ ^ 15(7 i - 4 j - 4 k ) 5 = (7 i - 4 j - 4 k ). 9 3

SSS Mathematics for Class 12 1047

Cross, or Vector, Product of Vectors ®

EXAMPLE 6

^

^

®

^

^

^

1047 ®

^

^

^

^

Let a = ( i + 4 j + 2 k ), b = ( 3 i - 2 j + 7 k ) and c = ( 2 i - j + 4 k ). ®

®

®

Find a vector d which is perpendicular to both a and b such that ® ®

c × d = 18.

[CBSE 2010]

®

SOLUTION

®

®

®

Since d is perpendicular to both a and b , it follows that d is ®

®

parallel to ( a ´ b ). ^

^

^

Now, ( a ´ b ) = 1

4

2

3

-2

7

i

®

j

®

k

^

^

^

^

^

^

= ( 28 + 4) i - (7 - 6) j + ( -2 - 12) k = ( 32 i - j - 14 k ).

®

® ®

®

®

®

Since d is parallel to ( a ´ b ), we have d = l( a ´ b ) for some scalar l. \

®

^

^

^

^

^

^

d = l( 32 i - j - 14 k ) = ( 32l i - l j - 14l k )

... (i)

® ®

Since c × d = 18, we have ^

^

^

^

^

^

( 2 i - j + 4 k ) × ( 32l i - l j - 14l k ) = 18 Þ

( 64l + l - 56l) = 18 Þ 9l = 18 Þ l = 2. ®

^

^

^

Hence, d = ( 64 i - 2 j - 28 k ) [putting l = 2 in (i)]. EXAMPLE 7

Find a vector of magnitude 5 units, perpendicular to each of the vectors ®

®

®

®

®

^

^

®

^

^

^

^

( a + b ) and ( a - b ), where a = ( i + j + k ) and b = ( i + 2 j + 3 k ). [CBSE 2008C] SOLUTION

We have ®

®

^

^

^

^

^

^

^

®

®

^

^

^

^

^

^

^

^

^

( a + b ) = ( i + j + k ) + ( i + 2 j + 3 k ) = ( 2 i + 3 j + 4 k ) and ^

( a - b ) = ( i + j + k ) - ( i + 2 j + 3 k ) = ( - j - 2 k ). ^

^

^

( a + b ) ´( a - b ) = 2

3

4

i

\

®

®

®

j

®

0

k

-1 -2 ^

^

^

= ( -6 + 4) i - ( -4 - 0) j + ( -2 - 0) k ^

^

^

= ( -2 i + 4 j - 2 k ). \

®

®

®

®

|( a + b ) ´ ( a - b )| = ( -2) 2 + 42 + ( -2) 2 = 24 = 2 6.

SSS Mathematics for Class 12 1048

1048

Senior Secondary School Mathematics for Class 12

So, the vectors of magnitude 5 units and perpendicular to each of ®

®

®

®

the vectors ( a + b ) and ( a - b ) are: ®

±

®

®

®

5{( a + b ) ´ ( a - b )} ®

®

®

®

^

=±

|( a + b ) ´ ( a - b )| ®

^

^

®

^

^

^

^

^

®

^

If a = 4 i + 3 j + 2 k and b = 3 i + 2 k , find| b ´ 2 a |.

SOLUTION

We have ^

®

^

^

®

EXAMPLE 8

®

^

5( -2 i + 4 j - 2 k ) 5( - i + 2 j - k ) =± × 2 6 6

^

^

^

b = ( 3 i + 2 k ) and 2 a = ( 8 i + 6 j + 4 k ). ^

^

^

( b ´2 a ) = 3

0

2

8

6

4

i

\

®

®

j

k

^

^

^

= ( 0 - 12) i + (16 - 12) j + (18 - 0) k ^

^

^

^

^

^

= ( -12 i + 4 j + 18 k ). \

®

®

| b ´ 2 a | = |-12 i + 4 j + 18 k | = ( -12) 2 + 42 + (18) 2 = 484 = 22. ®

®

Hence,| b ´ 2 a | = 22. ®

®

®

®

® ®

EXAMPLE 9

If| a | = 26 ,| b | = 7 and| a ´ b | = 35 , find a × b .

SOLUTION

Given that| a | = 26 and| b | = 7 , and| a ´ b | = 35.

®

\

®

®

®

®

®

®

| a ´ b | = 35 Þ | a || b |sin q = 35 Þ sin q =

35 ®

®

=

| a || b | Now, cos q = 1 - sin 2 q = 1 \

®

35 5 = × ( 26) ´ 7 26

25 1 = × 26 26

® ®

® ® 1 ö æ a × b = | a || b |cos q = ç 26 ´ 7 ´ ÷ = 7. 26 ø è ® ®

Hence, a × b = 7. ®

EXAMPLE 10

®

®

®

^

^

^

If | a | = 2, | b | = 7 and ( a ´ b ) = ( 3 i + 2 j + 6 k ), find the angle ®

®

between a and b .

SSS Mathematics for Class 12 1049

Cross, or Vector, Product of Vectors ®

SOLUTION

1049

®

Let q be the angle between a and b . Then, ®

®

®

®

®

®

^

^

^

a ´b =3 i +2j +6k

3 2 + 22 + 62 = 49 = 7

Þ | a ´ b |=

®

®

®

®

[Q | a ´ b | = | a|| b |sin q] Þ | a || b |sin q = 7 p 7 7 1 = Þ q= × Þ sin q = ® ® = ( 2 ´ 7) 2 6 | a || b | ® ® p Hence, the required angle between a and b is × 6 ®

EXAMPLE 11

^

^

^

Find the sine of the angle between the vectors a = ( 2 i - j + 3 k ) and ®

^

^

^

b = ( i + 3 j + 2 k ).

SOLUTION

We have ^

^

^

(a ´ b) = 2

-1

3

1

3

2

i

®

j

®

k

^

^

^

= ( -2 - 9) i - ( 4 - 3) j + ( 6 + 1) k ^

^

^

= ( -11 i - j + 7 k ). ®

®

| a ´ b |= ( -11) 2 + ( -1) 2 + 7 2 = 171 = 3 19 , ®

| a |= 22 + ( -1) 2 + 3 2 = 14 , ®

| b |= 12 + 3 2 + 22 = 14. ®

®

Let q be the angle between a and b . Then, sin q =

®

®

®

®

|a ´ b| | a || b | ®

EXAMPLE 12

=

3 19 3 = 19. ( 14)( 14) 14

®

®

®

If the vectors a and b are such that|a|= 3 ,|b|= ®

®

vector then write the angle between a and b . ®

SOLUTION

®

® ® 2 and a ´ b is a unit 3

Let q be the angle between a and b . Then, ® ® ® ® 2ö æ |a ´ b|= 1 Þ |a||b|sin q = 1 Þ ç 3 ´ ÷ sin q = 1 3ø è 1 Þ sin q = Þ q = 30°. 2 ®

®

Hence, the angle between a and b is 30°.

[CBSE 2014]

SSS Mathematics for Class 12 1050

1050 EXAMPLE 13

Senior Secondary School Mathematics for Class 12

Find the area of the parallelogram whose adjacent sides are represented by ^

^

^

^

^

^

the vectors ( 3 i + j - 2 k ) and ( i - 3 j + 4 k). ®

SOLUTION

^

^

®

^

^

^

^

Let a = ( 3 i + j - 2 k ) and b = ( i - 3 j + 4 k). ®

®

Then, vector area of the | |gm is ( a ´ b ). ^

^

^

Now, ( a ´ b ) = 3

1

-2

1

-3

4

i

®

j

®

k

^

^

^

= ( 4 - 6) i - (12 + 2) j + ( -9 - 1) k ^

^

^

= ( - 2 i - 14 j - 10 k ). ®

®

Required area = | a ´ b | = ( -2) 2 + ( -14) 2 + ( -10) 2 sq units = EXAMPLE 14

300 sq units = 10 3 sq units.

Find the area of the parallelogram whose diagonals are represented by the ®

^

^

®

^

^

^

^

vectors d 1 = ( 2 i - j + k ) and d 2 = ( 3 i + 4 j - k ). ®

SOLUTION

^

^

®

^

^

^

^

Given that d 1 = ( 2 i - j + k ) and d 2 = ( 3 i + 4 j - k ). Vector area of the | |gm is ^

1 ® ® ( d 1 ´ d 2 ). 2

^

^

Now, ( d1 ´ d 2 ) = 2

-1

1

3

4

-1

i

®

j

®

k

^

^

^

= (1 - 4) i - ( -2 - 3) j + ( 8 + 3) k ^

^

^

= ( - 3 i + 5 j + 11 k ). 1 ® ® Required area = | d 1 ´ d 2 | 2 1 = ( -3) 2 + 5 2 + (11) 2 sq units 2 1 = 155 sq units. 2 EXAMPLE 15

Find the area of the triangle whose adjacent sides are determined by the ®

^

^

®

^

^

^

vectors a = ( -2 i - 5 k ) and b = ( i - 2 j - k ).

SSS Mathematics for Class 12 1051

Cross, or Vector, Product of Vectors SOLUTION

1051

Two adjacent sides of the given triangle are represented by the ®

^

®

^

^

^

^

vectors a = ( -2 i - 5 k ) and b = ( i - 2 j - k ). 1 ® ® So, the area of the triangle is | a ´ b |. 2 ^

i

®

^

^

0

-5

j

®

Now, ( a ´ b ) = -2

k

-2 -1

1

^

^

^

= ( 0 - 10) i - ( 2 + 5) j + ( 4 - 0) k ^

^

^

= ( -10 i - 7 j + 4 k ). 1 ® ® \ required area = | a ´ b | 2 1 = × { ( -10) 2 + ( -7) 2 + 42} sq units 2 1 = 165 sq units. 2 EXAMPLE 16

SOLUTION

Using vectors find the area of C ABC whose vertices are A(1, 2, 3), [CBSE 2013] B( 2, - 1, 4) and C( 4, 5 , - 1). We have ^

^

^

Position vector of A = ( i + 2 j + 3 k ), ^

^

^

position vector of B = ( 2 i - j + 4 k ) and ^

^

^

position vector of C = ( 4 i + 5 j - k ). \

¾®

AB = (position vector of B) – (position vector of A) ^

^

^

^

^

^

^

^

^

= ( 2 i - j + 4 k ) - ( i + 2 j + 3 k ) = ( i - 3 j + k ). ¾®

AC = (position vector of C) – (position vector of A) ^

^

^

^

^

^

^

^

^

= ( 4 i + 5 j - k ) - ( i + 2 j + 3 k ) = ( 3 i + 3 j - 4 k ). \

½1 ¾® ¾® ½ area of C ABC = ½ ( AB ´ AC )½. ½2 ½ ^

^

^

Now, AB ´ AC = 1

-3

1

3

3

-4

i

¾®

¾®

j

k

SSS Mathematics for Class 12 1052

1052

Senior Secondary School Mathematics for Class 12 ^

^

^

= (12 - 3) i - ( -4 - 3) j + ( 3 + 9) k ^

^

^

= ( 9 i + 7 j + 12 k ). 1 ¾® ¾® \ area of C ABC = |AB ´ AC| 2 ^ ^ ^ 1 1 = |( 9 i + 7 j + 12 k )| = × ( 9) 2 + 7 2 + (12) 2 2 2 1 1 = × ( 81 + 49 + 144) = 274 sq units. 2 2 1 Hence, the area of DABC is 274 sq units. 2 EXAMPLE 17

^

^

^

Show that the points whose position vectors are (5 i + 6 j + 7 k ), ^

^

^

^

^

^

(7 i - 8 j + 9 k ) and ( 3 i + 20 j + 5 k ) are collinear. SOLUTION

Let the given points be A, B, and C respectively. Then, ¾®

AB = (position vector of B) – (position vector of A) ^

¾®

^

^

^

^

^

^

^

^

= (7 i - 8 j + 9 k ) - (5 i + 6 j + 7 k ) = ( 2 i - 14 j + 2 k ).

And, AC = (position vector of C) – (position vector of A) ^

^

^

^

^

^

= ( 3 i + 20 j + 5 k ) - (5 i + 6 j + 7 k ) ^

^

^

= ( -2 i + 14 j - 2 k ). ^

^

^

2 -14

2

i

\

¾®

¾®

( AB ´ AC ) =

-2

j

k

-2

14 ^

i

= 2 ´ ( -2) ×

¾®

^

j

^

k

® ®

1 -7

1 = ( -4) ´ 0 = 0

1 -7

1 R 2 and R 3 are identical].

[Q

¾®

So, AB and AC are parallel vectors, having a common end point, A. Hence, the given points A, B and C are collinear. EXAMPLE 18

Using vector method, show that the points A( 2, - 1, 3), B( 4, 3 , 1) and C( 3 , 1, 2) are collinear.

SOLUTION

Clearly, we have

^

^

^

position vector of A = ( 2 i - j + 3 k ),

SSS Mathematics for Class 12 1053

Cross, or Vector, Product of Vectors ^

^

1053

^

position vector of B = ( 4 i + 3 j + k ), and ^

^

^

position vector of C = ( 3 i + j + 2 k ). ¾®

\

AB = ( position vector of B) - ( position vector of A) ^

¾®

^

^

^

^

^

^

^

^

= ( 4 i + 3 j + k ) - ( 2 i - j + 3 k ) = ( 2 i + 4 j - 2 k ).

AC = ( position vector of C) - ( position vector of A) ^

^

^

^

^

^

^

^

^

= ( 3 i + j + 2 k ) - ( 2 i - j + 3 k ) = ( i + 2 j - k ). ^

^

2

4 -2

1

2 -1

i

¾®

¾®

( AB ´ AC ) =

\

¾®

k

^

^

1

2 -1 = 0 [Q R 2 and R 3 are identical].

1

2 -1

i

=2

^

j

^

j

k

®

¾®

Thus, AB and AC are parallel vectors, having a common end point, A. Hence, the given points A, B, and C are collinear. EXAMPLE 19

Show that the points having position vectors ®

®

®

®

®

®

®

®

( a - 2 b + 3 c ), ( -2 a + 3 b + 2 c ), ( - 8 a + 13 b ) ® ® ®

are collinear, whatever be a , b , c . SOLUTION

Let A , B , C be the given points whose position vectors are ®

®

®

®

®

®

®

®

( a - 2 b + 3 c ), ( -2 a + 3 b + 2 c ) and ( - 8 a + 13 b ) respectively. Then, ¾®

AB = (position vector of B) – (position vector of A) ®

®

®

®

®

®

®

®

®

= ( -2 a + 3 b + 2 c ) - ( a - 2 b + 3 c ) = ( -3 a + 5 b - c ), and ¾®

AC = (position vector of C) – (position vector of A) ®

®

®

®

®

®

®

®

= ( -8 a + 13 b ) - ( a - 2 b + 3 c ) = ( -9 a + 15 b - 3 c ). \

¾®

¾®

®

®

®

®

®

®

( AB ´ AC ) = ( -3 a + 5 b - c ) ´ ( -9 a + 15 b - 3 c ) ®

®

®

®

®

®

= p ´ 3 p , where ( -3 a + 5 b - c ) = p ®

= 0

®

®

®

[Q p ´ p = 0 ].

SSS Mathematics for Class 12 1054

1054

Senior Secondary School Mathematics for Class 12 ¾®

¾®

\ AB and AC are parallel vectors, having a common end point, A. Hence, the points A, B and C are collinear. ®

EXAMPLE 20 PROOF

®

®

®

®

®

Prove that ( a - b ) ´ ( a + b ) = 2( a ´ b ).

We have ®

®

®

®

( a - b ) ´( a + b ) ®

®

®

®

®

®

®

®

®

®

®

®

®

®

= a ´a + a ´b - b ´a - b ´b = a ´b - b ´a ®

®

®

®

®

®

®

®

®

®

®

®

®

®

[Q - b ´ a = ( a ´ b )]

®

®

®

[Q a ´ a = 0 and b ´ b = 0 ]

=(a ´ b) + (a ´ b) = 2( a ´ b ).

[by the distributive law]

®

®

®

®

®

Hence, ( a - b ) ´ ( a + b ) = 2( a ´ b ). ®

EXAMPLE 21

®

®

®

®

®

® ®

If a ´ b = c ´ d and a ´ c = b ´ d , show that ( a - d ) is parallel to ®

®

( b - c ), it being given that a ¹ d and b ¹ c. ®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

[CBSE 2009]

®

a ´ b = c ´ d , and a ´ c = b ´ d

SOLUTION

Þ

a ´b - a ´c = c ´d - b ´d

Þ

®

a ´b - a ´c + b ´d - c ´d = 0

Þ

®

®

®

®

®

®

®

®

®

®

a ´( b - c ) + ( b - c ) ´ d = 0

Þ

®

®

®

®

a ´( b - c ) - d ´( b - c ) = 0 ®

®

®

®

®

®

®

®

®

Þ ( a - d ) ´( b - c ) = 0 Þ ( a - d ) | | ( b - c ). ®

®

®

®

Hence, ( a - d ) is parallel to ( b - c ). EXAMPLE 22

Prove that

SOLUTION

We have ®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

a ´ ( b + c ) + b ´ ( c + a ) + c ´ ( a + b ) = 0. ®

®

a ´( b + c ) + b ´( c + a ) + c ´( a + b ) ®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

=(a ´ b) + (a ´ c) + (b ´ c) + (b ´ a) + ( c ´ a) + ( c ´ b) [by the distributive law] ®

®

®

®

=(a ´ b) + (b ´ c) + ( c ´ a) -(a ´ b) -(b ´ c) -( c ´ a) ®

®

®

®

®

®

®

®

®

= 0 [Q ( b ´ a ) = - ( a ´ b ), ( c ´ b ) = - ( b ´ c ) ®

®

®

®

and ( a ´ c ) = - ( c ´ a )].

SSS Mathematics for Class 12 1055

Cross, or Vector, Product of Vectors ®

®

®

®

®

®

®

1055

®

®

®

Hence, a ´ ( b + c ) + b ´ ( c + a ) + c ´ ( a + b ) = 0 . ®

EXAMPLE 23

®

®

®

®

®

®

®

®

®

If a + b + c = 0, prove that ( a ´ b ) = ( b ´ c ) = ( c ´ a ). [CBSE 2001, ’04C] ®

SOLUTION

®

®

®

®

®

®

a + b + c = 0 Þ a + b =-c ®

®

®

®

®

®

®

®

®

Þ ( a + b ) ´ b = (- c ) ´ b ®

®

®

®

Þ ( a ´ b ) + ( b ´ b ) = (- c ) ´ b [by the distributive law] ®

®

®

Þ (a ´ b) + 0 =(b ´ c) ®

®

®

®

®

®

®

[Q b ´ b = 0 and ( - c ) ´ b = b ´ c ] ®

®

®

®

Þ a ´b = b ´c Also,

®

®

®

®

®

… (i)

®

®

a + b + c = 0 Þ b + c =-a ®

®

®

®

®

®

®

®

®

Þ ( b + c ) ´ c = (- a ) ´ c ®

®

®

®

Þ ( b ´ c ) + ( c ´ c ) = (- a ) ´ c [by the distributive law] ®

®

®

®

®

Þ (b ´ c) + 0 = c ´ a ®

®

®

®

®

®

®

Þ b ´c = c ´a ®

®

®

®

c ´ c = 0 and ( - a ) ´ c = c ´ a ]

[Q

®

®

… (ii)

® ®

From (i) and (ii), we get a ´ b = b ´ c = c ´ a . ® ® ®

EXAMPLE 24

Prove that the points A, B, C with position vectors a , b , c are collinear ®

®

®

®

®

®

®

if and only if ( b ´ c ) + ( c ´ a ) + ( a ´ b ) = 0 . PROOF

We have ¾®

®

®

¾®

®

®

AB = (position vector of B) - (position vector of A) = ( b - a )

and BC = (position vector of C) - (position vector of B) = ( c - b ). Now, A , B , C are collinear ¾®

¾®

®

®

®

®

®

®

®

®

®

®

Û (b - a) ´ c -(b - a) ´ b = 0 ®

®

®

®

®

®

®

®

®

®

®

®

®

[by the distributive law] ®

Û b ´c - a ´c - b ´b + a ´b = 0 ®

®

Û ( b - a ) ´( c - b ) = 0

Û AB and BC are parallel

®

®

®

[by the distributive law] ®

®

®

®

®

®

Û ( b ´ c ) + ( c ´ a ) + ( a ´ b ) = 0 [Q b ´ b = 0 and - a ´ c = c ´ a]. ®

®

®

®

®

®

®

Thus, A , B , C are collinear Û ( b ´ c ) + ( c ´ a ) + ( a ´ b ) = 0 .

SSS Mathematics for Class 12 1056

1056

Senior Secondary School Mathematics for Class 12

EXAMPLE 25

® ®

®

® ®

PROOF

®

®

®

®

®

®

If a × b = 0 and a ´ b = 0, prove that a = 0 or b = 0. ®

®

®

Let a × b = 0 and a ´ b = 0. Then, ® ®

®

®

®

a × b = 0 and a ´ b = 0 ®

®

®

®

®

®

®

®

®

®

®

®

Þ

( a = 0 or b = 0 or a ^ b ) and ( a = 0 or b = 0 or a | | b )

Þ

a = 0

®

®

or

®

®

b = 0 ®

®

®

®

®

®

[Q a ^ b and a | | b can never hold simultaneously]. ® ®

®

®

®

®

®

Hence, ( a × b = 0 and a ´ b = 0 ) Þ ( a = 0 or b = 0 ). ® ®

EXAMPLE 26

® ® PROOF

® ® ®

®

®

®

®

®

®

®

®

®

If a × b = a × c , a ´ b = a ´ c and a ¹ 0 then prove that b = c . ® ®

®

®

a × b = a × c and a ¹ 0 ® ®

® ®

®

®

®

®

®

Þ a × b - a × c = 0 and a ¹ 0 Þ a × ( b - c ) = 0 and a ¹ 0 ®

®

®

®

®

®

®

®

®

®

®

®

®

Þ b - c = 0 or a ^ ( b - c ) Þ b = c or a ^ ( b - c ). ®

®

®

Again, a ´ b = a ´ c and a ¹ 0 ®

®

®

®

®

®

... (i)

®

®

®

®

®

®

®

®

®

®

Þ ( a ´ b ) - ( a ´ c ) = 0 and a ¹ 0 Þ a ´ ( b - c ) = 0 and a ¹ 0 ®

®

®

®

®

®

®

®

®

®

®

Þ ( b - c ) = 0 or a | | ( b - c ) Þ b = c or a | | ( b - c ).

... (ii)

From (i) and (ii), we get b = c ®

®

®

®

®

®

[Q a ^ ( b - c ) and a | | ( b - c ) both cannot hold simultaneously]. Lagrange’s Identity ®

EXAMPLE 27

®

® ®

® ®

® ®

® ®

a× a

Prove that |a ´ b|2 =

a× b

a× b

®

SOLUTION

×

[CBSE 2002C, ’04]

b× b

®

Let q be the angle betwen a and b . Then, ® ®

® ®

® ®

^

^

|a ´ b|2 = ( a ´ b ) × ( a ´ b ) = ( a b sin q) n × ( a b sin q) n ^ ^

= ( a 2b 2 sin 2 q)( n × n ) = a 2b 2 sin 2 q = a 2b 2(1 - cos2 q) = a 2b 2 - ( ab cos q) 2 ® ® ® ®

® ®

= ( a × a )( b × b ) - ( a × b ) 2 =

® ®

® ®

® ®

® ®

a× a

a× b

® ®

Hence,|a ´ b|2 =

® ®

® ®

® ®

® ®

a× a

a× b

a× b

b× b

×

a× b

b× b

×

SSS Mathematics for Class 12 1057

Cross, or Vector, Product of Vectors

1057

EXERCISE 24 1.

®

®

®

®

Find ( a ´ b ) and|a ´ b |, when ®

^

^

®

^

^

^

^

(i) a = i - j + 2 k and b = 2 i + 3 j - 4 k ®

^

^

^

®

^

^

^

^

^

®

^

^

^

^

^

®

^

(ii) a = 2 i + j + 3 k and b = 3 i + 5 j - 2 k ®

^

(iii) a = i - 7 j + 7 k and b = 3 i - 2 j + 2 k ®

^

®

^

^

(iv) a = 4 i + j - 2 k and b = 3 i + k ®

^

^

^

^

(v) a = 3 i + 4 j and b = i + j + k ^

^

^

^

^

®

^

2. Find l if ( 2 i + 6 j + 14 k ) ´ ( i - l j + 7 k ) = 0 . ®

^

^

®

^

^

^

[CBSE 2010] ® ®

^

3. If a = ( -3 i + 4 j - 7 k ) and b = ( 6 i + 2 j - 3 k ), find ( a ´ b ). ®

® ®

®

® ®

Verify that (i) a and ( a ´ b ) are perpendicular to each other and (ii) b and ( a ´ b ) are perpendicular to each other. 4. Find the value of: ^ ^ ^

^ ^

^

(i) ( i ´ j ) × k + i × j ^

^

^

^

^ ^

^ ^

(ii) ( k ´ j ) × i + j × k ^

^

^

^

[CBSE 2012]

^

(iii) i ´ ( j + k) + j ´ ( k + i ) + k ´ ( i + j )

®

[CBSE 2014]

®

5. Find the unit vectors perpendicular to both a and b when ®

^

^

^

®

^

®

^

®

^

®

^

^

^

(i) a = 3 i + j - 2 k and b = 2 i + 3 j - k ®

^

®

^

^

^

^

^

(ii) a = i - 2 j + 3 k and b = i + 2 j - k ^

^

^

(iii) a = i + 3 j - 2 k and b = - i + 3 k ®

^

^

^

^

^

(iv) a = 4 i + 2 j - k and b = i + 4 j - k. 6. Find the unit vectors perpendicular to the plane of the vectors ®

^

^

®

^

^

^

^

a = 2 i - 6 j - 3 k and b = 4 i + 3 j - k.

7. Find a vector of magnitude 6 which is perpendicular to both the vectors ®

^

^

®

^

^

^

^

a = 4 i - j + 3 k and b = -2 i + j - 2 k.

8. Find a vector of magnitude 5 units, perpendiuclar to each of the vectors ®

®

®

®

®

^

^

^

®

^

^

^

( a + b ) and ( a - b ), where a = ( i + j + k ) and b = ( i + 2 j + 3 k ). [CBSE 2008C] ®

®

9. Find the angle between two vectors a and b with magnitudes 1 and ® ®

2 respectively and|a ´ b|=

3.

[CBSE 2009]

SSS Mathematics for Class 12 1058

1058

Senior Secondary School Mathematics for Class 12 ®

^

®

^

^

®

^

^

®

^

10. If a = ( i - j ), b = ( 3 j - k ) and c = (7 i - k ), find a vector d which is ®

®

® ®

perpendicular to both a and b and for which c × d = 1. ®

^

^

®

^

^

^

®

^

^

[CBSE 2010]

^

®

^

11. If a = ( 4 i + 5 j - k ), b = ( i - 4 j + 5 k ), and c = ( 3 i + j - k ), find a vector d ®

®

® ®

which is perpendicular to both a and b and for which c × d = 21. ®

®

® ®

®

®

12. Prove that| a ´ b |= ( a × b ) tan q, where q is the angle between a and b . ®

^

^

®

^

^

^

^

13. Write the value of p for which a = ( 3 i + 2 j + 9 k ) and b = ( i + p j + 3 k ) are parallel vectors. [CBSE 2009] 14.

®

®

®

®

^ ®

^

®

®

®

Verify that a ´ ( b + c ) = ( a ´ b ) + ( a ´ c ), when ®

^

^

^

®

^

^

^

^

(i) a = i - j - 3 k , b = 4 i - 3 j + k and c = 2 i - j + 2 k ®

^

^ ®

^

^

^

®

^

^

^

^

(ii) a = 4 i - j + k , b = i + j + k and c = i - j + k. 15. Find the area of the parallelogram whose adjacent sides are represented by the vectors ®

^

^

®

^

^

^

^

(i) a = i + 2 j + 3 k and b = - 3 i - 2 j + k ®

^

^

®

^

^

^

^

(ii) a = ( 3 i + j + 4 k ) and b = ( i - j + k ) ®

^

®

^

^

®

^

^

^

(iii) a = 2 i + j + 3 k and b = i - j ®

^

(iv) a = 2 i and b = 3 j . 16. Find the area of the parallelogram whose diagonals are represented by the vectors ®

^

^

®

^

^

^

^

^

^

(i) d 1 = 3 i + j - 2 k and d 2 = i - 3 j + 4 k ®

^

^

®

^

[CBSE 2004]

^

(ii) d1 = 2 i - j + k and d 2 = 3 i + 4 j - k ®

^

^

®

^

^

^

(iii) d1 = i - 3 j + 2 k and d 2 = - i + 2 j . 17. Find the area of the triangle whose two adjacent sides are determined by the vectors ®

^

^

®

^

^

^

(i) a = - 2 i - 5 k and b = i - 2 j - k ®

^

^

®

^

^

(ii) a = 3 i + 4 j and b = - 5 i + 7 j . 18. Using vectors, find the area of C ABC whose vertices are (i) A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) (ii) A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1) (iii) A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1) (iv) A(1, –1, 2), B(2, 1, –1) and C(3, –1, 2).

[CBSE 2011] [CBSE 2013]

SSS Mathematics for Class 12 1059

Cross, or Vector, Product of Vectors

1059

19. Using vector method, show that the given points A, B, C are collinear: (i) A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6) (ii) A(6, –7, –1), B(2, –3, 1) and C(4, –5, 0). ^

^

^

20. Show that the points A , B , C with position vectors ( 3 i - 2 j + 4 k ), ^

^

^

^

^

^

( i + j + k ) and ( - i + 4 j - 2 k ) respectively are collinear. ® ®

®

®

®

21. Show that the points having position vectors a , b , ( c = 3 a - 2 b ) are ® ® ®

collinear, whatever be a , b , c . 22. Show ®

that

the

®

points

®

®

having

®

®

®

vectors ( -2 a + 3 b + 5 c ),

position

®

®

® ®

( a + 2 b + 3 c ) and (7 a - c ) are collinear, whatever be a , b , c . 23. Find a unit vector perpendicular to the plane ABC, where the points A , B , C are ( 3 , - 1, 2), (1, -1, - 3) and ( 4, - 3 , 1) respectively. ®

^

^

®

^

^

®

^

®

24. If a = ( i - 2 j + 3 k ) and b = ( i - 3 k ) then find|b ´ 2 a |. ®

®

®

®

® ®

25. If|a|= 2,|b|= 5 and|a ´ b |= 8, find a × b . ®

®

®

®

^

^

^

26. If |a|= 2,|b|= 7 and ( a ´ b ) = ( 3 i + 2 j + 6 k ), find the angle between ®

®

a and b .

ANSWERS (EXERCISE 24)

1.

®

®

^

^

®

^

®

(i) a ´ b = ( -2 i + 8 j + 5 k ) and| a ´ b | = 93 ®

®

®

®

®

®

®

®

^

^

®

^

®

(ii) a ´ b = ( -17 i + 13 j + 7 k ) and| a ´ b | = 13 3 ^

®

^

®

(iii) a ´ b = 19( j + k ) and| a ´ b | = 19 2 ^

®

^

®

®

(iv) a ´ b = ( i - 10 j - 3 k ) and| a ´ b | = 110 ^

^

®

^

®

(v) a ´ b = ( 4 i - 3 j - k ) and| a ´ b | = 26 ^

^

^

3. ( 2 i - 51 j - 30 k )

2.

l = -3

5.

(i) ±

^ ^ ^ 1 (5 i - j + 7 k ) 5 3

(ii) ±

(iii) ±

^ ^ ^ 1 (9 i - j + 3 k) 91

(iv) ±

4. (i) 1 (ii) –1 (iii) 0 ^ ^ ^ 1 (- i + j + k ) 3 ^ ^ ^ 1 ( 2 i + 3 j + 14 k ) 209

SSS Mathematics for Class 12 1060

1060

Senior Secondary School Mathematics for Class 12

^ ^ ^ 1 6. ± ( 3 i - 2 j + 6 k ) 7

9.

®

p 3

10. d =

^

®

^ 1 ^ ^ ( i + j + 3 k) 4

^ ^ ^

11. d = 7( i - j - k ) (ii)

^

^

13. p =

2 3

42 sq units

(iii) 3 3 sq units

(iv) 6 sq units 1 (ii) (iii) 16. (i) 5 3 sq units 155 sq units 2 1 41 17. (i) sq units 165 sq units (ii) 2 2 1 1 (ii) (iii) 18. (i) 61 sq units 274 sq units 2 2 (iv) 13 sq units ^

23.

^

^

-i -2 j +4k 21

25. 6

24. 4 29

1 2

21 sq units

1 165 sq units 2

p 6

26.

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 24) 4.

^

^

^

^ ^

^

^

^

^ ^

^ ^

^

^

^

^ ^

(i) ( i ´ j ) × k + i × j = k × k + 0 = ( 1 + 0 ) = 1 [Q i ´ j = k and i × j = 0]. ^ ^

(ii) ( k ´ j ) × i + j × k = - i × i + 0 = ( -1 + 0 ) = -1. ®

5. Required vector =

®

±(a´b ) ®

®

×

| a ´ b| ®

6. Required vector =

®

±( a ´ b ) ®

×

®

| a ´ b| ®

7. Required vector =

®

±6( a ´ b ) ®

®

×

| a ´ b| ®

8. Required vector =

®

®

®

±5[( a + b ) ´ ( a - b )] ®

®

®

®

×

|( a + b ) ´ ( a - b )| ®

®

® ®

9. | a ´ b|=| a|| b|sin q = ( 1 ´ 2 ´ sin q) \ 2 sin q =

3 Þ sin q =

3 p Þ q= × 2 3

®

®

®

^

^

®

®

10. Since d is perpendicular to both a and b , so it is parallel to ( a ´ b ). \

®

®

®

^

^

^

^

d = l( a ´ b ) = l( i + j + 3 k ) = ( l i + l j + 3 l k )

^

5( - i + 2 j - k ) 8. ± 6

^

7. ± 2( - i + 2 j + 2 k )

(i) 6 5 sq units

15.

^

SSS Mathematics for Class 12 1061

Cross, or Vector, Product of Vectors ® ®

^

^

^

^

1061

^

c × d = 1 Þ (7 i - k ) × ( l i + l j + 3 l k ) = 1 Þ 7 l - 3l = 1 Þ 4l = 1 Þ l =

\ ®

®

d=

1 × 4

^ 1 ^ ^ ( i + j + 3 k ). 4

®

®

®

^

12. ( a ´ b ) =| a|| b|sin q × n ®

®

®

®

Þ | a ´ b|= ab sin q, where| a|= a and| b|= b ® ®

= ( ab cos q)(tan q) = ( a × b ) tan q. ®

®

®

®

®

13. a and b are parallel vectors Û a ´ b = 0. ®

®

^

^

^

^

^

^

^

^

15. (iv) Area =| a ´ b|=|2 i ´ 3 j |=|6 ( i ´ j )|=|6 k|= 6 [Q ( i ´ j ) = k]. 1 ® ® 16. Area of a||gm = | d 1 ´ d 2|. 2 ® ® 1 ® ® 17. Area of a triangle = | a ´ b|, where a and b are adjacent sides. 2 23. Required vector =

¾®

¾®

¾®

¾®

( AB ´ AC )

×

| AB ´ AC| ®

®

®

®

25. | a ´ b|=| a|| b|sin q Þ 2 ´ 5 ´ sin q = 8 Þ sin q = Þ cos q = 1 \ ®

16 = 25

4 5

9 3 = × 25 5

® ®

® ® 3ö æ a × b =| a|| b|cos q = ç 2 ´ 5 ´ ÷ = 6. 5ø è

®

^

^

^

26. a ´ b = ( 3 i + 2 j + 6 k ) ®

®

Þ | a ´ b |= Þ sin q =

®

7 ®

®

3 2 + 2 2 + 6 2 = 7 Þ | a || b |sin q = 7 ®

| a || b |

=

7 1 = Þ q = 30 °. (2 ´7 ) 2

SSS Mathematics for Class 12 1062

25. PRODUCT OF THREE VECTORS ® ® ®

SCALAR TRIPLE PRODUCT ® ® ®

The scalar triple product of three vectors a , b , c is defined ® ®®

as ( a ´ b ) × c , and it is denoted by [ a b c ]. ® ®®

®

®

®

Thus [ a b c ] = ( a ´ b ) × c . ®

®

®

®

®

Since ( a ´ b ) is a vector, ( a ´ b ) × c is a scalar. ® ®®

Consequently, [ a b c ] is a scalar quantity. Some Theorems on Scalar Triple Product ®

®

®

( a ´ b ) × c represents the volume of a

THEOREM 1 (Geometrical interpretation)

® ® ®

parallelepiped whose coterminous edges are represented by a , b , c . PROOF

Let us consider a parallelepiped having coterminous edges OA, OB and OC respectively. ®

¾®

®

¾®

®

¾®

Let OA = a , OB = b and OC = c . ®

®

Then, ( a ´ b ) is a vector perpendi®

®

cular to the plane of a and b . ®

®

Let q be the angle between ( a ´ b ) ®

and c . \

® ®®

®

®

®

®

®

®

[ a b c ] = ( a ´ b ) × c. = | a ´ b || c |cos q

®

®

®

= (area of | |gm OADB)[projection of c along ( a ´ b )] = (area of | |gm OADB) × OL = (area of the base of the parallelepiped) × (height) = volume of the parallelepiped with coterminous

® ® ®

edges a , b , c . ® ®®

Hence [ a b c ] represents the volume of the parallelepiped with ® ® ®

coterminous edges a , b , c forming a right-handed system. 1062

SSS Mathematics for Class 12 1063

Product of Three Vectors ® ® ®

THEOREM 2

1063 ®

®

®

®

®

®

For any three vectors a , b , c prove that ( a ´ b ) × c = a × ( b ´ c ). ® ® ®

PROOF

Let a , b , c form a right-handed system, representing the coterminous edges of a parallelepiped of volume V. ®

®

®

Then, V = ( a ´ b ) × c . ®

®

®

Again, b , c , a form a right-handed system, representing the coterminous edges of the same parallelepiped. \

®

®

®

®

®

®

® ®

V =(b ´ c) × a = a ×(b ´ c) ®

®

®

®

®

® ®

[B A × B = B × A].

®

Hence, ( a ´ b ) × c = a × ( b ´ c ). THEOREM 3

The scalar triple product of three vectors remains unchanged so long as their cyclic order remains unchanged, ®

®

®

®

®

®

®

®

®

(a ´ b) × c =(b ´ c ) × a =( c ´ a) × b.

i.e., ® ® ®

PROOF

Let a , b , c form a right-handed system, representing coterminous edges of a rectangular parallelepiped of volume V. ®

®

the

®

Then, V = ( a ´ b ) × c . ® ® ®

® ® ®

Clearly, b , c , a as well as c , a , b form a right-handed system, representing the coterminous edges of the same parallelepiped. ®

®

®

®

®

®

®

®

\

V = ( b ´ c ) × a and V = ( c ´ a ) × b .

Thus,

(a ´ b) × c =(b ´ c) × a =( c ´ a) × b

®

® ® ®

®

®

®

® ® ®

®

®

®

[each equal to V].

® ® ®

Hence, [ a b c ] = [ b c a ] = [ c a b ]. THEOREM 4

The scalar triple product changes in sign but not in magnitude when the cyclic order of vectors is changed. ® ® ®

PROOF

For any three vectors a , b , c , we know that ® ® ®

® ® ®

® ® ®

[ a b c ] = [ b c a ] = [ c a b ]. \

® ® ®

®

®

®

®

®

®

® ® ®

® ® ®

® ® ®

®

®

®

®

®

®

® ® ®

® ® ®

[ c b a ] = ( c ´ b ) × a = - ( b ´ c ) × a = - [ b c a ] = - [ a b c ].

And, [ a c b ] = ( a ´ c ) × b = - ( c ´ a ) × b = - [ c a b ] = - [ a b c ]. \ THEOREM 5

® ® ®

® ® ®

® ® ®

® ® ®

[ c b a ] = - [ a b c ] and [ a c b ] = - [ a b c ]. The scalar triple product vanishes if any two of its vectors are equal, i.e., ® ® ®

® ® ®

® ® ®

[ a a b ] = 0, [ a b a ] = 0 and [ b a a ] = 0. PROOF

We have

® ® ®

®

®

®

® ®

[ a a b ] = ( a ´ a ) × b = 0 × b = 0;

SSS Mathematics for Class 12 1064

1064

Senior Secondary School Mathematics for Class 12 ® ® ®

® ® ®

[ a b a ] = [ a a b ] [cyclic-ordered property] ®

®

®

® ®

= ( a ´ a ) × b = 0 × b = 0; ® ® ®

® ® ®

and [ b a a ] = [ a b a ] [cyclic-ordered property] = 0. ® ® ®

THEOREM 6

® ® ®

® ® ®

[ a a b ] = [ a b a ] = [ b a a ] = 0.

Hence,

The scalar triple product vanishes if any two of its vectors are parallel or collinear. ® ® ®

PROOF

®

®

®

®

Let a , b , c be three vectors such that a || b , or a and b are collinear. ®

®

Then, a = m b for some scalar m. \

® ® ®

®

®

[ a b c] = [m b

®

®

®

®

® ®

c ] = ( m b ´ b ) × c = 0 × c = 0.

b

THEOREM 7 (Scalar triple product in terms of components) ® ^ ^ ^ ® ^ ^ ^ Let a = a1 i + a2 j + a 3 k , b = b1 i + b2 j + b 3 k ® ^ ^ ^

and c = c1 i + c2 j + c 3 k. ½ a1 ® ® ® Then, [ a b c ] = ½ b1 ½ ½ c1

PROOF

½^ ½i ® ® ½ a ´ b = a1 ½ ½ ½b1 ½

We have

^

j

a2 b2

a3 ½ b 3 ½. ½ c3 ½

a2 b2 c2 ^½

k½ ½ a3 ½ b 3½ ½ ½ ^

^

^

= ( a2b 3 - a 3b2) i + ( a 3b1 - a1b 3) j + ( a1b2 - a2b1) k . \

® ® ®

®

®

®

[a b c ] = ( a ´ b ) × c

= ( a2b 3 - a 3b2) c1 + ( a 3b1 - a1b 3) c2 + ( a1b2 - a2b1) c 3 = a1( b2c 3 - b 3c2) + a2( b 3c1 - b1c 3) + a 3( b1c2 - b2c1) ½a1 = ½b1 ½ ½c1

a 3½ b 3 ½× ½ c3 ½

a2 b2 c2 ® ®

THEOREM 8

®

For any three vectors a , b and c , prove that ®

®

®

®

®

®

® ® ®

[ a + b , b + c , c + a ] = 2[ a , b , c ]

[CBSE 2014]

SSS Mathematics for Class 12 1065

Product of Three Vectors PROOF

We have ®

®

[a + b

®

®

®

b + c

®

®

®

®

®

®

1065

®

c + a]

®

®

®

®

®

®

= ( a + b ) × [( b + c ) ´ ( c + a )] ®

®

®

®

®

®

= ( a + b ) × [ b ´ c + b ´ a + c ´ c + c ´ a ] [by the distributive law] ®

®

®

®

®

®

= ( a + b ) × [( b ´ c ) - ( a ´ b ) + ( c ´ a )] ®

®

®

®

®

®

®

[Q c ´ c = 0 , and b ´ a = - a ´ b ] ®

®

®

®

®

®

®

®

®

®

®

®

= a ×( b ´ c ) - a ×( a ´ b ) + a ×( c ´ a ) + b ×( b ´ c ) ®

®

®

®

®

®

- b × ( a ´ b ) + b × ( c ´ a ) [by the distributive law] ® ® ®

® ® ®

® ® ®

® ® ®

® ® ®

® ® ®

= [a b c ] -[a a b ] + [a c a ] + [b b c ] -[b a b ] + [b c a ] ® ® ®

® ® ®

= [a b c ] + [b c a ] [Q scalar triple product with two equal vectors is 0] ® ® ®

® ® ®

® ® ®

= 2[ a b c ] {Q [ b c a ] = [ a b c ]}. Hence, THEOREM 9

®

®

®

[a + b

®

®

b + c

®

®

c + a ] = 2[ a

®

®

b

c ].

The necessary and sufficient condition for three nonzero, noncollinear ® ® ®

® ® ®

vectors a , b , c to be coplanar is that [ a b c ] = 0. ® ® ®

PROOF

Let a , b , c be three nonzero, noncollinear and coplanar vectors. Then, ®

®

®

®

b ´ c is perpendicular to the plane of b and c

Þ Þ

®

®

®

®

(b ´ c) ^ a ®

®

®

®

[Q a , b , c are coplanar]

®

® ® ®

a × ( b ´ c ) = 0 Þ [ a b c ] = 0. ®

®

®

® ® ®

Thus, whenever a , b , c are coplanar, we have [ a b c ] = 0. ®

®

®

Conversely, let a , b , c be three nonzero, noncollinear vectors such ® ® ®

®

that [ a b c ] = 0 . Then, ® ® ®

®

®

®

®

®

[ a b c ] = 0 Þ a ×( b ´ c ) = 0 ®

®

®

®

®

®

Þ a = 0 , or ( b ´ c ) = 0 , or ( b ´ c ) ^ a ®

®

®

®

®

®

®

®

®

Þ ( b ´ c ) = 0 , or ( b ´ c ) ^ a

®

®

[Q a ¹ 0 ]

Þ (b ´ c) ^ a ®

®

®

®

® ®

®

®

®

[Q b ¹ 0 , c ¹ 0 , and b , c are noncollinear Þ b ´ c ¹ 0 ].

SSS Mathematics for Class 12 1066

1066

Senior Secondary School Mathematics for Class 12 ®

®

®

Thus, ( b ´ c ) is perpendicular to a . ®

®

®

®

But, ( b ´ c ) is perpendicular to the plane of b and c . ®

\

®

®

a must lie in the plane of b and c . ® ® ®

Hence a , b , c must be coplanar. ® ® ®

® ® ®

NOTE: a , b , c are coplanar Û [ a b c ] = 0. ® ® ®

THEOREM 10

®

®

®

®

®

( c - a ) are coplanar. PROOF

®

For any three vectors a , b , c show that the vectors ( a - b ), ( b - c ), [CBSE 2001]

We know that ®

®

®

®

®

®

®

®

®

®

®

®

( a - b ), ( b - c ), ( c - a ) are coplanar Û

[a - b ®

b - c

®

®

Now, [ a - b ®

®

®

®

c - a ]=0

®

b - c

®

®

®

®

®

c - a]

®

= ( a - b ) × [( b - c ) ´ ( c - a )] ®

®

®

®

®

®

®

®

= ( a - b ) × [ b ´ c - b ´ a - c ´ c + c ´ a ] [by the distributive law] ®

®

® ®

® ®

®

®

®

®

®

®

® ®

®

= ( a - b ) × [ b ´ c + a ´ b + c ´ a ] [Q - b ´ a = a ´ b , and c ´ c = 0] ®

® ®

® ® ®

®

®

®

= a ×( b ´ c) + a ×( a ´ b ) + a ×[ c ´ a ] ®

® ®

®

®

®

®

®

®

- b ×( b ´ c ) - b ×( a ´ b ) - b ×( c ´ a ) ® ®®

® ® ®

® ® ®

® ® ®

® ® ®

® ® ®

= [a b c ] + [a a b ] + [a c a ] -[b b c ] -[b a b ] -[b c a ] ® ® ®

® ® ®

® ® ®

® ® ®

= [a b c ] - [b c a ]

[Q scalar triple product with two equal vectors is 0]

{Q

= [a b c ] - [a b c ] = 0 ®

®

®

®

®

® ® ®

® ® ®

[ b c a ] = [ a b c ]} .

®

Hence, ( a - b ), ( b - c ), ( c - a ) are coplanar. ® ® ®

THEOREM 11

®

®

®

®

®

( b + c ), ( c + a ) are coplanar. ®

PROOF

®

Show that the vectors a , b , c are coplanar if and only if ( a + b ), ®

®

®

®

®

®

®

®

®

®

®

®

®

[CBSE 2013C, ’14]

( a + b ), ( b + c ), ( c + a ) are coplanar Û [a + b

®

b + c ®

®

c + a] = 0

®

®

®

Û ( a + b ) × {( b + c ) ´ ( c + a )} = 0 ®

®

®

®

®

®

®

®

Û (a + b) ×{ b ´ c + b ´ a + c ´ c + c ´ a } = 0

SSS Mathematics for Class 12 1067

Product of Three Vectors ®

®

®

®

®

®

®

1067

®

Û (a + b) ×{ b ´ c + b ´ a + c ´ a } = 0 ®

®

®

®

®

®

®

®

[Q

®

®

®

®

®

c ´c = 0]

®

Û a ×(b ´ c) + a ×(b ´ a) + a ×( c ´ a) ®

®

®

®

®

®

®

+ b ×(b ´ c) + b ×(b ´ a) + b ×( c ´ a) = 0 ® ® ®

® ® ®

® ® ®

® ® ®

® ® ®

® ® ®

Û [a b c ] + [a b a ] + [a c a ] + [b b c ] + [b b a ] + [b c a ] = 0 ® ® ®

® ® ®

Û [a b c ] + [b c a ] = 0 [Q scalar triple product with two equal vectors is 0] ® ® ®

® ® ®

{Q

Û 2[ a b c ] = 0 ® ® ®

® ® ®

[ b c a ] = [ a b c ]}

Û [a b c] = 0 ®

®

®

Û a , b , c are coplanar. ® ®

®

®

®

®

®

®

®

Hence, a , b and c are coplanar if and only if ( a + b ), ( b + c ) and ®

®

( c + a ) are coplanar. ® ® ®

THEOREM 12 PROOF

®

For any three vectors a , b , c , prove that [ a

We have ®

®

®

®

b + c

[a

®

®

®

b+ c

®

®

a + b + c ] = 0.

®

a + b + c]

®

®

®

®

®

= { a ´ ( b + c )} × ( a + b + c ) ®

®

®

®

®

®

®

= {( a ´ b ) + ( a ´ c )} × ( a + b + c ) [by the distributive law] ®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

®

=(a ´ b) × a + (a ´ b) × b + (a ´ b) × c + (a ´ c) × a + (a ´ c) × b + (a ´ c) × c ® ® ®

® ® ®

® ® ®

[by the distributive law]

® ® ®

=[a b a] + [a b b] + [a b c] + [a c a] ® ® ®

® ® ®

+ [a c b] + [a c c] ® ® ®

® ® ®

® ® ®

® ® ®

=[a b c] + [a c b] [B scalar triple product with two equal vectors is 0] =[a b c] -[a b c] = 0 ®

®

THEOREM 13

®

®

b + c

Hence, [ a

®

®

®

®

{Q

® ® ®

® ® ®

[ a c b ] = - [ a b c ] }.

a + b + c ] = 0.

®

If a , b , c are the position vectors of points A, B, C, prove that ®

®

®

®

®

®

( a ´ b + b ´ c + c ´ a ) is a vector perpendicular to the plane of triangle ABC. [CBSE 2001C]

SSS Mathematics for Class 12 1068

1068 PROOF

Senior Secondary School Mathematics for Class 12

In order to prove the required result, we have to show that ®

®

®

®

®

®

¾® ¾®

( a ´ b + b ´ c + c ´ a ) is perpendicular to each of the vectors AB , BC ¾®

and CA . ¾®

®

®

¾®

®

®

¾®

®

®

We have, AB = ( b - a ), BC = ( c - b ) and CA = ( a - c ). ®

®

®

®

®

®

®

®

®

®

®

®

Now, ( a ´ b + b ´ c + c ´ a ) × ( b - a ) ®

®

®

®

®

®

®

®

®

®

®

®

= ( a ´ b) ×( b - a) + ( b ´ c) ×( b - a) + ( c ´ a) ×( b - a) ®

®

®

®

®

®

®

®

=( a ´ b) × b -( a ´ b) × a +(b ´ c) × b -(b ´ c) × a ®

®

®

®

®

®

+ ( c ´ a) × b - ( c ´ a)× a ® ® ®

® ® ®

® ® ®

® ® ®

® ® ®

® ® ®

= [a b b ] -[ a b a ] + [ b c b ] -[ b c a ] + [ c a b ] -[ c a a ] ®

®

=[c

®

®

®

®

b] - [ b c a] [Q scalar triple product with two equal vectors is 0]

a

®

®

®

®

®

®

= 0 (B [ c

a

®

b] =[b ®

®

c

®

®

a ] ).

®

®

®

®

®

Similarly, ( a ´ b + b ´ c + c ´ a ) × ( c - b ) = 0. ®

®

®

®

®

®

And, ( a ´ b + b ´ c + c ´ a ) × ( a - c ) = 0. ®

®

®

®

®

®

Thus, ( a ´ b + b ´ c + c ´ a ) is perpendicular to each one of the ¾®

¾®

¾®

vectors AB , BC and CA , and therefore, it is perpendicular to the plane of CABC.

EXAMPLE 1 SOLUTION

SOLVED EXAMPLES ^ ^ ^ ^ ^ ^

Prove that [ i j k ] = 1, and [ i k j ] = - 1. We have ^ ^ ^

^

^

^

^ ^ ^

^

^

^

^ ^

[ i j k ] = ( i ´ j ) × k = k × k = 1 and ^ ^

[ i k j ] = ( i ´ k ) × j = - j × j = - 1. ®

EXAMPLE 2

^

^

^ ®

^

^

^

®

^

^

^

If a = 2 i + j + 3 k , b = - i + 2 j + k , and c = - 3 i + j + 2 k , find ® ® ®

[ a b c ]. SOLUTION

We have ½ 2 1 3½ ½ 0 5 5½ ® ® ® [ a b c ] = ½ -1 2 1½= ½-1 2 1½ ½ ½ ½ ½ ½-3 1 2½ ½ 0 -5 -1½ [R1 ® R1 + 2R 2 , R 3 ® R 3 - 3 R 2] = - ( -1) × [-5 + 25] = 20.

SSS Mathematics for Class 12 1069

Product of Three Vectors EXAMPLE 3

1069

Find the volume of the parallelepiped whose coterminous edges are represented by the vectors ®

^

^

®

^

^

^

®

^

^

^

^

a = 2 i - 3 j + k , b = i - j + 2 k and c = 2 i + j - k . [CBSE 2000C]

SOLUTION

We have 2 -3 1½ ½0 -1 -3½ ® ® ® ½ 2½ [ a b c ] = ½1 -1 2½= ½1 -1 ½ ½ ½ ½ 1 -1½ ½0 3 -5½ ½2 [R1 ® R1 - 2R 2 , and R 3 ® R 3 - 2R 2] = ( -1) × (5 + 9) = - 14. \ volume of the parallelepiped ® ® ®

= |[ a b c ]| = | - 14| = 14 cubic units. EXAMPLE 4

Show that the vectors ^

^

^

^

^

^

^

^

^

i - 3 j + 4 k , 2 i - j + 2 k and 4 i - 7 j + 10 k are coplanar. ®

SOLUTION

^

^

^

®

^

^

®

^

^

^

^

Let a = i - 3 j + 4 k , b = 2 i - j + 2 k and c = 4 i - 7 j + 10 k . 1 ® ® ® ½ [ a b c ] = ½2 ½ ½4

\

-3 -1 -7

4½ ½1 2½ = ½0 ½ ½ 10½ ½0

-3 5 5

4½ ½R 2 ® R 2 - 2R1 ½ -6½½ ½ ½ R ® R 3 - 4R1½ -6½½ 3

= ( -30 + 30) = 0. Hence, the given vectors are coplanar. EXAMPLE 5

Find the value of l so that the vectors ®

^

^

^ ®

^

^

^

®

^

^

a = 2 i - 3 j + k , b = i + 2 j - 3 k and c = j + l k are coplanar. ® ® ®

SOLUTION

The given vectors will be coplanar if [ a b c ] = 0. 1½ 7½ ½0 -7 ½2 -3 ®®® ½ ½ ½ Now, [ a b c ] = 0 Û 1 2 -3 = 0 Û 1 2 -3½ = 0 ½ ½ ½ ½ 1 l½ 1 l½ ½0 ½0 [R1 ® R1 - 2R 2] Û ( -1)( -7 l - 7) = 0 Û 7 l + 7 = 0 Û l = - 1. Hence, the given vectors are coplanar when l = - 1.

EXAMPLE 6

^

^

^

^

^

^

^

^

( 3 i + 9 j + 4 k ) and 4( - i + j + k ) are coplanar. SOLUTION

^

^

^

Show that the four points with position vectors ( 4 i + 5 j + k ), ( - j - k ), [CBSE 2014]

Let the given points be A , B , C , D respectively. ¾® ¾®

¾®

Points A, B, C, D are coplanar Û AB , AC and AD are coplanar ¾®

Û [ AB

¾®

¾®

AC AD ] = 0.

SSS Mathematics for Class 12 1070

1070

Senior Secondary School Mathematics for Class 12 ¾®

AB = (p.v. of B) - (p.v. of A)

Now,

^

^

^

^

^

^

^

^

= ( - j - k ) - ( 4 i + 5 j + k ) = ( -4 i - 6 j - 2 k )

¾®

AC = (p.v. of C) - (p.v. of A) ^

¾®

^

^

^

^

^

^

^

^

= ( 3 i + 9 j + 4 k) - (4 i + 5 j + k) = (- i + 4 j + 3 k)

AD = (p.v. of D) - (p.v. of A) ^

\

^

^

= ( -4 i + 4 j + ½-4 ¾® ¾® ¾® [ AB AC AD ] = ½-1 ½ ½-8 ½ 0 = ½-1 ½ ½ 0

^

^

^

^

^

^

4 k ) - ( 4 i + 5 j + k ) = ( -8 i - j + 3 k ). -6 -2½ 4 3½ ½ -1 3½ -22 -14½ ìR1 ® R1 - 4R 2 ü 4 3½ í ½ îR 3 ® R 3 - 8R 2 þý -21 -33½

= - ( -1)[462 - 462] = 0.

¾® ¾®

¾®

\ AB , AC and AD are coplanar. Hence, the points A, B, C, D are coplanar. EXAMPLE 7

Find the value of l so that the four points with position vectors ^

^

^

^

^

^

^

^

^

( -6 i + 3 j + 2 k ), ( 3 i + l j + 4 k ), (5 i + 7 j + 3 k ) ^

^

^

and ( -13 i + 17 j - 2 k ) are coplanar. SOLUTION

[CBSE 2000]

Let the given points be A , B , C , D respectively. Then, ¾®

AB = (p.v. of B) - (p.v. of A) ^

^

^

^

^

^

= ( 3 i + l j + 4 k ) - ( -6 i + 3 j + 2 k ) ^

¾®

^

^

= 9 i + ( l - 3) j + 2 k

AC = (p.v. of C) - (p.v. of A) ^

¾®

^

^

^

^

^

^

^

^

= (5 i + 7 j + 3 k ) - ( -6 i + 3 j + 2 k ) = (11 i + 4 j + k )

AD = (p.v. of D) - (p.v. of A) ^

^

^

^

^

^

= ( -13 i + 17 j - k ) - ( -6 i + 3 j + 2 k ) ^

^

^

= ( -7 i + 14 j - 3 k ) Now, A , B , C , D are coplanar

2½ ½ 9 l-3 ¾® ¾® ¾® Û [ AB AC AD ] = 0 Û ½ 11 4 1½ = 0 ½ ½ 14 -3½ ½-7

SSS Mathematics for Class 12 1071

Product of Three Vectors

1071

Û 9( -12 - 14) - ( l - 3)( -33 + 7) + 2(154 + 28) = 0 Û - 234 + 26l - 78 + 364 = 0 Û 26l = - 52 Û l = - 2. Hence, the required value of l is -2. EXAMPLE 8

Show that the points A( -1, 4, - 3), B( 3 , 2, - 5), C( -3 , 8, - 5) and D( -3 , 2, 1) are coplanar.

SOLUTION

Clearly, the position vectors of A, B, C, D are ( - i + 4 j - 3 k ),

^

^

^

^

^

^

^

^

^

^

^

^

( 3 i + 2 j - 5 k ), ( -3 i + 8 j - 5 k ) and ( -3 i + 2 j + k ) respectively. ¾®

AB = (p.v. of B) - (p.v. of A)

\

^

^

^

^

^

^

^

^

^

= ( 3 i + 2 j - 5 k) - (- i + 4 j - 3 k) = (4 i - 2 j - 2 k) ¾®

AC = (p.v. of C) - (p.v. of A) ^

^

^

^

^

^

^

^

^

= ( -3 i + 8 j - 5 k ) - ( - i + 4 j - 3 k ) = ( -2 i + 4 j - 2 k ) ¾®

AD = (p.v. of D) - (p.v. of A) ^

^

^

^

^

^

^

^

\

¾® ¾® ¾®

Þ Þ

AB , AC , AD are coplanar the points A , B , C , D are coplanar.

EXERCISE 25A 1.

Prove that ^ ^ ^

^ ^ ^

^ ^ ^

^ ^ ^

^ ^ ^

(i) [ i j k ] = [ j k i ] = [ k j i ] = 1 ^ ^ ^

(ii) [ i k j ] = [ k j i ] = [ j i k ] = 1 2.

^

= ( - 3 i + 2 j + k ) - ( - i + 4 j - 3 k ) = ( - 2 i - 2 j + 4 k ). 4 -2 -2½ ½ 2 -1 -1 ½ ¾® ¾® ¾® ½ [ AB AC AD] = ½-2 4 -2½ = 2 ´ ( -2) ´ ( -2) ×½ 1 -2 1 ½ ½ ½ ½ ½ 1 -2 ½ ½-2 -2 4½ ½1 ½0 -3 3½ = 8 ×½0 -3 3½ ½ ½ 1 -2½ ½1 [R1 ® R1 - 2R 3 , R 2 ® R 2 - R 3] [Q R1 and R 2 are identical] = ( 8 ´ 0) = 0

® ® ®

Find [ a b c ], when ®

^

®

^

^ ®

^

^

^

^

®

^

^

^

^

^

^

®

^

^

^

(i) a = 2 i + j + 3 k , b = - i + 2 j + k and c = 3 i + j + 2 k ^

^ ®

(ii) a = 2 i - 3 j + 4 k , b = i + 2 j - k and c = 3 i - j + 2 k

SSS Mathematics for Class 12 1072

1072

Senior Secondary School Mathematics for Class 12 ®

^ ®

^

^

^

®

^

^

^

(iii) a = 2 i - 3 j , b = i + j - k and c = 3 i - k

3. Find the volume of the parallelepiped whose coterminous edges are represented by the vectors ®

^

^

®

^

^

^

®

^

^

^

^

(i) a = i + j + k , b = i - j + k , c = i + 2 j - k ®

^

^ ®

^

^

^ ®

^

^

^

^

(ii) a = - 3 i + 7 j + 5 k , b = - 5 i + 7 j - 3 k , c = 7 i - 5 j - 3 k ®

^

^ ®

^

^

^ ®

^

^

^

(iii) a = i - 2 j + 3 k , b = 2 i + j - k , c = j + k ®

^ ®

^ ®

^

(iv) a = 6 i , b = 2 j , c = 5 i

® ® ®

4. Show that the vectors a , b , c are coplanar, when ®

^

®

^

^ ®

^

^

^

®

^

^

^

^

(i) a = i - 2 j + 3 k , b = - 2 i + 3 j - 4 k and c = i - 3 j + 5 k ^ ®

^

^

^

®

^

^

^

(ii) a = i + 3 j + k , b = 2 i - j - k and c = 7 j + 3 k ®

^

^ ®

^

^

^

®

^

^

^

^

(iii) a = 2 i - j + 2 k , b = i + 2 j - 3 k and c = 3 i - 4 j + 7 k ® ® ®

5. Find the value of l for which the vectors a , b , c are coplanar, where ®

^

®

^

^

®

^

^

^

®

^

^

^

^

(i) a = ( 2 i - j + k ), b = ( i + 2 j + 3 k and c = ( 3 i + l j + 5 k ) ^ ®

^

^

^

[CBSE 2004]

®

^

^

^

^

^

(ii) a = l i - 10 j - 5 k , b = - 7 i - 5 j and c = i - 4 j - 3 k ®

^

^

^ ®

^

^

^

®

^

(iii) a = i - j + k , b = 2 i + j - k and c = l i - j + l k ®

^

^

^

®

^

^

®

^

^

^

[CBSE 2004C] ® ® ®

^

6. If a = ( 2 i - j + k ), b = ( i - 3 j - 5 k ) and c = ( 3 i - 4 j - k ), find [ a b c ] and interpret the result. ^

^

^

^

7. The volume of the parallelepiped whose edges are ( -12 i + l k ), ( 3 j - k ) ^

^

^

and ( 2 i + j - 15 k ) is 546 cubic units. Find the value of l. ®

^

^

^ ®

^

^

^

[CBSE 2004] ®

^

^

8. Show that the vectors a = ( i + 3 j + k ), b = ( 2 i - j - k ) and c = (7 j + 3 k ) ®®®

are parallel to the same plane. { HINT: Show that [ a b c ] = 0} ^

^

^

^

^

^

^

^

9. If the vectors ( a i + a j + c k ), ( i + k ) and ( c i + c j + b k ) be coplanar, show that c2 = ab.

[CBSE 2005] ^

^

^

10. Show that the four points with position vectors ( 4 i + 8 j + 12 k ), ^

^

^

^

^

^

^

^

^

( 2 i + 4 j + 6 k ), ( 3 i + 5 j + 4 k ) and (5 i + 8 j + 5 k ) are coplanar.

SSS Mathematics for Class 12 1073

Product of Three Vectors

11. Show

that

^

the

^

four

^

^

^

points

^

with

^

1073

position ^

^

^

vectors ( 6 i - 7 j ),

(16 i - 19 j - 4 k ), ( 3 j - 6 k ) and ( 2 i - 5 j + 10 k ) are coplanar. [CBSE 2004, ‘05]

12. Find the value of l for which the four points with position vectors ^

^

^

^

^

^

^

^

^

^

^

^

( i + 2 j + 3 k ), ( 3 i - j + 2 k ), ( -2 i + l j + k ) and ( 6 i - 4 j + 2 k ) are coplanar. [CBSE 2000] ^

^

13. Find the value of l for which the four points with position vectors ( - j + k ), ^

^

^

^

^

^

^

^

( 2 i - j - k ), ( i + l j + k ) and ( 3 j + 3 k ) are coplanar.

[CBSE 2000]

14. Using vector method, show that the points A( 4, 5 , 1), B( 0, - 1, - 1), C( 3 , 9, 4) and D( -4, 4, 4) are coplanar. 15. Find the value of l for which the points A( 3 , 2, 1), B( 4, l , 5), C( 4, 2, - 2) and D( 6, 5 , - 1) are coplanar. [CBSE 2004C]

ANSWERS (EXERCISE 25A)

2. (i) –10

(ii) –7

(iii) 4

3. (i) 4 cubic units (ii) 264 cubic units (iv) 60 cubic units

(iii) 12 cubic units

5. (i) l = - 4 (ii) l = - 3 (iii) l = 1 6. 0, the given vectors are coplanar 7. l = -3 12. l = 3

13. l = 1

15. l = 5

EXERCISE 25B Very-Short-Answer Questions ®

^

^

®

^

^

^

^

1. If a = x i + 2 j - z k and b = 3 i - y j + k are two equal vectors then x+ y+z =? [CBSE 2013] 2. Write a unit vector in the direction of the sum of the vectors ®

^

^

^

®

^

^

^

a = ( 2 i + 2 j - 5 k ) and b = ( 2 i + j - 7 k ).

[CBSE 2014] ®

^

^

^

3. Write the value of l so that the vectors a = ( 2 i + l j + k ) and ®

^

^

^

b = ( i - 2 j + 3 k ) are perpendicular to each other. ®

[CBSE 2013C] ^

^

^

4. Find the value of p for which the vectors a = ( 3 i + 2 j + 9 k ) and ®

^

^

^

b = ( i - 2p j + 3 k ) are parallel.

[CBSE 2014]

SSS Mathematics for Class 12 1074

1074

Senior Secondary School Mathematics for Class 12 ®

^

^

^

5. Find the value of l when the projection of a = ( l i + j + 4 k ) on ®

^

^

^

b = ( 2 i + 6 j + 3 k ) is 4 units. ®

[CBSE 2012]

®

®

®

®

6. If a and b are perpendicular vectors such that|a + b|= 13 and|a|= 5 , find ®

the value of |b|.

[CBSE 2014]

®

®

®

®

®

®

7. If a is a unit vector such that ( x - a ) × ( x + a ) = 15 , find|x|. 8. Find ®

the ^

sum ^

of

the

®

^

®

^

a = ( i - 3 k ),

vectors

[CBSE 2013] ^

^

b = (2 j - k)

^

c = ( 2 i - 3 j + 2 k ).

and

[CBSE 2012] ®

^ ®

^

^

®

^

^

^

9. Find the sum of the vectors a = ( i - 2 j ), b = ( 2 i - 3 j ) and c = ( 2 i + 3 k ). [CBSE 2012] ^

^

^

^

10. Write the projection of the vector ( i + j + k ) along the vector j . [CBSE 2014] ^

^

^

11. Write the projection of the vector (7 i + j - 4 k ) on the vector ^

^

^

( 2 i + 6 j + 3 k ). 12. Find ®

®

®

[CBSE 2013C]

®

a ×(b ´ c)

^

^

when

®

^

^

^

a = ( 2 i + j + 3 k ),

®

^

^

^

b = (- i + 2 j + k)

^

c = ( 3 i + j + 2 k ).

and

[CBSE 2014] ^

^

^

13. Find a vector in the direction of ( 2 i - 3 j + 6 k ) which has magnitude 21 units. [CBSE 2014] ®

^

^

^

®

^

^

®

^

^

^

14. If a = ( 2 i + 2 j + 3 k ), b = ( - i + 2 j + k ) and c = ( 3 i + j ) are such that ®

®

®

( a + l b ) is perpendicular to c then find the value of l.

[CBSE 2009C] ^

^

^

15. Write a vector of magnitude 15 units in the direction of vector ( i - 2 j + 2 k ). [CBSE 2010] ®

^

^

^

®

^

^

®

^

^

^

^

16. If a = ( i + j + k ), b = ( 4 i - 2 j + 3 k ) and c = ( i - 2 j + k ), find a vector of ®

®

®

magnitude 6 units which is parallel to the vector ( 2 a - b + 3 c ). [CBSE 2010] ^

^

^

^

17. Write the projection of the vector ( i - j ) on the vector ( i + j ). [CBSE 2011] ®

®

18. Write the angle between two vectors a and b with magnitudes ® ®

respectively having a × b = 6.

3 and 2

[CBSE 2011]

SSS Mathematics for Class 12 1075

Product of Three Vectors ®

^

^

®

^

^

^

1075 ®

^

®

19. If a = ( i - 7 j + 7 k ) and b = ( 3 i - 2 j + 2 k ) then find|a ´ b|. [CBSE 2008C] ®

®

20. Find the angle between two vectors a and b with magnitudes 1 and 2 ®

®

respectively, when|a ´ b|=

3.

[CBSE 2009C] ®

®

®

®

®

21. What conclusion can you draw about vectors a and b when a ´ b = 0 and ® ®

a × b = 0?

®

^

^

®

of l when the vectors

22. Find the value

^

^

^

a = ( i + l j + 3 k ) and

^

b = ( 3 i + 2 j + 9 k ) are parallel. ^

^

^

^

^

^

^

^

^

23. Write the value of i × ( j ´ k ) + j × ( i ´ k ) + k × ( i ´ j ). 24. Find the volume of the parallelepiped whose edges are represented by the ®

^

^ ®

^

^

^

®

^

^

^

^

vectors a = ( 2 i - 3 j + 4 k ), b = ( i + 2 j - k ) and c = ( 3 i - 2 j + 2 k ). ®

^

^

^

®

^

^

®

^

^

^

^

a = ( -2 i - 2 j + 4 k ), b = ( -2 i + 4 j - 2 k ) and c = ( 4 i - 2 j - 2 k ) then

25. If

® ®

®

prove that a , b and c are coplanar. ®

^

^

®

^

^

^

®

^

®

®

26. If a = ( 2 i + 6 j + 27 k ) and b = ( i + l j + m k ) are such that a ´ b = 0 then find the values of l and m. ®

®

® ®

®

®

27. If q is the angle between a and b , and |a × b|=|a ´ b| then what is the value of q? ®

®

®

®

28. When does|a + b| = |a| + |b|hold? 29. Find the direction cosines of a vector which is equally inclined to the x-axis, y-axis and z-axis. 30. If P (1, 5, 4) and Q(4, 1, –2) be the position vectors of two points P and Q, ¾®

find the direction ratios of PQ . ®

^

^

^

31. Find the direction cosines of the vector a = ( i + 2 j + 3 k ). ^

^

^

^

32. If a and b are unit vectors such that ( a + b ) is a unit vector, what is the ^

^

angle between a and b ? ANSWERS (EXERCISE 25B)

1. 0

2.

^ ^ 1 ^ ( 4 i + 3 j - 12 k ) 13

3. l =

5 2

4. p =

-1 3

5. l = 5

®

6. |b| = 12

SSS Mathematics for Class 12 1076

1076

Senior Secondary School Mathematics for Class 12 ®

^

7. |x| = 4

^

^

^

^

13. ( 6 i - 9 j + 18 k ) p 4

18.

23. 1

24. 7 cubic untis

^

19. 19 2

2 , 14

3 14

10. 1

^

®

p 3

20.

^

®

®

26. l = 3 , m = 29.

27 2

27.

11.

8 7

12. –10

^

^

22. l =

2 3

®

p 4

1 1 1 , , 3 3 3

30. 3, –4, –6

2p 3

32.

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 25B) ®

®

^

^

^

^

^

^

1. a = b Þ ( x i + 2 j - z k ) - ( 3 i - y j + k ) Þ x = 3 , - y = 2 and -z = 1 Þ x = 3 , y = -2 and z = -1 Þ ( x + y + z) = [ 3 + ( -2 ) + ( -1)] = 0. ®

2. Required unit vector =

®

(a´b ) ®

®

^

=

®

®

®

^

16 + 9 + 144

| a ´ b| ®

^

( 4 i + 3 j - 12 k )

=

® ®

3. a ^ b Û a × b = 0 Û ( 2 - 2 l + 3 ) = 0 Û 2 l = 5 Û l = ®

®

^ ^ ^ 1 ( 4 i + 3 j - 12 k ). 13

5 × 2

®

4. a || b Û a ´ b = 0 ^

^

i j Û 3 2 1 -2 p

^

k ^ ^ ^ ® 9 = 0 Û ( 6 + 18 p ) i - ( 9 - 9 ) j + ( -6 p - 2 ) k = 0 3 ^

^

^

^

Û ( 6 + 18 p ) i + 0 × j + ( -6 p - 2 ) k = 0

Û 6 + 18 p = 0 and -6 p - 2 = 0 Û p = ® ®

a× b

5. Clearly,

®

=4Þ

| b| ®

( 2 l + 6 + 12 ) 4 + 36 + 9

-1 × 3

= 4 Þ ( 2 l + 18 ) = 28 Þ 2 l = 10 Þ l = 5.

®

® ®

6. Since a and b are perpendicular vectors, we have a × b = 0. ®

®2

®

®

®

®

Now,| a + b| = 169 Þ ( a + b ) × ( a + b ) = 169 ® ®

® ®

® ®

Þ a × a + 2 a × b + b × b = 169

^

16. ( 2 i - 4 j + 4 k )

21. a = 0 or b = 0

®

1 , 14

^

15. (5 i - 10 j + 10 k )

28. a and b are parallel or collinear 31.

^

9. (5 i - 5 j + 3 k )

14. l = 8

17. 0

®

^

^

8. ( 3 i - j - 2 k )

SSS Mathematics for Class 12 1077

Product of Three Vectors ®2

®2

1077

® ®

Þ | a| + | b| = 169 [Q a × b = 0] ®2

®2

®

Þ 25 + | b| = 169 Þ | b| = 144 Þ | b| = 12 . ®

®

®

®

®2

®2

7. ( x - a ) × ( x + a ) = 15 Þ | x| - | a| = 15 ®2

®2

®

®

Þ | x| - 1 = 15 Þ | x| = 16 Þ | x|= 4 {Q | a|= 1}. ®

®

®

^

^

^

8. ( a + b + c ) = ( 1 + 0 + 2 ) i + ( 0 + 2 - 3 ) j + [( -3 ) + ( -1) + 2] k ^

^

^

= ( 3 i - j - 2 k ). ®

®

®

^

^

^

^

^

^

9. ( a + b + c ) = ( 1 + 2 + 2 ) i + ( -2 - 3 + 0 ) j + 3 k = (5 i - 5 j + 3 k ). ^

10. Required projection =

^

^

^

( i + j + k )× j

( 0 + 1 + 0) = 1. 1

=

^

|j | ^

11. Required projection =

^

^

^

^

^

(7 i + j - 4 k ) × ( 2 i + 6 j + 3 k ) ^

^

^

|2 i + 6 j + 3 k| ( 14 + 6 - 12 )

= ^

^

8 8 = × 49 7

=

22 + 62 + 32

^

i j k ^ ^ ^ 12. ( b ´ c ) = -1 2 1 = ( 3 i + 5 j - 7 k ) 3 1 2 ®

®

®

®

®

^

^

^

^

^

^

a × ( b ´ c ) = ( 2 i + j + 3 k ) × ( 3 i + 5 j - 7 k ) = ( 6 + 5 - 21) = -10.

\

^

13. Required vector =

^

^

21( 2 i - 3 j + 6 k ) 2 2 + ( -3 ) 2 + 6 2 ^

^

^

=

^

^

21( 2 i - 3 j + 6 k ) 49

^

^

^

^

= 3( 2 i - 3 j + 6 k ) = ( 6 i - 9 j + 18 k ). ®

®

®

®

^

^

^

14. ( a + l b ) = ( 2 - l) i + ( 2 + 2 l) j + ( 3 + l) k ®

^

^

^

^

^

( a + l b ) × c = 0 Þ [( 2 - l) i + ( 2 + 2 l) j + ( 3 + l) k] × ( 3 i + j ) = 0 Þ 3( 2 - l) + 1 × ( 2 + 2 l) = 0 Þ 6 - 3 l + 2 + 2 l = 0 Þ l = 8. 15. Required vector = ®

®

^

^

^

2

2

2

15 ( i - 2 j + 2 k ) 1 ( -2 ) + 2

®

^

^

^

^

^

^

^

^

^

= 5 ( i - 2 j + 2 k ) = (5 i - 10 j + 10 k ). ^

^

^

^

^

^

16. ( 2 a - b + 3 c ) = ( 2 i + 2 j + 2 k ) - ( 4 i - 2 j + 3 k ) + ( 3 i - 6 j + 3 k ) ^

^

^

= ( i - 2 j + 2 k ).

SSS Mathematics for Class 12 1078

1078

Senior Secondary School Mathematics for Class 12

\

^

^

^

^

^

^

^

^ ^ ^ 6( i - 2 j + 2 k ) = =(2 i - 4 j + 4 k ) 2 2 2 3 1 + ( -2 ) + 2

required vector =

^

^

( i - j )×( i + j )

17. Required projection =

^

6( i - 2 j + 2 k )

^

^

^ ^

=

^ ^

^ ^

12 + 12

| i + j| = ®

^ ^

(i × i + i × j - j × i - j × j ) ( 1 + 0 - 0 - 1)

= 0.

2

®

18. Let q be the angle between a and b . Then, ® ®

a×b

cos q =

®

6 ´ 3 ´2

=

®

| a|| b|

3´ 2´ 3 6

3 = 3

2 1 p = Þ q= × 2 4 2

= ^

^

^

i j k ^ ^ ^ 19. ( a ´ b ) = 1 -7 7 = ( -14 + 14 ) i - ( 2 - 21) j + ( -2 + 21) k 3 -2 2 ®

®

^

^

= ( 19 j + 19 k ) \

®

®

| a ´ b|= 0 2 + ( 19 ) 2 + ( 19 ) 2 = 2 ´ ( 19 ) 2 = 19 2 .

®

®

®

®

®

®

20. | a ´ b|= | a|| b|sin q Þ 1 ´ 2 ´ sin q = ®

® ®

®

3 p Þ q= × 2 3

3 Þ sin q = ®

®

®

21. ( a ´ b = 0 and a × b = 0) only when a = 0 or b = 0 . ®

®

®

®

®

22. Clearly, a || b Û a ´ b = 0 . ^

^

^

i j (a ´b)= 1 l 3 2 ®

\

k ^ ^ ^ 3 = ( 9 l - 6 ) i - 0 j + ( 2 - 3 l) k 9

®

®

®

®

( a ´ b ) = 0 Û 9 l - 6 = 0 and 2 - 3 l = 0 Û l = ^

^

^

^

^

^

^

^

^

^

^

^

^

2 × 3

^

23. Clearly, ( j ´ k ) = i , ( i ´ k ) = - j and ( i ´ j ) = k . \

^

^

^

^

^ ^

^ ^

^ ^

i ×( j ´ k ) + j ×( i ´ k ) + k ×( i ´ j ) = ( i × i - j × j + k × k )

= ( 1 - 1 + 1) = 1. 2 -3 4 ® ® ® 24. [ a b c ] = 1 2 -1 = 2( 4 - 1) + 3( 2 + 3 ) + 4( -1 - 6 ) 3 \

-1

2

= ( 6 + 15 - 28 ) = -7 . volume of the parallelepiped =|- 7|= 7 cubic units.

SSS Mathematics for Class 12 1079

Product of Three Vectors

1079

-2 -2 4 ® ® ® 25. [ a b c ] = -2 4 -2 = ( -2 )( -8 - 4 ) + 2( 4 + 8 ) + 4( 4 - 16 ) 4 -2 -2 = ( 24 + 24 - 48 ) = 0. ® ® ®

Hence a , b , c are coplanar. ^

^

^

i j k ^ ^ ^ 26. ( a ´ b ) = 2 6 27 = ( 6m - 27 l) i - ( 2m - 27 ) j + ( 2 l - 6 ) k . 1 l m ®

®

®

®

®

27 × 2 ® ® ® ® ® ® ® ® p 27. | a × b|=| a ´ b| Þ | a|| b|cos q = | a|| b|sin q Þ tan q = 1 Þ q = × 4 \

®

( a ´ b ) = 0 Û 2 l - 6 = 0 and 2m - 27 = 0 Û l = 3 and m =

®

®

®

®

®2

®

®

®

®

28. | a + b|=| a| + | b| Þ | a + b| = {| a| +| b|} ®

®

2

®2

®2

®

®

Þ ( a + b ) × ( a + b ) =| a| + | b| + 2| a|| b| ® ®

® ®

® ®

®2

®2

®2

®2

®

®

Þ a × a + 2 a × b + b × b =| a| +| b| + 2| a|| b| ®2

® ®

®2

®

®

Þ | a| + 2 a × b + | b| = | a| + | b| + 2 | a|| b| ® ®

®

®

® ®

®

®

Þ 2 ( a × b ) = 2 (| a|| b|) Þ a × b = | a|| b| ®

®

®

®

Þ | a|| b|cos q = | a|| b| Þ cos q = 1 Þ q = 0 °. \

®

®

either a and b are parallel or collinear.

29. cos 2 a + cos 2 a + cos 2 a = 1 Þ 3 cos 2 a = 1 Þ cos a = So, the direction cosines of the vector are

1 1 1 , , × 3 3 3

¾®

30. D. R.’s of PQ are ( 4 - 1), ( 1 - 5 ), ( -2 - 4 ), i.e., 3, –4, –6. ®

31. Direction ratios of a are 1, 2, 3. And, 12 + 2 2 + 3 2 = 14 . \ ^

®

direction cosines of a are ^

^

^

1 2 3 , , × 14 14 14 ^ ^

32. ( a + b ) × ( a + b ) = 1 Þ a2 + b 2 + 2 a × b = 1 ^ ^

^ ^

Þ 1 + 1 + 2 a × b = 1 Þ 2 a × b = -1 ^ ^ -1 -1 Þ |a ||b |cos q = 2 2 -1 2p Þ cos q = Þ q= × 2 3 ^ ^

Þ a× b =

1 × 3

SSS Mathematics for Class 12 1080

1080

Senior Secondary School Mathematics for Class 12

OBJECTIVE QUESTIONS Mark ( ) against the correct answer in each of the following. ®

^

^

^

1. A unit vector in the direction of the vector a = ( 2 i - 3 j + 6 k ) is ^ö æ^ 3 ^ (a) çç i - j + 3 k ÷÷ 2 ø è

æ 2 ^ 3 ^ 6 ^ö (b) çç i - j + k ÷÷ 5 5 ø è5

æ 2 ^ 3 ^ 6 ^ö (c) çç i - j + k ÷÷ 7 7 ø è7

(d) none of these ®

^

^

^

2. The direction cosines of the vector a = ( -2 i + j - 5 k ) are 1 -1 , , 3 6 -2 (d) , 30

(a) –2, 1, –5 (c)

2 , 30

(b) 1 , 30

5 30

-5 6 1 , 30

-5 30 ¾®

3. If A(1, 2, –3) and B(–1, –2, 1) are the end points of a vector AB then the ¾®

direction cosines of AB are -1 (a) –2, –4, 4 (b) , - 1, 1 2

(c)

-1 -2 2 , , 3 3 3

(d) none of these

4. If a vector makes angles a , b and g with the x-axis, y-axis and z-axis respectviely then the value of (sin 2 a + sin 2 b + sin 2 g) is (a) 1

(b) 2

(c) 0

^

(d) 3

^

^

5. The vector (cos a cosb) i + (cos a sin b) j + (sin a ) k is a (a) null vector (c) a constant vector

(b) unit vector (d) none of these ^

^

^

6. What is the angle which the vector ( i + j + 2 k ) makes with the z-axis? p p p 2p (b) (c) (d) (a) 4 3 6 3 ®

®

®

®

7. If a and b are vectors such that |a| = ®

®

angle between a and b is p p (a) (b) 6 3 ®

® ®

3 ,|b| = 2 and a × b = 6 then the

®

(c)

p 4

®

(d) ®

2p 3

® ®

8. If a and b are two vectors such that|a| = |b| = 2 and a × b = -1 then the ®

®

angle between a and b is p p (a) (b) 6 4

(c)

p 3

(d)

2p 3

SSS Mathematics for Class 12 1081

Product of Three Vectors ®

^

^

1081 ®

^

^

^

^

9. The angle between the vectors a = i - 2 j + 3 k and b = 3 i - 2 j + k is (a) cos-1 ®

5 7

(b) cos-1

^

^

3 5

®

^

3 14

(c) cos-1 ^

^

(d) none of these ®

^

®

10. If a = ( i + 2 j - 3 k ) and b = ( 3 i - j + 2 k ) then the angle between ( a + b ) ®

®

and ( a - b ) is p 3

(a)

p 4

(b)

®

^

^

(c) ®

^

^

p 2

(d)

^

2p 3

^

11. If a = ( i + 2 j - 3 k ) and b = ( 3 i - j + 2 k ) then the angle between ®

®

®

®

( 2 a + b ) and ( a + 2 b ) is æ 21 ö (a) cos-1 ç ÷ è 40 ø ®

^

æ 31 ö (b) cos-1 ç ÷ è 50 ø

^

®

^

^

æ 11 ö (c) cos-1 ç ÷ è 30 ø ^

(d) none of these ®

^

®

12. If a = ( 2 i + 4 j - k ) and b = ( 3 i - 2 j + l k ) be such that a ^ b then l = ? (a) 2

(b) –2

(c) 3 ®

^

^

(d) –3 ®

^

^

^

^

13. What is the projection of a = ( 2 i - j + k ) on b = ( i - 2 j + k )? 2 3

(a) ®

4 5

(b) ®

®

(c)

5 6

(d) none of these

®

14. If|a + b |=|a - b |, then ®

®

®

(a) |a|=|b | 15. If

®

and

a

®

®

®

b

®

®

®

®

(b) a || b

(c) a ^ b

are

perpendicular

mutually

®

(d) none of these unit

vectors

then

( 3 a + 2 b ) × (5 a - 6 b ) = ? (a) 3

(b) 5 ®

^

(c) 6 ^

®

^

(d) 12 ^

^

^

16. If the vectors a = 3 i + j - 2 k and b = i + l j - 3 k are perpendicular to each other then l = ? (a) –3 (b) –6 (c) –9 (d) –1 ^ ^ 1 ^ ^ 17. If q is the angle between two unit vectors a and b then |a - b|= ? 2 q q q (b) sin (c) tan (d) none of these (a) cos 2 2 2 ®

^

^

^

®

^

^

^

®

®

18. If a = ( i - j + 2 k ) and b = ( 2 i + 3 j - 4 k ) then|a ´ b |= ? (a) 174

(b) 87

(c) 93

(d) none of these

SSS Mathematics for Class 12 1082

1082

Senior Secondary School Mathematics for Class 12 ®

^

^

®

^

^

®

^

®

19. If a = ( i - 2 j + 3 k ) and b = ( i - 3 k ) then|b ´ 2 a |= ? (b) 5 17

(a) 10 3 ®

®

®

(c) 4 19

®

^

^

(d) 2 23 ®

^

20. If |a|= 2,|b|= 7 and ( a ´ b ) = ( 3 i + 2 j + 6 k ) then the angle between a ®

and b is p (a) 6

(b)

®

p 3

®

(c) ®

2p 3

®

(d)

3p 4

® ®

21. If|a| = 26 ,|b| = 7 and|a ´ b| = 35 then a × b = ? (a) 5

(b) 7

(c) 13

(d) 12

22. Two adjacent sides of a ||gm are represented by the vectors ®

^

^

®

^

^

^

^

a = ( 3 i + j + 4 k ) and b = ( i - j + k ). The area of the||gm is

(a) 42 sq units

(b) 6 sq units

(c) 35 sq units

(d) none of these ®

^

^

^

23. The diagonals of a||gm are represented by the vectors d1 = ( 3 i + j - 2 k ) ®

^

^

^

and d 2 = ( i - 3 j + 4 k ). The area of the||gm is (a) 7 3 sq units

(b) 5 3 sq units

(c) 3 5 sq units

(d) none of these ®

^

^

24. Two adjacent sides of a triangle are represented by the vectors a = 3 i + 4 j ®

^

^

and b = -5 i + 7 j . The area of the triangle is (a) 41 sq units

(b) 37 sq units

(c)

41 sq units 2

(d) none of these ®

^

^

^

25. The unit vector normal to the plane containing a = ( i - j - k ) and ®

^

^

^

b = ( i + j + k ) is ^

^

^

(a) ( j - k ) ®

26. If a , ® ®

^

(b) ( - j + k )

®

b and

® ®

(c)

^ ^ 1 ( - j + k) 2

®

(d) ®

^ ^ 1 ( - i + k) 2 ®

®

®

c are unit vectors such that a + b + c = 0

then

® ®

(a × b + b × c + c × a) =? (a)

1 2

® ®

(b)

-1 2

(c)

3 2

(d)

-3 2

®

®

®

®

27. If a , b and c are mutually perpendicular unit vectors then|a + b + c| = ? (a) 1

(b) 2

(c) 3

(d) 2

SSS Mathematics for Class 12 1083

Product of Three Vectors

1083

^ ^ ^

28. [ i j k ] = ? (a) 0 ®

(b) 1

^

^

(c) 2 ®

^

^

^

(d) 3 ®

^

^

^

^

29. If a = ( 2 i - 3 j + 4 k ), b = ( i + 2 j - k ) and c = ( 3 i - j - 2 k ) be the coterminous edges of a parallelepiped then its volume is (a) 21 cubic units (b) 14 cubic units (c) 7 cubic units (d) none of these 30. If

the

®

^

volume ^

of

a ®

^

®

^

^

^

^

^

having a = (5 i - 4 j + k ),

parallelepiped ^

b = ( 4 i + 3 j + l k ) and c = ( i - 2 j + 7 k ) as coterrminous edges, is 216 cubic units then the value of l is 5 4 2 1 (a) (b) (c) (d) 3 3 3 3 31. It is given that the vectors ®

^

®

^

®

^

a = ( 2 i - 2 k ),

^

^

b = i + ( l + 1) j

and

^

c = ( 4 i + 2 k ) are coplanar. Then, the value of l is 1 1 (a) (b) (c) 2 2 3

(d) 1

32. Which of the following is meaningless? ®

®

®

®

®

(a) a × ( b ´ c ) ®

®

® ®

®

®

®

(b) a ´ ( b × c )

(c) ( a ´ b ) × c

(b) 1

(c) a 2 b

(d) none of these

33. a × ( a ´ b ) = ? (a) 0

® ® ®

(d) meaningless ®

®

34. For any three vectors a , b , c the value of [ a - b ® ® ®

(a) 2[ a b c ]

(b) 1

®

®

b -c

(c) 0

®

®

c - a ] is

(d) none of these

ANSWERS (OBJECTIVE QUESTIONS)

1. (c) 8. (d) 15. (a) 22. (a) 29. (c)

2. (d) 9. (a) 16. (c) 23. (b) 30. (a)

3. (c) 10. (c) 17. (b) 24. (c) 31. (d)

4. (b) 11. (b) 18. (c) 25. (c) 32. (b)

5. (b) 12. (b) 19. (c) 26. (d) 33. (a)

6. (a) 13. (c) 20. (a) 27. (c) 34. (c)

HINTS TO THE GIVEN OBJECTIVE QUESTIONS ®

1. | a| = 2 2 + ( -3 ) 2 + 6 2 = 49 = 7 . æ 2 ^ 3 ^ 6 ^ö \ required unit vector is ç i - j + k ÷ . 7 7 ø è7

7. (c) 14. (c) 21. (b) 28. (b)

SSS Mathematics for Class 12 1084

1084

Senior Secondary School Mathematics for Class 12

®

2. | a| = ( -2 ) 2 + 12 + ( -5 ) 2 =

30 .

®

-2 1 -5 , , × 30 30 30

\ direction cosines of a are ¾®

3. AB = (p.v. of B) – (p.v. of A) ^

^

^

^

^

^

^

^

^

= ( - i - 2 j + k ) - ( i + 2 j - 3 k ) = ( -2 i - 4 j + 4 k ). ¾®

| AB |= ( -2 ) 2 + ( -4 ) 2 + 4 2 = ¾®

\ direction cosines of AB are

36 = 6. -2 -4 4 -1 -2 2 , , , i.e., , , × 6 6 6 3 3 3

4. We have cos 2 a + cos 2 b + cos 2 g = 1 Þ ( 1 - sin 2 a ) + ( 1 - sin 2 b ) + ( 1 - sin 2 g ) = 1 Þ (sin 2 a + sin 2 b + sin 2 g ) = 2 . ®

^

^

^

5. Let a = (cos a cos b ) i + (cos a sin b ) j + (sin a ) k . Then, ®2

| a| = (cos a cos b ) 2 + (cos a sin b ) 2 + (sin a ) 2 = cos 2 a cos 2 b + cos 2 a sin 2 b + sin 2 a = cos 2 a(cos 2 b + sin 2 b ) + sin 2 a = (cos 2 a + sin 2 a ) = 1. ®

Hence, the given vector a is a unit vector. ®

^

^

®

^

2 k ) Þ | a|= 12 + 12 + ( 2 ) 2 = 4 = 2 .

6. a = ( i + j +

®

Direction ratios of a are 1, 1, 2. ® 1 1 2 1 1 1 Direction cosines of a are , , , i.e., , , × 2 2 2 2 2 2 1 p \ cos g = Þ g= × 4 2 ® ®

a×b

7. cos q =

®

®

=

p 6 1 = Þ q= × 4 3 ´2 2

=

-1 -1 2p = Þ q= × 2 3 2´ 2

| a|| b| ® ®

a×b

8. cos q =

®

®

| a|| b| ® ®

9. cos q = ®

a×b

®

®

[1 ´ 3 + ( -2 ) ´ ( -2 ) + 3 ´ 1]

=

2

2

2

^

®

2

2

2

^

^

| a|| b| [ 1 + ( -2 ) + 3 ][ 3 + ( -2 ) + 1 ]

®

^

^

®

^

=

10. ( a + b ) = ( 4 i + j - k ) and ( a - b ) = ( -2 i + 3 j - 5 k ). \

cos q =

®

®

®

®

®

®

®

®

( a + b )×( a - b ) | a + b|×| a - b|

=

p -8 + 3 + 5 =0 Þ q= × 2 18 ´ 38

10 5 5 = Þ q = cos -1 × 14 7 7

SSS Mathematics for Class 12 1085

Product of Three Vectors ®

®

^

^

®

^

®

^

^

1085 ^

11. ( 2 a + b ) = (5 i + 3 j - 4 k ) and ( a + 2 b ) = (7 i + 0 j + k ). (5 ´ 7 + 3 ´ 0 - 4 ´ 1) 31 æ 31 ö \ cos q = = Þ q = cos -1 ç ÷ . 50 50 ´ 50 è 50 ø ®

®

® ®

12. a ^ b Þ a × b = 0 Þ 2 ´ 3 + 4 ´ ( -2 ) + ( -1) ´ l = 0 Þ l = ( 6 - 8 ) = -2 . ®

® ®

®

( a × b)

13. Projection of a on b =

®

2 ´ 1 + ( -1) ´ ( -2 ) + 1 ´ 1

=

2

®

®

®

®

®2

®

®

2

1 + ( -2 ) + 1

| a| ®

2

®

5 × 6

=

®2

14. | a + b| = | a - b| Þ | a + b| = | a - b| ®

®

®

®

®

®

Þ ( a + b )×( a + b ) = ( a - b )×( a - b ) ® ®

® ®

®

®

Þ 4( a × b ) = 0 Þ a × b = 0 Þ a ^ b . ® ®

15. Given a2 = b 2 = 1 and a × b = 0. \

®

®

®

®

® ®

( 3 a + 2 b ) × (5 a - 6 b ) = 15 a2 - 12 b 2 - 8 a × b

= ( 15 ´ 1) - ( 12 ´ 1) - ( 8 ´ 0 ) = ( 15 - 12 - 0 ) = 3. ®

®

® ®

16. a ^ b Þ a × b = 0 Þ 3 ´ 1 + 1 ´ l + ( -2 ) ´ ( -3 ) = 0 Þ l = -9. ^

^2

^

^

^

^

^ ^

^ ^

^ ^

17. |a - b| = ( a - b ) × ( a - b ) = a × a + b × b - 2 a × b

= ( 1 + 1 - 2 ´ 1 ´ 1 ´ cos q) = 2( 1 - cos q) = 4 sin 2 \

sin 2

q 1 ^ ^2 1 ^ ^ q = × |a - b| Þ |a - b|= sin × 2 4 2 2 ^

^

i ®

®

18. ( a ´ b ) =

q × 2

^

j

k

^

1 -1 2

^

^

2 = ( 4 - 6 ) i - ( -4 - 4 ) j + ( 3 + 2 ) k

3 -4 ^

^

^

= ( -2 ) i + 8 j + 5 k . ®

®

®

®

\

| a ´ b |2 = {( -2 ) 2 + 8 2 + 5 2 } = ( 4 + 64 + 25 ) = 93 Þ | a ´ b| = 93 .

®

^

^

®

^

^

^

^

19. b = ( i + 0 j - 3 k ) and 2 a = ( 2 i - 4 j + 6 k ). ^

^

1

0 -3 = ( 0 - 12 ) i - ( 6 + 6 ) j + ( -4 - 0 ) k

i \

®

®

( b ´ 2 a) =

2

j

^

k

^

-4 ^

6 ^

^

= ( -12 i - 12 j - 4 k ).

^

^

SSS Mathematics for Class 12 1086

1086 \

Senior Secondary School Mathematics for Class 12 ®

® 2

| b ´ 2 a | = {( -12 ) 2 + ( -12 ) 2 + ( -4 ) 2 } = ( 144 + 144 + 16 ) = 304 . ®

®

Hence,| b ´ 2 a | = ®

®

3 2 + 2 2 + 6 2 = 49 = 7

20. | a ´ b|= ®

304 = 4 19 .

®

Þ | a || b |sin q = 7 Þ sin q = ®

®

®

7 1 p = Þ q= × 2 ´7 2 6

®

21. | a ´ b |= 35 Þ | a|| b |sin q = 35 Þ sin q = \ \

25 = 26

cos q = 1 - sin 2 q = 1 ® ®

®

1 × 26

®

1 = 7. 26

a × b = | a|| b|cos q = 26 ´ 7 ´ ^

^

35 5 = × 26 ´ 7 26

^

i j k ^ ^ ^ ^ ^ ^ 22. ( a ´ b ) = 3 1 4 = ( 1 + 4 ) i - ( 3 - 4 ) j + ( -3 - 1) k = 5 i + j - 4 k . 1 -1 1 ®

®

®

®

Required area =| a ´ b|= 25 + 1 + 16 = 42 sq units. ^

^

^

i j k ^ ^ ^ 23. ( d1 ´ d2 ) = 3 1 -2 = ( 4 - 6 ) i - ( 12 + 2 ) j + ( -9 - 1) k 1 -3 4 ®

®

^

^

^

= ( -2 i - 14 j - 10 k ) ®

®

Þ |d1 ´ d2|= ( -2 ) 2 + ( -14 ) 2 + ( -10 ) 2 = Þ

300 = 10 3

1 ® ® area of the||gm = |d1 ´ d2|= 5 3 sq units. 2 ^

^

^

i j k ^ ^ ^ 24. ( a ´ b ) = 3 4 0 = 0 i - 0 j + ( 21 + 20 ) k -5 7 0 ®

®

Þ C=

1 ® ® 41 sq units. | a ´ b|= 2 2 ^

^

^

1

1

1

i j k ^ ^ 25. ( a ´ b ) = 1 -1 -1 = ( -2 j + 2 k ) ®

®

®

®

Þ | a ´ b|= ( -2 ) 2 + 2 2 = 8 = 2 2 ^

Þ

required vector =

^

2( - j + k ) 2 2

=

^ ^ 1 ( - j + k ). 2

SSS Mathematics for Class 12 1087

Product of Three Vectors ®

®

®

®

®

1087

®

26. ( a + b + c ) × ( a + b + c ) = 0 ®2

®2

®2

® ®

® ®

® ®

Þ | a| +| b| +| c| + 2 ( a × b + b × c + c × a ) = 0 Þ

® ®

® ®

® ®

(a× b + b× c + c× a )=

®

-3 2

®

®

[Q| a| = | b |=| c|= 1].

® ®

® ®

® ®

27. Given a2 = b 2 = c 2 = 1 and a × b = b × c = c × a = 0. ®

®

®2

®

®

®

®

®

®

\ | a + b + c| = ( a + b + c ) × ( a + b + c ) ® ®

® ®

® ®

® ®

® ®

® ®

= a × a + b × b + c × c + 2( a × b + b × c + c × a ) = a2 + b 2 + c 2 + 2 ´ 0 = ( 1 + 1 + 1 + 0 ) = 3. ®

®

®

Hence,| a + b + c|= ^ ^ ^

28. [ i

^

^

3.

^

^ ^

j k] = ( i ´ j ) × k = k × k = 1.

29. We have 2 -3 4 ® ® ® ( a ´ b ) × c = 1 2 -1 = 2( 4 - 1) + 3( 2 + 3 ) + 4 ( -1 - 6 ) 3 -1 2 = ( 6 + 15 - 28 ) = -7 . \ volume of the parallelopiped ®

®

®

= |( a ´ b ) × c | = |- 7|= 7 cubic units. 5 -4 1 ® ® ® 30. ( a ´ b ) × c = 4 3 l = 5( 21 + 2 l) + 4 ( 28 - l) + 1 × ( -8 - 3 ) 1 -2 7

\

= ( 105 + 112 - 11) + ( 10 l - 4 l) = ( 206 + 6 l). 10 5 206 + 6 l = 216 Þ 6 l = 10 Þ l = = × 6 3 ® ® ®

® ® ®

31. Since a , b , c are coplanar, we must have [ a b c ] = 0. ® ® ®

Now, [ a b c ] =

2 1 0

0 l+ 1 4

-2 0 2

= 2 ( 2 l + 2 - 0 ) - 2( 4 - 0 ) = 4 l + 4 - 8 = 4 l - 4. \

® ® ®

[ a b c ] = 0 Û 4 l - 4 = 0 Û 4 l = 4 Û l = 1. ®

® ®

32. Clearly, a ´ ( b × c ) is meaningless. ®

®

®

®

®

®

® ®

33. a × ( a ´ b ) = ( a ´ a ) × b = 0 × b = 0. ®

®

34. We have proved in the text that [ a - b

®

®

b - c

®

®

c - a ] = 0.

SSS Mathematics for Class 12 1088

26. FUNDAMENTAL CONCEPTS OF 3-DIMENSIONAL GEOMETRY This chapter consists of some important concepts of three-dimensional geometry. Though we have studied these in Class 11, yet we shall review here these fundamental concepts for ready reference. Coordinates of a Point in Space Let O be the origin, and let OX, OY and OZ be three mutually perpendicular lines, taken as the x-axis, y-axis and z-axis respectively in such a way that they form a right-handed system. The planes YOZ, ZOX and XOY are respectively known as the yz-plane, the zx-plane and the xy-plane. These planes, known as the coordinate planes, divide the space into eight parts called octants. Let P be a point in space. Through P, draw three planes PLAN, PNBM and PLCM parallel to the yz-plane, the zx-plane and the xy-plane respectively, and meeting the x-axis, y-axis and z-axis at the points A , B , C respectively. Complete the parallelepiped whose coterminous edges are OA , OB and OC. Let OA = x , OB = y and OC = z. We say that the coordinates of P are ( x , y , z). It is clear from the figure alongside that (i) x = distance of P from the yz-plane (ii) y = distance of P from the zx-plane (iii) z = distance of P from the xy-plane. Also, we can say that (i) the equation of the xy-plane is z = 0 (ii) the equation of the xz-plane is y = 0 (iii) the equation of the yz-plane is x = 0. ^ ^ ^

POSITION VECTOR OF A POINT IN SPACE

Let i , j , k be unit vectors along OX, OY

and OZ respectively. 1088

SSS Mathematics for Class 12 1089

Fundamental Concepts of 3-Dimensional Geometry

1089

If P(a, b, c) is a point in space, we say that the position vector (or, p.v.) of P is ^

^

^

( a i + b j + c k ). Some Results on Points in Space 1. DISTANCE BETWEEN TWO POINTS

The distance between two points P( x1 , y1 , z1)

and Q( x 2 , y 2 , z2) is given by PQ = ( x 2 - x1) 2 + ( y 2 - y1) 2 + (z2 - z1) 2 . 2. SECTION FORMULAE

(i) If P( x , y , z) divides the join of A( x1 , y1 , z1) and B( x 2 , y 2 , z2) in the ratio m : n then (mx 2 + nx1) (my 2 + ny1) (mz2 + nz1) x= , y= , z= × (m + n) (m + n) (m + n) (ii) The mid-point of the line joining A( x1 , y1 , z1) and B( x 2 , y 2 , z2) is given by æ x + x 2 y1 + y 2 z1 + z2 ö , , Mç 1 ÷× 2 2 2 ø è (iii) The centroid of C ABC with vertices A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) is given by æ x + x 2 + x 3 y1 + y 2 + y 3 z1 + z2 + z 3 ö Gç 1 , , ÷× 3 3 3 è ø Some Results on Lines in Space If a line makes angles a , b , g with the x-axis, y-axis and z-axis respectively then

1. DIRECTION COSINES OF A LINE

l = cos a , m = cos b , n = cos g are called the direction cosines (or, d.c.’s) of the line. We always have l REMARKS

2

+m

2

+ n 2 = 1.

(i) d.c.’s of the x-axis are 1, 0, 0. (ii) d.c.’s of the y-axis are 0, 1, 0. (iii) d.c.’s of the z-axis are 0, 0, 1.

2. DIRECTION RATIOS OF A LINE Any three numbers a , b , c, proportional to the direction cosines l , m , n respectively of a line, are called the direction ratios of the line. l m n Clearly, we have = = × a b c

Some Important Facts (i) If a , b , c are the direction ratios of a line then its direction cosines are a b c l= , m= , n= × 2 2 2 2 2 2 2 a +b +c a +b +c a + b2 + c2 ®

^

^

^

®

(ii) If r = a i + b j + c k then the direction ratios of r are a , b , c.

SSS Mathematics for Class 12 1090

1090

Senior Secondary School Mathematics for Class 12

(iii) Let PQ be a line joining P( x1 , y1 , z1) and Q( x 2 , y 2 , z2). Then, the direction ratios of the line PQ are ( x 2 - x1), ( y 2 - y1), (z2 - z1). If q is the angle between two lines L1 and L2 whose d.c.’s are l1 , m1 , n1 , and l2 , m2 , n2 then the following hold true. 3. ANGLE BETWEEN TWO LINES

(i) cos q = l1l2 + m1m2 + n1n2 (ii) sin q = S(m1n2 - m2n1)

2

(iii) lines L1 and L2 are perpendicular Û l1l2 + m1m2 + n1n2 = 0 (iv) lines L1 and L2 are parallel Û REMARKS

l1 m1 n1 = = × l2 m2 n2

In fact, there are two angles q and (p - q) between two lines. SOLVED EXAMPLES

EXAMPLE 1

Find the direction cosines of a line whose direction ratios are 2, - 6, 3.

SOLUTION

Here, a = 2, b = - 6, c = 3. \

a 2 + b 2 + c2 = 22 + ( -6) 2 + 3 2 = 4 + 36 + 9 = 49 = 7.

Hence, the direction cosines of the given line are EXAMPLE 2

Find the direction cosines of each of the following vectors: ^

^

^

(i) 2 i + j - 2 k SOLUTION

2 -6 3 , , × 7 7 7

^

^

(ii) - i - k

^

(iii) - j ^

^

^

(i) Direction ratios of the vector ( 2 i + j - 2 k ) are 2, 1, –2. And, 22 + 12 + ( -2) 2 = 4 + 1 + 4 = 9 = 3. \ d.c.’s of the given vector are ^

2 1 -2 , , × 3 3 3 ^

(ii) Direction ratios of the vector ( - i - k ) are –1, 0, –1. And, ( -1) 2 + 02 + ( -1) 2 = 1 + 0 + 1 = 2 . -1 -1 \ d.c.’s of the given vector are , 0, × 2 2 ^

(iii) Direction ratios of the vector - j are 0, –1, 0. And, 02 + ( -1) 2 + 02 = 1 = 1. \ d.c.’s of the given vector are 0, –1, 0.

SSS Mathematics for Class 12 1091

Fundamental Concepts of 3-Dimensional Geometry EXAMPLE 3

SOLUTION

EXAMPLE 4

SOLUTION

1091

Find the direction cosines of a line which makes angles 90°, 135° and 45° with the positive directions of the x-, y-, and z-axis respectively. The direction cosines of the given line are -1 1 cos 90° , cos 135° , cos 45°, i.e., 0, , × 2 2 If a line makes angles a , b and g with the coordinate axes, prove that (sin 2 a + sin 2 b + sin 2 g) = 2. Let the direction cosines of the given line be l, m, n. Then, l = cos a , m = cos b and n = cos g. \

( l 2 + m 2 + n2) = 1 Þ cos2 a + cos2 b + cos2 g = 1 Þ (1 - sin 2 a ) + (1 - sin 2 b) + (1 - sin 2 g) = 1 Þ sin 2 a + sin 2 b + sin 2 g = 2.

Hence, (sin 2 a + sin 2 b + sin 2 g) = 2. EXAMPLE 5

SOLUTION

If a line makes angles a , b and g with the coordinate axes, prove that (cos 2a + cos 2b + cos 2g) = -1. Let the direction cosines of the given line be l, m, n. Then, l = cos a , m = cos b and n = cos g. \

( l 2 + m 2 + n2) = 1 Þ cos2 a + cos2 b + cos2 g = 1 Þ 2 cos2 a + 2 cos2 b + 2 cos2 g = 2 Þ (1 + cos 2a ) + (1 + cos 2b) + (1 + cos 2g) = 2 Þ (cos 2a + cos 2b + cos 2g) = -1.

Hence, (cos 2a + cos 2b + cos 2g) = -1. EXAMPLE 6

Find the direction cosines of a line that makes equal angles with the coordinate axes. [CBSE 2011]

SOLUTION

Let the given line make angle a with each of the coordinate axes and let the direction cosines of this line be l, m, n. Then, l = m = n = cos a . \

( l 2 + m 2 + n2) = 1 Þ cos2 a + cos2 a + cos2 a = 1 1 Þ 3 cos2 a = 1 Þ cos2 = 3 1 Þ cos a = ± . 3

Hence, the d.c.’s of the given line are 1 1 1 -1 -1 -1 or , , , , × 3 3 3 3 3 3

SSS Mathematics for Class 12 1092

1092

Senior Secondary School Mathematics for Class 12

EXAMPLE 7

A line makes angle 60° and 45° with the positive direction of x-axis and y-axis respectively. What acute angle does it make with the z-axis?

SOLUTION

Let the given line make an angle g with the z-axis. Then, cos2 60° + cos2 45 ° + cos2 g = 1 2

2

Þ

æ 1 ö æ1ö 2 ÷ + cos g = 1 ç ÷ +ç è 2ø è 2ø

Þ

3ö 1 æ æ1 1ö 2 2 ç + ÷ + cos g = 1 Þ cos g = ç1 - ÷ = 4ø 4 è è 4 2ø

Þ

cos g = ±

Þ

g = 60° or 120°.

1 2

Hence, the acute angle which the given line makes with the z-axis is 60°. ®

EXAMPLE 8

^

^

^

Find the direction cosines of the vector r = ( 6 i + 2 j - 3 k ). ®

SOLUTION

The direction ratios of r are 6, 2, –3. And, 62 + 22 + ( -3) 2 =

36 + 4 + 9 = 49 = 7.

® 6 2 -3 Hence, the direction cosines of r are , , × 7 7 7 EXAMPLE 9

SOLUTION

Find the direction cosines of the line segment joining the points A(7 , - 5 , 9) and B(5 , - 3 , 8). D.r.’s of AB are (5 - 7), ( -3 + 5), (8 – 9), i.e., –2, 2, –1. |AB|= ( -2) 2 + 22 + ( -1) 2 = 4 + 4 + 1 = 9 = 3. Hence, the d.c.’s of the line AB are

-2 2 -1 , , × 3 3 3

®

EXAMPLE 10

^

^

^

Find the angles made by the vector r = ( i + j - k ) with the coordinate axes. ®

SOLUTION

D.r.’s of the given vector r are 1, 1, –1. ®

|r | = 12 + 12 + ( -1) 2 = \

®

d.c.’s of r are ®

1 , 3

3.

1 -1 , × 3 3

Let r make angles a, b and g with the x-axis, y-axis and z-axis respectively. Then, d.c.’s of the x-axis are 1, 0, 0.

SSS Mathematics for Class 12 1093

Fundamental Concepts of 3-Dimensional Geometry

1093

ìæ 1 1 öü ö æ 1 ö æ 1 ´ 0÷ ý = ´ 0÷ + ç cosa = íç ´ 1÷ + ç 3 øþ ø è 3 îè 3 ø è 3 1 æ ö Þ a = cos-1 ç ÷× è 3ø 1 æ 1 ö Similarly, cos b = Þ b = cos-1 ç ÷× 3 è 3ø -1 æ -1 ö And, cos g = Þ g = cos-1 ç ÷× 3 è 3ø

\

®

Hence, the angles made by the given vector r with the x-axis, y-axis æ 1 ö -1 æ 1 ö -1 æ -1 ö and z-axis are cos-1 ç ÷ , cos ç ÷ and cos ç ÷ respectively. è 3ø è 3ø è 3ø EXAMPLE 11

Find the angle between the lines whose direction ratios are 3 , 2, - 6 and 1, 2, 2.

SOLUTION

Let the given lines be L1 and L2 respectively. Then, D.r.’s of L1 are 3, 2, –6. And, \

3 2 + 22 + ( -6) 2 = 9 + 4 + 36 = 49 = 7.

3 2 -6 , , × 7 7 7 D.r.’s of L2 are 1, 2, 2.

d.c.’s of L1 are

And, 12 + 22 + 22 = 9 = 3. 1 2 2 \ d.c.’s of L2 are , , × 3 3 3 Let q be the angle between L1 and L2. Then, cos q =| l1l2 + m1m2 + n1n2| æ 3 1 ö æ 2 2 ö æ -6 2 ö = ç ´ ÷+ç ´ ÷+ç ´ ÷ è 7 3 ø è7 3 ø è 7 3 ø 1 4 4 -5 5 + - = = × 7 21 7 21 21 æ5 ö q = cos-1 ç ÷ × \ è 21 ø Find the angle between the vectors =

EXAMPLE 12

® r2

^

^

^

^

^

®

Let q be the angle between r 1 and r 2. Then, cos

® ® r ×r q = ® 1 ®2 = |r 1||r 2|

^

^

^

^

^

= ( 4 i - 3 j + 5 k) and

= ( 3 i + 4 j + 5 k). ®

SOLUTION

® r1

^

^

( 4 i - 3 j + 5 k) × ( 3 i + 4 j + 5 k) { 42 + ( -3) 2 + 5 2}{ 3 2 + 42 + 5 2}

SSS Mathematics for Class 12 1094

1094

Senior Secondary School Mathematics for Class 12

= \

(12 - 12 + 25) 25 25 1 = = = × { 16 + 9 + 25}{ 9 + 16 + 25} { 50 ´ 50} 50 2

æ1ö p q = cos-1 ç ÷ = × è 2ø 3

Hence, the angle between the given vectors is

p × 3

EXAMPLE 13

Find the direction cosines of the line which is perpendicular to each of the lines having direction ratios 1, - 2, - 2 and 0, 2, 1 respectively.

SOLUTION

Let the direction ratios of the required line be a, b, c. Then, it being perpendicular to each of the lines having d.r.’s 1, –2, –2 and 0, 2, 1 we have a - 2b - 2c = 0

... (i)

0a + 2b + c = 0.

... (ii)

On solving (i) and (ii) by cross multiplication, we have a b c a b c = = Þ = = × ( -2 + 4) ( 0 - 1) ( 2 - 0) 2 -1 2 \

d.r.’s of the required line are 2, –1, 2.

And, 22 + ( -1) 2 + 22 = 9 = 3. \ EXAMPLE 14

d.c.’s of the required line are

2 -1 2 , , × 3 3 3

If A( 8, 2, 0), B( 4, 6, - 7), C( -3 , 1, 2) and D( -9, - 2, 4) are four given ¾®

¾®

point then find the angle between AB and CD . ¾®

SOLUTION

D.r.’s of AB are ( 4 - 8), ( 6 - 2), ( -7 - 0), i.e., –4, 4, –7. ¾®

|AB|= ( -4) 2 + 42 + ( -7) 2 = 16 + 16 + 49 = 81 = 9. \

¾®

d.c.’s of AB are ¾®

-4 4 -7 , , × 9 9 9

D.r.’s of CD are ( -9 + 3), ( -2 - 1), ( 4 - 2), i.e., –6, –3, 2. ¾®

|CD|= ( -6) 2 + ( -3) 2 + 22 = \

36 + 9 + 4 = 49 = 7.

¾®

-6 -3 2 d.c.’s of CD are , , × 7 7 7

¾®

¾®

Let q be the acute angle between AB and CD . Then, æ -4 ö æ -6 ö æ 4 ö æ -3 ö æ -7 ö æ 2 ö cos q = ç ÷ ´ ç ÷ + ç ÷ ´ ç ÷ + ç ÷ ´ ç ÷ è 9 ø è 7 ø è 9ø è 7 ø è 9 ø è 7 ø

SSS Mathematics for Class 12 1095

Fundamental Concepts of 3-Dimensional Geometry

= \

1095

24 12 14 -2 2 = = × 63 63 63 63 63

æ 2ö q = cos-1 ç ÷. è 63 ø

¾® ¾® æ 2ö Hence, the required angle between AB and CD is cos-1 ç ÷. è 63 ø EXAMPLE 15

Find the angles of C ABC whose vertices are A( -1, 3 , 2), B( 2, 3 , 5) and C( 3 , 5 , - 2).

SOLUTION

D.r.’s of AB are ( 2 + 1), ( 3 - 3), (5 - 2), i.e. 3, 0, 3. D.r.’s of AC are ( 3 + 1),(5 - 3),( -2 - 2), i.e., 4, 2, –4. \

cos A =

\

ÐA =

{ 3 ´ 4 + 0 ´ 2 + 3 ´ ( -4)}

{

3 + 02 + 3 2} × { 42 + 22 + ( -4) 2} 2

= 0.

p × 2

D.r.’s of BA are ( -1 - 2),( 3 - 3),( 2 - 5), i.e., –3, 0, –3. D.r.’s of BC are ( 3 - 2),(5 - 3),( -2 - 5), i.e., 1, 2, –7. {( -3) ´ 1 + 0 ´ 2 + ( -3) ´ ( -7)} \ cos B = { ( -3) 2 + 02 + ( -3) 2} × { 12 + 22 + ( -7) 2} = \

18 1 = × 18 ´ 54 3

æ 1 ö ÐB = cos-1 ç ÷× è 3ø

D.r.’s of CB are ( 2 - 3), ( 3 - 5), (5 + 2), i.e., –1, –2, 7. D.r.’s of CA are ( -1 - 3), ( 3 - 5), ( 2 + 2), i.e., –4, –2, 4. {( -1) ´ ( -4) + ( -2) ´ ( -2) + 7 ´ 4 } \ cos C = { ( -1) 2 + ( -2) 2 + 7 2} × { ( -4) 2 + ( -2) 2 + 42} = \

36 = 54 ´ 36

36 = 54

36 = 54

2 × 3

æ 2ö ÷× C = cos-1 çç ÷ è 5ø

Hence, ÐA =

p æ 1 ö -1 æ 2 ö÷ , ÐB = cos-1 ç ÷ and ÐC = cos çç ÷× 2 è 3ø è 3ø

SSS Mathematics for Class 12 1096

1096

Senior Secondary School Mathematics for Class 12

EXAMPLE 16

Show that the points A( 2, 3 , - 4), B(1, - 2, 3) and C( 3 , 8, - 11) are collinear.

SOLUTION

D.r.’s of the line AB are (1 - 2), ( -2 - 3), ( 3 + 4), i.e., –1, –5, 7. D.r.’s of the AC are ( 3 - 2), ( 8 - 3), ( -11 + 4), i.e., 1, 5, –7, i.e., –1, –5, 7. \ d.r.’s of AB and AC are the same and so they are parallel. But, the lines AB and AC have a common point A. Hence, the points A , B and C are collinear.

EXAMPLE 17

Find the coordinates of the foot of the perpendicular drawn from the point A(1, 2, 1) to the line joining B(1, 4, 6) and C(5 , 4, 4).

SOLUTION

Draw AN^ BC. Let N divide BC in the ratio k : 1. Then, the coordinates of N are æ 5 k + 1 4k + 4 4k + 6 ö , , ÷. ç è k+1 k+1 k+1 ø

\

d.r.’s of AN are 4k + 4 4k + 6 ö æ 5k + 1 æ 4k 2k + 2 3 k + 5 ö , , - 1, - 2, - 1÷ , i.e., ç ÷ ç k + 1 k + 1 k + 1 ø è èk+1 k+1 k+1 ø And, the d.r.’s of BC are (5 - 1), ( 4 - 4), ( 4 - 6), i.e., 4, 0, –2.

Since AN ^ BC , we have 3k + 5 ö 2k + 2 ö æ 4k ö æ æ ÷=0 ÷ - ç2 ´ ÷ + ç0 ´ ç4 ´ k + 1ø è k+1 ø è k+1 ø è Þ 16k + 0 - 6k - 10 = 0 Þ 10k = 10 Þ k = 1. Putting k = 1, we get the coordinates of N as N(3, 4, 5). EXAMPLE 18

SOLUTION

If l1 , m1 , n1 and l2 , m2 , n2 be the direction ratios of two lines then show that the direction ratios of a line which is perpendicular to each of the given lines are: (m1n2 - m2n1), (n1l2 - n2l1), ( l1m2 - l2m1). Let a, b, c be the direction ratios of the required line. Then, this line being perpendicular to each of the given two lines, we have: l1a + m1b + n1c = 0

... (i)

l2a + m2b + n2c = 0.

... (ii)

SSS Mathematics for Class 12 1097

Fundamental Concepts of 3-Dimensional Geometry

1097

On solving (i) and (ii) for a , b , c by cross multiplication, we have a b c = = × (m1n2 - m2n1) (n1l2 - n2l1) ( l1m2 - l2m1) Hence, the d.r.’s. of the required line are (m1n2 - m2n1), (n1l2 - n2l1), ( l1m2 - l2m1). EXAMPLE 19

Find the angle between the two lines whose direction cosines are connected by the relations l + m + n = 0 and l 2 + m 2 - n2 = 0.

SOLUTION

The given relations are l+m+n = 0 2

2

... (i)

2

l + m - n = 0.

... (ii)

From (i), we get l = - (m + n). Putting l = - (m + n) in (ii), we get {- (m + n)} 2 + m 2 - n2 = 0 Þ 2m 2 + 2mn = 0 Þ 2m (m + n) = 0 Þ m = 0 or (m + n) = 0 Þ m = 0 or m = -n. Putting m = 0 in (i), we get l = - n. Putting m = -n in (i), we get l = 0. \ d.r.’s of these lines are -n, 0, n and 0, -n, n, i.e., –1, 0, 1 and 0, –1, 1. If q is the acute angle between these lines, we have 1 1 |( -1) ´ 0 + 0 ´ ( -1) + 1 ´ 1| = = × cos q = 2 2 2 2 2 2 2 ( 2 ´ 2 ) { ( -1) + 0 + 1 } × { 0 + ( -1) + 1 } \

q=

p × 3

Hence, the angle between the given lines is EXAMPLE 20

SOLUTION

p × 3

Find the direction cosines of the lines which are connected by the relations l - 5m + 3n = 0 and 7 l 2 + 5m 2 - 3n 2 = 0. The given equations are l - 5m + 3n = 0

… (i)

7 l 2 + 5m 2 - 3n2 = 0.

… (ii)

Putting l = (5m - 3n) from (i) in (ii), we get 7(5m - 3n) 2 + 5m 2 - 3n2 = 0 Þ

6m 2 - 7mn + 2n2 = 0

SSS Mathematics for Class 12 1098

1098

Senior Secondary School Mathematics for Class 12 2

Þ

m æmö æmö 6 ç ÷ - 7 ç ÷ + 2 = 0 Þ 6p 2 - 7 p + 2 = 0, where = p n ènø ènø

Þ

( 3 p - 2)( 2p - 1) = 0 2 1 m 2 m 1 or p = or Þ = = × Þ p= 3 2 n 3 n 2 m 2 m n 5m - 3n l Now, = Þ = = = n 3 2 3 5 ´2- 3 ´ 3 1 Þ

l m n = = = 1 2 3

Þ

l=

Again,

l 2 + m 2 + n2 2

2

1 +2 +3

2

=

1 14

1 2 3 ,m= ,n= × 14 14 14 m 1 m n 5m - 3n l = Þ = = = n 2 1 2 5 ´ 1 - 3 ´ 2 -1

l m n l 2 + m 2 + n2 1 = = = = 2 2 2 -1 1 2 6 ( -1) + 1 + 2 -1 1 2 ,m= ,n= × Þ l= 6 6 6 Hence, the direction cosines of the lines are 2 3 ö 1 2 ö æ 1 æ -1 , , , , ç ÷ and ç ÷× 14 14 14 6 6 6ø è ø è Þ

EXAMPLE 21

If a variable line in two adjacent positions has direction cosines (l, m, n) and ( l + dl , m + dm , n + dn), show that the small angle dq between the two positions is given by ( dq) 2 = ( dl) 2 + ( dm) 2 + ( dn) 2.

SOLUTION

Clearly, we have l 2 + m 2 + n2 = 1 2

… (i) 2

2

and ( l + dl) + (m + dm) + (n + dn) = 1.

… (ii)

Subtracting (i) from (ii), we get ( l + dl) 2 + (m + dm) 2 + (n + dn) 2 - ( l 2 + m 2 + n2) = 0 Þ

( dl) 2 + ( dm) 2 + ( dn) 2 = -2( l × dl + m × dm + n × dn)

\

cos dq = l × ( l + dl) + m × (m + dm) + n × (n + dn)

… (iii)

= ( l 2 + m 2 + n2) + ( l × dl + m × dm + n × dn) =1-

1 {( dl) 2 + ( dm) 2 + ( dn) 2} 2

[using (i) and (iii)].

SSS Mathematics for Class 12 1099

Fundamental Concepts of 3-Dimensional Geometry

\

1099

( dl) 2 + ( dm) 2 + ( dn) 2 = 2(1 - cos dq) = 4 sin

2

dq æ dq ö = 4× ç ÷ 2 è 2ø dq dq dq ù é êëQ 2 being small, sin 2 = 2 úû 2

= ( dq) 2. 2

Hence, ( dq) = ( dl) + ( dm) 2 + ( dn) 2. EXAMPLE 22

SOLUTION

2

Prove that the straight lines whose direction cosines are given by the relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are f g h perpendicular to each other if + + = 0, and a b c parallel if a 2 f 2 + b 2g 2 + c2h2 - 2bcgh - 2cahf - 2abfg = 0. The given equations are al + bm + cn = 0 fmn + gnl + hlm = 0. - ( al + bm) Putting n = from (i) in (ii), we get c ì - ( al + bm) ü ì - ( al + bm) ü f m ×í ý + hlm = 0 ý + gl × í c c þ î þ î Þ

agl 2 + ( af + bg - ch) lm + bfm 2 = 0

Þ

æ lö æ lö ag ç ÷ + ( af + bg - ch) × ç ÷ + bf = 0 m è ø èmø

… (i) … (ii)

2

… (iii)

æ lö Now, equation (iii), being a quadratic equation in ç ÷ , will have èmø æ l1 ö æ l2 ö two roots, say çç ÷÷ and çç ÷÷ × è m1 ø è m2 ø l1 l bf ll mm ´ 2 = Þ 12 = 1 2 \ m1 m2 ag bf ag l1l2 m1m2 n1n2 Þ = = = k [by symmetry]. æfö ægö æ hö ç ÷ ç ÷ ç ÷ è aø è bø è cø g hö æf \ l1l2 + m1m2 + n1n2 = k ç + + ÷ . è a b cø Thus, the given lines will be perpendicular to each other f g h Û l1l2 + m1m2 + n1n2 = 0 Û + + = 0. a b c The lines will be parallel only when the roots of (iii) are equal.

SSS Mathematics for Class 12 1100

1100

Senior Secondary School Mathematics for Class 12

\

( af + bg - ch) 2 - 4 abgf = 0

Û a 2 f 2 + b 2g 2 + c 2h2 - 2bcgh - 2cahf - 2abfg = 0. EXAMPLE 23

Show that the straight lines whose direction cosines are given by the equations al + bm + cn = 0 and ul 2 + vm 2 + wn2 = 0 are mutually perpendicular if a 2(v + w) + b 2(u + w) + c2(u + v) = 0, and parallel if a 2 b 2 c2 + + = 0. u v w

SOLUTION

The given equations are al + bm + cn = 0 2

2

2

ul + vm + wn = 0.

… (i) … (ii)

- ( bm + cn) Putting l = from (i) in (ii), we get a u( bm + cn) 2

Þ

+ vm 2 + wn2 = 0 a2 ( b 2u + a 2v)m 2 + 2ubcmn + ( c2u + a 2w)n2 = 0 2

æmö æmö ( b 2u + a 2v) ç ÷ + 2ubc ç ÷ + ( c2u + a 2w) = 0. ènø ènø m1 m2 Let and be the roots of (iii). n1 n2

Þ

… (iii)

m1 m2 c2u + a 2w × = 2 n1 n2 b u + a 2v m1m2 n1n2 l1l2 = = = k [by symmetry]. c2u + a 2w b 2u + a 2v b 2w + c2v l1l2 + m1m2 + n1n2 = k( b 2w + c2v + c2u + a 2w + b 2u + a 2v).

Then, Þ \

The given lines are mutually perpendicular Û l1l2 + m1m2 + n1n2 = 0 Û a 2(v + w) + b 2(w + u) + c2(u + v) = 0. For the given lines to be parallel, the direction cosines must be equal. \ the roots of (iii) must be equal. a 2 b 2 c2 + + = 0. \ 4u2b 2c2 - 4( b 2u + a 2v)( c2u + a 2w) = 0 Û u v w EXAMPLE 24

If the edges of a rectangular parallelepiped are a, b, c, prove that the æ ± a 2 ± b 2 ± c2 ö ÷× angles between the four diagonals are given by cos-1 ç 2 ç a + b 2 + c2 ÷ è ø

SSS Mathematics for Class 12 1101

Fundamental Concepts of 3-Dimensional Geometry SOLUTION

1101

Let OA , OB, OC be the coterminous edges of the parallelepiped, taken along the axes in such a way that OA = a , OB = b and OC = c. Then, the coordinates of the vertices are O( 0, 0, 0), A( a , 0, 0), B( 0, b , 0), C( 0, 0, c), P( a , b , c), L( 0, b , c), M( a , 0, c) and N ( a , b , 0). The direction ratios of the diagonals OP, AL, BM and CN are ( a , b , c), ( - a , b , c), ( a , - b , c) and ( a , b , - c) respectively.

\ the direction cosines of OP, AL, BM and CN are æ a ç , ç 2 2 2 è a +b +c

b a 2 + b 2 + c2

æ -a ç , ç 2 2 2 a + b + c è æ a ç , ç 2 2 2 è a +b +c æ a ç , ç 2 2 2 è a +b +c

ö ÷, ÷ a 2 + b 2 + c2 ø c

,

b a 2 + b 2 + c2 -b a 2 + b 2 + c2 b a 2 + b 2 + c2

,

,

ö ÷, 2 2 2÷ a +b +c ø

,

ö ÷, ÷ a 2 + b 2 + c2 ø

c

c

ö ÷ respectively. ÷ a 2 + b 2 + c2 ø -c

Let q1 be the angle between OP and AL. Then, cos q1 =

( - a 2 + b 2 + c2 ) 2

2

2

(a + b + c )

or

æ - a 2 + b 2 + c2 ö ÷× q1 = cos-1 ç 2 ç a + b 2 + c2 ÷ è ø

Again, let q2 be the angle between OP and BM. Then, æ a 2 - b 2 + c2 ö ÷× q2 = cos-1 ç 2 ç a + b 2 + c2 ÷ (a + b + c ) è ø Similarly, the angles between the other pairs of diagonals can be obtained. cos q2 =

( a 2 - b 2 + c2 ) 2

2

2

or

SSS Mathematics for Class 12 1102

1102

Senior Secondary School Mathematics for Class 12

Clearly, the angles between the four diagonals can be given by æ ± a 2 ± b 2 ± c2 ö ÷. cos-1 ç 2 ç a + b 2 + c2 ÷ è ø EXAMPLE 25 SOLUTION

æ1ö Show that the angle between any two diagonals of a cube is cos-1 ç ÷. è 3ø Let OA , OB, OC be the coterminous edges of a cube, taken along the axes in such a way that OA = OB = OC = a. Then, the coordinates of the vertices of the cube are O( 0, 0, 0), A( a , 0, 0,), B( 0, a , 0), C( 0, 0, a), P( a , a , a ,), L( 0, a , a), M( a , 0, a) and N ( a , a , 0). The direction ratios of the diagonals OP, AL, BM and CN are ( a , a , a), ( - a , a , a), ( a , - a , a) and ( a , a , - a) respectively. Thus, direction cosines of OP , AL, BM and CN are 1 1 ö æ -1 1 1 ö æ 1 1 ö -1 æ 1 , , , , , , ç ÷, ç ÷, ç ÷ and 3 3ø è 3 3 3ø è 3 3 3ø è 3 æ 1 , ç è 3

-1 ö 1 , ÷ respectively. 3 3ø

If q1 be the angle between OP and AL then ì 1 æ -1 ö 1 1 1 1 ü 1 cos q1 = í ×ç × + × ÷+ ý= 3 3 3 3þ 3 î 3 è 3ø Þ

æ1ö q1 = cos-1 ç ÷ × è 3ø

Similarly, the angle between each one of the other pairs is æ1ö cos-1 ç ÷ × è 3ø Hence, the angle between any two diagonals of the cube is æ1ö cos-1 ç ÷ × è 3ø EXAMPLE 26

A line makes angles a , b, g, d with the four diagonals of a cube. Prove that 4 cos 2 a + cos 2 b + cos 2 g + cos 2 d = × [CBSE 2007] 3

SOLUTION

Let OA , OB, OC be the coterminous edges of a cube, taken along the axes in such a way that OA = OB = OC = a.

SSS Mathematics for Class 12 1103

Fundamental Concepts of 3-Dimensional Geometry

1103

Then, the coordinates of the vertices of the cube are O( 0, 0, 0), A( a , 0, 0), B( 0, a , 0), C( 0, 0, a), P( a , a , a), L( 0, a , a), M( a , 0, a) and N ( a , a , 0). The direction ratios of the diagonals OP, AL, BM and CN are ( a , a , a), ( - a , a , a), ( a , - a , a) and ( a , a , - a) respectively. \

the direction cosines of OP, AL, BM and CN are 1 ö æ -1 , ÷,ç 3ø è 3

æ 1 , ç è 3

1 , 3

æ 1 , ç è 3

-1 ö 1 , ÷ respectively. 3 3ø

1 , 3

-1 1 ö æ 1 , , ÷,ç 3ø è 3 3

1 ö ÷ and 3ø

Let ( l , m , n) be the direction cosines of a line which makes angles a , b , g , d with the four diagonals of the cube. Then, 1 1 ö ( l + m + n) æ 1 cos a = ç l × + m× + n× , ÷= 3 3 3ø 3 è ì æ -1 ö 1 1 ü ( - l + m + n) + n× cos b = íl × ç , ÷ + m× ý= 3 3 3þ 3 ø è î ì 1 1 ü ( l - m + n) æ -1 ö cos g = íl × + m×ç , ÷ + n× ý= 3 3 3þ 3 ø è î ì 1 1 æ -1 ö ü ( l + m - n) cos d = íl × + m× + n× ç × ÷ý = 3 3 3 è 3 øþ î On squaring and adding, we get cos2 a + cos2 b + cos2 g + cos2 d 1 = × {( l + m + n) 2 + ( - l + m + n) 2 + ( l - m + n) 2 + ( l + m - n) 2} 3 4 = × 3

EXERCISE 26 1. Find the direction cosines of a line segment whose direction ratios are: (i) 2, –6, 3 (ii) 2, –1, –2 (iii) –9, 6, –2 2. Find the direction ratios and the direction cosines of the line segment joining the points: (i) A(1, 0, 0) and B(0, 1, 1) (ii) A(5, 6, –3) and B(1, – 6, 3) (iii) A(–5, 7, –9) and B(–3, 4, – 6)

[CBSE 2011]

SSS Mathematics for Class 12 1104

1104

Senior Secondary School Mathematics for Class 12

3. Show that the line joining the points A(1, –1, 2) and B(3, 4, –2) is perpendicular to the line joining the points C(0, 3, 2) and D(3, 5, 6). 4. Show that the line segment joining the origin to the point A(2, 1, 1) is perpendicular to the line segment joining the points B(3, 5, –1) and C(4, 3, –1). 5. Find the value of p for which the line through the points A(4, 1, 2) and B(5 , p , 0) is perpendicular to the line through the points C( 2, 1, 1) and D( 3 , 3 , - 1). 6. If O be the origin and P( 2, 3 , 4) and Q(1, -2, 1) be any two points, show that OP ^ OQ. 7. Show that the line segment joining the points A(1, 2, 3) and B( 4, 5 , 7) is parallel to the line segment joining the points C( -4, 3 , - 6) and D( 2, 9, 2). 8. If the line segment joining the points A(7 , p , 2) and B( q , -2, 5) be parallel to the line segment joining the points C( 2, - 3 , 5) and D( -6, -15 , 11), find the values of p and q. 9. Show that the points A( 2, 3 , 4), B( -1, - 2, 1) and C(5 , 8, 7) are collinear. 10. Show that the points A( -2, 4, 7), B( 3 , - 6, - 8) and C(1, - 2, - 2) are collinear. 11. Find the value of p for which the points A( -1, 3 , 2), B( -4, 2, - 2), and C(5 , 5 , p) are collinear. 12. Find the angle between the two lines whose direction cosines are: 2 -1 -2 3 2 6 and , , × , , 3 3 3 7 7 7 13. Find the angle between the two lines whose direction ratios are: a, b, c and ( b - c), ( c - a), ( a - b). 14. Find the angle between the lines whose direction ratios are: 2, –3, 4 and 1, 2, 1. 15. Find the angle between the lines whose direction ratios are: 1, 1, 2 and ( 3 -1), ( - 3 -1), 4. 16. Find ® r2

the ^

angle ^

between

the

vectors

® r1

^

^

^

= ( 3 i - 2 j + k)

and

^

= ( 4 i + 5 j + 7 k ).

17. Find the angles made by the following vectors with the coordinate axes: ^

^

^

(i) ( i - j + k )

^

^

(ii) ( j - k )

^

^

^

(iii) ( i - 4 j + 8 k)

18. Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1). [CBSE 2004, ’05]

SSS Mathematics for Class 12 1105

Fundamental Concepts of 3-Dimensional Geometry ANSWERS (EXERCISE 26)

2 -1 -2 -9 6 -2 , , (iii) , , 3 3 3 11 11 11 -1 1 1 (i) d.r.’s are –1, 1, 1 and d.c.’s are 2. , , × 3 3 3 2 6 -3 (ii) d.r.’s are 2, 6, –3 and d.c.’s are , , × 7 7 7 2 3 -3 (iii) d.r.’s are 2, –3, 3 and d.c.’s are , , × 22 22 22 -3 æ -8 ö 8. p = 4, q = 3 11. p = 10 12. cos-1 ç ÷ 5. p = 2 è 21 ø (i)

1.

13.

p 2

2 -6 3 , , 7 7 7

14.

(ii)

p 2

15.

p 3

æ 1 ö -1 æ -1 ö -1 æ 1 ö 17. (i) cos-1 ç ÷ , cos ç ÷ ÷ , cos ç è 3ø è 3ø è 3ø æ1ö (iii) cos-1 ç ÷ , cos-1 è 9ø

æ 4ö -1 ç ÷ , cos è 9ø

æ 3 ö 16. cos-1 ç ÷ è 2 35 ø (ii)

p p 3p , , 2 4 4

æ 8ö ç ÷ è 9ø

æ -5 2 19 ö 18. ç , , ÷ 3 3ø è 3 HINTS TO SOME SELECTED QUESTIONS 1. (ii) D.r.’s of AB are ( 1 - 5 ), ( -6 - 6 ), ( 3 + 3 ), i.e., –4, –12, 6, i.e., 2, 6, –3. 2 2 + 6 2 + ( -3 ) 2 = 4 + 36 + 9 = 49 = 7 . \ d.c.’s of AB are

2 6 -3 , , × 7 7 7

7. a1 = ( 4 - 1) = 3 , b1 = (5 - 2 ) = 3 , c1 = (7 - 3 ) = 4. a2 = ( 2 + 4 ) = 6 , b 2 = ( 9 - 3 ) = 6 , c2 = ( 2 + 6 ) = 8. a1 b1 c1 1 \ = = = × Hence, AB||CD. a2 b 2 c2 2 8. a1 = ( q - 7 ), b1 = ( -2 - p ), c1 = (5 - 2 ) = 3. a2 = ( -6 - 2 ) = -8 , b 2 = ( -15 + 3 ) = -12 , c2 = ( 11 - 5 ) = 6. a b c Since AB ||CD , we have 1 = 1 = 1 × a2 b 2 c2 \

q - 7 -2 - p 3 1 = = × Find p and q. = -12 6 2 -8

9. D.r.’s of AB are ( -1 - 2 ), ( -2 ,-3 ), ( 1 - 4 ), i.e., –3, –5, –3, i.e., 3, 5, 3. D.r.’s. of AC are (5 - 2 ), ( 8 - 3 ), (7 - 4 ), i.e., 3, 5, 3. \ AB|| AC and both have a common end point A. Hence, A , B and C are collinear.

1105

SSS Mathematics for Class 12 1106

1106

Senior Secondary School Mathematics for Class 12

10. D.r.’s of AB are ( 3 + 2 ), ( -6 - 4 ), ( -8 - 7 ), i.e., 5, –10, –15, i.e., 1, –2, –3. D.r.’s of AC are ( 1 + 2 ), ( -2 - 4 ), ( -2 - 7 ), i.e., 3 , - 6 , - 9 , i.e., 1, –2, –3. \ AB|| AC and both have a common end point A. Hence, A , B and C are collinear. 11. D.r.’s of AB are ( - 4 + 1), ( 2 - 3 ), ( -2 - 2 ), i.e., -3 , - 1, - 4 , i.e., 3, 1, 4. D.r.’s of AC are (5 + 1), (5 - 3 ), ( p - 2 ), i.e., 6, 2, p - 2. Since A , B and C are collinear, we must have AB ||AC . 3 1 4 \ = = Þ p - 2 = 8 Þ p = 10. 6 2 p-2 é æ 2 3 ö æ -1 2 ö æ -2 6 ö ù é 6 2 12 ù ( 6 - 14 ) -8 12. cos q = ê ç ´ ÷ + ç - ú= = × ´ ÷+ ç ´ ÷ú = ê 21 21 ë è 3 7 ø è 3 7 ø è 3 7 ø û ë 21 21 21 û \

æ -8 ö q = cos -1 ç ÷. è 21 ø

13. cos q = 14.

Sa 2

S( b - c ) 2

=0 Þ q=

p × 2

2 2 + ( -3 ) 2 + 4 2 = 4 + 9 + 16 = 29 and 12 + 2 2 + 12 = 6 . \

\ 15.

a( b - c ) + b( c - a) + c( a - b )

1 ö æ -3 2 ö æ 4 1 ö æ 2 ÷+ ç ÷+ ç ÷ cos q = ç ´ ´ ´ 6 ø è 29 6 ø è 29 6ø è 29 2 6 4 0 = + = = 0. 174 174 174 174 p q= × 2

12 + 12 + 2 2 = 6 and ( 3 - 1) 2 + ( - 3 - 1) 2 + 4 2 = 24 . \

ì 1 ( 3 - 1) 1 ( - 3 - 1) 2 4 ü cos q = í ´ + ´ + ´ ý 6 24 6 24 6 24 þ î ( 3 - 1) ( - 3 - 1) 8 6 1 + + = = 12 12 12 12 2

= \

q=

p × 3 ®

®

^

r 1× r

16. cos q =

2 ®

®

{

|r 1||r 2| ®

^

^

^

^

^

^

^

( 3 i - 2 j + k )×(4 i + 5 j + 7 k )

=

2

2

}{

2

3 + ( - 2) + 1

2

2

4 +5 +7

2

^

^

}

=

3 × 2 35

^

17. (i) Let a = i - j + k . ^

cos a =

^

^

^

(i - j + k )× i ^

^

^

^

=

|i - j + k ||i | ^

and cos g =

^

^

^

(i - j + k )× k ^

^

^

^

|i - j + k ||k |

^

^

(i - j + k )× j 1 -1 , cos b = = ^ ^ ^ ^ 3 3 |i - j + k || j | =

1 × 3

SSS Mathematics for Class 12 1107

27. STRAIGHT LINE IN SPACE Equation of a Line Passing through a Given Point and Parallel to a Given Vector Vector Form THEOREM 1

The vector equation of a straight line passing through a given point with ®

®

®

®

®

position vector r1 and parallel to a given vector m is r = r1 + l m , where l is a scalar. PROOF

Let L be the line, passing through a given point A with position vector ® r1 and

®

parallel to a given vector m. ¾®

®

Let O be the origin. Then, OA = r1 . ®

Let P be an arbitrary point on L, and let the position vector of P be r . ¾®

®

Then, OP = r . ¾®

®

Clearly, AP | | m

®

¾®

Þ

AP = l m , for some scalar l

Þ

(p.v. of P) - (p.v. of A) = l m

®

¾®

®

¾®

Þ OP - OA = l m Þ Þ

®

®

®

®

®

®

r - r1 = l m

r = r1 + l m .

… (i)

Clearly, every point on the line L satisfies (i), and for any value of l, (i) gives the position vector of a point P on the line. ®

®

®

Hence, r = r1 + l m is the desired equation. COROLLARY

The vector equation of a straight line passing through the origin and ®

®

®

parallel to a given vector m is r = l m . PROOF

®

®

®

®

Taking r1 = 0 in (i), we get the desired equation, r = l m .

Cartesian Form THEOREM 2

The equations of a straight line with direction ratios a, b, c, and passing through a point A( x1 , y1 , z1) are 1107

SSS Mathematics for Class 12 1108

1108

Senior Secondary School Mathematics for Class 12

x - x1 y - y1 z - z1 = = × a b c PROOF

We know that the vector equation of a straight line passing through a fixed point ®

A( x1 , y1 , z1) with position vector r1 and ®

parallel to a given vector m is given by ®

®

®

r = r1 + l m

… (i), where l is a scalar. ®

Let P( x , y , z) be the given point on line with position vector r . ®

^

^

®

^

^

^

^

Then, r = x i + y j + z k and r 1 = x1 i + y1 j + z1 k. Since the direction ratios of the given line are a , b , c, and this line is ®

®

parallel to m , the direction ratios of m are a , b , c. ®

^

^

^

®

^

^

\ m = a i + b j + c k. ^ ®

^

^

®

^

^

^

^

Putting r = x i + y j + z k , r1 = x1 i + y1 j + z1 k and m = a i + b j + c k in (i), we get the equation of the line as ^

^

^

^

^

^

^

^

^

^

^

^

( x i + y j + z k ) = ( x1 i + y1 j + z1 k ) + l( a i + b j + c k ) ^

^

^

Þ ( x i + y j + z k ) = ( x1 + la) i + ( y1 + l b) j + (z1 + l c) k Þ

x = x1 + l a , y = y1 + l b and z = z1 + l c x - x1 y - y1 z - z1 Þ = = = l. a b c Hence, the equations of a line having direction ratios a , b , c and passing through A( x1 , y1 , z1) are x - x1 y - y1 z - z1 = = × a b c The equations of a line having direction cosines l, m, n and passing through ( x1 , y1 , z1) are x - x1 y - y1 z - z1 = = × l m n Since the direction cosines of a line are proportional to the direction ratios of the line, the result follows.

COROLLARY

PROOF

Equation of a Line Passing through Two Given Points Vector Form THEOREM 3

The vector equation of a straight line passing through two points with ®

®

®

®

®

®

position vectors r1 and r2 is given by r = r1 + l( r2 - r1 ).

SSS Mathematics for Class 12 1109

Straight Line in Space PROOF

1109

Let L be the given line, passing through two

®

given points A and B with position vectors r1 ®

and r2 respectively. Let O be the origin. ¾®

®

®

¾®

Then, OA = r1 and OB = r2 . ® ®

¾®

\ AB = (p.v. of B) - (p.v. of A) = ( r2 - r1 ). ®

Let P be an arbitrary point on L, having the position vector r . ®

¾®

Then, OP = r . ¾®

\ AP = (p.v. of P) - (p.v. of A) ¾®

® ®

¾®

= (OP - OA) = ( r - r1 ). ¾®

¾®

Since AP and AB are collinear vectors, we have ¾®

¾®

AP = l(AB), for some scalar l ®

®

®

®

Þ ( r - r1 ) = l( r2 - r1 ) ®

®

®

®

Þ r = r1 + l( r2 - r1 ). Hence, the vector equation of a line L, passing through two given ®

®

points A and B with position vectors r1 and r2 , is given by ®

®

® ®

r = r1 + l( r2 - r1 ).

Cartesian Form THEOREM 4

PROOF

The equations of a line passing through two given points A( x1 , y1 , z1) and B( x 2 , y 2 , z2) are given by x - x1 y - y1 z - z1 = = × x 2 - x1 y 2 - y1 z2 - z1

We know that the vector equation of a line passing through two points ®

®

A and B with position vectors r1 and r2 is given by ®

®

® ®

r = r1 + l( r2 - r1 ).

Let A( x1 , y1 , z1) and B( x 2 , y 2 , z2) be the points on the given line with position ®

®

vectors r1 and r2 respectively. Let P( x , y , z) be an arbitrary point on this ®

line with position vector r .

... (i)

SSS Mathematics for Class 12 1110

1110

Senior Secondary School Mathematics for Class 12 ®

^

^ ®

^

^

^

®

^

^

^

^

Then, r1 = x1 i + y1 j + z1 k , r2 = x 2 i + y 2 j + z2 k and r = x i + y j + z k. Substituting these values in (i), we get ^

^

^

^

^

^

^

x i + y j + z k = ( x1 i + y1 j + z1 k ) + l[( x 2 - x1) i

^

^

+ ( y 2 - y1) j + (z2 - z1) k ] Þ

^

^

^

( x - x1) i + ( y - y1) j + (z - z1) k ^

^

^

= l[( x 2 - x1) i + ( y 2 - y1 j + (z2 - z1) k ] Þ Þ

( x - x1) = l( x 2 - x1), ( y - y1) = l( y 2 - y1) and (z - z1) = l(z2 - z1) ( x - x1) ( y - y1) (z - z1) = = ( = l), ( x 2 - x1) ( y 2 - y1) (z2 - z1)

which are the required equations. SUMMARY

®

1. (i) If a®line passes through a point with p.v. r1 and it is parallel to m then its vector equation is: ®

®

®

r = r1 + l m .

(ii) If a line passes through a point A( x1 , y1. z1) and it has d.r.’s a, b, c then its Cartesian equations are: x - x1 y - y1 z - z1 = = × a b c ®

®

2. (i) If a line passes through two points having p.v.’s r1 and r2 , then its vector equation is: ®

®

®

®

r = r1 + l ( r2 - r1 ).

(ii) If a line passes through two points A( x1 , y1 , z1) and B( x 2 , y 2 , z2) then its Cartesian equations are: x - x1 y - y1 z - z1 = = × x 2 - x1 y 2 - y1 z2 - z1

SOLVED EXAMPLES EXAMPLE 1

Find the vector equation of a line which is parallel to the vector ^

^

^

( 2 i - j + 3 k ) and which passes through the point (5 , - 2, 4). Also find its Cartesian equations. [CBSE 2007] SOLUTION

Vector equation of the given line: The given line passes through the point A(5 , - 2, 4) and it is ®

^

^

^

parallel to the vector m = ( 2 i - j + 3 k ).

SSS Mathematics for Class 12 1111

Straight Line in Space

1111

®

^

^

^

The position vector of A is given by r1 = (5 i - 2 j + 4 k ). Hence, the vector equation of the given line is ®

®

®

®

^

^

^

^

^

^

r = r1 + l m Þ r = (5 i - 2 j + 4 k ) + l( 2 i - j + 3 k ).

... (i)

Cartesisan equation of the given line: ®

^

^

^

Taking r = ( x i + y j + z k ), equation (i) becomes: ^

^

^

^

^

^

^

^

^

^

^

^

( x i + y j + z k ) = (5 i - 2 j + 4 k ) + l( 2 i - j + 3 k ) Þ

^

^

^

( x i + y j + z k ) = (5 + 2l) i - ( 2 + l) j + ( 4 + 3 l) k

Þ

x = 5 + 2l , y = - ( 2 + l) and z = 4 + 3l x -5 y + 2 z - 4 = = = l. Þ 2 -1 3 x -5 y + 2 z - 4 Hence, are the required equations of the given = = 2 -1 3 line in Cartesian form. EXAMPLE 2

The Cartesian equations of a line are 3 x - 3 = 2y + 1 = 5 - 6z. (a) Write these equations in standard form and find the direction ratios of the given line. (b) Write the equation of the given line in vector form.

SOLUTION

(a) The given equations may be written as 1ö æ 5ö æ 3( x - 1) = 2ç y + ÷ = -6çz - ÷ 6ø 2ø è è

Þ

Þ

1ö æ 5 ö æ ç y + ÷ çz - ÷ ( x - 1) è 6ø 2ø è = = æ1ö æ1ö æ -1 ö ç ÷ ç ÷ ç ÷ è 3ø è 2ø è 6ø 1ö æ 5ö æ ç y + ÷ çz + ÷ ( x - 1) è 2ø è 6ø = = 2 3 -1

[on dividing each by 6].

This is the standard form of the given equations in Cartesian form. Clearly, its direction ratios are 2, 3, –1. æ -1 5 ö (b) The given line passes through the point A ç1, , ÷ and it is è 2 6ø ®

^

^

^

parallel to the vector m = ( 2 i + 3 j - k ). ® æ^ 1 ^ 5 ^ ö The position vector of the point A is r1 = çç i - j + k ÷÷ . 2 6 ø è

SSS Mathematics for Class 12 1112

1112

Senior Secondary School Mathematics for Class 12

So, the vector equation of the given line is ® ® ® æ ^ 1 ^ 5 ^ö ^ ^ ^ r = r1 + l m Þ r = çç i - j + k ÷÷ + l( 2 i + 3 j - k ). 2 6 è ø

®

EXAMPLE 3

The Cartesian equations of a line are 6x - 2 = 3 y + 1 = 2z - 2. (a) Write these equations in standard form and find the direction cosines of the given line. (b) Write down the Cartesian and vector equations of a line, passing [CBSE 2013C] through ( 2, - 1, 1) and parallel to the given line.

SOLUTION

(a) The given equations may be written as 1ö 1ö æ æ 6 ç x - ÷ = 3 ç y + ÷ = 2(z - 1) 3ø 3ø è è

Þ

Þ

1ö æ 1ö æ çx - ÷ çy + ÷ 3ø è 3 ø (z - 1) è = = æ1ö æ1ö æ1ö ç ÷ ç ÷ ç ÷ è 3ø è 6ø è 2ø 1ö æ 1ö æ çx - ÷ çy + ÷ 3ø è 3 ø (z - 1) è = = 1 2 3

[on dividing each by 6]

This is the standard form of the given equations in Cartesian form. The direction ratios of the given are 1, 2, 3. Also, 12 + 22 + 3 2 = 14. So, the direction cosines of the given line are

1 2 3 , , × 14 14 14

(b) The required line passes through the point A( 2, - 1, 1) and it is parallel to the line having direction ratios 1,2, 3. The equations of this line in Cartesian form are given by x - 2 y + 1 z -1 = = × 1 2 3

... (i) ®

^

^

^

The position vector of the point A is r1 = ( 2 i - j + k ). ®

^

^

^

The required line is parallel to the vector m = ( i + 2 j + 3 k ). So, its equation in vector form is ®

®

®

®

^

^

^

^

^

^

r = r1 + l m Þ r = ( 2 i - j + k ) + l( i + 2 j + 3 k ). ... (ii)

Thus, the Cartesian and vector forms of equations of the desired line are given by (i) and (ii) respectively.

SSS Mathematics for Class 12 1113

Straight Line in Space

1113

EXAMPLE 4

Find the vector and Cartesian equations of the line passing through the point (1, 2, - 4) and perpendicular to each of the lines x - 8 y + 19 z - 10 x - 15 y + 29 z - 5 and [CBSE 2012] = = = = × 3 -16 7 3 8 -5

SOLUTION

The given lines are x - 8 y + 19 z - 10 ... (i) = = 3 -16 7 x - 15 y + 29 z - 5 and ... (ii) = = × 3 8 -5 Let a, b, c be the direction ratios of the required line. Then, it being perpendicular to each of the lines (i) and (ii), we have 3 a - 16b + 7 c = 0 and 3 a + 8b - 5 c = 0. On solving these equations by cross multiplication, we get a b c = = ( 80 - 56) ( 21 + 15) ( 24 + 48) a b c a b c = = Þ = = × Þ 24 36 72 2 3 6 Thus, the desired line has direction ratios 2, 3, 6. So, we have to find the equations of a line passing through the point A(1, 2, –4) and having 2, 3, 6 as its direction ratios. So, the required equations in Cartesian form are ( x - 1) ( y - 2) (z + 4) = = × 2 3 6 ®

^

^

^

The position vector of point A is r1 = ( i + 2 j - 4 k ). Also, the required line has direction ratios 2, 3, 6 and so it is ®

^

^

^

parallel to the vector m = ( 2 i + 3 j + 6 k ). So, its equation in vector form is ®

®

®

®

^

^

^

^

^

^

r = r1 + l m Þ r = ( i + 2 j - 4 k) + l( 2 i + 3 j + 6 k ).

EXAMPLE 5

Find the equation of a line passing through the point P( 2, - 1, 3) and perpendicular to the lines ®

^

^

^

^

^

^

r = ( i + j - k ) + l( 2 i - 2 j + k )

®

SOLUTION

^

^

^

^

^

^

r = (2 i - j - 3 k) + m( i + 2 j + 2 k)

and

[CBSE 2012]

The given lines are ®

®

®

®

®

®

®

^

^

®

^

^

^

^

r = a1 + l b1 ... (i), where a1 = ( i + j - k ) and b1 = ( 2 i - 2 j + k ) and ®

^

^

^

®

^

^

^

r = a2 + l b2 ... (ii), where a2 = ( 2 i - j - 3 k ) and b2 = ( i + 2 j + 2 k ).

The required line is perpendicular to (i) as well as (ii).

SSS Mathematics for Class 12 1114

1114

Senior Secondary School Mathematics for Class 12 ®

®

Also (i) is parallel to b1 and (ii) is parallel to b2. ®

®

So, the required line is perpendicular to both b1 and b2. ® ®

Consequently, this line must be parallel to ( b1 ´ b2). ^

^

^

Now, ( b1 ´ b2) = 2

-2

1 = ( -6 i - 3 j + 6 k ).

1

2

2

i

® ®

j

k

^

^

^

So, we have to find the equation of a line passing through the ® ®

point P( 2, - 1, 3) and parallel to ( b1 ´ b2). Hence, the required equation is ®

^

^

^

^

^

^

r = ( 2 i - j + 3 k ) + t( -6 i - 3 j + 6 k ),

where t is an arbitrary constant. EXAMPLE 6

Find the vector equation of the line passing through the point A( 2, - 1, 1), and parallel to the line joining the points B( -1, 4, 1) and [CBSE 2003] C(1, 2, 2). Also, find the Cartesian equations of the line.

SOLUTION

Vector equation of the given line: ^

^

^

^

^

^

The p.v. of B = ( - i + 4 j + k ) and p.v. of C = ( i + 2 j + 2 k ). \

¾®

BC = (p.v. of C ) - (p.v. of B) ^

^

^

^

^ ^

^

^

^

^

^

^

= ( i + 2 j + 2 k ) - ( - i + 4 j + k ) = ( 2 i - 2 j + k ). ®

The p.v. of A is r1 = 2 i - j + k . \

the vector equation of the given line is ®

®

¾®

r = r1 + l( BC)

Û

®

^

^

^

^

^

^

r = ( 2 i - j + k ) + l( 2 i - 2 j + k )

… (i)

Cartesian equations of the given line: ®

^

^

^

Taking r = x i + y j + z k , equation (i) becomes ^

^

^

^

^

^

^

^

^

^

^

^

( x i + y j + z k ) = ( 2 i - j + k ) + l( 2 i - 2 j + k ) ^

^

^

Û ( x i + y j + z k ) = ( 2 + 2l) i + ( -1 - 2l) j + (1 + l) k Û x = 2 + 2l, y = - 1 - 2l and z = 1 + l

SSS Mathematics for Class 12 1115

Straight Line in Space

1115

x - 2 y + 1 z -1 = = = l. 2 -2 1 x - 2 y + 1 z -1 are the required equations of the Hence, = = 2 -2 1 given line in the Cartesian form. Û

EXAMPLE 7

Find the vector and Cartesian equations of the line passing through the points A( 2, - 1, 4) and B(1, 1, - 2).

SOLUTION

Vector equation of the given line: ®

®

Let the position vectors of A and B be r1 and r2 respectively. Then, ®

^

^

^

®

^

^

^

^

^

r1 = 2 i - j + 4 k and r2 = i + j - 2 k . ®

®

^

^

^

^

^

^

^

\

( r2 - r1 ) = ( i + j - 2 k ) - ( 2 i - j + 4 k ) = ( - i + 2 j - 6 k ).

\

the vector equation of the line AB is ®

®

®

®

r = r1 + l( r2 - r1 ) for some scalar, l,

®

®

®

®

®

®

®

i.e., r = ( 2 i - j + 4 k ) + l( - i + 2 j - 6 k ).

… (i)

Cartesian equations of the given line: ®

^

^

^

Taking r = x i + y j + z k , equation (i) becomes ^

^

^

^

^

^

^

^

^

^

^

^

( x i + y j + z k ) = ( 2 i - j + 4 k ) + l( - i + 2 j - 6 k ) ^

^

^

Û ( x i + y j + z k ) = ( 2 - l) i + ( 2l - 1) j + ( 4 - 6l) k Û x = 2 - l , y = 2l - 1 and z = 4 - 6l x-2 y+1 z-4 = = = l. Û -1 2 -6

x-2 y+1 z-4 are the Cartesian equations of the = = -1 2 -6 given line. x+1 y+ 3 z+5 x-2 y-4 z-6 Show that the lines and = = = = 3 5 7 1 3 5 intersect. Also find their point of intersection. [CBSE 2014]

Hence,

EXAMPLE 8

SOLUTION

The given lines are x+1 y+ 3 z+5 = = = l (say) 3 5 7 x-2 y-4 z-6 and = = = m (say). 1 3 5 The general point on (1) is P( 3 l - 1, 5 l - 3 , 7 l - 5).

... (1) ... (2)

SSS Mathematics for Class 12 1116

1116

Senior Secondary School Mathematics for Class 12

The general point on (2) is Q(m + 2, 3m + 4, 5m + 6). The given lines will intersect only when they have a common point. This happens when P and Q coincide for some particular values of l and m. So, the given lines will intersect only when 3 l - 1 = m + 2, 5 l - 3 = 3m + 4 and 7 l - 5 = 5m + 6 ... (iii) Þ 3 l - m = 3 ... (i), 5 l - 3m = 7 ... (ii) and 7 l - 5m = 11. 1 -3 On solving (i) and (ii), we get l = and m = × 2 2 Clearly, these values of l and m also satisfy (iii). Hence, the given lines intersect. 1 -3 in Q, we get the point of intersection of Putting l = in P or m = 2 2 æ 1 -1 -3 ö the given lines as ç , , ÷× è2 2 2 ø EXAMPLE 9

SOLUTION

Show that the lines x -1 y + 1 z -1 x + 2 y -1 z + 1 and = = = = 3 2 5 4 3 -2 do not intersect.

[CBSE 2002]

The given lines are x -1 y + 1 z -1 … (1) = = = l (say) 3 2 5 x + 2 y -1 z + 1 … (2) = = = m (say). 4 3 -2 The general point on (1) is P( 3 l + 1, 2l - 1, 5 l + 1). The general point on (2) is Q( 4m - 2, 3m + 1, - 2m - 1). If possible, let the given lines intersect. Then, P and Q coincide for some particular values of l and m. In that case, we have 3 l + 1 = 4m - 2, 2l - 1 = 3m + 1 and 5 l + 1 = - 2m - 1 Û

3 l - 4m = - 3 … (i), 2l - 3m = 2 … (ii), 5 l + 2m = - 2 … (iii).

Solving (i) and (ii), we get l = - 17 and m = - 12. However, these values of l and m do not satisfy (iii). Hence, the given lines do not intersect. EXAMPLE 10

Show that the lines ®

^

^

^

^

^

®

^

^

^

^

r = ( i + j - k ) + l( 3 i - j ) and r = ( 4 i - k ) + m ( 2 i + 3 k ) intersect each other. Find their point of intersection. [CBSE 2013, ’14]

SSS Mathematics for Class 12 1117

Straight Line in Space SOLUTION

1117

The given lines are ®

^

^

^

^ ^

r = ( i + j - k ) + l( 3 i - j )

®

^

^

^

… (1)

^

… (2) r = ( 4 i - k ) + m ( 2 i + 3 k ). These lines will intersect if for some particular values of l and m , ®

the values of r given by (1) and (2) are the same. So, we must have ^

^

^

^

^

^

^

^

^

^

^

( i + j - k ) + l( 3 i - j ) = ( 4 i - k ) + m ( 2 i + 3 k ) ^

^

^

Þ (1 + 3 l) i + (1 - l) j - k = ( 4 + 2m ) i + ( 3m - 1) k Þ 1 + 3 l = 4 + 2m ,1 - l = 0 and 3m - 1 = -1 Þ 3 l - 2m = 3 ... (i), l = 1 ... (ii) and m = 0 ... (iii). Clearly, l = 1 and m = 0 also satisfy (i). Hence, the given lines intersect. ®

^

^

^

Putting l = 1 in (1), we get r = ( 4 i + 0 j - k ). Hence, the point of intersection of the given lines is (4, 0, –1). EXAMPLE 11

Show that the lines ®

^

^

^

^ ^

®

^

^

^

^

^ ^

^

r = ( i + 2 j + k ) + l( i - j + k ) and r = ( i + j + k ) + m ( i - j + 2 k ) do not intersect. SOLUTION

The given lines are ®

^

®

^

^

^

^

^

^

r = ( i + 2 j + k ) + l( i - j + k ) ^

^

^

^

… (1)

^

… (2) r = ( i + j + k) + m( i - j + 2 k) These lines will intersect if for some particular values of l and m , ®

the values of r given by (1) and (2) are the same. This happens, when ^

^

^

^ ^

^

^

^

^

^

^

^

( i + 2 j + k ) + l( i - j + k ) = ( i + j + k ) + m ( i - j + 2 k ) Þ

^

^

^

^

^

^

(1 + l) i + ( 2 - l) j + (1 + l) k = (1 + m ) i + (1 - m ) j + (1 + 2m ) k

Þ 1 + l = 1 + m , 2 - l = 1 - m and 1 + l = 1 + 2m Þ l - m = 0 … (i), l - m = 1 … (ii), l - 2m = 0 … (iii). From (ii) and (iii), we get l = 2 and m = 1. And, these values of l and m do not satisfy (i). Hence, the given lines do not intersect. EXAMPLE 12

Find the points on the line from the point (1, 3 , 3).

x+ 2 y+1 z- 3 at a distance of 5 units = = 3 2 2 [CBSE 2010]

SSS Mathematics for Class 12 1118

1118 SOLUTION

Senior Secondary School Mathematics for Class 12

The given line is x+ 2 y+1 z- 3 = = = r (say). ... (i) 3 2 2 The general point on this line is P( 3r - 2, 2r - 1, 2r + 3). The given point is A(1, 3, 3). Now, PA = 5 Þ ( PA) 2 = 25 Þ ( 3r - 2 - 1) 2 + ( 2r - 1 - 3) 2 + ( 2r + 3 - 3) 2 = 25 Þ ( 3r - 3) 2 + ( 2r - 4) 2 + ( 2r) 2 = 25 Þ 17r 2 - 34r = 0 Þ 17r(r - 2) = 0 Þ r = 0 or r = 2. r = 0 Þ the required point is P(–2, –1, 3). r = 2 Þ the required point is P(4, 3, 7). Hence, the rquired points are (–2, –1, 3) and (4, 3, 7).

EXAMPLE 13

Find the equations of the perpendicular from the point ( 3 , - 1, 11) to the x y-2 z- 3 line = = × Also find the coordinates of the foot of the 2 3 4 perpendicular and the length of the perpendicular. [CBSE 2011C]

SOLUTION

The given line is x y-2 z- 3 ... (i) = = = r (say). 2 3 4 The general point on this line is ( 2r , 3r + 2, 4r + 3). Let N be the foot of the perpendicular drawn from the point P( 3 , - 1, 11) on the given line. Then, this point is N ( 2r , 3r + 2, 4r + 3) for some fixed value of r. D.r.’s of PN are ( 2r - 3),( 3r + 3),( 4r - 8). D.r.’s of the given line (i) are 2, 3, 4. Since PN is perpendicular to the given line (i), we have 2( 2r - 3) + 3( 3r + 3) + 4( 4r - 8) = 0 Þ 29r = 29 Þ r = 1. \ D.r.’s of PN are –1, 6, –4 [putting r = 1 in (ii)]. So, the required equations of perpendicular PN are x - 3 y + 1 z - 11 = = × -1 6 -4 Now, the coordinates of N are N(2, 5, 7) [Putting r = 1] Length of perpendicular PN = ( 2 - 3) 2 + (5 + 1) 2 + (7 - 11) 2 = ( -1) 2 + 62 + ( -4) 2 = 1 + 36 + 16 = 53.

... (ii)

SSS Mathematics for Class 12 1119

Straight Line in Space EXAMPLE 14

1119

Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from the point P(5 , 4, 2) to the line ®

^

^

^

^

^

^

r = ( - i + 3 j + k ) + l( 2 i + 3 j - k ). Also find the image of P in this line. [CBSE 2012] SOLUTION

The vector equation of the given line is ®

^

^

^

^

^

^

r = ( - i + 3 j + k ) + l( 2 i + 3 j - k ).

Clearly, it passes through the point ( -1, 3 , 1) and it has direction ratios 2, 3, –1. So, its Cartesian equations are x + 1 y - 3 z -1 ... (i) = = = r (say). 2 3 -1 The general point on this line is ( 2r - 1, 3r + 3 , - r + 1). Let N be the foot of the perpendicular drawn from the point P(5 , 4, 2) on the given line. Then, this point is N ( 2r - 1, 3r + 3 , - r + 1) for some fixed value of r. D.r.’s of PN are ( 2r - 6, 3r - 1, - r - 1). D.r.’s of the given line are 2, 3, –1. Since PN is perpendicular to the given line (i), we have 2( 2r - 6) + 3( 3r - 1) - 1 × ( -r - 1) = 0 Þ 14r = 14 Þ r = 1. So, the point N is given by N(1, 6, 0). Hence, the foot of the perpendicular from the given point P(5 , 4, 2) on the given line is N(1, 6, 0). Let Q( a , b , g) be the image of P(5, 4, 2) in the given line. Then, N(1, 6, 0) is the midpoint of PQ. 5+a 4+b 2+ g \ = 1, = 6 and = 0 Þ a = -3 , b = 8 and g = -2. 2 2 2 Hence, the image of P(5, 4, 2) in the given line is Q( -3 , 8, - 2). EXAMPLE 15

Find the image of the point P(1, 6, 3) in the line

SOLUTION

The equations of the given line are x y -1 z - 2 ... (i) = = = r (say). 1 2 3 The general point on this line is (r , 2r + 1, 3r + 2). Let N be the foot of the perpendicular drawn from the point P(1, 6, 3) on the given line.

x y -1 z - 2 = = × 1 2 3

[CBSE 2003C, ’08C, ’10C]

SSS Mathematics for Class 12 1120

1120

Senior Secondary School Mathematics for Class 12

Then, this point is N (r , 2r + 1, 3r + 2) for some fixed value of r. D.r.’s of PN are (r - 1, 2r - 5 , 3r - 1). D.r.’s of the given line are 1, 2, 3. Since, PN is perpendicular to the given line (i), we have 1 × (r - 1) + 2( 2r - 5) + 3( 3r - 1) = 0 Þ 14r = 14 Þ r = 1. So, the point N is given by N(1, 3, 5). Let Q( a , b , g) be the image of P(1, 6, 3) in the given line. Then, N is the midpoint of PQ. a +1 b+ 6 g+ 3 \ = 1, = 3, = 5 Þ a = 1, b = 0 and g = 7. 2 2 2 Hence, the image of the point P(1, 6, 3) in the given line is Q(1, 0, 7). EXAMPLE 16

Find the equations of the line passing through the points A( 0, 6, - 9) and B( -3 , - 6, 3). If D is the foot of the perpendicular drawn from a point C(7 , 4, - 1) on the line AB then find the coordinates of D and the [CBSE 2010C] equations of the line CD.

SOLUTION

The equations of line AB are x-0 y-6 z + 9 é x - x1 y - y1 z - z1 ù = = = = ê -3 - 0 -6 - 6 3 + 9 ë x 2 - x1 y 2 - y1 z2 - z1 úû Þ Þ

x y-6 z+ 9 = = -3 -12 12 x y-6 z+ 9 = = = a (say). 1 4 -4

... (i)

Thus, the required equations of line AB are given by (i). From (i), we get x = a , y = 4a + 6 and z = -4a - 9. So, the coordinates of D are D( a , 4a + 6, - 4a - 9) for some particular value of a. D.r.’s of AB are 1, 4, –4. D.r.’s of CD are ( a - 7), ( 4a + 2), ( -4a - 8). Since AB ^ CD , we have 1 × ( a - 7) + 4( 4a + 2) - 4( - 4a - 8) = 0 Þ 33 a = -33 Þ a = -1. Putting a = -1, we get the coordinates of D as D( -1, 2, - 5).

SSS Mathematics for Class 12 1121

Straight Line in Space

1121

The equations of the line CD are x -7 y-4 z+1 x -7 y - 4 z + 1 = = Þ = = -1 - 7 2 - 4 -5 + 1 -8 -2 -4 x -7 y - 4 z + 1 Þ = = × 4 1 2 EXAMPLE 17

The points A( 4, 5 , 10), B( 2, 3 , 4) and C(1, 2, - 1) are three vertices of a parallelogram ABCD. Find the vector equations of the sides AB and BC [CBSE 2010] and also find the coordinates of point D.

SOLUTION

D.r.’s of side AB are ( 2 - 4), ( 3 - 5), ( 4 - 10), i.e., -2, - 2, - 6. Thus, side AB passes through the point A( 4, 5 , 10) and it has D.r.’s -2, - 2, - 6.

So, the vector equation of side AB is ®

^

^

^

^

^

^

r = ( 4 i + 5 j + 10 k ) + l( -2 i - 2 j - 6 k ).

D.r.’s. of side BC are (1 - 2), ( 2 - 3), ( -1 - 4), i.e., –1, –1, –5. Thus, side BC passes through the point B(2, 3, 4) and it has d.r.’s –1, –1, –5. So, the vector equation of side BC is ®

^

^

^

^

^

^

r = ( 2 i + 3 j + 4 k ) + m ( - i - 3 j - 5 k ). Let the coordinates of D be D( x , y , z). We know that midpoint of BD coincides with the midpoint of AC. 2+ x 4+1 3 + y 5 + 2 4 + z 10 + ( -1) and = , = = \ 2 2 2 2 2 2 Þ 2 + x = 5 , 3 + y = 7 and 4 + z = 9 Þ x = 3 , y = 4 and z = 5. \ coordinates of D are D(3, 4, 5).

EXERCISE 27A 1. A line passes through the point (3, 4, 5) and is parallel to the vector ^

^

^

( 2 i + 2 j - 3 k ). Find the equations of the line in the vector as well as Cartesian forms. 2. A line passes through the point (2, 1, –3) and is parallel to the vector ^

^

^

( i - 2 j + 3 k ). Find the equations of the line in vector and Cartesian forms.

SSS Mathematics for Class 12 1122

1122

Senior Secondary School Mathematics for Class 12

3. Find the vector equation of the line passing through the point with ^

^

^

^

^

^

position vector ( 2 i + j - 5 k ) and parallel to the vector ( i + 3 j - k ). Deduce the Cartesian equations of the line. ^

^

^

4. A line is drawn in the direction of ( i + j - 2 k ) and it passes through a point ^ ^

^

with position vector ( 2 i - j - 4 k ). Find the equations of the line in the vector as well as Cartesian forms. 5. The Cartesian equations of a line are x-3 y+2 z-6 = = × 2 -5 4 Find the vector equation of the line. 6. The Cartesian equations of a line are 3 x + 1 = 6y - 2 = 1 - z. Find the fixed point through which it passes, its direction ratios and also its vector equation. [CBSE 2004] 7. Find the Cartesian equations of the line which passes through the point (1, 3 , - 2) and is parallel to the line given by x+1 y-4 z+ 3 = = × 3 5 -6 Also, find the vector form of the equations so obtained. 8. Find the equations of the line passing through the point (1, –2, 3) x-6 y-2 z+7 and parallel to the line = = × 3 -4 5 Also find the vector form of this equation so obtained. 9. Find the Cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line - x - 2 y + 3 2z - 6 [CBSE 2004, ’07] = = × 1 7 3 10. Find the equations of the line passing through the point (–1, 3, –2) and x y z x + 2 y -1 z + 1 perpendicular to each of the lines = = and = = × 1 2 3 -3 2 5 [CBSE 2005, ’12]

11. Find the Cartesian and vector equations of the line passing through the point (1, 2, –4) and perpendicular to each of the lines x - 8 y + 19 z - 10 x - 15 y + 29 z - 5 [CBSE 2012] and = = = = × 8 -16 7 3 8 -5 12. Prove that the lines x-4 y+ 3 z+1 x - 1 y + 1 z + 10 and = = = = 1 7 2 8 -4 -3 intersect each other and find the point of their intersection.

SSS Mathematics for Class 12 1123

Straight Line in Space

13.

1123

Show that the lines x -1 y - 2 z - 3 x - 4 y -1 and = = = =z 2 3 4 5 2

[CBSE 2004, ’05]

intersect each other. Also, find the point of their intersection. 14. Show that the lines x -1 y + 1 x+1 y-2 = = z and = , z=2 2 3 5 1 do not intersect each other. 15. Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line x - 6 y -7 z -7 [CBSE 2003C] = = × 3 2 -2 Also, find the length of the perpendicular from the given point to the line. 16. Find the length and the foot of the perpendicular drawn from the point ( 2, - 1, 5) to the line x - 11 y + 2 z + 8 [CBSE 2008, ‘09C] = = × 10 -4 -11 17. Find the vector and Cartesian equations of the line passing through the points A(3, 4, –6) and B(5, –2, 7). 18. Find the vector and Cartesian equations of the line passing through the points A(2, –3, 0) and B(–2, 4, 3). 19. Find the vector and Cartesian equations of the line joining the points ^

^

^

^

^

^

whose position vectors are ( i - 2 j + k ) and ( i + 3 j - 2 k ). 20. Find the vector equation of a line passing through the point A( 3 , - 2, 1) and parallel to the line joining the points B( -2, 4, 2) and C( 2, 3 , 3). Also, find the Cartesian equations of the line. 21. Find the vector equation of a line passing through the point having the ^

^

^

position vector ( i + 2 j - 3 k ) and parallel to the line joining the points ^

^

^

^

^

^

with position vectors ( i - j + 5 k ) and ( 2 i + 3 j - 4 k ). Also, find the Cartesian equivalents of this equation. 22. Find the coordinates of the foot of the perpendicular drawn from the point A(1, 2, 1) to the line joining the points B(1, 4, 6) and C(5, 4, 4). 23. Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B( 0, - 1, 3) and C( 2, - 3 , - 1). [CBSE 2005]

x + 3 y -1 z + 4 24. Find the image of the point (0, 2, 3) in the line = = × 5 2 3 x -1 y - 2 z - 3 25. Find the image of the point (5, 9, 3) in the line = = × 2 3 4

SSS Mathematics for Class 12 1124

1124

Senior Secondary School Mathematics for Class 12

26. Find the image of the point (2, –1, 5) in the line ®

^

^

^

^

^

^

r = (11 i - 2 j - 8 k ) + l (10 i - 4 j - 11 k ). ANSWERS (EXERCISE 27A)

x - 3 y - 4 z -5 = = 2 2 -3 ^ ^ ^ ^ ^ ^ x - 2 y -1 z + 3 ® 2. r = ( 2 i + j - 3 k ) + l( i - 2 j + 3 k ), = = 1 -2 3 ^ ^ ^ ^ ^ ^ x - 2 y -1 z + 5 ® 3. r = ( 2 i + j - 5 k ) + l( i + 3 j - k ), = = 1 3 -1 ^ ^ ^ ^ ^ ^ x-2 y+1 z-4 ® 4. r = ( 2 i - j + 4 k ) + l( i + j - 2 k ), = = 1 1 -2 ®

^

^

^

^

^

^

®

^

^

^

^

^

^

1. r = ( 3 i + 4 j + 5 k ) + l( 2 i + 2 j - 3 k ),

5. r = ( 3 i - 2 j + 6 k ) + l( 2 i - 5 j + 4 k ) ^ ^ ^ ® æ -1 ^ 1 ^ ^ ö ö æ -1 1 6. ç , , 1÷ , d.r.’s are ( 2, 1, - 6), r = çç i + j + k ÷÷ + l( 2 i + j - 6 k ) 3 3 ø è 3 è 3 ø ^ ^ ^ ^ ^ ^ ® x -1 y - 3 z + 2 7. , r = ( i + 3 j - 2 k ) + l( 3 i + 5 j - 6 k ) = = 3 5 -6

8. 9. 10. 11. 12.

^ ^ ^ ^ ^ x -1 y+ 2 z- 3 ® ^ = , r = ( i - 2 j + 3 k ) + l( 3 i - 4 j + 5 k ) = 3 -4 5 ^ ^ ^ ^ ^ x -1 y-2 z- 3 ® ^ = = , r = ( i + 2 j + 3 k ) + l( -2 i + 14 j + 3 k ) -2 14 3 ^ ^ ^ ^ ^ ^ x +1 y-3 z+2 ® = , r = ( - i + 3 j - 2 k ) + l( 2 i - 7 j + 4 k ) = 2 -7 4 ^ ^ ^ ^ ^ x -1 y - 2 z + 4 ® ^ , r = ( i + 2 j - 4 k ) + l( 2 i + 3 j + 6 k ) = = 2 3 6 (5, –7, 6) 13. (–1, –1, –1)

15. (3, 5, 9), 7 units ®

^

^

®

^

^

16. ^

^

^

^

14 units, (1, 2, 3)

17. r = ( 3 i + 4 j - 6 k ) + l( 2 i - 6 j + 13 k ), ^

^

x-2 y+ 3 z = = -4 7 3

^

18. r = ( 2 i - 3 j ) + l( -4 i + 7 j + 3 k ), ®

^

^

^

^

^

x -1 y + 2 z -1 = = 0 5 -3

^

^

19. r = ( i - 2 j + k ) + l(5 j - 3 k ), ®

^

^

^

^

20. r = ( 3 i - 2 j + k ) + l( 4 i - j + k ), ®

^

^

^

^

x- 3 y-4 z+ 6 = = 2 -6 13

^

^

x - 3 y + 2 z -1 = = 4 1 1

21. r = ( i + 2 j - 3 k ) + l( i + 4 j - 9 k ),

x -1 y - 2 z + 3 = = 1 4 -9

SSS Mathematics for Class 12 1125

Straight Line in Space

22. (3, 4, 5)

æ -5 2 19 ö 23. ç , , ÷ è 3 3 3ø

25. (1, 1, 11)

26. (0, 5, 1)

1125

24. (4, 4, –5)

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 27A) 9. The given equation may be written as x+ 2 y+ 3 z- 3 x+ 2 y+ 3 z- 3 = = Þ = = × 3 -1 7 -2 14 3 2 x+ 1 y- 3 z+ 2 10. Let the required equations be = = × Then, a b c a + 2b + 3c = 0 -3 a + 2 b + 5 c = 0 On solving (i) and (ii) by cross multiplication, we get a b c a b c a b c = = Þ = = Þ = = ( 10 - 6 ) ( -9 - 5 ) ( 2 + 6 ) 4 -14 8 2 -7 4

... (i) ... (ii)

x+ 1 y- 3 z+ 2 = = × 2 -7 4 x+ 1 y- 2 z- 2 14. The second equation is = = × 5 1 0 Hence, the required equations are

17. The given line passes through the point A(3, 4, –6) and it has d.r.’s (5 - 3 ), ( -2 - 4 ), (7 + 6 ), i.e., 2, –6, 13. \

®

^

^

^

^

^

^

its vector equation is r = ( 3 i + 4 j - 6 k ) + l( 2 i - 6 j + 13 k ).

Its Cartesian equations are

( x - 3 ) ( y - 4 ) (z + 6 ) = = × 2 13 -6

19. The given line passes through the point A( 1, - 2 , 1) and it has d.r.’s ( 1 - 1), ( 3 + 2 ), ( -2 - 1), i.e., 0, 5, –3. ®

^

^

^

^

^

So, its vector equation is r = ( i - 2 j + k ) + l(5 j - 3 k ). And, its Cartesian equations are

x- 1 y+ 2 z- 1 = = × 0 5 -3

20. The given line passes through the point A(3, –2, 1) and it has d.r.’s ( 2 + 2 ), ( 3 - 4 ), ( 3 - 2 ), i.e., 4, –1, 1. 21. The required line passes through the point, A( 1, 2 , - 3 ) and it is parallel to the line joining B( 1, - 1, 5 ) and C( 2 , 3 , - 4 ). \ its d.r.’s are ( 2 - 1), ( 3 + 1), ( -4 - 5 ), i.e., 1, 4, –9. x-0 y+ 1 z- 3 y+ 1 z- 3 x 23. The equations of BC are = = Þ = = -2 0 - 2 -1 + 3 3 + 1 2 4 Þ

y+ 1 z- 3 x x y+ 1 z- 3 = = Þ = = = r (say). -2 2 4 -1 1 2

Any general point on BC is ( -r , r - 1, 2r + 3 ). Let AP ^ BC . Then P( -r , r - 1, 2r + 3 ) for some fixed value of r.

SSS Mathematics for Class 12 1126

1126

Senior Secondary School Mathematics for Class 12 d.r.’s of BC are –1, 1, 2.

d.r.’s of AP are ( -r - 1), (r - 1 - 8 ), ( 2r + 3 - 4 ), i.e., ( -r - 1), (r - 9 ), ( 2r - 1). Since AP ^ BC , we have -1 × ( -r - 1) + 1 × (r - 9 ) + 2( 2r - 1) = 0. 5 6r = 10 Þ r = × \ 3 10 æ -5 5 æ -5 2 19 ö ö Hence, the coordinates of P are ç + 3 ÷ , i.e., ç , - 1, , , ÷× 3 è 3 3 è 3 3 3 ø ø ^

^

^

^

^

^

x i + y j + z k = ( 11 + 10 l) i - ( 2 + 4 l) j - ( 8 + 11 l) k

26.

Û x = 11 + 10 l, y = - ( 2 + 4 l) and z = - ( 8 + 11 l) x - 11 y + 2 z + 8 are the equations of the given line. Û = = 10 -4 -11

Collinearity of Three Given Points When the coordinates of Three Points are Given The condition for three given points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) to be collinear is that x 3 - x1 y 3 - y1 z 3 - z1 = = × x 2 - x1 y 2 - y1 z2 - z1 The equations of the line AB are x - x1 y - y1 z - z1 … (i) = = × x 2 - x1 y 2 - y1 z2 - z1 Clearly, A , B , C will be collinear only when C lies on the line AB. This happens when C( x 3 , y 3 , z 3) lies on (i). x 3 - x1 y 3 - y1 z 3 - z1 \ = = × x 2 - x1 y 2 - y1 z2 - z1

THEOREM 1

PROOF

When Position Vectors of Three Points are Given THEOREM 2

® ®

®

®

®

®

d1 a + d 2 b + d 3 c = 0 and d1 + d 2 + d 3 = 0. PROOF

®

Three points A, B and C with position vectors a , b and c respectively are collinear if and only if there exist scalars d1 , d 2 , d 3 not all zero such that ® ®

Let A , B and C be three collinear points having position vectors a , b ® and c respectively. ¾®

Then, the vector equation of line AB is ®

®

®

®

r = a + l( b - a ) for some scalar l

SSS Mathematics for Class 12 1127

Straight Line in Space ®

Þ

®

®

®

c = a + l( b - a ) ®

®

[QA , B and C being collinear, C lies on AB]

®

®

®

®

Þ

(1 - l) a + l b - c = 0

Þ

d1 a + d 2 b + d 3 c = 0 , where d 3 = -1 ¹ 0 and

®

®

1127

d1 + d 2 + d 3 = (1 - l) + l - 1 = 0. Conversely, let A , B, and C be the given points having position vectors ® ®

®

a , b and c respectively and let d1 , d 2 , d 3 be scalars, not all zero such ®

®

®

®

that d1 a + d 2 b + d 3 c = 0 , and d1 + d 2 + d 3 = 0. Let d 3 ¹ 0. Then, ®

®

®

d1 a + d 2 b + d 3 c = 0, and d1 + d 2 + d 3 = 0 Þ Þ Þ Þ Þ Þ

æ d1 ö çç ÷÷ è d3 ø æ d1 ö çç ÷÷ è d3 ø

æd ö ® ® ® æd ö æd ö a + çç 2 ÷÷ b + c = 0 and çç 1 ÷÷ + çç 2 ÷÷ + 1 = 0 è d3 ø è d3 ø è d3 ø ® ® ® ® æ d1 ö æd ö a - l b + c = 0 and çç ÷÷ - l + 1 = 0, where çç 2 ÷÷ = - l è d3 ø è d3 ø ®

ù ® ® ® ® é æd ö ( l - 1) a - l b + c = 0 ê Q çç 1 ÷÷ = ( l - 1) ú è d3 ø ë û ® ® ® c = (1 - l) a + l b ¾® the point C lies on the line AB the points A , B and C are collinear. SOLVED EXAMPLES

EXAMPLE 1

Show that the points A( 2, 0, 3), B( 3 , 2, - 1) and C(1, - 2, - 5) are collinear.

SOLUTION

The equations of the line AB are x - 2 y -0 z - 3 x-2 y z-3 ... (i) = = Þ = = × 3 - 2 2 - 0 -1 - 3 1 2 -4 Putting x = 1, y = -2 and z = -5 in (i), we get 1 - 2 -2 -5 - 3 = = , which is clearly true. 1 2 -4 Thus, the point C(1, - 2, - 5) satisfies the equation of line AB. \ C lies on line AB. Hence, the given points A , B and C are collinear.

EXAMPLE 2

Find the value of l for which the points A( -1, 3 , 2), B( - 4, 2, - 2) and C(5 , 5 , l) are collinear.

SOLUTION

The equations of the line AB are x+1 y-3 z-2 = = - 4 + 1 2 - 3 -2 - 2

SSS Mathematics for Class 12 1128

1128

Senior Secondary School Mathematics for Class 12

Þ

x+1 y-3 z-2 = = × -3 -1 -4

... (i)

Since the points A, B and C are collinear, so the point C(5 , 5 , l) lies on (i). 5+1 5-3 l-2 l-2 = = Þ = - 2 Þ l - 2 = 8 Þ l = 10. \ -3 -1 -4 -4 Hence, the required value of l is 10. EXAMPLE 3

^

^

^

Show that the points whose position vectors are ( -2 i + 3 j + 5 k ), ^

^

^

^

^

( i + 2 j + 3 k ) and (7 i - k ) are collinear. SOLUTION

The coordinates of the given points are A( -2, 3 , 5), B(1, 2, 3) and C(7 , 0, - 1). The equations of line AB are x + 2 y - 3 z -5 x + 2 y - 3 z -5 = = Þ = = × 1 + 2 2 - 3 3 -5 3 -1 -2

... (i)

Putting x = 7 , y = 0 and z = -1 in (i), we get 7 + 2 0 - 3 -1 - 5 = = , which is clearly true. 3 -1 -2 Thus, the point C(7 , 0, - 1) satisfies the equation of line AB. \

C lies on line AB.

Hence, the given points A , B and C are collinear. EXAMPLE 4

Using the vector method, find the values of l and m for which the points A( 3 , l , m ), B( 2, 0, - 3) and C(1, - 2, - 5) are collinear.

SOLUTION

Let a , b and c be the position vectors of the given points A , B and C respectively. Then,

® ®

®

®

^

^

®

^

^

®

^

^

^

^

a = 3 i + l j + m k , b = 2 i - 3 k and c = i - 2 j - 5 k .

Now, the vector equation of the line BC is given by ®

®

®

®

r = b + a( c - b ) for some scalar a

Þ Þ Þ Þ

®

®

®

r = (1 - a ) b + a c

®

^

^

^

^

^

r = (1 - a ) ( 2 i - 3 k ) + a ( i - 2 j - 5 k )

®

^

^

^

r = ( 2 - 2a + a ) i - 2a j + ( -3 + 3 a - 5 a ) k

®

^

^

^

r = ( 2 - a ) i - 2a j + ( -3 - 2a ) k .

SSS Mathematics for Class 12 1129

Straight Line in Space

1129

If this line passes through the point A then we must have ^

^

^

^

^

^

3 i + l j + m k = ( 2 - a ) i - 2a j + ( -3 - 2a ) k Û 2 - a = 3 , - 2a = l and -3 - 2a = m Û a = -1, l = 2 and m = ( -3 + 2) = -1. Hence, l = 2 and m = - 1.

EXERCISE 27B 1. Show that the points A(2, 1, 3), B(5, 0, 5) and C(–4, 3, –1) are collinear. 2. Show that the points A(2, 3, –4), B(1, –2, 3) and C(3, 8, –11) are collinear. [CBSE 2006]

3. Find the value of l for which the points A(2, 5, 1), B(1, 2, –1) and C( 3 , l , 3) are collinear. 4. Find the values of l and m so that the points A(3, 2, –4), B(9, 8, –10) and C( l , m , - 6) are collinear. 5. Find the values of l and m so that the points A( -1, 4, - 2), B( l , m , 1) and C( 0, 2, - 1) are collinear. ^

^

^

6. The position vectors of three points A , B and C are ( -4 i + 2 j - 3 k ), ^

^

^

^

^

^

( i + 3 j - 2 k ) and ( -9 i + j - 4 k ) respectively. Show that the points A , B and C are collinear.

ANSWERS (EXERCISE 27B)

3.

l=8

4. l = 5 , m = 4

5. l = 2, m = -2

Angle between Two Lines (i) When the Given Lines are in Vector Form ®

®

®

Let the equations of the given lines in vector form be r = r1 + l m1 and ®

®

®

r = r2 + m m2 , where l and m are scalars.

Let q be the angle between these lines. ®

®

Since the given lines are parallel to m1 and m2 respectively, the angle between ®

®

the given lines must be equal to the angle between m1 and m2.

SSS Mathematics for Class 12 1130

1130

Senior Secondary School Mathematics for Class 12 ®

®

|m1 × m2|

\ cos q =

®

×

®

| m1 | × | m2 | (ii) When the Given Lines are in Cartesian Form Let the Cartesian equations of two given lines be x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 and = = = = × a1 b1 c1 a2 b2 c2 Then, the direction ratios of these lines are a1 , b1 , c1 , and a2 , b2 , c2 respectively. Let q be the angle between these lines. Then, | a1a2 + b1b2 + c1c2| cos q = × 2 ( a1 + b12 + c12 ) ( a22 + b22 + c22 ) SUMMARY

®

®

®

(i) If q is the angle between the lines r = r1 + l m1 and ®

®

®

r = r2 + m m2 then ®

cos q =

®

|m1 × m2|

× ® ® |m1|×|m2|

p Û m1 × m2 = 0. 2 (ii) If q is the angle between the lines whose Cartesian equations are x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 and then = = = = a1 b1 c1 a2 b2 c2

REMARK:

q=

|a1a2 + b1b2 + c1c2|

cos q = ( REMARK:

q=

a12

+ b12 + c12 )( a22 + b22 + c22 )

×

p Û a1a2 + b1b2 + c1c2 = 0. 2

SOLVED EXAMPLES EXAMPLE 1

Find the angle between the lines ®

^

^

®

^

^

^

^

^

^

^

^

r = ( 2 i - 5 j + k ) + l ( 3 i + 2 j + 6 k ) and ^

r = (7 i - 6 k ) + m ( i + 2 j + 2 k ).

SOLUTION

[CBSE 2014]

The given lines are of the form ®

®

®

®

®

®

r = r1 + l m1 and r = r2 + l m2, where

®

^

^

^

®

^

^

^

m1 = ( 3 i + 2 j + 6 k ) and m2 = ( i + 2 j + 2 k ).

SSS Mathematics for Class 12 1131

Straight Line in Space

1131

Let q be the angle between the given lines. Then, cos q =

®

®

®

®

|m1 × m2| |m1||m2| ^

\

^

^

^

^

^

=

|( 3 i + 2 j + 6 k ) × ( i + 2 j + 2 k )| ( 3 + 4 + 12) = ( 3 2 + 22 + 62 ) × ( 12 + 22 + 22 ) ( 49)( 9)

=

19 19 = × (7 ´ 3) 21

æ 19 ö q = cos-1 ç ÷ × è 21 ø

æ 19 ö Hence, the angle between the given lines is cos-1 ç ÷ × è 21 ø EXAMPLE 2

Find the angle between the lines -x + 2 y - 1 z + 3 x + 2 2y - 8 z - 5 and = = = = × [CBSE 2011] -2 7 -3 -1 4 4

SOLUTION

The given equations in standard form are x - 2 y -1 z + 3 x + 2 y - 4 z -5 and = = = = × 2 7 -3 -1 2 4 Here a1 = 2, b1 = 7 , c1 = -3 and a2 = -1, b2 = 2, c2 = 4. Let q be the angle between the given lines. Then, |a1a2 + b1b2 + c1c2| cos q = 2 ( a1 + b12 + c12 )( a22 + b22 + c22 ) = =

SOLUTION

{

2 + 7 2 + ( -3) 2} { ( -1) 2 + 22 + 42}

|( -2) + 14 + ( -12)| = 0. ( 62) ( 21)

p × 2 p Hence, the angle between the given lines is × 2 Find the value of l so that the following lines are perpendicular to each other: x -5 2 - y 1 -z x 2y + 1 1 - z and = = = = × [CBSE 2009] 5l + 2 5 -1 1 4l -3

\

EXAMPLE 3

|2 ´ ( -1) + 7 ´ 2 + ( -3) ´ 4| 2

q=

The given equations in standard form are 1 y+ x -5 y - 2 z -1 x 2 = z -1 × and = = = 5l + 2 -5 1 1 2l 3

SSS Mathematics for Class 12 1132

1132

Senior Secondary School Mathematics for Class 12

Here a1 = 5 l + 2, b1 = -5 , c1 = 1 and a2 = 1, b2 = 2l and c2 = 3. Since the given lines are perpendicular to each other, we must have a1a2 + b1b2 + c1c2 = 0. \ (5 l + 2) ´ 1 + ( -5) ´ ( 2l) + 1 ´ 3 = 0 Þ 5l + 2 - 10l + 3 = 0 Þ 5l = 5 Þ l = 1. Hence, the required value of l is 1. EXAMPLE 4

Find the angle between the lines x + 1 2y - 3 z - 6 x-4 y+ 3 and = = = , z = 5. 1 3 2 3 -2

SOLUTION

The given equations may be written as 3 3 yyx+1 z-6 x+1 2 2 = z-6 = = Þ = 3 1 2 2 3 4 2 x - 4 y + 3 z -5 and = = × 3 -2 0 Here a1 = 2, b1 = 3 , c1 = 4 and a2 = 3 , b2 = -2, c2 = 0. Let q be the angle between the given lines. Then, |a1a2 + b1b2 + c1c2| cos q = 2 ( a1 + b12 + c12 )( a22 + b22 + c22 ) = \

q=

|( 2 ´ 3) + 3 ´ ( -2) + 4 ´ 0| 0 = = 0. ( 4 + 9 + 16)( 9 + 4 + 0) ( 29 ´ 13 )

p × 2

Hence, the angle between the given lines is

p × 2

EXAMPLE 5

Find the angle between the lines 5-x y+ 3 x 1-y z -6 = , z = 7 and = = × 3 -4 1 2 2

SOLUTION

The given lines in standard form are x -5 y + 3 z -7 x y -1 z - 6 and = = = = -3 -4 0 1 -2 2 Here a1 = -3 , b1 = -4, c1 = 0 and a2 = 1, b2 = -2, c2 = 2. Let q be the angle between the given lines. Then, |a1a2 + b1b2 + c1c2| cos q = 2 { a1 + b12 + c12}{ a22 + b22 + c22} =

|( -3) ´ 1 + ( -4) ´ ( -2) + 0 ´ 2| 5 1 = = × 5´3 3 { 9 + 16 + 0}{ 1 + 4 + 4}

… (i)

… (ii)

SSS Mathematics for Class 12 1133

Straight Line in Space

\

1133

æ1ö q = cos -1 ç ÷ × è 3ø

æ1ö Hence, the angle between the given lines is cos -1 ç ÷ × è 3ø EXAMPLE 6

Find the value of k so that the lines 1 - x 7 y - 14 z - 3 7 - 7x 5 - y 6 - z and = = = = 3 2k 2 3k 1 5 are at right angles.

SOLUTION

The given equation in standard form are x -1 y - 2 z - 3 x -1 y - 2 z - 3 = = Þ = = -3 æ 2k ö 2 -21 2k 14 ç ÷ è7 ø x -1 y -5 z - 6 x -1 y -5 z - 6 = = Þ = = × k 3 æ ö -1 -5 -3 k -7 -35 ç ÷ è 7 ø

… (i)

… (ii)

Here, a1 = -21, b1 = 2k , c1 = 14 and a2 = -3 k , b2 = -7 , c2 = -35. Since the given lines are at right angles, we have a1a2 + b1b2 + c1c2 = 0 Þ

( -21) ´ ( -3 k) + ( 2k) ´ ( -7) + 14 ´ ( -35) = 0

Þ

63 k - 14k - 490 = 0 Þ 49k = 490 Þ k = 10.

Hence, k = 10. EXAMPLE 7

Prove that the lines x = ay + b , z = cy + d , and x = a ¢ y + b ¢, z = c¢ y + d ¢ are perpendicular if aa ¢+ cc¢+ 1 = 0.

SOLUTION

The equations of the first line are x = ay + b , z = cy + d x-b z-d = y, =y Û a c x-b y z-d Û = = × a 1 c Similarly, the equations of the second line are x - b¢ y z - d ¢ = = × a¢ 1 c¢

… (i)

… (ii)

The given lines are perpendicular to each other Û aa ¢ + 1 ´ 1 + cc¢ = 0 Û aa ¢ + cc¢ + 1 = 0. EXAMPLE 8

Find the angle between the two lines, one of which has direction ratios 2, 2, 1, and the other is obtained by joining the points ( 3 , 1, 4) and (7 , 2, 12).

SSS Mathematics for Class 12 1134

1134

Senior Secondary School Mathematics for Class 12

Let L1 and L2 be the given lines. Then, d.r.’s of L1 are 2, 2, 1.

SOLUTION

D.r.’s of L2 are (7 - 3), ( 2 - 1), (12 - 4), i.e., 4, 1, 8. Let q be the angle between the given lines. Then, |( 2 ´ 4 + 2 ´ 1 + 1 ´ 8)| 18 2 cos q = = = 2 2 2 2 2 2 ( 2 + 2 + 1 ) ( 4 + 1 + 8 ) 27 3 æ 2ö q = cos -1 ç ÷ × è 3ø

Þ

æ 2ö Hence, the angle between the given lines is cos -1 ç ÷ × è 3ø

EXERCISE 27C Find the angle between each of the following pairs of lines: ®

^

®

^

^

^

^

^

®

^

^

^

^

^

^

^

1. r = ( 3 i + j - 2 k ) + l( i - j - 2 k ) and r = ( 2 i - j - 5 k ) + m ( 3 i - 5 j - 4 k ) ^

^

^

®

^

^

^

^

^

2. r = ( 3 i - 4 j + 2 k ) + l( i + 3 k ) and r = 5 i + m ( - i + j + k ) ®

^

^

^

^

^

®

^

^

^

^

3. r = ( i - 2 j ) + l( 2 i - 2 j + k ) and r = 3 k + m ( i + 2 j - 2 k ) Find the angle between each of the following pairs of lines: x + 1 y - 4 z -5 x+ 3 y-2 z+5 4. and = = = = 1 1 2 3 5 4 x-4 y+1 z-6 x -5 2y + 5 z - 3 5. and = = = = 3 4 5 1 -2 1 3 - x y + 5 1 -z x 1-y z + 2 and = 6. = = = -2 1 3 3 -2 -1 x z x y z 7. = , y = 0 and = = 1 -1 3 4 5 5-x y+ 3 x -1 1 - y z -5 8. = , z = 5 and = = 3 -2 1 3 2 x- 3 y+1 z-2 x+ 2 y-4 z+5 9. Show that the lines and are = = = = 2 -3 4 2 4 2 perpendicular to each other. x -1 y - 2 z - 3 x -1 y -1 6 -z and are perpendicular to 10. If the lines = = = = -3 2l 2 3l 1 5 each other then find the value of l. [CBSE 2009] 11. Show that the lines x = - y = 2 z and perpendicular to each other. HINT: The given lines are

x + 2 = 2 y - 1 = -z + 1

y z x x + 2 y - 12 z - 1 = = and = = × 2 -2 1 2 1 -2

are

SSS Mathematics for Class 12 1135

Straight Line in Space

1135

12. Find the angle between two lines whose direction ratios are (i) 2, 1, 2 and 4, 8, 1 (ii) 5, –12, 13 and –3, 4, 5 (iii) 1, 1, 2 and ( 3 - 1), ( - 3 - 1), 4 (iv) a , b , c and ( b - c), ( c - a), ( a - b) 13. If A(1, 2, 3), B(4, 5, 7), C(–4, 3, –6) and D(2, 9, 2) are four given points then find the angle between the lines AB and CD.

ANSWERS (EXERCISE 27C)

æ8 3ö ÷ 1. cos-1 çç ÷ è 15 ø æ2 6ö ÷ 5. cos-1 çç ÷ è 15 ø 10. l =

æ 30 ö ÷ 2. cos -1 çç ÷ è 15 ø æ 11 ö 6. cos-1 ç ÷ è 14 ø

æ8 3ö ÷ 4. cos-1 çç ÷ è 15 ø æ 3 ö 8. cos-1 ç ÷ è 182 ø

æ - 4ö 3. cos-1 ç ÷ è 9 ø æ1ö 7. cos-1 ç ÷ è5 ø

-10 p p æ 2ö æ 1 ö 12. (i) cos-1 ç ÷ (ii) cos-1 ç ÷ (iii) (iv) 7 3 2 è 3ø è 65 ø

13. 0°

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 27C) 5. Given lines in standard form are: 5 y+ x- 4 y+ 1 z- 6 x-5 2 = z- 3× and = = = 3 4 5 1 -1 1 6. Given lines in standard form are: x- 3 y+ 5 z- 1 x y- 1 z+ 2 and = = = = × 2 1 -3 3 2 -1 7. Given lines in standard form are: x y-0 z x y z and = = × = = 1 0 -1 3 4 5 8. Given lines in standard form are: x-5 y+ 3 z-5 x- 1 y- 1 z-5 and = = = = × -3 -2 0 1 -3 2 9. Given lines in standard form are: 1 yy z x x+ 2 z- 1 2 = = and = = × 2 -2 1 2 1 -2 ¾®

^

¾®

^

^

^

^

^

^

^

^

^

13. AB = (p.v. of B) - (p.v. of A) = ( 4 i + 5 j + 7 k ) - ( i + 2 j + 3 k ) = ( 3 i + 3 j + 4 k ) ^

^

^

^

^

^

^

^

CD = (p.v. of D) - (p.v. of C) = ( 2 i + 9 j + 2 k ) - ( -4 i + 3 j - 6 k ) = ( 6 i + 6 j + 8 k ) cos q =

¾®

¾®

¾®

¾®

| AB × CD| | AB||CD|

=

=

( 3 ´ 6 + 3 ´ 6 + 4 ´ 8) ( 9 + 9 + 16 ) ( 36 + 36 + 64 )

68 68 = = 1. 34 ´ 136 34 ´ 2

SSS Mathematics for Class 12 1136

1136

Senior Secondary School Mathematics for Class 12

Shortest Distance between Two Lines COPLANAR LINES

Two lines lying in the same plane are called coplanar lines.

Coplanar lines are either parallel or intersecting. Two lines in space which are not coplanar are called skew lines.

SKEW LINES

Skew lines are neither parallel nor intersecting. LINE OF SHORTEST DISTANCE BETWEEN TWO SKEW LINES If L1 and L2 are two skew lines then there is a unique line which is perpendicular to both the lines L1 and L2. This line is called the line of shortest distance between L1 and L2. SHORTEST DISTANCE BETWEEN TWO SKEW LINES ¾®

The length of the line segment

PQ , intercepted by two skew lines L1 and L2 on the common perpendicular to both the lines, is called the shortest distance (SD) between L1 and L2.

If two lines in space intersect at a point then the shortest distance between them is zero.

REMARK

To Find the Shortest Distance between Two Skew Lines Vector Form THEOREM

®

®

®

The shortest distance between two skew lines r = a1 + l b1 and ®

®

®

r = a2 + m b2 is given by

PROOF

½ ® ® ® ®½ ( a - a1 ) × ( b1 ´ b2) ½ d =½ 2 × ½ ½ ® ® b ´ b | | 1 2 ½ ½ be two skew lines whose vector equations are

Let L1 and L2 respectively ®

®

®

®

®

r = a1 + l b1

... (i)

®

and r = a2 + m b2 .

... (ii) ®

Then, L1 is parallel to b1 and passes through a point A , whose position ®

vector is a1 .

SSS Mathematics for Class 12 1137

Straight Line in Space

1137

®

And, L2 is parallel to b2 and passes through a point B, whose position ®

vector is a2. ¾®

Let PQ be the shortest-distance vector between L1 and L2. ®

¾®

®

¾®

Then, PQ ^ b1 and PQ ^ b2. ®

¾®

®

\

PQ | | ( b1 ´ b2).

\

PQ = |projection of AB along ( b1 ´ b2)|

®

¾®

®

½ ® ® ® ®½ ( a - a1) × ( b1 ´ b2)½ =½ 2 ® × ® ½ ½ | b1 ´ b2| ½ ½ CONDITION FOR TWO GIVEN LINES TO INTERSECT

®

®

®

®

®

®

Suppose that the lines

r = a1 + l b1 and r = a2 + m b2 intersect. Then, the shortest distance between them is zero. ®

® ® ®

\ [( a2 - a1 ) b1 b2] = 0. REMARK

Two lines intersect only when the shortest distance between them is zero. SOLVED EXAMPLES

EXAMPLE 1

Find the shortest distance between the lines whose vector equations are ®

^

^

^

^

^

^

r = ( 6 i + 2 j + 2 k) + l ( i - 2 j + 2 k )

®

^

^

^

^

^

and r = ( -4 i - k ) + m(3 i - 2 j - 2 k ). SOLUTION

[CBSE 2013C]

Comparing the given equations with the standard equations

®

®

®

®

^

®

®

®

r = a1 + l b1 and r = a2 + l b2 , we have ^

^

®

^

^

^

a1 = ( 6 i + 2 j + 2 k ), b1 = ( i - 2 j + 2 k ), ®

^

®

^

^

^

^

a2 = ( -4 i - k ) and b2 = ( 3 i - 2 j - 2 k ). \

®

®

^

^

^

^

^

^

^

^

( a2 - a1 ) = ( -4 i - k ) - ( 6 i + 2 j + 2 k ) = ( -10 i - 2 j - 3 k ) ^ ® ®

and, ( b1 ´ b2 ) =

^

^

i j k ^ ^ ^ 1 -2 2 = ( 4 + 4) i - ( -2 - 6) j + ( -2 + 6) k 3 -2 -2 ^

^

^

= ( 8 i + 8 j + 4 k ).

SSS Mathematics for Class 12 1138

1138

Senior Secondary School Mathematics for Class 12

\

®

®

|b1 ´ b2|= 82 + 82 + 42 = 64 + 64 + 16 = 144 = 12. ®

\

®

® ®

( a2 - a1 ) × ( b1 ´ b2)

SD =

® ®

|b1 ´ b2| ^

EXAMPLE 2

^

^

^

^

^

=

( -10 i - 2 j - 3 k ) × ( 8 i + 8 j + 4 k ) -80 - 16 - 12 = 12 12

=

-108 = - 9 = 9 units. 12

Find the shortest distance between the lines L1 and L2 whose vector equations are given below: ®

^

^

^

^

^

L1 : r = i + j + l ( 2 i - j + k ) ®

^

^

^

^

^

^

L2 : r = 2 i + j - k + m ( 3 i - 5 i + 2 k ) SOLUTION

[CBSE 2008C, ’14]

Comparing the given equations with the standard equations

®

®

®

®

®

®

r = a1 + l b1 and r = a2 + m b2 , we have ®

^

®

^

^

^

^

a1 = ( i + j ), b1 = ( 2 i - j + k ) ®

^

^

^

a2 = ( 2 i + j - k ) ®

®

^

^

½^ ½i ® ® and, ( b1 ´ b2 ) = ½ 2 ½ ½3 ½

^

\

®

^

^

^

b2 = ( 3 i - 5 j + 2 k )

and ^

^

^

^

^

( a2 - a1 ) = ( 2 i + j - k ) - ( i + j ) = ( i - k ) ^½

j -1 -5

k½ ^ ^ ^ 1 ½= ( -2 + 5) i - ( 4 - 3) j + ( -10 + 3) k 2½ ½ ½ ^

^

^

= ( 3 i - j - 7 k ). \ \

EXAMPLE 3

®

®

3 2 + ( -1) 2 + ( -7) 2 = 59.

| b1 ´ b2 | =

^ ^ ^ ^ ^ ½ ® ® ® ®½ ( a - a1 ) × ( b1 ´ b2)½ |( i - k ) × ( 3 i - j - 7 k )| = SD = ½ 2 ½ ½ ® ® 59 | b1 ´ b2 | ½ ½ |3 - 0 + 7| 10 59 units. = = 59 59

Find the shortest distance between the lines whose vector equations are ®

^

®

^

^

^

r = (1 - t) i + (t - 2) j + ( 3 - 2t) k , and ^

^

r = (s + 1) i + ( 2s - 1) j - ( 2s + 1) k.

[CBSE 2011]

SSS Mathematics for Class 12 1139

Straight Line in Space SOLUTION

1139

The given equations can be written as ®

^

®

^

^

^

^

^

^

r = ( i - 2 j + 3 k ) + t ( - i + j - 2 k ), and ^

^

^

^

^

r = ( i - j - k ) + s( i + 2 j - 2 k ).

Comparing the given equations with the standard equations ®

®

®

®

®

®

r = a1 + t b1 and r = a2 + s b2 , we get ®

^

®

^

^

®

^

^

^

^

a1 = ( i - 2 j + 3 k ), b1 = ( - i + j - 2 k ) ^

®

^

^

^

^

a2 = ( i - j - k ) and b2 = ( i + 2 j - 2 k ). \

®

®

^

^

^

^

^

^

^

^

( a2 - a1 ) = ( i - j - k ) - ( i - 2 j + 3 k ) = ( j - 4 k )

½ ^ ^ ^½ k½ ½ i j and, ( b1 ´ b2 ) = ½- 1 1 -2 ½ ½ ½ 1 2 -2 ½ ½ ½ ½ ^ ^ ^ = ( -2 + 4) i - ( 2 + 2) j + ( -2 - 1) k ®

®

^

^

^

= ( 2 i - 4 j - 3 k ). ®

®

\

| b1 ´ b2 | = 22 + ( - 4) 2 + ( -3) 2 = 29.

\

® ® ½ ½ ® ® ( a - a1 ) × ( b1 ´ b2 ) ½ SD = ½ 2 ½ ½ ® ® | b1 ´ b2 | ½ ½ ^

^

^

^

^

|( j - 4 k ) × ( 2 i - 4 j - 3 k)| |0 - 4 + 12| = 29 29 8 29 units. = 29 =

EXAMPLE 4

Show that the lines ®

^

^

^

^

®

^

^

^

^

^

r = ( i + j - k ) + l ( 3 i - j ) and r = ( 4 i - k ) + m ( 2 i + 3 k ) [CBSE 2014] intersect. Find their point of intersection. SOLUTION

Comparing the given equations with the standard equations ®

®

®

®

®

®

r = a1 + l b1 and r = a2 + m b2 , we get ®

^

^

^

®

^

®

^

^

a1 = ( i + j - k ), b1 = ( 3 i - j ), ®

^

^

^

a2 = ( 4 i - k ) and b2 = ( 2 i + 3 k ). \

®

®

^

^

^

^

^

^

^

( a2 - a1 ) = ( 4 i - k ) - ( i + j - k ) = ( 3 i - j ).

SSS Mathematics for Class 12 1140

1140

Senior Secondary School Mathematics for Class 12

½^ ½i And, ( b1 ´ b2) = ½ 3 ½ ½2 ®

^½

^

j

®

k½ ^ ^ ^ 0 ½= ( -3 - 0) i - ( 9 - 0) j + ( 0 + 2) k 3½ ½

-1 0

½

½

^

^

^

= ( -3 i - 9 j + 2 k ). ®

®

\

| b1 ´ b2 | = ( -3) 2 + ( -9) 2 + 2 2 = 94.

\

½ ® ® ® ®½ ( a - a1 ) × ( b1 ´ b2)½ SD = ½ 2 ½ ½ ® ® |b1 ´ b2 | ½ ½ ^

^

^

^

^

|( 3 i - j ) × ( -3 i - 9 j + 2 k )| 94 | - 9 + 9 + 0| = = 0. 94 Thus, the shortest distance between the given lines is 0. =

Hence, the given lines intersect. Thus, for some particular values of l and m , we have ^

^

^

^

^

^

^

^

^

( i + j - k) + l ( 3 i - j ) = (4 i - k) + m (2 i + 3 k) Þ

^

^

^

^

^

(1 + 3 l) i + (1 - l) j - k = ( 4 + 2m ) i + ( 3m - 1) k

Þ 1 + 3 l = 4 + 2m , 1 - l = 0 and 3m - 1 = - 1 Þ l = 1 and m = 0. Thus, the position vector of the point of intersection of the given lines is given by ®

^

^

^

^

®

^

^

^

[putting l = 1], i.e., r = ( 4 i - k ).

r = ( i + j - k) + ( 3 i - j )

Hence, the point of intersection of the given lines is P(4, 0, –1). EXAMPLE 5

Show that the lines ®

^ ^

^

®

^

^

^

^

^

^

r = ( i - j ) + l ( 2 i + k ) and r = ( 2 i - j ) + m ( i + j - k ) [CBSE 2012C] do not intersect. SOLUTION

Comparing the given equations with the standard equations

®

®

®

®

®

®

r = a1 + l b1 and r = a2 + m b2 , we get ®

^

®

^

^

^

a1 = ( i - j ), b1 = ( 2 i + k ) ®

^

^

®

^

^

^

^

^

^

^

a2 = ( 2 i - j ) and b2 = ( i + j - k ). \

®

®

^

( a2 - a1 ) = ( 2 i - j ) - ( i - j ) = i .

SSS Mathematics for Class 12 1141

Straight Line in Space

½^ ½i And, ( b1 ´ b2 ) = ½2 ½ ½1 ®

^½

^

j

®

k½ ^ ^ ^ 1 ½= ( 0 - 1) i - ( -2 - 1) j + ( 2 - 0) k -1 ½ ½

0 1

½

1141

½

^

^

^

= ( - i + 3 j + 2 k ). ®

®

\

| b1 ´ b2 | = ( -1) 2 + 3 2 + 2 2 = 14.

\

½ ® ® ® ®½ ( a - a1 ) × ( b1 ´ b2 )½ SD = ½ 2 ½ ½ ® ® | b1 ´ b2 | ½ ½ ^ ^½ ½^ ^ i × ( - i + 3 j + 2 k )½ |- 1| = =½ ½ ½ 14 14 ½ ½

1 ´ 14 14 = ¹ 0. 14 14 Since the shortest distance between the given lines is not zero, the given lines do not intersect. =

DISTANCE BETWEEN PARALLEL LINES

Let L1 and L2 be two parallel lines. Then,

these lines are clearly coplanar. Let the equations of these lines be ®

®

®

®

®

®

r = a1 + l b

... (i)

r = a2 + m b .

®

... (ii)

Let A be a point on L1 with position vector a1 and let B be a point on L2 ® with position vector a2. Draw BM ^ L1. Then, ¾®

distance between L1 and L2 = | BM|. ®

¾®

Let q be the angle between AB and b . Then, ®

®

¾®

¾®

^

( b ´ AB) = {| b || AB |× sin q } n , ^

where n is a unit vector, perpendicular to the plane of L1 and L2. \ Þ

®

®

®

®

®

®

®

®

¾®

^

b ´ ( a2 - a1 ) = {| b || AB |× sin q } n ®

b ´ ( a2 - a1 ) = | b |( BM) n ®

®

®

®

¾®

Þ | b ´ ( a2 - a1 )| = | b || BM |× 1

{Q ( AB) sin q = BM} ^

[Q |n| = 1]

SSS Mathematics for Class 12 1142

1142

Senior Secondary School Mathematics for Class 12

½® ® ® ½ b ´ ( a2 - a1 )½ Þ |BM | = ½ × ½ ½ ® | b| ½ ½ ¾®

EXAMPLE 6

Find the shortest distance between the lines L1 and L2 , given by ®

^

^

^

^

®

^

^

^

^

^

^

^

r = i + j + l ( 2 i - j + k ) and r = 2 i + j - k + m ( 4 i - 2 j + 2 k ).

SOLUTION

The given lines are ®

^

^

^

^

^

L1 : r = ( i + j ) + l ( 2 i - j + k ) ®

^

^

^

^

... (i) ^

^

®

®

L2 : r = ( 2 i + j - k ) + 2m ( 2 i - j + k ).

... (ii)

These equations are of the form: ®

®

®

^

®

®

®

®

r = a1 + l b and r = a2 + 2m b = a2 + m ¢ b , where ®

^

^

^

®

^

^

^

^

^

^

a1 = ( i + j ), a2 = ( 2 i + j - k ), b = ( 2 i - j + k ) and m ¢ = 2m .

Clearly, the given lines are parallel. ®

®

^

^

^

^

^

Now, ( a2 - a1 ) = ( 2 i + j - k ) - ( i + j ) = ( i - k ) ^

\

^

^

i j k { b ´ ( a2 - a1 )} = 2 -1 1 1 0 -1 ®

®

®

^

^

^

= (1 - 0) i - ( -2 - 1) j + ( 0 + 1) k ^

®

^

^

= ( i + 3 j + k) ®

®

Þ | b ´ ( a2 - a1 )| = 12 + 3 2 + 12 = 11 ®

and | b | = 22 + ( -1) 2 + 12 = 6 Þ

shortest distance between L1 and L2 = distance between L1 and L2 ®

=

®

®

| b ´ ( a2 - a1 )| ®

|b|

=

11 6

æ 11 6ö 66 ÷= units. = çç ´ ÷ 6 6ø è 6 EXAMPLE 7

Write the vector equations of the following lines and hence find the shortest distance between them: x+1 y+1 z+1 x - 3 y -5 z -7 [CBSE 2008, ’14] and = = = = 2 -6 1 1 -2 1

SSS Mathematics for Class 12 1143

Straight Line in Space

1143

The given lines are x+1 y+1 z+1 ... (i) = = 2 -6 1 x - 3 y -5 z -7 and ... (ii) = = × 1 -2 1 The line (i) passes through the point (–1, –1, –1) and has d.r.’s. 7, –6, 1. So, its vector equation is

SOLUTION

®

®

®

r = a1 + l b1 , ®

^

... (iii)

^

®

^

^

^

^

where a1 = ( - i - j - k ) and b1 = 7 i - 6 j + k . The line (ii) passes through the point (3, 5, 7) and has d.r.’s. 1, –2, 1. So, its vector equation is ®

®

®

r = a2 + l b2 , ®

... (iv)

^

^

®

^

^

^

^

where a2 = ( 3 i + 5 j + 7 k ) and b2 = i - 2 j + k . ®

\

®

^

^

^

^

^

^

^

^

^

( a2 - a1 ) = ( 3 i + 5 j + 7 k ) - ( - i - j - k ) = ( 4 i + 6 j + 8 k ) ^

^

i

®

j

^

k

®

^

^

^

and ( b1 ´ b2) = 7 -6 1 = ( -6 + 2) i - (7 - 1) j + ( -14 + 6) k 1 -2 1 ^

^

^

= ( -4 i - 6 j - 8 k ). ®

®

\

|b1 ´ b2|= ( -4) 2 + ( -6) 2 + ( -8) 2 = 16 + 36 + 64 = 116.

\

SD =

®

®

® ®

( a2 - a1 ) × ( b1 ´ b2) ® ®

^

=

|b1 ´ b2| =

^

^

^

^

^

( 4 i + 6 j + 8 k ) × ( -4 i - 6 j - 8 k ) 116

|( -16 - 36 - 64)| |- 116| 116 116 = = ´ 116 116 116 116

= 116 = 4 ´ 29 = 2 29 units.

EXERCISE 27D In problems 1–8, find the shortest distance between the given lines. ®

^

^

^

^

^

1. r = ( i + j ) + l ( 2 i - j + k ), ®

^

^

^

^

^

^

r = ( 2 i + j - k ) + m ( 3 i - 5 j + 2 k ).

[CBSE 2007, ’08C, ’09, ’14]

SSS Mathematics for Class 12 1144

1144

Senior Secondary School Mathematics for Class 12 ®

^

^

®

^

^

^

^

^

^

2. r = ( - 4 i + 4 j + k ) + l ( i + j - k ), ^

^

^

^

r = ( -3 i - 8 j - 3 k ) + m ( 2 i + 3 j + 3 k ).

®

^

^

^

^

^

^

3. r = ( i + 2 j + 3 k ) + l ( i - 3 j + 2 k ), ®

^

^

^

^

^

^

r = ( 4 i + 5 j + 6 k ) + m ( 2 i + 3 j + k ).

®

^

^

^

^

^

^

^

[CBSE 2006, ’11]

^

4. r = ( i + 2 j + k ) + l ( i - j + k ), ®

^

^

^

^

r = ( 2 i - j - k ) + m ( 2 i + j + 2 k ).

®

^

^

^

^

^

[CBSE 2008, ’11] ^

5. r = ( i + 2 j - 4 k ) + l ( 2 i + 3 j + 6 k ), ®

^

^

®

^

^

^

^

^

^

r = ( 3 i + 3 j - 5 k ) + m ( -2 i + 3 j + 8 k ). ^

^

^

6. r = ( 6 i + 3 k ) + l ( 2 i - j + 4 k ), ®

^

^

^

^

^

^

r = ( -9 i + j - 10 k ) + m ( 4 i + j + 6 k ).

®

^

^

®

^

^

®

^

^

®

^

^

7. r = ( 3 - t) i + ( 4 + 2t) j + (t - 2) k , ^

r = (1 + s) i + ( 3s - 7) j + ( 2s - 2) k. ^

8. r = ( l - 1) i + ( l + 1) j - ( l + 1) k , ^

^

r = (1 - m ) i + ( 2m - 1) j + (m + 2) k .

9. Compute the shortest distance between the lines ®

^

^

^

®

^

^

^

^

^

^

r = ( i - j ) + l ( 2 i - k ) and r = ( 2 i - j ) + m ( i - j - k ).

Determine whether these lines intersect or not.

[CBSE 2012C]

10. Show that the lines ®

^

®

^

^

^

^

^

^

r = ( 3 i - 15 j + 9 k ) + l ( 2 i - 7 j + 5 k ), and ^

^

^

^

^

r = (- i + j + 9 k) + m (2 i + j - 3 k) do not intersect. 11. Show that the lines ®

^

^

^

^

®

^

^

^

^

^

^

^

r = ( 2 i - 3 k ) + l ( i + 2 j + 3 k ) and r = ( 2 i + 6 j + 3 k ) + m ( 2 i + 3 j + 4 k )

intersect. Also, find their point of intersection. 12. Show that the lines ®

^

^

^

^

^

^

®

^

^

^

^

^

r = ( i + 2 j + 3 k ) + l ( 2 i + 3 j + 4 k ) and r = ( 4 i + j ) + m (5 i + 2 j + k ) intersect. Also, find their point of intersection.

SSS Mathematics for Class 12 1145

Straight Line in Space

1145

13. Find the shortest distance between the lines L1 and L2 whose vector equations are [CBSE 2014] ®

^

^

^

^

^

^

r = ( i + 2 j - 4 k ) + l ( 2 i + 3 j + 6 k ) and

®

^

^

^

^

^

^

r = ( 3 i + 3 j - 5 k ) + m ( 2 i + 3 j + 6 k ).

HINT: The given lines are parallel.

14. Find the distance between the parallel lines L1 and L2 whose vector equations are ®

^

^

^

^

^

®

^

^

^

^

^

^

^

r = ( i + 2 j + 3 k ) + l ( i - j + k ), and r = ( 2 i - j - k ) + m ( i - j + k ). 15. Find the vector equation of a line passing through the point (2, 3, 2) and ®

^

^

^

^

^

^

^

parallel to the line r = ( -2 i + 3 j ) + l ( 2 i - 3 j + 6 k ). Also, find the distance between these lines. HINT:

The given line is

®

^

^

^

^

^

^

^

L1 : r = ( -2 i + 3 j ) + l( 2 i - 3 j + 6 k ). ®

^

^

^

^

L2 : r = ( 2 i + 3 j + 2 k) + m ( 2 i - 3 j + 6 k ). Now, find the distance between the parallel lines L1 and L2. 16. Write the vector equation of each of the following lines and hence determine the distance between them: [CBSE 2010] x -1 y - 2 z + 4 x- 3 y- 3 z+5 and = = = = × 2 3 6 4 6 12 The required line is

HINT:

The given lines are

®

^

^

^

^

L1 : r = ( i + 2 j - 4 k ) + l( 2 i + 3 j + 6 k ) ®

^

^

^

^

^

^

L2 : r = ( 3 i + 3 j - 5 k) + 2m ( 2 i + 3 j + 6 k ). Now, find the distance between the parallel lines L1 and L2. 17. Write the vector equations of the following lines and hence find the shortest distancebetween them: [CBSE 2010C] x -1 y - 2 z - 3 x - 2 y - 3 z -5 and = = = = × 2 3 4 3 4 5 Find the shortest distance between the lines given below: x -1 y + 2 z - 3 x -1 y + 1 z + 1 and 18. = = = = × -1 1 -2 1 2 -2 x - 12 y - 1 z - 5 x - 23 y - 19 z - 25 and 19. = = = = × -9 4 2 -6 -4 3 HINT:

Change the given equations in vector form. ANSWERS (EXERCISE 27D)

10 units 59 14 241 5. units 241 1.

2.

62 units

3.

3 19 units 19

6.

38 units

7.

35 units

3 2 units 2 5 2 8. units 2 4.

SSS Mathematics for Class 12 1146

1146

Senior Secondary School Mathematics for Class 12

14 11. (2, 6, 3) ¹ 0, the given lines do not intersect 14 293 units 14. 26 units ( -1, - 1, - 1) 13. 7 ® ^ ^ ^ ^ ^ ^ 580 units L2 : r = ( 2 i + 3 j + 2 k ) + m ( 2 i - 3 j + 6 k ), 7 ü ® ^ ^ ^ ^ ^ ^ 293 ï L1 : r = ( i + 2 j - 4 k ) + l( 2 i + 3 j + 6 k ) units ý, SD = ® ^ ^ ^ ^ ^ ^ 7 L2 : r = ( 3 i + 3 j - 5 k ) + 2m ( 2 i + 3 j + 6 k ) ïþ ü ® ^ ^ ^ ^ ^ ^ 6 L1 : r = ( i + 2 j + 3 k ) + l( 2 i + 3 j + 4 k ) ï units ý, SD = ® ^ ^ ^ ^ ^ ^ 6 L2 : r = ( 2 i + 3 j + 5 k ) + m ( 3 i + 4 j + 5 k ) ïþ 8 29 units 19. 26 units 29

9. SD = 12. 15. 16.

17.

18.

Shortest Distance between Two Skew Lines in the Cartesian Form The shortest distance between the skew lines x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 and = = = = a1 b1 c1 a2 b2 c2 is given by ½ x 2 - x1 ½ a1 ½ a2 SD = ½

y 2 - y1 b1 b2 D

z2 - z1 ½ c1 ½ ½ c2 ½ ,

2

where D = {( a1 b2 - a2 b1) + ( b1 c2 - b2 c1) 2 + ( c1 a2 - c2 a1) 2}. CONDITION FOR TWO GIVEN LINES TO INTERSECT

Let L1 and L2 be the given lines

whose equations are x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 and = = = = × a1 b1 c1 a2 b2 c2 1.

L1 and L2 intersect Û SD between them is 0 ½ x 2 - x1 y 2 - y1 z2 - z1 ½ Û ½ a1 b1 c1 ½ = 0. ½ ½ a b c ½ 2 ½ 2 2

2.

L1 and L2 do not intersect Û they are skew lines ½ x 2 - x1 y 2 - y1 z2 - z1 ½ Û ½ a1 b1 c1 ½ ¹ 0. ½ ½ b2 c2 ½ ½ a2

SSS Mathematics for Class 12 1147

Straight Line in Space

1147

SOLVED EXAMPLES EXAMPLE 1

Find the shortest distance between the lines x+ 3 y-6 z x + 2 y z -7 = = and = = × -4 3 2 -4 1 1

SOLUTION

Comparing the given equations with x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 and , we get = = = = a1 b1 c1 a2 b2 c2 ( x1 = - 3 , y1 = 6, z1 = 0), ( x 2 = - 2, y 2 = 0, z2 = 7), ( a1 = - 4, b1 = 3 , c1 = 2) and ( a2 = - 4, b2 = 1, c2 = 1). Now, D = ( a1 b2 - a2 b1) 2 + ( b1 c2 - b2 c1) 2 + ( c1 a2 - c2 a1) 2 = ( - 4 + 12) 2 + ( 3 - 2) 2 + ( - 8 + 4) 2 = ( 64 + 1 + 16) = 81. \

SD =

=

½x 2 - x1 1 ½ a1 × D ½ a ½ 2 ½ -2 + 3 1 ½ × -4 81 ½ ½ -4

y 2 - y1 z2 - z1½ b1 c1 ½ ½ b2 c2 ½ 0-6 3 1

7 - 0½ ½ 1 -6 7½ 1 2 ½ = ×½-4 3 2½ ½ ½ 9 ½ 1 ½ ½-4 1 1½

1 × {1 × ( 3 - 2) + 6 × ( - 4 + 8) + 7 × ( - 4 + 12)} 9 81 = = 9 units. 9 =

Hence, the shortest distance between the given lines is 9 units. EXAMPLE 2

Find the length and the equations of the line of shortest distance between the lines x +1 y +1 z+1 x - 3 y -5 z -7 and = = = = × [CBSE 2008, ’14] 7 -6 1 1 -2 1

SOLUTION

The equations of the given lines are x +1 y +1 z+1 … (i) = = = l (say) 7 -6 1 x - 3 y -5 z -7 … (ii) = = = m (say). 1 -2 1 Any point on (i) is P(7 l - 1, - 6l - 1, l - 1). Any point on (ii) is Q(m + 3 , - 2m + 5 , m + 7). The direction ratios of PQ are (m - 7 l + 4, - 2m + 6l + 6, m - l + 8). Now, PQ will be the shortest distance between (i) and (ii) only when PQ is perpendicular to both (i) and (ii).

SSS Mathematics for Class 12 1148

1148

Senior Secondary School Mathematics for Class 12

ì 7(m - 7 l + 4) - 6( -2m + 6l + 6) + 1 × (m - l + 8) = 0 ï \ í and ï 1 × (m - 7 l + 4) - 2( -2m + 6l + 6) + 1 × (m - l + 8) = 0 î ì20m - 86l = 0 ì10m - 43 l Þ í Þ í Þ 6 m 20 l = 0 î î 3m - 10l

l = 0 and m = 0.

\ PQ will be the line of shortest distance when l = 0 and m = 0. Putting l = 0 and m = 0, we get the points P( -1, - 1, - 1) and Q( 3 , 5 , 7). \

SD = PQ = ( 3 + 1) 2 + (5 + 1) 2 + (7 + 1) 2 = ( 4) 2 + ( 6) 2 + ( 8) 2 = 16 + 36 + 64 = 116 = 2 29 units.

Clearly, the equation of the line of shortest distance is the equation of line PQ given by x - 3 y -5 z -7 = = 3+ 1 5+1 7+1 x - 3 y -5 z -7 Û = = 4 6 8 x - 3 y -5 z -7 = = × Û 2 3 4 Hence, the equation of the line of shortest distance is x - 3 y -5 z -7 = = × 2 3 4 EXAMPLE 3

Find the length and the equations of the line of shortest distance between the lines x - 8 y + 9 z - 10 x - 15 y - 29 z - 5 and = = = = × 3 -16 7 3 8 -5

SOLUTION

The equations of the given lines are x - 8 y + 9 z - 10 = = = l (say) 3 -16 7 x - 15 y - 29 z - 5 = = = m (say) 3 8 -5 Any point on the line (i) is P( 3 l + 8, - 16l - 9, 7 l + 10).

... (i) ... (iii)

Any point on the line (ii) is Q( 3m + 15 , 8m + 29, - 5m + 5). The direction ratios of PQ are ( 3m - 3 l + 7 , 8m + 16l + 38, - 5m - 7 l - 5). Now, PQ will be the shortest distance between (i) and (ii) only when PQ is perpendicular to each one of (i) and (ii).

SSS Mathematics for Class 12 1149

Straight Line in Space

1149

ì 3( 3m - 3 l + 7) - 16( 8m + 16l + 38) + 7( -5m - 7 l - 5) = 0 ï \ í and ï 3( 3m - 3 l + 7) + 8( 8m + 16l + 38) - 5 ( -5m - 7 l - 5) = 0 î ...(iii) ì 314l + 154m + 622 = 0 ì157 l + 77m + 311 = 0 Þ í Þ í 154 l + 98 m + 350 = 0 77 l + 49 m + 175 = 0 . ...(iv) î î On multiplying (iii) by 7 and (iv) by 11 and subtracting, we get (1099l - 847 l) + ( 2177 - 1925) = 0 Þ 252l = -252 Þ l = -1. Putting l = -1 in (iv), we get 49m + (175 - 77) = 0 Þ 49m + 98 = 0 Þ 49m = -98 Þ m = -2. Now, l = -1 gives P(5, 7, 3) and m = -2 gives Q(9, 13, 15). \

SD = PQ = ( 9 - 5) 2 + (13 - 7) 2 + (15 - 3) 2 = 16 + 36 + 144 = 196 = 14 units.

Equation of the line of shortest distance is the equation of PQ, given by x -5 y -7 z-3 = = 9 - 5 13 - 7 15 - 3 x -5 y -7 z - 3 x -5 y -7 z - 3 Þ = = Þ = = × 4 6 12 2 3 6 Hence, the equation of the line of shortest distance is x -5 y -7 z - 3 = = × 2 3 6 EXAMPLE 4

SOLUTION

Show that the lines x -1 y - 2 z - 3 x - 4 y -1 and = = = =z 2 3 4 5 2 intersect each other. Find their point of intersection.

[CBSE 2004]

The equations of the given lines are x -1 y - 2 z - 3 ... (i) = = = l (say) 2 3 4 x - 4 y -1 z - 0 ... (ii) = = = m. 5 2 1 Any point on the line (i) is P( 2l + 1, 3 l + 2, 4l + 3). Any point on the line (ii) is Q(5m + 4, 2m + 1, m ). If the lines (i) and (ii) intesect then P and Q must coincide for some particular values of l and m. This gives 2l + 1 = 5m + 4, 3 l + 2 = 2m + 1 and 4l + 3 = m ... ( i) ì2l - 5m = 3 ï Þ í 3 l - 2m = -1 ... ( ii) ï4l - m = -3. ... ( iii) î

SSS Mathematics for Class 12 1150

1150

Senior Secondary School Mathematics for Class 12

On solving (i) and (ii), we get l = -1 and m = -1. These values of l and m also satisfy (iii). Hence, the given lines intersect. Putting l = -1, we get P( -1, - 1, - 1). Note that putting m = -1, we get Q( -1, - 1, - 1). Hence, the point of intersection of the given lines is ( -1, - 1, - 1). EXAMPLE 5

SOLUTION

Show that the lines x -1 y + 1 x+1 y-2 = = z and = , z = 2. 2 3 5 2 do not intersect each other. The equations of the given lines are x -1 y + 1 z - 0 = = = l (say) 2 3 1 x+1 y-2 z-2 = = = m (say) 5 1 0

[CBSE 2010]

... (i) ... (ii)

Any point on the line (i) is P( 2l + 1, 3 l - 1, l). Any point on the line (ii) is Q(5m - 1, m + 2, 2). If the lines (i) and (ii) intersect, then P and Q must coincide for some particular values of l and m. This gives 2l + 1 = 5m - 1, 3 l - 1 = m + 2 and l = 2 Þ

ì2l - 5m = -2 ï í 3l - m = 3 ï l = -3. î

... ( iii) ... ( iv) ... ( v)

Putting l = 2 in (iv), we get m = 3. Clearly, l = 2 and m = 3 do not satisfy (iii). Hence, the given lines do not intersect each other.

EXERCISE 27E Find the length and the equations of the line of shortest distance between the lines given by x- 3 y-8 x + 3 y +7 z-6 = = z - 3 and = = × 3 -1 -3 2 4 x- 3 y-4 z+ 2 x -1 y + 7 z + 2 and 2. = = = = × -1 2 1 1 3 2 x + 1 y -1 z - 9 x - 3 y + 15 z - 9 3. and = = = = × 2 1 -3 2 -7 5 1.

SSS Mathematics for Class 12 1151

Straight Line in Space

4.

1151

x - 6 y -7 z - 4 x y+ 9 z-2 and = = = = × 3 -1 1 -3 2 4

5. Show that the lines x y-2 z+ 3 x-2 y-6 z- 3 and = = = = 1 2 3 2 3 4 intersect and find their point of intersection. 6. Show that the lines x - 1 y + 1 z -1 x - 2 y -1 z + 1 and = = = = 3 2 5 2 3 -2 do not intersect each other. ANSWERS (EXERCISE 27E)

1. 3 30 units,

x- 3 y-8 z- 3 = = 2 5 -1

3. 4 3 units, x = y = z

x-4 y-2 z+ 3 = = -1 -3 5 x- 3 y-8 z- 3 4. 3 30 units, = = 2 5 -1

2.

35 units,

5. (2, 6, 3)

EXERCISE 27F Very-Short-Answer Questions 1. If a line has direction ratios 2, –1, –2 then what are its direction cosines? [CBSE 2012]

4 - x y 1 -z 2. Find the direction cosines of the line [CBSE 2013C] = = × 2 6 3 3-x y+ 2 z+ 2 3. If the equations of a line are = = , find the direction cosines -3 -2 6 of a line parallel to the given line. [CBSE 2012] 4. Write the equations of a line parallel to the line passing through the point (1, –2, 3).

x-2 y+ 3 z+5 and = = -3 2 6 [CBSE 2009C]

5. Find the Cartesian equations of the line which passes through the point x + 3 4-y z + 8 = = × ( -2, 4, - 5) and which is parallel to the line 3 5 6 [CBSE 2013] 6. Write the vector equation of a line whose Cartesian equations are x -5 y + 4 6 -z [CBSE 2010, ’11] = = × 3 7 2 3 - x y + 4 2z - 6 7. The Cartesian equations of a line are = = × Write the 5 7 4 vector equation of the line. [CBSE 2014]

SSS Mathematics for Class 12 1152

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Senior Secondary School Mathematics for Class 12

8. Write the vector equation of a line passing through the point (1, –1, 2) and x - 3 y -1 z + 1 parallel to the line whose equations are = = × [CBSE 2013C] 1 2 -2 9. If P(1, 5 , 4) and Q( 4, 1, - 2) be two given points, find the direction ratios of PQ. [CBSE 2008] 4-x y + 3 z + 2 = = × Find the direction cosines 2 2 1 of a line parallel to this line. [CBSE 2013]

10. The equations of a line are

11. The Cartesian equations of a line are

x -1 y + 2 5 -z = = × Find its vector 2 3 1

equation. 12. Find the vector equation of a line passing through the point (1, 2, 3) and ^

^

^

parallel to the vector ( 3 i + 2 j - 2 k ). ®

^

^

^

^

^

^

13. The vector equation of a line is r = ( 2 i + j - 4 k ) + l( i - j - k ). Find its Cartesian equation. 14. Find the Cartesian equation of a line which passes through the point x+ 3 y-4 z+ 8 ( -2, 4, - 5) and which is parallel to the line = = × 3 5 6 15. Find the Cartesian equation of a line which passes through the point ^

^

^

having position vector ( 2 i - j + 4 k ) and is in the direction of the vector ^

^

^

( i + 2 j - k ). ®

^

^

^

^

^

^

16. Find the angle between the lines r = ( 2 i - 5 j + k ) + l( 3 i + 2 j + 6 k ) and ®

^

^

^

^

^

r = (7 i - 6 k ) + m ( i + 2 j + 2 k ).

17. Find the angle between the lines =

z -5 × 2

18. Show that the lines

x + 3 y -1 z + 3 x+1 y-4 and = = = 3 5 4 1 1

x -5 y + 2 z x y z = = and = = are at right angles. 7 -5 1 1 2 3

19. The direction ratios of a line are 2, 6, –9. What are its direction cosines? 20. A line makes angles 90°, 135° and 45° with the positive directions of x-axis, y-axis and z-axis respectively. What are the direction cosines of the line? 21. What are the direction cosines of the y-axis? ^

^

^

22. What are the direction cosines of the vector ( 2 i + j - 2 k )? ®

^

^

^

23. What is the angle between the vector r = ( 4 i + 8 j + k ) and the x-axis?

SSS Mathematics for Class 12 1153

Straight Line in Space

1153

ANSWERS (EXERCISE 27F)

2 -1 -2 , , 3 3 3 x -1 y + 2 z - 3 4. = = -3 2 6

-2 6 -3 , , 7 7 7 x+ 2 y-4 z+5 5. = = 3 -5 6 2.

1.

®

^

^

®

^

^

®

^ ^

^

^

^

3.

3 -2 6 , , 7 7 7

^

6. r = (5 i - 4 j + 6 k ) + l( 3 i + 7 j - 2 k ) ^

^

^

^

7. r = ( 3 i - 4 j + 3 k ) + l( -5 i + 7 j + 2 k ) ^

^

^

^

8. r = ( i - j + 2 k ) + l( i + 2 j - 2 k ) 10.

®

-2 2 1 , , 3 3 3

®

^

9. 3, –4, –6

^

^

^

^

^

^

^

x - 2 y -1 z + 4 = = 1 -1 -1 x-2 y+1 z-4 -1 æ 19 ö 15. 16. cos ç ÷ = = 1 2 -1 è 21 ø ^

^

^

12. r = ( i + 2 j + 3 k ) + l( 3 i + 2 j - 2 k ) 14.

^

11. r = ( i - 2 j + 5 k ) + l( 2 i + 3 j - k )

x+ 2 y-4 z+5 = = 3 5 6

æ8 3ö ÷ 17. cos-1 çç ÷ è 15 ø 2 1 -2 22. , , 3 3 3

19.

2 6 -9 , , 11 11 11

13.

20. 0,

-1 1 , 2 2

21. 0, 1, 0

æ 4ö 23. cos-1 ç ÷ è 9ø

HINTS TO THE GIVEN QUESTIONS (EXERCISE 27F) 1. D.r.’s of the given line are 2, –1, –2 and 2 2 + ( -1) 2 + ( -2 ) 2 = 9 = 3. 2 -1 -2 , , × 3 3 3 x- 4 y z- 1 2. The given equations are = = × -2 6 -3 \ d.c.’s of the given line are

\ d.r.’s of the given line are –2, 6, –3 and ( -2 ) 2 + 6 2 + ( -3 ) 2 = 49 = 7 . -2 6 -3 , , × 7 7 7 x- 3 y+ 2 z+ 2 3. The given equations are = = × 3 -2 6 \ d.c.’s of the given line are

\ d.r.’s of this line are 3, –2, 6. \ d.r.’s of a line parallel to this line are 3, –2, 6 and \ d.c.’s of a line parallel to the given line are

3 2 + ( -2 ) 2 + 6 2 = 49 = 7 .

3 -2 6 , , × 7 7 7

4. D.r.’s of the given line are –3, 2, 6. D.r.’s of a line parallel to the given line are –3, 2, 6. x- 1 y+ 2 z- 3 Required equations of the line are = = × -3 2 6

SSS Mathematics for Class 12 1154

1154

Senior Secondary School Mathematics for Class 12

5. The equations of the given line are

x+ 3 y- 4 z+ 8 = = × 3 -5 6

D.r.’s of this line are 3, –5, 6. D.r.’s of a line parallel to this line are 3, –5, 6. x+ 2 y- 4 z+ 5 = = × 3 -5 6 x-5 y+ 4 z- 6 6. The given equations of the line are = = × 3 7 -2 Required equations of a line through (–2, 4, –5) are

Clearly, this line passes through the point (5, –4, 6) and it has d.r.’s 3, 7, –2. ®

^

^

^

^

^

^

\ its vector equation is r = (5 i - 4 j + 6 k ) + l( 3 i + 7 j - 2 k ). 7. The given equations of the line are

x - 3 y - ( -4 ) z - 3 = = × -5 7 2

Clearly, this line passes through the point (3, –4, 3) and it has d.r.’s –5, 7, 2. ®

^

^

^

^

^

^

\ its vector equation is r = ( 3 i - 4 j + 3 k ) + l( -5 i + 7 j + 2 k ). 8. Clearly, the requried line passes through the point (1, –1, 2) and its d.r.’s are 1, 2, –2. ®

^

^

^

^

^

^

Hence, its vector equation is r = ( i - j + 2 k ) + l( i + 2 j - 2 k ). 9. D.r.’s of PQ are ( 4 - 1), ( 1 - 5 ), ( -2 - 4 ), i.e., 3, –4, –6. 10. The given equations are

x- 4 y+ 3 z+ 2 = = × -2 2 1

The d.r.’s of this line are –2, 2, 1. The d.r.’s of a line parallel to this line are –2, 2, 1 and ( -2 ) 2 + 2 2 + 12 = 9 = 3. The d.c.’s of a line parallel to the given line are 11. The given equations are

-2 2 1 , , × 3 3 3

x - 1 y - ( -2 ) z - 5 = = × 2 3 -1

Clearly, this line passes thorugh the point (1, –2, 5) and it has d.r.’s 2, 3, –1. ®

^

^

^

^

^

^

So, its vector equation is r = ( i - 2 j + 5 k ) + l( 2 i + 3 j - k ). ®

^

^

^

^

^

^

12. Clearly, the required equation is r = ( i + 2 j + 3 k ) + l( 3 i + 2 j - 2 k ). 13. The given line passes through the point (2, 1, –4) and it is parallel to a line whose direction ratios are 1, –1, –1. x- 2 y- 1 z+ 4 So, its Cartesian equation is = = × 1 -1 -1 14. Clearly, the line passes through the point (–2, –4, –5) and it has direction ratios 3, 5, 6. x+ 2 y- 4 z+ 5 So, its equation is = = × 3 5 6 15. The required line passes through the point (2, –1, 4) and it has direction ratios 1, 2, –1. x- 2 y+ 1 z- 4 \ its equation is = = × 1 2 -1

SSS Mathematics for Class 12 1155

Straight Line in Space ®

1155 ®

®

®

and r = a1 + lb1 and r = a2 + l b 2 is given by

16. The angle between the lines ® ®

b × b2 cos q = ®1 ® × |b1||b 2| ^

^

^

^

^

^

|( 3 i + 2 j + 6 k ) × ( i + 2 j + 2 k )| ( 3 + 4 + 12 ) 19 = = 21 (7 ´ 3 ) { 3 2 + 2 2 + 6 2}{ 12 + 2 2 + 2 2}

\

cos q =

Þ

æ 19 ö q = cos -1 ç ÷ . è 21 ø

17. Here ( a1 ) = 3 , b1 = 5 , c1 = 4 ) and ( a2 = 1, b 2 = 1, c2 = 2 ). \

cos q = = =

Þ

| a1 a2 + b1 b 2 + c1 c2| { a12 + b12 + c12 }{ a22 + b 22 + c22 } |( 3 ´ 1) + (5 ´ 1) + ( 4 ´ 2 )| { 3 2 + 5 2 + 4 2 }{ 12 + 12 + 2 2 } 16 16 8 = = × ( 50 ´ 6 ) 10 3 5 3

æ 8 q = cos -1 çç ´ è5 3

æ8 3ö 3ö ÷ = cos -1 ç ÷ ç 15 ÷ . 3 ÷ø è ø

18. Here ( a1 = 7 , b1 = -5 , c1 = 1) and ( a2 = 1, b 2 = 2 , c2 = 3 ). ( a1 a2 + b1 b 2 + c1 c2 ) = (7 ´ 1) + ( -5 ) ´ 2 + ( 1 ´ 3 ) = 0. Hence, the given lines are at right angles. 19. We have 2 2 + 6 2 + ( -9 ) 2 = 121 = 11. \

d.c.’s of the given line are

2 6 -9 , , × 11 11 11

-1 1 ö æ 20. D.c.’s of the line are cos 90°, cos 135° and cos 45°, i.e., ç 0 , , ÷. 3 2ø è 21. Clearly, the y-axis makes an angle of 90°, 0°, 90° with the x-axis, y-axis and z-axis respectively. So, its d.c.’s are cos 90°, cos 0°, cos 90°, i.e., 0, 1, 0. 22. D.r.’s of the given vector are 2, 1, –2 and 2 2 + 12 + ( -2 ) 2 = 9 = 3. \ d.c.’s of the given vector are

2 1 -2 , , × 3 3 3

23. D.r.’s of the given vector are 4, 8, 1 and 4 2 + 8 2 + 12 = 81 = 9. \ d.c.’s of the given vector are

4 8 1 , , × 9 9 9

Let a be the angle between the given vectors and the x-axis. 4 æ4ö Then, cos a = Þ a = cos -1 ç ÷ . 9 è9ø

SSS Mathematics for Class 12 1156

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Senior Secondary School Mathematics for Class 12

OBJECTIVE QUESTIONS Multiple-Choice Questions 1. The direction ratios of two lines are 3, 2, –6 and 1, 2, 2 respectively. The acute angle between these lines is æ5ö æ 3ö (b) cos-1 ç ÷ (a) cos-1 ç ÷ è 18 ø è 20 ø æ5 ö æ 8ö (c) cos-1 ç ÷ (d) cos-1 ç ÷ è 21 ø è 21 ø 2. The direction ratios of two lines are a , b , c and ( b - c), ( c - a), ( a - b) respectively. The angle between these lines is p p p 3p (b) (c) (d) (a) 3 2 4 4 x - 2 y -1 z + 3 x + 2 y - 4 z -5 3. The angle between the lines and is = = = = 2 7 -3 -1 2 4 p p p æ 3ö (b) (c) (d) cos-1 ç ÷ (a) 6 3 2 è 8ø x -1 y - 2 z - 3 x -1 y -1 z - 6 4. If the lines and are perpendicular to = = = = -3 2k 2 3k 1 -5 each other then k = ? -5 5 10 -10 (b) (c) (d) (a) 7 7 7 7 5. A line passes through the points A(2, –1, 4) and B(1, 2, –2). The equations of the line AB are x-2 y+1 z-4 x+ 2 y+1 z-4 (b) (a) = = = = -1 2 -6 -1 2 6 x-2 y+1 z-4 (c) (d) none of these = = 1 2 6 x y z x -5 y - 2 z - 3 is 6. The angle between the lines = = and = = 2 2 1 4 1 8 p æ 3ö æ5 ö æ 2ö (b) cos-1 ç ÷ (c) cos-1 ç ÷ (d) (a) cos-1 ç ÷ 4 6 3 3 è ø è ø è ø 7. The angle between the lines ®

^

^

^

^

^

^

®

^

^

^

^

^

^

r = ( 3 i + j - 2 k ) + l( i - j - 2 k ) and r = ( 2 i - j - 5 k ) + m ( 3 i - 5 j - 4 k ) is æ8 3ö æ6 2ö æ5 3 ö æ5 2 ö ÷ (b) cos-1 ç ÷ ÷ ÷ (a) cos-1 çç (c) cos-1 çç (d) cos-1 çç ÷ ç ÷ ÷ ÷ è 15 ø è 5 ø è 8 ø è 6 ø 8. A line is perpendicular to two lines having direction ratios 1, –2, –2 and 0, 2, 1. The direction cosines of the line are -2 1 2 2 1 -1 2 -1 2 (b) , , (c) , (d) none of these (a) , , , 3 3 3 3 3 3 3 3 3

SSS Mathematics for Class 12 1157

Straight Line in Space

1157

9. A line passes through the point A(5, –2. 4) and it is parallel to the vector ^

^

^

( 2 i - j + 3 k ). The vector equation of the line is ®

^

®

^

^

^

^

^

^

^

^

(a) r = ( 2 i - j + 3 k ) + l(5 i - 2 j + 4 k ) ^

^

^

(b) r = (5 i - 2 j + 4 k ) + l( 2 i - j + 3 k ) ®

^

^

^

(c) r × (5 i - 2 j + 4 k ) = 14 (d) none of these 10. The Cartesian equations of a line are equation is ®

^

^

®

^

^

^

^

^

x -1 y + 2 z -5 = = × Its vector 2 3 -1

^

(a) r = ( - i + 2 j - 5 k ) + l( 2 i + 3 j - k ) ^

^

^

^

(b) r = ( 2 i + 3 j - k ) + l( i - 2 j + 5 k ) ®

^

^

^

^

^

^

(c) r = ( i - 2 j + 5 k ) + l( 2 i + 3 j - 4 k ) (d) none of these 11. A line passes through the point A(–2, 4, –5) and is parallel to the line x+ 3 y-4 z+ 8 = = × The vector equation of the line is 3 5 6 ®

^

^

^

^

^

^

®

^

^

^

^

^

^

(a) r = ( -3 i + 4 j - 8 k ) + l( -2 i + 4 j - 5 k ) (b) r = ( -2 i + 4 j - 5 k ) + l( -3 i + 5 j + 6 k ) ®

^

^

^

^

^

^

(c) r = ( 3 i + 5 j + 6 k ) + l( -2 i + 4 j - 5 k ) (d) none of these 12. The coordinates of the point where the line through the points A(5, 1, 6) and B( 3 , 4, 1) crosses the yz-plane is æ -17 13 ö (a) (0, 17, –13) (b) ç 0, , ÷ 2 2ø è æ 17 -13 ö (d) none of these (c) ç 0, , ÷ 2 2 ø è 13. The vector equation of the x-axis is given by ®

®

^

(a) r = i

^

®

^

(b) r = j + k

^

(c) r = l i (d) none of these x-2 y+1 z- 3 14. The Cartesian equations of a line are = = × What is its vector 2 3 -2 equation? ®

^

®

^

®

^

^

^

^

^

^

(a) r = ( 2 i + 3 j - 2 k ) + l( 2 i - j + 3 k ) ^

^

^

^

^

(b) r = ( 2 i - j + 3 k ) + l( 2 i + 3 j - 2 k ) ^

^

(c) r = ( 2 i + 3 j - 2 k ) (d) none of these

SSS Mathematics for Class 12 1158

1158

Senior Secondary School Mathematics for Class 12

15. The angle between two lines having direction ratios 1, 1, 2 and ( 3 - 1), ( - 3 - 1), 4 is p p p p (a) (b) (c) (d) 6 2 3 4 x-2 y- 3 z+1 is 16. The straight line = = 3 1 0 (a) parallel to the x-axis (b) parallel to the y-axis (c) parallel to the z-axis (d) perpendicular to the z-axis 17. If a line makes angles a , b and g with the x-axis, y-axis and z-axis respectively then (sin 2 a + sin 2 b + sin 2 g) =? 3 2 18. If ( a1 , b1 , c1) and ( a2 , b2 , c2) be the direction ratios of two parallel lines then a b c (b) 1 = 1 = 1 (a) a1 = a2 , b1 = b2 , c1 = c2 a2 b2 c2 (a) 1

(b) 3

(c) 2

(c) a12 + b12 + c12 = a22 + b22 + c22

(d)

(d) a1 a2 + b1 b2 + c1 c2 = 0

19. If the points A(–1, 3, 2), B(–4, 2, –2) and C(5, 5, l) are collinear then the value of l is (a) 5 (b) 7 (c) 8 (d) 10 ANSWERS (OBJECTIVE QUESTIONS)

1. (c) 9. (b) 17. (c)

2. (b) 10. (c) 18. (b)

3. (c) 11. (b) 19. (d)

4. (d) 12. (c)

5. (a) 13. (c)

6. (c) 14. (b)

7. (a) 15. (c)

8. (c) 16. (d)

HINTS TO THE GIVEN OBJECTIVE QUESTIONS 1. (c) cos q = = Þ

|( 3 ´ 1) + ( 2 ´ 2 ) + ( -6 ) ´ 2| { 3 2 + 2 2 + ( -6 ) 2 } { 12 + 2 2 + 2 2 }

=

|- 5| ( 49 ´ 49 )

=

5 5 = (7 ´ 3 ) 21

1 - 51 5 5 = = ( 49 ´ 9 ) (7 ´ 3 ) 21

æ5 ö q = cos -1 ç ÷ . è 21 ø

2. (b) cos q =

a( b - c ) + b( c - a) + c( a - b ) { a 2 + b 2 + c 2 } { ( b - c ) 2 + ( c - a) 2 + ( a - b ) 2 }

=0 Þ q=

3. (c) The d.r.’s of these lines are 2, 7, –3 and –1, 2, 4. |2 ´ ( -1) + 7 ´ 2 + ( -3 ) ´ 4| p \ cos q = =0 Þ q= × 2 { 2 2 + 7 2 + ( -3 ) 2 } { ( -1) 2 + 2 2 + 4 2 }

p × 2

SSS Mathematics for Class 12 1159

Straight Line in Space

1159

4. (d) Since the given lines are perpendicular to each other, we have -10 ( -3 )( 3 k ) + ( 2 k ´ 1) + 2 ´ ( -5 ) = 0 Þ 7 k = -10 Þ k = × 7 5. (a) The given line passes through the point A( 2 , -1, 4 ) and its direction ratios are ( 1 - 2 ), ( 1 + 1), ( -2 - 4 ), i.e., –1, 2, –6. x-2 y+ 1 x-4 \ required equations of the line are = = × -1 2 -6 6. (c) The d.r.’s of the given lines are 2, 2, 1 and 4, 1, 8. |( 2 ´ 4 ) + ( 2 ´ 1) + ( 1 ´ 8 )| 18 2 18 \ cos q = = = = 2 2 2 2 2 2 ( 3 ´ 9 ) 3 81 ) ( ´ 9 { 2 + 2 + 1 }{ 4 + 1 + 8 } Þ

æ2ö q = cos -1 ç ÷ . è 3ø ®

®

®

®

®

®

7. (a) The given lines are r = a1 + l b1 and r = a2 + l b 2 . ® ®

\

= Þ

^

^

^

^

^

^

|b × b 2| |( i - j - 2 k ) × ( 3 i - 5 j - 4 k )| cos q = ®1 ® = 2 + { 1 ( -1) 2 + ( -2 ) 2 }{ 3 2 + ( -5 ) 2 + ( -4 ) 2 } |b1||b 2| ( 3 + 5 + 8) ( 6 ´ 50 )

=

16 8 3 8 = ´ = 3 15 10 3 5

3

æ8 3ö ÷. q = cos -1 çç ÷ è 15 ø

8. (c) Let the direction ratios of the required line be a, b , c. Then, a - 2b - 2c = 0

... (i)

0 a + 2 b + c = 0. On solving (i) and (ii) by cross multiplication, we get a b c a b c = = Þ = = × ( -2 + 4 ) ( 0 - 1) ( 2 - 0 ) 2 -1 2

... (ii)

\

d.r.’s of the required line are 2, –1, 2 and 2 2 + ( -1) 2 + 2 2 = 9 = 3.

\

d.c.’s of the required line are

2 -1 2 , , × 3 3 3

9. (b) Clearly, the required vector equation of the line is ®

^

^

^

^

^

^

r = (5 i - 2 j + 4 k ) + l( 2 i - j + 3 k ).

10. (c) The line passes through the point (1, –2, 5) and it has direction ratios 2, 3, –1. ®

^

^

^

^

^

^

So, its vector equation is r = ( i - 2 j + 5 k ) + l( 2 i + 3 j - k ). ^

^

^

11. (b) The required line passes through the point ( -2 i + 4 j + 5 k ) and it is parallel to ^

^

^

the vector ( 3 i + 5 j + 6 k ). Hence, the required equation of the line is ®

^

^

^

^

^

^

r = ( -2 i + 4 j - 5 k ) + l( 3 i + 5 j + 6 k ).

SSS Mathematics for Class 12 1160

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Senior Secondary School Mathematics for Class 12

12. (c) Equation of the line AB is x-5 y- 1 z- 6 x-5 y- 1 z- 6 = = Þ = = = l. 5 - 3 1- 4 6 - 1 2 -3 5 Any point on this line is ( 2 l + 5 , - 3 l + 1, 5 l + 6 ). -5 × 2 -25 æ 17 -13 ö æ 15 ö + 1, + 6 ÷ , i.e., ç 0 , , \ the required point is ç 0 , ÷. 2 2 2 2 ø è è ø x- 0 y- 0 z- 0 and its vector 13. (c) Note that the Cartesian equation of x-axis is = = 1 0 0 If this point lies on the yz-plane, we have 2l + 5 = 0 Þ l =

®

^

equation is r = l i . ^

^

^

14. (b) The line passes through the point ( 2 i - j + 3 k ) and it is parallel to the vector ^

^

^

( 2 i + 3 j - 2 k ). ®

^

^

^

^

^

^

So, its vector equation is r = ( 2 i - j + 3 k ) + l( 2 i + 3 j - 2 k ). 15. (c) cos q = =

1 ´ ( 3 - 1) + 1 ´ ( - 3 - 1) + 2 ´ 4 2

{ 1 + 12 + 2 2 } { ( 3 - 1) 2 + ( - 3 - 1) 2 + 4 2 } ( 3 - 1-

3 - 1 + 8)

( 6 ´ 24 )

=

6 1 p = Þ q= × 3 24 2

16. (c) If a line is perpendicular to the z-axis then cos g = cos

p = 0. 2

So, the given line is perpendicular to the z-axis. 17. (c) We have cos 2 a + cos 2 b + cos 2 g = 1. \

( 1 - sin 2 a ) + ( 1 - sin 2 b ) + ( 1 - sin 2 g ) = 1

Þ

sin 2 a + sin 2 b + sin 2 g = 2.

18. (b) If the given lines are parallel, then

a1 b1 c1 = = × a2 b 2 c2

19. (c) Equations of the line AB are x+ 1 y- 3 z- 2 x+ 1 y- 3 z- 2 = = Þ = = -4 + 1 2 - 3 -2 - 2 -3 -1 -4 x+ 1 y- 3 z- 2 = = × Þ 3 1 4 If the point C(5 , 5 , l) lies on AB, we have 5+ 1 5- 3 l- 2 l- 2 = = Þ = 2 Þ l = ( 8 + 2 ) = 10. 3 1 4 4

SSS Mathematics for Class 12 1161

28. THE PLANE A plane is a surface such that a line segment joining any two points on it lies wholly on it.

PLANE

NORMAL TO A PLANE A straight line which is perpendicular to every line lying on a plane is called a normal to the plane.

All the normals to a plane are parallel to each other. General Equation of a Plane in the Cartesian Form Every equation ax + by + cz + d = 0 of the first degree in x , y , z always represents a plane. Also, a, b, c are the direction ratios of the normal to this plane. Let us consider a surface represented by the equation … (i) ax + by + cz + d = 0. Let A( x1 , y1 , z1) and B( x 2 , y 2 , z2) be any two points on the surface represented by (i). Then, … (ii) ax1 + by1 + cz1 + d = 0 … (iii) and, ax 2 + by 2 + cz2 + d = 0. Multiplying (iii) by l and adding it to (ii), we get a( lx 2 + x1) + b( ly 2 + y1) + c( lz2 + z1) + d( l + 1) = 0 æ lx + x1 ö æ ly + y1 ö æ lz + z1 ö ÷÷ + b çç 2 ÷÷ + c çç 2 ÷÷ + d = 0 a çç 2 Þ l l + 1 + 1 è ø è ø è l+1 ø

THEOREM 1

PROOF

Þ

æ lx 2 + x1 ly 2 + y1 lz2 + z1 ö çç ÷ lies on surface (i), when l ¹ - 1. , , l+1 l + 1 ÷ø è l+1

But, these are the general coordinates of a point which divides AB in the ratio l : 1. Since l may take any real value other than -1, it follows that every point of AB lies on (i). Hence, ax + by + cz + d = 0 represents a plane. To Show that a, b, c are the Direction Ratios of the Normal to the Plane Subtracting (ii) from (iii), we get a( x 2 - x1) + b( y 2 - y1) + c(z2 - z1) = 0 Þ a line with direction ratios a , b , c is perpendicular to an arbitrary line AB taken on plane (i) [Q ( x 2 - x1), ( y 2 - y1), (z2 - z1) are d.r.’s of AB] Þ a line with d.r.’s a , b , c is perpendicular to the plane (i) 1161

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Senior Secondary School Mathematics for Class 12

Þ a , b , c are the direction ratios of the normal to the plane (i). Hence, ax + by + cz + d = 0 represents a plane, and a , b , c are the direction ratios of the normal to this plane. Equation of a Plane Passing through a Given Point The equation of a plane passing through a point P( x1 , y1 , z1) is a( x - x1) + b( y - y1) + c(z - z1) = 0, where a , b , c are constants. The general equation of a plane is

THEOREM 2 PROOF

ax + by + cz + d = 0. If this plane passes through the point P( x1 , y1 , z1) then ax1 + by1 + cz1 + d = 0.

… (i) … (ii)

Subtracting (ii) from (i), we get a( x - x1) + b( y - y1) + c(z - z1) = 0. This is the general equation of a plane passing through the point P( x1 , y1 , z1). Equation of a Plane in the Intercept Form THEOREM 3

If a plane makes intercepts of lengths a, b, c with the x-axis, y-axis and x y z z-axis respectively, the equation of the plane is + + = 1. a b c [CBSE 2004C]

PROOF

Let O be the origin, and let the plane meet the coordinate axes at A , B , C respectively such that OA = a , OB = b and OC = c. So, the coordinates of these points are A( a , 0, 0), B( 0, b , 0) and C( 0, 0, c). Let the equation of the given plane be Ax + By + Cz + D = 0. … (i) Since the given plane does not pass through O( 0, 0, 0), D ¹ 0. Also, since (i) passes through A( a , 0, 0), B( 0, b , 0) and C( 0, 0, c), we have -D Aa + D = 0 Þ A = , a -D Bb + D = 0 Þ B = , b -D Cc + D = 0 Þ C = × c Putting these values in (i), we get

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The Plane

1163

- Dx Dy Dz +D=0 a b c x y z + + = 1 [on dividing throughout by -D]. Þ a b c This is the required equation of the plane in the intercept form. Equation of a Plane Passing through Three Given Points Suppose that a plane passes through the points A( x1 , y1 , z1), B( x 2 , y 2 , z2)and C( x 3 , y 3 , z 3). We know that the equation of a plane passing through the point A( x1 , y1 , z1) is given by a( x - x1) + b( y - y1) + c(z - z1) = 0.

... (i)

If this plane passes through the points B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3), we have a( x 2 - x1) + b( y 2 - y1) + c(z2 - z1) = 0

... (ii)

and a( x 3 - x1) + b( y 3 - y1) + c(z 3 - z1) = 0.

... (iii)

On eliminating a , b , c from (i), (ii) and (iii), we get x - x1 y - y1 z - z1 x 2 - x1 y 2 - y1 z2 - z1 = 0. x 3 - x1

y 3 - y1

z 3 - z1

This is the required equation of the plane passing through three points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and ( x 3 , y 3 , z 3). SUMMARY OF THE RESULTS

1. The general equation of a plane is ax + by + cz + d = 0. The d.r.’s of the normal to the plane are a, b, c. 2. The equation of the plane passing through the point P( x1 , y1 , z1) is a ( x - x1) + b ( y - y1) + c(z - z1) = 0. 3. If a plane makes intercepts a , b and c with the x-axis, y-axis and z-axis respectively, then its equation is x y z + + =1× a b c 4. The equation of a plane passing through three points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) is given by x - x1 y - y1 z - z1 x 2 - x1 y 2 - y1 z2 - z1 = 0. x 3 - x1

y 3 - y1

z 3 - z1

SSS Mathematics for Class 12 1164

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Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Find the equation of the plane passing through the points A( 2, 5 , - 3), B( -2, - 3 , 5) and C(5 , 3 , - 3).

SOLUTION

The equation of the plane passing through the point A( 2, 5 , - 3) is given by ... (i) a( x - 2) + b( y - 5) + c(z + 3) = 0. If this plane passes through the points B( -2, - 3 , 5) and C(5 , 3 , - 3), then we have a( -2 - 2) + b( -3 - 5) + c(5 + 3) = 0 Þ - 4a - 8b + 8c = 0 ... (ii) ... (iii) a(5 - 2) + b( 3 - 5) + c( -3 + 3) = 0 Þ 3 a - 2b + 0c = 0. On solving (ii) and (iii) by cross multiplication, we get a b c a b c = = Þ = = ( 0 + 16) ( 24 - 0) ( 8 + 24) 16 24 32 a b c = = = k (say) Þ a = 2k , b = 3 k and c = 4k. 2 3 4 Putting these values of a , b and c in (i), we get 2k( x - 2) + 3 k( y - 5) + 4k(z + 3) = 0 Þ 2( x - 2) + 3( y - 5) + 4(z + 3) = 0 Þ 2x + 3 y + 4z - 7 = 0. Hence, the required equation of the plane is 2x + 3 y + 4z - 7 = 0.

Þ

Alternate method We know that equation of the plane passing through the points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) is given by x - x1 y - y1 z - z1 x 2 - x1 y 2 - y1 z2 - z1 = 0. x 3 - x1

y 3 - y1

z 3 - z1

Here, ( x1 = 2, y1 = 5 , z1 = -3), ( x 2 = -2, y 2 = -3 , z2 = 5) and ( x 3 = 5 , y 3 = 3 , z 3 = -3). \ the required equation of the plane is x-2 y -5 z+ 3 x-2 -2 - 2 -3 - 5 5 + 3 = 0 Þ -4 5-2 3 - 5 -3 + 3 3

y -5 z + 3 -8 8 = 0. -2 0

Þ ( x - 2)( 0 + 16) - ( y - 5)( 0 - 24) + (z + 3)( 8 + 24) = 0 Þ 16( x - 2) + 24( y - 5) + 32(z + 3) = 0 Þ 16x + 24y + 32z - 56 = 0 Þ 2x + 3 y + 4z - 7 = 0. Hence, the required equation of the plane is 2x + 3 y + 4z - 7 = 0.

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1165

EXAMPLE 2

Show that the four points A(1, - 1, 1), B( 2, 3 , 1), C(1, 2, 3) and D( 0, - 2, 3) are coplanar. Find the equation of the plane containing them.

SOLUTION

We know that the equation of the plane passing through the points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) is given by x - x1 y - y1 z - z1 x 2 - x1 y 2 - y1 z2 - z1 = 0. x 3 - x1

y 3 - y1

z 3 - z1

Here, ( x1 = 1, y1 = -1, z1 = 1), ( x 2 = 2, y 2 = 3 , z2 = 1) and ( x 3 = 1, y 3 = 2, z 3 = 3). \ the equation of the plane passing through the points A, B and C is given by x -1 y + 1 z -1 x -1 y + 1 z -1 2 -1 1 -1 Þ Þ

3 + 1 1 -1 = 0 Þ 2 + 1 3 -1

1 0

4 3

0 2

=0

( x - 1)( 8 - 0) - ( y + 1)( 2 - 0) + (z - 1)( 3 - 0) = 0 8( x - 1) - 2( y + 1) + 3(z - 1) = 0 Þ 8x - 2y + 3z = 13.

... (i)

Putting x = 0, y = -2 and z = 3 in (i), we get LHS = ( 8 ´ 0) - 2 ´ ( -2) + 3 ´ 3 = ( 0 + 4 + 9) = 13 = RHS.

Thus, the point D( 0, - 2, 3) lies on the plane (i). Hence, the given four points are coplanar and the equation of the plane containing them is 8x = 2y + 3z = 13. EXAMPLE 3

Find the equation of the plane which cuts off intercepts 3 , 6 and -4 from the axes of coordinates.

SOLUTION

We know that the equation of a plane which cuts off intercepts a , b , c from the x-axis, y-axis and z-axis respectively, is x y z + + = 1. a b c Here a = 3 , b = 6 and c = -4. Hence, the required equation of the plane is x y z + + = 1 Þ 4x + 2y - 3z = 12. 3 6 -4

EXAMPLE 4

Reduce the equation of the plane 2x - 3 y + z = 6 to intercept form and find its intercepts on the coordinate axes.

SOLUTION

We have 2x 3 y z x y z + =1 Þ + + = 1. 6 6 6 3 -2 6 Hence, the equation of the given plane in intercept form is x y z + + = 1. 3 -2 6 2x - 3 y + z = 6 Þ

SSS Mathematics for Class 12 1166

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Senior Secondary School Mathematics for Class 12

The intercepts of the given plane on x-axis, y-axis and z-axis are 3, –2 and 6 respectively. EXAMPLE 5

SOLUTION

A plane meets the coordinate axes in A, B, C such that the centroid of x y z CABC is (p, q, r). Show that the equation of the plane is + + = 3. p q r x y z ... (i) + + = 1. a b c Then, clearly the plane meets the coordinate axis at A( a , 0, 0), B( 0, b , 0) and C( 0, 0, c). Since the centroid of CABC is G( p , q , r), we have a + 0+ 0 0+ b+ 0 0+ 0+ c = p, = q and =r 3 3 3 Þ a = 3 p , b = 3 q and c = 3r. Putting these values of a , b , c in (i) we get x y z x y z + + =1 Þ + + = 3. 3 p 3 q 3r p q r Let the required equation of the plane be

Hence, the required equation of the plane is

x y z + + = 3. p q r

EXAMPLE 6

A variable plane moves in such a way that the sum of the reciprocals of its intercepts on the coordinate axes is constant. Show that the plane passes through a fixed point.

SOLUTION

Let the equation of the variable plane be x y z … (i) + + = 1. a b c Then, it makes intercepts a , b , c with the coordinate axes. 1 1 1 + + = k , where k is a constant (given) \ a b c 1 1 1 1 æ1ö 1 æ1ö 1 æ1ö + + =1 Þ ç ÷ + ç ÷ + ç ÷ =1 Þ ka kb kc a è kø b è kø c è kø æ1 1 1ö Þ ç , , ÷ satisfies (i). è k k kø æ1 1 1ö Hence, the given plane passes through a fixed point ç , , ÷ × è k k kø

EXERCISE 28A 1. Find the equation of the plane passing through each group of points: (i) A( 2, 2, - 1), B( 3 , 4, 2) and C(7 , 0, 6) (ii) A( 0, - 1, - 1), B( 4, 5 , 1) and C( 3 , 9, 4) [CBSE 2006] (iii) A( -2, 6, - 6), B( -3 , 10, - 9) and C( -5 , 0, - 6). 2. Show that the four points A( 3 , 2, - 5), B( -1, 4, - 3), C( -3 , 8, - 5) and D( -3 , 2, 1) are coplanar. Find the equation of the plane containing them.

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1167

3. Show that the four points A( 0, - 1, 0), B( 2, 1, - 1), C(1, 1, 1) and D( 3 , 3 , 0) are coplanar. Find the equation of the plane containing them. 4. Write the equation of the plane whose intercepts on the coordinate axes are 2, – 4 and 5 respectively. 5. Reduce the equation of the plane 4x - 3 y + 2z = 12 to the intercept form, and hence find the intercepts made by the plane with the coordinate axes. 6. Find the equation of the plane which passes through the point (2, –3, 7) and makes equal intercepts on the coordinate axes. 7. A plane meets the coordinate axes at A , B and C respectively such that the centroid of C ABC is (1, -2, 3). Find the equation of the plane. 8. Find the Cartesian and vector equations of a plane passing through the point (1, 2, 3) and perpendicular to a line with direction ratios 2, 3, –4. [CBSE 2005C]

9. If O be the origin and P(1, 2, - 3) be a given point, then find the equation of the plane passing through P and perpendicular to OP. ANSWERS (EXERCISE 28A)

1. (i) 5 x + 2y - 3z = 17 (ii) 5 x - 7 y + 11z + 4 = 0 (iii) 2x - y - 2z = 2 2. x + y + z = 0 3. 4x - 3 y + 2z = 3 4. 10x - 5 y + 4z = 20 x y z 5. + + = 1, Intercepts on the axes are 3, –4, 6. 3 -4 6 x y z 6. x + y + z = 6 7. + + =1 3 -6 9 ®

^

^

^

8. 2x + 3 y - 4z + 4 = 0, r × ( 2 i + 3 j - 4 k ) + 4 = 0 9. x + 2y - 3z - 14 = 0 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28A) 5. 4 x - 3 y + 2z = 12 Þ

y x z + + = 1 [on dividing each term by 12]. 3 -4 6

6. Let it make intercept a on each of the coordinate axes. x y z Then, its equation is + + = 1 Þ x + y + z = a. a a a Putting x = 2 , y = -3 , z = 7 , we get a = 2 + ( -3 ) + 7 = 6. So, the required equation of the plane is x + y + z = 6. 7. Let the plane meet the coordinate axes at A( a, 0 , 0 ), B( 0 , b , 0 ) and C ( 0 , 0 , c ). Then, a b c = 1, = -2 , = 3 Þ a = 3 , b = -6 , c = 9. 3 3 3 y x z \ required equation of the plane is + + = 1. 3 -6 9 8. Required equation is 2( x - 1) + 3( y - 2 ) - 4(z - 3 ) = 0 Þ 2 x + 3 y - 4z + 4 = 0.

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Senior Secondary School Mathematics for Class 12

Its vector equation is ^

^

^

^

^

®

^

^

^

^

( x i + y j + z k ) × ( 2 i + 3 j - 4 k ) + 4 = 0 , i.e., r × ( 2 i + 3 j - 4 k ) + 4 = 0. 9. Let the required equation of the plane passing through the point A( 1, 2 , - 3 ) be a( x - 1) + b( y - 2 ) + c(z + 3 ) = 0 D.r.’s of OP are ( 1 - 0 ), ( 2 - 0 ), ( -3 - 0 ), i.e., 1, 2, –3.

... (i)

Since the plane is perpendicular to OP, so the normal to the plane is parallel to OP. a b c \ = = = k (say) Þ a = k , b = 2 k and c = -3 k. 1 2 -3 \ required equation of the plane is k( x - 1) + 2 k( y - 2 ) - 3 k(z + 3 ) = 0 Þ

( x - 1) + 2( y - 2 ) - 3(z + 3 ) = 0.

Þ Þ

( x + 2 y - 3z) + ( -1 - 4 - 9 ) = 0 x + 2 y - 3z - 14 = 0.

EQUATIONS OF A PLANE IN VARIOUS FORMS

Equation of a Plane in the Normal Form Vector Form THEOREM 1

PROOF

^

If n is a unit vector normal to a given plane, directed from the origin to the plane, and p is the length of the perpendicular drawn from the origin ® ^ to the plane then the vector equation of the plane is r × n = p.

Let O be the origin, and let ON be the perpendicular drawn from O to the given plane. Let ON = p. ¾®

Let n$ be a unit vector along ON . ¾®

^

Then, ON = pn. Let P be an arbitrary point on the plane, and ®

let the position vector of P be r . ®

¾®

Then, OP = r . ¾®

¾®

^

Since NP lies on the plane, NP is perpendicular to n. ¾®

^

\

NP × n = 0

Þ

(OP - ON ) × n = 0 Þ ( r - p n) × n = 0

Þ

¾®

® ^

¾®

^

^ ^

®

^

^

® ^

r × n - p n × n = 0 Þ r × n = p. ® ®

Hence, the required equation of the plane is r × n = p. REMARK

The equation of a plane which is at a distance p from the origin and ^

® ^

which is perpendicular to n is r × n = p.

SSS Mathematics for Class 12 1169

The Plane

1169 ® ^

®

If n is a vector normal to a given plane then r × n = q represents a plane.

COROLLARY ® ® PROOF

®

r ×n =q Þ r ×

®

n

®

=

® ^

q ®

Þ r × n = p , where

= p.

|n|

|n | |n |

® ^

q ®

But, r × n = p represents a plane. ® ®

Hence, r × n = q also represents a plane. Cartesian Forms (i) Normal Form: ^

Let l , m , n be the d.c.’s of n and P( x , y , z) be the given point. ®

^

^

^

^

^

^

^

Then, r = x i + y j + z k and n = l i + m j + n k . ® ^

On substituting these values in r × n = p, we get ^

^

^

^

^

^

( x i + y j + z k) × ( l i + m j + n k) = p Þ lx + my + nz = p. This is known as the normal form of the Cartesian equation of a plane. (ii) General Form: ®

Let a , b , c be the d.r.’s of n and P( x , y , z) be the given point. ®

^

^

^

®

^

^

^

Then, r = x i + y j + z k and n = a i + b j + ck . ® ®

On substituting these values r × n = q , we get ^

Þ

^

^

^

^

^

( x i + y j + z k ) × ( a i + b j + ck ) = q ax + by + cz = q Þ ax + by + cz + d = 0, where d = - q.

The equation ax + by + cz + d = 0 is called the General Form of the Cartesian equation of a plane. Equation of a Plane Passing through a Given Point and Perpendicular to a Given Vector Vector Form THEOREM 1

The vector equation of a plane passing through a point A with position ®

®

®

®

®

vector a and perpendicular to a given vector n is ( r - a ) × n = 0. PROOF

Let O be the origin, and let p be the given plane. ® Let A be a given point on the plane with position vector a . ®

Let P be an arbitrary point on the plane with position vector r .

SSS Mathematics for Class 12 1170

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Senior Secondary School Mathematics for Class 12 ®

¾®

®

¾®

Then, OA = a , and OP = r . ¾®

¾®

®

¾®

®

AP = (OP - OA) = ( r - a ).

\

¾®

®

Let ON = n be normal to the plane. ®

¾®

Now, AP lies in the plane and n is normal to the plane. ®

¾®

AP ^ n Þ

\

¾®

®

®

®

®

AP × n = 0 Þ ( r - a ) × n = 0.

Hence, the equation of the plane passing through the point with ®

®

®

®

®

position vector a and perpendicular to n is ( r - a ) × n = 0. COROLLARY

The vector equation of the plane passing through the origin and ®

® ®

perpendicular to n is r × n = 0. Cartesian Form THEOREM 2

The Cartesian equation of a plane passing through a point A( x1 , y1 , z1) and perpendicular to a line having d.r.’s a, b, c is a( x - x1) + b( y - y1)+ c(z - z1) = 0.

®

PROOF

^

^

^

®

^

^

®

^

^

^

^

Let r = x i + y j + z k , a = x1 i + y1 j + z1 k and n = A i + B j + C k . ®

®

®

Substituting these values in ( r - a ) × n = 0, we get ^

^

^

^

^

^

{( x - x1) i + ( y - y1) j + (z - z1) k } × ( A i + B j + C k ) = 0 Þ a( x - x1) + b( y - y1) + c(z - z1) = 0. This is known as the equation of a plane in one-point form. SUMMARY

^

1. (i) Let a plane be at a distance p form the origin and let n be a unit vector perpendicular to the plane. ® ^

Then, equation of the plane is r × n = p. (ii) The Cartesian form of equation of this plane is lx + my + nz = p , where l , m , n are the d.c.’s of normal to the plane. ® ®

(iii) The general equation of a plane is r × n = q. (iv) Its Cartesian form is ax + by + cz + d = 0, where a, b, c are the ®

d.r.’s of n . 2. (i) The vector equation of a plane passing through a point A with ®

®

®

®

®

position vector a and perpendicular to n is ( r - a ) × n = 0. (ii) The Cartesian equation of a plane passing through a point A( x1 , y1 , z1) and perpendicular to a line having d.r.’s a, b, c is a( x - x1) + b( y - y1) + c(z - z1) = 0.

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SOLVED EXAMPLES EXAMPLE 1

Find the vector equation of a plane which is at a distance of 5 units from ^

the origin and which has j as the unit vector normal to it. ® ^

SOLUTION

Clearly, the required equation of the plane is r × j = 5.

EXAMPLE 2

Find the vector equation of a plane which is at a distance of 6 units from ^

^

^

the origin and which is normal to the vector ( i + 2 j - 2 k ). ®

SOLUTION

^

^

^

^

^

^

Here n = ( i + 2 j - 2 k ) and p = 6. æ 1 ^ 2 ^ 2 ^ö = çç i + j - k ÷÷ × 3 3 ø 12 + 22 + ( -2) 2 è 3 So, the vector equation of the plane is ® ^ ® æ1^ 2 ^ 2 ^ö r × n = p Þ r × çç i + j - k ÷÷ = 6 3 3 ø è3 ( i + 2 j - 2k)

^

\

n=

®

^

^

^

Þ r = ( i + 2 j - 2 k ) = 18. Hence, the required vector equation of the plane is ®

^

^

^

r × ( i + 2 j - 2 k ) = 18.

EXAMPLE 3

Find the vector equation of a plane passing through a point having ^

^

^

^

vector ( 2 i - j + k )

position ^

and

perpendicular to

the

vector

^

( 4 i + 2 j - 3 k ). Also reduce it to Cartesian form. ®

SOLUTION

^

^

®

^

^

^

^

Here a = ( 2 i - j + k ) and n = ( 4 i + 2 j - 3 k ). ®

®

®

Clearly, the required vector equation is ( r - a ) × n = 0 Þ Þ

® ®

® ®

r ×n = a ×n

®

^

^

^

®

^

^

^

^

^

^

^

^

^

^

^

r ×(4 i + 2 j - 3 k) = (2 i - j + k) ×(4 i + 2 j - 3 k) = ( 8 - 2 - 3) = 3. Hence, the vector equation of the given plane is r ×(4 i + 2 j - 3 k) = 3 Reduction to Cartesian Form: ®

... (i)

^

Putting r = ( x i + y j + z k ), we get ^

^

^

^

^

^

( x i + y j + z k ) × ( 4 i + 2 j - 3 k ) = 3 Þ 4x + 2y - 3z = 3. Hence, the equation of the given plane in Cartesian form is 4x + 2y - 3z = 3. ®

^

^

^

EXAMPLE 4

Find a unit vector normal to the plane r × ( 2 i - 3 j + 6 k ) + 14 = 0.

SOLUTION

The equation of the given plane is

SSS Mathematics for Class 12 1172

1172

Senior Secondary School Mathematics for Class 12 ®

^

^

^

®

^

^

^

r × ( 2 i - 3 j + 6 k ) + 14 = 0

Û Þ Þ

®

r × ( 2 i - 3 j + 6 k ) = - 14 Þ ®

® ®

^

^

^

^

+ ( - 6)

2

r × ( -2 i + 3 j - 6 k ) = 14

^

^

r × n = 14, where n = ( -2 i + 3 j - 6 k )

®

r ×

®

n

=

®

®

14

, where| n | = ( -2)

®

2

+ 3

2

=7

|n | |n | ^

^

^

® æ 2^ ( -2 i + 3 j - 6 k ) 14 3 ^ 6 ^ö Þ r × çç - i + j - k ÷÷ = 2. = 7 7 7 ø 7 7 è Hence, the unit vector normal to the given plane is ^ æ 2^ 3 ^ 6 ^ö n = çç - i + j - k ÷÷ . 7 7 ø è 7

Þ

EXAMPLE 5

®

r ×

Find the direction cosines of the perpendicular from the origin to the ®

^

^

^

plane r × ( 6 i - 3 j - 2 k ) + 3 = 0. SOLUTION

Clearly, we have to find the direction cosines of the normal to the given plane. The given equation may be written as ®

^

^

®

^

^

^

^

r × ( 6 i - 3 j - 2 k ) = -3 Þ r × ( -6 i + 3 j + 2 k ) = 3

Þ Þ

®

® ®

®

®

r ×

n

=

®

| n| Þ

^

^

^

r × n = 3, where n = ( - 6 i + 3 j + 2 k )

®

r ×

®

3 ®

, where| n | = ( - 6)

2

+ 3

2

+ 22 =7

|n | ^

^

^

® æ 6^ (- 6 i + 3 j + 2 k) 3 3 ^ 2 ^ö 3 = Þ r × çç - i + j + k ÷÷ = × 7 7 ø 7 7 7 è 7

Hence, the direction cosines of the normal to the given plane are æ 6 3 2ö ç- , , ÷ . è 7 7 7ø EXAMPLE 6

Find the Cartesian equation of a plane whose vector equation is ®

^

^

^

r × ( 2 i + 5 j - 4 k ) = 3.

SOLUTION

We have ®

^

^

^

^

^

^

^

^

^

r × (2 i + 5 j - 4 k ) = 3 Û (x i + y j + z k) × (2 i + 5 j - 4 k) = 3 Û 2x + 5 y - 4z = 3. Hence, the required equation is 2x + 5 y - 4z = 3.

SSS Mathematics for Class 12 1173

The Plane

1173

EXAMPLE 7

Find the vectors equation of a plane whose Cartesian equation is 2x - 3 y + 4z + 6 = 0. Find the direction cosines of the normal to the plane and its distance from the origin.

SOLUTION

The given equation of the plane is 2x - 3 y + 4z = -6 Þ - 2x + 3 y - 4z = 6.

... (i)

®

If r is the position vector of any point P( x , y , z) on plane (i), then we may write it as ^

^

^

^

^

^

( x i + y j + z k ) × ( -2 i + 3 j - 4 k ) = 6 Þ Þ

®

^

^

^

r × ( -2 i + 3 j - 4 k ) = 6 ®

® ®

^

^

^

r × n = 6, where n = ( -2 i + 3 j - 4 k )

®

®

n

®

6

, where|n| = ( -2) 2 + 3 2 + ( -4) 2 = 29 |n| |n| ® æ -2 ^ 3 ^ 4 ^ö 6 Þ r × çç i + j k ÷÷ = × 29 29 ø 29 è 29 -2 3 -4 \ d.c.’s of normal to the given plane are , , , and its 29 29 29 6 distance from the origin is × 29 Þ

r ×

®

=

®

EXAMPLE 8

Find the vector equation of the plane whose Cartesian equation is 5 y + 8 = 0. Find the direction cosines of the normal to the plane and its distance from the origin.

SOLUTION

The given equation of the plane is 5 y = -8 Þ - 5 y = 8 Þ 0 × x - 5 y + 0z = 8.

... (i)

®

Let r be the position vector of any point P( x , y , z) on plane (i). Then, the above equation may be written as ®

^

^

® ®

^

®

^

^

^

r × ( 0 i - 5 j + 0 k ) = 8 Þ r × n = 8, where n = 0 i - 5 j + 0 k

Þ

Þ

®

r ×

®

n

8

®

, where|n|= 02 + ( -5) 2 + 02 = 5 |n| |n| ®

=

®

® ^ ^ ^ æ 0 ^ 5 ^ 0 ^ö 8 8 r × çç i - j + k ÷÷ = Þ r × ( 0 i - j + 0 k ) = × 5 5 ø 5 5 è5

®

\ d.c.’s of the normal to the plane are 0, –1, 0 and its distance 8 from the origin is units. 5

SSS Mathematics for Class 12 1174

1174 EXAMPLE 9

Senior Secondary School Mathematics for Class 12

Find the Cartesian form of the equation of the plane ®

^

^

^

r = (s - 2t) i + ( 3 - t) j + ( 2s + t) k .

SOLUTION

We have ®

^

^

^

r = (s - 2t) i + ( 3 - t) j + ( 2s + t) k ^

^

^

^

^

^

Û ( x i + y j + z k ) = (s - 2t) i + ( 3 - t) j + ( 2s + t) k Û x = s - 2t , y = 3 - t and z = 2s + t Û x - 2y = (s - 6) and y + z = ( 3 + 2s) [eliminating t] 1 [equating the values of s] Û x - 2y + 6 = ( y + z - 3) 2 Û 2x - 4y + 12 = y + z - 3 Û 2x - 5 y - z + 15 = 0. This is the required Cartesian form of the equation of the given plane. EXAMPLE 10

Find the vector and Cartesian equations of the plane which passes through the point (5 , 2, - 4) and perpendicular to the line with direction ratios 2, 3 , - 1.

SOLUTION

The plane passes through the point A(5, 2, –4) whose position vector

®

^

^

^

a = (5 i + 2 j - 4 k )

is

®

and

the

^

^

normal

vector

®

n

^

perpendicular to the plane is n = ( 2 i + 3 j - k ). So, the vector equation of the plane is ® ®

® ®

r ×n = a×n

Þ

®

^

^

^

^

^

^

^

^

^

r × ( 2 i + 3 j - k ) = (5 i + 2 j - 4 k ) × ( 2 i + 3 j - k ) = (10 + 6 + 4) = 20. Hence, the required vector equation of the plane is ®

^

^

^

r = ( 2 i + 3 j - k ) = 20.

... (i)

Cartesian Form: ®

^

^

^

Taking r = ( x i + y j + z k ), equation (i) may be written as ^

^

^

^

^

^

( x i + y j + z k ) × ( 2 i + 3 j - k ) = 20 Þ 2x + 3 y - z = 20. Hence, the required Cartesian equation is 2x + 3 y - z = 20. EXAMPLE 11

The foot of the perpendicular drawn from the origin to a plane is ( 4, - 2, - 5). Find the equation of the plane in (i) vector form, [CBSE 2007] (ii) Cartesian form.

SOLUTION

Let O( 0, 0, 0) be the origin and P( 4, - 2, - 5) be the foot of the perpendicular drawn from O to the given plane. (i) Clearly, the required plane passes through the point P and it is perpendicular to OP.

SSS Mathematics for Class 12 1175

The Plane ®

^

^

1175 ®

^

¾®

^

^

^

\ a = ( 4 i - 2 j - 5 k ) and n = OP ( 4 i - 2 j - 5 k ). So, the required equation of the plane is ® ®

® ®

r ×n = a×n

Þ

®

^

^

^

^

^

^

r = (4 i - 2 j - 5 k) ^

^

^

= (4 i - 2 j - 5 k) ×(4 i - 2 j - 5 k) = (16 + 4 + 25) = 45. Hence, the vector equation of the plane is ®

^

^

^

^

^

r × ( 4 i - 2 j - 5 k ) = 45 ®

... (i)

^

(ii) Putting r = ( x i + y j + z k ) in (i), we get ^

Þ

^

^

^

^

^

( x i + y j + z k ) × ( 4 i - 2 j - 5 k ) = 45 4x - 2y - 5z = 45 , which is the required equation of the plane in Cartesian form.

EXAMPLE 12

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 3 y + 4z - 6 = 0.

SOLUTION

The equation of the given plane is 0x + 3 y + 4z - 6 = 0.

... (i)

D.r.’s of normal to the given plane are 0, 3, 4. The equation of the line through the origin O and perpendicular to the plane (i) is given by x y z ... (ii) = = = l (say). 0 3 4 The general point on (ii) is given by N( 0, 3 l , 4l). If this point lies on plane (i), we have ( 0 ´ 0) + ( 3 ´ 3 l) + ( 4 ´ 4l) = 6 Þ 25 l = 6 Þ l =

EXAMPLE 13

SOLUTION

6 × 25

Hence, the foot of the perpendicular drawn from the origin to the æ 18 24 ö given plane is N ç 0, , ÷. è 20 25 ø x+1 y+ 2 z+ 3 Find the coordinates of the point where the line = = 2 3 4 meets the plane x + y + 4z = 6. [CBSE 2006, ’08] The equation of the given line is x+1 y+ 2 z+ 3 = = = l (say). 2 3 4

... (i)

SSS Mathematics for Class 12 1176

1176

Senior Secondary School Mathematics for Class 12

The equation of the given plane is ... (ii) x + y + 4z = 6 The general point on the line (i) is ( 2l - 1, 3 l - 2, 4l - 3). Let the given line (i) meet the given plane (ii) at the point P( 2l - 1, 3 l - 2, 4l - 3) for some value of l. Then, ( 2l - 1) + ( 3 l - 2) + 4( 4l - 3) = 6 Þ 21l = 21 Þ l = 1. Hence, the required point of intersection of the given line (i) and the given plane (ii) is P( 2 ´ 1 - 1, 3 ´ 1 - 2, 4 ´ 1 - 3), i.e. P(1, 1, 1). EXAMPLE 14

Find the coordinates of the point where the line through the points [CBSE 2012] A( 3 , 4, 1) and B(5 , 1, 6) crosses the XY-plane.

SOLUTION

The equation of the line through the points A( 3 , 4, 1) and B(5 , 1, 6) is given by x-3 y-4 z -1 x - 3 y - 4 z -1 ... (i) = = Þ = = × (5 - 3) (1 - 4) ( 6 - 1) 2 -3 5 The equation of the XY-plane is z = 0. Since the line (i) meets the XY-plane, we have x - 3 y - 4 0 -1 x - 3 -1 y - 4 -1 and = = Þ = = 2 -3 5 2 5 -3 5 2 ö 13 3 ö 23 æ æ and z = 0. , y = ç4 + ÷ = Þ x = ç3- ÷ = 5ø 5 5ø 5 è è Hence, the given line crosses the XY-plane at the point æ 13 23 ö Pç , , 0÷ . ø è5 5

EXAMPLE 15

Find the distance of the point P( -1, - 5 , - 10) from the point of intersection of the line joining the points A( 2, - 1, 2) and B(5 , 3 , 4) with [CBSE 2014] the plane x - y + z = 5.

SOLUTION

The equation of the given plane is ... (i) x-y+z =5 The equation of the line AB is x-2 y+1 z-2 x-2 y+1 z-2 = = Þ = = = l (say). ... (ii) (5 - 2) ( 3 + 1) ( 4 - 2) 3 4 2 For some value of l, let the point Q( 3 l + 2, 4l - 1, 2l + 2) be the point of intersection of the plane (i) and the line (ii). Then, ( 3 l + 2) - ( 4l - 1) + (2l + 2) = 5 Þ l = 0. So, the required point Q is Q( 2, - 1, 2). \

PQ = ( 2 + 1) 2 + ( -1 + 5) 2 + ( 2 + 10) 2 =

3 2 + 42 + (12) 2 = 9 + 16 + 144

= 169 = 13 units. Hence, the required distance is 13 units.

SSS Mathematics for Class 12 1177

The Plane

1177

EXAMPLE 16

Find the coordinates of the point where the line through ( 3 , - 4, - 5) and ( 2, - 3 , 1) crosses the plane passing through the points ( 2, 2, 1), ( 3 , 0, 1) [CBSE 2013] and ( 4, - 1, 0).

SOLUTION

We know that the equation of a plane passing through the points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) is given by x - x1 y - y1 z - z1 x 2 - x1 y 2 - y1 z2 - z1 = 0. x 3 - x1

y 3 - y1

z 3 - z1

\ the equation of the plane passing through the points A( 2, 2, 1), B( 3 , 0, 1) and C( 4, - 1, 0) is given by x - 2 y - 2 z -1 x - 2 y - 2 z -1 Þ

3-2 4-2

0-2 -1 - 2

1 -1 = 0 Þ 0 -1

1 2

-2 -3

0 -1

=0

Þ ( x - 2)( 2 - 0) - ( y - 2)( -1 - 0) + (z - 1)( -3 + 4) = 0 ... (i) Þ 2( x - 2) + 1 × ( y - 2) + 1 × (z - 1) = 0 Þ 2x + y + z = 7. The equation of the line passing through the points P( 3 , - 4, - 5) and Q( 2, - 3 , 1) is given by x-3 y+4 z+5 x- 3 y+ 4 z+5 = = Þ = = = l (say). ( 2 - 3) ( -3 + 4) (1 + 5) -1 1 6 Any general point on line PQ is given as ( - l + 3 , l - 4, 6l - 5). For some value of l, let the point R(-l + 3, l - 4, 6l - 5) lie on the plane (i). Then, 2( - l + 3) + ( l - 4) + (6l - 5) = 7 Þ 5l = 10 Þ l = 2. Putting l = 2, we get the point of intersection of the given line and the given plane as R( -2 + 3 , 2 - 4, 12 - 5), i.e., R(1, –2, 7). EXAMPLE 17

Find the distance of the point (1, - 2, 3) from the plane x - y + z = 5 x -1 y - 3 z + 2 measured parallel to the line [CBSE 2013C] = = × 2 3 -6

SOLUTION

The equation of the given plane is .... (i) x-y+z =5 The equation of the given line is x -1 y - 3 z + 2 ... (ii) = = × 2 3 -6 The equation of the line passing through the point P(1, - 2, 3) and parallel to line (ii) is given by x -1 y + 2 z - 3 = = = l (say). 2 3 -6 The general point on line (iii) is ( 2l + 1, 3 l - 2, 6l + 3).

... (iii)

SSS Mathematics for Class 12 1178

1178

Senior Secondary School Mathematics for Class 12

For some value of l, let the point Q(2l + 1, 3 l - 2, - 6l + 3) lie on plane (i). Then, we have 1 (2l + 1) - (3l - 2) + (-6l + 3) = 5 Þ -7 l = -1 Þ l = × 7 So, the coordinates of Q are 3 -6 ö æ2 æ 9 -11 15 ö Q ç + 1, - 2, + 3 ÷ , i.e., Q ç , , ÷. 7 7 7 ø è è7 7 7 ø 2

2

ö æ 15 ö æ -11 æ9 ö Distance PQ = ç - 1÷ + ç + 2÷ + ç - 3 ÷ 7 7 7 ø è ø è ø è 2

2

2

2

1 æ 2ö æ 3ö æ -6 ö 4 + 9 + 36 = ç ÷ +ç ÷ +ç ÷ = 7 è7 ø è7ø è7 ø =

1 æ1 ö ´ 49 = ç ´ 7 ÷ = 1 unit. 7 è7 ø

Hence, the required distance is 1 unit. EXAMPLE 18

Find the distance of the point ( 3 , 4, 5) from the plane x + y + z = 2, [CBSE 2012C] measured parallel to the line 2x = y = z.

SOLUTION

The equation of the given plane is x + y + z = 2. The equation of the given line is x y z 2x = y = z Þ = = × 1 2 2

... (i)

... (ii)

The equation of the line passing through the point P( 3 , 4, 5) and parallel to the given line (ii) is x - 3 y - 4 z -5 = = = l (say). 1 2 2

... (iii)

The general point on line (iii) is ( l + 3 , 2l + 4, 2l + 5). For some value of l, let the point Q( l + 3, 2l + 4, 2l + 5) lie on the plane (i). Then, (l + 3) + (2l + 4) + (2l + 5) = 2 Þ

5l = -10 Þ l = -2.

\

coordinates of Q are Q( -2 + 3 , - 4 + 4, - 4 + 5), i.e., Q(1, 0, 1).

Distance PQ = ( 3 - 1) 2 + ( 4 - 0) 2 + (5 - 1) 2 = 22 + 42 + 42 = 4 + 16 + 16 =

36 = 6 units.

Hence, the required distance is 6 units.

SSS Mathematics for Class 12 1179

The Plane EXAMPLE 19

SOLUTION

Find the distance of x + 2 2y + 3 3z + 4 = = , 3 4 5 4x + 12y - 3z + 1 = 0.

the

1179

point ( -2, 3 , - 4)

measured

parallel

from to

the

the

line plane

[CBSE 2008]

The equation of the given plane is 4x + 12y - 3z + 1 = 0.

... (i)

Let P( -2, 3 , - 4) be the given point. Let l be the given line whose equation is x + 2 2y + 3 3z + 4 ... (ii) = = = l (say). 3 4 5 4l - 3 5 l - 4 ö æ The general point on this line is ç 3 l - 2, , ÷. 2 3 ø è 4l - 3 5 l - 4 ö æ For some value of l, let the point Q ç 3 l - 2, , ÷ lie on 2 3 ø è the line (ii) such that PQ is parallel to the given plane (i). ö ö æ 4l - 3 æ 5l - 4 D.r.’s of PQ are ( 3l - 2 + 2), ç - 3÷ , ç + 4÷ , 2 3 ø ø è è 4l - 9 5 l + 8 i.e., 3l, , × 2 3 D.r.’s of the normal to the given plane are 4, 12, –3. Now, PQ is parallel to the given plane (i). Þ PQ is perpendicular to the normal to the the plane (i). ( 4l - 9) (5 l - 8) -3´ =0 Þ ( 4 ´ 3 l) + 12 ´ 2 3 Þ 12l + (24l - 54) - (5l - 8) = 0 Þ 31l = 62 Þ l = 2. æ 5 ö \ coordinates of Q are Q ç 4, , 2÷ è 2 ø 2

2

æ5 ö æ -1 ö Þ PQ = ( 4 + 2) 2 + ç - 3 ÷ + ( 2 + 4) 2 = 62 + ç ÷ + 62 è2 ø è 2ø = 72 +

1 = 4

289 17 units = 8.5 units. = 4 2

Hence, the required distance of the given point from the given line is 8.5 units.

SSS Mathematics for Class 12 1180

1180

Senior Secondary School Mathematics for Class 12

EXAMPLE 20

Find the length and the foot of the perpendicular from the point P(7 , 14, 5) to the plane 2x + 4y - z = 2. Also, find the image of the point [CBSE 2012] P in the plane.

SOLUTION

The equation of the given plane is ... (i) 2x + 4y - z = 2 The d.r.’s of the normal to this plane are 2, 4, –1. The equation of the line passing through the point P(7 , 14, 5) and perpendicular to the given plane (i) is given by x - 7 y - 14 z - 5 = = = l (say) 2 4 -1 A general point on this line is (2l + 7, 4l + 14, - l + 5). For some value of l, let the point Q(2l + 7, 4l + 14, - l + 5) lie on the plane (i). Then, 2( 2l + 7) + 4(4l + 14) - (-l + 5) = 2 Þ 21l = -63 Þ l = -3. \ the coordinates of the foot of the perpendicular PQ are Q[2 ´ ( -3) + 7 , 4 ´ ( -3) + 14, - ( -3) + 5], i.e., Q(1, 2, 8). \

PQ = (7 - 1) 2 + (14 - 2) 2 + (5 - 8) 2 = 62 + (12) 2 + ( -3) 2 =

36 + 144 + 9

= 189 = 3 21 units. Let R( x , y , z) be the image of P in the plane (i). Then, Q is the midpoint of PR. 7+x 14 + y 5+z \ = 1, = 2 and = 8 Þ x = -5 , y = -10, z = 11. 2 2 2 Hence, the image of P(7, 14, 5) in the given plane is R( -5 , - 10, 11). EXAMPLE 21

Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38.

SOLUTION

The equation of the given plane is ... (i) x + 2y + 4z = 38. The d.r.’s of the normal to this plane are 1, 2, 4. The equation of the line passing through the point P(1, 2, 3) and perpendicular to the given plane (i) is given by ( x - 1) ( y - 2) (z - 3) = = = l (say). 1 2 4 A general point on this line is ( l + 1, 2l + 2, 4l + 3). For some value of l, let the point Q(l + 1, 2l + 2, 4l + 3) lie on the plane (i). Then,

SSS Mathematics for Class 12 1181

The Plane

1181

(l + 1) + 2(2l + 2) + 4(4l + 3) = 38 Þ 21l = 21 Þ l = 1. \ the coordinates of the foot of the perpendicular PQ are Q(1 + 1, 2 + 2, 4 + 3), i.e., Q( 2, 4, 7). Let R( x , y , z) be the image of P(1, 2, 3) in the given plane. Then, Q is the midpoint of PR. 1+ x 2+ y 3+z = 2, = 4 and = 7 Þ x = 3 , y = 6, z = 11. \ 2 2 2 Hence, the image of P(1, 2, 3) in the given plane is R( 3 , 6, 11).

EXERCISE 28B 1. Find the vector and Cartesian equations of a plane which is at a distance of ^

5 units from the origin and which has k as the unit vector normal to it. 2. Find the vector and Cartesian equations of a plane which is at a distance of 7 units from the origin and whose normal vector from the origin is ^

^

^

( 3 i + 5 j - 6 k ). 3. Find the vector and Cartesian equations of a plane which is at a distance of 6 from the origin and whose normal vector from the origin is 29 ^

^

^

( 2 i - 3 j + 4 k ). 4. Find the vector and Cartesian equations of a plane which is at a distance of 6 units from the origin and which has a normal with direction ratios 2, –1, –2. 5. Find the vector and Cartesian equations of a plane which passes through ^

^

^

the point (1, 4, 6) and normal vector to the plane is ( i - 2 j + k ). 6. Find the length of perpendicular from the origin to the plane ®

^

^

^

r × ( 3 i - 12 j - 4 k ) + 39 = 0. Also write the unit normal vector from the origin to the plane. 7. Find the Cartesian equation of the plane whose vector equation is ®

^

^

^

r × ( 3 i + 5 j - 9 k ) = 8.

8. Find the vector equation of a plane whose Cartesian equation is 5 x - 7 y + 2z + 4 = 0. 9. Find a unit vector normal to the plane x - 2y + 2z = 6. 10. Find the direction cosines of the normal to the plane 3 x - 6y + 2z = 7. 11. For each of the following planes, find the direction cosines of the normal to the plane and the distance of the plane from the origin: (i) 2x + 3 y - z = 5 (ii) z = 3 (iii) 3 y + 5 = 0

SSS Mathematics for Class 12 1182

1182

Senior Secondary School Mathematics for Class 12

12. Find the vector and Cartesian equations of the plane passing through the point (2, –1, 1) and perpendicular to the line having direction ratios 4, 2, –3. 13. Find the coordinates of the foot of the perpendicular drawn from the origin to the plane (i) 2x + 3 y + 4z - 12 = 0 (ii) 5 y + 8 = 0. 14. Find the length and the foot of perpendicular drawn from the point (2, 3, 7) to the plane 3 x - y - z = 7. 15. Find the length and the foot of the perpendicular drawn from the point ®

^

^

^

(1, 1, 2) to the plane r × ( 2 i - 2 j + 4 k ) + 5 = 0. 16. From the point P(1, 2, 4), a perpendicular is drawn on the plane 2x + y - 2z + 3 = 0. Find the equation, the length and the coordinates of the [CBSE 2008] foot of the perpendicular. 17. Find the coordinates of the foot of the perpendicular and the perpendicular distance from the point P( 3 , 2, 1) to the plane 2x - y + z + 1 = 0. [CBSE 2010] Find also the image of the point P in the plane. 18. Find the coordinates of the image of the point P(1, 3 , 4) in the plane [CBSE 2008] 2x - y + z + 3 = 0. 19. Find the point where the line

x -1 y - 2 z + 3 meets the plane = = 2 -3 4

2x + 4y - z = 1. 20. Find the coordinates of the point where the line through (3, –4, –5) and ( 2, - 3 , 1) crosses the plane 2x + y + z = 7. 21. Find the distance of the point (2, 3, 4) from the plane 3 x + 2y + 2z + 5 = 0, x+ 3 y-2 z measured parallel to the line [CBSE 2007, ’09C] = = × 3 6 2 22. Find the distance of the point (0, –3, 2) from the plane x + 2y - z = 1, x+1 y+1 z measured parallel to the line = = × 3 2 3 23. Find the equation of the line passing through the point P( 4, 6, 2) and the x -1 y z + 1 point of intersection of the line and the plane x + y - z = 8. = = 3 2 7 [CBSE 2008]

24. Show that the distance of the point of intersection of the line x-2 y+1 z-2 and the plane x - y + z = 5 from the point ( -1, - 5 , - 10) = = 3 4 12 is 13 units. 25. Find the distance of the point A( -1, - 5 , - 10) from the point of intersection of ® ^

the ^

line ^

r × ( i - j + k ) = 5.

®

^

^

^

^

^

^

r = ( 2 i - j + 2 k ) + l( 3 i + 4 j + 2 k )

and

the

plane

[CBSE 2011, ’14]

HINT: Convert the equations of the line and the plane to Cartesian form.

SSS Mathematics for Class 12 1183

The Plane

1183

26. Prove that the normals to the planes 4x + 11y + 2z + 3 = 0 and 3 x - 2y + 5z = 8 are perpendicular to each other. ®

^

^

^

^

^

^

27. Show that the line r = ( 2 i - 2 j + 3 k ) + l( i - j + 4 k ) is parallel to the ® ^

^

^

plane r × ( i + 5 j + k ) = 7. 28. Find the equation of a plane which is at a distance of 3 3 units from the origin and the normal to which is equally inclined to the coordinate axes. ®

29. A vector n of magnitude 8 units is inclined to the x-axis at 45°, y-axis at 60° and an acute angle with the z-axis. If a plane passes through a point ®

( 2 , - 1, 1) and is normal to n , find its equation in vector form. ^

^

^

30. Find the vector equation of a line passing through the point ( 2 i - 3 j - 5 k ) ®

^

^

^

and perpendicular to the plane r × ( 6 i - 3 j + 5 k ) + 2 = 0. Also, find the point of intersection of this line and the plane. [CBSE 2006C]

ANSWERS (EXERCISE 28B) ® ^

®

1. r × k = 5 , z = 5 ®

^

®

^

®

^

^

^

^

2. r × ( 3 i + 5 j - 6 k) = 7 70, 3 x + 5 y - 6z = 7 70

^

^

3. r × ( 2 i - 3 j + 4 k ) = 6, 2x - 3 y + 4z = 6 ^

^

4. r × ( 2 i - j - 2 k ) = 18, 2x - y - 2z = 18 ^

^

5. r × ( - i + 2 j - k ) = 1, x - 2y + z + 1 = 0 ^ æ -3 ^ 12 ^ 4 ^ö 6. p = 3 , n = çç i + j+ k÷ 13 13 ÷ø è 13 ®

^

^

^

8. r × ( -5 i + 7 j - 2k ) = 4 æ 2 11. (i) ç , è 14 ®

^

9.

-1 ö 5 3 , ÷, 14 14 ø 14 ^

7. 3 x + 5 y + 9z = 8 1 ^ 2^ 2^ i - j+ k 3 3 3

(ii) (0, 0, 1), 3

10.

3 -6 2 , , 7 7 7

(iii) (0, –1, 0),

5 3

^

12. r × ( 4 i + 2 j - 3 k ) = 3 , 4x + 2y - 3z = 3 æ 24 36 48 ö 13. (i) ç , , ÷ è 29 29 29 ø 15.

æ -8 ö (ii) ç 0, , 0÷ è 5 ø

13 6 æ -1 25 -1 ö units, ç , , ÷ 12 è 12 12 6 ø

17. (1, 3, 0), 6 units, (–1, 4, –1)

16.

14.

11 units, (5, 2, 6)

x -1 y - 2 z - 4 1 æ 11 19 34 ö = = ; unit; ç , , ÷ 2 1 -2 3 è 9 9 9ø

18. (–3, 5, 2)

19. (3, –1, 1)

20. (1, –2, 7)

SSS Mathematics for Class 12 1184

1184

Senior Secondary School Mathematics for Class 12

21. 7 units

22. 10 units

x-4 y-6 z-2 = = 1 2 2

23.

®

^

^

25. 13 units

^

29. r × ( 2 i + j + k ) = 2

28. x + y + z = 9

® ^ ^ ^ ^ ^ ^ æ 76 -108 -34 ö 30. r × ( 2 i - 3 j - 5 k ) + l( 6 i - 3 j + 5 k ), ç , , ÷ 4 ø è 35 35

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28B) ®

^

^

^

6. Given equation in standard form is r × ( -3 i + 12 j + 4 k ) = 39. ®

® ®

^

^

^

This is r × n = p , where n = ( -3 i + 12 j + 4 k ) and p = 39. ®

|n|= ( -3 ) 2 + ( 12 ) 2 + 4 2 = 9 + 144 + 16 = 169 = 13. \

®

®

n

r ×

®

p

=

=

®

|n| |n|

39 = 3. 13

æ -3 ^ 12 ^ 4 ^ö r × çç i + j + k ÷ = 3. 13 13 ÷ø è 13

®

^

Hence, p = 3 and n =

-3 ^ 12 ^ 4 ^ i + j + k. 13 13 13 ®

^

^

^

9. Given equation in vector form is r × ( i - 2 j + 2 k ) = 6. ^ ^ ^ Here,|i - 2 j + 2 k|= 12 + ( -2 ) 2 + 2 2 = 9 = 3.

æ 1 ^ 2 ^ 2 ^ö 6 r × çç i - j + k ÷÷ = = 2. 3 3 ø 3 è3

®

æ 1 ^ 2 ^ 2 ^ö Required unit vector = çç i - j + k ÷÷ . 3 3 ø è3 ®

^

^

^

10. Given equation in vector form is r × ( 3 i - 6 j + 2 k ) = 7 . ^

^

^

| 3 i - 6 j + 2 k|= \ \

3 2 + ( -6 ) 2 + 2 2 = 49 = 7 .

æ 3 ^ 6 ^ 2 ^ö r × çç i - j + k ÷÷ = 1. 7 7 ø è7

®

d.c.’s of the normal to the plane are

3 -6 2 , , × 7 7 7

®

^

^

^

11. (i) Given equation in vector form is r × ( 2 i + 3 j - k ) = 5. ^

^

^

|2 i + 3 j - k|= 2 2 + 3 2 + ( -1) 2 = 14 . \

æ 2 ^ r × çç i + è 14

®

3 ^ j 14

1 14

^ö 5 k ÷÷ = × 14 ø

SSS Mathematics for Class 12 1185

The Plane ^

d.c.’s of n are

2 , 14

1185

5 3 -1 and p = × , 14 14 14 ®

^

^

^

(ii) Given equation in vector form is r × ( 0 i + 0 j + 1 k ) = 3. ® ^

This is r × n = p. ^

D.c.’s of n are 0, 0, 1 and p = 3. ®

^

^

^

(iii) Given equation in vector form is r × ( 0 i - 1 × j + 0 × k ) =

5 3

5ù é êëQ - y = 3 úû .

^ ^ ^ Clearly, |0 i - 1 × j + 0 × k |= 0 2 + ( -1) 2 + 0 2 = 1.

\

® ^

^

^

^

^

r × n = p, where n = ( 0 i - 1 × j + 0 k ) and p = ^

D.c.’s of n are 0, –1, 0 and p =

5 × 3

5 × 3

12. Let the equation of the plane be a( x - 2 ) + b( y + 1) + c (z - 1) = 0 , where a = 4, b = 2 and c = -3. So, the required equation is 4( x - 2 ) + 2( y + 1) - 3(z - 1) = 0 Þ 4 x + 2 y - 3z = 3. ®

^

^

^

In vector form, we have r × ( 4 i + 2 j - 3 k ) = 3. 13. (i) Given equation is 2 x + 3 y + 4z = 12.

... (i)

D.r.’s of normal to the plane are 2, 3, 4. Equation of the line through O( 0 , 0 , 0 ) and perpendicular to plane (i) is x y z = = = l (say). 2 3 4

... (ii)

A general point on (ii) is N( 2 l, 3 l, 4 l). If this point lies on plane (i), then ( 2 ´ 2 l) + ( 3 ´ 3 l) + ( 4 ´ 4 l) = 12 Þ 29 N = 12 Þ l =

12 × 29

æ 24 36 48 ö Hence, the required point is ç , , ÷. è 29 29 29 ø (ii) The given plane is 0 x - 5 y + 0 z = 8. D.r.’s of normal to the plane are 0, –5, 0. Equation of the line through O( 0 , 0 , 0 ) and perpendicular to plane (i) are y x z = = = l (say). 0 -5 0 A general point on this line is N( 0 , - 5 l, 0 ). If this point lies on plane (i), we have 8 -5 ´ ( -5 l) = 8 Þ 25 l = 8 Þ l = × 25 æ -8 ö Hence, the required point is N ç 0 , , 0 ÷. 5 è ø

... (i)

... (ii)

SSS Mathematics for Class 12 1186

1186

Senior Secondary School Mathematics for Class 12

21. Equation of a line through A( 2 , 3 , 4 ) and parallel to the given line is x- 2 y- 3 z- 4 = = = l (say). 3 6 2 A general point on line (i) is N( 3 l + 2 , 6 l + 3, 2 l + 4 ). For some value of l, it lies on the plane 3 x + 2 y + 2z + 5 = 0. \ 3( 3 l + 2) + 2(6l + 3) + 2(2l + 4) + 5 = 0 Þ 25l = -25 Þ l = -1. the coordinates of N are ( -3 + 2 , - 6 + 3 , - 2 + 4 ), i.e., ( -1, - 3 , 2 ). \ Now, find AN = 7 units. x- 2 y+ 1 z- 2 24. Given line is = = = l (say) 3 4 12 A general point on this line is P( 3 l + 2, 4 l - 1, 12l + 2). If this point lies on the plane x - y + z = 5, then ( 3l + 2) - (4l - 1) + (12l + 2) = 5 Þ 11l = 0 Þ l = 0. \ point P is P( 2 , - 1, 2 ) and the other point is Q( -1, - 5 , - 10 ) Find PQ = 13 units. x- 2 y+ 1 z- 2 25. The Cartesian equation of the plane is = = = l (say). 3 4 2 A general point on this line is P( 3 l + 2, 4 l - 1, 2l + 2). The Cartesian equation of the given plan is x - y + z = 5. If this point lies on the given plane, we have \ \

( 3l + 2) - (4l - 1) + (2l + 2) = 5 Þ l = 0. point P is P( 2 , - 1, 2 ) and the other point is Q( -1, - 5 , - 10 ). PQ = 13 units.

®

^

^

^

®

^

^

^

26. n1 = ( 4 i + 11 j + 2 k ) and n2 = ( 3 i - 2 j + 5 k ). \

®

®

n1 × n2 = ( 4 ´ 3 ) + 11 ´ ( -2 ) + 2 ´ 5 = 0. ®

®

Hence, n1 ^ n2. 27. D.r.’s of the line are 1, –1, 4. D.r.’s of normal to the plane are 1, 5, 1. \ ( 1 ´ 1) + ( -1) ´ 5 + ( 4 ´ 1) = 0. So, the given line is perpendicular to the normal of the given plane. Hence, the given line is parallel to the given plane. ® ^

28. Let the required equation of the plane be r × n = p , where p = 3 3. ^

^

^

^

Let n = (cos a ) i + (cos a ) j + (cos a ) k , where a is acute. Then, cos 2 a + cos 2 a + cos 2 a = 1 Þ 3 cos 2 a = 1 Þ cos 2 a = Þ cos a = \ the required equation is ® æ 1 ^ 1 ^ 1 r × çç i + j + 3 3 è 3

^ö k ÷÷ = 3 3 . ø

1 × 3

1 3

... (i)

SSS Mathematics for Class 12 1187

The Plane ®

^

^

^

^

^

^

1187 ^

^

^

Hence, r × ( i + j + k ) = 9 Þ ( x i + y j + z k ) × ( i + j + k ) = 9 Þ x + y + z = 9. ®

^

^

n

29. We know that n =

®

^

^

= ( l i + m j + n k ).

|n| 1 1 , m = cos 60 °= and let n = cos g . Then, 2 2 1 1 l 2 + m 2 + n2 = 1 Þ + + cos 2 g = 1 2 4 1 Þ cos 2 g = 4 1 Þ cos g = × 2 ® ® ® ^ ^ ^ ^ ^ ^ æ 1 ^ 1 ^ 1 ^ö n =|n|( l i + m j + n k ) Þ n = 8 çç i + j + k ÷÷ = ( 4 2 i + 4 j + 4 k ). 2 2 2 è ø

Here l = cos 45 °=

\

So, the equation of the plane is ® ®

® ®

®

^

^

^

^

^

^

^

^

^

r × n = a × n Þ r ×(4 2 i + 4 j + 4 k ) = ( 2 i - j + k )×(4 2 i + 4 j + 4k ) ®

^

^

^

Þ r × ( 4 2 i + 4 j + 4 k ) = ( 8 - 4 + 4) = 8 ®

^

^

^

Þ r × ( 2 i + j + k ) = 2. ^

^

^

30. Clearly, the required line passes through the point ( 2 i - 3 j - 5 k ) and is parallel to ^

^

^

the normal of the given plane, which is ( 6 i - 3 j + 5 k ). So, the required vector equation is ®

^

^

^

^

^

^

r = ( 2 i - 3 j - 5 k ) + l( 6 i - 3 j + 5 k ).

The Cartesian equation of the line is

x- 2 y+ 3 z+ 5 = = = k. 6 -3 5

A general point on this line is P( 6 k + 2 , - 3 k - 3 , 5 k - 5 ). For some particular value of k , let the line cut the plane 6 x - 3 y + 5z + 2 = 0. Then, 6( 6 k + 2 ) - 3( -3 k - 3 ) + 5(5 k - 5 ) + 2 = 0 Þ Þ

( 36 k + 9 k + 25 k ) = 2 Þ 70 k = 2 1 k= × 35

\ required point of intersection of the line and plane is -3 1 æ 6 æ 76 -108 -34 ö ö Pç + 2, - 3 , - 5 ÷ , i.e., P ç , , ÷. 35 7 7 ø è 35 è 35 35 ø

SSS Mathematics for Class 12 1188

1188

Senior Secondary School Mathematics for Class 12

DISTANCE OF A POINT FROM A PLANE

1.

VECTOR FORM

(i)

Let p be the length of perpendicular drawn from a point P with ®

® ®

position vector a to the plane r × n = q. Then, ® ®

p=

|a × n - q| × ® |n|

(ii) Let p be the length of perpendicular drawn from the origin to the ® ®

plane r × n = q. Then, p= 2.

|q|

× ® |n|

CARTESIAN FORM

(i)

Let p be the length of perpendicular drawn from a point P( x1 , y1 , z1) to the plane ax + by + cz + d = 0. Then, |ax1 + by1 + cz1 + d| p= × a 2 + b 2 + c2

(ii) Let p be the length of perpendicular drawn from the origin to the plane ax + by + cz + d = 0. Then, |d| p= × 2 a + b 2 + c2 3.

DISTANCE BETWEEN PARALLEL PLANES

Let ax + by + cz + d1 = 0 and ax + by + cz + d 2 = 0 be two parallel planes. Let P( x1 , y1 , z1) be any point on the first plane. Then, the length of perpendicular from P( x1 , y1 , z1) to the other plane is the required distance between these planes.

SOLVED EXAMPLES ^

EXAMPLE 1

^

^

Find the distance of the point ( i + 2 j - 3 k) from the plane ®

^

^

^

r × ( 2 i - 5 j - k) = 4.

SOLUTION

We know that the perpendicular distance of a point with position ®

® ®

® ®

vector a , from the plane r × n = q is given by p = ®

^

^

^

®

^

^

^

|a × n - q| × ® |n|

Here a = ( i + 2 j - 3 k ), n = ( 2 i - 5 j - k ) and q = 4.

SSS Mathematics for Class 12 1189

The Plane

1189

Hence, the required distance is given by ^

p=

^

^

^

^

^

|( i + 2 j - 3 k ) × ( 2 i - 5 j - k ) - 4| 22 + ( -5) 2 + ( -1) 2

|2 - 10 + 3 - 4| |- 9| 9 30 = = ´ 30 30 30 30 9 3 = 30 = 30 units. 30 10 3 Hence, the required distance is 30 units. 10 =

EXAMPLE 2

Find the distance of the point ( 2, 3 , 4) from the plane ®

^

^

^

r × ( 3 i - 6 j + 2 k ) + 11 = 0.

SOLUTION

We know that the perpendicular distance of a point with position ®

® ®

vector a , from the plane r × n = q is given by ® ®

p= ®

^

^

|a × n - q| × ® |n|

^

®

^

^

^

^

^

Here a = ( 2 i + 3 j + 4 k ), n = ( 3 i - 6 j + 2 k ) and q = -11. \ the required distance is given by ^

p=

^

^

^

|( 2 i + 3 j + 4 k ) × ( 3 i - 6 j + 2 k ) - ( -11)| 3 2 + ( -6) 2 + 22 ^

^

^

^

^

^

|( 2 i + 3 j + 4 k ) × ( 3 i - 6 j + 2 k ) + 11| = 9 + 36 + 4 |( 2 ´ 3) + 3 ´ ( -6) + ( 4 ´ 2) + 11| = 49 |6 - 18 + 8 + 11| 7 = = = 1 unit. 7 7 Hence, the required distance is 1 unit. EXAMPLE 3 SOLUTION

Find the distance of the point ( 2, 3 , - 5) from the plane x + 2y - 2z = 9. Required distance = Length of perpendicular from (2, 3, –5) to the plane x + 2y - 2z - 9 = 0 |2 + 2 ´ 3 - 2 ´ ( -5) - 9| |2 + 6 + 10 - 9| = = 9 12 + 22 + ( -2) 2 9 = 3 units. 3 Hence, the required distance is 3 units. =

SSS Mathematics for Class 12 1190

1190

Senior Secondary School Mathematics for Class 12

EXAMPLE 4

If a plane has intercepts a, b and c on the coordinate axes and is at a 1 1 1 1 distance of p untis from the origin, prove that 2 + 2 + 2 = 2 × a b c p

SOLUTION

We know that the equation of the plane which makes intercepts a, b and c on the axes is x y z x y z ... (i) + + =1 Þ + + - 1 = 0. a b c a b c It is given that p is the length of perpendicular drawn from O( 0, 0, 0) to the plane (i). |0 + 0 + 0 - 1| 1 = \ p= 1 1 1 1 1 1 + + + + a 2 b 2 c2 a 2 b 2 c2 Þ

1 a2

Hence EXAMPLE 5

SOLUTION

+

1 a2

1

+

b2

+

1 b2

1 c2

+

=

1 c2

=

1 1 1 1 1 Þ 2 + 2 + 2 = 2× p a b c p 1 p2

×

Find the distance of the point P( 6, 5 , 9) from the plane determined by the [CBSE 2009, ’10, ’12] points A( 3 , - 1, 2), B(5 , 2, 4) and C( -1, - 1, 6). Here ( x1 = 3 , y1 = -1, z1 = 2), ( x 2 = 5 , y 2 = 2, z2 = 4) and ( x 3 = -1, y 3 = -1, z 3 = 6). The equation of the plane ABC is given by x - x1 y - y1 z - z1 x 2 - x1 y 2 - y1 z2 - z1 = 0

Þ

x 3 - x1

y 3 - y1

z 3 - z1

x-3 5-3

y+1 2+1

z-2 x-3 4-2 = 0 Þ 2

-1 - 3

-1 + 1

6-2

-4

y+1 3

z-2 2 =0

0

Þ

( x - 3)(12 - 0) - ( y + 1)( 8 + 8) + (z - 2)( 0 + 12) = 0

Þ

12( x - 3) - 16( y + 1) + 12(z - 2) = 0

Þ

12x - 16y + 12z - 76 = 0 Þ 3 x - 4y + 3z - 19 = 0.

4

... (i)

Distance of the point P( 6, 5 , 9) from the plane 3 x - 4y + 3z - 19 = 0 = length of perpendicular from P( 6, 5 , 9) to the plane 3 x - 4y + 3z - 19 = 0 |( 3 ´ 6) - ( 4 ´ 5) + ( 3 ´ 9) - 19| |18 - 20 + 27 - 19| = = 34 3 2 + ( -4) 2 + 3 2

SSS Mathematics for Class 12 1191

The Plane

1191

6 34 6 3 ´ = 34 = 17 34 34 34

=

Hence, the required distance is

34 units.

3 17

34 units.

EXAMPLE 6

Find the distance between the parallel planes 2x - y + 3z + 4 = 0 and 6x - 3 y + 9z - 3 = 0.

SOLUTION

Let P( x1 y1 , z1) be any point on the plane 2x - y + 3z + 4 = 0. Then, 2x1 - y1 + 3z1 = -4.

... (i)

Length of perpendicular from P( x1 , y1 , z1) to 6x - 3 y + 9z = 0 is given by |6x1 - 3 y1 + 9z1 - 3| |3( 2x1 - y1 + 3z1) - 3| p= = 126 62 + ( -3) 2 + 92 =

| 3 ´ ( -4) - 3 | [using (i)] 3 14

=

| - 15 | 15 14 5 14 units. = ´ = 3 14 3 14 14 14

Hence, the distance between the given planes is ®

EXAMPLE 7

the

plane

5 14 units. 14 ^

^

^

Find the distance between the parallel planes r × ( 2 i - 3 j + 6 k ) = 5 ®

^

^

^

^

^

and r × ( 6 i - 9 j + 18 k ) + 20 = 0. ®

SOLUTION

^

Taking r = ( x i + y j + z k ), the Cartesian equations of the given planes are ^

^

^

^

^

^

^

^

^

^

^

^

(x i + y j + z k) ×(2 i - 3 j + 6 k) = 5 Þ 2x - 3 y + 6z - 5 = 0 and ( x i + y j + z k ) × ( 6 i - 9 j + 18 k ) + 20 = 0 Þ 6x - 9y + 18z + 20 = 0

... (i)

... (ii)

Let P( x1 , y1 , z1) be any point on (i). Then, 2x1 - 3 y1 + 6z1 = 5 ... (iii) Distance between the given planes = length of the perpendicular from P( x1 , y1 , z1) to the plane (ii) |6x1 - 9y1 + 18z1 + 20| |3( 2x1 - 3 y1 + 6z1) + 20| = = 36 + 81 + 324 62 + ( -9) 2 + (18) 2 |( 3 ´ 5) + 20| [using (iii)] 441 35 5 = = units. 21 3 =

SSS Mathematics for Class 12 1192

1192

Senior Secondary School Mathematics for Class 12

Hence, the distance between the given planes is

5 units. 3

EXAMPLE 8

Find the equations of planes parallel to the plane x - 2y + 2z = 3 which are at a unit distance from the point (1, 2, 3).

SOLUTION

The equation of the given plane is x - 2y + 2z - 3 = 0. Let the required plane parallel to the given plane be ... (i) x - 2y + 2z + l = 0 for some real number l Now, the plane (i) is at a unit distance from the point (1, 2, 3). |1 - ( 2 ´ 2) + ( 2 ´ 3) + l | = 1 Þ | l + 3 |= 3 \ 12 + ( -2) 2 + 22 Þ l + 3 = 3 or - ( l + 3) = 3 Þ l = 0 or l = -6. Hence, the equations of the required planes are x - 2y + 2z = 0 and x - 2y + 2z - 6 = 0.

EXAMPLE 9

Find the equation of the plane passing through the point ( 2, - 3 , 5) and parallel to the plane 3 x - 7 y - 2z = 5. Also, find the distance between the two planes.

SOLUTION

The given plane is 3 x - 7 y - 2z - 5 = 0. ... (i) Let the required plane parallel to the given plane be ... (ii) 3 x - 7 y - 2 z + l = 0 for some real number l. Since the plane (ii) passes through the point (2, –3, 5), we have ( 3 ´ 2) - 7 ´ ( -3) - ( 2 ´ 5) + l = 0 Þ ( 6 + 21 - 10) + l = 0 Þ l = -17. \ the equation of the required plane is 3 x - 7 y - 2z - 17 = 0. ... (iii) Now, the planes (i) and (ii) are parallel and the point (2, –3, 5) lies on (iii). \ distance between the two planes = distance between the point (2, –3, 5) and the plane (i) |( 3 ´ 2) - 7 ´ ( -3) - ( 2 ´ 5) - 5 | = 3 2 + ( -7) 2 + ( -2) 2 =

| 6 + 21 - 10 - 5 | 9 + 49 + 4

12 62 12 ´ = 62 62 62 62 6 = 62 units. 31 =

Hence, the distance between the two planes is

6 62 units. 31

SSS Mathematics for Class 12 1193

The Plane EXAMPLE 10

1193

Find the Cartesian as well as the vector equations of the planes through the ®

^

®

^

^

^

^

intersection of the planes r × ( 2 i + 6 j ) + 12 = 0 and r × ( 3 i - j + 4 k ) = 0 [CBSE 2005] which are at a unit distance from the origin. SOLUTION

The equations of the given planes are ®

^

^

r × ( 2 i + 6 j ) + 12 = 0

Þ Þ

^

^

^

^

^

( x i + y j + z k ) × ( 2 i + 6 j ) + 12 = 0 2x + 6y + 12 = 0 Þ x + 3 y + 6 = 0 ®

^

^

... (i)

^

and, r × ( 3 i - j + 4 k ) = 0 ^

^

^

^ ^

^

Þ (x i + y j + z k) ×( 3 i - j + 4 k) = 0 ... (ii) Þ 3 x - y + 4z = 0. Equation of any plane through the intersection of these planes is ( x + 3 y + 4) + l( 3 x - y + 4z) = 0 for some real number l ... (iii) Þ (1 + 3 l) x + ( 3 - l) y + 4lz + 6 = 0 Now, plane (iii) is at a unit distance from O( 0, 0, 0). |0 + 0 + 0 + 6| =1 \ (1 + 3 l) 2 + ( 3 - l) 2 + ( 4l) 2 Þ

(1 + 3 l) 2 + ( 3 - l) 2 + ( 4l) 2 = 36

Þ

(1 + 9l2 + 6l) + ( 9 + l2 - 6l) + 16l2 = 36

Þ

26l2 = 26 Þ l2 = 1 Þ l = 1 or l = -1.

Putting l = 1in (iii), we get 4x + 2y + 4 z + 6 = 0 Þ 2x + y + 2 z + 3 = 0. Putting l = -1in (iii) we get -2x + 4y - 4z + 6 = 0 Þ x - 2y + 2z - 3 = 0. Hence, the required equations of the planes in Cartesian form are 2x + y + 2z + 3 = 0 and x - 2y + 2z - 3 = 0. The vector forms of these equations are ®

^

^

®

^

^

^

^

r × ( 2 i + j + 2 k ) + 3 = 0 and r × ( i - 2 j + 2 k ) = 3.

EXAMPLE 11

If the points (1, 1, p) and ( -3 , 0, 1) be equidistant from the plane ®

^

^

^

r × ( 3 i + 4 j - 12 k ) + 13 = 0, find the values of p.

SOLUTION

The equation of the given plane in Cartesian form is ^

^

^

^

^

^

x i + y j + z k × ( 3 i + 4 j - 12 k ) + 13 = 0 ... (i) Þ 3 x + 4y - 12z + 13 = 0. It is being given that the distances of the plane (i) from each of the points A(1, 1, p) and B( -3 , 0, 1) are equal.

SSS Mathematics for Class 12 1194

1194

Senior Secondary School Mathematics for Class 12

\

|( 3 ´ 1) + ( 4 ´ 1) - (12 ´ p) + 13 | |( 3 ´ ( -3) + ( 4 ´ 0) - (12 ´ 1) + 13 | = 3 2 + 42 + ( -12) 2 3 2 + 42 + ( -12) 2 |20 - 12p| = |- 8| Þ |20 - 12p|= 8 Þ ( 20 - 12p) = 8 or -( 20 - 12p) = 8

Þ 12p = 2 or 12p = 28 7 Þ p = 1 or p = × 3 7 Hence, p = 1 or p = × 3 EXAMPLE 12

Find the equation of the plane mid-parallel 2x - 2y + z + 3 = 0 and 2x - 2y + z + 9 = 0.

to

the

planes

SOLUTION

Let the required equation of the plane be 2x - 2y + z + k = 0. This plane equidistant from each of the given planes. Let P( x1 , y1 , z1) be any point on the plane 2x - 2y + z + k = 0. ... (i) Then, 2x1 - 2y1 + z1 + k = 0 Þ 2x1 - 2y1 + z1 = - k. \ P( x1 , y1 , z1) is equidistant from the planes 2x - 2y + z + 3 = 0 and 2x - 2y + z + 9 = 0. | 2x1 - 2y1 + z1 + 3 | | 2x1 - 2y1 + z1 + 9| = \ 22 + ( -2) 2 + 12 22 + ( -2) 2 + 12 Þ |( 2x1 - 2y1 + z1) + 3| = |( 2x1 - 2y1 + z1) + 9| Þ |- k + 3| = |- k + 9| [using (i)] Þ

( - k + 3) = ( - k + 9) or - ( - k + 3) = - k + 9

Þ

k - 3 = - k + 9 Þ 2k = 12 Þ k = 6 [Q ( - k + 3) ¹ ( - k + 9)].

Hence, the required equation of the plane is 2x - 2y + z + 6 = 0. EXAMPLE 13

A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of CABC is x -2 + y -2 + z -2 = p -2.

SOLUTION

Let the equation of the variable plane be x y z … (i) + + = 1. a b c This plane meets the x-axis, y-axis and z-axis at the points A( a , 0, 0), B( 0, b , 0) and C( 0, 0, c) respectively. Let ( a , b , g) be the coordinates of the centroid of C ABC. a+ 0+ 0 0+ b+ 0 0+ 0+ c and g = Then, a = , b= 3 3 3 a b c Þ a = , b = and g = Þ a = 3 a , b = 3b and c = 3g. … (ii) 3 3 3 \ 3p = length of the perpendicular from (0, 0, 0) to the plane (i)

SSS Mathematics for Class 12 1195

The Plane

Þ

½ 0 + 0 + 0 - 1½ ½a b c ½ = 3p = 1 1 1 + + a 2 b 2 c2 1

Þ Þ Þ

a2 1

1

+

a2

+

b2

1 b2 +

1

+

c2

1

=

c2

=

1 a2

1195

+

1 1 b2

+

1 c2

1 3p

1 9p 2

1

Þ

9a 2

+

1 9b 2

+

1 9g2

=

1 9p 2

[using (ii)]

a -2 + b -2 + g -2 = p -2 .

Hence, the required locus is x -2 + y -2 + z -2 = p -2. EXAMPLE 14

A variable plane is at a constant distance p from the origin and meets the coordinate axes in A, B, C. Show that the locus of the centroid of the tetrahedron OABC is x -2 + y -2 + z -2 = 16p -2.

SOLUTION

Let the equation of the variable plane be x y z …(i) + + = 1. a b c This plane meets the x-axis, y-axis and z-axis at the points A( a , 0, 0), B( 0, b , 0) and C( 0, 0, c) respectively. Let ( a , b , g) be the coordinates of the centroid of the tetrahedron OABC. 0+ a+ 0+ 0 a 0+ 0+ b+ 0 b Then, a = = ,b = = and 4 4 4 4 0+ 0+ 0+ c c … (ii) g= = Þ a = 4a , b = 4b , c = 4g 4 4 \ Þ

Þ Þ Þ

p = distance of the plane (i) from (0, 0, 0) ½ 0 + 0 + 0 - 1½ ½a b c ½ p= 1 1 1 + + a 2 b 2 c2 1 = p 1 p2 1 a

2

1 a2

= +

+

1 16a 2 1 b

2

+

1 b2 + 1 g

2

1

+

c2

1 16b 2 =

16 p2

= +

1 p2

1 1ö æ 1 = ç 2 + 2 + 2÷ èa b c ø

1 16g 2

[using (ii)]

Þ a -2 + b -2 + g -2 = 16p -2.

Hence, the required locus is x -2 + y -2 + z -2 = 16p -2.

SSS Mathematics for Class 12 1196

1196

Senior Secondary School Mathematics for Class 12

EXERCISE 28C 1. Find

the

®

^

^

^

^

^

^

^

of

the

point ( 2 i - j - 4 k )

of

the

point ( i + 2 j + 5 k )

distance

from

the

plane

from

the

plane

^

r × ( 3 i - 4 j + 12 k ) = 9.

2. Find

the

®

^

^

distance

^

^

r × ( i + j + k ) + 17 = 0.

3. Find ®

the ^

distance

^

of

the

point

(3, 4, 5)

from

the

plane

of

the

point

(1, 1, 2)

from

the

plane

^

r × ( 2 i - 5 j + 3 k ) = 13.

4. Find ®

the ^

distance

^

^

r × ( 2 i - 2 j + 4 k ) + 5 = 0.

5. Find the distance of the point (2, 1, 0) from the plane 2x + y + 2z + 5 = 0. 6. Find the distance of the point (2, 1, –1) from the plane x - 2y + 4z = 9. 7. Show ®

^

that ^

the ^

point

®

(1, 2, 1) ^

is ^

equidistant

from

the

planes

^

r × ( i + 2 j - 2 k ) = 5 and r × ( 2 i - 2 j + k ) + 3 = 0.

8. Show that the points (–3, 0, 1) and (1, 1, 1) are equidistant from the plane 3 x + 4y - 12z + 13 = 0. 9. Find the distance between the parallel planes 2x + 3 y + 4z = 4 and 4x + 6y + 8z = 12. 10. Find the distance between the parallel planes x + 2y - 2z + 4 = 0 and x + 2y - 2z - 8 = 0. 11. Find the equations of the planes parallel to the plane x - 2y + 2z - 3 = 0, each one of which is at a unit distance from the point (1, 1, 1). 12. Find the equation of the plane parallel to the plane 2x - 3 y + 5z + 7 = 0 and passing through the point (3, 4, –1). Also, find the distance between the two planes. 13. Find the equation of the plane mid-parallel 2x - 3 y + 6z + 21 = 0 and 2x - 3 y + 6z - 14 = 0.

to

the

ANSWERS (EXERCISE 28C)

47 25 3 units 3. units 2. 3 13 10 13 5. units 6. 9. 21 units 3 21 11. x - 2y + 2z + 2 = 0, x - 2y + 2z - 4 = 0 2 12. 2x - 3 y + 5z + 11 = 0, 38 units 19 1.

6 38 units 19 2 29 units 29

13 6 units 12 5 10. 14 units 14 4.

13. 4x - 6y + 12z + 7 = 0

planes

SSS Mathematics for Class 12 1197

The Plane

1197

Equation of a Plane Parallel to a Given Plane VECTOR FORM ® ®

® ®

Any plane parallel to r × n = q1 is given by r × n = q2, where the constant q2 is determined by a given condition. CARTESIAN FORM

Let ax + by + cz + d = 0 be a given plane. Then, any plane parallel to this plane is of the form ax + by + cz + k = 0, where k is determined by a given condition. DISTANCE BETWEEN TWO PARALLEL PLANES

Let a1x + b1y + c1z + d1 = 0 and a1x + b1y + c1z + d 2 = 0 be any two parallel planes. Then, we take a point P( x1 , y1 , z1) on any of these planes and find the length of perpendicular drawn from P( x1 , y1 , z1) to the other plane.

SOLVED EXAMPLES EXAMPLE 1

Find the vector equation of ®

^

^

a plane which is parallel to the plane

^

r × ( 2 i - j + 2 k) = 5 and passes through the point whose position vector ^

^

^

is ( i + j + k). SOLUTION

®

^

^

^

Any plane parallel to the plane r × ( 2 i - j + 2 k) = 5 is given by ®

^

^

^

r × ( 2 i - j + 2 k ) = q for some constant q.

... (i) ^

^

^

Since it passes through a point having position vector ( i + j + k ), we have ^

^

^

^

^

^

( i + j + k ) × ( 2 i - j + 2 k ) = q Þ q = ( 2 - 1 + 2) = 3. Putting q = 3 in (i), we get the required equation of the plane as ®

^

^

^

r × ( 2 i - j + 2 k ) = 3.

EXAMPLE 2

Find the vector equation of a plane which is parallel to the plane ®

^

^

^

r × ( 2 i - 3 j + 5 k ) + 2 = 0 and passes through the point ( 3 , 4, - 1). ®

SOLUTION

^

^

^

Any plane parallel to the plane r × ( 2 i - 3 j + 5 k ) + 2 = 0 is given ®

^

^

^

by r × ( 2 i - 3 j + 5 k ) = q for some constant q.

... (i)

Since it passes through a point A(3, 4, –1) whose position vector is ^

^

^

( 3 i + 4 j - k ), we have

SSS Mathematics for Class 12 1198

1198

Senior Secondary School Mathematics for Class 12 ^

^

^

^

^

^

( 3 i + 4 j - k ) × ( 2 i - 3 j + 5 k ) = q Þ q = ( 6 - 12 - 5) = -11. Putting q = -11 in (i), we get the required equation of the plane as ®

^

^

®

^

^

^

^

r × ( 2 i - 3 j + 5 k ) = -11 Þ r × ( 2 i - 3 j + 5 k ) + 11 = 0.

EXAMPLE 3

Find the equation of the plane which is parallel to the plane 2x - 3 y + z + 8 = 0 and which passes through the point ( -1, 1, 2).

SOLUTION

Any plane parallel to the given plane is 2x - 3 y + z + k = 0 for some constant k.

... (i)

If it passes through the point A( -1, 1, 2), then 2 ´ ( -1) - 3 ´ 1 + 2 + k = 0 Þ

( -2 - 3 + 2) + k = 0 Þ - 3 + k = 0 Þ k = 3.

Hence, the required equation of the plane is 2x - 3 y + z + 3 = 0. EXAMPLE 4

Find the distance between the planes 2x - y + 3z + 4 = 0 and 6x - 3 y + 9z - 3 = 0.

SOLUTION

Clearly, we have

a1 b1 c1 = = × a2 b2 c2

\ the given planes are parallel. Let P( x1 , y1 , z1) be any point on the plane 2x - y + 3z + 4 = 0. Then, 2x1 - y1 + 3z1 + 4 = 0 Þ 2x1 - y1 + 3z1 = - 4.

... (i)

Let p be the distance of the point P( x1 , y1 , z1) from the plane 6x - 3 y + 9z - 3 = 0. Then, | 6x1 - 3 y1 + 9z1 - 3 | | 3( 2x1 - y1 + 3 z1) - 3 | p= = 36 + 9 + 81 62 + ( -3) 2 + 92 |3 ´ ( -4) - 3| [using (i)] 126 |- 15| 15 5 units. = = = 3 14 3 14 14 =

Hence, the distance between the given planes is

5 units. 14

EXERCISE 28D 1. Show that the planes 2x - y + 6z = 5 and 5 x - 2.5 y + 15z = 12 are parallel. ^

^

^

2. Find the vector equation of the plane through the point ( 3 i + 4 j - k ) and ®

^

^

^

parallel to the plane r × ( 2 i - 3 j + 5 k ) + 5 = 0.

SSS Mathematics for Class 12 1199

The Plane

1199

3. Find the vector equation of the plane passing through the point ( a , b , c) ® ^

^

^

and parallel to the plane r × ( i + j + k ) = 2. 4. Find the vector equation of the plane passing through the point (1, 1, 1) ®

^

^

^

and parallel to the plane r × ( 2 i - j + 2 k ) = 5. 5. Find the equation of the plane passing through the point (1, 4, –2) and parallel to the plane 2x - y + 3z + 7 = 0. 6. Find the equation of the plane passing through the origin and parallel to the plane 5 x - 3 y + 7z + 13 = 0. 7. Find the equation of the plane passing through the point (–1, 0, 7) and parallel to the plane 3 x - 5 y + 4z = 11. 8. Find the equations of planes parallel to the plane x - 2y + 2z = 3 which are at a unit distance from the point (1, 2, 3). 9. Find the distance 2x + 4y + 6z + 7 = 0.

between

the

planes

x + 2y + 3z + 7 = 0

and

ANSWERS (EXERCISE 28D) ®

^

®

^

^

^

2. r × ( 2 i - 3 j + 5 k ) + 11 = 0 ^

® ^

^

^

3. r × ( i + j + k ) = a + b + c

^

4. r × ( 2 i - j + 2 k ) = 3 6. 5 x - 3 y + 7z = 0

5. 2x - y + 3z + 8 = 0 7. 3 x - 5 y + 4z = 25 7 units 8. x - 2y + 2z = 0 or x - 2y + 2z - 6 = 0 9. 56 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28D) 1. The given planes are in the form a1 x + b1 y + c1z + d1 = 0 and a2 x + b 2 y + c2 z + d2 = 0 , a b c where 1 = 1 = 1 × Hence, the given planess are parallel. a2 b 2 c2 8. Let the required plane be x - 2 y + 2z + k = 0 for some constants k. Then, its distance from the point P(1, 2, 3) is |1 - 2 ´ 2 + 2 ´ 3 + k| | 3 + k| = = 1 Þ | 3 + k|= 3 3 12 + ( -2 ) 2 + 2 2 Þ 3 + k = 3 or 3 + k = -3 Þ k = 0 or k = -6. Hence, the required equations are x - 2 y + 2z = 0 or x - 2 y + 2z - 6 = 0. 9. Let P( x1 , y1 z1 ) be any point on the plane x + 2 y + 3z + 7 = 0. Then, x1 + 2 y1 + 3z1 = -7 . |2 x1 + 4 y1 + 6z1 + 7| |2( x1 + 2 y1 + 3z1 ) + 7| \ p= = 56 22 + 42 + 62 |2 ´ ( -7 ) + 7| 7 units. = = 56 56

... (i)

SSS Mathematics for Class 12 1200

1200

Senior Secondary School Mathematics for Class 12

PLANE PASSING THROUGH THE INTERSECTION OF TWO PLANES

The equation of a plane passing through the intersection of two planes

THEOREM 1

® ® r × n1

PROOF

® ®

® ®

®

= q1 and r × n2 = q2 is given by r × (n1 + l n2) = ( q1 + lq2).

® ®

Let p1 and p 2 be the two given planes with equations r × n1 = q1 and ® ® r × n2

= q2 respectivley.

®

Let t be the position vector of any point on their line of intersection. Then, it must satisfy both the equations. ® ® t × n1

\

®

Þ

® ®

= q1 and t × n2 = q2

®

®

t × (n1 + l n2) = ( q1 + lq2) for every real value l. ®

Since t is arbitrary, the required equation of a plane p 3 passing through the intersection of the given planes, is given by ®

®

®

r × (n1 + l n2) = ( q1 + lq2).

The equation of a plane passing through the intersection of two planes a1x + b1y + c1z = d1 and a2x + b2y + c2z = d 2 is given by

THEOREM 2

( a1x + b1y + c1z - d1) + l( a2x + b2y + c2z - dz ) = 0. PROOF

Taking

® n1

®

^

^

^

®

^

^

^

= ( a1 i + b1 j + c1 k ), n2 = ( a2 i + b2 j + c 3 k ), q1 = d1 , q2 = d 2 ^

^

^

and r = ( x i + y j + z k ), we may write the above vector equation as ^

^

^

^

^

^

^

^

^

( x i + y j + z k ) × [( a1 i + b1 j + c1 k ) + l( a2 i + b2 j + c2 k ) = ( d1 + ld 2)] Þ ( a1x + b1y + c1z = d1) + l( a2x + b2y + c2z - d 2) = 0, which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of l. SUMMARY

1. The vector equation of a plane passing through the intersection ® ®

® ®

of two planes r × n1 = q1 and r × n2 = q2 is given by ®

®

®

r × (n1 + l n2) = ( q1 + lq2).

2. The equation of a plane passing through the intersection of two planes a1x + b1y + c1z = d1 and a2x + b2y + c2z = d 2 is given by ( a1x + b1y + c1z - d1 + l( a2x + b2y + c2z - d 2) = 0.

SSS Mathematics for Class 12 1201

The Plane

1201

SOLVED EXAMPLES EXAMPLE 1

Find the equation of the plane through the intersection of the planes 3 x - y + 2z = 4 and x + y + z = 2 and passing through the point ( 2, 2, 1).

SOLUTION

Let the required equation of the plane be ( 3 x - y + 2z - 4) + l( x + y + z - 2) = 0 for some real value of l ... (i) Þ ( 3 + l) x + ( -1 + l) y + ( 2 + l)z - ( 4 + 2l) = 0. It is given that the point A(2, 2, 1) lies on (i). \ ( 3 + l) ´ 2 + ( -1 + l) ´ 2 + ( 2 + l) ´ 1 - ( 4 + 2l) = 0 Þ ( 6 + 2l) + ( -2 + 2l) + ( 2 + l) - 4 - 2l = 0 -2 × Þ 3 l = -2 Þ l = 3 -2 in (i), we get Putting l = 3 4 2ö 2ö 2ö æ æ æ ç 3 - ÷ x + ç -1 - ÷ y + ç 2 - ÷ z - 4 + = 0 3 3 3 3 ø è ø è ø è 7 x 5 y 4z 8 Þ + - = 0 Þ 7 x - 5 y + 4z - 8 = 0. 3 3 3 3 Hence, the required equation of the plane is 7 x - 5 y + 4z - 8 = 0.

EXAMPLE 2

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3 y + 4z = 5 , which is perpendicular to the plane x - y + z = 0. Also find the distance of the plane so obtained from [CBSE 2014] the origin.

SOLUTION

Let the required equation of the plane be ( x + y + z - 1) + l( 2x + 3 y + 4z - 5) = 0 for some real value of l Þ

(1 + 2l)x + (1+ 3l)y + (1 + 4l)z - (1 + 5 l) = 0.

... (i)

Since this plane is perpendicular to the plane x - y + z = 0, we have (1 + 2l) ´ 1 + (1 + 3 l) ´ ( -1) + (1 + 4l) ´ 1 = 0 Þ

(1 + 2l) - (1 + 3 l) + (1 + 4l) = 0 Þ 3 l = -1 Þ l =

-1 × 3

-1 in (i), we get 3 2ö 4ö 5 æ æ ç1 - ÷ x + (1 - 1) y + ç1 - ÷z - 1 + = 0 3ø 3ø 3 è è 1 1 2 Þ x - z + = 0 Þ x - z + 2 = 0. 3 3 3 Hence, the required equation of the plane is x - z + 2 = 0. Distance of the plane from the origin = Length of perpendicular from the origin to the plane |0 - 0 + 2| 2 = = = 2. 2 2 2 2 1 + 0 + ( -1) Putting l =

SSS Mathematics for Class 12 1202

1202

Senior Secondary School Mathematics for Class 12

EXAMPLE 3

Find the equation of the plane passing through the line of intersection of the planes 2x + y - z = 3 and 5 x - 3 y + 4z + 9 = 0 and parallel to the line x -1 y - 3 z -5 [CBSE 2011] = = × 2 4 5

SOLUTION

Let the required equation of the plane be ( 2x + y - z - 3) + l(5 x - 3 y + 4z + 9) = 0 Þ

( 2 + 5 l) x + (1 - 3 l) y + ( -1 + 4l)z - 3 + 9l = 0.

... (i)

The plane given by (i) is parallel to the line x -1 y - 3 z -5 = = × 2 4 5

... (ii)

So, the normal to the plane (i) is perpendicular to the line (ii). \

2( 2 + 5 l) + 4(1 - 3 l) + 5( -1 + 4l) = 0

Þ

( 4 + 10l) + ( 4 - 12l) + ( -5 + 20l) = 0 -3 -1 Þ l= × Þ 18l = -3 Þ l = 18 6 -1 Putting l = in (i), we get 6 1 4 ö 3ö 5ö æ æ æ ç 2 - ÷ x + ç 1 + ÷ y + ç -1 - z ÷ - 9 ´ = 0 6 6 ø 6ø 6ø è è è Þ

7 x 9y 10z 27 + =0 6 6 6 6

Þ

7 x + 9y - 10z - 27 = 0.

Hence, the required equation of the plane is 7 x + 9y - 10z - 27 = 0. EXAMPLE 4

Find the equation of the plane passing through the intersection of the ®

^

^

®

^

^

^

^

planes r × ( i + j + k ) = 1 and r × ( 2 i + 3 j - k ) + 4 = 0 and parallel to [CBSE 2011] the x-axis. SOLUTION

Converting the given equations of the planes in Cartesian form, we get ®

^

^

^

^

^

^

^

^

^

r × ( i + j + k) = 1 Þ (x i + y j + z k) ×( i + j + k) = 1 Þ x + y + z - 1 = 0,

®

^

^

^

^

^

... (i) ^

^

^

^

r × (2 i + 3 j - k) + 4 = 0 Þ (x i + y j + z k) ×(2 i + 3 j - k) + 4 = 0 Þ 2x + 3 y - z + 4 = 0.

... (ii)

Now, the equation of a plane passing through the intersection of the planes (i) and (ii) is given by ( x + y + z - 1) + l( 2x + 3 y - z + 4) = 0 for some real value of l Þ

(1 + 2l) x + (1 + 3 l) y + (1 - l)z + ( -1 + 4l) = 0.

... (iii)

SSS Mathematics for Class 12 1203

The Plane

1203

D.r.’s of the normal to the plane are (1 + 2l),(1 + 3 l),(1 - l). D.r.’s of the x-axis are 1, 0, 0. Plane (iii) is parallel to the x-axis means that the normal to this plane is perpendicular to the x-axis. -1 × \ 1 ´ (1 + 2l) + 0 ´ (1 + 3 l) + 0 ´ (1 - l) = 0 Þ 1 + 2l = 0 Þ l = 2 -1 in (iii), we get the required equation of the plane as Putting l = 2 1ö 3ö æ æ (1 - 1) x + ç1 - ÷ y + ç1 + ÷z + ( -1 - 2) = 0 2ø 2ø è è - y 3z Þ + - 3 = 0 Þ y - 3z + 6 = 0. 2 2 Hence, the required equation of the plane is y - 3z + 6 = 0. EXAMPLE 5

Find the equation of the plane passing through the line of intersection of ®

^

®

^

^

^

^

the planes r × ( i + 3 j ) - 6 = 0 and r × ( 3 i - j - 4 k ) = 0, whose perpendicular distance from the origin is unity. [CBSE 2013] ®

SOLUTION

^

^

^

Taking r = ( x i + y j + z k ), the equations of the given planes are ^

^

^

^

^

^

^

^

( x i + y j + z k ) × ( i + 3 j ) - 6 = 0 Þ x + 3 y - 6 = 0, ^

^

^

( x i + y j + z k ) × ( 3 i - j - 4 k ) = 0 Þ 3 x - y - 4z = 0.

... (i) ... (ii)

The equation of any plane passing through the line of intersection of the given planes is given by ( x + 3 y - 6) + l( 3 x - y - 4z) for some real number l. ... (iii) Þ (1 + 3 l) x + ( 3 - l) y - 4lz - 6 = 0. Length of perpendicular from origin to plane (iii) is given as 1. |0 + 0 - 0 - 6| =1 \ (1 + 3 l) 2 + ( 3 - l) 2 + ( -4l) 2 Þ

(1 + 3 l) 2 + ( 3 - l) 2 + ( -4l) 2 = 36

Þ

26l2 = 26 Þ l2 = 1 Þ l = ±1.

Putting l = 1 in (iii), we get 4x + 2y - 4z - 6 = 0 Þ 2x + y - 2z - 3 = 0. Putting l = -1 in (iii), we get -2x + 4y + 4z - 6 = 0 Þ x - 2y - 2z + 3 = 0. Hence, the required equations of the plane are 2x + y - 2z - 3 = 0 and x - 2y - 2z + 3 = 0. Note: The vector equations of these planes are ®

^

^

^

®

^

^

^

r × ( 2 i + j - 2 k ) = 3 and r × ( i - 2 j - 2 k ) + 3 = 0.

SSS Mathematics for Class 12 1204

1204 EXAMPLE 6

Senior Secondary School Mathematics for Class 12

Find the vector equation of the plane passing through the intersection of ®

^

^

^

®

^

^

^

®

^

^

^

the planes r × ( 2 i + 2 j - 3 k ) = 7 and r × ( 2 i + 5 j + 3 k ) = 9 and passing through the point ( 2, 1, 3). SOLUTION

The equations of the given planes are r × ( 2 i + 2 j - 3 k ) = 7 and ®

^

^

^

r × ( 2 i + 5 j + 3 k ) = 9. ® ®

® ®

These are of the form r × n1 = q1 and r × n2 = q2 , where ®

^

^

®

^

^

^

^

n1 = ( 2 i + 2 j - 3 k ), n2 = ( 2 i + 5 j + 3 k ), q1 = 7 and q2 = 9. The equation of the plane through the intersection of the given planes is given by ®

®

®

r × (n1 + l n2) = ( q1 + lq2) for some real number l

Þ Þ

®

^

^

^

^

^

^

r × [( 2 i + 2 j - 3 k ) + l( 2 i + 5 j + 3 k )](7 + 9l)

®

^

^

^

r × [( 2 + 2l) i ( 2 + 5 l) j + ( -3 + 3 l) k ] = (7 + 9l).

... (i)

If the plane (i) passes through the point A(2, 1, 3), then ®

^

^

^

r = ( 2 i + j + 3 k ) must satisfy it. ^

^

^

^

^

^

\

( 2 i + j + 3 k ) × [( 2 + 2l) i + ( 2 + 5 l) j + ( -3 + 3 l) k ] = (7 + 9l)

Þ

2( 2 + 2l) + 1 × ( 2 + 5 l) + 3( -3 + 3 l) = 7 + 9l

Þ

( 4 + 4l) + ( 2 + 5 l) + ( -9 + 9l) = 7 + 9l Þ 9l = 10 Þ l =

Putting l =

10 × 9

10 in (i), we get the required equation of the plane as 9

éæ 30 ö ^ ù 50 ö ^ æ 20 ö ^ æ r × ê ç 2 + ÷ i + ç 2 + ÷ j + ç -3 + ÷ k ú = (7 + 10) 9ø û 9ø 9ø è è ëè

®

Þ

®

^

^

^

r × ( 38 i + 68 j + 3 k ) = 153.

Hence, the required equation of the plane is ®

^

^

^

r × ( 38 i + 68 j + 3 k ) = 153.

EXAMPLE 7

Find the vector equation of the plane through the line of intersection of ®

^

^

®

^

^

^

^

the planes r × ( i + 2 j + 3 k ) - 4 = 0 and r × ( 2 i + j - k ) + 5 = 0 and ®

^

^

^

perpendicular to the plane r × (5 i + 3 j - 6 k ) + 8 = 0.

[CBSE 2011]

SSS Mathematics for Class 12 1205

The Plane SOLUTION

1205

The equations of the given planes are ®

^

^

®

^

^

^

^

r × ( i + 2 j + 3 k ) = 4 and r × ( -2 i - j + k ) = 5. ® ®

® ®

These equations are of the form r × n1 = q1 and r × n2 = q2, where ®

^

^

^

®

^

^

^

n1 = ( i + 2 j + 3 k ), n2 = ( -2 i - j + k ), q1 = 4 and q2 = 5. The equation of the plane through the intersection of the given planes is given by ®

®

®

^

®

r × (n1 + l n2) = ( q1 + l q2) for some real number l

Þ Þ

^

^

^

^

^

r × [( i + 2 j + 3 k ) + l( -2 i - j + k )] = ( 4 + 5 l)

®

^

^

^

r × [(1 - 2l) i + ( 2 - l) j + ( 3 + l) k ] = ( 4 + 5 l) ®

... (i) ^

^

^

This plane (i) is perpendicular to the plane r × ( -5 i - 3 j + 6 k ) = 8, so we have (1 - 2l) ´ ( -5) + ( 2 - l) ´ ( -3) + ( 3 + l) ´ 6 = 0 Þ

( -5 + 10l) + ( -6 + 3 l) + (18 + 6l) = 0 -7 × Þ 19l = -7 Þ l = 19 -7 Putting l = in (i), we get 19 ® é 7 ö ^ù æ 7 ö^ æ 35 ö 14 ö ^ æ æ r × ê ç1 + ÷ i + ç 2 + ÷ j + ç 3 - ÷ k ú = ç 4 ÷ 19 19 19 19 è ø è ø è ø è ø ë û ®

^

^

^

®

^

^

^

Þ r × ( 33 i + 45 j + 50 k ) = 41. Hence, the required equation of the plane is r × ( 33 i + 45 j + 50 k ) = 41.

EXERCISE 28E 1. Find the equation of the plane through the line of intersection of the planes x + y + z = 6 and 2x + 3 y + 4z + 5 = 0, and passing through the point (1, 1, 1). 2. Find the equation of the plane through the line of intersection of the planes x - 3 y + z + 6 = 0 and x + 2y + 3z + 5 = 0, and passing through the origin. 3. Find the equation of the plane passing through the intersection of the planes 2x + 3 y - z + 1 = 0 and x + y - 2z + 3 = 0, and perpendicular to the plane 3 x - y - 2z - 4 = 0. 4. Find the equation of the plane passing through the line of intersection of the planes 2x - y = 0 and 3z - y = 0, and perpendicular to the plane 4x + 5 y - 3z = 9.

SSS Mathematics for Class 12 1206

1206

Senior Secondary School Mathematics for Class 12

5. Find the equation of the plane passing through the intersection of the planes x - 2y + z = 1 and 2x + y + z = 8, and parallel to the line with direction ratios 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) [CBSE 2005] from the plane. 6. Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z - 5 = 0 and 3 x - 2y - z + 1 = 0 and cutting off equal intercepts on the x-axis and z-axis. 7. Find the equation of the plane through the intersection of the planes 3 x - 4y + 5z = 10 and 2x + 2y - 3z = 4 and parallel to the line x = 2y = 3z. HINT: The equations of the given line are

x y z = = × 6 3 2

8. Find the vector equation of the plane through the intersection of the planes ®

^

^

®

^

^

^

r × ( i + 3 j - k ) = 0 and r × ( j + 2 k ) = 0, and passing through the point (2, 1, –1). 9. Find the vector equation of the plane through the point (1, 1, 1), and passing through the intersection of the planes ®

^

^

®

^

^

^

^

r × ( i - j + 3 k ) + 1 = 0 and r × ( 2 i + j - k ) - 5 = 0.

10. Find the vector equation of the plane passing through the intersection of the planes ®

^

^

®

^

^

^

^

r × ( 2 i - 7 j + 4 k ) = 3 and r × ( 3 i - 5 j + 4 k ) + 11 = 0, and passing through the point (–2, 1, 3). [CBSE 2006] 11. Find the equation of the plane through the line of intersection of the planes ®

^

®

^

^

® ^

^

^

r × ( 2 i - 3 j + 4 k ) = 1 and r × ( i - j ) + 4 = 0 and perpendicular to the plane ^

^

r × ( 2 i - j + k ) + 8 = 0.

12. Find the Cartesian and vector equations of the planes through the line of ® ^

®

^

^

^

^

intersection of the planes r × ( i - j ) + 6 = 0 and r × ( 3 i + 3 j - 4 k ) = 0, which are at a unit distance from the origin. ANSWERS (EXERCISE 28E)

1. 20x + 23 y + 26z = 69 3. 7 x + 13 y + 4z = 9 13 5. 9x - 8y + 7z = 21, units 194 7. x - 20y + 27z = 14 ®

^

®

^

^

^

9. r × (11 i + j + 5 k ) - 17 = 0 ^

^

11. r × ( -5 i + 2 j + 12 k ) = 47

2. x + 27 y + 13z = 0 4. 28x - 17 y + 9z = 0 6. 5 x + 2y + 5z - 9 = 0 ®

^

^

^

8. r × ( i + 9 j + 11 k ) = 0 ®

^

^

^

10. r × (15 i - 47 j + 28 k ) = 7

SSS Mathematics for Class 12 1207

The Plane

1207

12. 2x + y - 2z + 3 = 0, x + 2y - 2z - 3 = 0, and ®

^

^

^

®

^

^

^

r × ( 2 i + j - 2 k ) + 3 = 0, r × ( i + 2 j - 2 k ) - 3 = 0

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28E) 5. Let the required plane be ( x - 2 y + z - 1) + l ( 2 x + y + z - 8 ) = 0 Û

( 1 + 2 l)x + ( l - 2 )y + ( 1 + l)z - ( 1 + 8 l) = 0.

… (i)

The d.r.’s of the normal to this plane are ( 1 + 2 l), ( l - 2 ), ( 1 + l). The normal to the plane (i) is perpendicular to the line with d.r.’s 1, 2, 1. 2 \ ( 1 + 2 l) + 2( l - 2 ) + ( 1 + l) = 0 Þ l = × 5 6. Let the required plane be ( x + 2 y + 3z - 5 ) + l ( 3 x - 2 y - z + 1) = 0. Then, ( 1 + 3 l)x + ( 2 - 2 l)y + ( 3 - l) z - 5 + l = 0. Þ Þ

( 1 + 3 l)x + ( 2 - 2 l)y + ( 3 - l)z = 5 - l ( 1 + 3 l) ( 2 - 2 l) ( 3 - l) x+ y+ z= 1 (5 - l) (5 - l) (5 - l) y x z + + = 1. æ 5-l ö æ 5-l ö æ5-lö ç ÷ ç ÷ ç ÷ è 1 + 3l ø è 2 - 2l ø è 3 - l ø

Since the intercepts on the x-axis and z-axis are equal, we have 5-l 5-l = Þ 3 - l = 1 + 3l 1 + 3l 3 - l 1 Þ 4l = 2 Þ l = × 2 7. Let the required plane be ( 3 x - 4 y + 5z - 10 ) + l( 2 x + 2 y - 3z - 4 ) = 0 Þ

( 3 + 2 l)x + ( 2 l - 4 )y + (5 - 3 l)z - 10 - 4 l = 0.

... (i)

D.r’s of normal to the plane are ( 3 + 2 l), ( 2 l - 4), (5 - 3 l). x y z Given line is = = × 6 3 2 D.r.’s of the line are 6, 3, 2. \

This line is perpendicular to the plane.

\

6( 3 + 2 l) + 3(2l - 4) + 2(5 - 3l) = 0 Þ 12l = -16 Þ l =

-4 × 3

SSS Mathematics for Class 12 1208

1208

Senior Secondary School Mathematics for Class 12

12. The equations of the given planes are ^

^

^

^

^

^

^

^

^

^

^

( x i + y j + z k ) × ( i - j ) + 6 = 0 and ( x i + y j + z k ) × ( 3 i + 3 j - 4 k ) = 0 Þ x - y + 6 = 0 and 3 x + 3 y - 4z = 0. Any plane through their intersection is … (i) ( x - y + 6 ) + l( 3 x + 3 y - 4z) = 0 Þ ( 1 + 3 l)x + ( 3 l - 1)y - 4 lz + 6 = 0. 6 = 1 Þ 34 l2 + 2 = 36 Þ l2 = 1 Þ l = ±1 \ ( 1 + 3 l) 2 + ( 3 l - 1) 2 + ( - 4 l) 2 So, the required planes are 2 x + y - 2z + 3 = 0 and x + 2 y - 2z - 3 = 0. In vector form, they are ®

^

^

®

^

^

^

^

r × ( 2 i + j - 2 k ) + 3 = 0 and r × ( i + 2 j - 2 k ) - 3 = 0.

Equation of a Plane Passing through Three Noncollinear Points Vector Form THEOREM 1

The vector equation of a plane passing through three noncollinear points ® ® ®

with position vectors a , b , c is ®

®

®

®

®

®

( r - a ) × [( b - a ) ´ ( c - a )] = 0. PROOF

Let A , B , C be three given noncollinear points having position vectors ® ® ®

a , b , c respectively.

Let P be an arbitrary point on the plane passing through the points ®

A , B , C , and let r be the position vector of P. Then, ¾®

® ®

¾®

®

®

¾®

®

®

AP = (p.v. of P) - (p.v. of A) = ( r - a ), AB = (p.v. of B) - (p.v. of A) = ( b - a ), AC = (p.v. of C) - (p.v. of A) = ( c - a ). ¾®

¾®

¾®

Since the points A , B , C , P lie on the plane, the vectors AP , AB , AC are coplanar. So, the scalar triple product of these vectors is 0. ¾® ¾® ¾®

\

[AP AB AC ] = 0

Þ

AP × ( AB ´ AC ) = 0

Þ

( r - a ) × [( b - a ) ´ ( c - a )] = 0,

¾®

®

¾®

®

¾®

®

®

®

®

which is the required equation of the plane.

SSS Mathematics for Class 12 1209

The Plane

1209

Cartesian Form THEOREM 2

The equation of a plane passing through three given noncollinear points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) is given by ½ x - x1 ½ x 2 - x1 ½ ½ x 3 - x1

PROOF

y - y1

z - z1 ½ y 2 - y1 z2 - z1 ½ = 0. ½ y 3 - y1 z 3 - z1 ½ Let P( x , y , z) be an arbitrary point on the given plane. Then, ¾®

AP = (p.v. of P) - (p.v. of A) ^

^

^

^

^

^

= ( x i + y j + z k ) - ( x1 i + y1 j + z1 k ) ^

^

^

= ( x - x1) i + ( y - y1) j + (z - z1) k . Similarly, we have ¾®

^

^

¾®

^

^

^

AB = ( x 2 - x1) i + ( y 2 - y1) j + (z2 - z1) k

^

and AC = ( x 3 - x1 ) i + ( y 3 - y1) j + (z 3 - z1) k. ¾®

¾®

¾®

¾®

¾®

Since AP , AB and AC are coplanar, we have ¾®

[AP , AB, AC] = 0. Hence, the required equation of the plane passing through three given points A, B, C is given by ½ x - x1 ½ x 2 - x1 ½ ½ x 3 - x1

y - y1 y 2 - y1 y 3 - y1

z - z1 ½ z2 - z1 ½ = 0. ½ z 3 - z1 ½

SUMMARY

(i) The vector equation of a plane passing through three noncollinear ® ® ®

points having position vectors a , b , c is given by ®

®

®

®

®

®

( r - a ) × [( b - a ) ´ ( c - a )] = 0. (ii) The Cartesian equation of a plane passing through three noncollinear points A( x1 , y1 , z1), B( x 2 , y 2 , z2) and C( x 3 , y 3 , z 3) is given by x - x1 y - y1 z - z1 x 2 - x1 y 2 - y1 z2 - z1 = 0. x 3 - x1 y 3 - y1 z 3 - z1

SSS Mathematics for Class 12 1210

1210

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Find the vector equation of a plane passing through the points A( 2, 5 , - 3), B( -2, - 3 , 5) and C(5 , 3 , - 3).

SOLUTION

The position vectors of the given points A, B, and C are ®

^

^

®

^

^

^

®

^

^

^

^

a = ( 2 i + 5 j - 3 k ), b = ( -2 i - 3 j + 5 k ) and c = (5 i + 3 j - 3 k ) respectively. Then, the vector equation of the required plane is ®

®

®

®

®

®

( r - a ) × [( b - a) ´ ( c - a )] = 0. ®

®

®

®

^

^

^

^

^

^

Now ( b - a ) = ( -2 - 2) i + ( -3 - 5) j + (5 + 3) k = ( -4 i - 8 j + 8 k ). ^

^

^

^

^

^

and ( c - a ) = (5 - 2) i + ( 3 - 5) j + ( -3 + 3) k = ( 3 i - 2 j + 0 k ). ^

\

®

®

®

^

i

®

^

j

k

( b - a ) ´ ( c - a ) = -4 -8

8

-2

0

3

^

^

^

= ( 0 + 16) i - ( 0 - 24) j + ( 8 + 24) k ^

^

^

= (16 i + 24 j + 32 k ). \ the required vector equation is ®

®

^

^

^

^

^

^

( r - a ) × (16 i + 24 j + 32 k ) = 0 Þ

®

®

^

^

^

^

^

r × (16 i + 24 j + 32 k ) = a × (16 i + 24 j + 32 k ) ^

^

^

^

= ( 2 i + 5 j - 3 k ) × (16 i + 24 j + 32 k ) = ( 32 + 120 - 96) = 56. Hence, the required vector equation of the plane is ®

^

^

^

®

^

^

^

r × (16 i + 24 j + 32 k ) = 56 Þ r × ( 2 i + 3 j + 4 k ) = 7.

EXAMPLE 2

SOLUTION

Find the equation of the plane passing through the points A(1, 1, 0), B(1, 2, 1) and C( -2, 2, - 1). Here ( x1 = 1, y1 = 1, z1 = 0), ( x 2 = 1, y 2 = 2, z2 = 1) and ( x 3 = -2, y 3 = 2, z 3 = -1). Hence, the required equation of the plane is

SSS Mathematics for Class 12 1211

The Plane

Þ

x - x1

y - y1

z - z1

x 2 - x1 x 3 - x1

y 2 - y1 y 3 - y1

z2 - z1 = 0 z 3 - z1

x -1

y -1

z-0

1 -1 -2 - 1

2 -1 2 -1

1-0 = 0 -1 - 0

x -1 y -1 Þ

1211

0 -3

1 1

z 1 =0 -1

Þ ( x - 1)( -1 - 1) - ( y - 1)( 0 + 3) + z( 0 + 3) = 0 Þ ( -2)( x - 1) - 3( y - 1) + 3z = 0 Þ -2x - 3 y + 3z + 5 = 0 Þ 2x + 3 y - 3z - 5 = 0. Hence, the required equation of the plane is 2x + 3 y - 3z - 5 = 0. ANGLE BETWEEN TWO PLANES

The angle between two given planes is the angle between their normals. If q is an angle between two planes, then (180° - q) is also an angle between them. An Important Note: We shall take the acute angle as the angle between two planes. Vector Form Let q be the angle between two planes whose vector equations are ® ®

® ®

r × n1 = q1 and r × n2 = q2. ®

®

Then, q is the angle between their normals n1 and n2. ® ®

\

cos q =

|n1 × n2|

× ® ® |n1||n2|

Two Important Results

1. Condition for two planes to be perpendicular to each other ® ®

® ®

Two planes r × n1 = q1 and r × n2 = q2 are perpendicular to each other ®

®

®

®

Û n1 ^ n2 Û n1 × n2 = 0. 2. Condition for two planes to be parallel to each other ® ®

® ®

Two planes r × n1 = q1 and r × n2 = q2 are parallel to each other

SSS Mathematics for Class 12 1212

1212

Senior Secondary School Mathematics for Class 12 ®

®

Û n1 | | n2 ®

®

Û n1 = l n2 for some scalar l. ® ®

® ®

REMARK 1

Any plane parallel to r × n = q is r × n = q1.

REMARK 2

The equation of the plane parallel to the plane r × n = q and passing ®

® ®

®

®

®

through the point with position vector a is ( r - a ) × n = 0. Cartesian Form Let q be the acute angle between the planes: a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0. Then, q is the angle between their normals. D.r.’s of the normals to the given planes are a1 , b1 , c1 and a2 , b2 , c2. a1a2 + b1b2 + c1c2 \ cos q = × 2 { a1 + b12 + c12} { a22 + b22 + c22} Two Important Results

1. Condition for two planes to be perpendicular to each other Two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0 are perpendicular to each other Û their normals are perpendicular to each other Û lines with direction ratios a1 , b1 , c1 , and a2 , b2 , c2 are perpendicular to each other Û a1a2 + b1b2 + c1c2 = 0. 2. Condition for two planes to be parallel to each other Two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0 are parallel to each other Û their normals are parallel to each other Û lines with direction ratios a1 , b1 , c1 , and a2 , b2 , c2 are parallel to each other a b c Û 1 = 1 = 1× a2 b2 c2 REMARK 1

The equation of any plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + l = 0.

REMARK 2

The equation of a plane passing through ( x1 , y1 , z1) and parallel to the plane ax + by + cz + d = 0 is given by a( x - x1) + b( y - y1) + c(z - z1) = 0.

REMARK 3

The equation of any plane parallel to the xy-plane is z = l. This plane is perpendicular to the z-axis. The equation of any plane parallel to the xz-plane is y = l.

SSS Mathematics for Class 12 1213

The Plane

1213

xz-plane is perpendiuclar to the y-axis. The equation of any plane parallel to the yz-plane is x = l. yz-plane is perpendicular to the x-axis. SUMMARY OF THE RESULTS

® ®

® ®

1. The acute angle q between the planes r × n1 = q1 and r × n2 = q2 is given by ® ®

cos q = ® ®

|n1 × n2|

× |n1||n2| ®

®

® ®

(i) Two planes r × n1 = q1 and r × n2 = q2 are perpendicular to each other ® ®

Û n1 × n2 = 0.

® ®

® ®

(ii) Two planes r × n1 = q1 and r × n2 = q2 are perpendicular to each other ®

®

Û n1 = l n2 for some scalar l.

® ®

® ®

(iii) Any plane parallel to the plane r × n = q is r × n = q1. ® ®

(iv) Any plane parallel to r × n = q and passing through a point with p.v. ®

®

®

®

a is ( r - a ) × n = 0.

2. The acute angle q between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0 is given by a1a2 + b1b2 + c1c2 cos q = × 2 { a1 + b12 + c12} { a22 + b22 + c22} (i) Two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0 are perpendicular Û a1a2 + b1b2 + c1c2 = 0. (ii) Two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0 are a b c parallel Û 1 = 1 = 1 × a2 b2 c2 (iii) The equation of a plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + l = 0. (iv) The equation of a plane passing through the point ( x1 , y1 , z1) and parallel to the plane ax + by + cz + d = 0 is given by a( x - x1) + b( y - y1) + c(z - z1) = 0. (v) Any plane parallel to the yz-plane is x = l. Any plane parallel to the xz-plane is y = l. Any plane parallel to the xy-plane is z = l.

SSS Mathematics for Class 12 1214

1214

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

Find the angle between the planes ®

^

^

®

^

^

^

^

r × ( i + j + 2 k ) = 5 and r × ( 2 i - j + k ) = 8.

SOLUTION

® ®

We know that the angle between the planes r × n1 = q1 and

® ®

r × n2 = q2 is given by ®

®

|n1 × n2| cos q = ® ® × |n1||n2| ®

^

^

®

^

^

^

^

Here, n1 = ( i + j + 2 k ) and n2 = ( 2 i - j + k). ^

\

= Þ

^

^

^

^

^

|( i + j + 2 k ) × ( 2 i - j + k )| ^ ^ ^ ^ ^ ^ |i + j + 2 k||2 i - j + k|

cos q =

1 ´ 2 + 1 ´ ( -1) + 2 ´ 1 2

2

2

2

2

=

2

{ 1 + 1 + 2 } { 2 + ( -1) + 1 }

3 1 = 6 2

p q= × 3

æ pö Hence, the angle between the given planes is ç ÷ × è 3ø EXAMPLE 2

Find the angle between the planes whose vector equations are ®

^

^

®

^

^

^

^

r × ( 2 i + 2 j - 3 k ) = 5 and r × ( 3 i - 3 j + 5 k ) = 3. ® ®

SOLUTION

We know that the acute angle q between the planes r × n1 = q1 and ® ®

r × n2 = q2 is given by

|n®1 × n®2| cos q = × |n®1||n®2| ®

^

^

^

®

^

^

^

Here, n1 = ( 2 i + 2 j - 3 k ) and n2 = ( 3 i - 3 j + 5 k ). ^

\

^

^

^

^

^

|2 i + 2 j - 3 k ) × ( 3 i - 3 j + 5 k )| ^ ^ ^ ^ ^ ^ |2 i + 2 j - 3 k||3 i - 3 j + 5 k|

cos q =

= =

2 ´ 3 + 2 ´ ( -3) + ( -3) ´ 5 2

{ 2 + 22 + ( -3) 2} { 3 2 + ( -3) 2 + 5 2} -15 ( 17 ) ( 43 )

=

15 731

SSS Mathematics for Class 12 1215

The Plane

Þ

1215

æ 15 ö q = cos -1 ç ÷× è 731 ø

æ 15 ö Hence, the angle between the given planes is cos-1 ç ÷× è 731 ø EXAMPLE 3

Find the value of l for which the planes ®

^

^

®

^

^

^

^

r × ( i + 2 j + 3 k ) = 13 and r × ( l i + 2 j - 7 k ) = 9 are perpendicular to each other. SOLUTION

We know that the planes

® ®

r × n1 = q1 ®

and ®

® ®

r × n2 = q2

are

perpendicular to each other only when n1 × n2 = 0. ®

^

^

®

^

^

^

^

Here, n1 = ( i + 2 j + 3 k ) and n2 = ( l i + 2 j - 7 k). \

the given planes are perpendicular to each other ®

®

Û n1 × n2 = 0 ^

^

^

^

^

^

Û ( i + 2 j + 3 k) × ( l i + 2 j - 7 k) = 0 Û 1 ´ l + 2 ´ 2 + 3 ´ ( -7) = 0 Û l = 17. Hence, the required value of l is 17. EXAMPLE 4

Find the angle between the planes whose Cartesian equations are 7 x + 5 y + 6z + 30 = 0 and 3 x - y - 10 z + 4 = 0.

SOLUTION

We know that the acute angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0 is given by a1a2 + b1b2 + c1c2

cos q = {

a12

+ b12 + c12} { a22 + b22 + c22}

×

Here a1 = 7 , b1 = 5 , c1 = 6 and a2 = 3 , b2 = -1. c2 = -10. \

cos q = = =

Þ

7 ´ 3 + 5 ´ ( -1) + 6( -10) 2

{ 7 + 5 2 = 62} { 3 2 + ( -1) 2 + ( -10) 2} 21 - 5 - 60 { 49 + 25 + 36}{ 9 + 1 + 100} 44 44 2 = = × { 110 ´ 110} 110 5

æ 2ö q = cos-1 ç ÷ × è5 ø

æ 2ö Hence, the acute angle between the given planes is cos-1 ç ÷ × è5 ø

SSS Mathematics for Class 12 1216

1216

Senior Secondary School Mathematics for Class 12

EXAMPLE 5

Find the value of l for which the planes 2x - 4y + 3z = 7 and x + 2y + lz = 18 are perpendicular to each other.

SOLUTION

The equations of the given planes are 2x - 4y + 3z - 7 = 0 and x + 2y + lz - 18 = 0. They are of the form a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d 2 = 0 where a1 = 2, b1 = -4, c1 = 3 , d1 = -7, and a2 = 1, b2 = 2, c2 = l, d 2 = -18. The given planes are perpendicular to each other Û a1a2 + b1b2 + c1c2 = 0 Û 2 ´ 1 + ( -4) ´ 2 + 3 ´ l = 0 Û 3 l = 6 Û l = 2. Hence, the required value of l is 2.

EXAMPLE 6

Find the equation of the plane passing through the point ( -1, 3 , 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3 x + 3 y + z = 0. [CBSE 2009, ’11]

SOLUTION

Any plane passing through the point (–1, 3, 2) is given by a( x + 1) + b( y - 3) + c(z - 2) = 0.

... (i)

Now, (i) being perpendicular to each of the planes x + 2y + 3z - 5 = 0 and 3 x + 3 y + z = 0, we have, ( a ´ 1) + ( b ´ 2) + ( c ´ 3) = 0 Þ a + 2b + 3 c = 0 ( a ´ 3) + ( b ´ 3) + ( c ´ 1) = 0 Þ 3 a + 3 b + c = 0. On solving (ii) and (iii) by cross multiplication, we get a b c = = ( 2 - 9) ( 9 - 1) ( 3 - 6)

... (ii) ... (iii)

a b c a b c = = Þ = = = l (say) -7 8 -3 7 -8 3 Þ a = 7 l , b = -8l and c = 3l. Putting a = 7 l , b = -8l and c = 3l in (i), we get 7 l( x + 1) - 8l( y - 3) + 3 l(z - 2) = 0 Þ 7( x + 1) - 8( y - 3) + 3(z - 2) = 0 Þ 7 x - 8y + 3z + 25 = 0, which is the required equation of the plane.

Þ

EXAMPLE 7

Find the vector equation of the plane through the points ( 2, 1, - 1) and ( -1, 3 , 4) and perpendicular to the plane x - 2y + 4z = 10. [CBSE 2013]

SOLUTION

The general equation of a plane passing through the point A( 2, 1, - 1) is given by

SSS Mathematics for Class 12 1217

The Plane

1217

a( x - 2) + b( y - 1) + c(z + 1) = 0.

... (i)

If the point B( -1, 3 , 4) lies on plane (i), then we have a( -1 - 2) + b( 3 - 1) + c( 4 + 1) = 0 Þ

-3 a + 2b + 5 c = 0.

... (ii)

If the plane (i) is perpendicular to the plane x - 2y + 4z = 10, then we have (1 ´ a) - ( 2 ´ b) + ( 4 ´ c) = 0 Þ

a - 2b + 4c = 0.

... (iii)

On solving (ii) and (iii) by cross multiplication, we have a b c = = ( 8 + 10) (5 + 12) ( 6 - 2) Þ

a b c = = = l (say) 18 17 4

Þ

a = 18l , b = 17 l and c = 4l.

Substituting these values of a, b and c in (i), we get 18l( x - 2) + 17 l( y - 1) + 4l(z + 1) = 0 Þ

18( x - 2) + 17( y - 1) + 4(z + 1) = 0

Þ

18x + 17 y + 4z = 49.

Required equation of the plane in vector form is given by ®

^

^

^

r × (18 i + 17 j + 4 k ) = 49.

EXAMPLE 8

Find the equation of the plane passing through the point (1, 3 , 2) and parallel to the plane 3 x - 2y + 2z + 33 = 0.

SOLUTION

The equation of a plane parallel to the given plane is of the form 3 x - 2y + 2z = k for some scalar k. Since it passes through the point A(1, 3, 2), we have ( 3 ´ 1) - ( 2 ´ 3) + ( 2 ´ 2) = k Þ k = ( 3 - 6 + 4) = 1. Hence, the required equation of the plane is 3 x - 2y + 2z - 1 = 0.

EXERCISE 28F 1. Find the acute angle between the following planes: ®

^

®

^

^

®

^

^

^

^

(i) r × ( i + j - 2 k ) = 5 and r × ( 2 i + 2 j - k ) = 9 ^

®

^

^

^

^

(ii) r × ( i + 2 j - k ) = 6 and r × ( 2 i - j - k ) + 3 = 0 ®

^

^

^

®

^

^

(iii) r × ( 2 i - 3 j + 4 k ) = 1 and r × ( - i + j ) = 4

... (i)

SSS Mathematics for Class 12 1218

1218

Senior Secondary School Mathematics for Class 12 ®

^

^

®

^

^

^

^

(iv) r × ( 2 i - 3 j + 6 k ) = 8 and r × ( 3 i + 4 j - 12 k ) + 7 = 0 2. Show that the following planes are at right angles: ®

^

^

^

®

^

^

^

®

^

^

^

(i) r × ( 4 i - 7 j - 8 k ) = 5 and r × ( 3 i - 4 j + 5 k) + 10 = 0 ®

^

^

^

(ii) r × ( 2 i + 6 j + 6 k ) = 13 and r × ( 3 i + 4 j - 5 k ) + 7 = 0 3. Find the value of l for which the given planes are perpendicular to each other: ®

^

®

^

^

®

^

^

^

^

^

^

^

(i) r × ( 2 i - j - l k ) = 7 and r × ( 3 i + 2 j + 2 k) = 9 ^

®

^

(ii) r × ( l i + 2 j + 3 k ) = 5 and r × ( i + 2 j - 7 k ) + 11 = 0 4. Find the acute angle between the following planes: (i) 2x - y + z = 5 and x + y + 2z = 7 (ii) x + 2y + 2z = 3 and 2x - 3 y + 6z = 8 (iii) x + y - z = 4 and x + 2y + z = 9 (iv) x + y - 2z = 6 and 2x - 2y + z = 11 5. Show that each of the following pairs of planes are at right angles: (i) 3 x + 4y - 5 z = 7 and 2x + 6y + 6z + 7 = 0 (ii) x - 2y + 4z = 10 and 18x + 17 y + 4z = 49 6. Prove that the plane 2x + 3 y - 4z = 9 is perpendicular to each of the planes x + 2y + 2z - 7 = 0 and 5 x + 6y + 7z = 23. 7. Show that the planes 2x - 2y + 4z + 5 = 0 and 3 x - 3 y + 6z - 1 = 0 are parallel. 8. Find the value of l for which the planes x - 4y + lz + 3 = 0 and 2x + 2y + 3z = 5 are perpendicular to each other. 9. Write the equation of the plane passing through the origin and parallel to the plane 5 x - 3 y + 7z + 11 = 0. 10. Find the equation of the plane passing through the point ( a , b , c) and ®

^

^

^

parallel to the plane r × ( i + j + k) = 2.. 11. Find the equation of the plane passing through the point (1, –2, 7) and parallel to the plane 5 x + 4y - 11z = 6. 12. Find the equation of the plane passing through the point A( -1, - 1, 2) and perpendicular to each of the planes 3 x + 2y - 3z = 1 and 5 x - 4y + z = 5. [CBSE 2008]

13. Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y - z = 1 and 3 x - 4y + z = 5. [CBSE 2004] 14. Find the equation of the plane that contains the point A(1, - 1, 2) and is perpendicular to both the planes 2x + 3 y - 2z = 5 and x + 2y - 3z = 8. Hence, find the distance of the point P( -2, 5 , 5) from the plane obtained [CBSE 2014] above.

SSS Mathematics for Class 12 1219

The Plane

1219

15. Find the equation of the plane passing through the points A(1, - 1, 2) and [CBSE 2005] B( 2, - 2, 2) and perpendicular to the plane 6x - 2y + 2z = 9. 16. Find the equation of the plane passing through the points A( -1, 1, 1) and B(1, - 1, 1) and perpendicular to the plane x + 2y + 2z = 5. 17. Find the equation of the plane through the points A( 3 , 4, 2) andB(7 , 0, 6) and perpendicular to the plane 2x - 5 y = 15. [CBSE 2012] The given plane is 2 x - 5 y + 0 z = 15.

HINT:

18. Find the equation of the plane through the points A( 2, 1, - 1) and B( -1, 3 , 4) and perpendicular to the plane x - 2y + 4z = 10. Also, show that the plane thus obtained contains the line ®

^

^

^

^

^

^

r = ( - i + 3 j + 4 k) + l( 3 i - 2 j - 5 k ).

[CBSE 2012]

ANSWERS (EXERCISE 28F)

æ 6ö 1 ÷ (ii) cos-1 æç ö÷ 1. (i) cos -1 çç ÷ 6ø 3 è ø è 3. (i) l = -2 (ii) l = 17 4. (i)

p 3

æ 8ö (ii) cos-1 ç ÷ è 21 ø

æ 5 ö (iii) cos-1 ç ÷ è 58 ø

æ 2ö ÷ (iii) cos-1 çç ÷ è 3 ø ®

^

^

æ 6ö (iv) cos-1 ç ÷ è7 ø

æ 2 ö (iv) cos-1 ç ÷ è 3 6ø

^

10. r × ( i + j + k) = a + b + c

9. 5 x - 3 y + 7z = 0 12. 5 x + 9y + 11z = 8

13. x + 2y + 5z = 0

8. l = 2

11. 5 x + 4y - 11z + 80 = 0

14. 5 x - 4y - z - 7 = 0, 42 units

15. x + y - 2z + 4 = 0 16. 2x + 2y - 3z + 3 = 0 18. 18x + 17 y + 4z = 49

17. 5 x + 2y - 3z - 17 = 0

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28F) ® ®

® ®

10. We know that any plane parallel to r × n = q is r × n = q1. ®

^

^

^

Let the required plane parallel to the given plane be r × ( i + j + k ) = q1 ^

^

^

^

^

^

^

^

^

^

Þ (x i + y j + zk )×( i + j + k ) = q

1

^

^

Þ (a i + b j + ck )×( i + j + k ) = q

1

Þ q1 = ( a + b + c ).

®

^

^

^

Hence, the required equation is r × ( i + j + k ) = ( a + b + c ). 14. Any plane through A( 1, - 1, 2 ) is given by ... (i) a( x - 1) + b( y + 1) + c(z - 2 ) = 0. Since it is perpendicular to each of the planes 2 x + 3 y - 2 z = 5 and x + 2 y - 3z = 8, we have

SSS Mathematics for Class 12 1220

1220

Senior Secondary School Mathematics for Class 12

2a + 3b - 2c = 0 a + 2 b - 3 c = 0. On solving (ii) and (iii) by cross multiplication, we have a b c = = = l Þ a = -5 l, b = 4 l, c = l. ( -9 + 4 ) ( -2 + 6 ) ( 4 - 3 )

... (ii) ... (iii)

Putting these value in (i), we get the required equation as -5 l( x - 1) + 4 l( y + 1) + l(z - 2 ) = 0 Þ 5( x - 1) - 4( y + 1) - (z - 2 ) = 0 Þ 5 x - 4 y - z - 7 = 0. Distance of the point P( -2 , 5 , 5 ) from this plane is given by 5 ´ ( -2 ) - 4 ´ 5 - 5 - 7 -42 42 d= = = = 42 units. 2 2 2 42 42 5 + ( -4 ) + ( -1) 18. Obtain the required equation of the plane as 18 x + 17 y + 4z = 49. ... (A) ®

^

^

^

The given line is r = ( 3 l - 1) i + ( 3 - 2 l) j + ( 4 - 5 l) k . Coordinates of any point on this line are ( 3 l - 1, 3 - 2 l, 4 - 5 l). These coordinates satisfy (A), as shown below: LHS = 18( 3 l - 1) + 17 ( 3 - 2 l) + 4( 4 - 5 l) = 49 = RHS.

So, ( 3 l - 1, 3 - 2 l, 4 - 5 l) lies on plane (A). Hence, the plane (A) obtained above contains the given line.

ANGLE BETWEEN A LINE AND A PLANE

The angle between a line and a plane is the complement of the angle between the line and normal to the plane.

Vector Form ®

THEOREM 1

®

®

® ®

If f is the angle between the line r = a + l b and the plane r × n = q , then prove that ® ®

|b × n|

sin f = ® ® × | b || n | ®

PROOF

®

®

®

We know that the line r = a + l b is parallel to b . ® ®

®

And, the plane r × n = q is normal to n . Let q be the angle between the line and the normal to the plane. Then, ® ®

|b × n|

cos q = ® ® × | b || n |

SSS Mathematics for Class 12 1221

The Plane

1221

Let f be the angle between the line and the plane Then, f = 90° - q Þ q = 90° - f. ® ®

® ®

|b × n| |b × n| cos( 90° - f) = ® ® Þ sin f = ® ® × | b || n | | b || n |

\

Two Important Results

1. Condition for a given line to be perpendicular to a given plane ®

®

®

® ®

The line r = a + l b is perpendicular to the plane r × n = q ®

® ®

Û b is perpendicular to the plane r × n = q ®

®

Û b is parallel to the normal n to the plane ®

®

Û b = t n , for some scalar t. 2. Condition for a given line to be parallel to a given plane ®

®

®

® ®

The line r = a + l b is parallel to the plane r × n = q ®

® ®

Û b is parallel to the plane r × n = q ®

®

Û b is perpendicular to the normal n to the plane ® ®

Û b × n = 0. Cartesian Form THEOREM 2

If f is the angle between the line

x - x1 y - y1 z - z1 and the plane = = a1 b1 c1

a2x + b2y + c2z + d = 0, then sin f =

PROOF

ì a2 í 1 î

a1a2 + b1b2 + c1c2 × + b12 + c12 üý ìí a22 + b22 + c22 üý þ þî

The direction ratios of the given line are a1 , b1 , c1. ®

^

^

^

So, the given line is parallel to b = ( a1 i + b1 j + c1 k ). ®

^

^

^

The normal to the given plane is parallel to n = a2 i + b2 j + c2 k . Let f be the angle between the line and the plane. Then, ® ®

|b × n|

sin f = ® ® | b || n |

SSS Mathematics for Class 12 1222

1222

Senior Secondary School Mathematics for Class 12

|( a1 ^i + b1 ^j + c1 ^k ) × ( a2 ^i + b2 ^j + c2 ^k )|

=

|( a1 ^i + b1 ^j + c1 ^k | |( a2 ^i + b2 ^j + c2 ^k|

=

ì í î

a12

a1a2 + b1b2 + c1c2 × + b12 + c12 üý ìí a22 + b22 + c22 üý þî þ

Important Results

1. Condition for the given line to be perpendicular to the given plane x - x1 y - y1 z - z1 The line is perpendicular to the plane = = a1 b1 c1 a2x + b2y + c2z + d = 0 x - x1 y - y1 z - z1 is parallel to the normal to the plane Û the line = = a1 b1 c1 a2x + b2y + c2z + d = 0 a1 b1 c1 Û = = × a2 b2 c2 2. Condition for the given line to be parallel to the given plane x - x1 y - y1 z - z1 The line is parallel to the plane = = a1 b1 c1 a2x + b2y + c2z + d = 0 x - x1 y - y1 z - z1 is perpendicular to the normal to the Û the line = = a1 b1 c1 plane a2x + b2y + c2z + d = 0 Û a1a2 + b1b2 + c1c2 = 0. 3. Distance between a line and a plane, parallel to each other ®

®

®

® ®

If the line r = a + l b

is parallel to the plane r × n = q then the

® ®

distance between them is

| a × n - q| ®

, which is the same as the distance

|n| of a point from the plane. 4. Length of perpendicular from P( x1 , y1 , z1) to the plane ax + by + cz = d is ax1 + by1 + cz1 - d a 2 + b 2 + c2

×

SSS Mathematics for Class 12 1223

The Plane

1223

SUMMARY ®

®

®

1. If f is the angle between the line r = a + l b and the plane ® ®

r × n = q then ® ®

|b × n| sin f = ® ® × | b || n| ®

®

®

® ®

2. The line r = a + l b is perpendicular to the plane r × n = q only ®

®

when b = t n for some scalar t. ®

®

®

® ®

3. (i) The line r = a + l b is parallel to the plane r × n = q only ® ®

when b × n = 0. ®

®

®

® ®

(ii) If the line r = a + l b is parallel to the plane r × n = q then the distance between them is ® ®

| a × n - q| ®

×

|n| x - x1 y - y1 z - z1 and the = = a1 b1 c1 plane a2x + b2y + c2z + d = 0 then a1a2 + b1b2 + c1 c 2 sin f = × 2 ( a1 + b12 + c12 ) ( a22 + b22 + c 22 )

4. If f is the angle between the line

5. The line

x - x1 y - y1 z - z1 is perpendicular to the plane = = a1 b1 c1

a2x + b2y + c2z + d = 0 only if 6. The line

a1 b1 c1 = = × a 2 b2 c 2

x - x1 y - y1 z - z1 is parallel to the plane = = a1 b1 c1

a2x + b2y + c2z + d = 0 only if a1a2 + b1b2 + c1 c 2 = 0. SOLVED EXAMPLES EXAMPLE 1

®

^

^

^

^

^

^

Find the angle between the line r = ( i + j - 3 k ) + l ( 2 i + 2 j + k ) ®

^

^

^

and the plane r × ( 6 i - 3 j + 2 k ) = 5. ®

SOLUTION

®

®

We know that the angle f between the line r = a + l b and the

SSS Mathematics for Class 12 1224

1224

Senior Secondary School Mathematics for Class 12 ® ®

plane r × n = q is given by ® ®

|b × n| sin f = ® ® × | b || n | ®

^

^

®

^

^

^

^

Here, b = ( 2 i + 2 j + k ) and n = ( 6 i - 3 j + 2 k ). \

sin f =

|( 2 ^i + 2 ^j + ^k ) × ( 6 ^i - 3 ^j + 2 ^k )| ì 22 + 22 + 12 ü ì 62 + ( -3) 2 + 22 ü ý í ý í þ î þ î

= Þ

(12 - 6 + 2)

=

{ 9 ´ 49}

8 8 = 3 ´ 7 21

æ 8ö f = sin -1 ç ÷ × è 21 ø

Hence, the angle between the given line and the given plane is æ 8ö sin -1 ç ÷ × è 21 ø ®

EXAMPLE 2

^

^

^

^

^

^

Find the value of m for which the line r = ( i + 2 j - k ) + l( 2 i + j + 2 k ) ®

^

^

^

is parallel to the plane r × ( 3 i - 2 j + m k ) = 12. ®

SOLUTION

®

®

We know that the line r = a + l b is parallel to the plane ® ®

® ®

r × n = q only when b × n = 0.

®

^

^

®

^

^

^

^

Here, b = ( 2 i + j + 2 k ) and n = ( 3 i - 2 j + m k ). So, the given line is parallel to the given plane Û

® ®

b×n =0 ^

^

^

^

^

^

Û (2 i + j + 2 k ) × ( 3 i - 2 j + m k) = 0 Û ( 2 ´ 3) + 1 ´ ( -2) + 2 ´ m = 0 Û 2m = - 4 Û m = - 2. Hence, the required value of m is –2. EXAMPLE 3

®

^

^

^

^

^

^

Show that the line r = ( 2 i - 2 j + 3 k ) + l( i - j + 4 k ) is parallel to ® ^

^

^

the plane r × ( i + 5 j + k ) = 5. Also, find the distance between the given line and the given plane. [CBSE 2010]

SSS Mathematics for Class 12 1225

The Plane ®

SOLUTION

1225 ®

®

The given line is in the form r = a + l b and the given plane is in ® ®

the form r × n = q, where ®

^

^

^

®

^

^

®

^

^

^

^

a = ( 2 i - 2 j + 3 k ) , b = ( i - j + 4 k ) , n = ( i + 5 j + k ) and q = 5. ®

®

®

We know that the line r = a + l b is parallel to the plane ® ®

® ®

r × n = q only when b × n = 0. ® ®

^

^

^

^

^

^

Here, b × n = ( i - j + 4 k ) × ( i + 5 j + k ) = (1 ´ 1) + ( -1) ´ 5 + 4 ´ 1 = 0. Hence, the given line is parallel to the given plane. Distance between the given line and the given plane ® ®

| a × n - q| |( 2 ^i - 2 ^j + 3 ^k ) × ( ^i + 5 ^j + ^k ) - 5| = ® |n | |^i + 5 ^j + ^k|

=

= =

( 2 ´ 1) + ( -2) ´ 5 + ( 3 ´ 1) - 5 12 + 5 2 + 12 2 - 10 + 3 - 5 27

=

10 units. 3 3

Hence, the distance between the given line and the given plane is 10 units. 3 3 EXAMPLE 4

Find the vector equation of a line passing through the point A(1, - 1, 2) ®

^

^

^

and perpendicular to the plane r × ( 2 i - j + 3 k ) = 5. ®

SOLUTION

^

^

^

The position vector of the given point is a = ( i - j + 2 k ). ® ®

®

^

^

^

The given plane is r × n = q ; where n = ( 2 i - j + 3 k ) and q = 5. Since the required line is perpendicular to the given plane, so it ®

must be parallel to its normal n . Thus, we need a line which passes through a point having position ®

®

vector a and which is parallel to vector n . So, its vector equation is ®

®

®

®

^

^

^

^

^

^

r = a + l n , i.e., r = ( i - j + 2 k ) + l ( 2 i - j + 3 k )

for some scalar l.

SSS Mathematics for Class 12 1226

1226

EXAMPLE 5

Senior Secondary School Mathematics for Class 12

Find the angle between the line 2x - 3 y + z = 5.

SOLUTION

The given line is

x-2 y+ 3 z+ 4 and the plane = = -1 2 3

x - x1 y - y1 z - z1 , where a1 = -1, b1 = 2, c1 = 3. = = a1 b1 c1

The given plane is a2x + b2y + c2z + d = 0, where a2 = 2, b2 = -3 , c2 = 1, d = -5. Let the required angle be f. Then sin f =

=

= Þ

| a1a2 + b1b2 + c1c2 | ì a 2 + b 2 + c 2 ü× ì a 2 + b 2 + c 2 ü í 1 1 1 ý í 2 2 2 ý þ î þî ( -1) ´ 2 + 2 ´ ( -3) + 3 ´ 1 ì ( -1) 2 + 22 + 3 2 ü × ì 22 + ( -3) 2 + 12 ü ý í ý í þ î þ î | - 2 - 6 + 3 | |- 5| 5 = = 14 14 { 14 ´ 14}

æ5ö f = sin -1 ç ÷ × è 14 ø

Hence, the angle between the given line and and the given plane is æ5ö sin -1 ç ÷ × è 14 ø EXAMPLE 6

Find the equation of the line passing through the point ( 3 , 0, 1) and [CBSE 2012] parallel to the planes x + 2y = 0 and 3 y - z = 0.

SOLUTION

Let the direction ratios of the required line be a , b , c. Then, its equation is given by x - 3 y - 0 z -1 ... (i) = = × a b c Since the line (i) is parallel to each of the planes x + 2y + 0z = 0 and 0x + 3 y - z = 0, so it must be perpendicular to the normal of each of these planes. … (ii) \ a ´ 1 + b ´ 2 + c ´ 0 = 0 Þ a + 2b + 0c = 0 and a ´ 0 + b ´ 3 + c ´ ( -1) = 0 Þ 0a + 3 b - c = 0 … (iii) On solving (ii) and (iii) by cross multiplication, we get a b c = = ( -2 - 0) ( 0 + 1) ( 3 - 0) a b c = = = l (say) Þ a = -2l , b = l and c = 3l. -2 1 3 Putting these values of a , b , c in (i), we get

Þ

SSS Mathematics for Class 12 1227

The Plane

1227

x - 3 y - 0 z -1 x - 3 y z -1 = = Þ = = × -2l l 3l -2 1 3 x - 3 y z -1 Hence, the required equation of the line is = = × -2 1 3 EXAMPLE 7

Find the equation of the plane passing through the point A(1, 2, 1) and perpendicular to the line joining the points P(1, 4, 2) and Q( 2, 3 , 5). Also find the distance of this plane from the line x + 3 y -5 z -7 [CBSE 2010C, ’11] = = × 2 -1 -1

SOLUTION

The general equation of a plane passing through the point A(1, 2, 1) is given by ... (i) a( x - 1) + b( y - 2) + c(z - 1) = 0. D.r.’s of noraml to the plane (i) are a, b, c. D.r.’s of line PQ are ( 2 - 1), ( 3 - 4), (5 - 2), i.e., 1, –1, 3. Since the required plane is perpendicular to line PQ, so the normal to this plane must be parallel to PQ. a b c = = = l (say) Þ a = l , b = - l and c = 3l. \ 1 -1 3 Putting these values in (i), we get the required equation of the plane as l( x - 1) - l( y - 2) + 3 l(z - 1) = 0 ... (ii) Þ ( x - 1) - ( y - 2) + 3(z - 1) = 0 Þ x - y + 3z = 2 Hence, the required equation of the plane is x - y + 3z = 2. x + 3 y -5 z -7 ... (iii) The given line is = = × 2 -1 -1 This line passes through the point (–3, 5, 7). Distance between the plane (ii) and the line (iii) = Distance of any point on this line from the plane = Length of perpendicular from (–3, 5, 7) on plane (ii) | - 3 - 5 + 21 - 2| 11 = = = 11 units. 11 12 + ( -1) 2 + 3 2 Hence, the distance between the desired plane and the given line is 11 units.

EXAMPLE 8

Find the equation of the plane passing through the points ( 0, 0, 0) and x-4 y+ 3 z+1 [CBSE 2011] = = × ( 3 , -1, 2) and parallel to the line 1 -4 7

SOLUTION

The general equation of the plane passing through the point (3, –1, 2) is given by … (i) a( x - 3) + b( y + 1) + c(z - 2) = 0. If this plane passes through the point (0, 0, 0), we have … (ii) a( 0 - 3) + b( 0 + 1) + c( 0 - 2) = 0 Þ - 3 a + b - 2c = 0.

SSS Mathematics for Class 12 1228

1228

Senior Secondary School Mathematics for Class 12

D.r.’s of the normal to the plane (i) are a, b, c. x-4 y+ 3 z+1 are 1, –4, 7. D.r.’s of the line = = 1 -4 7 The required plane is parallel to the given line only when the normal to this plane is perpendicular to this line. … (iii) \ ( a ´ 1) + b ´ ( -4) + c ´ 7 = 0 Þ a - 4b + 7 c = 0. On solving (ii) and (iii) by cross multiplication, we get: a b c = = (7 - 8) ( -2 + 21) (12 - 1) a b c = = = l ( say) Þ a = - l , b = 19l and c = 11l. -1 19 11 Putting these values of a , b , c in (i), we get: - l( x - 3) + 19l( y + 1) + 11l(z - 2) = 0 Þ ( x - 3) - 19( y + 1) - 11(z - 2) = 0 Þ x - 19y - 11z = 0. Hence, the required equation of the plane is x - 19y - 11z = 0. Þ

EXAMPLE 9

Find the equation of the plane which passes through the point ( 4, - 1, 2) x -1 y - 3 z - 4 and which is parallel to each of the lines and = = 1 2 3 x+ 2 y-2 z+1 = = × 3 -1 2

SOLUTION

The general equation of a plane passing through the point A( 4, - 1, 2) is given by ... (i) a( x - 4) + b( y + 1) + c(z - 2) = 0 This plane will be parallel to each of the given lines only when the normal to the plane is perpendicular to each of the given lines. ... (ii) \ (1 ´ a) + ( 2 ´ b) + ( 3 ´ c) = 0 Þ a + 2b + 3 c = 0. ... (iii) ( 3 ´ a) + ( -1) ´ b + ( 2 ´ c) = 0 Þ 3 a - b + 2c = 0 On solving (ii) and (iii) by cross multiplication, we get a b c = = ( 4 + 3) ( 9 - 2) ( -1 - 6) a b c a b c = = Þ = = = l (say) 7 7 -7 1 1 -1 Þ a = l , b = l and c = -l. Putting these values of a , b , c in (i), we get l( x - 4) + l( y + 1) - l(z - 2) = 0 Þ ( x - 4) + ( y + 1) - (z - 2) = 0 Þ x + y - z = 1. Hence, the required equation of the plane is x + y - z = 1.

Þ

EXAMPLE 10

Find the equation of the plane passing through ( 2, 3 , - 4) and (1, - 1, 3) and parallel to the x-axis.

SSS Mathematics for Class 12 1229

The Plane

1229

SOLUTION

The general equation of the plane passing through the point A( 2, 3 , - 4) is given by ... (i) a( x - 2) + b( y - 3) + c(z + 4) = 0. Since it passes through the point B(1, - 1, 3), we have ... (ii) a(1 - 2) + b( -1 - 3) + c( 3 + 4) = 0 Þ - a - 4b + 7 c = 0. If this plane is parallel to x-axis, then the normal to the plane is perpendicular to the x-axis. D.r.’s of normal to plane (i) are a , b , c and d.r.’s of the x-axis are 1, 0, 0. \ a ´ 1 + b ´ 0 + c ´ 0 = 0 Þ a = 0. Putting a = 0 in (ii), we get 7 c - 4b = 0. 4l Let b = l. Then, 7 c = 4l Þ c = × 7 4l Putting a = 0, b = l and c = in (i), we get 7 4l 0 ´ ( x - 2) + l( y - 3) + (z + 4) = 0 7 4 Þ ( y - 3) + (z + 4) = 0 Þ 7 y - 21 + 4z + 16 = 0 7 Þ 7 y + 4z - 5 = 0. Hence, the required equation of the plane is 7 y + 4z - 5 = 0.

EXAMPLE 11

Find the equation of the plane passing through the point ( 0, 7 , - 7) and x+1 y- 3 z+ 2 containing the line = = × -3 2 1

SOLUTION

The general equation of the plane passing through the point ( 0, 7 , - 7) is given by … (i) a( x - 0) + b( y - 7) + c(z + 7) = 0. If (i) contains the given line, then it must pass through the point ( -1, 3 , - 2) and must be parallel to the given line. If (i) passes through the point (–1, 3, –2), we have ... (ii) a( -1 - 0) + b( 3 - 7) + c( -2 + 7) = 0 Þ a + 4b - 5 c = 0. If (i) is parallel to the given line, then its normal should be perpendicular to this line. ... (iii) \ ( -3) a + 2b + 1 ´ c = 0 Þ - 3 a + 2b + c = 0. On solving (ii) and (iii) by cross multiplication, we get a b c a b c = = Þ = = ( 4 + 10) (15 - 1) ( 2 + 12) 14 14 14 a b c = = = l (say). 1 1 1 Then a = l, b = l and c = l.

Þ

SSS Mathematics for Class 12 1230

1230

Senior Secondary School Mathematics for Class 12

Putting a = l , b = l and c = l in, we get lx + l( y - 7) + l(z + 7) = 0 Þ x + ( y - 7) + (z + 7) = 0 Þ x + y + z = 0. Hence, the required equation of the plane is x + y + z = 0. EXAMPLE 12

Find the equation of the plane passing through the line of intersection of the planes 2x + y - z = 3 and 5 x - 3 y + 4z + 9 = 0 and parallel to the line x -1 y - 3 z -5 [CBSE 2011] = = × 2 4 5

SOLUTION

The equation of a plane passing through the intersection of the given planes is given by ( 2x + y - z - 3) + l(5 x - 3 y + 4z + 9) = 0 for some real number l Þ

... (i) ( 2 + 5 l) x + (1 - 3 l) y + ( 4l - 1)z + ( 9l - 3) = 0 x -1 y - 3 z -5 If this plane is parallel to the line = = , then the 2 4 5 normal to the plane is perpendicular to this line.

\

2( 2 + 5 l) + 4(1- 3l) + 5(4l - 1) = 0

Þ

(10l - 12l + 20l) + ( 4 + 4 - 5) = 0 -3 -1 = × Þ 18l = -3 Þ l = 18 6 -1 Putting l = in (i), we get 6 5ö 3ö ö æ -9 æ -4 ö æ æ - 3÷ = 0 - 1÷z + ç ç 2 - ÷ x + ç1 + ÷ y + ç 6ø 6ø ø è 6 ø è 6 è è Þ 7 x + 9y - 10z - 27 = 0. Hence, the required equation of the plane is 7 x + 9y - 10z - 27 = 0. EXAMPLE 13

Show that the equation by + cz + d = 0 represents a plane parallel to the x-axis. Find the equation of a plane which is parallel to the x-axis and passes through the points A( 2, 3 , 1) and B( 4. - 5 , 3).

SOLUTION

The given equation is 0 × x + by + cz + d = 0, which is of the form ax + by + cz + d = 0. Hence, the given equation represents a plane. D.r.’s of the normal to this plane are 0, b, c. D.r.’s of the x-axis are 1, 0, 0. Now 0 ´ 1 + b ´ 0 + c ´ 0 = 0. This shows that the given plane is parallel to the x-axis. Thus, the equation of a plane parallel to the x-axis is ... (i) by + cz + d = 0. If it passes through the points A( 2, 3 , 1) and B( 4, - 5 , 3), we have ... (ii) 3b + c + d = 0 ... (ii) -5 b + 3 c + d = 0

SSS Mathematics for Class 12 1231

The Plane

1231

On solving (ii) and (iii) by cross multiplication, we get b c d = = (1 - 3) ( -5 - 3) ( 9 + 5) Þ

b c d b c d = = Þ = = = k (say) -2 -8 14 1 4 -7

Þ

b = k , c = 4k and d = -7 k.

Putting these values in (i), we get ky + 4kz - 7 k = 0 Þ y + 4z - 7 = 0. Hence, the required equation of the plane is y + 4z - 7 = 0.

EXERCISE 28G ®

^

^

^

^

^

^

^

^

1. Find the angle between the line r = ( i + 2 j - k ) + l( i - j + k ) and the ®

^

^

^

plane r × ( 2 i - j + k ) = 4. ®

^

^

^

^

2. Find the angle between the line r = ( 2 i - j + 3 k ) + l( 3 i - j + 2 k ) and the ®

^

^

^

plane r × ( i + j + k ) = 3. ®

^

^

^

^

3. Find the angle between the line r = ( 3 i + k ) + l( j + k ) and the plane ®

^

^

^

r × ( 2 i - j + 2 k ) = 1. x-2 y+1 z- 3 and the plane = = 3 -1 2

4. Find the angle between the line 3 x + 4y + z + 5 = 0. 5. Find the angle between the line 10x + 2y - 11z = 3.

x+1 y z- 3 and the plane = = 2 3 6

6. Find the angle between the line joining the points A( 3 , - 4, - 2) and B(12, 2, 0) and the plane 3 x - y + z = 1. 7. If the plane 2x - 3 y - 6z = 13 makes an angle sin -1( l) with the x-axis, then find the value of l. ®

^

^

^

^

^

^

8. Show that the line r = ( 2 i + 5 j + 7 k ) + l( i + 3 j + 4 k ) is parallel to the ®

^

^

^

plane r × ( i + j - k ) = 7. Also, find the distance between them. ®

^

^

^

^

^

9. Find the value of m for which the line r = ( i + 2 k ) + l( 2 i - m j - 3 k ) is ®

^

^

^

parallel to the plane r × (m i + 3 j + k ) = 4.

SSS Mathematics for Class 12 1232

1232

Senior Secondary School Mathematics for Class 12

10. Find the vector equation of a line passing through the origin and ®

^

^

^

perpendicular to the plane r × ( i + 2 j + 3 k ) = 3. 11. Find the vector equation of the line passing through the point with position

vector

®

^

^

^

^

^

( i - 2 j + 5 k)

and

perpendicular

to

the

plane

^

r × ( 2 i - 3 j - k ) = 0.

12. Show that the equation ax + by + d = 0 represents a plane parallel to the z-axis. Hence, find the equation of a plane which is parallel to the z-axis and passes through the points A(2, –3, 1) and B(–4, 7, 6). 13. Find the equation of the plane passing through the points (1, 2, 3) and x -1 y + 2 z [CBSE 2006] = = × ( 0, - 1, 0) and parallel to the line 2 3 -3 14. Find the equation of a plane passing through the point (2, –1, 5), perpendicular to the plane x + 2y - 3z = 7 and parallel to the line x+5 y+1 z-2 = = × 3 -1 1 15. Find the equation of the plane passing through the intersection of the planes 4x - y + z = 10 and x + y - z = 4 and parallel to the line with direction ratios 2, 1, 1. Find also the perpendicular distance of (1, 1, 1) from this plane.

ANSWERS (EXERCISE 28G)

1. sin 3. sin

-1

-1

æ2 2ö ç ÷ ç 3 ÷ è ø

2. sin

æ 1 ö ç ÷ è 3 2ø

æ 7 ö 4. sin -1 ç ÷ è 2 91 ø

®

^

^

^

10. r = l ( i + 2 j + 3 k ) for some scalar l

9. m = -3 ^

æ 4 ö ç ÷ è 42 ø

æ 23 ö 6. sin -1 ç ÷ è 11 11 ø 7 8. units 3

æ 8ö 5. sin -1 ç ÷ è 21 ø 2 7. 7

®

-1

^

^

^

^

^

11. r = ( i - 2 j + 5 k ) + l( 2 i - 3 j - k ) for some scalar l 12. 5 x + 3 y - 1 = 0

13. 6x - 3 y + z = 3

14. x + 10y + 7z - 27 = 0

15. 5 y - 5z - 6 = 0,

3 2 5

SSS Mathematics for Class 12 1233

The Plane

1233

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28G) y+ 4 x- 3 z+ 2 x- 3 y+ 4 z+ 2 , i.e., = = = = × ( 12 - 3 ) ( 2 + 4 ) ( 0 + 2 ) 9 6 2

6. Equation of line AB is

7. D.r.’s of the x-axis are 1, 0, 0 and d.r.’s of normal to the plane are 2, –3, –6. Let f be the angle between the x-axis and the given plane. Then, | 1 ´ 2 + 0 ´ ( -3 ) + 0 ´ ( -6 )| 2 æ2ö sin f = = Þ f = sin -1 ç ÷. ì 12 + 0 2 + 0 2 ü ì 2 2 + ( -3 ) 2 + ( -6 ) 2 ü 7 è7 ø í ýí ý î þî þ 2 Hence, l = × 7 ®

®

®

® ®

8. We know that a line r = a + l b is parallel to the plane r × n = q only when this line is perpendicular to the normal to the plane. ® ®

So, we must have b × n = 0. ® ®

^

^

^

^

^

^

Here, ( b × n ) = ( i + 3 j + 4 k ) × ( i + j - k ) = ( 1 + 3 - 4 ) = 0. Hence, the given line is parallel to the given plane. Required distance between the line and the plane ® ®

^

^

^

^

^

^

| a × n - q| |( 2 i + 5 j + 7 k ) × ( i + j - k ) - 7 | = = ® ^ ^ ^ | n| | i + j - k| =

|2 + 5 - 7 - 7| 2

2

1 + 1 + ( -1)

2

=

|- 7| 3

=

7 units. 3

12. The given equation is ax + by + 0 × z + d = 0 which is of the form ax + by + cz + d = 0. Therefore, it represents a plane. D.r.’s of normal to the plane are a, b , 0. D.r.’s of the z-axis are 0, 0, 1. Now, a ´ 0 + b ´ 0 + 0 ´ 1 = 0. This shows that the given plane is parallel to the z-axis. Let the required plane be ax + by + d = 0. Since it passes through the points A( 2 , - 3 , 1) and B( -4 , 7 , 6 ), we have 2a - 3b + d = 0 -4 a + 7 b + d = 0 On solving (ii) and (iii) by cross multiplication, we get a b c = = ( -3 - 7 ) ( -4 - 2 ) ( 14 - 12 ) Þ

a b c a b c = = Þ = = = k (say). -10 -6 2 5 3 -1

\ a = 5 k , b = 3 k and c = - k. Putting these values in (i), we get 5 kx + 3 ky - k = 0 Þ 5 x + 3 y - 1 = 0 , which is the required equation of the plane.

... (i) ... (ii) ... (iii)

SSS Mathematics for Class 12 1234

1234

Senior Secondary School Mathematics for Class 12

13. Any plane through the point (1, 2, 3) is a( x - 1) + b( y - 2 ) + c(z - 3 ) = 0. Since it passes through the point (0, –1, 0), we have

... (i)

a( 0 - 1) + b( -1 - 2 ) + c( 0 - 3 ) = 0 Þ a + 3 b + 3 c = 0 x-1 y+ 2 z This plane is parallel to the line = = , so 2 3 -3

... (ii)

2a + 3b - 3c = 0 On solving (ii) and (iii), we get a b c a b c a b c = = Þ = = Þ = = = l (say). ( 9 + 9 ) ( -3 - 6 ) ( 6 - 3 ) 18 -9 3 6 -3 1

... (iii)

Hence, the required equation of the plane is 6 l( x - 1) - 3 l( y - 2 ) + l(z - 3 ) = 0 Þ 6 x - 3 y + z = 3.

Equation of a Plane Passing through a Given Point and Parallel to Two Given Lines Vector Form THEOREM 1

The vector equation of a plane passing through a given point with ®

®

®

position vector a and parallel to two given vectors b and c is ®

®

®

®

( r - a ) × ( b ´ c ) = 0. PROOF

®

®

The required plane is parallel to vectors b and c . ®

®

®

So, the vector n = ( b ´ c ) is perpendicular to this plane. Thus, we have to find the equation of a plane passing through the ®

®

point with position vector a and perpendicular to the vector n . So, its equation is ® ®

®

®

®

®

®

( r - a ) × n = 0, i.e., ( r - a ) × ( b ´ c ) = 0. ®

®

®

®

Hence, the required equation of the plane is ( r - a ) × ( b ´ c ) = 0. Cartesian Form THEOREM 2

PROOF

The equation of the plane passing through a given point A( x1 , y1 , z1) and parallel to two given lines having direction ratios b1 , b2 , b 3 and c1 , c2 , c 3 is ½ x - x1 y - y1 z - z1 ½ ½ b b2 b 3 ½ = 0. ½ 1 ½ c2 c3 ½ ½ c1

Let us consider a plane passing through a given point A( x1 , y1 , z1) and parallel to two given lines having direction ratios b1 , b2 , b 3 and c1 , c2 , c 3.

SSS Mathematics for Class 12 1235

The Plane

1235

Let P( x , y , z) be an arbitrary point on the plane. Then, ¾®

AP = (p.v. of P) - (p.v. of A) ^

^

^

^

^

^

= ( x i + y j + z k ) - ( x1 i + y1 j + z1 k ) ^

^

^

= ( x - x1) i + ( y - y1) j + (z - z1) k . It is given that the plane is parallel to two lines having direction ratios b1 , b2 , b 3 and c1 , c2 , c 3. So, the given plane is parallel to each of the vectors ®

^

^

®

^

^

^

^

b = b1 i + b2 j + b 3 k and c = c1 i + c2 j + c 3 k .

¾®

®

®

\ AP , b and c are coplanar, and therefore their scalar triple product must be zero. ½ x - x1 y - y1 z - z1 ½ ½ b b2 b 3 ½ = 0, ½ 1 ½ c2 c3 ½ ½ c1 which is the required equation. SUMMARY OF THE RESULTS

(i) The vector equation of a plane passing through a point having position ®

®

®

vector a and parallel to vectors b and c , is given by ®

®

®

®

( r - a) ×( b ´ c) = 0 (ii) The Cartesian equation of a plane passing through a point A( x1 , y1 , z1) and parallel to two nonparallel lines having d.r.’s b1 , b2 , b 3 and c1 , c2 , c 3 is given by ½ x - x1 y - y1 z - z1 ½ ½ b b2 b 3 ½ = 0. ½ 1 ½ c2 c3 ½ ½ c1

SOLVED EXAMPLES EXAMPLE 1

Find the vector and Cartesian equations of the plane passing through the point (1, 2, - 4) and parallel to the lines ®

^

^

®

^

^

^

^

^

^

r = ( i + 2 j + k ) - l( 2 i + 3 j + 6 k ) ^

^

^

^

and r = ( i - 3 j + 5 k ) + m ( i + j - k ). SOLUTION

We know that the vector equation of a plane passing through a ®

®

®

® ®

® ®

point a and parallel to b and c is given by ( r × a ) × ( b ´ c ) = 0.

SSS Mathematics for Class 12 1236

1236

Senior Secondary School Mathematics for Class 12 ®

^

^ ®

^

^

^

^

Here a = ( i + 2 j - 4 k ), b = ( 2 i + 3 j + 6 k ) ®

^

®

®

^

^

and c = ( i + j - k ).

\

^

^

^

b ´c = 2

3

1

1

6 -1

i

j

k

^

^

^

^

^

^

= ( -3 - 6) i - ( -2 - 6) j + ( 2 - 3) k = ( -9 i + 8 j - k ). So, the required vector equation is ®

®

®

®

®

®

r ×( b ´ c ) = a ×( b ´ c )

Þ

®

^

^

^

®

^

^

^

^

^

^

^

^

^

r × ( -9 i + 8 j - k ) = ( i + 2 j - 4 k ) × ( -9 i + 8 j - k ) = ( -9 + 16 + 4) = 11

Þ Þ

r × ( -9 i + 8 j - k ) - 11 = 0

®

^

^

^

r × ( 9 i - 8 j + k ) + 11 = 0.

Hence, the required vector equation of the plane is ®

^

^

^

r × ( 9 i - 8 j + k ) + 11 = 0.

Clearly, the required Cartesian equation of the plane is 9x - 8y + z + 11 = 0. EXAMPLE 2

Find the Cartesian and vector equation of the plane passing through the point ( 2, 0, - 1) and parallel to the lines 1-y x y-2 = 2z . = = z + 1 and x - 4 = 2 -3 4

SOLUTION

The given lines are x y-2 z+1 x - 4 y -1 z and = = = = , 1 -3 4 1 1 -2 2 x y-2 z+1 x - 4 y -1 z i.e., and = = = = × -3 4 1 2 -4 1 So, the required plane passes through the point A(2, 0, –1) and it is parallel to the lines having direction ratios –3, 4, 1 and 2, –4, 1. Here ( x1 = 2, y1 = 0, z1 = -1), ( a1 = -3 , b1 = 4, c1 = 1) and ( a2 = 2, b2 = -4, c2 = 1). The required equation of the plane is

SSS Mathematics for Class 12 1237

The Plane

½ x - x1 ½ a1 ½ ½ a2 ½x - 2 Þ ½ -3 ½ ½ 2

y - y1

z - z1 ½ c1 ½ = 0 ½ c2 ½

b1 b2 y -0 4 -4

1237

z + 1½ 1 ½= 0 ½ 1 ½

Þ

( x - 2)( 4 + 4) - y( -3 - 2) + (z + 1)(12 - 8) = 0

Þ

8( x - 2) + 5 y + 4(z + 1) = 0

Þ

8x + 5 y + 4z - 12 = 0.

Hence, the required Cartesian equation of the plane is 8x + 5 y + 4z = 12 and its corresponding vector equation is ®

^

^

^

r × ( 8 i + 5 j + 4 k ) = 12.

EXERCISE 28H 1. Find the vector and Cartesian equations of the plane passing through the ^

^

^

^

^

origin and parallel to the vectors ( i + j - k ) and ( 3 i - k ). 2. Find the vector and Cartesian equations of the plane passing through the ®

^

^

^

^

^

point (3, –1, 2) and parallel to the lines r = ( - j + 3 k ) + l( 2 i - 5 j - k ) and ®

^

^

^

^

^

r = ( i - 3 j + k ) + m ( -5 i + 4 j ).

3. Find the vector equation of a plane passing through the point (1, 2, 3) and parallel to the lines whose direction ratios are 1, –1, –2, and –1, 0, 2. 4. Find the Cartesian and vector equations of a plane passing through the x -1 y - 2 z + 1 and point (1, 2, –4) and parallel to the lines = = 2 3 6 x -1 y + 3 z = = × 1 1 -1 5. Find the vector equation of the plane passing through the point ^

^

^

^

^

^

^

^

^

( 3 i + 4 j + 2 k ) and parallel to the vectors ( i + 2 j + 3 k ) and ( i - j + k ).

ANSWERS (EXERCISE 28H) ®

^

^

^

1. r × ( i + 2 j + 3 k ) = 0, x + 2y + 3z = 0 ®

^

^

^

2. r × ( 4 i + 5 j - 17 k ) + 27 = 0, 4x + 5 y - 17z + 27 = 0

SSS Mathematics for Class 12 1238

1238

Senior Secondary School Mathematics for Class 12

®

^

®

^

®

^

3. r × ( 2 i + k ) = 5 ^

^

^

^

4. 9x - 8y + z + 11 = 0, r × ( 9 i - 8 j + k ) + 11 = 0 ^

5. r × (5 i + 2 j - 3 k ) = 17 Condition for the Coplanarity of Two Lines Vector Form THEOREM 1

®

®

®

®

®

®

The condition for two lines r = a1 + l b1 and r = a2 + m b2 to be coplanar is that ®

®

®

®

®

®

®

®

®

®

( a2 - a1 ) × ( b1 ´ b2 ) = 0, i.e., [ a2 - a1 b1 b2 ] = 0. Also, the equation of the plane containing both these lines is ®

®

®

®

®

®

( r - a1 ) × ( b1 ´ b2 ) = 0 or ( r - a2 ) × ( b1 ´ b2) = 0. PROOF

The equations of the given lines are ®

®

®

®

®

®

r = a1 + l b1

… (i)

r = a2 + m b2.

… (ii) ®

The line (i) passes through a point A with position vector a1 and is ®

parallel to b1 . ®

The line (ii) passes through a point B with position vector a2 and is ®

parallel to b2. ®

¾®

®

Now, AB = (p.v. of B) - (p.v. of A) = ( a2 - a1 ). \

the given lines are coplanar

Û

AB , b1 , b2 are coplanar

Û

AB× ( b1 ´ b2) = 0

Û

( a2 - a1 ) × ( b1 ´ b2 ) = 0.

® ®

¾®

¾®

®

®

®

®

®

®

Equation of the Plane Containing Both the Lines ®

®

®

®

®

®

If the lines r = a1 + l b1 and r = a2 + m b2 are coplanar then their common ®

®

plane is perpendicular to the vector ( b1 ´ b2 ), and this plane passes through ®

®

each of the points a1 and a2.

SSS Mathematics for Class 12 1239

The Plane

1239

So, its equation is ®

®

®

®

®

®

®

®

( r - a1 ) × ( b1 ´ b2 ) = 0, or ( r - a2) × ( b1 ´ b2 ) = 0. Cartesian Form THEOREM 2

The lines x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 , and = = = = a1 b1 c1 a2 b2 c2 ½ x 2 - x1 are coplanar Û ½ a1 ½ ½ a2

y 2 - y1 b1 b2

z2 - z1 ½ c1 ½ = 0. ½ c2 ½

And, the equation of the plane containing both these lines is ½ x - x1 y - y1 z - z1 ½ ½ x - x 2 y - y 2 z - z2 ½ ½ a ½ b1 c1 = 0 or ½ a1 b1 c1 ½ = 0. ½ 1 ½ ½ ½ b2 c2 ½ b2 c2 ½ ½ a2 ½ a2 PROOF

The given lines are x - x1 y - y1 z - z1 = = a1 b1 c1

… (i)

x - x 2 y - y 2 z - z2 = = × a2 b2 c2

… (ii)

The line (i) passes through a point A( x1 , y1 , z1) and is parallel to the ®

^

^

^

vector u1 = a1 i + b1 j + c1 k . Also, the line (ii) passes through a point B( x 2 , y 2 , z2) and is parallel to ®

^

^

^

the vector u2 = a2 i + b2 j + c2 k . ¾®

Now, AB = (p.v. of B) - (p.v. of A) ^ ^ ^ = ( x 2 i + y 2 j + z2 k ) - ( x1 i$ + y1 j$ + z1 k$ ) ^

^

^

= ( x 2 - x1) i + ( y 2 - y1) j + (z2 - z1) k . \ lines (i) and (ii) are coplanar Û there is a plane which passes through the points A and B, and ®

®

which is parallel to each of u1 and u2 ¾®

® ®

¾®

® ®

Û AB , u1 , u2 are parallel to the same plane Û AB , u1 , u2 are coplanar ¾®

® ®

Û [AB , u1 , u2] = 0

SSS Mathematics for Class 12 1240

1240

Senior Secondary School Mathematics for Class 12

½ x 2 - x1 Û ½ a1 ½ ½ a2

y 2 - y1 b1 b2

z2 - z1 ½ c1 ½ = 0. ½ c2 ½

Equation of the Plane Containing Both the Lines If the lines (i) and (ii) are coplanar then their common plane is the plane containing the line (i) and parallel to the line (ii) or it is the plane containing the line (ii) and parallel to the line (i). Hence, its equation is ½ x - x2 ½ x - x1 y - y1 z - z1 ½ ½ a1 ½ b1 c1 = 0 or ½ a1 ½ ½ ½ b2 c2 ½ ½ a2 ½ a2

y - y2

z - z2 ½ c1 ½ = 0. ½ c2 ½

b1 b2

SUMMARY

1.

®

®

®

®

®

®

(i) Two lines r = a1 + l b1 and r = a2 + l b2 are coplanar only when ®

®

®

®

( a2 - a1) × ( b1 ´ b2) = 0. ®

®

®

®

®

®

®

®

(ii) If two lines r = a1 + l b1 and r = a2 + l b2 are coplanar, then the equation of the plane containing both of these lines is given by ®

®

®

®

®

®

{( r - a1) × ( b1 ´ b2) = 0} or {( r - a2) × ( b1 ´ b2) = 0}. x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 2. (i) The lines and are = = = = a1 b1 c1 a2 b2 c2 ½ x 2 - x1 y 2 - y1 z2 - z1 ½ coplanar only when½ a1 b1 c1 ½ = 0. ½ ½ b2 c2 ½ ½ a2 (ii) The equation of the common plane is ½ x - x 2 y - y 2 z - z2 ½ ½ x - x1 y - y1 z - z1 ½ ½ a1 b1 c1 ½ = 0. b1 c1 ½ = 0 or ½ a1 ½ ½ ½ ½ b2 c2 ½ b2 c2 ½ ½ a2 ½ a2

SOLVED EXAMPLES EXAMPLE 1

Show that the lines ®

^

^

^

^

®

^

^

^

^

^

r = ( i + j - k ) + l ( 3 i - j ) and r = ( 4 i - k ) + m ( 2 i + 3 k ) are coplanar. Also, find the equation of the plane containing both these [CBSE 2013C] lines. ®

SOLUTION

®

®

®

®

®

We know that the lines r = a1 + l b1 and r = a2 + m b2 are coplanar

SSS Mathematics for Class 12 1241

The Plane ®

1241

®

®

®

Û ( a2 - a1 ) × ( b1 ´ b2) = 0. ®

^

^ ®

^ ^ ®

^

^

^

^

^

®

^

^

^

^

^

Here, a1 = i + j - k , b1 = 3 i - j , a2 = 4 i - k and b2 = 2 i + 3 k. \

®

®

^

^

^

^

^

( a2 - a1 ) = ( 4 i - k ) - ( i + j - k ) = ( 3 i - j ). ½^ ½i ( b1 ´ b2 ) = ½ 3 ½ ½2 ®

®

^½

^

j

k½ 0½ 3½ ½

-1 0

½

½

^

^

^

= ( -3 - 0) i - ( 9 - 0) j + ( 0 + 2) k = - 3 i - 9 j + 2 k . \

®

®

®

®

^

^

^

^

^

( a2 - a1 ) × ( b1 ´ b2 ) = ( 3 i - j ) × ( -3 i - 9 j + 2 k ) = [ 3 ´ ( -3) + ( -1) ´ ( -9) + 0 ´ 2] = 0.

Hence, the given lines are coplanar. The equation of the plane containing both the given lines is given by ®

®

®

®

® ®

®

® ®

®

^

^

( r - a1 ) × ( b1 ´ b2 ) = 0 Û r × ( b1 ´ b2) = a1 × ( b1 ´ b2) Û Û

®

^

^

^

®

^

^

^

^

^

^

^

r × ( -3 i - 9 j + 2 k ) = ( i + j - k ) × ( -3 i - 9 j + 2 k ) ®

^

^

^

r × ( -3 i - 9 j + 2 k ) = ( -3 - 9 - 2) Û r × ( 3 i + 9 j - 2 k ) = 14.

Hence, the required equation of the plane is ®

^

^

^

r × ( 3 i + 9 j - 2 k ) = 14.

EXAMPLE 2

x + 3 y -1 z -5 x + 1 y - 2 z -5 and are = = = = -3 1 5 -1 2 5 coplanar. Also, find the equation of the plane containing these lines. Show that the lines

[CBSE 2013C] SOLUTION

We know that the lines x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 , and = = = = a1 b1 c1 a2 b2 c2 ½ x 2 - x1 are coplanar only when½ a1 ½ ½ a2

y 2 - y1 b1 b2

z2 - z1 ½ c1 ½ = 0. ½ c2 ½

Here, x1 = -3 , y1 = 1, z1 = 5 and x 2 = -1, y 2 = 2, z2 = 5. a1 = -3 , b1 = 1, c1 = 5 and a2 = -1, b2 = 2, c2 = 5

SSS Mathematics for Class 12 1242

1242

Senior Secondary School Mathematics for Class 12

\

½ x 2 - x1 ½ a1 ½ ½ a2

y 2 - y1 b1 b2

z2 - z1 ½ ½-1 + 3 c1 ½ = ½ -3 ½ ½ c2 ½ ½ -1

2 - 1 5 - 5½ 1 5 ½ ½ 2 5 ½

½ 2 1 0½ ½ 2 1 0½ = ½-3 1 5½ = ½-2 -1 0½ [R 2 ® ( R 2 - R 3)] ½ ½ ½ ½ ½-1 2 5½ ½-1 2 5½ = 5( -2 + 2) = (5 ´ 0) = 0 [expanding by c 3]. Hence, the given lines are coplanar. Equation of the plane containing both these lines is given by ½x + 3 y -1 z -5 ½ ½ x - x1 y - y1 z1 ½ ½ a1 1 5 ½= 0 b1 c1 ½ = 0 Û ½ -3 ½ ½ ½ ½ 2 5 ½ b2 c2 ½ ½ -1 ½ a2 Û ( x + 3)(5 - 10) - ( y - 1)( -15 + 5) + (z - 5)( -6 + 1) = 0 Û ( -5)( x + 3) + 10( y - 1) - 5(z - 5) = 0 Û -5 x + 10y - 5z = 0 Û x - 2y + z = 0. Hence, the required equation plane is x - 2y + z = 0. EXAMPLE 3

Show that the lines x-a+ d y-a z-a-d x-b+ c y-b z-b-c = = , and = = a -d a a+d b-g b b+ g are coplanar. Also find the equation of the plane containing them.

SOLUTION

The given lines are x - ( a - d) y - a z - ( a + d) = = a-d a a+d

… (i)

x - ( b - c) y - b z - ( b + c) = = × b-g b b+g

… (ii)

We know that the lines x - x1 y - y1 z - z1 x - x 2 y - y 2 z - z2 , and = = = = a1 b1 c1 a2 b2 c2 ½ x 2 - x1 are coplanar Û ½ a1 ½ ½ a2

y 2 - y1 b1 b2

z2 - z1 ½ c1 ½ = 0. ½ c2 ½

Here, x1 = ( a - d), y1 = a , z1 = ( a + d); x 2 = ( b - c), y 2 = b , z2 = ( b + c); a1 = ( a - d), b1 = a , c1 = ( a + d); and a2 = (b - g), b2 = b , c2 = (b + g).

SSS Mathematics for Class 12 1243

The Plane

\

1243

½ x 2 - x1 y 2 - y1 z2 - z1 ½ ½ a1 b1 c1 ½ ½ ½ b2 c2 ½ ½ a2 ½ ( b - c) - ( a - d) b - a ( b + c) - ( a + d) ½ ½ =½ a-d a a+d ½ ½ b-g b b+g ½ ½ ½ 2( b - a) b - a b + c - a + d ½ ½ {C1 ® C1 + C 3} = ½ 2a a a+d ½ ½ + 2 b b b g ½ ½ = 0 [Q C1 and C 2 are proportional].

Hence, the given lines are coplanar. Equation of the plane containing the given lines is given by x - ( a - d) y - a z - ( a + d) a -d b-g

a b

a+d b+ g

= 0.

Applying C1 ® C1 + C 3 - 2C 2 , we get x - z - 2y y - a z - ( a + d) 0 a a+d =0 0

b

b+ g

Þ

( x + z - 2y) - [a (b + g) - b( a + d)] = 0

Þ

( x + z - 2y)( ag - bd) = 0 Þ x + z - 2y = 0.

Hence, the required equation of the plane is x + z - 2y = 0. PLANE CONTAINING PARALLEL LINES EXAMPLE 4

Find the equation of the plane which contains the two parallel lines x - 3 y + 4 z -1 x+1 y-2 z = = , and = = × 3 2 1 3 2 1

SOLUTION

Clearly, the plane which contains the two given parallel lines must pass through the points A(3, – 4, 1) and B(–1, 2, 0), and it must be parallel to the line having direction ratios 3, 2, 1. Any plane passing through the point A(3 , – 4 , 1) is given by … (i) a( x - 3) + b( y + 4) + c(z - 1) = 0. If this plane passes through the point B( -1, 2, 0), we have a( -1 - 3) + b( 2 + 4) + c( 0 - 1) = 0 Þ 4a - 6b + c = 0

… (ii)

If the plane (ii) is parallel to the line having direction ratios 3, 2, 1 then 3 a + 2b + c = 0. On solving (ii) and (iii) by cross multiplication, we get

… (iii)

SSS Mathematics for Class 12 1244

1244

Senior Secondary School Mathematics for Class 12

a b c a b c = = Þ = = = l (say) ( - 6 - 2) ( 3 - 4) ( 8 + 18) -8 -1 26 Þ a = -8l , b = - l and c = 26l. Putting these values of a , b and c in (i), we get -8l( x - 3) - l ( y + 4) + 26l(z - 1) = 0 Þ

8( x - 3) + ( y + 4) - 26(z - 1) = 0 Þ 8x + y - 26z + 6 = 0.

Hence, the required equation of the plane is 8x + y - 26z + 6 = 0.

EXERCISE 28I 1. Show that the lines ®

^

^

^

^

®

^

^

^

^

^

^

^

r = ( 2 j - 3 k ) + l( i + 2 j + 3 k ) and r = ( 2 i + 6 j + 3 k) + m ( 2 i + 3 j + 4 k)

are coplanar. Also find the equation of the plane containing these lines. 2. Find the vector and Cartesian forms of the equations of the plane containing the two lines ®

^

^

^

^

^

^

r = ( i + 2 j - 4 k ) + l( 2 i + 3 j + 6 k )

®

^

^

^

^

^

^

and r = ( 9 i + 5 j - k) + m ( -2 i + 3 j + 8 k). 3. Find the vector and Cartesian equations of a plane containing the two lines ®

^

®

^

^

^

^

^

^

r = ( 2 i + j - 3 k ) + l( i + 2 j + 5 k ) ^

^

^

^

^

and r = ( 3 i + 3 j + 2 k) + m ( 3 i - 2 j + 5 k). ®

^

^

^

^

^

^

Also show that the line r = ( 2 i + 5 j + 2 k) + p( 3 i - 2 j + 5 k) lies in the plane. [CBSE 2011C] x y-2 z+ 3 x-2 y-6 z- 3 and are coplanar. = = = = 1 2 3 2 3 4 Also find the equation of the plane containing these lines. x-2 y-4 z-6 x+1 y+ 3 z+5 5. Prove that the lines and are = = = = 1 4 7 3 5 7 coplanar. Also find the equation of the plane containing these lines. 5 - x y -7 z + 3 x - 8 2y - 8 z - 5 and 6. Show that the lines are = = = = -4 4 -5 7 2 3 coplanar. Find the equation of the plane containing these lines. [CBSE 2014]

4. Prove that the lines

x+1 y- 3 z+ 2 x y -7 z + 7 and = are coplanar. = = = -3 2 1 1 -3 2 Find the equation of the plane containing these lines.

7. Show that the lines

SSS Mathematics for Class 12 1245

The Plane

1245

x -1 y - 3 z x - 4 y -1 z -1 and are coplanar. = = = = 2 -1 -1 3 -2 -1 Also find the equation of the plane containing these lines.

8. Show that the lines

9. Find the equation of the plane which contains two parallel lines given by x-3 y+2 z x-4 y- 3 z-2 = = and = = × 1 -4 5 1 -4 5 ANSWERS (EXERCISE 28I) ®

^

^

^

1. r × ( i - 2 j + k) + 7 = 0 ®

^

^

^

2. r × ( 6 i - 28 j + 12 k) + 98 = 0, 6x - 28y + 12z + 98 = 0 ®

^

^

^

3. r × (10 i + 5 j - 4 k) = 37 , 10x + 5 y - 4z = 37 4. x - 2y + z + 7 = 0

5. x - 2y + z = 0

6. 17 x - 47 y - 24z + 172 = 0

7. x + y + z = 0

8. 2x - 5 y - 16z + 13 = 0

9. 11x - y - 3 z = 35

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28I) ®

®

®

®

®

®

3. The given lines are r = a1 + l b1 and r = a2 + l b 2 , where ®

^

^

^

®

^

^

^

^

^

®

^

^

^

a1 = ( 2 i + j - 3 k ), a2 = ( 3 i + 3 j + 2 k ), ®

^

b1 = ( i + 2 j + 5 k ), b 2 = ( 3 i - 2 j + 5 k ).

\

^ i ® ® ( b1 ´ b 2 ) = 1

^

^

j k ^ ^ ^ 2 5 = ( 10 + 10 ) i - (5 - 15 ) j + ( -2 - 6 ) k 3 -2 5 ^

^

^

= ( 20 i + 10 j - 8 k ). Vector equation of the required plane is ®

®

®

®

®

^

^

®

r × ( b1 ´ b 2 ) = a1 × ( b1 ´ b 2 )

®

^

®

^

^

^

^

^

^

^

Þ r × ( 20 i + 10 j - 8 k ) = ( 2 i + j - 3 k ) × ( 20 i + 10 j - 8 k ) = ( 40 + 10 + 24 ) = 74 ^

^

Þ r × ( 10 i + 5 j - 4 k ) = 37 .

... (i)

The Cartesian equation is ^

^

^

^

^

^

( x i + y j + z k ) × ( 10 i + 5 j - 4 k ) = 37 Þ 10 x + 5 y - 4z = 37 . ®

^

^

^

^

^

^

The third line is r = ( 2 i + 5 j + 2 k ) + p( 3 i - 2 j + 5 k ).

... (ii) ... (iii)

SSS Mathematics for Class 12 1246

1246

Senior Secondary School Mathematics for Class 12 ^

^

^

Now, the line (iii) will lie in the plane (ii) if (2, 5, 2) lies on (ii) and ( 3 i - 2 j + 5 k ) is perpendicular to the normal of (ii). Now, 10 ´ 2 + 5 ´ 5 - 4 ´ 2 = 37 shows that (2, 5, 2) lies on (ii). ^

^

^

Also 10 ´ 3 + 5 ´ ( -2 ) - 4 ´ 5 = 0 shows that ( 3 i - 2 j + 5 k ) is perpendicular to the normal of (ii). Hence, the line (iii) lies in plane (ii). 6. The given equations are:

x-5 y-7 z+ 3 x- 8 y- 4 z-5 and = = = = × 4 4 -5 7 1 3

EXERCISE 28J Very-Small-Answer Questions 1. Find the direction ratios of the normal to the plane x + 2y - 3z = 5. 2. Find the direction cosines of the normal to the plane 2x + 3 y - z = 4. 3. Find the direction cosines of the normal to the plane y = 3. 4. Find the direction cosines of the normal to the plane 3 x + 4 = 0. 5. Write the equation of the plane parallel to XY-plane and passing through the point (4, –2, 3). 6. Write the equation of the plane parallel to YZ-plane and passing through the point (–3, 2, 0). 7. Write the general equation of a plane parallel to the x-axis. 8. Write the intercept cut off by the plane 2x + y - z = 5 on the x-axis. [CBSE 2011] 9. Write the intercepts made by the plane 4x - 3 y + 2z = 12 on the coordinate axes. 10. Reduce the equation 2x - 3 y + 5z + 4 = 0 to intercept form and find the intercepts made by it on the coordinate axes. 11. Find the equation of a plane passing through the points A( a , 0, 0), B( 0, b , 0) and C( 0, 0, c). 12. Write the value of k for which the planes 2x - 5 y + kz = 4 and x + 2y - z = 6 are perpendicular to each other. 13. Find the angle between the planes 2x + y - 2z = 5 and 3 x - 6y - 2z = 7. ®

^

®

^

^

^

14. Find the angle between the planes r × ( i + j ) = 1 and r × ( i + k ) = 3. 15. Find ®

the ^

^

angle

between

the

planes

®

^

^

^

r × ( 3 i - 4 j + 5 k) = 0

and

^

r × ( 2 i - j - 2 k) = 7.

16. Find the angle between the line 10x + 2y - 11z = 3.

x+1 y z- 3 and the plane = = 2 3 6

SSS Mathematics for Class 12 1247

The Plane ®

^

1247 ^

^

^

^

^

17. Find the angle between the line r = ( i + j - 2 k) + l( i - j + k) and the plane ®

^

^

^

r × ( 2 i - j + k) = 4. x - 2 y -1 z + 5 is perpendicular = = 6 l 4

18. Find the value of l such that the line to the plane 3 x - y - 2z = 7.

[CBSE 2010C]

19. Write the equation of the plane passing through the point ( a , b , c) and ®

^

^

^

parallel to the plane r × ( i + j + k) = 2.

[CBSE 2014]

20. Find the length of perpendicular drawn from the origin to the plane [CBSE 2013] 2x - 3 y + 6z + 21 = 0. 21. Find the direction cosines of the perpendicular from the origin to the plane ®

^

^

^

r × ( 6 i - 3 j - 2 k) + 1 = 0. ®

^

^

^

^

^

22. Show that the line r = ( 4 i - 7 k) + l( 4 i - 2 j + 3 k) is parallel to the plane ®

^

^

^

r × (5 i + 4 j - 4 k) = 7.

23. Find the length of perpendicular from the origin to the plane ®

^

^

^

r × ( 2 i - 3 j + 6 k) + 14 = 0.

24. Find the value of l for which the line ®

^

^

x -1 y -1 z -1 is parallel to the = = 2 3 l

^

plane r × ( 2 i + 3 j + 4 k) = 4. 25. Write the angle between the line x + y + 4 = 0.

x -1 y - 2 z + 3 and the plane = = 2 1 -2

26. Write the equation of a plane passing through the point (2, –1, 1) and parallel to the plane 3 x + 2y - z = 7. ANSWERS (EXERCISE 28J)

2 , 14

1. 1, 2, –3

2.

5. z = 3

6. x = -3

9. 3, –4, 6

3 -1 , 14 14

4. –1, 0, 0

7. by + cz + d = 0

8.

5 2

4 -4 , 3 5 p 14. 3

x y z 12. k = -8 + + =1 a b c p æ 8ö 15. 16. sin -1 ç ÷ 2 è 21 ø

18. l = -2

19. r × ( i + j + k) + a + b + c

10. -2,

æ 4ö 13. cos-1 ç ÷ è 21 ø æ2 2ö ÷ 17. sin -1 çç ÷ è 3 ø

3. 0, 1, 0

11.

®

^

^

^

SSS Mathematics for Class 12 1248

1248

Senior Secondary School Mathematics for Class 12

20. 3 units 25.

p 4

21.

-6 3 2 , , 7 7 7

23. 2 units

24. l =

-13 4

26. 3 x + 2y - z = 3

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 28J) 1. The direction ratios of the normal to the plane x + 2 y - 3z = 5 are 1, 2, –3. 2. The given plane is 2 x + 3 y - z = 4. Direction ratios of the normal to the given plane are 2, 3, –1 and 2 2 + 3 2 + ( -1) 2 = 14 . Hence, the required direction cosines are

2 , 14

3 -1 , × 14 14

3. Direction ratios of the normal to the plane are 0, 1, 0 and 0 2 + 12 + 0 2 = 1. Hence, the required direction cosines are 0, 1, 0. 4 4. 3 x + 4 = 0 Þ - x = × 3 Direction ratios of the normal to this plane are –1, 0, 0 and ( -1) 2 + 0 2 + 0 2 = 1. Hence, the required direction cosines are –1, 0, 0. 5. Any plane parallel to XY-plane is z = k. Since it passes through (4, –2, 3), we have 3 = k. Hence, the required equation of the plane is z = 3. 6. Any plane parallel to YZ-plane is x = k. Since it passes through (–3, 2, 0), we have -3 = k. Hence, the required equation of the plane is x = -3. 7. Let the required equation of the plane be ax + by + cz + d = 0. The d.r.’s of this plane are a, b , c. The d.r.’s of the x-axis are 1, 0, 0. Normal of the required plane is perpendicular to the x-axis. \ ( a ´ 1) + ( b ´ 0 ) + ( c ´ 0 ) = 0 Þ a = 0. Hence, the required equation is by + cz + d = 0. y x z 8. 2 x + y - z = 5 Þ + + = 1. æ 5 ö 5 -5 ç ÷ è2ø 5 × 2 y 4 x ( -3 y ) 2z x z 9. 4 x - 3 y + 2z = 12 Þ + + =1Þ + + = 1. 12 12 12 3 -4 6 \ intercept cut off by the given plane on the x-axis is

Hence, the required intercepts are 3, –4, 6. 10. The given equation may be written as -2 x + 3 y - 5z = 4

SSS Mathematics for Class 12 1249

The Plane Þ

\

1249

y ( -2 x ) 3 y ( -5z) x z + + =1Þ + + = 1. -4 4 4 4 -2 4 3 5 4 -4 the required intercepts are -2 , , × 3 5

11. Clearly, the required equation of the plane is

x y z + + = 1. a b c

12. Clearly, the normals of the given planes are perpendicular to each other. \ ( 2 ´ 1) + ( -5 ) ´ 2 + k ´ ( -1) = 0 Þ k = ( 2 - 10 ) = -8. 13. D.r.’s of normals to the given planes are 2, 1, –2 and 3, –6, –2. \

cos q =

|( 2 ´ 3 ) + 1 ´ ( -6 ) + ( -2 ) ´ ( -2 )| ì 2 2 + 12 + ( -2 ) 2 ü ì 3 2 + ( -6 ) 2 + ( -2 ) 2 ü í ýí ý î þî þ

=

4 4 4 æ 4 ö = = Þ q = cos -1 ç ÷ . ( 9 )( 49 ) ( 3 ´ 7 ) 21 è 21 ø

14. Given planes are x + 1 = 1 and x + z = 3. The d.r.’s of normals to these planes are 1, 1, 0 and 1, 0, 1. |( 1 ´ 1) + ( 1 ´ 0 ) + ( 0 ´ 1)| \ cos q = ì 12 + 12 + 0 2 ü ì 12 + 0 2 + 12 ü í ýí ý î þî þ =

1 1 p = Þ q= × 3 ( 2 ´ 2) 2

15. The given planes are 3 x - 4 y + 5z = 0 and 2 x - y - 2z = 7 . The d.r.’s of normals to these planes are 3, –4, 5 and 2, –1, –2. |( 3 ´ 2 ) + ( -4 ) ´ ( -1) + 5 ´ ( -2 )| p \ cos q = =0 Þ q= × 2 ì 3 2 + ( -4 ) 2 + 5 2 ü ì 2 2 + ( -1) 2 + ( -2 ) 2 ü í ýí ý î þî þ 16. D.r.’s of the given line are 2, 3, 6. D.r.’s of the normal to the given plane are 10, 2, –11. |( 2 ´ 10 ) + ( 3 ´ 2 ) + 6 ´ ( -11)| \ sin q = ì 2 2 + 3 2 + 6 2 ü ì ( 10 ) 2 + 2 2 + ( -11) 2 ü í ýí ý î þî þ =

40 40 8 æ 8 ö = = Þ q = sin -1 ç ÷ . { 49 } ´ { 225 } (7 ´ 15 ) 21 è 21 ø

® ® ® ® ® 17. Given line is r = a + l b and given plane is r × n = p. ® ®

sin q =

| b × n|

^ ^ ^ ^ ^ ^ |( i - j + k ) × ( 2 i - j + k )|

=

® ® ì 2 2 2üì 2 2 2ü | b || n| í 1 + ( -1) + 1 ý í 2 + ( -1) + 1 ý î þî þ

=

|( 2 ´ 1) + ( -1) ´ ( -1) + 1 ´ 1| ( 3 ´ 6)

SSS Mathematics for Class 12 1250

1250

Senior Secondary School Mathematics for Class 12

=

æ 4 æ2 2ö 4 2ö 2 2 ÷= ÷. = çç ´ Þ q = sin -1 çç ÷ ÷ 3 18 è 3 2 2ø è 3 ø

18. D.r.’s of the given line are 6 , l, - 4. D.r.’s of normal to the given plane are 3, –1, –2. Given line is parallel to the normal of the plane. \

6 l -4 = = Þ l = -2. 3 -1 -2

^ ^ ^ ^ ^ ^ 19. Given plane is ( x i + y j + z k ) × ( i + j + k ) = 2 Þ

x + y + z = 2.

Let the required plane be x + y + z = k , where k is a constant. Since it passes through ( a, b , c ), we have k = ( a + b + c ). So, the required plane is x + y + z = a + b + c. ® ^ ^ ^ In vector form, it is given by r × ( i + j + k ) = a + b + c. 20. We have p =

2 ´ 0 - 3 ´ 0 + 6 ´ 0 + 21 2 2 + ( -3 ) 2 + 6 2

=

21 21 = = 3 units. 49 7

® ^ ^ ^ 21. The given equation is r × ( -6 i + 3 j + 2 k ) = 1.

D.r.’s of normal to the plane are –6, 3, 2 and ( -6 ) 2 + 3 2 + 2 2 = 49 = 7 . \ d.c.’s of normal to the plane are

-6 3 2 , , × 7 7 7

® ® ® ® ^ ^ ^ 22. Given line is r = a + l b , where b = ( 4 i - 2 j + 3 k ).

D.r.’s of the line are 4, –2, 3. ® ® ® ^ ^ ^ Given plane is r × n = q, where n = (5 i + 4 j - 4 k ). D.r.’s of the normal to the given plane are 5, 4, –4. So, the given line will be parallel to the given plane when this line is perpendicular to the normal to the plane. Hence, we must have ( 4 ´ 5 ) + ( -2 ´ 4 ) + 3 ´ ( -4 ) = 0 , which is true. ® ^ ^ ^ 23. We have r × ( -2 i + 3 j - 6 k ) = 14 . ® ® ^ ^ ^ Here n = ( -2 i + 3 j - 6 k ) and| n|= ( -2 ) 2 + 3 2 + ( -6 ) 2 = 7 .

\

®

r ×

®

n

®

=

14 ®

| n| | n|

® ^ 14 Þ r ×n = = 2. 7

Hence, the length of perpendicular from origin to the given plane is 2 units. 24. Clearly, the given line must be perpendicular to the normal to the given plane. D.r.’s of the given line are 2, 3, l. D.r.’s of the normal to the given plane are 2, 3, 4.

SSS Mathematics for Class 12 1251

The Plane

1251

\ ( 2 ´ 2 ) + ( 3 ´ 3 ) + ( l ´ 4 ) = 0 Þ 4 l = -13 Þ l =

-13 × 4

25. D.r.’s of the given line are 2, 1, –2. D.r.’s of the normal to the given plane are 1, 1, 0. ( 2 ´ 1) + ( 1 ´ 1) + ( -2 ) ´ 0 ( 2 + 1 + 0) \ sin q = = ì 2 2 + 12 + ( -2 ) 2 ü ì 12 + 12 + 0 2 ü ( 9 )( 2 ) í ý ýí î þ þî 3 1 p = = Þ q= × 4 3 2 2 26. Let the required equation of the plane be a( x - 2 ) + b( y + 1) + c(z - 1) = 0. Here a = 3 , b = 2 and c = -1. So, the required equation of the plane is 3( x - 2 ) + 2( y + 1) - 1 × (z - 1) = 0 Þ 3 x + 2 y - z = 3.

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 1. The direction cosines of the perpendicular from the origin to the plane ®

^

^

^

r × ( 6 i - 3 j + 2 k ) + 1 = 0 are

6 3 -2 (a) , , 7 7 7

6 -3 2 (b) , , 7 7 7

(c)

-6 3 2 , , 7 7 7

(d) none of these

2. The direction cosines of the normal to the plane 5 y + 4 = 0 are -4 (a) 0, (b) 0, 1, 0 (c) 0, –1, 0 (d) none of these ,0 5 3. The ®

length ^

^

of

perpendicular

from

the

origin

to

the

plane

^

r × ( 3 i - 4 j - 12 k ) + 39 = 0 is

(a) 3 units

(b)

13 units 5

(c)

5 units 3

(d) none of these

4. The equation of a plane passing through the point A(2, –3, 7) and making equal intercepts on the axes, is (a) x + y + z = 3 (b) x + y + z = 6 (c) x + y + z = 9 (d) x + y + z = 4 5. A plane cuts off intercepts 3, –4, 6 on the coordinate axes. The length of perpendicular from the origin to this plane is 5 8 6 12 units (b) units (c) units (d) units (a) 29 29 29 29 x+1 y-2 z+ 6 is parallel to the plane 2x - 3 y + kz = 0, then 6. If the line = = 3 4 5 the value of k is 5 6 3 4 (a) (b) (c) (d) 6 5 4 5

SSS Mathematics for Class 12 1252

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Senior Secondary School Mathematics for Class 12

7. If O is the origin and P(1, 2, –3) is a given point, then the equation of the plane through P and perpendicular to OP is (a) x + 2y - 3z = 14 (b) x - 2y + 3z = 12 (c) x - 2y - 3z = 14 (d) none of these x-4 y-2 z-k lies in the plane 2x - 4y + z = 7 , then the 8. If the line = = 1 1 2 value of k is (a) –7 (b) 7 (c) 4 (d) –4 9. The plane 2x + 3 y + 4z = 12 meets the coordinate axes in A , B and C. The centroid of C ABC is æ 4 ö (d) none of these (a) (2, 3, 4) (b) (6, 4, 3) (c) ç 2, , 1÷ è 3 ø 10. If a plane meets the coordinate axes in A , B and C such that the centroid of C ABC is (1, 2, 4), then the equation of the plane is (a) x + 2y + 4z = 6 (b) 4x + 2y + z = 12 (c) x + 2y + 4z = 7 (d) 4x + 2y + z = 7 11. The equation of a plane through the point A(1, 0, –1) and perpendicular to x+1 y+ 3 z+7 the line is = = 2 4 -3 (a) 2x + 4y - 3z = 3 (b) 2x - 4y + 3z = 5 (c) 2x + 4y - 3z = 5 (d) x + 3 y + 7z = -6 x -1 y - 2 z - 3 meets the plane 2x + 3 y - z = 14 in the point 12. The line = = 2 4 4 (a) (2, 5, 7) (b) (3, 5, 7) (c) (5, 7, 3) (d) (6, 5, 3) 13. The equation of the plane passing through the points A( 2, 2, 1) and B( 9, 3 , 6) and perpendicular to the plane 2x + 6y + 6z = 1, is (a) x + 2y - 3z + 5 = 0 (b) 2x - 3 y + 4z - 6 = 0 (c) 4x + 5 y - 6z + 3 = 0 (d) 3 x + 4y - 5z - 9 = 0 14. The equation of the plane passing through the intersection of the planes 3 x - y + 2z - 4 = 0 and x + y + z - 2 = 0 and passing through the point A( 2, 2, 1) is given by (a) 7 x + 5 y - 4z - 8 = 0 (b) 7 x - 5 y + 4z - 8 = 0 (c) 5 x - 7 y + 4z - 8 = 0 (d) 5 x + 7 y - 4z + 8 = 0 15. The equation of the plane passing through the points A( 0, - 1, 0), B( 2, 1, - 1) and C(1, 1, 1) is given by (a) 4x + 3 y - 2z - 3 = 0 (b) 4x - 3 y + 2z + 3 = 0 (c) 4x - 3 y + 2z - 3 = 0 (d) none of these 2x - 1 2 - y z + 1 16. If the plane 2x - y + z = 0 is parallel to the line = = , then 2 2 a the value of a is (a) –4 (b) –2 (c) 4 (d) 2

SSS Mathematics for Class 12 1253

The Plane

17. The angle between the line x - y + z = 0 is

1253

x + 1 y z -1 and a normal to the plane = = 1 2 1

(a) 0°

(b) 30°

(c) 45° (d) 90° x -1 y + 2 z - 3 and the plane 18. The point of intersection of the line = = 3 4 -2 2x - y + 3z - 1 = 0, is (a) (–10, 10, 3)

(b) (10, 10, –3)

(c) (10, –10, 3)

(d) (10, –10, –3)

19. The equation of a plane passing through the points A( a , 0, 0), B( 0, b , 0) and C( 0, 0, c) is given by (a) ax + by + cz = 0

(b) ax + by + cz = 1

x y z (c) + + = 0 a b c

(d)

x y z + + =1 a b c

20. If q is the angle between the planes 2x - y + 2z = 3 and 6x - 2y + 3z = 5 , then cos q = ? 11 12 17 20 (a) (b) (c) (d) 20 23 25 21 21. The angle between the planes 2x - y + z = 6 and x + y + 2z = 7, is p p p p (a) (b) (c) (d) 6 4 3 2 22. The ®

angle ^

^

between

the

planes

®

^

^

^

r ×( 3 i - 6 j + 2 k) = 4

and

^

r × ( 2 i - j + 2 k ) = 3 , is

æ 16 ö (a) cos-1 ç ÷ è 21 ø

æ 4ö (b) cos-1 ç ÷ è 21 ø

æ 3ö (c) cos-1 ç ÷ è 4ø

æ1ö (d) cos-1 ç ÷ è 4ø

23. The equation of the plane through the points A(2, 3, 1) and B(4, –5, 3), parallel to the x-axis, is (a) x + y - 3z = 2 (b) y + 4z = 7

(c) y + 3z = 6

(d) x + 5 y - 3z = 4

24. A variable plane moves so that the sum of the reciporcals of its intercepts on the coordinate axes is (1/2). Then, the plane passes through the point æ1 1 1ö (d) (2, 2, 2) (a) (0, 0, 0) (b) (1, 1, 1) (c) ç , , ÷ è 2 2 2ø ^

^

^

25. The equation of a plane which is perpendicular to ( 2 i - 3 j + k ) and at a distance of 5 units from the origin is (a) 2x - 3 y + z = 5 (c)

x y z - + =5 2 3 1

(b) 2x - 3 y + z = 5 14 (d)

x y z 5 - + = 2 3 1 14

SSS Mathematics for Class 12 1254

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Senior Secondary School Mathematics for Class 12

26. The equation of the plane passing through the point A(2, 3, 4) and parallel to the plane 5 x - 6y + 7z = 3 , is (a) 5 x - 6y + 7z = 20 (b) 7 x - 6y + 5z = 72 (c) 20x - 18y + 14z = 11 (d) 10x - 18y + 28z = 13 27. The foot of the perpendicular from the point A(7, 14, 5) to the plane 2x + 4y - z = 2 is (a) (3, 1, 8) (b) (1, 2, 8) (c) (3, –3, 5) (d) (5, –3, –4) 28. The equation of the plane which makes with the coordinate axes, a triangle with centroid ( a , b , g) is given by (a) ax + by + gz = 1 (b) ax + by + gz = 3 x y z x y z (d) + + = 3 (c) + + = 1 a b g a b g ®

^

^

^

29. The intercepts made by the plane r × ( 2 i - 3 j + 4 k ) = 12 are (a) 2, –3, 4

(b) 2, –3, –6

30. The angle between the line 2x - 3 y + z = 5 is æ5ö æ5ö (a) cos-1 ç ÷ (b) sin -1 ç ÷ è 14 ø è 14 ø ® ^

(c) –6, –4, 3 (d) –6, 4, 3 x-2 y+ 3 z+ 4 and the plane = = 1 -2 -3 æ 3ö (c) cos-1 ç ÷ è7ø ^

^

æ 3ö (d) sin -1 ç ÷ è7ø ^

^

^

31. The angle between the line r × ( i + j - 3 k ) + l( 2 i + 2 j + k ) and the plane ®

^

^

^

r × ( 6 i - 3 j + 2 k ) = 5 , is

æ 8ö (a) cos-1 ç ÷ è 21 ø 32. The

distance

® ^

^

æ5 ö (b) cos-1 ç ÷ è 21 ø of

the

point

æ5 ö (c) sin -1 ç ÷ è 21 ø ^

^

æ 8ö (d) sin -1 ç ÷ è 21 ø ^

( i + 2 j + 5 k)

from

the

plane

^

r × ( i + j + k ) + 17 = 0, is

(a)

25 units 2

(b)

25 units 3

(c) 25 2 units

33. The distance between the parallel planes 6x - 9y + 18z + 20 = 0, is 5 8 (c) units (b) 5 3 units (a) units 3 5

(d) 25 3 units 2x - 3 y + 6z = 5

and

(d) 8 5 units

34. The distance between the planes x + 2y - 2z + 1 = 0 and 2x + 4y - 4z + 5 = 0, is 1 1 (d) units (a) 4 units (b) 2 units (c) units 2 4 35. The image of the point P(1, 3, 4) in the plane 2x - y + z + 3 = 0, is (a) (3, –5, 2) (b) (3, 5, –2) (c) (3, 5, 2) (d) (–3, 5, 2)

SSS Mathematics for Class 12 1255

The Plane

1255

ANSWERS (OBJECTIVE QUESTIONS)

1. (c)

2. (c)

3. (a)

4. (b)

5. (d)

6. (b)

7. (a)

8 (b)

9. (c) 10. (b)

11. (c) 12. (b) 13. (d) 14. (b) 15. (c) 16. (a) 17. (d) 18. (b) 19. (d) 20. (d) 21. (c) 22. (a) 23. (b) 24. (d) 25. (b) 26. (a) 27. (b) 28. (d) 29. (c) 30. (b) 31. (d) 32. (b) 33. (a) 34. (c) 35. (d)

HINTS TO GIVEN OBJECTIVE QUESTIONS ®

^

^

^

1. (c) Given plane is r × ( -6 i + 3 j + 2 k ) = 1. ® ®

®

^

^

^

\ r × n = p. where n = ( -6 i + 3 j + 2 k ). D.r.’s of the normal to the given plane are –6, 3, 2 and ( -6 ) 2 + 3 2 + 2 2 = 49 = 7 . -6 3 2 , , × 7 7 7 4 4 2. (c) 5 y + 4 = 0 Þ - 5 y = 4 Þ - y = Þ 0 x - 1 × y + 0 × z = × 5 5 \ d.c.’s of the normal to the given plane are

D.r.’s of the normal to this plane are 0, –1, 0 and 0 2 + ( -1) 2 + 0 2 = 1 = 1. \

d.c.’s of the normal to the given plane are 0, –1, 0. ®

^

^

^

3. (a) Given plane is r × ( -3 i + 4 j + 12 k ) = 39. ® ®

®

r × n = 39 Þ r ×

®

n

®

=

39 ®

® ^

Þ r ×n =

| n| | n|

39 ®

×

| n|

®

And,| n|= ( -3 ) 2 + 4 2 + ( 12 ) 2 = 169 = 13. ® ^

39 = 3 units. Hence, p = 3 units. 13 x y z 4. (b) Let the required equation of the plane be + + = 1, i.e., x + y + z = a. a a a \ r ×n =

Since, it passes through the point A( 2 , - 3 , 7 ), we have 2 + ( -3 ) + 7 = a Þ a = 6. Hence, the required equation of the plane is x + y + z = 6. y x z 5. (d) The given plane is + + = 1 Þ 4 x - 3 y + 2 z = 12 3 -4 6 \

p=

4 ´ 0 - 3 ´ 0 + 2 ´ 0 - 12 2

2

4 + ( -3 ) + 2

2

=

12 units. 29

6. (b) D.r.’s of the given line are 3, 4, 5. D.r.’s of the normal to the given plane are 2, –3, k. Since the given line is perpendicular to the normal of the plane, we have

SSS Mathematics for Class 12 1256

1256

Senior Secondary School Mathematics for Class 12 ( 3 ´ 2 ) + 4 ´ ( -3 ) + 5 ´ k = 0 Þ 5 k = 6 Þ k =

6 × 5

7. (a) Let the required equation of the plane through P( 1, 2 , - 3 ) be a( x - 1) + b( y - 2 ) + c(z + 3 ) = 0. D.r.’s of OP are ( 1 - 0 ), ( 2 - 0 ), ( -3 , 0 ), i.e., 1, 2, –3.

... (i)

\ a = 1, b = 2 , c = -3. Hence, the required equation of the plane is 1( x - 1) + 2( y - 2 ) - 3(z + 3 ) = 0 Þ x + 2 y - 3z = 14. 8. (b) Clearly, the given line passes through the point (4, 2, k ). Since, the given line lies in the plane 2 x - 4 y + z = 7 , so the above point lies in this plane. \

( 2 ´ 4) - ( 4 ´ 2) + k = 7 Þ k = 7. x y c 9. (c) The given equation is + + = 1. 6 4 3 This plane meets the coordinate axes in A( 6 , 0 , 0 ), B( 0 , 4 , 0 ) and C( 0 , 0 , 3 ). æ6+ 0+ 0 0+ 4+ 0 0+ 0+ 3ö æ 4 ö centroid of C ABC is G ç \ , , ÷ , i.e., G ç 2 , , 1 ÷. 3 3 3 è ø è 3 ø 10. (b) Let the required equation of the plane be

x y z + + = 1. a b c

Then, it meets the coordinate axes in A( a, 0 , 0 ), B( 0 , b , 0 ), C ( 0 , 0 , c ). æa+ 0+ 0 0+ b + 0 0+ 0+ cö æa b cö , , \ centroid of C ABC is G ç ÷ , i.e., G ç , , ÷ . 3 3 3 è ø è 3 3 3ø b c æa ö \ ç = 1, = 2 , = 4 ÷ Þ a = 3 , b = 6 , c = 12. 3 3 è3 ø Hence, the required equation of the plane is x y z + + = 1 Þ 4 x + 2 y + z = 12. 3 6 12 11. (c) D.r.’s of the given line are 2, 4, –3. Let the required equation of plane through (1, 0, –1) be a( x - 1) + b( y - 0 ) + c(z + 1) = 0. Since the line is perpendicular to the plane, so it is parallel to the normal to the plane. \

a = 2 , b = 4 and c = -3.

Hence, the required equation of the plane is 2( x - 1) + 4( y - 0 ) - 3(z + 1) = 0 Þ 2 x + 4 y - 3z = 5. x- 1 y- 2 z- 3 12. (b) Let = = = l (say). 2 3 4 A general point on this line is ( 2 l + 1, 3 l + 2 , 4 l + 3 ). For some value of l, let the given line meet the plane 2 x + 3 y - z = 14 at a point P( 2 l + 1, 3 l + 2 , 4 l + 3 ). Then, 2( 2 l + 1) + 3( 3 l + 2 ) - ( 4 l + 3 ) = 14 Þ 9 l = 9 Þ l = 1. So, the required point is P( 2 + 1, 3 + 2 , 4 + 3 ), i.e., P( 3 , 5 , 7 ).

SSS Mathematics for Class 12 1257

The Plane

1257

13. (d) The equation of a plane passing through the point A( 2 , 2 , 1) is a( x - 2 ) + b( y - 2 ) + c(z - 1) = 0.

... (i)

Since it passes through the point B( 9 , 3 , 6 ), we have a( 9 - 2 ) + b( 3 - 2 ) + c( 6 - 1) = 0 Þ 7 a + b + 5 c = 0.

... (ii)

Also, it being perpendicular to the plane 2 x + 6 y + 6z = 1, we have 2 a + 6 b + 6 c = 0 Þ a + 3 b + 3 c = 0.

... (iii)

On solving (ii) and (iii) by cross multiplication, we have a b c a b c a b c = = Þ = = Þ = = =k ( 3 - 15 ) (5 - 21) ( 21 - 1) -12 -16 20 3 4 -5 Þ

a = 3 k , b = 4 k and c = -5 k

Substituting these values in (i), we get 3 k( x - 2 ) + 4 k( y - 2 ) - 5 k(z - 1) = 0 Þ

3( x - 2 ) + 4( y - 2 ) - 5(z - 1) = 0 Þ 3 x + 4 y - 5z - 9 = 0.

14. (b) Let the required equation of the plane be ( 3 x - y + 2z - 4 ) + l( x + y + z - 2 ) = 0 Þ ( 3 + l)x + ( l - 1)y + ( 2 + l)z - ( 4 + 2 l) = 0. Since it passes through the point A( 2 , 2 , 1), we have Þ Þ

... (ii)

( 3 + l) ´ 2 + ( l - 1) ´ 2 + ( 2 + l) ´ 1 - ( 4 + 2 l) = 0 ( 6 + 2 l) + ( 2 l - 2 ) + ( 2 + l) - ( 4 + 2 l) = 0 -2 3l + 2 = 0 Þ l = × 3

So, the required equation of the plane is 2ö 2ö 4ö æ æ -2 ö æ æ - 1÷ y + ç 2 - ÷ z - ç 4 - ÷ = 0 ç 3 - ÷x + ç 3ø 3ø 3ø è è 3 ø è è 7 x 5 y 4z 8 + - = 0 Þ 7 x - 5 y + 4z - 8 = 0. Þ 3 3 3 3 15. (c) Here ( x1 = 0 , y1 = -1, z1 = 0 ), ( x2 = 2 , y2 = 1, z2 = -1) and ( x 3 = 1, y 3 = 1, z 3 = 1). So, the required equation of the plane is x - x1 y - y1 z - z1 x2 - x1 y2 - y1 z2 - z1 = 0 x 3 - x1 y 3 - y1 z 3 - z1 Þ

x- 0 y+ 1 z- 0 2 - 0 1 + 1 -1 - 0 = 0 Þ 1- 0 1+ 1 1- 0

Þ Þ

x y+ 1 z 2 2 -1 = 0 1 2 1

x( 2 + 2 ) - ( y + 1)( 2 + 1) + z( 4 - 2 ) = 0 4 x - 3( y + 1) + 2z = 0 Þ 4 x - 3 y + 2 z - 3 = 0. 1 x2 = y - 2 = z - ( -1) × 16. (a) The given line is 1 -2 a The d.r.’s of the line are 1, –2, a. The given plane is 2 x - y + z = 0.

SSS Mathematics for Class 12 1258

1258

Senior Secondary School Mathematics for Class 12 The d.r.’s of the normal to this plane are 2, –1, 1. Since the given plane is parallel to the given line, so the normal to this plane is perpendicular to the given line. \

( 2 ´ 1) - 1 ´ ( -2 ) + 1 ´ a = 0 Þ a = -4

17. (d) D.r.’s of the given line are 1, 2, 1. D.r.’s of the normal to the given plane are 1, –1, 1. |( 1 ´ 1) + 2 ´ ( -1) + ( 1 ´ 1)| cos q = \ = 0 Þ q = 90 °. ì 12 + 2 2 + 12 ü ì 12 + ( -1) 2 + 12 ü í ýí ý î þî þ 18. (b) The given line is

x- 1 y+ 2 z- 3 = = = l (say). 3 4 -2

A general point on this line is ( 3 l + 1, 4 l - 2 , - 2 l + 3 ). For some value of l, let the point P( 3 l + 1, 4 l - 2 , - 2 l + 3 ) lie on the plane 2 x - y + 3z - 1 = 0. Then, 2( 3 l + 1) - ( 4 l - 2 ) + 3( -2 l + 3 ) - 1 = 0 Þ 4 l = 12 Þ l = 3. So, the required point is P( 9 + 1, 12 - 2 , - 6 + 3 ), i.e., P( 10 , 10 , - 3 ). x y z 19. (d) The required equation is + + = 1. a b c 20. (d) Angle between the planes means the angle between their normals. |( 2 ´ 6 ) + ( -1) ´ ( -2 ) + 2 ´ 3| 20 20 20 \ cos q = = = = × ì 2 2 + ( -1) 2 + 2 2 ü ì 6 2 + ( -2 ) 2 + 3 2 ü ( 9 )( 49 ) ( 3 ´ 7 ) 21 í ýí ý î þî þ 21. (c) Angle between the planes means the angle between their normals. \

cos q =

=

|( 2 ´ 1) + ( -1) ´ 1 + ( 1 ´ 2 )| ì 2 2 + ( -1) 2 + 12 ü ì 12 + 12 + 2 2 ü í ý ýí î þ þî 3 3 1 p = = Þ q= × 3 ( 6 ´ 6) 6 2

22. (a) The given planes are 3 x - 6 y + 2 z = 4 and 2 x - y + 2 z = 3. \

cos q =

= Þ

|( 3 ´ 2 ) + ( -6 ) ´ ( -1) + ( 2 ´ 2 )| ì 3 2 + ( -6 ) 2 + 2 2 ü ì 2 2 + ( -1) 2 + 2 2 ü í ýí ý î þî þ 16 16 16 = = ( 49 )( 9 ) (7 ´ 3 ) 21

æ 16 ö q = cos -1 ç ÷. è 21 ø

23. (b) Let the required equation be by + cz + d = 0.

... (i)

Since it passes through the points A( 2 , 3 , 1) and B( 4 , - 5 , 3 ), we have 3b + c + d = 0 -5 b + 3 c + d = 0.

... (ii) ... (iii)

SSS Mathematics for Class 12 1259

The Plane

1259

On solving (ii) and (iii) by cross multiplication, we have b c d = = ( 1 - 3 ) ( -5 - 3 ) ( 9 + 5 ) Þ Þ

® ®

r × n = 5 14.

b c d = = = k (say). 1 4 -7

Then, b = k , c = 4 k and d = -7 k. Substituting these values in (i), we get the required equation as ky + 4 kz - 7 k = 0 Þ y + 4z - 7 = 0 Þ y + 4z = 7. 24. (d) Let the required equation of the plane be Then,

x y z + + = 1. a b c

1 1 1 1 2 2 2 + + = Þ + + = 1. a b c 2 a b c

It means that the plane passes through the point (2, 2, 2). ®

25. (b) We have| n|= 2 2 + ( -3 ) 2 + 12 = 14 . \

® ^

®

r ×n =5 Þ r ×

®

n =5 14

® ®

Þ r × n = 5 14. \

^

^

^

^

^

^

( x i + y j + z k ) × ( 2 i - 3 j + k ) = 5 14 Þ 2 x - 3 y + z = 5 14 .

26. (a) Let the required equation of the plane be 5 x - 6 y + 7 z = k. Since it passes through the point A( 2 , 3 , 4 ), we have (5 ´ 2 ) - 6 ´ 3 + 7 ´ 4 = k Þ k = ( 10 - 18 + 28 ) = 20. Hence, the required equation of the plane is 5 x - 6 y + 7 z = 20. 27. (b) Let N be the foot of the perpendicular drawn from the point A(7 , 14 , 5 ) and perpendicular to the plane 2 x + 4 y - z = 2. x - 7 y - 14 z - 5 Then, the equation of the line PN is = = = l (say). 2 4 -1 Let the coordinates of N be N( 2 l + 7 , 4 l + 14 , - l + 5 ). Since N lies on the plane 2 x + 4 y - z = 2 , so 2( 2 l + 7 ) + 4( 4 l + 14 ) - ( - l + 5 ) = 2 Þ 21l = -63 Þ l = -3. \ required foot of the perpendicular is N( -6 + 7 , - 12 + 14 , 3 + 5 ), i.e., N( 1, 2 , 8 ). 28. (d) Let the equation of the plane be

x y z + + = 1. a b c

Then, it meets the axes at A( a, 0 , 0 ), B( 0 , b , 0 ) and C ( 0 , 0 , c ).

SSS Mathematics for Class 12 1260

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Senior Secondary School Mathematics for Class 12

æa+ 0+ 0 0+ b + 0 0+ 0+ cö æa b cö \ centroid of C ABC is G ç , , ÷ , i.e., G ç , , ÷ × 3 3 3 è ø è 3 3 3ø b c æa ö \ ç = a , = b and = g ÷ Þ a = 3 a , b = 3b and c = 3g. 3 3 è3 ø \ the required equation of the plane is y x z x y z + + =1Þ + + = 3. 3 a 3b 3 g a b g y x z + + = 1. 6 -4 3

29. (c) The given plane is 2 x - 3 y + 4 z = 12 Þ \ required intercepts are 6, –4, 3. 30. (b) D.r.’s of the given line are 1, –2, –3.

D.r.’s of the noraml to the given plane are 2, –3, 1. |( 1 ´ 2 ) + ( -2 ) ´ ( -3 ) + ( -3 ) ´ 1| \ sin q = ì 12 + ( -2 ) 2 + ( -3 ) 2 ü ì 2 2 + ( -3 ) 2 + 12 ü í ýí ý î þî þ =

5 æ 5 ö Þ q = sin -1 ç ÷ . 14 è 14 ø ® ®

31. (d) sin q =

| b × n| ®

×

®

| b || n | ®

^

^

®

^

^

^

^

Here b = ( 2 i + 2 j + k ) and n = ( 6 i - 3 j + 2 k ). ^

\

^

^

^

^

^

|( 2 i ´ 2 j + k ) × ( 6 i - 3 j + 2 k )| ì 2 2 + 2 2 + 12 ü ì 6 2 + ( -3 ) 2 + 2 2 ü í ý ýí î þ þî | 12 - 6 + 2| 8 8 æ 8 ö = = = Þ q = sin -1 ç ÷ × ( 9 )( 49 ) ( 3 ´ 7 ) 21 è 21 ø

sin q =

®

32. (b) We know that the perpendicular distance of a point having position vector a ® ®

from the plane r × n = q is given by ® ®

p=

| a × n - q| ®

×

| n| ®

^

^

^

^

^

^

®

^

^

^

^

^

Here a = ( i + 2 j + 5 k ), n = ( - i - j - k ) and q = 17. \

p=

^

|( i + 2 j + 5 k ) × ( - i - j - k ) - 17 | ^

^

^

|- i - j - k | =

|( -1) + ( -2 ) + ( -5 ) + ( -17 )| 2

2

( -1) + ( -1) + ( -1)

2

=

25 units. 3

SSS Mathematics for Class 12 1261

The Plane

1261

33. (a) The given parallel planes are 2 x - 3 y + 6z = 5 6 x - 9 y + 18z + 20 = 0. Let P( x1 , y1 , z1 ) be a point on (i). Then, 2 x1 - 3 y1 + 6z1 = 5. Distance between the given planes

... (i) ... (ii) ... (iii)

= Length of perpendicular from P( x1 , y1 , z1 ) on (ii) =

| 6 x1 - 9 y1 + 18z1 + 20| | 3( 2 x1 - 3 y1 + 6z1 ) + 20| = 441 6 2 + ( -9 ) 2 + ( 18 ) 2

=

|( 3 ´ 5 ) + 20| 35 5 units = units [using (iii)] = 3 21 21

34. (c) Clearly the given planes are parallel. These are x + 2y - 2 z + 1 = 0

... (i)

2 x + 4 y - 4 z + 5 = 0.

... (ii)

Let P( x1 , y1 , z1 ) be a point on (i). Then, x1 + 2 y1 - 2z1 = -1. | 2 x1 + 4 y1 - 4z1 + 5| p= \ 2 2 + 4 2 + ( -4 ) 2 =

| 2( x1 + 2 y1 - 2z1 ) + 5| 36

| 2 ´ ( -1) + 5| 3 1 = = = units. 6 6 2 35. (d) The equation of a line through the point P( 1, 3 , 4 ) and perpendicular to the plane 2 x - y + z + 3 = 0 is x- 1 y- 3 z- 4 = = = k (say) 2 -1 1 A general point on this line is ( 2 k + 1, - k + 3 , k + 4 ). For some value of k let N ( 2 k + 1, - k + 3 , k + 4 ) be a point of the line lying on the plane. Then, 2( 2 k + 1) - ( - k + 3 ) + ( k + 4 ) + 3 = 0 Þ 6 k = -6 Þ k = -1. \

coordinates of N are N( -2 + 1, 1 + 4 , - 1 + 4 ), i.e., N( -1, 4 , 3 ).

Let Q( a , b , g ) be the image of P in the given plane. Then 1+ a 3+b 4+ g = -1, = 4 and = 3. So, a = -3 , b = 5 and g = 2. 2 2 2 Hence, the required image of P(1, 3, 4) in the given plane is Q(–3, 5, 2).

SSS Mathematics for Class 12 1262

29. PROBABILITY

CONDITIONAL PROBABILITY AND PROBABILITY OF INDEPENDENT EVENTS The concept of probability and various results on it were discussed in class XI. In this chapter, we shall be dealing with problems based on conditional probability and probability of independent events.

CONDITIONAL PROBABILITY Let A and B be the two events associated with the same random experiment. Then, the probability of occurrence of A under the condition B has already occurred and P( B) ¹ 0, is called conditional probability, denoted by P( A / B). We define: P( A / B) = and P( B/A) =

P( A Ç B) , where P( B) ¹ 0 P( B) P( B Ç A) P( A Ç B) = , where P( A) ¹ 0. P( A) P( A)

SOLVED EXAMPLES EXAMPLE 1

A die is rolled. If the outcome is an odd number, what is the probability that it is prime?

SOLUTION

When a die is rolled, the sample space is given by S = {1, 2, 3 , 4, 5 , 6 }. Let A = event of getting a prime number, and B = event of getting an odd number. Then, A = {2, 3 , 5}, B = {1, 3 , 5} and A Ç B = { 3 , 5}. n( A) 3 1 n( B) 3 1 = = , P( B) = = = \ P( A) = n( S) 6 2 n( S) 6 2 and P( A Ç B) =

n( A Ç B) 2 1 = = × n( S) 6 3

Suppose B has already occurred and then A occurs. So, we have to find P( A / B). 1262

SSS Mathematics for Class 12 1263

Probability

1263

P( A Ç B) (1/ 3) æ 1 2 ö 2 = =ç ´ ÷= × P( B) (1/ 2) è 3 1 ø 3 P( A Ç B) (1/ 3) æ 1 2 ö 2 = =ç ´ ÷= × P( A / B) = (1/ 2) è 3 1 ø 3 P( B)

Now, P( A / B) =

Hence, the required probability is

2 × 3

EXAMPLE 2

Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

SOLUTION

Clearly, the sample space is S = {1, 2, 3 , 4, 5 , 6, 7 , 8, 9, 10}. Let A = event that the number on the drawn card is even, and B = event that the number on the drawn card is more than 3. Then A = {2, 4, 6, 8, 10}, B = {4, 5 , 6, 7 , 8, 9, 10} and A Ç B = {4, 6, 8, 10}. n( A) 5 1 n( B) 7 and \ P( A) = = = , P( B) = = n( S) 10 2 n( S) 10 P( A Ç B) =

n( A Ç B) 4 2 = = × n( S) 10 5

Suppose B has already occurred and then A occurs. So, we have to find P( A / B). P( A Ç B) ( 2/5) æ 2 10 ö 4 Now, P( A / B) = = =ç ´ ÷= × P( B) (7 /10) è 5 7 ø 7 Hence, the required probability is

4 × 7

EXAMPLE 3

A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

SOLUTION

We know that when a die is thrown twice, then the sample space has 36 possible outcomes. Let A = event that 4 appears at least once, and B = event that the sum of the numbers appearing is 6. Then, A = {( 4, 1), ( 4, 2), ( 4, 3), ( 4, 4), ( 4, 5), ( 4, 6), (1, 4), ( 2, 4), ( 3 , 4), (5 , 4), ( 6, 4)} and B = {(1, 5), ( 2, 4), ( 3 , 3), ( 4, 2), (5 , 1)}. \ A Ç B = {( 2, 4), ( 4, 2)}. n( A) 11 n( B) 5 So, P( A) = = , P( B) = = n( S) 36 n( S) 36

SSS Mathematics for Class 12 1264

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Senior Secondary School Mathematics for Class 12

and

P( A Ç B) =

n( A Ç B) 2 1 = = × n( S) 36 18

Suppose B has already occurred and then A occurs. So, we have to find P( A / B). P( A Ç B) (1/18) æ 1 36 ö 2 Now, P( A / B) = = =ç ´ ÷= × P( B) (5 / 36) è 18 5 ø 5 2 Hence, the required probability is × 5 EXAMPLE 4

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest child is a girl, (ii) at least one of the children is a girl?

SOLUTION

We may write the sample space as S = {G1G2 , G1 B2 , B1G2 , B1 B2}, where the youngest child appears later. (i) Let A = event that both the children are girls, and B = event that the youngest child is a girl. Then, A = {G1G2}, B = {G1G2 , B1G2} and A Ç B = {G1G2}. n( A) 1 n( B) 2 1 \ P( A) = = , P( B) = = = n( S) 4 n( S) 4 2 and P( A Ç B) =

n( A Ç B) 1 = × n( S) 4

Suppose B has already occurred and then A occurs. So, we have to find P( A / B). P( A Ç B) (1/ 4) æ 1 2 ö 1 Now, P( A / B) = = =ç ´ ÷= × P( B) (1/ 2) è 4 1 ø 2 1 Hence, the required probability is × 2 (ii) Let A = event that both the children are girls, and E = event that at least one of the children is a girl. Then, A = {G1G2}, E = {G1 B2 , B1G2 , G1G2} and A Ç E = {G1G2}. n( A) 1 n(E) 3 \ P( A) = = , P(E) = = n( S) 4 n( S) 4 n( A Ç E) 1 and P( A Ç E) = = × n( S) 4 Suppose E has already occurred and then A occurs. So, we have to find P( A /E). P( A Ç E) (1/ 4) æ 1 4 ö 1 Now, P( A /E) = = =ç ´ ÷= × P(E) ( 3 / 4) è 4 3 ø 3

SSS Mathematics for Class 12 1265

Probability

1265

Hence, the required probability is

1 × 3

EXAMPLE 5

An instructor has a question bank consisting of 300 easy true/false questions; 200 difficult true/false questions; 500 easy multiple-choice questions and 400 difficult multiple-choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple-choice question?

SOLUTION

Clearly, the sample space consists of 1400 questions. \ n( S) = 1400. Let A = event of selecting an easy question, and B = event of selecting a multiple-choice question. Then, A Ç B = event of selecting an easy multiple-choice question. \ n( A) = ( 300 + 500) = 800, n( B) = (500 + 400) = 900 and n( A Ç B) = 500. n( A) 800 4 n( B) 900 9 So, P( A) = = = , P( B) = = = n( S) 1400 7 n( S) 1400 14 and P( A Ç B) =

n( A Ç B) 500 5 = = × n( S) 1400 14

Suppose B has already occurred and then A occurs. Thus we have to find P( A / B). P( A Ç B) (5 /14) æ 5 14 ö 5 Now, P( A / B) = = =ç ´ ÷= × ( 9/14) è 14 9 ø 9 P( B) Hence, the required probability is

5 × 9

EXAMPLE 6

Two numbers are selected at random from the integers 1 through 9. If the sum is even, find the probability that both the numbers are odd.

SOLUTION

Out of the numbers from 1 to 9, there are 5 odd numbers and 4 even numbers. Let A = event of choosing two odd numbers, and B = event of choosing two numbers whose sum is even. Then, n( A) = number of ways of choosing 2 odd numbers out of 5 = 5C 2 . n( B) = number of ways of choosing 2 numbers whose sum is even = ( 4C2 + 5 C2)

[2 out of 4 even and 2 out of 5 odd].

n( A Ç B) = number of ways of choosing 2 odd numbers out of 5 = 5 C2. Suppose B has already occurred and then A occurs.

SSS Mathematics for Class 12 1266

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Senior Secondary School Mathematics for Class 12

Then, we have to find P( A / B). Now, P( A / B) = =

5 P( A Ç B) n( A Ç B) C2 = = 4 P( B) n( B) C 2 + 5C 2

5 ´4 2 20 5 ´ = = × 2 ( 4 ´ 3 + 5 ´ 4) 32 8

Hence, the required probability is

5 × 8

Properties of Conditional Probability Let A and B be the events of a sample space S of an experiment. Then, prove that P( S/ B) = P( B/ B) = 1. We know that: P( S Ç B) P( B) P( S/ B) = = = 1 [B S Ç B = B]. P( B) P( B)

THEOREM 1

PROOF

And, P( B/ B) =

P( B Ç B) P( B) = = 1. P( B) P( B)

Hence, P( S/ B) = P( B/ B) = 1. THEOREM 2

PROOF

Let A and B be the two events of a sample space S and let E be an event such that P(E) ¹ 0. Then, prove that P[( A È B) /E ] = P( A /E) + P( B/E) - P[( A Ç B)/E].

We have P[( A È B) /E] =

P[( A È B ) Ç E] P(E)

=

P[( A Ç E) È ( B Ç E)] P(E)

=

P( A Ç E) + P( B Ç E) - P[( A Ç E) Ç ( B Ç E)] P(E)

=

P( A Ç E) + P( B Ç E) - P( A Ç B Ç E) P(E)

=

P( A Ç E) P( B Ç E) P[( A Ç B ) Ç E] + P(E) P(E) P(E)

= P( A /E) + P( B /E) - P[( A Ç B)/E]. Hence, P[( A È B ) /E] = P( A /E) + P( B/E) - [( A Ç B)/E]. COROLLARY

PROOF

If A and B are disjoint events, prove that P[( A È B ) /E] = P( A /E) + P( B /E).

For any events A and B, we have P[( A È B) /E] = P( A /E) + P( B /E) - P[( A Ç B ) /E].

SSS Mathematics for Class 12 1267

Probability

1267

If A and B are disjoint, then P[( A Ç B) /E ] = 0. Hence, in this case, we have P[( A È B) /E] = P( A /E) + P( B /E). THEOREM 3

PROOF

For any events A and B of a sample space S, prove that P( A / B) = 1 - P( A / B), where A denotes ‘not A’.

We know that P( S/ B) = 1 Þ P[( A È A ) / B] = 1 [B S = A È A ] Þ P( A / B) + P( A / B) = 1 [B A Ç A = j] Þ P( A / B) = 1 - P( A / B). Hence, P( A / B) = 1 - P( A / B).

EXAMPLE 7

If A and B are the two events such that P( A) = P( A È B) =

SOLUTION

3 7 and , P( B) = 5 10

9 , then find (i) P( A Ç B) (ii) P( A / B) (iii) P( B/A). 10

(i) We know that P( A Ç B) = P( A) + P( B) - P( A È B) 9 ö ( 6 + 7 - 9) 4 2 æ3 7 =ç + - ÷= = = × 10 10 5 è 5 10 10 ø (ii) P( A / B) =

( A Ç B) ( 2/5) æ 2 10 ö 4 = =ç ´ ÷= × P( B) (7/10) è 5 7 ø 7

(iii) P( B /A) =

( B Ç A) P( A Ç B) æ 2 / 5 ö æ 2 5 ö 2 = =ç ÷=ç ´ ÷= × P( A) P( A) è 3 / 5 ø è5 3 ø 3

EXAMPLE 8

Evaluate P( A È B), if 2P( A) = P( B) =

SOLUTION

2P( A) = P( B) = \

6 1 and P( A / B) = × 13 3

6 3 6 and P( B) = Þ P( A) = × 13 13 13 P( A Ç B) P( A/B) = P( B)

2 æ1 6 ö × Þ P( A Ç B) = P( A/B) × P( B) = ç ´ ÷ = è 3 13 ø 13 So, P( A È B) = P( A) + P( B) - P( A Ç B) 6 2 ö ( 3 + 6 - 2) 7 æ 3 =ç + - ÷= = × 13 13 è 13 13 13 ø Hence, P( A È B) =

7 × 13

SSS Mathematics for Class 12 1268

1268

EXAMPLE 9

Senior Secondary School Mathematics for Class 12

Let A and B be the events such that P( A) =

1 1 and , P( B) = 3 4

1 × 5 Find: (i) P( A / B) (ii) P( B / A) (iii) P( A È B) (iv) P( B / A) P( A Ç B) =

SOLUTION

We have: (i) P( A / B) =

P( A Ç B) (1/5) æ 1 4 ö 4 = =ç ´ ÷= × P( B) (1/4) è 5 1 ø 5

(ii) P( B / A) =

P( B Ç A) P( A Ç B) (1/5) æ 1 3 ö 3 = = =ç ´ ÷= × P( A) P( A) (1/ 3) è 5 1 ø 5

(iii) P( A È B) = P( A) + P( B) - P( A Ç B) æ 1 1 1 ö ( 20 + 15 - 12) 23 = × =ç + - ÷= 60 60 è 3 4 5ø (iv) P( B / A) =

P( B Ç A ) P( A Ç B) P( A È B) = = P( A ) P( A ) P( A )

23 ö æ 37 ö æ ç1 - ÷ ç ÷ 1 - P( A È B) è 60 ø è 60 ø æ 37 3 ö 37 = × =ç ´ ÷= = = 1ö 1 - P( A) æ æ 2ö è 60 2 ø 40 1 ÷ ç ç ÷ 3ø è è 3ø

EXERCISE 29A 1. Let A and B be the events such that 7 9 4 P( A) = , P( B) = and P( A Ç B) = × 13 13 13 Find (i) P( A / B) (ii) P( B /A) (iii) P( A È B) (iv) P( B /A). 2. Let A and B be the events such that 5 6 7 P( A) = , P( B) = and P( A È B) = × 11 11 11 Find (i) P( A Ç B) (ii) P( A /B) (iii) P( B /A) (iv) P(A / B). 3. Let A and B be the events such that 3 1 2 P( A) = , P( B) = and P( B/A) = × 10 2 5 Find (i) P( A Ç B) (ii) P( A È B) (iii) P( A /B). 4. Let A and B be the events such that 5 2 and P( A / B) = × 2P( A) = P( B) = 13 5 Find (i) P( A Ç B) (ii) P( A È B). 5. A die is rolled. If the outcome is an even number, what is the probability that it is a number greater than 2?

SSS Mathematics for Class 12 1269

Probability

1269

6. A coin is tossed twice. If the outcome is at most one tail, what is the probability that both head and tail have appeared? 7. Three coins are tossed simultaneously. Find the probability that all coins show heads if at least one of the coins shows a head. 8. Two unbiased dice are thrown. Find the probability that the sum of the numbers appearing is 8 or greater, if 4 appears on the first die. 9. A die is thrown twice and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once? 10. Two dice were thrown and it is known that the numbers which come up were different. Find the probability that the sum of the two numbers was 5. 11. A coin is tossed and then a die is thrown. Find the probability of obtaining a 6, given that a head came up. 12. A couple has 2 children. Find the probability that both are boys if it is known that (i) one of the children is a boy, and (ii) the elder child is a boy. 13. In a class, 40% students study mathematics; 25% study biology and 15% study both mathematics and biology. One student is selected at random. Find the probability that (i) he studies mathematics if it is known that he studies biology (ii) he studies biology if it is known that he studies mathematics. 14. The probability that a student selected at random from a class will pass in 4 1 Hindi is and the probability that he passes in Hindi and English is × 5 2 What is the probability that he will pass in English if it is known that he has passed in Hindi? 15. The probability that a certain person will buy a shirt is 0.2, the probability that he will buy a coat is 0.3 and the probability that he will buy a shirt given that he buys a coat is 0.4. Find the probability that he will buy both a shirt and a coat. 16. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random. (i) Find the probability that he reads neither Hindi nor English newspaper. (ii) If he reads Hindi newspaper, what is the probability that he reads English newspaper? (iii) If he reads English newspaper, what is the probability that he reads Hindi newspaper? 17. Two integers are selected at random from integers 1 through 11. If the sum is even, find the probability that both the numbers selected are odd.

SSS Mathematics for Class 12 1270

1270

Senior Secondary School Mathematics for Class 12 ANSWERS (EXERCISE 29A)

1. (i) 3. (i) 1 2 5 14. 8 8.

4 4 12 1 4 2 4 4 (ii) (iii) (iv) 2. (i) (ii) (iii) (iv) 9 7 13 6 11 3 5 5 3 17 6 2 11 2 2 1 (ii) (iii) 4. (i) (ii) 5. 6. 7. 25 25 25 13 26 3 3 7 2 2 1 1 1 3 3 9. 10. 11. 12. (i) (ii) 13. (i) (ii) 5 15 6 3 2 5 8 1 1 1 3 15. 0.12 16. (i) (ii) (iii) 17. 5 3 2 5 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 29A)

8. Let A = event of getting the sum 8 or greater, and B = event of getting a 4 on the first die. \ A = {( 2 , 6 ),( 3 , 5 ), ( 3 , 6 ),( 4 , 4 ), ( 4 , 5 ),( 4 , 6 ), (5 , 3 ),(5 , 4 ), (5 , 5 ),(5 , 6 ), ( 6 , 2 ),( 6 , 3 ), ( 6 , 4 ),( 6 , 5 ), ( 6 , 6 )} B = {( 4 , 1), ( 4 , 2 ),( 4 , 3 ), ( 4 , 4 ), ( 4 , 5 ), ( 4 , 6 )}. ( A Ç B) = {( 4 , 4 ), ( 4 , 5 ), ( 4 , 6 )}. \ n( A ) = 15 , n( B) = 6 and n( A Ç B) = 3. Hence, the required probability = P( A/ B) =

n( A Ç B ) 3 1 = = × n( B ) 6 2

10. Let A = event that the sum of two numbers appeared is 5, and B = event that the two dice show different numbers. Then, A = {( 2 , 3 ), ( 3 , 2 ), ( 1, 4 ), ( 4 , 1)}. Thus, n( A ) = 4 , n( B) = 30 , A Ç B = A and so n( A Ç B) = n( A ) = 4. n( A Ç B ) 4 2 = = × \ the required probability = P( A/ B) = n( B ) 30 15 11. Let A = event that the die shows a 6, and B = event that a head comes up. Then, A = {( H , 6 ), ( T , 6 )} B = {( H , 1), ( H , 2 ), ( H , 3 ), ( H , 4 ), ( H , 5 ), ( H , 6 )}. n( A Ç B ) 1 Then, P( A / B) = = × n( B ) 6 and

13. Let M = event of studying mathematics, and B = event of studying biology. 40 2 25 1 15 3 Then P( M ) = = , P( B) = = and P( M Ç B) = = × 100 5 100 4 100 20 P( M Ç B) (i) P( M / B) = × P( B) (ii) P( B / M ) =

P( M Ç B) × P( M )

SSS Mathematics for Class 12 1271

Probability

1271

4 1 and P( H Ç E) = × 5 2 P( E Ç H ) P( H Ç E) \ P( E / H ) = = × P( H ) P( H )

14. Given, P( H ) =

15. P( S) = 0. 2 , P(C ) = 0. 3 , P( S / C ) = 0.4. P( S Ç C ) \ P( S/ C ) = Þ P( S Ç C ) = P(C ) × P( S/ C ). P(C ) 17. Let A = event of choosing both odd numbers, B = event that sum of chosen numbers is even. In integers from 1 to 11, there are 5 even and 6 odd integers. 6 ö 5 C 3 ÷= , P( A Ç B) = 11 2 = × ÷ C2 C 2 11 ø 11 P( A Ç B) × \ required probability = P( A / B) = P( B)

P( A ) =

6

C2

11

=

æ 6C + 5C 2 3 , P( B) = çç 211 11 C2 è

Multiplication Theorem on Probability Let A and B be the two events associated with a sample space S. Then, the simultaneous occurrence of two events A and B is denoted by ( A Ç B) and also written as AB. Let A and B be the two events associated with a sample space S. Then, prove that P( AB) = P( A Ç B) = P( A) × P( B /A) = P( B) × P( A / B), provided P( A) ¹ 0 and P( B) ¹ 0. For any events A and B, we have P( A Ç B) P( A / B) = , where P( A) ¹ 0. P( B)

MULTIPLICATION THEOREM

PROOF

\ P( A Ç B) = P( B) × P( A / B). P( B Ç A) P( A Ç B) Again, P( B /A) = = , where P( B) ¹ 0. P( A) P( A)

... (i)

\ P( A Ç B) = P( A) × P( B/A). From (i) and (ii), we get P( A Ç B) = P( B) × P( A / B) = P( A) × P( A / B), where P( A) ¹ 0.

... (ii)

Multiplication Rule for Three Events For any three events A , B , C of the same sample space, we have P( A Ç B Ç C) = P( A) × P( B / A) × P[C /( A Ç B)] = P( A) × P( B / A) × (C /AB). This rule can be extended for four or more events.

SSS Mathematics for Class 12 1272

1272

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

An urn contains 8 white and 4 red balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are white?

SOLUTION

Let A and B denote respectively the events that first and second balls both drawn are white. Then, we have to find P( A Ç B). Now, P( A) = P (white ball in the first draw) =

8 × 12

After the occurrence of event A , we are left with 7 white and 4 red balls. The probability of drawing second white ball, given that the first ball drawn is white, is clearly the conditional probability of occurrence of B, given that A has occurred. 7 \ P( B /A) = × 11 By multiplication rule of probability, we have æ 8 7 ö 14 P( A Ç B) = P( A) × P( B /A) = ç ´ ÷ = × è 12 11 ø 33 Hence, the required probability is

14 × 33

EXAMPLE 2

Three cards are drawn successively without replacement from a pack of 52 well-shuffled cards. What is the probability that first two cards are queens and the third card drawn is a king?

SOLUTION

Let Q denote the event that the card drawn is a queen and K be the event that the card drawn is a king. Then, we have to find P(QQK). 4 Probability of drawing first queen is P(Q) = × 52 Now, there are 3 queens in remaining 51 cards. Let P(Q /Q) be the probability of getting the second queen with the condition that one queen has already been drawn. 3 \ P(Q /Q) = × 51 Lastly, P( K /QQ) is probability of third drawn card to be a king, with the condition that two queens have already been drawn. Now, there are 4 kings in remaining 50 cards. 4 \ P( K /QQ) = × 50

SSS Mathematics for Class 12 1273

Probability

1273

By multiplication law of probability, we have P(QQK) = P(Q Ç Q Ç K) = P(Q) × P(Q /Q) × P( K /QQ) 2ö 2 æ 4 3 4ö æ 1 1 × =ç ´ ´ ÷=ç ´ ´ ÷= è 52 51 50 ø è 13 17 25 ø 5525 Hence, the required probability is

2 × 5525

Independent Events Two events A and B are said to be independent if P( A / B) = P( A), where P( B) ¹ 0 and P( B / A) = P( B), where P( A) ¹ 0. i.e., the event A does not depend on the occurence of event B and vice versa. Condition for Independence of Two Events For any two events A and B, we have P( A Ç B) = P( A) × P( B /A).

... (i)

If A and B are independent, we have P( B /A) = P( B). \ (i) becomes P(A Ç B) = P(A) ´ P(B). Thus, two events A and B associated with the same random experiment are said to be independent if P( A Ç B) = P( A) ´ P( B). NOTE: Two events A and B are said to be dependent if they are not independent, i.e., if P( A Ç B) ¹ P( A) ´ P( B). Difference between Two Mutually Exclusive and Independent Events Two events A and B are said to be mutually exclusive if A Ç B = j and in this case P( A Ç B) = P( j) = 0. Also we know that two events A and B are independent if P( A Ç B) = P( A) ´ P( B). Clearly, two independent events with nonzero probabilities cannot be mutually exclusive. Also, two mutually exclusive events with nonzero probabilities cannot be mutually independent. Three events A , B and C are said to be mutually independent, if P( A Ç B) = P( A) ´ P( B), P( A Ç C) = P( A) ´ P(C), P( B Ç C) = P( B) ´ P(C) and P( A Ç B Ç C) = P( A) ´ P( B) ´ P(C). If at least one of the above is not true for three given events A , B and C, then we say that these events are not independent.

SSS Mathematics for Class 12 1274

1274

Senior Secondary School Mathematics for Class 12

EXAMPLE 3

Let E1 and E2 be two events such that P(E1 ) = 0. 3 , P(E1 È E2 ) = 0.4 and P(E2 ) = x. Find the value of x such that (i) E1 and E 2 are mutually exclusive, (ii) E1 and E 2 are independent.

SOLUTION

(i) Let E1 and E 2 be mutually exclusive. Then, E1 Ç E 2 = j. \ P(E1 È E 2 ) = P(E1 ) + P(E 2 ) Þ 0.4 = 0. 3 + x Þ x = 01 .. Thus, when E1 and E 2 mutually exclusive, then x = 01 .. (ii) Let E1 and E 2 be two independent events. Then, P(E1 Ç E 2 ) = P(E1 ) ´ P(E 2 ) = 0. 3 ´ x = 0. 3 x. \ P(E1 È E 2 ) = P(E1 ) + P(E 2 ) - P(E1 Ç E 2 ) Þ 0.4 = 0. 3 + x - 0. 3 x Þ 07 . x = 01 . 01 . 1 = × Þ x= 07 . 7 1 Thus, when E1 and E 2 are independent, then x = × 7

EXAMPLE 4

Let E1 and E 2 are the two independent events such that P(E1 ) = 0. 35 and P(E1 È E 2 ) = 0. 60, find P(E 2 ).

SOLUTION

Let P(E 2 ) = x. Then, E1 and E 2 being independent events, we have P(E1 Ç E 2 ) = P(E1 ) ´ P(E 2 ) = 0. 35 ´ x = 0. 35 x. Now, P(E1 È E 2 ) = P(E1 ) + P(E 2 ) - P(E1 Ç E 2 ) Þ 0. 60 = 0. 35 + x - 0. 35 x Þ 0. 65 x = 0.25 0. 25 25 5 = = × Þ x= 0. 65 65 13 Hence, P(E 2 ) =

5 × 13

EXAMPLE 5

A coin is tossed thrice. Let the event E be ‘the first throw results in a head’, and the event F be ‘the last throw results in a tail’. Find whether the events E and F are independent.

SOLUTION

When a coin is tossed three times, the sample space is given by S = {HHH , HHT , HTH , THH , TTH , THT , HTT , TTT}. Now, E = event that the first throw results in a head. \ E = {HHH , HHT , HTH , HTT}. And, F = event that the last throw results in a tail. \ F = {HHT , THT , HTT , TTT}.

SSS Mathematics for Class 12 1275

Probability

1275

So, (E Ç F) = {HHT , HTT}. Clearly, n(E) = 4, n( F) = 4, n(E Ç F ) = 2 and n( S) = 8. n(E) 4 1 n( F ) 4 1 \ P(E) = = = , P( F ) = = = n( S) 8 2 n( S) 8 2 and P(E Ç F ) =

n(E Ç F ) 2 1 = = × n( S) 8 4

Thus, P(E Ç F ) = P(E) ´ P( F ). Hence, E and F are independent events. EXAMPLE 6

An unbiased die is tossed twice. Find the probability of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss.

SOLUTION

In each case, the sample space is given by S = {1, 2, 3 , 4, 5 , 6}. Let E = event of getting a 4, 5 or 6 on the first toss. And, F = event of getting a 1, 2, 3 or 4 on the second toss. 3 1 4 2 Then, P(E) = = and P( F ) = = × 6 2 6 3 Clearly, E and F are independent events. \ required probability = P(E Ç F ) = P(E) ´ P( F ) [B E and F are independent] æ1 2ö 1 =ç ´ ÷= × è2 3ø 3

EXAMPLE 7

Ramesh appears for an interview for two posts, A and B, for which the selection is independent. The probability for his selection for Post A is (1/ 6) and for Post B, it is (1/ 7). Find the probability that Ramesh is selected for at least one post. [CBSE 2001]

SOLUTION

Let E1 = event that Ramesh is selected for the post A, and E 2 = event that Ramesh is selected for the post B. 1 1 Then, P(E1 ) = and P(E2 ) = × 6 7 Clearly, E1 and E 2 are independent events. æ1 1ö 1 × \ P(E1 Ç E 2 ) = P(E1 ) ´ P(E 2 ) = ç ´ ÷ = è 6 7 ø 42 \ P(Ramesh is selected for at least one post) = P(E1 È E 2 ) = P(E1 ) + P(E 2 ) - P(E1 Ç E2 ) æ 1 1 1 ö 12 2 = × =ç + - ÷= è 6 7 42 ø 42 7 2 Hence, the required probability is × 7

SSS Mathematics for Class 12 1276

1276

Senior Secondary School Mathematics for Class 12

EXAMPLE 8

A can solve 90% of the problems given in a book, and B can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book?

SOLUTION

Let E1 = event that A solves the problem, and E 2 = event that B solves the problem. 90 9 70 7 Then, P(E1 ) = and P(E 2 ) = = = × 100 10 100 10 Clearly, E1 and E 2 are independent events. æ 9 7 ö 63 × \ P(E1 Ç E 2 ) = P(E1 ) ´ P(E 2 ) = ç ´ ÷ = è 10 10 ø 100 \ P(at least one of them will solve the problem) = P(E1 È E 2 ) = P(E1 ) + P(E 2 ) - P(E1 Ç E 2 ) 63 ö ( 90 + 70 - 63) 97 æ 9 7 =ç + = × ÷= 100 100 è 10 10 100 ø Hence, the required probability is 0.97.

EXAMPLE 9

The probability that A hits a target is (1/3) and the probability that B hits it is (2/5). What is the probability that the target will be hit if both A and B shoot at it?

SOLUTION

Let E1 = event that A hits the target, and E 2 = event that B hits the target. 1 2 Then, P(E1 ) = and P(E 2 ) = × 3 5 Clearly, E1 and E 2 are independent events. 2 æ 1 2ö × \ P(E1 Ç E 2 ) = P(E1 ) ´ P(E 2 ) = ç ´ ÷ = è 3 5 ø 15 \ P(target is hit) = P(A hits or B hits) = P(E1 È E 2 ) = P(E1 ) + P(E 2 ) - P(E1 Ç E 2 ) 9 3 æ1 2 2 ö = × =ç + - ÷= è 3 5 15 ø 15 5 3 Hence, the required probability is × 5

EXAMPLE 10

A and B appear for an interview for two posts. The probability of A‘s selection is (1/3) and that of B‘s selection is (2/5). Find the probability that only one of them will be selected.

SOLUTION

Let E1 = event that A is selected, and E 2 = event that B is selected.

SSS Mathematics for Class 12 1277

Probability

1277

1 2 and P(E 2 ) = 3 5 1ö 2 2ö 3 æ æ P( E1 ) = ç1 - ÷ = and P( E 2 ) = ç1 - ÷ = × 3ø 3 5ø 5 è è

Then, P(E1 ) = Þ

\ P(event that only one of them is selected) = P[(E1 and not E 2 ) or (E 2 and not E1 )] = P[(E1 Ç E 2 ) or (E 2 Ç E1 )] = P(E1 Ç E 2 ) + P(E 2 Ç E1 )

[B (E1 Ç E 2 ) Ç (E 2 Ç E1 ) = f]

= P(E1 ) × P( E 2 ) + P(E 2 ) × P( E1 ) é B E1 and E 2 are independent, ù ê ú ë and E 2 and E1 are independent û æ 1 3ö æ2 2ö æ1 4 ö 7 =ç ´ ÷+ç ´ ÷=ç + ÷= × è 3 5 ø è 5 3 ø è 5 15 ø 15 EXAMPLE 11

SOLUTION

A speaks the truth in 60% of the cases, and B in 90% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact? [CBSE 2001] Let E1 = event that A speaks the truth, and E 2 = event that B speaks the truth. Then, E1 = event that A tells a lie, and E 2 = event that B tells a lie. Clearly, E1 and E 2 are independent events. Also, (E1 and E 2 ) as well as ( E1 and E 2 ) are independent. 60 3 90 9 Now, P(E1 ) = = ; P(E 2 ) = = ; 100 5 100 10 9ö 1 3ö 2 æ æ P( E1 ) = ç1 - ÷ = and P( E 2 ) = ç1 - ÷ = × 5ø 5 è 10 ø 10 è \ P(A and B contradict each other) = P[(A speaks the truth and B tells a lie) or (A tells a lie and B speaks the truth)] = P[(E1 Ç E 2 ) È ( E1 Ç E 2 )] = P(E1 Ç E 2 ) + P( E1 Ç E 2 ) [B (E1 Ç E 2 ) Ç ( E1 Ç E 2 ) = f] = {P(E1 ) ´ P( E 2 )} + {P( E1 ) ´ P(E 2 )} æ 3 1 ö æ 2 9 ö æ 3 18 ö 21 × =ç ´ ÷+ç ´ ÷=ç + ÷= è 5 10 ø è 5 10 ø è 50 50 ø 50 Percentage of cases in which A and B contradict each other æ 21 ö = ç ´ 100÷ % = 42%. è 50 ø

SSS Mathematics for Class 12 1278

1278 EXAMPLE 12

Senior Secondary School Mathematics for Class 12

The probabilities of a specific problem being solved independently by A and B are 1 2 and 1 3 respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

SOLUTION

Let E1 = event that A solves the problem, and E 2 = event that B solves the problem. 1 1 Then, P(E1 ) = and P(E 2 ) = 2 3 1 1 1ö 2 æ ö æ Þ P( E1 ) = ç1 - ÷ = and P( E 2 ) = ç1 - ÷ = × 2ø 2 3ø 3 è è Clearly, E1 and E 2 are independent events. æ1 1 ö 1 \ P(E1 Ç E 2 ) = P(E1 ) ´ P(E 2 ) = ç ´ ÷ = × è2 3ø 6 (i) P(the problem is solved) = P(at least one of A and B solves the problem) = P(E1 or E 2 ) = P(E1 È E 2 ) = P(E1 ) + P(E 2 ) - P(E1 Ç E 2 ) æ1 1 1ö 4 2 =ç + - ÷= = × è 2 3 6ø 6 3 (ii) P(exactly one of them solves the problem) = P[(E1 and not E 2 ) or (E 2 and not E1 )] = P(E1 and not E 2 ) + P(E 2 and not E1 ) = P(E1 Ç E 2 ) + P(E 2 Ç E1 ) = P(E1 ) ´ P( E 2 ) + P(E 2 ) ´ P( E1 ) æ1 2 ö æ 1 1ö æ 1 1ö 3 1 =ç ´ ÷+ç ´ ÷=ç + ÷= = × è 2 3 ø è 3 2ø è 3 6ø 6 2

EXAMPLE 13

Amit and Nisha appear for an interview for two vacancies in a company. The probability of Amit’s selection is 15 and that of Nisha’s selection is 1 . What is the probability that 6 (i) both of them are selected? (ii) only one of them is selected? (iii) none of them is selected?

SOLUTION

Let E1 = event that Amit is selected, and E 2 = event that Nisha is selected. 1 1 Then, P(E1 ) = and P(E 2 ) = × 5 6 Clearly, E1 and E 2 are independent events.

SSS Mathematics for Class 12 1279

Probability

1279

(i) P(both are selected) = P(E1 Ç E 2 ) = P(E1 ) ´ P(E 2 ) [B E1 and E 2 are independent] 1 æ1 1ö =ç ´ ÷= × è 5 6 ø 30 (ii) P(only one of them is selected) = P[(E1 and not E 2 ) or (E 2 and not E1 )] = P(E1 and not E 2 ) + P(E 2 and not E1 ) = P(E1 ) × P(not E 2 ) + P(E 2 ) × P(not E1 ) = P(E1 ) × [1 - P(E 2 )] + P(E 2 ) × [1 - P(E1 )] 1 æ 1 ö 1 æ 1 ö æ 1 5 ö æ 1 4ö = × ç1 - ÷ + × ç1 - ÷ = ç ´ ÷ + ç ´ ÷ 5 è 6ø 6 è 5 ø è 5 6ø è 6 5 ø 9 3 æ1 2 ö = × =ç + ÷= è 6 15 ø 30 10 (iii) P (none of them is selected) = P(not E1 and not E 2 ) = P(not E1 ) and P(not E 2 ) = [1 - P(E1 )] × [1 - P(E 2 )] æ 1ö æ 1ö æ 4 5 ö 2 = ç1 - ÷ × ç1 - ÷ = ç ´ ÷ = × 6ø è 5 6ø 3 è 5ø è EXAMPLE 14

Three groups of children contain 3 girls and 1 boy; 2 girls and 2 boys; and 1 girl and 3 boys. One child is selected at random from each group. Find the chance that the three children selected comprise 1 girl and 2 boys.

SOLUTION

Let G1 , G2 , G 3 be the events of selecting a girl from the first, second and third group respectively, and let B1 , B2 , B 3 be the events of selecting a boy from the first, second and third group respectively. Then, 3 2 1 1 P(G 1 ) = ; P(G2 ) = = ; P(G 3 ) = × 4 4 2 4 1 2 1 3 P( B1 ) = ; P( B2 ) = = and P( B 3 ) = × 4 4 2 4 \ P(selecting 1 girl and 2 boys) = P[(G 1 B2 B 3 ) or ( B1G2 B 3 ) or ( B1 B2G 3 )] = P(G 1 B2 B 3 ) + P( B1G2 B 3 ) + P( B1 B2G 3 ) = {P(G 1 ) ´ P( B2 ) ´ P( B 3 )} + {P( B1 ) ´ P(G2 ) ´ P( B 3 )} + {P( B 1 ) ´ P( B2 ) ´ P(G 3 )} 3 1 ö 13 æ 3 1 3 ö æ1 1 3 ö æ1 1 1ö æ 9 + × + =ç ´ ´ ÷+ç ´ ´ ÷+ç ´ ´ ÷=ç ÷= è 4 2 4 ø è 4 2 4 ø è 4 2 4 ø è 32 32 32 ø 32

SSS Mathematics for Class 12 1280

1280

Senior Secondary School Mathematics for Class 12

Hence, the chances of selecting 1 girl and 2 boys are EXAMPLE 15

13 × 32

A problem is given to three students whose chances of solving it are 2 3 3 , 7 and 8 . What is the probability that the problem will be solved?

1

SOLUTION

Let the three students be named A, B, and C respectively. Let E1 , E 2 , E 3 be the events that the problem is solved by A, B, C respectively. Then, 1 2 3 P(E1 ) = , P(E 2 ) = , P(E3 ) = × 3 7 8 2ö 5 1ö 2 æ æ \ P( E1 ) = ç1 - ÷ = ; P( E 2 ) = ç1 - ÷ = and 7 3 3 ø 7 è ø è 3ö 5 æ P( E3 ) = ç1 - ÷ = × 8ø 8 è \ P(none solves the problem) = P[(not E1 ) and (not E 2 ) and (not E3 )] = P( E1 Ç E 2 Ç E3 ) = P( E1 ) ´ P( E 2 ) ´ P( E3 ) [B E1 , E 2 , E3 are independent] æ 2 5 5 ö 25 × =ç ´ ´ ÷= è 3 7 8 ø 84 \ P(that the problem is solved) = 1 - P(none solves the problem) 25 ö 59 æ × = ç1 - ÷ = 84 ø 84 è 59 Hence, the required probability is × 84

EXAMPLE 16

A problem in mathematics is given to three students whose chances of solving it correctly are 1 2 , 1 3 and 1 4 respectively. What is the probability [CBSE 1999C] that only one of them solves it correctly?

SOLUTION

Let A, B, C be the given students and let E1 , E2 and E3 be the events that the problem is solved by A, B, C respectively. Then, E1 , E 2 and E 3 are the events that the given problem is not solved by A, B, C respectively. Then, 1 1 1 P(E1 ) = ; P(E 2 ) = ; P(E3 ) = ; 2 3 4 1ö 2 æ 1ö 1 æ æ 1ö 3 P( E1 ) = ç1 - ÷ = ; P( E 2 ) = ç1 - ÷ = and P( E3 ) = ç1 - ÷ = × 2ø 2 3ø 3 4ø 4 è è è P(exactly one of them solves the problem) = P[(E1 Ç E 2 Ç E3 ) or ( E1 Ç E 2 Ç E3 ) or ( E1 Ç E 2 Ç E3 )] = P(E1 Ç E 2 Ç E3 ) + P( E1 Ç E 2 Ç E3 ) + P( E1 Ç E 2 Ç E3 )

SSS Mathematics for Class 12 1281

Probability

1281

= {P(E1 ) ´ P( E 2 ) ´ P( E3 )} + {P( E1 ) ´ P(E 2 ) ´ P( E3 )} + {P( E1 ) ´ P( E 2 ) ´ P(E3 )} æ1 2 3 ö æ1 1 3 ö æ1 2 1ö =ç ´ ´ ÷+ç ´ ´ ÷+ç ´ ´ ÷ è 2 3 4 ø è 2 3 4 ø è 2 3 4ø æ 1 1 1 ö 11 × =ç + + ÷= è 4 8 12 ø 24 11 Hence, the required probability is × 24 EXAMPLE 17

Three critics review a book. For the three critics, the odds in favour of the book are (5 : 2), ( 4 : 3) and ( 3 : 4) respectively. Find the probability that the majority is in favour of the book.

SOLUTION

Let A , B , C denote the events that the book be favoured by the first, second and third critic respectively. Then, 5 4 3 P( A) = ; P( B) = ; P(C) = ; 7 7 7 3ö 4 4ö 3 æ 5ö 2 æ æ P( A ) = ç1 - ÷ = ; P( B ) = ç1 - ÷ = and P(C ) = ç1 - ÷ = × 7ø 7 è 7ø 7 è 7ø 7 è Required probability = P(2 critics favour the book or 3 critics favour the book) = P(2 critics favour the book) + P(3 critics favour the book) = P[{A and B and not C} or {A and C and not B} or {B and C and not A}] + P(A and B and C) = P( A Ç B Ç C ) + P( A Ç B Ç C) + P( A Ç B Ç C) + P( A Ç B Ç C) = {P( A) ´ P( B) ´ P(C )} + {P( A) ´ P( B ) ´ P(C )} + {P( A ) ´ P( B) ´ P(C)} + {P( A) ´ P( B) ´ P(C)} æ 5 4 4ö æ 5 3 3 ö æ 2 4 3 ö æ 5 4 3 ö =ç ´ ´ ÷+ç ´ ´ ÷+ç ´ ´ ÷+ç ´ ´ ÷ è7 7 7 ø è7 7 7 ø è7 7 7 ø è7 7 7 ø 45 24 60 ö 209 æ 80 × =ç + + + ÷= è 343 343 343 343 ø 343 209 Hence, the required probability is × 343

EXAMPLE 18

SOLUTION

The odds against a man who is 45 years old, living till he is 70 are 7 : 5, and the odds against his wife who is now 36, living till she is 61 are 5 : 3. Find the probability that (i) the couple will be alive 25 years hence (ii) at least one of them will be alive 25 years hence. [CBSE 2007] Let E1 = event that the husband will be alive 25 years hence, and E 2 = event that the wife will be alive 25 years hence.

SSS Mathematics for Class 12 1282

1282

Senior Secondary School Mathematics for Class 12

5 3 and P(E 2 ) = × 12 8 5ö 7 3ö 5 æ æ and P( E 2 ) = ç1 - ÷ = × \ P( E1 ) = ç1 - ÷ = 12 ø 12 8ø 8 è è Then, P(E1 ) =

Clearly, E1 and E 2 are independent events. (i) P(the couple will be alive 25 years hence) = P(E1 and E 2 ) = P(E1 Ç E 2 ) 3ö 5 æ5 × = P(E1 ) × P(E 2 ) = ç ´ ÷ = 12 8 ø 32 è (ii) P(at least one of them will be alive 25 years hence) = 1 - P (none will be alive 25 years hence) = 1 - P[(not E1 ) and (not E 2 )] = 1 - P( E1 Ç E 2 ) = 1 - [P( E1 ) × P( E 2 )] [Q E1 and E 2 are independent] 35 ö 61 æ 7 5ö æ = 1 - ç ´ ÷ = ç1 × ÷= 96 ø 96 12 8 è ø è EXAMPLE 19

A, B and C shoot to hit a target. If A hits the target 4 times in 5 trials; B hits it 3 times in 4 trials and C hits it 2 times in 3 trials, what is the probability that the target is hit by at least 2 persons?

SOLUTION

Let E1 , E 2 and E3 be the events that A hits the target, B hits the target and C hits the target respectively. Then, 4 3 2 P(E1 ) = , P(E 2 ) = , P(E3 ) = ; 5 4 3 3ö 1 4ö 1 2ö 1 æ æ æ P( E1 ) = ç1 - ÷ = , P( E 2 ) = ç1 - ÷ = and P( E3 ) = ç1 - ÷ = × 4ø 4 3ø 3 è è 5ø 5 è Case I

Case II

A, B, C all hit the target In this case, P(A, B and C all hit the target) = P(E1 and E 2 and E3 ) = P(E1 ) × P(E 2 ) × P(E3 ) [Q E1 , E 2 , E3 are independent] æ4 3 2ö 2 =ç ´ ´ ÷= × è5 4 3 ø 5 A and B hit but not C In this case, P(A and B hit but not C) = P(E1 and E 2 and not E3 ) = P(E1 Ç E 2 Ç E3 )

SSS Mathematics for Class 12 1283

Probability

1283

= P(E1 ) × P(E 2 ) × P( E3 ) [B E1 , E 2 , E 3 are independent] æ4 3 1ö 1 =ç ´ ´ ÷= × è5 4 3 ø 5 Case III

A and C both hit but not B In this case, P(A and C hit but not B) = P(E1 and E3 and E 2 ) = P(E1 ) × P(E3 ) × P( E 2 ) [B E1 , E3 , E 2 are independent] 2 æ 4 2 1ö × =ç ´ ´ ÷= è 5 3 4 ø 15

Case IV

B and C both hit but not A In this case, P(B and C hit but not A) = P(E 2 and E3 and E1 ) = P(E 2 ) × P(E3 ) × P( E1 ) [Q E 2 , E3 , E1 are independent] æ 3 2 1ö 1 =ç ´ ´ ÷= × è 4 3 5 ø 10

Clearly, all these are mutually exclusive. 1ö 5 æ2 1 2 Hence, required probability = ç + + + ÷= × è 5 5 15 10 ø 6

EXERCISE 29B 1. A bag contains 17 tickets, numbered from 1 to 17. A ticket is drawn and then another ticket is drawn without replacing the first one. Find the probability that both the tickets may show even numbers. 2. Two marbles are drawn successively from a box containing 3 black and 4 white marbles. Find the probability that both the marbles are black, if the first marble is not replaced before the second draw. 3. A card is drawn from a well-shuffled deck of 52 cards and without replacing this card, a second card is drawn. Find the probability that the first card is a club and the second card is a spade. 4. There is a box containing 30 bulbs of which 5 are defective. If two bulbs are chosen at random from the box in succession without replacing the first, what is the probability that both the bulbs chosen are defective? 5. A bag contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that the first ball is white and the second is black? 6. An urn contains 5 white and 8 black balls. Two successive drawings of 3 balls at a time are made such that the balls drawn in the first draw are not replaced before the second draw. Find the probability that the first draw gives 3 white balls and the second draw gives 3 black balls.

SSS Mathematics for Class 12 1284

1284

Senior Secondary School Mathematics for Class 12

7. Let E1 and E 2 be the events such that P(E1 ) =

1 3 and P(E2 ) = × 3 5

Find: (i) P(E1 È E 2 ), when E1 and E 2 are mutually exclusive, (ii) P(E1 Ç E 2 ), when E1 and E 2 are independent. 8. If E1 and E 2 are the two events such that P(E1 ) = P(E1 È E 2 ) =

1 1 and , P(E 2 ) = 4 3

1 , show that E1 and E 2 are independent events. 2

9. If E1 and E 2 are independent events such that P(E1 ) = 0. 3 and P(E 2 ) = 0.4, find (i) P(E1 Ç E 2 ) (ii) P(E1 Ç E 2 ) (iii) P(E 1 Ç E 2 ) (iv) P(E 1 Ç E 2 ). 10. Let A and B be the events such that 1 7 1 and P(not A or not B) = × P( A) = , P( B) = 2 12 4 State whether A and B are (i) mutually exclusive, (ii) independent. 11. Kamal and Vimal appeared for an interview for two vacancies. The probability of Kamal’s selection is 1 3 and that of Vimal’s selection is 15 × Find the probability that only one of them will be selected. 12. Arun and Ved appeared for an interview for two vacancies. The probability of Arun’s selection is 1 4 and that of Ved’s rejection is 2 3 × Find the probability that at least one of them will be selected. 13. A and B appear for an interview for two vacancies in the same post. The probability of A’s selection is 1 6 and that of B’s selection is 1 4 × Find the probability that (i) both of them are selected (ii) only one of them is selected (iii) none is selected (iv) at least one of them is selected. 14. Given the probability that A can solve a problem is 2 3 , and the probability that B can solve the same problem is 35 , find the probability that (i) at least one of A and B will solve the problem (ii) none of the two will solve the problem. 15. A problem is given to three students whose chances of solving it are 1 , 1 and 1 respectively. Find the probability that the problem is solved. 4 5 6 16. The probabilities of A, B, C solving a problem are 1 3 , 1 4 and 1 6 respectively. If all the three try to solve the problem simultaneously, find the probability that exactly one of them will solve it. 17. A can hit a target 4 times in 5 shots, B can hit 3 times in 4 shots, and C can hit 2 times in 3 shots. Calculate the probability that (i) A, B and C all hit the target (ii) B and C hit and A does not hit the target.

SSS Mathematics for Class 12 1285

Probability

1285

18. Neelam has offered physics, chemistry and mathematics in Class XII. She estimates that her probabilities of receiving a grade A in these courses are 0.2, 0.3 and 0.9 respectively. Find the probabilities that Neelam receives (i) all A grades (ii) no A grade (iii) exactly 2 A grades. 19. An article manufactured by a company consists of two parts X and Y. In the process of manufacture of part X, 8 out of 100 parts may be defective. Similarly, 5 out of 100 parts of Y may be defective. Calculate the probability that the assembled product will not be defective. 20. A town has two fire-extinguishing engines, functioning independently. The probability of availability of each engine when needed is 0.95. What is the probability that (i) neither of them is available when needed? (ii) an engine is available when needed? 21. A machine operates only when all of its three components function. The probabilities of the failures of the first, second and third components are 0.14, 0.10 and 0.05 respectively. What is the probability that the machine [CBSE 2005C] will fail? 22. An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that at least one shot hits the plane? 23. Let S1 and S2 be the two switches and let their probabilities of working be given by P( S1 ) = 45 and P( S2 ) = 910 × Find the probability that the current flows from the terminal A to terminal B when S1 and S2 are installed in series, shown as follows: S1

S2

A

B

24. Let S1 and S2 be two the switches and let their probabilities of working be given by P( S1 ) = 2 3 and P( S2 ) = 3 4 . Find the probability that the current flows from terminal A to terminal B, when S1 and S2 are installed in parallel, as shown below: S1

A

S2

B

25. A coin is tossed. If a head comes up, a die is thrown but if a tail comes up, the coin is tossed again. Find the probability of obtaining (i) two tails (ii) a head and the number 6 (iii) a head and an even number.

SSS Mathematics for Class 12 1286

1286

Senior Secondary School Mathematics for Class 12 ANSWERS (EXERCISE 29B)

1.

7 34

2.

1 7

3.

13 204

4.

2 87

9. (i) 0.12 (ii) 0.58 (iii) 0.42 (iv) 0.28

5.

1 4

6.

7 429

10. (i) No (ii) No

1 1 5 3 13 2 (ii) (iii) (iv) 14. (i) (ii) 24 3 8 8 15 15 2 1 17. (i) (ii) 18. (i) 0.054 (ii) 0.056 (iii) 0.348 5 10 1 19 (ii) 21. 0.2647 22. 0.6976 20. (i) 400 200 1 1 3 (ii) (iii) 25. (i) 8 8 8 13. (i)

14 1 (ii) 15 5 2 1 11. 12. 5 2 1 31 15. 16. 2 72 437 19. 500 18 11 23. 24. 25 12 7. (i)

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 29B) 8 ì 5C C ü æ 5´4´ 3 8 ´7 ´ 6 ö 6. Required probability = í 13 3 ´ 10 3 ý = ç ´ ÷ C 3 þ è 13 ´ 12 ´ 11 10 ´ 9 ´ 8 ø î C3

10. P(not A or not B) = P( A or B ) = P( A È B ) = P( A Ç B) = 1 - P( A Ç B) Þ 1 - P( A Ç B) =

1 3 Þ P( A Ç B) = × 4 4

(i) Since P( A Ç B) ¹ 0 , so A and B are not mutually exclusive. æ1 7 ö 7 (ii) P( A ) ´ P( B) = ç ´ ÷ = ¹ P( A Ç B) è 2 12 ø 24 Þ A and B are not independent. 11. Let E 1 = event that Kamal is selected, and E 2 = event that Vimal is selected. Then, 1 1 1ö 2 1ö 4 æ æ P( E 1 ) = , P( E 2 ) = , P( E 1 ) = ç 1 - ÷ = and P( E 2 ) = ç 1 - ÷ = × 3 5 3 3 5 è ø 5 è ø \ required probability = P[( E 1 and not E 2 ) or ( E 2 and not E 1 )] = P[( E 1 and E 2 ) or ( E 2 and E 1 )] = P( E 1 Ç E 2 ) + P( E 2 Ç E 1 ) = {P( E 1 ) ´ P( E 2 )} + {P( E 2 ) ´ P( E 1 )}. 12. Let E 1 = event that Arun is selected, and E 2 = event that Ved is selected. 1 2ö 1 æ Then, P( E 1 ) = and P( E 2 ) = ç 1 - ÷ = × 4 3ø 3 è Clearly, E 1 and E 2 are independent events.

SSS Mathematics for Class 12 1287

Probability

1287

æ 1 1ö 1 \ P( E 1 Ç E 2 ) = P( E 1 ) ´ P( E 2 ) = ç ´ ÷ = × è 4 3 ø 12 Required probability = P( E 1 or E 2 ) = P( E 1 È E 2 )

1 ö 6 1 æ1 1 = P( E 1 ) + P( E 2 ) - P( E 1 Ç E 2 ) = ç + = × ÷= è 4 3 12 ø 12 2

1 1 , P( B) = × 6 4 1ö 5 1ö 3 æ æ \ P( A ) = 1 - P( A ) = ç 1 - ÷ = and P( B ) = 1 - P( B) = ç 1 - ÷ = × 6ø 6 4ø 4 è è

13. We have P( A ) =

(i) P(both are selected) = P( A Ç B) = P( A ) ´ P( B). (ii) P(only one of them is selected) = P[( A and not B) or ( B and not A )] = P( A Ç B ) + P( B Ç A ) = {P( A ) ´ P( B )} + {P( B) ´ P( A )}. (iii) P(none is selected) = P(not A and not B) = P( A Ç B ) = P( A ) ´ P( B ). (iv) P(at least one is selected) = 1 - P(none is selected). 14. Let E 1 = event that A can solve the problem, and E 2 = event that B can solve the problem. Then, E 1 and E 2 are clearly independent events. æ 2 3ö 2 \ P( E 1 Ç E 2 ) = P( E 1 ) ´ P( E 2 ) = ç ´ ÷ = × è3 5ø 5 (i) P(at least one of A and B can solve the problem) æ 2 3 2 ö 13 = P( E 1 È E 2 ) = P( E 1 ) + P( E 2 ) - P( E 1 Ç E 2 ) = ç + - ÷ = × è 3 5 5 ø 15 (ii) P(none can solve the problem) = 1 - P(at least one can solve the problem) 13 ö 2 æ = ç1× ÷= 15 ø 15 è 15. Let A , B, C be the events of solving the problem by the 1st, 2nd and 3rd student respectively. Then, 1 1 1 , P( B) = and P(C ) = 4 5 6 1ö 4 1ö 5 1ö 3 æ æ æ Þ P( A ) = ç 1 - ÷ = ; P( B ) = ç 1 - ÷ = and P(C ) = ç 1 - ÷ = × 4ø 4 5ø 5 6ø 6 è è è P( A ) =

\ P(none solves the problem) = P[(not A ) and (not B) and (not C )] = P( A and B and C ) = P( A ) ´ P( B ) ´ P(C ). 16. Given P( A ) =

1 1 1 , P( B) = and P(C ) = 3 4 6

1ö 2 1ö 3 1ö 5 æ æ æ Þ P( A ) = ç 1 - ÷ = , P( B ) = ç 1 - ÷ = and P(C ) = ç 1 - ÷ = × 6ø 6 3ø 3 4ø 4 è è è

SSS Mathematics for Class 12 1288

1288

Senior Secondary School Mathematics for Class 12

\ required probability = P(exactly one of them solves the problem) = P[( A Ç B Ç C ) or ( A Ç B Ç C ) or ( A Ç B Ç C )] = P( A Ç B Ç C ) + P( A Ç B Ç C ) + P( A Ç B Ç C ) = {P( A ) ´ P( B ) ´ P(C )} + {P( A ) ´ P( B) ´ P(C )} + {P( A ) ´ P( B ) ´ P(C )}. 4 3 2 , P(B hits) = and P(C hits) = × 5 4 3 (i) P(A hits and B hits and C hits) = P(A) ´ P( B) ´ P( C). (ii) P(B hits and C hits and A does not hit) = P( B) ´ P( C) ´ P( A ).

17. P(A hits) =

18.

(i) P(getting all A grades) = P( A1 A 2 A 3 ) = P( A1 ) ´ P( A 2 ) ´ P( A 3 ). (ii) P(getting no A grade) = P( A1 A 2 A 3 ) = P( A1 ) ´ P( A 2 ) ´ P( A 3 ). (iii) P(getting exactly two A grades) = P[( A1 A 2 A 3 ) or ( A1 A 2 A 3 ) or ( A1 A 2 A 3 )] = P( A1 A 2 A 3 ) + P( A1 A 2 A 3 ) + P( A1 A 2 A 3 ) = {P( A1 ) ´ P( A 2 ) ´ P( A 3 )} + {P( A1 ) ´ P( A 2 ) ´ P( A 3 )} + {P( A1 ) ´ P( A 2 ) ´ P( A 3 )}. 8 2 = × 100 25 5 1 P(Y is defective) = = × 100 20 2 ö 23 æ P(X is not defective) = ç 1 × ÷= 25 ø 25 è 1 ö 19 æ P(Y is not defective) = ç 1 × ÷= 20 ø 20 è Required probability = P(assembled part is not defective)

19. P(X is defective) =

= P(X is not defective and Y is not defective) æ 23 19 ö 437 =ç ´ ÷= × è 25 20 ø 500 20. Let E 1 = event of availability of the first engine. And, E 2 = event of availability of the second engine. Then, P( E 1 ) = P( E 2 ) = 0.95 and P( E 1 ) = P( E 2 ) = ( 1 - 0.95 ) = 0.05. (i) P(neither of them is available when needed) = P( E 1 and E 2 ) = P( E 1 ) ´ P( E 2 ). (ii) P(an engine is available when needed) = P[( E 1 and not E 2 ) or ( E 2 and not E 1 )] = P[( E 1 and E 2 ) or ( E 2 and E 1 )] = P( E 1 Ç E 2 ) + P( E 2 Ç E 1 ) = {P( E 1 ) ´ P( E 2 )} + {P( E 2 ) ´ P( E 1 )}.

SSS Mathematics for Class 12 1289

Probability

1289

21. Let E 1 , E2 , E3 be the respective events that the 1st, 2nd and 3rd components function. Then, P( E 1 ) = 0.14 , P( E 2 ) = 0.10 and P( E 3 ) = 0.05 Þ Þ

P( E 1 ) = ( 1 - 0.14 ) = 0.86 , P( E 2 ) = ( 1 - 0.10 ) = 0.90 and P( E3 ) = ( 1 - 0.05 ) = 0.95 P(machine fails) = 1 - P(machine functions)

= 1 - P[( E 1 and E 2 and E 3 )] = 1 - [P( E 1 ) ´ P( E 2 ) ´ P( E 3 )]. 22. Let E 1 , E 2 , E3 , E 4 be the respective events that the plane is hit in the 1st, 2nd, 3rd and 4th shot. Then, P( E 1 ) = 0.4 , P( E 2 ) = 0. 3 , P( E3 ) = 0. 2 and P( E4 ) = 0.1 \

P( E 1 ) = ( 1 - 0.4 ) = 0.6 , P( E 2 ) = ( 1 - 0. 3 ) = 0.7 , P( E3 ) = ( 1 - 0. 2 ) = 0.8 , P( E4 ) = ( 1 - 0.1) = 0.9.

\

P(at least one shot hits the plane) = 1 - P( E 1 and E2 and E3 and E4 ) = 1 - {P( E 1 ) ´ P( E2 ) ´ P( E3 ) ´ P( E4 )}.

23. P(current flows from A to B) = P( S1 is closed and S2 is closed) = P( S1 and S2 ) = P( S1 ) ´ P( S2 ). 24. P(the current flows) = P( S1 or S2 ) = P( S1 È S2 ) = P( S1 ) + P( S2 ) - P( S1 Ç S2 ) = P( S1 ) + P( S2 ) - P( S1 ) ´ P( S2 ). 25. S = {H 1, H 2 , H 3 , H 4 , H 5 , H 6 , TT , TH } ® n( S) = 8. 1 (i) P(two tails) = × 8 1 (ii) P(head and the number 6) = × 8 3 (iii) P(head and an even number) = × 8

SSS Mathematics for Class 12 1290

30. BAYES’S THEOREM AND ITS APPLICATIONS Theorem of Total Probability THEOREM

Let E1 , E2 , ¼, E n be mutually exclusive and exhaustive events associated with a random experiment and let E be an event that occurs with some Ei . Then, prove that P(E) =

PROOF

n

S

i =1

P(E/E i ) × P(E i ).

Let S be the sample space. Then, S = E1 È E2 È ¼ È E n and E i Ç Ej = f for i ¹ j. \

E = E Ç S = E Ç (E1 È E 2 È ¼ È En)

Þ

P(E) = P{(E Ç E1 ) È (E Ç E 2 ) È ¼ È (E Ç E n)}

= (E Ç E1 ) È (E Ç E 2 ) È ¼ È (E Ç E n) = P(E Ç E1 ) + P(E Ç E 2 ) + ¼ + P(E Ç E n) {Q (E Ç E1 ), (E Ç E 2 ), ¼, (E Ç E n) are pairwise disjoint} = P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) + ¼ + P(E/E n) × P(E n) [by multiplication theorem] =

n

S P(E/E i ) × P(E i ).

i =1

EXAMPLE

There are three urns containing 3 white and 2 black balls; 2 white and 3 black balls; 1 black and 4 white balls respectively. There is equal probability of each urn being chosen. One ball is drawn from an urn chosen at random. What is the probability that a white ball is drawn?

SOLUTION

Let E1 , E 2 and E3 be the events of choosing the first, second and third urn respectively. Then, 1 P(E1 ) = P(E 2 ) = P(E3 ) = × 3 Let E be the event that a white ball is drawn. Then, 3 2 4 P(E/E1 ) = , P(E/E2 ) = and P(E/E3 ) = × 5 5 5 By the theorem of total probability, we have P(E) = P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) + P(E/E3 ) × P(E3 ) 2 4ö 9 3 æ 3 1 2 1 4 1 ö æ1 =ç ´ + ´ + ´ ÷=ç + + = × ÷= è 5 3 5 3 5 3 ø è 5 15 15 ø 15 5 1290

SSS Mathematics for Class 12 1291

Bayes’s Theorem and Its Applications

1291

BAYES’S THEOREM Let E1 , E 2 , ¼ , E n be mutually exclusive and exhaustive events, associated with a random experiment, and let E be any event that occurs with some E i . Then, P(E/E i ) × P(E i ) P(E i /E) = n ; i = 1, 2, 3 , ¼ , n.

S

i =1

PROOF

P(E/E i ) × P(E i )

By the theorem of total probability, we have P(E) = \

n

S

i =1

P(E i /E) = = =

P(E/E i ) × P(E i ) P(E Ç E i ) P(E)

… (i) [by multiplication theorem] é êQ êë

P(E/E i ) × P(E i ) P(E) P(E/E i ) × P(E i ) n

P(E/E i ) =

P(E Ç E i ) ù ú P(E i ) úû

[using (i)].

S P(E/E i ) × P(E i )

i =1

Hence, P(E i /E) =

P(E/E i ) × P(E i ) n

×

S P(E/E i ) × P(E i )

i =1

SOLVED EXAMPLES EXAMPLE 1

A factory has three machines, X, Y and Z, producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% defective bolts and Z produces 2% defective bolts. At the end of the day, a bolt is drawn at random and it is found to be defective. What is the probability that this defective bolt has been [CBSE 2002] produced by the machine X?

SOLUTION

Total number of bolts produced in a day = (1000 + 2000 + 3000) = 6000. Let E1 , E 2 and E3 be the events of drawing a bolt produced by machines X, Y and Z respectively. Then, 1000 1 2000 1 3000 1 P(E1 ) = = ; P(E2 ) = = and P(E3 ) = = × 6000 6 6000 3 6000 2 Let E be the event of drawing a defective bolt. Then, P(E/E1 ) = probability of drawing a defective bolt, given that it is produced by the machine X 1 = × 100 P(E/E 2 ) = probability of drawing a defective bolt, given that it is produced by the machine Y

SSS Mathematics for Class 12 1292

1292

Senior Secondary School Mathematics for Class 12

=

1.5 15 3 = = × 100 1000 200

P(E/E3 ) = probability of drawing a defective bolt, given that it is produced by the machine Z 2 1 = = × 100 50 Required probability = P(E1 /E) = probability that the bolt drawn is produced by X, =

given that it is defective P(E1 ) × P(E/E1 ) P(E1 ) × P(E/E1 ) + P(E 2 ) × P(E/E 2 ) + P(E 3 ) × P(E/E 3 )

1 ö æ1 ÷ ç ´ 6 100 ø è = 1 ö æ1 3 ö æ1 1 ö æ1 ÷+ç ´ ÷ ÷+ç ´ ç ´ è 6 100 ø è 3 200 ø è 2 50 ø 600 ö 1 æ 1 =ç ´ = 0.1. ÷= è 600 10 ø 10 Hence, the required probability is 0.1. EXAMPLE 2

In a bolt factory, three machines, A, B, C, manufacture 25%, 35% and 40% of the total production respectively. Of their respective outputs, 5%, 4% and 2% are defective. A bolt is drawn at random from the total product and it is found to be defective. Find the probability that it was [CBSE 2006, ‘10] manufactured by the machine C.

SOLUTION

Let E1 , E 2 and E3 be the events of drawing a bolt produced by machine A , B and C respectively. Then, 25 1 35 7 40 2 P(E1 ) = , and P(E3 ) = = , P(E 2 ) = = = × 100 4 100 20 100 5 Let E be the event of drawing a defective bolt. Then, P(E/E1 ) = probability of drawing a defective bolt, given that it is produced by the machine A 5 1 = = × 100 20 P(E/E 2 ) = probability of drawing a defective bolt, given that it is produced by the machine B 4 1 = = × 100 25 P(E/E3 ) = probability of drawing a defective bolt, given that it is produced by the machine C

SSS Mathematics for Class 12 1293

Bayes’s Theorem and Its Applications

=

1293

2 1 = × 100 50

Probability that the bolt drawn is manufactured by C, given that it is defective = P(E3 /E) =

P(E / E3 ) × P(E3 ) P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) + P(E/E3 ) × P(E3 ) [by Bayes’s theorem]

æ 1 2ö ç ´ ÷ 2000 ö 16 æ 1 è 50 5 ø =ç ´ × = ÷= 7 ö æ 1 2 ö è 125 æ 1 1ö æ 1 69 ø 69 ´ ÷+ç ´ ÷ ´ ÷+ç ç è 20 4 ø è 25 20 ø è 50 5 ø Hence, the required probability is

16 × 69

EXAMPLE 3

A company has two plants to manufacture bicycles. The first plant manufactures 60% of the bicycles and the second plant, 40%. Also, 80% of the bicycles are rated of standard quality at the first plant and 90% of standard quality at the second plant. A bicycle is picked up at random and found to be of standard quality. Find the probability that it comes [CBSE 2003] from the second plant.

SOLUTION

Let E1 and E2 be the events of choosing a bicycle from the first plant and the second plant respectively. Then, 60 3 40 2 P(E1 ) = = , and P(E2 ) = = × 100 5 100 5 Let E be the event of choosing a bicycle of standard quality. Then, P(E/E1 ) = probability of choosing a bicycle of standard quality, given that it is produced by the first plant 80 4 = = × 100 5 P(E/E2 ) = probability of choosing a bicycle of standard quality, given that it is produced by the second plant 90 9 = = × 100 10 The required probability P(E 2 /E ) = probability of choosing a bicycle from the second plant, given that it is of standard quality P(E 2 ) × P(E/E 2 ) [by Bayes’s theorem] = P(E1 ) × P(E/E1 ) + P(E 2 ) × P(E/E 2 )

SSS Mathematics for Class 12 1294

1294

Senior Secondary School Mathematics for Class 12

æ2 9 ö ç ´ ÷ 3 è 5 10 ø = × = 3 4 2 9 ö 7 ö æ æ ç ´ ÷+ç ´ ÷ è 5 5 ø è 5 10 ø EXAMPLE 4

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a 1 3 3 and respectively. One of the scooter, a car and a truck is , 100 100 20 insured persons meets with an accident. What is the probability that he is [CBSE 2000, ’02, ’08] a scooter driver?

SOLUTION

Total number of persons insured = ( 2000 + 4000 + 6000) = 12000. Let E1 , E 2 and E3 be the events of choosing a scooter driver, a car driver and a truck driver respectively. Then, P(E1 ) =

2000 1 4000 1 6000 1 = , P(E 2 ) = = and P(E3 ) = = × 12000 6 12000 3 12000 2

Let E be the event of an insured person meeting with an accident. Then, P(E/E1 ) = probability that an insured person meets with an accident, given that he is a scooter driver =

1 × 100

Similarly, P(E/E 2 ) =

3 3 and P(E/E3 ) = × 100 20

Required probability = P(E1 /E)

[by Bayes’s theorem]

= probability of choosing a scooter driver, given that he meets with an accident =

=

P(E/E1 ) × P(E1 ) P(E/E1 ) × P(E1 ) + P(E/E2 ) × P(E2 ) + P(E/E3 ) × P(E3 )

1ö æ 1 ´ ÷ ç è 100 6 ø

1ö æ 1 ´ ÷ ç 1 è 100 6 ø = × 3 1 3 1 æ ö æ ö 52 +ç ´ ÷+ç ´ ÷ è 100 3 ø è 20 2 ø

Hence, the required probability is

1 × 52

SSS Mathematics for Class 12 1295

Bayes’s Theorem and Its Applications

1295

EXAMPLE 5

A doctor is to visit a patient. From past experience, it is known that the probabilities that he will come by train, bus, scooter or by car are 3 1 1 2 1 and ×The probabilities that he will be late are , respectively , , 10 5 10 5 4 1 1 and , if he comes by train, bus and scooter respectively; but if he 3 12 comes by car, he will not be late. When he arrives, he is late. What is the [CBSE 2008C] probability that he has come by train?

SOLUTION

Let E1 , E 2 , E3 and E4 be the events that the doctor comes by train, bus, scooter and car respectively. Then, 3 1 1 2 and P(E4 ) = × P(E1 ) = , P(E 2 ) = , P(E3 ) = 10 5 10 5 Let E be the event that the doctor is late. Then, P(E/E1 ) = probability that the doctor is late, given that he comes by train 1 = × 4 P(E/E2 ) = probability that the doctor is late, given that he comes by bus 1 = × 3 P(E/E3 ) = probability that the doctor is late, given that he comes by scooter 1 = × 12 P(E/E 4 ) = probability that the doctor is late, given that he comes by car = 0. Probability that he comes by train, given that he is late = P(E1 /E) P(E1 ) × P(E/E1 ) = P(E1 ) × P(E/E1 ) + P(E2 ) × P(E/E2 ) + P(E3 ) × P(E/E3 ) + P(E4 ) × P(E/E4 ) æ 3 1ö ç ´ ÷ è 10 4 ø = 1 ö æ2 æ 3 1ö æ1 1 ö æ 1 ö ç ´ ÷ + ç ´ ÷ + ç ´ ÷ + ç ´ 0÷ è 10 4 ø è 5 3 ø è 10 12 ø è 5 ø æ 3 120 ö 1 =ç ´ ÷= × è 40 18 ø 2 1 Hence, the required probability is × 2

[by Bayes’s theorem]

SSS Mathematics for Class 12 1296

1296 EXAMPLE 6

Senior Secondary School Mathematics for Class 12

A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six. [CBSE 2005, ‘11]

SOLUTION

In a throw of a die, let E1 = event of getting a six, E 2 = event of not getting a six, and E = event that the man reports that it is a six. 1ö 5 1 æ Then, P(E1 ) = , and P(E 2 ) = ç1 - ÷ = × 6ø 6 6 è P(E/E1 ) = probability that the man reports that six occurs, when six has actually occurred = probability that the man speaks the truth 3 = × 4 P(E/E 2 ) = probability that the man reports that six occurs, when six has not actually occurred = probability that the man does not speak the truth 3ö 1 æ = ç1 - ÷ = × 4ø 4 è Probability of getting a six, given that the man reports it to be six = P(E1 /E) P(E/E1 ) × P(E1 ) [by Bayes’s theorem] = P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) æ 3 1ö ç ´ ÷ ö 3 æ1 è 4 6ø = ç ´ 3÷ = × = 3 1 1 5 ö è8 ö æ æ ø 8 ç ´ ÷+ç ´ ÷ è 4 6ø è 4 6ø 3 Hence, the required probability is × 8

EXAMPLE 7

In an examination, an examinee either guesses or copies or knows the answer to a multiple-choice question with four choices. The probability that he makes a guess is (1/ 3) and the probability that he copies the answer is (1/6). The probability that his answer is correct, given that he copied it, is (1/8). The probability that his answer is correct, given that he guessed it, is (1/4). Find the probability that he knew the answer to the question, given that he correctly answered it.

SOLUTION

Let E1 = event that the examinee guesses the answer, E 2 = event that he copies the answer, E3 = event that he knows the answer, and E = event that he answers correctly.

SSS Mathematics for Class 12 1297

Bayes’s Theorem and Its Applications

1297

1 1 æ 1 1ö 1 , P(E 2 ) = , and P(E3 ) = 1 - ç + ÷ = 3 6 è 3 6ø 2 [Q E1 , E2 , E3 are mutually exclusive and exhaustive]. P(E/E1 ) = probability that he answers correctly, given that he guesses 1 = × 4 P(E/E 2 ) = probability that he answers correctly, given that he copies 1 = × 8 P(E/E3 ) = probability that he answers correctly, given that he knew the answer

Then, P(E1 ) = \

= 1. Required probability = P(E3 /E) =

P(E/E3 ) × P(E3 ) P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) + P(E/E3 ) × P(E3 )

[by Bayes’s theorem] 1ö æ ç1 ´ ÷ 24 2ø è × = = 1 1 1 1 1 ö 29 ö æ ö æ æ ç ´ ÷ + ç ´ ÷ + ç1 ´ ÷ 2ø è 4 3 ø è 8 6ø è 24 Hence, the required probability is × 29 EXAMPLE 8

By examining the chest X-ray, the probability that a person is diagnosed with TB when he is actually suffering from it, is 0.99. The probability that the doctor incorrectly diagnoses a person to be having TB, on the basis of X-ray reports, is 0.001. In a certain city, 1 in 1000 persons suffers from TB. A person is selected at random and is diagnosed to have TB. What is the chance that he actually has TB?

SOLUTION

Let E = event that the doctor diagnoses TB, E1 = event that the person selected is suffering from TB, and E2 = event that the person selected is not suffering from TB. 1 ö 999 1 æ and P(E2 ) = ç1 Then, P(E1 ) = × ÷= 1000 ø 1000 1000 è P(E/E 1) = probability that TB is diagnosed, when the person actually has TB 99 = × 100

SSS Mathematics for Class 12 1298

1298

Senior Secondary School Mathematics for Class 12

P(E/E 2 ) = probability that TB is diagnosed, when the person has no TB 1 = × 1000 Using Bayes’s theorem, we have P(E1 /E) = probability of a person actually having TB, if it is known that he is diagnosed to have TB P(E/E1 ) × P(E1 ) = P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) 1 ö æ 99 ´ ÷ ç 110 100 1000 ø è = × = 1 ö æ 1 999 ö 221 æ 99 ´ ´ ÷+ç ç ÷ è 100 1000 ø è 1000 1000 ø Hence, the required probability is

110 × 221

EXAMPLE 9

Bag A contains 2 white and 3 red balls, and bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag B.

SOLUTION

Let E1 = event of choosing bag A,

[CBSE 2004C]

E2 = event of choosing bag B, and E = event of drawing a red ball. 1 1 Then, P(E1 ) = and P(E 2 ) = × 2 2 3 , and 5 5 P(E/E 2 ) = event of drawing a red ball from bag B = × 9

Also, P(E/E1 ) = event of drawing a red ball from bag A =

Probability of drawing a ball from B, it being given that it is red = P(E 2 /E) P(E/E 2 ) × P(E 2 ) [by Bayes’s theorem] = P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) æ5 1ö ç ´ ÷ 25 è 9 2ø = = × æ 3 1 ö æ 5 1 ö 52 ç ´ ÷+ç ´ ÷ è 5 2ø è 9 2ø Hence, the required probability is

25 × 52

SSS Mathematics for Class 12 1299

Bayes’s Theorem and Its Applications

1299

EXAMPLE 10

There are 5 bags, each containing 5 white balls and 3 black balls. Also, there are 6 bags, each containing 2 white balls and 4 black balls. A white ball is drawn at random. Find the probability that this white ball is from a bag of the first group.

SOLUTION

Let E1 = event of selecting a bag from the first group, E2 = event of selecting a bag from the second group, and E = event of drawing a white ball. 5 6 and P(E2 ) = Then, P(E1 ) = × 11 11 P(E/E1 ) = probability of getting a white ball, given that it is from a bag of the first group 5 = × 8 P(E/E 2 ) = probability of getting a white ball, given that it is from a bag of the second group 2 1 = = × 6 3 Probability of getting the ball from a bag of the first group, given that it is white = P(E1 /E) P(E/E1 ) × P(E1 ) [by Bayes’s theorem] = P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) æ5 5 ö ç ´ ÷ 75 è 8 11 ø = × = æ 5 5 ö æ 1 6 ö 123 ç ´ ÷+ç ´ ÷ è 8 11 ø è 3 11 ø

EXAMPLE 11

Urn A contains 1 white, 2 black and 3 red balls; urn B contains 2 white, 1 black and 1 red ball; and urn C contains 4 white, 5 black and 3 red balls. One urn is chosen at random and two balls are drawn. These happen to be one white and one red. What is the probability that they come from [CBSE 2009] urn A?

SOLUTION

Let E1 , E 2 , E3 be the events that the balls are drawn from urn A, urn B and urn C respectively, and let E be the event that the balls drawn 1 are one white and one red. Then, P(E1 ) = P(E 2 ) = P(E3 ) = × 3 P(E/E1 ) = probability that the balls drawn are one white and one red, given that the balls are from urn A =

1

C1 ´ 3C1 6

C2

=

3 1 = × 15 5

P(E/E 2 ) = probability that the balls drawn are one white and one red, given that the balls are from urn B

SSS Mathematics for Class 12 1300

1300

Senior Secondary School Mathematics for Class 12

=

2

C1 ´ 1C1 4

=

C2

2 1 = × 6 3

P(E/E3 ) = probability that the balls drawn are one white and one red, given that the balls are from urn C =

4

C1 ´ 3C1 12

C2

=

12 2 = × 66 11

Probability that the balls drawn are from urn A, it being given that the balls drawn are one white and one red = P(E1 /E) P(E/E1 ) × P(E1 ) = P(E/E1 ) × P(E1 ) + P(E/E 2 ) × P(E 2 ) + P(E/E3 ) × P(E3 ) [by Bayes’s theorem] æ1 1 ö ç ´ ÷ è5 3 ø = æ1 1 ö æ 1 1 ö æ 2 1 ö ç ´ ÷+ç ´ ÷+ç ´ ÷ è 5 3 ø è 3 3 ø è 11 3 ø æ 1 495 ö 33 =ç ´ × ÷= è 15 118 ø 118 33 Hence, the required probability is × 118 EXAMPLE 12

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both spades. Find the [CBSE 2002] probability of the lost card being a spade.

SOLUTION

Let E1 , E 2 , E3 and E4 be the events of losing a card of spades, clubs, hearts and diamonds respectively. 13 1 Then, P(E1 ) = P(E 2 ) = P(E3 ) = P(E4 ) = = × 52 4 Let E be the event of drawing 2 spades from the remaining 51 cards. Then, P(E/E1 ) = probability of drawing 2 spades, given that a card of spades is missing =

12

C2

51

C2

=

(12 ´ 11) 2! 22 ´ = × 2! (51 ´ 50) 425

P(E/E 2 ) = probability of drawing 2 spades, given that a card of clubs is missing =

13

C2

51

C2

=

(13 ´ 12) 2! 26 ´ = × 2! (51 ´ 50) 425

P(E/E3 ) = probability of drawing 2 spades, given that a card of hearts is missing

SSS Mathematics for Class 12 1301

Bayes’s Theorem and Its Applications

=

13

C2

51

C2

=

1301

26 × 425

P(E/E4 ) = probability of drawing 2 spades, given that a card of diamonds is missing = \

=

13

C2

51

C2

=

26 × 425

P(E1 /E) = probability of the lost card being a spade, given that 2 spades are drawn from the remaining 51 cards P(E1 ) × P(E/E1 ) P(E1 ) × P(E/E1 ) + P(E2 ) × P(E/E2 ) + P(E3 ) × P(E/E3 ) + P(E4 ) × P(E/E4 )

æ 1 22 ö ÷ ç ´ è 4 425 ø = æ 1 22 ö æ 1 26 ö æ 1 26 ö æ 1 26 ö ÷+ç ´ ÷ ÷+ç ´ ÷+ç ´ ç ´ è 4 425 ø è 4 425 ø è 4 425 ø è 4 425 ø 22 = = 0.22. 100 Hence, the required probability is 0.22.

EXERCISE 30 Long-Answer Questions 1. In a bulb factory, three machines, A , B , C , manufacture 60%, 25% and 15% of the total production respectively. Of their respective outputs, 1%, 2% and 1% are defective. A bulb is drawn at random from the total product and it is found to be defective. Find the probability that it was manufactured by machine C. 2. A company manufactures scooters at two plants, A and B. Plant A produces 80% and plant B produces 20% of the total product. 85% of the scooters produced at plant A and 65% of the scooters produced at plant B are of standard quality. A scooter produced by the company is selected at random and it is found to be of standard quality. What is the probability that it was manufactured at plant A? 3. In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Furthermore, 60% of the students are girls. If a student is selected at random and is taller than 1.75 metres, what is the probability that the selected student is a girl? 4. In a class, 5% of the boys and 10% of the girls have an IQ of more than 150. In this class, 60% of the students are boys. If a student is selected at random and found to have an IQ of more than 150, find the probability that the student is a boy.

SSS Mathematics for Class 12 1302

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Senior Secondary School Mathematics for Class 12

5. Suppose 5% of men and 0.25% of women have grey hair. A grey-haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females. [CBSE 2011] 6. Two groups are competing for the positions on the board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and when the second group wins, the corresponding probability is 0.3. Find the probability that the new product introduced was by the second group. 7. A bag A contains 1 white and 6 red balls. Another bag contains 4 white and 3 red balls. One of the bags is selected at random and a ball is drawn from it, which is found to be white. Find the probability that the ball [CBSE 2005] drawn is from the bag A. 8. There are two bags I and II. Bag I contains 3 white and 4 black balls, and bag II contains 5 white and 6 black balls. One ball is drawn at random from one of the bags and is found to be white. Find the probability that it [CBSE 2005C] was drawn from bag I. 9. A box contains 2 gold and 3 silver coins. Another box contains 3 gold and 3 silver coins. A box is chosen at random, and a coin is drawn from it. If the selected coin is a gold coin, find the probability that it was drawn from [CBSE 2004C] the second box. 10. Three urns A, B and C contain 6 red and 4 white; 2 red and 6 white; and 1 red and 5 white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the [CBSE 2004] ball was drawn from the urn A. 11. Three urns contain 2 white and 3 black balls; 3 white and 2 black balls, and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random and it was found to be white. Find the probability that it was drawn from the first urn. 12. There are three boxes, the first one containing 1 white, 2 red and 3 black balls; the second one containing 2 white, 3 red and 1 black ball and the third one containing 3 white, 1 red and 2 black balls. A box is chosen at random and from it two balls are drawn at random. One ball is red and the other, white. What is the probability that they come from the second box? 13. Urn A contains 7 white and 3 black balls; urn B contains 4 white and 6 black balls; urn C contains 2 white and 8 black balls. One of these urns is chosen at random with probabilities 0.2, 0.6 and 0.2 respectively. From the chosen urn, two balls are drawn at random without replacement. Both the balls happen to be white. Find the probability that the balls drawn are from the urn C. 14. There are 3 bags, each containing 5 white and 3 black balls. Also, there are 2 bags, each containing 2 white and 4 black balls. A white ball is drawn at random. Find the probability that this ball is from a bag of the first group.

SSS Mathematics for Class 12 1303

Bayes’s Theorem and Its Applications

1303

15.

There are four boxes, A , B , C and D, containing marbles. A contains 1 red, 6 white and 3 black marbles; B contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from the box A? 16. A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of the cars and plant Y manufactures 30%. At plant X, 80% of the cars are rated of standard quality and at plant Y, 90% are rated of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plantX. [CBSE 2005, ’06C]

17. An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcycle is 0.02. An insured vehicle met with an accident. Find the [CBSE 2005] probability that the accidented vehicle was a motorcycle. 18. In a bulb factory, machines A , B and C manufacture 60%, 30% and 10% bulbs respectively. Out of these bulbs 1%, 2% and 3% of the bulbs produced respectively by A , B and C are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the [CBSE 2008] probability that this bulb was produced by the machine A.

ANSWERS (EXERCISE 30)

1. 9. 17.

2 68 3 2. 3. 25 81 11 5 36 2 10. 11. 9 61 9 3 2 18. 4 5

3 7 6 12. 11 4.

5. 13.

2 3 1 40

6. 14.

2 1 7. 9 5 45 1 15. 61 15

8. 16.

33 68 56 83

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 30) 2. Let E 1 = event that the selected scooter is produced at plant A , and E2 = event that the selected scooter is produced at plant B. 80 4 20 1 Then, P( E 1 ) = = and P( E2 ) = = × 100 5 100 5 Let E be the event of choosing a scooter which is of standard quality. 85 17 65 13 Then, P( E/ E 1 ) = , and P( E/ E2 ) = = = × 100 20 100 20 Probability that the selected scooter was produced at plant A , given that it is of standard quality = P( E 1 / E) =

P( E/ E 1 ) × P( E 1 )

P( E/ E 1 ) × P( E 1 ) + P( E/ E 2 ) × P( E 2 )

×

SSS Mathematics for Class 12 1304

1304

Senior Secondary School Mathematics for Class 12

3. Let E 1 and E2 be the events of selecting a boy and a girl respectively. 40 2 60 3 Then, P( E 1 ) = = , and P( E2 ) = = × 100 5 100 5 Let E = event that the student selected is taller than 1.75 m. 4 1 1 Then, P( E/ E 1 ) = and P( E/ E2 ) = = × 100 25 100 Probability that the selected student is a girl, given that she is taller than 1.75 m = P( E2 / E) =

P( E/ E2 ) × P( E2 ) × P( E/ E 1 ) × P( E 1 ) + P( E/ E2 ) × P( E2 )

4. Let E 1 and E2 be the events of selecting a boy and a girl respectively. Then, 60 3 40 2 P( E 1 ) = = , and P( E2 ) = = × 100 5 100 5 Let E be the event of selecting a student having an IQ of more than 150. 5 1 10 1 Then, P( E/ E 1 ) = , and P( E/ E2 ) = = = × 100 20 100 10 Required probability = P( E 1 / E) P( E/ E 1 ) × P( E 1 ) = × P( E/ E 1 ) × P( E 1 ) + P( E/ E2 ) × P( E2 ) 5. Let there be 1000 males and 1000 females. Let E 1 and E2 be the events of choosing a male and a female respectively. Then, 1000 1 1000 1 P( E 1 ) = = , and P( E2 ) = = × 2000 2 2000 2 Let E be the event of choosing a grey haired person. Then, 50 1 25 1 P( E/ E 1 ) = , and P( E/ E2 ) = = = × 1000 20 1000 40 Probability of selecting a male person, given that the person selected is a grey haired = P( E1 / E) P( E/ E 1 ) × P( E 1 ) = × P( E/ E 1 ) × P( E 1 ) + P( E/ E2 ) × P( E2 ) 6. Let E 1 = event that the first group wins, E2 = event that the second group wins, and E = event that a new product is introduced. Then, P( E 1 ) = 0.6 , P( E2 ) = 0.4 , P( E/ E 1 ) = 0.7 , P( E/ E2 ) = 0. 3. Required probability = P( E2 / E) P( E/ E2 ) × P( E2 ) = × P( E/ E 1 ) × P( E 1 ) + P( E/ E2 ) × P( E2 ) 10. Let E 1 , E2 and E3 be the events of choosing the urns A, B and C respectively, and let E be the event of drawing a red ball. Then, 1 P( E 1 ) = P( E2 ) = P( E3 ) = × 3 æ E P çç è E1

ö 6 æ E ÷÷ = , P çç ø 10 è E2

ö 2 æ E ÷÷ = , and P çç ø 8 è E3

ö 1 ÷÷ = × ø 6

SSS Mathematics for Class 12 1305

Bayes’s Theorem and Its Applications æ E1 Required probability = P çç è E

1305

ö ÷ ÷ ø æ E P( E1 ) ´ P ç ç E1 è

=

æ E ö ÷ + P( E ) ´ P æç E P( E 1 ) ´ P ç 2 ç ç E1 ÷ è E2 ø è æ1 6 ö ç ´ ÷ è 3 10 ø = = æ 1 6 ö æ 1 2ö æ 1 1ö ç ´ ÷+ ç ´ ÷+ ç ´ ÷ è 3 10 ø è 3 8 ø è 3 6 ø

ö ÷ ÷ ø ö ÷÷ + P( E3 ) ´ P ø

æ E çç è E3

ö ÷÷ ø

36 × 61

13. P( A ) = 0.2 , P( B) = 0.6 and P(C ) = 0.2. Let E be the event that 2 white balls are drawn. Then, P( E/ A ) = \

7

C2

10

C2

; P( E/ B) =

4

C2

10

C2

; P( E/C ) =

2

C2

10

C2

×

required probability = P(C / E) =

P( E/C ) × P(C ) × P( E/ A ) × P( A ) + P( E/ B) × P( B) + P( E/C ) × P(C )

14. Let E 1 = event of selecting a bag from the first group, and E2 = event of selecting a bag from the second group. 3 2 Then, P( E 1 ) = and P( E2 ) = × 5 5 Let E = event that the ball drawn is white. Then, 5 2 1 P( E/ E 1 ) = , P( E/ E2 ) = = × 8 6 3 P( E/ E 1 ) × P( E1 ) × \ P( E 1 / E) = P( E/ E 1 ) × P( E 1 ) + P( E/ E2 ) × P ( E2 ) 15. Let E 1 , E2 , E3 , E4 be the events of selecting boxes A , B, C , D respectively. Then, 1 P( E 1 ) = P( E2 ) = P( E3 ) = P( E4 ) = × 4 Let E = event that the marble drawn is red. Then, 1 6 3 8 4 P( E/ E 1 ) = , P( E/ E2 ) = = , P( E/ E3 ) = = , P( E/ E4 ) = 0. 10 10 5 10 5 P( E/ E1 ) × P( E1 ) \ P( E 1 / E) = P( E/ E 1 ) × P( E 1 ) + P( E/ E2 ) × P( E2 ) + P( E/ E3 ) × P( E3 ) + P( E/ E4 ) × P( E4 ) 16. Let E 1 and E2 be the events that the car is manufactured by plant X and Y respectively. Let E be the event that the car is of standard quality. Then, 70 7 30 3 P( E 1 ) = , P( E2 ) = ; = = 100 10 100 10 80 4 90 9 P( E/ E 1 ) = = , P( E/ E2 ) = = × 100 5 100 10

SSS Mathematics for Class 12 1306

1306

\

Senior Secondary School Mathematics for Class 12

P( E 1 / E) =

P( E 1 ) ´ P( E/ E 1 ) P( E 1 ) ´ P( E/ E 1 ) + P( E2 ) ´ P( E/ E2 )

æ 7 4ö ´ ÷ ç 56 è 10 5 ø = = × 4ö æ 3 9 ö 83 æ7 ´ ÷+ ç ´ ÷ ç è 10 5 ø è 10 10 ø 17. Let E 1 and E 2 be the events that an insured vehicle is a scooter and a motorcycle respectively. Let E be the event that the insured vehicle meets an accident. 2000 2 3000 3 P( E 1 ) = = , P( E2 ) = = , ( 2000 + 3000 ) 5 5000 5 \

P( E/ E 1 ) = 0.01 and P( E/ E 2 ) = 0.02. P( E 2 ) × P( E/ E 2 ) P( E 2 / E) = P( E1 ) × P( E/ E1 ) + P( E 2 ) × P( E/ E 2 ) æ3 ö ç ´ 0.02 ÷ 3 è5 ø = = × æ2 ö æ3 ö 4 ç ´ 0.01 ÷ + ç ´ 0.02 ÷ è5 ø è5 ø

SSS Mathematics for Class 12 1307

31. PROBABILITY DISTRIBUTION RANDOM VARIABLE Let S be the sample space associated with a given random experiment. A real-valued function X which assigns a unique real number X(w) to each w Î S, is called a random variable.

A random variable which can assume only a finite number of values is called a discrete random variable. Example

Suppose that a coin is tossed twice. Then, sample space S = {TT , HT , TH , HH }. Consider a real-valued function X on S, defined by X : S ® R : X(w) = number of heads in w, for all w Î S. Then, X is a random variable such that X( TT ) = 0, X( HT ) = 1, X( TH ) = 1 and X( HH) = 2. Clearly, range (X) = {0, 1, 2}.

Probability Distribution of a Random Variable A description giving the values of a random variable along with the corresponding probabilities is called the probability distribution of the random variable. If a random variable X takes the values x1 , x 2 , ¼, x n with respective probabilities p1 , p 2 , ¼, p n then the probability distribution of X is given by

REMARK

X

x1

x2

x3

…

…

xn

P(X)

p1

p2

p3

…

…

pn

The above probability distribution of X is defined only when n

(i) each p i ³ 0

(ii)

S

i =1

pi = 1

Mean and Variance of Random Variables Let a random variable X assume values x1 , x 2 , ¼ , x n with probabilities p1 , p 2 , ¼ , p n respectively such that each p i ³ 0 and

n

S

i =1

p i = 1. Then mean of X,

denoted by m [or expected value of X, denoted by E(X)], is defined as m = E(X) =

n

S

i =1

xi pi .

1307

SSS Mathematics for Class 12 1308

1308

Senior Secondary School Mathematics for Class 12

And, the variance, denoted by s 2 , is defined as s 2 = (Sx i2 p i - m 2 ). Standard deviation, s, is given by s = variance.

SOLVED EXAMPLES EXAMPLE 1

Find the mean, variance and standard deviation of the number of tails in two tosses of a coin.

SOLUTION

In two tosses of a coin, the sample space is given by S = {HH , HT , TH , TT }. \ n( S) = 4. 1 So, every single outcome has a probability × 4 Let X = number of tails in two tosses. In two tosses, we may have no tail, 1 tail or 2 tails. So, the possible values of X are 0, 1, 2. P(X = 0) = P(getting no tail) = P( HH ) =

1 × 4

P(X = 1) = P(getting 1 tail) = P( HT or TH ) =

2 1 = × 4 2

1 × 4 Hence, the probability distribution of X is given by P(X = 2) = P(getting 2 tails) = P( TT ) =

X = xi pi \

0 1 4

1 1 2

2 1 4

1ö æ 1ö æ 1ö æ mean, m = Sx i p i = ç 0 ´ ÷ + ç1 ´ ÷ + ç 2 ´ ÷ = 1. 4ø è 2ø è 4ø è

Variance, s 2 = Sx i2 p i - m

2

éæ 1ö æ 1ö æ 1öù = ê ç 0 ´ ÷ + ç1 ´ ÷ + ç 4 ´ ÷ ú - 12 4ø û 2ø è 4ø è ëè =

1 × 2

Standard deviation, s =

1 2 2 1.414 ´ = = = 0.707. 2 2 2 2

SSS Mathematics for Class 12 1309

Probability Distribution

1309

EXAMPLE 2

Find the mean, variance and standard deviation of the number of heads when three coins are tossed.

SOLUTION

Here, S = {TTT , TTH , THT , HTT , THH , HTH , HHT , HHH}. \ n( S) = 8. 1 So, every single outcome has a probability × 8 Let X = number of heads in tossing three coins. The number of heads may be 0, 1, 2, or 3. So, the possible values of X are 0, 1, 2, 3. 1 P(X = 0) = P(getting no head) = P( TTT) = × 8 P(X = 1) = P(getting 1 head) = P( TTH or THT or HTT ) = P(X = 2) = P(getting 2 heads) = P( THH , HTH , HHT ) =

3 × 8

3 × 8

1 × 8 Thus, we have the following probability distribution: P(X = 3) = P(getting 3 heads) = P( HHH ) =

X = xi pi \

0 1 8

1 3 8

2 3 8

3 1 8

1ö 3 3ö æ 3ö æ 1ö æ æ mean, m = Sx i p i = ç 0 ´ ÷ + ç1 ´ ÷ + ç 2 ´ ÷ + ç 3 ´ ÷ = × 8ø 2 8 8 8 è ø è ø è ø è

Variance, s 2 = Sx i2 p i - m

2

1ö æ 3ö æ 3ö æ 1 ö 9ù 3 éæ = ê ç 0 ´ ÷ + ç1 ´ ÷ + ç 4 ´ ÷ + ç 9 ´ ÷ - ú = × 8ø è 8ø è 8ø è 8ø 4û 4 ëè Standard deviation, s = EXAMPLE 3

3 × 2

A die is tossed once. If the random variable X is defined as ì 1, if the die results in an even number X =í î 0, if the die results in an odd number then find the mean and variance of X.

SOLUTION

In tossing a die once, the sample space is given by S = {1, 2, 3, 4, 5, 6}. 3 1 \ P(getting an even number) = = , 6 2 3 1 P(getting an odd number) = = × 6 2

SSS Mathematics for Class 12 1310

1310

Senior Secondary School Mathematics for Class 12

As given, X takes the value 0 or 1. 1 × 2 1 P(X = 1) = P(getting an even number) = × 2 P(X = 0) = P(getting an odd number) =

Thus, the probability distribution of X is given by X = xi

0 1 2

pi \

1 1 2

1ö 1 1ö æ æ mean, m = Sx i p i = ç 0 ´ ÷ + ç1 ´ ÷ = × 2ø 2 2ø è è

Variance, s 2 = Sx i2 p i - m

2

1ö æ 1ö æ1ö æ = ç 0 ´ ÷ + ç1 ´ ÷ - ç ÷ 2ø è 2ø è 2ø è EXAMPLE 4

SOLUTION

2

æ1 1ö 1 =ç - ÷= × è 2 4ø 4

Find the mean, variance and standard deviation of the number of sixes in two tosses of a die. In a single toss, we have probability of getting a six =

1 , and 6

æ probability of getting a non-six = ç1 è

1ö 5 ÷= × 6ø 6

Let X denote the number of sixes in two tosses. Then, clearly X can assume the value 0, 1, or 2. P(X = 0) = P[(non-six in the 1st draw) and (non-six in the 2nd draw)] = P(non-six in the 1st draw) ´ P(non-six in the 2nd draw) æ 5 5 ö 25 × =ç ´ ÷= è 6 6 ø 36 P(X = 1) = P[(six in the 1st draw and non-six in the 2nd draw) or(non-six in the 1st draw and six in the 2nd draw)] = P(six in the 1st draw and non-six in the 2nd draw) + P(non-six in the 1st draw and six in the 2nd draw) 1 5 ö æ5 1ö æ 5 5 ö 10 5 æ =ç ´ ÷+ç ´ ÷=ç + = × ÷= è 6 6 ø è 6 6 ø è 36 36 ø 36 18 P(X = 2) = P[six in the 1st draw and six in the 2nd draw] = P(six in the 1st draw) ´ P(six in the 2nd draw)

SSS Mathematics for Class 12 1311

Probability Distribution

1311

1 æ1 1ö × =ç ´ ÷= è 6 6 ø 36 Hence, the probability distribution is given by X = xi pi \

0 25 36

1 5 18

2 1 36

25 ö æ 5ö æ 1 ö 6 1 æ mean, m = Sx i p i = ç 0 ´ ÷ + ç1 ´ ÷ + ç 2 ´ ÷ = = × 36 ø è 18 ø è 36 ø 18 3 è

Variance, s 2 = Sx i2 p i - m 2 1 ö 1ù 5 5ö æ 25 ö æ éæ × = ê ç 0 ´ ÷ + ç1 ´ ÷ + ç 4 ´ ÷ - ú = 36 ø 9 û 18 36 ø è 18 ø è ëè Standard deviation, s = EXAMPLE 5

5 1 5 = × × 18 3 2

Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the mean and variance of the number of kings. [CBSE 2005C]

SOLUTION

Let X be the random variable. Then, X = number of kings obtained in two draws. Clearly, X can assume the value 0, 1 or 2. 4 1 P(drawing a king) = = × 52 13 1 ö 12 æ × P(not drawing a king) = ç1 - ÷ = 13 ø 13 è P(X = 0) = P(not a king in the 1st draw and not a king in the 2nd draw) æ 12 12 ö 144 =ç ´ ÷= × è 13 13 ø 169 P(X = 1) = P(a king in the 1st draw and not a king in the 2nd draw) or P(not a king in the 1st draw and a king in the 2nd draw) æ 1 12 12 1 ö 24 × =ç ´ + ´ ÷= è 13 13 13 13 ø 169 P(X = 2) = P(a king in the 1st draw and a king in the 2nd draw) 1ö 1 æ 1 =ç ´ ÷= × 13 13 169 è ø

SSS Mathematics for Class 12 1312

1312

Senior Secondary School Mathematics for Class 12

Hence, the probability distribution is given by X = xi pi \

0 144 169

1 24 169

2 1 169

1 ö 24 ö æ 144 ö æ 2 æ mean, m = Sx i p i = ç 0 ´ × ÷= ÷ + ç2 ´ ÷ + ç1 ´ 169 ø 13 169 ø è 169 ø è è

Variance, s 2 = Sx i2 p i - m 2 1 ö 4 ù 24 24 ö æ 144 ö æ éæ = × = êç0 ´ ÷ + ç1 ´ ÷ + ç4 ´ ÷169 ø 169 úû 169 169 ø è 169 ø è ëè EXAMPLE 6

Two cards are drawn simultaneously (or successively without replacement) from a well-shuffled pack of 52 cards. Find the mean and variance of the number of aces. [CBSE 2001, ’12]

SOLUTION

Let X be the random variable. Then, X denotes the number of aces in a draw of 2 cards. \ X can assume the value 0, 1 or 2. Number of ways of drawing 2 cards out of 52 = C(52, 2). P(X = 0) = P(both non-aces, i.e., 2 non-aces out of 48) 48 æ 48 ´ 47 C 2 ö 188 ÷= ´ = 52 2 = çç × 52 ´ 51 ÷ø 221 C2 è 2 ´ 1 P(X = 1) = P[(one ace out of 4) and (one non-ace out of 48)] 4 ö C ´ 48C1 æ 4 ´ 48 32 = 152 = çç ´ 2÷÷ = × ´ 52 51 221 C2 è ø P(X = 2) = P(both aces) =

4

æ4´ 3 2 ´1 ö 1 ÷÷ = ´ × = çç ´ ´ 2 1 52 51 221 C2 è ø

C2

52

Thus, we have the following probability distribution: X = xi pi \

0 188 221

1 32 221

2 1 221

188 ö æ 32 ö æ 1 ö 2 æ mean, m = Sx i p i = ç 0 ´ × ÷ + ç1 ´ ÷ + ç2 ´ ÷= 221 ø è 221 ø è 221 ø 13 è

Variance, s 2 = Sx i2 p i - m 2 1 ö 4 32 ö æ 188 ö æ æ = ç0 ´ ÷÷ + ç4 ´ ÷ + ç1 ´ 221 169 221 221 ø ø è ø è è 4 ö 400 æ 36 =ç × ÷= è 221 169 ø 2873

SSS Mathematics for Class 12 1313

Probability Distribution

1313

EXAMPLE 7

Three defective bulbs are mixed with 7 good ones. Let X be the number of defective bulbs when 3 bulbs are drawn at random. Find the mean and variance of X.

SOLUTION

Let X denote the random variable showing the number of defective bulbs. Then, X can take the value 0, 1, 2 or 3. \ P(X = 0) = P(none of the bulbs is defective) = P(all the 3 bulbs are good ones) 7 æ7 ´ 6 ´5 C 3 ´ 2 ´1 ö 7 ÷= = 10 3 = çç ´ × C 3 è 3 ´ 2 ´ 1 10 ´ 9 ´ 8 ÷ø 24 P(X = 1) = P(1 defective and 2 non-defective bulbs) 3 C1 ´ 7C 2 æ 7 ´ 6 3 ´ 2 ´ 1 ö 21 ÷= ´ × = 10 = çç 3 ´ 2 ´ 1 10 ´ 9 ´ 8 ÷ø 40 C3 è P(X = 2) = P(2 defective and 1 good one) 3 C 2 ´ 7C1 æ 3 ´ 2 3 ´ 2 ´1 ö 7 ÷= × ´7 ´ = 10 = çç 10 ´ 9 ´ 8 ÷ø 40 C3 è 2 ´1 P(X = 3) = P(3 defective bulbs) 3 æ C 3 ´ 2 ´1 ö 1 ÷÷ = = 10 3 = çç1 ´ × 10 ´ 9 ´ 8 120 C3 è ø Thus, the probability distribution is given by X = xi pi

0 7 24

1 21 40

2 7 40

3 1 120

1 ö 9 7 ö æ 7 ö æ 21 ö æ æ \ mean, m = Sx i p i = ç 0 ´ ÷ + ç1 ´ ÷ + ç 2 ´ ÷ + ç 3 ´ ÷= × 120 ø 10 40 ø è 24 ø è 40 ø è è Variance, s 2 = Sx i2 p i - m 2 1 ö 81 7 ö æ 21 ö æ 7 ö æ æ = ç 0 ´ ÷ + ç1 ´ ÷ + ç 4 ´ ÷ + ç 9 ´ ÷120 40 40 24 è ø è ø è ø è ø 100 81 ö 49 æ 13 × =ç ÷= è 10 100 ø 100 EXAMPLE 8

An urn contains 4 white and 3 red balls. Let X be the number of red balls in a random draw of 3 balls. Find the mean and variance of X.

SOLUTION

When 3 balls are drawn at random, there may be no red ball, 1 red ball, 2 red balls or 3 red balls. Let X denote the random variable showing the number of red balls in a draw of 3 balls. Then, X can take the value 0, 1, 2 or 3.

SSS Mathematics for Class 12 1314

1314

Senior Secondary School Mathematics for Class 12

P(X = 0) = P(getting no red ball) = P(getting 3 white balls) 4 æ 4 ´ 3 ´ 2 3 ´ 2 ´ 1ö C 4 ÷= ´ × = 7 3 = çç C 3 è 3 ´ 2 ´ 1 7 ´ 6 ´ 5 ÷ø 35 P(X = 1) = P(getting 1 red and 2 white balls) 3 C1 ´ 4C 2 æ 3 ´ 4 ´ 3 3 ´ 2 ´ 1 ö 18 ÷= = = çç ´ × 7 2 7 ´ 6 ´ 5 ÷ø 35 C3 è P(X = 2) = P(getting 2 red and 1 white ball) 3 C 2 ´ 4C1 æ 3 ´ 2 3 ´ 2 ´ 1 ö 12 ÷= ´4´ × = = çç 7 7 ´ 6 ´ 5 ÷ø 35 C3 è 2 ´1 P(X = 3) = P(getting 3 red balls) =

3 7

C3

C3

=

1 ´ 3 ´ 2 ´1 1 = × 7 ´ 6 ´5 35

Thus, the probability distribution of X is given below. X = xi pi

0 4 35

1 18 35

2 12 35

3 1 35

4 ö æ 18 ö æ 12 ö æ 1 ö 9 æ \ mean, m = Sx i p i = ç 0 ´ ÷ + ç1 ´ ÷ + ç 2 ´ ÷ + ç 3 ´ ÷ = × 35 ø è 35 ø è 35 ø è 35 ø 7 è Variance, s 2 = Sx i2 p i - m 2 1 ö 81 ù 12 ö æ 18 ö æ 4 ö æ éæ = êç0 ´ ÷÷ + ç1 ´ ÷ + ç4 ´ ÷ + ç9 ´ 35 ø 49 úû 35 ø è 35 ø è 35 ø è ëè æ 15 81 ö 24 × =ç - ÷= 49 ø 49 è7 EXAMPLE 9

In a game, 3 coins are tossed. A person is paid ` 5 if he gets all heads or all tails; and he is supposed to pay ` 3 if he gets one head or two heads. What can he expect to win on an average per game?

SOLUTION

In tossing 3 coins, the sample space is given by S = {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT }. \ n( S) = 8. 2 1 P(getting all heads or all tails) = = × 8 4 6 3 P(getting one head or 2 heads) = = × 8 4 Let X = number of rupees the person gets. Then, possible values of X are 5 and -3. 1 3 P(X = 5) = and P(X = - 3) = × 4 4

SSS Mathematics for Class 12 1315

Probability Distribution

1315

Thus, we have X = xi

5 1 4

pi \

–3 3 4

the required expectations = mean, m = Sx i p i 3 æ 1ö = ç5 ´ ÷ + ( -3) ´ = -1, 4ø 4 è

i.e., he loses ` 1 per toss.

EXERCISE 31 1. Find the mean (m ), variance ( s 2 ) and standard deviation (s) for each of the following probability distributions: (i)

0 1 6

1 1 2

2 3 10

3 1 30

xi

1

2

3

4

pi

0.4

0.3

0.2

0.1

x p( x)

(ii)

(iii)

(iv)

[CBSE 2007]

xi

–3

–1

0

2

pi

0.2

0.4

0.3

0.1

xi

–2

–1

0

1

2

pi

0.1

0.2

0.4

0.2

0.1

2. Find the mean and variance of the number of heads when two coins are tossed simultaneously. 3. Find the mean and variance of the number of tails when three coins are tossed simultaneously. 4. A die is tossed twice. ‘Getting an odd number on a toss’ is considered a success. Find the probability distribution of number of successes. Also, find the mean and variance of the number of successes. 5. A die is tossed twice. ‘Getting a number greater than 4’ is considered a success. Find the probability distribution of number of successes. Also, find the mean and variance of the number of successes. 6. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes. Also, find the mean and variance of number of successes. [CBSE 2008]

SSS Mathematics for Class 12 1316

1316

Senior Secondary School Mathematics for Class 12

7. A coin is tossed 4 times. Let X denote the number of heads. Find the probability distribution of X. Also, find the mean and variance of X. [CBSE 2005]

8. Let X denote the number of times ‘a total of 9’ appears in two throws of a pair of dice. Find the probability distribution of X. Also, find the mean, variance and standard deviation of X. 9. There are 5 cards, numbered 1 to 5, one number on each card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Find the mean and variance of X. 10. Two cards are drawn from a well-shuffled pack of 52 cards. Find the probability distribution of number of kings. Also, compute the variance [CBSE 2007] for the number of kings. 11. A box contains 16 bulbs, out of which 4 bulbs are defective. Three bulbs are drawn at random from the box. Let X be the number of defective bulbs drawn. Find the mean and variance of X. 12. 20% of the bulbs produced by a machine are defective. Find the probability distribution of the number of defective bulbs in a sample of 4 [CBSE 2004C] bulbs chosen at random. 13. Four bad eggs are mixed with 10 good ones. Three eggs are drawn one by one without replacement. Let X be the number of bad eggs drawn. Find the mean and variance of X. 14. Four rotten oranges are accidentally mixed with 16 good ones. Three oranges are drawn at random from the mixed lot. Let X be the number of rotten oranges drawn. Find the mean and variance of X. 15. Three balls are drawn simultaneously from a bag containing 5 white and 4 red balls. Let X be the number of red balls drawn. Find the mean and variance of X. 16. Two cards are drawn without replacement from a well-shuffled deck of 52 cards. Let X be the number of face cards drawn. Find the mean and variance of X. 17. Two cards are drawn one by one with replacement from a well-shuffled deck of 52 cards. Find the mean and variance of the number of aces. 18. Three cards are drawn successively with replacement from a well-shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Find the mean and variance of X. 19. Five defective bulbs are accidently mixed with 20 good ones. It is not possible to just look at a bulb and tell whether or not it is defective. Find [CBSE 2007] the probability distribution from this lot.

ANSWERS (EXERCISE 31)

1.

(i) Mean = 1.2, variance = 0.56, SD =0.74

SSS Mathematics for Class 12 1317

Probability Distribution

(ii) Mean = 2, variance = 1, SD = 1 (iii) Mean = - 0.8, variance = 2.6, SD = 1.612 (iv) Mean = 0, variance = 1.2, SD = 1.095 2. Mean = 1, variance = 0.5 4.

X = xi

0

1

2

pi

1 4

1 2

1 4

1317

3. Mean =1.5 variance = 0.75

Mean = 1, variance = 0.5 5.

X = xi

0

1

2

pi

4 9

4 9

1 9

Mean = 6.

X = xi

0

1

2

3

4

pi

625 1296

125 324

25 216

5 324

1 1296

Mean = 7.

2 4 , variance = 3 9

2 5 , variance = 3 9

X = xi

0

1

2

3

4

pi

1 16

1 4

3 8

1 4

1 16

Mean = 2, variance = 1 8.

X = xi

0

1

2

pi

64 81

16 81

1 81

Mean = 9.

2 16 4 , variance = , SD = 9 81 9

X = xi

3

4

5

6

7

8

9

pi

1 10

1 10

1 5

1 5

1 5

1 10

1 10

Mean = 6, variance = 3

SSS Mathematics for Class 12 1318

1318

10.

Senior Secondary School Mathematics for Class 12

X = xi

0

1

2

pi

188 221

32 221

1 221

Variance = 11. 14. 16. 18. 19.

400 2873

3 39 , variance = 4 80 3 68 Mean = , variance = 5 125 20 1000 Mean = , variance = 13 2873 3 9 Mean = , variance = 4 16 Mean =

13. 15. 17.

X

0

1

2

3

4

P(X)

969 2530

114 253

38 253

4 253

1 2530

6 30 , variance = 7 49 4 5 Mean = , variance = 3 9 2 24 Mean = , variance = 13 169

Mean =

HINTS TO SOME SELECTED QUESTIONS (EXERCISE 31) 2. S = {HH , HT , TH , TT }. Let X be the number of heads. Then, X = 0, 1 or 2. 1 P(X = 0 ) = P (getting no head) = × 4 2 1 P(X = 1) = P (getting 1 head) = = × 4 2 1 P(X = 2 ) = P (getting 2 heads) = × 4

3.

X = xi

0

1

2

pi

1 4

1 2

1 4

X = xi

0

1

2

3

pi

1 8

3 8

3 8

1 8

4. In a single toss, P(success) = P(getting an odd number) =

3 1 = , and 6 2

1ö 1 æ P(non-success) = ç 1 - ÷ = × 2ø 2 è Let X be the number of successes. Then, X = 0, 1 or 2. P(X = 0 ) = P [(non-success in the 1st toss and non-success in the 2nd)]

SSS Mathematics for Class 12 1319

Probability Distribution

1319

æ 1 1ö 1 =ç ´ ÷= × è2 2ø 4 P(X = 1) = P [(success in the 1st toss and non-success in the 2nd) or (non-success in 1st toss and success in the 2nd)] æ 1 1ö æ 1 1ö 1 =ç ´ ÷+ ç ´ ÷= × è2 2ø è2 2ø 2 P(X = 2 ) = P [(success in the 1st toss and success in the 2nd)] æ 1 1ö 1 =ç ´ ÷= × è2 2ø 4 X = xi

0

1

2

pi

1 4

1 2

1 4

5. In a single toss, P(success) =

2 1 1ö 2 æ = and P(non-success) = ç 1 - ÷ = × 6 3 3ø 3 è

P(X = 0 ) = P (non-success in the 1st draw and non-success in the second) æ2 2ö 4 =ç ´ ÷= × è 3 3ø 9 P(X = 1) = P (success in the 1st toss and non-success in the 2nd) or ( non-success in the 1st toss and success in the 2nd)] æ 1 2ö æ 2 1ö 4 =ç ´ ÷+ ç ´ ÷= × è 3 3ø è 3 3ø 9 P(X = 2 ) = P [(success in the 1st toss and success in the 2nd)] æ 1 1ö 1 =ç ´ ÷= × è 3 3ø 9 X = xi

0

1

2

pi

4 9

4 9

1 9

6. In a single throw, P(doublet) =

6 1 1ö 5 æ = , and P(non-doublet) = ç 1 - ÷ = × 36 6 6ø 6 è

Let X be the number of doublets. Then, X = 0 , 1, 2 or 3. P(X = 0 ) = P (non-doublet in each case) æ 5 5 5 5 ö 625 = P( D1 D2 D3 D4 ) = ç ´ ´ ´ ÷ = × è 6 6 6 6 ø 1296 P(X = 1) = P (one doublet) = P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) æ 1 5 5 5 ö æ5 1 5 5 ö æ5 5 1 5 ö æ5 5 5 1ö =ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷ è6 6 6 6ø è6 6 6 6ø è6 6 6 6ø è6 6 6 6ø

SSS Mathematics for Class 12 1320

1320

Senior Secondary School Mathematics for Class 12 125 ö 125 æ = ç4 ´ × ÷= 1296 ø 324 è

P(X = 2 ) = P (two doublets) = P( D1 D2 D3 D4 )

or P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 )

or P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) æ 1 1 5 5 ö æ 1 5 1 5 ö æ 1 5 5 1ö =ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷ è6 6 6 6ø è6 6 6 6ø è6 6 6 6ø æ5 1 1 5 ö æ5 1 5 1ö æ5 5 1 1ö + ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷ è6 6 6 6ø è6 6 6 6ø è6 6 6 6ø 25 ö 25 æ = ç6 ´ × ÷= 1296 ø 216 è P(X = 3 ) = P (three doublets) = P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) or P( D1 D2 D3 D4 ) æ 1 1 1 5 ö æ 1 1 5 1ö æ 1 5 1 1ö æ5 1 1 1ö =ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷+ ç ´ ´ ´ ÷ è6 6 6 6ø è6 6 6 6ø è6 6 6 6ø è6 6 6 6ø 5 ö 5 æ = ç4 ´ × ÷= 1296 ø 324 è P(X = 4 ) = P (four doublets) = P( D1 D2 D3 D4 ) 1 æ 1 1 1 1ö =ç ´ ´ ´ ÷= × è 6 6 6 6 ø 1296 Thus, we have X = xi

0

1

2

3

4

pi

625 1296

125 324

25 216

5 324

1 1296

Now, find m and s 2 . 7. In a single throw of a coin, P( H ) =

1 1 and P( H ) = P( T ) = × 2 2

Let X show the number of heads. Then, X = 0 , 1, 2 , 3 or 4. æ 1 1 1 1ö 1 P(X = 0 ) = P (no head) = P( H1 H 2 H 3 H 4 ) = ç ´ ´ ´ ÷ = × è 2 2 2 2 ø 16 P(X = 1) = P (one head) = P( H1 H 2 H 3 H 4 ) or ( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) 1 ö 1 æ 1 1 1 1ö æ = 4 ç ´ ´ ´ ÷ = ç4 ´ ÷ = × 16 ø 4 è2 2 2 2ø è P(X = 2 ) = P (two heads) = P( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 )

SSS Mathematics for Class 12 1321

Probability Distribution

1321

1 ö 3 æ = ç6 ´ ÷ = × 16 ø 8 è P(X = 3 ) = P (three heads) = P( H1 H 2 H 3 H 4 ) or ( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) or P( H1 H 2 H 3 H 4 ) 1 ö 1 æ 1 1 1 1ö æ = 4 ´ç ´ ´ ´ ÷ = ç4 ´ ÷ = × 16 ø 4 è2 2 2 2ø è æ 1 1 1 1ö 1 × P(X = 4 ) = P (four heads) = P( H1 H 2 H 3 H 4 ) = ç ´ ´ ´ ÷ = è 2 2 2 2 ø 16 Thus, we have X = xi

0

1

2

3

4

pi

1 16

1 4

3 8

1 4

1 16

8. Let E = {( 3 , 6 ), ( 6 , 3 ), ( 4 , 5 ), (5 , 4 )}. So, n( E) = 4. \

P( E) =

4 1 1ö 8 æ = , and P( E ) = ç 1 - ÷ = × 36 9 9 è ø 9

Let X be the number of times ‘a total of 9’ appears in 2 throws. Then, X = 0, 1 or 2. æ 8 8 ö 64 P(X = 0 ) = P( E1 E2 ) = ç ´ ÷ = × è 9 9 ø 81 P(X = 1) = P[( E1 E2 ) or ( E1 E2 )] = P( E1 E2 ) + P( E1 E2 ) æ 1 8 ö æ 8 1 ö 16 =ç ´ ÷+ ç ´ ÷= × è 9 9 ø è 9 9 ø 81 æ 1 1ö 1 P(X = 2 ) = P( E1 E2 ) = P( E1 ) ´ P( E2 ) = ç ´ ÷ = × è 9 9 ø 81 X = xi

0

1

2

pi

64 81

16 81

1 81

9. S = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)} Then X = 3, 4, 5, 6, 7, 8 and 9. 2 1 2 1 4 1 P(X = 3 ) = = ; P(X = 4 ) = = ; P(X = 5 ) = = ; 20 10 20 10 20 5 P(X = 6 ) =

4 1 4 1 2 1 = ; P(X = 7 ) = = ; P(X = 8 ) = = ; 20 5 20 5 20 10

P(X = 9 ) =

2 1 = × 20 10

SSS Mathematics for Class 12 1322

1322

Senior Secondary School Mathematics for Class 12

Thus, we have X = xi

3

4

5

6

7

8

9

pi

1 10

1 10

1 5

1 5

1 5

1 10

1 10

10. Let X be the number of kings. Then, X = 0, 1 or 2. P(X = 0 ) = P (none is a king) = P(both are non-kings) =

48

C2

52

C2

æ 48 ´ 47 2 ö 188 ÷= = çç ´ × 2 52 ´ 51 ÷ø 221 è

P(X = 1) = P (one king and one non-king) =

4

C 1 ´ 48C 1 52

C2

æ 2 ö 32 ÷= = çç 4 ´ 48 ´ × 52 ´ 51 ÷ø 221 è

P(X = 2 ) = P (both are kings) =

4

æ4´3 2´1 ö 1 ÷÷ = = çç ´ × è 2 ´ 1 52 ´ 51 ø 221

C2

52

C2

Thus, we have X = xi

0

1

2

pi

188 221

32 221

1 221

11. Three bulbs drawn one by one without replacement is the same as drawing 3 bulbs simultaneously. Let X = number of defective bulbs in a lot of 3 bulbs drawn. Then, X = 0 , 1, 2 or 3. P(X = 0 ) = P (none of the bulbs is defective) =

12

C3

16

C3

æ 12 ´ 11 ´ 10 3 ´ 2 ´ 1 ö 11 ÷= = çç × ´ 16 ´ 15 ´ 14 ÷ø 28 è 3´2´1

P(X = 1) = P (1 defective bulb and 2 nondefective bulbs) =

4

C 1 ´ 12C 2 16

C3

æ 4 ´ 12 ´ 11 3 ´ 2 ´ 1 ö 33 ÷= = çç ´ × 2 1 16 ´ ´ 15 ´ 14 ÷ø 70 è

P(X = 2 ) = P (2 defective bulbs and 1 nondefective bulb) =

( 4 C 2 ´ 12C 1 ) 16

C3

æ4´3 3´2´1 ö 9 ÷= = çç × ´ 12 ´ 2 1 ´ 16 ´ 15 ´ 14 ÷ø 70 è

P(X = 3 ) = P (3 defective bulbs) =

4

C3

16

C3

æ4´ 3´2 3´2´1 ö 1 ÷÷ = = çç ´ × 3 2 1 16 15 14 140 ´ ´ ´ ´ è ø

SSS Mathematics for Class 12 1323

Probability Distribution

1323

X = xi

0

1

2

3

pi

11 28

33 70

9 70

1 140

12. Let there be 100 bulbs in all and let X be the number of defective bulbs. Then, P(X = 0 ) = P( none is defective) =

80

C4

× C4 P(X = 1) = P (1 defective and 3 nondefective) = P(X = 2 ) = 15. P(X = 0 ) = P(X = 2 ) =

( 20C 1 ´ 80C 3 ) 100

C4

×

( 20C 2 ´ 80C 2 ) 100

C4

5 9

C3 C3

; P(X = 3 ) =

( 20C 3 ´ 80C 1 ) 100

C4

; P(X = 4 ) =

( 4 C 1 ´ 5C 2 ) 10 5 , P(X = 1) = = , 9 42 21 C3

=

4

( C 2 ´ 5C 1 ) 9

100

C3

=

4 1 5 C , P(X = 3 ) = 9 3 = × 14 C 3 21

16. There are 12 face cards (4 kings, 4 queens and 4 jacks). Clearly, X = 0 or 1 or 2. P(X = 0 ) = P (no face card) = P(drawing 2 cards out of 40 non-face cards) 40

æ 40 ´ 39 2 ´ 1 ö 10 ÷= = çç ´ × 52 ´ 51 ÷ø 17 è 2´1

C2

=

52

C2

P(X = 1) = P (1 face card and 1 non-face card) =

( 12 C 1 ´ 40 C 1 ) 52

C2

æ 2 ´ 1 ö 80 ÷= = çç 12 ´ 40 ´ × 52 ´ 51 ÷ø 221 è

P(X = 2 ) = P (2 face cards) =

40

C2

52

C2

æ 40 ´ 39 2 ´ 1 ö 10 ÷= = çç ´ × 52 ´ 51 ÷ø 17 è 2´1

Thus, we have X = xi

0

1

2

pi

10 17

80 221

10 17

Now, find the mean and variance. 17. See Example 5. 18. There are 13 hearts and 39 other cards. Let E = event of drawing a heart. 13 1 1ö 3 æ and P( E ) = ç 1 - ÷ = × Then, P( E) = = 52 4 4ø 4 è Let X = number of hearts in a draw.

20

C4

100

C4

×

SSS Mathematics for Class 12 1324

1324

Senior Secondary School Mathematics for Class 12

Then, X = 0 , 1, 2 or 3. æ 3 3 3 ö 27 P(X = 0 ) = P( E E E ) = P( E ) ´ P( E ) ´ P( E ) = ç ´ ´ ÷ = × è 4 4 4 ø 64 P(X = 1) = P[( E E E ) or ( E E E ) or ( E E E)] = P( E E E ) + P( E E E ) + P( E E E) æ 1 3 3 ö æ 3 1 3 ö æ 3 3 1 ö 27 =ç ´ ´ ÷+ ç ´ ´ ÷+ ç ´ ´ ÷= × è 4 4 4 ø è 4 4 4 ø è 4 4 4 ø 64 P(X = 2 ) = P[( E E E ) or ( E E E) or ( E E E)] = P( E E E ) + P( E E E) + P( E E E) æ 1 1 3ö æ 1 3 1ö æ 3 1 1ö 9 =ç ´ ´ ÷+ ç ´ ´ ÷+ ç ´ ´ ÷= × è 4 4 4 ø è 4 4 4 ø è 4 4 4 ø 64 æ 1 1 1ö 1 P(X = 3 ) = P( E E E) = P( E) ´ P( E) ´ P( E) = ç ´ ´ ÷ = × è 4 4 4 ø 64 X = xi

0

1

2

3

pi

27 64

27 64

9 64

1 64

Find the mean and variance.

SSS Mathematics for Class 12 1325

32. BINOMIAL DISTRIBUTION SUCCESS AND FAILURE IN AN EXPERIMENT There are certain kinds of experiments which have two possible outcomes. One of these two outcomes is called a success, while the other is called a failure.

For example, in tossing a coin, we get either a head or a tail. If getting head is taken as a success then getting a tail is a failure. BERNOULLI’S THEOREM Let there be n independent trials in an experiment and let the random variable X denote the number of successes in these trials. Let the probability of getting a success in a single trial be p and that of getting a failure be q so that p + q = 1. Then, P(X = r) = nC r × p r × q (n - r ). PROOF

Let us denote a success by S and a failure by F. Number of ways of getting r successes in n trials = nC r . ì ï \ P(X = r) = nC r × í ïî

SSS ... S

ü ï ý (n - r) times ïþ FFF ... F

and r times

n

= C r × {P( S) × P( S) ¼ r times} ´ {P( F) × P( F) ¼ (n - r) times} = nC r × ( p × p × p ¼ r times) ´ [q × q × q ¼ (n - r) times] = nC r × p r × q (n - r ). Hence, P(X = r) = nC r × p r × q (n - r ). REMARK

We have P(X = 0) = q n ; P(X = 1) = npq (n - 1 ) ; P(X = 2) = nC 2 × p 2 × q (n - 2 ) , etc. The probability distribution of X may be expressed as 0 1 ... r ö æX: çç n (n -1 ) r (n - r ) ÷ ÷. ... C r × p × q ø è P(X): q npq This distribution is called a binomial distribution.

Conditions for the Applicability of a Binomial Distribution (i) The experiment is performed for a finite and fixed number of trials. (ii) Each trial must give either a success or a failure. (iii) The probability of a success in each trial is the same. 1325

SSS Mathematics for Class 12 1326

1326

Senior Secondary School Mathematics for Class 12 SOLVED EXAMPLES

EXAMPLE 1

SOLUTION

A coin is tossed 4 times. If X is the number of heads observed, find the probability distribution of X. When a coin is tossed, we have S = {H , T}. 1 P(getting a head) = , and 2 æ 1ö 1 P(not getting a head) = ç1 - ÷ = × 2ø 2 è Let X be the random variable denoting the number of heads. In 4 trials, we may get 0 or 1 or 2 or 3 or 4 heads. So, X may assume the values 0, 1, 2, 3, 4. 0

æ1ö æ1ö P(X = 0) = 4C 0 × ç ÷ × ç ÷ è 2ø è 2ø 1

æ1ö æ1ö P(X = 1) = 4C1 × ç ÷ × ç ÷ è 2ø è 2ø

(4 - 0)

=

1 × 16

=

1 × 4

(4 - 1 )

2

æ1ö æ1ö P(X = 2) = 4C 2 × ç ÷ × ç ÷ è 2ø è 2ø 3

(4 - 2 )

æ1ö æ1ö P(X = 3) = 4C 3 × ç ÷ × ç ÷ è 2ø è 2ø 4

æ1ö æ1ö P(X = 4) = 4C 4 × ç ÷ × ç ÷ è 2ø è 2ø

=

3 × 8

=

1 × 4

=

1 × 16

(4 - 3 )

(4 - 4 )

Hence, the required probability distribution is given by 0 1 2 3 4 ö æX : ÷ ç 1 1 3 1 1 ç P(X) : ÷× ç ÷ è 16 4 8 4 16 ø EXAMPLE 2

SOLUTION

Find the probability distribution of the number of sixes in three tosses of a die. When a die is tossed, we have S = {1, 2, 3 , 4, 5 , 6}. 1 æ 1ö 5 \ P(getting a six) = and P(not getting a six) = ç1 - ÷ = × 6ø 6 6 è Let X be the random variable denoting the number of sixes. In 3 trials, the number of sixes may be 0 or 1 or 2 or 3. So, X may assume the values 0, 1, 2, 3. 0

æ 1ö æ5 ö P(X = 0) = 3C 0 × ç ÷ × ç ÷ è 6ø è 6ø 1

æ 1ö æ5 ö P(X = 1) = 3C1 × ç ÷ × ç ÷ è 6ø è 6ø

(3 - 0)

=

125 × 216

=

25 × 72

(3 - 1 )

SSS Mathematics for Class 12 1327

Binomial Distribution

1327

(3 - 2 )

2

æ 1ö æ5 ö P(X = 2) = 3C 2 × ç ÷ × ç ÷ è 6ø è 6ø

=

3

æ 1ö æ5 ö P(X = 3) = 3C 3 × ç ÷ × ç ÷ è 6ø è 6ø

5 × 72

(3 - 3 )

=

1 × 216

The required probability distribution of X is given below: 0 1 2 3 ö æ X: ç ÷ 125 25 5 1 ç P(X) : ÷× ç ÷ è 216 72 72 216 ø EXAMPLE 3

Find the probability distribution of the number of doublets in four throws of a pair of dice.

SOLUTION

When a pair of dice is thrown, there are 36 possible outcomes. \ n( S) = 36. All possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). 6 1 P(getting a doublet) = = , and 36 6 1ö 5 æ P(not getting a doublet) = ç1 - ÷ = × 6ø 6 è Let X denote the number of doublets. In 4 throws, we can have 0 or 1 or 2 or 3 or 4 doublets. 0

4

625 æ 1ö æ5 ö P(X = 0) = 4C 0 × ç ÷ × ç ÷ = × 1296 è 6ø è 6ø 1

3

125 æ 1ö æ5 ö P(X = 1) = 4C1 × ç ÷ × ç ÷ = × 6 6 324 è ø è ø 2

æ 1ö æ5 ö P(X = 2) = 4C 2 × ç ÷ × ç ÷ è 6ø è 6ø

2

=

3

1

4

0

25 × 216

5 æ 1ö æ5 ö P(X = 3) = 4C 3 × ç ÷ × ç ÷ = × 324 è 6ø è 6ø 1 æ 1ö æ5 ö P(X = 4) = 4C 4 × ç ÷ × ç ÷ = × 1296 è 6ø è 6ø The required probability distribution is given below: 0 1 2 3 4 ö æ ç X: ÷ 625 125 25 5 1 ç P(X) : ÷× ç ÷ è 1296 324 216 324 1296 ø EXAMPLE 4

An unbiased coin is tossed 6 times. Find, using binomial distribution, the [CBSE 2000] probability of getting at least 5 heads.

SOLUTION

In a single throw of a coin, we have S = {H , T}. 1 æ 1ö 1 P(getting a head) = , and P(not getting a head) = ç1 - ÷ = × 2ø 2 2 è

SSS Mathematics for Class 12 1328

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Senior Secondary School Mathematics for Class 12

\

p=

1 1 and q = × 2 2 r

æ1ö æ1ö P(X = r) = nC r × p r × q (n - r ) = 6C r × ç ÷ × ç ÷ è 2ø è 2ø \

(6 - r )

æ1ö = 6C r × ç ÷ è 2ø

6

P(getting at least 5 heads) = P(X ³ 5) = P(X = 5) + P(X = 6) 6

6

1ö 7 æ1ö æ1ö æ 3 = 6 C5 × ç ÷ + 6 C 6 × ç ÷ = ç + ÷= × è 2ø è 2ø è 32 64 ø 64 7 Hence, the required probability is × 64 EXAMPLE 5

An unbiased coin is tossed 8 times. Find, by using binomial distribution, [CBSE 2000] the probability of getting at least 3 heads.

SOLUTION

In a single throw of a coin, we have S = {H , T}. 1ö 1 1 æ P(getting a head) = , and P(not getting a head) = ç1 - ÷ = × 2ø 2 2 è 1 1 \ p = and q = × 2 2 r

æ1ö æ1ö P(X = r) = nC r × p r × q (n - r ) = 8C r × ç ÷ × ç ÷ è 2ø è 2ø

(8 - r )

æ1ö = 8C r ç ÷ è 2ø

8

\

P(getting at least 3 heads) = P(X ³ 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] 8 8 8 é æ1ö ù æ1ö æ1ö = 1 - ê 8 C 0 × ç ÷ + 8C1 × ç ÷ + 8C 2 × ç ÷ ú è 2 ø úû è 2ø è 2ø êë 37 ö 219 1 æ × =1× (1 + 8 + 28) = ç1 ÷= 256 256 ø 256 è 219 Hence, the required probability is × 256

EXAMPLE 6

Six coins are tossed simultaneously. Find the probability of getting (i) 3 heads (ii) no head (iii) at least one head [CBSE 2003] (iv) not more than 3 heads.

SOLUTION

The experiment may be taken as throwing a single coin 6 times. In a single throw of a coin, we have S = {H , T }. 1 æ 1ö 1 P(getting a head) = , and P(not getting a head) = ç1 - ÷ = × 2 2ø 2 è Let X be the random variable showing the number of heads.

SSS Mathematics for Class 12 1329

Binomial Distribution

1329 r

æ1ö æ1ö P(X = r) = nC r × p r × q (n - r ) = 6C r × ç ÷ × ç ÷ è 2ø è 2ø

6- r

æ1ö = 6C r × ç ÷ è 2ø

6

6

5 æ1ö (i) P(getting 3 heads) = P(X = 3) = 6C 3 × ç ÷ = × 2 16 è ø 6

1 æ1ö (ii) P(getting no head) = P(X = 0) = 6C 0 × ç ÷ = × 64 è 2ø (iii) P(getting at least 1 head) 6 é æ1ö ù = 1 - P(X = 0) = 1 - ê 6 C 0 × ç ÷ ú è 2 ø úû êë 1 ö 63 æ = ç1 - ÷ = × 64 ø 64 è (iv) P(getting not more than 3 heads) = P(no head or 1 head or 2 heads or 3 heads) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) 6

6

6

æ1ö æ1ö æ1ö = 6C 0 × ç ÷ + 6C1 × ç ÷ + 6C 2 × ç ÷ + 6C 3 è 2ø è 2ø è 2ø 6 æ1ö æ 1 ö 21 = ç ÷ × (1 + 6 + 15 + 20) = ç ´ 42÷ = × è 2ø è 64 ø 32

æ1ö ×ç ÷ è 2ø

6

EXAMPLE 7

A die is thrown 5 times. If getting an odd number is a success, find the probability of getting at least 4 successes.

SOLUTION

When a die is thrown, we have S = {1, 2, 3 , 4, 5 , 6}. 3 1 \ P(getting an odd number) = = × 6 2 1 1ö 1 æ \ P(a success) = , and P(not a success) = ç1 - ÷ = × 2ø 2 2 è Let X be the random variable showing the number of successes. r

æ1ö æ1ö P(X = r) = nC r × p r × q (n - r ) = 5C r × ç ÷ × ç ÷ è 2ø è 2ø

(5 - r )

5

æ1ö = 5C r × ç ÷ . è 2ø

P(at least 4 successes) = P(4 successes or 5 successes) = P(X = 4) + P(X = 5) 5

5

1 ö 6 3 æ1ö æ1ö æ 5 = 5C 4 × ç ÷ + 5C5 × ç ÷ = ç + = × ÷= 2 2 32 32 32 16 è ø è ø è ø EXAMPLE 8

In 4 throws with a pair of dice, what is the probability of throwing doublets at least twice?

SOLUTION

In a single throw of a pair of dice, the number of all possible outcomes is 36. All doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

SSS Mathematics for Class 12 1330

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Senior Secondary School Mathematics for Class 12

6 1 = , and 36 6 1ö 5 æ P(not getting a doublet) = ç1 - ÷ = × 6ø 6 è

P(getting a doublet) =

Let X be the random variable denoting the number of doublets. r

æ 1ö æ5 ö Then, P(X = r) = nC r × p r × q (n - r ) = 4C r × ç ÷ × ç ÷ è 6ø è 6ø

(4 - r )

.

P(at least 2 doublets) = P(X = 2) + P(X = 3) + P(X = 4) 2

æ 1ö æ5 ö = 4C 2 × ç ÷ × ç ÷ è 6ø è 6ø 2

(4 - 2 )

3

æ 1ö æ5 ö + 4C 3 × ç ÷ × ç ÷ è 6ø è 6ø

2

3

1

(4 - 3 )

4

æ 1ö æ5 ö + 4C 4 × ç ÷ × ç ÷ è 6ø è 6ø

4

æ 1ö æ5 ö æ 1ö æ5 ö æ 1ö æ5 ö = 6× ç ÷ × ç ÷ + 4× ç ÷ × ç ÷ + ç ÷ × ç ÷ è 6ø è 6ø è 6ø è 6ø è 6ø è 6ø 25 5 1 171 ö æ × =ç + + ÷= è 216 324 1296 ø 1296 EXAMPLE 9

SOLUTION

(4 - 4 )

0

The bulbs produced in a factory are supposed to contain 5% defective bulbs. What is the probability that a sample of 10 bulbs will contain not more than 2 defective bulbs? 5 1 = , and 100 20 1 ö 19 æ × P(getting a nondefective bulb) = ç1 - ÷ = 20 ø 20 è 1 19 Then, p = and q = × 20 20 Let X denote the number of defective bulbs. P(getting a defective bulb) =

r

æ 1 ö æ 19 ö P(X = r) = nC r × p r × q (n - r ) = 10C r × ç ÷ × ç ÷ è 20 ø è 20 ø

(10 - r)

P(getting not more than 2 defective bulbs) = P(X = 0 or X = 1 or X = 2) = P(X = 0) + P(X = 1) + P(X = 2) 0

10

æ 1 ö æ 19 ö = 10C 0 × ç ÷ × ç ÷ + è 20 ø è 20 ø æ 19 ö =ç ÷ è 20 ø

10

1

10

9

+

9

æ 1 ö æ 19 ö C1 × ç ÷ × ç ÷ + è 20 ø è 20 ø

1 æ 19 ö 9 æ 19 ö ×ç ÷ + ×ç ÷ 2 è 20 ø 80 è 20 ø

8

8

2

10

æ 1 ö æ 19 ö C2 × ç ÷ × ç ÷ è 20 ø è 20 ø

æ 19 ö æ 149 ö =ç ÷ ×ç ÷× è 20 ø è 100 ø

8

æ 19 ö æ 149 ö Let A = ç ÷ × ç ÷ . Then, è 20 ø è 100 ø log A = 8(log 19 - log 20) + log 149 - log 100

8

SSS Mathematics for Class 12 1331

Binomial Distribution

1331

= 8(1.2788 - 1. 3010) + 2.1732 - 2 = - 0.0044 = 1.9956. \ A = antilog ( 1.9956) = 0.99. Hence, the required probability =

99 . 100

EXAMPLE 10

If on an average, out of 10 ships, one gets drowned then what is the probability that out of 5 ships at least 4 reach the shore safely?

SOLUTION

Probability of a ship to reach the shore safely =

9 . 10

9ö 1 æ Probability that a ship gets drowned = ç1 - ÷ = × 10 ø 10 è Let X be the random variable showing the number of ships reaching the shore safely. \ P(at least 4 reaching safely) = P(4 reaching safely or 5 reaching safely) = P(4 reaching safely) + P(5 reaching safely) = P(X = 4) + P(X = 5) 4

æ 9ö æ1ö = 5C 4 × ç ÷ × ç ÷ è 10 ø è 10 ø 4

=

(5 - 4 )

5

æ 9ö æ1ö + 5C5 × ç ÷ × ç ÷ è 10 ø è 10 ø

5

0

4

4

1 æ 9ö 9ö 7 æ 9ö æ 9ö æ 9 ö æ1 ×ç ÷ + ç ÷ =ç ÷ ç + ÷ = ×ç ÷ × 2 è 10 ø 10 10 2 10 è ø è ø è ø 5 è 10 ø 4

Let A = Þ

7 æ 9ö 7 ´ ( 9) 4 ×ç ÷ = 5 è 10 ø 5 ´ (10) 4

log A = log 7 + 4 ´ log 9 - log 5 - 4 ´ log 10 = ( 0.8451 + 4 ´ 0.9542 - 0.6990 - 4) = - 0.0371 = 1.9629

Þ

A = antilog ( 1.9629) = 0.9181.

Hence, the required probability is 0.9181.

Mean and Variance of a Binomial Distribution If a random variable X assumes the values x1 , x 2 , ¼ , x n with probabilities p1 , p 2 , ¼ , p n respectively then the mean of X is defined by

MEAN

m =

n

S

i =1

xi pi.

SSS Mathematics for Class 12 1332

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Senior Secondary School Mathematics for Class 12

To Find the Mean of a Binomial Distribution For the binomial distribution P(X = r) = P(r) = nC r × p r × q (n - r ) , where r = 0, 1, 2, ¼ , n the mean, m , is given by n

n

m = S r × p(r) = S r × nC r × p r × q (n - r ) r=0

r=0

n

= C1 × p × q

n -1

+ 2 × nC 2 × p 2 × q (n - 2 ) + ¼ + n × nC n × p nq 0

= 1 × np × q n - 1 + n(n - 1) × p 2 × q (n - 2 ) + ¼ + np n = np × [(n - 1 )C 0 × p 0 × q (n - 1 ) +

(n - 1 )

C1 × p1 × q (n - 2 ) + ... +

= (np) × ( q + p) n - 1 = (np)

(n - 1 )

C (n - 1 ) × p n - 1 × q 0]

[Q q + p = 1].

Hence, the mean is given by m = np. The variance = s 2 is given by n

s 2 = S r 2 × p(r) - (mean) 2 r=0 n

= S r 2 × nC r × p r q (n - r ) - (np) 2

[Q mean = np]

r=0 n

= S {r + r(r - 1)} × nC r p r q (n - r ) - (np) 2 r=0 n

n

= S r × nC r p r q (n - r ) + S r(r - 1) × nC r × p r × q (n - r ) - (np) 2 r=0

r=0

n

= np + S r(r - 1) × r=2

n(n - 1) × r(r - 1)

(n - 2 )

C (r - 2 ) × p 2 × p (r - 2 ) × q (n - r ) - (np) 2

n é ù n r (n - r ) = mean = np ú ê Q r S= 0 r × C r × p × q ë û

æ n = np + n(n - 1) × p 2 × ç S èr = 0

ö C (r - 2 ) × p (r - 2 ) × q (n - r ) ÷ - n 2 p 2 ø

(n - 2 )

= np + n(n - 1) p 2 ( q + p) (n - 2 ) - n 2 p 2 = np + n(n - 1) p 2 - n 2 p 2 = np - np 2 = np(1 - p) = npq. Hence, variance = npq. \ standard deviation = npq.

[Q q + p = 1]

SSS Mathematics for Class 12 1333

Binomial Distribution

1333

Recurrence Relation for a Binomial Distribution We have P(r) = nC r × p r × q (n - r ) and P(r + 1) = nC (r + 1 ) × p (r + 1 ) × q n - r - 1 . \

n C (r + 1 ) × p (r + 1 ) × q n - r - 1 P(r + 1) = n P(r) C r × p r × q (n - r )

= \

(n !) (r) ! × (n - r) ! p (n - r) p × × = × (r + 1) ! × (n - r - 1) ! (n) ! q (r + 1) q

P(r + 1) =

EXAMPLE 11

SOLUTION

(n - r) p × × P(r) × (r + 1) q

If X follows a binomial distribution with mean 3 and variance (3/2), find [CBSE 2001C] (i) P(X ³ 1) (ii) P(X £ 5). We know that mean = np and variance = npq. 3 3 1 np = 3 and npq = Þ 3q = Þ q= × 2 2 2 1 1 ö æ \ p = (1 - q) = ç1 - ÷ = × 2ø 2 è 1 1 Now, np = 3 and p = Þ n ´ = 3 Þ n = 6. 2 2 So, the binomial distribution is given by 6 r (6 - r ) æ1ö æ1ö æ1ö P(X = r) = nC r × p r × q (n - r ) = 6C r × ç ÷ × ç ÷ = 6C r ç ÷ . è 2ø è 2ø è 2ø

\

(i) P(X ³ 1) = 1 - P(X = 0) 6

1 ö 63 æ1ö æ = 1 - 6C 0 × ç ÷ = ç 1 - ÷ = × 64 2 è ø è ø 64 (ii) P(X £ 5) = 1 - P(X = 6) 6

1 ö 63 æ1ö æ = 1 - 6C 6 ç ÷ = ç 1 - ÷ = × 64 ø 64 è 2ø è EXAMPLE 12

If X follows a binomial distribution with mean 4 and variance 2, find [CBSE 2001C] P(X ³ 5).

SOLUTION

We know that mean = np and variance = npq. \ np = 4 and npq = 2. 1 Now, np = 4 and npq = 2 Þ 4q = 2 Þ q = × 2 1ö 1 æ \ p = (1 - q) = ç1 - ÷ = × 2ø 2 è 1 1 Now, np = 4 and p = Þ n = 4 Þ n = 8. 2 2

SSS Mathematics for Class 12 1334

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Senior Secondary School Mathematics for Class 12

So, the binomial distribution is given by r

æ1ö æ1ö P(X = r) = nC r × p r × q (n - r ) = 8C r × ç ÷ × ç ÷ è 2ø è 2ø \

(8 - r )

8

æ1ö = 8C r × ç ÷ . è 2ø

P(X ³ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) æ1ö = 8C5 × ç ÷ è 2ø

8

æ1ö + 8C 6 × ç ÷ è 2ø

8

æ1ö + 8C 7 × ç ÷ è 2ø

8

æ1ö + 8C 8 × ç ÷ è 2ø

8

8

æ1ö = [8 C 3 + 8C 2 + 8C1 + 1] ç ÷ è 2ø 1 93 = (56 + 28 + 8 + 1) × = × 256 256 EXAMPLE 13

Find the binomial distribution for which the mean and variance are 12 and 3 respectively. [CBSE 2004C]

SOLUTION

Let X be a binomial variate for which mean = 12 and variance = 3. 1 Then, np = 12 and npq = 3 Û 12 ´ q = 3 Û q = × 4 1ö 3 æ \ p = (1 - q) = ç1 - ÷ = × 4ø 4 è 12 æ 4ö = ç12 ´ ÷ = 16. 3ø p è 3 1 Thus, n = 16, p = and q = × 4 4 Hence, the binomial distribution is given by And, np = 12 Û n =

r

æ 3 ö æ1ö P(X = r) = 16C r × ç ÷ × ç ÷ è 4 ø è 4ø

(16 - r )

, where r = 0, 1, 2, 3 , ¼, 15.

EXAMPLE 14

If the sum of the mean and variance of a binomial distribution for 5 trials [CBSE 2004] is 1.8, find the distribution.

SOLUTION

We know that mean = np and variance = npq. It is being given that n = 5 and mean + variance = 1.8. \ np + npq = 1.8, where n = 5 Û 5 p + 5 pq = 1.8 Û p + p(1 - p) = 0. 36 [Q q = (1 - p)] Û p 2 - 2p + 0. 36 = 0 Û 100p 2 - 200p + 36 = 0 Û 25 p 2 - 50p + 9 = 0 Û 25 p 2 - 45 p - 5 p + 9 = 0

SSS Mathematics for Class 12 1335

Binomial Distribution

1335

Û 5 p(5 p - 9) - (5 p - 9) = 0 Û (5 p - 9)(5 p - 1) = 0 1 Û p = = 0.2 [Q p cannot exceed 1]. 5 Thus, n = 5, p = 0.2, and q = (1 - p) = (1 - 0.2) = 0.8. Let X denote the binomial variate. Then, the required distribution is P(X = r) = nC r × p r × q (n - r ) = 5C r × ( 0.2) r × ( 0.8) (5 - r ) , where r = 0, 1, 2, 3 , 4, 5. EXAMPLE 15

The sum and the product of the mean and variance of a binomial distribution are 24 and 128 respectively. Find the distribution.

SOLUTION

We have np + npq = 24 and np ´ npq = 128 Û (np)(1 + q) = 24 and n 2 p 2 ´ q = 128 Û n 2p 2 = Û

576 and n 2 p 2 ´ q = 128 (1 + q) 2

576 128 = Û 2(1 + q 2 + 2q) = 9q 2 q (1 + q)

Û 2q 2 - 5 q + 2 = 0 Û ( 2q - 1)( q - 2) = 0 1 [Q q ¹ 2]. 2 1ö 1 æ p = (1 - q) = ç1 - ÷ = × 2ø 2 è

Û q= \

Now, np(1 + q) = 24 Û n ´

1ö 1æ ç1 + ÷ = 24 Û n = 32. 2ø 2è

Hence, the required probability distribution is given by 32

P(X = r) =

32

æ1ö Cr × ç ÷ . è 2ø

EXAMPLE 16

In a binomial distribution, prove that mean > variance.

SOLUTION

Let X be a binomial variate with parameters n and p. Then, mean = np and variance = npq. \ (mean) - (variance) = (np - npq) = np(1 - q) = np 2 > 0 [Q (1 - q) = p and np 2 > 0 as n Î N ] Þ [(mean) - (variance)] > 0 Þ mean > variance. Hence, mean > variance.

SSS Mathematics for Class 12 1336

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Senior Secondary School Mathematics for Class 12

EXAMPLE 17

A die is tossed thrice. Getting an even number is considered a success. What is the variance of the binomial distribution?

SOLUTION

Here, n = 3. Let p = probability of getting an even number in a single throw 3 1 Þ p= = × 6 2 1ö 1 æ Þ q = (1 - p) = ç1 - ÷ = × 2ø 2 è \

1 1ö 3 æ variance = npq = ç 3 ´ ´ ÷ = × 2 2ø 4 è

EXAMPLE 18

A die is rolled 20 times. Getting a number greater than 4 is a success. Find the mean and variance of the number of successes.

SOLUTION

In a single throw of a die, we have p = probability of getting a number greater than 4 = \

2 1 = × 6 3

1ö 2 æ q = (1 - p) = ç1 - ÷ = × 3ø 3 è

Also, n = 20 (given). \

1ö æ mean = np = ç 20 ´ ÷ = 6.67 , and 3ø è 1 2ö æ variance = npq = ç 20 ´ ´ ÷ = 4.44. 3 3ø è

EXAMPLE 19

A die is tossed 180 times. Find the expected number (m ) of times the face with the number 5 will appear. Also, find the standard deviation ( s), and variance ( s 2 ).

SOLUTION

In a single throw of a die, S = {1, 2, 3, 4, 5, 6}. 1 p = (probability of getting the number 5) = × 6 1ö 5 æ \ q = (1 - p) = ç1 - ÷ = × 6ø 6 è 1 5 and q = × 6 6 1ö æ m = np = ç180 ´ ÷ = 30. 6ø è

Thus, n = 180, p = \

1 5ö æ Variance = s 2 = npq = ç180 ´ ´ ÷ = 25. 6 6ø è Standard deviation = s = 25 = 5.

SSS Mathematics for Class 12 1337

Binomial Distribution

1337

EXERCISE 32 1. A coin is tossed 6 times. Find the probability of getting at least 3 heads. 2. A coin is tossed 5 times. What is the probability that a head appears an even number of times? 3. 7 coins are tossed simultaneously. What is the probability that a tail appears an odd number of times? 4. A coin is tossed 6 times. Find the probability of getting (i) exactly 4 heads (ii) at least 1 head (iii) at most 4 heads. 5. 10 coins are tossed simultaneously. Find the probability of getting (i) exactly 3 heads (ii) not more than 4 heads (iii) at least 4 heads. 6. A die is thrown 6 times. If ‘getting an even number’ is a success, find the probability of getting (i) exactly 5 successes (ii) at least 5 successes (iii) at most 5 successes. 7. A die is thrown 4 times. ‘Getting a 1 or a 6’ is considered a success. Find the probability of getting (i) exactly 3 successes (ii) at least 2 successes (iii) at most 2 successes. 8. Find the probability of a 4 turning up at least once in two tosses of a fair die. 9. A pair of dice is thrown 4 times. If ‘getting a doublet’ is considered a success, find the probability of getting 2 successes. 10. A pair of dice is thrown 7 times. If ‘getting a total of 7’ is considered a success, find the probability of getting (i) no success (ii) exactly 6 successes (iii) at least 6 successes

(iv) at most 6 successes.

11. There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item. 12. In a box containing 60 bulbs, 6 are defective. What is the probability that out of a sample of 5 bulbs (i) none is defective, (ii) exactly 2 are defective? 13. The probability that a bulb produced by a factory will fuse after 6 months of use is 0.05. Find the probability that out of 5 such bulbs (i) none will fuse after 6 months of use (ii) at least one will fuse after 6 months of use (iii) not more than one will fuse after 6 months of use. 14. In the items produced by a factory, there are 10% defective items. A sample of 6 items is randomly chosen. Find the probability that this sample contains (i) exactly 2 defective items, (ii) not more than 2 defective items, (iii) at least 3 defective items.

SSS Mathematics for Class 12 1338

1338

Senior Secondary School Mathematics for Class 12

15. Assume that on an average one telephone number out of 15, called between 3 p.m. and 4 p.m. on weekdays, will be busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy? 16. Three cars participate in a race. The probability that any one of them has an accident is 0.1. Find the probability that all the cars reach the finishing line without any accident. 17. Past records show that 80% of the operations performed by a certain doctor were successful. If he performs 4 operations in a day, what is the probability that at least 3 operations will be successful? 18. The probability of a man hitting a target is (1/4). If he fires 7 times, what is the probability of his hitting the target at least twice? 19. In a hurdles race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is (5/6). What is the probability that he will knock down fewer than 2 hurdles? 20. A man can hit a bird, once in 3 shots. On this assumption he fires 3 shots. What is the chance that at least one bird is hit? 21. If the probability that a man aged 60 will live to be 70 is 0.65, what is the probability that out of 10 men, now 60, at least 8 will live to be 70? 22. A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that (i) none is white, (ii) all are white, (iii) at least one is white? 23. A policeman fires 6 bullets at a burglar. The probability that the burglar will be hit by a bullet is 0.6. What is the probability that the burglar is still unhurt? 24. A die is tossed thrice. A success is 1 or 6 on a toss. Find the mean and variance of successes. 25. A die is thrown 100 times. Getting an even number is considered a success. Find the mean and variance of successes. 26. Determine the binomial distribution whose mean is 9 and variance is 6. [CBSE 2006]

27. Find the binomial distribution whose mean is 5 and variance is 2.5. 28. The mean and variance of a binomial distribution are 4 and (4/3) respectively. Find P(X ³ 1). [CBSE 2004] 29. For a binomial distribution, the mean is 6 and the standard deviation is 2. Find the probability of getting 5 successes. 30. In a binomial distribution, the sum and the product of the mean and the variance are (25/3) and (50/3) respectively. Find the distribution. 31. Obtain the binomial distribution whose mean is 10 and standard deviation is 2 2.

SSS Mathematics for Class 12 1339

Binomial Distribution

1339

32. Bring out the fallacy, if any, in the following statement: ‘The mean of a binomial distribution is 6 and its variance is 9’.

ANSWERS (EXERCISE 32)

1.

21 32

2.

1 2

1 2 15 193 53 5. (i) (ii) (iii) 128 512 64 8 11 8 7. (i) (ii) (iii) 81 27 9 3.

15 63 57 (ii) (iii) 64 64 64 3 7 63 6. (i) (ii) (iii) 32 64 64 4. (i)

8.

11 36

9.

æ5 ö 10. (i) ç ÷ è 6ø

25 216

7

12.

5

æ 19 ö æ 19 ö 13. (i) ç ÷ (ii) 1 - ç ÷ è 20 ø è 20 ø (i)

3 æ 9ö ´ç ÷ 20 è 10 ø

5

4

4547 8192

r

27.

10

29.

9

31.

50

é 3 æ 19 ö 4 ù (iii) 1 - ê ´ ç ÷ ú êë 2 è 20 ø úû

16.

729 1000

17.

512 625

19.

5 10 (2 ´ 6 9)

20.

19 27

22. (i)

0.004096 27

4

81 1 175 (ii) (iii) 256 256 256 2 2 25. m = 50, s 2 = 25 24. m = 1, s = 3

21. 0.2615

26.

729 æ 9ö (i) ç ÷ (ii) 10000 è 10 ø

4

4

23.

1ö 35 1 æ (iii) 5 (iv) ç1 - 7 ÷ 6 ø 67 6 è

æ 19 ö æ 6ö (iii) ç ÷ ´ ç ÷ è 20 ø è5 ø

3 æ 19 ö ´ç ÷ 2 è 20 ø

(ii)

æ 14 ö æ 59 ö 15. 1 - ç ÷ × ç ÷ è 15 ø è 45 ø 18.

(ii) 5

æ 47 ö æ 71 ö 11. ç ÷ ´ ç ÷ è 50 ø è 50 ø

14.

7

æ 1ö æ 2ö Cr × ç ÷ × ç ÷ è 3ø è 3ø r

æ1ö æ1ö Cr × ç ÷ × ç ÷ è 2ø è 2ø 5

r

, where r = 0, 1, 2, 3 , ¼, 27

(10 - r )

, 0 £ r £ 10

æ 2ö æ 1ö C5 × ç ÷ × ç ÷ è 3ø è 3ø æ 1 ö æ 4ö Cr × ç ÷ × ç ÷ è5 ø è5 ø

(27 - r )

28. r

4

30.

15

728 729

æ 1ö æ 2ö Cr × ç ÷ × ç ÷ è 3ø è 3ø

(15 - r )

(50 - r )

, 0 £ r £ 50

32. The probability of getting a failure (i.e., q) cannot be greater than 1.

SSS Mathematics for Class 12 1340

1340

Senior Secondary School Mathematics for Class 12 HINTS TO SOME SELECTED QUESTIONS (EXERCISE 32)

1. In a single toss, P( H ) =

1 1 and P(not H ) = × 2 2

1 1 , q = and n = 6. 2 2 Let X show the number of heads. Then,

\

p=

6

æ 1ö P(X = r ) = nC r × p r × q( n - r) = 6C r ç ÷ . è2ø Required probability = P(X = 3 ) + P(X = 4 ) + P(X = 5 ) + P(X = 6 ). 1 1 2. Here, p = , q = and n = 5. 2 2 Let X show the number of heads. Then, 5

æ 1ö P(X = r ) = nC r × p r × q( n - r) = 5C r ç ÷ . è2ø Required probability = P(X = 0 ) + P(X = 2 ) + P(X = 4 ). 3. 7 coins being tossed simultaneously is the same as one coin being tossed 7 times. 1 1 \ p = , q = and n = 7. 2 2 Let X show the number of tails. Then, 7

æ 1ö P(X = r ) = nC r × p r × q n - r = 7C r × ç ÷ × è2ø Required probability = P(X = 1) + P(X = 3 ) + P(X = 5 ) + P(X = 7 ). 1 1 4. P(a head) = and P(not a head) = × 2 2 1 1 \ p = , q = and n = 6. 2 2 6 æ 1ö \ P(X = r ) = nC r × p r × q( n - r) = 6C r × ç ÷ × è2ø 6 1 æ ö (i) P(exactly 4 heads) = 6C 4 × ç ÷ . è2ø (ii) P(at least 1 head) = 1 - P(no head) 6 é æ 1ö ù = 1 - P( 0 head) = ê 1 - 6C 0 × ç ÷ ú × è 2 ø úû êë (iii) P(at the most 4 heads) = P(4 or less than 4 heads) = 1 - P[5 heads or 6 heads] 6 é æ 1ö ù = 1 - ê( 6C 5 + 6C 6 ) ç ÷ ú × è 2 ø úû êë 5. 10 coins being tossed simultaneously is the same as one coin being tossed 10 times. 10

æ 1ö P(X = r ) = nC r × p r × q( n - r) = 10C r × ç ÷ . è2ø 10

æ 1ö (i) P(exactly 3 heads) = 10C 3 × ç ÷ . è2ø

SSS Mathematics for Class 12 1341

Binomial Distribution (ii) P(not more than 4 heads) = P(X £ 4 ) = P(X = 0 ) + P(X = 1) + P(X = 2 ) + P(X = 3 ) + P(X = 4 ) 10 æ 1ö = ( 10C 0 + 10C 1 + 10C 2 + 10C 3 + 10C 4 ) ç ÷ × è2ø (iii) P(at least 4 heads) = P(4 heads or 5 heads or … or 10 heads) = 1 - P(0 head or 1 head or 2 heads or 3 heads) = 1 - [P(X = 0 ) + P(X = 1) + P(X = 2 ) + P(X = 3 )] 10 æ 1ö = 1 - ( 10C 0 + 10C 1 + 10C 2 + 10C 3 ) ç ÷ × è2ø 6. p =

3 1 1ö 1 æ = , q = ç 1 - ÷ = and n = 6. 6 2 2ø 2 è 6

æ 1ö P(X = r ) = nC r × p r × qn - r = 6C 5 ç ÷ × è2ø 6

æ 1ö (i) P(exactly 5 successes) = 6C 5 ç ÷ × è2ø (ii) P(at least 5 successes) = P[(5 successes) or (6 successes)] 6 æ 1ö = ( 6C 5 + 6C 6 ) ç ÷ × è2ø (iii) P(at most 5 successes) = P[0 or 1 or 2 or 3 or 5 successes] 6 é æ 1ö ù = 1 - P( 6 successes) = ê 1 - 6C 6 × ç ÷ ú × è 2 ø úû êë 2 1 1ö 2 æ = , q = ç 1 - ÷ = and n = 4. 6 3 3ø 3 è r ( 4 - r) æ 1ö æ 2ö n r P(X = r ) = C r × p × q ( n - r) = 4C r × ç ÷ × ç ÷ × è 3ø è 3ø

7. p =

3

1

æ 1ö æ 2ö (i) P(exactly 3 successes) = 4C 3 × ç ÷ × ç ÷ × è 3ø è 3ø (ii) P(at least 2 successes) = [P(X = 2 ) or P(X = 3 ) or P(X = 4 )] = 1 - [P(X = 0 ) + P(X = 1)]. (iii) P(at most 2 successes) = P[(X = 0 ) or (X = 1) or (X = 2 )]. 8. p =

1 1ö 5 æ , q = ç 1 - ÷ = and n = 2. 6 6ø 6 è r

æ 1ö æ5 ö P(X = r ) = nC r × p r × q ( n - r) = 2 C r × ç ÷ × ç ÷ è6ø è6ø

( 2 - r)

.

P[at least one 4] = P(X = 1 or X = 2 ) = P(X = 1) + P(X = 2 ). 6 1 1ö 5 æ = , q = ç 1 - ÷ = and n = 4. 36 6 6ø 6 è r ( 4 - r) æ 1ö æ5 ö P(X = r ) = nC r × p r × q( n - r) = 4C r × ç ÷ × ç ÷ . è6ø è6ø

9. p =

1341

SSS Mathematics for Class 12 1342

1342

Senior Secondary School Mathematics for Class 12 2

\

2

æ 1ö æ5 ö P(X = 2 ) = 4C 2 × ç ÷ × ç ÷ . è6ø è6ø

10. Let E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}. 6 1 1ö 5 æ \ p= = , q = ç 1 - ÷ = , n = 7. 36 6 6ø 6 è r

æ 1ö æ5 ö P(X = r ) = nC r × p r × q( n - r) = 7C r × ç ÷ × ç ÷ è6ø è6ø

( 7 - r)

7

æ5 ö (i) P(X = 0 ) = 7C 0 × ç ÷ . è6ø 6

æ 1ö æ5 ö (ii) P(X = 6 ) = 7C 6 × ç ÷ × ç ÷ × è6ø è6ø (iii) P(at least 6 successes) = P(6 successes or 7 successes) = P(X = 6 ) + P(X = 7 ) 7

6

æ 1ö æ5 ö æ 1ö = 7C 6 × ç ÷ × ç ÷ + 7C 7 × ç ÷ × è6ø è6ø è6ø (iv) P(at most 6 successes) = P(X £ 6 ) 7

æ 1ö = [1 - P(X = 7 )] = 1 - 7C 7 × ç ÷ . è6ø 11. p =

6 3 3 ö 47 æ and n = 8. = , q= ç1- ÷ = 100 50 50 ø 50 è r

æ 3 ö æ 47 ö P(X = r ) = nC r × p r × q( n - r) = 8C r × ç ÷ × ç ÷ è 50 ø è 50 ø

( 8 - r)

Required probability = P(0 defective or 1 defective) = P(X = 0 ) + P(X = 1) 7

8

æ 47 ö æ 3 ö æ 47 ö = 8C 0 × ç ÷ + 8C 1 × ç ÷ ç ÷ × è 50 ø è 50 ø è 50 ø 12. p =

6 1 1 ö 9 æ and n = 5. = ,q= ç1÷= 60 10 10 ø 10 è r

æ 1 ö æ 9 ö P(X = r ) = nC r × p r × qn - r = 5C r × ç ÷ × ç ÷ è 10 ø è 10 ø

( 5 - r)

5

æ 9 ö (i) P(none is defective) = P(X = 0 ) = 5C 0 × ç ÷ × è 10 ø 2

3

æ 1 ö æ 9 ö (ii) P(exactly 2 are defective) = P(X = 2 ) = 5C 2 × ç ÷ × ç ÷ × è 10 ø è 10 ø 13. p =

1 1 ö 19 æ and n = 5. , q= ç1÷= 20 20 ø 20 è r

æ 1 ö æ 19 ö P(X = r ) = nC r × p r × q( n - r) = 5C r × ç ÷ × ç ÷ è 20 ø è 20 ø 5

æ 19 ö (i) P(X = 0 ) = 5C 0 × ç ÷ × è 20 ø

( 5 - r)

SSS Mathematics for Class 12 1343

Binomial Distribution

1343

(ii) P(X ³ 1) = 1 - P(X < 1) = 1 - P(X = 0 ) 5

æ 19 ö = 1 - 5C 0 × ç ÷ × è 20 ø (iii) P(X £ 1) = P(X = 0 ) + P(X = 1) 4

5

æ 19 ö æ 1 ö æ 19 ö = 5C 0 × ç ÷ + 5C 1 × ç ÷ ç ÷ × è 20 ø è 20 ø è 20 ø 14. p =

10 1 1 ö 9 æ and n = 6. = , q= ç1÷= 100 10 10 ø 10 è r

\

æ 1 ö æ 9 ö P(X = r ) = nC r × p r × q( n - r) = 6C r × ç ÷ × ç ÷ è 10 ø è 10 ø 2

( 6 - r)

×

4

æ 1 ö æ 9 ö (i) P(X = 2 ) = 6C 2 × ç ÷ × ç ÷ × è 10 ø è 10 ø 5

6

4

1 æ 9 ö 1 æ 9 ö æ 9 ö (ii) P(X £ 2 ) = 6C 0 × ç ÷ + 6C 1 × ´ ç ÷ + 6C 2 × ´ç ÷ × 10 è 10 ø 100 è 10 ø è 10 ø (iii) P(X ³ 3 ) = 1 - P(X < 3 ) = 1 - P(X £ 2 ). 15. p =

1 1 ö 14 æ and n = 6. , q= ç1÷= 15 15 ø 15 è r

æ 1 ö æ 14 ö P(X = r ) = nC r × p r × q( n - r) = 6C r × ç ÷ × ç ÷ è 15 ø è 15 ø

( 6 - r)

×

P(X ³ 3 ) = 1 - P(X < 3 ) = 1 - [P(X = 0 ) + P(X = 1) + P(X = 2 )] 6 5 2 4 é æ 1 ö æ 14 ö æ 14 ö æ 1 ö æ 14 ö ù = 1 - ê 6C 0 × ç ÷ + 6C 1 × ç ÷ × ç ÷ + 6C 2 × ç ÷ × ç ÷ ú è 15 ø è 15 ø è 15 ø è 15 ø è 15 ø úû êë

æ 14 ö = 1- ç ÷ è 15 ø 16. p =

4

æ 59 ö ç ÷× è 45 ø

1 1 ö 9 æ and n = 3. , q= ç1÷= 10 10 ø 10 è r

æ 1 ö æ 9 ö P(X = r ) = nC r × p r × q( n - r) = 3C r × ç ÷ × ç ÷ è 10 ø è 10 ø

( 3 - r)

×

3

æ 9 ö P(X = 0 ) = 3 C 0 × ç ÷ × è 10 ø 17. p =

80 4 4ö 1 æ = , q = ç 1 - ÷ = and n = 4. 100 5 5ø 5 è r

æ 4 ö æ 1ö P(X = r ) = nC r × p r × q( n - r) = 4C r × ç ÷ × ç ÷ è 5 ø è5 ø

( 4 - r)

P(X ³ 3 ) = P(X = 3 ) + P(X = 4 ) 3 4 é æ 4 ö æ 1ö æ4ö ù = ê 4 C 3 × ç ÷ × ç ÷ + 4C 4 × ç ÷ ú × è 5 ø è5 ø è 5 ø úû êë

×

SSS Mathematics for Class 12 1344

1344 18. p =

Senior Secondary School Mathematics for Class 12 1 1ö 3 æ and n = 7. , q= ç1- ÷ = 4 4ø 4 è r

æ 1ö æ 3ö P(X = r ) = nC r × p r × q( n - r) = 7C r × ç ÷ × ç ÷ è4ø è 4ø P(X ³ 2 ) = 1 - P(X < 2 ) = 1 - P[(X = 0 ) or (X = 1)]

( n - r)

= 1 - [P(X = 0 ) + P(X = 1)]. 5ö 1 5 æ 19. p = P (knocking down 1 hurdle) = ç 1 - ÷ = , q = and n = 10. 6ø 6 6 è r

( 10 - r)

æ 1ö æ5 ö P(X = r ) = nC r × p r × q( n - r) = 10C r × ç ÷ × ç ÷ è6ø è6ø P(X < 2 ) = P(X = 0 ) + P(X = 1) 10 1 9 é æ5 ö æ 1ö æ5 ö ù = ê 10C 0 × ç ÷ + 10C 1 × ç ÷ × ç ÷ ú × è6ø è 6 ø è 9 ø úû êë 20. p =

×

1 1ö 2 æ , q = ç 1 - ÷ = and n = 3. 3 3ø 3 è r

( 3 - r)

æ 1ö æ 2ö P(X = r ) = nC r × p r × q( n - r) = 3C r × ç ÷ × ç ÷ × è 3ø è 3ø P(X ³ 1) = P(X = 1) + P(X = 2 ) + P(X = 3 ) 2 2 é 1 æ2ö ù é 2ù æ 1ö = ê 3C 1 ´ ´ ç ÷ ú + ê 3C 2 ´ ç ÷ ´ ú + 3 è 3 ø úû êë 3 úû è 3ø êë æ 4 2 1ö 7 =ç + + ÷= × è9 9 9ø 9

3 é 3 æ 1ö ù ê C3 ´ ç ÷ ú è 3 ø úû êë

21. P(X ³ 8 ) = P(X = 8 ) + P(X = 9 ) + P(X = 10 ) 8 2 9 10 é æ 13 ö æ 7 ö æ 13 ö ù æ 13 ö æ 7 ö = ê 10C 8 × ç ÷ × ç ÷ + 10C 9 × ç ÷ × ç ÷ + 10C 10 × ç ÷ ú è 20 ø è 20 ø è 20 ø úû è 20 ø è 20 ø êë 8 9 10 é 49 æ 13 ö æ 13 ö 7 æ 13 ö ù = ê 10C 2 × ç ÷ × + 10C 1 × ç ÷ × + ç ÷ ú× è 20 ø 400 è 20 ø 20 è 20 ø úû êë 8 8 821 ù æ 13 ö é 441 91 169 ù é æ 13 ö \ P = ç ÷ ×ê = êç ÷ ´ + + ú× ú 100 úû 40 400 û êë è 20 ø è 20 ø ë 80 Þ log P = [8(log 13 - log 20 ) + log 821 - log 100] = [-1.4968 + 2.9143 - 2] = - 0.5825

= 1.4175 Þ P = antilog ( 1.4175 ). 5 1 3 = , P(nonwhite) = × 20 4 4 1 3 \ p = , q = and n = 4. 4 4 r ( 4 - r) æ 1ö æ 3ö P(X = r ) = nC r × p r × q ( n - r) = 4C r × ç ÷ × ç ÷ × è4ø è 4ø

22. P(white) =

SSS Mathematics for Class 12 1345

Binomial Distribution

1345

4

æ 3ö (i) P(X = 0 ) = 4C 0 × ç ÷ × è4ø 4

æ 1ö (ii) P(X = 4 ) = 4C 4 × ç ÷ × è4ø (iii) P(X ³ 1) = 1 - P(X < 1) = 1 - P(X = 0 ). 23. p =

6 3 3ö 2 æ = , q = ç 1 - ÷ = and n = 6. 10 5 5ø 5 è r

æ 3ö æ2ö P(X = r ) = n C r × p r × q( n - r) = 6 C r × ç ÷ × ç ÷ è5 ø è5 ø

( 6 - r)

×

6

\ 24. p =

æ2ö P(X = 0 ) = 6 C 0 × ç ÷ . è5 ø 2 1 1ö 2 æ and n = 3. = , q= ç1- ÷ = 6 3 3ø 3 è

\ m = np and s 2 = npq. 4ö 1 æ 28. ç np = 4 and npq = ÷ Þ q = × 3 3 è ø \

1ö 2 4 æ 3ö æ p = ( 1 - q) = ç 1 - ÷ = and n = = ç 4 ´ ÷ = 6. p è 3ø 3 2ø è

\

P(X = r ) = n C r × p r × q( n - r) r

æ 2ö æ 1ö = 6Cr × ç ÷ × ç ÷ è 3ø è 3ø

( 6 - r)

×

P(X ³ 1) = 1 - P(X = 0 ) 0

æ 2ö æ 1ö = 1 - 6C 0 × ç ÷ × ç ÷ è 3ø è 3ø

6

1 ö 728 æ = ç1- 6 ÷ = × 3 ø 729 è 29. np = 6 and npq = 2 Þ q =

1 2 and p = × 3 3

2ö 3ö æ æ ç n ´ ÷ = 6 Þ n = ç 6 ´ ÷ = 9. 3ø 2ø è è 5

\

4

æ 2ö æ 1ö P(5 successes) = 9C 5 × ç ÷ × ç ÷ × è 3ø è 3ø

32. np = 6 and npq = 9 Þ q =

npq np

=

But, q cannot be greater than 1.

9 3 = × 6 2

SSS Mathematics for Class 12 1346

1346

Senior Secondary School Mathematics for Class 12

OBJECTIVE QUESTIONS Mark (3) against the correct answer in each of the following: 1. If A and B are mutually exclusive events such that P( A) = 0.4, P( B) = x and P( A È B) = 0.5, then x = ? 4 (d) none of these (a) 0.2 (b) 0.1 (c) 5 2. If A and B are independent events such that P( A) = 0.4, P( B) = x and P( A È B) = 0.5 , then x = ? 4 1 (b) 0.1 (c) (d) none of these (a) 5 6 3. If P( A) = 0. 8, P( B) = 0.5 and P( B/A) = 0.4, then P( A/B) = ? (a) 0.32

(b) 0.64 (c) 0.16 (d) 0.25 6 5 7 and P( A È B) = , then P( A/B) = ? 4. If P( A) = , P( B) = 11 11 11 5 5 6 4 (b) (c) (d) (a) 6 7 7 5 1 7 1 and P( A ¢ È B ¢) = , 5. If A and B are events such that P( A) = , P( B) = 2 12 4 then A and B are (a) independent

(b) mutually exclusive

(c) both ‘a’ and ‘b’

(d) none of these

6. It is given that the probability that A can solve a given problem is the probability that B can solve the same problem is that at least one of A and B can solve a problem is 2 1 13 (b) (c) (a) 5 15 15

3 and 5

2 × The probability 3

(d)

2 15

1 1 1 , and 6 5 3 respectively. What is the probability that the problem is solved? 4 5 1 (b) (c) (d) none of these (a) 9 9 3

7. The probabilities of A , B and C of solving a problem are

8. A can hit a target 4 times in 5 shots, B can hit 3 times in 4 shots, and C can hit 2 times in 3 shots. The probability that B and C hit and A does not hit is 1 2 (b) (a) 10 5 7 (c) (d) none of these 12

SSS Mathematics for Class 12 1347

Binomial Distribution

1347

9. A machine operates only when all of its three components function. The probabilities of the failures of the first, second and third component are 0.2, 0.3 and 0.5 respectively. What is the probability that the machine will fail? (a) 0.70 (b) 0.72 (c) 0.07 (d) none of these 10. A die is rolled. If the outcome is an odd number, what is the probability that it is prime? 2 3 5 (b) (c) (d) none of these (a) 3 4 12 11. If A and B are events such that P( A) = 0. 3 , P( B) = 0.2 and P( A Ç B) = 01 . , then P( A Ç B) = ? (a) 0.2 (b) 0.1 (c) 0.4 (d) 0.5 1 1 1 12. If P( A) = , P( B) = and P( A Ç B) = , then P( B /A ) = ? 4 3 5 11 11 23 37 (a) (b) (c) (d) 15 45 60 45 13. If A and B are events such that P( A) = 0.4, P( B) = 0.8 and P( B/A) = 0.6, then P( A/B) = ? (a) 0.2 (b) 0.3 (c) 0.4 (d) 0.5 14. If A and B are independent events, then P( A / B ) = ? (a) 1 - P( A) (b) 1 - P( B) (c) 1 - P( A/ B )

(d) -P( A /B) 5 1 15. If A and B are two events such that P( A È B) = , P( A Ç B) = and 6 3 1 P( B ) = , then the events A and B are 2 (a) independent (b) dependent (c) mutually exclusive (d) none of these 16. A die is thrown twice and the sum of the numbers appearing is observed to be 7. What is the conditional probability that the number 2 has appeared at least once? 1 1 2 3 (b) (c) (d) (a) 6 3 7 5 17. Two numbers are selected at random from integers 1 through 9. If the sum is even, what is the probability that both numbers are odd? 1 2 4 5 (a) (b) (c) (d) 6 3 9 8 18. In a class, 60% of the students read mathematics, 25% biology and 15% both mathematics and biology. One student is selected at random. What is the probability that he reads mathematics, if it is known that he reads biology? 2 3 3 5 (b) (c) (d) (a) 5 5 8 8

SSS Mathematics for Class 12 1348

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Senior Secondary School Mathematics for Class 12

19. A couple has 2 children. What is the probability that both are boys, if it is known that one of them is a boy? 1 2 3 1 (b) (c) (d) (a) 3 3 4 4 20. An unbiased die is tossed twice. What is the probability of getting a 4, 5 or 6 on the first toss and a 1, 2, 3 or 4 on the second toss? 1 2 3 5 (b) (c) (d) (a) 3 3 4 6 21. A fair coin is tossed 6 times. What is the probability of getting at least 3 heads? 11 21 1 3 (b) (c) (d) (a) 16 32 18 64 22. A coin is tossed 5 times. What is the probability that tail appears an odd number of times? 3 2 1 1 (b) (c) (d) (a) 5 15 2 3 23. A coin is tossed 5 times. What is the probability that head appears an even number of times? 2 3 4 1 (b) (c) (d) (a) 5 5 15 2 24. 8 coins are tossed simultaneously. The probability of getting 6 heads is 7 57 37 249 (b) (c) (d) (a) 64 64 256 256 25. A die is thrown 5 times. If getting an odd number is a success, then what is the probability of getting at least 4 successes? 4 7 3 3 (b) (c) (d) (a) 5 16 16 20 26. In 4 throws of a pair of dice, what is the probability of throwing doublets at least twice? 7 17 19 (b) (c) (d) none of these (a) 36 144 144 27. A pair of dice is thrown 7 times. If getting a total of 7 is considered a success, what is the probability of getting at most 6 successes? æ5 ö (a) ç ÷ è 6ø

7

æ1ö (b) ç ÷ è 6ø

7

1ö æ (c) ç1 - 7 ÷ 6 ø è

(d) none of these

3 × He tries five times. What is 4 the probability that he will hit the target at least 3 times? 459 291 371 (b) (c) (d) none of these (a) 512 364 464

28. The probability that a man can hit a target is

SSS Mathematics for Class 12 1349

Binomial Distribution

1349

29. The probability of the safe arrival of one ship out of 5 is probability of the safe arrival of at least 3 ships? 1 3 181 (b) (c) (a) 31 52 3125

(d)

1 × What is the 5

184 3125

30. The probability that an event E occurs in one trial is 0.4. Three independent trials of the experiment are performed. What is the probability that E occurs at least once? (a) 0.784 (b) 0.936 (c) 0.964 (d) none of these

ANSWERS (OBJECTIVE QUESTIONS)

1. (b)

2. (c)

3. (b)

4. (d)

5. (d)

6. (c)

7. (b)

8 (a)

9. (b) 10. (a)

11. (b) 12. (d) 13. (b) 14. (a) 15. (a) 16. (b) 17. (d) 18. (b) 19. (a) 20. (a) 21. (b) 22. (c) 23. (d) 24. (c) 25. (c) 26. (c) 27. (c) 28. (a) 29. (c) 30. (a)

HINTS TO THE GIVEN OBJECTIVE QUESTIONS 1. We have A Ç B = f Þ P( A Ç B) = 0. \

P( A È B) = P( A ) + P( B) Þ 0 . 5 = 0.4 + x Þ x = 0.1.

2. We have P( A Ç B) = P( A ) ´ P( B) = 0.4 ´ x. \

P( A È B) = P( A ) + P( B) - P( A Ç B)

Þ 0 . 5 = 0.4 + x - 0.4 x Þ 0.6 x = 0.1 Þ x = 3. P( B/ A ) =

1 × 6

P( B Ç A ) P( A Ç B) = × P( A ) P( A )

Þ P( A Ç B) = P( B/ A ) × P( A ) = ( 0.4 ´ 0.8 ) = 0. 32. P( A Ç B) 0. 32 Þ P( A/ B) = = = 0.64. P( B) 0 .5 4. P( A È B) = P( A ) + P( B) - P( A Ç B). 5 7 ö 4 æ 6 Þ P( A Ç B) = ç + - ÷= × è 11 11 11 ø 11 Þ P( A/ B) = 5. P{( A Ç B)¢} =

P( A Ç B) ( 4 11 ) 4 = 5 = = 0.8. P( B) ( 11 ) 5

1 1ö 3 æ Þ P( A Ç B) = ç 1 - ÷ = × 4 4ø 4 è

Since P( A Ç B) ¹ 0 , so A and B are not mutually exclusive. æ1 7 ö 7 P( A ) ´ P( B) = ç ´ ÷ = ¹ P( A Ç B). è 2 12 ø 24 Þ A and B are not independent.

SSS Mathematics for Class 12 1350

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Senior Secondary School Mathematics for Class 12

6. Let E1 be the event that A can solve the problem and E2 be the event that B can solve the problem. Then, E1 and E2 are independent. \

æ3 2ö 2 P( E1 Ç E2 ) = P( E1 ) ´ P( E2 ) = ç ´ ÷ = × è5 3ø 5

\

æ 3 2 2 ö 13 P( E1 È E2 ) = P( E1 ) + P( E2 ) - P( E1 Ç E2 ) = ç + - ÷ = × 3 5 ø 15 è5

1ö 5 1ö 4 1ö 2 æ æ æ 7. P( A ¢ ) = ç 1 - ÷ = , P( B¢ ) = ç 1 - ÷ = and P(C ¢ ) = ç 1 - ÷ = × 6ø 6 5ø 5 3ø 3 è è è P(that the problem is not solved) = P( A ¢ and B¢and C ¢) æ5 4 2 ö 4 = P( A ¢) ´ P( B¢) ´ P(C ¢) = ç ´ ´ ÷ = × è6 5 3ø 9 4ö 5 æ P(that the problem is solved) = ç 1 - ÷ = × 9ø 9 è 8. Let E1 , E2 and E3 be the events that A , B and C can hit respectively. 4 3 2 Then, P( E1 ) = , P( E2 ) = and P( E3 ) = × 5 4 3 Required probability = P( E2 and E3 but not E1 ) = P( E2 ) ´ P( E3 ) ´ P( E1¢ ) ì3 2 æ 4 öü æ 1 1 ö 1 × = í ´ ´ ç 1 - ÷ý = ç ´ ÷ = 5 ø þ è 2 5 ø 10 î4 3 è 9. Let E1 , E2 and E3 be the events that the 1st, 2nd and 3rd component function. Then, P( E1 ) = ( 1 - 0.2 ) = 0.8 , P( E2 ) = ( 1 - 0. 3 ) = 0.7 and P( E3 ) = ( 1 - 0.5 ) = 0.5. P(machine fails) = 1 - P(machine functions) = 1 - P( E1 , E2 and E3 ) = 1 - {P( E1 ) ´ P( E2 ) ´ P( E3 )} = 1 - ( 0.8 ´ 0.7 ´ 0.5 ) = ( 1 - 0.280 ) = 0.72. 10. Here S = {1, 2 , 3 , 4 , 5 , 6}. Let A = {1, 3 , 5}, B = {2 , 3 , 5}. Then, A Ç B = { 3 , 5}. 3 1 3 1 2 1 \ P( A ) = = , P( B) = = and P( A Ç B) = = × 6 2 6 2 6 3 P( A Ç B) ( 1 3 ) 2 = 1 = × \ P( B/ A ) = P( A ) ( 2) 3 11. B = ( A È A ) Ç B = ( A Ç B) È ( A Ç B) Þ P( B) = P( A Ç B) + P( A Ç B) Þ P( A Ç B) = P( B) - P( A Ç B) = ( 0.2 - 0.1) = 0.1. æ 1 1 1 ö 23 12. P( A È B) = P( A ) + P( B) - P( A Ç B) = ç + - ÷ = × è 4 3 5 ø 60 \

23 ö 37 æ P( A Ç B ) = P( A È B) = 1 - P( A È B) = ç 1 × ÷= 60 ø 60 è P( B/ A ) =

P( A Ç B ) P( A )

æ 37 4 ö 37 =ç ´ ÷= × è 60 3 ø 45

SSS Mathematics for Class 12 1351

Binomial Distribution 13. P( B/ A ) = \

P( A Ç B) Þ P( A Ç B) = 0.6 ´ 0.4 = 0.24. P( A )

P( A/ B) =

14. P( A / B ) =

1351

P( A Ç B) 0.24 24 3 = = = = 0. 3. P( B) 0.8 80 10

P( A Ç B ) P( B )

=

P( A ) × P( B ) P( B )

= P( A ) = 1 - P( A ).

1ö 1 æ 15. P( B) = 1 - P( B ) = ç 1 - ÷ = × 2ø 2 è P( A È B) = P( A ) + P( B) - P( A Ç B) Þ P( A ) =

2 3

æ 2 1ö 1 P( A ) × P( B) = ç ´ ÷ = = P( A Ç B). è 3 2ø 3 16. Let A = {( 1, 6 ), ( 2 , 5 ), ( 3 , 4 ), ( 4 , 3 ), (5 , 2 ), ( 6 , 1)} and B = {( 1, 2 ), ( 2 , 1), ( 2 , 2 ), ( 3 , 2 ), ( 2 , 3 ), ( 4 , 2 ), ( 2 , 4 ), (5 , 2 ), ( 2 , 5 ), ( 6 , 2 ), ( 2 , 6 )} A Ç B = {( 2 , 5 ), (5 , 2 )} n( A Ç B) 2 1 P( B/ A ) = = = × n( A ) 6 3

\

17. Out of given 9 numbers, 4 are even and 5 odd. Let A = event of choosing odd number and B be the event of getting the sum even. n( B) = ( 4 C 2 + 5C 2 ) = 16 and n( A Ç B) = 5C 2 = 10. \

P( A/ B) =

n( A Ç B) 10 5 = = × n( B) 16 8

18. Let A = event of reading mathematics and B of reading biology. 60 3 25 1 15 3 P( A ) = = , P( B) = = and P( A Ç B) = = × 100 5 100 4 100 20 P( A Ç B) æ 3 4 ö 3 \ P( A/ B) = =ç ´ ÷= × P( B) è 20 1 ø 5 19. Let S = {B1 B2 , B1G 2 , G1 B2 , G1G 2}. Let A = event that both are boys and B = event that one of the two is a boy. Then, A = {B1 B2}, B = {B1 B2 , B1G 2 , G1 B2} and A Ç B = {B1 B2}. n( A Ç B) 1 \ P( A/ B) = = × n( B) 3 20. Here S = {1, 2 , 3 , 4 , 5 , 6}. Let A = {4 , 5 , 6} and B = {1, 2 , 3 , 4}. 3 1 4 2 \ P( A ) = = and P( B) = = × 6 2 6 3 Clearly, A and B are independent events. æ1 2ö 1 \ P( A Ç B) = P( A ) ´ P( B) = ç ´ ÷ = × è2 3ø 3 1 1 21. In a single throw, we have P( H ) = and P( not H ) = × 2 2 1 1 \ p = , q = and n = 6. 2 2

SSS Mathematics for Class 12 1352

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Senior Secondary School Mathematics for Class 12

Required probability = P(3 heads or 4 heads or 5 heads or 6 heads) 3

3

4

æ 1ö æ 1ö æ 1ö = 6C 3 × ç ÷ × ç ÷ + 6C 4 × ç ÷ è2ø è2ø è2ø 1 1 1 æ = ç 20 ´ + 15 ´ + 6´ + 64 64 64 è 22. In a single throw, we have P( T ) =

5

2

æ 1ö æ 1ö æ 1ö æ 1ö × ç ÷ + 6C 5 × ç ÷ × ç ÷ + 6C 6 × ç ÷ è2ø è2ø è2ø è2ø 1 ö 42 21 = × ÷= 64 ø 64 32

1 1 and P(not T ) = × 2 2

1 1 , q = and n = 5. 2 2 Required probability = P(X = 1) or P(X = 3 ) or P(X = 5 ) = P(X = 1) + P(X = 3 ) + P(X = 5 )

\

p=

1

æ 1ö = 5C 1 × ç ÷ è2ø 10 æ 5 =ç + è 32 32

3

4

2

5

æ 1ö æ 1ö æ 1ö æ 1ö × ç ÷ + 5C 3 × ç ÷ × ç ÷ + 5C 5 × ç ÷ è2ø è2ø è2ø è2ø 1 ö 16 1 + = × ÷= 32 ø 32 2

23. In a single throw, we have P( H ) =

1 1 and P(not H ) = × 2 2

1 1 , q = and n = 5. 2 2 Required probability = P(X = 0 ) or P(X = 2 ) or P(X = 4 )

\

p=

= P(X = 0 ) + P(X = 2 ) + P(X = 4 ) 0

æ 1ö = 5C 0 × ç ÷ è2ø 10 æ 1 =ç + è 32 32

2

5

24. In a single throw, we have P( H ) = \

p=

3

4

1

æ 1ö æ 1ö æ 1ö æ 1ö æ 1ö × ç ÷ + 5C 2 × ç ÷ × ç ÷ + 5C 4 × ç ÷ × ç ÷ è2ø è2ø è2ø è2ø è2ø 5 ö 16 1 + = × ÷= 32 ø 32 2 1 1 and P(not H ) = × 2 2

1 1 , q = and n = 8. 2 2

Required probability = P( 6 heads or 7 heads or 8 heads) = P(6 heads) + P(7 heads) + P(8 heads) 6

7

2

1

æ 1ö æ 1ö æ 1ö æ 1ö æ 1ö = 8C 6 × ç ÷ × ç ÷ + 8C 7 × ç ÷ × ç ÷ + 8C 8 × ç ÷ è2ø è2ø è2ø è2ø è2ø 8 1 ö 37 æ 28 =ç + + × ÷= è 256 256 256 ø 256 25. In a single throw of a die, P(getting an odd number) = \

p=

8

3 1 = × 6 2

1 1 , q = ( 1 - p ) = and n = 5. 2 2

Required probability = P(4 successes or 5 successes) 4

1

5

1 ö 6 3 æ 1ö æ 1ö æ 1ö æ 5 = 5C 4 × ç ÷ × ç ÷ + 5C 5 × ç ÷ = ç = × + ÷= è2ø è2ø è2ø è 32 32 ø 32 16

6

SSS Mathematics for Class 12 1353

Binomial Distribution

1353

26. In a single throw of a pair of dice, we have n( S) = ( 6 ´ 6 ) = 36. Let E = {( 1, 1), ( 2 , 2 ), ( 3 , 3 ), ( 4 , 4 ), (5 , 5 ), ( 6 , 6 )}. n( E ) 6 1 1ö 5 æ Then, P( E) = = = , P (not E) = ç 1 - ÷ = × n( S) 36 6 6ø 6 è 1 5 , q = and n = 4. 6 6 Required probability = P( 2 or 3 or 4 successes)

\

p=

2

3

2

æ 1ö æ5 ö æ 1ö æ 1ö æ5 ö = 4C 2 × ç ÷ × ç ÷ + 4C 3 × ç ÷ × ç ÷ + 4C 4 × ç ÷ è6ø è6ø è6ø è6ø è6ø 5 1 ö 171 19 æ 25 =ç + + = × ÷= è 216 324 1296 ø 1296 144

4

27. In a single throw of a pair of dice, we have n( S) = ( 6 ´ 6 ) = 36. Let E = {( 1, 6 ), ( 6 , 1), ( 2 , 5 ), (5 , 2 ), ( 3 , 4 ), ( 4 , 3 )} 6 1 1ö 5 æ = , q = ç 1 - ÷ = and n = 7. \ p= 36 6 6ø 6 è 7

7

æ 1ö æ 1ö P(7 successes) = 7C 7 × ç ÷ = ç ÷ × è6ø è6ø 7

1 ö æ 1ö æ P(at most 6 successes) = 1 - P(7 successes) = 1 - ç ÷ = ç 1 - 7 ÷ × 6 ø è6ø è 28. Here p =

3 3ö 1 æ , q = ç 1 - ÷ = and n = 5. 4 4ø 4 è

P(3 or 4 or 5 successes) = P(3 successes) + P(4 successes) + P(5 successes) 3

4

2

5

æ 3ö 1 æ 3ö æ 3ö æ 1ö = 5C 3 × ç ÷ × ç ÷ + 5C 4 × ç ÷ × + 5C 5 × ç ÷ è 4ø è4ø è4ø 4 è4ø 135 405 243 918 459 ö æ =ç + + = × ÷= è 512 1024 1024 ø 1024 512 29. Here p = \

1 1ö 4 æ , q = ç 1 - ÷ = and n = 5. 5 5ø 5 è

P(3 or 4 or 5 successes) = P(3 successes) + P(4 successes) + P(5 successes) 3

2

æ 1ö æ 4 ö = 5C 3 × ç ÷ × ç ÷ + 5C 4 è5 ø è 5 ø 4 1 ö æ 32 =ç + + ÷= è 625 625 3125 ø

4

1

5

æ 1ö æ 1ö æ 4 ö × ç ÷ × ç ÷ + 5C 5 × ç ÷ è5 ø è5 ø è 5 ø 181 × 3125

30. p = 0.4 , q = ( 1 - 0.4 ) = 0.6 and n = 3. Required probability = P( E occuring at least once) = 3C 1 × ( 0.4 )1 ´ ( 0.6 ) 2 + 3C 2 × ( 0.4 ) 2 ´ ( 0.6 )1 + 3C 3 × ( 0.4 ) 3 36 8 ö 98 æ 54 =ç + + = 0.784. ÷= è 125 125 125 ø 125

SSS Mathematics for Class 12 1354

33. LINEAR PROGRAMMING Linear Inequations in Two Variables The inequalities of the form ax + by £ c, ax + by < c, ax + by ³ c and ax + by > c are called linear inequations in two variables x and y. The points ( x , y) for which the inequation is true, constitute its solution set. Graph of a Linear Inequation Let us consider an inequation, ax + by £ c. For drawing its graph to obtain a solution set, we proceed as under. STEP 1

Consider the equation ax + by = c, and plot the resulting line. In case of strict inequalities < or >, draw the line as dotted, otherwise mark it thick.

STEP 2

Choose a point [if possible (0, 0)], not lying on this line. Substitute its coordinates in the inequation. If the inequation is satisfied then shade the portion of the plane which contains the chosen point; otherwise shade the portion which does not contain this point.

The shaded portion represents the solution set. The dotted line is not a part of the shaded region while the thick line is a part of it. EXAMPLE

Graph the solution set of the inequation 2x - y ³ 1.

SOLUTION

Consider the equation 2x - y = 1. x y We may write it as + = 1. ( 1 2) -1 This shows that the line 2x - y = 1 1 makes intercepts of and -1 on the 2 axes. Thus, the line meets the æ1 ö x-axis at A ç , 0÷ and the y-axis at è2 ø B( 0, - 1). We plot these points and join them by a thick line. Consider O( 0, 0). Clearly, (0, 0) does not satisfy the given inequation. So, out of the portions divided by this line, the one not containing O(0, 0), together with the points on the line, forms the solution set. 1354

SSS Mathematics for Class 12 1355

Linear Programming

1355

Simultaneous Inequations The solution set of a system of linear inequations in two variables is the set of all points ( x , y) which satisfy all the inequations in the system simultaneously. So, we find the region of the plane common to all the portions comprising the solution sets of the given inequations. When there is no region common to all the solutions of the given inequations, we say that the solution set of the system is empty. The linear inequations are also known as linear constraints.

SOLVED EXAMPLES EXAMPLE 1

Draw the graph of the solution set of the system of inequations 2x + 3 y ³ 6, x + 4y £ 4, x ³ 0 and y ³ 0.

SOLUTION

Consider the equations 2x + 3 y = 6, x + 4y = 4, x = 0 and y = 0. x y Now, 2x + 3 y = 6 Þ + = 1. 3 2

This line meets the axes at A(3, 0) and B(0, 2). Join these points and draw a thick line. Clearly, the portion not containing (0, 0) represents the solution set of the inequation 2x + 3 y ³ 6. x y Again, x + 4y = 4 Þ + = 1. 4 1 This line meets the axes at C(4, 0) and D(0, 1). Join these points and draw a thick line. Clearly, the portion containing (0, 0) represents the solution set of the inequation x + 4y £ 4. Clearly, x ³ 0 is represented by the y-axis and the portion on its right-hand side. Also, y ³ 0 is represented by the x-axis and the portion above the x-axis. Hence, the shaded region represents the solution set of the given inequations.

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Senior Secondary School Mathematics for Class 12

EXAMPLE 2

Exhibit graphically the solution set of the system of linear inequations x + y ³ 1, 7 x + 9y £ 63 , y £ 5 , x £ 6, x ³ 0 and y ³ 0.

SOLUTION

x + y = 1 meets the axes at A(1, 0) and B(0, 1). Join these points by a thick line. Clearly, the portion not containing O(0, 0) is the solution set of x + y ³ 1. x y 7 x + 9y = 63 Þ + = 1. 9 7 This line meets the axes at C(9, 0) and D(0, 7). Join these points by a thick line. Clearly, the portion containing (0, 0) is the solution set of 7 x + 9y £ 63. y = 5 is a line parallel to the x-axis at a distance 5 from the x-axis and the portion containing O(0, 0) is the solution set of the inequation y £ 5. x = 6 is a line parallel to the y-axis at a distance 6 from the y-axis and the portion containing (0, 0) is the solution set of x £ 6. Clearly, x ³ 0 has a solution represented by the y-axis and the portion on its right. Also, y ³ 0 has a solution represented by the x-axis and the portion above it. The shaded region represents the solution set of the given system of inequations.

EXAMPLE 3

Find the linear constraints for which the shaded area in the figure below is the solution set.

SSS Mathematics for Class 12 1357

Linear Programming SOLUTION

1357

Consider the line 3 x + 4y = 18. Clearly, O(0, 0) satisfies 3 x + 4y £ 18. Clearly, the shaded area and (0, 0) lie on the same side of the line 3 x + 4y = 18. So, we must have 3 x + 4y £ 18. Consider the line x - 6y = 3. Clearly, (0, 0) satisfies the inequation x - 6y £ 3. Also, the shaded area and (0, 0) lie on the same side of the line x - 6y = 3. So, we must have x - 6y £ 3. Consider the line 2x + 3 y = 3. Clearly, (0, 0) satisfies the inequation 2x + 3 y £ 3. But, the shaded region and the point (0, 0) lie on the opposite sides of the line 2x + 3 y = 3. So, we must have 2x + 3 y ³ 3. Consider the line -7 x + 14y = 14. Clearly, (0, 0) satisfies the inequation -7 x + 14y £ 14. Also, the shaded region and the point (0, 0) lie on the same side of the line -7 x + 14y = 14. So, we must have -7 x + 14y £ 14. The shaded region is above the x-axis and on the right-hand side of the y-axis, so we have y ³ 0 and x ³ 0. Thus, the linear constraints for which the shaded area in the given figure is the solution set, are 3 x + 4y £ 18, x - 6y £ 3 , 2x + 3 y ³ 3 , -7 x + 14y £ 14, x ³ 0 and y ³ 0.

EXERCISE 33A Graph the solution sets of the following inequations: 1. x + y ³ 4 4. 2x - 3 y < 4

2. x - y £ 3 5. x ³ y - 2

3. x + 2y > 1 6. y - 2 £ 3 x

Solve each of the following systems of simultaneous inequations: 7. 2x + y > 1 and 2x - y ³ - 3 8. x - 2y ³ 0, 2x - y £ - 2 9. 3 x + 4y ³ 12, x ³ 0, y ³ 1 and 4x + 7 y £ 28 10. Show that the solution set of the following linear constraints is empty: x - 2y ³ 0, 2x - y £ - 2, x ³ 0 and y ³ 0. 11. Find the linear constraints for which the shaded area in the figure given is the solution set.

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Senior Secondary School Mathematics for Class 12

ANSWERS (EXERCISE 33A)

11.

x ³ 0, y ³ 0, 2x + y ³ 2, x - y £ 1 and x + 2y £ 8

Linear Programming Linear programming is the method used in decision making in business for obtaining the maximum or minimum value of a linear expression, subject to satisfying certain given linear inequations. The linear expression is known as an objective function and the linear inequations are known as linear constraints. LINEAR CONSTRAINTS In business or industry we want to make the best use of our limited resources like money, labour, time, materials, etc.

The limitations on the resources can often be expressed in the form of linear inequations, known as linear constraints. OBJECTIVE FUNCTION A linear function of the involved variables, which we want to maximize or minimize, subject to the given linear constraints, is known as an objective function. OPTIMAL VALUE OF AN OBJECTIVE FUNCTION The maximum or minimum value of an objective function is known as its optimal value. FEASIBLE SOLUTION A set of values of the variables satisfying all the constraints is known as a feasible solution of the system of inequations. OPTIMAL SOLUTION A feasible solution which leads to the optimal value of an objective function is known as an optimal solution of the system of inequations. OPTIMIZATION TECHNIQUES The processes of obtaining the optimal values of a system of inequations are called optimization techniques.

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Linear Programming

1359

A Linear Programming Problem (LPP) A general linear programming problem consists of maximizing or minimizing an objective function, subject to certain given constraints. Formulation of a Linear Programming Problem (LPP) Working rules STEP 1

Identify the unknowns in the given LPP. Denote them by x and y.

STEP 2

Formulate the objective function in terms of x and y. Be sure whether it is to be maximized or minimized.

STEP 3

Translate all the constraints in the form of linear inequations.

STEP 4

Solve these inequations simultaneously. Mark the common area by a shaded region. This is the feasible region.

STEP 5

Find the coordinates of all the vertices of the feasible region.

STEP 6

Find the value of the objective function at each vertex of the feasible region.

STEP 7

Find the values of x and y for which the objective function Z = ax + by has maximum or minimum value (as the case may be).

Graphical Solution of an LPP We shall restrict ourselves to the case of an LPP in two variables. We shall consider at least three constraints or inequations. Each inequation gives rise to a line in the plane. For a simultaneous solution of these inequations, we consider the region common to their solution sets. In each case we obtain such a region, a convex polygon, i.e., a closed region bounded by straight lines with the property that the line joining any two points of the region lies wholly in the region. The maximum or minimum value of a linear function over a convex polygon occurs at some vertex of the polygon. So, we look at the values of the objective function at the vertices of the set of feasible solutions. The largest of these values is the maximum value of the objective function and the smallest of these values is the minimum. Graphical Method It will be clear from the following solved examples. SOLVED EXAMPLES EXAMPLE 1

Solve the following problem graphically: Minimize and maximize z = 3 x + 9y , subject to the constraints x + 3 y £ 60, x + y ³ 10, x £ y , x ³ 0 and y ³ 0.

SOLUTION

Region represented by x + 3y £ 60 Consider the equation x + 3 y = 60. x = 0 Þ 3 y = 60 Þ y = 20; y = 0 Þ x = 60 Plot the points A( 0, 20) and B( 60, 0). Join AB and produce it both ways.

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Putting x = 0 and y = 0, we get 0 + 3 ´ 0 = 0 £ 60. \ O( 0, 0) lies in the region x + 3 y £ 60. So, the region containing the origin is the solution set of x + 3 y £ 60. Region represented by x + y ³ 10 Consider the equation x + y = 10. x = 0 Þ y = 10; y = 0 Þ x = 10. Plot the points C( 0, 10) and D(10, 0). Join CD and produce it both ways. Now, x = 0, y = 0 Þ 0 + 0 ³ 10 is not true. \ O( 0, 0) does not lie in the region x + y ³ 10. Region represented by x £ y, i.e., x - y £ 0 Consider the equation x = y , i.e., x - y = 0. Clearly, x = 10 Þ y = 10. And, x = 20 Þ y = 20. Plot the points E(10, 10) and F( 20, 20). Join EF and produce it both ways. Clearly, O( 0, 0) satisfies x - y £ 0. \ O( 0, 0) lies in the region x - y £ 0. We know that: x ³ 0 is the y-axis and the region on its RHS. y ³ 0 is the x-axis and the region above the x-axis. On solving x = y and x + y = 10, we get the point G(5 , 5). On solving x = y and x + 3 y = 60, we get H(15 , 15). Thus, the feasible region is ACGH, as shown in the figure.

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Value of z = 3 x + 9y : (i) At A( 0, 20) it is ( 3 ´ 0 + 9 ´ 20) = 180. (ii) At C( 0, 10) it is ( 3 ´ 0 + 9 ´ 10) = 90. (iii) At G(5 , 5) it is ( 3 ´ 5 + 9 ´ 5) = 60. (iv) At H(15 , 15) it is ( 3 ´ 15 + 9 ´ 15) = 180. So, the minimum value of z is 60 and its maximum value is 180. EXAMPLE 2

A furniture dealer deals in only two items: tables and chairs. He has ` 5000 to invest and a space to store at most 60 pieces. A table costs him ` 250 and a chair, ` 50. He can sell a table at a profit of ` 50 and a chair at a profit of ` 15. Assuming that he can sell all the items that he buys, how should he invest his money in order that he may maximize his [CBSE 2006C] profit?

SOLUTION

This problem can be formulated as under. Let x and y be the required numbers of tables and chairs respectively. Then, clearly we have x ³ 0, y ³ 0; x + y £ 60; 250x + 50y £ 5000, i.e., 5 x + y £ 100. Let P be the profit function. Then, P = 50x + 15 y. Now, we have to maximize P. x y Now, x + y = 60 Þ + = 1. 60 60 This line meets the axes at (60, 0) and (0, 60). Plot these points and join them to get the line x + y = 60.

x y + = 1. 20 100 This line meets the axes at (20, 0) and (0, 100). Plot these points and

Also, 5 x + y = 100 Þ

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join them to get the line 5 x + y = 100. Also, the line x = 0 is the y-axis and the line y = 0 is the x-axis. These four straight lines enclose the quadrilateral OABC. The coordinates of the points O, A , B, C are (0, 0), (20, 0), (10, 50) and (0, 60) respectively. At these points, the corresponding values of P = 50x + 15 y are 0, 1000, 1250 and 900 respectively. Clearly, it is maximum at B(10, 50). So, for a maximum profit, the dealer should purchase 10 tables and 50 chairs. EXAMPLE 3

If a young man rides his motorcycle at 25 km per hour, he has to spend ` 2 per kilometre on petrol; if he rides it at a faster speed of 40 km per hour, the petrol cost increases to ` 5 per kilometre. He has ` 100 to spend on petrol and wishes to find the maximum distance he can travel within one hour. Express this as a linear programming problem and then [CBSE 2013C] solve it.

SOLUTION

Suppose that the young man rides x km at 25 km per hour and y km at 40 km per hour. Then, we have to maximize P = x + y. Clearly, x ³ 0, y ³ 0, 2x + 5 y £ 100. Since the available time is at most one hour, we have x y + £ 1 or 8x + 5 y £ 200. 25 40

Now, we solve the system of the inequations. x y 2x + 5 y = 100 Þ + = 1. 50 20 This line meets the axes at (50, 0) and (0, 20). Plot these points and join them to get the line 2x + 5 y = 100.

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x y + = 1. 25 40 This line meets the axes at (25, 0) and (0, 40). Plot these points and obtain the line 8x + 5 y = 200. x = 0 is the y-axis and y = 0 is the x-axis. We find that the solution set of the above system is the shaded region OABC. æ 50 40 ö The coordinates of O , A , B , C are (0, 0), (25, 0), ç , ÷ and è 3 3ø (0, 20) respectively. The values of P = x + y at these points are 0, 25, 30 and 20 respectively. 50 40 and y = So, P = x + y is maximum when x = × 3 3 Thus, the young man can cover the maximum distance of 30 km, if 50 40 km at 25 km/h and km at 40 km/h. he rides 3 3 Also, 8x + 5 y = 200 Þ

EXAMPLE 4

SOLUTION

Suppose every gram of wheat provides 01 . g of proteins and 0.25 g of carbohydrates, and the corresponding values for rice are 0.05 g and 0.5 g respectively. Wheat costs ` 5 and rice ` 20 per kilogram. The minimum daily requirements of proteins and carbohydrates for an average man are 50 g and 200 g respectively. In what quantities should wheat and rice be mixed in the daily diet to provide the minimum daily requirements of proteins and carbohydrates at minimum cost, assuming that both wheat and rice are to be taken in the diet? Let x g of wheat and y g of rice be mixed to fulfil the requirements. Then, we have to minimize the cost function 5x 20y x y … (i) Z= + , i.e., Z = + × 1000 1000 200 50 x g of wheat and y g of rice must give at least 50 g of proteins. So, we must have 01 . x + 0.05 y ³ 50 or 2x + y ³ 1000. Similarly, x g of wheat and y g of rice must give at least 200 g of carbohydrates. So, we must have 0.25 x + 05 . y ³ 200 or x + 2y ³ 800. x y Thus, we have to minimize Z = + , subject to the constraints 200 50 x > 0, y > 0, 2x + y ³ 1000 and x + 2y ³ 800. x y 2x + y = 1000 Þ + = 1. 500 1000

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This line meets the axes at A(500, 0) and B(0, 1000). Plot these points and join them to obtain the line 2x + y = 1000. Clearly, (0, 0) does not satisfy 2x + y ³ 1000. x y Again, x + 2y = 800 Þ + = 1. 800 400 This line meets the axes at C(800, 0) and D(0, 400). Plot these points and join them to obtain the line x + 2y = 800. Clearly, (0, 0) does not satisfy x + 2y ³ 800. x = 0 is the y-axis and y = 0 is the x-axis. We obtain the solution set of the above system, as shown by the shaded region.

Solving 2x + y = 1000 and x + 2y = 800, we get the point of intersection of AB and CD, given by E(400, 200). x y would be at some vertex of The minimum value of Z = + 200 50 the unbounded feasible region BEC. Clearly, at B we have x = 0, and at C we have y = 0. 400 200 Also, the value of Z at E(400, 200) = + = 6. 200 50 So, we must have 400 g of wheat and 200 g of rice. EXAMPLE 5

A firm manufactures two types of products, A and B, and sells them at a profit of ` 5 per unit of type A and ` 3 per unit of type B. Each product is processed on two machines, M1 and M2 . One unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2 ; whereas one unit of type B requires one minute of processing time on M1 and one minute on M2 . Machines M1 and M2 are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce a day in order to [CBSE 2000] maximize the profit. Solve the problem graphically.

SOLUTION

Let x units of A and y units of B be produced in order to have a maximum profit.

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Then, clearly x ³ 0 and y ³ 0. x units of A and y units of B will take ( x + y) minutes on M1 . \ x + y £ 300. x units of A and y units of B will take ( 2x + y) minutes on M2 . \ 2x + y £ 360. Let Z be the profit function. Then, Z = 5 x + 3 y. We have to maximize Z = 5 x + 3 y, subject to the constraints x ³ 0, y ³ 0, x + y £ 300 and 2x + y £ 360. x y Now, x + y = 300 Þ + = 1. 300 300 This line meets the axes in (300, 0) and (0, 300). Joining these points, we get the line x + y = 300. Since (0, 0) satisfies the inequation x + y £ 300, the region below the line x + y = 300 containing O(0, 0) represents x + y £ 300. x y Again, 2x + y = 360 Þ + = 1. 180 360 This line meets the axes in (180, 0) and (0, 360). Joining these points, we get the line 2x + y = 360. Since (0, 0) satisfies 2x + y £ 360, the region below the line 2x + y = 360 containing (0, 0) represents 2x + y £ 360. Also x = 0 is the y-axis and y = 0 is the x-axis. On drawing these lines and shading the feasible region, we obtain a figure, given below.

On solving x = 0 and x + y = 300, we get the point R(0, 300).

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On solving x + y = 300 and 2x + y = 360, we get the point Q(60, 240). \ the vertices of the feasible region are O( 0, 0), P(180, 0), Q( 60, 240) and R( 0, 300). Values of Z = 5 x + 3 y at O, P, Q, R are 0, 900, 1020, 900 respectively. \ Z is maximum when x = 60 and y = 240. EXAMPLE 6

An aeroplane of an airline can carry a maximum of 200 passengers. A profit of ` 400 is made on each first-class ticket and a profit of ` 300 is made on each economy-class ticket. The airline reserves at least 20 seats for first class. However, at least 4 times as many passengers prefer to travel by economy class than by first class. Determine how many of each type of tickets must be sold in order to maximize the profit for the airline. What is the maximum profit?

SOLUTION

Let x tickets of first class and y tickets of economy class be sold to maximize the profit. Then, x ³ 20, y ³ 4x , y ³ 80 and x + y £ 200. The profit function is given by Z = 400x + 300y. Draw the graphs of the lines x = 20, y = 4x , y = 80 and x + y = 200 as shown below.

Graph of the inequation x ³ 20

Since (0, 0) does not satisfy x ³ 20, the line x = 20 together with the region to its right-hand side, not containing (0, 0), represents the region x ³ 20. Graph of the inequation y ³ 4 x

Clearly, since (20, 0) does not satisfy the inequation y ³ 4x , the line

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y = 4x together with the region to its left, not containing (20, 0), represents y ³ 4x. Graph of the inequation y ³ 80

Clearly, the line y = 80 and the region above this line represents y ³ 80. Graph of the inequation x + y £ 200

Clearly, (0, 0) satisfies x + y £ 200. So, the line x + y = 200 together with the region containing O(0, 0) represents x + y £ 200. Thus, the shaded region in the given figure is the feasible region, whose vertices are A , B and C. A is the point of intersection of x = 20 and y = 80. So, its coordinates are A(20, 80). On solving y = 4x and x + y = 200, we get B(40, 160). On solving x = 20 and x + y = 200, we get C(20, 180). The values of Z = 400x + 300y at A( 20, 80), B( 40, 160) and C( 20, 180) are respectively ` 32000, ` 64000 and ` 62000. \ Z is maximum at x = 40, y = 160. EXAMPLE 7

A chemical industry produces two compounds, A and B. The following table gives the units of ingredients C and D (per kg) of compounds A and B as well as minimum requirements of C and D, and costs per kg of A and B. Compound (inunits)

Minimum requirement (inunits)

A

B

Ingredient C (per kg)

1

2

80

Ingredient D (per kg)

3

1

75

Cost per kg (in `)

4

6

Find the quantities of A and B which would minimize the cost. SOLUTION

Let x kg of A and y kg of B be produced. Then, x ³ 0, y ³ 0, x + 2y ³ 80 and 3 x + y ³ 75. Then cost function is given by Z = 4x + 6y. Thus, we have to minimize Z = 4x + 6y, subject to the constraints: x ³ 0, y ³ 0, x + 2y ³ 80 and 3 x + y ³ 75. Draw the graphs of the lines x = 0, y = 0, x + 2y = 80 and 3 x + y = 75.

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Since (0, 0) does not satisfy the inequation x + 2y ³ 80, the line x + 2y = 80 together with the region not containing (0, 0) represents x + 2y ³ 80. Since (0, 0) does not satisfy the inequation 3 x + y ³ 75, the line 3 x + y = 75 together with the region not containing (0, 0) represents 3 x + y ³ 75. Thus, the shaded region is the feasible region. The vertices of this region are P , Q and R. On solving x = 0 and 3 x + y = 75, we get the point P(0, 75). On solving x + 2y = 80 and 3 x + y = 75, we get the point Q(14, 33). On solving y = 0 and x + 2y = 80, we get the point R(80, 0). The values of Z = 4x + 6y at the points P( 0, 75), Q(14, 33) and R( 80, 0) are 450, 254 and 320 respectively. Thus, Z is minimum at Q(14, 33). Hence, for a minimum cost, 14 kg of A and 33 kg of B must be taken. EXAMPLE 8

A company makes two types of belts, A and B; profits on these belts being ` 4 and ` 3 each respectively. Each belt of type A requires twice as much time as a belt of type B, and if all belts were of type B, the company could make 1000 belts per day. The supply of leather is sufficient for only 800 belts per day (both A and B combined). At the most 400 buckles for belts of type A and 700 for those of type B are available per day. How many belts of each type should the company make per day so as to maximize the profit?

SOLUTION

Let x belts of type A and y belts of type B be made. Then, x ³ 0, y ³ 0, x £ 400, y £ 700 and x + y £ 800. Now, 1000 belts of type B can be made in 1 day.

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\

500 belts of type A can be made in 1 day.

\

time taken to make x belts of type A and y belts of type B y ö æ x =ç + ÷ days. 500 1000 ø è

\

x y + £ 1, i.e., 2x + y £ 1000. 500 1000

1369

We have to maximize Z = 4x + 3 y , subject to the constraints x ³ 0, y ³ 0, x £ 400, y £ 700, x + y £ 800 and 2x + y £ 1000. We draw the graphs of the lines x = 0, y = 0, x = 400, y = 700, x + y = 800, 2x + y = 1000 as shown below.

Since (0, 0) satisfies x £ 400, the line x = 400 together with the region containing O(0, 0) represents x £ 400. Since ( 0, 0) satisfies y £ 700, the line y = 700 together with the region containing O(0, 0) represents y £ 700. Since (0, 0) satisfies x + y £ 800, the line x + y = 800 together with the region containing O(0, 0) represents x + y £ 800. Since (0, 0) satisfies 2x + y £ 1000, the line 2x + y = 1000 together with the region containing O(0, 0) represents 2x + y £ 1000. The y-axis and the region to its right-hand side represents x ³ 0. The x-axis and the region above it represents y ³ 0. Thus, the shaded region represents the feasible region, whose vertices are P , Q , R , S, T and U. Clearly, the coordinates of P and Q are (0, 0) and (400, 0) respectively. On solving x = 400 and 2x + y = 1000, we get R(400, 200).

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On solving x + y = 800 and 2x + y = 1000, we get S(200, 600). On solving y = 700 and x + y = 800, we get T(100, 700). On solving x = 0 and y = 700, we get U(0, 700). The values of Z = 4x + 3 y at the points P , Q , R , S, T and E are respectively 0, 1600, 2200, 2600, 2500 and 2100. The maximum of these values is 2600 occurring at S(200, 600). \ Z is maximum when x = 200 and y = 600. Thus, the company should make 200 belts of type A and 600 belts of type B to have a maximum profit. EXAMPLE 9

A company has factories located at each of the two places P and Q. From these locations, a certain commodity is delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 7, 6 and 4 units of the commodity while the weekly production capacities of the factories at P and Q are respectively 9 and 8 units. The cost of transportation per unit is given below. Cost (in `)

To From

A

B

C

P

16

10

15

Q

10

12

10

How many units should be transported from each factory to each depot in order that the transportation cost is minimum? Formulate the above LPP mathematically and then solve it. SOLUTION

This problem can be explained diagramatically as follows:

Let x units and y units of the commodity be transported from the factory at P to the depots at A and B respectively. Then, ( 9 - x - y) units will be transported from the factory at P to the depot at C. \ x ³ 0, y ³ 0, and 9 - x - y ³ 0 Þ x + y £ 9. The weekly requirement of the depot at A is 7 units. So, (7 - x) units will be transported to A from the factory at Q. Similarly, ( 6 - y) units will be transported to B from the factory at Q.

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And, 8 - (7 - x + 6 - y) = ( x + y - 5) units will be transported to C from the factory at Q. \ 7 - x ³ 0, 6 - y ³ 0 and x + y - 5 ³ 0, i.e., x £ 7 , y £ 6 and x + y ³ 5. The total cost of transportation is Z = 16x + 10(7 - x) + 10y + 12( 6 - y) + 15( 9 - x - y) + 10( x + y - 5) Þ Z = x - 7 y + 227. Now, we have to find the values of x and y which minimize Z = x - 7 y + 227, subject to the constraints x ³ 0, y ³ 0, x + y £ 9, x £ 7 , y £ 6 and x + y ³ 5. We draw the graphs of the lines x = 0, y = 0, x + y = 9, x = 7 , y = 6 and x + y = 5 as shown below.

Since (0, 0) satisfies x + y £ 9, the line x + y = 9 together with the region containing O(0, 0) represents x + y £ 9. Since (0, 0) does not satisfy x + y ³ 5, the line x + y = 5 together with the region not containing O(0, 0) represents x + y ³ 5. Since (0, 0) satisfies x £ 7, the line x = 7 together with the region containing O(0, 0) represents x £ 7. Since (0, 0) satisfies y £ 6, the line y = 6 together with the region containing O(0, 0) represents y £ 6. The y-axis and the region to its right-hand side represents x ³ 0. The x-axis and the region above it represents y ³ 0. Thus, the shaded region represents the feasible region whose vertices are R , S, T , U , V and W. On solving y = 0 and x + y = 5, we get R(5, 0). On solving y = 0 and x = 7, we get S(7, 0).

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On solving x = 7 and x + y = 9, we get T(7, 2). On solving x + y = 9 and y = 6, we get U(3, 6). On solving x = 0 and y = 6, we get V(0, 6). On solving x = 0 and x + y = 5, we get W(0, 5). The values of Z = x - 7 y + 227 at R , S, T , U , V and W are 232, 234, 220, 188, 185 and 192 respectively. Thus, Z is minimum at V(0, 6), i.e., when x = 0 and y = 6. \ from the factory at P there must be a delivery of 0, 6 and 3 units to A, B, C respectively. And, the factory at Q must deliver 7, 0 and 1 units to A, B, C respectively. EXAMPLE 10

A factory owner purchases two types of machines A and B for his factory. The requirements and the limitations for the machines are as follows: Machine

Area occupied

Labour force on each machine

Daily output (in units)

A

1000 m 2

12 men

60

B

1200 m 2

8 men

40

He has maximum area of 9000 m 2 available and 72 skilled labourers who can operate both the machines. How many machines of each type should [CBSE 2008] he buy to maximize the daily output? SOLUTION

Let x machines of type A and y machines of type B be bought and let z be the daily output. Then, z = 60 x + 40 y. … (i) Maximum area available = 9000 m 2 . \ 1000 x + 1200 y £ 9000 Þ 5 x + 6 y £ 45. Maximum labour available = 72 men. \ 12 x + 8 y £ 72 Þ 3 x + 2 y £ 18. Now, we have to maximize z = 60 x + 40 y , subject to the constraints 5 x + 6 y £ 45, 3 x + 2 y £ 18, x ³ 0 and y ³ 0. x y Now, 5 x + 6 y = 45 Þ + = 1. 9 (15 2)

… (ii) … (iii)

æ 15 ö This line meets the axes at A( 9, 0) and B ç 0, ÷ × 2ø è Plot these points and join them to obtain the line 5 x + 6 y = 45.

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Clearly, (0, 0) satisfies 5 x + 6 y £ 45. So, the region below AB represents 5 x + 6 y £ 45. x y Again, 3 x + 2y = 18 Þ + = 1. 6 9 This line meets the axes at C( 6, 0) and D( 0, 9). Plot these points and join them to obtain the line 3 x + 2 y = 18. Clearly, (0, 0) satisfies 3 x + 2 y £ 18. So, the region below CD represents 3 x + 2 y £ 18. x ³ 0 is the region to the right of the y-axis. And, y ³ 0 is the region above the x-axis. On solving 5 x + 6 y = 45 and 3 x + 2 y = 18 simultaneously, we get 9 45 x = and y = × 4 8

æ 9 45 ö So, the lines AB and CD intersect at E ç , ÷× è4 8 ø Thus, the corner points of the feasible region are æ 15 ö æ 9 45 ö O( 0, 0), C( 6, 0), E ç , ÷ and B ç 0, ÷ × 2ø è è4 8 ø Value of daily output z = 60 x + 40 y :

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(i) At O( 0, 0) it is z = ( 60 ´ 0 + 40 ´ 0) = 0. (ii) At C( 6, 0) it is z = ( 60 ´ 6 + 40 ´ 0) = 360. 45 ö 9 æ 9 45 ö æ ÷ it is z = ç 60 ´ + 40 ´ ÷ = 360. (iii) At E ç , 8ø 4 8 4 è ø è 15 ö æ 15 ö æ (iv) At B ç 0, ÷ it is z = ç 60 ´ 0 + 40 ´ ÷ = 300. 2ø 2 ø è è Thus, either (6 machines of type A and no machine of type B) or (2 machines of type A and 6 machines of type B) be used to have maximum output. 9 45 [NOTE machines º 2 machines and machines º 6 machines.] 4 8 EXAMPLE 11

SOLUTION

A retired person has ` 70000 to invest and two types of bonds are available in the market for investment. First type of bond yields an annual income of 8% on the amount invested and the second type of bond yields 10% per annum. As per norms, he has to invest minimum of ` 10000 in the first type and not more than ` 30000 in the second type. How should he plan his investment, so as to get maximum return, after [CBSE 2007] one year of investment? Let bonds A be at 8% and bonds B be at 10%. Suppose he plans to invest ` x in bonds A and ` y in bonds B. Then, clearly x + y = 70000. ... (i) He invests minimum of ` 10000 in bonds A. \

x ³ 10000.

... (ii)

Also, he invests not more than ` 30000 in bonds B. \

y £ 30000

Let z be the annual return on these investments. Then, 8x 10y z= + Þ z = 0.08x + 0.1y. 100 100

... (iii)

... (iv)

Thus, we have to maximize z, subject to the conditions x + y = 70000 ü ï x ³ 10000 ý y £ 30000 ïþ x y Now, x + y = 70000 Þ + = 1. 70000 70000 This line meets the axes at A(70000, 0) and B( 0, 70000). Plot these points and join them to obtain the line x + y = 70000. Clearly, (0, 0) satisfies x + y £ 70000.

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So, the region below AB represents x + y £ 70000. x ³ 10000 is the region parallel to the y-axis and to the right of it beyond the line x = 10000. Thus, the region to the right of line CD represents x ³ 10000. y £ 30000 is the region parallel to the x-axis and above it, but below the line y = 30000. Thus, the region below the line EF and above the x-axis represents y £ 30000.

Thus, the corner points of the feasible region are D(10000, 0), A(70000, 0), E( 30000, 40000) and F(10000, 40000). Value of annual return z = 0.08x + 0.1y : (i) At D(10000, 0) it is z = ( 0.08 ´ 10000 + 0.1 ´ 0) = 800. (ii) At A(70000, 0) it is z = ( 0.08 ´ 70000 + 0.1 ´ 0) = 5600. (iii) At E( 30000, 40000) it is z = ( 0.08 ´ 30000 + 0.1 ´ 40000) = 6400. (iv) At F(10000, 40000) it is z = ( 0.08 ´ 10000 + 0.1 ´ 40000) = 4800. So, in order to get a maximum annual return, he should invest ` 30000 in bond A and ` 40000 in bond B.

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EXERCISE 33B Long-Answer Questions 1. Find the maximum value of Z = 7 x + 7 y , subject to the constraints x ³ 0, y ³ 0, x + y ³ 2 and 2x + 3 y £ 6. 2. Maximize Z = 4x + 9y, subject to the constraints x ³ 0, y ³ 0, x + 5 y £ 200, 2x + 3 y £ 134. 3. Find the maximum value of Z = 3 x + 5 y, subject to the constraints -2x + y £ 4, x + y ³ 3 , x - 2y £ 2, x ³ 0 and y ³ 0. 4. Minimize Z = 2x + 3 y, subject to the constraints x ³ 0, y ³ 0, x + 2y ³ 1 and x + 2y £ 10. 5. Maximize Z = 3 x + 5 y, subject to the constraints x + 2y £ 2000, x + y £ 1500, y £ 600, x ³ 0 and y ³ 0. 6. Find the maximum and minimum values of Z = 2x + y, subject to the constraints x + 3 y ³ 6, x - 3 y £ 3 , 3 x + 4y £ 24, -3 x + 2y £ 6, 5 x + y ³ 5 , x ³ 0 and y ³ 0. 7. Mr Dass wants to invest ` 12000 in Public Provident Fund (PPF) and in National bonds. He has to invest at least ` 1000 in PPF and at least ` 2000 in bonds. If the rate of interest on PPF is 12% per annum and that on bonds is 15% per annum, how should he invest the money to earn maximum annual income? Also find the maximum annual income. 8. A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes 1 hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ` 100 and that on a bracelet is ` 300, how many of each should be produced daily to maximize the profit? It is being given that at least one of each must be produced. 9. A man has ` 1500 to purchase rice and wheat. A bag of rice and a bag of wheat cost ` 180 and ` 120 respectively. He has a storage capacity of 10 bags only. He earns a profit of ` 11 and ` 8 per bag of rice and wheat respectively. How many bags of each must he buy to make maximum profit? [CBSE 2008C, ’09C]

10. A manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour of work on machine A and 3 hours on machine B to produce a packet of nuts while it takes 3 hours on machine A and 1 hour on machine B to produce a packet of bolts. He earns a profit of ` 17.50 per packet on nuts and ` 7 per packet on bolts. How many packets of each should be produced each day so as to maximize his profit if he operates his machines for at the most 12 hours a day? Also find the maximum profit. [CBSE 2009C, ’12]

SSS Mathematics for Class 12 1377

Linear Programming

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11. Two tailors, A and B, earn ` 300 and ` 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. How many days should each of them work if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost? 12. A dealer wishes to purchase a number of fans and sewing machines. He has only ` 5760 to invest and has space for at most 20 items. A fan costs him ` 360 and a sewing machine, ` 240. He expects to gain ` 22 on a fan and ` 18 on a sewing machine. Assuming that he can sell all the items he can buy, how should he invest the money in order to maximize the profit? [CBSE 2009, ’09C]

13. A firm manufactures two types of products, A and B, and sells them at a profit of ` 2 on type A and ` 2 on type B. Each product is processed on two machines, M1 and M2. Type A requires one minute of processing time on M1 and two minutes on M2. Type B requires one minute on M1 and one minute on M2. The machine M1 is available for not more than 6 hours 40 minutes while M2 is available for at most 10 hours a day. Find how many products of each type the firm should produce each day in order to get maximum profit. 14. A manufacturer produces two types of soap bars using two machines, A and B. A is operated for 2 minutes and B for 3 minutes to manufacture the first type, while it takes 3 minutes on machine A and 5 minutes on machine B to manufacture the second type. Each machine can be operated at the most for 8 hours per day. The two types of soap bars are sold at a profit of ` 0.25 and ` 0.50 each. Assuming that the manufacturer can sell all the soap bars he can manufacture, how many bars of soap of each type should be manufactured per day so as to maximize his profit? 15. A manufacturer of a line of patent medicines is preparing a production plan on medicines A and B. There are sufficient ingredients available to make 20000 bottles of A and 40000 bottles of B but there are only 45000 bottles into which either of the medicines can be put. Furthermore, it takes 3 hours to prepare enough material to fill 1000 bottles of A and it takes 1 hour to prepare enough material to fill 1000 bottles of B, and there are 66 hours available for this operation. The profit is ` 8 per bottle for A and ` 7 per bottle for B. How should the manufacturer schedule the production in order to maximize his profit? Also, find the maximum profit. 16. A toy company manufactures two types of dolls, A and B. Each doll of type B takes twice as long to produce as one of type A, and the company would have time to make a maximum of 2000 per day, if it produces only type A. The supply of plastic is sufficient to produce 1500 dolls per day (both A and B combined). Type B requires a fancy dress of which there are only 600 per day available. If the company makes profits of ` 3 and ` 5 per doll respectively on dolls A and B, how many of each should be produced per day in order to maximize the profit? Also, find the maximum profit.

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Senior Secondary School Mathematics for Class 12

17. A small manufacturer has employed 5 skilled men and 10 semiskilled men and makes an article in two qualities, a de luxe model and an ordinary model. The making of a de luxe model requires 2 hours’ work by a skilled man and 2 hours’ work by a semiskilled man. The ordinary model requires 1 hour by a skilled man and 3 hours by a semiskilled man. By union rules, no man can work more than 8 hours per day. The manufacturer gains ` 15 on the de luxe model and ` 10 on the ordinary model. How many of each type should be made in order to maximize his total daily profit? Also, find the maximum daily profit. 18. A company producing soft drinks has a contract which requires a minimum of 80 units of chemical A and 60 units of chemical B to go in each bottle of the drink. The chemicals are available in a prepared mix from two different suppliers. Supplier X has a mix of 4 units of A and 2 units of B that costs ` 10, and the supplier Y has a mix of 1 unit of A and 1 unit of B that costs ` 4. How many mixes from X and Y should the company purchase to honour the contract requirement and yet minimize [CBSE 2012] the cost? 19. A small firm manufactures gold rings and chains. The combined number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and half an hour for a chain. The maximum number of hours available per day is 16. If the profit on a ring is ` 300 and that on a chain is ` 190, how many of each should be manufactured daily so as to maximize the profit? 20. A manufacturer makes two types, A and B, of teapots. Three machines are needed for the manufacture and the time required for each teapot on the machines is given below. Machine Type A B

I 12 6

Time (in minutes) II 18 0

III 6 9

Each machine is available for a maximum of 6 hours per day. If the profit on each teapot of type A is 75 paise and that on each teapot of type B is 50 paise, show that 15 teapots of type A and 30 of type B should be manufactured in a day to get the maximum profit. 21. A manufacturer makes two products, A and B. Product A sells at ` 200 1 each and takes hour to make. Product B sells at ` 300 each and takes 1 2 hour to make. There is a permanent order for 14 of product A and 16 of product B. A working week consists of 40 hours of production and the weekly turnover must not be less than ` 10000. If the profit on each of the product A is ` 20 and on product B, it is ` 30 then how many of each should be produced so that the profit is maximum? Also, find the maximum profit.

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Linear Programming

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22. A man owns a field of area 1000 m 2 . He wants to plant fruit trees in it. He has a sum of ` 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 m 2 of ground per tree and costs ` 20 per tree, and type B requires 20 m 2 of ground per tree and costs ` 25 per tree. When full grown, a type-A tree produces an average of 20 kg of fruit which can be sold at a profit of ` 2 per kg and a type-B tree produces an average of 40 kg of fruit which can be sold at a profit of ` 1.50 per kg. How many of each type should be planted to achieve maximum profit when trees are fully grown? What is the maximum profit? 23. A publisher sells a hardcover edition of a book for ` 72 and a paperback edition of the same for ` 40. Costs to the publisher are ` 56 and ` 28 respectively in addition to weekly costs of ` 9600. Both types require 5 minutes of printing time although the hardcover edition requires 10 minutes of binding time and the paperback edition requires only 2 minutes. Both the printing and binding operations have 4800 minutes available each week. How many of each type of books should be produced in order to maximize the profit? Also, find the maximum profit per week. 24. A gardener has a supply of fertilizers of the type I which consists of 10% nitrogen and 6% phosphoric acid, and of the type II which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type-I fertilizer costs 60 paise per kg and the type-II fertilizer costs 40 paise per kg, determine how many kilograms of each type of fertilizer should be used so that the nutrient requirements are met [CBSE 2008] at a minimum cost. What is the minimum cost? 25. Two godowns, A and B, have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shops, D , E and F, whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table. Cost of transportation (in ` per quintal) From A

B

6.00 3.00 2.50

4.00 2.00 3.00

To D E F

How should the supplies be transported in order that the transportation cost is minimum?

SSS Mathematics for Class 12 1380

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Senior Secondary School Mathematics for Class 12

26. A brick manufacturer has two depots, P and Q, with stocks of 30000 and 20000 bricks respectively. He receives orders from three buildings A, B, C, for 15000, 20000 and 15000 bricks respectively. The cost of transporting 1000 bricks to the buildings from the depots are given below. Cost of transportation (in ` per 1000 bricks) To A

B

C

P

40

20

30

Q

20

60

40

From

How should the manufacturer fulfil the orders so as to keep the cost of transportation minimum? 27. A medicine company has factories at two places, X and Y. From these places, supply is made to each of its three agencies situated at P , Q and R. The monthly requirements of the agencies are respectively 40 packets, 40 packets and 50 packets of medicines, while the production capacity of the factories at X and Y are 60 packets and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given as follows. Transportation cost per packet (in `) From X

Y

P

5

4

Q

4

2

R

3

5

To

How many packets from each factory should be transported to each agency so that the cost of transportation is minimum? Also, find the minimum cost. 28. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E, F, whose requirements are 4500 L, 3000 L and 3500 L respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:

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Linear Programming

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Distance (in km) From A

B

To D

7

3

E

6

4

F

3

2

Assuming that the transportation cost per km is Re 1 per litre, how should the delivery be scheduled in order that the transportation cost is minimum? 29. A firm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. They need certain nutrients, named as X, Y , Z. The pigs are fed on two products, A and B. One unit of product A contains 36 units of X, 3 units of Y and 20 units of Z, while one unit of product B contains 6 units of X, 12 units of Y and 10 units of Z. The minimum requirements of X, Y , Z are 108 units, 36 units and 100 units respectively. Product A costs ` 20 per unit and product B costs ` 40 per unit. How many units of each product must be taken to minimize the cost? Also, find the minimum cost. 30. A dietician wishes to mix two types of food, X and Y, in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food X contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C, while food Y contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs ` 5 per kg to purchase the food X and ` 7 per kg to purchase the food Y. Determine the minimum cost of such a mixture. [CBSE 2011C, ’12]

31. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 calories. Two foods, A and B, are available at a cost of ` 4 and ` 3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of minerals and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost. [CBSE 2008]

32. A housewife wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of [CBSE 2009C] vitamin B and 8 units of vitamin C. The vitamin contents of 1 kg of each food are given below. Vitamin A

Vitamin B

Vitamin C

Food X

1

2

3

Food Y

2

2

1

If 1 kg of food X costs ` 6 and 1 kg of food Y costs ` 10, find the minimum cost of the mixture which will produce the diet.

SSS Mathematics for Class 12 1382

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Senior Secondary School Mathematics for Class 12

33. A firm manufactures two types of products, A and B, and sells them at a profit of ` 5 per unit of type A and ` 3 per unit of type B. Each product is processed on two machines, M1 and M2. One unit of type A requires one minute of processing time on M1 and two minutes of processing time on M2 ; whereas one unit of type B requires one minute of processing time on M1 and one minute on M2. Machines M1 and M2 are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximize the [CBSE 2000] profit. Solve the problem graphically. 34. A small firm manufactures items A and B. The total number of items that it can manufacture in a day is at the most 24. Item A takes one hour to make while item B takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item A be ` 300 and that on one unit of item B be ` 160, how many of each type of item should be produced to maximize the profit? Solve the problem graphically. [CBSE 2000, ’04] 35. A manufacturer produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of ` 30 and ` 25 per trunk of the first type and second type respectively. How many trunks of each type must he make each day to make the maximum profit? [CBSE 2005, ’12] 36. A company manufactures two types of toys, A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is ` 50 each on type A and ` 60 each on type B. How many toys of each type should the company manufacture in a [CBSE 2001] day to maximize the profit? 37. Kellogg is a new cereal formed of a mixture of bran and rice, that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if [CBSE 2002] bran costs ` 5 per kilogram and rice costs ` 4 per kilogram. 38. A dealer wishes to purchase a number of fans and sewing machines. He has only ` 5760 to invest and has space for at most 20 items. A fan costs him ` 360 and a sewing machine ` 240. He expects to sell a fan at a profit of ` 22 and a sewing machine at a profit of ` 18. Assuming that he can sell all the items that he buys, how should he invest his money to maximize the profit? Solve graphically and find the maximum profit. [CBSE 2002, ’03, ’06]

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Linear Programming

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39. Anil wants to invest at the most ` 12000 in bonds A and B. According to rules, he has to invest at least ` 2000 in Bond A and at least ` 4000 in Bond B. If the rate of interest of Bond A is 8% per annum and on Bond B, it is 10% per annum, how should he invest his money for maximum interest? [CBSE 2004C] Solve the problem graphically. 40. Maximize z = 60x + 15 y , subject to the constraints x + y £ 50, 3 x + y £ 90, x , y ³ 0.

[CBSE 2005]

41. A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of ` 50 each on type A and ` 60 each on type B. How many toys of each type should the company manufacture [CBSE 2007] in a day to maximize the profit? 42. One kind of cake requires 200 g of flour and 25 g of fat, another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat, assuming that there is no shortage of the other ingredients used in making the cakes. [CBSE 2014C] Make it an LPP and solve it graphically. 43. A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of ` 80 on each piece of type A and ` 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum [CBSE 2014] profit per week?

ANSWERS (EXERCISE 33B)

1. Maximum Z = 21 at x = 3 , y = 0 2. Maximum Z = 382 when x = 10 and y = 38 3. Minimum Z = 9

2 æ 8 1ö at ç , ÷ 3 è 3 3ø

4. Minimum Z =

3 æ 1ö at ç 0, ÷ 2 è 2ø

5. Maximum Z = 5500 at x = 1000 and y = 500 1 æ 84 15 ö 1 æ 9 25 ö 6. Maximum Z = 14 at ç at ç , , ÷ and minimum Z = 3 ÷ 3 è 13 13 ø 14 è 14 14 ø 7. Rs 1000 in PPF and ` 11000 in bonds; ` 1770 8. x = 16 and y = 8

9. 5 bags of each

SSS Mathematics for Class 12 1384

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Senior Secondary School Mathematics for Class 12

10. Three packets each of nuts and bolts; maximum profit = ` 73.50 11. 5 days and 3 days 12. 24 sewing machines only

13. 200, 200

14. For maximum profit, 96 soap bars of 2nd type must be manufactured. 15. 10500 bottles of A and 34500 bottles of B, and maximum profit is ` 325500 16. 1000 of type A and 500 of type B, maximum profit = ` 5500 17. 10 de luxe, 20 ordinary; maximum daily gain = ` 400 18. 10 mixes from X and 40 mixes from Y 19. Number of rings = 8, number of chains = 16 21. A = 48, B = 16; maximum profit = ` 1440 22. Type A = 20, type B = 40; maximum profit = ` 3200 23. 360 and 600; ` 2880 24. 100 kg of type I and 80 kg of type II; minimum cost = ` 92 25. 10 q, 50 q and 40 q from A to D , E, F respectively, and 50 q, 0 q and 0 q from B to D , E, F respectively 26. From P: 0, 20000, 10000 bricks to A , B , C respectively From Q: 15000, 0, 5000 bricks to A , B , C respectively 27. From X: 10 pkt, 0 pkt and 50 pkt to P , Q , R respectively From Y: 30 pkt, 40 pkt and 0 pkt to P , Q , R respectively Minimum cost = ` 400 28. From A: 500 L, 3000 L, 3500 L to D , E, F respectively From B: 4000 L, 0 L, 0 L to D , E, F respectively 29. 2 units of A, 4 units of B; minimum cost = ` 160 30. x = 2, y = 4; ` 38 31. 5 units of A and 30 units of B

32. ` 52

33. Z is maximum at P( 60, 240) and its maximum value is ` 1020. 34. Z is maximum at (8, 16) and its maximum value is ` 4960. 35. For getting a maximum profit of ` 165, 3 trunks of each type should be manufactured. 36. For getting a maximum profit of ` 1500, 12 toys of type A and 15 toys of type B should be manufactured. 37. The minimum cost of producing this cereal is ` 4.60 per kg. 38. The profit is maximum at E(8, 12) and it is ` 392. 39. The interest is maximum at E(2000, 10000) and it is ` 1160. 40. The maximum value of z is at C(30, 0). 41. 12 toys of type A and 15 toys of type B to earn maximum profit ` 1500 42. Max. no. of cakes: 1st kind—20, 2nd kind—10 43. For max. profit, 12 pieces of type A and 8 pieces of type B; max. profit = ` 1680

SSS Mathematics for Class 12 1385

Linear Programming

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HINTS TO SOME SELECTED QUESTIONS (EXERCISE 33B) 12 x 15 y subject to x ³ 1000 , y ³ 2000 , x + y £ 12000. + 100 100 1 8. Maximize P = 100 x + 300 y subject to x > 0 , y > 0 , x + y £ 24 , x + y £ 16. 2 7. Maximize Z =

10. Maximize P = 17 . 5 x + 7 y subject to x ³ 0 , y ³ 0 , x + 3 y £ 12 and 3 x + y £ 12. 11. Let A and B work for x and y days respectively. Minimize Z = 300 x + 400 y subject to x ³ 0, y ³ 0, 6 x + 10 y ³ 60 and 4 x + 4 y ³ 32. 12. Maximize P = 22 x + 18 y subject to x ³ 0, y ³ 0, x + y £ 20, 360 x + 240 y £ 5760. 13. Maximize Z = 2 x + 2 y subject to x ³ 0, y ³ 0, x + y £ 400, 2 x + y £ 600. x y 14. Maximize P = + subject to x ³ 0, y ³ 0, 2 x + 3 y £ 480 and 3 x + 5 y £ 480. 4 2 15. Maximize Z = 8 x + 7 y subject to 3 x + y £ 66000 , x + y £ 45000, x £ 20000, y £ 40000, x ³ 0 and y ³ 0. 16. Maximize Z = 3 x + 5 y subject to x + 2 y £ 2000, x + y £ 1500, y £ 600, x ³ 0 and y ³ 0. 17. Suppose x de luxe models and y ordinary models of articles be made each day. Then, maximize Z = 15 x + 10 y, subject to the constraints x ³ 0 , y ³ 0 , 2 x + y £ 40 and 2 x + 3 y £ 80. 18. Let the company purchase x mixes from X and y mixes from Y. Then, minimize Z = 10 x + 4 y, subject to the constraints x ³ 0 , y ³ 0 , 4 x + y ³ 80 and 2 x + y ³ 60. 20. Let x teapots of type A and y teapots of type B be manufactured. Then, x ³ 0 , y ³ 0 , 12 x + 6 y £ 6 ´ 60 , 18 x + 0 y £ 6 ´ 60 , 6 x + 9 y £ 6 ´ 60 Þ x ³ 0 , y ³ 0 , 2 x + y £ 60 , x £ 20 , 2 x + 3 y £ 120. 75 50 3 1 Profit function is Z = x+ y Þ Z = x + y. 100 100 4 2 Maximize Z, subject to the constraints x ³ 0 , y ³ 0 , x £ 20 , 2 x + y £ 60 , 2 x + 3 y £ 120. Then x = 15 and y = 30 will give the maximum value of Z. 21. Let the number of articles produced per week be x of A and y of B. 1 Then, x + y £ 40 , 200 x + 300 y ³ 10000 , x ³ 14 , y ³ 16. 2 Profit function is Z = 20 x + 30 y. Maximize Z = 20 x + 30 y, subject to the constraints x + 2 y £ 80 , 2 x + 3 y ³ 1000 , x ³ 14 , y ³ 16. 22. Let x plants of type A and y plants of type B be planted. Then, x ³ 0 , y ³ 0 , 20 x + 25 y £ 1400 , 10 x + 20 y £ 1000. 3 Profit function is Z = 2 ´ 20 x + ´ 40 y, i.e., Z = 40 x + 60 y. 2

SSS Mathematics for Class 12 1386

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Senior Secondary School Mathematics for Class 12

23. Let x copies of the hardcover edition and y copies of the paperback edition be prepared. Then, C = 9600 + 56 x + 28 y and S = 72 x + 40 y. Profit function is Z = ( S - C ), i.e., Z = 16 x + 12 y - 9600. Maximize Z subject to x ³ 0 , y ³ 0 , (5 x + 10 x ) + (5 y + 2 y ) £ 4800. 24. Let x kg of type I and y kg of type II be used. Then, 6 10 10 5 x ³ 0, y ³ 0, x+ y ³ 14 , x+ y ³ 14. 100 100 100 100 60 40 3 2 Cost function is Z = x+ y , i.e., Z = x + y. 100 100 5 5 25. The entire problem can be explained diagramatically as shown below.

Let x q and y q of grains be transported from A to shops D and E respectively. Then, [100 - ( x + y )] q will be transported from A to F. \

x ³ 0 , y ³ 0 , 100 - ( x + y ) ³ 0 Þ x + y £ 100.

D , E, F require 60 q, 50 q and 40 q respectively. From B, there will be a supply of ( 60 - x )q to D; (50 - y )q to E and [50 - ( 60 - x + 50 - y )] q to F. \ ( 60 - x ) ³ 0 , (50 - y ) ³ 0 , ( x + y - 60 ) ³ 0 Þ x £ 60 , y £ 50 and x + y ³ 60. \ cost of transportation is given by 5 Z = 6 x + 4( 60 - x ) + 3 y + 2(50 - y ) + + ( 100 - x - y ) + 3( x + y - 60 ) 2 5 3 Þ Z = x + y + 410. 2 2 28. Let x and y litres be transported from A to D and E. Then, (7000 - x - y ) litres will be transported to F from A. From B, ( 4500 - x ) litres, ( 3000 - y ) litres, [4000 - ( 4500 - x + 3000 - y )] litres will be transported to D , E and F respectively. Cost of transportation is Z = 7 x + 6 y + 3(7000 - x - y ) + 3( 45000 - x ) + 4( 3000 - y ) + 2( x + y - 3500). Minimize Z = 3 x + y + 39500, subject to x ³ 0 , y ³ 0 , x + y £ 7000 , x £ 4500 , y £ 3000 , x + y ³ 3500 and x = 500.

SSS Mathematics for Class 12 1387

Linear Programming

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29. Let x units of A and y units of B be taken. Then, minimize C = 20 x + 40 y , subject to x ³ 0 , y ³ 0 , 36 x + 6 y ³ 108 , 3 x + 12 y ³ 36 , 20 x + 10 y ³ 100. 30. Let x kg of the food X and y kg of the food Y be taken. Then, x ³ 0 , y ³ 0 , 2 x + y ³ 8 and x + 2 y ³ 10. The cost function is Z = 5 x + 7 y. 31. Let x units of the food A and y units of the food B be taken. Then, x ³ 0 , y ³ 0 , 200 x + 100 y ³ 4000 , x + 2 y ³ 50 , 40 x + 40 y ³ 1400. The cost function is Z = 4 x + 3 y. 32. Let x kg of the food X and y kg of the food Y be taken. Then, x ³ 0 , y ³ 0 , x + 2 y ³ 10 , 2 x + 2 y ³ 12 and 3 x + y ³ 8. The cost function is Z = 6 x + 10 y. 33. Let x units of the type A and y units of the type B be produced. Then, x1 + x2 £ 300 ; 2 x1 + x2 £ 360 ; x1 ³ 0 and x2 ³ 0. Maximize Z = 5 x + 3 y. 34. Let x items of A and y items of B be produced. Then, 1 x + y £ 24 , 1 × x + × y £ 16 ; x ³ 0 and y ³ 0. 2 Maximize P = 300 x + 160 y. 35. Let x trunks of the first type and y trunks of the second type be manufactured. Then, 3 x + 3 y £ 18 ; 3 x + 2 y £ 15 ; x ³ 0 and y ³ 0. Maximize Z = 30 x + 25 y. 36. Let x and y be the number of toys of type A and type B respectively. Then, 5 x + 8 y £ 180 , 10 x + 8 y £ 240 and x ³ 0 , y ³ 0. Maximize Z = 50 x + 60 y. 37. Let the cereal contain x kg of bran and y kg of rice. Maximize Z = 5 x + 4 y , subject to the conditions 80 ö æ 100 ö 88 æ ; çx ´ ÷ + çy ´ ÷³ 1000 1000 1000 è ø è ø 40 ö æ 30 ö 36 æ ; çx ´ ÷ + çy ´ ÷³ 1000 ø è 1000 ø 1000 è x ³ 0 and y ³ 0. 38. Let the number of fans and sewing machines bought be x and y respectively. Maximize Z = 22 x + 18 y , subject to x + y £ 20 , 360 x + 240 y £ 5760 and x ³ 0 , y ³ 0. y ö 10 y æ 2 x 8x ÷ subject to x ³ 2000 , y ³ 4000 and x + y £ 12000. + =ç + 100 100 è 25 10 ø y x Draw the graphs of x = 2000 , y = 4000 and + = 1. 12000 12000 y x Shade x ³ 2000 , y ³ 4000 and + £ 1 to give C BEF. 12000 12000

39. Maximize Z =

SSS Mathematics for Class 12 1388

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Senior Secondary School Mathematics for Class 12

y ö æ 2x ÷ at B, E and F are respectively 560, 1160 and 1040. Values of ç + 25 10 è ø So, it is maximum at E(2000, 10000). 41. Maximize z = 50 x + 60 y , subject to the constraints 5 x + 8 y £ 180 , 5 x + 4 y £ 120 , x ³ 0 and y ³ 0. The corner points of the feasible region are O ( 0 , 0 ), C ( 24 , 0 ), D( 0 , 22.5 ) and E( 12 , 15 ). z is maximum at E( 12 , 15 ). 42. Let the number of cakes of the first and second type be x and y respectively. Then, we have to maximise z = x + y subject to the constraints 200 x + 100 y £ 5000 Þ 2 x + y £ 50 25 x + 50 y £ 1000 Þ x + 2 y £ 40 and

x ³ 0 , y ³ 0.

First we draw the line 2 x + y = 50. For this, we plot the points A( 25 , 0 ), B( 0 , 50 ) and C( 10 , 30 ). Draw the line ACB.

... (i) ... (ii)

SSS Mathematics for Class 12 1389

Linear Programming

1389

The region below this line represents 2 x + y £ 50. Now, we draw the line x + 2 y = 40. For this, we plot the points D( 40 , 0 ), E( 0 , 20 ) and F( 20 , 10 ). Draw the line DFE. The region below this line represetns x + 2 y £ 40. The feasible region contains the points A( 25 , 0 ), F( 20 , 10 ) and E( 0 , 20 ). Value of z at A( 25 , 0 ) = 25 + 0 = 25. Value of z at F( 20 , 10 ) = 20 + 10 = 30. Value of z at E( 0 , 20 ) = 0 + 20 = 20. For maximum value of z, we have x = 20 and y = 10. Maximum number of cakes required for first and second type are 20 and 10 respectively. 43. Let the number of pieces of type A and type B manufactured per week be x and y respectively. Then, we have to maximize P = 80 x + 120 y subject to the constraints 9 x + 12 y £ 180 Þ 3 x + 4 y £ 60, x + 3 y £ 30

... (i) ... (ii)

x ³ 0 , y ³ 0. We leave it to the reader to draw the graphs. and

For maximum profit, we shall have 12 pieces of type A and 6 pieces of type B; and the maximum profit is ` 1680.

SSS Mathematics for Class 12 1390

Sample Question Paper MATHEMATICS CLASS 12

Time : 3 hrs.

Max. Marks: 100

General Instructions: 1. All questions are compulsory. 2. There are 29 questions in all. 3. Section A contains 4 questions of 1 mark each. Section B contains 8 questions of 2 marks each. Section C contains 11 questions of 4 marks each. Section D contains 6 questions of 6 marks each. SECTION A

Question numbers 1 to 4 carry 1 mark each. 1. If A is a square matrix of order 3 and|3A|= k|A|then find the value of k. 2. If * is a binary operation on the set R of all real numbers defined by a * b = a + b - 3 then find the identity element for *. 3. Give an example to show that the relation R = {( a , b) : a £ b 2} on the set R of all real numbers is not reflexive. ®

®

®

®

® ®

4. If a and b are two nonzero vectors such that|a ´ b|= a × b then find the ®

®

angle between a and b . SECTION B

Question numbers 5 to 12 carry 2 marks each. 5. Prove that each diagonal element of a skew-symmetric matrix is zero. -1 1 æ 5x ö 6. If y = tan -1 ç then prove that