Krishna's Remedial Mathematics [5 ed.]

Table of contents :
Remedial Mathematics
Dedication 1
Dedication 2
Preface to the Revised Edition
Preface to the First Edition
Syllabus
Conversion Tables
Brief Contents
Detailed Contents 1
Detailed Contents 2
Detailed Contents 3
Chapter 1: Linear and Quadratic Equations
Chapter 2: Determinants
Chapter 3: Matrices
Chapter 4: Trigonometry
Chapter 5: Logarithms
Chapter 6: System of Coordinates
Chapter 7: The Straight Line
Chapter 8: Functions and Limits
Chapter 9: Differentiation
Chapter 10: Linear Differential Equations
Logarithms 1
Logarithms 2
Antilogarithms 1
Antilogarithms 2
Appendix (Unit-IV) Integration

Citation preview

Krishna's

Remedial Mathematics (For B.Pharm. First Semester Students of All Pharmacy Colleges Affiliated to U.P. Technical University, Lucknow and other Indian Universities.)

By

A. R. Vasishtha

Hemlata Vasishtha M.Sc., Ph.D.

Retired Head, Dep’t. of Mathematics

C.C.S. University, Meerut (U.P.)

Meerut College, Meerut (U.P.)

A. K. Vasishtha M.Sc., Ph.D. C.C.S. University, Meerut (U.P.)

KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

Jai Shri Radhey Shyam

Dedicated to

Lord

Krishna Publishers

Dedicated To Pt. Pt. Pt.

TULSI RAM RAM RAKHA RAMJI DAS

MAL

who served the cause of education with a true missionary zeal

Remembering: Rai Sahib Vikas Gautam

P reface to the R evised E dition The authors feel great pleasure in presenting the Thoroughly Revised Edition of the book REMEDIAL MATHEMATICS and wish to record thanks to the teachers and students for their warm reception to the previous edition. The present edition has been specially designed, made up-to-date and well organised in a systematic order according to the latest syllabus prescribed by Uttar Pradesh Technical University, Lucknow for B.Pharm. students of first semester. The entire text has been thoroughly revised, improved and modified wherever necessary keeping in view the current trends and latest thoughts on the subject. A number of new problems selected from various university examination papers have been added in this new edition, thus making it more useful. The authors have always endeavoured to keep the text update in the best interests of the students community — a gesture which the authors hope would be appreciated by the students and teachers alike. In the last the authors wish to express their thanks to the Publishers for their painstaking in bringing out this book in the present nice form. Suggestions for the improvement of the book will be thankfully received.

September, 2012

— Authors

(v)

P reface to the F irst E dition The authors feel great pleasure in presenting the First Edition of the book REMEDIAL MATHEMATICS. This book on mathematics has been primarily written according to the unified syllabus of “Remedial Mathematics” of first semester of all pharmacy colleges affiliated to U.P. Technical University, Lucknow and other Indian Universities. The subject matter has been discussed in such a simple way that the students will find no difficulty to understand it. Each chapter of this book contains complete self-explanatory theory and a fairly large number of solved examples. Sufficient problems have also been selected from various university examination papers. We have tried to solve each problem in an elegant and more interesting way. Every effort has been made to explain the subject matter in such a simple way that the students can easily understand and feel encouraged to solve themselves the questions given in unsolved exercises. In each chapter, a number of unsolved questions with answers have been given so that students may have a lot of practice on that topic. We have tried our best to keep the book free from misprints. The authors shall be grateful to the readers who point out errors and omissions which, inspite of all care, might have been there. The authors in general hope that the present book will be warmly received by the students and teachers. We shall indeed be very thankful to our colleagues for their recommending this book to their students. The authors wish to express their thanks to Mr. S.K. Rastogi (M.D.) and Mr. Sugam Rastogi (E.D.) and entire team of KRISHNA Prakashan Media (P) Ltd., Meerut for bringing out this book in the present nice form. The authors will feel amply rewarded if the book serves the purpose for which it is meant. The suggestions for the improvement of the book are always welcome and will be highly appreciated.

July, 2007

— Authors

(vi)

Syllabus

Remedial Mathematics Pharmacy- Ist Semester Uttar Pradesh Technical University, Lucknow

PHARM- III M

Unit-I: Algebra: Equations reducible to quadratics, Cramer's Rule Algebra of matrics. 1.

Addition

2.

Subtraction

3.

Multiplication

4.

Inverse of matrices

Simultaneous equation by matrices. Unit-II: Certain co-ordinates, distance between two parts, area of triangle, locus of points, straight line, intercept form in normal. Unit-III: Differential Calculus 1.

Limit and function

2.

Definition and formulation of differential calculus.

3.

Rules of standard form of differential calculus.

4.

Chain Rule, Parametric rule.

Unit-IV: 1.

Standard form of Integral calculus

2.

Partial fraction of Integral

3.

Trigonometric function of Integral calculus.

Unit-V: Linear differential equation of order greater than one with constant coefficient complimentary function and particular Integral (xn, eax ) Trignometric (sin/cos ex).

(vii)

CONVERSION TABLES LENGTH (LINEAR MEASURE)

MEASURE OF CAPACITY

1 centimetre = 10 millimetres

1 centilitre

= 10 millilitres

1 decimetre

= 10 centimetres

1 decilitre

= 10 centilitres

1 metre

= 10 decimetres

1 litre

= 10 decilitres

1 decametre

= 10 metres

1 decalitre

= 10 litres

1 hectometre = 10 decametres

1 hectolitre

= 10 decalitres

1 kilometre

1 kilolitre

= 10 hectolitres

= 10 hectometres

WEIGHT MEASURE

AREA (SQUARE MEASURE)

1 centigram

= 10 milligrams

1 square decimetre = 100 square centimetres

1 decigram

= 10 centigrams

1 square metre

= 100 square decimetres

1 gramme

= 10 decigrams

1 are

= 100 square metres

1 decagram

= 10 grams

1 hectare

= 100 ares

1 hectogram = 10 decagrams 1 kilogram

1 square kilometre = 100 hectares

= 10 hectograms

VOLUME (CUBIC MEASURE)

1 myriagram = 10 kilograms

1 cubic decimetre

1 quintal

= 100 kilograms

1 cubic metre or stere = 1000 cubic decimetres

1 tonne

= 1000 kilograms

= 1000 cubic centimetres

ABBREVIATIONS mm

millimetre

ml millilitre

g

cm centimetre

l

kg kilogram

m metre

kl kilolitre

a

km kilometre

mg milligram

ha hectare

litre

(viii)

[I]

gramme are

B rief C ontents Dedication.....................................................................(III-IV) Preface to the Revised Edition...............................................(v) Preface to the First Edition..................................................(vi) Syllabus...........................................................................(vii) Conversion Tables .....................................................(viii) Brief Contents...................................................................(ix) Detailed Contents..........................................................(x-xII)

Unit-I Chapter 1: Linear and Quadric Equations...........................................(01-48) Chapter 2: Determinants....................................................................(49-86) Chapter 3: Matrices..........................................................................(87-130)

Unit-II Chapter 4: Trigonometry................................................................(131-194) Chapter 5: Logarithms....................................................................(195-214) Chapter 6: System of Coordinates..................................................(215-234) Chapter 7: The Straight Line...........................................................(235-262)

Unit-III Chapter 8: Functions and Limits.....................................................(263-302) Chapter 9: Differentiation...............................................................(303-340)

Unit-V Chapter 10: Linear Differential Equations.........................................(341-392) Tables.............................................................................................(393-396)

Unit-IV Appendix: Integration...........................................................................(01-32) (ix)

D etailed C ontents Chapter 1: Linear and Quadric Equations................................................................(01-48) 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16

Linear Equation 01 Linear Equations in Two Variables 03 Simultaneous Linear Equations in Two Variables 03 Consistent and Inconsistent Systems of Linear Equations 03 Graphical Method of Solving Simultaneous Linear Equations 03 Algebraic Methods of Solving Simultaneous Linear Equations 06 General Solution and Conditions for Solvability 08 Homogeneous Linear Equations 09 Quadratic Equation 12 Solving a Quadratic Equation 12 Nature of the Roots of a Quadratic Equation 15 Relations between Roots and Coefficients 17 Formation of a Quadratic Equation with Given Roots 18 Solution of Equations Reducible to Quadratic Forms 21 Simultaneous Equations in Two Variables 36 Applied Problems on Quadratic Equations 40

Chapter 2: Determinants ............................................................................................(49-86) 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

Determinants of Order 2 49 Determinants of Order 3 50 Determinants of Order 4 51 Minors and Cofactors 51 Properties of Determinants 54 Product of Two Determinants of The Same Order 57 Working Rule for Finding the Value of a Determinant 57 Symbols and Notations to be Employed for Finding the Values of a Determinant 58 Applications of Determinants to Coordinate Geometry 71 Applications of Determinants in Solving a System of Linear Equations 74 System of Linear Non-homogeneous Equations in Two Unknowns (Cramer’s Rule) 75 System of Linear Non-homogeneous Equations in Three Unknowns (Cramer’s Rule) 77

Chapter 3: Matrices...................................................................................................(87-130) 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12

Matrix 87 Special Types of Matrices 88 Submatrices of a Matrix 90 Equality of Two Matrices 90 Addition of Matrices 93 Properties of Matrix Addition 93 Multiplication of a Matrix by a Scalar 94 Properties of Multiplication of a Matrix by a Scalar Multiplication of Matrices 99 Properties of Matrix Multiplication 100 Transpose of a Matrix 107 Properties of Transpose of a Matrix 107

(x)

94

3.13 3.14 3.15 3.16

Symmetric and Skew-symmetric Matrices 107 Adjoint of a Square Matrix 112 Inverse or Reciprocal of a Matrix 113 Solving Systems of Linear Equations using Inverse of a Matrix

121

Chapter 4: Trigonometry........................................................................................(131-194) 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18

Angles 131 Quadrants 132 Systems of Measurement of Angles 132 Sexagesimal System (English System or Degree Measure) 132 Circular System (or Radian Measure) 133 Relation between Degrees and Radians 133 Centesimal System (French System or Grade Measure) 134 Trigonometric Ratios or Functions 139 Fundamental Trigonometric Identities 141 Values of Trigonometric Functions at Some Special Angles 148 Signs of the Values of Trigonometric Functions in Different Quadrants Trigonometric Ratios of Allied Angles 149 Trigonometric Ratios of Compound Angles 156 Transformation Formulae 165 Trigonometric Ratios of Multiple Angles 171 Trigonometric Ratios of Sub-multiple Angles 172 Trigonometric Ratios of Angles of 18°, 36°, 54° and 72° 173 Some Important Trigonometric Identities Involving the Angles of a Triangle

149

185

Chapter 5: Logarithms............................................................................................(195-214) 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12

Logarithm 195 Properties of Logarithms 196 Fundamental Laws of Logarithms 196 Transformation of Logarithm from One Base to Another Common Logarithms 197 Standard Form of Decimal 202 Characteristic and Mantissa of a Common Logarithm To find the Characteristic 203 To find the Mantissa 204 Use of Log Tables 205 Anti-logarithm 205 Applications of Logarithms 209

197

203

Chapter 6: System of Coordinates .......................................................................(215-234) 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

Introduction 215 Rectangular Cartesian Coordinates 215 Sign Convention 216 Distance between Two Points 216 Properties of some Geometrical Figures 217 Section Formulae 221 Area of a Triangle 225 Area of a Quadrilateral 225 Locus and its Equation 229

Chapter 7: The Straight Line .................................................................................(235-262) 7.1 7.2 7.3 7.4

Straight Line 235 Equations of Lines Parallel to the Coordinate Axes Slope or Gradient of a Non-vertical Line 236 Intercepts of a Line on the Axes 236

(xi)

235

7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14

Various Forms of Equations of a Line 236 Reduction of General Form to Standard Forms 244 Angle between Two Lines 247 Condition for Two given Lines to be Parallel 248 Condition for Two given Lines to be Perpendicular 248 Parallel Lines 249 Perpendicular Lines 249 The Length of Perpendicular 253 Distance between Two Parallel Lines 254 Equations of the Bisectors of the Angles between Two given Lines

257

Chapter 8: Functions and Limits...........................................................................(263-302) 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11

Function 263 Types of Functions 268 Composite Functions 269 Graphs of Functions 272 Limit of a Function at a Point 275 Algebra of Limits 276 Some Important Expansions 277 Some Important Properties of Limits Some Standard Limits 278 Direct Substitution Method 278 Factorisation Method 280

277

Chapter 9: Differentiation......................................................................................(303-340) 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14

The Differential Coefficient 303 Some Standard Results 304 List of Standard Results to be Committed to Memory 307 Differential Coefficient of the Sum of Two Functions 307 Differential Coefficient of the Product of Two Functions 309 Differential Coefficient of the Quotient of Two Functions 309 Differential Coefficient of a Function of a Function (Chain Rule of Differentiation) 312 Differential Coefficients of Inverse Trigonometric Functions 318 Logarithmic Differentiation 320 Implicit Functions 323 Parametric Equations 327 Differentiation of a Function with Respect to a Function 328 Differentiation of Infinite Recurring Expressions 331 Derivatives of Second and Higher Order (Successive Differentiation) 333

Chapter 10: Linear Differential Equations ...........................................................(341-392) 10.1 10.2 10.3 10.4 10.5

Differential Equation and its Solution 341 Linear Differential Equation 342 Determination of Complementary Function (C.F.) The Particular Integral (P.I.) 349 Particular Integral in some Special Cases 349

10.6

To Find P.I. when Q = e ax V , where V is any Function of x

10.7

To Find P.I. when Q = e ax and F ( a ) = 0

10.8

To Find P.I. when Q = sin ax or cos ax and F ( − a2 ) = 0

343

363

368

10.9

To Find P.I. when Q = xV , where V is any Function of x 1 1 and , α being a Constant 10.10 The Operator D−α D+α

373 379 386

Tables ........................................................................................................................(393-396) Appendix: Integration..................................................................................................(01-32) (xii)

1

1 L inear a nd Q uadratic E quations

1.1 Linear Equation n equation which contains variables of degree one only is called a linear equation. z −5 z −3 1 For Example : (i) 3 ( x − 1) = 8 (ii) − = 2 5 2 2+n 6y −5 7 (iii) (iv) =5 = 3+n 2y 9

A

Operations used for Solving a Linear Equation. (i)

We can add or subtract the same number on both sides of the linear equation.

(ii)

We can multiply both sides of the linear equation by the same number.

(iii) We can divide both sides of the equation by the same non-zero number.

2

(iv) Cross-multiplication : If

ax + b cx + d

l then m (ax + b) = l (cx + d). This m

=

operation is called cross-multiplication. (v)

Transposition : Any term of a linear equation can be taken to other side by changing its sign. This operation is called transposition.

Illustration 1 : Solution : or or or

Solve :

We have

x+6

4 x+6

+

x−3

5 x−3

2 (9 x + 18) = 5 (5 x − 4)

or

18 x + 36 = 25 x − 20

or

25 x − 18 x = 36 + 20 7 x = 56 56 x= = 8. 7

or ∴

or

Solve :

We have

3x + 5 4x + 2

3x + 5 4x + 2

=

=

3x + 4 4x + 7

[By cross-multiplication] [By transposition]

.

4x + 7

(3 x + 5) (4 x + 7) = (4 x + 2) (3 x + 4) 2

12 x

or

41x + 35 = 22 x + 8

or

41x − 22 x = 8 − 35

or

19 x = − 27 27 . x=− 19 27 . x=− 19

∴

.

3x + 4

or

or

8 5x − 4

x = 8.

Illustration 2 : Solution :

5x − 4

+ = 4 5 8 5 ( x + 6) + 4 ( x − 3) 5 x − 4 = 20 8 5 x + 30 + 4 x − 12 5 x − 4 = 5 2 9 x + 18 5 x − 4 = 5 2

or

or

=

+ 21x + 20 x + 35 = 12 x

2

[By cross-multiplication]

+ 16 x + 6 x + 8 [By transposition]

3

1.2 Linear Equations in Two Variables An equation which contains two variables of degree one only is called a linear equation in two variables. An equation of the form, ax + by + c = 0, where a, b, c are real numbers (a ≠ 0 , b ≠ 0) is called a linear equation in two variables x and y. For example : (i) 5 x + 4 y − 9 = 0 (iii) 0.3 m + 0.2 n = 27

(ii) 8 x + 2 y = 1 (iv) 3 p − 2q = 0

1.3 Simultaneous Linear Equations in Two Variables When two or more than two linear equations in two unknowns x and y are satisfied by the same pair of values of x and y then these linear equations are called simultaneous linear equations. For example : The system of linear equations x+ y=2 x− y=0 is satsified by the values x = 1 and y = 1.

1.4 Consistent and Inconsistent Systems of Linear Equations Any two simultaneous linear equations in two variables x and y are said to be consistent if both equations can be satisfied by the same set of values of x and y. If there does not exist any pair of values of x and y which satisfy both the equations then the two equations are said to be inconsistent.

1.5 Graphical Method of Solving Simultaneous Linear Equations We know that every linear equation represents a straight line. On solving two simultaneous linear equations graphically the following three cases are possible : Case I. When the lines intersect then the given equations are consistent and there exists a unique solution of two linear equations. The co-ordinates of the point of intersection give the unique solution. Case II. When the lines are parallel then the given equations are inconsistent and have no solution. Case III. When the lines are coincident then the given equations are said to be dependent equations and there are infinitely many solutions.

4

Illustration 1 :

Solve graphically the following system of linear equations : 2 x − y = 2 , 4 x − y = 8.

Also find the coordinates of the points where the lines meet the axis of x. The equation 2 x − y = 2 ⇒ y = 2 x − 2

...(I)

Here

x=0 ⇒ y=2×0 −2=−2

and

x = 2 ⇒ y = 2 × 2 − 2 = 4 − 2 = 2. Y

y

−2

2

Plot the points A (0 , − 2) and B (2, 2). Join AB and produce the line on both sides to get the graph of the equation 2 x − y = 2.

7 6 5 4 3 2 X′

Further, equation 4 x − y = 8 ...(2)

⇒

y = 4x − 8

Here

x=0 ⇒ y=4×0 −8=−8

and

x = 1 ⇒ y = 4 × 1 − 8 = 4 − 8 = − 4.

(2, 2)

1 0

1 2 3 4 5 6 –4 –3 –2 –1 –1 (0, –2) –2 A –3 D(1, –4) –4 –5 –6 –7 –8 C(0, –8) –9 –10 –11

X

2x

For (2), the table is

(3, 4) 8

2

=2

0

–y

x

y=

For (1), the table is

4x –

Solution :

x

0

1

y

−8

−4

Y′

Plot the points C (0 , − 8) and D (1 , − 4). Join CD and produce the line on both sides to get the graph of 4 x − y = 8. From the graph we see that the two lines intersect at the point (3, 4). Hence x = 3 , y = 4 is the solution of the equations 2 x − y = 2 and 4 x − y = 8. Also, from the graph of 2 x − y = 2 we see that it intersects the x-axis at the point (1 , 0). Similarly the graph of 4 x − y = 8 shows that this line intesects the x-axis at the point (2, 0). Illustration 2 :

Solve graphically the following system of equations 2 x + y = 6 , x − 2 y = − 2.

Also, find the coordinates of the points where the lines meet the axis of x.

5 Y

Solution :

9 8 7 6 5 4 3 2

Table for 2 x + y = 6 x

1

3

y

4

0

Table for x − 2 y = − 2 4

y

1

3

1 (0, 1) (3, 0) 0 X 3 1 2 –3 –2 –1 4 5 –7 –6 –5 –4 –1 2 =– –2 2y – x –3 –4 –5 –6 (–2, 0)

X′

+y

0

=6

The graph shows that the lines intersect at the point (2, 2). Hence x = 2 , y = 2 is the required solution. Further, the line 2 x + y = 6 meets the x-axis at (3, 0). And, the line x − 2 y = − 2 meets the x-axis at (− 2 , 0). Illustration 3 :

(4, 3) (2, 2)

2x

x

(1, 4)

Y′

Solve the following system of equations graphically :

x − 2 y = 6 , 3 x − 6 y = 0. x−6 We have x − 2 y = 6 ⇒ y = 2

Solution :

Here x=0 ⇒ y=−3 x=6 ⇒ y=0

and

Y 5 4 3 2

Thus, the table is x

0

6

y

−3

0

X′

Plot the points A (0, − 3) and B (6, 0). Join the points A and B and extend the line on both sides to get the graph of x − 2 y = 6. x Again, 3x − 6 y = 0 ⇒ y = 2

C(4, 2)

1 0

B X –4 –3 –2 –1 (0, 0)1 2 3 4 5 6 7 8 –1 (6, 0) –2 –3 A(0, –3) –4 –5 Y′

Here x = 0 ⇒ y = 0 and x = 4 ⇒ y = 2. x

0

4

y

0

2

Plot the points O (0 , 0) and C (4 , 2). Join the points O and C and extend the line on both sides to get the graph of the line 3 x − 6 y = 0. From the graph we see that the two lines are parallel to each other and do not intersect. Hence the given system of equations is inconsistent and there is no solution.

6

1.6 Algebraic Methods of Solving Simultaneous Linear Equations (1) Substitution Method Step 1. Find the value of any variable in terms of the other variable from any one equation. Step 2. Substitute this value in the other equation and solve the simple equation obtained and find the value of one variable. Step 3. Substitute the value of the variable obtained in any one equation to get the value of the other variable.

Illustration 1 : Solve for x and y x+ y=3 and

2 x + 5 y = 12.

Solution : We have x + y = 3

...(1)

2 x + 5 y = 12 From equation (1), we have

...(2) ...(3)

x=3− y Putting this value of x in equation (2), we get 2 (3 − y) + 5 y = 12 or

6 − 2 y + 5 y = 12

or

6 + 3 y = 12 6 y = = 2. 3

or

or

3 y = 12 − 6 = 6

Putting y = 2 in equation (3), we get x = 3 − 2 = 1. Hence

x = 1 , y = 2.

(2) Elimination Method. Step 1. Multiply the equations by suitable non-zero constants so as to make equal the coefficients of the variable to be eliminated. Step 2. Add the equations obtained if the terms with equal coefficients are of opposite sign and subtract if they are of the same sign. Step 3. Solve the equation obtained after eliminating one variable to get the value of the other variable . Step 4. Substitute the value of the variable obtained above in any one of the given equations and find the value of the other variable.

7

Illustration 1 : Solve for x and y 3 x + 4 y = 17 2 x + 9 y = 5.

and Solution : We have

...(1)

3 x + 4 y = 17 2x + 9 y = 5 Multiplying equation (1) by 2 and equation (2) by 3, we get

...(2) ...(3)

6 x + 8 y = 34 6 x + 27 y = 15 Subtracting equation (4) from equation (3), we get

...(4)

− 19 y = 19 or y = − 1 Putting y = − 1 in equation (1), we get 3 x − 4 = 17 or or

x = 7.

Hence

x = 7 , y = − 1.

3 x = 17 + 4 = 21

(3) Special Case Method When the two given equations are such that the coefficients of the two variables in one equation are interchanged in the other equation then we add and subtract the given equations to obtain two new equations of the type x + y = a and x − y = b. We can now solve x + y = a and x − y = b easily.

Illustration 1 : Solve for x and y 47 x + 31 y = 63 and

31x + 47 y = 15.

Solution : We have and

...(1)

47 x + 31 y = 63 31x + 47 y = 15.

...(2)

Adding equations (1) and (2), we get 78 x + 78 y = 78 or

78 ( x + y ) = 78

or

x + y =1 Subtracting equation (2) from equation (1) , we get 16 x − 16 y = 48 or 16 ( x − y) = 48

...(3)

8

or

...(4)

x− y=3 Adding equations (3) and (4),we get 2 x = 4 or

x = 2.

Putting x = 2 in equation (3), we get 2 + y = 1 or Hence

y = − 1.

x = 2 , y = − 1.

Illustration 2 : Solve for x and y 1 1 − = −1 2x y and

1 1 + = 8. x 2y

Solution : We have 1 1 − = −1 2x y

...(1)

1 1 + =8 x 2y

...(2)

Multiplying equation (2) by 2, we get 2 1 + = 16 x y Adding equations (1) and (3), we get 1 2 or + = 15 2x x Putting x =

or Hence

...(3)

5 = 15 2x

or

x=

5 1 = . 2 × 15 6

1 in equation (1), we get 6 1 1 1 1 − = − 1 or = +1 1 1 y y 2× 2× 6 6 1 1 = 3 +1= 4 or y= . y 4 x=

1 1 , y= ⋅ 6 4

1.7 General Solution and Conditions for Solvability Let a1 x + b1 y = c1 and a2 x + b2 y = c 2 be the simultaneous equations. a b Case I. If 1 ≠ 1 , then the given system has a unique solution. a2 b2 x=

b2 c1 − b1 c 2 a1 b2 − a2 b1

and

y=

c 2 a1 − c1 a2 a1 b2 − a2 b1

9

Thus solution is given by y x −1 = = b1 c 2 − b2 c1 c1 a2 − c 2 a1 a1 b2 − a2 b1 In this case the two equations are said to be consistent. Cross Multiplication Method : The above solution can be easily obtained with the help of the following diagram : x

y

−1

b1

c1

a1

b1

b2

c2

a2

b2

The arrows between two numbers mean that the numbers are to be multiplied. First multiply the numbers with downward arrow and from their product subtract the product of the numbers with upward arrow. a b c Case II. If 1 = 1 = 1 , then the given equations have infinite number of a2 b2 c2 solutions and the two equations are said to be depended equations. a b c Case III. If 1 = 1 ≠ 1 , then the system has no solution and the two equations a2 b2 c2 are said to be inconsistent.

1.8 Homogeneous Linear Equations If c1 = c 2 = 0, then the equations become a1 x + b1 y = 0 and a2 x + b2 y = 0. a b (i) 1 ≠ 1 ⇒ x = 0 , y = 0 is only solution, and a2 b2 (ii)

a1 a2

=

b1 b2

⇒

the system has infinitely many solutions.

Illustration 1 : In each of the following system of equations, determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it. (i)

3 x − 7 y = 10 , 2 x − y = 3 9 15 (ii) − 3 x + 4 y = 5 , x −6y = 2 2 35 (iii) 4 x + 7 y = 10 , 10 x + y = 25 2

10

Solution :

(i) We have 3 x − 7 y = 10

or

3 x − 7 y − 10 = 0

2 x − y = 3 or 2 x − y − 3 = 0 These equations are of the form

...(1) ...(2)

a1 x + b1 y + c1 = 0 , a2 x + b2 y + c 2 = 0 Here

a1 = 3 , b1 = − 7 , c1 = − 10 , a2 = 2 , b2 = − 1, c 2 = − 3 a1 3 b −7 7 c1 −10 10 . = , 1 = = , = = a2 2 b2 −1 1 c 2 −3 3

∴ Since

a1 a2

≠

b1 b2

, therefore, a unique solution exists. Now by cross multiplica-

tion, we have x

y

−1

−7

10

3

−7

−1

3

2

−1

∴

y x −1 = = (− 7) × 3 − (− 1) × 10 10 × 2 − 3 × 3 3 × (− 1) − 2 × (− 7)

or

y x −1 = = − 21 + 10 20 − 9 − 3 + 14

or

y −1 x = = −11 11 11 x −1 11 = or x= = 1, −11 11 11 y −1 −11 = or y = = −1 11 11 11

Hence,

x = 1 , y = − 1.

or

(ii) We have − 3 x + 4 y = 5 9x 15 −6y = 2 2 or

or or

− 3x + 4 y − 5 = 0 , 9 x − 12 y = 15

9 x − 12 y − 15 = 0

These equations are of the form a1 x + b1 y + c1 = 0 and a2 x + b2 y + c 2 = 0 Here ∴ ⇒

a1 = − 3 , b1 = 4 , c1 = − 5 , a2 = 9 , b2 = − 12 , c 2 = − 15 a1 −3 1 b 4 1 c 5 1 = =− , 1 = =− , 1 = = a2 9 3 b2 −12 3 c 2 15 3 a1 a2

=

b1 b2

≠

c1 c2

11

Hence, the given equations are inconsistent haiving no solution. (iii) We have 4 x + 7 y = 10 or 4 x + 7 y − 10 = 0 , 35 10 x + y = 25 or 20 x + 35 y = 50 2 or

20 x + 35 y − 50 = 0

Here

a1 = 4 , b1 = 7 , c1 = − 10 , a2 = 20 , b2 = 35 , c 2 = − 50 a1 4 1 b 7 1 c 10 1 = = , 1 = = , 1 = = a2 20 5 b2 35 5 c 2 50 5

∴ ⇒

a1 a2

=

b1 b2

=

c1 c2

.

Hence the given system of equations has infinitely many solutions and thus has no unique solution.

Comprehensive Exercise 1 1. Solve the following system of simultaneous linear equations graphically : (i) x + y = 3 , 2 x + 5 y = 12 (ii) 3 x − 5 y = − 1 , 2 x − y = − 3 (iii) x − 2 y = 5 , 2 x − 4 y = 10 35 (iv) 2 x + 7 y = 11 , 5 x + y = 25 2 (v) 3 x + 2 y = 14 , − x + 4 y = 7 (vi) 4 x + 6 y = 9 , 4 x − 2 y = − 11 2.

Solve the following system of equations : (i) 3 x − 5 y = − 1, x − y = − 1 (ii) 11x + 15 y + 23 = 0 , 7 x − 2 y − 20 = 0 y y y x x 5x (iii) + (iv) + = 3 , 2x − y = 4 =3, − =4 2 4 3 4 6 8 x+ y x− y x+ y x− y 8 3 6 5 (v) + (vi) = 1, − =8 − = 1, + = 12 x y x y 4 3 2 6

3. In each of the following system of equations, find whether it has a unique, infinite or no solution. (i) 2 x − y = 3 , 6 x − 3 y = 2 (ii) 7 x + 6 y = 13 , 2 x + 3 y = 4 (iii) x − 5 y + 3 = 0 , 3 x − 15 y + 9 = 0 8 (iv) 5 x + 2 y = 12 , 4 x + y=3 5 5 (v) 3 x + y = 13 , 2 x − 3 y = 5 8

12

(vi) 2 x −

10 y = 4 , 3x − 5 y = 9 3

A nswers 1.

(i) 1 , y = 2

(ii)

2.

(iii) infinite solutions 5 (v) x = 3 , y = 2 (i) x = − 2 , y = − 1

(iv) no solution −3 5 (vi) x = , y= 2 2 (ii) x = 2 , y = − 3 11 14 (iv) x = , y= 2 3 (vi) x = 16 , y = 4 (ii) unique solution (iv) no solution (vi) no solution

(iii) x = 4 , y = 4 3.

(v) x = 2 , y = − 1 (i) no solution (iii) infinite solutions (v) unique solution

x = − 2, y = −1

1.9 Quadratic Equation An equation of the form ax 2 + bx + c = 0 , where a, b, c are real numbers and a ≠ 0 , is called a quadratic equation with real coefficients. Here, x is a variable and a, b, c are given real numbers which are called coefficients of the equation. The coefficient of x 2 is a, the coefficient of x is b and c is called the constant term of the equation. Roots of an Equation. The values of the variable satisfying a given equation are called its roots. Thus, x = α is a root of the equation f ( x) = 0 if f (α) = 0. Solution Set of a Given Equation. The set of all roots of an equation , in a given domain, is called the solution set of the equation.

1.10 Solving a Quadratic Equation (I)

Solving a Pure Quadratic Equation. The equation of the form ax 2 + c = 0,

where a ≠ 0, is known as a pure quadratic equation. Obviously, ax 2 + c = 0

⇒ ax 2 = − c ⇒ x 2 = −

c a

⇒ x=±

−

c ⋅ a

The equation ax 2 + bx + c = 0, where a ≠ 0, and b ≠ 0, is called a complete quadratic equation.

13

Illustration 1 : Solution :

Solve : 16 x 2 − 25 = 0.

We have 16 x 2 − 25 = 0 16 x 2 = 25

⇒

x2 =

⇒

25 5 =± ⋅ 16 4 5 Hence, the solution set is  , − 4 x=±

⇒

(II)

25 16

5 ⋅ 4

Method of Factorization for the Solution of a Quadratic Equation.

Illustration 2 :

Solve : (i) x 2 − 5 x + 4 = 0, (ii) abx 2 − (a2 + b 2 ) x + ab = 0,

(iii) a ( x 2 + 1) = x (a2 + 1). Solution :

(i) We have x 2 − 5 x + 4 = 0

⇒

x2 − 4 x − x + 4 = 0

⇒

x ( x − 4) − 1 ( x − 4) = 0

⇒

( x − 4) ( x − 1) = 0

⇒

x−4=0

or

x −1= 0

⇒ x = 4 or x = 1. Hence, the roots of the given equation are 4, 1. (ii) We have abx 2 − (a2 + b 2 ) x + ab = 0 ⇒

abx 2 − a2 x + b 2 x + ab = 0

⇒

ax (bx − a) − b (bx − a) = 0

(bx − a) (ax − b) = 0 or ⇒ bx − a = 0 ax − b = 0 a b or ⇒ x= ⋅ b a a b Hence, the roots of the given equation are , ⋅ b a ⇒

(iii) We have ax 2 − (a2 + 1) x + a = 0 ⇒

ax 2 − a2 x − x + a = 0

⇒

ax ( x − a) − ( x − a) = 0

⇒

( x − a) (ax − 1) = 0 or ( x − a) = 0 1 or x=a ⋅ a

⇒ ⇒

(ax − 1) = 0

14

1 ⋅ a (III) Method of Completing the Square for the Solution of a Quadratic Equation (Formula for the Solution of a Quadratic Equation). Hence, the roots of the given equation are a,

(Shreedharacharyya’s Method). Let the given quadratic equation be ax 2 + bx + c = 0 ,where a, b, c ∈ R and a ≠ 0. Then roots of the equation ax 2 + bx + c = 0, where a ≠ 0, are x=

−b±

b 2 − 4ac 2a

i. e. , x =

b 2 − 4ac

−b+

2a

(i) Solve : 2 x 2 − 4 x + 3 = 0,

Illustration 1 :

and x =

b 2 − 4ac

−b−

2a

⋅

(ii) 25 x 2 + 30 x + 7 = 0,

(iii) 4 x 2 − 4ax + (a2 − b 2 ) = 0. Solution ax

2

:

(i) Comparing

the

given

equation

2 x2 − 4 x + 3 = 0

with

+ bx + c = 0, we have a = 2, b = − 4 and c = 3.

Using the formula for the solution of a quardratic equation, we have x=

=

b 2 − 4ac

−b±

2a 4 ± 16 − 24 4

=1± i

2 2

=1±

− (− 4) ± (− 4)2 − 4 × 2 × 3

=

2×2

4±

=

−8 4

1

=

2±

−2 2

=

2±i 2 2

i.

2 1

Hence, the roots of the given equation are 1 +

i and 1 −

2

1

i.

2

(ii) Comparing the given equation 25 x 2 + 30 x + 7 = 0 with ax 2 + bx + c = 0, we have a = 25, b = 30 and c = 7. Using the formula for the solution of a quadratic equation, we have x=

= =

−b±

b 2 − 4ac 2a

=

− 30 ± 900 − 700 50 − 30 ± 10 2 50

=

− 30 ± (30)2 − 4 × 25 × 7 2 × 25 =

− 30 ± 200

−3± 2 5

50 ⋅

15

 − 3 + 2 − 3 − 2  Hence, the solution set of the given equation is  , ⋅ 5 5   (iii) Comparing Ax

2

the

given

4 x 2 − 4ax + (a2 − b 2 ) = 0

equation

2

with

2

+ Bx + C = 0, we have A = 4, B = − 4a and C = (a − b ).

Using the formula for the solution of a quardratic equation, we have x=

B2 − 4 AC

−B ±

2A 2

=

=

− (− 4a) ± (− 4a)2 − 4 × 4 × (a2 − b 2 )

2

2×4 2

4a ± 16a − 16(a − b ) 8

=

4a ± 16b 2

=

4a ± 4b

8 8  a + b a − b Hence, the solution set of the given equation is  , . 2   2

⋅

Comprehensive Exercise 2 Solve the following equations : 1.

x 2 + 25 = 0.

2. 16 x 2 − 49 = 0.

3.

x 2 − 8 x − 48 = 0.

4. 8 x 2 − 22 x − 21 = 0.

5.

70 x − 63 = 7 x 2 .

6. 2 x 2 − 11x = − 5.

7.

6 x 2 + 7 x − 10 = 0.

8. 2 x 2 − 10 x = 3 x − 15.

9.

x2 + 5 x + 4 = 0

10.

1 1 1 1 = + + p+ q + x p q x

A nswers 1.

± 5i.

2.

7 ± . 4

6.

1 , 5. 2

7.

− 2,

5 ⋅ 6

3.

12, − 4.

4.

7 3 ,− ⋅ 2 4

8.

3 , 5. 2

9.

− 4,−1

5.

1, 9.

10. − p , − q

1.11 Nature of the Roots of a Quadratic Equation Let the quadratic equation be ax 2 + bx + c = 0, where a, b, c ∈ R and a ≠ 0.

...(1)

16

We know that if α, β are the roots of (1), then α=

−b+

b 2 − 4ac 2a

and β =

−b−

b 2 − 4ac 2a

⋅

Discriminant of a Quadratic Equation. The expression ‘ b 2 − 4ac’ upon which the nature of the roots of the quadratic equation ax 2 + bx + c = 0 depends is called the discriminant of this equation and is denoted by the Greek letter ∆ . Thus, ∆ = b 2 − 4ac = discriminant of ax 2 + bx + c = 0. The roots of the equation ax 2 + bx + c = 0 are (i) real if and only if ∆ ≥ 0, (ii) real and equal if and only if ∆ = 0, and (iii) imaginary and unequal if and only if ∆ < 0.

Illustration 1: solving them :

Find the nature of the roots of the following equations without actually

(i ) 3 x 2 − 8 x − 3 = 0

(ii ) 9 x 2 − 6 x + 1 = 0 (iii ) 2 x 2 + x + 5 = 0.

(i) The given equation is 3 x 2 − 8 x − 3 = 0.

Solution :

Its discriminant ∆ = (− 8)2 − 4 × 3 × (− 3) = 64 + 36 = 100 > 0. ∴

The roots of the given equation are real and unequal.

Since ∆ = 100 = (10)2 is a perfect square and the coefficients of the equation are all rational, therefore the roots are rational and unequal. The given equation is 9 x 2 − 6 x + 1 = 0. Its discriminant

(ii)

∆ = (− 6)2 − 4 × 9 × 1 = 36 − 36 = 0. ∴

The roots of the equation are real and equal. The given equation is 2 x 2 + x + 5 = 0.

(iii)

Its discriminant ∆ = 12 − 4 × 2 × 5 = − 39 < 0. ∴

The roots of the given equation are imaginary and unequal.

Illustration 2: x Solution : Now, ⇒

2

For what value of m will the equation

− 2 (1 + 3m) x + 7 (3 + 2m) = 0 have equal roots ?

The given equation will have equal roots if its discriminant ∆ = 0. ∆ = 0 ⇒ 4 (1 + 3m)2 − 4 × 7 (3 + 2m) = 0 (1 + 3m)2 − 7 (3 + 2m) = 0

17

⇒

9m2 − 8m − 20 = 0

⇒

9m2 − 18m + 10 m − 20 = 0

⇒

9m (m − 2) + 10 (m − 2) = 0

⇒

(m − 2) (9m + 10) = 0 10 m = 2 or m = − ⋅ 9

⇒

Hence, the roots will be equal for m = 2 or Illustration 3 :

m=−

10 ⋅ 9

The roots of the quadratic equation 5 x 2 − px + 1 = 0 are real and

distinct. Find the possible values of p. Solution :

The roots of the quadratic equation 5 x 2 − px + 1 = 0 are real and

distinct ⇒

its discriminant ∆ = (− p)2 − 4 × 5 × 1 > 0

⇒

p2 − 20 > 0

⇒

p2 > 20

⇒

| p | > 20

⇒

p > 20

⇒ or

| p |2 > 20 p < − 20 .

Hence, p may have any value less than − 20 or any value greater than 20 . Illustration 4 : Determine the positive values of k for which the equations x 2 + kx + 64 = 0 and x 2 − 8 x + k = 0 will both have real roots. Solution : i.e.,

The equation x 2 + kx + 64 = 0 will have real roots if its discriminant ≥ 0 if k 2 − 4 × 1 × 64 ≥ 0

The equation x

2

i. e., if

k 2 ≥ 256 i.e.,

k ≥ 16.

[∵ k is + ve]

− 8 x + k = 0 will have real roots if its discriminant ≥ 0

i.e.,

if 64 − 4k ≥ 0 i.e., if 4k ≤ 64 i.e., if k ≤ 16. Thus, both the equations will have real roots if k ≥ 16 and k ≤ 16 i.e., if k = 16. Hence, the only +ve value of k for which both the given equations has real roots is 16.

1.12 Relations between Roots and Coefficients Let α and β be the roots of the equation ax 2 + bx + c = 0, where a, b, c ∈ R and a ≠ 0. Then,

α+β=−

and

αβ =

 coefficient of x  b  = −  2 a  coefficient of x 

c  constant term   ⋅ = a  coefficient of x 2 

...(1)

18

1.13 Formation of a Quadratic Equation with Given Roots The quadratic equation whose roots are α and β is given by ( x − α) ( x − β) = 0 i.e., x 2 − (α + β) x + αβ = 0 i.e.,

x 2 − (sum of the roots) x + product of the roots = 0.

Illustration 1 :

Without solving the equation 3 x 2 − 5 x − 8 = 0 , find the sum and the

product of its roots. Solution : Then, and

Let α, β be the roots of the equation 3 x 2 − 5 x − 8 = 0.  coefficient of x  (− 5) 5  =− α + β = −  = 2 3 3  coefficient of x  constant term −8 8 αβ = = =− ⋅ 2 3 3 coefficient of x

Illustration 2 :

If c , d are the roots of the equation ( x − a) ( x − b) − k = 0 , prove that

a, b are the roots of the equation ( x − c ) ( x − d ) + k = 0. Solution :

We have c , d are the roots of ( x − a) ( x − b) − k = 0

⇒

( x − a) ( x − b) − k = ( x − c ) ( x − d)

⇒

( x − c ) ( x − d ) + k = ( x − a) ( x − b ) a, b are the roots of ( x − c ) ( x − d) + k = 0. [∵ a and b are the roots of ( x − a) ( x − b) = 0 ]

⇒

Illustration 3 :

The roots α, β of the equation x 2 + kx + 12 = 0 are such that α − β = 1 ,

find k. Solution :

We have α + β = − k and αβ = 12.

Also, it is given that α − β = 1. Now, (α + β)2 − (α − β)2 = 4αβ ⇒

(− k )2 − 1 = 4 × 12 ⇒

⇒

k 2 = 49

Illustration 4 :

⇒

k 2 − 1 = 48

k = ± 7.

If α, β are the roots of the equation x 2 − 3 x + k = 0 , find k such that

α = 2β. Solution : and

We have α + β = − αβ =

k = k. 1

(− 3) 1

=3

19

Now,

α+β=3 ⇒

2β + β = 3

⇒ Also,

3β = 3

β = 1.

∴

k = αβ = 2 × 1 = 2.

⇒

[∵ α = 2β]

α + β = 3 and β = 1 ⇒ α + 1 = 3

Illustration 5 :

⇒ α = 2.

If 4 is a root of the equation x 2 − 5 x + k = 0, find the value of k and the

other root. Solution : Since 4 is a root of x 2 − 5 x + k = 0, therefore 42 − 5 × 4 + k = 0 Now, let β be the other root of x

⇒ k = 20 − 16 = 4. 2

...(1)

− 5 x + k = 0.

Since 4 and β are the roots of (1), therefore k 4 4 . β = = = 4 ⇒ β = 1. 1 1 Hence, k = 4 and the other root is 1. Illustration 6 : If α be a root of the equation 4 x 2 + 2 x − 1 = 0, prove that 4α 3 − 3α is the other root. Solution : Then

[UPTU 2002]

Given α be a root of the equation 4 x

2

+ 2 x − 1 = 0.

4α 2 + 2α − 1 = 0.

Let β be the other root of the given equation. 1 1 Then α+β=− ⇒ β = − −α. 2 2

...(1)

...(2)

We have to show that β = 4α 3 − 3α . Now

Hence,

1 1 (4α 2 + 2α − 1) − α − 2 2 1 1 [From (1)] = α ⋅0 − ⋅ 0 − α − , 2 2 1 [From (2)] =−α− =β. 2

4α 3 − 3α = α (4α 2 + 2α − 1) −

4α 3 − 3α is the other root of the given equation.

Illustration 7 :

Find the value of k for which the roots α, β of the equation x 2 − 6 x + k = 0

satisfy the relation 3α + 2β = 20 . Solution :

Since α, β are the roots of x 2 − 6 x + k = 0, therefore, α+β=6

...(1)

and

αβ = k.

...(2)

Now,

3α + 2β = 20

⇒

α + 2 × 6 = 20

⇒ α + 2 (α + β) = 20 [Using (1)]

⇒ α = 20 − 12 = 8. Putting α = 8 in (1), we get β = 6 − 8 = − 2. Hence, from (2) we have

k = αβ = 8 × (− 2) = − 16.

20

Illustration 8 : Solution :

Find the quadratic equation whose roots are 8 and − 6.

We have, sum of the roots = 8 + (− 6) = 8 − 6 = 2

and product of the roots = 8 × (− 6) = − 48. Hence, the required equation is x 2 − (sum of the roots) x + product of the roots = 0 x 2 − 2 x − 48 = 0.

i. e.,

Illsutration 9 :

Form a quadratic equation with rational coefficients whose one root is

2+ 5 . Solution :

Let the roots of the required equation be α and β.

Let

α=2+ 5 .

Then

β=2− 5

[∵ In an equation with rational coefficients, irrational roots occur in conjugate pairs]

∴

α + β = (2 + 5 ) + (2 − 5 ) = 4

and

αβ = (2 + 5 ) (2 − 5 ) = 22 − ( 5 )2 = 4 − 5 = − 1.

Hence, the required equation is x 2 − 4 x − 1 = 0. Illustration 10 : Solution :

Form a quadratic equation with real coefficients whose one root is 2 − 3i..

Let the roots of the required equation be α and β.

Let

α = 2 − 3i.

Then

β = 2 + 3i.

∴ and

α + β = (2 − 3i ) + (2 + 3i ) = 4

[∵ In an equation with real coefficients, complex roots occur in conjugate pairs]

αβ = (2 − 3i ) (2 + 3i ) = 22 + 32 = 13.

Hence, the required equation is x 2 − 4 x + 13 = 0.

Comprehensive Exercise 3 1. Find the nature of the roots of the following equations without actually solving them : (i) 2 x 2 − 9 x + 7 = 0 (ii) 4 x 2 − 20 x + 25 = 0 (iii) 2 x 2 − x + 2 = 0 (v) 3 x

2

+ 7x + 2 = 0

(iv) x 2 + 2 3 (vi) 3 x

2

x −1= 0

− 2 x + 5 = 0.

2. For what values of m will the following equations have equal roots ? (i)

x 2 − 2 (5 + 2m) x + 3 (7 + 10 m) = 0

(ii) (m + 1) x 2 + 2 (m + 3) x + (m + 8) = 0

21

(iii) (2m + 1) x 2 + 2 (m + 3) x + (m + 5) = 0 (iv) mx 2 + mx + 1 = − 4 x 2 − x. 2 3. Without solving the equation 7 x − 8 x + 49 = 0, find the sum and the product of its roots.

4.

Find a quadratic equation whose roots are : (i)

3, 8

(ii) 9, − 11

(iii) 5 + 7 , 5 − 7

(iv) − 3 + i 5 , − 3 − i 5 .

5. Find a quadratic equation with rational coefficients whose one root is −4+ 7 . 6.

Find a quadratic equation with real coefficients whose one root is

1 ⋅ 2 − 3i

7. If α, β are the roots of the equation x 2 − 2 x + 3 = 0, find the equation whose roots are α + 2 and β + 2.

A nswers 1.

2.

(i) Rational and unequal (iii) Imaginary and unequal (v) Rational and unequal 1 (i) 2 , 2 (iii)

− 5 ± 41 2 8 , product = 7. 7 x 2 − 11x + 24 = 0

(ii) Real and equal (iv) Real and unequal (vi) Imaginary and unequal. 1 (ii) 3 (iv) 5, − 3.

3. Sum = 4.

(i)

(iii) x 2 − 10 x + 18 = 0 5.

x 2 + 8 x + 9 = 0.

7.

x 2 − 6 x + 11 = 0.

(ii)

x 2 + 2 x − 99 = 0

(iv) x 2 + 6 x + 14 = 0. 6. 13 x 2 − 4 x + 1 = 0.

1.14 Solution of Equations Reducible to Quadratic Forms Type I.

Algebraic Equations Reducible to Quadratic Forms.

The general form of such an equation is px 2 n + qx

n

+ r = 0, where p, q, r are

constant coefficients. Putting x

n

= y, so that x 2 n = ( x n )2 = y 2 , the given equation reduces to

py 2 + qy + r = 0, which is a quadratic in y. Now, we solve this equation for y and then we derive from it the values of x.

22

Illustration 1 : Solution :

Solve : x 4 − 8 x 2 − 9 = 0.

We have x 4 − 8 x 2 − 9 = 0

⇒

y2 − 8 y − 9 = 0

⇒

y2 − 9 y + y − 9 = 0

⇒

y ( y − 9) + 1 ( y − 9) = 0

⇒

[Putting x 2 = y]

( y − 9) ( y + 1) = 0

⇒

y−9=0

⇒ Now,

y = 9 or y = − 1. y=9

y +1= 0

or

⇒ x2 = 9

and

y = −1 ⇒ x

⇒

x=±

2

⇒ x=±3

= −1

− 1 = ± i.

Hence, the roots of the given equation are − 3, 3, i and − i. Illustration 2 : Solution :

Solve : 6 x 3 /4 = 7 x1 /4 − 2 x − 1 /4 .

We have 6 x 3 /4 = 7 x1 /4 − 2 x − 1 /4 2

⇒

6 x 3 /4 = 7 x1 /4 −

⇒

6 x = 7 x1 /2 − 2

[Multiplying both sides by x1 /4 ]

⇒

6 y2 = 7 y − 2

[Putting x1 /2 = y so that x = ( x1 /2 )2 = y 2 ]

⇒

6 y2 − 7 y + 2 = 0

⇒

6 y2 − 4 y − 3 y + 2 = 0

⇒

2 y (3 y − 2) − (3 y − 2) = 0

⇒

(3 y − 2) (2 y − 1) = 0 2 1 y= or y = ⋅ 3 2

⇒ Now, and Thus,

y=

2 3

x1 /4

⇒ x1 /2 =

2 3

1 1 ⇒ x1 /2 = 2 2 1 4 x = or x = ⋅ 4 9 y=

2 2 4 ⇒ x =   =  3 9 ⇒

1 2 1 x =   = ⋅  2 4

Hence, the roots of the given equation are

1 4 and ⋅ 4 9

23

Illustration 3 : Solution : ⇒ ⇒ Now, and

Solve : x 2 /3 + x1 /3 − 2 = 0.

We have x 2 /3 + x1 /3 − 2 = 0 y2 + y − 2 = 0

[Putting x1 /3 = y so that x 2 /3 = ( x1 /3 )2 = y 2 ] ⇒ y = − 2 or y = 1.

( y + 2) ( y − 1) = 0 y=−2

1 /3

⇒ x

x = (− 2)3 = − 8

=−2 ⇒

y = 1 ⇒ x1 /3 = 1

x = 13 = 1.

⇒

Thus,

x = − 8 or x = 1. Hence, the solution set of the given equation is {− 8, 1}. Illustration 4 : Solution :

Solve : ( x 2 − 5 x + 7)2 − ( x − 2) ( x − 3) = 1.

We have ( x 2 − 5 x + 7)2 − ( x − 2) ( x − 3) = 1

⇒

( x 2 − 5 x + 7)2 − ( x 2 − 5 x + 6) − 1 = 0

⇒

( x 2 − 5 x + 7)2 − ( x 2 − 5 x + 7) = 0

⇒

y2 − y = 0

[Putting x 2 − 5 x + 7 = y ]

⇒ Now,

y ( y − 1) = 0

⇒

x=

Also,

y = 1 ⇒ x2 − 5 x + 7 = 1

⇒

x2 − 5 x + 6 = 0

⇒

( x − 3) ( x − 2) = 0

⇒

x = 3 or

or y = 1.

⇒ y=0

y=0

⇒

5 ± 25 − 28 2

=

x

2

5±

− 5x + 7 = 0 −3

2

=

Solution : ⇒ ⇒ ⇒ ⇒

2

⋅

x = 2.

Hence, the roots of the given equation are 2, 3, Illustration 5 :

5±i 3

5+i 3 2

1 2 3 1 Solve :  x +  −  x −  = 4, x ≠ 0.    x 2 x

1 We have  x +   x

2

−

and

5−i 3 2

⋅

[UPTU 2004]

3 1  x −  = 4  2 x

  3 1 2 1  x −  + 4 −  x −  = 4 x 2 x   3 y2 + 4 − y=4 2 3 3 y2 − y = 0 ⇒ y  y −  = 0  2 2 3 y = 0 or y− =0 2

Putting x − 1 = y   x 

24

⇒

y=0

Now,

y=0 ⇒

⇒

x2 − 1 = 0

Again,

y=

3 2

3 ⋅ 2

y=

or x−

1 =0 x

⇒ x 2 = 1 ⇒ x = ± 1.

⇒ x−

1 3 = x 2

⇒

x2 − 1 x

⇒

2 x2 − 2 = 3 x ⇒ 2 x2 − 3 x − 2 = 0

⇒

2 x2 − 4 x + x − 2 = 0

⇒

2 x ( x − 2) + ( x − 2) = 0

⇒

( x − 2) (2 x + 1) = 0

⇒

x−2=0

⇒

x = 2 or

=

3 2

2x + 1 = 0 1 x=− ⋅ 2 or

Hence, the solution set of the given equation is 1, − 1, 2, −   2 x + 1 Illustration 6 : Solve    x − 1  2 x + 1 Solution : We have    x − 1 2

4

4

 2 x + 1 − 10    x − 1  2 x + 1 − 10    x − 1

y

⇒

y2 − 9 y − y + 9 = 0

⇒

y ( y − 9) − 1 ( y − 9) = 0

⇒

2

+ 9 = 0. [UPTU 2005]

2

+9=0 2    2 x + 1 Putting   = y  x − 1  

⇒

⇒

− 10 y + 9 = 0

( y − 1) ( y − 9) = 0 or y =1 y = 9. 2

Now, ⇒

2x + 1  2 x + 1 y =1 ⇒  = ±1  =1 ⇒ x − 1 x −1   2x + 1 2x + 1 or =1 = −1 x −1 x −1

⇒

2x + 1 = x − 1

⇒

x = −2

Again ⇒

 2 x + 1 y = 9⇒   x − 1 2x + 1 or =3 x −1

1 . 2

or or

2x + 1 = − x + 1 x = 0.

2

=9

⇒

2x + 1 x −1

2x + 1 x −1 =−3

=±3

25

⇒

2x + 1 = 3x − 3

⇒

x=4

or 2 x= ⋅ 5

or

2x + 1 = − 3x + 3

2 Hence, the solution of the given eqaution is  − 2, 0, 4,  . 5  Type II.

Exponential Equations Reducible to Quadratic Forms.

Consider the equation ap 2

x

+ bp

= y so that p 2

x

= ( p x )2 = y 2 , the given equation reduces to

Putting p

x

x

+ c = 0.

ay 2 + by + c = 0, which is a quadratic in y. Now, we solve this equation for y and then we derive from it the values of x.

Solve : 4 x − 3 . 2 x + 3 = − 128.

Illustration 1 : Solution :

We have 4 x − 3 . 2 x + 3 = − 128

⇒

(22 )

⇒

(2 x )2 − 24 . 2 x + 128 = 0

⇒ ⇒ ⇒ Now, and

x

− 3 . 23 . 2 x = − 128

y 2 − 24 y + 128 = 0

[Putting 2 x = y ]

( y − 16) ( y − 8) = 0 y = 16 or y = 8. y = 16 y=8

2 x = 16 ⇒

⇒ ⇒

x

2 =8

2 x = 24 x

⇒ 2 =2

3

⇒ x=4 ⇒

x = 3.

Thus,

x = 4 or x = 3. Hence, the roots of the given equation are 3 and 4. Illustration 2 : Solve 7 x + 1 + 71 − x = 50. Solution : We have 7 x + 1 + 71 − x = 50 ⇒ ⇒ ⇒

x

[UPTU 2007, 08]

−x

7 ⋅7 + 7⋅7 = 50 7 7 x ⋅ 7 + x = 50 7 7 7y + = 50 y

⇒

7 y 2 + 7 = 50 y

⇒

7 y 2 − 50 y + 7 = 0

⇒

7 y 2 − 49 y − y + 7 = 0

⇒

7 y ( y − 7) − ( y − 7) = 0

⇒

(7 y − 1) ( y − 7) = 0

[Putting 7 x = y]

26

⇒ Now, ⇒

1 7 1 y= 7 y=

7

x

=7

or

y = 7.

⇒

7

−1

x

=

1 7

x = −1

⇒ x

and

y=7 ⇒ 7 =7 ⇒ x = 1. Thus, x = − 1 or x = 1. Hence, the solution set of the given equation is {−1, 1}. Illustration 3 : Solve [ 5 + 2 6 ] x + [ 5 − 2 6 ] x = 10. 2

Solution : We have (5 + 2 6 ) (5 − 2 6 ) = 5 − (2 6 )

[UPTU 2003] 2

=1

1 = (5 + 2 6 ) − 1 5+2 6

⇒

5−2 6 =

∴

[ 5+2

⇒

(5 + 2 6) x /2 +

6] x + [ 5 − 2

6 ] x = 10

1 (5 + 2 6) x /2

= 10

1 = 10 y

[Putting (5 + 2 6 )

⇒

y+

⇒

y 2 − 10 y + 1 = 0

Now,

y=5+2 6

⇒ and

x / 2 = 1 ⇒ x = 2.

⇒

x /2

= y]

y = 5 ± 2 6.

⇒ (5 + 2 6 ) x /2 = (5 + 2 6 )1

y = 5 − 2 6 ⇒ (5 + 2 6 ) x /2 = (5 + 2 6 ) − 1

⇒ x / 2 = − 1 ⇒ x = − 2. Hence, the solution set of the given equation is {2, –2}. Type III.

Equations of the Form ( x + a) ( x + b) ( x + c) ( x + d) + k = 0.

Such type of equations are easily solvable if the sum of any two of the numbers a, b, c , d is equal to the sum of the other two. We take such pairs together and solve as illustrated below :

Illustration 1 : Solution :

Solve : ( x + 9) ( x − 3) ( x − 7) ( x + 5) = 385.

We have ( x + 9) ( x − 3) ( x − 7) ( x + 5) = 385

⇒

[( x + 9) ( x − 7)] [( x + 5) ( x − 3)] = 385 [Note that − 7 + 9 = − 3 + 5]

⇒

( x 2 + 2 x − 63) ( x 2 + 2 x − 15) = 385

⇒

( y − 63) ( y − 15) = 385

[Putting x 2 + 2 x = y ]

27

⇒

y 2 − 78 y + 945 = 385

⇒

y 2 − 78 y + 560 = 0

⇒

y 2 − 70 y − 8 y + 560 = 0

⇒

y ( y − 70) − 8 ( y − 70) = 0

⇒

( y − 70) ( y − 8) = 0

⇒ Now,

y = 70

⇒

x=

y = 70

⇒ ⇒

⇒ x

2

+ 2 x = 70 ⇒ x 2 + 2 x − 70 = 0

− 2 ± 4 + 280 2 − 2 ± 2 71

= Again,

y = 8.

or

− 2 ± 284 2

= − 1 ± 71 .

2

y=8

=

⇒ x2 + 2 x = 8 ⇒ x2 + 2 x − 8 = 0

( x + 4) ( x − 2) = 0 x = − 4 or x = 2.

Hence, the roots of the given equation are 2, − 4, − 1 ± 71 . Illustration 2 : Solution :

Solve : (2 x + 7) ( x 2 − 9) (2 x − 5) = 91.

We have (2 x + 7) ( x 2 − 9) (2 x − 5) = 91

⇒

(2 x + 7) ( x − 3) ( x + 3) (2 x − 5) = 91

⇒

[(2 x + 7) ( x − 3)] [( x + 3) (2 x − 5)] = 91

⇒

(2 x 2 + x − 21) (2 x 2 + x − 15) = 91

⇒

( y − 21) ( y − 15) = 91

[Putt ing 2 x 2 + x = y ]

⇒

y 2 − 36 y + 315 = 91 ⇒ y 2 − 36 y + 224 = 0

⇒

y 2 − 28 y − 8 y + 224 = 0

⇒

y ( y − 28) − 8 ( y − 28) = 0

⇒ ⇒ Now,

( y − 28) ( y − 8) = 0 y = 28

y = 8.

or

y = 28 ⇒ 2 x

2

⇒ 2 x 2 + x − 28 = 0

+ x = 28

⇒

2 x 2 + 8 x − 7 x − 28 = 0

⇒

2 x ( x + 4) − 7 ( x + 4) = 0

⇒

( x + 4) (2 x − 7) = 0 7 x = − 4 or x = ⋅ 2

⇒ Again,

y=8

⇒

x=

⇒ 2 x2 + x = 8 ⇒

− 1 ± 1 + 64 4

=

− 1 ± 65 4

2 x2 + x − 8 = 0 ⋅

28

Hence, the roots of the given equation are − 4,

Evaluate 6 + 6 + 6 + … ∞ .

Illustration 3 : Solution :

− 1 − 65 7 − 1 + 65 and , ⋅ 2 4 4 [UPTU 2011]

Let x = 6 + 6 + 6 + … ∞ .

Then

x= 6+ x

⇒

x2 = 6 + x x

⇒

2

− x−6=0

[Squaring both sides] ⇒ ( x − 3) ( x + 2) = 0

⇒ x = 3 or x = − 2. But, the given expression is +ve. So, we reject the value x = − 2. Hence, the value of the given expression is 3. 1 Illustration 4 : Find the value of : 2 + 1 2+ 2 +…∞ Solution :

Let the given expression be equal to x. Then, 1 x=2+ ⇒ x2 = 2 x + 1 ⇒ x2 − 2 x − 1 = 0 x x=

⇒

− (− 2) ± (− 2)2 − 4 × 1 × (− 1) 2

=

2± 8 2

=

2±2 2 2

=1± 2 . But, the given expression is +ve. So, we reject the value x = 1 − 2 . Hence, the value of the given expression is 1 + 2 . b = c, where X is a function of x. X b Putting X = y, the given equation becomes ay + = c or ay 2 − cy + b = 0, which y Type IV.

Equations of the Form a X +

is a quadratic in y. Now, we solve this equation for y and then we derive from it the values of x.

Illustration 1: Solution :

Solve :

We have

1 − x 13 x + = ⋅ 1− x x 6 1 − x 13 x + = 1− x x 6

29

⇒

y+

y2 + 1

⇒

 Putting 

1 13 = y 6

y

=

13 6

 x = y 1− x 

6 y 2 + 6 = 13 y

⇒

⇒

6 y 2 − 13 y + 6 = 0

⇒

3 y (2 y − 3) − 2 (2 y − 3) = 0 ⇒ (2 y − 3) (3 y − 2) = 0 3 2 y= or y= ⋅ 2 3 3 x 3 x 9 y= ⇒ = ⇒ = 2 1− x 2 1− x 4

⇒ Now, ⇒ Again,

4x = 9 − 9 x y=

2 3

⇒

⇒ 6 y2 − 9 y − 4 y + 6 = 0

9 ⋅ 13 x 4 ⇒ = 1− x 9

⇒ 13 x = 9 ⇒ x = x 2 = 1− x 3

4 ⋅ 13 9 4 Hence, the solution set of the given equation is  ,  ⋅ 13 13 ⇒

9x = 4 − 4x

⇒ ⇒

x=

[UPTU 2009]

 4 x + 1  x + 1  5 We have   +  =  x + 1   4 x + 1 2 y+

y2 + 1 y 2

 Putting 

1 5 = y 2 =

5 2

2y

−4y − y +2 =0

⇒

( y − 2) (2 y − 1) = 0

Now,

y=2

⇒

Again,

y=

1 2

⇒

4x + 1 x +1 4x + 1 x +1

4x + 1 x +1

2 y2 + 2 = 5 y

⇒

⇒

⇒

⇒

 4 x + 1  x + 1  5 Solve :   +  = ⋅  x + 1   4 x + 1 2

Illustration 2 : Solution :

⇒ 13 x = 4

⇒ 2 y ( y − 2) − ( y − 2) = 0

=2

7x = − 1 ⇒ x = −

so that

x +1 4x + 1

⇒ 2 y2 − 5 y + 2 = 0

⇒ y = 2 or

=

= y

y=

1 ⋅ 2

⇒ 2 x + 2 = 4x + 1 ⇒

1 ⇒ 8x + 2 = x + 1 2

1 ⋅ 7

Hence, the roots of the given equation are 1 and −

1 ⋅ 7

1 x= . 2

=

1  y

30

Type V. Equations of the form 1 1 1 1   (i) a  x 2 + 2  + b  x +  + c = 0 (ii) a  x 2 + 2  + b  x −  + c = 0.     x x x  x  We have

2 2  x + 1 = x 2 + 1 + 2 ⇒ x 2 + 1 =  x + 1 − 2   x x x2 x2

and

2 2  x − 1 = x 2 + 1 − 2 ⇒ x 2 + 1 =  x − 1 + 2.   x x x2 x2

Thus, the given equations (i) and (ii) can be easily solved by putting x + case (i) and x −

1 = y in case (ii). x

Illustration 1 : Solve by reducing to the quadratic equation 1 1  3  x 2 + 2  − 16  x +  + 26 = 0.    x x 1 1  Solution : The given equation is 3  x 2 + 2  − 16  x +  + 26 = 0.    x x Put

x+

1 1 1 2 = y i. e.,  x +  = y 2 i. e., x 2 + 2 = y 2 − 2.  x x x

∴

3 ( y 2 − 2) − 16 y + 26 = 0

⇒

3 y 2 − 16 y + 20 = 0

⇒

( y − 2) (3 y − 10) = 0 10 y = 2 or ⋅ 3 1 y=2 ⇒ x+ =2 x

⇒ Now, ⇒

x2 − 2 x + 1 = 0

and

y=

⇒ ⇒

10 1 10 ⇒ x+ = 3 x 3

x2 + 1 x

⇒ ( x − 1)2 = 0 ⇒ x = 1

=

10 ⇒ 3 x 2 − 10 x + 3 = 0 3

( x − 3) (3 x − 1) = 0 ⇒ x = 3 or 1/3.

Hence, the required solutions of the given equation are x = 3, Illustration 2 : Solve by reducing to the quadratic equation 1 1  4  x 2 + 2  + 8  x +  + 3 = 0, x ≠ 0, x ∈ R.    x x

1 , 1. 3

1 = y in x

31

1 1  Solution : The given equation is 4  x 2 + 2  + 8  x +  + 3 = 0.   x x  1 1 1 2 = y i. e.,  x +  = y 2 i. e., x 2 + 2 = y 2 − 2.  x x x

Put

x+

∴

4 ( y 2 − 2) + 8 y + 3 = 0 ⇒ 4 y 2 + 8 y + 3 − 8 = 0

⇒

4 y2 + 8 y − 5 = 0

⇒

(2 y − 1) (2 y + 5) = 0 ⇒ y =

Now, ⇒

y=

1 2

x2 + 1 x

⇒ x+ =

1 2

1 or 2

−

5 ⋅ 2

1 1 = x 2

⇒

2 x 2 − x + 2 = 0.

Here, discriminant = (−1)2 − 4 (2) (2) = 1 − 16 = − 15. Since the discriminant is negative, therefore real roots are not possible in this case. 5 1 5 Further y=− ⇒ x+ =− 2 x 2 ⇒ ⇒

x2 + 1 x

=−

5 2

⇒ 2 x2 + 5 x + 2 = 0

( x + 2) (2 x + 1) = 0 ⇒ x = − 2 or −

Hence, the required solutions are x = − 2, −

1 ⋅ 2

1 ⋅ 2

Illustration 3 : Solve by reducing to the quadratic equation 1 1  6  x 2 + 2  − 25  x −  + 12 = 0.   x x  1  Solution : The given equation is 6  x 2 + 2  − 25  x 

1 1 1 2 = y i. e.,  x −  = y 2 i. e., x 2 + 2 = y 2 + 2.  x x x

Put

x−

∴

6 ( y 2 + 2) − 25 y + 12 = 0

⇒

6 y 2 − 25 y + 24 = 0

⇒

(2 y − 3) (3 y − 8) = 0 or 8 / 3. y =3/2 3 1 3 y= ⇒ x− = 2 x 2

⇒ Now, ⇒ ⇒

x2 − 1 x

 x − 1 + 12 = 0.  x

=

3 2

⇒

( x − 2) (2 x + 1) = 0

2 x2 − 3 x − 2 = 0 ⇒

x = 2 or –1/2

32

and ⇒

y=

8 3

x2 − 1 x

x−

⇒ =

8 3

1 8 = x 3 3 x2 − 8 x − 3 = 0

⇒

x = 3 or −1 / 3. 1 1 Hence, the solutions of the given equation are x = 2, − , 3 and − ⋅ 2 3 ⇒

( x − 3) (3 x + 1) = 0

⇒

Illustration 4 : Solve by reducing to the quadratic equation 1 1  2  x 2 + 2  − 3  x −  − 4 = 0.    x x 1 1  Solution : The given equation is 2  x 2 + 2  − 3  x −  − 4 = 0.   x x  1 1 1 2 = y i. e.,  x −  = y 2 i. e., x 2 + 2 = y 2 + 2.  x x x

Put

x−

∴

2 ( y 2 + 2) − 3 y − 4 = 0

⇒

2 y2 − 3 y = 0

3 or y= ⋅ 2 1 ⇒ x− =0 x

⇒

y=0

Now,

y=0

⇒

x2 − 1 = 0

and

y=

⇒ ⇒

3 2

x2 − 1 x

y (2 y − 3) = 0

⇒

x−

⇒ =

x2 = 1

⇒

3 2

x = ±1

⇒

1 3 = x 2 2 x2 − 3 x − 2 = 0

⇒

( x − 2) (2 x + 1) = 0

⇒

x = 2 or −

1 ⋅ 2

Hence, the solutions of the given equation are x = 1, − 1, 2 or − Type VI.

1 ⋅ 2

Irrational Equations Reducible to Quadratics:

Methods of solving such equations are illustrated by means of the following :

Solution by Introducing a New Variable : Illustration 1 : Solution :

Solve :

We have

3 x−2 2

3x − 2 2

+ 2 x2 − 5 x + 3 =

+ 2 x2 − 5 x + 3 =

( x + 1)2 3

( x + 1)2 3

⋅

33

⇒

3 (3 x − 2) + 6 2 x 2 − 5 x + 3 = 2 ( x + 1)2 = 2 ( x 2 + 2 x + 1)

⇒

2 x2 − 5 x + 8 − 6 2 x2 − 5 x + 3 = 0

⇒

(2 x 2 − 5 x + 3) + 5 − 6 2 x 2 − 5 x + 3 = 0

⇒

y2 + 5 − 6 y = 0 [Putting 2 x 2 − 5 x + 3 = y so that 2 x 2 − 5 x + 3 = y 2 ]

⇒ ⇒ Now,

y2 − 6 y + 5 = 0 ( y − 5) ( y − 1) = 0 y=5

2 x 2 − 5 x + 3 = 25

⇒

2 x 2 − 5 x − 22 = 0

⇒

x=

Again,

y =1 ⇒

⇒

or

y = 1.

2 x2 − 5 x + 3 = 5

⇒

⇒

⇒

⇒ y=5

5 ± 25 + 176

=

4

5 ± 201 4

⋅

2 x2 − 5 x + 3 = 1 ⇒ 2 x2 − 5 x + 3 = 1

2 x 2 − 5 x + 2 = 0 ⇒ (2 x − 1) ( x − 2) = 0 1 x= or x = 2. 2

Hence, the roots of the given equation are

1 , 2 and 2

5 ± 201 4

⋅

Solution by Applying the Method of Identity : Illustration 1 : Solution :

x 2 − 3 x + 27 +

Solve :

x 2 − 3 x + 11 = 8.

x 2 − 3 x + 27 and M =

Let

L=

Then

L+ M =8 2

2

x 2 − 3 x + 11 ...(1)

2

Also

L − M

or

( L + M) ( L − M) = 16 or 8 ( L − M) = 16

= (x

− 3 x + 27) − ( x

2

− 3 x + 11) = 16

L− M =2

Now from (1) and (2), we get 2 L = 10 or or

[UPTU 2008]

The given equation is x 2 − 3 x + 27 +

or

x 2 − 3 x + 11 = 8.

or

x 2 − 3 x + 27 = 25

L=5 or

or

L2 = 25

x2 − 3 x + 2 = 0

or ( x − 1) ( x − 2) = 0 x = 1 , 2. Hence, the roots of the given equation are 1 and 2.

...(2)

34

Illustration 2 : Solution :

Solve : 3 x 2 − 7 x − 30 + 2 x 2 − 7 x − 5 = x + 5.

The given equation is 3 x 2 − 7 x − 30 + 2 x 2 − 7 x − 5 = x + 5.

Let

L = 3 x 2 − 7 x − 30 , M = 2 x 2 − 7 x − 5

Then

L+ M = x+5

Also

L2 − M 2 = (3 x 2 − 7 x − 30) − (2 x 2 − 7 x − 5) = x 2 − 25

or

( L + M) ( L − M) = ( x + 5) ( x − 5)

or

( x + 5) ( L − M) = ( x + 5) ( x − 5)

...(1)

or

...(2)

L− M = x−5 Now from (1) and (2), we get or

2L = 2x or

3 x 2 − 7 x − 30 = x 2

or

2 x 2 − 7 x − 30 = 0

or

( x − 6) (2 x + 5) = 0

or

x=6

or

x=−

L2 = x 2

or

L= x

5 ⋅ 2

Hence, the roots of the given equation are 6 and −

5 ⋅ 2

Solution by the Method of Rationalisation : Suppose we have to solve equations of the form

or

ax + b

=k

ax + b

+

ax + b

or cx + d

=

ex + f

+

cx + d

=k

.

In such equations, we square both sides and then transpose so that the expression under radical sign alone is on one side. To remove the radical sign, we again square both sides. Thus, we get a quadratic equation in x and we solve it. However, in the process of squaring, sometimes extraneous roots are obtained. So, before writing the solution set, we check whether the roots obtained actually satisfy the given equation or not. Illustration 3 : Solution :

Solve : x + 2 = 2 x + 7 .

We have x + 2 = 2 x + 7

⇒

( x + 2)2 = 2 x + 7

⇒

x2 + 2 x − 3 = 0

[Squaring both sides]

⇒ ( x + 3) ( x − 1) = 0

⇒ x = − 3 or x = 1. Rejecting the extraneous root x = − 3 because it does not satisfy the given equation, we get x = 1.

35

Hence, the only root of the given equation is x = 1. Illustration 4 : Solution :

Solve : 1 − 5 x + 1 − 3 x = 2.

We have 1 − 5 x + 1 − 3 x = 2 1 − 5x

⇒

= 2 − 1 − 3x

⇒

1 − 5x = 4 − 4 1 − 3x

⇒

4 1 − 3x

= 4 + 2x

[Squaring both sides]

+ (1 − 3 x) ⇒ 2 1 − 3x

2

⇒

4 (1 − 3 x ) = (2 + x ) = 4 + 4 x + x

⇒

x 2 + 16 x = 0

=2+ x

2

[Squaring both sides]

x ( x + 16) = 0

⇒

⇒ x = 0 or x = − 16. We see that x = 0 satisfies the given equation. But, x = − 16 does not satisfy the given equation and so we reject this extraneous root. Hence, the only root of the given equation is x = 0. Illustration 5 : Solution : ⇒

Solve :

We have

x+5 x+5

x + 21 = 6 x + 40 .

+ +

x + 21 = 6 x + 40

( x + 5) + ( x + 21) + 2

x+5

x + 21 = 6 x + 40 [Squaring both sides]

⇒ ⇒ ⇒ ⇒

2

x+5 x+5

x + 21 = 4 x + 14 x + 21 = 2 x + 7

( x + 5) ( x + 21) = (2 x + 7)2 3x

2

+ 2 x − 56 = 0

[Squaring both sides]

⇒ (3 x + 14) ( x − 4) = 0

14 or x = 4. 3 14 Check : For x = − , we have 3 14 14 14 − +5 + − + 21 = 6 ×  −  + 40  3 3 3 ⇒

x=−

⇒

1 + 3

49 = 3

36 3

⇒

1 7 6 + = 3 3 3

1 + 7 = 6, which is false. 14 is an extraneous root because it does not satisfy the given equation ∴ x=− 3 and so we reject it. ⇒

36

For x = 4, we have 4+5

+ 4 + 21

= (6 × 4) + 40

⇒

9 + 25 = 64

⇒

3 + 5 = 8, which is true.

∴ x = 4 satisfies the given equation. Hence, the only root of the given equation is x = 4.

1.15 Simultaneous Equations in Two Variables Method of solving such equations are illustrated by means of the following :

Illustration 1 : (i)

x + y

y x

=

Solve : 5 , x + y = 10 2

(ii) x 2 + y 2 = 185 , x + y = 19

(iii) x 3 + y 3 = 91 , x + y = 7 Solution :

(i) We have y 5 x + = y x 2

and

x + y = 10 From (1), we have y 5 x + = 2 y x or or or or

x+ y xy

=

...(1) ...(2)

5 2

10 5 = xy 2 xy = 4 or xy = 16 or x (10 − x) = 16 10 x − x 2 − 16 = 0 or x 2 − 10 x + 16 = 0

[using (2)] [using (2)]

or

x = 2 ,8 Now if x = 2 , from (2), y = 10 − x = 10 − 2 = 8 or if x = 8 , from (2), y = 10 − x = 10 − 8 = 2 Hence solution is x = 2 , y = 8 ; x = 8 , y = 2. (ii) We have x 2 + y 2 = 185

...(1)

and

...(2)

x + y = 19

37

From (2), we get ∴

y = 19 − x from (1), we get x 2 + (19 − x)2 = 185

or

x 2 + 361 + x 2 − 38 x = 185

or

2 x 2 − 38 x + 176 = 0

or

x − 19 x + 88 = 0

or

( x − 8) ( x − 11) = 0

or

x = 8 , 11 Now if x = 8 then y = 19 − x = 19 − 8 = 11 or if x = 11 then y = y − x = 19 − 11 = 8 Hence the solution is x = 8 , y = 11 ; x = 11 , y = 8 (iii) We have x 3 + y 3 = 91

...(1)

and

...(2)

x+ y=7

From (2), we get x = 7 − y ∴ from (1), we get (7 − y)3 + y 3 = 91 {343 − y 3 − 21 y (7 − y)} + y 3 = 91 or

343 − 147 y + 21 y 2 = 91

or

21 y 2 − 147 y + 252 = 0 y 2 − 7 y + 12 = 0 or y = 3 , 4

or Now

if y = 3 then x = 7 − y = 7 − 3 = 4 if y = 4 then x = 7 − y = 7 − 4 = 3

Hence the solution is x =4, y =3; x =3, y =4

Comprehensive Exercise 4 Solve the following equations : 1.

(i) x 2 /3 + 2 x1 /3 − 8 = 0.

(ii) 3 x1 /2 + 3 x − 1 /2 = 10.

(iii) 8 x 3 /2 − 8 x − 3 /2 = 63.

(iv) x − 4 − 144 = 25 x − 2 .

(v) x + 2.

x =

6 25

(i) 32 x + 9 = 10 . 3 x .

(ii) 5 x + 1 + 52 − x = 53 + 1.

(iii) 4 x − 5 . 2 x + 4 = 0.

(iv) 22 x + 8 + 1 = 32 . 2 x .

38

(v) 22

x +3

− 57 = 65 (2 x − 1).

(vi) 3 x + 2 + 3 − x = 32 + 1.

2

(viii) 51 + x + 51 − x = 26

(vii) 2 x + 4 x = 8 + 1 3.

 2 x + 3 (i)   +6  x +1

 x +1   − 7 = 0.  2 x + 3

 x + 2  x − 2 5 (ii)   −  = ⋅ x − 2 x + 2 6     (iii) 10 (iv) 4.

x+3

x − x+3

4x − 1 4x + 1

+

x

4x + 1 4x − 1

=

= 3. 8 ⋅ 3

(i) ( x + 1) ( x + 2) ( x + 3) ( x + 4) = 120.

[UPTU 2006]

2

(ii) x ( x + 1) ( x + 2) = 72. (iii) 16 x ( x + 1) ( x + 2) ( x + 3) = 9. (iv) ( x 2 + x − 6) ( x 2 − 3 x − 4) = 24. (v) ( x + 1) ( x + 2) ( x + 3) ( x + 4) + 1 = 0. 5.

(i) 12 + 12 + 12 + … ∞ .

1

(ii) 4 +

1

4+ 4+ (iii) 20 + 20 + 20 + ... ∞ 6.

(i) (5 + 2 6 ) x

2

−3

+ (5 − 2 6 )

(iv) 8 − 8 − 8 − ... ∞ x 2 −3

= 10.

1 1  (ii) 2  x 2 + 2  − 3  x +  − 1 = 0.   x x  1 1  (iii) 8  x 2 + 2  − 42  x −  + 29 = 0.    x x (iv) 3 x 2 − 4 x + 3 x 2 − 4 x − 6 = 18. 7.

(i) 2 x 2 + 5 x − 2 − 2 x 2 + 5 x − 9 = 1. (ii)

x = ( x − 2)

(iii) 5 x 2 + 7 x + 2 + 4 x 2 + 7 x + 18 = x + 4. (iv)

x +1

(v)

3x + 1 −

(vi)

2x + 1 + 3x + 2

+ 2x

= 7.

x −1

= 2. = 5x

1 4 +…∞

+ 3.

⋅

39

8.

Solve the following simultaneous equations : y 10 x a b (i) (ii) x + y = a + b , + + = , x + y = 10 =2 y x 3 x y (iii)

1 1 5 + = , 3 x + 4 y = 18 x y 6

(iv) x + y = 20 , xy = 64

(v) x + 2 y = 1 , x 2 + y 2 = 10

(vi)

1 1 3 1 1 5 + = , 2 + 2 = x y 4 x 16 y

A nswers 1 ⋅ 9

(iii) 4 ,

1 ⋅ 4

(iv) ±

1 1 ,± ⋅ 3 4

(i) 8, − 64.

(ii) 9 ,

1 . 5 (i) 0, 2.

(ii) −1, 2.

(iii) 0, 2.

(iv) − 4.

(vi) 0, − 2. 2 (ii) 10, − ⋅ 5

(vii) − 1 , 3

3.

(v) 3, − 3. 3 (i) − 2, − ⋅ 4

(viii) − 1 , 1 5 (iv) − ⋅ 16

4.

(i) − 6, 1,

1.

(v) 2.

(iii) − (v) 5.

6.

−5± 5 2

⋅

(ii) 2, − 4, − 1 ± i 8 . (iv) 0, 1,

1 ± 57

(ii) 2 + 5.

(iii) 5

(iv)

(i) ± 2, ± 2 .

(i) 2, −

1 1 , 4, − ⋅ 2 4

9 ⋅ 2 8.

2

⋅

⋅

(i) 4.

(iv) 8.

⋅

2

3 − 3 ± 10 , 2 2

(iii) 2, − 7.

− 5 ± i 39

(iii) 1.

33 − 1

2 −1± i 3 1 (ii) , 2, ⋅ 2 2 5 1 (iv) 3, − , (2 ± 70 ). 3 3 − 7 ± 17

(ii) 4.

(iii)

(v) 1, 5.

(vi) −

(i) x = 1 , y = 9 ; x = 9 , y = 1 a+b a+b (ii) x = a , y = b ; x = , y= 2 2 18 9 (iii) x = 2 , y = 3 ; x = , y= 5 5 (iv) x = 16 , y = 4 ; x = 4 , y = 16

8 1 2 ,− ⋅ 2 3

⋅

40

13 9 , y= 5 5 (vi) x = 2 , y = 4 ; x = 4 , y = 2. (v) x = 3 , y = − 1 ; x = −

1.16 Applied Problems on Quadratic Equations In this section we shall discuss applications of quadratic equations in solving applied problems.

Illustration 1 : Solution :

Divide 33 into two parts whose product is 242 .

[UPTU 2001]

Let the two parts be x and 33 − x. Then, according to the question x (33 − x ) = 242 ⇒ x 2 − 33 x + 242 = 0

⇒

x 2 − 22 x − 11x + 242 = 0

⇒

x ( x − 22) − 11 ( x − 22) = 0

⇒

( x − 22) ( x − 11) = 0

⇒ x = 22 or x = 11. Hence, the required parts are 22 and 11. Illustration 2 : Solution :

Find two consecutive numbers whose squares have the sum 85.

Let the two consecutive numbers be x and x + 1.

Then,

x 2 + ( x + 1)2 = 85

⇒

x 2 + x 2 + 2 x + 1 = 85

⇒

2 x 2 + 2 x − 84 = 0

⇒

x 2 + x − 42 = 0

( x + 7) ( x − 6) = 0 or 6. ⇒ x=−7 When x = − 7, we have x + 1 = − 7 + 1 = − 6 and when x = 6, we have x + 1 = 6 + 1 = 7. Hence, the two required numbers are 6 and 7 or – 7 and – 6. ⇒

Illustration 3 : numbers. Solution :

The product of two successive positive multiples of 3 is 810. Find the

Let the required multiples of 3 be x and x + 3.

Then,

x ( x + 3) = 810

⇒

x 2 + 3 x − 810 = 0

⇒

( x + 30) ( x − 27) = 0 or – 30. x = 27

⇒

41

Since only +ive multiple is required therefore – 30 is not acceptable. ∴ x = 27 and x + 3 = 30. Hence, the required numbers are 27 and 30. Illustration 4 : Solution :

A number exceeds its positive square root by 12. Find the number.

Let the number be x. Then, according to the question x−

x = 12

⇒

x − 12 =

x

⇒

( x − 12)2 = x [Squaring both sides]

⇒

x

2

− 25 x + 144 = 0 ⇒ ( x − 16) ( x − 9) = 0

⇒ x = 16 or x = 9. Since 9 cannot exceed its positive square root by 12, therefore the required number is 16. Illustration 5 : A two digit number is such that the product of its digits is 8. When 63 is subtracted from the number the digits interchange their places. Find the number. Solution :

Let the number be 10 x + y.

Then,

xy = 8.

and

10 x + y − 63 = 10 y + x

...(1)

⇒ 9 x − 9 y = 63 ⇒ x − y = 7. Putting y = x − 7 from (2) in (1), we get x ( x − 7) = 8 ⇒

⇒

...(2)

x 2 − 7 x − 8 = 0.

( x − 8)( x + 1) = 0 x = 8, neglecting the negative value.

⇒ Putting x = 8 in (1), we get y = 1.

Hence, the required number is 10 × 8 + 1 = 81. Illustration 6: The sum of the ages of a man and his son is 35 years and the product of their ages is 150. Find their ages. Solution : Given that

Let the man’s age be x years. Then the son’s age is 35 − x years. x (35 − x) = 150

⇒

x 2 − 35 x + 150 = 0

⇒

( x − 30)( x − 5) = 0 x = 30 or 5.

⇒

35 x − x 2 = 150

⇒ Since the father’s age cannot be 5 years, therefore we have father’s age = 30 years and son’s age = 5 years. Illustration 7 : B is a point on the line segment AC such that it lies between A and C. If AC = 9 cm and AB × AC = 12( BC )2 , find BC. Solution :

Let BC = x cm, then AB = 9 − x cm.

Given that AB × AC = 12( BC )2

42

⇒

9 (9 − x) = 12 ( x)2

⇒

81 − 9 x = 12 x 2

⇒

4 x 2 + 3 x − 27 = 0

⇒

x=

9 4

or

27 − 3 x = 4 x 2

⇒ ⇒

− 3.

Since length cannot be –ive, therefore x = Illustration 8 :

( x + 3) (4 x − 9) = 0

9 = 2 ⋅ 25 cm. Hence BC = 2 ⋅ 25 cm. 4

The area of a right angled triangle is 40 cm2 . If its base is 2 cm. less than

its altitude, find its base and altitude. Solution :

Let the base of the triangle = x cm .

Then, its altitude = ( x + 2) cm. 1 ∴ Area of the triangle = x ( x + 2) cm2 . 2 1 According to the question, x ( x + 2) = 40 2 ⇒

x ( x + 2) = 80 ⇒ x 2 + 2 x − 80 = 0

( x + 10) ( x − 8) = 0 [Neglecting x = − 10] ⇒ x = 8. Hence, the base of the triangle = 8 cm. and its altitude = 10 cm. ⇒

Illustration 9 : A passenger train running at uniform speed, takes 2 hours less for a journey of 300 km , if its speed is increased by 5 km per hour from its usual speed. Find the usual speed. Solution :

Let the usual speed of the train be x km/hr.

Then, time taken to describe a distance of 300 km at x km/hr 300  =   hours.  x  Also, time taken to describe a distance of 300 km at ( x + 5) km/hr  300  =  hours.  x + 5 According to question,

300 300  =   −2 x+5  x 

⇒

1 1  300  −  =2 x x + 5 

⇒

( x + 5) − x  150   =1  x ( x + 5) 

⇒

x ( x + 5) = 750

⇒

x 2 + 5 x − 750 = 0

43

⇒

x 2 + 30 x − 25 x − 750 = 0

⇒

x ( x + 30) − 25 ( x + 30) = 0

⇒

( x + 30) ( x − 25) = 0 x = 25 or x = − 30.

⇒ Rejecting x = − 30 because x cannot be −ve, we have x = 25. Hence, the usual speed of the train = 25 km/hr. Illustration 10 :

A piece of cloth costs Rs. 35. If the piece were 4 metres longer and each

metre costs Re. 1. less, the total cost would remain unchanged. How long is the piece ? Solution :

Let the length of the cloth piece be x metres. 35 Then, rate per metre = Rs.   ⋅  x

If the piece were 4 metres longer, then the new length = ( x + 4) metres. Since the total cost remains unchanged, therefore, new rate per metre  35  = Rs.  .  x + 4 35 35 According to the question, − =1 x x+4 [∵ Difference in rates = Re 1 per metre] ⇒

1 1  35  −  =1  x x + 4

⇒

 x + 4 − x 35   =1  x ( x + 4)

⇒

x ( x + 4) = 140

⇒

x 2 + 4 x − 140 = 0

⇒

( x + 14) ( x − 10) = 0 ⇒ x = 10. Neglecting x = − 14 because the length of the cloth cannot be −ve]

Hence, the required length of the piece of cloth = 10 metres. Illustration 11 : Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hrs they are 50 km apart, find the average speed of each. Solution : Let the speed of the second train be x km/hr, then the speed of the first train is ( x + 5) km/hr. Let the two trains start from the station A, then distance travelled by the first train in 2 hrs = 2( x + 5) km = AB, say and distance travelled by the second train in 2 hrs = 2x km = AC, say.

44

Given that BC = 50 km. By Pythagoras Theorem, we have AB2 + AC2 = BC2 ⇒

[2( x + 5)]2 + [2 x]2 = (50)2

⇒

4 ( x + 5)2 + 4 x 2 = 2500

⇒

x 2 + 10 x + 25 + x 2 = 625

⇒

2 x 2 + 10 x − 600 = 0

⇒

x 2 + 5 x − 300 = 0

( x + 20) ( x − 15) = 0 ⇒ x = 15 or −20. Since speed cannot be negative, therefore x = 15 and x + 5 = 20. ⇒

Hence, the speed of the first train is 20 km/hr and the speed of the second train is 15 km/hr. Illustration 12 : The current of a stream runs at the rate of 2 km/hr. A motor boat goes10 km upstream and back again to the starting point in 55 minutes. Find the speed of the motor boat in still water. Solution :

Let the speed of the motor boat in still water be x km/hr.

Then, its speed down stream = ( x + 2) km/hr and its speed upstream = ( x − 2) km / hr.  10  ∴ Time taken to cover 10 km upstream =   hrs.  x − 2  10  and the time taken to cover 10 km down stream =   hrs.  x + 2  10   10  55 According to the question,   +  =  x − 2  x + 2 60 ⇒

 1 1  55 10  + =  x − 2 x + 2  60

⇒

 x+2+ x−2  55 11 10  =  = 60 12 ( x − 2) ( x + 2)

⇒

11 ( x 2 − 4) = 240 x

⇒

11x 2 − 240 x − 44 = 0

⇒

11x 2 − 242 x + 2 x − 44 = 0

⇒

11x ( x − 22) + 2 ( x − 22) = 0

⇒

( x − 22) (11x + 2) = 0 2   x = 22.  Neglecting x = − 11 because the speed cannot be −ve   

⇒

45

Hence, the speed of the motor boat in still water = 22 km/hr.

Comprehensive Exercise 5 1.

Divide 55 into two parts whose product is 516.

2.

The sum of two numbers is 8 and the sum of their squares is 34. Find the numbers.

3.

The sum of the squares of two consecutive even natural numbers is 340. Find the numbers. 1 The sum of a number and its reciprocal is 2 ⋅ Find the number. 20 6 The sum of a number and its positive square root is ⋅ Find the number. 25

4. 5. 6.

The product of two consecutive even integers is 528. Find the integers.

7.

The sum of the squares of three consecutive positive integers is 77. Find the numbers.

8.

Two numbers are such that their sum is 54 and the product is 629. Find the numbers.

9.

The sum of the squares of two consecutive odd positive integers is 290. Find the integers.

10. The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number. Find the numbers. 11. The sum of the squares of two numbers is 130. The sum of the smaller number and twice the larger number is 25. Find the numbers. 12. A two digit number is four times the sum and three times the product of its digits. Find the number. 13. The area of a right angled triangle is 54 cm2 . If its base is three times its altitude, find its base. 14. The perimeter of a rectangular plot is 42 m and its area is 108 m2 . Find the dimensions of the plot. 15. A man sold a table for Rs. 75. If the percentage of his profit in this transaction is equal to the cost price of the table, find the cost price of the table. 16. A person covers a distance of 300 km by car in a certain time with a certain fixed uniform speed. If he increases his speed by 5 km/hr, he takes 2 hours less to cover the same distance. Find his original speed.

46

A nswers 1.

43 and 12.

3.

12 and 14.

2. 3 and 5. 5 4 4. or ⋅ 4 5

1 ⋅ 25 6. (− 24 and − 22) or (22 and 24).

5.

7. 9.

4, 5 and 6. 11 and 13. 71 28 10.  − and −  or (13 and 8).  5 5 11. (3 and 11) or (7 and 9). 12. 24. 14. Length 12 m and breadth 9 m.

8. 37 and 17.

13. 18 cm.

15. Rs. 50. 16. 25 km/hr.

Comprehensive Exercise 6 Fill in the Blanks Fill in the blanks ‘‘..........’’, so that the following statements are complete and correct. 1. The roots of the equation x 2 − 8 = 0 are .......... . 2. If the discriminant D = b 2 − 4ac = 0 then the two roots are ............ and ............ . 3. The sum and product of the roots of the quadratic equation 3 x 2 − 7 x − 5 = 0 are .......... and .......... respectively. 4. The quadratic polynomial, whose zeros are 3 and − 2 is .......... . 5. A quadratic equation whose roots have the sum 5 and the product 6 is .......... . 6. If the product of the two roots of the equation 4 x 2 + 3 x + k = 0 is 3, the value of k is .......... . 7. If one root of the equation 6 x 2 − 12 x + k = 0 be the reciprocal of the other, then the value of k is .......... .

47

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 8. The solution of the equation x 2 − 10 x + 9 = 0 is (a) 9 or 1

(b) − 9 or − 1

(c) − 9 or 1

(d) 9 or − 1

9. The roots of the quadratic equation 15 x 2 − 11x + 3 = 0 are (a) real and equal

(b) real and unequal

(c) imaginary

(d none of these

10. The value of k for which kx 2 − 8 x + 4 = 0 has two equal roots, is (a) − 8

(b) 25 / 16 (d) 16 / 25

(c) 4

11. If the equation x 2 + 4 x + k = 0 has real roots, the value of k is (a) k ≤ 4 (c) k ≤ 0

(b) k ≥ 4 (d) k ≥ 0.

12. The value of k for which x 2 − 8 x + k = 0 has coincident roots is (a) 16

(b) 4

(c) − 4 (d) 60 13. If sum and product of the roots of a quadratic equation is 9, the equation is (a) x 2 + 9 x − 18 = 0

(b) x 2 − 18 x + 9 = 0

(c) x 2 + 9 x − 9 = 0

(d) x 2 − 9 x + 9 = 0

14. The sum of a positive number and its reciprocal is 10/3. Then the number is (a) 3 or 1/3

(b) 4 or 1/4

(c) − 3 or − 1/3

(d) none of these

True or False Write ‘T’ for true and ‘F’ for false statement. 2 15. The roots of equation x − = 1 are 1 and 3. x 2

16. If the discriminant D = b

2

[UPTU 2009]

− 4ac is perfect square then the roots are

rational and equal. 17. The nature of the roots of the equation 3 x 2 + 13 x + 5 = 0 are imaginary. 18. A quadratic equation whose one root is (5 + 3) is x 2 − 10 x + 22 = 0. 19. The sum of the roots of the equation 2 x 2 − 12 x + 5 = 0 is 7. 2 20. If one of the roots of the equation 4 x + ax + 20 = 0 is 4, then the value of

a is − 21.

48

A nswers 1. ± 2 2 4. ( x − 3) ( x + 2) 7. 10. 13. 16. 19.

6 (c) (d) T F

2. real, equal 5. x 8. 11. 14. 17. 20.

2

(a) (a) (a) F T

− 5x + 6 = 0

3. 7/3, − 5/3 6. 12 9. 12. 15. 18.

(c) (a) F T

49

2 D eterminants

2.1 Determinants of Order 2

D

efinition.

Let a11 , a12 , a21 , a22 be any four numbers (real or complex). The

symbol a11  a21

a12  a22

represents the number a11 a22 − a12 a21 and is called a determinant of order 2. The numbers a11 , a12 , a21 , a22 are called the elements of the determinant and the number a11 a22 − a12 a21 is called the value of the determinant. Thus the value of a determinant of order two is equal to the product of the elements along the principal diagonal minus the product of the off-diagonal elements. A determinant of order two has two rows and two columns.

50

Note. In a determinant the number of rows is equal to the number of columns.

2.2 Determinants of Order 3 a11 The symbol ∆ = a21  a31

a12

a13 a23is called a determinant of order 3 and its value  a33

a22 a32

is the number a22 a11  a32

a23 a − a   21 a33 12 a31

a23 a + a   21 a33 13 a31

a22 ⋅ a32

This is called the expansion of the determinant along its first row. In this expansion each element in the first row is multiplied by that determinant of the second order which is obtained by deleting the row and the column containing that element. Starting from the first the signs of the products are alternately positive and negative. A determinant of order 3 has three rows and three columns. It has 3 × 3 i. e., 9 elements. We can find the value of a determinant of order 3 by expanding it along any of its rows or along any of its columns. For example, we have a22 ∆ = a11  a32

a23 a  − a12   21 a33 a31

a23 a  + a13   21 a33 a31

a22  a32

= a11 (a22 a33 − a23 a32 ) − a12 (a21 a33 − a31 a23 ) + a13 (a21 a32 − a31 a22 ) = − a12 (a21 a33 − a31 a23 ) + a22 (a11 a33 − a31 a13 ) a21 = − a12  a31

a22 a  + a22   11 a33 a  31

a21 = (−1)1 + 2 a12  a31

− a32 (a11 a23 − a21 a13 ) a13 a a13  − a32   11  a33 a a  21 23

a23 a  + (−1)2 + 2 a22   11 a33 a31

a13  a33

a11 + (−1)3 + 2 a32  a21

a13  a23

and this is the expansion of ∆ along its second column. In the expansion of ∆ if a ij is the element occurring in the ith row and the jth column, then to fix the positive or negative sign before it we should multiply it by (−1) i + j .

51

2.3 Determinants of Order 4 a11 a21 The symbol ∆ =  a  31 a41 its value is the number a22 a11 a32  a42

a12 a22

a13 a23

a32

a33

a42

a43

a23

a24 a21 a34 − a12 a31   a44 a41

a33 a43

a21 + a13 a31  a41

a14 a24 is called a determinant of order 4 and a34  a44

a22 a32 a42

a23 a33 a43

a24 a34  a44

a24 a21 a34 − a14 a31   a44 a41

a22 a32 a42

a23 a33⋅  a43

This is the expansion of ∆ along its first row. The determinant of order 4 has 4 rows and 4 columns. We can expand it along any of its rows or columns as is the case of a determinant of order 3. Proceeding in the same fashion we can evaluate the determinant of order greater than 4.

2.4 Minors and Cofactors Minors. Consider the determinant a12 a13 a11 ∆ = a21 a22 a23  a32 a33 a31

 ⋅  

If we leave the row and the column passing through the element a ij , then the second order determinant thus obtained is called the minor of the element a ij . We shall denote the minor of the element a ij by M ij . For example, a12 the minor of the element a21 =  a32

a13  = M21 , a33

a11 the minor of the element a32 =  a21

a13  = M32 , and so on. a23

In terms of the notation of minors, if we expand ∆ along the first row then ∆ = (−1)1 + 1 a11 M11 + (−1)1 + 2 a12 M12 + (−1)1 + 3 a13 M13 = a11 M11 − a12 M12 + a13 M13 . Cofactors. The minor M ij multiplied by (−1) i + j is called the cofactor of the

52

element a ij . We shall denote the cofactor of an element by the corresponding capital letter. Then cofactor of a ij = Aij = (−1) i + j M ij . For example, a12 the cofactor of the element a21 = A21 = (−1)2 + 1 M21 = −  a32 a11 the cofactor of the element a32 = A32 = (−1)3 + 2 M32 = −  a21

a13 , a33 a13 ⋅ a23

Thus the cofactor of any element a ij = (−1) i + j × the determinant obtained by leaving the row and the column passing through that element. In terms of notation of cofactors, we have ∆ = a11 A11 + a12 A12 + a13 A13 or

∆ = a31 A31 + a32 A32 + a33 A33 and so on.

Therefore, in a determinant the sum of the products of the elements of any row or column with the corresponding cofactors is equal to the value of the determinant.

Illustration 1 :

2 Evaluate the determinant ∆ =  1  3

−1 2 1

5 1 by expanding it along  3

the third row. Solution :

We have −1 ∆ = (−1)3 + 1 3 . 2

5 2  + (−1)3 + 2 1 .  1 1

5  1 2 + (−1)3 + 3 3 . 1

−1  2

( By definition) = 3 (−1 − 10) − (2 − 5) + 3 (4 + 1) = − 33 + 3 + 15 = − 15. Illustration 2 : Find the order and value of the following determinants: 1 1 1 4 − 3     (i) (ii) 2 1 − 3 ⋅   − 1 2 2 − 2 1   [UPTU 2008] a (iii)h  g

h b f

g f ⋅  c

53

4 Solution : (i) The determinant 2

− 3 has 2 rows and 2 columns, so the order − 1

of this determinant is 2. − 3 4  = (4) (− 1) − (− 3) (2) = − 4 + 6 = 2. Let ∆ = − 1 2 1 (ii) The determinant2  2 this determinant is 3. 1 Let ∆ = 2  2

1 1 −2 1 1

1 −3  has 3 rows and 3 columns, so the order of  1 1  1 −3  = 1.  −2 1

−2

−3 2  − 1.  1 2

−3 2  + 1.  1 2

1 , −1

expanding the determinant along the first row = 1 [(1) . (1) − (3) . (−2)] − 1 [(2) . (1) − (−3) . (2)] + 1 [(2) . (−1) − (1) . (2)]

(iii) Let

= (1 − 6) − (2 + 6) − (−2 − 2) = − 5 − 8 + 4 = − 9. h g a ∆ = h b f   f c g f f b b h h  − h  + g , = a c c f f g g on expanding the determinant along the first row 2

= a (bc − f ) − h (hc − gf ) + g (hf − bg) = abc + 2 fgh − af 2 − bg 2 − ch2 . Illustration 3 :

Solution :

Find the minors of the elements of the determinant 2 3 ⋅ ∆ = 0 −1

M11 = minor of the element a11 (= 3) is 0. M12 = minor of the element a12 (= 2) is −1. M21 = minor of the element a21 (= − 1) is 2. M22 = minor of the element a22 (= 0) is 3.

Illustration 4 :

Write the cofactors of the elements of the determinant −1 1 3 ∆ = 1 4 7 ⋅   2 3 4

Solution : A11 = cofactor of the element a11 (= 3)

54

4 = (−1)1 + 1  2

7  = 12 − 14 = − 2. 3

A12 = cofactor of the element a12 (= − 1) 7 1  = − (3 − 28) = 25. = (−1)1 + 2  3 4 1 A13 = cofactor of the element a13 (= 1) = (−1)1 + 3  4

4  2

= 2 − 16 = − 14. −1 A21 = cofactor of the element a21 (= 1) = (−1)2 + 1  2

1  3

= − (− 3 − 2) = 5. 3 A22 = cofactor of the element a22 (= 4) = (−1)2 + 2  4

1  3

= 9 − 4 = 5. 3 A23 = cofactor of the element a23 (= 7) = (−1)2 + 3  4

−1  2

= − (6 + 4) = − 10. −1 A31 = cofactor of the element a31 (= 4) = (−1)3 + 1  4

1  7

= − 7 − 4 = − 11. 3 A32 = cofactor of the element a32 (= 2) = (−1)3 + 2  1

1  7

= − (21 − 1) = − 20. 3 A33 = cofactor of the element a33 (= 3) = (−1)3 + 3  1

−1  4

= − (12 + 1) = − 13.

2.5 Properties of Determinants Property 1. The value of a determinant does not change when rows and columns are interchanged,

i. e.,

a1 a2  a3

b1 b2 b3

c1  a1 c 2  = b1   c 3  c1

a2 b2 c2

a3  b3 ⋅  c3 

Property 2. If any two rows (or two columns) of a determinant are intechanged the value of the determinant is multiplied by −1,

55

a1 a2  a3

i. e.,

b1 b2 b3

c1  c2  =  c3 

a3 − a2  a1

b3

c3  c 2 ⋅  c1 

b2 b1

Corollary. If any row (or column) of a determinant ∆ be passed over n rows (or columns), the resulting determinant will be (−1) n ∆. Property 3. If all the elements of any row (or any column) of a determinant are multiplied by the same number k, the value of the new determinant is k times the value of the given determinant, b1 c1  b1 c1   a1 a1 ka2 i. e., kb2 kc 2  = k a2 b2 c2  ⋅     b3 c3  b3 c3   a3 a3 Corollary 1.

If A be an n-rowed square matrix and k be any scalar then |k A| = k n | A|.

Corollary 2. If all the elements of any row (or any column) of a determinant are zero, the value of the determinant is zero. i. e.,

0 a2  a3

0 b2 b3

0 0   c 2 = 0 or 0    c3  0

a2 b2 c2

a3   b3 = 0.   c3 

Property 4. If two rows (or two columns) of a determinant are identical, the value of the determinant is zero, a1 c1  b1 c1  a1 a1    i. e., a a2 c 2 = 0 or a1 b1 c1  = 0 2    a2 c2  b3 c3  a2 a3 Property 5. In a determinant the sum of the products of the elements of any row (or any column) with the cofactors of the corresponding elements of any other row (column) is zero. b1 c1  a1 Let ∆ = a2 b2 c 2  be a determinant of order 3.   b3 c3  a3 Expanding ∆ along the first row, we get ∆ = a1 (b2 c 3 − b3 c 2 ) − b1 (a2 c 3 − a3 c 2 ) + c1 (a2 b3 − a3 b2 ) c 2 a c 2 a b2 b2  + b1 (−1)1 + 2  2  + c1 (−1)1 + 3  2  = a1 (−1)1 + 1  c 3 c 3 b3 b3 a3 a3 = a1 A1 + b1 B1 + c1 C1 . Similarly, we have a2 A2 + b2 B2 + c 2 C2 = ∆ ; a3 A3 + b3 B3 + c 3 C3 = ∆ ; a1 A1 + a2 A2 + a3 A3 = ∆ and so on. Again take the elements of the first row and cofactors of the corresponding elements of the third row then we have

56

a1 A3 + b1 B3 + c1 C3 c1 b1 a1  + b1 (−1)3 + 2  = a1 (−1)3 + 1  c 2 b2 a2

c1  c 2 a1 + c1 (−1)3 + 3  a2

b1  b2

= a1 (b1 c 2 − b2 c1 ) − b1 (a1 c 2 − c1 a2 ) + c1 (a1 b2 − a2 b1 ) = 0. Thus the sum of the products of the elements of any row (or column) with the cofactors of the corresponding elements of any other row (or column) is zero. Similarly we have a2 A1 + b2 B1 + c 2 C1 = 0, a1 B1 + a2 B2 + a3 B3 = 0 etc. Property 6. If in a determinant each element in any row (or column) consists of the sum of two terms then the determinant can be expressed as the sum of two determinants of the same order b1 c1   a1 + α1 a2 + α 2 i. e., b2 c2    b3 c3  a3 + α 3 b1 c1  α1 b1 c1  a1    = a2 b2 c2 + α2 b2 c 2 ⋅     b3 c 3  α 3 b3 c3  a3 Property 7. If to the elements of a row (or column) of a determinant are added m times the corresponding elements of another row (or column), the value of the determinant thus obtained is equal to the value of the original determinant. b1 c1  b1 c1  a1  a1 + mb1    Let and ∆′ = a2 + mb2 ∆ = a2 b2 c2 b2 c2      b3 c3  b3 c3  a3 a3 + mb3 be the determinant obtained by adding to the elements of first column of ∆ m times the elements of the second column. b1 c1   a1 + mb1  Now ∆′ = a2 + mb2 b2 c2    b3 c3  a3 + mb3 a1 = a2  a3

b1 b2

a1 = a2  a3

b2

b3 b1 b3

c1  mb1 c 2  + mb2   c 3  mb3

b1 b2

c1  b1 c 2  + mb2   c3  b3

b2

b3 b1 b3

c1  c 2 , by prop. 6  c3  c1  c 2 , by prop. 3  c3 

57

a1 = a2  a3

b1 b2 b3

c1  c 2  + 0,  c3  since two columns of the second det. are identical

= ∆. Property 8. If a determinant ∆ becomes zero when we put x = α in it, then ( x − α) is a factor of ∆.

2.6 Product of Two Determinants of The Same Order Rule For The Multiplication Of Two Determinants Of The Third Order. β1 γ1   α1 a1 b1 c1 Let and ∆1 = a2 b2 c 2 ∆ 2 = α2 β2 γ2      γ3  β3 a3 b3 c 3  α3 be two determinants of the third order. a1 α1 + b1 α 2 + c1 α 3 Then ∆1 . ∆ 2 = a2 α1 + b2 α 2 + c 2 α 3

a2 β1 + b2 β 2 + c 2 β 3

a3 α1 + b3 α 2 + c 3 α 3

a3 β1 + b3 β 2 + c 3 β 3

a1 β1 + b1 β 2 + c1 β 3

a1 γ 1 + b1 γ 2 + c1 γ 3 a2 γ 1 + b2 γ 2 + c 2 γ 3 . a3 γ 1 + b3 γ 2 + c 3 γ 3 This is the row-by-column multiplication rule for writing down the product of two determinants of the third order in the form of a determinant of the third order. This rule is applicable for writing down the product of two determinants of the n th order, where n is arbitrary. It should be noted that this multiplication rule is the same as the rule for multiplication of two matrices. Other ways of multiplying the determinants. The value of a determinant does not change by interchanging the rows and columns. Therefore while writing down the product of two determinants of the same order, we can also follow the row-by-row multiplication rule, or the column-by-row multiplication rule or column-by-column multiplication rule.

2.7 Working Rule for Finding the Value of a Determinant If the determinant is of order 2, we can at once find its value. But to find the value of a determinant of order ≥ 3, we should always try to make zeros at maximum number of places in any particular row (or column) and then to expand the determinant along that row (or column). The property 7 helps us to make zeros.

58

2.8 Symbols and Notations to be Employed for finding theValues of a Determinant For convenience we shall denote first, second, third rows of a determinant by R1 , R2 , R3 and columns by C1 , C2 , C3 etc. If we change the ith row of a determinant by interchanging it with jth row, then we shall denote this operation by writing Ri ↔ R j . If we change the ith column (or row) by multiplying its elements by some number k then we shall denote this operation by writing Ci → kCi . If we change the ith row by adding to it m times the corresponding elements of the jth row then we shall denote this operation by writing Ri → Ri + mR j . It should be noted that in this operation only Ri will change while R j will remain as it is.

Illustration 1 :

Solution :

1 Show that1  1

1 Let ∆ = 1  1

a b c

a

b + c c + a  = 0.  a + b

b c

b + c c + a ⋅  a + b

Applying C3 → C3 + C2 , we get a a + b + c 1  ∆=1 b b + c + a   c a + b + c 1 1 = (a + b + c )1  1

a b c

1 1, taking a + b + c common from C3  1

= 0, since C1 and C3 are identical. Illustration 2 :

Without expanding the determinant, show that: b − c  0 − b 0 a  = 0.   −a 0  c

Solution :

Let

 0 ∆ = − b   c

b 0 −a

− c a ⋅  0

[UPTU 2002]

59

By property 1, changing columns into rows −b c  0 ∆ = b 0 − a   a 0 − c  0 = (−1) − b   c

b

3

0 −a

− c a ,  0 taking (–1) common from each column

3

= (−1) ∆ = − ∆ or

Illustration 3 :

Solution :

or

2∆ = 0

∆ = 0.

b 2 c 2  Without expanding show thatc 2 a2 2 2 a b

b 2 c 2  Let ∆ = c 2 a2 2 2 a b

bc ca ab

b + c  c + a  = 0. a + b 

b + c  c + a ⋅ a + b 

bc ca ab

Multiplying the 1st, 2nd and 3rd rows by a, b, c respectively, we get ab 2 c 2 1  2 2 ∆= bc a abc   2 2 ca b =

bc ca abc  ab

abc bca cab 1

(abc )2

ab + ac   bc + ba  ca + cb   ab + ac  bc + ba ,  ca + cb 

1 1

taking abc common from 1st and 2nd columns bc = abc ca  ab

1 1 1

ab + bc + ca  ab + bc + ca , applying C3 → C3 + C1  ab + bc + ca 

bc = abc (ab + bc + ca)ca  ab

1 1 1

1 1,  1

taking ab + bc + ca common from third column = 0, since C2 and C3 are identical.

60

Illustration 4 :

Solution :

b − c Show thatc − a  a − b

b − c Let ∆ = c − a  a − b

c −a

a − b b − c  = 0.  c − a

a−b b−c

c −a

a − b b − c ⋅  c − a

a−b b−c

Applying R1 → R1 + R2 + R3 , we get 0 0   0 ∆ = c − a a−b b − c  = 0,   b−c c − a a − b since all elements of R1 are zero.

Illustration 5 :

ω2  1  = 0, where ω is a complex cube root of   ω 

ω

1 Show that  ω   2 ω

ω

2

1

unity. Solution :

1 ∆ = ω   2 ω

Let

ω2  1 ⋅   ω 

ω ω

2

1

Applying C1 → C1 + C2 + C3 , we get 1 + ω + ω 2  ∆ = 1 + ω + ω 2  2 1 + ω + ω

ω 2  0  1  = 0  ω  0 

ω ω

2

1

ω ω

2

1

ω2  1   ω 

[ ∵ 1 + ω + ω 2 = 0] since all entries in C1 are 0.

= 0, Illustration 6 : 1 1  1

Solution :

Without expanding the determinant, show that a b c

1 Let ∆ = 1  1

bc  1  ca  = 1  ab  1  a b c

a b c

a2   b 2  and evaluate it.  c2 

[UPTU 2001]

bc  ca  ⋅  ab 

Multiplying the 1st, 2nd and 3rd rows by a, b, c respectively, we get

61

 a a2 1  ∆= b b2 abc  2 c c  a a2 abc  = b b2 abc  2 c c

abc  bca  cab 1 1 , taking abc common from 3rd column  1

 a 1 a2  = −  b 1 b 2  , applying C2 ↔ C3  2 c 1 c  1 1 a2  = 1 b b 2  , applying C1  C2 .  2 1 c c   applying R2 → R2 − R1 and R3 → R3 − R1 , we get 1  ∆ = 0 0 

a2

  b −a   c 2 − a2 

a

2

b−a c −a

2

b 2 − a2 , on expanding the determinant along C1 c 2 − a2 b + a 1 , = (b − a) (c − a) c + a 1 b − a = 1 c −a 

taking (b − a) common from R1 and (c − a) common from R2 = (b − a)(c − a){c + a − (b + a)} = (b − a)(c − a)(c − b) = (a − b)(b − c )(c − a). Illustration 7 : a  ∆ = b c 

Prove that a2 b2 c2

bc  1   ca  = 1 ab  1  

a2 b2 c2

a3   b3   c3 

= (a − b)(b − c )(c − a)(ab + bc + ca). Solution : Multiplying the first, second and third rows of ∆ on the left hand side by a, b and c respectively, we get  a2 1  2 ∆= b abc   2 c

a3 b

3

c

3

 a2 abc   abc  2 abc  = b abc   2 abc   c

a3 b

3

c3

1  1, 1 

taking abc common from C3

62

 a2  = − b2  2 c 1  = 1 1  1  ∆ = 0 0 

a3   b 3 ,  c3 

1 1 1 a2

a3   b 3 ,  c3 

b2 c

2

a2 b

2

−a

applying C2 ↔ C3

applying C1 ↔ C2 .

a3

2

c 2 − a2

  b − a ,  c 3 − a3  3

3

applying R2 → R2 − R1 , R3 → R3 − R1 b 2 − a2 = 1 2 2 c − a

b 3 − a3 , expanding along first column c 3 − a3

b + a = (b − a)(c − a) c +a  b + a = (b − a)(c − a) c −b 

b 2 + a2 + ab   c 2 + a2 + ca 

c2

b 2 + a2 + ab  , − b 2 + ca − ab  applying R2 → R2 − R1

b + a = (b − a)(c − a)(c − b)  1

b 2 + a2 + ab  , c + b + a taking (c − b) common from R2

= (b − a)(c − a)(c − b) {(a + b)(a + b + c ) − (b 2 + a2 + ab)} = (a − b)(b − c )(c − a)(ab + bc + ca). Illustration 8 : Prove that 1 1 + a ≡ 1 1+ b  1  1 = abc + ab + bc + ca. Solution :

1   1 1 1 1  = abc 1 + + +    a b c   1+ c [UPTU 2004, 06]

Taking a, b, c common from first, second and third columns

respectively, we get

63

1 + 1 a  1 ∆ = abc  a  1   a 1 + 1 +  a  1 = abc 1 + + a  1 1 + +  a

1 b

1 c 1 c

    1  + 1  c 1 1  b c  1 1  +1 , b c   1 1  +1  b c

1 +1 b 1 b 1 1 + b c 1 1 + b c 1 1 + b c

applying C1 → C1 + C2 + C3 1  1  c  1 1 1  1 1   = abc  1 + + +  1 +1 ,  a b c  b c   1 1  1 + 1   b c 1 1 1 taking  1 + + +  common from C1  a b c  1 b

1 b 1 0

1  1 1 1 = abc  1 + + +  0  a b c  0 

1 c 0 , 1  

applying R2 → R2 − R1 , R3 → R3 − R1 1 1 1 = abc  1 + + +  = abc + ab + bc + ca.  a b c  Illustration 9 :

Prove that x − y − z ∆ ≡ 2 y   2z

2x y− x−z 2z

2x 2y

  = ( x + y + z )3 .  z − x − y

Solution : Applying R1 → R1 + R2 + R3 , we get x + y + z ∆ = 2 y   2z 1 = ( x + y + z )2 y  2 z

x+ y+z y− x−z 2z 1 y− x−z 2z

x + y + z  2y  z − x − y 1

   z − x − y 2y

64

 1 = ( x + y + z ) 2 y  2 z

0

0

 ,  − (x + y + z)

− (x + y + z)

0

0

applying C2 → C2 − C1 , C3 → C3 − C1 = ( x + y + z )3 . Illustration 10 :

Prove that

1 + a2 − b 2 ∆ ≡  2ab   2b  Solution :

2ab 2

1− a + b

  = (1 + a2 + b 2 )3 .   1 − a2 − b 2 − 2b 2a

2

− 2a

Applying C1 → C1 − bC3 and C2 → C2 + aC3 , we get 0  1 + a2 + b 2 − 2b    2 2 ∆= 0 1+ a + b 2a     2 2 2 2 2 2 b ( 1 + a + b ) − a ( 1 + a + b ) 1 − a − b   1 0 − 2b   , = (1 + a + b ) 0 1 2a     b − a 1 − a2 − b 2   2

2 2

taking (1 + a2 + b 2 ) common from C1 and C2 . 1 0  0  , = (1 + a + b ) 0 1 2a     b − a 1 − a2 + b 2   2

2 2

2

2 2

2

2 3

applying C3 → C3 + 2bC1 2

= (1 + a + b ) (1 − a + b

2

2

+ 2a ), expanding along first row

= (1 + a + b ) . Illustration 11 :

Without expanding the determinant show that (a + b + c ) is a factor of

the following determinant : a ∆ ≡ b  c

b c a

c a ⋅  b

[UPTU 2003]

If a, b, c are positive and unequal, show that the value of ∆ is always negative. Solution :

Applying C1 → C1 + C2 + C3 , we get a + b + c ∆ = a + b + c  a + b + c

b c a

c 1  a = (a + b + c )1   b 1

b c a

c a  b

65

1 = (a + b + c )0  0

b c −b a−b

c

 a − c ,  b − c

applying R2 → R2 − R1 , R3 → R3 − R1 a − c  b − c

c − b = (a + b + c ) a − b

= (a + b + c ) {− (b − c )2 − (a − b) (a − c )} = (a + b + c ) (− a2 − b 2 − c 2 + ab + bc + ca). Thus (a + b + c ) is a factor of ∆. Now we shall prove the next part. We have

∆ = (a + b + c ) (− a2 − b 2 − c 2 + ab + bc + ca) 1 (a + b + c ) (2a2 + 2b 2 + 2c 2 − 2ab − 2bc − 2ca) 2 1 = − (a + b + c ) {(a − b)2 + (b − c )2 + (c − a)2 } 2 =−

< 0, since a, b, c are positive and unequal. Therefore the given determinant is always negative if a, b, c are positive and unequal. Illustration 12 :

Prove that

(b + c )2  ∆ ≡ b2  2  c Solution :

a2

a2

2

(c + a) c2

  3 b  = 2abc (a + b + c ) .  (a + b)2  2

Applying C1 → C1 − C2 , C2 → C2 − C3 , we get (b + c + a)(b + c − a)  ∆ = 0 (c + a + b)(c − a − b)  b + c − a  = (a + b + c )2  0 c − a − b 

a2

0 (c + a + b)(c + a − b) (c + a + b)(c − a − b) 0 c +a−b c −a−b

  b2   (a + b)2 

a2

  b 2 ,  (a + b)2 

taking (a + b + c ) common from each of C1 and C2 b + c − a 2 = (a + b + c )  0  − 2b 

0 c +a−b − 2a

a2   b2  ; 2ab  

by R3 → R3 − R2 − R1

66

b + c a2  by C → C + 1 C a2 / b 1 1 3   a = (a + b + c )2 b 2 / a c +a b 2 , 1  0 0 2ab  C2 → C2 + C3 b   2 b + c a / b , expanding along R3 = (a + b + c )2 2ab  2 c + a b / a = (a + b + c )2 2ab {(b + c )(c + a) − ab } = 2abc (a + b + c )3 . Illustration 13 :

x − 2 x − 4  x − 8

Solve the equation

2x − 3 2x − 9 2 x − 27

3x − 4  3 x − 16  = 0.  3 x − 64  [UPTU 2007]

Solution :

or

Applying R2 → R2 − R1 , R3 → R3 − R1 , we get x − 2  −2   −6

2 x−3

x − 2  1   1

2 x−3

3 x − 4 −12  = 0  −60 

−6 −24

3x − 4 6  = 0.  10 

3 4

Expanding the determinant along the first row, we get ( x − 2) . 6 − (2 x − 3) . 4 + (3 x − 4) . 1 = 0 or 6 x − 12 − 8 x + 12 + 3 x − 4 = 0 or x − 4 = 0 or x = 4. Illustration 14 :

If a + b + c = 0, solve the equation a − x  c   b

Solution :

c b− x a

Applying C1 → C1 + C2 + C3 , we get a + b + c − x a + b + c − x  a + b + c − x

or

or

b  a  = 0.  c − x c

b− x a

1 (a + b + c − x)1  1 1 − x 0  0

b  a = 0  c − x

c b−c − x a−c

c

b  a = 0  c − x

b− x a b

 a − b  = 0,  c − b − x by R2 → R2 − R1 , R3 → R3 − R1

67

or

− x {(b − c − x) (c − b − x) − (a − c ) (a − b)} = 0

or

x ( x 2 − b 2 − c 2 + 2bc − a2 + ab + ca − bc ) = 0

or

x ( x 2 − a2 − b 2 − c 2 + ab + bc + ca) = 0

or

x=0

or

x 2 = a2 + b 2 + c 2 − (ab + bc + ca) = a2 + b 2 + c 2 − =

∴

x=0

or

x=±

Illustration 15 :

3 2 (a + b 2 + c 2 ). 2

 3 (a2 + b 2 + c 2 ) ⋅   2 

Without expanding evaluate

1  1 1  Solution :

1 {(a + b + c )2 − (a2 + b 2 + c 2 )} 2

a2 − bc   b 2 − ca  ⋅  c 2 − ab 

a b c

[UPTU 2008]

Let 1  ∆ = 1 1  1  = 0 0 

a2 − bc  1   b 2 − ca  = 1   c 2 − ab  1

a b c a b−a c −a

a2

a b c

 1  2 2   b − a + 0   c 2 − a 2  0

a2  1  b 2  + 1   c 2  1 a b−a c −a

a

− bc  − ca   − ab 

b c − bc

 c (b − a)  ,  b (c − a) 

applying R2 → R2 − R1 and R3 → R3 − R1 in each determinant 1 = (b − a) (c − a)0  0

a 1 1

a2  1  b + c + (b − c ) (c − a)0   c + a 0

a 1 1

− bc  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 ) = (b − a) (c − a) (c − b) − (b − a) (c − a) (c − b) = 0.

68

Illustration 16 :

If x, y, z are all different and if x2

x  y z  Solution :

1 + x3   1 + y 3  = 0,  1 + z3 

y2 z

2

prove that xyz = − 1.

We have x2

x  y z 

1 + x 3  x   1 + y3  = y   1 + z 3  z

y2 z2

x2

1  x   1 +  y 1 z  

y2 z2

x  = − y z 

1 x2   2  y  + xyz 1  1 z2  

1 1 1

x2 y2 z2 x y z

x3   y3   z3  x2   y 2 ,  z2 

applying C2 ↔ C3 in first det. and taking x, y, z common from R1 , R2 , R3 respectively of the second determinant 1  = 1 1 

x y z

1 x2    y 2  + xyz 1  1 z2  

x y z

x2   y 2 ,  z2 

applying C1 ↔ C2 in first determinant 1  = (1 + x y z )1 1  1  = (1 + x y z )0 0 

x2   y2   z2 

x y z x y− x z − x

x2

  y 2 − x 2 ,  z 2 − x2 

applying R2 → R2 − R1 , R3 → R3 − R1 y2 − x2   z 2 − x2  y+ 1 = (1 + x y z ) ( y − x)(z − x) z + 1 y − x = (1 + x y z ) z − x 

x  x

= (1 + xyz ) ( y − x)(z − x) { z + x − ( y + x)} = (1 + xyz ) ( x − y)( y − z )(z − x). Since x, y, z are all different therefore x − y ≠ 0, y − z ≠ 0, z − x ≠ 0. Hence

(1 + xyz )( x − y)( y − z )(z − x) = 0

⇒

1 + xyz = 0 → xyz = − 1.

69

Comprehensive Exercise 1 1. Evaluate the following determinants : b x + y  a  (i)  (ii) 3 a − b  x 2.

3. 4.

x

 . − xy + y  2

Evaluate each of the following determinants : 3 − 1 2 3  1 1 (i) − 2 (ii)2 1 2 3 6⋅     5 1 y z  4 x sin 10 ° Show that  sin 80 °

− cos 10 °   = 1. cos 80 ° 

Find the cofactors of each element 1 0 5 7     (i) (ii) 1 4  − 1 3 5 3  1 (iii)1  1

5.

1

2

a

of the following determinants : 2 1  7

bc  ca⋅  ab

b c

Using the method of cofactors find the value of the determinant a b + c 1 1 b c + a⋅   c a + b 1

Evaluate the following determinants : b−c c − a a − b  6. b−c c −a a − b ⋅   a−b b − c c − a a 7.  h  g 1 8. 1  1  1 9.  α  β + γ

h

g f ⋅  c

b f a b c

bc  ca⋅  ab 1 β γ +α

1  γ ⋅  α + β

70

1 1 1 10. Without expanding the determinant, show that 1 + + +  is a  a b c  1 1 1 + a factor of the determinant 1 1+ b 1 ⋅   1 1 + c  1 11. Using properties of determinants show that a b a + b + 2c    = 2 (a + b + c )3 . c 2a + b + c b   c a a + 2b + c   12. Using properties of determinants, prove the following results : (i)

x + 9  x   x

x + 4 (ii)  x   x x + 4 (iii)  2 x  2 x b + c (iv) c + a  a + b

x x+9 x x x+4 x 2x x+4 2x c +a a+b b+c

x  x  = 243 ( x + 3).  x + 9 x  x  = 16 (3 x + 4).  x + 4 2x  2 x  = (5 x + 4)(4 − x)2 .  x + 4 a + b a b + c  = 2b   c + a c

b c a

c a⋅  b

Prove the following identities : y z  x 13.  x 2 y2 z 2 = xyz ( x − y)( y − z )(z − x).    3  3 3 x y z   1 a a3  14. 1 b b 3  = (a − b)(b − c )(c − a)(a + b + c ).  3 1 c c  a + b + c 15.  − c   −b 1  16. 1 1 

b+c c +a a+b

−c a+b+c −a

−b

[UPTU 2005]

 − a  = 2 (a + b)(b + c )(c + a).  a + b + c

b 2 + c 2  c 2 + a2 = (a − b)(b − c )(c − a).  a2 + b 2

71

− a2 17.  ba   ac

ac  bc  = 4a2 b 2 c 2 .  − c 2

ab −b bc

2

Solve the following equations : 3 5  x + 1  18. 2 x+2 5  = 0.   3 x + 4  2 x + a 19.  a   a

b

c  c  = 0.  x + c

x+b

3 x − 8 20.  3   3

b 3

3

 3  = 0.  3 x − 8

3x − 8 3

A nswers 1.

(i) a2 + b 2

4.

(i)

(ii) y 3 .

2.

(i) 34

(ii) 3x − z .

A11 = − 1, A12 = − 3, A21 = − 7, A22 = 5.

(ii) A11 = 25, A12 = − 2, A13 = − 17, A21 = − 1, A22 = − 10, A23 = 5, A31 = − 7, A32 = 2, A33 = − 1. (iii) A11 = a (b 2 − c 2 ), A12 = a (c − b), A13 = c − b, A21 = b (c 2 − a2 ), A22 = b (a − c ), A23 = a − c , A31 = c (a2 − b 2 ), A32 = c (b − a), A33 = b − a. 5. 7.

0. abc + 2 fgh − af

2

6. 0. − bg 2 − ch2 .

8. (a − b) (b − c ) (c − a). 19.

x = 0, − (a + b + c ).

9. 0.

18. 1, 1, − 9. 2 11 11 20. x = , , ⋅ 3 3 3

2.9 Applications of Determinants to Coordinate Geometry Area of a triangle. We know in coordinate geometry the area of the triangle, whose vertices are ( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ), is given by 1 ∆ = [ x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )]. 2 Since we can express x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) as a third order determinant

72

 x1 x2  x3

y1

1 1  1

y2 y3

therefore the area of the triangle with vertices ( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ) is y1 1  x1 1 ∆= x2 y2 1⋅  2 y3 1 x3 Since the area has to be a positive quantity, we always take the absolute value of the above determinant for the area. Condition for three points to be collinear. The three points A ( x1 , y1 ), B ( x2 , y2 ) and C ( x3 , y3 ) are collinear if and only if area of the triangle ABC = 0 iff

 x1 1 x2 2 x3

y1 y2 y3

1 1 = 0  1

 x1 iff x2  x3

y1 y2 y3

1 1 = 0.  1

Illustration 1 : Using determinants find the area of the triangle whose vertices are (1, − 1), (2, 4), (−3, 5). Are the given points collinear ? Solution :

We know the area of the triangle whose vertices are ( x1 , y1 ), ( x2 , y2 )

and ( x3 , y3 ) is  x1 1 x2 2 x3

y1 y2 y3

1 1⋅  1

Thus the area of the triangle with vertices (1, − 1), (2, 4), (−3, 5) is −1 1  1 1 1 2 4 1 = {(4 − 5) + (2 + 3) + (10 + 12)}   2 2 5 1 − 3 1 = (−1 + 5 + 22) = 13 sq. units. 2 Since the area of the triangle is non-zero therefore the given three points are non-collinear. Illustration 2 :

Using determinants, show that the points (− 2, 5), (− 6, − 7) and

(− 5, − 4) are collinear. Solution :

We know that the three points are collinear if the area of the triangle

formed by these points is zero.

73

Now area of the triangle formed by the given points 5 1 −2 1 1 = −6 −7 1 = {−2 (−7 + 4) − 5 (−6 + 5) + (24 − 35)}  2 2 −5 −4 1 1 = (6 + 5 − 11) = 0. 2 Hence the given points are collinear. Illustration 3 :

If the points (2, − 3), (λ , − 1) and (0, 4) are collinear, find the value of λ.

Solution : Since the given points are collinear therefore the area of the triangle formed by these points must be zero. −3 1 2 1 ∴ λ −1 1 = 0  2 4 1 0 or

2 (−1 − 4) + 3 (λ − 0) + (4λ − 0) = 0

or

−10 + 3λ + 4λ = 0 10 λ = ⋅ 7

∴

Illustration 4 :

or

7λ = 10.

For what value of k, are the points (1, 5), (k, 1) and (11, 7) collinear ? (Use

determinants).

[UPTU 2003]

Solution : If the three points are collinear then the area of the triangle formed by these points must be zero. 5 1 1 1 ∴ k 1 1 = 0  2 7 1 11 or

1 (1 − 7) + 5 (k − 11) + 1 (7k − 11) = 0

or

− 6 + 5k − 55 + 7k − 11 = 0

or

12k − 72 = 0

or

12k = 72 72 k= = 6. 12

or

k = 6.

∴

Comprehensive Exercise 2 1.

Using determinants find the area of the triangle with vertices (i)

(− 2, − 3), (3, 2), (− 1, − 8) (ii) (2, − 7), (1, 3), (10, 8)

(iii) (5, 4), (− 2, 4), (2, − 6)

(iv) (− 1, − 8), (− 2, − 3), (3, 2).

74

2.

Using determinants, show that the following points are collinear: (i) (3, 8), (− 4, 2), (10, 14)

(ii) (11, 7), (5, 5), (− 1, 3)

(iii) (1, − 1), (2, 1), (4, 5)

(iv) (2, 3), (− 1, − 2), (5, 8).

3.

For what value of k, the points (5, 5), (k, 1), (11, 7) are collinear?

4.

Find the value of λ so that the points (1, − 5), (− 4, 5) and ( λ , 7) are collinear.

5. If the points (a, b), (a′ , b ′ ) and (a − a′ , b − b ′ ) are collinear, prove that ab ′ = a′ b. 6.

[UPTU 2006]

Prove that the points (a, b + c ), (b, c + a), (c , a + b) are collinear. [UPTU 2001, 07]

A nswers 1.

(i) 15 sq. units

3.

(iii) 35 sq. units k = − 7.

101 sq. units 2 (iv) 15 sq. units. 4. λ = − 5. (ii)

2.10 Application of Determinants in solving a System of Linear Equations The equation ax + by + cz + d = 0 is a linear non-homogeneous equation in unknowns x, y and z. If we have d = 0, the above equation becomes homogeneous linear equation in unknowns x, y and z. Consider the system of n linear simultaneous equations in n unknowns x1 , x2 , …, x n a11 x1 + a12 x2 + … + a1n x n = b1 a21 x1 + a22 x2 + … + a2 n x n = b2 …

…

…

…

…

…

…

…

a n1 x1 + a n2 x2 + … + a nn x n = b n . If b’s are all zero, this system is called homogeneous and if at least one of b’s is non-zero then the above system is called non-homogeneous. Any set of values of x1 , x2 , …, x n (real or complex) which simultaneously satisfy all these equations is called a solution of this system. When the system of equations has one or more solutions the equations are said to be consistent otherwise they are said to be inconsistent. If the system of equations has a unique solution it is said to be determinate and it is said to be indeterminate if it has more than one solution :

75

2.11 System of Linear Non-homogeneous Equations in Two Unknowns. (Cramer’s Rule) The solution of the system of non-homogeneous linear equations a1 x + b1 y = c1 , a2 x + b2 y = c 2 D1 D is given by x = and y = 2 , where D D a1 b1 c b1 a  , D1 =  1  and D2 =   1 D = b2 b2 a2 c 2 a2

c1 , D ≠ 0. c 2

Proof. Multiplying first equation by b2 and second by b1 and subtracting, we get

∴

(a1 b2 − a2 b1 ) x = b2 c1 − b1 c 2 . b c − b1 c 2 x= 2 1 ⋅ a1 b2 − a2 b1

Substituting for x in the first equation, we get y =

a1 c 2 − a2 c1 a1 b2 − a2 b1

⋅

We can express the above solution in terms of determinants as follows :

x=

c1  c 2 a   1 a2

b1  b2 b1  b2

=

D1 D

and y =

a1  a2

c1  c 2

a   1 a2

b1  b2

=

D2 D

, provided D ≠ 0.

Here D is the determinant of the coefficients of x and y and D1 , D2 are the determinants obtained by replacing the coefficients of x and y respectively by constant terms in D. A system of non-homogeneous linear equations in two unknowns is (i) Consistent and has a unique solution given by x = D1 / D and y = D2 / D, if D ≠ 0. (ii) Inconsistent, if D = 0 but at least one of D1 , D2 is non-zero. (iii) Consistent and indeterminate if D = 0 = D1 = D2 . Homogeneous Linear Equations In Two Unknowns. Consider the system a1 x + b1 y = 0, a2 x + b2 y = 0. This system is always consistent because the trivial solution (zero solution) x = 0, y = 0 is always there. The above system has a non-trivial solution iff D = 0 and then the equations have an infinite number of solutions. If D ≠ 0, the only solution of this system is x = 0, y = 0.

76

Illustration 1 :

(i) Solve by Cramer’s rule x − 2 y = 4, −3 x + 5 y = − 7.

[UPTU 2001]

(ii) Using determinants solve the system of equations 2 x − 4 y = − 3, 4 x + 2 y = 9. (iii) Using determinants solve the system of equations 2 x − 2 y = 1 , x + 2 y = 2. Solution : Also

∴ Hence,

 1 (i) We have D =  −3  4 D1 =  −7

−2  = 20 − 14 = 6, 5

 1 D2 =  −3

4  = − 7 + 12 = 5. −7

x=

D1 D

=

and

y=

D2 D

=

5 = − 5. −1

− 4  = 4 + 16 = 20 ≠ 0. 2

−3 D1 =   9 2 D2 =  4

∴

x=

Hence

x=

D1 D

−4  = − 6 + 36 = 30, 2 −3  = 18 + 12 = 30. 9

=

D 30 3 30 3 = and y = 2 = = ⋅ 20 2 D 20 2

3 3 , y= ⋅ 2 2

2 (iii) We have D =  1 Also

6 =−6 −1

x = − 6, y = − 5.

2 (ii) We have D =  4 Also

−2  = 5 − 6 = − 1 ≠ 0. 5

−2  = 4 + 2 = 6 ≠ 0. 2

1 D1 =  2

−2 = 2 + 4 = 6 , 2

2 D2 =  1

1  = 4 − 1 = 3. 2

77

D1

6 = 1 and 6

∴

x=

Hence

x = 1 , y = 1 / 2.

Illustration 2 :

Solution :

D

=

y=

D2 D

=

3 1 = ⋅ 6 2

Using determinants, solve the system of equations

3 x − 5 y = 0, 3 x + 8 y = 0 . − 5 3  = 24 + 15 = 39 ≠ 0. We have D =  8 3

Since D ≠ 0, therefore the given system of homogeneous equations has only zero solution : − 5 3 0 0  = 0, D2 =    = 0. Now D1 =  0 5 3 0    ∴

x=

D1 D

= 0, y =

D2 D

= 0.

2.12 System of Linear Non-homogeneous Equations in Three Unknowns. (Cramer’s Rule) The solution of the system of non-homogeneous linear equations a1 x + b1 y + c1 z = d1 , a2 x + b2 y + c 2 z = d2 , a3 x + b3 y + c 3 z = d3 D D D is given by x = 1 , y = 2 , z = 3 , where D D D a b c b1 c1  1 d1 1 1 D = a2 b2 c 2, D1 = d2 b2 c 2,     b3 c 3 b3 c 3 a3 d3 a1 D2 = a2  a3 Proof.

Then

a1 Let D = a2  a3  xa1 x D = xa2  xa3

d1 d2 d3 b1

c1 a1 c 2, D3 = a2   c 3 a3

b1 b2 b3

d1 d2, D ≠ 0.  d3

c1 c 2⋅  c 3

b2 b3 b1 b2 b3

c1  a1 x + b1 y + c1 z c 2 = a2 x + b2 y + c 2 z   c 3 a3 x + b3 y + c 3 z

b1 b2 b3

c1 c 2  c 3

applying C1 → C1 + yC2 + zC3 d1 = d2  d3

b1 b2 b3

c1 c 2 = D1 , say.  c 3

78

Similarly,

∴

a1 yD = a2  a3 x=

D1 D

, y=

d1 d2 d3 D2 D

c1 a1 c 2 = D2 , zD = a2   c 3 a3 ,z =

D3 D

b1 b2 b3

d1 d2 = D3 .  d3

, provided D ≠ 0.

This method of solving a system of three non-homogeneous linear equations in three unknowns can be used in exactly the same way to solve a system of n simultaneous linear non-homogeneous equations in n unknowns. A system of 3 simultaneous linear non-homogeneous equations in three unknowns is (i) consistent and has a unique solution given by D D D x = 1 , y = 2 , z = 3 if D ≠ 0. D D D (ii) consistent and has infinitely many solutions if D = 0 = D1 = D2 = D3 . (iii) inconsistent if D = 0 but at least one of D1 , D2 and D3 is non-zero. System Of Homogeneous Linear Equations In Three Unknowns. Consider the system a1 x + b1 y + c1 z = 0 a2 x + b2 y + c 2 z = 0 a3 x + b3 y + c 3 z = 0. This system is always consistent because x = 0, y = 0, z = 0 is always a solution of this system. If D = 0, this system has a non-zero solution and in this case infinite number of solutions exist. If D ≠ 0, zero solution x = 0, y = 0, z = 0 is the only solution of the system.

Illustration 1 :

Solution :

Solve using Cramer’s rule

x + y = 5, 1 Let D = 0  1

[UPTU 2001, 07] y + z = 3, z + x = 4. 1 0 1 1 = 1 (1 − 0) − 1 (0 − 1) = 1 + 1 = 2.  0 1

Since D ≠ 0, therefore the given system has a unique solution given by y x z 1 = = = ⋅ D1 D2 D3 D

79

5 D1 = 3  4

Now

and

1

0 1 = 5 (1 − 0) − 1 (3 − 4) = 5 + 1 = 6,  1

1 0

1 D2 = 0  1

5

1 D3 = 0  1

1

0 1 = 1 (3 − 4) − 5 (0 − 1) = − 1 + 5 = 4  1

3 4

5 3  4

1 0

= 1 (4 − 0) − 1 (0 − 3) + 5 (0 − 1) = 4 + 3 − 5 = 2. ∴

The solution of the given system is x=

D1 D

=

D D 6 4 2 = 3, y = 2 = = 2, z = 3 = = 1. 2 D 2 D 2

Hence, the required solution is x = 3, y = 2, z = 1. Illustration 2 :

Solve the following by using Cramer’s rule

x − 2 y + 3z = 2, 2 x − 3z = 3, x + y + z = 6. Solution :

1 Let D = 2  1 4 = 3

−2 0 1

3 1 −3 = 2   1 1

0 4 3

[UPTU 2002]

0 −9 , by R2 + 2 R1 , R3 − 3 R1  −2

− 9  = − 8 + 27 = 19 ≠ 0. − 2

Since D ≠ 0, therefore the given system has a unique solution given by y x z 1 = = = ⋅ D1 D2 D3 D Now,

2 D1 = 3  6 3 = 2 7

−2 0 1

3 2 −3 = 3   1 6

0 3  + 0 + 5  7 6

0 3 7

5 0 , by R2 + R1 , R3 + R1  7

3  7

= 2 (21 − 0) + 5 (21 − 18) = 42 + 15 = 57, 1 D2 = 2  1 −1 = 1 4

2 3 6

3  1 −3 = 2   1  1

0 −1 4

−9  = 2 + 36 = 38, −2

0 −9 , by R2 − 2 R1 , R3 − 3 R1  −2

80

1 D3 = 2  1

and

4 = 1 3

−2 0 1

2  1 3 = 2   6  1

0

0 −1 , by R2 + 2 R1 , R3 − 2 R1  4

4 3

−1  = 16 + 3 = 19. 4

∴ The solution of the given system is D D D 57 38 19 x= 1 = = 3, y = 2 = = 2, z = 3 = = 1. D 19 D 19 D 19 Hence, the required solution is x = 3, y = 2, z = 1. Illustration 3 : rule :

Solve the following system of linear equations with the help of Cramer’s

x + 2 y + 3z = 6, 2 x + 4 y + z = 7, 2 x + 2 y + 9z = 14. [UPTU 2008] 2 3 1 2 3 1 by R2 − 2 R1 ,    Solution : Let D = 2 4 1 = 0 0 −5     R3 − 2 R1 2 9 0 −2 3 2 −5 0  = − 10 ≠ 0. = − 2 3  Since D ≠ 0, therefore the given system has a unique solution given by y x z 1 = = = ⋅ D1 D2 D3 D Now

6 D1 =  7  14

2 4 2

−5 = − 2  8 1 D2 = 2  2 −5 =  2 and

1 D3 = 2  2 0 = −2

∴

6 7 14

3  6 2 3 by R2 → R2 − R1 , 1 = −5 0 −5    R3 → R3 − R1 9  8 0 6 −5  = − 2 (−30 + 40) = − 20, 6 3 1 1 = 0   9 0

6 −5 2

3 by R2 → R2 − 2 R1 , −5  R3 → R3 − 2 R1 3

−5  = − 15 + 10 = − 5, 3 2 4 2

6 1 7 = 0   14 0

2 0 −2

6 by R2 → R2 − 2 R1 , −5  R3 → R3 − 2 R1 2

−5  = − 10. 2

The solution of the given system is D D D −20 −5 1 −10 x= 1 = = 2, y = 2 = = ,z = 3 = = 1. D −10 D −10 2 D −10

81

Hence, the required solution is x = 2, y = Illustration 4 : rule :

Solution :

1 , z = 1. 2

Solve the following system of linear equations with the help of Cramer’s

3 x − 4 y + 5z = − 6, x + y − 2z = − 1, 2 x + 3 y + z = 5.[UPTU 2004] −4 5 3  Let D = 1 1 −2 = 3 (1 + 6) + 4 (1 + 4) + 5 (3 − 2)   3 1 2 = 3 × 7 + 4 × 5 + 5 × 1 = 21 + 20 + 5 = 46 ≠ 0.

Since D ≠ 0, therefore the given system has a unique solution given by y x z 1 = = = ⋅ D1 D2 D3 D Now,

−6 D1 = −1   5 −10 =  −1   8

and

∴

−4 1 3 0 1 0

3 D2 =  1  2

−6

0 = 1  0

−3

3 D3 =  1  2

−4

0 = 1  0

−7

−1 5 −1 7 1 3 1 1

5 −2 , by R1 → R1 + 4 R2 , R3 → R3 − 3 R2  1 −3 −10 −2 =    8 7

−3  = − 70 + 24 = − 46, 7

5 −2 , R1 → R1 − 3 R2 , R3 → R3 − 2 R2  1 11 −3 −2 = −    7 5

11  = − (−15 − 77) = 92 5

− 6 −1 , R1 → R1 − 3 R2 , R3 → R3 − 2 R2  5 −3 −1 =  7

−7 −  1

−3  = − (− 49 + 3) = 46. 7

The solution of the given system is − 46 D D D 92 46 x= 1 = = − 1, y = 2 = = 2, z = 3 = = 1. D 46 D 46 D 46

Hence, the required solution is x = − 1, y = 2, z = 1. Illustration 5 : If the system of equations x = cy + bz , y = az + cx, z = bx + ay are consistent, having non-zero solution, show that a2 + b 2 + c 2 + 2abc = 1. Solution :

The given system of eqaution can be written as x − cy − bz = 0, cx − y + az = 0, bx + ay − z = 0.

82

If the system of equations are consistent and have non-zero solution, then −c − b 1 c or D=0 −1 a = 0   a −1 b or

1 (1 − a2 ) + c (− c − ab) − b (ac + b) = 0

or

1 − a2 − c 2 − abc − abc − b 2 = 0

or

a2 + b 2 + c 2 + 2abc = 1.

Illustration 6 :

If a, b, c are all different, solve the system of equations:

x + y + z = 1, ax + by + cz = k , a2 x + b 2 y + c 2 z = k 2 .

Solution :

1 Let ∆ =  a   2 a 1 = a   2 a

1 b b2

1 c   c 2 0  by C2 → C2 − C1 , c −a   C3 → C3 − C1  c 2 − a2

0 b−a b 2 − a2

 b−a = 2 2 b − a

c −a  1   = (b − a)(c − a) 2 c −a b + a 2

1   c + a

= (b − a)(c − a) {c + a − (b + a)} = (b − a)(c − a)(c − b) = (a − b)(b − c )(c − a) ≠ 0, since a, b, c are all different. ∴ The given system has a unique solution given by y x z 1 = = = ⋅ D1 D2 D3 D

Now

1 D1 =  k   2 k 1 D2 =  a   2 a

and

1 D3 =  a   2 a

1 b b2 1 k k2 1 b b2

1 c  = (k − b)(b − c )(c − k),   c 2 1 c  = (a − k)(k − c )(c − a),   c 2 1 k  = (a − b)(b − k)(k − a).   k 2

83

Hence

x= y=

(k − b) (b − c ) (c − k) (a − b) (b − c ) (c − a) (a − k) (k − c ) (a − b) (b − c )

, z =

=

(k − b) (c − k) (a − b) (c − a)

(b − k) (k − a) (b − c ) (c − a)

,

⋅

Comprehensive Exercise 3

1.

Solve the following systems of linear equations by Cramer’s rule : 5 x + 7 y = − 2, 4 x + 6 y = − 3.

2.

x + 2 y = 1, 3 x + y = 4.

3.

2 x + 3 y = 10, x + 6 y = 4.

4.

2 x − y = 1, 7 x − 2 y = − 7. Solve the following systems of linear equations in three unknowns by Cramer’s rule :

5.

5 x − 7 y + z = 11, 6 x − 8 y − z = 15, 3 x + 2 y − 6z = 7.

6.

3 x + y + z = 2, 2 x − 4 y + 3z = − 1, 4 x + y − 3z = − 11.

7.

x − 4 y − z = 11, 2 x − 5 y + 2z = 39, − 3 x + 2 y + z = 1.

8.

2 x − 3 y + 4z = − 9, − 3 x + 4 y + 2 z = − 12, 4 x − 2 y − 3z = − 3.

A nswers

5.

9 7 , y=− ⋅ 2 2 16 2 x= , y=− ⋅ 3 9 x = 1, y = − 1, z = − 1.

7.

x = − 1, y = − 5, z = 8.

1. 3.

x=

2. x =

7 1 , y=− ⋅ 5 5

3. x = − 3, y = − 7. 6. x = − 1, y = 2, z = 3. 342 321 189 8. x = − , y=− , z =− ⋅ 53 53 53

Comprehensive Exercise 4 Fill in the Blanks Fill in the blanks ‘‘..........’’, so that the following statements are complete and correct.

3 x 1. If  2

7  = 10, then the value of x is .......... . 4

84

2. Let AX = B be a system of linear equations in n unknowns. Then, the system has a consistent and unique solution if .......... . 3. A determinant of order two has .......... rows and .......... columns. 4. The solution of the system of linear equations; x − 2 y = 4, 3 x + 4 y = 5 is x = .......... and y = .......... . 5. If two rows or two columns of a determinant are identical, then the value of the determinant is .......... . 6. The points (1, 5) , (k, 1) and (11, 7) are collinear if k = .......... . x + y 7. The value of a determinant 3  x

1

x

  is .......... . − xy + y 

2

2

8. If all the elements of any row or any column of a determinant are zero, the value of the determinant is .......... . b + c 1 a  9. The value of a determinant 1 b c + a is .......... .   a + b 1 c

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d).

5 7  is 10. The minor of the element 5 in the determinant 2 4 (a) 4

(b) − 4

(c) 7

(d) 2

6 11. The cofactor of the element 2 in the determinant 2 (a) 7

(b) − 7

(c) 4

(d) 6 − cos 10 °  is cos 80 °

 sin 10 ° 12. The value of the determinant sin 80 ° (a) 1

(b) − 1

(c) 0 1 13. The value of the determinant ω   2 ω

7  is 4

(d) none of these ω ω 1

2

ω 2 1  , where ω is a complex   ω 

cube root of unity is (a) 1

(b) 0

(c) ω 2

(d) ω

85

a11 14. The value of the determinant a21

a12  is a22

(a) a12 a21 − a11 a22

(b) a11 a22 − a12 a21

(c) a11 a21 − a12 a22

(d) a11 a12 − a21 a22

 cos θ 15. The order and value of − sin θ

sin θ  are respectively cos θ

(a) 2, 1

(b) 1, 2

(c) 0, 2

(d) 1, 0 2ω  , then ∆ is ω 2

1 16. If ω is a cube root of unity and ∆ =  ω (a) ω 2

(b) − ω 2

(c) 1

(d) ω

True or False Write ‘T’ for true and ‘F’ for false statement. 4 17. The order and value of 2

− 3  are 2 × 2 and 2 respectively. − 1 [UPTU 2009]

18. The value of a determinant does not change when rows and columns are interchanged. 19. If a determinant ∆ becomes zero when we put x = α in it, then ( x − α) is a factor of ∆.

1 20. If ∆1 =  a

0 1  and ∆ 2 =   b c

0  then ∆1 ∆ 2 = bd. d

 0 21. The value of the determinant− a   b

a 0 −c

− b c  = 0.  0

22. The system of equations 8 x + 6 y = 1, 4 x + 3 y = 5 is inconsistent. 23. In a determinant the number of rows and columns may be different. 24. The co-factors of the elements in the 3rd row of the determinant 1 2  3

2 3 −3

− 3 2are 13, − 8, − 1.  − 4

86

A nswers 1. 2 5. zero

2. | A| ≠ 0 6. 6

3. 2, 2 7. y

3

4. − 6, − 5 8. zero

9. 0 13. (b)

10. (a) 11. (b) 14. a11 a22 − a12 a21

12. (a) 15. (a)

16. (b) 20. T 24. T

17. T 21. T

19. T 23. F

18. T 22. T

87

3 Matrices

3.1 Matrix efinition. A set of mn numbers (real or complex) arranged in the form of a rectangular array having m rows and n columns is called an m × n matrix [to be read as ‘m by n’ matrix].

D

Thus a rectangular array of mn numbers of the form …  a11 a12 a1n  a … a22 a2 n   21  A =  a31 a32 … a3 n   … … … …   a a m2 … a mn  m1  is called an m × n matrix. The numbers a11 , a12 , …, a mn etc. are called the elements of the matrix. The horizontal lines are called rows or row vectors and the vertical lines are called columns or column vectors of the matrix. The element a ij belongs to the i th row

88

and the jth column and is called the (i, j)th element of the matrix. Thus in the element a ij the first suffix iwill always denote the number of the row and the second suffix j, the number of the column in which the element occurs. In a compact form the above matrix is represented by A = [ a ij ], i = 1, 2, …, m, j = 1, 2, …, n or simply by A = [ a ij ] m × n . Besides the square brackets [ ], sometimes we also use parentheses ( ) or double bars || || to represent a matrix. We use capital bold or itallic letters to denote matrices. In a matrix, the number of rows and columns need not be equal. Further, a matrix has no numerical value. It is just an ordered collection of numbers arranged in the form of a rectangular array.

3.2 Special Types of Matrices Square Matrix. An m × n matrix for which m = n (i.e., the number of rows is equal to the number of columns) is called a square matrix of order n. A square matrix of order n is also called an n-rowed square matrix. Thus in a square matrix, we have the same number of rows and columns. The elements a ij of a square matrix A = [a ij ] n × n for which i = j i. e., the elements a11 , a22 , a33 , …, a nn are called the diagonal elements and the line along which they lie is called the principal diagonal of the matrix. 0 1 2 3 4 1 7 0  Example. The matrix A =  0 5 1 0 1 3 9 1 4 × 4  is a square matrix of order 4. The elements 3, 1, 5, 1 constitute the principal diagonal of the matrix A. Diagonal Matrix. A square matrix each of whose non-diagonal elements is equal to zero is called a diagonal matrix. Thus the square matrix A = [ a ij ] n × n is a diagonal matrix if a ij = 0 for i ≠ j. A diagonal matrix of order n is denoted by diag [a11 , a22 , … , a nn ]. 0 3 0 1  For example,  and 0 8  4 0 0 0 and 3 respectively.

0 0 are the diagonal matrices of order 2  1

Scalar Matrix. A square matrix each of whose non-diagonal elements is equal to zero and all the diagonal elements are equal is called a scalar matrix.

89

3 For example,  0

−1 0  , 0 3   0

0 −1 0

0 0  are the scalar matrices of order 2 and 3  −1 

respectively. Unit Matrix. A square matrix each of whose diagonal elements is 1 and each of whose non-diagonal elements is equal to zero is called a unit matrix or an identity matrix and is denoted by I . The unit matrix of order n is denoted by I n . Thus a square matrix A matrix if a ij = 1 when i = j and a ij = 0 when i ≠ j. 0 0 0  1 0 1 0 1 0 0  and I 3 = 0 For example, I 4 =  1  0 1 0 0 0 0 0 0 0 1 

= [a ij ] is a unit 0 0  1

are identity matrices of order 4 and 3 respectively. Row Matrix. Any 1 × n matrix which has only one row and n columns is called a row matrix or a row vector. For example, [1 2 7] and [0 3 1 4 9] are the row matrices of order 1 × 3 and 1 × 5 respectively. Column Matrix. Any m × 1matrix which has m rows and only one column is a column matrix or a column vector.  4 3 For example, −1  and      1  0  respectively.

are column matrices of order 3 × 1 and 2 × 1

Null Matrix. The m × n matrix whose elements are all 0 is called the null matrix or a zero matrix of the type m × n. The null matrix of order m × n is usually denoted by O or by O m, n . 0 0 0 0 0 0  For example, 0 0 0 0 and    0 0 0 0 0 0 are zero matrices of order 3 × 4 and 2 × 2 respectively. Upper Triangular Matrix.

A square matrix A = [a ij ] is called an upper triangular

matrix if a ij = 0 whenever i > j. Thus in an upper triangular matrix all the elements below the principal diagonal are zero.

90

5 0  0 0 

For example,

0 4

1 −3

0

1

0

0

2 1 1  and 0  7 0 −2

7 6 0

4 0  2

are upper triangular matrices of order 4 × 4 and 3 × 3 respectively. Lower Triangular Matrix.

A square matrix A = [a ij ] is called a lower triangular

matrix if a ij = 0 whenever i < j. Thus in a lower triangular matrix all the elements above the principal diagonal are zero. 1 For example,  7

2 3 0  and 2 0 4 

0 1

0 0

−2

9

5

1

0 0  0 1

are the lower triangular matrices of order 2 and 4 respectively.

3.3 Submatrices of a Matrix Definition. Any matrix obtained by omitting some rows and columns from a given matrix A is called a submatrix of A. The matrix A itself is a submatrix of A as it can be obtained from A by omitting no rows or columns.  1 1 For example, 7 and    0 0 1 A = 7  0

2 2

3 are the submatrices of the matrix 1

2

3

11

6

2

1

9 5 ⋅  8

3.4 Equality of Two Matrices Two matrices A = [a ij ] m × n and B = [b ij ] p × q are said to be equal if (i)

they are of the same order {i.e., if number of rows in A equals the number of rows in B (m = p) and the number of columns in A equals the number of columns in B (n = q)}

(ii)

their corresponding elements are equal i.e., a ij = b ij for each pair of subscripts i and j.

91

Illustration 1 :

(i) Construct a 2 × 3 matrix A whose elements in the ith row and the jth

column are given by a ij = 2i − j (ii) Construct a 2 × 2 matrix A = [a ij ] whose elements are given by i− j a ij = i+ j Solution :

(i) We have a ij = 2i − j , 1 ≤ i ≤ 2 and 1 ≤ j ≤ 3. a11 = 2 × 1 − 1 = 2 − 1 = 1, a12 = 2 × 1 − 2 = 2 − 2 = 0

∴

a13 = 2 − 3 = − 1 , a21 = 4 − 1 = 3 , a22 = 4 − 2 = 2 , a23 = 4 − 3 = 1 a11 a12 a13  Hence the required matrix A =   a21 a22 a23   1 0 −1 = . 3 2 1 (ii) We have a ij = Now

i− j 1+ j

a11 = a21 =

,1 ≤ i ≤ 2 and 1 ≤ j ≤ 2.

1−1 1+1 2 −1 2 +1

= 0 , a12 = =

1− 2 1+ 2

=−

1 , 3

2−2 1 , a22 = =0 3 2+2

 a11 a12  Hence the required matrix A =   a21 a22  −1 / 3  0 . = 0  1 / 3 Illustration 2 : If a matrix has 18 elements, what are the possible order it can have ? What, if it has 11 elements ? Solution : Given

Let the given matrix A be of order m × n. mn = 18

Factorising the possible values of (m , n) are (1, 18) , (2 , 9) , (3 , 6) , (6 , 3) , (9 , 2) , (1 , 8). ∴

The possible orders matrix containing 18 elements are 1 × 18 , 2 × 9 , 3 × 6 , 6 × 3 , 9 × 2 , 1 × 18.

Let the matrix B = [b ij ] m × n contain 11 elements. ∴

m × n = 13 , (a prime number)

Thus the possible orders of B are 1 × 11 or 11 × 1.

92

Illustration 3 : where

Find the values of a, b, c and d so that the matrices A and B may be equal,

a A = c Solution :

b d

 1  , B = 0  

−1 ⋅ 3

The matrices A and B are of the same order 2 × 2. If A = B, the

corresponding elements of A and B must be same. Hence, we have Illustration 4 :

a = 1, b = − 1, c = 0, d = 3. 2  6 2 a + b If matrices  = , find the values of a and b.  ab  5 8  5

Solution : We know the corresponding elements of two equal matrices are equal. Therefore we have a+b=6

…(1)

ab = 8. Substituting for b from (2) in (1), we get 8 a + = 6 or a2 + 8 = 6a or a2 − 6a + 8 = 0 a or (a − 4)(a − 2) = 0 From (2) we have

or

…(2)

a = 2 , 4.

b = 4, when a = 2 and b = 2 when a = 4. 2 x + z  −1 5  x− y Illustration 5 : If  =0  , find x, y, z and w. 2 x − y 3 z + w 13     Solution. We know the corresponding elements of equal matrices are equal. Therefore we have x − y = − 1, 2 x + z = 5, 2 x − y = 0, 3 z + w = 13. Solving first and third equations simultaneously, we get x = 1, y = 2. Substituting x = 1 in 2 x + z = 5, we get z = 3. Again putting z = 3 in 3z + w = 13, we get w = 4. Hence x = 1, y = 2 , z = 3, w = 4. Illustration 6 :

For what values of x and y are the following matrices equal 3y 2 x + 1  x + 3 y 2 + 2 A = , B=  . 2 y −5y  −6   0  0

Solution. For matrices A and B to be equal their corresponding elements must be same. Thus we have 2 x + 1 = x + 3, 3 y = y 2 + 2, y 2 − 5 y = − 6. From first equation, we get x = 2. From second equation we get y 2 − 3 y + 2 = 0

93

or

or ( y − 2)( y − 1) = 0 y = 1, 2 . Substituting these values of y in the third equation we observe that only y = 2 satisfies it. Therefore y = 1 is inadmissible. Hence we have x = 2, y = 2.

3.5 Addition of Matrices Definition. Let A and B be two matrices of the same type m × n. Then their sum (to be denoted by A + B) is defined to be the matrix of the type m × n obtained by adding the corresponding elements of A and B. Thus if A = [a ij ] m × n and B = [b ij ] m × n then A + B = [a ij + b ij ] m × n . We observe that addition is defined only for the matrices which are of the same order. Two matrices A and B are said to be conformable for addition if they are of the same type. …  a11 a12 a1n    a21 a22 … a2 n  If A =  … … … …   a a a … mn  m2  m1  m×n

and

then

 b11  b21 B=  …  b m1

b12

…

b22 …

…

b m2

… …

 a11 + b11  a21 + b21 A +B=  …………  a m1 + b m1

Important Note. define A + B.

b1n   b2 n  …  b mn 

m × n,

a12 + b12

…

a22 + b22 ………… a m2 + b m2

… … …

a1n + b1n   a2 n + b2 n  …  a mn + b mn 

m × n.

If the matrices A and B are not of the same order, we cannot

3.6 Properties of Matrix Addition (i) Matrix Addition Is Commutative. If A and B be two m × n matrices then A + B = B + A . (ii) Matrix Addition Is Associative. If A, B, C be three matrices each of the type m × n, then A + (B + C) = (A + B) + C.

94

(iii) Existence Of Additive Identity. If O be the m × n matrix each of whose elements is zero, then A + O = A = O + A , for every m × n matrix A. (iv) Existence Of The Additive Inverse. Negative Of A Matrix. Definition. Let A = [ a ij ] m × n . Then the negative of the matrix A is defined as the matrix [− a ij ] m × n and is denoted by − A. If A = [ a ij ] m × n then there exists a matrix (− A) = [− a ij ] m × n such that A + (− A) = O = − A + A . Subtraction Of Two Matrices. Definition. If A and B are two m × n matrices, then we define A − B = A + (− B). The difference A − B is obtained by subtracting from each element of A the corresponding element of B. (v) Cancellation Laws Hold Good In The Case Of Addition Of Matrices. If A, B, C are three m × n matrices, then and

A+B=A+C → B=C

(Left cancellation law)

B + A = C + A → B = C.

(Right cancellation law)

3.7 Multiplication of a Matrix by a Scalar Definition. Let A be any m × n matrix and k any complex number called scalar. The m × n matrix obtained by multiplying every element of the matrix A by k is called the scalar multiple of A by k and is denoted by k A or A k. Thus if A = [ a ij ] m × n and k is any scalar, then …  k a11 k a12 k a1n  k a21 k a22 … k a2 n kA =  … … … …  k a m2 k a mn … k a m1 2 −1 3 Example : If k = 2 and A =  −3 1 2 × 3 , 4 then

2 × 3 2A = 2 × 4

2×2 2 × −3

2 × − 1 6 = 2 × 1 8

   = [k a ] ij m × n .   

4 −6

−2 2 2 × 3 .

3.8 Properties of Multiplication of a Matrix by a Scalar If A = [a ij ] m × n and B = [b ij ] m × n , k and l are two scalars, then (i) k (A + B) = k A + kB (ii) (k + l) A = k A + lA

95

(iii) (kl ) A = k (lA) = l (k A) (v) (−1) A = − A .

Illustration 1 : 2 (ii) If A =  5

Now

5 3 2 − 4 (i) If A =  and B =  , find 3A − 2B .  3 7 − 9 5 1 4 2 3 1 A , B=   , find 2A − 3B and B − 2 ⋅ −3 7 2 − 4 6  

3 × 5  9 15 =  3 × − 9 21 −27 −2 × − 4  − 4 8 = ⋅  − 2 × 3 − 10 − 6 15 − 8  5 7  9−4 3A − 2B = 3A + (−2B) =  = ⋅  −27 − 6 11 −33 21 − 10

Solution : and

(iv) (− k) A = − (k A) = k (− A)

3 × 3 (i) We have 3A =  3 × 7 − 2 × 2 − 2B =  − 2 × 5

2 ×1 2 × 4  4 2 8 2 × 2 (ii) We have 2A =  = ,  2 × (−3) 2 × 7 10 −6 14 2 × 5 −3 × 2 −3 × 3 −3 −6  −3 × 1 − 3B =  =  −3 × (− 4) −3 × 6 −6 12 −3 × 2 1 4 1 − 2 −2 − −   −1 −    A  2 2 2 2 and − = 7 . = 5 5 3 7 3 − − 2  − −   2 2 2 2  2  2  Now

2A − 3B = 2A + (− 3B) 2−6  4−3 = − 6 + 12 10 − 6

−9 −18

8 − 9  1 −4 − 1 =  14 − 18 4 6 − 4 1  1−1 3 − 2 2−   A A 2 and B− = B +  −  =  5 7   3 6−  2 2 2 − 2 −4+ 2 2   0 3 / 2 1   = ⋅ − 1 / 2 − 5 / 2 5 / 2   −3 2 −1 −1 1 2 Illustration 2 : If A =  , find the matrix C  , B = 1 2 0 2 0 −1    such that A + B + C is a zero matrix. Solution :

A + B + C = O ⇒ A + B + C + (− C) = O + (− C)

⇒

A + B + O = − C ⇒ A + B = − C ⇒ C = − (A + B).

96

Now

1 + 2 A +B= 2 + 1

− 3 −1 0 +0

∴

− 3 C = − (A + B) =  − 3 Find x, y, z, t if z   x 1 2 +3  t  y 0

2 − 1 3 −4 1 = ⋅  2 − 1 3 0 1 − 1 is the required matrix. − 1

4 0

Illustration 3 :

Solution :

We have  x 2  y  2x 2 y  2 x+3  2 y 

or or

z  1 +3 t  0 2 z  3 + 2 t  0

−1 3 =3  2 4

5 ⋅ 6

−1 =3 2

5 6

3 4 

−3  9 = 6 12 2z −3  9 15 =  2 t + 6  12 18

15 18

or

2 x + 3 = 9, 2 z − 3 = 15, 2 y = 12, 2 t + 6 = 18

or

x = 3, z = 9, y = 6, t = 6.

Illustration 4 : Solution :

or or Now ⇒

7 Find X and Y if X + Y =  2

0 ⋅ 3

We have 7 X + Y + (X − Y ) =  2 0 10 2X= 2 8  X=

1 2

10  2 

7 X+Y = 2 − 5 Y = −1 2 = 1

0 3 + 5 0

0 7 + 3 = 3 2 + 0

0 + 0 5 + 3

0 5 0 = ⋅  8  1 4 0 0 7 → Y =−X+  5 5 2 0 7 0 − 5 + 7 0 + 0 + =   − 4 2 5  − 1 + 2 − 4 + 5 0 ⋅ 1

4 If A + 2B =  2 determine the matrices A and B. Illustration 5 :

Solution :

0 3 and X − Y =  5 0

2 1

3 5

7 and 2A − B =  2

6 3

4 4 + 1  2

3 5

We have 7 2 (2A − B) + A + 2B = 2  2

6 3

2 1

4 , 1

97

12 + 2 8 + 3 14 + 4 5A =  6 +1 2 + 5  4+2 14 11 1 18 A =  ⋅ 7 7 5 6

or or

7 2A − B − 2 (A + 2B) =  2 6−4 7 − 8 − 5B =  3−2 2 − 4

Now or or

1 − 1 5 − 2

B=−

2 1

6 3

4 −2 1

4 2 

2 1

3 5

4 − 6 1 − 10 − 2 ⋅ − 9

Comprehensive Exercise 1 1.

A matrix has 24 elements. What are the possible order or dimensions it can have ? What if it has 13 elements ?

2.

Construct a m × n matrix A = [a ij ] whose element a ij is given by (i) a ij =

3i − j

(ii) a ij = (iii) a ij = 3.

2

,m=2,n=3

(i + 2 j)2 2 (i + j)2 2

,m=3,n=2 ,m=3,n=4

For what values of x and y are the following matrices equal?  5 (i) A =  3 2y

49 x2   5  , B = 128 −17 −17   2y  2 x +1  x + 3 (ii) A =  , B=  2 y −5y   0  0 4.

Find a, b, c , d if 2a − 3b (i)   1 3a + 4b (ii)  c +d

5.

y 2 + 2 ⋅ −6 

c −d a + 4b 2 2c − d

3  1 = 3c + 4d  1 a − 2b  2 = −1  5

−2 6

3 29

2 −5

4 ⋅ −1

Find the values of x and y from the following equation : x 2 7

5  3 + y − 3  1

− 4  7 = 2 15

6 ⋅ 14

98

6.

Find x, y and z so that A = B, where x − 2 A =  18 z

7.

1 If A =  5

3 y+2

2 6

2 0 , B= 4 −1

z  y , B=  z 6 y

3  2 , C= 7 − 4

z x

6 ⋅ 2 y

9 , find 7

A + B , A − C , 2 A + 3B , B − 5C . 8.

 2 If A = − 3   4

2 6 1 , B =  1   0 0

2 3 , find the matrix C such that  4

A + B + C is a zero matrix. 9.

1 If A = 2  7

0 −1 1

3 3  2 , B = 1   5 0

1 2 4

5 4  3 , C = 0   7 0

6 1 2

1 7 , verify that  1

A + (B + C) = (A + B) + C. 5 10. Find matrices A and B if A + B =  0

2 3 , and A − B =  9 0

6 ⋅ −1

A nswers 1. 1 × 24 , 2 × 12 , 3 × 8 , 4 × 6 , 6 × 4 , 8 × 3 , 12 × 2 , 24 × 1 ; 1 × 13 , 13 × 1  9 / 2 25 / 2 0   1 1/ 2 2. (i)  (ii)  8 18     5 / 2 2 3 / 2   25 / 2 49 / 2 9/2 8 25 / 2  2  (iii) 9 / 2 8 25 / 2 18    25 / 2 18 49 / 2  8 3. (i) x = ± 7, y = 4. 4.

(ii) There do not exist such x and y for which A = B. (i) a = 2, b = 1, c = 3, d = 5.(ii) a = 2, b = − 1, c = 0, d = 5.

5.

x = 2, y = 9.

6.

x = 11, y = 9, z = 3. 5 −7 1 −1 A+ B=  , A− C=  ,  11 −3 4  9 13 − 42 2 − 10 2 A + 3B =  ⋅  , B − 5C =  19 7 29 − 28   

7.

99

− 8 8. C =  2  − 4 4 10. A =  0

− 4 − 4 ⋅  − 4 4 1 ,B=  4 0

− 2 ⋅ 5

3.9 Multiplication of Matrices Definition. Let A = [a ij ] m × n and B = [b jk ] n × p be two matrices such that the number of columns in A is equal to the number of rows in B. Then the m × pmatrix C = [c ik ] m × p such that n

c ik = Σ a ij b jk j=1

is called the product of the matrices A and B in that order and we write C = AB. Thus the (i, k )th element c ik of the matrix AB is obtained by multiplying the corresponding elements of ith row of A and kth column of B and then adding the products. The rule of multiplication is row by column multiplication. In the product AB, A is called the pre-factor (pre-multiplier) and B the post-factor (post-multiplier). It is important to note that the product AB of two matrices A and B exists if and only if the number of columns in A is equal to the number of rows in B. Two such matrices are said to be conformable for multiplication. a12   a11 b12   b11  For example, if A = a21 a22  ,B =    b22  2 × 2 b21 a32  3 × 2 a31  a11 b11 + a12 b21 then AB = a21 b11 + a22 b21  a31 b11 + a32 b21

a11 b12 + a12 b22  a21 b12 + a22 b22   a31 b12 + a32 b22  3 × 2 .

If the product AB exists, then it is not necessary that the product BA will also exist. In the above example the number of columns in B is not equal to the number of rows in A therefore BA does not exist.

Illustration 1 :

0 If A =  1   2

1 2 3

0 1  1 and B = 3   0 4

2 0 , find AB. Does BA exist ?  1

100

Solution : The matrix A has 3 columns and the matrix B has 3 rows therefore the product AB is defined. By row by column rule of multiplication, we have 1 0  1 2 0    AB = 1 2 1 3 0    3 0 4 1 2 . + 13 . + 0.4 0.2 + 10 . + 01 .  3 0  01    = 11 . + 23 . + 14 . 1.2 + 2.0 + 11 . = 11 3     . + 3 .3 + 0.4 2 .2 + 30 . + 01 .  11 4 3 × 2 . 21 The matrix BA does not exist since the number of columns in B is not equal to the number of rows in A .  1 0 3 5 Illustration 2 : If A =   , B = 2 1 , find AB and BA . Show that AB ≠ BA. 3 2     Solution : Since A and B are square matrices of same order 2, therefore both AB and BA are defined. 0 3 5 1.3 + 0.2 1.5 + 0.1  3 5 1 We have AB =  = =     2 2 1 3.3 + 2.2 3.5 + 2.1 13 17 3 and

3 BA =  2

5 1

1 3 

0 3.1 + 5.3 = 2  2.1 + 1.3

3.0 + 5.2 18 = 2.0 + 1.2  5

10 ⋅ 2

Both AB and BA are square matrices of the type 2 × 2 but the corresponding elements of these matrices are not equal, therefore AB ≠ BA.

3.10 Properties of Matrix Multiplication (i)

Matrix Multiplication Is Associative.

If A , B , C are three matrices conformable for multiplication, then A (BC) = (AB) C. (ii) Multiplication Of Matrices Is Distributive With Respect To Addition Of Matrices. If A , B and C are matrices of the type m × n, n × p, n × p respectively then A (B + C) = AB + AC. (iii) Matrix Multiplication Is Not Always Commutative. (iv) The Product Of Two Matrices May Be A Zero Matrix When Neither Of Them Is A Zero Matrix.

Illustration 1 :

Find the product matrix of the matrices

101

2 A = 1 Solution :

1 1

 2 2 and B =  0  1 −2

−2 1 3

0 1 ⋅  0

The matrix A is of the type 2 × 3 and B is of the type 3 × 3. Therefore the

product AB is defined and it will be a matrix of the type 2 × 3. −2 0  2 1 2  2 AB =  1 1 × 0  1 1 1   3 0 −2 .  2 . 2 + 1. 0 + 2. (−2) 2. (−2) + 1. 1 + 2.3 2.0 + 1.1 + 20 = 1. (−2) + 1.1 + 1. 3 1.0 + 1.1 + 10 .   1. 2 + 1. 0 + 1. (−2) 0 = 0

Illustration 2 :

1 If A = 3  2

3 2

1 ⋅ 1 2 −1 6

3 1 4 and B = 2   1 3

1 −2 2

−1 2 , find AB and  1

BA. Solution : Since A and B are square matrices of same order 3, therefore both AB and BA are defined. 2 3  1 1 −1 1 We have AB = 3 −1 4 2 −2 2     6 1 3 2 1 2 1.1 + 2.(−2) + 3 .2 1.(−1) + 2.2 + 3 .1   1 .1 + 2.2 + 3.3  = 3.1 + (−1).2 + 4.3 3.1 + (−1).(−2) + 4.2 3.(−1) + (−1). 2 + 4.1   2. 1 + 6.(−2) + 1. 2 2.(−1) + 6. 2 + 1.1   2.1 + 6. 2 + 1 .3 3 6 14  = 13 13 −1   −8 11 17 1 −1  1 2 3 1    and BA = 2 −2 2 3 −1 4     2 1 2 6 1 3 1.2 + 1.(−1) + (−1).6 13 . + 1.4 + (−1).1 1.1 + 1.3 + (−1).2  = 2.1 + (−2).3 + 2.2 2.2 + (−2).(−1) + 2.6 23 . + (−2).4 + 2.1   3.2 + 2.(−1) + 1.6 33 . + 2.4 + 1.1   3.1 + 2.3 + 1.2 −5 6 2  = 0 18 0 ⋅   10 18 11

102

−1 1 −1 If A =  ,B =   2 0  2 verify that A ( B + C) = AB + AC. Illustration 3 :

Solution :

0 0 and C =   3 3

−1 , 1

Obviously both A ( B + C) and AB + AC are square matrices of the

same order 2. Now ∴

Again

and

∴ Thus,

−1 + 0 B+C=  2+3 1 A ( B + C) =  0

0 − 1 −1 −1 = ⋅  3 + 1  5 4 −1 −1 −1 ×  2  5 4 1. (−1) + (−1).4 − 5 1. (−1) + (−1).5 − 6 = =  ⋅  0. (−1) + 24 .  8  0. (−1) + 2.5  10 0 1. (−1) + (−1).2 10 . + (−1).3  1 −1 −1 AB =  × =   2  2 3  0. (−1) + 2.2 0.0 + 23 .  0 −3 −3 = 6  4  1 −1 0 AC =  × 2 3 0 −3 −2 = ⋅ 2  6 −3 − 3 AB + AC =   4+6

−1 1.0 + (−1).3 = 1  0.0 + 23 .

− 3 − 2 − 6 = 6 + 2  10

1. (−1) + (−1).1 0. (−1) + 21 . 

− 5 ⋅ 8

A ( B + C) = AB + AC.

Illustration 4 :

Find the value of x such that 0 2 1 1  [1 1 x] 0 2 1 1 = O.    1 0 1 2

Solution :

The given matrix equation is 0 2 1 1  [1 1 x] 0 2 1 1 = O    1 0 1 2

or

[1 + 0 + 2 x

or

[1 + 2 x

or

[1 + 2 x + 2 + x + 3] = O

0 +2+ x

2+ x

1 2 + 1 + 0] 1 = O  1

1 3] 1 = O  1

[UPTU 2004, 06]

103

or

3x + 6 = 0

or

x = − 2.

Illustration 5 :

Find the value of x such that 3 2  1   1  [1 x 1] 2 5 1  2  = O.    3 2  x  15

Solution :

or

or

The given matrix equation is 3 2  1 [1 x 1]  2 5 1   3 2 15

 1 2 =O    x 

 1 [1 + 2 x + 15 3 + 5 x + 3 2 + x + 2]  2  = O    x   1 [2 x + 16 5 x + 6 x + 4]  2  = O    x 

or

[2 x + 16 + 10 x + 12 + x 2 + 4 x] = O

or

x 2 + 16 x + 28 = 0

or

( x + 2)( x + 14) = 0 or x = − 2, − 14. −1 1  1 a Illustration 6 : If A =  and (A + B)2 = A2 + B2 , find a , B =  b 2 − 1 − 1    and b. Solution :

Given (A + B)2 = A2 + B2

or

(A + B)(A + B) = A2 + B2

or

A (A + B) + B (A + B) = A2 + B2

or

A2 + AB + BA + B2 = A2 + B2

or

AB + BA = O −1  a 1  a 1  1  1 + 2    −1  b −1  b −1  2  2  a + 2 − a − 1  a−b + =O  2a − b  3  b − 2 − b + 1  − a + 1  2a − b + 2 =O  2a − 2 − b + 4 

or or or or

−1 =O −1

2a − b + 2 = 0, − a + 1 = 0, 2a − 2 = 0, − b + 4 = 0

or

a = 1, b = 4.  1 Illustration 7 : If A =   −1

0  1 and I =   7 0

0 , find k so that A2 = 8A + kI . 1

104

Solution :

and

We have  1 A2 = A A =   −1  1 8A + kI = 8   −1

0  1 7  −1 0  1 +k  7 0

0  1 = 7  −8 0  8 = 1  −8

0  ⋅ 56 + k  0  8 + k  1 = 8 A + kI ⇒  = 49  − 8  −8

0 49 0  k + 56  0

0 k

8+k =  −8

Now

A2

⇒

1 = 8 + k, 49 = 56 + k

⇒

0  ⋅ 56 + k

k = − 7.

Illustration 8 : Prove that the product of two matrices cos θ sin θ  cos2 θ  cos2 φ and    sin2 θ   cos θ sin θ  cos φ sin φ

cos φ sin φ  sin2 φ 

is a zero matrix when θ and φ differ by an odd multiple of π / 2. Solution : We have  cos 2 θ   cos θ sin θ

cos θ sin θ  cos 2 φ  sin2 θ   cos φ sin φ

cos φ sin φ  sin2 φ 

cos 2 θ cos 2 φ + cos θ sin θ cos φ sin φ cos 2 θ cos φ sin φ + cos θ sin θ sin2 φ =  cos θ sin θ cos 2 φ + sin2 θ cos φ sin φ cos θ sin θ cos φ sin φ + sin2 θ sin2 φ cos θ sin φ (cos θ cos φ + sin θ sin φ)  cos θ cos φ (cos θ cos φ + sin θ sin φ) = sin θ sin φ (cos θ cos φ + sin θ sin φ)  sin θ cos φ (cos θ cos φ + sin θ sin φ)  cos θ cos φ cos (θ − φ) =  sin θ cos φ cos (θ − φ) 0 0 = , 0 0

cos θ sin φ cos (θ − φ) sin θ sin φ cos (θ − φ)

since θ and φ differ by an odd multiple of π / 2 therefore cos (θ − φ) = 0. 0 − tan (α / 2)   Illustration 9 : Let A =   and I be the identity matrix of tan ( α / 2 ) 0   cos α − sin α   order 2. Show that I + A = (I − A)  ⋅ sin α cos α   Solution : Then

Let

tan

0 A=  t sin α =

α = t. 2 −t , 0  2 tan (α / 2) 2

1 + tan (α / 2)

=

2t 1+ t 2

105

and

cos α =

1 − tan2 (α / 2) 1 + tan2 (α / 2)

1 I + A=  0 1 I − A=  0

Now and

0 0 + 1  t 0 0 − 1  t

cos α (I − A)   sin α

∴

=

1− t 2 1+ t 2

⋅

− t 1 = 0 t − t  1 = 0 − t

− sin α  1 = cos α − t

t 1

− t ⋅ 1 t ⋅ 1 1 − t 2  2 1 + t  2t  2 1 + t

−

2t  1+ t 2   1− t 2   1 + t 2 

 1− t 2 t (1 − t 2 ) 2t 2 2t + − +   1+ t 2 1+ t 2 1+ t 2 1+ t 2   =  2 1− t 2  2t 2  − t (1 − t ) + 2t +  1+ t 2 1+ t 2 1 + t 2 1 + t 2   − t 1 = = I + A. 1 t − sin α cos α (I − A)  = I + A. cos α  sin α

Thus

Illustration 10 : 1 I = 0 Solution :

4 If A =  3

1 0 , then prove that A 2 − 6A + 5I =   2 0

0 ⋅ 1

[UPTU 2002]

4 We have A 2 = A A =  3

1 4 2 3

1 19 = 2 18

6 ⋅ 7

Now 19 A 2 − 6A + 5I =  18

6 − 24 + 7  − 18

− 6 5 + − 12 0

0 0 = 5 0

0 ⋅ 0

Comprehensive Exercise 2 1.

For the following matrices verify A (BC) = (AB) C i. e., matrix multiplication is associative 1 (i) A =  2

0 where 0

3 −1

 1 1 , B = −1  0  0

1 3 3 and C =   1 5

2 ⋅ −2

106

2.

3.

4.

5.

6. 7. 8.

5 (ii) A = 3  0 3 If A =  1  2

1 4 −1

0 0  −1 , B =  1   2  0

−1  2 0  ,B =   1 1 

−1 2 3

3 2  9 and C = 0   7  3

7 0 and C =   1 −1

verify that A ( B + C) = AB + AC.  2 −3 −5  2   If A = − 1 4 5 and B = − 1     1 − 3 − 4   1 AB = A and BA = B. 1 0 −1 0   If A = 3 −3 3 ,B =  2    5   5 −5 −1

4 4 4

−2 3 −2

1 −1  ⋅  5 

1 , 3 

−4 4  , show that  − 3 

3 6  , show that A 2 B 2 = A2 .  4  [UPTU 2005]

2 Let f ( x) = x 2 − 5 x + 6, find f (A) i. e., A2 − 5A + 6I, if A = 2   1 2  4 If A =  , find (A − 2 I )(A − 3I ). 1 −1

0 1 −1

1 3 ⋅  0

2 If A =  3

−1 4 0 2 and B =    , find 3A − 2B. 2 − 1 7   2 3 3 −2  1 2 (i) If A =  ,B= , and C =    , verify 1 4 −1 2 3 4 (A + B) C = AC + BC 0  1  (ii) If A = 3 −1  1 −2

[UPTU 2010]

−2 5 −4  0   0 , B= −2 1 3    1  0 2   − 1 5 2  1  and C = −1 1 0  , verify that A ( B − C) = AB − AC.   −1 1   0 3 2 9. Show that the matrix A =  satisfies the equation 1 2  A3 − 4 A2 + A = O. 1 0  3 1 2 10. If A =  and I =    , then find λ so that A = 5A + λI. − 1 2 0 1      3 1 2 11. If A =  , show that A − 5 A + 7I = O. −1 2

107

A nswers  1 5.  −1  −5  3 7.  38

−1 −1 4 −20 ⋅ −11

−3 −10 ⋅  4

0 6.  0

0 ⋅ 0

10. λ = − 7.

3.11 Transpose of a Matrix Definition. Let A be an m × n matrix. Then the n × m matrix obtained from A by interchanging its rows into columns and columns into rows is called the transpose of A and is denoted by A′ or AT . Thus if A = [ a ij ] m × n then A′ = [ b ji ] n × m , where b ji = a ij i. e., the ( j, i)th element of A′ = (i, j)th element of A. 3 2 5  1  For example, if A = 0 then 1 1 1   −1 3 0 3 × 4  2  1 3 A′ =  2 5 

0 1 1 1

2 −1  3 0 4 × 3 .

3.12 Properties of Transpose of a Matrix If A′ and B′ are the transposes of A and B respectively then (i)

(A ′ )′ = A,

(ii) (A + B)′ = A ′ + B′ , A and B being of the same size (order), (iii) (k A) ′ = k A ′ , k being any complex number, (iv) (AB)′ = B′ A ′ , A and B being conformable for multiplication.

3.13 Symmetric and Skew-symmetric Matrices Symmetric Matrix:

A square matrix A = [ a ij ] is said to be symmetric if its (i, j)th

element is the same as its ( j, i)th element, i.e., if a ij = a ji for all i and j.

108

Thus a square matrix A is symmetric if A = A′ . h g   a1 a2 a3  a 0 Example:  h b f , a2 b1 b2 ,      1 f c  a3 b2 c1    g matrices. Skew-symmetric Matrix :

1 2 

are

symmetric

A square matrix A = [ a ij ] is said to be skew-symmetric if

the (i, j)th element of A is the negative of the ( j, i)th element of A i.e., if a ij = − a ji for all i and j. Thus a square matrix A is skew-symmetric if A = − A′ . Note: For a skew-symmetric matrix, we have ∴

a ij = − a ji , for all i, j. a ii = − a ii , for all values of i

or

2a ii = 0 or a ii = 0. Thus the diagonal elements of a skew-symmetric matrix are all zero. Example :  0 −h  − g

h 0 −f

g  0 f , − i   0   3

i 0 1

−3  −1  are skew-symmetric matrices.  0 

Illustration 1 : If A, B are symmetric (skew-symmetric), then so is also A + B. Solution : Let A and B be symmetric matrices of the same order. Then, we have A = A′ and B = B′ . Now ( A + B)′ = A′ + B′ = A + B. ∴ A + B is a symmetric matrix. Again suppose A and B are skew-symmetric matrices of the same order. Then we have A′ = − A, B′ = − B. Now Since

( A + B)′ = A′ + B′ = − A − B = − ( A + B). ( A + B)′ = − ( A + B), therefore A + B is a skew-symmetric matrix.

Illustration 2 : If A and B are symmetric matrices and AB = BA, then show that AB is symmetric. Solution : Suppose A and B are symmetric matrices. Then we have A = A′ and B = B′. Also suppose that AB = BA. Now ( AB)′ = B′ A′ = BA

[ ∵ A = A′ , B = B′ ]

= AB.

[ ∵ AB = BA]

109

Since

( AB)′ = AB, therefore AB is a symmetric matrix.

Illustration 3 : Show that every square matrix is uniquely expressible as the sum of a symmetric matrix and a skew-symmetric matrix. Solution : 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 We have

∴

′ 1 1 P ′ =  ( A + A′ ) = ( A + A′ )′ 2 2   1 1 = { A′ + ( A′ )′ } = ( A′ + A) 2 2

[ ∵ (k A)′ = k A′ ] [( A′ )′ = A]

= P. P is a symmetric matrix.

′ 1 1 1 Q′ =  ( A − A′ ) = { A − ( A′ )}′ = { A′ − ( A′ )′ } 2 2 2  1 1 = ( A′ − A) = − ( A − A′ ) = − Q 2 2 ∴ Q is a skew-symmetric matrix. Thus we have expressed the square matrix A as the sum of a symmetric and a skew-symmetric matrix. To prove that the representation is unique, let A = R + S be another such representation of A, where R is symmetric and S is skew-symmetric.

Again

Then we have to show that R = P, S = Q. We have A′ = ( R + S)′ = R′ + S ′ = R − S. [ ∵ R = R′ , S = − S ′ ] and ∴ A + A′ = 2 R A − A′ = 2S, 1 1 so that R = ( A + A′ ) and S = ( A − A′ ). 2 2 Thus

R = P and S = Q. Hence every square matrix is uniquely expressible as the sum of a symmetric and a skew-symmetric matrix. Illustration 4 : Show that every square matrix is uniquely expressible as the sum of a symmetric matrix and a skew-symmetric matrix. Solution : Let A be any square matrix. 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 1 1 We have P′ =  (A + A ′ ) ′ = (A + A ′ ) ′ 2 2 

[∵ (kA)′ = kA ′ ]

110

1 {A ′ + (A ′ )′ } 2 1 = (A ′ + A) 2 1 = (A + A ′ ) = P. 2

[∵ (A + B)′ = A ′ + B′ ]

=

[∵(A ′ )′ = A]

Therefore P is a symmetric matrix. 1 1 1 Again Q′ =  (A − A ′ ) ′ = (A − A ′ ) ′ = {A ′ − (A ′ )′ } 2 2 2  1 1 = (A ′ − A) = − (A − A ′ ) = − Q. 2 2 Therefore Q is a skew-symmetric matrix. Thus we have expressed the square matrix A as the sum of a symmetric and a skew-symmetric matrix. To prove that the representation is unique, let A = R + S be another such representation of A, where R is symmetric and S skew-symmetric. Then to prove that R = P and S = Q. We have

A ′ = (R + S)′ = R ′ + S ′ = R − S

∴

A + A ′ = 2R and A − A ′ = 2S. 1 1 R = (A + A ′ ) and S = (A − A ′ ). 2 2

This gives Thus

[∵ R ′ = R and S ′ = − S]

R = P and S = Q.

Therefore the representation is unique. 3 2 3 Illustration 5 : If A =  ,B =   1 0 2

4 , verify that 1

(A + B )′ = A ′ + B′ and ( AB )′ = B′ A ′ . Solution :

We have 2 A′ =  0

3 T 0 2 = and B ′ =  1 1 3 +3 3 + 4 5 7 = ⋅  +2 1 + 1 2 2

Now

2 A +B= 0

∴

5 (A + B)′ =  2

and Thus, Again

7 T 2 5 =  2 2 7 0 + 2 5 2 + 3 A ′+B′= = 1 + 1 7 3 + 4

(A + B)′ = A ′ + B′ . 3 3 4 2.3 + 3.2 2 AB =  =   1 2 1 0.3 + 1.2 0

3 2 

4 T 3 =  1 4

2 ⋅ 1

2 ⋅ 2 2.4 + 3.1 = 0.4 + 1.1

12  2 

11 ⋅ 1

111

12 (AB)′ =  11 3 B′ A ′ =  4

∴ Also Thus,

2 ⋅ 1 2 1

2 3 

0 3.2 + 2.3 = 1 4.2 + 1.3

3.0 + 2.1 12 = 4.0 + 1.1 11

2 ⋅ 1

(AB)′ = B′ A ′ .

Comprehensive Exercise 3 1.

3 If A =  0

1 1

−1 , find AA′ and A ′ A. 2

Also show that AA′ and A ′ A are both symmetric matrices. 2.

3. 4.

Express the following matrices as the sum of a symmetric and a skew symmetric matrix. 2 −1 5 −1  2 (i) A = 3 0 1 , (ii)  ⋅   4 2   1 1 0 sin α   cos α If A =  , show that AA ′ = I = A ′ A. cos α  − sin α For the matrices A and B verify that (AB)′ = B ′ A ′ , where 1 A = 3

5.

 3 If A =  10

2 4

2 1

5  4  ,B =  1 2    5  3 and B =  −2 −2

4 0

7 , show that 5

(A + B) ′ = A ′+ B′ . 6.

 −1 Show that (AB)′ = B′ A ′ , when A =  −7

7.

1 If A =  2

3 4

 1  , B = 2  

4 5

3 2

  , verify that (AB) ′ = B′ A ′ . 

A nswers 11 1. AA ′ =  −1

−5 0  0 , B =  8  1

 9 −1  , A′A =  3  5  −3

3 2 1

−3  1 ⋅  5 

0 3⋅  −8 

112

2.

2 1 (i) A = P + Q =  1 0  3 1  2 (ii) A = P + Q =  3 / 2

3  0 −2 1 +  2 0   0 −2 0 3 / 2  2 + 2  5 / 2

2 0  0 −5 / 2 . 0 

3.14 Adjoint of a Square Matrix Definition. Let A = [a ij ] n × n be any n × n matrix. The transpose B′ of the matrix B = [A ij ] n × n , where Aij denotes the cofactor of the element a ij in the determinant |A|, is called the adjoint of the matrix A and is denoted by the symbol Adj A. Thus the adjoint of a matrix A is the transpose of the matrix formed by the cofactors of A. Thus if …  a11 a12 a1n    a a … a 21 22 2n  A = ,  … … … …   a n2 a nn  …  a n1 then  A11  A21 adj A =   …   An1

A12

…

A22 …

…

An2

… …

A1n   A2 n  …  Ann 

T

 A11  A12 =  …   A1n

A21

…

A22 …

…

A2 n

… …

An1   An2  ⋅ …  Ann 

The adjoint of a matrix is also called the adjugate of the matrix. Theorem. If A be any n-rowed square matrix, then (adj A) A = A(adj A) = | A | I n , where I n is the n-rowed unit matrix.

Illustration 1 : Solution :

α If A =  γ

β , then find adj A . δ 

In |A|, the cofactor of α is δ and the cofactor of β is − γ.

The cofactor of γ is − β and the cofactor of δ is α. Therefore the matrix B formed of the cofactors of the elements of |A| is − γ −β  δ  δ T ⋅ B=  ⋅ Now adj A = B = − γ α α  − β   

113

1 Find the adjoint of the matrix A =  1  2 A (adj A) = |A| I3 = (adj A) A . Illustration 2 :

Solution :

1 2 0

1 −3 and verify that  1

We have |A| = 1 . (2 − 0) − 1 . (1 + 6) + 1 . (0 − 4) = − 9.

Let us find the cofactors A11 , A12 etc. of the elements of |A|. 1 − 3 1 2 2 − 3  = 2, A 12 = −    = − 7, A 13 =    = − 4, We have A 11 =  1 1 0 2 2 0 1 1 1 1 1 1  = − 1, A 22 =    = − 1, A 23 = −  = 2 , A 21 = − 0 1 2 1 2 0 1 1 1 1 1 1  = − 5, A 32 = −   = 4, A 33 =    = 1. A 31 =  2 − 3 1 − 3     1 2 The matrix B formed of the cofactors of the elements of |A| is −7 − 4  2 B =  −1 −1 2 ⋅   4 1 −5 −1 −5  2 T  Now adj A = B = −7 −1 4 ⋅   2 1 −4 1 1  2 −1 −5 −9 1    ∴ A (adj A) = 1 2 −3 −7 −1 4 =  0     0 1 −4 2 1  0 2 1 = − 9 0  0 Also

 2 (adj A) A = −7  −4 −9 = 0   0

0 1 0 −1 −1 2 0 −9 0

0 0 = |A| I 3 ,  1 −5  1 1   4 1 2  1 2 0 0 0 = |A| I 3 .  −9

0 −9 0

0 0  −9

since | A| = − 9. 1 −3  1

Hence, A (adj A) =|A| I3 = (adj A) A .

3.15 Inverse or Reciprocal of a Matrix Definition. Let A be any n-rowed square matrix. Then a matrix B, if it exists, such that AB = BA = I n

114

is called the inverse of A i.e., A −1 = B . The inverse of a matrix A is usually denoted by A −1 . Theorem 1. (Uniqueness of the inverse of a matrix). Every invertible matrix possesses a unique inverse. Note. For the products AB, BA to be both defined and be equal, it is necessary that A and B are both square matrices of the same order. Thus non-square matrices cannot possess inverse. Theorem 2. (Existence of the Inverse). The necessary and sufficient condition for a square matrix A to possess the inverse is that |A| ≠ 0. OR The necessary and sufficient condition for a matrix A to be invertible is that it is non-singular. Important. A

−1

If A be an invertible matrix then 1 = adj A. |A|

Theorem 3. (Reversal law). If A, B be two n-rowed non-singular matrices then AB is also non-singular and (AB) −1 = B −1 A −1 , i.e., the inverse of a product is the product of the inverses taken in the reverse order. Theorem 4. If A be an n × n non-singular matrix, then (A ′ ) −1 = (A −1 )′ .

Illustration 1 : Solution :

 1 Find the inverse of the matrix A =  1   2

2

5 −1 ⋅  −1 [UPTU 2001]

−1 3

We have | A | = (1 + 3) − 2 (− 1 + 2) + 5 (3 + 2) = 4 − 2 + 25 = 27.

Since | A | ≠ 0, therefore A −1 exists. Let Aij be the cofactor of a ij in | A |. Then we have A11 = 4, A12 = − 1, A13 = 5, A21 = 17, A22 = − 11, A23 = 1, A31 = 3, A32 = 1, A33 = − 3. Now

∴

 4 adj A = 17   3 A

−1

−1 −11

5 1  1 −3  4 1 1  = adj A = −1 |A | 27   5

T

 4 = −1   5 17 −11 1

17 −11 1 3 1 ⋅  −3

3 1 ⋅  −3

115

1 Find the inverse of A = 1  1

Illustration 2 : Solution :

3

3 3 and verify that A −1 A = I3 .  4

4 3

We have |A| = (16 − 9) − 3 (4 − 3) + 3 (3 − 4) = 7 − 3 − 3 = 1.

Since |A| ≠ 0, therefore A −1 exists. Let Aij be the cofactor of a ij in |A|. Then we have A11 = 7,

A12 = − 1,

A31 = − 3, Now

∴

Now

 7 adj A = −3  −3 A

−1

−1 1 0

 7 1 = adj A = −1  |A| −1

 7 A A = −1  −1 7−3−3 = −1 + 1 + 0  −1 + 0 + 1 −1

Illustration 3 : Solution :

A32 = 0,

A13 = − 1, A21 = − 3,

A22 = 1,

A23 = 0,

A33 = 1. −1 0  1

T

 7 = −1  −1

−3 0

−3 0  0 1 21 − 12 − 9 −3 + 4 + 0 −3 + 0 + 3

1 1  1

0

−3 0 ⋅  1

−3 0 ⋅  1 3 4

1

−3 1

−3 1

3 3  3 4 21 − 9 − 12 − 3 + 3 + 0 =  − 3 + 0 + 4

1 Find the inverse of the matrix A = 2  3

2 3 1

1 0  0

0 1 0

0 0 = I 3 .  1

3 1 ⋅  2 [UPTU 2001]

We have | A | = 1 (6 − 1) − 2 (4 − 3) + 3 (2 − 9) = 5 − 2 − 21 = − 18

Since | A | ≠ 0, therefore A −1 exists. Let Aij be the cofactor of a ij in | A |. Then we have A11 = 5, A12 = − 1, A13 = − 7, A21 = − 1, A22 = − 7, A23 = 5, A31 = − 7, A32 = 4, A33 = − 1. Now

∴

 5 adj A =  −1  −7 A −1

−1 −7

−7 5  5 −1  5 1 −1  = adj A = −1 |A | 18  −7

T

 5 =  −1  −7 −1 −7 5

−1 −7 5 −7 5 ⋅  −1

−7 5 ⋅  −1

116

2 Illustration 4 : Find the inverse of the matrix A = 4   1

3

4 1 ⋅  4 [UPTU 2002]

3 2

Solution : We have | A | = 2 (12 − 2) − 3 (16 − 1) + 4 (8 − 3) = 20 − 45 + 20 = − 5. Since | A | ≠ 0, Therefore A −1 exists. Let Aij be the cofactor of a ij in | A |. Then we have A11 = 10, A12 = − 15, A13 = 5, A21 = − 4, A22 = 4, A23 = − 1, A31 = − 9, A32 = 14, A33 = − 6 Now

∴

 10 adj A = − 4   −9 A −1 =

−15 4

1 adj A |A |

1 Illustration 5 : If A = 1  1

1 2 4

T

5  10 −1 = −15   14 −6  5 −4  10 1 = − −15 4 5 −1  5 1 3 , find A −1 .  9

−4 4 −1

−9 14 ⋅  −6

−9 14 ⋅  −6

Solution : We have |A| = 1 (18 − 12) − 1 (9 − 3) + 1 (4 − 2) = 6 − 6 + 2 = 2. Since |A| ≠ 0, therefore A −1 exists. Let A ij be the cofactor of a ij in |A|. Then we have A11 = 6 , A12 = − 6 , A13 = 2 , A21 = − 5, A22 = 8 , A23 = − 3 , A31 = 1 , A32 = − 2, A33 = 1.  6 adj A = − 5   1

T

2 −5 1  6   Now −3 = −6 8 − 2    −2 1 −3 1  2 −5 1  6 1 1 −1 ∴ A = adj A = −6 8 − 2 ⋅  |A| 2 −3 1  2 2 2 1  Illustration 6 : If A = 2 1 2 ,using A2 − 4A − 5I = O and hence find A −1 .   2 1 2 [UPTU 2003; 08] 1 Solution : We have 2  2

−6 8

2 1 2

2 2 ,  1

117

2 2  1 2 1    A =A A = 2 1 2 2 1   2 1 2 2 2 8 8 2 9 1 2    A − 4A − 5I = 8 9 8 −4 2 1    8 9 2 8 2 8−8−0 8−8 9 − 4 − 5  = 8−8−0 9−4−5 8−8  8−8−0 9−8 8 − 8 − 0 2

Now

2 9 8   2 = 8 9   1 8 8 2 0 1   2 −5 0 1   1 0 0 − 0 0 0   −0 ≡ 0 0   − 0 0 0

8 8 ⋅  9 0 0  1 0 0 ⋅  0

Now we have to find A −1 . Multiplying both sides of A2 − 4A − 5I = O by A −1 , we get A −1 (A2 − 4A − 5I ) = O A −1 A 2 − 4A −1 A − 5A −1 I = O

or

2 1 5A −1 = A − 4I = 2 1  2 2 2 2 −3  = 2 −3 2 ⋅   2 −3  2 2 −3 1 −1 A = 2 −3 5 2  2

or

∴

−8 Show that A =   2 −1 + 4 x − 42 = 0. Hence find A .

Solution : and

−8 We have A =   2 −8 A2 = A A =   2

2 2 − 4  1

A − 4I − 5 A −1 = O

1 0  0

0

0 0  1

1 0

2 2 ⋅  −3

5 satisfies the equation 4

Illustration 7 : x2

or

5 4 5 4

−8  2 

5 74 = 4 −8

−20 ⋅ 26

In order to show that the matrix A satisfies the given equation, we have to show that A2 + 4A − 42 I = O. Now

−20 74 A2 + 4A − 42 I =  +4 26 −8 74 − 32 − 42 =  −8 + 8 + 0

−8  2 

5 − 42 4

1 0 

−20 + 20 0 = 26 + 16 − 42 0

Thus the matrix A satisfies the equation A2 + 4A − 42I = O.

0 1 0 = O. 0

118

Now we have to find A −1 . Multiplying A2 + 4A − 42I = O by A −1 , we get A −1 (A2 + 4A − 42 I ) = A −1 O or

A −1 A2 + 4A −1 A − 42 A −1 I = O

or

A + 4I − 42 A −1 = O

or ∴

− 8 42 A −1 = A + 4I =   2 5 1 − 4 A −1 = ⋅  8 42  6

Illustration 8 :

5 +4 4

3 For the matrix A =  7

1 0 

0 − 4 = 1  6

5 ⋅ 8

1 , find x and y, so that A2 + xI = yA. 5

Hence find A −1 . Solution :

3 Given A =  7

1 ⋅ 5

3 A2 = A A =  7

1 3 5 7 16 + xI = yA ⇒  56

1 16 = 5 56 8 1 + x 32 0

8 ⋅ 32 0 3 = y 1 7

We have

A2

⇒

16 + x  56 

⇒

16 + x = 3 y, 8 = y, 56 = 7 y, 32 + x = 5 y

8 32 +

 3 y = x 7 y

1 5

y 5 y

⇒ y = 8, x = 8. Substituting these values in the given equation, we ge t A2 + 8 I = 8 A. Pre-multiplying both sides by A −1 , we get A −1 (A2 + 8 I ) = A −1 (8 A) or

A −1 A 2 + 8 A −1 I = 8 A −1 A

or

A + 8 A −1 = 8 I

or ∴

1 8 A −1 = 8 I − A = 8  0 −1 1 5 A −1 =  ⋅ 3 8 −7

[ ∵ A −1 A2 = (A −1 A) A = IA = A ] 0 3 − 1 7

 2 Illustration 9 : Show that the matrix A = −1   1 3 2 −1 A − 6A + 9A − 4I = O. Hence find A .

1  5 = 5 −7

−1 2 −1

−1 ⋅ 3

1 −1 satisfies the equation  2 [UPTU 2004]

119

Solution :

 2 We have A = −1   1  2 2 A = A A = −1   1  6 = −5   5 A

Now

3

−1

1 2 −1 ,  −1 2 −1 1  2 2 −1 −1  −1 2  1 −5 5 6 −5  ,  −5 6  −1 1  6  2 2  = A A = −1 2 −1 −5   −1 2  5  1 −21 21  22  = −21 22 −21 ⋅   −21 22  21

−1 2 −1

1 −1  2

−5

5 −5  6

6 −5

A 3 − 6A 2 + 9A − 4I  22 = −21   21

−21

21  6  22 −21 − 6 −5   −21 22  5 −1  2  + 9 −1 2  1 − 1  0 0 0 0 = O.  0 0

0 = 0  0

−5

5 6 −5  −5 6 1 0 1   −1 − 4 0 1   2 0 0 

0 0  1

Now we have to find A −1 . Multiplying both sides of A3 − 6A2 + 9A − 4I = O by A −1 , we get A −1 (A3 − 6A2 + 9A − 4 I ) = O or

A −1 A 3 − 6A −1 A 2 + 9A −1 A − 4 A −1 I = O

or

A 2 − 6 A + 9I − 4A −1 = O

or

4 A −1 = A 2 − 6 A + 9I  6 = −5   5

−5 6 −5

5 −5 − 6  6

 2 −1   1

−1 2 −1

1 −1 + 9  2

1 0  0

0 1 0

0 0  1

120

 3 = 1  −1

∴

A −1

1

−1 3 1 ⋅  1 3 1 −1  3 1 = 1 3 1 ⋅  4 1 3 −1

Comprehensive Exercise 4 1.

2.

3.

4. 5.

6.

Find the adjoint of the following matrices : sin α  cos α 2 (i)  (ii)   cos α − sin α 5 1 tan (α / 2)  (iii)  ⋅ 1 − tan (α / 2) 

3 1

Find the adjoint of the following matrices : 1 1 2 0  1    (i) 1 (ii) 2 0 2 3    3 − 1 4 − 1 1    −2 3  1 For the matrix A =  0 2 −1 find A (adj A).   5 2 − 4

5 1 ⋅  1

Also verify that A (adj A) = |A| I 3 = (adj A) A . −3 −3 − 4  If A = 1 0 1 , show that adj A = A .   4 3  4 Find the inverse of each of the following matrices : b 3 a 2 (i)  (ii) A =   c d 5 − 2     Find the inverse of each of the following matrices : 0 −1 2 3  0 1 (i)  3 (ii) 2 4 5 3 1     −4 −7 1 2 −2 3 3 4 2  (iii) 4 3 1 ⋅   2 4  1

 2 (iii)  −3

5 ⋅ 1

[UPTU 2006]

121

7.

4 If A =  2

3 , find x and y such that A2 − xA + yI = O. Hence evaluate 5

A −1 .

A nswers − sin α cos α  1 (ii)   sin α  cos α  −5 1 − tan ( α / 2 )   (iii)  ⋅ tan ( α / 2 ) 1  

1. (i)

−5 − 2  8  2. (i) − 4 −3 1   3 − 1 − 7 3. |A| I3 , where |A| = 25.

 2 (ii) −3   5

−b  d 1  a  (ad − bc ) − c −5  1 1 (iii) ⋅  2  17 3 4 4 − 8 1 6. (i) 11 −2 − 3  4 0 0 − 4 −4 −9  10 1 (iii) − − 15 4 14 5 −1 −6  5 5 − 3  1  7. ⋅ 4  14 − 2 5.

(i)

(ii)

−3 2

3 6 −3

1 2 19 5

 5 1  (ii) − −1 18  −  7

−13 9 ⋅  −1

3 A −2 

−1 −7 5

−7 5  − 1 

 ⋅  

3.16 Solving Systems of Linear Equations using Inverse of a Matrix Consider a system of n linear equations in n unknowns x1 , x2 , …, x n a11 x1 + a12 x2 + … + a1n x n = b1 a21 x1 + a22 x2 + … + a2 n x n = b2 …

…

…

…

…

…

…

…

…

…

a n1 x1 + a n2 x2 + … + a nn x n = b n . These equations can be written in the form of a single matrix equation

122

where

AX=B a12  a11 a a22  21 A = … …  … …  a n2  a n1

…(1) … … … … …

a1n  a2 n   … …  a nn 

n × n,

 x1  x   2 X =  …  …    x n 

n ×1 ,

 b1  b   2 B =  …  …    b n 

n ×1 .

Suppose A is a non-singular matrix i. e.,|A| ≠ 0. Then A −1 exists. Therefore pre-multiplying (1) by A −1 , we get A −1 (A X ) = A −1 B ⇒ (A −1 A) X = A −1 B ⇒

⇒ I X = A −1 B

X = A −1 B, which gives the solution of the given equations.

This solution of the given system will be unique as shown below. Let x1 , x2 be two solutions of AX = B. Then AX1 = B and AX2 = B. Now A X1 = A X2 ⇒ A −1 (A X1 ) = A −1 (A X2 ) ⇒

(A −1 A) X1 = (A −1 A) X2 ⇒ I X1 = I X2

⇒ X1 = X2 .

Hence the solution is unique. Thus, if A is a non-singular matrix, then the system AX = B, of linear equations, has a unique solution given by X = A −1 B. Criterion For Consistency Of A System Of Linear Equations. Let AX = B be a system of n linear equations in n unknowns. (i) if |A| ≠ 0 i. e., A is non-singular, then the given system is consistent and has a unique solution given by X = A −1 B. (ii) If|A| = 0 and (adj A) B = 0, the system is consistent and has infinite number of solutions. (iii) If |A| = 0 and (adj A) B ≠ 0, the given system is inconsistent.

Illustration 1 :

Solve the following system of equations using matrix method : 5 x + 2 y = 3, 3 x + 2 y = 5.

Solution. The given system of equations can be written in matrix form as

where

5 A = 3

AX = B, 2  x 3 , X =   ,B =   ⋅  2  y 5

We have |A| = 10 − 6 = 4 ≠ 0. Therefore A −1 exists. Cofactors of the elements of the first row of |A| are 2, − 3. Cofactors of the elements of the second row of |A| are −2, 5.

…(1)

123

Now ∴ ∴ Hence

−3 T −2  2 = ⋅  5 5 −3 −2 1 1 2 A −1 = adj A =  ⋅ 5 |A| 4 −3 −2 3 −1 1 2 X = A −1 B =  = − 3 5 5  4 4

 2 adj A =  −2

or

 x  −1  y  =  4 ⋅    

x = − 1, y = 4.

Illustration 2 : Solve for x and y by inverting the matrix in the following 1  x  4 2 = ⋅ 1 3  y  7  [UPTU 2001] Solution :

We have | A | = 6 − 1 = 5 ≠ 0. Therefore A −1 exists.

Cofactors of the elements of the first row of |A| are 3, –1. Cofactors of the elements of the second row of |A| are –1, 2. Now, ∴ ∴

−1 T −1  3 = ⋅  2 2 −1 −1 1 1 3 A −1 = adj A =  ⋅ 2 |A| 5 −1 −1 4  1 1 3 X = A −1B =  = 2 7 2 5 −1

 3 adj A =  −1

or

 x   1  y  = 2 ⋅    

Hence x = 1, y = 2. a Find A −1 , if possible, where A =  c ax + by + 1 = 0 and cx + dy + 1 = 0. Illustration 3 :

Solution :

where

b and hence solve the equations d

The given system of equations can be written in matrix form as AX = B a A = c

...(1) b  x −1 , X =   ,B =   ⋅  d  y −1

We have |A| = ad − bc ≠ 0. Therefore A −1 exists. Cofactors of the elements of the first row of |A| are d, − c. Cofactors of the elements of the second row of |A| are − b, a. Now

 d adj A =  − b

∴

A −1 =

∴

− c T  d =  a − c

 d 1 1 adj A = |A| (ad − bc ) − c − b  d 1 X = A −1B =  a (ad − bc ) − c

− b ⋅ a − b ⋅ a −1 −1  

124

or Hence,

 x − d + 1  y  = (ad − bc )  c −    b−d c −a x= , y= ⋅ ad − bc ad − bc

b ⋅ a

1 1 1  Illustration 4 : If A = 1 −1 1 , find A −1 . Hence solve the equations   1 −1 2 [UPTU 2007] x + y + z = 6, x − y + z = 2, 2 x + y − z = 1. Solution :

We have

|A| = 1 (1 − 1) − 1 (− 1 − 2) + 1 (1 + 1) = 0 + 3 + 3 = 6. Since, |A| ≠ 0, therefore A is invertible. Let Aij be the cofactor of a ij in |A|. Then we have 1 − 1  = 1 − 1 = 0, A11 = (− 1)1 + 1   1 − 1 1 A12 = (− 1)1 + 2  2

1  = − (− 1 − 2) = 3 − 1

A13 = 3, A21 = 2, A22 = − 3, A23 = 1, A31 = 2, A32 = 0, A33 = − 2. Now,

∴

0 adj A = 2  2 A

−1

3 −3

3 1  0 − 2 0 1 1 = adj A = 3 |A| 6 3

T

0 = 3  3 2 −3 1

2 −3 1

2 0 ⋅  − 2

2 0 ⋅  − 2

Writing the given system as a single matrix equation, AX = B, we get 1 1  x  6 1 1 − 1 1  y  = 2 ⋅      1 − 1  z   1 2 2 2 6 0  x   1 1 −1     y  = 2 ⋅ or ∴ X=A B= 3 −3 0 2       6 1 − 2  1 3  z  3 Hence,

x = 1, y = 2, z = 3.

Illustration 5 :

Solve using matrix method, the following equations

x + y = 0, y + z = 1, z + x = 3. Solution :

[UPTU 2005]

The given system of equations can be written in matrix form as AX = B

125

1 A = 0   1

where

1 1 0

0  x 0    1 , X = y , B =  1 ⋅      1  z  3

We have |A| = 1 (1 − 0) − 1 (0 − 1) = 1 + 1 = 2 ≠ 0 Let A ij be the cofactor of a ij in |A|. Then we have A11 = 1, A12 = 1, A13 = − 1, A21 = − 1, A22 = 1, A23 = 1, A31 = 1, A32 = − 1, A33 = 1.  1 adj A = −1   1

Now

A

∴

We have

−1

−1 1  −1 1  1 1 1 = adj A = 1 |A| 2 −1

X=A

Hence,

−1

1 1

 1 1 B= 1 2 −  1

T

−1 1 1

 1 = 1  −1 −1 1 1 1 −1  1

−1 1

1 −1 ⋅  1

1 1 −1 ⋅  1

0  2  1 = 1 −2 or   2  3  4

x = 1, y = − 1, z = 2.

Illustration 6 :

Solve the following equations by matrix method.

2 x − y + 3z = 1, x + 2 y − z = 2, 5 y − 5z = 3. Solution :

 x  1  y = −1 ⋅      z   2

[UPTU 2003, 06]

The given system can be written as a single matrix equation AX = B,

where 2 A = 1  0

−1 2 5

3  x  1 −1 , X =  y , B = 2 ⋅      −5  z  3

We have |A| = 2 (− 10 + 5) + 1 (− 5 + 0) + 3 (5 − 0) = − 10 − 5 + 15 = 0. Since |A| = 0, therefore to judge whether the given system is consistent or incosistent, we shall calculate (adj A) B. Let A ij be the cofactor of a ij in |A|. Then we have A11 = − 5, A12 = 5, A13 = 5, A21 = 10, A22 = − 10, A23 = − 10, A31 = − 5, A32 = 5, A33 = 5. Now

−5 adj A = 10  −5 −5 (adj A ) B =  5   5

5 −10

T

5 −10  5 5 10 −5 −10 5  −10 5

−5 = 5   5  1 −5 2 =  5    3  5

10 −10 −10 + 20 − 20 − 20

−5 5 ,  5 − 15 0 + 15 = 0 ⋅    + 15 0

126

Since |A| = 0, (adj A) B = 0, therefore the given system of linear equations is consistent and possesses infinite number of solutions. Taking z = k, where k is real. Substituting for z in the first and second equations of the given system, we get 2 x − y = 1 − 3k , x + 2 y = 2 + k . Writing these equations as a single matrix equation A X = B, we get −1  x 1 − 3k 2 = ⋅ 1 2  y  2 + k  Here |A| = 4 + 1 = 5. Cofactors of the elements of the first row of |A| are 2, –1. Cofactors of the elements of the second row of |A| are 1, 2. −1 T 1  2 = ⋅  2 2 −1 1 1 1 2 A −1 = adj A =  ⋅ 2 |A| 5 −1 1 1 − 3k 1 2 X = A −1B =  2  2 + k 5 −1

2 adj A =  1

Now ∴ We have,

1  2 − 6k + 2 + k 1 4 − 5k = ⋅ 5 −1 + 3k + 4 + 2k 5 3 + 5k 1 1 ∴ x = (4 − 5k), y = (3 + 5k). 5 5 1 1 Since x = (4 − 5k), y = (3 + 5k), z = k, k is real, also satisfy the third equation 5 5 1 1 of the given system, therefore x = (4 − 5k), y = (3 + 5k), z = k form infinite 5 5 solutions of the given system for all real values of k. =

Comprehensive Exercise 5 Solve the following systems of equations using matrix method : 1.

3 x + 4 y − 5 = 0, x − y + 3 = 0.

2.

3 x + y = 7, 5 x + 3 y = 12.

3.

x + 2 y + z = 7, x + 3z = 11, 2 x − 3 y = 1.

4. 3 x − 4 y + 2 z = − 1, 2 x + 3 y + 5z = 7, x + z = 2. Show that each one of the following systems of linear equations is inconsistent :

5. 4 x − 2 y = 3, 6 x − 3 y = 5. 6. 2 x + 5 y = 7, 6 x + 15 y = 13.

127

7.

x + y − 2 z = 5, x − 2 y + z = − 2, − 2 x + y + z = 4. Show that each of the following systems of linear equations is consistent. Also find their solutions :

8. 2 x + 3 y = 5, 6 x + 9 y = 15. 9.

x + y + z = 6, x + 2 y + 3z = 14, x + 4 y + 7z = 30.

10. 2 x − y + 3z = 5, 3 x + 2 y − z = 7, 4 x + 5 y − 5z = 9. 11. Find A

−1

1 if A = 2  3

2 3 −3

− 3 2 ⋅  − 4

Hence solve the system of linear equations x + 2 y − 3z = − 4, 2 x + 3 y + 2 z = 2, 3 x − 3 y − 4z = 11.

A nswers 9 1 , y= ⋅ 4 4 4. x = 3, y = 2, z = − 1.

1.

x = − 1, y = 2.

3.

x = 2, y = 1, z = 3. 1 x = (5 − 3k), y = k, where k is any real number. 2 x = − 2 + k, y = 8 − 2k, z = k, k is real. 1 1 x = (17 − 5k), y = (11k − 1), z = k, k is real. 7 7 17 13  −6 1  A −1 = − 14 5 − 8 ; x = 3, y = 2, z = 1.  67  9 − 1 − 15

8. 9. 10. 11.

2. x =

Comprehensive Exercise 6 Fill in the Blanks Fill in the blanks ‘‘..........’’, so that the following statements are complete and correct.

6 1. If A =  x

3 y and B =   8 7

z are equal matrices, then the respective values 8

of x, y and z are .......... . 2 6 5 3 2. If X + Y =  and X − Y =  , then the matrix X is .......... .  9 − 1 0 0 2 0 3  1 3. If Y =  and 2X + Y =  then X = .......... .  4 2 1 − 3

128

3 4. The order of the matrix   is .......... . 8 5. A square matrix each of whose non-diagonal elements is equal to zero is called a .......... matrix. 6. The necessary and sufficient condition for a square matrix A to possess the inverse is that .......... . β α 7. If A =  , then (adj A) = .......... . δ γ 8. A square matrix A = [a ij ] is called an .......... triangular matrix if a ij = 0 whenever i > j. 9. If A is any matrix and O is a null matrix, then A + O is .......... .

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d).

2 10. If A =  9 1 (a)  9 9 (c)  5

7 1 ,B =   8 0

2 , then A − B is 8

5 5

9 (b)  1 5 (d)  5

5 1

5 5 9 1

x − y 11. The values of x and y such that  x + y (a) x = 2, y = 4 (c) x = 4, y = 4 2 − 1 2 12. If A =  and B =   1 0 6 4 − 4 2 4 − 3 (a)  2 6 0 3 0 (c) does not exist

4 2 = 7 6

(b) x = 2, y = 2 (d) x = 4, y = 2 4 − 3 , then 2 A + B is 0 3 2 (b)  6

4 0

− 3 3

(d) none of these.

cos α − sin α 13. The inverse of the matrix A =  is cos α  sin α sin α sin α − cos α  cos α (a)  (b)   sin α cos α − sin α cos α    cos α (c)   sin α

− sin α cos α

4 is 7

cos α (d)   sin α

sin α − cos α

129

 cos α 14. The value of cos α  − sin α 0 0 (a)  1 1 1 (c)  0 1 15. If  0

0 1 y 3 +2  1 5

(a) x = 8, y = 2, (c) x = 2, y = − 8

4  7 = x 10

sin α + sin α cos α 1 (b)  0 0 (d)  1

 sin α cos α 

− cos α is sin α

1 0 1 0

0 , then x and y is 5 (b) x = − 8, y = 2 (d) x = − 2, y = − 8

16. The inverse of a square matrix A, i. e., A −1 is (a) (adj A) −1

(b)

(adj A) |A|

(c) |A| (adj A)

(d) none of these 1 5  6  17. The product of two matrices  is 3  − 1 2  9 (a)    1  1 (c)    9 2 18. If A =  1

(b) [1 9] (d) [9 1] 1 0 3 and B =   0 1 − 2

2 , then AB is 3

(a) undefined

7 (b)  3

 7 (c) − 2   7

(d) none of these

3 0  2

−2 0

4 1 3 19. The inverse of a matrix  is 2 0 5  4 1 3 (a)  (b)   5 2 0 (c) does not exist

7 2

1 2

(d) none of these

[UPTU 2011]

True or False Write ‘T’ for true and ‘F’ for false statement. 20. In a matrix the number of rows and columns may be different. 21. Matrix multiplication is always commutative.

130

2  6 a + b 22. If  = a − b 5  5 respectively.

2 , then the values of a and b are 7 and − 1 8

5 23. The inverse of the matrix  7

2  3 is 3 − 7

− 2 ⋅ 5

24. If the matrices A and B are not of the same order then we cannot define A + B. 25. If A is a square matrix and |A| ≠ 0, then A is a singular matrix. [UPTU 2010]

26. A square matrix A is invertible only if |A| = 1. 27. Let AX = B be a system of n linear equations in n unknowns. If|A| ≠ 0, then the given system is consistent and has infinite number of solutions. 2 y 8 10 1 2 28. If 2  +3 = , then the respective values of x and   x 5 6 21 3 0 y are 3 and 2. 0 3 29. If A =  2 4 2 30. If  1

2 4 and B =   1 2 5  3 , then (adj A) is   3 − 1

2 0

1 3 , then A + B is   − 1 0

0 0

2 ⋅ − 1

− 5 ⋅ 2

A nswers 1. 7, 6, 3

2.

4 0 

5. diagonal

6.

|A | ≠ 0

9. 13. 17. 21. 25. 29.

A (b) (c) F F F

10. 14. 18. 22. 26. 30.

(a) (c) (b) T F T.

4 4

−1 3.  − 2  δ 7.  − γ 11. (d) 15. (c) 19. (c) 23. T 27. F

− 1 − 1

4. 3 × 1

− β α

8. upper 12. 16. 20. 24. 28.

(c) (b) T T T

131

4 T rigonometry

4.1 Angles

A

ngle.

→

If a ray OA starting from its initial

position OA rotates about its end point O and takes the final position OB,we say that the ∠ AOB has been generated. The revolving ray is called the generating line of the angle. Its initial position OA is called the initial side and the final position OB is called the terminal side of the angle. The end point O about which the ray rotates is called the vertex of the angle. Measure of an Angle. The measure of an angle is the amount of rotation of the revolving line from the initial side to the terminal side. Positive and Negative Angles. An angle described by a revolving ray is said to be positive or negative according as its rotation from the initial side to the terminal side is anticlockwise or clockwise respectively.

132

Right Angle. If the revolving ray starting from its initial position to the final position describes one quarter of a circle, we say that the measure of the angle so formed is a right angle.

4.2 Quadrants Let X ′ OX and Y ′ OY be two coplanar lines at right angles to each other. These lines divide the plane into four equal parts, called quadrants. The lines X ′ OX and Y ′ OY are known as x-axis and y-axis respectively. The regions XOY , YOX ′ , X ′ OY ′ and Y ′ OX are known as the first, the second, the third and the fourth quadrants respectively.

4.3 Systems of Measurement of Angles There are three systems for measuring angles, viz., (i) Sexagesimal or English system (ii) Circular system and (iii) Centesimal or French system.

4.4 Sexagesimal System (English System or Degree Measure) The number of degrees on the circumference of the circle between the initial and terminal sides of the angle is its degree measure. Each degree is subdivided into 60 equal parts, called minutes and each minute is subdivided into 60 equal parts, called seconds. The symbol 1′ is used to denote one minute and the symbol 1′ ′ is used to denote one second. Thus, 60 °45′ 25′ ′ means 60 degrees, 45 minutes and 25 seconds. We have 1 right angle = 90 degrees (= 90 ° ) 1° = 60 minutes (= 60 ′ ) and 1′ = 60 seconds (= 60 ′ ′ ). Remark 1. If there is no rotation, i. e., the initial and terminal sides of the angle coincide, the measure of the angle is 0° .

133 →

Remark 2. If a ray OA after having made n complete rotations arrives at the final position →

OB, then the measure of the angle so formed is (n . 360 + θ)°, where ∠ AOB = θ°. In the →

adjoining figure, the ray OA makes two complete rotations and then further moves through 60°. The measure of the angle so formed is (2 × 360 + 60)° = 780 ° . Different Types of Angles. An angle θ is said to be : (i) an acute angle, if 0 ° < θ < 90 ° (ii) a right angle, if θ = 90 ° (iii) an obtuse angle, if 90 ° < θ < 180 ° (iv) a straight angle, if θ = 180 ° (v) a reflex angle, if 180 ° < θ < 360 ° .

4.5 Circular System (or Radian Measure) One radian, written as 1c , is the measure of an angle subtended at the centre of a circle of radius r by an arc of length r. Theorem 1. Radian is a constant angle. Theorem 2. The number of radians in an angle subtended arc s by an arc of a circle at the centre = i. e.,θ = , where θ c is radius r the angle subtended at the centre of a circle of radius r by an arc of length s.

4.6 Relation between Degrees and Radians →

Suppose a ray OA revolves about its end point O and generates a circle of radius r. Then the radian measure of the angle formed by one complete revolution of the revolving ray 2 πr s 2 π r  = = 2 π. ∵ θ= =  r r r   But one complete revolution of the revolving ray generates an angle of 360° . 2 π radians = 360 ° or

∴ Corollary 1.

π radians = 180 ° .

Since π radians = 180 ° , therefore

1 radian =

180 °  7 o = 180 ×  = 57° 16′ 45′ ′ (approx.)  π 22

134

Corollary 2.

Since 180° = π radians, therefore  22  π radians =   radians = 0 ⋅ 01746 radian. 180  7 × 180 

1° =

4.7 Centesimal System (French System or Grade Measure) In this system a right angle is divided into 100 equal parts, called grades. Each grade is subdivided into 100 equal parts, called minutes and each minute is subdivided into 100 equal parts, called seconds. Thus,

1 right angle = 100 grades = 100

g

1 grade = 100 minutes = 100 ′ and 1 minute = 100 seconds = 100 ′ ′ . Obviously, π radians = 180 ° = 200 g . Some Useful Facts About the Hands of a Clock. (i) The angle between two consecutive digits in a clock is 30°, i. e.,

π radians. 6

(ii) The minute hand rotates through an angle of 6° in one minute. (iii) The hour hand rotates through an angle of 30° in one hour.

Illustration 1 : 7π (i)    12 

Find the degree measure corresponding to the following radian measures : c

π c (ii)    8

1 c (iii)    4 Solution :

(iv) − 2 c .

180  ° We know that π radians = 180 °. Therefore,1c =   ⋅  π 

Hence, c

7π 180  ° =  ×  = 105° .  12 π 

(i)

 7π     12   π    8

c

(ii)

π 180  °  45 °  =  ×  =   = 22 8 2  π 

 1    4

c

(iii)

1 180  °  1 7  °  315 ° =  ×  =  × 180 ×  =  4 4  22  π  22

1 °  = 22° 30 ′ . 2

7 °  7 × 60 ′ = 14° 19 1  ′ = 14  = 14°      22  22   11 1 = 14° 19′  × 60 ′ ′ = 14° 19′ 5′ ′ . 11 

135 ° ° 180 7 − 2 c =  × (− 2) = 180 × × (−2)  π    22

(iv)

6 ° =  − 114  = − 114° 32′ 44′ ′ .  11 Illustration 2 :

Find the radian measures corresponding to the following degree measures :

(i) 240° Solution :

(ii) 75°

(iii) − 22° 30 ′

(iv) 5° 37′ 30 ′ ′ .

c

We know that 180° = π radians = π . c

Therefore,

π  1° =   ⋅ 180 

Hence, (i)

π 240 ° =  × 240 180 

(ii)

π 75° =  × 75 180 

(iii)

−22° 30 ′ =  −22 

c

c

c

4π =   ⋅ 3 c

5π =   ⋅  12 

° c c 1 °  45  π × 45 = −  π  ⋅  =−  =−    2 180  8 2 2

30 ′ 1 ′ (iv) We have, 30 ′ ′ =   = ⋅  60  2 1 ′  75 ′  75 1  °  5 ° ×  =  =  =  ⋅ 2 2  8 2 60 

∴

37′ 30 ′ ′ = 37 

So,

5° 37′ 30 ′ ′ = 5 

Illustration 3 : 15° .

5 °  45 °  45 π c  π c ×  =  =  =  ⋅ 8 8  32  8 180 

Find the length of an arc of a circle of radius 5 cm subtending a central angle [UPTU 2007] c

Solution : Let s be the length of the arc subtending an angle θ at the centre of a circle of radius r. Then, θ = s / r. Here, ∴

π c  πc r = 5 cm and θ = 15° = 15 ×  =  ⋅  12 180  s π 5π θ= → s = θ × r =  × 5 cm = cm. 12  r 12

Illustration 4 : of 187 cm. Solution : Here, Now

Find the radius of a circle in which a central angle of 45° intercepts an arc

Let r be the radius of the circle. c π π c s = 187 cm and θ = 45° =  × 45 =   ⋅ 180   4 s s 4 7 θ= → r = = 187 ×  cm = 187 × 4 ×  cm = 238 cm.    r θ π 22

136

Illustration 5 :

Find in degrees the angle subtended at the centre of a circle of radius 10 ft

by an arc of length 20 ft. Solution :

Here, r = 10 ft and s = 20 ft. s θ =    r

∴

[UPTU 2005]

c

20 =    10 

c

180  = 2 ×   π 

°

° 7 1260  °  6° = 2 × × 180 =   = 114     11   22 11 ′ 6 8 ′ = 114°  × 60 = 114° 32  11   11

8 = 114° 32′  × 60 11 

′′

= 114° 32′43′ ′ .

Illustration 6 : A horse is tied to a post by a rope. If the horse moves along a circular path always keeping the rope tight and describes 88 metres when it has traced out 72° at the centre, find the length of the rope. Solution : Let P be the position of the post and let PA be the length of the rope in tight position. Suppose the horse moves along the arc AB so that ∠APB = 72° and arc AB = 88 m. Let r be the length of the rope i. e., PA = r metres. π θ = 72° =  × 72 180 

Now, and We have, ⇒

c

c 2 π =   5 

s = arc AB = 88 metres. 2 π 88 s θ= ⇒ = r 5 r  5  5 7 r = 88 ×  m = 88 × ×  m = 70 metres.  2 π 2 22 

Illustration 7 : The minute hand of a big clock is 70 cm long. How many centimetres does its extremity move in 6 minutes time ? Solution : ∴

Angle traced by the minute hand in 60 minutes = 360 ° .

360 Angle traced by it in 6 minutes =  × 6  60 

Thus,

π θ = 36° =  × 36 180 

c

°

= 36° .

π c =   and r = 70 cm.  5

Let s be the required arc length moved by the tip of the minute hand. s π Then, θ= ⇒ s = r θ = 70 ×  cm  r 5 22 1 = 70 × ×  cm = 44 cm.  7 5

137

Illustration 8 : Find the angle between the minute hand and the hour hand of a clock when the time is 7 ⋅ 20. Solution :

The angle traced by the hour hand in 12 hours = 360 °.

∴ The angle traced by it in 7 hours 20 minutes, i. e.,

22 hrs 3

360 22 ° =  ×  = 220 ° .  12 3 Also, the angle traced by the minute hand in 60 minutes = 360 °. 360 ∴ The angle traced by it in 20 minutes =  × 20  60 

°

= 120 ° .

Hence, the required angle between the two hands = 220 °−120 ° = 100 °. Illustration 9 : The moon’s distance from the earth is 360000 kms and its diameter subtends an angle of 31′ at the eye of the observer. Find the diameter of the moon. Solution : Let AB be the diameter of the moon and let E be the eye of the observer. Since the distance between the earth and the moon is quite large, so we may take diameter AB of the moon as arc AB of a circle whose centre is E. If the diameter of the moon is d kilometres, then arc AB = d km. Now, and ∴ ⇒ ⇒

31 θ = 31′ =    60 

°

π 31 c =  ×  180 60 

r = 360000 km. arc θ= radius π 31 d × = 180 60 360000 31 22 1  d = 360000 × × ×  km = 3247 ⋅ 62 km.  60 7 180 

Hence, the diameter of the moon is 3247 ⋅ 62 km. Illustration 10 :

The angles of a triangle are in A.P. The number of degrees in the least is to

the number of radians in the greatest as 60 : π. Find the angles in degrees. Solution :

Let the angles be (a − d )° , a° and (a + d )° .

Then,

(a − d ) + a + (a + d ) = 180

⇒

3a = 180 ⇒ a = 60.

∴ The angles are (60 − d )° , 60 ° and (60 + d )° . Obviously, the least angle is (60 − d )° and the greatest is (60 + d )°.

138

π (60 + d )° =  × (60 + d ) 180 

Now

It is given that,

⋅

Number of degrees in the least angle Number of radians in the greatest angle 60 − d

∴

c

π (60 + d ) ⋅ 180

=

=

60 ⋅ π

60 π

⇒

3 (60 − d ) = 60 + d

⇒

120 = 4d ⇒ d = 30.

Hence, the angles are (60 − 30)° , 60 ° and (60 + 30)° i. e., 30 ° , 60 ° and 90° .

Comprehensive Exercise 1 1. Find the degree measure corresponding to the following radian measures : (i)

c 2 π    15 

c 18 π  (iii)    5 

(v)

5π  c (ii)  −   6 (iv) 11c

− 3c .

2. Find the radian measure corresponding to the following degree measures : (i)

340°

(ii) 15°

(iii) 420°

(iv) −270 °

(v) −300 °

(vi) 7° 30 ′ .

3. The difference between the two acute angles of a right angled triangle is 2π radians. Express the angles in degrees. 5 4. The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians. 5. The angles of a triangle are in A.P. The number of grades in the least is to the number of radians in the greatest as 40 : π. Find the angles in degrees. 6. The circular measure of an angle of a triangle is

3π , the number of grades 10

in the second angle is 70. Find the number of degrees in the third angle.

139

7.

The minute hand of a clock is 10 cm long. How far does the tip of the hand move in 20 minutes ?

8.

A wheel makes 180 revolutions in 1 minute. Through how many radians does it turn in 1 second ?

9.

A railway train is travelling on a circular curve of 1500 metres radius at the rate of 66 km per hour. Through what angle has it turned in 10 seconds ?

10. A railroad curve is to be laid out on a circle. What radius should be used if the track is to change direction of 25° in a distance of 40 metres ? 11. A wire 121 cm long is bent so as to lie along the arc of a circle of radius 180 cm. Find in degrees the angle subtended at the centre by the arc. 12. The number of sides in two regular polygons are 5 : 4 and the difference between their angles is 9°. Find the number of sides in the polygons.

A nswers 1. (i) 24° (ii) −150 ° (v) − (171° 49′ 5′ ′ ) c c 17 π   π 2. (i)  (ii)    12  9  (v)

c  − 5π     3

(iii) 648° 7π (iii)   3

c

3π  c (iv)  −   2

π c (vi)   ⋅  24

π c π c 5π c 4.   ,   ,   ⋅  9  3  9  20 π 6. 63°. 7. cm. 3 10. 91⋅ 64 metres. 11. 38° 30 ′. 3. 81° , 9°.

(iv) 630°

5. 20 ° , 60 ° and 100°. 8. (6π) c .

c

11 9.   ⋅  90 

12. 10 and 8.

4.8 Trigonometric Ratios or Functions Let a revolving ray start from OX and move about O so as to trace out an angle θ = ∠ XOA. Let P ( x , y) be any point other than O on the terminal side OA of this angle θ. From P draw PM perpendicular to x-axis. Let length OP = r , OM = x and MP = y. We take the length OP = r always positive while x and y can be positive or negative depending upon the position of the terminal side OA of ∠ XOA . In the right angled triangle OMP (known as the triangle of reference) we have base OM = x, perpendicular PM = y and hypotenuse OP = r .

140

We define the six trigonometric ratios which are also known as trigonometric functions as given below.

(i)

Perpendicular

y

, and is written as sin θ ; Hypotenuse r Base x (ii) Cosine θ = = , and is written as cos θ ; Hypotenuse r Perpendicular y (iii) Tangent θ = = , and is written as tan θ ; Base x Hypotenuse r (iv) Cosecant θ = = , and is written as cosec θ ; Perpendicular y Sine θ =

Hypotenuse

=

r , and is written as sec θ ; Base x Base x (vi) Cotangent θ = = , and is written as cot θ. Perpendicular y (v) Secant θ =

=

Remark 1. When the terminal side coincides with x-axis, we have y = 0. So in this case cosec θ and cot θ are not defined. When the terminal side coincides with y-axis, we have x = 0. So in this case sec θ and tan θ are not defined. Remark 2. The trigonometric ratios may be positive or negative depending upon x and / or y. Remark 3. The following relations are obvious from the definitions of trigonometric ratios :

141

(i) sin θ cosec θ = 1 → sin θ = (ii) cos θ sec θ = 1 → cos θ =

1 1 and sec θ = sec θ cos θ

(iii) tan θ cot θ = 1 → tan θ = (iv) tan θ =

sin θ cos θ

and cot θ =

1 1 and cosec θ = cosec θ sin θ

1 1 and cot θ = cot θ tan θ cos θ sin θ

⋅

4.9 Fundamental Trigonometric Identities For all values of θ for which the trigonometric functions involved are defined, we have sin2 θ + cos2 θ = 1

(i)

(iii) 1 + cot 2 θ = cosec

2

(ii) 1 + tan2 θ = sec (iv)

θ

(v)

2

θ

tan θ . cot θ = 1

(vi)

sec θ . cos θ = 1 sin θ (vii) tan θ = cos θ

sin θ . cosec θ = 1 cos θ (viii) cot θ = ⋅ sin θ

Remark 1. For convenience, we write (sin θ)2 as sin2 θ, (cos θ)2 as cos 2 θ, (tan θ)2 as tan2 θ, etc. Remark 2. 1 + tan2 θ = sec 2 θ → sec 2 θ − tan2 θ = 1 (sec θ − tan θ) (sec θ + tan θ) = 1 1 sec θ − tan θ = sec θ + tan θ

⇒ ⇒ and

sec θ + tan θ =

Illustration 1 : (i)

1 ⋅ sec θ − tan θ

Prove the following identities :

4

tan θ + tan2 θ = sec 4 θ − sec 8

8

2

2

θ

2

(ii) sin θ − cos θ = (sin θ − cos θ)(1 − 2 sin2 θ cos2 θ) (iii) tan2 θ − sin2 θ = tan2 θ sin2 θ (iv) (sin θ + cosec θ)2 + (cos θ + sec θ)2 = tan2 θ + cot 2 θ + 7. (v) tan2 θ + cot 2 θ + 2 = sec 2 θ . cosec 2 θ 4

[UPTU 2001 (Special)] 2

Solution. (i) We have L.H.S. = tan θ + tan θ = tan2 θ (tan2 θ + 1) = tan2 θ sec 2 θ

[ ∵ sec 2 θ = 1 + tan2 θ]

142

= (sec 2 θ − 1) sec 2 θ

[ ∵ sec 2 θ = 1 + tan2 θ ⇒ tan 2 θ = sec 2 θ − 1]

= sec 4 θ − sec 2 θ = R. H. S. (ii) We have L.H.S. = sin8 θ − cos 8 θ = (sin4 θ)2 − (cos 4 θ)2 = (sin4 θ − cos 4 θ)(sin4 θ + cos 4 θ) = (sin2 θ − cos 2 θ) (sin2 θ + cos 2 θ) [(sin2 θ + cos 2 θ)2 − 2 sin2 θ cos 2 θ] = (sin2 θ − cos 2 θ) (1 − 2 sin2 θ cos 2 θ) = R. H. S. [ ∵ sin2 θ + cos 2 θ = 1] (iii) We have L.H.S. = tan2 θ − sin2 θ =

=

sin2 θ cos 2 θ

− sin2 θ =

sin2 θ − sin2 θ cos 2 θ

sin2 θ (1 − cos 2 θ) 2

cos θ

cos 2 θ =

sin2 θ 2

cos θ

⋅ sin2 θ = tan2 θ sin2 θ = R. H. S.

(iv) We have L.H.S. = (sin θ + cosec θ)2 + (cos θ + sec θ)2 = sin2 θ + cosec2 θ + 2 sin θ cosec θ + cos 2 θ + sec2 θ + 2 cos θ sec θ 2

2

= (sin θ + cos θ) + cosec

2

2

2

θ + sec

2

θ+2+2

= 1 + (1 + cot θ) + (1 + tan θ) + 2 + 2 = tan2 θ + cot 2 θ + 7 = R. H. S. (v) We have L.H.S. = tan2 θ + cot 2 θ + 2 = =

sin2 θ cos 2 θ

+

cos 2 θ sin2 θ

+2=

sin4 θ + cos 4 θ sin2 θ cos 2 θ

(sin2 θ + cos 2 θ)2 − 2 sin2 θ cos 2 θ sin2 θ cos 2 θ

+2

+2

[∵ a2 + b 2 = (a + b)2 − 2ab] = Illustration 2 :

1 − 2 sin2 θ cos 2 θ 2

2

sin θ cos θ

+2=

1 2

sin θ cos 2 θ

= sec2 θ cosec2 θ.

Prove the following identities :

(i) sin6 θ + cos6 θ = 1 − 3 sin2 θ cos2 θ (ii) 2 sec

2

θ − sec

4

θ − 2 cosec

2

θ + cosec

[UPTU 2007] 4

(iii) (1 + cot θ − cosec θ)(1 + tan θ + sec θ) = 2 tan θ + sec θ − 1 1 + sin θ (iv) = ⋅ tan θ − sec θ + 1 cos θ

4

4

θ = cot θ − tan θ

[UPTU 2004]

143

Solution :

(i) We have L.H.S. = sin6 θ + cos 6 θ = (sin2 θ)3 + (cos 2 θ)3 = (sin2 θ + cos 2 θ)(sin4 θ + cos 4 θ − sin2 θ cos 2 θ) [ ∵ a3 + b 3 = (a + b)(a2 + b 2 − ab ] = sin4 θ + cos 4 θ − sin2 θ cos 2 θ

[ ∵ sin2 θ + cos 2 θ = 1]

= [(sin2 θ + cos 2 θ)2 − 2 sin2 θ cos 2 θ] − sin2 θ cos 2 θ = 1 − 3 sin2 θ cos 2 θ = R. H. S. (ii) We have L.H.S. = 2 sec 2 θ − sec 4 θ − 2 cosec 2 θ + cosec 4 θ = 2 (1 + tan2 θ) − (sec 2 θ)2 − 2 (1 + cot 2 θ) + (cosec 2 θ)2 = 2 + 2 tan2 θ − (1 + tan2 θ)2 − 2 − 2 cot 2 θ + (1 + cot 2 θ)2 = 2 tan2 θ − (1 + tan4 θ + 2 tan2 θ) − 2 cot 2 θ + (1 + cot 4 θ + 2 cot 2 θ) = cot 4 θ − tan4 θ = R. H. S. (iii) We have L.H.S. = (1 + cot θ − cosec θ)(1 + tan θ + sec θ)  cos θ sin θ 1   1  = 1 + − +  1 +  sin θ sin θ  cos θ cos θ  =

= =

sin θ + cos θ − 1 cos θ + sin θ + 1 ⋅ sin θ cos θ (sin θ + cos θ)2 − 1 sin θ cos θ 1 + 2 sin θ cos θ − 1 sin θ cos θ

(iv) We have L.H.S. =

= = = =

= =

sin2 θ + cos 2 θ + 2 sin θ cos θ − 1 sin θ cos θ 2 sin θ cos θ sin θ cos θ

= 2 = R. H. S.

tan θ + sec θ − 1 tan θ − sec θ + 1

(tan θ + sec θ) − (sec 2 θ − tan2 θ) tan θ − sec θ + 1

[ ∵ sec 2 θ − tan2 θ = 1]

(sec θ + tan θ) − (sec θ + tan θ)(sec θ − tan θ) tan θ − sec θ + 1 (sec θ + tan θ) [1 − (sec θ − tan θ)] tan θ − sec θ + 1 (sec θ + tan θ)(tan θ − sec θ + 1) tan θ − sec θ + 1

= sec θ + tan θ =

sin θ 1 + sin θ 1 + = = R. H. S. cos θ cos θ cos θ

144

Illustration 3 :

Prove that the expression 2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is

independent of the angle θ. Solution :

[UPTU 2005] 6

6

4

4

We have 2 (sin θ + cos θ) − 3 (sin θ + cos θ) = 2 [(sin2 θ)3 + (cos 2 θ)3 ] − 3 (sin4 θ + cos 4 θ) = 2 [(sin2 θ + cos 2 θ) (sin4 θ + cos 4 θ − sin2 θ cos 2 θ)] − 3 (sin4 θ + cos 4 θ) [∵ a3 + b 3 = (a + b) (a2 + b 2 − ab)] = 2 sin4 θ + 2 cos 4 θ − 2 sin2 θ cos 2 θ − 3 sin4 θ − 3 cos 4 θ [∵ sin2 θ + cos 2 θ = 1] = − (sin4 θ + cos 4 θ + 2 sin2 θ cos 2 θ) = − (sin2 θ + cos 2 θ)2 = − 1, which is independent of the angle θ.

Illustration 4 : Prove the following identities : tan θ cot θ (i) + = sec θ cosec θ + 1 1 − cot θ 1 − tan θ (ii)

1 + cos θ 1 − cos θ

Solution :

= cosec θ + cot θ.

(i) L.H.S. =

tan θ 1 − cot θ

sin θ =

= =

=

cot θ 1 − tan θ

cos θ

cos θ sin θ + cos θ sin θ 1− 1− sin θ cos θ sin θ sin θ cos θ cos θ ⋅ + ⋅ cos θ sin θ − cos θ sin θ cos θ − sin θ sin2 θ cos θ (sin θ − cos θ) 3

=

+

+

cos 2 θ sin θ (cos θ − sin θ)

3

sin θ − cos θ cos θ sin θ (sin θ − cos θ) (sin θ − cos θ)(sin2 θ + cos 2 θ + sin θ cos θ) cos θ sin θ (sin θ − cos θ) [ ∵ a3 − b 3 = (a − b) (a 2 + b 2 + ab)]

=

1 + sin θ cos θ sin θ cos θ

(ii) We have L.H.S. =

=

1 + cos θ 1 − cos θ

1 + 1 = sec θ cosec θ + 1 = R. H. S. sin θ cos θ

145

= = = = Illustration 5 :

1 + cos θ 1 − cos θ

1 + cos θ

×

1 + cos θ

1 + cos θ (1 − cos θ)(1 + cos θ) 1 + cos θ

=

sin2 θ

=

1 + cos θ 1 − cos 2 θ

1 + cos θ sin θ

cos θ 1 + = cosec θ + cot θ = R. H. S. sin θ sin θ

If tan θ + sin θ = m and tan θ − sin θ = n, show that m2 − n2 = 4 mn.

Solution : and

Given that tan θ + sin θ = m

…(1) …(2)

tan θ − sin θ = n

Adding (1) and (2), we have 2 tan θ = m + n m+n 2 or or cot θ = tan θ = ⋅ 2 m+n Subtracting (2) from (1), we have or

2 sin θ = m − n or

sin θ =

m−n 2

2 cosec θ = ⋅ m−n

Now we know that cosec 2 θ − cot 2 θ = 1. 4

∴

or or

2

(m − n)

−

4 [(m + n)2 − (m − n)2 ] (m − n)2 (m + n)2 4 (4mn) (m2 − n 2 )2

(m + n)2

=1

=1

=1

or

16mn = (m2 − n 2 )2

or

m2 − n 2 = 4 mn.

Illustration 6 :

4

If cos θ + sin θ = 2 cos θ, show that cos θ − sin θ = 2 sin θ. [UPTU 2003]

Solution : ⇒ ⇒

We have cos θ + sin θ = 2 cos θ sin θ = ( 2 − 1) cos θ ( 2 + 1) sin θ = ( 2 + 1)( 2 − 1) cos θ, [Multiplying both sides by ( 2 + 1)]

⇒

( 2 + 1) sin θ = (2 − 1) cos θ = cos θ

146

⇒

cos θ − sin θ = 2 sin θ.

Illustration 7 :

If a cos θ − b sin θ = c , show that

a sin θ + b cos θ = ± Solution :

a2 + b 2 − c

2

.

We have a cos θ − b sin θ = c

⇒

(a cos θ − b sin θ)2 = c

⇒

a2 cos 2 θ − 2 ab cos θ sin θ + b 2 sin2 θ = c

⇒

a2 (1 − sin2 θ) − 2 ab cos θ sin θ + b 2 (1 − cos 2 θ) = c

⇒

a2 sin2 θ + b 2 cos 2 θ + 2 ab cos θ sin θ = a2 + b 2 − c

⇒

(a sin θ + b cos θ)2 = a2 + b 2 − c

⇒

a sin θ + b cos θ = ±

Illustration 8 :

2

a2 + b 2 − c

2 2 2

2 2

.

If a cos3 θ + 3a cos θ sin2 θ = m and a sin3 θ + 3a cos2 θ sin θ = n,

then prove that (m + n)2 /3 + (m − n)2 /3 = 2a2 /3 . Solution :

We have a cos 3 θ + 3a cos θ sin2 θ = m

and

a sin3 θ + 3a cos 2 θ sin θ = n

⇒

a cos 3 θ + 3a cos θ sin2 θ + a sin3 θ + 3a cos 2 θ sin θ = m + n

and

a cos 3 θ + 3a cos θ sin2 θ − a sin3 θ − 3a cos 2 θ sin θ = m − n

⇒

a (cos 3 θ + 3 cos 2 θ sin θ + 3 cos θ sin2 θ + sin3 θ) = m + n

and

a (cos 3 θ − 3 cos 2 θ sin θ + 3 cos θ sin2 θ − sin3 θ) = m − n

⇒

a (cos θ + sin θ)3 = m + n

and

a (cos θ − sin θ)3 = m − n

⇒

m + cos θ + sin θ =   a

n 1 /3  

and

m − cos θ − sin θ =   a

n 1 /3  

⇒

2 /3 2 /3  m + n  m − n (cos θ + sin θ)2 + (cos θ − sin θ)2 =  +     a   a 

⇒

2 /3 2 /3  m + n  m − n 2 (cos 2 θ + sin2 θ) =  +     a   a 

⇒

(m + n)2 /3 + (m − n)2 /3 = 2 a2 /3 .

Illustration 9 : If

ax cos θ

+

by sin θ

= a2 − b 2 and

a x sin θ 2

cos θ

−

by cos θ sin2 θ

(a x)2 /3 + (by)2 /3 = (a2 − b 2 )2 /3 .

= 0,prove that

147

Solution :

a x sin θ

We have

2

cos θ

a x sin θ

⇒

cos 2 θ sin3 θ

⇒

3

cos θ

⇒

tan θ =

⇒

sin θ =

and

cos θ =

=

=

−

by cos θ

by cos θ sin2 θ

by

by

⇒ tan3 θ =

ax

ax

(by)1 /3 (a x)1 /3 (by)1 /3 (a x)2 /3 + (by)2 /3 (a x)1 /3 (a x)2 /3 + (by)2 /3

Putting these values of sin θ and cos θ in (a x)

=0

sin2 θ

[(a x)2 /3 + (by)2 /3 ]1 /2 1 /3

(a x) 2 /3

2 /3 1 /2

⋅ ax cos θ

+ (by) 2 /3

+

(by) 2 /3

or

[(a x)

or

(a x)2 /3 + (by)2 /3 = (a2 − b 2 )2 /3 .

[(a x)

= a2 − b 2 , we get

1 /3

[(a x)

]

sin θ

[(a x)2 /3 + (by)2 /3 ]1 /2

or

+ (by)

by

+ (by)

= a2 − b 2

] = a2 − b 2

2 /3

+ (by)2 /3 ]3 /2 = a2 − b 2

Comprehensive Exercise 2 Prove the following trigonometric identities : 1. cot 4 θ + cot 2 θ = cosec 4 θ − cosec 2 θ. 2. (cosec θ − sin θ ) (sec θ − cos θ ) (tan θ + cot θ ) = 1. 3. cosec θ (sec θ − 1) − cot θ (1 − cos θ) = tan θ − sin θ. 4. sec

2

θ + cosec

2

θ = sec

2

θ cosec

2

θ.

5. sec 4 θ (1 − sin 4 θ ) − 2 tan 2 θ = 1. 6. 8.

1 − tan θ 1 + tan θ

=

cot θ − 1 cot θ + 1

⋅

1 1 + = 2 sec 2 θ. 1 + sin θ 1 − sin θ

7. 9.

1 + cos θ 1 − cos θ

= (cosec θ + cot θ)2 .

tan2 θ 1 + tan2 θ

+

cot 2 θ 1 + cot 2 θ

= 1.

148

10.

1 − tan2 θ

= cos 2 θ − sin2 θ.

2

1 + tan θ

11.

cot A cosec A + 1

+

cosec A + 1 cot A

= 2 sec A .

 sec θ − 1  sin θ − 1 12. cot 2 θ   + sec 2 θ   = 0.  1 + sin θ  1 + sec θ 13. (sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A . 14. cos θ (tan θ + 2) (2 tan θ + 1) = 2 sec θ + 5 sin θ . sin2 θ

15. 1 − 16.

1 + cot θ

−

tan A + tan B cot A + cot B

cos 2 θ 1 + tan θ

= sin θ cos θ.

= tan A tan B.

17. (1 + tan α tan β)2 + (tan α − tan β)2 = sec 2

18. sin

A cos

2

B − cos

2

2

A sin

2

B = sin

2

α sec 2

A − sin

2

β.

B.

19. a cos θ + b sin θ = x and a sin θ − b cos θ = y, prove that a2 + b 2 = x 2 + y 2 . 20. If cosec θ − sin θ = a3 , sec θ − cos θ = b 3 , then prove that a2 b 2 (a2 + b 2 ) = 1. 21. If cot θ (1 + sin θ) = 4n and cot θ (1 − sin θ) = 4n, prove that (m2 − n 2 )2 = mn.

4.10 Values of Trigonometric Functions at some Special Angles Measure of θ sin θ In degrees In radians 0° 30° 45° 60° 90° 180° 270°

cos θ

tan θ

cot θ

cosec θ

sec θ

1

0 1

undefined

undefined

1 2

0 π 6

0 1 2

π 4

1

π 3

3 2

2 1 2

π 2

1

0

undefined

0

1

undefined

0

−1

0

undefined

undefined

−1

−1

0

undefined

0

−1

undefined

π 3π 2

2

3 2 1

3 1 3

3 1 1 3

2

3 2 2

2 2

3

149

4.11 Signs of the Values of Trigonometric Functions in Different Quadrants (i) In I quadrant, all trigonometric ratios are positive ; (ii) In II quadrant sin θ and cosec θ are positive while cos θ, sec θ, tan θ and cot θ are negative ; (iii) In III quadrant tan θ and cot θ are positive while sin θ, cosec θ, cos θ and sec θ are negative ; (iv) In IV quadrant cos θ and sec θ are positive while sin θ, cosec θ, tan θ and cot θ are negative.

I

Quadrant:

II

All

Trigonometric ratios which are positive

sin and cosec

III

IV

tan and cot

cos and sec

4.12 Trigonometric Ratios of Allied Angles The angles − θ, 90 ° ± θ, 180 ° ± θ, 270 ° ± θ, 360 ° ± θ, etc., are called allied angles of θ. Remember the following relations : −θ

90° − θ 90° + θ

180° − θ

180° + θ

270° − θ

270° + θ

sin

− sin θ

cos θ

cos θ

sin θ

− sin θ

− cos θ

− cos θ

cos

cos θ

sin θ

− sin θ

− cos θ

− cos θ

− sin θ

sin θ

tan

− tan θ

cot θ

− cot θ

− tan θ

tan θ

cot θ

− cot θ

cot

− cot θ

tan θ

− tan θ

− cot θ

cot θ

tan θ

− tan θ

cosec

− cosec θ sec θ

sec θ

cosec θ

− cosec θ − sec θ

sec

sec θ

− sec θ

− sec θ

cosec θ − cosec θ

− sec θ

− cosec θ cosec θ

Trigonometric ratios for 360° + θ° and n × 360 ° + θ° are the same as those for θ° . Also, the trigonometric ratios for 360° − θ° and n × 360 ° − θ° are the same as those for − θ° .

150

Illustration 1 :

If sin θ = −

3 and tan θ is positive, find all other trigonometric ratios. In 5

which quadrant does θ lie ? Solution :

Since sin θ is negative and tan θ is positive, therefore θ lies in the third

quadrant. So, cos θ is negative. ∴

cos θ = − 1 − sin2 θ 3 2 = − 1 −  −   5 = − 1−

9 25

16 4 =− ⋅ 25 5 1 5 1 5 cosec θ = = − , sec θ = =− , sin θ 3 cos θ 4 =−

Now,

and

sin θ

tan θ =

=  − cos θ 

cot θ =

1 4 = ⋅ tan θ 3

Illustration 2 :

Find the trigonometric ratios of

(i) 135° Solution :

(i)

3  5 3  −  = 5  4 4

(ii) 180°

We have sin 135° = sin (180 ° − 45° ) = sin 45° =

cos 135° = cos (180 ° − 45° ) = − cos 45° = −

∴ and

1 , 2

1 2

[UPTU 2001] tan 135° = tan (180 ° − 45° ) = − tan 45° = − 1. 1 1 cosec 135° = = 2 , sec 135° = =− 2 sin 135° cos 135°

cot 135° =

1 1 = = − 1. tan 135° −1

(ii) We have sin 180 ° = sin (180 ° − 0 ° ) = sin 0 ° = 0, cos 180 ° = cos (180 ° − 0 ° ) = − cos 0 ° = − 1, ∴

tan 180 ° = tan (180 ° − 0 ° ) = − tan 0 ° = 0. 1 1 sec 180 ° = = = − 1. cos 180 ° −1

Obviously, cosec 180° and cot 180° are not defined because sin 180 ° = 0.

151

Illustration 3 : Solution :

If sec θ + tan θ = x , prove that tan θ =

x 2 −1 2x

and sin θ =

x 2 −1 x 2 +1

⋅

We have sec 2 θ − tan2 θ = 1 (sec θ − tan θ)(sec θ + tan θ) = 1 1 sec θ − tan θ = ⋅ sec θ + tan θ

⇒ ⇒

It is given that sec θ + tan θ = x . 1 ∴ sec θ − tan θ = ⋅ x

…(1) …(2)

Subtracting (2) from (1), we get 2 tan θ = x −

2 x 2 −1 1 x −1 = ⇒ tan θ = ⋅ x x 2x

Adding (1) and (2), we get 2 sec θ = x + Now,

sin θ =

Illustration 4 :

tan θ sec θ

2 x 2 +1 1 x +1 = ⇒ sec θ = ⋅ x x 2x

=

x 2 −1 2x

2x x 2 +1

=

x 2 −1 x 2 +1

⋅

Find the values of

(i) cos 855° Solution :

⋅

(ii) tan (−1575° )

(i) We have, cos 855° = cos (2 × 360 ° + 135° ) = cos 135° = cos (180 ° − 45° ) = − cos 45° = −

1 ⋅ 2

(ii) We have tan (−1575° ) = − tan 1575° = − tan (4 × 360 ° + 135° ) = − tan 135° = − tan (180 ° − 45° ) = − (− tan 45° ) = tan 45° = 1. Illustration 5 :

Show that 1 3 sin2 30 ° + cosec 4

Solution :

2

We have 3 sin2 30 ° + 1 = 3 ⋅    2

2 5 sin2 60 ° = − ⋅ 3 12 2 60 ° − 2 tan2 45° + sin2 60 ° 3

60 ° − 2 tan2 45° + 1 cosec 2 4 2

2

1  2 2  3 ⋅   − 2 ⋅ (1)2 + ⋅   4  3 3  2 3 1 1 9 + 4 − 24 + 6 5 = + −2+ = =− ⋅ 4 3 2 12 12 Illustration 6 : (i)

2

+

Prove that

tan 225° cot 405° + tan 765° cot 675° = 0.

(ii) sin θ + sin (90 ° + θ) + sin (180 ° + θ) + sin (270 ° + θ) = 0.

152

Solution :

(i) We have tan 225° cot 405° + tan 765° cot 675° = tan (180 ° + 45° ) cot (360 ° + 45° ) + tan (2 × 360 ° + 45° ) cot (2 × 360 ° − 45° ) = tan 45° cot 45° + (tan 45° ) (− cot 45° ) = 1 . 1 + 1 . (−1) = 1 − 1 = 0.

(ii) We have sin θ + sin (90 ° + θ) + sin (180 ° + θ) + sin (270 ° + θ) = sin θ + cos θ − sin θ − cos θ = 0. Illustration 7 :

Find the values of θ for which sec θ and cosec θ are equal, where

0 ° ≤ θ ≤ 360 ° . Solution :

We have sec θ = cosec θ

⇒

sin θ 1 1 = ⇒ = 1 ⇒ tan θ = 1 cos θ sin θ cos θ

⇒

θ = 45° or 180 ° + 45°

⇒

θ = 45° or 225° .

[ ∵ tan θ is +ve in I and III quad rants]

Hence, the required values of θ are 45°and 225°. Illustration 8 : 1 (i) sin θ = 2

Find all values of θ between 0° and 360° for which 2 (ii) cosec θ = − ⋅ 3

Solution :

(i) We know that sin θ is positive in I and II quadrants. 1 Now, sin θ = = sin 45° . 2 So, θ = 45° . Also, sin 45° = sin (180 ° − 45° ) = sin 135° . [ ∵ sin (180° − θ) = sin θ] Hence, θ = 45° , 135° . 2 3 (ii) We have cosec θ = − ⇒ sin θ = − ⋅ 2 3 Now, sin θ is negative in III and IV quadrants. 3 We have sin θ = − = − sin 60 ° 2 = sin (180 ° + 60 ° ) or = sin (360 ° − 60 ° ) [ ∵ sin (180 ° + θ) = − sin θ and sin (360 ° − θ) = − sin θ] = sin 240 ° or = sin 300 ° . Hence,

θ = 240 ° or 300 ° .

Illustration 9 : (i) tan θ = 2 sin θ Solution : ⇒ ⇒

Find the acute angle θ such that (ii) cot θ + tan θ = 2 cosec θ.

(i) We have tan θ = 2 sin θ sin θ = 2 sin θ ⇒ sin θ = 2 sin θ cos θ cos θ sin θ − 2 sin θ cos θ = 0

153

⇒

sin θ (1 − 2 cos θ) = 0

⇒

sin θ = 0

⇒

θ = 0 ° or θ = 60 ° .

(ii) ⇒

or cos θ =

1 2

We have cot θ + tan θ = 2 cosec θ 1 + tan θ = 2 cosec θ tan θ

⇒

1 + tan2 θ = 2 cosec θ tan θ

⇒

sec 2 θ = 2 sec θ

⇒

sec θ (sec θ − 2) = 0

⇒

sec θ = 2 i. e., cos θ =

Illustration 10 : that

1 i. e., θ = 60 ° . 2

[ ∵ sec θ = 0 is not pos sible]

If A, B, C and D are angles of a cyclic quadrilateral taken in order, prove cos A + cos B + cos C + cos D = 0.

Solution : We know that the opposite angles of a cyclic quadrilateral are supplementary i. e., A + C = 180 ° and B + D = 180 ° . We have

cos A + cos B + cos C + cos D = cos A + cos B + cos (180 ° − A ) + cos (180 ° − B ) [ ∵ C = 180 ° − A and D = 180 ° − B ] = cos A + cos B − cos A − cos B = 0. [ ∵ cos (180° − θ) = − cos θ]

Comprehensive Exercise 3 1. If θ is acute, find all other trigonometric ratios when 24 1 (i) sin θ = (ii) cos θ = 25 2 9 17 (iii) cot θ = (iv) cosec θ = ⋅ 40 8 2. In which quadrant does θ lie, if 5 5 (i) cos θ = − (ii) tan θ = 12 4 (iii) cosec θ = − 4. 3. Find the values of other five trigonometric ratios, when 8 (i) sin θ = − and θ lies in the third quadrant ; 17 4 (ii) cos θ = − and θ lies in the second quadrant ; 5 1 (iii) tan θ = − and θ lies in the fourth quadrant. 3

154

5 and θ lies in the fourth quadrant, find the value of 13 2 − 3 cot θ ⋅ 4 + 9 sec2 θ − 1

4. If cos θ =

5. Evaluate 6.

4 1 1 cot 2 30 ° + 3 sin2 60 ° − 2 cosec 2 45° + tan2 60 ° − cos 0 ° . 3 3 4

Find x from the following equations : (i) cosec (90 ° + θ) + x cos θ cot (90 ° + θ) = sin (90 ° + θ) (ii) x cot (90 ° + θ) + tan (90 ° + θ) sin θ + cosec (90 ° + θ) = 0

7. Find the trigonometric ratios of (i) 225°

(ii) 330°

8. Verify that : (i) cos 2 θ = 1 − 2 sin2 θ, when θ = 30 ° (ii) sin 3θ = 3 sin θ − 4 sin3 θ, when θ = 60 ° (iii) cos 2 θ =

1 − tan2 θ 1 + tan2 θ

, when θ = 30 °

(iv) cos 2 θ = cos 2 θ − sin2 θ, when θ = 60 ° . 9. Show that the value of (sin2 θ + cosec 2 θ) can never be less than 2 . 10. Prove that : (i) cos 510 ° cos 330 ° + sin 390 ° cos 120 ° = − 1 (ii) sin

8π 23π 13π 35π 1 cos + cos sin = 3 6 3 6 2

(iii) sin 180 ° + 2 cos 180 ° + 3 sin 270 ° + 4 cos 270 ° − 5 sec 180 ° − 6 cosec 270 ° = 6. 11. Prove that (i) sin (270 ° − θ) sin (90 ° − θ) − cos (270 ° − θ) cos (90 ° + θ) = − 1 tan (90 ° − θ) sec (180 ° − θ) sin (− θ) (ii) =1 sin (180 ° + θ) cot (360 ° − θ) cosec (90 ° − θ) 3π 5π   5π + θ tan θ − 3π  = − 1 (iii) sec  − θ sec θ −  + tan     2   2   2 2 sin (180 ° + θ) cos (90 ° + θ) tan (270 ° − θ) cot (360 ° − θ) (iv) = 1. sin (360 ° − θ) cos (360 ° + θ) cosec (−θ) sin (270 ° + θ) 12. Prove that (i) tan 720 ° − cos 270 ° − sin 150 ° cos 120 ° = (ii) sin 780 ° cos 30 ° + cos 120 ° sin 390 ° =

1 4

1 2

(iii) sin 600 ° cos 390 ° + cos 480 ° sin 150 ° = − 1 sin 300 ° tan 330 ° sec 420 ° 2 (iv) =− ⋅ cot 135° cos 210 ° cosec 315° 3

155

13. In any quadrilateral ABCD, prove that (i) sin ( A + B) + sin (C + D) = 0 (ii) cos ( A + B) = cos (C + D). 14. In a ∆ ABC , prove that (i) cos ( A + B) + cos C = 0 C  A + B (ii) cos   = sin  2  2 C  A + B (iii) tan  ⋅  = cot  2  2 15. If A, B, C , D be the angles of a cyclic quadrilateral, taken in order, prove that cos (180 ° − A ) + cos (180 ° + B ) + cos (180 ° + C ) − sin (90 ° + D ) = 0.

A nswers 1.

2. 3.

4. 5. 6. 7.

7 24 7 25 25 , tan θ = , cot θ = , sec θ = , cosec θ = ⋅ 25 7 24 7 24 1 (ii) sin θ = , tan θ = 1, cot θ = 1, cosec θ = 2 , sec θ = 2 . 2 41 41 40 9 40 (iii) cosec θ = and tan θ = , sec θ = , sin θ = , cos θ = ⋅ 40 9 41 41 9 8 15 8 15 17 (iv) sin θ = and sec θ = , cot θ = , tan θ = , cos θ = ⋅ 17 8 15 17 15 (i) II or III (ii) I or III (iii) III or IV 15 17 17 8 15 (i) cos θ = − and cot θ = , sec θ = − , cosec θ = − , tan θ = ⋅ 17 15 8 15 8 3 5 5 3 4 (ii) sin θ = , cosec θ = , sec θ = − , tan θ = − and cot θ = − ⋅ 5 3 4 4 3 10 3 (iii) cot θ = − 3, cosec θ = − 10 , sec θ = , cos θ = 3 10 1 and sin θ = − ⋅ 10 65 ⋅ 512 3. (i) tan θ (ii) sin θ. 1 1 (i) sin 225° = − , cos 225° = − , cosec 225° = − 2 , 2 2 sec 225° = − 2 , tan 225° = 1 and cot 225° = 1. 1 3 (ii) sin 330° = − , cos 330° = , cosec 330° = − 2 , 2 2 2 1 and cot 330 ° = − 3 . sec 330° = , tan 330° = 3 3 (i) cos θ =

156

4.13 Trigonometric Ratios of Compound Angles (a) Trigonometric Ratios of the Sum of Two Angles For any two angles A and B, we have (i) sin ( A + B) = sin A cos B + cos A sin B, (ii) cos ( A + B) = cos A cos B − sin A sin B , tan A + tan B and (iii) tan ( A + B) = ⋅ 1 − tan A tan B (b) Trigonometric Ratios of the Difference of Two Angles For any two angles A and B , we have (i) sin ( A − B) = sin A cos B − cos A sin B , (ii) cos ( A − B) = cos A cos B + sin A sin B , tan A − tan B and (iii) tan ( A − B) = ⋅ 1 + tan A tan B (c) For any two angles A and B, we have (i) sin ( A + B) sin ( A − B) = sin2 A − sin2 B and (ii) cos ( A + B) cos ( A − B) = cos

2

2

A − sin

[UPTU 2001]

B.

Proof. (i) We have sin ( A + B) sin ( A − B) = (sin A cos B + cos A sin B) (sin A cos B − cos A sin B) = sin2 A cos 2 B − cos 2 A sin2 B = sin2 A (1 − sin2 B) − (1 − sin2 A) sin2 B = sin2 A − sin2 A sin2 B − sin2 B + sin2 A sin2 B = sin2 A − sin2 B. (ii) We have cos ( A + B) cos ( A − B) = (cos A cos B − sin A sin B) (cos A cos B + sin A sin B) = cos 2 A cos 2 B − sin2 A sin2 B = cos 2 A (1 − sin2 B) − (1 − cos 2 A) sin2 B = cos 2 A − cos 2 A sin2 B − sin2 B + cos 2 A sin2 B = cos 2 A − sin2 B. (d) For any three angles A, B and C, we have (i) sin ( A + B + C ) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C , (ii) cos ( A + B + C ) = cos A cos B cos C − cos A sin B sin C − sin A cos B sin C − sin A sin B cos C , tan A + tan B + tan C − tan A tan B tan C and (iii) tan ( A + B + C ) = ⋅ 1 − tan A tan B − tan B tan C − tan C tan A

157

Illustration 1 :

Evaluate :

(i) sin 78° cos 18° − cos 78° sin 18°

[UPTU 2001]

(ii) cos 68° cos 8° + sin 68° sin 8° Solution. (i) We have sin 78° cos 18° − cos 78° sin 18° 3 ⋅ 2 [ ∵ sin A cos B − cos A sin B = sin ( A − B)]

= sin (78° − 18° ) = sin 60 ° =

(ii)

We have cos 68° cos 8° + sin 68° sin 8° 1 = cos (68° − 8° ) = cos 60 ° = ⋅ 2 [ ∵ cos A cos B + sin A sin B = cos ( A − B)]

Illustration 2 : (i) cos

Prove that

5π π 5π π (1 − 3) cos + sin sin = 3 4 3 4 2 2

π π π π 3 cos + cos sin = 4 12 4 12 2 7π π 7π π 3 (iii) sin cos − cos sin = ⋅ 12 4 12 4 2 5π π 5π π Solution : (i) We have cos cos + sin sin 3 4 3 4 π π π π = cos 2 π −  cos + sin 2 π −  sin     3 4 3 4 π π π π = cos cos − sin sin 3 4 3 4 (ii) sin

1 1 ⋅ 2 2 π π π π π π We have sin cos + cos sin = sin  +   4 12 4 12 4 12 =

(ii)

[ ∵ cos (2 π − θ) = cos θ and sin (2 π − θ) = − sin θ] 3 1 (1 − 3) − ⋅ = ⋅ 2 2 2 2

[ ∵ sin A cos B + cos A sin B = sin ( A + B)] π 3 = ⋅ 3 2 7π π 7π π 7π π  (iii) We have sin cos − cos sin = sin  −   12 4  12 4 12 4 = sin

[ ∵ sin A cos B − cos A sin B = sin ( A − B)] = sin

π 3 = ⋅ 3 2

158

Illustration 3 : Solution :

Prove that cos 105° + cos 15° = sin 75° − sin 15° .

We have cos 105° + cos 15° = cos (90 ° + 15° ) + cos (90 ° − 75° ) = − sin 15° + sin 75° [ ∵ cos (90 ° + θ) = − sin θ and cos (90 ° − θ) = sin θ] = sin 75° − sin 15° .

Illustration 4 :

Obtain the values of

(i) sin 15° [UPTU 2004]

(ii) cos 15°

(iii) sin 75°

(iv) cos 75°

(v) cos 105°

(vi) cot 75°

(vii) tan 15°. [UPTU 2006] Solution :

(i) We have sin 15° = sin (45° − 30 ° ) = sin 45° cos 30 ° − cos 45° sin 30 ° 3 −1 1 3 1 1 = ⋅ − ⋅ = ⋅ 2 2 2 2 2 2

(ii) We have cos 15° = cos (45° − 30 ° ) = cos 45° cos 30 ° + sin 45° sin 30 ° 3 +1 1 3 1 1 = ⋅ + ⋅ = ⋅ 2 2 2 2 2 2 (iii) We have sin 75° = sin (45° + 30 ° ) = sin 45° cos 30 ° + cos 45° sin 30 ° 3 +1 1 3 1 1 = ⋅ + ⋅ = ⋅ 2 2 2 2 2 2 (iv) We have cos 75° = cos (45° + 30 ° ) = cos 45° cos 30 ° − sin 45° sin 30 ° 3 −1 1 3 1 1 = ⋅ − ⋅ = ⋅ 2 2 2 2 2 2 (v) We have cos 105° = cos (60 ° + 45° ) = cos 60 ° cos 45° − sin 60 ° sin 45° 1− 3 3 −1 1 1 3 1 = ⋅ − ⋅ = =− ⋅ 2 2 2 2 2 2 2 2 (vi) We have cot 75° = cot (45° + 30 ° ) cot 45° cot 30 ° − 1 1⋅ 3 − 1 = = = cot 45° + cot 30 ° 1+ 3 (vi) We have tan 15° = tan (45° − 30 ° ) = 1 3 = = 1 1+ 3 1−

3 −1 3 +1

⋅

3 −1 3 +1

⋅

tan 45° − tan 30 ° 1 + tan 45° tan 30 °

[UPTU 2002]

159

5 1 and tan B = , prove that A + B = 45° . 6 11 tan A + tan B We have tan ( A + B) = 1 − tan A tan B

Illustration 5 : Solution :

If tan A =

55 + 6 5 1 + 6 × 11 61 = 6 11 = = = 1 = tan 45° . 5 1 66 − 5 61 1− × 6 11 6 × 11 ∴

A + B = 45° .

Illustration 6 :

If A + B = π / 4, prove that (1 + tan A) (1 + tan B) = 2 . [UPTU 2005]

Solution : ∴

It is given that A + B = π / 4. tan ( A + B) = tan (π / 4) = 1 ⇒

⇒

tan A + tan B = 1 − tan A tan B

⇒

tan A + tan B + tan A tan B = 1

⇒

1 + tan A + tan B + tan A tan B = 2

tan A + tan B 1 − tan A tan B

(1 + tan A) (1 + tan B) = 2 . cos 8° + sin 8° Illustration 7 : Prove that = cot 37° . cos 8° − sin 8°

=1

⇒

Solution :

We have = =

[UPTU 2002]

cos 8° + sin 8° cos 8° − sin 8°

1 + tan 8° 1 − tan 8°

[Dividing the Nr and the Dr by cos 8°]

tan 45° + tan 8°

[ ∵ tan 45° = 1]

1 − tan 45° tan 8°

= tan (45° + 8° ) = tan 53°   tan A + tan B = tan ( A + B) ∵ 1 − tan A tan B   = tan (90 ° − 37° ) = cot 37° . Illustration 8 : Solution :

Prove that tan 70 ° = tan 20 ° + 2 tan 50 ° .

We have tan 70 ° = tan (20 ° + 50 ° ) tan 20 ° + tan 50 °

⇒

tan 70 ° =

⇒

tan 70 ° − tan 70 ° tan 20 ° tan 50 ° = tan 20 ° + tan 50 °

⇒

tan 70 ° − tan (90 ° − 20 ° ) tan 20 ° tan 50 ° = tan 20 ° + tan 50 °

⇒

tan 70 ° − cot 20 ° tan 20 ° tan 50 ° = tan 20 ° + tan 50 °

1 − tan 20 ° tan 50 °

[ ∵ tan (90° − θ) = cot θ]

160

[ ∵ cot A tan A = 1]

⇒

tan 70 ° − tan 50 ° = tan 20 ° + tan 50 °

⇒

tan 70 ° = tan 20 ° + tan 50 ° + tan 50 ° = tan 20 ° + 2 tan 50 ° .

Illustration 9 :

Prove that tan 13θ − tan 9θ − tan 4 θ = tan 13θ tan 9θ tan 4 θ. [UPTU 2005]

Solution :

We have tan 13 θ = tan (9 θ + 4 θ) tan 9θ + tan 4 θ

⇒

tan 13 θ =

⇒

tan 13 θ (1 − tan 9 θ tan 4 θ) = tan 9 θ + tan 4 θ

⇒

tan 13 θ − tan 13 θ tan 9 θ tan 4 θ = tan 9 θ + tan 4 θ

1 − tan 9 θ tan 4 θ

tan 13 θ − tan 9 θ − tan 4 θ = tan 13 θ tan 9 θ tan 4 θ. 3 Illustration 10 : Prove that cos2 45° − sin2 15° = ⋅ 4 ⇒

Solution :

We have cos 2 45° − sin2 15° = cos (45° + 15° ) cos (45° − 15° ) [ ∵ cos 2 A − sin2 B = cos ( A + B ) cos ( A − B )]

1 3 3 ⋅ = ⋅ 2 2 4 4 5 Illustration 11 : If sin A = and cos B = , find the values of sin ( A − B) and 5 13 π π cos ( A + B), where 0 < A < and 0 < B < ⋅ [UPTU 2001] 2 2 4 π Solution : We have sin A = and 0 < A < 5 2 = cos 60 ° cos 30 ° =

4 2 16 A = 1 −   = 1 − =  5 25 π and 0 < B< 2

⇒

cos A = 1 − sin2

Again

cos B =

⇒

5  sin B = 1 − cos 2 B = 1 −   13

5 13

2

=

144 12 = ⋅ 169 13

∴

sin ( A − B) = sin A cos B − cos A sin B 4 5 3 12 20 36 16 =  ×  −  ×  = − =− ⋅  5 13  5 13 65 65 65

Also

cos ( A + B) = cos A cos B − sin A sin B 3 5 4 12 15 48 33 =  ×  −  ×  = − =− ⋅  5 13  5 13 65 65 65

Illustration 12 :

If sin A =

9 3 = ⋅ 25 5

3 π 12 3π and cos B = − ,0 < A< , π < B< , find the 5 2 13 2

following : (i)

(ii) cos ( A + B) 3 π We have sin A = , where 0 < A < ⋅ 5 2

sin ( A − B)

Solution :

(iii) tan ( A − B).

161

Since, A lies in the first quadrant, therefore sin A is positive and cos A is positive. ∴

cos A = 1 − sin2

Also,

tan A =

Again,

cos B = −

sin A cos A

=

3 2 9 16 4 A = 1 −   = 1 − = = ⋅  5 25 25 5 3 5 3 × = ⋅ 5 4 4

12 3π , π< B< ⋅ 13 2

Since, B lies in the third quadrant, therefore cos B is negative and sin B is also negative. 12 sin B = − 1 − cos 2 B = − 1 −  −   13

∴

2

144 25 5 =− =− ⋅ 169 169 13 sin B  5   13 5 tan B = = −  × −  = ⋅     cos B 13 12 12 = − 1−

Also,

Now, (i) sin ( A − B) = sin A cos B − cos A sin B 3 12 4 5 36 20 −16 = ×  −  −    −  = − + = ⋅ 5  13  5  13 65 65 65 (ii) cos ( A + B) = cos A cos B − sin A sin B 4 12 3 5 48 15 33 = ×  −  − ×  −  = − + =− ⋅     5 13 5 13 65 65 65 3 5 − 16 4 12 (iii) tan ( A − B) = = = ⋅ 1 + tan A tan B 1 + 3 × 5 63 4 12 tan A − tan B

Illustration 13 : that tan 2 α = Solution : 0 0 to a given base a > 0 and a ≠ 1is the

index (power) to which the base must be raised in order to become equal to the given number. Thus if a

x

= y, then x is called the logarithm of y to the base a. The logarithm of y to the base a is usually written as log a y. Thus the following equations express the same meaning a x = y and x = log a y. The first equation is called the exponential form, whereas the second equation is called the corresponding logarithmic form. Thus, we have 1. 43 = 64, 2. 2 −3 =

1 = 0 ⋅ 125, 8

∴

log

∴

log 2 0 ⋅ 125 = − 3.

4

64 = 3

196

5.2 Properties of Logarithms (i)

The logarithm of 1 is 0 at every base.

We have a0 = 1,for all values of a ;therefore log 1 = 0, whatever the base may be. (ii)

Logarithm of a to the base a itself is 1.

We have a1 = a. ∴

log a a = 1.

Note. Since 11 = 1,12 = 1, 13 = 1, so by definition of logarithm, we have log 1 1 = 1, log 1 1 = 2 , log 1 1 = 3 1 = 2 = 3 = …… ; which is not possible.

⇒

log 1 1 is meaningless.

∴ (iii) a Let Then

log a n

= n.

log a

x

a

n = x. a log a n = n.

or

=n

[∵

x = log

a

n]

5.3 Fundamental Laws of Logarithms (i) Logarithm of a product. If m and n are any two numbers, then log a mn = log a m + log a n. i. e., the log of the product of two numbers is equal to the sum of their logs. Proof. Let log

m = x and log

a

a

n = y.

Then, by definition of logarithm, m = a Now,

mn = a

∴

log

a

x

.a

y

=a

x

and n = a y .

x+ y

mn = x + y = log

[By law of indices] m + log

a

n.

a

Similarly, if m1 , m2 , …… , mn are n numbers, then log

a

(m1 m2 … mn ) = log

a

m1 + log

a

m2 + … + log

(ii) Logarithm of ratio or quotient of two numbers. numbers, then m log a = log a m − log a n n

a

mn .

If m and n are any two

i. e., the log of the quotient of two numbers is equal to the difference of their logs. Proof. Let log

m = x and log

a

a

n = y.

Then, by definition of logarithm, m = a Now ∴

m a = n a log

x y a

=a

x

and n = a y .

x− y

m = x − y = log n

[By law of indices] a

m − log

a

n.

197

(iii) Logarithm of power of a number. If m and n are any two numbers, then log a m n = n log a m i. e.,the logarithm of the power of a number is the product of the logarithm of the number and the index. Proof. Let

log

a

m = x.

Then, by definition of logarithm, m = a x . m n = (a x ) n = a n x

Now

log

∴

a

[By law of indices]

n

m = n x = n log

a

m.

5.4 Transformation of Logarithm from one Base to another Formula :

log b a = (log c a) ×

log c a 1 = ⋅ log c b log c b

Proof. Let y = log b a. Then b log

∴

y = log

or

log b a = log

Corollary.

c

c

a⋅

c

b.

1 ⋅ log c b

…(1)

Putting c for a in (1), we get 1 1 log b c = log c c ⋅ = 1⋅ log c b log c b

or ∴

a = log c

= a.

(b y ) = y log 1 a⋅ log c b

∴

c

y

(log b c) (log c b) = 1. (1) can also be written as log b a = log c a . log b c.

5.5 Common Logarithms Logarithms to the base 10 are called common logarithms. We shall denote log 10 m as log m only i. e., if no base is mentioned, the base is to be taken as 10.

Illustration 1 : (i) 10 Solution :

−1

Write the following in the logarithmic form : 1 (ii) 9 −5 /2 = = 0 ⋅1 ⋅ 243

(i) By the definition of logarithm, 10 −1 = 0 ⋅ 1 ⇒ log 10 0 ⋅ 1 = − 1.

198

1 ⇒ log 243

(ii) By the definition of logarithm, 9 −5 /2 =

9

1 5 =− ⋅ 243 2

Illustration 2 : (i)

log10

Solution :

Write the following in the exponential form : 1 (ii) log2 0 ⋅ 01 = − 2 = − 6. 64 (i) By the definition of logarithm, log 10 0 ⋅ 01 = − 2 ⇒ 10 −2 = 0 ⋅ 01.

(ii) By the definition of logarithm, 1 1 log 2 = − 6 ⇒ 2 −6 = ⋅ 64 64 Illustration 3 :

Find the value of each of the following :

(i) log 2 8 Solution : (i) Let log Then ∴ (ii)

x

( 2)

(ii) log9 3 8 = x.

2

x

= 3 ⇒ (32 )

9

∴

log

(iii) Let

log 10

Then

10

x

= 3 ⇒ 32

1 ⋅ 2 0 ⋅ 000001 = x.

Illustration 4 :

x

= 0 ⋅ 000001 =

6

= 10 −6 ⇒ x = − 6.

3

(iii) log16 x = 1⋅ 5.

x=6

3

1 10

3

=

1 = 0 ⋅ 001. 1000

x=6

x = ( 3)6 = (31 /2 )6 = 33 = 27. We have log 16 x = 1⋅ 5

⇒

x = (16)1⋅5 = (16)3 /2 = (42 )3 /2 = 43 = 64 .

Illustration 5 : Solution : Now

1 ⋅ 2

(i) We have log 10 x = − 3

We have log

⇒

and

1

(ii) log

x = 10 −3 =

⇒

= 31 ⇒ 2 x = 1 ⇒ x =

Find the value of x in each of the following :

(i) log10 x = − 3 Solution :

x

3=

9

10 log 10 0 ⋅ 000001 = − 6.

∴

(iii)

x = 3 ⇒ x = 6. 2

= 8 ⇒ 2 x / 2 = 23 ⇒

log 2 8 = 6. Let log 9 3 = x.

Then

(ii)

(iii) log 10 0 ⋅ 000001.

x

If log2 m = x and log3 n = x, find 108

We have log 2 m = x ⇒ 2 log 3 n = x ⇒ 3 108

x

= (4 × 27) = (2

x

x

=m

x

= n.

x

= (22 × 33 )

)2 × (3

x

in terms of m and n.

x

= 22

)3 = m2 n3 .

x

× 33

x

199

Illustration 6 :

Simplify the following : 34 × 21 /5 (i) log a 81 (ii) log a ⋅ 7 Solution : (i) We have log a 81 = log a 34 = 4 log 4

(ii) We have log

1 /5

3 ×2 a

7

= log

a

a

3.

(34 × 21 /5 ) − log

a

71 /2

1 log a 7 2 1 1 = 4 log a 3 + log a 2 − log a 7. 5 2 15 2 3 Illustration 7 : Prove that log + log − log = 0. 16 5 8 15 2 3 Solution : We have log + log − log 16 5 8 = log

a

34 + log

a

21 /5 −

= log 15 − log 16 + log 2 − log 5 − (log 3 − log 8) = log (3 × 5) − log 24 + log 2 − log 5 − log 3 + log 8 = log 3 + log 5 − 4 log 2 + log 2 − log 5 − log 3 + log 23 = − 3 log 2 + 3 log 2 = 0. Alternative Method. 15 2 3 We have log + log − log 16 5 8 15 2 8  15 2 3 = log   ×  ÷  = log  × ×  16 5 3  16 5 8 Illustration 8 : Solution :

Express log 972 in terms of log 2 and log 3.

We have log

972 = log (972)1 /2

= log (9 × 9 × 3 × 4)1 /2 = log (35 × 22 )1 /2 = log (35 /2 × 2) = log 35 /2 + log 2 =

5 log 3 + log 2 . 2

Illustration 9 :

Solution :

Prove that 15 25 4 2 log − log + 3 log = log 2 . 8 162 9 15 25 4 We have 2 log − log + 3 log 8 162 9 15 2 25 4 3 = log   − log + log   8  9 162  15 2 15 2 4 3 25 4 3 25  = log   + log   − log = log    ×    ÷  8  9    9 162  162  8 225 64 162 162 = log  × × = log 2 .  = log  64 729 25  81

200

Illustration 10 :

Prove that log b a × log c b × log a c = 1.

Solution :

We have log =

log

m

a

log

m

b

×

a × log

b

log

m

b

log

m

c

×

b × log

c

log

m

c

log

m

a

c

a

, changing all logarithms to the base m

= 1. Illustration 11 : Solution :

If x = log2 a a, y = log3 a 2a, z = log4 a 3a,prove that xyz + 1 = 2 yz .

Taking log, we get log (2a)

x

Similarly,

y=

log 2a

= =

log a

Solve :

×

log 2a log a log 4a

log 2a log 3a

log 2a

= ×

×

⋅

log 3a

z =

log 3a

2

log 4a

log a

log 2a

+1=

log 4a

=2 Illustration 12 :

and

log 3a

x y z +1=

∴

= a.

= log a

x log 2a = log a ⇒ x =

⇒

Solution :

x

We have x = log 2 a a ⇒ (2a)

log 4a log 3a log 4a

⋅ +1

log a + log 4a log 4a

log (2a)2 log 4a log 3a log 4a

=

2 log 2a log 4a

= 2 yz .

log (4 x − 3) − log ( x + 1) = log 3.

We have log (4 x − 3) − log ( x + 1) = log 3

⇒

4 x−3  4 x − 3 log  =3  = log 3 ⇒ x +1  x +1 

⇒

4 x − 3 = 3 x + 3 ⇒ 4 x − 3 x = 3 + 3 ⇒ x = 6.

Comprehensive Exercise 1 1. Write the following in the logarithmic form : (i) 43 = 64

(ii) 35 = 243

(iii) 5 −2 = 0 ⋅ 04

(iv) 10 −3 = 0 ⋅ 001

(v) x

m

= n.

2. Write the following in the exponential form : (i) log 2 64 = 6

(ii) log 2 0 ⋅ 125 = − 3.

201

3. Find the value of each of the following : (i) log 2 16 (ii) log 5 5 −2 (iii) log 4 256 4. Find the value of log 2

(iv) log 3 (3 3). 3

144 .

5. If log 5 x = y, find the value of 53

y

6. If log 10 x = a, find the value of 10

2 a−1

in terms of x. in terms of x.

7. If log 2 x = a and log 5 y = a, find the value of 2000

2 a−1

in terms of x

and y. 8. Simplify each of the following : (i) log 24 − log 8 (ii) log 16 − log 6 + log 3 a b c (iii) log + log + log (iv) 2 + log 5. b c a a2 b1 /3 9. Express log in terms of logarithms of a, b, c . c 2 /3 10. Express the following in terms of log 2 and log 3 : (i) log 108 3

(iii) log ( 54 × 243)

(ii) log 144 16 (iv) log ⋅ 27

11. Prove the following : (i) log 10 25 = 2 (1 − log 10 2). (ii) log (1 + 2 + 3) = log 1 + log 2 + log 3. 10 65 25 (iii) log + log − log = 0. 13 68 34 16 25 81 (iv) 7 log + 5 log + 3 log = log 2 . 15 24 80 12. Simplify the following : 65 133 11 (i) log + log + log ⋅ 19 143 35 75 5 32 (ii) log − 2 log + log ⋅ 16 9 243 16 25 81 (iii) log + log + log ⋅ 15 24 80 13. Prove that : log (m n p q) = log m + log n + log p + log q. a 14. If log (a + b) = log a + log b, show that b = ⋅ a −1 1 15. If log 10 a + log 10 b = 1, then prove that ab 2 = 100. 2  a + b 1 16. If log   = (log a + log b), show that a = b.  2  2 a + 17. If log   3

b 1 2 2  = (log a + log b), prove that a + b = 7ab.  2

202

A nswers 1. (i) log 4 64 = 3

(ii) log 3 243 = 5

(iii) log 5 0 ⋅ 04 = − 2 (v) log

(iv) log 10 0 ⋅ 001 = − 3

n = m.

x

6

(ii) 2 −3 = 0 ⋅ 125.

2. (i) 2 = 64 3. (i) 4

(ii) −2

(iii) 4

3 ⋅ 2

5. x 3 .

4. 4. 6.

(iv)

x2 ⋅ 10

7.

8. (i) log 3

x8 y6 2000

⋅

(ii) log 8

(iii) 0

(iv) log 500.

1 2 9. 2 log a + log b − log c . 3 3 10. (i) 2 log 2 + 3 log 3 (iii)

1 19 log 2 + log 3 2 6

12. (i) 0.

(ii) 4 log 2 + 2 log 3 (iv) 4 log 2 − 3 log 3. (ii) log 2 .

(iii) 2 log 3 − 3 log 2 .

5.6 Standard Form of Decimal We can express any number in decimal form as the product of a number between 1 and 10 and an integral power of 10. Thus, any positive decimal number n can be written in this form as n = m × 10 p , where p is an integer (positive, zero or negative) and 1 ≤ m < 10. This is known as the standard form of the decimal number. Working Rule to convert a given decimal number into standard form. Move the decimal point to the left or to the right, as may be necessary, to bring one non-zero digit to the left of decimal point. If the decimal is shifted p places to the left, multiply by 10 p . If the decimal is shifted p places to the right, multiply by 10 −p . If the decimal is not moved at all, multiply by 10 0 . Illustration :

Write each of the following in standard form :

(i) 1⋅ 234

(ii) 12 ⋅ 34

(iii) 123 ⋅ 4

(iv) 1234

(v) 0 ⋅ 1234

(vi) 0 ⋅ 01234

(vii) 0 ⋅ 001234. Solution :

(i) 1⋅ 234 = 1⋅ 234 × 10 0

203

(ii) 12 ⋅ 34 = 1⋅ 234 × 101

(iii) 123 ⋅ 4 = 1⋅ 234 × 10 2

(iv) 1234 = 1⋅ 234 × 10 3

(v) 0 ⋅ 1234 = 1⋅ 234 × 10 −1

(vi) 0 ⋅ 01234 = 1⋅ 234 × 10 −2

(vii) 0 ⋅ 001234 = 1⋅ 234 × 10 −3 .

5.7 Characteristic and Mantissa of a Common Logarithm Let the standard form of n be n = m × 10 p , where 1 ≤ m < 10 and p is an integer. Then log n = log (m × 10 p ) = log m + log 10

p

= log m + p log 10 = p + log m.

[ ∵ log 10 10 = 1]

Since 1 ≤ m < 10, so 0 ≤ log m < 1 i. e., log m lies between 0 and 1. Thus, the logarithm of a positive real number n has two parts : (i) the integral part p known as the characteristic and (ii) the fractional part log m known as the mantissa. Hence, log n = characteristic + mantissa. It follows that characteristic is an integer which may be positive, negative or zero and mantissa is always a positive number less than 1.

5.8 To find the Characteristic Put the number n in the standard form as n = m × 10 p .Then, the characteristic is p. Alternative Method (Computation of Characteristic by Inspection). (a) 10 0 = 1, 101 = 10, 10

2

= 100,

10 3 = 1000,

∴

log 10 1 = 0,

∴

log 10 10 = 1,

∴

log 10 100 = 2 ,

∴

log 10 1000 = 3,

……………

…………………

…………… ………………… It is clear from the above that logarithms of all the numbers lying between 1 and 10 will lie between 0 and 1 i. e., the integral part of the logarithm is zero. Logarithms of all the numbers lying between 10 and 100 will lie between 1 and 2 i. e., the integral part of the logarithm is 1. Similarly, the integral part of the logarithm of a number lying between 100 and 1000 is 2 ; the integral part of the logarithm of a number lying between 1000 and 10,000 is 3, etc. Thus, we obtain the following law : Law 1 : The characteristic of the logarithm of a number greater than unity is less by one than the number of digits in its integral part, and is positive i.e., > 0.

204

Example :

The characteristics of log 312 , log 64 ⋅ 372 and log 2250 are 2 , 1and 3

respectively. (b) 10 0 = 1,

∴ log 10 1 = 0, 1 10 = = 0 ⋅ 1, ∴ log 10 0 ⋅ 1 = − 1, 10 1 10 −2 = = 0 ⋅ 01, ∴ log 10 0 ⋅ 01 = − 2 , 100 1 10 −3 = = 0 ⋅ 001, ∴ log 10 0 ⋅ 001 = − 3, 1000 … … … … … … … … … … … … … … … … … … … … Thus, we observe that the logarithm of a number lying between 0 ⋅ 1 and 1 lies between −1 and 0 and hence it can be written as −1 + decimal part. Thus, the characteristic of the logarithm of a number smaller than unity and having no zero after the decimal point, is −1and we shall write it as 1 in place of −1. The logarithm of a number lying between 0 ⋅ 01and 0 ⋅ 1lies between −2 and −1and it can be written as −2 + decimal part. Thus, the characteristic of the logarithm of such number is −2 and we shall write it as 2 . −1

Similarly, the characteristic of the logarithm of a number lying between 0 ⋅ 001and 0 ⋅ 01 is −3 i. e., 3. Thus, we obtain the following law : Law 2 : The characteristic of the logarithm of a positive decimal fraction less than one is negative and one more than the number of consecutive zeros immediately after the decimal point. Example : The characteristics of log 0 ⋅ 532 , log 0 ⋅ 0417 and log 0 ⋅ 000187 are −1, − 2 and −4 respectively.

5.9 To find the Mantissa Example :

If log 7513 = 3 ⋅ 8758 , then find log 75 ⋅ 13, log 751⋅ 3, log 0 ⋅ 7513 and

log 0 ⋅ 007513. Solution : Now

We have log 7513 = 3 ⋅ 8758. log 75 ⋅ 13 = log (7513 × 10 −2 ) = log 7513 + log 10 −2 = 3 ⋅ 8758 − 2 = 1⋅ 8758 log 751⋅ 3 = log (7513 × 10 −1 ) = log 7513 + log 10 −1 = 3 ⋅ 8758 − 1 = 2 ⋅ 8758 log 0 ⋅ 7513 = (log 7513 × 10 −4 ) = log 7513 + log 10 −4 = 3 ⋅ 8758 − 4 = 1 ⋅ 8758 log 0 ⋅ 007513 = log (7513 × 10 −6 ) = log 7513 + log 10 −6 = 3 ⋅ 8758 − 6 = 3 ⋅ 8758.

205

For the above example, we conclude the following law : Law 3 : The mantissa of logarithms of all the numbers having the same digits in the same order is the same. The mantissa is found from the log tables. For this purpose the position of the decimal is immaterial. We consider the first four significant digits from the left side of the number.

5.10 Use of Log Tables (a) To find the logarithm of a number of one digit. Suppose we have to find log 7. See the log table in the end of the book. We find that the number in the row headed by 70 and under the column of 0 is 8451. So mantissa of log 70 is 0 ⋅ 8451. According to Law 3, the mantissa of log 7 will be the same. Again, since 7 is a number of one digit, so by Law 1 the characteristic of log 7 will be zero. ∴ log 7 = 0 ⋅ 8451. (b) To find the logarithm of a number of two digits. Suppose we have to find log 72. We find that in the log table the number in the row headed by 72 and under the column of 0 is 8573. So mantissa of log 72 is 0 ⋅ 8573. Since 72 is a number of two digits, so the characteristic of its logarithm will be 1. ∴ log 72 = 1⋅ 8573. (c) To find the logarithm of a number of three digits. Suppose we have to find log 734. We find that the number in the row headed by 73 and under the column of 4 is 8657. ∴ log 734 = 2 ⋅ 8657. Similarly, log 73 ⋅ 4 = 1⋅ 8657, log ⋅ 00734 = 3 ⋅ 8657, etc. (d) To find the logarithm of a number of four digits. Suppose we have to find log 71⋅ 56. As explained in (c), we have log 71⋅ 5 = 1⋅ 8543. Now locate the number in the row headed by 71 and under the column headed by 6 of mean differences. We find the number 4 there. Thus log 71⋅ 50 = 1⋅ 8543 difference of 6 = 4 [On adding] ∴ log 71⋅ 56 = 1⋅ 8547.

5.11 Anti-logarithm If log n = m, then n = antilog of m. For example , log 71⋅ 56 = 1⋅ 8547. antilog 1⋅ 8547 = 71⋅ 56. ∴ To find the Antilog of a Number from the table. First take just the decimal part (mantissa) of the number. In the first column of antilogarithm table the first two digits in the decimal part are given. Thus, in the antilog table locate the number formed by the first two digits in the decimal part. In the row against this number, find the number in the column headed by the third digit and to this number add the mean difference under the

206

column headed by the fourth digit. To find the position of the decimal point, we consider the characteristic. If the characteristic is non-negtive and is equal to n, then insert the decimal point after (n + 1) digits from the left in the number seen from the antilog table. If n > 4 , then put zeros on the right of the number seen from the antilog table so that the total number of digits becomes n + 1. If the characteristic is negative and is equal to −n i. e., n, then write (n − 1) zeros between the decimal point and the number seen from the antilog table. For example, suppose we have to find that number whose logarithm is 1⋅ 3156. See 31 in the first column of the antilogarithm table. In the row against 31, the number in the column headed by 5 is 2065. In the same row, the mean difference of 6 is 3. Adding this to 2065, we get 2068. Now since the characteristic of logarithm is 1, so in the required number the decimal point will be inserted after two digits. Thus, the number whose logarithm being 1⋅ 3156 is 20 ⋅ 68. If we have to find the antilog of 2 ⋅ 9875,then we shall use the table in this way : antilog 2 ⋅ 987 = 0 ⋅ 09705 difference of 5 = 11 [On adding] ∴ antilog 2 ⋅ 9875 = 0 ⋅ 09716.

Illustration 1 : How many digits are there in the integral part of a number whose logarithm is 13 ⋅ 7993 ? Solution : Since the characteristic of the logarithm of the number is 13, so the number of digits in the integral part of the number = 13 + 1 = 14. Illustration 2 : The logarithm of a number is 5 ⋅ 5798. How many consecutive zeros are in the number after the decimal point ? Solution : Since the characteristic of the logarithm of the number is 5, so the number of consecutive zeros immediately after the decimal point in the number = 5 − 1 = 4. Illustration 3 : Find the number of digits in 2100 if log10 2 = 0 ⋅ 30103. Solution : Let x = 2100 . Then log 10 x = 100 log 10 2 = 100 × 0 ⋅ 30103 = 30 ⋅ 103. Since the characteristic of log x is 30, so the number of digits in 2100 = 30 + 1 = 31. 3 ⋅ 274 × 0 ⋅ 0059 Illustration 4 : Evaluate , correct to 4 decimal places. 14 ⋅ 83 × 0 ⋅ 077 3 ⋅ 274 × 0 ⋅ 0059 Solution : Let x = ⋅ 14 ⋅ 83 × 0 ⋅ 077 Then log x = log 3 ⋅ 274 + log 0 ⋅ 0059 − log 14 ⋅ 83 − log 0 ⋅ 077 Using log tables, we have log 3⋅ 27

=

0⋅ 5145

diff. of 4 =

5

log 14 ⋅ 8

=

1⋅ 1703

diff. of 3 =

9

…(1)

207

∴ log 3⋅ 274 = Again,

0⋅ 5150

∴

log 14 ⋅ 83 =

1⋅ 1712

log 0 ⋅ 0059 = 3 ⋅ 7709 and log 0 ⋅ 077 = 2 ⋅ 8865

Putting these values in (1), we get log x = 0 ⋅ 5150 + 3 ⋅ 7709 − 1⋅ 1712 − 2 ⋅ 8865 = 2 ⋅ 2859 − 0 ⋅ 0577 = 2 ⋅ 2282 . Again from antilog table, we have antilog 2 ⋅ 2282 = 0 ⋅ 01691. x = 0 ⋅ 0169.

∴

Illustration 5 : Solution :

Find the cube root of 11, correct to four decimal places. x = 111 /3 .

Let

Then

log x =

1 1 log 11 = × 1⋅ 0414 , from table 3 3

= 0 ⋅ 3471. Now

antilog 0 ⋅ 347 = 2 ⋅ 223, from the table diff. of

1 =

1

∴

antilog 0 ⋅ 3471 = 2 ⋅ 224

∴

(11)1 /3 = 2 ⋅ 224.

Illustration 6 :

If log 2 = 0 ⋅ 3010 and log 3 = 0 ⋅ 4771, then find the number of digits in

17

6 . Solution :

Let x = 617 .

Then

log x = 17 log 6 = 17 log (2 × 3) = 17 (log 2 + log 3) = 17 (0 ⋅ 3010 + 0 ⋅ 4771) = 17 × 0 ⋅ 7781 = 13 ⋅ 2277.

Since the characteristic of log x is 13, so the number of digits in 617 = 13 + 1 = 14. Illustration 7 :

If log 7 = 0 ⋅ 8451, then how many consecutive zeros are after the decimal

1 11 point in converting   into a decimal fraction ?  49 Solution : Then

Let

1 11 x =   = (7 −2 )11 = 7 −22 .  49

log x = − 22 log 7 = − 22 × 0 ⋅ 8451 = − 18 ⋅ 5922 = − 18 − 1 + (1 − 0 ⋅ 5922) = − 19 + 0 ⋅ 4078 = 19 ⋅ 4078.

Since the characteristic of log x is −19, so number of consecutive zeros after the decimal point in x = 19 − 1 = 18. Illustration 8 : Solution :

Solve : 2 x = 25, if log 2 = 0 ⋅ 30103.

We have 2 x = 25.

208

100 100 = log 2 4 2

∴

log 2 x = log 25 = log

or

x log 2 = log 100 − 2 log 2

or

x × 0 ⋅ 30103 = 2 − 2 × 0 ⋅ 30103 = 1⋅ 39794 1⋅ 39794 x= = 4 ⋅ 64386. 0 ⋅ 30103

or

Comprehensive Exercise 2 1. Find, only by inspection, the characteristic of logarithm of the following numbers : (i) 19

(ii) 4173

(iv) 0 ⋅ 0205

(v) 0 ⋅ 000137

(iii) 375 ⋅ 57

2. If log 10 2 = 0 ⋅ 3010, then find the value of log 10 200. 3. If log 6732 = 3 ⋅ 8281, then find logarithms of the following numbers : (i) 67 ⋅ 32

(ii) 6732000

(iii) 0 ⋅ 6732

(iv) 6 ⋅ 7320. 4. If log 8 ⋅ 057 = 0 ⋅ 9062 , then find the numbers having following logarithms: (i) 4 ⋅ 9062

(ii) 1⋅ 9062

(iii) 1 ⋅ 9062 .

5. How many digits are there in the integral part of a number whose logarithm is 17 ⋅ 3212 ? 6. How many digits are there in 21000 , if log 10 2 = 0 ⋅ 30103 ? 7. If log 2 = ⋅ 3010, log 3 = ⋅ 4771, then find the number of digits in the following : 230 , 264 , 342 , (540)10 , (1296)8 . 8. Solve the following equations : (i) 6

x −2

= 10

(ii) 2 x × 32 − x = 5.

9. If log 3 = 0 ⋅ 47712 , then find the value of 2 log

4 25 9 + log + log ⋅ 5 24 2

10. If log 3 = 0 ⋅ 4771, find the number of digits in 315 and the position of the first significant digit in 3 −15 . 11. Evaluate the following : (i) (ii)

215 ⋅ 9 if log 2159 = 3 ⋅ 3342 3

0 ⋅ 0007652 if log 76 ⋅ 52 = 1⋅ 8838. 35 ⋅ 8 × 375 ⋅ 2 + 12 ⋅ 8 12. Evaluate by using log tables ⋅ 29 × 0 ⋅ 0015

209

13. Using the log tables evaluate the following : (53 ⋅ 2)2 (0 ⋅ 126)1 /3 1 − (0 ⋅ 234)2

⋅

A nswers 1. (i) 1

(ii) 3

(iii) 2

(ii) 6 ⋅ 8281

(iii) 1 ⋅ 8281

(ii) 80 ⋅ 57

(iii) 0 ⋅ 8057.

(iv) 2

(v) 4 2. 2 ⋅ 3010. 3. (i) 1⋅ 8281 (iv) 0 ⋅ 8281. 4. (i) 80570 5. 18.

6. 302.

7. 10, 20, 21, 28, 25.

8. (i) 3 ⋅ 2851

9. 0 ⋅ 47712 . 11. (i) 14 ⋅ 51

(ii) 1⋅ 4496.

10. 8, 8th. (ii) 0 ⋅ 09147.

12. 16210.

13. 1501.

5.12 Applications of Logarithms (I) Determination of Half Life. The half-life period of a radioactive element is the time during which half the amount of a given sample of the element disintegrates. It is denoted by t1 /2 . log 2 0 ⋅ 693 t1 /2 = = , λ λ where λ is the decay constant or disintegration constant.

Illustration 1 : Solution : or or

Half life of radium is 1600 years. Calculate the disintegration constant. log 2 We have t1 /2 = λ log 2 λ = t1 /2 λ =

0 ⋅ 693 = 1⋅ 3736 × 10 1600 × 365 × 24 × 60 × 60 s

−11

s −1 .

Illustration 2 : Calculate the half life period of a nucleus if at the end of 4 ⋅ 2 days, N = 0 ⋅ 798 N 0 . Given N = N 0 e − λ t .

210

Solution :

Given N = N 0 e − λ t

⇒

λ =−

Half life,

t1 /2 =

Illustration 3 : 14 6

log (N / N 0 ) t

=−

log (0 ⋅ 798) 4 ⋅ 2 days

= 0 ⋅ 054 day

−1

.

0 ⋅ 693 0 ⋅ 693 = = 12 ⋅ 83 days λ 0 ⋅ 054 day −1

Find the age of death of an organism from the following data : t1 /2 of

8 C = 5600 years, and ratio of amounts of 14 6 C at the death and present time is 10 .

The formula used for carbon dating is N 0 ⋅ 693 0 ⋅ 693 2 ⋅ 303 t= log 0 , where λ = = year λ N t1 /2 5600 Solution : ⇒

−1

.

N0

= 10 8 N 2 ⋅ 303 × 5600 2 ⋅ 303 t= log 10 8 = × 8 log 10 0 ⋅ 693 / 5600 0 ⋅ 693 2 ⋅ 303 × 5600 = × 8 = 1⋅ 488 × 10 5 years. 0 ⋅ 693

We have

Illustration 4 : A piece of wood recorded in an excavation has 25 ⋅ 6% as much C14 (half-life = 5760 years) as ordinary wood today has. When did the piece get burried ? Solution : Also ∴ ⇒

Here N 0 = 100, N t = 25 ⋅ 6, t1 /2 = 5760 years. 0 ⋅ 693 2 ⋅ 303 100 and λ = log ⋅ 5760 t 25 ⋅ 6 0 ⋅ 693 2 ⋅ 303 100 = log 5760 t 25 ⋅ 6 2 ⋅ 303 × 5760 100 t= log = 11330 years. 0 ⋅ 693 25 ⋅ 6 λ =

Illustration 5 :

A piece of wood was found to have C14 / C12 ratio 0 ⋅ 7 times that in a

living plant. Calculate the period when the plant died. Half-life of C14 = 5760 years. Solution : We have, ∴ ⇒

Here λ =

N 0 = 1, N t = 0 ⋅ 70, t1 /2 = 5760 years. N 2 ⋅ 303 log 0 t Nt

and

λ =

0 ⋅ 693 0 ⋅ 693 = ⋅ t1 /2 5760

0 ⋅ 693 2 ⋅ 303 1 = log 5760 t 0 ⋅ 70 2 ⋅ 303 × 5760 × 0 ⋅ 155 t= = 2970 years. 0 ⋅ 693

Illustration 6 : One microgram of Na −24 is injected into the blood of a patient. How long will it take the radioactivity to fall to 10% of initial value ? (t1 /2 for Na − 24 is14 − 8 hours). 0 ⋅ 693 Solution : We know that t1 /2 = and t1 /2 = 14 ⋅ 8 hours (given). λ

211

∴ Also ∴

0 ⋅ 693 ⋅ 14 ⋅ 8 N N 0 100 1 t = × 2 ⋅ 303 log 0 and = = 10. λ N N 10 14 ⋅ 8 14 ⋅ 8 t= × 2 ⋅ 303 log 10 = × 2 ⋅ 303 × 1 0 ⋅ 693 0 ⋅ 693 = 49 ⋅ 2 hrs = 2 ⋅ 05 days. λ =

(II) Computation of Equilibrium Constant. Illustration 7 :

Calculate K p for the reaction

3 / 2 O2 ( g)

∆G ° for the reaction is 163 ⋅ 43 kJ mol Solution :

O3 ( g) at 298 K

⇔

−1

. Given ∆G ° = − RT log K p .

∆G ° = − RT log K p 163 ⋅ 43 × 10 3 J mol − 1 ∆G ° =− RT (8 ⋅ 314 JK − 1 mol − 1 ) (298 K )

⇒

log K p = −

∴

K p = 2 × 10

− 29

.

(III) Computation of Avogadro’s Number. Illustration 8 :

Using Stirling’s approximation,

log N A ! ≈ N A log N A − N A calculate log N A ! , where N A is Avogadro’s Number. (N A = 6 ⋅ 022 × 10 23 ). Solution :

By Stirling’s approximation log N A ! ≈ N A log N A − N A = (6 ⋅ 022 × 10 23 ) log (6 ⋅ 022 × 10 23 ) − (6 ⋅ 022 × 10 23 ) = 323 ⋅ 77 × 10 23 = 3 ⋅ 2377 × 10 25 .

(IV) Chemical Kinetics Problems. Illustration 9 : In a first order reaction, it takes the reactant 40 ⋅ 5 minutes to be 25% decomposed. Calculate the rate constant of the reaction. 1 a Given k1 = log ⋅ t a− x Solution : The reactant gets 25% decomposed in 40 ⋅ 5 min, thus the remaining reactant is 75%. Now at t = 40 ⋅ 5 min, we have a − x = 0 ⋅ 75 a. 1 a 1 a ∴ k1 = log = log = 7 ⋅ 11 × 10 t a − x 40 ⋅ 5 min 0 ⋅ 75 a Illustration 10 :

Show that for a first order reaction, t =

−3

min

−1

.

1 a log , the time required k1 a− x

for 99 ⋅ 9% completion of the reaction is 10 times that required for 50% completion. 1 a Solution :. We have t = log k1 a− x

212

1 100 log t (99 ⋅ 9%) k1 100 − 99 ⋅ 9 3 ⋅ 00 = = ≈ 10. 1 100 t (50%) 0 ⋅ 3010 log k1 100 − 50

⇒

Illustration 11 : Decomposition of diazobenzene chloride was followed at constant temperature by measuring the volume of nitrogen evolved at suitable intervals. The readings thus, obtained are given below : Time from start (min) 0 20 50 70 ∞ Volume of N 2 (ml) 0 10 25 33 162 Calculate the order of the reaction. Solution :

Here a = V∞ = 162 ml and a − x = V∞ − Vt .

For the first order reaction V∞ 2 ⋅ 303 a 2 ⋅ 303 k1 = log = log ⋅ t a− x t V∞ − Vt t 20 50 70

V∞ − Vt

k1 =

V∞ 2 ⋅ 303 log t V∞ − Vt

2 ⋅ 303 162 log = 3 ⋅ 22 × 10 − 3 / min 20 152 2 ⋅ 303 162 162 − 25 = 137 k1 = log = 3 ⋅ 36 × 10 − 3 / min 50 137 2 ⋅ 303 162 162 − 33 = 129 k1 = log = 3 ⋅ 26 × 10 − 3 / min 70 129 162 − 10 = 152 k1 =

Thus, constant value of k1 shows that the reaction is of the first order. Illustration 12 :

Half change time for the first order decomposition.

CH3 — N == N — CH3 → C2 H6 + N 2 is 30 minutes. Calculate : (i) What fraction of azo-methane will be decomposed in 1⋅ 5 hours ? (ii) How long will it take to be 60% completed ? Solution :

(i) The first order reaction equation is k1 =

2 ⋅ 303 a log ⋅ t a− x

For half change, x = a / 2, then 2 ⋅ 303 a 2 ⋅ 303 0 ⋅ 693 k= log = log 2 = = 0 ⋅ 0231 / min. t1 /2 a − (a / 2) 30 30 Let y% of azo-methane decompose in 1⋅ 5 hours. Then x = ∴

or

k1 =

2 ⋅ 303 log t

a a−

ya

=

y 100

a.

2 ⋅ 303 100 log t 100 − y

100 2 ⋅ 303 100 0 ⋅ 0231 = log 90 100 − y

or

y = 87 ⋅ 4.

213

So, the fraction decomposed is 0 ⋅ 874. 60 (ii) For 60% decomposition, x = a. 100 2 ⋅ 303 a Here k1 = log 60 a 90 a− 100 2 ⋅ 303 100 or 0 ⋅ 0231 = log ⋅ t 40 ∴ t = 39 ⋅ 7 minutes.

[ ∵ k1 = 0 ⋅ 0231]

Comprehensive Exercise 3 1. The half life of

238 92 U

is 4 ⋅ 51 × 10 9 years. What percentage of

238 92 U

that

10

existed 10 years ago still survives ? 2. Calculate the age in the following cases : (i) A piece of hair in which C-14 activity is 60% of the activity found today (t1 /2 = 5760 years). (ii) A vegetarian beverage whose tritium content is only 5% of level in living plants (t1 /2 = 12 ⋅ 3 years). (iii) A uranium rock has uranium-238 and lead-206 in mass ratio of1⋅ 50 to 1⋅ 00 ( λ = 1⋅ 52 × 10 − 10 year − 1 ). 3. 100 mg of Co − 60 is stocked in a laboratory as a γ-ray source. It has a halflife of 5 ⋅ 26 years. Calculate the percentage decrease in its activity after 1 year. 4. Following observations were made during hydrolysis of methyl acetate at 25° C , using 0 ⋅ 05 N HCl as catalysts : ∞

t (sec.) :

0

75

119

183

Volume of the alkali used :

9⋅ 62

12⋅ 10

13⋅ 10

14 ⋅ 75

21⋅ 05

Show that the reaction is of the first order. 5. The decomposition of H2 O2 was studied by titrating it at different intervals of time with potassium permanganate. Calculate the velocity constant from the following data, if the reaction is of the first order. t (min.) : 0 10 30 KMnO4 (ml. ) : 25⋅ 0 20⋅ 0 12⋅ 5

40 9⋅ 6

A nswers 1. 21⋅ 5%. 2. (i) 4245 years (ii) 53 ⋅ 17 years 3. 12 ⋅ 36%.

5. 0 ⋅ 0231 per min.

(iii) 3 ⋅ 76 × 10 9 years.

214

Comprehensive Exercise 4 Fill in the Blanks Fill in the blanks ‘‘..........’’, so that the following statements are complete and correct. 1. The value of log 3 27 is .......... . 2. The logarithmic form of 25 = 32 is .......... . 3. If log (k 2 − 4k + 5) = 0, then the value of k is .......... . 4. If log m x + log m y + log m z = 0 , then xyz = .......... .

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 5.

The value of log b a log c b log a c is (a) 0

6.

(b) log (abc ) (b) 2

(c) 1

(d) 0

(c) 1.8854

(d) 3.8854

(c) 3

(d) 4

If x = 0.0768 , then log x is (a) 0.8854

8.

(d) 10

If x = log 6 216 , then the value of x is (a) 3

7.

(c) 1

(b) 2.8854

The characteristic of log 37295 is (a) 1

(b) 2

True or False Write ‘T’ for true and ‘F’ for false statement. 9.

The logarithm of 1 is 0 at every base.

10. The logarithm of a to the base a itself is 0. 11. log b a =

log c a log c b

.

12. The mantissa of 21.34 is 0.3292.

A nswers 1. 3 5. (c) 9. T

2. log 2 32 = 5. 6. (a) 10. F

3.

2

7. (b) 11. T

4. 1 8. (d) 12. T

215

6 S ystem O f C oordinates

6.1 Introduction he French Mathematician and Philosopher Descartes (1596 –1650) was the first, who established relation between Geometry and Algebra. He invented Analytical Geometry or Coordinate Geometry. In this branch of Geometry, the first important relation is established between the points in a plane and the ordered pairs of real numbers.

T

6.2 Rectangular Cartesian Coordinates Let X ′ OX and Y ′ OY be two mutually perpendicular lines, intersecting at O. We fix up a convenient unit of length and starting from O, mark off distances on x-axis as well as y-axis. The two lines X ′ OX , Y ′ OY are called axes of reference. X ′ OX is called the x-axis and Y ′ OY is called the y-axis. The point of intersection of axes, O, is called the origin.

216

Let P be any point in the plane of lines. Draw PM perpendicular from P on OX . Then the lengths OM and MP are called the rectangular cartesian coordinates or in short only coordinates and generally denoted by x and y respectively. The length OM or x is called the x-coordinate or abscissa of the point P and the length MP or y is called the y-coordinate or ordinate of the point P. The point P is represented by ( x , y). The lines X ′ OX and Y ′ OY are infinite. They divide the plane into 4 regions, called the quadrants. The regions XOY , YOX ′ , X ′ OY ′ and Y ′ OX are called the first, the second, the third and the fourth quadrants respectively.

6.3 Sign Convention The distances measured in the direction OX are taken positive while distances measured in the opposite direction OX ′ are taken negative. Similarly, distances measured in the direction OY are taken positive and in the opposite direction OY ′ are taken negative. From the above sign convention, it is clear that if a point ( x , y) lies in the first quadrant then both x and y are positive; if a point lies in the second quadrant then x is negative and y is positive; if a point lies in the third quadrant then both x and y are negative and if the point lies in the fourth quadrant then x is positive and y is negative.

6.4 Distance between two Points Let P ( x1 , y1 ) and Q ( x2 , y2 ) be two given points and let PQ = d . Draw PM and QN perpendiculars from P and Q respectively on the x-axis. Draw PR perpendicular from P on QN. Then OM = x1 , MP = y1 , ON = x2 , NQ = y2 . ∴

PR = MN = ON − OM = x2 − x1 ,

and RQ = NQ − NR = NQ − MP = y2 − y1 .

217

Now, in ∆ PRQ , ∠ R is right angle, so PQ 2 = PR 2 + RQ 2 , d 2 = ( x2 − x1 )2 + ( y2 − y1 )2

i. e., ∴

d = ( x2 − x1 )2 + ( y2 − y1 )2

= (difference of x - coordinates ) 2 + (difference of y - coordinates ) 2 . Corollary.

The distance of the point ( x, y) from the origin is x 2 + y 2 .

6.5 Properties of some Geometrical Figures (i) Equilateral triangle. All sides are equal and each angle is 60°. (ii) Isosceles triangle. Two sides (or two angles) are equal. (iii) Parallelogram. Opposite sides are parallel, or a pair of opposite sides are parallel and equal, or the diagonals bisect each other. (iv) Rectangle. Opposite sides are equal and each angle is a right angle. The diagonals are equal. (v) Rhombus. All the four sides are equal. The diagonals bisect each other at right angles and are in general not equal. (vi) Square. All the four sides are equal and each angle is a right angle. The diagonals are also equal. (vii) Circum-centre of a triangle. It is the centre of the circle which passes through the vertices of the triangle and so it is equidistant from the three vertices of the triangle. It is the point of intersection of the right bisectors of the sides of the triangle (i. e.,the lines through the mid-point of each side and perpendicular to it).

Illustration 1. Find the distance between the points (i) A (3, − 1) and B (−2 , 4) (ii) A (a cos α , a sin α) and B (a cos β , a sin β). Solution :

(i) The distance between the given points = AB = {−2 − 3}2 + {4 − ( −1)}2 = 25 + 25 = 5 2.

(ii) The distance between the given points = AB = (a cos β − a cos α)2 + (a sin β − a sin α)2 =

a2 (cos β − cos α)2 + a2 (sin β − sin α)2

218

= a (cos β − cos α)2 + (sin β − sin α)2 = a (cos 2 β + sin2 β) + (cos 2 α + sin2 α) − 2 (cos α cos β + sin α sin β) = a 2 − 2 cos (α − β) = a 2 {1 − cos (α − β)} α − β α − β = a 4 sin2   = 2 a sin  ⋅  2   2  Illustration 2 : Find the value of a if the distance between the points (3, a) and (4, 1) is 10. [UPTU 2003]

Solution :

Here, (4 − 3)2 + (1 − a)2 = 10

or

1 + (1 − a)2 = 10

∴

a = 1 ± 3 = − 2, 4.

Illustration 3 :

or (1 − a)2 = 9

or

1 − a = ± 3.

Show that the points A (2, − 2), B (8, 4), C (5, 7), D (− 1, 1) are the

vertices of a rectangle. Solution :

[UPTU 2007]

Here, AB = (8 − 2)2 + (4 + 2)2 = 62 + 62 = 72, BC = (5 − 8)2 + (7 − 4)2 = (− 3)2 + 32 = 18, CD = (− 1 − 5)2 + (1 − 7)2 = (− 6)2 + (− 6)2 = 72, DA = (2 + 1)2 + (− 2 − 1)2 = 32 + (− 3)2 = 18, AC = (5 − 2)2 + (7 + 2)2 = 32 + 92 = 90 ,

and

BD = (− 1 − 8)2 + (1 − 4)2 = (− 9)2 + (− 3)2 = 90 .

∴

AB = CD , BC = DA and AC = BD .

Thus, the opposite sides of the quadrilateral ABCD are equal and its diagonals are also equal. Hence, it is a rectangle. Illustration 4 :

Show that the points (12 , 8), (−2 , 6) and (6, 0) are the vertices of a right

angled triangle. Solution : Now

Let A , B and C be the points (12 , 8), (−2 , 6) and (6, 0) respectively. AB 2 = (12 + 2)2 + (8 − 6)2 = 142 + 22 = 200, BC 2 = (−2 − 6)2 + (6 − 0)2 = (−8)2 + 62 = 100

and

CA 2 = (12 − 6)2 + (8 − 0)2 = 62 + 82 = 100.

Thus,

BC 2 + CA 2 = AB 2 ,

i. e., the sum of the squares of two sides is equal to the square of the third side. This shows that ABC is a right angled triangle. Hence, the given points are the vertices of a right angled triangle.

219

Illustration 5 :

The vertices A, B, C of a triangle are (2 , 1), (5, 2) and (3, 4) respectively.

Find the coordinates of the circum-centre and also the radius of the circum-circle. Solution :

Let ( x , y) be the coordinates of the centre P of the circum-circle of the

triangle ABC . Then PA 2 = PB 2 ⇒

( x − 2)2 + ( y − 1)2 = ( x − 5)2 + ( y − 2)2

⇒

− 4 x + 4 − 2 y + 1 = − 10 x + 25 − 4 y + 4

⇒

6 x + 2 y = 24

⇒

3 x + y = 12 . 2

…(1)

2

Also,

PB

⇒

( x − 5)2 + ( y − 2)2 = ( x − 3)2 + ( y − 4)2

⇒

−10 x + 25 − 4 y + 4 = − 6 x + 9 − 8 y + 16

⇒

−4 x + 4 y + 4 = 0

⇒

x − y = 1.

= PC

…(2)

13 9 Solving the equations (1) and (2), we get the centre P as the point  ,  ⋅  4 4 Now if r be the radius of the circum-circle, then 2 2 13 9 25 25 50 r 2 = PA 2 =  − 2 +  − 1 = + = 4  4  16 16 16

∴

r=

Illustration 6 :

5 2 4

⋅

The vertices of a triangle ABC are A (0, 0), B (2 , − 1) and C (9, 2).

Evaluate cos B . Solution :

We know that cos B =

Now

a2 + c

2

−b2

2 ac

=

BC 2 + AB 2 − AC 2 2 BC . AB

BC = (9 − 2)2 + (2 + 1)2 = 49 + 9 = 58 , AC = (9 − 0)2 + (2 − 0)2 = 81 + 4 = 85 ,

and

AB = (2 − 0)2 + (−1 − 0)2 = 4 + 1 = 5.

Substituting these values in (1), we get cos B =

58 + 5 − 85 2 ( 58 )( 5)

=

−11 ⋅ 290

⋅

…(1)

220

Comprehensive Exercise 1 1. Find the distance between the points : (i) (−6, 7) and (−1, − 5) (ii) (a + b , a − b) and (a − b , a + b) (iii) (cos α , − sin α) and (− cos α , sin α) (iv) (at12 , 2 at1 ) and (at2 2 , 2 at2 ). 2. Find a, if the distance between the points (a , 2) and (3, 4) is 8. 3. If the points (2 , 1) and (1, − 2) are equidistant from the point ( x , y), show that x + 3 y = 0. 4. Find the values of x , y if the distance of the point ( x , y) from (−3, 0) as well as from (3, 0) is 4. 5. Use distance formula to show that the points A (−2 , 3), B (1, 2) and C (7, 0) are collinear. 6. Show that the points A (a , a), B (− a , − a) and C (− 3 a , 3 a) are the vertices of an equilateral triangle. 7. Show that four points (0, − 1), (6, 7), (−2 , 3) and (8, 3) are the vertices of a rectangle. 8. Show that the points A (1, − 2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram. 9. Show that the points A (0, 1), B (1, 4), C (4, 3) and D (3, 0) are the vertices of a square. Show that (−2 , 3), (8, 3) and (6, 7) are the vertices of a right angled triangle. 10. Show that the quadrilateral whose vertices are A (2, − 1), B (3, 4), C (−2 , 3) and D (−3, − 2) is a rhombus. 11. Find the coordinates of the circum-centre of the triangle whose vertices are (8, 6), (8, − 2) and (2 , − 2). Also, find its circum-radius. 12. The vertices of a triangle are A (1, 1), B (4, 5) and C (6, 13). Find cos A. −2 a   a 13. P and Q are two points whose coordinates are (at 2 , 2 at) and  2 ,  t t  1 1 respectively and S is the point (a , 0). Show that is independent of t . + SP SQ

A nswers 1. (i) 13

(ii) 2 2b

(iv) a (t1 − t2 ) (t1 + t2 )2 + 4.

(iii) 2

221

2. 3 ± 60 .

4. x = 0, y = ± 7. 63 12. ⋅ 65

11. (5, 2) ; 5.

6.6 Section Formulae. (i) The coordinates of a point which divides the line joining the two given points ( x1 , y1 ) and ( x2 , y2 ) internally in the ratio m1 : m2 is m1 y 2 + m2 y 1 m x + m2 x 1 x= 1 2 , y= ⋅ m1 + m2 m1 + m2 Corollary.

(i) The coordinates of the mid-point of the line joining the points  x + x2 y1 + y2   ⋅ ( x1 , y1 ) and ( x2 , y2 ) are  1 , 2  2  (ii) The coordinates of a point which divides the line joining the two given points ( x1 , y1 ) and ( x2 , y2 ) externally in the ratio p : q is px2 − q x1 py2 − q y1 x= , y= ⋅ p− q p− q

Illustration 1 : Find the coordinates of the points which divide internally and externally the line segment joining the points (1, 7) and (6, − 3) in the ratio 2 : 3. Solution : Internal division. Let the coordinates of the required point be ( x1 , y1 ). Then 2 . 6 + 3 .1 2 . (−3) + 3 . 7 x1 = = 3, y1 = = 3. 2+3 2+3 ∴

The required point is (3, 3).

External division. x2 = ∴

If the coordinates of the required point be ( x2 , y2 ), then 2 . 6 − 3 .1 2−3

= − 9, y2 =

2 . (−3) − 3 . 7 2−3

= 27.

The required point is (−9, 27).

Illustration 2 :

If the coordinates of the ends of a diameter of a circle are (− 4, 7) and (2 , 5),

then find the centre of the circle. Solution :

Let ( x , y) be the centre of the circle. Then ( x , y) is the mid-point of

(− 4, 7) and (2 , 5). ∴

x=

−4+2 2

= − 1 and

y=

7+5 2

Hence, the coordinates of the centre are (−1, 6 ).

= 6.

222

Illustration 3 :

In what ratio the line x + y = 4 divides the line joining the points (−1, 1)

and (5, 7) ? Solution :

[UPTU 2008]

Suppose the line x + y = 4

…(1)

divides the line joining the points (–1, 1) and (5, 7) in the ratio m : n . If ( x1 , y1 ) are the coordinates of the point dividing the line joining (−1, 1) and (5, 7) in the ratio m : n , then m . 5 + n . (−1) 5m − n x1 = = m+n m+n and

y1 =

m. 7 + n . 1 m+n

=

7m + n m+n

⋅

Since the point ( x1 , y1 ) lies on the line (1), therefore, x1 + y1 = 4 5m − n 7m + n or + =4 m+n m+n or or or

5m − n + 7m + n = 4m + 4n m 4 1 8m = 4n or = = n 8 2 m : n = 1: 2 .

Illustration 4 :

The extremities of the diagonal of a parallelogram are the points (3, − 4)

and (− 6, 5). Third vertex is the point (−2 , 1. ) Find the coordinates of the fourth vertex. Solution :

Let ( x , y) be the coordinates of the

fourth vertex D. If M is the point of intersection of the diagonals AC and BD , then M is the middle point of both the diagonals AC and BD . x−2 3−6 ∴ = 2 2 y +1 −4+5 and = 2 2 or

x − 2 = − 3 and

or

x = − 1 and

y +1=1

y = 0.

Hence, the fourth vertex is the point (−1, 0). Illustration 5 :

If A (−1, 3), B (1, − 1) and C (5, 1) are the vertices of a triangle ABC , find

the length of the median through A . Solution :

Let D be the middle point of BC .

Then AD is the median through A, and D is the point 1 + 5 −1 + 1 ,   i.e. (3, 0)  2 2  ∴

AD = {3 − (−1)}2 + {0 − 3}2

[UPTU 2005]

223

= 42 + (−3)2 = 5. Illustration 6 : The coordinates of the mid-points of the sides of a triangle are (1, 2), (0, − 1) and (2 , − 1). Find the coordinates of its vertices. [UPTU 2007] Solution :

Let (1, 2), (0, − 1) and (2 , − 1) be the

mid-points of AB , BC and CA respectively. Let the coordinates of A , B , C be ( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ) respectively. Then x1 + x2 2 and or and

y1 + y2

= 1,

=2,

2

x2 + x3 2 y2 + y3 2

= 0, = − 1,

x3 + x1 2

=2

y3 + y1 2

= −1

x1 + x2 = 2 , x2 + x3 = 0, x3 + x1 = 4

…(1)

y1 + y2 = 4, y2 + y3 = − 2 , y3 + y1 = − 2

…(2)

From (1) and (2), we get 2 ( x1 + x2 + x3 ) = 6 and

2 ( y1 + y2 + y3 ) = 0

or

x1 + x2 + x3 = 3

and

y1 + y2 + y3 = 0.

∴

x1 = 3, x2 = − 1, x3 = 1 ; y1 = 2 , y2 = 2 , y3 = − 4.

Hence, the vertices A , B , C of the triangle are (3, 2), (−1, 2) and (1, − 4) respectively. Illustration 7 :

A quadrilateral has the vertices at the points (− 4, 2), (2 , 6), (8, 5) and

(9, − 7). Show that the mid-points of the sides of this quadrilateral are the vertices of a parallelogram. [UPTU 2004] Solution : Let ABCD be the given quadrilateral with vertices A (− 4, 2), B (2, 6), C (8, 5) and D (9, − 7). Let E, F, G and H be the mid-points of the sides AB, BC, CD  − 4 + 2 2 + 6 and AD respectively. Then the co-ordinates of E are  ,  i. e., (− 1, 4);  2 2   2 + 8 6 + 5  11 the coordinates of F are  ,  i. e., 5,  ;  2  2 2  8 + 9 5 − 7 17  the coordinates of G are  ,  i. e.,  , − 1 ;  2 2  2  − 4 + 9 2 − 7 5 and the coordinates of H are  ,  i. e.,  , −  2 2  2 Now the coordinates of the mid-point of EG are  − 1 + 17    4 − 1 15 3 2 ,   i. e.,  ,   4 2 2 2    

5 ⋅ 2

224

 5 + 5 11 − 5   2 , 2 2  i. e., 15 , 3 ⋅ and the coordinates of the mid-point of FH are   4 2 2   2   Thus, we see that the diagonals EG and FH of the quadrilateral EFGH bisect each other. Hence EFGH is a parallelogram.

Comprehensive Exercise 2 1. Find the coordinates of the point which divides the line segment joining the points (6, 3) and (− 4, 5) in the ratio 3 : 2 : (i) internally and (ii) externally. 2. Find the coordinates of the mid-point of the line segment joining the points (−2 , − 5) and (3, − 1). 3. Find the points on the line through A (5, − 4) and B (− 3, 2), which are, twice as far from A as from B . 4. Find the coordinates of the point which divides the line segment joining the points (5, − 2) and (9, 6) in the ratio 3 : 1. 5. The coordinates of one end point of a diameter of a circle are (3, 5). If the coordinates of the centre be (6, 6), find the coordinates of the other end point of the diameter. 6. Find the ratio in which the point (2 , y) divides the join of (− 4, 3) and (6, 3), and hence, find the value of y. 7. In what ratio does the point (−1, − 1) divide the line segment joining the points (4, 4) and (7, 7) ? 8. A (1, 1) and B (2 , − 3) are two points and D is a point on AB produced such that AD = 3 AB . Find the coordinates of D . 9. If the point C (−1, 2) divides externally the line segment joining A (2 , 5) and B in the ratio 3 : 4, find the coordinates of B . 10. Find the lengths of the medians of a triangle whose vertices are A (−1, 3), B (1, − 1) and C (5, 1. ) 11. The three vertices of a parallelogram taken in order are (−1, 0), (3, 1) and (2 , 2) respectively. Find the coordinates of the fourth vertex. 12. Show that the points A (−2 , − 1), B (1, 0), C (4, 3) and D (1, 2) are the vertices of a parallelogram. 13. If the points (−2 , − 1), (1, 0), ( x, 3), (1, y) form a parallelogram, find the values of x and y. 14. The three vertices of a rhombus, taken in order are (2 , − 1), (3, 4) and (−2 , 3). Find the fourth vertex. 15. Show that the points (− 4, − 1), (− 2 , − 4), (4, 0) and (2 , 3) are the vertices of a rectangle.

225

A nswers 1.

21 (i) 0,   5  1 , − 3 .   2 

(ii) (−24, 9).

4.

(8, 4).

 − 1 , 0 , (−11, 8).    3  5. (9, 7).

6.

3 : 2 ; 3.

7.

8.

(4, − 11).

9. (3, 6).

2.

10.

3.

externally 5 : 8.

AD = 5, BE = 10 , CF = 5.

11. (− 2 , 1).

13. x = 4, y = 2 .

14. (− 3, − 2).

6.7 Area of a Triangle. The area of a triangle when the coordinates of its three vertices A, B and C are ( x1 , y1 ), ( x2 , y2 ) and ( x3 , y3 ) respectively is given by 1 area of ∆ ABC = { x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 )} 2 1 = {( x 1 y 2 + x 2 y 3 + x 3 y 1 ) − ( y 1 x 2 + y 2 x 3 + y 3 x 1 )} 2 y1 1 x1 1 = x2 y2 1⋅  2 y3 1 x3 Condition for collinearity of three points. The points ( x1 , y1 ), ( x2 , y2 ), ( x3 , y3 ) will be collinear if x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0. Corollary.

If one vertex of a triangle is (0, 0) and the other two vertices are 1 ( x1 , y1 ) and ( x2 , y2 ), then its area = ( x1 y2 − x2 y1 ). 2 Note : If the area of ∆ ABC is zero, then the three points A , B , C will be collinear.

6.8 Area of a Quadrilateral If ( x1 , y1 ), ( x2 , y2 ), ( x3 , y3 ) and ( x4 , y4 ) be the four vertices A, B, C and D respectively of a quadrilateral, then area of quadrilateral ABCD 1 = {( x 1 y 2 + x 2 y 3 + x 3 y 4 + x 4 y 1 ) − ( y 1 x 2 + y 2 x 3 + y 3 x 4 + y 4 x 1 )}. 2

226

Illustration 1 :

Find the area of the triangle whose vertices are (0 , 5),(2, 3) and (4, 5). [UPTU 2002, 09]

Solution : ∴

Here, x1 = 0 , y1 = 5 , x2 = 2, y2 = 3, x3 = 4, y3 = 5.

area of the given triangle 1 = [( x1 y2 + x2 y3 + x3 y1 ) − ( y1 x2 + y2 x3 + y3 x1 )] 2 1 = [{(0).(3) + (2).(5) + (4).(5)} − {(5).(2) + (3).(4) + (5).(0)}] 2 1 1 1 = [(0 + 10 + 20) − (10 + 12 + 0)] = [30 − 22] = ⋅ 8 = 4 sq. units. 2 2 2

Illustration 2 :

Prove that the points (a, b + c ), (b, c + a), (c , a + b) are collinear. [UPTU 2001, 07]

Solution :

Here x1 = a, y1 = b + c , x2 = b, y2 = c + a, x3 = c , y3 = a + b.

Now, area of the triangle formed by the given points 1 = [( x1 y2 + x2 y3 + x3 y1 ) − ( y1 x2 + y2 x3 + y3 x1 )] 2 1 = [{a (c + a) + b (a + b) + c (b + c )} − {(b + c ) b + (c + a) c + (a + b) a}] 2 1 = [ac + a2 + ba + b 2 + cb + c 2 − b 2 − cb − c 2 − ac − a2 − ba] = 0. 2 Hence, the given points are collinear. Illustration 3 :

For what value of k are the points (k, 2 − 2k), (1 − k, 2k) and

( − 4 − k, 6 − 2k) collinear ? Solution : If the points A (k, 2 − 2k), B (1 − k, 2k) and C (− 4 − k, 6 − 2k) are collinear, then the area of ∆ ABC must be zero, 1 i. e., [k {2k − (6 − 2k)} + (1 − k) {(6 − 2k) − (2 − 2k)} 2 + (−4 − k) {(2 − 2k) − 2k}] = 0 2

or

2k

or

(2k − 1)(k + 1) = 0. 1 k = − 1 or ⋅ 2

∴

Illustration 4 :

+ k −1= 0

Find the area of a quadrilateral whose vertices are (1, 1),(3, 4), (5, − 2) and

(4, − 7). Solution : Here, x1 = 1, y1 = 1, x2 = 3, y2 = 4, x3 = 5, y3 = − 2 , x4 = 4, y4 = − 7. Required area

227

1 [( x1 y2 + x2 y3 + x3 y4 + x4 y1 ) 2 − ( y1 x2 + y2 x3 + y3 x4 + y4 x1 )] 1 = [{1 . 4 + 3 . (−2) + 5 . (−7) + 4 . 1} 2 =

− {1 . 3 + 4 . 5 + (−2) . 4 + (−7) . 1}] 1 1 = [{4 − 6 − 35 + 4} − {3 + 20 − 8 − 7}] = [(−33) − 8] 2 2 41 = , numerically. 2 Illustration 5 :

If the four points (1, 2), (−5, 6), (7, − 4) and (a, − 2) are collinear, find a. [UPTU 2008]

Solution :

Let A , B , C and D be the points (1, 2), (−5, 6), (7, − 4) and (a, − 2)

respectively. If these points are collinear, then the area of the quadrilateral ABCD is zero. 1 ∴ [{1 . 6 + (−5) . (−4) + 7 . (−2) + a . 2} 2 − {(−5) . 2 + 76 . + a . (−4) + 1 . (−2)}] = 0 or

[(12 + 2 a) − (30 − 4a)] = 0

∴

a = 3.

Illustration 6 :

Find the area of a triangle formed by the lines

y = 2 x, Solution :

or 6a − 18 = 0

y= x

and

y = 3 x + 4.

[UPTU 2001]

Let the equations of the sides AB , BC and CA of ∆ ABC be

y − x = 0, y − 2 x = 0 and y − 3 x − 4 = 0 respectively. Solving these equations in pairs, the coordinates of A , B and C are (−2 , − 2),(0, 0) and (− 4, − 8) respectively. 1 ∴ Area of ∆ ABC = { x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 )} 2 1 1 = {(− 2) (0 + 8) + 0 (− 8 + 2) + (− 4) (− 2 − 0)} = {− 16 + 8} 2 2 1 = (− 8) = 4 sq. units, neglecting the negative sign. 2 Illustration 7 : Four points A (6, 3), B (− 3, 5), C (4, − 2) and D ( x, 3 x) are given in ∆ DBC 1 such a way that [UPTU 2006] = , find x. ∆ ABC 2 Solution :

⇒

1 [ x (5 + 2) − 3 (− 2 − 3 x) + 4 (3 x − 5)] = 2 ⋅ 1 Area of ∆ ABC [6 (5 + 2) − 3 (− 2 − 3) + 4 (3 − 5)] 2 1 7 x + 6 + 9 x + 12 x − 20 = 2 42 + 15 − 8 Area of ∆ DBC

228

1 28 x − 14 = 2 49 1 4x − 2 = 2 7 7 =4 x−2 2 11 x= ⋅ 8

⇒ ⇒ ⇒ ⇒

Comprehensive Exercise 3 1. Find the area of a triangle whose vertices are : (i) (4, 4), (3, − 16) and (3, − 2) (ii) (3, 2), (11, 8) and (8, 12) (iii) (a, c + a), (a, c ) and (− a, c − a) (iv) (at12 , 2at1 ), (at2 2 , 2at2 ) and (at3 2 , 2 at3 ). 2. Find the area of the quadrilateral whose vertices are : (i) (1, 1), (7, − 3), (12 , 2) and (7, 21) (ii) (1, 2), (−1, 4), (− 5, 1) and (0, − 3). 3. If (1, − 1), (6, − 6) and (4, k) are the vertices of a triangle whose area is 10 square units, find k . 4. Find whether the following points are collinear or not : (−2 , 1), (0, 5), (−1, 2). 5. Show that the points (a2 , 0), (0, b 2 ) and (1, 1) are collinear, if 1 a

2

+

1 b2

= 1.

6. If a ≠ b ≠ c , prove that the points (a, a2 ), (b, b 2 ), (c , c 2 ) can never be collinear. 7. Find the condition that the point ( x , y) may lie on the line joining the points (3, 4) and (–5, – 6). 8. For what value of k , the following points are collinear : (i) (k , − 1), (2 , 1) and (4, 5) (UPTU 2010) (ii) (2 , 3), (5, k) and (6, 7). ) (2 p + 1, 3) and (2 p + 2 , 2 p) are 9. Find the value of p if the points ( p + 1, 1, collinear. 10. Find whether the points (0, 0), (a, b), (− a, − b), (a2 , b 2 ) are collinear or not. 11. Find the area of a triangle ABC the coordinates of the mid-points of whose sides are D (−1, − 2), E (6, 1) and F (3, 5) respectively.

229

A nswers 1. (i) 7 sq. units (iii) a2

(ii) 25 sq. units (iv) a2 (t1 − t2 ) (t2 − t3 ) (t3 − t1 ).

2. (i) 132 sq. units 3. 0. 7. 5 x − 4 y + 1 = 0. 1 9. 2 , − ⋅ 2

(ii) 21⋅ 5 sq. units. 4. not. 8. (i) 1 (ii) 6. 10. Not.

11. 74 sq. units.

6.9 Locus and its Equation Definition. The locus of a moving point is the path traced out by it under certain geometrical condition or conditions. To find the equation of the locus of a point, we should proceed as follows : Take the coordinates of the moving point whose locus is to be found as (h, k). Use the given geometrical condition or conditions and find a relation between h, k by eliminating the variable quantity for different values of which the locus of (h, k) is to be found. Finally, generalise (h, k), i. e., put x for h and y for k to obtain the required equation of the locus of (h, k).

Illustration 1 :

Find the locus of the moving point P, such that 2 PA = 3 PB , where A is

(0, 0) and B is (4, − 3). Solution : Let the coordinates of P be (h, k). Given

2 PA = 3 PB .

∴

4 PA2 = 9 PB 2

or

4 (h2 + k 2 ) = 9 [(h − 4)2 + (k + 3)2 ]

or

4 (h2 + k 2 ) = 9 [h2 + k 2 − 8h + 6k + 25]

or

5h2 + 5k 2 − 72 h + 54k + 225 = 0.

∴

locus of P (h, k) on generalising, is 5 x 2 + 5 y 2 − 72 x + 54 y + 225 = 0.

Illustration 2 : A point moves so that its distance from (3, 0) is twice the distance from (– 3, 0). Find the equation of the locus. [UPTU 2005] Solution :

Let A represent the point (3, 0); B the point (– 3, 0) and P (h, k) be the

moving point. According to the question, PA = 2 PB

or

( PA)2 = 4 ( PB)2

230

or

[(h − 3)2 + (k − 0)2 ] = 4 [(h + 3)2 + (k − 0)2 ]

or

h2 + 9 − 6h + k 2 = 4h2 + 36 + 24h + 4k 2

or

3h2 + 3k 2 + 30 h + 27 = 0.

Generalising, the required locus is 3 x 2 + 3 y 2 + 30 x + 27 = 0. Illustration 3 : Find the locus of a point such that the line segments having end points (2, 0) and (– 2, 0) subtend a right angle at that point. [UPTU 2006] Solution : Let A (2, 0) and B (− 2, 0) be the given points and P (h, k) be the variable point. According to the question, ∠ APB = 90 ° , ∆ APB is a right triangle

i. e., ∴

AB2 = PA2 + PB2

⇒

{2 − (− 2)}2 + (0 − 0)2 = {(2 − h)2 + (0 − k)2 } + {(− 2 − h)2 + (0 − k)2 }

⇒

16 = (2 − h)2 + k 2 + (− 2 − h)2 + k 2

⇒

16 = 4 + h2 − 4h + 2k 2 + 4 + h2 + 4h

⇒

16 = 2h2 + 2k 2 + 8

⇒

h2 + k 2 = 4.

Generalising, the required locus is x 2 + y 2 = 4. Illustration 4 : A point moves so that the sum of the squares of its distances from two fixed points A (a, 0) and B (− a, 0) is constant and equal to 2c 2 ; find the locus of the point. Solution :

Let P (h, k) be the moving point. According to the question, PA2 + PB 2 = 2c

2

[(h − a)2 + (k − 0)2 ] + [(h + a)2 + (k − 0)2 ] = 2c

or

h2 + k

or

2

=c

2

2

− a2 .

∴ the required locus of the point P (h, k) is x 2 + y 2 = c

2

− a2 .

Illustration 5 : Find the equation to the locus of a point which moves so that the sum of its distances from (3, 0) and (– 3, 0) is less than 9. [UPTU 2003, 04] Solution :

Let A (3, 0) and B (− 3, 0) be the two given points and (h, k) be the

coordinates of the moving point P whose locus is to be found. According to question PA + PB < 9 or

(h − 3)2 + k 2 + (h + 3)2 + k 2 < 9

or

(h − 3)2 + k 2 < 9 − (h + 3)2 + k 2

231

Squaring both the sides, we get (h − 3)2 + k 2 < 81 − 18 (h + 3)2 + k 2 + (h + 3)2 + k 2 or

− 12h − 81 < − 18 (h + 3)2 + k 2

or

4h + 27 > 6 (h + 3)2 + k 2 .

Again squaring both the sides, we get 16h2 + 729 + 216h > 36 [h2 + 9 + 6h + k 2 ] or ∴

20 h2 + 36k 2 < 405. the required locus of the point (h, k) is 20 x 2 + 36 y 2 < 405.

Comprehensive Exercise 4 1. Find the equation to the locus of a point equidistant from the points A (1, 3) and B (−2 , 1). 2. Find the locus of a point, so that the join of (−5, 1) and (3, 2) subtends a right angle at the moving point. 3. A point moves so that its distance from the axis of y is half its distance from the origin. Find the equation of its locus. [UPTU 2008] 4. A point P moves in such a way that the area of the triangle formed with A (1, − 1) and B (5, 2) is of magnitude 5 units. Find the locus of P. 5. Find the locus of a point such that the sum of its distances from the points (0, 2) and (0, − 2) is 6. 6. Find the equation of the locus of a point which moves so that its distance from (a, 0) is equal to its distance from the y-axis. [UPTU 2007] 7. Find the locus of the point of intersection of the lines x cos α + y sin α = a and x sin α − y cos α = b, where α is a variable. 8. A rod of length l slides with its ends on two perpendicular lines. Find the locus of its mid-point. 9. Find the locus of the mid-point of the portion of the line x cos α + y sin α = p which is intercepted between the axes. 10. If O is the origin and Q is a variable point on x 2 = 4 y, find the locus of the mid-point of OQ . 11. If O is the origin and Q is a variable point on y 2 = x , find the locus of the mid-point of OQ .

232

A nswers 1. 6 x + 4 y = 5. 2.

x 2 + y 2 + 2 x − 3 y − 13 = 0.

3.

x 2 − 3 y 2 = 0.

4. 3 x − 4 y + 3 = 0, 3 x − 4 y − 17 = 0. 5. 9 x 2 + 5 y 2 = 45. 6.

y 2 − 2ax + a2 = 0.

7.

x 2 + y 2 = a2 + b 2 .

8. 4 x 2 + 4 y 2 = l 2 . 9. 10.

p 2 (x2 + y2 ) = 4 x2 y2 . x 2 = 2 y.

11. 2 y 2 = x .

Comprehensive Exercise 5 Fill in the Blanks Fill in the blanks ‘‘.........’’, so that the following statements are complete and correct. 1. If (3, a) is the mid-point of the line joining the points (4, 3) and (2, 5), the value of a is .......... . 2. The distance of the point ( x, y) from the origin is .......... . 3. The condition for collinearity of three points ( x1 , y1 ), ( x2 , y2 ), ( x3 , y3 ) is .......... . 4. If one vertex of a triangle is (0, 0) and the other two vertices are ( x1 , y1 ) and ( x2 , y2 ), then its area is .......... . 5. If the points (− 2, − 5), (2, − 2), (8, a) are collinear, then the value of a is .......... . 6. The area of a triangle whose vertices are (4, 4), (3, − 16) and (3, − 2) is .......... . 7. The co-ordinates of the points which divides the join of (3, − 4) and (− 5, − 3) externally in the ratio 3 : 4 are .......... .

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 8. The area of the triangle whose vertices are (2, 2), (− 4, 0) and (1, − 1) is (a) 8 sq. units

(b) 6 sq. units

233

(c) 7 sq. units

(d) 5 sq. units

9. If the four points (1, 2), (− 5, 6), (7, − 4) and (a, − 2) are collinear, then a is (a) 0

(b) 1

(c) 2

(d) 3

10. The co-ordinates of a point which divides the line joining the two given points ( x1 , y1 ) and ( x2 , y2 ) externally in the ratio p : q is py − qy1 px − qx1 (a) x = 2 , y= 2 p− q p− q (b) x = (c) x = (d) x =

px2 − qx1 p− q px2 + qx1 p− q px2 − qx1 p+ q

, y=

py2 − qy1 p− q py2 + qy1

, y= , y=

p− q py2 − qy1 p+ q

11. The distance between the points (− 1, − 4) and (3, 5) is (a) 97

(b) 97

(c) 13

(d) 39

12. The points (− 2, 2), (8, − 2) and (− 4, − 3) are the vertices of (a) a right angled triangle (b) an isosceles triangle (c) an equilateral triangle (d) none of these

True or False Write ‘T’ for true and ‘F’ for false statement. 13. If the area of ∆ABC is one, then the three points A, B, C will be collinear. 14. The points (− 5, 7), (− 4, 5) and (1, − 5) are collinear. 15. The coordinates of a point which divides the line joining the two given points ( x1 , y1 ) and ( x2 , y2 ) internally in the ratio m1 : m2 are m x + m2 x1 m y + m2 y1 x= 1 2 , y= 1 2 ⋅ m1 + m2 m1 + m2 16. The distance between the points (3, − 1) and (− 2, 4) is 5 2. 17. The points (6, 2), (3, − 1) and (− 2, 4) are the vertices of a right angle triangle. 18. If the points ( x, − 1), (2, 1) and (4, 5) are collinear, then x is 2. 19. The co-ordinates of the point which divides the join of (2, 3) and (5, − 3) internally in the ratio 1 : 2 are (1, 1).

234

A nswers 1. 4. 3. 4. 8. 12. 16.

2.

(x2 + y2 )

x1 ( y2 − y3 ) + x2 ( y3 − y1 ) + x3 ( y1 − y2 ) = 0 1 5. 5/2 6. − 7 ( x1 y2 − x2 y1 ) 2 (a) 9. (d) 10. (b) (a) 13. F 14. T T 17. T 18. F

7. (27, − 7) 11. (b) 15. T 19. F

235

7 T he S traight L ine

7.1 Straight Line

A

straight line is a curve such that all the points of the line segment joining any two points on it lie on it.

7.2 Equations of Lines Parallel to the Coordinate Axes (i) Equation of a line parallel to the x-axis. The equation of the straight line parallel to the x-axis and at a distance b from it is y = b. If the line parallel to x-axis lies below it at a distance b,then its equation is y = − b. Corollary.

The equation of the x-axis is y = 0.

(ii) Equation of a line parallel to the y-axis.

236

The equation of the straight line parallel to the y-axis and at a distance a from it is x = a. If the line parallel to y-axis lies to the left of it at a distance a, then its equation is x = − a. Corollary.

The equation of the y-axis is x = 0.

7.3 Slope or Gradient of a Non-vertical Line The tangent of the angle θ which the line AB makes with the positive direction of the x-axis is called the slope or gradient of the line AB and is denoted by m. slope m = tan θ. The angle θ is positive if it is measured in anticlockwise direction from OX and it is negative if measured in clockwise direction. Note. The angle of inclination of a line with the positive direction of x-axis in anticlockwise direction always lies between 0° and 180°. The slope of x-axis or a line parallel to x-axis is 0. The slope of y-axis or a line parallel to y-axis is not defined.

7.4 Intercepts of a Line on the Axes If a line intersects the x-axis at A , then OA is called the intercept of the line on x-axis or x-intercept of the line. Similarly, if a line intersects the y-axis at B , then OB is called the intercept of the line on y-axis or y-intercept of the line. The intercepts are positive or negative according as the line meets with positive or negative directions of the coordinate axes. If a line intersects the axes at A and B , then AB is called the portion of the line intercepted between the axes.

7.5 Various Forms of Equations of a Line (i) The slope-intercept form of a line. To find the equation of a straight line which cuts off an intercept c on y-axis and whose slope is m.

237

Let the line AB make angle BHX = θ with OX . Then m = tan θ. Suppose that AB cuts off from the y-axis an intercept OK , equal to c . Take any point P ( x, y) on AB . To find the equation of AB , we have to find a relation between x, y, m and c . From P draw PM perpendicular to the x-axis and from K draw KL perpendicular to PM . Then, in ∆ PKL ∠ PKL = ∠ BHX = θ, LK = x, LP = MP − ML = MP − OK = y − c and ∴

∠ PLK = 1 right angle. y−c LP m = tan θ = = LK x

or

y = mx + c ,

which is the required equation of the line AB . Corollary.

If the straight line passes through the origin then c = 0. Hence, the

equation of the line passing through the origin is y = mx . Note. The equation y = mx + c represents any straight line except those whose equations are of the form x = a, i. e., which are parallel to the y-axis. (ii) The double intercept form of a line. To find the equation of a straight line which cuts off intercepts of lengths a and b from x-axis and y-axis respectively. Let the line AB meet x-axis and y-axis at A and B respectively. Then OA = a and OB = b. Let P ( x, y) be any point on AB . To find the equation of AB , we have to find a relation between x, y, a, b . From P, draw PM and PN perpendiculars on the axes. Join OP. Evidently, in area ∆OAB = ∆OAP + ∆OPB 1 1 1 ∴ OA . OB = OA . MP + OB . NP 2 2 2 1 1 1 or ab = ay + bx . 2 2 2 1 Dividing both sides by ab, we get 2 y x x y or 1= + + = 1, b a a b which is the required equation of the line AB .

238

(iii) The normal (perpendicular) form of a line. To find the equation of a straight line upon which the length of the perpendicular from the origin is p and this perpendicular makes an angle α with positive direction of x-axis. Let the required line be AB and OL be the perpendicular on it from O. Then OL = p and ∠ LOX = α . To find the equation of AB , we have to find a relation between x, y, p and α . Let P ( x, y) be any point on AB . From P, draw PM perpendicular on the x-axis. From M , draw MK perpendicular on OL and lastly draw PN perpendicular on MK . Then

∠ PMN = 90 ° − ∠ KMO = ∠ MOK = α ,

and

OM = x, MP = y, PN = LK .

Now

OL = OK + KL = OK + NP. OK NP = cos α and = sin α. OM MP

But ∴

OL = OM cos α + MP sin α,

i. e.,

p = x cos α + y sin α .

Hence, the required equation is x cos α + y sin α = p . Note. The convention is to regard the length of the perpendicular from the origin on the line as positive, so that p is positive. The value of α is taken positive or negative according as the line making an angle α with OX takes the position OL after rotating anticlockwise or clockwise respectively. The angle α will be taken to lie between − π and π. (iv) The two-points form of a line. To find the equation of a straight line which passes through two given points ( x1 , y1 ) and ( x2 , y2 ). Let A and B be two points whose coordinates are ( x1 , y1 ) and ( x2 , y2 ). To find the equation of AB , we have to find a relation between x, y, x1 , y1 , x2 and y2 . Let P ( x, y) be any point on AB . From A , P and B draw AL, PM and BN respectively perpendiculars on OX . From A , draw AK perpendicular to BN which meets PM at the point H. Then triangles AHP and AKB are similar triangles. AH HP …(1) ∴ = ⋅ AK KB Now

AH = LM = OM − OL = x − x1 , AK = LN = ON − OL = x2 − x1 ,

239

HP = MP − MH = MP − LA = y − y1 and

KB = NB − NK = NB − LA = y2 − y1 .

Substituting these values in (1), we get y − y1 y2 − y1 x − x1 or y − y 1 = = ( x − x 1 ), x2 − x1 y2 − y1 x2 − x1 which is the required equation of the line AB . Note. The slope of the line joining the two points ( x1 , y1 ) and ( x2 , y2 ) is y2 − y1 x2 − x1

=

difference of ordinates difference of abscissae

⋅

Corollary 1. The equation of the straight line having slope m and passing through the point ( x1 , y1 ) is y − y 1 = m ( x − x 1 ). Corollary 2. The equation of any straight line joining the origin and any point y x ( x1 , y1 ) is = ⋅ x1 y1

Illustration 1 : Find the equation of a straight line which cuts off an intercept of 3 units on negative direction of y-axis and makes an angle of 120° with the positive direction of x-axis. Solution : and

We have, m = tan 120 ° = tan (180 ° − 60 ° ) = − tan 60 ° = − 3 c = − 3.

So the equation of the line is y = mx + c i. e.,

y = − 3x − 3

Illustration 2 :

or

3 x + y + 3 = 0.

Find the equation of the line which makes intercepts of 2 and − 3 with x axis

and y-axis respectively. Solution :

Here, a = 2 and b = − 3.

Hence, the required equation of the line is y y x x or + =1 − = 1. 2 (−3) 2 3 Illustration 3 : Find the equation of straight line which makes equal intercepts on the axes and passes through the point (3, − 5). [UPTU 2002] Solution :

Let the equation of the straight line be y x + = 1. a b

The line (1) makes equal intercepts on the axes i. e., a = b. y x or ∴ + =1 x + y = a. a a If this line passes through the point (3, − 5), then

...(1)

240

3−5=a

or

a=−2

Hence, the required equation is x+ y=−2 Illustration 4 :

or

x + y + 2 = 0.

A line passes through (3, 4) and the sum of its intercepts on the axes is 14.

Find the equation of the line. Solution :

[UPTU 2009]

Let the intercept made by the line on x-axis be a.

Then the intercept made by the line on y-axis is 14 − a. ∴

The equation of the line in intercept form is y x + = 1. a 14 − a

…(1)

If the line (1) passes through the point (3, 4), then 3 4 or + =1 42 − 3a + 4a = 14a − a2 a 14 − a or

a2 − 13a + 42 = 0

⇒

a = 7 or 6.

Hence, the required equation is y x or + =1 7 7 i. e.,

x+ y=7

or

or

(a − 7)(a − 6) = 0

y x + =1 6 8 4 x + 3 y = 24.

Illustration 5 : Find the equation of the straight line the portion of which intercepted between the axes is divided by the point (−2, 6) in the ratio 3 : 2 . [UPTU 2007] Solution :

Let the equation of the straight line be y x + = 1. a b

…(1)

The line (1) meets x-axis at the point A (a, 0) and y-axis at the point B (0, b). According to question, the point (− 2, 6) divides the line segment AB in the ratio 3 :2 . 2 . a + 3 .0 2 a or ∴ −2= = a=−5 2+3 5 and

6=

2 .0 + 3 . b 2+3

=

3b 5

or

b = 10 .

Putting the values of a and b in (1), the required equation of the line is y x or + =1 y − 2 x = 10. − 5 10 π with 4 + ive x-axis and intersects another line x + 2 y + 1 = 0 at point B. Find the length AB. Illustration 6 :

A straight line, drawn through the point A (2, 1), makes an angle

[UPTU 2003]

241

Solution :

The equation of any line passing through the given point A (2, 1) and π making an angle with x-axis is 4 x−2 y −1 = = r (say), ...(1) cos 45° sin 45° where r represents the distance of any point B on this line from the given point A (2, 1. ) The coordinates ( x, y) of any point B on the line (1) are 1 1  (2 + r cos 45° 1 + r sin 45° ) i. e., 2 + r ⋅ ,1 + r ⋅ ⋅  2 2 If the point B lies on the line x + 2 y + 1 = 0, then 1 1   2 + r ⋅  + 2 1 + r ⋅  +1= 0   2 2 3 5 or or 5+r =0 r=− 2. 3 2 Hence, the length AB = 5 2 / 3. Illustration 7 :

Find the equation of the line passing through the points (4, 3) and (7 , 8). [UPTU 2001]

Solution : Using

The two points are ( x1 , y1 ) = (4, 3) and ( x2 , y2 ) = (7 , 8). y − y1 = y−3=

or

y−3=

y2 − y1 x2 − x1

8−3 7−4

( x − x1 ), the required equation is

( x − 4)

5 ( x − 4) 3

or

5 x − 3 y − 11 = 0.

Illustration 8 : Find the slope and the equation of the straight line joining the points [UPTU 2002] (2, − 5) and (4, 1). Solution :

The slope of the line joining the points (2, − 5) and (4, 1) is 1 − (− 5) 6 = = = 3. 4−2 2

Now the equation of the straight line joining the points (2, − 5) and (4, 1) and whose slope is 3 is or y − (− 5) = 3 ( x − 2) y + 5 = 3x − 6 or 3 x − y = 11. Illustration 9 : Find the equation of the straight line which divides the line joining points (2, 3) and (– 5, 8) in the ratio 3 : 4 and is also perpendicular to it. [UPTU 2006] Solution :

The equation of the line joining the points (2, 3) and (– 5, 8) is 8−3 5 or y−3= ( x − 2) y−3= ( x − 2) −5−2 −7

242

or

− 7 y + 21 = 5 x − 10

or

5 x + 7 y = 31. ...(1) 5 The slope of line (1) is − and so the slope of the line perpendicular to it will be 7 / 5. 7 The coordinates (h, k) of the point dividing line (1) in the ratio 3 : 4 are given as 3 × (− 5) + 4 × 2 h= 3+4 3×8+4×3

and

k=

i. e.,

h = −1

3+4 and

k=

36 ⋅ 7

Hence the equation of the line passing through (h, k) and having slope

7 is 5

7 ( x − h) 5 36 7 y− = [ x − (− 1)] 7 5 y−k=

or or

35 y − 180 = 49 x + 49

or

49 x − 35 y + 229 = 0.

Comprehensive Exercise 1 1. Write the equation of a straight line (i) which is parallel to x-axis and lies at a distance –5 from it. (ii) which is parallel to y-axis and lies at a distance 3 from it. (iii) which is parallel to x-axis and passes through the point (3, − 5). (iv) which is parallel to y-axis and passes through the point (−4, 7). (v) which is equidistant from the lines x = − 4 and x = 8. (vi) which is equidistant from the lines y = 10 and y = − 2 . 2. Find the equation of a straight line (i) with slope 2 and y-intercept 3; 1 (ii) with slope − and y-intercept − 4. 3 3. Find the equation of a line with slope –1 and cutting off an intercept of 4 units on negative direction of y-axis. −1 4. Find the equation of a line which makes an angle of tan (3) with the

x-axis and cuts off an intercept of 4 units on negative direction of y-axis. 5. Find the equation of a straight line cutting off an intercept –1 from y-axis and being equally inclined to the axes.

243

6. Find the equation of a straight line which cuts off intercepts (i) 2 and –3 from x-axis and y-axis respectively; (ii) –5 and 6 from x-axis and y-axis respectively. 7. Find the equation of the line which cuts off an intercept 4 on the positive direction of x-axis and an intercept 3 on the negative direction of y-axis. 8. Find the equation of a line which passes through (1, − 2) and cuts off equal intercepts on the axes. 9. Find the equation of the line which passes through (3, − 5) and cuts off intercepts on the axes which are equal in magnitude but opposite in sign. 10. Find the equation of a line which passes through the point (−3, 8) and cuts off positive intercepts on the coordinate axes whose sum is 7. 11. Find the equation of the line which passes through the point (− 4, 3) and is such that the portion of it between the axes is divided by the point in the ratio 5 : 3. 12. Find the equation of a line which passes through the point (22 , − 6) and is such that the intercept on x-axis exceeds the intercept on y-axis by 5. 13. Find the equation of a straight line (i) which passes through (5, − 7) and having slope 4; (ii) which passes through (−2 , 3) and having slope –3. 14. Determine the equation of the line through the point (3, − 4) and parallel to x-axis. 15. Find the equation of the line passing through : (i) (−1, 3) and (4, − 2) (iii) (1, 2) and (7, 8).

(ii) (0, − 4) and (− 6, 2) [UPTU 2001 (Special)]

16. Show that the points (5, 1), (1, − 1) and (11, 4) are collinear. Also find the equation of the straight line on which these points lie. 17. Find the equation of the straight line which passes through the points (3, 3) and (7, 6) and find the length of the intercept cut off by the axes. 18. The vertices of a quadrilateral are A (−2 , 6) ; B (1, 2) ; C (10, 4) and D (7, 8). Find the equations of its diagonals. 19. Find the equations of the sides of a triangle whose vertices are the points (0, 0), (2 , 4) and (6, 4). ) (− 5, 7) and (− 5, − 5). 20. The mid-points of the sides of a triangle are (2 , 1, Find the equations of the sides.

A nswers 1. (i) y = − 5

(ii) x = 3

(iii) y = − 5

(iv) x = − 4 2. (i) y = 2 x + 3

(v) x = 2 (vi) y = 4. (ii) x + 3 y + 12 = 0.

244

3.

x + y + 4 = 0.

5.

x − y −1= 0

6.

(i) 3 x − 2 y = 6

7.

3 x − 4 y = 12 .

9.

x − y = 8.

4. or

y = 3 x − 4.

x + y + 1 = 0. (ii) 8.

− 6 x + 5 y = 30. x + y = − 1.

10. 4 x + 3 y = 12 .

11. 9 x − 20 y + 96 = 0.

12. 6 x + 11 y − 66 = 0.

13. (i) 4 x − y − 27 = 0

(ii)

3 x + y + 3 = 0.

(ii)

x+ y+4=0

14.

y + 4 = 0.

15. (i) x + y − 2 = 0

18.

(iii) x − y + 1 = 0. 5 17. 3 x − 4 y + 3 = 0 ; ⋅ x − 2 y − 3 = 0. 4 x + 6 y − 34 = 0 ; x − y + 1 = 0.

19.

y = 4, 2 x − 3 y = 0, 2 x − y = 0.

20.

x = 2 , 7 y − 6 x − 79 = 0, 7 y + 6 x + 65 = 0.

16.

7.6 Reduction of General Form to Standard Forms Let the general equation of the line be ax + by + c = 0. (i) Slope-intercept form. The equation (1) can be written as

…(1)

a c x − , if b ≠ 0. b b This is the equation of a line in slope-intercept form. a c Here, m = slope = − and y-intercept = − ⋅ b b (ii) Intercept form. The equation (1) can be written as by ax or ax + by = − c + = 1, −c −c y x if c ≠ 0 or + = 1. c c − − a b This is the equation of a line in intercept form. c c Here, x-intercept = − and y-intercept = − ⋅ a b (iii) Normal form. First, write the equation (1) so that the constant term c is negative ( if necessary, by = − ax − c

or

y=−

then multiply by –1). Dividing (1) by a2 + b 2 , we get a a2 + b 2

x+

b a2 + b 2

y=

−c a2 + b 2

,

…(2)

245

where the term on the R.H.S. is positive. Since,

 a   a2 + b 2 

   

2

 b +  a2 + b 2 

   

2

=

a2 + b 2 a2 + b 2

= 1,

we can find an angle α such that a cos α = 2 a +b2 and

b

sin α =

2

a +b2

⋅

Putting these values in (2), we get x cos α + y sin α = −

c 2

a + b2

⋅

This is the equation of a line in normal form. a b Here, cos α = , sin α = , p= − a2 + b 2 a2 + b 2

Illustration 1 :

c a2 + b 2

⋅

Reduce the equation 3 x + y + 2 = 0 to

(i) slope-intercept form and find its slope and y-intercept; (ii) intercept form and find intercepts on the axes; (iii) normal form and find p and α . Solution :

The given equation of the line is …(1)

3 x + y + 2 = 0. (i) The equation (1) can be written as y = − 3 x − 2 , which is the slope-intercept form. Here, slope = − 3 and y-intercept = − 2 . (ii) The equation (1) can be written as 3x + y = − 2 or −

x 2

+

y −2

y 3x + =1 −2 −2

or

= 1, which is the intercept form.

3 Here, x-intercept = −

2

and y-intercept = − 2 .

3 (iii) The equation (1) can be written as − 3x − y − 2 = 0

or

− 3 x − y = 2.

246

Dividing both sides by (− 3)2 + (−1)2 = 2 , we get 3 1 x− y = 1, which is the normal form. 2 2 3 1 Here, cos α = − , sin α = − and p = 1. 2 2 −

Since sin α and cos α both are negative, therefore, α is in the third quadrant. But α 5π lies between − π and π so we take α = − ⋅ 6 5π Hence, α = − and p = 1. 6 Illustration 2 :

Find the equation of a line at a distance of 3 units from the origin such that 3 the perpendicular from the origin to the line makes an angle tan −1   with the positive  4 direction of x-axis. Solution : ∴ ⇒

[UPTU 2006]

3 We have p = 3 and α = tan − 1   ⋅  4 3 tan α = 4 4 3 and sin α = ⋅ cos α = 5 5

Hence, the equation of the line in normal form is or or

x cos α + y sin α = p 4 3 x× + y× =3 5 5 4 x + 3 y = 15.

Illustration 3 :

Reduce 4 x + 3 y − 9 = 0 to the normal form and find the distance

(perpendicular distance p) from origin. Solution :

We have 4 x + 3 y − 9 = 0

[UPTU 2004]

or

4 x + 3 y = 9.

Dividing both sides by (4)2 + (3)2 = 5, we get 4 3 9 x+ y = , which is the normal form . 5 5 5 Hence, the length of perpendicular from the origin to the line is p =

9 ⋅ 5

Illustration 4 : Find the angle made by the line x cos 30 ° + y sin 30 ° + sin 120 ° = 0 with the positive direction of x-axis.

[UPTU 2007]

Solution : or

The equation of the given line is x cos 30 ° + y sin 30 ° + sin 120 ° = 0 cos 30 ° x sin 120 ° y=− − sin 30 ° sin 30 °

...(1)

247

3 /2 1/ 2

or

y = − cot 30 ° x −

or

y = tan 120 ° x − 3, which is the slope-intercept form .

Hence, the angle made by the given line with the positive direction of x-axis is 120°.

Comprehensive Exercise 2 1. Reduce the equation 3 x + y − 8 = 0 to : (i) slope-intercept form and find slope and y-intercept; (ii) intrcept form and find intercepts on the axes; (iii) the normal form and find p and α . 2. Find the angle made by the line x + 3 y − 6 = 0 with the positive direction of x-axis. 3. Find the intercepts cut off by the line 2 x − y + 16 = 0 on the coordinate axes. 4. Find the intercepts of the line x sin α + y cos α = sin 2 α on the coordinate axes and obtain the mid-point of the line-segment intercepted between the axes. 5. Find the length of perpendicular from origin to the line x + y = 2 . Also, find the inclination of this perpendicular with the x-axis. 6. Which of the lines 2 x − y + 3 = 0 and x − 4 y − 7 = 0 is farther from the origin?

A nswers 1. (i) y = − 3 x + 8 ; slope = − 3 and y-intercept = 8 y 8 x (ii) and y-intercept = 8 + = 1 ; x-intercept = 8 8/ 3 3 3 1 π and p = 4. x+ y = 4 ;α = 2 2 6 2. 150°. 3. x-intercept = − 8, y-intercept = 16. (iii)

4. 2 cos α, 2 sin α ; (cos α, sin α). 6.

5. p = 1, α = 45° .

x − 4 y − 7 = 0.

7.7 Angle between two Lines. Prove that the angle θ between two lines whose slopes are m1 and m2 is given by tan θ =

m1 − m2 1 + m1 m2

⋅

248

Let y = m1 x + c1 and y = m2 x + c 2 be the equations of two given lines AB and CD respectively which intersect at the point K . Then, we have to find the ∠ BKD . Let ∠ BKD = θ. If the lines AB and CD make angles θ1 and θ2 with the x-axis, then tan θ1 = m1 and tan θ2 = m2 . It is clear from the figure, that ∠ θ = ∠ θ1 − ∠ θ2 ∴

i. e.,

tan θ = tan (θ1 − θ2 ) tan θ1 − tan θ2 m − m2 = = 1 1 + tan θ1 tan θ2 1 + m1 m2 the angle between the lines y = m1 x + c 1 and y = m2 x + c 2 is given by θ = tan −1

m1 − m2 1 + m1 m2

⋅

7.8 Condition for two given Lines to be Parallel Two lines y = m1 x + c1 and y = m2 x + c 2 are parallel if the angle θ between them is 0° or 180° i. e., if tan θ = 0, i. e., if m1 − m2 = 0 i.e., if m1 = m2 , i. e. , their slopes are equal. Thus, the lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c 2 = 0 are parallel if a a a1 b − 1 = − 2 , i. e., = 1 ⋅ b1 b2 a2 b2 i. e.,

coefficients of x and y in their equations are proportional.

7.9 Condition for two given Lines to be Perpendicular Two lines y = m1 x + c1 and y = m2 x + c 2 are perpendicular if the angle θ between them is 90° i. e.,

if tan θ = tan 90 ° = ∞,

i. e.,

if m1 m2 = − 1

i. e.,

i. e.,

if 1 + m1 m2 = 0 (UPTU 2011)

the product of their slopes is equal to –1.

If the slope of a line is m, then the slope of any line perpendicular to it is −

1 ⋅ m

The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c 2 = 0 are perpendicular if

249

 a1   a2  −  ⋅ −  = −1  b1   b2  or i. e.,

a1 a2 + b1 b2 = 0 product of the coefficients of x + product of the coefficients of y = 0.

7.10 Parallel Lines The equation of any straight line parallel to a given line a x + by + c = 0 is a x + by + λ = 0. Rule. In order to write the equation of any line parallel to a given line, we do not change the terms of x and y but we only change the constant term. The changed constant will be found by an additional given condition.

7.11 Perpendicular Lines The equation of any straight line perpendicular to a given line a x + by + c = 0 is b x − ay + λ = 0. Rule. In order to write the equation of any line perpendicular to a given line, we interchange the coefficients of x and y in the given equation and change the sign of one of these coefficients. Also we change the constant term. The value of the new constant is to be found by an additional given condition.

Illustration 1 :

Find the angle between the lines

(i) x − y 3 − 5 = 0 and 3 x + y − 7 = 0 ;

[UPTU 2001 (Special)]

(ii) y = (2 − 3) x + 5 and y = (2 + 3) x − 7. Solution :

(i) The equations of two lines are …(1)

x− y 3 −5=0 and

…(2)

3 x + y − 7 = 0.

Here,

1 m1 = slope of the line (1) = − = − 3 3

and

m2 = slope of the line (2) = −

Clearly,

1

m1 m2 = − 1.

Hence, the two lines are at right angle.

3 = − 3. 1

250

(ii) The equations of two lines are y = (2 − 3) x + 5

…(1)

and

y = (2 + 3) x − 7.

…(2)

Here,

m1 = slope of the line (1) = 2 − 3

and

m2 = slope of the line (2) = 2 + 3.

Let θ be the acute angle between the lines. Then  (2 − 3) − (2 + 3)  = 3. tan θ =  1 + (2 − 3)(2 + 3) ∴

θ = 60 ° .

Hence, the acute angle between the lines is 60°. Illustration 2 :

Show that the line passing through the points (3, − 4) and (− 2 , 6 ) is

parallel to the line passing through (− 3, 6 ) and (9, − 18). Solution : Here, m1 = slope of the line joining the points (3, − 4) and (− 2 , 6 ) 6 − (− 4) 10 = = =−2 −2−3 −5 and

m2 = slope of the line joining the points (−3, 6) and (9, − 18) − 18 − 6 − 24 = = = −2. 9 − (− 3) 12

Since

m1 = m2 , therefore, the two lines are parallel.

Illustration 3 :

The line joining (− 5, 7 ) and (0, − 2 ) is perpendicular to the line joining

(1, 3) and (4, x), then find x. Solution :

Here, m1 = slope of line joining the points (− 5, 7) and (0, − 2) =

and

[UPTU 2003]

−2−7 0 − (− 5)

=−

9 , 5

m2 = slope of the line joining the points (1, 3) and (4, x). x−3 x−3 = = ⋅ 4 −1 3

If the given two lines are perpendicular, then

⇒

m1 m2  − 9    5

= −1  x − 3   =−1  3 

⇒

− 9 ( x − 3) = − 15

⇒

− 9 x + 27 = − 15 14 x= ⋅ 3

⇒

Illustration 4 :

⇒

− 9 x = − 42

Find the equation of the straight line which passes through (−1, 2) and is

parallel to the line 3 x − 4 y + 5 = 0.

251

Solution :

The equation of any straight line parallel to the line 3 x − 4 y + 5 = 0 is …(1)

3 x − 4 y + λ = 0. If the line (1) passes through the point (−1, 2), then or

−3−8+λ =0

λ = 11.

Putting λ = 11 in (1), the required equation of the line is 3 x − 4 y + 11 = 0. Illustration 5 :

Find the equation of the straight line which passes through (1, 2) and is

perpendicular to the line 4 x − 3 y = 8.

[UPTU 2001]

Solution : The equation of any straight line perpendicular to the line 4 x − 3 y − 8 = 0 is …(1)

3 x + 4 y + λ = 0. If the line (1) passes through the point (1, 2), then 3+8+λ =0

or

λ = − 11.

Putting λ = − 11 in (1), the required equation of the line is 3 x + 4 y − 11 = 0. Illustration 6 : Prove that the equation to the straight line passing through the point (a cos3 θ, a sin3 θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ − y sin θ = a cos 2 θ. Solution :

The slope of the given line x sec θ + y cosec θ = a is −

∴

[UPTU 2007]

sec θ cosec θ

, i. e., −

sin θ cos θ

⋅

the slope of a line perpendicular to the given line =

cos θ sin θ

⋅

Now the equation of the straight line which passes through the point cos θ is (a cos 3 θ, a sin3 θ) and whose slope is sin θ cos θ y − a sin3 θ = ( x − a cos 3 θ) sin θ or

x cos θ − y sin θ = a (cos 4 θ − sin4 θ)

or

x cos θ − y sin θ = a (cos 2 θ + sin2 θ)(cos 2 θ − sin2 θ)

or

x cos θ − y sin θ = a cos 2 θ.

Illustration 7 : Find the equation of the line which is perpendicular to the line 3 x − 2 y + 4 = 0 and passes through the point of intersection of the lines x + 2 y + 1 = 0 and [UPTU 2009] y = x + 7. Solution :

The point of intersection of the lines x + 2 y + 1 = 0 and y = x + 7 is

(− 5, 2). The equation of any straight line perpendicular to the line 3 x − 2 y + 4 = 0 is 2x + 3 y + λ = 0

...(1)

252

If the line (1) passes through the point (− 5, 2), then 2(− 5) + 3(2) + λ = 0

or

λ = 4.

Putting λ = 4 in (1), the required equation of the line is 2 x + 3 y + 4 = 0. Illustration 8 : Find the equation of perpendicular bisector of the line segment joining the points A (2, 3) and B (6, − 5). [UPTU 2004] Solution :

The slope of the line joining the points A (2, 3) and B (6, − 5) is =

∴

−5−3 6−2

=−

8 = − 2. 4

slope of a line perpendicular to the line AB =

1 ⋅ 2

The coordinates of the middle point M of AB are  2 + 6 3 + (− 5) , i. e., (4, − 1).    2  2 Hence, the equation of the perpendicular bisector of AB, i. e., the equation of the line passing through M and perpendicular to AB is 1 or y + 1 = ( x − 4) x − 2 y = 6. 2

Comprehensive Exercise 3 1. Find the angle between the lines whose equations are : (i) y − 3 x − 5 = 0 and 3 y − x + 6 = 0; (ii) x − 2 y + 3 = 0 and 3 x + y − 1 = 0; (iii) 2 x − y + 3 = 0 and x + y + 2 = 0; y y x x (iv) + = 1 and − = 1. a b b a 2. Show that (i) the line joining (2 , − 3) and (− 5, 1) is parallel to the line joining (7, − 1) and (0, 3) ; (ii) the line through (2 , − 5) and (−2 , 5) is perpendicular to the line through (6, 3) and (1, 1). 3. What is the value of y so that the line through (3, y) and (2 , 7) is parallel to the line through (−1, 4) and (0, 6) ? [UPTU 2006] 4. Without using Pythagoras theorem, show that the points A (0, 4), B (1, 2) and C (3, 3) are the vertices of a right angled triangle. 5. Show that the points A (2 , − 1), B (0, 2), C (2 , 3) and D (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

253

6. Show that A (4, 3), B (6, 4), C (5, 6) and D (3, 5) are the angular points of a square. 7. Find the angles of a triangle whose vertices are A (3, 4), B (4, 4) and C (4, 5). 8. Find the equation of the perpendicular bisector of the line joining the points A (2 , 3) and B (6, − 5). 9. Find the equation of a straight line passing through the origin and parallel to the line 3 x − 2 y + 1 = 0. 10. Find the equation of a line passing through (1, 1) and perpendicular to the straight line 3 x − 4 y = 6. [UPTU 2001 (Special)] 11. Find the equation of the line passing through (−2 , − 4) and perpendicular to the line 3 x − y + 5 = 0. [UPTU 2007] 12. Find the equation of a line passing through (7, 1) and parallel to the line joining the points (2 , − 2) and (4, 8). 13. Let A (1, 1), B (2 , − 1) and C (−1, 3) be the vertices of a ∆ ABC . Find the equations of the perpendiculars drawn from vertices to the opposite sides. 14. Find the equation of a line which is perpendicular to the line 3 x − y + 5 = 0 and cuts off – 4 units from the y-axis. 15. Find the equation of a line perpendicular to the line 3 x − y + 5 = 0 and at a distance of 3 units from the origin.

A nswers 1. (i) 30° (iii) tan −1 (3)

(ii)

tan −1 (7)

(iv) 90°.

3. 9.

5. tan −1 2 .

7.

8.

45°, 90°, 45°.

9. 3 x − 2 y = 0. 11.

x + 3 y + 14 = 0.

x − 2 y − 6 = 0.

10. 4 x + 3 y − 7 = 0. 12. 5 x − y − 34 = 0.

13. 3 x − 4 y + 1 = 0, x − y − 3 = 0, x − 2 y + 7 = 0. 14.

x + 3 y + 4 3 = 0.

15.

x + 3 y ± 6 = 0.

7.12 The Length of Perpendicular The length of the perpendicular drawn from the point ( x 1 , y 1 ) to the line x cos α + y sin α = p is x 1 cos α + y 1 sin α − p. The length of the perpendicular drawn from the point ( x 1 , y 1 ) to the line ax 1 + by 1 + c ax + by + c = 0 is ⋅ a2 + b 2

254

In particular, the length of perpendicular from origin (0, 0) to the line c ax + by + c = 0 is ⋅ 2 a + b2 Rule. Bring all the terms in the equation of the line to the L.H.S. and make the R.H.S. 0, substitute for x and y the coordinates of the point ( x1 , y1 ) and divide by the square root of the sum of the squares of the coefficients of x and y. Note. If the length of perpendicular obtained from the above formula is negative, we can ignore the sign and give the positive value in answer, unless there is some special reason.

7.13 Distance between two Parallel Lines First method. Choose a convenient point on any of the lines (put x = 0 and find the value of y or put y = 0 and find the value of x). Now the perpendicular of this point from the other line will give the required distance between the given parallel lines. Second method. Write the equations of the two parallel lines in such a way that coefficients of x are of the same sign. Then the two lines will lie on the same side of the origin if c1 and c 2 are of the same sign and on opposite sides of the origin if c1 and c 2 are of opposite signs. Find the perpendicular distances p1 and p2 of both the lines from the origin and retain their signs. The required distance between the parallel lines is | p1 − p2|.

Illustration 1 : If p is the length of perpendicular from origin on the line whose intercepts on the axes are a and b, show that 1 1 1 = 2 + 2 ⋅ 2 p a b Solution : axes is

The equation of the line making intercepts a and b on the coordinate y x + =1 a b

or

y x + −1= 0 a b

…(1)

If pbe the length of the perpendicular drawn from origin (0, 0) to the line (1), then −1 p= 2 2  1 +  1      a  b or

1 p2

=

1 a2

+

1 b2

⋅

255

Illustration 2 :

Find the equations of two lines through (0, a) which are at a distance a from

the point (2 a, 2 a). Solution :

Equation of any line through (0, a) is y − a = m ( x − 0)

or

…(1)

mx − y + a = 0.

If the length of perpendicular from (2 a, 2 a) to the line (1) is a, then m (2 a) − 2 a + a a =    2 m + 1   or

m2 + 1.

2m − 1 = ±

Squaring, we get (2m − 1)2 = m2 + 1 3m2 − 4m = 0 ⇒ m = 0,

or

4 ⋅ 3

Putting these values of m in (1), the required equations are y−a=0 Illustration 3 :

and

4 x − 3 y + 3a = 0.

Find the distance between the parallel lines 3 x + 4 y = 12 and

3 x + 4 y = 3. Solution :

[UPTU 2004]

The given lines are 3 x + 4 y = 12

and

3 x + 4 y = 3.

...(1) ...(2)

Putting x = 0 in (1), we get y = 3. Thus (0, 3) is a point on the line (1). The perpendicular distance between the lines (1) and (2) is = the length of perpendicular from the point (0, 3) to the line (2) 3⋅0 + 4⋅3 − 3 9 = = ⋅ 5 9 + 16 Illustration 4 :

Show that the lines 2 x + 3 y = 19 and 2 x + 3 y = − 7 are equidistant

from the line 2 x + 3 y = 6. Solution :

All the three lines are parallel. Putting x = 0 in the equation

2 x + 3 y = 6 of the third line, we get y = 2 . Thus, (0, 2) is a point on the third line. The perpendicular distances of this point from the first and second lines are 0 + 6 − 19 = − 13 13 0 +6+7 and = 13. 13 These two perpendiculars are equal in magnitude and opposite in sign. Hence, the first two lines are equidistant from the third line on opposite sides of it. Illustration 5 : 3x − 4 y = 8 ?

Are the points (3, 4) and (2 , − 6) on the same or opposite sides of the line

256

Solution :

The equation of the given line is …(1)

3 x − 4 y − 8 = 0. Putting x = 3, y = 4 on the L.H.S. of (1), we get 3 . (3) − 4 . (4) − 8 = 9 − 16 − 8 = − 15, which is negative. Again putting x = 2 , y = − 6 on the L.H.S. of (1), we get 3 . (2) − 4 . (−6) − 8 = 6 + 24 − 8 = 22 , which is positive.

Since the values of the L.H.S. of the equation of the line (1) are of the opposite signs when the coordinates of the points (3, 4) and (2 , − 6) are substituted in it, therefore, these points lie on the opposite sides of the line (1).

Comprehensive Exercise 4 1. Find the distance of the point (4, 5) from the line 3 x − 5 y + 7 = 0. 2. Find the length of perpendicular drawn from (2 , − 1) to the line 4 ( x − 3) = 3 ( y + 4). 3. Find the length of perpendicular drawn from the origin to the line a ( x + a) + b ( y + b) = 0. 4. If the length of the perpendicular form the point (1, 1) to the line 1 1 1 c ax − by + c = 0 be unity, show that + − = ⋅ c a b 2 ab 5. If p and p′ be the perpendiculars from the origin to the straight lines x sec θ − y cosec θ = a and x cos θ − y sin θ = a cos 2 θ, prove that 4 p 2 + p′ 2 = a2 . 6. Find the distance between the parallel lines : (i) 3 x − 4 y = 8 and 3 x − 4 y + 12 = 0 ; (ii) 8 x + 15 y + 20 = 0 and 8 x + 15 y = 14 ; (iii) 2 x − 3 y + 9 = 0 and 4 x − 6 y + 1 = 0.

[UPTU 2005]

7. Are the points (2 , 5) and (−3, 7) on the same or opposite sides of the line 2 x −3y =4? 8. Show that the origin and the point (−2 , 5) lie on the opposite sides of the line 4 x − 3 y + 9 = 0.

A nswers 1.

6 ⋅ 34

6. (i) 4

2.

13 ⋅ 5

(ii)

2

3. (iii) −

a2 + b 2 . 17 ⋅ 2 13

7. same side.

257

7.14 Equations of the Bisectors of the Angles between two given Lines The equations of the bisectors of the angles between the lines a1 x + b1 y + c1 = 0 and are

a2 x + b2 y + c 2 = 0 a1 x + b1 y + c1 a x + b2 y + c 2 =± 2 ⋅ a12 + b12 a2 2 + b2 2

Particular Cases : The equations of the lines should be so written that c1 and c 2 are either both positive or both negative. (i) Equation of the bisector of the angle in which the origin lies is a1 x + b1 y + c1 a x + b2 y + c 2 =+ 2 a12 + b12 a2 2 + b2 2 and the equation of the bisector of the angle which does not contain the origin is a1 x + b1 y + c1 a x + b2 y + c 2 =− 2 ⋅ a12 + b12 a2 2 + b2 2 (ii) Bisectors of acute and obtuse angles. If θ be the angle between one of the given lines and one bisector, then if the absolute value of tan θ, i. e.,|tan θ| < 1, then it is the bisector of acute angle and if |tan θ| > 1, then it is the bisector of obtuse angle. (iii) If the constant terms c1 and c 2 in the two equations a1 x + b1 y + c1 = 0 and

a2 x + b2 y + c 2 = 0

are of the same sign, then the angle between these two lines in which the origin lies is acute if a 1 a 2 + b 1 b 2 < 0 and obtuse if a 1 a 2 + b 1 b 2 > 0.

Illustration 1 : Find the equations of the bisectors of the angles between the lines 3 x − 4 y + 7 = 0 and 12 x − 5 y − 8 = 0. Solution : The equations of the bisectors of the angles between the lines 3 x − 4 y + 7 = 0 and 12 x − 5 y − 8 = 0 are

258

3x − 4 y + 7 2

12 x − 5 y − 8

=±

2

122 + (−5 2 )

3 + (− 4 ) or or

3x − 4 y + 7 5

=±

12 x − 5 y − 8 13

39 x − 52 y + 91 = ± (60 x − 25 y − 40).

Taking positive sign, we get one bisector as 21x + 27 y − 131 = 0. Taking negative sign, we get the other bisector as 99 x − 77 y + 51 = 0. Illustration 2 :

Find the equation of the bisector of the acute angle between the lines

3 x − 4 y + 7 = 0 and 12 x + 5 y − 2 = 0. Solution :

Making the constant terms of the same sign, the equations of the

given lines are

and

3x − 4 y + 7 = 0

…(1)

−12 x − 5 y + 2 = 0.

…(2)

The equations of the bisectors of the angles between the lines (1) and (2) are 3x − 4 y + 7 32 + 42

=±

−12 x − 5 y + 2 122 + 5 2

…(3)

i. e.,

13 (3 x − 4 y + 7) = ± 5 (−12 x − 5 y + 2)

i. e.,

11 x − 3 y + 9 = 0

…(4)

and

21 x + 77 y − 101 = 0.

…(5)

Let θ be the acute angle between the line (1) and the bisector (4). Then  11 − 3  35 tan θ =  3 4  = 1 5 − 2 ( 5 − 2)( 5 + 2)

= i. e.,

θ > 45° or 2 θ > 90 ° .

∴ the line (4) bisects the obtuse angle between the given lines.

Comprehensive Exercise 5 1. For the lines 4 x + 3 y − 6 = 0 and 5 x + 12 y + 9 = 0, find the equation of the bisector of the angle which contains the origin. 2. Find the equation of the obtuse angle bisector of the lines 12 x − 5 y + 7 = 0 and 3 y − 4 x − 1 = 0. 3. Find the bisector of the acute angle between the lines 3 x + 4 y − 11 = 0 and 12 x − 5 y − 2 = 0.

A nswers 1. 7 x + 9 y − 3 = 0. 2. 4 x + 7 y + 11 = 0. 3. 11 x + 3 y − 17 = 0.

260

Comprehensive Exercise 6 Fill in the Blanks Fill in the blanks ‘‘..........’’, so that the following statements are complete and correct. 1. The slope of the line joining the points (2, − 5) and (4, 1) is .......... . [UPTU 2009]

2. The slope of the line y =

1 x 3 + 3 is .......... . 3

3. The equation of the line passing through the points (4, 2) and (− 2, 8) is .......... . 4. The equation of the line with slope 4 and passing through (4, 3) is .......... . 5. The line passing through the point (2, 1) and parallel to the join of the points (1, 3) and (− 3, 1) is .......... . 6. The angle θ between two lines whose slopes are m1 and m2 is given by .......... . 7. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c 2 = 0 are parallel if ..... . 8. The lines a1 x + b1 y + c1 = 0 and a2 x + b2 y + c 2 = 0 are perpendicular if ........... .

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 9. The slope of the line passing through the points (3, 4) and (− 3, 6) is 1 3

(b)

2 (c) 3

(d)

(a)

−1 3 −2 3

10. The equation of a straight line which cuts off intercepts of lengths a and b from x-axis and y-axis respectively is y y x x (a) (b) + =1 + =1 a b b a (c) x + y = a

(d) x + y = b

261

11. The equation of the line passing through the points (1, 1) and (5, 4) is (a)

y = 3x + 1

(c) 4 y = 3 x − 1

(b) 4 y = 3 x + 1 (d) 4 y = 4 x − 3

12. The equation of the straight line having slope m and passing through the point ( x1 , y1 ) is (a)

y1 − y = m ( x − x1 )

(c) x − x1 = m ( y − y1 )

(b) y − y1 = m ( x − x1 ) (d) none of these

13. The equation of a line passing through (2, − 3) and inclined at an angle of 135° with the positive direction of x-axis is (a) x + y + 1 = 0

(b) x + y − 1 = 0

(c) x + y = 1 / 2

(c) x + y = 0

14. Condition for two given lines to be parallel is m 1 (a) m1 m2 = 1 (b) 1 = m2 2 (c) m1 = m2

(d) m1 m2 = − 1

15. Condition for two given lines to be parpendicular is (a) m1 = m2

(b) m1 = − m2

(c) m1 m2 = 1

(d) m1 m2 = − 1

[UPTU 2011]

True or False Write ‘T’ for true and ‘F’ for false statement 16. The equation of a straight line upon which the length of the perpendicular from the origin is p and this perpendicular makes an angle α with positive direction of x-axis is p = x cos α + y sin α 17. The slope of the line joining the two points ( x1 , y1 ) and ( x2 , y2 ) is y2 − y1 x2 − x1

⋅

18. The length of the intercept on x-axis cut by the line 4 x + 3 y = 20 is 4. 19. The length of the intercept of the line 2 x + 3 y = 4 from x-axis is 3. 20. The angle between the lines joining the points (0, 0) and (2, 3) and the 11 points (2, − 2) and (3, 5) is θ = tan −1 ⋅ 23 21. The equation of the line cutting equal intercepts on the axes and passing through the point (6, 9) is x + y = 6.

262

22. The equation of any straight line perpendicular to a given line ax + by + c = 0 is bx + ay + λ = 0. 23. The equation of any straight line parallel to a given line ax + by + c = 0 is ax + by + λ = 0. 24. The length of the perpendicular drawn from the point ( x1 , y1 ) to the line x cos α + y sin α = p is x1 cos α − y1 sin α − p.

A nswers 1. 3 5.

x −2y =0

2. 30°

3. x + y − 6 = 0 4. y = 4 x − 3 m1 − m2 a b 6. tan θ = 7. 1 = 1 1 + m1 m2 a2 b2

8. a1 a2 + b1 b2 = 0 9. (b) 12. (b) 17. T 22. F

13. (a) 18. F 23. T

10. (a)

11. (b)

14. (c) 19. F 24. F

15. (d) 20. T

16. T 21. F

263

8 F unctions a nd L imits

8.1 Function

A

constant is a quantity which, during any set of mathematical operations, retains the same value.

A variable is a quantity, or a symbol representing a number, which is capable of assuming different values. A continuous variable is a variable that can take all the numerical values between two given numbers. An independent variable is one which may take up any arbitrary value that may be assigned to it. A dependent variable is a variable which can assume its value as a result of some other variable taking some assigned value. A set is any well-defined class or collection of objects. The objects constituting the set are also called elements or members of the set.

264

Set Theoretic Definition Of A Function. Let A and B be any two given non-empty sets. Suppose there exists a correspondence denoted by f, which associates to each member of A a unique member of B. Then f is called a function or a mapping from A to B. The mapping f of A to B is denoted by f : A → B. The set A is called the domain of the function f , and B is called the co-domain of f . The element y ∈ B which the mapping f associates to an element x ∈ A is symbolically denoted by y = f ( x) and is called the f-image of x or the value of the function f at x or y is a function of x or y equals f of x. Each element of A has a unique image and each element of B need not appear as the image of an element in A. We define the range of f to consist of those elements in B which appear as the image of at least one element in A. It is denoted by f ( A). This symbolic representation was invented by Leonhard Euler. Thus, range f = f ( A) = { f ( x) : x ∈ A}. The range of a mapping f : A → B is always a subset of the co-domain B of f i. e., f ( A) ⊆ B. How to find the domain of a function ? The domain of a function f ( x) is the set of those real numbers x for which f ( x) is defined i. e., for which f ( x) is a unique real number. Sometimes a function is described only by a formula and the domain of the function is not explicitly stated. In such cases we consider the set of all real numbers x for which f ( x) is meaningful i. e., for which f ( x) is a unique real number, as the domain of f . How to find the range of a function ? The range of a function f is the set of all values taken by f ( x) when x takes values in the domain of f . Thus, range f = { y ∈ R : y = f ( x) for some x ∈ domain f }. So to find the range of a function f ( x) we put y = f ( x) and try to solve this equation to get x in terms of y in the form x = φ ( y). Then we find those real numbers y for which φ ( y) is in the domain of f . The set of these real numbers y constitute the range of f .

Illustration 1 : Solution : Now

State the domain of the function f ( x) =

x2 + 3 x + 1

⋅ x 2 − 5 x + 6 [UPTU 2007]

Obviously f ( x) is not defined when x 2 − 5 x + 6 = 0. x 2 − 5 x + 6 = 0 ⇒ ( x − 2)( x − 3) = 0 ⇒ x = 2 or x = 3.

So f ( x) is not defined at x = 2, 3. Hence, domain f = R − {2, 3}. Illustration 2 :

Find the domain of the function f ( x) =

x 2 − 9.

265

Solution :

Obviously f ( x) is a real number if and only if x 2 − 9 ≥ 0.So the domain

of f ( x) is the set of those real numbers x for which x 2 − 9 ≥ 0. Now

x 2 − 9 ≥ 0 ⇒ x 2 ≥ 9 ⇒ | x|2 ≥ 9

⇒

| x| ≥ 3 ⇒ x ≥ 3 or

x ≤ − 3 ⇒ x ∈ (− ∞, − 3] ∪ [3, ∞).

Hence, domain f = (− ∞, − 3] ∪ [3, ∞). Illustration 3 : Solution :

Find the domain and range of the function f ( x) =

1+ x 1− x

⋅

Obviously f ( x) is not defined when 1 − x = 0, i. e., when x = 1 and at

every other real number x, f ( x) is a unique real number. So domain f = R − {1} = (− ∞, 1) ∪ (1, ∞). Now to find range f ( x), we put y = f ( x). 1+ x Then y= ⋅ ⇒ y (1 − x) = 1 + x 1− x x=

⇒

y −1 y +1

⇒

x ( y + 1) = y − 1

, provided y + 1 ≠ 0.

Obviously when y + 1 = 0 i. e., when y = − 1, x is not a real number and so it is not in the domain of f . For all real numbers y except y = − 1, x is in the domain of f . Hence, range f = R − {−1. } Illustration 4 : Solution :

If f ( x) = x 2 + 3 x + 2,find that value of x for which f ( x + 1) = f ( x).

We have f ( x) = x 2 + 3 x + 2.

∴

f ( x + 1) = ( x + 1)2 + 3 ( x + 1) + 2 = x 2 + 5 x + 6.

Now

f ( x + 1) = f ( x) ⇒ x 2 + 5 x + 6 = x 2 + 3 x + 2

2 x = − 4 ⇒ x = − 2. 1− x θ Illustration 5 : If f ( x) = , show that f (cos θ) = tan2 ⋅ 1+ x 2 ⇒

Solution :

We have f ( x) =

1− x 1+ x

⋅

1 θ θ 2 ∴ f (cos θ) = = = tan2 ⋅ 1 + cos θ 2 cos 2 1 θ 2 2  2x  1 + x   = 2 f ( x). Illustration 6 : If f ( x) = log   , show that f  2 1 − x  1 + x  1 − cos θ

Solution :

1 + We have f ( x) = log  1 −

2 sin2

x ⋅ x

266

∴

1 + 2 x   2x  1 + x2  = log  f  2 2x 1 + x  1 −  1 + x2 = log

Illustration 7 :

(1 + x)2 2

(1 − x)

1 + x  = log   1 − x 

  1 + x2 + 2 x  = log 1 + x2 − 2 x   2

1 + x  = 2 log   = 2 f ( x). 1 − x 

1 + x   x + y If f ( x) = log   show that f ( x) + f ( y) = f  ⋅ 1 − x  1 + xy  [UPTU 2005]

Solution :

We have

1 + x  1 + y  Given f ( x) = log   and f ( y) = log  ⋅ 1 − x  1 − y    x + y 1 +   1 + xy    x + y f   = log  1 + xy   1 −  x + y    1 + xy  

      

1 + xy + x + y  1 + x + y + xy  = log    = log  1 + xy − x − y 1 − x − y + xy    and

...(1)

1 + x  1 + y  f ( x) + f ( y) = log   + log   1 − x  1 − y  1 + = log  1 −

x  x

1 + y   1 + x + y + xy     = log  ⋅ 1 − y   1 − x − y + xy 

...(2)

From (1) and (2), we get  x + y f ( x) + f ( y) = f  ⋅ 1 + xy  Illustration 8 : Find the domain of the following functions : 1 (i) f ( x) = (ii) f ( x) = ( x − 2)( x + 5). 9 − x2 Solution :

(i) We have f ( x) =

1 9 − x2

⋅

Obviously f ( x) is not defined for those real numbers x for which 9 − x 2 ≤ 0 i. e., x 2 ≥ 9. Thus, f ( x) is defined only when x 2 < 9. Now

x2 < 9

∴

Dom f = ] − 3, 3[.

⇒

| x|2 < 9

⇒

| x| < 3

⇒

− 3 < x < 3.

267

(ii) We have f ( x) = ( x − 2)( x + 5). Obviously f ( x) is defined only for those real numbers x for which ( x − 2)( x + 5) ≥ 0. Now

( x − 2)( x + 5) ≥ 0

⇒

x ≤ − 5 or

Hence,

dom f = ] − ∞, − 5] ∪ [2, ∞[.

⇒

[ x − (−5)]( x − 2) ≥ 0

x ≥ 2 ⇒ x ∈ ] − ∞, − 5] ∪ [2, ∞[.

Comprehensive Exercise 1 sin x

π  , find f  ⋅  2  1 + sin x

1.

If f ( x) =

2.

If f ( x) = ( x − a)2 ( x − b)2 , find f (a + b).

3.

If f ( x) = x 2 − 3 x + 4, find the values of x for which f ( x) = f (2 x + 1).

4.

If f ( x) = x +

5. 6.

1 , prove that [ f ( x)]3 = f ( x 3 ) + 3 x 1 1 If f ( x) = x 3 − 3 , find the value of f ( x) + f   x x 2x If f ( x) = , prove that f (tan θ) = sin 2θ. 1 + x2 1 − x2

1 f   ⋅  x   ⋅ 

7.

If f ( x) =

, prove that f (tan θ) = cos 2θ.

8.

 x   x + 1 If f ( x) = log   , show that f ( x) + f ( x + 1) = log  ⋅  x − 1  x − 1

9.

Find the domain and the range of the constant function f ( x) = 5.

1 + x2

10. For what value or values of x, if any, are the following functions not defined ? sin x 1 (i) (ii) f ( x) = sin f ( x) = x x 1 1 (iii) f ( x) = (iv) f ( x) = x−7 (1 − x)( x − 2) (v)

f ( x) =

1 x

3

+ x

(vi) f ( x) = 1 − x 2 .

11. Find the domain and range of the following functions : 1 x (i) f ( x) = 2 (ii) f ( x) = 1− x x −1 (iii) f ( x) =

x2 1+ x

2

(iv) f ( x) =

1 ⋅ 2 − cos 3 x

268

12. Find the domain and the range of the following functions. | x − 3| 3x − 5 (i) f ( x) = (ii) f ( x) = x−3 4 x−7 (iii) f ( x) = 3 sin (π 2 / 16) − x 2 .

A nswers 1. 9.

1 2. a2 b 2 . 3. ⋅ 2 Dom f = R , range f = {5}.

10. (i)

x = 0 (ii)

x=0

(iii)

−1, 2 / 3.

5.

0.

x≤7

(iv) x ≤ 1 or x ≥ 2 (v) x = 0 (vi) 11. (i) dom f = R − {−1, 1}, range f = R − ] − 1, 0]

x < − 1 or x > 1.

(ii) dom f = R − {1}, range f = R − {−1} (iii) dom f = R , range f = [0, 1[  1  (iv) dom f = R , range f =  , 1 ⋅  3  12. (i)

dom f = R − {3}, range f = {−1, 1}.

(ii) dom f = R − {7 / 4}, range f = R − {3  3 π π (iii) dom f = − ,  , range f = 0,  4 4  2 

/ 4}.  ⋅ 

8.2 Types of Functions Even and odd functions. A function f ( x) is said to be an even function, if f (− x) = f ( x) for all x ∈ domain f. A function f ( x) is said to be an odd function, if f (− x) = − f ( x) for all x ∈ domain f. In the above definitions it is implied that if x is in domain f , then − x is also in domain f . For example, f ( x) = cos x is an even function because f (− x) = cos (− x) = cos x = f ( x) for all x ∈ R . Similarly f ( x) = sin2 x and f ( x) = x 2 + 1 are also even functions. The function f ( x) = sin x is an odd function because f (− x) = sin (− x) = − sin x for all x ∈ R . Similarly f ( x) = x 3 is an odd function. Bounded functions.

A function f ( x) defined on a domain D is said to be bounded on

D, if there exists a positive real number k such that | f ( x)| ≤ k for all x ∈ D. For example, the function f ( x) = sin x is bounded on R because |sin x| ≤ 1 for all x ∈ R .

269

‘Increasing’ or ‘decreasing’ functions. A function f ( x) is said to be an increasing function on a domain D, if x1 > x2 ⇒ f ( x1 ) ≥ f ( x2 ) for all x1 , x2 ∈ D. Similarly, a function f ( x) is said to be a decreasing function on a domain D if x1 > x2 ⇒ f ( x1 ) ≤ f ( x2 ) for all x1 , x2 ∈ D.

8.3 Composite Functions Let f and g be two functions with domains D1 and D2 respectively. If range f ⊆ domain g, then the composite mapping g o f of the mappings f and g is defined by the formula ( g o f ) ( x) = g ( f ( x)) for all x ∈ D1 . Also, if range g ⊆ domain f , the composite mapping f o g of the mappings f and g is defined by the formula ( f o g) ( x) = f ( g ( x)) for all x ∈ D2 .

Illustration 1 : Prove that Solution :

Let f ( x) = sin x, g ( x) = 2 x and h ( x) = cos x.

fog = go( f h). Obviously dom ( fog) = dom g = R

and

dom [ go ( f h)] = dom ( f h) = dom f ∩ dom h = R ∩ R = R .

Thus,

dom ( fog) = dom [ go( f h)] = R .

Now for all x ∈ R , we have ( fog) ( x) = f ( g ( x)) = f (2 x) = sin 2 x and

…(1)

[ go( f h)] ( x) = g [( f h) ( x)] = g [ f ( x) h ( x)]

...(2) = g (sin x cos x) = 2 (sin x cos x) = sin 2 x. From (1) and (2), we have ( fog) ( x) = [ go( f h)] ( x), ∀ x ∈ R . So by definition of equality of two functions, we have f o g = g o ( f h). Illustration 2 : Let f ( x) = x 2 and g ( x) = cos x for all x ∈ R .Show that gof ≠ fog. Solution :

Here obviously dom ( gof ) = dom f = R

and We have

dom ( fog) = dom g = R . ( gof ) ( x) = g ( f ( x)) = g ( x 2 ) = cos x 2

and

( fog) ( x) = f ( g ( x)) = f (cos x) = (cos x)2 = cos 2 x.

Since there exist real numbers x for which cos x 2 ≠ cos 2 x i. e., ( gof ) ( x) ≠ ( fog) ( x), therefore gof ≠ fog. 1 + x, 0 ≤ x ≤ 2 Illustration 3 : Let f ( x) =  Find fof. 3 − x, 2 < x ≤ 3. Solution :

When 0 ≤ x ≤ 1, we have ( fof ) ( x) = f ( f ( x)) = f (1 + x) = 1 + (1 + x) = 2 + x [ ∵ 0 ≤ x ≤ 1 ⇒ 1 ≤ 1 + x ≤ 2]

270

When 1 < x ≤ 2 , we have ( fof ) ( x) = f ( f ( x)) = f (1 + x) = 3 − (1 + x) = 2 − x [ ∵ 1 < x ≤ 2 ⇒ 2 < 1 + x ≤ 3] and when 2 < x ≤ 3, we have ( fof ) ( x) = f ( f ( x)) = f (3 − x) = 1 + (3 − x) = 4 − x. [ ∵ 2 < x ≤ 3 ⇒ 0 ≤ 3 − x < 1] Hence,

2 + x, 0 ≤ x ≤ 1  ( fof ) ( x) = 2 − x, 1 < x ≤ 2 4 − x, 2 < x ≤ 3. 

Illustration 4 : (i)

f ( x) = x

5

(iii) f ( x) = x + (v)

f ( x) =

Solution : ∴

| x| x

Test the nature of the following functions for even or odd: + 7 x − 4 sin3 x

(ii) f ( x) = e

1 ,x≠0 x

(iv) f ( x) = x

x

− e − x + sin x e

x

−1

e

x

+1

(vi) f ( x) = log ( x +

,x≠0

x 2 + 1).

(i) We have f ( x) = x 5 + 7 x − 4 sin3 x. f (− x) = (− x)5 + 7 (− x) − 4 sin3 (− x) = − x 5 − 7 x + 4 sin3 x = − ( x 5 + 7 x − 4 sin3 x) = − f ( x).

Hence, f ( x) is an odd function. (ii) We have f ( x) = e ∴

x

− e − x + sin x.

f (− x) = e − x − e − (− x) + sin (− x) = e− x − e

x

− sin x = − (e

x

− e − x + sin x) = − f ( x).

Hence, f ( x) is an odd function. 1 (iii) We have f ( x) = x + , x ≠ 0. x 1 1 1 ∴ f (− x) = − x + = − x − = −  x +  = − f ( x).  (− x) x x Hence, f ( x) is an odd function. (iv) We have f ( x) = x ∴

e

x

e

x

−1 +1

f (− x) = (− x) = x

= x

e

− x /2

e

− x /2

e

x /2

− e − x /2

e

x /2

+ e − x /2

−e

x /2

+e

x /2

e

x /2

− e − x /2

e

x /2

+ e − x /2

Hence, f ( x) is an even function. | x| (v) We have f ( x) = , x ≠ 0. x

⋅

= (− x)(−1)

= f ( x).

e

x /2

− e − x /2

e

x /2

+ e − x /2

271

f (− x) =

∴

|− x| − x

| x|

=

− x

=−

| x| x

= − f ( x).

Hence, f ( x) is an odd function. (vi) We have

x 2 + 1)

f (− x) = log (− x +

∴

x 2 + 1).

f ( x) = log ( x +

 − x + x2 + 1 = log  ⋅  x +  x + x2 + 1   ( x 2 + 1) − x 2 = log   x + x2 + 1 

 x 2 + 1     

   = log      x +

1 x2

   + 1 

x 2 + 1) = − f ( x).

= − log ( x + Hence, f ( x) is an odd function.

Comprehensive Exercise 2 1. If f ( x) = e

x

and g ( x) = log

e

(i) ( f + g) (1)

(ii) ( fg) (1)

(iv) ( g o f ) (1)

(v) (4 f ) (1).

2. If f ( x) = cos

2

Let f ( x) = x

2

(iii) ( fog) (1)

x and g ( x) = 3 sin x, find

(i) ( f + g) (π / 3) 3.

x, find

(ii) ( f g) (π / 3).

+ x and g ( x) = 3 − x. Describe

(i) f + g

(ii) f − g

(iii) fg

(iv) f / g .

Find the domain in each case. 4. Test the nature of the following functions for even or odd : (i)

(ii) f ( x) = x

f ( x) = x sin x

(iii) f ( x) = sin2 x cos x

x2 − 1

(iv) f ( x) = x 2 − | x|.

A nswers 1.

(i) e

2.

(i)

3.

6 3 +1

(ii) 0

(iii) 1 (ii)

3 3

⋅ 4 8 (i) ( f + g) ( x) = x 2 + 3, dom ( f + g) = R (ii) ( f − g) ( x) = x 2 + 2 x − 3, dom ( f − g) = R

(iv) 1

(v) 4 e.

272

(iii) ( fg) ( x) = 3 x + 2 x 2 − x 3 , dom ( fg) = R x2 + x  f   f  (iv)  , dom   ( x) =  = R − {3}. 3− x  g   g  4.

(i) even (iii) even

(ii) odd (iv) even.

8.4 Graphs of Functions To draw the graph of a function f ( x), we put y = f ( x). Then the aggregate of the points ( x, y) satisfying the equation y = f ( x) gives us the graph of the function f ( x). In practice we plot some of the points and join them by a smooth free hand curve to get the graph. An important property of the graph of an even function. The graph of an even function f ( x) is symmetrical about the y-axis. Since f (− x) = f ( x), therefore the equation y = f ( x) remains unchanged on replacing x by − x. So, if the point ( x, y) lies on the curve y = f ( x), the point (− x, y) also lies on it and consequently the curve is symmetrical about y-axis. An important property of the graph of an odd function. The graph of an odd function f ( x) is symmetrical in opposite quadrants i.e., symmetrical about the origin. Since f (− x) = − f ( x), therefore the equation y = f ( x) remains unchanged on replacing x by − x and y by − y. So, if the point ( x, y) lies on the curve y = f ( x), the point (− x, − y) also lies on it and consequently the curve is symmetrical in opposite quadrants.

Illustration 1 : (i)

f ( x) = 1

Solution :

Draw the graphs of the constant functions (ii) f ( x) = 0.

(i) Putting y = f ( x),the graph

of the function f ( x) = 1 is the straight line y = 1. It is parallel to x-axis and is at a distance 1 above x-axis. (ii) The graph of the function f ( x) = 0 is the straight line y = 0 which is x-axis. Remark. The graph of the constant function f ( x) = c is always a straight line parallel to x-axis. It is above x-axis if c > 0 and is below x-axis if c < 0. If c = 0, it coincides with x-axis. Illustration 2 : Draw the graph of the modulus function f ( x) = | x|.  x, x ≥ 0 Solution : We have f ( x) = | x| =   − x, x ≤ 0.

273

Putting y = f ( x),the graph of the function f ( x) = | x| consists of the straight lines y = x, when x ≥ 0 and

y = − x, when x ≤ 0.

We have f (− x) = |− x| = | x| = f ( x) and so f ( x) is an even function of x. Hence, the graph is symmetrical about y-axis. Also | x| ≥ 0 for all x ∈ R and so the graph does not lie below x-axis. The line y = x has been drawn in the region where x ≥ 0. It passes through the points (0, 0) and (1, 1). Its slope is 1. The line y = − x has been drawn in the region where x ≤ 0. It passes through the points (0, 0) and (−1, 1) and its slope is −1so that it makes an angle of 135° with the positive direction of x-axis. Illustration 3 :

Draw the graph of the polynomial functions f ( x) = 1 − x 2 .

Solution : Putting y = f ( x), the graph of the function f ( x) = 1 − x 2 is the curve y = 1 − x 2 which is the parabola x 2 = − ( y − 1). Since f ( x) is an even function of x, the graph is symmetrical about y-axis. The vertex of the parabola is the point (0, 1) and its axis is the straight line x = 0 i. e., the y-axis. The curve passes through the points (1, 0) and (–1, 0). If y > 1, then x is not real and so the curve does not lie above the straight line y = 1. Remark. If the function f ( x) = ax 2 + bx + c , a ≠ 0, i. e., f ( x) is a quadratic polynomial, then its graph is always a parabola. 1 2 x over the interval [–3, 3]. 2 1 Putting y = f ( x), the graph of the function f ( x) = x 2 is the curve 2

Illustration 4 : Solution :

Draw the graph of the function f ( x) =

1 2 x which is a parabola. Since f ( x) is an even function of x, the graph is 2 symmetrical about y-axis. y=

The vertex of the parabola is at the orgin and its axis coincides with the y-axis. Since y = f ( x) ≥ 0 for all x ∈ R , therefore the graph does not lie below x-axis. In the first quadrant as x increases, y also increases.

y=

x

–3

–2

–1

0

1

2

3

1 2 x 2

4⋅5

2

0⋅ 5

0

0⋅ 5

2

4⋅5

274

Plotting the points ( x, y) and joining them with a smooth curve, the required graph is

Comprehensive Exercise 3 1. Draw the graph of the constant function f ( x) = − 1. 2. Draw the graph of the identity function f ( x) = x. 3. Draw the graphs of the linear functions (i) f ( x) = 2 + x (ii) f ( x) = 3 − 4 x. 4. Draw the graph of the polynomial function f ( x) = x 2 .

A nswers

1.

3.

2.

275

4.

8.5 Limit of a Function at a Point Consider the function y =

x2 − 1 x −1

⋅ The value of this function at x = 1is of the form

0 which is meaningless. In this case we cannot divide the numerator by the 0 denominator since x − 1is zero. Now suppose x is not actually equal to 1 but tends to 1. Then x − 1is not equal to zero. Hence, in this case we can divide the numerator by the denominator. ∴

x2 − 1 x −1

=

( x − 1)( x + 1) x −1

= x + 1.

If x is little greater or smaller than 1, then the value of y will be little greater or smaller than 2 and as x gets nearer to 1, y comes nearer to 2. Now the difference between y and 2 is x 2 − 1  x 2 − 2 x + 1 ( x − 1)2 − 2 =    =  = | x − 1|.  x −1   x −1   x −1  This difference | x − 1| can be made as small as we please by letting x tend to 1. Thus, we see that when x has a fixed value 1, the value of y is meaningless but when x tends to 1, y tends to 2 and we say that the limit of y is 2 when x tends to 1. Symbolically we write lim x 2 − 1 = 2. x →1 x −1 Definition Of Limit Of A Function At A Point. Let f be a function defined on some neighbourhood of a point a except possibly at a itself so that x may tend to a. Then a real number l is said to be the limit of f at x = a or the limit of f as x tends to a if for any arbitrarily chosen positive number ε, there exists a corresponding number δ > 0 such that | f ( x) − l| < ε i. e., l − ε < f ( x) < l + ε for all values of x for which 0 < | x − a| < δ i. e., a − δ < x < a or a < x < a + δ.

276

Symbolically we write

lim

f ( x) = l,

x→ a

which is read as limit of f ( x), as x tends to a, is l.

8.6 Algebra of Limits Theorem : Let (i)

lim

(iii)

x→ a

f ( x) = l and

{ f ( x) + g ( x)} = l + m =

x→ a

(ii)

lim

lim

x→ a

{ f ( x) g ( x)} = lm =

x→ a

lim x→ a

{ f ( x) − g ( x)} = l − m =

x→ a lim

lim

f ( x) +

lim

lim

lim x→ a

f ( x) −

x→ a

x→ a

g ( x) = m. Then

f ( x) .

g ( x).

lim x→ a

lim x→ a

g ( x).

g ( x).

In particular, 2 lim  lim  2 { f ( x)} =  f ( x) , x→ a  x→ a  3 lim  lim  { f ( x)}3 =  f ( x) , etc. x→ a  x→ a  lim f ( x) f ( x) lim x→ a l (iv) = = , provided m ≠ 0. lim x → a g ( x) m g ( x) x→ a

(v)

lim x→ a

lim

c f ( x) = c l = c

x→ a

f ( x), where c is some scalar i. e., c is any real

number. lim  lim  (vi) | f ( x)| = | l | =  f ( x)⋅ x→ a x→ a  lim lim (vii) | f ( x)| = 0 ⇔ f ( x) = 0. x→ a x→ a (viii)

lim x→ a

f ( x)

g ( x)

=l

m

.

8.7 Some Important Expansions (i) Binomial Expansion. (1 + x) n = 1 + nx +

n (n − 1) 2!

x2 +

n (n − 1)(n − 2) 3!

x 3 + ……

277

If n is a positive integer, the number of terms in the expansion is n + 1. If n is not a positive integer, the expansion is an infinite series and in that case we must have | x| < 1. (ii) e

x

x x2 x3 + + + ……, ∀ x ∈ R . 1! 2 ! 3!

=1+

(iii) e − x = 1 − (iv) a

x

x x2 x3 + − + ...... , ∀ x ∈ R . 1! 2 ! 3!

= 1 + x (log a) +

(v) log (1 + x) = x −

x2 (log a)2 + ……, ∀ x ∈ R . 2!

x2 x3 x4 + − + ……, − 1 < x ≤ 1. 2 3 4

(vi) log (1 − x) = − x −

x2 x3 x4 − − − ......, − 1 ≤ x ≤ 1. 2 3 4

(vii) sin x = x −

x3 x5 x7 + − + ……, ∀ x ∈ R . 3! 5! 7!

(viii) cos x = 1 −

x2 x4 x6 + − + ……, ∀ x ∈ R . 2! 4! 6!

(ix) tan x = x +

x3 2 5 + x + …… . 3 15

Some Important Factors. (i) a2 − b 2 = (a − b) (a + b) (ii) a3 − b 3 = (a − b)(a2 + ab + b 2 ) (iii) a3 + b 3 = (a + b)(a2 − ab + b 2 ) (iv) a4 − b 4 = (a2 − b 2 )(a2 + b 2 ) = (a − b)(a + b)(a2 + b 2 ) (v) a n − b n = (a − b)(a n − 1 + a n − 2 b + a n − 3 b 2 + …… + ab n − 2 + b n − 1 ) (vi) If f ( x) is a polynomial, then f (a) = 0 ⇔ ( x − a) is a factor of f ( x). For example, if f ( x) = x 3 − 6 x 2 + 11x − 6, then x − 1 is a factor of f ( x) because f (1) = 0.

8.8 Some Important Properties of Limits (i) (ii) (iii)

lim x→ a lim x→ a

f ( x) = f ( x) =

lim h→0 lim h→0

f (a + h), where h may be positive or negative. f (a − h), where h may be positive or negative.

If f and g are two functions such that lim lim lim g ( x) = l and f ( y) = m, then f ( g ( x)) = m. x→ a y→ l x→ a

278

8.9 Some Standard Limits (i)

lim x n − a n = n a n−1 , where n is a positive integer. x→ a x−a

(ii)

lim x n − a n = na n−1 , where a > 0 and n is any real number. x→ a x−a

(iii)

lim e x − 1 = 1. x→0 x

lim e ax − 1 = 1, a ≠ 0. x→0 ax lim (v) e x = 1. x→0 (iv)

(vi) (vii)

lim a x − 1 = log e a. x→0 x lim

a bx − 1

= log e a, b ≠ 0. bx lim log (1 + x) (viii) = 1. x→0 x lim log (1 + ax) (ix) = 1, a ≠ 0. x→0 ax lim (x) log (1 + x) = 0. x→0 (xi) (xii)

x→0

lim x→0 lim x→0

(1 + x)1 / x = e. (1 + ax)1 / x = e a .

8.10 Direct Substitution Method

Illustration 1 : (iii)

lim x → −1

Evaluate (i)

lim x→3

(2 x 3 − 3 x 2 + 4 x − 1).

x2 ,

(ii)

lim x→ a

(4 x 2 − 6 x + 7),

279

Solution :

(i) We have

lim x→3

x2 =

lim x→3

( x . x) =

lim x→3

x.

lim x→3

x

[ ∵ limit of a product = product of the limits] lim   = 3 . 3 = 9. ∵ x→ c x = c    (ii) We have

lim x→ a

(4 x 2 − 6 x + 7)

lim  lim   lim  =4 x2  − 6  x + 7 = 4a2 − 6a + 7.  x→ a   x→ a  x→ a lim (iii) We have (2 x 3 − 3 x 2 + 4 x − 1) x → −1 lim  lim   lim   lim  =2 x3  − 3  x2  + 4  x − 1  x → −1   x → −1   x → −1  x → −1 = − 2 − 3 − 4 − 1 = − 10. lim x − 2 Illustration 2 : Evaluate (i) , x →1 x +3

(ii)

lim

x2 + 2

x →1

x−2

⋅

lim

Solution :

( x − 2) x →1 (i) We have = lim x →1 x +3 ( x + 3) x →1 lim

x−2

lim x →1 = lim x →1

x− x+

lim x →1 lim x →1

2 = 3

1− 2 1+ 3

=−

1 ⋅ 4

lim ( x 2 + 2) lim x 2 + 2 x → 1 (ii) We have = lim x →1 x −2 ( x − 2) x →1 lim lim x2 + 2 x →1 x →1 1+ 2 = = = − 3⋅ lim lim 1− 2 x− 2 x →1 x →1 Illustration 3 : Solution :

If

lim x n − 2 n = 80 and if n is a positive integer, find n. x→2 x−2

We have lim x n − 2 n = n . 2 n−1 = 80 = 5.25 −1 ⇒ n = 5. x→2 x−2

280

8.11 Factorisation Method Suppose we are to evaluate such that

lim

lim

f ( x)

x→ a

g ( x) lim

f ( x) = 0 and

x→ a

x→ a

, where f ( x) and g ( x) are polynomials g ( x) = 0. Here the method of direct

substitution gives the indeterminate form 0/0, which is meaningless. So we proceed as follows : Since f (a) = 0 and g (a) = 0, therefore x − a is a common factor of f ( x) and g ( x). So we factorize the numerator and the denominator. Now x → a ⇒ x − a ≠ 0 and so x − a can be cancelled from the numerator and the denominator. After cancelling x − a, we proceed to find the limit by direct substitution and if we get a meaningful value the solution is complete. But if we still get the indeterminate form 0/0, we repeat the above process again till we get a meaningful value.

Illustration 1 :

(iii)

1+ x −1

lim x→0

Solution :

x

=

lim

x → 1 x2 − 3 x + 2

, (ii)

lim

x 2 − 16

x→ −4

x+4

⋅

[UPTU 2005]

(i) We have =

x2 − 1

lim

Evaluate (i)

x2 − 1

lim

 form 0   0 

x → 1 x2 − 3 x + 2

( x − 1)( x + 1)

x → 1 ( x − 1)( x − 2) lim x + 1 x →1 x −2

[ ∵ x → 1 ⇒ x − 1 ≠ 0 and so x − 1 can be

cancelled from the numerator and the denominator] = (ii) We have

1+1 1− 2

= − 2.

lim

x 2 − 16

x→ −4

x+4

= =

=

lim

x 2 − (4)2

x→ −4

x+4

lim

( x + 4) ( x − 4)

x→ −4

( x + 4)

lim x→ −4

( x − 4) =

= (− 4) − 4 = − 8.

lim x→ −4

x−

lim x→ −4

4

281

(iii) We have

=

= = = Illustration 2 : Solution :

Illustration 3 :

 form 0   0 

x [ (1 + x) − 1 ] [ (1 + x) + 1]

lim x→0

x [ (1 + x) + 1]

lim

[ (1 + x)]2 − (1)2

x→0

x [ (1 + x) + 1]

lim

1+ x −1

x → 0 x [ (1 + x) + 1] lim

1 = (1 + x) + 1

x→0 Evaluate

We have =

Solution :

(1 + x) − 1

lim x→0

lim

cos x − cos a

x→ a

x−a

lim

cos x − cos a

x→ a

x−a

lim

− sin x − 0

x→ a

1− 0

Evaluate

We have

1 1 = ⋅ (1 + 0) + 1 2

=

⋅

[UPTU 2004]

 form 0   0 

lim x→ a

− sin x = − sin a.

lim

 1 2   ⋅ − x → 1  x − 1 x 2 − 1

lim

 1 2    − x → 1  x − 1 x 2 − 1

[ form ∞ − ∞]

lim

 1  2 −   x →1  x −1 ( x − 1)( x + 1) lim lim ( x + 1) − 2 x −1 = = x → 1 ( x − 1)( x + 1) x → 1 ( x − 1)( x + 1) =

= Illustration 4 : Solution :

lim 1 1 = ⋅ x →1 x +1 2 Evaluate

We have = =

 form 0  0  

lim

( x + 2)5 /3 − (a + 2)5 /3

x→ a

x−a

( x + 2)5 /3 − (a + 2)5 /3

lim x→ a

lim x→ a

⋅

x−a 5 /3

( x + 2)

− (a + 2)5 /3

( x + 2) − (a + 2)

lim

y 5 /3 − b 5 /3

y→ b

y−b

, putting x + 2 = y and a + 2 = b ; obviously x → a ⇒ y → a + 2 i. e., y → b

282

=

5 5 5 ⋅ b (5 /3) − 1 = ⋅ b 2 /3 = ⋅ (a + 2)2 /3 . 3 3 3

Illustration 5 : Evaluate the following limits : lim e4 x − 1 lim 2 x − 1 (i) (ii) x→0 x→0 x x (iv)

lim a x→0

Solution :

x

−b

x

(v)

x

(i) We have =

lim e x→0

x

(iii)

+ e− x − 2 x2

lim

e4 x − 1

x→0

x

lim a x − 1 x→0 x [UPTU 2006]

⋅

[UPTU 2007]

e − 1  lim e y − 1   =4 ⋅ 4 ,  4x  y→0 y   4x

lim x→0

putting 4x = y ; obviously y → 0 when x → 0  lim e y − 1  = 1 ∵ y→0 y  

= 41 . = 4.

x

(ii)

lim 2 x − 1 lim e log 2 − 1 lim e x log 2 − 1 = = x→0 x→0 x→0 x x x

We have =

 e x log 2 − 1  ⋅ log x → 0  x log 2 lim

 lim e y − 1 2 = (log 2) , y→0 y 

putting x log 2 = y ; obviously y → 0 when x → 0 = ( log 2 ) . 1 = log (iii) We have

lim

a

x

x→0 =

lim

−1 x

e

=

x log a

x→0

e

2.

x→0 −1

x lim

= (log a) .

e log

lim

e

y→0

−1

x

 e x log a − 1  ⋅ log x → 0  x log a lim

=

y

a x

−1 y

 a 

,

putting x log a = y, obviously x → 0 ⇒ y → 0 = (log a) . 1 = log (iv) We have

lim a x→0

x

−b x x

=

e

x

=

a. lim (a x→0

x

− 1) − (b x

x

lim  a − 1 b − 1   − x → 0  x x 

 lim a x − 1  lim b x − 1  −  =  x   x → 0 x   x→0 = log a − log b = log (a / b).

x

− 1)

283

(v) We have

lim e x→0 =

x

+e x

−x

−2

2

=

x→0

2   e x − 1  1 ⋅  x → 0  e x  x  

lim

 1   lim  lim =   x→0 e x  x→0   1 =   ⋅ (1)2 = 1.  1

e2 x − 2e x + 1

lim

x

2

e

 ∵ e− x = 1   e x 

x

   

 e x − 1    x 

2

   

 lim lim e x − 1  e x = 1 and = 1 ∵ x→0 x→0 x  

Illustration 6 : Evaluate the following limits : lim x − 1 lim log x − log 5 (i) (ii) x → 1 log x x→5 x−5 (iii)

lim e x→0

Solution :

x

−e

−x

(iv)

x

(i) We have = = =

(ii) We have

log (a + x) − log (a − x)

x→0

x

lim x − 1 x → 1 log x

lim (1 + h) − 1 h → 0 log (1 + h)

lim lim   ∵ x → a f ( x) = h → 0 f (a + h)  

lim

h h → 0 log (1 + h) lim

1 = h → 0 log (1 + h) h

=

lim

x→5

1 log (1 + h)

h→0

h lim log (1 + x)   = 1 ∵ x→ 0 x  

1 = 1. 1

lim

lim

log x − log 5 x−5

log (5 + h) − log 5  lim lim  ∵ f ( x) = f (a + h)  x → a h → 0 h→0 (5 + h) − 5   5 + h   h log  log 1 +   lim lim  5   5 = = h→0 h → 0 5 . (h / 5) h =

=

lim

1 lim log {1 + (h / 5)} 1 1 = ⋅1= ⋅ 5 h→0 h/5 5 5 lim log [1 + (h / 5)] lim log (1 + y)   = = 1 ∵ h→ 0 y → 0 h / 5 y  

284

(iii) We have

lim e x→0 =

=

(iv) We have

lim

e

= = =

2x

−1

xe

x

1  ∵ e− x = x  e 

 1 e2 x − 1  1  ⋅ ⋅ 2 = ⋅ 1 . 2 = 2 . x → 0  e x 2x  1 lim

lim e2 x − 1 lim e y − 1   ∵ = = 1  x→0 y→0 2x y   log (a + x) − log (a − x)

x→0

=

− e− x x

x→0

lim

=

x

x log [a {1 + ( x / a)}] − log [a {1 − ( x / a)}]

lim x→0 lim

x log a + log {1 + ( x / a)} − log a − log {1 − ( x / a)}

x→0

x

lim

log {1 + ( x / a)}

x→0

x log {1 + ( x / a)}

lim x→0

1 lim a x→0

a . ( x / a)

− +

log {1 + ( x / a)} x/a

log {1 − ( x / a)}

lim x→0 lim

x log {1 + (− x / a)}

x→0

a (− x / a)

+

1 lim log {1 + (− x / a)} a x→0 (− x / a)

1 1 2 = ⋅1+ ⋅1= ⋅ a a a

Comprehensive Exercise 4 Evaluate the following limits : lim 1. (i) {( x − 1)2 + 7} x →1 2.

(i)

3.

(i)

4.

(i)

lim

x −1

x →1 x +1 lim

x 2 − 9 x + 20

x→5

x2 − 6 x + 5

lim

x 5 − 3125

x→5

x−5

(ii) (ii) (ii) (ii)

lim x→0

{( x − 1)2 + 5}.

lim

x 2 − a2

x→ a

x−a

⋅

lim

2 x2 + 9 x − 5

x→ −5

x+5

lim

x 3 − 64

x → 4 x 2 − 16

⋅

285

lim

(iii) 5. (i) 6. (i) 7. If 8. If

x−2

(iv)

x → 2 x 3 / 2 − 23 / 2

lim e − x − 1 x→0 x lim x →1

log

e

(ii)

x

(ii)

x −1

lim

x 5 /7 − a5 /7

⋅

x → a x 2 /7 − a2 /7 lim e x − e ⋅ x →1 x −1 lim

log (5 + x) − log (5 − x)

x→0

x

⋅

lim x n − 3 n = 108, and n is a positive integer, find the value of n. x→3 x−3 lim

x 9 + a9

x→ −a

x+a

= 9, find all possible values of a.

A nswers 1. (i) 7 2. (i) 0 1 3. (i) 4 4. (i) 3125 2 (iii) 3 5. (i) –1 6. (i) 1 7. 4

(ii) 6. (ii) 2a. (ii) −11. (ii) 6 5 (iv) a3 /7 . 2 (ii) e. (ii) 2/5. 8. ± 1.

Some useful results. lim lim x 1 1 1 1. = = = = 1. sin x lim x → 0 sin x x → 0 sin x 1 x→0 x x 2. 3. 4.

lim tan x lim  sin x lim sin x lim 1 1  1 = 1 ⋅ = 1. = ⋅ ⋅   = x→0 x x→0  x 1 cos x  x → 0 x x → 0 cos x lim

lim  x lim lim  x x = ⋅ cos x = ⋅ cos x = 1 . 1 = 1.  x → 0 tan x x → 0  sin x  x → 0 sin x x → 0 lim sin ( x − a) lim sin (a + h − a) = x→ a h→0 x−a a+h−a lim lim    ∵ x → a f ( x) = h → 0 f (a + h)   =

lim

sin h

h→0

h

= 1.

286

5. 6. 7. 8.

lim

tan ( x − a)

x→ a

x−a

lim x→0

=

lim

tan (a + h − a)

h→0

a+h−a

sin ax = 0 and

lim x→0

=

lim

tan h

h→0

h

= 1.

cos ax = 1.

lim sin ax lim tan ax = 1, = 1, a ≠ 0. x→0 x→0 ax ax lim

sin −1 x

x→0

x

=

lim

θ = 1, θ → 0 sin θ

putting sin −1 x = θ or x = sin θ, obviously θ → 0 when x → 0. 9.

lim lim sin θ x = = 1, x → 0 sin −1 x θ → 0 θ putting sin −1 x = θ or x = sin θ ; obviously θ → 0 when x → 0.

10.

lim

tan −1 x

x→0

x

=

lim

θ , θ → 0 tan θ

putting tan −1 x = θ or x = tan θ ; obviously θ → 0 when x → 0. 11.

lim

x

x → 0 tan

−1

x

=

lim

tan θ

θ→0

θ

,

putting tan −1 x = θ or x = tan θ ; obviously θ → 0 when x → 0 = 1. All the above results can be directly applied while evaluating limits of trigonometric functions. However, these results can themselves be asked in the form of short questions.

Illustration 1 : Evaluate the following limits : lim sin 3 x lim x3 (i) (ii) x → 0 5x x → 0 sin x 2 (iii) (iv)

lim sin a θ , where a and b are fixed non-zero real numbers θ → 0 sin b θ lim

sin2 a θ

(v)

2

θ → 0 sin b θ

Solution :

(i) We have =

lim

sin 3 x

x→0

5x

lim

sin 5 x

x → 0 tan 3 x

⋅

lim  sin 3 x 3 3 lim sin 3 x 3 3 ⋅  = = ⋅1 = ⋅  x → 0  3x x → 0  5 5 3x 5 5

287

(ii) We have

lim

x3

x → 0 sin x 2

=

 x2   ⋅ x   2 x → 0  sin x  lim

 lim x 2  lim ⋅ =  x = 1 . 0 = 0. 2  x → 0 sin x  x → 0  lim x2 2  ∵ x → 0 ⇒ x → 0 and so x → 0 sin x 2  lim sin a θ lim  sin a θ bθ a (iii) We have = ⋅ ⋅  θ → 0 sin b θ θ → 0  a θ sin b θ b =

=

lim

x2

2

x → 0 sin x

2

 = 1 

  

bθ a lim sin a θ lim a a ⋅ = ⋅ 1.1 = ⋅ θ → 0 θ → 0 b aθ sin b θ b b [ ∵ θ → 0 ⇒ a θ → 0 and b θ → 0] 2

(iv) We have

lim sin aθ lim  sin a θ =   2 θ → 0 sin b θ θ → 0  sin b θ  lim sin a θ =   θ → 0 sin b θ

2

2

 lim  sin aθ bθ a  = ⋅ ⋅   θ → 0 sin b θ b    aθ 

2

a 2 a2 =  1 . 1⋅  = 2 ⋅  b  b lim sin 5 x lim  sin 5 x 3x 5 (v) We have = ⋅ ⋅   x → 0 tan 3 x x → 0  5 x tan 3 x 3 5 lim sin 5 x lim 3x 5 5 = ⋅ = ⋅ 1.1 = ⋅ x → 0 tan 3 x 3 3 x→0 5x 3 Illustration 2 : Evaluate the following limits : lim 1 − cos mθ (i) , where m and n are fixed non-zero real numbers. θ → 0 1 − cos nθ (ii) (iv)

lim

1 − cos 5 x

x → 0 1 − cos 6 x lim

tan x − sin x

x→0

3

Solution :

x

(i) We have

=

(iii) (v)

lim

cos mx − cos nx

lim

x2 cosec x − cot x

x→0

x

x→0

lim 1 − cos mθ θ → 0 1 − cos nθ

lim 2 sin2 (mθ / 2) lim sin (mθ / 2) = θ → 0 2 sin2 (nθ / 2) θ → 0  sin (nθ / 2) 

lim sin (mθ / 2) (nθ / 2) m = . . θ → 0  (mθ / 2) sin (nθ / 2) n 

2

2

m 2 m2 = 1.1  = 2 .  n n

[ ∵ θ → 0 ⇒ (mθ / 2) → 0 and (nθ / 2) → 0]

288

lim

(ii) We have

1 − cos 5 x

x → 0 1 − cos 6 x

=

lim

2 sin2 (5 x / 2)

x→0

2 sin2 3 x

lim sin (5 x / 2) = x → 0  sin 3 x 

2

lim sin (5 x / 2) 3x 5 1 = ⋅ ⋅ ⋅   x → 0  (5 x / 2) sin 3 x 2 3

2

5 2 25 = 1 . 1⋅  = ⋅  6 36 (iii)

lim

1 1 (m + n) x sin (n − m) x 2 2 = x→0 x2 x2 1 1   lim sin 2 (m + n) x sin 2 (n − m) x (m + n)(n − m) =2 ⋅ ⋅  1 x→0  1 4 (m + n) x (n − m) x   2  2  lim

cos mx − cos nx

x→0

= 2⋅ (iv) We have

n2 − m2

⋅ 1.1 =

lim

4 tan x − sin x

x→0

3

= =

x

2 sin

n2 − m2

=

2 lim

⋅ sin x (1 − cos x) x 3 cos x

x→0

lim

sin x (1 − cos x)(1 + cos x)

x→0

x 3 cos x (1 + cos x) sin x . sin2 x

lim

x → 0 x 3 cos x (1 + cos x)

  sin x  3  1     ⋅ x→0   x  cos x (1 + cos x)  1 1 = (1)3 ⋅ = ⋅ 1 (1 + 1) 2 =

(v) We have

lim

lim

cosec x − cot x

x→0 lim

x 1 − cos x

x→0

x sin x

= = =

lim

=

lim

(1 − cos x)(1 + cos x)

x→0

x sin x (1 + cos x)

sin2 x

x → 0 x sin x (1 + cos x) lim  sin x  1 1 1 ⋅   = 1⋅ = ⋅ x→0  x 2 2 1 + cos x 

289

Illustration 3 : Evaluate the following limits : lim 1 − cos 2 x lim 1 − cos θ (i) (ii) x→0 θ→0 x 2θ2 lim lim tan 3 x − 2 x sin 3 x + 7 x (iii) (iv) x → 0 4 x + sin 2 x x → 0 3 x − sin2 x (v)

lim

1 + cos x 2

x→ π

Solution :

tan

x

(i) We have =

(ii) We have

lim

tan x − sin x

x→0

sin3 x

lim

1 − cos 2 x

x→0

x

lim

2 sin2 x

x→0

x

lim

=2

lim

 sin x    ⋅ sin x → 0  x 

1 − cos θ

lim

2θ2 (1 − cos θ)(1 + cos θ)

θ→0

2θ2 (1 + cos θ)

θ→0 =

(vi)

⋅ form 0  0  

 x = 210 . . = 0.  form 0  0  

 sin θ 2  1    ⋅  θ  1 + cos θ 1 1 1 = ⋅ (1)2 ⋅ = ⋅ 2 1+1 4 =

(iii) We have

1 lim 2 θ→0

lim

sin 3 x + 7 x x → 0 4 x + sin 2 x sin 3 x =

lim x→0

x 4+

form 0  0  

+7

sin 2 x x

[Dividing the numerator and the denominator by x] sin 3x   ⋅ 3 + 7  lim  3 x (1.3) + 7 10 5  = = = = ⋅ sin 2 x x→0 4 + 2.1 6 3 4 +2⋅ 2x lim tan 3 x − 2 x form 0  (iv) We have x → 0 3 x − sin2 x 0   tan 3 x =

lim x→0

x 3−

−2

sin2 x x

[Dividing the numerator and the denominator by x ]

290

=

(v) We have

x→0

lim

= =

3−

tan 3 x 3x sin x x

= ⋅ sin x

2

tan

lim

cos

x (1 + cos x) sin2 x

x→ π

x → π (1 + cos x)(1 − cos x) (−1)2

1 ⋅ 3

1 − (−1)

lim

sin

cos 2 x (1 + cos x)

x→ π

(1 − cos 2 x)

=

lim

cos 2 x

x → π 1 − cos x

form 0   0 

x

sin x (1 − cos x) 3

x→0

lim

1 ⋅ 2

=

3

=

=

cos 2 x (1 + cos x)

lim

tan x − sin x

=

3 − 1. 0

=

x 2

lim

=

3 .1 − 2

form 0  0  

x→0 =

−2

1 + cos x

x→ π =

(vi) We have

3⋅

lim

cos x sin

lim

x

=

lim

1 − cos x

x → 0 cos x sin2 x

1 − cos x

x → 0 cos x (1 − cos 2 x) lim

1 − cos x

x → 0 cos x (1 − cos x)(1 + cos x) lim

1 1 1 = = ⋅ x → 0 cos x (1 + cos x) 1 (1 + 1) 2

Illustration 4 : Evaluate the following limits : lim sin x ° lim (i) (ii) (sec x − tan x) x→0 x→ π /2 x (iii) (v)

lim x→ π lim

1 + sec 3 x 2

tan

x

( x + y) sec ( x + y) − x sec x

y→0

Solution :

y

=

sin 2 x + sin 6 x x → 0 sin 5 x − sin 3 x

lim

⋅

sin x °

x→0

lim

x sin (πx / 180)

form 0   0 

x→0

x

 ∵ x ° = π x radians  180 

(i) We have =

lim

(iv)

lim

sin (πx / 180) π  ⋅  x → 0  (πx / 180) 180 

291

=

lim sin (πx / 180) π π π = ⋅1= ⋅ 180 x → 0 (πx / 180) 180 180 [ ∵ x → 0 ⇒ (πx / 180) → 0]

(ii) We have

lim

= = = = (iii) We have

lim lim

(1 − sin x)(1 + sin x)

x→ π /2

cos x (1 + sin x)

lim x→ π /2

=

cos cos x

1 − sin2 x

lim

x → π / 2 cos x (1 + sin x)

x

=

x → π / 2 1 + sin x

0 0 = = 0. 1+1 2

1 + sec 3 x 2

tan

form 0   0 

x

lim

(1 + sec x)(1 − sec x + sec 2 x)

x→ π

sec 2 x − 1

lim

(sec x + 1)(1 − sec x + sec 2 x)

x→ π

(sec x + 1)(sec x − 1)

lim

1 − sec x + sec 2 x

x→ π

sec x − 1

lim

=

1 − (−1) + (−1)2 −1 − 1

=−

sin 2 x + sin 6 x

lim

2 sin 4 x cos 2 x

x→0

2 cos 4 x sin x sin 4 x

=

lim

4x

x→0

⋅ 4 x ⋅ cos 2 x

cos 4 x ⋅ sin 4 x

=4

lim x→0

lim

sin x x

⋅x

⋅ cos 2 x

cos 4 x ⋅

sin x

=4⋅

1.1 = 4. 1.1

x ( x + y) sec ( x + y) − x sec x

y→0 =

4x

3 ⋅ 2 form 0   0 

x → 0 sin 5 x − sin 3 x =

(v) We have

2

=

cos x (1 + sin x)

lim

x→ π

=

[ form ∞ − ∞]

lim sin x  1 − sin x  1 −   = x → π / 2 cos x  cos x cos x 

x→ π /2

lim

=

(iv) We have

(sec x − tan x)

x→ π /2

y

lim

x {sec ( x + y) − sec x} + y sec ( x + y)

y→0

y

form 0  0  

292

=

lim y→0

x⋅

sec ( x + y) − sec x y

+

lim y→0

sec ( x + y)

cos x − cos ( x + y)   lim = x  + sec x y cos x cos ( x + y)   y→0 y  y     sin  2 sin  x +  lim  2 2  1  = sec x + x⋅ ⋅ y→0  ( y / 2) . 2 cos x cos ( x + y)     y  lim sin ( y / 2)   x = sec x + ⋅ sin  x + ⋅   y → 0  ( y / 2) 2  cos x cos ( x + y)  x = sec x + 1. sin x ⋅ = sec x + x sec x tan x. cos x. cos x Illustration 5 : Evaluate the following limits. lim lim cot θ (i) (ii) θ → π / 2 (π / 2) − θ x→ π /2 (iii) (v)

lim

sin x − cos x

x→ π /4

x − (π / 4)

(iv)

lim πx (1 − x) tan x →1 2

Solution :

(vi)

(i) We have

=

lim

cos x

−1

cos x

lim

2x − π

x→ π /2

cos x

lim 1 + cos 2 x ⋅ x → π / 2 (π − 2 x)2

lim

cot θ

θ→ π /2

(π / 2) − θ

 1  cot  π + h  2 

e

lim lim    ∵ x → a f ( x) = h → 0 f (a + h)  

h→0 1  1  π −  π + h 2  2  lim − tan h lim tan h = = = 1. h→0 h →0 −h h (ii) We have

lim x→ π /2 =

lim y→0

e

cos x

−1

cos x e

y

−1 y

,

putting cos x = y, obviously y → 0 when x → π / 2 (iii) We have

= 1. lim x→ π /4

sin x − cos x x − (π / 4)

293

=

lim x→ π /4

2 {(1 / 2) sin x − (1 / 2) cos x} x − (π / 4)

1  sin  x −  4 = 2⋅ 1 x→ π /4 x− π 4 1 putting x − π = 4 lim

 π 

= 2⋅

lim y→0

sin y y

,

y ; obviously y → 0 when x → π / 4

= ( 2) . 1 = 2.

(iv) We have

lim x→ π /2

2x − π cos x

=

lim h→0

 1  2  π + h − π  2   1  cos  π + h  2 

lim lim    ∵ x → a f ( x) = h → 0 f (a + h)   lim lim 2h h = =−2 = (− 2) . 1 = − 2. h → 0 − sin h h → 0 sin h

(v)

lim x →1

lim πx π = {1 − (1 + h)} tan  (1 + h) h→0 2 2   lim lim    ∵ x → a f ( x) = h → 0 f (a + h)   lim lim 1 = − h tan  π + (πh / 2) = h cot (πh / 2) h→0 2  h→0

(1 − x) tan

=

lim

 (πh / 2) ⋅ h → 0  tan (πh / 2)

lim lim 1 + cos 2 x (vi) = 2 x → π / 2 (π − 2 x) h→0

2 2 2  = 1⋅ = ⋅ π π π   1  1 + cos 2  π + h   2   2

  1   π − 2  π + h   2   lim lim    ∵ x → a f ( x) = h → 0 f (a + h)   lim 1 + cos (π + 2h) lim 1 − cos 2h = = 2 h→0 h →0 4h 4h2 =

lim

2 sin2 h

4h2 1 1 = ⋅ (1)2 = ⋅ 2 2 h→0

=

2 1 lim  sin h   2 h→0  h 

294

Illustration 6 : (i)

lim e x→0

(iii) (v)

x

e

x→0

Solution :

− sin x − 1

(ii)

x

lim e x→0 lim

Evaluate the following limits :

tan x

−1

x

2x

−e

lim

log (1 + x 3 )

x→0

sin3 x

(UPTU 2011) (iv)

x

(vi)

sin 2 x (i) We have

lim e x→0

x

lim

23 x − 3 x

x→0

sin 3 x

lim x (2 x − 1) ⋅ x → 0 1 − cos x

− sin x − 1 x

x

= (ii)

lim

log (1 + x 3 )

x→0

sin3 x =

(iii)

lim e − 1 lim sin x − = 1 − 1 = 0. x→0 x→0 x x

lim

e

lim

log (1 + x 3 )

x→0

x3

⋅

x3 sin3 x

3 3 lim log (1 + x ) lim  x  ⋅ = 1 . (1)3 = 1.   x→0 x → 0  sin x  x3

tan x

x→0

−1

x =

=

lim

= e

x→0

 e tan x − 1 tan x  ⋅   x → 0  tan x x   lim

tan x

−1

tan x

⋅

lim

tan x

x→0

x

 lim e y − 1  ⋅ 1, =  y   y→0

in the first limit putting tan x = y ; obviously y → 0 when x → 0 = 1 . 1 = 1. (iv)

lim x→0

23 x − 3

x

sin 3 x

=

(23 x − 1) − (3  x→0  x  lim

x

− 1)

⋅

3x 1 ⋅  sin 3 x 3

 lim  23 x − 1  lim 3 x − 1 1 lim 3x   − ⋅ ⋅ 3   x→0 3 x → 0 sin 3 x  x → 0  3 x x     1 1 = ⋅ 1 . (3 log 2 − log 3) = log 2 − log 3. 3 3 2x x x x lim e − e lim e (e − 1) (v) We have = x → 0 sin 2 x x→0 sin 2 x =

=

(vi)

lim

lim  x e x − 1 2 x 1 1 1 . .  = 1⋅ 1⋅ 1⋅ = ⋅ e . x→0  x sin 2 x 2 2 2 

x (2

x

− 1)

x → 0 1 − cos x

=

2 x − 1  x2 ⋅ ⋅ (1 + cos x)  2 x→0  x 1 − cos x   lim

295

=

lim 2 x − 1  x  ⋅ x→0  x  sin 

  x

2

 ⋅ (1 + cos x) = (log 2) ⋅ (1)2 ⋅ (1 + 1) 

= 2 log 2 = log 22 = log 4. Illustration 7 : Evaluate the following limits : 1+ x − 1− x lim lim tan −1 2 x (i) (ii) x→0 x → 0 sin 3 x sin −1 x (iii)

1 − x 2    1 + x 2   

lim

1 cos −1 x→0 x

(iv)

lim 1− x x → 1 (cos −1 x)2

 x − cos (sin −1 x) lim 1− x (vi) ⋅   x → 1 π − 2 sin −1 x x → 1 / 2 1 − tan (sin −1 x) 1+ x − 1− x lim Solution : (i) We have x→0 sin −1 x (v)

lim

=

form 0  0  

(1 + x) − (1 − x)  1   ⋅ x→0  sin −1 x 1 + x + 1 − x   lim

[Rationalising the numerator]  lim  x 1 2 ⋅  = ⋅ x → 0  sin −1 x 1 + x + 1 − x    =2

(ii) We have

x

⋅

lim

x → 0 sin −1 x x → 0

lim

tan −1 2 x

x→0

sin 3 x

= (iii) We have

lim

=

1

= 2 ⋅ 1⋅

1+ x + 1− x

1 = 1. 2

 tan −1 2 x 3x 2 ⋅ ⋅   x → 0  2x sin 3 x 3  lim

−1 2 lim tan 2 x lim 3x 2 2 ⋅ = ⋅ 1⋅ 1 = ⋅ x → 0 x → 0 3 2x sin 3 x 3 3

lim

1 cos −1 x→0 x

1 − x 2  lim  1 −1    1 + x 2  = x → 0  x ⋅ 2 tan x   2  −1 1 − x ∵ cos = 2 tan −1  2 1 + x 

=2 (iv) We have

lim

tan −1 x

x→0

x

 x 

= 21 . = 2.

lim lim 1− x = − 1 2 x → 1 (cos x → 1 (1 + x)

1− x x ) (cos −1 x)2 [Rationalizing the numerator]

296

=

lim

lim 1− x 1 ⋅ x → 1 1 + x x → 1 (cos −1 x)2

=

1 lim 1 − cos θ , 2 θ→0 θ2 putting cos −1 x = θ or x = cos θ ; obviously θ → 0 when x → 1

=

 sin2 θ  1    θ2 ⋅ 1 + cos θ  

1 lim 2 θ→0

2  1 lim  sin θ 1    ⋅ 2 θ → 0  θ  1 + cos θ 1 1 1 = ⋅ (1)2 ⋅ = ⋅ 2 2 4 lim  x − cos (sin −1 x) (v) We have   x → 1 / 2 1 − tan (sin −1 x)  lim sin θ − cos θ = , θ→ π /4 1 − tan θ

=

= (vi) We have

lim θ→ π /4

lim

putting sin −1 x = θ or x = sin θ ;

obviously θ → π / 4 when x → 1 / 2 lim  sin θ − cos θ 1 (− cos θ) = − ⋅   ⋅ cos θ = θ→ π /4 2  cos θ − sin θ

1− x

x → 1 π − 2 sin

−1

x

=

lim

1 − sin θ

θ→ π /2

π − 2θ

,

putting sin −1 x = θ or x = sin θ ; obviously θ → π / 2 when x → 1 1 1 − sin ( π + h) lim lim   2 = ∵ f ( x) = f (a + h)  1 x → a h → 0 h→0   π − 2 ( π + h) 2 lim 1 − cos h 1 lim (1 − cos h)(1 + cos h) = =− h→0 − 2h 2 h→0 h (1 + cos h) lim

=−

 1 1 1 lim  sin h 1  ⋅ sin h ⋅   = − (1) ⋅ 0 ⋅ = 0. h → 0  2 2 2 1 + cos h  h

Illustration 8 : Evaluate the following limits : lim lim 3 x 2 + 5 x − 4 3x − 7 (i) (ii) 2 2 x → ∞ 5x + 8 x − 5 x → ∞ 7x − 8x + 3 (iii)

lim x→ ∞

7 x2 − 6 x + 5

(iv)

5x + 8

Solution : (i) We have

lim x→ ∞

3x

2

+ 5x − 4

x → ∞ 7x

2

− 8x + 3

lim

x−

x2 + x. [UPTU 2008]

297

=

= (ii)

x → ∞ 7 − 8 . (1 / x) + 3 . (1 / x 2 ) 3 + 5.0 − 4.0 7 − 8.0 + 3.0 lim

We have

2

+8 x−5

x [3 − 7 . (1 / x)]

lim

=

2

x → ∞ x [5 + 8 . (1 / x) − 5 . (1 / x 2 )]

3 − 7 . (1 / x) 1 ⋅ x → ∞ x 5 + 8 . (1 / x) − 5 . (1 / x 2 ) 3 − 7 .0 5 + 8 .0 − 5 .0

=0⋅

lim 1 lim 3 1   = 0.∵ = 0, = 0 2 x → ∞ x → ∞ 5 x x  

7 x2 − 6 x + 5

lim x→ ∞

=

3 ⋅ 7

=

lim

=0⋅

=

[Dividing the Nr. and the Dr. by x 2 ] lim 1 lim 1    ∵ x → ∞ x = 0, x → ∞ 2 = 0 x  

3x − 7

x → ∞ 5x =

(iii) We have

3 + 5 . (1 / x) − 4 . (1 / x 2 )

lim

5x + 8 2

lim

x [7 − 6 . (1 / x) + 5 . (1 / x 2 )]

x→ ∞

x [5 + 8 . (1 / x)]

lim x→ ∞

x⋅

7 − 6 . (1 / x) + 5 . (1 / x 2 ) 5 + 8 . (1 / x)

= ∞.

lim lim 1 lim 1    ∵ x → ∞ x = ∞, x → ∞ x = 0, x → ∞ 2 = 0 x   (iv) We have

lim x→ ∞ =

=

= Illustration 9 : Solution :

x→ ∞

x+

lim

x→ ∞

x2 + x − x

x→ ∞

x + x 1+

−1 1+ 1+ 0

=

Show that lim x→ ∞

lim x→ ∞

−1

1 x

x2 + x) (x +

(x −

lim

x 2 − ( x 2 + x)

lim

We have

=

x2 + x =

x−

=

− x

x→ ∞

=−

x2 + x)

(x +

lim

x2 + x

x+

lim x→ ∞

x2 + x)

1 1+ 1+

1 x

⋅

2 lim

x→ ∞

( x 2 + x + 1 − x) ≠

lim x→ ∞

( x 2 + x + 1 − x)

( x 2 + x + 1 − x)( x 2 + x + 1 + x) ( x 2 + x + 1 + x)

( x 2 + 1 − x).

298

=

=

=

( x 2 + x + 1) − x 2

lim

x → ∞ ( x 2 + x + 1 + x) lim

x +1

x→ ∞

x

2

+ x +1+ x

lim

1 + (1 / x)

x→ ∞

1 + (1 / x) + (1 / x)2 + 1 [Dividing the Nr. and the Dr. by x]

1+ 0

1 = = ⋅ 2 1+ 0 + 0 +1 lim

Again

x→ ∞ =

=

=

( x 2 + 1 − x) =

lim x→ ∞

( x 2 + 1 − x)( x 2 + 1 + x)

x→ ∞

x2 + 1 + x

lim

( x 2 + 1) − x 2

x→ ∞

x2 + 1 + x

lim

1

x→ ∞

2

x +1+ x

lim

1 1 [Dividing the Nr. and the Dr. by x] ⋅ x → ∞ x 1 + (1 / x 2 ) + 1

=0⋅ Hence,

lim

1 1+ 0 +1

=0⋅

( x 2 + x + 1 − x) ≠

1 = 0. 2 lim

x→ ∞

( x 2 + 1 − x).

Comprehensive Exercise 5 Evaluate the following limits. lim sin 7 x 1. (i) x→0 5x (iii) 2.

(i)

lim

tan 3 x

(iv)

x → 0 sin 2 x lim

3 sin x − 4 sin3 x

x→0

x

(iii)

(ii)

lim

cos 3 x − cos 7 x

x→0

2

x

(ii) (iv)

lim

sin 3θ

θ → 0 sin 2θ lim

sin x cos x

x→0

3x

lim

x2

x → 0 sin x 2 lim 1 − cos 4θ ⋅ θ → 0 1 − cos 5θ

⋅

299

3. (i)

lim

(i)

1 + cos π x

(iv)

2

(1 − x)

lim e x→0

x

(ii)

+ sin x − 1

(iv)

x

x+3

(ii)

x→ ∞ x−3

(iv) (i)

x →1

lim

(iii)

6.

lim

lim sin 2 x − 1 x → π / 4 x − (π / 4)

(iii) 5. (i)

(ii)

x→ π x−π

(iii) 4.

sin x

lim

cos θ

θ→ π /2

(π / 2) − θ

lim

1 − tan x

x → π / 4 x − (π / 4) lim x→ π /3 lim x→0 lim

e

⋅

3 − tan x π − 3x sin x

−1

sin x

⋅

8 x2 − 9 x + 5

x → ∞ 4 x2 + 6 x − 7

lim 3 x 3 − 7 x 2 + 5 x − 1 x → ∞ 4 x3 + x2 − 6 x + 8 lim

(3 x − 1)(4 x − 2)

x→ ∞

( x + 8)( x − 1)

lim

2 x2 − 5 x + 7

⋅

x → ∞ 3 x3 − 6 x2 + x − 4

(ii) (iii) (iv)

lim

4 x 2 − 7 x + 11

x→ ∞

3 − 5x

lim x→ ∞ lim x→ ∞

[ x2 + x + 1 − [ x +1 −

x 2 + 1]

x ].

A nswers 1. (i)

7 5

(ii)

3 2

(iii)

3 2

2. (i) 3

(ii) 1

(iii) 20

3. (i) − 1

(ii) 1

(iii)

4. (i) 0

(ii)

5. (i) 1

(ii) 2

6. (i) 0

(ii) − ∞

4 3

π2 2

(iii) 2 3 4 1 (iii) 2 (iii)

1 ⋅ 3 16 (iv) ⋅ 25 (iv)

(iv) − 2. (iv) 1. (iv) 12. (iv) 0.

300

Comprehensive Exercise 6 Fill in the Blanks Fill in the blanks ‘‘..........’’, so that the following statements are complete and correct. 1− x 1. If f ( x) = , then f (cos θ) = .......... . 1+ x 2. If f ( x) =

sin x 1 + sin x

, then f (π / 2) = .......... .

3. The value of lim

x →1

x2 − 1 x −1

is .......... . [UPTU 2009, 10]

x

4. The value of lim e is .......... . x→0

5. The value of lim

x→3

4 x 2 − 17 x + 15 x2 − x − 6

is .......... .

6. The domain of the real function f ( x) = 7. The value of lim

x→4

8. The value of lim

x→0

9. The value of lim

x→ a

10. The value of lim

x→ ∞

x 2 − 16 x−4 ax + b cx + d

1 − x2

is .......... .

is .......... .

is .......... .

x n − an x−a

1

= ........., where a > 0 and n is any real number.

3 x −1 + 4 x −2 5 x −1 + 6 x −2

= .......... .

a x 11. The value of lim 1 +  is .......... . x→ ∞  x

[UPTU 2010]

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 12. The value of lim log (1 + x) is x→0

(a) 1

(b) 0

(c) − 1

(d) ∞

301

13. The value of lim (1 + x)4 x is x→0

(a) e

2

(b) e −1

(c) e

(d) − e

 sin 2 x  14. The value of lim   is x→0  x  (a) 0

(b) 1

(c) 1/2

(d) 2

[UPTU 2011]

x

e −1

15. The value of lim

x→0

x

is

(a) 1

(b) − 1 1 (d) 2

(c) 0 a bx − 1

16. The value of lim

x→0

bx

is ........., where b ≠ 0

(a) log e b b (c) log e a 17. If lim

x→ a

(b) log e a a (d) log e b

x 9 − a9 x−a

= 9, the all possible values of a are

(a) 1

(b) − 1

(c) ± 1

(d) 0

True or False Write ‘T’ for true and ‘F’ for false statement. 18. The value of lim

x→4

19. The value of lim

x 2 − 16 x−4

is ( x + 4).

ax −1

= log e a. x sin x 20. The value of lim = 1. x→0 x x −1 21. The value of lim is 0. x→ ∞ x +1 x→0

22. The domain of the real function f ( x) = 23. The value of lim

θ→0

24. The value of lim

x→ ∞

sin aθ sin bθ

=

[UPTU 2010]

x2 1 + x2

is R.

a ⋅ b

x2 + 2 3x

2

1 is . −4 3

[UPTU 2011]

302

25. The value of lim

x→ ∞

26. The value of lim

x→ ∞

27. The value of lim

θ→0

x 2 + ax + b x 2 + cx + d

is 1.

ax 2 + bx + c dx

2

+ ex + f

sin 4θ tan 3θ

=

=

d ⋅ a

4 ⋅ 3

A nswers 1. tan2 (θ / 2) 5. 9. 13. 17. 21. 25.

7 5 na n −1 (c) (c) F T

2.

1 2

6. R − {− 1, 1} 10. 14. 18. 22. 26.

3/5 (d) T T F

3. 2

4. 1

7. 8

8. b/d

11. 15. 19. 23. 27.

ea (a) T T T

12. 16. 20. 24.

(b) (b) T F

303

9 D ifferentiation

9.1 The Differential Coefficient he differential coefficient of a function y = f ( x) with respect to x is defined as δy f ( x + δx) − f ( x) lim = lim δx → 0 δ x δx → 0 δx provided the limit exists. The differential coefficient is also called the derivative, or the derived function. The differential coefficient of y = f ( x) with respect to x

T

may be denoted by any of the symbols dy d d y, , y ′ , Dy , f ( x), f ′ ( x) , Df ( x). dx dx dx The process of finding the differential coefficient is called differentiation. The Differential Coefficient At A Point. lim

h→ 0

f (a + h) − f (a) h

If y = f ( x) is a function of x, then

304

is called the differential coefficient of f ( x) for x = a, provided the above limit  dy  exists. It is denoted by   , ( y ′ ) a , or f ′ (a). It gives us the rate of change of  dx  x = a y with respect to x at x = a.

9.2 Some Standard Results (I) Differential Coefficient Of x n (n being real number). Let y = x n . Then y + δy = ( x + δx) n . Therefore δy = ( x + δx) n − x n . δy ( x + δx) n − x n ∴ = ⋅ δx δx Taking limit when δx → 0, we get dy δy = lim = lim δx → 0 δx → 0 dx δx n

x =

lim

δx → 0

x =

n

lim

δx → 0

δx   1 +   x δx

( x + δx) n − x

n

δx n

− x

x

n

=

n

lim

δx → 0

n   δx  1 +  − 1 x   δx

2    δx  n (n − 1)  δx  1 + n +     + … − 1   x  x 2!   δx

[Expanding by binomial theorem since δx / x is numerically less than unity, δx being numerically small]   δx  n (n − 1)  δx  2  x n n   +   + .....  x 2!   x   = lim δx → 0 δx = Thus

lim

δx → 0

x

n

 n n (n − 1) δx  + ...... = x  + 2 2! x x 

n

.

n = nx x

n −1

.

d x n = nx n - 1 . dx

(II) Differential Coefficient Of sin x. sin ( x + δx) − sin x d We have sin x = lim , by definition δ x → 0 dx δx

=

lim

δx → 0

δx  δx  2 cos  x +  sin  2 2 = lim δx → 0 δx

= cos x, since

lim

δx → 0

sin (δx / 2) δx / 2

= 1.

δx   cos  x +   2

δx 2 δx 2

sin

305

Thus,

d sin x = cos x. dx d cos x = − sin x. dx

Similarly, it can be shown that

(III) Differential Coefficient of e x . lim e x + δx − e d x We have e = δx → 0 dx δx

=

lim

e

= Thus

−e

, by definition

x

x

lim

lim

=

δx

(e

δx

− 1)

δx

 δx (δx)2  (δx)3 + + + ..... − 1 1 + 2! 3!  1 ! 

x

2

δx (δx)  (δx)3 + + + ....  2! 3! 1 ! 

δx → 0 δx → 0

x

δx

lim lim

e

δx → 0

δx → 0 e

=

δx

e

δx → 0 e

=

x

x

δx x

e

2   δx (δx) 1 + + + … = e x .  3!  2 ! 

d (e x ) = e x . dx

(IV) Differential Coefficient of a x log e a. We have

d a dx

x

=

lim

x+h

a

h→0

= lim a

−a

x

h x

h→ 0

=

h

 a − 1   = a  h 

lim

a

x

ah − a

h→0 x

h h

lim

x

a −1 h

h→ 0

=a

x

. log

e

a

 lim a h − 1 ∵ = log  h→0 h  Thus,

d x a = a x log e a = a x log a. dx

(V) Differential Coefficient of log e x. We have

lim d log x = δx → 0 dx

=

lim δx → 0

log ( x + δx) − log x δx  x + δx  log    x  δx

=

, by definition

lim δx → 0

δx   log 1 +   x δx

e

 a 

306

=

lim δx → 0

2 (δx)3 δx (δx) − + − ..... x 2 x2 3 x3 δx

(the expansion is justified since δx / x is numerically less than unity, δx being numerically small) 1  1 lim (δx)2 δx = − ...... = ⋅  − 2 + 3 δx → 0  x 2 x 3x  x d 1 (log e x) = ⋅ dx x

Thus

(VI) Differential Coefficient of a Constant. Let f ( x) = c , where c is a constant. lim f ( x + δx) − f ( x) d Then f ( x) = , by definition δx → 0 dx δx lim c −c = , since f ( x) = c for every value of x δx → 0 δx lim 0 = = 0. δx → 0 δx Thus, the differential coefficient of a constant is zero. (VII) Differential coefficient of the product of a constant and a function. Let c be a constant and f ( x) be a function of x. Then by definition lim c f ( x + δx) − c f ( x) d { c f ( x)} = δx → 0 dx δx lim f ( x + δx) − f ( x) = c⋅ δx → 0 δx lim f ( x + δx) − f ( x) d =c⋅ =c f ( x). δx → 0 δx dx Thus the differential coefficient of the product of a constant and a function is equal to the product of the constant and the differential coefficient of the function. (VIII) Differential coefficient of log a x. We have log constant. ∴

a

x = log d (log dx

a

e

x log

a

x) = log

e = log

a

e

a

d (log dx =

Thus,

e log

e

e

x, where log

x) = (log

1 x log

d 1 1 (log a x) = = ⋅ dx x log e a x log a

e

a

a

e) .

a

e is simply a

1 1 = ⋅ log x x

, since log

a

e . log

e

a

e

a = 1.

307

9.3 List of Standard Results to be Committed to Memory d n x = nx n −1 dx d 1 log e x = dx x

d x e =ex dx d 1 log a x = log dx x

d dx d dx d dx d dx d dx d dx d dx

d dx d dx d dx d dx d dx d dx d dx

sin x = cos x tan x = sec 2 x sec x = sec x tan x sinh x = cosh x tanh x = sech 2 x . sech x = − sech x tanh x sin − 1 x =

1 2

√ (1 − x )

a

e or

1 x log e a

cos x = − sin x cot x = − cosec 2 x cosec x = − cosec x cot x cosh x = sinh x coth x = − cosech 2 x cosech x = − cosech x coth x cos − 1 x = −

1 √ (1 − x 2 )

d 1 tan − 1 x = dx 1 + x2

d 1 cot − 1 x = − dx 1 + x2

d 1 sec − 1 x = dx x √ ( x 2 − 1)

d 1 cosec − 1 x = − dx x √ ( x 2 − 1)

d 1 sinh − 1 x = dx √ ( x 2 + 1)

d 1 cosh − 1 x = dx √ ( x 2 − 1)

d 1 tanh − 1 x = dx 1 − x2

d a dx

x

=a

x

log

e

a.

9.4 Differential Coefficient of the Sum of two Functions Let Then

f ( x) = f1 ( x) + f 2 ( x). lim f ( x + δx) − f ( x) d f ( x) = δ x → 0 dx δx lim { f1 ( x + δx) + f 2 ( x + δx)} − { f1 ( x) + f 2 ( x)} = δx → 0 δx lim { f1 ( x + δx) − f1 ( x)} + { f 2 ( x + δx) − f 2 ( x)} = δx → 0 δx lim f ( x + δx) − f 2 ( x)  f1 ( x + δx) − f1 ( x) = + 2   δx → 0  δx δx 

308

=

lim

f1 ( x + δx) − f1 ( x)

δx → 0

δx

+

lim

f 2 ( x + δx) − f 2 ( x)

δx → 0

δx

d d f1 ( x) + f 2 ( x). dx dx d d d [ f1 ( x) + f 2 ( x)] = f1 ( x) + f 2 ( x). dx dx dx d { f 1 ( x) + f 2 ( x) + . . . . + f n ( x)} dx d d d = f 1 ( x) + f 2 ( x) + . . . . + f n ( x) dx ds dx d { f 1 ( x) − f 2 ( x) − . . . . − f n ( x)} dx d d d = f 1 ( x) − f 2 ( x) − . . . . − f n ( x). dx dx dx =

∴ Similarly,

and

Illustration 1 :

Find the derivatives of the following functions with respect to (w.r.t.) x. (ii) x − 8

(i) 1 / x (iii) 5

cot x

(iv) log2 x. d n Solution : We know that x = n x n−1 . Therefore dx d  1 d −1 1 (i) x = − 1 x −1−1 = − x −2 = − 2 ⋅   = dx  x  dx x d −8 − 8 −1 −9 (ii) x =−8 x = − 8x . dx d cot x (iii) 5 = 5 cot x log e5 . dx d 1 1 (iv) ( log 2 x) = log 2 e or ⋅ dx x x log e 2 Illustration 2 :

Differentiate the following functions with respect to x:

(i) 4 x 3 − 7 − 6 x 2

(ii) 5 sin x − 2 log x + 7 sec x .

d d d d (4 x 3 − 7 − 6 x 2 ) = (4 x 3 ) − (7) − (6 x 2 ) dx dx dx dx d 3 d 2 =4 x −0 −6 x = 4 . (3 x 2 ) − 0 − 6 . (2 x) = 12 x 2 − 12 x. dx dx d d d d (5 sin x − 2 log x + 7 sec x) = 5 sin x − 2 log x + 7 sec x dx dx dx dx 1 2 = 5 cos x − 2 ⋅ + 7 sec x tan x = 5 cos x − + 7 sec x tan x. x x

Solution :

(ii)

[UPTU 2001]

(i)

309

Comprehensive Exercise 1 Differentiate the following functions with respect to x. 1 x x 1. x + ⋅ 2. 3 sin − 4 sin3 ⋅ x 3 3 3.

5 x −7 .

4. log

5.

tan −1 x + cot −1 x.

6. tan x + 2 sin x + 3 cos x −

7.

e

x

+ 4 sec x + 5 cos −1 x.

8. 2

x

e

x. 1 log x − e x . 2

+ x 2 + 22 .

A nswers 1. 1 − 5.

0.

7.

e

x

1 x

2

2. cos x.

⋅

3. − 35 x − 8 .

6. sec 2 x + 2 cos x − 3 sin x − 5

+ 4 sec x tan x −

1− x

2

⋅

8. 2

x

4.

1 ⋅ 2x

1 − e x. 2x

log

e

2 + 2 x.

9.5 Differential Coefficient of the Product of two Functions Let

f ( x) = f1 ( x) f 2 ( x).

Then

d d d [ f1 ( x) f 2 ( x)] = f1 ( x) f 2 ( x) + f 2 ( x) f1 ( x). dx dx dx (Product Rule)

9.6 Differential Coefficient of the Quotient of two Functions f1 ( x)

Let

f ( x) =

Then

d  f 1 ( x)    = dx  f 2 ( x) 

f 2 ( x)

⋅ f 2 ( x) .

d d f1 ( x) − f1 ( x) . f 2 ( x) dx dx ⋅ [ f 2 ( x)]2 (Qoutient Rule)

310

Illustration 1 : (i)

e

x

Find the derivatives of the following functions with respect to x cos x [UPTU 2004]

(iii) x log x − x

(iv) (1 + x 2 ) tan −1 x

(v)

(vi) e

sin x . log x

Solution :

(i)

d (e dx

x

cos x) = e =e

(ii)

(ii) x 3 log x

d 3 (x dx

x

x

+e

x

log x.

d d cos x + cos x e dx dx

x

(− sin x) + cos x . e d 3 3 d log x) = x log x + log x x dx dx 1 = x 3 . + ( log x) . 3 x 2 x

x

[UPTU 2007] x

=e

x

(cos x − sin x).

= x 2 + 3 x 2 log x = x 2 (1 + 3 log x). d d d ( x log x − x) = ( x log x) − ( x) dx dx dx d d = x⋅ ( log x) + ( log x) ⋅ ( x) − 1 dx dx 1 = x ⋅ + ( log x) . 1 − 1 = 1 + log x − 1 = log x. x d d d 2 −1 (iv) {(1 + x ) tan x} = (1 + x 2 ) tan −1 x + (tan −1 x) (1 + x 2 ) dx dx dx 1 = (1 + x 2 ) ⋅ + (tan −1 x) . (0 + 2 x) 1 + x2 (iii)

= 1 + 2 x tan −1 x. d d d (sin x log x) = sin x (log x) + log x (sin x) dx dx dx 1 = sin x . + log x . (cos x) x sin x = + log x (cos x). x d d d (vi) (e x + e x log x) = (e x ) + (e x log x) dx dx dx d d =ex +ex ( log x) + log x (e dx dx 1 = e x + e x ⋅ + log x ⋅ e x x 1 = e x 1 + + log x ⋅   x (v)

x

)

311

Illustration 2 : (i) (iii)

3x

2

Find the derivatives of the following functions with respect to x.

−2

x2 + 7 x2 + 2 x + 4

(iv)

x+4

Solution :

ex 1 + sin x

(ii)

d (i) dx

e ax − e − ax e ax + e − ax

d d 2 2 2 2  3 x 2 − 2 ( x + 7) dx (3 x − 2) − (3 x − 2) dx ( x + 7)   = 2 ( x 2 + 7)2  x +7 [By quotient rule for differentiation] = =

(x

6 x 3 + 42 x − 6 x 3 + 4 x (x

=

(iii)

d dx

2

2

+ 7)

(1 + sin x)

=

46 x (x

2

+ 7)2

⋅

d x d e −ex (1 + sin x) dx dx (1 + sin x)2

(1 + sin x) e

x

−e

x

cos x

2

(1 + sin x)

=

e

x

(1 + sin x − cos x) (1 + sin x)2

⋅

d 2 d 2  x 2 + 2 x + 4 ( x + 4) dx ( x + 2 x + 4) − ( x + 2 x + 4) dx ( x + 4)   = x + 4  ( x + 4)2  = =

(iv)

+ 7)(6 x) − (3 x 2 − 2)(2 x) ( x 2 + 7)2

d  ex    = dx 1 + sin x 

(ii)

2

ax − ax d e − e   ax = dx  e + e − ax 

= = = =

( x + 4) (2 x + 2) − ( x 2 + 2 x + 4) (1) ( x + 4)2 2 ( x + 4) ( x + 1) − ( x + 2)2 ( x + 4)2

(e ax + e − ax )

⋅

d ax d ax (e − e − ax ) − (e ax − e − ax ) (e + e − ax ) dx dx (e ax + e − ax )2

(e ax + e − ax ) (ae ax + ae − ax ) − (e ax − e − ax ) (ae ax − ae − ax ) (e ax + e − ax )2 a (e ax + e − ax )2 − a (e ax − e − ax )2 (e ax + e − ax )2 a {(e2 ax + e − ax + 2) − (e2 ax + e −2 ax − 2) (e ax + e − ax )2 4a (e ax + e − ax )2

312

Comprehensive Exercise 2 Differentiate the following functions with respect to x. 1.

x2 e x .

2. x 3 sin x. 4. (1 + x 4 ) cos x.

3. sec x tan x.

Find the derivatives of the following functions with respect to x. x + sin x 2 x+3 5. 6. 2 ⋅ ⋅ x + cos x x −5 7. 9.

x ⋅ 1 + tan x log x x

8.

e

1 + x2

⋅

x2 + 5 x − 6

10.

⋅

x

⋅

4 x2 − x + 3

A nswers x

2. x 2 (3 sin x + x cos x).

1.

x ( x + 2) e

3.

sec x (sec 2 x + tan2 x).

5. 7. 9.

⋅

x (cos x + sin x) + cos x − sin x + 1 ( x + cos x)2 1 + tan x − x sec 2 x (1 + tan x)2 1 − log x x2

⋅

⋅

4. 4 x 3 cos x − (1 + x 4 ) sin x. ⋅

6. 8. 10.

−2 ( x 2 + 3 x + 5) ( x 2 − 5)2 e

x

(1 − x)2

(1 + x 2 )2

⋅

⋅

9 + 54 x − 21 x 2 (4 x 2 − x + 3)2

⋅

9.7 Differential Coefficient of a Function of a Function (Chain Rule of Differentiation) Consider the function log sin x. Here log (sin x) is a function of sin x whereas sin x is itself a function of x. Thus we have case of a function of a function. If y is a function of t and t is a function of x,then y is also a function of x and we have dy dy dt = ⋅ ⋅ dx dt dt Similarly, if y is a function of u, u is a function of v and v is a function of x, then y is also a function of x and we have,

313

dy dy du dv = ⋅ ⋅ dx du dv dx derivative of y w.r.t. x

i. e.,

=(derivative of y w.r.t. u).(derivative of u w.r.t v).(derivative of v w.r.t. x).

Illustration 1 : (i) cosec

5

Find the derivatives of the following functions with respect to x. (ii) e

x

(iii) tan (sin x) Solution :

(iv) e

sin x tan −1 x

(i) Let y = cosec 5 x.

Put cosec x = t. Then y = t 5 and t = cosec x. ∴ and Now

dy

d 5 = (t ) = 5t 4 dt dt dt d = cosec x = − cosec x cot x. dx dx dy dy dt = ⋅ = 5t 4 . (− cosec x cot x) dx dt dx = 5 cosec4 x. (− cosec x cot x) = − 5 cosec 5 x cot x.

Hence,

d cosec 5 x = − 5 cosec 5 x cot x. dx

(ii) Let y = esin x . Put sin x = t. Then y = e t and t = sin x. ∴ Now

dy dt dy dx

=e =

t

and

dt = cos x. dx

dy dt ⋅ = e t . cos x = esin x . cos x. dt dx

(iii) Let y = tan (sin x). Put sin x = t. Then y = tan t. ∴ Now

dy dt dy dx

(iv) Let y = e tan ∴

Now

dy dt

= sec 2 t and = −1

dy dt x

⋅

dt = cos x. dx

dt = (sec 2 t) . (cos x) = (cos x) sec 2 (sin x). dx

. Put tan −1 x = t. Then y = e t and t = tan −1 x.

=e

t

dy

and

dt 1 = ⋅ dx 1 + x 2 −1

dt 1 e tan x = ⋅ =et ⋅ = ⋅ dx dt dx 1 + x2 1 + x2

dy

314

Illustration 2 : (i) sin

−1

(ii) log5 (log5 x)

( x)

1 (iii) log e  x +   x Solution :

Find the derivatives of the following functions with respect to x

1 1 ⋅ x 1− x 2

d 1 log 5 ( log 5 x) = dx log 5 x . log =

(iii)

[UPTU 2001 (Special)]

d 1 d 1 /2 (i) sin −1 ( x ) = ⋅ x 2 dx dx 1 − ( x) =

(ii)

(iv) log3 ( x 2 + 3 x).

[UPTU 2002]

1 log 5 x . log

d 1 1 log e  x +  =   dx x  x + 1  x

=

e

e

−1 /2

=

2

d ⋅ log 5 dx 1 ⋅ 5 x log

e

5

5

1 ⋅ x (1 − x) x =

1 (log

e

2

5) . x log

5

x

⋅

d  1  x +   dx x

1

  1  1 + − 2  =  1  x  x+  x

1  1 − 2   x  x2 + 1

=

( x 2 − 1) x (x 2 + 1)

⋅

x d 1 d (iv) log 3 ( x 2 + 3 x) = 2 ( x 2 + 3 x) dx ( x + 3 x) log e 3 dx =

1 (x

2

+ 3 x) log e 3

(2 x + 3).

Illustration 3 : (i)

Find the derivatives of the following functions with respect to x. 1 − tan x x (ii) log tan    2 1 + tan x

(iii) log log log x 3 [UPTU 2007]

(iv) sin ( sin x + cos x )

(v) log sin ( x 2 + 1). [UPTU 2004]

(vi) sin (a tan −1 x).

Solution :

[UPTU 2005]

d x 1 d x (i) log tan   = ⋅ tan      dx 2 tan ( x / 2) dx 2 =

1 d  x  ⋅ sec 2 ( x / 2) ⋅   tan ( x / 2) dx  2 

=

1 1 d ⋅ ⋅ ( x) sin ( x / 2) cos ( x / 2) 2 dx

=

1 1 = = cosec x. 2 sin ( x / 2) cos ( x / 2) sin x

315

(ii)

d 1 − tan x    dx 1 + tan x 

1 /2

=

= = (iii)

1 1 − tan x    2 1 + tan x 

1 1 + tan x    2 1 − tan x 

⋅

1 /2

d 1 − tan x    dx 1 + tan x 

(1 + tan x).(− sec 2 x) − (1 − tan x).sec 2 x

⋅

(1 + tan x)2

1 /2 − 2 sec 2 x − sec 2 x 1 (1 + tan x) ⋅ = ⋅ 2 (1 − tan x)1 /2 (1 + tan x)2 (1 − tan x)1 /2 (1 + tan x)3 /2

d 1 d log log log x 3 = ⋅ log log x 3 3 dx dx log log x =

1

⋅

log log x 3 1

=

log log x

3

⋅

1 log x 3

⋅

d log x 3 dx

1 d ⋅ (3 log x) 3 log x dx

1

= (iv)

−1 / 2

log log x

3

⋅

1 1 1 ⋅3⋅ = ⋅ 3 log x x ( x log x)(log log x 3 )

d d sin ( sin x + cos x ) = cos ( sin x + cos x ) ⋅ sin x + cos x dx dx 1 d = cos ( sin x + cos x ) ⋅ (sin x + cos x) −1 /2 ⋅ (sin x + cos x) 2 dx =

(cos x − sin x) cos ( sin x + cos x )

⋅

2 sin x + cos x (v)

d 1 d log sin ( x 2 + 1) = sin ( x 2 + 1) 2 dx dx sin ( x + 1) =

(vi)

1 sin ( x

2

+ 1)

cos ( x 2 + 1)

d 2 ( x + 1) = 2 x cot ( x 2 + 1). dx

d d sin (a tan −1 x) = cos (a tan −1 x) a (tan −1 x) dx dx a = cos (1 tan −1 x). a + x2

Illustration 4 :

Find

dy dx

by using chain rule for the following functions

y = u2 + u + 2, u = v2 + 5 , v = 6 x + 7 Solution :

We have y = u2 + u + 2 , u + v2 + 5 , v = 6 x + 7

∴

dy du

= 2u + 1 ,

du dv = 2v , =6 dv dx

Now by chain rule, dy dy du dv = . . = (2u + 1) . (2v) . (6) dx du dv dx

316

= {2 (v2 + 5) + 1} .12v = (2u2 + 11) . 12v = {2 (6 x + 7)2 + 11} . 12 (6 x + 7) = 12 (6 x + 7) (72 x 2 + 168 x + 109) Illustration 5 : Solution :

If y = z + 1 , z = w + 2 , w = u2 , u = 4 x + 5, find

dy dw dz dy , , and . dw dx dx dx

We have y = z + 1, z = w + 2, w = u2 , u = 4 x + 5. dy

dz dw du = 1, = 2u , =4 dw du dx dy dz = = 1⋅ 1 = 1. dw dz dz dw dw du = ⋅ = 2u ⋅ 4 = 8u dx du dx

∴

dz dy

Now

= 1,

= 8 (4 x + 5). dz dz dw du = ⋅ ⋅ = 1⋅ 2 u ⋅ 4 dx dw du dx = 8u = 8 (4 x + 5) dy dz dw du = ⋅ ⋅ ⋅ dx dz dw du dx

dy

= 1⋅ 1⋅ 2u ⋅ 4 = 8u = 8 (4 x + 5) Illustration 6 :

If y = u2 + 1and u = x 3 + x + 4,find

dy dx

(i) Without using chain rule

(ii) Using chain rule. Verify that the derivatives are same. Solution :

We have y = u2 + 1 and u = x 3 + x + 4

(i) Without using chain rule Now

y = u2 + 1 = ( x 3 + x + 4)2 + 1

[∵ u = x 3 + x + 4]

= x 6 + x 2 + 16 + 2 x 4 + 8 x + 8 x 3 + 1 = x 6 + 2 x 4 + 8 x 3 + x 2 + 8 x + 17 ∴

dy dx

= 6 x 5 + 8 x 3 + 24 x 2 + 2 x + 8.

...(1)

(ii)

Using chain rule dy du Now = 2u , = 3 x2 + 1 du dx dy dy du ∴ = ⋅ = (2u) ⋅ (3 x 2 + 1) dx du dx = 2 ( x 3 + x + 4) (3 x 2 + 1) = 6 x 5 + 8 x 3 + 24 x 2 + 2 x + 8 From (1) and (2) we see that derivatives are same.

...(2)

317

Comprehensive Exercise 3 1. Differentiate the following functions with respect to x. (i) cos (ax + b).

(ii) tan (3 x / 2).

2

(iii) sin x .

(iv) e cos x .

(v) tan −1 (5x).

(vi) sin −1 e x .

(vii) sin [cos (tan x)]. 2.

Find

dy dx

[UPTU 2006]

by using chain rule for the following functions :

(i) y = u2 + u + 5, u = x 2 − x + 1 (ii) y =

z , z = 5 x4 + 3

(iii) y = u2 , u = x 2 + 4 (iv) y = v2 + v + 1 , v = u2 + 1 , u = 6 x (v) y = v3 + 2v2 + 4 , v = 3u + 7 , u = 9 x + 3 (vi) y = 2u2 + 3 , u = 12v2 + v , v = 3w + 8 , w = 2 x 3.

If y = u2 and u = 1 + x 3 , find the derivative of y w.r. to x : (i) without using chain rule (ii) using chain rule.

A nswers 1.

(i) − a sin (a x + b).

(ii)

(iii) 2 x cos x 2 . (v)

5 1 + 25 x

2

3 sec 2 (3 x / 2). 2 (iv) − sin x . e (vi)

⋅

e

cos x

.

x

1 − e2

x

⋅

(vii) − [cos {cos (tan x)}] [sin (tan x)] [sec2 x]. 2.

(i) (2 x − 1) (2 x 2 − 2 x + 3)

(ii) 10 x 3 / (5 x 4 + 3)

(iii) 4 x ( x 2 + 4)

(iv) 72 x (72 x 2 + 3)

(v) 27 (2187 x 2 + 2700 x + 832) (vi) 24 (62208 x 3 + 250128 x 2 + 335238 x + 149768) 3.

6 x 2 (1 + x 3 ).

318

9.8 Differential Coefficients Of Inverse Trigonometric Functions d 1 d 1 d 1 sin − 1 x = , cos − 1 x = − ; tan − 1 x = ; 2 2 dx √ (1 − x ) dx √ (1 − x ) dx 1 + x2 d 1 d 1 cot − 1 x = − ; sec − 1 x = ; 2 dx dx 1+ x x √ ( x 2 − 1) d 1 cosec − 1 x = − ⋅ dx x √ ( x 2 − 1)

Illustration 1 : (i) tan

−1

Find the derivatives of the following functions with respect to x. (ii) e

log x

(iii) (cot −1 x)2 Solution :

(i)

(iv) cos −1 (cot x). d 1 d {tan −1 (log x)} = ⋅ log x 2 dx dx 1 + ( log x) =

(ii)

d { e sin dx

−1

( x + 1)

} = e sin =e

(iii)

−1

1 2

1 + ( log x)

( x + 1)

sin −1 ( x + 1)

⋅

⋅

⋅

1 1 = ⋅ x x [1 + ( log x)2 ]

d sin −1 ( x + 1) dx 1 1 − ( x + 1)2

−1

d e sin ( x + 1) ⋅ ( x + 1) = ⋅ dx 1 − ( x + 1)2

 d d 1   (cot −1 x)2 = 2 cot −1 x ⋅ cot −1 x = 2 cot −1 x ⋅  − 2 dx dx  1+ x  =

(iv)

sin −1 ( x + 1)

− 2 cot −1 x 1 + x2

⋅

  d d 1 ⋅ {cos −1 (cot x)} =  − (cot x)  2 dx 1 − cot x  dx    1  ⋅ (− cosec 2 x) = −  2 1 − cot x   =

cosec 2 x 1 − cot 2 x

⋅

319

Illustration 2 : (i) tan (sin

−1

Find the derivatives of the following functions with respect to x (ii) sin (tan −1 x).

x)

(i) Let y = tan (sin −1 x).

Solution :

Putting sin −1 x = t, we get y = tan t and t = sin −1 x. dy

∴

= sec 2 t and

dt dy

Now

=

dx

dt = dx

1

⋅

1− x

2

dy dt 1 1 1 ⋅ = sec 2 t ⋅ = ⋅ 2 2 dt dx cos t 1 − x 2 1− x =

1

1

⋅

1 − sin2 t

1− x

=

2

1 1 − x2

⋅

1 1 − x2

[ ∵ t = sin −1 x ⇒ sin t = x ] = (ii)

1 (1 − x 2 )3 /2

⋅

Let y = sin (tan −1 x).

Putting

tan −1 x = t, we get y = sin t and t = tan −1 x. dy

∴

dt dy

Now

dx

=

dt 1 = ⋅ dx 1 + x 2

and

= cos t

dy dt 1 ⋅ = cos t ⋅ dt dx 1 + x2 = =

1 1 1 1 ⋅ = ⋅ 2 2 2 sec t 1 + x 1 + tan t 1 + x 1 1+ x

=

2

⋅

1

[ ∵ t = tan −1 x ⇒ tan t = x ]

1 + x2

1 (1 + x 2 )3 /2

⋅

Comprehensive Exercise 4

Find the derivatives of the following functions with respect to x. 1. 3. 5.

tan −1 (cot x). sec

−1

4. cot

x.

log (tan

−1

2. sin −1

x).

−1

6. (cot

−1

x. 2

x . x 2 )3 .

320

A nswers 1. –1.

1

2. 2

−2 x

4.

1 + x4

x (1 − x)

2x

1

5.

⋅

1

3.

⋅

(1 + x 2 ) tan −1 x

⋅

x −1

− 6 x (cot −1 x 2 )2

⋅ 6.

1 + x4

⋅

9.9 Logarithmic Differentiation Whenever we are required to differentiate a function of x in which a function of x is raised to a power which itself is a function of x, neither the formula for a x nor that for x n is applicable. In such cases we first take logarithm of the function and then differentiate. This process is called logarithmic differentiation. It is also helpful in the cases where we are to differentiate a function which consists of the product or the quotient of a number of functions.

Illustration 1 : (i)

x

(iii) x

Differentiate the following functions with respect to x.

x sin x

Solution :

(ii) x . [UPTU 2004] (i) Let y = x x .

sin −1 x

(iv) (sin x)

[UPTU 2003]

log x

Then log y = log x

x

= x . log x.

Differentiating both sides w.r.t. x, we get 1 dy 1 = x ⋅ + ( log x) ⋅ 1 = 1 + log x. y dx x ∴ Thus,

dy

= y (1 + log x) = x x (1 + log x). dx d x x = x x (1 + log x). dx

(ii) Let y = x sin Then

−1

x

.

log y = log ( x sin

−1

x

) = sin −1 x . log x.

Differentiating both sides w.r.t. x, we get  1 dy 1 1 = (sin −1 x) ⋅ + ( log x) ⋅   1 − x2 y dx x 

 ⋅  

321

∴

sin −1 x log x   = y + 2  dx  x 1 − x  

dy

= x sin

−1

x

sin −1 x log x   ⋅ + 2   x 1 − x  

(iii) Let

y = x sin x .

Then

log y = log ( x sin x ) = sin x log x.

Differentiating both sides w.r.t. x, we get 1 dy 1 = sin x . + log x . (cos x). y dx x ∴

dy

sin x  = y + (cos x) log x dx  x  sin x  = x sin x  + (cos x) log  x

 x ⋅ 

(iv) Let

y = (sin x) log x .

Then

log y = log {(sin x) log x } = log x . log sin x.

Differentiating both sides w.r.t. x, we get  1  1 dy 1 = ( log x) ⋅  ⋅ cos x + (log sin x) ⋅ ⋅ y dx x  sin x  ∴

log sin x   = y cot x log x +  dx x   log sin x   = (sin x) log x cot x log x +  ⋅ x 

dy

Illustration 2 : Solution : ∴

If y = x ( x

x

)

We have y = x ( x x

log y = x

, find

x

)

dy dx

⋅

[UPTU 2008]

.

. log x.

Differentiating both sides w.r.t. x, we get 1 dy 1 d = x x . + ( log x) ⋅ x y dx x dx ∴

dy

x

 = y x  dx

x −1

+ ( log x).  x . x 

= y[x

x −1

+ ( log x) . ( x

= x( x

x

)

[x

x −1

+ x

x

x

[By the product rule] x −1

+ x

x

⋅

d ( x) + x dx log x)]

(1 + log x) log x].

x

log x ⋅

d  ( x) dx  

322

Illustration 3 : Solution : ∴

dy

= Illustration 4 : Let

x

x

x

x

x

x

d ⋅  x dx  x

+ cosec 2 x ⋅

x

+ cosec 2 x

x

x

+ cosec 2 x)

+

d cosec 2 dx

x 

×

d ( x) + x dx

1 x

d (x dx x ⋅

1 2

x), find (dy / dx).

+ cosec 2 x).

+ cosec

+ cosec

x −1

2

2

1

 x . x 

∴

x

1

=

dx

=

Then

+ cosec

We have y = log ( x

=

Solution :

x

If y = log ( x

x

⋅ {x

log x ⋅ x

 d ( x) + 2 cosec x . (− cosec x cot x)  dx 

(1 + log x) − 2 cosec 2 x cot x}.

If y = (tan x) cot x + (cot x) tan x , find (dy / dx). u = (tan x)

cot x

and v = (cot x)

tan x

[UPTU 2009]

.

y = u + v. dy du dv = + ⋅ dx dx dx

…(1)

Now

u = (tan x) cot

⇒

log u = cot x . log tan x  1  1 du = cot x ⋅  ⋅ sec 2 x + ( log tan x) ⋅ (− cosec 2 x) u dx  tan x 

⇒

⇒

x

[Differentiating both sides w.r.t. x] du 2 2 = u [cosec x − cosec x log tan x ] dx = (tan x) cot x . cosec 2 x (1 − log tan x).

Again

v = (cot x) tan x .

∴

log v = tan x . log cot x.

Differentiating both sides w.r.t. x, we get  1  1 dv = (tan x) ⋅  ⋅ (− cosec 2 x) + ( log cot x) . sec 2 x v dx  cot x  dv 2 2 or = v [− sec x + sec x log cot x ] dx = (cot x) tan

x

sec 2 x {( log cot x) − 1}.

Putting the values of du / dx and dv / dx in (1), we get dy = (tan x) cot x . cosec 2 x (1 − log tan x) dx + (cot x) tan

x

. sec2 x {( log cot x) − 1}.

323

9.10 Implicit Functions If y is a function of x given by a relation of the type y = f ( x), then y is said to be an explicit function of x. On the other hand, if the relation between x and y is given by an equation involving both x and y s.t. y cannot be expressed as y = f ( x), then y is said to be an implicit function of x. If we are given y implicitly in terms of x, we can find dy / dx without first expressing y explicitly in terms of x. Thinking of y as a function of x, we differentiate both sides of the given equation with respect to x and then solve the resulting relation for dy / dx.

Illustration 1 : Solution :

If x

y

=e

x− y

, prove that

We are given that x

y

=e

dy dx

x− y

=

log x (1 + log x)2

⋅

.

Taking logarithm of both sides, we get y log x = ( x − y) or or ∴

y (1 + log x) = x x y= ⋅ 1 + log x dy dx

= =

Illustration 2 :

(1 + log x)2 1 + log x − 1 2

(1 + log x)

=

log x (1 + log x)2

⋅

If sin y = x sin (a + y), prove that

dy dx Solution :

(1 + log x).1 − x . {0 + (1 / x)}

=

sin (a + y) cos y − x cos (a + y)

⋅ [UPTU 2008]

Given that sin y = x sin (a + y).

Differentiating both sides of (1) with respect to x, we get dy d d cos y ⋅ = x . cos (a + y) ⋅ (a + y) + sin (a + y) ⋅ ( x) dx dx dx dy dy or cos y = x cos (a + y) + sin (a + y) dx dx dy or {cos y − x cos (a + y)} = sin (a + y) dx sin (a + y) dy or = ⋅ dx cos y − x cos (a + y)

…(1)

324

Illustration 3 :

If x (1 + y) + y (1 + x) = 0, show that

dy

=−

dx Solution :

1 ( x + 1)2

⋅

Given that x (1 + y) + y (1 + x) = 0

or

x (1 + y) = − y (1 + x)

or

x 2 (1 + y) = − y 2 (1 + x)

or

x 2 + x 2 y = y 2 + xy 2

or

x 2 − y 2 = xy 2 − x 2 y

or

( x + y) ( x − y) = − xy ( x − y)

or

x + y = − xy or x + y + xy = 0 x x + y (1 + x) or y = − 1+ x

or ∴

dy dx

(1 + x) y

= ex +

Given that e x + e

or

ex + e

or

e

y

=

2

If e x + e

Illustration 4 : Solution :

(1 + x) − x

=−

y

= ex e

y

y

[On squaring both sides]

−1 (1 + x)2 y

.

, show that

= ex +

[∵ ( x − y) ≠ 0]

dy dx

+e

= 0.

y

or e x = e

y

(e x − 1)

ex

=

y− x

....(1)

ex −1

Differentiating both sides of (1) w.r. to x, we get e

y

dy

=

dx

dy dx

e

y

(e x − 1)2

− ex −

=

=

− ex (e x − 1)2

− ex −

y

(e x − 1)2

y

 x ex  ∵ e − 1 = y  e  

(e x / e y )2

=−

or

(e x − 1)2 − ex

=

=

(e x − 1)2 e2 x − e x − e2 x

= or

(e x − 1)e x − e x (e x − 0)

ex − (e

y

x− y 2

)

=−

= − e y− x dy + e y − x = 0. dx

1 e

x− y

= − e− (x −

y)

325

Illustration 5 : Solution :

Find

dy dx

for the following implicit function xy + xe − y + ye x = x 2 .

Given that xy + xe − y + ye x = x 2

Differentiating both sides w.r. to x, we get dy dy   dy     + y ⋅ 1 +  x (− e − y ) + e − y  +  ye x + e x x  = 2x  dx     dx dx  dy or ( x − xe − y + e x ) = 2 x − y − e − y − ye x dx dy 2 x − y − e − y − ye x or . = dx x − xe − y + e x Illustration 6 : Solution :

Find

dy dx

for the following implicit function sin ( xy) +

Given that sin ( xy) +

x = x2 − y y

x = x2 − y y

Differentiating both sides w.r.t. x, we get  dy  cos ( xy)  x + y ⋅ 1 +  dx  dy

y ⋅1 − x y

2

dy dx = 2 x − dy dx

dy 1 x dy − 2 = 2x − y dx y dx

or

x cos ( xy)

or

 dy  x 1 − y cos ( xy)  x cos ( xy) − 2 + 1 = 2 x − dx  y y 

or

dy  xy 2 cos ( xy) − x + y 2  2 xy − 1 − y 2 cos ( xy)  = dx  y y2  dy

or

dx

=

dx

+ y cos ( xy) +

y (2 xy − 1 − y 2 cos ( xy)) xy 2 cos ( xy) − x + y 2

Comprehensive Exercise 5 1. Find ( dy / dx ) when x and y are connected by the following relations: (i)

x2 a2

+

y2 b2

= 1.

(ii) y 2 = 4a {x + a sin ( x / a)}. (iii) x 2 /3 + y 2 /3 = a2 /3 (iv) ax 2 + 2hxy + by 2 = 1.

326

(v) 3 x 2 y − y 3 − 2 = 0. (vi) cos ( x + y) = y sin x. (vii) x 3 + y 3 = 3axy (viii) y sec x + tan y + xy = 0 2. Find the derivatives of the following functions with respect to x. (i) (sin x) x . (ii) x1 / x . (iii) (2 x + 3)

(iv) ( x + 1) x .

.

cos −1 x

(v) ( log x) x .

(vi) x

(vii) x sin x + cos x . (ix) e x cos x + x sin x . [UPTU 2004]

(viii) x

(x) x log 3.

x −5

x

x

.

.

+ (log x) x

Find ( dy / dx) of the following functions : (i) y tan x − y 2 cos x + 2 x = 0 (ii) sin y + log y = x 2 + 18 x + 3 (iii) tan ( x + y) + tan ( x − y) = 1 (iv) y 2 sin x + y tan x + (1 + x 2 ) cos x = 0 ( x 2 + y 2 ) = tan −1

(v) log 4.

y x

(i) If y = x sin y , show that x (ii) If xy = tan ( xy), show that

dy

=

dx dy

y 1 − x cos y y

=−

dx dy (iii) If sec y = x 3 , show that x = dx

x

.

. 3

. x6 − 1

A nswers 1.

(i) −

b2 x a

2

(iii) − (v) (vii)

y

y1 /3

x1 /3 2x y

y2 − x2 ay − x 2 y

2

(ii)

⋅

− ax

2a [1 + cos ( x / a)]

⋅

(iv) −

⋅

(vi) −

y ax + hy hx + by

⋅

⋅

sin ( x + y) + y cos x

(viii) −

sin x + sin ( x + y) y sec x tan x + y sec x + sec 2 y + x

⋅

327 x

2. (i) (sin x) 1/ x

(ii)

x

[ x cot x + log sin x].

(1 − log x) x2

(iii) (2 x + 3) x

(iv) ( x + 1) (v) ( log x)

(vi) x

2 ( x − 5)  + log (2 x + 3) ⋅   2x + 3 

 x  + log ( x + 1) ⋅   x +1   1  + log log x ⋅  log x  

cos −1 x log x   ⋅ − 2  x  1 − x   sin x + cos x x + cos x  + (cos x − sin x) log  x

cos −1 x

(vii) x sin (viii) x (ix) e

x

x −5

⋅

x

 x ⋅ 

 x log x  +  ⋅ 2 x   x

x

(cos x − sin x) + x sin

x

sin x  x + (cos x) log

 x ⋅ 

x  2 log x   x 1 (x) x log  + log (log x)  + (log x)   x   log x 

3.

(i) (ii)

y sec 2 x + y 2 sin x + 2 2 y cos x − tan x 2 y ( x + 9) y cos y + 1

(iii) − (iv) (v)

sec 2 ( x + y) + sec 2 ( x − y) sec 2 ( x + y) − sec 2 ( x − y)

(1 + x 2 ) sin x − (2 x + y 2 ) cos x − y sec 2 x 2 y sin x + tan x x+ y x− y

.

9.11 Parametric Equations If x and y are both expressed in terms of a third variable, say t, then t is usually called a parameter. In the case of parametric equations we can always find dy / dx, without first eliminating the parameter.

328

Thus, if the parametric equations are x = φ ( t ), y = ψ ( t ), then dy dy dt dy dy dx or = ⋅ = ⋅ dx dt dx dx dt dt

9.12 Differentiation of a Function with Respect to a Function Suppose we are to find the differential coefficient of the function u = f ( x) with respect to another function, say, v = φ ( x). It means we are to find du / dv, where u and v are both given in terms of a third variable x. Therefore, as in the case of parametric equations, we have du du / dx = , dv dv / dx

i.e.,

dφ ( x ) ⋅ dx

df ( x) df ( x) = dφ( x) dx

(i) If x = at 2 and y = 2at, find dy / dx ⋅

Illustration 1 :

[UPTU 2005]

1 (ii) If x = a (cos t + log tan t ), y = a sin t, find dy / dx. 2 dy dx Solution : (i) Here = 2at and = 2a. dt dt dy dy / dt 2a 1 Now, = = = ⋅ dx dx / dt 2at t (ii) Here

 dx 1 t 1 = a  − sin t + ⋅ sec2  ⋅    2 dt tan ( t / 2 ) 2    1 = a  − sin t + 1 1  2 sin t cos t 2 2 

Also Now

 1 − sin2 t  cos 2 t    = a = a ⋅   sin t  sin t    

dy / dt = a cos t. dy dy / dt a cos t = = = tan t. dx dx / dt (a cos 2 t) / sin t 3

θ and y = a tan3 θ, find dy / dx at θ = π / 4.

Illustration 2 :

If x = a sec

Solution :

dx = 3a sec 2 θ . sec θ tan θ = 3a sec 3 θ tan θ dθ

and ∴

Here dy dθ dy dx

= 3a tan2 θ . sec 2 θ = 3a sec 2 θ tan2 θ. =

dy / dθ dx / dθ

=

3a sec 2 θ tan2 θ 3a sec

3

θ tan θ

=

tan θ sec θ

= sin θ.

329

∴

 dy  π 1 = sin = ⋅   4 2  dx  at θ = π /4

Illustration 3 : Solution :

Let

Differentiate x u = x sin

−1

sin− 1 x

with respect to sin −1 x.

and v = sin − 1 x.

x

Then

log u = sin −1 x. log x.

∴

1 du 1 1 = log x + sin − 1 x 2 u dx √ (1 − x ) x

or

−1 du = x sin x dx

Again

dy dx

=

 log x sin − 1 +  2 x  √ (1 − x )

1 √ (1 − x 2 )

⋅ x sin

Now

x ⋅ 

−1

x

du du / dx = = dv dv / dx

 log x sin − 1 x  +   2 x   √ (1 − x ) 1 √ (1 − x 2 )

= x sin Illustration 4 : Solution :

−1

x

 x log x + √ (1 − x 2 ) sin − 1  x 

x ⋅ 

Differentiate log sin x with respect to cos x.

Let u = log sin x and v = cos x . Then to find du / dv. du 1 = ⋅ cos x = cot x dx sin x

We have

u = log sin x ⇒

and

v = cos x ⇒

∴

du du / dx = = (cot x) ÷ dv dv / dx

− sin x dv 1 = ⋅ (− sin x) = ⋅ dx 2 cos x 2 cos x  − sin x  2 cos x   = cot x ⋅ (− sin x)  2 cos x 

= − 2 cos x . cosec x cot x. Illustration 5 : Differentiate sin 2 x with respect to cos x. Solution : Let u = sin 2 x and v = cos x. du Then to find . dv du We have u = sin 2 x ⇒ = 2 cos 2 x, dx dv and v = cos x ⇒ = − sin x dx du du / dx 2 cos 2 x ∴ = = dv dv / dx − sin x = − 2 cos 2 x cosec x.

330

Comprehensive Exercise 6 1. Find dy / dx, when (i) x = (ii) x =

a (1 − t 2 ) 1+ t

2

3 1 + t3

, y=

, y=

2bt 1 + t2

3t 2 1 + t3

.

.

(iii) x = a cos t, y = b sin t. (iv) x = a sec θ, y = b tan θ. (v) x = sin 2t, y = 2 cos t. (vi) x = log t, y = sin t. (vii) x = a (θ − sin θ), y = a (1 − cos θ). (viii) x =

3at 1 + t2

2. (i) x = a e

θ

, y=

3at 2 1 + t2

⋅

(sin θ − cos θ), y = a e

θ

(sin θ + cos θ).

(ii) x = a (cos θ + θ sin θ), y = a (sin θ − θ cos θ).

[UPTU 2009]

(iii) x = 4 sin θ , y = 7 cos θ (iv) x = a cos 3 θ , y = a sin3 θ 3.

(i) Differentiate x 2 with respect to x 3 .  2x   with respect to tan −1 x. (ii) Differentiate tan −1  2 1 − x   (iii) Differentiate sin −1

2x 1 + x2

(iv) Differentiate tan −1

with respect to tan −1 x.

1 + x2 − 1 x

with respect to tan −1

2x 1 − x2

A nswers 1.

(i)

b (t 2 − 1)

2at b (iv) cosec θ. a (vii) cot θ / 2 .

t2 2 − 3 3t − sin t (v) ⋅ cos 2 t (ii)

(viii)

2t 1 − t2

(iii) − (vi) ⋅

b cot t . a

t cos t .

.

331

2.

(i) cot θ.

3. (i)

2 ⋅ 3x

−7 tan θ 4

(ii) tan θ.

(iii)

(ii) 2

(iii) 2

(iv) − tan θ (iv) 1/4.

9.13 Differentiation of Infinite Recurring Expressions If from an infinite collection of terms we take out a single term or a finite number of terms, the collection still remains infinite. We use this fact to find the derivatives of infinite recurring expressions.

Illustration 1 : prove that Solution :

dy dx

If y = =

sec

2

x

2y −1

+

tan x

tan x + … ∞ ,

+

⋅

It is given that y=

∴

tan x

tan x

y=

tan x

+

+

tan x + … ∞ .

tan x + y .

Squaring both sides, we get y 2 = tan x + y. Now differentiating both sides w.r.t. x, we get 2y or or

dy dx dy dx

Illustration 2 : Solution : ∴

dy dx

= sec 2 x +

dy dx

(2 y − 1) = sec 2 x =

sec 2 x 2y −1

⋅

If y = ( x )(

x )( x) … ∞

It is given that y = ( x )(

, prove that x )( x) … ∞

y = ( x) y .

Taking logarithm of both sides, we get log y = y log x1 /2 or

log y =

y 2

log x.

.

dy dx

=

y2 x (2 − y log x)

⋅

332

Now differentiating both sides w.r.t. x, we get 1 dy 1  dy 1 =  . ( log x) + y .  y dx 2  dx x or

y 1 1  dy =  − log x y 2  dx 2 x

or

y 2 − y log x  dy =   2y   dx 2 x dy

or

dx

Illustration 3 : Solution :

=

y2 x (2 − y log x)

If y = e

⋅

x + … to ∞ x+e x+e

, show that

dy

=

dx

From the given relation, we have y = e

y 1− y

x+ y

⋅

[UPTU 2003]

.

Taking logarithm of both sides, we get log y = x + y. Differentiating both sides w.r.t. x, we get dy 1 dy =1+ y dx dx or

dy  1   − 1 = 1 dx  y 

or

dy 1 − y ⋅ =1 dx y

dy

or

dx

=

y 1− y

⋅

Comprehensive Exercise 7

1. If y = x

x x… ∞

2. If y = sin x

, prove that

+

x)(sin x)… ∞

4. If y =

x

x

5. If y =

log x +

+

+

dx

=

y2 x (1 − y log x)

⋅

+ sin x + … to ∞ , prove that

sin x

3. If y = (sin x)(sin

dy

, prove that

dy dx

=

y 2 cot x 1 − y log sin x

x + … to ∞ , prove that

log x +

dy dx

=

dy dx

=

cos x 2y −1

⋅

⋅

1 ⋅ 2y −1

log x + … to ∞ , prove that (2 y − 1)

dy dx

=

1 ⋅ x

333

9.14 Derivatives of Second and Higher Order (Successive Differentiation) Let y = f ( x) be a differentiable function of x ; then its derivative y ′ = the first derivative of y with respect to x. If the first derivative differentiable, then its derivative i. e.,

dx dy dx

is called itself is

d  dy    is called the second derivative of y or dx  dx 

the derivative of y of second order and is denoted by y ′ ′ or derivative of

dy

d2 y dx 2

⋅ Similarly, the

d2 y

d3 y

is called the third derivative of y and is written as y ′ ′ ′ or ⋅ dx 2 dx 3 In general, the nth derivative of y or the differential coefficient of y of order n is denoted by dny and is obtained by differentiating y, n times successively with respect y (n) = dx n to x. The method of obtaining successive derivatives is called successive differentiation. The successive derivatives of y = f ( x) are also denoted as : Dy, D2 y, D3 y, …, D n y, … y1 , y2 , y3 , … , y n , … (n)

f ′ ( x), f ′ ′ ( x), f ′ ′ ′ ( x), …, f

( x), …

The value of the nth derivative of y = f ( x) at x = a is denoted by ( y n ) x = a or by ( y n )(a) or by f (n) (a).

Illustration 1 : Solution :

dy

=

dx d2 y dx

2

Illustration 2 : Solution :

x

We have y = e

∴ ∴

If y = e

=

+ sin x, find x

d2 y dx 2

[UPTU 2006]

+ sin x.

d d (e x ) + (sin x) = e dx dx

d dx

⋅

d  dy  (e   =  dx  dx

x

x

+ cos x.

+ cos x) =

d d (e x ) + (cos x) = e dx dx

Find the second derivative of e3 x sin 4 x.

Let y = e3 x sin 4 x.

x

− sin x.

334

Then ∴

dy

= 3e3 x sin 4 x + 4e3 x cos 4 x = e3 x (3 sin 4 x + 4 cos 4 x).

dx d2 y dx

2

=

d dx

 dy  d 3x {e (3 sin 4 x + 4 cos 4 x)}   =  dx  dx

= 3e3 x (3 sin 4 x + 4 cos 4 x) + e3 x (12 cos 4 x − 16 sin 4 x) = e3 x (24 cos 4 x − 7 sin 4 x). Illustration 3 :

(i) If y = A cos nx + B sin nx , show that

(ii) If y = cosec x + cot x , show that sin x Solution : ∴ and

d2 y dx 2

d2 y dx 2

+ n2 y = 0.

= y2 .

(i) We have y = A cos nx + sin nx. dy dx

= − A n sin nx + B n cos nx

d2 y dx 2

= − A n ⋅ n cos nx − B n ⋅ n sin nx = − n2 ( A cos nx + B sin nx) = − n2 y

Hence,

d2 y dx

2

+ n2 y = 0.

(ii) We have y = cosec x + cot x dy ∴ = − cosec x cot x − cosec 2 x. dx and

d2 y dx 2

= − cosec x (− cosec 2 x) − ( cosec x cot x) cot x − 2 cosec x (− cosec x cot x) 3

= cosec x + cosec x cot ∴

sin x

d2 y dx

2

2

x + 2 cosec 2 x cot x

= sin x cosec x (cosec 2 x + cot 2 x + 2 cosec x cot x)

= (cosec x + cot x)2 = y 2 Hence,

sin x

Illustration 4 :

d2 y dx

2

= y2 .

If y = (sin −1 x)2 , prove that (1 − x 2 )

d2 y dx

2

− x

dy dx

− 2 = 0. [UPTU 2011]

Solution : ∴

We have y = (sin dy dx

= 2 sin −1 x

−1

2

...(1)

x) . −1

2 sin x d (sin −1 x) = dx 1 − x2

335

Squaring both sides, we get  dy  (1 − x 2 )    dx 

2

 dy  (1 − x 2 )    dx 

2

or

 dy  (1 − x 2 )    dx 

2

or

= 4 (sin −1 x)2 [∵ from (1)]

=4y

...(2)

−4y = 0

Differentiating both sides again w.r.t. x, we have (1 − x 2 ) 2 or

2 (1 − x 2 )

or

2

or

(1 − x 2 )

Illustration 5 : Solution :

2 dy d  dy   dy  =0   +   (− 2 x) − 4 dx dx  dx   dx  dx

dy

⋅

dy d2 y ⋅ − 2x dx dx 2

 dy     dx 

2

−4

dy dx

=0

2  dy  dy 2 d y ( 1 − x ) − x − 2 = 0  2 dx  dx dx 

d2 y dx

2

− x −1

If y = e a sin

x

We have y = e sin

−1

dy

−1

= e a sin

∴

y1 =

or

y1 1 − x 2 = ay

dx

dy dx

dy   ∵ 2 dx ≠ 0

− 2 = 0.

, prove that (1 − x 2 ) y2 − xy1 − a2 y1 = 0. x x

...(1) .

a 1 − x2

=

ay 1 − x2 ...(2)

Squaring both sides of (2), we get (1 − x 2 ) y12 = a2 y 2 or

(1 − x 2 ) y12 − a2 y 2 = 0

...(3)

Differentiating both sides of (3) w.r.t. x, we get (1 − x 2 ) 2 y1 y2 + y12 (− 2 x) − a2 . 2 yy1 = 0 or

2 y1 [(1 − x 2 ) y2 − xy1 − a2 y] = 0

or

(1 − x 2 ) y2 − xy1 − a2 y = 0

Illustration 6 : Solution : ∴ ∴

Find

d

2

dx

y

2

, if x = at 2 and y = 2at.

We have x = at 2 and y = 2at dy dx = 2at and = 2a dt dt dy dy / dt 2a 1 = = = . dx dx / dt 2at t

[∵ 2 y1 ≠ 0]

336

d2 y

∴

dx

=

2

=

d dx −1 t2

=− Illustration 7 : Solution :

d  1  d  1   dy    =     =  dx  dx  t   dt  t   1 ∵ .  2at 1

dt dx dx dt 1  = 2at ⇒ = dt dx 2at 

2at 3

If x = a cos 3 θ , y = a sin3 θ ,then find 3

d2 y dx 2

. [UPTU 2010]

3

We have x = a cos θ , y = a sin θ dy dx = − 3a cos 2 θ sin θ and = 3a sin2 θ cos θ dθ dθ

∴

dy

∴

dx

dy / dθ

=

d2 y

dx / dθ

=

3a sin2 θ cos θ − 3a cos 2 θ sin θ

= − tan θ

d d dθ  dy  (− tan θ) =  (− tan θ) .   =    dx  dx  dθ  dx dx 1 = − sec 2 θ . − 3a cos 2 θ sin θ

∴

2

=

d dx

 dx  dθ 1 = − 3a cos 2 θ sin θ ⇒ = ∵  2 dx − 3a cos θ sin θ  dθ =

1 sec 4 θ cosec θ. 3a

Comprehensive Exercise 8 Find the second derivatives of the following functions. 1. (i) x sin x.

(ii) log x − x. x

(iii) log log x.

(iv) e

sin x.

(v) e4 x sin 3 x.

(vi) cos −1 x.

(vii) sin 3 x sin 5 x. d2 y

dy

2.

If y = tan −1 x, show that (1 + x 2 )

3.

If y = e m cos −1 x, show that (1 − x 2 ) y2 − xy1 − m2 y = 0.

4.

If y =

sin −1 x 1 − x2

dx

2

+ 2x

, then show that (1 − x 2 )

dx

d2 y dx

2

= 0.

− 3x

dy dx

− y=0

337

d2 y

cos x

5.

If y = tan x + sec x , prove that

6.

If x = a cos θ , y = b sin θ , show that

7.

If x = a (t − sin t) and y = a (1 + cos t) , prove that

8.

If x = a (cos θ + θ sin θ) and y = a (sin θ − θ cos θ) , show that d2 y dx 2

9.

=

sec 2 θ aθ

dx

2

=

(1 − sin x)2

d2 y dx

2

=−

b4 2

a y3

.

d2 y dx

2

=

1 cosec 4 4a

 t   2

.

If x = a (θ + sin θ) and y = a (1 − cos θ) , find

d2 y dx

2

at θ =

π . 2

A nswers 1. (i) 2 cos x − x sin x. (iii) −

1 + log x

⋅

( x log x)2

(ii) −1 / x 2 . (iv) 2e

x

cos x.

(v) e4 x (7 sin 3 x + 24 cos 3 x). (vi) −

9.

x (1 − x 2 )3 /2

⋅

(vii) 32 cos 8 x − 2 cos 2 x. 1 . a

Comprehensive Exercise 9 Fill in the Blanks Fill in the blanks ‘‘............’’, so that the following statements are complete and correct. dy 1. If y = a x , then = .......... . dx dy 2. If y = cos x , then = .......... . dx [UPTU 2011] dy 3. If y = log a x, then = .......... . dx

338

4. If y = x 2 + 2 x , then 5. If 6. If 7. If 8. If 9. If 10. If 11. If

dy

= .......... . dx dy y = sec −1 x, then = .......... . dx dy y = (2 x + 3)5 , then = .......... . dx dy y = cos 2 x 2 , then = .......... . dx dy y = cos −1 2 x, then = .......... . dx dy x 3 + y 3 = 3axy, then = .......... . dx dy y = cosh −1 x, then = .......... . dx dy = .......... . y = e cos x , then dx

[UPTU 2009]

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). dy 12. If y = e − 3 x , then is dx 1 (a) 3e − 3 x (b) e − 3 x 3 − 1 −3 x −3 x (c) − 3e (d) e 3 dy 13. If y = log sec x, then is dx (a) cot x

(b) tan x

(c) tan x sec x 14. If y = 3 x + 2 , then

(d) sec x dy dx

is

1 x 3 log 3 a 1 (c) (3 x log x) 9 dy x 15. If y = sin −1 , then is a dx 1 (a) 2 x − a2 (a)

(c)

1 2

2

(a − x )

(b) 9 (3 x log 3) (d) 9 (3 x log x)

(b) −

1 2

(a − x 2 ) 1

(d) − (x

2

− a2 )

339

16. If y = cot −1 (1 / x), then (a) (c)

dy dx

is

1

(b) x 2

x2 1 1+ x

17. If y = e cot (a) e cot

x

(c) e − cot

(d)

2 x

, then

dy dx

1 1 − x2

is

(cosec2 x)

(b) − e cot

x

(d) − e − cot

(cosec2 x)

18. If y = sin x ° then

dy dx

x

(cosec2 x) x

(cosec2 x)

is

(a) cos x° π (c) cos x° 180

(b) − cos x ° (d) None of these

[UPTU 2011]

True or False Write ‘T’ for true and ‘F’ for false statement. 19.

The derivative of the function (a2 − x 2 )5 /2 with respect to x is 5 x (a2 − x 2 )1 /2 .

20. The differentiation of ( x − 1) ( x − 2) with respect to x is 2 x − 3. 21. If y = log sin 3 x, then

dy dx

= 3 cot 3 x.

22. If y is a function of t and t is a function of x, then 23. If y = sin (tan −1 x), then 24. If y = e x cot x, then 25. If y = log sin x, then

dy

dx

=

cos (tan −1 x) 1 − x2

⋅

= e x (cot x − cosec2 x).

dx dy dx

26. If y = sin (log x), then

dy

= tan x.

dy dx

=

sin (log x) x

⋅

dy dx

=

dy dt ⋅ ⋅ dt dx

340

A nswers 1. a x log a 1

5. x x 8.

2

2. −

sin

x

2 x

6. 10 (2 x + 3)4

3.

1 x log a

4. 2 x + 2 x log e 2

7. − 4 x cos x 2 sin x 2

−1

−2 (1 − 4 x 2 )

9.

ay − x 2 y

2

− ax

10.

1 x2 − 1

11. − e cos x . sin x

12. (c)

13. (b)

14. (b)

15. (c)

16. (c)

17. (b)

18. (c)

19. F

20. T

21. T

22. T

23. F

24. T

25. F

26. F

341

10 L inear D ifferential E quations

10.1 Differential Equation and its Solution differential equation is an equation containing the dependent and independent variables and different derivatives of the dependent variables w.r.t. one or more independent variables.

A

The order of a differential equation is the order of the highest derivative (or differential coefficient) occurring in the equation. The degree of a differential equation is the degree of the highest derivative (or diff. coeff.) which occurs in it, after the differential equation has been rationalized (i.e,. made free from radicals and fractions so far as derivatives are concerned). A differential equation is called ordinary, if the unknown function depends on only one argument (independent variable).

342

A differential equation is said to be partial if there are two or more independent variables. A differential equation is said to be linear if the dependent variable, say, ‘ y ' and all its derivatives occur in the first degree, otherwise it is non-linear. A function y = f ( x) is called a Solution (or the primitive) of a differential equation if, when substituted into the equation, it reduces the equation to an identity and the process of finding all the solutions is called integrating (or solving) the differential equation. General solution. A solution of a differential equation, containing independent arbitrary constants equal in number to the order of the differential equation is called its general solution. Particular solution. A solution obtained by giving particular values to the arbitrary constants in the general solution is called a particular solution or particular integral. Arbitrary Constants. The solution of a differential equation may contain as many arbitrary constants as is the order of the differential equation i. e., the solution of an nth order differential equation may contain n arbitrary constants.

10.2 Linear Differential Equation A linear differential equation is an equation in which the dependent variable and its derivatives appear only in the first degree. A linear differential equation of order n of the form d

n

dx

y n

+ a1

d

n −1

dx

y

n −1

+ a2

d

n−2

dx

n−2

y

+ … + an −1

dy dx

+ an y = Q ,

...(1)

where a1 , a2 , … , a n − 1 , a n are constants and Q is any function of x is called a linear differential equation with constant coefficients. For convenience, the operators

d d2 d3 dn , 2 , 3 , … , n are also denoted by D, D2 , dx dx dx dx

D3 , … , D n respectively. Thus the equation (1) can also be written as D n y + a1 D n − 1 y + … + a n − 1 Dy + a n y = Q or

[ D n + a1 D n − 1 + … + a n − 1 D + a n ] y = Q .

...(1)

If y = f ( x) is the general solution of [ D n + a1 D n − 1 + … + a n − 1 D + a n ] y = 0,

...(2)

and y = φ ( x) is any particular solution of the equation (1) not containing any arbitrary constant, then

343

y = f ( x) + φ ( x) , is the general solution of (1). Thus the method of solving a linear equation is divided into two parts : First, we find the general solution of the equation (2). It is called the complementary function (C.F.). It must contain as many arbitrary constants as is the order of the given differential equation. Next, we find a solution of (1) which does not contain an arbitrary constant. This is called the particular integral (P.I.). If we add (C.F.) and (P.I.), we get the general solution of (1). Thus the general solution of (1) is y = C. F. + P. I.

10.3 Determination of Complementary Function (C.F.) Consider a linear nth order differential equation with constant coefficients of the form f ( D) y = 0 i. e., [ D n + a1 D n − 1 + a2 D n − 2 + … + a n ] y = 0.

...(1)

This is equivalent to ...(2)

[( D − m1 ) ( D − m2 ) ...... ( D − mn )] y = 0. The solution of any one of the equations ( D − m1 ) y = 0, ( D − m2 ) y = 0, … , ( D − mn ) y = 0

...(3)

is also a solution of (2) and we know that the general solution of ( D − m1 ) y = 0 is y = Ae m1 x . Hence we can assume that a solution of the equation (2) is of the form Then, substituting for in y y = e mx . e mx mx 2 2 mx n n mx Dy = me , D y = m e , ...., D y = m e , we get e mx

so

that

(m n + a1 m n − 1 + a2 m n − 2 + … + a n ) = 0

m n + a1 m n − 1 + a2 m n − 2 + … + a n = 0 , because e

or Hence e

mx

(1),

mx

≠ 0.

will be a solution of (1) if m has the value obtained from the equation m n + a1 m n − 1 + … + a n = 0.

...(4)

The equation (4) is called the auxiliary equation (A.E.) and is obtained by putting D = m in f ( D) = 0. It will give in general n roots, say, m1 , m2 , m3 , ..., mn . Case I.

If all the roots of the Auxiliary equation (A.E.) are distinct.

If the roots m1 , m2 , m3 , … , mn are all distinct, then e

m1 x

, e m2 x , ......, e mn

x

are all

distinct and linearly independent. So the general solution of (1) in this case is y = c1 e m1 x + c 2 e m2

x

+ … + c n e mn x .

...(5)

344

Case II. Auxiliary equation having equal roots : If two roots are equal say m1 = m2 , then the solution (5) becomes or

y = c1 e m1 x + c 2 e m1 x + c 3 e

m3 x

y = (c1 + c 2 ) e m1 x + c 3 e m3

+ … + c n e mn x .

x

+ … + c n e mn

x

Now (c1 + c 2 ) can be replaced by single constant say c . Similarly if three roots of the auxiliary equation are equal say, m1 = m2 = m3 , the general solution of f ( D) y = 0 will be y = (c1 + c 2 x + c 3 x 2 ) e m1 x + c 4 e m4

x

+ … + c n e mn

x

and so on. Case III. Auxiliary equation having complex roots. Let the two roots of the auxiliary equation be complex, say m1 = α + iβ and m2 = α − iβ, (where i = √ − 1). The solution corresponding to these two roots will be y=e

αx

(c1 cos βx + c 2 sin βx)

Note. The expression e c1 e

αx

αx

(c1 cos βx + c 2 sin βx) can also be written as

sin ( βx + c 2 ) or

c1 e

αx

cos ( βx + c 2 ) .

Case IV. If the imaginary roots are repeated, say α + iβ and α − iβ occur twice then the solution will be y=e

αx

[(c1 + c 2 x) cos βx + (c 3 + c 4 x) sin βx],

and so on. Case V.

If a pair of the roots of the auxiliary equation are irrational i. e., they are α ± √ β,

where β is positive, then the corresponding term in the C.F. will be e or

αx

c1 e

(c1 cosh √ βx + c 2 sinh √ βx) , αx

sinh (√ βx + c 2 ) or

c1 e

αx

cosh (√ βx + c 2 ) .

If these irrational roots are repeated twice, then the corresponding portion of the solution will be e

Illustration 1 : Solution : ∴

αx

{(c1 + c 2 x) cosh √ βx + (c 3 + c 4 x) sinh √ βx} .

Solve

d2 y dx

2

−7

dy dx

+ 12 y = 0.

[UPTU 2001]

The given differential equation is ( D2 − 7 D + 12) y = 0.

the auxiliary equation is m2 − 7m + 12 = 0

or

(m − 3) (m − 4) = 0.

∴

m = 3, 4. Hence the solution is

345

y = c1 e3 x + c 2 e4 x . Illustration 2 :

Solve ( D 2 ± µ 2 ) y = 0. 2

Solution : The auxiliary equation is m − µ

[UPTU 2008] 2

2

= 0 or m + µ

2

= 0.

When m2 − µ 2 = 0, we have m = ± µ and

when m2 + µ 2 = 0, we have m = 0 ± µi.

Hence the solution of ( D 2 − µ 2 ) y = 0 is y = c1 e µx + c 2 e −µx and the solution of ( D 2 + µ 2 ) y = 0 is y = e0 or

(c 3 cos µx + c 4 sin µx)

y = c 3 cos µx + c 4 sin µx.

Illustration 3 : Solution :

x

Solve ( D3 + 6 D2 + 11D + 6) y = 0.

[UPTU 2001]

The auxiliary equation is m3 + 6m2 + 11m + 6 = 0

or

(m + 1) (m2 + 5m + 6) = 0

or

(m + 1) (m + 2) (m + 3) = 0.

∴

m = − 1, − 2, − 3.

Hence the solution is y = c1 e − x + c 2 e − 2 Illustration 4 :

Solve

d2 x dt

2

−3

x

+ c3 e − 3 x .

dx + 2 x = 0, given that when t = 0, x = 0 and dt

dx / dt = 0. Solution : or

The auxiliary equation is m2 − 3m + 2 = 0 (m − 1) (m − 2) = 0

Hence the solution is x = c1 e

t

or

+ c2 e

m = 1, 2 . 2t

,

...(1)

where c1 and c 2 are arbitrary constants. Now x = 0 when t = 0; ∴

...(2)

0 = c1 + c 2 . t

Also

dx / dt = c1 e

∴

0 = c1 + 2c 2 .

+ 2c 2 e

2t

, and dx / dt = 0 when t = 0 . ...(3)

Solving (2) and (3), we get c1 = 0, c 2 = 0. Now putting the values of c1 and c 2 in (1), we get the required solution as x = 0. Illustration 5 : 1 x = π. 2

Solve (d2 y / dx 2 ) + y = 0, given y = 2 for x = 0, and y = − 2 for

Solution : The auxiliary equation is m2 + 1 = 0. ∴

m = ± i = 0 ± i.

346

Hence the solution is y = c1 cos x + c 2 sin x.

...(1)

Now if x = 0, y = 2; c1 = 2, from (1). 1 Also when x = π, y = − 2; 2

∴

∴

c 2 = − 2, from (1).

Substituting these values of c1 and c 2 in (1), the solution is y = 2 cos x − 2 sin x = 2 (cos x − sin x) or

y = 2 √ 2 {(1 / √ 2) cos x − (1 / √ 2) sin x} = 2 √ 2 cos ( x +

Illustration 6 : Solution :

Solve ( D2 + 1) ( D − 1) y = 0.

Given differential equation is, ( D2 + 1) ( D − 1) y = 0

The A.E. is,

(m2 + 1) (m − 1) = 0

...(1)

⇒

m = ± i, 1.

Hence the solution of (1) is, y = C. F. i. e.,

y = c1 cos x + c 2 sin x + c 3 e x

Illustration 7 :

Solve

d2 y dx

2

− 2p

dy dx

+ ( p2 + q 2 ) y = 0.

Solution : The given differential equation is d2 y dx or

2

− 2p

dy dx

+ ( p2 + q 2 ) y = 0.

[ D2 − 2 pD + ( p2 + q 2 )] y = 0

The A.E. is, m2 − 2mp + p2 + q 2 = 0 or →

or

m − p = ± qi

(m − p)2 = − q 2 m = p ± qi.

Hence the required solution is, y = C. F. i. e.,

y = e px (c1 cos qx + c 2 sin qx)

Illustration 8 : Solution :

1 π). 4

Solve ( D3 − 3 D + 2) y = 0.

The auxiliary equation is m3 − 3m + 2 = 0

or

(m − 1) (m2 + m − 2) = 0

or

(m − 1) {(m − 1) (m + 2)} = 0.

∴

m = 1, 1, − 2 .

Hence the solution is y = (c1 + c 2 x) e

x

+ c3 e − 2 x .

347

Illustration 9 : Solution :

Solve

d3 y dx 3

− 8 y = 0.

[UPTU 2008] 3

The auxiliary equation is m − 8 = 0

or

(m − 2) (m2 + 2m + 4) = 0 i. e., m − 2 = 0 and m2 + 2m + 4 = 0.

∴

m = 2 and

m = − 1 ± i √ 3.

Hence the solution is y = e − x (c1 cos √ 3 x + c 2 sin √ 3 x) + c 3 e2 x . Illustration 10 : Solution :

Solve

d4 y dx

4

−2

d3 y dx

dy

−2

3

dx

− y = 0.

The auxiliary equation is m4 − 2m3 − 2m − 1 = 0

or (m4 − 1) − 2m (m2 + 1) = 0

or

(m2 + 1) (m2 − 1) − 2m (m2 + 1) = 0

or

(m2 + 1) (m2 − 2m − 1) = 0 i. e., m2 = − 1 or m2 − 2m − 1 = 0.

∴

m = 0 ± i, 1 ± √ 2 .

Hence the solution is y = e0 or

x

(c1 cos x + c 2 sin x) + e

y = c1 cos x + c 2 sin x + e

Illustration 11 :

x

x

(c 3 cosh √ 2 x + c 4 sinh √ 2 x)

{c 3 cosh √ 2 x + c 4 sinh √ 2 x} .

Write the complementary function for ( D + 1)2 ( D2 + 4) y = 0.

Solution :

Given differential equation is ( D + 1)2 ( D2 + 4) y = 0

The A.E. is, (m + 1)2 (m2 + 4) = 0 or

or

m + 1 = 0, m + 1 = 0, m2 + 4 = 0 (Note)

m = − 1, − 1, − 1, ± 2i

Hence the required solution is, y = C. F. i. e.,

y = (c1 + xc 2 ) e − x + c 3 cos 2 x + c 4 sin 2 x

Illustration 12 : Solution :

Solve (i) ( D2 + 1)2 ( D − 1) y = 0 (ii) ( D2 − 4) y = e x + sin 2 x

(i) Here the auxiliary equation is (m2 + 1)2 (m − 1) = 0.

∴

m = ± i, ± i, 1.

∴

C.F. = c1 e x + (c 3 + c 4 x) cos x + (c 5 + c 6 x) sin x.

(ii) The auxiliary equation is m2 − 4 = 0 or m = 2, − 2. ∴

C.F. = c1 e2 x + c 2 e − 2 x .

348

And P.I. =

1 2

D −4

ex +

1 2

D −4

sin 2 x =

sin 2 x ex 1 1 + = − e x − sin 2 x. 1 − 4 − 22 − 4 2 8

Hence the complete solution is y = C.F. + P.I. = c1 e2 x + c 2 e − 2 x − Illustration 13 : Solution :

1 x 1 e − sin 2 x. 3 8

Solve ( D4 + k 4 ) y = 0.

The auxiliary equation is m4 + k 4 = 0

or

(m2 + k 2 )2 − 2 k 2 m2 = 0

or (m2 + k 2 )2 − (√ 2 km)2 = 0

or

(m2 + k 2 − √ 2 km) (m2 + k 2 + √ 2 km) = 0

or

m2 − √ 2 km + k 2 = 0

or

m=

and m2 + √ 2 km + k 2 = 0

√ 2k ± √ (2k 2 − 4k 2 )

and m =

− √ 2 k ± √ (2 k 2 − 4k 2 )

2 k k k k and − m= ±i ±i ⋅ √2 √2 √2 √2

or

2

Hence the solution is y=e

k x / √2

{ c1 cos (k x / √ 2) + c 2 sin (k x / √ 2)} + e − k x / √2 { c 3 cos (k x / √ 2) + c 4 sin (k x / √ 2)} .

Illustration 14 :

Solve ( D 4 + 8 D 2 + 16) y = 0.

Solution : The auxiliary equation is m4 + 8m2 + 16 = 0 or

(m2 + 4)2 = 0

∴

m = 0 ± 2i, 0 ± 2i.

or

m2 + 4 = 0 and m2 + 4 = 0.

Thus the imaginary roots are repeated twice. Hence the solution is y = (c1 + c 2 x) cos 2 x + (c 3 + c 4 x) sin 2 x, since e 0 = 1.

Comprehensive Exercise 1 Solve the following differential equations : 1. 2. 3.

d2 y dx

2

d2 y dx

2

d2 y dx

2

+ (a + b) −3 +5

dy dx dy dx

dy dx

+ aby = 0.

− 4 y = 0. + 4 y = 0.

(Note)

349

4. 5.

d3 y dx

3

d2 y dx

2

+6 −4

d2 y dx dy dx

2

+3

dy dx

− 10 y = 0.

+ 4 y = 0.

6. ( D3 − 4 D2 + 5 D − 2) y = 0. 7. 8. 9.

d4 y dx

4

d4 y dx

4

d4 y dx 4

+2 −

d3 y dx

d3 y dx

3

3

−3

−9

d2 y dx

2

d2 y dx

2

−4

− 11

dy dx

dy dx

+ 4 y = 0.

− 4 y = 0.

− k 4 y = 0.

A nswers 1.

y = c1 e − ax + c 2 e − bx .

2. y = c1 e − x + c 2 e4 x .

3.

y = c1 e − x + c 2 e −4 x .

4. y = c1 e

5.

y = (c1 + c 2 x) e2 x .

6. y = (c1 + c 2 x) e

7.

y = (c1 + c 2 x) e

8.

y = (c1 + c 2 x + c 3 x 2 ) e − x + c 4 e4 x .

9.

y = c1 e

kx

x

x

+ c2 e − 2 x

x

+ c3 e − 5 x .

+ c 3 e2 x .

+ (c 3 + c 4 x) e − 2 x .

+ c 2 e − k x + c 3 cos k x + c 4 sin k x .

10.4 The Particular Integral (P.I.) As already shown in 10.1 the complete solution of ( D n + a1 D n − 1 + a2 D n − 2 + …… + a n − 1 D + a n ) y = Q or

...(1)

F ( D) y = Q is y = C. F. + P. I. ,

where the C.F. consists of the general solution of the differential equation F ( D) y = 0. 1 The particular integral of the differential equation F ( D) y = Q is Q. It is F ( D) obviously a function of x which when operated by F ( D) gives Q . Note. If D stands for differentiation then 1 / D will stand for integration.

10.5 Particular Integral in some Special Cases Case I. To find P.I. when Q is of the form e ax , where a is any constant and F ( a) ≠ 0.

350

P.I. =

1 1 e ax = e ax , provided F ( a) ≠ 0. F ( D) F ( a)

Working Rule : If P.I.= {1 / F ( D)} e ax , then put a for D in F ( D) and we get P.I. provided F ( a) ≠ 0.

Illustration 1: Solution :

d2 y

Solve

dx

−3

2

dy dx

+ 2 y = e5 x .

The given equation can be written as ( D2 − 3 D + 2) y = e5 x , where d / dx ≡ D .

Here

F ( D) = D2 − 3 D + 2 and

Q = e5 x .

The auxiliary equation is m2 − 3m + 2 = 0 ∴

(m − 1) (m − 2) = 0;

m = 1, 2 .

∴ The C. F. = c1 e And P. I. =

or

x

1 e F ( D)

+ c 2 e2 x , the roots of the A.E. being distinct. ax

1

=

D2 − 3 D + 2

e5 x =

e5 x 52 − 3 . 5 + 2

=

e5 x ⋅ 12

[We have put 5 for D in F ( D), because here a = 5 ] Hence the complete solution is y = (C. F. ) + (P. I. ) 1 5x or y = c1 e x + c 2 e2 x + e . 12 Illustration 2 : Solution :

Solve

d2 x dt

+λ

2

dx + µ x = et . dt

The given equation can be written as, ( D2 + λD + µ) x = e t .

The auxiliary equation is m2 + λm + µ = 0 or

m=

−λ ±

λ2 − 4µ 2

∴

C. F. = c1 e

and

P.I. =

 − λ ± λ2 − 4µ    t   2  

1 D2 + λD + µ

et =

+ c2 e

 − λ − λ2 − 4µ    t   2  

et ⋅ 1+ λ + µ

Hence the general solution is y = (C. F. ) + (P. I. )

[UPTU 2007]

351

or

y = c1 e

Illustration 3 : Solution :

 − λ +  

λ2 − 4 x  t  2 

d2 y

Solve

dx

2

+ 31

dy

+ c2 e

 − λ −  

λ2 − 4µ  t  2 

+

1 et . 1+ λ + µ

+ 240 y = 272 e − x .

dx

The auxiliary equation is m2 + 31m + 240 = 0

or

(m + 15) (m + 16) = 0.

∴

m = − 15, − 16.

∴

C. F. = c1 e − 15 x + c 2 e − 16 x .

And

P. I. =

1 2

D + 31D + 240

= 272 ⋅ =

(272 e − x )

1 2

D + 31D + 240

e − x = 272 ⋅

1 2

(− 1) + 31 (− 1) + 240

e− x

272 − x 136 − x e = e . 210 105

Hence the general solution is y = (C. F. ) + (P. I. ) 136 − x i. e., y = c1 e − 15 x + c 2 e − 16 x + e . 105 Illustration 4 : Solution :

Solve

d2 y dx

2

−5

dy dx

+ 6 y = e4 x

[UPTU 2004, 06]

The given differential equation can be written as ( D2 − 5 D + 6) y = e4 x .

The auxiliary equation is m2 − 5m + 6 = 0

or

or

m = 2, 3.

The

C.F. = c1 e2 x + c 2 e3 x .

And

P.I. =

1 2

D − 5D + 6

e4 x =

(m − 2) (m − 3) = 0

e4 x 2

4 − 54 . +6

=

1 4x e . 2

Hence the complete solution is y = (C.F.) + (P.I.) 1 or y = c1 e2 x + c 2 e3 x + e4 x . 2 Illustration 5 :

Obtain the complete solution of the differential equation

d2 y dx

2

−7

dy dx

+ 6 y = e2 x ,

and determine the constant so that y = 0 when x = 0. Solution : ∴

The given equation is [ D2 − 7 D + 6] y = e2 x .

auxiliary equation is m2 − 7m + 6 = 0

352

or

or

(m − 1) (m − 6) = 0

∴

C. F. = c1 e

And

P. I. =

x

+ c2 e

6x

1 D2 − 7 D + 6

m = 1, 6.

.

e2 x =

1 22 − 7 . 2 + 6

e2 x = −

e2 x ⋅ 4

∴ the general solution is y = (C. F. ) + (P. I. ) 1 i. e., y = c1 e x + c 2 e6 x − e2 x . 4

...(1)

Now when y = 0, x = 0 ; 1 , from (1) 4

∴

0 = c1 + c 2 −

so that

1 c 2 =  − c1  ⋅ 4 

Hence from (1), we have 1 1 1 y = c1 e x +  − c1  e6 x − e2 x = c1 (e x − e6 x ) + (e6 x − e2 x ) 4  4 4 1 y = c1 (e x − e6 x ) + e2 x (e4 x − 1) is the required solution. 4

or

Comprehensive Exercise 2 Solve the following differential equations : 1. 3. 5.

d2 y dx

2

d2 y dx

2

d3 y dx

3

dy

− 2k +

dx

dy dx

+6

+ k2 y = e x .

+ y = e x.

d2 y dx

2

+ 11

dy dx

2.

d2 y dx

2

+2

dy dx

+ y = 2 e2 x .

4. [ D2 + D + 1] y = e − x . + 6 y = e2 x .

A nswers + e x / (1 − k)2 . 2 = (c1 + c 2 x) e − x + e2 x . 9 1 1 1 − x /2 =e { c1 cos ( x √ 3) + c 2 sin ( x √ 3)} + e x . 2 2 3 = c1 e − x /2 cos { x (√ 3 / 2) + c 2 } + e − x . 1 2x = c1 e − x + c 2 e − 2 x + c 3 e − 3 x + e . 60

1.

y = (c1 + c 2 x) e

2.

y

3.

y

4.

y

5.

y

kx

353

Case II. To find P.I. when Q is of the form sin ax or cos ax and F ( − a 2 ) ≠ 0. 1

sin ax =

2

F (D ) 1

1 F (− a 2 ) 1

cos ax =

2

2

F (− a )

F (D )

sin ax , provided F (− a2 ) ≠ 0. cos ax , provided F (− a2 ) ≠ 0.

Working Rule. If P. I. = {1 / F ( D)} sin ax or cos ax , put − a2 for D2 , − a2 D for D3 , (− a2 )2 i. e., a4 for D4 , a4 D for D5 , − a6 for D6 etc. in F ( D) and calculate the P.I. Note. Linear factors in D of the form ( pD ± q) appearing in the denominator are removed by first multiplying the Nr. and Deno. by the conjugate factors ( pD + q) and then putting − a2 for D2 in the denominator. The operation left in the numerator can be easily worked out because the operator D stands for differentiation with respect to x .

Illustration 1: Solution : ∴ ∴

Solve

d2 y dx

2

−

dy dx

− 2 y = sin 2 x .

[UPTU 2001]

The given equation is [ D2 − D − 2] y = sin 2 x .

auxiliary equation is m2 − m − 2 = 0

or (m + 1) (m − 2) = 0.

m = − 1, 2 .

∴

C. F. = c1 e − x + c 2 e2 x .

And

P. I. =

1 2

D − D−2

sin 2 x =

1 sin 2 x , −4− D−2 putting − 22 i. e., − 4 for D2

=− =−

( D − 6) ( D − 6) 1 sin 2 x = − sin 2 x = − 2 sin 2 x D+6 ( D + 6) ( D − 6) D − 36 ( D − 6) − 4 − 36

sin 2 x , putting − 4 for D2

1 1 = ( D − 6) sin 2 x = {D (sin 2 x) − 6 sin 2 x} 40 40 1 1 3 = (2 cos 2 x − 6 sin 2 x) = cos 2 x − sin 2 x . 40 20 20 ∴ or

the complete solution is y = (C. F. ) + (P. I. ) 1 3 y = c1 e − x + c 2 e2 x + cos 2 x − sin 2 x . 20 20

Illustration 2 :

Solve ( D2 + D + 1) y = sin 2 x.

[UPTU 2001 (Special)]

354

Solution :

The auxiliary equation is m2 + m + 1 = 0, giving − 1 ± 1 − 4 .1

1 i 3 ± , (complex roots). 2 2 2 1 1 C.F. = e (−1 /2) x { c1 cos ( x 3) + c 2 sin ( x 3)}. 2 2 1 1 1 P.I. = sin 2 x = 2 sin 2 x = sin 2 x, F (D ) − 4 + D +1 D + D +1 m=

∴

=−

putting − 22 i. e., − 4 for D2 = = =

( D + 3) D+3 1 sin 2 x = sin 2 x = 2 sin 2 x D−3 ( D − 3) ( D + 3) D −9 D+3 −4 − 9 D+3

(sin 2 x) = −

− 13

=−

putting − 22 i. e., − 4 for D2

sin 2 x,

1 [ D (sin 2 x) + 3 sin 2 x] 13

1 [2 cos 2 x + 3 sin 2 x]. 13

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1   y = e (−1 /2) x  c1 cos  x 3 + c 2 sin  x 3   2   2   1 − [2 cos 2 x + 3 sin 2 x]. 13

or

Illustration 3 : Solution :

Solve ( D2 − 5 D + 6) y = sin 3 x .

Here the auxiliary equation is m2 − 5m + 6 = 0

∴

m = 2 , 3.

∴

C. F. = c1 e2

And

P. I. =

+ c 2 e3 x .

1 2

sin 3 x =

D − 5D + 6 1 = sin 3 x − 9 − 5D + 6

= = = ∴

x

or (m − 2) (m − 3) = 0 .

−1 5D + 3

sin 3 x =

− (5 D − 3) 2

25 . (− 3 ) − 9

1 2

− 3 − 5D + 6

− (5 D − 3) (5 D + 3) (5 D − 3)

sin 3 x =

sin 3 x

sin 3 x =

− (5 D − 3) 25 D2 − 9

1 {5 D (sin 3 x) − 3 sin 3 x} 234

1 1 (5 . 3 cos 3 x − 3 sin 3 x) = (5 cos 3 x − sin 3 x) . 234 78

the complete solution is y = (C. F. ) + (P. I. )

sin 3 x

355

y = c1 e2 x + c 2 e3 x + (1 / 78) (5 cos 3 x − sin 3 x) .

or

Illustration 4 : Solution :

d2 y

Solve

+ 9 y = cos 2 x + sin 2 x .

dx 2

The auxiliary equation is m2 + 9 = 0. m = ± 3i.

∴

C. F. = c1 cos 3 x + c 2 sin 3 x or 1 P. I. = (cos 2 x + sin 2 x) 2 ( D + 9)

∴ And

= =

1 2

D +9

cos 2 x +

cos 2 x 2

−2 +9

+

1 2

D +9

sin 2 x 2

−2 +9

=

= c1 cos (3 x + c 2 ) .

sin 2 x

cos 2 x 5

+

sin 2 x 5

=

1 (cos 2 x + sin 2 x) . 5

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 or y = c1 cos (3 x + c 2 ) + (cos 2 x + sin 2 x) . 5 Illustration 5 : Solution :

Solve

d2 y

− 4 y = e x + sin 2 x.

dx 2

The given differential equation can be written as ( D2 − 4) y = e x + sin 2 x.

The auxiliary equation is m2 − 4 = 0. ∴

(m + 2) (m − 2) = 0 or m = 2, − 2.

∴

C.F. = c1 e2 x + c 2 e −2 x

and

P.I. = = =

1 2

D −4 1 2

D −4 ex 12 − 4

+

[e x + sin 2 x] ex +

1 2

D −4

sin 2 x − 22 − 4

sin 2 x

=−

ex 1 − sin 2 x. 3 8

Comprehensive Exercise 3 Solve the following differential equations : 1. (d2 y / dx 2 ) + 9 y = cos 4 x .

2. (d2 y / dx 2 ) + y = cos 2 x .

3. ( D2 − 3 D + 2) y = sin 3 x .

4. ( D2 − 2 D + 5) y = sin 3 x .

5. ( D2 − 3 D + 2) y = cos 3 x .

6. ( D3 + D2 − D − 1) y = cos 2 x .

356

7. 9.

d2 y dx

2

d2 y dx

2

−8 +2

dy dx dy dx

8.

+ 9 y = 40 sin 5 x .

d2 y dx

2

−4

dy dx

+ y = a sin 2 x .

+ 10 y + 37 sin 2 x = 0.

A nswers 1.

y = c1 cos 3 x + c 2 sin 3 x −

2.

y = c1 cos x + c 2 sin x −

3.

y = c1 e

4.

y

5.

y

6.

y

7.

y

8.

y

9.

y

1 cos 4 x . 7

1 cos 2 x . 3

1 (9 cos 3 x − 7 sin 3 x) . 130 = e2 x (c1 cos 2 x + c 2 sin 2 x) + (1 / 26) (3 cos 3 x − 2 sin 3 x) . 1 = c1 e x + c 2 e2 x − (7 cos 3 x + 9 sin 3 x) . 130 = (c1 + c 2 x) e − x + c 3 e x − (1 / 25) (2 sin 2 x + cos 2 x) . 5 = c1 e4 x cosh ( x √ 7 + c 2 ) + (5 cos 5 x − 2 sin 5 x) . 29 = c1 e2 x cosh (√ 3 x + c 2 ) + (1 / 73) a (8 cos 2 x − 3 sin 2 x) . 37 = e − x (c1 cos 3 x + c 2 sin 3 x) + (2 cos 2 x − 3 sin 2 x) . 26 x

+ c 2 e2

x

+

Case III. To find P.I. when Q is of the form x m , where m is a positive integer. Consider first {1 / ( D − a)} x m . We have 1 1 xm =− x ( D − a) (a − D) =−

 1 D D2 + 2 + … x 1 + a a  a

=−

1 x a

+

=−

1 D − 1 x 1 −  a a

=−

m

m

1 mx a

m −1

+

m

1 a

2

1 x a {1 − ( D / a)}

m

m

, expanding by the binomial theorem m (m − 1) x

m −2

 + … ⋅ 

Here we observe that in the expansion by the binomial theorem, the terms of the expansion beyond the mth power of D need not be written since D m + 1 x m = 0, Dm + 2 x

m

= 0, etc.

Working Rule. In order to evaluate {1 / F ( D)} x

m

, bring out common the

lowest degree term in D from F ( D) so that remaining factor in the denominator is of the form [1 + f ( D)] or [1 − f ( D)] which is taken in the numerator with a

357

negative index. Next we expand [1 ± f ( D)] − 1 in powers of D by the binomial m

theorem and operate upon x

with the expansion obtained.

This expansion should be done upto the term D m , since D m + 1 x higher differential coefficients of x the following examples.

m

m

= 0 and all

are zero. The whole process will be clear from

The following binomial expansions should be remembered well. (i)

(1 − D) − 1 = 1 + D + D2 + D3 + D4 + …

(ii)

(1 + D) − 1 = 1 − D + D2 − D3 + D4 − …

(iii) (1 − D) − 2 = 1 + 2 D + 3 D2 + 4 D3 + … (iv) (1 + D) − 2 = 1 − 2 D + 3 D2 − 4 D3 + … (v)

(1 − D) −3 = 1 + 3 D + 6 D2 + 10 D3 + …

(vi) (1 + D) −3 = 1 − 3 D + 6 D2 − 10 D3 + … n (n + 1)

(vii) (1 + x) − n = 1 − nx + (viii) (1 − x) − n = 1 + nx +

Illustration 1 : Solution :

2! n (n + 1)

x2 −

2!

n (n + 1) (n + 2)

x2 +

3! n (n + 1) (n + 2) 3!

x3 + … x3 + …

Solve (d2 y / dx 2 ) − 4 y = x 2 .

The auxiliary equation is m2 − 4 = 0 or m = ± 2 .

∴

C. F. = c1 e2 x + c 2 e − 2 x .

And

P.I. =

1 D2 − 4

=−

x2 =

1 1 − 4 [1 − D2 ] 4

x2 = −

1 4

1 − 1 D2   4 

−1

x2

1 1 [1 + D2 + ...] x 2 , 4 4

expanding by binomial theorem upto the terms containing D2 1 1 = − [ x 2 + D2 ( x 2 )] , because all the remaining terms vanish 4 4 1 2 1 1 1 = − [ x + ⋅ 2] = − [ x 2 + ] ⋅ 4 4 4 2 Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1 e2 x + c 2 e − 2 x − ( x 2 + ) . 4 2 Illustration 2 : Solution :

Solve

d3 y dx

3

−

d2 y dx

2

−6

The auxiliary equation is

dy dx

= 1 + x2 .

358

m3 − m2 − 6m = 0 ∴

m = 0, − 2, 3.

∴

C. F. = c1 e0

And

P. I. =

x

or

+ c 2 e − 2 x + c 3 e3 x = c1 + c 2 e − 2 x + c 3 e3 x . 1

4

2

D − D − 6D

1 6D 1 =− 6D =−

m (m + 2) (m − 3) = 0.

(1 + x 2 ) =

1 (1 + x 2 ) 1 1 2 − 6 D [1 + D − D ] 6 6

−1 1 − 1 (− D + D2 ) (1 + x 2 )  6  1 + 1 (− D + D2 ) + 1 (− D + D2 )2 + … (1 + x 2 ) ,  6  36

the terms in the expansion being needed only upto D2 1 6D 1 =− 6D 1 =− 6D =−

1 − 1 D +  6 1 − 1 D +  6

1 2 1 D + D2 + … (1 + x 2 )  6 36 7 D2 + … (1 + x 2 )  36 (1 + x 2 ) − 1 D (1 + x 2 ) + 7 D2 (1 + x 2 ) ,   6 36

because all other terms vanish 1  1 7 1  25 1 2 =− − x + x 2  1 + x − x +  = −   6D  3 18 6 D  18 3 1 25 1 = − ∫  − x + x 2  dx [ ∵ 1 / D ≡ ∫ dx]   6 18 3 1 25 1 1 =−  x − x2 + x3  ⋅   6 18 6 3 Hence the complete solution is y = (C. F. ) + (P. I. ) 1 25 1 1 or y = c1 + c 2 e − 2 x + c 3 e3 x −  x − x2 + x3  ⋅  6 18 6 3 Illustration 3 : Solution :

Solve ( D3 + 8) y = x 4 + 2 x + 1.

The auxiliary equation is m3 + 8 = 0

or (m + 2) (m2 − 2m + 4) = 0.

∴

m = − 2, 1 ± i √ 3.

Hence

C. F. = c1 e − 2

And

P. I. =

=

x

1 3

D +8

+e

x

{ c 2 cos ( x √ 3) + c 3 sin ( x √ 3)} .

( x 4 + 2 x + 1) =

1 ( x 4 + 2 x + 1) 1 3 8 [1 + D ] 8

1 1 [1 + D3 ] − 1 ( x 4 + 2 x + 1) 8 8

359

=

1 1 [1 − D3 + ...] ( x 4 + 2 x + 1) , 8 8

the other terms in the expansion being of no need 1 1 = [( x 4 + 2 x + 1) − D3 ( x 4 + 2 x + 1)] 8 8 1 4 1 = [ x + 2 x + 1 − 3 x] = ( x 4 − x + 1) . 8 8 Hence the complete solution is y = (C. F. ) + (P. I. ) y = c1 e − 2

or

+e

{ c 2 cos ( x √ 3) + c 3 sin ( x √ 3)} +

1 4 ( x − x + 1) . 8

The auxiliary equation is m2 + m − 2 = 0

∴

m = 1, − 2 .

∴

C. F. = c1 e

And

x

Solve ( D2 + D − 2) y = x + sin x .

Illustration 4 : Solution :

x

P. I. = =

1 2

D + D−2

x

or (m − 1) (m + 2) = 0.

+ c2 e − 2

x

( x + sin x) =

. 1 2

( D + D − 2)

x+

1 2

D + D−2

sin x

1 1 x+ sin x 2 1 1 2 −1 + D − 2 − 2 [1 − D − D ] 2 2

−1 ( D + 3) 1 2   1 1 − D + D x+ sin x     2   2 ( D − 3) ( D + 3) ( D + 3) D+3 1 1 1 1 = − 1 + D + … x + 2 sin x = −  x +  + sin x     2 2 2 2 −1− 9  D −9

=−

1 2

1 1 1 (x + ) − {D (sin x) + 3 sin x} 2 2 10 1 1 1 =− x− − {cos x + 3 sin x} . 2 4 10 =−

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 1 3 or y = c1 e x + c 2 e − 2 x − x − − cos x − sin x . 2 4 10 10 Illustration 5 :

Solve ( D2 − 5 D + 6) y = x + sin 3 x.

[UPTU 2003, 10]

2

Solution : The auxiliary equation is m − 5m + 6 = 0. ∴

m = 2, 3.

Hence

C. F. = c1 e2 x + c 2 e3 x .

And

P. I. = =

1 2

( D − 5 D + 6) 1 2

D − 5D + 6

( x + sin 3 x)

x+

1 2

( D − 5 D + 6)

sin 3 x.

...(1)

360

1

Now

2

D − 5D + 6

x=

1 x 5 1 6 [1 − D + D2 ] 6 6

= (1 / 6) [1 − {(5 / 6) D − (1 / 6) D2 }] −1 x = (1 / 6) [1 + (5 / 6) D +...] x = (1 / 6) ( x + 5 / 6). 1 sin 3 x = sin 3 x, putting −32 for D2 2 2 [ D − 5 D + 6] −3 − 5 D + 6 1

And

=

(3 − 5 D) 1 sin 3 x = − sin 3 x −(3 + 5 D) (3 + 5 D) (3 − 5 D)

=−

(3 − 5 D) 2

9 − 25 D

sin 3 x = −

(3 − 5 D) 9 − 25.(−9)

sin 3 x

1 [3 sin 3 x − 5 D (sin 3 x)] 234 1 =− [3 sin 3 x − 15 cos 3 x]. 234 =−

∴

from (1), P. I. =

1 5 1 (x + ) − [3 sin 3 x − 15 cos 3 x]. 6 6 234

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 5 1 or y = c1 e2 x + c 2 e3 x + ( x + ) − (3 sin 3 x − 15 cos 3 x). 6 6 234 Illustration 6 : Solution :

Solve

d2 y dx 2

C.F. = c1 e

∴ P.I. = = Now

And

[UPTU 2003]

The auxiliary equation is m2 − 4 = 0

And

− 4 y = cos2 x.

1 2

( D − 4)

2x

or

m = ± 2.

+ c 2 e −2 x .

cos 2 x =

1

1 1 1 (2 cos 2 x) = (1 + cos 2 x) 2 ( D − 4) 2 ( D − 4) 2 2

1 1 1 1 ⋅1+ cos 2 x 2 2 2 ( D − 4) 2 ( D − 4) 1 1 1 1 1 1 2  −1 1 = 1 = − 1 − D 1  2 ( D2 − 4) 2 −4 (1 − 1 D2 ) 8  4 4 1 1 2 1 = − [1 + D + …] 1 = − ⋅ 8 4 8 1 1 1 1 1 cos 2 x = cos 2 x = − cos 2 x. 2 2 2 ( D − 4) 2 (−2 − 4) 16

Substituting these values in (1), the

…(1)

361

P.I. = −

1 1 − cos 2 x. 8 16

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1 e2 x + c 2 e −2 x − − cos 2 x. 8 16 Illustration 7 : Solution :

d2 y

Solve

dx

−4

2

dy

+ 4 y = x2 + e

dx

m = 2 ,2 .

Hence

C. F. = (c1 + c 2 x) e2 x .

And

P. I. =

where

P1 =

1 2

( D − 4 D + 4) 1 D2 − 4 D + 4

(x2 + e

x2 =

+ cos 2 x .

m2 − 4m + 4 = 0.

The auxiliary equation is

∴

x

x

+ cos 2 x) = P1 + P2 + P3 , (say),

1 ( D − 2)2

x2 =

1 4

1 − 1 D  2 

−2

x2

1 3 1 3 [1 + D + D2 + ...] x 2 = [ x 2 + 2 x + ] , 4 4 4 2 x x 1 e e P2 = ex = 2 = = e x, 1 ( D2 − 4 D + 4) 1 − 4 .1 + 4 =

and P3 =

1 2

( D − 4 D + 4) =−

1 4

∫

cos 2 x =

1 2

− 2 − 4D + 4

cos 2 x dx = −

cos 2 x = −

1 cos 2 x 4D

1 1 1 ⋅ sin 2 x = − sin 2 x . 4 2 8

Hence the complete solution is y = C. F. + P. I. or or

y = C. F. + P1 + P2 + P3 1 3 y = (c1 + c 2 x) e2 x + ( x 2 + 2 x + ) + e 4 2

Illustration 8 : Solution :

x

−

1 sin 2 x . 8

Solve ( D2 − 3 D + 2) y = 6e −3 x + sin 2 x

(UPTU 2010)

The auxiliary equation is m2 − 3m + 2 = 0

∴

m = 1, 2 C.F. = C1 e x + C2 e2 x

And

P.I. = =

Now

1 2

D − 3D + 2 1 2

D − 3D + 2 1

2

D − 3D + 2

(6e −3 x + sin 2 x) 6e −3 x +

6e −3 x =

1 2

D − 3D + 2 6

2

(− 3) − 3 (− 3) + 2

sin 2 x

...(1)

e −3 x , putting (− 3) for D

362

= and

1 2

D − 3D + 2

sin 2 x = = =

3 −3 x e 10 1

2

− 2 − 3D + 2

sin 2 x , putting − 22 for D2

− (2 − 3 D) 1 sin 2 x = sin 2 x − (2 + 3 D) (2 + 3 D) (2 − 3 D) − (2 − 3 D) 2

4 − 9D

sin 2 x =

− (2 − 3 D) 4 − 9 (− 4)

sin 2 x

1 (2 − 3 D) sin 2 x 40 1 =− [2 sin 2 x − 3 D (sin 2 x)] 40 1 =− [2 sin 2 x − 6 cos 2 x] 40 =−

∴

from (1), we have 3 −3 x 1 P.I. = e − (2 sin 2 x − 6 cos 2 x) 10 40

Hence the complete solution is y = C.F. + P.I. y = C1 e x + C2 e2 x +

or

3 −3 x 1 e − (2 sin 2 x − 6 cos 2 x). 10 40

Comprehensive Exercise 4 Solve the following differential equations 1. ( D2 + D − 6) y = x .

2. (d2 y / dx 2 ) + (dy / dx) − 2 y = 2 x .

3. (d3 y / dx 3 ) + 3 (d2 y / dx 2 ) + 2 (dy / dx) = x 2 . 4.

d3 y dx

3

−4

d2 y dx

2

+5

dy dx

− 2 = 0.

6. ( D2 + 3 D + 2) y = x 2 . 8.

d2 y dx 2

5. ( D2 + 2 D + 1) y = 2 x + x 2 . 7. ( D4 + D2 + 16) y = 16 x 2 + 256.

+ 4 y = cos 3 x + sin 3 x.

9. ( D2 − 4 D + 3) y = e − x + 5.

10. (d2 y / dx 2 ) − 4 y = sin2 x .

11. ( D2 − 2 D + 3) y = cos x + x 2 . 12. 13.

d3 y dx

3

+2

d2 y dx

2

+

dy dx

= e2

Solve the equation y ′ ′ + y ′ − 2 y = x, provided that y = x = 0.

x

+ x2 + x .

dy dx

= 0, when

363 2 2 2 14. Solve the equation (d y / dx ) = a + bx + cx , given that (dy / dx) = 0,

when x = 0 and y = d, when x = 0.

A nswers 1. 2. 3. 4.

1 (6 x + 1) . 36 1 y = c1 e x + c 2 e − 2 x − x − ⋅ 2 1 y = c1 + c 2 e − x + c 3 e − 2 x + x (2 x 2 − 9 x + 21) . 12 2 y = c1 + c 2 e2 x cos ( x + c 3 ) + x . 5 y = c1 e2 x + c 2 e − 3 x −

5.

y = (c1 + c 2 x) e − x + ( x 2 − 2 x + 2) .

6.

y = c1 e − x + c 2 e −2 x +

7. 8. 9. 10. 11. 12. 13. 14.

1 2 3 7 x − x+ ⋅ 2 2 4

3 3 127 y = c1 e − (1 /2) √7. x sin ( x + c 2 ) + c 3 e(1 /2) √7 x sin  x + c 4  + x 2 + ⋅ 2  2 8 1 c1 cos 2 x + c 2 sin 2 x − (cos 3 x + sin 3 x). 5 1 5 y = c1 e x + c 2 e3 x + e − x + ⋅ 8 3 1 1 2x −2 x y = c1 e + c 2 e − + cos 2 x . 8 16 1 4 2 1 + (cos x − sin x) . y = e x [c1 cos √ 2 x + c 2 sin √ 2 x] + x 2 + x + 3 9 27 4 1 2x 1 3 3 2 −x y = c1 + (c 2 + c 3 x) e + e + x − x + 4x . 18 3 2 1 x 1 −2 x 1 1 y= e − e − x− ⋅ 3 12 2 4 1 2 1 3 1 4 y = d + ax + bx + cx . 2 6 12

10.6 To Find P. I. when Q = e a x V , whereV is any Function of x. 1 (e F ( D)

ax

V ) =e

ax

1 V. F ( D + a)

Working Rule : Replace D by ( D + a) and take out e

a x

before the operator

1 / F ( D). Then determine {1 / F ( D + a)} V by the methods discussed in 10.4.

364

Illustration 1: Solution :

Solve ( D2 − 2 D + 1) y = x 2 e3 x .

The auxiliary equation is m2 − 2m + 1 = 0 or (m − 1)2 = 0.

∴

m = 1, 1.

∴

C. F. = (c1 + c 2 x) e x . 1 P.I. = 2 x 2 e3 D − 2D + 1

And

= e3

1

x

2

{( D + 3) − 1}

x

=

1 ( D − 1)2

e3

x

x2

x2 ,

putting D + 3 for D and bringing e3 = e3

1

x

2

( D + 2)

x 2 = e3

x

x

before the operator

1 x2 1 2 4 (1 + D) 2

1 3x  1  −2 2 e x 1 + D  4 2  1 1 1 = e3 x [1 − 2 ⋅ D + 3 ⋅ D2 + ...] x 2 , 4 2 4 =

1 = e3 4 1 = e3 4

x

x

expanding by binomial theorem 3 2 2 [1 − D + D + ...] x 4 3 1 3 2 {x − 2 x + ⋅ 2} = e3 x {x 2 − 2 x + } . 4 4 2

Hence the required solution is y = (C. F. ) + (P. I. ) 1 3 or y = (c1 + c 2 x) e x + e3 x { x 2 − 2 x + } . 4 2 Illustration 2: Solution :

Solve ( D2 − 4 D + 4) y = x 3 e2 x .

[UPTU 2010]

The auxiliary equation is m2 − 4m + 4 = 0 or (m − 2)2 = 0.

∴

m = 2, 2.

∴

C. F. = (c1 + c 2 x) e 2 x .

And

P. I. = =

1 2

D − 4D + 4 1 ( D − 2)2

x 3 e2 x

x 3 e2 x = e2 x

1 ( D + 2 − 2)2

x 3 , by 10.5

365

1

= e2 x = e2 x

D2

1  x4    D 4

x 3 = e2 x

x5 20

Hence the complete solution is y = (C.F.) + (P.I.) or

y = (c1 + c 2 x) e2 x + e2 x

Illustration 3 :

Solve

d2 y dx

2

+4

dy dx

x5 . 20

− 12 y = ( x − 1) e 2 x .

Solution : The auxiliary equation is m2 + 4m − 12 = 0 or

(m − 2) (m + 6) = 0.

∴

m = 2, − 6.

∴

C. F. = c1 e 2 x + c 2 e −6 x .

And

P.I. =

1 2

D + 4 D − 12

= e 2x = e 2x

1 2

( D + 2) + 4 ( D + 2) − 12 1 D2 + 8 D

1 2x e 8 1 = e 2x 8 1 2x = e 8 1 = e 2x 8 =

=

e 2 x ( x − 1)

( x − 1) = e 2 x

( x − 1) 1

8 D (1 +

1 D) 8

( x − 1)

1  1  −1 1 + D ( x − 1) D 8  1  1  1 − D +... ( x − 1)   D 8 1  1  x − 1 −  , because all other terms vanish  D 8 1  9 1 2 x 9   x −  = e ∫  x −  dx    D 8 8 8

1 2 x  x2 9  1 2 2x 9 e − x = x e − xe 2 x .  8 8  16 64  2

Hence the general solution is y = c1 e 2 x + c 2 e −6 x + (1 / 16) x 2 e 2 x − (9 / 64) xe 2 x . Illustration 4 :

Solve ( D2 + 2 D + 1) y = e − x / ( x + 2).

Solution : The auxiliary equation is m2 + 2m + 1 = 0. ∴

m = − 1, − 1.

∴

C. F. = (c1 + c 2 x) e − x .

366

 e− x  1 e− x =   D2 + 2 D + 1  x + 2 ( D + 1)2 x + 2 1 1 1 1 = e− x = e− x ( D − 1 + 1)2 ( x + 2) D2 ( x + 2)

And

1

P.I. =

= e− x

1 D

∫

1 1 dx = e − x log ( x + 2) x+2 D

= e − x ∫ log ( x + 2). 1 dx = e − x [ x log ( x + 2) −

∫

{x / ( x + 2)} dx],

(integrating by parts taking 1 as second function) =e

−x

{x log ( x + 2) − x + 2 log ( x + 2)}

=e

−x

{x log ( x + 2) + 2 log ( x + 2}, omitting the term − xe − x

because such a term has already been included in the C.F. in c 2 x e − x . Hence complete solution is y = (C. F. ) + (P. I. ) y = (c1 + c 2 x) e − x + e − x {x log ( x + 2) + 2 log ( x + 2)}.

or

Illustration 5 : Solution :

d2 y dx 2

+ y = x 2 e x , find y.

The auxiliary equation is (m2 + 1) = 0. m = ± i.

∴

C.F. = c1 cos x + c 2 sin x. 1 1 1 P.I. = 2 x2 e x = e x ⋅ x2 = e x ⋅ x2 2 2 D +1 ( D + 1) + 1 ( D + 2 D + 1 + 1)

∴ And

= ex

If

1 2

D + 2D + 2

⋅ x2 = e x

1

⋅ x2 =

  D2   2 1 +  D +  2    

ex 2

 1 + 

 D2   D +  2   

−1

x2

2   ex   D2   D2  ex  D2 1 −  D + = + D2  x 2  + D +  − … x 2 = 1 − D − 2   2   2  2  2    

=

ex  2 1 x − D ( x 2 ) + D2 ( x 2 )   2  2

=

ex  2 1 ex 2 x − 2 x + (2) = [ x − 2 x + 1]. 2  2  2

Hence the complete solution is y = (C.F.) + (P.I.) ex 2 or y = c1 cos x + c 2 sin x + ( x − 2 x + 1). 2 Illustration 6 : Solution :

Solve ( D2 − 2 D + 5) y = e2 x sin x . 2

The auxiliary equation is m − 2m + 5 = 0.

∴

m = {2 ± √ (4 − 20)} / 2 = 1 ± 2 i .

Hence

C. F. = e

x

(c1 cos 2 x + c 2 sin 2 x) .

[UPTU 2008]

367

And

P. I. =

1 2

D − 2D + 5

e2 x sin x 1

= e2 x

2

( D + 2) − 2 ( D + 2) + 5

= e2 x

1 2

(− 1 + 2 D + 5)

sin x = e2 x

1 2

D + 2D + 5

sin x , putting − 12 for D2

=

D−2 1 2x 1 1 e sin x = e2 x sin x 2 D+2 2 ( D − 2) ( D + 2)

=

D−2 1 2x D − 2 1 e sin x = e2 x sin x 2 2 2 − 1− 4 ( D − 4)

1 2x 1 2x e ( D − 2) sin x = − e {D (sin x) − 2 sin x} 10 10 1 2x =− e (cos x − 2 sin x) . 10 =−

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 2x or y = e x (c1 cos 2 x + c 2 sin 2 x) − e (cos x − 2 sin x) . 10

Comprehensive Exercise 5 Solve the following differential equations : 1. ( D + 1)3 y = x 2 e − x .

2. ( D2 − 2 D + 1) y = x 2 e x .

3. ( D2 − 2 D + 1) y = x 2 e2 x .

4. ( D2 − 3 D + 2) y = xe x .

5. ( D2 + 2 D + 2) y = xe x .

6. ( D2 − 4 D + 1) y = e2 x sin x .

7. ( D2 − 4 D + 4) y = e2 x sin 3 x . 8. ( D4 − 1) y = e 2

x

2

x

cos x .

9. ( D − 2 D + 4) y = e cos x . 10. ( D − 5 D + 6) y = e 2 x sin 2 x .

A nswers 1.

y = (c1 + c 2 x + c 3 x 2 ) e − x +

2.

y = (c1 + c 2 x) e

3. 4. 5.

1 5 −x x e . 60

1 4 x x e . 12 y = (c1 + c 2 x) e x + e2 x ( x 2 − 4 x + 6) . 1 y = c1 e x + c 2 e2 x − e x ( x 2 + x) . 2 1 4 y = e − x (c1 cos x + c 2 sin x) + e x ( x − ) . 5 5 x

+

sin x

368

6.

y = e2 x (c1 cosh √ 3 x + c 2 sinh √ 3 x) −

7.

y = (c1 + c 2 x) e2 x −

8.

y = c1 e

9.

y=e

10.

x

x

1 2x e sin x . 4

1 2x e sin 3 x . 9

+ c 2 e − x + c 3 cos x + c 4 sin x −

(c1 cos √ 3 x + c 2 sin √ 3 x) +

y = c1 e2 x + c 2 e3 x +

1 e 2

x

1 e 5

x

cos x .

cos x .

1 2x e (cos 2 x − 2 sin 2 x). 10

10.7 To Find P.I. when Q = e a x and F (a )=0. We have

P. I. =

1 1 e ax = e a x . 1, if F (a) = 0. F ( D) F ( D)

Now this is of the form e a x V . Hence, by putting ( D + a) for D in the operator and by making e a x free from the operator, we have 1 P. I. = e a x 1. F ( D + a) Now this can be evaluated by using the method for finding P.I. in the case of x

Illustration 1 : Solution :

Solve

d2 y dx

2

−3

dy dx

.

+ 2 y = e x.

The auxiliary equation is m2 − 3m + 2 = 0

∴

m = 1, 2.

Hence

C. F. = c1 e

And

m

x

+ c 2 e2 x .

1

P. I. =

e

x

1 e (1 − 2) ( D − 1)

x

2

D − 3D + 2

=

or (m − 1) (m − 2) = 0.

=

1 e ( D − 2) ( D − 1)

x

[putting 1 for D in the factor D − 2 because it does not vanish by doing so]

1 =− e D −1

x

.1

[Note that D − 1 becomes zero

by putting 1 for D; so here we shall apply the method for e a x V by taking 1 for V .] 1 1 = − ex 1= − ex 1= − ex x . ( D + 1) − 1 D

369

Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 e

x

+ c 2 e2

x

− x e x.

Note. While finding P.I. in the case of e

a x

if F ( D) becomes zero by putting a for

D, we factorise F ( D) . Then we put D = a in the factors which do not vanish by doing so. The remaining operator is then dealt with by using the method for e a x V on taking 1 for V . Illustration 2 : Solution :

Solve ( D3 − 7 D + 6) y = e2 x .

[UPTU 2002]

The auxiliary equation is m3 − 7m + 6 = 0

∴

m = 1, 2, − 3.

∴

C. F. = c1 e

And

P. I. =

x

or (m − 1) (m − 2) (m + 3) = 0.

+ c 2 e2 x + c 3 e − 3 x . 1

3

( D − 7 D + 6)

e2 x =

1 e2 ( D − 1) ( D − 2) ( D + 3)

x

1 e2 x , (2 − 1) ( D − 2) (2 + 3)

=

putting 2 for D in all the factors except D − 2 1 1 1 1 1 1 2x = e . 1 = e2 x 1 = e2 x 1 5 D−2 5 ( D + 2) − 2 5 D =

1 2x xe . 5

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 or y = c1 e x + c 2 e2 x + c 3 e − 3 x + xe2 x . 5 Illustration 3 : Solution :

Solve ( D2 − 5 D + 6) y = e 3 x .

[UPTU 2002]

2

The auxiliary equation is m − 5m + 6 = 0.

∴

m = 2, 3.

Thus,

C.F. = c1 e 2

and

P.I. = =

x

+ c2 e 3 x

1 2

( D − 5 D + 6)

e 3x

1 1 e 3x = e3 x , ( D − 2) ( D − 3) (3 − 2) ( D − 3)

putting 3 for D except ( D − 3) 1 1 1 3x 3x = e .1 = e 1= e 3x 1 ( D − 3) ( D + 3) − 3 D = xe 3 x . Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 e 2 x + c 2 e 3 x + xe 3 x .

370

Illustration 4 : Solution :

Solve ( D2 + 4 D + 4) y = e2 x − e − 2 x .

The auxiliary equation is (m + 2)2 = 0.

∴

m = − 2, − 2 .

Hence

C. F. = (c1 + c 2 x) e − 2 x .

And

P. I. =

Now Also

1 2

( D + 2)

1 2

( D + 2) 1

2

( D + 2) = e− 2

e2 x =

e

2x

=

2

(2 + 2)

e− 2 x =

1 2

( D + 2)

1

x

1

(e2 x − e − 2 x ) =

2

{( D − 2) + 2}

2

( D + 2)

e2 x −

1 2

( D + 2)

1 2x e . 16

[ ∵ here, F (a) ≠ 0]

e − 2 x . 1,

1 = e− 2

x

e− 2 x .

1 2

D

[ ∵ here F (a) = 0] 1 = e− 2

x

1 1 x = x2 e − 2 x . D 2

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 2 x 1 2 −2 x or y = (c1 + c 2 x) e − 2 x + e − x e . 16 2 Illustration 5 : Solution :

Solve D2 y − 3 Dy + 2 y = cosh x .

The auxiliary equation is m2 − 3m + 2 = 0

∴

m = 1, 2 .

∴

C. F. = c1 e

And

P. I. = =

Now

+ c 2 e2 x . 1

1 1 e 2 2 D − 3D + 2 x

=

1 1 e 2 ( D − 1) (1 − 2)

=−

1 e 2

x

x

x

1 1 e 2 ( D − 1) ( D − 2) =−

1 1 e 2 D −1

1 1 1= − e D +1−1 2

x

x

x

.1

1 1 1= − x e x. D 2

1 1 1 1 e− x = e − x , putting − 1 for D 2 2 2 D − 3D + 2 2 (− 1) − 3.(− 1) + 2 =

∴

 e x + e− x    2 ( D2 − 3 D + 2)   1 1 + e− x . 2 2 D − 3D + 2 1

cosh x =

D2 − 3 D + 2

1 1 e 2 D2 − 3 D + 2 =

Also

x

or (m − 1) (m − 2) = 0 .

P. I. = −

1 xe 2

x

+

1 −x e . 12

1 −x e . 12

Hence the complete solution is y = (C. F. ) + (P. I. )

371

or

x

y = c1 e

Illustration 6 : Solution :

+ c 2 e2 x −

1 xe 2

1 −x e . 12

The auxiliary equation is m = 2, 2 .

Hence

C. F. = (c1 + c 2 x) e2 x .

And

P. I. =

1 2

( D − 2)

1 2

( D − 2)

e2 x =

e2 x +

1 D − 4D + 4

1 2

D − 4D + 4

1 2

( D − 2)

= e2 x 2

(m − 2)2 = 0.

or

∴

and

+

Solve ( D2 − 4 D + 4) y = e2 x + sin 2 x .

m2 − 4m + 4 = 0

Now

x

1 D2

sin 2 x =

sin 2 x .

e2 x . 1 = e2 x

(1) = e2 x

1 ( D + 2 − 2)2

1

1 x2 1 ( x) = e2 x . = x 2 e2 x , D 2 2

1 2

− 2 − 4D + 4

sin 2 x , putting − 22 for D2

=

1 1 sin 2 x = − − 4D 4

∫

sin 2 x dx = −

1 1 1 ⋅ (− cos 2 x) = cos 2 x . 4 2 8

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = (c1 + c 2 x) e2 x + x 2 e2 x + cos 2 x . 2 8 Illustration 7 :

Solve ( D2 − 4 D + 4) y = 8 ( x 2 + e 2

x

+ sin 2 x).

[UPTU 2003]

Solution : Proceeding as in illustration 6 above, here

And

Now

and

C. F. = (c1 + c 2 x) e 2 x . 1 P. I. = 2 {8 ( x 2 + e 2 x + sin 2 x)} D − 4D + 4 1 1 1 =8 2 x2 + 8 2 e 2x + 8 2 sin 2 x. D − 4D + 4 D − 4D + 4 D − 4D + 4 1 1 1 8 2 x2 = 8 x2 = 8 ⋅ x2 1 2 D − 4D + 4 ( D − 2)2 4 (1 − D) 2 1 1 1 2 −2 2 = 2 (1 − D) x = 2 {1 + 2 ⋅ D + 3 ⋅ D +...} x 2 2 2 4 3 2 3 3 2 2 = 2 (1 + D + D + ...) x = 2 ( x + 2 x + . 2) = 2( x 2 + 2 x + ), 4 4 2 1 1 2x 2x 2 2x 8 2 e =8 e .1 = 4 x e , D − 4D + 4 ( D − 2)2 8

1 2

D − 4D + 4

sin 2 x = 8 ⋅

1 cos 2 x. 8

[See Illustration 6]

372

3 P. I. = 2 ( x 2 + 2 x + ) + 4 x 2 e2 x + cos 2 x. 2

∴

Hence the complete solution is y = (C. F. ) + (P. I. ) 3 y = (c1 + c 2 x) e 2 x + 2 ( x 2 + 2 x + ) + 4 x 2 e 2 x + cos 2 x. 2

or

Illustration 8 :

Solve ( D3 − 5 D2 + 7 D − 3) y = e 2 x cosh x.

Solution : The auxiliary equation is m3 − 5m2 + 7m − 3 = 0 or

(m − 1) (m2 − 4m + 3) = 0 or (m − 1) {(m − 1) (m − 3)} = 0.

∴

m = 1, 1, 3.

Hence

C. F. = (c1 + c 2 x) e

And

P. I. =

x

+ c3 e 3 x .

1

e2

x

cosh x

e e 2 x  2 ( D − 1) ( D − 3) 

x

=

+ e− x   , 2 

=

1 e 3 x + 1 e x     2 ( D − 1) ( D − 3)  2

=

1 1 e3 2 2 ( D − 1) ( D − 3)

3

2

D − 5D + 7D − 3 1

x

+ e − x   2 

1

2

1 1 e3 2 2 ( D − 1) ( D − 3)

Now

 e  ∵ cosh x = 

x

=

x

+

1 1 e x. 2 2 ( D − 1) ( D − 3)

...(1)

1 1 e 3 x, 2 2 (3 − 1) ( D − 3) putting 3 for D in the factor ( D − 1)2

= Again

1 1 1 1 1 1 1 e 3 x .1 = e 3 x 1= e 3x 1 = xe 3 x . 8 D−3 8 D+3−3 8 D 8 1 2

2 ( D − 1) ( D − 3)

e

x

=

1 2 ( D − 1)2 (1 − 3)

1 1 1 =− e x .1 = − e 2 4 ( D − 1) 4 1 x 1 e 1= − 4 D2 1 1 = − e x . x2 = − 4 2 1 1 from (1), P. I. = xe 3 x − 8 8 =−

∴

x

e x,

putting 1 for D in the factor D − 3 1 1 {( D + 1) − 1}2

1 x 1 e x 4 D 1 2 x x e . 8 x2 e x .

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = (c1 + c 2 x) e x + c 3 e 3 x + x e 3 x − x 2 e x . 8 8

373

Comprehensive Exercise 6 Solve the following differential equations : 1. ( D2 + D − 2) y = e x .

2. ( D2 − 2 D + 1) y = e x .

3. ( D2 + D − 6) y = e2 x .

4. ( D2 − 4 D + 3) y = 2e3 x .

5. ( D3 + 3 D2 + 3 D + 1) y = e − x . 6. (d2 y / dx 2 ) − 5 (dy / dx) + 6 y = e

x

+ e2 x .

7. ( D2 − a2 ) = cosh ax . 8. (2 D + 1)2 ( D + 2) ( D2 + 4) y = 51 . e − x /2 .

A nswers 1. 3. 5. 6.

1 1 xe x . 2. y = (c1 + c 2 x) e x + x 2 e x . 3 2 1 2x 2 x −3 x x 3x y = c1 e + c2 e + xe . 4. y = c1 e + c 2 e + xe3 x . 5 1 2 y = (c1 + c 2 x + c 3 x ) e − x + x 3 e − x . 6 1 y = c1 e2 x + c 2 e3 x + e x − xe2 x . 2 y = c1 e

x

+ c2 e − 2

x

+

7.

y = c1 e ax + c 2 e − ax + ( x / 2a) sinh ax .

8.

y = c1 e − 2

x

+ (c 2 + c 3 x) e − x /2 + c 4 cos 2 x + c 5 sin 2 x + x 2 e − x /2 .

10.8 To Find P.I. when Q =sin ax or cosax and F (- a 2 )=0. To find out these types of particular integrals, it is convenient to replace sin ax or cos ax by the exponential value and then apply 10.6. Thus the particular integral of cos ax 1 i. e., cos ax , when F (− a2 ) = 0, can be written as F ( D2 ) =

1 F ( D2 )

(Real part of e

= Real Part of

1 2

e

iax

),

ia x

F (D )

Similarly, the particular integral of sin ax

[∵ e

iax

= cos ax + i sin ax]

. 1, which is now of the form of 10.5.

374

i. e.,

1 2

F (D ) =

sin ax , when F (− a2 ) = 0

1

(Imaginary part of e

2

F (D )

iax

),

e

iax

1

= Imaginary part of

F (D 2 )

[∵ e

iax

= cos ax + i sin ax]

. 1,

which is now of the form

1 e F ( D)

ax

V of 10.5.

Now the method of 10.5 is to be followed.

Illustration 1 (a) : Solution :

Solve

d2 y dx 2

+ 9 y = cos 3 x .

[UPTU 2002]

The auxiliary equation is m2 + 9 = 0

∴

C. F. = e0

And

P. I. =

x

i. e.,

i. e., m = 0 ± 3i .

(c1 cos 3 x + c 2 sin 3 x) = c1 cos 3 x + c 2 sin 3 x . 1

2

m2 + 32 = 0

D + 32

cos 3 x

= the real part in = the real part in

1 2

D + 32 1 2

D + 32

(cos 3 x + i sin 3 x)

[Note]

e 3 ix ,

by Euler’s theorem of trigonometry. Now

1 D2 + 32

e 3 ix =

1 e 3 ix (D + 3 i ) (D − 3 i )

=

1 e 3 ix , putting 3i for D in the factor D + 3i (3 i + 3 i ) ( D − 3 i )

=

1 e 3 ix . 1 6 i ( D − 3i)

=

1 3 ix 1 1 3 ix 1 e 1= e 1 2i ( D + 3i ) − 3i 6i D

=

1 3 ix x e .x= (cos 3 x + i sin 3 x) , 6i 6i

=−

[Note]

[∵ e

iax

= cos ax + i sin ax]

ix (cos 3 x + i sin 3 x) , 6

  1 i i = 2 = = − i ∵ i i −1  

x x cos 3 x + sin 3 x , 6 6

[∵ i 2 = − 1]

=−i

375

1

x x cos 3 x = the real part in  sin 3 x − i cos 3 x   6 6 D +3

∴

2

2

x sin 3 x . 6

= Hence the complete solution is

y = (C. F. ) + (P. I. ) 1 y = c1 cos 3 x + c 2 sin 3 x + x sin 3 x. 6

or

Remember : While solving the differential equation ( D2 + a2 ) y = cos ax , the P.I. 1 2

D +a

cos ax =

2

Illustration 1 (b) :

Solve

x x sin ax = 2a 2

d2 y dx

2

∫

cos ax dx .

+ a2 y = cos ax.

Solution : Proceed exactly as in Illustration 1 (a). Ans.

y = c1 cos ax + c 2 sin ax + ( x / 2a) sin ax .

Illustration 2 : Solution : Also

Solve ( D2 + a2 ) y = sin ax .

Here as in Illustration 1, C. F. = c1 cos ax + c 2 sin ax . P. I. =

1 2

D + a2

sin ax

= the coefficient of i in = the coefficient of i in Now

1 2

D +a

2

e

iax

=−i

1 2

D + a2 1 D2 + a2

(cos ax + i sin ax) e

iax

[Note]

.

x x cos ax + sin ax . 2a 2a [Proceed as in Illustration 1]

∴

1 D2 + a2

sin ax

x x = the coefficient of i in  − i cos ax + sin ax  2a  2a x =− cos ax . 2a Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 cos ax + c 2 sin ax − ( x / 2a) cos ax . 1 x x Remember : sin ax = − cos ax = ∫ sin ax dx . 2 2 2a 2 D +a Illustration 3 : Solution :

Solve ( D2 + 4) y = sin2 x .

The auxiliary equation is m2 + 2 = 0.

376

∴

m = 0 ± 2i .

Hence

C. F. = c1 cos 2 x + c 2 sin 2 x .

And

P. I. =

1 D2 + 4

sin2

[ ∵ e0

x

= 1]

1

1 x= 2 (2 sin2 x) D +4 2

=

1 1 (1 − cos 2 x) 2 2 ( D + 4)

=

1 1 1 1 1− cos 2 x . 2 D2 + 4 2 D2 + 4

Now

1 1 1 1 2  −1 1 1 1 1 = 1 + D  1 = 1 − D2 + … 1 = ⋅  2     2 D +4 8 4 8 4 8

Again

1 1 1 1 cos 2 x = ⋅ x sin 2 x. 2 D2 + 4 2 4

[Refer Illustration 1.]

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1 cos 2 x + c 2 sin 2 x + − x sin 2 x . 8 8 Illustration 4 : Solution : ∴ And

The auxiliary equation is m2 + 1 = 0 or m = ± i . C. F. = c1 cos x + c 2 sin x . 1 1 1 P. I. = 2 (sin x sin 2 x) = 2 (2 sin x sin 2 x) 2 D +1 D +1 =

Now

Solve ( D2 + 1) y = sin x sin 2 x .

1 1 1 1 1 1 (cos x − cos 3 x) = cos x − cos 3 x . 2 D2 + 1 2 D2 + 1 2 D2 + 1 1 2

D +1

cos x = the real part in

1 2

D +1

= the real part in

1 e ( D + i) ( D − i)

= the real part in

1 e (i + i) ( D − i)

e

ix

ix

ix

= the real part in

1 e 2 i ( D − i)

= the real part in

1 e 2i

ix

1 1 D+ i− i

= the real part in

1 e 2i

ix

1 1 1 = the real part in e D 2i

= the real part in

x (cos x + i sin x) 2i

= the real part in −

ix

.1

1 xi (cos x + i sin x) 2

ix

x

377

1 1 1 x cos x + x sin x = x sin x . 2 2 2 1 1 1 cos 3 x = cos 3 x = − cos 3 x . 8 D2 + 1 − 32 + 1

= the real part in − i Again ∴

the P. I. =

1 1 1 1 1 1 ⋅ x sin x − ⋅ (− cos 3 x) = x sin x + cos 3 x . 2 2 2 8 4 16

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1 cos x + c 2 sin x + x sin x + cos 3 x . 4 16 Illustration 5 :

Solve the differential equation 1 ( D − 1)2 ( D2 + 1)2 y = sin2 x + e x. 2

[UPTU 2002]

Solution : Here the auxiliary equation is (m − 1)2 (m2 + 1)2 = 0. ∴

m = 1, 1, ± i, ± i.

∴

C. F. = (c1 + c 2 x) e x + (c 3 + c 4 x) cos x + (c 5 + c 6 x) sin x. 1 e x + sin2 1 x P.I. = 2 2 2  2  ( D − 1) ( D + 1) 

Also

= =

1 2

2

e 

2

2

1 1 + e 2 2 ( D − 1) ( D2 + 1)2

( D − 1) ( D + 1) 1 2

1 (1 − cos x)  2

2

( D − 1) ( D + 1)

x

+

− Now

1 2

2

1 2

e

2

( D − 1) ( D + 1)

and

−

1 e 4

1

x

( D + 1 − 1) 1 2

x

1=

2

2

2

( D − 1) ( D + 1)

=− =

2

2

( D − 1) ( D + 1)

1 cos x. 2

1 1 1 1 1 e0 x = e0 x = , 2 2 2 2 2 2 ( D − 1) ( D + 1) 2 (0 − 1) (0 + 1) 2 2

=

1 2

1 1 1 = 1 2 2 ( D − 1)2 ( D2 + 1)2

2

( D − 1) ( D + 1) =

x

=

1 2

e x .1

2

(1 + 1) ( D − 1)

1 e 4

x

1 2

D

1=

1 e 4

x

⋅

1 2 1 2 x = x e 2 8

x

,

1 1 1 cos x = − cos x 2 2 2 ( D − 2 D + 1) ( D2 + 1)2

1 1 1 1 cos x = cos x 2 2 2 2 (−1 − 2 D + 1) ( D + 1) 4 ( D + 1)2 D

1 1 sin x 4 ( D2 + 1)2

 ∵ 1 cos x = sin x    D

378

= the coefficient of i in = the coeff. of i in = the coeff. of i in

∴

1 2

e

2

4 ( D + 1) 1 2

2

4 ( D + i) ( D − i) 1

1 2

4 (i + i)

ix

e

ix

e ix . 1

( D − i)2

= the coeff. of i in −

1 e 16

ix

= the coeff. of i in −

1 e 16

ix

= the coeff. of i in −

x2 x2 (cos x + i sin x) = − sin x. 32 32

P. I. =

1 1 2 + x e 2 8

x

−

1 ( D + i − i)2 1 D2

1

1

1 2 x sin x. 32

Hence the general solution is y = (C. F. ) + (P. I. ) or

y = (c1 + c 2 x) e

Illustration 6 :

x

+ (c 3 + c 4 x) cos x + (c 5 + c 6 x) sin x 1 1 1 2 + + x2 e x − x sin x. 2 8 32

Solve ( D3 + a2 D) y = sin ax.

Solution : The auxiliary equation is m3 + a2 m = 0 or

m (m2 + a2 ) = 0, i. e., m = 0, ± ai.

∴

C. F. = c1 + c 2 cos ax + c 3 sin ax .

And

P. I. = =

1 3

2

D +a D

sin ax =

[ ∵ e 0 x = 1]

1 2

( D + a2 ) D

sin ax

1  cos ax   1 sin ax  = − ,   2 2  D a    (D + a ) D +a  1

2

2

[∵ (1 / D) sin ax = =− Now

D +a =

∴

2

cos ax = the real part in

x sin ax. 2a

the P. I. = −

1 2

D + a2

(cos ax + i sin ax)

[Proceed as in Illustration 1 (a)] 1 x x ⋅ sin ax = − 2 sin ax. a 2a 2a

Hence the complete solution is y = (C. F. ) + (P. I. ) or

sin ax dx ]

1 1 cos ax. a D2 + a2

1 2

∫

y = c1 + c 2 cos ax + c 3 sin ax − (1 / 2a) x sin ax.

379

Comprehensive Exercise 7 Solve the following differential equations : 1. ( D2 + 4) y = cos 2 x . 3.

2. ( D2 + 4) y = sin 2 x .

( D2 + 9) ( D2 + 1) y = cos 3 x . 4.

d2 y dx

2

+

dy dx

= sin x . sin 2 x.

A nswers 1. 2.

1 x sin 2 x . 4 1 y = c1 cos 2 x + c 2 sin 2 x − x cos 2 x . 4 y = c1 cos 2 x + c 2 sin 2 x +

1 x sin 3 x . 48 1 1 1 1 + sin x − cos x − sin 3 x + cos 3 x. 4 4 60 20

3.

y = c1 cos 3 x + c 2 sin 3 x + c 3 cos x + c 4 sin x −

4.

y = c1 + c 2 e − x

10.9 To Find P.I. when Q = xV , whereV is any Function of x F ′ ( D) 1 1 ( xV ) = x V − V F ( D) F ( D) { F ( D )} 2 or

[Remember]

 d 1 1 1  (x V ) = x V + V. F ( D) F ( D)  dD F ( D)

Proceeding by repeated application of the above formula {1 / F ( D)} x determined.

m

V can be

Important Note. The method of 10.8 cannot be used when V is of the form cos ax or sin ax and F (− a2 ) = 0 i. e., F ( D2 ) vanishes by putting − a2 for D2 . In that case P.I. is found by the method discussed in 10.7.

Illustration 1 : Solution : ∴

Solve

d2 y dx

2

−2

dy dx

+ y = x sin x .

The auxiliary equation is m2 − 2m + 1 = 0 . m = 1, 1.

380

Hence And

C. F. = (c1 + c 2 x) e x . 1 1 2 P. I. = x sin x = x sin x − sin x , by 10.8 2 2 ( D − 1) ( D − 1) ( D − 1)3 1

= x

2

D − 2D + 1

2

sin x −

3

D − 3 D2 + 3 D − 1

sin x

1 2 sin x − sin x , − 2D − D + 3 + 3D − 1

= x⋅

putting − 12 i. e., − 1 for D2 x 1 2 x 1 ⋅ sin x − sin x = cos x − sin x 2 D 2 ( D + 1) 2 D +1

=− =

( D − 1) x cos x D −1 x cos x − sin x = − 2 sin x 2 ( D + 1) ( D − 1) 2 D −1

=

1 1 1 1 x cos x + ( D − 1) sin x = x cos x + (cos x − sin x) . 2 2 2 2

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = (c1 + c 2 x) e x + x cos x + (cos x − sin x) . 2 2 Illustration. 2 : Solution :

And

d2 y dx 2

+ 4 y = x sin x .

The auxiliary equation is m2 + 4 = 0.

Therefore ∴

Solve

m = 0 ± 2 i. [ ∵ e0

C. F. = c1 cos 2 x + c 2 sin 2 x . P. I. = = =

1

x sin x = x

2

D +4 x sin x

−

2

−1 + 4

1 2

D +4

2D 2

2

(− 1 + 4)

sin x −

2D 2

( D + 4)2

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 2 or y = c1 cos 2 x + c 2 sin 2 x + x sin x − cos x . 3 9 Illustration 3 : Solution :

Solve ( D4 − 1) y = x sin x .

The auxiliary equation is m4 − 1 = 0

or (m2 − 1) (m2 + 1) = 0.

∴

m = 1, − 1, 0 ± i .

∴

C. F. = c1 e

And

P. I. =

x

+ c 2 e − x + e0

1 4

D −1

x sin x .

x

(c 3 cos x + c 4 sin x) .

= 1]

sin x , by 10.8

sin x , putting − 12 for D2

1 2 1 2 x sin x − D (sin x) = x sin x − cos x . 3 9 3 9

x

381

Here we cannot use the method given in 10.8 because D4 − 1vanishes by putting − 12 i. e., − 1 for D2 . So here we shall proceed by the method given in 10.7. ∴

1

P.I. = Imaginary part of = I. P. of e

ix

= I. P. of e

ix

= I. P. of e

ix

= I. P. of e

ix

x e ix , [∵ e ix = cos x + i sin x]

4

D −1

1 4

{( D + i ) − 1}

x , by using the method for e 1

4

3

2

1 3

2

D + 4iD − 6 D − 4iD

x

V

x

( D + 4iD + 6i D2 + 4i3 D + i4 − 1) 4

ax

[ ∵ i 2 = − 1]

1 − 4iD [1 + (3 D / 2 i ) − D2 − ( D3 / 4i )]

x

−1

= I. P. of

e ix 1   3 D − …  ⋅  1 +   − 4i D  2i

= I. P. of

 e ix 1  3D ⋅ + … x 1 − −4i D  2i 

= I. P. of

e ix 1  3 x −  − 4i D  2 i

= I. P. of

e ix  x 2 3 x −   − 4i  2 2i 

x

∵ D ≡ d   dx 

[∵ (1 / D) stands for integration w.r.t. x ] 1 1 3 = I. P. of i e ix ( x 2 + i x) [∵ i2 = − 1 ⇒ 1/ i = − i] 4 2 2 1 1 3 = I. P. of i (cos x + i sin x) ( x 2 + ix) 4 2 2 1 1 2 3 1 2 = ( x cos x − x sin x) = ( x cos x − 3 x sin x) . 4 2 2 8 Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 e

Illustration 4 : Solution :

x

+ c 2 e − x + c 3 cos x + c 4 sin x +

1 2 ( x cos x − 3 x sin x) . 8

Solve ( D4 + 2 D2 + 1) y = x 2 cos x . 4

[UPTU 2007]

2

The auxiliary equation is m + 2m + 1 = 0

or

(m2 + 1)2 = 0 giving m = ± i, ± i .

∴

C. F. = (c1 + c 2 x) cos x + (c 3 + c 4 x) sin x .

And

P. I. =

1 4

2

D + 2D + 1

x 2 cos x

[ ∵ e0

x

= 1]

382

= Real part of = R. P. of e

ix

= R. P. of e

ix

= R. P. of e

ix

1 2

2

( D + 1)

x2 e

1 2

2

{( D + i ) + 1} 1 2

( D + 2iD)2

ix

[∵ e

[By 10.5]

x2

[ ∵ i 2 = − 1]

2

2

4i D [1 + ( D / 2i)]

1 = R. P. of − e 4

= cos x + i sin x ]

x2

1 2

ix

x2

−2

ix

1 ix e 4 1 = R.P. of − e ix 4 = R. P. of −

1 = R.P. of − e 4

ix

= R. P. of −

1 e 4

ix

= R. P. of −

1 e 4

ix

1  D 1+  x2 [ ∵ i 2 = − 1] 2  D  2 i 1 1  ∵ 1 = − i [1 − iD] − 2 x 2 2   2 i D 1  1 1 1 + 2 ⋅ iD + 3 ⋅ i2 D2 + … x 2 , 2   2 4 D  expanding by binomial theorem 1  3 2 1 + iD − D + … x 2 2   4 D  i 3  1  2 2  D2 + D − 4 + terms in D, D , and so on x 1 x 4  1 3 + i x 3 − x 2 + terms in x1 , x 0  ,  3 4 3 4 

(∵ 1 / D stands for integration w.r.t. x) 1 1 4 1 3 3 2 = R. P. of − (cos x + i sin x) ( x + ix − x + terms in x1 , x 0 ) 4 12 3 4 1 1 4 3 2 1 1 3 =− ( x − x ) cos x + ( x ) sin x 4 12 4 4 3 + terms already included in the C.F. 1 4 1 3 2 =− ( x − 9 x ) cos x + x sin x , 48 12 neglecting the terms already included in the C.F. Hence the complete solution is y = (C. F. ) + (P. I. ) or

Note.

y = (c1 + c 2 x) cos x + (c 3 + c 4 x) sin x 1 4 1 3 − ( x − 9 x 2 ) cos x + x sin x . 48 12 1 1 1 (e ax . x . sin bx) = e ax . ( x sin bx) = e ax x sin bx F( D) F( D) F( D + a)    d  1 1 = e ax  x ⋅ sin bx +  ⋅  sin bx ⋅  F( D + a)   dD F( D + a)

383

Illustration 5 : Solution :

Solve ( D2 − 2 D + 1) y = xe

sin x .

The auxiliary equation is m2 − 2m + 1 = 0

∴

m = 1, 1.

∴

C. F. = (c1 + c 2 x) e x . 1 P. I. = 2 xe D − 2D + 1

And

x

1

x

=e

x

=e

x

=e

x

=e

x

∫

=e

x

[∫

=e

x

[− x sin x +

=e

x

(− x sin x − 2 cos x) .

{( D + 1) − 1}2 1 D

( D − 1)2

x

e

x sin x

x sin x , by 10.5

( x sin x) = e

2

1

sin x =

or (m − 1)2 = 0.

x

1 D

⋅

∫

x sin x dx ,

[∵ (1 / D) stands for integration w.r.t. x] 1 (− x cos x + sin x) , integrating by parts D (− x cos x + sin x) dx − x cos x dx +

sin x dx ]

∫

sin x dx +

∫

∫

sin x dx ]

Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = (c1 + c 2 x) e

Illustration 6 :

Solve

d

2

dx

y

x

−e

x

( x sin x + 2 cos x) .

− y = x sin x + (1 + x 2 ) e x .

2

[UPTU 2001 (Special)]

Solution : The auxiliary equation is m2 − 1 = 0. ∴

m = ± 1.

Hence

C. F. = c1 e x + c 2 e − x . 1 P. I. = {x sin x + (1 + x 2 ) e x } 2 ( D − 1)

And

= where

P1 =

1 ( D2 − 1) 1 2

( D − 1)

= I. P. of e

ix

x sin x +

1 ( D2 − 1)

(1 + x 2 ) e

x sin x = Imaginary part of 1 2

( D + i) − 1

x = I. P. of e

= I. P. of e ix . (1 / −2) [1 − (iD +

x

= P1 + P2 (say), 1

2

D −1

xe

ix

1

ix 2

D + 2iD − 2

1 2 −1 D )] x 2

x

...(1)

384

1 ix 1 e [1 + iD +...] x = I. P. of − e ix [ x + i] 2 2 1 1 = I. P. of − (cos x + i sin x) ( x + i) = − (cos x + x sin x), 2 2 1 1 P2 = 2 (1 + x 2 ) e x = e x (1 + x 2 ) 2 D −1 {( D + 1) + 1} = I. P. of −

and

=e

∴

1

x 2

D + 2D

(1 + x 2 ) = e

1

x

1 2 D (1 + D) 2

(1 + x 2 )

=

 ex 1  D  −1 ex 1  D D2 2 ⋅ ⋅ + −... (1 + x 2 ) 1 − 1 +  (1 + x ) =   2 D 2 2 D 2 4 

=

ex 1 1 1 ex 1  2 3 ⋅ [(1 + x 2 ) − (2 x) + ⋅ 2] = ⋅ x − x+  2 D 2 4 2 D  2 

=

1 e 2

x

1 x 3 − 1 x 2 + 3 3 2 2

1 x = e  12

x

(2 x 3 − 3 x 2 + 9 x).

from (1), P. I. = −

1 (cos x + x sin x) + (1 / 12) e 2

x

(2 x 3 − 3 x 2 + 9 x).

Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 e

x

+ c2 e − x −

1 (cos x + x sin x) + (1 / 12) e 2

x

(2 x 3 − 3 x 2 + 9 x).

Illustration 7 : Solve ( D2 − 4 D + 4) y = 8 x 2 e 2 x sin 2 x.

(UPTU 2005)

Solution : The auxiliary equation is m2 − 4m + 4 = 0 or (m − 2)2 = 0. ∴ m = 2, 2. ∴

C. F. = (c1 + c 2 x) e 2 x .

And

P. I. =

1 2

D − 4D + 4

= 8e 2 x = 8e 2 x

8 x 2 e 2 x sin 2 x = 8

1 2

{( D + 2) − 2} 1 D2

1 2

( D − 2)

e 2 x x 2 sin 2 x

x 2 sin 2 x, by 10.5

x 2 sin 2 x

= I.P. of 8e 2 x

1 D2

= I.P. of 8e 2 x e

x2 e

i2 x

i2 x

[∵ e

,

1 2

( D + 2i)

i2 x

= cos 2 x + i sin 2 x ]

x 2 , by 10.5

385

= I.P. of 8e 2 x e

1

i2 x

= I.P. of −2e 2 x e

2

2

4i [1 + ( D / 2i)] i2 x

x2

[ 1 − (i / 2) D ] −2 x 2 , [ ∵ i 2 = − 1, (1 / i) = − i ]

= I.P. of −2e 2 x e

i2 x

[1 + 2 (i / 2) D + 3. (i 2 / 4) D2 +...] x 2

= I.P. of −2e 2 x e

i2 x

[1 + iD − (3 / 4) D2 +...] x 2

= I.P. of −2e 2 x (cos 2 x + i sin 2 x) {x 2 + i2 x − (3 / 2)} = − 2e 2 x [2 x cos 2 x + {x 2 − (3 / 2)} sin 2 x] = e 2 x [ − 2 x 2 sin 2 x + 3 sin 2 x − 4 x cos 2 x]. Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = (c1 + c 2 x) e 2 x + e 2 x (−2 x 2 sin 2 x + 3 sin 2 x − 4 x cos 2 x).

Comprehensive Exercise 8 Solve the following differential equations : 1.

d2 y / dx 2 + 9 y = x sin x .

2. ( D2 + 2 D + 1) y = x cos x . 3. ( D2 + m2 ) y = x cos mx . 4. ( D2 + 4) y = x sin 2 x . 5. ( D2 + D) y = x cos x . 6. ( D2 − 1) y = x 2 cos x . 7. ( D2 + 1) y = x 2 sin 2 x . 8. ( D4 − 1) y = x sin 2 x. 9. ( D2 + 4) x = t 2 + sin 2t.

A nswers 1 1 x sin x − cos x . 8 32 1 1 1 + x sin x − sin x + cos x . 2 2 2

1.

y = c1 cos 3 x + c 2 sin 3 x +

2.

y = (c1 + c 2 x) e − x

3.

y = c1 cos mx + c 2 sin mx + ( x / 4m2 ) cos mx + ( x 2 / 4m) sin mx .

386

1 2 1 x cos 2 x + x sin 2 x . 8 16 1 1 + c 2 e − x − x (cos x − sin x) + cos x + sin x . 2 2 1 e x + c 2 e − x − ( x 2 − 1) cos x + x sin x . 2 1 cos x + c 2 sin x − [24 x cos 2 x − (9 x 2 − 26) sin 2 x] . 27 1  32 e x + c 2 e − x + c 3 cos x + c 4 sin x + cos 2 x + x sin 2 x ⋅   15  15

4.

y = c1 cos 2 x + c 2 sin 2 x −

5.

y = c1

6.

y = c1

7.

y = c1

8.

y = c1

9.

y = c1 cos 2t + c 2 sin 2t +

10.10 The Operator

1 2 1 1 t − − t cos 2t. 4 8 4

1 1 and , α being a Constant D−α D +α

If Q is any function of x, then 1 1 Q = e αx ∫ e − αx Q dx and Q = e− α D−α D+α

Solution : ∴ And

Solve

∫

eα

x

Q dx.

d2 y

+ a2 y = sec ax . dx 2 The auxiliary equation is m2 + a2 = 0,

Illustration 1 :

x

or

m = ± ia .

C. F. = c1 cos ax + c 2 sin ax . 1 1 P. I. = 2 sec ax = sec ax 2 ( D + ia) ( D − ia) D +a =

1  1 1  −   sec ax 2 ia  D − ia D + ia

[By resolving into partial fractions]  1  1 1 = sec ax − sec ax  2 ia  D − ia D + ia  1 iax − iax − iax iax = [e ∫ e sec ax dx − e ∫ e sec ax dx] , by 10.9 2 ia =

1 2 ia

 e 

iax

∫

cos ax − i sin ax cos ax

dx − e − iax ∫

[ ∵ e − iax = cos ax − i sin ax and e

cos ax + i sin ax cos ax iax

 dx , 

= cos ax + i sin ax]

387

=

1 2 ia

 e

 x + i log cos ax − e − iax  x − i log cos ax      a a    

iax

 e − e − iax  1  + 2 (log cos ax)  2i  a  x 1 = sin ax + 2 (log cos ax) cos ax . a a =

 e  

x a

iax

iax

+ e − iax   2 

Hence the complete solution is or

y = (C. F. ) + (P. I. ) x 1 y = c1 cos ax + c 2 sin ax + sin ax + 2 cos ax . log (cos ax) . a a

Illustration 2 :

Solve

d2 y dx 2

+ 9 y = sec 3 x.

[UPTU 2002]

Solution : Proceeding as in illustration 1, we have

And

C. F. = c1 cos 3 x + c 2 sin 3 x. 1 1 P. I. = 2 sec 3 x = sec 3 x ( D + 3i) ( D − 3i) D +9 =

1 6i

 1 1  −   sec 3 x,  D − 3i D + 3i  by resolving into partial fractions

Now putting

 1  1 sec 3 x − sec 3 x   D + 3i  D − 3i 

=

1 6i

=

1 3 ix [e ∫ e −3 ix sec 3 x dx − e −3 ix ∫ e3 ix sec 3 x dx], by 10.9 6i

=

1  3 ix e 6i 

=

1  3 ix  i i −3 ix    e  x + log cos 3 x − e  x − log cos 3 x    6i  3 3    

=

x e ⋅ 3

3 ix

e3 ix − e −3 ix 2i P. I. =

∫

cos 3 x − i sin 3 x

− e −3 ix 2i

cos 3 x

+

dx − e −3 ix

∫

cos 3 x + i sin 3 x cos 3x

(log cos 3 x) e3 ix + e −3 ix ⋅ ⋅ 9 2

= sin 3 x and

e3 ix + e −3 ix 2

 dx 

(Note)

= cos 3 x, we get

1 1 x sin 3 x + cos 3 x log (cos 3 x). 3 9

Hence the complete solution is y = (C. F. ) + (P. I. ) 1 1 or y = c1 cos 3 x + c 2 sin 3 x + x sin 3 x + cos 3 x log (cos 3 x). 3 9

388

Illustration 3 :

Solve ( D2 + a2 ) y = cosec ax.

Solution : The auxiliary equation is m2 + a2 = 0 or m = ± ai. ∴ And

C. F. = c1 cos ax + c 2 sin ax. 1 1 P. I. = 2 cosec ax = cosec ax 2 ( D + ia ) ( D − ia) D +a =

1 2ai

 1 1  −   cosec ax D − ia D + ia 

=

1 2ai

 1  1 cosec ax − cosec ax   D − ia D + ia  

=

1 [e 2ai

=

1 2ai

 e 

iax

∫

iax

∫

e − iax cosec ax dx − e − iax ∫ e cos ax − i sin ax sin ax

iax

cosec ax dx] , [by 10.9] cos ax + i sin ax

dx − e − iax ∫

sin ax

[ ∵ e − iax = cos ax − i sin ax and e =

1 2ai

=

e  (log sin ax )  a2 

=

 e

iax

a

2

= cos ax + i sin ax]

 1 log sin ax − ix − e − iax  1 log sin ax + ix      a  a 

1 1

iax

 dx, 

iax

− e − iax  x  e  −   a 2i  

sin ax (log sin ax) −

iax

+ e − iax    2 

x cos ax. a

Hence the complete solution is y = (C. F. ) + (P. I. ) or

y = c1 cos ax + c 2 sin ax + (1 / a2 ) sin ax log sin ax − ( x / a) cos ax.

Illustration 4 :

Solve ( D2 + a2 ) y = tan ax.

Solution : Here the C. F. = c1 cos ax + c 2 sin ax, as in Illustration 3. And

Now

P. I. =

1 2

D +a

2

tan ax =

1 tan ax ( D + ia) ( D − ia)

=

1  1 1  −   tan ax 2ai  D − ia D + ia

=

1 2ai

 1  1 tan ax − tan ax ⋅  D − ia D + ia  

1 tan ax = e D − ia

iax

∫

e − iax tan ax dx,

=e

iax

∫

(cos ax − i sin ax).

[by 10.9] sin ax cos ax

dx

389

Again

∫

 sin2 ax sin ax − i  cos ax 

iax

∫

 1 − cos 2 ax sin ax − i  cos ax 

=e

iax

∫

{sin ax − i sec ax + i cos ax} dx

=e

iax

=e

iax

=e

 cos ax  −  − ie  a 

iax

1 tan ax = e − iax ∫ e D + ia

∫

  dx     dx  

sec ax dx + ie

iax

= e − iax ∫ (cos ax + i sin ax) ⋅

iax

 sin ax   ⋅  a 

...(1)

[by 10.9]

tan ax dx, sin ax cos ax

dx

 sin2 ax   dx = e − iax ∫ sin ax + i cos ax    1 − cos 2 ax = e − iax ∫ sin ax + i cos ax 

  dx  

= e − iax ∫ {sin ax + i sec ax − i cos ax} dx  cos ax  − iax − iax = e − iax  −  + ie ∫ sec ax dx − ie  a 

 sin ax     a 

...(2)

Subtracting (2) from (1) and dividing by 2ai, we find that the P.I.

=−

e

iax

− e − iax cos ax 1 e ⋅ − 2i a a2

iax

+ e − iax 2

∫

sec ax dx +

=−

sin ax cos ax a2

−

e

iax

+ e − iax sin ax ⋅ 2 2 a

1 1 1 1 (cos ax) ⋅ log tan  π + ax 4 a a 2  +

cos ax sin ax a2

1 1 = (−1 / a2 ) cos ax log tan ( π + ax). 4 2 Hence the complete solution is y = (C. F. ) + (P. I. ) or

1 1 y = c1 cos ax + c 2 sin ax − (1 / a2 ) cos ax. log tan ( π + ax). 4 2

390

Comprehensive Exercise 9 Fill in the Blanks Fill in the blanks “……” so that the following statements are complete and correct. 1. The general solution of the equation f ( D) y = 0 is called the …… of the equation f ( D) y = Q. 2. The solution of the differential equation

d2 y dx 2

+ a2 y = 0 is …… .

3. The complementary function (C.F.) of the differential equation d2 y dx

2

−5

dy

+ 4 y = 0 is …… .

dx

4. The particular integral (P.I.) of the differential equation d3 y dx

3

−4

d2 y dx

2

+5

dy dx

− 2 = 0 is …… .

5. The particular integral of the differential equation f ( D) y = Q will be …..… . 6.

The number of arbitrary constants in general solution of second order differential equation is .........

[UPTU 2011]

Multiple Choice Questions Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 7. The solution of differential equation (a) c1 e − x + c 2 e

d2 y dx 2

+ y = 0 is

x

(b) c1 cos x + c 2 sin x (c) (c1 + c 2 x) cos x + (c 3 + c 4 x) sin x (d) None of these. 8. P.I. of the differential equation ( D2 + D + 1) y = e (a)

1 e 3

(c) e

x

x

(b) 3 e

x

(d) None of these.

x

is

391

9. For the differential equation F ( D) y = e P.I. =

1 e F ( D)

1 e F (a)

(a)

(c) e

ax

, if F (a) = 0, then

is given by

a x

(b)

1 1 F ( D + a)

a x

a x

1 e F (− a)

a x

(d) None of these.

10. For the particular integral

1 f ( D2 )

sin ax when f (− a2 ) = 0, which one of the

following is correct 1 1 (a) sin ax = sin ax 2 f (D ) f (− a2 ) (b)

1 2

D +a

(c)

2

sin ax = −

1 2

D +a

sin ax =

2

x cos ax 2a

x cos ax 2a

(d) None of these. 11. The solution of the differential equation (a) y = c1 e

x

d2 y dx 2

−3

dy dx

+2 y=e

x

is

+ c2 e 3 x + x

(b) y = (c1 + c 2 ) e x − xe x (c) y = c1 e x + c 2 e 2 x − xe (d) y = c1 x + c 2 e

x

x

− e 2 x.

12. The solution of the differential equation

d2 y dx

2

−2

dy dx

(a) y = (c1 + c 2 x) e − x

(b) y = (c1 − c 2 x) e − x

(c) (c1 + c 2 x)`e x

(d) None of these

+ y = 0 is

[UPTU 2011]

True or False Write ‘‘ T ’’ for true and ‘‘F’’ for false statement. 13. A linear differential equation is an equation in which the dependent variable and its derivatives appear only in the first degree. [UPTU 2011]

14.

d4 y dx 4

−5

d3 y dx 3

−2

dy dx

+ 4 xy = 0 is a linear differential equation with

constant coefficients. 15. e − x (c1 cos √ 3 x + c 2 sin √ 3 x) + c 3 e 2 x is the solution of the differential equation

d3 y dx 3

− 8 y = 0.

392

A nswers 1. 4.

C.F. 2 x. 5

8. (a). 12. (d)

y = c1 cos ax + c 2 sin ax. 1 6. Two 5. Q. F ( D) 2.

9. (c). 13. T.

10. (b). 14. F.

3. c1 e

x

7. (b). 11. (c). 15. T.

+ c2 e 4 x .

393

LOGARITHMS 0

1

2

3

4

10

0000 0043 0086 0128 0170

5

6

7

8

9

11

0414 0453 0492 0531 0569

12

0792 0828 0864 0899 0934

13

1139 1173 1206 1239

14

1461 1492 1523 1553

15

1761 1790 1818 1847

16

2041 2068 2095 2122

17

2304 2330 2355 2380

18

2553 2577 2601 2625

19

2788 2810 2833 2856

2878 2900 2923 2945 2967 2989

123 5 9 13 4 8 12 4 8 12 4 7 11 3 7 11 3 7 10 3 6 10 3 7 10 3 6 9 3 6 9 3 6 9 3 6 8 3 6 8 3 5 8 3 5 8 3 5 8 2 5 7 2 4 7 2 4 7 2 4 6

20

3010 3032 3054 3075 3096 3118 3139 3160 3181 3201

2 4 6

4 5 6 17 21 26 16 20 24 16 20 23 15 18 22 14 18 21 14 17 20 13 16 19 13 16 19 12 15 19 12 14 17 11 14 17 10 14 17 11 14 16 10 13 16 10 13 15 10 12 15 9 12 14 9 11 14 9 11 13 8 11 13 8 11 13

21

3222 3243 3263 3284 3304 3324 3345 3365 3385 3404

2 4 6

8 10 12

14 16 18

22

3424 3444 3464 3483 3502 3522 3541 3560 3579 3598

2 4 6

8 10 12

14 15 17

23

3617 3636 3655 3674 3692 3711 3729 3747 3766 3784

2 4 6

7 9 11

13 15 17

24

3802 3820 3838 3856 3874 3892 3909 3927 3945 3962

2 4 5

7 9 11

12 14 16

25

3979 3997 4014 4031 4048 4065 4082 4099 4116 4133

2 3 5

7 9 10

12 14 15

26

4150 4166 4183 4200 4216 4232 4249 4265 4281 4298

2 3 5

7 8 10

11 13 15

27

4314 4330 4346 4362 4378 4293 4409 4425 4440 4456

2 3 5

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28

4472 4487 4502 4518 4533 4548 4564 4579 4594 4609

2 3 5

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29

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1 3 4

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30

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1 3 4

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31

4914 4928 4942 4955 4969 4983 4997 5011 5024 5038

1 3 4

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32

5051 5065 5079 5092 5105 5119 5132 5145 5159 5172

1 3 4

5 7 8

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33

5185 5198 5211 5224 5237 5250 5263 5276 5289 5302

1 3 4

5 6 8

9 10 12

34

5315 5328 5340 5353 5366 5378 5391 5403 5416 5428

1 3 4

5 6 8

9 10 11

35

5441 5453 5456 5478 5490 5502 5514 5527 5539 5551

1 2 4

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36

5563 5575 5587 5599 5611 5623 5635 5647 5658 5670

1 2 4

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37

5682 5694 5705 5717 5729 5740 5752 5763 5775 5786

1 2 3

5 6 7

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38

5798 5809 5821 5832 5843 5855 5866 5877 5888 5899

1 2 3

5 6 7

8 9 10

39

5911 5922 5933 5944 5955 5966 5977 5988 5999 6010

1 2 3

4 5 7

8 9 10

40

6021 6031 6042 6053 6064 6075 6085 6096 6107 6117

1 2 3

4 5 6

8 9 10

41

6128 6138 6149 6160 6170 6180 6191 6201 6212 6222

1 2 3

4 5 6

7 8 9

42

6232 6243 6253 6263 6274 6284 6294 6304 6314 6325

1 2 3

4 5 6

7 8 9

43

6335 6345 6355 6365 6375 6385 6395 6405 6415 6425

1 2 3

4 5 6

7 8 9

44

6435 6444 6454 6464 6474 6484 6493 6503 6513 6522

1 2 3

4 5 6

7 8 9

45

6532 6542 6551 6561 6571 6580 6590 6599 6609 6618

1 2 3

4 5 6

7 8 9

46

6628 6637 6646 6656 6665 6675 6684 6693 6702 6712

1 2 3

4 5 6

7 7 8

47

6721 6730 6739 6749 6758 6767 6776 6785 6794 6803

1 2 3

4 5 6

6 7 8

48

6812 6821 6830 6839 6848 6857 6866 6875 6884 6893

1 2 3

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49

6902 6911 6920 6928 6937 6946 6955 6964 6972 6981

1 2 3

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6 7 8

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1271 1303 1335 1367 1399 1430 1584 1614 1644 1673 1703 1732 1875 1903 1932 1959 1987 2014 2148 2175 2201 2227 2253 2279 2405 2430 2455 2480 2504 2529 2648 2672 2695 2718 2742 2765

7 30 28 27 26 25 24 23 22 22 20 20 19 19 18 18 17 17 16 16 15 15

8 34 32 31 29 28 27 26 25 25 23 23 22 22 21 20 20 19 18 18 17 17

9 38 36 35 33 32 31 29 29 28 26 26 25 24 23 23 22 21 21 20 19 19

394

LOGARITHMS 0

1

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5

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1

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7160 7168 7177 7185 7193 7202 7210

7218 7226 7235 1

2 2

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53

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7300 7308 7316 1

2 2

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6 6 7

54

7324 7332 7340 7348 7356 7364 7372

7380 7388 7396 1

2 2

3 4

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6 6 7

55

7404 7412 7419 7427 7435 7443 7451

7459 7466 7474 1

2 2

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56

7482 7490 7497 7505 7513 7520 7528

7536 7543 7551 1

2 2

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7612 7619 7627 1

2 2

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58

7634 7642 7649 7657 7664 7672 7679

7686 7694 7701 1

1 2

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59

7709 7716 7723 7731 7738 7745 7752

7760 7767 7774 1

1 2

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5 6 7

60

7782 7789 7796 7803 7810 7818 7825

7832 7839 7846 1

1 2

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61

7853 7860 7868 7875 7882 7889 7896

7903 7910 7917 1

1 2

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62

7924 7931 7938 7945 7952 7959 7966

7973 7980 7987 1

1 2

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63

7993 8000 8007 8014 8021 8028 8035

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1 2

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67

8261 8267 8274 8280 8287 8293 8299

8306 8312 8319 1

1 2

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1 2

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8432 8439 8445 1

1 2

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1 2

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8633 8639 8645 8651 8557 8663 8669

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1 2

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1 2

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1 2

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8921 8927 8932 8938 8943 8949 8954

8960 8965 8971 1

1 2

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8976 8982 8987 8993 8998 9004 9009

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1 2

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1 2

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1 2

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9395 9400 9405 9410 9415 9420 9425

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1 1

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9445 9450 9455 9460 9465 9469 9474

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1 1

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9494 9499 9504 9509 9513 9518 9523

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1 1

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9542 9547 9552 9557 9562 9566 9571

9576 9581 9586 0

1 1

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9590 9595 9600 9605 9609 9614 9619

9624 9628 9633 0

1 1

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92

9638 9643 9647 9652 9657 9661 9666

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1 1

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93

9685 9689 9694 9699 9703 9708 9713

9717 9722 9727 0

1 1

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9731 9736 9741 9745 9750 9754 9759

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1 1

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9777 9782 9786 9791 9795 9800 9805

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1 1

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96

9823 9827 9832 9836 9841 9845 9850

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1 1

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9868 9872 9877 9881 9886 9890 9894

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395

ANTILOGARITHMS 0

1

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1274

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1291

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1297

1300

1303

1306

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1312

1315

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1318

1321

1324

1327

1330

1334

1337

1340

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1950

1954

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ANTILOGARITHMS 0

1

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1

Unit IV

A ppendix Integration

§ 1. Integration The process inverse to differentiation is defined as integration. d F ( x) = f ( x), we say that F ( x) is an integral or a primitive of f ( x), and, in dx symbols, we write ∫ f (x) dx = F (x) .The letter x in dx denotes that the integration Thus if

is to be performed with respect to the variable x. The letter x in dx denotes that the integration is to be performed with respect to the variable x. The process of determining an integral of a function is called integration and the function to be integrated is called integrand. § 2. Constant of Integration As the differential coefficient of a constant is zero, if d [ F ( x) + c ] = f ( x), dx therefore

∫f

d F ( x) = f ( x) ;we have dx

(x) dx = F(x) + c.

This constant c is called a constant of integration and can take any constant value. Also ∫ f ( x) dx is called the indefinite integral of f ( x) w.r.t. `x' ; for by giving different values to the constant of integration, the indefinite nature is preserved. § 3. Some Properties of Integral (i)

d { f ( x) dx} = f ( x). dx ∫

(ii)

The integral of the product of a constant and a function is equal to the product of the constant and the integral of the function.

2

Thus if λ is a constant, then

∫λ f

(x) dx = λ

∫

f (x) dx.

(iii) The integral of a sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the functions. Symbolically.

∫[ ∫

f1 ( x) ± f 2 ( x) ± . . . ± f n ( x)] dx =

∫

f1 ( x) dx ± ∫ f 2 ( x) dx ± . . . ±

f n ( x) dx.

§ 4. Standard Results. (i)

d  x n +1    = x n, n √ − 1 dx  n + 1 

→

∫

(ii)

d 1 ( log | x |) = , x √ 0 dx x

→

∫

→

∫e

x

dx = e

a

x

(iii) d (e x ) = e dx

x

n

x

dx =

x n +1 + c, n +1

1 dx = log ∞x∞ + c , x x

+ c,

(iv)

d  ax    = a x , a > 0, a √ 1 dx  log a

→

∫

(v)

d (sin x) = cos x dx

→

∫ cos x dx = sin x + c,

(vi)

d (− cos x) = sin x dx

→

∫ sin x dx = − cos x + c,

(vii) d (tan x) = sec2 x dx

→

∫ sec

(viii) d (− cot x) = cosec 2 x dx

→

∫ cosec

2

dx =

ax + c, log a

x dx = tan x + c , 2

x dx = − cot x + c ,

(ix)

d (sec x) = sec x tan x dx

→

∫sec x tan x dx = sec x + c,

(x)

d (− cosec x) = cosec x cot x dx

→

∫ cosecx cot x dx = –cosec x + c,

(xi)

d (sin − 1 x) = dx

→

∫

1 1 − x2

1 1 − x2

dx = sin − 1 x + c ,

3

1 (xii) d (tan − 1 x) = dx 1 + x2

→

∫

(xiii) d (sec − 1 x) = dx

→

∫

Illustration 1:

1 x2 − 1

x

1+ x

2

dx = tan − 1 x + c ,

1

dx = sec − 1 x + c .

x2 − 1

x

Evaluate :

1 5  3 −5 ∫ 5 x + 2 x − 7 x + x + x  dx (iii) ∫ (3 sin x − 2 cos x + 4 sec 2 x − 5 cosec (i)

(iv)

1

(ii) 2

∫

x3 + 5 x2 + 4 x + 1 x2

dx

x) dx

∫ (1 − x) (2 + 3 x) (5 − 4 x) dx.

Solution : (i)

Let Then

I =

∫

I =5 =5⋅

=

1 5  3 −5 − 7x + +  dx. 5 x + 2 x  x x

∫

x 3 dx + 2

∫

x − 5 dx − 7

x dx +

∫

x − 1 /2 dx + 5

∫

x4 x–4 x2 x1 /2 +2⋅ −7⋅ + + 5 log | x | + c 4 (– 4) 2 (1 / 2)

5 x4 1 7 x2 − − +2 4 2 2 x4

x + 5 log | x | + c .

x3 + 5 x2 + 4 x + 1

(ii) Let

I =

∫

Then

I =

∫

=

∫

=

x2 1 + 5 x + 4 log | x | − + c . 2 x

(iii) Here

∫

 x 

dx. x2 4 1 + 5 + + 2  dx, dividing each term by x 2 x x 

x dx + 5

∫

dx + 4

∫

1 dx + x

∫

x − 2 dx

I = ∫ (3 sin x − 2 cos x + 4 sec2 x − 5 cosec2 x) dx = 3∫ sin x dx − 2 ∫ cos x dx + 4 ∫ sec2 x dx − 5 ∫ cosec 2 x dx = − 3 cos x − 2 sin x + 4 tan x + 5 cot x + c .

1 dx x

4

I = ∫ (1 − x) (2 + 3 x) (5 − 4 x) dx = ∫ (10 − 3 x − 19 x 2 + 12 x 3 ) dx

(iv) Here

= 10

∫ dx − 3 ∫ x dx − 19 ∫ x

= 10 x − 3 ⋅

2

dx + 12

∫x

3

dx

x2 x3 x4 3 x 2 19 x 3 − 19 ⋅ + 12 ⋅ + c = 10 x − − 2 3 4 2 3 + 3 x 4 + c.

Illustration 2 : 2

(i)

∫ tan

(iii)

∫

(v)

∫

(vii)

∫

Evaluate :

x dx

cos x

dx [UPTU 2001(Special)] sin2 x 1 ` dx 2 sin x cos2 x sec x dx. (sec x + tan x)

2

(ii)

∫ cot

(iv)

∫ (tan x + cot x)

(vi)

∫

x dx 2

cos 2 x 2

sin

x cos2 x

dx dx

Solution : (i)

Here

I =

∫

tan2 x dx =

= ∫ sec 2 x dx −

∫

(sec 2 x − 1) dx

∫ dx = tan x − x + c.

I = ∫ cot 2 x dx = ∫ (cosec2 x − 1) dx

(ii) Here

= ∫ cosec2 x dx − ∫ dx = − cot x − x + c . cos x

 cos x

(iii) Here

I =

(iv) Here

I = ∫ (tan x + cot x)2 dx

∫ sin2

dx =

x = − cosec x + c.

1 

∫  sin x ⋅ sin x  dx = ∫ cot x cosec x dx

= ∫ (tan2 x + cot 2 x + 2 tan x .cot x) dx = ∫ (tan2 x + cot 2 x + 2) dx = ∫ {(1 + tan2 x) + (1 + cot 2 x)} dx = ∫ (sec 2 x + cosec 2 x) dx = ∫ sec 2 x dx + ∫ cosec 2 x dx = tan x − cot x + c . (v) Here

I =

∫ sin2

1 x cos 2 x

dx =

 sin2 x + cos 2 x  1   1 ∫  sin2 x cos2 x  dx = ∫  cos2 x + sin2 x  dx

= ∫ sec 2 x dx + ∫ cosec 2 x dx = tan x − cot x + c .

5

(vi) Here

I =

cos 2 x

∫ sin2

2

x cos x

dx =

∫

cos 2 x − sin2 x 2

2

sin x cos x



dx =

1

∫  sin2

x

−

1

  dx cos x  2

= ∫ cosec 2 x dx − ∫ sec 2 x dx = − cot x − tan x + c . (vii) Here

I =

=

sec x

sec x (sec x − tan x)

∫ sec x + tan x dx = ∫ (sec x + tan x) (sec x − tan x) dx ∫

(sec 2 x − sec x tan x) sec 2 x − tan2 x

dx = ∫ (sec 2 x − sec x tan x) dx

= ∫ sec 2 x dx − ∫ sec x tan x dx = tan x − sec x + c . Illustration 3: (i)

∫

Evaluate : (ii)

1 + sin 2 x dx

dx ⋅ 1 + sin x

∫

Solution : (i)

Let

I =

∫

1 + sin 2 x dx.

Then

I =

∫

cos 2 x + sin2 x + 2 sin x cos x dx =

(cos x + sin x)2 dx

∫

= ∫ (cos x + sin x) dx = ∫ cos x dx + ∫ sin x dx = sin x − cos x + c (ii) Here

I = =

dx

∫ 1 + sin x 1 − sin x

∫ 1 − sin2

x

=

1

(1 − sin x)

∫ (1 + sin x) ⋅ (1 − sin x) dx

dx =

∫

1 − sin x cos 2 x

dx =



1

∫  cos2

x

−

sin x   dx cos 2 x 

= ∫ (sec 2 x − sec x tan x) dx = ∫ sec 2 x dx − ∫ sec x tan x dx = tan x − sec x + c .

Comprehensive Exercise 1 Integrate the following functions with respect to x : 1.

(i)

x4

(ii) x − 7

(iii) x 5 /3

(iv) 2

(v)

(vi)

3

x2

x

1 4

(vii) 32 log 3

x

(viii) log

x3 x

x

6

2.

(ix) x 3 + 5 x 2 − 4 + 7 + 2 x x

(x)

(i) (1 − x)

(ii)

(iii)

x

x m + + x m x

m

x

+m

3.

(vi)

x2 – 1

(i) 3 sin x − 4 cos x + (ii) 1 +

1 1+ x

2

5 cos

2

2

−

cot x

+ 2

− tan2 x −

−

x

1− x (iii) sec x (sec x + tan x) (iv)

a 2

x

tan x

sin x cos x (v) cosec x (cosec x − cot x)

+

6 sin2 x 5 x

2

+

b +c x

x (1 + x)2 x x6 – 1

(viii)

x2 + 1

5 1 − ⋅ x x1 /3

x (ax 2 + bx + c )

(iv)

(v) (2 − 5 x) (3 + 2 x) (1 − x) (vii)

2x +

x2 + 1

+ tan2 x − cot 2 x +3

x

−1

2 cos 2 x

(vi) 4 − 5 cos x sin2 x

(vii)

1 + 2 sin x cos 2 x

⋅

Evaluate the following integrals : 4.

(i)

∫

(iii)

∫ (1 + cos 2 x) dx

(v)

∫ 1 − sin x dx

1 + cos 2 x dx 1

(vii) ∫

sin x

cosec x cosec x − cot x

∫ (1 − cos x) dx

(iv)

∫ 1 + cos x dx

(vi)

∫ sec x + tan x dx

cos x

(viii) ∫

dx

A

1.

1

(ii)

tan x

cot x cosec x − cot x

dx.

A nswers (i)

x5 + c. 5

(ii) −

1 6 x6

+c.

(iii)

3 5 /3 x + c . (vi) 4 x1 /4 + c . (vii) 5 x4 5 3 (ix) + x − 4 x + 7 log| x| + 4 x + c . 4 3 (v)

2x + c. log 2

3 8 /3 x + c. 8

(iv)

x3 + c. 3

(viii) x + c .

7

2.

(x)

2x 3 + 5 log | x | − x 2 /3 + c . log 2 2

(i)

2 3 /2 2 5 /2 x − x + c. 3 5

(iii) −

a + b log | x | + cx + k . x

(v) 6 x −

17 x 2 x3 5 x4 + + + c. 2 3 2

(vii) x − 2 tan − 1 x + c . 3.

(i)

2ax 7 /2 2bx 5 /2 2cx 3 /2 + + +k. 7 5 3

(ii) (iv)

x2 x m +1 mx + m log | x |+ + + c. 2m m + 1 log m

(vi) 2 (viii)

x +

4 3 /2 2 5 /2 x + x + c. 3 5

x5 x3 − + x − 2 tan − 1 x + c . 5 3

− 3 cos x − 4 sin x + 6 tan x + 7 cot x + c .

(ii) x + tan − 1 x − 2 sin − 1 x + 5 sec − 1 x +

3x + c. log 3

(iii) tan x + sec x + c .

(iv) − cosec x + tan x + x − sec x + c .

(v)

(vi) − 4 cot x + 5 cosec x + c .

− cot x + cosec x + c .

(vii) tan x + 2 sec x + c . 4.

(i) (iii)

2 sin x + c . 1 tan x + c . 2

(v) sec x + tan x − x + c . (vii) − cot x − cosec x + c .

(ii) − cot x − cosec x + c . (iv) x + cot x − cosec x + c . (vi) sec x − tan x + x + c . (viii) − cosec x − cot x − x + c .

§ 5. Extended Forms of Fundamental Formulae. n +1 1 (ax + b) ⋅ + c , n √ − 1. a n +1 1 1 (ii) ∫ dx = ⋅ log | ax + b | + c . ax + b a 1 ax + b ax + b (iii) ∫ e dx = e + c. a 1 a bx + c (iv) ∫ a bx + c dx = ⋅ + k, a > 0 and a √ 1. b log a 1 (v) ∫ sin (ax + b) dx = − cos (ax + b) + c . a 1 (vi) ∫ cos (ax + b) dx = sin (ax + b) + c . a 1 2 (vii) ∫ sec (ax + b) dx = tan (ax + b) + c . a

(i)

n ∫ (ax + b) dx =

8

1 cot (ax + b) + c . a 1 (ix) ∫ sec (ax + b) tan (ax + b) dx = sec (ax + b) + c . a 1 (x) ∫ cosec (ax + b) cot (ax + b) dx = − cosec (ax + b) + c . a § 6. Methods of Integration. (viii) ∫ cosec 2 (ax + b) dx

(i)

=−

Integration by substitution,

(ii) Integration by parts, (iii) Integration by decomposition into sum, (iv) Integration by successive reduction. § 7. Integration By Substitution. A change in the variable of integration often reduces an integral to one of the fundamental integrals. The method in which we change the variable to some other variable is called the Method of substitution. Rule to remember : To evaluate ∫ f { φ (x)} . φ′ (x) dx, put φ (x) = t and φ′ (x) dx = dt, where φ′ ( x) is the differential coefficient of φ ( x) w.r.t. x. Important. The success of the method of substitution depends on choosing the substitution x = φ (t ) so that the new integrand f {φ (t )}. φ′ (t ) is of a form whose integral is known. This is done by guess rather than in according with some rule. However, try to put that expression of x equal to t whose differential coefficient is multiplied with dx. Three important forms of integrals. f ′ (x) 1. ∫ dx = log | f (x) | + c. f (x) Remember : The integral of a fraction whose numerator is the exact derivative of its denominator is equal to the logarithm of modulus of its denominator. For example

∫

4 x3 1+ x

4

dx = log |1 + x 4 | + c , as in this case numerator is the exact

derivative of the denominator. Integrals of tan x, cot x, sec x and cosec x. (i)

∫

tan x dx =

∫

sin x cos x

dx = −

∫

(− sin x) cos x

dx, adjusting the numerator as

the exact diff. coeffi. of the denominator (ii) Similarly, ∫

= − log | cos x | + c = log | sec x | + c. cos x cot x dx = ∫ dx = log | sin x | + c. sin x

[UPTU 2002]

9

(iii)

∫

cosec x dx =

dx = sin x

∫

∫

2 sin

1 x dx 2 , 1 2 tan x 2

sec 2

dx 1 1 x cos x 2 2

=

∫

dividing Nr. and Dr. by cos 2 1 x  2  1 = logtan x + c. 2   ...  2.

[ f (x)] n f ′ (x) dx =

1 1 1 sec 2 x is the diff. coeffi. of tan x 2 2 2 

[ f (x)] n + 1

+ c, when n √ - 1. n +1 Remember : If the integrand consists of the product of a constant power of a function f ( x)

∫

and the derivative f ′ ( x) of f ( x), to obtain the integral we increase the index. This is known as Power formula. f (ax + b) 3. ∫ f ′ (ax + b) dx = + c. a Note. To evaluate integrals of the type dx dx dx dx ∫ a + b cos x , ∫ a + b sin x , ∫ a sin x + b cos x , ∫ a sin x + b cos x + c ; we write

sin x =

2 tan ( x / 2)

and

1+ tan2 ( x / 2)

cos x =

1– tan2 ( x / 2) 1+ tan2 ( x / 2)

x x by sec2 ⋅ 2 2 x 2 x Now put tan = t, so that sec dx = 2 dt . 2 2 In numerator replace 1 + tan2

Illustration 1: (i)

∫

tan

−1

x 2

1+ x log x (iv) ∫ dx x Solution: (i) ∴

Evaluate : dx

(ii) (v)

∫ ∫

e

tan − 1 x 2

1+ x sec 2 (log x) x

Put tan −1 x = t, so that

∫

tan −1 x 1 + x2

dx =

∫

dx

(iii) dx

(vi)

1

∫

sin e

x

∫ex

x dx x − sin x + cos x

dx = dt. 1 + x2 (tan −1 x)2 t2 t dt = +c = + c. 2 2

(ii) Put tan − 1 x = t, so that

1 1 + x2

dx = dt.

dx [UPTU 2001]

10

∴

e

∫

1+ x

(iii) Put ∴

tan − 1 x

dx =

2

x = t, so that sin

∫

x x

dx =

∫

e

∫

t

t

dt = e

+ c = e tan

−1

x

+ c.

1 − 1 /2 1 x dx = dt or dx = 2 dt. 2 x 2 sin t dt = 2 (− cos t ) + c = − 2 cos

x + c.

1 dx = dt. x log x 1 2 1 2 ∴ ∫ x dx = ∫ t dt = 2 t + c = 2 ( log x) + c. 1 (v) Put log x = t, so that dx = dt. x sec 2 ( log x) ∴ dx = ∫ sec 2 t dt = tan t + c = tan ( log x) + c . ∫ x (iv) Put log x = t, so that

(vi) Put e

x

+ cos x = t, so that (e

e

x

− sin x

e

x

∴

∫

+ cos x

Illustration 2: (i)

(iii)

(v)

2

x

∫

dx

1 + x6

∫

− sin x) dx = dt.

1 dt = log t + c = log (e t

∫

(ii)

[UPTU 2001]

log sec x

∫

(iv)

x6 − 1

x

x

+ cos x) + c .

Evaluate :

dx

tan x

∫

dx =

x

dx.

∫

x8 (1 − x 3 )1 /3 (1 + log x)2 x

dx

dx

[UPTU 2002]

[UPTU 2003]

Solution: (i)

Put x 3 = t, so that 3 x 2 dx = dt.

∴

1 3

3 x 2 dx

∫

1 + x6

1 3

=

dt

∫

1+ t 2

=

1 1 tan − 1 t + c = tan − 1 x 3 + c . 3 3

(ii) Put (1 − x 3 ) = t, so that x 3 = 1 − t and 3 x 2 dx = − dt. ∴

∫

x8 (1 − x 3 )1 /3

=− =−

1 3

∫

dx =

∫

(1 − 2t + t 2 ) t 1 /3

x6 . x2 (1 − x 3 )1 /3

dt = −

1 3

∫t

dx = −

− 1 /3

1 2 /3 2 5 /3 1 8 /3 t + t − t +c 2 5 8

dt +

1 3

∫

2 3

∫t

(1 − t )2 dt t 1 /3 2 /3

dt −

1 3

∫t

5 /3

dt

11

1 2 1 (1 − x 3 )2 /3 + (1 − x 3 )5 /3 − (1 − x 3 )8 /3 + c . 2 5 8 1 = t, so that 3 x 2 dx = dt or x 2 dx = dt. 3 x 2 dx dx [Multiplying Nr. and Dr. by x 2 ] =∫ 6 3 6 x −1 x x −1 =−

(iii) Put x 3 ∴

∫

x

=

1 3

dt

∫

=

t 2 −1

t

1 1 sec − 1 t + c = sec − 1 x 3 + c . 3 3

1 dx = dt x (1 + log x)2 dx t3 1 2 ∴ = t dt = + c = (1 + log x)3 + c . ∫ ∫ x 3 3 1 (v) Put slog sec x = t, so that sec x tan x dx = dt or tan x dx = dt sec x tan x 1 ∴ ∫ log sec x dx = ∫ t dt = log t + c = log ( log sec x) + c. (iv) Put 1 + log x = t, so that

Illustration 3: 5

Evaluate :

(i)

∫ cos

(iii)

∫

sin x cos x dx

(v)

∫

dx ⋅ (5 + 4 cos x)

(vi) ∫

x dx

2

∫ sin

(iv)

∫ sec

4

x dx x tan x dx.

[UPTU 2001]

dx 2

7

(ii)

a cos x + b 2 sin2 x

⋅

[UPTU 2008]

Solution: (i)

Let I = ∫ cos 5 x dx = ∫ cos 4 x . cos x dx = ∫ (1 − si n2 x)2 cos x dx. Put sin x = t, so that cos x dx = dt. I = ∫ (1 − t 2 )2 dt = ∫ (1 − 2t 2 + t 4 ) dt = t − 2

∴

= sin x − (ii) Let

t3 t5 + +c 3 5

2 1 sin3 x + sin5 x + c . 3 5

I = ∫ sin7 x dx = ∫ sin6 x . sin x dx = ∫ (1 − cos 2 x)3 sin x dx.

Put cos x = t, so that (− sin x) dx = dt or sin x dx = − dt.

12

I = − ∫ (1 − t 2 )3 dt = ∫ (t 6 − 3t 4 + 3t 2 − 1) dt

∴

= (iii) Let

I =

t7 t5 cos 7 x 3 −3 + t3 − t + c = − cos 5 x + cos 3 x − cos x + c 7 5 7 5

∫

sin x cos x dx.

Put sin x = t, so that cos x dx = dt. t dt =

2 3 /2 2 t + c = (sin x)3 /2 + c . 3 3

∴

I =

(iv) Let

I = ∫ sec 4 x tan x dx = ∫ sec 2 x . sec 2 x . tan x dx

∫

= ∫ (1 + tan2 x) sec 2 x tan x dx. Put tan x = t, so that sec 2 x dx = dt. ∴ I = ∫ (1 + t 2 ) t dt = ∫ (t + t 3 ) dt = (v) Let I =

∫

Writing cos x = I =

(vi)

∫

dx ⋅ 5 + 4 cos x 1 − tan2 ( x / 2)

=

1 − tan2 ( x / 2) 5+4  2 1 + tan ( x / 2) 2 dt =∫ , 9 + t2 Here I =

, we have

1 + tan2 ( x / 2) dx

t2 t4 1 1 + + c = tan2 x + tan4 x + c . 2 4 2 4

1 + tan2 ( x / 2)

∫ 9 + tan2

putting tan dx

∫

( x / 2)

a2 cos 2 x + b 2 sin2 x

=

∫

dx =

sec2 ( x / 2)

∫

9 + tan2 ( x / 2)

dx

1 x x = t, so that sec2 dx = 2 dt 2 2 sec2 x dx

a2 + b 2 tan2 x

dividing the numerater and the denominator by cos 2 x. Put tan x = t, so that sec2 x dx = dt. ∴

I =

=

=

dt

∫

2

a +b t

1 b

2 2

2

.

=

1 b

2

∫

dt 2

2

(a / b ) + t

2

1 t 1 tan −1 +c = tan −1 a/b a/b ab

1  b tan x  tan −1   + c.  a  ab

=

1 b

2

∫

 tb  + c    a

dt (a / b)2 + t 2

13

Illustration 4: (i)

∫

Evaluate :

sin ( x − a)

dx

sin x

(ii)

∫

sin x

dx.

sin ( x − a)

[UPTU 2007]

Solution: (i)

Here

sin ( x − a)

I =

sin x cos a − cos x sin a

dx ∫ sin x dx = ∫ sin x = ∫ cos a dx − ∫ sin a cot x dx = cos a∫ dx − sin a ∫ cot x dx = x cos a − sin a . log | sin x | + c .

(ii) Put ( x − a) = t, so that x = t + a and dx = dt. ∴

∫

sin x sin ( x − a)

dx =

∫

sin (t + a) sin t

dt =

∫

sin t cos a + cos t sin a sin t

dt

= cos a∫ dt + sin a∫ cot t dt = cos a . t + sin a . log |sin t | + c = ( x − a) cos a + sin a . log |sin ( x − a)| + c .

Comprehensive Exercise 2

Evaluate the following integrals : 1.

(i) (iii)

(v)

2.

(i)

∫x ∫

∫e

∫

cos x 4 dx

(ii)

sin x (3 + 4 cos x)2 sin − 1 x

∫

(iii) ∫ (v)

3

dx

dx

(iv)

sec 2 x dx

cos

x

dx

x e

∫

(vi)

∫

(ii)

∫

(iv)

∫

(vi)

∫

(1 − x 2 ) tan x

∫ sin (ax + b) cos (ax + b) dx

m tan − 1 x

1+ x

2

dx

sin (2 tan − 1 x) 1 + x2 x tan − 1 x 2 1 + x4 e

x

(1 + e2 x )

dx

cos ( log x) x 1 x

2

dx.

dx

e − 1 / x dx.

dx

14

3.

(i)

∫

(iii)

∫

(v)

∫

sin x

(ii)

∫

dx

(iv)

∫

dx

(vi)

∫ sin

1 + cos x cot x log sin x 3 x2 (1 + x 6 )

4x − 5

dx

2x

2

− 5x + 1

(4 x + 2) 3

dx

x 2 + x + 1 dx

x cos x dx.

A nswers (ii)

(iii)

1 +c 4 (3 + 4 cos x)

(iv)

−

(v)

1 (sin − 1 x)2 + c 2

(vi)

1 (tan − 1 x 2 )2 + c . 4

e tan

(ii)

tan − 1 (e x ) + c

(iv)

sin ( log x) + c

(vi)

e − 1 / x + c.

2. (i)

x

(iii) 2 sin (v) 3.

sin2 (ax + b)

1 sin x 4 + c 4

1. (i)

1 e m

+c x +c

m tan

−1

x

+c

(i) − log |1 + cos x | + c

(ii)

(iii) log | log sin x | + c

(iv)

(v) tan −1 x 3 + c

(vi)

2a

+c

1 cos (2 tan − 1 x) + c 2

log | 2 x 2 − 5 x + 1| + c 4 2 ( x + x + 1)3 /2 + c 3 sin4 x 4

+ c.

§ 8. Integral of the Product of two Functions. Integration by Parts. Let f1 ( x) and f 2 ( x) be two functions of x. Then  d

∫ f 1 (x) f 2 (x) dx = f 1 (x) . ∫ f 2 (x) dx – ∫  dx f 1

(x) ⋅ ∫ f 2 (x) dx dx  

i.e., the integral of the product of two functions = first function × integral of second function − integral of {diff. coeffi. of first function × integral of second function}.

15

Note 1. Care must be taken in choosing the first function and the second function. Obviously we must take that function as the second function whose integral is well known to us. Thus, to evaluate  x log x dx we shall take x as the second function because we so far do not know the integral of log x. But to evaluate  x sin x dx we must take sin x as the second function and x as the first function. Here, if we take x as the second function, then the new integral will become more complicated. Thus, to evaluate integrals of the type ∫ x 2 e x dx,  x 3 cos x dx etc., the function of the type x n must be taken as the first function. In certain cases we can take unity (i. e., 1) as the second function. Thus, to evaluate  log x dx we shall take 1 as the second function. To evaluate ∫ e x sin x dx we can take either e x or sin x as the second function. Note 2. The formula of integration by parts can be applied more than once if necessary.

Illustration 1:

Evaluate : 2

(i)

∫ x sin 2 x dx

(ii)

∫x

(iii)

∫ log x dx

(iv)

∫ sin

[UPTU 2009]

e −1

x

dx x dx.

Solution: (i)

Integrating by parts, taking x as the first function, we have d

∫ x sin 2 x dx = x ∫ sin 2 x dx − ∫  dx ( x) . ∫ sin 2 x dx

 dx  

− cos 2 x   − cos 2 x  dx = − x cos 2 x + 1 cos 2 x dx = x   − 1.    2  ∫  2  2 2∫ x cos 2 x

=−

2

+

(ii) Integrating by parts taking e

∫x

2 x

e dx = x 2

∫e

x

x cos 2 x 1 1 sin 2 x ⋅ +c =− + sin 2 x + c . 2 2 2 4 x

as the second function, we have

d dx –  x 2 . ∫ e x dx dx   dx

( )

= x 2 e x – ∫ 2 x. e x dx = x 2 e x – 2∫ x e x dx d = x 2 e x – 2 x ∫ e x dx – ∫  ( x ). ∫ e x dx dx     dx  [Again integrating by parts]

16

= x2 e

x

− 2 {x e

x

x

−  1. e

dx} = x 2 e

x

− 2 {x e

x

− e x } + c.

(iii) As there is only one function here, unity should be taken as the second function. We have ∫ log x dx =

∫ ( log

x) . 1 dx 1 = ( log x ). x − ∫  . x dx = x log x − ∫ dx x  = x log x − x + c = x ( log x − 1) + c .

(iv) As there is only one function here, unity should be taken as the 2nd function. We have ∫ sin − 1 x dx = ∫ (sin − 1 x) . 1 dx. Integrating by parts regarding 1 as the second function, we have

∫ (sin

–1

)

(

)

x .1dx = sin –1 x . ∫ 1dx – = (sin − 1 x) . x −

–1

1

∫

1 2

= x sin − 1 x +

d

∫  dx (sin

x . ∫ 1dx dx 

)

⋅ x dx

1 − x2 (1 − x 2 ) − 1 /2 (− 2 x) dx [Note]

∫

(

1 /2

)

2 1 1– x –1 = x sin x + 2 1/ 2

[By power formula]

+c

= x sin − 1 x + 1 − x 2 + c . Illustration 2: (i) (iii)

∫x ∫

2

sin

−1

Evaluate : x dx

x 2 tan −1 x 1 + x2

(ii)

dx .

[UPTU 2004]

∫x

(iv)

3

∫− e

sin x 2 dx x

[UPTU 2004]

sin x dx

[UPTU 2008]

Solution: (i)

Integrating by parts taking x 2 as the second function, we have

∫

x 2 sin − 1 x dx = (sin − 1 x) .

=

=

x 3 sin − 1 x 1 − 3 3 x 3 sin − 1 x 3 3

=

x3 − 3

x sin 3

–1

x

+

+ 1 3

∫

1 3

∫

∫

1

⋅

1 − x2 x3

x3 dx 3

dx =

1− x

2

t (1 − t 2 ) t

x 3 sin − 1 x 1 − 3 3

1− x

dt, where (1 − x 2 ) = t 2 .3 3

[∫ dt – ∫ t dt] = x sin3 12

∫

x . x2

–1

x

+

1  t3  t –  + c 3 3

2

dx

17

=

∴

3

1 1 1 − x 2 − (1 − x 2 )3 /2 + c . 3 9

+

∫x

sin x 2 dx =

2

⋅ x sin x 2 dx = ∫ x 2 sin x x dx. 1 Put x 2 = t so that 2x dx = dt or x dx = dt. 2 1 1 2 2 ∫ x sin x x dx = 2 ∫ t sin t dt = 2 [− t cos t + ∫ cos t dt] 1 1 = [− t cos t + sin t] + c = [− x 2 cos x 2 + sin x 2 ] + c . 2 2

(ii) We have

3

x 3 sin − 1 x

∫x

2

(iii) Put tan −1 x = t so that x = tan t. Also dx = sec2 t dt. ∴

x 2 tan −1 x

∫

1 + x2 =

∫ t (sec

=t

∫

2

dx =

t − 1) dt =

sec2 t dt −

= t tan t −

∫

∫

∫

tan2 t ⋅ t ⋅ sec2 t dt 1 + tan2 t

∫ t sec d t  dt

tan t dt −

2

∫

t dt − ∫ t dt t2 sec2 t dt dt − +c 2 

t2 t2 + c = t tan t − log | sec t | − +c 2 2

= t tan t − log | 1 + tan2 t | − = x tan −1 x − log | 1 + x 2 | −

t2 +c 2 (tan −1 x)2 2

+ c.

(iv) We have ∫ − e x sin x dx = − [e x (− cos x) − ∫ e x (− cos x) dx] , integrating by parts taking sin x as the second function = + e x cos x − ∫ e x cos x dx = e x cos x − [e x sin x − ∫ e x sin x dx] , again integrating by parts taking cos x as the second function = e x cos x − e x sin x + ∫ e x sin x dx Transposing the last term on the right hand side to the left and dividing by 2, we get 1 x x ∫ − e sinx dx = 2 e (cos x − sin x) + c.

Comprehensive Exercise 3 Evaluate the following integrals : 1.

∫ x cos 2 x dx.

2.

∫xe

x

dx.

18

x

3.

∫ x sin 3 x dx.

4.

∫ sin2

5.

∫ x sin x dx. 2 ∫ x cos x dx. 3 ∫ cosec x dx. −1 ∫ sec x dx.

2

6.

∫ x cos x dx. 2 3x ∫ x e dx. 2 ∫ log (1 + x ) dx. −1 ∫ tan x dx

7. 9. 11. 13.

∫

x sin −1 x 1− x

dx.

8. 10. 12.

3

14.

[UPTU 2006]

dx.

x

∫

2

x tan –1 x 3 /2

(

)

1 + x2

dx.

A nswers 1. 3. 5. 6. 7.

1 1 2. ( x − 1) e x + c . x sin 2 x + cos 2 x + c . 2 4 − x cos 3 x sin 3 x 4. + + c. − x cot x + log ∞ sin x∞ + c . 3 9 x sin 2 x cos 2 x x2 − − + c. 4 4 8 1 1 3 3 x sin 3 x + cos 3 x + x sin x + cos x + c . 12 36 4 4  x2 2 x 2 3x   e 8. x 2 sin x + 2 x cos x − 2 sin x + c . − + + c. 9 27  3 1 1 x cosec x cot x + log tan  + c . 2 2 2 

9.

−

10.

x log (1 + x 2 ) − 2 x + 2 tan − 1 x + c .

11.

x sec −1 x − log | x +

x 2 − 1| + c .

13. − (sin −1 x) 1 − x 2 + x + c .

12. 14.

x tan −1 x − x − tan −1 x 1+ x

1 log |1 + x 2 | + c . 2 + c.

2

§ 9. Integration by Parts as Applied to the Functions of the type e x [ f ( x) + f ′ ( x)] . Let I =

∫e

x

[ f ( x) + f ′ ( x)] dx =

∫e

x

f ( x) dx + ∫ e

Integrating the first integral by parts regarding e have I = f ( x) . e

x

− ∫ f ′ ( x) . e

= e x f (x) + c.

x

dx + ∫ e

x

x

x

f ′ ( x) dx.

as the 2nd function, we

f ′ ( x) dx

19

[Note that we have left the other integral unchanged because the last two integrals cancel each other.]

§ 10. Integrals of e

∫e

ax

ax

sin bx and e

sin bx dx =

ax ∫ e cos bx dx =

and

ax

cos bx.

ax

e

(a sin bx - b cos bx) + c

a2 + b2 e ax

(a cos bx + b sin bx) + c.

a2 + b2

Alternative forms of ∫ e ax sin bx dx and∫ e ax cos bx dx. e ax

ax ∫ e sin bx dx =

and

∫

e

ax

b sin  bx – tan - 1  + c  a

a2 + b2 ax

e

cos bx dx =

a2 + b2

Illustration 1: Evaluate : 1 1 (i) ∫ e x  − 2  dx x x  (iii)

∫e

x

(cot x + log sin x) dx

Solution: (i)

We have

=

x

x

(ii)

∫e

(iv)

∫ [sin ( log

(1 + x)2

dx [UPTU 2007]

x) + cos ( log x)] dx.

1 1  − 2  dx x x  1 { f ( x) + f ′ ( x)} dx, where f ( x) = x

x

e

∫ =∫

b  + c. a

cos  bx – tan - 1 

e

x

x

f ( x) dx + ∫ e x f ′ ( x) dx = e x f ( x) − ∫ e x f ′ ( x) dx + ∫ e x f ′ ( x) dx,

∫e

applying integration by parts to the first integral taking e =e

x

f ( x) + c =

(ii) We have

∫e

x

=

∫e

x

x 2

( x + 1)

x

1 e x

dx =

x

x

as the second function

+ c.

∫e

x

( x + 1) − 1 2

( x + 1)

dx =

[ f ( x) + f ′ ( x)] dx, where f ( x) =

∫e

x

 1 1  − dx  2  x + 1 ( x + 1) 

1 x +1

x

∫ e f ( x) dx + ∫ e f ′ ( x) dx = ∫ e x f ( x) − ∫ e x f ′ ( x) dx + ∫ e x f ′ ( x) dx, =

applying integration by parts to the first intergal taking e

x

as the second function

20

=e

x

x

f ( x) + c = e

1 + c. x +1

(iii) We have ∫ e x (cot x + log sin x) dx =

∫e

x

( log sin x + cot x) dx =  e

x

[ f ( x) + f ′ ( x)] dx, where f ( x) = log sin x

so that f ' ( x ) = (1 / sin x ) cos x = cot x =e (iv) Let

x

x

f ( x) + c = e

log sin x + c .

I = ∫ [sin ( log x) + cos ( log x)] dx.

Put log x = t or x = e t so that dx = e I =

∴

∫e

t

[sin t + cos t ] dt =

∫e

t

t

dt.

[ f (t) + f ′ (t )] dt,

where f (t ) = sin t so that f ′ (t) = cos t =e Illustration 2: (i)

∫e

(iii)

∫e

x

2x

t

f (t ) + c = e

log x

sin ( log x) + c = x sin ( log x) + c .

Evaluate :

cos2 x dx

(ii)

(− sin x + 2 cos x) dx .

∫ sin ( log x) dx

[UPTU 2004]

Solution: (i)

We have ∫ =

1 2

e

∫

x

cos 2 x dx = e

x

dx +

1 2

∫

∫ e

e x

x

1 + cos 2 x    dx   2

cos 2 x dx =

1 e 2

x

+

1 I, 2

where I =

∫e

x

cos 2 x dx.

Integrating by parts taking cos 2 x as the second function, we get  sin 2 x  x sin 2 x I =ex  dx  −∫ e ⋅  2  2 1 1 = e x sin 2 x − ∫ e x sin 2 x dx 2 2 1 1  cos 2 x   x  cos 2 x  = e x sin 2 x − e x  −  − ∫ e −  dx ,   2 2 2  2  

or

[Again integrating by parts taking sin 2x as the 2nd function] 1 x 1 1 = e sin 2 x + e x cos 2 x − ∫ e x cos 2 x dx 2 4 4 1 1 1 = e x sin 2 x + e x cos 2 x − I 2 4 4 1 1 x 1 x 1 I + I = e sin 2 x + e cos 2 x or I = e x (2 sin 2 x + cos 2 x) + c . 4 2 4 5 1 1 x Thus, ∫ e x cos 2 x dx = e x + e (2 sin 2 x + cos 2 x) + c . 2 10

21

(ii) Let

I = ∫ sin ( log x) dx.

Put log x = t or x = e t so that dx = e t dt. Then I = ∫ e t sin t dt = e t ( − cos t) − ∫ e t (− cos t) dt, [Integrating by parts taking sin t as the second function] = − e t cos t + ∫ e t cos t dt = − e t cos t + e t . sin t − ∫ e t sin t dt, [Again integrating by parts taking cos t as the second function]

= − e t cos t + e t sin t − I 2I = − e t cos t + e t sin t or

or

Thus,

∫

(iii) We have =

sin ( log x) dx =

∫e

2 x

−e 2

I =

1 t e (sin t − cos t ) + c . 2

x {sin ( log x) − cos ( log x)} + c . 2

(− sin x + 2 cos x) dx = ∫ (− e2 x sin x + 2e2

2 x

2

2 +1

(2 sin x − cos x) +

2e

x

cos x) dx

2 x

22 + 12

(2 cos x + sin x) + c

  e ax ax (a sin bx – b cos bx)+c  ∵ ∫ e sin bx dx = 2 2 a +b     e ax ax (a cos bx + b sin bx)+c and e cos bs dx = 2 2   a +b =

e2 x [−2 sin x + cos x + 4 cos x + 2 sin x] + c = e2 5

x

cos x + c .

Comprehensive Exercise 4 Evaluate the following integrals : x 1. ∫ e (sin x + cos x) dx.

∫e

(cot x − cosec2 x) dx.

4.

∫e

 2− x  dx.  2  (1 − x) 

6.

∫e

8.

∫ e cos x dx. 2 x ∫ e cos (3 x + 4) dx.

3.

∫e

x

5.

∫e

x

7.

∫ e sin 4 x dx. a x ∫ e sin (bx + c) dx.

9.

3 x

An

1.

x

2.

10.

(tan x + log sec x) dx.

x

x

2  1  2 − 3  dx. x x  1   log x + 2  dx.  x 

−x

A nswers e

x

sin x + c .

2.

e

x

log (sec x) + c .

22

3.

e

x

x

e

4.

cot x + c .

x2

+ c.

5.

ex + c. 1− x

6.

e

7.

e3 x 4 sin 4 x − tan − 1  + c .  5 3

8.

e− x (− cos x + sin x) + c . 2

ea x

9.

 log x − 1 + c .    x

b sin  bx + c − tan − 1  + k. a 

a2 + b 2 10.

x

e2 x {2 cos (3 x + 4) + 3 sin (3 x + 4)} + c . 13

§ 11. Three Special Integrals.

∫ (i)

dx a

2

+x

;

2

dx

∫

To evaluate

x

2

- a dx

, ( x > a) ;

2

∫

dx a

2

- x2

, ( x < a).

⋅ a2 + x2 Put x = a tan θ; so that dx = a sec 2 θ dθ .

∫

[UPTU 2003]

Also a2 + x 2 = a2 + a2 tan2 θ = a2 (1 + tan2 θ) = a2 sec 2 θ. ∴

∫

dx 2

a + x

Thus,

∫

2

=

∫

a

We know that

∴

∫ x2

2

a sec θ 2

+x dx

=

1 x

−a

1 a

∫

dθ =

1 x 1 θ + c = tan − 1 + c . a a a

1 x tan - 1 + c . a a

x2 - a 2

2

=

2

,

=

( x > a) . 1 ( x − a) ( x + a)

1 ( x + a) − ( x − a) 1  1 1  = − ⋅ 2a  ( x − a) ( x + a)  2a ( x − a) ( x + a)

dx –a

2

dx 2

(ii) To evaluate ∫

=

a sec2 θ dθ

2

=

1 

1



=

=

1



∫ 2a ( x – a) – ( x + a) dx 1  1 dx – ∫ 2a  ( x – a )



1



∫ ( x + a) dx 

1 1 x–a [log| x – a|–log| x +a|] + c = 2a log x + a + c 2a

23

Thus,

∫ x2

1 -a

dx =

2

1  x - a  + c, x > a. log 2a x + a

dx

(iii) To evaluate ∫

a

2

- x2

,

( x < a).

We know that 1 2

a − x ∴

∫ a2

2

dx – x

2

Thus,

=

1 1 (a + x) + (a − x) 1  1 1  = = + ⋅    (a − x) (a + x) 2a  (a − x) (a + x)  2a (a − x) (a + x)

=

∫ 2a ( a – x ) + ( a + x ) dx

1 

1

1



 

=

 1  1 1 dx + ∫ dx ∫ 2a  ( a – x ) (a + x) 

=

1 1 a+ x – log|a – x|+ log |a + x|] + c = log +c [ 2a 2a a– x

∫ a2

dx -x

2

=

1 a + x  + c, x < a. log 2a a - x 

§ 12. Integration by Partial Fractions. The following table gives an idea what kind of partial fractions are to be taken for what kind of factors in the denominator : S.No.

Factor in the denominator

Form of the partial fraction

(i)

( x − a) ( x − b)

A B + ( x − a) ( x − b)

(ii)

( x − a)2 ( x − b)

A B C + + ( x − a) ( x − a)2 ( x − b)

(iii)

( x − a)3

A B C + + ( x − a) ( x − a)2 ( x − a)3

(iv)

(ax 2 + bx + c )

Ax + B ax

(v)

(ax

2

2

Ax + B

2

+ bx + c )

ax

2

+ bx + c

+ bx + c +

Cx + D (ax

2

+ bx + c )2

Note. There are as many constants to be determined as the degree of the denominator.

Illustration 1:

Resolve

( x − 1) ( x − 3) ( x − 2)

into partial fractions.

24

Solution: Clearly

Let =

x −1 ( x − 3) ( x − 2)

=

A ( x − 2) + B ( x − 3) A B + = ⋅ ( x − 3) ( x − 2) ( x − 3) ( x − 2) …(1)

x − 1 ≡ A ( x − 2) + B ( x − 3) .

Comparing the coefficients of x and the constant terms on both sides of (1), we get …(2)

1= A + B and

…(3)

− 1 = − 2 A − 3B Solving (2) and (3), we get A = 2, B = − 1. =

∴

( x − 1) ( x − 3) ( x − 2)

Important Note.

=

2 1 − ⋅ ( x − 3) ( x − 2)

An easy way to find the constants A and B etc. corresponding

to linear non-repeated factors is like this : The factor below A is ( x − 3) . The equation x − 3 = 0 gives x = 3. Now suppress ( x − 3) in the given fraction ( x − 1) ( x − 1) and put x = 3 in the remaining fraction to get A. Thus, ( x − 3) ( x − 2) ( x − 2) 3 −1 A= =2. 3−2 2 −1 = − 1. 2−3 Evaluate :

Similarly B = Illustration 2: (i)

∫

dx x+ x

[UPTU 2005]

2

(ii)

∫

dx x − x3

⋅

[UPTU 2004]

Solution: (i)

Here

∴

∫

1

1 1 1 = − , resolving into partial fractions x (1 + x) x 1 + x x+ x 1 dx 1  1 1 =∫  − dx − ∫ dx  dx = ∫ 2 x 1 + x x 1 + x   x+ x 2

=

 x  = log | x | − log |1 + x | + c = log   + c. 1 + x 1 1 1 A B C (ii) Here = = ≡ + + , (say). 3 2 x (1 − x) (1 + x) x− x x (1 − x ) x (1 − x) (1 + x) To find A suppress x in the given fraction and put x = 0 in the remaining fraction. ∴

A=

1 = 1. (1 − 0) (1 + 0)

To find B suppress (1 − x) in the given fraction and put x = 1 in the remaining fraction.

25

B=

∴ 1

Thus, ∴

∫

x− x

3

(x − x )

Illustration 3:

Similarly,

=

1 1 1 + + ⋅ x 2 (1 − x) 2 (1 + x)

=

∫

3

dx

1 1 = ⋅ 1 (1 + 1) 2

1 1 dx + x 2

∫

1 1 dx + (1 − x) 2

∫

C=

1 ⋅ 2

1 dx (1 + x)

= log x −

1 1 log (1 − x) + log (1 + x) + c 2 2

= log x +

(1 + x) x 2 (1 + x) 1 1 log + c = log + c. 2 (1 − x) 2 (1 − x) x3 dx . ( x − 1) ( x − 2) ( x − 3)

Evaluate ∫

[UPTU 2006]

Solution: Here since the numerator is not of a lower degree than the denominator, we divide the numerator by the denominator till the remainder is of lesser degree than the denominator. We orally see that the quotient is 1. We need not find out the actual value of the remainder because ultimately we have to break the fraction into partial fractions. Note that the denominators of the partial fractions depend only upon the denominator of the given fraction. So let x3 A B C ≡1+ + + ⋅ ( x − 1) ( x − 2) ( x − 3) ( x − 1) ( x − 2) ( x − 3) We have A =

13 1 23 = , B= = − 8, (1 − 2) (1 − 3) 2 (2 − 1) (2 − 3)

and

33 27 = ⋅ (3 − 1) (3 − 2) 2

∴

C=

x3 1 8 27 =1+ − + ⋅ ( x − 1) ( x − 2) ( x − 3) 2 ( x − 1) ( x − 2) 2 ( x − 3)

Hence,

∫

x 3 dx ( x − 1) ( x − 2) ( x − 3) =

1 . dx +

∫

= x+ Illustration 4: Solution:

∫

dx − 2 ( x − 1)

∫

8 dx ( x − 2)

+

∫

27 dx 2 ( x − 3)

1 27 log | x − 1| − 8 log | x − 2| + log | x − 3| + c . 2 2

Evaluate ∫

Let y = x 2 .

x2 (x

2

+ 2) ( x 2 + 3)

dx .

26

x2

Then

(x

2

+ 2) ( x

2

+ 3)

y

=

We have A = the value of

( y + 2) ( y + 3) y

≡

A B + , (say). ( y + 2) ( y + 3)

, when y is − 2 ,

y+3

=−2 and

B = the value of x2

Thus, ∴

∫

y y+2

(x

2

x (x

2

+ 2) ( x

2

+ 3)

, when y is − 3 , = 3. −2

=

y+2

+

2

+ 2) ( x

2

+ 3)

dx = − 2

−2 3 3 = + ⋅ y + 3 x2 + 2 x2 + 3 dx

∫

x

2

+2

+3

dx

∫

x

2

+3

1 x 1 x tan − 1 +3⋅ tan − 1 +c 2 2 3 3 x x = − 2 tan − 1 + 3 tan − 1 +c. 2 3 = −2⋅

Note. In the above example, the substitution was made only for the partial fraction part and not for the integration part. Illustration 5:

Evaluate

(3 x + 1)

∫

( x − 2)2 ( x + 2)

(3 x + 1)

dx .

A B C + + ( x − 2) ( x − 2)2 x+2

Solution:

Let

or

(3 x + 1) ≡ A ( x − 2) ( x + 2) + B ( x + 2) + C ( x − 2)2

Then

( x − 2)2 ( x + 2)

B = the value of C = the value of

≡

3x + 1 x+2

, when x is 2 =

3x + 1 ( x − 2)2

…(1)

7 , 4

, when x is − 2 = −

5 ⋅ 16

Comparing the coefficients of x 2 on both sides of (1), we get 5 A+ C=0 → A= − C= ⋅ 16 (3 x + 1) 5 7 5 Thus, = + − ⋅ 2 2 16 ( x − 2 ) 16 ( x + 2) ( x − 2) ( x + 2) 4 ( x − 2) ∴

∫

(3 x + 1) 2

( x − 2) ( x + 2)

dx =

5 16

∫

dx 7 + x−2 4

∫

dx 2

( x − 2)

5 7 5 = log ∞x − 2∞ − − log ∞x + 2∞ + c . 16 4 ( x − 2) 16 Illustration 6:

Evaluate

∫

cos3 x + cos5 x sin2 x + sin4 x

dx.

−

5 16

∫

dx x+2

27

Solution:

(cos 3 x + cos 5 x)

∫ =

(sin2 x + sin4 x)

dx =

∫

(1 − sin2 x) (2 − sin2 x)

∫

sin2 x (1 + sin2 x) (1 − t 2 ) (2 − t 2 )

=

∫

=

∫ 1 + t 2

=

∫

t 2 (1 + t 2 ) 

2

sin2 x (1 + sin2 x)

cos x dx

cos x dx [Putting sin x = t so that cos x dx = dt]

dt

6  dt 1 + t 2  dt dt + 2 ∫ 2 − 6 t

cos 2 x (1 + cos 2 x)

[Resolving into partial fractions]

–

∫

1 1 + t2

dt

2 − 6 tan −1 t + c t 2 = sin x − − 6 tan −1 (sin x) + c sin x =t−

= sin x − 2 cosec x − 6 tan −1 (sin x) + c . A special Type of Integral. Here we shall give a very interesting method for evaluating the integral of a fraction in which the denominator is of degree 4 and the numerator is either of degree 2 or is constant. Moreover the odd powers of x occur neither in the numerator nor in the denominator. Method. Divide the numerator and denominator by x 2 and put x +

1 = t or x

1 = t, whichever on differentiation gives the numerator of the resulting x integrand. x−

Illustration 7: (i)

∫

x2 + 1 x4 + 1

Evaluate : dx

[UPTU 2005]

(ii)

∫

x2 − 1 x4 + x2 + 1

dx .

Solution: (i)

Let

I =

∫

x2 + 1 x4 + 1

dx .

Here both the numerator and the denominator do not contain odd powers of x . Also the numerator is of degree 2 and the denominator is of degree 4. So dividing the numerator and the denominator by x 2 , we get 1 + (1 / x 2 ) ( ) I =∫ dx = ∫ 2 x 2 + (1 / x 2 ) [ x – (1 / x )] + 2 1 + 1 / x2

Note that d  x − 1 = 1 + 1     dx  x x 2 

28

{ (

)} dx = dt

Now put x + (1 / x ) = t, so that 1 + 1 / x 2 dt

I =

∴

∫ t2

 x – (1 / x ) 1 1  t  tan –1   + c = tan –1  +c  2 2 2 2  

=

+t

 x 2 − 1    x 2 +c.   2 x −1 dx , 4 x + x2 + 1

1 tan − 1 2

=

(ii) We have I = =

∫

1 − (1 / x 2 )

∫

x 2 + 1 + (1 / x 2 )

[Note the form of the integrand]

dx ,

dividing the numerator and the denominator by x 2

(

1 – 1 / x2 =

∫ x + (1 / x )2

)

Note that d {x + (1 / x )} = 1 – 1 / x 2    dx

(

dx,

–1

{ (

Now put x + (1 / x ) = t, so that 1 – 1 / x 2 dt

I =

∴

∫ t2

=

`

x + (1 / x ) – 1 1 t –1 1 log + c = log 2 t +1 2 x + (1 / x ) + 1

Evaluate

We have

∫ I =

( tan θ

I =

∫

∫

= 2 tan

Illustration 9: Solution:

   dθ = tan θ  2 x dx θ dθ = 2 x dx or dθ = ⋅ 1 + x4

x 2 + 1 2 x dx ⋅ =2 x 1 + x4 −1

Evaluate

Let I =

∫

  tan θ − 1   2 tan θ 

∫

+ cot θ ) dθ .

   tan θ  

Put tan θ = x 2 , so that sec 2

∴

)} dx = dt

1 x2 – x + 1 log 2 +c 2 x + x +1

Illustration 8: Solution:

–1

=

)

∫

+

x2 + 1 x4 + 1

1

∫

tan θ + 1

dθ .

tan θ

dx

   + c . [Proceeding as in illustration 7 (i)]  

cot x dx .

cot x dx .

Put cot x = t 2 , so that − cosec 2 x dx = 2t dt

29

or

dx =

− 2t dt cosec

I =

∴

2

x

=

t( –2t )dt

∫

1+ t

4

− 2t dt 1 + cot =–

2

x

=

2t 2 dt

∫ 1 + t4

− 2t dt 1 + t4 =–

∫ t2

⋅ 2dt

(

+ 1 / t2

)

dividing the numerator and the denominator by t 2

{1 + (1 / t )} + {1 – (1 / t )} dt 2

I =

∫

2

(

t2 + 1 1 / t2

{1 + (1 / t )}

)

2

I =

∫

{t – (1 / t )}2

{1 – (1 / t )} 2

+2

dt – ∫

{t + (1 / t )}2

–2

{ ( )} dt = dz. In the second integral put t + (1 / t ) = u, so that {1 – (1 / t )} dt = du. In the first integral put t – (1 / t ) = z , so that 1 – 1 / t 2

2

Hence, I = −

∫

dz z

2

+2

−

∫

du 2

u −2

=−

u − 2 1 z 1 tan − 1 − log  + c 2 2 2 2 u + 2

=−

t − (1 / t) t + (1 / t) − 2 1 1 tan − 1  log   +c  − 2 2  2 2  t + (1 / t) + 2

=−

 t 2 − 1 t 2 − 2 t + 1 1 1 tan − 1  log  2  +c  − 2  t 2 2 2 t + 2 t + 1

=−

 cot x − 2 cot x  cot x − 1 1 1 tan − 1  log   −  2  2 cot x  2 2 cot x + 2 cot x

 +1 + c .  + 1

Comprehensive Exercise 5 Evaluate the following integrals : (1 − x 2 )

1.

∫

x (1 − 2 x)

3.

∫

5.

∫

(2 x + 1)

2.

∫

( x + 1) ( x − 2)

dx ⋅ x ( x + 2)

4.

∫

x dx. ( x + 2) ( x + 3)

x dx . ( x + 2) (3 − 2 x)

6.

∫

dx .

2x + 5 x

2

− x−2

dx .

dx .

30

x3 − x − 2

7.

∫

9.

∫

x ( x n + 1)

11.

∫

sin 4 x

1− x

∫

(x

⋅

10.

∫

(1 − e x )

dx .

12.

∫

dx sin x

2x − 3

8.

2

dx .

An

2

− 1) (2 x + 3)

dx

dx .

⋅

tan x dx .

A nswers

1.

1 3 x + log | x | − log |1 − 2 x | + c . 2 4

2.

1 5 log | x + 1| + log | x − 2 |+ c . 3 3

4.

3 log | x + 3 | − 2 log | x + 2 | + c .

5.

−

6.

3 log | x − 2 | − log | x + 1| + c .

8.

5 1 12 log | x + 1| − log | x − 1| − log | 2 x + 3 | + c . 2 10 5

9.

1 1 log | x n ∞ − log | x n n

 x  1 log  + c . 2 x + 2

3.

2 3 log | x + 2 | − log | 3 − 2 x | + c . 7 14

n

7.

−

1 − x x2 + log  + c . 2 1 + x

+ 1| + c .

10. log |e x| − log |1 − e x| + c . 11. −

12.

1 + sin 1 log  8 1 − sin

x 1 + 2 sin x 1 log  +  +c. x 4 2 1 − 2 sin x

   tan x − 2 tan x 1 1  tan x − 1  tan − 1  log   +  2  2 tan x  2 2   tan x + 2 tan x

 +1  +c.  + 1

Comprehensive Exercise 6 Fill in the Blank(s). Fill in the blanks ‘’..........’’, so that the following statements are complete and correct. 1.

The value of ∫ cosec x cot x dx = .......... + c .

2.

The value of ∫ a x dx = .......... + c .

31

3.

The value of ∫ tan2 x dx = .......... + c .

4. 5.

The value of ∫ e(1 − 3 x) dx = .......... + c . [ f ( x)] n +1 The value of ∫ [ f ( x)] n f ′ ( x) dx = + c , when n ≠ − 1 is known as n +1 ....... formula.

6.

The value of ∫ sin 2x dx = .......... + c .

7.

The value of ∫ log x dx = .......... + c .

[UPTU 2009] [UPTU 2009]

2

8.

The value of ∫

sec (log x) x

dx = .......... + c .

Multiple Choice Questions. Indicate the correct answer for each question by writing the corresponding letter from (a), (b), (c) and (d). 9.

The value of ∫ sec (ax + b) tan (ax + b) dx is (a) (c)

10.

∫ (a)

1 sec (ax + b) + c a

1 tan (ax + b) + c a 1 dx is equal to 2 (a − x 2 ) sin −1 (a − x) + c

sin −1 ( x / a) + c (tan x) 11. The value of ∫ dx is sin x cos x (a) 2 (tan x) + c (c)

(c) 2 / (tan x) + c

(b)

1 sec (ax + b) + c b

(d)

tan (ax + b) + c

(b)

sin −1 ( x − a) + c

(d)

sin −1 (a / x) + c

(b)

2 (sec x) + c

(d)

none of these

(b)

2 (sin x / 2 + cos x / 2) + c

(d)

none of these.

12. The value of ∫ (1 + sin x) dx is (a) 2 (sin x / 2 − cos x / 2) + c (c) sin x / 2 − cos x / 2 + c cos x 13. The value of ∫ dx is x (a) 2 cos x + c (c)

sin

x +c

(b) (d)

14. The value of ∫ e ax sin bx dx is (a)

e bx

a2 + b 2

(a sin bx − b cos bx) + c (b)

{(cos x) / x} + c 2 sin

x +c

e ax a2 + b 2

(a sin bx − b cos bx) + c

32

(c)

e ax 2

a +b

2

(a sin bx + b cos bx) + c (d)

e ax 2

a + b2

(b cos bx − a sin bx) + c

True or False. Write ‘T’ for true and ‘F’ for false statement. 2( x + 3 ) + c. log 2 f1 (x) . f 2 ( x) dx = f1 ( x) . ∫ f 2 ( x) dx +

15.

( x + 3) dx = ∫2

16.

∫ dx ∫ a2 + x 2

17.

=

(a x + b x )2

1 x tan −1 + c. a a (a / b) x

∫

19.

∫ 1 − sin 2 x dx = sin x − cos x + c. 1 2 x –3 dx = e2 x − 3 + c. ∫e 2

20.

dx =

log (a / b)

+

 d f ( x) . f ( x) dx dx.  dx 1  ∫ 2 

(b / a) x

18.

ax bx

∫

log (b / a)

+ 2 x + c.

A nswers 1.

− cosec x.

2.

ax log a

3.

tan x − x.

4.

−1 (1 − 3 x) . e 3

5.

power

6.

 cos 2 x  − ⋅  2 

7.

x (log x − 1. )

8.

tan (log x).

9.

(a)

10.

(c)

11. (a)

12.

(a)

13. (d)

14.

(b)

15. T

16.

F

17. T

18.

T

19. F

20.

T

❍❍❍