43 Years Chapterwise Topicwise Solved Papers 2021-1979 Arihant IIT JEE Mathematics Amit M Agarwal 9789325796133

Table of contents :
Cover
Title
Copyrights
CONTENTS
SYLLABUS
1 : Complex Numbers
2 : Theory of Equations
3 : Sequences and Series
4 : Permutations and Combinations
5 : Binomial Theorem
6 : Probability
7 : Matrices and Determinants
8 : Functions
9 : Limit, Continuity and Differentiability
10 : Application of Derivatives
11 : Indefinite Integration
12 : Definite Integration
13 : Area
14 : Differential Equations
15 : Straight Line and Pair of Straight Lines
16 : Circle
17 : Parabola
18 : Ellipse
19 : Hyperbola
20 : Trigonometrical Ratios and Identities
21 : Trigonometrical Equations
22 : Inverse Circular Functions
23 : Properties of Triangles
24 : Vectors
25 : 3D Geometry
26 : Miscellaneous
JEE ADVANCED Solved Paper 2021

Citation preview

Chapterwise Topicwise

Solved Papers 2021-1979

IITJEE JEE Main & Advanced

Mathematics Amit M Agarwal

Arihant Prakashan (Series), Meerut

Arihant Prakashan (Series), Meerut All Rights Reserved

© Author

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CONTENTS 1. Complex Numbers

1-28

2. Theory of Equations

29-50

3. Sequences and Series

51-75

4. Permutations and Combinations

76-88

5. Binomial Theorem

89-102

6. Probability

103-134

7. Matrices and Determinants

135-170

8. Functions

171-187

9. Limit, Continuity and Differentiability

188-238

10. Application of Derivatives

239-277

11. Indefinite Integration

278-293

12. Definite Integration

294-328

13. Area

329-353

14. Differential Equations

354-378

15. Straight Line and Pair of Straight Lines

379-404

16. Circle

405-439

17. Parabola

440-457

18. Ellipse

458-473

19. Hyperbola

474-485

20. Trigonometrical Ratios and Identities

486-498

21. Trigonometrical Equations

499-514

22. Inverse Circular Functions

515-525

23. Properties of Triangles

526-547

24. Vectors

548-579

25. 3D Geometry

580-605

26. Miscellaneous

606-632

JEE Advanced Solved Paper 2021

1-16

SYLLABUS JEE MAIN UNIT I Sets, Relations and Functions

UNIT V Mathematical Induction

Sets and their representation, Union, intersection and complement of sets and their algebraic properties, Power set, Relation, Types of relations, equivalence relations, functions, one-one, into and onto functions, composition of functions.

Principle of Mathematical Induction and its simple applications.

UNIT VI Binomial Theorem and its Simple Applications

UNIT II Complex Numbers and Quadratic Equations

Binomial theorem for a positive integral index, general term and middle term, properties of Binomial coefficients and simple applications.

Complex numbers as ordered pairs of reals, Representation of complex numbers in the form a+ib and their representation in a plane, Argand diagram, algebra of complex numbers, modulus and argument (or amplitude) of a complex number, square root of a complex number, triangle inequality, Quadratic equations in real and complex number system and their solutions. Relation between roots and co-efficients, nature of roots, formation of quadratic equations with given roots.

UNIT III Matrices and Determinants Matrices, algebra of matrices, types of matrices, determinants and matrices of order two and three. Properties of determinants, evaluation of deter-minants, area of triangles using determinants. Adjoint and evaluation of inverse of a square matrix using determinants and elementary transformations, Test of consistency and solution of simultaneous linear equations in two or three variables using determinants and matrices.

UNIT VII Sequences and Series Arithmetic and Geometric progressions, insertion of arithmetic, geometric means between two given numbers. Relation between AM and GM Sum upto n terms of special series: ∑ n, ∑ n2, ∑n3. Arithmetico Geometric progression.

UNIT VIII Limit, Continuity and Differentiability Real valued functions, algebra of functions, polynomials, rational, trigonometric, logarithmic and exponential functions, inverse functions. Graphs of simple functions. Limits, continuity and differenti-ability. Differentiation of the sum, difference, product and quotient of two functions. Differentiation of trigonometric, inverse trigonometric, logarithmic, exponential, composite and implicit functions; derivatives of order upto two. Rolle's and Lagrange's Mean Value Theorems. Applications of derivatives: Rate of change of quantities, monotonic increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals.

UNIT IV Permutations and Combinations

UNIT IX Integral Calculus

Fundamental principle of counting, permutation as an arrangement and combination as selection, Meaning of P(n,r) and C (n,r), simple applications.

Integral as an anti - derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by

parts and by partial fractions. Integration using trigonometric identities. Evaluation of simple integrals of the type dx

dx

,

x2 ± a2

x ±a

dx , ax 2 + bx + c (px + q) dx ax 2 + bx + c

,

2

2

dx

dx , a2 – x2

dx

,

,

2

a – x2 (px + q) dx

,

ax 2 + bx + c

ax 2 + bx + c

circle when the end points of a diameter are given, points of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to a circle, equation of the tangent. Sections of cones, equations of conic sections (parabola, ellipse and hyperbola) in standard forms, condition for y=mx + c to be a tangent and point (s) of tangency.

UNIT XII Three Dimensional Geometry ,

2

2

a ± x dx

and

2

2

x – a dx

Integral as limit of a sum. Fundamental Theorem of Calculus. Properties of definite integrals. Evaluation of definite integrals, determining areas of the regions bounded by simple curves in standard form.

UNIT X Differential Equations Ordinary differential equations, their order and degree. Formation of differential equations. Solution of differential equations by the method of separation of variables, solution of homogeneous and linear differential equations of the type

UNIT XI Coordinate Geometry Cartesian system of rectangular coordinates in a plane, distance formula, section formula, locus and its equation, translation of axes, slope of a line, parallel and dy lines, perpendicular of a line on the coordinate + p(x)yintercepts = q(x) dx axes. Straight lines Various forms of equations of a line, intersection of lines, angles between two lines, conditions for concurrence of three lines, distance of a point from a line, equations of internal and external bisectors of angles between two lines, coordinates of centroid, orthocentre and circumcentre of a triangle, equation of family of lines passing through the point of intersection of two lines. Circles, Conic sections Standard form of equation of a circle, general form of the equation of a circle, its radius and centre, equation of a

Coordinates of a point in space, distance between two points, section formula, direction ratios and direction cosines, angle between two intersecting lines. Skew lines, the shortest distance between them and its equation. Equations of a line and a plane in different forms, intersection of a line and a plane, coplanar lines.

UNIT XIII Vector Algebra Vectors and scalars, addition of vectors, components of a vector in two dimensions and three dimensional space, scalar and vector products, scalar and vector triple product.

UNIT XIV Statistics and Probability Measures of Dispersion: Calculation of mean, median, mode of grouped and ungrouped data. Calculation of standard deviation, variance and mean deviation for grouped and ungrouped data. Probability: Probability of an event, addition and multiplication theorems of probability, Baye's theorem, probability distribution of a random variate, Bernoulli trials and Binomial distribution.

UNIT XV Trigonometry Trigonometrical identities and equations. Trigonometrical functions. Inverse trigonometrical functions and their properties. Heights and Distances.

UNIT XVI Mathematical Reasoning Statements, logical operations And, or, implies, implied by, if and only if. Understanding of tautology, contradiction, converse and contra positive.

JEE ADVANCED Algebra Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and principal argument, triangle inequality, cube roots of unity, geometric interpretations. Quadratic equations with real coefficients, relations between roots and coefficients, formation of quadratic equations with given roots, symmetric functions of roots. Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural numbers. Logarithms and their Properties, Permutations and combinations, Binomial theorem for a positive integral index, properties of binomial coefficients.

Matrices Matrices as a rectangular array of real numbers, equality of matrices, addition, multiplication by a scalar and product of matrices, transpose of a matrix, determinant of a square matrix of order up to three, inverse of a square matrix of order up to three, properties of these matrix operations, diagonal, symmetric and skew-symmetric matrices and their properties, solutions of simultaneous linear equations in two or three variables.

Probability Addition and multiplication rules of probability, conditional probability, independence of events, computation of probability of events using permutations and combinations.

Trigonometry Trigonometric functions, their periodicity and graphs, addition and subtraction formulae, formulae involving multiple and sub-multiple angles, general solution of trigonometric equations. Relations between sides and angles of a triangle, sine rule, cosine rule, half-angle formula and the area of a triangle, inverse trigonometric functions (principal value only).

Analytical Geometry Two Dimensions Cartesian oordinates, distance between two points, section formulae, shift of origin. Equation of a straight line in various forms, angle between two lines, distance of a point from a line. Lines through the point of intersection of two given lines, equation of the bisector of the angle between two lines, concurrency of lines, centroid, orthocentre, incentre and circumcentre of a triangle. Equation of a circle in various forms, equations of tangent, normal and chord. Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the points of intersection of two circles and those of a circle and a straight line. Equations of a parabola, ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric equations, equations of tangent and normal.

Locus Problems, Three Dimensions Direction cosines and direction ratios, equation of a straight line in space, equation of a plane, distance of a point from a plane.

Differential Calculus Real valued functions of a real variable, into, onto and one-to-one functions, sum, difference, product and quotient of two functions, composite functions, absolute value, polynomial, rational, trigonometric, exponential and logarithmic functions. Limit and continuity of a function, limit and continuity of the sum, difference, product and quotient of two functions, l'Hospital rule of evaluation of limits of functions. Even and odd functions, inverse of a function, continuity of composite functions, intermediate value property of continuous functions. Derivative of a function, derivative of the sum, difference, product and quotient of two functions, chain rule, derivatives of polynomial, rational, trigonometric, inverse trigonometric, exponential and logarithmic functions. Derivatives of implicit functions, derivatives up to order two, geometrical interpretation of the derivative, tangents and normals, increasing and decreasing functions, maximum and minimum values of a function, applications of Rolle's Theorem and Lagrange's Mean Value Theorem.

Integral Calculus Integration as the inverse process of differentiation, indefinite integrals of standard functions, definite integrals and their properties, application of the Fundamental Theorem of Integral Calculus. Integration by parts, integration by the methods of substitution and partial fractions, application of definite integrals to the determination of areas involving simple curves. Formation of ordinary differential equations, solution of homogeneous differential equations, variables separable method, linear first order differential equations.

Vectors Addition of vectors, scalar multiplication, scalar products, dot and cross products, scalar triple products and their geometrical interpretations.

1 Complex Numbers Topic 1 Complex Number in Iota Form Objective Questions I (Only one correct option) 2z − n 1. Let z ∈ C with Im (z ) = 10 and it satisfies = 2i − 1 2z + n for some natural number n, then (2019 Main, 12 April II) (a) n = 20 and Re(z ) = − 10 (b) n = 40 and Re(z ) = 10 (c) n = 40 and Re(z ) = − 10 (d) n = 20 and Re(z ) = 10

 α + i 2. All the points in the set S =  : α ∈ R (i = −1 ) lie  α − i on a (2019 Main, 9 April I) (a) circle whose radius is 2. (b) straight line whose slope is −1. (c) circle whose radius is 1. (d) straight line whose slope is 1.

3. Let z ∈ C be such that|z|< 1. If ω =

5 + 3z , then 5(1 − z )

3

1 x + iy (i = −1 ), where x and y are real  3  27 (2019 Main, 11 Jan I) numbers, then y − x equals 

(c) – 85

(d) – 91

 π  3 + 2i sin θ , π : is purely imaginary   2  1 − 2i sin θ 

5. Let A = θ ∈  −

 Then, the sum of the elements in A is (2019 Main, 9 Jan I)

(a)

3π 4

(b)

5π 6

(c) π

(b)

6i 7. If 4

–3 i 3i

20

3

 3 −1  1  (c) sin −1   (d) sin    3 4  

π 6

1 –1 = x + iy, then i

(1998, 2M)

(a) x = 3, y = 1 (b) x = 1, y = 1 (c) x = 0, y = 3 (d) x = 0, y = 0 13

(a) i

(b) 5 Re (ω) > 1 (d) 5 Re(ω) > 4

(b) 85

π 3

2 + 3i sin θ is purely imaginary, is 1 − 2i sin θ (2016 Main)

∑ (i n + i n + 1 ), where i =

−1, equals

n =1

4. Let  −2 − i = (a) 91

(a)

8. The value of sum

(2019 Main, 9 April II)

(a) 4 Im(ω) > 5 (c) 5 Im (ω) < 1

6. A value of θ for which

(d)

2π 3

(b) i − 1

(1998, 2M)

(c) − i

(d) 0

n

 1 + i  = 1, is 1 − i

9. The smallest positive integer n for which  (a) 8 (c) 12

(b) 16 (d) None of these

(1980, 2M)

Objective Question II (One or more than one correct option) 10. Let a , b, x and y be real numbers such that a − b = 1 and y ≠ 0. If the complex number z = x + iy satisfies  az + b Im   = y, then which of the following is(are)  z+1 possible value(s) of x?

(2017 Adv.)

(a) 1 − 1 + y2

(b) − 1 − 1 − y2

(c) 1 + 1 + y2

(d) − 1 +

1 − y2

Topic 2 Conjugate and Modulus of a Complex Number Objective Questions I (Only one correct option) 1. The equation|z − i| = |z − 1|, i = −1, represents 1 (2019 Main, 12 April I) 2 (b) the line passing through the origin with slope 1 (c) a circle of radius 1 (d) the line passing through the origin with slope − 1 (a) a circle of radius

2. If a > 0 and z = equal to 1 3 (a) − i 5 5 1 3 (c) − + i 5 5

(1 + i )2 2 , has magnitude , then z is a−i 5 (2019 Main, 10 April I)

1 (b) − − 5 3 (d) − − 5

3 i 5 1 i 5

2 Complex Numbers 3. Let z1 and z2 be two complex numbers satisfying| z1 | = 9 and | z2 − 3 − 4i | = 4. Then, the minimum value of (2019 Main, 12 Jan II) | z1 − z2|is (a) 1

(b) 2

(c)

(d) 0

2

z −α 4. If (α ∈ R) is a purely imaginary number and z+α (2019 Main, 12 Jan I) |z| = 2, then a value of α is (a) 2

1 (b) 2

(c) 1

(where i = − 1). Then,| z |is equal to 34 3

(b)

(2019 Main, 11 Jan II)

5 3

41 4

(c)

(d)

5 4

6. A complex number z is said to be unimodular, if z ≠ 1. z – 2 z2 is If z1 and z2 are complex numbers such that 1 2 – z1z2 unimodular and z2 is not unimodular. Then, the point z 1 lies on a

(2015 Main)

(a) straight line parallel to X-axis

(d) circle of radius 2

7. If z is a complex number such that |z| ≥ 2, then the 1 2

(c) is strictly greater than 5/2

(x − x0 )2 + ( y − y0 )2 = r 2 and (x − x0 )2 + ( y − y0 )2 = 4r 2, respectively. If z0 = x0 + iy0 satisfies the equation 2|z0|2 = r 2 + 2, then (2013 Adv.) |α |is equal to (c)

1 7

(d)

1 3

9. Let z be a complex number such that the imaginary part

of z is non-zero and a = z + z + 1 is real. Then, a cannot take the value (2012) 2

(b)

1 3

(c)

1 2

(d)

3 4

10. Let z = x + iy be a complex number where, x and y are integers. Then, the area of the rectangle whose vertices are the root of the equation zz3 + zz3 = 350, is (2009) (a) 48 (c) 40

 1  1 2 (c) (d) ⋅ 2 z + 1 |z + 1|2   |z + 1|

1 |z + 1|

2

14. For all complex numbers z1 , z2 satisfying |z1| = 12 and |z2 − 3 − 4i| = 5, the minimum value of|z1 − z2|is (a) 0 (c) 7

(b) 2 (d) 17

(2002, 1M)

15. If z1 , z2 and z3 are complex numbers such that 1 1 1 + = 1, then |z1 + z2 + z3|is |z1| = |z2| = |z3| =  + z1 z2 z3 (a) equal to 1 (c) greater than 3

(b) less than 1 (d) equal to 3

(2000, 2M)

(a) n1 = n2 + 1 (c) n1 = n2

(b) n1 = n2 − 1 (d) n1 > 0, n2 > 0

complex numbers sin x + i cos 2x cos x − i sin 2x are conjugate to each other, for (b) x = 0 (d) no value of x

and

(1988, 2M)

vertices of a parallelogram taken in order, if and only if

8. Let complex numbers α and 1 /α lies on circles

(a) − 1

(b)

18. The points z1 , z2, z3 and z4 in the complex plane are the

(d) is strictly greater than 3/2 but less than 5/2

1 2

(a) 0

z −1 (where, z ≠ − 1), then Re (w) is z+1 (2003, 1M)

(a) x = nπ (c) x = (n + 1/2) π

(b) lies in the interval (1, 2)

(b)

13. If|z| = 1 and w =

17. The (2014 Main)

(a) is equal to 5/2

1 2

(b)| z | = 1 and z ≠ 1 (d) None of these

(1 + i )n 1 + (1 + i3 )n1 + (1 + i5 )n 2 + (1 + i7 )n 2 , here (1996, 2M) i = −1 is a real number, if and only if

(c) circle of radius 2

(a)

(a)| z | = 1, z ≠ 2 (c) z = z

16. For positive integers n1 , n2 the value of expression

(b) straight line parallel toY -axis

minimum value of z +

 w − wz  condition that   is purely real, then the set of  1−z  values of z is (2006, 3M)

(d) 2

5. Let z be a complex number such that | z | + z = 3 + i

(a)

12. If w = α + iβ, where β ≠ 0 and z ≠ 1, satisfies the

(b) 32 (d) 80

11. If|z|= 1 and z ≠ ± 1, then all the values of (a) a line not passing through the origin (b)|z|= 2 (c) the X-axis (d) the Y-axis

(a) z1 + z4 = z2 + z3 (c) z1 + z2 = z3 + z4

(b) z1 + z3 = z2 + z4 (1983, 1M) (d) None of these

19. If z = x + iy and w = (1 − iz ) / (z − i ), then |w| = 1 implies that, in the complex plane

(1983, 1M)

(a) z lies on the imaginary axis (b) z lies on the real axis (c) z lies on the unit circle (d) None of these

20. The inequality |z − 4| < |z − 2| represents the region given by

(1982, 2M)

(a) Re (z ) ≥ 0 (c) Re (z ) > 0

(b) Re (z ) < 0 (d) None of these 5

5

 3 i  3 i 21. If z =  +  + −  , then 2 2    2 2 (a) Re (z ) = 0 (c) Re (z ) > 0, Im (z ) > 0

(1982, 2M)

(b) Im (z ) = 0 (d) Re (z ) > 0, Im (z ) < 0

22. The complex numbers z = x + iy which satisfy the z lie on 1 − z2 (2007, 3M)

z − 5i   = 1, lie on equation z + 5i  (a) the X-axis (b) the straight line y = 5 (c) a circle passing through the origin (d) None of the above

(1981, 2M)

Complex Numbers 3 29. Let z be any point in A ∩ B ∩ C and let w be any point

Objective Questions II (One or more than one correct option)

satisfying |w − 2 − i| < 3. between

23. Let S be the set of all complex numbers z satisfying

(a) − 6 and 3 (c) − 6 and 6

| z 2 + z + 1| = 1. Then which of the following statements is/are TRUE? (2020 Adv.)

1 1 ≤ for all z ∈S 2 2 1 1 (c) z + ≥ for all z ∈S 2 2

(a) z +

solutions z = x + iy (x, y ∈ R, i = − 1 ) of the equation sz + tz + r = 0, where z = x − iy. Then, which of the following statement(s) is (are) TRUE? (2018 Adv.)

(a) If L has exactly one element, then| s|≠ |t | (b) If|s|=|t |, then L has infinitely many elements (c) The number of elements in L ∩ {z :| z − 1 + i| = 5} is at most 2 (d) If L has more than one element, then L has infinitely many elements

25. Let z1 and z2 be complex numbers such that z1 ≠ z2 and

30. Let z be any point in A ∩ B ∩ C. The|z + 1 − i|2 + |z − 5 − i|2 lies between (a) 25 and 29 (c) 35 and 39 (a) 0 (c) 2

that |z1| = |z2| = 1 and Re (z1z2) = 0, then the pair of complex numbers w1 = a + ic and w2 = b + id satisfies (1985, 2M)

Passage Based Problems

32. Match the statements of Column I with those of Column II. Here, z takes values in the complex plane and Im (z ) and Re (z ) denote respectively, the imaginary part and (2010) the real part of z

A.

The set of points z satisfying | z − i| z|| = | z + i | z|| is contained in or equal to

p.

an ellipse with eccentricity 4/5

B.

The set of points z satisfying | z + 4| + | z − 4| = 0 is contained in or equal to

q.

the set of points z satisfying Im ( z) = 0

C.

If| w| = 2 , then the set of 1 points z = w − is contained w in or equal to

r.

the set of points z satisfying|Im( z) |≤ 1

D.

If| w| = 1, then the set of points s. 1 z = w + is contained in or t. w equal to

Passage I

27. min|1 − 3i − z|is equal to z ∈s

2− 3 (a) 2

2+ 3 (b) 2

3− 3 (c) 2

3+ 3 (d) 2

28. Area of S is equal to (a)

10 π 3

(b)

20 π 3

(c)

16 π 3

(d)

32 π 3

Column II

Column I

Read the following passages and answer the questions that follow. Let A, B, C be three sets of complex number as defined below A = { z : lm (z ) ≥ 1} B = { z :|z − 2 − i| = 3} (2008, 12M) C = { z : Re((1 − i )z ) = 2 }

(b) 1 (d) ∞

Match the Columns

(b) real and positive (d) purely imaginary

26. If z1 = a + ib and z2 = c + id are complex numbers such

(b) 30 and 34 (d) 40 and 44

31. The number of elements in the set A ∩ B ∩ C is

|z1| = |z2|. If z1 has positive real part and z2 has negative z + z2 imaginary part, then 1 may be (1986, 2M) z1 − z2

(b)|w2| = 1 (d) None of these

(b) − 3 and 6 (d) − 3 and 9

Let S = S1 ∩ S 2 ∩ S3 , where   z − 1 + 3 i  S1 = { z ∈ C :|z | < 4}, S 2 = z ∈ C : lm   > 0  1− 3i    and S3 : { z ∈ C : Re z > 0} (2008)

(d) The set S has exactly four elements

(a)|w1| = 1 (c) Re (w1 w2 ) = 0

lies

Passage II

(b)|z|≤ 2 for all z ∈S

24. Let s, t, r be non-zero complex numbers and L be the set of

(a) zero (c) real and negative

Then, |z | − |w| + 3

the set of points satisfying|Re( z)|≤ 2 the set of points z satisfying| z| ≤ 3

Fill in the Blanks 33. If α , β, γ are the cube roots of p, p < 0, then for any x, y and z then

xα + yβ + zγ = ... . xβ + yγ + zα

(1990, 2M)

34. For any two complex numbers z1 , z2 and any real numbers a and b,|az1 − bz2|2+ |bz1 + az2|2 = K . (1988, 2M)

4 Complex Numbers    x  x sin  2 + cos  2 − i tan (x)   is real, 35. If the expression   x  1 + 2 i sin  2    then the set of all possible values of x is… . (1987, 2M)

41. If z1 and z2 are two complex numbers such that |z1| < 1 < |z2|, then prove that

1 − z1z2 < 1. z1 − z2

(2003, 2M)

42. For complex numbers z and w, prove that

| z |2 w − |w|2 z = z − w, if and only if z = w or z w = 1. (1999, 10M)

43. Find all non-zero complex numbers z satisfying

True/False

z = iz 2.

36. If three complex numbers are in AP. Then, they lie on a circle in the complex plane

(1985 M)

37. If the complex numbers, z1 , z2 and z3 represent the vertices of an equilateral triangle | z1 | = | z2| = | z3 |, then z1 + z2 + z3 = 0.

such

that

(1984, 1M)

38. For complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, we write z1 ∩ z2, if x1 ≤ x2 and y1 ≤ y2. Then, for all complex 1−z numbers z with 1 ∩ z, we have ∩ 0. (1981, 2M) 1+ z

Analytical & Descriptive Questions 39. Find the centre and radius of the circle formed by all the

points represented by z = x + iy satisfying the relation z − α    = k (k ≠ 1), where α and β are the constant z −β complex numbers given by α = α 1 + iα 2, β = β1 + iβ 2. (2004, 2M)

40. Prove that there exists no complex number z such that |z| < 1 /3 and

n

∑

44. If

(1996, 2M)

iz + z − z + i = 0, 3

2

then

show

that

|z| = 1.

(1995, 5M)

45. A relation R on the set of complex numbers is defined z1 − z2 is real. z1 + z2 Show that R is an equivalence relation.

by z1 R z2, if and only if

(1982, 2M)

46. Find the real values of x and y for which the following equation is satisfied (1 + i ) x − 2i (2 − 3i ) y + i + = i. 3+ i 3−i

47. Express

( 1980, 2M)

1 in the form A + iB. (1 − cos θ ) + 2i sin θ (1979, 3M)

48. If x + iy =

a + ib a 2 + b2 , prove that (x2 + y2)2 = 2 c + id c + d2 (1978, 2M)

Integer & Numerical Answer Type Question 49. If z is any complex number satisfying | z − 3 − 2i | ≤ 2, then the maximum value of|2z − 6 + 5i |is …… (2011)

a rz r = 1, where|a r|< 2.

r =1

(2003, 2M)

Topic 3 Argument of a Complex Number Objective Questions I (Only one correct option) 1. If z1 , z2 are complex numbers such that Re(z1 ) = |z1 − 1|, Re(z2) = |z2 − 1| and arg(z1 − z2) =

π , then Im(z1 + z2) is 6

equal to (a)

(2020 Main, 3 Sep II)

3 2

(b)

1 3

(c)

2 3

(d) 2 3

2. Let z1 and z2 be any two non-zero complex numbers such that 3|z1| = 4|z2|. If z = (a) |z| =

1 17 2 2

3z1 2z2 + , then 2z2 3z1 (2019 Main, 10 Jan I)

(d) |z| =

5 2

(a) − θ

(b)

π −θ 2

(c) θ

(2000, 2M)

(b) − π (d) π /2

5. Let z and w be two complex numbers such that|z| ≤ 1, |w| ≤ 1 and|z + i w|= |z − iw| = 2 , then z equals (1995, 2M)

(a) 1 or i (c) 1 or −1

(b) i or −i (d) i or −1

6. Let z and w be two non-zero complex numbers

such that |z| = |w| and arg (z ) + arg (w) = π, then z (1995, 2M) equals (b) −w (d) −w

7. If z1 and z2 are two non-zero complex numbers such

3. If z is a complex number of unit modulus and argument  1 + z θ, then arg   is equal to 1 + z

(a) π (c) − π/2

(a) w (c) w

(b) Im(z ) = 0

(c) Re(z) = 0

4. If arg (z ) < 0 , then arg (−z ) − arg (z ) equals

(2013 Main)

(d) π − θ

that|z1 + z2| = |z1| + |z2|, then arg (z1 ) − arg (z2) is equal to (a) − π (c) 0

π (b) − 2 π (d) 2

(1987, 2M)

Complex Numbers 5 8. If a , b, c and u , v, w are the complex numbers

10. Let z1 and z2 be two distinct complex numbers and let

representing the vertices of two triangles such that c = (1 − r ) a + rb and w = (1 − r ) u + rv, where r is a (1985, 2M) complex number, then the two triangles

z = (1 − t ) z1 + tz2 for some real number t with 0 < t < 1. If arg (w) denotes the principal argument of a non-zero (2010) complex number w, then

(a) have the same area (c) are congruent

(a) |z − z1| + |z − z2|= |z1 − z2|(b) arg (z − z1 ) = arg (z − z2 )

(b) are similar (d) None of these

(c)

Objective Questions II (One or more than one correct option) principal argument with − π < arg(z ) ≤ π . Then, which of the following statement(s) is (are) FALSE ? (2018 Adv.) π , where i = 4

z − z1

z2 − z1

z2 − z1

=0

(d) arg (z − z1 ) = arg (z2 − z1 )

Match the Columns

9. For a non-zero complex number z, let arg(z ) denote the

(a) arg (−1 − i ) =

z − z1

11. Match the conditions/expressions in Column I with statement in Column II (z ≠ 0 is a complex number) Column II

Column I

−1

Re ( z) = 0 π arg ( z) = 4

A.

(b) The function defined by f : R → (− π, π], f (t ) = arg (−1 + it ) for all t ∈ R, is continuous at all points of R, where i = −1. (c) For any two non-zero complex numbers z1 and z2, z  arg  1  − arg (z1 ) + arg (z2 ) is an integer multiple of  z2 

B.

p.

Re ( z2 ) = 0

q.

Im ( z2 ) = 0

r.

Re ( z2 ) = Im ( z2 )

Analytical & Descriptive Questions 12. |z| ≤ 1,|w|≤ 1, then show that |z − w|2 ≤ (|z| − |w|)2 + (arg z − arg w)2

2π. (d) For any three given distinct complex numbers z1 , z2 and z3 , the locus of the point z satisfying the condition  (z − z1 ) (z2 − z3 )  arg   = π, lies on a straight line.  (z − z3 ) (z2 − z1 ) 

(1995, 5M)

13. Let z1 = 10 + 6i and z2 = 4 + 6i. If z is any complex number such that the argument of (z − z1 ) / (z − z2) is (1991, 4M) π /4, then prove that|z − 7 − 9i| = 3 2.

Topic 4 Rotation of a Complex Number Objective Questions I (Only one correct option) 1. Let S be the set of all complex numbers z satisfying | z − 2 + i | ≥ 5. If the complex number z0 is such that   1 1 is the maximum of the set  : z ∈ S , then | z0 − 1 | | z − 1 |   the principal argument of (a)

π 4

(b)

3π 4 5

4 − z0 − z0 is z0 − z0 + 2 i (c) −

π 2

(2019 Adv.)

(d)

π 2

5

 3 i  3 i +  + −  . If R(z ) and I (z ) 2 2  2  2

2. Let z = 

respectively denote the real and imaginary parts of z, then (2019 Main, 10 Jan II) (a) R (z ) > 0 and I (z ) > 0 (c) R (z ) < 0 and I (z ) > 0

(b) I (z ) = 0 (d) R (z ) = − 3

3. A particle P starts from the point z0 = 1 + 2 i, where i = −1. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z1. From z1 the particle moves 2 units in the direction of the vector $i + $j and then it moves π in anti-clockwise direction on a through an angle 2 circle with centre at origin, to reach a point z2. The point (2008, 3M) z2 is given by (a) 6 + 7i

(b) −7 + 6i

(c) 7 + 6i

(d) − 6 + 7i

4. A man walks a distance of 3 units from the origin towards the North-East (N 45° E) direction. From there, he walks a distance of 4 units towards the North-West (N 45° W) direction to reach a point P. Then, the position of P in the Argand plane is (2007, 3M) (a) 3ei π/ 4 + 4 i (b) (3 − 4 i ) ei π / 4 (c) (4 + 3 i )ei π / 4 (d) (3 + 4 i ) ei π / 4

5. The shaded region, where P = (−1, 0), Q = (−1 + 2 , 2 ) R = (−1 + 2 , − 2 ), S = (1, 0) is represented by (2005, 1M) π 4 π (b)|z + 1|< 2,| arg (z + 1)|< 2 π (c)|z + 1|> 2,| arg (z + 1)|> 4 π (d)|z − 1|< 2,| arg (z + 1)|> 2

Y

(a)|z + 1|> 2,| arg (z + 1)|
 2  1 −   and H 2 = z ∈ C : Re (z ) < , where C is the set of all  2  complex numbers. If z1 ∈ P ∩ H 1, z2 ∈ P ∩ H 2 and O represents the origin, then ∠ z1Oz2 is equal to

9. Let W =

π 6

(c)

14. Let bz + bz = c, b ≠ 0, be a line in the complex plane,

(1997C, 5M)

15. Let z1 and z2 be the roots of the equation z + pz + q = 0, 2

  1 Suppose S = z ∈ C : z = , t ∈ R, t ≠ 0, where a + i bt   i = − 1. If z = x + iy and z ∈ S, then (x, y) lies on

(b)

circle |z − 1| = 2 is 2 + 3i. Find the other vertices of (2005, 4M) square.

that c = z1b + z2b.

8. Let a , b ∈ R and a 2 + b2 ≠ 0.

π 2

13. If one of the vertices of the square circumscribing the

where b is the complex conjugate of b. If a point z1 is the reflection of the point z2 through the line, then show

Objective Questions II (One or more than one correct option)

(a)

Analytical & Descriptive Questions

2π 3

(2013 JEE Adv.)

(d)

5π 6

Fill in the Blanks 10. Suppose z1 , z2, z3 are the vertices of an equilateral triangle inscribed in the circle |z| = 2. If z1 = 1 + i 3, then z2 = K, z3 = … . (1994, 2M)

11. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2 AC. If the points D and M represent the complex numbers 1 + i and 2 − i respectively, then A represents the complex number …or… (1993, 2M)

where the coefficients p and q may be complex numbers. Let A and B represent z1 and z2 in the complex plane. If ∠ AOB = α ≠ 0 and OA = OB, where O is the origin prove α that p2 = 4q cos 2  .  2 (1997, 5M)

16. Complex numbers z1 , z2, z3 are the vertices A , B, C respectively of an isosceles right angled triangle with right angle at C. Show that (z1 − z2)2 = 2(z1 − z3 ) (z3 − z2). (1986, 2 1 M) 2

17. Show that the area of the triangle on the argand

diagram formed by the complex number z , iz and z + iz 1 is |z|2. 2 (1986, 2 1 M) 2

18. Prove that the complex numbers z1 , z2 and the origin form an equilateral triangle only if z12 + z22 − z1z2 = 0. (1983, 2M)

19. Let the complex numbers z1 , z2 and z3 be the vertices of an equilateral triangle. Let z0 be the circumcentre of the triangle. Then, prove that z12 + z22 + z32 = 3z02. (1981, 4M)

Integer & Numerical Answer Type Questions 20. For a complex number z, let Re(z ) denote the real part of z. Let S be the set of all complex numbers z satisfying z 4 − |z|4 = 4 i z 2, where i = −1. Then the minimum possible value of|z1 − z2|2, where z1 , z2 ∈ S with Re(z1 ) > 0 and Re(z2) < 0 is …… . (2020 Adv.) kπ   kπ   + i sin   , where  7  7 i = −1. The value of the expression

21. For any integer k, let α k = cos  12

∑|α k + 1 − α k|

k =1

12. If a and b are real numbers between 0 and 1 such that the points z1 = a + i , z2 = 1 + bi and z3 = 0 form an equilateral triangle, then a = K and b = K . (1990, 2M)

3

∑|α 4k − 1 − α 4k − 2|

k =1

is (2016 Adv.)

Complex Numbers 7

Topic 5 De-Moivre’s Theorem, Cube Roots and nth Roots of Unity Objective Questions I (Only one correct option)

9. Let z1 and z2 be nth roots of unity which subtend a right angled at the origin, then n must be of the form (2001, 1M) (where, k is an integer)

3

2π 2π   + i cos   1 + sin 9 9  is 1. The value of   1 + sin 2π − i cos 2π   9 9 

(a) 4k + 1

[2020 Main, 2 Sep I]

1 (a) − ( 3 − i ) 2 1 (c) ( 3 − i ) 2

1 (b) − (1 − i 3 ) 2 1 (d) (1 − i 3 ) 2

π , then 2

3. If z = to

(2019 Main, 10 April II)

(a) 1

(c) −1

(d) 0

(d)

5. Let z = cos θ + i sin θ . Then, the value of

π 3

∑ Im (z

θ = 2° is

2m − 1

) at

6. The minimum value of |a + bω + cω |, where a, b and c are all not equal integers and ω (≠ 1) is a cube root of unity, is (2005, 1M)

(c) 1

(d) 0

7. If ω (≠ 1) be a cube root of unity and (1 + ω 2)n = (1 + ω 4 )n, then the least positive value of n is (a) 2

(b) 3

(c) 5

(2004, 1M)

(d) 6

1 3 , then value of the determinant +i 2 2 1 1 1 1 −1 − ω 2 ω 2 is (2002, 1M) ω2 ω 1

8. Let ω = −

(a) 3 ω

(a) – 1

(b) 3 ω (ω − 1) (c) 3 ω2

3

(d) −128 ω2

B

are

respectively

(b) 1, 1



∑ sin

(c) 1, 0

(d) 3 ω (1 − ω)

(d) −1, 1

2 πk 2 πk  – i cos  is 7 7 

(b) 0

(c) – i

(1998, 2M)

(d) i

Match the Columns 2 kπ   2 kπ   ; k = 1, 2, …9.  + i sin   10   10 

14. Let zk = cos 

Column II

P.

For each zk, there exists a z j such that zk ⋅ z j = 1

(i)

True

Q.

There exists a k ∈ { 1, 2, … , 9 } such that z1 ⋅ z = zk has no solution z in the set of complex numbers

(ii)

False

R.

|1 − z1||1 − z2| … |1 − z9| equal 10 9 2 kπ  1 − ∑ cos   equals  10  k =1

(iii)

1

(iv)

2

S.

2

1 2

and

A

k =1

(2009)

1 (b) 3 sin 2° 1 (d) 4 sin 2°

(b)

(d) −i

(1998, 2M)

(c) 128 ω2

Column I

m =1

1 (a) sin 2° 1 (c) 2 sin 2°

3

12. If ω (≠ 1) is a cube root of unity and (1 + ω )7 = A + Bω,

(2019 Main, 9 Jan II)

(c) 0

(b) −128 ω

6

15

(a) 3

(a) 128 ω

13. The value of

z = 3 + 6iz081 − 3iz093 , then arg z is equal to π (b) 6

(c) i

(1995, 2M)

4. Let z0 be a root of the quadratic equation, x2 + x + 1 = 0, If π (a) 4

365

(1999, 2M)

(b) −1 + i 3

3

(a) 0, 1

(2019 Main, 8 April II)

(b) (−1 + 2i )

 1 i 3 + 3 − +  2   2

(1 + ω − ω 2)7 is equal to

then

3 i + (i = −1 ), then (1 + iz + z5 + iz 8 )9 is equal 2 2 9

(d) 4k

11. If ω is an imaginary cube root of unity, then

(b) zw =

(c) zw = i

i 3  2 

334

is equal to

1− i 2 − 1+ i (d) zw = 2

(a) zw = − i

 1  2

(c) 4k + 3

10. If i = −1, then 4 + 5  − + (a) 1 − i

2. If z and w are two complex numbers such that | zw| = 1 and arg(z ) − arg(w) =

(b) 4k + 2

(2011)

Codes P (a) (i) (b) (ii) (c) (i) (d) (ii)

Q (ii) (i) (ii) (i)

R (iv) (iii) (iii) (iv)

S (iii) (iv) (iv) (iii)

Fill in the Blanks 2π 2π + i sin . Then 3 3 the number of distinct complex number z satisfying

15. Let ω be the complex number cos

ω ω2 z+1 2 ω 1 = 0 is equal to ... . z+ω ω2 1 z+ω

(2010)

8 Complex Numbers 16. The value of the expression

19. If 1, a1 , a 2, ... , a n − 1 are the n roots of unity, then show

1 ( 2 − ω ) (2 − ω 2) + 2(3 − ω ) (3 − ω 2) + ... + (n − 1) ⋅ (n − ω ) (n − ω 2) , where, ω is an imaginary cube root of unity, is….

that (1 − a1 ) (1 − a 2) (1 − a3 ) K (1 − a n − 1 ) = n

(1984, 2M)

20. It is given that n is an odd integer greater than 3, but n

(1996, 2M)

True/False

is not a multiple of 3. Prove that x3 + x2 + x is a factor of (1980, 3M) (x + 1)n − xn − 1 .

21. If x = a + b, y = aα + bβ, z = aβ + bα , where α , β are

17. The cube roots of unity when represented on Argand diagram form the vertices of an equilateral triangle.

complex cube xyz = a3 + b3 .

roots

of

unity,

then

show

that

(1979, 3M)

(1988, 1M)

Integer & Numerical Answer Type Questions

Analytical & Descriptive Questions 18. Let a complex number α , α ≠ 1, be a root of the equation z p + q − z p − zq + 1 = 0 where, p and q are distinct primes. Show that either 1 + α + α 2 + ... + α p − 1 = 0 or 1 + α + α 2 + ... + α q − 1 = 0 but not both together.

(2002, 5M)

22. Let ω ≠ 1 be a cube root of unity. Then the minimum of the set {| a + bω + cω 2|2 : a , b, c distinct non-zero integers} equals ............ (2019 Adv.)

23. Let ω = eiπ /3 and a , b, c, x, y, z be non-zero complex numbers such that a + b + c = x, a + bω + cω 2 = y, a + bω 2 + cω = z. | x|2 + | y|2 + | z |2 Then, the value of (2011) is …… | a |2 + | b|2 + | c|2

Answers Topic 1 1. (c) 5. (d) 9. (d)

2. (c) 6. (d) 10. (b, d)

3. (b) 7. (d)

4. (a) 8. (b)

1 cot (θ / 2 ) −i θ   1 + 3 cos2 (θ / 2 ) 2 1 + 3 cos2   2

49. 5

Topic 3 1. (d)

Topic 2 1. (b)

47. A + iB =

2. (b)

3. (d)

4. (d)

5. (c) 9. (a, b, d)

2. (*) 6. (d) 10. (a, c, d)

3. (c)

4. (a)

7. (c)

8. (b)

11. A → q ; B → p

5. (b)

6. (c)

7. (b)

8. (c)

9. (d)

10. (a)

11. (d)

12. (b)

13. (a)

14. (b)

15. (a)

16. (d)

1. (c)

2. (b)

3. (d)

4. (d)

17. (d)

18. (b)

19. (b)

20. (d)

5. (a)

6. (d)

7. (c)

8. (a,c,d)

21. (b)

22. (a)

23. (b,c)

24. (a, c, d)

25. (a,d)

26. (a, b, c)

27. (c)

28. (b)

29. (d)

30. (c)

31. (b)

32. A → q, r ; B → p; C → p, s, t ; D → q, r, s, t 33. ω

2

34. (a + b )(| z1| + | z 2| ) 2

2

2

2

−1

35. x = 2nπ + 2α , α = tan k, where k ∈(1, 2 ) or x = 2nπ 36. False

37. True

38. True k (α − β ) α − k 2β  39. Centre = , Radius =  2 2 1 −k  1 −k 

Topic 4

9. (c, d) 10. z 2 = − 2, z 3 = 1 − i 3 i 3i 11. 3 − or 1 − 12. a = b = 2 ± 3 2 2 13. z 2 = − 3 i , z 3 = (1 − 3 ) + i and z 4 = (1 + 3 ) − i 20. (8)

21. (4)

Topic 5 1. (a)

2. (a)

3. (c)

3. (a)

5. (d)

6. (c)

7. (b)

8. (b)

9. (d)

10. (c)

11. (d)

12. (b)

14. (c)

15. (1)

13. (d)

 3 i 43. z = i , ± –  2 2 

 n (n + 1 ) 16.   −n  2 

46. (x = 3 and y = −1)

23. (3)

2

17. True

22. (3)

Hints & Solutions Topic 1 Complex Number in Iota Form 1. Let z = x + 10i, as Im (z ) = 10 (given). Since z satisfies, 2z − n = 2i − 1, n ∈ N , 2z + n ∴ (2x + 20i − n ) = (2i − 1) (2x + 20i + n ) ⇒ (2x − n ) + 20i = (− 2x − n − 40) + (4x + 2n − 20)i On comparing real and imaginary parts, we get 2x − n = − 2x − n − 40 and 20 = 4x + 2n − 20 ⇒ 4x = − 40 and 4x + 2n = 40 ⇒ x = − 10 and − 40 + 2n = 40 ⇒ n = 40 So, n = 40 and x = Re (z ) = − 10 α+i 2. Let x + iy = α −i (α + i )2 (α 2 − 1) + (2α )i α 2 − 1  2α  ⇒ x + iy = 2 = + = 2 i α +1 α2 + 1 α + 1  α 2 + 1

3

 3 + 2i sin θ   1 + 2 i sin θ   ×   1 − 2i sin θ   1 + 2 i sin θ 

5. Let z = 

(rationalising the denominator) =

2α α2 −1 and y = 2 α +1 α2 + 1 2

2  α 2 − 1  2α  Now, x + y =  2  + 2   α + 1  α + 1 2

=

2

α 4 + 1 − 2α 2 + 4α 2 (α 2 + 1)2 = 2 =1 (α 2 + 1)2 (α + 1)2

⇒ x2 + y2 = 1 Which is an equation of circle with centre (0, 0) and radius 1 unit.

3 − 4 sin θ + 8i sin θ 1 + 4 sin 2 θ 2

[Q a 2 − b2 = (a + b)(a − b) and i 2 = − 1]  3 − 4 sin θ   8 sin θ  =  + i  1 + 4 sin 2 θ   1 + 4 sin 2 θ 

On comparing real and imaginary parts, we get x=

2

As z is purely imaginary, so real part of z = 0 3 − 4 sin 2 θ = 0 ⇒ 3 − 4 sin 2 θ = 0 ∴ 1 + 4 sin 2 θ 3 ⇒ sin 2 θ = 4 3 ⇒ sin θ = ± 2 Y 1 √3/2

 α + i So, S =  ; α ∈ R lies on a circle with radius 1.  α − i X′

3. Given complex number

5 ω − 5 ω z = 5 + 3z (3 + 5 ω )z = 5 ω − 5 …(i) |3 + 5 ω||z| = |5 ω − 5| [applying modulus both sides and |z1z2| = |z1||z2|] Q |z| < 1 [from Eq. (i)] ∴ |3 + 5 ω| > |5 ω − 5| 3  ⇒ ω + > |ω − 1| 5  2 3  Let ω = x + iy, then  x +  + y2 > (x − 1)2 + y2  5 9 6 2 2 ⇒ x + + x > x + 1 − 2x 25 5 1 16x 16 ⇒ > ⇒ x > ⇒ 5x > 1 5 5 25 ⇒ 5 Re( ω ) > 1

y=sin θ

–π/2 –π/3 O π/3

5 + 3z ω= 5(1 − z )

⇒ ⇒ ⇒

3

1   –1 x + iy   =  − 2 − i = (6 + i )   27 3   3 1 x + iy ⇒ =− (216 + 108i + 18i 2 + i3 ) 27 27 1 =− (198 + 107i ) 27 [Q (a + b)3 = a3 + b3 + 3a 2b + 3ab2, i 2 = − 1, i3 = − i] On equating real and imaginary part, we get x = − 198 and y = − 107 ⇒ y − x = − 107 + 198 = 91

4. We have,

−1

2π/3

π

X

–√3/2

Y′

 π π 2π  θ ∈ − , ,   3 3 3  2π . Sum of values of θ = 3 2 + 3i sin θ 6. Let z = is purely imaginary. Then, we have 1 − 2i sin θ Re(z ) = 0 2 + 3i sin θ Now, consider z = 1 − 2i sin θ (2 + 3i sin θ) (1 + 2i sin θ ) = (1 − 2i sin θ ) (1 + 2i sin θ) ⇒

=

2 + 4i sin θ + 3i sin θ + 6i 2 sin 2 θ 12 − (2i sin θ) 2

10 Complex Numbers =

2 + 7i sin θ − 6 sin 2 θ 1 + 4 sin 2 θ

=

2 − 6 sin 2 θ 7 sin θ +i 1 + 4 sin 2 θ 1 + 4 sin 2 θ

Re(z ) = 0 2 − 6 sin 2 θ = 0 ⇒ 2 = 6 sin 2 θ 1 + 4 sin 2 θ 1 sin 2 θ = 3 1 sin θ = ± 3 1 −1  −1  1  θ = sin  ±  = ± sin      3 3

Q ∴ ⇒ ⇒ ⇒

7. Given,

 6i 4 20

 6i −3 i  4  20

⇒

−3 i 3i 3

1 −1  = x + iy i

1 −1 i

1 −1  = x + i y i

⇒

x = 0, y = 0

13

8.

∑ (i

+i

n

n=1

n+1

13

) = ∑ i (1 + i ) = (1 + i ) n

n=1

1. Let the complex number z = x + iy | z − i | =| z − 1|

Also given, ⇒

| x + iy − i | = | x + iy − 1|

⇒

x2 + ( y − 1)2 = (x − 1)2 + y2 [Q| z | = (Re(z ))2 + (Im(z ))2 ]

On squaring both sides, we get x2 + y2 − 2 y + 1 = x2 + y2 − 2x + 1 ⇒ y = x, which represents a line passing through the origin with slope 1. (1 + i )2 a−i (1 − 1 + 2i ) (a + i ) = a2 + 1 2i (a + i ) −2 + 2ai = = a2 + 1 a2 + 1

2. The given complex number z =

x + iy = 0 [Q C 2 and C3 are identical]

⇒

Topic 2 Conjugate and Modulus of Complex Number

13

∑

i

n

n =1

 i − (1 − i13 )  = (1 + i ) (i + i 2 + i3 + K + i13 ) = (1 + i )     1−i  i (1 − i )  = (1 + i )   = (1 + i ) i = i − 1  1−i  Alternate Solution

4 + 4a 2 2 2 2 ⇒ = = 2 5 5 (a 2 + 1)2 1+ a 4 2 = ⇒ a 2 + 1 = 10 ⇒ a 2 = 9 ⇒ a = 3 [Qa > 0] ⇒ 1 + a2 5 –2 + 6i [From Eq. (i)] ∴ z= 10 1 3  −2 + 6 i   1 3  So, z =   =  − + i ⇒ z = − − i  10   5 5  5 5

+ (i 2 + i3 + K + i14 ) = i + i 2 = i − 1 n

 1 + i   =1 1 − i

n

 1 + i 1 + i × ⇒   =1  1 − i 1 + i

n

⇒

 2i    =1  2

⇒ in = 1 The smallest positive integer n for which i n = 1 is 4. ∴ n =4 az + b ax + b + aiy (ax + b + aiy)((x + 1) − iy) 10. = = z+1 (x + 1) + iy (x + 1)2 + y2 ∴

 az + b − (ax + b) y + ay(x + 1) Im  =  z+1 (x + 1)2 + y2

⇒

(a − b) y =y (x + 1)2 + y2

Q ∴

a − b =1 (x + 1)2 + y2 = 1

∴

[Qif z = x + iy, then z = x − iy]

3. Clearly|z1|= 9, represents a circle having centre C1 (0, 0)

n=1

9. Since,

x = − 1 ± 1 − y2

[given]

⇒

13

∑ (i n + i n + 1 ) = (i + i 2 + K + i13 )

…(i)

z = 2 /5

Q

Since, sum of any four consecutive powers of iota is zero. ∴

[Q i 2 = − 1]

and radius r1 = 9. and |z2 − 3 − 4i|= 4 represents a circle having centre C 2(3, 4) and radius r2 = 4. The minimum value of |z1 − z2| is equals to minimum distance between circles|z1|= 9 and|z2 − 3 − 4i|= 4. Q

C1C 2 = (3 − 0)2 + (4 − 0)2 = 9 + 16 = 25 = 5

and |r1 − r2|=|9 − 4|= 5 ⇒ C1C 2 =|r1 − r2| ∴ Circles touches each other internally. Hence, |z1 − z2|min = 0 z −α 4. Since, the complex number (α ∈ R) is purely z+α imaginary number, therefore z −α z −α [Qα ∈ R] + =0 z+α z+α ⇒

zz − αz + αz − α 2 + zz − αz + αz − α 2 = 0

⇒

2 z

⇒

α2 = z

⇒

α=±2

2

− 2 α2 = 0 2

=4

2

[Qzz = z ] [| z | = 2 given]

Complex Numbers 11 5. We have,|z | + z = 3 + i ∴

x2 + y2 + x + iy = 3 + i

⇒

(x + x + y ) + iy = 3 + i

⇒

x + x2 + y2 = 3 and y = 1

2

⇒

∴ ⇒

= |z1|2 − z1z2 − z2z1 + |z2|2 Here, (x − x0 ) + ( y − y0 ) = r 2 2

|z − z0|2 = r 2 and |z − z0|2 = 4r 2 Since, α and ∴ ⇒

(α − z0 ) (α − z0 ) = r 2

⇒

|α|2 − z0α − z0α + |z0|2 = r 2

⇒ (z1 − 2z2)(z1 − 2z2) = (2 − z1z2) (2 − z1z2)

[zz = |z|2 ]

⇒ |z1|2+4|z2|2−2z1z2 − 2z1z2

⇒

  1 1 2  − z0   − z0  = 4 r   α α

⇒

z z 1 − 0 − 0 + |z0|2 = 4r 2 α |α|2 α |α|2 = α ⋅ α

Since,

= 4+|z1|2|z2|2−2z1z2 − 2z1z2 ⇒ (|z2|2−1)(|z1|2−4) = 0 Q |z2|≠ 1 ∴ |z1|= 2 Let z1 = x + iy ⇒ x2 + y2 = (2)2 ∴ Point z1 lies on a circle of radius 2.

⇒

z ⋅α z 1 − 0 − 0 ⋅ α + |z0|2 = 4r 2 |α|2 |α|2 |α|2 1 − z0α − z0α + |α|2|z0|2 = 4r 2|α|2

⇒

(|α|2 − 1) + |z0|2 (1 − |α|2 ) = r 2 (1 − 4|α|2 )

(0, 0) and radius is 2. 1 is distance of z, which lie on circle Minimum z + 2 | z | = 2 from (−1 / 2, 0). 1  1  = Distance of  − , 0 from (−2, 0) ∴ Minimum z +  2  2

⇒

(|α|2 − 1) (1 − |z0|2 ) = r 2(1 − 4|α|2 )

⇒

 r 2 + 2 (|α|2 − 1) 1 −  = r 2(1 − 4|α|2 ) 2   |z0|2 =

Given,

2

1 3 3  −1  + 2 + 0 =  +0= =   2  2 2 2

⇒

|α|2 − 1 = − 2 + 8|α|2 ⇒ 7|α|2 = 1 |α| = 1 / 7

∴

9. A 1 , (0,0) (– — ) 2 0

(2,0)

1 = AD 2 1 Hence, minimum value of z + lies in the interval 2 (1, 2).

Geometrically Min z +

PLAN If ax 2 + bx + c = 0 has roots α, β, then α, β =

X

Y′

r2 + 2 2

 − r 2 (|α|2 − 1) ⋅   = r 2(1 − 4|α|2 )  2 

⇒

Y

D (–2,0)

…(ii)

On subtracting Eqs. (i) and (ii), we get

7. |z| ≥ 2 is the region on or outside circle whose centre is

X′

…(i)

1 − z0 = 4 r 2 α

and

z1 − 2z2 = 1 ⇒ |z1 − 2z2|2 =|2 − z1z2|2 2 − z1z2

2

1 lies on first and second respectively. α 2 1 − z0 = 4 r 2 |α − z0|2 = r 2 and α

2

Given, z2 is not unimodular i.e.|z2|≠ 1 z − 2 z2 is unimodular. and 1 2 − z1z2

 =  −2 + 

2

(x − x0 )2 + ( y − y0 )2 = 4r 2 can be written as,

and

PLAN If z is unimodular, then| z| = 1. Also, use property of modulus i.e. z z =| z|2

⇒

|z|2 = z ⋅ z |z1 − z2|2 = (z1 − z2) (z1 − z2)

and

x2 + 1 = 9 − 6x + x2 4 6x = 8 ⇒ x = 3 4 z= +i 3 5 16 25 |z | = +1= ⇒ |z | = 9 9 3

⇒

PLAN Intersection of circles, the basic concept is to solve the equations simultaneously and using properties of modulus of complex numbers.

Formula used

2

x2 + 1 = 3 − x

Now,

6.

8.

z = x + iy

Let

− b ± b 2 − 4ac 2a

For roots to be real b 2 − 4ac ≥ 0.

Description of Situation z = x + iy is non-zero. ⇒ y ≠0

As imaginary part of

Method I Let z = x + iy ∴

a = (x + iy)2 + (x + iy) + 1

⇒

(x2 − y2 + x + 1 − a ) + i (2xy + y) = 0

⇒

(x2 − y2 + x + 1 − a ) + iy (2x + 1) = 0,

…(i)

12 Complex Numbers It is purely real, if y (2x + 1) = 0

On squaring both sides, we get 1 + |w|2 − 2|w| Re (w) = 1 + |w|2 + 2|w| Re (w)

but imaginary part of z, i.e. y is non-zero.

[using|z1 ± z2|2 = |z1|2 + |z2|2 ± 2|z1||z2| Re (z1z2)]

⇒

2x + 1 = 0 or x = − 1 / 2 1 1 From Eq. (i), − y2 − + 1 − a = 0 4 2 3 2 3 a =− y + ⇒ a< ⇒ 4 4 Method II

4|w|Re|w| = 0

⇒

Re (w) = 0

14. We know, |z1 − z2| = |z1 − (z2 − 3 − 4i ) − (3 + 4i )| ≥ |z1| − |z2 − 3 − 4i | − |3 + 4i|

Here, z 2 + z + (1 − a ) = 0

∴

z=

− 1 ± 1 − 4 (1 − a ) 2 ×1

⇒

z=

− 1 ± 4a − 3 2

For z do not have real roots, 4 a − 3 < 0 ⇒

≥ 12 − 5 − 5 ∴

Alternate Solution

a
|z − z2| represents the region on right side of perpendicular bisector of z1 and z2. ∴ |z − 2| > |z − 4| ⇒ Re (z ) > 3 and Im (z ) ∈ R Y

X′

O

(2, 0) (3, 0) (4, 0)

5

 3 i  3 i +  + −  2 2  2  2

X

⇒

1 3 − ≤1 2 4 2

⇒

−

1 1 7 ≤ z+ ≤ 4 2 4

 −1 + i 3  3+i = −i   = − iω 2 2    −1 − i 3  3−i =i  = iω 2 2 2   z = (− iω )5 + (iω 2)5 = − iω 2 + iω = i(ω − ω 2) = i (i 3 ) = − 3 Re(z ) < 0 and lm (z ) = 0

0≤ z+

1 7 ≤ 2 4

⇒

0≤ z+

1 7 ≤ 2 2

Q

z+

1 7 ≤ 2 2

5

 −1 + i 3 −1 − i 3  and ω 2 =  Q ω = 2 2  

∴

−1 ≤ z +

2

21. Given, z = 

and

1 3 − ≤1 2 4 2

⇒

⇒

Y′

Now,

2

1 3  + =1  2 4

Q ∴

|z1 + z2| ≥ ||z1| − |z2|| 1 1 z + ≥ |z| − 2 2 1 1 7 ≤ z+ ≤ 2 2 2

⇒

|z| −

⇒

7 1 7 ≤ |z| − ≤ 2 2 2 1− 7 7+1 ≤ |z| ≤ 2 2 1+ 7 |z| ≤ 2 |z| ≤ 2

⇒ ⇒ ∴

−

{Q|z| ≥ 0} … (i)

14 Complex Numbers  z + 

Q

2

1≤ z+

2

⇒

|w2| = b2 + d 2 = a 2 + b2 = 1

1 3 1 3 + ≥1 + ⇒ z+ 2 4 2 4

1 1 z+ ≥ 2 4

|w1| = a 2 + c2 = a 2 + b2 = 1

Now,

2

2

⇒

Also, given w1 = a + ic and w2 = b + id

2

1 3 1 3 +  + ≤ z+ 2 4 2 4

⇒

1 1 z+ ≥ 2 2

… (ii)

and Re(w1 w2) = ab + cd = (bλ )b + c(− λc) [from Eq. (i)] = λ (b2 − c2) = 0

27. min|1 − 3 i − z|= perpendicular distance of point (1, − 3) Z ∈S

From Eqs. (i) and (ii), we get 1 1 7 ≤ z+ ≤ 2 2 2

from the line

3x + y = 0 ⇒

28. Since, S = S1 ∩ S 2 ∩ S3

24. We have, sz + tz + r = 0 On taking conjugate sz + tz + r = 0 On solving Eqs. (i) and (ii), we get rt − rs z= 2 |s| − |t|2

Y

…(i) …(ii)

150°

X′

For unique solutions of z |s|2 − |t|2 ≠ 0 ⇒ |s| ≠ |t| It is true (b) If|s| = |t|, then rt − rs may or may not be zero. So, z may have no solutions. ∴ L may be an empty set. It is false. (c) If elements of set L represents line, then this line and given circle intersect at maximum two point. Hence, it is true. (d) In this case locus of z is a line, so L has infinite elements. Hence, it is true.

O

25. Given,|z1| = |z2|

y = –3√x

∴

S=

1 2 1 5π 20π r θ = × 42 × = 2 2 3 6

|w − (2 + i )| < 3 ⇒ |w| − |2 + i| < 3

29. Since, ⇒

−3 + 5 < |w| < 3 + 5

⇒

−3 − 5 < − |w| < 3 − 5

…(i)

|z − (2 + i )| = 3

Also,

−3 + 5 ≤ |z| ≤ 3 + 5

⇒ ∴

…(ii)

−3 < |z| − |w| + 3 < 9 = (x + 1)2 + ( y − 1)2 + (x − 5)2 + ( y − 1)2 = 2(x2 + y2 − 4x − 2 y) + 28 = 2(4) + 28 = 36

[Q|z1|2 = |z2|2 ]

[Q x2 + y2 − 4x − 2 y = 4]

31. Let z = x + iy Set A corresponds to the region y ≥ 1

…(i)

Set B consists of points lying on the circle, centre at (2, 1) and radius 3. i.e. …(ii) x2 + y 2 − 4 x − 2 y = 4

As, we know z − z = 2i Im (z ) ∴

∴

30. |z + 1 − i| + |z − 5 − i|2

|z |2 + (z2 z1 − z1 z2) − |z2|2 = 1 |z1 − z2|2 z2z1 − z1z2 |z1 − z2|2

Y′

Clearly, the shaded region represents the area of sector

2

z1 + z2 z1 − z2 z1z1 − z1z2 + z2z1 − z2 z2 = × z1 − z2 z1 − z2 |z1 − z2|2

=

X (4, 0)

(a)

Now,

| 3 − 3| 3 − 3 = 2 3+1

z2z1 − z1z2 = 2i Im (z2z1 ) z1 + z2 2i Im (z2 z1 ) = z1 − z2 |z1 − z2|2

Set C consists of points lying on the x + y = 2

…(iii)

Y

which is purely imaginary or zero.

26. Since, z1 = a + ib and

P

z2 = c + id

⇒ |z1|2 = a 2 + b2 = 1 and |z2|2 = c2 + d 2 = 1

(0,√2)

…(i)

[Q|z1|=|z2| = 1] Also, Re (z1z2) = 0 ⇒ ac + bd = 0 a d ⇒ =− =λ b c From Eqs. (i) and (ii), b2λ2 + b2 = c2 + λ2c2 ⇒ b2 = c2 and a 2 = d 2

X′

(√2,0)

(2,1)

y=1

X

[say]…(ii) Y′

Clearly, there is only one point of intersection of the line x + y = 2 and circle x2 + y2 − 4x − 2 y = 4.

Complex Numbers 15 z = x + iy

32. A. Let

y x 2 + y2 = 0

⇒ we get ⇒

y = 0 ⇒ Im (z ) = 0

B. We have 2ae = 8, 2a = 10 ⇒ 10e = 8 4 16  e = ⇒ b2 = 25 1 −  = 9  5 25

⇒

C. Let ∴

z = 2 (cos θ + i sin θ ) − = 2 (cos θ + i sin θ ) − =

⇒

2

⇒

x2 y2 + =1 9 / 4 25 / 4 e= 1−

⇒ Let and Then, and

9 /4 4 = 25 / 4 5 1 cos θ + i sin θ

x = 2 cos θ , y = 0

ω 2 (x + yω + z ω 2) ω 2 (xω + yω 2 + z ) ω 2 (x + yω + zω 2) = ω2 x + yω + zω 2

34. |az1 − bz2| + |bz1 + az2|

2

= [a 2|z1|2 + b2|z2|2 − 2ab Re (z1z2)] + [b2|z1|2 + a 2|z2|2 + 2ab Re (z1z2)] = (a 2 + b2) (|z1|2 + |z2|2 )

35.

... (i)

x     2 x 2 x  tan + 1 1 − tan  + 1 + tan  = 0    2 2  2

xα + yβ + z γ x( p)1/3 + y( p)1/3 ω + z ( p)1/3 ω 2 33. = xβ + yγ + zα x( p)1/3 ω 2 + y( p)1/3 ω3 + z ( p)1/3 ω

2

x = 2 nπ

x x  x  2x 2 x sin + cos   cos − sin  + cos = 0     2 2 2 2 2

x On dividing by cos3 , we get 2

= 2 cos θ

=

x =0 2

1 (cos θ − i sin θ ) 2

z = x + iy = cos θ + i sin θ +

⇒

sin

or

w = cos θ + i sin θ

Then,

x  x x  x 2x 2 x sin + cos   cos − sin  + cos  = 0 2  2 2  2 2 2

1 2 (cos θ + i sin θ )

2

⇒

D. Let

⇒ sin

3 5 cos θ + i sin θ 2 2

 2x  2 y   +   =1  3 5

∴

2 sin

⇒

z = x + iy 3 5 x = cos θ and y = sin θ 2 2

Let

x x x x x sin + cos  cos x + 2 sin cos = 0 2 2 2 2 2

⇒

∴

x2 y2 + =1 25 9 w = 2 (cos θ + i sin θ )

∴

Since, it is real, so imaginary part will be zero. x x x −2 sin sin + cos  − tan x = 0 ∴ 2 2 2

x x  sin + cos  − i tan x  2 2 ∈R x 1 + 2 i sin 2 x x x   sin + cos − i tan x 1 − 2i sin     2 2 2 = x 1 + 4 sin 2 2

tan3

x x − tan − 2 = 0 2 2 x tan = t 2 f (t ) = t3 − t − 2 f (1) = − 2 < 0 f (2) = 4 > 0

Thus, f (t ) changes sign from negative to positive in the interval (1, 2). ∴ Let t = k be the root for which f (k) = 0 and k ∈(1, 2) x ∴ t = k or tan = k = tan α 2 ⇒ x/2 = nπ + α x = 2nπ + 2α , α = tan −1 k, where k ∈ (1, 2) ⇒  or x = 2nπ 

36. Since, z1 , z2, z3 are in AP. ⇒ 2z2 = z1 + z3 i.e. points are collinear, thus do not lie on circle. Hence, it is a false statement.

37. Since, z1 , z2, z3 are vertices of equilateral triangle and |z1| = |z2| = |z3| ⇒ z1 , z2, z3 lie on a circle with centre at origin. ⇒ Circumcentre = Centroid z + z2 + z3 ⇒ 0= 1 3 ∴ z1 + z2 + z3 = 0

38. Let z = x + iy ⇒ 1 ∩ z gives 1 ∩ x + iy or Given, ⇒

1 ≤ x and 0 ≤ y 1−z 1 − x − iy ∩0 ⇒ ∩0 1+ z 1 + x + iy (1 − x − iy) (1 + x − iy) ∩ 0 + 0i (1 + x + iy) (1 + x − iy)

…(i)

16 Complex Numbers ⇒

1 − x2 − y2 2iy − ∩ 0 + 0i 2 2 (1 + x) + y (1 + x)2 + y2

⇒

⇒

x2 + y2 ≥ 1 and −2 y ≤ 0

⇒

⇒

or x2 + y2 ≥ 1 and y ≥ 0 which is true by Eq. (i).

39. As we know,|z| = z ⋅ z |z − α|2 = k2 |z − β|2

⇒

1 − |z1|2 − |z2|2+ | z1|2|z2|2 < 0

2

2

 1 − z1z2 < 1  z1 − z2 

⇒

⇒ |z| − αz − αz + |α| = k (|z| − βz − β z + |β| ) 2

2

2

Hence proved.

42. Given,|z|2 w − |w|2 z = z − w

⇒ |z|2 (1 − k2) − (α − k2β )z − (α − β k2) z + (|α|2 − k2|β|2 ) = 0 |α|2 − k2|β|2 (α − k2β ) (α − β k2) z+ z− = 0 …(i) 2 2 (1 − k2) (1 − k ) (1 − k )

⇒

zz w − ww z = z − w

[Q |z|2 = zz ] …(i)

Taking modulus of both sides, we get |zw||z − w| = |z − w|

On comparing with equation of circle,

⇒

|zw||z − w| = |z − w|

|z| + az + az + b = 0 whose centre is (− a ) and radius = |a|2 − b

⇒

| zw|| z − w | = |z − w |

2

∴ Centre for Eq. (i) =

α −kβ 1 − k2 2

 α − k β  α − k β  αα − k ββ and radius =   −   1 − k2  1 − k2   1 − k2  2

2

2

∴ ⇒

1 3 |a1z + a 2z 2 + a3 z3 + K + a nz n| = 1 |a1z| + |a 2z 2| + |a3 z3| + K + |a nz n| ≥ 1

⇒

2{(|z| + |z| + |z| + K + |z|n } > 1

|z|
1 1 − |z|

[using sum of n terms of GP]

2|z| − 2|z|n + 1 > 1 − |z| 3|z| > 1 + 2|z|n + 1 1 2 |z| > + |z|n + 1 3 3 1 |z| > , which contradicts 3

|zw| − 1 = 0

⇒

|z − w| = 0

or

|zw| = 1

⇒

z − w=0

or

|z w|= 1

⇒

z=w

or

|zw| = 1

   z1  |z1| using z = |z |   2 2   …(ii)

(1 − z1z2)(1 − z1z2) < (z1 − z2)(z1 − z2) [using|z|2 = zz ]

1 iφ e r

On putting these values in Eq. (i), we get 1 1  1 r 2 ei φ  − 2 (reiθ ) = reiθ − eiφ r  r r 1 iθ 1 iφ iφ iθ re − e = re − e ⇒ r r 1  iφ  1  iθ  ⇒ = + r + e r  e      r r

NOTE

…(i)

[say]

z = reiθ and w =

Let

⇒

r =1

|1 − z1z2| < |z1 − z2| ⇒ On squaring both sides, we get,

or

…(ii)

n

 1 − z1z2 < 1  z1 − z2 

|z − w| = 0

Therefore,

|z| < 1 / 3 and ∑ a rz r = 1

Then, to prove

|z − w|(|zw| − 1) = 0

⇒

∴ There exists no complex number z such that

41. Given,|z1| < 1 and |z2| > 1

⇒ ⇒

Then,|zw| = 1 or|z||w| = 1 1 |z| = =r ⇒ |w|

40. Given, a1z + a 2z 2 + K + a nz n = 1

2

[∴ |z| = |z| ]

Now, suppose z ≠ w

k(α − β )   = 2 1−k 

and

…(iii)

∴ Eq. (iii) is true whenever Eq. (ii) is true.

(z − α )(z − α ) = k2(z − β ) (z − β )

⇒ |z|2 −

1 + |z1|2|z2|2 1 ∴ (1 − |z1|2 ) > 0 and (1 − |z2|2 ) < 0

2

Given,

1 − z1z2 − z1z2 + z1z1z2z2 < z1z1 − z1z2 − z2z1 + z2z2

eiφ = eiθ ⇒ φ = θ 1 z = reiθ and w = eiθ r 1 zw = reiθ . e−iθ = 1 r

‘If and only if ’ means we have to prove the relation in both directions.

Conversely Assuming that z = w or z w = 1 If z = w, then LHS = zz w − w wz = |z|2⋅z − |w|2⋅z = |z|2⋅z − |z|2⋅z = 0 and RHS = z − w = 0 If zw = 1, then zw = 1 and LHS = zz w − ww z = z ⋅ 1 − w ⋅ 1 = z − w = z − w = 0 = RHS Hence proved.

Complex Numbers 17 Alternate Solution We have, |z|2 w − |w|2 z = z − w 2 ⇔ |z| w − |w|2 z − z + w = 0 ⇔

45. Here, z1Rz2 ⇔

(i) Reflexive z1Rz1 ⇔

(|z|2 + 1)w − (|w|2 + 1)z = 0

⇔

(|z|2 + 1)w = (|w|2 + 1)z z |z|2 + 1 = w |w|2 + 1

⇔ z is purely real. ∴ w ⇔ Again, ⇔ ⇔

z z = w w

⇒

zw = zw

(z − w)(zw − 1) = 0

⇔

z = w or zw = 1

(ii) Symmetric z1Rz2 ⇔

…(i)

⇒ ⇒ ⇒ NOTE

[from Eq. (i)]

Here, let z1 = x1 + iy1 , z2 = x2 + iy2 and z3 = x3 + iy3 z − z2 (x − x2) + i ( y1 − y2) is real ⇒ 1 is real ∴ 1 z1 + z2 (x1 + x2) + i ( y1 + y2) ⇒

It is a compound equation, therefore we can generate from it more than one primary equations.

⇒

x + 2xy = 0 and x2 − y2 + y = 0 x(1 + 2 y) = 0 x = 0 or y = − 1 / 2

Similarly,

0 − y2 + y = 0 ⇒ y = 0 or y = 1

3 i − 2 2 44. Given, iz3 + z 2 − z + i = 0 ⇒

z = i, ±

⇒

iz3 − i 2z 2 − z + i = 0

⇒

iz 2(z − i ) − 1(z − i ) = 0

⇒

(iz 2 − 1)(z − i ) = 0

⇒ ⇒ If

z − i = 0 or iz 2 − 1 = 0 1 z = i or z 2 = = − i i z = i, then|z| = |i| = 1

If z 2 = − i, then |z 2| = |− i| = 1 ⇒

|z|2 = 1 ⇒ |z| = 1

2 x2y1 − 2 y2x1 = 0 x1 x2 = y1 y2

⇒

⇒ When x = 0, x2 − y2 + y = 0 ⇒

Therefore, z = 0 + i 0 , 0 + i ; ±

{(x1 − x2) + i ( y1 − y2)}{(x1 + x2) − i ( y1 + y2)} (x1 + x2)2 + ( y1 + y2)2

⇒ ( y1 − y2) (x1 + x2) − (x1 − x2) ( y1 + y2) = 0

and − y = x2 − y2

x2 =

z1Rz2 z1 − z2 is real z1 + z2 z2Rz3 z2 − z3 is real z2 + z3

⇒

(x + iy) = i (x + i y)2 x − iy = i (x2 − y2 + 2i xy) x − iy = − 2xy + i (x2 − y2)

⇒ y(1 − y) = 0 When, y = − 1 / 2 , x2 − y2 + y = 0 1 1 ⇒ x2 − − = 0 ⇒ 4 2 3 ⇒ x=± 2

z1Rz2 ⇒ z2Rz1

and

On equating real and imaginary parts, we get ⇒ ⇒

∴

⇒

z = iz 2

x = − 2xy

− (z2 − z1 ) is real ⇒ z2Rz1 z1 + z2

Therefore, it is symmetric.

z = x + iy.

Given,

[purely real]

z1 − z2 is real z1 + z2

⇒

(iii) Transitive

Therefore, |z|2 w − |w|2 z = z − w if and only if z = w or zw = 1.

43. Let

z1 − z1 =0 z1 + z2

∴ z1Rz1 is reflexive.

|z|2 w − |w|2 z = z − w z ⋅ zw − w ⋅ wz = z − w z (zw − 1) − w (zw − 1) = 0

⇔

z1 − z2 is real z1 + z2

⇒

z2Rz3 x2 x3 = y2 y3

From Eqs. (i) and (ii), we have

3 4

... (i)

... (ii)

x1 x3 = ⇒ z1Rz3 y1 y3

Thus, z1Rz2 and z2Rz3 ⇒ z1Rz3 .

[transitive]

Hence, R is an equivalence relation.

3 i − 2 2

46. [Q z ≠ 0]

[Q i 2 = − 1]

(1 + i ) x − 2i (2 − 3i ) y + i + =i 3+ i 3−i ⇒ (1 + i ) (3 − i ) x − 2i (3 − i ) + (3 + i ) (2 − 3i ) y + i (3 + i ) = 10i ⇒

4x + 2ix − 6i − 2 + 9 y − 7iy + 3i − 1 = 10i

⇒

4x + 9 y − 3 = 0 and 2x − 7 y − 3 = 10

⇒

x = 3 and y = − 1

1 1 47. Now, = θ (1 − cos θ ) + 2i sin θ 2 sin 2 + 4i sin θ cos θ 2 2 2 θ θ sin − 2 i cos 1 2 2 × = θ θ θ  θ θ 2 sin sin + 2 i cos  sin − 2 i cos  2 2 2  2 2

18 Complex Numbers θ θ − 2 i cos 2 2 = θ  2θ θ + 4 cos 2  2 sin sin  2 2 2 θ θ sin − 2 i cos 2 2 = θ θ 2 sin 1 + 3 cos 2  2 2

y1 + y2 = 2 3 ⇒ Im(z1 + z2) = 2 3 ⇒ Hence, option (d) is correct. |z | 4 2. (*) Given, 3|z1| = 4|z2|⇒ 1 = [Q z2 ≠ 0 ⇒|z2| ≠ 0] |z2| 3

sin

⇒ A + iB =

1 θ  2 1 + 3 cos 2   2

48. Since, (x + iy)2 = ⇒

|x + iy|2 =

⇒

(x2 + y2) =

⇒

(x2 + y2)2 =

−i

cot

∴

θ 2

1 + 3 cos 2

[Q z =|z|(cos θ + i sin θ) = |z| eiθ ] z1 4 iθ z 3 ⇒ = e and 2 = e−iθ z2 3 z1 4 3 z1 2 z2 1 −iθ = e = 2eiθ and ⇒ 2 z2 3 z1 2

θ 2

a + ib c + id  z1 |z1| Q z = |z | 2 2 

|a + ib| |c + id| a 2 + b2 c2 + d 2 a + b2 c2 + d 2 2

On adding these two, we get 1 3 z1 2 z2 z= = 2eiθ + e−iθ + 2 2 z2 3 z1 1 1 = 2 cos θ + 2i sin θ + cos θ − i sin θ 2 2 [Q e± iθ = (cos θ ± i sin θ)] 5 3 = cos θ + i sin θ 2 2 2

Hence proved.

49. Given,|z − 3 − 2 i| ≤ 2

…(i)

To find minimum of|2z − 6 + 5 i| 5 or 2 z − 3 + i , using triangle inequality 2

= (z − 3 − 2 i ) +

9 i 2

≥ |z − 3 − 2 i| −

9 9 5 ≥ 2− ≥ 2 2 2

Note that z is neither purely imaginary and nor purely real. ‘*’ None of the options is correct. z = e–iθ ⇒ z =

∴

   1 + z  1 + z arg   = arg (z ) = θ  = arg  1 + z 1 + 1  z

4. Since, arg (z ) < 0 ⇒

arg (z ) = − θ Y

(–z)

π−θ

r X′

X

–θ

O

Topic 3 Argument of a Complex Number

r

1. Let the complex numbers

From Eqs. (i) and (ii), we get y12 − y22 = 2(x1 − x2) y1 − y2 1 ⇒ ( y1 + y2) = 2 ( y1 + y2) = 2 ⇒ x1 − x2 3

1 z

∴

5 5 z − 3 + i ≥ or|2z − 6 + 5 i| ≥ 5 2 2

z1 = x1 + iy1 and z2 = x2 + iy2 Now, it is given that Re (z1 ) = |z1 − 1| ⇒ x12 = (x1 − 1)2 + y12 ⇒ y12 + 1 = 2x1 and Re(z2) =|z2 − 1| ⇒ x22 = (x2 − 1)2 + y22 ⇒ y22 + 1 = 2x2 π y − y2 1 and arg (z1 − z2) = ⇒ 1 = 6 x1 − x2 3

2

34 17  3  5 = |z| =   +   =  2  2 4 2

⇒

3. Given,|z| = 1 , arg z = θ∴z = eiθ

i.e. ||z1| − |z2|| ≤ |z1 + z2| 5 5 ∴ z −3 + i = z −3 −2i + 2i + i 2 2

⇒

z1 z1 iθ z z e and 2 = 2 e−iθ = z2 z2 z1 z1

Y′

⇒

(z)

z = r cos (−θ ) + i sin (−θ ) = r (cos θ − i sin θ )

…(i) …(ii) …(iii)

and

− z = − r [cos θ − i sin θ ] = r [cos (π − θ ) + i sin (π − θ )]

∴ arg (− z ) = π − θ Thus, arg (− z ) − arg (z ) = π − θ − (− θ ) = π Alternate Solution

 − z arg (− z ) − arg z = arg   = arg (− 1) = π  z   z  and also arg z − arg (− z ) = arg   = arg (− 1) = π  − z Reason

Complex Numbers 19 5. Given,

|z + iw| = |z − iw |= 2

⇒

|z − (− iw)| = |z − (iw )| = 2

⇒

|z − (− iw)| = |z − (− iw )|

Consider

∴ z lies on the perpendicular bisector of the line joining − iw and − iw. Since, − iw is the mirror image of − iw in the X-axis, the locus of z is the X-axis. Let

z = x + iy and y = 0.

Now,

|z | ≤ 1 ⇒ x2 + 02 ≤ 1

⇒

−1 ≤ x ≤ 1.

|z + iw| ≤ 2

⇒

arg z − arg i w = 0 z arg =0 iw

|z − i w| = 2, then

This function is discontinuous at t = 0. z is purely w

imaginary Now, given relation |z + iw| = |z − iw| = 2 Put w = i, we get

⇒

Put w = − i , we get |z − i 2| = |z − i 2| = 2 ⇒ |z + 1| = 2 ⇒ z = 1

z  arg  1  − arg (z1 ) + arg (z2)  z2 

∴

|z − 1| = 2 z = −1

(c) We have,

z  Now,arg  1  = arg (z1 ) − arg (z2) + 2nπ  z2 

|z + i 2| = |z + i 2| = 2 ⇒

z = − 1 − i and arg(z) = θ

(b) We have, f (t ) = arg (−1 + it )  π − tan −1 t , t ≥ 0 arg (−1 + it ) =  −1 − (π + tan t ), t < 0

z is purely real. ⇒ iw z is purely imaginary. ⇒ w Similarly, when

[from Eq. (i)]

Now,

≤1 +1 =2 |z + i w| = 2 holds when

1 1 = 0 0 

 im (z )    −1  =1 tan θ =  =  Re(z )   −1  π θ= ⇒ 4 Since, x < 0, y < 0 π 3π  arg (z ) = −  π −  = − ∴  4 4

|z + i w| ≤ |z| + |iw|= |z| + |w|

⇒

u v 0

9. (a) Let

Alternate Solution

∴

Applying R3 → R3 – {(1 − r ) R1 + rR2} a u 1 = b v 1 c − (1 − r ) a − rb w − (1 − r ) u − rv 1 − (1 − r ) − r a = b 0

∴ z may take values given in option (c).

 a u 1 b v 1  c w 1 

[Q|z| ≤ 1]

[Q|z| ≤ 1]

∴ z = 1 or − 1 is the correct option. w = re iθ , then w = re–iθ

∴

z = rei ( π − θ ) = re iπ ⋅ e−iθ = − re−iθ = − w

= arg (z1 ) − arg (z2) + 2nπ − arg (z1 ) + arg (z2) = 2nπ So, given expression is multiple of 2π.  (z − z1 ) (z2 − z3 ) (d) We have, arg   =π  (z − z3 ) (z2 − z1 )  z − z1   z2 − z3  ⇒   is purely real  z − z3   z2 − z1 

6. Since,|z|=|w|and arg (z ) = π − arg (w) Let

z  arg  1  − arg (z1 ) + arg (z2)  z2 

Thus, the points A (z1 ), B(z2), C (z3 ) and D (z ) taken in order would be concyclic if purely real. Hence, it is a circle.

7. Given,|z1 + z2| = |z1| + |z2|

C(z3)

On squaring both sides, we get |z1|2 + |z2|2 + 2|z1||z2| cos (arg z1 − arg z2) = |z1|2 + |z2|2 + 2|z1||z2| ⇒ ⇒ ⇒

D(z)

2|z1||z2|cos (arg z1 − arg z2) = 2|z1||z2| cos (arg z1 − arg z2) = 1 arg (z1 ) − arg (z2) = 0

8. Since a , b, c and u , v, w are the vertices of two triangles. Also,

c = (1 − r ) a + rb

and

w = (1 − r ) u + rv

…(i)

B(z2)

A(z1)

∴(a), (b), (d) are false statement.

20 Complex Numbers (1 − t ) z1 + t z2 (1 − t ) + t P z t : (1 - t)

B z2

≤ |r1 − r2|2 + |θ1 − θ 2|2 ⇒

Clearly, z divides z1 and z2 in the ratio of t : (1 − t ), 0 < t 0. 4 − z0 − z0 4 − x − iy − x + iy So, = z0 − z0 + 2i x + iy − x + iy + 2i

5. Since, |PQ | = |PS | = |PR| = 2 ∴ Shaded part represents the external part of circle having centre (−1, 0) and radius 2. As we know equation of circle having centre z0 and radius r, is|z − z0| = r ∴ |z − (−1 + 0i )| > 2 ⇒ |z + 1| > 2

2(2 − x)  2 − x = = −i    y + 1 2i ( y + 1) Q

2−x is a positive real number, so y+1 4 − z0 − z0 is purely negative imaginary number. z0 − z0 + 2 i

5

5

∴

π i( π / 6 )  =e 6

i

5π

−i

3 X′

π/4

X

O Y′

π ∴ I (z ) = 0 and R(z ) = −2 cos = − 3 < 0 6 5π π π   Q cos 6 = cos  π − 6  = − cos 6   

7.

z1 − z3 1 − i 3 (1 − i 3 )(1 + i 3 ) = = z2 − z3 2 2 (1 + i 3 ) 1 − i 23 2 (1 + i 3 ) 4 2 z3 = = 2 (1 + i 3 ) (1 + i 3 ) z2 − z3 1 + i 3 π π = = cos + i sin z1 − z3 2 3 3   − z z − π z z   3 3 2 = 1 and arg  2  =   − z z − 3 z z 1 3 1 3

z2

=

⇒ ⇒

Imaginary axis

…(ii)

represented by e i( α − θ ) Now, rotation about a line with angle α is given by e θ → e (α − θ ). Therefore, Q is obtained from P by reflection in the line making an angle α /2.

[Q eiθ = cos θ + i sin θ]

5π = 2 cos 6

− π / 4 ≤ arg (z + 1)

6. In the Argand plane, P is represented by e i0 and Q is

5π

z = (eiπ / 6 )5 + (e−iπ / 6 )5 = e 6 + e 6 5π 5π   5π 5π   =  cos + i sin  +  cos − i sin     6 6 6 6

3.

3e iπ/4 A

From Eqs. (i) and (ii), |arg (z + 1)|≤ π / 4

3 i  π  −π  and − = cos   + i sin  −  = e−iπ / 6  6   2 2 6 So,

4

and argument of z + 1 in anticlockwise direction is −π /4.

2. Given, z = 

Q Euler’s form of 3 i  π + =  cos + i sin 2 2  6

Y

Also, argument of z + 1 with respect to positive direction of X-axis is π/4. π …(i) arg (z + 1) ≤ ∴ 4

 4 − z0 − z0  π ⇒ arg   =−  z0 − z0 + 2 i  2  3 i  3 i +  + −  2 2    2 2

P

π/3

Hence, the triangle is an equilateral. z2

z'2 (7,6)

(1

,2

1

)

90° z 0

5

z2′ = (6 + 2 cos 45° , 5 + By rotation about (0, 0),  iπ  z2 = ei π/ 2 ⇒ z2 = z2′  e 2    z2′   π  = (7 + 6i )  cos + i sin  2

3 1 (6,2) Real axis

2 sin 45° ) = (7, 6) = 7 + 6i

Alternate Solution z1 − z3 1 − i 3 = ∴ z2 − z3 2 z2 − z3 2 1 +i 3 π π ⇒ = cos + i sin = = z1 − z3 1 − i 3 2 3 3  z2 − z3  π z2 − z3 and also arg  =1 ⇒  =  z1 − z3  3 z1 − z3 Therefore, triangle is equilateral.

8. Here, x + iy = π  = (7 + 6i ) (i ) = − 6 + 7i 2

4. Let OA = 3, so that the complex number associated with A is 3e iπ / 4 . If z is the complex number associated with P, then z − 3eiπ / 4 4 − iπ / 2 4i = e =− iπ / 4 3 3 0 − 3e ⇒ 3z − 9eiπ / 4 = 12 ieiπ / 4 ⇒ z = (3 + 4 i ) eiπ / 4

∴ Let ∴ ⇒

x + iy =

a − ibt 1 × a + ibt a − ibt a − ibt a 2 + b2t 2

a ≠ 0, b ≠ 0 − bt a and y = 2 x= 2 a + b2t 2 a + b2t 2 ay y − bt = ⇒t = bx x a

z1

22 Complex Numbers On putting x =

a , we get a 2 + b2t 2

So,

 a 2 y2  x  a 2 + b2 ⋅ 2 2  = a ⇒ bx   x2 + y2 −

or

Since, |z2| = |z3| = 2 2

2

x =0 a

z1

… (i) z2

1 1  2 x −  + y =   2a 4a 2

P (2, 0) O

P (–1, 0)

∴Option (a) is correct. For a ≠ 0 and b = 0, 1 1 x + iy = ⇒ x = , y = 0 a a ⇒ z lies on X-axis. ∴ Option (c) is correct. 1 1 For a = 0 and b ≠ 0, x + iy = ⇒ x = 0, y = − ibt bt ⇒ z lies on Y-axis. ∴ Option (d) is correct.

9.

[given]

Y-axis

a (x + y ) = ax 2

2

or

z1 = 2 (cos π / 3 + sin π / 3)

X-axis

z3

PLAN It is the simple representation of points on Argand plane and to find the angle between the points.

Now, the triangle z1 , z2 and z3 being an equilateral and the sides z1z2 and z1z3 make an angle 2π / 3 at the centre. π 2π Therefore, ∠POz2 = + =π 3 3 π 2π 2π 5π and ∠POz3 = + + = 3 3 3 3 Therefore, z2 = 2 (cos π + i sin π ) = 2 (− 1 + 0) = − 2  1 5π 5π  3  =2  − i and z3 = 2  cos + i sin  =1 − i 3 3 3  2 2 

n

nπ nπ π π  Here, P = W n =  cos + i sin  = cos + i sin  6 6 6 6 1  H 1 = z ∈ C : Re(z ) >  2  ∴ P ∩ H 1 represents those points for which cos

nπ is + ve. 6

Hence, it belongs to I or IV quadrant. π π 11π 11π + i sin ⇒ z1 = P ∩ H 1 = cos + i sin or cos 6 6 6 6 ∴

z1 =

3 i 3 i + or − 2 2 2 2

…(i)

Similarly, z2 = P ∩ H 2 i.e. those points for which nπ cos

5 3

4 5 1 5 (c) λ ∈  ,  (d) λ ∈  ,   3 3  3 3

28. If one root is square of the other root of the equation x2 + px + q = 0, then the relation between p and q is

(a) p3 (b) p3 (c) p3 (d) p3

− q(3 p − 1) + q2 = 0 − q(3 p + 1) + q2 = 0 + q(3 p − 1) + q2 = 0 + q(3 p + 1) + q2 = 0

(2004, 1M)

is

(2002, 1M)

(a) (− ∞ , − 2) ∪ (2, ∞ ) (c) (− ∞ , − 1) ∪ (1, ∞ )

(b) (− ∞ , − 2 ) ∪ ( 2 , ∞ ) (d) ( 2 , ∞ )

30. The number of solutions of log 4 (x − 1) = log 2(x − 3) is (a) 3 (c) 2

(b) 1 (d) 0

(2001, 2M)

31. For the equation 3x2 + px + 3 = 0, p > 0, if one of the root is square of the other, then p is equal to (b) 1

(c) 3

(2000, 1M)

(d) 2 /3

x2 + bx + c = 0, where c < 0 < b, then (a) 0 < α < β (c) α < β < 0

a10 − 2a 8 is 2a 9 (d) 4

(2011)

(2000, 1M)

(b) α < 0 < β < |α| (d) α < 0 < |α|< β

33. The equation x + 1 − x − 1 = 4x − 1 has

(1997C, 2M)

(a) no solution (b) one solution (c) two solutions (d) more than two solutions 3

24. Let α and β be the roots of x2 − 6x − 2 = 0, with α > β. If

(a) 1

2 (q − p ) (2 p − q) 9 2 (d) (2 p − q) (2q − p ) 9

(b)

32. If α and β (α < β) are the roots of the equation

1 1 + = 4, then the α β

value of|α − β|is (a)

2 ( p − q) (2q − p ) 9 2 (c) (q − 2 p ) (2q − p ) 9

(a) 1 /3

23. Let α and β be the roots of equation px2 + qx + r = 0, p ≠ 0. If p, q and r are in AP and

α , 2 β be the roots of the equation x2 − qx + r = 0 . Then, 2 (2007, 3M) the value of r is

29. The set of all real numbers x for which x2 − |x + 2| + x > 0

21. Let α and β be the roots of equation x2 − 6x − 2 = 0. If

(a) 6

( p3 + q) x2 − ( p3 + 2q) x + ( p3 + q) = 0 ( p3 + q) x2 − ( p3 − 2q) x + ( p3 + q) = 0 ( p3 − q) x2 − (5 p3 − 2q) x + ( p3 − q) = 0 ( p3 − q) x2 − (5 p3 + 2q) x + ( p3 − q) = 0

(d) 5

π π 20. Let − < θ < − . Suppose α 1 and β1 are the roots of the 6 12 equation x2 − 2x secθ + 1 = 0 , and α 2 and β 2 are the roots of the equation x2 + 2x tan θ − 1 = 0. If α 1 > β1 and (2016 Adv.) α 2 > β 2, then α 1 + β 2 equals (a) 2(secθ − tan θ) (c) −2tanθ

(a) (b) (c) (d)

(a)

18. For a positive integer n, if the quadratic equation,

(a) 12

p3 ≠ − q. If α and β are non-zero complex numbers satisfying α + β = − p and α 3 + β3 = q, then a quadratic α β equation having and as its roots is (2010) β α

26. Let α, β be the roots of the equation x2 − px + r = 0 and

17. If α , β ∈ C are the distinct roots of the equation (a) −1

25. Let p and q be real numbers such that p ≠ 0, p3 ≠ q and

34. The equation x 4

(log 2 x )2 + log 2 x −

5 4

= 2 has

(a) atleast one real solution (b) exactly three real solutions (c) exactly one irrational solution (d) complex roots

(1989; 2M)

Theory of Equations 31 35. If α and β are the roots of x2 + px + q = 0 and α 4 , β 4are the roots of x − rx + s = 0, then x2 − 4qx + 2q2 − r = 0 has always 2

the

equation (1989, 2M)

(a) two real roots (b) two positive roots (c) two negative roots (d) one positive and one negative root

2 2 36. The equation x − has =1 − x−1 x−1 (a) no root (c) two equal roots

(1984, 2M)

(b) one root (d) infinitely many roots

37. For real x, the function

(x − a )(x − b) will assume all real (x − c)

values provided (a) a > b > c

(1984, 3M)

(b) a < b < c

(c) a > c < b

(d) a ≤ c ≤ b

38. The number of real solutions of the equation |x|2−3|x| + 2 = 0 is (a) 4

(1982, 1M)

(b) 1

(c) 3

(d) 2

39. Both the roots of the equation (x − b) (x − c) + (x − a ) (x − c) + (x − a ) (x − b) = 0 are always

(1980, 1M)

(a) positive (c) real

(b) negative (d) None of these

equation ax + bx + c = 0

p and q are real, then ( p, q) = (…,…).

(1979, 1M)

(a) are real and negative (b) have negative real parts (c) have positive real parts (d) None of the above

45. The coefficient of x in the polynomial (x − 1)(x − 2)... (x − 100) is....

(1982, 2M)

True/False 46. If P (x) = ax2 + bx + c and Q (x) = − ax2 + bx + c, where ac ≠ 0, then P (x) Q (x) has atleast two real roots. (1985, 1M)

47. The equation 2x + 3x + 1 = 0 has an irrational root. 2

(1983, 1M)

Analytical & Descriptive Questions 48. If x2 − 10ax − 11b = 0 have roots c and d. x2 − 10cx − 11d = 0 have roots a and b, then find a + b + c + d.

(2006, 6M)

49. If α , β are the roots of ax + bx + c = 0, (a ≠ 0) and 2

α + δ , β + δ are the roots of Ax2 + Bx + C = 0, ( A ≠ 0) for some constant δ, then prove that b2 − 4ac B2 − 4 AC (2000, 4M) = a2 A2 numbers. prove that if f (x) is an integer whenever x is an integer, then the numbers 2 A , A + B and C are all integers. Conversely, prove that if the numbers 2 A , A + B and C are all integers, then f (x) is an integer (1998, 3M) whenever x is an integer.

51. Find the set of all solutions of the equation

Assertion and Reason For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows : (a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

41. Let a , b, c, p, q be the real numbers. Suppose α , β are the roots of the equation x + 2 px + q = 0. 1 and α , are the roots of the equation ax2 + 2bx + c = 0, β 2

2|y| − |2y − 1 − 1| = 2y − 1 + 1

2x 1 > x +1 2x + 5x + 2

(1987, 3M)

53. Solve| x2 + 4x + 3 | + 2x + 5 = 0

(1987, 5M)

2

54. For a ≤ 0, determine all real roots of the equation x2 − 2a|x − a|− 3a 2 = 0

55. Solve for x : (5 + 2 6 )x

Statement I ( p − q) (b − ac) ≥ 0 Statement II b ∉ pa or c ∉ qa .

−3

+ (5 − 2 6 )x

2

(1986, 5M) −3

= 10 (1985, 5M)

56. If one root of the quadratic equation ax + bx + c = 0 is equal to the nth power of the other, then show that 1 n n +1

1

+ (a c) n

n +1

+ b =0

(1983, 2M)

57. If α and β are the roots of x + px + q = 0 and γ , δ are the 2

(2008, 3M)

Fill in the Blanks

roots of x2 + rx + s = 0, then evaluate (α − γ ) (β − γ ) (1979, 2M) (α − δ ) (β − δ ) in terms of p, q, r and s.

58. Solve 2 log x a + log ax a + 3 logb a = 0 ,

42. The sum of all the real roots of the equation |x − 2|2 + |x − 2| − 2 = 0 is…… .

2

2

(ac ) 2

(1997 C, 3M)

52. Find the set of all x for which

where β 2 ∉ { −1, 0, 1}. 2

(1982, 2M)

99

50. Let f (x) = Ax2 + Bx + C where, A , B, C are real

40. Let a > 0, b > 0 and c > 0. Then, both the roots of the 2

44. If 2 + i 3 is a root of the equation x2 + px + q = 0, where

(1997, 2M)

43. If the products of the roots of the equation x2 − 3kx + 2e2log k − 1 = 0 is 7, then the roots are real for (1984, 2M) k =K .

where a > 0, b = a 2 x

59. If α

(1978 , 3M )

and β are the roots of the equation x2 + px + 1 = 0 ; γ , δ are the roots of x2 + qx + 1 = 0, then (1978, 2M) q2 − p2 = (α − γ )(β − γ )(α + δ )(β + δ )

32 Theory of Equations Passage Type Questions

60. a12 =

Let p, q be integers and let α , β be the roots of the equation, x2 − x − 1 = 0 where α ≠ β. For n = 0, 1, 2, …… , let a n = pα n + qβ n. FACT : If a and b are rational numbers and a + b 5 = 0, then (2017 Adv.) a = 0 = b.

61. If a 4 = 28, then p + 2q =

(a) a11 + 2a10 (c) a11 − a10

(b) 2a11 + a10 (d) a11 + a10

(a) 14 (c) 21

(b) 7 (d) 12

Topic 2 Common Roots Objective Questions I (Only one correct option) 1. If α , β and γ are three consecutive terms of a non-constant GP such that the equations αx2 + 2βx + γ = 0 and x2 + x − 1 = 0 have a common root, then, α(β + γ) is equal to (2019 Main, 12 April II)

3. A value of b for which the equations x2 + bx − 1 = 0, x2 + x + b = 0 have one root in common is (a) −

2

(b) −i 3

(c) i 5

(2011)

(d) 2

Fill in the Blanks the quadratic equations x2 + ax + b = 0 and x + bx + a = 0 (a ≠ b) have a common root, then the numerical value of a + b is… . (1986, 2M)

(a) 0 (b) αβ (c) αγ (d) βγ

4. If

2

2. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0,

5. If x − r is a factor of the polynomial f (x) = a n xn,

a , b, c ∈ R have a common root, then a : b : c is (a) 1 : 2 : 3 (c) 1 : 3 : 2

(b) 3 : 2 : 1 (d) 3 : 1 : 2

True/False

(2013 Main)

+ a n − 1 x n−1 + K + a 0 repeated m times (1 < m ≤ n ), then (1983, 1M) r is a root of f ′ (x) = 0 repeated m times.

Topic 3 Transformation of Roots Objective Question I (Only one correct option) 1. Let α , β be the roots of the equation,(x − a )(x − b) = c, c ≠ 0. Then the roots of the equation (x − α ) (x − β ) + c = 0 are

(a) a , c

(b) b , c

(c) a , b (d) a + c, b + c

(1992, 2M)

Analytical & Descriptive Question 2. Let a, b and c be real number s with a ≠ 0 and let α , β be the roots of the equation ax2 + bx + c = 0. Express the roots of a3 x2 + abcx + c3 = 0 in terms of α , β. (2001, 4M)

Topic 4 Graph of Quadratic Expression Objective Questions I (Only one correct option) 1. Let P(4, − 4) and Q(9, 6) be two points on the parabola, y2 = 4x and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of ∆PXQ is maximum. Then, this (2019 Main, 12 Jan I) maximum area (in sq units) is 125 2 625 (c) 4

(a)

75 2 125 (d) 4 (b)

2. Consider the quadratic equation, (c − 5) x2− 2cx + (c − 4) = 0, c ≠ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then, the number of elements in S is (2019 Main, 10 Jan I) (a) 11 (c) 12

(b) 10 (d) 18

3. If

both the roots of the quadratic equation x2 − mx + 4 = 0 are real and distinct and they lie in the interval [1, 5] then m lies in the interval (2019 Main, 9 Jan II)

(b) (−5, − 4) (d) (3, 4)

(a) (4, 5) (c) (5, 6)

4. If a ∈ R and the equation −3 (x − [x])2 + 2 (x − [x]) + a 2 = 0 (where, [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval (2014 Main) (a) (−1, 0) ∪ (0, 1) (c) (−2 , − 1)

(b) (1, 2) (d) (−∞ , − 2) ∪ (2 , ∞ )

5. For all ‘x’, x2 + 2ax + (10 − 3a ) > 0, then the interval in which ‘a’ lies is (a) a < − 5

(b) −5 < a < 2 (c) a > 5

(2004, 1M)

(d) 2 < a < 5

Theory of Equations 33 6. If b > a, then the equation (x − a ) (x − b) − 1 = 0 has (a) both roots in (a , b) (b) both roots in ( − ∞ , a ) (c) both roots in (b, + ∞ ) (d) one root in (−∞ , a ) and the other in (b, ∞ )

(2000, 1M)

9. If x2 + (a − b)x + (1 − a − b) = 0 where a , b ∈ R, then find the values of a for which equation has unequal real (2003, 4M) roots for all values of b.

7. If the roots of the equation x2 − 2ax + a 2 + a − 3 = 0 are real and less than 3, then

Analytical & Descriptive Questions

(1999, 2M)

(a) a < 2 (b) 2 ≤ a ≤ 3 (c) 3 < a ≤ 4 (d) a> 4

10. Let a , b, c be real. If ax2 + bx + c = 0 has two real roots α and β, where α < − 1 and β > 1, then show that c b 1 + +  < 0 (1995, 5M) a a 

11. Find all real values of x which satisfy x2 − 3x + 2 > 0 and x2 − 2x − 4 ≤ 0.

(1983, 2M)

8. Let f (x) be a quadratic expression which is positive for all real values of x. If g (x) = f (x) + f ′ (x) + f ′ ′ (x), then for (1990, 2M) any real x (a) g (x) < 0 (c) g (x) = 0

(b) g (x) > 0 (d) g (x) ≥ 0

Integer & Numerical Answer Type Question 12. The smallest value of k, for which both the roots of the

equation x2 − 8kx + 16 (k2 − k + 1) = 0 are real, distinct (2009) and have values atleast 4, is …… .

Topic 5 Some Special Forms Objective Questions I (Only one correct option)

x12 − x9 + x4 − x + 1 > 0 is

1. The number of real roots of the equation 5 + |2 − 1| = 2 (2 − 2) is x

x

x

(a) 1 (c) 4

(2019 Main, 10 April II)

(b) 3 (d) 2

2. All the pairs (x, y) that satisfy the inequality 2

sin 2 x − 2 sin x + 5

⋅

1 sin 2 y

≤ 1 also satisfy the equation

4

(a) 2|sin x| = 3 sin y (c) sin x = 2 sin y

(2019 Main, 10 April I)

(b) sin x = |sin y| (d) 2 sin x = sin y

(b) 12 (d) 10

real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1] (2013 Main)

(b) lies between 2 and 3 (d) does not exist

5. Let a , b, c be real numbers, a ≠ 0. If α is a root of a 2x2 + bx + c = 0, β is the root of a 2x2 − bx − c = 0 and 0 < α < β, then the equation a 2x2 + 2bx + 2c = 0 has a root (1989, 2M) γ that always satisfies β 2 (d) α < γ < β (b) γ = α +

6. If a + b + c = 0, then the quadratic equation 3ax2 + 2bx + c = 0 has (a) at least one root in (0, 1) (b) one root in (2, 3) and the other in (−2, − 1) (c) imaginary roots (d) None of the above

8

x)(ax2 + bx + c)dx

(1981, 2M)

0

4. The

α+β 2 (c) γ = α

1

∫0 (1 + cos

Then, the quadratic equation ax2 + bx + c = 0 has

(2019 Main, 8 April I)

(a) γ =

8. Let a , b, c be non-zero real numbers such that

= ∫ (1 + cos 8 x)(ax2 + bx + c)dx

| x − 2| + x ( x − 4) + 2 = 0 (x > 0) is equal to

(a) lies between 1 and 2 (c) lies between − 1and 0

(1982, 2M)

(a) −4 < x ≤ 0 (b) 0 < x < 1 (c) −100 < x < 100 (d) −∞ < x < ∞

2

3. The sum of the solutions of the equation

(a) 9 (c) 4

7. The largest interval for which

(a) no root in (0, 2) (b) atleast one root in (1, 2) (c) a double root in (0, 2) (d) two imaginary roots

Objective Questions II (One or more than one correct option) 9. Let S be the set of all non-zero real numbers α such that

the quadratic equation αx2 − x + α = 0 has two distinct real roots x1 and x2 satisfying the inequality x1 – x2 < 1. Which of the following interval(s) is/are a subset of S? (2015 Adv.)

1 1  (a)  – , –   2 5 1  (c)  0,   5

1 (b)  – , 0  5  1 1 (d)  ,   5 2

10. Let a ∈ R and f : R → R be given by f (x) = x5 − 5x + a. (1983, 1M)

Then, (a) (b) (c) (d)

f (x) has three real roots, if a > 4 f (x) has only one real root, if a > 4 f (x) has three real roots, if a < − 4 f (x) has three real roots, if −4 < a < 4

34 Theory of Equations Passage Based Problems Read the following passage and answer the questions. Passage I Consider the polynomial f ( x ) = 1 + 2x + 3x 2+ 4x3 . Let s be the sum of all distinct real roots of f ( x ) and let t =| s|. (2010)

11. The real numbers s lies in the interval 1 (a)  − , 0  4 

(d)  0, 

3 3 1 (b)  − 11, −  (c)  − , −    4 4 2

1  4

12. The area bounded by the curve y = f (x) and the lines x = 0, y = 0 and x = t , lies in the interval

3 (a)  , 3 4 

21 11 (b)  ,   64 16 

21 (d)  0,   64 

(c) (9, 10)

13. The function f ′ (x) is 1 1 (a) increasing in  − t, −  and decreasing in  − , t    4  4 1 1 (b) decreasing in  − t, −  and increasing in  − , t    4  4 (c) increasing in (−t , t ) (d) decreasing in (−t , t )

Passage II If a continuous function f defined on the real line R, assumes positive and negative values in R, then the equation f ( x ) = 0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum values is negative, then the

equation f ( x ) = 0 has a root in R. Consider f ( x ) = kex − x for all real x where k is real constant. (2007, 4M)

14. The line y = x meets y = kex for k ≤ 0 at (a) no point (b) one point (c) two points (d) more than two points

15. The positive value of k for which kex − x = 0 has only one root is 1 e (c) e

(b) 1

(a)

(d) log e 2

16. For k > 0, the set of all values of k for which kex − x = 0 has two distinct roots, is 1 (a)  0,   e 1 (c)  , ∞  e 

1 (b)  , 1 e  (d) (0, 1)

True/False 17. If a < b < c < d, then the roots of the equation (x − a ) (x − c) + 2 (x − b) (x − d ) = 0 are real and distinct. (1984, 1M)

Analytical & Descriptive Question 18. Let −1 ≤ p < 1. Show that the equation 4x3 − 3x − p = 0 has a unique root in the interval [1/2, 1] and identify it. (2001, 4M)

Answers Topic 1

Topic 2

1. (d)

2. (d)

5. (b) 9. 13. 17. 21. 25. 29. 33. 37. 41.

6. (b) 10. 14. 18. 22. 26. 30. 34. 38. 42.

(c) (d) (c) (c) (b) (b) (a) (d) (b)

45. –5050

(c) (d) (d) (d) (d) (b) (b) (a) 4

46. True

3. (d) 7. (c) 11. 15. 19. 23. 27. 31. 35. 39. 43.

4. (a)

1. (d)

8. (a)

5. False

12. 16. 20. 24. 28. 32. 36. 40. 44.

(d) (c) (a) (d) (a) (c) (a) (c) k =2

(b) (c) (c) (c) (a) (b) (a) (b) ( −4, 7 )

51. y ∈ { −1 } ∪ [1, ∞ )

48. 1210 1  2 52. x ∈ ( −2, − 1 ) ∪  − , −   3 2

53. −4 and ( − 1 − 3 )

54. x = {a (1 − 2 ), a ( 6 − 1 )}

55. ±2, ± 2 58. x = a 61. (d)

−1/ 2

47. False

57. (q − s ) 2 − rqp − rsp + sp 2 + qr 2 or a

− 4 /3

59. q − p 2

2

60. (d)

2. (a)

3. (b)

4. –1

2. (a)

3. (a)

4. (a)

6. (d)

7. (a)

8. (b)

Topic 3 1. (c)

2. x = α 2β, αβ 2

Topic 4 1. (d) 5. (b) 9. a > 1

11. x ∈ [1 − 5, 1 ) ∪ [1 + 5, 2 )

12. k = 2

Topic 5 1. (a)

2. (b)

3. (d)

4. (d)

5. (d)

6. (a)

7. (d)

8. (b)

10. (b, d)

11. (c)

12. (a)

15. (a) 1  18. x = cos cos−1 p 3 

16. (a)

9. (a,d) 13. (b) 17. True

14. (b)

Hints & Solutions Topic 1 Quadratic Equations 1. Given quadratic polynomials, x + 20x − 2020 and 2

x2 − 20x + 2020 having a , b distinct real and c, d distinct complex roots respectively. So, a + b = − 20, ab = − 2020 and c + d = 20, cd = 2020 Now, ac(a − c) + ad (a − d ) + bc(b − c) + bd (b − d ) = a 2(c + d ) − a (c2 + d 2) + b2(c + d ) − b(c2 + d 2) = (c + d ) (a 2 + b2) − (c2 + d 2)(a + b) = (c + d )[(a + b)2 − 2ab] − (a + b) [(c + d )2 − 2cd ] = 20 [(20)2 + 4040] + 20 [(20)2 − 4040] = 2 × 20 × (20)2 = 40 × 400 = 16000

2. It is given that α is a common roots of given quadratic equations x2 – x + 2λ = 0 and 3x2 – 10x + 27λ = 0 2 ∴ 3 α – 10 α + 27λ = 0 3 α 2 − 3 α + 6λ = 0 – + – 0 – 7α + 21λ = 0 ⇒ α = 3λ So, 9λ2 – 3λ + 2λ = 0 1 1 [Qλ ≠ 0] λ= ⇒α= ⇒ 9 3 1 2× 2λ 9 =2 As αβ = 2λ ⇒ β = = 1 α 3 3 1 9× 9λ 9 =3 and αγ = 9λ ⇒ γ = = 1 α 3 2 ×3 βγ 3 = = 18 ∴ 1 λ 9

3. Given quadratic equations f (x) = (λ2 + 1)x2 − 4λx + 2 = 0 have exactly one root in the interval (0, 1). So, D > 0 ⇒ 16λ2 − 4(λ2 + 1)2 > 0 2 ⇒ 8λ − 8 > 0 ⇒ λ2 > 1 …(i) ⇒ λ ∈ (−∞ , − 1) ∪ (1, ∞ ) and f (0) f (1) < 0 ⇒ 2(λ2 + 1 − 4λ + 2) < 0 ⇒ λ2 − 4λ + 3 < 0 …(ii) ⇒ (λ − 3)(λ − 1) < 0 ⇒ λ ∈ (1, 3) From Eqs, (i) and (ii), we get λ ∈ (1, 3) And if λ = 3, then the quadratic equation is 10x2 − 12x + 2 = 0 ⇒ 5 x2 − 6 x + 1 = 0 1 x = 1, ⇒ 5

Q The root x =

1 ∈ (0, 1) for λ = 3 5

∴ λ ∈ (1 , 3]. Hence, option (d) is correct.

4. Given quadratic equation is  x2 + x sin θ − 2 sin θ = 0, θ ∈ 0, 

π  2

and its roots are α and β. So, sum of roots = α + β = − sin θ and product of roots = αβ = − 2 sin θ ⇒ αβ = 2(α + β ) α 12 + β12 Now, the given expression is −12 (α + β −12)(α − β)24 =

…(i)

α 12 + β12 α 12 + β12 = 12  1  β + α 12 1 24  12 12  (α − β)24  12 + 12 (α − β) α β   α β  12

 αβ    αβ = =  2 2  (α + β ) − 4αβ  (α − β ) 

12

12

  2(α + β) =  2  (α + β ) − 8 (α + β)    2 =   (α + β ) − 8 =

12

  2 =   − sin θ − 8

[from Eq. (i)] 12

[Q α + β = − sin θ]

212 (sin θ + 8)12

5. Given quadratic equation is x2 + px + q = 0, where p, q ∈R having one root 2 − 3 , then other root is 2 + 3 (conjugate of 2 − 3 ) [Q irrational roots of a quadratic equation always occurs in pairs] So, sum of roots = − p = 4 ⇒ p = −4 and product of roots = q = 4 − 3 ⇒ q = 1 Now, from options p2 − 4q − 12 = 16 − 4 − 12 = 0

6. Given quadratic equation is (m2 + 1)x2 − 3x + (m2 + 1)2 = 0

…(i)

Let the roots of quadratic Eq. (i) are α and β, so 3 and αβ = m2 + 1 m2 + 1

α+β=

According to the question, the sum of roots is greatest and it is possible only when ‘‘(m2 + 1) is minimum’’ and ‘‘minimum value of m2 + 1 = 1, when m = 0’’. ∴α + β = 3 and αβ = 1, as m = 0 Now, the absolute difference of the cubes of roots = |α 3 − β3| = |α − β||α 2 + β 2 + αβ| = (α + β )2 − 4αβ |(α + β )2 − αβ| = 9 − 4 |9 − 1|= 8 5

36 Theory of Equations 7. Given, α and β are the roots of the quadratic equation, x − 2x + 2 = 0 ⇒ (x − 1)2 + 1 = 0 ⇒ (x − 1)2 = − 1 ⇒ x−1 = ± i ⇒ x = (1 + i ) or (1 − i ) Clearly, if α = 1 + i, then β = 1 − i 2

10. Let the given quadratic equation in x, [where i = −1]

n

n

 1 + i α According to the question   = 1 ⇒   =1  β 1 − i n  (1 + i )(1 + i ) [by rationalization] ⇒   =1  (1 − i )(1 + i )  n

⇒

 1 + i 2 + 2i   =1⇒   1 − i2 

n

 2i  n   =1⇒ i =1  2

So, minimum value of n is 4.

8.

From inequalities Eqs. (iii) and (iv), the integral values of m are 0, 1, 2, 3, 4, 5, 6 Hence, the number of integral values of m is 7.

⇒ ⇒

α 2 + β 2 = αβ (α + β )2 = 3αβ 2 2 m (m − 4)2 =3 3m2 9m4 (m − 4)2 = 18 m − 4 = ±3 2 m =4±3 2

⇒

[Q i = 1] 4

Key Idea (i) First convert the given equation in quadratic equation. (ii) Use, Discriminant, D = b 2 − 4 ac < 0

Given quadratic equation is (1 + m2)x2 − 2(1 + 3m)x + (1 + 8m) = 0 Now, discriminant D = [−2(1 + 3m)]2 − 4(1 + m2)(1 + 8m) = 4 [(1 + 3m)2 − (1 + m2)(1 + 8m)] = 4 [1 + 9m2 + 6m − (1 + 8m + m2 + 8m3 )] = 4 [−8m3 + 8m2 − 2m] = − 8m(4m2 − 4m + 1) = − 8m(2m − 1)2

3m2x2 + m(m − 4)x + 2 = 0, m ≠ 0 have roots α and β, then m(m − 4) 2 and αβ = α +β = − 2 3m 3m2 α Also, let =λ β α β 1 Then, (given) λ + =1 ⇒ + =1 λ β α

⇒ ⇒ ⇒

[Qm ≠ 0]

The least value of m = 4 − 3 2

11. Given quadratic equation is …(i)

According to the question there is no solution of the quadratic Eq. (i), then D 0

81x2 + kx + 256 = 0 Let one root be α, then other is α 3 . k 256 and α ⋅ α 3 = Now, α + α 3 = − 81 81 b a c and product of roots = ] a

[Q for ax2 + bx + c = 0, sum of roots = −

4

 4 α4 =    3 4 α=± 3 k = − 81 (α + α 3 )

⇒ ⇒ ∴

So, there are infinitely many values of ‘m’ for which, there is no solution of the given quadratic equation.

= − 81 α (1 + α 2) 16  4  = − 81  ±  1 +  = ± 300  3  9

9. The quadratic expression ax2 + bx + c, x ∈ R is always positive, if a > 0 and D < 0. So, the quadratic expression (1 + 2m) x2 − 2 (1 + 3m)x + 4(1 + m), x ∈ R will be always positive, if 1 + 2m > 0 …(i) and D = 4(1 + 3m)2 − 4(2m + 1) 4(1 + m) < 0 …(ii) From inequality Eq. (i), we get 1 …(iii) m>− 2 From inequality Eq. (ii), we get 1 + 9m2 + 6m − 4 (2m2 + 3m + 1) < 0 ⇒ m2 − 6m − 3 < 0 ⇒ [m − (3 + 12 )][m − (3 − 12 )] < 0 6 ± 36 + 12 = 3 ± 12] [Q m2 − 6m − 3 = 0 ⇒ m = 2 …(iv) 3 − 12 < m < 3 + 12 ⇒

12. Let a = 5, b = 5r and c = 5r 2 We know that, in a triangle sum of 2 sides is always greater than the third side. ∴ a + b > c; b + c > a and c + a > b Now,

a+ b>c

⇒ 5 + 5r > 5r 2

⇒ 5r 2 − 5r − 5 < 0

⇒ r − r −1 5 ⇒ r2 + r − 1 > 0   −1 + 5    − 1 − 5    >0   r −  r −  2 2        2 −1 ± 1 + 4 − 1 ± 5  =  Q r + r − 1 = 0 ⇒ r = 2 2     −1 + 5  −1 − 5  r ∈  − ∞, , ∞  ∪ 2 2    

⇒

+

–1+ √ 5 2

⇒ x=

c+ a >b 5r 2 + 5 > 5r r2 − r + 1 > 0

and ⇒ ⇒ 2

and

… (iii)

b2 − 4ac ] 2a

1+ √ 5 2

= − [(1 − i )15 + (1 + i )15 ] 15 15   1   1 i  i   + = −  2  −     +  2 2  2      2   2   = −   + 

∞

15  π π   2  cos − i sin     4 4   15 π π      2  cos + i sin    4 4   

15π 15π   15π 15π    = − ( 2 )15  cos − i sin + i sin  +  cos  4 4   4 4   

7 is the only value that does not satisfy. 4

[using De’ Moivre’s theorem (cos θ ± i sin θ )n = cos nθ ± i sin nθ, n ∈ Z ]

13. Given quadratic equation is x2 + (3 − λ )x + 2 = λ 2 x + (3 − λ )x + (2 − λ ) = 0

−b±

Then, α 15 + β15 = (−1 + i )15 + (− 1 − i )15

2

–1+ √ 5 2

[Q roots of ax2 + bx + c = 0 are

⇒ x = −1 ± i Let α = − 1 + i and β = − 1 − i.

2

1 3  ⇒ r −  + > 0  2 4 ⇒ r ∈R From Eqs. (i), (ii) and (iii), we get  −1 + 5 1 + 5 r ∈ ,  2 2  

1– √ 5 2

−2 ± 4 −8 2

given by x =

1  1  1 ⇒ r2 − 2 ⋅ r +   + 1 −   > 0  2  2 2

–∞ –1– √ 5 2

rational D = (b2 − 4ac) should be perfect square. In the equation 6x2 − 11x + α = 0 a = 6, b = − 11 and c = α ∴For roots to be rational D = (− 11)2 − 4(6) (α) should be a perfect square. ⇒ D(α) = 121 − 24α should be a perfect square Now, D(1) = 121 − 24 = 97 is not a perfect square. D(2) = 121 − 24 × 2 = 73 is not a perfect square. D(3) = 121 − 24 × 3 = 49 is a perfect square. D(4) = 121 − 24 × 4 = 25 is a perfect square. D(5) = 121 − 24 × 5 = 1 is a perfect square. and for α ≥ 6, D(α) < 0, hence imaginary roots. ∴For 3 values of α (α = 3, 4, 5), the roots are rational.

15. We have, x2 + 2x + 2 = 0

+

–

–1– √ 5 2

…(ii)

14. For the roots of quadratic equation ax2 + bx + c = 0 to be

… (i)

Let Eq. (i) has roots α and β, then α + β = λ − 3 and αβ = 2 − λ b [Q For ax2 + bx + c = 0, sum of roots = − a c and product of roots = ] a Now, α 2 + β 2 = (α + β )2 − 2αβ = (λ − 3)2 − 2(2 − λ ) = λ2 − 6λ + 9 − 4 + 2λ = λ2 − 4λ + 5 = (λ2 − 4λ + 4) +1 = (λ − 2)2 + 1 2 Clearly, a + β 2 will be least when λ = 2.

15π  1    = − ( 2 )15 2 cos = − ( 2 )15 2 × 4  2    15π π π 1    Q cos 4 = cos 4π − 4  = cos 4 = 2    = − ( 2 )16 = − 28 = − 256. Alternate Method α 15 + β15 = (−1 + i )15 + (−1 − i )15 = − [(1 − i )15 + (1 + i )15 ]  (1 − i )16 (1 + i )16  =− + 1 + i   1−i  [(1 − i )2]8 [(1 + i )2]8  =− + 1 + i   1−i

38 Theory of Equations  [1 + i 2 − 2i ]8 [1 + i 2 + 2 i ]8  =− +  1−i 1+ i    (−2 i )8 (2 i )8  =− + 1 + i   1−i  1 1  [Q i 4n = 1, n ∈ Z ] = − 28  +  − + 1 i 1 i    2  2  = − 256  = − 256 = − 256 2 2  1 − (i )  16. We have, 2| x − 3| +

x ( x − 6) + 6 = 0

Let x − 3 = y x = y+3 ⇒ ∴ 2| y| + ( y + 3)( y − 3) + 6 = 0 ⇒ 2| y| + y2 − 3 = 0 ⇒ | y|2 + 2| y| − 3 = 0 ⇒ (| y| + 3)(| y| − 1) = 0 ⇒ | y| ≠ − 3 ⇒ | y| = 1 ⇒ y= ±1 ⇒ x −3 = ±1 x = 4, 2 ⇒ x = 16, 4 ⇒ 17. We have, α , β are the roots of x2 − x + 1 = 0 Q Roots of x2 − x + 1 = 0 are −ω ,−ω 2 ∴ Let α = − ω and β = − ω 2 ⇒ α 101 + β107 = (− ω )101 + (− ω 2)107 = − (ω101 + ω 214 ) = − (ω 2 + ω ) [Q ω3 = 1] [Q1 + ω + ω 2 = 0] = − (−1) =1

or III. x2 − 5x + 5 = − 1 and x2 + 4x − 60 = Even integer. Case I When x2 + 4x − 60 = 0 ⇒ x2 + 10x − 6x − 60 = 0 ⇒ x(x + 10) − 6(x + 10) = 0 ⇒ (x + 10) (x − 6) = 0 ⇒ x = − 10or x = 6 Note that, for these two values of x, x2 − 5x + 5 ≠ 0 Case II When x2 − 5 x + 5 = 1 ⇒ x2 − 5 x + 4 = 0 2 ⇒ x − 4x − x + 4 = 0 ⇒ x(x − 4) − 1 (x − 4) = 0 ⇒ (x − 4) (x − 1) = 0 ⇒ x = 4or x = 1 Case III When x2 − 5 x + 5 = − 1 ⇒ x2 − 5 x + 6 = 0 2 ⇒ x − 2x − 3x + 6 = 0 ⇒ x(x − 2) − 3(x − 2) = 0 ⇒ (x − 2) (x − 3) = 0 ⇒ x = 2 or x = 3 Now, when x = 2, x2 + 4x − 60 = 4 + 8 − 60 = − 48, which is an even integer. When x = 3, x2 + 4x − 60 = 9 + 12 − 60 = − 39, which is not an even integer. Thus, in this case, we get x = 2. Hence, the sum of all real values of x = − 10 + 6 + 4 + 1 + 2 = 3

18. Given quadratic equation is x(x + 1) + (x + 1)(x + 2) + ... + (x + n − 1) (x + n ) =10n ⇒ (x2 + x2 + ... + x2) + [(1 + 3 + 5 + ... + (2n − 1)]x + [(1 ⋅ 2 + 2 ⋅ 3 + ... + (n − 1)n ] = 10n n (n 2 − 1) 2 2 nx + n x + − 10n = 0 ⇒ 3 2 n −1 ⇒ x2 + nx + − 10 = 0 3 ⇒ 3x2 + 3nx + n 2 − 31 = 0 Let α and β be the roots. Since, α and β are consecutive. ∴ |α − β| = 1 ⇒ (α − β )2 = 1 2 2 Again, (α − β ) = (α + β ) − 4αβ 2  n 2 − 31  − 3n  ⇒ 1=   − 4  3   3  4 ⇒ 1 = n 2 − (n 2 − 31) ⇒ 3 = 3n 2 − 4n 2 + 124 3 ⇒ n 2 = 121 ⇒ n = ± 11 [Qn > 0] ∴ n = 11

19. Given, (x − 5x + 5) 2

x 2 + 4 x − 60

=1

Clearly, this is possible when I. x2 + 4x − 60 = 0 and x2 − 5x + 5 ≠ 0 or II. x2 − 5x + 5 = 1

20. Here, x − 2x secθ + 1 = 0 has roots α 1 and β1. 2

2 sec θ ± 4 sec2 θ − 4 2 ×1 2 sec θ ± 2|tan θ| = 2 π  π Since, θ ∈  − ,−  ,  6 12 2 sec θ m 2 tan θ i.e. θ ∈ IV quadrant = 2 ∴ α 1 = sec θ − tan θ and β1 = sec θ + tan θ [as α 1 > β1] and x2 + 2x tan θ − 1 = 0 has roots α 2 and β 2 .

∴

i.e. ∴ and Thus,

α 1 , β1 =

−2 tan θ ± 4 tan 2 θ + 4 2 α 2 = − tan θ + sec θ [as α 2 > β 2] β 2 = − tan θ − sec θ α 1 + β 2 = −2 tan θ α 2, β 2 =

21. Given, α and β are the roots of the equation x2 − 6x − 2 = 0. ∴

a n = α n − β n for n ≥ 1 a10 = α 10 − β10 a 8 = α 8 − β8 a 9 = α 9 − β9

Theory of Equations 39 Now, consider a10 − 2a 8 α = 2a 9

10

8

8

and

α 8 (α 2 − 2) − β 8 (β 2 − 2) = 2(α 9 − β 9 ) =

−q 4r 4r 4 = = =− p p −9 r 9 r r 1 αβ = = = p −9 r −9 16 4 16 + 36 (α − β )2 = (α + β )2 − 4 αβ = + = 81 9 81 52 (α − β )2 = 81 2 |α − β| = 13 9 α+β=

Now,

− β − 2(α − β ) 2(α 9 − α 9 ) 10

∴

α 8 ⋅ 6 α − β 86 β 6 α 9 − 6 β 9 6 = =3 = 2(α 9 − β 9 ) 2(α 9 − 6 β 9 ) 2

⇒

 Q α and β are the roots of x2 − 6x − 2 = 0 or x2 = 6x + 2   2 2 ⇒ α = 6 α + 2 ⇒ α − 2 = 6 α and β 2 = 6 β + 2 ⇒ β 2 − 2 = 6 β 

⇒

24.

a10 − 2a 8 (α 10 − β10 ) − 2 (α 8 − β 8 ) = 2a 9 2 (α 9 − β 9 )

Alternate Solution =

Since, α and β are the roots of the equation x2 − 6x − 2 = 0.

Qα is root of x2 − 6 x − 2 = 0 ⇒ α 2 − 2 = 6α

x2 = 6 x + 2

or ∴

α 2 = 6α + 2

⇒

α 10 = 6 α 9 + 2 α 8

[and β is root of x2 − 6 x − 2 = 0 ⇒ β 2 − 2 = 6 β]

…(ii)

25. Sum of roots =

On subtracting Eq. (ii) from Eq. (i), we get α 10 − β10 = 6(α 9 − β 9 ) + 2(α 8 − β 8 ) ⇒

=

...(i)

β10 = 6 β 9 + 2 β 8

Similarly,

(Q a n = α n − β n)

a10 = 6a 9 + 2a 8

α 2 + β2 and product = 1 αβ

(α + β ) (α 2 − αβ + β 2) = q −q α 2 + β 2 − αβ = p

∴

22. If quadratic equation has purely imaginary roots, then

α 8 (6 α ) − β 8 (6 β ) 6 (α 9 − β 9 ) =3 = 2 (α 9 − β 9 ) 2 (α 9 − β 9 )

Given, α + β = − p and α 3 + β3 = q ⇒

⇒ a10 − 2a 8 = 6a 9 a − 2a 8 ⇒ 10 =3 2a 9

⇒

Let p(x) = ax + b with a, b of same sign and a , b ∈ R.

From Eqs. (i) and (ii), we get

Then,

p[ p(x)] = a (ax2 + b)2 + b

α 2 + β 2 + 2αβ = p2 α 2 + β2 =

p(x) has imaginary roots say ix. Then, also ax2 + b ∈ R and (ax2 + b)2 > 0 ∴ Thus, PLAN

If ax 2 + bx + c = 0 has roots α and β, then α + β = − b / a c and α β = . Find the values of α + β and αβ and then put a in (α − β )2 = (α + β )2 − 4αβ to get required value.

Given, α and β are roots of px + qx + r = 0, p ≠ 0. −q r , αβ = α+β= ∴ p p 2

2 (−4r ) = p + r ⇒

p = − 9r

p3 + q 3p

( p3 − 2q) x +1 =0 ( p3 + q)

…(i)

...(ii)

[from Eq. (i)]

26. The equation x2 − px + r = 0 has roots α, β and the equation x2 − qx + r = 0 has roots ⇒

α , 2β. 2

r = αβ and α + β = p, α 2q − p 2 (2 p − q) and and α = + 2β = q ⇒ β = 2 3 3 2 ⇒ αβ = r = (2q − p) (2 p − q) 9

27. Since, roots are real, therefore D ≥ 0 ⇒

4 (a + b + c)2 − 12λ (ab + bc + ca ) ≥ 0

⇒

(a + b + c)2 ≥ 3λ (ab + bc + ca )

⇒ ⇒

On putting the value of q in Eq. (ii), we get ⇒

and αβ =

...(ii)

⇒ ( p3 + q) x2 − ( p3 − 2q) x + ( p3 + q) = 0

p [ p(x)] ≠ 0, ∀ x

Since, p, q and r are in AP. ∴ 2q = p + r 1 1 α+β Also, + =4 ⇒ =4 α β αβ −q 4r = ⇒ α + β = 4 αβ ⇒ p p ⇒ q = − 4r

p3 − 2q 3p

∴ Required equation is, x2 −

a (ax2 + b)2 + b ≠ 0, ∀ x

...(i)

(α + β )2 = p2

and

coefficient of x must be equal to zero. 2

23.

α 8 (α 2 − 2) − β 8 (β 2 − 2) 2(α 9 − β 9 )

Also,

a 2 + b2 + c2 ≥ (ab + bc + ca ) (3λ − 2) a 2 + b2+ c2 3λ − 2 ≤ ab + bc + ca cos A =

b2 + c2 − a 2 0 ⇒ x2 − 2 > 0

⇒

x < − 2 or x > 2

− 1 = − p/3 p=3 c 0 ⇒ − b < 0 ⇒ α + β < 0 ⇒ the sum is negative. ⇒ Modulus of negative quantity is > modulus of positive quantity but α < β is given. Therefore, it is clear that α is negative and β is positive and modulus of α is greater than modulus of β ⇒ α < 0 < β < |α| NOTE This question is not on the theory of interval in which root lie, which appears looking at first sight. It is new type and first time asked in the paper. It is important for future. The actual type is interval in which parameter lie.

x + 1 − x − 1 = 4x − 1

33. Since, ⇒

(x + 1) + (x − 1) − 2 x2 − 1 = 4x − 1

Case II When x + 2 < 0

1 − 2 x = 2 x2 − 1 ⇒ 1 + 4 x2 − 4 x = 4 x 2 − 4 5 ⇒ 4x = 5 ⇒ x = 4 But it does not satisfy the given equation.

∴

x2 + x + 2 + x > 0

Hence, no solution exists.

⇒

x2 + 2 x + 2 > 0

⇒

(x + 1)2 + 1 > 0

⇒

x ∈ [−2, − 2 ) ∪ ( 2 , ∞ )

⇒ …(ii)

34. Given, ⇒

which is true for all x. ∴

x ≤ − 2 or x ∈ (−∞ , − 2)

…(iii)

From Eqs. (ii) and (iii), we get x ∈ (−∞ , − 2 ) ∪ ( 2 , ∞ )

30. Given, log 4 (x − 1) = log 2(x − 3) = log 41/ 2 (x − 3) ⇒

log 4 (x − 1) = 2 log 4 (x − 3)

⇒

log 4 (x − 1) = log 4 (x − 3)2

⇒ ⇒

(x − 3)2 = x − 1 x2 + 9 − 6x = x − 1

⇒ ⇒

x2 − 7x + 10 = 0 (x − 2)(x − 5) = 0 x = 2, or x = 5

⇒ ⇒ x = 5 [Q x = 2 makes log (x − 3) undefined]. Hence, one solution exists.

31. Let α , α 2 be the roots of 3x2 + px + 3 = 0 Now, ⇒ Now, ⇒

S = α + α 2 = − p /3, P = α3 = 1 α = 1, ω , ω 2 α + α 2 = − p/3 ω + ω 2 = − p/3

⇒

x

3 5 (log 2 x )2 + log 2 x − 4 4

= 2

3 5 (log 2 x)2 + log 2 x − = log x 2 4 4 3 5 1 2 (log 2 x) + log 2 x − = 4 4 2 log 2 x

⇒

3(log 2 x)3 + 4(log 2 x)2 − 5(log 2 x) − 2 = 0

Put ∴ ⇒ ⇒ ⇒

log 2 x = y 3 y3 + 4 y2 − 5 y − 2 = 0 ( y − 1) ( y + 2) (3 y + 1) = 0 y = 1, −2 , −1 / 3 log 2 x = 1, − 2, − 1 / 3 1 1 x = 2, 1/3 , 4 2

⇒

35. Since, α , β are the roots of x2 + px + q = 0 and α 4 , β 4 are the roots of x2 − rx + s = 0. ⇒ α + β = − p ; α β = q; α 4 + β 4 = r and α 4β 4 = s Let roots of x2 − 4qx + (2q2 − r ) = 0 be α ′ and β′ Now, α ′ β ′ = (2q2 − r ) = 2 (αβ )2 − (α 4 + β 4 ) = − (α 4 + β 4 − 2α 2β 2) = − (α 2 − β 2)2 < 0 ⇒ Roots are real and of opposite sign. 2 2 36. Given, x − =1 − ⇒ x=1 x−1 x−1 But at x = 1, the given equation is not defined. Hence, no solution exist.

Theory of Equations 41 37. Let y = ⇒

x2 − (a + b)x + ab x−c

41. Given, ∴

yx − cy = x2 − (a + b)x + ab

⇒ x2 − (a + b + y)x + (ab + cy) = 0

And

For real roots, D ≥ 0 ⇒

(a + b + y)2 − 4(ab + cy) ≥ 0

⇒ (a + b)2 + y2 + 2(a + b) y − 4ab − 4cy ≥ 0 ⇒

x2 + 2 px + q = 0

y2 + 2(a + b − 2c) y + (a − b)2 ≥ 0

and

... (i)

αβ = q

... (ii)

ax2 + 2bx + c = 0 1 2b α+ =− β a α c = β a

D ≤0  α + β 2  =   − αβ     −2 

4 (a + b − 2c) − 4 (a − b) ≤ 0 2

2

⇒ 4 (a + b − 2c + a − b)(a + b − 2c − a + b) ≤ 0 ⇒

(2a − 2c)(2b − 2c) ≤ 0

⇒

(a − c)(b − c) ≤ 0

⇒

(c − a )(c − b) ≤ 0

⇒ c must lie between a and b

⇒

| x|2 − 3|x| + 2 = 0 (|x| − 1) (|x| − 2) = 0

⇒

=

(α − β )2 16

x = 1, − 1, 2, − 2

39. (x − a ) (x − b) + (x − b) (x − c) + (x − c) (x − a ) = 0 3x2 − 2 (a + b + c) x + (ab + bc + ca ) = 0

Now, discriminant = 4 (a + b + c) − 12 (ab + bc + ca ) 2

= 4 { a 2 + b2 + c2 − ab − bc − ca } = 2 {(a − b)2 + (b − c)2 + (c − a )2} which is always positive. Hence, both roots are real.

40. Since, a , b, c > 0 and ax2 + bx + c = 0 −b x= ± 2a

⇒

b2 − 4ac 2a

Case I When b2 − 4ac > 0 x=

⇒ and

−b + 2a

−b − 2a

2

1  2 α −  . a ≥ 0  β

a 1 α +   β 2

Since,

pa ≠ b

⇒ α+

⇒

β 2 ≠ 1, β ≠ { −1, 0, 1}, which is correct.

Hence, four real solutions exist. ⇒

2   1   α +  α β  −  a2 β  2       

b=−

and

|x| = 1, 2

∴

... (iv)

∴ Statement I is true. a  α + β Again, now pa = −   a = − (α + β )  2  2

i.e. a ≤ c ≤ b or b ≤ c ≤ a

38. Since,

... (iii)

Now, ( p2 − q) (b2 − ac)

which is true for all real values of y. ∴

∴

α + β = −2 p

b2 − 4ac 2a

b2 − 4ac both roots, are negative. 2a

1 ≠α +β β

Similarly, if c ≠ qa α 1  ⇒ a ≠ a α β ⇒ α β −  ≠ 0  β β ⇒

α ≠ 0 and β −

⇒

β ≠ { −1 , 0 , 1 }

Statement II is true. Both Statement I and Statement II are true. But Statement II does not explain Statement I.

42. Given,|x − 2|2 + |x − 2| − 2 = 0 Case I When x ≥ 2 ⇒ (x − 2)2 + (x − 2) − 2 = 0 ⇒

x2 + 4 − 4 x + x − 2 − 2 = 0

⇒

x2 − 3 x = 0

⇒

x (x − 3) = 0

Case II When b − 4ac = 0 −b , i.e. both roots are equal and negative ⇒ x= 2a

⇒

x = 0, 3

⇒

x=3

Case III When b2 − 4ac < 0

⇒

{ − (x − 2)}2 − (x − 2) − 2 = 0

⇒

(x − 2)2 − x + 2 − 2 = 0

⇒

x2 + 4 − 4 x − x = 0

have negative real part.

⇒

x2 − 4x − (x − 4) = 0

∴ From above discussion, both roots have negative real parts.

⇒

x(x − 4) − 1 (x − 4) = 0

⇒

(x − 1) (x − 4) = 0

2

⇒

x=

4ac − b2 −b ±i 2a 2a

1 ≠0 β

Case II When x < 2

[0 is rejected] …(i)

42 Theory of Equations ⇒

x = 1, 4

⇒

x=1

[4 is rejected] …(ii)

Hence, the sum of the roots is 3 + 1 = 4.

⇒ (| x − 2| + 2) (| x − 2| − 1) = 0 | x − 2| = − 2, 1

⇒ ⇒

| x − 2| = 1

43. Since, x − 3kx + 2e ⇒ ⇒ ⇒ ⇒

⇒ [neglecting –2]

⇒ x = 3, 1

Sum of the roots = 4 2

2 log k

− 1 = 0 has product of roots 7.

2e2 log k − 1 = 7 e2 log e k = 4 k2 = 4 k =2

[neglecting −2]

D1 = b2 − 4ac and D2 = b2 + 4ac D1 + D2 = 2b2 ≥ 0

∴ Atleast one of D1 and D2 is (+ ve). Hence, atleast two real roots. Hence, statement is true.

…(i) …(ii)

…(i)

2y + (2y − 1 − 1) = 2y − 1 + 1 2y = 2 y = 1 ∈ (0, 1] 2y − 2y − 1 + 1 = 2y − 1 + 1 2y − 2 ⋅ 2y − 1 = 0

⇒

2y − 2y = 0 true for all y > 1

From Eqs. (i), (ii) and (iii), we get …(iii) …(iv) [from Eq. (i)]

…(ii)

Case III When y ∈ (1, ∞ )

⇒

On subtracting Eq. (iv) from Eq. (ii), we get

= 10 (c + a ) = 1210

y = − 1 ∈ (−∞ , 0]

∴

a 2 − 10ca − 11d = 0

c + a = 121

2−y = 2

⇒

x − 10cx − 11d = 0

a + b + c + d = 10c + 10a

2− y + (2y − 1 − 1) = 2y − 1 + 1

⇒

a + b = 10c and c + d = 10a

2

⇒

Case I When y ∈ (−∞ , 0] ∴

∴

Similarly, ‘a’ is the root of

∴

51. Given, 2| y | − |2y − 1 − 1| = 2y − 1 + 1

Case II When y ∈(0, 1]

∴ Roots are rational. Hence, statement is false.

(c + a ) (c − a ) = 11 × 11 (c − a )

Let n be any integer. We have,  n (n − 1)  + ( A + B) n + C f (n ) = An 2 + Bn + C = 2 A 2  

⇒

Here, D = (3)2 − 4 ⋅ 2 ⋅ 1 = 1 which is a perfect square.

∴

C , A + B + C , A − B + C are integers. C , A + B, A − B are integers. C , A + B, ( A + B) − ( A − B) = 2 A are integers.

⇒

47. Given, 2x2 + 3x + 1 = 0

(c2 − a 2) = 11 (b − d )

is an integer.

Since, n is an integer, n (n − 1) / 2 is an integer. Also, 2 A , A + B and C are integers. We get f (n ) is an integer for all integer n.

46. P (x) ⋅ Q (x) = (ax2 + bx + c) (−ax2 + bx + c)

⇒

50. Suppose f (x) = Ax2 + Bx + C is an integer, whenever x

Conversely, suppose 2 A , A + B and C are integers.

= − (1 + 2 + 3 + ... + 100) 100 =− (1 + 100) = −50(101) = −5050 2

⇒ (a − c) + (b − d ) = 10 (c − a ) ⇒ (b − d ) = 11 (c − a ) Since, ‘c’ is the root of x2 − 10ax − 11b = 0 ⇒ c2 − 10ac − 11b = 0

2

b2 4c B2 4C − = 2− A a2 a A 2 2 b − 4ac B − 4 AC = a2 A2

⇒

⇒ ⇒ ⇒

45. The coefficient of x99 in (x − 1)(x − 2)... (x − 100)

48. Here,

2

4c  B  4C  b = −  − −  −  a   a A A

⇒

∴ f (0), f (1), f (−1) are integers.

other root is 2 − i 3. ⇒ − p =2 + i 3 + 2 − i 3 = 4 and q = (2 + i 3 ) (2 − i 3 ) = 7 ⇒ ( p, q) = (−4, 7)

Clearly,

(α + β )2 − 4αβ = (α + δ + β + δ )2 − 4(α + δ ) (β + δ )

⇒

44. If 2 + i 3 is one of the root of x2 + px + q = 0. Then,

Now,

B C , (α + δ ) (β + δ ) = A A α − β = (α + δ ) − (β + δ ) (α − β )2 = [(α + δ ) − (β + δ )]2

Now, ⇒

Given,| x − 2|2 + | x − 2| − 2 = 0

b c , αβ = a a

α+δ+β+δ=−

and

Alternate Solution

∴

49. Since, α + β = −

52. Given,

y ∈{ −1} ∪ [1, ∞ ). 2x 1 > 2 x2 + 5 x + 2 x + 1

⇒

2x 1 − >0 (2x + 1) (x + 2) (x + 1)

⇒

2x (x + 1) − (2x + 1) (x + 2) >0 (2x + 1) (x + 2) (x + 1)

…(iii)

Theory of Equations 43 ⇒

− (3x + 2) > 0 ; using number line rule (2x + 1) (x + 1) (x + 2) –

+

–

–2

+

–

–1

1  2 x ∈ (−2, − 1) ∪  − , −   3 2

∴

Given, ⇒

– 1 2

– 2 3

56. Let α , β are roots of ax2 + bx + c = 0 α = βn αβ =

c c  c ⇒ β=  ⇒ βn + 1 =  a a a

1/( n + 1 )

It must satisfy ax2 + bx + c = 0 i.e.

53. Given,| x + 4x + 3|+ 2x + 5 = 0 2

Case I x2 + 4x + 3 > 0 ⇒ (x < − 3 or x > − 1) 2 ∴ x + 4x + 3 + 2x + 5 = 0 ⇒ x2 + 6x + 8 = 0 ⇒ (x + 4) (x + 2) = 0 [but x < − 3 or x > − 1] ⇒ x = − 4, − 2 …(i) ∴ x = − 4 is the only solution. Case II x2 + 4x + 3 < 0 ⇒ (−3 < x < − 1) ∴ − x2 − 4 x − 3 + 2 x + 5 = 0 ⇒ x2 + 2x − 2 = 0 ⇒ (x + 1)2 = 3 ⇒ | x + 1| = 3 ⇒

x = − 1 − 3, −1 + 3

∴

x = − 1 − 3 is the only solution.

[but x ∈ (−3, − 1)] …(ii)

 c a   a

⇒ ⇒

c1/( n + 1) a1/( n + 1)

2 /( n + 1 )

 c + b   a

1 /( n + 1 )

+ c=0

a . c2/( n + 1) b. c1/( n + 1) + 1/( n + 1) + c = 0 a 2/( n + 1) a + 1 /( n 1 )  a. c c. a1/( n + 1)   1/( n + 1) + b + 1/( n + 1)  = 0 c  a 

⇒

a n/( n + 1)c1/( n + 1) + b + cn/( n + 1)a1 / ( n + 1) = 0

⇒

(a nc)1/( n + 1) + (cna )1/( n + 1) + b = 0

57. Since, α , β are the roots of x2 + px + q = 0 and γ , δ are the roots of x2 + rx + s = 0 ∴

α + β = − p, αβ = q and γ + δ = − r, γδ = s

Now, (α − γ )(α − δ )(β − γ )(β − δ )

From Eqs. (i) and (ii), we get

= [α 2 − (γ + δ ) α + γδ ][β 2 − (γ + δ ) β + γδ ]

x = − 4 and (−1 − 3) are the only solutions.

= (α 2 + rα + s)(β 2 + rβ + s)

a ≤0 Given, x2 − 2 a | x − a | − 3 a 2 = 0 Case I When x ≥ a

= (αβ )2 + r (α + β )αβ + s(α 2 + β 2) + αβr 2 + rs(α + β ) + s2

54. Here,

⇒

x2 − 2a (x − a ) − 3a 2 = 0

⇒

x2 − 2ax − a 2 = 0

= q2 − rqp + s( p2 − 2q) + qr 2 − rsp + s2 = (q − s)2 − rqp − rsp + sp2 + qr 2

58. The given equation can be rewritten as 2 1 3 + + = 0 [Q b = a 2x, given] log a x log a ax log a a 2x

x = a ± 2a

⇒

[as a (1 + 2 ) < a and a (1 − 2 ) > a] ⇒

∴ Neglecting x = a (1 + 2 ) as x ≥ a ⇒

x = a (1 − 2 )

Case II ⇒

…(i) ⇒

When x < a ⇒ x + 2a (x − a ) − 3a = 0 2

2

x2 + 2 ax − 5a 2 = 0 ⇒ x = − a ± 6a [as a ( 6 − 1) < a and a (−1 − 6 ) > a]

∴ Neglecting x = a (−1 − 6 ) ⇒ x = a ( 6 − 1)

...(ii)

From Eqs. (i) and (ii), we get

55. Given, (5 + 2 6 )x

2

Put y = (5 + 2 6 ) From Eq. (i), y +

−3

+ (5 − 2 6 )x

x2 − 3

= 10 x2 − 3

…(i) 1 = y

1 = 10 y

y2 − 10 y + 1 = 0 ⇒

⇒

(5 + 2 6 )x

or

(5 + 2 6 )x x2 − 3 = 1

−3

⇒ (5 − 2 6 )

⇒

⇒

2

or

2

⇒

6 t 2 + 11 t + 4 = 0

⇒

(2 t + 1) (3 t + 4) = 0 1 4 t=− or − 2 3 1 4 log a x = − or log a x = − 2 3

∴ ⇒

x = a −1/ 2 or

x =a

−4/3

59. Since, α + β = − p, αβ = 1 and γ + δ = − q, γδ = 1 y=5±2 6

2

2 1 3 + + = 0, where t = log a x t 1+ t 2+ t

⇒ 2 (1 + t ) (2 + t ) + 3 t (1 + t ) + t (2 + t ) = 0

⇒

x = { a (1 − 2 ), a ( 6 − 1)}

2 1 3 + + =0 log a x 1 + log a x 2 + log a x

−3

=5 + 2 6

−3

=5 −2 6

x2 − 3 = − 1

⇒

x = ± 2 or x = ± 2

⇒

x = ± 2, ± 2

Now, (α − γ )(β − γ )(α + δ )(β + δ ) = {αβ − γ (α + β ) + γ 2}{αβ + δ (α + β ) + δ 2} = {1 − γ (− p) + γ 2}{1 + δ ( − p) + δ 2} = (1 + γ 2 + γp)(1 − δp + δ 2) = (− qγ + γp)(−δp − δq) [Q γ 2 + qγ + 1 = 0 and δ 2 + qδ + 1 = 0] = (q2 − p2)(γδ ) = q2 − p2

[Q γ δ = 1]

44 Theory of Equations α2 = α + 1

60.

β =β + 1 a n = pα + qβ n

and a 2x2 + b2x + c2 = 0 have a common real root, then

n

= p(α n − 1 + α n − 2) + q(β n − 1 + β n − 2)

⇒

= an − 1 + an − 2 α=

1+ 5 1− 5 ,β = 2 2

a 4 = a3 + a 2 = 2a 2 + a1 = 3a1 + 2a 0

⇒

⇒

(1 + b)2 = (b2 + 1) (1 − b)

⇒

b2 + 2b + 1 = b2 − b3 + 1 − b

⇒

b3 + 3b = 0

∴

b (b2 + 3) = 0

x2 + bx + a = 0 have common root

p− q =0 7 ( p + q) × = 28 2

On subtracting above equations, we get (a − b) x + (b − a ) = 0 x=1

⇒

∴ x = 1 is the common root.

p+ q =8

⇒

p= q =4

⇒

∴

p + 2q = 12

⇒

1 Given α, β and γ are three consecutive terms of a non-constant GP. Let α = α, β = αr , γ = αr 2, { r ≠ 0, 1} and given quadratic equation is αx2 + 2 βx + γ = 0 On putting the values of α,β, γ in Eq. (i), we get αx2 + 2αrx + αr 2 = 0 ⇒ x2 + 2rx + r 2 = 0 ⇒ (x + r )2 = 0 ⇒ x=−r

…(i)

f (x) = a nxn + a n − 1xn − 1 + ... + a 0 Then, x = r is root of f ′ (x) = 0 repeated (m − 1) times.

Topic 3 Transformation of Roots 1. Given, α , β are the roots of (x − a )(x − b) − c = 0 ⇒ ⇒

(x − a )(x − b) − c = (x − α ) (x − β ) (x − a )(x − b) = (x − α )(x − β ) + c

⇒ a , b are the roots of equation (x − α )(x − β ) + c = 0

2. Given equations are x2 + 2x + 3 = 0

…(i)

ax + bx + c = 0

…(ii)

Since, Eq. (i) has imaginary roots, so Eq. (ii) will also have both roots same as Eq. (i). a b c Thus, = = 1 2 3 Hence, a : b : c is 1 : 2 : 3.

a + b = −1

Hence, statement is false.

Q The quadratic equations αx2 + 2 βx + γ = 0 and x2 + x − 1 = 0 have a common root, so x = − r must be root of equation x2 + x −1 = 0, so …(ii) r2 − r − 1 = 0 Now, α (β + γ ) = α (αr + αr 2) = α 2 (r + r 2) From the options, βγ = αr ⋅ αr 2 = α 2r3 = α 2 (r + r 2) [Q r 2 − r − 1 = 0 ⇒ r3 = r + r 2] ∴ α (β + γ ) = βγ 2

1 + a + b =0

5. Since, (x − r ) is a factor of the polynomial

Topic 2 Common Roots

and

b = 0, ± 3 i

4. Given equations are x2 + ax + b = 0 and

3 5 3  28 = ( p + q) + 2 + ( p − q)  2   2  and

x2 + bx − 1 = 0  have a common root. x2 + x + b = 0 

⇒

28 = p(3α + 2) + q(3β + 2)

∴

(a1c2 − a 2c1 )2 = (b1c2 − b2c1 ) (a1b2 − a 2b1 )

∴

∴ a12 = a11 + a10

61.

a1x2 + b1x + c1 = 0

3. If

2

2. Since, ax2 + bx + c = 0 has roots α and β. ⇒

α + β = − b /a αβ = c / a

and

Now, a x + abcx + c3 = 0 3 2

On dividing the equation by c2, we get a3 2 abcx c3 x + 2 + 2 =0 c c c2 2

⇒ ⇒ ⇒

 ax  ax a   + b   + c=0  c  c ax = α , β are the roots c c c x = α , β are the roots a a

⇒

x = α β α , α β β are the roots

⇒

x = α 2β , αβ 2 are the roots

…(i)

Theory of Equations 45 1 [4(2t − 6) + 4(t 2 − 9) + 1(6t 2 − 18t ] 2 1 = |[8t − 24 + 4t 2 − 36 + 6t 2 − 18t ]| 2

Divide the Eq. (i) by a3 , we get

=

3

x2 +

b c  c ⋅ ⋅ x+   =0  a a a

⇒

x2 − (α + β ) ⋅ (αβ ) x + (αβ )3 = 0

⇒

x2 − α 2βx − αβ 2 x + (αβ )3 = 0

⇒

x (x − α 2β ) − αβ 2 (x − α 2β ) = 0

⇒

= |5t 2 − 5t − 30| = |5(t + 2) (t − 3)|

(x − α 2β )(x − αβ 2) = 0

⇒ x = α β , αβ which is the required answer. 2

2

Alternate Solution a3 x2 + abcx + c3 = 0

Since, ⇒ x= ⇒ x= ⇒ x=

− abc ±

(abc)2 − 4 . a3 ⋅ c3

Now, as X is any point on the arc POQ of the parabola, therefore ordinate of point X, 2t ∈ (− 4, 6) ⇒ t ∈ (− 2, 3). ∴ Area of ∆PXQ = − 5(t + 2) (t − 3) = − 5t 2 + 5t + 30 [Q| x − a | = − (x − a ), if x < a] The maximum area (in square units) 25 − 4(− 5) (30)  125 =− = 4 4(− 5)   [Q Maximum value of quadratic expression D ] ax2 + bx + c, when a < 0 is − 4a

2 a3 − (b/a )(c/a ) ±

(b/a ) (c/a ) − 4(c/a ) 2 2

(α + β ) (α β ) ±

2

3

(α + β ) (α β ) − 4(α β ) 2 2

2

3

2. Let f (x) = (c − 5)x2 − 2 cx + (c − 4) = 0. Then, according to problem, the graph of y = f (x) will be either of the two ways, shown below.

(α + β )(αβ ) ± α β (α + β )2 − 4 αβ 2  (α + β ) ± (α − β )2  ⇒ x = αβ   2    (α + β ) ± (α − β )  ⇒ x = αβ   2 ⇒ x=

α + β + α − β α + β − α + β  , ⇒ x = αβ   2 2 2α 2β  , ⇒ x = αβ  2 2  ⇒

x = α 2 β , α β 2 which is the required answer.

Topic 4 Graph of Quadratic Expression

O

O

X (t 2,2t)

Q(9,6)

y 2=4x

2

f (0) f (2) < 0

Now, consider ⇒

(c − 4) [4 (c − 5) − 4c + (c − 4)] < 0

⇒

(c − 4) (c − 24) < 0

⇒

c ∈ (4, 24) +

– 4

⇒ P(4,–4) Y′

1 ∴ Area of ∆PXQ = 2

4 2

t 9

−4 1 2t 6

1 1

24

–

49/4

X

O

… (i)

+

Similarly, f (2) ⋅ f (3) < 0 ⇒ [4 (c − 5) − 4c + (c − 4)] [9(c − 5) − 6c + (c − 4)] < 0 ⇒ (c − 24) (4c − 49) < 0 +

X′

3

In both cases f (0). f (2) < 0 and f (2) f (3) < 0

1. Given parabola is y = 4x,

Y

3

Or

2

Since, X lies on the parabola, so let the coordinates of X be (t 2, 2t ). Thus, the coordinates of the vertices of the triangle PXQ are P (4,–4), X (t 2,2t ) and Q (9, 6).

2

+ 24

 49  c ∈ , 24 4 

From Eqs. (i) and (ii), we get  49  c ∈ , 24 4  ∴Integral values of c are 13, 14, ……, 23. Thus, 11 integral values of c are possible.

…(ii)

46 Theory of Equations 3. According to given information, we have the following

5. As we know, ax2 + bx + c > 0 for all x ∈ R, iff

graph

a > 0 and D < 0. Given equation is x2 + 2ax + (10 − 3a ) > 0, ∀ x ∈ R

Y

D 0 ⇒ b2 − 4ac > 0 ⇒ m2 − 4 × 1 × 4 > 0 ⇒ m2 − 16 > 0 ⇒ (m − 4) (m + 4) > 0 ⇒ m ∈ (− ∞ , − 4) ∪ (4, ∞ ) (ii) The vertex of the parabola should lie between x = 1and x = 5 b m − ∈ (1, 5) ⇒1 < < 5 ⇒ m∈ (2, 10) ∴ 2a 2 (iii) f (1) > 0 ⇒1 − m + 4 > 0 ⇒ m < 5 ⇒ m ∈ (−∞ , 5)

y = (x – a)(x – b) –1

6.

α a

b β 1

From graph, it is clear that one of the roots of (x − a )(x − b) − 1 = 0 lies in (− ∞ , a ) and other lies in (b, ∞ ).

7. Let f (x) = x2 − 2ax + a 2 + a − 3 Since, both root are less than 3.

(iv) f (5) > 0 ⇒ 25 − 5m + 4 > 0 29  ⇒ 5m < 29 ⇒ m ∈  − ∞,   5

⇒

α < 3, β < 3

⇒

From the values of m obtained in (i), (ii), (iii) and (iv), we get m ∈ (4, 5).

⇒

Sum, S = α + β < 6 α+β 0

∴ [Q{ x } > 0]

1 + 3a 2 < 2 ⇒ 1 + 3a 2 < 4 a 2 − 1 < 0 ⇒ (a + 1) (a − 1) < 0

∴ a ∈ (−1, 1), for no integer solution of a, we consider (−1, 0) ∪ (0, 1)

a + a −3 0, ∀ x ∈ R ⇒ and

a >0 b2 − 4ac < 0

…(i)

Theory of Equations 47 ∴ ⇒ ⇒

g (x) = f (x) + f ′ (x) + f ′ ′ (x) g (x) = ax2 + bx + c + 2ax + b + 2a g (x) = ax2 + x (b + 2a ) + (c + b + 2a )

11. Since, x2 − 3x + 2 > 0 and x2 − 2x − 4 ≤ 0 ⇒ ⇒

whose discriminant = (b + 2a )2 − 4a (c + b + 2a ) = b2 + 4a 2 + 4ab − 4ac − 4ab − 8a 2 [from Eq. (i)] = b2 − 4a 2 − 4ac = (b2 − 4ac) − 4a 2 < 0 ∴ g (x) > 0 ∀ x, as a > 0 and discriminant < 0. Thus, g (x) > 0, ∀ x ∈ R.

9. Given, x2 + (a − b) x + (1 − a − b) = 0 has real and unequal roots. ⇒ D >0 ⇒ (a − b)2 − 4(1) (1 − a − b) > 0 ⇒ a 2 + b2 − 2ab − 4 + 4a + 4b > 0 Now, to find the values of ‘a’ for which equation has unequal real roots for all values of b. i.e. Above equation is true for all b. or b2 + b(4 − 2a ) + (a 2 + 4a − 4) > 0, is true for all b. ∴ Discriminant, D < 0 ⇒

(4 − 2a )2 − 4 (a 2 + 4a − 4) < 0

⇒

16 − 16a + 4a − 4a − 16a + 16 < 0 2

⇒

2

−32a + 32 < 0

10.

⇒

a >1

Y a0 y = ax2 + bx + c α

–1

0 1

β

X

From figure, it is clear that, if a > 0, then f (−1) < 0 and f (1) < 0 and if a < 0, f (−1) > 0 and f (1) > 0. In both cases, af (−1) < 0 and af (1) < 0. ⇒

a (a − b + c) < 0

and

a (a + b + c) < 0

On dividing by a 2, we get b c 1 − + < 0 and a a On combining both, we get b c 1 ± + 0

x ∈ [1 − 5 , 1) ∪ [1 + 5 , 2)

(i) Given, x2 − 8kx + 16 (k2 − k + 1) = 0 Now, D = 64 { k2 − (k2 − k + 1)} = 64 (k − 1) > 0 k >1 b 8k (ii) − >4 ⇒ >4 ⇒ k >1 2a 2 (iii) f (4) ≥ 0 ⇒

16 − 32k + 16 (k2 − k + 1) ≥ 0

⇒

k2 − 3k + 2 ≥ 0

⇒

(k − 2) (k − 1) ≥ 0

⇒

k ≤1

or k ≥ 2 k =2

Hence,

Topic 5 Some Special Forms 1. Given equation 5 + |2x − 1| = 2x (2x − 2) Case I If 2x − 1 ≥ 0 ⇒ x ≥ 0 , then 5 + 2x − 1 = 2x (2x − 2) Put 2x = t, then 5 + t − 1 = t 2 − 2t ⇒ t 2 − 3t − 4 = 0 2 ⇒ t − 4t + t − 4 = 0 ⇒ t (t − 4) + 1(t − 4) = 0 ⇒ t = 4 or − 1 ⇒ t = 4 (Q t = 2x > 0) x ⇒ 2 =4⇒x=2 >0 ⇒ x = 2 is the solution. Case II If 2x − 1 < 0 ⇒ x < 0 , then 5 + 1 − 2x = 2x (2x − 2) Put 2x = y, then 6 − y = y2 − 2 y ⇒ y2 − y − 6 = 0 ⇒ y2 − 3 y + 2 y − 6 = 0 ⇒ ( y + 2) ( y − 3) = 0 ⇒ y = 3 or − 2 ⇒ y = 3(as y = 2x > 0) ⇒ 2x = 3 ⇒ x = log 2 3 > 0 So, x = log 2 3 is not a solution. Therefore, number of real roots is one.

2. Given, inequality is 2

sin 2 x − 2sin x + 5

⋅

1 sin 2 y

≤1

4 ⇒2

b c 1 + + 2) and (1 − 5 ≤ x ≤ 1 + 5 )

∴

12.

and

⇒2 ⇒

(sin x − 1 )2 + 4 2

(sin x − 1 ) + 4

⋅ 2−2sin

2

≤2

2

y

≤1

2sin y

(sin x − 1)2 + 4 ≤ 2 sin 2 y [if a > 1 and am ≤ a n ⇒ m ≤ n]

Q Range of (sin x − 1)2 + 4 is [2, 2 2 ] and range of 2 sin 2 y is [0, 2].

48 Theory of Equations ∴ By Rolle’s theorem, f ′ (α ) = 0 for 0 < α < 1

∴The above inequality holds, iff (sin x − 1)2 + 4 = 2 = 2 sin 2 y ⇒ sin x = 1 and sin 2 y = 1 ⇒ sin x = |sin y| 3.

[from the options]

Key Idea Reduce the given equation into quadratic equation.

Now,

f ′ (x) = 3ax2 + 2bx + c

⇒

f ′ (α ) = 3aα 2 + 2bα + c = 0

∴ Eq. (i) has exist atleast one root in the interval (0, 1). Thus, f ′ (x) must have root in the interval (0, 1) or 3ax2 + 2bx + c = 0 has root ∈ (0, 1).

7. Given, x12 − x9 + x4 − x + 1 > 0

Given equation is | x − 2| + x ( x − 4) + 2 = 0 ⇒ | x − 2| + x − 4 x + 4 = 2

Here, three cases arises: Case I When x ≤ 0 ⇒ x12 > 0, − x9 > 0, x4 > 0, − x > 0 …(i) ∴ x12 − x9 + x4 − x + 1 > 0, ∀ x ≤ 0

⇒ | x − 2| + ( x − 2)2 = 2 ⇒ (| x − 2|)2 + | x − 2| − 2 = 0 Let| x − 2| = y, then above equation reduced to y2 + y − 2 = 0 ⇒ y2 + 2 y − y − 2 = 0 ⇒ y( y + 2) − 1( y + 2) = 0 ⇒ ( y + 2)( y − 1) = 0 ⇒ y = 1, − 2 [Q y = | x − 2| ≥ 0] ∴ y=1 ⇒ | x − 2| = 1 x −2 = ±1 ⇒ x = 3 or 1 ⇒ ⇒ x = 9 or 1 ∴ Sum of roots = 9 + 1 = 10

4. Let f (x) = 2x3 + 3x + k

Case II When 0 < x ≤ 1 x9 < x4 and x < 1 ⇒ − x9 + x4 > 0 ∴

9

4

f ′ (x) = 6x2 + 3 > 0, ∀ x ∈ R ⇒ f (x) = 0 has only one real root, so two roots are not possible.

and β is a root of

a x − bx − c = 0

⇒

a 2β 2 − bβ − c = 0 f (x) = a 2x2 + 2bx + 2c

∴

f (α ) = a 2α 2 + 2bα + 2c

... (i)

8. Consider, 1

f (x) = ∫ (1 + cos 8 x)(ax2 + bx + c) dx 0

f (1) = f (2)

... (ii)

f ′ (x) = (1 + cos 8 x)(ax2 + bx + c)

Now,

f (β ) = α 2β 2 + 2bβ + 2c f (α ) f (β ) < 0

f (x) must have a root lying in the open interval (α , β ). ∴ α < γ 1 ⇒ x > x and x > x x12 − x9 + x4 − x + 1 > 0, ∀ x > 1 12

Obviously, f (x) is continuous and differentiable in the interval [1, 2].

On differentiating w.r.t. x, we get

2 2

and 1 − x > 0

x − x + x − x + 1 > 0, ∀ 0 < x ≤ 1 12

k ∈ (1, 2)

9. Given, x1 and x2 are roots of αx2 − x + α = 0. ∴ Also, ⇒ ⇒ or ⇒ ⇒

1 and x1x2 = 1 α x1 − x2 < 1 |x1 − x2|2 < 1 (x1 − x2)2 < 1 (x1 + x2)2 − 4x1x2 < 1 1 1 − 4 < 1 or 0 or ( 5 α − 1) ( 5 α + 1) > 0 + –1/√5

∴

–

+ 1/√5

1  1   α ∈  −∞ , − , ∞  ∪  5  5 

…(i)

Theory of Equations 49 D >0 1 − 4α 2 > 0  1 1 α ∈ − ,   2 2

Also, ⇒ or

1  3 f (x) changes its sign in  − , −   4 2

From Eqs. (i) and (ii), we get  1 −1   1 1  ,  α ∈ − ,  ∪   2 5   5 2

10.

 3 1 Hence, f (x) = 0 has a root in  − , −  .  4 2

…(ii)

PLAN (i) Concepts of curve tracing are used in this question.

12.

1/ 2

∫0

t

3/ 4

0

0

f (x) dx < ∫ f (x) dx < ∫

f (x) dx

Now, ∫ f (x) dx = ∫ (1 + 2x + 3x2 + 4x3 ) dx = x + x2 + x3 + x4

(ii) Number of roots are taken out from the curve traced.

Let y = x5 − 5x (i) As x → ∞, y → ∞ and as x → − ∞, y → − ∞ (ii) Also, at x = 0, y = 0, thus the curve passes through the origin. dy (iii) = 5x4 − 5 = 5 (x4 − 1) = 5 (x2 − 1) (x2 + 1) dx = 5 (x − 1) (x + 1) (x2 + 1) +

–

+ –1

1

dy > 0 in (− ∞ , − 1) ∪ (1, ∞ ), thus f (x) is dx increasing in these intervals. dy Also, < 0 in (− 1, 1), thus decreasing in (− 1, 1). dx (iv) Also, at x = − 1, dy /dx changes its sign from + ve to –ve. ∴ x = − 1 is point of local maxima. Similarly, x = 1 is point of local minima. Local maximum value, y = (− 1)5 − 5 (−1) = 4 Local minimum value, y = (1)5 − 5(1) = − 4

1 1 t 3 2 4

f (x) dx =

15 3 > , 16 4

3/ 4

∫0

f (x) dx =

1 and 4 1 f ′ (x) < 0, when x < − . 4 f ′ (x) > 0, when x > −

∴ It could be shown as 1

S –3 4

–1 2

1 2

3 4

14. Let y = x intersect the curve y = kex at exactly one point when k ≤ 0 .

–1

Y

(1, – 4)

Now, let y = − a As evident from the graph, if − a ∈ (− 4, 4) i.e. a ∈ (− 4, + 4) Then, f (x) has three real roots and if − a > 4 or − a < − 4, then f (x) has one real root. i.e. for a < − 4 or a > 4, f (x) has one real root.

11. Given, f (x) = 4x3 + 3x2 + 2x + 1 D = 9 − 24 < 0

Hence, f (x) = 0 has only one real root. 3 4  1 f −  = 1 −1 + − > 0  2 4 8 6 27 108  3 − f −  = 1 − +  4 4 16 64 64 − 96 + 108 − 108 = 0) ln k < −1 ⇒ k < Hence,

1 e

∴ f (x) has atleast one root in

 1 k ∈ 0,   e

Now, f ′ (x) = 12x2 − 3 = 3 (2x − 1) (2x + 1) 3 1  1 1  =  x −   x +  > 0 in ,1 2  4 2  2

17. Let f (x) = (x − a ) (x − c) + 2 (x − b) (x − d ) Here,

⇒ f (x) is an increasing function in [1 /2, 1]

f (a ) = + ve

Therefore, f (x) has exactly one root in [1 /2, 1] for any p ∈ [−1, 1].

f (b) = − ve f (c) = − ve

Now, let x = cos θ

f (d ) = + ve ∴ There exists two real and distinct roots one in the interval (a , b) and other in (c, d ). Hence, statement is true.

18. Let f (x) = 4x3 − 3x − p

…(i) 3

Now,

4 3  1  1  1 f   =4   −3   − p= − − p  2  2  2 8 2 = − (1 + p) f (1) = 4(1)3 − 3(1) − p = 1 − p

⇒

1  ,1 . 2 

 1 f   . f (1) = − (1 + p)(1 − p)  2 = ( p + 1)( p − 1) = p − 1 2

∴

x∈

1 , 2

1

 

 π ⇒ θ ∈ 0,  3 

From Eq. (i), 4 cos3 θ − 3 cos θ = p ⇒ ⇒ ⇒ ⇒ ⇒

cos 3θ = p 3θ = cos −1 p 1 θ = cos −1 p 3 1  cos θ = cos  cos −1 p 3   1 x = cos  cos −1 p  3

3 Sequences and Series Topic 1 Arithmetic Progression (AP) Objective Questions I (Only one correct option) 1. If a1 , a 2, a3 , ... , a n are in AP and a1 + a 4 + a7 + ... + a16 = 114 , then a1 + a 6 + a11 + a16 is equal to

(2019 Main, 10 April I)

(a) 64 (c) 98

(b) 76 (d) 38

Analytical and Descriptive Question 5. If a1 , a 2 ..... , a n are in arithmetic progression, where ai > 0, ∀ i, then show that 1 1 + + ... a1 + a 2 a 2 + a3 +

2. If 19th term of a non-zero AP is zero, then its (49th term) : (29th term) is (a) 1 : 3 (c) 2 : 1

(2019 Main, 11 Jan II)

(b) 4 : 1 (d) 3 : 1

3. For any three positive real numbers a , b and c, if 9 (25a 2 + b2) + 25 (c2 − 3ac) = 15b (3a + c), then (2017 Main) (a) b, c and a are in GP (b) b, c and a are in AP (c) a , b and c are in AP (d) a , b and c are in GP

1 an − 1 +

an

=

n −1 a1 + a n (1982, 2M)

True/False 6. n1 , n2, K , n p are p positive integers, whose sum is an even number, then the number of odd integers among them is odd. (1985, 1M)

Integer & Numerical Answer Type Questions

4. If Tr is the r th term of an AP, for r = 1, 2, 3, .... . If for

7. Let AP (a ; d ) denote the set of all the terms of an

1 some positive integers m and n, we have Tm = and n 1 Tn = , then Tmn equals m (1998, 2M)

infinite arithmetic progression with first term a and common difference d > 0. If

1 mn (c) 1

(a)

1 1 + m n (d) 0 (b)

AP (1 ; 3) ∩ AP (2 ; 5) ∩ AP (3 ; 7) = AP (a ; d ) then a + d equals ................

(2019 Adv.)

8. The sides of a right angled triangle are in arithmetic progression. If the triangle has area 24, then what is the length of its smallest side? (2017 Adv.)

Topic 2 Sum of n Terms of an AP Objective Questions I (Only one correct option) 1. If a1 , a 2, a3 , ... are in AP such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this AP is (2019 Main, 12 April II)

(a) 200 (c) 120

(b) 280 (d) 150

S 4 = 16 and S 6 = − 48, then S10 is equal to (2019 Main, 12 April I)

(b) − 410 (d) − 380

sum of the series 1   1  1 + + − − −  3   3 100 

2   1 +…+ − −  3 100 

 1 99  is − −  3 100 

(2019 Main, 12 April I)

2. Let S n denote the sum of the first n terms of an AP. If (a) − 260 (c) − 320

3. For x ∈ R, let [x] denote the greatest integer ≤ x, then the

(a) − 153 (c) − 131

(b) − 133 (d) − 135

4. If the sum and product of the first three terms in an AP are 33 and 1155, respectively, then a value of its 11th term is (2019 Main, 9 April II) (a) 25

(b) –36

(c) –25

(d) –35

52 Sequences and Series 5. Let the sum of the first n terms of a non-constant AP n (n − 7) A, where A is a constant. a1 , a 2, a3 .....be 50n + 2 If d is the common difference of this AP, then the ordered pair (d , a50 ) is equal to (2019 Main, 9 April I)

(a) (A, 50 + 46A) (c) (50, 50 + 46A)

(b) (50, 50 + 45A) (d) (A, 50 + 45A)

6. The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is (2019 Main, 10 Jan I)

(a) 1256

(b) 1465

(c) 1356

(d) 1365

7. Let a1 , a 2, ..... a30 be an AP, S = ∑ ai and

∑ a( 2 i − 1). If a5 = 27 and S − 2T = 75,

i =1

(2019 Main, 9 Jan I)

(a) 42 (c) 52

(b) 57 (d) 47

... , log e b101 are in AP with the common difference log e 2 . Suppose a1 , a 2, ... , a101 are in AP, such that a1 = b1 and If and a51 = b51. t = b1 + b2 + ... + b51 (2016 Adv.) s = a1 + a 2 + ... + a51, then (a) s > t and a101 > b101

(b) s > t and a101 < b101

(c) s < t and a101 > b101

(d) s < t and a101 < b101

(a) Q1 , Q2 , Q3 ,... are in an AP with common difference 5 (b) Q1 , Q2 , Q3 ,... are in an AP with common difference 6 (c) Q1 , Q2 , Q3 ,... are in an AP with common difference 11 (d) Q1 = Q2 = Q3 = ...

of squares of these n terms is

15. Let p and q be the roots of the equation x2 − 2x + A = 0 and let r and s be the roots of the equation x2 − 18x + B = 0. If p < q < r < s are in arithmetic (1997, 2M) progression, then A = … and B = … .

16. The sum of the first n terms of the series n (n + 1)2 , when 2 n is even. When n is odd, the sum is .... . (1988, 2M)

12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + K is

9. If the sum of first n terms of an AP is cn 2, then the sum (2009)

n (4n 2 + 1) c2 (b) 3 n (4n 2 + 1) c2 (d) 6

17. The sum of integers from 1 to 100 that are divisible by 2 or 5 is ……

(1984, 2M)

Analytical & Descriptive Questions 18. The fourth power of the common difference of an

10. If the sum of the first 2n terms of the AP series 2,5,8,..., is equal to the sum of the first n terms of the AP series 57, 59, 61,..., then n equals (2001, 1M) (a) 10 (c) 11

(b) an even number (d) a composite number

Fill in the Blanks

8. Let bi > 1 for i = 1, 2, ... , 101 . Suppose log e b1 , log e b2,

n (4n 2 − 1) c2 (a) 6 n (4n 2 − 1) c2 (c) 3

13. Tr is always

14. Which one of the following is a correct statement ?

i =1

15

then a10 is equal to

1 n (n + 1) (3n 2 − n + 1) 12 1 (b) n (n + 1) (3n 2 + n + 2) 12 1 (c) n (2n 2 − n + 1) 2 1 (d) (2n3 − 2n + 3) 3 (a)

(a) an odd number (c) a prime number

30

T=

12. The sum V1 + V 2 + ... + V n is

arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that resulting sum is the square of an integer.(2000, 4M)

(b) 12 (d) 13

19. The real numbers x1 , x2, x3 satisfying the equation

Objective Question II (One or more than one correct option)

20. The interior angles of a polygon are in arithmetic

4n

11. If S n = ∑

k( k + 1 ) (−1) 2 k2. Then, S

n

x3 − x2 + βx + γ = 0 are in AP. Find the intervals in (1996, 3M) which β and γ lie.

can take value(s) (2013 Adv.)

2

(a) 1056 (c) 1120

(b) 1088 (d) 1332

Passage Based Problems Read the following passage and answer the questions. Passage Let V r denotes the sum of the first r terms of an arithmetic progression (AP) whose first term is r and the common difference is (2r − 1). Let Tr = V r + 1 − V r and (2007, 8M) Qr = Tr + 1 − Tr for r = 1, 2, . . .

progression. The smallest angle is 120° and the common difference is 5°. Find the number of sides of the polygon. (1980, 3M)

Integer & Numerical Answer Type Questions 21. Suppose that all the terms of an arithmetic progression are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in between 130 and 140, then the common difference of this AP is (2015 Adv.)

22. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k − 20 is equal to (2013 Adv.)

Sequences and Series 53 23. Let a1 , a 2, a3 , K , a100 be an arithmetic progression with p

a1 = 3 and S p =

∑

24. Let a1 , a 2, a3 , ... , a11 be real numbers satisfying a1 = 15, 27 − 2a 2 > 0 and a k = 2a k − 1 − a k − 2 for k = 3, 4, ... , 11. 2 a 2 + a 22 + K + a11 If 1 = 90, then the value of 11 a1 + a 2 + K + a11 (2010) is…… 11

ai , 1 ≤ p ≤ 100. For any integer n with

i =1

1 ≤ n ≤ 20, let m = 5n. If

Sm does not depend on n, then a 2 Sn

is equal to ……

(2011)

Topic 3 Geometric Progression (GP) Objective Questions I (Only one correct option) 1. Let a , b and c be in GP with common ratio r, where a ≠ 0 1 and 0 < r ≤ . If 3a, 7b and 15care the first three terms of 2 an AP, then the 4th term of this AP is (2019 Main, 10 April II)

(a) 5a

2 (b) a 3

7 (d) a 3

(c) a

2. If three distinct numbers a , b and c are in GP and the equations ax + 2bx + c = 0 and dx + 2ex + f = 0 have a common root, then which one of the following (2019 Main, 8 April II) statements is correct? 2

2

(a) d , e and f are in GP (c) d , e and f are in AP

d e f , and are in AP a b c d e f (d) , and are in GP a b c

(b)

3. The product of three consecutive terms of a GP is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an AP. Then, the sum of the original three terms of the given GP is (2019 Main, 12 Jan I)

(a) 36

(b) 28

(c) 32

4. Let a1 , a 2, .... , a10 be a GP. If a9 equals a5 (a) 53

(d) 24

(c) 4(52 )

(d) 54

5. Let a , b and c be the 7th, 11th and 13th

terms respectively of a non-constant AP. If these are also the a three consecutive terms of a GP, then is equal to c (2019 Main, 9 Jan II)

(a) 2

(b)

7 13

(c) 4

(d)

1 2

6. If a , b and c be three distinct real numbers in GP and a + b + c = xb, then x cannot be (a) 4

(b) 2

(c) −2

(2019 Main, 9 Jan I)

(d) −3

7. If the 2nd, 5th and 9th terms of a non-constant AP are in GP, then the common ratio of this GP is 8 (a) 5

4 (b) 3

(c) 1

α 2 + β 2 and α 3 + β3 are in GP, then (a) ∆ ≠ 0

(b) b∆ = 0

(c) c∆ = 0

(2005, 1M)

(d) bc ≠ 0

9. Let a , b, c be in an AP and a 2, b2, c2 be in GP. If a < b < c 3 and a + b + c = , then the value of a is 2 (a)

1 2 2

(b)

1 2 3

(c)

1 1 − 2 3

(2002, 1M)

(d)

(2016 Main)

7 (d) 4

1 1 − 2 2

10. Let α , β be the roots of x2 − x + p = 0 and γ, δ be the roots of x2 − 4x + q = 0. If α , β , γ, δ are in GP, then the integer (2001, 1M) values of p and q respectively are (a) − 2, − 32

(b) − 2,3

(c) − 6, 3

(d) − 6, − 32

11. If a , b, c, d and p are distinct real numbers such that (a 2 + b2 + c2) p2 − 2 (ab + bc + cd ) p + (b2 + c2 + d 2) ≤ 0, then a , b, c, d (a) are in AP (c) are in HP

(b) are in GP (d) satisfy ab = cd

(1987, 2M)

12. If a , b, c are in GP, then the equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root, if in

a3 = 25, then a1 (2019 Main, 11 Jan I)

(b) 2(52 )

8. Let f (x) = ax2 + bx + c, a ≠ 0 and ∆ = b2 − 4ac. If α + β,

(a) AP (c) HP

d e f , , are a b c (1985, 2M)

(b) GP (d) None of these

13. The third term of a geometric progression is 4. The product of the first five terms is (a) 43 (c) 44

(1982, 2M)

(b) 45 (d) None of these

Analytical & Descriptive Questions 14. Find three numbers a , b, c between 2 and 18 such that (i) their sum is 25. (ii) the numbers 2, a , b are consecutive terms of an AP. (iii) the numbers b, c, 18 are (1983, 2M) consecutive terms of a GP.

15. Does there exist a geometric progression containing 27,8 and 12 as three of its term? If it exists, then how many such progressions are possible? (1982, 2M)

16. If the mth, nth and pth terms of an AP and GP are equal and are x, y, z, then prove that xy − z ⋅ yz − x ⋅ z x − y = 1.

(1979, 3M)

54 Sequences and Series

Topic 4 Sum of n Terms & Infinite Terms of a GP Objective Questions I (Only one correct option) 1. If|x|< 1,| y|< 1 and x ≠ y, then the sum to infinity of the following series (x + y) + (x2 + xy + y2) (2020 Main, 2 Sep I) + (x3 + x2y + xy2 + y3 ) + … is

x + y + xy (1 + x) (1 + y) x + y + xy (c) (1 − x) (1 − y)

x + y − xy (1 − x) (1 − y) x + y − xy (d) (1 + x) (1 + y)

(a)

20

2. The sum

(2019 Main, 8 April II)

k =1

(a) 2 − (c) 2 −

(b) 1 −

19

2 3

(d) 2 −

217

11

∞

(a)

n

2

n

(b) 202 (d) 299

terms is 3 and the sum of the cubes of its terms is

27 . 19

Then, the common ratio of this series is (2019 Main, 11 Jan I)

4 9

(b)

2 3

(c)

2 9

(d)

1 3

2 3

(2014 Main)

6. If (10)9 + 2(11)1 (10)8 + 3(11)2(10)7 + ... + 10(11)9 = k(10)9, then k is equal to (a)

121 10

(b)

441 100

(b) bn = α n + β n for all n ≥ 1 (c) a1 + a2 + a3 + K + an = an + ∞ 10 a (d) ∑ nn = 89 n = 1 10

2

− 1for all n ≥ 1

(2014 Main)

(c) 100

length of a side of S n equals the length of a diagonal of S n + 1. If the length of a side of S1 is 10 cm, then for which of the following values of n is the area of S n less than (1999, 3M) 1 sq cm? (a) 7

(b) 8

(c) 9

2

middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is (b) 3 + (d) 2 +

(2019 Adv.)

n =1

(d) 10

Analytical & Descriptive Questions

5. Three positive numbers form an increasing GP. If the (a) 2 + 3 (c) 2 − 3

8

b

∑ 10nn = 89

12. Let S1 , S 2, ... be squares such that for each n ≥ 1 the

4. The sum of an infinite geometric series with positive

(a)

all positive integers n, define α n − βn , n ≥ 1, an = α −β Then which of the following options is/are correct?

220

 q + 1  q + 1  q + 1 Tn = 1 +   , where q is a  +K+  +  2   2   2  real number and q ≠ 1. If 101 C1 + 101C 2 ⋅ S1 + K + 101C101 ⋅ S100 = αT100, then α is (2019 Main, 11 Jan II) equal to (a) 2100 (c) 200

(1988, 2M)

(c) n + 2 − n − 1 (d) 2n + 1

b1 = 1 and bn = a n − 1 + a n + 1, n ≥ 2

220 21

3. Let S n = 1 + q + q + K + q and 2

is equal to (a) 2n − n − 1 (b) 1 − 2−n

11. Let α and β be the roots of x2 − x − 1 = 0, with α > β. For

1

11

3 7 15 + + + ... 4 8 16

Objective Question II (One or more than one correct option)

(b)

∑ k 2k is equal to

1 2

10. Sum of the first n terms of the series +

(d) 110

3

n

3 3 3 3 13. Let An =   −   +   + ... + (− 1)n − 1   ,         4

4

4

4

Bn = 1 − An. Find a least odd natural number n0 , so that (2006, 6M) Bn > An , ∀ n ≥ n0.

14. If S1 , S 2, S3 , ... , S n are the sums of infinite geometric series, whose first terms are 1, 2, 3,..., n and whose 1 1 1 1 common ratios are , , ,... , respectively, then 2 3 4 n+1 find the values of S12 + S 22 + S32 + ... + S 22n − 1. (1991, 4M)

7. The sum of first 20 terms of the sequence

15. The sum of the squares of three distinct real numbers,

(2013 Main)

which are in GP, is S 2. If their sum is a S, then show that 1  a 2 ∈  , 1 ∪ (1, 3) 3  (1986, 5M)

0.7, 0.77, 0.777,… , is 7 (a) (179 − 10− 20 ) 81 7 (c) (179 + 10− 20 ) 81

7 (b) (99 − 10− 20 ) 9 7 (d) (99 + 10− 20 ) 9

8. An infinite GP has first term x and sum 5, then x belongs to (a) x < − 10

(2004, 1M)

(b) −10 < x < 0 (c) 0 < x < 10

(d) x > 10

Integer & Numerical Answer Type Question 16. Let S k , where k = 1, 2, , K , 100, denotes the sum of the infinite geometric series whose first term is

9. Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3 /4, then (2000, 2M) (a) a = 4/7, r = 3/7 (c) a = 3/2, r = 1/2

(b) a = 2, r = 3 /8 (d) a = 3, r = 1/4

the

common 2

100 + 100 !

ratio

is

1 . k

Then,

the

k −1 and k!

value

of

100

∑|(k2 − 3k + 1) Sk |is ……

k =1

(2010)

Sequences and Series 55

Topic 5 Harmonic Progression (HP) Objective Questions I (Only one correct option)

5. Suppose four distinct positive numbers a1 , a 2, a3 , a 4 are in GP. Let b1 = a1 , b2 = b1 + a 2, b3 = b2 + a3 and b4 = b3 + a 4. Statement I The numbers b1 , b2, b3 , b4 are neither in AP nor in GP.

1. If a1 , a 2, a3 ,… are in a harmonic progression with a1 = 5 and a 20 = 25. Then, the least positive integer n for which (2012) a n < 0, is (a) 22 (c) 24

(b) 23 (d) 25

Statement II The numbers b1 , b2, b3 , b4 are in HP. (2008, 3M)

2. If the positive numbers a , b, c,d are in AP. Then, abc, abd , acd , bcd are

(2001, 1M)

(a) not in AP / GP / HP (c) in GP

(b) in AP (d) in HP

Fill in the Blank 6. If cos (x − y), cos x and cos (x + y)’ are in HP. Then  y cos x ⋅ sec   = K .  2

3. Let a1 , a 2, ..., a10 be in AP and h1 , h2, equal to ..... , h10 be in HP. If a1 = h1 = 2 and a10 = h10 = 3, then a 4h7 is (1999, 2M)

(a) 2

(b) 3

(c) 5

(d) 6

4. If x > 1, y > 1, z > 1 are in GP, then 1 are in 1 + ln z

1 1 , , 1 + ln x 1 + ln y

(1997C, 2M)

Analytical & Descriptive Questions 7. If a , b, c are in AP, a 2, b2, c2 are in HP, then prove that either a = b = c or a , b, −

(1998, 2M)

(a) AP (c) GP

(b) HP (d) None of these

(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

(2003, 4M)

8. Let a and b be positive real numbers. If a , A1 , A2, b are in arithmetic progression, a , G1 , G2, b are in geometric progression and a , H 1 , H 2, B are in harmonic progression, then show that G1G2 A + A2 (2a + b)(a + 2b) = 1 = 9ab H 1H 2 H 1 + H 2 (2002, 5M)

Assertion and Reason For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows:

c form a GP. 2

9.

(i) The value of x + y + z is 15. If a , x, y, z , b are in AP 1 1 1 5 while the value of + + is . If a , x, y, z , b are in x y z 3 HP, then find a and b. (ii) If x, y, z are in HP, then show that log (x + z ) + log (x + z − 2 y) = 2 log (x − z ). (1978, 3M)

Topic 6 Relation between AM, GM, HM and Some Special Series Objective Questions I (Only one correct option) 1 +2 1 +2 +3 + + ... 1+2 1+2+3 13 + 23 + 33 + K + 153 1 + − (1 + 2 + 3 + K + 15) is 1 + 2 + 3 + K + 15 2 (2019 Main, 10 April II) equal to

1. The sum of series 1 +

(a) 620

(b) 660

3

3

3

(c) 1240

3

3

(d) 1860

3 × 13 5 × (13 + 23 ) + 12 12 + 22 3 3 3 7 × (1 + 2 + 3 ) + .......... + upto 10th term, is + 12 + 22 + 32 (2019 Main, 10 April I)

2. The sum of series

(a) 680

(b) 600

(c) 660

(d) 620

3. The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +... upto 11th term is (a) 915

(2019 Main, 9 April II)

(b) 946

(c) 916

(d) 945

4. If the sum of the first 15 terms of the series 3

3

3

3

 3  1  1  3 3   + 1  +  2  + 3 +  3  + ...  4  4  2  4 is equal to 225 k, then k is equal to (2019 Main, 12 Jan II)

(a) 108

(b) 27

(c) 54

(d) 9

5 1 + 2 + 3 + ... + k 2 5. Let S k = . If S12 + S 22 + ... + S10 A, = k 12 (2019 Main, 12 Jan I) then A is equal to (a) 156

(b) 301

(c) 283

(d) 303

6. Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression xm yn is 2m (1 + x ) (1 + y2n ) (2019 Main, 11 Jan II) (a)

1 2

(b) 1

(c)

1 4

(d)

m+ n 6mn

56 Sequences and Series 18. If a , b and c are distinct positive numbers, then the

7. The sum of the following series

expression (b + c − a ) (c + a − b) (a + b − c) − abc is

9 (1 + 2 + 3 ) 12 (1 + 2 + 3 + 4 ) + 7 9 2 2 2 15 (1 + 2 + ... + 5 ) + + ... up to 15 terms is 11 2

1+6+

2

2

2

2

2

2

(a) positive (c) non-positive

(2019 Main, 9 Jan II)

(a) 7510

(b) 7820

(c) 7830

(d) 7520

∑ a 4k + 1 = 416

k=0

2 and a 9 + a 43 = 66. If a12 + a 22 + … + a17 = 140 m, then m is equal to (2018 Main)

(a) 66

(b) 68

(c) 34

(d) 33

9. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + … If B − 2 A = 100λ, then λ is equal to (a) 232

(b) 248

(c) 464

(2018 Main)

(d) 496

10. If the sum of the first ten terms of the series 2

2

2

2

16  4  3  2  1 2 m, then 1  + 2  + 3  + 4 + 4  + K, is  5  5  5  5 5 (2016 Main) m is equal to (a) 102

(b) 101

(c) 100

(d) 99

11. If m is the AM of two distinct real numbers l and

n (l, n > 1) and G1 , G2 and G3 are three geometric means between l and n, then G14 + 2G24 + G34 equals (2015)

(a) 4l2mn

(b) 4lm2n

(c) lmn 2

12. The

sum of first 9 terms 13 13 + 23 13 + 23 + 33 + + + ... is 1 1+3 1+3+5

(a) 71

(b) 96

13. If α ∈ 0, 

π  , then 2

(c) 142

x2 + x +

(d) l2m2n 2

of

the

(2015)

(d) 192

tan α 2

x2 + x

is always greater

than or equal to (a) 2 tan α

(b) 1

series

(2003, 2M)

(c) 2

(d) sec2 α

14. If a1 , a 2,... , a n are positive real numbers whose product is a fixed number c, then the minimum value of (2002, 1M) a1 + a 2 + ... + a n − 1 + 2a n is (b) (n + 1)c1/ n (d) (n + 1) (2c)1/ n

(a) n (2c)1/ n (c) 2nc1/ n

a + b + c + d = 2 , then M = (a + b) (c + d ) satisfies the (2000, 2M) relation (a) 0 < M ≤ 1

(b) 1 ≤ M ≤ 2

(c) 2 ≤ M ≤ 3

(d) 3 ≤ M ≤ 4

16. The harmonic mean of the roots of the equation (5 + 2 ) x2 − (4 + 5 ) x + 8 + 2 5 = 0 is (a) 2 (c) 6

(1999, 2M)

(b) 4 (d) 8

17. The product of n positive numbers is unity, then their sum is (a) a positive integer 1 (c) equal to n + n

integer, then   (a) n ∑ xi2 <  ∑ xi    i =1  i =1 n

n

(c) n ∑

i =1

xi2

(1991, 2M)

(b) divisible by n (d) never less than n

(1982, 1M) 2

 n  ≥ n  ∑ xi    i =1 

  (b) n ∑ xi2 ≥  ∑ xi    i =1  i =1 n

n

2

2

(d) None of these

Passage Based Problems Passage Let A1 , G1 , H 1 denote the arithmetic, geometric and harmonic means, respectively, of two distinct positive numbers. For n ≥ 2, let An − 1 and H n − 1 has arithmetic, geometric and harmonic means as An , Gn , H n , respectively. (2007, 8M)

20. Which one of the following statements is correct? (a) G1 > G2 > G3 > ... (b) G1 < G2 < G3 < ... (c) G1 = G2 = G3 = ... (d) G1 < G3 < G5 < ... and G2 > G4 > G6 >...

21. Which of the following statements is correct? (a) A1 > A2 > A3 >... (b) A1 < A2 < A3 A3 > A5 >... and A2 < A4 < A6 A6 >...

22. Which of the following statements is correct ? (a) H1 > H 2 > H3 >... (b) H1 < H 2 < H3 H3 > H5 >... and H 2 < H 4 < H 6 H 6 >...

Objective Questions II (One or more than one correct option) 23. For a positive integer n let a (n ) = 1 +

15. If a , b, c are positive real numbers such that

(1991, 2M)

19. If x1 , x2,... , xn are any real numbers and n is any positive n

12

8. Let a1 , a 2, a3 , …, a 49 be in AP such that

(b) negative (d) non-negative

1 1 1 1 + + + ... + n , then 2 3 4 (2 ) − 1

(a) a (100) ≤ 100 (c) a (200)≤ 100

(1999, 3M)

(b) a (100) > 100 (d) a (200) > 100

24. If the first and the (2n − 1)th term of an AP, GP and HP are equal and their nth terms are a , b and c (1988, 2M) respectively, then (a) a = b = c (c) a + c = b

(b) a ≥ b ≥ c (d) ac − b2 = 0

Fill in the Blanks 25. If x be is the arithmetic mean and y, z be two geometric means between any two positive numbers, then y3 + z3 = ... xyz (1997C, 2M)

Sequences and Series 57 26. If the harmonic mean and geometric mean of two

32. If a > 0, b > 0 and c > 0, then prove that

positive numbers are in the ratio 4 : 5. Then, the two numbers are in the ratio … . (1992, 2M)

True/False

(a + b + c)

(1984, 2M)

Integer & Numerical Answer Type Questions

27. If x and y are positive real numbers and m, n are any

33. Let

m be the minimum possible value of log3 (3y1 + 3y2 + 3y3 ), where y1 , y2, y3 are real numbers for which y1 + y2 + y3 = 9. Let M be the maximum possible value of (log3 x1 + log3 x2 + log3 x3 ), where x1 , x2, x3 are positive real numbers for which x1 + x2 + x3 = 9. Then the (2020 Adv.) value of log 2(m3 ) + log3 (M 2) is ……… .

n m

positive

1 1 1 + + ≥9 a b c

integers,

1 x y > . (1 + x2n )(1 + y2m ) 4

then

(1989, 1M)

28. For 0 < a < x, the minimum value of function log a x + log x a is 2.

34. Let a1 , a 2, a3 , .... be a sequence of positive integers in arithmetic progression with common difference 2. Also, let b1 , b2, b3 , .... be a sequence of positive integers in geometric progression with common ratio 2. If a1 = b1 = c, then the number of all possible values of c, for which the equality

Analytical & Descriptive Questions 29. If a , b, c are positive real numbers, then prove that {(1 + a ) (1 + b) (1 + c)}7 > 77 a 4b4c4

(2004, 4M)

2(a1 + a 2 + K + a n ) = b1 + b2 + K + bn holds for some positive integer n, is ……

30. Let a1 , a 2,.. be positive real numbers in geometric

(2020 Adv.)

progression. For each n, if An , Gn , H n are respectively, the arithmetic mean, geometric mean and harmonic mean of a1 , a 2, .... , a n. Then, find an expression for the geometric mean of G1 , G2, ... , Gn in terms of (2001, 5M) A1 , A2, ... , An , H 1 , H 2, ... , H n.

35. If m arithmetic means (AMs) and three geometric means

31. If p is the first of the n arithmetic means between two

If a,b,c are in geometric progression and the arithmetic a 2 + a − 14 mean of a,b,c is b + 2, then the value of is a+1

numbers and q be the first on n harmonic means between the same numbers. Then, show that q does not 2  n + 1 lie between p and  (1991, 4M)  p.  n − 1

(GMs) are inserted between 3 and 243 such that 4th AM is equal to 2nd GM, then m is equal to [2020 Main, 3 Sep II]

36. Let a,b,c be positive integers such that b /a is an integer.

(2014 Adv.)

37. The minimum value of the sum of real numbers a − 5 , a − 4 , 3a − 3 , 1, a 8 and a10 with a > 0 is ……

(2011)

Answers Topic 1 1. (b) 6. False

Topic 4 2. (d) 7. (157.00)

3. (b) 8. (6)

4. (c)

(a) 2. (a) 6. (c) 10. (d) 14. 2 n (n + 1 )  16.   17. 2   1  19. β ∈ – ∞, and γ  3  22. (5) 23.

(c) (c) (c) (b)

3. 7. 11. 15.

(b) 4. (c) (c) 8. (b) (a, d) 12. (b) A = – 3 , B = 77

3050  1  ∈ – , ∞ 20. (9)  27  (9) 24. (0)

(c) (c) (d) (b)

3. (a)

4. (b)

7. (c)

8. (c)

2. (b) 6. (b) 10. (a)

14. (a = 5 ) (b = 8 ) (c = 12 )

3. (b) 7. (b) 11. (b) 15. Yes, infinite

13. (7)

10. (c) 11. (b, c, d) 1 14. (2n )(2n + 1 )( 4n + 1 ) – 1 6

12. (b,c,d) 16. (4)

Topic 5 1. (d) 5. (c)

2. (d) 6. ± 2

3. (d) 4. (b) 9. (i) a = 1, b = 9

Topic 6 21. (9)

Topic 3 1. 5. 9. 13.

2. (a) 6. (c)

9. (d)

Topic 2 1. 5. 9. 13.

1. (b) 5. (d)

4. (d) 8. (c) 12. (a)

1. 5. 9. 13. 17. 21. 25. 33.

(a) (d) (b) (a) (d) (a) 2 (8)

36. (4)

2. 6. 10. 14. 18. 22. 26. 34.

(c) (c) (b) (a) (b) (b) 4:1 C =12 for n

37. (8)

3. 7. 11. 15. 19. 23. 27. =3

(b) (b) (b) (a) (b) (a, d) False

4. 8. 12. 16. 20. 24. 28. 35.

(b) (c) (b) (b) (c) (a, b, d) False (39)

Hints & Solutions Topic 1 Arithmetic Progression (AP) 1.

Key Idea Use nth term of an AP i.e. an = a + ( n − 1) d , simplify the given equation and use result.

Given AP is a1 , a 2, a3 , … , a n Let the above AP has common difference ‘d’, then a1 + a 4 + a7 + … + a16 = a1 + (a1 + 3d ) + (a1 + 6d ) + … + (a1 + 15d ) = 6a1 + (3 + 6 + 9 + 12 + 15)d (given) ∴ 6a1 + 45d = 114 …(i) ⇒ 2a1 + 15d = 38 Now, a1 + a 6 + a11 + a16 = a1 + (a1 + 5d ) + (a1 + 10d ) + (a1 + 15d ) = 4a1 + 30d = 2(2a1 + 15d ) [from Eq. (i)] = 2 × 38 = 76

2. Let tn be the nth term of given AP. Then, we have t19 = 0 ⇒ a + (19 − 1)d = 0 ⇒ a + 18d = 0 t49 a + 48d Now, = t29 a + 28d − 18d + 48d = − 18d + 28d 30d = = 3 :1 10d

[Qtn = a + (n − 1)d ] …(i)

[using Eq. (i)]

225a 2 + 9b2 + 25c2 − 75ac − 45ab − 15bc = 0 ⇒ (15a )2 + (3b)2 + (5c)2 − (15a )(5c) − (15a )(3b) − (3b)(5c) = 0 1 2 2 2 ⇒ [(15a − 3b) + (3b − 5c) + (5c − 15a ) ] = 0 2 ⇒ 15a = 3b, 3b = 5c and 5c = 15a a b c = = =λ ∴ 15a = 3b = 5c ⇒ 1 5 3

(say)

⇒

a = λ , b = 5λ , c = 3λ ∴ b, c, a are in AP.

and

1 n 1 Tn = a + (n − 1) d = m

Tm = a + (m − 1) d =

On subtracting Eq. (ii) from Eq. (i), we get 1 1 m−n (m − n ) d = − = n m mn 1 d= ⇒ mn Again, Tmn = a + (mn − 1) d = a + (mn − n + n − 1) d

5. Since, a1 , a 2, ... , a n are in an AP. ∴ (a 2 − a1 ) = (a3 − a 2) = ... = (a n − a n − 1 ) = d 1 1 1 Thus, + + ... + a1 + a 2 a 2 + a3 a n −1 + a n  a 2 − a1   a3 − a 2  +  + ... + = d d     =

1 1 (a n − a1 ) ( a n − a1 ) = = d d a n + a1

 a n − a n −1      d   (n − 1) a n + a1

6. Since, n1 , n2,... , n p are p positive integers, whose sum is even and we know that, sum of any two odd integers is even. ∴Number of odd integers must be even. Hence, it is a false statement.

7. Given that, AP (a ; d ) denote the set of all the terms of an

3. We have,

4. Let

= a + (n − 1) d + (mn − n ) d (m − 1) 1 1 = Tn + n (m − 1) = + =1 mn m m

…(i) …(ii)

infinite arithmetic progression with first term ‘a’ and common difference d > 0. Now, let mth term of first progression …(i) AP(1 ; 3) = 1 + (m − 1)3 = 3m − 2 and nth term of progression …(ii) AP (2; 5) = 2 + (n − 1)5 = 5n − 3 and rth term of third progression AP (3; 7) …(iii) = 3 + (r − 1)7 = 7r − 4 are equal. Then 3m − 2 = 5n − 3 = 7r − 4 Now, for AP ( 1 ; 3 ) ∩ AP (2 ; 5 ) ∩ AP (3; 7), the common 5n − 1 terms of first and second progressions, m = 3 ⇒ n = 2, 5, 11, … and the common terms of second and 5n + 1 the third progressions, r = ⇒ n = 4, 11,… 7 Now, the first common term of first, second and third progressions (when n = 11), so a = 2 + (11 − 1)5 = 52 and d = LCM(3, 5, 7) = 105 So, AP (1 ; 3) ∩ AP (2 ; 5) ∩ AP (3 ; 7) = AP (52 ; 105) So, a = 52 and d = 105 ⇒ a + d = 157.00

8. Let the sides are a − d , a and a + d. Then, a (a − d ) = 48 and a − 2ad + d 2 + a 2 = a 2 + 2ad + d 2 ⇒ a 2 = 4ad ⇒ a = 4d Thus, a = 8, d = 2 Hence, a − d =6 2

Sequences and Series 59 = − 2 − 2 − 2 − 2 K 33 times

Topic 2 Sum of n Terms of an AP

= (−2) × 33 = − 66 1   1 2   1  1 + − − + − − +K+ ∴ −  3   3 100   3 100 

1. Let the common difference of given AP is ‘d’. Since, a1 + a7 + a16 = 40 ∴ a1 + a1 + 6d + a1 + 15d = 40 [Q a n = a1 + (n − 1) d ] …(i) ⇒ 3a1 + 21d = 40 Now, sum of first 15 terms is given by 15 S15 = [2a1 + (15 − 1) d ] 2 15 = [2a1 + 14d ] = 15 [a1 + 7d ] 2 From Eq. (i), we have 40 a1 + 7d = 3 40 So, S15 = 15 × 3 = 5 × 40 = 200

= (− 67) + (− 66) = − 133. Alternate Solution Q [− x] = − [x] − 1, if x ∉Integer, 1  2 n − 1   and [x] + x + + x+ +K+ x+ = [nx], n   n  n    n ∈N. So given series 1   1  1 − + − − +  3   3 100 

∴

S 4 = 2[2a + 3d ] = 16

(given) n   Q S n = [2a + (n − 1)d ]   2 … (i) ⇒ 2a + 3d = 8 and [given] S 6 = 3[2a + 5d ] = − 48 … (ii) ⇒ 2a + 5d = − 16 On subtracting Eq. (i) from Eq. (ii), we get 2d = − 24 ⇒ d = − 12 So, [put d = −12 in Eq. (i)] 2a = 44 Now, S10 = 5[2a + 9d ] = 5[44 + 9(− 12)] = 5[44 − 108] = 5 × (− 64) = − 320

2   99   1   1 − 1 − 1 + K +  − + + − +  3 100    3 100  

1   1 2   1  1  1 99  − + − − + − − + K ... + − −  3   3 100   3 100   3 100  [where, [x] denotes the greatest integer ≤ x] Now, 1   1 2   1  1  1 66  , …+ − − − , − − , − −  3   3 100   3 100   3 100  all the term have value − 1  1 67   1 68   1 99  and − − , − − , …, − − all the term  3 100   3 100   3 100  have value − 2. 1   1 2   1  1  1 66  So, − + − − + − − + ... + − −  3   3 100   3 100   3 100  = − 1 − 1 − 1 − 1 K 67 times. = (− 1) × 67 = − 67  1 67  and − − +  3 100 

1  × 100 = − 100 − 33 = − 133. 3  4. Let first three terms of an AP as a − d, a, a + d. = (− 1) × 100 −

3a = 33 ⇒ a = 11 [given sum of three terms = 33 and product of terms = 1155] [given] ⇒ (11 − d )11(11 + d ) = 1155 ⇒ 112 − d 2 = 105 ⇒ d 2 = 121 − 105 = 16 ⇒ d = ±4 So the first three terms of the AP are either 7, 11, 15 or 15, 11, 7. So, the 11th term is either 7 + (10 × 4) = 47 or 15 + (10 × (−4)) = − 25. So,

5.

3. Given series is

 1 68  − − +K+  3 100 

 1 99  − −  3 100 

2   1  − 1 99  − − − +…K+  3 100   3 100 

1    1    1 − 1 − 1 +  − + = −   3    3 100 

2. Given S n denote the sum of the first n terms of an AP. Let first term and common difference of the AP be ‘a’ and ‘d’, respectively.

 1 99  − −  3 100 

Key Idea Use the formula of sum of first n terms of AP, i.e Sn =

n [2 a + ( n − 1) d ] 2

Given AP, is a1 , a 2, a3 ,… having sum of first n-terms n = [2a1 + (n − 1)d ] 2 [where, d is the common difference of AP] n (n − 7) (given) = 50n + A 2 1 n−7 ⇒ [2a1 + (n − 1)d ] = 50 + A 2 2 1 7  n  [2a1 + nd − d ] = 50 − A + A ⇒  2 2  2 d  nd  7  n  ⇒ = 50 − A + A  a1 −  +   2 2 2  2 On comparing corresponding term, we get d 7 d = A and a1 − = 50 − A 2 2

60 Sequences and Series A 7 = 50 − A 2 2 ⇒ a1 = 50 − 3 A So a50 = a1 + 49d = (50 − 3 A ) + 49 A = 50 + 46 A Therefore, (d , a50 ) = ( A , 50 + 46 A ) ⇒

a1 −

[Qd = A]

Also, a1 , a 2, ... , a101 are in AP.

[Q d = A]

⇒ ⇒ Now,

12 [16 + 93] = 654 2 n   QS n = ( a + l )   2

Similarly, the two digit number which leaves remainder 5 when divided by 7 is of the form N = 7k + 5

For k = 1, N = 12 k = 2, N = 19 M k = 13, N = 96 ∴13 such numbers are possible and these numbers also forms an AP.

Now,

Total sum = S + S′ = 654 + 702 = 1356

s = a1 + a 2 + K + a51 51 = (2a1 + 50 D ) 2

…(iv)

t = a1 (251 − 1) [Q a1 = b1 ] …(v) t = 251 a1 − a1 < 251 a1 51 [from Eq. (ii)] and s= [a1 + (a1 + 50 D )] 2 51 = [a1 + 250 a1 ] 2 51 51 50 = a1 + 2 a1 2 2 …(vi) ∴ s > 251 a1 From Eqs. (v) and (vi), we get s > t Also, a101 = a1 + 100 D and b101 = 2100 b1  250 a1 − a1  ∴ a101 = a1 + 100   and b101 = 2100 a1 50  

∴ 12 such numbers are possible and these numbers forms an AP.

13 [12 + 96] = 702 S′ = 2 n  QS n = ( a +  2

…(iii)

∴ or

k = 2, N = 16 k = 3, N = 23 M M k = 13, N = 93

S=

[Q a1 = b1 ] …(ii)

t = b1 + b2 + K + b51 (251 − 1) t = b1 2 −1

and

when divided by 7 is of the form N = 7k + 2 [by Division Algorithm]

Now,

a1 + 50 D = 250 b1 a1 + 50 D = 250 a1

⇒

6. Clearly, the two digit number which leaves remainder 2

For,

a1 = b1 and a51 = b51

Given,

⇒ ⇒ ⇒

a101 = a1 + 251 a1 − 2a1 = 251 a1 − a1 a101 < 251 a1 and b101 > 251 a1 b101 > a101

9. Let S n = cn 2 S n − 1 = c (n − 1)2 = cn 2 + c − 2cn

 l ) 

∴

Tn2

…(i) = 15[2a1 + 29d ] (where d is the common difference) n   Q S n = [2a + (n − 1)d ]   2 T = a1 + a3 + … + a 29 15 = [2a1 + 14 × 2d )] 2 (Q common difference is 2d) …(ii) ⇒ 2T = 15[2a1 + 28d ] From Eqs. (i) and (ii), we get [QS − 2T = 75] S − 2T = 15d = 75 ⇒ d =5 Now, a10 = a5 + 5d = 27 + 25 = 52

2

2 2

4c2 ⋅ n (n + 1) (2n + 1) + nc2 − 2c2n (n + 1) 6

2 c2n (n + 1) (2n + 1) + 3nc2 − 6c2n (n + 1) 3 nc2(4n 2 + 6n + 2 + 3 − 6n − 6) nc2(4n 2 − 1) = = 3 3 =

10. According to given condition, S 2n = S′n 2n n [2 × 2 + (2n − 1) × 3] = [2 × 57 + (n − 1) × 2] ⇒ 2 2 1 ⇒ (4 + 6n − 3) = (114 + 2n − 2) 2 ⇒ 6n + 1 = 57 + n − 1 ⇒ 5n = 55 ∴ n = 11

11.

PLAN

Convert it into differences and use sum ofn terms of an AP, n S n = [2a + ( n − 1 )d ] 2

i.e.

8. If log b1 , log b2, ... , log b101 are in AP, with common difference log e 2 , then b1 , b2, ... , b101 are in GP, with common ratio 2. …(i) ∴ b1 = 20 b1 , b2 = 21 b1 , b3 = 22b1,…, b101 = 2100 b1

[Q Tn = S n − S n − 1 ]

= (2 cn − c) = 4c n + c2 − 4c2n

∴ Sum = Σ Tn2 =

7. We have, S = a1 + a 2 + … + a30

and

Tn = 2cn − c

Now,

Sn =

4n

∑

k =1

(−1)

k( k + 1) 2

⋅ k2

Sequences and Series 61 = − (1)2 − 22 + 32 + 42 − 52 − 62 + 72 + 82 + K

∴

p = a − 3d = 5 − 6 = − 1

= (3 − 1 ) + (4 − 2 ) + (7 − 5 ) + (8 − 6 ) + K

q = a − d =5 −2 =3

= 2{(4 + 6 + 12 + K ) + (6 + 14 + 22 + K )} 144424443 144424443

r = a + d =5 + 2 = 7

2

2

2

2

2

2

n terms

2

2

s = a + 3d = 5 + 6 = 11

and

n terms

n n  =2 {2 × 4 + (n − 1) 8} + {2 × 6 + (n − 1) 8}  2  2 = 2 [n (4 + 4n − 4) + n (6 + 4n − 4)]

Therefore, A = pq = − 3 and B = rs = 77

16. Here, 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + ... upto n terms =

= 2 [4n + 4n + 2n ] = 4n (4n + 1) 2

2

Here, 1056 = 32 × 33, 1088 = 32 × 34, 1056 and 1332 are possible answers. r 1 12. Here, V r = [ 2r + (r − 1) ( 2r − 1)] = ( 2r3 − r 2 + r ) 2 2 1 ∴ ΣV r = [ 2 Σr3 − Σr 2 + Σr ] 2 2 1   n (n + 1)  n (n + 1) ( 2n + 1) n (n + 1)  + = 2   −  2  2 6 2  

13. V r + 1

n (n + 1) [ 3n (n + 1) − ( 2n + 1) + 3] 12 1 = n (n + 1) ( 3n 2 + n + 2) 12 1 1 − V r = (r + 1)3 − r3 − [(r + 1)2 − r 2] + = 3r 2 + 2r − 1 2 2 =

∴

which is a composite number. Tr = 3r 2 + 2r − 1

∴

Qr = Tr+1 − Tr = 3 [ 2r + 1] + 2 [1]

⇒

Qr = 6 r + 5

⇒

(n + 1) n −1  = n2  + 1 = n 2  2  2 ∴ 12 + 2⋅ 22 + 32 + 2 ⋅ 42 + ... upto n terms, when n is odd =

17. Integers divisible by 2 are {2,4,6,8,10, ...,100}. Integers divisible by 5 are {5,10,15, ...,100}. Thus, sum of integers divisible by 2 50 = (2 + 100) = 50 × 51 = 2550 2

∴ Sum of integers from 1 to 100 divisible by 2 or 5 = 2550 + 1050 − 550

Qr+ 1 = 6(r + 1) + 5

= 2550 + 500 = 3050

Common difference = Qr+ 1 − Qr = 6

18. Let four consecutive terms of the AP are a − 3d , a − d ,

15. Given, p + q = 2, pq = A and

a + d , a + 3d, which are integers. Again, required product

r + s = 18, rs = B

and it is given that p, q, r , s are in an AP.

P = (a − 3d )(a − d )(a + d )(a + 3d ) + (2d )4 [by given condition]

Therefore, let p = a − 3d , q = a − d , r = a + d s = a + 3d

and We have,

d >0

Now,

2 = p + q = a − 3d + a − d = 2a − 4d

⇒ Again,

= (a 2 − 9d 2)(a 2 − d 2) + 16d 4

p 0 and 0 < r < 1. Then, according the problem, we have a 3= 1−r 27 and = a3 + (ar )3 + (ar 2)3 + (ar3 )3 + ... 19  27 a3 a  = ⇒ Q S∞ =  3 19 1 − r 1 − r   3  a 27 (3 (1 − r ))  ⇒ = Q3 = ⇒ a = 3 (1 − r ) 1−r 19 1 − r3  

27 27 (1 − r ) (1 + r 2 − 2r ) = 19 (1 − r ) (1 + r + r 2) [Q (1 − r )3 = (1 − r ) (1 − r )2] r 2 + r + 1 = 19 (r 2 − 2r + 1) 18r 2 − 39r + 18 = 0 6r 2 − 13r + 6 = 0 (3r − 2) (2r − 3) = 0 2 3 [Q0 < r < 1] r = or r = (reject) 3 2

5. Let a , ar , ar 2 are in GP, where (r > 1). On multiplying middle term by 2, we have a , 2ar , ar 2 are in an AP. ⇒ 4ar = a + ar 2 2 ⇒ r − 4r + 1 = 0 4 ± 16 − 4 ⇒ r= =2 ± 3 2 [since, AP is increasing] r =2 + 3 ⇒

6. Given, k ⋅ 109 = 109 + 2 (11)1 (10)8 + 3(11)2(10)7 + ... + 10(11)9 2

 11  11  11 k = 1 + 2   + 3  + ... + 10    10  10  10

⇒

 11  k =1  10

9

2

9

 11  11  11  11   + 2   + ... + 9   + 10    10  10  10  10

...(i) 10

…(ii)

On subtracting Eq. (ii) from Eq. (i), we get 2

9

11 11  11  11  11  +   + ... +   − 10   k 1 −  = 1 +  10  10  10 10  10

10

 11 10  1   − 1 10    10 − 11  10  − 10  11 ⇒ k  =    10   10  11  − 1   10    a (r n − 1) Q In GP,sum of n terms = r − 1 , when r > 1   10    11 10  11 ⇒ − k = 10 10   − 10 − 10     10  10   ∴

k = 100

7. Let S = 0.7 + 0.77 + 0.777 + … 7 77 777 + + + … upto 20 terms 10 102 103 11 111 1  =7 + + + … upto 20 terms 10 102 103  79 99 999  = + + +… upto 20 terms  9 10 100 1000 =

=

=

7 9

1  1   1   1 − 10 + 1 − 102 + 1 − 103  

+ … + upto 20 terms] 7 [(1 + 1 +…+ upto 20 terms) 9 1 1 1  −  + 2 + 3 + … + upto 20 terms   10 10  10

Sequences and Series 67 20  1   1    1 −     10  10   7 = 20 −  1 9 1−   10       20 Q ∑ = 20 and sum of n terms of    i =1 a (1 − r n )   GP, S n = 1 − r when (r < 1)   

=

20 7  1  1   20 − 1 −     10 9  9  

=

7 9

179 1  1  20  7 +   = [179 + (10)− 20 ]   10 81 9 9  

8. We know that, the sum of infinite terms of GP is

∴ or ⇒ i.e. ⇒

 a , |r| < 1 S ∞ = 1 − r  ∞ ,|r| ≥ 1 x S∞ = =5 1−r

[|r| < 1]

x 5 5−x exists only when|r| < 1. r= 5 5−x −1 < < 1 or −10 < − x < 0 5

1−r=

0 < x < 10 3 4

9. Since, sum = 4 and second term = . It is given first term a and common ratio r. 3 3 a ⇒ = 4, ar = ⇒ r = 4 4a 1−r ⇒ ⇒

a 1−

3 4a

=4 ⇒

4a 2 =4 4a − 3

(a − 1) (a − 3) = 0 ⇒ a = 1 or 3

When a = 1, r = 3 / 4 and when a = 3, r = 1 / 4

10. Sum of the n terms of the series

1 3 7 15 + + + + ... 2 4 8 16

upto n terms can be written as 1  1  1  1  1 −  + 1 −  + 1 −  + 1 −  + ... upto n terms  2  4  8  16 1 1 1  = n −  + + + ... + n terms 2 4 8  1 1 1 − n   2 2 = n + 2−n − 1 =n− 1 1− 2

11. Given quadratic equation x2 − x − 1 = 0 having roots α

and β, (α > β ) 1+ 5 1− 5 So, α = and β = 2 2 and α + β = 1, αβ = −1 α n − βn , n ≥1 an = Q α −β α n+ 1 − β n+ 1 So, a n+ 1 = = α n + α n−1 β + α n− 2 β 2+ .... α −β + αβ n−1 + β n n n− 2 n−3 n−2 n = α −α −α β −....−β +β [as αβ = −1] = α n + β n − (α n− 2 + α n−3β + ...+ β n− 2) = α n + β n − a n−1   α n−1 − β n−1 = α n− 2 + α n−3 β + ...+ β n− 2  as a n−1 = − α β   ⇒ a n+ 1 + a n−1 = α n + β n = bn , ∀n ≥ 1 So, option (b) is correct. ∞ ∞ α n + βn b Now, ∑ nn = ∑ [as, bn = α n + β n] n 10 10 n=1 n=1 n

∞ ∞ α β =∑  +∑   10  10 n=1

n

n=1

β  α  Q 10 < 1 and 10 < 1  

α β α β 10 10 = + = + α β 10 − α 10 −β 1− 1− 10 10 10 α − αβ + 10 β − αβ 10(α + β ) − 2 αβ = = (10 − α ) (10 − β ) 100 − 10(α + β ) + αβ 10(1) − 2(−1) 12 [as α + β = 1 and αβ = −1] = = 100 − 10(1) − 1 89 So, option (a) is not correct. Q α 2 = α + 1 and β 2 = β + 1 n+ 2 ⇒ α = α n+ 1 + α n and β n+ 2 = β n+ 1 + β n n+ 2 n+ 2 ⇒ (α + β ) = (α n+ 1 + β n+ 1 ) + (α n + β n ) ⇒ a n+ 2 = a n+ 1 + a n Similarly, a n+ 1 = a n + a n−1 a n = a n−1 + a n−2 ……… ……… a3 = a 2 + a1 On adding, we get a n+ 2 = (a n + a n−1 + a n− 2+ ....+ a 2 + a1 ) + a 2   α 2 − β2 Q a = = α + β = 1 2  α −β   So, a n+ 2 − 1 = a1 + a 2 + a3 + .....+ a n So, option (c) is also correct. ∞ ∞ an α n − βn And, now ∑ =∑ n n n=1 (α − β ) 10 n=1 10 =

n n ∞ 1 ∞ α β  ∑   − ∑    α − β n=1 10 10  n=1

68 Sequences and Series β   α 1  10   α 10 − = , as  α β α −β 1 − 1 −   10   10 10  1  α β  1 = −   = α − β  10 − α 10 − β α − β

β  < 1 and < 1 10 

 10 α − αβ − 10 β + αβ  100 − 10 (α + β ) + αβ    10(α − β ) 10 10 = = = (α − β ) [100 − 10 (α + β ) + αβ ] 100 − 10 − 1 89 Hence, options (b), (c) and (d) are correct.

∴

S1 =

1 =2 1 − 1 /2

S2 =

2 =3 1 − 1 /3

M

S 2n − 1 ∴

M M 2n − 1 = = 2n 1 − 1 / 2n

S12 + S 22 + S32 + ... + S 22n − 1 = 22 + 32 + 42 + ... + (2n )2 1 = (2n ) (2n + 1) (4n + 1) − 1 6

12. Let a n denotes the length of side of the square S n. We are given, a n = length of diagonal of S n + 1. a ⇒ an = 2 an + 1 ⇒ an + 1 = n 2 This shows that a1 , a 2, a3 , K form a GP with common ratio 1 / 2. n −1  1 Therefore, a n = a1    2  1 a n = 10    2

⇒

 1 ⇒ a n2 = 100    2

2( n − 1)

⇒

100 ≤1 2n − 1

[Q a n2 ≤ 1, given]

13. Bn = 1 − An > An  3 1 −  −    4  1 1 3 An < ⇒ < 3 2 2 4 1+ 4 n

n

1  3 −  > −  4 6

a 2 (1 + r 2 + r 4 ) S2 (1 + r 2)2 − r 2 1 = 2 2 ⇒ = 2 2 2 a (1 + r + r ) aS (1 + r + r 2)2 a 2 1 r+ +1 (1 + r 2 − r ) 1 2 r ⇒ a = = 1 (1 + r 2 + r ) a 2 r + −1 r 1 y+ 1 2 r+ =y ⇒ =a r y−1

⇒

⇒

y + 1 = a 2y − a 2 a2 + 1 a2 − 1

⇒

y=

⇒

a2 + 1 >2 a2 −1

⇒

1   Q | y | = r + r > 2   [where , (a 2 − 1) ≠ 0]

|a 2 + 1| > 2|a 2 − 1|

⇒ (a + 1) − {2 (a 2 − 1)}2 > 0 2

2

⇒ {(a 2 + 1) − 2 (a 2 − 1)}{(a 2 + 1) + 2 (a 2 − 1)} > 0 ⇒

But for

∴

3 1

4 5 19 × 25 ⇒

n >1 +

95 or n > 24.75 4

∴ Least positive value of n = 25

2. Since, a , b, c, d are in AP. a b c d are in AP. , , , abcd abcd abcd abcd 1 1 1 1 are in AP. , , , ⇒ bcd cda abd abc ⇒ bcd , cda , abd , abc are in HP. ⇒ abc, abd , cda , bcd are in HP. ⇒

3. Since, a1 , a 2, a3 , K , a10 are in AP. Now, ⇒ ⇒

a10 = a1 + 9d ⇒ 3 = 2 + 9d d = 1 / 9 and a 4 = a1 + 3d a 4 = 2 + 3(1 / 9) = 2 + 1 / 3 = 7 / 3

Also, h1 , h2, h3 , K , h10 are in HP. 1 1 1 1 are in AP. ⇒ , , ,K, h1 h2 h3 h10 Given, h1 = 2, h10 = 3 1 1 1 1 1 ∴ = + 9d1 ⇒ = + 9d1 ⇒ − = 9d1 6 h10 h1 3 2 ⇒ ⇒ ⇒ ∴

1 1 1 and = + 6d1 h7 h1 54 1 1 6 ×1 = + h7 2 − 54 1 1 1 18 = − ⇒ h7 = h7 2 9 7 7 18 a 4h7 = × =6 3 7 d1 = −

1 1 1 = = 1 + ln z 1 + ln x + 2 ln r A + 2D

and

Therefore,

1 1 1 are in HP. , , 1 + ln x 1 + ln y 1 + ln z a1 = 1, a 2 = 2 , ⇒ a3 = 4 , a 4 = 8

5. Let ∴

b1 = 1, b2 = 3, b3 = 7, b4 = 15

Clearly, b1 , b2, b3 , b4 are not in HP. Hence, Statement II is false. Statement I is already true.

6. Since, cos (x − y), cos x and cos (x + y) are in HP. ∴

cos x =

2 cos (x − y) cos (x + y) cos (x − y) + cos (x + y)

⇒ cos x (2 cos x ⋅ cos y) = 2 {cos 2 x − sin 2 y} ⇒ ⇒

cos 2 x ⋅ cos y = cos 2 x − sin 2 y cos x (1 − cos y) = sin 2 y y y y cos 2 x ⋅ 2 sin 2 = 4 sin 2 ⋅ cos 2 2 2 2 y 2 2 y cos x ⋅ sec = 2 ⇒ cos x ⋅ sec = ± 2 2 2 2

⇒ ⇒

7. Since, a , b, c are in an AP. ∴ 2b = a + c and a 2, b2, c2 are in HP. ⇒

b2 =

⇒

2a 2c2 ⇒ a 2 + c2

2

2a 2c2  a + c   = 2 2  2  a +c

(a 2 + c2)(a 2 + c2 + 2ac) = 8a 2c2

⇒

(a 2 + c2) + 2ac(a 2 + c2) = 8a 2c2

⇒ (a + c ) + 2ac(a 2 + c2) + a 2c2 = 9a 2c2 2

2

⇒

(a 2 + c2 + ac)2 = 9a 2c2

⇒

a 2 + c2 + ac = 3ac

⇒ ⇒

a 2 + b2 – 2ac = 0 (a – c)2 = 0 ⇒ a = c

and if a = c ⇒ b = c or a 2 + c2 + ac = – 3ac ⇒

a 2 + c2 + 2ac = –2ac

⇒

(a + c)2 = –2ac ac 4b2 = –2ac ⇒ b2 = – ⇒ 2 c Hence, a , b, – are in GP. 2 c ∴ Either a = b = c or a , b, − are in GP. 2

8. Since, a , A1 , A2, b are in AP. ⇒

A1 + A2 = a + b a , G1 , G2, b are in GP ⇒ G1G2 = ab

⇒ ln y = ln x + ln r and ln z = ln x + 2 ln r

and

A = 1 + ln x, D = ln r 1 1 1 1 1 Then, = , = = 1 + ln x A 1 + ln y 1 + ln x + ln r A + D

⇒

a , H 1 , H 2, b are in HP. 3ab 3ab H1 = , H2 = b + 2a 2b + a

4. Let the common ratio of the GP be r. Then, y = xr and z = xr Let

2

∴

1 1 1 1 + = + H1 H 2 a b

70 Sequences and Series H 1 + H 2 A1 + A2 1 1 = = + H 1H 2 G1G2 a b

⇒

G1G2 ab = H 1H 2  3ab   3ab       2b + a   b + 2a 

Now,

=

(2a + b) (a + 2b) 9ab

9. (i) Now,

a + b = 10

…(i)

1 1 1 1 1 Since, a , x, y, z , b are in HP, then , , , , a x y z b are in AP. Now,

a + b 10 5  1 1 5 =  +  − ⇒ ab 9 2  a b 3 9 × 10 [from Eq. (i)] ab = 10 =

⇒

ab = 9

…(ii)

On solving Eqs. (i) and (ii), we get a = 1, b = 9 (ii) LHS = log (x + z ) + log (x + z − 2 y)   2 xz   = log (x + z ) + log x + z − 2    x + z   = log (x + z ) + log

n =1

1 15 × 16 × 31 15 × 16  + 2  6 2 

1 [(5 × 8 × 31) + (15 × 8)] 2 = (5 × 4 × 31) + (15 × 4) = 620 + 60 = 680 1 15 × 16 1 and S 2 = (1 + 2 + 3 + K + 15) = × = 60 2 2 2 =

Therefore, S = S1 − S 2 = 680 − 60 = 620.

1 1  1 1 1 1 1   1 1 1 + = + + + +  − + +  a b  a x y z b  x y z 

⇒

15

∑ (n 2 + n )

 n n (n + 1) (2n + 1)  2  Q ∑ r = 6   r = 1

[since, a , x, y, z are in AP]

⇒

=

n( n + 1) 1 = 2 2 n =1 15

∑

5 (a + b) − 15 2

5 Sum = (a + b) 2

∴

=

a + b = (a + x + y + z + b) − (x + y + z ) =

2

2 n  n n (n + 1)   n (n + 1) Q ∑ r3 =   and ∑ r =    2 2  r = 1  r =1

…(ii)

From Eqs. (i) and (ii), we get G1G2 A + A2 ( 2a + b) (a + 2b) = = 1 9ab H 1H 2 H 1 + H 2

2xz   Qy=  x + z 

(x − z )2 (x + z )

= 2 log (x − z ) = RHS

Topic 6 Relation between AM, GM, HM and Some Special Series 1. Given series, S =1+

 n( n + 1)  15 15    13 + 23 + K + n3 2 = ∑ = ∑ + n n ( ) 1 + + + K n 1 2 n =1 n =1 2

…(i)

13 + 23 13 + 23 + 33 + ... + + 1+ 2 1+ 2+ 3

13 + 23 + 33 + K + 153 1 − (1 + 2 + 3 + K + 15) 1 + 2 + 3 + K + 15 2 = S1 − S 2 (let) where, 13 + 23 13 + 23 + 33 S1 = 1 + + +K+ 1+ 2 1+ 2+ 3 13 + 23 + 33 + K + 153 1 + 2 + 3 + K + 15

2. Given series is 3 × 13 5 × (13 + 23 ) 7 × (13 + 23 + 33 ) + ... + + 2 2 2 1 1 +2 12 + 22 + 33 So, nth term (3 + (n − 1)2)(13 + 23 + 33 ... + n3 ) Tn = 12 + 22 + 32 + K + n 2  n (n + 1) (2n + 1) ×     2 = n (n + 1)(2n + 1) 6

2

n  n 3  n (n + 1)  2 n (n + 1)(2n + 1)  and Σ r 2 = Q rΣ= 1 r =    r =1 2 6    3n (n + 1) 3 2 So, Tn = = (n + n ) 2 2 Now, sum of the given series upto n terms 3 S n = ΣTn = [Σn 2 + Σn ] 2 3  n (n + 1)(2n + 1) n (n + 1)  = +  2  6 2 3 10 × 11 × 21 10 × 11  + S10 = ∴ 2  6 2 

3 [(5 × 11 × 7) + (5 × 11)] 2 3 3 = × 55(7 + 1) = × 55 × 8 = 3 × 55 × 4 2 2 = 12 × 55 = 660 =

3. (b) Given series is 1 + (2 × 3) + (3 × 5) + (4 × 7) + …upto 11 terms. Now, the rth term of the series is a r = r (2r − 1) ∴Sum of first 11-terms is

Sequences and Series 71 S11 =

11

11

11

11

r =1

r =1

r =1

r =1

( x m + x − m ) ≥ 2 and ( y n + y − n ) ≥ 2

∑ r (2r − 1) = ∑ (2r 2 − r ) = 2 ∑ r 2 − ∑ r

11 × (11 + 1)(2 × 11 + 1) 11 × (11 + 1) − 6 2 n  n n (n + 1)(2n + 1) n (n + 1)  2 and ∑ r = Q ∑ r =  2 6  r = 1  r =1  11 × 12 × 23  11 × 12 =   −   2   3 =2

= (11 × 4 × 23) − (11 × 6) = 11(92 − 6) = 11 × 86 = 946

4. Given series is 3

3

3

3

 3  1  1  3 3   + 1  +  2  + 3 +  3  + ...  4  4  2  4 3

3

3

3

 12  9  6  3 Let S =   +   +   +    4  4  4  4 3  15 +   + … + upto 15 terms  4 3

 3 =   [13 + 23 + 33 + 43 + 53 + ... + 153 ]  4 3 2  3  15 × 16 =     4  2  2  3   n ( n + 1) 3 3 3 Q1 + 2 + 3 + ... + n =   , n ∈N   2   27 225 × 256 = 27 × 225 = × 64 4

⇒ S = 27 × 225 = 225 k ⇒ k = 27. 1 + 2 + 3 + ... + k 5. Since, S k = k

=

[given]

k ( k + 1) k + 1 = 2k 2

2

1  k + 1 2 S 2k =   = (k + 1)  2  4

So,

10 5 2 A = S 12 + S 22 + S 23 + ... S 10 = ∑ S 2k 12 k =1

Now, ⇒

… (i)

5 1 A= 12 4

10

1

∑ (k + 1)2 = 4 [22 + 32 + 42 + ... 112]

k =1

1 11 × (11 + 1) (2 × 11 + 1) 2 = −1 4  6 

[Q ∑ n 2 = 1 4 1 = 4 =

n ( n + 1) ( 2n + 1) ] 6

11 × 12 × 23  1 − 1 = [(22 × 23) − 1]   4 6 1 5 505 [506 − 1] = [505] ⇒ A= ⇒ A = 303 4 12 4

1 xm yn 6. Consider, = m 2m 2n −m (1 + x )(1 + y ) (x + x )( yn + y−n )

By using AM ≥ GM (because x , y ∈ R + ), we get

[Q If x > 0, then x + ( x m + x − m )( y n + y − n ) ≥ 4 1 1 ⇒ ≤ ( x m + x − m )( y n + y − n ) 4 1 ∴ Maximum value = . 4

1 ≥ 2] x

⇒

7. General term of the given series is 3r (12 + 22 + K + r 2) 3r [r (r + 1) (2r + 1)] = 2r + 1 6(2r + 1) 1 3 2 = (r + r ) 2 15 1 15 Now, required sum = ∑ Tr = ∑ (r3 + r 2) 2 r =1 r =1

Tr =

 n (n + 1)  2 n (n + 1) (2n + 1)  +    2 6 n = 15  1  n (n + 1)  n 2 + n 2n + 1  =  +   2 2 2 3   n = 15 1  n (n + 1) (3n 2 + 7n + 2)  =   2 2 6 n = 15 1 15 × 16 (3 × 225 + 105 + 2) = × × = 7820 2 2 6 8. We have, a1 , a 2, a3 , … a 49 are in AP. =

1 2

12

∑ a 4k + 1 = 416 and

a 9 + a 43 = 66

k=0

Let a1 = a and d = common difference Q a1 + a5 + a 9 + L + a 49 = 416 ∴ a + (a + 4d ) + (a + 8d ) + … (a + 48d ) = 416 13 …(i) (2a + 48d ) = 41633a + 24d = 32 ⇒ 2 Also a 9 + a 43 = 66 ⇒ a + 8d + a + 42d = 66 …(ii) ⇒ 2a + 50d = 66 ⇒ a + 25d = 33 Solving Eqs. (i) and (ii), we get a = 8and d = 1 2 2 2 2 Now, a1 + a 2 + a3 + L + a17 = 140m 2 2 2 8 + 9 + 10 + … + 242 = 140m ⇒ (12 + 22 + 32 + … + 242) − (12 + 22 2 2 + 3 + … + 7 ) = 140m 24 × 25 × 49 7 × 8 × 15 − = 140m ⇒ 6 6 3 × 7 ×8 ×5 ⇒ (7 × 5 − 1) = 140m 6 ⇒ 7 × 4 × 5 × 34 = 140m ⇒ 140 × 34 = 140m ⇒ m = 34

9. We have, 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + … A = sum of first 20 terms B = sum of first 40 terms

72 Sequences and Series ∴A = 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + … + 2 ⋅ 202 A = (12 + 22 + 32 + … + 202) + (22 + 42 + 62 + … + 202) A = (12 + 22 + 32 + … + 202) + 4(12 + 22 + 32 + …+102) 20 × 21 × 41 4 × 10 × 11 × 21 A= + 6 6 20 × 21 20 × 41 × 63 A= (41 + 22) = 6 6 Similarly B = (12 + 22 + 32 + … + 402) + 4(12 + 22 +… + 202) 40 × 41 × 81 4 × 20 × 21 × 41 B= + 6 6 40 × 41 40 × 41 × 123 B= (81 + 42) = 6 6 Now, B − 2 A = 100λ 40 × 41 × 123 2 × 20 × 21 × 63 ∴ − = 100λ 6 6 40 40 ⇒ (5043 − 1323) = 100λ ⇒ × 3720 = 100λ 6 6 40 × 620 = 248 ⇒ 40 × 620 = 100λ ⇒ λ = 100

10. Let S10 be the sum of first ten terms of the series. Then, we have 2 2 2 2  3  2  1  4 S10 = 1  + 2  + 3  + 42 + 4   5  5  5  5 + ... to 10 terms 2 2 2 2  8  12  16  24 =   +   +   + 42 +   + ... to 10 terms  5 5 5 5 1 2 2 2 2 2 = 2 (8 + 12 + 16 + 20 + 24 + ... to 10 terms) 5 42 = 2 (22 + 32 + 42 + 52 + ... to 10 terms) 5 42 = 2 (22 + 32 + 42 + 52 + ... + 112) 5 16 = ((12 + 22 + ... + 112) − 12) 25 16  11 ⋅ (11 + 1) (2 ⋅ 11 + 1)  = − 1   25  6 16 16 16 16 m= × 505 = 101 = × 505 ⇒ (506 − 1) = 5 25 25 25

Write the nth term of the given series and simplify it to get its lowest form. Then, apply, S n = ∑ Tn

PLAN

13 13 + 23 13 + 23 + 33 + + + ... 1 1+3 1+3+5 Let Tn be the nth term of the given series. 13 + 23 + 33 + ... + n3 ∴ Tn = 1 + 3 + 5 + ... + upto n terms Given series is

2

 n (n + 1)    2 2  = (n + 1) = 2 4 n 9 (n + 1)2 1 2 S9 = ∑ = (2 + 32 + ... + 102) + 12 – 12] 4 4 n =1 =

1 10(10 + 1)(20 + 1)  384 = 96 –1 =  4  6 4 π 2

13. Here, α ∈ (0, ) ⇒ tan α > 0 x2 + x + ∴

⇒

tan 2 α x2 + x

2 x +x+ 2

tan 2 α x2 + x

l + n = 2m

[using AM ≥ GM] ≥ 2 tan α

15. Since, AM ≥ GM, then

Also,

(a + b) + (c + d ) ≥ (a + b)(c + d ) ⇒ M ≤ 1 2 (a + b) + (c + d ) > 0 [Q a , b, c, d > 0]

∴

0 < M ≤1

α+β=

4+ 5 8+2 5 and α β = 5+ 2 5+ 2

Let H be the harmonic mean between α and β, then

∴ G1 = lr, G2 = lr 2, G3 = lr3 , n − lr 4

⇒

 n r=   l

= (lr ) + 2(lr ) + (lr ) 4

2 4

2

H =

1 4

3 4

= l × r 4 (1 + 2r 4 + r 6 ) = l4 × r 4 (r 4 + 1)2 = l4 ×

x2 + x

…(i) a1 a 2 a3 ... (a n − 1 )(2a n ) = 2c a1 + a 2 + a3 + ... + 2a n 1/ n ≥ (a1 ⋅ a 2 ⋅ a3 ... 2a n ) ∴ n [using AM ≥ GM] [from Eq. (i)] ⇒ a1 + a 2 + a3 + ... + 2a n ≥ n (2c)1/ n ⇒ Minimum value of a1 + a 2 + a3 + ... + 2a n = n (2c)1/ n

Let r be the common ratio of this GP.

+ G34 4

tan 2 α

⇒

... (i)

l, G1 , G2, G3 , n are in GP.

+ 2G24

x2 + x ⋅

14. Given, a1 a 2 a3 ... a n = c

and G1 , G2, G3 are geometric means between l and n.

Now, G14

≥

16. Let α , β be the roots of given quadratic equation. Then,

11. Given, m is the AM of l and n. ∴

12.

n  n + l 2 2   = ln × 4 m = 4lm n l  l 

2 αβ 16 + 4 5 =4 = α+β 4+ 5

17. Since, product of n positive numbers is unity. ⇒

... (i) x1 ⋅ x2 ⋅ x3 ... xn = 1 x1 + x2 + ... + xn 1/ n Using AM ≥ GM, ≥ (x1 ⋅ x2 ... xn ) n ⇒

x1 + x2 + ... + xn ≥ n (1)1/ n

[from Eq. (i)]

Hence, sum of n positive numbers is never less than n.

Sequences and Series 73 18. Since, AM > GM

Again, a (n ) = 1 +

(b + c − a ) + (c + a − b) > (b + c − a )(c + a − b)1/ 2 ∴ 2 ⇒

c > [(b + c − a )(c + a − b)]1/ 2

…(i)

Similarly

b > [(a + b − c)(b + c − a )]1/ 2

…(ii)

and

a > [(a + b − c)(c + a − b)]

1/ 2

1  1 1  1 1 +  +  +  +K+  2  3 4  5 8 1 1 1  + K +  n −1 + K + n − n 2 2  2 +1

>1 +

…(iii)

1  1 1  1 1 1 +  +  +  + +K+     2 4 4 8 8 8 1 1  1 + K +  n + K + n − n 2 2  2

On multiplying Eqs. (i), (ii) and (iii), we get abc > (a + b − c)(b + c − a )(c + a − b)

1 2 4 2n − 1 1 + + +K+ n − n 2 4 8 2 2 1 1 1 1 1  1 n = 1 + + + + K + − n = 1 − n  +   2 2 2 2 2 2 2 14442444 3 n times

Hence, (a + b − c)(b + c − a )(c + a − b) − abc < 0

=1 +

19. Since, x1 , x2 ,…, xn are positive real numbers. ∴ Using nth power mean inequality x12 + x22 + ... + xn2  x1 + x2 + ... + xn  ≥    n n ⇒

2

2

 n   n  n  n 2  n   ∑ xi2 ≥  ∑ xi  x x ⇒ n ≥ i  ∑ i     n  i∑     i =1   i =1  =1 i =1 2

2

20. Let a and b are two numbers. Then, a+b 2ab ; G1 = ab ; H 1 = 2 a+b An − 1 + H n − 1 , G n = An − 1 H n − 1 , An = 2 2 An − 1 H n − 1 Hn = An − 1 + H n − 1 A1 =

Clearly, G1 = G2 = G3 = ... = ab .

21. A2 is AM of A1 and H 1 and A1 > H 1 ⇒ A1 > A2 > H 1 A3 is AM of A2 and H 2 and A2 > H 2 ⇒ A2 > A3 > H 2 M M M A1 > A2 > A3 > ...

∴

22. As above, A1 > H 2 > H 1 , A2 > H 3 > H 2 ∴

H 1 < H 2 < H 3 < ... 1 1 1 1 + + +K+ n 2 3 4 2 −1 1  1 1  1 1  1 =1 +  +  +  + K +  +  + K+   2 3  4   7 8 15  1 1  + K +  n −1 + K + n  2 2 − 1 1  1 1 1  1 1  1 1 1 − 200  + > 100  2 2  Therefore, (d) is also the answer.

24. Since, first and (2n − 1)th terms are equal. Let first term be x and (2n − 1) th term be y, whose middle term is tn. x+ y Thus, in arithmetic progression, tn = =a 2 In geometric progression, tn = xy = b 2xy In harmonic progression, tn = =c x+ y [using AM > GM > HM] ⇒ b2 = ac and a > b > c Here, equality holds (i. e. a = b = c) only if all terms are same. Hence, options (a), (b) and (d) are correct.

25. Let the two positive numbers be a and b. ∴ and

a+b [since, x is AM between a and b] … (i) 2 a y z [since, y, z are GM’s between a and b] = = y z b

x=

∴

a=

y2 z

and b =

z2 y

On substituting the values of a and b in Eq. (i), we get 2x =

y2 z 2 y3 + z3 y3 + z3 + ⇒ = 2x ⇒ =2 z y yz xyz

26. Let the two positive numbers be ka and a , a > 0. Then, G = ka ⋅ a = k ⋅ a and H = Again,

⇒

[given]

⇒

2 k 4 = k+1 5

5 k = 2k + 2 ⇒ 2k − 5 k + 2 = 0

⇒ ⇒

H 4 = G 5 2ka k+1 4 = ka 5

2(ka )a 2ka = ka + a k + 1

k=

5 ± 25 − 16 5 ± 3 1 = = 2, ⇒ k = 4, 1 / 4. 4 4 2

Hence, the required ratio is 4 : 1.

74 Sequences and Series 27. Using AM ≥ GM,

= a12n ⋅ r1 + 2 + K + ( n − 1) = a12nr

1 + x2 n 1 + x2n ≥ 1 ⋅ x2 n ⇒ ≥ xn 2 2 1 1 xn xn ⋅ ym ≤ ≤ ⇒ 2n 2n 2 (1 + x ) (1 + y2m ) 4 1+x

⇒

⇒

Gm = ( A1 A2 K AnH 1H 2 K H n )1/ 2n

31. Let two numbers be a and b and A1 , A2, ... , An be n arithmetic means between a and b. Then, a , A1 , A2, ... , An , b are in AP with common difference b−a b−a d= ⇒ p = A1 = a + d = a + n+1 n+1 p=

⇒

29. Here, (1 + a )(1 + b)(1 + c) …(i) = 1 + a + b + c + ab + bc + ca + abc a + b + c + ab + bc + ca + abc 4 4 4 1/7 Since, ≥ (a b c ) 7 [using AM ≥ GM] a + b + c + ab + bc + ca + abc ≥ 7(a 4b4c4 )1/7 1 + a + b + c + ca + abc > 7(a 4b4c4 )1/7

…(ii)

From Eqs. (i) and (ii), we get {(1 + a )(1 + b)(1 + c)}7 > 77 (a 4b4c4 )

30. Let Gm be the geometric mean of G1 , G2, ... , Gn. Gm = (G1 ⋅ G2 K Gn )1/ n K (a1 ⋅ a1r ⋅ a1r K a1r

∴

1 1 1 1 (a − b) = +D ⇒ = + q a q a (n + 1) ab

⇒

( n + 1) ab 1 nb + a ... (ii) = ⇒ q= nb + a q (n + 1) ab

From Eq. (i),

b = (n + 1) p − na.

q { n (n + 1) p − n 2a + a } = (n + 1) a {(n + 1) p − na }

n − 1 1/ n 1/ n

) ]

⇒ n (n + 1) a 2 − {(n + 1)2 p + (n 2 − 1)q}a

where, r is the common ratio of GP a1 , a 2, K , a n. = [(a1 ⋅ a1. K n times ) (r1/ 2 ⋅ r3/3 ⋅ r 6/ 4 K r =

( n − 1 )n 2n

+ n (n + 1) pq = 0 )]1/ n

{(n + 1) p + (n − 1)q}2 − 4n 2pq > 0 ⇒ (n + 1) p + (n − 1)2q2 + 2 (n 2 − 1) pq − 4n 2pq > 0 2 2

1/ n

 1  ( n − 1)n    n −1  = a1 r 2  2   = a1 r 4        An = Hn =

and

=

∴ ⇒

An ⋅ H n =

…(i)

a1 + a 2 + ... + a n a1 (1 − r n ) = n n (1 − r ) 1 1 1 + +K+    a1 a 2 an n 1  1 1  1 + + K + n − 1   a1  r r

=

a1n (1 − r ) r n − 1 1 − rn

a1 (1 − r n ) a1n (1 − r )r n − 1 = a12r n − 1 × n (1 − r ) (1 − r n ) n

k =1

k =1

⇒

∏ AkH k = ∏ (a12r n − 1 ) = (a12 ⋅ a12 ⋅ a12K n times ) × r 0 ⋅ r1 ⋅ r 2K r n − 1

(n + 1)2 p2 + (n − 1)2q2 − 2 (n 2 + 1) pq > 0 2

⇒ ⇒

n

n

⇒ na − {(n + 1) p + (n − 1)q} a + npq = 0 2

Since, a is real, therefore

n −1 1 3 + 1 + + L+ 2 2 ]1/ n [a1n ⋅ r 2

Now,

…(i)

Putting it in Eq. (ii), we get

= [(a1 ) ⋅ (a1 ⋅ a1r )1/ 2 ⋅ (a1 ⋅ a1r ⋅ a1r 2)1/3 2

na + b n+1

Let H 1 , H 2, ... , H n be n harmonic means between a and b. 1 1 1 1 1 is an AP with common ∴ , , , ... , , a H1 H 2 Hn b (a − b) difference, D = . (n + 1) ab

(1 + a )(1 + b)(1 + c) > 7(a 4b4c4 )1/7

⇒

[from Eq. (i)]

 n  Gm =  ∏ AkH k  k = 1 

Hence, it is a false statement.

or

n −1 4 ]2 n

1/ 2 n

Here, equality holds only when x = a which is not possible. So, log a x + log x a is greater than 2.

⇒

= [a1r

= [Gm ]2n

Hence, it is false statement. 1 log a x + log a x 28. Since, > 1, using AM > GM 2

⇒

n ( n − 1) 2

⇒

q2 −

 n + 1 2 (n 2 + 1) 2 pq +   p >0  n − 1 (n − 1)2

2 2   n + 1   n + 1) 2 q2 − 1 +    pq +   p >0  n − 1   n −1   2   n + 1  (q − p) q −   p > 0  n − 1    2

⇒

Q

 n + 1 q < p or q >   p  n − 1  n + 1 2   p > p   n − 1   2

 n + 1 Hence, q cannot lie between p and   p.  n − 1

Sequences and Series 75 ⇒ 2n 2 + 1 ≥ 2n ⇒ n ≤ 6 and, also C >0⇒n >2 ∴The possible values of n are 3, 4, 5, 6 (2 × 9) − 6 So, at n = 3, C = = 12 8 − 6 −1 32 − 8 24 8 at, n = 4, C = = = ∉N 16 − 8 − 1 9 3 50 − 10 40 at, n = 5, C = = ∉N 32 − 10 − 1 21 72 − 12 60 and at, n = 6, C = = ∉N 64 − 12 − 1 51

32. Since a , b, c > 0 ⇒

Also,

(a + b + c) > (abc)1/3 3

…(i)

[using AM ≥ GM] 1 1 1 + + 1/3 a b c ≥  1 ⋅ 1 ⋅ 1 …(ii)    a b c 3 [using AM ≥ GM]

On multiplying Eqs. (i) and (ii), we get  1 1 1 (a + b + c)  + +   a b c 1 ≥ (abc)1/3 9 (abc)1/3 ∴

 1 1 1 (a + b + c)  + +  ≥ 9  a b c

∴ The required value of C = 12 for n = 3.

35. The 4th AM out of m AMs inserted between 3 and 243 is y1

33. For real numbers y1 , y2, y3 , the quantities 3 , 3

A4 = 3 + 4

y2

and 3y3 are positive real numbers, so according to the AM-GM inequality, we have

2

1     243 3 + 1  1/ 2 G2 = 3     = 3(81) = 27  3   

1

1

3y1 + 3y2 + 3y3 ≥ 3(3y1 . 3y2 . 3y3 )3

 240  ⇒ 4   = 24 ⇒ m + 1 = 40 ⇒ m = 39  m + 1 Hence, answer is 39.00

36.

Plan (i) If a, b, c are in GP, then they can be taken as a, ar, ar 2 where r, (r ≠ 0) is the common ratio. x + x2 + K + xn (ii) Arithmetic mean of x1, x2, K , xn = 1 n

1

x1 + x2 + x3 ≥ (x1x2x3 )3 3 On applying logarithm with base ‘3’, we get  x + x2 + x3  1 log3  1  ≥ (log3 x1 + log3 x2 + log3 x3 )  3  3 1 ⇒ 1 ≥ (log3 x1 + log3 x2 + log3 x3 ) 3 {Q x1 + x2 + x3 = 9} ∴ M =3 Now, log 2(m3 ) + log3 (M 2) = 3 log 2(4) + 2 log3 (3 ) =6 + 2 =8

34. Given arithmetic progression of positive integers terms a1 , a 2, a3 , ..... having common difference ‘2’ and geometric progression of positive integers terms b1 , b2, b3 , .... having common ratio ‘2’ with a1 = b1 = c, such that 2 (a1 + a 2 + a3 + ... + a n ) = b1 + b2 + b3 + .... + bn  2 n − 1 n 2 × [2C + (n − 1)2] = C  ⇒  2  2 −1 ⇒ ⇒ Q

2nC + 2n − 2n = 2 . C − C C [2n − 2n − 1] = 2n 2 − 2n C ∈ N ⇒ 2n 2 − 2n ≥ 2n − 2n − 1 2

n

 240  A4 = G 2 ⇒ 3 + 4   = 27  m + 1

Q

On applying logarithm with base ‘3’, we get 1   log3 (3y1 + 3y2 + 3y3 ) ≥ 1 + ( y1 + y2 + y3 )   3 =1 + 3 =4 {Q y1 + y2 + y3 = 9} ∴ m =4 Now, for positive real numbers x1 , x2 and x3 , according to AM-GM inequality, we have

…(i)

and the 2nd GM out of three GMs inserted between 3 and 243 is

3y1 + 3y2 + 3y3 ≥ (3y1 . 3y2 . 3y3 )3 3 ⇒

243 − 3 m+1

Let a , b, c be a , ar , ar 2, where r ∈ N a+ b+ c Also, =b+2 3 ⇒

a + ar + ar 2 = 3 (ar ) + 6

⇒

ar 2 − 2ar + a = 6 6 (r − 1)2 = ⇒ a Since, 6 /a must be perfect square and a ∈ N . So, a can be 6 only. ⇒

r −1 = ± 1 ⇒ r =2 a + a − 14 36 + 6 − 14 = =4 a+1 7 2

and

37. Using AM ≥ GM, a −5 + a −4 + a −3 + a −3 + a −3 + 1 + a 8 + a10 8 1

≥ (a − 5 ⋅ a − 4 ⋅ a − 3 ⋅ a − 3 ⋅ a − 3 ⋅ 1 ⋅ a 8 ⋅ a10 ) 8 ⇒ a −5 + a −4 + 3a −3 + 1 + a 8 + a10 ≥ 8 ⋅ 1 Hence, minimum value is 8.

4 Permutations and Combinations Topic 1 General Arrangement Objective Questions I (Only one correct option) 1. The number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is (2019 Main, 8 April II) (a) 306 (c) 360

(b) 310 (d) 288

are there, for which the sum of the diagonal entries of (2017 Adv.) M T M is 5 ? (b) 162

(c) 126

(d) 135

3. The number of integers greater than 6000 that can be formed using the digits 3, 5, 6, 7 and 8 without repetition is (2015 Main) (a) 216 (c) 120

(b) 192 (d) 72

4. The number of seven-digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is (2009) (a) 55

(b) 66

(c) 77

(d) 88

5. How many different nine-digit numbers can be formed from the number 22 33 55 888 by rearranging its digits so that the odd digits occupy even positions? (2000, 2M) (a) 16 (c) 60

(b) 36 (d) 180

6. An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2,5 and 7. The smallest value of n for which this is possible, is (1998, 2M) (a) 6

(b) 7

(c) 8

5 newspapers and every newspaper is read by 60 students. The number of newspapers is (1998, 2M) (a) atleast 30 (b) atmost 20 (c) exactly 25 (d) None of the above

8. A five digits number divisible by 3 is to be formed using

2. How many 3 × 3 matrices M with entries from {0, 1, 2}

(a) 198

7. In a collage of 300 students, every student reads

(d) 9

the numbers 0, 1 , 2, 3 , 4 and 5, without repetition. The total number of ways this can be done, is (1989, 2M) (a) 216 (c) 600

(b) 240 (d) 3125

9. Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 and then the men select the chairs from amongst the remaining. The number of possible arrangements is (a) 6C3 × 4 C2 (c) 4 C2 + 4 P3

(b) 4 P2 × 4 P3 (d) None of these

(1982, 2M)

10. The different letters of an alphabet are given. Words with five letters are formed from these given letters. Then, the number of words which have at least one letter repeated, is (1980, 2M) (a) 69760 (c) 99748

(b) 30240 (d) None

Analytical & Descriptive Question 11. Eighteen guests have to be seated half on each side of a long table. Four particular guests desire to sit on one particular side and three other on the other side. Determine the number of ways in which the sitting arrangements can be made. (1991, 4M)

Permutations and Combinations 77 Match the Columns Match the conditions/expressions in Column I with statement in Column II.

12. Consider all possible permutations of the letters of the word ENDEANOEL. Column I

(2008, 6M)

Column II

A.

The number of permutations containing the word ENDEA, is

p.

5!

B.

The number of permutations in which the letter E occurs in the first and the last positions, is

q.

2 × 5!

C.

The number of permutations in which none of the letters D, L, N occurs in the last five positions, is

r.

7 × 5!

D.

The number of permutations in which the letters A, E, O occur only in odd positions, is

s.

21 × 5!

Topic 2 Properties of Combinational and General Selections Objective Questions I (Only one correct option)

6. If n C 4 , n C 5 and n C 6 are in AP, then n can be (2019 Main, 12 Jan II)

1. There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is [2020 Main, 5 Sep II]

(a) 3000 (c) 2255

(b) 1500 (d) 2250

objects of which 10 are identical and the remaining 21 are distinct, is (2019 Main, 12 April I) (a) 2

−1

21

(b) 2

20

(c) 2

20

(d) 2

+1

3. Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is (2019 Main, 10 April II) (a) 180

(b) 210

(c) 170

(d) 190

4. Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then, the number of balls used to form the equilateral triangle is (2019 Main, 9 April II)

(a) 262

(b) 190

(c) 225

(d) 157

5. There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is (2019 Main, 12 Jan II) (a) 12

(b) 11

(c) 9

(b) 11

25

7. If

∑{

50

Cr ⋅

(d) 7

(c) 14

(d) 12

50 − r

C 25 − r } = K (50C 25 ),

r = 0

then, K is equal to (a) 224

8. If

(2019 Main, 10 Jan II)

(b) 225 − 1

(c) 225

(d) (25)2

3

20   Ci − 1 k   = , then k equals ∑ 20 20 Ci + Ci − 1  21 i=1 (2019 Main, 10 Jan I) 20

2. The number of ways of choosing 10 objects out of 31 20

(a) 9

(a) 100

(b) 400

(c) 200

(d) 50

9. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then, the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is (2017 Main) (a) 485

(b) 468

(c) 469

(d) 484

10. Let S = {1, 2, 3, …… , 9}. For k = 1, 2 , …… 5, let N k be the number of subsets of S, each containing five elements out of which exactly odd. Then k are (2017 Adv.) N1 + N 2 + N 3 + N 4 + N 5 = (a) 210

(b) 252

(c) 126

(d) 125

11. A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include atmost one boy, the number of ways of selecting the team is (2016 Adv.) (a) 380

(b) 320

(c) 260

(d) 95

12. Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If (2013 Main) Tn + 1 − Tn = 10, then the value of n is (a) 7 (c) 10

(b) 5 (d) 8

78 Permutations and Combinations 13. If r , s, t are prime numbers and p, q are the positive

17. Let n ≥ 2 be an integer. Take n distinct points on a circle

integers such that LCM of p, q is r 2s4 t 2 ,then the number of ordered pairs ( p, q) is (2006, 3M)

and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is (2014 Adv.)

(a) 252

(b) 254

(c) 225

(d) 224

14. The value of the expression 47 C 4 +

5

∑

52− j

C 3 is

j =1

(1980, 2M)

47

(a) C5 (b) 52C5 (c) 52C4 (d) None of these

Fill in the Blanks 18. Let A be a set of n distinct elements. Then, the total number of distinct functions from A to A is…and out of these… are onto functions. (1985, 2M)

19. In a certain test, a i students gave wrong answers to at least i questions, where i = 1, 2, K , k. No student gave more that k wrong answers. The total number of wrong answers given is … . (1982, 2M)

Match the Columns 15. In a high school, a committee has to be formed from a group of 6 boys M1 , M 2, M 3, M 4 , M 5, M 6 and 5 girls G1 , G2, G 3, G 4 , G 5. (i) Let α1 be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls. (ii) Let α 2 be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls. (iii) Let α 3 be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls. (iv) Let α 4 be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls such that both M1 and G1 are NOT in the committee together. (2018 Adv.)

List-I

List-II

P.

The value of α1 is

1.

136

Q.

The value of α 2 is

2.

189

R.

The value of α 3 is

3.

192

S.

The value of α 4 is

4.

200

5.

381

6.

461

The correct option is (a) P → 4; Q → 6; R → 2; S → 1 (b) P → 1; Q → 4; R → 2; S → 3 (c) P → 4; Q → 6; R → 5; S → 2 (d) P → 4; Q → 2; R → 3; S → 1

Integer & Numerical Answer Type Questions 16. Five persons A , B, C , D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ……… (2019 Adv.)

True/False 20. The product of any r consecutive natural numbers is always divisible by r !.

(1985, 1M)

Analytical & Descriptive Questions 21. A committee of 12 is to be formed from 9 women and 8 men. In how many ways this can be done if at least five women have to be included in a committee ? In how many of these committees (i) the women are in majority? (ii) the men are in majority?

(1994, 4M)

22. A student is allowed to select atmost n books from n

collection of (2n + 1) books. If the total number of ways in which he can select at least one books is 63, find the value of n. (1987, 3M)

23. A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw ? (1986, 2 12 M)

24. 7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife has also 7 relatives ; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife's relatives? (1985, 5M)

25. m men and n women are to be seated in a row so that no

two women sit together. If m > n, then show that the number of ways in which they can be seated, is m ! (m + 1) ! . (m − n + 1) ! (1983, 2M)

26. mn squares of equal size are arranged to form a rectangle of dimension m by n where m and n are natural numbers. Two squares will be called ‘neighbours’ if they have exactly one common side. A natural number is written in each square such that the number in written any square is the arithmetic mean of the numbers written in its neighbouring squares. Show that this is possible only if all the numbers used are (1982, 5M) equal.

27. If n C r −1 = 36, n C r = 84 and n C r +1 = 126, then find the values of n and r.

(1979, 3M)

Permutations and Combinations 79

Topic 3 Multinomial, Repeated Arrangement and Selection Objective Question I (Only one correct option)

Integer & Numerical Answer Type Questions

1. The number of 6 digits numbers that can be formed

8. An engineer is required to visit a factory for exactly four

using the digits 0, 1, 2,5, 7 and 9 which are divisible by 11 and no digit is repeated, is (2019 Main, 10 April I)

days during the first 15 days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1-15 June 2021 is ……… (2020 Adv.)

(a) 60 (c) 48

(b) 72 (d) 36

2. A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with atleast 3 females, then (2019 Main, 9 April I) (a) m = n = 68 (c) m = n = 78

(b) m + n = 68 (d) n = m − 8

3. Consider three boxes, each containing 10 balls labelled 1, 2, …, 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni , the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that (2019 Main, 12 Jan I) n1 < n2 < n3 is (a) 82 (c) 240

(b) 120 (d) 164

4. The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to (2019 Main, 9 Jan II) (a) 374 (c) 372

(b) 375 (d) 250

5. Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is (2019 Main, 9 Jan I) (a) 350 (c) 200

(b) 500 (d) 300

6. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is (2016 Main) (a) 46th (c) 52nd

(b) 59th (d) 58th

7. The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN, is (2007, 3M) (a) 360 (c) 96

(b) 192 (d) 48

9. If the letters of the word ‘MOTHER’ be permuted and all the words so formed (with or without meaning) be listed as in dictionary, then the position of the word ‘MOTHER’ is………… (2020 Main, 2 Sep I)

10. The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of (2018 Adv.) digits is allowed, is ..................... .

11. Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice y and no other letter is repeated. Then, = (2017 Adv.) 9x

12. Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively m in the queue. Then, the value of is n (2015 Adv.)

13. Let n1 < n2 < n3 < n4 < n5 be positive integers such that n1 + n2 + n3 + n4 + n5 = 20. The number of such distinct (2014 Adv.) arrangements (n1 , n2 , n3 , n4 , n5 ) is

Fill in the Blanks 14. Let n and k be positive integers such that n ≥ The number of solutions (x1 , x2 ,... , xk ), x1 ≥ 1, x2 ≥ 2, ... , xk ≥ k for all integers x1 + x2 + ... + xk = n is …

k(k + 1) . 2

satisfying (1996, 2M)

15. Total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two ‘–’signs occur together is… . (1988, 2M)

Analytical & Descriptive Question 16. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty? (1981, 4M)

80 Permutations and Combinations

Topic 4 Distribution of Object into Group Objective Questions I (Only one correct option)

4. The total number of ways in which 5 balls of different

1. A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is (2019 Main, 12 April II) 1750, then n is equal to (a) 28 (c) 25

(b) 27 (d) 24

(a) 75

having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is (2019 Main, 9 Jan II)

(b) 32 (d) 9

3.. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf, so that the dictionary is always in the middle. The number of such arrangements is (2018 Main)

(a) atleast 1000 (b) less than 500 (c) atleast 500 but less than 750 (d) atleast 750 but less than 1000

(b) 150

(c) 210

(d) 243

5. The number of arrangements of the letters of the word BANANA in which the two N’s do not appear adjacently, is (2002, 1M) (a) 40

2. Let S be the set of all triangles in the xy-plane, each

(a) 36 (c) 18

colours can be distributed among 3 persons so that each person gets at least one ball, is (2012)

(b) 60

(c) 80

(d) 100

Analytical & Descriptive Questions 6. Using permutation or otherwise, prove that integer, where n is a positive integer.

n2 ! is an (n !)n (2004, 2M)

7. In how many ways can a pack of 52 cards be (i) divided equally among four players in order (ii) divided into four groups of 13 cards each (iii) divided in 4 sets, three of them having 17 cards each and the fourth just one card? (1979, 3M)

Integer & Numerical Answer Type Question 8. In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ……… (2020 Adv.)

Topic 5 Dearrangement and Number of Divisors Objective Question I (Only one correct option)

Fill in the Blank

1. Number of divisors of the form (4n + 2), n ≥ 0 of the integer 240 is

(1998, 2M)

(a) 4 (c) 10

(b) 8 (d) 3

2. There are four balls of different colours and four boxes of colours, same as those of the balls. The number of ways in which the balls, one each in a box, could be placed such that a ball does not go to a box of its own colour is.... . (1992, 2M)

Answers Topic 1 1. (b)

Topic 3 2. (a)

5. (c)

6. (b)

9. (d)

10. (a)

3. (b)

4. (c)

7. (c)

8. (a)

11.

9

P4 × 9 P3 (11 )!

13. (7)

Topic 2 (d) (a) (a) (c)

2. (c)

3. (b)

6. (d)

7. (c)

10. (625) 11. (5) 1 2 14. (2n − k + k − 2 ) 2

4. (a) 8. (495) 12. (5) 15. (35 ways)

16. (300) 2. 6. 10. 14.

3. 7. 11. 15.

(c) (c) (c) (c) n

17. (5)

5. (d) 9. (309)

12. ( A → p; B → s; C → q ; D → q )

1. 5. 9. 13.

1. (a)

18. n , ∑ ( −1 ) n

n −r n

4. 8. 12. 16.

(c) (c) (a) (c) n

Cr (r )

(b) (a) (b) (30)

19. 2 − 1

21. 6062, (i) 2702 (ii) 1008

23. (64)

24. (485)

1. (c) 5. (a)

2. (a)

3. (a) 4. (b) (52 )! (52 )! (52 )! 7. (i) (ii) (iii) 4 4 (13 !) 4 ! (13 !) 3 ! (17 ) 3

n

r =1

20. (True)

Topic 4

22. n = 3

27. (n = 9 and r = 3 )

8. (1080)

Topic 5 1. (a)

2. (9)

Hints & Solutions Topic 1 General Arrangement 1. Following are the cases in which the 4-digit numbers strictly greater than 4321 can be formed using digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) Case-I 4

3

∴ Total number of numbers = 42 + 35 = 77

5. X  X  X  X  X . The four digits 3, 3, 5, 5 can be 2

4! = 6 ways. 2 !2 ! The five digits 2, 2, 8, 8, 8 can be arranged at (X ) places 5! in ways = 10 ways. 2 !3 ! Total number of arrangements = 6 × 10 = 60 [since, events A and B are independent, therefore A ∩ B = A × B] arranged at () places in

2/3/4/5

4 ways

4 numbers

Case-II 4

Case II Four 1’s, three 2’s 7! Number of numbers = = 35 4!3!

3 3/4/5 0/1/2/3/4/5 3 ways

3×6=18 numbers

6 ways

6. Distinct n-digit numbers which can be formed using

Case-III 4 4/5

0/1/2/3/4/5

2 ways

6 ways

2×6×6=72 numbers

Case-IV

digits 2, 5 and 7 are 3n . We have to find n, so that 3n ≥ 900 n− 2 ⇒ 3 ≥ 100 ⇒ n −2 ≥5 ⇒ n ≥ 7, so the least value of n is 7.

7. Let n be the number of newspapers which are read by

5

6×6×6=216 numbers

the students. Then,

0/1/2/3/4/5

⇒

6 ways

So, required total numbers = 4 + 18 + 72 + 216 = 310

2. Sum of diagonal entries of M M is ∑ a . T

9

∑a

2 i

2 i

=5

i =1

Possibilities 9! matrices 7! 9! matrices II. 1, 1, 1, 1, 1, 0, 0, 0, 0, which gives 4! × 5! I. 2, 1, 0, 0, 0, 0, 0, 0, 0, which gives

Total matrices = 9 × 8 + 9 × 7 × 2 = 198

3. The integer greater than 6000 may be of 4 digits or 5 digits. So, here two cases arise. Case I When number is of 4 digits. Four-digit number can start from 6, 7 or 8. Thus, total number of 4-digit numbers, which are greater than 6000 = 3 × 4 × 3 × 2 = 72 Case II When number is of 5 digits. Total number of five-digit numbers which are greater than 6000 = 5 ! = 120 ∴ Total number of integers = 72 + 120 = 192

4. There are two possible cases Case I Five 1’s, one 2’s, one 3’s 7! Number of numbers = = 42 5!

60n = (300) × 5 n = 25

8. Since, a five-digit number is formed using the digits {0, 1, 2, 3, 4 and 5} divisible by 3 i.e. only possible when sum of the digits is multiple of three. Case I Using digits 0, 1, 2, 4, 5 Number of ways = 4 × 4 × 3 × 2 × 1 = 96 Case II Using digits 1, 2, 3, 4, 5 Number of ways = 5 × 4 × 3 × 2 × 1 = 120 ∴ Total numbers formed = 120 + 96 = 216

9. Since, the first 2 women select the chairs amongst 1 to 4 in 4 P2 ways. Now, from the remaining 6 chairs, three men could be arranged in 6 P3. ∴ Total number of arrangements = 4 P2 × 6P3.

10. Total number of five letters words formed from ten

different letters = 10 × 10 × 10 × 10 × 10 = 105 Number of five letters words having no repetition = 10 × 9 × 8 × 7 × 6 = 30240 ∴ Number of words which have at least one letter repeated = 105 − 30240 = 69760

11. Let the two sides be A and B. Assume that four particular guests wish to sit on side A. Four guests who wish to sit on side A can be accommodated on nine chairs in 9 P4 ways and three guests who wish to sit on side B can be accommodated in 9 P3 ways. Now, the remaining guests are left who can sit on 11 chairs on both the sides of the table in (11!) ways. Hence, the total number of ways in which 18 persons can be seated = 9P4 × 9P3 × (11)!.

82 Permutations and Combinations 12.

A. If ENDEA is fixed word, then assume this as a single letter. Total number of letters = 5 Total number of arrangements = 5 !. B. If E is at first and last places, then total number of permutations = 7 !/ 2 ! = 21 × 5 ! C. If D, L, N are not in last five positions ← D, L, N, N → ← E, E, E, A, O → 4! 5! Total number of permutations = × = 2 × 5 ! 2! 3! D. Total number of odd positions = 5 5! Permutations of AEEEO are . 3! Total number of even positions = 4 4! ∴ Number of permutations of N, N, D, L = 2! 5! 4! ⇒ Total number of permutations = × = 2 × 5 ! 3! 2!

Topic 2 Properties of Combinational and General Selections 1. As each section has 5 questions, so number of ways to select 5 questions are S1 S 2 S 3 1 1 3 1 3 1 1 2 2 2 1 2 2 2 1

0 identical + 10 distincts, number of ways = 1 ×

21

1 identical + 9 distincts, number of ways = 1 ×

C9

21

2 identicals + 8 distincts, number of ways = 1 × . . . . . . . . . . . .

C9 +

⇒ C11 +

21

21

C8 + K +

C12 +

21

21

21

C 0 = x (let)

C13 + K +

21

C 21 = x

⇒

2x = 2

⇒ x=2

20

two

games

with

every

other

and the number of games played between the men and the women = 2 × mC1 × 2C1 Now, according to the question, ⇒

C8

… (i) … (ii) [Q nC r = nC n − r ]

On adding both Eqs. (i) and (ii), we get 2x = 21C 0 + 21C1 + 21C 2 + K + 21C10 + 21C11 + 21C12 + K + 21

triangle.

According to the question, we have n ( n + 1) + 99 = ( n − 2 )2 2 ⇒ n 2 + n + 198 = 2 [ n 2 − 4 n + 4 ] ⇒ n 2 − 9n − 190 = 0 ⇒ n 2 − 19n + 10n − 190 = 0 ⇒ ( n − 19 )( n + 10 ) = 0 ⇒ n = 19, − 10 ⇒ n = 19 [Qnumber of balls n > 0]

C10

So, total number of ways in which we can choose 10 objects is 21

4. Let there are n balls used to form the sides of equilateral

∴ Number of games played by the men between themselves = 2 × mC 2

remaining 21 are distinct, so in following ways, we can choose 10 objects.

21

Q 20C 2 is selection of any two vertices of 20-sided polygon which included the sides as well. So, required number of total beams = 20C 2 − 20 [Q the number of diagonals in a n-sided closed polygon = nC 2 − n] 20 × 19 = − 20 = 190 − 20 = 170 2

participant plays participant.

2. Given that, out of 31 objects 10 are identical and

C10 +

Now, the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then total number of beams = number of diagonals of 20-sided polygon.

5. Since, there are m-men and 2-women and each

∴Total number of selection of 5 questions = 3 × ( 5C1 × 5C1 × 5C 3 ) + 3 × ( 5C1 × 5C 2 × 5C 2 ) = 3(5 × 5 × 10) + 3(5 × 10 × 10) = 750 + 1500 = 2250.

21

have been erected along the boundary of a circular stadium.

Now, number of balls used to form an equilateral n (n + 1) triangle is 2 19 × 20 = = 190. 2

3 1 1 and

3. It is given that, there are 20 pillars of the same height

21

C 21

⇒ ⇒ ⇒ ⇒ ⇒ ⇒

2 mC 2 = 2 mC1 2C1 + 84 m! = m × 2 + 42 2 !(m − 2)! m(m − 1) = 4m + 84 m2 − m = 4m + 84 m2 − 5m − 84 = 0 m2 − 12m + 7m − 84 = 0 m(m − 12) + 7 (m − 12) = 0 m = 12 n

n

[Q m > 0]

n

6. If C 4 , C 5 and C 6 are in AP, then 2 ⋅n C 5 = n C 4 + n C 6 [If a , b, c are in AP , then 2b = a + c] n! n! n! = + ⇒ 2 5 !(n − 5)! 4 !(n − 4)! 6 !(n − 6)! n!  n  Q C r = r !(n − r )! 

Permutations and Combinations 83 ⇒

2 5 ⋅ 4 !(n − 5) (n − 6)!

9. Given, X has 7 friends, 4 of them are ladies and 3 are men while Y has 7 friends, 3 of them are ladies and 4 are men.

1 1 + 4 !(n − 4) (n − 5) (n − 6)! 6 ⋅ 5 ⋅ 4 ! (n − 6)! 2 1 1 = + 5(n − 5) (n − 4) (n − 5) 30 2 30 + (n − 4) (n − 5) = 5(n − 5) 30 (n − 4) (n − 5) =

⇒ ⇒

∴ Total number of required ways = 3C 3 × 4C 0 × 4C 0 × 3C 3 + 3C 2 × 4C1 × 4C1 × 3C 2 + 3C1 × 4C 2 × 4C 2 × 3C1 + 3C 0 × 4C 3 × 4C 3 × 3C 0 = 1 + 144 + 324 + 16 = 485

10. N i = 5C k × 4C 5 − k

⇒ 12 (n − 4) = 30 + n − 9n + 20 ⇒ n 2 − 21n + 98 = 0 2 ⇒ n − 14n − 7n + 98 = 0 ⇒ n (n − 14) − 7(n − 14) = 0 ⇒ (n − 7) (n − 14) = 0 ⇒ n = 7 or 14 2

N1 = 5 × 1 N 2 = 10 × 4 N 3 = 10 × 6 N4 = 5 × 4 N5 = 1 N 1 + N 2 + N 3 + N 4 + N 5 = 126

25

7. Given, Σ { 50C r .50− r C 25− r } = K 50C 25 r=0

50 ! (50 − r )!   50 ⇒ Σ  ×  = K C 25 r = 0  r !(50 − r )! (25 − r )! 25 ! 25  50 ! 25 !  50 Σ  × ⇒  = K C 25 r = 0  25 ! 25 ! r !(25 − r )! [on multiplying 25 ! in numerator and denominator.] 25

⇒

50

25

C 25 Σ

Cr = K

25

r = 0

 50 50 !  Q C 25 = 25 ! 25 !   

50

C 25 25

⇒

K= Σ

Cr = 2

25

r = 0

K =2

∑

i=1

⇒

∴

C 3 − C 3 = 10

[given]

[Q C r + C r

n

n

C3

n

n

n

+1

=

n +1

Cr

+1

]

C 2 = 10 ⇒ n = 5

∴ Selection of p and q are as under

3

  Ci − 1 k  20  = 20 21  Ci + Ci − 1  20

∑ 20

3

 20C i − 1  k  21  = 21 C  i 

(Q nC r + nC r − 1 =

∑

   20C i − 1  k   = 21 21 20  Ci − 1   i  3 20 k  i   = ∑   21 21 i=1 1 (21)3

⇒

⇒

− Tn =

n +1

13. Since, r , s, t are prime numbers.

i=1

⇒

+1

n

⇒

n +1

Cr )

3

⇒

Tn

n +1

⇒ C 2 + C 3 − C 3 = 10

25

20

i=1

⇒

12. Given, Tn = nC 3 ⇒ Tn + 1 = n

8. Given, 20

(atmost one boy) i.e. (1 boy and 3 girls) or (4 girls) = 6 C 3 ⋅4 C1 + 6C 4 …(i) Now, selection of captain from 4 members = 4 C1 …(ii) ∴ Number of ways to select 4 members (including the selection of a captain, from these 4 members) = ( 6C 3 ⋅4 C1 + 6C 4 ) 4C1 = (20 × 4 + 15) × 4 = 380 ∴

25

[Q nC 0 + nC1 + n C 2 + ....+ nC n = 2n ] ⇒

11. We have, 6 girls and 4 boys. To select 4 members

1 (21)3

20

∑i i=1

2

3

=

n  n Q C r =  r

n−1

 Cr − 1  

k 21

k  n (n + 1)  =   2  n = 20 21 2  3  n (n + 1)   3 3 Q 1 + 2 + K + n =    2    2 21  20 × 21 k=   = 100 (21)3  2  k = 100

p r0 r1 r2

q r2 r2 r 0 , r1 , r 2

Number of ways 1 way 1 way 3 ways

∴ Total number of ways to select, r = 5 Selection of s as under s0 s1 s2 s3 s4

s4 s4 s4 s4

1 way 1 way 1 way 1 way 5 ways

∴ Total number of ways to select s = 9 Similarly, the number of ways to select t = 5 ∴ Total number of ways = 5 × 9 × 5 = 225 5

14. Here, 47 C 4 + ∑

52 − j

C3

j =1

= 47C 4 + = ( C4 + 47

51

C 3 + 50C 3 + 49C 3 + 48C 3 +

47

47

C3 ) + C3 + C3 + C3 + 48

49

50

C3 51

C3

[using C r + C r − 1 = n +1C r ] n

n

84 Permutations and Combinations = (48C 4 + 48C 3 ) + 49C 3 + 50C 3 + 50

= ( C4 + C3 ) + C3 +

51

= ( C4 + C3 ) +

51

51

C3 = C4 +

49 50

49 50

51

18. Let A = { x1 , x2 , ... , xn }

C3

∴ Number of functions from A to A is n n and out of these

C3 51

C3 = C4 52

15. Given 6 boys M1 , M 2 , M 3 , M 4 , M 5 , M 6 and 5 girls G1 , G2 , G3 , G4 , G5

(ii) α 2 → Total number of ways selecting at least 2 member and having equal number of boys and girls 5 5 5 5 5 i.e., 6C1 C1 + 6C 2 C 2 + 6C 3 C 3 + 6C 4 C 4 + 6C 5 C 5 = 30 + 150 + 200 + 75 + 6 = 461 ⇒ α 2 = 461 (iii) α 3 → Total number of ways of selecting 5 members in which at least 2 of them girls 6 6 6 6 i.e., 5C 2 C 3 + 5C 3 C 2+ 5C 4 C1 + 5C 5 C 0 = 200 + 150 + 30 + 1 = 381 α 3 = 381 (iv) α 4 → Total number of ways of selecting 4 members in which at least two girls such that M1 and G1 are not included together. G1 is included → 4 C1 ⋅ 5C 2 + 4C 2 ⋅ 5C1 + 4C 3 = 40 + 30 + 4 = 74 M1 is included → 4 C 2 ⋅ 5C1 + 4C 3 = 30 + 4 = 34 G1 and M1 both are not included 4 C 4 + 4 C 3 ⋅ 5C1 + 4C 2 ⋅ 5C 2 1 + 20 + 60 = 81 ∴ Total number = 74 + 34 + 81 = 189 α 4 = 189 Now, P → 4; Q → 6; R → 5; S → 2 Hence, option (c) is correct.

16. Given that, no two persons sitting adjacent in circular arrangement, have hats of same colour. So, only possible combination due to circular arrangement is 2 + 2 + 1. So, there are following three cases of selecting hats are 2R + 2B + 1G or 2B + 2G + 1R or 2G + 2R + 1B. To distribute these 5 hats first we will select a person which we can done in 5C1 ways and distribute that hat which is one of it’s colour. And, now the remaining four hats can be distributed in two ways. So, total ways will be 3 × 5C1 × 2 = 3 × 5 × 2 = 30 PLAN Number of line segment joining pair of adjacent point = n Number of line segment obtained joining n points on a circle = nC 2 Number of red line segments = nC 2 − n

Number of blue line segments = n ∴ ⇒ ⇒

n−r n

C r (r )n are onto functions.

r =1

19. The

(i) α 1 → Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls. i..e, 6C 3 × 5C 2 = 20 × 10 = 200 ∴ α 1 = 200

17.

n

∑ (−1)

C2 − n = n n (n − 1) = 2n 2 n

n =5

number of students answering exactly k (1 ≤ k ≤ n − 1) questions wrongly is 2n − k − 2n − k − 1 . The number of students answering all questions wrongly is 20. Thus, total number of wrong answers = 1 (2n − 1 − 2n − 2 ) + 2 (2n − 2 − 2n − 3 ) + K + (n − 1) (21 − 20 ) + 20 ⋅ n = 2n − 1 + 2n − 2 + 2n − 3 + K + 21 + 20 = 2n − 1

20. Let r consecutive integers be x + 1, x + 2, K , x + r. ∴ (x + 1) (x + 2) K (x + r ) = (x + r )! r ! ⋅ = (x)! r !

=

(x + r )(x + r − 1) K (x + 1) x ! x! x + r

C r ⋅ (r )!

Thus, (x + 1) (x + 2) K (x + r ) = x + rC r ⋅ (r )! , which is clearly divisible by (r )! . Hence, it is a true statement.

21. Given that, there are 9 women and 8 men, a committee of 12 is to be formed including at least 5 women. This can be done in = (5 women and 7 men) + (6 women and 6 men) + (7 women and 5 men) + (8 women and 4 men) + (9 women and 3 men) ways Total number of ways of forming committee = (9C 5 . 8C7 ) + (9C 6. 8C 6 ) + (9C7 . 8C 5 ) + (9C 8 . 8C 4 ) + (9C 9. 8C 3 ) = 1008 + 2352 + 2016 + 630 + 56 = 6062 (i) The women are in majority = 2016 + 630 + 56 = 2702 (ii) The man are in majority = 1008 ways

22. Since, student is allowed to select at most n books out of (2n + 1) books. 2 n +1 ∴ C1 + We know ⇒ 2( ⇒

2n +1

2n +1

2 n +1

C0 +

C0 +

2n + 1

2n +1

C1 +

C 2 + .... +

2n +1

C1 +

2n +1

2 n +1

C1 + ..... + 2n + 1

C 2 + ... +

C 2 + ... +

2n +1

C n = 63

2n +1

C2 n + 1 = 2

2n +1

... (i) 2n +1

C n ) = 22 n + 1

C n = (22 n − 1)

.... (ii)

From Eqs. (i) and (ii), we get 22 n − 1 = 63 ⇒

22 n = 64

⇒

2n = 6

⇒

n =3

23. Case I

When one black and two others balls are drawn. Number of ways = 3C1 ⋅6 C 2 = 45 ⇒ Case II When two black and one other balls are drawn

⇒

Number of ways = 3C 2 ⋅6 C1 = 18

Permutations and Combinations 85 n

Case III When all three black balls are drawn ⇒ ∴

Also given,

Number of ways = C 3 = 1 Total number of ways = 45 + 18 + 1 = 64 3

⇒

24. The possible cases are

⇒

Case I A man invites 3 ladies and women invites 3 gentlemen. Number of ways = 4C 3 ⋅4 C 3 = 16 Case II A man invites (2 ladies, 1 gentleman) and women invites (2 gentlemen, 1 lady). Number of ways = ( C 2 ⋅ C1 ) ⋅ ( C1 ⋅ C 2 ) = 324 4

3

3

4

Case III A man invites (1 lady, 2 gentlemen) and women invites (2 ladies, 1 gentleman). Number of ways = (4 C1 ⋅3 C 2 ) ⋅ (3C 2 ⋅ 4C1 ) = 144 Case IV A man invites (3 gentlemen) and women invites (3 ladies). Number of ways = 3C 3 ⋅3 C 3 = 1 ∴ Total number of ways, = 16 + 324 + 144 + 1 = 485

25. Since, m men and n women are to be seated in a row so that no two women sit together. This could be shown as × M1 × M 2 × M 3 × ... × M m × which shows there are (m + 1) places for n women. ∴ Number of ways in which they can be arranged (m)! ⋅ (m + 1)! = (m) ! m + 1 Pn = (m + 1 − n )!

26. Let mn squares of equal size are arrange to form a rectangle of dimension m by n. Shown as, from figure. neighbours of x1 are {x2 , x3 , x4 , x5} x5 are {x1 , x6 , x7 } and x7 are {x5 , x4 }. x + x3 + x4 + x5 x + x6 + x7 x1 = 2 , x5 = 1 ⇒ 4 3 x + x5 and x7 = 4 2 x + x6 + x7 ∴ 4x1 = x2 + x3 + x4 + 1 3 x + x5 ⇒ 12x1 = 3x2 + 3x3 + 3x4 + x1 + x6 + 4 2 ⇒

24x1 = 6x2 + 6x3 + 6x4 + 2x1 + 2x6 + x4 + x5

⇒ 22x1 = 6x2 + 6x3 + 7x4 + x5 + 2x6 where, x1 , x2 , x3 , x4 , x5 , x6 are all the natural numbers and x1 is linearly expressed as the sum of x2 , x3 , x4 , x5 , x6 where sum of coefficients are equal only if, all observations are same. ⇒ x2 = x3 = x4 = x5 = x6 ⇒ All the numbers used are equal. Cr n−r+1 = C r −1 r

n

27. We know that, ⇒ ⇒

n

84 7 n − r + 1 = = 36 3 r 3n − 10r + 3 = 0

[given] …(i)

Cr 84 = C r +1 126 r+1 2 = n−r 3

n

2n − 5r − 3 = 0

…(ii)

On solving Eqs. (i) and (ii), we get r = 3 and n = 9

Topic 3 Multinomial, Repeated Arrangement and Selection 1.

Key Idea Use divisibility test of 11 and consider different situation according to given condition.

Since, the sum of given digits 0 + 1 + 2 + 5 + 7 + 9 = 24 Let the six-digit number be abcdef and to be divisible by 11, so the difference of sum of odd placed digits and sum of even placed digits should be either 0 or a multiple of 11 means|(a + c + e) − (b + d + f )|should be either 0 or a multiple of 11. Hence, possible case is a + c + e = 12 = b + d + f (only) Now, Case I set { a , c, e} = {0, 5, 7} and set { b, d , f } = {1, 2, 9} So, number of 6-digits numbers = (2 × 2 !) × (3 !) = 24 [Q a can be selected in ways only either 5 or 7]. Case II Set { a , c, e} = {1, 2, 9} and set { b, d , f } = {0, 5, 7} So, number of 6-digits numbers = 3 ! × 3 ! = 36 So, total number of 6-digits numbers = 24 + 36 = 60

2. Since there are 8 males and 5 females. Out of these 13 members committee of 11 members is to be formed. According to the question, m = number of ways when there is at least 6 males = (8C 6 × 5C 5 ) + (8C7 × 5 C 4 ) + (8C 8 × 5C 3 ) = (28 × 1) + (8 × 5)+ (1 × 10) = 28 + 40 + 10 = 78 and n = number of ways when there is at least 3 females = ( 5C 3 × 8 C 8 ) + ( 5C 4 × 8 C7 ) + ( 5C 5 × 8 C 6 ) = 10 × 1 + 5 × 8 + 1 × 28 = 78 So, m = n = 78

3. Given there are three boxes, each containing 10 balls labelled 1, 2, 3, … , 10. Now, one ball is randomly drawn from each boxes, and ni denote the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is same as selection of 3 different numbers from numbers {1, 2, 3, … , 10} = 10C 3 = 120.

4. Using the digits 0, 1, 3, 7, 9 number of one digit natural numbers that can be formed = 4,

86 Permutations and Combinations number of two digit natural numbers that can be formed = 20,

4×5

Number of words starting with CH, CI, CN is 4! each. Similarly, number of words before the first word starting with CO = 4! + 4! + 4! + 4! = 96. The word starting with CO found first in the dictionary is COCHIN. There are 96 words before COCHIN.

(Q 0 can not come in Ist box)

8. Let the engineer visits the factory first time after x1 days

number of three digit natural numbers that can be formed = 100

to 1 June, second time after x2 days to first visit and so on. ∴ x1 + x2 + x3 + x4 + x5 = 11 where x1 , x5 ≥ 0 and x2 , x3 , x4 ≥ 1 according to the requirement of the question. Now, let x2 = a + 1, x3 = b + 1 and x4 = c + 1 where a , b, c ≥ 0 ∴New equation will be x1 + a + b + c + x5 = 8 Now, the number of all possible ways in which the engineer can made visits is equals to the non-negative integral solution of equation x1 + a + b + c + x5 = 8, and it is equal to 12 × 11 × 10 × 9 8 + 5−1 C 5 − 1 = 12C 4 = = 495. 4 ×3 ×2

4×5× 5

and number of four digit natural numbers less than 7000, that can be formed = 250

2×5× 5×5

(Q only 1 or 3 can come in Ist box) ∴Total number of natural numbers formed = 4 + 20 + 100 + 250 = 374

5. Number of girls in the class = 5 and number of boys in the class = 7

9. Given word in MOTHER, now alphabetical order

Now, total ways of forming a team of 3 boys and 2 girls = 7C 3 ⋅5 C 2 = 350 But, if two specific boys are in team, then number of ways = 5C1 ⋅5 C 2 = 50 Required ways, i.e. the ways in which two specific boys are not in the same team = 350 − 50 = 300. Alternate Method Number of ways when A is selected and B is not = 5C 2 ⋅5 C 2 = 100 Number of ways when B is selected and A is not = 5C 2 ⋅5 C 2 = 100 Number of ways when both A and B are not selected = 5C 3 ⋅5 C 2 = 100 ∴ Required ways = 100 + 100 + 100 = 300. 4! 6. Clearly, number of words start with A = = 12 2! Number of words start with L = 4 ! = 24 4! Number of words start with M = = 12 2! 3! Number of words start with SA = =3 2! Number of words start with SL = 3 ! = 6 Note that, next word will be “SMALL”. Hence, the position of word “SMALL” is 58th.

7. Arrange the letters of the word COCHIN as in the order of dictionary CCHINO. Consider the words starting from C. There are 5! such words. Number of words with the two C’s occupying first and second place = 4 ! .

of letters is EHMORT, so number of words start with letter. E ------ is 5! H ------ is 5! M E ------ is 4! M H ------ is 4! M O E ------ is 3! M O H ------ is 3! M O R ------ is 3! M O T E ------ is 2! M O T H E R is 1 So, position of the word ‘MOTHER’ is 5! + 5! + 4! + 4! + 3! + 3! + 3! + 2! + 1 = 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1 = 309

10. A number is divisible by 4 if last 2 digit number is divisible by 4. ∴ Last two digit number divisible by 4 from (1, 2, 3, 4, 5) are 12, 24, 32, 44, 52 ∴ The number of 5 digit number which are divisible by 4, from the digit (1, 2, 3, 4, 5) and digit is repeated is 5 × 5 × 5 × (5 ×1) = 625

11. x = 10 ! y = 10C1 × 9C 8 ×

12. Here,

B1

10 ! 10 ! y 10 = 10 × 9 × ⇒ = =5 2 2! 9x 2

B2

B3

B4

B5

Out of 5 girls, 4 girls are together and 1 girl is separate. Now, to select 2 positions out of 6 …(i) positions between boys = 6C 2 …(ii) 4 girls are to be selected out of 5 = 5C 4 Now, 2 groups of girls can be arranged in 2 !ways. …(iii) Also, the group of 4 girls and 5 boys is arranged in 4 ! × 5 ! ways . …(iv)

Permutations and Combinations 87 Now, total number of ways ∴ and ⇒

13.

= 6C 2 × 5C 4 × 2 ! × 4 ! × 5 ! [from Eqs. (i), (ii), (iii) and (iv)] m = 6C 2 × 5C 4 × 2 ! × 4 ! × 5 ! n = 5! × 6! m 6C 2 × 5C 4 × 2 ! × 4 ! × 5 ! 15 × 5 × 2 × 4 ! = = =5 n 6! × 5! 6 × 5 × 4!

PLAN Reducing the equation to a newer equation, where sum of variables is less. Thus, finding the number of arrangements becomes easier.

As, n1 ≥ 1, n2 ≥ 2 , n3 ≥ 3, n4 ≥ 4, n5 ≥ 5 Let n1 − 1 = x1 ≥ 0, n2 − 2 = x2 ≥ 0, ..., n5 − 5 = x5 ≥ 0 ⇒ New equation will be x1 + 1 + x2 + 2 + ... + x5 + 5 = 20 ⇒

x1 + x2 + x3 + x4 + x5 = 20 − 15 = 5 x1

x2

x3

x4

x5

0 0 0 0 0 0 1

0 0 0 0 0 1 1

0 0 0 1 1 1 1

0 1 2 1 2 1 1

5 4 3 3 2 2 1

So, 7 possible cases will be there.

14. The number of solutions of x1 + x2 + ... + xk = n = Coefficient of t n in (t + t 2 + t 3 + ... )(t 2 + t 3 + ... )... (t k + t k +1 + ... ) n 1 + 2 + ... + k 2 = Coefficient of t in t (1 + t + t + ... )k k(k + 1) [say] Now, 1 + 2 + ... + k = =p 2 1 and 1 + t + t 2 + ... = 1−t Thus, the number of required solutions = Coefficient of t n − p in (1 − t )− k = Coefficient of t n − p in [1 + k C1 t + k +1 C 2t 2 + k + 2 C 3t 3 + ... ] = k + n − p −1 C n − p = r C n − p where, r = k + n − p − 1 = k + n − 1 −

1 k(k + 1) 2

1 = (2k + 2n − 2 + k2 − k) 2 1 = (2n − k2 + k − 2) 2

15. Since, six ‘+’ signs are + + + + + + ∴ 4 negative sign has seven places to be arranged in 7 C 4 ways = 35 ways ⇒

16. Since, each box can hold five balls. ∴ Number of ways in which balls could be distributed so that none is empty, are (2, 2, 1) or (3, 1, 1). i.e.

( 5C 2 3C 2 1C1 + 5C 3 C1 1C1 ) × 3 ! 2

= (30 + 20) × 6 = 300

1. It is given that a group of students comprises of 5 boys and n girls. The number of ways, in which a team of 3 students can be selected from this group such that each team consists of at least one boy and at least one girls, is = (number of ways selecting one boy and 2 girls) + (number of ways selecting two boys and 1 girl) = ( C1 × nC 2 ) ( C 2 × nC1 ) = 1750 [given] n (n − 1)  5 × 4   ⇒ 5 × × n = 1750  +    2  2 2 ⇒ n (n − 1) + 4n = × 1750 ⇒ n 2 + 3n = 2 × 350 5 5

5

⇒ n 2 + 3n − 700 = 0 ⇒ n 2 + 28n − 25n − 700 = 0 ⇒ n (n + 28) − 25(n + 28) = 0 ⇒ (n + 28) (n − 25) = 0 ⇒ n = 25

x1 ≤ x2 ≤ x3 ≤ x4 ≤ x5

Now,

Topic 4 Distribution of Object into Group

[Q n ∈ N ]

2. According to given information, we have the following figure.

Y B

(0, b)

O

A (a, 0)

X

(Note that as a and b are integers so they can be negative also). Here O(0, 0), A (a , 0) and B(0, b) are the three vertices of the triangle. Clearly, OA =| a |and OB =| b|. 1 ∴Area of ∆OAB = | a ||b|. 2 But area of such triangles is given as 50 sq units. 1 |a || b| = 50 ∴ 2 ⇒ |a || b| = 100 = 22 ⋅ 52 Number of ways of distributing two 2’s in|a |and| b| = 3 | a| 0 1 2

| b| 2 1 0

⇒ 3 ways Similarly, number of ways of distributing two 5’s in| a | and|b|= 3 ways. ∴ Total number of ways of distributing 2’s and 5’s = 3 × 3 = 9 ways Note that for one value of | a | , there are 2 possible values of a and for one value of|b|, there are 2 possible values of b. ∴Number of such triangles possible = 2 × 2 × 9 = 36. So, number of elements in S is 36. 3. Given 6 different novels and 3 different dictionaries. Number of ways of selecting 4 novels from 6 novels is 6! 6 C4 = = 15 2 !4 !

88 Permutations and Combinations Number of ways of selecting 1 dictionary is from 3 3! dictionaries is 3C1 = =3 1 !2 ! ∴ Total number of arrangement of 4 novels and 1 dictionary where dictionary is always in the middle, is 15 × 3 × 4 ! = 45 × 24 = 1080

4. Objects Distinct Distinct

Groups Distinct Identical

Objects Identical Identical

Groups Identical Distinct

Description of Situation Here, 5 distinct balls are distributed amongst 3 persons so that each gets at least one ball. i.e. Distinct → Distinct So, we should make cases A B C  A B C  Case I Case II   1 1 2  1 2 2  Number of ways to distribute 5 balls 3 !  3 !  =  5C1 ⋅4 C1 ⋅3 C 3 ×  +  5C1 ⋅4 C 2 ⋅2 C 2 ×   2 !  2 ! = 60 + 90 = 150

5. Total number of arrangements of word BANANA 6! = = 60 3!2! The number of arrangements of words BANANA in 5! which two N’s appear adjacently = = 20 3! Required number of arrangements = 60 − 20 = 40

6. Here, n 2 objects are distributed in n groups, each group containing n identical objects. ∴ Number of arrangements =

n

2

Cn .

n − n 2

Cn . n

2

− 2n

Cn .

n − 3n 2

Cn .

n − 2n 2

C n K nC n

(n 2 )! (n 2 − n )! n! (n 2 )! = . K = 2 2 n ! (n − n )! n ! (n − 2n )! n ! ⋅ 1 (n !)n ⇒ Integer (as number of arrangements has to be integer).

7.

(i) The number of ways in which 52 cards be divided equally among four players in order (52)! = 52C13 × 39C13 × 26C13 × 13C13 = (13 !)4 (ii) The number of ways in which a pack of 52 cards can be divided equally into four groups of 13 cards 52 (52)! C13 × 39C13 × 26C13 × 13C13 each = = 4! 4 !(13 !)4 (iii) The number of ways in which a pack of 52 cards be divided into 4 sets, three of them having 17 cards each and the fourth just one card =

52

(52)! C17 × 35C18 × 18C17 × 1C1 = 3 !(17)3 3!

8. The groups of persons can be made only in 2, 2, 1, 1 ∴ So the number of required ways is equal to number of ways to distribute the 6 distinct objects in group sizes 1, 1, 2 and 2 6! = (4 !) (2 !)2 (1 !)2 (2 !) (2 !) = 360 × 3 = 1080.

Topic 5 Dearrangement and Number of Divisors 1. Since, 240 = 24 .3.5 ∴ Total number of divisors = (4 + 1)(2)(2) = 20 Out of these 2, 6, 10, and 30 are of the form 4n + 2.

2. The number of ways in which the ball does not go its 1 1 1 1  own colour box = 4 ! 1 − + − +   1 ! 2 ! 3 ! 4 ! 1 1 1 = 4!  − +   2 6 24  12 − 4 + 1 = 24   =9   24

5 Binomial Theorem Topic 1 Binomial Expansion and General Term Objective Questions I (Only one correct option) 1. If the constant term in the binomial expansion of k   x − 2  x 

10

is 405, then|k|equals

(a) 9

(b) 1

(2020 Main, 6 Sep II)

(c) 3

(d) 2

2. If α and β be the coefficients of x and x respectively in 4

the expansion of (x +

2

x2 − 1 )6 + (x − x2 − 1 )6, then (2020 Main, 8 Jan II)

(a) α + β = − 30 (c) α + β = 60

(b) − 126

(b) 104

(a) 100

(d) 103

(c) 10

10. The sum of the coefficients of all even degree terms is x in the expansion of

(2019 Main, 8 April I)

x − 1 ) + (x − x − 1 ) , (x > 1) is equal to 3

6

(a) 29

3. The coefficient of x18 in the product (a) 84

6

  1  1   1+ log10 x   12 + x  is equal to 200, and x > 1, then the  x     (2019 Main, 8 April II) value of x is

(x +

(b) α − β = − 132 (d) α − β = 60

(1 + x)(1 − x)10 (1 + x + x2)9 is

9. If the fourth term in the binomial expansion of

3

(b) 32

6

(c) 26

(d) 24

11. The total number of irrational terms in the binomial (2019 Main, 12 April I)

(c) − 84

(d) 126

expansion of (71/5 − 31/10 )60 is

(a) 49

(b) 48

(2019 Main, 12 Jan II)

(c) 54

(d) 55

4. If the coefficients of x2 and x3 are both zero, in the

12. The ratio of the 5th term from the beginning to the 5th

(2019 Main, 10 April I)

term from the end in the binomial expansion of 10  1   3 1  is 2 + (2019 Main, 12 Jan I) 1   3   2(3)

expansion of the expression (1 + ax + bx2) (1 − 3x)15 in powers of x, then the ordered pair (a , b) is equal to (b) (− 21, 714) (d) (− 54, 315)

(a) (28, 315) (c) (28, 861)

5. The term independent of x in the expansion of 8

1 x 3  −  . 2x2 − 2 is equal to   x   60 81

(a) − 72

6

(c) − 36

(b) 36

(2019 Main, 12 April II)

(d) − 108

6. The smallest natural number n, such that the n

1  coefficient of x in the expansion of  x2 + 3  is nC 23 , is  x  (2019 Main, 10 April II)

(a) 35

(b) 23

(c) 58

(d) 38

7. If some three consecutive coefficients in the binomial expansion of (x + 1) in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficients is n

(2019 Main, 9 April II)

(a) 964 (c) 232

(b) 227 (d) 625

1

1

(a) 1 : 2(6)3

(b) 1 : 4(16)3

1

8

 x3 3 term in the binomial expansion of  +  equals x 3 5670 is (2019 Main, 11 Jan I) (a) 4

(b) 0

(c) 6

(b) 83 (d) 82

(d) 8

14. The positive value of λ for which the coefficient of x2 in λ  the expression x2  x + 2  x 

(a) 3

(b) 5

10

is 720, is (2019 Main, 10 Jan II)

(c) 2 2

(d) 4

15. If the third term in the binomial expansion of (1 + xlog 2 x )5 equals 2560, then a possible value of x is

(2019 Main, 10 Jan I)

(a) 4 2

(b)

1 4

(c)

1 8

6

(a) 8−2 (c) 8

(d) 2(36)3 : 1

13. The sum of the real values of x for which the middle

8. If the fourth term in the binomial expansion of log x  2 7  + x 8  (x > 0) is 20 × 8 , then the value of x is x  (2019 Main, 9 April I)

1

(c) 4(36)3 : 1

(d) 2 2 3

 1 − t6   is 1−t

16. The coefficient of t 4 in the expansion of 

(2019 Main, 9 Jan II)

(a) 12

(b) 10

(c) 15

(d) 14

90 Binomial Theorem 17. The sum of the coefficients of all odd degree terms in the expansion of  x + (a) −1

5

5

x3 − 1 +  x − x3 − 1 , (x > 1) is

(2018 Main)

(b) 0

(c) 1

value of (21C1 − 10C1 ) + (21C 2 − 10C 2) 21 10 21 + ( C3 − C3 ) + ( C 4 − 10C 4 ) + ... + (21C10 − 10C10 ) is (2017 Main)

(c) 220 − 29

(d) 220 − 210

(d) 729

20. The sum of coefficients of integral powers of x in the binomial expansion (1 − 2 x )

50

(a)

1 50 1 (3 + 1) (b) (350 ) 2 2

(c)

is

(2015 Main)

1 50 (3 − 1) 2

(d)

1 50 (2 + 1) 2

(1 + x ) (1 + x ) (1 + x ) is (a) 1051

22. The

term

4 12

(b) 1106

 x+1 x−1  −  2/ 3  1 /3 x − x + 1 x − x1 / 2 (a) 4

(2014 Adv.)

(c) 1113

independent

(b) 120

of

x

(d) 1120

in

expansion

is

(2013 Main)

(b)

n−4 5

r =1

2n

2n

r=0

r=0 2n + 1

∑ a r (x − 2)r = ∑ br (x − 3)r

show that bn = (c) 210

(d) 310

5 n−4

(d)

Cn+ 1

(1992, 6M)

r   1 3r 7r 15 r n C ( − 1 ) + + + ... upto m terms .  r ∑ r 2r 3r 4r 2 2 2 2 r=0 

(1985, 5M)

37. Given, sn = 1 + q + q + K + q 2

(2001, 1M)

(c)

and a k = 1 , ∀ k ≥ n, then

n

(d) 12C6 + 2

(c) 12C6

(1993, 5M)

36. Find the sum of the series

the 5th and 6th terms is zero. Then, a / b equals n−5 6

34. Prove that ∑ (−3)r − 1 3 nC 2r − 1 = 0, where k = (3n )/ 2 and n

35. If

10

24. In the binomial expansion of (a − b)n , n ≥ 5 the sum of

(a)

33. The larger of 9950 + 10050 and 10150 is ... .

of

(2003, 1M)

(b) 12C6 + 1

+ (− 1)n−113 = K

is an even positive integer.

23. Coefficient of t 24 in (1 + t 2)12 (1 + t12) (1 + t 24 ) is (a) 12C6 + 3

32. For any odd integer n ≥ 1, n − (n − 1) + K 3

k

21. Coefficient of x in the expansion of 3 7

(1983, 2M) 3

Analytical & Descriptive Questions

11

2 4

Fill in the Blanks

31. If (1 + a x)n = 1 + 8x + 24x2 + … , then a = … and n = K .

(2016 Main)

(c) 243

272  (d)  16,   3 

and 4th terms in the expansion of (1 + x)n are in AP, then the value of n is… . (1994, 2M)

n

2 4  1 − + 2 , x ≠ 0, is 28, then the sum of the  x x  coefficients of all the terms in this expansion, is (b) 2187

272  (c)  14,   3 

30. Let n be a positive integer. If the coefficients of 2nd, 3rd,

19. If the number of terms in the expansion of

(a) 64

(1 + ax + bx2)(1 − 2x)18 in powers of x are both zero, then (a , b) is equal to

251 251  (a)  16,  (b)  14,    3  3 

(d) 2

18. The

(a) 221 − 211 (b) 221 − 210

29. If the coefficients of x3 and x4 in the expansion of

6 n−5

25. If in the expansion of (1 + x)m (1 − x)n, the coefficients of x and x are 3 and −6 respectively, then m is euqal to 2

n 2

Sn = 1 + Prove that

n+ 1

n

q + 1  q + 1  q + 1 +  + ... +   ,q ≠1  2   2  2 C1 +

n+ 1

C 2 s1 +

n+ 1

C3 s2

+ ... +

n+ 1

C n+ 1 sn = 2nS n (1984, 4M)

(1999, 2M)

(a) 6

(b) 9

(c) 12

(d) 24

26. The expression [x + (x3 − 1) 1/ 2]5 + [x − (x3 − 1)1/ 2]5 is a polynomial of degree (a) 5

(b) 6

(1992, 2M)

(c) 7

x 3 27. The coefficient of x in  − 2 2 x  4

405 256 450 (c) 263

(a)

(b)

10

(1983, 1M)

504 259

(d) None of these

28. Given positive integers r > 1, n > 2 and the coefficient of (3r )th and (r + 2)th terms in the binomial expansion of (1980, 2M) (1 + x)2n are equal. Then, (a) n = 2r (c) n = 3r

(b) n = 2r + 1 (d) None of these

38. The natural number m, for which the coefficient of x in 1  the binomial expansion of  xm + 2  x 

(d) 8

is

Integer & Numerical Answer Type Questions 22

is 1540, is …… . [2020 Main, 5 Sep I]

39. Let m be the smallest positive integer such that the coefficient of x2 in the expansion of (1 + x)2 + (1 + x)3 + K + (1 + x)49 + (1 + mx)50is (3n + 1) 51 C3 for some positive integer n. Then, the value of n is (2016 Adv.)

40. The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3 ) ... (1 + x100 ) is

(2015 Adv.)

41. The coefficients of three consecutive terms of (1 + x)n + 5 are in the ratio 5 : 10 : 14. Then, n is equal to (2013 Adv.)

Binomial Theorem 91

Topic 2 Properties of Binomial Coefficient Objective Questions I (Only one correct option)

10. If C r stands for nC r, then the sum of the series

1. Let (x + 10)50 + (x − 10)50 = a 0 + a1x + a 2x2 + K + a50x50, for all x ∈ R; then

a2 is equal to a0

(a) 12.25 (c) 12.00

(2019 Main, 11 Jan II)

where n is an even positive integer, is

(b) 12.50 (d) 12.75

Cr

C0 +

20

20

Cr−1

C1 +

20

Cr− 2 20C2 + .... + 20C 020C r

20

is maximum, is

(2019 Main, 11 Jan I)

(a) 15 (c) 11

(b) 10 (d) 20

2 k is , then k is 15 15

equal to

(2019 Main, 9 Jan I)

(a) 14 (c) 4

(b) 6 (d) 8

4. For r = 0, 1, ... , 10, if Ar, Br and C r denote respectively the coefficient of xr in the expansions of (1 + x)10, (1 + x)20 10

and (1 + x)30. Then,

∑

Ar (B10Br − C10 Ar ) is equal to

r =1

(a) B10 − C10 (c) 0

(b) − C10 A10 ) (d) C10 − B10 2 A10 (B10

(2010)

30 30  30 30 30 30 30 5. 30 0  10 −  1  11 +  2  12 + K + 20 30 equal to

6. If

n −1

(b) (d)

C10 65 C55

m

i=0

maximum when m is equal to

is

(2002, 1M)

 n + 1 (b) 2    r + 1

 n + 2 (c) 2    r 

 n + 2 (d)    r 

9. If a n = ∑

r=0

1 , then Cr

n

(a) (n − 1) an (c)

1 n an 2

n

r

∑ nC

r=0

 = 0 holds for some  ∑ C k3  k=0  n n Ck positive integer n. Then ∑ equals ……… k +1 k=0 k=0 n

k

n

(2019 Adv.)

+ ... + 10(10C10 )2, where 10C r, r ∈{1, 2,... , 10} denote binomial coefficients. 1 Then, the value of X is .......... . 1430 (2018 Adv.) 2

10

2

10

2

14. The sum of the coefficients of the polynomial (1 + x − 3x2)2163 is …. .

(1982, 2M)

Analytical & Descriptive Questions 15. Prove that

r

(d) None of these

∑ nC kk2 

Fill in the Blank

equals

(b) n an



n

13. Let X = ( C1 ) + 2( C 2) + 3( C3 )

 n   n  +  is equal to  r − 1  r − 2 (2000, 2M)

 n + 1 (a)    r − 1

 n  ∑k k=0 det  n  n  ∑ C kk k = 0

10

8. For 2 ≤ r ≤ n,   + 2 

n

(a) g (m, n ) = g (n , m) for all positive integers m, n (b) g (m, n + 1) = g (m + 1, n ) for all positive integers m, n (c) g (2m, 2n ) = 2 g (m, n ) for all positive integers m, n (d) g (2m, 2n ) = ( g (m, n ))2 for all positive integers m, n

12. Suppose

(b) 10 (d) 20

 n  r

(2020 Adv.)

Then which of the following statements is/are TRUE?

(2004, 1M)

10 20  , where  p = 0 if p > q,    i  m − i  q

(a) 5 (c) 15

where for any non-negative integer p, p  m  n + i  p + n f (m, n , p) = Σ      . i =0  i   p   p − i 

Integer & Numerical Answer Type Questions

(b) [2, ∞ ) (d) ( 3 , 2]

∑

sum

s!   s  if r ≤ s,   =  r ! (s − r )!  r  0 if r > s  For positive integers m and n, let m+ n f (m, n , p) g (m, n ) = Σ p = 0  n + p    p 

60

C r = (k2 − 3) nC r + 1, then k belongs to

(a) (− ∞ , − 2] (c) [ − 3 , 3 ]

7. The

is

(2005, 1M)

(a) 30 C11 (c) 30 C10

Objective Question II (One or more than one correct option) 11. For non-negative integers s and r, let

403

3. If the fractional part of the number

(1986, 2M)

(b) (−1)n (n + 1) (d) None of these

(a) (−1)n/ 2 (n + 2) (c) (−1)n/ 2 (n + 1)

2. The value of r for which 20

 n  n 2  !  !  2  2 [C 02 − 2 C12 + 3 C 22 − ... + (−1)n (n + 1) C n2 ], n!

(1998, 2M)

 n  n  n  n − 1 k − 2  n  n − 2 2k     − 2k −1     − ...  +2     2  k − 2  0  k  1  k − 1  n  n − k  n + (−1)k     =   (2003, 4 M)  k  0   k

92 Binomial Theorem 16. For any positive integers m, n (with n ≥ m),

18. If n is a positive integer and (1 + x + x2)n = a 0 + a1x + ... + a 2n x2n.

 n If   = nCm. Prove that m  n   n − 1  n − 2 m  n + 1  + +   + ... +   =   m  m   m  m m + 1

Then, show that, a 02 − a12 + ... + a 22n = a n.

(1994, 5M)

19. Prove that C 0 − 2 2 ⋅ C1 + 3 2 ⋅ C 2 − ... + (−1)n (n + 1) 2⋅ C n = 0 , n > 2, where C r = nC r.

(1989, 5M)

20. If (1 + x) = C 0 + C1x + C 2 x + ... + C nx , then show that

or

2

n

Prove that  n  n − 1  n − 2   +2  +3  + ... + (n − m + 1) m  m   m  m  n + 2 (IIT JEE 2000, 6M)   =  m m + 2

n

the sum of the products of the Ci’s taken two at a time represented by Σ Σ CiC j is equal to (2n !) (1983, 3M) . 0 ≤ i < j ≤ n 22 n −1 − 2 (n !)2

21. Prove that C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ...−2n ⋅ C 22n = (−1)n n ⋅ C n (1979, 4M)

n  nC  3! 17. Prove that = ∑ (−1)r  r + 3 r  . 2(n + 3) r = 0 Cr  

22. Prove that ( C 0 ) − ( C1 ) + ( C 2) − ... + (2 nC 2 n )2 2n

2

2n

2

2n

2

= (−1)n ⋅2n C n.

(1997C, 5M)

(1978, 4M)

Answers Topic 1

32.

1. (c)

2. (b)

3. (a)

4. (a)

5. (c)

6. (d)

7. (c)

8. (d)

9. (c)

10. (d)

11. (c)

12. (c)

13. (b)

14. (d)

15. (b)

16. (c)

17. (d)

18. (d)

19. (d)

20. (a)

21. (c) 25. (c)

22. (c) 26. (c)

23. (d) 27. (a)

24. (b) 28. (a)

29. (d)

30. (n = 7)

1 (n + 1 ) 2 (2n − 1 ) 4

33. (101 ) 50

 2mn − 1  36.  mn n  2 (2 − 1 ) 

38. (13)

39. (5)

40. (8)

41. (n = 6 )

2. (d) 6. (d) 10. (a)

3. (d) 7. (c) 11. (a,b,d)

4. (d) 8. (d) 12. (6.20)

Topic 2 1. (a) 5. (c) 9. (c)

31. (a = 2, n = 4)

13. (646)

14. ( − 1 )

Hints & Solutions Topic 1 Binomial Expansion and General Term 1. Since, the general term in the expansion of binomial k   x − 2  x 

10

Tr + 1 =

is 10

Cr x

 10 − r     2 

10 − 5 r 2

(− k)r x− 2r = 10C r (− k)r x

Now, coefficient of x18 in the product (1 + x) (1 − x)10 (1 + x + x2)9 = coefficient of x18 in the product (1 − x2) (1 − x3 )9

x2 − 1 )6 + (x − x2 − 1 )6

= 2 [ C 0x + C 2 x (x − 1) + C 4x (x − 1) + 6C 6 (x2 − 1)3 ] 4 The coefficients of x in above expansion α = 2 [6C 2(−1) + 6C 42C1 (−1) + 6C 63C1 (−1)] 6

6

4

3. Given expression is (1 + x) (1 − x)10 (1 + x + x2)9 = (1 − x2) (1 − x3 )9

2. Given expression is 6

∴ α − β = − 96 − 36 = − 132 Hence, option (b) is correct. = (1 + x) (1 − x) [(1 − x) (1 + x + x2)]9

Q Term is constant, so r = 2. 10 × 9 2 k = 405 ∴ 10C 2(− k)2 = 405 ⇒ 2 ⇒ k2 = 9 ⇒ | k| = 3 (x +

= 2 [−15 − (15 × 2) − (1 × 3)] = − 96 and the coefficient of x2 in the expansion β = 2 [6C 44C 0 + 6C 63C 2] = 2 [15 + 3] = 36

2

6

2

2

2

= coefficient of x18 in (1 − x3 )9 − coefficient of x16 in (1 − x3 )9 Since, (r + 1)th term in the expansion of (1 − x3 )9 is 9C r (− x3 )r = 9C r (− 1)r x3 r

Binomial Theorem 93 = nC rx2n − 2r − 3 r = nC rx2n − 5 r For the coefficient of x , …(i) 2n − 5r = 1 ⇒ 2n = 5r + 1 As coefficient of x is given as nC 23 , then either r = 23 or n − r = 23 . If r = 23, then from Eq. (i), we get 2n = 5(23) + 1 ⇒ 2n = 115 + 1 ⇒ 2n = 116 ⇒ n = 58.

Now, for x18, 3r = 18 ⇒ r = 6 16 and for x16, 3r = 16 ⇒ r = ∉N. 3 9 ×8 × 7 9! = = 84 ∴Required coefficient is 9C 6 = 3 ×2 6 !3 !

4. Given expression is (1 + ax + bx2)(1 − 3x)15 . In the expansion of binomial (1 − 3x)15 , the (r + 1) th term is Tr + 1 = 15C r (−3x)r = 15C r (−3)r xr Now, coefficient of x2, (1 + ax + bx2)(1 − 3x)15 is 15

⇒

in

the

expansion

C 2(−3) + a C1 (−3) + b C 0 (−3) = 0 (given) (105 × 9) − 45 a + b = 0 ⇒ 45a − b = 945 2

15

1

15

⇒ 2n = 5n − 115 + 1 ⇒ 3n = 114 ⇒ n = 38 So, the required smallest natural number n = 38.

0

…(i)

Similarly, the coefficient of x3 , in the expansion of (1 + ax + bx2)(1 − 3x)15 is C3 (−3)3 + a 15C 2(−3)2 + b 15C1 (−3)1 = 0 ⇒ − 12285 + 945a − 45b = 0 15

⇒

If n − r = 23, then from Eq. (i) on replacing the value of ‘r’, we get 2n = 5(n − 23) + 1

of

(given)

63a − 3b = 819 ⇒ 21a − b = 273

…(ii)

7.

Key Idea Use general term of Binomial expansion ( x + a) n i.e. T r + 1 = nC r 1 x n − r a r

Given binomial is (x + 1)n, whose general term, is Tr + 1 = nC r xr According to the question, we have C r − 1 : nC r : nC r + 1 = 2 : 15 : 70

n

From Eqs. (i) and (ii), we get

n

24a = 672 ⇒ a = 28 b = 315 ⇒ (a , b) = (28, 315)

So,

Key Idea Use the general term (or (r + 1)th term) in the

5.

⇒

expansion of binomial (a + b) n Tr + 1 = C r a n

i.e.

n−r

b

r

⇒

6

3  Let a binomial 2x2 − 2 , it’s (r + 1)th term  x  = Tr + 1 = C r (2x )

 3  − 2  x 

6−r

12 − 2r − 2r

2 6−r

6

⇒

r

Now, the 1 x8  −    60 81

r

…(i)

term independent of x in the expansion of 6  2 3  2x − 2  x 

= the term independent of x in the expansion of 6 1  2 3 2x − 2 + the term independent of x in the 60  x  6 x8  2 3  expansion of −  2x − 2 81  x  =

6

C3 (− 3)3 (2)6 − 3 x12 − 4 (3 ) 60  1 +  −  6C5 (− 3)5 (2)6 − 5 x12 − 4 (5 ) x8  81

[put r = 3] [put r = 5]

1 35 × 2(6) = (− 3)3 23 + = 36 − 72 = − 36 3 81

Tr + 1

n

1 th  , its (r + 1) term, is  x3  r 1  1 = nC r (x2)n − r  3  = nC rx2n − 2r 3 r x  x

6. Given binomial is  x2 +

n

Cr

⇒ ⇒

2 15

=

n! (r − 1)!(n − r + 1)! 2 = n! 15 r !(n − r )! r 2 = ⇒ 15r = 2n − 2r + 2 n − r + 1 15 2n − 17r + 2 = 0

n C Similarly, n r Cr + 1

= C r (− 3) (2) x 6 r 6 − r 12 − 4 r = C r (−3) (2) x 6

Cr − 1

Now,

…(i)

n! 3 15 r !(n − r )! = ⇒ = n! 14 70 (r + 1)!(n − r − 1)!

r+1 3 = ⇒ 14r + 14 = 3n − 3r n − r 14 3n − 17r − 14 = 0

…(ii)

On solving Eqs. (i) and (ii), we get n − 16 = 0 ⇒ n = 16 and r = 2 C 2 + 16C3 3 16 + 120 + 560 696 = = = 232 3 3

Now, the average =

16

C1 +

log 8 x 

8. Given binomial is  + x 2 x

16

6

 

Since, general term in the expansion of (x + a )n is Tr+ 1 = n C r xn− ra r 6 −3  2 (xlog 8 x )3 = 20 × 87 (given) T4 = T3 + 1 = 6 C3   ∴  x 3  2 [Q 6C3 = 20] ⇒ 20   x3 log 8 x = 20 × 87  x  3  log 2 x −3  

⇒ 23 x [3(log 8 x )−3 ]= (23 )7 ⇒ x 3

= (23 )6

94 Binomial Theorem 1   Q log an (x) = log a x for x > 0; a > 0, ≠ 1   n

(x +

x3 − 1 )6 + (x − x3 − 1 )6 = 2 [6C 0 − 6C 2 + 6 C 4 + 6C 4 − 1 − 3]

log x − 3 ) ⇒ x( 2 = 218

= 2 [1 − 15 + 15 + 15 − 1 − 3] = 2(15 − 3) = 24

On taking log 2 both sides, we get

11. The general term in the binomial expansion of (a + b)n

(log 2 x − 3) log 2 x = 18 ⇒

is Tr + 1 = nC r a n − rbr.

(log 2 x) − 3 log 2 x − 18 = 0 2

So, the general term in the binomial expansion of (71/5 − 31/10 )60 is

⇒ (log 2 x)2 − 6 log 2 x + 3 log 2 x − 18 = 0 ⇒ log 2 x(log 2 x − 6) + 3 (log 2 x − 6) = 0 ⇒

log 2 x = −3, 6 1 ⇒ x = 2 −3 , 2 6 ⇒ x = , 8 2 8 6

⇒ ⇒

=

60

6

Cr 7

3  2

1 1 3  C3  x1 + log10 x   x12 = 200        3 1 +    2 (1 + log10 x ) 4 

20 × x = 200 ⇒ x  3 1 2(1 + log x) + 4  log10 x = 1   10

3 1 + 2(1 + log10 x ) 4

= 10

⇒ [6 + (1 + log10 x)] log10 x = 4(1 + log10 x) t 2 + 7t = 4 + 4t t + 3t − 4 = 0 t = 1 , −4 = log10 x

10.

[let log10 x = t]

2

⇒ Since,

x = 10, 10−4 x = 10

x>1

Given expression is (x +

binomial (x + a )n is same as the (n − r + 2)th term from the beginning in the expansion of same binomial. T4 + 1 T5 T = 5 = ∴Required ratio = T10 − 5 + 2 T7 T6 + 1 ⇒

T5 = T10 − 5 + 2

x3 − 1 )6 + (x − x3 − 1 )6

= 2 [nC 0a n + nC 2a n − 2b2 + nC 4a n − 4b4 + …]} = 2 [ C 0x + C 2x4 (x3 − 1) + 6C 4x2(x3 − 1)2 + 6C 6 (x3 − 1)3 ] 6

= 2 [ C 0x + C 2(− x ) + C 4x + C 4x + C 6 (−1 − 3x )] 4

6

8

6

2

6

6

= 2 [ C 0x − C 2x + C 4x + C 4x − 1 − 3x ] 6

6

4

6/3

[QTr + 1 = n Cr x n − r a r ]

1/3 6

2 (2(3) ) 24/3 (2(3)1/3 )4

[Q 10C 4 = 10C 6 ]

= 26/3 − 4/3 (2(3)1/3 )6 − 4 / / = 223 ⋅ 22 ⋅ 323 23 / = 4(6) = 4(36)1/3 So, the required ratio is 4(36)1/3 : 1. 8

6

8

6

2

 x3   3 4 8 ⋅ 7 ⋅ 6 ⋅ 5 8 x ∴ 5670 = 8C4     = 1⋅ 2⋅ 3⋅ 4  3   x 8− r   x3  QTr + 1 = 8Cr    3   8 4 ⇒ x = 3 ⇒x= ± 3

r  3     x  

So, sum of all values of x i.e + 3 and − 3 = 0

The sum of the terms with even power of x 6

 1  C 6 (21/3 )10− 6    2(3)1/3 

6

10

4

{Q (a + b)n + (a − b)n

6

 1  C 4 (21/3 )10− 4    2(3)1/3 

4

10

[Q Here, n = 8, which is even, therefore middle term  n + 2 =  th term]  2 

+ 6C 4x2( x3 − 1 )4 + 6C 6 ( x3 − 1 )6 ]

6

r

310

3 +  , the middle term is T4 + 1. x 3

= 2 [6C 0x6 + 6C 2x4 ( x3 − 1 )2

6

r 5

12. Since, rth term from the end in the expansion of a

 x3

Key Idea Use formula :

6

12 −

Cr 7

13. In the expansion of 

( a + b) n + ( a − b) n = 2 [ n C 0 a n + nC 2a n − 2b 2 + nC 4 a n − 4 b 4 + ...... ]

6

60

∴There are 7 rational terms in the binomial expansion and remaining 61 − 7 = 54 terms are irrational terms.

=

(7 + log10 x) log10 x = 4 + 4 log10 x

⇒ ⇒ ⇒

r

310 = (−1)r

The possible non-negative integral values of ‘r’ for r r and are integer, where r ≤ 60, are 5 10 r = 0, 10, 20, 30, 40, 50, 60.

[applying log10 both sides] ⇒

60 − r 5 (−1 )r

which

     1 1   9. Given binomial is   1 + log10 x   + x12  x   Since, the fourth term in the given expansion is 200. ∴

60

(log 2 x − 6) (log 2 x + 3) = 0

⇒

C r (71/5 )60 − r (−31/10 )r

Tr + 1 =

6

Now, the required sum of the coefficients of even powers of x in

14. The general term in the expansion of binomial expression (a + b)n is Tr+ 1 = nC r a n− rbr, so the general term in the expansion of binomial expression 10 λ  x2 x + 2 is  x 

Binomial Theorem 95 10 − r r   λ  Tr+ 1 = x2  10C r ( x )10− r  2  =10C r x2 ⋅ x 2 λr x−2r x   

= 10C r λr x

2+

10 − r − 2r 2

Now, for the coefficient of x2, put 2 +

10 − r − 2r = 2 2

10 − r − 2r = 0 2 ⇒ 10 − r = 4r ⇒ r = 2 So, the coefficient of x2 is 10C 2 λ2 = 720 [given] 10 ! 2 10 ⋅ 9 ⋅ 8 ! 2 λ = 720 ⇒ λ = 720 ⇒ 2!8! 2⋅ 8! ⇒ 45 λ2 = 720 ⇒ λ2 = 16 ⇒ λ = ± 4 ∴ λ =4 [λ > 0] ⇒

15. The (r + 1)th term in the expansion of (a + x)n is given by Tr + 1 = nC ra n − rxr ∴ 3rd term in the expansion of (1 + xlog 2 x )5 is 5

⇒ 5 C 2(1)5 − 2(xlog 2 x )2 = 2560 (given)

log 2 x2log 2 x = log 2 256 (taking log 2 on both sides) (Q log 2 256 = log 2 28 = 8) ⇒ 2(log 2 x)(log 2 x) = 8 (log 2 x)2 = 4 ⇒ log 2 x = ± 2 ⇒ log 2 x = 2 or log 2 x = − 2 1 ⇒ x = 4 or x = 2−2 = 4 3 6 1 − t  16. Clearly,   = (1 − t 6 )3 (1 − t )− 3 1−t ∴ Coefficient of t 4 in (1 − t 6 )3 (1 − t )−3 = Coefficient of t 4 in (1 − t18 − 3t 6 + 3t12) (1 − t )− 3 = Coefficient of t 4 in (1 − t )− 3 = 3 + 4 − 1C 4 = 6C 4 = 15 (Q coefficient of xr in (1 − x)− n = n + r − 1C r)

We have, (x +

18. ( 21C1 − 10C1 ) + (21C 2 − 10C 2) + (21C3 − 10C3 ) + ... + ( C10 − 21

C10 ) − (10C1 +

21

Alternate Solution We have, (1 − 2 x )50 = C o − C12 x + C 2( 2x)2 + ... + C50 (2 x )50 ...(i) (1 + 2 x )50 = C o + C12 x + C 2(2 x )2 + ... + C50 (2 x )50...(ii) On adding Eqs. (i) and (ii), we get (1 − 2 x )50 + (1 + 2 x )50 = 2 [C 0 + C 2(2 x )2 + ... + C50 (2 x )50 ] ...(iii) (1 − 2 x )50 + (1 + 2 x )50 2

(−1)50 + (3)50 = C 0 + C 2(2)2 + ... + C50 (2)50 2 1 + 350 = C 0 + C 2(2)2 + ... + C50 (2)50 2

21. Coefficient of xr in (1 + x)n is nC r.

2 (1 − 10 + 5 + 5) = 2

C 2 + ... +

r= 0

1 1 = [(1 + 2)50 + (1 − 2)50 ] = (350 + 1) 2 2

⇒

Sum of coefficients of all odd degree terms is

21

25

∴ Sum of coefficients = ∑ 50C 2r (2)2r

⇒

= 2(x5 + 10x6 − 10x3 + 5x7 − 10x4 + 5x)

C1 +

For the integral power of x, r should be even integer.

(1 − 2 1 )50 + (1 + 2 1 )50 = C 0 + C 2 + ... + C50 (2)50 2

5

= 2(x5 + 10x3 (x3 − 1) + 5x(x3 − 1)2)

21

Tr+ 1 = 50C r (1)50− r (−2x1/ 2)r = 50C r2r xr/ 2(−1)r

On putting x = 1, we get

x − 1 ) + (x − x − 1 ) , x > 1 3

= 2(5 C 0x5 + 5C 2x3 ( x3 − 1 )2 + 5C 4x( x3 − 1 )4 )

=(

∴

= C 0 + C 2(2 x )2 + ... + C50 (2 x )50

= 2 ( n C 0 a n + nC 2a n − 2b 2 + nC 4 a n − 4 b 4 + ...) 5

1 1 and 2 are functions of same variables, therefore x x number of dissimilar terms will be 2 n + 1, i.e. odd, which is not possible. Hence, it contains error.

⇒

Key Idea Use formula : = ( a + b) n + ( a − b) n

3

Hence, sum of coefficients = (1 − 2 + 4)6 = 36 = 729

(1 − 2 x )50

x( 2log 2 x ) = 256

⇒

17.

n

(n + 2) (n + 1) 2 4  or n + 2C 2. 1 − + 2 is  2 x x  1 1 [assuming and 2 distinct] x x (n + 2) (n + 1) = 28 ∴ 2 ⇒ (n + 2) (n + 1) = 56 = (6 + 1) (6 + 2) ⇒ n = 6

20. Let Tt +1 be the general term in the expension of

10 (xlog 2 x )2 = 2560

⇒

19. Clearly, number of terms in the expansion of

Note As

C 2(1)5 − 2(xlog 2 x )2

⇒

1 21 ( C1 + 21C 2 + ... + 21C 20 ) − (210 − 1) 2 1 = (21C1 + 21C 2 + ... + 21C 21 − 1) − (210 − 1) 2 1 = (221 − 2) − (210 − 1) = 220 − 1 − 210 + 1 = 220 − 210 2 =

C 2 + ... +

10

10

C10 )

10

C10 )

In this type of questions, we find different composition of terms where product will give us x11. Now, consider the following cases for x11 in (1 + x2)4 (1 + x3 )7 (1 + x4 )12.

96 Binomial Theorem Coefficient of x0 x3 x8; Coefficient of x2 x9 x0 Coefficient of x4 x3 x4; Coefficient of x8 x3 x0 = 4C 0 × 7C1 × 12C 2 + 4C1 × 7C3 × 12C 0 + 4C 2 × 7C1 × 12C1 + 4C 4 × 7C1 × 12C 0 = 462 + 140 + 504 + 7 = 1113



22. 

x+1

x

2/3

− x1/ 3 + 1

−

( x − 1)   x − x1/ 2 

10

Therefore, the given expression is a polynomial of degree 7.

27. The general term in  − x 2

10

C r (x )

(− x

− 1/ 2 r

) =

C r (− 1)

10

r

tr + 1 = (−1)r

23. Here, Coefficient of t

10 − r r − 2 x 3

12

2 12

= Coefficient of t 24 = (12C12 +

12

C6 +

12

C0 ) = 2 +

12

36

12

C6

24. Given, T5 + T6 = 0 ⇒ ⇒

C 4a n − 4b4 = nC5 a n − 5 b5 ⇒

n

3r = 6 r =2 10

32 28 45 × 9 405 = = 256 256

= (−1)2. 10C 2.

tr + 2 =

C r + 1 (x)r + 1

2n

Since, binomial coefficients of t3 r and tr + 2 are equal. ∴

C3 r −1 = 2nC r + 1

2n

⇒ ⇒ ⇒ But

3r − 1 = r + 1 or 2n = (3r − 1) + (r + 1) or 2n = 4r 2r = 2 or r =1 n = 2r r >1

In expansion of (1 + ax + bx2)(1 − 2x)18.

term containing power of x ≥ 3.

⇒ ⇒ ⇒ ⇒

⋅ x10 − 3 r

coefficient of x r in (1 − x)n is (−1)r nCr and then simplify.

m (m − 1) 2  x +K  2  n (n − 1) 2   1 − nx + x − ...   2  m(m − 1) n (n − 1) = 1 + (m − n ) x + + − mn x2 + K   2 2

and

210 − r

29. To find the coefficient of x3 and x4, use the formula of

a nC5 n − 4 = = 5 b nC 4

25. (1 + x)m (1 − x)n = 1 + mx +

m − n =3

C r.

∴ We take, n = 2r

C 4a n − 4b4 − nC5 a n − 5 b5 = 0

n

Now,

⇒ ⇒

and 24

24

3r

10

28. In the expansion (1 + x)2n, t3 r = 2nC3 r − 1 (x)3 r − 1

= Coefficient of t in {(1 + t ) ⋅ (1 + t + t + t )} = Coefficient of t 24 in {(1 + t 2)12 + t12(1 + t 2)12 + t 24 (1 + t 2)12}; [neglecting t36 (1 + t 2)12] 24

10 − r

x 3 ∴ Coefficient of x4 in  − 2 2 x 

in {(1 + t ) (1 + t )(1 + t )} 2 12

 x Cr    2

10

10

3  is x2  r  3 r  2 = (−1) x 

For coefficient of x4, we put 10 − 3r = 4

For independent of x , put 10 − r r − = 0 ⇒ 20 − 2r − 3r = 0 3 2 ⇒ 20 = 5r ⇒ r = 4 10 × 9 × 8 × 7 ∴ T5 = 10C 4 = = 210 4 ×3 ×2 ×1 24

= 2 [a5 + 10a3 b2 + 5ab4 ] 1/ 2 5

= 2 [x5 + 10x3 (x3 − 1) + 5x (x3 − 1)2]

 (x1/ 3 + 1)(x2/ 3 + 1 − x1/3 ) {( x )2 − 1}  − =  / x ( x − 1)  x23 − x1/3 + 1  10  ( x + 1)  1/3 − 1/ 2 10 = (x1/3 + 1) −  = (x − x ) x   ∴ The general term is Tr + 1 =

+ 5C3 a 2b3 + 5C 4ab4 + 5C5 b5 + 5C 0a5 − 5C1a 4b + 5C 2a3 b2 − 5C3 a 2b3 + 5C 4ab4 − 5C5 b5 ∴ [x + (x − 1) ] + [x − (x3 − 1)1/ 2]5

 (x1/ 3 )3 + 13 {( x )2 − 1}  − =  23  1/3 / x ( x − 1)  x − x + 1

1/3 10 − r

(a + b)5 + (a − b)5 = 5C 0a5 + 5C1a 4b + 5C 2a3 b2

3

10

10

26. We know that,

…(i) [Q coefficient of x = 3, given]

1 1 m(m − 1) + n (n − 1) − mn = − 6 2 2 m(m − 1) + n (n − 1) − 2mn = − 12 m2 − m + n 2 − n − 2mn = − 12 (m − n )2 − (m + n ) = − 12 m + n = 9 + 12 = 21

On solving Eqs. (i) and (ii), we get m = 12

Coefficient of x3 = Coefficient of x3 in (1 − 2x)18 + Coefficient of x2 in a (1 − 2x)18 + Coefficient of x in b(1 − 2x)18 =

C3 ⋅ 23 + a 18C 2 ⋅ 22 − b 18C1 ⋅ 2

Given, coefficient of x3 = 0 C3 ⋅ 23 + a 18C 2 ⋅ 22 − b 18C1 ⋅ 2 = 0 18 × 17 × 16 18 × 17 2 ⋅8 + a ⋅ ⋅ 2 − b ⋅ 18 ⋅ 2 = 0 ⇒ − 3 ×2 2 34 × 16 ⇒ 17a − b = 3 Similarly, coefficient of x4 = 0 ⇒

⇒ ∴ …(ii)

18

18

18

..(i)

C 4 ⋅ 24 − a ⋅18 C3 23 + b ⋅18 C 2 ⋅ 22 = 0 32a − 3b = 240

On solving Eqs. (i) and (ii), we get a = 16, b =

272 3

…(ii)

Binomial Theorem 97 30. Let the coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x) is C1, C 2, C3 . According to given condition, n

n

n

n

2 (nC 2) = nC1 + nC3 n (n − 1) n (n − 1) (n − 2) 2 =n+ ⇒ 1 ⋅2 1 ⋅2 ⋅3 (n − 1) (n − 2) ⇒ n −1 =1 + 6 n 2 − 3n + 2 n −1 =1 + ⇒ 6 ⇒ 6n − 6 = 6 + n 2 − 3n + 2 ⇒ n 2 − 9n + 14 = 0 ⇒ (n − 2) (n − 7) = 0 ⇒ n =2 ⇒ n = 7 n But C3 is true for n ≥ 3, therefore n = 7 is the answer.

3m

Also, k =

(1 + ax) = 1 + 8x + 24x + ... n (n − 1) 2 2 a x + ... = 1 + 8x + 24x2 + ... ⇒ 1 + anx + 2! n (n − 1) = 24 ⇒ 8 (8 − a ) = 48 ∴ an = 8 and a 2 2 ⇒ 8 − a =6 ⇒ a =2 a =2

and

n =4

32. Since, n is an odd integer, (− 1)n−1 = 1

C1 + (−3)

(1 + x)6 m =

C0 +

6m

(1 − x)

6m

=

C0 +

6m

[Q n − 1, n − 3, ... , are even integers] 3   n − 1  = Σ n3 − 16 Σ     2    2

1  n − 1  n − 1   n (n + 1)  + 1  − 16   =         2 2 2 2 

2

1 1 (n + 1)2 [n 2 − (n − 1)2] = (n + 1)2(2n − 1) 4 4

33. Consider, (101)50 − (99)50 − (100)50

= (100)50 {2 [50C3 (0.01)3 + ∴ ⇒

50

(101)

− {(99)

50

C3 (0.01)3 + K ] − 1}

50

C5 (0.01)5 + ... ]}

50

+ (100) } > 0 50

(101)50 > (99)50 + (100)50

34. Since, n is an even positive integer, we can write n = 2 m ,m = 1, 2, 3, K

+

6m

C 6 m x6 m

…(ii)

C 2(− x) + K

6m

2

C 6 m (− x)6 m

6m

+ (1 + x)

6m

…(iii)

− (1 − x) 2x

6m

6m

6m

C3 x3

C5 x5 + K + 6 m C 6 m − 1x6 m − 1 ]

6m

= 6mC1 +

C3 x2 +

6m

x =y (1 +

⇒

C5 x4 + ...

6m

+ 6 mC 6 m − 1 x6 m − 2

2

Let

y )6 m − (1 − 2 y

y )6 m +

=

C1 +

6m

6m

C3 y

C5 y2 + K + 6 m C 6 m − 1 y 3 m − 1

6m

For the required sum we have to put y = − 3 in RHS. S= =

(1 +

−3 )6m − (1 − −3 )6m 2 −3

(1 + i 3 )6m − (1 − i 3 )6m 2i 3

…(iv)

z = 1 + i 3 = r (cos θ + i sin θ )

Let ⇒

r = |z| = 1 + 3 = 2

and

θ = π /3 z 6 m = [r (cos θ + i sin θ )]6 m

Now,

= r 6 m (cos 6m θ + i sin 6m θ ) z = r (cos θ − i sin θ )

Again,

(z )6 m = r 6 m (cos 6m θ − i sin 6m θ )

and ⇒

z 6 m − z 6 m = r 6 m (2i sin 6m θ )

…(v)

From Eq. (i), S=

z 6 m − z 6 m r 6 m (2i sin 6 m θ ) 26 m sin 6 m θ = = 2i 3 2i 3 3

= 0 as m ∈ z , and θ = π /3

35. Let y = (x − a )m, where m is a positive integer, r ≤ m

= (100 + 1)50 − (100 − 1)50 − (100)50 = (100)50 {2 ⋅ [50C1 (0.01) +

C1 (− x) +

6m

x2 + K

(1 + x)6 m − (1 − x)6 m = 2 [6 mC1x +

Now,

= {(100)50 (1 + 0.01)50 − (1 − 0.01)50 − 1)}

…(i)

C3m − 1

On subtracting Eq. (iii) from Eq. (ii), we get

1 16 (n − 1)2(n + 1)2 = n 2 (n + 1)2 − 4 4 ×4 ×4 =

6m C2 6m − 1

C 6 m − 1 (− x)6 m − 1 +

+

3

− 2 [(n − 1)3 + (n − 3)3 + K + 23 )]   n − 1 3  n − 3 3  3 = Σ n3 − 2 × 23   +K+1   +  2   2   

⋅

C 6 m − 1x

∴

= n + (n − 1) + (n − 2) + K + 1 3

C1x +

6m

6m

n3 − (n − 1)3 + (n − 2)3 − (n − 3)3 + K + (− 1)n−1 ⋅ 13 3

6m

+ (−3)

and n − 1, n − 3, n − 5, etc., are even integers, then 3

C3 + K 3 m − 1 6m

6m

From the binomial expansion, we write

2

n

Hence,

S = (−3)0

i.e.

⇒

31. Given,

3n 3(2m) = = 3m ∴ S = ∑ (−3)r − 1 ⋅6mC 2r − 1 2 2 r =1

⇒

dy d 2y = m(m − 1) (x − a )m − 2 = m(x − a )m − 1 ⇒ dx dx2 d3 y = m(m − 1)(m − 2)(m − 3)(x − a )m − 4 dx3 ………………………………… …………………………………

On differentiating r times, we get dr y = m(m − 1) ... (m − r + 1)(x − a )m − r dxr m! = (x − a )m − r = r !(mC r )(x − a )m − r (m − r )!

98 Binomial Theorem and for r > m, 2n

n +1

dr y =0 dxr

2n

∑ a r (x − 2)r = ∑ br (x − 3)r

Now,

r=0

[given]

r=0

On differentiating both sides n times w.r.t. x, we get 2n

∑ a r (n !) C n (x − 2) r

r−n

2n

∑ br (n !) C n (x − 3)

=

r=n

r

∑ a r (n !)rC n = (bn )n !

r=n

[since, all the terms except first on RHS become zero] ⇒ bn = nC n + n + 1C n + n + 2C n + K + 2nC n

n+3

[Q a r = 1, ∀ r ≥ n ] C n ) + K + 2n C n

=

2n

n

∑ (−1)r nC r

r=0

n

=

∑ (−1)

Cn + 1 +

Cn =

2n

= ....

2n

Cn 2n + 1

Cn + 1

 1 3 7 15  r + 2r + 3 r + 4r + ... upto m terms  2 2 2 2   r

r

r

 1 Cr   +  2

r n

r=0

n+ 2

Cn + 1 + K +

r

r

n

∑ (−1)

 3 Cr   +  4

r n

r=0

n

r

 7 ∑ (−1)r nC r  8  + ... upto m terms

r=0 n

n

n

1 3 7    = 1 −  + 1 −  + 1 −  + ... upto m terms    2 4 8  ∑ (−1)r nC rxr = (1 − x)n   r= 0

 using  n

n

n

n

 1  1  1 =   +   +   + ... upto m terms  2  4  8 m   1  1 −  n   mn  1 2  = 2 −1 =   mn n 1  2  1 − n  2 (2 − 1)   2   n+ 1

C1 +

n+ 1

C 2 s1 +

n+ 1

C n+ 1sn

 1 − qr 

1

1 − qn + 1 1−q

 n +1

n+ 1



 r =1

r =1



∑ n + 1C r  1 − q  = 1 − q  ∑ n + 1C r − ∑ n + 1C r qr

r =1

C3 s2 + ... +

n+ 1

C n + 1sn = 2nS n

22

is

Tr + 1 = 22C r (xm )22 − r x−2r Q The coefficient 22C r = 1540 ⇒ r = 3 or 19 and 22m − mr − 2r = 1 at r = 3, 22m − 3m − 6 = 1 ⇒ 18m = 7 7 m= ∉N ⇒ 16 Now, at r = 19, 22m − 19m − 38 = 1 ⇒ 3m = 39 ⇒ m = 13 ∈ N So, the natural value of m = 13

[given]

39. Coefficient of x2 in the expansion of {(1 + x)2 + (1 + x)3 + K + (1 + x)49 + (1 + mx)50} ⇒ 2C 2 + 3C 2 + 4C 2 + K + 49C 2 + 50C 2 ⋅ m2 = (3n + 1) ⋅51 C3 50 ⇒ C3 + 50C 2m2 = (3n + 1) ⋅51 C3 [QrC r + r + 1C r + K+ nC r = n + 1C r + 1] 51 × 50 × 49 50 × 49 × 48 50 × 49 ⇒ + × m2 = (3n + 1) 3 ×2 ×1 3 ×2 ×1 2 ⇒ m2 = 51n + 1 ∴ Minimum value of m2 for which (51n + 1) is integer (perfect square) for n = 5. ∴ m2 = 51 × 5 + 1 ⇒ m2 = 256 ⇒ m = 16 and n = 5 Hence, the value of n is 5.

n+ 5

where sn = 1 + q + q2 + ... + qn = n+ 1

n+ 1

tr , tr + 1 , tr + 2 having coefficients

∑ n + 1C rsr −1,

r =1

∴

C 2 s1 +

41. Let the three consecutive terms in (1 + x)n + 5 be

n+ 1

=

n+ 1

(1 + x)(1 + x2)(1 + x3 ) K (1 + x100 ) =Terms having x9 = [199 ⋅ x9 , 198 ⋅ x ⋅ x8 , 198 ⋅ x2 ⋅ x7 , 198 ⋅ x3 ⋅ x6 , 198 ⋅ x4 ⋅ x5 , 197 ⋅ x ⋅ x2 ⋅ x6 , 197 ⋅ x ⋅ x3 ⋅ x5 ,197 ⋅ x2 ⋅ x3 ⋅ x4] ∴ Coefficient of x9 = 8

n +1

C3 s2 + ... +

...(ii)

40. Coefficient of x9 in the expansion of

n

37.

C1 +

1  binomial  xm + 2  x 

2n

=

2n + 1 − (q + 1)n + 1 2n (1 − q)

38. The general term (i.e. (r + 1)th term) in the expansion of

r=n

= (n + 2C n + 1 +

=

From Eqs. (i) and (ii), n+ 1

r−n

On putting x = 3, we get

36.

 q + 1 1−   2  =  q + 1 1−   2 

1 [(1 + 1)n + 1 − (1 + q)n + 1 ] 1 −q 1 …(i) = [2n + 1 − (1 + q)n + 1 ] 1 −q =

2

 q + 1  q + 1  q + 1 Also, S n = 1 +   + ... +   +  2   2   2 

n

C r − 1 , n + 5C r , n + 5C r + 1. Given, n + 5 C r − 1 : n + 5C r : n + 5C r + 1 = 5 : 10 : 14 n+ 5 n+ 5 C r + 1 14 10 Cr and n + 5 = ∴ = n+ 5 5 10 C r −1 Cr ⇒

n + 5 − (r − 1) n−r+5 7 = 2 and = r r+1 5

⇒ n − r + 6 =2r and 5n − 5 r + 25 = 7r + 7 ⇒ n + 6 = 3 r and 5n + 18 = 12r n + 6 5n + 18 ∴ = 3 12 ⇒

4n + 24 = 5n + 18 ⇒ n = 6

Binomial Theorem 99 ∴ When (16)100 is divided by 15, gives remainder 1100 = 1

Topic 2 Properties of Binomial Coefficient

and when 8(16)100 is divided by 15, gives remainder 8.

1. We have,

2403  8  = .  15  15

(x + 10)50 + (x − 10)50 = a 0 + a1x + a 2x2 + … + a50x50 ∴a 0 + a1x + a 2x2 + … + a50x50

∴

= [(50C 0x50 +

(where {⋅} is the fractional part function)

C1x49 10 +

50

C 2 x48 ⋅ 102 + … +

50

+ ( C 0x − C1x 10 + C 2 x 10 − … + = 2 [50C 0 x50 + 50C 2x48 ⋅ 102 + 50C 4x46 ⋅ 104 50

50

50

49

50

48

2

50

C50 1050 )

50

50

⇒

C50 10 )]

+…+

C50 ⋅ 1050 ]

50

4. A r = Coefficient of xr in (1 + x)10 = 10C r B r = Coefficient of xr in (1 + x)20 =

By comparing coefficients, we get

∴

a 50 ⋅ 49 (10)48 2(50C 2)(10)48 =2 ∴ 2 = 50 a0 1 ⋅ 2 2 ⋅ (10)50 2 (10) C 48 = C 2] 50

=

20

C0 +

C1x +

C r − 1 xr − 1 +

C rxr + ... +

20

20

20

C 20x20

∴ (1 + x)20 ⋅ (1 + x)20 = (20C 0 + 20

C 2x2 + ... +

× (20C 0 +

C r − 1 xr − 1 +

20

20

20

C rxr

+ ....+ C 20x ) ⇒ (1 + x)

= ( C0 .

Cr +

20

20

20

C1

C120C r − 1 + ... +

20

40

The maximum value of

C r − 1 ...

∴

C r20C 0 ) xr + ...

20

Cr

20

C0 =

C10

Cr

C r is possible only when r = 20

C 2(15)2 + … +

100

C10 ( 30C10 − 1) − 30C10 (20C10 − 1)

20

100

C100 (15)100 )

C 2(15)2 + … +

∴

Alternate Method 2403 = 8 ⋅ 2400 = 8(16)100 Note that, when 16 is divided by 15, gives remainder 1.

C 20 ⋅30 C30

30

∑ (−1)r

30

C r (x2)r

[for coefficient of x20, put r = 10]

30

n −1

6. Given , ⇒

C10

n −1

C r = (k2 − 3) nC r + 1

C r = (k2 − 3)

n r+1

n −1

C r ⇒ k2 − 3 =

r+1 n

⇒

r+1 ≤ 1 and n , r > 0] n 0 < k2 − 3 ≤ 1 ⇒ 3 < k2 ≤ 4

⇒

k ∈ [−2, − 3 ) ∪ ( 3 , 2]

[since, n ≥ r ⇒

99

k =8

C 2 ⋅30 C12

30

= 30C10

C100 (15)100 )

(where {⋅} is the fractional part function)

30 30 20 30

r=0

100

C1 +......+ C100 (15) ∈ N 2403  8 2403 8 + 8 × 15λ 8 ∴ = = 8λ + ⇒  = 15 15 15  15  15

30 30  2  12

30

= 8 + 8 × 15λ where λ =

30 30  1  11

A = 30C 0 ⋅30 C10 − 30C1 ⋅30 C11 +

(1 + x)n = nC 0 + nC1x + nC 2x2 + … nC nxn , n ∈ N ] 100

C10 = C10 − B10

20

= Coefficient of x20 in

[By binomial theorem,

100

r =1

= (−1)10

2403 = 2400 + 3 = 8 ⋅ 2400 = 8 ⋅ (24 )100 = 8 (16)100= 8(1 + 15)100

100

r =1

= Coefficient of x20 in (1 − x2)30

3. Consider,

= 8 + 8 (100C1 (15) +

10

∑ 10C10 − r ⋅ 20C r − 30C10 ∑ 10C10 − r 10C r

= Coefficient of x20 in (1 + x)30 (1 − x)30

40

Thus, required value of r is 20.

C1 (15) +

20

−K +

[Q C n/2 is maximum when n is even]

100

r =1

10

30 30  0  10

n

= 8 (1 +

10

5. Let A =     −     +     − ... +    

On comparing the coefficient of xr of both sides, we get C 020C r +

10

∑ 10C10 − r 20C10 20C r − ∑ 10C10 − r 30C10 10C r

20

20

20

20

r =1

∑ 10C r 20C10 20C r − ∑ 10C r 30C10 10C r

= 30C10 −

20

20

40

=

c20x20 )

r =1

r =1

C1x +

C rxr + ... +

r =1

10

20

C r − 1 xr − 1 +

C1x + ...+

20

=

20

Cr 10

10

r =1

C 2x2 + ... +

20

Cr

30

∑ Ar (B10 Br − C10 Ar ) = ∑ Ar B10 Br − ∑ Ar C10 Ar

=

2. We know that,

=

20

10

10

r =1 50

[Q 50 × 49 5 × 49 245 = = = 12 .25 = 20 2 ⋅ (10 × 10) 20 20

(1 + x)

C r = Coefficient of x in

30

r

a 2 = 2 50C 48 (10)48 ; a 0 = 2 50C50 (10)50 = 2(10)50

(1 + x)20 =

k =8

m

7.

10  20 

∑  i  m − i is the coefficient of xm in the expansion of

i=0

(1 + x)10 (x + 1)20, m

⇒

∑

i=0

10  20  m    is the coefficient of x in the  i  m − i expansion of (1 + x)30

100 Binomial Theorem m

10  20  30 30    = Cm =    i  m − i  m

∑

i.e.

i=0

p

p=0 m+ n n+ p

n

[Q C r + C r − 1 =

9. Let b =

∑

n

r=0 n

=n

∑

=

n

∑

r=0

n

Σ

n an 2

m+ n m+ n

∑

C p = 2m + n

p=0

= [C 02 − C12 + C 22 − C32 + ... + (−1)nC n2 ] − [ C12 − 2 C 22 + 3 C32 − ... + (−1)n nC n2 ] n −1 n n! n! = (−1)n/ 2 − (−1) 2 2  n  n  n  n   !  !  !  !  2  2  2  2 n  1 +   2

n

Σ

n

Σ

n

C k k2

k=0 n

n

Ck . k

=0

C k3k

k=0

n n n (n + 1)   , Σ nC k k = n.2 n − 1 ,  Q k Σ= 0 k = k=0 2   n n  Σ nC k k2 = n (n + 1)2n − 2 and Σ nC k3k = 4n  = k 0 0 = k   n (n + 1) n 2 2n − 3 =0 4 − n (n + 1) 2 2

⇒

C 02 − 2C12 + 3C 22 − 4C32 + ... + (−1)n (n + 1) C n2

n!  n  n  !  !  2  2

=

n (n + 1) n (n + 1)2 n − 2 =0 2 4n n.2 n − 1

[Q nC r = nC n − r ]

10. We have,

= (−1)n/ 2

Cp

n

Σ k

⇒

= na n − b ⇒ 2b = na n ⇒ b =

C p × m + nC p

k=0 k=0

n−r n Cn − r

Cp

12. It is given that

Cr ]

n

= na n −

m+ n

, and ∴ g (m, n ) = g (n , m) = 2 g (m, n + 1) = 2m + n + 1 = g (m + 1, n ), and g (2m; 2n ) = 22(m + n) = (2(m + n) )2 = ( g (m, n ))2.

n+1

n−r 1 − ∑ n C r r = 0 nC r

r=0

Cp

m+ n

r n − (n − r ) = ∑ n n Cr Cr r=0

∑

n+ p

f (m, n , p)  n + p    p 

n+ p

p=0

 n   n   n  n   + =  +   r − 1  r − 2  r   r − 1   n   n    n + 1  n + 1  n + 2 +  + + =   =   r − 1  r − 2   r   r − 1   r  n

∑

Since, g (m, n ) =

8.   + 2 

n

(n + p)! ∑ mCi × nC p − i = n ! p! i = 0 m+ n

 n and we know that,   is maximum, when  r n  if n ∈ even.  n  r=2, =   n±1  r  max  .  r = 2 , if n ∈ odd 30 Hence,   is maximum when m = 15.  m  n  r

=

…(i)

4n 4n−1 −n =0⇒n =4 2 2

⇒

4 4 Ck Ck 1 4 5 1 Σ C k + 1 = (25 − 1) = Σ = k + 1 k=0 k + 1 5 k=0 5 n

n

∴Σ

k=0

=

1 31 = 6.20 (32 − 1) = 5 5

13. We have, X = (10C1 )2 + 2(10C 2)2+ 3(10C3 )2 + ... + 10 (10C10 )2

 n  n 2 !  !  2  2 [C 02 − 2 C12 + 3 C 22 − ... + (− 1)r (n + 1) C n2 ] ∴ n!  n  n 2  !  !  2  2 n! (n + 2) (−1)n/ 2 = = (−1)n/ 2(n + 2) n!  n  n 2  !  !  2  2

⇒

X=

10

∑ r (10C r )2 ⇒ X =

r =1

⇒

X=

10

∑r×

r =1

⇒

10 9 Cr − 1 r

X = 10

10

∑ r 10C r 10C r

r =1

n  n Q Cr = r 

10

Cr

n −1

Cr − 1

 

10

∑ 9C r − 1 10C r

r =1

11. Since, m  n + i  p + n ∑  i   p   p − i  i=0 p

f (m, n , p) = p

=

(n + i )!

m!

⇒ ( p + n )!

i=0

=

10

∑ 9C r − 1 10C10 − r

[Q nC r = nC n − r ]

r =1

∑ i !(m − i )! × p !(n + i − p)! × ( p − i )! (n + i )! p

X = 10

(n + p)! 1 ∑ mCi ( p − i )! (n + i − p)! p! i = 0

⇒ X = 10 ×

19

[Q

C9

C r − 1nC n − r =

C9 C9 1 10 × C 9 = = X= 143 11 × 13 1430 1430 19 × 17 × 16 = = 19 × 34 = 646 8 19

Now,

n−1

19

19

2n − 1

Cn − 1 ]

Binomial Theorem 101  n   n − 1  n − 2 m  =  +  +  + ... +    m  m   m  m  n 2 − n 1 −     m  + +  + ... +     m   m  m  m  n − m + 1 rows  n − 2 ... + + +    m  m   ...... m  +   m 

14. Sum of coefficients is obtained by putting x = 1 i.e. (1 + 1 − 3)2163 = − 1 Thus, sum of the coefficients of the polynomial (1 + x − 3x2)2163 is −1.

15. To show that

2k.n C 0.n C k − 2k − 1.n C1.n − 1 C k − 1 + 2k − 2.n C 2.n − 2 C k − 2 − K + (−1)k nC kn − kC 0 = nC k Taking LHS 2k.n C 0.n C k − 2k −1.n C1 ⋅n − 1 C k − 1 + K + (−1)k.n C k.n − k C 0 k

=

∑ (−1)

r. k − r . n

∑ (−1)

r

∑ (−1)

r. k − r.

2

r=0 k

=

r=0 k

=

2

k − r.

2

r=0 k

=

 n + 1 Now, sum of the first row is  . m + 1

C r.n − r C k − r

 n  Sum of the second row is  . m + 1

n! (n − r )! . r !(n − r )! (k − r )!(n − k)!

 n − 1 Sum of the third row is  , m + 1 ………………………… m m + 1 Sum of the last row is   =  . m m + 1

n! k! ⋅ (n − k)!. k ! r !(k − r )!

∑ (−1)r. 2k − r nC k.kC r

r=0

 k 1 = 2k.n C k  ∑ (−1)r. r .k C r 2 r = 0 =2

k.n

1  C k 1 −   2

  

m + 1  n + 1 + 1  n + 2 +K+ =   =  m + 1  m + 2  m + 2

k

[from Eq. (i) replacing n by n + 1 and m by m + 1]

= n C k = RHS  n m

 n − 1  n − 2 m  n + 1 +  + ... +   =   …(i)  m   m  m m + 1

16. Let S =   + 

It is obvious that, n ≥ m. NOTE

 n + 1  n   n − 1  S= + +  m + 1 m + 1 m + 1

Thus,

[given]

This question is based upon additive loop.

m + 1 m + 1  m + 2  n =  + +   + K        m + m 1 m m    m m + 1  Q m = 1 = m + 1    m + 2 m + 2  n = +  + ... +   m + 1  m  m

[Q nC r + nC r + 1 =

= ...............................  n   n   n + 1 =  +   =  , which is true. m + 1 m m + 1

17.

∑ (−1)r

r=0

n

= ∑ (−1)r r=0

m m + 1 m + 2  n Now, S =   +  +  + ... +   m  m   m  m

m + 3  n =  + ... +   m + 1 m

n

n+1

Cr + 1 ]

n

Cr Cr

r +3

n n !⋅ 3 ! n! = 3 ! ∑ (−1)r (n − r ) ! (r + 3) ! (n − r ) ! (r + 3) ! r=0 n

(−1)r. (n + 3)! ∑ (n − r )!(r + 3)! r=0

=

3! (n + 1)(n + 2)(n + 3)

=

n 3! ⋅ ∑ (−1)r ⋅n + 3C r + 3 (n + 1)(n + 2)(n + 3) r = 0

=

3 ! (− 1)3 (n + 1)(n + 2)(n + 3)

n+3

∑ (−1)s ⋅n + 3 C3

s =3

 n + 3 −3!  ∑ (−1)s ⋅n + 3 C s =  (n + 1)(n + 2)(n + 3)  s =` 0  C1 − n + 3C 2 (n + 3)(n + 2)    0 − 1 + (n + 3) − 2!   −

…(ii)

Again, we have to prove that  n  n − 1  n − 2 m  n + 2   +2  +3  + ... + (n − m + 1)   =   m  m   m  m m + 2  n  n − 1  n − 2 m Let S1 =   + 2   +3  + ... + (n − m + 1)   m  m   m  m

n+3

C0 +

n+3

=

−3! (n + 1)(n + 2)(n + 3)

=

3! −3! (n + 2)(2 − n − 3) = ⋅ 2(n + 3) (n + 1)(n + 2)(n + 3) 2

18. (1 + x + x2)n = a 0 + a1x + K + a 2nx2n

…(i)

Replacing x by −1 / x, we get n a 2n a a a 1 1  1 − + 2 = a 0 − 1 + 22 − 33 + K + 2n  x x  x x x x

…(ii)

102 Binomial Theorem Now, a 02 − a12 + a 22 − a32 + K + a 22n = coefficient of the term independent of x in [a 0 + a1x + a 2x + K + a 2nx ] a 2n   a a × a 0 − 1 + 22 − K + 2n  x x x   2

2n

= Coefficient of the term independent of x in n

1 1  (1 + x + x2)n 1 − + 2  x x  n 1 1   Now, RHS = (1 + x + x2)n 1 − + 2  x x  (1 + x + x2)n (x2 − x + 1)n [(x2 + 1)2 − x2]n = = x2 n x2 n 2 4 2 n 2 4 n (1 + 2x + x − x ) (1 + x + x ) = = 2n x x2n 2 2 2 2 2 Thus, a 0 − a1 + a 2 − a3 + K+ a 2n = Coefficient of the term independent of x in 1 (1 + x2 + x4 )n x2n = Coefficient of x2n in (1 + x2 + x4 )n = Coefficient of t n in (1 + t + t 2)n = a n

= =

n

∑ (−1) r(r + 1)2 nC r =

r=0 n

∑ (−1)r (r 2 + 2r + 1) nC r n

∑ (−1)

n

r=0

r (r − 1) ⋅ C r + 3 ⋅ ∑ (−1) r ⋅ nC r n

r.

r=0

+

n

∑ (−1)

r n

Cr

r=0

=

n

∑ (−1)r n (n − 1) n − 2C r − 2

r=2

n

+ 3 ∑ (−1)r n . n − 1C r − 1 r =1

+

n

∑ (−1)r nC r

r=0

= n (n − 1){ n − 2C 0 − n − 2C1 + n − 2C 2−... + (−1)n n − 2C n − 2} + 3n { − n−1C 0 + n − 1C1 − n −1C 2 + ...+ (−1)n n − 1C n − 1} + { nC 0 − nC1 + nC 2 + ... + (−1)n nC n } = n (n − 1) . 0 + 3n . 0 + 0, ∀n > 2 = 0, ∀n > 2

20. We know that, n

2 ∑ ∑ Ci C j = ∑ 0 ≤i < j ≤ n

i=0

=

n

n

n

∑ Ci C j − ∑ ∑ CiC j

j=0

i = 0 j=0

n

n

n

i=0

j=0

i=0

∑ Ci ∑ C j − ∑ Ci2

0 ≤i< j≤n

(2n )! 22 n − 2 nC n = 22 n−1 − 2 2 (n !)2

21. We know that, (1 + x)2 n = C 0 + C1x + C 2 x2 + ... + C 2 n x 2 n On differentiating both sides w.r.t. x, we get 2n (1 + x)2 n−1 = C1 + 2 ⋅ C 2 x + 3 ⋅ C3 x2 + ... + 2nC 2n x2 n−1 …(i) 2n 1 1 1 1  and 1 −  = C 0 − C1 ⋅ + C 2 ⋅ 2 − C3 ⋅ 3   x x x x 1 …(ii) + ... + C 2n ⋅ 2 n x On multiplying Eqs. (i) and (ii), we get 1  2n (1 + x)2 n−1 1 −   x

2n

= [C1 + 2 ⋅ C 2x + 3 ⋅ C3 x2 + ... + 2n ⋅ C 2n x2n−1 ]   1  1  1  × C 0 − C1   + C 2 2 − ..... + C 2 n  2 n       x  x x 

1  1  in 2n  2 n  (1 + x)2 n−1 (x − 1)2 n x  x

= 2n (−1)n−1 ⋅ (2n − 1) C n−1 (−1) (2n )! (2n − 1)! = (−1)n (2n ) = (−1)n n ⋅n (n − 1)! n ! (n !)2

r=0

r=0

r.

CiC j =

= Coefficient of x2 n−1 in 2n (1 − x2)2 n−1 (1 − x)

r=0

∑ (−1)r r 2 ⋅ nC r + 2 ∑ (−1)r r ⋅ nC r + ∑ (−1)r. nC r

r=0 n

∑∑

= Coefficient of

n

n

∴

 1 Coefficient of   on the LHS  x

19. C 0 − 22 ⋅ C1 + 32 ⋅ C 2 − ... + (−1)n (n + 1)2 ⋅ C n =

= 2n 2n − (2 nC n ) = 22 n − 2 nC n

= − (−1)n n ⋅ C n  1 Again, the coefficient of   on the RHS  x = − (C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ... − 2n C 22n )

…(iii)

…(iv)

From Eqs. (iii) and (iv), C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ... − 2n ⋅ C 22n = (−1)n n ⋅ C n 1 22. (1 + x)2 n 1 −  

2n

x

= [ C 0 + (2nC1 )x + (2nC 2)x2 + ...+ (2 nC 2 n )x2 n ] 1 1 1   × 2 nC 0 − (2 nC1 ) + (2 nC 2) 2 + ... + (2 nC 2n ) 2n  x x x  2n

Independent terms of x on RHS = (2nC 0 )2 − (2nC1 )2 + (2nC 2)2 − ...+ (2nC 2n )2  x − 1 LHS = (1 + x)2n    x 

2n

=

1 (1 − x2)2n x2n

Independent term of x on the LHS = (−1)n ⋅2n C n.

6 Probability Topic 1 Classical Probability Objective Questions I (Only one correct option) 1. A person throws two fair dice. He wins ` 15 for throwing a doublet (same numbers on the two dice), wins ` 12 when the throw results in the sum of 9, and loses ` 6 for any other outcome on the throw. Then, the expected gain/loss (in `) of the person is (2019 Main, 12 April II)

(a)

1 gain 2

(b)

1 loss 4

(c)

1 loss 2

(d) 2 gain

2. In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is (2019 Main, 12 Jan I) equal to (a) (c)

175

(b)

65 200

(d)

65

225 65 150

(2019 Main, 12 April I)

3 (c) 10

3 (d) 20

4. Let S = {1, 2, K , 20}. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then, the probability that a randomly chosen subset of S is ‘‘nice’’, is (2019 Main, 11 Jan II) (a)

6 220

(b)

4 220

(c)

7 220

(d)

5 220

5. If two different numbers are taken from the set {0, 1, 2, 3, …, 10}, then the probability that their sum as well as absolute difference are both multiple of 4, is (2017 Main)

(a)

6 55

(b)

12 55

(c)

14 45

(a)

55  2    3  3

11

2 (b) 55    3

10

1 (c) 220    3

12

1 (d) 22    3

11

8. Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is atleast one more that the number of girls ahead of her, is (2014 Adv.) (a) 1 /2

(b) 1 /3

(c) 2 /3

(d) 3 /4

9. Four fair dice D1 , D2, D3 and D4 each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of (2012) D1 , D2 and D3 , is 91 216

(b)

108 216

(c)

125 216

(d)

127 216

10. Let ω be a complex cube root of unity with ω ≠ 1. A fair die is

65

chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is 1 (b) 5

then the probability that one of the boxes contains excatly (2015 Main) 3 balls, is

(a)

3. If there of the six vertices of a regular hexagon are

1 (a) 10

7. If 12 identical balls are to be placed in 3 different boxes,

(d)

7 55

thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ω r 1 + ω r2 + ω r3 = 0, is (2010)

(a) 1/18

(b) 1/9

(c) 2/9

(d) 1/36

11. If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3, is (2004, 1M) (a)

4 55

(b)

4 35

(c)

4 33

(d)

4 1155

12. Two numbers are selected randomly from the set

S = {1, 2, 3, 4, 5, 6} without replacement one by one. The probability that minimum of the two numbers is less than (2003, 1M) 4, is (a) 1/15

(b) 14/15

(c) 1/5

(d) 4/5

13. If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form (1999, 2M) 7m + 7n is divisible by 5, equals (a)

1 4

(b)

1 7

(c)

1 8

(d)

1 49

6. Three randomly chosen non-negative integers x, y

14. Seven white balls and three black balls are randomly

and z are found to satisfy the equation x + y + z = 10. (2017 Adv.) Then the probability that z is even, is

placed in a row. The probability that no two black balls are placed adjacently, equals (1998, 2M)

(a)

1 2

(b)

36 55

(c)

6 11

(d)

5 11

(a)

1 2

(b)

7 15

(c)

2 15

(d)

1 3

104 Probability 15. Three of the six vertices of a regular hexagon are chosen at rondom. The probability that the triangle with three vertices is equilateral, equals (1995, 2M) (a) 1/2

(b) 1/5

(c) 1/10

(d) 1/20

16. Three identical dice are rolled. The probability that the same number will appear on each of them, is (1984, 2M) (a)

1 6

(b)

1 36

(c)

1 18

(d)

3 28

17. Fifteen coupons are numbered 1, 2, ..., 15, respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9, is 9 (a)    16 

6

8 (b)    15 

7

3 (c)    5

7

(d) None of these

For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows. (a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

18. Consider the system of equations ax + by = 0, cx + dy = 0, where a , b, c, d ∈ {0, 1}. Statement I The probability that the system of equations has a unique solution, is 3/8. Statement II The probability that the system of equations has a solution, is 1. (2008, 3M)

Passage Based Problems Passage Box I contains three cards bearing numbers 1, 2, 3 ; box II contains five cards bearing numbers 1, 2, 3, 4, 5; and box III contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the i th box i = 1, 2, 3. (2014 Adv.)

19. The probability that x1 + x2 + x3 is odd, is 29 105

(b)

53 105

(c)

57 105

(d)

1 2

20. The probability that x1 , x2 and x3 are in an arithmetic progression, is 9 (a) 105

10 (b) 105

21. Three faces of a fair die are yellow, two faces red and one face blue. The die is tossed three times. The probability that the colours, yellow, red and blue, appear in the first, second and the third tosses respectively, is…… . (1992, 2M) 1 + 3p 1 − p 1 − 2p 22. If are the probabilities of three and , 3 4 2 mutually exclusive events, then the set of all values of p is… . (1986, 2M)

23. A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of the determinant chosen is positive, is… . (1982, 2M)

True/False

Assertion and Reason

(a)

Fill in the Blanks

11 (c) 105

7 (d) 105

24. If the letters of the word ‘ASSASSIN’ are written down at random in a row, the probability that no two S’s occur together is 1/35.

Analytical and Descriptive Questions 25. An unbiased die, with faces numbered 1, 2, 3, 4, 5 and 6 is thrown n times and the list of n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5 and 6 only three numbers appear in (2001, 5M) this list?

26. If p and q are chosen randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9 and 10} with replacement, determine the probability that the roots of the equation x2 + px + q = 0 (1997, 5M) are real.

27. In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats? Now, if all the seating arrangements are equally likely, what is the probability of 3 girls sitting together in a back row on adjacent (1996, 5M) seats?

28. A box contains 2 fifty paise coins, 5 twenty five paise coins and a certain fixed number n (≥ 2) of ten and five paise coins. Five coins are taken out of the box at random. Find the probability that the total value of these 5 coins is less than one rupee and fifty paise. (1988, 3M)

29. Six boys and six girls sit in a row at random. Find the probability that (i) the six girls sit together. (ii) the boys and girls sit alternatively.

(1978, 3M)

Probability 105

Topic 2 Addition and Subtraction Law of Probability Objective Questions I (Only one correct option) 1. The probabilities of three events A , B and C are given by P ( A ) = 0.6, P (B) = 0.4 and P (C ) = 0.5. If P ( A ∪ B) = 0.8, P ( A ∩ C ) = 0.3, P ( A ∩ B ∩ C ) = 0.2, and where P (B ∩ C ) = β P(A ∪ B ∪ C ) = α , 0.85 ≤ α ≤ 0.95, then β lies in the interval (2020 Main, 6 Sep II)

(a) [ 0.35, 0.36] (c) [0.20, 0.25]

(b) [0.25, 0.35] (d) [0.36, 0.40]

2. For three events A , B and C, if P (exactly one of A or B occurs) = P(exactly one of B or C occurs) = P (exactly 1 one of C or A occurs) = and P (all the three events 4 1 occur simultaneously) = , then the probability that 16 atleast one of the events occurs, is (2017 Main)

(a)

7 32

(b)

7 16

3 4 then P (B ∩ C ) is equal to

(c)

3. If P (B) = , P ( A ∩ B ∩ C ) = (a)

1 12

(b)

1 6

7 64

3 16

(d)

1 1 and P ( A ∩ B ∩ C ) = , 3 3 (2002, 3M)

(c)

1 15

(d)

1 9

4. If E and F are events with P (E ) ≤ P (F ) and P (E ∩ F ) > 0, then which one is not correct? (1998, 2M) (a) occurrence of E ⇒ occurrence of F (b) occurrence of F ⇒ occurrence of E (c) non-occurrence of E ⇒ non-occurrence of F (d) None of the above

3p + 2p 2 p + 3p2 (c) 2

p + 3p 4 3p + 2p2 (d) 4 (b)

6. If 0 < P ( A ) < 1, 0 < P (B) < 1 and P ( A ∪ B) = P ( A ) (1995, 2M)

(a) P (B / A ) = P (B ) − P (A ) (b) P (A ′ − B ′ ) = P (A ′ ) − P (B ′ ) (c) P (A ∪ B ) ′ = P (A ) ′ P (B )′ (d) P (A / B ) = P (A ) − P (B )

occurs is 0.6. If A and B occur simultaneously with probability 0.2, then P ( A ) + P (B ) is equal to (1987, 2M)

(b) 0.8

(c) 1.2

Objective Questions II (One or more than one correct option) 9. For two given events A and B, P ( A ∩ B) is

(1988, 2M)

(a) not less than P (A ) + P (B ) − 1 (b) not greater than P (A ) + P (B ) (c) equal to P (A ) + P (B ) − P (A ∪ B ) (d) equal to P (A ) + P (B ) + P (A ∪ B )

10. If M and N are any two events, then the probability that exactly one of them occurs is

(1984, 3M)

(a) P (M ) + P (N ) − 2 P (M ∩ N ) (b) P (M ) + P (N ) − P (M ∪ N ) (c) P (M ) + P (N ) − 2 P (M ∩ N ) (d) P (M ∩ N ) − P (M ∩ N )

Fill in the Blanks 11. Three

numbers are chosen at random without replacement from {1, 2,…, 10}. The probability that the minimum of the chosen number is 3, or their maximum is 7, is … . (1997C, 2M) (1985, 2M)

13. If the probability for A to fail in an examination is 0.2 and that of B is 0.3, then the probability that either A or B fails is 0.5. (1989, 1M)

Analytical and Descriptive Questions 14. In a certain city only two newspapers A and B are published, it is known that 25% of the city population reads A and 20% reads B, while 8% reads both A and B. It is also known that 30% of those who read A but not B look into advertisements and 40% of those who read B but not A look into advertisements while 50% of those who read both A and B look into advertisements. What is the percentage of the population reads an advertisement? (1984, 4M)

15. A, B, C are events such that Pr ( A ) = 0.3, Pr (B) = 0.4, Pr (C ) = 0.8, Pr ( AB) = 0.08, Pr ( AC ) = 0.28 and Pr ( ABC ) = 0.09

7. The probability that at least one of the events A and B

(a) 0.4

(b) 0.25 (d) None of these

True/False

2

+ P (B) − P ( A ) P (B), then

(a) 0.39 (c) 0.11

P ( A ) and P (B) is… .

events A or B occurs) = P(exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p and P(all the three events occurs 1 simultaneously) = p2, where 0 < p < . Then, the 2 probability of atleast one of the three events A, B and C occurring is (1996, 2M)

(a)

respectively. The probability that both A and B occur simultaneously is 0.14. Then, the probability that neither A nor B occurs, is (1980, 1M)

12. P ( A ∪ B) = P ( A ∩ B) if and only if the relation between

5. For the three events A, B and C, P(exactly one of the

2

8. Two events A and B have probabilities 0.25 and 0.50,

(d) 1.4

If Pr ( A ∪ B ∪ C ) ≥ 0.75, then show that Pr (BC ) lies in the (1983, 2M) interval [ 0.23, 0.48 ].

16. A and B are two candidates seeking admission in IIT. The probability that A is selected is 0.5 and the probability that both A and B are selected is atmost 0.3. Is it possible that the probability of B getting selected is 0.9? (1982, 2M)

106 Probability Pragraph Based Questions There are five students S1 , S 2, S3 , S 4 and S5 in a music class and for them there are five seats R1 , R2, R3 , R4and R5 arranged in a row, where initially the seat Ri is allotted to the student Si , i = 1, 2, 3, 4, 5. But, on the examination day, the five students are randomly allotted the five seats. (There are two questions based on Paragraph, the question given below is one of them) (2018 Adv.)

17. The probability that, on the examination day, the

NONE of the remaining students gets the seat previously allotted to him/her is (a)

3 40

(b)

1 8

(c)

7 40

(d)

1 5

18. For i = 1, 2, 3, 4, let Ti denote the event that the students Si and Si+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1 ∩ T2 ∩ T3 ∩ T4 is (a)

student S1 gets the previously allotted seat R1, and

1 15

(b)

1 10

(c)

7 60

(d)

1 5

Topic 3 Independent and Conditional Probability Objective Questions I (Only one correct option) 1. Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls; is (2019 Main, 10 April I) (a)

1 17

(b)

1 12

(c)

1 10

(d)

1 11

2. Four persons can hit a target correctly with

1 1 1 1 , , and respectively. If all hit at the 2 3 4 8 target independently, then the probability that the target would be hit, is (2019 Main, 9 April I)

probabilities

(a)

1 192

(b)

25 32

(c)

7 32

(d)

25 192

3. Let A and B be two non-null events such that A ⊂ B. Then, which of the following statements is always correct. (2019 Main, 8 April I) (a) P (A /B ) = P (B ) − P (A ) (c) P (A/B ) ≤ P (A )

(b) P (A/B ) ≥ P (A ) (d) P (A/B ) = 1

4. Two integers are selected at random from the set { 1, 2, …… , 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is (2019 Main, 11 Jan I) (a)

2 5

(b)

1 2

(c)

7 10

(d)

3 5

5. An unbiased coin is tossed. If the outcome is a head, then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail, then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, …, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is (2019 Main, 10 Jan I)

(a)

15 72

(b)

13 36

(c)

19 72

(d)

19 36

6. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E 2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is not true? (2016 Main)

(a) E1 and E2 are independent (b) E2 and E3 are independent (c) E1 and E3 are independent (d) E1 , E2 and E3 are independent

1 6

7. Let A and B be two events such that P ( A ∪ B) = , 1 1 and P ( A ) = , where A stands for the 4 4 complement of the event A. Then , the events A and B (2014 Main) are

P ( A ∩ B) =

(a) independent but not equally likely (b) independent and equally likely (c) mutually exclusive and independent (d) equally likely but not independent

8. Four persons independently solve a certain problem 1 3 1 1 , , , . Then, the 2 4 4 8 probability that the problem is solved correctly by (2013 Adv.) atleast one of them, is

correctly with probabilities

235 256 3 (c) 256

21 256 253 (d) 256

(a)

(b)

9. An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. If A consists of 4 outcomes, then the number of outcomes that B must have, so that A and B are independent, is (a) 2, 4 or 8 (c) 4 or 8

(b) 3, 6 or 9 (d) 5 or 10

(2008, 3M)

10. Let E c denotes the complement of an event E. If E, F, G

are pairwise independent events with P (G ) > 0 and P (E ∩ F ∩ G ) = 0 . Then, P (E c ∩ F c|G ) equals(2007, 3M)

(a) P (E c ) + P (F c ) (c) P (E c ) − P (F )

(b) P (E c ) − P (F c ) (d) P (E ) − P (F c )

11. One Indian and four American men and their wives are to be seated randomly around a circular table. Then, the conditional probability that Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife, is (2007, 3M) (a)

1 2

(b)

1 3

(c)

2 5

(d)

1 5

Probability 107 12. A fair die is rolled. The probability that the first time 1 occurs at the even throw, is (a) 1/6

(b) 5/11

(2005, 1M)

(c) 6/11

(d) 5/36

13. There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then, the probability that only two tests are needed, is (1998, 2M) (a)

1 3

(b)

1 6

(c)

1 2

(d)

1 4

14. A fair coin is tossed repeatedly. If tail appears on first four tosses, then the probability of head appearing on fifth toss equals (1998, 2M) (a)

1 2

(b)

1 32

(c)

31 32

(d)

1 5

15. If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black balls will be drawn, is (1998, 2M)

(a)

13 32

(b)

1 4

(c)

1 32

(d)

3 16

16. The probability of India winning a test match against West Indies is 1/2. Assuming independence from match to match the probability that in a 5 match series India’s second win occurs at third test, is (1995, 2M)

(a) 1/8

(b) 1/4

(c) 1/2

(d) 2/3

17. An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5, is (1993, 1M) (a) 16/81

(b) 1/81

(c) 80/81

(d) 65/81

18. A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in 1 tests I, II and III are p, q and , respectively. If the 2 1 probability that the student is successful, is , then 2 1 2

(a) p = q = 1

(b) p = q =

(c) p = 1, q = 0

(d) p = 1, q =

(1986, 2M)

1 2

P ( A ) > 0, and P (B) ≠ 1, then P ( A / B ) is equal to (c)

1 − P (A ∪ B ) P (B )

(b) 1 − P (A / B ) (d)

(1982, 2M)

P (A ) P (B )

20. The probability that an event A happens in one trial of an experiment, is 0.4. Three independent trials of the experiments are performed. The probability that the event A happens atleast once, is (1980, 1M) (a) 0.936 (c) 0.904

(b) 0.784 (d) None of these

1 3

21. Let X andY be two events such that P (X ) = , P (X /Y ) = 2 and P (Y /X ) = . Then 5 (a) P (Y ) =

4 15

(c) P (X ∪Y ) =

22. If

(2017 Adv.)

(b) P (X ′/Y ) = 2 5

1 2

1 2

(d) P (X ∩ Y ) =

1 5

and Y are two events such that 1 1 1 , P (Y /X ) = and P (X ∩ Y ) . Then, which of 2 3 6 (2012) the following is/are correct? X

P (X / Y ) =

(a) P (X ∪ Y ) = 2/3 (b) X and Y are independent (c) X and Y are not independent (d) P (X c ∩ Y ) = 1/3

23. Let E and F be two independent events. The probability that exactly one of them occurs is

11 and the probability of 25

2 . If P (T ) denotes the 25 (2011) probability of occurrence of the event T, then none of them occurring is

4 3 , P (F ) = 5 5 2 1 (c) P (E ) = , P (F ) = 5 5

(a) P (E ) =

1 , P (F ) = 5 3 (d) P (E ) = , P (F ) = 5 (b) P (E ) =

2 5 4 5

24. The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the students has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly two. Which of the following relations are true? (1999, 3M) (2011) (a) p + m + c = (c) pmc =

19 20

1 10

(b) p + m + c = (d) pmc =

27 20

1 4

25. If E and F are the complementary events of E and F respectively and if 0 < P (F ) < 1, then

(a) P (E / F ) + P (E / F ) = 1 (c) P (E / F ) + P (E / F ) = 1

(1998, 2M)

(b) P (E / F ) + P (E / F ) = 1 (d) P (E / F ) + P (E / F ) = 1

26. Let E and F be two independent events. If the probability

19. If A and B are two independent events such that (a) 1 − P (A / B )

Objective Questions II (One or more than one correct option)

that both E and F happen is 1/12 and the probability that neither E nor F happen is 1/2. Then, (a) P (E ) = 1 / 3, P (F ) = 1 / 4 (b) P (E ) = 1 / 2, P (F ) = 1 / 6 (c) P (E ) = 1 / 6, P (F ) = 1 / 2 (d) P (E ) = 1 / 4, P (F ) = 1 / 3

(1993, 2M)

27. For any two events A and B in a sample space (1991, 2M)

P (A ) + P (B ) − 1 A , P (B ) ≠ 0 is always true (a) P   ≥  B P (B ) (b) P (A ∩ B ) = P (A ) − P (A ∩ B ) does not hold (c) P (A ∪ B ) = 1 − P (A )P (B ), if A and B are independent (d) P (A ∪ B ) = 1 − P (A )P (B ), if A and B are disjoint

108 Probability 28. If E and F are independent events such that 0 < P (E ) < 1 and 0 < P (F ) < 1, then

(1989, 2M)

(a) E and F are mutually exclusive (b) E and F c (the complement of the event F) are independent (c) E c and F c are independent (d) P (E / F ) + P (E c / F ) = 1

result is a tail, a card from a well-shuffled pack of eleven cards numbered 2, 3, 4, …, 12 is picked and the number on the card is noted. What is the probability that the noted number is either 7 or 8? (1994, 5M)

39. A lot contains 50 defective and 50 non-defective bulbs.

Fill in the Blanks 29. If two events A and B are such that P ( A c ) = 0.3, P (B) = 0.4 and P ( A ∩ Bc ) = 0.5, then P [B / ( A ∪ Bc )] = K . (1994, 2M)

30. Let A and B be two events such that P ( A ) = 0.3 and P ( A ∪ B) = 0.8. If A and B are independent events, (1990, 2M) then P (B) = … .

31. A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then, the probability that 5 comes before 7, is… . (1989, 2M)

32. Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then, one ball is drawn at random from urn B and placed in urn A. If one ball is drawn at random from urn A, the probability that it is found to be red, is…. (1988, 2M)

33. A box contains 100 tickets numbered 1, 2, …,100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The maximum number on them is 5 with (1985, 2M) probability… .

Analytical and Descriptive Questions 34. If A and B are two independent events, prove that P ( A ∪ B) ⋅ P ( A′ ∩ B ′ ) ≤ P (C ), where C is an event defined that exactly one of A and B occurs. (2004, 2M)

35. A is targeting to B, B and C are targeting to A.

2 1 Probability of hitting the target by A, B and C are , 3 2 1 and , respectively. If A is hit, then find the probability 3 that B hits the target and C does not. (2003, 2M)

36. For a student to qualify, he must pass atleast two out of three exams. The probability that he will pass the 1st exam is p. If he fails in one of the exams, then the p probability of his passing in the next exam, is 2 otherwise it remains the same. Find the probability (2003, 2M) that he will qualify.

37. A coin has probability p of showing head when tossed. It is tossed n times. Let pn denotes the probability that no two (or more) consecutive heads occur. Prove that p1 = 1, p2 = 1 − p2 and pn = (1 − p). pn − 1 + p(1 − p) pn − 2, ∀ n ≥ 3. (2000, 5M)

38. An unbiased coin is tossed. If the result in a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the

Two bulbs are drawn at random, one at a time, with replacement. The events A, B, C are defined as : A = ( the first bulb is defective) B = (the second bulb is non-defective) C = (the two bulbs are both defective or both non-defective). Determine whether (i) A, B, C are pairwise independent. (1992, 6M) (ii) A, B, C are independent.

40. In a multiple-choice question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks the correct answers. The candidates decide to tick the answers at random, if he is allowed upto three chances to answer the questions, find the probability that he will get (1985, 5M) marks in the question.

41. A and B are two independent events. The probability that both A and B occur is

1 and the probability that 6

1 neither of them occurs is . Find the probability of the 3 occurrence of A. (1984, 2M)

42. Cards are drawn one by one at random from a well shuffled full pack of 52 playing cards until 2 aces are obtained for the first time. If N is the number of cards required to be drawn, then show that (n − 1)(52 − n )(51 − n ) Pr { N = n } = 50 × 49 × 17 × 13 where, 2 < n ≤ 50. (1983, 3M)

43. An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2, and 0.1, respectively. What is the probability that the gun hits the plane? (1981, 2M)

44. A box contanis 2 black, 4 white and 3 red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball is drawn at random and kept beside the first. This process is repeated till all the balls are drawn from the box. Find the probability that the balls drawn are in the sequence of 2 black, 4 white and 3 red. (1979, 2M)

Integer & Numerical Answer Type Questions 45. Let S be the sample space of all 3 × 3 matrices with entries from the set {0, 1}. Let the events E1 and E 2 be given by E1 = { A ∈ S : det A = 0} and E 2 = { A ∈ S : sum of entries of A is 7}. If a matrix is chosen at random from S, then the conditional probability P (E1 | E 2) equals ...........

Probability 109 46. Of the three independent events E1 , E 2 and E3 , the probability that only E1 occurs is α , only E 2 occurs is β and only E3 occurs is γ. Let the probability p that none of events E1 , E 2 or E3 occurs satisfy the equations (α − 2 β ), p = αβ and (β − 3γ ) p = 2 βγ. All the given probabilities are assumed to lie in the interval (0, 1). Then,

1 1 1 losing a game against T2 are , and , respectively. Each 2 6 3 team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X andY denote the total points scored by (2016 Adv.) teams T1 and T2, respectively, after two games.

47. P (X > Y ) is 1 4 1 (c) 2

probability of occurrence of E1 is equal to probability of occurrence of E3

(a)

Passage Type Questions

5 12 7 (d) 12

(b)

48. P (X = Y ) is

Passage

11 36 13 (c) 36 (a)

Football teams T1 and T2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1 winning, drawing and

1 3 1 (d) 2

(b)

Topic 4 Law of Total Probability and Baye’s Theorem Objective Question I (Only one correct option) 1. A pot contain 5 red and 2 green balls. At random a ball is drawn from this pot. If a drawn ball is green then put a red ball in the pot and if a drawn ball is red, then put a green ball in the pot, while drawn ball is not replace in the pot. Now we draw another ball randomnly, the probability of second ball to be red is (2019 Main, 9 Jan II) (a)

27 49

(b)

26 49

(c)

21 49

(d)

32 49

2. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is (2018 Main) 3 10 1 (c) 5 (a)

2 5 3 (d) 4 (b)

3. A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective, given that it is produced in plant T1) = 10P (computer turns out to be defective, given that it is produced in plant T2), where P (E ) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then, the probability that it is produced in plant T2, is (2016 Adv.) 36 73 78 (c) 93

(a)

47 79 75 (d) 83 (b)

4. A signal which can be green or red with probability and

transmitted to station B. The probability of each station 3 receiving the signal correctly is . If the signal received 4 at station B is green, then the probability that the (2010) original signal green is 3 5 20 (c) 23

(a)

6 7 9 (d) 20 (b)

Objective Question II (One or more than one correct option) 5. There are three bags B1, B2 and B3 . The bag B1 contains 5 red and 5 green balls, B2 contains 3 red and 5 green balls, and B3 contains 5 red and 3 green balls. Bags B1, 3 3 4 and respectively B2 and B3 have probabilities , 10 10 10 of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct? (a) Probability that the chosen ball is green, given that 3 the selected bag is B3 , equals . 8 (b) Probability that the selected bag is B3 , given that the 5 chosen ball is green, equals . 13 39 (c) Probability that the chosen ball is green equals . 80 (d) Probability that the selected bag is B3 and the chosen 3 ball is green equals . 10

6. A ship is fitted with three engines E1 , E 2 and E3 . The

4 5

1 respectively, is received by station A and then 5

engines function independently of each other with respective probabilities 1/2, 1/4 and 1/4. For the ship to be operational atleast two of its engines must function. Let X denotes the event that the ship is operational and let X1, X 2 and X3 denote, respectively the events that the engines E1, E 2 and E3 are functioning.

110 Probability Which of the following is/are true? (a) P

[X1c| X ] =

(2012)

is

3 / 16

(b) P [exactly two engines of the ship are functioning] = (c) P [X | X 2 ] =

10. The probability of the drawn ball fromU 2 being white,

5 16

(d) P [X | X1 ] =

7 8

7 16

(a)

(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

P (H i ) > 0, i = 1, 2,... , n . Let E be any other event with 0 < P (E ) < 1. Statement I P (H i/E ) > P (E/H i ) ⋅ P (H i ) for

19 30

11 30

(d)

17 23

(b)

11 23

(c)

15 23

(d)

12 23

Passage III

12. The probability that X = 3 equals (a)

25 216

(b)

25 36

(c)

5 36

(d)

125 216

(d)

25 216

13. The probability that X ≥ 3 equals (a)

125 216

(b)

25 36

(c)

5 36

14. The conditional probability that X ≥ 6 given X > 3 equals (a)

i = 1, 2, . . . , n

125 216

(b)

25 216

(c)

5 36

(d)

25 36

Passage IV (2007, 3M)

i =1

Passage Based Problems Passage I Let n1 and n2 be the number of red and black balls, respectively in box I. Let n3 and n4 be the number of red and black balls, respectively in box II. (2015 Adv.)

8. One of the two boxes, box I and box II was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this 1 red ball was drawn from box II, is , then the correct 3 option(s) with the possible values of n1 , n2, n3 and n4 is/are (a) n1 = 3, n2 = 3, n3 = 5, n4 = 15 (b) n1 = 3, n2 = 6, n3 = 10, n4 = 50 (c) n1 = 8, n2 = 6, n3 = 5, n4 = 20 (d) n1 = 6, n2 = 12, n3 = 5, n4 = 20

9. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, 1 after this transfer, is , then the correct option(s) with 3 the possible values of n1 and n2 is/are

(a) n1 = 4 and n2 = 6 (c) n1 = 10 and n2 = 20

(c)

A fair die is tossed repeatedly until a six is obtained. Let X (2009) denote the number of tosses required.

7. Let H 1 , H 2,... , H n be mutually exclusive events with

∑ P (Hi ) = 1

23 30

probability that head appeared on the coin is

For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.

Statement II

(b)

11. Given that the drawn ball from U 2 is white, the (a)

Assertion and Reason

n

13 30

(b) n1 = 2 and n2 = 3 (d) n1 = 3 and n2 = 6

Passage II LetU 1 andU 2 be two urns such thatU 1 contains 3 white and 2 red balls andU 2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U 1 and put intoU 2. However, if tail appears then 2 balls are drawn at random from U 1 and put into U 2. Now, 1 ball is drawn at random from U 2. (2011)

There are n urns each containing (n + 1) balls such that the ith urn contains ‘i’white balls and (n + 1 − i) red balls. Let ui be the event of selecting ith urn, i = 1, 2, 3, ... , n and W (2006, 5M) denotes the event of getting a white balls.

15. If P (ui ) ∝ i, where i = 1, 2, 3,... , n , then lim P (W ) is n→ ∞

equal to (a) 1

2 (b) 3

1 (c) 4

(d)

3 4

16. If P (ui ) = c , where c is a constant, then P (un / W ) is equal to 2 n+1 n (c) n+1

(a)

1 n+1 1 (d) 2 (b)

17. If n is even and E denotes the event of choosing even 1  numbered urn P (ui )= , then the value of P ( W /E ) is  n 

(a) n + 2 2n + 1 n (c) n+1

n+ 2 2 (n + 1) 1 (d) n+1 (b)

Analytical and Descriptive Questions 18. A person goes to office either by car, scooter, bus or 1 3 2 1 and , , , 7 7 7 7 respectively. Probability that he reaches offices late, if 2 1 4 1 he takes car, scooter, bus or train is , , and , 9 9 9 9 respectively. Given that he reached office in time, then what is the probability that he travelled by a car ? train probability of which being

(2005, 2M)

Probability 111 19. A bag contains 12 red balls and 6 white balls. Six balls are drawn one by one without replacement of which at least 4 balls are white. Find the probability that in the next two drawn exactly one white ball is drawn. (Leave the answer in nC r). (2004, 4M)

20. A box contains N coins, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed, is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair? (2002, 5M)

21. An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white? (2001, 5M)

22. Eight players P1 , P2, K , P8 play a knock-out tournament. It is known that whenever the players Pi and Pj play, the player Pi will win if i < j. Assuming that the players are paired at random in each round, what is the probability (1999, 10M) that the player P4 reaches the final?

23. Three players, A, B and C, toss a coin cyclically in that order (i.e. A, B, C, A, B, C, A, B, …) till a head shows. Let p be the probability that the coin shows a head. Let α, β and γ be, respectively, the probabilities that A, B and C gets the first head. Prove that β = (1 − p) α . Determine α, β and (1998, 8M) γ (in terms of p).

24. Sixteen players S1 , S 2, K , S16 play in a tournament. They are divided into eight pairs at random from each pair a winner is decided on the basis of a game played between

the two players of the pair. Assume that all the players are of equal strength. (i) Find the probability that the player S1 is among the eight winners. (ii) Find the probability that exactly one of the two players S1 and S2 is among the eight winners. (1997C, 5M)

25. In a test an examinee either guesses or copies of knows the answer to a multiple choice question with 1 four choices. The probability that he make a guess is 3 1 and the probability that he copies the answer is . The 6 probability that his answer is correct given that he 1 copied it, is . Find the probability that he knew the 8 answer to the question given that he correctly answered it. (1991, 4M)

26. An urn contains 2 white and 2 blacks balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. Find the probability that the third ball drawn is black. (1987, 4M)

27. A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.6. Articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing? (1986, 5M)

Topic 5 Probability Distribution and Binomial Distribution Objective Questions I (Only one correct option) 1. Let C1 and C 2 be two biased coins such that the 2 1 and , 3 3 respectively. Suppose α is the number of heads that appear when C1 is tossed twice, independently, and suppose β is the number of heads that appear when C 2 is tossed twice, independently. Then the probability that the roots of the quadratic polynomial x2 − ax + β are real (2020 Adv.) and equal, is probabilities of getting head in a single toss are

40 81 1 (c) 2

(a)

20 81 1 (d) 4 (b)

2. Four fair dice are thrown independently 27 times. Then, the expected number of times, at least two dice show up a three or a five, is …… . [2020 Main, 5 Sep I]

3. In a workshop, there are five machines and the probability of any one of them to be out of service on a day 1 is . If the probability that at most two machines will be 4

3

 3 out of service on the same day is   k, then k is equal  4 to

[2020 Main, 7 Jan II]

(a) 4

(b)

17 4

17 (c) 8

(d)

17 2

4. For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem 4 is , then the probability that he is unable to solve less 5 than two problem is (2019 Main, 12 April II) (a) (c)

201  1    5  5 54  4    5  5

49

48

(b)

316  4    25  5 

(d)

164  1    25  5 

49

48

5. Let a random variable X have a binomial distribution with mean 8 and variance 4. If P (X ≤ 2) = equal to (a) 17

k , then k is 216

(2019 Main, 12 April I)

(b) 121

(c) 1

(d) 137

112 Probability 6. Minimum number of times a fair coin must be tossed so that the probability of getting atleast one head is more than 99% is (2019 Main 10 April II) (a) 8

(b) 6

(c) 7

coin so that the probability of observing atleast one head (2019 Main, 8 April II) is atleast 90% is (b) 3

15. If the mean and the variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than one is equal to… . (1991, 2M)

(d) 5

7. The minimum number of times one has to toss a fair

(a) 2

Fill in the Blanks

(c) 5

16. For a biased die the probabilities for the different faces to turn up are given below

(d) 4

8. In a game, a man wins ` 100 if he gets 5 or 6 on a throw of a fair die and loses ` 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is (2019 Main, 12 Jan II)

(a)

400 400 loss (b) loss 3 9

(c) 0

(d)

400 gain 3

(a) 6

(b) 3

(c) 5

(d) 4

10. Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then, P (X = 1) + P (X = 2) equals (2019 Main, 9 Jan I)

25 (a) 169

52 (b) 169

49 (c) 169

24 (d) 169

11. A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn one-by-one with replacement, then the variance of the number of green balls drawn is (2017 Main)

12 (a) 5

(b) 6

6 (d) 25

(c) 4

1

2

3

4

5

6

Probability

0.1

0.32

0.21

0.15

0.05

0.17

This die is tossed and you are told that either face 1 or face 2 has turned up. Then, the probability that it is face 1, is… . (1981, 2M)

Analytical & Descriptive Questions 17. Numbers are selected at random, one at a time, from the two-digit numbers 00, 01, 02, …, 99 with replacement. An event E occurs if and only if the product of the two digits of a selected number is 18. If four numbers are selected, find probability that the event E occurs at least (1993, 5M) 3 times.

9. If the probability of hitting a target by a shooter in any

1 shot, is , then the minimum number of independent 3 shots at the target required by him so that the probability of hitting the target at least once is greater 5 than , is (2019 Main, 10 Jan II) 6

Face

18.

A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q of A is again chosen at random. Find the probability that P and Q (1991, 4M) have no common elements.

19. Suppose the probability for A to win a game against B is 0.4. If A has an option of playing either a ‘best of 3 games’ or a ‘best of 5 games’’ match against B, which option should choose so that the probability of his winning the match is higher? (no game ends in a (1989, 5M) draw).

20. A man takes a step forward with probability 0.4 and backwards with probability 0.6. Find the probability that at the end of eleven steps he is one step away from the starting point. (1987, 3M)

12. A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is (2013 Main) (a)

17 3

5

(b)

13 3

5

(c)

11 3

5

(d)

10 3

5

13. India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.50, respectively. Assuming that the outcomes are independent. The probability of India getting at least 7 points, is (1992, 2M) (a) 0.8750

(b) 0.0875

(c) 0.0625

(d) 0.0250

14. One hundred identical coins, each with probability p, of

showing up heads are tossed once. If 0 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is (1988, 2M)

(a) 1/2

(b) 49/101

(c) 50/101

(d) 51/101

Integer & Numerical Answer Type Questions 21. Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14 p is ……… .

22. The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than 0.95, is ……… .

23. The minimum number of times a fair coin needs to be tossed, so that the probability of getting atleast two (2015 Adv.) heads is atleast 0.96, is

Probability 113

Answers Topic 1

39. (i) A, B and C are pairwise independent

1. (c)

2. (a)

5. 9. 13. 17.

(a) 6. (c) (a) 10. (c) (a) 14. (b) (d) 18. (b) 1 1 1 21. 22. ≤ p ≤ 36 3 2 (3n − 3. 2n + 3 ) × 6C 3 25. 6n 10 (n + 2 ) 28. 1 − n + 7 C5

3. (a) 7. 11. 15. 19.

(a) (d) (c) (b) 3 23. 16

8. 12. 16. 20.

26. 0.62

27.

29. (i)

(a) (d) (b) (c)

24. False 1 91

1 1 (ii) 132 462

41.

2. (b)

5. (a)

6. (c)

9. (a, b, c) 13. False

3. (a)

4. (c)

7. (c) 11 40

11.

12. P ( A ∩ B )

14. 13.9%

16. No

17. (a)

18. (c)

Topic 3

1. (d) 5. (a,c) 9. (c,d)

2. (b) 6. (d)

3. (b) 7. (a)

4. (a) 8. (a)

9. (d)

10. (c)

11. (c)

12. (b)

13. (b)

14. (a)

15. (a)

16. (b)

17. (a)

18. (c)

19. (b)

20. (b)

21. (a, b)

22. (a,b)

23. (a, d)

24. (b, c)

25. (a, d) 1 29. 4 1 33. 9

26. (a, d) 5 30. 7 1 35. 2

27. (a, c) 2 31. 5

28. (b, c, d) 32 32. 55 193 38. 792

36. 2 p 2 – p 3

44.

47. (b)

48. (c)

2. (b)

3. (c)

45. 0.50

6. (b, d) 10. (b)

4. (c)

7. (d) 11. (d)

8. (a,b) 12. (a)

13. (b)` 17. 19. 21. 23. 24.

1. (d) 5. (c)

1 1260

43. 0.6976

1 5

Topic 4

8. (a)

10. (a, c)

1 1 or 3 2

46. (6)

Topic 2 1. (b)

40.

4. (d)

14. (d) 15. (b) 16. 1 (b) 18. 7 10 2 12 6 C 2 ⋅ C 4 C1 ⋅ C1 12C1 ⋅6 C 5 11C1 ⋅1 C1 20. ⋅ 12 + 18 ⋅ 12 18 C6 C2 C6 C2 m 4 22. m+n 35 p p (1 − p ) p − 2p 2 + p 3 , , = α= β = γ 1 − (1 − p ) 3 1 − (1 − p ) 3 1(1 − p ) 3 1 8 24 23 (i) (ii) 25. 26. 27. 2 15 29 30

(a)

9m 8N + m

99 1900

Topic 5 1. (b)

2. (11)

3. (c)

4. (c)

5. (d)

6. (c)

7. (d)

8. (c)

9. (c)

10. (a)

13. (b)

14. (d)

11. (a) 11 15. 16

12. (c) 5 16. 21

17.

97 25 4

20.

11

 3 18.    4

C 6( 0 . 24 ) 5

21. (8)

n

19. Best of 3 games 22. (6)

23. (8)

Hints & Solutions Topic 1 Classical Probability 1. It is given that a person wins `15 for throwing a doublet (1, 1) (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) and win `12 when the throw results in sum of 9, i.e., when (3, 6), (4, 5), (5, 4), (6, 3) occurs. Also, losses ` 6 for throwing any other outcome, i.e., when any of the rest 36 − 6 − 4 = 26 outcomes occurs. Now, the expected gain/loss = 15 × P (getting a doublet) + 12 × P (getting sum 9) − 6 × P (getting any of rest 26 outcome) 6  4  26  = 15 ×  + 12 ×  − 6 ×   36  36  36

5 4 26 15 + 8 − 26 + − = 2 3 6 6 23 − 26 3 1 1 = = − = − , means loss of ` 6 6 2 2 =

2. Since, the experiment should be end in the fifth throw of the die, so total number of outcomes are 65 . Now, as the last two throws should be result in two 4 4 fours (i) (ii) (iii) (iv) (v) So, the third throw can be 1, 2, 3, 5 or 6 (not 4). Also, throw number (i) and (ii) can not take two fours in succession, therefore number of possibililites for throw (i) and (ii) = 62 − 1 = 35 [Q when a pair of dice is thrown then (4, 4) occur only once].

114 Probability Hence, the required probability =

5 × 35 175 = 5 65 6

7. We have mentioned that boxes are different and one particular box has 3 balls.

3. Since, there is a regular hexagon, then the number of 6

ways of choosing three vertices is C3 . And, there is only two ways i.e. choosing vertices of a regular hexagon alternate, here A1, A3 , A5 or A2, A4, A6 will result in an equilateral triangle. A2

A5

A3 A4

∴Required probability 2 2 2 ×3 ×2 ×3 ×2 1 = 6 = = = 6 ! 6 × 5 × 4 × 3 × 2 × 1 10 C3 3 !3 ! 4. Number of subset of S = 220 20(21) Sum of elements in S is 1 + 2 + .....+20 = = 210 2 n (n + 1)   Q 1 + 2+ ...... + n =   2 Clearly, the sum of elements of a subset would be 203, if we consider it as follows S − { 7}, S − {1, 6} S − {2, 5}, S − {3, 4} S − { 1, 2, 4 } ∴ Number of favourables cases = 5 5 Hence, required probability = 20 2

5. Total number of ways of selecting 2 different numbers from {0, 1, 2, ..., 10} = 11C 2 = 55 Let two numbers selected be x and y. Then, x + y = 4m and x − y = 4n ⇒ 2x = 4(m + n ) and 2 y = 4(m − n ) ⇒ x = 2(m + n )and y = 2(m − n )

∴x and y both are even numbers. x

y

0

4, 8

2

6, 10

4

0, 8

6

2, 10

8

0, 4

10

2, 6

∴Required probability =

6. Sample space → 12C 2

6 55

Number of possibilities for z is even. z = 0 ⇒ 11C1; z = 2 ⇒ 9C1 z = 4 ⇒ 7C1; z = 6 ⇒ 5C1 z = 8 ⇒ 3C1; z = 10 ⇒ 1C1 Total = 36 36 6 ∴ Probability = = 66 11

…(i) …(ii)

12

According to given condition, following cases may arise. B G G B B G G B B B G B G B B G B B G B B G B G B So, number of favourable ways = 5 × 3 ! × 2 ! = 60 60 1 Required probability = = ∴ 120 2

A1

A6

11

C3 × 29 55  2 =   3  3 312 8. Total number of ways to arrange 3 boys and 2 girls are 5!. Then, number of ways =

9.

PLAN As one of the dice shows a number appearing on one of P1, P2 and P3. Thus, three cases arise. (i) All show same number. (ii) Number appearing on D 4 appears on any one of D1, D 2 and D 3. (iii) Number appearing on D 4 appears on any two of D1, D 2 and D 3.

Sample space = 6 × 6 × 6 × 6 = 64 favourable events = Case I or Case II or Case III Case I First we should select one number for D4 which appears on all i.e. 6C1 × 1. Case II For D4 there are 6C1 ways. Now, it appears on any one of D1 , D2 and D3 i.e. 3 C1 × 1. For other two there are 5 × 5 ways. 6 ⇒ C1 × 3C1 × 1 × 5 × 5 Case III For D4 there are 6C1 ways now it appears on any two of D1 , D2 and D3 3 ⇒ C 2 × 12 For other one there are 5 ways. 6 ⇒ C1 × 3C 2 × 12 × 5 6 C1 + 6C1 × 3C1 × 52 + 6C1 × 3C 2 × 5 Thus, probability = 64 6 (1 + 75 + 15) = 64 91 = 216

10. Sample space A dice is thrown thrice, n (s) = 6 × 6 × 6. r

r

Favorable events ω r 1 + ω 2 + ω 3 = 0 i.e. (r1 , r2, r3 ) are ordered 3 triples which can take values, (1, 2 , 3), (1, 5, 3), (4, 2 , 3), (4, 5, 3)  i.e. 8 ordered pairs (1, 2 , 6), (1, 5, 6), (4, 2 , 6), (4, 5, 6) and each can be arranged in 3 ! ways = 6 8 ×6 2 ∴ = n (E ) = 8 × 6 ⇒ P (E ) = 6 ×6 ×6 9

Probability 115 Only two equilateral triangles can be formed ∆AEC and ∆BFD.

11. Since, three distinct numbers are to be selected from first 100 natural numbers. ⇒ n (S ) = 100C3

D

E(favourable events) = All three of them are divisible by both 2 and 3 . ⇒ Divisible by 6 i.e. {6, 12, 18, …, 96} Thus, out of 16 we have to select 3. ∴ n (E ) = 16C3 16 C 4 ∴ Required probability = 100 3 = C3 1155

13. 71 = 7, 72 = 49, 73 = 343, 74 = 2401, … Therefore, for 7r, r ∈ N the number ends at unit place 7, 9, 3, 1, 7, …

∴ Total number of cases S = 6 × 6 × 6 = 216 and the same number appear on each of them = 6C1 = 6 6 1 ∴ Required probability = = 216 36

17. Since, there are 15 possible cases for selecting a coupon and seven coupons are selected, the total number of cases of selecting seven coupons = 157 It is given that the maximum number on the selected coupon is 9, therefore the selection is to be made from the coupons numbered 1 to 9. This can be made in 97 ways. Out of these 97 cases, 87 does not contain the number 9. Thus, the favourable number of cases = 97 − 87. 97 − 87 ∴ Required probability = 157

For this m and n should be as follows n 4r − 2 4r − 3 4r 4r − 1

18. The number of all possible determinants of the form

For any given value of m, there will be 25 values of n. Hence, the probability of the required event is 100 × 25 1 = 100 × 100 4

a

b

c

d

0 1 0 0 , 0 0 1 0

14. The number of ways of placing 3 black balls without any restriction is 10C3 . Since, we have total 10 places of putting 10 balls in a row. Now, the number of ways in which no two black balls put together is equal to the number of ways of choosing 3 places marked ‘—’ out of eight places.

Vanish and remaining six determinants have non-zero 6 3 values. Hence, the required probability = = 16 8 Statement I is true. Statement II is also true as the homogeneous equations have always a solution and Statement II is not the correct explanation of Statement I.

—W—W—W—W —W—W—W— This can be done in 8C3 ways. 8

C3 8 × 7 ×6 7 = = C3 10 × 9 × 8 15

10

15. Three vertices out of 6 can be chosen in 6C3 ways. So, total ways = 6C3 = 20

= 24 = 16

Out of which only 10 determinants given by 1 1 0 0 1 1 0 0 0 1 1 0 1 0 , , , , , , , 1 1 0 0 0 0 1 1 0 1 1 0 0 0

NOTE Power of prime numbers have cyclic numbers in their unit place.

∴ Required probability =

2 1 = 20 10

16. Since, three dice are rolled.

Also for end at 0.

4r 4r − 1 4r − 2 4r − 3

B

So, required probability =

But it cannot end at 5.

m

F

∴ Favourable ways = 2

∴ 7m + 7n will be divisible by 5 if it end at 5 or 0.

1 2 3 4

C

A

12. Here, two numbers are selected from {1, 2, 3, 4, 5, 6} ⇒ n (S ) = 6 × 5 {as one by one without replacement} Favourable events = the minimum of the two numbers is less than 4. n (E ) = 6 × 4 {as for the minimum of the two is less than 4 we can select one from (1, 2, 3, 4) and other from (1, 2, 3, 4, 5, 6) n (E ) 24 4 = = ∴ Required probability = n (S ) 30 5

E

19.

PLAN Probability =

Number of favourable outcomes Number of total outcomes

As, x1 + x2 + x3 is odd. So, all may be odd or one of them is odd and other two are even.

116 Probability ∴ Required probability 2 C1 × 3C1 × 4C1 + 1C1 × 2C1 × 4C1 + 2C1 × 2C1 × 3C1 + 1C1 × 3C1 × 3C1 = 5 7 3 C1 × C1 × C1 24 + 8 + 12 + 9 105 53 = 105 =

∴ x1 + x3 = 2x2 So, x1 + x3 should be even number.

B: [1, 2, 3], where r1 , r2, K , rn are the readings of n throws and 1, 2, 3 are the numbers that appear in the n throws. Number of such functions, M = N − [n (1) − n (2) + n (3)]

C1 × 4C1 + 1C1 × 3C1 3 C1 × 5C1 × 7C1

2

where, N = total number of functions and n (t ) = number of function having exactly t elements in the range. Now, N = 3n, n (1) = 3 . 2n, n(2) = 3, n(3) = 0

11 = 105

21. According to given condition, 3 1 = 6 2 2 1 P (red at the second toss) = = 6 3 1 and P (blue at the third toss) = 6 Therefore, the probability of the required event 1 1 1 1 = × × = 2 3 6 36 1 + 3p 1 − p 1 − 2p 22. Since, are the probability of , and 3 4 2 mutually exclusive events. 1 + 3p 1 − p 1 −2p + + ≤1 ∴ 3 4 2 P ( yellow at the first toss) =

4 + 12 p + 3 − 3 p + 6 − 12 p ≤ 12

⇒

13 − 3 p ≤ 12 1 ⇒ p≥ 3 1 + 3p 1− p 1 −2p and 0 ≤ ≤ 1, 0 ≤ ≤ 1, 0 ≤ ≤1 3 4 2 ⇒ ⇒

0 ≤ 1 + 3 p ≤ 3, 0 ≤ 1 − p ≤ 4, 0 ≤ 1 − 2 p ≤ 2 1 2 1 1 − ≤ p ≤ , 1 ≥ p ≥ −3 , ≥ p ≥ − 3 3 2 2

...(i)

...(ii)

From Eqs. (i) and (ii), 1 / 3 ≤ p ≤ 1 / 2

23. Since, determinant is of order 2 × 2 and each element is 0 or 1 only. ∴

n (S ) = 24 = 16

and the determinant is positive are 1 0 1 1 1 0 , , 0 1 0 1 1 1 ∴

n (E ) = 3

Thus, the required probability =

First we fix the position ⊗ A ⊗ A ⊗ I ⊗ N ⊗. Number of ways in which no two S’s occur together 4! 5 = × C4 2! 4! × 5 × 4! × 2! 1 ∴ Required probability = = 2! × 8! 14

25. Let us define a onto function F from A : [ r1 , r2, K , rn] to

Either both x1 and x3 are odd or both are even.

⇒

8! . 4 !⋅ 2 !

Hence, it is a false statement.

20. Since, x1 , x2, x3 are in AP.

∴ Required probability =

24. Total number of ways to arrange ‘ASSASSIN’ is

3 16

⇒

M = 3n − 3 . 2n + 3

Hence, the total number of favourable cases = (3n − 3 . 2n + 3) . 6C3 (3n − 3 . 2n + 3) × 6C3 ∴ Required probability = 6n

26. The required probability = 1 − (probability of the event that the roots of x2 + px + q = 0 are non-real). The roots of x2 + px + q = 0 will be non-real if and only if p2 − 4q < 0, i.e. if p2 < 4 q The possible values of p and q can be possible according to the following table. Value of q

Value of p

Number of pairs of p, q

1

1

1

2

1, 2

2

3

1, 2, 3

3

4

1, 2, 3

3

5

1, 2, 3, 4

4

6

1, 2, 3, 4

4

7

1, 2, 3, 4, 5

5

8

1, 2, 3, 4, 5

5

9

1, 2, 3, 4, 5

5

1, 2, 3, 4, 5, 6

6

10

Therefore, the number of possible pairs = 38 Also, the total number of possible pairs is 10 × 10 = 100 38 ∴ The required probability = 1 − = 1 − 0.38 = 0.62 100

27. We have 14 seats in two vans and there are 9 boys and 3 girls. The number of ways of arranging 12 people on 14 seats without restriction is 14 ! 14 P12 = = 7(13 !) 2!

Probability 117 Now, the number of ways of choosing back seats is 2. and the number of ways of arranging 3 girls on adjacent seats is 2(3!) and the number of ways of arranging 9 boys on the remaining 11 seats is 11 P9 ways. Therefore, the required number of ways 4 ⋅ 3 ! 11 ! = 12 ! = 2. (2 .3 !).11 P9 = 2! Hence, the probability of the required event 12 ! 1 = = 7 ⋅ 13 ! 91

28. There are (n + 7) coins in the box out of which five coins can be taken out in

n+7

C5 ways.

The total value of 5 coins can be equal to or more than one rupee and fifty paise in the following ways. (i) When one 50 paise coin and four 25 paise coins are chosen. (ii) When two 50 paise coins and three 25 paise coins are chosen. (iii) When two 50 paise coins, 2 twenty five paise coins and one from n coins of ten and five paise. ∴ The total number of ways of selecting five coins so that the total value of the coins is not less than one rupee and fifty paise is (2C1 ⋅5 C5 ⋅n C 0 ) + (2C 2 ⋅5 C3 ⋅n C 0 ) + (2C 2 ⋅5 C 2 ⋅n C1 ) = 10 + 10 + 10n = 10 (n + 2)

2. We have, P (exactly one of A or B occurs) = P ( A ∪ B) − P ( A ∩ B) = P ( A ) + P (B) − 2P ( A ∩ B) According to the question, 1 …(i) P ( A ) + P (B) − 2P ( A ∩ B) = 4 1 …(ii) P (B) + P (C ) − 2P (B ∩ C ) = 4 1 …(iii) and P (C ) + P ( A ) − 2P (C ∩ A ) = 4 On adding Eqs. (i), (ii) and (iii), we get 2 [P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C ) 3 − P (C ∩ A )] = 4 ⇒ P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C ) 3 − P (C ∩ A ) = 8 ∴P (atleast one event occurs) = P(A ∪ B ∪ C ) = P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C ) 3 1 7 1  = + = Q P(A ∩ B ∩ C ) =  16  8 16 16 3 1 3. Given, P (B) = , P ( A ∩ B ∩ C ) = 4 3 B

A

So, the number of ways of selecting five coins, so that the total value of the coins is less than one rupee and fifty paise is n + 7C5 − 10(n + 2)

(A ∩ B ∩ C)

n+7

(B ∩ C)

C5 − 10(n + 2) n+7 C5 10 (n + 2) =1 − n+7 C5

∴ Required probability =

29. (i) The total number of arrangements of six boys and six girls = 12 !

6! × 7! 1 = (12)! 132 [since, we consider six girls at one person] 2 ×6! ×6! 1 (ii) Required probability = = (12)! 462 ∴ Required probability =

Topic 2 Addition and Subtraction Law of Probability 1. As, we know that P ( A ∩ B) = P ( A ) + P (B) − P ( A ∪ B) ⇒ P ( A ∩ B) = 0.6 + 0.4 − 0.8 ⇒ P ( A ∩ B) = 0.2 and, as P ( A ∪ B ∪ C ) = P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C ) ⇒ α = 0.6 + 0.4 + 0.5 − 0.2 − β − 0.3 + 0.2 ⇒ α = 1.2 − β Q 0.85 ≤ α ≤ 0.95 ⇒ 0.85 ≤ 1.2 − β ≤ 0.95 ⇒ 0.25 ≤ β ≤ 0.35

(A ∩ B ∩ C)

C

and

P(A ∩ B ∩ C ) =

1 3

which can be shown in Venn diagram. ∴ P (B ∩ C ) = P (B) − { P ( A ∩ B ∩ C + P ( A ∩ B ∩ C ))} 3  1 1 3 2 1 = − +  = − = 4  3 3 4 3 12

4. It is given that, P (E ) ≤ P (F ) ⇒ E ⊆ F

…(i)

P (E ∩ F ) > 0 ⇒ E ⊂ F

…(ii)

(a) occurrence of E ⇒ occurrence of F

[from Eq. (i)]

(b) occurrence of F ⇒ occurrence of E

[from Eq. (ii)]

and

(c) non-occurrence of E ⇒ occurrence of F Hence, option (c) is not correct.

[from Eq. (i)]

5. We know that, P (exactly one of A or B occurs) = P ( A ) + P (B) − 2P ( A ∩ B) ∴

P ( A ) + P (B) − 2P ( A ∩ B) = p

…(i)

Similarly,

P (B) + P (C ) − 2P (B ∩ C ) = p

…(ii)

and

P (C ) + P ( A ) − 2P (C ∩ A ) = p

…(iii)

118 Probability On adding Eqs. (i), (ii) and (iii), we get

= P (M ∪ N ) − P (M ∩ N ) = P (M ) + P (N ) − 2P (M ∩ N )

2 [P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C ) − P (C ∩ A )] = 3 p ⇒ P ( A ) + P (B) + P (C ) − P ( A ∩ B) 3p …(v) − P (B ∩ C ) − P (C ∩ A ) = 2 …(v) It also given that, P ( A ∩ B ∩ C ) = p2 ∴ P(at least one of the events A, B, and C occurs) = P ( A ) + P (B) + P (C ) − P ( A ∩ B)

Hence, (a) and (c) are correct answers.

− P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C ) 3p [from Eqs. (iv) and (v)] = + p2 2 =

3 p + 2 p2 2

6. Since, P ( A ∩ B) = P ( A ) ⋅ P (B) It means A and B are independent events, so A ′ and B ′ are also independent. ∴ P ( A ∪ B) ′ = P ( A ′∩ B ′ ) = P ( A )′ ⋅ P (B)′

11. Let E1 be the event getting minimum number 3 and E 2 be the event getting maximum number 7. Then, P (E1 ) = P (getting one number 3 and other two from numbers 4 to 10) 1 C1 × 7C 2 7 = 10 = 40 C3 P (E 2) = P(getting one number 7 and other two from numbers 1 to 6) 1 C1 × 6C 2 1 = 10 = 8 C3 and P (E1 ∩ E 2) = P(getting one number 3, second number 7 and third from 4 to 6) 1 C1 × 1C1 × 3C1 1 = = 10 40 C3 ∴

Alternate Solution P ( A ∪ B)′ = 1 − P ( A ∪ B) = 1 − { P ( A ) + P (B) − P ( A ) ⋅ P (B)} = {1 − P ( A )}{1 − P (B)} = P ( A )′ P (B)′

7. Given,

P ( A ∪ B) = 0.6 , P ( A ∩ B) = 0.2

∴ P ( A ) + P (B ) = [1 − P ( A )] + [1 − P (B)] = 2 − [P ( A ) + P (B)] = 2 − [P ( A ∪ B) + P ( A ∩ B)] = 2 − [0.6 + 0.2] = 1.2

8. Given, P ( A ) = 0.25, P (B) = 0.50, P ( A ∩ B) = 0.14 P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B) = 0.25 + 0.50 – 0.14 = 0.61 Now, P ( A ∪ B) = 1 − P ( A ∪ B) = 1 − 0.61 = 0.39

P (E1 ∪ E 2) = P (E1 ) + P (E 2) − P (E1 ∩ E 2) 7 1 1 11 = + − = 40 8 40 40

12. P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B) If

P ( A ∪ B) = P ( A ∩ B),

then P ( A ) and P (B) are equals. Since, P ( A ∪ B) = P ( A ∩ B) ⇒ A and B are equals sets Thus, P ( A ) and P (B) is equal to P ( A ∩ B).

13. Given, P (A fails in examination) = 0.2 and

P (B fails in examination) = 0.3 P ( A ∩ B) = P ( A )P (B) = (0.2) (0.3)

∴

∴

P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B) = 0.2 + 0.3 − 0.06 = 0.44

Hence, it is a false statement.

9. We know that, P ( A ∩ B) = P ( A ) + P (B) − P ( A ∪ B)

14. Let P ( A ) and P (B) denote respectively the percentage of

Also,

P ( A ∪ B) ≤ 1

city population that reads newspapers A and B.

∴

P ( A ∩ B)min. , when P ( A ∪ B)max = 1

Then,

⇒

P ( A ∩ B) ≥ P ( A ) + P (B) − 1

25 1 20 1 = , P (B) = = , 100 4 100 5 8 2 P ( A ∩ B) = = , 100 25 1 2 17 P ( A ∩ B) = P ( A ) − P ( A ∩ B) = − = , 4 25 100 1 2 3 P ( A ∩ B) = P (B) − P ( A ∩ B) = − = 5 25 25 Let P (C ) be the probability that the population who reads advertisements. ∴ P (C ) = 30% of P ( A ∩ B) + 40% of P ( A ∩ B)

∴ Option (a) is true. Again,

P ( A ∪ B) ≥ 0

∴

P ( A ∩ B)max. , when P ( A ∪ B)min. = 0

⇒

P ( A ∩ B) ≤ P ( A ) + P (B)

∴ Option (b) is true. Also, P ( A ∩ B) = P ( A ) + P (B) − P ( A ∪ B), Thus, (c) is also correct. Hence, (a), (b), (c) are correct options.

10. P(exactly one of M, N occurs) = P{(M ∩ N ) ∪ (M ∩ N )} = P (M ∩ N ) + P (M ∩ N ) = P (M ) − P (M ∩ N ) + P (N ) − P (M ∩ N ) = P (M ) + P (N ) − 2P (M ∩ N ) Also, P(exactly one of them occurs) = {1 − P (M ∩ N )}{1 − P (M ∪ N )}

P ( A) =

+ 50% of P ( A ∩ B) [since, A ∩ B, A ∩ B and A ∩ B are all mutually exclusive] 3 17 2 3 1 2 139 ⇒ P (C ) = × + × + × = = 13 . 9% 10 100 5 25 2 25 1000

Probability 119  1    2

15. We know that, P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C ) = P ( A ∪ B ∪ C ) ⇒ 0.3 + 0.4 + 0.8 – {0.08 + 0.28 + P (BC )} + 0.09 = P(A ∪ B ∪ C ) ⇒ 1.23 − P (BC ) = P ( A ∪ B ∪ C ) where, 0.75 ≤ P ( A ∪ B ∪ C ) ≤ 1 ⇒ 0.75 ≤ 1.23 − P (BC ) ≤ 1 ⇒ − 0.48 ≤ − P (BC ) ≤ − 0.23 ⇒ 0.23 ≤ P (BC ) ≤ 0.48

16. Given, P ( A ) = 0.5 and P ( A ∩ B) ≤ 0.3 ⇒

P ( A ) + P (B) − P ( A ∪ B) ≤ 0.3

⇒

P (B) ≤ 0.3 + P ( A ∪ B) − P ( A ) ≤ P ( A ∪ B) − 0.2 [since, P ( A ∪ B) ≤ 1 ⇒ P ( A ∪ B) − 0.2 ≤ 0.8 ]

∴

P (B) ≤ 0.8

⇒

P (B) cannot be 0.9.

17. Here, five students S1 , S 2, S3 , S 4 and S5 and five seats R1 , R2, R3 , R4 and R5 ∴ Total number of arrangement of sitting five students is 5 ! = 120 Here, S1 gets previously alloted seat R1 ∴S 2, S3 , S 4 and S5 not get previously seats. Total number of way S 2, S3 , S 4 and S5 not get previously seats is 1 1 1 1 1 1 1   4 ! 1 − + − +  = 24 1 − 1 + − +    1 ! 2 ! 3 ! 4 ! 2 6 24  12 − 4 + 1 = 24   =9   24 ∴ Required probability =

9 3 = 120 40

18. Here, n (T1 ∩ T2 ∩ T3 ∩ T4 ) Total = − n (T1 ∪ T2 ∪ T3 ∪ T4 ) ⇒ n (T1 ∩ T2 ∩ T3 ∩ T4 ) = 5 ! − [4C1 4 ! 2 ! − (3 C1 3 ! 2 ! + 3C1 3 ! 2 ! 2 !) + (2C1 2 ! 2 ! + 4C1 2 ⋅ 2 !) − 2] ⇒ n (T1 ∩ T2 ∩ T3 ∩ T4 ) = 120 − [192 − (36 + 72) + (8 + 16) − 2] ∴

= 120 − [192 − 108 + 24 − 2] = 14 14 7 Required probability = = 120 60

Topic 3 Independent and Conditional Probability 1. Let event B is being boy while event G being girl. 1 2 Now, required conditional probability that all children are girls given that at least two are girls, is All 4 girls = (All 4 girls ) + (exactly 3 girls + 1 boy) + (exactly 2 girls + 2 boys) According to the question, P (B) = P (G ) =

=

2.

4

3

4

2

 1  1  1 4  1  1 4   + C3     + C 2     2  2  2  2  2

2

=

1 1 = 1 + 4 + 6 11

Key Idea Use P ( A) = 1 − P ( A) and condition of independent events i.e P ( A ∩ B) = P ( A) ⋅ P ( B)

Given that probability of hitting a target independently by four persons are respectively 1 1 1 1 P1 = , P2 = , P3 = and P4 = 2 3 4 8 Then, the probability of not hitting the target is 1  1  1  1  = 1 −  1 −  1 −  1 −   2  3  4  8 [Q events are independent] 1 2 3 7 7 × × × = 2 3 4 8 32 Therefore, the required probability of hitting the target = 1 − (Probability of not hitting the target) 7 25 =1− = 32 32 P ( A ∩ B) 3. We know that, P( A / B) = P (B) [by the definition of conditional probability] Q A⊂B ⇒ A∩B= A P ( A) …(i) P( A / B) = ∴ P (B) As we know that, 0 ≤ P (B) ≤ 1 1 P ( A) 1≤ < ∞ ⇒ P ( A) ≤ −1 P ( A ) + P (B) − P ( A ∪ B) > P ( A ) + P (B) − 1 P ( A ) + P (B) − P ( A ∪ B) P ( A ) + P (B) − 1 > P (B) P (B)  A  P ( A ) + P (B) − 1 P  >  B P (B)

Hence, option (a) is correct. The choice (b) holds only for disjoint i.e. P ( A ∩ B) = 0 Finally, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B) = P ( A ) + P (B) − P ( A ) ⋅ P (B), if A , B are independent = 1 − {1 − P ( A )} {1 − P (B)} = 1 − P ( A ) ⋅ P (B ) Hence, option (c) is correct, but option (d) is not correct.

28. Since, E and F are independent events. Therefore,

P (E ∩ F ) = P (E ) ⋅ P (F ) ≠ 0, so E and F are not mutually exclusive events.

= 1 − [1 − P (E ) ⋅ P (F )] [Q E and F are independent] = P (E ) ⋅ P (F ) So, E and F as well as E and F are independent events. P (E ∩ F ) + P (E ∩ F ) Now, P (E / F ) + P (E / F ) = P (F ) P (F ) = =1 P (F )

29. P ( A c ) = 0.3

[given]

⇒

P ( A ) = 0.7 [given] P (B) = 0.4 [given] ⇒ P (Bc ) = 0.6 and P ( A ∩ Bc ) = 0.5 Now, P ( A ∪ Bc ) = P ( A ) + P (Bc ) − P ( A ∩ Bc ) = 0.7 + 0.6 − 0.5 = 0.8 P{ B ∩ ( A ∪ Bc )} c ∴ P [B / ( A ∪ B ] = P ( A ∪ Bc ) P{(B ∩ A ) ∪ (B ∩ Bc )} P{(B ∩ A ) ∪ φ } P (B ∩ A ) = = 0.8 0.8 0.8 1 c = [P ( A ) − P ( A ∩ B )] 0.8 0.7 − 0.5 0.2 1 = = = 0.8 0.8 4 =

30. P ( A ∪ B) = P ( A ) + P (B) − P ( A ) P (B), as A and B are independent events. ⇒ 0.8 = (0.3) + P (B) − (0.3) P (B) 5 ⇒ 0.5 = (0.7) P (B) ⇒ P (B) = 7

31. 5 can be thrown in 4 ways and 7 can be thrown in 6 ways, hence number of ways of throwing neither 5 nor 7 is 36 − (4 + 6) = 26 ∴ Probability of throwing a five in a single throw with a 4 1 pair of dice = = and probability of throwing neither 36 9 26 13 5 nor 7 = = 36 18 Hence, required probability 1 2 2  1  13  1  13  1 =   +     +     + ... = 9 = 13 5  9  18  9  18  9 1− 18

32. Let R be drawing a red ball and B for drawing a black ball, then required probability = RRR + RBR + BRR + BBR 5 6 6 6 5 6 × ×  = × × +  10 11 10  10 11 10 4 7  4 7 6 4 + × × + × ×   10 11 10  10 11 10 640 32 = = 1100 55

124 Probability 33. Let A be the event that the maximum number on the two chosen tickets is not more than 10, and B be the event that the minimum number on them is 5 5 C ∴ P ( A ∩ B) = 100 1 C2 10

C P ( A ) = 100 2 C2

and

[Q P ( A′ ) ≤ 1 and P (B ′ ) ≤ 1] . ⇒ P ( A ∪ B) P ( A ′ ∩ B ′ ) ≤ P ( A ) . P (B ′ ) + P (B) . P ( A′ ) ⇒ P ( A ∪ B) . P ( A′ ∩ B ′ ) ≤ P (C ) [Q P (C ) = P ( A ) . P (B ′ ) + P (B). P ( A′ )] 2 3

P (B) = probability that B will hit A =

1.2 P (B) . P (C ) 2 3 1 = P (B ∩ C / E ) = = = 2 2 P (E ) 3 36. Let Ei denotes the event that the students will pass the ith exam, where i = 1, 2, 3 and E denotes the student will qualify. ∴

P (E ) = [P (E1 ) × P (E 2 / E1 )] + [P (E1 ) × P (E 2′ /E1 ) × P (E3 / E 2′ )] + [P (E1′ ) × P (E 2 / E ′1 ) × P (E3 / E 2)] p p 2 = p + p(1 − p) . + (1 − p) . . p 2 2

⇒

P (E ) =

⇒

pn = (1 − p) pn − 1 + p(1 − p) pn − 2 Hence proved.

∴ By law of total probability,

⇒ { P ( A ) + P (B) − P ( A ∩ B)}{ P ( A′ ) . P (B ′ )} [since A, B are independent, so A ′ , B ′ are independent] ∴ P ( A ∪ B) . P ( A ′ ∩ B ′ ) ≤ { P ( A ) + P (B)}. { P ( A′ ) . P (B′ )} = P ( A ) . P ( A′ ) . P (B′ ) + P (B) . P ( A′ ) . P (B′ ) …(i) ≤ P ( A ) . P (B′ ) + P (B) . P ( A′ )

1 2 1 P (C ) = probability that C will hit A = 3 P (E ) = probability that A will be hit 1 2 2 P (E ) = 1 − P (B ) ⋅ P (C ) = 1 − ⋅ = ⇒ 2 3 3 Probability if A is hit by B and not by C

For n ≥ 3, pn = pn − 1 (1 − p) + pn − 2(1 − p) p

E 2 = the event noted number is 8 H = getting head on coin T = be getting tail on coin

34. Here, P ( A ∪ B) . P ( A′ ∩ B ′ )

35. Given, P ( A ) = probability that A will hit B =

= 1 − p(HH ) = 1 − pp = 1 − p2

38. Let, E1 = the event noted number is 7

 B  P ( A ∩ B) P  =  A P ( A) 5 C1 1 = 10 = C2 9

Then

For n = 2, p2 = 1 − p (two heads simultaneously occur).

2 p2 + p2 − p3 + p2 − p3 = 2 p2 − p3 2

37. Since, pn denotes the probability that no two (or more) consecutive heads occur. ⇒ pn denotes the probability that 1 or no head occur. For n = 1 , p1 = 1 because in both cases we get less than two heads (H, T).

P (E1 ) = P (H ) ⋅ P (E1 / H ) + P (T ) ⋅ P (E1 / T ) and P (E 2) = P (H ) ⋅ P (E 2 / H ) + P (T ) ⋅ P (E 2 / T ) where, P (H ) = 1 / 2 = P (T ) P (E1/H ) = probability of getting a sum of 7 on two dice Here, favourable cases are ∴

{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}. 6 1 = P (E1 / H ) = 36 6

Also,

P (E1 / T ) = probability of getting 7 numbered card out of 11 cards 1 = 11

P (E 2 / H ) = probability of getting a sum of 8 on two dice Here, favourable cases are {(2, 6), (6, 2), (4, 4), (5, 3), (3, 5)}. 5 P (E 2 / H ) = 36

∴

P (E 2 / T ) = probability of getting ‘8’ numbered card out of 11 cards = 1 / 11 1 1 17  1 1  1 1  P (E1 ) =  ×  +  ×  = + =  2 6  2 11 12 22 132

∴ and

1 5  1 1  P (E 2) =  ×  +  ×   2 36  2 11 =

1 2

91  91    =  396 729

Now, E1 and E 2 are mutually exclusive events. Therefore, P (E1 or E 2) = P (E1 ) + P (E 2) =

17 91 193 + = 132 792 792

39. Let D1 denotes the occurrence of a defective bulb in Ist draw. Therefore, P (D1 ) =

50 1 = 100 2

and let D2 denotes the occurrence of a defective bulb in IInd draw. 50 1 Therefore, P (D2) = = 100 2 and let N 1 denotes the occurrence of non-defective bulb in Ist draw.

Probability 125 50 1 = 100 2 Again, let N 2 denotes the occurrence of non-defective bulb in IInd draw. 50 1 Therefore, P (N 2) = = 100 2 Now, D1 is independent with N 1and D2 is independent with N 2 . According to the given condition, A = {the first bulb is defective} = { D1D2, D1N 2} B = {the second bulb is non-defective} = { D1N 2, N 1N 2} and C = {the two bulbs are both defective} = { D1D2, N 1N 2} Again, we know that, Therefore, P (N 1 ) =

⇒

6 x2 − 5 x + 1 = 0

⇒

(3x − 1)(2x − 1) = 0 1 1 x = and 3 2 1 1 P ( A ) = or 3 2

⇒ ∴

42. P (N th draw gives 2nd ace) = P{ 1 ace and (n − 2) other cards are drawn in (N − 1) draws} × P { N th draw is 2nd ace} 4 ⋅ (48)! ⋅ (n − 1)! (52 − n )! 3 = ⋅ (52)! ⋅ (n − 2)! (50 − n )! (53 − n ) 4(n − 1)(52 − n )(51 − n ) ⋅ 3 52 ⋅ 51 ⋅ 50 ⋅ 49 (n − 1) (52 − n ) (51 − n ) = 50 ⋅ 49 ⋅ 17 ⋅ 13 =

A ∩ B = { D1N 2}, B ∩ C = { N 1N 2}. C ∩ A = { D1D2} and A ∩ B ∩ C = φ P ( A ) = P{ D1D2} + P{ D1N 2}

Also,

= P (D1 )P (D2) + P (D1 )P (N 2)  1  1  1  1 1 =   +   =  2  2  2  2 2 1 1 Similarly, P (B) = and P (C ) = 2 2  1  1 1 Also, P ( A ∩ B) = P (D1N 2) = P (D1 )P (N 2) =     =  2  2 4 1 1 Similarly, P (B ∩ C ) = , P (C ∩ A ) = 4 4 and P ( A ∩ B ∩ C ) = 0. Since, P ( A ∩ B) = P ( A )P (B), P (B ∩ C ) = P (B)P (C ) and P (C ∩ A ) = P (C )P ( A ). Therefore, A, B and C are pairwise independent. Also, P ( A ∩ B ∩ C ) ≠ P ( A )P (B)P (C ) therefore A, B and C cannot be independent.

40. The total number of ways to answer the question = C1 + C 2 + C3 + C 4 = 2 − 1 = 15 4

4

4

4

4

P(getting marks) = P( correct answer in I chance) + P(correct answer in II chance) + P( correct answer in III chance) 1 3 1  14 1   14 13 1  = + ⋅  + ⋅ ⋅ = =     15 15 14 15 14 13 15 5 1 6

41. Given, P ( A ) ⋅ P (B) = , P ( A ) ⋅ P (B ) = ∴ Let ⇒ ⇒ ⇒ ⇒

[1 − P ( A )] [1 − P (B)] =

1 3

1 3

P ( A ) = x and P (B) = y 1 1 and xy = (1 − x)(1 − y) = 3 6 1 1 and xy = 1 − x − y + xy = 3 6 5 1 and xy = x+ y= 6 6 5  1 x  − x = 6  6

43. Let P (H 1 ) = 0.4, P (H 2) = 0.3, P (H 3 ) = 0.2, P (H 4 ) = 0.1 P (gun hits the plane) = 1 − P(gun does not hit the plane) = 1 − P (H 1 ) ⋅ P (H 2) ⋅ P (H 3 ) ⋅ P (H 4 ) = 1 − (0.6) (0.7) (0.8) (0.9) = 1 − 0.3024 = 0.6976

44. Since, the drawn balls are in the sequence black, black, white, white, white, white, red, red and red. Let the corresponding probabilities be

Then,

p1 , p2,... , p9 2 1 4 3 2 p1 = , p2 = , p3 = , p4 = , p5 = 9 8 7 6 5 1 3 2 p6 = , p7 = , p8 = , p9 = 1 4 3 2

∴ Required probabilitie p1 . p2 . p3 ⋅ K ⋅ p9 1  2  1  4   3   2  1   3  2  =                 (1) =  9  8  7  6   5  4   3  2  1260

45. Given sample space (S) of all 3 × 3 matrices with entries from the set {0, 1} and events E1 = { A ∈ S : det( A ) = 0} and E 2 = { A ∈ S : sum of entries of A is 7}. For event E 2, means sum of entries of matrix A is 7, then we need seven 1s and two 0s. 9! ∴Number of different possible matrices = 7!2! ⇒ n (E 2) = 36 For event E1 ,| A|= 0, both the zeroes must be in same row/column. ∴ Number of matrices such that their determinant is zero 3! = 6 × = 18 = n (E1 ∩ E 2) 2!  E  n (E1 ∩ E 2) ∴Required probability, P  1  =  E 2 n (E 2) =

18 1 = = 0.50 36 2

126 Probability 46.

3. Let x = P (computer turns out to be defective, given that

PLAN Forthe events to be independent,

it is produced in plant T2)

P( E1 ∩ E 2 ∩ E 3 ) = P( E1 ) ⋅ P( E 2 ) ⋅ P( E 3 ) P( E1 ∩ E 2 ∩ E 3 ) = P(only E1 occurs) = P( E1 ) ⋅ (1 − P( E 2 )) (1 − P( E 3 ))

Let x, y and z be probabilities of E1 , E 2 and E3 , respectively. …(i) ∴ α = x (1 − y) (1 − z ) …(ii) β = (1 − x) ⋅ y(1 − z ) …(iii) γ = (1 − x) (1 − y)z …(iv) ⇒ p = (1 − x) (1 − y) (1 − z ) Given, (α − 2 β ) p = αβ and ( β − 3γ ) p = 2 βγ

…(v)

From above equations, x = 2 y and y = 3z ∴

x = 6z x =6 z

⇒

47. Here, P (X > Y ) = P (T1win ) P (T1 win ) + P (T1 win ) P (draw ) + P (draw ) P (T1 win )  1 1  1 1  1 1 5 = ×  + ×  + ×  =  2 2  2 6  6 2 12

48. P [X = Y ] = P (draw ) ⋅ P (draw ) + P (T1 win ) P (T2 win ) + P (T2 win ) ⋅ P (T1 win ) = (1 / 6 × 1 / 6) + (1 / 2 × 1 / 3) + (1 / 3 × 1 / 2) = 13 / 36

Topic 4 Law of Total Probability and Baye’s Theorem 1. Let A be the event that ball drawn is given and B be the event that ball drawn is red. 2 5 P ( A ) = and P (B) = ∴ 7 7 Again, let C be the event that second ball drawn is red. ∴

2.

 D x= P    T2

⇒

where, D = Defective computer ∴ P (computer turns out to be defective given that is produced in plant T1) = 10x  D …(ii) i.e. P   = 10x  T1  20 80 and P (T2) = 100 100 7 Given, P (defective computer) = 100 7 i.e. P (D ) = 100 P (T1 ) =

Also,

Using law of total probability,  D  D P (D ) = 9(T1 ) ⋅ P   + P (T2) ⋅ P    T1   T2 ∴ ⇒ ∴

7  20   80  =  ⋅ 10x +  ⋅x  100 100  100 1 7 = (280)x ⇒ x = 40  D  D  10 1 and P   = P  =  T2 40  T1  40

Using Baye’s theorem, P (T2 ∩ D ) T  P  2 =  D  P (T1 ∩ D ) + P (T2 ∩ D )  D P (T2) ⋅ P    T2 =  D  D P (T1 ) ⋅ P   + P (T2) ⋅ P    T2  T1  80 39 ⋅ 78 100 40 = = 20 30 80 39 93 ⋅ + ⋅ 100 40 100 40

Key idea Use the theorem of total probability

⇒

P (E 2) =

=

4. From the tree diagram, it follows that S

4 6 6 4 24 + 24 48 2 × + × = = = 10 12 10 12 120 120 5

1 5

4 5

 A 4 6 , P  = 10  E 2 12

By law of total probability  A  A P ( A ) = P (E1 ) × P   + P (E 2) × P    E1   E 2

…(iii)

 D  D 1 39 10 30 and P   = 1 − …(iv) = = ⇒ P   =1− 40 40 40 40  T2  T1 

P (C ) = P ( A ) P (C / A ) + P (B) P (C / B) 2 6 5 4 = × + × 7 7 7 7 12 + 40 32 = = 49 49

Let E1 = Event that first ball drawn is red E 2 = Event that first ball drawn is black A = Event that second ball drawn is red  A 4 6 P (E1 ) = , P   = 10  E1  12

…(i)

G 3 4

R 1 4

AR

AR

AG

1 4

3 4

3 4

1 3 4 4

11 44

BG

BR

BG

1 AG 3 4 4 BR

3 4 BG

Probability 127 46 80 10 5 P (BG|G ) = = 16 8 5 4 1 P (BG ∩ G ) = × = 8 5 2 1 1 80 20 P (G|BG ) = 2 = × = P (BG ) 2 46 23 P (BG ) =

∴

5.

= 1 /4 Now, (a) P

5 R 5 G B1

3 R 5 G

5 R 3 G

B2

B3

Now, probability that the chosen ball is green, given G 3 that selected bag is B3 = P   =  B3  8 Now, probability that the selected bag is B3 , given that B  the chosen ball is green = P  3  G

=

G P   P (B3 )  B3  G G G P   P (B1 ) + P   P (B2) + P   P (B3 )  B3   B1   B2

[by Baye’s

theorem] =

3 5 ×    10 10

3 4  1  ×   8 10 4 2 = =  5 3   3 4  1 + 5 + 1 13 + ×  + ×   8 10  8 10 2 8 2

Now, probability that the chosen ball is green G G G = P (G ) = P (B1 )P   + P (B2)P   + P (B3 )P    B1   B2  B3  [By using theorem of total probability] 5   3 5  4 3 3 = ×  + ×  + ×   10 10  10 8  10 8 3 3 3 12 + 15 + 12 39 = + + = = 20 16 20 80 80 Now, probability that the selected bag is B3 and the G 4 3 3 chosen ball is green = P (B3 ) × P   = × =  B3  10 8 20 Hence, options (a) and (c) are correct. PLAN It is based on law of total probability and Bay’s Law.

Description of Situation It is given that ship would work if atleast two of engines must work. If X be event that the ship works. Then, X ⇒ either any two of E1 , E 2, E3 works or all three engines E1 , E 2, E3 works.

(X1c

/X)

1 1 1  X1c ∩ X  P (E1 ∩ E 2 ∩ E3 ) 2 ⋅ 4 ⋅ 4 1 = =P = = 1 P (X ) 8  P (X )  4

Key Idea Use conditional probability, total probability and Baye’s theorem.

It is given that there are three bags B1 , B2 and B3 and probabilities of being chosen B1 , B2 and B3 are respectively 3 3 4 and P (B3 ) = . ∴ P (B1 ) = , P (B2) = 10 10 10

6.

1 1 1 Given, P (E1 ) = , P (E 2) = , P (E3 ) = 2 4 4 P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 )  P(X) =  ∴  + P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 )   1 1 3 1 3 1 1 1 1  1 1 1 = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅   2 4 4 2 4 4 2 4 4  2 4 4

(b) P (exactly two engines of the ship are functioning) =

P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) P (X )

1 1 3 1 3 1 1 1 1 ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ 7 =2 4 4 2 4 4 2 4 4= 1 8 4  X  P (X ∩ X 2) (c) P   =  X 2 P (X 2) =

P (ship is operating with E 2 function ) P (X 2)

P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) P (E 2) 1 1 3 1 1 1 1 1 1 ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ 5 = 2 4 4 2 4 4 2 4 4 = 1 8 4 P (X ∩ X1 ) (d) P (X / X1 ) = P (X1 ) 1 1 1 1 3 1 1 1 3 ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =2 4 4 2 4 4 2 4 4 1 /2 7 = 16 =

7. Statement I If P (H i ∩ E ) = 0 for some i, then  E H  P  i = P   = 0  E  Hi  If P (H i ∩ E ) ≠ 0, ∀ i = 1, 2, K , n , then  H  P (H i ∩ E ) P (H i ) × P  i =  E P (H i ) P (E ) =

 E P   × P (H i )  Hi  P (E )

 E > P   ⋅ P (H i )  Hi 

[Q0 < P (E ) < 1]

Hence, Statement I may not always be true. Statement II Clearly, H 1 ∪ H 2 ∪ . . . ∪ H n = S [sample space] ⇒

P (H 1 ) + P (H 2) + . . . + P (H n ) = 1

Hence, Statement II is ture.

128 Probability Head appears

Passage I

2W

2W

8. n1 n2

Red

n3

Black

n4

Box I

2R

Red

or Black

1R

∴

1 3 P (B2) ⋅ P (B2 ∩ A ) 1 = P ( A) 3

⇒

3W

1W 2W

2R

U2

1W

2W

2W

0R

2R

1R

1R

U1

U2

U1

U2

3W

3 Cases

10. Now, probability of the drawn ball from U 2 being white

⇒

1  n3    2  n3 + n4  1 = 3     1 n1 1 n3   +   2  n1 + n2 2  n3 + n4 

⇒

n3 (n1 + n2) 1 = n1 (n3 + n4 ) + n3 (n1 + n2) 3

is 1 2 2  3C C C C  P (white / U 2) = P (H ) ⋅  5 1 × 2 1 + 5 1 × 2 1  C1 C1 C1   C1

Now, check options, then clearly options (a) and (b) satisfy.

1 2 3 3  3C C C C C1 ⋅ 2C1 + P (T )  5 2 × 3 2 + 5 2 × 3 1 + 5 C2 C2 C2 C2  C2 1 3 2 1 =  ×1 + ×  2 5 5 2

+

9.

1 Red (n1 – 1)

n2

Red

(n3 + 1)

Black

Red

n1

= P (H ) .

Box II

Red

(n2 – 1) Black

(n4 + 1) Black

Box I

Box II

1 3  n1 − 1   n1   n2    1 n1 ⇒   +   =  n1 + n2 − 1  n1 + n2  n1 + n2  n1 + n2 − 1 3

∴ P (drawing red ball from B1) =

n12 + n1n2 − n1 1 = (n1 + n2) (n1 + n2 − 1) 3

⇒

Clearly, options (c) and (d) satisfy. Passage II 3W

1W

2R U1

U2

Initial

1 2  3C1 2C1 C C  × 2 + 5 1 × 2 1 5 C1 C1 C1   C1

23 / 30 2 1 3  ×1 + ×  1 5 5 2 =  20 / 30 2 12 = 23

Red

n3

C1   C2 

2 3

11. P (Head appeared/white from U 2)

1 Black or

×

13 1 1 6 2  23 × + × =  ×1 + 2 10 10 3 10 3  30

Black

n4

Box I

1R U2

Tail appears

U1

P (B2 / A ) =

Given,

1W 1R

U1

P ( A ) = P (B1 ) ⋅ P ( A / B1 ) + P (B2) ⋅ P ( A / B2) 1  n1  1  n3  =  +   2  n1 + n2 2  n3 + n4 

2 Cases

U2

U1 3W

Box II

A = Drawing red ball

Let

1W

Passage III 5 5 1 25 6 6 6 216 5 5 25 13. P (X ≥ 3) = ⋅ ⋅ 1 = 6 6 36 P {(X > 3) / (X ≥ 6)} ⋅ P (X ≥ 6) 14. P {(X ≥ 6) / (X > 3)} = P (X > 3) 5   5  1  5 6  1  1 ⋅   ⋅   +   ⋅   + ... ∞   6  6  6  6  = 25 =  36   5 3 1  5 4 1    ⋅ +   ⋅ + ... ∞  6 6 6  6 

12. P (X = 3) = ⋅ ⋅ =

Probability 129 Passage IV 2 15. Here, P (ui ) = ki, Σ P (ui ) = 1 ⇒ k = n (n + 1) lim P (W ) = lim

n→ ∞

n→ ∞

n

2i

∑ n (n + 1)2 i=1

n 2 u +1 n 16. P  n  = = Σi W  n+1 n+1 W  2 + 4 + 6 + ... n+2 =  = n (n + 1)  E 2 (n + 1) 2

17. P 

18. As, the statement shows problem is to be related to Baye’s law. Let C , S , B, T be the events when person is going by car, scooter, bus or train, respectively. 1 3 2 1 ∴ P (C ) = , P (S ) = , P (B) = , P (T ) = 7 7 7 7 Again, L be the event of the person reaching office late. ∴ L be the event of the person reaching office in time.  L 7  L 8  L 5 Then, P   = , P   = , P   = C 9 S 9  B 9

∴

12

=

2

2n (n + 1)(2n + 1) = 2 /3 = lim n→∞ 6n (n + 1)2

and

 B  B P (B) = P   P ( A1 ) + P   P ( A2)  A1   A2 

∴

 L 8 P  = T 9  L P   ⋅ P (C ) C C P  =  L  L  L  L P   ⋅ P (C ) + P   ⋅ P (S ) + P   ⋅ P (B) S C  B  L + P   ⋅ P (T ) T 7 1 × 1 9 7 = = 7 1 8 3 5 2 8 1 7 × + × + × + × 9 7 9 7 9 7 9 7

19. Let A1 be the event exactly 4 white balls have been drawn. A2 be the event exactly 5 white balls have been drawn. A3 be the event exactly 6 white balls have been drawn. B be the event exactly 1 white ball is drawn from two draws. Then,  B  B  B P (B) = P   P ( A1 ) + P   P ( A2) + P   P ( A3 )  A1   A2   A3   B But P   = 0  A3  [since, there are only 6 white balls in the bag]

C 2.6 C 4 18

C6

10

.

C1.2 C1 + 12 C2

C1.6 C5 . 11C1.1 C1 12 18 C2 C6

12

20. Let E be the event that coin tossed twice, shows head at first time and tail at second time and F be the event that coin drawn is fair. P (E / F ) ⋅ P (F ) P (F / E ) = P (E / F ) ⋅ P (F ) + P (E / F ′ ) ⋅ P (F ′ ) 1 1 m ⋅ ⋅ 2 2 N = 1 1 m 2 1 N −m ⋅ ⋅ + ⋅ ⋅ 2 2 N 3 3 N m 9m 4 = = m 2 (N − m) 8N + m + 4 9

21. Let W1 = ball drawn in the first draw is white. B1 = ball drawn in the first draw in black. W 2 = ball drawn in the second draw is white. Then , P (W 2) = P (W1 ) P (W 2 / W1 ) + P (B1 )P (W 2 / B1 )  m   m+ k   n    m =    +     m + n   m + n + k  m + n   m + n + k =

m(m + k) + mn m (m + k + n ) m = = (m + n ) (m + n + k) (m + n ) (m + n + k) m + n

22. The number of ways in which P1 , P2, K , P8 can be paired in four pairs 1 8 = [( C 2)(6C 2)(4C 2)(2C 2)] 4! 1 8! 6! 4! = × × × ×1 4 ! 2 !6 ! 2 !4 ! 2 !2 ! 1 8×7 6 ×5 4 ×3 = × × × 4 ! 2 ! ×1 2 ! × 1 2 ! × 1 8 × 7 ×6 ×5 = = 105 2 .2 .2 .2 Now, atleast two players certainly reach the second round between P1, P2 and P3 and P4 can reach in final if exactly two players play against each other between P1, P2, P3 and remaining player will play against one of the players from P5 , P6, P7, P8 and P4 plays against one of the remaining three from P5 …P8. This can be possible in 3 C 2 × 4C1 × 3C1 = 3 . 4 . 3 = 36 ways ∴ Probability that P4 and exactly one of P5 ... P8 reach 36 12 second round = = 105 35 If P1 , Pi , P4 and Pj , where i = 2 or 3 and j = 5 or 6 or 7 reach the second round, then they can be paired in 2 1 4 pairs in ( C 2) (2C 2) = 3 ways. But P4 will reach the 2! final, if P1 plays against Pi and P4 plays against Pj .

130 Probability Hence, the probability that P4 will reach the final round 1 from the second = 3 12 1 4 × = . ∴ Probability that P4 will reach the final is 35 3 35

∴

The probability of any one wining in the pairs of S1 , S 2 = P (certain event) = 1 ∴ The pairs of S1 , S 2 being in two pairs separately and S1 wins, S 2 loses + The probability of S1 , S 2 being in two pairs separately and S1 loses, S 2 wins.

α = probability of A getting the head on tossing firstly = P (H 1 or T1T2T3 H 4 or T1T2T3T4T5T6H 7 or …) = P (H ) + P (H )P (T )3 + P (H )P (T )6 + … P (H ) p = = 3 1 − P (T ) 1 − q3

(14)!  (14)!    7    1 1 2 ( !)7 ⋅ 7 !  1 1 (2 !) ⋅ 7 ! × × + 1 − = 1 − × ×  (16)! (16)!  2 2  2 2     (2 !)8 ⋅ 8 !  (2 !)8 ⋅ 8 !  1 14 × (14)! 7 = × = 2 15 × (14)! 15

Also, β = probability of B getting the head on tossing secondly = P (T1H 2 or T1T2T3T4H 5 or T1T2T3T4T5T6T7H 8 or …) = P (H ) [P (T ) + P (H )P (T )4 + P (H )P (T )7 + K ] = P (T )[P (H ) + P (H )P (T )3 + P (H )P (T )6 + ... ] p(1 − p) = q α = (1 − p) α = 1 − q3

∴ Required probability =

p + p(1 − p) =1 − 1 − (1 − p)3 =

1 − (1 − p)3 − p − p(1 − p) 1 − (1 − p)3

1 − (1 − p)3 − 2 p + p2 p − 2 p2 + p3 = γ= 1 − (1 − p)3 1 − (1 − p)3 Also,

α=

p(1 − p) p , β= 1 − (1 − p)3 1 − (1 − p)3

24. (i) Probability of S1 to be among the eight winners = (Probability of S1 being a pair ) × (Probability of S1 winning in the group) 1 1 [since, S1 is definitely in a group] =1 × = 2 2 (ii) If S1 and S 2 are in the same pair, then exactly one wins. If S1 and S 2 are in two pairs separately, then exactly one of S1 and S 2 will be among the eight winners. If S1 wins and S 2 loses or S1 loses and S 2 wins. Now, the probability of S1 , S 2 being in the same pair and one wins = (Probability of S1 , S 2 being the same pair) × (Probability of anyone winning in the pair). and the probability of S1 , S 2 being the same pair n (E ) = n (S ) where, n (E ) = the number of ways in which 16 persons can be divided in 8 pairs.

1 7 8 + = 15 15 15

25. Let E1 , E 2, E3 and A be the events defined as E1 = the examinee guesses the answer

p + p(1 − p) γ = 1 − (α + β ) = 1 − 1 − q3

⇒

(14)! (16)! and n (S ) = (2 !)7 ⋅ 7 ! (2 !)8 ⋅ 8 !

∴ Probability of S1 and S 2 being in the same pair (14)! ⋅ (2 !)8 ⋅ 8 ! 1 = = (2 !)7 ⋅ 7 !⋅ (16)! 15

23. Let q = 1 − p = probability of getting the tail. We have,

Again, we have α + β + γ =1

n (E ) =

E 2 = the examinee copies the answer E3 = the examinee knows the answer and A = the examinee answer correctly 1 1 We have, P (E1 ) = , P (E 2) = 3 6 Since, E1 , E 2, E3 are mutually exclusive and exhaustive events. 1 1 1 ∴ P (E1 ) + P (E 2) + P (E3 ) = 1 ⇒ P (E3 ) = 1 − − = 3 6 2 If E1 has already occured, then the examinee guesses. Since, there are four choices out of which only one is correct, therefore the probability that he answer correctly given that he has made a guess is 1/4. 1 i.e. P ( A / E1 ) = 4 1 It is given that, P ( A / E 2) = 8 and P ( A / E3 ) = probability that he answer correctly given that he know the answer = 1 By Baye’s theorem, we have P (E3 ) ⋅ P ( A / E3 ) P (E3 / A ) =   P (E1 ) ⋅ P ( A / E1 ) + P (E 2) ⋅ P ( A / E 2)   + P (E3 ) ⋅ P ( A / E3 )   1 ×1 24 2 = ∴ P (E3 / A ) =  1 1  1 1  1  29  ×  +  ×  +  × 1  3 4  6 8  2 

26. Let and

Bi = ith ball drawn is black. Wi = ith ball drawn is white, where i = 1, 2 A = third ball drawn is black.

Probability 131 We observe that the black ball can be drawn in the third draw in one of the following mutually exclusive ways. (i) Both first and second balls drawn are white and third ball drawn is black. i.e. (W1 ∩ W 2) ∩ A (ii) Both first and second balls are black and third ball drawn is black. i.e. (B1 ∩ B2) ∩ A (iii) The first ball drawn is white, the second ball drawn is black and the third ball drawn is black. i.e. (W1 ∩ B2) ∩ A (iv) The first ball drawn is black, the second ball drawn is white and the third ball drawn is black. i.e. ∴

(B1 ∩ W 2) ∩ A P ( A ) = P [{(W1 ∩ W 2) ∩ A } ∪{(B1 ∩ B2) ∩ A } ∪ {(W1 ∩ B2) ∩ A } ∪ {(B1 ∩ W 2) ∩ A }] = P{(W1 ∩ W 2) ∩ A } + P{(B1 ∩ B2) ∩ A } + P{(W1 ∩ B2) ∩ A } + P{(B1 ∩ W 2) ∩ A } = P (W1 ∩ W 2) ⋅ P ( A / (W1 ∩ W 2)) + P (B1 ∩ B2)

∴

P ( A / (B1 ∩ B2)) + P (W1 ∩ B2) ⋅ P ( A / (W1 ∩ B2)) + P (B1 ∩ W 2) ⋅ P ( A / (B1 ∩ W 2))  2 3 4  2 1 =  ×  ×1 +  ×  ×  4 5 6  4 3  2 2 3  2 2 3 + ×  × + ×  ×  4 3 4  4 5 4 1 1 1 3 23 = + + + = 6 5 4 20 30

27. The testing procedure may terminate at the twelfth testing in two mutually exclusive ways. I : When lot contains 2 defective articles. II : When lot contains 3 defective articles. Let A = testing procedure ends at twelth testing A1 = lot contains 2 defective articles A2 = lot contains 3 defective articles ∴ Required probability = P ( A1 ) ⋅ P ( A / A1 ) + P ( A2) ⋅ P ( A / A2) Here, P ( A / A1 ) = probability that first 11 draws contain 10 non-defective and one-defective and twelfth draw contains a defective article. 18 C10 × 2C1 1 …(i) = × 20 9 C11 P ( A / A2) = probability that first 11 draws contains 9 non-defective and 2-defective articles and twelfth draw 17 C 9 × 3C 2 1 … (ii) contains defective = × 20 9 C11 ∴ Required probability = (0.4)P ( A / A1 ) + 0.6 P ( A / A2) =

0.4 × 18C10 × 2C1 1 0.6 × 17C 9 × 3C 2 1 99 × + × = 20 20 9 1900 9 C11 C11

Topic 5 Probability Distribution and Binomial Distribution 1. It is given that α is the number of heads that appear when C1 is tossed twice, the probability distribution of random variable α is α

0

P(α)

 1    3

2

1

2

2 1 2      3  3

2    3

2

Similarly, it is given that β is the number of heads that appear when C 2 is tossed twice, so probability distribution of random variable β is β P(β)

0 2    3

2

1

2

4 9

1 9

Now, as the roots of quadratic polynomial x2 − αx + β are real and equal, so D = α 2 − 4β = 0 and it is possible if (α,β) = (0, 0) or (2, 1) 2

2

2

 2  4  1  2 ∴ Required probability =     +      3  9  3  3 4 16 20 = + = 81 81 81

2. As we know that the probability of show up a three or a five on a throw of a dice is 2 1 P (E ) = = 6 3 2 ∴ P (E ) = 3 Now, as four fair dice are thrown independently 27 times, so probability of getting at least two 3’s or 5’s in one trial 2 2 3 4  1  2  1  2  1 p = 4C 2    + 4C3     + 4C 4    3  3  3  3  3 2 1  4 = 6  4 + 4 4 + 4 3  3 3 24 + 8 + 1 33 11 = = 4 = 27 34 3 Therefore, the expected number of times in 27 trials = 27 p = 11

3. It is given that out of the five machines in a workshop, the probability of any one of them to be out of service on 1 a day is . 4 So, the probability that at most two machines will be out of service on the same day = probability that no machine is out of service + probability that exactly one machine is out of service + probability that exactly two of machines are out of service 3 5 4 2 3 1 1 1 1  3   1  5 5 1 −  + C1 1 −  + C 2  1 −  =   k  4   4  4 4 4 4 (given)

132 Probability 5

3

4

5  3 33  3  3 ⇒   +   + 10 5 =   k  4  4 4  4 4

According to the question, n n 99 99  1  1 1−  > ⇒  100 ⇒   <  2 100 [for minimum] ⇒ n=7

2

15 10  3 k=  + 2 + 2  4 4 4 9 + 15 + 10 34 = = 16 16 17 k= 8

⇒

⇒

7. The required probability of observing atleast one head

4. Given that, there are 50 problems to solve in an admission test and probability that the candidate can 4 solve any problem is = q (say). So, probability that the 5 4 1 candidate cannot solve a problem is p = 1 − q = 1 − = . 5 5 Now, let X be a random variable which denotes the number of problems that the candidate is unable to solve. Then, X follows binomial distribution with 1 parameters n = 50 and p = . 5 Now, according to binomial probability distribution concept r 50 − r  1  4 , r = 0, 1, ... , 50 P (X = r ) = 50C r      5  5 ∴Required probability = P (X < 2) = P (X = 0) + P (X = 1) 49 49 50 449  4  4  4 50 54  4 = 50C 0   + 50C1 =   +  =   50  5  5  5 5  5  5 (5)

5. Let for the given random variable ‘X’ the binomial probability distribution have n-number of independent trials and probability of success and failure are p and q respectively. According to the question, Mean = np = 8 and variance = npq = 4 1 1 q = ⇒ p =1−q = ∴ 2 2 1 Now, n × = 8 ⇒ n = 16 2  1 P (X = r ) = 16C r    2 ∴

16

 1 + 16 C1    2

16

 1 + 16 C 2   2

k 1 + 16 + 120 137 = = 16 = 16 216 2 2 ⇒

16

failure in a trial respectively. Then, 2 1 4 2 p = P (5 or 6) = = and q = 1 − p = = . 6 3 6 3 Now, as the man decides to throw the die either till he gets a five or a six or to a maximum of three throws, so he can get the success in first, second and third throw or not get the success in any of the three throws. So, the expected gain/loss (in `) = ( p × 100) + qp(− 50 + 100) + q2p(− 50 − 50 + 100) + q3 (− 50 − 50 − 50) 3

2

1   2 1  2  1  2 =  × 100 +  ×  (50) +     (0) +   (− 150) 3   3 3  3  3  3 100 100 1200 = + +0− 3 9 27 900 + 300 − 1200 1200 − 1200 = = =0 27 27

9. The probability of hitting a target at least once

[by using binomial distribution]

Here, (given)

k = 137

6. As we know probability of getting a head on a toss of a fair coin is P (H ) =

8. Let p and q represents the probability of success and

= 1 − (probability of not hitting the target in any trial) = 1 − nC 0 p0qn where n is the number of independent trials and p and q are the probability of success and failure respectively.

16

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)  1 =16 C 0    2

= 1 − P (no head) 1 [let number of toss are n] =1 − n 2 1  Q P (Head) = P (Tail) =  2  1 90 According to the question, 1 − n ≥ 100 2 1 1 ⇒ 2n ≥ 10 ⇒ n ≥ 4 ⇒ n ≤ 10 2 So, minimum number of times one has to toss a fair coin so that the probability of observing atleast one head is atleast 90% is 4.

1 = p (let) 2

Now, let n be the minimum numbers of toss required to get at least one head, then required probability = 1 − (probability that on all ‘n’ toss we are getting tail) n 1   1 =1 −   Q P (tail) = P (Head ) =   2 2 

1 p= 3

and

q =1− p =1− 0

1 2 = 3 3 n

5  1  2 According to the question, 1 − nC 0     >  3  3 6 n n 1 5  2  2 ⇒   1) = 1 − { P (X = 0) + P (X = 1)} 0

4

1

Clearly, X follows binomial distribution with parameter 1 n = 2 and p = . 13 x 2− x  1   12 Now, P (X = x) = 2C x     , x = 0, 1, 2  13  13 ∴

17. Let E be the event that product of the two digits is 18,

P (X = 1) + P (X = 2) 1

2

 1   12  1   12 = C1     + 2C 2     13  13  13  13

0

therefore required numbers are 29 , 36, 63 and 92. 4 Hence, p = P (E ) = 100

2

1  12  =2  +  169 169 24 1 25 = + = 169 169 169

and probability of non-occurrence of E is 4 96 q = 1 − P (E ) = 1 − = 100 100

11. Given box contains 15 green and 10 yellow balls. ∴ Total number of balls = 15 + 10 = 25 15 3 P(green balls) = = = p = Probability of success 25 5 10 2 P(yellow balls) = = = q = Probability of unsuccess 25 5 and n = 10 = Number of trials. 3 2 12 ∴Variance = npq = 10 × × = 5 5 5 1 12. Probability of guessing a correct answer, p = and 3 probability of guessing a wrong answer, q = 2 /3 ∴ The probability of guessing a 4 or more correct 4 5 2 1 11  1 2  1 answers = 5C 4   ⋅ + 5C5   = 5 ⋅ 5 + 5 = 5  3 3  3 3 3 3

13. India play 4 matches and getting at least 7 points. It can only be possible in WWWD or WWWW position, where W represents two points and D represents one point. Therefore, the probability of the required event = [4(0.05) + 0.5] (0.5)

C50 p50 (1 − p)50 = 100C51 ( p)51 (1 − p)49 (100) ! (51 !) × (49 !) p p 51 ⋅ = ⇒ = (50 !) (50 !) 100 ! 1− p 1 − p 50 100

51 101

15. For Binomial distribution, mean = np and ∴ ⇒

variance = npq np = 2 and npq = 1 q = 1 / 2 and p + q = 1

4

18. Since, set A contains n elements. So, it has 2n subsets. ∴ Set P can be chosen in 2n ways, similarly set Q can be chosen in 2n ways. ∴ P and Q can be chosen in (2n )(2n ) = 4n ways. Suppose, P contains r elements, where r varies from 0 to n. Then, P can be chosen in nC r ways, for 0 to be disjoint from A, it should be chosen from the set of all subsets of set consisting of remaining (n − r ) elements. This can be done in 2n − r ways. ∴ P and Q can be chosen in nC r ⋅ 2n − r ways. But, r can vary from 0 to n. ∴ Total number of disjoint sets P and Q n

∑ nC r2n − r = (1 + 2)n = 3n

Hence, required probability =

binomial variate with parameters n = 100 and p. Since, P (X = 50) = P (X = 51)

p=

3

97  4   96   4  =4   = 4  +    100  100  100 25

r=0

14. Let X be the number of coins showing heads. Let X be a

⇒

P = 4C3 p3 q + 4C 4 p4

3

= 0.0875

⇒

Out of the four numbers selected, the probability that the event E occurs atleast 3 times, is given as

=

= 4C3 (0.05) (0.5)3 + 4C 4 (0.5)4

⇒

 1  1      2  2

3

 1  1 = 1 − 4C 0     − 4C1  2  2 1 4 11 =1 − − = 16 16 16 0 .1 0 .1 5 16. Probability (face 1) = = = 0 .1 + 0 .32 0 .42 21

[given]

3 n  3 =  4 n  4

n

19. Case I When A plays 3 games against B. In this case, we have n = 3, p = 0.4 and q = 0.6 Let X denote the number of wins. Then, P (X = r ) = 3C r (0.4)r (0.6)3 − r; r = 0, 1, 2, 3 ∴ P1 = probability of winning the best of 3 games = P (X ≥ 2) = P (X = 2) + P (X = 3) = 3C 2(0.4)2(0.6)1 + 3C3 (0.4)3 (0.6)0 = 0.288 + 0.064 = 0.352 Case II When A plays 5 games against B. In this case, we have n = 5, p = 0.4

and

q = 0.6

134 Probability Let X denotes the number of wins in 5 games. Then, P (X = r ) = 5C r (0.4)r (0.6)5 − r , where r = 0, 1, 2K , 5 ∴ P2 = probability of winning the best of 5 games = P (X ≥ 3) = P (X = 3) + P (X = 4) + P (X = 5) = 5C3 (0.4)3 (0.6)2 + 5C 4 (0.4)4 (0.6) + 5C5 (0.4)5 (0.6)0 = 0.2304 + 0.0768 + 0.1024 = 0.31744 Clearly, P1 > P2. Therefore, first option i.e. ‘best of 3 games’ has higher probability of winning the match.

20. The man will be one step away from the starting point, if (i) either he is one step ahead or (ii) one step behind the starting point. The man will be one step ahead at the end of eleven steps, if he moves six steps forward and five steps backward. The probability of this event is 11 C 6 (0.4)6 (0.6)5 . The man will be one step behind at the end of eleven steps, if he moves six steps backward and five steps forward. The probability of this event is 11C 6 (0.6)6 (0.4)5 . ∴ Required probability = 11C 6 (0.4)6 (0.6)5 +

C 6 (0.6)6 (0.4)5 = 11C 6 (0.24)5

11

21. Let an event E of sum of outputs are perfect square (i.e., 4 or 9), so E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)} and an event F of sum of outputs are prime numbers (i.e., 2, 3, 5, 7, 11) so F = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)} and event T of sum of outputs are odd numbers (i.e., 3, 5, 7, 9, 11) T = {(1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (3, 6), (4, 5), (5, 4), (6, 3), (5, 6), (6, 5)} Now, required probability p = P (T / E ) P (T ∩ E ) = P (E ) where, P (T ∩ E ) = probability of occuring perfect square odd number before prime  4   14  4  =  +   +  36  36  36 4 4 2 36 = = = 14 22 11 1− 36

2

 14  4      +…∞  36  36

and P (E ) = probability of occuring perfect square before prime 2

 7   14  7   14  7  =   +     +     +…∞  36  36  36  36  36 7 7 36 = = 14 22 1− 36 2 4 ∴ P (T / E ) = 11 = = p 7 7 22 ⇒ 14 p = 8

22. It is given that the probability, a missile hits a target 3 successfully p = , so the probability to not hits the 4 1 target is q = . 4 And it is also given that to destroy the target completely, at least three successful hits are required. Now, according to the question, let the minimum number of missiles required to fired is n, so n C3 p3 qn − 3 + nC 4 p4qn − 4 + … + nC n pn ≥ 0.95 ⇒ n − 2 2 n n −1   1  3  1  3  1 1 −  nC 0   + nC1     + nC 2     ≥ 0.95  4  4  4  4  4   95 1 3n n (n − 1) 9 ⇒ ≥ + + 1− 100 4n 4n 2 4n n 2 4 2 + 6n + 9n − 9n ≥ ⇒ 20 2 ⇒ 10(9n 2 − 3n + 2 ) ≤ 4n Now, at n = 3, LHS = 720, RHS = 64 at n = 4, LHS = 1340, RHS = 256 at n = 5, LHS = 2120, RHS = 1024 at n = 6, LHS = 3080, RHS = 4096 Hence, n = 6 missiles should be fired.

23. Using Binomial distribution, P (X ≥ 2) = 1 − P (X = 0) − P (X = 1) n n −1   1  1  1 = 1 −   −  nC1 ⋅   ⋅     2  2  2   1 1  1 + n = 1 − n − nC1 ⋅ n = 1 −  n   2  2 2 Given, P (X ≥ 2) ≥ 0. 96 n+1 1 (n + 1) 24 ∴ ≤ ⇒ n =8 1− ≥ ⇒ 25 25 2n 2n

7 Matrices and Determinants Topic 1 Types of Matrices, Addition, Subtraction, Multiplication and Transpose of a Matrix Objective Questions I (Only one correct option) 1. If A is a symmetric matrix and B is a skew-symmetric 2 3  matrix such that A + B =  , then AB is equal to 5 −1

(2019 Main, 12 April I)

 −4 −2 (a)    −1 4   4 −2 (c)    1 −4

4 (b)   −1  −4 (d)  1

−2 −4

(a) 52 (c) 201

2 4

 0 2y 1 

(x, y ∈ R, x ≠ y) for which AT A = 3I3 is

  2x − y 1 

(2019 Main, 9 April II)

(b) 4 (d) 6

0 −1 cos α − sin α  3. Let A =  , (α ∈ R) such thatA32 =  .  1 0  sin α cos α  (2019 Main, 8 April I) Then, a value of α is (a)

π 32

(b) 0

(c)

π 64

(d)

π 16

1 0 0

4. Let P = 3 1 0 and Q = [qij ] be two 3 × 3 matrices such   9 3 1

(b) 135 (d) 15

0 2q r  5. Let A =  p q −r . If AAT = I3 , then| p|is    p − q r  (2019 Main, 11 Jan I) (a)

1 5

(b)

1 2

(c)

1 3

(d)

   c 2 b  AAT = 9 I, where, I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to (2015 Main)

(a) (2, − 1) (c) (2, 1)

1 6

(b) (−2, 1) (d) (−2, − 1)

 3 /2

8. If P = 

 −1 /2 PTQ 2005 P is

1 1 1 /2  T , A = 0 1 and Q = PAP , then 3 / 2  

 1 2005 (a)  1  0 1 0  (c)    2005 1

 A 2 = B,is

(2019 Main, 12 Jan I)

2

7. If A = 2 1 −2 is a matrix satisfying the equation

9. If A = α 1

q + q31 that Q − P 5 = I3 . Then, 21 is equal to q32 (a) 10 (c) 9

(b) 103 (d) 205

1 2

2. The total number of matrices A = 2x y −1 ,

(a) 2 (c) 3

0 0 1 0 and I be the identity matrix of order   16 4 1 3. If Q = [qij ] is a matrix, such that P50 − Q = I, then q31 + q32 equals q21 (2016 Adv.) 1

6. Let P =  4

1 0 and B =  1 5

(a) 1 (c) 4

(2005, 1M)

2005  1 (b)  2005 1   1 0   (d)    0 1

0 , then value of α for which 1 (2003, 1M)

(b) –1 (d) no real values

10. If A and B are square matrices of equal degree, then which one is correct among the following? (a) A + B = B + A (b) A + B = A − B (c) A − B = B − A (d) AB = BA

(1995, 2M)

136 Matrices and Determinants 15. If the point P (a , b, c), with reference to Eq. (i), lies on

Objective Questions II One or more than one correct option) 1 0 0 11. Let P1 = I = 0 1 0,   0 0 1

0 P3 = 1  0 0 P5 = 1  0 and

the plane 2x + y + z = 1, then the value of 7a + b + c is (a) 0

1 0 0 P2 = 0 0 1,   0 1 0

0 1 P4 = 0 0  1 0 0 0 P6 = 0 1  1 0 2 1 3 6 X = ∑ Pk 1 0 2 PkT   k=1 3 2 1 1 0 0 0,  0 1 0 1 0 0,  1 0

0 1,  0 1 0  0

(a) 6

(c)

6 7

(d) ∞

with b and c satisfying Eq. (i) then the value of 3 1 3 + b + c is a ω ω ω (a) − 2 (c) 3

(b) 2 (d) − 3

Passage II Let p be an odd prime number and T p be the following set of 2 × 2 matrices a b   Tp =  A =  ; a , b, c ∈ { 0, 1, 2, K , p − 1}  c a  (2010) 

18. The number of A in T p such that det (A) is not divisible by p, is (b) p3 − 5 p (d) p3 − p 2

(a) 2 p 2 (c) p3 − 3 p

12. Let X and Y be two arbitrary, 3 × 3, non-zero, skew-symmetric matrices and Z be an arbitrary, 3 × 3, non-zero, symmetric matrix. Then, which of the following matrices is/are skew-symmetric? (2015 Adv.)

(b) X 44 + Y 44 (d) X 23 + Y 23

19. The number of A in T p such that the trace of A is not divisible by p but det ( A ) is divisible by p is (a) ( p − 1) ( p 2 − p + 1) (c) ( p − 1)2

(b) p3 − ( p − 1)2 (d) ( p − 1) ( p 2 − 2)

20. The number of A in T p such that A is either symmetric or skew-symmetric or both and det (A) is divisible by p is

13. For 3 × 3 matrices M and N , which of the following (2013 Adv.)

(a) N T M N is symmetric or skew-symmetric, according as M is symmetric or skew-symmetric (b) MN − NM is symmetric for all symmetric matrices M and N (c) M N is symmetric for all symmetric matrices M and N (d) (adj M ) (adj N ) = adj (MN ) for all invertible matrices M and N

14. Let ω be a complex cube root of unity with ω ≠ 0 and

P = [ pij ] be an n × n matrix with pij = ωi+ j . Then, P 2 ≠ 0 when n is equal to (2013 Adv.) (b) 55 (d) 56

(b) 2 ( p − 1) (d) 2 p − 1

(a) ( p − 1)2 (c) ( p − 1)2 + 1 NOTE

The trace of a matrix is the sum of its diagonal entries.

Analytical and Descriptive Questions   , where a , b, c are real   c positive numbers, abc = 1 and AT A = I , then find the (2003, 2M) value of a3 + b3 + c3 . a

21. If matrix A =  b

b c a

c a b

Integer & Numerical Answer Type Questions 22. The trace of a square matrix is defined to be the sum of

its diagonal entries. If A is a 2 × 2 matrix such that the trace of A is 3 and the trace of A3 is − 18, then the value (2020 Adv.) of the determinant of A is ……… .

Passage Based Problems

−1 + 3 i , where i = −1, and r , s ∈ {1, 2, 3}. Let 2 (− z )r z 2s  P =  2s  and I be the identity matrix of order 2. zr   z Then, the total number of ordered pairs (r , s) for which (2016 Adv.) P 2 = − I is

23. Let z =

Passage I Let a , b and c be three real numbers satisfying

1 9 7 [a b c] 8 2 7 = [0 0 0]   7 3 7

(b) 7

17. Let ω be a solution of x3 − 1 = 0 with Im (ω ) > 0. If a = 2

(a) X is a symmetric matrix (b) The sum of diagonal entries of X is 18 (c) X − 30 I is an invertible matrix 1 1 (d) If X 1 = α 1, then α = 30   1 1

(a) 57 (c) 58

(d) 6

the roots of the quadratic equation ax2 + bx + c = 0, then n ∞  1 1 ∑  α + β is equal to n= 0

where, denotes the transpose of the matrix Pk. Then which of the following option is/are correct? (2019 Adv.)

statement(s) is/are not correct ?

(c) 7

16. Let b = 6, with a and c satisfying Eq. (i). If α and β are

PkT

(a)Y 3 Z 4 − Z 4Y 3 (c) X 4 Z 3 − Z 3 X 4

(b) 12

...(i) (2011)

Matrices and Determinants 137

Topic 2 Properties of Determinants Objective Questions I (Only one correct option) 1. If the minimum and the maximum values of the function f :

π π  , → R,  4 2 

If det( A ) ∈ [2, 16], then c lies in the interval

− sin 2 θ −1 − sin 2 θ defined by f (θ ) = − cos 2 θ −1 − cos 2 θ 12

1 1 are m and M −2

10

respectively, then the ordered pair (m, M ) is equal to (2020 Main, 5 Sep I)

(a) (0, 2 2 )

(b) (−4, 4)

(d) (−4, 0)

(c) (0, 4)

2. A value of θ ∈ (0, π / 3), for which

π 9

(b)

π 18

(c)

7π 24

(2019 Main, 12 April II)

(d)

7π 36

3. The sum of the real roots of the equation −6 −1 − 3x x − 3 = 0, is equal to − 3 2x x + 2 x 2

(b) − 4

(a) 0

4. If

(2019 Main, 10 April II)

(c) 6

(d) 1

sin θ cos θ x ∆1 = − sin θ − x 1 cos θ 1 x

sin 2θ cos 2θ x and ∆ 2 = − sin 2θ , x ≠ 0, −x 1 cos 2θ 1 x  π then for all θ ∈ 0,   2

(2019 Main, 10 April I)

1 n − 1 1 78 = ,then the 1  0 1 

5. If  . .  ...  0 1 0 1 0 1 0 1 n  inverse of   is 0 1 

 1 0 (a)   12 1

(2019 Main, 9 April I)

 1 −12 (d)   0 1 

 1 0  1 −13 (b)   (c) 13 1 0 1    

6. Let α and β be the roots of the equation x2 + x + 1 = 0. Then, for y ≠ 0 in R, α β y+1 α 1 is equal to y+β β 1 y+α (a) y( y2 − 1) (c) y3 − 1

(2019 Main, 9 April I)

(b) y ( y2 − 3) (d) y3

(b) (2 + 23 / 4 , 4) (d) [2, 3)

sin θ 1   3π 5π  1 , , sin θ ; then for all θ ∈   4 4   − sin θ 1   − 1 (2019 Main, 12 Jan II) det( A ) lies in the interval 

1

8. If A = − sin θ

5 (b)  , 4  2 

3 (c)  0,   2 

5 (d)  1,   2 

2a 2a a−b−c 9. If 2b 2b b−c−a 2c 2c c−a −b = (a + b + c) (x + a + b + c)2, x ≠ 0 and a + b + c ≠ 0, then (2019 Main, 11 Jan II) x is equal to (a) − (a + b + c) (b) − 2(a + b + c) (d) abc (c) 2(a + b + c)

10. Let a1 , a 2, a3 ..... , a10 be in GP with ai > 0 for i = 1, 2, ..... ,10 and S be the set of pairs (r , k), r , k ∈ N (the set of natural numbers) for which log e a1r a 2k

log e a 2ra3k

log e a 4r a5k log e a7r a 8k

log e a5r a 6k log e a 8r a 9k

log e a3r a 4k log e a 6r a7k = 0 k log e a 9r a10

Then, the number of elements in S, is (2019 Main, 10 Jan II) (a) 4 (b) 2 (c) 10 (d) infinitely many

(a) ∆1 + ∆ 2 = − 2(x3 + x − 1) (b) ∆1 − ∆ 2 = − 2x3 (c) ∆1 + ∆ 2 = − 2x3 (d) ∆1 − ∆ 2 = x(cos 2θ − cos 4θ)

1 1 1 2 1 3

(2019 Main, 8 April II)

(a) [3, 2 + 23 / 4 ] (c) [4, 6]

3 (a)  , 3  2 

sin 2 θ 1 + cos 2 θ 4 cos 6θ 2 cos θ = 0, is 1 + sin 2 θ 4 cos 6θ 2 2 cos θ sin θ 1 + 4 cos 6θ (a)

1 1 1  7. Let the numbers 2, b, c be in an AP and A = 2 b c .   2 2 4 b c 

b 1 2 2  11. Let A = b b + 1 b, where b > 0. Then, the minimum   b 2  1 det ( A ) is value of b (2019 Main, 10 Jan II) (a) − 3

(b) −2 3

(c) 2 3

(d)

3

12. Let d ∈ R, and

(sin θ ) − 2 4+ d   −2  , θ ∈ [θ , 2π ]. If  (sin θ ) + 2 A= 1 d    5 (2 sin θ ) − d (− sin θ ) + 2 + 2d  the minimum value of det(A) is 8, then a value of d is (2019 Main, 10 Jan I)

(a) −5

(b) −7

(c) 2( 2 + 1)

(d) 2( 2 + 2)

2x  x − 4 2x 2  13. If  2x x − 4 2x   = ( A + Bx)(x − A ) , 2x x − 4  2x ordered pair ( A , B) is equal to (a) (−4, − 5)

(b) (−4, 3)

(c) (−4, 5)

then

the

(2018 Main)

(d) (4, 5)

138 Matrices and Determinants 14. Let ω be a complex number such that 2ω + 1 = z, where 1 1 1 2 z = − 3. If 1 −ω − 1 ω 2 = 3 k, then k is equal to ω2 ω7 1 (2017 Main) (a) − z

(b) z

(c) − 1

(d) 1

15. If α, β ≠ 0 and f (n ) = α n + β n and 1 + f (1) 1 + f (2)

3

1 + f (1) 1 + f (2) 1 + f (3) 1 + f (2) 1 + f (3) 1 + f (4) = K (1 − α )2(1 − β )2 (α − β )2, then K is equal to (2014 Main)

1 (b) αβ

(a) αβ

(d) −1

(c) 1

16. Let P = [aij ] be a 3 × 3 matrix and let Q = [bij ], where

bij = 2i + j aij for 1 ≤ i , j ≤ 3. If the determinant of P is 2, (2012) then the determinant of the matrix Q is

(a) 210

(b) 211

(c) 212

(d) 213

2 17. If A = α and| A3| = 125, then the value of α is 2 α 



(a) ± 1

(b) ± 2

(c) ± 3

(d) ± 5 (2001, 1M)

cos x cos x  π π sin x cos x  = 0 in the interval − ≤ x ≤ is 4 4 cos x sin x 

(a) 0 (c) 1

(b) 2 (d) 3

matrix with real entries? 1 0 0  (a)  0 1 0     0 0 −1

(2017 Adv.)

1 0 0  (b)  0 − 1 0     0 0 −1

− 1 0 0  (c)  0 − 1 0     0 0 −1

 1 0 0 (d)  0 1 0    0 0 1

24. Which of the following values of α satisfy the equation (1 + α )2 (1 + 2α )2 (1 + 3 α )2

(2 + α )2 (2 + 2α )2 (2 + 3 α )2 = − 648 α ? (3 + α )2 (3 + 2α )2 (3 + 3 α )2 (a) −4

(c) −9

(b) 9

(2015 Adv.)

(d) 4

25. Let M and N be two 3 × 3 matrices such that MN = NM . Further, if M ≠ N 2 and M 2 = N 4, then

(2014 Adv.)

(a) determinant of (M 2 + MN 2 ) is 0 (b) there is a 3 × 3 non-zero matrixU such that (M 2 + MN 2 ) U is zero matrix (c) determinant of (M 2 + MN 2 ) ≥ 1 (d) for a 3 × 3 matrix U, if (M 2 + MN 2 ) U equals the zero matrix, then U is the zero matrix



x x+1 1  x (x − 1) (x + 1) x 2x ,  3x (x − 1) x (x − 1) (x − 2) (x + 1) x (x − 1)  

19. If f (x) = 

then f (100) is equal to (a) 0 (c) 100

23. Which of the following is(are) NOT the square of a 3 × 3

(2004, 1M)

18. The number of distinct real roots of  sin x  cos x  cos x

Objective Questions II (One or more than one correct option)

(1999, 2M)

(b) 1 (d) –100

a

26. The determinant  b

aα + b

aα + b bα + c  0 

b c bα + c

is equal to zero, then

(1986, 2M)

(a) a, b, c are in AP (b) a, b, c are in GP (c) a, b, c are in HP (d) (x − α ) is a factor of ax2 + 2bx + c

20. The parameter on which the value of the determinant

Integer & Numerical Answer Type Questions

  1 a a2 cos ( p d ) x cos cos ( px − p + d) x     sin ( p − d ) x sin px sin ( p + d ) x  does not depend upon, is (a) a

(b) p

(1997, 2M)

(c) d

xp + y x 21. The determinant yp + z y 0 xp + y (a) x, y, z are in AP (c) x, y, z are in HP

27. Let P be a matrix of order 3 × 3 such that all the entries in

(d) x

y = 0, if z yp + z (1997C, 2M)

(b) x, y, z are in GP (d) xy, yz , zx are in AP

22. Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of A consisting of all determinants with value 1. Let C be the subset of A consisting of all determinants with value –1. Then, (a) C is empty (b) B has as many elements as C (c) A = B ∪ C (d) B has twice as many elements as C

(1981, 2M)

P are from the set { − 1, 0, 1}. Then, the maximum possible value of the determinant of P is ......... .

Fill in the Blanks 28. For positive numbers x, y and z, the numerical value of the log x y log x z 1 determinant log y x log y z is…… . 1 log z x log z y 1 1

29. The value of the determinant  1  1 

a b c

(1993, 2M)

a − bc b2 − ca is … .  c2 − ab  2

(1988, 2M)

x 30. Given that x = − 9 is a root of 2 7 roots are... and... .

3 x 6

7 2  = 0, the other two x (1983, 2M)

Matrices and Determinants 139 1

4

1

2x

20  5   = 0 is… . 5x2 (1981, 2M)

31. The solution set of the equation 1 − 2  λ2 + 3λ 32. Let pλ + qλ + rλ + sλ + t =   λ+1  λ −3 be an identity in λ , where p,q,r,s and Then, the value of t is…. . 4

3

2

λ − 1 λ + 3 − 2λ λ − 4   λ + 4 3λ  t are constants. (1981, 2M)

True/False 1 a

1 a

bc

a2

33. The determinants 1 b ca and 1 b b 1

c

1

ab

c

2

are not

2

(1983, 1M)

Analytical and Descriptive Questions 34. If M is a 3 × 3 matrix, where M T M = I and det (M ) = 1, then prove that det (M − I ) = 0

(2004, 2M)

35. Let a , b, c be real numbers with a + b + c = 1 . Show 2

2

 cos ( A − P )   cos (B − P )  cos (C − P )

2

that the equation bx + ay cx + a  ax − by − c  cy + b  = 0 represents − ax + by − c  bx + ay cy + b − ax − by + c   cx + a (2001, 6M) a straight line.

cos θ

2π   sin θ +   3 2π   sin θ −   3

2π   cos θ +   3 2π   cos θ −   3

(n + 2)!  (n + 3)! , (n + 4)! 

 D  then show that  − 4 is divisible by n. 3  (n !) 

(1992, 4M)

 p b c

a b r p q r Then, find the value of . (1991, 4M) + + p−a q−b r−c

43. Let the three digit numbers A28, 3B9 and 62C, where A, B and C are integers between 0 and 9, be divisible by a fixed integer k. Show that the determinant  A 3 6 (1990, 4M)  8 9 C is divisible by k. 2 B 2   a −1

n 2n 2 3n3

44. Let ∆ a =   (a − 1)2 3  (a − 1)

 6 4n − 2   3n 2 − 3n 

n

sin 2θ 4π   sin 2θ +  =0  3 4π   sin 2 θ −   3  (2000, 3M)

(a) f (0) = 2, f (1) = 1 (b) f has a minimum value at x = 5 / 2, and 2ax − 1  2ax (c) for all x, f ′ (x) =  b b+ 1   2(ax + b) 2ax + 2b + 1

∑ ∆ a = c ∈ constant.

(1989, 5M)

a =1

conditions

45. Show that  xCr  yC z r C  r

x

x

x+1

x+ 2

y

y

y+1

y+ 2

Cr + 2   xCr C r + 2 =  y C r  z z Cr + 2   Cr

Cr + 1 Cr + 1 z Cr + 1

Cr + 1 Cr + 1 z +1 Cr + 1

C r + 2 C r + 2  z+2 C r + 2 (1985, 3M)

46. If α be a repeated root of a quadratic equation f (x) = 0 2ax + b + −1 2ax + b

1   

where a, b are some constants. Determine the constants a, b and the function f (x). (1998, 3M) bc ca ab 38. Find the value of the determinant p q r , where 1 1 1 a , b and c are respectively the pth , qth and rth terms of a harmonic progression. (1997C, 2M)

39. Let a > 0, d > 0. Find the value of the determinant 1 a (a + d ) 1 (a + d ) (a + 2d ) 1 (a + 2d ) (a + 3d )

(n + 1)! (n + 2)! (n + 3)!

 n! D =  (n + 1)!  (n + 2)!

Show that

37. Suppose, f (x) is a function satisfying the following

1 a 1 (a + d ) 1 (a + 2d )

(1994, 4M)

cos ( A − R)  cos (B − R)  = 0 cos (C − R) 

41. For a fixed positive integer n, if

36. Prove that for all values of θ sin θ

cos ( A − Q ) cos (B − Q ) cos (C − Q )

42. If a ≠ p, b ≠ q, c ≠ r and a q c  = 0

c

identically equal.

40. For all values of A , B, C and P , Q , R, show that

1 (a + d ) (a + 2d ) 1 (a + 2d ) (a + 3d ) 1 (a + 3d ) (a + 4d )

and A (x), B (x) and C (x) be polynomials of degree 3, 4 and 5 respectively, then show that B (x) C (x)   A (x)  A (α ) B (α ) C (α )   A′ (α ) B′ (α ) C′ (α )  is divisible by derivatives.

f (x), where prime denotes the (1984, 3M)

47. Without expanding a determinant at any stage, show that  x2 + x  2x2 + 3x − 1  2  x + 2x + 3

x+1 3x 2x − 1

x−2  3x − 3  = xA + B  2x − 1 

where A and B are determinants of order 3 not involving x. (1982, 5M)

48. Let a, b, c be positive and not all equal. Show that the  a b c value of the determinant  b c a is negative.  c a b (1981, 4M)

140 Matrices and Determinants

Topic 3 Adjoint and Inverse of a Matrix Objective Questions I (Only one correct option) − 1 − sin 2 θ  −1  = α I + βM , 2 cos 4 θ  1 + cos θ  sin 4 θ

1. Let M = 

where α = α (θ ) and β = β (θ ) are real numbers, and I is the 2 × 2 identity matrix. If α * is the minimum of the set {α (θ ): θ ∈ [0, 2π )} and β * is the minimum of the set { β(θ ) : θ ∈ [0, 2π) }, then the value of α * + β * is (2019 Adv.)

17 16 37 (c) − 16

31 16 29 (d) − 16

(a) −

(b) −

(2019 Main, 12 April I)

 et e− t sin t e− t cos t  t −t −t −t −t 3. If A = e −e cos t − e sin t − e sin t + e cos t  then A  et −2e− t cos t 2e− t sin t   (2019 Main, 9 Jan II)

(a) invertible only when t = π (b) invertible for every t ∈ R (c) not invertible for any t ∈ R π (d) invertible only when t = 2

4. Let A and B be two invertible matrices of order 3 × 3. If

det( ABAT ) = 8 and det( AB− 1 ) = 8, then det(BA − 1BT ) is equal to (2019 Main, 11 Jan II) (a) 1 (c)

(b)

1 16

1 4

cos θ − sin θ  , then the matrix cos θ   π when θ = , is equal to 12 (2019 Main, 9 Jan I)

 1  (a)  2 − 3  2  3  (c)  2 − 1  2

3 2  1  2  1  2  3 2 

  (b)      (d)   

− b and A adj A = AAT , then 5a + b is equal 2  (2016 Main)

(b) 5 (d) 13

8. If A is a 3 × 3 non-singular matrix such that AAT = AT A and B = A −1 AT , then BBT is equal to

(a) I + B (c) B −1

(2014 Main)

(b) I (d) (B −1 )T

1 α 3

9. If P = 1 3 3 is the adjoint of a 3 × 3 matrix A and   2 4 4 | A | = 4 , then α is equal to (a) 4 (c) 5

(2013 Main)

(b) 11 (d) 0

10. If P is a 3 × 3 matrix such that PT = 2P + I, where PT is the transpose of P and I is the 3 × 3 identity matrix, then  x  0 there exists a column matrix, X =  y ≠ 0 such that      z  0 (2012)  0 (a) PX =  0    0 (c) PX = 2X

(b) PX = X (d) PX = − X

11. Let ω ≠ 1 be a cube root of unity and S be the set of all

(d) 16

5. If A =  sin θ A −50

5a 3

7. If A = 

 51 63 (b)    84 72  72 − 63 (d)    − 84 51 

to

(b) −1 (d) 2

is

 72 − 84 (a)    − 63 51   51 84 (c)    63 72

(a) − 1 (c) 4

5 2α 1  2. If B = 0 2 1  is the inverse of a 3 × 3 matrix A, then   α 3 −1 the sum of all values of α for which det ( A ) + 1 = 0, is (a) 0 (c) 1

 2 −3  2  , then adj (3 A + 12 A ) is equal to  −4 1  (2017 Main)

6. If A = 

3 2 1 2 1 2 3 2

1 −  2 3 2  3 − 2  1  2 

 1 a b non-singular matrices of the form  ω 1 c , where  2  ω ω 1  each of a , b and c is either ω or ω 2. Then, the number of (2011) distinct matrices in the set S is (a) 2 (c) 4

(b) 6 (d) 8

12. Let M and N be two 3 × 3 non-singular skew-symmetric matrices such that MN = NM . If PT denotes the transpose of P, then (2011) M 2N 2(M T N )−1 (MN −1 )T is equal to (a) M 2 (c) −M 2

(b) −N 2 (d) MN

Matrices and Determinants 141 1 0 0 13. If A = 0 1 1, 6 A −1 = A 2 + cA + dI , then (c, d ) is   0 −2 4 (2005, 1M) (a) (− 6, 11) (c) (11, 6)

(b) (− 11, 6) (d) (6, 11)

identity matrix of order 3. If q23 = − then

(2016 Adv.)

(a) α = 0, k = 8 (c) det (P adj (Q )) = 29

(b) 4α − k + 8 = 0 (d) det (Q adj (P )) = 213

17. Let M be a 2 × 2 symmetric matrix with integer entries.

Objective Questions II (One or more than one correct option)

Then, M is invertible, if

let I denote the 3 × 3 identity matrix. If M −1 = adj(adj M ), then which of the following (2020 Adv.) statements is/are ALWAYS TRUE? (b) det M = 1 (d) (adj M )2 = I

0 1 a  − 1 1 − 1 15. Let M = 1 2 3  and adj M =  8 − 6 2      3 b 1  − 5 3 − 1

(2014 Adv.)

(a) the first column of M is the transpose of the second row of M (b) the second row of M is the transpose of the first column of M (c) M is a diagonal matrix with non-zero entries in the main digonal (d) the product of entries in the main diagonal of M is not the square of an integer

14. Let M be a 3 × 3 invertible matrix with real entries and

(a) M = I (c) M 2 = I

k k2 and det (Q ) = , 8 2

1 4 4 

18. If the adjoint of a 3 × 3 matrix P is 2 1 7, then the   1 1 3  possible value(s) of the determinant of P is/are

(a) − 2

where a and b are real numbers. Which of the following (2019 Adv.) options is/are correct?

(b) − 1

(c) 1

(d) 2

Integer & Numerical Answer Type Question

(a) det (adj M 2 ) = 81 α   1 (b) If M β  =  2, then α − β + γ = 3      γ   3

19. Let k be a positive real number and let  2k − 1 2 k 2 k    A= 2 k − 2k  and 1  − 2 k 2k − 1 

(c) (adj M )− 1 + adj M − 1 = − M (d) a + b = 3

 0 2k − 1 k  0 2 k B =  1 − 2k  − k − 2 k 0

3 −1 −2 16. Let P = 2 0 α , where α ∈ R. Suppose Q = [qij ] is a   3 −5 0  matrix such that PQ = kI , where k ∈ R, k ≠ 0 and I is the

   

If det (adj A ) + det (adj B) = 106, then [k] is equal to…… (2010)

Topic 4 Solving System of Equations Objective Questions I (Only one correct option) 1. If [x] denotes the greatest integer ≤ x , then the system of liner equations [sin θ ]x + [− cos θ ] y = 0, [cot θ ]x + y = 0 (2019 Main, 12 April II)

π 2π  (a) have infinitely many solutions if θ ∈  ,  2 3 7π  and has a unique solution if θ ∈  π , .  6 (b) has a unique solution if π 2π   7π  θ ∈  ,   ∪  π, 2 3  6 π 2π  (c) has a unique solution if θ ∈  ,  2 3 7π  and have infinitely many solutions if θ ∈  π ,   6 (d) have infinitely many solutions if 7π  π 2π θ ∈  ,  ∪  π ,  2 3  6

2. Let λ be a real number for which the system of linear equations x + y + z = 6, 4x + λy − λz = λ − 2 and 3x + 2 y − 4z = − 5 has infinitely many solutions. Then λ is a root of the quadratic equation (2019 Main, 10 April II) (b) λ2 + 3λ − 4 = 0 (a) λ2 − 3λ − 4 = 0 (c) λ2 − λ − 6 = 0

(d) λ2 + λ − 6 = 0

3. If the system of linear equations x+ y+ z =5 x + 2 y + 2z = 6 x + 3 y + λz = µ,(λ , µ ∈ R), has infinitely many solutions, then the value of λ + µ is (2019 Main, 10 April I)

(a) 7

(b) 12

(c) 10

(d) 9

142 Matrices and Determinants 4. If the system of equations 2x + 3 y − z = 0, x + ky − 2z = 0 and 2x − y + z = 0 has a non-trivial solution (x, y, z ), then x y z + + + k is equal to y z x (2019 Main, 9 April II) (a) −4

(b)

1 2

(c) −

1 4

(d)

3 4

5. If the system of linear equations

(b) 3x − 4 y − 1 = 0 (d) 4x − 3 y − 1 = 0

6. The greatest value of c ∈ R for which the system of linear equations x −cy − cz = 0, cx − y + cz = 0, cx + cy − z = 0 has a non-trivial solution, is

(2019 Main, 8 April I)

(a) −1

1 (b) 2

(c) 2

(d) 0

7. The set of all values of λ for which the system of linear equations x − 2 y − 2z = λx, − x − y = λz

x + 2 y + z = λyand

has a non-trivial solution

(2019 Main, 12 Jan II)

(a) (b) (c) (d)

contains exactly two elements. contains more than two elements. is a singleton. is an empty set. (2019 Main, 12 Jan I)

(1 + α )x + βy + z = 2 αx + (1 + β ) y + z = 3 ax + βy + 2z = 2 has a unique solution, is

(b) a + b + c = 0 (d) b + c − a = 0

10. The number of values of θ ∈ (0, π ) for which the system

(c) four

(d) one

(2019 Main, 10 Jan I)

(c) 21

has a non-zero solution ( x , y , z ), then (a) −10

(b) 10

(c) −30

xz

y2

is equal to (2018 Main)

(d) 30

15. The system of linear equations x + λy − z = 0; λx − y − z = 0; x + y − λz = 0 has a non-trivial solution for

(2016 Main)

(a) infinitely many values of λ (b) exactly one value of λ (c) exactly two values of λ (d) exactly three values of λ

equations 2x1 − 2x2 + x3 = λx1, 2x1 − 3x2 + 2x3 = λx2 and (2015 Main) − x1 + 2x2 = λx3 has a non-trivial solution

(a) is an empty set (b) is a singleton set (c) contains two elements 9d) contains more than two elements

(b) 1

(c) 2

(d) 3

 x  1 0 or 1 and for which the system A  y = 0 has exactly      z  0 two distinct solutions, is (2010)

(a) 0

(b) 29 − 1

(c) 168

(d) 2

19. Given, 2x − y + 2z = 2, x − 2 y + z = − 4, x + y + λz = 4, (b) 1

(c) 0

(d) –3

20. The number of values of k for which the system of (2019 Main, 10 Jan II)

x+ y+z=5 x + 2 y + 3z = 9 x + 3 y + αz = β has infinitely many solutions, then β − α equals (b) 18

x + ky + 3z = 0, 3x + ky − 2z = 0, 2x + 4 y − 3z = 0

(a) 3

11. If the system of equations

(a) 8

14. If the system of linear equations

then the value of λ such that the given system of (2004, 1M) equations has no solution, is

of linear equations

(b) three

(2019 Main, 9 Jan I)

has infinitely many solutions for a = 4 is inconsistent when a = 4 has a unique solution for|a| = 3 is inconsistent when|a| = 3

18. The number of 3 × 3 matrices A whose entries are either

where a , b, c are non-zero real numbers, has more than one solution, then (2019 Main, 11 Jan I)

(a) two

(a) (b) (c) (d)

(a) infinite

2x + 2 y + 3z = a 3x − y + 5z = b x − 3 y + 2z = c

x + 3 y + 7z = 0, − x + 4 y + 7z = 0, (sin 3θ )x + (cos 2θ ) y + 2z = 0 has a non-trivial solution, is

x + y + z = 2, 2x + 3 y + 2z = 5 2x + 3 y + ( a 2 − 1)z = a + 1

equation (k + 1) x + 8 y = 4 y ⇒ kx + (k + 3) y = 3k − 1 has no solution, is (2013 Main)

9. If the system of linear equations

(a) b − c − a = 0 (c) b − c + a = 0

(b) g + 2h + k = 0 (d) g + h + 2k = 0

17. The number of value of k, for which the system of

(b) (− 4, 2) (d) (−3, 1)

(a) (2, 4) (c) (1, − 3)

(a) 2 g + h + k = 0 (c) g + h + k = 0

16. The set of all values of λ for which the system of linear

8. An ordered pair (α , β) for which the system of linear equations

x − 4 y + 7z = g, 3 y − 5z = h, − 2x + 5 y − 9z = k is consistent, then (2019 Main, 9 Jan II)

13. The system of linear equations

x − 2 y + kz = 1 , 2x + y + z = 2 ,3x − y − kz = 3 has a solution (x, y, z ), z ≠ 0, then (x, y) lies on the straight line whose equation is (2019 Main, 8 April II) (a) 3x − 4 y − 4 = 0 (c) 4x − 3 y − 4 = 0

12. If the system of linear equations

(d) 5

equations (k + 1) x + 8 y = 4k and kx + (k + 3) y = 3k − 1 has infinitely many solutions, is/are (2002, 1M)

(a) 0 (c) 2

(b) 1 (d) ∞

21. If the system of equations x + ay = 0, az + y = 0 and ax + z = 0 has infinite solutions, then the value of a is (a) –1 (c) 0

(b) 1 (d) no real values

Matrices and Determinants 143 22. If the system of equations x − ky − z = 0, kx − y − z = 0, x + y − z = 0 has a non-zero solution, then possible values (2000, 2M) of k are (a) –1, 2

(b) 1, 2

(c) 0, 1

(d) –1, 1

Assertion and Reason

(a) x + 2 y + 3z = b1 , 4 y + 5z = b2 and x + 2 y + 6z = b3 (b) x + y + 3z = b1 , 5x + 2 y + 6z = b2 and − 2x − y − 3z = b3 (c) − x + 2 y − 5z = b1 , 2x − 4 y + 10z = b2 and x − 2 y + 5z = b3 (d) x + 2 y + 5z = b1 , 2x + 3z = b2 and x + 4 y − 5z = b3

Fill in the Blank

For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows. (a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false. (d) Statement I is false; Statement II is true.

x − 2 y + 3 z = −1 , x − 3 y + 4z = 1 and − x + y − 2z = k Statement I The system of equations has no solution for k ≠ 3 and 1 3 −1 Statement II The determinant −1 −2 k ≠ 0 , for

23. Consider the system of equations

1 k ≠ 0.

4

1

26. The system of equations λx + y + z = 0 , − x + λy + z = 0 and − x − y + λz = 0 will have a non-zero solution, if (1982, 2M) real values of λ are given by ...

Analytical and Descriptive Questions a 27. A = 1 1

0 c d

1 a b, B = 0  f b

1 d g

a 2 1 f  c , U =  g , V = 0  0  h  h   

If there is a vector matrix X, such that AX = U has infinitely many solutions, then prove that BX = V cannot have a unique solution. If a f d ≠ 0. Then, prove that BX = V has no solution. (2004, 4M)

28. Let λ and α be real. Find the set of all values of λ for which the system of linear equations

(2008, 3M)

λx + (sin α ) y + (cos α )z = 0, x + (cos α ) y + (sin α )z = 0

Objective Questions II (One or more than one correct) and

24. Let x ∈ R and let 1 1 1 2 x x P = 0 2 2, Q = 0 4 0 and R = PQP −1,     0 0 3 x x 6 the which of the following options is/are correct? (2019 Adv.)

(a) There exists a real, number x such that PQ = QP  1  1 (b) For x = 0, if R  a  = 6  a , then a + b = 5      b  b (c) For x = 1, there exists a unit vector α $i + β$j + γk$ for α   0 which R β  =  0      γ   0  2 x x (d) det R = det  0 4 0 + 8, for all x ∈ R    x x 5

 b1  25. Let S be the set of all column matrices  b2  such that b1,   b3  b2, b3 ∈R and the system of equations (in real variables)

− x + 2 y + 5z = b1 2x − 4 y + 3z = b2 x − 2 y + 2z = b3

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one  b1  solution for each  b2  ∈ S ?    b3 

− x + (sin α ) y − (cos α )z = 0

has a non-trivial solution. For λ = 1, find all values of α.

(1993, 5M)

29. Let α 1 , α 2, β1 , β 2 be the roots of ax + bx + c = 0 and 2

px2 + qx + r = 0, respectively. If the system of equations α 1 y + α 2z = 0 and β1 y + β 2z = 0 has a non-trivial solution, b2 ac then prove that 2 = . (1987, 3M) pr q

30. Consider the system of linear equations in x, y, z (sin 3θ ) x − y + z = 0, (cos 2θ ) x + 4 y + 3z = 0 and 2x + 7 y + 7z = 0 Find the values of θ for which this system has non-trivial solution. (1986, 5M)

31. Show that the system of equations, 3x − y + 4z = 3, x + 2 y − 3z = − 2 and 6x + 5 y + λz = − 3 has atleast one solution for any real number λ ≠ − 5. Find the set of solutions, if λ = − 5. (1983, 5M)

32. For what values of m, does the system of equations 3x + my = m and 2x − 5 y = 20 has a solution satisfying (1979, 3M ) the conditions x > 0, y > 0?

33. For what value of k, does the following system of equations possess a non-trivial solution over the set of rationals x + y − 2z = 0, 2x − 3 y + z = 0, and x − 5 y + 4z = k Find all the solutions. (1979, 3M )

34. Given, x = cy + bz , y = az + cx, z = bx + ay, where x, y, z are not all zero, prove that a 2 + b2 + c2 + 2ab = 1. (1978, 2M)

144 Matrices and Determinants Integer & Numerical Answer Type Questions

37. For a real number α , if the system  1 α α 2  x   1        α 1 α   y = −1 2 α α 1   z   1   

35. If the system of equations x − 2 y + 3z = 9 2x + y + z = b x − 7 y + az = 24, has infinitely many solutions, then a − b is equal to ............. . (2020 Main, 4 Sep I)

36. If the system of linear equations,

of linear equations, has infinitely many solutions, then (2017 Adv.) 1 + α + α2 = 0

 −1 

  0

   3 

38. Let M be a 3 × 3 matrix satisfying M 1 =  2 ,

x+ y+ z =6 x + 2 y + 3z = 10

1 1 M −1 =  1 , and M      0  −1

3 x + 2 y + λz = µ has more than two solutions, then µ − λ2 is equal to (2020 Main, 7 Jan II) ………… .

1  0  1 =  0  ,     1 12

Then, the sum of the diagonal entries of M is … (2011)

Answers Topic 3

Topic 1 1. (b) 5. (b)

2. (b) 6. (b)

3. (c) 7. (d)

4. (a) 8. (a)

1. (d)

2. (c)

3. (b)

4. (c)

5. (c)

6. (b)

7. (b)

8. (b)

9. (b)

9. (d)

10. (a)

11. (a,b,d)

12. (c, d)

10. (d)

11. (a)

12. (c)

13. (c, d)

14. (b, c, d)

15. (d)

16. (b)

13. (a)

14. (b,c,d)

15. (b,c,d)

16. (b,c)

17. (a)

18. (d)

19. (c)

20. (d)

17. (c, d)

18. (a,d)

19. (4)

21. (4)

22. (5)

23. (1)

2. (a)

3. (a)

2. 6. 10. 14. 18. 22. 26.

3. 7. 11. 15. 19. 23.

Topic 4

Topic 2

21. (b)

22. (b)

23. (a, c)

24. (b, c)

1. 5. 9. 13. 17. 21. 25.

25. (a, b)

26. (b,d)

27. (4)

28. (0)

28. − 2 ≤ λ ≤ 2 , α = nπ , nπ +

29. (0)

30. (2 and 7) 31. {–1,2} 32. (0) 1 5 1 2 5   37. a = , b = − and f ( x ) = x − x + 2   4 4 4 4

1. (d)

4. (c)

5. (b)

6. (d)

7. (c)

8. (a)

9. (b)

10. (d)

11. (c)

12. (a)

13. (c)

14. (a)

15. (c)

16. (d)

17. (c)

18. (c)

19. (a)

20. (b)

33. False 38. (0) 42. (2)

  4d 4 39.   2 3 2  a (a + d ) (a + 2d ) (a + 3d ) (a + 4d )

(a) (c) (a) (d) (d) (a) (a,d)

(c) (b) (a) (b) (a) (d) λ=0

(c) (c) (a) (d) (b) (a)

4. 8. 12. 16. 20. 24.

(b) (a) (a) (c) (b) (b,d)

π 4

π , n ∈Z 6 4 − 5k 13k − 9 15 31. x = ,y = , z = k 32. m < − or m > 30 7 7 2 33. (k = 0, the given system has infinitely many solutions) 30. θ = nπ , nπ + ( −1 )n

35. (5)

36. (13)

37. (1)

38. (9)

Hints & Solutions Topic 1 Types of Matrices, Addition, Subtraction and Transpose of a Matrix 1. Given matrix A is a symmetric and matrix B is a skew-symmetric. ∴

A = A and B = − B 2 3  Since, A + B =   5 − 1 T

T

(given) …(i)

T

2 3  ( A + B)T =   5 − 1 2 5  AT + BT =   3 − 1 T T Given, A = A and B = − B 2 5  ⇒ A−B=  3 − 1

… (ii)

On solving Eqs. (i) and (ii), we get 2 4  0 − 1 A=  and B = 1 0  − 4 1     2 4  0 − 1  4 − 2 So, AB =   =  4 − 1 1 0  − 1 − 4

2. Given matrix  0 2y 1  A = 2x y −1 , (x, y ∈ R, x ≠ y)   2x − y 1  for which AT A = 3I3 ⇒

 0 2x 2x   0 2 y 1  3 0 0 2 y y − y 2x y −1 = 0 3 0       1 −1 1  2x − y 1  0 0 3

⇒

8x2 0 0 3 0 0    2   0 6 y 0 = 0 3 0  0 0 3 0 0 3 

cos α ∴ A2 =  sin α

 cos2 α − sin 2 α = sin cosα + cosα sin α α 

cos 2 α = sin 2 α

4. Given matrix 1 P = 3  9 ⇒ P=X + Now, P5 = (I +

and

3 8 1 y=± 2

Q There are 2 different values of x and y each. So, 4 matrices are possible such that AT A = 3I3 .

0 0 0 0 0 1 0 = 3 0 0 +    3 1 9 3 0 I (let) X )5

1 0 0 0 1 0   0 0 1

= I + 5C1 (X ) + 5C 2(X 2) + 5C3 (X 3 ) + … [Q I n = I , I ⋅ A = A and (a + x)n = nC 0a n + n C1a n − 1x + ...+ nC nxn] 0 0 0 0 0 0 0 0 0 Here, X 2 = 3 0 0 3 0 0 = 0 0 0      9 3 0 9 3 0 9 0 0 0 0 0 0 0 0 0 0 0 and X 3 = X 2 ⋅ X = 0 0 0 3 0 0 = 0 0 0      9 0 0 9 3 0 0 0 0 0 0 0 ⇒ X 4 = X 5 = 0 0 0   0 0 0 0 0 0 0 0 0 So, P5 = I + 5 3 0 0 + 10 0 0 0     9 3 0 9 0 0

8x2 = 3 and 6 y2 = 3 x=±

− sin 2 α  cos 2 α 

cos(32 α ) − sin(32 α ) 0 −1 ⇒ A32 =  =  (given) sin(32 α ) cos(32 α )  1 0  So, cos(32 α ) = 0 and sin(32 α ) = 1 π π 32 α = ⇒ α = ⇒ 2 64

Here, two matrices are equal, therefore equating the corresponding elements, we get

⇒

− cosα sin α − sin α cosα   − sin 2 α + cos2 α 

Similarly, cos(nα ) − sin(nα ) An =  , n ∈ N sin(nα ) cos(nα ) 

On taking transpose both sides, we get

⇒

cos α − sin α   sin α cos α  − sin α  cos α − sin α  cos α  sin α cos α 

3. Given, matrix A = 

and ⇒

0 0  1  = 15 1 0   135 15 1 0 0  2 Q = I + P5 =  15 2 0 = [q ij ]   135 15 2

q21 = 15, q31 = 135 and q 32 = 15 q + q31 15 + 135 150 Hence, 21 = = = 10 q32 15 15

146 Matrices and Determinants 5. Given, AAT = I p p  1 0 0 0 2q r   0    ⇒ p q − r 2q q − q = 0 1 0       p − q r   r − r r  0 0 1 0 + 4q2 + r 2 0 + 2q2 − r 2 0 − 2q2 + r 2  1 0 0   ⇒ 0 + 2q2 − r 2 p2 + q2 + r 2 p2 − q2 − r 2  = 0 1 0   2 2 2 2 2 2 2 2 0 − 2q + r p −q −r p + q + r  0 0 1  We know that, if two matrices are equal, then corresponding elements are also equal, so

4q2 + r 2 = 1 = p2 + q2 + r 2, 2q2 − r 2 = 0 ⇒ r 2 = 2q2 and p2 − q 2 − r 2 = 0 Using Eqs. (ii) and (iii), we get

…(i) …(ii) …(iii)

…(iv) p2 = 3 q 2 Using Eqs. (ii) and (iv) in Eq. (i), we get 4q2 + 2q2 = 1 ⇒ 6q2 = 1 [using Eq. (iv)] ⇒ 2 p2 = 1 1 1 2 p = ⇒ | p| = 2 2  1 0 0 6. Here, P =  4 1 0   16 4 1  1 0 0  1 0 0  1 2  ∴ P = 4 1 0  4 1 0 =  4 + 4     16 4 1 16 4 1 16 + 32 1 0 0   4 ×2 1 0 =   16 (1 + 2) 4 × 2 1 1 0 0  1 0 0  and P3 =  4 × 2 1 0  4 1 0    16 (1 + 2) 4 × 2 1 16 4 1 1 0 0  4 ×3 1 0 =   16 (1 + 2 + 3) 4 × 3 1 From symmetry, 1 0  50  P = 4 × 50 1  16 (1 + 2 + 3 + ... + 50) 4 × 50 P50 − Q = I

1 2

2

2 a 1 2    2 b  2 −2 b  2 2  1 2 a  1 −2 2 1 2    2 b  2 −2 b  a + 4 + 2b  9 0 0 9 2a + 2 − 2b  + 4 + 2b 2a + 2 − 2b a 2 + 4 + b2  1

7. Given, A = 2 1 −2, AT = 2  a 1 T and AA = 2  a

 =  a

It is given that, AAT = 9I 9 0 a + 4 + 2b   1 0 0 ⇒  0 9 2a + 2 − 2b = 9 0 1 0     2 2 a + 4 + 2b 2a + 2 − 2b a + 4 + b  0 0 1 a + 4 + 2b  9 0 0 9 0  ⇒  0 9 2a + 2 − 2b = 0 9 0     2 2 a + 4 + 2b 2a + 2 − 2b a + 4 + b  0 0 9 On comparing, we get a + 4 + 2b = 0 ⇒ a + 2b = − 4

...(i)

2a + 2 − 2b = 0 ⇒ a − b = − 1 0 1 0  4 + 4 1 0

…(i)

a = − 2, b = − 1 This satisfies Eq. (iii) (a , b) ≡ (−2,−1)

Hence,

 3 /2 −1 /2   3 /2 PT P =   3 /2  −1 /2  1 /2 1 0 PT P =   0 1

⇒

1 /2   3 /2

PT P = I PT = P −1

⇒

Q = PAPT

Since, ∴ P Q T

0 0  1

…(iii)

On solving Eqs. (i) and (ii), we get

⇒

...(ii)

…(ii)

and a 2 + 4 + b2 = 9

8. Now,

[given]   − q12 − q13  1 0 0 1 − q11  ∴  − q23  = 0 1 0 200 − q21 1 − q22     50 (51) − q31 200 − q32 1 − q33  0 0 1 16 ×   2 16 × 50 × 51 ⇒ 200 − q21 = 0, − q31 = 0, 2 200 − q32 = 0 ∴ q21 = 200, q32 = 200, q31 = 20400 Q

q31 + q32 20400 + 200 20600 = = 103 = 200 q21 200

Thus,

2005

P = PT [(PAPT )(PAPT ) K 2005 times ] P

= (PT P ) A (PT P ) A (PT P ) K (PT P ) A (PT P ) 1444444442444444443 2005 times

∴

= IA 2005 = A 2005 1 1 A=  0 1 1 1 1 1 1 2 A2 =   =  0 1 0 1 0 1 1 2 1 1 1 3 A3 =   =  0 1 0 1 0 1 …………………… ……………………

Matrices and Determinants 147

∴

1 2005 A 2005 =  1  0 1 2005 PTQ 2005 P =  1  0

9. Given, A = α 1 

⇒

α A2 =  1

1 0 , B= 1 5

0 1

⇒

Which is not possible at the same time.

10. If A and B are square matrices of equal degree, then A+ B=B+ A 0 0 0 0 1, P3 = 1   1 0 0 0 1 0 0 0, P6 = 0   1 0 1

1 0 0 0  0 1 0 1 1 0  0 0

2 1 3 and X = Σ Pk 1 0 2 PKT   K =1 3 2 1 Q P1T = P1 , P2T = P2, P3T = P3 , P4T = P5 , P5T = P4 and 2 1 3 P6T = P6 and Let Q = 1 0 2 and Q QT = Q   3 2 1 Now, X = (P1QP1T ) + (P2QP2T ) + (P3QP3T ) + (P4QP4T ) + (P5QP5T ) + (P6QP6T ) T T T T T So, X = (P1QP1 ) + (P2QP2 ) + (P3QP3T )T + (P4QP4T )T + (P5QP5T )T + (P6QP6T )T 6

= P1QP1T + P2QP2T + P3QP3T + P4QP4T + P5QP5T + P6QP6T [Q ( ABC )T = CT BT AT and ( AT )T = A and QT = Q] T ⇒X = X ⇒ X is a symmetric matrix. The sum of diagonal entries of X = Tr (X ) 6

= Σ Tr (Pi Q PiT ) i =1

[QTr ( ABC ) = Tr (BCA )]

6

= Σ Tr (QI ) i =1 6

[QPi’s are orthogonal matrices]

= Σ Tr ( Q ) = 6 Tr (Q ) = 6 × 3 = 18 i =1

[Q PKT R = R]

6 2 2 2 6 6 6 6 = Σ PK 3 = Σ PK 3 = 2 2 2 3      K =1   K =1 6 2 2 2 6 6 30 1 1     XR = 30 ⇒ XR = 30R ⇒ X 1 = 30 1       30 1 1

⇒ (X − 30 I ) R = 0 ⇒|X − 30 I| = 0 So, (X − 30I ) is not invertible and value of α = 30. Hence, options (a), (b) and (d) are correct.

12. Given,

11. Given matrices,

i =1

K =1

K =1

0  1

∴ No real values of α exists.

6

K =1

= Σ (PK Q R)

⇒ α 2 = 1 and α + 1 = 5

= Σ Tr (Q PiT Pi )

6

6

0 1

1 1 0 0 P1 = I = 0 1 0, P2 = 0    0 0 0 1 0 0 1 0 P4 = 0 0 1, P5 = 1    0 1 0 0

6

XR = Σ (PK Q PKT )R = Σ (PK QPKT R)

0  α 2 = 1 α + 1

0  α 1 1

Also, given, A 2 = B α 2 0 1 = ⇒ α + 1 1 5

1 Now, Let R = 1, then   1

X T = − X , Y T = −Y , Z T = Z

P = Y 3 Z 4 − Z 4Y 3 PT = (Y 3 Z 4 )T − (Z 4 Y 3 )T = (Z T )4 (Y T )3 − (Y T )3 (Z T )4 = − Z 4Y 3 + Y 3 Z 4 = P ∴ P is symmetric matrix. (b) Let P = X 44 + Y 44 Then, PT = (X T )44 + (Y T )44 = X 44 + Y 44 = P ∴ P is symmetric matrix. (c) Let P = X 4Z 3 − Z 3 X 4 Then, PT = (X 4Z 3 )T − (Z 3 X 4 )T = (Z T )3 (X T )4 − (X T )4 (Z T )3 = Z 3 X 4 − X 4Z 3 = − P ∴ P is skew-symmetric matrix. (d) Let P = X 23 + Y 23 Then, PT = (X T )23 + (Y T )23 = − X 23 − Y 23 = − P ∴ P is skew-symmetric matrix. (a) Let Then,

13. (a) (N T MN )T = N T M T (N T )T = N T M T N , is symmetric if M is symmetric and skew-symmetric, if M is skew-symmetric. (b) (MN − NM )T = (MN )T − (NM )T = NM − MN = − (MN − NM ) ∴ Skew-symmetric, when M and N are symmetric. (c) (MN )T = N T M T = NM ≠ MN ∴ Not correct. (d) (adj MN ) = (adj N ) ⋅ (adj M ) ∴ Not correct. 14. Here,

P = [ pij ]n × n with pij = wi + j

∴ When n = 1 ⇒ ∴ When

P = [ pij ]1 × 1 = [ω 2] P 2 = [ω 4 ] ≠ 0 n =2

148 Matrices and Determinants  p11 p12  ω 2 ω3  ω 2 1  = P = [ pij ]2× 2 =   4 = 3  p21 p22 ω ω   1 ω  ω 2 1  ω 2 1  P2 =     1 ω  1 ω ω 4 + 1 ω 2 + ω  ≠0 ⇒ P2 =  2 2 ω + ω 1 + ω  When n = 3 P = [ pij ]3 × 3

ω 2 ω3 ω 4  ω 2 1 ω      = ω3 ω 4 ω5  =  1 ω ω 2 ω 4 ω5 ω 6   ω ω 2 1     

ω 2 1 ω  ω 2 1 ω  0 0 0    2 P =  1 ω ω 2  1 ω ω 2 = 0 0 0 = 0   2 2 ω ω 1  0 0 0 1   ω ω  ∴

P 2 = 0, when n is a multiple of 3. P 2 ≠ 0, when n is not a multiple of 3.

⇒

n = 57 is not possible.

∴

n = 55, 58, 56 is possible.

15. As (a , b, c) lies on 2x + y + z = 1 ⇒ 2a + b + c = 1 ⇒

2a + 6a − 7a = 1

⇒

a = 1, b = 6, c = − 7

∴ 7a + b + c = 7 + 6 − 7 = 6

a c

20. Given, A = 

If A is skew-symmetric matrix, then a = 0, b = − c ∴ | A|= − b2 Thus, P divides| A|, only when b = 0. ...(i) Again, if A is symmetric matrix, then b = c and | A|= a 2 − b2 Thus, p divides| A|, if either p divides (a − b) or p divides (a + b). p divides (a − b), only when a = b, a = b ∈ {0, 1, 2 ,... , ( p − 1)}

i.e. i.e. p choices

ax2 + bx + c = 0 ⇒ x2 + 6x − 7 = 0

⇒

(x + 7) (x − 1) = 0

∴

x = 1, − 7 ∞

⇒ p choices, including a = b = 0 included in Eq. (i). ∴ Total number of choices are ( p + p − 1) = 2 p − 1 a

21. Given, A = b c

Now, AT A = I a b ⇒ b c c a

1 6  6  1 1 ⇒∑  −  =1 + +   + L+ ∞ = 6  1 7  7 7 n= 0 1− 7 1 = =7 1 /7

17. If a = 2, b = 12, c = − 14

c  a a  b b  c

 a 2 + b2 + c2 ab + bc + ca ab + bc + ca 

⇒

1 = 0 0

0 1 0

3 1 3 + + ω a ωb ω c 3 3 1 3 = + 1 + 3ω 2 = 3ω + 1 + 3ω 2 + + ⇒ ω 2 ω12 ω −14 ω 2 = 1 + 3 (ω + ω 2) = 1 − 3 = − 2

18. The number of matrices for which p does not divide Tr ( A ) = ( p − 1) p2 of these ( p − 1)2 are such that p divides| A|. The number of matrices for which p divides Tr ( A ) and p does not divides| A|are ( p − 1)2.

| A|= a 2 − bc, for (a 2 − bc) to be divisible by p. There are exactly ( p − 1) ordered pairs (b, c) for any value of a. ∴ Required number is ( p − 1)2.

c  1 0 0 a  = 0 1 0 b  0 0 1 ab + bc + ca ab + bc + ca  a 2 + b2 + c2 ab + bc + ca  ab + bc + ca a 2 + b2 + c2  b c a

0 0 1 and 3

ab + bc + ca = 0

…(ii)

3

= (a + b + c)(a 2 + b2 + c2 − ab − bc − ca ) ⇒ a3 + b3 + c3 = (a + b + c) (1 − 0) + 3 [from Eqs. (i) and (ii)] ∴ a3 + b3 + c3 = (a + b + c) + 3 Now,

…(iii)

(a + b + c)2 = a 2 + b2 + c2 + 2(ab + bc + ca ) =1

…(iv)

From Eq. (iii), a3 + b3 + c3 = 1 + 3 ⇒ a3 + b3 + c3 = 4 22. Let a square matrix ‘A’ of order 2 × 2 , such that tr ( A ) = 3, is y  x A=  z 3 − x So,

Required number = ( p − 1) p2 − ( p − 1)2 + ( p − 1)2 = p3 − p2

19. Trace of A = 2a, will be divisible by p, iff a = 0.

…(i)

We know, a + b + c − 3abc 3

∴

∴

c a , abc = 1 and AT A = I b 

b c a

⇒ a 2 + b2 + c2 = 1 2

n

...(ii)

p divides (a + b).

16. If b = 6 ⇒ a = 1 and c = − 7 ∴

b , a, b, c ∈ {0, 1, 2 ,... , p − 1} a 

∴ Q

y  x y  x A2 =     z 3 − x z 3 − x  x2 + yz xy + 3 y − xy  =  xz z xz yz + (3 − x)2 + − 3   x x2 + yz y  3y A3 =   z 3 − x 2 yz + 9 + x − 6x    + 3z tr ( A3 ) = x3 + xyz + 3 yz + 3 yz + 3 yz − xyz + 27 − 9x + 3x2 − x3 − 18x + 6x2

Matrices and Determinants 149 = 9 yz + 27 − 27x + 9x2 = − 18 yz + 3 − 3x + x2 = − 2 3x − x2 − yz = 5 x y = 3x − x2 − yz = 5. | A| = z 3−x

⇒ ⇒ Q

23. Here, Q

(given)

 π f (θ ) minimum = f   = − 4 = m  2  π and f (θ ) maximum = f   = 0 = M  4

∴

∴The ordered pair (m, M ) is (−4, 0).

−1 + i 3 z= =ω 2 (−ω )r ω 2s  P =  2s  ωr   ω (− ω )r ω 2s  (− ω )r ω 2s  P 2 =  2s   ω r   ω 2s ωr   ω  ω 2r + ω 4s ω r + 2s [(− 1)r + 1] =  r + 2s  ω 4s + ω 2r [(− 1)r + 1]  ω

Given, P = − I ∴ ω 2r + ω 4s = − 1 and ω r + 2s [(− 1)r + 1] = 0 Since, r ∈{1, 2, 3} and (− 1)r + 1 = 0 ⇒ r = {1, 3} Also, ω 2r + ω 4s = − 1 If r = 1, then ω 2 + ω 4s = − 1 which is only possible, when s = 1. As, ω2 + ω4 = − 1 ∴ r = 1, s = 1 Again, if r = 3, then ω 6 + ω 4s = − 1 [never possible] ⇒ ω 4 s = −2 ∴ r ≠3 ⇒ (r , s) = (1, 1) is the only solution. Hence, the total number of ordered pairs is 1. 2

Topic 2 Properties of Determinants 1. Given function

 − sin 2 θ −1 − sin 2 θ 1  f (θ ) =  − cos 2 θ −1 − cos 2 θ 1    10 −2   12 − sin 2 θ −1 − sin 2 θ 1  = 2− cos 2 θ −1 − cos 2 θ 1    5 −1   6 On applying R1 → R1 + R3 and R2 → R2 + R3 , we get

sin 2 θ 1 + cos 2 θ 4 cos 6 θ 2 2. Let ∆ = cos θ 1 + sin 2 θ 4 cos 6 θ =0 2 2 1 + 4 cos 6 θ cos θ sin θ Applying C1 → C1 + C 2, we get 4 cos 6 θ sin 2 θ 2 ∆ = 2 1 + sin θ 4 cos 6 θ =0 1 + 4 cos 6 θ 1 sin 2 θ 2

Applying R1 → R1 − 2R3 and R2 → R2 − 2R3 , we get − sin 2 θ

− 2 − 4 cos 6 θ ∆ = 0 1 − sin 2 θ − 2 − 4 cos 6 θ = 0 1 + 4 cos 6 θ 1 sin 2 θ 0

On expanding w.r.t. C1, we get ⇒ sin 2 θ (2 + 4 cos 6 θ ) + (2 + 4 cos 6 θ ) (1 − sin 2 θ ) = 0 1 2π ⇒ 2 + 4 cos 6 θ = 0 ⇒ cos 6 θ = − = cos 2 3 2π π   π  ⇒ 6θ = ⇒θ = Q θ ∈ 0, 3   9 3   −6 −1 x 3. Given equation 2 − 3x x − 3 = 0 − 3 2x x + 2 On expansion of determinant along R1, we get x [(− 3x) (x + 2) − 2x(x − 3)] + 6 [2(x + 2) + 3(x − 3)] − 1 [2(2x) − (− 3x) (− 3)] = 0 ⇒ x [− 3x2 − 6x − 2x2 + 6x] + 6[2x + 4 + 3x − 9] − 1 [4x − 9x] = 0 ⇒ x(− 5x2) + 6(5x − 5) − 1(− 5x) = 0 ⇒ −5x3 + 30x − 30 + 5x = 0 ⇒ 5x3 − 35x + 30 = 0 ⇒ x3 − 7x + 6 = 0. Since all roots are real ∴ Sum of roots = −

coefficient of x2 coefficient of x3

=0

4. Given determinants are sin θ cos θ x ∆1 = − sin θ − x 1 cos θ 1 x

6 − sin θ 4 − sin θ 0  f (θ ) = 26 − cos 2 θ 4 − cos 2 θ 0    6 5 −1  

= − x3 + sin θ cos θ − sin θ cos θ + x cos 2 θ − x + x sin 2 θ

= 2(−1)[(6 − sin 2 θ )(4 − cos 2 θ ) − (4 − sin 2 θ )(6 − cos 2 θ )]

= − x3

2

2

= − 2 [24 − 6 cos 2 θ − 4 sin 2 θ + sin 2 θ cos 2 θ −

x

24 + 4 cos θ + 6 sin θ − sin θ cos θ ] 2

2

= − 2 [−2 cos θ + 2 sin θ ] = 4 cos 2 θ π π  π  As θ ∈ , ⇒2θ ∈ ,π  4 2   2  2

2

2

2

and ∆ 2 = − sin 2θ cos 2θ

sin 2θ cos 2θ −x 1

= − x3 (similarly as ∆1)

1 x

,x≠0

So, according to options, we get ∆1 + ∆ 2 = − 2x3

150 Matrices and Determinants = y [ y2 − (α + β )2 + 2αβ + 2αβ − 1 + (α + β ) − αβ ]

5. Given 1 1 1 2 1 3 ... 1 n − 1 1 78 = 0 1 0 1 0 1 0 1  0 1       1 1 1 2 1 2 + 1 , 0 1 0 1 = 0 1     

Q

1 2 + 1 0 1  

1 3 1 3 + 2 + 1 0 1 = 0 , 1    

: : : : : : 1 1 1 2 1 3 1 n − 1 ∴     ...  1  0 1 0 1 0 1 0 1 (n − 1) + (n − 2)+ ...+3 + 2 + 1 =  1 0  n (n − 1)  1 78   = 2  0 1    1

 1 = 0

Since, both matrices are equal, so equating corresponding element, we get n (n − 1) = 78 ⇒ n (n − 1) = 156 = 13 × 12 = 13(13 − 1) 2 ⇒

n = 13 1 13 1 −13 A= = A −1 =    0 1  0 1 

So,

 d − b  a b −1 [Qif|A|= 1and A =   , then A =  − c a      c d

6. Given, quadratic equation is x2 + x + 1 = 0 having roots α , β.

Then, α + β = −1 and αβ = 1 Now, given determinant ∆=

y+1

α

β

α β

y+β 1

1 y+α

On applying R1 → R1 + R2 + R3 , we get y+1+α +β y+1+α +β y+1+α +β ∆= α y+β 1 β y = α β

1

y

y

y+β 1

1 y+α

y+α [Qα + β = −1]

On applying C 2 → C 2 − C1 and C3 → C3 − C1, we get 0 0 y 1 −α ∆ = α y + β −α β

1 −β

y + α −β

= y[( y + (β − α )) ( y − (β − α )) − (1 − α ) (1 − β )] [expanding along R1] = y [ y2 − (β − α )2 − (1 − α − β + αβ )] = y [ y2 − β 2 − α 2 + 2αβ − 1 + (α + β ) − αβ ]

= y[ y2 − 1 + 3 − 1 − 1] = y3 [Qα + β = −1 and αβ = 1] 1 1 1  7. Given, matrix A = 2 b c , so   2 2 4 b c  1 1 1 det( A ) = 2 b c   2 2 4 b c  On applying, C 2 → C 2 − C1 and C3 → C3 − C1, 0 0  1 b − 2 c − 2  we get det( A ) = 2 b − 2 c − 2=  2  b − 4 c2 − 4  2 2 4 b − 4 c − 4 b −2 c−2 = (b − 2)(b + 2) (c − 2)(c + 2) 1   1  = (b − 2)(c − 2) + + 2 2 b c  [taking common (b − 2) from C1 and (c − 2) from C 2] = (b − 2)(c − 2)(c − b) Since, 2, b and c are in AP, if assume common difference of AP is d, then b = 2 + d and c = 2 + 2d [given] So, | A| = d (2d )d = 2d3 ∈ [2, 16] ⇒ d3 ∈ [1, 8] ⇒ d ∈ [1, 2] ∴ 2 + 2d ∈ [2 + 2, 2 + 4] = [4, 6] ⇒ c ∈ [4, 6] 1  sin θ  1  8. Given matrix A = − sin θ 1 sin θ    − sin θ 1   −1 1 sin θ 1 1 sin θ ⇒ det( A ) =| A|= − sin θ −1 − sin θ 1 = 1(1 + sin 2 θ ) − sin θ (− sin θ + sin θ ) + 1(sin 2 θ + 1) …(i) ⇒| A| = 2 (1 + sin 2 θ )  3π 5π  As we know that, for θ ∈  ,   4 4 1  1  1 2 sin θ ∈  − ,  ⇒ sin θ ∈ 0,    2 2  2 1   3  ⇒ 1 + sin 2 θ ∈ 0 + 1, + 1 ⇒ 1 + sin 2 θ ∈ 1,   2   2 3 ⇒ 2(1 + sin 2θ ) ∈ [2, 3) ⇒| A| ∈ [2, 3) ⊂  , 2 2a 2a a−b−c 9. Let ∆ = 2b 2b b−c−a 2c 2c c−a −b Applying R1 → R1 + R2 + R3 , we get a+ b+ c a+ b+ c a+ b+ c 2b b−c−a 2b ∆= 2c

2c

c−a −b

 3 

Matrices and Determinants 151 1 1 1 2b = (a + b + c) 2b b − c − a 2c 2c c−a −b (taking common (a + b + c) from R1) Applying C 2 → C 2 − C1 and C3 → C3 − C1 , we get 1 0 0 ∆ = (a + b + c) 2b − (a + b + c) 0 2c

− (a + b + c)

0

Now, expanding along R1, we get ∆ = (a + b + c) 1. {(a + b + c)2 − 0 } = (a + b + c)3 = (a + b + c)(x + a + b + c)2 (given) ⇒ (x + a + b + c)2 = (a + b + c)2 ⇒ x + a + b + c = ± (a + b + c) ⇒ x = − 2(a + b + c) [Q x ≠ 0]

10. Given,

log e a1ra 2k

log e a 2ra3k

log e a3r a 4k

log e a 4r a5k log e a7ra 8k

log e a5r a 6k log e a 8r a 9k

log e a 6r a7k = 0 k log e a 9r a10

On applying elementary operations C 2 → C 2 − C1 and C3 → C3 − C1, we get log e a1ra 2k log e a 4r a5k log e a7ra 8k

log e a 2ra3k log e a5r a 6k log e a 8r a 9k

− − −

log e a1ra 2k log e a 4r a5k log e a7ra 8k

log e a3r a 4k − log e a1ra 2k log e a 6r a7k − log e a 4r a5k = 0 k − log e a7ra 8k log e a 9r a10 log e a1ra 2k ⇒ log e a 4r a5k log e a7ra 8k

 a ra k   arak  log e  2r 3k  log e  3r k4   a1 a 2   a1 a 2  r k a a   arak  log e  5r 6k  log e  6r 7k  = 0  a 4a5   a 4a5  k   a 8r a 9k   a 9r a10 log e  r k  log e  r k   a7 a 8   a7 a 8 

  m  Q log e m − log e n = log e  n     [Q a1 , a 2, a3 ....... , a10 are in GP, therefore put a1 = a , a 2 = aR, a3 = aR2, ... , a10 = aR9 ] log e a r + kRk ⇒ log e a r + kR3 r + 4k log e a r + kR6r + 7k

 a r + kRr + 2k  log e    a r + k Rk   a r + k R4 r + 5 k  log e  r + k 3 r + 4k  R a   a r + kR7r + 8k  log e  r + k 6r + 7k  R  a  a r + kR2r + 3 k  log e    a r + k Rk   a r + kR5 r + 6k  log e  r + k 3 r + 4k  = 0 R  a  a r + k R8 r + 9 k  log e  r + k 6r + 7k  R  a

log e (a r + kRk ) log e Rr + k log e R2r + 2k ⇒ log e a r + kR3 r + 4k log e Rr + k log e R2r + 2k = 0 log e a r + kR6r + 7k log e Rr + k log e R2r + 2k log e (a r + kRk ) log e Rr + k 2 log e Rr + k ⇒ log e (a r + kR3 r + 4k ) log e Rr + k 2 log e Rr + k = 0 log e (a r + kR6r + 7k ) log e Rr + k 2 log e Rr + k [Q log mn = n log m and here log e R2r + 2k = log e R2( r + k) = 2 log e Rr + k ] Q Column C 2 and C3 are proportional, So, value of determinant will be zero for any value of (r , k), r , k ∈ N . ∴ Set ‘S’ has infinitely many elements. b 1 2 11. Given matrix, A = b b2 + 1 b, b > 0   b 2  1 2 b 1 So, det ( A ) =| A| = b b2 + 1 b b 1 2 = 2 [2(b2 + 1) − b2] − b(2b − b) +1(b2 − b2 − 1) 2 2 2 = 2[2b + 2 − b ] − b − 1 = 2b2 + 4 − b2 − 1 = b2 + 3 det( A ) b2 + 3 3 = = b+ b b b Now, by AM ≥ GM, we get 3 1/ 2 b+ b ≥  b × 3   b 2 3 ⇒ b+ ≥2 3 b det ( A ) So, minimum value of =2 3 b

⇒

{Q b > 0 }

12. Given, (sin θ ) − 2  4+ d  −2  A =  1 (sin θ ) + 2 d    5 (2 sin θ ) − d (− sin θ ) + 2 + 2d  4+ d −2 (sin θ ) − 2 ∴ | A| = 1 (sin θ ) + 2 d 5 (2 sin θ ) − d (− sin θ ) + 2 + 2d (sin θ ) − 2 −2 4+ d d = 1 (sin θ ) + 2 1

0 (R3 → R3 − 2R2 + R1 ) = 1 [(4 + d )d − (sin θ + 2) (sin θ − 2)] (expanding along R3 ) = (d 2 + 4d − sin 2 θ + 4) = (d 2 + 4d + 4) − sin 2 θ = (d + 2)2 − sin 2 θ Note that| A|will be minimum if sin 2 θ is maximum i.e. if sin 2 θ takes value 1. Q

0

| A|min = 8,

152 Matrices and Determinants therefore

(d + 2)2 − 1 = 8

⇒ ⇒ ⇒

15.

(d + 2)2 = 9 d+2=±3 d = 1, − 5

PLAN Use the property that, two determinants can be multiplied column-to-row or row-to-column, to write the given determinant as the product of two determinants and then expand.

Given, f (n ) = α n + β n, f (3) = α 3 + β3 , 3 1+ Let ∆ = 1 + f (1) 1 + 1 + f (2) 1 +

13. Given, 2x  x − 4 2x  2x x − 4 2x  = ( A + Bx)(x − A )2   2x x − 4  2x ⇒ Apply C1 → C1 + C 2 + C3 2x   5x − 4 2x  5x − 4 x − 4 2x  = ( A + Bx)(x − A )2    5x − 4 2x x − 4

3 ⇒

∆= 1+α +β 1 + α 2 + β2

2ω + 1 = z 2ω + 1 = − 3 − 1 + 3i ⇒ ω= 2 Since, ω is cube root of unity. − 1 − 3i and ω3 n = 1 ω2 = ∴ 2 1 1 1 2 Now, 1 − ω − 1 ω 2 = 3k

14. Given, ⇒

1

⇒

ω 3

⇒ ⇒ ⇒ ∴

0

0

ω ω 2 = 3k 1 ω2 ω 1

3(ω 2 − ω 4 ) = 3k (ω 2 − ω ) = k  − 1 − 3i   − 1 + 3i  k=  −  = − 3i = − z 2 2    

1 ⋅ 1 + α 2 ⋅ α + β2 ⋅ β 1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2 1 ⋅ 1 + α ⋅ α 2 + β ⋅ β2

1

α β 1 α 2 β2

1

1

α 1 α2

1

1 1 1 β = 1 α β 1 α 2 β2 β2

2

On expanding, we get ∆ = (1 − α )2(1 − β )2(α − β )2 ∆ = K (1 − α )2(1 − β )2(α − β )2

Hence, K (1 − α )2(1 − β )2(α − β )2 = (1 − α )2(1 − β )2(α − β )2 ∴ K =1

16.

PLAN It is a simple question on scalar multiplication, i.e. k a1 k a2 k a3 a1 a2 a3 b1 b 2 b 3 = k b1 b 2 b 3 c1 c 2 c 3 c1 c 2 c 3 Description of Situation Construction of matrix, i.e.

ω7

ω2

1

=1

[Q z = − 3]

[Q 1 + ω + ω 2 = 0 and ω7 = (ω3 )2 ⋅ ω = ω] On applying R1 → R1 + R2 + R3 , we get 3 1 + ω + ω2 1 + ω + ω2 ω ω2 = 3k 1 1

1

But given,

1 1 1 1 ω ω 2 = 3k 1 ω2 ω

1 ⋅1 + 1 ⋅α + 1 ⋅β 1 ⋅1 + α ⋅α + β ⋅β

1 ⋅ 1 + α 2 ⋅ α 2 + β2 ⋅ β2

(5x − 4)(x + 4)2 = ( A + Bx)(x − A )2 Equating, we get, A = − 4 and B = 5

ω2

1 + α + β 1 + α 2 + β2 1 + α 2 + β 2 1 + α 3 + β3 1 + α 3 + β3 1 + α 4 + β 4

x − 4

2x

Apply R2 → R2 − R1 and R3 → R3 − R1 2x 0  1 2 ∴ (5x − 4) 0 4 0  − x −  = ( A + Bx)(x − A )  −x − 4 0 0 Expanding along C1, we get

1

f (3) 1 + f (4)

1 ⋅1 + 1 ⋅1 + 1 ⋅1 = 1 ⋅1 + 1 ⋅α + 1 ⋅β 1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2

Taking common (5x − 4) from C1, we get 2x  1 2x 2 x − (5x − 4) 1 4 2x   = ( A + Bx)(x − A )  1

f (1) = α + β, f (2) = α 2 + β 2, f (4) = α 4 + β 4 f (1) 1 + f (2) f (2) 1 + f (3)

Here,

 a11 a12 a13  if a = [aij ]3 × 3 =  a21 a22 a23   a31 a32 a33 

 a11 a12 a13  P = [aij ]3 × 3 =  a 21 a 22 a 23   a31 a32 a33   b11 b12 b13  Q = [bij ]3 × 3 =  b21 b22 b23   b31 b32 b33 

where, bij = 2i + j aij 4 a11 8 a12 16 a13 ∴ Q = 8 a 21 16 a 22 32 a 23 16 a31 32 a32 64 a33 a11 a12 a13 = 4 × 8 × 16 2 a 21 2 a 22 2 a 23 4 a31 4 a32 4 a33 a11 a12 = 29 × 2 × 4 a 21 a 22 a31 a32 = 212 ⋅ P = 212 ⋅ 2 = 213

a13 a 23 a33

Matrices and Determinants 153 17. We know,

xp + y

| A n| = | A|n

| A | = 125 ⇒ | A| = 125 α 2 = 5 ⇒ α2 − 4 = 5 ⇒ α = ± 3 2 α  3

Since, ⇒

3

 sin x cos x cos x  18. Given,  cos x sin x cos x  = 0  cos x cos x sin x  Applying C1 → C1 + C 2 + C3  sin x + 2 cos x cos x =  sin x + 2 cos x sin x  sin x + 2 cos x cos x 1 = (2 cos x + sin x) 1 1

0

cos x sin x − cos x 0

− (xp + yp + yp + z ) xp + y ⇒

cos x  cos x  sin x 

cos x sin x cos x

cos x  cos x  = 0 sin x 

=0

yp + z

− (xp + 2 yp + z ) (xz − y2) = 0 2

∴ Either xp + 2 yp + z = 0 or y2 = xz 2

⇒ x, y, z are in GP.

22. Since, A is the determinant of order 3 with entries 0 or 1 only. Also, B is the subset of A consisting of all determinants with value 1.

cos x  0 = 0 sin x − cos x 

(2 cos x + sin x) (sin x − cos x)2 = 0 2 cos x + sin x = 0 or sin x − cos x = 0 2 cos x = − sin x or sin x = cos x π π ⇒ cot x = − 1 / 2 gives no solution in − ≤ x ≤ 4 4 and sin x = cos x ⇒ tan x = 1 ⇒ x = π /4

[since, if we interchange any two rows or columns, then among themself sign changes] Given, C is the subset having determinant with value −1. ∴ B has as many elements as C. 23. For a matrix to be square of other matrix its determinant should be positive. (a) and (c) → Correct, (b) and (d) → Incorrect

24. Given determinant could be expressed as product of two

19. Given,

Applying C3 → C3 − (C1 + C 2) x 1  x (x − 1) 2x =  3x (x − 1) x (x − 1)(x − 2)

determinants. (1 + α )2 (1 + 2α )2 (1 + 3α )2 i.e. (2 + α )2 (2 + 2α )2 (2 + 3α )2 = − 648 α (3 + α )2 (3 + 2α )2 (3 + 3α )2

x+1  (x + 1) x  (x + 1) x (x − 1) 

1 x  f (x) = 2x x (x − 1)  3x (x − 1) x (x − 1) (x − 2)

20. Let

y z

2

⇒ ⇒ ⇒

∴

y =0 z yp + z

Applying C1 → C1 − ( p C 2 + C3 ) x 0 y ⇒ 0

Applying R2 → R2 − R1 , R3 → R3 − R1 1 ⇒ (2 cos x + sin x) 0 0

x y xp + y

21. Given, yp + z

0 0 = 0 0

1 + 2α + α2 ⇒ 4 + 4α + α2

  a2 a 1 p px cos ( p d ) x cos cos ( ∆ = − + d) x    sin ( p − d ) x sin px sin ( p + d) x 

⇒

1 + a2 a a2 ⇒ ∆ = cos ( p − d ) x + cos ( p + d ) x cos px cos ( p + d )x sin( p − d ) x + sin( p + d ) x sin px sin ( p + d )x

1 1 1 1 1 1 ⇒ α 3 4 2 1 ⋅ 2 4 6 = − 648 α 9 3 1 1 4 9

1 + a2 a a2 ⇒ ∆ = 2 cos px cos dx cos px cos ( p + d ) x 2 sin px cos dx sin px sin ( p + d ) x

a cos px sin px

 a2 cos ( p + d ) x   sin ( p + d ) x 

= − 648 α α α2 1 1 1 4 2 α α 2 ⋅ 2 4 6 = − 648 α 9 3α α2 1 4 9 1

Applying C1 → C1 + C3

1 + a 2 − 2a cos dx ⇒ ∆ = 0  0 

1 + 6α + 9 α2 4 + 12 α + 9 α 2

9 + 6 α + α 2 9 + 12 α + 4 α 2 9 + 18 α + 9 α 2

f (x) = 0 ⇒ f (100) = 0

Applying C1 → C1 − 2 cos dx C 2

1 + 4α + 4α2 4 + 8α + 4α2

⇒ ⇒ ∴

25.

−8 α 3 = − 648 α ⇒ α 3 − 81α = 0 α (α − 81) = 0 α = 0, ± 9 2

PLAN (i) If A and B are two non-zero matrices and AB = BA, then ( A − B)( A + B) = A 2 − B 2. (ii) The determinant of the product of the matrices is equal to product of their individual determinants, i.e. | A B | = | A || B |.

Given, M 2 = N 4

⇒ ∆ = (1 + a 2 − 2a cos dx) [sin ( p + d ) x cos px − sin px cos ( p + d ) x] ⇒ ∆ = (1 + a 2 − 2a cos dx) sin dx

⇒

which is independent of p.

⇒

⇒

M2 − N 4 = 0

(M − N ) (M + N 2) = 0

[as MN = NM ]

2

M ≠N

Also, M + N2 =0

2

154 Matrices and Determinants ⇒

det (M + N 2) = 0

Also, det (M 2 + MN 2) = (det M) (det M + N 2)

1 a a 2 − bc 1 a a2 1 a bc 2 2 1 b b − ca = 1 b b − 1 b ca 1 c c2 − ab 1 c c2 1 c ab

29.

= (det M) (0) = 0 As,

det (M 2 + MN 2) = 0

Thus, there exists a non-zero matrix Usuch that (M 2 + MN 2) U = 0 b aα + b  a 26. Given, c bα + c  = 0  b 0   aα + b bα + c

Now,

1 a 1 b 1

c

a a 2 abc 1 b b2 abc ca = abc c c2 abc ab bc

Applying R1 → aR1 , R2 → bR2, R3 → cR3 1 a a2 a a2 1 1 2 = ⋅ abc b b 1 = 1 b b2 abc c c2 1 1 c c2

Applying C3 → C3 − (α C1 + C 2)  a  b 0 c 0    b = 0 2  aα + b bα + c − (aα + 2bα + c) ⇒ ⇒

− (aα 2 + 2bα + c) (ac − b2) = 0 aα 2 + 2bα + c = 0 or b2 = ac

∴

1 a a 2 − bc 1 b b2 − ca = 0 1 c c2 − ab

⇒ x − α is a factor of ax2 + 2bx + c or a , b, c are in GP. a1 27. Let Det (P ) = a 2

b1 b2

c1 c2

a3

b3

c3

7 6 x

= a1 (b2c3 − b3 c2) − a 2 (b1c3 − b3 c1 ) + a3 (b1c2 − b2c1 ) Now, maximum value of Det (P ) = 6 If a1 = 1, a 2 = − 1, a3 = 1, b2c3 = b1c3 = b1c2 = 1 and b3 c2 = b3 c1 = b2c1 = − 1 But it is not possible as (b2c3 ) (b3 c1 ) (b1c2) = − 1 and (b1c3 ) (b3 c2) (b2c1 ) = 1 i.e., b1b2b3 c1c2c3 = 1 and − 1 Similar contradiction occurs when a1 = 1, a 2 = 1, a3 = 1, b2c1 = b3 c1 = b1c2 = 1 and b3 c2 = b1c3 = b1c2 = − 1 Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5 1 1 1 Now, −1 1 1 =4 1

−1 1

Hence, maximum value is 4. log x y log x z 1 28. Let log y z ∆ = log y x 1 log z x log z y 1 =

log x log y log x log z

log y log x 1 log y log z

x 3 7 2 x 2 =0

30. Given,

1 log z log x log z log y 1

On dividing and multiplying R1 , R2, R3 by log x, log y, log z, respectively. log x log y log z 1 log x log y log z = 0 = log x log y log z log x log y log z

Applying R1 → R1 + R2 + R3 x+9 x+9 x+9 1 1 1 2 2 = 0 ⇒ (x + 9) 2 x 2 = 0 ⇒ x 7 6 x 7 6 x Applying C 2 → C 2 − C1 1 0 0

and C3 → C3 − C1

0 = 0 ⇒ (x + 9) (x − 2) (x − 7) = 0 ⇒ (x + 9) 2 x − 2 7 −1 x − 7 ⇒

x = − 9, 2, 7 are the roots.

∴ Other two roots are 2 and 7. 1

4

31. Given, 1 −2

20 =0

5

1 2x 5x

2

⇒ 1 (− 10 x2 − 10x) − 4 (5x2 − 5) + 20 (2x + 2) = 0 ⇒

−30x2 + 30x + 60 = 0

⇒

(x − 2) (x + 1) = 0

⇒

x = 2, − 1

Hence, the solution set is { −1, 2}. λ 2 + 3λ

32. Given,

λ+1 λ −3

λ −1

λ+3

−2 λ λ − 4 λ + 4 3λ = pλ 4 + qλ 3 + rλ 2 + sλ + t

Thus, the value of t is obtained by putting λ = 0. 0 −1 3 1 0 −4 = t ⇒ −3 4 0 ⇒ t =0 [Q determinants of odd order skew-symmetric matrix is zero]

Matrices and Determinants 155 1 [(x2 + y2 + 1) (acx + a 2x2 + abxy)] = 0 ax 1 ⇒ [ax(x2 + y2 + 1) (c + ax + by)] = 0 ax ⇒ (x2 + y2 + 1)(ax + by + c) = 0 ⇒ ax + by + c = 0 which represents a straight line. sin 2θ sin θ cos θ 4π  2π  2π     sin θ +   sin 2θ +  cos θ +      3 3 3 36. Let ∆ =

a a 2 abc  1 a bc  1 2  33. Let ∆ =   1 b ca  = abc b b abc 2 c c abc  1 c ab 

⇒

Applying R1 → aR1 , R2 → bR2, R3 → cR3 a a2 1 1 a a2 1 = ⋅ abc b b2 1 = 1 b b2 abc c c2 1 1 c c2 1 a 1 b

∴

1

c

1 a

a2

ca = 1 b ab 1 c

b2 c2

bc

2π  2π    sin θ −   cos θ −   3 3

4π   sin 2θ −   3

Applying R2 → R2 + R3

Hence, statement is false.

34. Since, M M = I and|M | = 1 T

∴

|M − I | = |IM − M T M |

[Q IM = M ]

⇒ |M − I | = |(I − M T )M |= |(I − M )T ||M |= | I − M | = (− 1)3 |M − I |[Q I − M is a 3 × 3 matrix] = − |M − I | ⇒

2|M − I| = 0

⇒

|M − I | = 0

bx + ay cx + a  ax − by − c 35. Given,  bx + ay −ax + by − c cy + b cy + b − ax − by +  cx + a

 = 0 c

a x − aby − ac cx + a bx + ay − ax + by − c cy + b abx + a 2y =0 cy + b − ax − by + c acx + a 2 2

⇒

1 a

⇒

 (a 2 + b2 + c2) x cx + a  ay + bx 1 2 b + cy  = 0 (a + b2 + c2) y by − c − ax  a 2 2 2 c ax − by − + b cy  a +b +c 

⇒

ay + bx cx + a  1x b + cy  = 0  y by − c − ax a 1 b + cy c − ax − by  [Q a 2 + b2 + c2 = 1]

Applying C 2 → C 2 − bC1 and C3 → C3 − cC1 ay a  1x b ⇒ = 0  y − c − ax a 1 cy − ax − by  x2 axy ax 1 y − c − ax b =0 ax 1 cy − ax − by

Applying R1 → R1 + yR2 + R3 ⇒

  0 0 1 x + y +1 y b − c − ax    = 0 ax 1 cy − ax − by  

⇒

1 [(x2 + y2 + 1) {(− c − ax)(− ax − by) − b(cy)}] = 0 ax

⇒

1 [(x2 + y2 + 1) (acx + bcy + a 2x2 + abxy − bcy)] = 0 ax

2

cos θ 2π   cos θ +   3 2π   + cos θ −   3 2π   cos θ −   3

sin 2θ 4π   sin 2θ +   3 4π   + sin 2θ −   3 4π   sin 2θ −   3

2π  2π    Now, sin θ +  + sin θ −    3 3 2π 2π  2π 2π    +θ− −θ +   θ + θ + 3 3  cos  3 3  = 2 sin  2 2         π 2π  = 2 sin θ cos  π −   3 3 π = − 2 sin θ cos = − sin θ 3 2 π 2π     and cos θ +  + cos θ −     3 3 = 2 sin θ cos

Applying C1 → C1 + bC 2 + cC3

⇒

sin θ 2π   sin θ +   3 = 2π   + sin θ −   3 2π   sin θ −   3

2

2π 2π  2π 2π    −θ + +θ−  θ +  θ + 3 3 3 3  = 2 cos   cos  2 2          2π  = 2 cos θ cos    3  1 = 2 cos θ  −  = − cos θ  2 4π  4π    and sin 2θ +   + sin 2θ −   3 3 4π 4π    + 2θ −  2θ +   2θ + 3 3 = 2 sin   cos  2       4π  = 2 sin 2θ cos = 2 sin 2θ cos  π +  3 π = − 2 sin 2θ cos = − sin 2θ 3

4π 4π  − 2θ +  3 3  2   π  3

156 Matrices and Determinants     sin 2 θ cos θ sin θ ∴ ∆ =  − sin θ − cos θ − sin 2θ = 0  sin θ − 2π  cos θ − 2π  sin 2θ − 4π       3 3 3  

39. Given, a > 0, d > 0 and let 1 a 1 ∆= (a + d ) 1 (a + 2d )

[since, R1 and R2 are proportional] 2ax − 1 b+1

2ax

37. Given, f ′ (x) =

b

2ax + b + 1 −1

2 (ax + b) 2ax + 2b + 1

2ax + b

 2ax 2ax − 1 2ax + b + 1  f ′ (x) =  b b+1 −1  1 0  0 2ax − 1   2ax −1  = b+1   b 1

⇒

[C 2 → C 2 − C1]

a   (a + d )(a + 2d ) (a + 2d )  (a + 2d )(a + 3d ) (a + 3d ) (a + d )   (a + 3d )(a + 4d ) (a + 4d ) (a + 2d )  1 ⇒ ∆= ∆′ a (a + d )2(a + 2d )3 (a + 3d )2(a + 4d )

f ′ (x) = 2ax + b

On integrating, we get f (x) = ax + bx + c, 2

where c is an arbitrary constant. Since, f has maximum at x = 5 / 2. ⇒

f ′ (5 / 2) = 0 ⇒ 5a + b = 0

Also,

…(i)

f (0) = 2

⇒

c = 2 and f (1) = 1

⇒

a + b + c=1

…(ii)

On solving Eqs. (i) and (ii) for a , b, we get 1 5 a = ,b = − 4 4 1 2 5 Thus, f (x) = x − x + 2 4 4

1 1 1 , , are in an AP. a b c 1  = A + ( p − 1) D  a  1 = A + ( q − 1) D  b  1 = A + (r − 1) D  c 

⇒

bc ca Let ∆ = p 1

q 1

1 a r = abc p 1 1

1 b q 1

ab

A + ( p − 1) D = abc p

…(i) ∴

1 c r 1

40. Let [from Eq. (i)] ⇒ A + (r − 1) D r

1

1

Applying R1 → R1 − ( A − D ) R3 − DR2 0 0 0 = abc p q r = 0 ⇒ 1 1 1

ab r =0

1

1

∆′ = (2d 2)(d )(a + 2d − a ) = 4d 4 4d 4 ∆= 2 a (a + d ) (a + 2d )3 (a + 3d )2(a + 4d ) cos ( A − P ) ∆ = cos (B − P ) cos (C − P ) cos A cos P ∆ = cos B cos P cos C cos P

cos ( A − Q ) cos ( A − R) cos (B − Q ) cos (B − R) cos (C − Q ) cos (C − R) + sin A sin P cos ( A − Q ) + sin B sin P cos (B − Q ) + sin C sin P cos (C − Q ) cos ( A − R) cos (B − R) cos (C − R)

⇒

bc ca p q 1

Applying R2 → R2 − R1 , R3 → R3 − R2 (a + d ) (a + 2d ) (a + 2d ) a ⇒ d d ∆′ = (a + 2d ) (2d ) d d (a + 3d ) (2d )

Expanding along R3 , we get a + 2d a  ∆′ = 2d 2 d  d

A + (q − 1) D q

1

a   (a + d )(a + 2d ) (a + 2d ) where, ∆′ =  (a + 2d )(a + 3d ) (a + 3d ) (a + d )   (a + 3d )(a + 4d ) (a + 4d ) (a + 2d ) 

Applying R3 → R3 − R2 (a + d )(a + 2d ) (a + 2d ) a ∆′ = (a + 2d )2d d d 2d 2 0 0

38. Since, a , b, c are pth , qth and rth terms of HP. ⇒

1 common from R1 , a (a + d ) (a + 2d ) 1 from R2, (a + d )(a + 2d )(a + 3d ) 1 from R3 (a + 2d ) (a + 3d )(a + 4d ) 1 ⇒ ∆= a (a + d )2(a + 2d )3 (a + 3d )2(a + 4d ) Taking

Applying R3 → R3 − R1 − 2R2, we get

2ax =  b

1 1 (a + d ) (a + 2d ) a (a + d ) 1 1 (a + d ) (a + 2d ) (a + 2d ) (a + 3d ) 1 1 (a + 2d ) (a + 3d ) (a + 3d ) (a + 4d )

 cos A cos P cos ( A − Q ) cos ( A − R)  ∆ =  cos B cos P cos (B − Q ) cos (B − R)   cos C cos P cos (C − Q ) cos (C − R)   sin A sin P cos ( A − Q ) cos ( A − R)  +  sin B sin P cos (B − Q ) cos (B − R)   sin C sin P cos (C − Q ) cos (C − R) 

Matrices and Determinants 157 ⇒

 cos A cos ( A − Q ) cos ( A − R)  ∆ = cos P  cos B cos (B − Q ) cos (B − R)   cos C cos (C − Q ) cos (C − R)   sin A + sin P  sin B  sin C

cos ( A − Q ) cos (B − Q ) cos (C − Q )

cos ( A − R)  cos (B − R)  cos (C − R) 

Applying C 2 → C 2 − C1 cos Q , C3 → C3 − C1 cos R in first determinant and C 2 → C 2 − C1 sin Q and in second determinant ⇒

cos A sin A sin Q sin A sin R ∆ = cos P cos B sin B sin Q sin B sin R cos C sin C sin Q sin C sin R  sin A cos A cos Q cos A cos R  + sin P sin B cos B cos Q cos B cos R   sin C cos C cos Q cos C cos R   cos A sin A sin A  ∆ = cos P sin Q sin R cos B sin B sin B   cos C sin C sin C   sin A cos A cos A  + sin P cos Q cos R sin B cos B cos B   sin C cos C cos C  

n!

 (n + 2)!

(n + 1)! (n + 2)! (n + 3)!

(n + 2)!  (n + 3)!  (n + 4)! 

Taking n !, (n + 1)! and (n + 2)! common from R1 , R2 and R3 , respectively. ∴

a

1 D = n ! (n + 1)! (n + 2) ! 1 1

(n + 1) (n + 2) (n + 3)

(n + 1) (n + 2)  (n + 2) (n + 3)  (n + 3) (n + 4) 

Applying R2 → R2 − R1 and R3 → R3 − R2, we get 1 D = n !(n + 1)!(n + 2)! 0 0

(n + 1) 1 1

(n + 1) (n + 2)  2n + 4  2n + 6 

Expanding along C1 , we get D = (n !)(n + 1)!(n + 2)![(2n + 6) − (2n + 4)] D = (n !)(n + 1)! (n + 2)! [2] On dividing both side by (n !)3 D (n !)(n !)(n + 1)(n !)(n + 1)(n + 2)2 = ⇒ 3 (n !) (n !)3 D = 2(n + 1)(n + 1)(n + 2) ⇒ (n !)3 D = 2(n3 + 4n 2 + 5n + 2) ⇒ (n !)3

Applying R1 → R2 − R1 and R3 → R3 − R1 , we get p b c 0 ∆= a− p q−b a−p 0 r−c =c

a− p q−b a−p

D − 4 = 2n (n 2 + 4n + 5) (n !)3

 D  which shows that  − 4 is divisible by n. 3 ( n !)  

0

+ (r − c)

p

b

a− p q−b

= − c (a − p) (q − b) + (r − c) [ p(q − b) − b(a − p)] = − c (a − p) (q − b) + p(r − c) (q − b) − b(r − c)(a − p) Since, ∆ = 0 ⇒ − c (a − p) (q − b) + p(r − c) (q − b) − b(r − c) (a − p) = 0 c p b + + =0 ⇒ r−c p−a q−b ⇒

[on dividing both sides by (a − p)(q − b)(r − c)] p b c + +1+ + 1 =2 p−a q−b r−c p q r + + =2 p−a q−b r−c

43. We know, A 28 = A × 100 + 2 × 10 + 8 3B9 = 3 × 100 + B × 10 + 9 and

62 C = 6 × 100 + 2 × 10 + C

Since, A 28, 3B 9 and 62 C are divisible by k, therefore there exist positive integers m1 , m2 and m3 such that, 100 × A + 10 × 2 + 8 = m1k , 100 × 3 + 10 × B + 9 = m2k and 100 × 6 + 10 × 2 + C = m3 k … (i) A 3 6 ∴

∆= 8 2

9 C B 2

Applying R2 → 100R1 + 10R3 + R2 3 A ⇒ ∆ = 100 A + 2 × 10 + 8 100 × 3 + 10 × B + 9 2 B 6 100 × 6 + 10 × 2 + C 2 3 6 A = A28 3B9 62C 2 2 B A

= 2n (n 2 + 4n + 5) + 4 ⇒

b r

⇒

∆ =0 + 0 =0

41. Given, D =  (n + 1)!

p b c

42. Let ∆ = a q c

3

6

[from Eq. (i)] A

3

6

= m1k m2k m3 k = k m1 m2 m3 2 B 2 2 B 2 ∴

∆ = mk

Hence, determinant is divisible by k.

158 Matrices and Determinants  a −1

n

 (a − 1)

3n

 6 4n − 2   3n 2 − 3n 

2 2 44. Given, ∆ a =   (a − 1) 2n 3 3

46. Since, α is repeated root of f (x) = 0. ∴ f (x) = a (x − α )2, a ∈ constant (≠ 0) B(x) C (x)   A (x) Let φ (x) =  A (α ) B(α ) C (α )   A ′ (α ) B ′ (α ) C ′ (α ) 

 n  n  ∑ (a − 1)  6  an=1  n 2 2   = a − n n − ∆ 1 2 4 2 ( ) ∑ a ∑  a=1 a=1  n  2 3  ∑ (a − 1)3 3n 3n − 3n   a=1 

∴

To show φ (x) is divisible by (x − α )2, it is sufficient to show that φ (α ) and φ ′ (α ) = 0. B(α ) C (α )   A (α ) ∴ φ (α ) =  A (α ) B(α ) C (α )   A ′ (α ) B ′ (α ) C ′ (α ) 

n (n − 1) 6 n 2 n (n − 1)(2n − 1) 2n 2 4n − 2 = 6 n 2(n − 1)2 3n3 3n 2 − 3n 4

=0

[Q R1 and R2 are identical]  A ′ (x) B ′ (x) C ′ (x)  Again, φ ′ (x) =  A (α ) B(α ) C (α )   A ′ (α ) B ′ (α ) C ′ (α )   A ′ (α ) B ′ (α ) C ′ (α )  φ ′ (α ) =  A (α ) B(α ) C (α )   A ′ (α ) B ′ (α ) C ′ (α ) 

1 1 6 n 2(n − 1) (2n − 1) = 2n 4n − 2 2 3 n (n − 1) 3n 2 3n 2 − 3n 2 1 1 6 n3 (n − 1) = 2n − 1 6n 12n − 6 12 n − 1 6n 6n − 6

=0

Thus, α is a repeated root of φ (x) = 0. Hence, φ (x) is divisible by f (x).  x2 + x x+1 x−2 47. Let ∆ =  2x2 + 3x − 1 3x 3x − 3  2 2x − 1 2x − 1  x + 2x + 3 Applying R2 → R2 − (R1 + R3 ), we get

Applying C3 → C3 − 6 C1 1

1

 x2 + x x+ 1 x−2  0 4 0  ∆ = −   2  x + 2x + 3 2x − 1 2x − 1 

0

n3 (n − 1) 2n − 1 6n 0 = 0 = 12 n − 1 6n 0 ⇒

n

∑ ∆a = c

[c = 0, i.e. constant]

Applying R1 → R1 +

a=1

 xCr 45. Let ∆ =  y C r z C  r

x

Cr + 1 y Cr + 1 z Cr + 1

x2 R2, we get 4 x+ 1 x−2  x  ∆= 0 0  −4 − −1 2 2 x 1 x  2x + 3

Applying C3 → C3 + C 2 x

x+1

y

y+1

Cr + 1 Cr + 1 z Cr + 1

Cr + 2  Cr + 2   z +1 Cr + 2 

[Q nC r + nC r − 1 =

 x + 0 x + 1 x − 2 Applying R3 → R3 − 2R1 =  − 4 0 0  −3 3   3 x x  0 1 −2   x =  −4 0 0 0 0  +  −4 3 3   3 −3  3 −3 1 1 − 1 0 1 2     = x − 4 0 0  +  −4 0 0 3  3 −3 3   3 −3

n+1

Cr ]

Applying C 2 → C 2 + C1  xCr ∆ = y C r z C  r

x+1

x+1

y+1

y+1

Cr + 1 Cr + 1 z+1 Cr + 1

Cr + 2  Cr + 2   z+1 Cr + 2 

Applying C3 → C3 + C 2  xCr ⇒ ∆ = y C r z C  r

x+1

x+ 2

y+1

y+ 2

Cr + 1 Cr + 1 z +1 Cr + 1

x2 R2 4

and R3 → R3 +

Cr + 2  y Cr + 2   z Cr + 2  x

 xCr ∆ = y C r z C  r

[Q R1 and R3 are identical]

Cr + 2  Cr + 2   z+ 2 Cr + 2 

Hence proved.

⇒

∆ = Ax + B 1 1  1 where, A =  −4 0 0  3 −3 3 

and

 0 B =  −4  3

1 0 −3

−2  0 3

Matrices and Determinants 159 a

b

c

c

a

b

2. Given matrix B is the inverse matrix of 3 × 3 matrix A,

48. Let ∆ =  b c a

Applying C1 → C1 + C 2 + C3 1 b c  a + b + c b c ∆ =  a + b + c c a = (a + b + c) 1 c a  a + b + c a b 1 a b Applying R2 → R2 − R1 and R3 → R3 − R1 , we get b c  1 = (a + b + c) 0 c − b a − c  0 a − b b − c  = (a + b + c) [− (c − b)2 − (a − b) (a − c)] = − (a + b + c) (a 2 + b2 + c2 − ab − bc − ca ) 1 = − (a + b + c) (2a 2 + 2b2 + 2c2 − 2ab − 2bc − 2ca ) 2 1 = − (a + b + c)[(a − b)2 + (b − c)2 + (c − a )2] 2 which is always negative.

Topic 3 Adjoint and Inverse of a Matrix 1. It is given that matrix.0  sin 4 θ −1 − sin 2 θ  −1 M =  = α I + βM , where 2 4 cos cos 1 θ θ +   α = α (θ ) and β = β (θ ) are real numbers and I is the 2 × 2 identity matrix. Now, det(M ) =|M|= sin 4 θ cos 4 θ + 1 + sin 2 θ + cos 2 θ + sin 2 θ cos 2 θ 4 4 2 2 = sin θ cos θ + sin θ cos θ + 2  sin 4 θ −1 − sin 2 θ  α 0  β and   = 0 α  + |M| (adj M ) 2 4 1 + θ θ cos cos     adj M   Q M −1 =  |M|   sin 4 θ −1 − sin 2 θ  α 0  ⇒  = 2 cos 4 θ  0 α  1 + cos θ a 1 + sin 2 θ   β  cos 4 θ + Q adj    2 4 |M| −1 − cos θ sin θ   c

b   d −b =  d  − c a 

⇒ β = −|M|and α = sin 4 θ + cos 4 θ 1 ⇒ α = α (θ) = 1 − sin 2(2 θ ), and 2 2  1 7 β = β (θ ) = − sin 2 θ cos 2 θ +  +    2 4   2  sin 2(2 θ ) 1 7   = −  +  +  4 2 4    1 37 and β * = β min = − 2 16 2 Q α is minimum at sin (2 θ ) = 1 and β is minimum at sin 2(2 θ ) = 1 1 37 29 So, α * + β * = − =− 2 16 16 Now, α * = α min =

5 2α where B = 0 2  α 3

1  1   − 1

We know that, det( A ) =

1   −1 Q det( A ) = det( A ) 

1 det(B)

det( A ) + 1 = 0 1 + 1 =0 det(B)

Since,

(given)

⇒

det(B) = − 1

⇒

5(− 2 − 3) − 2α (0 − α ) + 1 (0 − 2α ) = − 1

⇒

− 25 + 2α 2 − 2α = − 1

⇒

2α 2 − 2α − 24 = 0

⇒

α 2 − α − 12 = 0

⇒

(α − 4) (α + 3) = 0

⇒

α = − 3, 4

So, required sum of all values of α is 4 − 3 = 1 et 3. | A | = et et

−t

−e

e− t cos t cos t − e− t sin t

−e

2e− t sin t 1

−t

e− t sin t sin t + e− t cos t

− 2e− t cos t cos t

= (et ) (e− t ) (e− t ) 1 − cos t − sin t 1 2 sin t

sin t − sin t + cos t − 2 cos t

(taking common from each column) Aplying R2 → R2 − R1 and R3 → R3 − R1, we get [Qet − t = e0 = 1] cos t sin t 1 = e− t 0 − 2 cos t − sin t − 2 sin t + cos t 0 2 sin t − cos t − 2 cos t − sin t = e− t ((2 cos t + sin t )2 + (2 sin t − cos t )2) (expanding along column 1) = e− t (5 cos 2 t + 5 sin 2 t ) (Qcos 2 t + sin 2 t = 1) = 5e− t −t for all t ∈ R ⇒| A | = 5e ≠ 0 ∴ A is invertible for all t ∈ R [QIf| A | ≠ 0, then A is invertible]

4. Given,| ABAT| = 8 ⇒ | A||B|| AT| = 8 ∴ | A|2|B| = 8 Also, we have | AB−1| = 8 ⇒ | A||B−1| = 8 | A| ⇒ =8 |B|

[Q|XY | = |X ||Y |] …(i) [Q| AT| = | A|]

1   …(ii) Q| A −1|=| A|−1 =  | A|

On multiplying Eqs. (i) and (ii), we get | A|3 = 8 ⋅ 8 = 43 ⇒ | A| = 4

160 Matrices and Determinants ⇒ Now,

| A| 4 1 = = 8 8 2 1 −1 T |BA B | = |B| |B| | A| 1  1 1  1 =    =  2 4  2 16 |B| =

Now,

cos θ − sin θ   sin θ cos θ 

5. We have, A = 

| A| = cos 2 θ + sin 2 θ = 1  cos θ sin θ  and adj A =   − sin θ cos θ  a b   d − b [Q If A =  , then adj A = − c a  ] c d     cos θ sin θ    adj A  −1 ⇒ A −1 =  Q A =   − sin θ cos θ  | A|   

51 63  ∴ adj (3 A 2 + 12 A ) =   84 72

∴

− b and A adj A = AAT 2  Clearly, A (adj A ) = A I 2 [Q if A is square matrix of order n, then A (adj A ) = (adj A ) ⋅ A = A I n ] 5a − b I2 = 3 2 5a 3

7. Given, A = 

Note that, A −50 = ( A −1 )50 Now, A −2 = ( A −1 )( A −1 )  cos θ sin θ   cos θ sin θ  ⇒ A −2 =    − sin θ cos θ  − sin θ cos θ   cos 2 θ − sin 2 θ cos θ sin θ + sin θ cos θ  =  θ θ θ θ − cos sin − cos sin − sin 2 θ + cos 2 θ    cos 2 θ sin 2 θ  =  − sin 2 θ cos 2 θ 

= (10a + 3b) I 2 1 0 = (10a + 3b)   0 1 0 10a + 3b  =  + 0 10 a 3 b   and

Also, A −3 = ( A −2)( A −1 )  cos 2 θ sin 2 θ   cos θ sin θ  A −3 =    − sin 2 θ cos 2 θ  − sin θ cos θ   cos 3 θ sin 3 θ  =  − sin 3 θ cos 3 θ 

Q ∴

 2 − 3  − 4 1   2 − 3  2 − 3 A2 = A ⋅ A =    − 4 1  − 4 1   4 + 12 − 6 − 3  16 − 9 =  = − 8 − 4 12 + 1  − 12 13 

∴

5a − b  5a 3 AAT =  2  − b 2 3 25a 2 + b2 15a − 2b ...(ii) =  13   15a − 2b A (adj A ) = AAT 0 10a + 3b  25a 2 + b2 15a − 2b =   0 10a + 3b  15a − 2b 13  

⇒

1  2   3 2 

6. We have, A = 

...(i)

[using Eqs. (i) and (ii)] 15a − 2b = 0 2b ...(iii) a= ⇒ 15 and ...(iv) 10a + 3b = 13 On substituting the value of ‘a ’ from Eq. (iii) in Eq. (iv), we get  2b 10 ⋅   + 3b = 13  15  20b + 45b = 13 ⇒ 15 65b ⇒ = 13 ⇒ b = 3 15 Now, substituting the value of b in Eq. (iii), we get 5a = 2 Hence, 5a + b = 2 + 3 = 5

 cos 50 θ sin 50 θ  Similarly, A =  − sin 50 θ cos 50 θ  25 25   π sin π cos π   6 6  =  when θ =  25 25    12 π cos π − sin  6 6  π π   Q cos  25π  = cos 4π + π  = cos π       cos sin   6    6 6  6 6 =   π π 25π  π π  cos  and sin   = sin 4π +  = sin  − sin  6    6 6   6 6  −50

 3  = 2  −1  2

 16 − 9  2 − 3 3 A 2 + 12 A = 3   + 12 − 4 1  − 12 13      48 − 27  24 − 36 = +   − 36 39  − 48 12   72 − 63 =  − 84 51 

8.

PLAN Use the following properties of transpose ( AB)T = BT AT , ( AT )T = A and A −1 A = I and simplify. If A is non-singular matrix, then| A | ≠ 0.

Given,

AAT = AT A and B = A −1 AT BBT = ( A −1 AT )( A −1 AT )T = A −1 AT A ( A −1 )T

[Q ( AB)T = BT AT ]

Matrices and Determinants 161 = A −1 AAT ( A −1 )T = IAT ( A −1 )T = AT ( A −1 )T = ( A −1 A )T

[Q AAT = AT A] [Q A −1 A = I ] [Q ( AB)T = BT AT ]

= IT = I

1 α

3

∴

A3 + cA 2 + dA − 6I = O c = − 6 and d = 11

14. It is given that matrix M be a 3 × 3 invertible matrix,

| P | = 1(12 − 12) − α (4 − 6) + 3 (4 − 6) = 2 α − 6 P = adj ( A )

[given] [Q|adj A| = | A|n −1 ]

|P| = |adj A|= | A|2 = 16

such that ⇒

M −1 = adj(adj M ) M −1 = |M| M (Q for a matrix A of order ‘n’ adj(adj A ) = | A|n − 2 A}

∴

2 α − 6 = 16

⇒

2 α = 22

⇒

M −1M =|M|M 2 ⇒ M 2|M|= I

Q

⇒

α = 11

det(M 2|M|) = det(I ) = 1

⇒

10. Given, PT = 2 P + I ∴

…(i)

(PT )T = (2 P + I )T = 2 PT + I

⇒

P = 2 PT + I

⇒

P = 2 (2 P + I ) + I

⇒

P = 4P + 3I

⇒

PX = − IX = –X

or 1

11. | A| ≠ 0, as non-singular ω ω2 ⇒

a

b

⇒

(1 − cω ) (1 − aω ) ≠ 0 1 1 a ≠ ,c≠ ω ω

12. Given, M T = − M , N T = − N and MN = NM

...(i)

∴ M 2N 2 (M T N )− 1 (MN −1 )T ⇒ M 2 N 2N −1 (M T )−1 (N −1 )T ⋅ M T

⇒ ⇒

|M| = −1(6 − 6) − 1(−8 + 10) − 1(24 − 30) = −2 + 6 = 4 |M|= ±2

M = |M|(adj M )−1

⇒ Q

⇒ − M2

13. Every square matrix satisfied its characteristic

−2

0 1

So

=0

4−λ

⇒

(1 − λ ) {(1 − λ ) (4 − λ ) + 2} = 0

⇒

λ3 − 6λ2 + 11λ − 6 = 0

⇒

A − 6 A 2 + 11 A − 6I = O 3

⇒ …(i)

−2 −6  0 1  −2 −4 −2   |adj M|  −4 −6 −2  0 −2 −4  1 = −2 −4 −6   4 −6 −2 −2 −4  −2 a 0 |M|  −6  −2 −4 3 =   4  −2 1  −6 −2

(adj M )−1 =

Here, non-singular word should not be used, since there is no non-singular 3 × 3 skew-symmetric matrix.

0

…(i) T

⇒ − MN (NM −1 )N −1 M = − M (N N −1 )M

i.e. | A − λI | = 0 ⇒

−1

As we know A (adj A ) = | A| I

M 2 N (NN −1 )(− M )−1 (N T )−1 (− M ) M 2 N I (− M −1 ) ( − N )−1 (− M )

1−λ 0 0 1−λ

3

2

∴det (adj M 2) =|M 2|2 =|M|4 = 16

⇒ − M 2 NM −1N −1 M ⇒ − M ⋅ (MN )M −1N −1 M = − M (NM )M −1N −1 M

⇒

(adj M )2 = M 2

−5

⇒ a = ω , c = ω and b ∈{ω , ω 2} ⇒ 2 solutions

equation,

… (ii)

(adj M )2 = I 0 1 a  15. Given square matrix M = 1 2 3    3 b 1  1 −1   −1 and 2 adj (M ) =  8 −6   3 −1 −5 −1 1 −1 2 Q adj (M ) =|M|2 = 8 −6

1 c ≠0 ω 1

1 − cω − aω + acω 2 ≠ 0

NOTE

⇒ |M| = 1 from Eqs. (i) and (ii), we get M2 = I −1 As, adj M =|M|M = M ⇒

⇒

⇒ ⇒

… (i)

|M|3 |M|2 = 1

⇒

3P = − 3I

1 (1 − cω ) − a (ω − cω 2) + b (ω 2 − ω 2) ≠ 0

⇒

…(ii)

On comparing Eqs. (i) and (ii), we get

  2 4 4

Q

6I = A3 + cA 2 + dA ⇒

9. Given, P = 1 3 3 ∴

Given, 6 A −1 = A 2 + cA + dI , multiplying both sides by A, we get

0 1 1 2  3 b

|M|= −2, a = 2and b = 1 0 1 2 M = 1 2 3   3 1 1

162 Matrices and Determinants 0 1 2 α  1 α  1 M β  = 2 ⇒ 1 2 3 β  = 2          3 1 1  γ  3  γ  3 ⇒ B + 2 γ = 1, α + 2 β + 3γ = 2 and 3 α + β + γ = 3 ⇒ α = 1, β = −1and γ = 1 ⇒ α −β + γ = 3 And (adj M )−1 + adj (M −1 ) [Qadj (M −1 ) = (adj M )−1] = 2(adj M )−1 M  M [Q(adj M )−1 = from Eq. (i)] = 2 −  = −M  2 |M|

17.

and Q a = 2 and b = 1, so a + b = 3 Hence, options (b), (c) and (d) are correct. 3 − 1 − 2 16. Here, α  P = 2 0   3 − 5 0  Now, |P| = 3(5α ) + 1(− 3α ) − 2(− 10) …(i) = 12α + 20 T 2α − 10  5α 6 12  ∴ adj (P ) = − 10   2   − α − (3α + 4) −α   5α − 10 …(ii) 6 =  2α − 3α − 4   10 12 2 −   As, PQ = kI ⇒ |P||Q| = |kI| ⇒ |P||Q| = k3   k2 k2  ⇒ |P|  = k3 given,|Q| =   2  2  ⇒ Q

…(iii) |P| = 2k PQ = kI ⇒ Q = kp−1I adjP k(adj P ) [from Eq. (iii)] = k⋅ = |P| 2k −α   5 α − 10 adj P 1  = = − 3α − 4 2α 6  2 2 2  − 10 12 k − 3α − 4  q23 = given, q23 = − ∴ 8  2  (3α + 4) k − =− ⇒ 2 8 ⇒ (3α + 4) × 4 = k …(iv) ⇒ 12α + 16 = k From Eq. (iii), |P|= 2k [from Eq. (i)] …(v) ⇒ 12α + 20 = 2k On solving Eqs. (iv) and (v), we get …(vi) α = − 1 and k = 4 ∴ 4α − k + 8 = − 4 − 4 + 8 = 0 ∴ Option (b) is correct. Now, |P adj (Q )| = |P||adj Q| 2  k2 k5 210 = = 29 = 2k  = 2 2  2 ∴ Option (c) is correct.

PLAN A square matrix M is invertible, iff dem (M) or| M| ≠ 0.

a b M =   b c a  b (a) Given,   =   ⇒ a = b = c = α [let]  b  c α α  ⇒ M =  ⇒ |M |= 0 ⇒ M is non-invertible. α α 

Now, If

Let

(b) Given, [b c] = [a b] ⇒

a = b = c=α

[let]

Again,|M | = 0 ⇒ M is non-invertible. a 0 (c) As given M =   ⇒|M |= ac ≠ 0  0 c [Q a and c are non-zero] ⇒ M is invertible. a b (d) M =   ⇒|M |= ac − b2 ≠ 0  b c Q ac is not equal to square of an integer. M is invertible.

18.

PLAN If| A n × n| = ∆, then|adj A| = ∆A − 1

1 4 4  Here, adj P3 × 3 = 2 1 7   1 1 3  ⇒

|adj P | = | P |2 1 4 4

∴

| adj P | = 2 1 7 = 1 (3 − 7) − 4 (6 − 7) + 4 (2 − 1) 1 1 3 = − 4 + 4 + 4 = 4 ⇒ |P |= ± 2

19. | A | = (2k + 1) ,| B| = 0 3

But det (adj A) + det (adj B) = 106 ⇒

(2k + 1)6 = 106 9 k= ⇒ 2

⇒

[k] = 4

Topic 4 Solving System of Equations 1. Given system of linear equations is [sin θ ] x + [− cos θ ] y = 0

…(i)

[cot θ ] x + y = 0

…(ii)

and

where, [x] denotes the greatest integer ≤ x. [sin θ ] [− cos θ ] Here, ∆= [cot θ ] 1 ⇒ ∆ = [sin θ ] − [− cos θ ] [cot θ ]  π 2π  When θ ∈ ,  2 3   3  sin θ ∈  , 1  2 

Matrices and Determinants 163 ⇒

[sin θ ] = 0  1 − cos θ ∈ 0,   2

⇒

[− cos θ ] = 0  1  cot θ ∈  − , 0  3 

and

…(iii)

4. Given system of linear equations 2x + 3 y − z = 0, x + ky − 2z = 0

…(iv)

and 2x − y + z = 0 has a non-trivial solution (x, y, z ). 2 3 −1  ∴ ∆ = 0 ⇒ 1 k −2 = 0

⇒ So,

…(v) [cot θ ] = − 1 ∆ = [sin θ ] − [− cos θ ] [cot θ ] − (0 × (− 1)) = 0 [from Eqs. (iii), (iv) and (v)]  π 2π  Thus, for θ ∈  ,  , the given system have infinitely 2 3  many solutions.  7π   1  When θ ∈  π,  , sin θ ∈  − , 0 ⇒ [sin θ ] = − 1   2  6

2 −1

2(k − 2) − 3(1 + 4) − 1(−1 − 2k) = 0 ⇒

So, system of linear equations is 2x + 3 y − z = 0 2x + 9 y − 4z = 0 and 2x − y + z = 0

cot θ ∈ ( 3 , ∞ ) ⇒ [cot θ ] = n , n ∈ N .

∆ = − 1 − (0 × n ) = − 1  7π  Thus, for θ ∈  π,  , the given system has a unique  6 solution.

From Eqs. (i) and (iii), we get x 1 4x + 2 y = 0 ⇒ = − y 2

2. Given, system of linear equations

1 1 ⇒ 4 λ 3 2

… (i) …(ii) …(iii)

1 − λ =0

So, ∴

x x y z 1 = × = − ⇒ = −4 z y z x 4

x 1  y 1 Q z = 2 and y = − 2 

x y z 1 1 9 1 + + + k= − + −4 + = . y z x 2 2 2 2

5. Given system of linear equations …(i) x − 2 y + kz = 1 …(ii) 2x + y + z = 2 and …(iii) 3x − y − kz = 3 has a solution (x, y, z ), z ≠ 0. On adding Eqs. (i) and (iii), we get x − 2 y + kz + 3x − y − kz = 1 + 3 4x − 3 y = 4 ⇒ 4x − 3 y − 4 = 0 This is the required equation of the straight line in which point (x, y) lies.

−4

⇒ 1(− 4λ + 2λ ) − 1(− 16 + 3λ) + 1(8 − 3λ) = 0 ⇒ − 8λ + 24 = 0 ⇒ λ = 3 From, the option λ = 3, satisfy the quadratic equation λ2 − λ − 6 = 0.

3. Given system of linear equations x+ y+ z =5 x + 2 y + 2z = 6 x + 3 y + λz = µ

…(i) …(ii) …(iii) (λ, µ ∈ R) The above given system has infinitely many solutions, then the plane represented by these equations intersect each other at a line, means (x + 3 y + λz − µ ) = p(x + y + z − 5) + q (x + 2 y + 2z − 6) = ( p + q)x + ( p + 2q) y + ( p + 2q)z − (5 p + 6q) On comparing, we get p + q = 1, p + 2q = 3, p + 2q = λ and 5 p + 6q = µ So, ( p, q) = (−1, 2) ⇒ λ = 3 and µ = 7 ⇒ λ + µ = 3 + 7 = 10

…(i) …(ii) …(iii)

From Eqs. (i) and (ii), we get y 1 6 y − 3z = 0, = z 2

So,

x+ y+ z =6 4 x + λy − λz = λ − 2 and 3x + 2 y − 4z = − 5 has infinitely many solutions, then ∆ = 0

2k − 4 − 15 + 1 + 2k = 0 9 4 k = 18 ⇒ k = 2

⇒

 3  − cos θ ∈  , 1 ⇒ [cos θ ] = 0  2  and

1

6.

Key Idea A homogeneous system of linear equations have non-trivial solutions iff ∆ = 0

Given system of linear equations is x − cy − cz = 0, cx − y + cz = 0 and cx + cy − z = 0 We know that a homogeneous system of linear equations have non-trivial solutions iff ∆ =0 1 − c − c c −1 c ⇒ = 0  c c − 1  

164 Matrices and Determinants ⇒ 1(1 − c2) + c(− c − c2) − c(c2 + c) = 0 ⇒ 1 − c2 − c2 − c3 − c3 − c2 = 0 ⇒ −2c3 − 3c2 + 1 = 0 ⇒ 2c3 + 3c2 − 1 = 0 ⇒ (c + 1)[2c2 + c − 1] = 0 ⇒ (c + 1)[2c2 + 2c − c − 1] = 0 ⇒ (c + 1)(2c − 1)(c + 1) = 0 ⇒

has more than one solution, then D1 = D2 = D3 = 0. In the given problem, a 2 3 D1 = 0 ⇒ b − 1 5 = 0 c −3 2

c = − 1 or

1 2

10. We know that,

7. The given system of linear equations is

x − 2 y − 2 z = λx x + 2 y + z = λy − x − y − λz = 0, which can be rewritten as (1 − λ )x − 2 y − 2z = 0 ⇒ x + (2 − λ ) y + z = 0 x + y + λz = 0 Now, for non-trivial solution, we should have 1 − λ −2 −2 1 2 − λ 1 =0 1

λ

[Q If a1x + b1 y + c1z = 0; a 2x + b2y + c2z = 0 a3 x + b3 y + c3 z = 0]  has a non-trivial solution, then a 2 b2 c2 = 0  a3 b3 c3  ⇒ (1 − λ ) [(2 − λ )λ − 1] + 2 [λ − 1] − 2 [1 − 2 + λ ] = 0 ⇒ (λ − 1)[λ2 − 2λ + 1 + 2 − 2] = 0 ⇒ (λ −1)3 = 0 ⇒ λ =1 a1

b1

c1

8. Given system of linear equations, (1 + α )x + βy + z = 2 αx + (1 + β ) y + z = 3 αx + βy + 2z = 2 has a unique solution, if 1+α β 1 α α

(1 + β ) 1 ≠ 0 β 2

Apply R1 → R1 − R3 and R2 → R2 − R3 1 0 −1 0 1 −1 ≠0 α β

2

⇒ 1(2 + β ) − 0(0 + α ) − 1(0 − α ) ≠ 0 ⇒ α + β + 2 ≠ 0 … (i) Note that, only (2, 4) satisfy the Eq. (i).

9. We know that, if the system of equations a1x + b1 y + c1z = d1 a 2x + b2y + c2z = d2 a3 x + b3 y + c3 z = d3

and

⇒ a (− 2 + 15) − 2(2b − 5c) + 3(− 3b + c) = 0 ⇒ 13a − 4b + 10c − 9b + 3c = 0 ⇒ 13a − 13b + 13c = 0 ⇒ a − b + c=0⇒b − a − c=0

1 Clearly, the greatest value of c is . 2

1

D =0

the system of linear equations a1x + b1 y + c1z = 0 a 2x + b2y + c2z = 0 a3 x + b3 y + c3 z = 0 has a non-trivial solution, if a1 b1 c1 a 2 b2 c2 = 0 a3 b3 c3 Now, if the given system of linear equations x + 3 y + 7z = 0 − x + 4 y + 7z = 0, and (sin 3 θ )x + (cos 2 θ ) y + 2z = 0 has non-trivial solution, then 1 3 7 −1 4 7 =0 sin 3 θ cos 2 θ 2 ⇒ 1(8 − 7 cos 2 θ ) − 3 (− 2 − 7 sin 3 θ ) + 7 (− cos 2 θ − 4 sin 3 θ ) = 0 ⇒ 8 − 7 cos 2 θ + 6 + 21 sin 3 θ − 7 cos 2 θ − 28 sin 3 θ = 0 ⇒ − 7 sin 3 θ − 14 cos 2 θ + 14 = 0 ⇒ − 7 (3 sin θ − 4 sin3 θ ) − 14 (1 − 2 sin 2 θ ) +14 = 0 [ Q sin 3 A = 3 sin A − 4 sin3 A and cos 2 A = 1 − 2 sin 2 A] 3 2 ⇒ 28 sin θ + 28 sin θ − 21 sin θ − 14 + 14 = 0 ⇒ 7 sin θ [4 sin 2 θ + 4 sin θ − 3] = 0 ⇒ sin θ [4 sin 2 θ + 6 sin θ − 2 sin θ − 3] = 0 ⇒ sin θ [2 sin θ (2 sin θ + 3) − 1 (2 sin θ + 3)] = 0 ⇒ (sin θ ) (2 sin θ − 1) (2 sin θ + 3) = 0 1 Now, either sin θ = 0 or 2 3   Q sin θ ≠ − as − 1 ≤ sin θ ≤ 1   2 In given interval (0, π ), 1 sin θ = 2 π 5π θ= , ⇒ 6 6 Hence, 2 solutions in (0, π )

[Q sin θ ≠ 0, θ ∈ (0, π )]

Matrices and Determinants 165 11. Since, the system of equations has infinitely many solution, therefore D = D1 = D2 = D3 = 0 Here, 1 1 1 D = 1 2 3 = 1 (2α − 9) − 1 (α − 3) + 1(3 − 2) = α − 5 1 3 α 1 1 5 and D3 = 1 2 9 = 1 (2 β − 27) − 1(β − 9) + 5 (3 − 2) 1 3 β = β − 13 Now, ⇒ and ∴

12. Here,

D =0 α −5 =0 ⇒ α =5 D3 = 0 ⇒ β − 13 = 0 ⇒ β = 13 β − α = 13 − 5 = 8 1 −4 7 3 −5 D= 0 −2

−9

5

= 1(− 27 + 25) + 4(0 − 10) + 7(0 + 6) [expanding along R1] = − 2 − 40 + 42 = 0 ∴The system of linear equations have infinite many solutions. [Q system is consistent and does not have unique solution as D = 0] ⇒ D1 = D2 = D3 = 0 g −4 7 Now, D1 = 0 ⇒ h 3 − 5 = 0 k

5

−9

⇒ g (− 27 + 25) + 4(− 9h + 5k) + 7(5h − 3k) = 0 ⇒ − 2 g − 36h + 20k + 35h − 21k = 0 ⇒ − 2g − h − k = 0 ⇒2g + h + k = 0

13 According to Cramer’s rule, here 1 1

1

1 0

0

2 0 D= 2 3 = 2 1 2 3 a2 − 1 2 1 a2 − 3 (Applying C 2 → C 2 − C1 and C3 → C3 − C1) (Expanding along R1) = a −3 2 1 1 2 1 0 and D1 = 5 3 2 = 5 3 −1 2

a + 1 3 a2 − 1

a + 1 3 a2 − 1 −3 (Applying C3 → C3 − C 2)

2

0

0 5 = 5 −1 3− 2 a + 1 3 − (a + 1) a 2 − 1 − 3 2 1 (Applying C 2 → C 2 − C1) 2

2 5

0 1 2

−1

a+1

5 a − 2 2

a2 − 4

=

0

1 5 a  [Expanding along R1] = 2  (a 2 − 4) +  −   2 2  2  a2 5 a =2 − 2 + −  = a2 − 4 + 5 − a = a2 − a + 1 2 2 2 Clearly, when a = 4, then D = 13 ≠ 0 ⇒ unique solution and when|a| = 3, then D = 0 and D1 ≠ 0. ∴When |a| = 3, then the system has no solution i.e. system is inconsistent. 14. We have, x + ky + 3z = 0; 3x + ky − 2z = 0; 2x + 4 y − 3z = 0 System of equation has non-zero solution, if 1 k 3   3 k −2 = 0    2 4 −3 ⇒ (−3k + 8) − k(−9 + 4) + 3(12 − 2k) = 0 ⇒ − 3k + 8 + 9k − 4k + 36 − 6k = 0 ⇒ − 4k + 44 = 0 ⇒ k = 11 Let z = λ , then we get x + 11 y + 3λ = 0 3x + 11 y − 2λ = 0 and 2x + 4 y − 3λ = 0 Solving Eqs. (i) and (ii), we get 5λ −λ 5λ2 xz , y= ,z=λ ⇒ 2= x= = 10 2 2 2 y  λ 2 × −   2

…(i) …(ii) …(iii)

15. Given, system of linear equation is x + λy − z = 0; λx − y − z = 0; x + y − λz = 0 Note that, given system will have a non-trivial solution only if determinant of coefficient matrix is zero, i.e.

1 λ 1

λ −1 −1 −1 = 0 1 −λ

⇒ 1 (λ + 1) − λ(− λ2 + 1) − 1(λ + 1) = 0 ⇒ λ + 1 + λ3 − λ − λ − 1 = 0 ⇒ λ3 − λ = 0 ⇒ λ(λ2 − 1) = 0 ⇒ λ = 0or λ = ± 1 Hence, given system of linear equation has a non-trivial solution for exactly three values of λ.

16. Given system of linear equations 2x1 − 2x2 + x3 = λx1 ⇒

(2 − λ )x1 − 2x2 + x3 = 0 2x1 − 3x2 + 2x3 = λx2

…(i) …(ii) …(iii)

166 Matrices and Determinants ⇒ 2x1 − ( 3 + λ )x2 + 2x3 = 0 − x1 + 2x2 = λx3 ⇒ − x1 + 2x2 − λx3 = 0 Since, the system has non-trivial solution. −2 1 2 − λ  2 ∴ − (3 + λ ) 2  = 0   2 λ   −1 ⇒ (2 − λ )(3λ + λ2 − 4) + 2(−2λ + 2) + 1(4 − 3) − λ ) = 0 ⇒ (2 − λ )(λ2 + 3λ − 4) + 4(1 − λ ) + (1 − λ ) = 0 ⇒ (2 − λ )(λ + 4)(λ − 1) + 5(1 − λ ) = 0 ⇒ (λ − 1)[(2 − λ )(λ + 4) − 5] = 0 ⇒ (λ − 1)(λ2 + 2λ − 3) = 0 ⇒ (λ − 1)[(λ − 1)(λ + 8)] = 0 ⇒ (λ − 1)2(λ + 3) = 0 ⇒ λ = 1, 1, − 3 Hence, λ contains two elements.

17. Given equations can be written in matrix form AX = B where, A =  

k+1 k

4k x 8  , X =   and B = k + 3  3k − 1  y

For no solution, A = 0 and (adj A) B ≠ 0 8  k+1 Now, A =  = 0 ⇒ (k2 + 1)(k + 3) − 8k = 0 k + 3   k

⇒

k = 1, 3 8 4.1 If k –1, then , false ≠ 1+3 2 And, if k = 3, then

(k − 1)(k − 3) = 0 ⇒ k = 1, k = 3, k + 3 −8  adj A =  k + 1   − k

Now

k + 3 − 8   4k  (adj A)B =  k + 1   3 k + 1   − k  (k + 3)(4k) − 8 (3k − 1) =  − 4k2 + (k + 1)(3k − 1) 

Now,

 4k2 − 12k + 8  = 2   − k + 2k − 1 

Hence, only one value of k exist.  x

Put k = 3 36 − 36 + 8  8  (adj A) B =  ≠ 0 true = −9 + 6 − 1 −4 Hence, required value of k is 3.

     z  0 and that could have only unique, no solution or infinitely many solution.

∴It is not possible to have two solutions. Hence, number of matrices A is zero.

19. Since, given system has no solution. ∴ ∆ = 0 and any one amongst ∆ x , ∆ y , ∆ z is non-zero. 2  2 −1  2 −1 2  Let 1 −2 1  = 0 and ∆ z =  1 −2 −4  = 6 ≠ 0 1 λ 1 4 1 1 ⇒

8 4k k+1 = = ⇒ k =1 k k + 3 3k − 1

21. Given equations x + ay = 0, az + y = 0, ax + z = 0 has infinite solutions. 1 a 0  ∴ 0 1 a = 0 a 0 1  ⇒

∴

−k −1 1

−1  −1  = 0 −1  − C 2, C 2 → C 2 + C3 −1  −1  = 0 −1 

⇒

2(k + 1) − (k + 1)2 = 0

⇒

(k + 1)(2 − k − 1) = 0

⇒

8 k+1 = ⇒ k2 + 4k + 3 = 8k k k+3

1 k 1

Applying C1 → C1  1 + k −k − 1 −2 ⇒ 1 + k 0  0

NOTE

Take

1 + a3 = 0 or a = − 1

22. Since, the given system has non-zero solution.

Condition for the system of equations has no solution is a1 b1 c1 = ≠ a 2 b2 c2 8 4k k+1 = ≠ k k + 3 3k − 1

λ =1

20. For infinitely many solutions, we must have

Alternate Solution

∴

1

18. Since, A  y = 0 is linear equation in three variables

Put k = 1 4 − 12 + 8 0 not true (adj A) B =  = −1 + 2 − 1  0

8 4.3 , true ≠ 6 9 −1

Therefore, k = 3

k2 + 4k + 3 − 8k = 0 ⇒ k2 − 4k × 3 = 0 ⇒

k2 − 4k + 3 ⇒ (k –1) (k –3) = 0

k=±1 There is a golden rule in determinant that n one’s ⇒ ( n − 1) zero’s or n (constant) ⇒ ( n − 1) zero’s for all constant should be in a single row or a single column.

23. The given system of equations can be expressed as 3  x   1 −2  1 −3 4  y =    −1 1 −2  z 

 −1   1    k 

Matrices and Determinants 167 2 2 + 2x 2 + 5x  = 0 8 8    3x 3x + 18 x

Applying R2 → R2 − R1 , R3 → R3 + R1 1 −2 3 ~ 0 −1 1   0 −1 1

 x   −1   y =  2       z  k − 1

Applying R3 → R3 − R2 1 −2 3  x   −1  ~ 0 −1 1  y =  2        0 0 0  z  k − 3 When k ≠ 3 , the given system of equations has no solution. ⇒ Statement I is true. Clearly, Statement II is also true as it is rearrangement of rows and columns of  1 −2 3   1 −3 4 .   −1 1 −2

24. It is given, that matrices

∴

2 x x 1 1 1 P = 0 2 2, Q = 0 4 0     x x 6 0 0 3 adj (P ) P −1 = |P|

6 as|P| = 6 and adj P = −3   0 6 −3 1 P −1 = 0 3 ⇒ 6 0 0 ∴ ⇒

T

0 3 0  −2 2 0 −2   2  0

−1

−1

[QR = PQP (given] |R| = |P Q P | −1 [Q|P||P −1| = |I| = 1] |R| = |P||Q||P | = |Q| 2 x 0 2 x x 2 x x = 0 4 0 = 0 4 0 + 0 4 0 x x 6

x x 5

x x 1

2 x x = 0 4 0 + 2 (4 − 0) − x (0 − 0) + 0(0 − 4x) x x 5 2 x x = 0 4 0 + 8, for all x ∈ R x x 5 Q

and

1 1 PQ = 0 2  0 0 2 + x =  2x   3x

1 2 x x 2 0 4 0   3 x x 6 4 + 2x x + 6 2x + 8 12   3x 18 

2 x x 1 1 1 QP = 0 4 0 0 2 2    x x 6 0 0 3

There is no common value of ‘x’, for which each corresponding element of matrices PQ and QP is equal. 2 0 0 For x = 0, Q = 0 4 0   0 0 6 1  1  then, if R a  = 6 a       b   b  1  1  [Q R = PQP −1 ] ⇒ PQP −1 a  = 6 a       b   b  1  1 1 1 2 0 0 6 −3 0  1  1 ⇒ 0 2 2 0 4 0 0 3 −2 a  = 6 a        6 2   b   b  0 0 3 0 0 6 0 0 2 4 6 6 3 0 1 − 1       1 0 8 12 0 3 −2 a  = 6 a  ⇒      6 2   b  0 0 18 0 0  b  12 6 4  1  1   0 24 8  a  = 36 a  ⇒       0 0 36  b   b  + + a b 36 12 6 4     0 + 24a + 8b = 36a  ⇒      0 + 0 + 36b  36b  ⇒ ⇒

6a + 4b = 24 and 12a = 8b 3a + 2b = 12 and 3a = 2b

⇒ a = 2 and b = 3 So a + b = 5. α  0 Now, R β  = 0 and αi$ + β$j + γk$ is a unit vector, so det      γ  0

(R) = 0 ⇒

det(Q ) = 0 [Q R = PQP −1 So,|R|=|Q|] 2 x x

⇒

0 4 0 =0 x x 6

⇒ ⇒

2 (24 − 0) − x (0 − 0) + x(0 − 4x) 48 − 4x2 = 0

⇒ ⇒

x2 = 12 x=±2 3

So, for x = 1, there does not exist a unit vector α  0 αi$ + β$j + γk$ , for which R β  = 0      γ  0 Hence, options (b) and (d) are correct.

168 Matrices and Determinants 26. Given system λx + y + z = 0, − x + λy + z = 0

25. We have, − x + 2 y + 5z = b1 2x − 4 y + 3z = b2 x − 2 y + 2z = b3 has at least one solution. −1 2 5 D= 2 1

∴ and ⇒

will have non-zero solution, if λ 1

⇒ λ (λ 2 + 1) − 1 (− λ + 1) + 1(1 + λ ) = 0 ⇒ λ3 + λ + λ − 1 + 1 + λ = 0 ⇒ λ 3 + 3λ = 0 ⇒ λ (λ 2 + 3) = 0 ⇒

D1 = D2 = D3 = 0 b1 2 5 D1 = b2 − 4 3 b3 − 2 2 ...(i)

= 1(− 6 + 6) − 1 (− 15 + 12) + 3 (− 5 + 4) = 0 For atleast one solution D1 = D2 = D3 = 0 b1 1 3 Now, D1 = b2 2 6

5

⇒

a | A2| = 1 1

and

f | A1| = g h

−2

3

BX = V a |B| = 0 f

1 d g

= 1(0 − 12) − 2 (− 10 − 3) + 5 (8 − 0)

1 4 −5 = 54 D ≠0 It has unique solution for any b1, b2, b3 .

1 b =0 ⇒ b

g=h

1 b =0 ⇒ b

g=h …(i)

1 c =0 h

[from Eq. (i)]

[since, C 2 and C3 are equal] BX = V has no solution. a2 1 1 [from Eq. (i)] |B1|= 0 d c =0 0 g h a |B2| = 0 f

Since,

a2 0 0

[since, c = d and g = h] 1 [Q c = d ] c = a 2cf = a 2df h

adf ≠ 0 ⇒ |B2| ≠ 0 |B| = 0

5

= − 1(− 20 + 20) − 2(10 − 10) − 5(− 4 + 4) = 0 Here, b2 = − 2b1 and b3 = − b1 satisfies the Eq. (i) Planes are parallel. 1 2 5 (d) D = 2 0

∴

f g h 0 c d

g = h , c = d and ab = 1

Now,

−3

= − b1 (− 15 + 12) + b2(− 3 + 6) − b3 (6 − 15) = 3b1 + 3b2 + 9b3 = 0 ⇒ b1 + b2 + 3b3 = 0 not satisfies the Eq. (i) It has no solution. −1 2 −5 (c) D = 2 − 4 10 1

∴

6

b2

1 b =0 b

a (bc − bd ) + 1(d − c) = 0 ⇒ (d − c)(ab − 1) = 0 ab = 1 or d = c a 0 f Again, | A3|= 1 c g =0 ⇒ g = h 1 d h

−1 −3

= b1 (− 6 + 6) − b2(− 3 + 3) + b3 (6 − 6) = 0 1 b1 3

0 c d

⇒ ∴

−2 −1 −3

− 2 b3

a 1 1

⇒| A| = 0 ⇒

= 14 − 2 = 12 ≠ 0 Here, D ≠ 0 ⇒ unique solution for any b1, b2, b3 . 1 1 3 (b) D = 5 2 6

D2 =

λ =0

27. Since, AX = U has infinitely many solutions.

1 2 6

b3

1

−1 λ 1 = 0 −1 −1 λ

−4 3 −2 2

= − 2b1 − 14b2 + 26b3 = 0 ⇒ b1 + 7b2 = 13b3 1 2 3 (a) D = 0 4 5 = 1(24 − 10) + 1(10 − 12)

− x − y + λz = 0

and

∴

and

|B2| ≠ 0

BX = V has no solution.

28. Given, λx + (sin α ) y + (cos α ) z = 0 x + (cos α ) y + (sin α ) z = 0 and − x + (sin α ) y − (cos α ) z = 0 has non-trivial solution. ∴ ⇒

λ 1

sin α cos α

−1 sin α

∆ =0 cos α sin α = 0 − cos α

Matrices and Determinants 169 ⇒

λ (− cos 2 α − sin 2 α ) − sin α (− cos α + sin α ) + cos α (sin α + cos α ) = 0

⇒

− λ + sin α cos α + sin α cos α − sin 2 α + cos 2 α = 0

⇒

λ = cos 2α + sin 2α Q − a 2 + b2 ≤ a sin θ + b cos θ ≤ a 2 + b2   

∴

− 2≤λ≤ 2

…(i)

Again, when λ = 1, cos 2α + sin 2α = 1 1 1 1 cos 2α + sin 2α = ⇒ 2 2 2 ⇒ ⇒

2α − π / 4 = 2n π ± π / 4 2α = 2nπ − π / 4 + π / 4 or 2α = 2nπ + π / 4 + π / 4

∴

⇒

⇒

31. The given system of equations

has atleast one solution, if ∆ ≠ 0. 3 −1 4 ∴ ∆ = 1 2 −3 ≠ 0

α = nπ or nπ + π /4

6

29. Since, α 1 , α 2 are the roots of ax + bx + c = 0. 2

⇒

sin θ (4 sin 2 θ + 4 sin θ − 3) = 0 sin θ (2 sin θ − 1) (2 sin θ + 3) = 0 1 sin θ = 0, sin θ = 2 [neglecting sin θ = − 3 / 2] π θ = nπ , nπ + (−1)n , n ∈ Z 6 3x − y + 4z = 3 x + 2 y − 3z = − 2 6 x + 5 y + λz = − 3

cos (2α − π / 4) = cos π / 4

∴

⇒ ⇒

b α1 + α 2 = − a

c and α 1α 2 = a

Also, β1 , β 2 are the roots of px2 + qx + r = 0. q r and β1β 2 = ⇒ β1 + β 2 = − p p

...(i)

...(ii)

Given system of equations α 1 y + α 2z = 0

⇒ 3 (2λ + 15) + 1 (λ + 18) + 4 (5 − 12) ≠ 0 ⇒ 7 (λ + 5) ≠ 0 ⇒ λ ≠ −5 Let z = − k, then equations become 3x − y = 3 − 4k and x + 2 y = 3k − 2 On solving, we get 4 − 5k 13k − 9 x= ,y= ,z=k 7 7

32. Given system of equations are

and β1 y + β 2 z = 0, has non-trivial solution. α1 α 2 α 1 β1 = ∴ =0 ⇒ β1 β 2 α 2 β2 Applying componendo-dividendo,

α 1 + α 2 β1 + β 2 = α 1 − α 2 β1 − β 2

3x + my = m and 2x − 5 y = 20 3 m Here, ∆= = −15 − 2m 2 −5 ∆x =

and

⇒ (α 1 + α 2) (β1 − β 2) = (α 1 − α 2) (β1 + β 2)

∆y =

⇒ (α 1 + α 2)2 {(β1 + β 2)2 − 4 β 2 β 2} = (β1 + β 2)2{(α 1 + α 2)2 − 4 α 1α 2} From Eqs. (i) and (ii), we get b2  q2 4r  q2  b2 4c − = −    a 2  p2 p  p2  a 2 a  ⇒ ⇒

b2q2 4b2r b2q2 4q2c − = − a 2p2 a 2p a 2p2 ap2 b2r q2c b2 ac = ⇒ 2= a p pr q

30. The system of equations has non-trivial solution, if ∆ = 0.

sin 3θ −1 1 cos 2θ 4 3 = 0

y=

and For x > 0 ,

∆y ∆

=

sin 3θ ⋅ (28 − 21) − cos 2 θ (−7 − 7) + 2 (−3 − 4) = 0

⇒

⇒

7

7

7 sin 3θ + 14 cos 2θ − 14 = 0

⇒ sin 3θ + 2 cos 2 θ − 2 = 0 ⇒ 3 sin θ − 4 sin3 θ + 2 (1 − 2 sin 2 θ ) − 2 = 0

m

3 m 2 20

= −25m = 60 − 2m

or

60 − 2m 2m − 60 = − 15 − 2m 15 + 2m

25m >0 ⇒ m >0 15 + 2m

Expanding along C1 , we get

2

m

20 −5

If ∆ = 0, then system is inconsistent, i.e. it has no solution. 15 If ∆ ≠ 0, i.e. m ≠ , the system has a unique solution 2 for any fixed value of m. ∆ − 25m 25m We have, x= x = = ∆ − 15 − 2 m 15 + 2m

15 2 2m − 60 and y > 0, >0 2m + 15

⇒

λ

5

m 30 or m < −

…(i)

15 2

From Eqs. (i) and (ii), we get m < −

…(ii) 15 or m > 30 2

170 Matrices and Determinants …(ii) x + 2 y = 10 and …(iii) 3x + 2 y = µ On solving Eqs. (i) and (ii), we are getting y = 4 and x = 2 From Eq. (iii), we get µ = 3(2) + 2(4) = 14 Now, on putting y = 0 and µ = 14, in the given system of linear Eq., we get …(iv) x+ z =6 …(v) x + 3z = 10 and …(vi) 3x + λz = 14 On solving Eqs. (iv) and (v), we get z = 2 and x = 4 From Eq. (vi), we get 3(4) + λ (2) = 14 ⇒ λ =1 ∴ µ − λ2 = 14 − 1 = 13

33. Since, the given system of equations posses non-trivial 0

1

solution, if 0 −3 k −5

−2 1 4

=0 ⇒ k =0

On solving the equations x = y = z = λ

[say]

∴ For k = 0, the system has infinite solutions of λ ∈R.

34. Given systems of equations can be rewritten as − x + cy + by = 0, cx − y + az = 0 and bx + ay − z = 0 Above system of equations are homogeneous equation. Since, x, y and z are not all zero, so it has non-trivial solution. Therefore, the coefficient of determinant must be zero. −1 c b c −1 a = 0 ∴ b

a

−1

⇒ − 1 (1 − a 2) − c (− c − ab) + b (ca + b) = 0 ⇒

a 2 + b2 + c2 + 2abc − 1 = 0

⇒

a 2 + b2 + c2 + 2abc = 1

35. System of given linear equations is

x − 2 y + 3z = 9 2x + y + z = b and x − 7 y + az = 24, has infinitely many solution, so ∆ = 0 = ∆1 = ∆ 2 = ∆3 So, ∆ =0 1 −2 3 2 1 1= 0 ⇒   1 −7 a

⇒ 1(a + 7) + 2(2a − 1) + 3(−14 − 1) = 0 ⇒ 5a = 40 ⇒ a = 8 1 −2 9  and b ∆3 = 0 ⇒ = 0 2 1 1 −7 24

⇒ 1(24 + 7b) + 2(48 − b) + 9(−14 − 1) = 0 ⇒ 5b + 120 − 135 = 0 ⇒ b = 3 ∴ a − b =8 −3=5 Hence, answer is 5.00.

36. The given system of linear equations x + y + z = 6, x + 2 y + 3z = 10, and 3x + 2 y + λz = µ has more than two solutions, so system must have infinite number of solutions. Now, on putting z = 0 in above equation, we get …(i) x+ y=6

1

37.

α α2

α α2 1 α =0 α 1

⇒ α 4 − 2α 2 + 1 = 0 ⇒ α2 = 1 ⇒α = ± 1 But α = 1 not possible [Not satisfying equation] ∴ α = −1 Hence, 1 + α + α 2 = 1 a1 a 2 a3  38. Let M =  b1 b2 b3     c1 c2 c3  0 − 1  1  1 ∴ M 1 =  2 ⇒ M − 1 =  1         0  3  0 − 1 0 1 a 2 − 1 a1 − a 2  1         M ⋅ 1 = 0 ⇒  b2  =  2,  b1 − b2  =  1,              c2   3  c1 − c2  − 1 1 12 a1 + a 2 + a3   0  b + b + b  =  0 2 3     1  c1 + c2 + c3  12 ⇒

a 2 = − 1, b2 = 2, c2 = 3, a1 − a 2 = 1, b1 − b2 = 1, c1 − c2 = − 1

⇒ a1 + a 2 + a3 = 0, b1 + b2 + b3 = 0 c1 + c2 + c3 = 12 ∴

a1 = 0, b2 = 2, c3 = 7

⇒ Sum of diagonal elements = 0 + 2 + 7 = 9

8 Functions Topic 1 Classification of Functions, Domain and

Range and Even, Odd Functions Objective Questions I (Only one correct option)

7. The domain of definition of f (x) =

1. If R = {(x, y): x, y ∈ Z , x2 + 3 y2 ≤ 8} is a relation on the set of integers Z, then the domain of R−1 is

(2020 Main, 2 Sep I)

(a) {−1, 0,1} (c) {−2, − 1, 0,1, 2}

(b) {− 2, − 1,1, 2} (d) {0,1}

2. The domain of the definition of the function f (x) =

1 + log10 (x3 − x) is 4 − x2

(a) (−1, 0) ∪ (1, 2) ∪ (3, ∞ ) (c) (−1, 0) ∪ (1, 2) ∪ (2, ∞ )

(2019 Main, 9 April II)

(b) (−2, − 1) ∪ (−1, 0) ∪ (2, ∞ ) (d) (1, 2) ∪ (2, ∞ )

where f1 (x) is an even function and f2(x) is an odd function. Then f1 (x + y) + f1 (x − y) equals (2019 Main, 8 April II)

(b) 2f1 (x + y) ⋅ f1 (x − y) (d) 2f1 (x) ⋅ f1 ( y)

4. Domain of definition of the function f (x) = sin −1 (2x) + 1 1 (a)  − ,   4 2  1 1 (c)  − ,   2 9

π for real valued x, is 6

(2003, 2M)

1 1 (b)  − ,   2 2  1 1 (d)  − ,   4 4 

minimum value of f (x). As b varies, the range of m (b) is

(2000, 1M)

(b) 0 ≤ x ≤ 1 (d) −∞ < x < 1

9. Let f (θ ) = sin θ (sin θ + sin 3 θ ). Then, f (θ ) (a) ≥ 0, only when θ ≥ 0 (c) ≥ 0, for all real θ

(2000, 1M)

(b) ≤ 0, for all real θ (d) ≤ 0, only when θ ≤ 0

10. The domain of definition of the function y=

1 + log10 (1 − x)

x + 2 is (1983, 1M)

(a) (− 3, − 2) excluding − 2. 5 (b) [0, 1] excluding 0.5 (c) (−2, 1) excluding 0 (d) None of these

Match the conditions / expressions in Column I with statement in Column II.

11. Let f (x) =

x2 − 6 x + 5 . x2 − 5 x + 6

(2007, 6M)

Column I

Column II

A.

If −1 < x < 1, then f ( x ) satisfies

p.

0 < f (x ) < 1

1 (b)  0,   2 

B.

If 1 < x < 2 , then f ( x ) satisfies

q.

f (x ) < 0

C.

If 3 < x < 5, then f ( x ) satisfies

r.

f (x ) > 0

(d) (0, 1]

D.

If x > 5, then f ( x ) satisfies

s.

f (x ) < 1

(2001, 1M)

1 (c)  , 1  2 

(a) 0 < x ≤ 1 (c) − ∞ < x ≤ 0

(b) (1, 11/7) (d) (1, 7/5)

6. Let f (x) = (1 + b2) x2 + 2bx + 1 and let m (b) be the

(a) [0, 1]

the equation 2x + 2y = 2 , is

Match the Columns

x2 + x + 2 5. Range of the function f (x) = 2 ; x ∈ R is x + x+1 (2003, 2M) (a) (1, ∞) (c) (1, 7/3]

(2001, 1M)

(a) R / {− 1, − 2} (b) (− 2, ∞ ) (c) R / {− 1, − 2, − 3} (d) (− 3, ∞ ) / {− 1, − 2}

8. The domain of definition of the function y (x) is given by

3. Let f (x) = a x (a > 0) be written as f (x) = f1 (x) + f2(x),

(a) 2f1 (x + y) ⋅ f2 (x − y) (c) 2f1 (x) ⋅ f2 ( y)

log 2(x + 3) is x2 + 3x + 2

172 Functions True/False

Objective Question II (One or more than one correct option) 2x − 1 is 12. If S is the set of all real x such that 3 2 x + 3 x2 + x positive, then S contains (1986, 2M) 3 (a)  − ∞ , −   2

3 1 (b)  − , −   2 4

1 1 (c)  − ,   4 2

1 (d)  , 3 2 

16. If f1 (x) and f2(x) are defined on domains D1 and D2 respectively, then f1 (x) + f2(x) is defined on D1 ∩ D2. (1988, 1M)

Analytical & Descriptive Questions 17. Find the range of values of t for which 2 sin t =

Fill in the Blanks

y=

18. Let

 4 − x2   , then the domain of  1−x   

13. If f (x) = sin log  f (x) is ……… .

1 − 2 x + 5 x2  π π . , t∈ − ,  2 2  3 x2 − 2 x − 1

(2005, 2M)

(x + 1) (x − 3) . (x − 2)

Find all the real values of x for which y takes real values. (1980, 2M) (1985, 2M)

 x2  14. The domain of the function f (x) = sin −1  log 2  is 2  given by ... (1984, 2M)

 π2  15. The values of f (x) = 3 sin  − x2 lie in the   16  interval ……… (1983, 2M)

Integer & Numerical Answer Type Question 19. The value of the limit 4 2 (sin 3x + sin x) 3x 5x  3x  + cos  −  2 + 2 cos 2x + cos  2 sin 2x sin  2 2  2 (2020 Adv.) is ……… lim

π x→ 2

Topic 2 Composite of Functions Objective Questions I (Only one correct option)

10

4. Let

1. For a suitable chosen real constant a, let a functin

a −x . Further a+x suppose that for any real number x ≠ − a and f (x) ≠ − a,  1 ( fof )(x) = x. Then, f  −  is equal to  2 [2020 Main, 6 Sep II] f : R − { a } → R be defined by f (x) =

1 (b) − 3 (d) 3

1 (a) 3 (c) −3

3 2. For x ∈ 0,  , let f (x) = x , g (x) = tan x and

 2 1 − x2  π . If φ(x) = ((hof )og )(x), then φ   is equal to h (x) =  3 1 + x2 (2019 Main, 12 April I)

π (a) tan 12 7π (c) tan 12

11π (b) tan 12 5π (d) tan 12

For any define A ⊆ R, f (x) = x , x ∈ R. g ( A ) = { x ∈ R : f (x) ∈ A }. If S = [0, 4], then which one of the following statements is not true?

3. Let

∑ f (a + k) = 16(210 − 1),

where the function f

k =1

satisfies f (x + y) = f (x) f ( y) for all natural numbers x, y and f (1) = 2. Then, the natural number ‘a’ is (2019 Main, 9 April I)

(a) 2

(b) 4

(c) 3

 1 − x  ,|x| < 1, then  1 + x

5. If f (x) = log e 

 2x  f  is equal to  1 + x2 

(b) 2f (x2 ) (d) −2f (x)

(a) 2f (x) (c) (f (x))2

x ∈ R − {0, 1},

6. For

(d) 16

let

f1 (x) =

(2019 Main, 8 April I)

1 , f2(x) = 1 − x x

and

1 be three given functions. If a function, J (x) 1 −x satisfies ( f2° J ° f1 )(x) = f3 (x), then J (x) is equal to f3 (x) =

(2019 Main, 9 Jan I)

(a) f2 (x)

(b) f3 (x) 1 (d) f3 (x) x

(c) f1 (x)

2

(2019 Main, 10 April I)

(a) f ( g (S)) = S (c) g (f (S)) = g (S)

(b) g (f (S)) ≠ S (d) f(g(S)) ≠ f (S)

7. Let a , b, c ∈ R. If f (x) = ax2+ bx + c be such that a + b + c = 3 and f (x + y) = f (x) + f ( y) + xy, ∀ x, y ∈ R, 10

then

∑ f (n ) is

equal to

n =1

(a) 330

(b) 165

(2017 Main)

(c) 190

(d) 255

Functions 173 8. Let f (x) = x2 and g (x) = sin x for all x ∈ R. Then, the set of all x satisfying ( fog )(x) = f ( g (x)), is (a) (b) (c) (d)

( fogogof )(x) = ( gogof )(x) ,

where (2011)

± nπ, n ∈ {0, 1, 2, K } ± nπ, n ∈ {1, 2, K } π /2 + 2nπ, n ∈ {...,− 2, − 1, 0, 1, 2, K } 2nπ, n ∈ {..., − 2, − 1, 0, 1, 2, K }

αx , x ≠ − 1. Then, for what value of α is x+1 (2001, 1M) f [ f (x)] = x ? (b) −

2

(d) −1

(c) 1

− 1 , x < 0  10. Let g (x) = 1 + x − [x] and f (x) =  0, x = 0 , then for all  1, x > 0  (2001, 1M) x, f [ g (x)] is equal to (a) x

(b) 1

(c) f (x)

(d) g (x)

11. If g { f (x) } = |sin x| and f { g (x) } = (sin x )2, then (1998, 2M)

(a) f (b) f (c) f (d) f

(x) = sin 2 x, g (x) = x (x) = sin x, g (x) = | x| (x) = x2 , g (x) = sin x and g cannot be determined



1   x





has the value

(1983, 1M)

1 2 (d) None of these (b)

(c) − 2

13. Let f (x) = | x − 1|. Then, (a) f (x ) = {f (x) } (c) f (| x|) = | f (x)| 2

2

(1991, 2M)

( π / 2) = − 1 ( π) = 1 (− π ) = 0 ( π / 4) = 1

the equilateral triangle with two of its vertices at (0, 0) and [x, g (x)] is 3 / 4, then the function g (x) is (1989, 2M)

(a) g (x) = ±

1 − x2

(b) g (x) = 1 − x2

(c) g (x) = − 1 − x2

(d) g (x) = 1 + x2

17. If y = f (x) =

x+2 , then x−1

(1984, 3M)

(a) x = f ( y) (b) f (1) = 3 (c) y increases with x for x < 1 (d) f is a rational function of x

Fill in the Blank π π   f (x) = sin 2 x + sin 2 x +  + cos x cos  x +  and   3 3  5 g   = 1, then (g o f ) (x) = ... .  4 (1996, 2M)

18. If

12. If f (x) = cos(log x), then f (x) ⋅ f ( y) −  f   + f (xy) 2  y (a) −1

greatest integer function, then (a) f (b) f (c) f (d) f

16. Let g (x) be a function defined on [− 1, 1]. If the area of

9. Let f (x) = (a) 2

15. If f (x) = cos [π 2] x + cos [− π 2] x, where [x] stands for the

True/False 19. If f (x) = (a − xn )1/ n, where a > 0 and n is a positive integer, then f [ f (x)] = x.

(1983, 1M)

(b) f (x + y) = f (x) + f ( y) (d) None of the above

Objective Questions II (One or more than one correct option) π π  f (x) = sin  sin  sin x  for all x ∈ R and   2 6 π g (x) = sin x for all x ∈ R. Let ( fog )(x) denotes f { g (x)} 2 and (gof ) (x) denotes g{f (x)}. Then, which of the following is/are true? (2015 Adv.)

14. Let

1 1 1 1 (a) Range of f is  − ,  (b) Range of fog is  − ,   2 2   2 2  f (x ) π (c) lim = x → 0 g (x ) 6 (d) There is an x ∈ R such that ( gof ) (x) = 1

(1983, 1M)

Analytical & Descriptive Question 20. Find the natural number a for which n

∑

f (a + k) = 16 (2n − 1),

k =1

where the function f satisfies the relation f (x + y) = f (x) f ( y) for all natural numbers x , y and further f (1) = 2. (1992, 6M)

Integer & Numerical Answer Type Question 21. Let the function f : [0, 1] → R be defined by f (x) =

4x 4x + 2

Then the value of  39 3 2 1  1 f   + f   + f   + ... + f   − f   is ………  40  40  40  40  2

(2020 Adv.)

174 Functions

Topic 3 Types of Functions Objective Questions I (Only one correct option) 1. If the function f : R → R is defined by f (x) =|x|(x − sin x), then which of the following statements is TRUE? (a) f is one-one, but NOT onto (b) f is onto, but NOT one-one (c) f is BOTH one-one and onto (d) f is NEITHER one-one NOR onto

2. If the function f (x) =

(2020 Adv.)

f : R − {1, − 1} → A

defined

by

 2 2

x , is surjective, then A is equal to 1 − x2 (2019 Main, 9 April I)

(2017 Main)

9. The function f: [0, 3] → [1, 29], defined by f (x) = 2x3 − 15 x2 + 36x + 1, is

0, if x is rational x, if x is rational , g (x) =  0 if is irrational , x x, if x is irrational 

1 . Then, f is x

(2019 Main, 11 Jan II)

injective only both injective as well as surjective not injective but it is surjective neither injective nor surjective

2, 3, … , 20} such that f (k) is a multiple of 3, whenever k (2019 Main, 11 Jan II) is a multiple of 4, is (b) 56 × 15 (d) 65 × (15)!

5. Let f : R → R be defined by f (x) = x ∈ R. Then, the range of f is

1 1 (a)  − ,   2 2  1 1 (c) R −  − ,   2 2 

x , 1 + x2 (2019 Main, 11 Jan I)

(b) (−1, 1) − {0} (d) R − [−1, 1]

6. Let N be the set of natural numbers and two functions f and g be defined as f , g : N → N such that n + 1 ; if n is odd  f (n ) =  2 n  if n is even ;  2 and g (n ) = n − (−1)n. Then, fog is (2019 Main, 10 Jan II) (a) one-one but not onto (c) both one-one and onto

Then, f − g is

(2005, 1M)

(a) one-one and into (c) many one and onto

(b) onto but not one-one (d) neither one-one nor onto

(b) neither one-one nor onto (d) one-one and onto

x , then f is 1+ x (2003, 2M)

11. If f : [0, ∞ ) → [0, ∞ ) and f (x) = (a) one-one and onto (c) onto but not one-one

4. The number of functions f from {1, 2, 3, … , 20} onto {1, (a) (15)! × 6! (c) 5! × 6!

(2012)

(a) one-one and onto (b) onto but not one-one (c) one-one but not onto (d) neither one-one nor onto

10. f (x) = 

3. Let a function f : (0, ∞ ) → (0, ∞ ) be defined by f (x) = 1 −

x is 1 + x2

(a) invertible (b) injective but not surjective (c) surjective but not injective (d) neither injective nor surjective

2

(a) R − {−1} (b) [0, ∞ ) (c) R − [−1, 0) (d) R − (−1, 0)

(a) (b) (c) (d)

1 1 8. The function f : R → − ,  defined as f (x) =

(b) one-one but not onto (d) neither one-one nor onto

12. Let function f : R → R be defined by f (x) = 2 x + sin x for x ∈ R . Then, f is

(2002, 1M)

(a) one-to-one and onto (b) one-to-one but not onto (c) onto but not one-to-one (d) neither one-to-one nor onto

13. Let E = {1, 2, 3, 4} and F = {1, 2}. Then, the number of onto functions from E to F is (a) 14

(b) 16

(2001, 1M)

(c) 12

(d) 8

Match the Columns Match the conditions/expressions in Column I with statement in Column II.

14. Let f1 : R → R, f2 : [0, ∞ ] → R, f3 : R → R and f4 : R → [0, ∞ ) be defined by

(2014 Adv.)

sin x, if x < 0 |x|, if x < 0 ; f2(x) = x2; f3 (x) =  f1 (x) =  x if e , x ≥ 0  x, if x ≥ 0   f [ f (x)], if x < 0 and f4 (x) =  2 1  f2[ f1 (x)] − 1, if x ≥ 0 Column I

Column II

A.

f4 is

p.

onto but not one-one

7. Let A = { x ∈ R : x is not a positive integer}. Define a

B.

f3 is

q.

neither continuous nor one-one

2x , then f is x−1 (2019 Main, 9 Jan II)

C.

f2of1 is

r.

differentiable but not one-one

D.

f2 is

s.

continuous and one-one

function f : A → R as f (x) =

(a) injective but not surjective (b) not injective (c) surjective but not injective (d) neither injective nor surjective

Codes A B C D (a) r p s q (c) r p q s

A B C D (b) p r s q (d) p r q s

Functions 175 15. Let the functions defined in Column I have domain (−π /2, π /2) and range (− ∞ , ∞ ) Column I A. 1 + 2 x B. tan x

(1992, 2M)

18. The function f (x) =

Column II p. q. r. s.

True/False

onto but not one-one one-one but not onto one-one and onto neither one-one nor onto

x2 + 4x + 30 is not one-to-one. x2 − 8x + 18 (1983, 1M)

Analytical & Descriptive Question 19. A function f : IR → IR, where IR, is the set of real

Objective Question II (One or more than one correct option)

numbers, is defined by f (x) =

π π 16. Let f :  − ,  → R be given by

αx2 + 6x − 8 . α + 6 x − 8 x2

Find the interval of values of α for which is onto. Is the functions one-to-one for α = 3 ? Justify your answer.

 2 2 f (x) = [log(sec x + tan x)]3 . Then,

(1996, 5M)

(a) f (x) is an odd function (b) f (x) is a one-one function (c) f (x) is an onto function (d) f (x) is an even function

Fill in the Blanks 17. There are exactly two distinct linear functions, …, and… which map {– 1, 1} onto {0, 2}.

20. Let A and B be two sets each with a finite number of elements. Assume that there is an injective mapping from A to B and that there is an injective mapping from B to A. Prove that there is a bijective mapping from A to B. (1981, 2M)

(1989, 2M)

Topic 4 Inverse and Periodic Functions Objective Questions I (Only one correct option)

whose graph is reflection of the graph of f (x) with respect to the line y = x, then g (x) equals

1. The inverse function of f (x) =

−2x

8 −8 , x ∈ (− 1, 1) is 82x + 8−2x 2x

(a)

1− x 1 (log 8 e) log e   4 1 + x

(b)

1− 1 log e  4 1+

(2002, 1M) [2020 Main 8 Jan I]

(a) −

x − 1, x ≥ 0

(c) x + 1 , x ≥ − 1

(b)

1 (x + 1)2

,x> −1

(d) x − 1, x ≥ 0

1 x

x  x

1 + x 1 (c) (log 8 e) log e   4 1− x

4. Suppose f (x) = (x + 1)2 for x ≥ − 1. If g (x) is the function

5. If f : [1, ∞ ) → [2, ∞ ) is given by f (x) = x + , then f −1(x) equals

1 + x 1 (d) log e   4 1− x

(a)

2. If X and Y are two non-empty sets where f : X → Y , is function is defined such that f (C ) = { f (x) : x ∈ C } for C ⊆ X (2005, 1M)

3. If f (x) = sin x + cos x, g (x) = x2 − 1, then g { f (x) } is (2004, 1M)

π (a)  0,   2 

π π (b)  − ,   4 4 

π π (c)  − ,   2 2 

(d) [0 , π ]

x−

x2 − 4 2

(b)

x 1 + x2

(d) 1 +

x2 − 4

f : [1, ∞ ) → [1, ∞ ) is f (x) = 2x ( x − 1), then f −1 (x) is

(a) f −1 {f (A )} = A (b) f −1 {f (A )} = A, only if f (X ) = Y (c) f {f −1 (B )} = B, only if B ⊆ f (x) (d) f {f −1 (B )} = B

invertible in the domain

x −4 2

6. If the function

and f −1 (D ) = { x : f (x) ∈ D } for D ⊆ Y , for any A ⊆ Y and B ⊆ Y , then

(c)

x+

(2001, 1M) 2

x ( x − 1)

1 (a)    2 1 (c) (1 − 1 + 4 log2 x ) 2

(b)

1 (1 + 2

defined

by

(1999, 2M)

1 + 4 log2 x )

(d) not defined

7. If f (x) = 3x − 5, then f −1 (x) 1 (a) is given by 3x − 5 x+ 5 (b) is given by 3 (c) does not exist because f is not one-one (d) does not exist because f is not onto

(1998, 2M)

176 Functions 8. Which of the following functions is periodic?

(1983, 1M)

(a) f (x) = x − [ x ], where [x] denotes the greatest integer less than or equal to the real number x (b) f (x) = sin (1 /x) for x ≠ 0, f (0) = 0 (c) f (x) = x cos x (d) None of the above

Objective Question II (One or more than one correct option) 9. Let f : (0, 1) → R be defined by f (x) = constant such that 0 < b < 1. Then,

b−x , where b is a 1 − bx

(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I. (c) Statement I is true; Statement II is false. (d) Statement I is false; Statement II is true.

10. Let F (x) be an indefinite integral of sin 2 x. Statement I The function F (x) satisfies F (x + π ) = F (x) for all real x. Because Statement II sin 2(x + π ) = sin 2 x, for all real x.

(2011)

(2007, 3M)

(a) f is not invertible on (0, 1) 1 f ′ (0) 1 on (0, 1) and f ′ (b) = f ′ (0)

Analytical & Descriptive Question

(b) f ≠ f −1 on (0, 1) and f ′ (b) = (c) f = f −1 (d) f

−1

11. Let f be a one-one function with domain { x, y, z } and range {1, 2, 3}. It is given that exactly one of the following statements is true and the remaining two are false f (x) = 1, f ( y) ≠ 1, f (z ) ≠ 2 determine f −1 (1).

is differentiable on (0, 1)

(1982, 2M)

Assertion and Reason For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows. (a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I.

12. If f is an even function defined on the interval (− 5, 5), then four real values of x satisfying the equation  x + 1 f (x) = f   are ………. .  x + 2 (1996, 1M)

Answers Topic 1

Topic 3

1. (a)

2. (c)

3. (d)

4. (a)

1. (c)

2. (c)

3. (d)

4. (a)

5. (c)

6. (d)

7. (d)

8. (d)

5. (a)

6. (b)

7. (a)

8. (c) 12. (a)

9. (c) 12. (a,d) 15.

 3  0, 2  

10. (c)

11. A→ p; B→ q; C→ q; D→ p

9. (b)

10. (d)

11. (b)

13. (–2,1)

14. Domain ∈ [ −2,−1 ] ∪ [1, 2 ]

13. (a)

14. (d)

15. A → q; B → r

16. True

 π π  3π π  17. t ∈ − , ∪ ,  2 10   10 2 

16. (a, b, c)

19. (8)

18. True

18. x ∈ [ −1, 2 ) ∪ [3, ∞ )

19. 2 ≤ α ≤ 14, No

Topic 2 1. (d)

2. (b)

3. (c)

4. (c)

5. (a)

6. (b)

7. (a)

8. (b)

9. (d)

10. (b)

11. (a)

12. (d)

13. (d)

14. (a,b,c)

15. (a, c)

16. (b, c)

17. (a, d)

18. 1

19. True

20. (a = 3)

21. (19)

17. y = x + 1 and y = − x + 1

Topic 4 1. (c)

2. (c)

3. (b)

4. (d)

5. (a)

6. (b)

7. (b)

8. (a)

9. (b) 11. f

−1

(1 ) = y

10. (d)  ± 3 ± 5 12.   2  

Hints & Solutions –1 1 π  – π ≤ 2x ≤ sin  ⇒  ≤ 2x ≤ sin  6  2 2 2 –1 1 ≤x≤ 4 2 – 1 1 x∈ , Q  4 2 

Topic 1 Classification of Functions, Domain and Range 1. Given relation R = {(x, y) : x, y ∈ Z , x2 + 3 y2 ≤ 8} For, y2 = 0, x2 = 0, 1, 4 For, y2 = 1, x2 = 0, 1, 4 For, y2 = 4, x2 ∈ φ ∴ Range of R is possible values of y = { − 1, 0, 1}

5. Let y = f (x) =

∴ Domain of R−1 = Range of R = { − 1, 0, 1} 1 2. Given function f (x) = + log10 (x3 − x) 4 − x2 For domain of f (x) 4 − x2 ≠ 0 ⇒ x ≠ ± 2 and x3 − x > 0 ⇒ x(x − 1)(x + 1) > 0 From Wavy curve method, + –∞

–

–1

+ 0

–

+1

x ∈ (−1, 0) ∪ (1, ∞ )

x2 + x + 2 x2 + x + 1 1 y=1 + 2 x + x+1

∴

…(i)

+∞

…(ii)

From Eqs. (i) and (ii), we get the domain of f (x) as (−1, 0) ∪ (1, 2) ∪ (2, ∞ ).

y=

So, f1 (x + y) + f1 (x − y) 1 1 = [a x + y + a − ( x + y ) ] + [a x − y + a − ( x − y ) ] 2 2 x  1 1 a ay  = a x a y + x y + y + x  2 a a a a 

2

1 is positive 1 + b2 and m (b) varies from 1 to 0, so range = (0, 1] log (x + 3) log 2 (x + 3) 7. Given, f (x) = 2 2 = (x + 3x + 2) (x + 1) (x + 2) m (b) = minimum value of f (x) =

For numerator, x + 3 > 0 ⇒

x> −3

⇒

From Eqs. (i) and (ii), Domain is (− 3 , ∞ ) /{ − 1, − 2}

π , to find domain we must 6

have, π sin (2x) + ≥ 0 6 π π − ≤ sin −1 (2x) ≤ 6 2 −1

π π  but − ≤ sin −1 θ ≤  2 2 

…(ii)

 b  b2 = (1 + b2)  x +  +1− 2  1+ b  1 + b2

1 1  1 = ax + x  ay + y  2 a  a 

4. Here, f (x) = sin −1 (2x) +

…(i)

6. Given, f (x) = (1 + b2) x2 + 2bx + 1

1  1 1  1  = a x  a y + y  + x  y + a y    2  a  a a

 a x + a −x   a y + a −y  =2  = 2 f1 (x) ⋅ f1 ( y)  2 2   

[i.e. y > 1]

⇒ yx2 + yx + y = x2 + x + 2 2 ⇒ x ( y − 1) + x ( y − 1) + ( y − 2) = 0, ∀ x ∈ R Since, x is real, D ≥ 0 ⇒ ( y − 1 )2 − 4 ( y − 1 ) ( y − 2 ) ≥ 0 ⇒ ( y − 1) {( y − 1) − 4 ( y − 2)} ≥ 0 ⇒ ( y − 1) (− 3 y + 7) ≥ 0 7 1≤ y≤ ⇒ 3  7 From Eqs. (i) and (ii), Range ∈ 1 ,  3 

3. Given, function f (x) = a x , a > 0 is written as sum of an even and odd functions f1 (x) and f2(x) respectively. a x + a −x a x − a −x and f2(x) = Clearly, f1 (x) = 2 2

x2 + x + 2 , x ∈R x2 + x + 1

…(i)

and for denominator, (x + 1) (x + 2) ≠ 0 x ≠ − 1, − 2

8. Given, 2x + 2y = 2, ∀ x , y ∈ R 2x , 2y > 0, ∀ x , y ∈ R

But

Therefore, 2x = 2 − 2y < 2 ⇒ 0 < 2x < 2 Taking log on both sides with base 2, we get log 2 0 < log 2 2x < log 2 2 ⇒ − ∞ < x < 1

9. It is given, f (θ ) = sin θ (sin θ + sin 3 θ ) = (sin θ + 3 sin θ − 4 sin3 θ ) sin θ

…(ii)

178 Functions = (4 sin θ − 4 sin3 θ ) sin θ = sin 2 θ (4 − 4 sin 2 θ ) = 4 sin 2 θ cos 2 θ = (2 sin θ cos θ )2 = (sin 2θ )2 ≥ 0 which is true for all θ.

⇒

1 x2 ≤ ≤2 2 2

⇒

1 ≤ x2 ≤ 4

⇒

1 ≤ |x| ≤ 2

⇒

10. For domain of y, and 1 − x > 0, 1 − x ≠ 1 and ⇒ x < 1, x ≠ 0 ⇒ −2 < x < 1 excluding 0 ⇒ x ∈ (−2, 1) − {0} (x − 1) (x − 5) 11. Given, f (x) = (x − 2) (x − 3)

x+ 2 >0 x > −2

Domain ∈ [−2, − 1] ∪ [1, 2]

15. Given, f (x) = 3 sin

 π π ⇒ Domain ∈ − ,  4 4   π2  − x2  ⋅ ∴ For range, f ′ (x) = 3 cos   16   

The graph of f (x) is shown as : Y

Where, y=1 X'

0

1

2

X

5

3

Y'

⇒

C. If 3 < x < 5 ⇒ f (x) < 0

(2x − 1) >0 x(2x2 + 3x + 1)

⇒

(2x − 1) >0 x (2x + 1) (x + 1)

−∞

−1

−1/2

0

+ 1/2

∞

x ∈ (−∞ , − 1) ∪ (−1/2, 0) ∪ (1 / 2, ∞ ) Hence, (a) and (d) are the correct options.  4 − x2   13. Given, f (x) = sin log   1−x   

and

x 0 , 4 − x2 > 0 and 1 − x ≠ 0 1−x (1 − x) > 0

 3  range ∈ 0,  2 

Since, x ∈ R − {1, − 1 / 3}

2

⇒

π 3 = 4 2

Thus, domain of [ f1 (x) + f2(x)] is D1 ∩ D2 . Hence, given statement is true.

Put

⇒

x=0

16. Since, domains of f1 (x) and f2(x) are D1 and D2 .

−

+

=0

 π  π f −  = f   =0  4  4

Hence,

Hence, the solution set is,

For domain,

f (0) = 3 sin

and

D. If x > 5 ⇒ 0 < f (x) < 1 2x − 1 12. Since, >0 3 2 x + 3 x2 + x ⇒

or

π2 2 − x2 16

x =0

Thus,

B. If 1 < x < 2 ⇒ f (x) < 0

−

 π2  cos  − x2  = 0  16   

1(−2x)

  π2  π2 π2 − x2  = 0 ⇒ − x2 = neglecting cos    16 4   16    3π 2 2 , never possible ⇒ x = −    16

A. If − 1 < x < 1 ⇒ 0 < f (x) < 1

+

π2 − x2 16

4 − x >0 2

| x| < 2 ⇒ −2 < x < 1

∴

D ≥0

⇒ ⇒

4( y − 1) + 4(3 y − 5) ( y + 1) ≥ 0 y2 − y − 1 ≥ 0

⇒

1 5  y−  − ≥0  2 4

⇒

 1 5  1 5 y− −  y− +  ≥0 2 2  2 2 

2

2

⇒

y≤

1− 5 2

Functions 179 1+ 5 2 1− 5 2 sin t ≤ 2 1+ 5 2 sin t ≥ 2  π sin t ≤ sin  −   10

Topic 2 Composite of Functions and Even, Odd Functions

y≥

or ⇒ or ⇒

1. For a given function f : R − { −a } → R defined by f (x) =

⇒

 3π  sin t ≥ sin    10 

or

π 3π t≤− t≥ or ⇒ 10 10 π  3π π   π . Hence, range of t is − , − ∪ ,  2 10   10 2 

18. Since, y =

(x + 1) (x − 3) ≥0 (x − 2)

when −

−∞

⇒

−1 ≤ x < 2

∴

− 2

or

+ 3

x≥3

19. The limit π x→ 2

4 2 (sin 3x + sin x) 3x 3x 5x   + cos  −  2 + 2 cos 2x + cos  2 sin 2x sin  2 2 2 

= lim

π x→ 2

4 2 (2 sin 2x cos x) 3x  5x 3x 2 sin 2x sin +  cos − cos  − 2 (1 + cos 2x)  2 2 2

8 2 sin 2x cos x x 3x − 2 sin 2x sin − 2 (2 cos 2 x) 2 sin 2x sin 2 2 4 2 sin 2x cos x = lim π 3x x  x → sin 2 x sin − sin  − 2 cos 2 x  2  2 2 = lim

π x→ 2

4 2 sin 2x cos x x 2 sin 2x cos x sin − 2 cos 2 x 2 4 2 sin 2x = lim π x x → 2 sin 2 x sin − 2 cos x 2 2 8 2 sin x = lim π x x → 4 sin x sin − 2 2 2 8 2 16 = = = 8. 4 −2 4 − 2 2 = lim

π x→ 2

a−x a+x =x a−x a+ a+x a−

a 2 + ax − a + x =x a 2 + ax + a − x

⇒

a 2 + ax − a + x = a 2x + ax2 + ax − x2

⇒ So,

∞

x ∈ [−1, 2) ∪ [3, ∞ ).

lim

 a − x f  =x ⇒  a + x

a (a − 1) = (a 2 − 1)x + x2(a − 1) (a − 1) [x2 + (a + 1)x − a ] = 0 ⇒ a = 1 1−x  1 1 + (1 /2) f (x) = =3 ∴ f −  =  2 1 − (1 /2) 1+ x

2. Given, for x ∈ (0, 3 / 2), functions

+ −1

( fof ) (x) = x

⇒

⇒

(x + 1) (x − 3) takes all real values only (x − 2)

a−x . Q a+x

and

f (x) = x g (x) = tan x 1 − x2 h (x) = 1 + x2

… (i) … (ii) … (iii)

Also given, φ(x) = ((hof )og )(x) = (hof ) ( g (x)) = h ( f ( g (x))) = h ( f (tan x)) 1 − ( tan x )2 = h ( tan x ) = 1 + ( tan x )2 1 − tan x π  = = tan  − x 4  1 + tan x Now,

 π  π π φ   = tan  −   3  4 3  3π − 4π   π = tan   = tan  −   12   12 π π   11π  = − tan   = tan  π −  = tan    12   12  12

3. Given, functions f (x) = x2, x ∈ R and g ( A ) = { x ∈ R : f (x) ∈ A }; A ⊆ R Now, for S = [0, 4] g (S ) = { x ∈ R : f (x) ∈ S = [0, 4]} = { x ∈ R : x2 ∈ [0, 4]} = { x ∈ R: x ∈ [−2, 2]} ⇒ g (S ) = [−2, 2] So, f ( g (S )) = [0, 4] = S Now, f (S ) = { x2 : x ∈ S = [0, 4]} = [0, 16] and g ( f (S )) = { x ∈ R : f (x) ∈ f (S ) = [0, 16]} = { x ∈ R : f (x) ∈ [0, 16]} = { x ∈ R: x2 ∈ [0, 16]} = { x ∈ R : x ∈ [−4, 4]} = [−4 ,4] From above, it is clear that g ( f (S )) = g (S ).

180 Functions 4. Given,

−1 J (X ) = X 1 1− X −1 1 = = X −1 1 − X

f (x + y) = f (x) ⋅ f ( y) f (x) = λ f (1) = 2

[where λ > 0] (given)

x

Let Q ∴ λ =2

10  10  Σ f (a + k) = Σ λa+ k = λa  Σ λk  k=1  k =1 k =1 10

So,

= 2a [21 + 22 + 23 + ......+210 ] 10 2(2 − 1)  = 2a    2 −1  [by using formula of sum of n-terms of a GP having first term ‘a’ and common ratio ‘r’, is  a (r n − 1) Sn = , where r > 1 r −1  ⇒ ⇒

7. We have, f (x) = ax2 + bx + c Now, f (x + y) = f (x) + f ( y) + xy Put y = 0 ⇒ f (x) = f (x) + f (0) + 0 ⇒ f (0) = 0 ⇒ c=0 Again, put y = − x ∴ f (0) = f (x) + f (− x) − x2 ⇒ 0 = ax2 + bx + ax2 − bx − x2 ⇒

2ax2 − x2 = 0 1 a= ⇒ 2 Also, a + b + c = 3 5 1 ⇒ + b + 0 =3⇒ b = 2 2 x2 + 5 x ∴ f (x) = 2

 1 − x  ,|x| < 1, then  1 + x

5. Given, f (x) = log e 

  2x    < 1 Q 2  1 + x  

 1 + x2 − 2 x    2  (1 − x)2   1 − x 1 + x2  = log e  = = log log     e e  1 + x2 + 2x   1 + x  (1 + x)2   2  1+ x 

∑

f (n ) =

n 2 + 5n 1 2 5 = n + n 2 2 2

f (n ) =

1 2

[Q f2(x) = 1 − x and f3 (x) = 1  1 1 − J  =  x 1 − x 1  1 J  = 1 −  x 1−x = 1 = X , then x

1 − x−1 −x = 1−x 1−x

1 ] 1−x

1 [Q f1 (x) = ] x

∑ n2 +

n =1

5 2

10

∑n

n =1

1 10 × 11 × 21 5 10 × 11 ⋅ + × 2 6 2 2 385 275 660 = + = = 330 2 2 2 f (x) = x2, g (x) = sin x

8.

1 1 f1 (x) = , f2(x) = 1 − x and f3 (x) = x 1−x Also, we have ( f2 o J o f1 )(x) = f3 (x) ⇒ f2((J o f1 )(x)) = f3 (x) ⇒ f2(J ( f1 (x)) = f3 (x) 1 ⇒ 1 − J ( f1 (x)) = 1−x

10

=

  1 − x  Q f (x) = log e  1 + x   

6. We have,

Now, put

10

∴

[Q log e| A|m = m log e| A|]

= 2 f (x)

⇒

Now,

n =1

 1 − x = 2 log e    1 + x

⇒

J (X ) = f3 (X ) or J (x) = f3 (x)

⇒

2a+ 1 (210 − 1) = 16 (210 − 1) (given) 2a+ 1 = 16 = 24 ⇒ a + 1 = 4 ⇒ a = 3

2x   1 −   2x  1 + x2  f  = log e  2  1 + 2x  1 + x     1 + x2 

1  Qx=  X 

( gof )(x) = sin x2 go( gof ) (x) = sin (sin x2) ( fogogof ) (x) = (sin (sin x2))2

…(i)

Again, ( gof ) (x) = sin x2 ( gogof ) (x) = sin (sin x2) Given, ⇒

…(ii)

( fogogof ) (x) = ( gogof ) (x) (sin (sin x2))2 = sin (sin x2)

⇒ sin (sin x2) {sin (sin x2) − 1} = 0 ⇒ sin (sin x2) = 0 or sin (sin x2) = 1 π ⇒ sin x2 = 0 or sin x2 = 2 ∴

x2 = n π [sin x2 = x=±

π is not possible as − 1 ≤ sin θ ≤ 1] 2

nπ

Functions 181 9. Given, f (x) =

αx x+1

 αx  α   x + 1  αx  f [ f (x) ] = f   = αx  x + 1 +1 x+1 α 2x α 2x x+1 = = = x [given] …(i) α x + (x + 1) (α + 1) x + 1 x+1 2 ⇒ α x = (α + 1) x2 + x ⇒ x [α 2 − (α + 1) x − 1] = 0 ⇒ x(α + 1)(α − 1 − x) = 0 ⇒ α − 1 = 0 and α + 1 = 0 ⇒ α = −1 But α = 1 does not satisfy the Eq. (i).

10. g (x) = 1 + x − [x] is greater than 1 since x − [x] > 0 f [ g (x)] = 1, since f (x) = 1 for all x > 0 f (x) = sin 2 x and g (x) = x

11. Let Now,

fog (x) = f [ g (x)] = f ( x ) = sin 2 x

and

gof (x) = g [ f (x)] = g (sin 2 x) = sin 2 x = |sin x|

Again, let f (x) = sin x , g (x) = | x| fog (x) = f [ g (x)] = f (| x|) = sin| x|≠ (sin x )2 f (x) = x2, g (x) = sin x

When

fog (x) = f [ g (x)] = f (sin x ) = (sin x )2 ( gof ) (x) = g [ f (x)] = g (x2) = sin x2

and

= sin| x|≠|sin x|

12. Given, f (x) = cos (log x) ∴ f (x) ⋅ f ( y) −

 1   x f   + f (xy) 2   y 

1 [cos (log x − log y) 2 + cos(log x + log y)] 1 = cos (log x) ⋅ cos (log y) − [(2 cos (log x) ⋅ cos (log y)] 2 = cos (log x) ⋅ cos (log y) − cos (log x) ⋅ cos (log y) = 0 = cos (log x) ⋅ cos(log y) −

13.

Given, ∴ and ⇒

f (x) = |x − 1| f (x ) = |x − 1| 2

2

{ f (x)}2 = (x − 1)2 f (x2) ≠ ( f (x))2, hence (a) is false.

Also, f (x + y) = |x + y − 1| and ⇒

f (x) = |x − 1|, f ( y) = | y − 1| f (x + y) ≠ f (x) + f ( y), hence (b) is false.

f (|x|) = ||x| − 1| and

| f (x)| = ||x − 1|| = |x − 1|

∴

f (|x|) ≠| f (x)|, hence (c) is false. π 6

π 2

 

14. f (x) = sin  sin  sin x  , x ∈ R π π   π π = sin  sin θ , θ ∈ − , , where θ = sin x  2 2  6  2 π  π π ,where α = sin θ = sin α, α ∈ − ,  6 6  6  1 1 f (x) ∈ − , ∴  2 2   1 1 Hence, range of f (x) ∈ − ,  2 2  So, option (a) is correct.  π π  1 1 (b) f { g (x)} = f (t ), t ∈ − , ⇒ f (t ) ∈ − ,  2 2   2 2  ∴

Option (b) is correct. π  π sin  sin  sin x   2 f (x) 6  (c) lim = lim π x → 0 g (x) x→ 0 (sin x) 2 π  π π  π sin  sin  sin x  sin  sin x  6   2 6 2  = lim ⋅ x→ 0 π  π π  sin  sin x  sin x  2 2  6 π π =1 × ×1 = 6 6 ∴Option (c) is correct. (d) g{ f (x)} = 1 π sin { f (x)} = 1 ⇒ 2 2 ...(i) ⇒ sin { f (x)} = π  1 1  π π  But f (x) ∈ − , ⊂ − ,  2 2   6 6 

 1 1 sin { f (x)} ∈ − ,  2 2  2 ⇒ sin { f (x)} ≠ , π i.e. No solution. ∴ Option (d) is not correct. ∴

...(ii) [from Eqs. (i) and (ii)]

15. Since, f (x) = cos [π 2] x + cos [−π 2] x ⇒ ∴

f (x) = cos (9) x + cos (−10) x [using [π 2] = 9 and [− π 2] = − 10] 9π  π f   = cos + cos 5π = − 1  2 2 f (π ) = cos 9π + cos 10π = − 1 + 1 = 0 f (− π ) = cos 9π + cos 10π = − 1 + 1 = 0 9π 10π 1 1  π f   = cos + cos = +0=  4 4 4 2 2

Hence, (a) and (c) are correct options.

182 Functions 16. Since, area of equilateral triangle = ⇒

1 π   = 2 sin (2x + π / 3) ⋅ − sin 2x +  = 0   2  3

3 (BC )2 4

3 3 = ⋅ [x2 + g 2(x)] ⇒ g 2(x) = 1 − x2 4 4

⇒ f (x) = c, where c is a constant. But f (0) = sin 2 0 + sin 2(π / 3) + cos 0 cos π / 3

A

2

 3 1 3 1 5 =  + = + = 2 4 2 4  2 Therefore, ( gof ) (x) = g [ f (x)] = g(5 / 4) = 1 C (x,g(x))

B (0,0)

⇒

g (x) = 1 − x2 or − 1 − x2

Hence, (b) and (c) are the correct options. x+2 17. Given , y = f (x) = x−1 ⇒ ⇒

yx − y = x + 2 ⇒ x( y − 1) = y + 2 y+2 x= ⇒ x = f ( y) y−1

f (x) = (a − xn )1/ n

19. Given, ⇒

f [ f (x)] = [a − {(a − xn )1/ n }n ]1/ n = (xn )1/ n = x

∴

f [ f (x)] = x

Hence, given statement is true.

20. Let f (n ) = 2n for all positive integers n. Now, for n = 1, f (1) = 2 = 2 ! ⇒ It is true for n = 1. Again, let f (k) is true. ⇒

f (k) = 2k, for some k ∈ N .

Here, f (1) does not exist, so domain ∈ R − {1} dy (x − 1) ⋅ 1 − (x + 2) ⋅ 1 3 = =− dx (x − 1)2 (x − 1)2

Again, f (k + 1) = f (k) ⋅ f (1)

⇒ f (x) is decreasing for all x ∈ R − {1}. Also, f is rational function of x. Hence, (a) and (d) are correct options.

Therefore, the result is true for n = k + 1. Hence, by principle of mathematical induction,

18. f (x) = sin 2 x + sin 2(x + π / 3) + cos x cos (x + π / 3) ⇒ ⇒

sin 2 x 3 cos 2 x 2 ⋅ 3 + + sin x cos x 4 4 4

3 cos 2 x − cos x sin x ⋅ 2 2 sin 2 x 3 cos 2 x cos 2 x = sin 2 x + + + 4 4 2 5 5 5 2 2 = sin x + cos x = 4 4 4 +

and

gof (x) = g { f (x)} = g (5 / 4) = 1

Alternate Solution f (x) = sin x + sin (x + π / 3) + cos x cos (x + π / 3) 2

2

⇒ f ′ (x) = 2 sin x cos x + 2 sin (x + π / 3) cos (x + π / 3) − sin x cos (x + π / 3) − cos x sin (x + π / 3) = sin 2x + sin (2x + 2π / 3) − [sin (x + x + π / 3)]  2x − 2x − 2π / 3  2x + 2x + 2π / 3 = 2 sin    ⋅ cos      2 2 − sin (2x + π / 3) = 2 [sin (2x + π / 3) ⋅ cos π / 3] − sin (2x + π / 3)

[from induction assumption]

f (n ) = 2n , ∀ n ∈ N Now,

+ cos x cos (x + π / 3) 2 .  sin x ⋅ 1 cos x 3  f (x) = sin 2 x +  +  2  2 

f (x) = sin 2 x +

[by definition]

= 2k + 1

f (x) = sin 2 x + (sin x cos π / 3 + cos x sin π / 3)2

+ cos x (cos x cos π / 3 − sin x sin π / 3) ⇒

=2 ⋅2 k

n

n

k =1

k =1

n

∑ f (a + k) = ∑ f (a ) f (k) = f (a ) ∑ 2k = f (a ) ⋅

k =1

2 (2 − 1) 2 −1 n

= 2a ⋅ 2 (2n − 1) = 2a + 1 (2n − 1) n

But

∑ f (a + k) = 16 (2n − 1) = 24 (2n − 1)

k =1

Therefore,

a + 1 =4 ⇒ a =3

21. The given function f : [0, 1] → R be define by f (x) = ∴ So, ∴

2 41 − x 4x − = ⇒ ( 1 ) f x = 1−x x + 2 2 + 4x 4 4 +2

f (x) + f (1 − x) =

4x 2 4x + 2 + = 4x + 2 2 + 4x 4x + 2

f (x) + f (1 − x) = 1 1 2     3  39  1 f  + f  + f  +… + f  − f   40  40  40  40  2  1  39  = f   + f    +    40  40    18 + … + f     40

… (i)

 2  38   f  40 + f  40     22    19  21  + f    + f   + f     40    40  40  20  1   + f  − f   2  40  1  1 = {1 + 1 + K + 1 + 1} + f   − f   {from Eq. (i)}  2  2 19 − times = 19.

Functions 183

Topic 3 Types of Functions 1. The given function f : R → R is f (x) = |x|(x − sin x)

… (i)

Q The function ‘f’ is a odd and continuous function and as lim f (x) = ∞ and lim f (x) = − ∞, so range is R, x→ ∞

x→ −∞

∴ f (x) is neither injective nor surjective

4. According to given information, we have if k ∈{4, 8, 12, 16, 20} Then, f (k) ∈ {3, 6, 9, 12, 15, 18} [Q Codomain ( f ) = {1, 2, 3, …, 20}]

therefore ‘f’ is a onto function.  x(x − sin x), x ≥ 0 Q f (x) =  − x (x − sin x), x < 0 ∴

Clearly, f (x) is not injective because if f (x) < 1, then f is many one, as shown in figure. Also, f (x) is not surjective because range of f (x) is [0, ∞ [ and but in problem co-domain is (0, ∞ ), which is wrong.

 2x − sin x − x cos x, x > 0 f ′ (x) =  − 2x + sin x + x cos x, x < 0

 (x − sin x) + x(1 − cos x), x > 0 (− x + sin x) − x(1 − cos x), x < 0  Q for x > 0, x − sin x > 0 and x (1 − cos x) > 0 ∴ f ′ (x) > 0 ∀ x ∈ (0, ∞ ) ⇒ f is strictly increasing function, ∀ x ∈ (0, ∞ ). Similarly, for x < 0, − x + sin x > 0 and (− x) (1 − cos x) > 0, therefore, f ′ (x) > 0 ∀ x ∈ (− ∞ , 0) ⇒ f is strictly increasing function, ∀ x ∈ (0, ∞ )

Now, we need to assign the value of f (k) for k ∈{4, 8, 12, 16, 20} this can be done in 6C5 ⋅ 5 ! ways = 6 ⋅ 5 ! = 6 ! and remaining 15 element can be associated by 15 ! ways. ∴Total number of onto functions = = 15 ! 6 ! x 5. We have, f (x) = , x ∈R 1 + x2 Ist Method f (x) is an odd function and maximum occur at x = 1 Y

x2 =y 1 − x2 x2 = y(1 − x2)

f (x) = ⇒ ⇒ Q

(let) [Q x2 ≠ 1]

2

Since, for surjective function, range of f = codomain ∴ Set A should be R − [−1, 0).  (x − 1) |x − 1| − x , if 0 < x ≤ 1 3. We have, f (x) = = x−1 x  if x > 1 ,  x 1 − 1, if 0 < x ≤ 1  = x 1 1 − , if x > 1  x Now, let us draw the graph of y = f (x) Note that when x → 0, then f (x) → ∞, when x = 1, then f (x) = 0, and when x → ∞, then f (x) → 1 Y

y=1

y=–

1

y=0

X

1 2

 1 1 From the graph it is clear that range of f (x) is − ,  2 2  IInd Method f (x) =

1 x+

1 x

If x > 0, then by AM ≥ GM, we get x + 1

⇒

1 x+ x

≤

1

⇒

1 x+ x

1 ≥2 x

1 1 ⇒ 0 < f (x) ≤ 2 2

If x < 0, then by AM ≥ GM, we get x + ≥−

1 2

⇒–

1 ≤ −2 x

1 ≤ f (x) < 0 2

0 1 1 = 0. Thus, − ≤ f (x) ≤ 1+0 2 2  1 1 Hence, f (x) ∈ − ,  2 2  If x = 0, then f (x) =

IIIrd Method x Let y = ⇒ yx2 − x + y = 0 1 + x2

+

– O

X

O1

Q x ∈ R, so D ≥ 0 ⇒ 1 − 4 y2 ≥ 0  1 1 ⇒ (1 − 2 y) (1 + 2 y) ≥ 0 ⇒ y ∈ − ,  2 2 

x=0

1 2

(–1, 1/2)

y [provided y ≠ −1] x (1 + y) = y ⇒ x = 1+ y y ≥ 0 ⇒ y ∈ (−∞ , − 1) ∪ [0, ∞ ) x2 ≥ 0 ⇒ 1+ y 2

y=

–1

Therefore ‘f’ is a strictly increasing function for x ∈ R and it implies that f is one-one function.

2. Given, function f : R – {1, − 1} → A defined as

(1, 1/2)

–1/2

– 1/2

184 Functions  1 1 So, range is − , .  2 2 

∴ (− 1) − 4 ( y)( y) ≥ 0 ⇒ 1 − 4 y2 ≥ 0 ⇒ y ∈ 2

n + 1 , if n is odd 

6. Given, f (n ) =  n 2

 , if n is even, 2  n + 1 , if n is odd and g (n ) = n − (−1)n =  n − 1, if n is even 1 if f ( n + ), n is odd  Now, f ( g (n )) =   f (n − 1), if n is even n + 1 , if n is odd  =  n 2− 1 + 1 n = , if n is even   2 2 = f (x) [Q if n is odd, then (n + 1) is even and if n is even, then (n − 1) is odd] Clearly, function is not one-one as f (2) = f (1) = 1 But it is onto function. [Q If m ∈ N (codomain) is odd, then 2m ∈ N (domain) such that f (2m) = m and if m ∈ N codomain is even, then

2m − 1 ∈ N (domain) such that f (2m − 1) = m] ∴Function is onto but not one-one 2x 7. We have a function f : A → R defined as, f (x) = x −1 One-one Let x1, x2 ∈ A such that 2x1 2x2 = f (x1 ) = f (x2) ⇒ x1 − 1 x2 − 1 ⇒

2x1x2 − 2x1 = 2x1x2 − 2x2 ⇒ x1 = x2

Thus, f (x1 ) = f (x2) has only one solution, x1 = x2 ∴ f (x) is one-one (injective) 2 ×2 Onto Let x = 2, then f (2) = =4 2 −1 But x = 2 is not in the domain, and f (x) is one-one function ∴f (x) can never be 4. Similarly, f (x) can not take many values. Hence, f (x) is into (not surjective). ∴f (x) is injective but not surjective. x 8. We have, f (x) = 1 + x2 1 x  1 f  = x = ∴ = f (x) 1 1 + x2  x 1+ 2 x  1  1 f   = f (2)or f   = f (3)and so on. ∴  2  3 So, f (x) is many-one function. x Again, let y = f (x) ⇒ y = 1 + x2 ⇒

y + x2y = x ⇒ yx2 − x + y = 0

x ∈R

As,

∴ Range = Codomain =

− 1 1 ,  2 2 

− 1 1 ,  2 2 

So, f (x) is surjective. Hence, f (x) is surjective but not injective.

9.

PLAN To check nature of function. (i) One-one To check one-one, we must check whether f ′ ( x )> 0 or f ′ ( x )< 0 in given domain. (ii) Onto To check onto, we must check Range = Codomain

Description of Situation To find range in given domain [a , b], put f ′ (x) = 0 and find x = α 1, α 2, …, α n ∈[a , b] Now, find { f (a ), f (α 1 ), f (α 2), K , f (α n ), f (b)} its greatest and least values gives you range. Now, f : [0, 3] → [1, 29] f (x) = 2x3 − 15x2 + 36x + 1 ∴

f ′ (x) = 6x2 − 30x + 36 = 6 (x2 − 5x + 6) = 6 (x − 2) (x − 3) −

+

+ 3

2

For given domain [0, 3], f (x) is increasing as well as decreasing ⇒ many-one Now, put

f ′ (x) = 0 ⇒ x = 2, 3

Thus, for range

f (0) = 1, f (2) = 29, f (3) = 28

⇒

Range ∈[1, 29]

∴ Onto but not one-one.  x, x ∈ Q − x, x ∉ Q

10. Let φ (x) = f (x) − g (x) = 

Now, to check one-one. Take any straight line parallel to X-axis which will intersect φ(x) only at one point. ⇒ φ(x) is one-one. To check onto As

 x, x ∈ Q , which shows f (x) =  − x, x ∉ Q

y = x and

y = − x for rational and irrational values

⇒ y ∈ real numbers. ∴

Range = Codomain ⇒ onto

Thus, f − g is one-one and onto.

11. Given, f : [0, ∞ ) → [0, ∞ ) Here, domain is [0, ∞ ) and codomain is [0, ∞ ). Thus, to check one-one x 1 Since, f (x) = > 0, ∀ x ∈ [0, ∞ ) ⇒ f ′ (x) = 1+ x (1 + x)2 ∴ f (x) is increasing in its domain. Thus, f (x) is one-one in its domain. To check onto (we find range)

Functions 185 x ⇒ y + yx = x 1+ x

Again,

y = f (x) =

⇒

y y x= ⇒ ≥0 1− y 1− y

Y y = 1 + 2x

X′

Since, x ≥ 0, therefore 0 ≤ y < 1

−π 2

O

π 2

X

i.e. Range ≠ Codomain ∴ f (x) is one-one but not onto.

Y′

f (x) = 2x + sin x

12. Given, ⇒

f ' (x) = 2 + cos x

⇒

f ' (x) > 0 , ∀x ∈ R

which shows f (x) is one-one, as f (x) is strictly increasing. Since, f (x) is increasing for every x ∈ R, ∴ f (x) takes all intermediate values between (−∞ , ∞ ). Range of f (x) ∈ R. Hence, f (x) is one-to-one and onto.

13. The number of onto functions from

It is clear from the graph that y = tan x is one-one and onto, therefore (B) → (r). 16. PLAN (i) If f ′ ( x ) > 0, ∀x ∈ ( a, b ), then f( x ) is an increasing function in ( a, b ) and thus f( x ) is one-one function in ( a, b ) . (ii) If range of f( x ) = codomain of f( x ) , then f( x ) is an onto function. (iii) A function f( x ) is said to be an odd function, if f( − x ) = − f( x ), ∀x ∈ R, i.e. f( − x ) + f( x ) = 0, ∀ x ∈ R

E = {1, 2, 3, 4} to F = {1, 2} = Total number of functions which map E to F

f (x) = [ln (sec x + tan x)]3 3 [ln (sec x + tan x)]2 (sec x tan x + sec 2x) f ′ (x) = (sec x + tan x)

− Number of functions for which map f (x) = 1 and f (x) = 2 for all x ∈ E = 24 − 2 = 14

 −π π  ,  f ′ (x) = 3 sec x [ln (sec x + tan x)]2 > 0, ∀x ∈   2 2

14. PLAN (i) For such questions, we need to properly define the functions and then we draw their graphs. (ii) From the graphs, we can examine the function for continuity, differentiability, one-one and onto.

f (x) is an increasing function. ∴ f (x) is an one-one function.  π x  π π (sec x + tan x) = tan  +  , as x ∈  − ,  , then  4 2  2 2  π x 0 < tan  +  < ∞  4 2

− x, x < 0 f1 (x) =  x e , x ≥ 0 f2(x) = x2, x ≥ 0 sin x, x < 0 f3 (x) =  x≥0 x,

0 < sec x + tan x < ∞ ⇒ − ∞ < ln (sec x + tan x) < ∞ − ∞ < [ln (sec x + tan x)]3 < ∞ ⇒ −∞ < f (x) < ∞ Range of f (x) is R and thus f (x) is an ont function.

x 0 (46 + α 2)2 − 4 (9 + 8α )2 ≤ 0

⇒

α > − 9 /8

and [46 + α − 2 (9 + 8α )][46 + α + 2 (9 + 8α )] ≤ 0 2

2

⇒ and

α > − 9 /8 α > − 9 /8

and [(α − 2) (α − 14)] (α + 8)2 ≤ 0 ⇒ α > − 9 /8 and ⇒

(α − 2) (α − 14) ≤ 0

[Q (α + 8)2 ≥ 0]

α > − 9 / 8 and 2 ≤ α ≤ 14 ⇒ 2 ≤ α ≤ 14

Thus, f (x) =

αx2 + 6x − 8 will be onto, if 2 ≤ α ≤ 14 α + 6 x − 8 x2

Again, when α = 3 3 x2 + 6 x − 8 , in this case f (x) = 0 f (x) = 3 + 6 x − 8 x2 ⇒ ⇒

3 x2 + 6 x − 8 = 0 − 6 ± 36 + 96 − 6 ± 132 1 x= = = (− 3 ± 33 ) 6 6 3

This shows that 1  f (− 3 + 33) = f 3 

 1 + y  1 + y 1 1 log 8    = (log 8 e) log e   1 − y  1 − y 4 4

By interchanging the variables x and y, we get the inverse function of f (x) and it is  1 + x 1 f −1 (x) = (log 8 e) log e  .  1 − x 4 Hence, option (c) is correct.

2. Since, only (c) satisfy given definition f { f −1 (B)} = B

i.e.

B ⊆ f (x)

Only, if

3. By definition of composition of function,

(α 2 − 16α + 28) (α 2 + 16α + 64) ≤ 0

⇒

x=

{by base change property of logarithm log a b = log a e ⋅ log e b}

⇒ 9 (1 − 2 y + y2) + [8α + (64 + α 2) y + 8 αy2] ≥ 0 ⇒

82x − 8−2x 84x − 1 , x ∈ ( − 1 , 1 ) = y (let) = 82x + 8−2x 84x + 1

1  (− 3 − 33) = 0 3 

Therefore, f is not one-to-one.

20. Since, there is an injective mapping from A to B, each element of A has unique image in B. Similarly, there is also an injective mapping from B to A, each element of B has unique image in A or in other words there is one to one onto mapping from A to B. Thus, there is bijective mapping from A to B.

g ( f (x) ) = (sin x + cos x)2 − 1, is invertible (i.e. bijective) ⇒ g { f (x) } = sin 2x is bijective.  π π We know, sin x is bijective, only when x ∈ − , .  2 2  π π Thus, g { f (x) } is bijective, if − ≤ 2x ≤ 2 2 π π − ≤x≤ ⇒ 4 4

4. It is only to find the inverse. y = f (x) = (x + 1)2, for x ≥ − 1

Let ±

y = x + 1,

⇒

y = x+1

⇒

x=

⇒

f

−1

x≥ −1 ⇒

y ≥ 0, x + 1 ≥ 0

y − 1 ⇒ f −1 ( y) =

(x) = x − 1 ⇒

y −1

x≥0

x2 + 1 ⇒ y= ⇒ xy = x2 + 1 x y ± y2 − 4 ⇒ x2 − xy + 1 = 0 ⇒ x = 2

1 5. Let y = x + x

⇒

f −1 ( y) =

y±

y2 − 4 x ± x2 − 4 ⇒ f −1 (x) = 2 2

Since, the range of the inverse function is [1, ∞), then we take

f −1 (x) =

x+

x2 − 4 2

Functions 187 x − x2 − 4 , then f −1 (x) > 1 2 This is possible only if (x − 2)2 > x2 − 4 ⇒ x2 + 4 − 4x > x2 − 4 ⇒ 8 > 4x ⇒ x < 2, where x > 2 Therefore, (a) is the answer. If we consider

f −1 (x) =

log 2 y = log 2 2x ( x − 1) ⇒ log 2 y = x (x − 1)

1 + 4 log 2 y ≥ 1 ⇒ − 1 + 4 log 2 y ≤ − 1

⇒

1 − 1 + 4 log 2 y ≤ 0

Let ⇒

Case III When f (z ) ≠ 2 is true. If f (z ) ≠ 2 is true, then remaining statements are false. ∴ If Thus, we have [given]

⇒ ⇒

For function to be invertible, it should be one-one onto. ∴ Check Range : b−x Let f (x) = y ⇒ y= 1 − bx ⇒

y − bxy = b − x ⇒ x (1 − by) = b − y b− y , where 0 < x < 1 ⇒ x= 1 − by

1 b (b − 1) ( y + 1) 1 0 and

f (x) = 2, f ( y) = 1 and f (z ) = 3 f −1 (1) = y

then f (− x) = f (x), ∀ x ∈ (− 5, 5)  x + 1 Given , f (x) = f    x + 2

1 , x cos x are non-periodic functions. x b−x 9. Here, f (x) = , where 0 < b < 1, 0 < x < 1 1 − bx

y < b or

Hence,

12. Since, f is an even function,

And sin

⇒

f (x) ≠ 1 and f ( y) = 1

But f is injective.

8. Clearly, f (x) = x − [x] = { x} which has period 1.

b− y 0 . If L is finite, then

(1991, 2M)

(b) −1 (d) None of these

(a) 1 (c) 0

23. Let L = lim

a − a 2 − x2 −

x→ 0

x −1  .  2 x2 − 7 x + 5 

34. Evaluate lim  x→1

(1979, 3M)

(1978, 3M)

190 Limit, Continuity and Differentiability Integer & Numerical Answer Type Questions x2 sin (βx) 35. Let α , β ∈ R be such that lim = 1 . Then, x → 0 αx − sin x (2016 Adv.) 6 (α + β ) equals

36. Let m and n be two positive integers greater than 1. If  ecos ( α n ) − e  = −  e , then the value of m is lim  m   2 α→ 0  n α  

(2015 Adv.)

Topic 2 1∞ Form, RHL and LHL Objective Questions I (Only one correct option) 1. Let f : R → R be a differentiable function satisfying 1

 1 + f (3 + x) − f (3) x f ′ (3) + f ′ (2) = 0. Then lim  is equal x→ 0  1 + f (2 − x) − f (2 )  to (2019 Main, 8 April II) −1

(a) e

2.

(b) e

(c) e

π − 2 sin − 1 x is equal to 1−x

lim

x→1 −

(a)

2

π 2

(b)

2 π

(c)

(d) 1

(2019 Main, 12 Jan II)

(d)

1 2π

3. Let [x] denote the greatest integer less than or equal to

(a) equals π (c) equals 0

(2019 Main, 11 Jan I)

4. For each t ∈ R, let [t ] be the greatest integer less than or equal to t. Then,

π  (1 − |x| + sin|1 − x|) sin  [1 − x] 2  lim x → 1+ |1 − x|[1 − x] (2019 Main, 10 Jan I) (b) does not exist (d) equals 1

(a) 0 (c) − sin 1

(2019 Main, 9 Jan II)

  1  2 15 lim x   +   + … +     x  x  x 

x→ 0 +

(2018 Main)

(b) is equal to 15 (d) does not exist (in R)

 1  1 − x (1 + |1 − x|) cos    1 − x |1 − x|

(b) limx→ 1 − f (x) does not exist

(c) limx→ 1 − f (x) = 0 (d) limx→ 1 + f (x) does not exist

1 and −1 2 9 (d) − and 3 2

(b) −

1

10. If lim [1 + x log (1 + b2)] x = 2b sin 2 θ, b > 0 and θ ∈ (− π , π ], then the value of θ is

π 4 π (c) ± 6

π 3 π (d) ± 2

(a) ±

(2011)

(b) ±



11. For x > 0, lim (sin x)1/ x +    x x→0  1

(a) 0 (c) 1

sin x 

 is 

(2006, 3M)

(b) – 1 (d) 2 1/ x

 f (1 + x)  lim x→ 0  f (1) 

equals

(2002, 2M)

(b) e 2 (d) e3

(a) 1 (c) e2 x

equal to t. Then,

for x ≠ 1. Then (a) limx→ 1 + f (x) = 0

a→ 0 +

5 (a) − and 1 2 7 (c) − and 2 2

x−3   is equal to x→ ∞ x + 2 

13. For x ∈ R , lim 

6. For each t ∈R, let [t ] be the greatest integer less than or

7. Let f (x) =

( 3 1 + a − 1) x2 − ( 1 + a − 1) x + (6 1 + a − 1) = 0, where (2012) a > − 1. Then, lim α (a ) and lim β (a ) are

1

(b) sin 1 (d) 1

(a) is equal to 0 (c) is equal to 120

9. Let α (a ) and β (a ) be the roots of the equation

12. Let f : R → R be such that f (1) = 3 and f ′ (1) = 6. Then,

5. For each x ∈ R, let [x] be the greatest integer less than or equal to x. Then, x([x] + |x|) sin [x] is equal to lim |x| x → 0−

(2016 Main)

(b) 1 1 (d) 4

x→ 0

(b) equals π + 1 (d) does not exist

(a) equals 0 (c) equals − 1

x→ 0

(a) 2 1 (c) 2

a→ 0+

π

x. Then, tan(π sin 2 x) + (|x| − sin(x[x]))2 lim x→ 0 x2

8. Let p = lim+ (1 + tan 2 x )1/ 2x , then log p is equal to

(2000, 2M)

(b) e−1 (d) e5

(a) e (c) e−5

Fill in the Blanks  1 + 5 x2  14. lim   x → 0  1 + 3 x2   x + 6  + 1

15. lim 

x → ∞ x

1/ x 2

=K . (1996, 1M)

x+ 4

= …… . (1991, 2M)

Analytical & Descriptive Question 

16. Find lim tan   x→ 0 

1/ x

π   + x  .   4

(1993, 2M)

Limit, Continuity and Differentiability 191 Integer & Numerical Answer Type Questions 17. Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit 1

(1 − x) x − e−1 is equal to a non zero real number, is ……… . lim + x→ 0 xa

(2020 Adv.) 1−x

 − ax + sin(x − 1) + a 1−  x→1  x + sin(x − 1 ) − 1 

18. The largest value of the non-negative integer a for which lim 

x

=

1 is 4

(2014 Adv.)

Topic 3 Squeeze, Newton-Leibnitz’s Theorem and Limit Based on Converting infinite Series into Definite Integrals Objective Question II (One more than one correct option)

Objective Questions I (Only one correct option) 1. If α and β are the roots of the equation 375x2 − 25x − 2 = 0, then lim

n→ ∞

n

n

∑ α r + nlim ∑ βr is equal →∞

r =1

r =1

to

(2019 Main, 12 April I)

21 346 1 (c) 12

29 358 7 (d) 116

(a)

(b)

sec 2 x

2.

∫ lim 2

π x→ 4

1 (a) f   ≥ f (1)  2

f (t ) dt

x2 −

(2007, 3M)

8 (a) f (2) π 2 (b) f (2) π 2  1 (c) f   π  2

n→ ∞

1 n

(a) 1 +

2n

∑

r =1

5

(2016 Adv.)

Integer & Numerical Answer Type Question 5. For each positive integer n, let

1

1 ((n + 1) (n + 2) ... (n + n )) n . n For x ∈ R, let [x] be the greatest integer less than or equal to x. If lim yn = L, then the value of [L ] is n→ ∞ ................. . (2018 Adv.) yn =

(d) 4f (2)

3. lim

x

n n n    ...  x +   n 2  , n 2  2 n 2   ...  x + 2  4  n  

1 2 (b) f   ≤ f    3  3 f ′ (3) f ′ (2) (d) ≥ f (3) f (2)

(c) f ′ (2) ≤ 0

equals

π2 16

   n n (x + n )  x +   4. Letf (x) = lim n→ ∞   2 2 2  n ! (x + n )  x +   for all x = 0. Then

r n2 + r2

Fill in the Blank equals

(b) 5 − 1

(1999, 2M)

(c) −1 +

(d) 1 +

2

2

x2

6. lim

∫0

x→ 0

cos 2t dt

x sin x

=K (1997C, 2M)

Topic 4 Continuity at a Point Objective Questions I (Only one correct option) π π 1. If the function f defined on  ,  by

2. The function f (x) = [x]2 − [x2] (where, [x] is the greatest integer less than or equal to x), is discontinuous at

 6 3

 2 cos x − 1 , x≠  f (x) =  cot x − 1  k, x=  then k is equal to (a)

1 2

(c) 1

π 4 is continuous, π 4 (2019 Main, 9 April I)

(b) 2 (d)

1 2

(1999, 2M)

(a) all integers

(b) all integers except 0 and 1

(c) all integers except 0

(d) all integers except 1

3. Let [.] denotes the greatest integer function and f (x) = [tan 2 x], then

(a) lim f (x) does not exist x→ 0

(b) f (x) is continuous at x = 0 (c) f (x) is not differentiable at x = 0 (d) f ′ (0) = 1

(1993, 1M)

192 Limit, Continuity and Differentiability 2x − 1  π, [⋅] denotes the  2  greatest integer function, is discontinuous at

4. The function f (x) = [x] cos  (a) all x (b) all integer points (c) no x (d) x which is not an integer

5. If f (x) = x ( x + (x + 1), then

(1993, 1M)

(1985, 2M)

(a) f (x) is continuous but not differentiable at x = 0 (b) f (x) is differentiable at x = 0 (c) f (x) is not differentiable at x = 0 (d) None of the above

log (1 + ax) − log (1 − bx) x is not defined at x = 0. The value which should be assigned to f at x = 0, so that it is continuous at x = 0, is

6. The function f (x) =

(a) a − b (b) a + b (c) log a + log b (d) None of the above

(1994, 4M)

  1 − cos 4x , x0   16 + x − 4 Determine the value of a if possible, so that the function (1990, 4M) is continuous at x = 0.

(a) x = − 1 (b) x = 1 (c) x = 0 (d) x = 2

 x + a 2 sin x, 0 ≤ x ≤ π /4  π /4 ≤ x ≤ π /2 f (x) =  2x cot x + b, a cos 2x − b sin x, π / 2 < x ≤ π 

x≤0  g (x), 1/ x  defined by f (x) =  (1 + x)   (2 + x)  , x > 0   Find the continuous f ′ (1) = f (− 1).

8. For every pair of continuous function f , g : [0, 1] → R such that max { f (x): x ∈ [0, 1]} = max { g (x): x ∈ [0,1]}. The correct statement(s) is (are)

(1989)

is continuous for 0 ≤ x ≤ π.

Then, at which of the following point(s) the function (2017 Adv.) f (x) = x cos (π (x + [x])) is discontinuous ?

(2014 Adv.)

[f (c)] + 3f (c) = [ g (c)] + 3 g (c) for some c ∈[0,1] [f (c)]2 + f (c) = [ g (c)]2 + 3 g (c) for some c ∈[0,1] [f (c)]2 + 3f (c) = [ g (c)]2 + g (c) for some c ∈[0,1] [f (c)]2 = [ g (c)]2 for some c ∈[0,1] 2

9. For every integer n, let a n and bn be real numbers. Let function f : R → R be given by a n + sin πx, for x ∈ [2n , 2n + 1] f (x) =  , bn + cos πx, for x ∈ (2n − 1, 2n ) for all integers n. If f is continuous, then which of the following hold(s) for (2012) all n ? (a) an−1 − bn−1 = 0 (b) an − bn = 1 (c) an − bn + 1 = 1 (d) an − 1 − bn = −1

Fill in the Blank 10. A discontinuous function y = f (x) satisfying x2 + y2 = 4 is given by f (x) = .... .

Determine a and b such that f (x) is continuous at x = 0.

14. Let g (x) be a polynomial of degree one and f (x) be

7. Let [x] be the greatest integer less than or equals to x.

(a) (b) (c) (d)

π  {1 + |sin x|}a/|sin x |, < x 5 (2019 Main, 9 April II) x = 5, then the value of a − b is

3. If the function f (x) =  −2 π+ 5 2 (c) π −5

2 π+ 5 2 (d) 5− π (b)

(a)

x 4. If f (x) = [x] −  , x ∈ R where [x] denotes the greatest 4  (2019 Main, 9 April II) integer function, then (a) lim f (x) exists but lim f (x) does not exist x→ 4 +

x→ 4 −

(b) f is continuous at x = 4

x→ 4 −

x→ 4 +

(d) lim f (x) exists but lim f (x) does not exist x→ 4 −

x→ 4 +

5. Let f : [−1, 3] → R be defined as |x| + [x], −1 ≤ x < 1  f (x) =  x + |x|, 1 ≤ x < 2  x + [x], 2 ≤ x ≤ 3 , 

(2019 Main, 8 April II)

where, [t ] denotes the greatest integer less than or equal to t. Then, f is discontinuous at

is continuous at x = 0 , then the ordered pair ( p, q) is equal to (2019 Main, 10 April I) 3 1 (a)  − , −   2 2 5 1  (c)  ,   2 2

(c) Both lim f (x) and lim f (x) exist but are not equal

(a) (b) (c) (d)

four or more points only two points only three points only one point

6. Let f: R → R be a function defined as  5, a + bx,  f (x) =   b + 5x,  30, Then, f is (a) (b) (c) (d)

if

x≤1

if 1 < x < 3 if 3 ≤ x < 5 if

x≥5 (2019 Main, 9 Jan I)

continuous if a = − 5 and b = 10 continuous if a = 5 and b = 5 continuous if a = 0 and b = 5 not continuous for any values of a and b

7. If f (x) = (a) tan [f (b) tan [f (c) tan [f (d) tan [f

1 x − 1, then on the interval [0, π ] 2

(1989, 2M)

(x)] and 1/ f (x) are both continuous (x)] and 1/ f (x) are both discontinuous (x)] and f −1 (x) are both continuous (x)] is continuous but 1/ f (x) is not continuous

194 Limit, Continuity and Differentiability the points of discontinuity of f in the domain are…… .

Objective Questions II (One or more than one correct option)

(1996, 2M)

Analytical & Descriptive Question

8. The following functions are continuous on (0, π) (a) tan x  1, 0 ≤ x≤ 3 π /4 (c)  2 3π < x< π 2 sin x,  9 4

1 (b) ∫ t sin dt 0 t xπ sin x, 0 < x ≤ (d)  sin ( π + x ),  2 x

(1991, 2M)

π /2 π < x< π 2

 x2 0 ≤ x 2 (d) f ′(x) is strictly decreasing in the interval [1, ∞ )

Analytical & Descriptive Questions  x + a , if x < 0 2. Let f (x) =  |x − 1|, if x ≥ 0 if x < 0  x + 1, and g (x) =  2 1 if x ≥ 0 ( x − ) + b , 

function. Prove that f (x) + g (x) is a discontinuous function. (1987, 2M) 1 + x, 0 ≤ x ≤ 2 3 − x, 2 < x ≤ 3

4. Let f (x) = 

Determine the form of g (x) = f [ f (x)] and hence find the points of discontinuity of g, if any (1983, 2M)

5. Let f (x + y) = f (x) + f ( y) for all x and y. If the function f (x) is continuous at x = 0, then show that f (x) is (1981, 2M) continuous at all x.

Topic 7 Differentiability at a Point Objective Questions I (Only one correct option) 1. Let f : R → R be differentiable at c ∈ R and f (c) = 0. If g (x) = | f (x)|, then at x = c, g is (a) (b) (c) (d)

not differentiable differentiable if f ′ (c) ≠ 0 not differentiable if f ′ (c) = 0 differentiable if f ′ (c) = 0

2. If f : R → R is a differentiable function and f ( x)

f (2) = 6, then lim

x→ 2

(2019 Main, 10 April I)

(a) 12f ′ (2)

(b) 0

∫

6

2t dt is (x − 2)

(2019 Main, 9 April II)

(c) 24f ′ (2)

(d) 2f ′ (2)

3. Let f (x) = 15 − x − 10 ; x ∈ R. Then, the set of all values of x, at which the function, g (x) = f ( f (x)) is not differentiable, is (2019 Main, 9 April I) (a) {5, 10, 15, 20} (c) {10}

(b) {5, 10, 15} (d) {10, 15}

Limit, Continuity and Differentiability 195 4. Let S be the set of all points in (− π , π ) at which the function, f (x) = min {sin x, cos x} is not differentiable. Then, S is a subset of which of the following? (2019 Main, 12 Jan I)

π π (a) − , 0,  4  4 3π π 3π π   (c) − ,− , ,  4 4 4  4

(b) −  (d) − 

π π π π , − , ,  2 4 4 2 3π π π 3π  ,− , ,  4 2 2 4

5. Let K be the set of all real values of x, where the function f (x) = sin| x| − | x| + 2(x − π ) cos| x|is not differentiable. (2019 Main, 11 Jan II) Then, the set K is equal to (b) φ (an empty set) (d) {0, π }

(a) {0} (c) { π }

−2 ≤ x < 0  −1 , and 2 1 0 ≤ x≤2 x − , 

6. Let f (x) = 

g (x) = | f (x)| + f (|x|). Then, in the interval (−2, 2), g is (2019 Main, 11 Jan I)

(a) (b) (c) (d)

not differentiable at one point not differentiable at two points differentiable at all points not continuous

7. Let f : (−1, 1) → R be a function defined by f (x) = max { − x , − 1 − x2 }. If K be the set of all points at which f is not differentiable, then K has exactly (2019 Main, 10 Jan II)

(a) three elements (c) two elements

(b) five elements (d) one element

max {|x|, x2}, |x| ≤ 2 2 < |x| ≤ 4  8 − 2|x|, Let S be the set of points in the interval (−4, 4) at which f is not differentiable. Then, S (2019 Main, 10 Jan I)

8. Let f (x) = 

(a) equals {−2, − 1, 0, 1, 2} (b) equals {−2, 2} (c) is an empty set (d) equals {−2,−1, 1, 2}

π



 2 13. Let f (x) = x cos x , x ≠ 0, x ∈ R, then f is  0,

(a) (b) (c) (d)

3

differentiable both at x = 0 and at x = 2 differentiable at x = 0 but not differentiable at x = 2 not differentiable at x = 0 but differentiable at x = 2 differentiable neither at x = 0 nor at x = 2

(x − 1)n ; 0 < x < 2, m and n are log cosm (x − 1) integers, m ≠ 0, n > 0 and let p be the left hand derivative of|x − 1| at x = 1 . If lim g (x) = p , then

14. Let

g (x) =

x→1 +

(a) n = 1, m = 1 (c) n = 2 , m = 2

(b) n = 1, m = −1 (d) n > 2, m = n

f 2(x) dx is equal to (b)

1 2

(2019 Main, 9 Jan II)

(c) 1

(d) 0

10. Let S = (t ∈ R : f (x) = |x − π |( ⋅ e − 1)sin| x| is not |x|

differentiable at t}.Then, the set S is equal to(2018 Main) (a) φ (an empty set) (c) { π }

(2008, 3M)

15. If f is a differentiable function satisfying  1 f   = 0, ∀ n ≥ 1, n ∈ I ,then  n

(2005, 2M)

(a) f (x) = 0, x ∈ (0, 1] (b) f ′ (0) = 0 = f (0) (c) f (0) = 0 but f ′ (0) not necessarily zero (d)|f (x)|≤ 1, x ∈ (0, 1]

16. Let f (x) = ||x|− 1|, then points where, f (x) is not differentiable is/are (a) 0, ± 1 (c) 0

(2005, 2M)

(b) ± 1 (d) 1

17. The domain of the derivative of the functions  tan −1 x , if | x| ≤ 1  is f (x) = 1 (| x| − 1), if | x| > 1 2

(a) R − {0} (c) R − {−1}

(2002, 2M)

(b) R − {1} (d) R − {−1, 1} (2001, 2M)

(a) cos (| x|) + | x| (c) sin (| x|) + | x|

(b) cos (| x|) − | x| (d) sin (| x|) − | x|

19. The left hand derivative of f (x) = [x] sin (π x) at x = k, k is an integer, is

1

(a) 2

(2012)

x =0

at x = 0 ?

| f (x) − f ( y)| ≤ 2|x − y|2 , for all x, y ∈ R. If f (0) = 1, then

0

(b) 2 f ′(c) = 3 g ′(c) (d) f ′(c) = 2 g ′(c)

18. Which of the following functions is differentiable

9. Let f be a differentiable function from R to R such that

∫

(a) 2f ′ (c) = g ′(c) (c) f ′(c) = g ′(c)

(b) {0} (d) {0, π }

11. For x ∈ R, f (x) = |log 2 − sin x|and g (x) = f ( f (x)), then (a) g is not differentiable at x = 0 (2016 Main) (b) g′ (0) = cos (log 2) (c) g′ (0) = − cos (log 2) (d) g is differentiable at x = 0 and g′ (0) = − sin (log 2)

12. If f and g are differentiable functions in (0, 1) satisfying f (0) = 2 = g (1), g(0) = 0 and f (1) = 6, then for some (2014 Main) c ∈] 0, 1 [

(a) (−1)k (k − 1) π (c) (−1)k kπ

(2001, 2M)

(b)(−1)k − 1 (k − 1) π (d) (−1)k − 1 kπ

20. Let f : R → R be a function defined by f (x) = max { x, x3 }. The set of all points, where f (x) is not differentiable, is (a) {−1,1 } (c) {0,1 }

(b) {−1, 0 } (d) {−1, 0,1 }

(2001, 2M)

21. Let f : R → R be any function. Define g : R → R by g (x) =| f (x)|, ∀ x. Then, g is

(2000, 2M)

(a) onto if f is onto (b) one-one if f is one-one (c) continuous if f is continuous (d) differentiable if f is differentiable

22. The function f (x) = (x2 − 1)| x2 − 3x + 2| + cos (| x|) is not differentiable at (a) −1

(b) 0

(1999, 2M)

(c) 1

(d) 2

196 Limit, Continuity and Differentiability 23. The set of all points, where the function f (x) = differentiable, is

x is 1 + | x| (1987, 2M)

(a) (− ∞ , ∞ ) (c) (− ∞ , 0) ∪ (0, ∞ )

(b) [0, ∞ ) (d) (0, ∞ )

24. There exists a function f (x) satisfying f (0) = 1, f ′ (0) = − 1, f (x) > 0, ∀ x and

(1982, 2M)

(a) f ′′ (x) < 0 , ∀ x

(b) − 1 < f ′′ (x) < 0 , ∀ x

(c) − 2 ≤ f ′′ (x) ≤ − 1 , ∀ x

(d) f ′′ (x) < − 2 , ∀ x

25. For a real number y, let [ y] denotes the greatest integer less than or equal to y. Then, the function tan π [(x − π )] (1981, 2M) is f (x) = 1 + [x]2 (a) discontinuous at some x (b) continuous at all x, but the derivative f ′ (x) does not exist for some x (c) f ′(x) exists for all x, but the derivative f ′ ′ (x) does not exist for some x (d) f ′(x) exists for all x

Objective Questions II (One or more than one correct option) 26. Let f : R → R and g : R → R be functions satisfying f (x + y) = f (x) + f ( y) + f (x) f ( y)and f (x) = xg (x)

29. For every twice differentiable function f : R → [−2, 2] with ( f (0))2 + ( f ′ (0))2 = 85, which of the following (2018 Adv.) statement(s) is (are) TRUE ? (a) There exist r , s ∈ R , where r < s, such that f is one-one on the open interval (r , s) (b) There exists x0 ∈ (−4, 0) such that|f ′ (x0 )|≤ 1 (c) lim f (x) = 1 x→ ∞

(d) There exists α ∈ (−4, 4) such that f (α ) + f ′′(α ) = 0 and f ′ (α ) ≠ 0

30. Let f : (0, π)→ R be a twice differentiable function such that lim

t→ x

f (x) sin t − f (t )sin x = sin 2 x for all x ∈ (0, π). t−x

π  π If f   = − , then which of the following statement(s)  6 12 is (are) TRUE? (2018 Adv.) π π (a) f   =  4 4 2 (b) f (x)
b (2013)

g (x) is continuous but not differentiable at a g (x) is differentiable on R g (x) is continuous but not differentiable at b g (x) is continuous and differentiable at either a or b but not both

π π  −x− , x≤−  2 2  π 37. If f (x) = − cos x, − < x ≤ 0, then 2  0 < x≤1  x − 1,  ln x, x>1 π 2 (b) f (x) is not differentiable at x = 0 (c) f (x) is differentiable at x = 1 3 (d) f (x) is differentiable at x = − 2

(2006, 3M)

(a) f (x) is continuous everywhere (b) f (x) is continuous and differentiable everywhere (c) f (x) is not differentiable at two points (d) f (x) is not differentiable at one point

40. Let h (x) = min { x, x2} for every real number of x, then (a) h is continuous for all x (b) h is differentiable for all x (c) h ′ (x) = 1, ∀ x > 1 (d) h is not differentiable at two values of x

| x − 3|, 3x 13  4 − 2 + 4 ,

41. The function f (x) =  x2

(1998, 2M)

x≥1 x b > 0, the

dx  π π  at  ,  is dy  4 4 

a−b (a) a+ b

a+ b (b) a−b

(2020 Main, 4 Sep I)

2a + b (c) 2a − b

(d)

a − 2b a + 2b

C1 + (22) 20C 2 + (32) 20C3 + ..... + (202)20C 20 = A (2β ) , then the ordered pair ( A , β) is equal to

2. If

20

(2019 Main, 12 April II)

(a) (420, 19) (b) (420, 18) (c) (380, 18) (d) (380, 19)

 sin x − cos x  x 3. The derivative of tan −1   , with respect to ,  sin x + cos x 2   π where  x ∈ 0,   is  2  (2019 Main, 12 April II) (a) 1

(b)

2 3

(c)

1 2

(d) 2

 dy d 2y 4. If e + xy = e, the ordered pair  , 2  at x = 0 is equal  dx dx  y

to

(2019 Main, 12 April I)

1 1 1 1 1 1 (a)  , − 2  (b)  − , 2  (c)  , 2   e e  e e e  e 

f (1) = 1, f′ (1) = 3, then f ( f ( f (x))) + ( f (x))2 at x = 1 is

5. If

(a) 12

(b) 9

1 1 (d)  − , − 2   e e 

the

derivative

of

(2019 Main, 8 April II)

(c) 15

(d) 33

2

  3 cos x + sin x  dy  π 6. If 2 y =  cot−1  is   , x ∈ 0,  then  2 dx  cos x − 3 sin x   equal to (a)

π −x 6

(b) x −

π 6

(c)

π −x 3

(2019 Main, 8 April I)

(d) 2x −

π 3

7. For x > 1, if (2x)2y = 4e2x − 2y , then (1 + log e 2x)2 equal to

(2019 Main, 12 Jan I)

x log e 2x + log e 2 (a) x (c) x log e 2x

x log e 2x − log e 2 (b) x (d) log e 2x

8. If x log e (log e x) − x2 + y2 = 4( y > 0), then equal to (a) (c)

dy is dx

e

(b)

4 + e2 (1+ 2e )

(d)

4 + e2

(2e − 1)

dy at x = e is dx

(2019 Main, 11 Jan I)

2 4 + e2 (1+ 2e ) 2 4 + e2

be a function such that f :R→ R f (x) = x3 + x2f ′ (1) + xf ′ ′ (2) + f ′′′ (3), x ∈ R. (2019 Main, 10 Jan I) Then, f (2) equals

9. Let

(b) − 4 (d) 8

(a) 30 (c) − 2

10. If x = 3 tan t and y = 3 sec t, then the value of t=

π , is 4

1 (a) 6

d 2y at dx2

(2019 Main, 9 Jan II)

(b)

1 6 2

(c)

1 3 2

(d)

3 2 2

1 11. For x ∈ 0,  , if the derivative of  4  6x x  tan −1   is x ⋅ g (x), then g (x) equals  1 − 9x3  (2017 Main) (a)

9 1 + 9x3

(b)

3x x 1 − 9x3

(c)

3x 1 − 9x3

(d)

3 1 + 9x3

200 Limit, Continuity and Differentiability 12. If g is the inverse of a function f and f ′ (x) =

1 , then 1 + x5

g′ (x) is equal to (a) 1 + x5

(b) 5 x4

13. If y = sec (tan (a)

(2015)

1 2

−1

(c)

1

(d) 1 + { g (x)}5

1 + { g (x)}5

dy at x = 1 is equal to x), then dx

1 (b) 2

(c) 1

(2013)

(d)

2

14. Let g (x) = log f (x), where f (x) is a twice differentiable positive function on (0, ∞ ) such that f (x + 1) = x f (x). 1  1  Then, for N = 1, 2, 3,... , g′′  N +  − g′′   is equal to  2  2

  1 1 1 (a) − 4 1 + + + ... +  9 25 (2 N − 1)2     1 1 1 (b) 4 1 + + + ... + 2 9 25 (2 N − 1)  

(2008, 3M)

 d 2y  (b) −  2   dx 

 d 2 y  dy − 2 (c)  2     dx   dx 

−1

 dy     dx 

−3

differentiable function and g (x) = f ′ (x).

(c) 10

(2006, 3M)

f (1) = 1, f (2) = 4, f (3) = 9, then

18. If y is a function of x and log (x + y) = 2xy, then the value of y′ (0) is

(2004, 1M)

(c) 2

19. If x2 + y2 = 1 , then

3

x

20. Let f (x) = 6 p Then,

(d) 0 (2000, 1M)

(a) yy′ ′ − 2 ( y ′ )2 + 1 = 0 (c) yy′ ′ + ( y ′ )2 − 1 = 0

(b) yy′ ′ + ( y ′ )2 + 1 = 0 (d) yy′ ′ + 2 ( y ′ )2 + 1 = 0

sin x cos x −1 p2

0 p3

d3 f (x) at x = 0 is dx3

22. If x exy = y + sin 2 x, then at x = 0,

dy = ...… . dx

(1996, 2M)

23. Let f (x) = x|x|. The set of points, where f (x) is twice differentiable, is … .

(1992, 2M)

24. If f (x) = |x − 2|and g (x) = f [ f (x)] , then g ′ (x) = K… for −1 

(1990, 2M)

that fr (a ) = gr (a ) = hr (a ), r = 1, 2, 3 f1 (x) f2(x) f3 (x) and F (x) = g1 (x) g2(x) g3 (x) , (1985, 2M)

 2x − 1 dy 28. If y = f  2 = ....... .  and f ′ (x) = sin 2 x, then dx  x + 1 (1982, 2M)

29. If y =

ax2 bx c + + 1, + (x − a ) (x − b) (x − c) (x − b) (x − c) (x − c)

Prove that

(2005, 2M)

(a) f ′ ′ (x) = 2, ∀ x ∈ (R ) (b) f ′ (x) = 5 = f ′ ′ (x), for some x ∈ (1, 3) (c) there exists atleast one x ∈ (1, 3) such that f ′ ′ (x) = 2 (d) None of the above

(b) −1

Fill in the Blanks

(d) 25

17. Let f be twice differentiable function satisfying

(a) 1

(b) P ′ ′ (x) ⋅ P ′ ′ ′ (x) (d) a constant

Analytical & Descriptive Questions

2

  x    x  If F (x) =  f    +  g    and F (5) = 5,   2    2  then F (10) is (b) 5

(a) P ′ ′ ′ (x) + P ′ (x) (c) P (x) P ′ ′ ′ (x)

then F ′ (x) at x = a is …… .

16. If f ′′ (x) = − f (x), where f (x) is a continuous double

(a) 0

(1988, 2M)

h1 (x) h2(x) h3 (x)

 d 2 y  dy − 3 (d) −  2     dx   dx 

2

d  3 d 2y  equals y dx  dx2 

27. If fr (x), gr (x), hr (x), r = 1, 2, 3 are polynomials in x such (2007, 3M)

−1

2

1  25. The derivative of sec  − 2  with respect to  2x − 1 1 (1986, 2M) 1 − x2at x = is …… . 2 26. If f (x) = log x (log x), then f ′ (x) at x = e is ...... . (1985, 2M)

d 2x equals dy2  d 2y  (a)  2   dx 

21. If y2 = P (x) is a polynomial of degree 3, then

x > 2.

  1 1 1 (c) − 4 1 + + + ... + 2 9 25 (2 N + 1)     1 1 1 (d) 4 1 + + + ... + 2 9 25 (2 N + 1)  

15.

(b) p + p 2 (d) independent of p

(a) p (c) p + p3

, where p is constant.

30. Find

y′ 1  a b c  =  + + . y x  a − x b − x c − x

(1998, 8M)

dy at x = − 1, when dx sin

(sin y)

π x 2

+

3 sec−1 (2x) + 2x tan ln (x + 2) = 0. 2 (1991, 4M)

31. If x = sec θ − cos θ and y = secn θ − cos nθ, then show that  dy (x2 + 4)    dx

2

= n 2 ( y2 + 4).

(1989, 2M)

32. If α be a repeated roots of a quadratic equation f (x) = 0 and A (x), B(x) and C (x) be polynomials of degree 3, 4 and A (x) B(x) C (x) 5 respectively, then show that A (α ) B(α ) C (α ) is A ′ (α ) B ′ (α ) C ′ (α ) divisible by f (x), where prime denotes the derivatives.

(1997, 2M)

(1984, 4M)

Limit, Continuity and Differentiability 201 33. Find the derivative with respect to x of the function   2x   y = (log cos x sin x) (log sin x cos x)−1 + sin −1     1 + x2    π at x = . (1984, 4M) 4

34. If (a + bx) ey/ x = x, then prove that

3

36. Let y = ex sin x + (tan x)x , find

dy . dx

(1981, 2M)

5x

dy 37. Given, y = + cos 2 (2x + 1), find . 2 dx 3 (1 − x)

(1980)

Integer & Numerical Answer Type Question 38. Let f : R → R be a continuous odd function, which

2

d 2y  dy  x = x − y .  dx2  dx

1 vanishes exactly at one point and f (1) = . 2

3

(1983, 3M)

Suppose that F (x) = ∫

35. Let f be a twice differentiable function such that (1983, 3M)

x –1

x

G (x) = ∫ t| f { f (t )}| dt

f ′ ′ (x) = − f (x) , f ′ (x) = g (x) and h (x) = [ f (x) ]2 + [ g (x)]2 Find h (10), if h (5) = 11.

for

–1

lim

x→1

f (t ) dt for all x ∈ [−1 , 2] and all

x ∈ [−1 , 2].

F (x) 1  1 = , then the value of f   is  2 G (x) 14

(2015 Adv.)

Answers Topic 1 1. 5. 9. 13. 17. 21. 25. 28. 32. 34.

3. (d) 7. (a) 11. (b) 15. (c) 19. (b) 23. (a, c) 1 2 1 26. h 2hr − h , 128r 2 7 29. 30. False π 33. 0 a 2 cosα + 2a sin α −1 35. (7) 36. (2) 3 (b) (c) (d) (d) (d) (a)

2. 6. 10. 14. 18. 22.

(c) (b) (c) (c) (d) (d)

4. 8. 12. 16. 20. 24.

(a) (b) (d) (c) (c) –1

27. − 1

(d) (c) (b) (c) (1)

2. 6. 10. 14. 18.

(b) (c) (d) e2 (2)

3. 7. 11. 15.

(d) (d) (c) e5

2. (a) 6. 1

3. (b)

4. 8. 12. 16.

(a) (c) (c) e2

4. (b,c)

2. (b) 6. (b)

1. (a)

2. (d)

3. (d)

4. (b)

5. (c)

6. (d)

7. (b)

8. (b, c)

9. (a,b, d)

10. x ∈ (– ∞,– 1 ) ∪ [ 0, ∞ ),[ − 1, 0 )

11. f and f ′′ are continuous and f ′ is discontinuous at x = {1, 2 }.

Topic 6 x + a + 1, ( x + a − 1 ) 2,  2. g{ f ( x )} =  2 x + b , ( x − 2 ) 2 + b,

if x < −a if a ≤ x < c if 0 ≤ x ≤ 1 if x > 1

a = 1, b = 0 gof is differentiable at x = 0 4 − x , 2 < x ≤ 3  4. g ( x ) = 2 + x, 0 ≤ x ≤ 1, discontinuous at x = {1, 2 } 2 − x, 1 < x ≤ 2  Discontinuity of g at x = {1, 2 }

Topic 7

Topic 4 1. (a) 5. (c)

16. (d)

Topic 5

1. (b, c, d)

Topic 3 1. (c) 5. (1)

–3 1 , c = and b ∈ R 2 2

12. (5) 31. loge 4

Topic 2 1. 5. 9. 13. 17.

15. a =

3. (b) 7. (a,b,d)

4. (c) 8. (a, d)

9. (b,d) 10. f ( x ) = 4 − x 2 2 π –π 11. a = , b = e 2/3 12. a = 8 13. a = ,b = 3 6 12 2   2 1 3  log  3 – 6 x, x ≤ 0 14. f ( x ) =  1/ x   1 + x  , x>0   2 + x 

1. 5. 9. 13. 17. 21. 25. 29. 32. 36. 40.

(b) (b) (c) (b) (d) (c) (d) (a,b,d) (a,b) (b, c) (a, c, d)

2. 6. 10. 14. 18. 22. 26. 30. 33. 37. 41.

(a) (a) (a) (c) (d) (d) (a,b,d) (b,c,d) (b,c) (a, b, c, d) (a, b)

3. 7. 11. 15. 19. 23. 27. 31. 34. 38. 42.

(b) (a) (b) (b) (a) (a) (b,c,d) (b,c) (a,d) (b, c) (b, d)

If

4. 8. 12. 16. 20. 24. 28.

(c) (a) (d) (a) (d) (a) (b,c)

35. (b,c) 39. (a, d) 43. (a, b, d)

202 Limit, Continuity and Differentiability 44. 45. 46. 48.

9. 13. 17. 21.

(b) (A) → p, q, r, s; (B) → p, s; (C) → r, s; (D) → p, s (A) → p; (B) → r 47. (24) + − f ′( 0 ) = 0, f ′( 0 ) = 1

49. x = 0

51. (a = 1 )

50. True

54. (1, 2 )

55. (i) Yes (ii) No

56. ( − 1 )

57. e 2x

58. { 0, 1, 2 )

2  52. 1 −   π 60. f ′( 0 ) = 0

61. g ( x ) is differentiable for all x ∈ ( − 2, 2 ) − { 0, 1 )

2. (b) 6. (b)

3. (d) 7. (b)

(a) (d) (b) x ∈ R − { 0}

12. 16. 20. 24.

(d) (b) (d) 1

27. 0

34. (11)

3

36. e x sin x (3 x 3 cos x 3 + sin x 3 ) + (tan x ) x [2 x cosec 2 x + log (tan x )] 5  – 2 sin ( 4 x + 2 ), x < 1 3 (1 – x ) 2 37.  –5 – 2 sin ( 4 x + 2 ), x > 1  2 3 ( x – 1 )

Topic 8 1. (b) 5. (d)

11. 15. 19. 23.

3

π π2 –3 −8 32 33. + loge 2 16 + π 2

68. p (2 ) = 0

67. (3)

(b) (a) (a) 1 1 25. –4 26. e –2 ( x 2 – x – 1 ) 2  2x – 1  28. ⋅ sin  2   x + 1 (x 2 + 1)2 30.

62. g ( x ) is continuous for all x ∈ ( 0, 2 ) − {1 ] and g ( x ) is differentiable for all x ∈ ( 0, 2 ) − {1 }  2 1 π 64.  +  63.  −  65. (4)  9 2 4 66. (2)

10. 14. 18. 22.

(c) (a) (c) (c)

4. (b) 8. (b)

38. (7)

Hints & Solutions Topic 1

0 ∞ and Form 0 ∞

1. Let P = lim

x→ 0

x + 2 sin x x + 2 sin x + 1 − sin x − x + 1 2

2

0  form 0 

On rationalization, we get (x + 2 sin x) P = lim 2 x → 0 x + 2 sin x + 1 − sin 2 x + x − 1 × ( x2 + 2 sin x + 1 + sin 2 x − x + 1 )

…(iii) ⇒ 1 − b =5 ⇒ b = −4 On putting value of ‘b’ from Eq. (iii) to Eq. (ii), we get a = −3 ⇒ a + b = − 7

= lim ( x2 + 2 sin x + 1 + sin 2 x − x + 1 ) x→ 0

× lim

x→ 0

= 2 × lim

x→ 0

x + 2 sin x x − sin 2 x + 2 sin x + x 2

x + 2 sin x x2 − sin 2 x + 2 sin x + x

0  form  0

Now applying the L′ Hopital’s rule, we get 1 + 2 cos x P = 2 × lim x → 0 2 x − sin 2 x +2 cos x + 1 (1 + 2) [on applying limit] =2 0 −0 + 2 + 1 3 =2 × =2 3 x + 2 sin x =2 ⇒ lim x→ 0 x2 + 2 sin x + 1 − sin 2 x − x + 1 x2 − ax + b =5 x→1 x−1

2. It is given that lim

Since, limit exist and equal to 5 and denominator is zero at x = 1 , so numerator x2 − ax + b should be zero at x = 1, So …(ii) 1 − a + b =0 ⇒ a =1 + b On putting the value of ‘a’ from Eq. (ii) in Eq. (i), we get x2 − (1 + b) x + b (x2 − x) − b(x − 1) lim = 5 ⇒ lim =5 x→1 x→1 x−1 x−1 (x − 1) (x − b) ⇒ lim = 5 ⇒ lim (x − b) = 5 x→1 x→1 x−1

…(i)

3. Given, lim

x→1

= lim

x→ k

(x − 1)(x + 1)(x2 + 1) x4 − 1 x3 − k3 ⇒ lim = lim 2 2 x→1 x −1 x − 1 x→ k x − k 8 3k2 (x − k)(x2 + k2 + xk) ⇒ k= ⇒ 2 ×2 = 2k (x − k)(x + k) 3

4. Given limit is lim x→ 0

= lim

sin 2 x

x x→ 0 2 − 2 cos 2 sin 2 x = lim x→ 0 x  2 1 − cos   2

sin 2 x 2 − 1 + cos x

0  form 0  x  Q 1 + cos x = 2 cos 2  2 

Limit, Continuity and Differentiability 203 sin 2 x

= lim

 x 2 × 2 sin 2   4

x→ 0

= lim x→ 0

x x  Q 1 − cos = 2 sin 2  2 4 

x2  x 2 2   4

16 = =4 2 2 2

2

[lim sin x = lim x] x→ 0

x→ 0

cot3 x − tan x x → π /4 π  cos  x +   4

5. Given, limit = lim

1 − tan 4 x

= lim

y→ 0

y ( 1 + 1 + y + 2 ) ( 1 + y + 1) 4

4

1 = lim ⋅ h→ 0 8

=

2 (sec2 x) cos 2 x − sin 2 x × x → π / 4 cos x − sin x cos 2 x tan3 x [Q 1 + tan 2 x = sec2 x]

1 = lim 8 h→ 0

2 sec4 x (cos x − sin x) (cos x + sin x) = lim × x → π /4 (cos x − sin x) tan3 x 2 2 [Q (a − b ) = (a − b) (a + b)] x → π /4

=

2 sec4 x (cos x + sin x) tan3 x

2 ( 2)  1 1 +    2 2 (1)3

=

2

×

1 + 1 + y4 + 2

11.

PLAN

1 + 1 + y4 + 2

[rationalising the numerator] = lim

y→ 0

= lim

y→ 0

(1 + 1 + y4 ) − 2 y4 ( 1 + 1 + y 4 + 2 ) 1 + y4 − 1 y4 ( 1 + 1 + y4 + 2 )

[Q (a + b) (a − b) = a − b ] 2

×

sin θ   Q lim =1   θ→ 0 θ

2 sin 2 x(3 + cos x) (1 – cos 2x)(3 + cos x) = lim tan 4x x→ 0 x→ 0 x tan 4x x× × 4x 4x (3 + cos x) 2 sin 2 x 1 = lim × lim × tan 4x x→ 0 x→ 0 4 x2 lim x→ 0 4x 4 sin θ tan   = 2 × ×1 Q lim = 1 and lim =1   θ θ → 0 → 0 4 θ θ  =2

y4

y

[Q sin (π − θ ) = sin θ ]

10. We have, lim

1 + 1 + y4 − 2

4

h3 cos h

 sin 2 x sin π sin 2 x × (π )  2  = π 2 x→ 0 π sin x  x 

x tan x   Q lim = 1 = lim  x→ 0 sin x x→ 0 x 

1 + 1 + y4 − 2

 h sin h ⋅ sin 2   2

sin(π cos 2 x) sin π (1 − sin 2 x) = lim 2 x→0 x→ 0 x x2 sin(π − π sin 2 x) = lim x→ 0 x2

= lim

 x   tan 2x 4   .  .  sin x  2x  1

y→ 0

y→ 0

cos h ⋅ h3

sin(π sin 2 x) x→ 0 x2

2

1 4 ⋅ 1 ⋅1 ⋅1 ⋅ = 1 4 1

= lim

1 = lim 4 h→ 0

= lim

1 4x 4 (tan 4x)

7. Clearly, lim

h  sin h 2 sin 2   2

h  sin  1  sin h   2 ⋅ 1 ⋅ 1 = 1 × 1 = 1 = lim    4 h → 0 h   h  cos h 4 4 4 16  2 

[on applying limit]

x cot 4x x 1 tan 2 2x = lim lim . 2 2 x→0 sin x. cot 2 x x→0 tan 4 x sin 2 x 1 2 2 1 4x x tan 2x = lim . x→0 4 (tan 4 x) sin 2 x x2 x→ 0

π  π  cos  − h 1 − sin  − h  2  2  3  π  π π sin  − h  − + h  2  2 2

9. lim

4

= lim

1 2 2 ×2

2

 2 = 4 2   = 8.  2

6.

=

sin h (1 − cos h ) 1 lim 8 h→ 0 cos h ⋅ h3

= lim

= lim

4

(by cancelling y4 and then by direct substitution). 1 . = 4 2 1 cos x(1 − sin x) cot x − cos x 8. lim = lim ⋅ 3 x → π/ 2 8 x→ π / 2 (π − 2 x)3 π  sin x − x 2 

1   Q cot x =  tan x 

1 × = lim 3 x → π /4 1 (cos x − sin x) tan x 2 2 (1 + tan 2 x) (1 − tan 2 x) = lim × x → π / 4 cos x − sin x tan3 x

y4

2

1 + y4 + 1 1 + y4 + 1

[again, rationalising the numerator]

 ∞  form   ∞ a0x + a1x n

lim

x→∞

n−1

+ K + an

b 0x m + b1x m − 1 + K + b m

 0,  a0  , =  b0  + ∞,  − ∞, 

if n< m if n = m if n> mand a0b 0 > 0 if n>m and a0b 0 < 0

Description of Situation As to make degree of numerator equal to degree of denominator.   x2 + x + 1 lim  − ax − b = 4 ∴ x→ ∞  x+1 

204 Limit, Continuity and Differentiability ⇒

lim

x→ ∞

x2 + x + 1 − ax2 − ax − bx − b =4 x+1

= lim

x→ 0

x2(1 − a ) + x(1 − a − b) + (1 − b) lim =4 x→ ∞ x+1

⇒

1 − 1 + tan 2 x  2x tan x   2  1 − tan x  = lim x→ 0 4 sin 4 x

Here, we make degree of numerator = degree of denominator ∴

3

1−a =0 ⇒ a =1 x (1 − a − b) + (1 − b) lim =4 x→ ∞ x+1

and ⇒

2x tan3 x 1 = lim x → 0 2 sin 4 x (1 − tan 2 x) x → 0 2 sin 4 x (1 − tan 2 x)

b = − 4 [Q (1 − a ) = 0]

12. Here, lim

h→ 0

x→ 0

 x x2 1 + x + + K 2 3!  2

 x 2   xn − 3  2

In trigonometry try to make all trigonometric functions in same angle. It is called 3rd Golden rule of trigonometry.

= lim

x→ 0

2 tan x − 2x tan x 1 − tan 2 x (2 sin 2 x)2

16. LHL = lim− x→1

= lim

x → 1−

1 − cos 2 (x − 1) x −1 2 sin 2 (x − 1) |sin (x − 1)| = 2 lim x −1 x −1 x → 1−

Put x = 1 − h , h > 0, for x → 1− , h → 0 sin h |sin (− h )| = 2 lim = 2 lim =− 2 h → 0 h→ 0 −h −h Again, RHL = lim x→1

+

= lim

x →1+

1 − cos 2 (x − 1) x−1 |sin (x − 1)| 2 x−1

x = 1 + h, h > 0 = lim 2

4

Above limit is finite, if n − 3 = 0, i.e. n = 3. x tan 2x − 2x tan x 15. lim x→ 0 (1 − cos 2x)2

x

1 1 ⋅ (1)3 1 lim = = 4 2 x → 0  sin x 4 2(1) (1 − 0) 2 2  (1 − tan x)   x 

For x → 1 , h → 0

   x x − ...   1 − + 2! 4 !    2 x   −2 sin      x2 x3 2  + + ...  − 1 + x +  2! 3!    = lim n x→ 0 x 2   x x x3 2   2 − − ...  − 2 sin   − x −  2  2! 3!  = lim 2 x→ 0  x 4   xn − 2  2

= lim

3

+

2

sin 2

 tan x    x 

Put

(cos x − 1) (cos x − ex ) x→ 0 xn

14. lim

NOTE

=

f (2h + 2 + h 2) − f (2) f (h − h 2 + 1) − f (1)

[Q f ′ (2) = 6 and f ′ (1) = 4, given] Applying L’Hospital’s rule, { f ′ (2h + 2 + h 2)} ⋅ (2 + 2h ) − 0 f ' (2) ⋅ 2 = = lim h → 0 { f ′ (h − h 2 + 1 )} ⋅ (1 − 2 h ) − 0 f ' (1) ⋅ 1 6 .2 [using f ′ (2) = 6 and f ′ (1) = 4] = =3 4 .1 {(a − n ) nx − tan x} sin nx 13. Given, lim =0 x→ 0 x2 tan x  sin n x  ⇒ lim (a − n ) n − ×n =0  x→ 0 x  nx 1 ⇒ {(a − n ) n − 1} n = 0 ⇒ (a − n ) n = 1 ⇒ a = n + n

 tan x 3 x  ⋅x  x 

= lim

1−a −b =4

⇒

1   −1 1 − tan 2 x  4 sin 4 x

2x tan x

h→ 0

∴

LHL ≠ RHL.

Hence,

17. lim x→ 0

At

and

|sin h | sin h = lim 2 = 2 h→ 0 h h

lim x→1

f (x) does not exist.

1 (1 − cos 2 x) 1 .|sin x| 2 = lim x→ 0 2 x x x=0 1 . sin h 1 RHL = lim = h→ 0 2 h 2 1 1 . sin h LHL = lim =− h→ 0 2 −h 2

Here, RHL ≠ LHL ∴ Limit does not exist. sin [x] , [x] ≠ 0  18. Since, f (x) =  [x] [x] = 0  0, ⇒

sin [x] , x ∈ R − [0, 1)  f (x) =  [x]  0, 0 ≤ x0

x4 1 1     x2  1 x2 2  2 − 1 x4 ⋅ 4 − ... − a − a ⋅ 1 − ⋅ 2 + 2 a  2 a  4   = lim x→ 0 x4

Now,

fn (x) = (tan − 1 (x + 1) − tan − 1 x) + (tan − 1 (x + 2) − tan − 1 (x + 1))

a − a 2 − x2 −

x→ 0

∴

n

j=1

⇒

x→ ∞

= 1 + lim tan 2( fn (x)) = 1 + 0 = 1

r B

h–r

and

D

C

Here, BD = r 2 − (h − r )2 = 2hr − h 2 1 ∴ A = . 2BD. h = ( 2hr − h 2) h 2

206 Limit, Continuity and Differentiability ∴

lim

h→ 0

h h  a 2 ⋅ 2 cos  a +  ⋅ sin  2 2 = lim + (2a + h ) sin (a + h ) h h→ 0 2⋅ 2 = a 2 cos a + 2a sin a

h 2hr − h 2 A = lim 3 h→ 0 P 8 ( 2hr − h 2 + 2hr )3 = lim

h→ 0

h3/ 2 ( 2r − h ) 8 h3/ 2 ( 2r − h + 2r )3

lim(x − sin x)1/ 2 x − sin x 33. lim = x→ 0 x→ 0 x + cos 2 x lim(x + cos 2 x)1/ 2

1 1 2r = = ⋅ 8 ( 2r + 2r )3 128 r

x→ 0

1/ 2

sin x    lim x 1 −  x→ 0   x   = lim(0 + 1)1/ 2

1 1  4 2 x4 sin + x2  x sin + x  x x 27. lim   = lim 1 − x3 1 + |x |3  x → −∞ x → −∞   

x→ 0

 x3 + x2 − 16x + 20  (x − 2)2  k

1 1 x −1  =−  = lim x → 1 ( )( ) ( x − ) 3 x x 2 5 − − 1 2 5  

34. lim  x→1

x2 sin (βx) =1 x → 0 αx − sin x

35. Here, lim

  (βx)3 (βx)5 + − K x2  β x − 3! 5!   lim =1 x→ 0   x3 x5 αx −  x − + − K 3! 5!  

⇒

, if x ≠ 2 , if x = 2

Since, continuous at x = 2. x3 + x2 − 16x + 20 ,[using L’Hospital’s rule] x→ 2 (x − 2)2

⇒ f (2) = lim

6x + 2 3x2 + 2x − 16 = lim =7 x→ 2 x→ 2 2 2 (x − 2)

= lim ∴

k=7

29. lim (1 − x) tan x→1

πx 2

x−1 = y π  π  ∴ − lim y tan ( y + 1) = − lim y − cot  y  2  y→ 0 y→ 0  2 π    2 2  y 2 = lim  ⋅ = y→ 0  tan π y π π  2  Put

x→ a

x→ 0

x→ 0

x→ a

(2x − 1)( 1 + x + 1) x

= log e (2) ⋅ (2) = 2 log e 2 = log e 4 (a + h )2 sin (a + h ) − a 2 sin a h→ 0 h a 2[sin (a + h ) − sin a ] = lim h→ 0 h h [2a sin (a + h ) + h sin (a + h )] + h

32. Here, lim

lim

x→ 0

(α − 1)x +

 ecos ( α n ) − e  e =− m α→ 0 2 α  

2x − 1 1+ x+1 × 1 + x −1 1+ x+1 = lim

=1 x3 x5 + −K 3! 5! Limit exists only, when α − 1 = 0 ⇒ α =1 3 2   x β β5 x4 + − K x3 β − 3! 5!   lim ∴ =1 2 x→ 0   x 1 3 x  − − K 3! 5!  ⇒

36. Given, lim 

may or may not exist. Hence, it is a false statement.

31. lim

  β3 x2 β 5 x4 + − K x3 β − 3! 5!  

⇒ 6β = 1 From Eqs. (i) and (ii), we get 6(α + β ) = 6α + 6 β = 6 + 1 = 7

30. If lim [ f (x) g (x)] exists, then both lim f (x) and lim g (x) x→ a

0 .0 =0 1



On dividing by x3 , we get sin (1 / x) 1 + 1 x 1+0 x = = −1 lim 1 x → −∞ 0 −1 −1 3 x

28. f (x) = 

=

n

⇒

lim

α→ 0

e { ecos( α ) −1 − 1} cos(α n ) − 1 − e ⋅ = 2 cos(α n ) − 1 αm

αn −2 sin 2  ecos( α n ) −1 − 1  2 = − e /2 ⇒ lim e   ⋅ lim m α→ 0  cos(α n ) − 1  α→ 0 α    α n sin 2   2  α 2n −e ⋅ m= e × 1 × (−2) lim ⇒ 2n α→ 0 2 4α α 4 α 2n − m − e ⇒ e × 1 × − 2 × 1 × lim = α→ 0 4 2 For this to be exists, 2n − m = 0 m ⇒ =2 n

…(i)

…(ii)

Limit, Continuity and Differentiability 207

Topic 2 1∞ Form, RHL and LHL  1 + f (3 + x) − f (3) 1. Let l = lim  x→ 0  1 + f (2 − x) − f (2 ) 

3.

[1 form]

 1 + f ( 2 − x ) − f ( 2) − 1 − f (3 + x ) + f (3 )  lim   x(1 + f ( 2 − x ) − f ( 2)) 

+

x→ 0

= lim

x→ 0 +

 f ( 2 − x ) − f (3 + x ) + f (3 ) − f ( 2)  lim   x(1 + f ( 2 − x ) − f ( 2)) 

x→ 0 

On applying L’Hopital rule, we get  lim 

 

− f ′( 2 − x )− f ′(3 + x )

x→ 0  1 − xf ′( 2 − x ) + f ( 2 − x ) − f ( 2) 

l=e

On applying limit, we get  − f ′( 2) − f ′(3 )     1 − 0 + f ( 2) − f ( 2) 

l=e

 1 + f (3 + x) − f (3) So, lim  x→ 0  1 + f (2 − x) − f (2 ) 

2. Let L = L=

π − 2 sin 1−x

lim x → 1−

lim x → 1−

1 x

= e0 = 1

=

lim x → 1−

−1

x

, then

π + 2 sin −1 x

=

lim 2 cos −1 x × − x→ 1 x → 1− 1−x

=

lim 2 cos −1 x 1 − 1−x 2 π x→ 1

lim

×

1 2 2 = 2 π

2 π

  tan(π sin 2 x) π sin 2 x . = lim  + 1 2 2 + π sin x x→ 0  x 

π + 2 sin −1 x π  Q sin −1 x + cos −1 x =  2  1 π + 2 sin −1 x lim  π −1 Q x → 1− sin x = 2   

lim   θ Q x → 0+ sin θ = 1  

+

sin 2 x tan (π sin 2 x) . lim +1 2 + x→ 0 x2 π sin x tan x  Q lim =1  x→ 0 x sin x  =1 and xlim →0 x 

=π+1

    

and LHL = lim x→ 0

−

x → 0−

tan (π sin 2 x) + (|x| − sin (x [x])2 x2 tan (π sin 2 x) + (− x − sin(x (− 1))2 x2  Q|x| = − x for x < 0  and [x] = − 1 for − 1 < x < 0  

1

Put x = cos θ, then as x → 1− , therefore θ → 0+ lim 2θ 1 Now, L = + 1 − cos θ 2 π θ→0 lim 2θ θ 1  = Q 1 − cos θ = 2 sin 2 +  θ   2  2 π θ→0  2 sin    2  θ 2⋅    2 1 = ⋅ 2 lim  θ 2 π θ→ 0 + sin    2 =

x → 0+

= lim

1

1−x

tan (π sin 2 x) + x2 x2

= lim

x→ 0

=1

[on rationalization]

π  π − 2 − cos −1 x 2 

tan(π sin 2 x) + (x − sin(x ⋅ 0))2 x2  Q|x| = x for x > 0  and [x] = 0 for 0 < x < 1  

= π lim

π − 2 sin −1 x π + 2 sin −1 x × 1−x π + 2 sin −1 x

lim π − 2 sin −1 x × = x → 1− 1−x

tan(π sin 2 x) + (|x| − sin(x [x]))2 x2

RHL = lim

x→ 0 

=e

x → a−

At x = 0,

l=e

=e

lim f ( x) = lim f ( x)

x → a+

∞

1 + f (3 + x ) − f (3 )  1  lim  1 − 1 + f ( 2 − x ) − f ( 2)  x→ 0 x 

⇒

Key Idea lim f ( x) exist iff x→a

1 x

= lim

x → 0−

tan(π sin 2 x) + (x + sin(− x))2 x2

tan(π sin 2 x) + (x − sin x)2 x → 0− x2 [Qsin (− θ ) = − sin θ] 2 2 2  tan(π sin x) + x + sin x − 2x sin x = lim   x → 0−  x2  = lim

 tan(π sin 2 x) sin 2 x 2x sin x = lim  +1+ −  2 x → 0−  x x2 x2   tan (π sin 2 x) π sin 2 x sin 2 x sin x . + 1+ = lim  −2  2 2 x  π sin x x2 x → 0−  x π sin 2 x tan(π sin 2 x) . lim + 2 − π sin x x→ 0 x→ 0 x2 sin 2 x sin x − 2 lim 1 + lim − x x→ 0 x2 x → 0− = π + 1 + 1 −2 = π = lim

−

Q RHL ≠ LHL ∴ Limit does not exist

208 Limit, Continuity and Differentiability  n  n n  = −  x  x  x  15 15   1 1  2 2  ∴Given limit = lim x −   + −   + … −   + x x  x  x→ 0 x x x  1  2  15  = lim (1 + 2 + 3+ ...+15) − x   +   + ... +   +  x→ 0  x  x x

4. Given,

Similarly,

 π (1 − |x| + sin|1 − x|) sin  [1 − x]  2

lim

|1 − x|[1 − x]

x →1+

Put x = 1 + h , then x → 1+ ⇒ h → 0+  π (1 − |x| + sin|1 − x|) sin  [1 − x]  2

∴ lim

= lim h→ 0

= 120 − 0 = 120

|1 − x|[1 − x]

x →1+

π  (1 − |h + 1| + sin|− h|) sin  [− h ] 2 

7.

|− h|[− h ]

+

π  (1 − (h + 1) + sin h ) sin  [− h ] 2  = lim + [ − ] h h h→ 0

 1  1 − x( 1 + |1 − x|) cos    1 − x |1 − x|

f (x) =

Now, lim f (x) = lim x→1 −

x→1 −

π  ( − h + sin h ) sin  (− 1) 2  = lim h (− 1) h→ 0 +

and

h→ 0 +

8. Given,

x→ 0

−

x(− 1 − x) sin (− 1) (Q lim [x] = − 1) −x x→ 0 x → 0− − x(x + 1) sin(− 1) = lim = lim (x + 1)sin(− 1) −x x → 0− x → 0−

= lim

−

= (0 + 1) sin (− 1) (by direct substitution) (Qsin(− θ) = − sin θ) = − sin 1 Key Idea Use property of greatest integer function [ x ] = x − { x }.

 1  lim x + +   x x→ 0  

15  2  + …+  x   x 

We know, [x] = x − { x} 1  1 1  ∴ = −  x  x  x 

 1  1 − x(1 − 1 + x) cos    1 − x x−1

x → 0+

=e

x([x] − x) sin [x] x([x] + | x|) sin [x] = lim − | x| −x x→ 0

+

p = lim (1 + tan 2 x ) lim

(Q| x| = − x, if x < 0)

6.

x→1

x → 0+

tan 2 x 2x

1

  sin h  h  sin h  = lim  = 1  − lim+   = 1 − 1 = 0 Q lim+ +     h h h h→ 0 h→ 0   h→ 0 lim

+

 1  = lim − (x + 1) ⋅ cos   , which does not exist. +  x + 1 x→1

 h  sin h  = lim   − lim+   h→ 0 +  h  h → 0  h

5.

lim f (x) = lim

x→1

(Q [x] = − 1 for − 1 < x < 0 and h → 0+ ⇒ − h → 0− ) (− h + sinh)  − π sin    2  −h

 1  1 − x(1 + 1 − x) cos    1 − x 1−x

 1  = lim (1 − x) cos   =0  1 − x x→1 −

(Q|− h| = h and|h + 1| = h + 1 as h > 0)

= lim

  n   Q 0 ≤  x  < 1, therefore   n  n  0 ≤ x  < x ⇒ lim+ x  = 0 x→ 0 x x  

∴

9.

PLAN

log p = log e2 =

1 2x

(1∞ form)

1  tan x  lim   2x → 0+  x 

2

=e

1

= e2

1 2

To make the quadratic into simple form we should eliminate radical sign.

Description of Situation As for given equation, when a → 0 the equation reduces to identity in x. i.e. ax2 + bx + c = 0, ∀ x ∈ R or a = b = c → 0 Thus, first we should make above equation independent from coefficients as 0. Let a + 1 = t 6. Thus, when a → 0, t → 1. ∴ (t 2 − 1) x2 + (t3 − 1)x + (t − 1) = 0 ⇒ (t − 1) {(t + 1) x2 + (t 2 + t + 1) x + 1} = 0, as t → 1 2 x2 + 3 x + 1 = 0 ⇒

2 x2 + 2 x + x + 1 = 0

⇒

(2x + 1) (x + 1) = 0 x = − 1, − 1 / 2

Thus, or and

lim α (a ) = − 1 / 2

a→ 0 +

lim β (a ) = − 1

a→ 0 +

Limit, Continuity and Differentiability 209 [1∞ form]

10. Here, lim {1 + x log (1 + b2)}1/ x x→ 0

1 lim {x log (1 + b )} ⋅ x x→ 0

 x + 6  x → ∞  x + 1

x+4

15. lim 

2

 5  = lim 1 +  x→ ∞  x + 1

=e

log (1 + b 2 )

=e

= (1 + b ) 2

⇒

1 + b2 2b

sin 2 θ =

…(ii)

1 1/ 2 b ≥  b ⋅ 1   b 2

b+

By AM ≥ GM,

b2 + 1 ≥1 2b

⇒

 π  lim tan  + x  4  x→ 0 1/ x π   1/ x tan + tan x 1 + tan x    4 = lim  = lim    π x → 0  1 − tan x  x→ 0 1 − tan tan x    4 [(1 + tan x)1/tan x ]tan x/ x e1 = lim = = e2 x → 0 [(1 − tan x)−1/tan x ]− tan x/ x e−1

17. The right hand limit lim

x →0 +

sin 2 θ = 1

(1 − x)1/ x − e−1 xa

π θ = ± , as θ ∈ (− π , π ] 2

1 11. Here, lim (sin x)1/ x + lim     x→ 0

x→ 0

1 log    x

sin x

= 0 + lim e

 1  x −   x2  x → 0 − cosec x cot x

e

=e

= lim = e0 = 1

18.

PLAN lim

x→0

[using L’ Hospital’s rule]

e

−1

xa

sin x =1 x (1 +

x ) (1 − x ) 1− x

1+

  sin (x − 1) −a   (x − 1) lim  sin(x − 1)  x→1  1 + (x − 1)  

lim y = e2 x

 x − 3 (1 − 3 / x)x e−3 = 2 = e−5  = lim x→∞ (1 + 2 / x)x  x + 2 e

13. For x ∈ R , lim   1 + 5x lim   x → 0  1 + 3 x2 

 x x2 x3  −  + + + ...  3 4 2 

sin (x − 1) + a (1 − x)  Given, lim   x → 1  (x − 1 ) + sin (x − 1 ) 

  ⇒ log  lim y = 2  x→ 0 

14.

x

− e−1

a

The above limit will be non-zero, if a = 1. And at a = 1, the value of the limit is 1  1 = e−1  −  = −  2 2e

f (1) 6 = = f (1) 3

2  1/ x

e−1 . e

x →0 +

1 ⇒ log y = [log f (1 + x) − log f (1)] x   1 ⋅ f ′ (1 + x) ⇒ lim log y = lim  x→ 0 x → 0  f (1 + x) 

x→ ∞

  x x2 x3  − −  2 3 − 4 − ......   

+

= e−1 lim

1/ x

x→ 0

xa

x →0 +

x →0

sin x tan x x→ 0 x lim

− e−1

e

= lim

 lim (sin x)1/ x → 0  x → 0  ∞ as, (decimal) → 0

 f (1 + x)  12. Let y =   f (1) 

⇒

x

 1  x2 x3 −x− − − K 2 3 x  

x

log (1/ x ) lim x → 0 cosec x

− e−1

a

x →0 +

Applying L’Hospital’s rule, we get lim

1   log e (1 − x ) 

e x

= lim

sin x

=e

x→ 0

= e5

…(iii)

From Eqs. (ii) and (iii), ⇒

5 ( x + 4) x +1

[1∞ form]

1/ x

16.

(1 + b2) = 2b sin 2 θ

∴

=e

…(i)

Given, lim {1 + x log (1 + b2)}1/ x = 2b sin 2 θ x→ 0

lim

x→ ∞

x+ 4

2 1/5 x 2 5

2

=

lim [(1 + 5x )

x→ 0

]

2 1/3 x 2 3

lim [(1 + 3x )

x→ 0

]

=

e5 = e2 e3

=

1 4

=

1 4

x

2

⇒

1 1 − a   =  2  4

⇒

a = 2 or 0

⇒ (a − 1)2 = 1

Hence, the maximum value of a is 2.

210 Limit, Continuity and Differentiability Topic 3 Squeeze, Newton-Leibnitz’s Theorem and Limit Based on Coverting infinite Serie s into Definite Integrals 1. Given α and β are roots of quadratic equation 375x2 − 25x − 2 = 0 25 1 = 375 15 2 αβ = − 375

∴

α+β=

and n

n

r =1

r =1

… (i) … (ii)

∑ α r + nlim ∑ βr n→ ∞ →∞

= (α + α 2 + α 3 + K + upto infinite terms)+ (β + β 2 + β3 + K + upto infinite terms)   a Q S ∞ = 1 − r for GP  

α β + 1 −α 1 −β

=

α (1 − β) + β (1 − α ) (1 − α ) (1 − β )

=

α − αβ + β − αβ 1 − α − β + αβ

=

(α + β ) − 2αβ 1 − (α + β ) + αβ

On substituting the value α + β =

1 −2 and αβ = from 15 375

Eqs. (i) and (ii) respectively, we get 1 4 + 1 29 29 = 15 375 = = = 1 2 − − 375 25 2 348 12 1− − 15 375  0 form 2  0 π x2 − 16 f (sec2 x) 2 sec x sec x tan x = lim x→ π /4 2x [using L’ Hospital’s rule] 2 f (2) 8 = = f (2) π /4 π 2

π x→ 4

= lim

n→ ∞

=∫

2 0

1 n

2n

∑

r =1 2n

∑

r =1

x 1 + x2

1 = lim 2 2 n→ ∞ n n +r r

2n

∑

r =1

r n 1 + (r / n )2

1 + (r / n )

∴ ⇒

∑

r =1

  r  ⋅x + 1   log  n2  r 2  2 ⋅ x + 1  n

x  1 + z  dz log e { f (x)} = x ∫ log   0  1 + z 2 x x  1+ z log e{ f (x)} = ∫ log   dz 0  1 + z 2

Using Newton-Leibnitz formula, we get  1 + x 1 ⋅ f ′ (x) = log   f (x)  1 + x2 

2

dx = [ 1 + x2 ]2 = 5 − 1 0

n

tx=z xdt = dz

Put, ⇒

Here, at x = 1 ,

r /n

x

Converting summation into definite integration, we get 1  xt + 1  log e{ f (x)} = x ∫ log  2 2  dt 0  x t + 1

∫ f (t ) dt

1 3. Let I = lim n→ ∞ n

x

n n n    … x +    2 n  ,x>0 2  n n 2   K  x2 + 2   4 n   

n  n n     n n (x + n )  x +  K  x +     2 n  log e { f (x)} = lim log  2  n →∞  n  n 2  2 2  2  K  x2 + 2    n !(x + n )  x + 4 n      n   1   Π x +     r =1  x r /n  = lim ⋅ log  n→ ∞ n 1  n   n  2 Π (r / n )  x +  rΠ 2 r = 1 =1   ( / ) r n     n   x+ 1 n r   log = x lim ∑ 2   n→ ∞ n  n r 2 r =1 x + 2  n r    

1 = x lim n→ ∞ n

sec 2 x

2. lim

   n n (x + n )  x +  f (x) = lim  n→ ∞  2 2  2  n ! (x + n )  x +  

Taking log on both sides, we get

Now, lim

=

4. Here,

∴

f ′ (1) = log (1) = 0 f (1) f′ (1) = 0

… (i)

Limit, Continuity and Differentiability 211 Now, sign scheme of f ′ (x) is shown below +

x2

–

6. lim

∫0

x→ 0

Applying L’Hospital’s rule, we get

∴ At x = 1 , function attains maximum. Since, f (x) increases on (0, 1). ∴ f (1) > f (1 / 2) ∴ Option (a) is incorrect. f (1 / 3) < f (2 / 3) ∴Option (b) is correct.

2 ⋅ cos 2(x2) 2 cos 2(x2) ⋅ 2x − 0 = =1 = lim sin x 1 + 1 x→ 0 x → 0 x cos x + sin x cos x + x

= lim

Topic 4 Continuity at a Point 1. Given function is  2 cos x − 1  f (x) =  cot x − 1  k 

Also, f ′ (x) < 0, when x > 1 ⇒ f′ (2) < 0 ∴ Option (c) is correct.  1 + x f ′ (x) Also, = log   f (x)  1 + x2

= log (2 / 3) < 0 f ′ (3) f ′ (2) < f (3) f (2)

4

lim π k= x→ 4

⇒

∴ Option (d) is incorrect.

5. We have, 1 yn = [(n + 1) (n + 2) … (n + n )]1/ n n and lim yn = L n→ ∞ 1 ⇒ L = lim [(n + 1) (n + 2) (n + 3) … (n + n )]1/ n n→ ∞ n

Put x =

2 cos x − 1 cot x − 1

π π + h, when x → , then h → 0 4 4  π 2 cos  + h − 1 lim  4 k= h→0  π cot + h − 1  4

1

⇒

1  2  3 n  n   L = lim 1 +  1 +  1 +  ... 1 +          n→∞  n n n n 

⇒ log L = lim

n→ ∞

1 n

=

n  1 2     log 1 + n  + log 1 + n  … log 1 + n    

⇒ log L = lim

n→ ∞

1 n

n

∑ log 1 + n 

r =1

=

1

⇒ log L = ∫ 1 × log (1 + x) dx 0 II

I

 (log(1 + x) ∫ dx dx  [by using integration by parts] 1 x 1 dx ⇒ log L = [x log(1 + x)]0 − ∫ 01 + x ⇒ log L = (x ⋅ log (1 + x))10 − ∫

1

d

0  dx

1 x + 1 1  ⇒ log L = log 2 − ∫  −  dx 0x+ 1 x + 1

⇒ log L = log 2 − [x]10 + [log(x + 1)]10 ⇒ log L = log 2 − 1 + log 2 − 0 4 4 ⇒ L= ⇒ log L = log 4 − log e = log e e 4  =1 [L ] =  e 

⇒

lim h→0

2

1  1 cos h − sin  2 2 cot h − 1 −1 cot h + 1

 h −1 

[Q cos (x + y) = cos x cos y − sin x sin y and cot x cot y − 1 ] cot (x + y) = cot y + cot x

r



π 4 π , x= 4 ,x≠

Q Function f (x) is continuous, so it is continuous at π x= . 4  π ∴ f   = lim f (x)  4  x→ π

f ′ (3) f ′ (2) 4  3 − = log   − log    10  5 f (3) f (2)

⇒

0  form 0 

x sin x

x=1

∴

cos 2tdt

=

lim cos h − sin h − 1 −2 h→0 1 + cot h  lim  (1 − cos h ) + sin h (sin h + cos h )  h→0  2 sin h 

h h   2h lim 2 sin 2 + 2 sin 2 cos 2  (sin h + cos h ) = h h h→0 4 sin cos     2 2 h h   lim sin 2 + cos 2  = × (sin h + cos h ) h h→0  2 cos    2 ⇒k=

1 2

212 Limit, Continuity and Differentiability 2.

NOTE All integers are critical point for greatest integer function.

6. For f (x) to be continuous, we must have

Case I When x ∈ I

f (0) = lim f (x) x→ 0 log (1 + ax) − log (1 − bx) = lim x→ 0 x a log (1 + ax) b log (1 − bx) = lim + x→ 0 − bx ax log (1 + x) [using lim = a ⋅1 + b ⋅1 = 1] x→ 0 x

f (x) = [x]2 − [x2] = x2 − x2 = 0 Case II When x ∉ I If

0 < x < 1, then [x] = 0

and

0 < x2 < 1, then [x2] = 0

Next, if

1 ≤ x2 < 2 ⇒ 1 ≤ x < 2

⇒

[x] = 1

and

[x2] = 1

=a+b

Therefore, f (x) = [x]2 − [x2] = 0, if 1 ≤ x < 2 Therefore,

f (x) = 0, if 0 ≤ x < 2

This shows that f (x) is continuous at x = 1. Therefore, f (x) is discontinuous in (−∞ , 0) ∪ [ 2 , ∞ ) on many other points. Therefore, (b) is the answer.

∴

7. f (x) = x cos(π (x + [x])) At x = 0 lim f (x) = lim x cos(π (x + [x]) = 0 x→ 0

3. Given, f (x) = [tan x] Now, − 45° < x < 45° tan (− 45° ) < tan x < tan (45° ) − tan 45° < tan x < tan (45° ) −1 < tan x < 1 0 < tan 2 x < 1 ⇒ [tan 2 x] = 0

8.

Case I Let f and g attain their common maximum value at p. ⇒

f ( p) = g ( p),

where p ∈ [0, 1] Case II Let f and g attain their common maximum value at different points.

, −1 ≤ x < 0

⇒

f (a ) = M and g (b) = M

⇒

f (a ) − g (a ) > 0 and f (b) − g (b) < 0

, 0 ≤ x0  2 + x 1 log f (x) = [log (1 + x) − log (2 + x)] x

1/ x

…(i)

⇒

(a + 1) + 1 = lim

⇒

a+2=

∴ and

x→ 0

bx 1 ⋅ =c bx 1 + bx + 1

1 =c 2 3 1 a =− ,c= 2 2

b ∈R

214 Limit, Continuity and Differentiability 2x sin  1  − cos  1  , x ≠ 0      f4 ′ (x) =   x  x  0, x=0

2

16. (i) Given, f1 : R → R and f1 (x) = sin ( 1 − e−x )

Thus,

∴ f1 (x) is continuous at x = 0 1

2

Now, f1 ′ (x) = cos 1 − e− x .

2

−x

2 1− e

2

(2xe− x )

1 1   Again, lim f ′ (x) = lim  2x sin   − cos    x→ 0  x→ 0  x  x 

At x = 0

does not exists.

f1 ′ (x) does not exists.

1 Since, lim cos   does not exists. x→ 0  x

∴ f1 (x) is not differential at x = 0

Hence, f ′ (x) is not continuous at x = 0.

Hence, option (2) for P.

∴ Option (3) for S.

 |sin x| , if x ≠ 0  (ii) Given, f2 (x) =  tan −1 x  if x = 0 1,

⇒

 − sin x  tan −1 x   sin x f2 (x) =  −1  tan x  1  

Topic 5 Continuity in a Domain

x< 0

1. Given ∫

x> 0

f ( x) 6

4t3 dt = (x − 2) g (x) f ( x)

∫ g (x) = 6

⇒

x=0

4t3 dt f ( x)

Clearly, f2 (x) is not continuous at x = 0.

So, lim g (x) = lim x→ 2

∫6

x→ 2

4t3 dt

x−2  0  Q form as x → 2 ⇒ f (2) = 6  0 

∴ Option (1)for Q. (iii) Given, f3 (x) = [sin (log e (x + 2))] , where [ ] is G.I.F. and f3 : (−1, eπ / 2 − 2) → R It is given

− 1 < x < eπ / 2 − 2

⇒ ⇒ ⇒

x2 sin  1  , if x ≠ 0    (iv) Given, f4 (x) =   x if x = 0 0, 

 1 Now, lim f4 (x) = lim x2 sin   = 0  x x→ 0 x→ 0  1  1 f4′ (x) = 2x sin   − cos    x  x f (0 + h ) − f (0) For x = 0, f4′ (x) = lim h→ 0 h 1  h 2 sin   − 0  h ⇒ f4′ (x) = lim h→ 0 h  1 f4′ (x) = lim h sin   = 0 ⇒  h h→ 0

4( f (x))3 f ′ (x) x→ 2 1

lim g (x) = lim

x→ 2

π/ 2

−1 + 2 < x+ 2 < e −2 + 2 1 < x + 2 < eπ/ 2 log e 1 < log e (x + 2) < log e eπ / 2 π ⇒ 0 < log e (x + 2) < 2 π ⇒ sin 0 < sin log e (x + 2) < sin 2 ⇒ 0 < sin log e (x + 2) < 1 [sin log e (x + 2)] = 0 ∴ ∴ f3 (x) = 0, f ′3 (x) = f3 ′ ′ (x) = 0 It is differentiable and continuous at x = 0. ∴Option (4) for R

[provided x ≠ 2]

(x − 2)

 d Q  dx

φ 2( x )

∫φ

1 ( x)

 f (t ) dt = f (φ 2(x)), φ 2′ (x) − f (φ1 (x)) ⋅ φ1′ (x) 

On applying limit, we get lim g (x) = 4( f (2))3 f ′ (2) = 4 × (6)3

x→ 2

1 4 × 216 = = 18 48 48

1  Q f (2) = 6 and f ′ (2) = 48  

2. Given function sin( p + 1)x + sin x  , x0  x3 / 2 is continuous at x =`0, then f (0) = lim f (x) = lim f (x) …(i) x→ 0 −

x→ 0 +

sin( p + 1)x + sin x x sin(ax)   Q lim =a = p+1+1=p+2  x→ 0  x

lim f (x) = lim

x→ 0 −

x→ 0 −

and lim f (x) = lim x→ 0 +

x→ 0 +

= lim

x→ 0 +

x + x2 − x x3/ 2 x [(1 + x)1/ 2 − 1] x x

Limit, Continuity and Differentiability 215 1 1     − 1    2 2 1 2 1 + x + x + .... − 1 2 2!       = lim + x x→ 0

5. Given function f : [−1, 3] → R is defined as |x| + [x], −1 ≤ x < 1  f (x) =  x + |x|, 1 ≤ x < 2  x + [x ], 2 ≤ x ≤ 3  − x − 1 , − 1 ≤ x < 0  x, 0 ≤ x 0 (d) a < 0, b < 0

Show that the abscissa of P1, P2, P3 , …, Pn, form a GP. Also, find the ratio of [area (∆P1P2P3 )]/[area (∆P2P3 P4 )]. (1993, 5M)

20. Find the equation of the normal to the curve y = (1 + x)y + sin −1 (sin 2 x) at x = 0.

(1993, 4M)

21. Find all the tangents to the curve y = cos (x + y), −2π ≤ x ≤ 2π , that are parallel to the line x + 2 y = 0.

Fill in the Blank

(1985, 5M)

16. Let C be the curve y − 3xy + 2 = 0. If H is the set of 3

points on the curve C, where the tangent is horizontal and V is the set of points on the curve C, where the tangent is vertical, then H = K and V = K . (1994, 2M)

Integer & Numerical Answer Type Question 22. The slope of the tangent to the curve ( y − x5 )2 = x (1 + x2)2 at the point (1, 3) is

(2014 Adv.)

Topic 2 Rate Measure, Increasing and Decreasing Functions Objective Questions I (Only one correct option) 1. A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3 /min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is (2020 Main, 9 Jan I) 1 9π 1 (c) 36 π

5 6π 1 (d) 18 π

(b)

(a)

of all x ∈ R, where the function h (x) = ( fog ) (x) is (2019 Main, 10 April II) increasing, is −1 1 (b)  −1,  ∪  , ∞   2   2  −1 (d)  , 0 ∪ [1, ∞ )  2 

(c) [0, ∞ )

3. A water tank has the shape of an inverted right circular −1  1 

cone, whose semi-vertical angle is tan   . Water is  2 poured into it at a constant rate of 5 cu m/min. Then, the rate (in m/min) at which the level of water is rising at the instant when the depth of water in the tank is 10 m is (2019 Main, 9 April II) (a)

2 π

(b)

1 5π

(c)

1 15π

that f ′ ′ (x) > 0, for all x ∈ (0, 2). If φ (x) = f (x) + f (2 − x) , then φ is (2019 Main, 8 April I) (a) increasing on (0, 1) and decreasing on (1, 2) (b) decreasing on (0, 2) (c) decreasing on (0, 1) and increasing on (1, 2) (d) increasing on (0, 2)

5. If the function f given by

2. Let f (x) = ex − x and g (x) = x2 − x, ∀ x ∈ R. Then, the set

1 (a)  0,  ∪ [1, ∞ )  2 

4. Let f : [0, 2] → R be a twice differentiable function such

(d)

1 10π

f (x) = x3 − 3(a − 2) x2 + 3ax + 7, for some a ∈ R is increasing in (0, 1] and decreasing in f (x) − 14 [1, 5), then a root of the equation, = 0 (x ≠ 1) is (x − 1)2 (2019 Main, 12 Jan II)

(a) − 7

(b) 6

(c) 7

x

d−x

(d) 5

, x ∈ R, where a, b − a 2 + x2 b2 + (d − x)2 and d are non-zero real constants. Then,

6. Let f (x) =

(a) (b) (c) (d)

(2019 Main, 11 Jan II) f is an increasing function of x f ′ is not a continuous function of x f is a decreasing function of x f is neither increasing nor decreasing function of x

π π 7. If the function g : (−∞ , ∞ ) →  − ,  is given by  2 2 π g (u ) = 2 tan (e ) − . Then, g is 2 −1

u

(2008, 3M)

Application of Derivatives 241 (a) (b) (c) (d)

even and is strictly increasing in (0, ∞ ) odd and is strictly decreasing in (−∞ , ∞ ) odd and is strictly increasing in (−∞ , ∞ ) neither even nor odd but is strictly increasing in (−∞ , ∞ )

18. If f : (0, ∞ ) → R be given by  1 x −  t +  t

f (x) = ∫ e 1/ x

3 sin x − 4 sin3 x is increasing, is (a)

π 3

(b)

π 2

(c)

(2002, 2M)

3π 2

Then,

(2001, 2M)

(2000, 1M)

(a) e < 1 + x (c) sin x > x

(b) log e (1+ x) < x (d) log ε x > x

Fill in the Blanks 20. The set of all x for which log (1 + x) ≤ x is equal to ..... .

12. Let f (x) = ∫ ex (x − 1) (x − 2) dx. Then, f decreases in the interval

(1987, 2M)

function y = 2x − log|x| is monotonically increasing for values of x ( ≠ 0), satisfying the inequalities… and monotonically decreasing for values of x satisfying the inequalities… . (1983, 2M)

(d) (2, ∞ )

(c) (1, 2)

13. The function f (x) = sin x + cos x increases, if (1999, 2M) 4

(a) 0 < x
2 f (x) for all x ∈ R, and f (0) = 1 then

(2017 Adv.)

(a) f (x) > e in (0, ∞ ) (b) f ′ (x) < e in (0, ∞ ) (c) f (x) is increasing in (0, ∞ ) (d) f (x) is decreasing in (0 ∞ ) 2x

x→ ∞

2x

(a) (I) (iii) (P) (c) (II) (iii) (P)

(2017 Adv.)

(b) (II) (iv) (Q) (d) (III) (i) (R)

23. Which of the following options is the only CORRECT combination? (a) (I) (ii) (R) (c) (II) (iii) (S)

(b) (III) (iv) (P) (d) (IV) (i) (S)

242 Application of Derivatives 24. Which of the following options is the only CORRECT combination?

27. Using the relation 2 (1 − cos x) < x2, x ≠ 0 or prove that sin (tan x) ≥ x , ∀ x ∈ [0, π / 4].

(a) (III) (iii) (R) (c) (II) (ii) (Q)

(b) (IV) (iv) (S) (d) (I) (i) (P)

statements in Column II. Let the functions defined in Column I have domain (−π / 2, π / 2). Column II

A. x + sin x

p. increasing

B. sec x

q. decreasing

−1 ≤ p ≤ 1, then show that the equation 4x3 − 3x − p = 0 has a unique root in the interval [1 / 2, 1] (2001, 5M) and identify it.

28. If

25. Match the conditions/expressions in Column I with

Column I

(2003, 4M)



x≤0 2 3 x ax x x >0 + − , 

29. Let f (x) = 

xeax ,

where, a is a positive constant. Find the interval in (1996, 3M) which f ′ (x) is increasing.

30. Show that 2 sin x + 2 tan x ≥ 3x, where 0 ≤ x < π /2.

r. neither increasing nor decreasing

(1990, 4M)

31. Show that 1 + x log (x + x2 + 1 ) ≥ 1 + x2 ∀x ≥ 0.

Analytical & Descriptive Questions 26. Prove that sin x + 2x ≥

3x (x + 1)  π , ∀ x ∈ 0,  2  π

(Justify the inequality, if any used).

(2004, 4M)

(1983, 2M)

π π 32. Given A = x : ≤ x ≤  and f (x) = cos x – x (1 + x). Find 3  6 (1980, 2M) f ( A ).

Topic 3 Rolle’s and Lagrange’s Theorem Objective Question I (Only one correct option) 1. If f : R → R is a twice differentiable function such that  1 1 f ′′ (x) > 0 for all x ∈ R, and f   = , f (1) = 1, then  2 2 (2017 Adv.) (a) f ′ (1) ≤ 0 (c) 0 < f ′ (1) ≤

1 2

(b) f ′ (1) > 1 1 < f ′ (1) ≤ 1 2

(d)

(a) Statement I is correct, Statement II is also correct; Statement II is the correct explanation of Statement I (b) Statement I is correct, Statement II is also correct; Statement II is not the correct explanation of Statement I (c) Statement I is correct; Statement II is incorrect (d) Statement I is incorrect; Statement II is correct

Analytical & Descriptive Question

Assertion and Reason 2. Let f (x) = 2 + cos x, for all real x. Statement I For each real t, there exists a point c in [t , t + π ], such that f ' (c) = 0. Because Statement II f (t ) = f (t + 2π ) for each real t. (2007, 3M)

3. If f (x) and g (x) are differentiable functions for 0 ≤ x ≤ 1, f (0) = 2, g (0) = 0

such that

f (1) = 6, g (1) = 2 Then, show that there exists c satisfying 0 < c < 1 and (1982, 2M) f ′ (c) = 2 g′ (c).

Topic 4 Maxima and Minima Objective Questions I (Only one correct option) 1. Consider the rectangles lying in the region π   (x, y) ∈ R × R : 0 ≤ x ≤ and 0 ≤ y ≤ 2 sin(2x) 2   and having one side on the X-axis. The area of the rectangle which has the maximum perimeter among all (2020 Adv.) such rectangles, is

3π (a) 2 π (c) 2 3

(b) π (d)

π 3 2

2. Let f (x) = 5 −|x − 2| and g (x) = |x + 1|, x ∈ R. If f (x) attains maximum value at α and g (x) attains minimum (x − 1) (x2 − 5x + 6) is equal to value of β, then lim x → − αβ x2 − 6x + 8 (2019 Main, 12 April II)

(b) − 3 / 2

(a) 1/2

(c) − 1 / 2

(d) 3/2

3. If the volume of parallelopiped formed by the vectors $i + λ$j + k $ , $j + λk $ and λ$i + k $ is minimum, then λ is equal to (2019 Main, 12 April I)

(a) −

1 3

(b)

1 3

(c)

3

(d) −

3

Application of Derivatives 243 4. If m is the minimum value of k for which the function

14. If 20 m of wire is available for fencing off a flower-bed in

f (x) = x kx − x2 is increasing in the interval [0, 3] and M is the maximum value of f in the interval [0, 3] when k = m, then the ordered pair (m, M ) is equal to

the form of a circular sector, then the maximum area (in sq. m) of the flower-bed is (2017 Main)

(2019 Main, 12 April I)

(a) (4, 3 2 ) (c) (3, 3 3 )

(b) (4, 3 3 ) (d) (5, 3 6 )

local extreme points at x = − 1, 0, 1, then the set S = { x ∈ R : f (x) = f (0)} contains exactly four rational numbers (2019 Main, 9 April I) two irrational and two rational numbers four irrational numbers two irrational and one rational number

6. The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is (2019 Main, 8 April II)

(a)

6

(b) 2 3

(c)

3

(d)

2 3 3

7. If S1 and S 2 are respectively the sets of local minimum and local maximum points of the function, f (x) = 9x4 + 12x3 − 36x2 + 25, x ∈ R, then (a) (b) (c) (d)

S1 = {−2} ; S2 = {0,1} S1 = {−2, 0} ; S2 = {1} S1 = {−2,1} ; S2 = {0} S1 = {−1} ; S2 = {0, 2}

(2019 Main, 8 April I)

(2019 Main, 8 April I)

(c)

7 4 2

11 (d) 4 2

9. The maximum area (in sq. units) of a rectangle having its base on the X-axis and its other two vertices on the parabola, y = 12 − x2 such that the rectangle lies inside the parabola, is (2019 Main, 12 Jan I) (a) 36

(b) 20 2

(c) 32

(d) 18 3

f (x) = 3x3 − 18x2 + 27x − 40 on the set S = { x ∈ R : x2 + 30 ≤ 11x} is (2019 Main, 11 Jan I) (b) − 122

(c) − 222

(d) 222

11. Let A (4, − 4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ∆ACB is maximum. Then, the area (in sq. units) of ∆ACB, is (2019 Main, 9 Jan II)

(a) 31

1 4

(b) 32

(c) 31

3 4

(d) 30

1 2

12. The maximum volume (in cu.m) of the right circular cone having slant height 3m is (a)

4 π 3

(b) 2 3π

13. Let f (x) = x2 +

(c) 3 3π

(2019 Main, 9 Jan I)

(d) 6π

1 1 and g (x) = x − , x ∈ R − { −1, 0, 1}. If x x2

f (x) , then the local minimum value of h (x) is h (x) = g (x) (2018 Main) (a) 3

(b) −3

(b) (4 − π )x = πr (d) 2x = r

16. The least value of α ∈ R for which 4αx2 + x > 0, is

(c) −2 2

(d) 2 2

1 ≥ 1 , for all x (2016 Adv.)

1 (a) 64

1 (b) 32

1 (c) 27

1 (d) 25

17. Let f (x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If

lim  x→ 0 1+



f (x)  = 3, then f (2) is x2 

equal to (a) −8 (c) 0

(b) −4 (d) 4

1 2 1 (c) α = 2, β = − 2

(a) α = − 6, β =

extreme

points

of

(2014 Main)

(b) α = − 6, β = − (d) α = 2, β =

1 2

1 2

19. The number of points in (− ∞ , ∞ ) for which x2 − x sin x − cos x = 0, is (a) 6

(2013 Adv.)

(b) 4

(c) 2

(d) 0

20. Let f, g and h be real-valued functions defined on the 2

2

2

2

interval [0, 1] by f (x) = ex + e− x , g (x) = x ex + e− x and 2

10. The maximum value of the function

(a) 122

(a) 2x = ( π + 4)r (c) x = 2r

x = − 1 and x = 2 are f (x) = α log|x| + βx2 + x, then

curve y2 = x − 2 is (a) 2

are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is (2016 Main) minimum, then

18. If

8. The shortest distance between the line y = x and the 7 (b) 8

(b) 10 (d) 30

15. A wire of length 2 units is cut into two parts which

5. If f (x) is a non-zero polynomial of degree four, having

(a) (b) (c) (d)

(a) 12.5 (c) 25

2

h (x) = x2ex + e− x . If a, b and c denote respectively, the absolute maximum of f, g and h on [0, 1], then (2010) (a) a = b and c ≠ b (c) a ≠ b and c ≠ b

(b) a = c and a ≠ b (d) a = b = c

21. The total number of local maxima and local minima of (2 + x)3 , − 3 < x ≤ −1 the function f (x) =  2 is 3 −1 < x < 2 x , (a) 0 (c) 2

(2008, 3M)

(b) 1 (d) 3

22. If f (x) = x2 + 2bx + 2c2 and g (x) = − x2 − 2cx + b2, such

that min f (x) > max g (x), then the relation between b and c, is (2003, 2M)

(a) No real value of b and c (b) 0 < c < b 2 (d)| c|> | b| 2 (c)| c|< | b| 2

23. If f (x) = | x|, for 0 < | x| ≤ 2 . Then, at x = 0, f has 1

, for

(a) a local maximum (c) a local minimum

x=0

(2000, 1M)

(b) no local maximum (d) no extremum

244 Application of Derivatives x2 − 1 , for every real number x, then the x2 + 1 (1998, 2M) minimum value of f

24. If f (x) =

(a) does not exist because f is unbounded (b) is not attained even though f is bounded (c) is 1 (d) is –1

25. The number of values of x, where the function f (x) = cos x + cos ( 2x) attains its maximum, is(1998, 2M) (a) 0

(b) 1

(c) 2

26. On the interval [0,1], the function x

(d) infinite 25

(1 − x)

75

maximum value at the point (a) 0

(b) 1/4

(c) 1/2

takes its (1995, 1M)

(d) 1/3

27. Find the coordinates of all the points P on the ellipse 2

2

x y + = 1, for which the area of the ∆PON is a 2 b2 maximum, where O denotes the origin and N is the foot of the perpendicular from O to the tangent at P. (1990, 10M)

 ± a2 ± b2   (a)  ,   a 2 + b2 a 2 + b2   ± a2 ± b2   (c)  ,  2 2  a +b a 2 – b2 

 ± a2 ± b2   (b)  ,   a 2 – b2 a 2 – b2   ± a2 ± b2   (d)  ,  2 2  a –b a 2 + b2 

28. If P (x) = a 0 + a1x2 + a 2x4 +... + a nx2n is a polynomial in a real variable x with 0 < a 0 < a1 < a 2 < K < a n. Then, the function P (x) has (1986, 2M) (a) neither a maximum nor a minimum (b) only one maximum (c) only one minimum (d) only one maximum and only one minimum

29. If y = a log x + bx2 + x has its extremum values at x = − 1 and x = 2 , then (a) a = 2, b = − 1 (c) a = − 2, b =

1 2

x5 + 5x4 + 10x3 + 10x2 + 3x + 1, x < 0;  2 0 ≤ x < 1; x − x + 1,  8 2 3 2 f (x) =  1 ≤ x < 3; x − 4x + 7x − , 3 3  10  x ≥ 3;  (x − 2) log e (x − 2) − x + 3 , Then which of the following options is/are correct? (a) f is increasing on (− ∞ , 0)

1 (b) a = 2, b = − 2

cos(2x) cos(2x) sin(2x) , then cos x − sin x   sin x cos x   sin x

33. If f (x) =  − cos x

(1982, 1M)

(a) max ( p , q) < max ( p , q, r ) 1 (b) min ( p , q) = ( p + q − | p − q|) 2 (c) max ( p , q) < min ( p , q, r ) (d) None of the above

34. Let f : R → (0, ∞ ) and g : R → R be twice differentiable functions such that f′ ′ and g′ ′ are continuous functions on R. Suppose f ′ (2) = g (2) = 0, f ′ ′(2) ≠ 0 and g′ (2) ≠ 0. f (x) g (x) If lim (2016 Adv.) = 1, then x → 2 f ′ (x) g′ (x) (a) f has a local minimum at x = 2 (b) f has a local maximum at x = 2 (c) f ′ ′ (2) > f (2) (d) f (x) − f ′ ′ (x) = 0, for atleast one x ∈ R

35. The function f (x) = 2|x| + |x + 2| − ||x + 2| − 2|x|| has a local minimum or a local maximum at x is equal to (b)

−2 3

(b) 32

 

x

Define F (x) = ∫ f (t )dt, x > 0 0

Then which of the following options is/are correct? (2019 Adv.)

F (x) ≠ 0 for all x ∈ (0, 5) F has a local maximum at x = 2 F has two local maxima and one local minimum in (0, ∞ ) F has a local minimum at x = 1

(d) 2/3

having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are (2013 Adv.) x

e

,

37. If f (x) = 2 − ex − 1 ,

31. Let f : R → R be given by f (x) = (x − 1)(x − 2)(x − 5).

(2013 Adv.)

(c) 2

36. A rectangular sheet of fixed perimeter with sides

(a) 24

Objective Questions II (One or more than one correct option)

(2017 Adv.)

(a) f (x) attains its minimum at x = 0 (b) f (x) attains its maximum at x = 0 (c) f ′ (x) = 0 at more than three points in (− π , π ) (d) f ′ (x) = 0 at exactly three points in (− π , π )

(a) −2

(d) None of the above

(2019 Adv.)

(b) f ′ is NOT differentiable at x = 1 (c) f is onto (d) f ′ has a local maximum at x = 1

(1983, 1M)

30. If p, q and r are any real numbers, then

(a) (b) (c) (d)

32. Let f : R → R be given by

 x−e ,  x ∈ [1, 3], then

(c) 45

0 ≤ x≤1 1 < x≤2 2 < x≤3

(d) 60

and g (x) = ∫

x 0

f (t ) dt , (2006, 3M)

(a) g (x) has local maxima at x = 1 + log e 2 and local minima at x = e (b) f (x) has local maxima at x = 1and local minima at x = 2 (c) g (x) has no local minima (d) f (x) has no local maxima

38. If f (x) is a cubic polynomial which has local maximum at x = − 1. If f (2) = 18, f (1) = − 1 minimum at x = 0, then

and f ′ (x) has local (2006, 3M)

Application of Derivatives 245 (a) the distance between (– 1, 2) and (a , f (a )), where x = a is the point of local minima, is 2 5 (b) f (x) is increasing for x ∈[1, 2 5 ] (c) f (x) has local minima at x = 1 (d) the value of f (0) = 5

39. The function f (x) = ∫

x

t (et − 1) (t − 1) (t − 2)3 (t − 3)5 dt has a local

−1

minimum at x equals (a) 0

(1999, 3M)

(b) 1

(c) 2

(d) 3

2 40. If f (x) = 3x + 12x − 1, − 1 ≤ x ≤ 2, then 37 2 < x≤3  − x,

(1993, 3M)

arc of the parabola y = 16x, 0 ≤ y ≤ 6 at the point F (x0 , y0 ). The tangent to the parabola at F (x0 , y0 ) intersects the Y -axis at G (0, y1 ). The slope m of the line L is chosen such that the area of the ∆EFG has a local maximum 2

Match List I with List II and select the correct answer using the codes given below the list. Column I m=

1.

1/2

Q.

Maximum area of ∆EFG is

2.

4

R.

y0 =

3.

2

S.

y1 =

4.

1

Analytical & Descriptive Questions 45. If f (x) is twice differentiable function such that f (a ) = 0,

f (b) = 2, f (c) = 1, f (d ) = 2, f (e) = 0, where a < b < c < d < e, then the minimum number of zeroes of g (x) = { f ′ (x)}2 + f ′′ (x) ⋅ f (x) in the interval [a, e] is

from the line x + y = 7, is minimum.

(2003, 2M)

48. Let f (x) is a function satisfying the following conditions (i) f (0) = 2, f (1) = 1 (ii) f (x) has a minimum value at x = 5 / 2 and 2 ax 2ax − 1 2 ax + b + 1 (iii) For all x, f ′ (x) = b b+ 1 −1 2 (ax + b) 2 ax + 2b + 1 2 ax + b

where, a and b are some constants. Determine the constants a , b and the function f (x). (1998, 8M)

49. Let C1 and C 2 be respectively, the parabolas x2 = y – 1

Codes S 3 4

(b) (d)

P 3 1

Q 4 3

R 1 4

S 2 2

Passage Based Problems Consider the function f : (−∞ , ∞ ) → (−∞ , ∞ ) defined by x2 − ax + 1 ; 0 < a < 2. f (x) = 2 (2008, 12M) x + ax + 1

42. Which of the following is true ? (a) (2 + a ) f ′ ′ (1) + (2 − a ) f ′ ′ (−1) = 0 (b) (2 − a )2 f ′ ′ (1) − (2 + a )2 f ′ ′ (−1) = 0 (c) f ′ (1) f ′ (−1) = (2 − a )2 (d) f ′ (1) f ′ (−1) = − (2 + a )2 2

(a) g ′ (x) is positive on (− ∞ , 0) and negative on (0, ∞ ) (b) g ′ (x) is negative on (− ∞ , 0) and positive on (0, ∞ ) (c) g ′ (x) changes sign on both (− ∞ , 0) and (0, ∞ ) (d) g ′ (x) does not change sign (− ∞ , ∞ )

Column II

P.

R 2 2

f ′ (t ) dt. Which of the following is true? 1 + t2

47. Find a point on the curve x2 + 2 y2 = 6 whose distance

41. A line L : y = mx + 3 meets Y -axis at E (0, 3) and the

Q 1 3

0

the area enclosed by the tangents drawn from the point P(6, 8) to the circle and the chord of contact is maximum. (2003, 2M)

Match the Columns

P 4 1

ex

46. For the circle x2 + y2 = r 2, find the value of r for which

(a) f (x) is increasing on [–1, 2] (b) f (x) is continuous on [–1, 3] (c) f ′(2) does not exist (d) f (x) has the maximum value at x = 2

(a) (c)

44. Let g (x) = ∫

2

43. Which of the following is true ? (a) f (x) is decreasing on (−1, 1) and has a local minimum at x=1 (b) f (x) is increasing on (−1, 1) and has a local maximum at x=1 (c) f (x) is increasing on (−1, 1) but has neither a local maximum nor a local minimum at x = 1 (d) f (x) is decreasing on (−1, 1) but has neither a local maximum nor a local minimum at x = 1

and y2 = x – 1. Let P be any point on C1 and Q be any point on C 2. If P1 and Q1 is the reflections of P and Q, respectively, with respect to the line y = x. Prove that P1 lies on C 2 Q1 lies on C1 and PQ ≥ min (PP1 , QQ1 ). Hence, determine points P0 and Q0 on the parabolas C1 and C 2 respectively such that P0Q0 ≤ PQ for all pairs of points (P , Q ) with P on C1 and Q on C 2.

50. If S is a square of unit area. Consider any quadrilateral which has one vertex on each side of S. If a , b, c and d denote the length of the sides of the quadrilateral, then prove that 2 ≤ a 2 + b2 + d 2 ≤ 4. (1997, 5M)

51. Determine the points of maxima and minima of the functionf (x) = constant.

1 ln x − bx + x2, x > 0, where b ≥ 0 is a 8 (1996, 5M)

52. Let (h , k) be a fixed point, where h > 0 , k > 0. A straight line passing through this point cuts the positive directions of the coordinate axes at the points P and Q. Find the minimum area of the ∆OPQ, O being the origin. (1995, 5M)

53. The circle x2 + y2 = 1 cuts the X-axis at P and Q. Another circle with centre at Q and variable radius intersects the first circle at R above the X-axis and the line segment PQ at S. Find the maximum area of the ∆QSR. (1994, 5M)

246 Application of Derivatives  3 (b3 − b2 + b − 1) − x + , 0 ≤ x≤1 54. Let f (x) =  (b2 + 3b + 2)  2x − 3, 1 ≤ x≤3  Find all possible real values of b such that f (x) has the smallest value at x = 1. (1993, 5M)

55. What normal to the curve y = x2 forms the shortest chord?

(1992, 6M)

56. A window of perimeter (including the base of the arch) is in the form of a rectangle surmounted by a semi-circle. The semi-circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass. The clear glass transmits three times as much light per square meter as the coloured glass does. What is the ratio for the sides of the rectangle so that the window transmits the maximum light? (1991, 4M)

57. A point P is given on the circumference of a circle of radius r. Chord QR is parallel to the tangent at P. Determine the maximum possible area of the ∆PQR. (1990, 4M)

58. Find the point on the curve 4x + a y = 4a , 4 < a 2 < 8 2

2 2

that is farthest from the point (0, –2).

2

(1987, 4M)

59. Let A ( p2, − p) B(q2, q), C (r 2, − r ) be the vertices of the triangle ABC. A parallelogram AFDE is drawn with vertices D, E and F on the line segments BC, CA and AB, respectively. Using calculus, show that maximum area 1 of such a parallelogram is ( p + q)(q + r )( p − r ). 4 (1986, 5M) π π 60. Let f (x) = sin3 x + λ sin 2 x, − < x < ⋅ Find the 2 2 intervals in which λ should lie in the order that f (x) has exactly one minimum and exactly one maximum. (1985, 5M)

x , 1 + x2 where the tangent to the curve has the greatest slope.

61. Find the coordinates of the point on the curve y =

66. For a polynomial g (x) with real coefficients, let mg denote the number of distinct real roots of g (x). Suppose S is the set of polynomials with real coefficients defined by S = {(x2 − 1)2(a 0 + a1x + a 2x2 + a3 x3 ) : a 0 , a1 , a 2, a3 ∈R}; For a polynomial f, let f′ and f′ ′ denote its first and second order derivatives, respectively. Then the minimum possible value of (mf ′ + mf ′ ′ ), where f ∈ S, is ……… (2020 Adv.)

67. Let AD and BC be two vertical poles at A and B

respectively on a horizontal ground. If AD = 8 m, BC = 11 m and AB = 10 m; then the distance (in meters) of a point M on AB from the point A such that (2020 Main, 6 Sep I) MD 2 + MC 2 is minimum is ........

68. A cylindrical container is to be made from certain solid material with the following constraints : It has a fixed inner volume of V mm3 , has a 2 mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum, when the inner radius of the container is V 10 mm, then the value of is (2015 Adv.) 250 π

69. A vertical line passing through the point (h , 0) x2 y 2 + = 1 at the points P and Q. If 4 3 the tangents to the ellipse at P and Q meet at the point R. intersects the ellipse

If ∆(h ) = area of the ∆ PQR, ∆1 = max ∆ (h ) and 1/ 2 ≤ h ≤ 1 8 ∆1 − 8∆ 2 is equal to (2013 Adv.) ∆ 2 = min ∆ (h ), then 1/ 2 ≤ h ≤ 1 5

70. Let f : R → R be defined as f (x) = | x| + | x2 − 1|. The total

(1984, 4M)

number of points at which f attains either a local (2012) maximum or a local minimum is

62. A swimmer S is in the sea at a distance d km from the

71. Let p(x) be a real polynomial of least degree which has a

closest point A on a straight shore. The house of the swimmer is on the shore at a distance L km from A. He can swim at a speed of u km/h and walk at a speed of v km/h (v > u ). At what point on the shore should be land so that he reaches his house in the shortest possible time? (1983, 2M)

63. If ax2 + b / x ≥ c for all positive x where a > 0 and b > 0, then show that 27ab ≥ 4c . 2

3

(1982, 2M)

64. If x and y be two real variables such that x > 0 and

xy = 1. Then, find the minimum value of x + y. (1981, 2M)

Integer & Numerical Answer Type Questions 65. Let the function f : (0, π) → R be defined by f (θ ) = (sin θ + cos θ ) + (sin θ − cos θ ) Suppose the function f has a local minimum at θ precisely when θ ∈{ λ 1π, ... , λ rπ }, where 0 < λ 1 < ... λ r < 1. (2020 Adv.) Then the value of λ 1 + ... + λ r is ....... . 2

4

local maximum at x = 1 and a local minimum at x = 3. If (2012) p(1) = 6 and p(3) = 2, then p′ (0) is equal to

72. The number of distinct real roots of x4 − 4x3 + 12x2 + x − 1 = 0 is……

73. Let f be a function defined on R (the set of all real numbers) such that f ′ (x) = 2010 (x − 2009) (x − 2010)2 (x − 2011)3 (x − 2012)4, ∀x ∈ R. If g is a function defined on R with values in the interval (0, ∞ ) such that f (x) = ln ( g (x)), ∀ x ∈ R, then the number of points in R at which g has a local maximum is…… (2010)

74. The maximum value of the expression 1 is …… sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ

(2010)

75. The maximum value of the function

f (x) = 2x3 − 15x2 + 36x − 48 on the set A = { x | x2 + 20 ≤ 9x} is ……… .

(2009)

Answers Topic 1

9. (c)

10. (a)

11. (a)

12. (b)

1. (a) 5. (c)

2. (c) 6. (b)

3. (b) 7. (d)

4. (d) 8. (b)

13. (d)

14. (c)

15. (c)

16. (c)

17. (c)

18. (c)

19. (c)

20. (d)

9. (b) 13. (c)

10. (d) 14. (b, d)

11. (d) 15. (b, c)

12. (d)

21. (c)

22. (d)

23. (a)

24. (d)

25. (b)

26. (b)

27. (a)

28. (c)

29. (b)

30. (b)

31. (a,b,d)

32. (b,c,d)

33. (b, c)

34. (a, d)

35. (a, b)

36. (a, c)

37. (a, b)

38. (b, c)

39. (b, d)

40. (a, b, c, d)

41. (a) P → 4 Q → 1 R → 2 S → 3

42. (a)

43. (a)

16. H = φ, V = {1, 1 } 1 3 18. a = – , b = – , c = 3 2 4

17. y − 2 = 0 19. 1 : 16

20. y + x − 1 = 0 π − 3π 21. x + 2y = and x + 2y = 2 2 22. (8)

Topic 2 1. (d)

2. (a)

3. (b)

4. (c)

5. (c)

6. (a)

7. (c)

8. (a)

9. (a)

10. (a)

11. (b)

12. (c)

13. (b) 17. (a, c)  1  21. x ∈  − , 0  2  22. (d)

14. (c) 15. (b) 18. (c, d) 19. (a, c) 1 1   ∪  , ∞ , x ∈  − ∞, −  ∪ 2   2

16. (d) 20. x > − 1  1  0,   2

54. b ∈ ( −2, − 1 ) ∪ [1, ∞ ] 55.

2. (b)

6. (b)

7. (c)

3 3 2 r sq units 4

58. (0, 2)

ud v − u2 2

64. (2)

65. 0.50

66. (5)

67. (5)

4. (b)

68. (4)

69. (9)

70. (5)

71. (9)

8. (c)

72. (2)

73. (1)

74. (2)

75. (7)

Topic 4 5. (d)

57.

 3 3 60. λ ∈  − ,  61. x = 0, y = 0  2 2 62.

3. (b)

2 x − 2y + 2 = 0, 2 x + 2y − 2 = 0

56. 6 : 6 + π

Topic 3

2. (a)

(b − b 2 − 1 ) 1 and minima at x = (b + b 2 − 1 ) 4 4 4 3 53. 9

52. 2hk

24. (c)  2 a 25. A → p, B → r 29. – ,  a 3  1 π  π 3 π  π  32.  − 1 +  , − 1 +      2 3 3 2 6 6 

1. (c)

45. 6

51. Maxima at x =

23. (c)

1. (b)

44. (a)

46. 5 units 47. (2, 1) 1 −5 1 5 48. a = , b = ; f ( x ) = x 2 − x + 2 4 4 4 4  1 5  5 1 49. P0 =  ,  and Q0 =  ,   2 4  4 2

Hints & Solutions Topic 1 Equations of Tangent and Normal

∴ Slope of this line = −

1. Equation of given curve is y=

x , x ∈ R, (x ≠ ± 3 ) 2 x −3

…(i)

On differentiating Eq. (i) w.r.t. x, we get dy (x2 − 3) − x(2x) (− x2 − 3) = 2 = dx (x − 3)2 (x2 − 3)2 It is given that tangent at a point (α , β ) ≠ (0, 0) on it is parallel to the line 2x + 6 y − 11 = 0.

⇒

−

2 dy = 6 dx ( α , β )

1 α2 + 3 =− 3 (α 2 − 3)2

⇒ 3α 2 + 9 = α 4 − 6 α 2 + 9 ⇒ α4 − 9 α2 = 0 ⇒ α = 0, − 3, 3 ⇒ α = 3 or − 3, Now, from Eq. (i), 3 α 1 1 −3 or ⇒ β= β= 2 = or − 9 −3 2 9 −3 2 α −3

[Q α ≠ 0]

248 Application of Derivatives According to the options,|6 α + 2 β | = 19 1  at (α , β ) =  ± 3, ±   2

2. Given curve is y = f (x) = x3 − x2 − 2x

...(i)

So, f (1) = 1 − 1 − 2 = −2 and f (−1) = −1 − 1 + 2 = 0 Since, slope of a line passing through (x1 , y1 ) and y −y (x2, y2) is given by m = tan θ = 2 1 x2 − x1 ∴Slope of line joining points (1, f (1)) and

5. The helicopter is nearest to the soldier, if the tangent to

f (1) − f (−1) −2 − 0 (−1, f (−1)) is m = = = −1 1 − (−1) 1+1 dy Now, = 3x2 − 2x − 2 [differentiating Eq. (i), w.r.t. ‘x’] dx According to the question, dy =m dx ⇒ 3 x2 − 2 x − 2 = − 1 ⇒ 3 x2 − 2 x − 1 = 0 1 ⇒ (x − 1) (3x + 1) = 0 ⇒ x = 1 ,− 3  1  Therefore, set S = − , 1.  3 

3. Given curve is y = x + ax − b 3

…(i)

passes through point P(1, − 5). …(ii) ∴ −5 =1+ a −b⇒b−a =6 and slope of tangent at point P(1, − 5) to the curve (i), is dy m1 = = [3x2 + a ](1, −5 ) = a + 3 dx (1, −5 ) Q The tangent having slope m1 = a + 3 at point P(1, − 5) is perpendicular to line − x + y + 4 = 0 , whose slope is m2 = 1. ∴ a + 3 = −1 ⇒ a = −4 [Q m1m2 = −1] Now, on substituting a = −4 in Eq. (ii), we get b = 2 On putting a = −2 and b = 2 in Eq. (i), we get y = x3 − 4x − 2 Now, from option (2, − 2) is the required point which lie on it.

4. The given curve is y = x2 − 5x + 5

Now, equation of tangent to the curve (i) at point  7 1  ,−  and having slope 2, is  2 4 1 7 1  y + = 2  x −  ⇒ y + = 2x − 7  4 2 4 29 …(iii) y = 2x − ⇒ 4  1 On checking all the options, we get the point  , − 7  8 satisfy the line (iii).

…(i)

Now, slope of tangent at any point (x, y) on the curve is dy …(ii) = 2x − 5 dx [on differentiating Eq. (i) w.r.t. x] Q It is given that tangent is parallel to line 2 y = 4x + 1 dy So, = 2 [Q slope of line 2 y = 4x + 1 is 2] dx 7 ⇒ 2x − 5 = 2 ⇒ 2x = 7 ⇒ x = 2 7 On putting x = in Eq. (i), we get 2 49 35 69 35 1 y= − +5 = − =− 4 2 4 2 4

the path, y = x3/ 2 + 7, (x ≥ 0) of helicopter at point (x, y) is perpendicular to the line joining (x, y) and the position 1  of soldier  , 7 . 2 

(x, y) y=x3/2+7 (1/2, 7)

Q Slope of tangent at point (x, y) is dy 3 1/ 2 = x = m1 (let ) dx 2 1  and slope of line joining (x, y) and  , 7 is 2  y−7 m2 = 1 x− 2 Now, m1 ⋅ m2 = −1 3 1/ 2  y − 7  x   = −1  x − (1 /2) 2

⇒

…(i)

…(ii)

[from Eqs. (i) and (ii)]

3 1/ 2 x3/ 2 [Q y = x3/ 2 + 7] = −1 x 1 2 x− 2 3 2 1 x = −x + ⇒ 2 2 ⇒ 3x2 + 2x − 1 = 0 ⇒ 3x2 + 3x − x − 1 = 0 ⇒ 3x(x + 1) − 1(x + 1) = 0 1 1 ⇒ x = , −1 ⇒ x ≥ 0 ⇒ x = 3 3 3/ 2  1 and So, [Q y = x3/ 2 + 7] y=  +7  3 ⇒

 1  1 3 / 2  Thus, the nearest point is  ,   + 7   3 3   Now, the nearest distance 2

2 3 2  3/ 2   1  1 1  1  1 =  −  +  7 −   − 7 =   +    3  6  2 3  3  

=

1 1 3+4 7 1 + = = = 36 27 108 108 6

7 3

Application of Derivatives 249 6.

Given equation of curves are y = 10 − x2 and y = 2 + x2

…(i) …(ii)

For point of intersection, consider 10 − x2 = 2 + x2 ⇒ 2 x2 = 8 ⇒ x2 = 4 ⇒ x=±2 Clearly, when x = 2 , then y = 6 (using Eq. (i)) and when x = − 2, then y = 6 Thus, the point of intersection are (2, 6) and (−2, 6). Let m1 be the slope of tangent to the curve (i) and m2 be the slope of tangent to the curve (ii) dy dy For curve (i) = −2x and for curve (ii) = 2x dx dx ∴ At (2, 6), slopes m1 = − 4 and m2 = 4, and in that case |tan θ| =

m2 − m1 4+4 8 = = 1 + m1m2 1 − 16 15

At (−2, 6), slopes m1 = 4 and m2 = − 4 and in that case |tan θ| =

m2 − m1 −4 − 4 8 = = 1 + m1m2 1 − 16 15

7. We have, y2 = 6x

⇒

2y

dy =6 dx

⇒

dy 3 = dx y

Slope of tangent at ( x1 , y1 ) is m1 = Also,

9x 2 + by 2 = 16 dy =0 18x + 2by dx

3 y1

dy −9x = dx by −9x1 Slope of tangent at ( x1 , y1 ) is m2 = by1 Since, these are intersection at right angle. 27x1 ∴ m1m2 = − 1 ⇒ =1 by12 27x1 [Q y12 = 6x1] =1 ⇒ 6bx1 9 b= ⇒ 2 ⇒

⇒

8. Given curve is

∴ Equation of normal at (0, 1) is given by −1 y−1 = (x − 0) ⇒ x + y − 1 = 0 1  1 1 which passes through the point  ,  .  2 2

9. We have, f (x) = tan − 1

f (x) = tan − 1

⇒

Put x = 0 in Eq. (i), we get y(− 2) (− 3) = 6 ⇒ y = 1 So, point of intersection is (0, 1). x+6 Now, y = (x − 2)(x − 3) dy 1 (x − 2)(x − 3) − (x + 6)(x − 3 + x − 2) = dx (x − 2)2(x − 3)2

…(i)

1 + sin x  π , x ∈ 0,   2 1 − sin x x x   cos + sin   2 2

2

x   cos − sin  2

2

x  2

x x   cos + sin  2 2 = tan    cos x − sin x   2 2 x x x π  Q cos > sin for 0 < <   2 2 2 4 x   1 + tan  −1 2 = tan    1 − tan x   2   π x  π x = tan − 1 tan  +   = +  4 2  4 2  1  π 1 f ′ (x) = ⇒ f′   = ⇒  6 2 2 π Now, equation of normal at x = is given by 6 π   π   y − f    = − 2  x −   6  6 −1

π π  π 4π π     π π ⇒  y −  = − 2  x −  Q f   = + = =    6  4 12 12 3  3 6   2π  which passes through 0,  .  3

10. Given equation of curve is x2 + 2xy − 3 y2 = 0

…(i)

On differentiating w.r.t x, we get 2x + 2xy′ + 2 y − 6 yy′ = 0 ⇒ y′ = At

y(x − 2)(x − 3) = x + 6

⇒

6 + 30 36  dy = =1 =    dx ( 0, 1) 36 4 ×9

⇒

Key Idea Angle between two curves is the angle between the tangents to the curves at the point of intersection.

i.e.

x+ y 3y − x

x = 1, y = 1, y′ = 1  dy =1    dx (1, 1)

Equation of normal at (1, 1) is 1 y − 1 = − (x − 1) ⇒ y − 1 = − (x − 1) 1 ⇒ x+ y=2 On solving Eqs. (i) and (ii) simultaneously, we get ⇒ x2 + 2x(2 − x) − 33(2 − x)2 = 0

…(ii)

250 Application of Derivatives ⇒ x2 + 4x − 2x2 − 3(4 + x2 − 4x) = 0 ⇒ − x2 + 4x − 12 − 3x2 + 12x = 0 ⇒ − 4x2 + 16x − 12 = 0 ⇒ 4x2 − 16x + 12 = 0 ⇒ x2 − 4 x + 3 = 0 ⇒ (1 − 1)(x − 3) = 0 ⇒ x = 1, 3 Now, when x = 1, then y = 1 and when x = 3, then y = − 1 ∴ P = (1, 1) and Q = (3, − 1) Hence, normal meets the curve again at (3, –1) in fourth quadrant. Alternate Solution Given, x2 + 2xy − 3 y2 = 0 ⇒ (x − y)(x + 3 y) = 0 ⇒ x − y = 0 or x + 3 y = 0 Equation of normal at (1, 1) is y − 1 = − 1(x − 1) ⇒ x + y − 2 = 0 It intersects x + 3 y = 0 at (3, –1) and hence normal meets the curve in fourth quadrant. x + y = 2 Y' x + 3y = 0

y=x

X

O (3,–1) Y'

11. Given,

y + 3x = 12 y 3

2

...(i)

On differentiating w.r.t. x, we get dy dy ⇒ 3 y2 + 6x = 12 dx dx dy 6x dx 12 − 3 y2 = ⇒ ⇒ = dx 12 − 3 y2 dy 6x dx For vertical tangent, =0 dy ⇒ 12 − 3 y2 = 0

⇒

y=±2

4 and again 3 2 putting y = − 2 in Eq. (i), we get 3x = − 16, no real solution.  4  So, the required point is  ± , 2 .  3  On putting, y = 2 in Eq. (i), we get x = ±

12. Slope of tangent y = f (x) is

dy = f ′ (x)(3 , 4) dx

Therefore, slope of normal 1 1 =− =− f ′ (x)(3 , 4) f ′ (3) But ⇒ ∴

−

1  3π  = tan    4 f ′ (3) −1  π π = tan  +  = − 1  2 4 f ′ (3) f ′ (3) = 1

y = a (sin θ − θ cos θ ) dx ∴ = a (− sin θ + sin θ + θ cos θ ) = a θ cos θ dθ dy and = a (cos θ − cos θ + θ sin θ ) dθ dy dy = a θ sin θ ⇒ = tan θ dθ dx Thus, equation of normal is y − a (sin θ − θ cos θ ) − cos θ = x − a (cos θ + θ sin θ ) sin θ and

⇒ − x cos θ + a θ sin θ cos θ + a cos 2 θ = y sin θ + θ a sin θ cos θ − a sin 2 θ ⇒ x cos θ + y sin θ = a whose distance from origin is |0 + 0 − a | =a cos 2 θ + sin 2 θ

14. Given, 4x2 + 9 y2 = 1

…(i)

On differentiating w.r.t. x, we get dy dy 8x 4x 8x + 18 y =0 ⇒ =− =− dx dx 18 y 9y

(1,1) X'

13. Given, x = a (cos θ + θ sin θ )

The tangent at point (h , k) will be parallel to 8x = 9 y, then 4h 8 − = ⇒ h = − 2k 9k 9 Point (h , k) also lies on the ellipse. ...(ii) ∴ 4h 2 + 9k2 = 1 On putting value of h in Eq. (ii), we get 4(− 2k)2 + 9 k2 = 1 ⇒ 16k2 + 9k2 = 1 1 ⇒ 25k2 = 1 ⇒ k2 = 25 1 ⇒ k=± 5 Thus, the point, where the tangents are parallel to 1 2  2 1 8x = 9 y are  − ,  and  , −  . 5  5 5 5 Therefore, options (b) and (d) are the answers. 1 15. Given, xy = 1 ⇒ y = x dy 1 ⇒ =− 2 dx x Thus, slope of normal = x2 (which is always positive) and a it is given ax + by + c = 0 is normal, whose slope = − . b a a ⇒ − > 0 or 0 ∀ x ∈ (0, 2) ⇒ f ′ (x) is a strictly increasing function ∀ x ∈ (0, 2). Now, for φ(x) to be increasing, φ′ (x) ≥ 0 [using Eq. (i)] ⇒ f ′ (x) − f ′ (2 − x) ≥ 0 ⇒ f ′ (x) ≥ f ′ (2 − x) ⇒ x > 2 − x [Q f′ is a strictly increasing function] ⇒ 2x > 2 ⇒ x > 1 Thus, φ(x) is increasing on (1, 2). Similarly, for φ(x) to be decreasing, φ′ (x) ≤ 0 ⇒ f ′ (x) − f ′ (2 − x) ≤ 0 [using Eq. (i)] ⇒ f ′ (x) ≤ f ′ (2 − x) [Q f′ is a strictly increasing ⇒ x 0, 2 3/ 2 (a + x ) (b + (d − x)2)3/ 2 2

11.

Option (a) Let f (x) = ex − 1 − x.

∀ x ∈R

⇒

g (x) decreases

for 0 < x < 1

⇒

g (x) < g (0)

for 0 < x < 1

⇒ log e (1 + x) − x < 0

for 0 < x < 1

log e (1 + x) < x

for 0 < x < 1

Option (c) sin x > x h (x) = sin x − x ⇒ h′ (x) = cos x − 1

Let

u

2e 1 + e 2u

For x ∈(0, 1), cos x − 1 < 0 [Q eu > 0]

So, g′ (u ) is increasing for all x ∈ R. f (x) = x3 + bx2 + cx + d f ′ (x) = 3x2 + 2bx + c

As we know that, if ax2 + bx + c > 0, ∀x, then a > 0 and D < 0. Now, D = 4b2 − 12c = 4(b2 − c) − 8c [where, b2 − c < 0 and c > 0] D = (–ve) or D < 0

⇒ f ′ (x) = 3x2 + 2bx + c > 0 ∀x ∈ (− ∞ , ∞ ) [as D < 0 and a > 0] Hence, f (x) is strictly increasing function.

9. Let f (x) = 3 sin x − 4 sin3 x = sin 3x The longest interval in which sin x is increasing is of length π . So, the length of largest interval in which f (x) = sin 3x is π increasing is . 3

10. Given, f (x) = xex (1 − x ) ⇒

f (x) > f (0) for 0 < x < 1

Therefore, option (b) is the answer.

g′ (u ) > 0, ∀ x ∈ R

∴

⇒

or

We have, g (u ) = 2 tan −1 (eu ) − π 2

⇒

f (x) increase in (0, 1)

Option (b) Let g (x) = log e (1 + x) − x, 0 < x < 1 x 1 g′ (x) = −1 = − < 0 for 0 < x < 1 1+ x 1+ x

⇒ g (u ) is an odd function.

8. Given,

⇒

⇒ ex − 1 − x > 0 or ex > 1 + x for 0 < x < 1

= π / 2 − 2 tan −1 (eu ) = − g (u ) g (− u ) = − g (u )

g′ (u ) =

f ′ (x) = ex − 1 > 0, ∀x ∈ (0, 1)

then

Hence, f (x) is an increasing function of x. π 7. Given, g (u ) = 2 tan −1 (eu ) − for u ∈ (−∞ , ∞ ) 2 π g (− u ) = 2 tan −1 (e− u ) − 2 π −1 u = 2 (cot (e )) − 2 π  π = 2  − tan −1 (eu ) − 2  2 ∴

PLAN Inequation based upon uncompatible function. This type of inequation can be solved by calculus only.

f ′ (x) = ex (1 − x ) + xex (1 − x ) (1 − 2x) = ex (1 − x ) [1 + x (1 − 2x)] = ex (1 − x ) (1 + x − 2x2) = − ex (1 − x ) (2x2 − x − 1)

⇒ h (x) is decreasing function. ⇒ h (x) < h (0) ⇒ sin x − x < 0 ⇒ sin x < x,which is not true. Option (d)

Therefore, p′ (x) is an increasing function. ⇒ p(0) < p(x) < p(1) ⇒

− ∞ < log x − x < −1

⇒

log x − x < 0

⇒

log x < x

Therefore, option (d) is not the answer. f (x) = ∫ ex (x − 1) (x − 2) dx

12. Let ⇒

f ′ (x) = ex (x − 1) (x − 2) + − 1 2

∴ f ′ (x) < 0 for 1 < x < 2 ⇒ f (x) is decreasing for x ∈(1, 2).

13. Given, f (x) = sin 4 x + cos 4 x On differentiating w.r.t. x, we get f ′ (x) = 4 sin3 x cos x − 4 cos3 x sin x

= − ex (1 − x ) (x − 1) (2x + 1)  1  which is positive in  − , 1 .  2   1  Therefore, f (x) is increasing in − , 1 .  2 

p(x) = log x − x 1 p′ (x) = − 1 > 0, ∀x ∈ (0, 1) x

= 4 sin x cos x (sin 2 x − cos 2 x) = 2 sin 2x (− cos 2x) = − sin 4x Now, ⇒

f ′ (x) > 0 , if sin 4x < 0 π < 4x < 2π

+

Application of Derivatives 255 π π 2dx ⇒ y

x

dy > 2∫ dx y 0

∫

1

ln( f (x)) > 2x ⇒ f (x) > e2x Also, as f ′ (x) > 2 f (x) ⇒ f ′ (x) > 2c2x > 0

18. Given, f (x) = ∫

x 1 x

As

 1 − t +   t

e

dt

t 1  − x +   x

e

x

1  − x +   x

e

x

+

1  − + x  

  −1  e x −  2  x  1 /x 1  − x +   x

e

1  − x +   x

2e

=

x

x

f ′ (x) > 0, ∀x ∈ (0, ∞ )

∴ f (x) is monotonically increasing on (0, ∞ ). ⇒ Option (a) is correct and option (b) is wrong.  1 − t +  t

x e   1 Now, f (x) + f   = ∫ 1/x  x t

dt + ∫

 1 − t +   t

e

1/ x

t

x

dt

= 0, ∀x ∈ (0, ∞ )

Therefore, option (c) is the answer. log (π + x) 15. Given, f (x) = log(e + x)

When

16. Let F (x) = h (x) − h (1) = f { g (x)} − f { g (1)}

=

⇒ f (x) increases in [0, 1]. x Thus, f (x) = increases in 0 < x ≤ 1. sin x

f ′ (x) =

From Eqs. (i) and (iv), f ′ (x) < 0

f ′ (x) = 1 ⋅

For 0 < x < 1, sin x − x cos x (tan x − x) cos x f ′ (x) = = > 0 for 0 < x < 1 sin 2 x sin 2 x

1 1 − log (π + x) ⋅ e+ x π+x [log (e + x)]2

…(iv)

∴ f (x) is decreasing for x ∈ (0, ∞ ).

0 < x≤1 for 0 < x≤1 for x=0

Now, f is continuous in [0, 1] and differentiable in ]0, 1[.

log (e + x) ⋅

On multiplying Eqs. (ii) and (iii), we get log (π + x) log (e + x) > π+x e+ x

17. f ′ (x) > 2 f (x) ⇒

⇒

Also,

... (iii)

Therefore, option (d) is the answer.

for 0 < x < 1

⇒

... (ii)

∴ f (x) is decreasing function.

0 < x < 1, H (x) = tan x − x sec2x, 0 ≤ x ≤ 1

⇒

and

Since,

Now, H (x) is continuous in [0, 1] and differentiable in ]0, 1[. For

log (π + x) > log (e + x) 1 1 > e+ x π + x

[since, f (x) is an increasing function f ′ ( g (x)) is + ve and g (x) is decreasing function g ′ ( f (x)) is −ve ]

H (x) = tan x − x sec2 x, 0 ≤ x ≤ 1

Again,

∴

Now, let

…(i)

g (x) = f (2x ) = ∫ −x

2x 2− x

g (− x) = f (2 ) = ∫ ∴ f (2x ) is an odd function.

 1 − t +   t

e

2− x 2x

dt

t

 1 − t +   t

e

t

dt = − g (x)

256 Application of Derivatives f ′ (x) > 0 for (0, 1)

19. Given, h (x) = f (x) − f (x)2 + f (x)3

f ′ (x) < 0 for (e, ∞ )

On differentiating w.r.t. x, we get h ′ (x) = f ′ (x) − 2 f (x) ⋅ f ′ (x) + 3 f 2(x) ⋅ f ′ (x)

∴ P and Q are correct, II is correct, III is incorrect. −1 1 f ′′ (x) = 2 − ⇒ f ′′ (x) < 0 for (0, ∞ ) x x

= f ′ (x)[1 − 2 f (x) + 3 f 2(x)] 2 1  = 3 f ′ (x) ( f (x))2 − f (x) +  3 3 

∴ S, is correct, R is incorrect.

 1 1 1 = 3 f ′ (x)  f (x) −  + −    3 3 9 

IV is incorrect.

 1 3 − 1 = 3 f ′ (x)  f (x) −  +    3 9  

∴ ii, iii, iv are correct.

2

lim f (x) = − ∞ ⇒ lim f ′ (x) = − ∞ ⇒ lim f ′′ (x) = 0

x→∞

2

 1 2 = 3 f ′ (x)  f (x) −  +    3 9  2

h ′ ( x ) < 0, if f ′ ( x ) < 0 and h ′ ( x ) > 0, if f ′ ( x ) > 0

NOTE

Therefore, h (x) is an increasing function, if f (x) is increasing function and h (x) is decreasing function, if f (x) is decreasing function. Therefore, options (a) and (c) are correct answers.

20. Let f (x) = log (1 + x) − x ⇒

f ′ (x) =

⇒ when when

x 1 −1 = − 1+ x 1+ x

f ′ (x) > 0 − 1 < x < 0 and f ′ (x) < 0 x>0

∴ f (x) is increasing for −1 < x < 0. ⇒

f (x) < f (0) ⇒ log (1 + x) < x

Again, f (x) is decreasing for x > 0. ⇒

f (x) < f (0) ⇒ log (1 + x) < x

∴

log (1 + x) ≤ x, ∀x > − 1

21. Here,

⇒

 2x2 − log x, x >0 y= 2 x − − x x 0 = dx 4x − 1 , x < 0  x

4x2 − 1 (2x − 1) (2x + 1) , x ∈ R − {0} = x x 1 1     Increasing when x ∈  − , 0 ∪  , ∞  2  2  =

∴

1  1  and decreasing when x ∈  −∞ , −  ∪ 0,  .  2  2 Solutions. (22-24) f (x) = x + ln x − x ln x ⇒ f (1) = 1 > 0 f (e2) = e2 + 2 − 2e2 = 2 − e2 < 0 ⇒ f (x) = 0 for some x ∈ (1, e2) ∴ I is correct f ′ (x) = 1 +

1 1 − ln x − 1 = − ln x x x

22. (d)

x→∞

x→∞

23. (c)

24. (c) d x π π (x + sin x) = 1 + cos x = 2 cos 2 > 0 for − < x < . dx 2 2 2 Therefore, x + sin x is increasing in the given interval. Therefore, (A)→ (p) is the answer. d Again, (sec x) = sec x tan x which is > 0 for 0 < x < π / 2 dx −π and < 0 for < x 0

[ Q x < π /2]

⇒

f ′ (x) > f ′ (π / 2)

∴ f (x) is increasing. Thus, when x ≥ 0, f (x) ≥ f (0) 3x(x + 1) 3x (x + 1) ⇒ sin x + 2x − ≥ 0 ⇒ sin x + 2x ≥ π π

27. Let

f (x) = sin (tan x) − x f ′ (x) = cos (tan x) ⋅ sec2x − 1 = cos (tan x) (1 + tan 2 x) − 1 = tan 2 x {cos (tan x)} + cos (tan x) − 1

tan 2 x 2  x2  2 Q2 1 0 1 ( cos ) , cos − < ≠ ⇒ > − x x x x  2  tan 2 x  cos (tan x) > 1 −   ⇒   2 1   f ′ (x) > tan 2 x cos (tan x) −  2  > tan 2 x cos (tan x) −

> tan 2 x [cos (tan x) − cos (π / 3)] > 0

Application of Derivatives 257 ⇒

 eax − 1  = lim aeax + x  x → 0− 

f (x) is increasing function, for all x ∈ [0, π / 4] f (0) = 0 ⇒ f (x) ≥ 0, for all x ∈ [0, π / 4]

As ⇒

sin (tan x) ≥ x −1 ≤ p ≤ 1

28. Given, Let Now,

= lim aeax + a ⋅ lim x → 0−

f (x) = 4x − 3x − p = 0 1 3 f (1 / 2) = − − p = − 1 − p ≤ 0 2 2

= ae0 + a (1) = 2a f ′ (x) − f ′ (0) Rf ′ ′ (0) = lim + x+0 x→ 0

3

f (1) = 4 − 3 − p = 1 − p ≥ 0

Also,

[Q p ≥ − 1]

and

[Q p ≤ 1]

∴ f (x) has atleast one real root between [1 /2, 1]. f ′ (x) = 12x − 3 > 0 on [1 / 2, 1]

⇒

f ′ (x) increasing on [1 /2, 1]

⇒

f has only one real root between [1 /2, 1].

To find a root, we observe f (x) contains 4x3 − 3x, which is multiple angle formula for cos 3θ. ∴ Put

x = cos θ −1

⇒

p = cos 3 θ ⇒ θ = (1 / 3) cos ( p)  1 ∴ Root is cos  cos −1 ( p) .  3 NOTE This type is asked in 1983 and repeat after 13 years.

At x = 0, LHL = lim f (x) = lim xeax = 0 and

x → 0−

x → 0−

+

+

RHL = lim f (x) = lim (x + ax2 − x3 ) = 0 x→ 0

x→ 0

Therefore, LHL = RHL = 0 = f (0) So, f (x) is continuous at x = 0. ax  ax Also, f ′ (x) = 1 ⋅ e + axe 2 , if x < 0 1 + 2ax − 3x , if x > 0 and

Lf ′ (0) = lim

x → 0−

= lim x→0

and

−

Rf ′ (0) = lim x→ 0

+

= lim

x →0 +

f (x) − f (0) x−0 xeax − 0 = lim eax = e0 = 1 x x → 0− f (x) − f (0) x+0 x + ax2 − x3 − 0 x

= lim 1 + ax − x = 1 2

Therefore, Lf ′ (0) = Rf ′ (0) = 1 ⇒ f ′ (0) = 1 (ax + 1) eax , if x < 0 Hence, f ′ (x) =  1, if x = 0 2 1 + 2ax − 3x , if x > 0 Now, we can say without solving that, f ′ (x) is continuous at x = 0 and hence on R. We have, ax ax f ′ ′ (x) = ae + a (ax + 1) e , if x < 0 2 6 , if x > 0 − a x  f ′ (x) − f ′ (0) and Lf ′ ′ (0) = lim x−0 x → 0− x → 0−

= lim

2ax − 3x2 = lim 2a − 3x = 2a x x → 0+

x→ 0

+

Therefore, Lf ′ ′ (0) = Rf ′ ′ (0) = 2a a (ax + 2) eax , Hence, f ′ ′ (x) = 2a , 2a − 6x, ⇒

if x < 0 if x = 0 if x > 0

For x < 0, f ′ ′ (x) > 0, if x > − 2 / a 2 f ′ (x) > 0, if − < x < 0 a

⇒

and for x > 0, f ′ ′ (x) > 0, ⇒ for x > 0,

(ax + 1)eax − 1 x

if 2a − 6x > 0

f ′ ′ (x) > 0, if x < a /3

Thus, f (x) increases on [–2/a, 0] and on [0, a/3].  2 a . Hence, f (x) increases on − ,  a 3  y = f (x) = 2 sin x + 2 tan x − 3x

30. Let ⇒

f ′ (x) = 2 cos x + 2 sec2 x − 3 0 ≤ x < π / 2, f ′ (x) > 0

For

Thus, f (x) is increasing. When x ≥ 0, f (x) ≥ f (0) ⇒ 2 sin x + 2 tan x − 3x ≥ 0 + 0 − 0 ⇒ 2 sin x + 2 tan x ≥ 3x

31. Let f (x) = 1 + x log (x + x2 + 1 ) − 1 + x2

∴ f ′ (x) = x ⋅

x → 0+

= lim

(1 + 2ax − 3x2) − 1 x

+

Now, for x < 0, f ′ ′ (x) > 0, if ax + 2 > 0

⇒ 4 cos θ − 3 cos θ − p = 0 3

29.

= lim x→ 0

2

Also,

x → 0−

eax − 1 ax

− ⇒

x x +1 2

 1 +  

  x2 + 1  x

x+ =

x +1 2

x x +1 2

f ′ (x) = log (x +

⇒

+ log (x +

+ log (x +

x2 + 1 )

x2 + 1 ) −

[Q log (x +

∴ f (x) is increasing for x ≥ 0. f (x) ≥ f (0)

⇒ 1 + x log (x + 1 + x2 ) − 1 + x2 ≥ 1 + 0 − 1 ⇒ 1 + x log (x + 1 + x2 ) ≥ 1 + x2 , ∀ x ≥ 0

32. Given, A = x : 

x +1

x2 + 1 )

f ′ (x) ≥ 0

⇒

x 2

π π ≤x≤  6 3

x2 + 1 ) ≥ 0]

258 Application of Derivatives ∴ Perimeter of the rectangle π  y=4 − x + sin 2x  4  dy For maximum, =0 dx ⇒ − 1 + 2 cos 2x = 0 1 cos 2x = ⇒ 2 π π ⇒ 2x = ⇒ x= 3 6

and f (x) = cos x − x − x2 ⇒

f ' (x) = − sin x − 1 − 2x = − (sin x + 1 + 2x) π π  which is negative for x ∈ , ⇒ f ' (x) < 0  6 3  or f (x) is decreasing.   π  π  Hence, f ( A ) =  f   , f      3  6  1 π  π 3 π  π  − 1 +   =  − 1 +  , 3 2 6  6  2 3 

and

Topic 3 Rolle’s and Lagrange’s Theorem

π ∴ At x = , the rectangle PQRS have maximum 6 perimeter. So length of sides  π π π PS = QR = 2  −  =  4 6 6

1  For some x in  , 1 2  f ′ (x) = 1 f′ (1) > 1

[LMVT]

2. Given, f (x) = 2 + cos x, ∀x ∈ R There exists a point ∈ [t , t + r ], where

Statement I f ′ (c) = 0

and ∴

Hence, Statement I is true. Statement II f (t ) = f (t + 2π ) is true. But statement II is not correct explanation for statement I. 0 ≤ x ≤ 1. f ′ (c) =

Using Lagrange’s Mean Value theorem, 6 −2 =4 1 −0 g (1) − g (0) 2 − 0 = =2 1 −0 1 −0

g ′ (c) =

⇒

f ′ (c) = 2 g′ (c)

Topic 4 Maxima and Minima π and 0 ≤ y ≤ 2 sin(2x) 2 On drawing the diagram, Let the side PS on the X-axis, such that P (x, 0), and π  Q (x, 2 sin(2x)), so length of the sides PS = QR = 2  − x 4  and PQ = RS = 2 sin 2x. {(x, y) ∈ R × R : 0 ≤ x ≤

Y 2 Q

O

P

R

π/4

S π/2

X

π π × 3= 6 2 3

Key Idea Volume of parallelopiped formed by the vectors a, b and c is V = [ a b c].

3. 1. Given region is

Required area =

and g (x) = | x + 1|, where x ∈ R. Clearly, maximum of f (x) occurred at x = 2, so α = 2. and minimum of g (x) occurred at x = − 1, so β = − 1. ⇒ αβ = − 2 (x − 1) (x2 − 5x + 6) Now, lim x → − αβ x2 − 6 x + 8 (x − 1) (x − 3) (x − 2) [Qαβ = − 2] = lim x→ 2 (x − 4) (x −2) (x − 1) (x − 3) (2 − 1) (2 − 3) 1 × (− 1) 1 = lim = = = x→ 2 (2 − 4) (x − 4) (− 2) 2

f (1) − f (0) 1 −0

and

 π PQ = RS = 2 sin   = 3  3

2. Given functions are f (x) = 5 − |x − 2|

3. Since, f (x) and g (x) are differentiable functions for ∴

d 2y = − 4 sin 2x| π < 0 x= dx2 x = π 6 6

1. f ′ (x) is increasing

∴

  π  Q x ∈ 0,   2  

$ , $j + λk $ and λ$i + k $ , which Given vectors are $i + λ$j + k forms a parallelopiped. ∴Volume of the parallelopiped is

⇒

1

λ

1

V = 0 λ

1 0

λ = 1 + λ3 − λ 1

V = λ3 − λ + 1

On differentiating w.r.t. λ, we get dV = 3 λ2 − 1 dλ dV For maxima or minima, =0 dλ 1 ⇒ λ=± 3

Application of Derivatives 259 1  2 3 > 0 , for λ =  d 2V 3 = 6λ =  1 dλ2 2 3 < 0 , for λ = − 3 

and

6.

1 d 2V , so volume ‘V ’ is minimum is positive for λ = 3 dλ2 1 for λ = 3

Key Idea (i) Use formula of volume of cylinder, V = πr 2h where, r = radius and h = height (ii) For maximum or minimum, put first derivative of V equal to zero

Q

4. Given function f (x) = x kx − x2

… (i) From the figure,

the function f (x) is defined if kx − x ≥ 0 2

⇒

x2 − kx ≤ 0

⇒

Let a sphere of radius 3, which inscribed a right circular cylinder having radius r and height is h, so

x ∈ [0, k]

… (ii)

⇒ and

h = 3 cos θ 2 h = 6 cos θ r = 3 sin θ

…(i)

r

because it is given that f (x) is increasing in interval x ∈ [0, 3], so k should be positive. Now, on differentiating the function f (x) w.r.t. x, we get x f ′ (x) = kx − x2 + × (k − 2x) 2 kx − x2 =

2(kx − x ) + kx − 2x 2

2 kx − x

2

2

=

3kx − 4x

h

2

2 kx − x

θ

h/2

2

r

as f (x) is increasing in interval x ∈ [0, 3], so f ′ (x) ≥ 0 ∀ x ∈ (0, 3) ⇒

3

3kx − 4x2 ≥ 0

⇒

4x2 − 3kx ≤ 0 3k  3k   (as k is positive) ⇒ 4x x −  ≤ 0 ⇒ x ∈ 0,  4   4

Q Volume of cylinder V = πr 2h = π (3 sin θ )2(6 cos θ ) = 54π sin 2 θ cos θ . dV For maxima or minima, =0 dθ 2 3 ⇒ 54π [2 sin θ cos θ − sin θ ] = 0 ⇒ sin θ [2 cos 2 θ − sin 2 θ ] = 0

3k ⇒k ≥4 4 ⇒ Minimum value of k = m = 4 and the maximum value of f in [0, 3] is f (3).

⇒

tan 2 θ = 2

⇒

tan θ = 2 ⇒ sin θ =

Q f is increasing function in interval x ∈ [0, 3]

and

So,

3≤

Q M = f (3) = 3 4 × 3 − 3 = 3 3 2

Therefore, ordered pair (m, M ) = (4, 3 3 )

5. The non-zero four degree polynomial

f (x) has extremum points at x = −1, 0, 1, so we can assume f ′ (x) = a (x + 1)(x − 0) (x − 1) = ax(x2 − 1) where, a is non-zero constant. f ′ (x) = ax3 − ax a a f (x) = x4 − x2 + C ⇒ 4 2 [integrating both sides] where, C is constant of integration. Now, since f (x) = f (0) a 4 a 2 x 4 x2 = ⇒ x − x +C =C ⇒ 4 2 4 2 ⇒ x2(x2 − 2) = 0 ⇒ x = − 2 , 0, 2 Thus, f (x) = f (0) has one rational and two irrational roots.

2 3

1 3 From Eqs. (i) and (ii), we get 1 h =6 =2 3 3 cos θ =

…(ii)

7. Given function is

f (x) = 9x4 + 12x3 − 36x2 + 25 = y (let) dy For maxima or minima put =0 dx dy = 36x3 + 36x2 − 72x = 0 ⇒ dx ⇒ x3 + x2 − 2x = 0 ⇒ x[x2 + x − 2] = 0 2 ⇒ x[x + 2x − x − 2] = 0 ⇒ x[x(x + 2) − 1(x + 2)] = 0 ⇒ x(x − 1)(x + 2) = 0 ⇒ x = − 2, 0, 1 By sign method, we have following –

+ –2

– 0

  π  Q θ ∈ 0, 2    

+ 1

260 Application of Derivatives dy changes it’s sign from negative to positive at dx x = ‘−2 ’ and ‘1’, so x = − 2, 1 are points of local minima. dy changes it’s sign from positive to negative at Also, dx x = 0, so x = 0 is point of local maxima.

Y

Since,

(0, 12) M

R

XN

∴ S1 = { −2, 1} and S 2 = {0}.

X y=12–x2

YN

y2 = x − 2 and the equation of line is

…(i)

y=x

…(ii) y=x

y2=x–2

M P(t2+2, t)

O

P(a, 0)

O

S

8. Given equation of curve is

Y

Q(a, 12–a2)

X

(2, 0)

Then, area of rectangle PQRS = 2 × (Area of rectangle PQMO) [due to symmetry aboutY -axis]

= 2 × [a (12 − a 2)] = 24a − 2a3 = ∆ (let). The area function ∆ a will be maximum, when d∆ = 0 ⇒ 24 − 6a 2 = 0 da ⇒ a2 = 4 ⇒ a = 2 [Q a > 0] So, maximum area of rectangle PQRS = (24 × 2) − 2 (2)3 = 48 − 16 = 32 sq units

10. We have, f (x) = 3x3 − 18x2 + 27x − 40 Consider a point P (t 2 + 2, t ) on parabola (i). For the shortest distance between curve (i) and line (ii), the line PM should be perpendicular to line (ii) and parabola (i), i.e. tangent at P should be parallel to y = x. ∴

dy = Slope of tangent at point P to curve (i) dx at point P =1

⇒

[Q tangent is parallel to line y = x]

1 =1 2yP [differentiating the curve (i), we get2 y

⇒

1 1 =1⇒t = 2 2t

dy =1] dx

[Q P (x, y) = P (t 2 + 2, t )]

⇒ f ′ (x) = 9x2 − 36x + 27 = 9(x2 − 4x + 3) = 9 (x − 1) (x − 3) Also, we have S = { x ∈ R : x2 + 30 ≤ 11 x} Clearly, x2 + 30 ≤ 11x ⇒ x2 − 11x + 30 ≤ 0 ⇒ (x − 5) (x − 6) ≤ 0 ⇒ x ∈ [5, 6] So, S = [5, 6] Note that f (x) is increasing in [5, 6]

…(i)

[Qf ′ (x) > 0 for x ∈[5, 6]

∴f (6) is maximum, where f (6) = 3(6)3 − 18(6)2 + 27(6) − 40 = 122

11. According to given information, we have the following figure. C(t2, 2t)

B(9, 6)

 9 1 So, the point P is  ,  .  4 2 9 − 1 4 2 Now, minimum distance = PM = 2 [Q distance of a point P (x1 , y1 ) from a line 7 |ax1 + by1 + c | units ax + by + c = 0 is = 2 2 a + b  4 2

9. Equation of parabola is given, y = 12 − x2 or

x2 = − ( y − 12).

Note that vertex of parabola is (0, 12) and its open downward. Let Q be one of the vertices of rectangle which lies on parabola. Then, the coordinates of Q be (a, 12− a 2)

A(4, –4)

For y2 = 4ax, parametric coordinates of a point is (at 2, 2at). ∴For y2 = 4x, let coordinates of C be (t 2, 2t). 1 Then, area of ∆ABC = 2

t2 9

2t 6

4

−4 1

1 1

1 = |t 2(6 − (− 4)) − 2t (9 − 4) + 1(− 36 − 24)| 2 1 10 2 = |10t 2 − 10t − 60| = |t − t − 6|= 5| t 2 − t − 6| 2 2

Application of Derivatives 261 Let, A (t ) = 5| t 2 − t − 6| Clearly, A (4, − 4) ≡ A (t12, 2t1 ) ⇒ 2t1 = − 4 ⇒ t1 = − 2

...(i)

and B(9, 6) ≡ B(t22,2t2) ⇒ 2t2 = 6 ⇒ t2 = 3 Since, C is on the arc AOB, the parameter ‘t’ for point C ∈ (− 2, 3). Let f (t ) = t 2 − t − 6 ⇒ f ′ (t ) = 2t − 1 1 Now, f ′ (t ) = 0 ⇒ t = 2 Thus, for A (t ), critical point is at t =

1 2

2

125 1 1  1  1 Now,A   = 5   − − 6 = = 31 [Using Eq. (i)]  2  2 4 4 2

12. Let h = height of the cone, r = radius of circular base = (3)2 − h 2 = 9−h

[Q l2 = h 2 + r 2]

2

…(i)

1 2 1  ∈ ( −∞ , 2 2 ] < 0,  x −  +  x x x − 1 x ∴ Local minimum value is 2 2. x−

14. Total length = 2r + r θ = 20

⇒

l= 3

Now, volume (V ) of cone = V (h ) =

13. We have,

f(x) = x2 +

1 x2

and g( x ) = x − 2

1 f(x) ⇒ h( x ) = g( x ) x

1  +2 2 x x = ∴ h( x ) = 1 1 x− x− x x 1 2  ⇒ h( x ) =  x −  +  x x − 1 x 1 2 1  ∈ [2 2 , ∞ ) x − > 0,  x −  +  x x − 1 x x x2 +

1

 x − 

r

r

rθ

For maxima or minima, put

dA = 0. dr

10 − 2r = 0 ⇒ r = 5 1 20 − 2 (5)  Amax = (5)2   2 5 1 = × 25 × 2 = 25 sq. m 2

15. According to given information, we have Perimeter of

1 2 πr h 3

1 [From Eq. (i)] π (9 − h 2)h 3 1 …(ii) = π [9h − h3 ] 3 For maximum volume V ′ (h ) = 0 and V ′′ (h ) < 0. Here, V ′ (h ) = 0 ⇒ (9 − 3h 2) = 0 [Q h 0. x i.e. to find the minimum value of α 1 y = 4αx2 + ; x > 0 attains minimum value of α. x dy 1 = 8αx − 2 ∴ dx x 4 αx2 +

when

…(i)

262 Application of Derivatives 2 d 2y …(ii) = 8α + 3 2 dx x dy When = 0, then 8x3α = 1 dx 1/3 d 2y  1 At x =   , = 8α + 16α = 24α, Thus, y attains  8α  dx2 1/3  1 minimum when x =   ; α > 0.  8α 

Now,

 1 ∴ y attains minimum when x =    8α  i.e.

 1 4α    8α 

PLAN The given equation contains algebraic and trigonometric functions called transcendental equation. To solve transcendental equations we should always plot the graph for LHS and RHS.

Here, x2 = x sin x + cos x

X′

Y′

Let f (x) = x2 and

23 /

+ (8α )1/3 ≥ 1

To plot,

PLAN Any function have extreme values (maximum or minimum) at its critical points, where f ′ ( x ) = 2.

∴

g (x) = x sin x + cos x g′ (x) = x cos x + sin x − sin x g′ (x) = x cos x

lim f (x) x→ 0 2

x

/2

–5 π/ 2

f (x) =2 x2 f (x) will be of the form ax4 + bx3 + 2x2.

–π –π/2 O

lim x→ 0

f (x) = ax4 + bx3 + 2x2 f ′ (x) = 4ax3 + 3bx2 + 4x f ′ (1) = 4a + 3b + 4 = 0 f ′ (2) = 32a + 12b + 8 = 0 8a + 3b + 2 = 0

π/2 π

⇒

Y′

...(i)

So, graph of f (x) and g (x) are shown as Y

...(ii)

f (x)

X′

x4 − 2x3 + 2x2 2 f (2) = 8 − 16 + 8 = 0

Y′

18. Here, x = − 1 and x = 2 are extreme points of f (x) = α log|x| + βx2 + x, then α f ′ (x) = + 2 βx + 1 x …(i) f ′ (−1) = − α − 2 β + 1 = 0 [at extreme point, f ′ (x) = 0] α …(ii) f ′ (2) = + 4 β + 1 = 0 2

g (x)

X

–π/2 O π/2

f (x) =

On solving Eqs. (i) and (ii), we get 1 α = 2, β = − 2

5π/2

3π 7π At x = 0, , ,... , f ′′ (x) > 0, so minimum 2 2 π 5π 9π At x = , , ,... , f ′ (x) < 0, so maximum 2 2 2

On solving Eqs. (i) and (ii), 1 we get a = , b = −2 2 ∴

…(ii)

g′ (x) = 0 ⇒ x cos x = 0 π 3π 5π 7π x = 0, , , , 2 2 2 2 g (x)

=3

[Q f (x) is of four degree polynomial] Let ⇒ ⇒ and ⇒

…(i)

g′′ (x) = − x sin x + cos x Put

Since, the function have extreme values at x = 1 and x = 2. ∴ f ′ (x) = 0 at x = 1 and x = 2 ⇒ f′ (1) = 0 and f′ (2) = 0

⇒

g (x) = x sin x + cos x

We know that, the graph for f (x) = x2

α 1/3 + 2α 1/3 ≥ 1 1 ⇒ 3α 1/3 ≥ 1 ⇒ α ≥ 27 1 Hence, the least value of α is . 27

⇒

X

O

.

Also, it is given that, f (x)  lim  =3 ⇒ 1 + x→ 0 1 + x2  

x2

Y

1/3

⇒

17.

19.

So, number of solutions are 2. 2

2

2

2

20. Given function, f (x) = ex + e− x , g (x) = xex + e− x and 2

2

h (x) = x2ex + e− x are strictly increasing on [0, 1]. Hence, at x = 1, the given function attains absolute maximum all equal to e + 1 / e . ⇒ a=b=c

21. Given,

(2 + x)3 , if − 3 < x ≤ −1 f (x) =  x 2 / 3 , if − 1 < x < 2 

Application of Derivatives 263

⇒

3 (x + 2)2 , if − 3 < x ≤ −1  f ′ (x) = 2 − 1 3 , if − 1 < x < 2 3 x

26. Let ⇒

f (x) = x25 (1 − x)75 , x ∈[0, 1] f ′ (x) = 25 x24 (1 − x)75 − 75x25 (1 − x)74 = 25x24 (1 − x)74 [(1 − x) − 3x]

Y

= 25x24 (1 − x)74 (1 − 4x) For maximum value of f (x), put f ′ (x) = 0

X′ (–3,0)

X (–1,0)

⇒

25x24 (1 − x)74 (1 − 4x) = 0 1 x = 0, 1, 4

⇒

Y′

x = 0, y = 0

Also, at

Clearly, f ′ (x) changes its sign at x = −1 from +ve to −ve and so f (x) has local maxima at x = −1 . Also, f ′ (0) does not exist but f ′ (0 − ) < 0 and f ′ (0 + ) < 0. It can only be inferred that f (x) has a possibility of a minima at x = 0 . Hence, the given function has one local maxima at x = −1 and one local minima at x = 0 .

22. Given f (x) = x2 + 2bx + 2c2 and g (x) = − x2 − 2cx + b2

At

x = 1, y = 0

and at

x = 1 / 4, y > 0

∴ f (x) attains maximum at x = 1 / 4.

27. Let the coordinates of P be (a cos θ , b sin θ ) Equations of tangents at P is Y

Then, f (x) is minimum and g (x) is maximum at  −b −D  and f (x) = x =  , respectively.  4a 4a  ∴ and

min f (x) =

− (4b2 − 8c2) = (2c2 − b2) 4

max g (x) = −

N

X'

O

(4c2 + 4b2) = (b2 + c2) 4(−1)

Y'

x y cos θ + sin θ = 1 a b

2c2 − b2 > b2 + c2

⇒

c2 > 2b2

⇒

| c| > 2| b|

Again, equation of normal at point P is

23. It is clear from figure that at x = 0, f (x) is not continuous. Y y=x

y = –x X′

X

O Y′

Here, f (0) > RHL at x = 0 and f (0) > LHL at x = 0

ax secθ − by cosec θ = a 2 − b2 Let M be foot of perpendicular from O to PK, the normal at P. 1 Area of ∆OPN = (Area of rectangle OMPN ) 2 1 = ON ⋅ OM 2 1 ab Now, ON = = cos 2 θ sin 2 θ b2 cos 2 θ + a 2 sin 2 θ + a2 b2 [perpendicular from O, to line NP]

So, local maximum at x = 0. 2 x2 − 1 24. Given, f (x) = 2 =1 − 2 x +1 x +1 f (x) will be minimum, when

X

M K

Now, min f (x) > max g (x) ⇒

P

2 is maximum, x +1 2

i.e.

when x2 + 1 is minimum,

i.e.

at x = 0.

∴ Minimum value of f (x) is f (0) = − 1

and OM =

a 2 sec2 θ + b2cosec2θ

Thus, area of ∆OPN =

Let f (θ ) = f ' (θ ) =

=

(a 2 – b2) ⋅ cos θ ⋅ sin θ a 2 sin 2 θ + b2 cos 2 θ

ab(a 2 − b2) cos θ ⋅ sin θ 2(a 2 sin 2 θ + b2 cos 2 θ )

=

25. The maximum value of f (x) = cos x + cos ( 2x) is 2 which occurs at x = 0. Also, there is no other value of x for which this value will be attained again.

a − b2 2

tan θ α 2 tan 2 θ + b2

ab(a 2 − b2) tan θ 2(a 2 tan 2 θ + b2) [0 < θ < π / 2]

sec2 θ( a 2 tan 2 θ + b2) − tanθ (2a 2 tanθ sec2θ) (a 2 tan 2 θ + b2)2

264 Application of Derivatives =

sec2 θ (a 2 tan 2 θ + b2 – 2a 2 tan 2 θ ) (a 2 tan 2 θ + b2)2

∴

sec2 θ( a tan θ + b)(b – a tan θ ) = (a 2 tan 2 θ + b2)2 For maximum or minimum, we put f′ (θ ) = 0 ⇒ b – a tan θ = 0 [sec2 θ ≠ 0, a tan θ + b ≠ 0, 0 < θ < π / 2 ] ⇒

tan θ = b / a

31. Given, f : R → R and f (x) = (x − 1) (x − 2 ) (x − 5 )

−1  Also, f ′ (θ )> 0, if 0 < θ−1< tan (b / a ) < 0, if tan (b / a ) < θ < π / 2

x

Since, F (x) = ∫ f (t ) dt , x > 0 0

So, F ′ (x) = f (x) = (x − 1)(x − 2 )(x − 5 ) According to wavy curve method

Therefore, f (θ ) has maximum, when b  b θ = tan −1   ⇒ tan θ =  a a Again,

sin θ =

b a 2 + b2

, cos θ =

-

2

…(i)

where, (a1 + 2a 2 x2 + 3a3 x4 + K + na nx2n − 2) > 0, ∀ x ∈ R. P ′ (x) > 0, when x > 0 Thus,  P ′ (x) < 0, when x < 0 i.e. P ′ (x) changes sign from (–ve) to (+ve) at x = 0. P (x) attains minimum at x = 0.

Hence, it has only one minimum at x = 0 .

29. y = a log x + bx2 + x has extremum at x = − 1 and x = 2.

∴ ⇒

dy = 0, at x = − 1 dx a x=2 ⇒ + 2bx + 1 = 0, at x = − 1 x x=2 − a − 2b + 1 = 0 and a = 2 and b = −

a + 4b + 1 = 0 2

1 2

 p, if p > q  q, if q > p

30. Since, max ( p, q) = 

 p, if p is greatest.  and max ( p, q, r ) =  q, if q is greatest.  r , if r is greatest.  ∴ max ( p, q) < max ( p, q, r ) is false.  p − q , if p ≥ q We know that, | p − q| =  q − p , if p < q

0

2

 t t t2 =  −8 + 17 − 10t  4 3 2 0  64 124 10 =4 − + 34 − 20 = 38 − =− 3 3 3 Q At the point of maxima x = 2 , the functional value 10 F (2 ) = − , is negative for the interval x ∈ (0, 5 ), so 3 F (x) ≠ 0 for any value of x ∈ (0, 5 ), Hence, options (a), (b) and (d) are correct. 4

⇒ P ′ (x) = 2a1x + 4a 2 x3 + ... + 2na nx2n − 1

and

5

2

0

where, a n > a n − 1 > a n − 2 > K > a 2 > a1 > a 0 > 0

and

2

+

Q F (2 ) = ∫ f (t ) dt = ∫ (t3 − 8t 2 + 17t − 10)dt

28. Given, P (x) = a 0 + a1x2 + a 2 x4 + K+ a nx2n

∴

-

F ′ (x) changes, it’s sign from negative to positive at x = 1 and 5, so, F (x) has minima at x = 1 and 5 and as F ′ (x) changes, it’s sign from positive to negative at x = 2 , so F (x) has maxima at x = 2 .

a 2 + b2

= 2x (a1 + 2a 2 x2 + K + na nx2n − 2)

+ 1

a

By using symmetry, we get the required points  ±a2 ± b2  ,    a 2 + b2 a 2 + b2 

∴

⇒

1 ( p + q − p + q), if p ≥ q 1 ( p + q −| p − q|) = 12 2  ( p + q + p − q), if p < q 2  q, if p ≥ q =  p, if p < q 1 { p + q − | p − q|} = min ( p, q) 2

3

32. Given function f : R → R is x5 + 5x4 + 10x3 + 10x2 + 3x + 1 , x 0 Since, f′ (2) = 0 and f ′′ (2) > 0 ∴ f (x) attains local minimum at x = 2. ⇒ Option (a) is correct.

35.

Graph for y = f (x) is shown as

… (i)

36. [from Eq. (i)]

PLAN The problem is based on the concept to maximise volume of cuboid, i.e. to form a function of volume, say f( x )find f ′( x )and f ′ ′( x ). Put f ′ ( x ) = 0 and check f ′ ′( x ) to be + ve or − ve for minimum and maximum, respectively.

Here,

a

PLAN x , if x ≥ We know that,| x| =  −  x , if x <  x − a, ⇒ | x − a| =   − ( x − a ),

l = 15x − 2a , b = 8x − 2a and h = a a

0 0

8x–2a

if x ≥ a if x < a

and for non-differentiable continuous function, the maximum or minimum can be checked with graph as Y

8x

15x – 2a

15x

∴ Volume = (8x − 2a ) (15x − 2a ) a

Y

V = 2a ⋅ (4x − a ) (15x − 2a )

…(i)

a O

x=a Minimum at x = a

X

x=a Maximum at x = a

Y

X

8x – 2 a 15x – 2a

On differntiating Eq. (i) w.r.t a, we get dv = 6a 2 − 46ax + 60x2 da Again, differentiating,

O

x=a

X

Neither maximum nor minimum at x = a

Here,

d 2v = 12a − 46x da 2  dv  2   = 0 ⇒ 6x − 23x + 15 = 0  da 

266 Application of Derivatives a = 5 ⇒ x = 3,

At

5 6

 d 2v   2 = 2 (30 − 23x)  da 

⇒

 d 2v  At x = 3,  2 = 2(30 − 69) < 0  da  ∴ Maximum when x = 3, also at x =

 d 2v  5 ⇒  2 > 0 6  da 

For maxima or minima, put f ′ (x) = 0 ⇒ x = 1, − 1 1 Now f ′ ′ (x) = (114x) 4 At x = 1 , f ′ ′ (x) > 0, minima At x = − 1, f ′ ′ (x) < 0 , maxima ∴ f (x) is increasing for [1, 2 5 ] . ∴ f (x) has local maxima at x = − 1 and f (x) has local minima at x = 1. Also, f (0) = 34 /4 Hence, (b) and (c) are the correct answers.

∴ At x = 5 /6, volume is minimum.

39. f (x) = ∫

Thus, sides are 8x = 24 and 15x = 45

x −1

37. Given,  ex ,  f (x) = 2 − ex − 1 ,  x−e ,  g (x) = ∫

and ⇒

x 0

– – –∞ 0

if 0 ≤ x ≤ 1 if 1 < x ≤ 2 if 2 < x ≤ 3

f ′ (x) =

Also,

 d Q  dx

g′ (x) = 0 ⇒ x = 1 + log e 2 and x = e.  ex , if 0 ≤ x ≤ 1  g′′ (x) =  − ex − 1 , if 1 < x ≤ 2  1 , if 2 < x ≤ 3

⇒ Let

g′′ (e) = 1 > 0, g (x) has a local minima. Q f (x) is discontinuous at x = 1, then we get local maxima at x = 1 and local minima at x = 2. Hence, (a) and (b) are correct answers.

38. Since, f (x) has local maxima at x = − 1 and f ′ (x) has local minima at x = 0. ∴ f ′ ′ (x) = λx

t (et − 1)(t − 1)(t − 2)3 (t − 3)5 dt

 f (t ) dt = f { ψ (x)}ψ′ (x) − f { φ (x)} φ′ (x) 

x = 0, 1, 2, 3 f ′ (x) = g (x) = x(ex − 1)(x − 1)(x − 2)3 (x − 3)5 +

− 1

+

− 2

3

This shows that f (x) has a local minimum at x = 1 and x = 3 and maximum at x = 2 . Therefore, (b) and (d) are the correct answers.

40. For −1 ≤ x ≤ 2 , we have f (x) = 3x2 + 12x − 1 ⇒

f ′ (x) = 6x + 12 > 0, ∀ − 1 ≤ x ≤ 2

2

x +c 2

[Q f ′ (− 1) = 0]

λ + c = 0 ⇒ λ = − 2c 2

Again, function is an algebraic polynomial, therefore it is continuous at x ∈ (−1, 2) and (2, 3). For continuity at x = 2, lim f (x) = lim (3x2 + 12x − 1) x → 2−

…(i)

x3 + cx + d 6  8 f (2) = λ   + 2c + d = 18  6 λ f (1) = + c + d = − 1 6

= lim [3 (4 + h 2 − 4h ) + 24 − 12h − 1] h→ 0

f (x) = λ

From Eqs. (i), (ii) and (iii), 1 f (x) = (19x3 − 57x + 34) 4 1 57 2 (x − 1) (x + 1) ∴ f ′ (x) = (57x − 57) = 4 4

x → 2−

= lim [3 (2 − h )2 + 12(2 − h ) − 1] h→ 0

Again, integrating on both sides, we get

and

∞

Hence, f (x) is increasing in [–1, 2].

On integrating, we get

⇒

3

Using sign rule,

Also, at x = e,

⇒

ψ( x )

∫ φ ( x)

g′′ (1 + log e 2) = − elog e 2 < 0, g (x) has a local maximum.

f ′ (x) = λ

x

∫ −1

2

For local minimum, f ′ (x) = 0

x = 1 + log e 2,

At

d dx

+

–

+ 1

= x(ex − 1)(x − 1)(x − 2)3 (x − 3)5 × 1

f (t ) dt

g′ (x) = f (x)

Put

t (et − 1) (t − 1)(t − 2)3 (t − 3)5 dt

= lim (12 + 3h 2 − 12h + 24 − 12h − 1) h→ 0

= lim (3h 2 − 24h + 35) = 35

…(ii)

h→ 0

and …(iii)

lim f (x) = lim (37 − x)

x → 2+

x → 2+

= lim [37 − (2 + h )] = 35 h→ 0

and

f (2) = 3 ⋅ 22 + 12 ⋅ 2 − 1 = 12 + 24 − 1 = 35

Therefore, LHL = RHL = f (2) ⇒ function is continuous at x = 2 ⇒ function is continuous in −1 ≤ x ≤ 3.

Application of Derivatives 267 Now, Rf ′ (2) = lim

x → 2+

f (x) − f (2) x−2

⇒

(x0 , y0 ) = (4t , 8t ) = (1, 4)

f (2 + h ) − f (2) = lim h→ 0 h

y0 = 4

37 − (2 + h ) − (3 × 22 + 12 × 2 − 1) = lim h→ 0 h −h = lim = −1 h→ 0 h f (x) − f (2) f (2 − h ) − f (2) and Lf ′ (2) = lim = lim h→ 0 x−2 −h x → 2−

f (x) =

42.

 (x2 − 1)   −2ax2 + 2a  = 2a  2 = 2 2 2  (x + ax + 1)   (x + ax + a ) 

3h 2 − 24h + 35 − 35 h→ 0 −h 3h − 24 = lim = 24 h→ 0 −1

2x (x2 + ax + 1) − 2 (x2 − 1) (2x + a )  = 2a   (x2 + ax + 1)3   4a (a + 2) 4a Now, f ′′ (1) = = (a + 2)3 (a + 2)2

Since, Rf ′ (2) ≠ Lf ′ (2), f ′ (2) does not exist. Again, f (x) is an increasing in [–1, 2] and is decreasing in (2, 3), it shows that f (x) has a maximum value at x = 2. Therefore, options (a), (b), (c), (d) are all correct.

and

f ′′ (−1) =

x2 < 1 ⇒ x2 − 1 < 0

(4t 2, 8t)

∴ f ′ (x) < 0, f (x) is decreasing. 4a Also, at >0 x = 1, f ′′ (1) = (a + 2)2

E (0, 3) X

O

fy

x = 0, y = at = 4 t

g′ (x) =

44.

Also, (4 t 2, 8 t ) satisfy y = mx + c . 4mt 2 − 8t + 3 = 0 0 3 1 1 1 Area of ∆ = 0 4t 1 = ⋅ 4t 2(3 − 4t ) 2 2 2 4t 8t 1 A = 2[3t 2 − 4t3 ] dA = 2[6t − 12t 2] = − 12t (12t − 1) dt + 0

∴ Maximum at t =

– 1 2

1 and 4mt 2 − 8t + 3 = 0 2

⇒

m −4 + 3 =0

⇒

m =1 G (0, 4t ) ⇒ G (0,2)

f ′ (ex ) x ⋅e 1 + (ex )2

  ex   e2x − 1 = 2a  2x   x 2 2x  (e + ae + 1)   1 + e 

8 t = 4mt 2 + 3

–

[Q0 < a < 2]

So, f (x) has local minimum at x = 1.

Tangent at F , yt = x + at 2

∴

4a (a − 2) − 4a = 3 (2 − a ) (a − 2)2

43. When x ∈ (−1, 1),

Y

∴

…(ii)

∴ (2 + a )2 f ′ ′ (1) + (2 − a )2 f ′ ′ (−1) = 4a − 4a = 0

41. Here, y2 = 16x,0 ≤ y ≤ 6

⇒

...(i)

 (x2 + ax + 1)2 (2x) − 2(x2 − 1)   (x2 + ax + 1) (2x + a )  f ′ ′ (x) = 2a   (x2 + ax + 1)4    

= lim

⇒

(x2 + ax + 1) − 2ax 2ax =1 − 2 x + ax + 1 x2 + ax + 1

 (x2 + ax + 1) ⋅ 2a − 2ax(2x + a )  f ′ (x) = −   (x2 + ax + a )2  

[12 + 3h 2 − 12h + 24 − 12h − 1] − 35 = lim h→ 0 −h

At

 3 1 1 Area = 2 −  =  4 2 2

∴

 [3(2 − h )2 + 12(2 − h ) − 1]  − (3 × 22 + 12 × 2 − 1) = lim h→ 0 −h

G

y1 = 2 2

g′ (x) = 0, if e2x − 1 = 0, i.e. x = 0 x < 0, e2x < 1 ⇒ g′ (x) < 0

If

45. Let g (x) =

d [ f (x) ⋅ f ′ (x)] dx

To get the zero of g (x), we take function h (x) = f (x) ⋅ f ′ (x) between any two roots of h (x), there lies atleast one root of h′ (x) = 0. ⇒ g (x) = 0 ⇒ h (x) = 0 ⇒ f (x) = 0 or f ′ (x) = 0 If

f (x) = 0 has 4 minimum solutions. f ′ (x) = 0 has 3 minimum solutions. h (x) = 0 has 7 minimum solutions.

⇒ h′ (x) = g (x) = 0 has 6 minimum solutions.

268 Application of Derivatives 46. To maximise area of ∆ APB, we know that, OP = 10 and sin θ = r /10 , where θ ∈ (0, π / 2)

… (i)

b 2(ax + b)

Y P(6,8) θ A r

θ Q

X′

X

O

2ax − 1 b+1

2ax + b + 1 −1

2ax + 2b + 1

2ax + b

2ax

48. Given, f ′ (x) =

Applying R3 → R3 − R1 − 2R2, we get 2ax 2ax − 1 2ax + b + 1 f ′ (x) = b b+1 −1 1 0 0 ⇒

f ′ (x) = 2ax + b

On integrating both sides, we get f (x) = ax2 + bx + c

B

Since, maximum at x = 5 /2 ⇒ f ′ (5 /2) = 0 Y′

⇒

1 ∴ Area = (2 AQ ) (PQ ) 2 = AQ . PQ = (r cos θ ) (10 − OQ ) = (r cos θ ) (10 − r sin θ ) = 10 sin θ cos θ (10 − 10 sin 2 θ )

[from Eq. (i)]

⇒

A = 100 cos3 θ sin θ dA ⇒ = 100 cos 4 θ − 300 cos 2 θ ⋅ sin 2 θ dθ dA Put =0 dθ ⇒ cos 2 θ = 3 sin 2 θ ⇒

5a + b = 0

…(i)

Also,

f (0) = 2

⇒

c=2

and

f (1) = 1

⇒

a + b + c=1

On solving Eqs. (i), (ii) and (iii), we get 1 5 a = ,b = − ,c=2 4 4 1 2 5 Thus, f (x) = x − x + 2 4 4

49. Let coordinates of P be (t , t 2 + 1) Reflection of P in y = x is P1 (t 2 + 1, t ) which clearly lies on y2 = x − 1 Similarly, let coordinates of Q be (s2 + 1, s)

tan θ = 1 / 3

⇒ θ = π /6 dA At which < 0, thus when θ = π /6, area is maximum dθ From Eq. (i), r = 10 sin

π = 5 units 6

x2 y2 47. Let us take a point P( 6 cos θ , 3 sin θ ) on + =1. 6 3 Now, to minimise the distance from P to given straight line x + y = 7, shortest distance exists along the common normal.

Its reflection in y = x is Q1 (s, s2 + 1), which lies on x2 = y − 1. We have,

PQ12 = (t − s)2 + (t 2 − s2)2 = P1Q 2

⇒

PQ1 = P1Q [Q both perpendicular to y = x]

Also PP1 || QQ1

Y

C1

x 2 = y −1 y=x

P1 Q1 (0, 1)

Q

O

X′

Y

(1, 0)

y

P y2− = 1 X

N P

X′

…(ii) …(iii)

O

C2

x+y=7

Y′

X

Thus, PP1QQ1 is an isosceles trapezium. Also, P lies on PQ1 and Q lies on P1Q , then Y′

Slope of normal at P = So, and

6 sec θ a 2 / x1 = = 2 tan θ = 1 6 cosec θ b2 / y1

2 3 1 sin θ = 3 cos θ =

Hence, required point is P(2, 1).

PQ ≥ min { PP1QQ1 } Let us take min { PP1QQ1 } = PP1 ∴

PQ 2 ≥ PP12 = (t 2 + 1 − t )2 + (t 2 + 1 − t 2) = 2(t 2 + 1 − t 2) = f (t )

we have,

f ′ (t ) = 4(t 2 + 1 − t )(2t − 1) = 4[(t − 1 / 2)2 + 3 / 4][2t − 1]

Now, ⇒

f ′ (t ) = 0 t = 1 /2

[say]

Application of Derivatives 269 Also, f ′ (t ) < 0 for t < 1 / 2

On adding Eqs. (i), (ii), (iii) and (iv), we get

and f ′ (t ) > 0 for t > 1 / 2

a 2 + b2 + c2 + d 2 = { x12 + (1 − x1 )2} + { y12 + (1 − y1 )2} + { x22 + (1 − x2)2} + { y22 + (1 − y2)2}

Thus, f (t ) is least when t = 1 / 2. Corresponding to t = 1 / 2, point P0 on C1 is (1/2, 5/4) and P1 (which we take as Q0) on C 2 are (5 / 4, 1 / 2). Note that P0Q0 ≤ PQ for all pairs of (P , Q ) with P on C 2.

50. Let the square S is to be bounded by the lines x = ±1 / 2 and y = ±1 / 2.

2

1  1  a 2 =  x1 −  +  − y1  2  2

We have,

2

⇒ 4x = 2

X'

Again,

1/2

−1/2

X B ( −1/2, y)

O c

b

C(x2 −1/2)

and maximum value is 1 + 1 + 1 + 1 = 4

−1/2 Y′

51. f (x) is a differentiable function for x > 0.

1 = − y1 + 2 1 2 2 2 b = x2 − y1 − x2 + y1 + 2 1 c2 = x22 − y22 + x2 + y2 + 2 1 d 2 = x12 − y22 + x1 − y2 + 2 x12 − y12 − x1

Similarly,

Therefore, for maxima or minima, f ′ (x) = 0 must satisfy. 1 Given, f (x) = ln x − bx + x2, x > 0 8 1 1 ⇒ f ′ (x) = . − b + 2x 8 x ⇒

a 2 + b2 + c2 + d 2 = 2(x12 + y12 + x22 + y22) + 2 1 Therefore, 0 ≤ x12, x22, y12, y22 ≤ 4

⇒

0 ≤ x12 + x22 + y12 + y22 ≤ 1

Alternate Solution c2 = x22 + y22

16x2 − 8bx + 1 = 0

⇒

(4x − b)2 = b2 − 1

⇒

(4x − b) = (b − 1) (b + 1)

... (i)

f ′ (x) =

Y A(x , 1)

=

a d (0, y2)D

X

b c X′

O

X

C(x2 , 0)

Y′

b = (1 − x2) + 2

2

y12

[b ≥ 0, given]

1 − 1 + 2x 8x 2 2 1 1 2 = x − x + x 2 16 x

1  x −   4

2

We have to check the sign of f ′ (x) at x = 1/4.

B (1, y1) 1/2

…(i)

2

Case I 0 ≤ b < 1 , has no solution. Since, RHS is negative in this domain and LHS is positive. 1 Case II When b = 1, then x = is the only solution. 4 When b = 1,

0 ≤ 2(x12 + x22 + y12 + y22) ≤ 2 2 ≤ 2(x12 + x22 + y12 + y22) + 2 ≤ 4

But

f ′ (x) = 0 1 − b + 2x = 0 8x

For

∴

⇒

⇒ x = 1 /2

d 2y =2+2 =4 dx2 1 Hence, y is minimum at x = and its minimum value is 2 1/4. Clearly, value is maximum when x = 1. 1 1 1 1 ∴Minimum value of a 2 + b2 + c2 + d 2 = + + + = 2 2 2 2 2

1/2

d

D( −1/2, 1/2)

Now, consider the function y = x2 + (1 − x)2, 0 ≤ x ≤ 1 dy differentiating ⇒ = 2x − 2(1 − x). For maximum or dx dy minimum = 0. dx ⇒ 2x − 2(1 − x) = 0 ⇒ 2x − 2 + 2x = 0

Y A(x , 1/2)

where x1 , y1 , x2, y2 all vary in the interval [0, 1].

...(ii)

a 2 = (1 − y1 )2 + (1 − x1 )2

...(iii)

d 2 = x12 + (1 − y2)2

...(iv)

Interval

Sign of f′(x)

Nature of f(x)

−∞, 0

−ve

↓

 0,  

1  4

+ ve

↑

 1, ∞   4 

+ ve

↑

From sign chart, it is clear that f ′ (x) has no change of sign in left and right of x = 1/4.

270 Application of Derivatives Case III When b > 1, then 2 1 1 1 f ′ (x) = − b + 2x =  x2 − bx +  x 2 16 8x

dA 1 = [− k2(− cosec2θ ) − h 2 sec2θ ] dθ 2 1 = [k2 cosec2θ − h 2 sec2θ ] 2 dA To obtain minimum value of A, =0 dθ

⇒

=

2  2  1 2 b − − x (b − 1)    4 16 x 

=

b 1 b 1 2     b 2 − 1  x − + b 2 − 1  x − −      x 4 4 4 4

⇒

2 (x − α ) (x − β ) x 1 where, α < β and α = (b − b2 − 1 ) and 4 1 β = (b + b2 − 1 ). From sign scheme, it is clear that 4 > 0, for 0 < x < α  f ′ (x) < 0, for α < x < β > 0, for x > β 

k2 cosec2 θ − h 2 sec2 θ = 0

⇒

k2 h2 = sin 2 θ cos 2 θ

⇒

tan θ = ±

=

By the first derivative test, f (x) has a maxima at x = α 1 = (b − b2 − 1 ) 4 1 and f (x) has a minima at x = β = (b + b2 − 1 ) 4

= − [k2(1 + cot2 θ ) cot θ + h 2(1 + tan 2 θ ) tan θ ]    d 2A  h 2  − h  = − k21 + 2     2 k   k  dθ  tan θ = − k/ h    k 2   − k  + h 2 1 + 2    h   h   2 2 2 2   k + h   h  h + k   k  = k2       + h 2 2    h2   h   k   k

⇒

… (i)

 (k2 + h 2) h (h 2 + k2)(k)  = +  k h   h k   = (k2 + h 2) + >0  k h 

Y

X′

O

(h, k)

P

1  − k   −h  2hk − k2  − h 2    h   k 2  1 = [2hk + kh + kh ] = 2hk 2 53. Since x2 + y2 = 1 a circle S1 has centre (0, 0) and cuts X-axis at P(–1, 0) and Q(1, 0) . Now, suppose the circle S 2, with centre at Q(1, 0) has radius r. Since, the circle has to meet the first circle, 0 < r < 2 . Again, equation of the circle with centre at Q(1, 0) and radius r is A=

Y′

k   The line (i) meets X-axis at P  h − , 0 and Y-axis at  m  Q (0, k − mh ). 1 Let A = area of ∆ OPQ = OP . OQ 2 1 k =  h −  (k − mh ) 2 m

Y

=

=− =

R S1 X′

[Qm = tan θ]

1 (k2 + h 2 tan 2 θ − 2hk tan θ ) 2 tan θ

1 (2kh − k2 cot θ − h 2 tan θ ) 2

[Q h, k > 0]

Therefore, A is least when tan θ = − k / h. Also, the least value of A is

X

1  mh − k  (k − mh )  2 m  1 =− (k − mh )2 2m 1 =− (k − h tan θ )2 2 tan θ

k h

[given] tan θ < 0 , k > 0, h > 0 k Therefore, tan θ = − (only possible value). h d 2A 1 2 Now, = [−2k cosec2 θ cot θ − 2h 2 sec2 θ tan θ ] dθ 2 2

For this line to intersect the positive direction of two axes, m = tan θ < 0 , since the angle in anti-clockwise direction from X-axis becomes obtuse.

Q

k2 = tan 2 θ h2

Q

52. Let equation of any line through the point (h , k) is y − k = m(x − h )

⇒

(–1,0)P

S2

r O S

Q(1,0)

X

Y′

(x − 1)2 + y2 = r 2

…(i)

Application of Derivatives 271 To find the coordinates of point R, we have to solve it with … (ii) x2 + y2 = 1

⇒

On subtracting Eq. (ii) from Eq. (i), we get (x − 1)2 − x2 = r 2 − 1 ⇒

x2 + 1 − 2 x − x 2 = r 2 − 1

⇒

1 − 2x = r 2 − 1

∴

2 − r2 x= 2

2 2 2  4 × 2  4 × 2 f ′ ′  = 48   − 30    3   3   3 10 × 64 640 256 = 16 × 8 − = 128 − =− 0 2

Therefore, L is minimum, when t = ± 1 / 2.For t = 1 / 2 , point A is (1 / 2 , 1 / 2) and point B is (− 2 , 2). When t = − 1 / 2 , A is (−1 / 2 , 1 / 2), B is ( 2 , 2). Again, when t = 1 / 2 , the equation of AB is y−2 x+ 2 = 1 1 + 2 −2 2 2  1  1  + 2  = (x + 2)  − 2 ⇒ ( y − 2 )      2 2   ⇒ ⇒

− 2 y + 4 = 2x + 2 2x + 2 y − 2 = 0

2b

2

1 2 1   1     = 2  1 + 2  4t −  = 4  1 + 2  2t −    t t 4t   4t   dL For maxima or minima, we must have =0 dt 1 1 ⇒ 2t − = 0 ⇒ t 2 = t 2 1 ⇒ t=± 2 1 1  1  d 2L  Now, = 8  1 + 2  − 3   2t −   t 4t   2t   dt 2

Then,

Q = 2ab (3L ) +

⇒

Q=

⇒ ⇒

1 πb2(L ) 2

L (πb2 + 12ab) 2 L Q = [πb2 + 6b (K − 4b − πb)] 2 L Q = (6Kb − 24b2 − 5πb2) 2

[from Eq. (i)]

On differentiating w.r.t. b, we get dQ L = (6K − 48b − 10πb) db 2 dQ For maximum, put =0 db 6K b= ⇒ 48 + 10 π Now,

…(ii)

d 2Q L = (− 48 − 10π ) < 0 db2 2

Thus, Q is maximum and from Eqs. (i) and (ii), we get (48 + 10π ) b = 6 {2a + 4b + πb} 2b 6 Ratio = ∴ = = 6 :6 + π a 6+ π

57. Since, the chord QR is parallel to the tangent at P. ∴

ON ⊥ QR

Consequently, N is the mid-point of chord QR. ∴

QR = 2QN = 2r sin θ

Also,

ON = r cos θ

Application of Derivatives 273 ∴

PN = r + r cos θ

Now,

P

d2 π , we have ( AP )2 = − (16 − 2a 2) < 0 2 dθ 2 π Thus, AP 2 i.e. AP is maximum when θ = .The point on 2 the curve 4x2 + a 2y2 = 4a 2 that is farthest from the point π π  A(0, − 2) is  a cos , 2 sin  = (0, 2)  2 2

r

For θ =

O r

r

Q

N

R

M

59. Let AF = x and AE = y, ∆ABC and ∆EDC are similar.

Let A denotes the area of ∆PQR. 1 Then, A = ⋅ 2r sin θ (r + r cos θ ) 2 ⇒ ⇒ ⇒ and

d2 ( AP )2 = − {(8 − 2a 2) sin θ + 8} sin θ dθ 2 + (8 − 2a 2) ⋅ cos 2 θ

A = r 2(sin θ + sin θ cos θ ) 1 A = r 2(sin θ + sin 2θ ) 2 dA 2 = r (cos θ + cos 2θ ) dθ

C

dA =0 dθ

Hence, the area of ∆PQR is maximum when θ =

A 2 x (p , – p )

3 3 2 r sq units 4

58. Let P (a cos θ , 2 sin θ ) be a point on the ellipse x2 y2 + =1 4 a2 Let A(0, − 2) be the given point. Then, ( AP )2 = a 2 cos 2 θ + 4 (1 + sin θ )2 d ⇒ ( AP )2 = − a 2 sin 2θ + 8 (1 + sin θ ) ⋅ cos θ dθ d ⇒ ( AP )2 = [(8 − 2a 2) sin θ + 8 ] cos θ dθ d For maximum or minimum, we put ( AP )2 = 0 dθ ⇒ [(8 − 2a 2) sin θ + 8] cos θ = 0 4 ⇒ cos θ = 0 or sin θ = 2 a −4 4x2 + a 2y2 = 4a 2, i.e.

4 > 1 ⇒ sin θ > 1, which is a2 − 4 impossible]

B (q 2, –q)

c b = x b− y

⇒ π . 3

F

c

⇒

The maximum area of ∆ PQR is given by  3 2π  3 π 1  + A = r 2 sin + sin  = r 2    3 2 3 4  2

[Q 4 < a 2 < 8 ⇒

a D

y

⇒ cos θ + cos 2θ = 0 ⇒ cos 2θ = − cos θ π ⇒ cos θ = cos (π − 2θ ) ⇒ θ = 3 d 2A π Clearly, < 0 for θ = 3 dθ 2

=

2

( r , – r)

b E

d 2A = r 2(− sin θ − 2 sin 2θ ) dθ 2

For maximum and minimum values of θ,we put

AB AC = ED CE

∴

bx = c (b − y) c x = (b − y) b

⇒

Let z denotes the area of par allelogram AFDE. z = xy sin A c z = (b − y) y ⋅ sin A b

Then, ⇒

…(i)

On differentiating w.r.t. y we get dz c = (b − 2 y) sin A dy b d 2z −2c = sin A b dy2

and

For maximum or minimum values of z, we must have dz =0 dy c ⇒ (b − 2 y) = 0 b b y= ⇒ 2 Clearly,

d 2z 2c =− < 0, ∀ y 2 b dy

Hence, z is maximum, when y = On putting y =

b . 2

b in Eq. (i), we get 2

274 Application of Derivatives the maximum value of z is 1 c b b z =  b −  ⋅ ⋅ sin A = bc sin A 2 2 4 b =

g′ (x) = 0 ⇒ x = 0, x = ± 3 Now,

1 area of ∆ABC 2

g′ ′ (x) =

p2 − p 1 1 1 2 = × q q 1 2 2 2 −r 1 r

At

∴

AC = x km

−p 1

1 ( p + q) (r − p) q − p 1 0 4 r + p −1 0

1 ( p + q) (r − p) (− q − r ) 4 1 = ( p + q) (q + r ) ( p − r ) 4 π π 3 60. Let y = f (x) = sin x + λ sin 2 x, − < x < 2 2 Let sin x = t

S

∴

∴

⇒ Let ∴

1−x dy (1 + x ) ⋅ 1 − x (2x) = = dx (1 + x2)2 (1 + x2)2 dy = g (x) dx

B

SC = x2 + d 2

CB = (L − x)

and

Distance Time = Speed

Time from S to B = Time from S to C + Time from C to B

x 1 + x2 2

(L – x)

C

L

For exactly one minima and exactly one maxima dy/dt must have two distinct roots ∈ (−1, 1). 2λ ⇒ t = 0 and t = − ∈ (−1, 1) 3 2λ −1 < − 0, ∀ x .

Application of Derivatives 275 b ≥ c, ∀ x > 0 ; a , b > 0 x b f (x) = ax2 + − c x

63. Given, ax2 + Let

b 2ax3 − b = x2 x2 2b ⇒ f ′ ′ (x) = 2a + 3 > 0 [since, a , b are all positive] x

∴

f ′ (x) = 2ax −

Now, put At

 b f ′ (x) = 0 ⇒ x =    2a   b x=   2a 

and

⇒

1/3

>0

[Q a , b > 0]

1/3

, f ′ ′ (x) = + ve

 b ⇒ f (x) has minimum at x =    2a 

1/3

.

2 /3   b  1/3  b  b + − c≥ 0 f     = a       2 2 a a ( / 2 b a )1/3  

 2a  =   b

1/3

 2a     b

1/3

π 5π 13π , , ,… 12 12 12 but θ ∈ { λ 1π , λ 2π , … λ rπ }, where 0 < λ 1 < … < λ r < 1. π 5π θ= , ∴ 12 12 1 5 So, λ 1 + … + λ r = + = 0.50 12 12 ⇒

⋅

3b − c≥0 2

⋅

3b ≥c 2

66. Given set S of polynomials with real coefficients S = {(x2 − 1)2(a 0 + a1x + a 2x2 + a3 x3 ) : a 0 , a1 , a 2, a3 ∈ R} and for a polynomial f ∈ S, Let f (x) = (x2 − 1)2 (a 0 + a1x + a 2x2 + a3 x3 ) it have − 1 and 1 as repeated roots twice, so graph of f (x) touches the X-axis at x = − 1 and x = 1, so f ′ (x) having at least three roots x = − 1, 1 and α . Where α ∈ (− 1, 1) and f ′ ′ (x) having at least two roots in interval (− 1, 1) So, mf ′ = 3 and mf ′ ′ = 2 ∴Minimum possible value of (mf ′ + mf ′ ′ ) = 5

67. On drawing the diagram of given situation MD 2 + MC 2 = 64 + x2 + 121 + (10 − x)2 = f (x) ⇒

f (x) = 2x − 20x + 285

D

27ab ≥ 4c 2

3

11

64. Let f (x) = x + y, where xy = 1 ⇒

[say]

2

C

On cubing both sides, we get 2a 27b3 ⋅ ≥ c3 8 b ⇒

θ=

f (x) = x +

8

1 x

⇒

1 x2 − 1 f ′ (x) = 1 − 2 = x x2

Also,

f ′ ′ (x) = 2 / x3

On putting f ′ (x) = 0, we get x = ± 1, but x > 0 [neglecting x = − 1] f ′ ′ (x) > 0, for x = 1 Hence, f (x) attains minimum at x = 1, y = 1

A

M x

B 10 – x

For minima, f ′ (x) = 0 ⇒ 4x − 20 = 0 ⇒ x = 5 As f ′ ′ (x) = 4 > 0 ∴ AM = 5 m

68. Here, volume of cylindrical container, V = πr 2h …(i) and let volume of the material used be T. r+2

⇒ (x + y) has minimum value 2.

O r

65. The given function f : R → R be defined by f ( θ ) = (sin θ + cos θ )2 + (sin θ − cos θ )4 = 1 + sin 2 θ + (1 − sin 2 θ )2 = 1 + sin 2 θ + 1 + sin 2 2 θ − 2 sin 2 θ = sin 2 2 θ − sin 2 θ + 2

2

h

2

1 7  = sin 2 θ −  +  2 4 The local minimum of function ‘ f ’ occurs when 1 sin 2 θ = 2 π 5π 13π 2θ = , , … ⇒ 6 6 6

∴ ⇒

T = π [(r + 2)2 − r 2] h + π (r + 2)2 × 2 V T = π [(r + 2)2 − r 2] ⋅ 2 + 2π (r + 2)2 πr [Q V = πr 2h ⇒ h =

V ] πr 2

276 Application of Derivatives 2

 r + 2 2 T =V   + 2π (r + 2) − V  r 

⇒

∴

On differentiating w.r.t. r, we get

or

69.

70.

1 9 = 2 2

PLAN (i) Local maximum and local minimum are those points at which f ′ ( x ) = 0, when defined for all real numbers. (ii) Local maximum and local minimum for piecewise functions are also been checked at sharp edges.

(h,0)

Also,

(x2 − 1), y =|x2 − 1|=  2  (1 − x ),

R (2 sec θ, 0)

∆ = Area of ∆PQR 1 = (2 3 sin θ ) (2 sec θ − 2 cos θ ) 2 = 2 3 ⋅ sin3 θ/cos θ 1 ≤ h ≤1 2 1 ≤ 2 cos θ ≤ 1 2 1 1 ≤ cos θ ≤ 4 2

∴ ⇒ ∴

d∆ 2 3 {cos θ ⋅ 3 sin θ cos θ − sin θ (− sin θ )} = dθ cos 2 θ 2

= =

When

2 3 ⋅ sin θ [3 cos 2 θ + sin 2 θ ] cos 2 θ 2

2 3 sin 2 θ ⋅ [2 cos 2 θ + 1] cos 2 θ

= 2 3 tan 2 θ (2 cos 2 θ + 1) > 0 1 1 ≤ cos θ ≤ , 4 2

if x ≤ − 1 if − 1 ≤ x ≤ 0 if 0 ≤ x ≤ 1 if x ≥ 1

 − x2 − x + 1 , if x ≤ − 1  2  − x − x + 1, if − 1 ≤ x ≤ 0 = 2 − x + x + 1, if 0 ≤ x ≤ 1  x2 + x − 1, if x≥1

x y cos θ + sin θ = 1 2 3 R(2 sec θ , 0)

Since,

if x ≤ − 1 or x ≥ 1 if − 1 ≤ x ≤ 1

 − x + 1 − x2 ,  2 − x + 1 − x , y =|x|+ |x2 − 1|=  2 x + 1 − x ,  x + x2 − 1 ,

(2 sec θ, –√3 sin θ)

⇒

x, if x ≥ 0 y =|x|=   − x, if x < 0

Description of Situation

(2 cos θ, √3 sin θ)

Q

∴

1 2

8 ∆1 − 8∆ 2 = 45 − 36 = 9 5

∴

Here, tangent at P(2 cos θ , 3 sin θ ) is

O

cos θ =

When

PLAN As to maximise or minimise area of triangle, we should find area in terms of parametric coordinates and use second derivative test.

P

1 45 5 = 8 4

 2 3 sin3 θ  =   cos θ 

r = 10 V =π 1000 V =4 250π

⇒

1  2 3 ⋅ sin3 θ  =  4  cos θ 

∆ 2 = ∆ min occurs at cos θ =

V =π r3

where

cos θ =

When

dT  r + 2   −2  = 2V ⋅   ⋅   + 4π (r + 2)  r   r2  dr dT At r = 10, =0 dr V  Now, 0 = (r + 2) ⋅ 4  π − 3   r  ⇒

∆1 = ∆ max occurs at cos θ =

which could be graphically shown as Y

…(i)

– x 2– x + 1

– x 2– x + 1 – x 2+ x +1

x 2+x –1

1 –1 –1/2 O

X 1/2 1

Thus, f (x) attains maximum at x = …(ii)

attains minimum at x = − 1, 0, 1. ⇒ Total number of points = 5

3

71.

1 −1 and f (x) , 2 2

PLAN If f( x ) is least degree polynomial having local maximum and local minimum at α and β.

Then,

f ′ (x) = λ (x − α ) (x − β )

Here,

p′ (x) = λ (x − 1) (x − 3) = λ (x2 − 4x + 3)

On integrating both sides between 1 to 3, we get 3

3

∫1 p′ (x) dx = ∫1 λ (x

2

− 4x + 3) dx 3

⇒

 x3  ( p(x))31 = λ  − 2x2 + 3x 3 1

Application of Derivatives 277 ⇒

  1 p(3) − p(1) = λ  (9 − 18 + 9) −  − 2 + 3   3  − 4 2 −6 = λ   3 

⇒ ⇒

λ =3

⇒

p′ (x) = 3 (x − 1) (x − 3)

∴

72.

p′ (0) = 9

f (x) = x − 4x3 + 12x2 + x − 1 4

f ′ (x) = 4x3 − 12x2 + 24x + 1 f ′′ (x) = 12x2 − 24x + 24 = 12 (x2 − 2x + 2) = 12 {(x − 1)2 + 1} > 0 ∀x ⇒

f ′ (x) is increasing.

Since, f ′ (x) is cubic and increasing.

74. Let

f (θ ) =

1 sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ

Again let, g (θ ) = sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ 1 − cos 2 θ  1 + cos 2 θ  3 = +5  + sin 2 θ  2  2 2 3 = 3 + 2 cos 2 θ + sin 2 θ 2 9 g (θ )min = 3 − 4 + ∴ 4 5 1 =3− = 2 2 1 ∴ Maximum value of f (θ ) = =2 12

75. Given, A = { x|x2 + 20 ≤ 9x} = { x|x ∈ [4, 5]}

⇒ f ′ (x) has only one real root and two imaginary roots.

Y

∴ f (x) cannot have all distinct roots. ⇒ Atmost 2 real roots.

O

Now, f (− 1) = 15, f (0) = − 1, f (1) = 9

2

3

4 5

∴ f (x) must have one root in (− 1, 0) and other in (0, 1). ⇒ 2 real roots.

X

–16 –20 –21

73. Let g (x) = e f ( x ), ∀ x ∈ R ⇒ g′ (x) = ef ( x ) ⋅ f ′ (x) ⇒ f ′ (x) changes its sign from positive to negative in the neighbourhood of x = 2009 ⇒ f (x) has local maxima at x = 2009. So, the number of local maximum is one.

Now,

f ′ (x) = 6(x2 − 5x + 6)

Put

f ′ (x) = 0 ⇒ x = 2, 3 f (2) = − 20, f (3) = − 21, f (4) = − 16, f (5) = 7

From graph, maximum value of f (x) on set A is f (5) = 7.

11 Indefinite Integration Topic 1 Some Standard Results Objective Questions I (Only one correct option) cos θ dθ = A log e|B(θ )| + C , where C is a 5 + 7 sin θ − 2 cos 2 θ B(θ ) can be constant of integration, then A

1. If ∫

(2020 Main, 5 Sep II)

2 sin θ + 1 sin θ + 3 5(sin θ + 3) (c) 2 sin θ + 1

(a) − (c)

1 and f (x) = 9 (x − 1) 27 1 (b) A = and f (x) = 3 (x − 1) 81 1 (c) A = and f (x) = 3 (x − 1) 54 1 (d) A = and f (x) = 9 (x − 1)2 54

x dx (x ≥ 0). Then f (3) − f (1) is equal to (1 + x)2 (2020 Main, 4 Sep I)

π 1 3 + + 6 2 4

π 1 3 + − 6 2 4

(b) − (d)

π 1 3 + + 12 2 4

π 1 3 + − 12 2 4

1

7. If ∫

∫

sin 2 α + C, where C is a constant of integration, then the functions A (x) and B (x) are respectively (2019 Main, 12 April II)

(a) x + α and log e|sin(x + α )| (b) x − α and log e|sin(x − α )| (c) x − α and log e|cos(x − α )| (d) x + α and log e|sin(x − α )|

2x3 − 1 dx is equal to x4 + x

|x + 1| 1 log e +C 2 x2 x3 + 1 (c) log e +C x (a)

(2019 Main, 12 April I)

(x + 1)2 1 +C log e 2 |x3| |x3 + 1| (d) log e +C x2 3

(b)

∫

1

(b) −

6x3 1

(d)

2x2

1

2x3 3

x2

5x 2 dx is equal to x sin 2

sin

(where, C is a constant of integration) (2019 Main, 8 April I)

(a) (b) (c) (d)

2x + sin x + 2 sin 2x + C x + 2 sin x + 2 sin 2x + C x + 2 sin x + sin 2x + C 2x + sin x + sin 2x + C

3x13 + 2x11 dx is equal to (where C (2x4 + 3x2 + 1)4 (2019 Main, 12 Jan II) is a constant of integration)

9. The integral ∫

(here C is a constant of integration) 3

(c) −

8.

tan x + tan α dx = A (x) cos 2α + B (x) tan x − tan α

5. The integral ∫

(a) −

(b) (1, 1 − tan θ) (d) (−1, 1 − tan θ)

4. Let α ∈ (0, π / 2) be fixed. If the integral

dx = xf (x)(1 + x6 )3 + C / x (1 + x6 )23 3

where, C is a constant of integration, then the function f (x) is equal to (2019 Main, 8 April II)

dθ 3. If ∫ = λ tan θ + 2 log e| f (θ )| + C 2 cos θ (tan 2 θ + sec 2 θ ) where C is a constant of integration, then the ordered pair (2020 Main, 9 Jan II) (λ , f (θ )) is equal to (a) (1, 1 + tan θ) (c) (−1, 1 + tan θ)

2

(a) A =

2 sin θ + 1 5(sin θ + 3) 5(2 sin θ + 1) (d) sin θ + 3 (b)

(a)

2. Let f (x) = ∫

dx (x − 2x + 10)2   f (x)  x − 1 = A  tan −1   + C , where, C is a  +  3  x2 − 2x + 10  (2019 Main, 10 April I) constant of integration, then

6. If ∫

(a) (c)

x4 6(2x4 + 3x2 + 1)3 x4 (2x + 3x + 1) 4

2

3

+C

+C

(b) (d)

x12 6(2x4 + 3x2 + 1)3 x12 (2x + 3x2 + 1)3 4

+C +C

Indefinite Integration 279 x+1 dx = f (x) 2x − 1 + C, where C is a constant 2x − 1 of integration, then f (x) is equal to (2019 Main, 11 Jan II)

10. If ∫

2 (x + 2) 3 2 (c) (x − 4) 3

is equal to

(c)

1

−1

(b)

9x4 −1

3 (1 + tan3 x) 1 (c) +C 1 + cot3 x

(2019 Main, 11 Jan I)

27x6

17.

(sin θ cos θ dθ is equal to sin n + 1 θ (where C is a constant of integration)

∫

(a)

1 n   1 −  sin n + 1 θ  n 2 − 1

(b)

1 1 +    n 2 − 1 sin n − 1 θ 

(c)

1 1 −    sin n − 1 θ  n 2 − 1

(d)

n

n

1 n   1 −  sin n − 1 θ  n 2 + 1

(2019 Main, 10 Jan I)

n+1 n

+C

n+1 n

+C

n+1 n

+C

n

+C

5x8 + 7x6 13. If f (x) = ∫ 2 dx, (x ≥ 0), and f (0) = 0, then (x + 1 + 2x7 )2 the value of f (1) is (2019 Main, 9 Jan I) (a) − (c)

1 2

1 4

(b) − (d)

1 4

1 2

14. For x ≠ nπ + 1, n ∈ N (the set of natural numbers), the 2

integral 2 sin(x2 − 1) − sin 2(x2 − 1) ∫ x 2 sin(x2 − 1) + sin 2(x2 − 1) dx is equal to (where C is a constant of integration ) (2019 Main, 9Jan I) 1 log e|sec(x2 − 1)| + C 2  x2 − 1  + C (b) log e sec   2  1 (c) log e sec2 (x2 − 1) + C 2  x2 − 1 1  + C (d) log e sec2  2  2  (a)

3 (1 + tan3 x) −1 (d) +C 1 + cot3 x

(2015 Main)

1

(b) (x4 + 1) 4 + c

(c) − (x +

 x4 + 1 4 (d) −  4  + c  x 

1 1) 4

1

+c

sec2 x ∫ (sec x + tan x)9/ 2 dx equals to (for some arbitrary constant K) −1 1 −  (sec x + tan x)11/ 2 11 1 1 − (b)  (sec x + tan x)11/ 2 11 −1 1 + (c)  (sec x + tan x)11/ 2 11 1 1 + (d)  (sec x + tan x)11/ 2 11

(a)

18. If I = ∫

n+1

+C

 x4 + 1 4 (a)  4  + c  x  4

π 2

1 − sin θ ) n

(2018 Main)

−1

(b)

1

12. Let n ≥ 2 be a natural number and 0 < θ < . Then, n

+C

(where C is a constant of integration) dx 16. The value of ∫ 2 4 is x (x + 1)3/ 4

3x3 1

(d)

27x9

1

(a)

dx = A (x)( 1 − x2 )m + C, x4 for a suitable chosen integer m and a function A (x), where C is a constant of integration, then ( A (x))m equals

dx

+ cos5 x)2

(b)

1 − x2

11. If ∫

sin 2 x cos 2 x

∫ (sin5 x + cos3 x sin 2 x + sin3 x cos2 x

1 (x + 4) 3 1 (d) (x + 1) 3

(a)

(a)

15. The integral

4x

e

(2012)

1 (sec x + tan x)2  + K 7  1 2 (sec x + tan x)  + K 7  1 2 (sec x + tan x)  + K 7  1 2 (sec x + tan x)  + K 7 

ex e− x dx, J = ∫ − 4x dx. 2x +e +1 + e−2x + 1 e

Then, for an arbitrary constant c, the value of J − I (2008, 3M) equals (a) (c)

e4 x − e2x + 1 1 +c log 4 x 2 e + e2x + 1 e2x − ex + 1 1 +c log 2x 2 e + ex + 1

19. If f (x) =

(b)

e2x + ex + 1 1 +c log 2x 2 e − ex + 1

(d)

e4 x + e2x + 1 1 log 4 x +c 2 e − e2x + 1

x for n ≥ 2 and g (x) = ( fofo ... of ) (x). 14243 (1 + xn )1/ n f occurs n times

Then, ∫ xn − 2g (x) dx equals 1 1− nxn ) n

(2007, 3M)

(a)

1 (1 + n (n − 1)

(c)

1+ 1+ 1 1 (1 + nxn ) n + c (d) (1 + nxn ) n + c n (n + 1) n+1

+ c (b)

1 (1 + n −1

1 1− nxn ) n

1

20. The value of ∫ (a) 2 2 − (c)

2 x2

+

1

(x2 − 1) dx 3

x 1 x4

2 x4 − 2 x2 + 1 +c

2 1 1 2− 2 + 4 + c 2 x x

+c

is

(b) 2 2 +

(2006, 3M)

2 x2

+

1 x4

(d) None of these

+ c

280 Indefinite Integration Analytical & Descriptive Questions

Objective Questions II (One or more than one correct option)

24. For any natural number m, evaluate

21. Let f : R → R and g : R → R be two non-constant differentiable functions. If f ′ (x) = (e( f ( x ) − g( x )) ) g′ (x) for all x ∈ R and f (1) = g (2) = 1, then which of the following statement(s) is (are) TRUE? (2018 Adv.) (a) f (2) < 1 − log e 2 (b) f (2) > 1 − log e 2 (c) g (1) > 1 − log e 2 (d) g (1) < 1 − log e 2

22. Let f : R → R be a differentiable function with f (0) = 1 and satisfying the equation + f ′ (x) f ( y) for all x, y ∈ R.

f (x + y) = f (x) f ′ ( y)

Then, the value of log e ( f (4)) is ....... .

(2018 Adv.)

+ x2m + xm ) (2 x2m + 3 xm + 6)1/m dx, x > 0. (2002, 5M)

25. Evaluate

 1 − x    1 + x

∫

1/ 2

⋅

1− x dx. 1+ x dx 27. Evaluate ∫ 2 4 . x (x + 1)3/ 4 28. Evaluate the following: 1  (i) ∫ 1 + sin  x dx 2 

dx x

(1997C, 3M)

29. Integrate

(1985, 2 1 M) 2

(1984, 2M) (1980, 4M)

(ii)

∫

x2 dx 1−x

x2 . (a + bx)2

(1979, 2M)

30. Integrate sin x ⋅ sin 2 x ⋅ sin 3x + sec2 x ⋅ cos 2 2x + sin 4 x ⋅ cos 4 x.

Fill in the Blank

(1979, 1M)

−x

4e + 6e dx = Ax + B log (9e2x − 4) + C, then A = K, 9ex − 4e− x (1989, 2M) B = ... and C = K .

23. If ∫

3m

26. Evaluate ∫

Integer & Numerical Answer Type Question

x

∫ (x

x . 31. Integrate the curve 1 + x4 1 sin x 32. Integrate or . 1 − cot x sin x − cos x

(1978, 1M)

(1978, 2M)

Topic 2 Some Special Integrals Objective Question I (Only one correct option) 1. The integral ∫ sec

23 /

4/3

x cosec x dx is equal to (here C is a

constant of integration) (a) 3 tan −1/3 x + C

(2017 Main)

1 1 (c)  , − 1 (d)  − , 0 5   5 

 1 3  x+

∫(

4

x

tan x +

+

ln (1 + 6 x )  dx. 3 x+ x 

(1992, 4M)

cot x ) dx.

(1988, 3M)

(cos 2x)1/ 2 (1987, 6M) dx. sin x 2 sin x − sin 2 x 6. If f (x) is the integral of , where x ≠ 0, then x3 (1979, 3M) find lim f ′ (x).

5. Evaluate

I 4 + I 6 = a tan5 x + bx5 + C, where C is a constant of integration, then the ordered pair (a , b) is equal to 1 (b)  , 0 5 

∫ 4. Evaluate

2. Let I n = ∫ tan n x dx (n > 1). If

1 (a)  − , 1  5 

3. Find the indefinite integral

(2019 Main, 9 April I)

(b) −3 tan −1/3 x + C 3 (d) − tan −4 /3 x + C 4

(c) −3 cot −1/3 x + C

Analytical & Descriptive Questions

∫

x→ 0

Topic 3 Integration by Parts Objective Questions (Only one correct option) 1. If

5 −x

∫xe

2

2

dx = g (x)e− x + C, where C is a constant of

integration, then g (− 1) is equal to (2019 Main, 10 April II) (a) − 1 1 (c) − 2

(b) 1 (d) −

5 2

2. If ∫ esec x (sec x tan x f (x) + (sec x tan x + sec2 x)) dx = esec x f (x) + C, then a possible choice of f (x) is (2019 Main, 9 April II)

(a) x sec x + tan x +

1 2

(b) sec x + tan x +

1 2

(c) sec x + x tan x −

1 2

(d) sec x − tan x −

1 2

Indefinite Integration 281

∫ cos (log e x) dx is equal to (where C is a

3. The integral

constant of integration)

6. If ∫ f (x) dx = ψ (x), then ∫ x5 f (x3 ) dx is equal to 1 3 [x ψ(x3 ) − ∫ x2ψ(x3 )dx] + c 3 1 3 (b) x ψ(x3 ) − 3 ∫ x3 ψ(x3 ) dx + c 3 1 (c) x3 ψ(x3 ) − ∫ x2ψ(x3 ) dx + c 3 1 (d) [x3 ψ(x3 ) − ∫ x3 ψ(x3 ) dx] + c 3

(2019 Main, 12 Jan I)

(a)

x (a) [cos(log e x) + sin(log e x)] + C 2 (b) x [cos(log e x) + sin(log e x)] + C

(c) x [cos(log e x) − sin(log e x)] + C x (d) [sin(log e x) − cos(log e x)] + C 2

1 −4 x 3 e f (x) + C , 48 where C is a constant of integration, then f (x) is equal to

(2013 Main)

3

4. If ∫ x5 e−4x dx =

(2019 Main, 10 Jan II)

5.

(a) − 4x3 − 1

(b) 4x3 + 1

(c) − 2x3 − 1

(d) − 2x3 + 1

∫

1 1 x + x e dx is

 1 + x −   x x+

(a) (x − 1) e x+

(b) x e

1 x

 2x + 2  dx.  4x2 + 8 x + 13    

7. Evaluate ∫ sin −1 

(c) (x + 1) e

1 x+ −x e x

1 x

(2000, 5M)

8. Find the indefinite integral  cos θ + sin θ 

∫ cos 2θ log cos θ − sin θ  dθ.

(2014 Main)

+ c

9. Evaluate ∫

+ c x+

(d)

1 x

equal to

Analytical & Descriptive Questions

10. Evaluate ∫

+ c

sin −1 x − cos −1 x dx. sin −1 x + cos −1 x (x − 1) ex dx. (x + 1)3

11. Evaluate ∫ (elog x + sin x) cos x dx.

+ c

(1994, 5M)

(1986, 2½ M)

(1983, 2M) (1981, 2M)

Topic 4 Integration, Irrational Function and Partial Fraction Objective Questions (Only one correct option) 1. The integral ∫ (a)

2x12 + 5x9 dx is equal to (x5 + x3 + 1)3

− x5 (x + x + 1) 5

3

2

(c)

x

2(x5 + x3 + 1)2 x5 2(x + x + 1) 5

3

2

−x

10

(d)

2(x5 + x3 + 1)2

(c) sin x − 2 (sin x)−1 − 6 tan −1 (sin x) + c (d) sin x − 2 (sin x)−1 + 5 tan −1 (sin x) + c

Analytical & Descriptive Questions 3.

+C

where, C is an arbitrary constant.

(1995, 2M)

(b) sin x − 2 (sin x)−1 + c

+C +C

cos3 x + cos5 x dx is sin 2 x + sin 4 x

(a) sin x − 6 tan −1 (sin x) + c

(2016 Main)

+C

10

(b)

2. The value of ∫

∫

x3 + 3x + 2 dx. (x + 1)2 (x + 1) 2

4. Evaluate ∫

(x + 1) dx. x (1 + xex )2

(1999, 5M)

(1996, 2M)

Answers Topic 1 1. 5. 9. 13. 17.

(d) (c) (b) (c) (c)

21. (b,c)

3. 2. 6. 10. 14. 18.

(d) (c) (b) (b) (c)

22. (2)

3. 7. 11. 15. 19.

4. (b) 8. (c) 12. (c) 16. (d) 20. (c) 3 35 23. A = − , B = and C ∈R 2 36 (c) (b) (c) (b) (a)

1 ⋅(2 x 3m + 3 x 2m + 6 xm ) (m + 1)/m + c 6 (m + 1 ) 1 25. 2 [cos−1 x − log| 1 + 1 − x | − log | x| ] + c 2 −1 26. −2 1 − x + cos x + x (1 − x ) + c 24.

29. 30.

31.

4.

 tan x − cot x  2 tan −1   +c 2  

5. − log| cot x +

cot 2 x − 1 | +

1 log 2

2 + 1 − tan 2 x

+c

2 − 1 − tan 2 x

Topic 3 1. (d)

( x 4 + 1 )1/4 +c x x x (i) 4 sin − 4 cos + c 4 4 2 1   (ii) − 2  1 − x − (1 − x ) 3/2 + (1 − x ) 5/2  + c 3 5    1  a2 + c a + bx − 2a log (a + bx ) − a + bx b3   cos 4 x cos 2 x cos 6 x − − + + sin 2 x + tan x − 2 x 16 8 24 3 x sin 4 x sin 8 x + − + 128 128 1024 1 1 x −1 2 32. log (sin x − cos x ) + + c tan ( x ) + c 2 2 2

Topic 2 1. (b)

+ 12 ln (1 + x1/2 ) − 3 [ ln(1 + x1/6 )]2 + c

6. (1)

27. − 28.

3 2/3 12 7/12 4 1/2 12 5/12 1 1/3 x − x + x − x + x − 4 x 1/ 4 − 7 x 1/ 6 2 7 3 5 2 − 12 x1/12 + (2 x1/2 − 3 x1/3 + 6 x1/6 + 11 ) ln (1 + x1/6 )

7. ( x + 1 ) tan −1 8.

3. (a)

4. (a)

6. (c) 2  x + 2 3 2  − log( 4 x + 8 x + 13 ) + c   3  4

 cosθ + sin θ  1 1 sin 2 θ ln   + ln(cos2 θ ) + c  cosθ − sin θ  2 2

2 [ x − x 2 − (1 − 2 x ) sin −1 x ] − x + c π cos2 x ex 10. +c + c 11. x sin x + cos x − 4 (x + 1)2 9.

Topic 4 1. (b) 2. (c) 1 1 3 x 3. − log| x + 1 | + log| x 2 + 1 | + tan −1 x + 2 +c 2 4 2 x +1 4. log

2. (b)

2. (b)

5. (b)

1 xe x + +c 1 + xe x 1 + xe x

Hints & Solutions Topic 1 Some Standard Results cos θ 1. Since, I = ∫ dθ 5 + 7 sin θ – 2 cos 2 θ cos θ =∫ dθ 3 + 7 sin θ + 2 sin 2 θ Let, sin θ = t ⇒ cos θ dθ = dt dt dt 1 = ∫ ∴ I=∫ 2 7 3 2 2t + 7t + 3 t2 + t + 2 2 dt 1 1 = ∫ = ∫ 2 2  7  3 49 2  t +  +  –  t +   2 16  4

7 5 t+ – 1 1 4 4 + C' log e = × 7 5 2 2×5 t+ + 4 4 4

dt 2

7  5  –   4 4

2

∴

=

1 2t + 1 + C' log e 5 2t + 6

=

1 1 2 sin θ + 1 + C, where C = C′– log e 2 log e 5 sin θ + 3 5

= A log e|B(θ )|+ C B(θ ) 5(2 sin θ + 1) . = A (sin θ + 3)

[given]

Indefinite Integration 283 Now, put x − α = t ⇒ dx = dt, so

2. It is given that, f (x) = ∫

x dx, (x ≥ 0) (1 + x)2

I=∫

Put x = tan 2 t ⇒ dx = 2 tan t sec2 tdt 2 tan 2 t sec2 t ∴ f (x) = ∫ dt = ∫ 2 sin 2 t dt sec4 t [Q2 sin 2 x = 1 − cos 2x] = ∫ (1 − cos 2t )dt sin 2t +C 2 tan t x So, =t− + C = tan −1 x − +C 2 1+ x 1 + tan t   3 ∴f (3) − f (1) =  tan −1 3 − + C 1+3  

sin t cos 2 α + sin 2α cos t dt sin t cos t   = ∫  cos 2 α + sin 2 α  dt  sin t 

=∫

= t (cos 2 α ) + (sin 2 α ) log e |sin t |+ C = (x − α ) cos 2 α + (sin 2 α ) log e |sin (x − α )| + C

=t−

  1 −  tan −1 (1) − + C   1+1 π 3 π 1 π 1 3 − − + = + − 3 4 4 2 12 2 4 Hence, option (d) is correct. dθ 3. Given integral, I = ∫ cos 2 θ (tan 2 θ + sec 2 θ ) =

=∫

sec2 θ cos 2 θ dθ (sin 2 θ + 1)

=∫

sec2 θ (1 − tan 2 θ ) dθ (2 tan θ + 1 + tan 2 θ )

=∫

sec2 θ (1 − tan θ ) dθ 1 + tan θ

= A (x) cos 2 α + B(x) sin 2 α + C (given) Now on comparing, we get A (x) = x − α and B(x) = log e |sin (x − α )|

5.

Key Idea (i) Divide each term of numerator and denominator by x 2. 1 (ii) Let x 2 + = t x

2x3 − 1 2 x − 1 / x2 dx = ∫ dx 4 1 x +x x2 + x [dividing each term of numerator and

Let integral is I = ∫

denominator by x2] 1  Put x2 + = t ⇒ 2x +  x I=∫

∴

= − t + 2 log e|1 + t| + C = − tan θ + 2 log e |1 + tan θ| + C = λ tan θ + 2 log e| f (θ )| + C (given) (on comparing) ∴ (λ , f (θ )) = (−1, 1 + tan θ ) Hence, option (c) is correct. tan x + tan α  π 4. Let I=∫ dx, α ∈ 0,   2 tan x − tan α sin x sin α + cos x cos α dx =∫ sin x sin α − cos x cos α sin x cos α + sin α cos x dx =∫ sin x cos α − sin α cos x sin (x + α ) =∫ dx sin (x − α )

= log e

6. Let I = ∫

 1   − 2  dx = dt  x 

dt = log e|(t )| + C t

 = log e  x2 + 

Let, tan θ = t ⇒ sec2 θdθ = dt 1 −t Then, I=∫ dt 1+t  2  = ∫  −1 +  dt  1 + t

sin (t + 2α ) dt sin t

1  +C x

x3 + 1 +C x

dx dx =∫ 2 ((x − 1)2 + 32)2 (x − 2x + 10) 2

Now, put x − 1 = 3 tan θ ⇒ dx = 3 sec2θ dθ 3sec2θ dθ 3sec2θ dθ =∫ 4 2 2 2 (3 tan θ + 3 ) 3 sec4θ 1 1 1 + cos 2θ = dθ cos 2θ dθ = 27 ∫ 27 ∫ 2 1 + cos 2θ   Q cos 2 θ =   2 1  sin 2θ  1 = (1 + cos 2θ ) dθ =  +C θ + 54  2  54 ∫

So,I = ∫

=

2

1 1  2 tan θ   x − 1 tan −1    +C  +   54 3 108  1 + tan 2 θ   2 tan θ  Q sin 2θ =  + tan 2 θ  1 

284 Indefinite Integration  x − 1    3 

1 1  x − 1 tan −1  =  +  3  54 54

+C 2  x − 1 1+    3    x−1   +C 2 2  (x − 1) + 3    x−1  2  +C  x − 2x + 10

=

1 1  x − 1 tan −1   +  3  18 54

=

1 1  x − 1 tan −1   +  3  18 54

=

1  3(x − 1)   x − 1 tan −1  +C  +  3  x2 − 2x + 10  54 

It is given, that   f (x)  x − 1 +C I = A tan −1   + 2  3  x − 2x + 10   On comparing, we get A =

1 and f (x) = 3(x − 1). 54

dx x (1 + x6 )2/ 3

7. Let I = ∫

3

dx

=∫

1  x3 ⋅ x4  6 + 1 x 

2/ 3

=∫

dx 1  x7  6 + 1 x 

2/ 3

1 + 1 = t3 x6

Now, put

6 dx t2 dx = 3t 2dt ⇒ 7 = − dt 7 2 x x 1 2 − t dt 1 2 = − ∫ dt 2 t2 −

⇒ So, I = ∫

1/3

1 1 1 1  3   Qt = 6 +1 t + C = −  6 + 1 + C    2 x 2 x 1 1 (1 + x6 )1/3 + C =− 2 x2 [given] = x ⋅ f (x) ⋅ (1 + x6 )1/3 + C On comparing both sides, we get 1 f (x) = − 3 2x 5x 5x x sin 2 sin cos 2 dx = 2 2 dx 8. Let I = ∫ ∫ x x x sin 2 sin cos 2 2 2 x [multiplying by 2 cos in numerator and 2 denominator] sin 3x + sin 2x =∫ dx sin x [Q2 sin A cos B = sin( A + B) + sin( A − B) and sin 2 A = 2 sin A cos A] (3 sin x − 4 sin3 x) + 2 sin x cos x =∫ dx sin x [Q sin 3x = 3 sin x − 4 sin3 x] 2 = ∫ (3 − 4 sin x + 2 cos x)dx =−

= ∫ [3 − 2(1 − cos 2x) + 2 cos x]dx = ∫ [3 − 2 + 2 cos 2x + 2 cos x]dx

[Q2 sin 2 x = 1 − cos 2x]

= ∫ [1 + 2 cos 2x + 2 cos x]dx = x + 2 sin x + sin 2x + C

9. Let I=∫

3 2 + 5 3 3x13 + 2x11 x x dx = ∫ dx 4 (2x4 + 3x2 + 1)4 3 1  2 + 2 + 4   x x 

[on dividing numerator and denominator by x16] 3 1 Now, put 2 + 2 + 4 = t x x ⇒

−6 4   3 − 5  dx = dt x x 

⇒

2 dt 3  3 + 5  dx = −  x 2 x 1 1 t− 4 + 1 − dt =− × +C= 3 +C 4 2 −4 + 1 2t 6t 3 1 1  = +C Qt =2 + 2 + 4 3   x x 3 1   6 2 + 2 + 4   x x 

So, I = ∫

=

x12 +C 6 (2x + 3x2 + 1)3 4

10. We have,

∫

x+1 dx = f (x) 2x − 1 + C 2x − 1

Let I = ∫

...(i)

x+1 dx 2x − 1

Put 2x − 1 = t 2 ⇒ 2dx = 2tdt ⇒ dx = tdt 2 t +1 +1 1 I=∫ 2 tdt = ∫ (t 2 + 3) dt t 2  t2 + 1  2 Q 2x − 1 = t ⇒ x = 2     1  t3 t 2 =  + 3t + C = (t + 9) + C 23 6  2x − 1 = [Q t = 2x − 1 ] (2x − 1 + 9) + C 6 2x − 1 = (2x + 8) + C 6 x+4 = 2x − 1 + C 3 On comparing it with Eq. (i), we get x+4 f (x) = 3

Indefinite Integration 285 11. We have,

∫

1 − x2 x4

dx = A (x) ( 1 − x2 )m + C

… (i)

1  x2 2 − 1 x 

1 − x2

LetI = ∫

13. We have, f (x) = ∫

=∫

dx = ∫ dx x4 x4 1 x 2 −1 1 1 x =∫ − 1 dx dx = ∫ 3 x4 x x2 1 −2 1 Put 2 − 1 = t 2 ⇒ 3 dx = 2t dt ⇒ 3 dx = − t dt x x x t3 2 ∴ I = − ∫ t dt = − +C 3 =−

1  1 − x2  .  3  x2 

3/ 2

1 1 ( 1 − x2 )3 + C 3 x3 On comparing Eqs. (i) and (ii), we get 1 A (x) = − 3 and m = 3 3x 1 m ( A (x)) = ( A (x))3 = − ∴ 27 x9 (sin θ − sin θ ) cos θ dθ sin n + 1 θ sin θ = t ⇒ cos θ dθ = dt 1/ n t   n − t 1     (t n − t )1/ n t n   dt = ∫  I=∫ dt n+1 n+1 t t t (1 − 1 / t n−1 )1/ n (1 − 1 / t n − 1 )1/ n dt = ∫ dt =∫ n+1 t tn 1 1 − n −1 = u t (n − 1) 1 − t −( n − 1) = u ⇒ dt = du tn dt du = n n −1 t

12. Let I = ∫ Put ∴

Put or ⇒

1

⇒

…(ii)

1/ n

n

I=∫

u1/ ndu = n −1

+1

un +C  1 (n − 1)  + 1  n n+1

1  n  n 1 − n−1   t  = +C (n − 1) (n + 1)

=

1   n 1 −   sin n − 1 θ  n2 − 1

 x6   x8  5 14  + 7 14  x  x   x2 1 2x7   7 + 7 + 7 x x  x

2

dx

(dividing both numerator and denominator by x14) 5 x − 6 + 7 x− 8 = ∫ −5 dx (x + x− 7 + 2)2

1/ 2    1 + C Q t =  2 − 1   x  

=−

5 x8 + 7 x 6 dx (x2 + 1 + 2x7 )2

n+1 n

Let x− 5 + x − 7 + 2 = t ⇒ (− 5x− 6 − 7x− 8 )dx = dt ⇒ (5x− 6 + 7x− 8 )dx = − dt dt ∴ f (x) = ∫ − 2 = − ∫ t −2dt t t− 2 + 1 t− 1 1 =− + C =− +C= +C −2 + 1 −1 t =

x

−5

1 x7 +C= 7 +C −7 +x +2 2 x + x2 + 1

Q f (0) = 0 ∴

0 + C ⇒C = 0 0+0+1

0=

∴ f (x) =

x7 2 x + x2 + 1

⇒

f (1) =

7

14. Let I = ∫ x

1 1 = 2 2(1) + 1 + 1 4 7

2 sin(x2 − 1) − sin 2(x2 − 1) dx 2 sin(x2 − 1) + sin 2(x2 − 1)

x2 − 1 =θ 2 ⇒ x2 − 1 = 2θ ⇒ 2x dx = 2 dθ ⇒ x dx = dθ 2 sin 2 θ − sin 4 θ Now, I = ∫ dθ 2 sin 2 θ + sin 4 θ Put

=∫

2 sin 2 θ − 2 sin 2 θ cos 2 θ dθ 2 sin 2 θ + 2 sin 2 θ cos 2 θ

=∫

2 sin 2 θ (1 − cos 2 θ ) dθ 2 sin 2 θ (1 + cos 2 θ )

=∫

1 − cos 2 θ dθ = ∫ 1 + cos 2 θ

(Qsin 2 A = 2 sin A cos A)

2 sin 2 θ dθ 2 cos 2 θ

[Q1 − cos 2 A = 2 sin 2 A and 1 + cos 2 A = 2 cos 2 A] +C

1   Q u = 1 − n − 1 and t = sin θ   t

=∫

tan 2 θ d θ = ∫ tan θd θ

 x2 − 1  = log e|sec θ| + C = log e sec   +C  2 

 x2 − 1  Q θ = 2   

286 Indefinite Integration 15. We have,

sin 2 x ⋅ cos 2 x I=∫ dx 5 3 (sin x + cos x ⋅ sin 2 x + sin3 x ⋅ cos 2 x + cos5 x)2 sin 2 x cos 2 x dx =∫ {sin3 x(sin 2 x + cos 2 x) + cos3 x(sin 2 x + cos 2 x)}2 =∫

sin 2 x cos 2 x sin 2 x cos 2 x = dx dx ∫ (sin3 x + cos3 x)2 cos 6 x(1 + tan3 x)2

=∫

tan 2 x sec2 x dx (1 + tan3 x)2

Put tan3 x = t ⇒ 3 tan 2 x sec2 xdx = dt 1 dt I= ∫ ∴ 3 (1 + t )2 −1 ⇒ I= +C 3 (1 + t ) −1 +C ⇒ I= 3 (1 + tan3 x)

16.

dx

1+

Put

18. Since, ∴ Put

1  x5 1 + 4   x 

3/ 4

1 = t4 x4

−4 dx = 4t3 dt x5 dx = − t3 dt x5

⇒ ⇒

Hence, the integral becomes

∫ 17.

1 − t3 dt  = − ∫ dt = − t + c = − 1 + 4   t3 x 

I=∫

J −I=∫

4x

e

ex e3 x dx and J = ∫ dx 2x + e +1 1 + e2x + e4x

(e3 x − ex ) dx 1 + e2x + e4x

ex = u ⇒ ex dx = du

1   1 − 2  (u 2 − 1) u  du ∴ du = ∫ J −I = ∫ 2 4 1 1+ u + u 1 + 2 + u2 u 1   1 − 2  u  du =∫ 2 1  u +  −1  u 1 Put u+ =t u 1   ⇒ 1 − 2 du = dt  u 

dx

∫ x2(x4 + 1)3/ 4 = ∫

1 1 dt t +  ⋅ 1  t t 1  1 2 = ∫  9/ 2 + 13/ 2 dt ⇒ I=∫ 2 t t 9/ 2 t  1 2 2  = −  7/ 2 + + K 2 7 t 11 t11/ 2    1 1 = − + +K 7/ 2 11/ 2  7 (sec x tan x ) 11 (sec x tan x ) + +   1 −1 1  + (sec x + tan x)2 + K = 11/ 2  11 7 (sec x + tan x)  

1/ 4

+c

=∫

PLAN Integration by Substitution i.e. Put ∴

I = ∫ f { g ( x )} ⋅ g ′ ( x )dx

=

u2 − u + 1 1 log 2 +c 2 u +u+1

I = ∫ f(t )dt

=

e2x − ex + 1 1 log 2x +c 2 e + ex + 1

g ( x ) = t ⇒ g ′ ( x )dx = dt

Description of Situation Generally, students gets confused after substitution, i.e. sec x + tan x = t. Now, for sec x, we should use sec2 x − tan 2 x = 1 ⇒ ⇒ Here, Put

(sec x − tan x) (sec x + tan x) = 1 1 sec x − tan x = t I=∫

sec2 dx (sec x + tan x)9/ 2

∴ ∴

x for n ≥ 2 (1 + xn )1/ n f (x) x ∴ = ff (x) = n 1/ n [1 + f (x) ] (1 + 2 xn )1/ n x and fff (x) = (1 + 3xn )1/ n x g (x) = ( fofo ... of ) (x) = ∴ 14243 (1 + n xn )1/ n

19. Given, f (x) =

n times

sec x + tan x = t

⇒ (sec x tan x + sec2 x) dx = dt ⇒

dt 1 t −1 = log +c t+1 t −1 2 2

dt t 1 1 1 sec x − tan x = ⇒ sec x =  t +  2 t t sec x ⋅ sec x dx I =∫ (sec x + tan x)9/ 2

Let

xn − 1 dx (1 + nxn )1/ n d (1 + nxn ) n 2 xn − 1 dx 1 dx = dx (1 + nxn )1/ n n 2 ∫ (1 + nxn )1/ n

I = ∫ xn − 2g (x) dx = ∫

sec x ⋅ t dx = dt ⇒ sec x dx =

=

1 n2 ∫

1

I=

1− 1 (1 + nxn ) n + c n (n − 1)

Indefinite Integration 287 20. Let I = ∫

Put ⇒ ∴ I=

(x2 − 1) dx

2x − 2x + 1 [dividing numerator and enominator by x5 ] 1 1  3 − 5  dx x x  =∫ 2 1 2− 2+ 4 x x 2 1 2− 2+ 4 =t x x 4 4  3 − 5  dx = dt x x  1 4

∫

3

4

x

ef ( x ) g′ (x) eg( x ) f ′ (x) g′ (x) = g( x ) ⇒ ef ( x ) e ⇒ e− f ( x ) f ′ (x) = e− g( x ) g′ (x) On integrating both side, we get

f (0) = 1, then A = 1 1

x

f (x) = e2 1 1 log e f (x) = x ⇒ log e f (4) = × 4 = 2 2 2

Hence,

∴

4ex + 6e− x dx = Ax + B log (9e2x − 4) + c 9ex − 4e− x 4e2x + 6 dx 9e2x − 4 4e2x + 6 = A (9e2x − 4) + B (18 e2x ) 9 A + 18B = 4 and − 4 A = 6 3 35 and A=− B= 2 36

∫

1 A (9e2x − 4) + B (18e2x ) dx = A ∫ 1 dx + B ∫ dt t 9e2x − 4

where t = 9e2x − 4

e− f (1) = e− g(1) + C [Q f (1) = 1] =e +C

= A x + B log (9e2x − 4) + c 3 35 =− x+ log (9e2x − 4) + c 2 36 3 35 A =− ,B= 2 36

…(i)

[Q g(2) = 1]

…(ii)

∴

c = any real number

and …(iii)

I = ∫ (x3 m + x2m + xm )

− f (2) < log e 2 − 1

⇒

e− g(1) + e− f (2) = 2e−1

⇒

− g(1 )

24. For any natural number m , the given integral can be written as (2x3 m + 3x2m + 6xm )1/ m dx x

⇒ I = ∫ (2 x3 m + 3x2m + 6 xm )

f (2) > 1 − log e 2 [from Eq. (iii)]

−1

2e−1 We know that, e− x is decreasing ∴

1

⇒

Let

e− f ( x ) = e− g( x ) + C

e At x = 2 e− f ( 2) = e− g( 2) + C ⇒ e− f ( 2) = e−1 + C From Eqs. (i) and (ii)

On integrating, we get 1 log f (x) = x + C 2

LHS = ∫

f ′ (x) =

−1

⇒

23. Given, ∫

21. We have, f ′ (x) = e( f ( x ) − g( x )) g′ (x) ∀ x ∈ R

At x = 1

f (x) =

⇒

dt 1 t1/ 2 = ⋅ +c t 4 1 /2 1 2 1 = 2− 2+ 4 + c 2 x x

⇒

1 f (x) + f ′ (x) 2 1 f ′ (x) 1 f ′ (x) = f (x) ⇒ = 2 f (x) 2

⇒

2

g (1) > 1 − log e 2

22. Given, f (x + y) = f (x) f ′ ( y) + f ′ (x) f ( y), ∀x, y ∈ R and f (0) = 1 Put x = y = 0, we get f (0) = f (0) f ′ (0) + f ′ (0) f (0) 1 ⇒ 1 = 2 f ′ (0) ⇒ f′ (0) = 2 Put x = x and y = 0, we get f (x) = f (x) f ′ (0) + f ′ (x) f (0)

1/ m

(x3 m − 1 + x2m − 1 + xm − 1 ) dx Put 2 x3 m + 3x2m + 6xm = t (6 mx3 m−1 + 6 mx2m−1 + 6 mxm −1 ) dx = dt

⇒

1

∴ I=∫ =

25. Let Put

+1

1 tm dt t1/m = ⋅ 1 6m 6m    + 1  m

1 ⋅ (2 x3m + 3x2m + 6 xm )(m + 1)/m + c 6 (m + 1) I=∫

 1 − x    1 + x

1/ 2

⋅

dx x

x = cos 2 θ ⇒ dx = − 2 cos θ sin θ dθ

288 Indefinite Integration ∴

 1 − cos θ  I=∫    1 + cos θ 

1/ 2

⋅

− 2 cos θ ⋅ sin θ dθ cos 2 θ

θ − 2 sin θ 2 =∫ ⋅ dθ θ cos θ cos 2 θ θ θ θ 2 sin ⋅ 2 sin ⋅ cos 2 sin 2 2 2 2 d θ −2 =−∫ ∫ cos θ 2 d θ θ cos ⋅ cos θ 2 1 − cos θ = −2∫ dθ cos θ = 2∫ (1 − sec θ ) dθ = 2 [θ − log|sec θ + tan θ|] + c  1 −1 + c x  1  x − log|1 + 1 − x|− log|x| + c  2

 1 + I = 2 cos −1 x − log x 

⇒

 I = 2 cos −1 

26. Let

I=∫

x2 dx 1−x

Put 1 − x = t 2 ⇒ − dx = 2 t dt

sin

⇒

I=∫

(ii) Let

(1 − t 2)2 ⋅ (−2t ) dt t 2 4 = − 2 ∫ (1 − 2t + t ) dt

I=∫

∴

 2t3 t5  = − 2 t − +  +c 3 5  2 1   = − 2  1 − x − (1 − x)3/ 2 + (1 − x)5/ 2 + c 3 5   I=

29. Let

x2 (a + bx)2

Put a + bx = t ⇒ b dx = dt 2  t − a    b  dt ∴ ⋅ I=∫ b t2  t 2 − 2 at + a 2 1 = 3 ∫   dt b t2  

1− x dx 1+ x

 2 a a 2 + 2  dt 1 − t t  

Put x = cos 2 θ ⇒ dx = − 2sin θ cos θ dθ

=

1 − cos θ ⋅ (− 2 sin θ cos θ ) dθ 1 + cos θ

1 b3

=

1  a 2  t − 2 a log t −  + c 3 t b 

=

1 b3

∴

I=∫

θ θ = − ∫ 2 tan ⋅ sin θ cos θ dθ = − 2 ∫ 2 sin 2 ⋅ cos θ dθ 2 2 = − 2∫ (1 − cos θ ) cos θ dθ = − 2 ∫ ( cos θ − cos θ ) dθ 2

= − 2∫ cos θ dθ + = − 2 sin θ + θ +

sin 2θ +c 2

dx = x2 (x4 + 1)3/ 4 ∫

Put 1 + x− 4 = t ⇒ − ∴

I=−

1 4

1 (sin 4x + sin 2 x − sin 6x) dx 4∫ cos 4x cos 2 x cos 6 x =− − + 16 8 24 =

x (1 − x) + c dx

1  x ⋅ x 1 + 4   x  2

I 2 = ∫ sec2 x ⋅ cos 2 2 x dx

3/ 4

3

= ∫ sec2 x (2 cos 2 x − 1)2dx

4 dx = dt x5

dt

1 t1/ 4

= ∫ (4 cos 2 x + sec2 x − 4) dx 

1

∫ t3/ 4 = − 4 ⋅ 1 / 4 + c = − 1 + x4 

(x4 + 1)1/ 4 =− +c x

28. (i) Let

I=∫

= ∫ (2 cos 2 x + sec2 x − 2) dx

1/ 4

+c

= sin 2 x + tan x − 2 x and I3 = ∫ sin 4 x cos 4 x dx 1 (3 − 4 cos 4x + cos 8x) dx 128 ∫ 3x sin 4x sin 8x = − + 128 128 1024 =

x 1 + sin dx 2

x x x x + sin 2 + 2 sin cos dx 4 4 4 4 x x  = ∫  cos + sin  dx  4 4 x x = 4 sin − 4 cos + c 4 4 =∫

cos 2

  a2 + c  a + bx − 2 a log (a + bx) − a + bx  

30. Let I1 = ∫ sin x sin 2 x sin 3x dx

∫ (1 + cos 2 θ ) dθ

= − 2 1 − x + cos −1 x +

27. Let I = ∫

∫

∴

I = I1 + I 2 + I3 cos 4x cos 2 x cos 6 x =− − + + sin 2 x + tan x − 2 x 16 8 24 3x sin 4x sin 8 x + − + 128 128 1024

Indefinite Integration 289 31. Let

I=∫

x dx 1 2x = dx 1 + x4 2 ∫ 1 + (x2)2

3. Let

Put x = u ⇒ 2 x dx = du 2

∴

du 1 1 1 I= ∫ = tan −1 (u ) + c = tan −1 (x2) + c 2 2 2 2 1+ u

∴

 1 where, I1 = ∫  3  x+

sin x dx sin x − cos x Again, let sin x = A (cos x + sin x) + B(sin x − cos x),

A=

∴

I2 = ∫ Now,

1 1 , B= 2 2 1 1 (cos x + sin x) + (sin x − cos x) 2 2 I=∫ dx (sin x − cos x) 1 cos x + sin x 1 = ∫ dx + ∫ 1 dx + c 2 sin x − cos x 2 1 1 = log (sin x − cos x) + x + c 2 2

⇒

Put ∴

2

4

∫

dx 4 x 3

 sin    cos x

4 cos3

2 x cos3

dx 4 tan3

x cos 2 x

sec2 x dx

1

t3

−3 1

dt t+1

+ 12 ln (t + 1)

∴

x = u6 ⇒ I2 = ∫

dx = 6 u5 du

ln (1 + u ) 5 ln (1 + u ) 6u du = ∫ 2 . 6 u5 du u (1 + u ) u 2 + u3 u3 ln (1 + u ) du (u + 1)

 1  = 6 ∫  u2 − u + 1 −  ln (1 + u ) du  u + 1 + C = −3 tan

−

1 3

x+C

(tan x)3

∴ I n + I n + 2 = ∫ tan n x dx +

∫ tan

n+ 2

x dx

= ∫ tan n x(1 + tan 2 x) dx = ∫ tan n x sec2 x dx =

n+1

tan x +C n+1

tan5 x +C 5 1 a = and b = 0 5

Put n = 4, we get I 4 + I 6 =

= 6 ∫ (u 2 − u + 1) ln (1 + u ) du − 6 ∫ II

I

ln (1 + u ) du (u + 1)

 u3 u 2  =6  − + u ln (1 + u ) 2  3 

2. We have, I n = ∫ tan n x dx

∴

t 8 dt t+1

 u3 − 1 + 1 =6 ∫   ln(1 + u ) du  u+1 

dt +C = −4 t 4/3 +1 3 +C =

t11 dt t 4 + t3

 ln (1 + 6 x )   dx  3  x+ x 

=6 ∫

4 x)3

−4 +1 t3

1

I1 = 12 ∫

+ 12∫

Put

Now, put tan x = t ⇒ sec x dx = dt

= −3

  dx x

4

x

(tan

4

x = t12 ⇒ dx = 12 t11dt

x sin3 x

2

∴I=∫

 1 I1 = ∫  3  x+

and I 2 = ∫

=∫

  dx, x

 t8 t7 t 6 t5 t 4 t3 t 2  = 12  − + − + − + − t 7 6 5 4 3 2 8 

[dividing and multiplying by cos 4/3 x in denominator] =∫

4

= 12∫ (t7 − t 6 + t5 − t 4 + t3 − t 2 + t − 1) dt

dx 2 cos3

x

ln (1 + 6 x )  dx 3 x+ x 

ln (1 + 6 x ) dx 3 x+ x

= 12 ∫

Topic 2 Some Special Integrals 1. Let I = ∫ sec3 x cos ec3 x dx = ∫

4

+

I = I1 + I 2

32. Let I = ∫

then A + B = 1 and A − B = 0

 1 I = ∫ 3  x+

−∫

2 u3 − 3u 2 + 6 u 1 du − 6 [ln (1 + u )]2 2 u+1

= (2 u3 − 3 u 2 + 6 u ) ln (1 + u )  11 u  2 −∫ 2 u 2 − 5 u +  du − 3 [ln (1 + u )]  u + 1 = (2 u3 − 3 u 2 + 6 u ) ln (1 + u )  2 u3 5 2  −  − u + 11u − 11 ln (u + 1) − 3 [ln (1 + u )]2 2  3 

290 Indefinite Integration ∴

I=

3 23 12 7/12 12 5/12 x/ − x + 2x1/ 2 − x + 3x1/3 − 4x1/ 4 2 7 5 − 6 x1/ 6 − 12 x1/12 + 12 ln (x1/12 + 1) + (2 x1/ 2 − 3x1/3 + 6 x1/ 6 ) ln (1 + x1/ 6 ) 2 1/ 2 5 1/3  − x − x 11 x1/ 6 − 11 ln (1 + x1/ 6 ) 2 3  −3 [ln (1 + x1/ 6 )]2 + c 12 7/12 4 1/ 2 12 5/12 − x + x − x 7 3 5 1 + x1/3 − 4x1/ 4 − 7x1/ 6 − 12 x1/12 2

3 = x2 / 3 2

= − ∫ secθ dθ + 2 ∫

cos θ dθ 1 + cos 2 θ

= − log|sec θ + tan θ | + 2 ∫ = − log|sec θ + tan θ | +

I = ∫ ( tan x +

4. Let

cot x ) dx = ∫

+

tan x + 1 dx tan x

I=∫

t2 + 1 t2

=2 ∫

t−

Put

⋅

t2 +

1 −2 + 2 t2

x  2 sin 2   sin x  2 lim f ′ (x) = lim 2     x   x2  x→ 0 x→ 0  

dt = 2 ∫

1+

1 t2

2

1  2 t −  + ( 2)  t

du u 2 + ( 2 )2

2  u tan −1   + c I=  2 2

⇒

 tan x − cot x  = 2 tan −1   +c 2  

5. Let I = ∫

cos 2 x dx = ∫ sin x =∫

cos 2 x − sin 2 x dx sin 2 x cot2 x − 1 dx

Put cot x = sec θ ∴

I=∫

 x   sin 2  2 =1 = 4 ⋅ 1 ⋅ lim  2 x→ 0   x  4 ×    2  

dt

1 1 = u ⇒ 1 + dt = du t t2 I =2 ∫

∴

1 t2

⇒ − cosec2x dx = sec θ tan θ dθ sec θ ⋅ tan θ sec2 θ − 1 ⋅ dθ − (1 + sec2 θ )

=−∫

sec θ ⋅ tan 2 θ dθ 1 + sec2 θ

=−∫

sin 2 θ dθ cos θ + cos3 θ

=−∫

1 − cos 2 θ dθ cos θ (1 + cos 2 θ )

=−∫

(1 + cos 2 θ ) − 2 cos 2 θ dθ cos θ (1 + cos 2 θ )



x

On differentiating w.r.t. x, we get 2 sin x − sin 2 x 2 sin x  1 − cos x f ′ (x) = =   x  x2  x3

t2 + 1 2t dt = 2 ∫ t 4 + 1 dt t4 + 1

1+

 2 + 1 − tan 2 x  1 log  + c 2 2  2 − 1 − tan x 

2 sin x − sin 2 x 6. Given, f (x) = ∫   dx 3 

Put tan x = t 2 ⇒ sec2x dx = 2t dt 2t ⇒ dx = dt 1 + t4 ∴

 2 + sin θ  1 + c log  2 2  2 − sin θ 

cot2 x − 1 |

+ (2 x1/ 2 − 3x1/3 + 6 x1/ 6 + 11) ln (1 + x1/ 6 ) + 12 ln (1 + x1/12) − 3 [ln (1 + x1/ 6 )]2 + c

dt

∫ 2 − t 2 , where sin θ = t

= − log|sec θ + tan θ |+ 2 ⋅ = − log|cot x +

cos θ dθ 2 − sin 2 θ

Topic 3 Integration by Parts 2

1. Let given integral, I = ∫ x5 e− x dx Put x2 = t ⇒ 2xdx = dt 1 So, I = ∫ t 2e− t dt 2 1 = [(− t 2e− t ) + ∫ e− t (2t ) dt ] [Integration by parts] 2 1 = [− t 2e− t + 2t (− e− t ) + ∫ 2e− t dt ] 2 1 = [− t 2e− t − 2te− t − 2e− t ] + C 2 e− t 2 (t + 2t + 2) + C =− 2 2

e− x (x4 + 2x2 + 2) + C [Q t = x2] 2 Q It is given that, =−

2

2

I = ∫ x5 e− x dx = g (x) ⋅ e− x + C By Eq. (i), comparing both sides, we get 1 g (x) = − (x4 + 2x2 + 2) 2 1 5 So, g(− 1) = − (1 + 2 + 2) = − 2 2

…(i)

Indefinite Integration 291 2. Given, ∫ esec x [(sec x tan x) f (x) + (sec x tan x + sec2 x)]dx = e sec x ⋅ f (x) + C On differentiating both sides w.r.t. x, we get esec x [(sec x tan x) f (x) + (sec x tan x + sec2 x)] = esec x f ′ (x) + esec x (sec x tan x) f (x) ⇒ esec x (sec x tan x + sec2 x) = esec x f ′ (x) ⇒ f ′ (x) = sec x tan x + sec2 x So, f (x) = ∫ f ′ (x)dx = ∫ (sec x tan x + sec2 x)dx

x+

=∫e

x+

=∫e

So, possible value of f (x) from options, is 1 f (x) = sec x + tan x + . 2

Put

∴

1 = x cos(log e x) − ∫ x(− sin(log e x)) ⋅ dx x [using integration by parts] = x cos(log e x) + ∫ sin(log e x) dx 1

[using integration by parts] 1  te−4t e−4t  +C =  + 3  −4 −16 

e−4x [4x3 + 1] + C 48

5.

[Q t = x3 ]

f (x) = −1 − 4x3 (comparing with given equation) 

1

x+

∫ 1 + x − x e

1 x dx

x+

1 x dx

+

x+

1 x dx

+ xe

=∫e =∫e

1

∫

1 x+  x 1 − 2 e x dx  x  x+

1 x

1

−∫

1 x

dx

x+

dx + xe

1 x

− ∫ ex

x+

1 x

x+

dx = xe

1 x

+c

I = ∫ x5 f (x3 ) dx x3 = t dt x2dx = 3

…(i)

1 t f (t ) dt 3∫ 1 d   = t ⋅ ∫ f (t ) dt − ∫  (t ) ∫ f (t ) dt  dt  3  dt  

I=

  2x + 2  dx I = ∫ sin −1   4x2 + 8 x + 13   

7. Let

=

∫ sin



  dx  (2x + 2)2 + 9   

−1 

2x + 2

Put 2x + 2 = 3 tan θ ⇒ 2 dx = 3 sec2θ dθ  3 tan θ  3 ∴ I = ∫ sin −1  ⋅ sec2 θ dθ  9 tan 2 θ + 9  2    3 tan θ  3 2 = ∫ sin −1   ⋅ sec θ dθ  3 secθ  2  sin θ  3 2 = ∫ sin −1   ⋅ sec θ dθ  cos θ ⋅ sec θ  2

3

∴

x+

−∫e

[integration by parts] 1 = [t ψ (t ) − ∫ ψ (t ) dt ] 3 1 3 [from Eq. (i)] = [x ψ (x3 ) – 3 ∫ x2ψ (x3 ) dx] + c 3 1 = x3 ψ (x3 ) − ∫ x2ψ (x3 ) dx + c 3

∫ x(cos(log e x)) x dx

[again, using integration by parts] ⇒ I = x cos(log e x) + x sin((log e x) − I x I = [cos(log e x) + sin(log e x)] + C. ⇒ 2 1 −4 x 3 5 −4 x 3 4. Given, ∫ x e dx = e f (x) + C 48 In LHS, put x3 = t ⇒ 3x2dx = dt 3 1 So, ∫ x5 e−4x dx = ∫ t e−4t dt 3 1  e−4t e−4t  dt  = t −∫ 3  −4 −4 

=−

1 x

⇒

3. Let I = ∫ cos(log e x)dx

1 −4 t e [4t + 1] + C 48

1 x

6. Given, ∫ f ( x ) dx = ψ( x ) Let

=−

x+

+ xe

1 1  x+  1 x+  Q ∫ 1 − 2 e x dx = e x   x   

= sec x + tan x + C

= x cos(log e x ) + x sin(log e x ) −

1 x dx

x+ d (x)e x dx dx

3 sin −1 (sin θ ) ⋅ sec2θ dθ 2∫ 3 3 = ∫ θ ⋅ sec2θ dθ = [θ ⋅ tan θ − ∫ 1 ⋅ tan θdθ] 2 2 3 = [θ tan θ − log sec θ ] + c 2 3  2 x + 2  2 x + 2 = tan −1    ⋅  3   3  2 =

2   2 x + 2 − log 1 +    + c1  3   

292 Indefinite Integration θ 1 cos 2θ + sin 2θ 2 4 1 1 = − θ (1 − 2 sin 2 θ ) + sin θ 1 − sin 2 θ 2 2 1 1 −1 = − sin x (1 − 2 x) + x 1−x 2 2

2   2x + 2 3  2 x + 2  = (x + 1) tan −1   − log 1 +    + c1  3  4  3   

=−

 2 x + 2 3 2 = (x + 1) tan −1   − log (4x + 8 x + 13) + c  3  4

3   let log 3 + c1 = c   2  cos θ + sin θ  cos 2 θ ln   dθ  cos θ − sin θ 

8. I = ∫

[given]

We integrate it by taking parts  cos θ + sin θ  ln   as first function  cos θ − sin θ  = − But

From Eqs. (i) and (ii), 4  1 1  I= − (1 − 2 x) sin −1 x + x − x2 − x + c π  2 2  2 = [ x − x2 − (1 − 2 x) sin −1 x ] − x + c π

10. Let I = ∫

 cos θ + sin θ  sin 2 θ ln    cos θ − sin θ  2

1 d   cos θ + sin θ   ln   sin 2 θdθ 2 ∫ dθ   cos θ − sin θ  

d [ln (cos θ + sin θ ) − ln (cos θ − sin θ )] dθ 1 (− sin θ − cos θ ) = . (− sin θ + cos θ ) − (cos θ + sin θ ) cos θ − sin θ (cos θ − sin θ ) (cos θ − sin θ ) − (cos θ + sin θ ) (− sin θ − cos θ ) = (cos θ + sin θ ) (cos θ − sin θ ) =

Applying integration by parts,   1 −2 = ⋅ ex − ∫ ex ⋅ dx 2 3 + + ( x 1 ) ( x 1 )   − 2 ∫ ex ⋅

= ∫ (x + sin x) cos x dx

=

2

=

= x sin x + cos x −

 cos θ + sin θ  1 1 sin 2 θ ln   + ln (cos 2 θ ) + c  cos θ − sin θ  2 2

9. Let I = ∫ =∫ 2 π

∫

sin −1 x − cos −1 x dx sin −1 x + cos −1 x

 π sin −1 x −  − sin −1 x  2 dx π 2 π 4  −1 x −  dx = ∫ sin −1 x dx − x + c …(i) 2 sin  2 π

Now, ∫ sin −1 x dx Put x = sin 2 θ ⇒ dx = sin 2θ θ cos 2 θ = ∫ θ ⋅ sin 2 θ dθ = − + 2

1 2

∫

cos 2 x +c 4

cos 2 x +c 4

Topic 4 Integration, Irrational Function and Partial Fraction 1. Let I = ∫ =∫

2x12 + 5x9 2x12 + 5x9 dx = ∫ x15 (1 + x− 2 + x− 5 ) 3 dx (x5 + x3 + 1) 3 2x− 3 + 5x− 6 dx (1 + x− 2 + x− 5 ) 3

Now, put 1 + x− 2 + x− 5 = t ⇒

(− 2x− 3 − 5x− 6 ) dx = dt

(2x− 3 + 5x− 6 ) dx = − dt dt ∴ I = − ∫ 3 = − ∫ t − 3 dt t ⇒

=− 1 cos 2θ dθ 2

∫ (sin 2x) dx

= (x ⋅ sin x − ∫ 1 ⋅ sin x dx) −

2

Therefore, on putting this value in Eq.(i), we get  cos θ + sin θ  1 1 2 I = sin 2θ ln  dθ sin 2 θ  −  cos θ − sin θ  2 ∫ 2 cos 2 θ

=

= ∫ x cos x dx +

2 2 (cos θ + sin θ ) = cos 2 θ cos 2θ 2

1 ex dx = +c (x + 1)3 (x + 1)2

11. Let I = ∫ (elog x + sin x) cos x dx

cos θ − cos θ sin θ − sin θ cos θ + sin θ + cos θ sin θ + cos 2 θ + sin 2 θ + cos θ ⋅ sin θ = cos 2 θ − sin 2 θ 2

(x − 1) ex dx (x + 1)3

x + 1 − 2 x  x  1 2 e dx = ∫  I=∫ − e dx 3  2 3 + x + + ( 1 ) ( x 1 ) ( x 1 )     1 1 = ∫ ex ⋅ dx − 2 ∫ ex ⋅ dx (x + 1)2 (x + 1)3

…(i)

d   cos θ + sin θ   ln   dθ   cos θ − sin θ  

…(ii)

=

t− 3 + 1 1 +C= 2+C −3 + 1 2t

x10 +C 2 (x + x3 + 1) 2 5

Indefinite Integration 293 I=∫

2. Let

=∫

cos3 x + cos5 x dx sin 2 x + sin 4 x

On putting x = 0, we get

(cos 2 x + cos 4 x) ⋅ cos x dx (sin 2 x + sin 4 x)

⇒

0=B+C x3 + 3x + 2 x+1 1 2 − + = (x + 1)2 (x + 1) 2 (x2 + 1) 2 (x + 1) (x2 + 1)2 2

Put sin x = t ⇒ cos x dx = dt ∴

I=∫

[(1 − t 2) + (1 − t 2)2] dt t2 + t4

⇒

I=∫

1 − t 2 + 1 − 2t 2 + t 4 dt t2 + t4

⇒

I=∫

2 − 3t 2 + t 4 dt t 2 (t 2 + 1)

⇒ ∴

x3 + 3x + 2 dx (x + 1)2 (x + 1)

∴ I=∫

2

=−

…(i)

Using partial fraction for y2 − 3 y + 2 A B =1 + + y ( y + 1) y y+1

B = − C = 1 /2

[where, y = t 2]

A = 2, B = − 6 y − 3y + 2 2 6 =1 + − y ( y + 1) y y+1 2

 2 6  Now, Eq. (i) reduces to, I = ∫ 1 + 2 −  dt  1 + t 2 t

1 2

3.

Again,

=

x(x + 1) + 2(x + 1) (x2 + 1)2(x + 1)

=

x 2 + 2 2 (x + 1)(x + 1) (x + 1)2

2

x Ax + B C + = (x + 1) (x + 1) (x2 + 1) (x + 1) 2

⇒ On putting

x = ( Ax + B) (x + 1) + C (x2 + 1) x = − 1, we get −1 = 2 C ⇒ C = − 1 / 2

On equating coefficients of x2, we get 0= A+C ⇒

A = − C = 1 /2

x+1

1

dx (x2 + 1)2

1 1 log|x + 1| + log|x2 + 1| 2 4 1 + tan −1 x + 2 I1 2 dx where, I1 = ∫ (x2 + 1)2 ⇒I = −

…(i)

x = tan θ

Put ⇒

dx = sec2θ dθ

∴ I1 = ∫

1 sec2θdθ = ∫ cos 2 θ dθ = ∫ (1 + cos 2 θ )dθ 2 2 2 (tan θ + 1) 1 1  θ + sin 2θ  2  2 1 1 tan θ = θ+ ⋅ 2 2 (1 + tan 2 θ) =

2 = t − − 6 tan −1 (t ) + c t 2 = sin x − − 6 tan −1 (sin x) + c sin x x3 + 3x + 2 x3 + 2x + x + 2 = 2 2 2 (x + 1) (x + 1) (x + 1)2(x + 1)

dx

∫ x + 1 + 2 ∫ x2 + 1 dx + 2 ∫

=

1 1 x tan −1 x + ⋅ 2 2 (1 + x2)

From Eq. (i), 1 3 1 x I = − log|x + 1| + log|x2 + 1|+ tan −1 x + 2 +c 2 2 4 x +1

4. Let

I=∫

(x + 1) ex (x + 1) dx dx = ∫ x 2 x (1 + xe ) x ex (1 + xex )2

Put 1 + xex = t ⇒ (ex + x ex ) dx = dt  1 dt 1 1 ∴ I=∫ = − − dt (t − 1)t 2 ∫  t − 1 t t 2  1 = log|t − 1| − log|t| + + c t t − 1 1  = log + +c  t  t  xex  1 + +c = log  x x 1 + xe  1 + xe

12 Definite Integration Topic 1 Properties of Definite Integral Objective Questions I (Only one correct option) 1.

π

∫−π|π −|x||dx is equal to

(2020 Main, 3 Sep I)

2π 2

(a)

(b) 2 π 2 π2 (d) 2

(c) π 2

8. The value of ∫ π −1 (a) 2

∫π / 6

2

2

If f (x + 5) = g (x), then ∫ f (t )dt equals

2

7 (b) 18

9 (d) 2

1 (c) − 18

3. A value of α such that

α

dx  9 = log e   is  8 (x + α ) (x + α + 1)

(a) − 2

4. If ∫

π /2 0

(b)

1 2

(c) −

1 (a) − 2 1 (c) 2

(2019 Main, 12 April II)

1 2

(d) 2

(b) 1

(d)

π−2 4

∫ g (t )dt

(b)

x+5

10. If f (x) =

(d) −1 π /3 π /6

(b) 37/ 6 − 35 / 6 (d) 34/3 − 31/3 2π

∫ [sin 2x (1 + cos 3x)] dx, where [t ] denotes 0

the greatest integer function, is

(2019 Main, 10 April I)

(c) − 2π

(b) 2π 1

−1

(d) π

7. The value of the integral ∫ x cot (1 − x + x )dx is 2

4

0

π 1 − log e 2 4 2 π (c) − log e 2 4

(2019 Main, 9 April II)

π 1 − log e 2 2 2 π (d) − log e 2 2

(b)

(c) 2

5

∫ g (t )dt

5

(d)

5

∫ g (t )dt

x+5

2 − x cos x and g (x) = log e x, (x > 0) then the 2 + x cos x π/ 4

−π/ 4

g ( f (x))dx is (2019 Main, 8 April I)

(a) log e 3 (c) log e 2

(b) log e e (d) log e 1

11. Let f and g be continuous functions on [0, a] such that f (x) = f (a − x) and g (x) + g (a − x) = 4, a then ∫ f (x) g (x) dx is equal to (2019 Main, 12 Jan I) a

(a) 4∫ f (x) dx

(b) ∫ f (x) dx

(c) 2∫ f (x) dx

(d) − 3∫ f (x) dx

0 a

sec 2/ 3 x cosec4/3 x dx is equal to

(a) 35 / 6 − 32/ 3 (c) 35 /3 − 31/3

(2019 Main, 8 April II)

x+5

0

(2019 Main, 10 April II)

(a)

∫ g (t )dt

(a) 5

a

5. The integral ∫

(a) − π

0

x+5

5

value of the integral ∫

cot x dx = m(π + n ), then m ⋅ n is equal to cot x + cosec x (2019 Main, 12 April I)

6. The value of

(2019 Main, 9 April I)

x

3

(2020 Main, 4 Sep II)

∫

π −1 (c) 4

0

+ 3 tan x ⋅ sin 6x) dx is equal to

α+1

π−2 (b) 8 x

tan x ⋅ sin 3x(2 sec x ⋅ sin 3x

1 (a) − 9

0

sin3 x dx is sin x + cos x

9. Let f (x) = ∫ g (t )dt, where g is a non-zero even function.

2. The integral π /3

π/ 2

0

a

0

0

e

12. The integral ∫   1  e  x

2x

x  e  −    log e x dx is  x 

equal to 3 1 (a) −e− 2 2 2e 1 1 (c) −e− 2 2 e

13. The integral ∫

(2019 Main, 12 Jan II)

1 1 1 + − 2 e 2e2 3 1 1 (d) − − 2 e 2e2 (b) −

π /4 π /6

dx equals sin 2x(tan5 x + cot5 x)

1 π 1  (a)  − tan − 1     5 4 3 3 (c)

1π −1 1   − tan    9 3 10  4

(2019 Main, 11 Jan II)

1 1  (b) tan − 1    20 9 3 (d)

π 40

Definite Integration 295 14. The value of the integral ∫

sin 2 x dx −2  x  1 +  π  2 2

24. The value of the integral ∫ is

(where, [x] denotes the greatest integer less than or equal to x) is (2019 Main, 11 Jan I) (a) 4 − sin 4 (b) 4 (c) sin 4 (d) 0

(2012)

π2 (b) −4 2

(a) 0

25. The value of

b

15. Let I = ∫ (x4 − 2x2) dx. If I is minimum, then the ordered

(a)

a

pair (a , b) is

(2019 Main, 10 Jan I)

(a) (− 2 , 0) (c) ( 2 , − 2 )

16. If ∫

π /3 0

(b) (0, 2 ) (d) (− 2 , 2 )

0

(2019 Main, 9 Jan II)

1 (b) 2 (d) 4

(a) 1 (c) 2 π

(2019 Main, 9 Jan I)

0

(a)

2 3

(b) −

18. The value of ∫ π (a) 8

19.

3 π/ 4

∫π / 4

4 3

(d)

sin 2 x dx is + 2x

π /2

π (d) 4

(c) 4π

(b) 2

x2 cos x dxis equal to −π / 2 1 + ex

π −2 4 (c) π 2 − e− π / 2 2

22. The integral ∫

∫0

π /2 π /4

(c) 1

log( 1 + 2)

(eu − e−u )17 du

∫0

(c)

∫0

log( 1 + 2)

∫0

π

0

(c) 4 3 − 4

1 0

(d) 1

1−x dx is 1+ x

(c) −1

(2004, 1M)

(d) 1

1 (d) log    2

(c) 1

cos 2 x dx, a > 0,is −π 1 + a x π

(2001, 1M)

(b) aπ (d) 2π

ecos x sin x , for | x|≤ 2 2 , otherwise 

then ∫

3 −2

(2000, 2M)

f (x) dx is equal to (b) 1

(c) 2

(d) 3

log x 31. The value of the integral ∫ −1  e  dx is e  x  (a) 3 / 2

(b) 5 / 2

2π − 4− 4 3 3 (d) 4 3 − 4 − π /3

(d) 5

than or equal to y, then the value of the integral 3π/2 (1999, 2M) ∫ π / 2 [2 sin x] dx is

3 π /4

∫ π /4

(a) 2

x dx is equal to 2 (2014 Main)

(c) 3

(2000, 2M)

32. If for a real number y, [ y] is the greatest integer less

(2014 Adv.)

−u 16

(b)

1 3 log 6 2

30. If f (x) = 

33.

x 2

(d)

  1 + x   [x] + log    dx equals (2002, 1M)  1 − x  

(a) − π π (c) − 2

23. The integral ∫ 1 + 4 sin 2 − 4 sin (a) π − 4

(d) 6

2(e − e ) du u

1/ 2 −1/ 2

(a) π π 2

−u 16

(eu + e−u )17 du

(b)

3 2

(2011)

e2

(2 cosec x)17dx is equal to

log( 1 + 2)

π −1 2

(b) 0

29. The value of ∫

2(e + e ) du u

(b)

1 2

(a) 0

log x2 dx is equal to log x2 + log(36 − 12x + x2) (2015, Main)

(b) 4

log( 1 + 2)

(a) −

(b)

21. The integral ∫

π +1 2

28. The integral ∫

(2016 Adv.)

π + 2 4 (d) π 2 + eπ / 2

(c) log

(c) 4

27. The value of the integral ∫

2

4

1 3 log 2 2

(b)

(b) 3

π /2

(a)

x sin x2 dx is log 2 sin x2 + sin (log 6 − x2 )

+ 3x2 + 3x + 3 + (x + 1) cos (x + 1)] dx is (2005, 1M)

3

(a) 0

(2017 Main)

(d) − 1

π2 (d) 2

log 3

(c) (c) 4

2

(a) 2

4 3

(2018 Main)

dx is equal to 1 + cos x

20. The value of ∫

(d)

(c) 0

−π / 2 1

π (b) 2

(a) − 2

(a)

∫ −2 [x

(a)

17. The value of ∫ |cos x|3 dx is

1 3 log 4 2

∫

π2 (c) + 4 2

26. The value of

tan θ 1 , (k > 0), then the value of k dθ = 1 − 2k sec θ 2

is

 2 π − x  x + log  cos x dx  π + x

π /2 − π /2

(b) 0 π (d) 2

dx is equal to 1 + cos x (b) −2

(1999, 2M)

1 (d) − 2

1 (c) 2

34. Let f (x) = x − [x] , for every real number x, where [x] is the integral part of x. Then,

1

∫ −1 f (x) dx is

(a) 1

(b) 2

(c) 0

(d) −

1 2

(1998, 2M)

296 Definite Integration x

35. If g (x) = ∫ cos 4 t dt , then g (x + π ) equals

(1997, 2M)

0

(a) g (x) + g ( π )

(b) g (x) − g ( π ) g (x ) g( π)

(c) g (x) g ( π )

(d)

36. Let f be a positive function. If I1 = ∫ I2 = ∫

k 1−k

(a) Statement I is correct; Statement II is correct; Statement II is a correct explanation for Statement I (b) Statement I is correct; Statement II is correct; Statement II is not a correct explanation for Statement I (c) Statement I is correct; Statement II is false (d) Statement I is incorrect; Statement II is correct

x f [x (1 − x)] dx and

k

Passage Based Questions

f [x (1 − x)] dx, where 2k − 1 > 0.

1−k

I Then, 1 is I2

Passage I

(1997C, 2M)

(a) 2 (c) 1/2

(b) k (d) 1

37. The value of

2π

∫0

44. The correct statement(s) is/are

[ 2 sin x ] dx, where [.] represents the

greatest integral functions, is

5π (a) − 3

(b) − π

Let F : R → R be a thrice differentiable function. Suppose that F (1) = 0, F (3) = − 4 and F ′ (x) < 0 for all x ∈ (1, 3). Let (2015 Adv.) f (x) = xF (x) for all x ∈ R.

(1995, 2M)

5π (c) 3

(d) −2π

πx 38. If f (x) = A sin   + B,  2  1 2A 1 f ′   = 2 and ∫ f (x) dx = , then constants 0  2 π

A and B are

(a) f ′ (1) < 0 (b) f (2) < 0 (c) f ′ (x) =/ 0 for any x ∈ (1, 3) (d) f ′ (x) = 0 for some x ∈ (1, 3) 3

45. If ∫ x2 F ′ (x) dx = − 12 and 1

39. The value of ∫

π /2 0

4 (c) 0 and − π

4 (d) and 0 π

dx is 1 + tan3 x

(a) 0 (c) π / 2

(b) ∫ f (x) dx = 12

(c) 9f ′ (3) – f ′ (1) + 32 = 0

(d) ∫ f (x) dx = − 12

Then, the value of the integral

(b) 1

F (c) =

(c) −1

cos 2 x

∫0 e

(1990, 2M)

(d) 0

a+b 2 b b−a ∫ a f (x) dx = 4 { f (a ) + f (b) + 2 f (c)} dx t

cos (2n + 1) x dx has the value 3

(a) π (c) 0

(1985, 2M)

(a) π / 4 (c) π

π /2 0

cot x dx is cot x + tan x

(b) π / 2 (d) None of these

46. If lim

∫ a f (x) dx −

t→ a

(t − a ) { f (t ) + f (a )} 2 = 0, (t − a )3

(1983, 1M)

(a) 0 (c) 3

(b) 1 (d) 2

47. If f ′ ′ (x) < 0, ∀x ∈ (a , b), and (c, f (c)) is point of maxima, where c ∈ (a , b), then f ′ (c) is f (b) − f (a ) b− a f (b) − f (a )   (c) 2  b − a 

(b) 3  

(a)

Assertion and Reason 43. Statement I The value of the integral π /3

∫π / 6

1+

Statement II

dx is equal to π /6 . tan x b

b

∫a f (x) dx = ∫a f (a + b − x) dx

(2006, 6M)

then degree of polynomial function f (x) atmost is

(b) 1 (d) None of these

42. The value of the integral ∫

c−a b−c [ f (a ) − f (c)] + [ f (b) − f (c)] 2 2

When c =

41. For any integer n, the integral π

 b − a  { f (a ) + f (b)}, 2 

∫ a f (x) dx = 

for more acurate results for c ∈ (a , b),

(b) 1 (d) π / 4

(a) π

3

1

Passage II

b

f (− x)] [ g (x) − g (− x)] dx is

1

For every function f (x) which is twice differentiable, these will be good approximation of

(1993, 1M)

∫ −π / 2 [ f (x) +

3

(a) 9f ′ (3) + f ′ (1) − 32 = 0

40. Let f : R → R and g : R → R be continuous functions. π/ 2

F ′ ′ (x) dx = 40, then the

correct expression(s) is/are

(1995, 2M)

π π 2 3 (b) and (a) and 2 2 π π

3 3

∫1 x

(2013 Main)

48. Good approximation of ∫ (a) π / 4 (c) π ( 2 + 1) / 8

f (b) − f (a )  b − a 

(d) 0 π /2 0

sin x dx, is

(b) π ( 2 + 1) / 4 π (d) 8

Definite Integration 297 Integer & Numerical Answer Type Questions

Objective Questions II (One or more than one correct option)

56. Let [t ] denote the greatest integer less than or equal to t.

49. Let f : R → (0, 1) be a continuous function. Then, which of the following function(s) has (have) the value zero at (2017 Adv.) some point in the interval (0, 1) ? x

(a) ex − ∫ f (t ) sin t dt (c) x −

0 π −x 2

∫ f (t ) cos t dt

(b) f (x) +

π 2

2

Then the value of ∫ |2x − [3x]|dx is ……… . 1

(2020 Main, 2 Sep II)

57. The value of the integral π /2

∫ f (t ) sin t dt

∫

0

0

(d) x9 − f (x)

3 cos θ d θ equals ………… ( cos θ + sin θ )5

58. If I =

0

k+1

50. If I = ∑ k = 1 98

∫k

k+1 dx , then x(x + 1)

(2017 Adv.)

(a) I > log e 99

(b) I < log e 99

49 (c) I < 50

(d) I >

 π π x ∈  – ,  . Then, the correct expression(s) is/are  2 2 π/4

1 12 π/4 1 (c) ∫ x f (x) dx = 0 6 0

x f (x) dx =

1

1/ 2

π/ 4

0

(d) ∫

f (x) dx = 0

π/ 4

0

(2015 Adv.)

f (x) dx = 1

3

192x  1 for all x ∈ R with f   = 0. If  2 2 + sin 4 πx

52. Let f ′ (x) = m≤∫

(b) ∫

f (x) dx ≤ M , then the possible values of m and M

are

(1 + e

sin x

− π/ 4

4π

equation

4

e4 π − 1 eπ − 1

(b) a = 2, L =

e4 π + 1 eπ + 1

(c) a = 4, L =

e4 π − 1 eπ − 1

(d) a = 4, L =

e4 π + 1 eπ + 1

0

22 −π 7

55. If I n = ∫

π −π

(b)

10

∑ I2 m = 0

m =1

x4 (1 − x)4 dx is (are) 1 + x2 2 105

(c) 0

(2010)

(d)

71 3 π − 15 2

sin nx dx , n = 0 , 1 , 2 ,... , then (1 + π x ) sin x

(a) I n = I n + 2

1+ 3 dx is ((x + 1)2 (1 − x)6 )1/ 4 (2018 Adv.)

where tan −1 x takes only principal values, then the 3π   value of  log e|1 + α | − (2015 Adv.)  is  4  d2  (1 − x2)5  dx is 2 dx  

(2014 Adv.)

Fill in the Blanks

(a) a = 2, L =

1

1/ 2 0

[x], x ≤ 2 60. Let f : R → R be a function defined by f (x) =  , 0 , x > 2 where [x] denotes the greatest integer less than or equal 2 xf (x2) to x. If I = ∫ dx, then the value of (4I − 1) −1 2 + f (x + 1 ) is (2015 Adv.) 2  12 + 9x  1 −1 61. If α = ∫ (e9x + 3 tan x )   dx, 0  1 + x2 

0

∫0 e (sin at + cos at )dt = L, is/are π t 6 4 ∫0 e (sin at + cos at )dt (2015 Adv.)

54. The value(s) of ∫

(2019 Adv.)

59. The value of the integral ∫

1

6

t

dx then 27I 2 equals ......... ) (2 − cos 2x)

62. The value of ∫ 4x3 

53. The option(s) with the values of a and L that satisfy the

(c)

∫

(2015 Adv.)

(a) m = 13, M = 24 1 1 (b) m = , M = 4 2 (c) m = − 11, M = 0 (d) m = 1, M = 12

(a)

π/ 4

.......

49 50

51. Let f (x) = 7 tan 8 x + 7 tan 6 x – 3 tan 4 x – 3 tan 2 x for all

(a) ∫

2 π

(2019 Adv.)

10

(b)

∑ I2 m + 1 = 10 π

m =1

(d) I n = I n + 1

(2009)

63. Let f : [1, ∞ ] → [2, ∞ ] be differentiable function such x

that f (1) = 2. If 6∫ f (t ) = dt = 3x f (x) − x3 , ∀ x ≥ 1 then 1

the value of f (2) is ....

(2011)

d e sin x 64. Let F (x) = , x > 0. dx x 2 4 2 e sin x If ∫ dx = F (k) − F (1), then one of the possible 1 x values of k is ..… . (1997, 2M) 37 π π sin (π log x) 65. The value of ∫ dx is …… . 1 x (1997, 2M)

66. For n > 0, ∫

x sin 2n x dx = …… . sin x + cos 2n x

2π

2n

0

(1996, 2M)

1 1 67. If for non-zero x, af (x) + bf   = − 5, where a ≠ b,  x x 2

then ∫ f (x) dx = …… . 1

68. The value of ∫

3 2

x 5−x+

(1996, 2M)

x

dx is …… . (1994, 2M)

298 Definite Integration 69. The value of ∫

3 π /4

x dx …… . 1 + sin x

π /4

76. Evaluate (1993, 2M)

π + 4x3 dx. π  2 − cos |x|+   3

π/ 3

∫ −π / 3

2

70. The value of ∫ |1 − x2| dx is … . −2

1.5

∫0

71. The integral

(1989, 2M)

2

[x ] dx, where [.] denotes the greatest

function, equals …… .

(1988, 2M)

77. If

is

f

π /2

∫0

an

Column I 1

dx

∫ −1 1 + x 2

A.

1

∫0

B.

3

∫2

C.

2

∫1

D.

P. Q.

2 2 log    3

dx 1− x 2 dx 1− x dx

R.

π 3

x x2 −1

S.

π 2

2

8

1

≤ 2, satisfying f( 0) = 0 and ∫ f( x ) dx = 1, is Q. The number of points in the interval (ii) [− 13, 13 ] at which f( x ) = sin( x 2 ) + cos( x 2 ) attains its maximum value, is

∫−2 1 + e x dx equals

(iii)

 1/ 2 1 + x    ∫ cos 2 x log   dx  − 1/ 2 1− x    equals  1/ 2 1 + x   dx   ∫ cos 2 x log   1− x    0

(iv)

2

4

0

(1998, 8M)

log (1 + tan x) dx. π

∫ −π

2x (1 + sin x) dx. (1995, 5M) 1 + cos 2 x

 x4   2x   cos −1    dx. 4  1 + x2  1 − x 

(1995, 5M)

2x5 + x4 − 2x3 + 2x2 + 1 dx. (x2 + 1) (x4 − 1)

(1993, 5M)

1/ 3

3

∫2

83. Evaluate

3

has relative minimum / maximum at x = − 1 and x = 1/3. 1 If ∫ f (x) dx = 14 / 3, find the cubic f (x). (1992, 4M)

85. Evaluate ∫

π

 π x sin (2x) sin  cos x  2

0

2x − π

86. Show that , π /2

∫0

f (sin 2x) sin x dx = 2

π /4

∫0

dx .

(1991, 4M)

f (cos 2x) cos x dx. (1990, 4M)

sin 2kx = 2 [cos x + cos 3x + K + cos(2k − 1) x] sin x Hence, prove that

π /2

∫0

sin 2kx ⋅ cot x dx = π /2.

(1990, 4M)

88. If f and g are continuous functions on [0, a ] satisfying f (x) = f (a − x) and g (x) + g (a − x) = 2, then show that a

π |cos x |

1  1   dx = 2 ∫ tan −1 x dx. 0  1 − x + x2 

(1997C, 2M)

∫ −1/

(5050) ∫ 1

∫0

1 0

(1 − x50 )100 dx

(1 − x50 )101 dx

a

∫ 0 f (x) g(x) dx = ∫ 0 f (x) dx.

Analytical & Descriptive Questions

∫ 0e

(1999, 3M)

87. Prove that for any positive integer k,

P Q R S (a) (iii) (ii) (iv) (i) (b) (ii) (iii) (iv) (i) (c) (iii) (ii) (i) (iv) (d) (ii) (iii) (i) (iv)

75. Evaluate

dx.

−1

Codes

74. The value of

(2003, 2M)

84. A cubic f (x) vanishes at x = − 2 and

0

S.

π /4 0

List II

3x 2

f (sin 2x) cos x dx.

82. Evaluate the definite integral

(2014)

(i)

that

−1 

81. Determine the value of

List I

2

prove

0

using codes given below the lists.

R.

1

80. Integrate ∫

The number of polynomials f( x ) with non-negative integer coefficients of degree

0

then

Hence or otherwise, evaluate the integral 1 −1 2 ∫ tan (1 − x + x ) dx.

73. Match List I with List II and select the correct answer

P.

+ e− cos x

∫ 0 tan

Column II 1 2 log    3 2

e

0 ecos x

79. Prove that

π /4

cos x

π

72. Match the conditions/expressions in Column I with statement in Column II.

function,

f (cos 2x) cos x dx = 2 ∫

78. Evaluate ∫

Match the Columns

even

(2004, 4M)

89. Prove is

(2006, 6M)

 1  1  2 sin  2 cos x + 3 cos  2 cos x  sin x dx.   (2005, 2M)

2a

∫0

that the value of the [ f (x) /{ f (x) + f (2a − x)}] dx is equal to a.

90. Evaluate

1

∫ 0 log[

91. Evaluate ∫

π 01

(1 − x) +

(1 + x)] dx.

x dx , 0 < α < π. + cos α sin x

(1989, 4M)

integral, (1988, 4M) (1988, 5M) 1 (1986, 2 M) 2

Definite Integration 299 92. Evaluate ∫ 93. Evaluate

π /2 0 1/ 2

∫0

94. Evaluate ∫

π /4 0

x sin x cos x dx. cos 4 x + sin 4 x x sin

−1

1−x

x 2

1 (1985, 2 M) 2

dx.

(1984, 2M)

sin x + cos x dx. 9 + 16 sin 2x π

95. (i) Show that ∫ x f (sin x) dx = 0

π 2

(1983, 3M) π

∫ 0 f (sin x) dx.(1982, 2M)

(ii) Find the value of

96. Evaluate

1

3/ 2

∫ −1 | x sin πx| dx.

(1982, 3M)

∫ 0 (tx + 1 − x) dx, n

where n is a positive integer and t is a parameter independent of x. Hence, show that 1 k 1 n−k . ∫ 0 x (1 − x) dx = nC k (n + 1), for k = 0, 1,... , n(1981, 4M)

Topic 2 Periodicity of Integral Functions Objective Questions I (Only one correct option)

(c)

π/ 2

dx 1. The value of ∫ , where [t ] denotes − π / 2 [x] + [sin x] + 4 the greatest integer less than or equal to t, is (2019 Main, 10 Jan II)

1 (a) (7 π − 5) 12 3 (c) (4 π − 3) 10

1 (b) (7 π + 5) 12 3 (d) (4 π − 3) 20

3 I 2 (c) 3I

f (2x) dx is

(2002,1M)

(d) 6I x

3. Let g (x) = ∫ f (t ) dt, where f is such that 0

t ∈ [0,1] and 0 ≤ f (t ) ≤

nπ + v 0

|sin x|dx = 2n + 1 − cos v, where n is a (1994, 4M)

every interval on the real line and f (t + x) = f (x), for every x and a real t, then show that the integral a+ t (1984, 4M) ∫ f (x) dx is independent of a.

Integer & Numerical Answer Type Question 6. For any real number x, let [x] denotes the largest

(b) I

(a)

4. Show that ∫

a

0

3

Analytical & Descriptive Questions

5. Given a function f (x) such that it is integrable over

continuous function such that for all T x ∈ R. f (x + T ) = f (x). If I = ∫ f (x) dx, 3 + 3T

(d) 2 < g(2) < 4

positive integer and 0 ≤ v < π .

2. Let T > 0 be a fixed real number. Suppose, f is a

then the value of ∫

3 < g(2) ≤ 5 / 2 2

1 ≤ f (t ) ≤ 1 for 2

1 for t ∈ [1, 2]. Then, g(2) satisfies 2

the inequality

integer less than or equal to x. Let f be a real valued function defined on the interval [− 10, 10] by if f (x) is odd  x − [x[, f(x) =  1 [ [ if + x − x, f (x) is even  π 2 10 (2010) Then, the value of f (x) cos πx dx is…… 10 ∫− 10

(2000, 2M)

1 3 (a) − ≤ g(2) < 2 2

(b) 0 ≤ g(2) < 2

Topic 3 Estimation, Gamma Function and Derivative of Definite Integration Objective Questions I (Only one correct option) x

1. If ∫ f (t ) dt = x2 + 0

 1 ∫x t f (t )dt , then f′  2 is 1 2

(2019 Main, 10 Jan II)

24 25 6 (c) 25 (a)

18 25 4 (d) 5

(b)

2. Let f : [0, 2] → R be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f (0) = 1.

Let F (x) = ∫

x2 0

f ( t ) dt, for x ∈ [0, 2]. If F ′ (x) = f ′ (x), ∀

x ∈ (0, 2), then F (2) equals (a) e2 − 1 (c) e − 1

(2014 Adv.)

(b) e4 − 1 (d) e4

3. The intercepts on X-axis made by tangents to the curve, x

y = ∫ | t | dt , x ∈ R, which are parallel to the line y = 2x, 0

are equal to (a) ± 1 (c) ± 3

(2013 Main)

(b) ± 2 (d) ± 4

300 Definite Integration 4. Let f be a non-negative function defined on the interval [0, 1].

x

x

∫0

If

1 − ( f ′ (t ))2 dt = ∫ f (t ) dt , 0 ≤ x ≤ 1 0

f (0) = 0 , then 1 1 (a) f   <  2 2 1 1 (b) f   >  2 2 1 1 (c) f   <  2 2 1 1 (d) f   >  2 2

5. If

1

∫ sin x

and (2009)

1 1 and f   >  3 3 1 1 and f   >  3 3 1 1 and f   <  3 3 1 1 and f   <  3 3

(1998, 2M)

1 (a) 2

(b) 0

(c) 1

(d) −

Then, the value of lim ∫ x→1

f ( x) 4

(a) 8 f ′(1) (c) 2 f ′(1)

1 2

2t dt is x −1

(b) 3 (d) None of these

6. If f (x) is differentiable and 4 f   equals  25

t2

(2005, 1M)

2

∫ 0 x f (x) dx = 5 t

5

,

then

(2004, 1M)

(1990, 2M)

(b) 4 f ′(1) (d) f ′(1) 1

is

2

13. The value of the definite integral ∫ (1 + e−x ) dx is 0

(a) −1 (b) 2 (c) 1+ e−1 (d) None of the above

(1981, 2M)

Objective Question II (One or more than one correct option) 14. Which of the following inequalities is/are TRUE? (2020 Adv.)

5 (b) − 2 5 (d) 2

2 (a) 5 (c) 1

7. If f (x) = ∫

1

x

∫ 0 f (t ) dt = x + ∫ x t f (t ) dt, then the value of f (1) is

12. Let f : R → R be a differentiable function and f (1) = 4.

 1 t 2 f (t ) dt = 1 − sin x, ∀ x ∈ (0, π / 2), then f    3

(a) 3 (c) 1/3

11.

x 2 + 1 −t 2

e

x2

3 (a) ∫ x cos x dx ≥ 0 8 1 2 1 (c) ∫ x cos x dx ≥ 0 2 1

dt, then f (x) increases in

(a) (2, 2 ) (c) (0, ∞ )

(2003, 1M)

(b) no value of x (d) (−∞ ,0) 1

8. If I (m, n ) = ∫ tm (1 + t )n dt, then the expression for

15. If g (x) = ∫

sin( 2x )

sin x

3 (b) ∫ x sin x dx ≥ 0 10 1 2 (d) ∫ x2 sin x dx ≥ 0 9 1

sin −1 (t ) dt, then

π (a) g′  −  = 2 π  2 π (c) g′   = 2 π  2

(2017 Adv.)

π (b) g′  −  = − 2 π  2 π (d) g′   = − 2 π  2

0

I (m, n ) in terms of I (m + 1, n − 1) is

(2003, 1M)

2n n (a) − I (m + 1, n − 1) m+1 m+1 n (b) I (m + 1, n − 1) m+1

Let f (x) = (1 − x)2 sin 2 x + x2, ∀ x ∈ R and x  2 (t − 1 )  g (x) = ∫  − ln t f (t ) dt ∀ x ∈ (1, ∞). 1  t+1 

16. Consider the statements

2n n + I (m + 1, n − 1) m+1 m+1 m (d) I (m + 1, n − 1) m+1

(c)

9. Let f (x) = ∫

x 1

2 − t 2 dt. Then, the real roots of the

equation x2 − f '(x) = 0 are (a) ±1 (c) ±

1 2

(2002, 1M)

1 (b) ± 2 (d) 0 and 1

P : There exists some x ∈ R such that, f (x) + 2x = 2 (1 + x2). Q : There exists some x ∈ R such that, 2 f (x) + 1 = 2x (1 + x). Then, (a) (b) (c) (d)

both P and Q are true P is true and Q is false P is false and Q is true both P and Q are false

17. Which of the following is true? x

10. Let f : (0, ∞ ) → R and F (x) = ∫ f (t ) dt. 0

If F (x2) = x2 (1 + x), then f (4) equals 5 (a) 4

Passage Based Questions

(b) 7

(c) 4

(2001, 1M)

(d) 2

(a) (b) (c) (d)

g is increasing on (1, ∞ ) g is decreasing on (1, ∞ ) g is increasing on (1, 2) and decreasing on (2, ∞ ) g is decreasing on (1, 2) and increasing on (2, ∞ )

Definite Integration 301 Fill in the Blank sec x cos x sec2x + cot x cosec x 18. f (x) = cos 2 x cos 2 x cosec2x . 2 2 1 cos x cos x Then, ∫

π /2 0

|x|→ ∞ , then show that every line y = mx intersects the curve y2 +

(1987, 2M)

Analytical & Descriptive Questions 1

(1988, 5M)

Integer & Numerical Answer Type Question 24. Letf : R → R be a differentiable function such that its derivative f′ is continuous and f (π) = − 6. x

dg differentiable function. If > 0, ∀ x prove that dx a b ∫ g(x) dx + ∫ g(x) dx increases as (b − a ) increases. 0

If F : [0, π ] → R is defined by F (x) = ∫ f (t )dt, and if 0

π

∫ ( f ′ (x) + F (x)) cos x dx = 2 0

(1997, 5M)

then the value of f (0) is ……… .

21. Determine a positive integer n ≤ 5, such that

∫ 0e

x

f (x) = ∫ [2(t − 1)(t − 2)3 + 3(t − 1)2(t − 2)2] dt.

x

20. Let a + b = 4, where a < 2 and let g (x) be a

1 x

(1991, 2M)

1

ln t dt. Find the function 1+ t f (x) + f (1 / x) and show that f (e) + f (1 / e) = 1 / 2, where (2000, 5M) ln t = log e t.

0

x

∫ 0 f (t ) dt = 2

23. Investigate for maxima and minima the function,

f (x) dx = K .

19. For x > 0, let f (x) = ∫

x

∫ 0 f (t ) dt → ∞ as

22. If ‘ f ’ is a continuous function with

(x − 1)n dx = 16 − 6e

(2020 Adv.)

(1992, 4M)

Topic 4 Limits as the Sum Objective Question I (Only one correct option)  (n + 1)1/3 (n + 2)1/3 (2n )1/3  + + ..... +  is equal to 4/3 4/3 n n 4/3   n

1. lim  n→ ∞

(2019 Main, 10 April I)

3 4/3 4 (2) − 3 4 4 3/ 4 (d) (2) 3

4 4/3 (2) 3 3 3 (c) (2)4/3 − 4 4 (a)

(b)



n n n 1 + 2 + 2 + ... +  is equal to 2 2 2 n→ ∞ n + 1 5n  n +2 n +3

2. lim  (a) (b) (c) (d)

2

(2019 Main, 12 Jan II)

tan − 1 (3) tan − 1 (2) π/4 π/2

(a) (b) (c)



18 e4 27 e2 9

e2 (d) 3 log 3 − 2

4. For a ∈ R,|a| > 1, let     3 3 1 + 2 +…+ n   = 54 lim n→∞    1 1 1   n7/3  + + … +    (an + 1)2 (an + 2)2 (an + n )2   Then the possible value(s) of a is/are (2020 Adv.) (a) −6

(b) 7

n



1/ n

(d) −9

(c) 8

5. For a ∈ R (the set of all real numbers), a ≠ − 1, (1a + 2a + K + n a ) 1 = . n→∞ (n + 1 ) [(na + 1) + (na + 2) + K + (na + n )] 60 Then, a is equal to lim

a−1

(a) 5

(n + 1)(n + 2) K 3n  3. lim   2n n→ ∞

Objective Questions II (One or more than one correct option)

(b) 7

(c)

−15 2

(d)

−17 2

n−1 n n and Tn = ∑ 2 , for 2 + + + + k2 n kn k n kn k=0 k=0 (2008, 4M) n = 1, 2, 3, ... , then n

is equal to (2016 Main)

6. Let S n = ∑ (a) Sn
3 3 3 3

(c) Tn


π 3 3

Analytical & Descriptive Question  1 1 1 + +K+  = log 6. n + 1 n + 2 6n  (1981, 2M)

7. Show that, lim  n→ ∞

Answers Topic 1

π (log 2 ) 8 1 1 83. log 6 − 2 10

π [ π + 3 log (2 + 3 ) − 4 3 ] 12 8 84. f ( x ) = x 3 + x 2 − x + 2 85. 2 π 81. π 2

80.

1. (c) 5. (b) 9. (d) 13. (c) 17. (d) 21. (c) 25. (a) 29. (c) 33. (a) 37. (a) 41. (c) 45. (c, d) 49. (c, d) 53. (a, c) 57. (0.5) 61. (9)

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62.

65. (2)

66. π 2

 1 68.    2

69. π ( 2 − 1 ) 70. (4)

(c) (a) (d) (d) (d) (a) (c) (c) (a) (d) (a) (b) (b, d) (a) (4.00) (2)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63.

(a) (a) (c) (d) (b) (d) (b) (b) (a) (d) (d) (a) (a, b) (a, b, c) (2) (8/3)

67.

1 a − b2 2

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 64.

(d) (c) (a) (c) (a) (b) (a) (c) (c) (d) (a, b, c) (c) (*) (1) (0) (k = 16)

7   a log 2 − 5a + b  2  71. (2 − 2 )

72. A → S; B → S; C → P; D → R 73. (d) 24 75. 5

74. 5051

78.

π 2

1 1 π log 2 − + 2 2 4

91.

απ sin α

92.

 3 1 3π + 1  1  93.  − π +  94. (log 3 ) 95. ( ii )  π 2  20  2  12

Topic 2 1. (d)

2. (c)

3. (b)

6. (4)

2. (b) 6. (a) 10. (c)

3. (a) 7. (d) 11. (a)

4. (c) 8. (a) 12. (a)

14. (a,b,d) 15. (*) 15 π 32 +   18. −    60 

16. (c) 1 19. 2

Topic 3 1. (a) 5. (a) 9. (a) 13. (d) 17. (b) 21. (n = 3 ) 7 23. At x = 1 and , f ( x ) is maximum and minimum respectively. 5

Topic 4

79. log 2

1. (c)

2. (b)

5. (b, d)

6. (a, d)

3. (b)

4. (c,d)

Hints & Solutions Topic 1 Properties of Definite Integral π

1. Given integral ∫ |π −|x|| dx

=∫

π /3 π/ 6

−π

π

π

0 π

0

= 2 ∫ |π −|x|| dx = 2∫ |π − x | dx = 2 ∫ (π − x) dx π

   π 2 x2  π2 = 2  πx −  = 2  π 2 − = 2   = π2  2 0 2  2   Hence, option (c) is correct.

=

1 1 1   (9 × 0) –  × 1  = − 9   2  18 α+1

3. Let I = =

π /3

π /3 π/ 6

= 3

2

∫

dx (x + α ) (x + α + 1)

∫

(x + α + 1) − (x + α ) dx (x + α ) (x + α + 1)

∫

 1  1 −   dx  x + α x + α + 1

α α+1

tan3 x ⋅ sin 2 3x(2 sec2 x sin 2 3x + 3 tan x ⋅ sin 6x)dx

π/ 6

=∫

π /3

 1 tan 4 x sin 4 3x  2 π/ 6

α α+1

2. Given integral

∫

 d  tan 4 x sin 4 3x   dx   2   dx 

= [Q x ∈ (0, π)]

0

I=

4

α

[2 tan x sec x sin 3x + 3 tan x(2 sin 3x cos 3x)]dx 4

π2 16

24. (4)

   1  1 e e cos  2 + 2 sin  2 − 1   

 4π  1  76.  tan −1     2   3

90.

82.

3

α +1

= [ log e (x + α ) − log e (x + α + 1)] α

Definite Integration 303 π /2

α+1

π 1 x  = x − tan = − 1 = (π − 2)   2 0 2 2

  x+ α  = log e    x + α + 1  α  2α + 1 2α = log e − log e 2α + 2 2α + 1  2α + 1 2α + 1  9 = log e  ×  = log e    8  2α + 2 2α 

(given)

⇒

(2 α + 1)2 9 = ⇒ 8 [4α 2 + 4α + 1] = 36 (α 2 + α ) 4α (α + 1) 8

⇒ ⇒

8α 2 + 8α + 2 = 9α 2 + 9α ⇒ α 2 + α − 2 = 0 (α + 2) (α − 1) = 0 ⇒ α = 1, − 2

π /3

5. Let

From the options we get α = − 2 π /2 cot x 4. Let I = ∫ dx 0 cot x + cosec x cos x π / 2 cos x π /2 sin x dx =∫ dx = ∫ 0 0 cos x 1 1 + cos x + sin x sin x π / 2 cos x(1 − cos x) =∫ dx 0 1 − cos 2 x =∫ =∫ =∫

0 π /2 0

=

cos 2/ 3

1 dx = x sin 4/3 x

π /3

sec2 x ∫ (tan x)4/ 3 dx π/6

and sec2 x dx = dt I=

So,

3

dt  t − 1/3  =  t 4/3  − 1 / 3 1/

∫

1/ 3

3

 1  = − 3  1/ 6 − 31/ 6 = 3 ⋅ 31/ 6 − 3 ⋅ 3− 1/ 6 3 

(cosec x cot x − cosec x + 1) dx 2

1 2 x 1 − sec  dx  2 2

x cosec4/3 x dx

[multiplying and dividing the denominator by cos 4/3 x] Put, tan x = t, upper limit, at x = π / 3 ⇒ t = 3 and lower limit, at x = π / 6 ⇒ t = 1 / 3

(cosec x cot x − cot2 x) dx

π /2

∫

π /6

= 37/ 6 − 35/ 6

6. Given integral 2π

I = ∫ [sin 2x ⋅ (1 + cos 3x)]dx 0 π

= ∫ [sin 2x ⋅ (1 + cos 3x)]dx 0

2π

+ ∫ [sin 2x ⋅ (1 + cos 3x)]dx π

= I1 + I 2 (let)

[given]

cos x π / 2 cos x π /2 sin x dx =∫ dx = ∫ 0 0 cos x 1 cos x + 1 + sin x sin x x 2 cos 2 − 1 π /2 2 =∫ dx 0 x 2 cos 2 2 θ θ  Q cos θ = 2 cos 2 − 1 and cos θ + 1 = 2 cos 2  2 2  0

2/3

3

= [− cosec x + cot x + x]π0 / 2 π /2   2 x  2 sin −   π /2   cos x − 1  2   = x+ = x +  x x sin x  0   2 sin cos  2 2  0  π /2 x π 1   = x − tan = − 1 = [π − 2]  2  0 2 2 = m[π + n ] 1 On comparing, we get m = and n = − 2 2 ∴ m⋅ n = − 1 Alternate Solution π /2 cot x Let I = ∫ dx 0 cot x + cosec x

=∫

∫ sec

π /6

cos 2 x dx sin 2 x

π /2

I=

π /3

π / 2 cos x − 0

I = m(π − n ) 1 ∴ m(π − n ) = (π − 2) 2 On comparing both sides, we get 1 m = and n = − 2 2 1 Now, mn = × − 2 = − 1 2 Since,

Now,

... (i)

2π

I 2 = ∫ [sin 2x ⋅ (1 + cos 3x)]dx π

let 2π − x = t, upper limit t = 0 and lower limit t = π and dx = − dt 0

I 2 = − ∫ [− sin 2x ⋅ (1 + cos 3x)]dx

So,

π

π

= ∫ [− sin 2x ⋅ (1 + cos 3x)]dx 0 π

…(ii)

I = ∫ [sin 2x ⋅ (1 + cos 3x)]dx

∴

0

+

π

∫0 [− sin 2x ⋅ (1 + cos 3x)]dx [from Eqs. (i) and (ii)] π

= ∫ (−1)dx] [Q [x] + [− x] = − 1, x ∉Integer] 0

= −π 1

7. Let I = ∫ x cot−1 (1 − x2 + x4 )dx 0

Now, put x2 = t ⇒ 2xdx = dt

304 Definite Integration Lower limit at x = 0, t = 0 Upper limit at x = 1, t = 1

x

9. Given, f (x) = ∫ g (t )dt 0

1

∴I=

1 cot−1 (1 − t + t 2)dt 2 ∫0

On replacing x by (−x), we get −x

  1 1 = ∫ tan −1   dt 2 20 1 − t + t  1

1  Q cot−1 x = tan −1  x 

1

=

 t − (t − 1)  1 tan −1   dt ∫  1 + t (t − 1) 20

=

1  1 −1 −1 ∫ ( tan t − tan (t − 1)dt  2 0 

 y 

1

Q ∫ tan −1 (t − 1)dt = ∫ tan −1 (1 − t − 1)dt = − ∫ tan −1 (t )dt 0 a

a

0

So, I =

⇒

f (− x) = − f (x)

⇒ f is an odd function. Now, it is given that f (x + 5) = g (x) ∴

f (5 − x) = g (− x) = g (x) = f (x + 5) f (5 − x) = f (x + 5)

0

Put t = u + 5 ⇒ t − 5 = u ⇒ dt = du

t dt t2 1 + 0

0

∴

[by integration by parts method] π 1 π 1 = − [log e (1 + t 2)]10 = − log e 2 4 2 4 2

π /2 0

5 −x

2I = ∫

0

=∫

π 2 0

=∫

π 2 0

∫

5

g (− t )dt =

b

∫ g(t )dt

5 −x

a

[Q − ∫ f (x)dx = ∫ f (x)dx and g is an even function] a

3

sin x dx sin x + cos x

…(i)

I=

b

5

∫ f ′ (t )dt

[by Leibnitz rule f ′ (x) = g (x)]

5 −x

b

b

∫a f (x)dx = ∫a f (a + b − x) dx,

= f (5) − f (5 − x) = f (5) − f (5 + x) 5

5

5+ x

5+ x

=

cos3 x dx cos x + sin x

…(ii)

On adding integrals (i) and (ii), we get π/ 2

−5

5

we get I=∫

−5

u = −t du = − dt, we get

Put ⇒

b

On applying the property,

x −5

∫ f (u + 5)du = ∫ g(u )du

I=−

∫a f ( x) dx = ∫a f ( a + b − x) dx π 2 0

x −5

I=

Key Idea Use property of definite integral.

I=∫

10. The given functions are g (x) = log e x, x > 0 2 − x cos x f (x) = 2 + x cos x

nd

(sin x + cos x) (sin 2 x + cos 2 x − sin x cos x) dx sin x + cos x

Let

I=∫

Then,

I=∫

π

1    1  1 − (2 sin x cos x) dx = ∫ 2 1 − sin 2x dx   2 0   2

π /4 −π / 4 π /4 −π / 4

g ( f (x))dx  2 − x cos x  log e   dx  2 + x cos x

π/ 2

π 1 π −1 I= − = 4 4 4

[from Eq. (i)]

∫ f ′ (t )dt = ∫ g(t )dt

sin3 x + cos3 x dx sin x + cos x

π 1 1 π  1   = x + cos 2x =  − 0 + (−1 − 1) = −     4 2 2 2 4 0 ⇒

…(i)

Let I = ∫ f (t )dt 1

1

Let

0

x

1 ( tan −1 t + tan −1 t )dt 2 ∫0

b

0

⇒

0

= ∫ tan −1 tdt = [t tan −1 t ]10 − ∫

8.

x

[Q g is an even function]

because ∫ f (x)dx = ∫ f (a − x)dx 0 1

x

[Q g is an even function]

1

0

0

Now, put t = − u, so f (− x) = − ∫ g (− u )du = − ∫ g (u )du = − f (x)

 −1 x − y −1 −1 Q tan 1 + xy = tan x − tan 

1

∫ g(t )dt

f (− x) =

Now, by using the property b

∫

a

b

f (x)dx = ∫ f (a + b − x)dx, we get a

…(i)

Definite Integration 305 I=∫

 2 + x cos x log e   dx −π / 4  2 − x cos x  π /4

⇒ 5 tan 4 x sec 2x dx = dt

…(ii)

  2 − x cos x   2 + cos x   log e   + log e   dx −π / 4   2 + x cos x  2 x cos x   π /4  2 − x cos x 2 + x cos x =∫ log e  ×  dx −π / 4  2 + x cos x 2 − x cos x 

2I = ∫

π /4

⇒ 2I = ∫

∴ I=

log e (1)dx = 0 ⇒ I = 0 = log e (1)

−π / 4 a

11. Let I = ∫ f (x) g (x) dx

… (i)

0 a

= ∫ f (a − x) g (a − x) dx 0

Q a f (x) dx =  ∫0

a

a

I = ∫ f (x) [4 − g (x)] dx

⇒

0



∫0 f (a − x) dx

[Q f (x) = f (a − x) and g (x) + g (a − x) = 4]

a

a

0

0

1 1 ⋅ 2 5

1

∫(1/

3 )5

a

I = 4 ∫ f (x) dx − I

[from Eq. (i)]

0 a

⇒ 2I = 4 ∫ f (x) dx ⇒ 0

e

12. Let I = ∫   1  e x

2x



⇒

1 π −1  1    − tan   9 3 10  4

14. Let I = ∫

sin 2 x dx −2 1 x + 2  π  sin 2 x 1 x + 2  π 

Also, let f (x) =

sin 2(− x) (replacing x by − x) 1  x + − 2  π 

Then, f (− x) =

a

I = 2 ∫ f (x) dx. 0

=

sin 2 x 1   x  + − 1 −  π  2  if   − [x], Q [− x] = − 1 − [x], if  

 x   = log t  e

x (log e x − log e e) = log t

1   1  x  x + (log e x − log e e) dx = t dt   1 1 ⇒ (1 + log e x − 1) dx = dt ⇒ (log e x) dx = dt t t Also, upper limit x = e ⇒ t = 1 and lower limit x = 1 ⇒ 1 t= e I=∫

1

⇒ I=∫

1

 2 1 1  t −  ⋅ dt t t

f (− x) = −

⇒

x ∈ I x ∉ I 

sin 2 x = − f (x) 1 x + 2  π 

i.e. f (x) is odd function ∴ I =0  a Q ∫ f (x) dx = −a 

0, if f (x) is odd function     a f (x)dx, f ( if x) is even function 2  ∫0 

b

15. We have, I = ∫ (x4 − 2x2)dx

1/ e 

1/ e

3 )5

2

⇒

∴

1 dt (tan −1 (t ))(11/ = t 2 + 1 10

=

 e −    log e x dx  x 

x

1

1  −1 −1  1    tan (1) − tan   9 3 10 

x

 x Now, put   = t ⇒ x log e  e

5

=

= ∫ 4 f (x) dx − ∫ f (x) g (x) dx ⇒

π 4

 1     3

t

[Q log e A + log e B = log e AB]

π /4

π 6

x

On adding Eqs. (i) and (ii), we get

a

(t − t −2) dt

Let

1

  t 2 1  1  1   1  3 I =  +   =  + 1 −  2 + e  = − e − 2     t 2 2 2 2 e 2 e   1    

f (x) = x4 − 2x2 = x2(x2 − 2) = x2(x − 2 ) (x + 2 )

Graph of y = f (x) = x4 − 2x2 is Y y=f(x)

e

13. Let I = ∫ =∫

π /4 π /6

π /4 π /6

dx sin 2x(tan5 x + cot5 x)

(1 + tan x) tan x dx 2 tan x (tan10 x + 1) 2

5

1 π / 4 tan 4 x sec 2x = ∫ dx 2 π / 6 (tan10 x + 1) Put tan5 x = t

 2 tan x  Q sin 2x =  1 + tan 2 x  

– √2

X

O

√2 ∴

f(x) < 0 for – √2 < x < √2 – – + + – √2

0

√2

b

[Qsec2 x = 1 + tan 2 x]

Note that the definite integral ∫ (x4 − 2x2)dx represent a

the area bounded byy = f (x) , x = a, b and the X -axis.

306 Definite Integration But between x = − 2 and x = 2 , f (x) lies below the X-axis and so value definite integral will be negative. Also, as long as f (x) lie below the X-axis, the value of definite integral will be minimum. ∴(a , b) = (− 2 , 2 ) for minimum of I. π /3 1 tan θ , (k > 0) 16. We have, ∫ dθ = 1 − 0 2 2k sec θ π /3 tan θ π /3 tan θ 1 Let I = ∫ dθ = dθ ∫ 0 0 sec θ 2k sec θ 2k π /3 π /3 sin θ (sin θ) 1 1 = dθ = dθ ∫ ∫ cos θ 2k 0 1 2k 0 (cos θ) cos θ Let cos θ = t ⇒ − sin θ dθ = dt ⇒ sin θ dθ = − dt for lower limit, θ = 0 ⇒ t = cos 0 = 1 π π 1 for upper limit, θ = ⇒ t = cos = 3 3 2 ⇒

I=

1 2k

1/ 2 −

∫1

dt −1 = 2k t

1/ 2 −

∫1

t

 1 1  1 =− [2 t ]12  =−  2k  − 1 + 1 2k 1  2  2  1  2 − 1 = 2k   1 I =1 − 2 2  1 1 ⇒ 1 −  =1 − 2k  2 2

Q ∴ ⇒

π

∴I=

1 2  1 −   2 2k (given)

2 2 ( cos 3x + 3 cos x) dx 4 ∫0 π

=

1 sin 3x 2 + 3 sin x 2  3  0

=

1  1 3π π  1  − sin 0 + 3 sin 0  + 3 sin  sin  2 3 2 2  3

=

1  1    (−1) + 3 − [0 + 0] 2 3   3π π π    Q sin 2 = sin  π + 2  = − sin 2 = − 1  

1 2dt

1 1 − +1 2 t 2 

=−

Now, as cos 3x = 4 cos3 x − 3 cos x 1 cos3 x = (cos 3x + 3 cos x) ∴ 4

=

1 1  4 − +3 =  3 2  3

π /2

Let

I=

⇒

I=

∫

−π/ 2

a

π π   − + − x   2 2

2

1+2

−

π π + −x 2 2

dx

b  b  Q ∫ f (x)dx = ∫ f (a + b − x)dx  a  a π /2

⇒

I=

⇒

I=

⇒

2I =

⇒

2I =

Y

O

a

sin x dx 1 + 2x −π/ 2

∫

π / 2 sin

2 = 2k ⇒ 2k = 4 ⇒ k = 2

X′

b

2

2 =1 2k

17. We know, graph of y = cos x is

b

18. Key idea Use property = ∫ f (x)dx = ∫ f (a + b − x)dx

sin 2 x ∫ 1 + 2−x dx −π / 2 π /2

π/2

π

X

2x sin 2 x dx 2x + 1 −π / 2

∫

π /2

 2 x + 1 sin 2 x x  dx  2 + 1 −π / 2

∫

π /2

Y′

∴ The graph of y =|cos x|is Y

∫ sin

2

x dx

−π / 2 π /2

⇒

y=|cos x|

2I = 2 ∫ sin 2 x dx [Qsin 2 x is an even function] 0 π /2

⇒ X′

π

O Y′

I = ∫ |cos x|3 = 2 ∫ 0

π/2

π

I=

π (Q y = |cos x|is symmetric about x = ) 2 π π     = 2∫ 2 cos3 x dx Q cos x ≥ 0 for x ∈ 0, 2  0  

2

0 π /2

X

π 2|cos x|3 dx 0

∫ sin xdx

⇒

I=

∫

cos 2xdx

0 π /2

∫ dx

⇒

2I =

⇒

2I = [x]0π / 2 π I= 4

0

⇒

a  a  Q ∫ f (x)dx = ∫ f (a − x)dx  0  0

Definite Integration 307 19. Let I = ∫

π /4

=∫ =∫

3 π / 4 1 − cos x dx =∫ dx π / 4 1 − cos 2 x 1 + cos x

3 π /4

3 π /4 π /4

22.

Given, I = ∫

1 − cos x dx sin 2 x

3 π /4 π /4

=∫

(cosec2 x − cosec x cot x)dx

= [− cot x + cosec

x]3π π/ 4/ 4

2

π /2

x cos x …(i) dx + ex Q b f (x) dx = b f (a + b − x) dx ∫a  ∫a 

− π /2 1

x2 cos (− x) ⇒ I=∫ dx −π / 2 1 + e− x π /2

…(ii)

On adding Eqs. (i) and (ii), we get π /2 2  1 1  x cos x  + 2I = ∫  dx x −π / 2 + e + e− x  1 1  =∫

π /2 2

x cos x ⋅ (1) dx

−π / 2

⇒ 2I = 2∫

  a a Q ∫ f (x) dx = 2 ∫ f (x) dx, when f (− x) = f (x) 0   − a

π /2 2 0

2I = 2 [x2(sin x) − (2x) (− cos x) + (2) (− sin x)] π0 / 2  π2 − 2 ⇒ 2I =2 4  

21.

⇒

(− cosec x ⋅ cot x − cosec2x) dx = dt

and

cosec x − cot x = 1 /t

⇒

2 cosec x = t + I=−∫

∴

⇒

1

217

2+1

⇒

u = ln ( 2 + 1 )

⇒

I=−∫ =2 ∫

log x2 dx I=∫ 2 log x2 + log(36 − 12 x + x2 ) 4 2 log x dx =∫ 2 2 log x + log(6 − x)2 4 2 log x dx =∫ 2 2 [log x + log(6 − x)] 4 log x dx …(i) I=∫ 2 [log x + log(6 − x)] 4 log(6 − x) …(ii) I=∫ dx 2 log(6 − x) + log x Q b f (x)dx = b f (a + b − x)dx ∫a  ∫a  4

On adding Eqs. (i) and (ii), we get 4 log x + log(6 − x) 2I = ∫ dx 2 log x + log(6 − x) 4

⇒

2I = 2 ⇒ I = 1

2

dt t

0 ln ( 2 + 1 )

ln ( 2 + 1 ) 0

2(eu + e− u )16

eudu eu

(eu + e− u )16 du

x − a, x ≥a  − ( x − a ), x < a to break given integral in two parts and then integrate separately. π

∫0

2

π x x  1 − 2 sin  dx = ∫ |1 − 2 sin |dx 0  2 2 π

π  x x  = ∫ 3 1 − 2 sin  dx − ∫ π 1 − 2 sin  dx 0    2 2 3

⇒ 2I = 2

π

x 3  x π  =  x + 4 cos  −  x + 4 cos  = 4 3 − 4 −  2 0  2 π 3

b

2I = ∫ dx = [x]42

16

and when t = 2 + 1, eu = 2 + 1

π

⇒

1  t +  t  2    

When t = 1, eu = 1 ⇒ u = 0

∫a f(x )dx = ∫a f(a + b − x )dx and then add.

⇒

1 t

Let t = eu ⇒ dt = eudu.

PLAN Apply the property

Let

π /4

217 (cosec x)16 cosec x (cosec x + cot x) dx (cosec x + cot x) cosec x + cot x = t

π2 I= −2 4 b

π /2

(2 cosec x)17 dx

23. PLAN Use the formula,| x − a | = 

x cos x dx

Using integration by parts, we get

∴

π /2 π /4

Let

= [(1 + 2 ) − (− 1 + 2 )] = 2

20. Let I = ∫

PLAN This type of question can be done using appropriate substitution.

3

24. I = ∫ As,

π /2 − π /2

 2  π − x  x + log  π + x  cos x dx  

a

∫− a f (x) dx = 0, when f (− x) = − f (x)

∴ I=∫

π /2 − π /2

x2 cos x dx + 0 = 2∫

= 2 {(x2 sin x)π0 / 2 − ∫

π /2 0

π /2 0

(x2 cos x) dx

2x ⋅ sin x dx}

π2 π /2 =2  − 2 {(− x ⋅ cos x)0π / 2 − ∫ 1 ⋅ (− cos x) dx} 0  4      π2 π2 π2 − 4 =2  − 2 (sin x)0π / 2 = 2  − 2 =      2 4 4

25. Put x2 = t ⇒ x dx = dt /2 ∴

dt 2 I=∫ log 2 sin t + sin (log 6 − t ) log 3

sin t ⋅

...(i)

308 Definite Integration b

b

∫a f (x) dx = ∫a f (a + b − x) dx

Using,

sin (log 2 + log 3 − t ) ∫log 2 sin (log 2 + log 3 − t ) + sin dt (log 6 − (log 2 + log 3 − t )) 1 log 3 sin (log 6 − t ) = ∫ dt 2 log 2 sin (log 6 − t ) + sin (t )

1 = 2

sin (log 6 − t ) dt sin (log 6 − t ) + sin t

log 3 log 2

2I = ∫ =∫

…(ii)

0 −2

=∫

−2

0

⇒

2I = π

30. Given,

[x3 + 3x2 + 3x + 3 + (x + 1) cos (x + 1)] dx

∴

I=∫ =∫

1 −1

t dt + 2 ∫ 3

−1

1 −1

dt +

1

∫ −1 t cos t dt

=∫

1 1−x 1−x dx = ∫ dx 0 1+ x 1 − x2

1

1

1

0

1−x

2

dx − ∫

1

x

0

1−x

2

= (sin

1 − sin

−1

0

dx = [sin −1 x] 10+

∫1

31.

e2

∫e

−1

−1/ 2

=∫

−1/ 2

=∫

1/ 2

0 −1/ 2

[x] dx +

0) + [t ]10 = π /2 − 1

∫ −1/ 2log 1 − x dx    1 + x Q log  1 − x  is an odd function   

[x]dx + 0 [x] dx +

1/ 2

∫0

 = − [x]−01/ 2 = − 0 + 

29. Let I = ∫ =∫

π −π −π π

[x] dx = ∫

0 −1/ 2

(−1) dx +

1/ 2

∫0

(0) dx

1 1  =− 2 2

cos 2(− x) d (− x) 1 + a −x

⇒ I = π /2

3

2 −2

ecos x sin x dx +

3

∫ 2 2 dx

= 0 + 2 [x]32

e 2 log x log e x e  dx + ∫  dx 1 x  x  2 1 1 = − [(log e x)2]1e−1 + [(log e x)2] 1e 2 2 1 1 5 = − {0 − (−1)2} + (22 − 0) = 2 2 2 1

e −1

figure. From the graph, it is clear that x = π /2  2, if  1, if π / 2 < x ≤ 5π /6  [2 sin x] =  0, if 5π / 6 < x ≤ π −1, if π < x ≤ 7π / 6  −2, if 7π / 6 < x ≤ 3π /2 Y 2 1 π/2

2

cos x dx 1 + ax

cos 2x dx = π + 0

32. The graph of y = 2 sin x for π /2 ≤ x ≤ 3π / 2 is given in

 1 + x

1/ 2

π /2 0

π

∫ 0 cos 2x dx

2 log e x  dx = 1  log e x  dx − e  log e x  dx − 1 ∫ e  x  ∫1  x   x   since, 1 is turning point for    log e x   for + ve and − ve values    x 

t dt t

  1 + x  28. ∫ [x] + log   dx −1/ 2  1 − x    1/ 2

π

0

= −∫

1/ 2

=∫

π

2

[where, t 2 = 1 − x2 ⇒ t dt = − x dx] −1

1 + cos 2x dx 2

[Q ecos x sin x is an odd function] Q 3 f (x) dx = 2 = 2 [3 − 2] = 2  ∫ −2 

[since, t3 and t cos t are odd functions] 0

π 0

0

=∫

= 0 + 2 ⋅ 2 [x]10 + 0 = 4

27. I = ∫

cos 2 x dx = 2 ∫

3

(t3 + 2 + t cos t ) dt

1

π −π

ecos x sin x, for | x| ≤ 2  f (x) =  2 otherwise ,  

Put x + 1 = t dx = dt

1 + ax    cos 2x dx −π  1 + a x 

∫ −2 f (x) dx = ∫ −2 f (x) dx + ∫ 2 f (x) dx

∴

[(x + 1)3 + 2 + (x + 1) cos (x + 1)] dx

⇒

…(ii)

π

= [x]π0 + 2∫

1 log 3 1 (t )log 2 = (log 3 − log 2) 2 2 1  3 I = log    2 4

26. Let I = ∫

cos 2 x dx 1 + ax

= ∫ (1 + cos 2x) dx = ∫ 1 dx +

2I =

∴

−π

ax

On adding Eqs. (i) and (ii), we get

On adding Eqs. (i) and (ii), we get 1 log 3 sin t + sin (log 6 − t ) 2I = dt 2 ∫log 2 sin (log 6 − t ) + sin t ⇒

π

log 3

I=∫

∴

I=∫

⇒

…(i)

–1 –2

5π/6

π

7π/6

3π/2

X

Definite Integration 309 Therefore, =∫

5 π /6 π /2

3 π /2

∫ π /2

dx + ∫

⇒

[2 sin x]dx

x

π 5π / 6

0 dx +

7 π /6

∫π

(−1) dx + ∫

3 π /2 7π / 6

∴

3 π /4

π /4 3 π /4

I=∫

π /4

k

⇒

2I = ∫ I=∫

⇒

dx 1 + cos (π − x)

π /4

dx 1 − cos x

…(ii)

π /4

1

1 −1

[x] dx

0

∫ −1 [x] dx = ∫ −1 [x] 0 −1

1

[Q x is an odd function]

I1 1 = I2 2

subdivide the interval π to 2π as under keeping in view that we have to evaluate [2 sin x ] Y 1,π/2 O 30°

30°

∴

⇒

∫ 0 [x] dx 1

∫0

11π π π 1  = sin 2π −  = − sin = −   6 6 6 2 9π 3π sin = sin = −1 6 6

sin

Hence, we divide the interval π to 2π as 7π   7π 11 π   11π   , , 2π π , ,  ,    6  6 6   6

0 dx 1

∫ −1 f (x) dx = 1

1  1  1   sin x = 0, −  ,  − 1, −  ,  − , 0  2  2  2 

x

⇒

0

⇒

4

cos t dt

π

0

π+x π

2 sin x = (0, − 1), (−2, − 1), (−1, 0) [2 sin x] = − 1 =∫

cos 4 t dt = I1 + I 2

7π / 6 π

[2 sin x] dx +

4

0

and Put

I2 = ∫

π+x π

t=π+ y

4

cos t dt

=∫

7π / 6 π

=−

(−1) dx +

11 π / 6

∫ 7π / 6

11 π / 6

∫ 7π / 6

+

π

I1 = ∫ cos t dt = g (π )

–1/2,11π/6

Y'

π 1 = 6 2 7π 1 π  sin  π +  = sin =−  6 6 2

1

(−1) dx +

= ∫ cos 4 t dt + ∫ where,

X (0,2π)

We know that, sin

x=1 dx +

–1,3π/2

–1/2,7π/6

⇒

g (x) = ∫ cos 4 t dt 0

⇒

xf (1 − x)]dx

37. It is a question of greatest integer function. We have,

−1 ≤ x < 0 0 ≤ x 0  1 1  1 1 sin   < and sin   <  2 2  3 3  1 1  1 1 f   < and f   <  2 2  3 3 1 sin x

4

⋅ 2x − e− ( x 2

+ 2x + 1 ) 2

∴

)

⋅ 2x

(1 − e2x

2x + 1

[where, e

2 2

2

+1

)

> 1, ∀ x and e− ( x

4

+ 2x 2 + 1 )

> 0, ∀ x]

f ′ (x) > 0 1

As we know that,

5. Since ∫

+ 1 )2

0

x = 0 ⇒ sin { f (0)} = c c = sin −1 (0) = 0 f (x) = ± sin x f (x) ≥ 0, ∀ x ∈ [0, 1] f (x) = sin x

⇒

2

8. Here, I (m, n ) = ∫ t m(1 + t )n dt reduce into I (m + 1, n − 1)

−1

∴

dt

which shows 2x < 0 or x < 0 ⇒ x ∈ (−∞ , 0)

∫ dx

⇒ sin −1 { f (x)} = ± x + c Put ⇒ ∴ but ∴

+ 1 )2 

= 2xe− ( x

0

2

e

x2

d 2   − ( x 2 )2  d (x2)   (x + 1) − e  dx   dx 

2

f ′ (x) = e− ( x

For x intercept put y = 0, we get 2

2  putting t =  5 

x 2 + 1 −t 2

f (x) = ∫

f (t 2) = t

On differentiating both sides using Newton’s Leibnitz’s formula, we get

Equation of tangent is

x

⇒

t 2 f (t ) dt = 1 − sin x, thus to find f (x).

On differentiating both sides using Newton Leibnitz formula d 1 d i.e. t 2 f (t ) dt = (1 − sin x) dx ∫ sin x dx ⇒ {12 f (1)} ⋅ (0) − (sin 2 x) ⋅ f (sin x) ⋅ cos x = − cos x 1 f (sin x) = ⇒ sin 2 x  1 For f   is obtained when sin x = 1 / 3  3  1 i.e. f   = ( 3 )2 = 3  3

[we apply integration by parts taking (1 + t )n as first and t m as second function] 1 1  tm + 1  tm + 1 ∴ I (m, n ) = (1 + t )n ⋅ − ∫ n (1 + t )( n − 1) ⋅ dt  0 m + 1 0 m+1  1 n 2n (1 + t )( n − 1) ⋅ tm + 1 dt = − ∫ m+1 m+1 0 n 2n ∴ I (m, n ) = − ⋅ I (m + 1, n − 1) m+1 m+1

9. Given,

f (x) = ∫

x 1

2 − t 2 dt ⇒

f ′ (x) = 2 − x2

Also, x2 − f ′ (x) = 0 ∴ ⇒

x2 = 2 − x 2

⇒ x4 = 2 − x 2

x4 + x2 − 2 = 0 ⇒ x = ± 1

10. Given, F (x) = ∫

x 0

f (t ) dt

By Leibnitz’s rule, F ′ (x) = f (x) But

…(i)

F (x2) = x2 (1 + x) = x2 + x3

⇒

F (x) = x + x3/ 2 ⇒

⇒

f (x) = F ′ (x) = 1 +

⇒

f (4) = 1 +

F ′ (x) = 1 +

3 1/ 2 x 2

[given] 3 1/ 2 x 2 [from Eq. (i)]

3 1/ 2 3 (4) ⇒ f (4) = 1 + × 2 = 4 2 2

Definite Integration 325 1

x

∫ 0 f (t )dt = x + ∫ x t f (t ) dt

11. Given,

On differentiating both sides w.r.t. x, we get f (x) 1 = 1 − x f (x) ⋅ 1 ⇒ (1 + x) f (x) = 1 1 1 1 ⇒ f (x) = ⇒ f (1) = = 1+ x 1+1 2 f ( x)

f ( x)

12. lim ∫ x→1

2t dt = lim x→1 x−1

∫4

2t dt

x−1 [using L’ Hospital’s rule] 2 f (x) ⋅ f ′ (x) = 2 f (1) ⋅ f ′ (1) = lim x→1 1 [Q f (1) = 4] = 8 f ′ (1)

4

13. If f (x) is a continuous function defined on [a , b], then b

m (b − a ) ≤ ∫ f (x) dx ≤ M (b − a ) where, M and m are maximum and minimum values respectively of f (x) in [a , b]. −x 2

Here, f (x) = 1 + e

is continuous in [0, 1]. x2

0 < x 0 ⇒ e 2

< e ⇒e

−x 2

> e ⇒e 0

e

∴

⇒

1

1



3

x

x

2

∫ 0 x cos x dx ≥ ∫ 0  x − 2 ! dx =  2

⇒ and,

−

sin 2x

sin x

sin −1 (t )dt

g′ (x) = 2 cos 2x sin −1 (sin 2x) − cos x sin −1 (sin x)  π  π g′   = − 2 sin −1 (0) = 0 ⇒ g′  −  = − 2 sin −1 (0) = 0  2  2 No option is matching. 16. Here, f (x) + 2x = (1 − x)2 ⋅ sin 2 x + x2 + 2x where, P : f (x) + 2x = 2 (1 + x)2

and,

4 1

1 1 3 x = − =  8 0 2 8 8

∫ 0 x cos x dx ≥ 8 1 2 x4  ∫ 0 x sin dx ≥ ∫ 0  x − 6  dx 1

 x3 x5  1 1 9 3 = −  = 3 − 30 = 30 = 10 3 30 0  1 3 ∫ 0 x sin x dx ≥ 10 1 2 1 3 x5  ∫ 0 x cos x dx ≥ ∫ 0  x − 2  dx

∴

6 1

x x 1 1 2 1 = −  = 4 − 12 = 12 = 6 4 12 0  1 2 1 ∫ 0 x cos x dx ≥ 6 4

…(ii)

2 (1 + x2) = (1 − x)2 sin 2 x + x2 + 2x (1 − x)2 sin 2 x = x2 − 2x + 2

⇒

(1 − x)2 sin 2 x = (1 − x)2 + 1 ⇒ (1 − x)2 cos 2 x = − 1

∴ P is false. Again, let Q : h (x) = 2 f (x) + 1 − 2x (1 + x) where,

h (0) = 2 f (0) + 1 − 0 = 1 h (1) = 2 f (1) + 1 − 4 = − 3, as h (0) h (1) < 0

⇒ h (x) must have a solution. ∴ Q is true. 17. Here, f (x) = (1 − x)2 ⋅ sin 2 x + x2 ≥ 0, ∀ x. x  2 (t − 1 )  and g (x) = ∫  − log t f (t ) dt 1 t + 1   2 (x − 1) g′ (x) =  (x) − log x ⋅ f { x ( ) 1 +  + ve 

…(i)

For g′ (x) to be increasing or decreasing, 2 (x − 1) let φ(x) = − log x (x + 1) φ′ (x) =

4 1 − (x − 1)2 − = 2 x x (x + 1)2 (x + 1)

φ′ (x) < 0, for x > 1 ⇒ φ (x) < φ (1) ⇒ φ(x) < 0

…(ii)

From Eqs. (i) and (ii), we get

1

⇒

…(i)

⇒

⇒

3

1

15. g (x) = ∫

which is never possible.

−x 2

x

∴

∴

a

1

1 3  x4 x6  x5  ∫ 0 x sin x dx ≥ ∫ 0  x − 6  dx =  4 − 36  0 1 1 8 2 = − = = 4 36 36 9 1 2 2 ∫ 0 x sin x dx ≥ 9 1 2

and,

g′ (x) < 0 for x ∈ (1, ∞ ) ∴g (x) is decreasing for x ∈ (1, ∞ ). sec x

cos x

18. Given, f (x) = cos 2 x cos 2 x 1

2

cos x

1 R3 , cos x sec x cos x

cosec x ⋅ cotx + sec2x cosec2x cos 2 x

Applying R3 →

f (x) = cos x cos x cos x sec x cos x 2

2

cosec x ⋅ cot x + sec2 x cosec2 x cos x

326 Definite Integration Applying R1 → R1 − R3 ⇒ f (x) cosec x ⋅ cot x + sec2x − cos x 0 0 = cos x cos 2 x cos 2 x sec x cos x

⇒

cosec2x

a

∴

∫0

3

2

f (x) dx = −   Q  

π /2

∫0

2

=∫

∫0

  π /2 ∫ 0 f (x) dx = −   

5

n +2 2 2 sinm x ⋅ cos n x dx = m +n+2 2 2 3 1 6 1  ⋅ ⋅ 2 2 2 2  +  2 2 7  2 2 

    

  2 π 8  1 /2 ⋅ π  15π + 32 π =− +   = −  = −  + 5 3 1 2 60  4 15    2⋅ ⋅ ⋅ ⋅ π    2 2 2

19. f (x) = ∫

x 1

ln t dt for x > 0 1+ t f (1 / x) = ∫

Now,

1/ x 1

[given]

ln t dt 1+ t

Now, =∫

 1 f (x) + f   = ∫  x

x

ln t dt + 1 (1 + t )

⇒ ⇒ ⇒

t =4 − a − a t = 4 − 2a t a =2 − 2

and

t = b − (4 − b)

⇒

t = 2b − 4 t = b −2 2 t b =2 + 2

⇒ ⇒

2 + t/ 2

g (x) dx

1  t 1  g 2 −  + g 2 +  2 2 2 

t  2

[given]

We get g (2 + t / 2) − g (2 − t / 2) > 0, ∀ t > 0 F ′ (t ) > 0, ∀ t > 0

So,

Hence, F (t ) increases with t, therefore F (t ) increases as (b − a ) increases.

21. Let I n = ∫

1 0

ex (x − 1)n dx

Put x − 1 = t In = ∫

⇒

dx = dt

0 t+1 −1

e

⋅ t ndt = e ∫

= e 0 − (−1)n e−1 − n 

ln t ∫ 1 t (1 + t ) dt x

= (−1)n + 1 − ne ∫ ⇒

0 n t −1

t e dt

0 −1

0

∫ −1 t

e dt 

n −1 t

t n − 1et dt

I n = (−1)n + 1 − nI n − 1

…(i)

1 x

For n = 1, I1 = ∫ e (x − 1) dx = [e (x x

0

− 1)]10

1

− ∫ ex dx 0

= e1 (1 − 1) − e0 (0 − 1) − [ex ] 10 = 1 − (e − 1) = 2 − e Therefore, from Eq. (i), we get I 2 = (−1)2 + 1 − 2I1 = − 1 − 2(2 − e) = 2e − 5

Put x = e,

20. Let t = b − a and a + b = 4

∫0

0 = e  [t net ] −01 − n ∫ t n − 1et dt   −1

x ln t 1 1 t ) ln t dt = ∫ dt = [(ln t )2]1x = (ln x)2 1 2 2 t (1 + t ) t

1  1 1 f (e) + f   = (ln e)2 =  e 2 2

g (x) dx +

g (x) dx

Now, 2 − t / 2 < 2 + t / 2 ∴ t > 0

x (1 + 1

∫0

dg > 0, ∀ x ∈ R dx

Again,

∴

x ln t ln u du = ∫ dt 1 u (u + 1 ) 1 t (1 + t )

x

0

2 + t/ 2

g (x) dx +

[using Leibnitz’s rule] 1  t 1  t = g 2 +  − g 2 −  2  2 2  2

Put t = 1 / u ⇒ dt = (−1 / u 2) du x ln (1 / u ) (−1 ) ⋅ 2 du ∴ f (1 / x) = ∫ 1 1 + 1 /u u =∫

0 2 − t/ 2

For t > 0, F ′ (t ) = −

(sin x + cos x) dx 2

F (x) = ∫

Let

2 − t/ 2

5

m+ 1

π /2

0

3

= − sin x − cos x (1 − sin x) = − sin x − cos x π /2

b

0

= (cosecx ⋅ cot x + sec x − cos x) ⋅ (cos x − cos x) ⋅ cos x sin 2 x + cos3 x − cos3 x ⋅ sin 2 x  2 2 =−  ⋅ cos x ⋅ sin x sin 2 x ⋅ cos 2 x   2

[given]

Now, ∫ g (x) dx + ∫ g (x) dx

cos x

2

a 0 ⇒ t >0 2 2

Again,

Hence proved. [given]

and

I3 = (−1)3 + 1 − 3I 2 = 1 − 3(2e − 5) = 16 − 6e

Hence, n = 3 is the answer.

22. Since, f is continuous function and ∫

x 0

f (t ) dt → ∞ ,

as| x|→ ∞. To show that every line y = mx intersects the x curve y2 + ∫ f (t ) dt = 2 0

Y (0,√2) A O B (0,–√2)

(Xp,0)

X

Definite Integration 327 At x = 0, y = ± 2 Hence, (0, 2 ), (0, − 2 ) are the point of intersection of the curve with the Y-axis. As x → ∞, ∫ x

∫0

x 0

f (t ) dt → ∞ for a particular x (say xn ), then

x

0

23. Given, f (x) = ∫ [2(t − 1) (t − 2) + 3 (t − 1) (t − 2) ] dt 2

1

=

lim 2n n ∑ n → ∞ r=1 n 2 + r 2

0

F ′ (π ) = f (π ) = − 6 π

0

= [cos x f (x)]0π +

… (i)

{by integration by parts} = (− 1) (− 6) − f (0) + [(sin x) F (x)]π0

⇒

π

∫ 0 F (x) cos x dx

= 6 − f (0) = 2 f (0) = 4

(given)

1 2 2n       log 1 + n  + log 1 + n  + ... + log 1 + n    

 (n + 1)1/3 (n + 2)1/3 (2n )1/3  1. Let p = lim  + +…+  4 3 4 3 / / n→ ∞  n n n 4/3  n→ ∞

n

(n + r )1/3 n 4/3 r =1

∑

⇒ ⇒

log l = lim

n→ ∞

1 n

2n



r

∑ log 1 + n 

r =1

2

log l = ∫ log (1 + x) dx 0

2

⇒

1/3

r  1/3  n n 1 +  1 n = lim ∑ = lim 4/3 n→ ∞ n→ ∞ n n r =1

pn   p 1  r Q lim ∑ f   = ∫ 0 f (x)dx n→∞ n n r =1  

Taking log on both sides, we get 1  1  2 2 n     log l = lim log 1 +  1 +  ... 1 +   n→ ∞ n  n  n n    1 ⇒ log l = lim n→ ∞ n

Topic 4 Limits as the Sum

= lim

2 dx 1 =∫ 0 n 1 + x2

  n + 1  n + 2  n + 2n   n = lim     K    n   n→ ∞   n   n 

π

π

⋅

1

∫ 0 f (x) sin x dx + ∫ 0 F (x) cos x dx

0

 r 1+    n

2

1

π

− ∫ F (x) cos x dx +

r =1

1

 (n + 1) ⋅ (n + 2) ... (n + 2n ) n = lim    n→ ∞  n 2n

∫ 0 F (x) cos xdx

π

n→ ∞

2n

∑

(n + 1) ⋅ (n + 2) K (3n ) n 3. Let l = lim    n→ ∞  n 2n

( f ′ (x) + F (x)) cos x dx

= ∫ f ′ (x) cos xdx +

lim

1

x

F (x) = ∫ f (t )dt , where f (π ) = − 6

∫0

0

= [tan −1 x]20 = tan −1 2

and F : [0, π ] → R ,

Now,

 3 (1 + x)4/3  4

lim  n  n n n + + + ....+ 2   n → ∞  n 2 + 12 n 2 + 22 n 2 + 32 n + (2n )2

=

7/5

24. It is given that, for functions f : R → R

π



r =1

=

+

∴ f (x) attains maximum at x = 1 and f (x) attains 7 minimum at x = . 5

⇒

n

r  ∑ 1 + n  r =1

Now, as per integration as limit of sum. r 1 Let = x and = dx n n



lim  n 1 n n + 2 + 2 + ...+   2 2 2 2  n→ ∞ n +1 5n  n +2 n +3

= (x − 1) (x − 2)2 (5x − 7)

1

1

2. Clearly,

= (x − 1) (x − 2)2 [2 (x − 2) + 3 (x − 1)] –

 r

∑ f  n  = ∫0 f (x) dx

3 3 3 = (24/3 − 1) = (2)4/3 − 4 4 4

f ′ (x) = [2 (x − 1) (x − 2)3 + 3 (x − 1)2(x − 2)2] ⋅ 1 − 0

+

n

1

=

2

3

 1 Q lim  n → ∞ n

1

So, p = ∫ (1 + x)1/3 dx

f (t ) dt = 2 and for this value of x, y = 0

The curve is symmetrical about X-axis. Thus, we have that there must be some x, such that f (xn ) = 2. Thus, y = mx intersects this closed curve for all values of m.

∴

Then, upper limit of integral is 1 and lower limit of integral is 0.

1/3

[Q n → ∞]

⇒

  1 log l = log (1 + x) ⋅ x − ∫ ⋅ x dx + x 1  0 2 x+ 1 −1 2 log l = [log (1 + x) ⋅ x]0 − ∫ dx 0 1+ x

⇒

2 1  log l = 2 ⋅ log 3 − ∫ 1 −  dx 0  1 + x

⇒

log l = 2 ⋅ log 3 − [x − log 1 + x ]20

328 Definite Integration ⇒ ⇒ ⇒

log l = 2 ⋅ log 3 − [2 − log 3] log l = 3 ⋅ log 3 − 2 log l = log 27 − 2

∴

l = elog 27 − 2 = 27 ⋅ e− 2 =

n

n

    3 3 3 1 + 2 + 3 + ... + n , lim  n→ ∞   1 1 1 7/3  + + ... +  n   (an + 1)2 (an + 2)2 (an + n)2   a ∈ R,|a| > 1 1/3 n n  r 1 Σ   Σ (r1/3 ) r = 1  n n r =1 = lim = lim n 1 1 1 n→ ∞ n → ∞ 7/3 n Σ Σ n 2 r =1  r = 1 (an + r )2 r n a +   n 1

1/3

=

∫

0

dx (a + x)2

= 54,

(given)

3 4/3 1 [x ]0 3 /4 4 = 54 = 54 ⇒ 1 1 1  1  − + a+1 a − x + a   0 3 1 = 4 × 54 a (a + 1)

⇒

⇒

⇒ a 2 + a = 72 2 ⇒ a + 9a − 8a − 72 = 0 ⇒ a (a + 9) − 8(a + 9) = 0 ⇒ (a − 8) (a + 9) = 0 ⇒ a = 8 or− 9 Hence, options (c) and (d) are correct. 5. PLAN

 r 2∑    n

Converting Infinite series into definite Integral h( n) i.e. lim n→ ∞ n h ( n) 1  r lim ∑ f  n  = ∫ f(x )dx n→ ∞ n r = g ( n) g ( n) n→ ∞ n lim

r is replaced with x. n Σ is replaced with integral.

where,

n→ ∞

1 1a + 2a + K + n a = a −1 (n + 1) {(na + 1) + (na + 2) + K + (na + n )} 60

1 60

a

n→ ∞

2 ⋅ [xa+ 1 ] 10 1 = (2a + 1) ⋅ (a + 1) 60 2 1 = ⇒ (2a + 1) (a + 1) = 120 (2a + 1) (a + 1) 60

⇒

⇒ 2a 2 + 3a + 1 − 120 = 0 ⇒ 2a 2 + 3a − 119 = 0 −17 ⇒ (2a + 17) (a − 7) = 0 ⇒ a = 7, 2 n

n 2 + + k2 n kn k=0

6. Given, S n = ∑

    1  1  < lim =∑ ⋅ n  k k2  n → ∞ k=0 1 + +    n n2  n

=∫

1 0

    1 1 ∑ n  k k  2  k=0 1 + +     n  n  n

1 dx 1 + x + x2 1

 2 1    2  tan −1  = x +     2   0 3 3  2  π π π ⋅ −  = 3  3 6 3 3 π Similarly, Tn > 3 3 =

i.e. S n
0), is 1 square unit. Then, k is (2019 Main, 10 Jan I)

(a)

3

(b)

1 3

(c)

2 3

(d)

3 2

3. The area (in sq units) of the region {(x, y) : y2 ≥ 2x and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is 4 3 4 2 (c) π − 3

(2016 Main)

8 3 π 2 2 (d) − 2 3

(a) π −

(b) π −

the parabola y2 = 8x touch the circle at the points P,Q and the parabola at the points R, S. Then, the area (in sq units) of the quadrilateral PQRS is (2014 Adv.)

(b) 6

(c) 9

end points of latusrectum to the ellipse (a) 27/4 sq units (c) 27/2 sq units

(d)15

5. The area of the equilateral triangle, in which three coins of radius 1 cm are placed, as shown in the figure, is

x2 y2 + = 1, is 9 5 (2003, 1M)

(b) 9 sq units (d) 27 sq units

7. The area (in sq units) bounded by the curves y = | x | − 1 and (2002, 2M) y = − | x | + 1 is (a) 1 (c) 2 2

(b) 2 (d) 4

8. The triangle formed by the tangent to the curve f (x) = x2 + bx − b at the point (1,1) and the coordinate axes, lies in the first quadrant. If its area is 2 sq units, then the value of b is (2001, 2M) (a) – 1

4. The common tangents to the circle x2 + y2 = 2 and

(a) 3

6. The area of the quadrilateral formed by the tangents at the

(b) 3

(c) – 3

(d) 1

Objective Question II (One or more than one correct option) 9. Let P and Q be distinct points on the parabola y2 = 2x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of ∆OPQ is 3 2, then which of the following is/are the (2015 Adv.) coordinates of P ? (a) (4 , 2 2 ) (b) (9 , 3 2 )

1 1  (c)  ,  4 2

(d) (1, 2 )

Integer & Numerical Answer Type Questions 10. A farmer F1 has a land in the shape of a triangle with

(2005, 1M)

(a) (6 + 4 3 ) sq cm (c) (7 + 4 3 ) sq cm

(b) (4 3 − 6) sq cm (d) 4 3 sq cm

vertices at P(0, 0), Q(1, 1) and R(2, 0). From this land, a neighbouring farmer F2 takes away the region which lies between the sides PQ and a curve of the form y = xn (n > 1). If the area of the region taken away by the farmer F2 is exactly 30% of the area of ∆PQR, then the value of n is .................... . (2018 Adv.)

330 Area y = f (x). If x ∈ (− 2, 2), the equation implicitly defines a unique real-valued differentiable function y = g (x), (2008, M) satisfying g (0) = 0.

Fill in the Blanks 11. The area of the triangle formed by the positive X-axis

and the normal and the tangent to the circle x2 + y2 = 4 (1989, 2M) at (1, 3 ) is … .

14. If f (− 10 2 ) = 2 2 , then f′′ (− 10 2 ) is equal to

12. The area enclosed within the curve|x| + | y| = 1 is ....... . (1981, 2M)

Analytical & Descriptive Question

(a)

4 2

4 2 73 32

(c)

4 2

(d) −

73 3

4 2 73 3

15. The area of the region bounded by the curve y = f (x), the X-axis and the lines x = a and x = b, where − ∞ < a < b < − 2, is

1 ) be the vertices of a 3 triangle. Let R be the region consisting of all those points P inside ∆ OAB which satisfy d (P , OA ) ≥ min { d (P , OB), d (P , AB)}, where d denotes the distance from the point to the corresponding line. Sketch the (1997C, 5M) region R and find its area.

13. Let O (0, 0), A (2, 0) and B (1,

(a) ∫

x

b

(b) − (c) ∫

x

b

∫a 3[{f (x)}2 − 1] dx + bf (b) − af (a) x

b

− 1]

a 3[{f (x)}2

(d) −

16.

dx + bf (b) − af (a )

− 1]

a 3[{f (x)}2

Passage Based Questions Consider the functions defined implicity by the equation y3 − 3 y + x = 0 on various intervals in the real line. If x ∈ (−∞ , − 2) ∪ (2, ∞ ), the equation implicitly defines a unique real-valued differentiable function

(b) −

73 32

dx − bf (b) + af (a )

x

b

∫a 3[{f (x)}2 − 1] dx − bf (b) + af (a)

1

∫− 1 g ′ (x) dx is equal to (a) 2 g(− 1)

(c) − 2 g(1)

(b) 0

(d) 2 g(1)

Topic 2 Area Using Integration Objective Questions I (Only one correct option) 1. Let the functions f : R → R and g : R → R be defined by 1 x −1 (e + e1 − x ). 2 Then the area of the region in the first quadrant bounded by the curves y = f (x), y = g (x) and x = 0 is f (x) = ex − 1 − e−|x − 1|and g (x) =

(2020 Adv.)

1 (a) (2 − 3 ) + (e − e−1 ) 2 1 (c) (2 − 3 ) + (e + e−1 ) 2

(b) (2 + (d) (2 +

1 3 ) + (e − e−1 ) 2 1 3 ) + (e + e−1 ) 2

2. Consider the region R = {(x, y) ∈ R : x ≤ y ≤ 2x }. If a 2

2

line y = α divides the area of region R into two equal parts, then which of the following is true? (2020 Main, 2 Sep II)

(a) α3 − 6α 2 + 16 = 0 (c) 3α 2 − 8α + 8 = 0

(b) 3α 2 − 8α3 / 2 + 8 = 0 (d) α3 − 6α3 / 2 − 16 = 0

3. The

area (in sq. units) of the region A = {(x, y) : (x − 1)[x] ≤ y ≤ 2 x, 0 ≤ x ≤ 2}, where [t ] denotes the greatest integer function, is (2020 Main, 5 Sep II)

1 8 (a) 2− 2 3 8 (c) 2−1 3

4 (b) 2+1 3 1 4 (d) 2− 2 3 (2020 Adv.)

7 (b) 8 log e 2 − 3 (d) 16 log e 2 − 6

1 y2 = 4λx and the line y = λx, λ > 0, is , then λ is equal to 9

(2019 Main, 12 April II)

(a) 2 6

(b) 48

(c) 24

(d) 4 3

6. If

the area (in sq units) of the region {(x, y): y2 ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0} is a 2 + b, then (2019 Main, 12 April I) a − b is equal to (a)

10 3

(b) 6

(c)

8 3

(d) −

2 3

7. The area (in sq units) of the region bounded by the curves y = 2x and y = | x + 1|, in the first quadrant is

(2019 Main, 10 April II)

(a)

3 2

(b) log e 2 +

3 1 (c) 2 2

(d)

3 1 − 2 log e 2

8. The area (in sq units) of the region   y2 A = (x, y) : ≤ x ≤ y + 4 is 2   53 (b) 3

(a) 30

(2019 Main, 9 April II)

(c) 16

(d) 18

9. The

area (in sq units) A = {(x, y) : x2 ≤ y ≤ x + 2} is

(a)

4. The area of the region {(x, y) : xy ≤ 8,1 ≤ y ≤ x2} is 14 (a) 8 log e 2 − 3 14 (c) 16 log e 2 − 3

5. If the area (in sq units) bounded by the parabola

13 6

(b)

9 2

(c)

of

the

region

(2019 Main, 9 April I)

31 6

(d)

10 3

10. Let S (α ) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α} and A(α ) is area of the region S(α ). If for λ, 0 < λ < 4, A (λ ) : A (4) = 2 : 5, then λ (2019 Main, 8 April II) equals 1

4 3 (a) 2   25 

1

2 3 (b) 4   5

1

4 3 (c) 4   25 

1

2 3 (d) 2   5

Area 331 11. The tangent to the parabola y2 = 4x at the point where it intersects the circle x2 + y2 = 5 in the first quadrant, passes through the point 1 (a)  , 4

3 (b)  , 4

3  4

7  4

1 (c)  − ,  3

1 (d)  − ,  4

4  3

1  2

53 (a) 6

(2019 Main, 8 April I)

59 (c) 6

(b) 8

(d)

26 3

13. The area (in sq units) of the region bounded by the

parabola, y = x + 2 and the lines, y = x + 1, x = 0 and (2019 Main, 12 Jan I) x = 3, is 2

(a)

15 2

(b)

17 4

(c)

21 2

(d)

15 4

14. The area (in sq units) in the first quadrant bounded by the parabola, y = x + 1, the tangent to it at the point (2, 5) and the coordinate axes is (2019 Main, 11 Jan II) 2

(a)

14 3

(b)

187 24

(c)

8 3

(d)

37 24

15. The area (in sq units) of the region bounded by the curve x2 = 4 y and the straight line x = 4 y − 2 is

(2019 Main, 11 Jan I)

7 (a) 8

9 (b) 8

5 (c) 4

3 (d) 4

16. The area of the region A = {(x, y); 0 ≤ y ≤ x| x| + 1 and − 1 ≤ x ≤ 1} in sq. units, is 4 (b) 3

(a) 2

(2019 Main, 9 Jan II)

1 (c) 3

2 (d) 3

y = x2 − 1, the tangent at the point (2, 3) to it and the (2019 Main, 9 Jan I) Y -axis is 8 3

(b)

56 3

(c)

32 3

(d)

14 3

18. Let g (x) = cos x2, f (x) = x and α , β (α < β) be the roots of the quadratic equation 18x2 − 9πx + π 2 = 0. Then, the area (in sq units) bounded by the curve y = ( gof )(x) and the lines x = α, x = β and y = 0, is (2018 Main) 1 ( 3 − 1) 2 1 (c) ( 3 − 2 ) 2

{(x, y) : x ≥ 0, x + y ≤ 3, x2 ≤ 4 y and y ≤ 1+ (b)

3 2

(c)

7 3

x } is (d)

5 2

(a)

1 6

of

the

(b)

4 3

(c)

3 2

(d)

5 3

21. The area (in sq units) of region described by (x, y) y ≤ 2x 2

and y ≥ 4x − 1 is (a)

7 32

(b)

(2015 JEE Main)

5 64

(c)

15 64

(c)

π 2 − 2 3

(2014 Main)

(d)

π 2 + 2 3

(2014 Adv.)

(b) 2 2 ( 2 − 1) (d) 2 2 ( 2 + 1)

24. The area (in sq units) bounded by the curves

y = x, 2 y − x + 3 = 0, X-axis and lying in the first (2013 Main, 03) quadrant, is

(a) 9

(b) 6

(c) 18

(d)

27 4

25. Letf : [−1, 2] → [0, ∞ ) be a continuous function such that f (x) = f (1 − x), ∀x ∈ [−1, 2]. If R1 = ∫

2 −1

xf (x) dx and R2 are

the area of the region bounded by y = f (x), x = − 1, x = 2 and the X-axis. Then, (2011) (a) R1 = 2R2 (c) 2R1 = R2

(b) R1 = 3R2 (d) 3R1 = R2

26. If the straight line x = b divide the area enclosed by y = (1 − x)2, y = 0 and x = 0 into two parts R1 (0 ≤ x ≤ b) 1 and R2(b ≤ x ≤ 1) such that R1 − R2 = . Then, b equals 4 3 4

(a)

(b)

1 2

(c)

1 3

(d)

1 4

(2011)

area of the region between the curves 1 + sin x 1 − sin x and bounded by the y= and y = cos x cos x π lines x = 0 and x = is 4 (2008, 3M) 2 −1

(a) ∫

0

(c) ∫

0

t

(1 + t ) 1 − t 2+1 4t 2

2

(1 + t 2 ) 1 − t 2

2 −1

4t

dt

(b) ∫

0

dt

(d) ∫

(1 + t ) 1 − t 2 2+1 t

0

(1 + t 2 ) 1 − t 2

2

dt dt

(d)

9 32

1 is 4

(2005, 1M)

1 2 sq unit (b) sq unit 3 3

(c)

1 sq unit 4

(d)

1 sq unit 5

29. The area enclosed between the curves y = ax2 and x = ay2 (a > 0) is 1 sq unit. Then, the value of a is (2004, 1M)

region {(x, y)} ∈ R2 : y ≥ |x + 3|, (2016 Adv.) 5 y ≤ (x + 9) ≤ 15} is equal to

20. Area

π 4 − 2 3

(a) 4( 2 − 1) (c) 2 ( 2 + 1)

(a) (2017 Main)

59 12

(b)

23. The area enclosed by the curves y = sin x + cos x and

and y =

19. The area (in sq units) of the region (a)

π 4 + 2 3

28. The area bounded by the curves y = (x − 1)2, y = (x + 1)2

1 ( 3 + 1) 2 1 (d) ( 2 − 1) 2

(b)

(a)

(a)

27. The

17. The area (in sq units) bounded by the parabola

(a)

A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 − x } is

 π is y = | cos x − sin x|over the interval 0,  2 

12. The area (in sq units) of the region A = {(x, y) ∈ R × R|0 ≤ x ≤ 3, 0 ≤ y ≤ 4, y ≤ x2 + 3x} is

22. The area (in sq units) of the region described by

1 (a) 3

1 (b) 2

(c) 1

1 (d) 3

30. The area bounded by the curves y = f (x),the X-axis and the ordinates x = 1 and x = b is (b − 1) sin (3b + 4). Then, (1982, 2M) f (x) is equal to (a) (x − 1) cos (3x + 4) (b) 8sin (3x + 4) (c) sin (3x + 4) + 3(x − 1) cos (3x + 4) (d) None of the above

332 Area 31. The slope of tanget to a curve y = f (x) at [x, f (x)] is 2x + 1. If the curve passes through the point (1, 2), then the area bounded by the curve, the X-axis and the line x = 1 is (a)

3 2

(b)

4 3

(c)

5 6

(d)

1 12

Objective Questions II (One or more than one correct option) divides the area of region R = {(x, y) ∈ R2 : x3 ≤ y ≤ x, 0 ≤ x ≤ 1} into two equal (2017 Adv.) parts, then (b) α 4 + 4α 2 − 1 = 0 1 (d) 0 < α ≤ 2

2

y = e− x , y = 0, x = 0 and x = 1. Then, 1 e 1 1 (c) S ≤  1 +  e 4

(b) S ≥ 1 − (d) S ≤

bounded by the Y-axis and the curve jπ ( j + 1)π . Then, show that xe = sin by, ≤ y≤ b (b) S 0 , S1 , S 2,... , S n are in geometric progression. Also, find their sum for a = − 1 and b = π . (2001, 5M) ay

2x, |x| ≤ 1 f (x) =  2 x + ax + b, |x| > 1 Then, find the area of the region in the third quadrant bounded by the curves x = −2 y2 and y = f (x) lying on the left on the line 8x + 1 = 0. (1999, 5M)

42. Let C1 and C 2 be the graphs of functions y = x2 and

33. If S be the area of the region enclosed by (a) S ≥

region

41. If f (x) is a continuous function given by

32. If the line x = α (a) 2α 4 − 4α 2 + 1 = 0 1 (c) < α < 1 2

40. Let b ≠ 0 and for j = 0,1,2..., n. If S j is the area of the

(2012)

1 e

1 1  1  + 1 −  e  2 2

y = 2x, 0 ≤ x ≤ 1, respectively. Let C3 be the graph of a function y = f (x), 0 ≤ x ≤ 1, f (0) = 0. For a point P on C1, let the lines through P, parallel to the axes, meet C 2 and C3 at Q and R respectively (see figure). If for every position of P(on C1) the areas of the shaded regions OPQ and ORP are equal, then determine f (x). (1998, 8M) Y

34. Area of the region bounded by the curve y = e and lines x

x = 0 and y = e is (a) e − 1 1

(c) e − ∫ ex dx 0

(d)

35. For which of the following values of m, is the area of the

region bounded by the curve y = x − x2 and the line 9 (1999, 3M) y = mx equals ? 2

(a) – 4

(b) –2

(c) 2

(d) 4

4a 2 4a 1  f (−1)  2 3a + 3a  36. If 4b2 4b 1  f (1)  = 3b2 + 3b ,   2    4c2 4c 1  f (2)  3c + 3c    f (x) is a quadratic function and its maximum value occurs at a point V. A is a point of intersection of y = f (x) with X-axis and point B is such that chord AB subtends a right angle at V. Find the area enclosed by f (x) and chord AB. (2005, 5M)

37. Find the area bounded by the curves x = y, x = − y and y2 = 4x − 3.

2

(2005, 4M)

38. A curve passes through (2,0) and the slope of tangent at (x + 1) + y − 3 . (x + 1) 2

point P (x, y) equals

(0,0) O C 3

(1,0)

X

R

43. Letf (x) = max { x , (1 − x) , 2x (1 − x)}, where 0 ≤ x ≤ 1. 2

2

Determine the area of the region bounded by the curves (1997, 5M) y = f (x), X-axis, x = 0 and x = 1. the bounded region enclosed between the parabolas x2 is maximum. y = x − bx2 and y = (1997C, 5M) b

45. If An is the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x =

π . 4

Then, prove that for n > 2 , An + An + 2 = and deduce

1 1 . < An < 2n + 2 2n − 2

1 n+1 (1996, 3M)

46. Consider a square with vertices at (1,1), (–1, 1), (–1, –1) and (1, –1). If S is the region consisting of all points inside the square which are nearer to the origin than to any edge. Then, sketch the region S and find its area. (1995, 5M)

Find the equation of the curve and area enclosed by the curve and the X-axis in the fourth quadrant. (2004, 5M)

39. Find the area of the region bounded by the curves y = x2, y = | 2 − x2| and y = 2 , which lies to the right of the line x = 1.

P

44. Find all the possible values of b > 0, so that the area of

Analytical & Descriptive Questions

2

C1

Q

e

∫1 ln (e + 1 − y) dy e ∫1 ln y dy

(1,1)

C2

(2009)

(b)

(1/2,1)

(0,1)

47. In what ratio, does the X-axis divide the area of the region bounded by the parabolas y = 4x − x2 and (1994, 5M) y = x2 − x ?

48. Sketch the region bounded by the curves y = x2 and (2002, 5M)

y = 2 / (1 + x2). Find its area.

(1992, 4M)

Area 333 49. Sketch the curves and identify the region bounded by x = 1 /2, x = 2, y = log x and y = 2x . Find the area of this (1991, 4M) region.

55. Sketch the region bounded by the curves y = 5 − x2 and y = |x − 1|and find its area.

56. Find the area of the region bounded by the X-axis

50. Compute the area of the region bounded by the curves log x , where log e = 1. y = ex log x and y = (1990, 4M) ex 51. Find all maxima and minima of the function y = x (x − 1)2, 0 ≤ x ≤ 2 . Also, determine the area bounded by the curve (1989, 5M) y = x (x − 1)2, the Y-axis and the line x = 2 .

and the curves defined by y = tan x, − y = cot x,

53. Find the area bounded by the curves

x2 + y2 = 25, 4 y = |4 − x2|and x = 0 above the X-axis. (1987, 6M)

54. Find the area bounded by the curves x2 + y2 = 4, x2 = − 2 y and x = y.

(1986, 5M)

π π ≤x≤ . 6 3

π π and ≤x≤ 3 3 (1984, 4M)

57. Find the area bounded by the X-axis, part of the curve

8  y = 1 + 2 and the ordinates at x = 2 and x = 4. If the  x  ordinate at x = a divides the area into two equal parts, then find a. (1983, 3M)

52. Find the area of the region bounded by the curve

C: y = tan x, tangent drawn to C at x = π / 4 and the X-axis. (1988, 5M)

( 1985, 5M)

58. Find the area bounded by the curve x2 = 4 y and the straight line x = 4 y − 2. et + e– t et – e– t is a point on the 59. For any real t, x = ,y= 2 2 hyperbola x2 – y2 = 1. Find the area bounded by this hyperbola and the lines joining its centre to the points corresponding to t1 and – t1. (1982, 3M)

Answers Topic 1 1. (a) 5. (a) 9. (a,d)

2. (b) 6. (d) 10. (4)

3. (b)

4. (d)

7. (b)

8. (c)

11. 2 3 sq units

12. 2 sq units

13. (2 − 3 ) sq unit

14. (b)

15. (a)

16. (d)

17 sq unit 27

47. 121 : 4

1. (a)

2. (b)

3. (a)

4. (c)

5. (c)

6. (b)

7. (d)

8. (d)

9. (b)

10. (c)

11. (b)

12. (c)

13. (a)

14. (d)

15. (b)

16. (a)

17. (a)

18. (a)

19. (d)

20. (c)

21. (d)

22. (a)

23. (b)

24. (a)

25. (c)

26. (b)

27. (b)

28. (a)

29. (a)

30. (c)

31. (c)

33. (b, d)

34. (b, c, d)

35. (b, d)

32. (a, c) 125 sq units 36. 3

38. y = x 2 – 2 x,

4 sq units 3

1 sq unit 3

 761 41.   sq units  192 43.

Topic 2

37.

 20 − 12 2  39.   sq units 3  

 π (1 + e ) (en + 1 − 1 )  40.  ⋅ 2 e − 1   (1 + π )

42. f ( x ) = x 3 − x 2, 0 ≤ x ≤ 1

44. b = 1

46.

2  48.  π −  sq units  3

4 − 2 5 49.  − log 2 + 2  log 2

1  (16 2 − 20 ) sq units 3 

 e 2 − 5 3  sq units  sq units 50.  2  4e 

4 10   51. y max = , y min = 0, sq units   27 3 1   52.  log 2 −  sq units   4   1  54.  − π  sq units 3  1   56.  loge 3 sq units  2 59.

  4  53. 4 + 25 sin −1    sq units  5    5π 1 55.  −  sq units  4 2 9 58. sq units 57. 2 2 8

e 2t1 − e −2t1 1 2t1 − (e − e −2t1 − 4t1 ) 4 4

Hints & Solutions Topic 1 Area Based on Geometrical Figures Without Using Integration

Now, the required area is the area of shaded region, i.e. Y

1. Equation of given parabola y = x2 – 1 According to

A (2, 2)

symmetry let, the coordinate of A (– a , 0), B(a , 0) C (a , a 2 – 1) and D (– a , a 2 – 1). ∴Area of rectangle P (a ) = 2a (a 2 – 1)

x 2 + y 2 = 4x X′

(0, 0)

y2 =

Y

B(a, 0)

A(–a, 0)

(1, 0)

O

D (–a, a2–1)

2 Area of a circle Required area = − ∫ 2x dx 0 4 2 2  x3/ 2  π (2)2 = − 2 ∫ x1/ 2dx = π − 2   0 4  3 /2  0 8 2 2  =π− [2 2 − 0] =  π −  sq unit  3 3

X

C(a, a2–1) (0, –1)

4.

Now, for maxima P ' (a ) = 0 ⇒ 2(a – 1) + 4a = 0 1 units ⇒ 3a 2 = 1 ⇒ a = 3 2

2

∴ Area of largest rectangle is 2 1 4  sq. units  – 1 =  3 3 3 3

2. We know that, area of region bounded by the parabolas 16 (ab) sq units. 3 On comparing y = kx2 and x = ky2 with above equations, 1 1 we get 4a = and 4b = k k 1 1 and b = ⇒ a= 4k 4k ∴ Area enclosed between y = kx2 and x = ky2 is 16  1   1  1    = 2     3 4k 4k 3k 1 = 1 [given, area = 1 sq.unit] ⇒ 3k2 1 1 ⇒ k2 = ⇒ k=± 3 3 1 [Q k > 0] k= ⇒ 3 x2 = 4ay and y2 = 4bx is

3. Given equations of curves are y2 = 2x which is a parabola with vertex (0, 0) and axis parallel to X-axis. . ..(i) And x2 + y 2 = 4 x which is a circle with centre (2, 0) and radius = 2 ...(ii) On substituting y2 = 2x in Eq. (ii), we get x2 + 2x = 4x ⇒ x2 = 2x ⇒ x = 0 or x = 2 [using Eq. (i)] ⇒ y = 0 or y = ± 2

2x

Y′

y=x2–1

(–1, 0)

X

B (2,0)

PLAN (i) y = mx + a / m is an equation of tangent to the parabola y 2 = 4ax . (ii) A line is a tangent to circle, if distance of line from centre is equal to the radius of circle. (iii) Equation of chord drawn from exterior point ( x 1, y1 ) to a circle/parabola is given by T = 0. 1 (iv) Area of trapezium = (Sum of parallel sides) 2

Let equation of tangent to parabola be y = mx +

2 m

It also touches the circle x2 + y2 = 2.   2  = 2 2 m 1 + m  m4 + m2 = 2 4 m + m2 − 2 = 0 2 (m − 1) (m2 + 2) = 0 [rejected m2 = − 2] m = ± 1,m2 = − 2

∴ ⇒ ⇒ ⇒ ⇒

So, tangents are y = x + 2, y = − x − 2. They, intersect at (−2, 0). R

Y P T(–2,0) X′

X

O Q Y′

S

Equation of chord PQ is −2x = 2 ⇒ x = − 1 Equation of chord RS is O = 4(x − 2) ⇒ x = 2 ∴ Coordinates of P, Q, R, S are P (−1, 1), Q (−1, − 1), R(2, 4), S (2, − 4) (2 + 8) × 3 ∴ Area of quadrilateral = = 15 sq units 2

Area 335 5. Since, tangents drawn from external points to the circle

7. The region is clearly square with vertices at the points

subtends equal angle at the centre.

(1, 0), (0, 1), (– 1, 0) and (0, – 1). Y

A

y = –IxI + 1 y = IxI –1

(0,1) O3

X′

X (1,0)

(–1,0) O1 B

(0,–1)

O2 1cm 1cm

30°

30° 2 cm E

3 cm D

Y′

C

∴ Area of square = 2 × 2 = 2 sq units

3 cm

8. Let y = f (x) = x2 + bx − b

∴

∠ O1BD = 30° OD In ∆O1BD, tan 30° = 1 BD

Y B

⇒

BD = 3 cm

Also,

DE = O1O2 = 2 cm and EC = 3 cm

Now,

BC = BD + DE + EC = 2 + 2 3

⇒

Area of ∆ABC = =

(1,1) P O

3 (BC )2 4

The equation of the tangent at P (1, 1) to the curve 2 y = 2x2 + 2bx − 2b is y + 1 = 2x ⋅ 1 + b (x + 1) − 2b ⇒ y = (2 + b) x − (1 + b) Its meet the coordinate axes at 1+ b xA = and yB = − (1 + b) 2+ b 1 ∴ Area of ∆ OAB = OA × OB 2 1 (1 + b)2 [given] =− × =2 2 (2 + b)

3 ⋅ 4 (1 + 3 )2 = (6 + 4 3 ) sq cm 4

x2 y2 + =1 9 5 To find tangents at the end points of latusrectum, we find ae.

6. Given,

ae = a 2 − b2 = 4 = 2

i.e.

4 5  b2(1 − e2) = 5 1 −  =  9 3

and

⇒ ⇒

By symmetry, the quadrilateral is a rhombus.

(1 + b)2 + 4(2 + b) = 0 (b + 3)2 = 0 Y

A

t 12 P 2 , t1

L(ae, √ b2(1 – e2)) D

O

B

⇒ b2 + 6b + 9 = 0 ⇒ b = −3

9. Since, ∠ POQ = 90°

Y

X′

X

A

X

L′ C

X′

X

O (0, 0)

Y′

So, area is four times the area of the right angled triangle formed by the tangent and axes in the Ist quadrant.  5 ∴ Equation of tangent at 2,  is  3 2 5 y x y x+ ⋅ =1 ⇒ + =1 9 3 5 9 /2 3 ∴ Area of quadrilateral ABCD [area of ∆ AOB] =4 1 9  = 4  ⋅ ⋅ 3 = 27 sq units 2 2 

Y′

⇒

Q ∴

⇒

t 22 Q 2 , t2

t1 − 0 t2 − 0 ⋅ 2 = − 1 ⇒ t1t2 = − 4 t12 t −0 2 −0 2 2 ar (∆OPQ ) = 3 2 0 0 1 1 2 t1 / 2 t1 1 = ± 3 2 2 2 t2 / 2 t2 1 1  t12t2 t1t22 −   =±3 2 2 2 2 

…(i)

336 Area ⇒

4 1 ( −4t1 + 4t2) = ± 3 2 ⇒ t1 + = 3 2 [Q t1 > 0 for P] t1 4

13. Let the coordinates of P be (x, y). Y

⇒ t12 − 3 2t 1 + 4 = 0 ⇒ (t1 − 2 2 ) (t1 − 2 ) = 0 t1 = 2 or 2 2 ⇒ ∴ P (1, 2 ) or P (4, 2 2 )

B

P

10. We have, y = xn , n > 1 Q P (0, 0) Q (1, 1) and R(2, 0) are vertices of ∆ PQR.

X′

O (0, 0)

A (2, 0)

y Y′

y=

x

Q(1,1) F2

F1 y = xn

x′

P(0,0) (1,0) R(2,0)

x

y′

∴ Area of shaded region = 30% of area of ∆ PQR 1 30 1 n ⇒ ∫0 (x − x ) dx = 100 × 2 × 2 × 1 1

 x2 xn + 1  1 3 1  3  −  =  2 − n + 1  = 10 ⇒   2 + 1 10 n 0  2 1 1 1 3 = = ⇒ n + 1 =5 ⇒ n =4 = − ⇒ n + 1 2 10 10 5

⇒

11. Equation of tangent at the point (1, 3 ) to the curve x2 + y2 = 4 is x + whose X-axis intercept (4, 0).

3y = 4

Y

Equation of line OA be y = 0. Equation of line OB be 3 y = x. Equation of line AB be 3 y = 2 − x. d (P , OA ) = Distance of P from line OA = y | 3 y − x| d (P , OB) = Distance of P from line OB = 2 | 3 y + x − 2| d (P , AB) = Distance of P from lineAB = 2 Given, d (P , OA ) ≤ min { d (P , OB), d (P , AB)}  | 3 y − x| | 3 y + x − 2| , y ≤ min   2 2   | 3 y − x| | 3 y + x − 2| and y ≤ 2 2 | 3 y − x| Case I When y ≤ [since, 3 y − x < 0] 2 x − 3y ⇒ (2 + 3 ) y ≤ x ⇒ y ≤ x tan 15° y≤ 2 | 3 y + x − 2| Case II When y ≤ , 2 2 y ≤ 2 − x − 3 y [since, 3 y + x − 2 < 0] ⇒ (2 + 3 ) y ≤ 2 − x ⇒ y ≤ tan 15°⋅ (2 − x) ⇒

y≤

Y

P (1,√3) X′

(0,0) O

A (4,0)

B (1, 1/ 3) X

P X′

Y′

Thus, area of ∆ formed by (0, 0) (1, 3 ) and (4, 0) 0 0 1 1 1 = 1 3 1 = |(0 − 4 3 )|= 2 3 sq units 2 2 4 0 1

12. The area formed by | x| + | y| = 1 is square shown as below :

Y

−x + y = 1 X'

−1

x+y=1 O

1

x+y=1

x−y=1

Y'

∴ Area of square = ( 2 )2 = 2 sq units

X

X

O (0, 0)

C (1, 0)

A (2, 0)

X

Y′

From above discussion, P moves inside the triangle as shown below : ⇒ Area of shaded region = Area of ∆OQA 1 = (Base) × (Height) 2 1 = (2) (tan 15° ) = tan 15° = (2 − 3 ) sq unit 2

14. Given, ⇒

y3 − 3 y + x = 0 dy dy 3 y2 −3 + 1 =0 dx dx

2  d 2y d 2y  dy ⇒ 3 y2 2  + 6 y   − 3 =0  dx dx2  dx 

At x = − 10 2 , y = 2 2

…(i) …(ii)

Area 337 On substituting in Eq. (i) we get dy dy 3(2 2 )2 ⋅ − 3⋅ + 1 =0 dx dx dy 1 =− ⇒ dx 21 Again, substituting in Eq. (ii), we get

So, required area is 1/ 2 log 3e + 1

∫0 =∫

2

2

=

2

d y d y  1 + 6 (2 2 ) ⋅  −  − 3 ⋅ 2 = 0 3(2 2 )2 2   21 dx dx ⇒

21 ⋅

⇒

1/ 2 log 3e + 1 0

b

b

a

a

[ f (x) ⋅ x]ba

− ∫ f ′ (x)x dx a

a b

xdx

∫a 3[{ f (x)}2 − 1]

y3 − 3 y + x = 0 y = g (x) { g (x)}3 − 3 g (x) + x = 0

y

…(iii)

g (1) + g (− 1) = 0 g (1) = − g ( − 1) I = g (1) − g (− 1) = g (1) − { − g (1)} = 2 g (1)

Topic 2 Area Using Integration 1. The given functions f : R → R and g : R → R be defined , x≥1

0, x 0

x=

6 ± 36 − 4 2

= 3 ± 2 2. Since, x-coordinate of P less than x-coordinate of point A(1, 0).

Area bounded by above two curve is, as per figure

∴ x=3 −2 2 Now, required area =∫

y2=4λx

X

y=λx

the intersection point A we will get on the solving Eqs. (i) and (ii), we get 4 λ2x2 = 4λx ⇒ x = , so y = 4. λ 4 So, A  , 4 λ 

3 −2 2 0

=2 =

O

x2 − 6 x + 1 = 0

⇒

…(i) …(ii)

A

(1 − x)2 = 4x ⇒ x2 + 1 − 2x = 4x ⇒

5. Given, equation of curves are

Y

X

Now, for point P, put value of y = 1 − x to y2 = 4x, we get

[Q Points P (1, 1), Q (2, 4)and

and

A(1,0)

x3/ 2 3 /2

2 x dx + 3 −2 2

0

1

∫3 − 2 2 (1 − x) dx 1

 x2  + x −  2 3 − 2 

2

(3 − 2 2 ) 2 4 1 (3 − 2 2 ) 3/ 2 +  1 −  − (3 − 2 2 ) +   3 2 2

4 1 1 [( 2 − 1)2]3/ 2 + − 3 + 2 2 + (9 + 8 − 12 2 ) 3 2 2 4 5 17 = ( 2 − 1)3 − + 2 2 + −6 2 3 2 2 4 = (2 2 − 3(2) + 3( 2 ) − 1) − 4 2 + 6 3 4 8 2 10 = (5 2 − 7) − 4 2 + 6 = − 3 3 3 =

=a 2+b

(given)

Area 339 8 10 and b = − 3 3 8 10 a−b= + =6 3 3

Now, the area enclosed by the region A

So, on comparing a = ∴

7. Given, equations of curves  x + 1 ,x ≥ − 1 y = 2x and y = | x + 1| =  − x − 1 , x < − 1

4

9. Given region is A = {(x, y) : x2 ≤ y ≤ x + 2}

Q The figure of above given curves is Y

4

  y2 y2  y3  ∫  ( y + 4) − 2 dy =  2 + 4 y − 6  −2 −2 16 64   4 8 8 − − + =  + 16 −     2 6  2 6 4 32 = 8 + 16 − − 2 + 8 − = 30 − 12 = 18 sq unit. 3 3 =

Now, the region is shown in the following graph

y=x+1

Y

(1,2) y=–x–1

y=x+2

x2=y

y=2x

B(2,4)

(0,1) X′

O

(–1,0)

A

X X'

In first quadrant, the above given curves intersect each other at (1, 2). 1

So, the required area = ∫ ((x + 1) − 2 ) dx x

0

1

x 2  = + x−  2 log e 2 0  2

x

  ax x Q ∫ a dx = log a + C  e  

1 2 1  = +1− +  2 log 2 log  e e 2 3 1 = − 2 log e 2 

8. Given region A = (x, y) : 

y2 =x 2 ⇒ y2 = 2x and x = y + 4 ⇒ y = x − 4

Y'

2

] dx

2

=8− …(i) …(ii)

1 9 1 1 9 − = 8 − − 3 = 5 − = sq units 2 3 2 2 2

10. Given, S (α ) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α } and A(α ) is area of the region S(α ) Y

Graphical representation of A is

y 2 =x

y2 =x 2 X

O

X

O

2

∫ [(x + 2) − x

 x2 x3  1 8 1 4 =  + 2x −  =  + 4 −  −  − 2 +     3 3 2 2 2 3  −1 

∴

X'

X'

2

−1

 y ≤ x ≤ y + 4 2 

Q

O

For intersecting points A and B Taking, x2 = x + 2 ⇒ x2 − x − 2 = 0 ⇒ x2 − 2x + x − 2 = 0 ⇒ x(x − 2) + 1(x − 2) = 0 ⇒ x = −1, 2 ⇒ y = 1, 4 So, A(−1, 1) and B (2, 4). Now, shaded area =

2

Y

(–2,0) –1

(0,2)

A(λ)

x=

y+

4

P Y'

On substituting y = x − 4 from Eq. (ii) to Eq. (i), we get (x − 4)2 = 2x 2 ⇒ x − 8x + 16 = 2x ⇒ x2 − 10x + 16 = 0 ⇒ (x − 2)(x − 8) = 0 ⇒ x = 2, 8 [from Eq. (ii)] ∴ y = − 2, 4 So, the point of intersection of Eqs. (i) and (ii) are P (2, − 2) and Q(8, 4).

x=λ λ

Clearly, A (λ ) = 2∫

0

Since, ⇒ ⇒

λ

 x3/ 2  4 x dx = 2  = λ3/ 2  3 / 2 0 3

A (λ ) 2 = , (0 < λ < 4) A (4) 5

3

λ3/ 2 2 = 43/ 2 5 λ 4 =  4  25

 2  λ ⇒   =   5  4 1/3

⇒

2

4 λ =4    25

1/3

340 Area 11. Given equations of the parabola y2 = 4x

…(i)

14. Given, equation of parabola is y = x2 + 1, which can be

and circle …(ii) x + y =5 So, for point of intersection of curves (i) and (ii), put y2 = 4x in Eq. (ii), we get x2 + 4 x − 5 = 0 2 ⇒ x + 5x − x − 5 = 0 ⇒ (x − 1)(x + 5) = 0 ⇒ x = 1, − 5 For first quadrant x = 1 , so y = 2 . Now, equation of tangent of parabola (i) at point (1, 2) is T = 0 ⇒ 2 y = 2(x + 1) ⇒ x − y + 1 = 0  3 7 The point  ,  satisfies, the equation of line  4 4

written as x2 = ( y − 1). Clearly, vertex of parabola is (0, 1) and it will open upward. y+5 Now, equation of tangent at (2, 5) is = 2x + 1 2

2

2

[QEquation of the tangent at (x1 , y1 ) is given by 1 T = 0. Here, ( y + y1 ) = xx1 + 1] 2 y = 4x − 3 y= 4x–3 Y P (2, 5)

x− y+ 1 =0

(0, 1)

12. Given, y ≤ x + 3x 2

2

2

3 9 3 9    ⇒ x +  ≥  y +  y ≤ x +  −    2 4 2 4

⇒

O

Since, 0 ≤ y ≤ 4 and 0 ≤ x ≤ 3 ∴The diagram for the given inequalities is Y

R Q (2, 0) 3, 0 4

X

Required area = Area of shaded region

y=x2+3x

2

= ∫ y(parabola) dx − (Area of ∆PQR) 0 2

= ∫ (x2 + 1) dx − (Area of ∆PQR)

y=4

0

2

–3/2 O 9 – 4

(–3, 0)

1

  x3 1 3 = + x − 2 −  ⋅ 5  3 2 4 0 

X

3 x=3

[QArea of a triangle =

and points of intersection of curves y = x2 + 3x and y = 4 are (1, 4) and (−4, 4) Now required area

1  5  8 =  + 2 − 0 −   5  3 2  4

1

 x3 3x2  3 4 dx =  3 + 2  + [4x]1 ∫   1 0 0 1 3 2+9 11 59 sq units = + + 4(3 − 1) = +8 = +8= 3 2 6 6 6 1

= ∫ (x2 + 3x)dx +

3

=

14 25 112 − 75 37 − = = 3 8 24 24

15. Given equation of curve is x2 = 4 y, which represent a parabola with vertex (0, 0) and it open upward.

y

x2 4 x+2 y= 4

Y

13. Given equation of parabola is y = x2 + 2, and the line is y = x+1

y=

y=x2 +2 y=x+1

(0,2)

1 × base × height] 2

X′

B

A –1 O

X 2

1 1

O

(3,0)

x

The required area = area of shaded region =∫

3 0

((x2 + 2) − (x + 1)) dx = ∫ 3

3 0

(x2 − x + 1) dx

 x3 x2   27 9  =  − + x =  − + 3 − 0   3 2 3 2  0 9 9 15 sq units = 9 − + 3 = 12 − = 2 2 2

Y′

Now, let us find the points of intersection of x2 = 4 y and 4y = x + 2 For this consider, x2 = x + 2 ⇒ x2 − x − 2 = 0 ⇒ (x − 2) (x + 1) = 0 ⇒ x = − 1, x = 2 1 When x = − 1, then y = 4 and when x = 2, then y = 1 1 Thus, the points of intersection are A  − 1,  and B (2, 1).  4

Area 341 Now, required area = area of shaded region =

∫−1 {y (line) − y (parabola)} dx 2

 x + 2 x2  x3  1  x2 ∫−1  4 − 4  dx = 4  2 + 2x − 3  −1 1  8  1 1  = 2 + 4 −  −  − 2 +  3  3  2 4  

=

1 4

1   1 8 − −3 =   4 2

1 9  = sq units. 5−  2  8

16. We have, A = {(x, y) : 0 ≤ y ≤ x| x|+ 1and − 1 ≤ x ≤ 1} When x ≥ 0, then 0 ≤ y ≤ x2 + 1 and when x < 0, then 0 ≤ y ≤ − x2 + 1

Now, the required region is the shaded region. y 2 y=–x2+1

y=x2+1

1

–1

x

1 y=0

[Q y = x2 + 1⇒ x2 = ( y − 1), parabola with vertex (0, 1) and y = − x2 + 1⇒ x2 = − ( y − 1) , parabola with vertex (0,1) but open downward] We need to calculate the shaded area, which is equal to 0

∫−1 (− x

2

y+3 = 2x − 1 ⇒ y = 4x − 5 2 Now, required area = area of shaded region

⇒

2

=

y1 = 3.

and

2

1

2

= ∫ (y(parabola) − y(tangent)) dx 0 2

= ∫ [(x2 − 1) − (4x − 5)] dx 0 2

2

0

0

= ∫ (x2 − 4x + 4) dx = ∫ (x − 2)2 dx =

(x − 2)3 3

2

(2 − 2)3 (0 − 2)3 8 − = sq units. 3 3 3

= 0

18. We have, ⇒ ⇒

18x2 − 9πx + π 2 = 0 18x − 6πx − 3πx + π 2 = 0 (6x − π )(3x − π ) = 0 π π x= , ⇒ 6 3 π Now, α < β α= , 6 π β= 3 2 Given, g(x) = cos x and f (x) = x 2

y = gof (x) ∴ y = g ( f (x)) = cos x Area of region bounded by x = α,x = β, y = 0 and curve y = g ( f (x)) is π /3

+ 1)dx + ∫ (x2 + 1) dx

A=

0

0

1

3

∫ cos x dx

π /6

 x   x + x = − + x +  3 3 0 − 1   3

A = [sin x]ππ //36 π π 3 1 A = sin − sin = − 3 6 2 2  3 − 1 A=   2 

  (− 1)3   1  =  0 − − + (− 1)  +   + 1 − 0    3    3   

1  4 2 4 = −  − 1 + = + = 2 3  3 3 3

19. Required area

17. Given, equation of parabola is y = x − 1, which can be 2

rewritten as x2 = y + 1 or x2 = ( y − (−1)). ⇒ Vertex of parabola is (0, − 1) and it is open upward.

=∫

1 0

(1 +

x )dx +

2

∫1

(3 − x)dx − ∫

2 0

x2 dx 4

Y y=1+√x (0, 3) (1, 2) 4y=x 2

y=x2–1 (2, 3)

(0, 1)

(2, 1) x+y=3

2 (0, –1)

y=4x–5

Equation of tangent at (2, 3) is given by T = 0 y + y1 ⇒ = x x1 − 1, where, x1 = 2 2

X′

X

(0, 0) (1, 0)(2, 0) (3, 0) Y′ 3/ 2  1

2

2

 x3    x x2  + 3x −  −   = x +  3 /2  0  2  1 12  0  2  = 1 +  +  3 5 3 2 = + − 3 2 3

1  8   6 − 2 − 3 +  −    2  12 3 5 = 1 + = sq units 2 2

342 Area and y ≥ 4x − 1 represents a region to the left of the line …(ii) y = 4x − 1 The point of intersection of the curves (i) and (ii) is (4x − 1)2 = 2x 2 ⇒ 16x + 1 − 8x = 2x ⇒ 16x2 − 10x + 11 = 0 1 1 ⇒ x= , 2 8 1  So, the points where these curves intersect are  , 1 2   1 1 and  ,  .  8 2

20. Here, {(x, y) ∈ R2 : y ≥ |x + 3|, 5 y ≤ (x + 9) ≤ 15} ∴

y ≥

x+3

 x + 3 , when x ≥ − 3 y≥   − x − 3 , when x ≤ − 3

⇒

 x + 3 , when x ≥ − 3 y2 ≥  − 3 − x, when x ≤ − 3

or Shown as

Y

y2=–x–3

y2=x+3

4x

−1

Y

1,1 2

X

0

–3

Also,

5 y ≤ (x + 9) ≤ 15

⇒ Shown as

(x + 9) ≥ 5 y and x ≤ 6

X′

1 2

−1 2

−1

−1 X

0

∴ Required area = ∫

x=6

–9

(–4,0) A

6

1  1 1  1  1  1  + 1 −  −   − 1 +  4 2 8 2  6 8 1 3 3  1 9  =  + −   4 2 8  6 8  1 15 3 9 sq units = × − = 4 8 16 32

X

22. Given, A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 − x}

Y′

∴ Required area = Area of trapezium ABCD − Area of ABE under parabola − Area of CDE under parabola 1 −3 1 = (1 + 2) (5) − ∫ − (x + 3) dx − ∫ (x + 3) dx 4 − − 3 2 −3

 y + 1 y2  −  dy  −1/ 2  4 2 1

=

(1,2)

(1,0) 0 E (–3,0) D

Y′

−1

Y C

Y

X′

(0,1)

(–1,0)

X

1

    15  (− 3 − x)3/ 2   (x + 3)3/ 2  = −  −  3 3 2  −      –3 − 4   2 2 15 2 15 2 16 15 18 3 2 = + [0 − 1] − [8 − 0] = − − = − = 2 3 2 3 3 2 3 2 3

21. Given region is {(x, y) : y2 ≤ 2x and y ≥ 4x − 1} y2 ≤ 2x repressents a region inside the parabola y2 = 2x

X

 1  y2 1 − ( y3 )1−1/ 2 =  + y 42  −1/ 2 6

∴ {(x, y) ∈ R2 : y ≥ |x + 3|, 5 y ≤ (x + 9) ≤ 15}

)

1

1 −1 , 8 2

−1 2

(0, 9/5)

4,1 B (–

1 2

O

Y

X′

y 2 = 2x

1 Y′

(–9, 0)

y=

X′

…(i)

Y′

Required area =

1 1 2 πr + 2∫ (1 − y2)dy 0 2 1

=

 1 y3  π (1)2 + 2  y −  2 3 0 

 π 4 =  +  sq units  2 3

Area 343 23.

PLAN To find the bounded area between y = f( x ) and y = g ( x ) between x = a to x = b. Y

Hence, required area

f(x)

a

O

3

g(x)

c

∴ Area bounded = =

2

25. R1 = ∫ x f (x) dx

b

b

∫a[g (x ) − f(x )]dx + ∫c [f(x ) − g (x )]dx ∫a | f(x ) − g (x )|dx

g (x) = y = | cos x − sin x| π  cos x − sin x, 0 ≤ x ≤  4 = π π sin x − cos x, ≤x≤  4 2 could be shown as

π 2

Y

π/4

O

∴ Area bounded = ∫ +

π /4

2 sin x dx +

π /2

∫π / 4 2 cos x dx …(i)

2y − x + 3 = 0

…(ii)

Y

x

2y –

x+

3

0 3= X

–3 2 Y'

On solving Eqs. (i) and (ii), we get 2 x − ( x )2 + 3 = 0

∴

b

∫0

∴

X

{(sin x + cos x) − (cos x − sin x)} dx

X'

…(iv)

f (x) dx

2R1 = R2

24. Given curves are y = x

⇒ ⇒

2 −1

(1 − x)2 dx −

1

∫b

(1 − x)2 dx =

b

y=

…(iii)

f (x) dx

From Eqs. (iii) and (iv), we get

= − 2 [cos x]π0 / 4 + 2 [sin x ⋅ n ]ππ // 24 = 4 − 2 2 = 2 2( 2 − 1) sq units and

−1

2R1 = ∫

∫π / 4 {(sin x + cos x) − (sin x − cos x)} dx 0

2

R2 = ∫

On adding Eqs. (i) and (ii), we get

0 π /2

=∫

…(ii)

26. Here, area between 0 to b is R1 and b to 1 is R2.

π/2

π /4

2

−1

∴

g(x)

g(x)

(1 − x) f (1 − x) dx

[f (x) = f (1 − x), given] Given, R2 is area bounded by f (x), x = − 1 and x = 2.

= √2 sin x + π 4

1

2 −1

R1 = ∫ (1 − x) f (x) dx

∴

y = sin x + cos x

f(x)

b

R1 = ∫

b

and

√2

b

∫ a f (x) dx = ∫ a f (a + b − x) dx

Using

c

…(i)

−1

X

f (x) = y = sin x + cos x, when 0 ≤ x ≤

Here,

3

0

 y3  = 9 + 9 − 9 = 9 sq units =  y2 + 3 y − 3  0 

g(x) f(x)

3

0

= ∫ (x2 − x1 ) dy = ∫ {(2 y + 3) − y2} dy

1

 (1 − x)3   (1 − x)3  1  −3  −  −3  = 4 b 0  

⇒

1 1 1 [(1 − b)3 − 1] + [0 − (1 − b)3 ] = 3 4 3 2 1 1 1 1 3 ⇒ − (1 − b) = − + = − ⇒ (1 − b)3 = 3 8 3 4 12 1 1 (1 − b) = ⇒ b= ⇒ 2 2 π / 4  1 + sin x 1 − sin x  27. Required area = ∫ −  dx  0 cos x cos x   ⇒−

 x 2 tan  2  1+ 2x  1 + tan π /4  2 − =∫  0 2x 1 − tan  2  x  1 + tan 2  2

( x )2 − 2 x − 3 = 0 ( x − 3) ( x + 1) = 0 ⇒ x =3 [since, x = − 1 is not possible] y=3

1 4

=∫

π /4 0

 1 + sin x Q  cos x x 2 tan 2 1−

1 − sin x  >0  cos x    x 1 + tan 2  2  dx  2x 1 − tan  2  x  1 + tan 2  2 

 x x 1 − tan   1 + tan 2 2   dx − x  1 − tan x 1 + tan    2 2

>

344 Area =∫

π /4 0

Put tan

π 8

tan x 1 x = t ⇒ sec2 dx = dt = ∫ 0 2 2 2

∫0

As

∴ Required area OABCO = Area of curve OCBDO – Area of curve OABDO 1/a  x  2 [given] ⇒ ∫ 0  a − ax  dx = 1

x x x − 1 + tan 2 tan 2 2 dx = π / 4 2 dx ∫0 x x 1 − tan 2 1 − tan 2 2 2

1 + tan

2 −1

4t dt

4t dt ⇒

π = 2 − 1] 8

⇒

[Q tan

(1 + t 2) 1 − t 2

1/ a

(1 + t 2) 1 − t 2

28. The curves y = (x − 1)2, y = (x + 1)2 and y = 1 /4 are

 1 x3/ 2 2 1 ax3  − =1 ⋅ − =1 ⇒   2 3 0 3a 3a 2  a 3 /2 1 1 [Q a > 0] ⇒ a= a2 = 3 3 b

30. Since, ∫ f (x) dx = (b − 1) sin (3b + 4) 1

shown as Y y = (x + 1)2

1/4

y = (x – 1)

2

P

R

Q O 1/2

–1 –1/2

y = 1/4 X

1

On differentiating both sides w.r.t. b, we get f (b) = 3(b − 1) ⋅ cos (3b + 4) + sin (3b + 4) ∴ f (x) = sin (3x + 4) + 3(x − 1) cos (3x + 4) dy 31. Given, = 2x + 1 dx On integrating both sides ∫ dy = ∫ (2x + 1) dx ⇒ ∴ ⇒ ∴

where, points of intersection are 1 1 ⇒ x= (x − 1)2 = 4 2 1 1 2 and (x + 1) = ⇒ x = − 4 2  1 1  1 1 i.e. Q  ,  and R  − ,   2 4  2 4 ∴ Required area = 2 ∫

1/ 2 0

y = x2 + x + C which passes through (1, 2) 2 =1+1+C C =0 y = x2 + x Y y = x (x + 1)

1  (x − 1)2 − dx  4 

X'

1/ 2

 (x − 1)3 1  =2  − x 4 0  3 1  1  1  8 1 = 2 − − −  − − 0  = = sq unit   24 3  8 ⋅3 8  3

B C

1

1  x3 x2  1 1 5 = ∫ (x2 + x)dx =  +  = + = sq unit 0 3 2  0 3 2 6

32.

O

1

∫0 (x − x

3

α

)dx = 2∫ (x − x3 )dx 0

α2 α4 1 = 2 −  4 4  2 x y 2= a

⇒

(1/a,1/a)

A X'

X

x=1

Y'

y = ax2

Y

O

Thus, the required area bounded by X-axis, the curve and x = 1

29. As from the figure, area enclosed between the curves is OABCO. Thus, the point of intersection of y = ax2 and x = ay2

−1

X

D

33.

2α 4 − 4α 2 + 1 = 0 4 − 16 − 8 α2 = 4 1 2 α =1 − 2

(Q α ∈ (0, 1))

PLAN (i) Area of region f( x ) bounded between x = a to x = b is

Y'

⇒ ⇒

x = a (ax2)2 1 x = 0, ⇒ a

y = 0,

y = f (x)

1 a

 1 1 So, the points of intersection are (0, 0) and  ,  ⋅  a a

a a 1 a 2 a3 b

D

∫a f(x )dx = Sum of areas of rectangle shown in shaded part.

Area 345 (ii) If f( x )≥ g ( x ) when defined in [a, b ], then b

∫a

OABCO and points of intersection are (0,0) and {1 − m, m(1 − m)}.

b

f( x )dx ≥ ∫ g ( x )dx a

2

Description of Situation As the given curve y = e − x cannot be integrated, thus we have to bound this function by using above mentioned concept.

Graph for y = e− x

∴ Area of curve OABCO = ∫

1 −m

[x − x2 − mx] dx

0

Y

2

Y

A

1

B (0, 0) O

1 — √e O

1 — √2

C

X

1

2

− x2 ≥ − x or e− x ≥ e− x 1 − x2

∫0 e

∴

Also,

∴

1

dx ≥ ∫ e− x dx 0

S ≥ − (e− x )10 = 1 −

⇒ 1 − x2

∫0 e

1 e

…(i)

dx ≤ Area of two rectangles

1  1 1  ≤ 1 × ×  + 1 −  e 2  2 1 1  1 …(ii) ≤ + 1 −  2 e 2 1 1  1 1 [from Eqs. (i) and (ii)] + 1 −  ≥S ≥1 − e 2 e 2

Also,

∫1 ln (e + 1 − y) dy

=∫

1

∴ ⇒ ⇒

(1 − m)3 = 27 1 −m =3

Y y = mx

[put e + 1 − y = t ⇒ − dy = dt]

e

e

1

1

1– m A

35. Case I When m = 0 In this case, …(i) y = x − x2 and …(ii) y=0 are two given curves, y > 0 is total region above X-axis. Therefore, area between y = x − x2 and y = 0 is area between y = x − x2 and above the X-axis

1

⇒

2

Hence, no solution exists. Case II When m < 0 In this case, area between y = x − x2 and y = mx is

0 1 −m

(x − x2 − mx) dx 0

X

 x2 x3  1 1 1 9 ∴ A = ∫ (x − x ) dx =  −  = − = ≠ 0 3 0 2 3 6 2 2 1

y = x − x2

∴ Area of shaded region = ∫

A B

X

O (0,0)

B

Y

O

y = mx

⇒ m = −2 Case III When m > 0 In this case, y = mx and y = x − x2 intersect in (0,0) and {(1 – m), m( 1 – m)} as shown in figure

ln t (− dt ) = ∫ ln t dt = ∫ ln y dy = 1

e

x – x2

 x2 x3  = (1 − m) −  2 3 0  1 1 1 = (1 − m)3 − (1 − m)3 = (1 − m)3 2 6 3 1 9 3 [given] (1 − m) = 6 2

1 34. Shaded area = e −  ∫ ex dx = 1 0 e

y=

1 −m

Since, x2 ≤ x when x ∈ [0, 1] ⇒

X {1 – m, m (1 – m)}

 x2 x3  = (1 − m) −  2 3 1−m  1 1 = − (1 − m) (1 − m)2 + (1 − m)3 2 3 1 3 = − (1 − m) 6 9 1 [given] = − (1 − m)3 2 6

⇒

(1 − m)3 = − 27

⇒

(1 − m) = − 3

⇒ m =3 + 1 =4 Therefore, (b) and (d) are the answers.

346 Area 4a 2 4a 1 f (−1) 3a 2 + 3a    2 36. Given, 4b 4b 1  f (1)  = 3b2 + 3b   2   4c2 4c 1  f (2)  3c + 3c    ⇒ 4a 2 f (−1) + 4a f (1) + f (2) = 3a 2 + 3a , 4b f (−1) + 4b f (1) + f (2) = 3b + 3b 2

2

4c2 f (−1) + 4cf (1) + f (2) = 3c2 + 3c

and

…(i) …(ii) …(iii)

where, f (x) is quadratic expression given by, f (x) = ax2 + bx + c and Eqs. (i), (ii) and (iii). ⇒

4x2 f (−1) + 4x f (1) + f (2) = 3x2 + 3x

or {4 f (−1) − 3} x2 + {4 f (1) − 3} x + f (2) = 0 As above equation has 3 roots a, b and c. So, above equation is identity in x. i.e. coefficients must be zero. ⇒ f (−1) = 3 / 4, f (1) = 3 / 4, f (2) = 0 Q f (x) = ax2 + bx + c ∴ a = − 1 / 4, b = 0 and c = 1, using Eq. (v) 4 − x2 Thus, shown as, f (x) = 4 Let A (−2, 0), B = (2t , − t 2 + 1)

 x3  1  (4x − 3)3/ 2 1  = 2   −     3  0  3 ⋅ 4 / 2  3/ 4  1 1  1 1 = 2  −  = 1 ⋅ = sq unit  3 6 6 3

…(iv)

…(v)

Since, AB subtends right angle at vertex V (0, 1). 1 −t 2 ⇒ ⋅ = −1 ⇒ t =4 2 2t ∴ B (8, − 15) − (3x + 6) So, equation of chord AB is y = . 2 8  4 − x2 3x + 6 ∴ Required area = ∫ +   dx −2  4 2 

38. Here, slope of tangent, dy (x + 1)2 + y − 3 dy ( y − 3) = (x + 1) + , ⇒ = dx dx (x + 1) (x + 1) Put x + 1 = X and y − 3 = Y dy dY ⇒ = dx dX dY Y ∴ =X + dX X dY 1 ⇒ − Y =X dX X ∫ IF = e

2  128   = 8 − + 48 + 24 −  −2 + + 3 − 6    3 3  =

125 sq units 3

37. The region bounded by the curves y = x2, y = − x2 and y2 = 4 x − 3 is symmetrical about X-axis, where y = 4x − 3 meets at (1, 1). ∴ Area of curve (OABCO ) 1 1 ( 4x − 3 ) dx = 2 ∫ x2 dx − ∫ 3/ 4  0  y=x

2

1 dX X

= e− log X =

∴ Solution is, Y ⋅

1 X

1 1 Y = X ⋅ dX + c ⇒ =X +c X ∫ X X

Y

y = x 2 _ 2x

X′

y − 3 = (x + 1)2 + c(x + 1), which passes through (2, 0). ⇒ − 3 = (3)2 + 3c ⇒ c = − 4 ∴ Required curve y = (x + 1)2 − 4(x + 1) + 3 ⇒ y = x2 − 2x 2   x3   2 − x2  ∴ Required area =  ∫ (x2 − 2x)dx=    0  3  0  8 4 = − 4 = sq units 3 3

39. The points in the graph are A (1, 1), B ( 2, 0), C (2, 2), D ( 2, 2) Y y = |2 – x 2 | y = x2 D (√2,2) C (2,2) y=2

Y

(1,1) A B (3/4, 0)

X'

O C

y = –x 2

Y′

X

2

O Y′

8

  x3 3x2 = x − + + 3x 12 4   −2

−

X

y 2 = 4x – 3

A X′

x = 1 B (√2,0) Y′

X

Area 347 ∴ Required area 2

=∫

1

=∫

{ x2 − (2 − x2)} dx +

2 1

(2x2 − 2) dx +

2x3  = − 2x 3  1

∫

2 2

∫

2 2

{2 − (x2 − 2)} dx ∴

(4 − x2) dx

Sj Sj −1

3 2

2

 x + 4x −  3 

a 2 + b2 a − π b

=e

= constant

⇒ S 0 , S1 , S 2, K , S j form a GP. For a = − 1 and b = π 1

. πj

π ⋅ eπ Sj = (1 + π 2)

− ay

40. Given, x = (sin by) e

Now, −1 ≤ sin by ≤ 1 ⇒ − e−ay ≤ e− ay sin by ≤ e− ay ⇒ − e− ay ≤ x ≤ e− ay

n

⇒

Y

∑ Sj =

j=0

j  1 ⋅π   eπ + 1 = π ⋅ e (1 + e)   (1 + π 2)  

π ⋅ (1 + e) (1 + π )2

n

π (1 + e)

∑ e j = (1 + π 2) (e0 + e1 + ...+ en )

j=0

π (1 + e) (en + 1 − 1) ⋅ = e−1 (1 + π 2)

x = e – ay S3

|x| ≤ 1  x + ax + b,|x| > 1

41. Given, f (x) =  2x2 ,

S2 S1

Y

S0

X′

a − jπ

e b a 2 + b2 = = a a − aπ   − ( j − 1 )π − ( j − 1 )π  e b + 1 e b be b    

2

4 2   8 2 2 2 = − 2 2 − + 2 + 8 − − 4 2 + 3 3 3 3      20 − 12 2  =  sq units 3  

x = e – ay

 − aπ   e b + 1    

a − jπ b

be

X

O

x = – 2y2

1 x=– 8 y = f (x)

–2

X′

Y′

–1 O

− ay

In this case, if we take a and b positive, the values − e and e− ay become left bond and right bond of the curve and due to oscillating nature of sin by, it will oscillate between x = e− ay and x = − e− ay Now,

Sj = ∫

( j + 1 ) π /b

jπ /b

∴

Y′

sin by ⋅ e− ay dy  since, I = sin by ⋅ e− ay dy ∫   − ay e −  I = a by b by ( sin + cos )   a 2 + b2

⇒

− a

Sj = −

− ( j + 1 )π 1 [e b 2 2 a +b

a  − b (− 1) j e b  =  a 2 + b2 

lim f (x) = lim f (x) = f (−1)

x → −1 −

{a sin ( j + 1)π + b cos ( j + 1) π} − ajπ e b

jπ

 (a sin jπ + b cos jπ )   {0 + b(−1) j + 1 } − e− ajπ / b{0 + b(− 1) j }]

  −aπ   e b + 1       [Q (−1) j + 2 = (−1)2 (−1) j = (−1) j ]

=

a − jπ be b

x2 + ax + b, if x < − 1  if − 1 ≤ x < 1 f (x) = 2x, x2 + ax + b, if x ≥ 1 

f is continuous on R, so f is continuous at –1 and 1.

 − a ( j + 1)π e b

 −1 S j = 2 2 a + b 

X

 − aπ   e b + 1 2 2   a +b  

and ⇒ ⇒ ∴

x → −1 +

lim f (x) = lim f (x) = f (1)

x → 1−

x →1+

1 − a + b = − 2 and 2 = 1 + a + b a − b = 3 and a + b = 1 a =2, b = −1

 x2 + 2 x − 1 ,  Hence, f (x) =  2x,  x2 + 2x − 1,

if if if

x < −1 −1 ≤ x < 1 x≥1

Next, we have to find the points x = − 2 y2 and y = f (x). The point of intersection is (–2, –1). −1/ 8  − x  Required area = ∫ − f (x) dx ∴ −2  2   −1/ 8 −1 −1/ 8 −x dx − ∫ (x2 + 2x − 1)dx − ∫ 2x dx =∫ −2 −2 −1 2

348 Area −1

 x3  =− −  + x2 − x  − [x2] −−11/ 8 [(− 3 3 2    −2  3/ 2   1 2   1  3/ 2 =−   − 2  −  − + 1 + 1 3 3 2 8    1  8 −1 +  − + 4 + 2 −   64  3 2 5 63 = [2 2 − 2−9/ 2] + + 3 3 64 63 509 761 sq units = + = 16 × 3 64 × 3 192 2

x)3/ 2]−−12/ 8

42. Refer to the figure given in the question. Let the coordinates of P be (x, x2), where 0 ≤ x ≤ 1. For the area (OPRO ), Upper boundary: y = x2 and lower boundary : y = f (x) Lower limit of x : 0 Upper limit of x : x

∴ Area (OPRO ) =

x

∫0

t 2 dt − ∫

x 0

Now, to get the point of intersection of y = x2 and y = 2x (1 − x), we get x2 = 2x (1 − x) ⇒ 3 x2 = 2 x ⇒ x (3x − 2) = 0 ⇒ x = 0, 2 / 3 Similarly, we can find the coordinate of the points of intersection of y = (1 − x2) and y = 2x (1 − x) are x = 1 / 3 and x = 1 From the figure, it is clear that,  (1 − x)2, if 0 ≤ x ≤ 1 / 3  if 1 / 3 ≤ x ≤ 2 / 3 f (x) = 2x (1 − x), 2 if  2 /3 ≤ x ≤ 1 x , ∴ The required area 1

A=∫

0

=∫

0

f (x) dx

1/3

(1 − x)2 dx +

x

x  t3  =   − ∫ f (t ) dt 0  3 0 x x3 = − ∫ f (t ) dt 0 3 For the area (OPQO ), The upper curve : x = y and the lower curve : x = y /2 Lower limit of y : 0 and upper limit of y : x2 x2 x2 t ∴ Area (OPQO ) = ∫ t dt − ∫ dt 0 0 2 2 2 2 1 2 x4 = [t3/ 2]x0 − [t 2]x0 = x3 − 3 4 3 4 According to the given condition, x 2 x3 x4 − ∫ f (t ) dt = x3 − 0 3 3 4 On differentiating both sides w.r.t. x, we get x2 − f (x) ⋅ 1 = 2x2 − x3 ⇒ f (x) = x3 − x2, 0 ≤ x ≤ 1

2x (1 − x) dx +

1

∫ 2/ 3

x

 1  = − (1 − x)3  3  0

1  2x 1 3  x +  x2 − +  3 1/ 3 3  2 / 3 

 1  2 3 1  = −   +  +   3  3 3

  2 2 2   − 3  3

2

3

2  1  2   −  +  3  3 3

19 13 19 17 + + = sq unit 81 81 81 27 x2 44. Eliminating y from y = and y = x − bx2, we get b x2 = bx − b2x2 b x = 0, ⇒ 1 + b2 =

Y y=

x2 b y = x – bx 2

X′

–1

O

X

b 2 1+ b

y = 2x(1 − x) in following figure Y′

(1, 1)

(0, 1)

y = (1 – x)

Thus, the area enclosed between the parabolas b/(1 + b )2  x2  A=∫  x − bx2 −  dx 0 b 

2 2

Q

y=x

(1/2, 1/2) A

B y = 2x (1 – x)

X′

O

1/3 Y′

2/3

1

X

dx

3  1      3 

1 1 +  (1) − 3 3

43. We can draw the graph of y = x2, y = (1 − x2) and Y

2

3 2 / 3

1/ 3

f (t ) dt

2 /3

∫ 1/ 3

b/(1 + b )2

 x2 x3 1 + b2  = − ⋅ 3 b  0 2

=

b2 1 ⋅ 6 (1 + b2)2

3  2      3 

Area 349 On differentiating w.r.t. b, we get dA 1 (1 + b2)2 ⋅ 2b − 2b2 ⋅ (1 + b2) ⋅ 2b = ⋅ db 6 (1 + b2)4 1 b (1 − b2) = ⋅ 3 (1 + b2)3 dA For maximum value of A, put =0 db ⇒ b = − 1, 0, 1, since b > 0 ∴ We consider only b = 1. dA Sign scheme for around b = 1 is as shown below : db –

+ 0

1 n −1 1 An < ⇒ 2n − 2 1 1 From Eqs. (i) and (ii), < An < 2n + 2 2n − 2 ⇒

AB : y = 1, BC : x = − 1, CD : y = − 1, DA : x = 1 Y B(–1,1)

– (0,1/2)

X′

π /4

∫0

(tan x)n+ 1 dx < ∫

(tan x)n dx

Y′

Let the region be S and (x, y) is any point inside it. Then, according to given conditions,

An + An + 2 = ∫ =∫

π /4 0 π /4 0

x2 + y2 < |1 − x|,|1 + x|,|1 − y|,|1 + y|

[( tan x)n + (tan x)n + 2] dx

⇒ ⇒

x2 + y2 < (1 − x)2, (1 + x)2, (1 − y)2, (1 + y)2 x2 + y 2 < x 2 − 2 x + 1 , x 2 + 2 x + 1 , y2 − 2 y + 1 , y 2 + 2 y + 1 2 2 2 ⇒ y < 1 − 2x, y < 1 + 2x, x < 1 − 2 y and x2 < 2 y + 1 Now, in y2 = 1 − 2x and y2 = 1 + 2x, the first equation represents a parabola with vertex at ( 1/2,0) and second equation represents a parabola with vertex ( –1/2, 0) and in x2 = 1 − 2 y and x2 = 1 + 2 y, the first equation represents a parabola with vertex at (0, 1/2) and second equation represents a parabola with vertex at (0, –1 /2) . Therefore, the region S is lying inside the four parabolas y2 = 1 − 2x, y2 = 1 + 2x, x2 = 1 + 2 y, x2 = 1 − 2 y

(tan x) (1 + tan x) dx 2

n

Y y = (tan x)

n

B X

O

Y 2

(tan x) sec x dx

2x

An + 2 < An + 1 < An, An + An + 2 < 2 An 1 < 2 An n+1 1 < An 2n + 2

Also, for n > 2 An + An < An + An − 2 =

G I

O

–1

π /4

 1  = (tan x)n + 1   (n + 1) 0 1 1 = (1 − 0) = (n + 1) n+1

⇒

A(1,1)

=

0

n

2

π /4

(0,1)

y

=∫

⇒

D(1,–1)

C(–1,–1)

π /4 0

X

(1/2,0)

O

(0,–1/2)

⇒ An+ 1 < An Now, for n > 2

Since, then

(–1/2,0)

(tan x)n dx

Since, 0 < tan x < 1, when 0 < x < π /4 We have, 0 < (tan x)n+ 1 < (tan x)n for each n ∈N ⇒

A(1,1)

1

π /4 0

…(ii)

46. The equations of the sides of the square are as follow :

From sign scheme, it is clear that A is maximum at b = 1.

45. We have, An = ∫

2 An
0.

∴

Both are positive when x > 1 and negative when 0 < x < 1. We know that, lim (log x) → −∞ x→ 0 +

log x Hence, lim → −∞. Thus, Y-axis is asymptote of ex x→ 0 + second curve. [(0) × ∞ form] And lim ex log x x→ 0 + e log x  ∞  = lim − form  ∞  x→ 0 + 1 / x  1 e   x [using L’Hospital’s rule] =0 = lim +  1 x→ 0  − 2  x  Thus, the first curve starts from (0, 0) but does not include (0, 0). Now, the given curves intersect, therefore log x ex log x = ex i.e. (e2x2 − 1) log x = 0 1 [Qx > 0] x = 1, ⇒ e Y y=

log x ex y = ex log x

X′

1/e

O

Y′

1

X

= (x − 1) ⋅ (2x + x − 1) = (x − 1) (3x − 1) +• − • + 1 /3 1 Maximum at x = 1 / 3 2 1  2 4 ymax =  −  =   3 3 27 Minimum at x = 1 ymin = 0

Now, to find the area bounded by the curve y = x (x − 1)2, the Y-axis and line x = 2 . Y B

C 2 4 27 X′

O

1

A x=2

X

Y′

∴ Required area = Area of square OABC − ∫ 2

= 2 × 2 − ∫ x (x − 1)2 dx 0

  x (x − 1)3  2 1 2 = 4 −  − ∫ (x − 1)3 ⋅ 1 dx  3   0 3 0 2 x (x − 1)4  = 4 −  (x − 1)3 − 12  0 3 1  10 2 1 sq units =4− − + = 3 12 12  3 dy 52. Given, = sec2x y = tan x ⇒ dx  dy ∴ =2    dx x = π 4

2 0

y dx

352 Area π  Hence, equation of tangent at A  , 1 is 4  y−1 =2 x − π /4 π ⇒ y − 1 = 2x − 2

 25  4  1  8  sin −1    − 8 − = 2 6 +  5  4  3  2  −

25   4 4 4 4  = 2 6 + sin −1   − − −   5 3 3 3  2 

y = tan x

Y

  4  = 4 + 25 sin −1    sq units  5  

A O

X′

1  64   8   − 16 −  − 8     3   4  3

1

X

B L 2

54. Given curves are x2 + y2 = 4, x2 = − 2 y and x = y.

−1

Y

Y′

y=x

2

 π (2x − y) =  − 1  2

⇒

x + y2 = 4

∴ Required area is OABO =∫

π /4

0

X′

( tan x) dx − area of ∆ ALB

1 ⋅ BL ⋅ AL 2 1  π π − 2 = log 2 −  −  ⋅1 2 4 4  1  =  log 2 −  sq unit  4

−2

−√ 2

√2

O

X

2

= [log|sec x|]π0 / 4 −

−2

Thus, the required area

curves, x2 + y2 = 25, 4 y = |4 − x2| could be sketched as below, whose points of intersection are (4 − x2)2 x2 + = 25 16

53. Given

=

X′

4y = 4 – x 2 4y =x 2 – 4

5

O –5 –4

2

–2

X

4

x 2 + y 2 = 25

∫−

=2 ∫

Y 4y = x 2 – 4

x 2 − √ 2y

Y′

2 2 2

0

4 − x2 dx −

0

∫−

2

x dx −

∫0

2

− x2 dx 2

  x2 0   x3  2  4 − x2 dx −     − 2− 2 3 2 0     2

4 x 2 x =2  4 − x2 − sin −1  − 1 − 2 2 0 3 2 5 1  = (2 − π ) − =  − π  sq units  3 3

55. Given curves y = 5 − x2 and y = |x − 1|could be sketched as shown, whose point of intersection are 5 − x2 = (x − 1)2 Y

–5 Y′

(x2 + 24) (x2 − 16) = 0 x=±4  4 2  4 − x2  ∴ Required area = 2 ∫ 25 − x2 dx − ∫   dx 0  0 4    4  x2 − 4 −∫   dx 2  4  

y = –x + 1

⇒ ⇒

4

 x 25  x  sin −1    25 − x2 + = 2   5  2 2 0 −

1 4

4 2    x3  1  x3 x − − − x 4 4    3  0 4  3   2 

X′

y=x–1

–1

1

X

2

Y′

⇒ 5 − x = x − 2 x + 1 ⇒ 2 x2 − 2 x − 4 = 0 ⇒ x = 2 , − 1 ∴ Required area 2

=∫

2 −1

2

5 − x2 dx − ∫

1 −1

2

( − x + 1) dx − ∫ (x − 1) dx 1

2

1

2

 − x2   x2  x 5  x  5 − x2 + sin −1    −  = + x −  − x   2 2 2 2 5  −1    −1  1

Area 353 5 5 2   −1  − 1   = 1 + sin −1  − −1 + sin     5   2 2 5   1 1     1 −  − + 1 + + 1 − 2 − 2 − + 1     2 2 2 5  −1 2 1 1  = sin + sin −1  − 2 5 5 2  2 5 = sin −1  2  5 =

58. The point of intersection of the curves x2 = 4 y and x = 4 y − 2 could be sketched are x = − 1 and x = 2. ∴ Required area  x2  2  x + 2  = ∫   −    dx −1  4   4  2

4 1 1−  − 5 2

1 1 1− + 5 5

5 1  5π 1 −  sq units sin −1 (1) − =  2 2  4 2

π π  tan x, − ≤ x ≤  3 3 56. Given, y =  π π  cot x, ≤x≤  6 2 which could be plotted as Y-axis.

1  x2 x3  2 + − x 4  2 3  −1

=

1 4

=

1 10  −7  1 9 9 −   = ⋅ = sq units 4  3  6   4 2 8

8  1 1   2 + 4 − 3 −  2 − 2 + 3   

 et1 + e– t1 et1 – e– t1  ,  2 2  

59. Let P = 

 e– t + et1 e– t1 – et  and Q =  ,  2 2  

Y y = cot x

y = tan x

X′

=

X

O

We have to find the area of the region bounded by the curve x2 – y2 = 1 and the lines joining the centre x = 0, y = 0 to the points (t1 ) and (– t1 ). Y

π/3

P (t1)

X′

Y′

∴ Required area = ∫

π /4 0

( tan x) dx +

π /3

3 1 = log − 2 log 2 2 3 1 1  = log − log =  log e 3 sq units  2 2 2

57. Here,

∫2

4 8 1 + 2 dx = ∫ a  x  a

⇒ ⇒ ⇒ ⇒ ⇒ ∴

A 1 N

C

∫ π / 4 ( cot x) dx

= [− log|cos x|]π0 / 4 + [log sin x] ππ //34 3 1 1    = −  log − 0 +  log − log     2 2 2

a

−1

8  1 + 2 dx  x  4

8 8   x− = x−  x  2  x  a 8 8    a −  − (2 − 4) = (4 − 2) −  a −    a a 8 8 a − + 2 =2 − a + a a 16 2a − = 0 ⇒ 2 (a 2 − 8) = 0 a [neglecting –ve sign] a = ±2 2 a =2 2

X

Q (−t1) Y′

Required area e t1 + e – t1   2 = 2 area of ∆PCN – ∫ ydx 1    

 1  et1 + e– t1   et1 – e– t1  t1 dy =2   ⋅ dt    –∫ y 1 dt 2 2    2   e2t1 – e–2t1 t1  et – e– t   –∫ =2  dt  8  2    0 =

e2t1 – e–2t1 1 t1 2t – ∫ (e + e–2t – 2)dt 4 2 0

=

 e2t1 – e–2t1 1  e2t e–2t –  – – 2t  4 22 2 

=

e2t1 – e–2t1 1 2t1 – (e – e–2t1 – 4t1 ) 4 4

14 Differential Equations Topic 1 Solution of Differential Equations by Variable Separation Method Objective Questions I (Only one correct option) 1. The solution curve of the differential equation, (1 + e− x )(1 + y2)

dy = y2, which passes through the point dx

(0, 1) is

(2020 Main, 3 Sep I)

 1+ e (a) y2 + 1 = y  log e   2 

−x

(c) 2e

2

(d) 2e

dy = (x − y)2, dx

(2019 Main, 11 Jan II)

1+ x − y =x+ y−2 1− x + y

differential  x 

−1

dx,

x>0

(2017 Adv.)

(c) 9

(d) 80

(2016 Adv.)

8. The differential equation

dy 1 − y2 determines a = dx y

family of circles with

(2007, 3M)

(a) variable radii and a fixed centre at (0, 1) (b) variable radii and a fixed centre at (0, – 1) (c) fixed radius 1 and variable centres along the X-axis (d) fixed radius 1 and variable centres along the Y-axis

(a) 1/3

2 + sin x y+1

(b) 2/3

 dy   = − cos x , y (0) = 1, then  dx (2004, 1M)

(c) − 1 / 3

(d) 1

10. A solution of the differential equation 2

x, y ∈ [0, 1] and

f (0) ≠ 0. If y = y (x) satisfies the dy differential equation, = f (x) with y(0) = 1, then dx  3  1 (2019 Main, 9 Jan II) y   + y   is equal to  4  4 (c) 2

and

(b) 2(3 − 3 ) (d) 2(2 + 3 )

 π y   equals  2

1− x + y = 2(x − 1) 1+ x − y

equation

1 is equal 1 π ( − ) π k   π kπ  sin  +  sin  +  4  4 6 6

(a) 3 − 3 (c) 2( 3 − 1)

9. If y = y (x) and

2− x =x− y 2− y

(b) 3

4 (d) 3

to

4. Let f : [0, 1] → R be such that f (xy) = f (x). f ( y), for all

(a) 5

∑

k =1

2− y = 2( y − 1) 2− x

(d) − log e

(b) 3 13

when y(1) = 1, is

(c) log e

the

 x ) dy =  4 + 9 + 

7. The value of

3. The solution of the differential equation,

(b) − log e

satisfies

1 (c) − 3

y(0) = 7, then y(256) =

f ′ (x) = f (x) for all x ∈ R. If h (x) = f ( f (x)), then h′ (1) is equal to (2019 Main, 12 Jan II)

(a) log e

y = y(x)

(a) 16

2. Let f be a differentiable function such that f (1) = 2 and

(b) 4e

(2017 Main)

2 (b) − 3

1 (a) 3

8 x( 9+

 1 + ex  (c) y2 = 1 + y log e    2   1 + e− x  (d) y2 = 1 + y log e    2 

(a) 4e

dy  π + ( y + 1) cos x = 0 and y(0) = 1, then y   2 dx

is equal to

6. If

   + 2  

   1 + ex  (b) y2 + 1 = y  log e   + 2  2   

2

5. If (2 + sin x)

(d) 4

dy  dy + y = 0 is   −x  dx dx (a) y = 2

(b) y = 2x

(1999, 2M)

(c)y = 2x − 4

(d) y = 2x2 − 4

11. The order of the differential equation whose general

solution is given by y = (c1 + c2) cos (x + c3 ) − c4 ex + c5 , where c1 , c2, c3 , c4 , c5 are arbitrary constants, is (1998, 2M)

(a) 5

(b) 4

(c) 3

(d) 2

Differential Equations 355 Integer & Numerical Answer Type Questions

Objective Questions II (One or more than one correct option)

17. Let f : R → R be a differentiable function with f (0) = 0.

12. Let b be a nonzero real number. Suppose f : R → R is a differentiable function such that f (0) = 1. If the derivative f′ of f satisfies the equation f (x) f ′ (x) = 2 b + x2 for all x ∈R , then which of the following statements (2020 Adv.) is/are TRUE? (a) If b > 0, then f is an increasing function (b) If b < 0, then f is a decreasing function (c) f (x)f (− x) =1for all x∈ R (d) f (x) − f (− x) = 0 for all x ∈ R x x− t

∫0 e

f (t ) dt for all x ∈ [0, ∞ ). Then,

which of the following statement(s) is (are) TRUE? (2018 Adv.)

(a) The curve y = f (x) passes through the point (1, 2) (b) The curve y = f (x) passes through the point (2, − 1) (c) The area of the region π−2 {(x, y) ∈ [0, 1] × R : f (x) ≤ y ≤ 1 − x2 } is 4 (d) The area of the region π −1 {(x, y) ∈ [0, 1] × R : f (x) ≤ y ≤ 1 − x2 } is 4

14. Let y(x) be a solution of the differential equation (1 + ex ) y′ + yex = 1. If y(0) = 2, then which of the following statement(s) is/are true? (2015 Adv.) (a) y (−4) = 0 (b) y (−2) = 0 (c) y(x) has a critical point in the interval (−1 , 0) (d) y(x) has no critical point in the interval (−1 , 0)

15. Consider the family of all circles whose centres lie on the straight line y = x.If this family of circles is represented by the differential equation Py′ ′+ Qy′ + 1 = 0, where P , Q are dy d 2y ), then the functions of x, y and y′ (here, y′ = , y′ ′ = dx dx2 which of the following statement(s) is/are true? (2015 Adv.)

f (t ) dt . Then, the value of f (ln 5) is … . (2009)

Assertion and Reason (a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I. (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I. (c) Statement I is true; Statement II is false. (d) Statement I is false; Statement II is true.

19. Let a solution y = y(x) of the differential equation x x2 − 1 dy − y y2 − 1 dx = 0 satisfy y(2) =

2 3

π  Statement I y(x) = sec sec−1 x −  and  6 1 2 3 1 Statement II y(x) is given by = − 1− 2 y x x

(2008, 3M)

Analytical & Descriptive Questions 20. If P(1) = 0 and P (x) > 0, ∀ x > 1.

dP (x) > P (x), ∀ x ≥ 1 , then prove that dx (2003, 4M)

21. Let y = f (x) be a curve passing through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2 unit. Form the differential equation and determine all such possible curves.

Passage

c), where c is a positive parameter, (1999, 3M)

(b) order 2 (d) degree 4

x 0

Passage Based Problems

16. The differential equation representing the family of (a) order 1 (c) degree 3

f (x) = ∫

(1995, 5M)

(a) P = y + x (b) P = y − x (c) P + Q = 1 − x + y + y′ + ( y′ )2 (d) P – Q = x + y – y′ – ( y′ )2

curves y2 = 2c (x + is of

18. Let f : R → R be a continuous function, which satisfies

For the following question, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.

13. Let f : [0, ∞ ) → R be a continuous function such that f (x) = 1 − 2x +

If y = f (x) satisfies the differential equation dy = (2 + 5 y) (5 y − 2), then the value of lim f (x) is ...... . x→− ∞ dx (2018 Adv.)

Let f : [0, 1] → R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f ( 0) = f (1) = 0 and satisfies f ′ ′ (x) − 2 f ′ (x) + f (x) ≥ ex , x ∈ [0, 1]

(2013 Adv.)

356 Differential Equations 22. If the function e−x f (x) assumes its minimum in the interval [0, 1] at x = 1 / 4, then which of the following is true? 1 3 < x< 4 4 1 (c) f ′ (x) < f (x), 0 < x < 4

(a) f ′ (x) < f (x),

1 4

(b) f ′ (x) > f (x), 0 < x < (d) f ′ (x) < f (x),

3 < x 1). If 2 y(2) = log e 4 − 1, then y(e) dx is equal to (2019 Main, 12 Jan I)

x

(a) −

e 2

(b) −

e2 2

(c)

e 4

(d)

e2 4

10. If y(x) is the solution of the differential equation

dy  2x + 1 −2x +  y = e , x > 0, dx  x  1 (2019 Main, 11 Jan I) where y (1) = e−2, then 2

1 (a) y(x) is decreasing in  , 1 2  (b) y(x) is decreasing in (0, 1) (c) y(log e 2) = log e 4 log e 2 (d) y(log e 2) = 4

Differential Equations 357 11. Let f

be a differentiable function such 3 f (x) f ′ (x) = 7 − , (x > 0) and f (1) ≠ 4. Then, lim x 4 x x→ 0 +

that  1 f   x

(2019 Main, 10 Jan II)

(a) does not exist

4 7 (d) exists and equals 4 (b) exists and equals

(c) exists and equals 0

dy 3 1  −π π   π 4 ,  and y  = , then + y= ,x ∈  2 2    4 3 3 3 dx cos x cos x π   y −  equals  4 (2019 Main, 10 Jan I)

12. If

1 (a) + e6 3

4 (b) − 3

1 (c) + e3 3

1 (d) 3

13. If y = y(x) is the solution of the differential equation, x

dy  1 + 2 y = x2 satisfying y(1) = 1, then y  is equal to  2 dx (2019 Main, 9 Jan I)

(a)

13 16

(b)

1 4

(c)

49 16

(d)

7 64

14. Let y = y(x) be the solution of the differential equation dy + y cos x = 4x, x ∈ (0, π ). dx  π  π If y  = 0, then y  is equal to  6  2

sin x

(a)

4 2 π 9 3

(b)

−8 2 π 9 3

(c) −

8 2 π 9

(2018 Main)

4 2 π 9

(d) −

satisfies the differential equation, y(1 + xy)dx = x dy,  1 then f  −  is equal to  2 (2016 Main) 2 5

(b) −

4 5

(c)

2 5

(d)

4 5

dy + y = 2x log x, (x ≥ 1). Then, y(e) is equal to dx (2015 Main)

(a) e

(b) 0

(c) 2

(d) 2e

17. The function y = f (x) is the solution of the differential equation

dy xy x4 + 2x + 2 = dx x − 1 1 − x2

f (0) = 0. Then, ∫ (a)

π 3 − 3 2

3 /2 −

(b)

3 2

in

(−1, 1) satisfying

f (x) dx is

π 3 − 3 4

(c)

(2014 Adv.)

π 3 − 6 4

(d)

π 3 − 6 2

18. Let f : [1 /2, 1] → R (the set of all real numbers) be a positive, non-constant and differentiable function such that f ′ (x) < 2 f (x) and f (1 / 2) = 1 . Then, the value of 1 ∫ f (x) dx lies in the interval (2013 Adv.)

1/ 2

(a) (2e − 1, 2e) e−1 (c)  , e − 1  2 

t→ x

t 2f (x) − x2f (t ) = 1 for each x > 0 . Then, t−x

f (x) is 1 (a) + 3x 1 (c) − + x

(2007, 3M)

2x2 3 2

1 4x2 (b) − + 3x 3 1 (d) x

x2

20. If x dy = y (dx + y dy), y (1) = 1 and y (x) > 0. Then, y (−3) is equal to

(2005, 1M)

(a) 3 (c) 1

(b) 2 (d) 0

21. If y (t ) is a solution of (1 + t )

dy − ty = 1 and y (0) = − 1, dt

then y (1) is equal to (a) −1 / 2 (c) e − 1 / 2

(2003, 1M)

(b) e + 1 / 2 (d) 1 / 2

Objective Questions II (One or more than one correct option)

(b) (e − 1, 2e − 1) e − 1 (d)  0,   2 

f ′ (x) = 2 −

f (x) for all x ∈ (0, ∞ ) and f (1) ≠ 1. Then x (2016 Adv.)

1 (a) lim f ′  = 1 x → 0+  x  1 (b) lim x f   = 2 x → 0+  x (c) lim x2f ′(x) = 0 x → 0+

(d)|f (x)|≤ 2 for all x ∈ (0, 2)

23. If

16. Let y(x) be the solution of the differential equation (x log x)

f (1) = 1, and lim

22. Let f : (0, ∞ ) → R be a differentiable function such that

15. If a curve y = f (x) passes through the point (1, − 1) and

(a) −

19. Let f (x) be differentiable on the interval (0, ∞) such that

satisfies the differential y(x) y′ − y tan x = 2 x sec x and y(0), then

π (a) y   =  4 π (c) y   =  3

π2 8 2 π2 9

equation (2012)

π π2 (b) y′   =  4  18 π 4π 2π 2  (d) y′   = +  3 3 3 3

Analytical & Descriptive Question 24. Let u (x) and v (x) satisfy the differential equations du dv + p (x) u = f (x) and + p (x) v = g (x), where dx dx p (x), f (x) and g (x) are continuous functions. If u (x1 ) > v (x1 ) for some x1 and f (x) > g (x) for all x > x1, prove that any point (x, y) where x > x1 does not satisfy (1997, 5M) the equations y = u (x) and y = v (x).

Integer & Numerical Answer Type Question 25. Let y′ (x) + y(x) g′ (x) = g (x) g′ (x), y(0) = 0, x ∈ R, where d f (x) and g (x) is a given non-constant dx differentiable function on R with g (0) = g (2) = 0. Then, the value of y(2) is …… (2011) f ′ (x) denotes

358 Differential Equations

Topic 3 Applications of Homogeneous Differential Equations Objective Questions I (Only one correct option) 1. If a curve y = f (x), passing through the point (1, 2), is the solution

of

equation,

7. Let Γ denote a curve y = y(x) which is in the first

(2020 Main, 2 Sep II)

quadrant and let the point (1, 0) lie on it. Let the tangent to Γ at a point P intersect the y-axis at YP . If PYP has length 1 for each point P on Γ, then which of the following options is/are correct? (2019 Adv.)

differential  1 2x dy = (2xy + y )dx, then f   is equal to  2 2

(a)

the

2

1 1 (b) (c) 1 + log e 2 1 + log e 2 1 − log e 2

(d)

−1 1 + log e 2

2. Given that the slope of the tangent to a curve y = y(x) at 2y . If the curve passes through the x2 centre of the circle x2 + y2 − 2x − 2 y = 0, then its (2019 Main, 8 April II) equation is any point (x, y) is

(a) x2 log e|y| = − 2(x − 1)

(b) x log e|y|= x − 1

(c) x log e|y| = 2(x − 1)

(d) x log e|y| = − 2(x − 1)

3. The curve amongst the family of curves represented by the differential equation, (x2 − y2)dx + 2xydy = 0, which passes through (1, 1), is (2019 Main, 10 Jan II) (a) a circle with centre on the Y-axis (b) a circle with centre on the X-axis (c) an ellipse with major axis along the Y-axis (d) a hyperbola with transverse axis along the X-axis.

4. Let the population of rabbits surviving at a time t be governed by the differential equation dp(t ) 1 = p(t ) − 200. If p(0) = 100, then p(t ) is equal to dt 2 (2014 Main)

(a) 400 −

t 300 e 2

(b) 300 − 200 e

(c) 600 −

t 500 e 2

(d) 400 − 300 e

−

t 2

−

t 2

y (b) cosec   = log x + 2  x 2y 1 (d) cos   = log x +  x  2

6. At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P with respect to additional number of workers x is given by dP = 100 − 12 x. If the firm employees 25 more dx workers, then the new level of production of items is (2013 Main)

(a) 2500

(b) 3000

(c) 3500

(a) xy′ +

1 − x2 = 0

(b) xy′ − 1 − x2 = 0

 1 + 1 − x2   − 1 − x2 (c) y = log e    x    1 + 1 − x2   + (d) y = − log e    x  

8. A

solution

1 − x2

curve

of the differential equation dy (x + xy + 4x + 2 y + 4) − y2 = 0, x > 0, passes through dx (2016 Adv.) the point (1, 3). Then, the solution curve 2

(a) intersects y = x + 2 exactly at one point (b) intersects y = x + 2 exactly at two points (c) intersects y = (x + 2)2 (d) does not intersect y = (x + 3)2

9. Tangent is drawn at any point P of a curve which passes through (1, 1) cutting X-axis and Y-axis at A and B, respectively. If BP : AP = 3 : 1, then (2006, 3M) dy + y =0 dx dy (b) differential equation of the curve is 3x − y=0 dx 1 (c) curve is passing through  , 2 8 

(a) differential equation of the curve is 3x

(d) normal at (1, 1) is x + 3 y = 4.

π 5. A curve passes through the point 1,  . Let the slope of  6 y  y the curve at each point (x, y) be + sec   , x > 0.  x x Then, the equation of the curve is (2013 Adv.) 1 y (a) sin   = log x +  x 2 2y   (c) sec   = log x + 2  x 

Objective Questions II (One or more than one correct option)

(d) 4500

Fill in the Blank 10. A spherical rain drop evaporates at a rate proportional to its surface area at any instant t. The differential equation giving the rate of change of the rains of the rain drop is …. . (1997C, 2M)

Analytical & Descriptive Questions 11. If length of tangent at any point on the curve y = f (x) intercepted between the point and the X-axis is of length 1. Find the equation of the curve. (2005, 4M)

12. A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Find the time after (2003, 4M) which the cone is empty.

Differential Equations 359 13. A hemispherical tank of radius 2 m is initially full of

15. A curve passing through the point (1, 1) has the property

water and has an outlet of 12 cm2 cross-sectional area at the bottom. The outlet is opened at some instant. The flow through the outlet is according to the law v (t ) = 0.6 2 gh (t ), where v (t ) and h (t) are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time t and g is the acceleration due to gravity. Find the time it takes to empty the tank. (2001, 10M)

that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the X-axis. Determine the equation of the (1999, 10M) curve.

16. A and B are two separate reservoirs of water. Capacity of reservoir A is double the capacity of reservoir B. Both the reservoirs are filled completely with water, their inlets are closed and then the water is released simultaneously from both the reservoirs. The rate of flow of water out of each reservoir at any instant of time is proportional to the quantity of water in the reservoir at the time.

Hint Form a differential equation by relating the decreases of water level to the outflow.

14. A country has food deficit of 10%. Its population grows continuously at a rate of 3% per year. Its annual food production every year is 4% more than that of the last year. Assuming that the average food requirement per person remains constant, prove that the country will become self- sufficient in food after n years, where n is the smallest integer bigger than or equal to ln 10 − ln 9 . ln (1.04) − (0.03) (2000, 10M)

One hour after the water is released, the quantity of 1 water in reservoir A is 1 times the quantity of water in 2 reservoir B. After how many hours do both the reservoirs have the same quantity of water? (1997, 7M)

17. Determine the equation of the curve passing through the

origin in the form y = f (x), which satisfies the dy differential equation = sin (10x + 6 y) (1996, 5M) dx

Match the Columns 18. Match the conditions/expressions in Column I with statements in Column II. Column I π /2

(2006, 6M)

Column II

A.

∫0

B.

Area bounded by − 4 y 2 = x and x − 1 = − 5 y 2

q.

0

C.

The angle of intersection of curves y = 3 x − 1 log x and y = x x − 1 is

r.

3e y / 2

D.

If

s.

4 3

cos x

(sin x )

{cos x cot x − log (sin x )

sin x

p.

}dx

dy 2 passing through (1, 0), then ( x + y + 2 ) is = dx x + y

1

Answers 21. (a)

Topic 1

22. (a)

23. (a, d)

25. (0)

2. (c)

3. (b)

4. (a)

6. (c)

7. (a,c)

1. (c)

2. (b)

3. (d)

4. (b)

5. (a)

6. (b)

7. (c)

8. (c)

1. (a)

9. (a)

10. (c)

11. (c)

12. (a,c)

5. (a)

14. (a, c)

15. (b, c)

8. (a, d)

9. (a, c)

13. (b, c) 16. (a, c)

17. (0.40) 18. (0) 19. (b) dy d 2y 21. Differential Equation: 2 = 0, x 2 +1 = 0 dx dx Curves : x + y = 2, xy = 1 22. (c) 23. (d) 24. (b) 25. (c)

Topic 2 1. (b)

2. (b)

3. (a)

4. (d)

5. (b)

6. (a)

7. (d)

8. (a)

9. (c) 13. (c)

10. (a) 14. (c)

11. (d) 15. (d)

12. (a) 16. (c)

17. (b)

18. (d)

19. (a)

20. (a)

Topic 3

  dr 10.  = − λ    dt

  1 + 1 −y2 11.  1 − y 2 − log = ± x + c 2   1 − 1 −y  

H  12. T =   k

  14 π × 10 5 13.  unit   27 g

 1  15. ( x 2 + y 2 = 2 x ) 16.  log 3 2   4 17.

1 4  tan −1  tan  4 x + tan −1  3 5

3  3  5x −  − 4 5  3

18. A → p; B → s; C → q; D → r

Hints & Solutions Topic 1 Solution of Differential Equations by Variable Separation Method

⇒

1. Given differential equation dy (1 + e ) (1 + y ) = y2 dx ex 1 + y2 ⇒ = dy ∫ y2 ∫ 1 + ex dx −x

2

which passes through (0, 1), so − 1 + 1 = log e 2 + C ⇒ C = − log e 2 So, equation of required curve is  1 + ex  y2 = 1 + y log e    2  Hence, option (c) is correct.

2. Given that, ⇒

∫

⇒

⇒ ∴ ⇒ ⇒ Q So, ⇒ ⇒

f ′ (x) = f (x) f ′ (x) =1 f (x) f ′ (x) dx = ∫ 1 ⋅ dx f (x)

⇒

[by integrating both sides w.r.t. x] Put f (x) = t ⇒ f ′ (x)dx = dt dt ∫ t = ∫ 1 dx dx   Q = ln|x|+ C ln|t|= x + C   ∫ x …(i) ln| f (x)|= x + C [Q t = f (x)] f (1) = 2 [using Eq. (i)] ln (2) = 1 + C C = ln 2 − ln e [Q ln e = 1]  2  A [Q ln A − ln B = ln   ] C = ln    e  B

From Eq. (i), we get

⇒ ⇒

Now,

ef (x) =x 2 e f (x) = ex 2 | f (x)|= 2ex −1

  a+x 1 dx log e = + C ∫ 2 2 a a x 2 −   a −x  1 + x − y 1 [Q t = x − y] log e   =x+C  1 − x + y 2

Since, y = 1 when x = 1, therefore  1 + 0 1 log e   =1 + C  1 + 0 2 ⇒ ∴ ⇒

C = −1 1  1 + x − y log e   = x −1  1 − x + y 2 − log e

[Q log 1 = 0]

1−x+ y = 2(x − 1) 1+ x− y 1 = log x−1 = − log x] x

4. Given, f (xy) = f (x) ⋅ f ( y), ∀ x, y ∈ [0, 1]

 2 ⇒ ln| f (x)|− ln   = x  e ln

⇒

[separating the variables]  1 + t 1 log e   =x+C  1 − t 2

[Q log

 2 ln| f (x)|= x + ln    e

⇒

h′ (x) = f ′ ( f (x)) ⋅ f ′ (x) [on differentiating both sides w.r.t. ‘x’] h′ (1) = f ′ ( f (1)) ⋅ f ′ (1) [Q f (1) = 2 (given)] = f ′ (2) ⋅ f ′ (1) = 2e2−1 ⋅ 2e1−1 [Q f ′ (x) = 2ex −1 or −2ex −1]

= 4e dy 3. We have, = (x − y)2 which is a differential equation of dx the form dy = f (ax + by + c) dx Put x − y = t dy dt dy dt =1 − 1− = ⇒ ⇒ dx dx dx dx dt dy [ Q = (x − y)2] ⇒ 1− = t2 dx dx dt dt 2 =1−t ⇒ ∫ = dx ⇒ dx 1 − t2 ∫

 1  x  − + y = log e (1 + e ) + C,  y 

⇒

⇒

[Q ln A − ln B = ln

A ] B

[Q ln a = b ⇒ a = eb , a > 0] e  e  Q 2 f (x) = 2 | f (x)|   

f (x) = 2ex −1 or −2ex −1 h (x) = f ( f (x))

Putting x = y = 0 in Eq. (i), we get f (0) = f (0) ⋅ f (0) ⇒ f (0) [ f (0) − 1] = 0 ⇒ f (0) = 1 as f (0) ≠ 0 Now, put y = 0 in Eq. (i), we get f (0) = f (x) ⋅ f (0) ⇒ f (x) = 1 dy dy So, = f (x) ⇒ =1 dx dx ⇒ ∫ dy = ∫ dx

...(i)

Differential Equations 361 ⇒ Q ∴ ⇒ ∴

y= x + C y(0) = 1 1 =0 + C C =1 y=x+1 5 7  1 1  3 3 Now, y   = + 1 = and y   = + 1 =  4 4  4 4 4 4  1  3 5 7 ⇒ y  + y  = + =3  4  4 4 4

13

= 2∑

k =1

13 π π   π π = 2 ∑ cot  + (k − 1)  − cot  + k   4 4 6 6     k =1

  π  π π  = 2 cot   − cot  +      4 6  4 

dy 5. We have, (2 + sin x) + ( y + 1) cos x = 0 dx dy cos x − cos x ⇒ y= + dx 2 + sin x 2 + sin x

  π 2π    π π + cot  +  − cot  +  4   4 6 6   π π    π π + K + cot  + 12  − cot  + 13      4 6 4 6   

which is a linear differential equation. ∫

∴ IF = e

cos x dx 2 + sin x

= elog ( 2 + sin x ) = 2 + sin x

π π   π = 2 cot − cot  + 13   4 4 6  

∴Required solution is given by − cos x y ⋅ (2 + sin x) = ∫ ⋅ (2 + sin x)dx + C 2 + sin x ⇒ ∴

y(0) = 1

1(2 + sin 0) = − sin 0 + C

⇒

= 2 (1 − 2 + 3 )

C =2 2 − sin x y= 2 + sin x

∴

 π y  =  2

⇒

6.

5 π     29 π    = 2 1 − cot    = 2 1 − cot 2π +     12  12     5π 5π    = 2 1 − cot Q cot = (2 − 3 )     12  12 

y(2 + sin x) = − sin x + C

Also,

dy = dx 8 x 9 + ⇒

= 2 ( 3 − 1)

8. Given,

π 2 =1 π 3 2 + sin 2 1 2 − sin

x 4+ 9+

⇒

⇒

c=0 y(256) = 4 + 9 + 16 = 4 + 5 = 3 13

∑

k =1

1  π (k − 1)π  sin  +  sin 6  4

 π kπ    + 4 6

Converting into differences, by multiplying and  π k π   π (k − 1)π  dividing by sin  +  − + , i.e. 6  4 6   4  π sin   .  6 ∴

13

∑

k =1

∫

y 1 − y2

dy = ∫ dx

− 1 − y2 = x + c ⇒ (x + c)2 + y2 = 1

Here, centre (– c, 0) and radius = 1 dy − cos x ( y + 1) 9. Given, = dx 2 + sin x

x

Now, y(0) = 7 + c

7. Here,

dy 1 − y2 = dx y

⇒

y= 4 + 9+ x + c

⇒

π    π kπ  π sin  4 + 6  cos  4 + (k − 1) 6    π π  π kπ    − sin  + (k − 1)  cos  + 4 6 6  4  π π π π sin  + (k − 1)  sin  + k  4 6 6 4

π  π π   π sin  + k  −  + (k − 1)    6 6  4  4 π  π π  π π sin sin  + (k − 1)  sin  + k   4 6  6 6  4

dy − cos x = dx y + 1 2 + sin x

On integrating both sides dy cos x ∫ y + 1 = − ∫ 2 + sin x dx ⇒

log ( y + 1) = − log (2 + sin x) + log c

When x = 0, y = 1 ⇒ c = 4 4 ⇒ y+1= 2 + sin x ∴

 π 4 y  = −1  2 3

⇒

 π 1 y  =  2 3

10. Given differential equation is 2

dy  dy + y=0   −x  dx dx

…(i)

362 Differential Equations dy =0 dx

(a) y = 2 ⇒

log e| f (x)| =

⇒

On putting in Eq. (i),

f (0) = 1 , so c = 0

Q

02 − x (0) + y = 0

1  x tan − 1    b b

∴

⇒ y = 0 which is not satisfied. dy (b) =2 y = 2x ⇒ dx

| f (x)| = e

1

On putting in Eq. (i),

⇒

f (x) = eb

Q

f (0) = 1

∴

f (x) = eb

1

(2) − x ⋅ 2 + y = 0 4 − 2x + y = 0 2

⇒

⇒ y = 2x which is not satisfied. dy (c) =2 y = 2x − 4 ⇒ dx

f ′ (x) =

and

On putting in Eq. (i) 4 − 2x + 2x − 4 = 0

1

and f (x) f (− x) = eb

On putting in Eq. (i),

11. Given, y = (c1 + c2) cos (x + c3 ) − c4 ex + c5

…(i)

y = (c1 + c2) cos (x + c3 ) − c4 e ⋅ e x

c5

c1 + c2 = A , c3 = B, c4ec5 = c y = A cos (x + B) − cex

…(ii)

On differentiating w.r.t. x, we get dy = − A sin ( A + B) − cex dx

…(iii)

Again, on differentiating w.r.t. x, we get d 2y = − A cos (x + B) − cex dx2

⇒

…(iv)

d 2y = − y − 2 cex dx2

⇒

…(v)

2

d y + y = − 2 cex dx2

d3 y dy d 2y + = + y dx2 dx dx2

…(vi) [from Eq. (v)]

which is a differential equation of order 3. f (x) b 2 + x2 f ′ (x) dx dx = ∫ 2 f (x) b + x2 f ′ (x) =

⇒

∫

 x tan − 1    b

−

⋅e

1  x tan − 1   b b

=1

x x−t

∫0 e

f (t ) dt

On multiplying e− x both sides, we get e− x f (x) = e− x − 2xe− x +

x −t

∫0 e

f (t ) dt

On differentiating both side w.r.t. x, we get e− x f ′ (x) − e− x f (x) = − e− x − 2e− x + 2xe− x + e− x f (x) ⇒ f ′ (x) − 2 f (x) = 2x − 3 [dividing both sides by e− x ] dy Let f (x) = y ⇒ f ′ (x) = dx dy − 2 y = 2x − 3 ∴ dx which is linear differential equation of the form dy + Py = Q. Here, P = −2 and Q = 2x − 3. dx P dx −2 dx Now, IF = e∫ = e∫ = e−2x

I

d3 y dy + = − 2 cex dx dx3

12. Given differential equation

 x tan − 1  

∴ Solution of the given differential equation is y ⋅ e−2x = ∫ (2x − 3) e−2x dx + C

Again, on differentiating w.r.t. x, we get

⇒

1 e 2 b2  x 1+    b

f (x) = 1 − 2x +

⇒ y = 0 which is not satisfied.

⇒

 x tan − 1    b

13. We have,

(4x)2 − x ⋅ 4x + y = 0

Now, let

 x tan − 1    b

∴ f ′ (x) > 0 ∀x ∈ R and b ∈ R0. Therefore, f (x) is an increasing function ∀ b ∈ R0.

y = 2 x2 − 4 dy = 4x dx

⇒

1

or − eb

b eb = b 2 + x2

[Q y = 2x − 4]

y = 2x − 4 is satisfied.

 x tan − 1    b

1  x tan − 1    b b

1

(2)2 − x − 2 + y

(d)

1  x tan − 1   + c  b b

II

− (2x − 3) ⋅ e e−2x + 2∫ dx + C 2 2 [by using integration by parts] − ( 2 x − 3 ) e−2x e−2x ⇒ y ⋅ e−2x = − +C 2 2 ⇒ y = (1 − x) + Ce2x On putting x = 0 and y = 1, we get 1 =1 + C ⇒ C =0 ∴ y=1 − x y = 1 − x passes through (2, − 1) y ⋅ e−2x =

−2x

Differential Equations 363 Now, area of region bounded by curve

y = 1 − x2 and

y = 1 − x is shows as

15. Since, centre lies on y = x. ∴ Equation of circle is

y

x2 + y2 − 2ax − 2ay + c = 0

B (0, 1)

On differentiating, we get

y= 1–x2

2x + 2 yy′ − 2a − 2ay′ = 0 x′

A (1,0)

O

–1,0

⇒

x y=1–x

x + yy′ − a − ay′ = 0 ⇒ a =

Again differentiating, we get

y′

∴ Area of shaded region = Area of 1st quadrant of a circle − Area of ∆ OAB π π 1 π −2 1 2 = (1) − × 1 × 1 = − = 4 4 2 4 2

0=

(1 + y′ )[1 + yy′ ′+ ( y′ )2] − (x + yy′ ) ⋅ ( y′ ′ ) (1 + y′ )2

⇒ (1 + y′ ) [1 + ( y′ )2 + yy′ ′ ] − (x + yy′ ) ( y′′ ) = 0 ⇒

1 + y′ [( y′ )2 + y′ + 1] + y′′ ( y − x) = 0 On comparing with Py′′ + Qy′ + 1 = 0, we get P = y − x and Q = ( y′ )2 + y′ + 1

Hence, options b and c are correct.

⇒ dy + ex dy + yex dx = dx dy + d (ex y) = dx

On putting this value of c in Eq. (i), we get dy  dy  y2 = 2 y x + y  dx  dx 

On integrating both sides, we get y + ex y = x + C y (0) = 2

Given, ⇒

C =4

∴

y (1 + e ) = x + 4 x+4 y= 1 + ex

Now at x = − 4, y = ∴

…(i) dy =0 dx

ex (x + 3) − 1 = 0 e− x = (x + 3) y = e–x

Y

y=x+3

(–1, e) (–1, 2) X′

–1

O

dy  dy =2 y    dx dx

dy   dy  y − 2x  = 4 y      dx dx

3/ 2

3/ 2

3

Therefore, order of this differential equation is 1 and degree is 3.

dy (1 + ex ) ⋅ 1 − (x + 4)ex =0 = dx (1 + ex )2

or

dy  dy ⋅ x + 2 y1/ 2    dx dx

2

y(−4) = 0

i.e.

y − 2x

⇒

−4 + 4 =0 1 + e−4

For critical points,

y=2

⇒

x

⇒

⇒

⇒

2 + e0 ⋅ 2 = 0 + C

⇒

…(i)

c)

On differentiating w.r.t. x, we get dy dy 2y = 2c ⇒ c = y dx dx

dy dy + ex ⋅ + yex = 1 dx dx

⇒

y2 = 2c (x +

16. Given,

14. Here, (1 + ex ) y′ + y ex = 1 ⇒

x + yy′ 1 + y′

X

Y′

Clearly, the intersection point lies between (− 1, 0). ∴ y(x) has a critical point in the interval (− 1, 0).

17. We have, dy = (2 + 5 y) (5 y − 2) dx   dy 1  dy  = dx ⇒ ⇒  = dx  25  y2 − 4  25 y2 − 4  25  On integrating both sides, we get 1 dy = dx 25 ∫ 2  2 2 ∫ y −   5 ⇒

⇒ ⇒

1 1 y − 2 /5 × log =x+C 2 25 2 × y + 2 /5 5 5y − 2 log = 20(x + C ) 5y + 2 5y − 2 = Ae20x 5y + 2

[Q e20C = A ]

364 Differential Equations when x = 0 ⇒ y = 0, then A = 1

dP (x) − P (x) > 0, ∀ x ≥ 1 dx On multiplying Eq. (i) by e− x , we get

20. Given, P(1) = 0 and

5 y −2 = e20x 5y + 2

∴ lim

x → −∞

⇒

lim

n → −∞

⇒

lim 5 f (x) −2 = 0

lim f (x) =

n→ − ∞

2 = 0.4 5

d (P (x) ⋅ e− x ) > 0 dx

⇒

P (x) ⋅ e− x is an increasing function.

⇒

P (x) ⋅ e− x > P (1) ⋅ e−1 , ∀ x ≥ 1

⇒

P (x) > 0, ∀ x > 1

A (x, y) is Y − y =

Also, differentiating the given integral equation w.r.t. x f (x) ≠ 0 f ′ (x) = 1 ⇒ log f (x) = x + c f (x)

(1, 1) Q

⇒

f (x) = e e

Q

f (0) = 0 ⇒

∴

f (x) = 0, ∀ x ∈ R

x ′'

ec = 0 , a contradiction

x

Given,

f (x) = ∫ f (t ) dt

⇒

f (0) = 0 and

dy  y-intercept 0, y − x   dx

0

f ′ (x) = f (x)

If f (x) ≠ 0 f ′ (x) = 1 ⇒ ln f (x) = x + c f (x)

⇒

f (x) = e ⋅ e

Q

f (0) = 0

c

y y −1 2

=∫

x x2 − 1

sec−1 y = sec−1 x + c 2 π π π At x = 2 , y = ; = + c ⇒ c=− 6 3 6 3 ⇒

 π 1 3  Now, y = sec sec−1 x −  = cos cos −1 − cos −1   6 x 2     3 1 = cos cos −1  + 1− 2 2 x  x  y=

3 1 1 + 1− 2 2x 2 x

3  1 −  4 

1   x − y  ( y − xp) = 4, where  p p2x2 − 2 pxy + 4 p + y2 = 0

⇒

( y − px)2 + 4 p = 0

⇒

dx

∆OPQ = 2 1  dx  dy ⋅ x − y   y − x  = 2 2  dy  dx

⇒ ∴

2 dy y y − 1 = 2 dx x x − 1

dy

⇒ ⇒

f (x) = 0, ∀ x ∈ R ⇒ f (ln 5) = 0

∫

Given,

x

⇒ ec = 0, a contradiction

19. Given,

x

P

dx   whose, x-intercept  x − y ⋅ , 0  dy 

Alternate Solution

∴

O

y = f(x)

y′

f (ln 5) = 0

⇒

(x, y) A

c x

⇒

dy (X − x) dx y

f ′ (x) = f (x) ⇒

[Q P (1) = 0 and e− x > 0]

21. Equation of tangent to the curve y = f (x) at point

18. From given integral equation, f (0) = 0 .

If

d d −x P (x) ⋅ e >0 dx dx

⇒

5 f (x) − 2 = 0 5 f (x) + 2

n→ − ∞

⇒

e− x ⋅

5 f (x) −2 = lim e20x x → −∞ 5 f (x) + 2

...(i)

p=

dy dx

y − px = 2 − p y = px + 2 − p

…(i)

On differentiating w.r.t. x, we get dp dp  1 p= p+ ⋅ x + 2 ⋅   (− p)−1/ 2 ⋅ (−1)  2 dx dx dp −1/ 2 { x − (− p) } = 0 ⇒ dx dp ⇒ = 0 or x = (− p)−1/ 2 dx dp If =0 ⇒ p= c dx On putting this value in Eq. (i), we get y = cx + 2 − c This curve passes through (1, 1). ⇒ ⇒ ∴

1 = c + 2 −c c = −1 y=−x+2

Differential Equations 365 ⇒

x+ y=2

25. Here, f (x) + 2x = (1 − x)2 ⋅ sin 2 x + x2 + 2x

−1/ 2

Again, if x = (− p) 1 ⇒ − p = 2 putting in Eq. (i) x −x 1 ⇒ xy = 1 y = 2 + 2⋅ x x

∴

Thus, the two curves are xy = 1 and x + y = 2.  1 φ′ (x) < 0, x ∈ 0,   4

⇒

d d { f ′ (x)e−x } − { f (x)e− x } ≥ 1 dx dx

⇒

d { f ′ (x)e−x − f (x)e− x } ≥ 1 dx

⇒

d 2 −x { e f (x)} ≥ 1, ∀x ∈ [0, 1] dx2

e

∴

( y2 − x3 ) dx − xy dy = 0, (x ≠ 0) dy ⇒ xy − y2 = − x3 dx dy dt dy 1 dt Now, put y2 = t ⇒ 2 y = ⇒y = dx dx dx 2 dx x dt dt 2 − t = − 2x2 − t = − x3 ⇒ ∴ 2 dx dx x which is the linear differential equation of the form dt + Pt = Q. dx 2 Here, P = − and Q = − 2x2. x

24. Here, f (x) = (1 − x)2 ⋅ sin 2 x + x2 ≥ 0, ∀ x x  2 (t − 1 )  and g (x) = ∫  − log t f (t )dt 1  t + 1 

…(i)

⇒

φ(x) < 0

From Eqs. (i) and (ii), g′ (x) < 0, x ∈ (1, ∞ ) ∴ g (x) is decreasing on x ∈ (1, ∞ ).

=

or

y2 + 2x3 + Cx2 = 0

[Q t = y2] [let C = − λ]

2. Given differential equation is 1  y2dx +  x −  dy = 0  y

φ′ (x) < 0, ∀ x > 1 φ (x) < φ (1)

2 dx x

⇒

⇒ ⇒

4 1 − (x − 1) − = (x + 1)2 x x (x + 1)2

IF = e

1  1  t  2 = − 2 ∫  x2 × 2 dx + λ x   x  t = − 2x + λ x2 y2 + 2x = λ x2 y2 + 2x3 − λx2 = 0

∴

For g′ (x) to be increasing or decreasing. 2(x − 1) Let φ(x) = − log x x+1

⇒

−∫

1 x2 Q Solution of the linear differential equation is (IF) t = ∫ Q (IF)dx + λ [where λ is integrating constant] Now,

f (x) < 0

φ′ (x) =

h (0) = 2 f (0) + 1 − 0 = 1

1. Given differential equation is

f (x) < 0

 2(x − 1) ⇒ g′ (x) =  (x) − log x ⋅ f{ ( x ) + 1  + ve 

(1 − x)2 cos 2 x = − 1

Topic 2 Linear Differential Equation and Exact Differential Equation

φ (x) < 0

⇒

⇒

∴ II is true.

φ(0) = 0 = f (1) −x

(1 − x)2 sin 2 x = (1 − x)2 + 1

h(1) = 2(1) + 1 − 4 = − 3 as [h (0)h (1) < 0]

f (0) = f (1) = 0 ⇒

⇒

⇒ h (x) must have a solution.

∴ φ (x) = e− x f (x) is concave function.

⇒

(1 − x)2 sin 2 x = x2 − 2x + 2

where,

23. Here, f ′′ (x) − 2 f ′ (x) + f (x) ≥ ex f ′′ (x)e− x − f ′ (x)e− x − f ′ (x)e− x + f (x)e− x ≥ 0

⇒

Again, let h (x) = 2 f (x) + 1 − 2x(1 + x)

 1 ⇒ e− x f ′ (x) − e− x f (x) < 0, x ∈ 0,   4 1 ⇒ f ′ (x) < f (x) ,0 < x < 4

⇒

2(1 + x2) = (1 − x)2 sin 2 x + x2 + 2x

∴I is false.

1  φ′ (x) > 0, x ∈  , 1 4 

and

…(ii)

which is never possible.

22. Let φ(x) = e−x f (x) Here,

…(i)

where, I: f (x) + 2x = 2(1 + x)2

…(ii)

dx 1 1 + x = 3 , which is the linear differential dy y2 y dx equation of the form + Px = Q. dy

⇒

366 Differential Equations Here, P =

2 1 π2 1 π2  π  π π and y  + y −  = + + + = + 2  4  4  16 4 2 2 16

1 1 and Q = 3 y2 y 1 ∫ 2 dy y

Now, IF = e

−

=e

1 y

2 1 1 π2  π  π π and y  − y −  = + − − =0  4  4  16 2 2 16

∴The solution of linear differential equation is

4. Given differential equation

x ⋅ (IF) ∫ Q (IF)dy + C ⇒ x e− 1/ y = ∫

dy = (tan x − y)sec2 x dx

1 − 1/ y e dy + C y3

∴ x e− 1/ y = ∫ (− t ) et dt + C = − tet +

[Qlet −

∫ e dt + C t

1 1 = t ⇒ + 2 dy = dt] y y

[integration by parts]

= − te + e + C 1 ⇒ x e− 1/ y = e− 1/ y + e− 1/ y + C y t

dy + (sec2 x) y = sec 2x tan x, dx which is linear differential equation of the form dy + Py = Q, dx where P = sec2 x and Q = sec2 x tan x sec 2 x dx IF = e∫ = etan x

⇒

t

… (i)

So, solution of given differential equation is y × IF = ∫ (Q × IF)dx + C

Now, at y = 1, the value of x = 1, so 1 1 ⋅ e− 1 = e− 1 + e− 1 + C ⇒ C = − e On putting the value of C, in Eq. (i), we get e1/ y 1 x= +1− y e

y(etan x ) = ∫ etan x ⋅ sec 2 x tan x dx + C Let

[using integration by parts method] = et (t − 1) + C tan x ⇒ y⋅ e = etan x (tan x − 1) + C [Q t = tan x] Q y(0) = 0 ⇒ 0 = 1(0 − 1) + C ⇒ C = 1 tan x ∴ y⋅ e = etan x (tan x − 1) + 1 π Now, at x = − 4 ye−1 = e−1 (−1 − 1) + 1 ⇒ ye−1 = − 2e−1 + 1 ⇒ y = e − 2

e1/ 2 3 1 1 So, at y = 2, the value of x = + 1 − = − e 2 2 e

3. Given differential equation is dy + y tan x = 2x + x2 tan x , which is linear differential dx dy equation in the form of + Py = Q . dx Here, P = tan x and Q = 2x + x2 tan x ∴IF = e∫ tan x dx = elog e (sec x ) = sec x Now, solution of linear differential equation is given as y × IF = ∫ (Q × IF)dx + C ∴ y(sec x) = ∫ (2x + x tan x) sec x dx + C 2

= ∫ (2x sec x) dx + Q

∫

∫x

2

sec x tan x dx + C

y sec x = 2∫ x sec x dx + x2 sec x − 2∫ x sec x dx + C Q

Key Idea (i) First convert the given differential equation into dy + Py = Q dx

(ii) Find IF

x2 sec x tan x dx = x2 sec x − ∫ (2x sec x) dx

y sec x = x2 sec x + C

5.

linear differential equation of the form

Therefore, solution is

⇒

tan x = t ⇒ sec 2x dx = dt yetan x = ∫ et ⋅ t dt + C = tet − ∫ et dt + C

…(i)

y(0) = 1 ⇒ 1(1) = 0(1) + C ⇒ C = 1

Now, [from Eq. (i)] y = x2 + cos x and y′ = 2x − sin x According to options, 1 1    π  − π   π  π y′   − y′   = 2  −  − 2 −  +  =π− 2  4    4   4 2 2   4 1 1    π  π  π   π and y′   + y′  −  = 2  −  + 2 −  +  =0  4  4   4 4 2 2 

(iii) Apply formula, y( IF) = ∫ Q( IF) dx + C

Given differential equation dy cos x − (sin x) y = 6x dx dy 6x , which is the linear ⇒ − (tan x) y = dx cos x differential equation of the form dy + Px = Q, dx 6x where P = − tan x and Q = cos x − tanx dx So, IF = e ∫ = e− log(sec x ) = cos x ∴Required solution of differential equation is cos x x2 y(cos x) = ∫ (6x) dx + C = 6 + C = 3 x2 + C cos x 2

Differential Equations 367  π y  = 0  3

Given,

⇒ y(1 + x2) = ∫ 2

⇒ y(1 + x2) = tan −1 (x) + C

π  π 0 = 3  + C ⇒ C = −  3 3

So, ∴ π Now, at x = 6

2

 3 π π π π y  = 3 − =− ⇒ y=− 36 3 4 2 3  2 2

2

y(0) = 0

Q

π2 y(cos x) = 3x2 − 3

2

dx +C 1 + x2

∴

C =0

∴

y(1 + x2) = tan −1 x tan x 1 + x2  tan −1 x ay = a    1 + x2 

⇒

y=

2

⇒

6. Given differential equation is

[multiplying both sides by a]

dy + 2 y = x2, (x ≠ 0) dx dy  2 +   y = x, dx  x x

⇒

Now,

which is a linear differential equation of the form dy + Py = Q dx 2 Here, P = and Q = x x ∫ IF = e

∴

2 dx x

y × IF = ∫ (Q × IF) dx + C x4 +C 4

∴ y(x2) = ∫ (x × x2) dx + C ⇒ yx2 =

∴

1 3 + C ⇒C = 4 4 3 x4 3 x2 2 + + ⇒y= yx = 4 4 x2 4 4

y(1) = 1, so 1 =

∴

the curve is

⇒

= eln(1 + x

dx

2

)

(given) …(i)

which is a linear differential equation of the form dy + P (x) ⋅ y = Q (x), dx 2 where P (x) = and Q (x) = x x Now, integrating factor 2

P ( x )dx ∫ dx (IF) = e∫ = e x = e2log e x 2

[Q m log a = log am ]

= (1 + x2)

y(IF) = ∫ Q (x)(IF)dx + C ⇒ y(x2) = ∫ x ⋅ x2 dx + C x4 …(ii) +C 4 Q The curve (ii) passes through the point (1, − 2), therefore 1 9 −2 = + C ⇒C = − 4 4 ∴ Equation of required curve is 4 yx2 = x4 − 9. Now, checking all the option, we get only ( 3 , 0) satisfy the above equation.

9. Given differential equation is dy + y = x log e x, (x > 1) dx dy 1 + y = log e x dx x

x

y ⋅ (IF) = ∫ Q (IF)dx + C 1 (1 + x2)dx + C (1 + x2)2

yx2 =

⇒

and required solution of differential Eq. (i) is given by

⇒ y(1 + x2) = ∫

dy x2 − 2 y = dx x dy 2 + y=x dx x

= x2 [Q elog e f ( x ) = f (x)] and the solution of differential Eq. (i) is

[dividing each term by (1 + x2)2] …(i) This is a linear differential equation of the form dy + P⋅ y =Q dx 2x 1 and Q = Here, P = 2 (1 + x ) (1 + x2)2 ∫ 2 ∴Integrating Factor (IF) = e 1 + x

1 1 ⇒a = 4 16

= elog e x

dy (x2 + 1)2 + 2x(x2 + 1) y = 1 dx dy  2x  1 + y= dx  1 + x2 (1 + x2)2

2x

a =

aπ π (given) = 8 32

8. We know that, slope of the tangent at any point (x, y) on

7. Given differential equation is

⇒

at x = 1

π  tan −1 (1) a y (1 ) = a   = a 4 = 2  1+1 

= e2log x = x2

Since, solution of the given differential equation is

Q

[Q C = 0]

−1

⇒

…(i)

368 Differential Equations Which is a linear differential equation. So, if =

1 ∫ dx e x

if

= elog e x = x

y × x = ∫ (log e x) x dx + C yx =

x2 x2 1 log e x − ∫ × dx + C 2 2 x [using integration by parts]

x x2 log e x − +C 2 4 Given that, 2 y(2) = log e 4 − 1 On substituting, x = 2, in Eq. (ii), we get 4 4 2 y(2) = log e 2 − + C, 2 4 ⇒

yx =

2

…(ii) …(iii)

2 y(2) = log e 4 − 1 + C

−2 log e 2 −2log e 2 1 e = log e 2 elog e 2 2 2 1 1 −2 = . log e 2 ⋅ 2 = log e 2 2 8 3 f (x) 11. Given, f ′ (x) = 7 − , (x > 0) 4 x dy On putting f (x) = y and f ′ (x) = , then we get dx dy 3 y = 7− dx 4 x dy 3 + y=7 ⇒ dx 4x

3

…(iv) [Q m log a = log am ]

From Eqs. (iii) and (iv), we get C =0 So, required solution is x2 x2 yx = log e x − 2 4 e2 e2 Now, at x = e, ey(e) = log e e − 2 4

∫ dx Now, integrating factor (IF) = e 4x 3

y(e) =

e 4

yx3/ 4 = ∫ 7x3/ 4dx + C 3

 1 + 2x   dx x 

∫ =e 

ln x + 2x

1

yx

⇒

y x3/ 4 = 4x 4 + C

y = 4 x + C x− 3 / 4 y = f (x) = 4x + C ⋅ x−3/ 4  1 4 Now, f   = + C ⋅ x3/ 4  x x  4  1 lim x f   = lim x + Cx3/ 4 ∴   x x→ 0 +  x x→ 0 + ⇒ So,



∫  + 2 dx = e x 

= lim (4 + Cx7/ 4 ) = 4 x→ 0 +

12. Given, differential equation is

=e = e . e = x. e and the solution of the given equation is ln x

2x

2x

y ⋅ (IF) = ∫ (IF) Q dx + C ⇒

y(xe2x ) = ∫ (x e2x . e−2x ) dx + C = ∫ x dx + C =

x2 +C 2

1 −2 e when x = 1 2 1 −2 2 1 ∴ e . e = + C ⇒ C = 0 (using Eq. (i)) 2 2 x x2 2x ⇒ y = e−2x y (xe ) = ∴ 2 2 dy 1 −2x x −2x 1  Now, = e + e (− 2) = e−2x  − x < 0, dx 2 2 2  Since, y =

+1

x4 =7 +C 3 +1 4

⇒

3/ 4

7

[Q log e e = 1].

dy  2x + 1 − 2x + y=e dx  x  dy which is of the form + Py = Q, where dx 2x + 1 and Q = e−2x P= x Pdx

3/ 4

y(IF) = ∫ (Q ⋅ (IF))dx + C

10. We have,

Now, IF = e∫

log x

= e4 = elog x = x3/ 4 and solution of differential Eq. (i) is given by

[where, y(e) represents value of y at x = e]

⇒

…(i)

which is a linear differential equation of the form dy 3 and Q = 7. + Py = Q, where P = dx 4x

[where, y(2) represents value of y at x = 2]

⇒

[by using product rule of derivative]

and y(log e 2) =

Now, solution of differential Eq. (i), is

⇒

1 < x 0,

y (1) = 1

Since, ⇒

∴ Option (d) is incorrect.

∴

c=2 x Now, Eq. (i) becomes, + y = 2 y

23. PLAN

Linear differential equation under one variable. Pdx dy + Py = Q; IF = e ∫ dx ∴ Solution is, y (IF) = ∫ Q ⋅ (IF) dx + C

Again, for x = − 3 ⇒

⇒

y − 2y − 3 = 0

⇒

( y + 1) ( y − 3) = 0

∴

As y > 0, take y = 3, neglecting y = − 1.

Solution is y ⋅ cos x = ∫ 2x sec x ⋅ cos x dx + C

dy  t  1 and y (0) = − 1 −  y= dt  1 + t  (1 + t )

Which represents linear differential equation of first order. 

t



∫ −  1 + t  dt = e− t + log (1 + t ) = e− t ⋅ (1 + t ) IF = e 

∴

⇒

ye

−t

2  π  2π y  =  3 9

y (0) = − 1 −1 ⋅ e0 (1 + 0) = − e0 + c c=0 y=−

∴

22. Here, f ′ (x) = 2 −

1 ⇒ (1 + t )

y (1) = −

1 2

f (x) x

1

∫ dx Integrating Factor, IF = e x = elog x = x ∴ Required solution is y ⋅ (IF) = ∫ Q (IF) dx + C y(x) = ∫ 2(x) dx + C

⇒

yx = x + C C y=x+ x

[QC ≠ 0, as f (1) ≠ 1]

 1 (a) lim f ′   = lim (1 − Cx2) = 1  x x → 0 + x → 0+ ∴ Option (a) is correct.  1 (b) lim x f   = lim (1 + Cx2) = 1  x x → 0 + x → 0+ ∴ Option (b) is incorrect. (c) lim x2f ′ (x) = lim (x2 − C ) = − C ≠ 0 x→ 0

+

x→ 0

24. Let

w (x) = u (x) − v (x)

and

h (x) = f (x) − g (x)

2  π  4π 2π y′   = +  3 3 3 3

…(i)

On differentiating Eq. (i) w.r.t. x dw du dv = − dx dx dx

+

∴ Option (c) is incorrect.

[given]

= { f (x) − g (x)} − p (x) [u (x) − v (x)] dw …(ii) ⇒ = h (x) − p (x) ⋅ w (x) dx dw + p (x) w (x) = h (x) which is linear differential ⇒ dx equation . The integrating factor is given by p ( x ) dx IF = e∫ = r (x)

2

∴

⇒

= { f (x) − p (x) ⋅ u (x)} − { g (x) − p (x) v (x)}

dy y + = 2 [i.e. linear differential equation in y] dx x

⇒

y ⋅ cos x = x2 + C y(0) = 0 ⇒ C = 0 y = x2 sec x π2  π y  =  4 8 2 π π2  π y′   = +  4 2 8 2

⇒

(1 + t ) = − e− t + c

Since, ⇒

⇒ As ∴ Now,

Required solution is, 1 ye− t (1 + t ) = ∫ ⋅ e− t (1 + t ) dt + c = ∫ e− t dt + c 1+ t

or

y ′ − y tan x = 2x sec x and y (0) = 0 dy − y tan x = 2x sec x dx IF = ∫ e− tan x dx = elog|cos x| = cos x

–3 + y2 = 2 y 2

21. Given,

lim f (x) = ∞

x → 0+

∴ Function is not bounded in (0, 2).

x = 1, y = 1

⇒

C ,C ≠0 x

(d) f (x) = x +

[let]

On multiplying both sides of Eq. (ii) of r (x), we get dw r (x) ⋅ + p (x) (r (x)) w (x) = r (x) ⋅ h (x) dx d  dr  [r (x) w (x)] = r (x) ⋅ h (x) ⇒ Q = p (x) ⋅ r (x)  dx  dx Now, and Thus,

r (x) = e∫

P ( x ) dx

> 0, ∀ x

h (x) = f (x) − g (x) > 0, for x > x1 d [r (x) w (x)] > 0, ∀ x > x1 dx

r (x) w (x) increases on the interval [x, ∞ [

372 Differential Equations Therefore, for all x > x1 r (x) w (x) > r (x1 ) w (x1 ) > 0 [Q r (x1 ) > 0 and u (x1) > v (x1)] ⇒

w (x) > 0 ∀ x > x1

⇒

u (x) > v (x) ∀ x > x1

[Q r (x) > 0]

Hence, there cannot exist a point (x, y) such that x > x1 and y = u (x) and y = v (x).

25.

dy + y ⋅ g′ (x) = g (x) g′ (x) dx g ′( x ) dx IF = e∫ = e g( x ) ∴ Solution is y (e g( x ) ) = ∫ g (x) ⋅ g′ (x) ⋅ e g( x ) dx + C Put

g (x) = t, g′ (x) dx = dt y(eg( x ) ) = ∫ t ⋅ et dt + C

= t ⋅ et − ∫ 1 ⋅ et dt + C = t ⋅ et − et + C …(i) y eg( x ) = ( g (x) − 1) eg( x ) + C Given, y(0) = 0, g (0) = g (2) = 0 ∴ Eq. (i) becomes, y(0) ⋅ eg( 0) = ( g (0) − 1) ⋅ eg( 0) + C ⇒ 0 = (− 1) ⋅ 1 + C ⇒ C = 1 ∴ y(x) ⋅ eg( x ) = ( g (x) − 1) eg( x ) + 1 ⇒ y(2) ⋅ eg( 2) = ( g (2) − 1) eg( 2) + 1, where g(2) = 0 ⇒ y(2) ⋅ 1 = (− 1) ⋅ 1 + 1 y(2) = 0

Topic 3 Applications of Homogeneous Differential Equations

⇒

Hence, option (a) is correct. dy 2 y 2. Given, = dx x2 dy 2 [integrating both sides] ⇒ ∫ y = ∫ x2dx 2 …(i) log e| y| = − + C ⇒ x Since, curve (i) passes through centre (1, 1) of the circle x2 + y2 − 2x − 2 y = 0 2 log e (1) = − + C ⇒ C = 2 ∴ 1 ∴ Equation required curve is 2 [put C = 2 in Eq. (i)] log e| y| = − + 2 x ⇒

x log e| y| = 2(x − 1)

3. Given differential equation is (x2 − y2)dx + 2xy dy = 0, which can be written as dy y2 − x2 = 2xy dx Put y = vx

[Q it is in homogeneous form] dy dv ⇒ = v+ x dx dx Now, differential equation becomes v+ x

1. Given differential equation is 2x2dy = (2xy + y2) dx dy 2xy + y2 ⇒ = dx 2 x2 [Homogeneous differential equation] dy dv Let y = vx = =v+ x dx dx ∴ The differential equation becomes dv 2vx2 + v2x2 = v+ x dx 2x2 dv 2v + v2 − 2v ⇒ x = dx 2 dv v2 dv dx ⇒2 2 = x = ⇒ dx 2 x v 2x  −1  …(i) + C =0 ⇒ 2   = log e x + C ⇒ log e x +  v y y as v= x Q The curve (i) passes through the point (1, 2). 2x So, c = − 1 ∴ log e x + =1 y 1 Now, at x= 2

1  1 1 log e   + = 1 ⇒ = 1 + log e 2  2 y y 1  1 y= f   =  2 1 + log e 2

⇒ ⇒ ⇒

x

dv v2x2 − x2 dv (v2 − 1)x2 = ⇒ v+ x = 2x(vx) dx dx 2vx2 dv v2 − 1 v2 − 1 − 2v2 = −v= dx 2v 2v

dv 1 + v2 dx 2v dv ⇒ ∫ = −∫ =− 2 dx 2v x 1+ v 2 ln (1 + v ) = − ln x − ln C f ′ (x)   Q ∫ f (x) dx ⇒ ln| f (x)|+ C  x

⇒ ln|(1 + v2)Cx|= 0 ⇒

[Q ln A + ln B = ln AB]

(1 + v )Cx = 1 [log e x = 0 ⇒ x = e0 = 1] y Now, putting v = , we get x  y2  1 + 2  Cx = 1 ⇒ C (x2 + y2) = x x   2

Q The curve passes through (1, 1), so 1 C (1 + 1) = 1 ⇒ C = 2 Thus, required curve is x2 + y2 − 2x = 0, which represent a circle having centre (1, 0) ∴ The solution of given differential equation represents a circle with centre on the X-axis.

Differential Equations 373 4. Given, differential equation is

dp 1 − p(t ) = −200 is a dt 2

linear differential equation. −1 Here, p(t ) = , Q (t ) = −200 2  1

Hence, solution is

t 2

t − p(t )⋅ e 2

−

t

+K

If p(0) = 100, then k = − 300

t

y =v x dv v+ x = v + sec (v) dx dv dx ∫ sec v = ∫ x

∴ ⇒

dv = sec (v) dx

∫ cos v dv = ∫

dx x

 y sin v = log x + log c ⇒ sin   = log(cx)  x

⇒

 As it passes through 1,  log c =

⇒

π  6

 π ⇒ sin   = log c  6

1 2

1  y sin   = log x +  x 2

∴

6. Given, ⇒

⇒ x ⇒

dP = (100 − 12 x ) dx dP = (100 − 12 x ) dx

On integrating both sides, we get

∫

dP = ∫ (100 − 12 x ) dx P = 100x − 8x3/ 2 + C

When x = 0, then P = 2000 ⇒ C = 2000 Now, when x = 25, then is P = 100 × 25 − 8 × (25)3/ 2 + 2000 = 2500 − 8 × 125 + 2000 = 4500 − 1000 = 3500

dy = ±

1 − sin 2 θ cos 2 θ dθ cos θdθ = ± sin θ sin θ = ± (cos ecθ − sin θ )dθ ⇒ y = ± [ln (cos ecθ − cot θ ) + cos θ ] + C   1 − cos θ   ⇒ y = ± ln   + cos θ  + C   sin θ     1 − 1 − sin 2 θ    + 1 − sin 2 θ  + C ⇒ y = ± ln    sin θ     2   1 − 1 − x   + 1 − x2  + C [Qx = sin θ] ⇒ y = ± ln    x       1 + 1 − x2 = ± − ln + 1 − x2  + C x   [on rationalization] Q The curve is in the first quadrant so y must be positive, so  1 + 1 − x2   − 1 − x2 + C y = ln    x   dy = ±

Here, slope of the curve at (x, y) is dy y  y = + sec    x dx x Put

[on replacing h by x]

1 − x2 dx x On putting x = sin θ, dx = cos θ dθ, we get

To solve homogeneous differential equation, i.e. substitute y =v x dy dv ∴ y = vx ⇒ =v +x dx dx

PLAN

dy 1 − x2 =± dx x

⇒

p(t ) = 400 − 300 e2

⇒

 dy h 2 + h 2  = 1  dx

⇒

p(t ) = 400 + ke−1/ 2

⇒

5.

2

⇒

= ∫ −200 ⋅ e 2dt t − = 400 e 2

…(i)

So, PY p = 1 (given)

p (t ) ⋅ IF = ∫ Q (t ) ⋅ IF dt −

tangent to the curve at point P is  dy y−k =   (x − h )  dx h, k

Now, the tangent (i) intersect the Y -axis at Y p, so dy  dy dy  coordinates Y p is 0, k − h =   , where  dx  dx ( h, k) dx

t

− ∫ −   dt IF = e  2 = e 2

p(t )⋅ e

7. Let a point P (h , k) on the curve y = y(x), so equation of

As curve passes through (1, 0), so 0 = 0 − 0 + c ⇒ c = 0, so required curve is  1 + 1 − x2   − 1 − x2 y = ln    x   and required differential equation is dy 1 − x2 =− dx x ⇒ xy′+ 1 − x2 = 0 Hence, options (a) and (c) are correct. dy 8. Given, (x2 + xy + 4x + 2 y + 4) − y2 = 0 dx dy ⇒ [(x2 + 4x + 4) + y(x + 2)] − y2 = 0 dx dy [(x + 2)2 + y(x + 2)] − y2 = 0 ⇒ dx

374 Differential Equations Put x + 2 = X and y = Y , then (X 2 + XY )

dY −Y 2 = 0 dX

⇒

X 2dY + XYdY − Y 2dX = 0

⇒

X 2dY + Y (XdY − YdX ) = 0 dY XdY − YdX − = Y X2 Y  − d (log|Y |) = d    X

⇒ ⇒

|x + 3|2  (x + 3)2 =0 log  +  3e  (x + 2)

i.e.

To check the number of solutions.

∴

On integrating both sides, we get Y − log|Y| = + C, where x + 2 = X and y = Y X y …(i) ⇒ − log| y| = +C x+ 2

g′ (x) = =

− log 3 = 1 + C

⇒

(x + 3)(x + 1) 2 + x+3 (x + 2)2

g (x) is increasing, when x > 0. when x > 0, then g (x) > g (0)  3 9 g (x) > log   + > 0  e 4

Thus,

Hence, there is no solution. Thus, option (d) is true.

C = − 1 − log 3 = − (log e + log 3)

9. Since, BP : AP = 3 : 1. Then, equation of tangent is

= − log 3e

Y − y = f ′ (x) (X − x) The intercept on the coordinate axes are   y A x − , 0  f ′ (x) 

∴ Eq. (i) becomes y − log (3e) = 0 x+2 y | y| log   + =0  3e  x + 2

log| y| + ⇒

 (x + 2) ⋅ 2 (x + 3) − (x + 3)2 ⋅ 1 2 +  −0 x+ 3  (x + 2)2 

Clearly, when x > 0, then, g′ (x) > 0 ∴

Since, it passes through the point (1, 3). ∴

…(ii)

Now, to check option (a), y = x + 2 intersects the curve. |x + 2| x + 2 |x + 2| log  = 0 ⇒ log  ⇒ +  = −1  3e  x + 2  3e  ⇒

|x + 2| 1 = e−1 = 3e e

⇒

|x + 2| = 3 or x + 2 = ± 3

∴ x = 1, − 5 (rejected), as x > 0

B [0, y − x f ′ (x)]

and

Since, P is internally intercepts a line AB,  y  3 x −  + 1 ×0  f ′ (x) x= 3+1

∴ Y B

(1, 1) 3

[given]

P

∴ x = 1 only one solution. To check option (c), we have y | y| y = (x + 2)2 and log   + =0  3e  x + 2 |x + 2|2  (x + 2 )2 |x + 2|2  + = 0 ⇒ log  ⇒ log    = − (x + 2) x+2  3e   3e  (x + 2) 3e = e−( x + 2) or (x + 2)2 ⋅ ex + 2 = 3e ⇒ex+ 2 = 3e (x + 2)2 2

Y

e x +2

O

y = f (x) A

X

dy y dy 1 = ⇒ =− dx dx − 3x y 3x

⇒

On integrating both sides, we get xy3 = c Since, curve passes through (1, 1), then c = 1. ∴

xy3 = 1

1 ⇒ y=2 8 Hence, (a) and (c) are correct answers. At

x=

10. Since, rate of change of volume ∝ surface area

e2 3e /4

(x, y)

1

Thus, (a) is the correct answer.

⇒

(x + 3)2 − log (3e) (x + 2)

Let g (x) = 2 log (x + 3) +

3e /( x + 2)2 X

Clearly, they have no solution. To check option (d), y = (x + 3)2

⇒ ⇒

dV ∝ SA dt dr 4 πr 2 ⋅ = − λ 4 πr 2 dt

dr = − λ is required differential equation. dt

Differential Equations 375  dx  dy

 y2 1 + 

⇒

 dx    dy

 =1  

dy =± dx

⇒

1 − y2 dy = ± y

⇒ ⇒

y

∫

dr = − k ∫

R cot θ = kT H T= k

⇒ x dx

T

dt

0

⇒ H = kT [from Eq. (iii)]

∴ Required time after which the cone is empty, T =

1− y dy = ± x + C y

H k

13. Let O be the centre of hemispherical tank. Let at any

instant t, water level be BAB1 and at t + dt, water level is B′ A′ B1. Let ∠ O1OB1 = θ.

Put y = sin θ ⇒ dy = cos θ dθ cos θ ∴ ∫ sin θ ⋅ cos θ dθ = ± x + C ⇒

0 R

cot θ (0 − R) = − k (T − 0)

⇒

1 − y2

∫

cot θ ∫

2

∫

⇒

=1

2

∴

∫

On substituting Eq. (iv) in Eq. (i), we get 1 dr cot θ ⋅ 3r 2 = − kπ r 2 3 dt

2

11. Since, the length of tangent = y 1 +  

O

cos 2 θ ⋅ sin θ dθ = ± x + C sin 2 θ

O1

θ

A

B B'

B1 B ′1

A′

Again put cos θ = t ⇒ − sin θ dθ = dt −∫

∴ ⇒

∫

t2 dt = ± x + C 1 − t2

O2

 1  1 −  dt = ± x + C  1 − t 2

⇒

t − log

⇒ 1 − y2 − log

⇒ AB1 = r cos θ and OA = r sin θ decrease in the water volume in time dt = π AB12 ⋅ d (OA ) [πr 2 is surface area of water level and d (OA ) is depth of water level]

1+ t =± x+C 1−t

1 + 1 − y2 1 − 1 − y2

= πr 2 ⋅ cos 2 θ ⋅ r cos θ dθ = πr3 ⋅ cos3 θ dθ

=±x+C

Also, h (t ) = O2A = r − r sin θ = r (1 − sin θ )

12. Given, liquid evaporates at a rate proportional to its

Now, outflow rate Q = A ⋅ v (t ) = A ⋅ 0.6 2 gr (1 − sin θ ) Where, A is the area of the outlet.

surface area. dV ∝ −S dt

⇒

…(i)

1 We know that, volume of cone = πr 2h 3

Thus, volume flowing out in time dt. ⇒

Q dt = A ⋅ (0.6) ⋅ 2 gr ⋅ 1 − sin θ dt

We have, πr3 cos3 θ dθ = A (0.6) ⋅ 2 gr ⋅ 1 − sin θ dt ⇒

R

πr3 ⋅ A (0.6) 2 gr

cos3 θ dθ = dt (1 − sin θ )

Let the time taken to empty the tank be T. r

Then,

H

h

= Let

and surface area = πr 2 1 or V = πr 2h and S = πr 2 3 R r Where, tan θ = = tan θ and H h From Eqs. (ii) and (iii), we get 1 V = πr3 cot θ and 3

T =∫

S = πr

2

π/ 2 0

πr3 cos3 θ ⋅ dθ 1 − sin θ A (0.6) ⋅ 2 gr

− πr3 A (0.6) 2 gr

π/ 2 1

∫0

t1 = 1 − sin θ − cos θ dt1 = dθ 1 − sin θ

…(ii)

⇒

…(iii)

∴

T=

−2πr3 A (0.6) 2 gr

∫1

⇒

T=

−2πr3 A (0.6) 2 gr

∫1

…(iv)

− sin 2 θ (− cos θ ) dθ 1 − sin θ

0

[1 − (1 − t12)2] dt1

0

[1 − (1 + t14 − 2t12)] dt1

376 Differential Equations ⇒

T=

−2πr3 A (0.6) 2 gr

∫1

⇒

T=

2πr3 A (0.6) 2 gr

∫1

⇒

 t15 2t13  2 πr T=  − 3  A (0.6) 2 gr  5 1

⇒

T=

⇒

T=

0

[1 − 1 − t14 + 2t12] dt1

0

(t14 − 2t12) dt1

Thus, the least integral values of the year n, when the country becomes self-sufficient is the smallest integer log 10 − log 9 greater than or equal to . log (1.04) − 0.03

15. Equation of normal at point (x, y) is

0

3

2π ⋅ r5/ 2 6 A   2 gr  10

Y − y=−

2π ⋅ 25/ 2 (102) 5/ 2 2 1  −   3 12 ⋅ ⋅ 2 ⋅ g 3 5  5

Also, distance between P and X-axis is |y|. dx y+ ⋅x dy ∴ = | y| 2  dx 1+    dy

2π × 105 × 7 14π × 105 unit = 3⋅3⋅ g ⋅3 27 g

14. Let X 0 be initial population of the country and Y 0 be its initial food production. Let the average consumption be a unit. Therefore, food required initially aX 0. It is given  90  …(i) Y p = aX 0   = 0.9 aX 0  100 Let X be the population of the country in year t. dX Then, = Rate of change of population dt 3 = X = 0.03 X 100 dX ⇒ = 0.03 dt X dX ⇒ ∫ X = ∫ 0.03 dt ⇒ ⇒

log X = 0.03 t + c X = A ⋅ e0. 03 t , where A = ec

At t = 0, X = X 0, thus X 0 = A ∴

X = X 0 e0. 03 t

Let Y be the food production in year t. t

Then,

4   t Y = Y 0 1 +  = 0.9aX 0 (1.04)  100

Q

Y 0 = 0.9 aX 0

[from Eq. (i)]

Food consumption in the year t is aX 0 e0. 03 t . Again, ⇒ ⇒

Y − X ≥0 0.9 X 0 a (1.04)t > a X 0 e0. 03 t t

(1.04) 1 10 > = . 9 e0. 03 t 0.9

Taking log on both sides, we get t[log (1.04) − 0.03] ≥ log 10 − log 9 log 10 − log 9 t≥ ⇒ log (1.04) − 0.03

…(i)

Distance of perpendicular from the origin to Eq. (i) dx ⋅x y+ dy = 2  dx 1+    dy

1 2  ⋅ 0 − −0 +  5 3 

2π × 105 ⋅ 4 ⋅ 5 10 − 3  = (12 × 3) g  15  =

dx (X − x) dy

[given]

⇒

y2 +

 dx 2 dx ⋅ x + 2xy = y2 1 + dy dy 

2  dx      dy  

2

⇒ ⇒

dx  dx 2 2 =0   (x − y ) + 2xy  dy dy dx dy

  dx 2 2  dy (x − y ) + 2xy = 0  

dx =0 dy dx But =0 dy

⇒

or

dy y2 − x2 = 2xy dx

⇒ x = c, where c is a constant. Since, curve passes through (1, 1), we get the equation of the curve as x = 1. dy y2 − x2 is a homogeneous equation. The equation = 2xy dx dy dv Put y = vx ⇒ =v+ x dx dx dv v2x2 − x2 = v+ x dx 2x2v 2 dv v − 1 v2 − 1 − 2v2 v2 + 1 ⇒x = −v= =− dx 2v 2v 2v dx −2 v ⇒ dv = x v2 + 1 ⇒

c1 − log (v2 + 1) = log|x|

⇒

log| x|(v2 + 1) = c1

  y2 ⇒ | x|  2 + 1 = ec1  x

⇒ x2 + y2 = ± ec1 x or x2 + y2 = ± ecx is passing through (1, 1). ∴ 1 + 1 = ± ec ⋅ 1 ⇒ ± ec = 2 Hence, required curve is x2 + y2 = 2x .

Differential Equations 377 16.

dV ∝ V for each reservoir. dt dV ∝ − VA ⇒ dx

Put tan t /2 = u 1 2 du sec2 t / 2 dt = du ⇒ dt = ⇒ 2 sec2 t / 2

dV A = − K 1V A dt

[K 1 is the proportional constant] V ′A

∫V

⇒ ⇒

log

A

t dV A = − K 1 ∫ dt 0 VA

V ′A = − K 1t VA

Similarly for B,

⇒ V A′ = V A ⋅ e−K1 t

VB′ = VB ⋅ e−K 2t

dt =

∴

I1 = ∫

…(ii) − K2) t

It is given that at t = 0, V A = 2 VB and at 3 3 t = , V A′ = VB′ 2 2 3 3 − (K1 − K 2 )t Thus, …(iii) = 2⋅ e ⇒ e− (K1 − K 2 ) = 2 4 Now, let at t = t0 both the reservoirs have some quantity of water. Then, V A′ = VB′ From Eq. (iii), 2e−(K − K 2 ) = 1  3 2⋅    4

⇒

17. Given, Let ⇒

t0

=

2 5

=

2 2 2 5∫  3  4 u +  +     5 5

dy 1 = dx 6

⇒

…(i)

 dt  − 10   dx 

On integrating both sides, we get 1 dt =x+ c 2 ∫ 3 sin t + 5 Let

I1 = ∫

dt = 3 sin t + 5 ∫

=∫

dt  2 tan t / 2  3  +5  1 + tan 2 t / 2 (1 + tan 2 t / 2) dt t  2 t 6 tan + 5 + 5tan   2 2

⇒

⇒ ⇒

Now, the given differential equation becomes 1  dt  sin t =  − 10  6  dx dt ⇒ 6 sin t = − 10 dx dt ⇒ = 6 sin t + 10 dx dt ⇒ = dx 6 sin t + 10

∫

∫

du 6 u2 + u + 1 5

du 6 9 9 − +1 u2 + u + 5 25 25 du

=

2 5  u + 3 / 5 ⋅ tan −1    4 /5  5 4

=

1 tan −1 2

=

1 5 tan t / 2 + 3  tan −1   2 4

5u + 3   4 

On putting this in Eq. (ii), we get

=1

t0 = log3/ 4 (1 / 2) dy = sin (10x + 6 y) dx 10x + 6 y = t dy  dt  10 + 6 =  dx  dx

2 (1 + u 2)du 2 = 2 2 (1 + u ) (5u + 6u + 5) 5

…(i)

On dividing Eq. (i) by Eq. (ii), we get V A′ V A − (K1 = ⋅e VB′ VB

2 du 2 du ⇒ dt = 1 + tan 2 t / 2 1 + u2

⇒

t   5 tan + 3 1  −1  2 tan  =x+ c 4 4     t   5 tan + 3   2 tan −1   = 4x + 4c 4     1 [5 tan (5x + 3 y) + 3] = tan (4x + 4c) 4 5 tan (5x + 3 y) + 3 = 4 tan (4x + 4c)

When x = 0, y = 0, we get ⇒ ⇒ Then,

5 tan 0 + 3 = 4 tan (4c) 3 = tan 4c 4 3 4c = tan −1 4  5 tan (5x + 3 y) + 3 = 4 tan 4x + tan −1 

⇒ tan (5x + 3 y) = …(ii)

4 3 3  tan 4x + tan  −  5 4 5

4 ⇒ 5x + 3 y = tan −1  5 4 3 y = tan −1  ⇒ 5 ⇒

y=

3  4

  −1 3   3  − tan 4x + tan 4  5   3   −1 3     −  − 5x tan 4x + tan  4  5 

4  1  tan −1  tan 4x + tan −1  3 5 

3  3  5x −  − 4  5  3

378 Differential Equations 18. A. I = ∫ =∫

π /2 0

π /2 0

x

(sin x)cos {cos x ⋅ cot x − log (sin x)sin x } dx

Hence,

d (sin x)cos x dx = 1 dx

∴

B. The point of intersection of − 4 y2 = x and x − 1 = − 5 y2 is (− 4, − 1) and (− 4, 1).

For ⇒

Y (– 4, 1)

∴

dy 3x −1 = + 3x −1 ⋅ log 3 ⋅ log x dx x  dy =1    dx (1, 0) y = xx − 1 dy = xx (1 + log x) dx  dy =1    dx (1, 0)

If θ is angle between the curves, then tan θ = 0. X′

(1, 0)

X

⇒ D.

(– 4, –1)

⇒ Y′

∴ Required area 1 1 = 2 ∫ (1 − 5 y2) dy − ∫ − 4 y2 dy 0   0 3 1

1  y 4 = sq units = 2 ∫ (1 − y2)dy = 2  y −  0 3 0 3 

C. The point of intersection y = 3x −1 log x and y = xx − 1 is (1, 0).

⇒ ⇒

θ = 0° dy 2 = dx x + y dx x y − = dy 2 2 1 xe− y/ 2 = ⋅ ∫ y ⋅ e− y/ 2dy 2  ye− y 2 1 e− y/ 2  +k xe− y/ 2 =  − 2  − 1 / 2 (1 / 2)2 

⇒ x + y + 2 = key/ 2 It passing through (1, 0). ⇒ k =3 ∴

x + y + 2 = 3e y/ 2

15 Straight Line and Pair of Straight Lines Topic 1 Various Forms of Straight Line Objective Questions I (Only one correct option) 1. The set of all possible values of θ in the interval (0, π) for which the points (1, 2) and(sin θ , cos θ) lie on the same (2020 Main, 2 Sep II) side of the line x + y = 1 is

π (a)  0,   2 3π  (c)  0,   4

π 3π  (b)  ,  4 4 π (d)  0,   4

(a) x + 3 y = 8 (b) ( 3 + 1) x + ( 3 − 1) y = 8 2 (c) 3x + y = 8 (d) ( 3 − 1)x + ( 3 + 1) y = 8 2

2x + a 2y = 1, (a ∈ R − {0, 1}) are perpendicular, then the distance of their point of intersection from the origin is (2019 Main, 9 April II)

2 (a) 5 2 (c) 5

2 (b) 5 2 (d) 5

intersecting the line, x + y = 7 at a distance of 4 units from P, is (2019 Main, 9 April I)

(2019 Main, 12 April I)

second and third quadrants only first, second and fourth quadrants first, third and fourth quadrants third and fourth quadrants only

1− 5 1+ 5 1− 7 (c) 1+ 7

(b)

(a)

(d)

7−1 7+1 5−1 5+1

8. If a point R(4, y, z ) lies on the line segment joining the

4. Lines are drawn parallel to the line 4x − 3 y + 2 = 0, at a 3 from the origin. Then which one of the 5 following points lies on any of these lines? distance

(2019 Main, 10 April I)

1 2 (b)  − ,   4 3 1 1 (d)  ,   4 3

(b) rhombus of area 8 2 sq units

7. Slope of a line passing through P (2, 3) and

3. The equation y = sin x sin(x + 2) − sin 2(x + 1) represents a

1 2 (a)  − , −   4 3 1 1  (c)  , −  4 3

(2019 Main, 10 April I)

(a) rhombus of side length 2 units

6. If the two lines x + (a − 1) y = 1 and

makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with the line x + y = 0. Then, an equation of the line L is (2019 Main, 12 April II)

(a) (b) (c) (d)

bounded by a

(c) square of side length 2 2 units (d) square of area 16 sq units

2. A straight line L at a distance of 4 units from the origin

straight line lying in

5. The region represented by|x − y| ≤ 2 and|x + y| ≤ 2 is

points P(2, − 3, 4) and Q(8, 0, 10), then the distance of R from the origin is (2019 Main, 8 April II) (a) 2 21

(b)

53

(c) 2 14

(d) 6

9. Suppose that the points (h , k), (1, 2) and (−3, 4) lie on the line L1. If a line L 2 passing through the points (h , k) and (4, 3) is perpendicular to L1, then k /h equals (2019 Main, 8 April II)

(a) − (c) 3

1 7

1 3 (d) 0

(b)

380 Straight Line and Pair of Straight Lines 10. A point on the straight line, 3x + 5 y = 15 which is equidistant from the coordinate axes will lie only in (2019 Main, 8 April I)

(a) IV quadrant (c) I and II quadrants

(b) I quadrant (d) I, II and IV quadrants

11. If a straight line passing through the point P(− 3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is (2019 Main, 12 Jan II)

(a) x − y + 7 = 0 (c) 3x − 4 y + 25 = 0

(b) 4x − 3 y + 24 = 0 (d) 4x + 3 y = 0

12. If the straight line, 2x − 3 y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals (2019 Main, 12 Jan I) (a)

35 3

(b) − 5

(c) −

35 3

(d) 5

13. If in a parallelogram ABDC, the coordinates of A , B and C are respectively (1, 2), (3, 4) and (2, 5), then the (2019 Main, 11 Jan II) equation of the diagonal AD is (a) 3x + 5 y − 13 = 0 (c) 5x − 3 y + 1 = 0

(b) 3x − 5 y + 7 = 0 (d) 5x + 3 y − 11 = 0

14. The tangent to the curve, y = xex passing through the point (1, e) also passes through the point (2019 Main, 10 Jan II)

(b) (3, 6e)

(c) (2, 3e)

5 (d)  , 2e 3 

15. Two sides of a parallelogram are along the lines, x + y = 3 and x − y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is (2019 Main, 10 Jan II)

(a) (3, 6) (c) (2, 1)

(b) (2, 6) (d) (3, 5)

16. The shortest distance between the point  , 0 and the 3 2

curve y = x , (x > 0), is 3 (a) 2

5 (b) 4



(2019 Main, 10 Jan I)

(c)

3 2

(d)

5 2

17. If the line 3x + 4 y − 24 = 0 intersects the X-axis at the point A and theY -axis at the point B, then the incentre of the triangle OAB, where O is the origin, is (2019 Main, 10 Jan I)

(a) (4, 3) (c) (4, 4)

(b) (3, 4) (d) (2, 2)

18. A point P moves on the line 2x − 3 y + 4 = 0. If Q(1, 4) and R(3, − 2) are fixed points, then the locus of the centroid of (2019 Main, 10 Jan I) ∆PQR is a line 2 (a) with slope 3 (c) parallel to Y -axis

3 (b) with slope 2 (d) parallel to X-axis

19. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is (2018 Main) (a) 3x + 2 y = 6 (c) 3x + 2 y = xy

(b) 2x + 3 y = xy (d) 3x + 2 y = 6xy

(k, − 3k), (5, k) and (− k, 2) has area 28 sq units. Then, the orthocentre of this triangle is at the point (2017 Main)

1 (a)  2, −   2

3 (b)  1,   4

3 (c)  1, −   4

1 (d)  2,   2

21. Let a , b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes, then (2014 Main) (a) 2bc − 3ad = 0 (c) 2ad − 3bc = 0

(b) 2bc + 3ad = 0 (d) 3bc + 2ad = 0

22. If PS is the median of the triangle with vertices P(2, 2), Q(6, – 1) and R(7, 3), then equation of the line passing (2014 Main, 2000) through (1, – 1) and parallel to PS is (a) 4x − 7 y − 11 = 0 (c) 4x + 7 y + 3 = 0

(b) 2x + 9 y + 7 = 0 (d) 2x − 9 y − 11 = 0

23. The x-coordinate of the incentre of the triangle that has the coordinates of mid-points of its sides as (0, 1), (1, 1) and (1, 0) is (2013 Main) (a) 2 +

2

4 (a)  , 2e 3 

20. Let k be an integer such that the triangle with vertices

2

(b) 2 −

2

(c) 1 +

2

(d) 1 −

2

24. A straight line L through the point (3, −2) is inclined at an angle 60° to the line 3x + y = 1. If L also intersects (2011) the X-axis, then the equation of L is

(a) y + 3x + 2 − 3 3 = 0 (c) 3 y − x + 3 + 2 3 = 0

(b) y − 3x + 2 + 3 3 = 0 (d) 3 y + x − 3 + 2 3 = 0

25. The locus of the orthocentre of the triangle formed by the lines (1 + p) x − py + p (1 + p) = 0 ,

(1 + q) x − qy + q (1 + q) = 0 and y = 0, where p ≠ q , is (2009) (a) a hyperbola (b) a parabola (c) an ellipse (d) a straight line

26. Let O(0, 0), P(3, 4) and Q(6, 0) be the vertices of a ∆OPQ.

The point R inside the ∆OPQ is such that the triangles OPR, PQR and OQR are of equal area. The coordinates of R are (2007, 3M)

4 (a)  , 3 3 

2 (b)  3,   3

4 (c)  3,   3

4 2 (d)  ,   3 3

27. Orthocentre of triangle with vertices (0, 0), (3, 4) and (4, 0) is 5 (a)  3,   4

(2003, 2M)

(b) (3, 12)

3 (c)  3,   4

(d) (3, 9)

28. The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4 y = 9 and y = mx + 1 is also an integer, is (2001, 1M) (a) 2

(b) 0

(c) 4

(d) 1

29. A straight line through the origin O meets the parallel

lines 4x + 2 y = 9 and 2x + y + 6 = 0 at points P and Q respectively. Then, the point O divides the segment PQ (2000, 1M) in the ratio

(a) 1 : 2

(b) 3 : 4

(c) 2 : 1

(d) 4 : 3

30. The incentre of the triangle with vertices (1, 3 ), (0, 0) and (2, 0) is  3 (a)  1,   2 

(2000, 2M)

2 1  (b)  ,   3 3

 2 3 (c)  ,  3 2 

(d)  1, 

1   3

Straight Line and Pair of Straight Lines 381 31. If A0 , A1 , A2, A3 , A4 and A5 be a regular hexagon inscribed in a circle of unit radius. Then, the product of the lengths of the line segments A0 A1 , A0 A2 and A0 A4 is (1998, 2M)

(a) 3 / 4

(b) 3 3

3 3 (d) 2

(c) 3

32. If the vertices P , Q , R of a ∆PQR are rational points, which of the following points of the ∆PQR is/are always (1998, 2M) rational point(s) (a) centroid (c) circumcentre

(b) incentre (d) orthocentre

(A rational point is a point both of whose coordinates are rational numbers)

33. If P (1, 2), Q (4, 6), R (5, 7) and S (a , b) are the vertices of a parallelogram PQRS, then (a) a = 2, b = 4 (c) a = 2, b = 3

(1998, 2M)

(b) a = 3, b = 4 (d) a = 3, b = 5

34. The diagonals of a parallelogram PQRS are along the lines x + 3 y = 4 and 6x − 2 y = 7. Then, PQRS must be a (a) rectangle (c) cyclic quadrilateral

(b) square (d) rhombus

(1998, 2M)

35. The graph of the function cos x cos (x + 2) − cos 2 (x + 1) (1997, 2M)

is

2 (a) a straight line passing through (0, − sin 1) with slope 2 (b) a straight line passing through (0, 0) (c) a parabola with vertex (1, − sin 2 1) π (d) a straight line passing through the point  , − sin 2 1 2  and parallel to the X-axis

36. The orthocentre of the triangle formed by the lines xy = 0 and x + y = 1,is

1 (a)  , 2

1  2

1 (b)  , 3

1  3

(1995, 2M)

1 1 (d)  ,   4 4

(c) (0, 0)

37. If the sum of the distance of a point from two perpendicular lines in a plane is 1, then its locus is (a) square (c) straight line

(b) circle (1992, 2M) (d) two intersecting lines

38. Line L has intercepts a and b on the coordinate axes. When, the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q, then (1990, 2M) (a) a 2 + b2 = p 2 + q2

(b)

(c) a 2 + p 2 = b2 + q2

(d)

1 a2 1 a

2

+ +

1 b2 1 p

2

= =

1 p2 1 2

b

+ +

1 q2 1 2

q

39. If P = (1, 0), Q = (−1, 0) and R = (2, 0) are three given

II. Transformation through a distance 2 units along the positive direction of X-axis. π III. Rotation through an angle about the origin in the 4 counter clockwise direction. Then, the final position of the point is given by the (1980, 1M) coordinates 1 7  (a)  ,   2 2 1 7  (c)  − ,   2 2

(b) (−

2, 7 2)

(d) ( 2, 7 2)

41. The points (−a , − b), (0, 0), (a , b) and (a 2, a3 ) are (a) collinear (b) vertices of a rectangle (c) vertices of a parallelogram (d) None of the above

(1979, 2M)

Objective Questions II (One or more than one correct option) 42. Let a, λ, µ ∈R. Consider the system of linear equations ax + 2 y = λ and 3x − 2 y = µ. Which of the following statement(s) is/are correct? (2016 Adv.)

(a) If a = − 3, then the system has infinitely many solutions for all values of λ and µ (b) If a ≠ − 3, then the system has a unique solution for all values of λ and µ (c) If λ + µ = 0, then the system has infinitely many solutions for a = − 3 (d) If λ + µ ≠ 0, then the system has no solution for a = − 3

43. For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and (2014 Adv.) bx + ay + c = 0 is less than 2 2. Then, (a) a + b − c > 0 (c) a − b + c > 0

(b) a − b + c < 0 (d) a + b − c < 0

44. All points lying inside the triangle formed by the points (1, 3), (5, 0) and (– 1, 2) satisfy (a) 3x + 2 y ≥ 0 (c) 2x − 3 y − 12 ≤ 0

(1986, 2M)

(b) 2x + y − 13 ≥ 0 (d) −2x + y ≥ 0

Fill in the Blanks 45. Let the algebraic sum of the perpendicular distance from the points (2, 0) , (0, 2) and (1, 1) to a variable straight line be zero, then the line passes through a fixed point whose coordinates are… . (1991, 2M)

points, then locus of the points satisfying the relation (1988, 2M) SQ 2 + SR2 = 2 SP 2, is

46. The orthocentre of the triangle formed by the lines

(a) a straight line parallel to X-axis (b) a circle passing through the origin (c) a circle with the centre at the origin (d) a straight line parallel to Y-axis

47. If a, b and c are in AP, then the straight line

40. The point (4, 1) undergoes the following three transformations successively I. Reflection about the line y = x.

x + y = 1, 2 x + 3 y = 6 and 4x − y + 4 = 0 lies in quadrant number… . (1985, 2M)

a x + by + c = 0 will always pass through a fixed point whose coordinates are (…) . (1984, 2M)

48. y = 10x is the reflection of y = log10 x in the line whose equation is .... .

(1984, 2M)

382 Straight Line and Pair of Straight Lines 61. Let ABC be a triangle with AB = AC. If D is mid point of

True/False 49. The lines 2x + 3 y + 19 = 0 and 9x + 6 y − 17 = 0 cut the coordinate axes in concyclic points.

(1988, 1M)

50. No tangent can be drawn from the point (5/2, 1) to the circumcircle of the triangle with vertices (1, 3 ), (1985, 1M) (1, − 3 ) and (3, 3 ).

51. The straight line 5x + 4 y = 0 passes through the point of intersection of the straight lines x + 2 y − 10 = 0 and (1983, 1M) 2x + y + 5 = 0.

Analytical & Descriptive Questions 52. A straight line L through the origin meets the line

BC, the foot of the perpendicular drawn from D to AC and F the mid-point of DE. Prove that AF is perpendicular to BE. (1989, 5M)

62. The equations of the perpendicular bisectors of the sides AB and AC of a ∆ABC are x − y + 5 = 0 and x + 2 y = 0, respectively. If the point A is (1, – 2), find the equation of the line BC. (1986, 5M)

63. One of the diameters of the circle circumscribing the rectangle ABCD 4 y = x + 7. If A and B are the points (−3, 4) and (5, 4) respectively, then find the area of (1985, 3M) rectangle.

x + y = 1 and x + y = 3 at P and Q respectively. Through P and Q two straight lines L1 and L 2 are drawn, parallel to 2x − y = 5 and 3x + y = 5, respectively. Lines L1 and L 2 intersect at R, show that the locus of R as L varies, is a straight line. (2002, 5M)

64. Two sides of a rhombus ABCD are parallel to the lines

53. A straight line L with negative slope passes through the

equations 7x − y + 3 = 0 and x + y − 3 = 0 and its third side passes through the point (1, −10). Determine the (1984, 4M) equation of the third side.

point (8, 2) and cuts the positive coordinate axes at points P and Q. Find the absolute minimum value of OP + OQ, as L varies, where O is the origin. (2002, 5M)

54. For points P = (x1 , y1 ) and Q = (x2, y2) of the coordinate plane, a new distance d (P , Q ) is defined by d (P , Q ) = | x1 − x2| + | y1 − y2|. Let O = (0, 0) and A = (3, 2). Prove that the set of points in the first quadrant which are equidistant (with respect to the new distance) from O and A consists of the union of a line segment of finite length and an infinite ray. Sketch this set in a labelled diagram. (2000, 10M)

55. A rectangle PQRS has its side PQ parallel to the line y = mx and vertices PQ and S on the lines y = a , x = b and x = − b, respectively. Find the locus of the vertex R.

y = x + 2 and y = 7x + 3. If the diagonals of the rhombus intersect at the point (1, 2) and the vertex A is on the Y-axis, find possible coordinates of A. (1985, 5M)

65. Two equal sides of an isosceles triangle are given by the

66. The vertices of a triangle are [at1t2, a (t1 + t2)], [at2 t3 , a (t2 + t3 )], [at3 t1 , a (t3 + t1 )]. Find the orthocentre of the triangle.

(1983, 3M)

67. The ends A , B of a straight line segment of constant length c slide upon the fixed rectangular axes OX , OY respectively. If the rectangle OAPB be completed, then show that the locus of the foot of the perpendicular drawn from P to AB is x2 / 3 + y2 / 3 = c2 / 3

(1983, 2M)

68. The points (1, 3) and (5, 1) are two opposite vertices of a

(1996, 2M)

rectangle. The other two vertices lie on the line y = 2x + c. Find c and the remaining vertices. (1981, 4M)

56. A line through A (−5, − 4) meets the line x + 3 y + 2 = 0,

69. Two vertices of a triangle are (5, − 1) and (−2, 3). If the

2x + y + 4 = 0 and x − y − 5 = 0 at the points B, C and D respectively. If (15 / AB)2 + (10 / AC )2 = (6 / AD )2, find the equation of the line. (1993, 5M)

57. Determine all values of α for which the point (α , α 2) lies inside the triangles formed by the lines 2x + 3 y − 1 = 0, x + 2 y − 3 = 0, 5x − 6 y − 1 = 0

(1992, 6M)

58. Find the equations of the line passing through the point (2, 3) and making intercept of length 3 unit between the lines y + 2x = 2 and y + 2x = 5. (1991, 4M)

59. Straight lines 3x + 4 y = 5 and 4x − 3 y = 15 intersect at the point A. Points B and C are chosen on these two lines such that AB = AC. Determine the possible equations of the line BC passing through the point (1990, 4M) (1, 2).

60. A line cuts the X-axis at A (7, 0) and the Y-axis at B (0, − 5). A variable line PQ is drawn perpendicular to AB cutting the X-axis in P and the Y-axis in Q. If AQ (1990, 4M) and BP inters at R, find the locus of R.

orthocentre of the triangle is the origin, find the (1978, 3M) coordinates of the third vertex.

70. One side of a rectangle lies along the line 4x + 7 y + 5 = 0. Two of its vertices are (−3, 1) and (1, 1). Find the equations of the other three sides. (1978, 3M)

Integer & Numerical Answer Type Questions 71. Let A(1, 0), B(6, 2) and C  , 6 be the vertices of a 3 2



triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q 1  7 is the point  − , −  , is (2020 Main, 7 Jan I)  6 3

72. For a point P in the plane, let d1 (P ) and d2(P ) be the distances of the point P from the lines x − y = 0 and x + y = 0, respectively. The area of the region R consisting of all points P lying in the first quadrant of the plane and satisfying 2 ≤ d1 (P ) + d2(P ) ≤ 4, is (2014 Adv.)

Topic 2 Angle between Straight Lines and Equation of Angle Bisector Objective Questions I (Only one correct option) 1. A ray of light coming from the point (2, 2 3 ) is incident at an angle 30° on the line x = 1 at the point A. The ray gets reflected on the line x = 1 and meets X-axis at the point B. Then, the line AB passes through the point (a)  3, −   (b)  4, − 

(2020 Main, 6 Sep I)

1   3 3  2 

For the following questions choose the correct answer from the codes (a), (b), (c) and (d) defined as follows. (a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

5. Lines L1 : y − x = 0 and L 2 : 2x + y = 0 intersect the line

(c) (3, − 3 ) (d) (4, − 3 )

L3 : y + 2 = 0 at P and Q, respectively. The bisector of the acute angle between L1 and L 2 intersects L3 at R.

Objective Questions II (One or more than one correct option)

Statement I The ratio PR : RQ equals 2 2 : 5 .

Because

2. A ray of light along x + 3 y = 3 gets reflected upon reaching X-axis, the equation of the reflected ray is (2013 Main)

(a) y = x + 3 (c) y = 3x − 3

Assertion and Reason

(b) 3 y = x − 3 (d) 3 y = x − 1

3. Consider three points P = {– sin ( β – α ) – cos β}, Q = {cos( β – α ),sin β} and R = {cos ( β – α + θ ) sin ( β – θ )}, π where 0 < α , β , θ < . Then, 4 (2008, 4M)

Statement II In any triangle, bisector of an angle divides the triangle into two similar triangles. (2007, 3M)

Fill in the Blank 6. The vertices of a triangle are A (− 1, − 7), B(5, 1) and C (1, 4). The equation of the bisector of the angle ABC (1993, 2M) is… .

Analytical & Descriptive Questions 7. The area of the triangle formed by the intersection of line parallel to X-axis and passing through (h, k) with the lines y = x and x + y = 2 is 4h 2. Find the locus of point P. (2005)

(a) P lies on the line segment RQ (b) Q lies on the line segment PR (c) R lies on the line segment QP (d) P, Q, R are non-collinear

8. Find the equation of the line which bisects the obtuse

4. Let P = (–1, 0), and Q (0, 0) and R = (3, 3 3 ) be three point. Then, the equation of the bisector of the angle (2001, 1M) PQR is 3 (a) x+ y = 0 2

(b) x +

3y = 0

(c) 3 x + y = 0

(d) x +

3 y=0 2

angle between the lines x − 2 y + 4 = 0 and 4x − 3 y + 2 = 0.

(1993, 2M)

L1 ≡ ax + by + c = 0 and L 2 ≡ lx + my + n = 0 intersect at the point P and makes an angle θ with each other. Find the equation of a line L different from L 2 which passes through P and makes the same angle θ (1988, 5M) with L1.

9. Lines

Topic 3 Area and Family of Concurrent Lines Objective Questions I (Only one correct option) 1. A triangle has a vertex at (1, 2) and the mid-points of

the two sides through it are (−1, 1) and (2, 3). Then, the (2019 Main, 12 April II) centroid of this triangle is

7 (a)  1,   3 1 (c)  , 1 3 

1 (b)  , 2 3  1 5 (d)  ,   3 3

2. Consider the set of all lines px + qy + r = 0 such that 3 p + 2q + 4 r = 0. Which one of the following statements (2019 Main, 9 Jan I) is true? (a) Each line passes through the origin.  3 1 (b) The lines are concurrent at the point  ,   4 2 (c) The lines are all parallel (d) The lines are not concurrent

384 Straight Line and Pair of Straight Lines 3. Two sides of a rhombus are along the lines, x − y + 1 = 0 and 7x − y − 5 = 0. If its diagonals intersect at (− 1, − 2), then which one of the following is a vertex of this (2016 Main) rhombus? (a) (− 3, − 9) 1 8 (c)  , −  3 3

Match the Columns Match the conditions/expressions in Column I with statement in Column II.

9. Consider the lines given by

(b) (− 3, − 8) 10 7 (d)  − , −   3 3

L1 : x + 3 y − 5 = 0, L3 : 5x + 2 y − 12 = 0

4. Area of the parallelogram formed by the lines

Column I

y = mx, y = mx + 1, y = nx and y = nx + 1 equals

(a)

(2001, 1M)

|m + n|

2 |m + n| 1 (d) |m − n| (b)

(m − n ) 1 (c) |m + n| 2

8 5. The points 0,  , (1, 3) and (82, 30) are vertices of 

(a) (b) (c) (d)

3

Column II

(A)

L1, L2, L3 are concurrent, if

(p)

k= −9

(B)

One of L1, L2, L3 is parallel to at least one of the other two, if

(q)

k=−

(C)

L1, L2, L3 form a triangle, if

(r)

k=

(D)

L1, L2, L3 do not form a triangle, if

(s)

k=5

6 5

5 6

(1986, 2M)

Fill in the Blank

an obtuse angled triangle an acute angled triangle a right angled triangle None of the above

10. The set of lines ax + by + c = 0, where 3a + 2b + 4c = 0 is concurrent at the point… .

6. The straight lines x + y = 0, 3x + y − 4 = 0, x + 3 y − 4 = 0 form a triangle which is

(1983, 1M)

(a) isosceles (b) equilateral (c) right angled (d) None of the above

x1

11. If

x + 2 y − 3 = 0, 3x + 4 y − 7 = 0, 2x + 3 y − 4 = 0,4x + 5 y − 6 = 0, then

x2 x3

a1 b1 1 y1 1 y2 1 = a 2 b2 1 a3 b3 1 y3 1

,then the two triangles

with vertices (x1 , y1 ), (x2, y2), (x3 , y3 ) and (a1 , b1 ), (a 2, b2), (1985, 1M) (a3 , b3 ) must be congruent. (1980, 1M)

(a) they are all concurrent (b) they are the sides of a quadrilateral (c) only three lines are concurrent (d) None of the above

Analytical & Descriptive Questions 12. Using coordinate geometry, prove that the three altitudes of any triangle are concurrent.

(1998, 8M)

13. The coordinates of A , B, C are (6, 3), (−3, 5), (4, − 2) respectively and P is any point (x, y). Show that the ratio of the areas of the triangles ∆ PBC and ∆ ABC is x+ y−2 (1983, 2M) . 7

Objective Question II (One or more than one correct option) 8. Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent, if

(1982, 2M)

True/False

7. Given the four lines with the equations

(a) p + q + r = 0 (c) p3 + q3 + r3 = 3 pqr

L 2 : 3x − ky − 1 = 0,

(1985, 2M)

14. A straight line L is perpendicular to the line in

(b) p 2 + q2 + r 2 = pr + rq (d) None of these

5x − y = 1. The area of the triangle formed by the line L and the coordinate axes is 5. Find the equation of the line L. (1980, 3M)

Topic 4 Homogeneous Equation of Pair of Straight Lines Objective Questions I (Only one correct option) 1. Let a and b be non-zero and real numbers. Then, the equation (ax2 + by2 + c) (x2 − 5xy + 6 y2) = 0 represents (a) four straight lines, when c = 0 and a , b are of the same sign (b) two straight lines and a circle, when a = b and c is of sign opposite to that of a (c) two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a (d) a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a

(2008, 3M)

Straight Line and Pair of Straight Lines 385 2. Area of triangle formed by the lines x + y = 3 and angle bisectors of the pair of straight lines x2 − y2 + 2 y = 1 is (2004, 2M)

(a) 2 sq units (c) 6 sq units

Analytical & Descriptive Question 3. Show that all chords of curve 3x2 − y2 − 2x + 4 y = 0,which subtend a right angle at the origin pass through a fixed (1991, 4M) point. Find the coordinates of the point.

(b) 4 sq units (d) 8 sq units

Topic 5 General Equation of Pair of Straight Lines Objective Question I (Only one correct option) 1. Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is (1999, 2M)

(a) 3x2 − 3 y2 + 8xy + 20x + 10 y + 25 = 0 (b) 3x2 − 3 y2 + 8xy − 20x − 10 y + 25 = 0 (c) 3x2 − 3 y2 + 8xy + 10 x + 15 y + 20 = 0 (d) 3x2 − 3 y2 − 8xy − 10 x − 15 y − 20 = 0

Answers 66. [( −a, a (t1 + t 2 + t 3 + t1t 2t 3 )]

Topic 1 1. (a)

2. (d)

3. (d)

68. c = − 4,( 4, 4 ),(2, 0 )

4. (b)

69. ( − 4, − 7 ) 70. 7 x − 4y + 25 = 0, 4 x + 7y = 11 = 0,7 x − 4y − 3 = 0

5. (c)

6. (d)

7. (c)

8. (c)

9. (c)

10. (c)

11. (b)

12. (d)

13. (c)

14. (a)

15. (a)

16. (d)

17. (d)

18. (a)

19. (c)

20. (d)

21. (c)

22. (b)

23. (b)

24. (b)

Topic 2

25. (d)

26. (c)

27. (c)

28. (a)

1. (c)

2. (b)

3. (d)

29. (b)

30. (d)

31. (c)

32. (a)

5. (c)

6. 7y = x + 2

7. 2 x = ± (y − 1 )

33. (c)

34. (d)

35. (d)

36. (c)

8. ( 4 + 5 ) x − (2 5 + 3 ) y + ( 4 5 + 2 ) = 0

37. (a)

38. (b)

39. (d)

40. (c)

9. 2 (al + bm )(ax + by + c ) − (a 2 + b 2 ) (lx + my + n ) = 0

41. (a)

42. (b, c, d)

43. (a, c)

44. (a, c)

45. (1, 1)

46. Ist

47. (1, − 2 )

48. (y = x )

49. True

50. True

51. True

53. (OP + OQ = 18) 55. (m 2 − 1 ) x − my + b (m 2 + 1 ) + am = 0 56. 2 x + 3y + 22 = 0

3 1 57. − < α < −1 ∪ < α < 1 2 2

58. x = 2 and 3 x + 4y = 18 59. x − 7y + 13 = 0 and 7 x + y − 9 = 0 60. x + y − 7 x + 5y = 0 2

2

63. 32 sq units

62. 14 x + 23y − 40 = 0

 5 64.  0,  or ( 0, 0 )  2

65. x − 3y − 31 = 0 or 3 x + y + 7 = 0

71. (5) 72. 6 sq units

4. (c)

Topic 3 1. (b) 5. (d)

2. (b) 6. (a)

3. (c) 7. (c)

9. A → s; B → p, q; C → r; D → p, q, s  3 1 10.  ,   4 2

11. False

14. x + 5y = ± 5 2

Topic 4 1. (b) 3. (1, − 2 )

Topic 5 1. (b)

2. (a)

4. (d) 8. (a, c)

Hints & Solutions Topic 1 Various Forms of Straight Line 1. According to the question, as points (1, 2) and

(sin θ , cos θ ), θ ∈ (0 , π ), lie on the same side of the line x + y = 1, so (1 + 2 − 1) (sin θ + cos θ − 1) > 0 ⇒ sin θ + cos θ > 1 1 1 1 1  π ⇒ sin θ + cos θ > ⇒ sin  + θ >  4 2 2 2 2 3π π π 2 nπ + < + θ < 2 nπ + , n ∈ I. ⇒ 4 4 4 π 2 nπ < θ < 2 nπ + ⇒ 2  π As θ ∈ (0, π ), So for n = 0, θ ∈ 0,   2

and as we know that y < 0, is in third and fourth quadrants only.

4. Since, equation of a line parallel to line ax + by + c = 0 is ax + by + k = 0 ∴ Equation of line parallel to line 4x − 3 y + 2 = 0 is 4x − 3 y + k = 0 Now, distance of line (i) from the origin is | k| 3 = 2 2 5 4 +3

[as per question’s requirement] ⇒ | k| = 3 ⇒ k=±3 So, possible lines having equation, either 4x − 3 y + 3 = 0 or 4x − 3 y − 3 = 0  1 2 Now, from the given options the point  − ,  lies on  4 3

Hence, option (a) is correct.

2. According to the question, we have the following figure.

the line 4x − 3 y + 3 = 0.

(0,b) M 60° O

5. The given inequalities are |x − y| ≤ 2 and|x + y| ≤ 2 . On drawing, the above inequalities, we get a square

(a,0)

α x+ y= 0

Y

x y — + — =1 a b

(0, 2)

Let θ be the inclination of the line x + y = 0. Then, tan θ = − 1 = tan (180° − 45° ) ⇒ tan θ = tan 135° ⇒ θ = 135° ⇒ α + 60° = 135° ⇒ α = 75° Since, line L having perpendicular distance OM = 4. So, equation of the line ‘L’ is x cos α + y sin α = 4 ⇒ x cos 75° + y sin 75° = 4 ⇒ x cos (45° + 30° ) + y sin (45° + 30° ) = 4  3  3 1  1  ⇒ x + − =4 + y 2 2 2 2  2 2 2 2  ⇒

3.

( 3 − 1) x + y ( 3 + 1) = 8 2

Key Idea Use formulae : 2sin A sin B = cos( A − B) − cos( A + B) and cos 2 θ = 1 − 2 sin2 θ

Given equation is y = sin x sin(x + 2) − sin 2(x + 1) 1 1 = [cos 2 − cos(2x + 2)] − [1 − cos(2x + 2)] 2 2 [Q2 sin A sin B = cos( A − B) − cos( A + B) and cos 2 θ = 1 − 2 sin 2 θ ⇒ 2 sin 2 θ = 1 − cos 2 θ] 1 1 1 1 = cos 2 − cos(2x + 2) − + cos(2x + 2) 2 2 2 2 1 1 = (cos(2) − 1) = − (2 sin 2(1)) 2 2 = − sin (1) < 0 ⇒ y < 0 2

…(i)

X′

(–2, 0)

(2, 0) O

X

(0, –2) Y′

Now, the area of shaded region is equal to the area of a square having side length (2 − 0)2 + (0 − 2)2 = 2 2 units.

6.

Key Idea (i) If lines are perpendicular to each other, then product of their slopes is −1, i.e. m1m2 = − 1 (ii) Distance between two points ( x 1, y 1) and (x 2, y 2) =

(x 2 − x 1) 2 + (y 2 − y 1) 2

Given, lines x + (a − 1) y = 1 and 2x + a 2y = 1, where a ∈ R − {0, 1} are perpendicular to each other  1   2 ∴ −  × −  = −1  a − 1  a 2

⇒ ⇒

[Q If lines are perpendicular, then product of their slopes is −1] a 2(a − 1) = − 2 ⇒ a3 − a 2 + 2 = 0 (a + 1)(a 2 − 2a + 2) = 0 ⇒ a = − 1

Straight Line and Pair of Straight Lines 387 ∴ Equation of lines are …(i) x − 2y = 1 and …(ii) 2x + y = 1 On solving Eq. (i) and Eq. (ii), we get 3 1 x = and y = − 5 5 ∴ Point of intersection of the lines (i) and (ii) is 1 3 P , −  . 5 5 1 3 Now, required distance of the point P  , −  from 5 5 9 1 10 2 origin = + = = 25 25 25 5

7. Let the slope of line is m, which is passing through P(2, 3).

x − x1 y − y1 z − z1  = = x2 − x1 y2 − y1 z2 − z1 

x − 8 y z − 10 …(i) = = 2 1 2 Q Points P , Q and R are collinear, so 4 − 8 y z − 10 = = 2 1 2 z − 10 ⇒ −2 = y = 2 ⇒ y = − 2 and z = 6 So, point R is (4, − 2, 6), therefore the distance of point R from origin is OR = 16 + 4 + 36 = 56 = 2 14 ⇒

9. Given, points (1, 2), (−3, 4) and (h , k) are lies on line L1, so slope of line L1 is

Y

4 −2 k −2 = −3 − 1 h − 1 −1 k − 2 m1 = = h −1 2

m1 =

8 7

⇒

6 5 4

R

d

3

P(2, 3

2

)

4

1 X′

1

2

3

4

θ

Q

5

6

7

8

9

X

x+y=7

Y′

Since, the distance of a point (x1 , y1 ) from the line ax + by1 + c . ax + by + c = 0 is d = 1 a 2 + b2 ∴The distance of a point P(2, 3) from the line x + y − 7 = 0, is |2 + 3 − 7| 2 d= = = 2 1+1 2 Now, in ∆PRQ,

⇒ 2(k − 2) = − 1(h − 1) ⇒ 2k − 4 = − h + 1 ⇒ h + 2k = 5 and slope of line L 2 joining points (h , k) and 3−k (4, 3), is m2 = 4−h

QR = 16 − d = 16 − 2 = 14 d 2 1 m+1  m2 − m1  = = = ∴ tan θ = Q tan θ = 1 + m m  QR 14 7 1 −m  1 2 m+1 1 =± ⇒ 1 −m 7 m+1 m+1 1 1 or ⇒ = =− 1 −m 1 −m 7 7 1− 7 −1 − 7 or m = m= ⇒ 1+ 7 7 −1

⇒ ⇒

3 − k = 8 − 2h 2h − k = 5

…(iii)

…(iv)

On solving Eqs. (ii) and (iv), we get (h , k) = (3, 1) k 3 = =3 h 1

10. Given equation of line is 3x + 5 y = 15

…(i)

Clearly, a point on the line (i), which is equidistance from X and Y -axes will lie on the line either y = x or y = − x. Y B

y=x (0, 3)

y=–x

A

8. Given points are P(2, − 3, 4), Q(8, 0, 10) and R(4, y, z ). Now, equation of line passing through points P and Q x − 8 y − 0 z − 10 is = = 6 3 6 [Since equation of a line passing through two points A (x1 , y1 , z1 ) and B(x2, y2, z2) is given by

…(ii)

Since, lines L1 and L 2 are perpendicular to each other. ∴ m1m2 = − 1   − 1 3 k   ⇒ −    = − 1[from Eqs. (i) and (iii)]  2  4 − h 

So, 2

…(i)

(5, 0) 0

X 3x+5y=15

In the above figure, points A and B are on the line (i) and are equidistance from the coordinate axes.

388 Straight Line and Pair of Straight Lines  15 15 On solving line (i) and y = x, we get A  , . 8 8

14. Given equation of curve is y = xex

Similarly, on solving line (i) and y = − x, we get  15 15 B − , .  2 2 So, the required points lie only in I and II quadrants.

11. Let the equation of required line having intercepts a and b with the axes is x y + =1 a b

…(i)

…(i)

Note that (1, e) lie on the curve, so the point of contact is (1, e). Now, slope of tangent, at point (1, e), to the curve (i) is 2 2 dy =  x(2x) ex + ex  = 2e + e = 3e dx (1 , e ) (1 , e )

Now, equation of tangent is given by ( y − y1 ) = m (x − x1 ) y − e = 3e(x − 1) ⇒ y = 3ex − 2e  4 On checking all the options, the option  , 2e satisfy  3 the equation of tangent.

Y

B (0,b)

15. According to given information, we have the following

P (–3,4)

b

2

figure X

a

O

x–y+3=0

Now, according to given information, P is the mid-point of AB  a b ∴ P =  ,  = (−3, 4)  2 2

[given]

12. Slope of the line 2x − 3 y + 17 = 0 is 2 = m1, (let) and the slope of line joining the points (7, 17) 3 β − 17 β − 17 and (15, β ) is = = m2 (let) 15 − 7 8 According to the question, m1m2 = − 1 2 β − 17 × = − 1 ⇒ β − 17 = − 12 ⇒ β = 5. ⇒ 3 8 13. According to given information, we have the following figure D P

A(1, 2)

B(3, 4)

We know that, diagonals of a parallelogram intersect at mid-point.  5 9 ∴ P = Mid-point of BC and so, P ≡  ,   2 2 Now, equation of AD is. 9 −2 5 ( y − 2) = 2 (x − 1) ⇒ y − 2 = (x − 1) 5 3 −1 2 ⇒ 3 y − 6 = 5x − 5 ⇒ 5x − 3 y + 1 = 0

M

(2,4) B

⇒ (a , b) = (−6, 8) On putting the value of a and b in Eq. (i), we get x y + = 1 ⇒ 8x − 6 y = −48 −6 8 ⇒ 4x − 3 y + 24 = 0

C(2, 5)

C (x2 , y2 )

D

A (a,0)

A

x+y=3

[Note that given lines are perpendicular to each other as m1 × m2 = −1] Clearly, point A is point of intersection of lines …(i) x+ y =3 and …(ii) x − y = −3 So, Eqs. (i) and (ii)] A = (0, 3[solving ) Now, as point M (2, 4) is mid-point of line joining the points A and C, so  0 + x2 3 + y2 (2, 4) =  ,   2 2    x1 + x2 y1 + y2  Qmid-point =  2 , 2     0 + x2 3 + y2 ⇒ 2= ;4 = 2 2 ⇒ x2 = 4 and y2 = 5 ∴Thus, C ≡ (4, 5) Now, equation of line BC is given by ( y − y1 ) = m (x − x1 ) y − 5 = 1(x − 4) [line BC is parallel to x − y + 3 = 0 and slope (−1) of x − y + 3 = 0 is = 1] (−1) …(iii) ⇒ y = x+1 and equation of line DC is y − 5 = −1 (x − 4) [line DC is parallel to x + y = 3 and −1 slope of x + y = 3 is = −1] 1 …(iv) ⇒ x+ y =9 On solving Eqs. (i) and (iii), we get B (1, 2) and on solving Eqs. (ii) and (iv), we get D (3, 6)

Straight Line and Pair of Straight Lines 389 Now, let the centroid of ∆PQR be G (h , k), then x +1+3 h= 1 3 …(ii) ⇒ x1 = 3h − 4 y1 + 4 − 2 and k= 3 2x1 + 4 +2 3 [from Eq. (i)] k= ⇒ 3 2x + 4 + 6 ⇒ 3k = 1 3 …(iii) ⇒ 9k − 10 = 2x1 Now, from Eqs. (ii) and (iii), we get 2(3h − 4) = 9k − 10

16. Let P (x1 , y1 ) be any point on the curve y = x . y1 = x1 ⇒ x1 =

Clearly,

∴ The point is

y12

[Q (x1 , y1 ) lies on y = x]

P ( y12, y1 )

3  Now, let the given point be A  , 0 , then 2  2

3  PA =  y12 −  + y12 =  2 =

y14 − 2 y12 +

y14 − 3 y12 +

9 + y12 4

9 5 = ( y12 − 1)2 + 4 4

Clearly, PA will be least when y12 − 1 = 0 ⇒

PAmin = 0 +

5 5 = 4 2

⇒ ⇒

17. Given equation of line is

6h − 8 = 9k − 10 6h − 9k + 2 = 0

Now, replace h by x and k by y. ⇒ 6x − 9 y + 2 = 0, which is the required locus and slope a 2  of this line is Q slope of ax + by + c = 0 is − b  3 

3x + 4 y − 24 = 0 For intersection with X-axis put y = 0 ⇒ 3x − 24 = 0 ⇒ x = 8 For intersection with Y -axis, put x = 0 ⇒ 4 y − 24 = 0 ⇒ y = 6 ∴ A(8, 0) and B (0, 6)

19. R(α, β)

Q(0, β)

B(0,6)

(2, 3) P(α, 0)

O

Equation of line PQ is

A(8,0)

Since this line is passes through fixed point (2, 3). 2 3 ∴ + =1 α β

Let AB = c = 82 + 62 = 10 OB = a = 6 and OA = b = 8 Also, let incentre is (h k), then ax + bx2 + cx3 (here, h= 1 a+ b+ c x1 = 8, x2 = 0, x3 = 0)

∴Locus of R is

2β + 3α = αβ

i.e.

2 y + 3x = xy

⇒

3x + 2 y = xy

20. Given, vertices of triangle are (k, − 3k), (5, k) and (− k, 2).

6 × 8 + 8 × 0 + 10 × 0 = 6 + 8 + 10 48 =2 24 ay + by2 + cy3 k= 1 a+ b+ c

∴

=

and

k − 3k 1 1 5 k 1 = ± 28 2 −k 2 1 k

(here, y1 = 0, y2 = 6, y3 = 0)

6 × 0 + 8 × 6 + 10 × 0 48 = = =2 6 + 8 + 10 24 ∴Incentre is (2, 2).

18. Let the coordinates of point P be (x1 , y1 ) Q P lies on the line 2x − 3 y + 4 = 0 ∴ 2x1 − 3 y1 + 4 = 0 2x + 4 ⇒ y1 = 1 3

x y + =1 α β

…(i)

⇒

− 3k 1

5

k

1 = ± 56

−k

2

1

⇒ k(k − 2) + 3k(5 + k) + 1(10 + k2) = ± 56 ⇒ k2 − 2k + 15k + 3k2 + 10 + k2 = ± 56 ⇒ 5k2 + 13k + 10 = ± 56 ⇒ 5k2 + 13k − 66 = 0 or 5k2 + 13k − 46 = 0 [Qk ∈ I] ⇒ k =2 Thus, the coordinates of vertices of triangle are A(2, − 6), B(5, 2) and C (− 2, 2).

390 Straight Line and Pair of Straight Lines Now, equation of altitude from vertex A is −1 y − (− 6) = (x − 2) ⇒ x = 2  2 −2     − 2 − 5

23. Given mid-points of a triangle are (0, 1), (1, 1) and …(i)

(1, 0). Plotting these points on a graph paper and make a triangle. So, the sides of a triangle will be 2, 2 and 22 + 22 i.e. 2 2.

Y

Y

C (–2, 2)

D

B (5, 2)

C (0,2)

(2, 1/2) E X′

X′

A (2, –6) Y′

B (0, 0)

(1, 0) 2

A (2, 0)

X

Y′

Equation of altitude from vertex C is −1 y−2 = [x − (− 2)] 2 − (− 6)   5 −2    ⇒

(1, 1)

2 (0, 1)

X

O

3x + 8 y − 10 = 0

x-coordinate of incentre = = …(ii)

1 On solving Eqs. (i) and (ii), we get x = 2 and y = . 2  1 ∴ Orthocentre = 2,   2

2 × 0 + 2 2 ⋅0 + 2 ⋅2 2+2+2 2 2 2− 2 =2 − 2 × 2+ 2 2− 2

24. A straight line passing through P and making an angle of α = 60º , is given by y − y1 = tan (θ ± α ) x − x1

21. Let coordinate of the intersection point in fourth

quadrant be (α, − α). Since, (α, − α) lies on both lines 4 ax + 2ay + c = 0 and 5bx + 2by + d = 0. −c …(i) ∴ 4 aα − 2aα + c = 0 ⇒ α = 2a and

5 bα − 2 bα + d = 0 ⇒ α =

−d 3b

60°

⇒ ⇒ ⇒

7 + 6 3 − 1  13  , , 1  =  2 2  2 

22. Coordinate of S = 

and

[QS is mid-point of line QR] ⇒

P (2,2)

and

S

R (7,3)

−2 Slope of the line PS is . 9 Required equation passes through (1 , − 1) and parallel to PS is −2 y+1= (x − 1) 9 ⇒ 2x + 9 y + 7 = 0

3 x+ y=1 y = − 3 x + 1, then tan θ = − 3 y+2 tan θ ± tan α = x − 3 1 m tan θ tan α y+2 − 3+ 3 = x − 3 1 − (− 3)( 3 )

2ad − 3bc = 0

Q (6,–1)

⇒ tan α = √3

√3x + y = 1

…(ii)

From Eqs. (i) and (ii), we get −c −d = ⇒ 3bc = 2ad 2a 3b ⇒

P (3, –2)

60°

y+2 − 3− 3 = x − 3 1 + (− 3)( 3 ) y+ 2 =0 y+ 2 −2 3 = = 3 x−3 1 − 3

y+ 2 = 3 x−3 3 Neglecting, y + 2 = 0, as it does not intersect Y-axis.

25. Given, lines are (1 + p) x − py + p (1 + p) = 0

... (i)

and ... (ii) (1 + q) x − qy + q (1 + q) = 0 On solving Eqs. (i) and (ii), we get C { pq , (1 + p) (1 + q)} ∴ Equation of altitude CM passing through C and perpendicular to AB is ... (iii) x = pq

Straight Line and Pair of Straight Lines 391  1 + q Q Slope of line (ii) is  .  q  ∴ Slope of altitude BN (as shown in figure) is

−q . 1+ q

Y

Lin

O

A

C

y=

 3 Required orthocentre = (3, y) = 3,   4

Now, for x to be an integer, 3 + 4m = ± 5 or ± 1 The integral values of m satisfying these conditions are −2 and −1.

−q (x + p) 1+ q

29. Now, distance of origin from 4x + 2 y − 9 = 0 is

−q (x + p) (1 + q)

| − 9|

... (iv)

26. Since, triangle is isosceles, hence centroid is the desired point. Y P (3, 4)

4 +2 2

30. Let the vertices of triangle be A(1, 3 ), B(0, 0) and C (2, 0). Here, AB = BC = CA = 2 . Therefore, it is an equilateral triangle. So, the incentre coincides with centroid. 0 + 1 + 2 0 + 0 + 3 I ≡ , ∴  3 3   I ≡ (1, 1 / 3 )

 1 3 31. Now, ( A0 A1 )2 = 1 −  + 0 − 

Q (6, 0)

X

2



2

2



2

 3 1 3  1 =   +   = + = 1 ⇒ A0 A1 = 1  2 4 4  2

Y'

 4 ∴ Coordinates of R 3,  .  3

 ( A 0 A 2 )2 =  1 + 

27. To find orthocentre of the triangle formed by (0, 0) (3, 4)

⇒ P

y)

and

H

Q

2

2

B(3, 4)

Y

2  1 3   + 0 −  2 2 

2  3 9 3 12  3 =3 =   + −  = + =  2 4 4 4  2

and (4, 0).

(0, 0)

9 / 20 3 = 4 6/ 5

2

(0, 0) O

O

9 20

=

Hence, the required ratio =

2

X'

2

and distance of origin from 2x + y + 6 = 0 is |6| 6 = 2 2 5 2 +1

⇒ R

(3,

5 3 + 4m

x=

Let orthocentre of triangle be H(h, k), which is the point of intersection of Eqs. (iii) and (iv). On solving Eqs. (iii) and (iv), we get x = pq and y = − pq ⇒ h = pq and k = − pq ∴ h+ k =0 ∴ Locus of H (h, k) is x + y = 0 .

X'

4 y = −1 3 3 y= 4

28. On solving equations 3x + 4 y = 9 and y = mx + 1, we get

H(h,k) (-p, 0) X M B Line (2)

∴ Equation of BN is y − 0 = ⇒

⇒ ∴

N 1) e(

−

⇒

A (4, 0)

Y'

Let H be the orthocentre of ∆OAB (slope of OP i.e. OH) ⋅(slope of BA) = −1 ∴  y − 0  4 − 0 ⇒  = −1  ⋅   3 − 0   3 − 4

A0 A2 = 3 2  1 3  ( A 0 A 4 )2 =  1 +  +  0 +   2 2 

2

2

 3 9 3 12  3 =3 =  +  = + =  4 4 4 4  2

X

⇒ Thus,

32. Since,

A0 A4 = 3 ( A0 A1 ) ( A0 A2) ( A0 A4 ) = 3

the coordinates of the centroid are  x1 + x2 + x3 y1 + y2 + y3  ,  , then the centroid is always    3 3 a rational point.

392 Straight Line and Pair of Straight Lines 33. PQRS is a parallelogram if and only if the mid point of

41. The point O(0, 0) is the mid-point of A (− a , − b) and

the diagonals PR is same as that of the mid-point of QS. That is, if and only if 1+5 4+ a 2+ 7 6+ b and = = 2 2 2 2 ⇒ a = 2 and b = 3.

B(a , b). Therefore, A , O , B are collinear and equation of line AOB is b y= x a Since, the fourth point D (a 2, ab) satisfies the above equation. Hence, the four points are collinear.

34. Slope of line x + 3 y = 4 is − 1 / 3 and slope of line 6x − 2 y = 7 is 3.  −1  Here, 3 ×   = −1  3

42. Here, ax + 2 y = λ and

Therefore, these two lines are perpendicular which show that both diagonals are perpendicular. Hence, PQRS must be a rhombus.

For a = − 3, above equations will be parallel or coincident, i.e. parallel for λ + µ ≠ 0 and coincident, if λ + µ = 0 and if a ≠ − 3, equations are intersecting, i.e. unique solution.

35. Let y = cos x cos (x + 2) − cos 2 (x + 1) = cos (x + 1 − 1) cos (x + 1 + 1) − cos 2 (x + 1) = cos 2 (x + 1) − sin 2 1 − cos 2 (x + 1) ⇒ y = − sin 2 1 This is a straight line which is parallel to X-axis. It passes through (π / 2, − sin 2 1).

36. Orthocentre of right angled triangle is at the vertex of right angle. Therefore, orthocentre of the triangle is at (0, 0).

37. By the given conditions, we can take two perpendicular lines as x and y axes. If (h , k) is any point on the locus, then |h| + |k| = 1.Therefore, the locus is |x| + | y| = 1. This consist of a square of side 1. Hence, the required locus is a square.

3x − 2 y = µ

43.

PLAN Application of inequality sum and differences, along with lengths of perpendicular. For this type of questions involving inequality we should always ckeck all options.

Situation analysis Check all the inequalities according to options and use length of perpendicular from the point (x1 , y1) to ax + by + c = 0 ax1 + by1 + c i.e. a 2 + b2 As

a > b > c>0 a − c > 0 and b > 0

⇒

a + b−c>0

38. Since, the origin remains the same. So, length of the

a − b > 0 and c > 0

perpendicular from the origin on the line in its position x y x y + = 1 and + = 1 are equal. Therefore, a b p q 1 1 1 1 1 1 = ⇒ + = + 1 1 1 1 a 2 b 2 p2 q 2 + + a 2 b2 p2 q 2

a + c− b > 0

39. Let the coordinate of S be (x, y). Q SQ 2 + SR2 = 2SP 2 2 ⇒ (x + 1) + y2 + (x − 2)2 + y2 = 2 [(x − 1)2 + y2] ⇒ x2 + 2x + 1 + y2 + x2 − 4x + 4 + y2 = 2(x2 − 2x + 1 + y2) 3 ⇒ 2x + 3 = 0 ⇒ x = − 2 Hence, it is a straight line parallel to Y-axis.

40. Let B, C , D be the position of the point A(4, 1) after the three operations I, II and III, respectively. Then, B is (1, 4), C (1 + 2, 4) i.e. (3, 4). The point D is obtained from C by rotating the coordinate axes through an angle π/4 in anti-clockwise direction. Therefore, the coordinates of D are given by 1 π π X = 3 cos − 4 sin = − 4 4 2 π π 7 and Y = 3 sin + 4 cos = 4 4 2 7  1 ∴ Coordinates of D are  − , .  2 2

…(i) ...(ii)

∴ Option (c) are correct. Also, the point of intersection for ax + by + c = 0 and bx + ay + c = 0 –c   −c i.e. ,    a + b a + b −c   −c The distance between (1, 1) and  ,   a + b a + b i.e. less than 2 2. 2

⇒ ⇒

2

c  c    1 +  + 1 +  0] [given] d (P , A ) = | X − 3| + |Y − 2| [given] d (P , 0) = d (P , A ) …(i) ⇒ x + y =| x − 3| + | y − 2| Case I When 0 < x < 3, 0 < y < 2 In this case, Eq. (i) becomes x+ y=3 − x+ 2 − y ⇒ 2x + 2 y = 5 or x + y = 5 /2 Case II When 0 < x < 3, y ≥ 2 Now, Eq. (i) becomes x+ y=3 − x+ y−2 ⇒ 2x = 1 ⇒ x = 1 /2 Case III When x ≥ 3, 0 < y < 2 Y 2

Infinite segment x = 1/2 y=2

1

O

1/2

, 5/2 ent y = egm x + ite s Fin

(1/2,2)

54.

Hence, the solution set is {(x, y)| x = 12, y ≥ 2 } ∪ {(x, y)}| x + y = 5 / 2, 0 < x < 3, 0 < y > 2} The graph is given in adjoining figure.

(5/2,0)

3

X

Now, Eq. (i) becomes x+ y= x−3 + 2 − y ⇒ 2 y = − 1 or y = − 1 / 2 Hence, no solution. Case IV When x ≥ 3, y ≥ 2 In this case, case I changes to x+ y= x−3 + y−2 ⇒ 0 = −5 which is not possible.

X′

O G

X

Q (– b,α) Y′ R(h,k)

Slope of RQ = −

1 m

⇒

k −α 1 =− h−b m

From Eq. (i), we get α − a = m (b + h ) ⇒ α = a + m (b + h ) and from Eq. (ii), we get 1 k −α = − (h − b) m 1 ⇒ α =k+ (h − b) m From Eqs. (iii) and (iv), we get 1 a + m (b + h ) = k + (h − b) m ⇒ am + m2 (b + h ) = km + (h − b) ⇒ (m2 − 1) h − mk + b (m2 + 1) + am = 0 Hence, the locus of vertex is (m2 − 1) x − my + b (m2 + 1) + am = 0

…(ii)

…(iii)

…(iv)

56. Let equation of line AC is

y+4 x+5 = =r sin θ cos θ

Let line AE make angle θ with X-axis and intersects x + 3 y + 2 = 0 at B at a distance r1 and line 2x + y + 4 = 0 at C at a distance r2 and line x − y − 5 = 0 at D at a distance r3 . ∴ AB = r1 , AC = r2, AD = r3 . r1 = −

−5 − 3 × 4 + 2 1 ⋅ cos θ + 3 ⋅ sin θ

  I′ Q r = − (a cos θ + b sin θ )   

Straight Line and Pair of Straight Lines 395 15 cos θ + 3 sin θ 2 × (−5) + 1 (−4) + 4 r2 = − 2 cos θ + 1 ⋅ sin θ

⇒

r1 =

Similarly,

…(i)

Dx–y –5= 0 r3 C r2 B

2x +

y+4

=0

x+ 3 y + 2 = 0

r1 A (–5,–4)

E

10 2 cos θ + sin θ −5 × 1 − 4 (−1) − 5 r3 = − cos θ − sin θ 6 r3 = cos θ − sin θ

⇒

r2 =

and ⇒

…(ii)

…(iii)

But it is given that, 2

2

 6   10   15    =  +   AD   AC   AB

2

2

2

 15  10  6 ⇒   +  =   r1   r2   r3 

2

⇒ (cos θ + 3 sin θ )2 + (2 cos + sin θ )2 = (cos θ − sin θ )2 [from Eqs. (i), (ii) and (iii)] ⇒ cos 2 θ + 9 sin 2 θ + 6 cos θ sin θ + 4 cos 2 θ + sin 2 θ + 4 cos θ sin θ = cos 2 θ + sin 2 θ − 2 cos θ sin θ ⇒ 4 cos 2 θ + 9 sin 2 θ + 12 sin θ cos θ = 0 ⇒ (2 cos θ + 3 sin θ )2 = 0 ⇒ 2 cos θ + 3 sin θ = 0 ⇒ cos θ = − (3 / 2) sin θ On substituting this in equation of AC, we get y+4 x+5 = 3 sin θ − sin θ 2 ⇒ −3 ( y + 4) = 2 (x + 5) ⇒ −3 y − 12 = 2x + 10 ⇒ 2x + 3 y + 22 = 0 which is the equation of required straight line.

57. Given lines are 2x + 3 y − 1 = 0

…(i)

x + 2y − 3 = 0 5x − 6 y − 1 = 0

…(ii) …(iii)

0 1= y–

2y –3

+3

x+ =

2x

Let P (α , α 2) be a point inside the ∆ABC. Since, A and P are on the same side of 5x − 6 y − 1 = 0, both 5 (−7) − 6 (5) − 1 and 5α − 6α 2 − 1 must have the same sign, therefore 5α − 6α 2 − 1 < 0 ⇒ 6 α 2 − 5α + 1 > 0 ⇒ (3α − 1) (2α − 1) > 0 1 1 …(iv) ⇒ α < or α > 3 2  5 7 Also, since P (α , α 2) and C  ,  lie on the same side of  4 8  5  7 2x + 3 y − 1 = 0, therefore both 2   + 3   − 1 and  4  8 2 α + 3α 2 − 1 must have the same sign. Therefore, 2α + 3α 2 − 1 > 0 1  ⇒ (α + 1) α −  > 0  3 ⇒

…(v) α < − 1 ∪ α > 1 /3  1 1 2 and lastly  ,  and P (α , α ) lie on the same side of the  3 9 1  1 line therefore, + 2   − 3 and α + 2α 2 − 3 must have  9 3

the same sign. Therefore, 2α 2 + α − 3 < 0 ⇒ 2α (α − 1) + 3 (α − 1) < 0 2 ⇒ (2α + 3) (α − 1) < 0 ⇒ − < α < 1 3 On solving Eqs. (i), (ii) and (iii), we get the common 3 1 answer is − < α < − 1 ∪ < α < 1. 2 2 58. Let l makes an angle α with the given parallel lines and intercept AB is of 3 units. l B 3

0

B(1/3,1/9) 5x – 6y – 1 = 0 C(5/4,7/8)

y + 2x = 2 3/√5

α

A

C

y + 2x = 5

(2,3)

Now, distance between parallel lines |5 − 2| 3 = = 2 2 5 1 +2 1 2 , cos α = 5 5 1 , and tan α = 2 ⇒ Equation of straight line passing through (2, 3 ) and making an angle α with y + 2x = 5 is y−3 = tan (θ + α ) x−2

∴

A(–7,5)

P (a, a2)

On solving Eqs. (i), (ii) and (iii), we get the vertices of a  1 1  5 7 triangle are A(−7, 5), B  ,  and C  ,  .  3 9  4 8

sin α =

396 Straight Line and Pair of Straight Lines y−3 tan θ + tan α = x − 2 1 − tan θ tan α y−3 tan θ − tan α = x − 2 1 + tan θ tan α y−3 y−3 1 3 and =− = x−2 x−2 0 4

⇒ and ⇒ ⇒

3x + 4 y = 18 and

x=2

1  h 1 1 −  = 5k  7  7h

⇒

59. Let m1 and m2 be the slopes of the lines 3x + 4 y = 5 and 4x − 3 y = 15, respectively. 3 4 Then, m1 = − and m2 = 4 3 Clearly, m1m2 = − 1. So, lines AB and AC are at right angle. Thus, the ∆ABC is a right angled isosceles triangle. Y

⇒ ⇒

k  1 +  [on eliminating λ]  5

h (7 − h ) = k(5 + k) h + k − 7h + 5k = 0 2

2

Hence, the locus of a point is x2 + y2 − 7x + 5 y = 0.

61. Let BC be taken as X-axis with origin at D, the mid-point of BC and DA will be Y-axis. Given, AB = AC Let BC = 2a, then the coordinates of B and C are (– a, 0) and (a , 0) let A (0, h ).

C

4x − 3y − 15 = 0

Let R(h , k) be their point of intersection of lines AQ and BP. h 5k Then, + =1 λ 7 7h k and − =1 5 λ

45°

Y A 45°

A (3,−1) 3x+ 4y – 5 = 0 B

X

E

Hence, the line BC through (1, 2) will make an angle of 45° with the given lines. So, the possible equations of BC are m ± tan 45° ( y − 2) = (x − 1) 1 m m tan 45° 3 where, m = slope of AB = − 4 3 − ±1 4 ⇒ ( y − 2) = (x − 1)  3 1 m −   4 −3 ± 4 ⇒ ( y − 2) = (x − 1) 4±3 1 ( y − 2) = (x − 1) ⇒ 7 and ( y − 2) = − 7 (x − 1) ⇒ x − 7 y + 13 = 0 and 7x + y − 9 = 0

60. The equation of the line AB is x y … (i) + =1 7 −5 ⇒ 5x − 7 y = 35 Equation of line perpendicular to AB is …(ii) 7x + 5 y = λ It meets X-axis at P(λ/7, 0) and Y-axis at Q(0, λ /5). x 5y The equations of lines AQ and BP are + = 1 and λ 7 7x y − = 1, respectively. 5 λ

X'

F B

D

C

X

Y'

Then, equation of AC is x y …(i) + =1 a h and equation of DE ⊥ AC and passing through origin is x y − =0 h a hy …(ii) ⇒ x= a On solving, Eqs. (i) and (ii), we get the coordinates of point E as follows hy y a 2h + = 1 ⇒ = y a2 h a 2 + h2  ah 2 a 2h  , 2 ∴Coordinate of E =  2  2  a + h a + h 2 Since, F is mid-point of DE.   ah 2 a 2h ∴Coordinate of F  , 2 2 2 2  2 (a + h ) 2 (a + h )  ∴ Slope of AF ,

a 2h 2 (a 2 + h 2) 2h (a 2 + h 2) − a 2h = m1 = − ah 2 ah 2 0− 2 2 2 (a + h ) h−

⇒

m1 =

− (a 2 + 2h 2) ah

…(iii)

Straight Line and Pair of Straight Lines 397 a 2h −0 a 2h a 2 + h2 and slope of BE , m2 = = ah 2 + a3 + ah 2 ah 2 +a 2 2 a +h ah …(iv) ⇒ m2 = 2 a + 2h 2 From Eqs. (iii) and (iv), m1m2 = − 1 ⇒ AF ⊥ BE

62. Let the coordinates of B and C be (x1 , y1 ) and (x2, y2)

 11 2 So, the coordinates of C are  , .  5 5 Thus, the equation of BC is 2 /5 − 6 y −6 = (x + 7) 11 / 5 + 7 ⇒ − 23 ( y − 6) = 14 (x + 7) ⇒ 14x + 23 y − 40 = 0

63. Let O be the centre of circle and M be mid-point of AB.

respectively. Let m1 and m2 be the slopes of AB and AC, respectively. Then, x_y+5=0 F

B (5, 4)

O

x + 2y = 0 E

D (a, b) C (x2, y2)

B (x1, y1)

y1 + 2 x1 − 1 y +2 m2 = slope of AC = 2 x2 − 1 m1 = slope of AB =

and

M

A (– 3, 4)

A(1,–2)

Let F and E be the mid point of AB and AC, respectively. Then, the coordinates of E and F are  x + 1 y2 − 2  x + 1 y1 − 2 , , E 2  and F  1  , respectively.  2  2 2  2  Now, F lies on x − y + 5 = 0. x1 + 1 y1 − 2 − = −5 ⇒ 2 2 ⇒ x1 − y1 + 13 = 0 Since, AB is perpendicular to x − y + 5 = 0. ∴ (slope of AB) ⋅ (slope of x − y + 5 = 0) = − 1. y1 + 2 ⋅ (1) = − 1 ⇒ x1 − 1

Then, OM ⊥ AB ⇒ M (1, 4) Since, slope of AB = 0 Equation of straight line MO is x = 1 and equation of diameter is 4 y = x + 7. ⇒ Centre is (1, 2). Also, O is mid-point of BD  α + 5 β + 4 ⇒ ,  = (1, 2)  2 2  ⇒

α = − 3, β = 0 AD = (−3 + 3)2 + (4 − 0)2 = 4

∴

and AB = 64 + 0 = 8 Thus, area of rectangle = 8 × 4 = 32 sq units

64. Let the coordinates of A be (0, α ). Since, the sides AB …(i)

and AD are parallel to the lines y = x + 2 and y = 7x + 3, respectively. Y D

C

E( 1,

…(ii)

A (0, α) X′

O

2)

⇒ y1 + 2 = − x1 + 1 ⇒ x1 + y1 + 1 = 0 On solving Eqs. (i) and (ii), we get x1 = − 7, y1 = 6. So, the coordinates of B are (−7, 6). Now, E lies on x + 2 y = 0. x2 + 1  y − 2 +2 2 ∴  =0  2  2

C

B X

Y′

⇒ x2 + 2 y2 − 3 = 0. Since, AC is perpendicular to x + 2 y = 0 ∴ (slope of AC) ⋅ (slope of x + 2 y = 0) = − 1 y2 + 2  1 ⇒ ⋅ −  = − 1 x2 − 1  2

…(iii)

⇒ 2x2 − y2 = 4 On solving Eqs. (iii) and (iv), we get 11 2 and y2 = x2 = 5 5

… (iv)

∴ The diagonal AC is parallel to the bisector of the angle between these two lines. The equation of the bisectors are given by x− y+2 7x − y + 3 =± 2 50 ⇒ 5 (x − y + 2) = ± (7x − y + 3) ⇒ 2x + 4 y − 7 = 0 and 12x − 6 y + 13 = 0. Thus, the diagonals of the rhombus are parallel to the lines 2x + 4 y − 7 = 0 and 12x − 6 y + 13 = 0. 2 12 or Slope of AE = − ∴ 4 6

398 Straight Line and Pair of Straight Lines 2 −α 1 2 −α or =− =2 1 −0 2 1 −0 5 or α = 0. α= ⇒ 2 Hence, the coordinates are (0, 5 /2) or (0, 0).

⇒ ⇒

⇒

65. The equation of any line passing through (1, − 10) is y + 10 = m (x − 1). Since, it makes equal angles, say θ, with the given lines, therefore 1 m−7 m − (−1) tan θ = = ⇒ m = or –3 3 1 + 7m 1 + m(−1) Hence, the equations of third side are 1 y + 10 = (x − 1) or y + 10 = − 3 (x − 1) 3 i.e. x − 3 y − 31 = 0 or 3x + y + 7 = 0

66. Let

ABC be a A [at1t2, a (t1 + t2)], C [at1t3 , a (t1 + t3 )].

triangle whose vertices B [at2t3 , a (t2 + t3 )]

are and

a (t 2 + t3 ) − a (t1 + t3 ) 1 Then, Slope of BC = = at 2t3 − at1t3 t3 a (t1 + t3 ) − a (t1 + t2) 1 Slope of AC = = at1t3 − at1t2 t1

So, the equation of a line through A perpendicular to BC … (i) is y − a (t1 + t2) = − t3 (x − at1t2) and the equation of a line through B perpendicular to AC is …(ii) y − a (t2 + t3 ) = − t1 (x − at2t3 ) The point of intersection of Eqs. (i) and (ii), is the orthocentre. On subtracting Eq. (ii) from Eq. (i), we get x = − a. On putting x = − a in Eq. (i), we get y = a (t1 + t2 + t3 + t1t2t3 ) Hence, the coordinates of the orthocentre are [− a , a (t1 + t2 + t3 + t1t2t3 )].

67. Let OA = a and OB = b. Then, the coordinates of A and B are (a, 0) and (0, b) respectively and also, coordinates of P are (a , b). Let θ be the foot of perpendicular from P on AB and let the coordinates of Q (h , k). Here, a and b are the variable and we have to find locus of Q. Given, AB = c ⇒ AB2 = c2 2 …(i) ⇒ OA + OB2 = c2 ⇒ a 2 + b2 = c2

bk − b2 = ah − a 2 …(ii) ah − bk = a 2 − b2 x y Equation of line AB is + = 1. a b Since, Q lies on AB, therefore h k + =1 a b …(iii) ⇒ bh + ak = ab On solving Eqs. (ii) and (iii), we get 1 h k = = ab2 + a (a 2 − b2) − b(a 2 − b2) + a 2b a 2 + b2 1 h k [from Eq. (i)] ⇒ = 3 = 2 3 a b c ⇒ a = (hc2)1/3 and b = (kc2)1/3 On substituting the values of a and b in a 2 + b2 = c2, we get h 2 / 3 + k2 / 3 = c2 / 3 Hence, locus of a point is x2 / 3 + y2 / 3 = c2 / 3 .

68. Since, diagonals of rectangle bisect each other, so mid point of (1, 3) and (5, 1) must satisfy y = 2x + c, i.e. (3, 2) lies on it. D

C(5,1)

y= B

A(1,3)

2x

+

c

⇒ 2 =6 + c ⇒ c = −4 ∴ Other two vertices lies on y = 2x − 4 Let the coordinate of B be (x, 2x − 4). ∴ Slope of AB ⋅ Slope of BC = − 1  2 x − 4 − 3  2 x − 4 − 1 ⇒   ⋅  = −1  x−1   x−5  ⇒ (x2 − 6x + 8) = 0 ⇒ x = 4, 2 ⇒ y = 4, 0 Hence, required points are (4, 4), (2, 0).

69. Let the coordinates of third vertex be C (a , b). C (a,b)

Since, PQ is perpendicular to AB. Y (0,b)

B

P(a,b)

H(

0, 0

)

Q (h ,k O

Y′

B (–2,3)

A (5,–1)

)

X′

A(a,0)

⇒ Slope of AB ⋅ Slope of PQ = − 1 0−b k−b ⋅ = −1 ⇒ a −0 h − a

X

Since, CH is ⊥ AB,  b  4  ∴     = −1  a   −7  ⇒

4b = 7a

...(i)

Straight Line and Pair of Straight Lines 399 Also, AH ⊥ BC  1  3 − b  ∴  = −1 −    5   −2 − a 

72.

PLAN Distance of a point ( x 1, y1 ) from ax + by + c = 0 is given by ax 1 + by1 + c

⇒ 3 − b = −10 − 5a On solving Eqs. (i) and (ii), we get a = −4 , b = −7

a2 + b 2

.

...(ii)

y=x

x 0 For obtuse angle bisector, we take negative sign. x −2y + 4 4x − 3 y + 2 ∴ =− 5 5 ⇒ ⇒

5 (x − 2 y + 4) = − (4x − 3 y + 2) (4 + 5 )x − (2 5 + 3) y + (4 5 + 2) = 0

Straight Line and Pair of Straight Lines 401 9. Since, the required line L passes through the intersection of L1 = 0 and L 2 = 0.

L2

P

L

So, point C is (3, 4). Now, centroid of ∆ABC is  1 + (− 3) + 3 2 + 0 + 4 , G  =G   3 3

L1

So, the equation of the required line L is L1 + λL 2 = 0. i.e. (ax + by + c) + λ (lx + my + n ) = 0

…(i)

where, λ is a parameter. Since, L1 is the angle bisector of L = 0 and L 2 = 0. ∴ Any point A (x1 , y1 ) on L1 is equidistant from L1 = 0 and L 2 = 0. | lx1 + my1 + n | ⇒ l2 + m 2 |(ax1 + by1 + c) + λ (lx1 + my1 + n )| …(ii) = (a + λl)2 + (b + λm)2 But, A (x1 , y1 ) lies on L1. So, it must satisfy the equation of L1 , ie, ax1 + by1 + c1 = 0. On substituting ax1 + by1 + c = 0 in Eq. (ii), we get | lx1 + my1 + n | |0 + λ (lx1 + my1 + n )| = (a + λl)2 + (b + λm)2 l2 + m 2 λ2(l2 + m2) = (a + λl)2 + (b + λm)2 λ=−

∴

3 p + 2q + 4r = 0 Consider, ⇒ ⇒

 3 1 ⇒  ,  satisfy px + qy + r = 0  4 2

 3 1 So, the lines always passes through the point  ,  .  4 2 3. As the given lines x − y + 1 = 0 and 7x − y − 5 = 0 are not parallel, therefore they represent the adjacent sides of the rhombus. On solving x − y + 1 = 0 and 7x − y − 5 = 0, we get x = 1 and y = 2. Thus, one of the vertex is A(1, 2).

(a 2 + b2) 2 (al + bm)

On substituting the value of λ in Eq. (i), we get

Topic 3 Area and Family of Concurrent Lines 1. Let a ∆ABC is such that vertices A (1, 2), B(x1 y1 ) and C (x2, y2). A(1,2)

B(x1,y1)

C(x2,y2)

It is given that mid-point of side AB is (− 1, 1). x1 + 1 So, = −1 2 y1 + 2 and =1 2

D

C (x, y) (–1, –2)

7x –

(a 2 + b2) (ax + by + c) − (lx + my + n ) = 0 2 (al + bm) ⇒ 2 (al + bm) (ax + by + c) − (a 2 + b2) (lx + my + n ) = 0 which is the required equation of line L.

3 p + 2q + 4r = 0 3 p 2q + + r =0 4 4 (dividing the equation by 4)  3  1 p  + q   + r = 0  4  2

=0

⇒

1   , 2 3 

2 Given, px + qy + r = 0 is the equation of line such that

y–5

A (x1,y1)

⇒ x1 = − 3 and y1 = 0 So, point B is (− 3, 0) Also, it is given that mid-point of side AC is (2, 3), so x2 + 1 y +2 = 2 and 2 =3 2 2 ⇒ x2 = 3 and y2 = 4

A x − y + 1=0 (1, 2)

B

Let the coordinate of point C be (x, y). x+1 y+2 and − 2 = Then, −1 = 2 2 ⇒ x + 1 = − 2 and y = − 4 − 2 ⇒ x= −3 and y= −6 Hence, coordinates of C = (− 3, − 6) Note that, vertices B and D will satisfy x − y + 1 = 0 and 7x − y − 5 = 0, respectively. Since, option (c) satisfies 7x − y − 5 = 0, therefore  1 − 8 coordinate of vertex D is  , . 3 3 

4. Let lines OB : y = mx CA : y = mx + 1 and

BA : y = nx + 1 OC : y = nx

402 Straight Line and Pair of Straight Lines The point of intersection B of OB and AB has x 1 coordinate ⋅ m−n Y

x y 1 = = −36 + 10 −25 + 12 2 − 15

B

D

X′

O

X

Y′

Now, area of a parallelogram OBAC = 2 × area of ∆ OBA 1 1 1 = 2 × × OA × DB = 2 × × 2 2 m−n 1 1 = = m − n |m − n | depending upon whether m > n or m < n.

5. Since, vertices of a triangle are (0, 8 / 3), (1, 3) and (82, 30) 0 8 /3 1 1 1 1 3 1 = 2 2 82 30 1 1 = 2 ∴ Points are collinear. Now,

 8  − (1 − 82) + 1(30 − 246)  3  [216 − 216] = 0

6. The points of intersection of three lines are A (1, 1), B(2, − 2), C (−2, 2). Now, | AB| = 1 + 9 = 10 , |BC| = 16 + 16 = 4 2, and |CA| = 9 + 1 = 10 ∴ Triangle is an isosceles.

7. Given lines, x + 2 y − 3 = 0 and 3x + 4 y − 7 = 0 intersect at (1,1), which does not satisfy 2x + 3 y − 4 = 0 and 4x + 5 y − 6 = 0. Also, 3x + 4 y − 7 = 0 and 2x + 3 y − 4 = 0 intersect at (5, –2) which does not satisfy x + 2 y − 3 = 0 and 4x + 5 y − 6 = 0 . Lastly, intersection point of x + 2 y − 3 = 0 and 2x + 3 y − 4 = 0 is (–1, 2) which satisfy 4x + 5 y − 6 = 0. Hence, only three lines are concurrent.

8. Given lines px + qy + r = 0, qx + ry + p = 0 and ∴

rx + py + q = 0 are concurrent. p q r q r p =0 r

p q

Applying R1 → R1 + R2 + R3 and taking common from R1 1 1 1 ( p + q + r) q r p = 0 r

Therefore, (a) and (c) are the answers.

9. (A) Solving equations L1 and L3 ,

A C

⇒ ( p + q + r ) ( p2 + q2 + r 2 − pq − qr − pr ) = 0 ⇒ p3 + q3 + r3 − 3 pqr = 0

p q

∴ x = 2, y = 1 L1 , L 2, L3 are concurrent, if point (2, 1) lies on L 2 ∴ 6 − k −1 =0 ⇒ k =5 (B) Either L1 is parallel to L 2,or L3 is parallel to L 2,then 1 3 3 −k or = = 3 −k 5 2 ⇒ k = −9 −6 or k= 5 (C) L1 , L 2, L3 form a triangle, if they are not concurrent, or not parallel. 6 ∴ k ≠ 5, − 9, − 5 5 k= ⇒ 6 (D) L1 , L 2, L3 do not form a triangle, if 6 k = 5, − 9, − ⋅ 5

10. The set of lines ax + by + c = 0, where 3a + 2b + 4c = 0 or

3 1 3 1  a + b + c = 0 are concurrent at  x = , y =  i.e.  4 2 4 2 comparing the coefficients of x and y.  3 1 Thus, point of concurrency is  ,  .  4 2 Alternate Solution

As, ax + by + c = 0, satisfy 3a + 2b + 4c = 0 which represents system of concurrent lines whose point of concurrency could be obtained by comparison as, 3a 2 ax + by + c ≡ + b+ c 4 4 3 1 ⇒ x = , y = is point of concurrency. 4 2  3 1 ∴  ,  is the required point.  4 2 x1 11. Since, x2 x3

a1 b1 1 y1 1 y2 1 = a 2 b2 1 a3 b3 1 y3 1

represents area of triangles are equal, which does not impies triangles are congrvent. Hence, given statement is false.

12. Let the vertices of a triangle be, O (0, 0) A (a , 0) and B (b, c) equation of altitude BD is x = b. c Slope of OB is . b b Slope of AF is − . c

Straight Line and Pair of Straight Lines 403 Now, the equation of altitude AF is b y − 0 = − (x − a ) c Suppose, BD and OE intersect at P.   (a − b)  Coordinates of P are b, b    c    a−b Let m1 be the slope of OP = c c m2 be the slope of AB = b−a

and Now,

 a − b  c  m1m2 =     = −1  c   b − a

Hence, the given equation, (ax2 + by2 + c)(x2 − 5xy + 6 y2) = 0 may represent two straight lines and a circle.

2. Given, ⇒ ⇒ and

(0,3)

3

(0, 1)

x + y −2 7 

X′

X

λ

Y′

1 λ  =5 λ⋅ 2 5 

Topic 4 Homogeneous Equation of Pair of Straight Lines 1. Let a and b be non-zero real numbers. Therefore the given equation (ax + by + c) (x − 5xy + 6 y ) = 0 implies either ⇒ ⇒ and

2

3

1

y=

–x

X

+1

2

From the graph, it is clear that equation of angle bisectors are y=1 and x=0 ∴ Area of region bounded by x + y = 3, x = 0 1 and y = 1 is ∆ = × 2 × 2 = 2 sq units 2

3. The given curve is

⇒ λ2 = 50 ⇒ |λ | = 5 2 ∴ Equation of the line L is, x + 5 y = ± 5 2

2

O

Y′ Angle bisector

Y

⇒

−1

Angle bisector

(2,1)

y=x+1

Since, area of triangle = 5

λ/5 O

x2 = ( y − 1 ) 2 x = y −1 x=−y+1 Y

14. A straight line perpendicular to 5x − y = 1 is x + 5 y = λ.

X′

x2 − y 2 + 2 y = 1

y=

= 

When, a = b and c is of sign opposite to that of a, ax2 + by2 + c = 0 represents a circle.

x+

We get, that the line through O and P is perpendicular to AB. 1 |x (5 + 2) + (−3) (−2 − y) + 4 ( y − 5)| Area of ∆PBC 2 13. = Area of ∆ABC 1 |6 (5 + 2) + (−3) (−2 − 3) + 4 (3 − 5)| 2 |7x + 7 y − 14| 7|x + y − 2| = = |42 + 15 − 8| 49

ax2 + by2 + c = 0, x=0 and y = 0. which is a point specified as the origin.

2

x2 − 5xy + 6 y2 = 0 (x − 2 y)(x − 3 y) = 0 x = 2y x = 3y

represent two straight lines passing through origin or ax2 + by2 + c = 0 when c = 0 and a and b are of same signs, then

3 x 2 − y2 − 2 x + 4 y = 0

…(i)

Let y = mx + c be the chord of curve (i) which subtend right angle at origin. Then, the combined equation of lines joining points of intersection of curve (i) and chord y = mx + c to the origin, can be obtained by the equation of the curve homogeneous, i.e.  y − mx  y − mx 3x2 − y2 − 2x   =0  + 4y   c   c  ⇒ ⇒

3cx2 − cy2 − 2xy + 2mx2 + 4 y2 − 4mxy = 0 (3c + 2m) x2 − 2 (1 + 2m) y + (4 − c) y2 = 0

Since, the lines represented are perpendicular to each other. ∴ Coefficient of x2 + Coefficient of y2 = 0 ⇒ 3c + 2m + 4 − c = 0 ⇒ c+ m + 2 =0 On comparing with y = mx + c ⇒ y = mx + c passes through (1, – 2).

404 Straight Line and Pair of Straight Lines Topic 5 General Equation of Pair of Straight Lines 1. Let S be the mid-point of QR and given ∆PQR is an isosceles. Therefore, PS ⊥ QR and S is mid-point of hypotenuse, therefore S is equidistant from P , Q , R. ∴ PS = QS = RS Since, ∠ P = 90° and ∠ Q = ∠ R But ∠ P + ∠ Q + ∠ R = 180° ∴ 90° + ∠ Q + ∠ R = 180° ⇒ ∠ Q = ∠ R = 45° Y Q 90°

P(2,1)

O

Y'

But QR ⊥ PS. ∴ Slope of PS is 1/2. Let m be the slope of PQ . ∴ ⇒

m − 1 /2 1 − m (−1 / 2) 2m − 1 ±1= 2+m

tan (± 45° ) =

⇒ m = 3, − 1 / 3 ∴ Equations of PQ and PR are y − 1 = 3 (x − 2) 1 and y − 1 = − (x − 2) 3 or 3 ( y − 1) + (x − 2) = 0 Therefore, joint equation of PQ and PR is [3 (x − 2) − ( y − 1)] [(x − 2) + 3 ( y − 1)] = 0 ⇒ 3 (x − 2)2 − 3 ( y − 1)2 + 8 (x − 2) ( y − 1) = 0

S X'

Now, slope of QR is −2 .

R

X

⇒

3x2 − 3 y2 + 8xy − 20x − 10 y + 25 = 0

[given]

16 Circle Topic 1 Equation of Circle Objective Questions I (Only one correct option) 1. A circle touching the X-axis at (3, 0) and making a intercept of length 8 on the Y -axis passes through the point (2019 Main, 12 April II) (a) (3, 10) (c) (2, 3)

(b) (3, 5) (d) (1, 5)

of a point P such that the perimeter of ∆AOP is 4, is (2019 Main, 8 April I)

(a) 8x2 − 9 y2 + 9 y = 18

(b) 9x2 − 8 y2 + 8 y = 16

(c) 9x2 + 8 y2 − 8 y = 16

(d) 8x2 + 9 y2 − 9 y = 18

3. If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is (2019 Main, 12 Jan II)

(a) (x2 + y2 )2 = 4R 2x2 y2 (b) (x2 + y2 )3 = 4R 2x2 y2 (c) (x2 + y2 )(x + y) = R 2xy (d) (x2 + y2 )2 = 4Rx2 y2

(2019 Main, 11 Jan II)

(c)

41

(d)

137

5. If the area of an equilateral triangle inscribed in the circle, x + y + 10x + 12 y + c = 0 is 27 3 sq units, then c is equal to (a) 20

2

(b) −25

(c) 13

(2019 Main, 10 Jan II)

(d) 25

6. Let the orthocentre and centroid of a triangle be A(−3, 5) and B(3, 3), respectively. If C is the circumcentre of this triangle, then the radius of the circle having line (2018 Main) segment AC as diameter, is (a) 10

(b) 2 10

(c) 3

5 2

(d)

3 5 2

7. The centre of circle inscribed in square formed by the lines x2 − 8x + 12 = 0 and y2 − 14 y + 45 = 0, is (2003, 1M)

(a) (4, 7) (c) (9, 4)

(b) a circle (d) a pair of straight lines

circle of area 154 sq units. Then, the equation of this circle is (1989, 2M) (a) x2 + y2 + 2x − 2 y = 62 (c) x2 + y2 − 2x + 2 y = 47

(b) x2 + y2 + 2x − 2 y = 47 (d) x2 + y2 − 2x + 2 y = 62

10. AB is a diameter of a circle and C is any point on the circumference of the circle. Then,

(1983, 1M)

(a) the area of ∆ ABC is maximum when it is isosceles (b) the area of ∆ ABC is minimum when it is isosceles (c) the perimeter of ∆ ABC is minimum when it is isosceles (d) None of the above

and touching the curve y = x2 at (2, 4) is

square is inscribed in the circle x2 + y2 − 6x + 8 y − 103 = 0 with its sides parallel to the coordinate axes. Then, the distance of the vertex of this square which is nearest to the origin is

2

(a) a parabola (c) an ellipse

11. The centre of the circle passing through the point (0, 1)

4. A

(b) 13

right angle at the centre. Then, the locus of the centroid (2001, 1M) of the ∆ PAB as P moves on the circle, is

9. The lines 2x − 3 y = 5 and 3x − 4 y = 7 are diameters of a

2. Let O(0, 0) and A(0, 1) be two fixed points, then the locus

(a) 6

8. Let AB be a chord of the circle x2 + y2 = r 2 subtending a

(b) (7, 4) (d) (4, 9)

16 27  (a)  − ,   5 10  16 53  (c)  − ,   5 10 

(1983, 1M)

16 53  (b)  − ,   7 10  (d) None of these

Objective Questions II (One or more than one correct option) 12. Circle(s) touching X-axis at a distance 3 from the origin and having an intercept of length 2 7 on Y-axis is/are (a) (b) (c) (d)

x2 + x2 + x2 + x2 +

y2 − 6x + 8 y + 9 = 0 y2 − 6x + 7 y + 9 = 0 y2 − 6x − 8 y + 9 = 0 y2 − 6x − 7 y + 9 = 0

(2013 Adv.)

13. Let L1 be a straight line passing through the origin and L 2 be the straight line x + y = 1. If the intercepts made by the circle x2 + y2 − x + 3 y = 0 on L1 and L 2 are equal, then which of the following equation can represent L1?

(a) x + y = 0 (c) x + 7 y = 0

(b) x − y = 0 (d) x − 7 y = 0

(1999, 1M)

406 Circle Integer & Numerical Answer Type Questions

Fill in the Blanks 14. The lines 3x − 4 y + 4 = 0 and 6x − 8 y − 7 = 0 tangents to the same circle. The radius of this circle is… .

22. For how many values of p, the circle x2 + y2 + 2x + 4 y − p = 0 and the coordinate axes have (2017 Adv.) exactly three common points?

(1984, 2M)

15. If A and B are points in the plane such that PA / PB = k (constant) for all P on a given circle, then the value of k cannot be equal to ……… . (1982, 2M)

23. The straight line 2x − 3 y = 1 divide the circular region into two parts. If x2 + y 2 ≤ 6  3   5 3   1 1   1 1   S = 2,  ,  ,  ,  ,−  ,  ,  , then the number of  4   2 4   4 4   8 4  

True/False 16. The line x + 3 y = 0 is a diameter of the circle x2 + y2 − 6x + 2 y = 0.

point (s) in S lying inside the smaller part is (2016 Adv.)

(1989, 1M)

Paragraph Based Questions

Analytical & Descriptive Questions

Let S be the circle in the XY -plane defined by the equation

17. Let C be any circle with centre (0, 2 ). Prove that

x2 + y2 = 4.

at most two rational points can be there on C.

(There are two questions based on above Paragraph, the question given below is one of them)

(A rational point is a point both of whose coordinates are rational numbers.) (1997, 5M)

18. Consider a curve ax2 + 2hxy + by2 = 1 and a point P not

24. Let E1E 2 and F1F2 be the chords of S passing through the point P0 (1, 1) and parallel to the X-axis and the Y -axis, respectively. Let G1G2 be the chord of S passing through P0 and having slope −1. Let the tangents to S at E1 and E 2 meet at E3 , then tangents to S at F1 and F2 meet at F3 , and the tangents to S at G1 and G2 meet at G3 . Then, the points E3 , F3 and G3 lie on the curve

on the curve. A line drawn from the point P intersect the curve at points Q and R. If the product PQ ⋅ QR is independent of the slope of the line, then show that the (1997, 5M) curve is a circle.

19. A circle passes through three points A, B and C with the line segment AC as its diameter. A line passing through A intersects the chord BC at a point D inside the circle. If angles DAB and CAB are α and β respectively and the distance between the point A and the mid-point of the line segment DC is d, prove that the area of the circle is πd 2 cos 2 α (1996, 5M) 2 2 cos α + cos β + 2 cos α cos β cos (β − α )

(2018 Adv.)

(a) x + y = 4 (b) (x − 4)2 + ( y − 4)2 = 16 (c) (x − 4) ( y − 4) = 4 (d) xy = 4

25. Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N . Then, the mid-point of the line segment MN must lie on the curve

20. If (mi , 1 / mi ), mi > 0, i = 1, 2, 3, 4 are four distinct points on a circle , then show that m1 m2 m3 m4 = 1. (1989, 2M)

(a) (x + y)2 = 3xy / / (b) x23 + y23 = 24/3 2 2 (c) x + y = 2xy (d) x2 + y2 = x2 y2

21. The abscissae of the two points A and B are the roots of the equation x2 + 2ax − b2 = 0 and their ordinates are the roots of the equation y2 + 2 py − q2 = 0. Find the equation and the radius of the circle with AB as diameter.

(1984, 4M)

Topic 2 Relation between Two Circles Objective Questions I (Only one correct option)

3. If a variable line, 3x + 4 y − λ = 0 is such that the two

the circles and x + y + 5Kx + 2 y + K = 0 2 (x2 + y2) + 2Kx + 3 y −1 = 0, (K ∈ R), intersect at the points P and Q, then the line 4x + 5 y − K = 0 passes (2019 Main, 10 April I) through P and Q, for

circles x2 + y2 − 2x − 2 y + 1 = 0 and x2 + y2 − 18x − 2 y + 78 = 0 are on its opposite sides, then the set of all values of λ is (2019 Main, 12 Jan I) the interval

(a) no values of K (b) exactly one value of K (c) exactly two values of K (d) infinitely many values of K

(a) [13, 23] (c) [12, 21]

1. If

2

2

2. If a tangent to the circle x2 + y2 = 1 intersects the coordinate axes at distinct points P and Q, then the locus (2019 Main, 9 April I) of the mid-point of PQ is (a) x2 + y2 − 2x2 y2 = 0

(b) x2 + y2 − 2xy = 0

(c) x2 + y2 − 4x2 y2 = 0

(d) x2 + y2 − 16x2 y2 = 0

(b) (2, 17) (d) (23, 31)

4 Let C1 and C 2 be the centres of the circles x2 + y2 − 2x − 2 y − 2 = 0 and x2 + y2 − 6x − 6 y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq units) of the (2019 Main, 12 Jan I) quadrilateral PC1QC 2 is (a) 8 (c) 6

(b) 4 (d) 9

Circle 407 5. If the circles x2 + y2 −16x − 20 y + 164 = r 2 and (x − 4)2 + ( y − 7)2 = 36 intersect at two distinct points, (2019 Main, 9 Jan II) then (a) 0 < r < 1 (c) 1 < r < 11

(b) r > 11 (d) r = 11

6. If one of the diameters of the circle, given by the equation, x2 + y2 − 4x + 6 y − 12 = 0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is (2016 Main)

(b) 5 3

(a) 5 2

(c) 5

(d) 10

7. If

one of the diameters of the circle x2 + y2 − 2x − 6 y + 6 = 0 is a chord to the circle with centre (2, 1), then the radius of the circle is (2004, 1M) (b) 2

(a) 3

(c) 3

(d) 2

8. The number of common tangents to the circles

x + y − 4x − 6 y − 12 = 0 and x + y + 6x + 18 y + 26 = 0 (2015) is 2

2

2

(a) 1

(b) 2

2

(c) 3

(d) 4

9. Let C be the circle with centre at (1, 1) and radius 1. If T is the circle centred at (0, y) passing through origin and touching the circle C externally, then the radius of T is (2014 Main) equal to 3 2

(a)

(b)

3 2

(c)

1 2

(d)

1 4

x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is (b) −2 or − 3 / 2 (d) −2 or 3 / 2

(2000, 2M)

11. The ∆ PQR is inscribed in the circle x2 + y2 = 25. If Q and R have coordinates (3, 4) and (− 4, 3) respectively, then (2000, 2M) ∠ QPR is equal to (a) π / 2

(b) π / 3

(c) π / 4

(d) π / 6

12. The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 − 6x − 8 y = 24 is

(a) 0

(b) 1

(c) 3

(1998, 2M)

(d) 4

13. The angle between a pair of tangents drawn from a point P to the circle x2 + y2 + 4 x − 6 y + 9 sin 2 α + 13 cos2 α = 0 is 2α. The equation of the locus of the point P is (1996, 1M)

(a) x + y + 4x − 6 y + 4 = 0 (b) x + y + 4x − 6 y − 9 = 0 (c) x2 + y2 + 4x − 6 y − 4 = 0 (d) x2 + y2 + 4x − 6 y + 9 = 0 2

2

2

2

and the two circles (x − 1) + ( y − 3) = r x2 + y2 − 8x + 2 y + 8 = 0 intersect in two distinct points, (1989, 2M) then

14. If

(a) 2 < r < 8 (b) r < 2

2

(c) r = 2

2

2

(d) r > 2

15. If a circle passes through the point (a, b) and cuts the circle x2 + y2 = k2 orthogonally, then the equation of the locus of its centre is (1988, 2M)

(a) 2ax + 2by − (a 2 + b2 + k 2 ) = 0 (b) 2ax + 2by − (a 2 − b2 + k 2 ) = 0 (c) x2 + y2 − 3ax − 4by + a 2 + b2 − k 2 = 0 (d) x2 + y2 − 2ax − 3by + (a 2 − b2 − k 2 ) = 0

16. Let T be the line passing through the points P(− 2, 7) and Q(2, − 5). Let F1 be the set of all pairs of circles (S1 , S 2) such that T is tangent to S1 at P and tangent to S 2 at Q, and also such that S1 and S 2 touch each other at a point, say M. Let E1 be the set representing the locus of M as the pair (S1 , S 2) varies in F1. Let the set of all straight line segments joining a pair of distinct points of E1 and passing through the point R(1, 1) be F2. Let E 2 be the set of the mid-points of the line segments in the set F2. Then, which of the following statement(s) is (are) TRUE? (2018 Adv.) (a) The point (− 2, 7) lies in E1 4 7 (b) The point  ,  does NOT lie in E2  5 5 1 (c) The point  ,1 lies in E2 2  3 (d) The point  0,  does NOT lie in E1  2

17. A circle S passes through the point (0, 1) and is orthogonal to the circles (x − 1)2 + y2 = 16 and x2 + y2 = 1. (2014 Adv.) Then,

10. If the circle x2 + y2 + 2x + 2ky + 6 = 0 and (a) 2 or − 3 / 2 (c) 2 or 3 / 2

Objective Questions II (One or more than one correct option)

(a) (b) (c) (d)

radius of S is 8 radius of S is 7 centre of S is (−7,1) centre of S is (−8,1)

Passage Based Problems Passage Let ABCD be a square of side length 2 unit. C 2 is the circle through vertices A , B, C , D and C1 is the circle touching all the sides of square ABCD. L is the line through A. (2006, 5M)

18. If P is a point of C1 and Q is a point on C 2 , then PA 2 + PB2 + PC 2 + PD 2 is equal to QA 2 + QB2 + QC 2 + QD 2 (a) 0.75 (c) 1

(b) 1.25 (d) 0.5

19. A circle touches the line L and the circle C1 externally such that both the circles are on the same side of the line, then the locus of centre of the circle is (a) ellipse (c) parabola

(b) hyperbola (d) parts of straight line

20. A line M through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If locus of S cuts M at T2 and T3 and AC at T1, then area of ∆T1T2T3 is 1 sq unit 2 (c) 1sq unit (a)

2 sq unit 3 (d) 2 sq units

(b)

408 Circle Match the Columns

Analytical & Descriptive Questions

21. Match the conditions/expressions in Column I with statement in Column II. Column I

Column II

A. Two intersecting circles

p.

have a common tangent

B. Two mutually external circles

q.

have a common normal

C. Two circles, one strictly inside r. the other

do not have a common tangent

D. Two branches of a hyperbola s.

do not have a common normal

22. Let C1 and C 2 be two circles with C 2 lying inside C1. A circle C lying inside C1 touches C1 internally and C 2 externally. Identify the locus of the centre of C. (2001, 5M)

23. Three circles touch one another externally. The tangents at their points of contact meet at a point whose distance from a point of contact is 4. Find the ratio of the product of the radii to the sum of the radii of the circles. (1992, 5M)

Topic 3 Equation of Tangent, Normal and Length of Tangents Objective Questions I (Only one correct option) 1. A circle touches the Y -axis at the point

8. Let ABCD be a quadrilateral with area 18, with side AB

(0, 4) and passes through the point (2, 0). Which of the following lines is not a tangent to this circle? (2020 Main, 9 Jan I)

(a) 4x − 3 y + 17 = 0 (c) 4x + 3 y − 8 = 0

(b) 3x + 4 y − 6 = 0 (d) 3x − 4 y − 24 = 0

2. The common tangent to the circles x + y = 4 and 2

2

x2 + y2 + 6x + 8 y − 24 = 0 also passes through the point (2019 Main, 9 April II )

(a) (6, − 2)

(b) (4, − 2)

(c) (−6, 4)

(d) (−4, 6)

3. A rectangle is inscribed in a circle with a diameter lying

along the line 3 y = x + 7. If the two adjacent vertices of the rectangle are (–8, 5) and (6, 5), then the area of the (2019 Main, 9 April II ) rectangle (in sq units) is

(a) 72

(b) 84

(c) 98

(d) 56

4. The tangent and the normal lines at the point ( 3 , 1) to the circle x + y = 4 and the X-axis form a triangle. The area of this triangle (in square units) is 2

2

(2019 Main, 8 April II )

(a)

1 3

(b)

4 3

(c)

2 3

(d)

1 3

5. The straight line x + 2 y = 1 meets the coordinate axes at A and B. A circle is drawn through A , B and the origin. Then, the sum of perpendicular distances from A and B on the tangent to the circle at the origin is (2019 Main, 11 Jan I )

(a) 2 5

(b)

5 4

(d)

(c) 4 5

5 2

6. If a circle C passing through the point

(4, 0) touches the circle x2 + y2 + 4x − 6 y = 12 externally at the point (1, − 1), then the radius of C is (2019 Main,10 Jan I )

(a) 5

(b) 2 5

(c)

(d) 4

57

7. If the tangent at (1, 7) to the curve x = y − 6 touches the 2

circle x2 + y2 + 16x + 12 y + c = 0, then the value of c is (2018 Main)

(a) 195

(b) 185

(c) 85

(d) 95

parallel to the side CD and AB = 2 CD . Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the sides, then its radius is (2007, 3M)

(a) 3

(b) 2

(c)

3 2

(d) 1

9. If the tangent at the point P on the circle

meets the straight line x2 + y 2 + 6 x + 6 y = 2 5x − 2 y + 6 = 0 at a point Q on the Y-axis, then the length of PQ is (2002, 1M) (a) 4

(b) 2 5

(c) 5

(d) 3 5

10. Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals (2001, 1M) (a) PQ ⋅ RS PQ + RS (b) 2 2PQ ⋅ RS (c) PQ + RS (d)

PQ 2 + RS 2 2

Objective Questions II (One or more than one correct option) 11. Let RS be the diameter of the circle x2 + y2 = 1, where S is the point (1, 0) . Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then, the locus of E passes through the point(s) (2016 Adv.)

1 1  (a)  ,   3 3 1 1  (c)  , −  3 3

1 1 (b)  ,   4 2 1 1 (d)  , −  4 2

Circle 409 12. The circle C1 : x2 + y2 = 3 with centre at O intersects the parabola x = 2 y at the point P in the first quadrant. Let the tangent to the circle C1 at P touches other two circles C 2 and C3 at R2 and R3 , respectively. Suppose C 2 and C3 have equal radii 2 3 and centres Q2 and Q3 , respectively. If Q2 and Q3 lie on the Y-axis, then 2

(a) Q2Q3 = 12 (b) R2R3 = 4 6 (c) area of the ∆OR2R3 is 6 2 (d) area of the ∆PQ2Q3 is 4 2

(2016 Adv.)

13. Tangents are drawn from the point (17, 7) to the circle x2 + y2 = 169. Statement I The tangents are mutually perpendicular. because Statement II The locus of the points from which a mutually perpendicular tangents can be drawn to the given circle is x2 + y2 = 338. (2007, 3M) (a) Statement I is true, Statement II is true; Statement II is correct explanation of Statement I (b) Statement I is true, Statement II is true, Statement II is not correct explanation of Statement I. (c) Statement I is true, Statement II is false. (d) Statement I is false, Statement II is true.

Passage Based Problems Passage 1 A tangent PT is drawn to the circle x2 + y2 = 4 at the point P ( 3 , 1). A straight line L, perpendicular to PT is a tangent to the circle (x − 3)2 + y2 = 1. (2012)

14. A possible equation of L is 3y = 1 3y = 5

15. A common tangent of the two circles is (b) y = 2 (d) x + 2 2 y = 6

Passage 2 A circle C of radius 1 is inscribed in an equilateral ∆PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F respectively. The line PQ is given by the  3 3 3 equation 3 x + y − 6 = 0 and the point D is  , .  2 2 Further, it is given that the origin and the centre of C (2008, 12M) are on the same side of the line PQ.

16. The equation of circle C is (a) (x − 2 3 )2 + ( y − 1)2 = 1 2 1 (b) (x − 2 3 )2 +  y +  = 1  2 (c) (x − 3 )2 + ( y + 1)2 = 1 (d) (x − 3 )2 + ( y − 1)2 = 1

 3 1 (b)  ,  , ( 3 , 0)  2 2

 3 3  3 1 (c)  , ,  ,   2 2  2 2

 3 3   3 1 (d)  , ,  ,   2 2   2 2

18. Equations of the sides QR, RP are 2 2 x + 1, y = − x−1 3 3 1 (b) y = x, y = 0 3 3 3 (c) y = x + 1, y = − x−1 2 2 (d) y = 3x, y = 0

Fill in the Blanks 19. A circle is inscribed in an equilateral triangle of side a. The area of any square inscribed in this circle is… . (1994, 2M)

20. If a circle passes through the points of intersection of

the coordinate axes with the lines λx − y + 1 = 0 and (1991,2M) x − 2 y + 3 = 0, then the value of λ is ... .

Analytical & Descriptive Questions 21. Find the equation of circle touching the line

2x + 3 y + 1 = 0 at the point (1, –1) and is orthogonal to the circle which has the line segment having end points (0, –1) and (–2, 3) as the diameter. (2004, 4M)

22. Find the coordinates of the point at which the circles x2 − y2 − 4x − 2 y + 4 = 0 and x2 + y2 − 12x − 8 y + 36 = 0 touch each other. Also, find equations of common tangents touching the circles the distinct points. (1993, 5M)

(b) x + (d) x −

(a) x = 4 (c) x + 3 y = 4

 3 3 (a)  ,  , ( 3 , 0)  2 2

(a) y =

Assertion and Reason

(a) x − 3 y = 1 (c) x − 3 y = − 1

17. Points E and F are given by

23. Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tangent is 4x + 3 y = 10, find the equations of the circles. (1991, 4M)

Integer & Numerical Answer Type Questions 24. Let the point B be the reflection of the point A( 2, 3) with respect to the line 8x − 6 y − 23 = 0. Let ΓA and ΓB be circles of radii 2 and 1 with centres A and B respectively. Let T be a common tangent to the circles ΓA and ΓB such that both the circles are on the same side of T. If C is the point of intersection of T and the line passing through A and B, then the length of the line segment AC is ........... (2019 Adv.)

25. The centres of two circles C1 and C 2 each of unit radius are at a distance of 6 units from each other. Let P be the mid-point of the line segment joining the centres of C1 and C 2 and C be a circle touching circles C1 and C 2 externally. If a common tangents to C1 and C passing through P is also a common tangent to C 2 and C , then the radius of the circle C is … (2009)

410 Circle

Topic 4 Radical Axis and Family of Circle Objective Questions I (Only one correct option) 1. The locus of the centres of the circles, which touch the

circle, x2 + y2 = 1 externally, also touch the Y-axis and lie in the first quadrant, is (2019 Main, 10 April II)

(a) y = 1 + 2x, x ≥ 0

(b) y = 1 + 4x, x ≥ 0

(c) x = 1 + 2 y, y ≥ 0

(d) x = 1 + 4 y, y ≥ 0

11. The equation of the circle passing through (1, 1) and

the points of intersection of x2 + y2 + 13x − 3 y = 0 and (1983, 1M) 2x2 + 2 y2 + 4x − 7 y − 25 = 0 is (a) 4x2 + 4 y2 − 30x − 10 y = 25 (b) 4x2 + 4 y2 + 30x − 13 y − 25 = 0 (c) 4 x2 + 4 y2 − 17x − 10 y + 25 = 0 (d) None of the above

2. The line x = y touches a circle at the point (1, 1). If the

12. Two circles x2 + y2 = 6 and x2 + y2 − 6x + 8 = 0 are given.

circle also passes through the point (1, − 3), then its radius is (2019 Main, 10 April I)

Then the equation of the circle through their points of (1980, 1M) intersection and the point (1, 1) is

(a) 3 2

(b) 2 2

(c) 2

(d) 3

3. Two circles with equal radii are intersecting at the

points (0, 1) and (0, −1). The tangent at the point (0,1) to one of the circles passes through the centre of the other circle. Then, the distance between the centres of these (2019 Main, 11 Jan I) circles is

(a)

2

(b) 2 2

(c) 1

(d) 2

4. Three circles of radii a , b, c(a < b < c) touch each other externally. If they have X-axis as a common tangent, (2019 Main, 9 Jan I) then (a) a , b, c are in AP

(b)

(c)

(d)

a , b , c are in AP

1 1 1 = + a b c 1 1 1 = + b a c

(b) (2, − 5)

(c) (5, − 2)

(d) (− 2, 5)

6. The circle passing through the point (− 1, 0) and touching the Y-axis at (0, 2) also passes through the (2011) point 3 5 (a)  − , 0 (b)  − , 2  2   2 

3 5 (c)  − ,   2 2

(d) (− 4, 0)

7. The locus of the centre of circle which touches ( y − 1)2 + x2 = 1 externally and also touches X-axis, is

(a) {x = 4 y, y ≥ 0} ∪ {(0, y), y< 0} (b) x2 = y (c) y = 4x2 (d) y2 = 4x ∪ (0, y), y∈R 2

(2005, 2M)

8. If two distinct chords, drawn from the point ( p, q) on the circle x2 + y2 = px + qy (where, pq ≠ 0) are bisected by the X-axis, then (1999, 2M) (a) p 2 = q2 (c) p 2 < 8q2

(b) p 2 = 8q2 (d) p 2 > 8q2

9. The locus of the centre of a circle, which touches

externally the circle x2 + y2 − 6x − 6 y + 14 = 0 and also touches the Y-axis, is given by the equation (1993, 1M)

(a) x2 − 6x − 10 y + 14 = 0 (c) y2 − 6x − 10 y + 14 = 0

(b) x2 − 10x − 6 y + 14 = 0 (d) y2 − 10x − 6 y + 14 = 0

10. The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle x + y = 9 is 2

(a) (3/2, 1/2) (c) (1/2, 1/2)

2

(b) (1/2, 3/2) (d) (1 / 2, − 21/ 2 )

(b) x2 + y2 − 3x + 1 = 0 (d) None of these

Fill in the Blanks 13. For each natural number k. Let C k denotes the circle with radius k centimetres and centre at origin. On the circle C k a particle moves k centimetres in the counter-clockwise direction. After completing its motion on C k the particle moves to C k + 1 in the radial direction. The motion of the particle continue in this manner. The particle starts at (1, 0). If the particle crosses the positive direction of the X-axis for the first time on the (1997, 2M) circle C n, then n = …

14. The intercept on the line y = x by the circle x2 + y2 − 2x = 0

5. The circle passing through (1, − 2) and touching the axis of x at (3, 0) also passes through the point (2013 Main) (a) (− 5, 2)

(a) x2 + y2 − 6 x + 4 = 0 (c) x2 + y2 − 4 y + 2 = 0

(1992, 2M)

is AB. Equation of the circle with AB as a diameter is… . (1996, 1M)

15. If the circle C1 : x + y = 16 intersects another circle C 2 2

2

of radius 5 in such a manner that the common chord is of maximum length and has a slope equal to 3/4, then (1988, 2M) the coordinates of the centre of C 2 are… .

16. The points of intersection of the line 4x − 3 y − 10 = 0 and the circle x2 + y2 − 2x + 4 y − 20 = 0 are…and….(1983, 2M)

Analytical & Descriptive Questions 17. Consider the family of circles x2 + y2 = r 2 , 2 < r < 5. If in the first quadrant, the common tangent to a circle of this family and the ellipse 4x2 + 25 y2 = 100 meets the coordinate axes at A and B, then find the equation of the (1999, 5M) locus of the mid-points of AB.

18. Consider a family of circles passing through two fixed points A (3, 7) and B (6, 5). Show that the chords in which the circle x2 + y2 − 4x − 6 y − 3 = 0 cuts the members of the family are concurrent at a point. Find (1993, 5M) the coordinates of this point.

19. A circle touches the line y = x at a point P such that OP = 4 2, where O is the origin. The circle contains the point (− 10, 2) in its interior and the length of its chord on the line x + y = 0 is 6 2. Determine the equation of (1990, 5M) the circle.

20. Let S ≡ x2 + y2 + 2 gx + 2 fy + c = 0 be a given circle. Find the locus of the foot of the perpendicular drawn from the origin upon any chord of S which subtends a right angle at the origin. (1988, 5M)

Circle 411 21. Let a given line L1 intersect the X and Y-axes at P and Q

22. Find the equations of the circles passing through (−4, 3)

respectively. Let another line L 2, perpendicular to L1, cut the X and Y-axes at R and S, respectively. Show that the locus of the point of intersection of the line PS and QR is a circle passing through the origin. (1987, 3M)

and touching the lines x + y = 2 and x − y = 2. (1982, 3M)

23. Find the equation of the circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of the lines 3x + 5 y = 1, (2 + c)x + 5c2y = 1 as c (1979, 3M) tends to 1.

Topic 5 Equation of Chord Bisected at a Point, Product of Pair of Tangents, Chord of Contact of Tangents, Pole and Equation of Polar Objective Questions I (Only one correct option)

7. The locus of the mid-point of a chord of the circle

line intersects the circle y = mx + 1 (x − 3)2 + ( y + 2)2 = 25 at the points P and Q. If the 3 midpoint of the line segment PQ has x-coordinate − , 5 then which one of the following options is correct?

1. A

(a) 6 ≤ m < 8 (c) 4 ≤ m < 6

(b) − 3 ≤ m < − 1 (d) 2 ≤ m < 4

(2019 Adv.)

2. If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90°, then the length (in cm) of their common chord is (2019 Main, 12 April I)

(a)

13 5

(b)

120 13

(c)

60 13

(d)

13 2

3. The sum of the squares of the lengths of the chords

intercepted on the circle, x2 + y2 = 16, by the lines, x + y = n, n ∈ N , where N is the set of all natural (2019, Main, 8 April I) numbers, is

(a) 320 (c) 160

(b) 105 (d) 210

4. The centres of those circles which touch the circle,

x + y − 8x − 8 y − 4 = 0, externally and also touch the (2016 Main) X-axis, lie on 2

2

(a) a circle (b) an ellipse which is not a circle (c) a hyperbola (d) a parabola

5. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4x − 5 y = 20 to the circle x2 + y2 = 9 is (2012) (a) (b) (c) (d)

20 (x + 20 (x2 + 36 (x2 + 36 (x2 + 2

y ) − 36x + 45 y = 0 y2 ) + 36x − 45 y = 0 y2 ) − 20 y + 45 y = 0 y2 ) + 20x − 45 y = 0 2

6. Tangents drawn from the point P (1, 8) to the circle

x2 + y2 − 6x − 4 y − 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the ∆PAB is

(a) (b) (c) (d)

x2 + y2 + 4x − 6 y + 19 = 0 x2 + y2 − 4x − 10 y + 19 = 0 x2 + y2 − 2x + 6 y − 29 = 0 x2 + y2 − 6x − 4 y + 19 = 0

(2009)

x2 + y2 = 4 which subtends a right angle at the origin, is

(a) x + y = 2 (b) x2 + y2 = 1 (c) x2 + y2 = 2 (d) x + y = 1

(1984, 2M)

Objective Question II (One or more than one correct option) 8. The equations of the tangents drawn from the origin to the circle x2 + y2 + 2rx + 2hy + h 2 = 0, are

(1988, 2M)

(a) x = 0 (b) y = 0 (c) (h 2 − r 2 ) x − 2rhy = 0 (d) (h 2 − r 2 ) x + 2rhy = 0

Assertion and Reason For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows. (a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I. (b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true

9. Consider

L1 : 2x + 3 y + p − 3 = 0 L2 : 2x + 3 y + p + 3 = 0

where, p is a real number and C : x2 + y2 − 6x + 10 y + 30 = 0 Statement I If line L1 is a chord of circle C, then line L 2 is not always a diameter of circle C. Statement II If line L1 is a diameter of circle C, then line L 2 is not a chord of circle C. (2008, 3M)

Fill in the Blanks 10. The chords of contact of the pair of tangents drawn from

each point on the line 2x + y = 4 to the circle x2 + y2 = 1 pass through the point… . (1997, 2M)

412 Circle 11. The equation of the locus of the mid-points of the

20. Find the intervals of values of a for which the line

12. The area of the triangle formed by the tangents from

y + x = 0 bisects two chords drawn from a point  1 + 2a 1 − 2a  ,  to the circle  2 2  

chords of the circle 4x2 + 4 y2 − 12x + 4 y + 1 = 0 that subtend an angle of 2π / 3 at its centre is … .(1993, 2M)

the point (4, 3) to the circle x2 + y2 = 9 and the line joining their points of contact is… . (1987, 2M)

13. From

the point on the circle A (0, 3) 2 2 x + 4x + ( y − 3) = 0, a chord AB is drawn and extended to a point M such that AM = 2 AB. The equation of the locus of M is….

2x2 + 2 y2 − (1 + 2a ) x − (1 − 2a ) y = 0.

(1996, 6M)

21. Let a circle be given by 2x (x − a ) + y (2 y − b) = 0, (a ≠ 0, b ≠ 0) Find the condition on a and b if two chords, each bisected by the X-axis can be drawn to the circle from (a , b /2).

(1986, 2M)

(1992, 6M)

14. The equation of the line passing through the points of

22. Lines 5x + 12 y − 10 = 0 and 5x − 12 y − 40 = 0 touch a circle

intersection of the circles 3x2 + 3 y2 − 2x + 12 y − 9 = 0 and x2 + y2 + 6x + 2 y − 15 = 0 is… . (1986, 2M)

15. Let x2 + y2 − 4x − 2 y − 11 = 0 be a circle. A pair of tangents from the point (4, 5) with a pair of radii form a quadrilateral of area... . (1985, 2M)

16. From the origin chords are drawn to the circle

(x − 1)2 + y2 = 1. The equation of the locus of the mid points of these chords is… . (1985, 2M)

Analytical & Descriptive Questions 17. Let 2x + y − 3xy = 0 be the equation of a pair of 2

2

tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the (2001, 5M) points of contact, find the length of OA.

18. Let T1 , T2 and be two tangents drawn from (−2, 0) onto the circle C : x2 + y2 = 1. Determine the circles touching C and having T1 , T2 as their pair of tangents. Further, find the equations of all possible common tangents to these circles when taken two at a time. (1999, 10M)

19. C1 and C 2 are two concentric circles, the radius of C 2 being twice that of C1. From a point P on C 2, tangents PA and PB are drawn to C1. Prove that the centroid of the ∆PAB lies on C1. (1998, 8M)

C1 of diameter 6. If the centre of C1 lies in the first quadrant, find the equation of the circle C 2 which is concentric with C1 and cuts intercepts of length 8 on these lines. (1986, 5M) 23. Through a fixed point (h , k) secants are drawn to the circle x2 + y2 = r 2. Show that the locus of the mid-points of the secants intercepted by the circle is x2 + y2 = hx + ky. (1983, 5M)

24. Let A be the centre of the circle x2 + y2 − 2x − 4 y − 20 = 0. Suppose that, the tangents at the points B (1, 7) and D (4, −2 ) on the circle meet at the point C. Find the area of the quadrilateral ABCD. (1981, 4M)

Integer & Numerical Answer Type Questions 5 . 2 Suppose PQ is a chord of this circle and the equation of the line passing through P and Q is 2x + 4 y = 5. If the centre of the circumcircle of the triangle OPQ lies on the line x + 2 y = 4, then the value of r is ……… . (2020 Adv.)

25. Let O be the centre of the circle x2 + y2 = r 2, where r >

26. Two parallel chords of a circle of radius 2 are at a distance

3 + 1 apart. If the chords subtend at the centre, angles of 2π , where k > 0, then the value of [k] is…… π /k and k (2010) NOTE

[k ] denotes the largest integer less than or equal to k]

Answers Topic 1 1. (a) 5. (d) 9. (c)

2. (c) 3. (b) 6. (c) 7. (a) 10. (d) 11. (c) 3 15. k ≠ 1 13. (b, c) 14. 4 21. x 2 + y 2 + 2ax + 2 py − (b 2 + q 2 ) = 0, radius = a + p + b + q 2

22. (2)

2

2

13. (d) 17. (b, c)

14. (a) 18. (a)

15. (a) 19. (c)

4. (c) 8. (b) 12. (a, c)

22. Ellipse having foci are (a, b) and (0, 0)

16. True

23. 16 : 1

16. (a,d) 20. (c)

21. A→p, q; B→p, q; C→q, r; D→q, r

Topic 3 1. (c)

2

23. (2)

24. (a)

25. (d)

1. (a)

2. (c)

3. (c)

4. (b)

5. (c)

6. (b)

7. (c)

8. (c)

9. (d)

10. (a)

11. (c)

12. (b)

Topic 2

2. (a)

3. (b)

4. (c)

5. (d)

6. (a)

7. (d)

9. (c)

10. (a)

11. (b, d)

12. (a, b, c)

13. (c)

14. (a)

16. (d)

17. (a)

18. (a)

15. (d) a2 19. sq unit 6

21. 2 x 2 + 2y 2 − 10 x − 5y + 1 = 0

8. (b)

20. λ = 2 or −

1 2

Circle 413 22. y = 0 and 7y − 24 x + 16 = 0

Topic 5

23. ( x − 5 ) 2 + (y − 5 ) 2 = 5 2 and ( x + 3 ) 2 + (y + 1 ) 2 = 5 2 24. (10)

1. (d) 5. (a)

25. (8)

2. (b) 6. (b)  1 1 10.  ,   2 4

9. (c)

Topic 4 1. (a)

2. (b)

3. (d)

4. (b)

5. (c) 9. (d) 13. n = 7

6. (d) 10. (b)

7. (a) 11. (b)

8. (d) 12. (b)

14. x 2 + y 2 − x − y = 0

 9 12  9 12 15.  − ,  and  ,−  5 5   5 5

16. ( −2, − 6 ) and ( 4, 2 )

17. 4 x + 25y = 4 x y

18. x = 2 and y = 23 / 3

19. x 2 + y 2 + 18 x − 2y + 32 = 0

2

2

2 2

c =0 2 22. x 2 + y 2 + 2 (10 ± 3 6 ) x + (55 ± 24 6 ) = 0 20. x 2 + y 2 + gx + f y +

23. 25 ( x 2 + y 2 ) − 20 x + 2y − 60 = 0

3. (d) 7. (c)

4. (d) 8. (a,c)

11. 16 x 2 + 16y 2 − 48 x + 16y + 31 = 0 13. x 2 + y 2 + 8 x − 6y + 9 = 0 15. 8 sq units

12.

192 sq units 25

14. 10 x − 3y − 18 = 0

16. x 2 + y 2 − x = 0

17. 3 (3 + 10 ) 2

2

4 5  4  1  18.  x +  + y 2 =   ; y = ± x +   3  3 5 39  20. a ∈ ( −∞, − 2 ) ∪ (2, ∞ )

21. a 2 > 2b 2

22. ( x − 5 ) 2 + (y − 2 ) 2 = 5 2

23. x 2 + y 2 = hx + ky

24. 75 sq units

26. (3)

25. (2)

Hints & Solutions Topic 1 Equation of Circle

⇒ (0 − 0)2 + (0 − 1)2 +

1. It is given that the circle touches the X-axis at (3, 0) and making an intercept of 8 on the Y -axis. Y

⇒1 +

(0 − x)2 + (1 − y)2 +

x2 + ( y − 1 ) 2 +

x2 + y2 = 4

⇒ x2 + y2 − 2 y + 1 +

x2 + y2 = 3

x2 + y2 = 4

⇒ x2 + y2 − 2 y + 1 = 3 − x2 + y2 B

⇒ M

8

r A O

[squaring both sides]

C

⇒ 1 − 2 y = 9 − 6 x2 + y2

r

(3,0)

x2 + y2 − 2 y + 1 = 9 + x2 + y2 − 6 x2 + y2

X

Let the radius of the circle is ‘r’, then the coordinates of centre of circle are (3, r ). From the figure, we have CM = 3 ⇒ CA = radius = r AB and AM = BM = =4 2 Then, r 2 = CM 2 + AM 2 = 9 + 16 = 25 ⇒ r=±5 Now, the equation of circle having centre (3, ± 5) and radius = 5 is (x − 3)2 + ( y ± 5)2 = 25 Now, from the options (3, 10) satisfy the equation of circle (x − 3)2 + ( y − 5)2 = 25

2. Given vertices of ∆AOP are O(0, 0) and A(0, 1) Let the coordinates of point P are (x, y). Clearly, perimeter = OA + AP + OP = 4 (given)

⇒ 6 x2 + y2 = 2 y + 8 ⇒

3 x2 + y2 = y + 4

⇒

9(x2 + y2) = ( y + 4)2

⇒

9x + 9 y = y + 8 y + 16 2

2

[squaring both sides]

2

⇒ 9x2 + 8 y2 − 8 y = 16 Thus, the locus of point P (x, y) is 9x2 + 8 y2 − 8 y = 16

3. Let the foot of perpendicular be P (h , k). Then, the slope of line OP =

k h Y

B P(h, k) A O

X

414 Circle Q Line AB is perpendicular to line OP, so slope of line h [Q product of slopes of two AB = − k perpendicular lines is (−1)] Now, the equation of line AB is h y − k = − (x − h ) ⇒ hx + ky = h 2 + k2 k x y or + =1  h 2 + k2  h 2 + k2      h   k   h 2 + k2  h 2 + k2  So, point A   , 0 and B0, k     h

∴

B (11, − 12) and D (− 5, 4)

Now,

OA = 25 + 144 = 169 = 13; OB = 121 + 144 = 265 OC = 121 + 16 = 137

and

OD = 25 + 16 = 41

5. Clearly, centre of the circumscribed circle is the centroid (G) of the equilateral triangle ABC. [Q in an equilateral triangle circumcentre and centroid coincide] C

Q ∆AOB is a right angled triangle, so AB is one of the diameter of the circle having radius R (given). ⇒ AB = 2R 2

⇒ ⇒ ⇒

r

2

 h 2 + k2  h 2 + k2   +  = 2R  h   k 

(–5,–6) G 120°

A

1  1 (h + k )  2 + 2 = 4R2 h k  2 (h + k2)3 = 4R2h 2k2 2

r B

2 2

On replacing h by x and k by y, we get (x2 + y2)3 = 4R2x2y2, which is the required locus.

4. Given equation of circle is x2 + y2 − 6x + 8 y − 103 = 0, which can be written as (x − 3)2 + ( y + 4)2 = 128 = (8 2 )2 ∴ Centre = (3, − 4) and radius = 8 2 Now, according to given information, we have the following figure. Y C

D

X

O

Also, we know that ∆AGB ≅ ∆BGC ≅ ∆CGA [by SAS congruence rule] ∴ ar (∆ABC ) = 3 ar (∆AGB) 1  = 3  r 2 sin 120° 2  1 [Q area of triangle = ab sin (∠C )] 2 [given] Q ar (∆ABC ) = 27 3 3 2 3 ∴ r = 27 3 2 2 3 ] [sin 120° = sin (180° − 60° ) = sin 60° = 2 ⇒ r2 = 4 × 9 ⇒ r = 6 Now, radius of circle, r=

45°

G (3,–4) A

[using distance (parametric) form of line, x − x1 y − y1 = = r] cos θ sin θ ⇒ x = 3 ± 8, y = − 4 ± 8 ∴ A(− 5, − 12) and C (11, 4) Similarly, for the coordinates of B and D, consider x−3 y+ 4 [in this case, θ = 135°] = = ±8 2 1 1 − 2 2 x = 3 m 8, y = − 4 ± 8

g2 + f 2 − c

⇒

6 = 25 + 36 − c [Q in the given equation of circle 2 g = 10 and 2 f = 12 ⇒ g = 5 and f = 6]

⇒ ⇒

36 = 25 + 36 − c c = 25

B

For the coordinates of A and C. x−3 y+ 4 Consider, = =±8 2 1 1 2 2

⇒

60°

6. Key idea Orthocentre, centroid and circumcentre are collinear and centroid divide orthocentre and circumcentre in 2 : 1 ratio. We have orthocentre and centroid of a triangle be A(−3, 5) and B(3, 3) respectively and C circumcentre.

Clearly, AB = (3 + 3)2 + (3 − 5)2 = 36 + 4 = 2 10 We know that, AB : BC = 2 : 1 BC = 10 ⇒ Now, AC = AB + BC = 2 10 + 10 = 3 10 Since, AC is a diameter of circle.

Circle 415 ∴

r=

AC 3 10 5 ⇒ r= =3 2 2 2

11. Let centre of circle be (h , k) . so that

7. Given, circle is inscribed in square formed by the lines

B (0,1)

x2 − 8x + 12 = 0 and y2 − 14 y + 45 = 0 ⇒ x = 6 and x = 2, y = 5 and y = 9 which could be plotted as Y

C (6,9)

D (2,9)

OA 2 = OB2

A (2,4) O (h,k )

y=9

y = x2 T

X

⇒ ⇒

h 2 + (k − 1)2 = (h − 2)2 + (k − 4)2 …(i) 4h + 6k − 19 = 0 k −4 Also, slope of OA = and slope of tangent at (2, 4) to h −2 y = x2 is 4. and (slope of OA) ⋅ (slope of tangent at A) = − 1 k −4 ∴ ⋅4 = − 1 h −2

y=5

B (6,5)

A (2,5)

X

O Y

where, ABCD clearly forms a square. ∴ Centre of inscribed circle = Point of intersection of diagonals = Mid-point of AC or BD  2 + 6  5 + 9 =  ,  = (4, 7)  2   2 

⇒

4k − 16 = − h + 2 h + 4k = 18 On solving Eqs. (i) and (ii), we get 53 16 and h = − k= 10 5  16 53 , . ∴ Centre coordinates are  −  5 10

⇒ Centre of inscribed circle is (4, 7).

8. Choosing OA as X-axis, A = (r , 0), B = (0, r ) and any point P on the circle is (r cos θ , r sin θ ). If (x, y) is the centroid of ∆ PAB, then Y B (0,r )

12.

P X′

O

A (r,0)

…(ii)

PLAN

X

(–g, –f )

Y′

3x = r cos θ + r + 0 and 3 y = r sin θ + 0 + r ∴ (3x − r )2 + (3 y − r )2 = r 2 Hence, locus of P is a circle.

9. Since, 2x − 3 y = 5 and 3x − 4 y = 7 are diameters of a circle. Their point of intersection is centre (1, − 1) . 7 Also given, πr 2 = 154 ⇒ r 2 = 154 × ⇒ r=7 22 ∴ Required equation of circle is (x − 1) + ( y + 1) = 7 2

⇒

2

Here, the length of intercept on Y-axis is ⇒ 2 f 2 − c and if circle touches X-axis ⇒ g2 = c 2 2 for x + y + 2 gx + 2 fy + c = 0 Here,

A

2

2√7

x + y − 2x + 2 y = 47 2

x2 + y2 + 2 gx + 2 fy + c = 0

2

B

10. Clearly, ∠C = 90° as angle in semi-circle is right angled.

A′

Now, area of triangle is maximum when AC = BC. i.e. Triangle is right angled isosceles.

(3, 0) 2√7

B′

C

passes through (3, 0). A

B

⇒ and

9 + 6g + c = 0

…(i)

g2 = c

…(ii)

2 f2 − c =2 7

416 Circle f2 − c = 7

…(iii)

On subtracting Eq. (ii) from Eq. (iii), we get p1 − 2 q1 = 0

From Eqs. (i) and (ii), we get g 2 + 6 g + 9 = 0 ⇒ ( g + 3 )2 = 0 ⇒

g = − 3 and c = 9

∴ ∴

p1 = x22 + y22 − x12 − y12,

where,

f 2 = 16 ⇒

q1 = y2 − y1

f =±4

On subtracting Eq. (ii) from Eq. (iv), we get p2 − 2q2 = 0

x2 + y2 − 6x ± 8 y + 9 = 0

13. Let equation of line L1 be y = mx. Intercepts made by L1 and L 2 on the circle will be equal i.e. L1 and L 2 are at the same distance from the centre of the circle; Centre of the given circle is (1/2, – 3/2). Therefore, m 3 + m+3 1 /2 – 3 /2 – 1 2 = 2 2 ⇒ = 2 1+1 2 m –1 2 m2 + 1 ⇒ ⇒

8m + 8 = m + 6m + 9 7m – 6 m – 1 = 0 ⇒ (7m + 1)(m – 1) = 0 1 ⇒ m = – , m =1 7 Thus, two chords are x + 7 y = 0 and x – y = 0. Therefore, (b) and (c) are correct answers. 7 14. Since, 3x − 4 y + 4 = 0 and 3x − 4 y − = 0 are two parallel 2 tangents. Thus, distance between them is diameter of circle 7 4+ 15 3 2 Diameter = = = ⇒ 32 + 42 2 . 5 2 2

2

2

and

3 radius = 4

15. Since, P lies on circle and A and B are points in plane PA = k, then the locus of P is perpendiular PB bisector of AB. Thus, the value of k ≠ 1.

such that,

Now, p1 , p2 , q1 , q2 are rational numbers. Also, either q1 ≠ 0 or q2 ≠ 0. If q1 ≠ 0, then 2 = p1 / q1 and if q2 ≠ 0, then 2 = p2 / q2. In any case 2 is a rational number. This is a contradiction.

18. The given circle is ax2 + 2hxy + by2 = 1

17. Equations of any circle C with centre at (0, 2 ) is given

Then, equation of line through P is x − x1 y − y1 = =r cos θ sin θ ⇒ x = x1 + r cos θ , y = y1 + r sin θ For points Q and R, above point must lie on Eq. (i). ⇒ a (x1 + r cos θ )2 + 2h (x1 + r cos θ ) ( y1 + r sin θ ) + b ( y1 + r sin θ )2 = 1 ⇒ (a cos 2 θ + 2h sin θ cos θ + b sin 2 θ )r 2 + 2 (ax1 cos θ + hx1 sin θ + hy1 cos θ + by1 sin θ )r + (ax12 + 2hx1 y1 + by12 − 1) = 0 It is quadratic in r, giving two values of r as PQ and PR. ∴ PQ ⋅ PR =

(x − 0) + ( y − 2 ) = r or

2

= a + 2 h sin θ cos θ + (b − a )sin 2 θ = a + sin θ {2h cos θ + (b − a ) sin θ } = a + sin θ ⋅ 4h 2 + (b − a )2 ⋅ (cos θ sin φ + sin θ cos φ ) where, tan θ =

b−a 2h

= a + 4h 2 + (b − a )2 sin θ sin (θ + φ )

2

x2 + y2 − 2 2 y + 2 = r 2

a x12 + 2hx1 y1 + by12 − 1 a cos θ + 2h sin θ cos θ + b sin 2 θ 2

Here, ax12 + 2hx1 y1 + by12 − 1 ≠ 0, as (x1 , y1 ) does not lie on Eq. (i), Also, a cos 2 θ + 2h sin θ cos θ + b sin 2 θ

by 2

…(i)

Let the point P not lying on Eq. (i) be (x1 , y1 ), let θ be the inclination of line through P which intersects the given curve at Q and R.

16. Since, centre of circle is (3, − 1) which lies on x + 3 y = 0 ⇒ x + 3 y = 0 is diameter of x2 + y2 − 6x + 2 y = 0 Hence, given statement is true.

p2 = x32 + y32 − x12 − y12, q2 = y3 − y1

where

which will be independent of θ, if …(i)

where, r > 0.

4h 2 + (b − a )2 = 0 ⇒

Let (x1 , y1 ), (x2, y2), (x3 , y3 ) be three distinct rational points on circle. Since, a straight line parallel to X-axis meets a circle in at most two points, either y1 , y2 or y1 , y3 .

which is a equation of circle.

19. Let the radius of the circle be r. Take X-axis along AC

On putting these in Eq. (i), we get x12 + y12 − 2 x22 + y22 − 2 x32 + y32 − 2

h = 0 and b = a 1 ∴ Eq. (i) reduces to x2 + y2 = a

2 y1 = r − 2

…(ii)

2 y2 = r 2 − 2

…(iii)

2 y3 = r 2 − 2

…(iv)

2

and the O (0, 0) as centre of the circle. Therefore, coordinate of A and C are (− r , 0) and (r , 0), respectively. Now, ∠ BAC = β , ∠ BOC = 2 β Therefore, coordinates of B are (r cos 2 β , r sin 2 β ).

Circle 417 And slope of AD is tan (β − α ). Let (x, y) be the coordinates of the point D. Equation of AD is Y

X′

A(−r, 0)

B (r cos 2 β, r sin 2 β) P D E O

C (r,0)

X

Y′

…(i) y = tan (β − α ) (x + r ) [Q slope = tan (β − α ) and point is (− r, 0)] Now, equation of BC is r sin 2 β − 0 y= (x − r ) r cos 2 β − r ⇒

y=

r ⋅ 2 sin β cos β (x − r ) r (− 2 sin 2 β )

⇒

y=

2 sin β cos β (x − r ) −2 sin 2 β

⇒

y = − cot β (x − r )

2

  cos β sin (β − α )   cos β cos (β − α ) d2 = r2  + 1 + r 2   cos α cos α     =

…(ii)

tan (β − α ) (x + r ) = − cot β (x − r )

⇒

x tan (β − α ) + r tan (β − α ) = − x cot β + r cot β

⇒

x [tan (β − α ) + cot β ] = r [cot β − tan (β − α )]

⇒

 cos β sin(β − α )  sin (β − α ) cos β  x − +   =r cos ( β − α ) β sin sin β cos (β − α )   

sin (β − α ) sin β + cos (β − α ) cos β  ⇒ x  cos (β − α ) sin β    cos β cos (β − α ) − sin β sin (β − α )  =r  sin β cos (β − α )   ⇒ x [cos (β − α − β )] = r [cos (β − α + β )] r cos (2 β − α ) ⇒ x= cos α On putting this value in Eq. (ii), we get   r cos (2 β − α ) y = − cot β  − r cos α   ⇒

cos β ⋅ r  cos (2β − α ) − cos α  y=−   sin β  cos α 

⇒

 2β − α + α α − 2β + α  2 sin   sin       r cos β  2 2   y=− sin β  cos α   

⇒

r cos β 2 sin β ⋅ sin (α − β )  y=−   sin β  cos α  = − 2r cos β sin (α − β ) / cos α

Thus, coordinates of E are  r cos (2 β − α ) + r cos α cos β sin (α − β ) ,−r   2 cos α cos α    2β − α + α   2β − α − α  2 cos   ⋅ cos       2 2 ⇒ r , 2 cos α cos β sin (β − α ) r cos α cos β ⋅ cos (β − α ) cos β sin (β − α ) r ⇒ ,r cos α cos α Since, AE = d, we get

To obtain the coordinate of D, solve Eqs. (i) and (ii) simultaneously ⇒

Therefore, coordinates of D are  r cos (2 β − α ) 2r cos β sin (α − β ) ,−   cos α cos α  

2

r2 [cos 2 β cos 2 (β − α ) + cos 2 α cos 2 α

+ 2 cos β cos (β − α ) cos α + cos 2 β sin 2 (β − α )] r2 [cos 2 β {cos 2 (β − α ) + sin 2 (β − α ) } + cos 2 α = cos 2 α + 2cos β cosα cos ( β − α )] r2 2 2 [cos β + cos α + 2 cos α cos β cos (β − α )] = cos 2 α ⇒ r2 =

d 2 cos 2 α cos β + cos α + 2 cos α cos β cos (β − α ) 2

2

Therefore, area of the circle π d 2 cos 2 α πr 2 = 2 2 cos β + cos α + 2 cos α cos β cos (β − α ) 1  ; i = 1, 2, 3, 4 mi   lie on a circle x2 + y2 + 2 gx + 2 fy + c = 0. 1 2f Then, mi2 + 2 + 2 gmi + + c=0; mi mi 

20. Let the points mi ,

Since, mi4 + 2 gmi3 + cmi2 + 2 fmi + 1 = 0 ; i = 1, 2, 3, 4 ⇒ m1 , m2, m3 and m4 are the roots of the equation m4 + 2 gm3 + cm2 + 2 fm + 1 = 0 1 m1m2 m3m4 = = 1 ⇒ 1

21. Let (x1 , y1 ) and (x2, y2) be the coordinates of points A and B, respectively. It is given that x1 , x2 are the roots of x2 + 2ax − b2 = 0 …(i) ⇒ x1 + x2 = − 2a and x1x2 = − b2 Also, y1 and y2 are the roots of y2 + 2 py − q2 = 0 …(ii) ⇒ y1 + y2 = − 2 p and y1 y2 = − q2 ∴ The equation of circle with AB as diameter is, (x − x1 ) (x − x2) + ( y − y1 ) ( y − y2) = 0 ⇒ x2 + y2 − (x1 + x2) x − ( y1 + y2) y + (x1x2 + y1 y2) = 0 ⇒ x2 + y2 + 2ax + 2 py − (b2 + q2) = 0 and radius = a 2 + p2 + b2 + q2

418 Circle 22. The circle and coordinate axes can have 3 common points, if it passes through origin. [ p = 0] If circle is cutting one axis and touching other axis. Only possibility is of touching X-axis and cutting

Y -axis. [ p = − 1]

23. x2 + y2 ≤ 6 and 2x − 3 y = 1 is shown as L

1/2

1/3

∴ For any point in shaded part L > 0 and for any point inside the circle S < 0.  3 Now, for 2,  L : 2x − 3 y − 1  4 9 3 L :4 − − 1 = > 0 4 4 9 2 2 and S : x + y − 6, S : 4 + −6 0  4 4 2 4 

E1(– 3,1)

Similarly, equation of tangent at F1 (1, − 3 ) and F2(1, 3 ) are x − 3 y = 4 and x + 3 y = 4, respectively and intersection point is (4, 0), i.e., F3 (4, 0) and equation of tangent at G1 (0, 2) and G2(2, 0) are 2 y = 4 and 2x = 4, respectively and intersection point is (2, 2) i.e., G3 (2, 2). Point E3 (0, 4), F3 (4, 0) and G3 (2, 2) satisfies the line x + y = 4.

25. We have,

For the point to lie in the shade part, origin and the point lie on opposite side of straight line L.

24.

Intersection point of tangent at E1 and E 2 is (0, 4). ∴ Coordinates of E3 is (0, 4)

Topic 2 Relation between Two Circles

E2( 3,1)

1. Equation of given circles G2(2, 0)

F1(1, – 3) (0, –2) Y′

Equation of tangent at E1 (− 3 , 1) is − 3x + y = 4 and at E 2( 3 , 1) is 3x + y = 4

x

x2 + y2 + 5Kx + 2 y + K = 0 and 2(x2 + y2) + 2Kx + 3 y − 1 = 0 3 1 ⇒ x2 + y2 + Kx + y − = 0 2 2 On subtracting Eq. (ii) from Eq. (i), we get 1 1 4Kx + y + K + = 0 2 2 ⇒ 8Kx + y + (2K + 1) = 0

…(i) …(ii)

…(iii)

[Q if S1 = 0 and S 2 = 0 be two circles, then their common chord is given by S1 − S 2 = 0.]

Circle 419 Eq. (iii) represents equation of common chord as it is given that circles (i) and (ii) intersects each other at points P and Q. Since, line 4x + 5 y − K = 0 passes through point P and Q. 8K 1 2K + 1 ∴ = = 4 5 −K 1 [equating first and second terms] ⇒ K = 10 − K = 10K + 5 [equating second and third terms] 5 ⇒ 11K + 5 = 0 ⇒ K = − 11 1 5 ≠ − , so there is no such value of K, for which Q 10 11 line 4x + 5 y − K = 0 passes through points P and Q.

3(9) + 4(1) − λ ≥ 81 + 1 − 78 5

and

⇒ |λ − 31 | ≥ 10 ⇒ λ ∈ (− ∞ , 21] ∪ [41, ∞ ) From Eqs. (iii), (iv) and (v), we get λ ∈ [ 12, 21 ]

4. Given circles, x2 + y 2 − 2 x − 2 y − 2 = 0

and

P

tangent at the point (cos θ ,sin θ ) on the given circle is

⇒

x2 + y2 − 4x2y2 = 0

... (i) x + y − 2x − 2 y + 1 = 0 and ... (ii) x + y2 − 18x − 2 y + 78 = 0, are on the opposite sides of the variable line 3x + 4 y − λ = 0. So, their centres also lie on the opposite sides of the variable line. ⇒ [3(1) + 4(1) − λ ] [3(9) + 4(1) − λ ] < 0 [Q The points P (x1 , y1 ) and Q (x2, y2) lie on the opposite sides of the line ax + by + c = 0, if (ax1 + by1 + c)(ax2 + by2 + c) < 0] ⇒ (λ − 7)(λ − 31) < 0 ... (iii) ⇒ λ ∈ (7, 31) 3(1) + 4(1) − λ Also, we have ≥ 1 + 1 −1 5 2

2

QDistance of centre from the given line is    ax1 + by1 + c ≥ r  greater than the radius,i.e.   a2 + b2 ⇒

|7 − λ|≥ 5 ⇒ λ ∈ (− ∞, 2] ∪ [12 , ∞ )

2 C1 (1,1)

2 (3,3) C2 Q

So, area of quadrilateral PC1QC 2 = 2 × ar (∆PC1C 2). 1  = 2 ×  × 2 × 2 = 4 sq units 2 

5. Circle I is x2 + y2 − 16x − 20 y + 164 = r 2 ⇒ (x − 8)2 + ( y − 10)2 = r 2 ⇒C1 (8, 10) is the centre of Istcircle and r1 = r is its radius Circle II is (x − 4)2 + ( y − 7)2 = 36 ⇒ C 2(4, 7) is the centre of 2nd circle and r2 = 6 is its radius. Two circles intersect if|r1 − r2| < C1C 2 < r1 + r2 ⇒| r − 6|< (8 − 4)2 + (10 − 7)2 < r + 6 ⇒| r − 6| < 16 + 9 < r + 6

3. The given circles, 2

… (i)

and … (ii) x + y2 − 6x − 6 y + 14 = 0 are intersecting each other orthogonally, because 2(1)(3) + 2(1)(3) = 14 − 2 [Q two circles are intersected orthogonally if 2 g1 g2 + 2 f1 f2 = c1 + c2] 2

2. Equation of given circle is x2 + y2 = 1, then equation of …(i) x cos θ + y sin θ = 1 [QEquation of tangent at the point P(cos θ , sin θ ) to the circle x2 + y2 = r 2 is x cos θ + y sin θ = r] Now, the point of intersection with coordinate axes are P(sec θ , 0) and Q(0, cos ec θ ). Q Mid-point of line joining points P and Q is  sec θ cos ecθ  M ,  = (h , k) (let)  2 2  1 1 So, cos θ = and sin θ = 2h 2k Q sin 2 θ + cos 2 θ = 1 1 1 1 1 ∴ + =1⇒ 2 + 2 =4 4h 2 4k2 h k Now, locus of mid-point M is 1 1 + =4 x2 y 2

... (v)

... (iv)

⇒| r − 6| < 5 < r + 6 Now as, 5 < r + 6 always, we have to solve only | r − 6| < 5 ⇒ − 5 < r − 6 < 5 ⇒

6 − 5 < r < 5 + 6 ⇒ 1 < r < 11

6. Given equation of a circle is x2 + y2 − 4x + 6 y − 12 = 0, whose centre is (2, − 3) and radius = 2 2 + (− 3) 2 + 12 = 4 + 9 + 12 = 5 Now, according to given information, we have the following figure. S A (–3, 2) O (2,–3) B

C

420 Circle x2 + y2 − 4x + 6 y − 12 = 0 Clearly, AO ⊥ BC, as O is mid-point of the chord. Now, in ∆AOB, we have

Y C C1 (1,1)

OA = (− 3 − 2) 2 + (2 + 3) 2 and

= 25 + 25 = 50 = 5 2 OB = 5

∴

AB = OA 2 + OB2

X′

7. Here, radius of smaller circle, AC = 12 + 32 − 6 = 2 Clearly, from the figure the radius of bigger circle r 2 = 22 + [(2 − 1)2 + (1 − 3)2] ⇒ r =3 A

O

C2

(1,0)

X

k + 1 = k2 + 2 − 2k

⇒ ⇒

k2 + 1 + 2k = k2 + 2 − 2k 1 ⇒ k= 4 1 So, the radius of circle T is k, i. e. . 4

10. Since, the given circles intersect orthogonally. 2

C

(1– k) 1

Y′

= 50 + 25 = 75 = 5 3

r2 = 9

T (0,k)

r

∴

2 (1) (0) + 2 (k) (k) = 6 + k [Q 2 g1 g2 + 2 f1 f2 = c1 + c2] 3 2k2 − k − 6 = 0 ⇒ k = − , 2 2

C1 (2,1)

(1,3) 2

⇒

B

11. Let O is the point at centre and P is the point at 8.

PLAN Number of common tangents depend on the position of the circle with respect to each other.

circumference. Therefore, angle QOR is double the angle QPR. So, it is sufficient to find the angle QOR. Y

(i) If circles touch externally ⇒ C1C 2 = r1 + r2, 3 common tangents. (ii) If circles touch internally ⇒ C1C 2 = r2 − r1,1 common tangent. (iii) If circles do not touch each other, 4 common tangents.

= 9 + 81 – 26 = 8(r2) Now,

C1C 2 = (2 + 3)2 + (3 + 9)2

⇒

C1C 2 = 5 + 12 2

...(i) ...(ii)

⇒

C1C 2 = 25 + 144 = 13 r1 + r2 = 5 + 8 = 13

Distance between their centre

⇒

k + 1 = 1 + k + 1 − 2k

Y′

slope of OQ , m1 = 4 / 3 , slope of OR, m2 = − 3 / 4 Here, m1 m2 = − 1 Therefore, ∠ QOR = π /2 which implies that ∠ QPR = π /4

Let the coordinate of the centre of T be (0, k).

2

X

Now,

PLAN Use the property, when two circles touch each other externally, then distance between the centre is equal to sum of their radii, to get required radius.

k + 1 = 1 + (k − 1)2

O (0,0)

P

[say]

Also, C1C 2 = r1 + r2 Thus, both circles touch each other externally. Hence, there are three common tangents.

9.

X′

[say]

2

∴

4 ,3 ) R (−

Given equations of circles are x2 + y2 − 4x − 6 y − 12 = 0 2 x + y2 + 6x + 18 y + 26 = 0 Centre of circle (i) is C1 (2, 3) and radius = 4 + 9 + 12 = 5(r1 ) Centre of circle (ii) is C 2(–3, – 9) and radius

Q (3,4)

[Q C1C 2 = k + 1]

12. Given, x2 + y2 = 4 Centre ≡ C1 ≡ (0, 0) and R1 = 2 Again, x2 + y2 − 6x − 8 y − 24 = 0, then C 2 ≡ (3, 4) and R2 = 7 Again, C1C 2 = 5 = R2 − R1 Therefore, the given circles touch internally such that, they can have just one common tangent at the point of contact.

13. Centre of the circle x2 + y2 + 4x − 6 y + 9 sin 2 α + 13 cos2 α = 0

Circle 421 is C (−2, 3) and its radius is = 13 − 13 cos2 α − 9 sin 2 α

∴ Locus of M is a circle having PQ as its diameter of circle. ∴ Equation of circle (x − 2) (x + 2) + ( y + 5) ( y − 7) = 0

= 13 sin 2 α − 9 sin 2 α = 4 sin 2 α = 2 sin α

⇒

(−2) + (3) − 9 sin α − 13 cos α 2

2

2

2

x2 + y2 − 2 y − 39 = 0

Hence, E1 : x2 + y2 − 2 y − 39 = 0, x ≠ ± 2 Locus of mid-point of chord (h , k) of the circle E1 is

A

xh + yk − ( y + k) − 39 = h 2 + k2 − 2k − 39 , −2 C(

P (h, k)

⇒ xh + yk − y − k = h 2 + k2− 2k Since, chord is passing through (1, 1). ∴ Locus of mid-point of chord (h , k) is

3)

h + k − 1 − k = h 2 + k2 − 2k

B

⇒

Let (h , k) be any point P and ∠ APC = α , ∠ PAC = π /2 That is, triangle APC is a right angled triangle. AC 2 sin α ∴ sin α = = PC (h + 2)2 + (k − 3)2 ⇒ ⇒

(h + 2)2 + (k − 3)2 = 4 2 h + 4 + 4h + k2 + 9 − 6k = 4

⇒

h 2 + k2 + 4h − 6k + 9 = 0

Locus is E 2 : x2 + y2 − x − 2 y + 1 = 0 Now, after checking options, (a) and (d) are correct.

17.

PLAN (i) The general equation of a circle is x 2 + y 2 + 2 gx + 2 f y + c = 0 where, centre and radius are given by ( − g , − f ) and g 2 + f 2 − c , respectively. (ii) If the two circles x 2 + y 2 + 2 g 1x + 2 f1 y + c1 = 0 and x 2 + y 2 + 2 g 2x + 2 f2 y + c 2 = 0 are orthogonal, then 2 g 1g 2 + 2 f1f2 = c1 + c 2.

Thus, required equation of the locus is x2 + y 2 + 4 x − 6 y + 9 = 0

Let circle be x2 + y2 + 2 gx + 2 fy + c = 0

14. As, the two circles intersect in two distinct points. ⇒ Distance between centres lies between | r1 − r2| and | r1 + r2|. | r − 3|
2

∴

2

−2 g ⋅ 0 − 2 f ⋅ 0 = c − k 2

⇒

c=k

c=1 ⇒

g = 7 and f = − 1

Centre is (− g , − f ) ≡ (− 7, 1)

2

…(i)

Also, x2 + y2 + 2 gx + 2 fy + k2 = 0 passes through (a , b). …(ii) ∴ a 2 + b2 + 2 ga + 2 fb + k2 = 0 ⇒ Required equation of locus of centre is −2ax − 2by + a + b + k = 0 2

c = 15 − 2 g

2

2

∴

Radius =

16. It is given that T is tangents to S1 at P and S 2 at Q and S1

= 49 + 1 − 1 = 7

18. Let the, equation of circles are C1 : (x − 1)2 + ( y − 1)2 = (1)2 and C 2 : (x − 1)2 + ( y − 1)2 = ( 2 )2 D(0, 2)

s1

P(–2,7) N

Q(2,–5)

MN = NP = NQ

P

(1, 1)

s2

M

C(2, 2) C1

and S 2 touch externally at M.

∴

g2 + f 2 − c

2ax + 2by − (a 2 + b2 + k2) = 0

or

A(0, 0)

T

…(ii)

x2 + y2 − 1 = 0

Orthogonal with

orthogonally. ⇒

…(i)

Orthogonal with x + y − 2x − 15 = 0 ⇒

15. Let x2 + y2 + 2 gx + 2 fy + c = 0, cuts x2 + y2 = k2 2 g1 g2 + 2 f1 f2 = c1 + c2

1 + 2f + c = 0 2

2 g (− 1) = c − 15

2 < r O k >O for first quadrant

Hence, the equations of the two circles are (x − 5)2 + ( y − 5)2 = 52 and (x + 3)2 + ( y + 1)2 = 52

X

O x2+y2=1

24. According to given informations the figure is as following ∴ OC = r + 1

Q q C

90° 2

P 1 B

[Q if circles touch each other externally, then C1C 2 = r1 + r2]

A (2,3)

M

⇒

and k > 0, for first quadrant.

8x-6y-23=0

From the figure, AC = Qsin θ =

2 sin θ

1 (from ∆CPB) CB

h 2 + k2 = h + 1, h > 0

…(i) …(ii)

⇒ h 2 + k2 = h 2 + 2h + 1 ⇒ k2 = 2h + 1 k = 1 + 2h , as k > 0 ⇒ Now, on taking locus of centre (h , k), we get y = 1 + 2x, x ≥ 0

Circle 429 2. Since, the equation of a family of circles touching line at their point of contact(x1 , y1 ) L =0 (x − x1 )2 + ( y − y1 )2 + λ L = 0, where λ ∈ R.

is

∴Equation of circle, touches the x = y at point (1, 1) is (x − 1)2 + ( y − 1)2 + λ (x − y) = 0 2 2 …(i) ⇒ x + y + (λ − 2)x + (− λ − 2) y + 2 = 0 Q Circle (i) passes through point (1, − 3). ∴ 1 + 9 + (λ − 2) + 3(λ + 2) + 2 = 0 ⇒ 4λ + 16 = 0 ⇒ λ = −4 So, equation of circle (i) at λ = − 4 , is x2 + y 2 − 6 x + 2 y + 2 = 0

∴In ∆BCF, BF = (b + c)2 − (c − b)2 = 2 bc AD + AE = BF

Q ∴

2 ab + 2 ac = 2 bc ⇒

5. Let the equation of circle be (x − 3)2 + ( y − 0)2 + λy = 0 Y A (3, 0)

X

Now, radius of the circle = 9 + 1 − 2 = 2 2.

Y

3. Clearly, circles are orthogonal because tangent at one point of intersection is passing through centre of the other.

X

P (1, –2)

As it passes through (1, − 2) ∴ (1 − 3)2 + (− 2)2 + λ (− 2) = 0 ⇒ 4 + 4 − 2λ = 0 ⇒ λ = 4 ∴ Equation of circle is (x − 3)2 + y2 + 4 y = 0

Y S2

S1

(0, 1) r

r

By hit and trial method, we see that point (5, − 2) satisfies equation of circle.

6. Equation of circle passing through a point (x1 , y1 ) and

X

(α, 0)

(–α, 0)

touching the straight line L, is given by

(0,–1)

(x − x1 )2 + ( y − y1 )2 + λ L = 0 ∴ Equation of circle passing through (0, 2) and touching x=0 …(i) ⇒ (x − 0)2 + ( y − 2)2 + λ x = 0

Let C1 (α, 0) and C 2(− α , 0) are the centres. Then, S1 ≡ (x − α )2 + y2 = α 2 + 1 ⇒ S1 ≡ x2 + y2 − 2 αx − 1 = 0

Also, it passes through (− 1, 0) ⇒ 1 + 4 − λ = 0

[Q radius, r = (α − 0) + (0 − 1) ] 2

2

and S 2 ≡ (x + α ) + y = α + 1 ⇒ S 2 ≡ x2 + y2 + 2αx − 1 = 0 Now, 2(α ) (− α ) + 2 ⋅ 0 ⋅ 0 = (− 1) + (− 1) ⇒ α = ± 1 2

1 1 1 + = c b a

2

2

[Q condition of orthogonality is 2 g1 g2 + 2 f1 f2 = c1 + c2]

∴ λ =5 Eq. (i) becomes, x2 + y 2 − 4 y + 4 + 5 x = 0 ⇒

x2 + y2 + 5x − 4 y + 4 = 0,

For x-intercept put y = 0

⇒ x2 + 5x + 4 = 0,

(x + 1) (x + 4) = 0

∴ C1 (1, 0) and C 2(− 1, 0) ⇒ C1C 2 = 2

4. According to given information, we have the following

∴

x = − 1, − 4

Hence, (d) option (–4, 0).

figure.

7. Let the locus of centre of circle be (h , k) touching ( y − 1)2 + x2 = 1 and X-axis shown as X

C B b

D

F

c

E A

a

(0, 1) O

where A , B, C are the centres of the circles Clearly, AB = a + b (sum of radii) and BD = b − a

X′

∴ AD = (a + b)2 − (b − a )2

1

|k| (h, k ) A |k| B

O

X

Y′

(using Pythagoras theorem in ∆ABD) = 2 ab Similarly, AC = a + c and CE = c − a

Clearly, from figure, Distance between C and A is always 1 + | k|,

∴In ∆ACE, AE = (a + c)2 − (c − a )2 = 2 ac

i.e.

Similarly, BC = b + c and CF = c − b

⇒

(h − 0)2 + (k − 1)2 = 1 + | k|, h 2 + k2 − 2k + 1 = 1 + k2 + 2| k|

430 Circle ⇒

h 2 = 2 | k | + 2 k ⇒ x2 = 2 | y | + 2 y  y , y≥0 | y| =  − y , y < 0

where ∴

x2 = 2 y + 2 y, y ≥ 0

and

x2 = − 2 y + 2 y, y < 0

10. Let C1 (h , k) be the centre of the required circle. Then,

⇒

x = 4 y, when y ≥ 0

(h − 0)2 + (k − 0)2 = (h − 1)2 + (k − 0)2

and

x = 0, when y < 0

2

2

∴

8.

h 2 + 9 − 6h + k2 + 9 − 6k = 4 + h 2 + 4h i.e. k2 − 10h − 6k + 14 = 0 Thus, the locus of (h, k) is y2 − 10x − 6 y + 14 = 0

NOTE

{(x, y): x2 = 4 y, when y ≥ 0} ∪ {(0, y): y < 0}

⇒ ⇒

h 2 + k2 = h 2 − 2h + 1 + k2 −2 h + 1 = 0 ⇒ h = 1 / 2

Since, (0, 0) and (1, 0) lie inside the circle x2 + y2 = 9. Therefore, the required circle can touch the given circle internally. i.e. C1 ⋅ C 2 = r1 ~ r2

In solving a line and a circle there oftengenerate a quadratic equation and further we have to apply condition of Discriminant so question convert from coordinate to quadratic equation.

From equation of circle it is clear that circle passes through origin. Let AB is chord of the circle.

⇒ ⇒

Y

⇒ ⇒

h 2 + k2 = 3 − h 2 + k2 1 + k2 = 3 4 1 3 1 9 + k2 = ⇒ + k2 = 4 2 4 4 k2 = 2 ⇒ k = ± 2

2 h 2 + k2 = 3 ⇒ 2

11. The required equation of circle is 1 25  (x2 + y2 + 13x − 3 y) + λ  11x + y +  =0  2 2

A (p,q) X′

C (h,0)

X

O

Its passing through (1, 1), ⇒ 12 + λ (24) = 0

B

1 2 On putting in Eq. (i), we get 11 1 25 x2 + y2 + 13x − 3 y − x− y− =0 2 4 4 ⇒

Y′

A ≡ ( p, q) ⋅ C is mid-point and coordinate of C is (h , 0) Then, coordinates of B are (− p + 2h , − q) and B lies on the circle x2 + y2 = px + qy, we have ⇒

λ=−

(− p + 2h )2 + (− q)2 = p (− p + 2h ) + q (− q)

⇒

4x2 + 4 y2 + 52x − 12 y − 22x − y − 25 = 0

p2 + 4h 2 − 4 ph + q2 = − p2 + 2 ph − q2

⇒

4x2 + 4 y2 + 30x − 13 y − 25 = 0

⇒

2 p2 + 2q2 − 6 ph + 4h 2 = 0

⇒

2h − 3 ph + p + q = 0 2

2

2

…(i)

12. The required equation of circle is, S1 + λ (S 2 − S1 ) = 0. …(i)

There are given two distinct chords which are bisected at X-axis then, there will be two distinct values of h satisfying Eq. (i). So, discriminant of this quadratic equation must be > 0. ⇒ D >0 ⇒ (− 3 p)2 − 4 ⋅ 2 ( p2 + q2) > 0 ⇒ 9 p2 − 8 p2 − 8 q 2 > 0 ⇒ p2 − 8 q 2 > 0 ⇒ p2 > 8 q 2

9. Let (h , k) be the centre of the circle which touches the circle x2 + y2 − 6x − 6 y + 14 = 0 and Y-axis. The centre of given circle is (3, 3) and radius is 32 + 32 − 14 = 9 + 9 − 14 = 2 Since, the circle touches Y-axis, the distance from its centre to Y-axis must be equal to its radius, therefore its radius is h. Again, the circles meet externally, therefore the distance between two centres = sum of the radii of the two circles. Hence, (h − 3)2 + (k − 3)2 = (2 + h )2

∴ (x2 + y2 − 6) + λ (−6x + 14) = 0 Also, passing through (1, 1). ⇒

− 4 + λ (8) = 0 1 λ= ⇒ 2 ∴ Required equation of circle is x2 + y2 − 6 − 3x + 7 = 0 or x2 + y 2 − 3 x + 1 = 0

13. It is given that, C1has centre (0, 0) and radius 1. Similarly, C 2 has centre (0, 0) and radius 2 and C k has centre (0, 0) and radius k. Now, particle starts it motion from (1, 0) and moves 1 radian on first circle then particle shifts from C1 to C 2. After that, particle moves 1 radian on C 2 and then particle shifts from C 2 to C3 . Similarly, particle move on n circles. Now, n ≥ 2π because particle crosses the X-axis for the first time on C n, then n is least positive integer. Therefore, n = 7.

Circle 431 14. Equation of any circle passing through the point of intersection of x2 + y2 − 2x = 0 and y = x is (x2 + y2 − 2x) + λ ( y − x) = 0 ⇒

x2 + y2 − (2 + λ )x + λy = 0 2 + λ − λ Its centre is  , .  2 2 

∴ ⇒

⇒ For AB to be the diameter of the required circle the centre must be on AB, i.e. 2 + λ = − λ ⇒ λ = −1 [Q y = x] Therefore, equation of the required circle is x2 + y2 − (2 − 1) x − 1 ⋅ y = 0 ⇒

x2 + y 2 − x − y = 0

15. Given,

Therefore,

and let C 2 : (x − h ) + ( y − k)2 = 25

∴

∴ Equation of common chords is S1 − S 2 = 0 ∴

2hx + 2ky = (h 2 + k2 − 9) h 3 ∴ Its slope = − = k 4

x1 =

r r and 2k = cos θ sin θ 25 2h

and

y1 =

4 2k

x2 y2 + = 1, we get 25 4 1  625 1  4  25 1 + 2 =1  +   =1 ⇒  2 25  4h 2  4  k2 4h k

[given]

or

25k2 + 4h 2 = 4h 2 k2

Therefore, required locus is 4x2 + 25 y2 = 4x2y2

18. The equation of the circle on the line joining the points

Also, the length of the chord is 2 r − p =2 4 − p 2

A(3, 7) and B (6, 5) as diameter is

2

(x − 3) (x − 6) + ( y − 7) ( y − 5) = 0

The chord will be of maximum length, if φ = 0 or 16 2 h =9 h 2 + k2 − 9 = 0 ⇒ h 2 + 9 9 ⇒ h=± 5 12 ∴ k= m 5 12 9  9 12 Hence, centres are  , −  and  − ,  5  5 5 5

and the equation of the line joining the points A (3, 7) 7 −5 (x − 3) 3 −6

⇒

S + λP = 0 ⇒ x2 − 6x − 3x + 18 + y2 − 5 y − 7 y + 35 + 2λx + 3λy − 27λ = 0 ⇒ S1 ≡ x2 + y2 + x (2λ − 9) + y (3λ − 12) + (53 − 27λ ) = 0 …(iii)

 3 y + 10  3 y + 10 2  + 4 y − 20 = 0  + y −2     4  4  ⇒ 25 y2 + 100 y − 300 = 0 ⇒ y2 + 4 y − 12 = 0 ⇒ ( y − 2) ( y + 6) = 0 ⇒ y = − 6, 2 When y = −6 ⇒ x = − 2 When y=2 ⇒ x=4 ∴ Point of intersection are (−2, − 6) and (4, 2) . ⇒

Again, the circle,which cuts the members of family of circles, is S 2 ≡ x 2 + y2 − 4 x − 6 y − 3 = 0

…(iv)

and the equation of common chord to circles S1 and S 2 is S1 − S 2 = 0 ⇒ x (2λ − 9 + 4) + y (3λ − 12 + 6) + (53 − 27λ + 3) = 0

17. Equation of any tangent to circle x2 + y2 = r 2 is

x2 y2 + = 1 at (x1 , y1 ) 25 4

…(ii)

⇒ (x − 3)(x − 6) + ( y − 7)( y − 5) + λ (2x + 3 y − 27) = 0

2

Suppose Eq. (i) is tangent to 4x2 + 25 y2 = 100

2x + 3 y − 27 = 0

Now, the equation of family of circles passing through the point of intersection of Eqs. (i) and (ii) is

3 y + 10 in x2 + y2 − 2x + 4 y − 20 = 0 x= 4

x cos θ + y sin θ = r

…(i)

and B (6, 5) is y − 7 =

16. For point of intersection, we put

or

25 cos θ 4 sin θ , y1 = r r

As (x1 , y1 ) lies on the ellipse

If p be the length of perpendicular on it from the centre h 2 + k2 − 9 . (0, 0) of C1 of radius 4, then p = 4h 2 + 4k2 2

2h =

2

2

2

x1 =

The line (i) meet the coordinates axes in A (r sec θ , 0) and β (0, r cosec θ ). Let (h , k) be mid-point of AB. r sec θ r cosecθ and k = Then, h= 2 2

C1: x + y = 16 2

xx1 yy + 1 = 1 are identical 25 4 y1 x1 / 25 1 = 4 = cos θ sin θ r

Then, Eq. (i) and

… (i)

⇒

2λx − 5x + 3λy − 6 y + 56 − 27λ = 0

⇒

(− 5x − 6 y + 56) + λ (2x + 3 y − 27) = 0

which represents equations of two straight lines passing through the fixed point whose coordinates are obtained by solving the two equations 5x + 6 y − 56 = 0 and

2x + 3 y − 27 = 0,

we get x = 2 and y = 23 /3

432 Circle 20. Let P (h , k) be the foot of perpendicular drawn from

19. The parametric form of OP is x−0 y−0 = cos 45° sin 45°

origin O(0, 0) on the chord AB of the given circle such that the chord AB subtends a right angle at the origin.

Since, OP = 4 2 So, the coordinates of P are given by x−0 y−0 = −4 2 = cos 45° sin 45°

x2 + y2 + 2g x + 2fy + c = 0 B

So, P(−4, − 4) Let C (h , k) be the centre of circle and r be its radius. Now, CP ⊥ OP

P (h, k) A

Y O

y=x

The equation of chord AB is h y − k = − (x − h ) ⇒ hx + ky = h 2 + k2 k The combined equation of OA and OB is homogeneous equation of second degree obtained by the help of the given circle and the chord AB and is given by, 2  hx + ky  hx + ky c =0 x2 + y2 + (2 gx + 2 fy)  2 +     h 2 + k2   h + k2 

B (x2, y2) M 3 3√2 A(x ,y ) 1 1 (h,k)C X P O

4 ,−

4)

4√2

(−

y = −x

Since, the lines OA and OB are at right angle. ⇒

k+4 ⋅ (1) = − 1 h+4

⇒

k + 4 = − h −4

⇒

h + k = −8

Also, ⇒

…(i)

CP = (h + 4) + (k + 4) 2

2

2

(h + 4)2 + (k + 4)2 = r 2

…(ii)

In ∆ACM , we have  h + k AC 2 = (3 2 )2 +    2  ⇒ ⇒ Also, ⇒ ⇒

2

∴ Required equation of locus is x2 + y2 + gx + fy +

r 2 = 18 + 32 r =5 2

…(iii)

CP = r h − k    =r  2 

21. Let the equation of L1 be x cos α + y sin α = p1. Y L2

h − k = ± 10

Q

…(iv)

S X′

Thus, the equation of the circles are (x + 9) + ( y − 1) = (5 2 ) 2

c =0 2

Then, any line perpendicular to L1 is

From Eqs. (i) and (iv), we get (h = − 9, k = 1) or (h = 1, k = − 9)

or

∴Coefficient of x2 + Coefficient of y2 = 0   ch 2 2 gh ⇒ 1 + 2 + 2 2 2 2 h +k (h + k )     2f k ck2 =0 + 1 + 2 + 2 2 2 2 h +k (h + k )   ⇒ 2 (h 2 + k2) + 2 ( g h + f k) + c = 0 c ⇒ h 2 + k2 + gh + fk + = 0 2

2

2

(x − 1)2 + ( y + 9)2 = (5 2 )2

⇒

x2 + y2 + 18x − 2 y + 32 = 0

or

x2 + y2 − 2x + 18 y + 32 = 0

Clearly, (−10, 2) lies interior of x2 + y2 + 18x − 2 y + 32 = 0 Hence, the required equation of circle, is x2 + y2 + 18x − 2 y + 32 = 0

O

R Y′

X

P L1

x sin α − y cos α = p2 where, p2 is a variable. Then, L1 meets X-axis at P ( p1 sec α ,0) and Y-axis at Q (0, p1cosec α). Similarly, L 2 meets X-axis at R ( p2 cosec α , 0) and Y-axis at S (0, − p2 sec α ). Now, equation of PS is, x y x y + =1 ⇒ − = secα p1 sec α − p2 sec α p1 p2

…(i)

Circle 433 Hence, the required equations of the circles are

Similarly, equation of QR is x y x y + =1 ⇒ + = cosec α p2cosec α p1cosec α p2 p1

x2 + y2 + 2 (10 ± 3 6 ) x + (55 ± 24 6 ) = 0

…(ii)

23. Given lines are

Locus of point of intersection of PS and QR can be obtained by eliminating the variable p2 from Eqs. (i) and (ii).  x  x y ∴ = cosec α − secα  +   y p1  p1 ⇒

and ∴

(x − p1 secα ) x + y = p1 y cosec α

⇒ x + y − p1x sec α − p1 y cosec α = 0 which is a circle through origin. 2

22. Let the equation of the required circle be x2 + y2 + 2 gx + 2 fy + c = 0

…(i)

It passes through (−4, 3). ∴

25 − 8 g + 6 f + c = 0

Since, circle touches the line x + y − 2 = 0 and x − y − 2 = 0.  −g − f − 2    −g + f − 2  = = g2 + f 2 − c ∴  2 2    

…(ii)

⇒

− g − f − 2 = ± (− g + f − 2)

⇒

− g − f −2 = − g + f −2

or

− g − f −2 = g − f + 2

⇒

⇒

g −c

( g + 2)2 = 2 ( g 2 − c) g − 4 g − 4 − 2c = 0

…(iv)

On putting f = 0 in Eq. (ii). we get 25 − 8 g + c = 0

…(v)

Eliminating c between Eqs. (iv) and (v), we get g 2 − 20 g + 46 = 0 g = 10 ± 3 6 and c = 55 ± 24 6 ⇒ On substituting the values of g , f and c in Eq. (i), we get x2 + y2 + 2 (10 ± 3 6 ) x + (55 ± 24 6 ) = 0

Thus, there is no circle in this case.

Key Idea Firstly find the centre of the given circle and write the coordinates of mid point ( A ), of line segment PQ, since AC ⊥ PQ therefore use (slope of AC) × (slope of PQ) = − 1

It is given that points P and Q are intersecting points of circle …(i) (x − 3)2 + ( y + 2 )2 = 25 Line …(ii) y = mx + 1 −3 so And, the mid-point of PQ is A having x-coordinate 5 3 y-coordinate is 1 − m. 5 3   3 So, A  − , 1 − m  5 5  From the figure, Q AC ⊥ PQ

Case II When g = − 2

This equation gives imaginary values of f.

25(x2 + y2) − 20x + 2 y − 60 = 0

Topic 5 Equation of Chord Bisected at a Point, Product of Pair of Tangents, Chord of Contact of Tangent, Pole and Equations of Polar 1.

2

2

2 1 1601   ∴ Equation of the circle is  x −  +  y +  =   5 25 625 4x 2 y 4 1 1601 ⇒ x2 + y 2 − + + + − =0 5 25 25 625 625 ⇒

2

From Eq. (iii), we get ⇒ f 2 = 2 (4 + f 2 − c) ⇒ f 2 − 2c + 8 = 0 On putting g = − 2 in Eq. (ii), we get c = − 6 f − 41 On substituting c in Eq. (vi), we get f 2 + 12 f + 90 = 0

2

2

Case I When f = 0

⇒

2

…(iii)

f = 0 or g = − 2

From Eq. (iii), we get  − g − 2   = 2  

x=

2 1 64 1 1601   ∴ Radius = 2 −  + 0 +  = + =   5 25 25 625 25

 −g − f − 2  =  −g + f − 2   2 2    

Now,

...(i) ...(ii)

1−c 5(c2 − 1) and y = 2 2 15c − 5c − 10 5(3c − c − 2) 2c lim x = lim ⇒ c→1 c→1 6 c − 1 −1 2 1 and lim y = lim ⇒ lim x = and lim y = − c→1 c→1 30 c − 5 c→1 c→1 5 25 1 2 ∴ Centre =  lim x, lim y =  , −   5 25 c→1 c→1 ⇒

2

2

3x + 5 y − 1 = 0 (2 + c)x + 5c2y − 1 = 0 x y 1 = = 2 2 ( ) 2 c 3 − + + 15c − 5c − 10 −5 + 5 c

Y

…(vi)

Q A

P

O

C (3,-2)

X

434 Circle ⇒ (slope of AC) × (slope of PQ) = −1 3    −2 − 1 + m  5  × m = −1 ⇒  3   3+   5 (3 /5 ) m − 3  3m − 15 m = −1 ⇒  ⇒  m = −1  18  18 / 5

1 C1C 2 × AM 2 1 AB = × 13 × 2 2

Also, area of ∆C1 AC 2 =

1 × 13 × AM = 30 cm 4 120 AM = cm 13

∴

⇒ 3m2 − 15m + 18 = 0 ⇒ m2 − 5m + 6 = 0 ⇒ m = 2 or 3

AB   Q AM = 2 

3. Given equation of line is x + y = n,n ∈ N

…(i) and equation of circle is x2 + y2 = 16 …(ii) Now, for intercept, made by circle (ii) with line (i)

2. Let the length of common chord = AB = 2 AM = 2x A 12 C1

M

12

x2+y2=16

5

(0, 0)

C2

4

5

B

Now, C1C 2 =

AC12

+

A

AC 22

… (i)

[Qcircles intersect each other at 90º] ⇒ C1C 2 = 122 − AM 2 + 52 − AM 2 From Eqs. (i) and (ii), we get

144 + 25 = 144 − x2 + 25 − x2

⇒

13 = 144 − x + 25 − x 2

2

⇒ n 0 6

π . 6

17. Given, y = mx − b 1 + m2 touches both the circles, so distance from centre = radius of both the circles. |ma − 0 − b 1 + m2 | m2 + 1

Y'

 12 9 On solving Eqs. (i) and (ii), we get M  − , .  5 5 Now, area of ∆ AOM =

1 27 sq units ⋅ OA × MN = 2 10

15. Let the point P ( 2 cos θ, sin θ) on

2

2

x y + = 1. 2 1

Y B(θ, cosec θ)

X′

⇒

P(√2 cos θ, sin θ) X

A (√2 sec θ, 0)

⇒ ⇒ ∴

m2 + 1

=b

|ma − b 1 + m2 | =|− b 1 + m2 |

ma − 2b 1 + m2 = 0 m2a 2 = 4b2 (1 + m2) 2b m= 2 a − 4b2

18. For ellipse, condition of tangency is c2 = a 2m2 + b2 Given line is y = 4x + c and curve ⇒ ⇒

Y′

|− b 1 + m2|

⇒ m2a 2 − 2abm 1 + m2 + b2 (1 + m2) = b2 (1 + m2)

(mid-point of AB) M

= b and

x2 + y2 = 1 4

c2 = 4 × 42 + 1 = 65 c = ± 65

So, there are two different values of c.

Ellipse 471 x2 y2 + = 1, (a > b) a 2 b2 x2 y2 E 2 : 2 + 2 = 1, (c < d ) and S : x2 + ( y − 1)2 = 2 c d as tangent to E1 , E 2 and S is x + y = 3.

20. Let the common tangent to x2 + y2 = 16 and

19. Here, E1:

and

)2 =

2

Y

–1

y = mx + 4 1 + m2

…(i)

y = mx + 25m + 4

…(ii)

2

Since, Eqs. (i) and (ii) are same tangent. R

∴ 4 1 + m2 = 25m2 + 4 ⇒ 16 (1 + m2) = 25 m2 + 4

(y

P

⇒

(0, 1)

S:

x2 +

be

O

m < 0 ⇒ m = − 2/ 3

∴

X E1

9m2 = 12 ⇒ m = ± 2 / 3

Since, tangent is in Ist quadrant.

Q X′

x+y=3

So, the equation of the common tangent is y=−

E2

Let the point of contact of tangent be (x1 , y1 ) to S. ∴ x ⋅ x1 + y ⋅ y1 − ( y + y1 ) + 1 = 2 or x x1 + y y1 − y = (1 + y1 ), same as x + y = 3. x1 y1 − 1 1 + y1 = = ⇒ 1 1 3 i.e. x1 = 1 and y1 = 2 ⇒ P = (1, 2)

∴

2 2 . Thus, by parametric form, 3 x −1 y −2 2 2 = =± 3 −1 / 2 1 / 2 5 4 1 8     x = , y =  and  x = , y =     3 3 3 3  5 4  1 8 Q =  ,  and R =  ,   3 3  3 3

21.

 7 AB = (2 7 − 0) + 0 − 4  3  

at

A(2 7 , 0) and

2

11 196 14 3 14 3 = = × = 3 3 3 3 3

Any point on the ellipse x2 y2 + 2 = 1 be P (a cos θ , b sin θ ) 2 a b The equation of tangent at point P is given by x cos θ y sin θ + =1 a b The equation of line perpendicular to tangent is x sin θ y cos θ − =λ b a

Now, equation of tangent at Q on ellipse E1 is x⋅5 y⋅4 =1 + a 2 ⋅ 3 b2 ⋅ 3 On comparing with x + y = 3, we get 4 1 b2 a 2 = 5 and b2 = 4 ⇒ e12 = 1 − 2 = 1 − = 5 5 a Also, equation of tangent at R on ellipse E 2 is x⋅1 y⋅8 + 2 =1 2 a ⋅3 b ⋅3 On comparing with x + y = 3, we get 1 7 a2 a 2 = 1, b2 = 8 ⇒ e22 = 1 − 2 = 1 − = 8 8 b 7 7 2 2 Now, e1 ⋅ e2 = ⇒ e1 e2 = 40 2 10 1 7 43 and e12 + e22 = + = 5 8 40 1 7 27 Also, e12 − e22 = − = 5 8 40

axes

2

= 28 +

Since, PR = PQ =

∴

2x 7 +4 3 3

which meets coordinate  7 . 0, 4 3 

Y′

⇒

x2 y2 + =1 25 4

…(i)

…(ii)

Since, it passes through the focus (ae, 0), then ae sin θ −0 = λ b ae sin θ ⇒ λ= b x sin θ y cos θ ae sin θ − = ∴ Equation is b a b

…(i)

Equation of line joining centre and point of contact P (a cos θ , b sin θ ) is b …(ii) y = (tan θ ) x a Point of intersection Q of Eqs. (i) and (ii) has x a coordinate, . Hence, Q lies on the corresponding e a directrix x = . e

472 Ellipse 22. Let the coordinates of A ≡ (a cos θ , b sin θ), so that the coordinates of B = { a cos (θ + 2π / 3), a sin (θ + 2π / 3)} and C = { a cos (θ + 4π / 3), a sin (θ + 4π / 3)} According to the given condition, coordinates of P are (a cos θ b sin θ ) and that of Q are { a cos (θ + 2π / 3), b sin (θ + 2π / 3)} and that of R are a cos (θ + 4π / 3), b sin (θ + 4π / 3) Y

Equation of second ellipse is Y

A (a cos θ, b sin θ)

B X'

x2 y2 …(i) + =1 4 1 Equation of any tangent to the ellipse on (i) can be written as x …(ii) cos θ + y sin θ = 1 2

23. Given, x2 + 4 y2 = 4 or

A

P

Q R C

P X′ −√6

Y'

[Q it is given that P, Q, R are on the same side of X-axis as A, B and C] Equation of the normal to the ellipse at P is ax by − = a 2 − b2 cos θ sin θ 1 or ax sin θ − by cos θ = (a 2 − b2) sin 2θ …(i) 2 Equation of normal to the ellipse at Q is 2π  2π    ax sin θ +   − by cos θ +    3 3 1 4π   …(ii) = (a 2 − b2)sin 2θ +   2 3 Equation of normal to the ellipse at R is a x sin (θ + 4π / 3) − by cos (θ + 4π / 3) 1 …(iii) = (a 2 − b2) sin (2θ + 8π / 3) 2 But sin (θ + 4π / 3) = sin (2π + θ − 2π / 3) = sin (θ − 2π / 3) and cos (θ + 4π / 3) = cos (2π + θ − 2π / 3) = cos (θ − 2π / 3) and sin (2θ + 8π / 3) = sin (4π + 2θ − 4π / 3 ) = sin (2θ − 4π / 3) Now, Eq. (iii) can be written as ax sin (θ − 2π / 3) − by cos (θ − 2π / 3) 1 …(iv) = (a 2 − b2) sin (2θ − 4π / 3) 2 For the lines (i), (ii) and (iv) to be concurrent, we must have the determinant a sin θ − b cos θ 2π  2π    ∆1 = a sin θ +  − b cos θ +     3 3 2π  2π    a sin θ −  − b cos θ −     3 3 1 2 (a − b2) sin 2 θ 2 1 2 (a − b2) sin (2 θ + 4π / 3) = 0 2 1 2 (a − b2) sin (2 θ − 4π / 3) 2 Thus, lines (i), (ii) and (iv) are concurrent.

Q

√3

X

O

1 −2

2

O

√6

X

−1 −√3 Y′

x2 + 2 y 2 = 6 ⇒

x2 y2 + =1 6 3

…(iii)

Suppose the tangents at P and Q meets at A (h , k). Equation of the chord of contact of the tangents through A (h , k) is hx ky …(iv) + =1 6 3 But Eqs. (iv) and (ii) represent the same straight line, so comparing Eqs. (iv) and (ii), we get h /6 k /3 1 = = cos θ / 2 sin θ 1 ⇒

h = 3 cos θ and k = 3 sin θ

Therefore, coordinates of A are (3 cos θ ,3 sin θ ). Now, the joint equation of the tangents at A is given by T 2 = SS1, 2    h 2 k2  x2 y2  hx ky  …(v) i.e.  + − 1 + − 1 =  + − 1   6  3 6 3 6 3    In Eq. (v), coefficient of x2 =

 h 2 1  h 2 k2 −  + − 1 36 6  6 3 

h 2 h 2 k2 1 1 k2 − − + = − 36 36 18 6 6 18  k2 1  h 2 k2 and coefficient of y2 = −  + − 1 9 3 6 3  =

=

k2 h 2 k2 1 h2 1 − − + =− + 9 18 9 3 18 3

Again, coefficient of x2 + coefficient of y2 1 1 1 =− (h 2 + k2) + + 18 6 3 1 1 =− (9 cos 2 θ + 9 sin 2 θ ) + 18 2 9 1 =− + =0 18 2

Ellipse 473 which shows that two lines represent by Eq. (v) are at right angles to each other.

24. Let the coordinates of point P be (a cos θ , b sin θ ). Then, equation of tangent at P is x y …(i) cos θ + sin θ = 1 a b We have, d = length of perpendicular from O to the tangent at P Y

∴

…(i) ⇒ m12 = 1 / 2 Also, tangent to P2 passes through (f1, 0) i.e. (2, 0). (−4) 4 ⇒ ⇒ 0 = 2m2 − T2 : y = m2x + m2 m2 ⇒

m22 = 2

∴

1 + m22 = 2 + 2 = 4 m12

P (a cosθ, b sin θ)

F1(−ae,0) O

X

F2(ae,0)

…(ii)

Topic 3 Equation of Chord of Contact, Chord Bisected at a Given Point and Diameter

d X'

2 2 ⇒ 0 = −4m1 + m1 m1

T1 : y = m1x +

1 3

1. Equation of AB is y − 0 = − (x − 3) Y'

cos 2 θ sin 2 θ 1 = + d= ⇒ 2 2 d a2 b2 cos θ sin θ + a2 b2 1 cos 2 θ sin 2 θ ⇒ = + 2 d a2 b2  b2  We have to prove (PF1 − PF2)2 = 4a 2 1 − 2 d   |0 + 0 − 1|

Now,

x + 3y −3 = 0 ⇒ |x + 3 y − 3|2 = 10 [(x − 3)2 + ( y − 4)2] On solving, we are getting 9x2 + y2 − 6xy − 54x − 62 y + 241 = 0

2. Equation of AB is P (3, 4)

(− 95 , 85(

 4a 2b2 b2  RHS = 4a 1 − 2 = 4a 2 − d  d2  2

 cos 2 θ sin 2 θ   = 4a 2 − 4a 2b2  + b2   a2 = 4a 2 − 4b2 cos 2 θ − 4a 2 sin 2 θ = 4a 2(1 − sin 2 θ ) − 4b2 cos 2 θ = 4a 2 cos 2 θ − 4b2 cos 2 θ 2 2 = 4 cos θ (a − b2) = 4 cos 2 θ ⋅ a 2e2 Q e = 1 − (b / a )2    Again, PF1 = e|a cos θ + a / e|= a | e cos θ + 1| = a (e cos θ + 1) [Q − 1 ≤ cos θ ≤ 1 and 0 < e < 1] Similarly, PF2 = a (1 − e cos θ ) Therefore, LHS = (PF1 − PF2)2 = [a (e cos θ + 1) − a (1 − e cos θ )]2 = (ae cos θ + a − a + ae cos θ )2 = (2ae cos θ )2 = 4a 2e2 cos 2 θ Hence, LHS = RHS

25.

Y

x + 3y = 3

...(i)

Equation of the straight line perpendicular to AB through P is 3x − y = 5 Equation of PA is 2x − 3 = 0 The equation of straight line perpendicular to PA 8  − 9 8 through B  ,  is y = .  5 5 5  11 Hence, the orthocentre is  , 5

8 . 5

F2(–2, 0)

P D

X

O F1(2, 0)

Y′

–9 , 8 5 5

y = 8x

Tangent to P1 passes through (2 f2, 0) i. e. (−4, 0).

(3, 4)

B F

2

y2 = –16x

⇒

T1

(– 4, 0) F2

8 8 1 y − 0 = 5 (x − 3) = (x − 3) ⇒ y = − (x − 3) 9 − 24 3 − −3 5

3. Figure is self-explanatory.

T2

X′

A (3, 0)

B

A (3, 0)

19 Hyperbola Topic 1 Equation of Hyperbola and Focal Chord Objective Questions I (Only one correct option) 1. A hyperbola having the transverse axis of length 2 has the same foci as that of the ellipse 3x2 + 4 y2 = 12, then this hyperbola does not pass through which of the following points? (2020 Main, 3 Sep I) 1  (a)  , 0  2 

 3  (b)  − ,1  2 

(c)  1, − 

 3 1  (d)  ,   2 2

1   2

to the parabola y2 = 12x and the hyperbola 8x2 − y2 = 8. If S and S′ denotes the foci of the hyperbola where S lies on the positive X-axis then P divides SS′ in a ratio (2019 Main, 12 April I)

(b) 14 : 13 (d) 2 : 1

16x − 9 y = 144, then its corresponding focus is 5 (a)  − , 0 (b) (− 5, 0)  3 

5 (c)  , 0 3 

(d) (5, 0)

4. If a directrix of a hyperbola centred at the origin and passing through the point (4, − 2 3 ) is 5x = 4 5 and its (2019 Main, 10 April I) eccentricity is e, then (a) 4e4 − 12e2 − 27 = 0 (c) 4e4 + 8e2 − 35 = 0

(b) 4e4 − 24e2 + 27 = 0 (d) 4e4 − 24e2 + 35 = 0

5. If the vertices of a hyperbola be at (−2, 0) and (2, 0) and one of its foci be at (−3, 0), then which one of the following points does not lie on this hyperbola? (2019 Main, 12 Jan I)

(a) (2 6 , 5) (c) (4, 15 )

(b) (6, 5 2 ) (d) (− 6, 2 10 )

6. If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is (2019 Main, 11 Jan II) 13 12 13 (c) 8

(b) 2 (d)

13 6

(2019 Main, 10 Jan II)

2 , when (a) a hyperbola whose eccentricity is 1−r 0 < r < 1. 2 (b) a hyperbola whose eccentricity is , when r +1 0 < r < 1.

(d) an ellipse whose eccentricity is

2 , when r > 1. r+1 1 , when r > 1. r+1

8. A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the X-axis. Then the eccentricity of the hyperbola is (a) 2

2

(2019 Main, 10 April II)

(a)

where r ≠ ± 1. Then, S represents

(2019 Main, 9 Jan II)

3. If 5x + 9 = 0 is the directrix of the hyperbola 2



 y2 x2 − = 1, 1+ r 1−r 

(c) an ellipse whose eccentricity is

2. Let P be the point of intersection of the common tangents

(a) 13 : 11 (c) 5 : 4



7. Let S = (x, y) ∈ R2 :

2 (b) 3

3 (c) 2

(d) 3

π 2

9. Let 0 < θ < . If the eccentricity of the hyperbola x2 y2 − = 1 is greater than 2, then the length of its 2 cos θ sin 2 θ (2019 Main, 9 Jan I) latus rectum lies in the interval 3 (a)  1,  2 

(b) (3,∞)

3  (c)  , 2  2 

(d) (2, 3]

10. The eccentricity of the hyperbola whose length of the latusrectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is (2017 Main)

4 (a) 3

4 (b) 3

2 (c) 3

(d) 3

11. Consider a branch of the hyperbola x2 − 2 y2 − 2 2x − 4 2 y − 6 = 0 with vertex at the point A. Let B be one of the end points of its latusrectum. If C is the focus of the hyperbola nearest to the point A, then the area of the ∆ ABC is (2008, 3M)

(a) 1 − 2 / 3 sq unit (c) 1 + 2 /3 sq unit

(b) 3 / 2 − 1 sq unit (d) 3 /2 + 1 sq unit

Hyperbola 475 12. A hyperbola, having the transverse axis of length 2 sin θ ,

17. An ellipse intersects the hyperbola 2x2 − 2 y2 = 1

is confocal with the ellipse 3x + 4 y = 12 . Then, its (2007, 3M) equation is

orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are (2009) along the coordinate axes, then

2

2

(a) x2cosec2 θ − y2sec2 θ = 1 (b) x2 sec2 θ − y2cosec2 θ = 1 (c) x2 sin 2θ − y2cos2θ = 1 (d) x2cos2 θ − y2sin 2 θ = 1

x2 y2 + = 1 and e2 is 16 25 the eccentricity of the hyperbola passing through the foci of the ellipse and e1 e2 = 1, then equation of the (2006, 3M) hyperbola is

13. If e1 is the eccentricity of the ellipse

x2 y2 − =1 9 16 x2 y2 (c) − =1 9 25

(b)

(a)

x2 y2 − = −1 16 9

(d) None of these

15. The equation

equation of ellipse is x2 + 2 y2 = 2 the foci of ellipse are (±1, 0) equation of ellipse is x2 + 2 y2 = 4 the foci of ellipse are (± 2 , 0)

Analytical & Descriptive Question 18. A variable straight line of slope 4 intersects the

hyperbola xy = 1 at two points. Find the locus of the point which divides the line segment between these two (1997, 5M) points in the ratio 1 : 2.

Match the List

x2 y2 14. For hyperbola − = 1, which of the 2 cos α sin 2 α following remains constant with change in ‘ α ’ ? (2003, 1M) (a) Abscissae of vertices (c) Eccentricity

(a) (b) (c) (d)

(b) Abscissae of foci (d) Directrix

x2 y2 − = 1, where a > b > 0, be a hyperbola in the a 2 b2 XY -plane whose conjugate axis LM subtends an angle of 60° at one of its vertices N . Let the area of the ∆ LMN be 4 3.

19. Let H :

x2 y2 − = 1,|r| < 1 represents 1− r 1 + r (1981, 2M)

(a) an ellipse (c) a circle

(b) a hyperbola (d) None of these

Objective Questions II (One or more than one correct option) x2 y2 16. Let the eccentricity of the hyperbola 2 − 2 = 1 be a b reciprocal to that of the ellipse x2 + 4 y2 = 4. If the hyperbola passes through a focus of the ellipse, then (a) the equation of the hyperbola is

x2 y2 − =1 3 2

(2011)

(b) a focus of the hyperbola is (2, 0) 5 3 (d) the equation of the hyperbola is x2 − 3 y2 = 3

(c) the eccentricity of the hyperbola is

List-I

List-II

P.

The length of the conjugate axis of H is

1. 8

Q.

The eccentricity of H is

2.

4 3

R.

The distance between the 3. foci of H is

2 3

S.

The length of the latus 4. 4 rectum of H is

The correct option is

(2018 Adv.)

(a) P → 4; Q → 2; R → 1; S → 3 (b) P → 4; Q → 3; R → 1; S → 2 (c) P → 4; Q → 1; R → 3; S → 2 (d) P → 3; Q → 4; R → 2; S → 1

Topic 2 Equation of Tangent and Normal Objective Questions I (Only one correct option) 1. If a hyperbola passes through the point P(10, 16) and it has vertices at (± 6, 0), then the equation of the normal to it at P is (2020 Main, 8 Jan II) (a) 3x + 4 y = 94 (c) 2x + 5 y = 100

(b) x + 2 y = 42 (d) x + 3 y = 58

2. The equation of a common tangent to the curves, y2 = 16x and xy = − 4, is

(a) x − y + 4 = 0 (c) x − 2 y + 16 = 0

(2019 Main, 12 April II)

(b) x + y + 4 = 0 (d) 2x − y + 2 = 0

3. If the line y = mx + 7 3 is normal to the hyperbola x2 y2 − = 1, then a value of m is 24 18 (a)

3 5

(b)

15 2

(c)

2 5

(2019 Main, 9 April I)

(d)

5 2

4. If the eccentricity of the standard hyperbola passing through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is (2019 Main, 8 April II) (a) 3x − 2 y = 0

(b) x − 2 y + 8 = 0

(c) 2x − y − 2 = 0

(d) 2x − 3 y + 10 = 0

476 Hyperbola 5. The equation of a tangent to the hyperbola 4x − 5 y = 20 parallel to the line x − y = 2 is 2

2

(2019 Main, 10 Jan I)

(a) x − y − 3 = 0 (c) x − y + 1 = 0

(b) x − y + 9 = 0 (d) x − y + 7 = 0

(a) a, 4, 1

6. Tangents are drawn to the hyperbola 4x2 − y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3), then the area (in sq units) of ∆PTQ is (a) 45 5 (c) 60 3

(b) 54 3 (d) 36 5

(2018 Main)

7. If a hyperbola passes through the point P( 2 , 3 ) and

has foci at (± 2, 0), then the tangent to this hyperbola at (2017 Main) P also passes through the point

(a) (3 2 , 2 3 ) (c) ( 3 , 2 )

(b) (2 2 , 3 3 ) (d) (− 2 , − 3 )

x2 y2 − = 1. If the a 2 b2 normal at the point P intersects the X-axis at (9, 0), then the eccentricity of the hyperbola is (2011)

8. Let P(6, 3) be a point on the hyperbola

(a)

5 2

(b)

3 2

(c)

2

(d)

3

x2 − 2 y2 = 4, then the point of contact is

(2004, 1M)

(d) (4, −

6)

10. Let P (a sec θ , b tan θ ) and Q (a sec φ , b tan φ ), where θ+φ=

x y π , be two points on the hyperbola 2 − 2 = 1. 2 a b 2

a 2 + b2 a

 a 2 + b2  a 2 + b2  (c) (b) −  b a  

 a 2 + b2   (d) −  b  

Objective Questions II (One or more than one correct option) 11. Let a and b be positive real numbers such that a > 1 and b < a. Let P be a point in the first quadrant that lies on x2 y 2 the hyperbola 2 − 2 = 1. Suppose the tangent to the a b hyperbola at P passes through the point (1, 0), and suppose the normal to the hyperbola at P cuts off equal intercepts on the coordinate axes. Let ∆ denote the area of the triangle formed by the tangent at P, the normal at P and the X-axis. If e denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE? (2020 Adv.) (a) 1 < e < 2 (c) ∆ = a 4

(b) 2 < e < 2 (d) ∆ = b4

(c) a, 4, 2

(d) 2a, 8, 1

centre N (x2 , 0). Suppose that H and S touch each other at a point P(x1 , y1 ) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the X-axis at point M. If (l , m) is the centroid of ∆PMN , then the correct (2015 Adv.) expression(s) is/are 1 dl = 1 − 2 for x1 > 1 dx1 3x1 x1 dm for x1 > 1 (b) = dx1 3( x2 − 1) 1 1 dl (c) for x1 > 1 = 1+ dx1 3x12 dm 1 (d) = for y1 > 0 dy1 3 (a)

x2 y2 − = 1, 9 4 parallel to the straight line 2x − y = 1. The points of (2012) contacts of the tangents on the hyperbola are

9 1  (a)  ,   2 2 2

9 1  (b)  − ,−   2 2 2

(c) (3 3 , − 2 2)

(d) (− 3 3 , 2 2 )

2

If (h , k) is the point of the intersection of the normals at P and Q, then k is equal to (1999, 2M) (a)

(b) 2a, 4, 1

13. Consider the hyperbola H: x2 − y2 = 1 and a circle S with

14. Tangents are drawn to the hyperbola

9. If the line 2x + 6 y = 2 touches the hyperbola 1 1  (a) ( − 2, 6) (b) ( − 5, 2 6 ) (c)  ,   2 6

x 2 y2 − =1 a 2 16 then which of the following CANNOT be sides of a right (2017 Adv.) angled triangle?

12. If 2x − y + 1 = 0 is a tangent to the hyperbola

Passage Based Problems The circle x 2 + y 2 − 8x = 0 and hyperbola

x2 y2 − =1 9 4

intersect at the points A and B.

(2010)

15. Equation of the circle with AB as its diameter is (a) (b) (c) (d)

x2 + y2 − 12 x + 24 = 0 x2 + y2 + 12 x + 24 = 0 x2 + y2 + 24 x − 12 = 0 x2 + y2 − 24 x − 12 = 0

16. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is (a) (b) (c) (d)

2 x − 5 y − 20 = 0 2x − 5y + 4 = 0 3x − 4y + 8 = 0 4x − 3y + 4 = 0

Integer & Numerical Answer Type Question 17. The line 2x + y = 1

is tangent to the hyperbola x2 y 2 − = 1. If this line passes through the point of a 2 b2 intersection of the nearest directrix and the X-axis, (2010) then the eccentricity of the hyperbola is……

Hyperbola 477

Topic 3 Equation of Chord of Contact, Chord Bisected Diameter, Asymptote and Rectangular Hyperbola Objective Question I (Only one correct option) 1. If x = 9

is the chord of contact of the hyperbola x − y = 9, then the equation of the corresponding pair (1999, 2M) of tangents is 2

2

(a) 9 x2 − 8 y2 + 18 x − 9 = 0 (b) 9 x2 − 8 y2 − 18 x + 9 = 0 (c) 9 x2 − 8 y2 − 18 x − 9 = 0 (d) 9 x2 − 8 y2 + 18 x + 9 = 0

Analytical & Descriptive Question

Objective Question II (Only one or more than one correct option)

3. Tangents are drawn from any point on the hyperbola

2. If the circle x + y = a intersects the hyperbola xy = c 2

2

(a) x1 + x2 + x3 + x4 = 0 (b) y1 + y2 + y3 + y4 = 0 (c) x1 x2 x3 x4 = c4 (d) y1 y2 y3 y4 = c4

2

2

in four points P (x1 , y1 ), Q (x2 , y2 ), R (x3 , y3 ), S (x4 , y4 ), (1998, 2M) then

x2 y2 − = 1 to the circle x2 + y2 = 9. Find the locus of 9 4 mid-point of the chord of contact. (2005, 4M)

Answers Topic 1 1. 5. 9. 13. 17.

2. 6. 10. 14. 18.

(d) (b) (b) (b) (a, b)

(c) 3. (b) (a) 7. (c) (c) 11. (b) (b) 15. (b) 2 2 16x + y + 10 xy = 2

4. 8. 12. 16. 19.

(d) (b) (a) (b, d) (b)

5. (c)

6. (a)

7. (b)

9. (d)

10. (d)

11. (a,d)

12. (a, c, d)

13. (a, b, d)

14. (a, b)

15. (a)

16. (b)

17. (2)

Topic 3 1. (b)

Topic 2 1. (c)

2. (a)

3. (c)

8. (b)

2. (a,b,c,d)

3.

4. (c)

x 2 y 2 (x 2 + y 2 )2 − = 9 4 81

Hints & Solutions  3 1 So, from the given options point  ,  does not pass  2 2

Topic 1 Equation of Hyperbola and Focal Chord 1. Equation of given ellipse is x2 y2 3x + 4 y = 12 ⇒ + =1 4 3 2

2

… (i)

3 1 ∴Eccentricity of ellipse (i) is e1 = 1 − = 4 2 ∴Coordinate of foci is (± 1, 0). Now, it is given that length of the transverse axis of 2 hyperbola is 2, so e2 = 1, where e2 is the eccentricity 2 of the hyperbola. So, e2 = 2 Given hyperbola is rectangular hyperbola and it’s ∴ equation is x2 y2 − =1 2 2  1  1      2  2 ⇒

x2 − y2 =

1 2

… (ii)

through the hyperbola (ii). Hence, option (d) is correct.

2. Equation of given parabola y2 = 12x

… (i)

and hyperbola 8x − y = 8 2

2

… (ii)

Now, equation of tangent to parabola y = 12x having 3 … (iii) m 2

slope ‘m’ is y = mx +

and equation of tangent to hyperbola x2 y2 − = 1 having slope ‘m’ is 1 8 …(iv) y = mx ± 12m2 − 8 Since, tangents (iii) and (iv) represent the same line 2  3 m2 − 8 =   ∴  m 4 ⇒ m − 8m2 − 9 = 0 2 ⇒ (m − 9) (m2 + 1) = 0 ⇒ m = ± 3.

478 Hyperbola Now, equation of common tangents to the parabola (i) and hyperbola (ii) are y = 3x + 1 and y = − 3x − 1 Q Point ‘P’ is point of intersection of above common tangents, ∴ P(− 1 / 3, 0) and focus of hyperbola S(3, 0) and S′ (− 3, 0). PS 3 + 1 / 3 10 5 Thus, the required ratio = = = = PS′ 3 − 1 / 3 8 4

3. Equation of given hyperbola is 16x2 − 9 y2 = 144 x2 y2 − =1 ⇒ 9 16 So, the eccentricity of Eq. (i) 16 5 e= 1+ = 9 3

…(i)

5. The vertices of hyperbola are given as (± 2, 0) and one of its foci is at (− 3, 0). ∴ (a , 0) = (2, 0) and (− ae, 0), = (− 3, 0) On comparing x-coordinates both sides, we get ⇒ a = 2 and − ae = − 3 3 ⇒ 2e = 3 ⇒ e = 2  2 b2  9 b2 Also, =1 + ⇒ b2 = 5 Q e = 1 + 2  4 4 a   So, equation of the hyperbola is x2 y2 − =1 4 5

The point (6, 5 2 ) from the given options does not satisfy the above equation of hyperbola.

x2 y 2 [Q the eccentricity (e) of the hyperbola 2 − 2 = 1 is a b 1 + (b / a )2 ]

x2 y 2 − = 1, where b2 = a 2 (e2 − 1), the a 2 b2 length of conjugate axis is 2b and distance between the foci is 2ae.

6. We know that in

∴According the problem, 2b = 5 and 2ae = 13 Now, b2 = a 2 (e2 − 1)

and given directrix is 5x + 9 = 0 ⇒ x = − 9 / 5   5  So, corresponding focus is  − 3   , 0 = (− 5, 0)  3  

2

⇒

4. Let the equation of hyperbola is x2 y 2 …(i) − =1 a 2 b2 Since, equation of given directrix is 5x = 4 5 a  a so 5  = 4 5 [Q equation of directrix is x = ]  e e a 4 …(ii) = ⇒ e 5 and hyperbola (i) passes through point (4, − 2 3 ) 16 12 so, …(iii) − =1 a 2 b2 The eccentricity e = 1 + ⇒

2

2

b b ⇒ e2 = 1 + 2 a2 a

a 2e2 − a 2 = b2

From Eqs. (ii) and (iv), we get 16 4 16 2 e − e = b2 5 5 From Eqs. (ii) and (iii), we get 16 5 12 12 − =1 ⇒ 2 − 2 =1 16 2 b2 e b e 5

⇒ ⇒

4(5e2 − e4 − 5 + e2 ) = 15 4e − 24e + 35 = 0 4

2

 5 2 2 2   =a e −a  2 25 (2ae)2 = − a2 4 4 25 169 = − a2 4 4 169 − 25 144 = = 36 a2 = 4 4

⇒ ⇒ ⇒ ⇒ Now, ⇒ ⇒

[Q 2ae = 13]

a =6 2ae = 13 2 × 6 × e = 13 13 e= 12 

7. Given, S = (x, y) ∈ R2: 

…(iv)

 y2 x2 − = 1 1+ r 1−r 

  y2 x2 = (x, y) ∈ R2 : + = 1 1 + − 1 r r  

…(v)

12 5 12 5 − e2 12e2 …(vi) = 2 −1 ⇒ 2 = ⇒ b2 = 2 2 5 − e2 b e b e From Eqs. (v) and (vi), we get  12e2  16e4 − 16e2 = 5  ⇒ 16(e2 − 1)(5 − e2 ) = 60  5 − e2 ⇒

…(i)

For r > 1,

y2 x2 + = 1, represents a vertical ellipse. 1 + r r −1 [Q for r > 1, r − 1 < r + 1 and r − 1 > 0]

Now, eccentricity (e) = 1 −

r −1 r+1

 a2  x2 y2 Q For 2 + 2 = 1, a < b, e = 1 − 2  a b b   =

(r + 1) − (r − 1) r+1

=

2 r+1

Hyperbola 479 8. Equation of hyperbola is given by

11. Given equation can be rewritten as focal chord

x2 y 2 − =1 a 2 b2 Q Length of transverse axis = 2a = 4 ∴ a=2 x2 y2 Thus, − = 1 is the equation of hyperbola 4 b2 Q It passes through (4, 2). 4 2 16 4 4 ∴ − 2 = 1 ⇒ 4 − 2 = 1 ⇒ b2 = ⇒ b = 4 3 3 b b Now, eccentricity, 4 b2 2 1 e= 1+ 2 = 1+ 3 = 1+ = 4 3 3 a

(x − 2 )2 ( y + 2 )2 − =1 4 2 For point A (x, y), e = 1 +

Y Conjugate axis B X′

Now,

X

and greatest value of latusrectum is < ∞ Hence, latusrectum length ∈ (3, ∞ ) 2b2 = 8 and 2b = ae a

AC = 6 + 2 − 2 − 2 = 6 − 2

b2 2 = =1 a 2 1 3 ∴ Area of ∆ ABC = × ( 6 − 2) × 1 = − 1 sq unit 2 2 BC =

and

sin 2 θ >2 cos 2 θ (Q a 2 = cos 2 θ and b2 = sin 2 θ) ⇒ 1 + tan 2 θ > 4 ⇒ tan 2 θ > 3 ⇒tan θ ∈ (− ∞ , − 3 ) ∪ ( 3 , ∞ ) [x2 > 3 ⇒|x| > 3 ⇒ x ∈ (−∞ , − 3 ) ∪ ( 3 , ∞ )]  π But θ ∈ 0,  ⇒ tan θ ∈ ( 3 , ∞ )  2  π π ⇒ θ ∈ ,   3 2 Now, length of latusrectum sin 2 θ 2b2 = 2 sin θ tan θ = =2 a cos θ Since, both sin θ and tan θ are increasing functions in  π π  , .  3 2 ∴ Least value of latusrectum is π π π 3  = 2 sin ⋅ tan = 2 ⋅ ⋅ 3 =3 at θ =   3 3 3 2

12. The given ellipse is x2 y2 + = 1 ⇒ a = 2, b = 3 4 3 1 3 = 4 (1 − e2 ) ⇒ e = 2 1 ae=2 × =1 2

∴ So,

Hence, the eccentricity e1 of the hyperbola is given by e1 = cosec θ

[Q ae = e sin θ ]

⇒ b = sin θ (cosec θ − 1) = cos 2 θ Hence, equation of hyperbola is 2

2

2

x2 y2 − = 1 or x2 cosec2 θ − y2 sec2 θ = 1. 2 sin θ cos 2 θ x2 y2 + = 1 is 16 25 16 3 e1 = 1 − = 25 5 5 e2 = ∴ 3 ⇒ Foci of ellipse (0, ± 3) x2 y 2 − = − 1. ⇒ Equation of hyperbola is 16 9

13. The eccentricity of

14. Given equation of hyperbola is

⇒ b2 = 4a and 2b = ae Consider, 2b = ae

⇒

C

For point C (x, y), x − 2 = ae = 6 ⇒ x = 6 + 2

e= 1+

4b2 = a 2e2 4a 2 (e2 − 1) = a 2e2 4e2 − 4 = e2 3 e2 = 4 2 e= 3

A

transverse axis

Y′

x2 y2 − = 1, a 2 b2 b2 e= 1+ 2 a ∴ For the given hyperbola,

⇒ ⇒ ⇒ ⇒

x− 2 =2 ⇒ x=2 + 2

⇒

9. For the hyperbola

10. We have,

2 3 = 4 2

Here,

x2 y2 − =1 2 cos α sin 2 α

a 2 = cos 2 α and b2 = sin 2 α

[i.e. comparing with standard equation [Q a ≠ 0] [Q e > 0]

We know that, foci = (± ae, 0) where, ae = a 2 + b2 = cos 2 α + sin 2 α = 1 ⇒

[Q e1 e2 = 1]

Foci = (± 1, 0)

where, vertices are (± cos α , 0).

x2 y 2 − = 1] a 2 b2

480 Hyperbola ae = 1 or e =

Eccentricity,

1 cos α

dy dx

Hence, foci remains constant with change in α. x2 y2 − = 1 , where| r | < 1 1−r 1+ r ⇒ 1 − r is (+ ve) and 1 + r is (+ ve) x2 y 2 ∴Given equation is of the form 2 − 2 = 1. a b Hence, it represents a hyperbola when| r | < 1.

Since, ellipse and hyperbola are orthogonal. 1 π ∴ − cosec2 θ = − 1 ⇒ cosec2 θ = 2 ⇒ θ = ± 4 2 1 1   ∴ A 1 ,  or 1 , −    2 2

x2 y2 + =1 4 1 b2 1 3 e2 = 1 − 2 = 1 − = 4 4 a

16. Here, equation of ellipse is

∴

2

⇒ b2 = 1 2 Equation of ellipse is x + 2 y2 = 2. 1   Coordinates of foci (± ae , 0) =  ± 2 ⋅ , 0 = (± 1 , 0)  2 

where, ∴

x2 y2 b2 − 2 = 1, e12 = 1 + 2 2 a b a e12

If major axis is along Y-axis, then

b2 4 ⇒ 1+ 2 = 3 a

1 4 = 2= 3 e b2 1 = a2 3

a2 1 = 1 − 2 ⇒ b2 = 2a 2 2 b …(i)

…(ii)

From Eqs. (i) and (ii), b = 1 2

…(iii) 2

x y − =1 3 1 2   Focus is (± a e1 , 0) ⇒  ± 3 ⋅ , 0 ⇒ (± 2, 0)  3 

∴ Equation of hyperbola is

Hence, (b) and (d) are correct answers.

17. Given, ⇒

∴ ⇒

and hyperbola passes through (± 3, 0) 3 = 1 ⇒ a2 = 3 ⇒ a2 2

 1 1 + 2   = 2b2  2

Form Eq. (ii),

3 and focus (± a e, 0) ⇒ (± 3 , 0) e= 2

For hyperbola

2 x 2 − 2 y2 = 1 x2 y2 − =1  1  1      2  2

... (i)

Eccentricity of hyperbola = 2

⇒

y′  

1 2

2 x2 + y2 = 2 a 2 2x Y′ = − y −2 = 1  sec θ , tan θ  sin θ 

2

As ellipse and hyperbola are orthogonal 2 ∴ − ⋅ cosec θ = − 1 sin θ π 1 cosec2 θ = ⇒ ⇒ θ=± 4 2 ∴ 2 x2 + y2 = 2 a 2 1 5 ⇒ 2 + = 2a 2 ⇒ a 2 = 2 4 5 2 2 ∴ 2x + y = , corresponding foci are (0, ± 1). 2 Hence, option (a) and (b) are correct.

18. Let y = 4x + c meets xy = 1 at two points A and B.

So, eccentricity of ellipse = 1 / 2

Y

Let equation of ellipse be x2 y2 + =1 a 2 b2

[where a > b]

∴

1 b2 1 b2 = 1− 2 ⇒ 2 = 2 2 a a

⇒

a 2 = 2b2 ⇒ x2 + 2 y2 = 2b2

Let ellipse and hyperbola intersect at 1   1 A sec θ , tan θ   2 2 On differentiating Eq. (i), we get dy dy x 4x − 4 y =0 ⇒ = dx dx y

at A

sec θ = cosec θ tan θ

and on differentiating Eq. (ii), we get 1 dy dy x 2x + 4 y =− = − cosec θ =0 ⇒ 2y 2 dx dx at A

15. Given equation is

⇒

=

y = 4x + c A

2:1 P

xy = 1 X

X′ B

... (ii)

Y′

i.e. A (t1 , 1 t1 ), B (t2 , 1 t2 ) ∴ Coordinates of P are 1 1  2⋅ + 1⋅   + 2 t t t t 2 1 2  1 , = (h , k) 2+1   2+1    

[say]

Hyperbola 481 ∴ h=

2t1 + t2 2t + t and k = 2 1 3 3t1 t2

...(i)

1. Since, the vertices of hyperbola on X-axis at (± 6, 0), so

  1 1 Also,  t1 ,  and  t2 ,  lie on y = 4 x + c.  t1   t2 ⇒

1 1 − 1 t2 t1 = =4 t2 − t1 t1 t2

or

t1 t 2 = − 1 4

k 4 k and t1 = h − 2 k  k 1  From Eqs. (ii) and (iii), − h −  2h +  = −     2 4 4

Topic 2 Equation of Tangent and Normal equation of hyperbola we can assume as

|a| = 6 and hyperbola passes through point P(10, 16) so (10)2 (16)2 − =1 62 b2

... (ii)

From Eq. (i), t1 = 2h +

⇒

1  2h + k  8h + k −  =−    2   4  4

⇒

( 2h + k) ( 8h + k) = 2

⇒

16h 2 + k2 + 10hk = 2

⇒ ...(iii) ⇒

x2 y2 − = 0, and the 36 144 equation of normal to hyperbola at point P is 36x 144 y + = 36 + 144 10 16 18 x + 9 y = 180 ⇒ 5 ⇒ 2x + 5 y = 100 Hence, option (c) is correct.

19. We have, Equation of hyperbola

Y L (0,–b) b X′

2.

y2 x2 – 2 =1 b a2 N(a,0)

X

M (0,–b) Y′

It is given, ∠ LNM = 60° and Area of ∆LMN = 4 3 Now, ∆LNM is an equilateral triangle whose sides is 2b [Q ∆LON ~ = ∆MOL ; ∴ ∠NLO = ∠NMO = 60°] ∴

Area of ∆LMN =

3 (2b)2 4

⇒

4 3 = 3b2 ⇒ b = 2 1 Also, area of ∆LMN = a (2b) = ab 2 4 3 = a(2) ⇒ a = 2 3 ⇒ (P) Length of conjugate axis = 2b = 2(2)= 4 b2 4 4 2 = 1+ = = 12 2 3 3 a2 2 (R) Distance between the foci = 2ae = 2 × 2 3 × =8 3

(Q) Eccentricity (e) = 1 +

(S) The length of latusrectum = P → 4; Q → 3; R → 1; S → 2

Key Idea An equation of tangent having slope ‘m’ to parabola y 2 = 4 ax is y = mx +

60° O b

(16)2 100 − 36 64 = = 36 36 b2 36 256 × = 144 b2 = 64

So, equation of hyperbola is

Hence, required locus is 16x2 + y2 + 10xy = 2.

x2 y 2 − =1 a 2 b2

2b2 2(4) 4 = = a 2 3 3

x2 y2 − = 1 and a 2 b2

a . m

Given equation of curves are y2 = 16x (parabola) and xy = − 4 (rectangular hyperbola)

…(i) …(ii)

Clearly, equation of tangent having slope ‘m’ to parabola 4 …(iii) (i) is y = mx + m Now, eliminating y from Eqs. (ii) and (iii), we get 4 4  x mx +  = − 4 ⇒ mx2 + x + 4 = 0,   m m which will give the points of intersection of tangent and rectangular hyperbola. 4 is also a tangent to the Since, line y = mx + m rectangular hyperbola. 4 ∴ Discriminant of quadratic equation mx2 + x + 4 = 0, m should be zero. [Q there will be only one point of intersection] 2

⇒ ⇒

 4 D =   − 4 (m) (4) = 0  m m3 = 1 ⇒ m = 1

So, equation of required tangent is y = x + 4. x2 y2 …(i) − =1 24 18 Since, the equation of the normals of slope m to the x2 y 2 m(a 2 + b2 ) hyperbola 2 − 2 = 1, are given by y = mx m a b a 2 − b2m2

3. Given equation of hyperbola, is

482 Hyperbola ∴ Equation of normals of slope m, to the hyperbola (i), are m(24 + 18) …(ii) y = mx ± 24 − m2 (18) Q Line y = mx + 7 3 is normal to hyperbola (i) ∴On comparing with Eq. (ii), we get m(42) ± =7 3 24 − 18m2 6m ⇒ ± = 3 24 − 18m2 36m2 [squaring both sides] =3 ⇒ 24 − 18m2 ⇒ 12m2 = 24 − 18m2 ⇒ 30m2 = 24 2 ⇒ 5m2 = 4 ⇒ m = ± 5

1+

Y T (0, 3)

(–3√5, –12)Q

(given)

⇒ a 2 + b2 = 4a 2 …(ii) ⇒ b2 = 3a 2 Since, hyperbola (i) passes through the point (4, 6) 16 36 …(iii) − =1 ∴ a 2 b2 On solving Eqs. (ii) and (iii), we get …(iv) a 2 = 4 and b2 = 12 Now, equation of tangent to hyperbola (i) at point (4, 6), is 4x 6 y − =1 a 2 b2 4x 6 y [from Eq. (iv)] ⇒ − =1 4 12 y ⇒ x − = 1 ⇒ 2x − y − 2 = 0 2

5. Given equation of hyperbola is 4x2 − 5 y2 = 20 which can be rewritten as x2 y2 ⇒ − =1 5 4 The line x − y = 2 has slope, m = 1 ∴ Slope of tangent parallel to this line = 1

having slope m is given by y = mx ±

a 2m2 − b2

Here, a = 5, b = 4 and m = 1 2

2

S(0, –12)

P(3√5, –12)

Clearly, P Q is chord of contact. ∴Equation of PQ is −3 y = 36 ⇒ y = − 12 Solving the curve 4x2 − y2 = 36 and y = − 12 , we get x= ±3 5 1 1 Area of ∆PQT = × PQ × ST = (6 5 × 15) = 45 5 2 2 x2 y 2 − = 1. a 2 b2 ae = 2 ⇒ a 2e2 = 4 2 a + b2 = 4 ⇒ b2 = 4 − a 2 x2 y2 − =1 a2 4 − a2

7. Let the equation of hyperbola be ∴ ⇒ ∴

Since, ( 2 , 3 ) lie on hyperbola. 2 3 − =1 ∴ a2 4 − a2 ⇒ ⇒ ⇒ ⇒ ∴

8 − 2a 2 − 3a 2 = a 2 (4 − a 2 ) 8 − 5a 2 = 4a 2 − a 4 4 a − 9a 2 + 8 = 0 4 (a − 8)(a 4 − 1) = 0 ⇒ a 4 = 8, a 4 = 1 a =1 x2 y2 Now, equation of hyperbola is − = 1. 1 3 ∴ Equation of tangent at ( 2 , 3 ) is given by 3y y 2x − =1 = 1 ⇒ 2x − 3 3 which passes through the point (2 2 , 3 3 ).

8. Equation of normal to hyperbola at (x1 , y1 ) is 2

We know equation of tangent to hyperbola

X

O

…(i)

b2 =2 a2

6. Tangents are drawn to the hyperbola 4x2 − y2 = 36 at the point P and Q. Tangent intersects at point T (0, 3)

4. Let the equation of standard hyperbola is x2 y 2 − =1 a 2 b2 Now, eccentricity of hyperbola is

∴ Required equation of tangent is y= x± 5 −4 ⇒ ⇒ y = x ± 1 ⇒ x − y ± 1 =0

a 2x b2 y + = (a 2 + b2 ) x1 y1

2

x y − 2 =1 2 a b

∴

At (6, 3) =

a 2x b2 y + = (a 2 + b2 ) 6 3

Q It passes through (9, 0). ⇒

a2⋅ 9 = a 2 + b2 6

Hyperbola 483 ⇒ ∴

3a 2 a2 − a 2 = b2 ⇒ 2 = 2 2 b 2 b 1 e2 = 1 + 2 = 1 + 2 a

⇒ e=

and equation of normal to the hyperbola at point P b sec θ having slope is , as normal cuts off equal a tan θ intercepts on the coordinate axes, so slope must be − 1. b sec θ Therefore, = −1 a tan θ {from Eq. (ii)} ⇒ b = − tan θ Q The eccentricity of hyperbola

3 2

9. The equation of tangent at (x1 , y1 ) is x x1 − 2 y y1 = 4, which is same as 2x + 6 y = 2. 2y 4 x1 =− 1 = ∴ 2 6 2 ⇒

x1 = 4

and

b2 = 1 + sin 2 θ a2  π 1 < e < 2 , as θ ∈ 0,   2

e= 1+

y1 = − 6

Thus, the point of contact is (4, − 6 ).

∴

x2 y 2 − = 1 at a 2 b2 (a sec θ , b tan θ ). On differentiating w.r.t. x, we get 2x 2 y dy dy b2 x = − 2 × =0 ⇒ 2 dx a 2 y dx a b

10. Firstly, we obtain the slope of normal to

And the area of required triangle is, which is isosceles is 1 1 ∆ = ( AP )2 = [(1 − sec2 θ )2 + tan 4 θ ] 2 2 Y

Slope for normal at the point (a sec θ , b tan θ ) will be −

a 2b tan θ a = − sin θ b b2a sec θ

P (a secθ, b tanθ)

(a sin θ ) x + by = (a 2 + b2 ) tan θ

⇒

a x + b cosecθ = (a 2 + b2 ) sec θ

=

…(i)

x2 y2 − = 1 at a 2 b2 (a sec φ, b tan φ ) is ax + b y cosec φ = (a 2 + b2 ) sec φ …(ii)

Similarly, equation of normal to

x2 y2 − =1 a 2 16 It point ≡ (a sec θ , 4 tan θ ), x sec θ y tan θ tangent ≡ − =1 a 4 On comparing, we get secθ = − 2a tan θ = − 4 ⇒ 4a 2 − 16 = 1 17 ∴ a= 2 Substitute the value of a in option (a), (b), (c) and (d). Hyperbola ≡

b (cosec θ − cosec φ ) y = (a 2 + b2 ) (sec θ − sec φ )

But

y=

a 2 + b2 sec θ − sec φ ⋅ b cosec θ − cosec φ

sec θ − sec (π / 2 − θ ) sec θ − sec φ = cosec θ − cosec φ cosec θ − cosec (π / 2 − θ )

13.

Y

,y

1

[Q φ + θ = π / 2] sec θ − cosec θ = = −1 sec θ − sec θ

1 (2 tan 4 θ ) = tan 4 θ = b4 . 2

12. Tangent ≡ 2x − y + 1 = 0

On subtracting Eq. (ii) from Eq. (i), we get

⇒

X

B

)

⇒

135º

A O (1, 0)

∴ Equation of normal at (a sec θ , b tan θ ) is a y − b tan θ = − sin θ (x − a sec θ ) b

P( x

1

 a 2 + b2  a 2 + b2 y=−  , i.e. k = −   b  b   

Thus,

2

(–1, 0)

2

x y − = 1 , a > b and a > 1 a 2 b2 Let point P (a sec θ, b tan θ) on the hyperbola in first  π quadrant i.e., Q ∈ 0,  .  2 Now equation of tangent to the hyperbola at point P is x sec θ y tan θ …(i) − =1 a b Q The tangent (i) passes through point A(1, 0) So, … (ii) a = secθ > 1 .

M

X′

11. Equation of given hyperbola is

(1, 0)

N(x2, 0)

X

x2 – y2 = 1 Y′

Equation of family of circles touching hyperbola at (x1 , y1 ) is (x − x1 )2 + ( y − y1 )2 + λ( x x1 − y y1 −1) = 0 Now, its centre is (x2 , 0).  − (λx1 − 2x1 ) – (−2 y1 − λy1 )  ∴ , = (x2 , 0)   2 2 ⇒ and

2 y1 + λy1 = 0 ⇒ λ = − 2 2x1 − λx1 = 2x2 ⇒ x2 = 2x1

484 Hyperbola ∴

P (x1 , x12 − 1 ) and N (x2 , 0) = (2x1 , 0)

15. The equation of the hyperbola is

1  As tangent intersect X-axis at M  ,0 .  x1 

circle is x2 + y2 − 8x = 0

Centroid of ∆PMN = (l, m) 1     3x1 + 0 0 y + + x 1  = (l, m)  , 1 ⇒ 3 3      

For their points of intersection,

1 3x1 + x1 l= 3

⇒

⇒

b

y = 2x ± 32

…(i)

The equation of tangent at (x1 , y1 ) is xx1 yy1 − =1 9 4 From Eqs. (i) and (ii),

13 not acceptable. 6

⇒

x2 − 12x + y2 + 24 = 0

⇒

x2 + y2 − 12x + 24 = 0

16. Equation of tangent to hyperbola having slope m is y = mx + 9m2 − 4

...(i)

Equation of tangent to circle is y = m (x − 4) + 16m2 + 16

...(ii)

2 satisfy. Eqs. (i) and (ii) will be identical for m = 5

2x – y = 1

x

…(ii)

Tangent

P (x1, y1) (3, 0)

a e

X



0=−

2a +1 e

a 1 = e 2

Also, y = − 2x + 1 is tangent to hyperbola. ∴

Y

12

13x (x − 6) + 6 (x − 6) = 0 13 x=6, x= − 6

⇒

y = 2x ± 9 (4) − 4

O

⇒

a1 b1 c1 = = a 2 b2 c2

Equation of tangent, parallel to y = 2x − 1 is

(–3, 0)

13x2 − 78x + 6x − 36 = 0

17. On substituting  , 0 in y = − 2x + 1, we get

and a 2x + b2 y + c2 = 0 are identical. Then,

x

⇒

∴ Equation of common tangent is 2x − 5 y + 4 = 0.

If two straight lines

a1 x + b1 y + c1 = 0

∴

13x2 − 72x − 36 = 0

Now, for x = 6, y = ± 2 3

2 2 14. PLAN Equation of tangent to x 2 − y 2 = 1 is y = mx ± a 2m2 − b 2

Description of Situation

⇒

Required equation is, (x − 6)2 + ( y + 2 3 ) ( y − 2 3 ) = 0

On differentiating w.r.t. x1 , we get dm 2x1 x1 , for x1 > 1 = = dx1 2 × 3 x12 − 1 3 x12 − 1 y Also, m= 1 3 dm 1 On differentiating w.r.t. y1 , we get = , for y1 > 0 dy1 3 a

4x2 + 9x2 − 72x = 36

x=−

x12 − 1 3

1 dl = 1 − 2 , for x1 > 1 and m = dx1 3x1

x2 x2 − 8 x + =1 9 4

⇒

⇒

1 3− 2 dl x1 On differentiating w.r.t. x1 , we get = dx1 3

x2 y2 − = 1 and that of 9 4

1 = 4a 2 − b2 1 = 4 − (e2 − 1) a2 4 = 5 − e2 e2

⇒ ⇒ ⇒

e4 − 5 e2 + 4 = 0

⇒

(e2 − 4) (e2 − 1) = 0

⇒

e=2, e=1

–1

Y

x2 – y2 =1 9 4

⇒ or

−1 ± 32 2 = = x1 − y1 1 9 4 9 1 and y1 = − x1 = − 2 2 2 9 1 , y1 = x1 = 2 2 2

F F1

X 1 , – 0 2

(

)

e = 1 gives the conic as parabola. But conic is given as hyperbola, hence e = 2.

Hyperbola 485 and

Topic 3 Equation of Chord of Contact, Chord Bisected Diameter, Asymptote and Rectangular Hyperbola

Similarly,

∴

hx − ky − 9 = 0

Again equation of pair of tangents is ⇒

T 2 = SS1 (x − 9)2 = (x2 − y2 − 9) (12 − 02 − 9)

⇒

x2 − 18x + 81 = (x2 − y2 − 9) (−8)

⇒

x2 − 18x + 81 = − 8x2 + 8 y2 + 72

2. It is given that, x 2 + y2 = a 2 xy = c

2

x + c /x = a

We obtain ⇒

4

2

…(i) …(ii)

2

x − a 2x2 + c4 = 0 4

Now, x1 , x2 , x3 , x4 will be roots of Eq. (iii). Therefore,

(3 sec θ ) x + (2 tan θ ) y = 9

Σ x1 = x1 + x2 + x3 + x4 = 0

x x1 + yy1 = x12 + y12

…(i)

…(ii)

Since, Eqs. (i) and (ii) are identically equal. 3 sec θ 2 tanθ ∴ = x1 y1 =

9x2 − 8 y2 − 18x + 9 = 0

2

∴ Chord of contact of the circle x2 + y2 = 9 with respect to the point (3 sec θ , 2 tanθ ) is,

⇒ Equation of chord in mid-point form is

This is possible when h = 1, k = 0 (by comparing both equations).

and

3. Let any point on the hyperbola is (3 sec θ , 2 tan θ ).

Let (x1 , y1 ) be the mid-point of the chord of contact.

But it is the equation of the line x = 9.

⇒

y1 y2 y3 y4 = c4

Hence, all options are correct.

to hyperbola x2 − y2 = 9 is x = 9. We know that, chord of contact of (h , k) with respect to hyperbola x2 − y2 = 9 is T = 0. h ⋅ x + k (− y) − 9 = 0

y1 + y2 + y3 + y4 = 0

and

1. Let (h , k) be a point whose chord of contact with respect

⇒

product of the roots x1 x2 x3 x4 = c4

9 x12 + y12

⇒

sec θ =

9x1 3 (x12 + y12 )

and

tanθ =

9 y1 2 (x12 + y12 )

Thus, eliminating ‘ θ ’ from above equation, we get 81 x12 81 y12 − =1 9 (x12 + y12 )2 4 (x12 + y12 )2

…(iii)

[Q sec2 θ − tan 2 θ = 1] ∴ Required locus is

x y (x + y2 )2 . − = 9 4 81 2

2

2

20 Trigonometrical Ratios and Identities Topic 1 Based on Trigonometric Formulae Objective Questions I (Only one correct option) π 3π π 3π 1. The value of cos3   ⋅ cos   + sin3   ⋅ sin   is  8

 8

 8

 8

(2020 Main, 9 Jan I)

(a)

1 4

(b)

1 2 2

(c)

1 2

(d)

1 2

2. The value of sin 10º sin 30º sin 50º sin 70º is (2019 Main, 9 April II)

(a)

1 36

(b)

1 32

(c)

1 16

(d)

1 18

3. The value of cos 2 10° − cos 10° cos 50° + cos 2 50° is (2019 Main, 9 April I)

3 (a) (1 + cos 20° ) 2 (c) 3 / 2

3 (b) + cos 20° 4 (d) 3 / 4

4. If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is (2019 Main, 8 April II) (a) 3 : 4 : 5

(b) 4 : 5 : 6

(c) 5 : 9 : 13 (d) 5 : 6 : 7

3 5 tan(2α ) is equal to

5. If cos(α + β) = , sin(α − β) = 63 (a) 52

63 (b) 16

5 π and 0 < α , β < , then 4 13 (2019 Main, 8 April I)

21 (c) 16

(d)

33 52

1 (sin k x + cos k x) for k = 1, 2, 3 ... . Then, for k all x ∈ R, the value of f4 (x) − f6 (x) is equal to

6. Let fk (x) =

(a)

1 12

(b)

5 12

(c)

−1 12

(2019 Main, 11 Jan I)

(d)

1 4

π π π π ⋅ cos 3 ....... cos 10 ⋅ sin 10 is 22 2 2 2

1 (a) 1024

(2019 Main, 9 Jan I)

(a) (b) (c) (d)

13 − 4 cos4 θ + 2 sin 2 θ cos2 θ 13 − 4 cos2 θ + 6 cos4 θ 13 − 4 cos2 θ + 6 sin 2 θ cos2 θ 13 − 4 cos6 θ

9. The expression

1 (b) 2

1 (c) 512

tan A cot A can be written as + 1 − cot A 1 − tan A (2013 Main)

(a) sin A cos A + 1 (c) tan A + cot A

(b) sec A cosec A + 1 (d) sec A + cosec A

10. The number of ordered pairs (α , β), where α , β ∈ (−π , π ) satisfying cos (α − β ) = 1 and cos (α + β ) =

1 is e (2005, 1M)

(a) 0

(d) 4

(b) 1

(c) 2

11. Given both θ and φ are acute angles and sin θ =

1 1 , cos φ = , then the value of θ + φ belongs to 2 3 (2004, 1M)

π π (a)  ,   3 6 

π 2π (b)  ,  2 3

2π 5π  (c)  ,  3 6 

5π  (d)  ,π  6 

12. Which of the following numbers is rational? (a) sin 15° (c) sin 15° cos 15°

equals

(1995, 2M)

(b) 12

(c) 13

14. The value of the expression (2019 Main, 10 Jan II)

1 (d) 256

(1998, 2M)

(b) cos 15° (d) sin 15° cos 75°

13. 3 (sin x − cos x)4 + 6 (sin x + cos x)2 + 4 (sin 6 x + cos 6 x) (a) 11

7. The value of cos

π π 4 2 3 (sin θ − cos θ )4 + 6 (sin θ + cos θ )2 + 4 sin 6 θ equals

8. For any θ ∈  ,  , the expression  

(d) 14

3 cosec 20° − sec 20° is

equal to (a) 2 (c) 4

(1988, 2M)

(b) 2 sin 20° /sin 40° (d) 4 sin 20° /sin 40°

Trigonometrical Ratios and Identities 487 π (a) f2   = 1  16  π (c) f4   = 1  64 

15. The expression    3π  3 sin 4  − α  + sin 4 (3π + α )   2    6 π   −2 sin  + α  + sin 6 (5π − α ) 2    is equal to

(1986, 2M)

(a) 0 (c) 3

(b) 1 (d) sin 4α + cos 6 α

π 3π 5π 7π 16. 1 + cos  1 + cos  1 + cos  1 + cos  is 8 

 8  equal to 1 2

(a)

Match the Column Match the conditions/expressions in Column I with values in Column II.

23. (sin 3 α ) / (cos 2 α ) is

8

8

Column I

(1984, 3M)

(b) cos

π 8

(c)

1 8

(d)

1+ 2 2 2

17. Given A = sin 2 θ + cos 4 θ , then for all real values of θ (a) 1 ≤ A ≤ 2 (c)

13 ≤ A≤1 16

4 18. If tan θ = − , then sin θ is 3 4 4 but not 5 5 4 4 (c) but not − 5 5 (a) −

p. (13 π / 48, 14 π / 48)

B. negative

q. (14 π / 48, 18 π / 48) r. (18 π / 48, 23 π / 48)

Fill in the Blanks

4 4 or 5 5

π 3π 5π 7π 9π 11π 13π ⋅ sin ⋅ sin ⋅ sin ⋅ sin ⋅ sin ⋅ sin 14 14 14 14 14 14 14 (1991, 2M) is equal to …… .

sin

Analytical & Descriptive Questions

2 for 19. Let f : (−1, 1) → R be such that f (cos 4 θ ) = 2 − sec2 θ  π π  1  π θ ∈  0 ,  ∪  ,  . Then, the value(s) of f   is/are  4 2  3  4 (2012) 3 2

(b) 1 +

3 2

(c) 1 −

2 3

(d) 1 +

π 4 π (c) 12

21. If

(2009)

π 6 5π (d) 12

(b)

2 (a) tan x = 3 1 2 (c) tan x = 3

(2009)

sin 8 x cos8 x 1 (b) + = 8 27 125 sin 8 x cos8 x 2 (d) + = 8 27 125

θ  fn (θ ) =  tan  (1 + sec θ )(1 + sec 2 θ )  2 (1 + sec 2nθ ), then

(1982, 2M)

29. Prove that sin 2 α + sin 2 β – sin 2 γ = 2 sinα sin β sin γ , where α + β + γ = π .

(1978, 4M)

2 sin α 1 1 − cos 2β 1  π , α , β ∈ 0,  , = and =  2 2 1 + cos 2α 7 10 then tan(α + 2β ) is equal to .............. .

30. If

22. For a positive integer n, let

(1 + sec 22 θ )...

1 (sin 12° ) (sin 48° ) (sin 54° ) = . 8

Integer & Numerical Answer Type Questions

sin 4 x cos 4 x 1 + = , then 2 3 5 2

27.

tan α + 2 tan 2α + 4 tan 4α + 8 cot 8α = cot α (1988, 2M)  2π   4π   8π   16π  Show that 16 cos   cos   cos   cos   =1  15   15   15   15 

28. Without using tables, prove that

π , the solution(s) of 2 6 (m − 1) π  mπ    ∑ cosec θ + 4  cosec θ + 4  = 4 2 is/are

m =1

26. Prove that

(1983, 2M)

2 3

20. For 0 < θ
t2 > t3 > t4 (c) t3 > t1 > t2 > t4

(b) t4 > t3 > t1 > t2 (d) t2 > t3 > t1 > t4

Fill in the Blank (2019 Main, 12 Jan I)

3. If A > 0, B > 0 and A + B = π /3, then the maximum value of tan A tan B is ...........

(1993)

(b) 34

Analytical & Descriptive Question

(d) 19

π 2. Let θ ∈ 0,  and t1 = (tan θ )tan θ , t2 = (tan θ )cot θ ,  4 t3 = (cot θ )tan θ and t4 = (cot θ )cot θ , then

(2006, 3M)

4. Prove that the values of the function lie between 1 / 3 and 3 for any real x.

sin x cos 3x do not sin 3x cos x (1997, 5M)

Topic 4 Height & Distance 1. The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30°, then the distance (in m) of the foot of the tower from the point A is (2019 Main, 12 April II)

(a) 15 (3 + 3 ) (c) 15 (3 − 3 )

(b) 15 (5 − 3 ) (d) 15 (1 + 3 )

2. A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/s, then the rate (in cm/s) at which the bottom of the ladder slides away from the wall on the horizontal

ground when the top of the ladder is 1 m above the ground is (2019 Main, 12 April I) (a) 25 3

(b)

25 3

(c)

25 3

(d) 25

3. ABC is a triangular park with AB = AC = 100 m. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot−1 (3 2 ) and cosec−1 (2 2 ) respectively, then the height of the tower (in m) is (2019 Main, 10 April I) (a) 25 (c) 10 5

(b) 20 100 (d) 3 3

Trigonometrical Ratios and Identities 489 4. Two poles standing on a horizontal ground are of

8. PQR is a triangular park with PQ = PR = 200 m. A TV

heights 5 m and 10 m, respectively. The line joining their tops makes an angle of 15º with the ground. Then, the distance (in m) between the poles, is

tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P , Q and R are respectively 45°, 30° and 30°, then the height of the tower (in m) is (2018 Main)

(2019 Main, 9 April II)

5 (b) (2 + 3 ) 2 (d) 5(2 + 3 )

(a) 5( 3 + 1) (c) 10( 3 − 1)

(a) 100 (c) 100 3

5. Two vertical poles of heights, 20 m and 80 m stand apart on a horizontal plane. The height (in m) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is (2019 Main, 8 April II)

(a) 15

(b) 16

(c) 12

(d) 18

6. If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30º and the angle of depression of reflection of the cloud in the lake from P be 60º, then the height of the cloud (in meters) from the surface of the (2019 Main, 12 Jan II) lake is (a) 50

(b) 60

(c) 45

(d) 42

7. Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and CA = 6 m. A vertical lamp-post at the mid-point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is (2019 Main, 10 Jan I) (a)

2 21 3

(b) 2 21

(c) 7 3

(d)

3 21 2

(b) 50 (d) 50 2

9. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2 AB. If ∠BPC = β, then tan β is equal to (2017 Main) 6 7 2 (c) 9

1 4 4 (d) 9

(a)

(b)

10. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 min from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then, the time taken (in minutes) by him, from B to reach the pillar, is (2016 Main) (a) 6 (c) 20

(b) 10 (d) 5

Answers Topic 1 1. 5. 9. 13. 17. 21.

(b) (b) (b) (c) (b) (a, b) 1 24. 8

2. 6. 10. 14. 18. 22.

(c) (a) (b) (c) (b) (a, b, c, d) 1 25. 64

3. 7. 11. 15. 19. 23.

(d) 4. (c) 8. (b) 12. (b) 16. (a, b) 20. A → r; B → p 31. θ =

30. (1)

(b) (d) (c) (c) (c, d) π 5π 9π , , 12 12 12

6. A = 45 °, B = 60 °, C = 75 ° 56 9. 33

Topic 3 1. (d)

3.

2. (b)

3. (b)

4. (d)

7. (a)

8. (a)

Topic 4 1. (a)

Topic 2

5. (b)

6. (a)

1. (c)

9. (c)

10. (d)

2. (a)

3. 6

4. True

1 3

2. (b)

Hints & Solutions Topic 1 Based on Trigonometric Formulae 1. Given trigonometric expression 3π  π  3π   π cos3   cos   + sin3   sin  8  8  8 8  π  π π  π  π π cos3   cos  −  + sin3   sin  −   8  2 8  8  2 8  π π 3π  Q − =  2 8 8 

 π  π  π  π = cos3   sin   + sin3   cos    8  8  8  8  π  π = sin   cos    8  8 =

 2 π  2 π   cos  8  + sin  8    

1 π 1  π 1  π × 2 sin   cos   = sin =  8 2  8 2 4 2 2

Hence, option (b) is correct.

490 Trigonometrical Ratios and Identities 2. We have, sin 10° sin 30° sin 50° sin 70° = sin(30° )[sin(10° )sin(50° )sin(70° )] 1 = [sin(10° )sin(60° − 10° )sin(60° + 10° )] 2 1 1  = sin(3(10° )) 2 4 

1 [Qsin θ sin(60° − θ )sin(60° + θ ) = sin 3 θ] 4 1 1 1 1 = sin 30° = × = 8 8 2 16

3. We have, cos 2 10º − cos 10º cos 50º + cos 2 50º 1 [2 cos 2 10º −2 cos 10º cos 50º +2 cos 2 50º ] 2 1 = [1 + cos 20º − (cos 60º + cos 40º ) + 1 + cos 100º ] 2 [Q 2 cos 2 A = 1 + cos 2 A and 2 cos A cos B = cos( A + B) + cos( A − B)] 1 1 1   = Q cos 60º = 2 + cos 20º + cos 100º − − cos 40º 2  2  2   =

3  + (cos 20º − cos 40º ) + cos 100º 2  20º + 40º 20°− 40° 3  − 2 sin + cos 100° sin 2 2 2  C + D C − D  Q cos C − cos D = −2 sin sin  2 2  1 3  = − 2 sin 30º sin(−10º ) + cos(90º +10º ) 2 2  1 3  = + sin 10º − sin 10º [Q cos (90º + θ ) = − sin θ ]  2 2 1 2 1 = 2 =

=

1 3 3 × = 2 2 4

4. Let a , b and c be the lengths of sides of a ∆ABC such that a < b < c. Since, sides are in AP. …(i) ∴ 2b = a + c Let ∠ A =θ Then, [according to the question] ∠ C = 2θ So, …(ii) ∠ B = π − 3θ On applying sine rule in Eq. (i), we get 2 sin B = sin A + sin C [from Eq. (ii)] ⇒ 2 sin(π − 3 θ ) = sin θ + sin 2 θ ⇒ 2 sin 3 θ = sin θ + sin 2 θ ⇒ 2 [3 sin θ − 4 sin3 θ ] = sin θ + 2 sin θ cos θ [Qsin θ can not be zero] ⇒ 6 − 8 sin 2 θ = 1 + 2 cos θ ⇒ 6 − 8(1 − cos 2 θ ) = 1 + 2 cos θ ⇒ 8 cos 2 θ − 2 cos θ − 3 = 0 ⇒ (2 cos θ + 1)(4 cos θ − 3) = 0 3 ⇒ cos θ = 4

1 (rejected). 2 Clearly, the ratio of sides is a : b : c = sin θ : sin 3 θ : sin 2 θ = sin θ : (3 sin θ − 4 sin3 θ ) : 2 sin θ cos θ or cos θ = −

= 1 : (3 − 4 sin 2 θ ) : 2 cos θ = 1 : (4 cos 2 θ − 1) : 2 cos θ 5 6 = 1 : : = 4 :5 :6 4 4 5 5. Given, sin(α − β) = 13 3  π and cos(α + β ) = , where α , β ∈ 0,   4 5 π π and 0 < β < 4 4 π π π 0 φ > cos −1    2

  π  the sign changed as cos x is decreasing between 0, 2     π π π 2π π and ⇒ 0.  48 48 

= sin 2

 8π π π  Q sin 7 = sin  π + 7  = − sin 7   

26. We know that,

23. In the interval 

25. sin

2

= 1 / 64

Therefore, (d) is the answer.

= cos A ⋅ cos 2 A ⋅ cos 22 A =

2

2

 1 sin 8π / 7 = − ⋅   8 sin π / 7 

Therefore, (b) is the answer. π   π π f4   = tan 24 ⋅  = tan   = 1    4  64 64 Therefore, (c) is the answer.  π   5 π   π f5   = tan 2 ⋅  = tan   = 1  128   4 128

⇒

2

π 2π 4π   =  − cos ⋅ cos ⋅ cos   7 7 7  sin 23 π / 7  = − 3   2 ⋅ sin π / 7

2

2π    32π  sin    sin 2π +   15  15  = =  2π   2π  sin   sin    15   15   2π  sin    15  =1 =  2π  sin    15 

28. sin 12° sin 48° sin 54° = 2

=

1 (2 sin 12° sin 48° ) sin 54° 2

1 [cos (36° ) − cos (60° )] sin 54° 2

494 Trigonometrical Ratios and Identities 1 1  cos 36° −  sin 54° 2 2 1 = (2 cos 36° sin 54° − sin 54° ) 4 1 = (sin 90° + sin 18° − sin 54° ) 4 1 5 −1 5 + 1 = 1 + −  4 4 4  =

=

1 5 − 1 − 5 − 1 1 +  4 4 

=

1 1 1 1 −  = 4 2 8

31. Given equations can be written as

cos 3 θ cos 3 θ − =0 y z 2 cos 3 θ 2 sin 3 θ x sin 3 θ − − =0 y z 2 1 and x sin 3 θ − cos 3 θ − (cos 3 θ + sin 3 θ ) = 0 y z Eqs. (ii) and (iii), implies 2 sin 3 θ = cos 3 θ + sin 3 θ x sin 3 θ −

⇒

sin 3 θ = cos 3 θ

∴

tan 3 θ = 1

⇒

29. LHS = sin 2 α + sin 2 β − sin 2 γ = sin 2 α + (sin 2 β − sin 2 γ ) = sin 2 α + sin ( β + γ ) sin ( β − γ ) = sin 2 α + sin ( π − α )sin (β − γ ) [Qα + β + γ = π] = sin 2 α + sin α sin ( β − γ ) = sin α [sin α + sin ( β − γ )] = sin α [sin (π − ( β + γ )) + sin ( β − γ )] = sin α [sin ( β + γ ) + sin ( β − γ )] = sin α [2 sin β cos γ ] = 2 sin α sin β cos γ = RHS π 30. It is given that, for α , β ∈ 0,   2

⇒ ⇒ ⇒ ⇒

π 5π 9π π 5π 9π , , or θ = , , 4 4 4 12 12 12

α = (π /2) − β tan α = tan (π /2 − β ) tan α = cot β tan α tan β = 1 β + γ =α γ = (α − β ) tan γ = tan (α − β ) tan α − tan β tan γ = 1 + tan α tan β

Again, ⇒ ⇒

⇒

2 sin α 1 = 2|cos α| 7

⇒

⇒

1 tan α = 7

Hence, answer is 1.

...(iii)

α + β = π /2

1. Given,

⇒

∴

1 − cos 2 β 1 and = 2 10 1 ⇒ |sin β| = 10 1  π sin β = Qβ ∈ 0,  ⇒  2 10 1 …(ii) ∴ tan β = 3 2 tan β tan 2 β = ∴ 1 − tan 2 β 2 6 3 = 3 = = 1 8 4 1− 9 tan α + tan 2 β Now, tan(α + 2 β ) = 1 − tan α tan 2 β 1 3 + 4 + 21 25 7 4 = = = =1  1 3 28 − 3 25 1− ×   7 4

...(ii)

Topic 2 Graph and Conditional Identities

2 sin α 1 = 1 + cos 2α 7

 π Q α ∈ 0,  …(i)  2

3θ =

...(i)

[given]

tan α − tan β 1+1 2 tan γ = tan α − tan β tan γ =

⇒

tan α = tan β + 2 tan γ α β  γ + = π −  2 2  2

2. Since,

γ   α β tan  +  = tan  π −    2 2 2 α β tan + tan 2 2 = − tan γ ⇒ α β 2 1 − tan tan 2 2 α β γ α β γ ⇒ tan + tan + tan = tan tan tan 2 2 2 2 2 2

∴

n

3. Given, sin3 x sin 3x = Σ Cm cos nx is an identity in x, m= 0

where, C 0 , C1 ,... , C n are constants. 1 sin3 x sin 3x = {3 sin x − sin 3x} ⋅ sin 3x 4 1 3  =  ⋅ 2 sin x ⋅ sin 3x − sin 2 3x  4 2 1 3 1   (cos 2x − cos x) − (1 − cos 6x) 4 2 2  1 = (cos 6x + 3 cos 2x − 3 cos x − 1) 8 ∴ On comparing both sides, we get n = 6 =

Trigonometrical Ratios and Identities 495 ⇒ tan A + tan B = − tan C + tan A tan B tan C ⇒ tan A + tan B + tan C = tan A tan B tan C 4 9. Since, cos(α + β) = 5 5 and sin(α − β ) = 13 3 tan(α + β ) = ∴ 4 5 and tan(α − β ) = 12 Now, tan 2 α = tan[(α + β ) + (α − β )] 3 5 + tan(α + β ) + tan(α − β ) 56 = = 4 12 = 1 − tan(α + β ) ⋅ tan(α − β ) 1 − 3 ⋅ 5 33 4 12

B 2 sin 2 1 − cos B 2 4. Since, tan A = = B B sin B 2 sin cos 2 2 tan A = tan B / 2 ⇒ tan 2 A = tan B Hence, it is a true statement. A+ B+C=π A B π C + = − ⇒ 2 2 2 2  A B  π C ⇒ cot  +  = cot  −  2   2 2 2 A B cot ⋅ cot − 1 C 2 2 ⇒ = tan B A 2 cot + cot 2 2 A B C C A B ⇒ cot ⋅ cot ⋅ cot − cot = cot + cot 2 2 2 2 2 2 A B C A B C cot + cot + cot = cot cot cot ⇒ 2 2 2 2 2 2

5. Since,

Topic 3 Maxima and Minima π 1. Given expression 3 cos θ + 5 sin θ −  

6 π π   = 3 cos θ + 5 sin θ cos − sin cos θ   6 6  3  1 = 3 cos θ + 5  sin θ − cos θ 2  2 

6. Given, in ∆ABC , A , B and C are in an AP. ∴ A + C = 2B Also, A + B + C = 180° ⇒ B = 60° and sin (2 A + B) = sin (C − A ) 1 = − sin (B + 2C ) = 2 ⇒ sin (2 A + 60° ) = sin (C − A ) = − sin (60° + 2C ) =

…(i) 1 2

⇒ 2 A + 60° = 30° , 150° [neglecting 30°, as not possible] ⇒ 2 A + 60° = 150° ⇒ A = 45° Again, from Eq. (i), sin (60° + 2C ) = − 1 / 2 ⇒ 60° + 2C = 210° , 330° ⇒ C = 75° or 135° Also, from Eq. (i), sin (C − A ) = 1 / 2 ⇒ C − A = 30° , 150° For A = 45° , C = 75° [not possible] C = 135° C = 75° A = 45° , B = 60° , C = 75°

and ∴ Hence,

7. LHS = sin 2 α + sin 2 β − sin 2 γ = sin 2 α + (sin 2 β − sin 2 γ ) = sin 2 α + sin (β + γ ) sin (β − γ ) = sin 2 α + sin (π − α )sin (β − γ ) [Qα + β + γ = π] = sin 2 α + sin α sin (β − γ ) = sin α [sin α + sin (β − γ )] = sin α [sin (π − (β + γ )) + sin (β − γ )] = sin α [sin (β + γ ) + sin (β − γ )] = sin α [2 sin β cos γ ] = 2 sin α sin β cos γ = RHS

8. Since, A + B = 180° − C ∴ ⇒

tan( A + B) = tan(180°− C ) tan A + tan B = − tan C 1 − tan A tan B

5 5 3 cos θ + sin θ 2 2 1 5 3 = cos θ + sin θ 2 2 = 3 cos θ −

Q The maximum value of a cos θ + b sin θ is a 2 + b2 1 5 3 So, maximum value of cos θ + sin θ is 2 2 2

2 5 3 1 75  1 =   + + =  =  2 4 4  2 

76 = 19. 4

π 2. As when θ ∈ 0,  , tan θ < cot θ 4



Since, tan θ < 1 and cot θ > 1 ∴ (tan θ )cot θ < 1 and (cot θ )tan θ > 1 ∴ t4 > t1 which only holds in (b). Therefore, (b) is the answer. π 3. Since, A + B = and, we know product of term is 3 maximum, when values are equal. ∴(tan A ⋅ tan B) is maximum. A = B = π /6 π π 1 i.e. y = tan tan = 6 6 3 sin x cos 3x tan x 4. Let y = = sin 3x cos x tan 3x When

⇒ y=

tan x tan x (1 − 3 tan 2 x) = tan 3x 3 tan x − tan3 x =

1 − 3 tan 2 x 3 − tan 2 x

[Q x ≠ 0]

496 Trigonometrical Ratios and Identities Put tan x = t

2. Given a ladder of length l = 2m leans against a vertical

⇒

1 − 3t 2 y= 3 − t2

wall. Now, the top of ladder begins to slide down the wall at the rate 25 cm/s.

⇒

3 y − t 2 y = 1 − 3t 2

Let the rate at which bottom of the ladder slides away dx from the wall on the horizontal ground is cm /s. dt

⇒

3 y − 1 = t 2y − 3t 2

⇒

3 y − 1 = t 2 ( y − 3)

wall l y

–

+

3

1/3

∴ NOTE

x

3y − 1 3y − 1 >0 = t2 ⇒ y−3 y−3

⇒

[by Pythagoras theorem]

It is a brilliant technique to convert equation into inequation and asked in IIT papers frequently. ⇒ y < 1 / 3 or y > 3. This shows that y cannot lie between 1 / 3 and 3.

Topic 4 Height & Distance 1. According to the question, we have the following figure.

⇒

x + y =4 2

From Eq. (i), when y = 1m, then x2 + 1 2 = 4 ⇒ x2 = 3 ⇒ x = 3 m On substituting x = 3m and dx 1  25  =− −  m /s dt 3  100

xm

30°

B

=

P 30 m

[Q x > 0]

y = 1m in Eq. (ii), we get dy   given = − 25 cm /sec   dt

25 cm /s 3

3. Given ABC is a triangular park with AB = AC = 100 m.

45° ym

A

A vertical tower is situated at the mid-point of BC. Let the height of the tower is h m.

Now, let distance of foot of the tower from the point A is y m. Draw BP ⊥ ST such that PT = x m. Then, in ∆TPB, we have x 1 …(i) tan 30° = ⇒ x = y y 3 x + 30 and in ∆TSA, we have tan 45° = y y = x + 30

Now, according to given information, we have the following figure. Q

C h

100

…(ii)

On the elimination of quantity x from Eqs. (i) and (ii), we get 1 y= y + 30 3 ⇒

1  y 1 −  = 30  3

⇒

30 3 30 3 ( 3 + 1) y= = 3 −1 3 −1 =

[Ql = 2m]… (i)

2

On differentiating both sides of Eq. (i) w.r.t. ‘t’, we get dx dy 2x + 2y =0 dt dt dx  y dy … (ii) ⇒ = −   x  dt dt

T

⇒

ground

x2 + y 2 = l 2

Q

t2 > 0

S

Ladder

+

30 3 ( 3 + 1) = 15 (3 + 3 ) 2

P l

β A

α B

1 00

From the figure and given information, we have β = cot−1 (3 2 ) and α = cosec−1 (2 2 ) Now, in ∆QPA, l cot β = h ⇒

l = (3 2 )h

…(i)

Trigonometrical Ratios and Identities 497 and in ∆BPQ, tan α = ⇒ cot α =

h BP

(100)2 − l2 BP = h h [Q p is mid-point of isosceles ∆ABC, AP ⊥ BC]

⇒ h 2 cot2 α = (100)2 − l2 2 ⇒ h (cosec2α − 1) = (100)2 − (3 2h )2 ⇒ h 2(8 − 1) = (100)2 − 18h 2 ⇒ 25h 2 = (100)2

[from Eq. (i)]

2

 100 h2 =   ⇒ h = 20 m  5 

⇒

4. Given heights of two poles are 5 m and 10 m. A

and MN = hm is the height of intersection point from the horizontal plane h 80 [in ∆MNB and ∆PQB] …(i) tan α = = Q x x+ y h 20 and tan β = = y x+ y [in ∆MNQ and ∆ABQ] From Eqs. (i) and (ii), we get y = 4 ⇒ y = 4x x From Eqs. (i) and (iii), we get 80 h 80 = 16 m = ⇒ h= 5 x x + 4x

…(ii) …(iii)

6. According to given information, we have the following figure,

5m 15°

B 10 m

Q Cloud

E

d

15°

C

D

d

xm M 25 m

30° P 60° 25 m

5m F

Surface

y

(25 +x)m

i.e. from figure AC = 10 m, DE = 5 m ∴ AB = AC − DE = 10 − 5 = 5 m Let d be the distance between two poles. Clearly, ∆ABE ~ ∆ACF [by AA- similarity criterion] ∴ ∠AEB = 15° In ∆ABE, we have 3 −1 5  AB 3 − 1 tan 15° = Q tan 15° = ⇒ =   BE 3 + 1 3+1 d  d=

⇒ ⇒

d =5

5( 3 + 1) ( 3 − 1)

3+1 3 + 1 5(3 + 2 3 + 1) × = 3 −1 3 −1 3+1

5(2 3 + 4) 2 × 5( 3 + 2) = = = 5(2 + 3 ) m 2 2

R Image of cloud

x y 25 + (25 + x) tan 60° = y

In ∆PQM ,

tan 30° =

In ∆PRM,

…(i) …(ii)

On eliminating ‘y’ from Eqs. (i) and (ii), we get 25 + (25 + x) 3= 3x ⇒ 3x = 50 + x ⇒ 2x = 50 ⇒ x = 25 m. ∴ Height of cloud from surface = x + 25 = 50 m.

7. According to given information, we have the following figure. E

A

5. Let a first pole AB having height 20 m and second pole c=7

PQ having height 80 m and ∠PBQ = α, ∠AQB = β

D

b=6

P

30° B 80 m A

M

20 m α B

Clearly,

x

h N

where, and

β y

Q

∴

a=5

C

1 2a 2 + 2c2 − b2, 2 (using Appollonius theorem) c = AB = 7, a = BC = 5 b = CA = 6 1 BD = 2 × 25 + 2 × 49 − 36 2 length

of

BD =

498 Trigonometrical Ratios and Identities 1 1 112 = 4 7 = 2 7 2 2 Now, let ED = h be the height of the lamp post. =

E

h 30°

B

D

h BD 1 h = 3 2 7 2 7 2 h= = 21 3 3

Then, in ∆BDE , tan 30° = ⇒ ⇒

P

8.

AB AP h 1 = = 2h 2 h AC 1 Also, in ∆ACP, tan α = = 2 = AP 2h 4 Now, in ∆ABP, tan (α + β ) =

Now, tan β = tan[(α + β ) − α ] tan(α + β ) − tan α = 1 + tan(α + β ) tan α 1 1 1 − 2 2 4 = =4 = 1 1 9 9 1+ × 2 4 8

10. According to given information, we have the following figure Now, from ∆ACD and ∆BCD, we have

45°

200 m

D

200 m

T 30°

Q

Pillar h

90°

30°

R

M

Let height of tower TM be h. TM In ∆PMT , tan 45° = PM h ⇒ ⇒ PM = h 1= PM h ; QM = 3h In ∆TQM, tan 30° = QM In ∆PMQ, PM 2 + QM 2 = PQ 2 h 2 + ( 3h )2 = ( 200)2 ⇒ 4h 2 = ( 200)2 ⇒ h = 100 m

9. Let AB = h, then AP = 2h AC = BC =

and

and ⇒

60º y

C

h x+ y

h y x+ y h= 3

tan 60° =

h= 3 y x+ y From Eqs. (i) and (ii), = 3 y 3

⇒

h/2

B

and

⇒

B

30º x

tan 30° =

⇒

h 2

Again, let ∠CPA = α

x +y = 3y x − 2y = 0 x y= 2

Q Speed is uniform C

h

A

h/2

and distance x covered in 10 min. x ∴ Distance will be cover in 5 min. 2

β α A

P 2h

∴ Distance y will be cover in 5 min.

...(i) ...(ii)

21 Trigonometrical Equations Topic 1 General Solution Objective Questions I (Only one correct option) 1. If the equation cos 4 θ + sin 4 θ + λ = 0 has real solutions for θ, then λ lies in the interval

5 (a)  − , − 1  4  1 1  (c)  − , −  2 4 

(2020 Main, 2 Sep II)

1 (b)  − 1, −   2  3 5  (d) − , −  4   2

(c) [3, 7]

(d) [2, 6]

3. The number of solutions of the equation

(b) 5

(c) 7

(d) 4

4. Let S = {θ ∈ [−2π , 2π ] : 2 cos 2 θ + 3 sin θ = 0}, then the sum of the elements of S is (a) 2π

(b) π

(c)

(2019 Main, 9 April I)

5π 3

(d)

13 π 6

5. If sin 4 α + 4 cos 4 β + 2 = 4 2 sin α cos β; α, β ∈ [0, π ], then cos(α + β ) − cos(α − β ) is equal to (2019 Main, 12 Jan II)

(a) − 1

(b)

2

(c) −

(d) 0

2

6. Let α and β be the roots of the quadratic equation x2 sin θ − x(sin θ cos θ + 1) + cos θ = 0 (0 < θ < 45º ) and ∞  (− 1)n  α < β. Then, ∑ α n +  is equal to βn  n = 0 (2019 Main, 11 Jan II) 1 − 1 − cosθ 1 + 1 (c) − 1 + cosθ 1 −

(a)

1 sin θ 1 sin θ

1 1 + 1 − cosθ 1 + sin θ 1 1 (d) + 1 + cosθ 1 − sin θ

(b)

π 7. The sum of all values of θ ∈ 0,  satisfying 

3 sin 2 2θ + cos 4 2θ = is 4 3π (a) 8

5π (b) 4

(d) 4

in [0, π ] is kπ, then k is equal to 2 (a) 3

13 (b) 9

(2018 Main)

8 (c) 9

20 (d) 9

10. If 5 (tan 2 x − cos 2 x) = 2 cos 2x + 9, then the value of

2

(2017 Main)

3 (a) − 5

1 (b) 3

(2019 Main, 10 Jan I)

(d) π

2 (c) 9

(d) −

7 9

11. If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0, is (2016 Main) (a) 3

(b) 5

(c) 7

(d) 9

(c) 0

(d)

π 12. Let S = x ∈ (− π , π ): x =/ 0, ± . The sum of all 2  distinct solutions of the equation 3 sec x + cosec x + 2(tan x − cot x) = 0 in the set S is equal to (2016 Adv.) (a) −

7π 9

(b) −

2π 9

5π 9

13. If P = {θ :sin θ − cos θ = 2 cos θ } and Q = {θ :sin θ + cos θ = 2 sin θ } be two sets. Then, (a) P ⊂ Q and Q − P ≠ φ (c) P ⊄ Q

value of θ, then (a) b0 = 1, b1 = 3 (c) b0 = − 1, b1 = n

(2011)

(b) Q ⊄ P (d) P = Q

14. Let n be an odd integer. If sin nθ =

n

∑ br sin r θ, for every

r=0

(1998, 2M)

(b) b0 = 0, b1 = n (d) b0 = 0, b1 = n 2 − 3n + 3

15. The general value of θ satisfying the equation 2 sin 2 θ − 3 sin θ − 2 = 0, is π (a) nπ + (−1) 6 n 5π (c) nπ + (−1) 6 n

π (c) 2

(2019 Main, 9 Jan II)

(c) 1

cos 4x is

 5π 5π  is , 1 + sin 4 x = cos 2 3x, x ∈ −  2 2  (2019 Main, 12 April I) (a) 3

(b) 3

 π  π  1 8 cos x ⋅  cos + x ⋅ cos − x −  = 1 6  6  2 

cos 2x + α sin x = 2α − 7 has a solution. Then, S is equal to (2019 Main, 12 April II) (b) [1, 4]

(a) 2

9. If sum of all the solutions of the equation

2. Let S be the set of all α ∈ R such that the equation,

(a) R

π 2 sin x − sin 2x + sin 3x = 0, is

8. If 0 ≤ x < , then the number of values of x for which

(1995,2M)

π (b) nπ + (−1) 2 n 7π (d) nπ + (−1) 6 n

500 Trigonometrical Equations 16. In a ∆ ABC , angle A is greater than angle B. If the measures of angles A and B satisfy the equation 3 sinx − 4 sin3 x − k = 0, 0 < k < 1, then the measure of ∠ C (1990, 2M) is (a)

π 3

(b)

π 2

(c)

2π 3

(d)

5π 6

17. The general solution of sin x − 3 sin 2x + sin 3x = cos x − 3 cos 2x + cos 3x is π 8 π n nπ (c) (−1) + 2 8

(a) nπ +

(b)

nπ π + 2 8

(d) 2nπ + cos−1

(1989, 2M)

23. The number of distinct solutions of the equation 5 cos2 2x + cos4 x + sin4 x + cos6 x + sin6 x = 2 in the 4 interval [0, 2π ] is (2015 Adv.)

Match the Columns 24. Let f (x) = sin(π cos x) and g (x) = cos(2π sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order : X = { x : f (x) = 0}, Y = { x : f ′ (x) = 0} Z = { x : g (x) = 0}, W = { x : g′ (x) = 0}

3 2

18. The general solution of the trigonometric equation sin x + cos x = 1 is given by

(1981, 2M)

(a) x = 2nπ ; n = 0, ± 1, ± 2, ... (b) x = 2 nπ + π / 2 ; n = 0, ± 1, ± 2, .... π π (c) x = nπ + (− 1)n − ; n = 0, ± 1, ± 2 , ... 4 4 (d ) None of the above

List-I contains the sets X , Y , Z and W . List-II contains (2019 Adv.) some information regarding these sets. List-I

π x 19. The equation 2 cos 2   sin 2 x = x2 + x−2, x ≤ has  2

9

(I)

X

(P)

 π 3π  ⊇ , , 4π , 7π  2 2 

(II)

Y

(Q)

an arithmetic progression

(III)

Z

(R)

NOT an arithmetic progression

(IV)

W

(S)

 π 7π 13π  ⊇ , ,  6 6 6 

(T)

  π 2π , π ⊇ , 3 3 

(U)

 π 3π  ⊇ ,  6 4 

(1980, 1M)

(a) no real solution (b) one real solution (c) more than one real solution (d) None of the above

Objective Questions II (One or more than one correct option) 20. Let α and β be non zero real numbers such that 2(cos β − cos α ) + cos α cos β = 1. Then which of the following is/are true? (2017 Adv.) α (a) 3 tan    2 α (b) tan   −  2 α (c) tan   +  2 α (d) 3 tan    2

β − tan   = 0  2 β 3 tan   = 0  2 β 3 tan   = 0  2 β + tan   = 0  2

functions defined for x > 0. Define the following sets whose elements are written in the increasing order : X = { x : f (x) = 0}, Y = { x : f ′ (x) = 0} Z = { x : g (x) = 0}, W = { x : g′ (x) = 0}

satisfying the equation cos 2 θ 1 + sin 2 θ 4 sin 4 θ 2 2 sin θ 1 + cos θ 4 sin 4θ = 0, is sin 2 θ cos 2 θ 1 + 4 sin 4θ (b) 5 π / 24

Which of the following is the only CORRECT combination? (a) (IV), (P), (R), (S) (b) (III), (P), (Q), (U) (c) (III), (R), (U) (d) (IV), (Q), (T)

25. Let f (x) = sin(π cos x) and g (x) = cos(2π sin x) be two

21. The values of θ lying between θ = 0 and θ = π /2 and

(a) 7 π / 24

List-II

(c) 11π / 24

List-I contains the sets X , Y , Z and W . List-II contains some information regarding these sets. (2019 Adv.) List-I

List-II

(I)

X

(P)

 π 3π  ⊇ , , 4π , 7π  2 2  

(II)

Y

(Q)

an arithmetic progression

(III)

Z

(R)

NOT an arithmetic progression

(IV)

W

(S)

 π 7π 13π  ⊇ , ,  6 6 6 

(T)

  π 2π , π ⊇ , 3 3 

(U)

 π 3π  ⊇ ,  6 4 

(1988, 3M)

(d) π / 24

Integer & Numerical Answer Type Questions 22. Let a , b, c be three non-zero real numbers such that the  π π equation 3 a cos x + 2b sin x = c, x ∈ − , , has two  2 2  π distinct real roots α and β with α + β = . Then, the 3 b (2018 Adv.) value of is .................. . a

Trigonometrical Equations 501 Which of the following is the only CORRECT combination? (a) (II), (Q), (T) (b) (II), (R), (S) (c) (I), (P), (R) (d) (I), (Q), (U)

29. If exp {(sin 2 x + sin 4 x + sin 6 x + ... ∞ ) log e 2}, satisfies the equation x2 − 9x + 8 = 0, π cos n x ,0 < x < . 2 cos x + sin x

find

the

value

of

(1991, 4M)

30. Consider the system of linear equations in x, y, z (sin 3 θ ) x − y + z = 0 ,

Fill in the Blank

(cos 2 θ ) x + 4 y + 3z = 0,

26. General value of θ satisfying the equation tan 2 θ + sec 2 θ = 1 is…… .

(1996, 1M)

True/False 27. There exists a value of θ between 0 and 2π that satisfies the equation sin θ − 2 sin θ + 1 = 0. 4

2

(1984, 1M)

2x + 7 y + 7z = 0 Find the values of θ for which this system has (1986, 4M) non-trivial solutions.

31. Find the values of x (− π , π ) which satisfy the equation 21 + | cos x | + | cos

2

x | + ...

=4

(1984, 2M)

32. Find all the solutions of 4cos x sin x − 2 sin x = 3 sin x. 2

Analytical & Descriptive Questions

2

(1983, 2M)

28. Determine the smallest positive value of x ( in degrees) for which tan (x + 100° ) = tan (x + 50° ) tan (x) tan (x − 50° ).

33. Solve 2 (cos x + cos 2x) + (1 + 2 cos x)sin 2x = 2 sin x, − π ≤ x ≤ π

(1978, 3M)

(1993, 5M)

Topic 2 Solving Equations with Graph Objective Question I (Only one correct option) 1. All x satisfying the inequality

(cot− 1 x)2 − 7(cot− 1 x) + 10 > 0, lie in the interval (2019 Main, 11 Jan II)

(a) (− ∞ , cot 5) ∪ (cot 2, ∞ ) (b) (cot 5, cot 4) (c) (cot 2, ∞ ) (d) (− ∞ , cot 5) ∪ (cot 4, cot 2)

4. Let θ, φ ∈ [0, 2π ] be such that 2 cos θ (1 − sin φ ) = sin 2 θ

2. The set of values of θ satisfying the inequation 2 sin 2 θ − 5 sin θ + 2 > 0, where 0 < θ < 2π , is

(2006, 3M)

π 5π (a)  0,  ∪  , 2 π    6  6

π 5π (b)  0,  ∪  , 2π   6   6

π 2π (c)  0,  ∪  , 2π   3   3

(d) None of these

Objective Question II (One or more than one correct option) 3. Let, f (x) =

sin πx , x>0 x2

θ θ   tan + cot  cos φ − 1, tan (2π − θ ) > 0  2 2 3 . Then, φ cannot satisfy and − 1 < sin θ < − 2

(2012)

π 2 π 4π (b) < φ< 2 3 4π 3π (c) < φ< 3 2 3π (d) < φ < 2π 2 (a) 0 < φ
1for every n

(b) xn + 1 − xn > 2 for every n (c) x1 < y1 1 (d) xn ∈  2n , 2n +  for every n  2

(2019 Adv.)

5. Find all values of θ in the interval  − 

π π ,  satisfying 2 2

the equation (1 − tan θ )(1 + tan θ )sec2 θ + 2tan

2

θ

= 0.

(1996, 2M)

502 Trigonometrical Equations

Topic 3 Problems Based on Maximum and Minimum Objective Questions I (Only one correct option) 1. For x ∈ (0, π ), the equation sin x + 2 sin 2x − sin 3x = 3 has (a) infinitely many solutions (b) three solutions (c) one solution (d) no solution

(2014 Adv.)

Integer & Numerical Answer Type Questions

2. The number of solutions of the pair of equations 2 sin θ − cos 2θ = 0 and interval [0, 2π] is

2 cos θ − 3 sin θ = 0

2

(a) 0

2

in

the

(2007, 3M)

(b) 1

(a) xyz = xz + y (b) xyz = xy + z (c) xyz = x + y + z (d) xyz = yz + x

(c) 2

(d) 4

12. Let f : [0, 2] → R be the function defined by π π   f (x) = (3 − sin(2πx))sin  πx −  − sin 3πx +  .   4 4

3. The number of integral values of k for which the

If α , β ∈ [0, 2] are such that { x ∈ [0, 2]: f (x) ≥ 0} = [α , β ], (2020 Adv.) then the value of β − α is ……

(2002, 1M)

13. The positive integer value of n > 3 satisfying the

equation 7 cos x + 5 sin x = 2k + 1 has a solution, is

(a) 4

(b) 8

(c) 10

(d) 12

equation

4. The number of values of x in the interval [0, 5π ] satisfying the equation 3 sin 2 x − 7 sin x + 2 = 0 is (1998, 2M)

(a) 0

(b) 5

(c) 6

(d) 10

5. Number

of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2π ] is (1993, 1M)

(a) 0

(b) 1

(c) 2

6. The

number of sin (ex ) = 5x + 5− x is

(d) 3

solutions

of

the

equation (1991, 2M)

(a) 0 (c) 2

(b) 1 (d) infinitely many

7. The equation (cos p − 1) x + (cos p) x + sin p = 0 in the variable x, has real roots. Then, p can take any value in (1990, 2M) the interval (b) ( − π , 0)

π π (c)  − ,   2 2

(d) (0, π )

8. The smallest positive root of the equation tan x − x = 0 lies in

(1987, 2M)

π (a)  0,   2

3π (d)  , 2 π   2 

3π (c)  π ,   2 

π (b)  , π  2 

9. The number of all possible triplets (a1 , a 2, a3 ) such that a1 + a 2 cos (2x) + a3 sin 2 (x) = 0, ∀ x is (a) 0

(b)1

(1987, 2M)

(d) ∞

(c) 3

Objective Questions II (One or more than one correct option)

11. For 0 < φ < π / 2, if x = z=

∞

∑ cos

n=0

∑ cos

n=0 2n

(1996, 1M)

φ sin

2n

φ , then

φ, y =

Fill in the Blanks 15. The set of all x in the interval [0, π ] for which (1987, 2M)

16. The

solution set of the system of equations 2π 3 x+ y= , cos x + cos y = , where x and y are real, 3 2 is…… . −π π 17. The larger of cos (log θ ) and log (cos θ ) if B and 0 < 3 A < π, 0 < 3B < π.  A − B Therefore, sin 3   ≠0  2  Hence, ⇒ ⇒ ⇒

 2

9

∴ The equation has no real solution.

20. We have, 2(cos β − cos α ) + cos α cos β = 1 or

4(cos β − cos α ) + 2 cos α cos β = 2

⇒ 1 − cos α + cos β − cos α cos β = 3 + 3 cos α − 3 cos β − 3 cos α cos β ⇒ ⇒ ⇒ ∴

(1 − cos α )(1 + cos β ) = 3(1 + cos α )(1 − cos β ) (1 − cos α ) 3(1 − cos β ) = (1 + cos α ) 1 + cos β α β = 3 tan 2 2 2 α β tan ± 3 tan = 0 2 2 tan 2

cos 2 θ 1 + sin 2 θ 4 sin 4 θ 2 2 21. Given, sin θ 1 + cos θ 4 sin 4 θ =0 sin 2 θ cos 2 θ 1 + 4 sin 4 θ Applying R3 → R3 − R1 and R2 → R2 − R1, we get 1 + sin 2 θ cos 2 θ 4 sin 4 θ

 A + B cos 3   =0  2 

−1 −1

 A + B π 3  =  2  2 π A+ B= 3

1 0

0 1

=0

Applying C1 → C1 + C 2, we get 2 0 −1

2π C = π − ( A + B) = 3

17. Given, sin 3x + sin x − 3 sin 2x = cos 3x + cos x − 3 cos 2x ⇒ 2 sin 2x cos x − 3 sin 2x = 2 cos 2x cos x − 3 cos 2x ⇒

sin 2x ( 2 cos x − 3) = cos 2x ( 2 cos x − 3)

⇒

sin 2x = cos 2x

⇒

tan 2x = 1

⇒

π x 19. Given equation is 2 cos 2   sin 2 x = x2 + x−2, x ≤ 1  x LHS = 2 cos 2  sin 2 x < 2 and RHS = x2 + 2 ≥ 2  2 x

( 2 sin θ + 1)(sin θ − 2) = 0

⇒

On dividing and multiplying each terms by 2, we get 1 1 1 sin x + cos x = 2 2 2 π π 1 ⇒ sin x cos = cos x sin = 4 4 2 π   π ⇒ sin  x +  = sin     4 4 π π x + = nπ + (−1)n ⇒ 4 4 π n π ⇒ x = nπ + (−1) − , n ∈I 4 4

[Q 2 cos x − 3 ≠ 0]

π nπ π 2 x = nπ + ⇒ x= + 4 2 8

cos 2 θ 4 sin 4 θ 1 0

0 1

⇒

2 + 4 sin 4 θ = 0 ⇒ sin 4 θ =

⇒

 π 4 θ = nπ + (−1)n  −   6

⇒ Clearly, θ = 0 and π /2.

θ=

=0 −1 2

nπ π + (−1)n+ 1    24 4

7π 11π are two values of θ lying between , 24 24

Trigonometrical Equations 507 22. We have, α , β are the roots of 3 a cos x + 2b sin x = c ∴ 3 a cos α + 2b sin α = c and 3 a cos β + 2b sin β = c On subtracting Eq. (ii) from Eq. (i), we get 3a (cos α − cos β ) + 2b(sin α − sin β ) = 0   α − β  α + β  3 a  − 2 sin  ⇒   sin   2   2   

…(i) …(ii)

  α − β  α + β  + 2b 2 cos   =0  sin   2   2     α + β  α + β 3 a sin    = 2b cos   2   2 

⇒

⇒ MM ⇒ ⇒ 23. Here,

⇒ (iv) → P, R, S Hence, option (a) is correct.

2b  α + β tan   =  2  3a

⇒

2b  π tan   =  6 3a 1 2b = 3 3a b = 0.5 a

3 1 1 3 , − , , are in an A.P. but 4 4 4 4 corresponding values of x are not in an AP so, (iii) → R. For W = { x : g′ (x) = 0}, x > 0 so, g′ (x) = − 2π cos x sin(2π sin x) = 0 ⇒ either cos x = 0 or sin(2π sin x) = 0 π ⇒ either x = (2n + 1) or 2π sin x = nπ , n ∈Integers. 2 Q 2π sin x = nπ n 1 1 ⇒ sin x = = − 1, − , 0, , 1 {Q sin x ∈ [− 1, 1)} 2 2 2 π  π x = nπ, (2n + 1) or nπ + (− 1)n  ±  ∴  6 2 here values of sin x, −

π   Q α + β = , given   3 b 1 ⇒ = a 2

5 cos 2 2x + (cos 4x + sin 4 x) + (cos 6x + sin 6 x) = 2 4

5 cot 2x + [(cos 2 x + sin 2 x)2 − 2 sin 2 x cos 2 x] 4 + (cos 2 x + sin 2 x)[(cos 2 x + sin 2 x)2 − 3 sin 2 x cos 2 x] = 2 5 ⇒ cos 2 2x + (1 − 2 sin 2 x cos 2 x) + (1 − 3 cos 2 x sin 2 x) = 2 4 5 cos 2 2x − 5 sin 2 x cos 2 x = 0 ⇒ 4 5 5 ⇒ cos 2 2x − sin 2 2x = 0 4 4 5 5 5 ⇒ cos 2 2x − + cos 2 2x = 0 4 4 4 1 5 5 2 ⇒ cos 2x = ⇒ cos 2 2x = 2 2 4 ⇒ 2 cos 2 2x = 1 ⇒ 1 + cos 4x = 1 ⇒ cos 4x = 0, as 0 ≤ x ≤ 2π  π 3π 5π 7π 9π 11π 13π 15π  , , , , , , ∴ 4x =  ,  2 2 2  2 2 2 2 2 as 0 ≤ 4x ≤ 8π  π 3π 5π 7π 9π 11π 13π 15π  ⇒ x= , , , , , , ,  8 8 8  8 8 8 8 8 Hence, the total number of solutions is 8. ⇒

24. For Z = { x : g (x) = 0}, x > 0 g (x) = cos(2π sin x) = 0 π ⇒ 2π sin x = (2n + 1) , n ∈Integer 2 2n + 1 sin x = ⇒ 4 3 1 1 3 ⇒ sin x = − , − , , 4 4 4 4

Q

25. For, X = { x : f (x) = 0}, x > 0 Now, f (x) = 0 ⇒ sin(π cos x) = 0, x > 0 ⇒ π cos x = nπ , n ∈ Integer. ⇒ cos x = n {Qcos x ∈ [− 1, 1]} ⇒ cos x = − 1, 0, 1 π ⇒ x = nπ or (2n + 1) , n is an integer. so, (i) → (P), (Q) 2 For, Y = { x : f ′ (x) = 0}, x > 0 Now, f ′ (x) = 0 ⇒ − π sin x cos(π cos x) = 0 ⇒ either sin x = 0 ⇒ x = nπ, n is an integer, or cos(π cos x) = 0 π ⇒ π cos x = (2n + 1) , n is an integer 2 2n + 1 ⇒ cos x = 2 1 ⇒ cos x = ± , {Q cos x ∈ [− 1, 1]} 2 π 2π , n is an integer. x = 2nπ ± or 2nπ ± ⇒ 3 3 So, (ii) → (Q), (T) Hence, option (a) is correct.

26. Given, ⇒ ⇒

tan 2 θ + sec 2θ = 1 tan 2 θ +

tan 2 θ +

1 =1 cos 2θ

1 + tan 2 θ =1 1 − tan 2 θ

⇒

tan 2 θ (1 − tan 2 θ ) + (1 + tan 2 θ ) = 1 − tan 2 θ

⇒

3 tan 2 θ − tan 4 θ = 0

⇒

tan 2 θ (3 − tan 2 θ ) = 0

⇒

tan θ = 0

or

tan θ = ± 3

Now, tan θ = 0 [Q sin x ∈ [− 1, 1]]

⇒

θ = mπ , where m is an integer.

508 Trigonometrical Equations  π tan θ = ± 3 = tan  ±   3 π θ = nπ ± 3

and ⇒ ∴ θ = mπ , nπ ±

π , where m and n are integers. 3

sin 4 θ − 2 sin 2 θ + 1 = 2

27. Given,

⇒ (sin θ − 1) = 2 2

2

⇒ sin θ = ± 2 + 1

⇒

x=nπ

and

⇒

x = nπ

and

Neglecting x = nπ as 0 < x < ⇒

2

which is not possible. Hence, given statement is false.

28. tan (x + 100° ) = tan (x + 50° ) tan x tan(x − 50° ) ⇒

tan(x + 100° ) = tan (x + 50° ) tan(x − 50° ). tan x sin(x + 100° ) cos x sin(x + 50° ) sin(x − 50° ) = ⋅ cos (x + 100° ) sin x cos(x + 50° ) cos (x − 50° )

⇒

sin(2x + 100° ) + sin 100° cos 100° − cos 2x = ⇒ sin(2x + 100° ) − sin 100° cos 100° + cos 2x ⇒

∴

⇒

⇒ sin(2x + 100° ) ⋅ cos 100° + sin(2x + 100° ) ⋅ cos 2x + sin 100° cos 100° + sin 100° cos 2x = cos 100° sin(2x + 100° ) − cos 100° sin 100° − cos 2x sin (2x + 100° ) + cos 2x sin 100°

∴

⇒

⇒

4x = n (180° ) + (− 1)n (− 40° ) − 100° 1 x = [n (180° ) + (− 1)n (− 40° ) − 100° ] 4

⇒

or

= e 1 − sin ⇒

2

tan 2 x

2

x

⋅log e 2

log e 2

=e

satisfies x − 9x + 8 = 0 2

⇒

x = 1, 8 2

2

∴

2tan

=1

and

2tan

⇒

tan 2 x = 0

and

tan 2 x = 3

x

x

=8

sin θ = 0 θ = nπ

...(i)

4 sin 2 θ + 4 sin θ − 3 = 0

∴ From Eqs. (i) and (ii), we get  π θ = nπ or nπ + (−1)n    6 2

x | + |cos 3 x| + ....

= 22

1

29. exp {(sin 2 x + sin 4 x + sin 6 x + ... ∞ ) log e 2} sin 2 x

sin θ (4 sin 2 θ + 4 sin θ − 3) = 0

31. Given, 2 1+ | cos x | + |cos

1 Therefore, x = (180° + 40° − 100° ) 4 1 x = (120° ) = 30° ⇒ 4

3 =0 7

⇒ ( 2 sin θ − 1) ( 2 sin θ + 3) = 0 1 3 [Qsin θ = − is not possible] sin θ = ⇒ 2 2 π   ...(ii) ∴ θ = nπ + (−1)n    6

The smallest positive value of x is obtained when n = 1.

4 7

sin 3 θ + 2 cos 2 θ − 2 = 0

sin(4x + 100° ) + 2 sin 150° cos 50° = 0 1 sin(4x + 100° ) + 2 ⋅ sin(90° − 50° ) = 0 2

4x + 100° = nπ + (− 1)n (− 40° )

cos 2 θ 2

⇒ 3 sin θ − 4 sin3 θ + 2 (1 − 2 sin 2 θ ) − 2 = 0

⇒

sin(4x + 100° ) = sin(− 40° )

−1 1

⇒

⇒

⇒

sin 3 θ

+ 1 (7 cos 2 θ − 8) = 0

sin(4x + 100° ) + sin 100° + sin 200° = 0

⇒

cos x 3 −1 = cos x + sin x 2

7 sin 3 θ + 14 cos 2 θ − 14 = 0

⇒

sin(4x + 100° ) + sin 40° = 0

π  π ∈ 0,  3  2 1 1 3 −1 cos x 2 = × = cos x + sin x 1 3 −1 3 1+ 3 + 2 2 x=

⇒

⇒

⇒

π 2

⇒ sin 3 θ (28 − 21) + 1 (7 cos 2 θ − 6)

⇒ 2 sin(2x + 100° ) cos 2x + 2 sin 100° cos 100° = 0

⇒

2

30. Since, the given system has non-trivial solution.

[sin(2x + 100° ) + sin 100° ] [cos 100° + cos 2x] = [cos 100° − cos 2x] × [sin(2x + 100° ) − sin 100° ]

π  3 π x = nπ ± 3

 tan 2 x =  tan 

sin 2 x

⇒

21 −|cos x| = 22

⇒

1 =2 1 − |cos x|

⇒

|cos x| =

cos 2 x

⇒

1 2

cos x = ±

1 2

π 2π π 2π , ,− ,− 3 3 3 3 π 2 π   Thus, the solution set is ± , ± . 3   3

∴

x=

[Q x ∈ (−π , π )]

Trigonometrical Equations 509 32. Given, 4 cos 2 x sin x − 2 sin 2 x = 3 sin x

Topic 2 Solving Equations with Graph

⇒ 4 (1 − sin 2 x) sin x − 2 sin 2 x − 3 sin x = 0

1. Given, (cot−1 x)2 − 7(cot−1 x) + 10 > 0

⇒ 4 sin x − 4 sin x − 2 sin x − 3 sin x = 0 3

2

⇒

− 4 sin3 x − 2 sin 2 x + sin x = 0

⇒

− sin x (4 sin 2 x + 2 sin x − 1) = 0

⇒ sin x = 0

⇒

x = nπ

or

sin x =

⇒ x = nπ or sin x = sin or

+ – + cot−1 x =2 cot−1 x =5

−2 ± 4 + 16 sin x = 2 (4)

or

∴ cot−1 x ∈ (−∞ , 2) ∪ (5, ∞ ) cot−1 x ∈ (0, 2) [Q Range of cot−1 x is (0, π )] ∴ x ∈ (cot 2, ∞ )

−1 ± 5 4

π 10

2. Since, 2 sin 2 θ − 5 sin θ + 2 > 0

 3π  sin x = sin  −   10 

⇒ x = nπ , nπ + (−1)n

(by factorisation)

By wavy curve method,

4 sin 2 x + 2 sin x − 1 = 0

or

⇒ sin x = sin 0

⇒ (cot−1 x − 2)(cot−1 x − 5) > 0 ⇒ cot−1 x < 2 or cot−1 x > 5

⇒

( 2 sin θ − 1) (sin θ − 2) > 0

∴

( 2 sin θ − 1) < 0

[where, (sin θ − 2) < 0, ∀ θ ∈ R]

π  − 3π  , nπ + (−1)n    10  10

Y

∴ General solution set is

y=1 2

π  { x : x = nπ} ∪ x : x = nπ + (−1)n  10  

X'

  −3 π   ∪ x : x = nπ + (−1)n    10   

∴

x = ( 2 n + 1)π

...(i)

...(ii)

sin θ
0 x2 x2π cos(πx) − 2x sin(πx) ⇒ f ′ (x) = x4  xπ  2x cos(πx) − tan(πx)  2  = 4 x  xπ  2 cos(πx) − tan(πx)  2  = 3 x Since, for maxima and minima of f (x), f ′ (x) = 0 πx ⇒ cos(πx) = 0 or tan(πx) = , (as x > 0) 2 πx Q cos(πx) ≠ 0 ⇒ tan(πx) = 2 y=tan (px)

Y

...(iii)

But given interval is [−π , π ] . π Put n = −1 in Eq. (i), x = − 2 π π ,π − ,− π 3 3 π π π Hence, the solution in [− π , π ] are − π , − , − , , π. 2 3 3

Put n = 0, 1, − 1, − 2 in Eq. (ii), x =

X

2π

3. (a, b, d) Given, f (x) =

1 + sin x = 0 3x x cos = 0 or cos = 0 2 2

If 1 + sin x = 0, then sin x = −1 3π x = 2 nπ + ∴ 2 3x 3x π If cos = 0, then = (2n + 1) 2 2 2 π x = (2 n + 1) ∴ 3 x x π And if cos = 0, then = ( 2 n + 1) 2 2 2

π

1 2  π  5π  ∴ From the graph, θ ∈ 0,  ∪  , 2π  6  6 

2 cos x + 2 cos 2x + sin 2x + sin 3x + sin x − 2 sin x = 0 ∴ 2 cos x + 2 cos 2x + 2 sin x cos x + (sin 3x − sin x) = 0 ⇒ 2 cos x + 2 cos 2x + 2 sin x cos x + 2 cos 2x sin x = 0 ⇒ 2 cos x(1 + sin x) + 2 cos 2x(1 + sin x) = 0 ⇒ 2 (1 + sin x)(cos x + cos 2x) = 0 x  3x ⇒ 4 (1 + sin x) cos   cos = 0  2 2

or

5π 6

Y'

⇒

33. Given that,

∴

O π 6

y=px 2

O

1 1 2

P1

3 2

2 P2

5 2

3 P3

7 2

X 4

9 2

510 Trigonometrical Equations  3 Q f ′ (P1− ) < 0 and f ′ (P1+ ) > 0 ⇒ x = P1 ∈ 1,  is point of  2

From Eq. (i),

local minimum.

⇒

Q f′

(P2− )

> 0 and f ′

(P2+ )

 5 < 0 ⇒ x = P2 ∈ 2,  is point of  2

local maximum. From the graph, for points of maxima x1 , x2, x3 …… it is clear that 5 9 13 17 − x1 > − x2 > − x3 > − x4 ...... 2 2 2 2 ⇒ xn + 1 − xn > 2, ∀ n. From the graph for points of minima y1 , y2, y3 ....., it is clear that 3 5 7 9 − y1 > − x1 > − y2 > − x2 ...... 2 2 2 2 |xn − yn|> 1, ∀ n and x1 > ( y1 + 1)  5  9  13 And x1 ∈ 2,  , x2 ∈ 4,  , x3 ∈ 6,  ........  2  2  2 1  ⇒ xn ∈ 2n , 2n +  , ∀ n.  2

∴

1 < 2 sin (θ + φ ) < 2 1 < sin (θ + φ ) < 1 2 π 5π 0

π

=0

2

tan θ = x

∴

As θ , φ ∈ [0, 2π ] and

3 tan (2π − θ ) > 0, − 1 < sin θ < − 2

O

θ

2

Put

[− 1,1] and the internal for a < x < b.

1

2

5. Given, (1 − tan θ )(1 + tan θ ) sec2 θ + 2tan

4. PLAN It is based on range of sin x, i.e.

Also,

…(iii)

2π   3π  3π 5π   2π 7π  ⇒ φ ∈− , −  or  ,  ,  , as θ ∈   2  2 3  3 6 3

Hence, options (a), (b) and (d) are correct.

⇒

2 cos θ + 1 ∈ (1, 2)

⇒

Description of Situation

3π 5π 0  2 2 2 ∴ f (x) = 0 has at least one root between π and

3π . 2

512 Trigonometrical Equations a1 + a 2 cos 2x + a3 sin 2 x = 0, ∀ x

9. Given,

Therefore, (b) is the answer from Eq. (i). [putting the value of xy]

 1 − cos 2x ⇒ a1 + a 2 cos 2x + a3   = 0, ∀ x   2

⇒

a   a    a1 + 3  +  a 2 − 3  cos 2x = 0, ∀ x  2  2 a3 a ⇒ a1 + = 0 and a 2 − 3 = 0 2 2 k k ⇒ a1 = − , a 2 = , a3 = k, where k ∈ R 2 2  k k  Hence, the solutions, are  − , , k , where k is any  2 2  real number. Thus, the number of triplets is infinite. ⇒

10. We know that, sec2 θ ≥ 1 4xy ≥1 (x + y)2

⇒ ⇒

4xy ≥ (x + y)2

⇒

(x + y)2 − 4xy ≤ 0

⇒

(x − y)2 ≤ 0

⇒

x− y=0 ⇒ x= y

Therefore, x + y = 2x

[add x both sides]

But x + y ≠ 0 since it lies in the denominator, ⇒

2x ≠ 0 ⇒ x ≠ 0

Hence, x = y, x ≠ 0 is the answer. Therefore, (a) and (b) are the answers.

12. The given function f : [0, 2] → R defined by π π   f (x) = (3 − sin(2πx))sin  πx −  − sin 3πx +    4 4 sin πx cos πx  sin 3πx cos (3πx)  = (3 − sin(2πx)) − − +   2 2   2 2 

[sin(πx) − cos(πx)] 1 − 2 2 [3 sin(πx) − 4 sin3 (πx) + 4 cos3 (πx) − 3 cos(πx)] sin(πx) − cos(πx) = [3 − sin(2πx) − 3 2 + 4 {sin 2(πx) + cos 2(πx) + sin(πx) cos(πx)}] sin(πx) − cos(πx) = [4 + sin(2πx)] 2 As, f (x) ≥ 0 ∀ ∈ [α , β ] , where α , β ∈ [0, 2] , so sin(π x) − cos(π x) ≥ 0 as 4 + sin(2π x) > 0 ∀x ∈ R.  π 5π  ⇒ π x∈ ,  4 4  1 5  x∈ , ⇒ 4 4  1 5 and β = ∴ α= 4 4 Therefore the value of (β − α ) = 1 = (3 − sin(2πx))

13. Given, n > 3 ∈ Integer

11. For 0 < φ < / π / 2, we have ∞

and

∑ cos2n φ = 1 + cos2 φ + cos4 φ + cos6 φ + K

x=

n=0

It is clearly a GP with common ratio of cos 2 φ which is ≤ 1.   1 1 a Hence, x = = Q S∞ = , − 1 < r < 1  2 2 1−r 1 − cos φ sin φ   Similarly, y = z=

and

Now, x + y = = Again, ⇒ ⇒

⇒

⇒

1 cos 2 φ 1 1 − sin 2 φ cos 2 φ

⇒

1 1 + sin 2 φ cos 2 φ 1 cos 2 φ + sin 2 φ = 2 2 2 cos φ sin φ cos φ sin 2 φ

⇒ ⇒

1 1 = 1 − sin 2 φ cos 2 φ = 1 − z xy 1 xy − 1 = ⇒ xy = xyz − z z xy

xy + z = xyz

xyz = x + y + z

Therefore, (c) is also the answer.

⇒ …(i)

⇒

1 1 1 + =  3π   2π   π sin   sin   sin    n  n  n 1

1 1 − = π 3π 2π sin sin sin n n n 3π π − sin sin n n = 1 π 3π 2π sin ⋅ sin sin n n n π 3π sin ⋅ sin π  2π  n n 2 cos   ⋅ sin = 2π  n n sin n 2π 2π 3π 2 sin ⋅ cos = sin n n n 4π 3π sin = sin n n 4π 3π =π− n n 7π =π ⇒ n=7 n

Trigonometrical Equations 513 14. Given, tan θ = cot 5 θ

18. Given, cos ( p sin x) = sin ( p cos x) , ∀ x ∈ [0, 2π ]

π  tan θ = tan  − 5θ 2 

⇒

π − 5 θ = nπ + θ 2 π 6 θ = − nπ 2 π nπ θ= − 12 6  π cos 4 θ = sin 2 θ = cos  − 2 θ  2

⇒ ⇒ ⇒ Also,

or

p sin x − p cos x = 2nπ − π / 2, n ∈ I

⇒

p (sin x + cos x) = 2nπ + π / 2

or ⇒ or

π 2 nπ π θ= + 3 12

⇒

6 θ = 2 nπ +

or

π 2

⇒ θ = nπ −

π 4

Above values of θ suggest that there are only 3 common solutions.

15. Given, ⇒ ⇒ ⇒ ⇒

(2 sin x − 1) (sin x − 1) ≥ 0 2 sin x − 1 ≤ 0 or sin x ≥ 1 1 or sin x = 1 sin x ≤ 2  π  π  5π  x ∈ 0, ∪ ∪ ,π  6   2   6 

2π 3 3 and cos x + cos y = 2 2 π 3   − x = ⇒ cos x + cos   2  3

16. Given,

x+ y=

 1  3 3 ⇒ cos x +  − cos x + sin x = 2  2  2 1 3 3 ⇒ cos x + sin x = 2 2 2 π  3 ⇒ sin  + x = , which is never possible. 6  2 Hence, no solution exists.

17. Since,

⇒

− 1 ≤ sin (x ± π / 4) ≤ 1 − p 2 ≤ p 2 sin (x ± π / 4) ≤ p 2 (4n + 1) ⋅ π −p 2≤ ≤ p 2, n ∈I 2 (4n − 1) π −p 2≤ ≤ p 2, n ∈I 2

⇒ or

2 sin 2 x − 3 sin x + 1 ≥ 0

2 sin 2 x − 2 sin x − sin x + 1 ≥ 0

⇒

p (sin x − cos x) = 2nπ − π / 2, n ∈ I π π π p 2 (cos sin x + sin cos x) = 2nπ + 4 4 2 π π π   p 2  cos sin x − sin cos x = 2nπ − , n ∈ I   4 4 2 (4n + 1)π p 2 [sin (x + π / 4)] = 2 π p 2 [sin (x − π / 4)] = (4n − 1) , n ∈ I 2

Now,

Taking negative sign, 2 θ = 2 nπ −

[Q cos θ = cos α ⇒ θ = 2nπ ± α , n ∈ I ] p sin x + p cos x = 2nπ + π / 2

⇒

Taking positive sign,

⇒

 π p sin x = 2nπ ±  − p cos x , n ∈ I  2

⇒

 π 4 θ = 2 nπ ±  − 2 θ  2

⇒

π  cos ( p sin x) = cos  − p cos x 2 

⇒

Second inequality is always a subset of first, therefore we have to consider only first. It is sufficient to consider n ≥ 0, because for n > 0, the solution will be same for n ≥ 0. If n ≥ 0, − 2 p ≤ (4n + 1) π / 2 ⇒

(4n + 1) π / 2 ≤ 2 p

For p to be least, n should be least. ⇒

n =0

⇒

2 p ≥ π /2 ⇒

Therefore, least value of p =

and

cos (log θ ) > 0 cos (log θ ) > log (cos θ )

1 =4 e sin x

e sin x −

⇒

(e sin x )2 − 4 (e sin x ) − 1 = 0

⇒

e sin x =

π 2 2

π 2 2

19. Given,

4 ± 16 + 4 =2 ± 5 2

But since, e ~ 2 .72 and we know, 0 < e sin x < e ∴ e sin x = 2 ± 5 is not possible. Hence, it does not exist any solution.

20. The point of intersection is given by π  sin 3x = cos x = sin  − x 2 

cos θ ≤ 1 ⇒ log (cos θ ) < 0

∴

p≥

⇒

 π 3x = nπ + (−1)n  − x  2

514 Trigonometrical Equations (i) Let n be even i.e. n = 2 m ⇒ ⇒

π 3x = 2 mπ + − x 2 mπ π n= + 2 8

⇒ ⇒ Now, ⇒

cos θ + sin θ
sin (cos θ )

π 22. Let f (θ ) = 5 cos θ + 3 cos θ +  + 3 

π 2

π   as, 2 = 1.4141, = 1.57 (approx) 2  

3

π π  = 5 cos θ + 3  cos θ cos − sin θ sin  + 3  3 3  3  1 = 5 cos θ + 3   cos θ − 3   sin θ + 3  2  2 13 3 3 cos θ − sin θ + 3 2 2 1 f (θ ) = (13 cos θ − 3 3 sin θ ) + 3 2 =

⇒

Put r cos α = 13, r sin α = 3 3 , then ∴

 π = 2 sin  + θ  4 ⇒ cos θ + sin θ ≤ 2
 , then x is equal to  3x  4x 2  4

5. If cos −1 

12. Let (x, y) be such that sin −1 (ax) + cos −1 ( y) + cos −1 (bxy) =

(2019 Main, 9 Jan I)

145 10

(a)

146 12

(b)

145 12

(c)

145 11

(d)

Column I

6 4 6. If α = 3sin −1   and β = 3 cos −1   , where the inverse    

11 9 trigonometric functions take only the principal values, then the correct option(s) is/are (2015 Adv.) (b) sin β < 0 (d) cos α < 0

(a) cos β > 0 (c) cos (α + β ) > 0

+ sin (cot−1 x)}2 − 1]1/ 2 is equal to x 1 + x2

p.

lies on the circle x 2 + y 2 = 1

B. If a = 1 and b = 1, then (x, y)

q.

lies on ( x 2 − 1)( y 2 − 1) = 0

C. If a = 1 and b = 2, then (x, y)

r.

lies on y = x

D. If a = 2 and b = 2, then (x, y)

s.

lies on ( 4x 2 − 1)( y 2 − 1) = 0

x > 0} and x−1     x  E 2 = x ∈ E1 : sin − 1  log e    is a real number   x − 1    

(d) 1 + x2

(Here, the inverse trigonometric function sin − 1 x  π π assumes values in − , ).Let f : E1 → R be the  2 2 

8. The value of x for which sin [cot−1 (1 + x)] = cos (tan −1 x) is

(2004, 1M)

1 (a) 2

(b) 1



9. If sin −1  x −

1 (d) − 2

(c) 0 3

4

6

 π = , for 0 0 ∀ x ∈R x2 + 1 x +1

Here, A = 2 tan −1 (2 2 − 1)

⇒ |x| + 5 ≤ x2 + 1 as x2 + 1 is positive ∀ x ∈ R ⇒ x2 − |x| − 4 ≥ 0 ⇒ |x|2 − |x| − 4 ≥ 0 [Q ∀ x ∈ R, x2 = |x|2 ] 2 Q For|x| − |x| − 4 = 0 1 ± 1 + 16 |x| = ⇒ 2 1 ± 17 1 + 17 as|x| ≥ 0 ⇒ |x| = ⇒|x| = 2 2 17 + 1 ∴ |x| ≥ 2  17 + 1   17 + 1  ⇒ x ∈  − ∞, − , ∞ ∪ 2 2     17 + 1 ∴ a= 2

∴ …(i)

x(x + 1) + sin −1

x2 + x + 1 =

π 2

Function is defined, if (i) x (x + 1) ≥ 0, since domain of square root function. (ii) x2 + x + 1 ≥ 0, since domain of square root function. 2 −1 (iii) x + x + 1 ≤ 1, since domain of sin function.

From (ii) and (iii), 0 ≤ x + x + 1 ≤ 1 ∩ x + x ≥ 0 2

⇒

2

0 ≤ x + x+ 1 ≤1∩x + x+ 1 ≥1 2

⇒

x2 + x + 1 = 1

⇒

x + x=0

⇒

x (x + 1) = 0

⇒

x = 0, x = − 1

2

2

To find the value of B, we first say 1 1 π sin −1 < sin −1 = 3 2 6 1 π so that 0 < 3 sin −1 < 3 2 1 1  23  1 Now, 3 sin −1 = sin −1 3 ⋅ − 4 ⋅  = sin −1    27  3 3 27  3 π = sin −1 (0.851) < sin −1   =  2 3

2. Given function is tan −1

= 2 tan −1 (2 × 1.414 − 1) = 2 tan −1 (1.828) π 2π A > 2 tan −1 ( 3 ) = 2 ⋅ = 3 3

 3 π  3 sin −1   = sin −1 (0.6) < sin −1   =  5  2 3 π π 2π B< + = ∴ 3 3 3 2π 2π Thus, A > and B < 3 3 Hence, greater angle is A.

Topic 2 Properties of Inverse Functions 1. We have, 4 5 16  + sin −1  2π − sin −1 + sin −1  5 13 65 4 5 16  = 2π −  tan −1 + tan −1 + tan −1   3 12 63 4 5   +   −1 3 −1 16 12 = 2π −  tan + tan  5 63  1−   9

Inverse Circular Functions 519   21 × 3 −1 16 = 2π −  tan −1    + tan  4 ×4  63 

 π  (n + 1) cos   +  n + 2

63 16  = 2π −  tan −1 + tan −1   16 63

=

16 16  = 2π −  cot−1 + tan −1   63 63

(n + 1) +

π 3π = 2π − = 2 2 Hence, option (c) is correct.

=

[Q cos(α ) + cos(α + β) + cos(α + 2 β ) + ...   nβ  sin    2 2 1 α ( n ) β + −   cos  + cos(α + (n − 1)β ) =    2  β sin    2   π  sin  π −   n + 2  π   π  (n + 1) cos  cos  π +  −   n + 2   π  n + 2 sin    n + 2  π  sin  π −   n + 2 (n + 1) − cos(π )  π  sin    n + 2

 π  sin    n + 2

 π  sin    n + 2  π  (n + 2 ) cos    n + 2  π  = = cos    n + 2 (n + 2 )

2. It is given, that for non-negative integers ‘n’, n k+1  k+2  Σ sin  π  sin  π k=0 n + 2  n + 2  f (n ) = n k+1  π Σ sin 2 k=0 n + 2  n   2k + 3   π π  Σ  cos − cos  k=0  n + 2  n+2 = n   2k + 2   π  Σ 1 − cos  k=0  n + 2  [Q2 sin A sin B = cos( A − B) − cos( A + B) and 2 sin 2 A = 1 − cos 2 A] 3π 5π 7π  cos + cos + cos    π  n n+2 n+2 n+2  cos   Σ 1 −   2n + 3  n + 2  k = 0   + ..... + cos   n+2  = 2π 4π 6π   + cos + cos + cos n   n+2 n+2 n+2 Σ 1−    + n 2 2 k=0  ..... + cos  π   n + 2     nπ  sin    n + 2 n + 3   π  (n + 1) cos  cos  π  − n + 2   n + 2  π  sin    n + 2 =  nπ  sin    n + 2 n + 2  cos  π (n + 1) − n + 2   π  sin    n + 2

 π  sin    n + 2  π  cos    n + 2  π  sin    n + 2

⇒

 π  f (n ) = cos    n + 2

 π Now, f (6) = cos    8

   π   

Q cos −1cos x = x   π  Qα = tan (cos −1 f ((6))) = tan     π   8 if x 0 ∈ ,     2   = 2 −1 ⇒ (α + 1) = 2 ⇒ (α + 1)2 = 2 ⇒ α 2 + 2α + 1 = 2 ⇒ α 2 + 2α − 1 = 0  π  3  π Now, f (4) = cos  ,  = cos   =    4 + 2 6 2 Now,

   π  sin (7 cos −1 f (5 )) = sin  7 cos −1  cos    5 + 2       π = sin  7   = sin π = 0   7  and Now, lim f (x) = lim cos n→ ∞

n→ ∞

π = cos 0 = 1 n+2

Hence, options (a), (b) and (c) are correct.

3. Given functions, f (x) = log e (sin x), (0 < x < π ) and g (x) = sin − 1 (e− x ), x ≥ 0. Now, fog (x) = f ( g (x)) = f (sin − 1 (e− x )) = log e (sin(sin − 1 (e− x ))) = log e (e− x )

{Qsin(sin − 1 x) = x, if x ∈ [− 1, 1]}

=−x

…(i) d …(ii) and ( fog )′ (x) = (− x) = − 1 dx According to the question, [from Eq. (ii)] Q a = ( fog )′ (α ) = − 1 and b = ( fog ) (α ) = − (α ) [from Eq. (i)] for a positive real value ‘α’. Since, the value of a = − 1 and b = − α, satisfy the quadratic equation (from the given options) aα 2 − bα − a = 1 .

520 Inverse Circular Functions 

19





n

⇒

x2(144x2 − 145) = 0 145 145 =± ⇒ x = 0 or x = ± 144 12 3, 145 But x > x= 4 12 6 1 6 4 6. Here, α = 3 sin −1   and β = 3 cos −1   as >  11  9 11 2

4. Consider, cot  ∑ cot−1 1 + Σ 2 p    p =1

 n =1

  19 = cot  ∑ cot−1(1 + n (n + 1))     n =1



 19 n (n + 1)   Q ∑ p = 2   n = 1

  19 = cot  ∑ cot−1 (1 + n + n 2)     n =1   19 1  = cot  ∑ tan −1  1 + n (n + 1)  n =1

⇒

[ Q cot−1 x = tan −1  19  (n + 1) − n   = cot  ∑ tan −1     1 + n (n + 1)   n =1

1 , if x > 0 ] x

[put 1 = (n + 1) − n]

n =1

∴

= cot [(tan

−1

−1

−1

 y 

−1

2 − tan 1) + (tan 3 − tan 2) + ......+ (tan −1 20 − tan − 1 19)] −1 = cot (tan 20 − tan −1 1)  π   π = cot  − cot−1 20 −  − cot− 1 1    2  2 −1

−1

∴

7. We have, Let

2  3 π We have, cos   + cos −1   =  3x  4x 2  2 3 9  π 4 ⇒ cos −1  ⋅ − 1− 2 1− =  3x 4x 16x2  2 9x −1 

[Q cos −1 x + cos −1 y = cos −1 (xy − 1 − x2 1 − y2 )]

⇒

 1 9x2 − 4 16x2 − 9  π cos −1  2 − =  2x  2 12x2   6 − 9x2 − 4 16x2 − 9 12x2

√1 + y2 θ

= cos

9x2 − 4 16x2 − 9 = 6

On squaring both sides, ⇒ (9x2 − 4)(16x2 − 9) = 36 4 ⇒ 144x − 81x2 − 64x2 + 36 = 36 ⇒ 144x4 − 145x2 = 0

π =0 2

x

B

⇒

cot θ = x

⇒

sin θ =

and

cos θ =

Key Idea Use the formula, cos −1 x + cos −1 y = cos −1( xy − 1 − x 2 1 − y 2 )

⇒

0 < x 0

1

[Q cot ( A − B) =

⇒

 4 β = 3 cos −1   > π  9

Now, α + β is slightly greater than

= cot (cot 1 − cot 20) cot (cot−1 1) cot (cot−1 20) + 1 = cot (cot−1 20) − cot (cot−1 1)

5.

4 1  4  1 π < ⇒ cos −1   > cos −1   =  9  2 3 9 2

∴ cos β < 0 and sin β < 0

[Q tan −1 x + cot−1 x = π / 2 ]

(1 × 20) + 1 21 = = 20 − 1 19

 4 β = 3 cos −1    9

Now, As

 −1 x − y −1 −1 Q tan 1 + xy = tan x − tan

6 π α = 3 sin −1   > ⇒ cos α < 0  11 2

∴

19

= cot ∑ (tan −1 (n + 1) − tan −1 n )

 1 π 6 sin −1   > sin −1   =  2 6  11

Now,

1 1+ x

2

x 1 + x2

A

= sin (cot−1 x) = cos (cot−1 x)

1 + x2 [{ x cos (cot−1 x) + sin (cot−1 x)}2 − 1]1/ 2 1/ 2

2   1  x − 1 = 1 + x  x +   1 + x2 1 + x2    1/ 2 2   1 + x2  − 1 = 1 + x2     1 + x2    2

= 1 + x2 [1 + x2 − 1]1/ 2 = x 1 + x2

8. Given, sin [cot−1 (1 + x)] = cos (tan −1 x)

… (i)

and we know that,   1  1  and tan −1 θ = cos −1  cot−1 θ = sin −1   1 + θ2   1 + θ2     

Inverse Circular Functions 521 From Eq. (i),    1 1   = cos  cos −1 sin sin −1   2 1 + (1 + x)  1 + x2    1

⇒

1 + (1 + x)2

⇒

=

1

π 2 Therefore, α should be equal in both functions.

9. We know that, sin −1 (α ) + cos −1 (α ) =

∴ ⇒ ⇒ ⇒ ⇒ ⇒

x2 x3 x4 x6 + − K = x2 − + −K 2 4 2 4 x x2 x x2 2x 2 x2 = = ⇒ = ⇒ 2 2 2+ x 2+ x x 2 + x 2 + x2 x 1+ 1+ 2 2 2 2 2 2 2x (2 + x ) = 2x (2 + x) 4x + 2x3 = 4x2 + 2x3 x (4 + 2x2 − 4x − 2x2) = 0 Either x = 0 or 4 − 4x = 0 x−

⇒

x=0

Q

0 < | x|< 2

∴

x=1

10. sin −1 sin 

and x ≠ 0

√1 + y2 θ

1

1/ 2

  1  cos (tan −1 y) + y sin(tan −1 y)  4  2⋅  + y  −1 −1   y  cot (sin y) + tan(sin y)  2

1/ 2

2      1 + y2  4  + y   y   2  1− y    1/ 2

1  =  2 ⋅ y2(1 − y4 ) + y4  y 

=1

y 1  =  2

 π  π  R. cos 2x ⋅  cos  − x − cos  + x  + 2 sin 2 x 4  4   = 2 sin x ⋅ cos x 2 ⇒ cos 2x ⋅ ( 2 sin x) + 2 sin x = 2 sin x ⋅ cos x ⇒

2 sin x [cos 2x + 2 sin x − 2 cos x] = 0

⇒ sin x = 0, (cos x − sin x) (cos x + sin x − 2 ) = 0 ⇒ sec x = 1 or tan x = 1 ⇒ sec x = 1 or 2 S. cot (sin −1 1 − x2 ) = sin(tan −1 (x 6 )) ⇒

⇒

x 1 − x2

=

x 6 1 + 6x2

1 + 6 x2 = 6 − 6 x2 12x2 = 5 ⇒

x=

5 5 = 12 2 3

(P) → 4, (Q) → 3, (R) → 2 or 4, (S) → 1

⇒ xy + 1 − x2 1 − y2 = xy ⇒ (x2 − 1)( y2 − 1) = 0

y = θ ⇒ tan θ = y

y2

y 1 x−  = ⇒ cos   2  2

⇒ sin −1 x = − cos −1 y ⇒ x2 + y2 = 1 B. If a = 1 and b = 1, then π sin −1 x + cos −1 y + cos −1 xy = 2 ⇒ cos −1 x − cos −1 y = cos −1 xy

y

  1   +  1  1 + y2 = 2  y  1 − y2 +   y   

x− ⇒ 2 cos 2   2

12. A. If a = 1, b = 0, then sin −1 x + cos −1 y = 0

11. P. Here, innermost function is inverse.

∴ Put tan

cos 2 x + sin 2 x + cos 2 y + sin 2 y + 2 cos x cos y

⇒

2

sin x + sin y = − sin z

On squaring and adding, we get

x=1

or

π π 2π  π   −1  −1   = sin sin  π −   = sin sin  =   3  3 3 3 

−1

and

cos x + cos y = − cos z

+ 2 sin x sin y = 1 1 ⇒ 2 + 2 [cos(x − y)] = 1 ⇒ cos (x − y) = − 2 1  x − y ⇒ 2 cos 2   −1 = −  2  2

1 + x2

1 + x2 + 2 x + 1 = x 2 + 1 1 x=− 2

⇒

Q. Given,

C. If a = 1, b = 2 , then

π 2 y = cos −1 (2xy)

sin −1 x + cos −1 y + cos −1 (2 xy) = ⇒ ⇒

cos −1 x – cos −1

xy + 1 − x2 1 − y2 = 2xy ⇒ x2 + y2 = 1

D. If a = 2 and b = 2, then

π 2 ( y) = cos −1 (2xy)

sin −1 (2x) + cos −1 ( y) + cos −1 (2xy) = ⇒ ⇒ ⇒

cos −1 (2x) − cos −1

2xy + 1 − 4x2 1 − y2 = 2xy (4x2 − 1) ( y2 − 1) = 0

Hence, A → p; B → q; C → p; D → s

522 Inverse Circular Functions 13. We have,

∴

  x E1 = x ∈ R : x ≠ 1 and > 0 x−1   x E1 = >0 x−1 +

–

+

0

1

E1 = x ∈ (− ∞ , 0) ∪ (1, ∞ ) and     x  E 2 = x ∈ E1 : sin −1  log e    is a real number    x − 1     x x E 2 = − 1 ≤ log e ≤ 1 ⇒ e−1 ≤ ≤e x−1 x−1 x x 1 Now, ≥ e−1 ⇒ − ≥0 x−1 x−1 e ex − x + 1 x (e − 1) + 1 ⇒ ≥0⇒ ≥0 e(x − 1) (x − 1) e +

–

⇒

x ≤e x−1 (e − 1)x − e ≥0 x−1 +

– 1

⇒

So,

 7π 2π   7π π  + tan  + +   − tan   12  12 2  2 . . .

1

 1  x ∈  − ∞, ∪ (1, ∞ )  1 − e 

Also,

π  7π (k + 1) π  7π kπ  π  − +  = and sin = 1 Q 12 +   2 12 2 2 2    7π (k + 1) π   7 π kπ  + + sin   cos    12   12 2 2  7π (k + 1) π   7 π kπ  − sin  + +   cos  10  12   12 2 2 = Σ k=0  7 π kπ   7π (k + 1) π  cos  + +   cos   12  12  2 2 10   7 π kπ    7π (k + 1) π  + = Σ tan  +   − tan   12   12 k=0  2 2    7π π   7π  = tan  +  − tan    12 2   12 

+

–1 e–1

⇒

 7π (k + 1)π   7π k π   sin  + +   −  12 2   12 2    = Σ k=0  7π (k + 1) π   7 π kπ  cos  + +   cos   12   12 2 2 10

+ e e–1

 e  x ∈ (− ∞ , 1) ∪  , ∞  e − 1   1  ∪ E 2 =  −∞ ,  1 − e 

 e   e − 1 , ∞ 

 7π 10 π   7π 11π  + + tan  +   − tan   12  12 2  2  7π π π  7π 11 π  = tan + cot = tan  +  − tan  12 2  12 12 12 1 2 = = =4 π π π sin cos sin 12 12 6 10 1 7 π kπ    7π (k + 1) π    So, sec−1  Σ sec  + +   sec     12   12 4 k = 0 2 2 = sec−1 (1) = 0

15. We have, i ∞  ∞  x  sin −1  Σ xi + 1 − x Σ    i = 1  2  i = 1

=

∴ The domain of f and g are  1  ∪  −∞ ,  1 − e  and Range of

 e   e − 1 , ∞ 

x is R+ − {1} x−1

⇒ Range of f is R − {0} or (−∞ , 0) ∪ (0, ∞ )  π  π π  π  Range of g is − , − {0} or − , 0 ∪ 0,  2 2   2   2  Now, P → 4, Q → 2, R → 1, S → 1 10 7 π kπ   7π (k + 1) π  14. Q Σ sec  + +   sec   12   12  k=0 2 2 10 1 = Σ k=0  7π (k + 1) π   7 π kπ  cos  + +   cos   12   12 2 2

∞  ∞  − x i  π − cos −1  Σ   − Σ (− x)i 2 i = 1  2  i = 1 

x    −x  x⋅ (− x)   π  x2 −1  2 2 − = − cos  − ⇒ sin  x 1 + x 1 − x 1 − x 2  1 +      2 2  ∞ i+1 x2  x = x2 + x3 + x4 + ... = Q i Σ =1 1 − x   using sum of infinite terms of GP   x2 x2  π x  −1  x − − ⇒ sin −1   = − cos  1 x 2 x 2 1 x 2 x  − − + +     x2  x x2  x  ⇒ sin −1  = sin −1  − −    1 + x 2 + x 1 − x 2 − x  −1

π   Q sin −1 x = − cos −1 x   2

Inverse Circular Functions 523  1  1 ⇒ f   = − sin sin −1 1 − 2   5  5 

⇒

x2 x2 x x − = − 1−x 2−x 1+ x 2+ x

⇒

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

⇒

x 1 or x = 0 = 2 − 3 x + x2 2 + 3 x + x 2

 2 6 2 6 = − sin sin −1  =− 5  5 

Topic 3 Sum and Difference Formulae

⇒ x3 + 3x2 + 2x = x2 − 3x + 2 ⇒ x3 + 2x2 + 5x − 2 = 0 or x = 0 Let f (x) = x3 + 2x2 + 5x − 2 f ′ (x) = 3x2 + 4x + 5 f ′ (x) > 0, ∀ x ∈ R ∴ x3 + 2x2 + 5x − 2 has only one real roots Therefore, total number of real solution is 2.

16.

1.

if if if if

x ∈ [0, π] x ∈ [π , 2π] x ∈ [2π, 3π] x ∈ [3π, 4π]

y = cos –1 (cos x)

π–

2π

2π

x

–π

(0, 1) π 2

π

4π

(ii) sin− 1 x = cos − 1 1 − x 2 and π (iii) sin− 1θ + cos − 1θ = 2

We have,  3  12 sin − 1   − sin − 1    5  13 2 2  12 3  12  3 1−  − 1−   = sin − 1   13  13   5 5  

[Qsin − 1 x − sin − 1 y = sin − 1 (x 1 − y2 − y 1 − x2), if x2 + y2 ≤ 1 or if xy > 0 and x2 + y2 > 1 ∀x, y ∈ [− 1, 1]]  12 4 3 5  = sin − 1  × − ×   13 5 5 13

Y π 10 – π y= 10 π/2

(i) sin− 1 x − sin− 1 y = sin− 1( x 1 − y 2 − y 1 − x 2 ) if x 2 + y 2 ≤ 1 or if xy > 0 and x 2 + y 2 > 1 ∀x , y ∈ [ − 1, 1]

PLAN (i) Using definition of f( x ) = cos −1( x ) , we trace the curve f( x ) = cos −1(cos x ) . (ii) The number of solutions of equations involving trigonometric and algebraic functions and involving both functions are found using graphs of the curves.

x, 2π − x,  We know that, cos −1 (cos x) =  − 2π + x, 4π − x,

Key Idea Use formulae

–x

3π 2π 5π 3π 10 4π 2 2 x 10 – x =1– y= 10 10

 48 − 15 − 1  33 = sin − 1   = sin    65   65 X

 33 = cos − 1 1 −    65

10 − x and 10 −1 y = cos (cos x) intersect at three distinct points, so number of solutions is 3.

= cos − 1

From above graph, it is clear that y =

3136 4225

 1   sin −1   1 + x2   

 1  = cos  tan −1  1 + x2   =

x +1 = RHS x2 + 2 2

18. Let f (x) = cos ( 2 cos −1 x + sin −1 x) π  = cos  cos −1 x +   2 = − sin (cos −1 x) ⇒

π  Q cos −1 x + sin −1 x =  2 

f (x) = − sin (sin −1 1 − x2 )

[Qsin − 1 x = cos − 1 1 − x2]

 56 π  56 = cos − 1   = − sin − 1    65 2  65

17. LHS = cos tan −1 [sin (cot−1 x)]  = cos tan −1 sin  

2

π  Q sin − 1 θ + cot− 1 θ =  2 

2. Given equation is y = α, where − 1 ≤ x ≤ 1, 2 y − 2 ≤ y ≤ 2 and x ≤ 2   y cos − 1  x + 1 − x2 1 − ( y / 2)2 = α   2 cos − 1 x − cos − 1

[Q cos − 1 x − cos − 1 y = cos − 1 (xy + 1 − x2 1 − y2), | x|,| y| ≤ 1 and x + y ≥ 0] xy + 1 − x2 1 − ( y / 2)2 = cos α ⇒ 2 ⇒

1 − x2 1 − ( y / 2)2 = cos α −

xy 2

∴

524 Inverse Circular Functions On squaring both sides, we get  y2  x2 y2 xy (1 − x2) 1 −  = cos 2 α + cos α −2 4 4 2  ⇒ 1 − x2 −

x2 y2 y2 x2y2 + = cos 2 α + − xy cos α 4 4 4

⇒

y2 − xy cos α = 1 − cos 2 α 4 ⇒ 4x2 − 4xy cos α + y2 = 4 sin 2 α 3 1 3. Given, α = cos −1   and β = tan −1    5  3

0≤x< 1 6 1 0≤x< 6

1 , − 1, 6 1 x= , 6

x=

⇒

1 6

0≤x
a, A >

2(c2 − a 2) . 3ac

(1980, 3M)

24. If in a triangle ABC, a = 1 + 3 cm, b = 2 cm and ∠C = 60° , then find the other two angles and the third side. (1978, 3M)

Topic 2 Applications of Area, Napier’s Analogy

and Solution of a Triangle Objective Questions I (Only one correct option) 1. With the usual notation, in ∆ABC, if

∠A + ∠B = 120°, a = 3 + 1 and b = 3 − 1, then the

ratio ∠A : ∠B, is (a) 7 : 1 (c) 9 : 7

(2019 Main, 10 Jan II)

(b) 3 : 1 (d) 5 : 3

2. If PQR is a triangle of area ∆ with a = 2, b =

7 and 2

5 c = , where a, b and c are the lengths of the sides of 2

the triangle opposite to the angles at P , Q and R, 2 sin P − sin 2P respectively. Then, equals (2012) 2 sin P + sin 2P (a)

3 4∆

(b)

45 4∆

 3  (c)    4∆ 

2

 45  (d)    4∆ 

5

3. In radius of a circle which is inscribed in a isosceles triangle one of whose angle is 2π / 3, is 3, then area of triangle (in sq units) is (2006, 2M)

(a) 4 3 (c) 12 + 7 3

(b) 12 − 7 3 (d) None of these

528 Properties of Triangles 4. The sides of a triangle are in the ratio 1 : 3 : 2 , then the angles of the triangle are in the ratio (a) 1 : 3 : 5

(b) 2 : 3 : 2

(c) 3 : 2 : 1

(2004, 1M)

(d) 1 : 2 : 3

Objective Questions II (One or more than one correct option)

12. Consider the following statements concerning a ∆ABC

5. Let x , y and z be positive real numbers. Suppose x , y and z are the lengths of the sides of a triangle opposite to its angles X , Y and Z, respectively. If X Z 2y tan + tan = , 2 2 x+ y+z then which of the following statements is/are TRUE? (a) 2Y = X + Z X x (c) tan = 2 y+ z

(b)Y = X + Z

(2020 Adv.)

2

(i) The sides a , b, c and area of triangle are rational. B C (ii) a , tan , tan are rational. 2 2 (iii) a , sin A , sin B, sin C are rational. Prove that (i) ⇒ (ii) ⇒ (iii) ⇒ (i) (1994, 5M)

13. In a triangle of base a, the ratio of the other two sides

is r(< 1). Show that the altitude of the triangle is less ar . (1991, 4M) 1 − r2

than or equal to

(d) x + z − y = xz 2

π and 4 tan B, tan C = p. Find all positive values of p such that A, B, C are the angles of triangle. (1997C, 5M)

11. Let A, B, C be three angles such that A =

2

14. If in a ∆ABC, cos A cos B + sin A sin B sin C = 1, then

Fill in the Blank

show that a : b : c = 1 : 1 : 2.

6. If the angle of a triangle are 30° and 45° and the

included side is ( 3 + 1) cm, then the area of the triangle is … . (1988, 2M)

7. The set of all real numbers a such that a 2 + 2a , 2a + 3 and a 2 + 3a + 8 are the sides of a triangle is …… (1985, 2M)

Analytical & Descriptive Questions 8. If ∆ is the area of a triangle with side lengths a , b, c, then show that 1 ( a + b + c) abc ∆≤ 4

(1986, 5M)

3 2

15. For a ∆ABC, it is given that cos A + cos B + cos C = . Prove that the triangle is equilateral.

(1984, 4M)

16. If p1 , p2 , p3 are the altitudes of a triangle from the vertices A, B, C and ∆ is the area of the triangle, then prove that C 1 1 1 2ab + − = cos 2 p1 p2 p3 (a + b + c)∆ 2

(1978, 3M)

17. If p1 , p2 , p3 are the perpendiculars from the vertices of a triangle to the opposite sides, then prove that p1 p2p3 =

Also, show that the equality occurs in the above inequality if and only if a = b = c. (2001, 6M)

9. Prove that a ∆ABC is equilateral if and only if tan A + tan B + tan C = 3 3.

(1998, 8M)

10. Show that for any triangle with sides a , b, c 3( ab + bc + ca ) ≤ ( a + b + c)2 ≤ 4( ab + bc + ca ).

(1979, 3M)

a 2b2c2 8R3

(1978, 3M)

Integer & Numerical Answer Type Question 18. Let ABC and ABC′ be two non-congruent triangles with sides AB = 4 , AC = AC ′ = 2 2 and angle B = 30° .The absolute value of the difference between the areas of these triangles is (2009)

Topic 3 Circumcircle, Incircle, Escribed, Orthocentre

and Centroid of a Triangle Objective Questions I (Only one correct option) 1. Two vertices of a triangle are (0, 2) and (4, 3). If its orthocentre is at the origin, then its third vertex lies in which quadrant? (2019 Main, 10 Jan II) (a) Fourth (c) Second

(b) Third (d) First

2. Let the equations of two sides of a triangle be

3x − 2 y + 6 = 0 and 4x + 5 y − 20 = 0. If the orthocentre of this triangle is at (1, 1) then the equation of its third side is (2019 Main, 9 Jan II)

(a) 122 y − 26x − 1675 = 0 (c) 122 y + 26x + 1675 = 0

(b) 26x − 122 y − 1675 = 0 (d) 26x + 61y + 1675 = 0

Properties of Triangles 529 3. In a triangle, the sum of two sides is x and the

product of the same two sides is y. If x 2 − c2 = y, where c is the third side of the triangle, then the ratio of the inradius to the circumradius of the triangle is (2014 Adv.)

3y 2x (x + c) 3y (c) 4x (x + c)

3y 2 c (x + c) 3y (d) 4 c (x + c) (b)

(a)

uniquely determine an acute angled ∆ABC (R being the radius of the circumcircle)? (2002, 1M) (b) a , b, c (d) a , sin A , R

the circumradius of the triangle, then 2 (r + R ) is equal to (2000, 2M) (b) b + c

(c) c + a

(d) a + b + c

Passage Based Problems Consider the circle x 2 + y 2 = 9 and the parabola y 2 = 8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the X-axis at R and tangents to the parabola at P and Q intersect the X-axis at S. (2007, 8M)

6. The radius of the incircle of ∆PQR is (a) 4

(b) 3

(c)

8 3

(d) 2

7. The radius of the circumcircle of the ∆PRS is (a) 5

(b) 3 3

(c) 3 2

(d) 2 3

8. The ratio of the areas of ∆PQS and ∆PQR is (a) 1 : 2 (c) 1 : 4

(b) 1 : 2 (d) 1 : 8

to the angles X , Y , Z respectively and 2s = x + y + z. s−x s− y s−z and area of incircle of the If = = 4 3 2 8π (2016 Adv.) ∆XYZ is , then 3

(b) the radius of circumcircle of the ∆XYZ is

35 6 6

X Y Z 4 sin sin = 2 2 2 35 X +Y 3 (d) sin 2   =  2  5 (c) sin

12. A straight line through the vertex P of a ∆PQR intersects the side QR at the point S and the circumcircle of the ∆PQR at the point T . If S is not the centre of the circumcircle, then (2008, 4M) 1 + PS 1 (b) + PS 1 (c) + PS 1 (d) + PS (a)

1 ST 1 ST 1 ST 1 ST

2 QS × SR 2 > QS × SR 4 < QR 4 > QR
5

Hints & Solutions Topic 1 Applications of Sine, Cosine, Projection and Half Angle Formulae 1. It is given that angles of a ∆ABC are in AP. So, ∠A + ∠B + ∠C = 180º ⇒ ∠B − d + ∠B + ∠B + d = 180º [if ∠A , ∠B and ∠C are in AP, then it taken as ∠B − d, ∠B, ∠B + d respectively, where d is common difference of AP] …(i) ⇒ 3∠B = 180º ⇒ ∠B = 60º a 1 and [given] = b 3 sin A 1 = ⇒ sin B 3 sin A sin B sin C   by sine rule a = b = c  sin A 1 = ⇒ 3 3 2 1 ⇒ sin A = ⇒ ∠A = 30º 2

So, ∠C = 90º ∴ From sine rule, a b c = = sin A sin B sin C a b 4 = = [Q c = 4 cm] ⇒ 1 3 1 2 2 ⇒ a = 2 cm, b = 2 3 cm 1 1 ∴ Area of ∆ABC = ab sin C = × 2 × 2 3 × 1 2 2 = 2 3 sq cm b+ c c+ a a + b 2. Given, = = = λ (say) 11 12 13 A

c

 3 Q sin B = sin 60° =  2   B

b

a

C

b + c = 11λ ,c + a = 12λ and a + b = 13λ

...(i)

Properties of Triangles 531 ⇒ 2(a + b + c) = 36λ ⇒ a + b + c = 18λ From Eqs. (i) and (ii), we get a = 7λ, b = 6λ, c = 5λ

4. Applying sine rule in ∆ABD, ...(ii)

q

D

C

θ 2

p

+

Now,

2

q

b2 + c2 − a 2 λ2[36 + 25 − 49] 12 1 = = = 60 5 2bc 60λ2 a 2 + c2 − b2 λ2[49 + 25 − 36] 19 cos B = = = 2ac 35 70λ2

P

cos A =

a 2 + b2 − c2 2ab λ2[49 + 36 − 25] = 84λ2 60 5 = = 84 7 1 7 Thus, cos A = = , 5 35 19 25 cos B = , cos C = 35 35 cos A cos B cos C 1 = = = 7 19 25 35 ⇒ (α , β , γ ) = (7, 19, 25) a b c 3. We know that, = = = 2R sin A sin B sin C and

π−(θ+α)

A

B

AB p2 + q 2 = sin θ sin { π − (θ + α )}

cos C =

⇒ ⇒

AB p2 + q 2 = sin θ sin(θ + α ) AB =

=

( p2 + q2)sin θ and sin α = p cos θ + q sin θ

q D

2

p

C

a

π−(θ+α) x−q

α q

⇒

a 2 + b2 − c2 = − 2ab + ab

⇒

⇒

a + b − c = − ab

⇒

2

2

2

2

a +b −c − ab 1 = =− 2ab 2ab 2 1 cos C = − 2 2

⇒

2

∴

a 2 + b2 − c2 ] 2ab

c = 2R sin C 1 c c 2 R= = 2 sin (120° ) 2 3 c R= 3

tan(θ + α ) =

B

p x−q

p q−x

q − x = p cot(θ + α ) x = q − p cot(θ + α )  cot θ cot α − 1 = q− p   cot α + cot θ    = q− p   

C = 120°

[using cosine rule, cos C =

⇒

p

M

In ∆DAM, tan(π − θ − α ) =

⇒

Now,

2

A

b

(a + b) − c = ab

∴

q

+

p

∴

⇒

C

α

x

B

2

p p2 + q 2

θ

R

2

2

AB = x

Let

A

O

  p + q  q

2

Alternate Solution

and given that, a + b = x, ab = y and x2 − c2 = y

c

 Q cos α = 

p2 + q2 sin θ sin θ cos α + cos θ sin α

q  Q cot α = p   

q  cot θ – 1  q cot θ − p  p  = q− p  q  q + p cot θ   + cot θ   p

 q cos θ − p sin θ  = q− p   q sin θ + p cos θ  ⇒

x=

⇒

AB =

q2 sin θ + pq cos θ − pq cos θ + p2 sin θ p cos θ + q sin θ ( p2 + q2)sin θ p cos θ + q sin θ

532 Properties of Triangles 8. We know that, A + B + C = 180°

5. Since, A, B, C are in AP. ⇒

2 B = A + C i.e. ∠ B = 60º a c ∴ (2 sin C cos C ) + (2 sin A cos A ) c a

⇒

= 2 k (a cos C + c cos A )   using, a = b = c = 1   sin A sin B sin C k  = 2 k (b)

= 2ac cos B =

2 ac ⋅ (a 2 + c2 − b2) 2ac

[by cosine rule]

= a 2 + c2 − b2

9. It is given that, tan (P / 2) and tan (Q / 2) are the roots of

= 2 sin B

[using b = a cos C + c cos A ]

= 3

the quadratic equation ax2 + bx + c = 0 and ∠R = π /2 ∴

6. Let a , b, c are the sides of ∆ABC.

tan (P / 2) + tan (Q / 2) = − b/a tan (P / 2) tan (Q / 2) = c / a

and

b + c k (sin B + sin C ) = a k sin A

Now,

A + C − B = 180 − 2B  1 Now, 2ac sin  ( A − B + C ) = 2ac sin (90° − B)  2

[by sine rule]

 B − C  B + C  B − C 2 sin  cos   cos     2   2   2  b+ c ⇒ = = A A A a sin 2 sin cos 2 2 2  B − C sin    2  b−c Also, = A a cos 2 A  B − C or (b − c) cos = a sin    2  2

⇒ ⇒ ⇒ ⇒ ⇒

7. Given, ratio of angles are 4 : 1: 1. ⇒

4 x + x + x = 180° ⇒ x = 30°

∴

∠ A = 120°, ∠B = ∠C = 30°

P + Q + R = 180°

Since, ⇒

⇒

P + Q = 90° P+Q = 45° 2  P + Q tan   = tan 45°  2  tan (P / 2) + tan (Q / 2) =1 1 − tan (P / 2) tan (Q / 2) − b/a =1 1 − c/a −b / a =1 a−c a −b =1 a−c

⇒ ⇒

C 30°

−b = a − c a+ b=c

10. By the law of sine rule, a

a b c = = =k sin P sin Q sin R

b 120 ° A

P

30° B

c

Thus, ratio of longest side to perimeter = Let

a a+ b+ c

a 2 = b2 + c2 − 2bc cos A

⇒

a 2 = 2x2 − 2x2 cos A

p2

[by cosine rule]

= 2x2(1 − cos A ) ⇒ ⇒

a 2 = 4x2 sin 2 A / 2 a = 2x sin A / 2 ⇒ a = 2x sin 60° = 3x

Thus, required ratio

b

c

b=c=x

⇒

[say]

Q

Also, ⇒ ⇒

p1 a

p3 R

1 ap1 = ∆ 2 2∆ = p1 a 2∆ p1 = k sin P 2∆ 2∆ and p3 = p2 = k sin Q k sin R

=

a 3x = a + b + c x + x + 3x

Similarly,

=

3 = 3 :2 + 3 2+ 3

Since, sin P, sin Q and sin R are in AP, hence p1 , p2, p3 are in HP.

Properties of Triangles 533 11. In ∆ABD, applying sine rule, we get

13. Using, cos C =

A

a 2 + b2 − c2 2ab

β

α

A

b=

30º C

3 AD = x sin α 2

and in ∆ACD, applying sine rule, we get AD 3x = sin π / 4 sin β 3 ⇒ AD = x sin β 2

…(i) ⇒

a = x2 + x + 1

…(ii)

PLAN Whenever cosine of angle and sides are given or to find out, we should always use Cosine law. b2 + c 2 − a2 a2 + c 2 − b2 i.e. cos A = , cos B = 2 bc 2 ac a2 + b2 − c 2 and cosC = 2 ab

3 (x2 + x + 1)2 + (x2 − 1)2 − (2x + 1)2 = 2 2 (x2 + x + 1) (x2 − 1)

⇒

x2 + 2x + (x2 − 1) = 3 (x2 + x + 1)

⇒

(2 − 3 ) x2 + (2 − 3 ) x − ( 3 + 1) = 0

⇒

x = − (2 + 3 )

and

x=1+ 3

But

x = − (2 + 3 )

⇒ c is negative. ∴ x = 1 + 3 is the only solution.

14. Given, cos B + cos C = 4 sin 2

n

b

B

(n + 2)

(n + 4)

Q

R

(n + 4) a

∴ ⇒

1 (2n + 4)2 + (2n + 2)2 − (2n + 6)2 = 3 2(2n + 4) (2n + 2)

⇒ ⇒ ⇒

4n 2 − 16 1 = 8(n + 1) (n + 2) 3 n2 − 4 1 = 2(n + 1) (n + 2) 3 (n − 2) 1 = 2(n + 1) 3 3n − 6 = 2n + 2 ⇒ n = 8

∴ Sides are (2n + 2), (2n + 4), (2n + 6), i.e. 18, 20, 22.

a

C

Fixed base

 B + C  B −C 2 A ⇒ 2 cos   cos   = 4 sin  2   2  2 ⇒ 2 sin

b2 + c2 − a 2 cos P = 2bc

1   Q cos p = 3 , given  ⇒

b

c

c

(n + 2)

A 2

A

P n

B

⇒ (x + 2) (x + 1) (x − 1) x + (x2 − 1)2 = 3 (x2 + x + 1) (x2 − 1)

3x 3x From Eqs. (i) and (ii), = 2 sin α 2 sin β sin α 1 = ⇒ sin β 6

12.

c

C

AD x = sin π / 3 sin α ⇒

1=

π/4 3x

D

+

x

x 21

2x

π/3 B

A A  B −C cos   − 2 sin  = 0  2  2  2

⇒

A  B + C  B −C cos   = 0 as sin ≠ 0  − 2 cos   2   2  2

⇒

− cos

⇒ ⇒ ⇒

B C B C cos + 3 sin sin = 0 2 2 2 2 B C 1 tan tan = 2 2 3

(s − a ) (s − c) (s − b) (s − a ) 1 . = s (s − b) s (s − c) 3 s− a 1 = s 3

⇒

2s = 3a

⇒

b + c = 2a

∴ Locus of A is an ellipse.

534 Properties of Triangles 15. Since, ∆ABC = ∆ABD + ∆ACD

17. In ∆ADC ,

1 1 A 1 A bc sin A = c AD sin + b AD sin 2 2 2 2 2

⇒

AD = sin 23° b

A

A

b

c

A 2

A 2

c

b E

B

23°

D

⇒ C

D a

AD = b sin 23° abc AD = 2 b − c2 abc = b sin 23° b2 − c2 sin 23° a = 2 2 c b −c

But ⇒ ⇒

F

AD =

⇒

Again, AE = AD sec ⇒

2bc A cos 2 b+ c

A 2bc = 2 b+ c

AE is HM of b and c. EF = ED + DF = 2DE = 2 AD tan =2

A 2

A 2bc A A 4bc sin cos tan = 2 b+ c 2 2 b+ c

Again, in ∆ABC, sin A sin 23° = a c sin A a = 2 ⇒ a b − c2 sin A =

a2 b2 − c2

⇒

sin A =

k2 sin 2 A k sin 2 B − k2 sin 2 C

⇒

sin A =

sin 2 A sin B − sin 2 C

⇒

sin A =

sin 2 A sin (B + C ) sin (B − C )

⇒

sin A =

sin 2 A sin A ⋅ sin (B − C )

Hence, (a), (b), (c), (d) are correct answers.

16. The sine formula is a b = sin A sin B ⇒

a sin B = b sin A

(a) b sin A = a ⇒ a sin B = a π ⇒ B= 2 π Since, ∠ A < , therefore the triangle is possible. 2 (b) and (c) b sin A > a ⇒ a sin B > a ⇒ sin B > 1 ∴ ∆ ABC is not possible. (d) b sin A < a ⇒ a sin B < a ⇒ sin B < 1 ⇒ ∠B exists. Now, b > a ⇒ B> A π Since, A < 2 ∴ The triangle is possible. Hence, (a) and (d) are the correct answers.

⇒ ⇒ ⇒

[given]

…(i)

[from Eq. (i)]

⇒

Since, AD ⊥ EF and DE = DF and AD is bisector. ⇒ ∆AEF is isosceles.

C

a

B

2

2

sin (B − C ) = 1

[Q sin A ≠ 0]

sin (B − 23° ) = sin 90° B − 23° = 90°

∴

B = 113° a b 2 cos A cos B 2 cos C 18. Given, = + + + bc ca a b c b2 + c2 − a 2 We know that, cos A = 2bc

and

cos B =

c2 + a 2 − b2 2ac

cos C =

a 2 + b2 − c2 2ab

…(i)

On putting these values in Eq. (i), we get 2 (b2 + c2 − a 2) c2 + a 2 − b2 + 2abc 2abc

+

b 2 ( a 2 + b2 − c2 ) a = + bc ca 2abc

Properties of Triangles 535 ⇒

2 (b2 + c2 − a 2) + c2 + a 2 − b2 + 2 (a 2 + b2 − c2) 2abc =

⇒

 π  2π  + sin   ⋅ sin    n  n

a 2 + b2 abc  2π  ⇒ sin    n

3b2 + c2 + a 2 = 2a 2 + 2b2 ⇒ b2 + c2 = a 2

Hence, the angle A is 90°.

19. Let O be the centre and r be the radius of the circle passing through the vertices A1 , A2, …, An. 2π Then, ∠A1OA2 = n

2 2 2  2π  OA1 + OA2 − A1 A2 cos   =  n 2(OA1 )(OA2)

⇒

 4π  sin   = sin  n

⇒

⇒ A2

⇒

2 2 2  2π  r + r − A1 A2 cos   =  n 2(r )(r )

n=7

20. Let ABC be the triangle such that the lengths of its

sides CA , AB and BC are (x − 1), x and (x + 1) respectively, where x ∈ N and x > 1. Let ∠B = α be the smallest angle and ∠ A = 2α be the largest angle.

 2π  2r 2 cos   = 2r 2 − A1 A22  n

A

⇒

 2π  A1 A22 = 2r 2 − 2r 2 cos    n

⇒

A1 A22

⇒

 π A1 A22 = 2r 2 ⋅ 2 sin 2    n

⇒

 π A1 A22 = 4r 2 sin 2    n

⇒

 π A1 A2 = 2r sin    n

Similarly,

 2π  A1 A3 = 2r sin    n

⇒

and

 3π  A1 A4 = 2r sin    n

⇒

2 cos α =

Since,

1 1 1 = + A1 A2 A1 A3 A1 A4

∴

cos α =

⇒ ⇒

⇒

 3π     n

4π 3π =π− n n 7π =π n

⇒

r

A1

⇒

 2π   2π   3π  2 sin   cos   = sin    n  n  n

⇒

O 2 n

 2π    3π + π   3 π − π   ⇒ sin   2 cos   sin    n    2n   2n  

 2π   2π   π  π  3π  ⇒ 2 sin   ⋅ cos   ⋅ sin   = sin   sin    n  n  n  n  n

Again, by cos formula, we know that,

r

  3π   π  3π   π  sin  n  − sin  n   = sin  n  ⋅ sin  n   

 π  3π  = sin   . sin    n  n

OA1 = OA2 = r

also

 2π   3π   π  3π  sin   ⋅ sin   = sin   sin    n  n  n  n

⇒

2

 2π   = 2r 1 − cos     n   2

(x−1)

x

B

(x +1)

C

Then, by sine rule, we have sin α sin 2 α = x−1 x+1

[given]

1 1 1 = + 2r sin (π / n ) 2r sin (2π / n ) 2r sin (3π / n )

Also, cos α =

sin 2α x + 1 = sin α x−1 x+1 x−1 x+1 2 (x − 1)

x2 + (x + 1)2 − (x − 1)2 2x (x + 1) cos α =

x+4 2 (x + 1)

1 1 1 = + sin (π / n ) sin (2π / n ) sin (3π / n )

⇒

 3π   2π  sin   + sin    n  n 1 = sin(π / n ) sin (2π / n ) sin (3π / n )

From Eqs. (i) and (ii), x+1 x+4 = 2 (x − 1) 2 (x + 1)

...(i) [using cosine law] ...(ii)

536 Properties of Triangles ⇒

(x + 1)2 = (x + 4) (x − 1)

⇒

x + 2x + 1 = x + 3x − 4

⇒

x=5

2

On solving Eqs. (i) and (ii), we get a = 7λ , b = 6λ and c = 5λ

2

cos A =

b2 + c2 − a 2 36λ 2 + 25λ 2 − 49λ 2 1 = = 2bc 5 2 (30) λ 2

21. Let AD be the median to the base BC = a of ∆ABC

cos B =

and let ∠ ADC = θ, then a a  a a  +  cot θ = cot 30° − cot 45°  2 2 2 2

a 2 + c2 − b2 49λ 2 + 25λ2 − 36λ 2 19 = = 2ac 35 70λ 2

cos C =

a 2 + b2 − c2 49λ 2 + 36λ 2 − 25λ 2 5 = = 2ab 7 84λ 2

Hence, the lengths of the sides of the triangle are 4, 5 and 6 units.

∴ cos A : cos B : cos C =

3 −1 2

cot θ =

⇒

1 19 5 : : = 7 : 19 : 25 5 35 7

23. In ∆ADC, we have

A

B 30° 45°

a/2 D

90°–A

c B

a/2

C

a/2

D

⇒ 1

a = 11 − 6 3 2

⇒

⇒ a=

( 3 + 1) 11 − 6 3

=

= =2

AC CD 2b cos C = a

=

cos C =

and

11 − 6 3

⇒

a 2 + b2 − c2 = 4b2

⇒

a 2 − c2 = 3b2

(4 + 2 3 ) (11 − 6 3 )

44 − 24 3 + 22 3 − 36 =2

⇒ (b + c) = 11λ , c + a = 12λ , a + b = 13λ ⇒

2 (a + b + c) = 36λ

⇒

a + b + c = 18λ

2

2

b2 + c2 − a 2 3b2 + 3 (c2 − a 2) = ac 3ac

=

(a 2 − c2) + 3 (c2 − a 2) 2 (c2 − a 2) = 3ac 3ac

24. Given that, a = 1 + 3 , b = 2 and ∠C = 60°

b+ c c+ a a + b 22. Let = = =λ 11 12 13

c2 = a 2 + b2 − 2ab cos C

We have, …(i) …(ii)

...(iii)

b + c − a 2b ⋅ 2bc a 2

=

2 8 −2 3

8 −2 3

...(ii)

From Eqs. (i) and (ii),

Now, cos A cos C =

2 8 −2 3

8 −2 3

a 2 + b2 − c2 2ab

a 2 + b2 − c2 2b = 2ab a

2 8 −2 3 ( 3 + 1 )2

…(i)

Applying cosine formula in ∆ABC, we have b2 + c2 − a 2 cos A = 2bc

 3 −1  2   +  8 −2 3 8 − 2 3  

2 8 −2 3

C

cos C =

a AD 2 = 1 sin (θ + 45° ) 2 a AD = (sin 45° cos θ + cos 45° sin θ ) 2 a  cos θ + sin θ  a (cos θ + sin θ ) AD =   =  2 2  2

⇒

b

A

Applying sine rule in ∆ ADC, we get AD DC = sin (π − θ − 45° ) sin 45° ⇒

a/2

⇒

c = (1 + 3 )2 + 4 − 2(1 + 3 ) ⋅ 2 cos 60°

⇒

c2 = 1 + 2 3 + 3 + 4 − 2 − 2 3

⇒

c2 = 6

⇒

c= 6

2

Properties of Triangles 537 Using sine rule, a b c = = sin A sin B sin C

∴

2 sin P − sin 2 P 2 sin P (1 − cos P ) = 2 sin P + sin 2 P 2 sin P (1 + cos P ) =

1+ 3 2 6 = = sin A sin B sin 60°

⇒

∴

sin B =

2 sin 60° = 6

2×

P

3 2 = 1 6 2

⇒

B = 45°

∴

A = 180° − (60°+45° ) = 75° ⇒

Topic 2 Applications of Area, Napier’s Analogy and Solution of a Triangle

Area of triangle =

D 60° 60° a

a

√3 O

15° 15° B

⇒

C

 1 + tan 45° tan 30° a =1 + 3    tan 45° − tan 30°   3 + 1 =1 + 3   =4 + 2 3  3 − 1

∴ Area of a triangle =

 3 1 (4 + 2 3 )2   = (12 + 7 3 ) sq units 2  2

4. Let a : b : c = 1 : 3 : 2 ⇒ c2 = a 2 + b2 ∴ Triangle is right angled at C. B

A

B

Then,

tan ( A / 2) =

where,

C

( s − b) ( s − a) s ( s − a)

a+ b+ c s= ⇒ s= 2

c

a

b a

2

A

PLAN If ∆ABC has sides a, b, c .

c

 3 =   4∆ 

3. Let AB = AC = a and ∠ A = 120° .

C

a

Clearly, ∠C = 60º [Q ∠A + ∠B + ∠C = 180º ] Now, by tangent law, we have A−B a−b C tan = cot 2 a+b 2 ( 3 + 1) − ( 3 − 1)  60º  cot =   2  ( 3 + 1) + ( 3 − 1) 2 cot (30º ) = 2 3 1 = × 3 =1 3  A − B ⇒tan   = 1 = tan 45º  2  A−B = 45º ⇒ 2 ⇒ ∠A − ∠B = 90º On solving ∠A − ∠B = 90º and ∠A + ∠B = 120º ,we get ∠A = 105º and ∠B = 15º So, ∠A : ∠B = 7 : 1

2.

2

1 2 a sin 120° 2 where, a = AD + BD = 3 tan 30° + 3 cot 15° 3 =1 + tan (45° − 15° )

b

B

(s − b) (s − c) (s − b) (s − c) × s (s − a ) (s − b) (s − c) 2

∴

c

R

a=2

7  5  4 −  4 −  [(s − b)2 (s − c)2]  2  2 = = ∆2 ∆2

1. For a ∆ABC, it is given that a = 3 + 1 , A

b = 7/2

c = 5/2 Q

b = 3 − 1 and ∠A + ∠B = 120º

2 sin 2 (P /2) = tan 2 (P /2) 2 cos 2 (P /2)

C

2+

7 5 + 2 2 =4 2

or and

b

∠C = 90° a 1 = b 3

A

538 Properties of Triangles tan A =

In ∆BAC ,

∴ Area of triangle

a 1 = b 3

⇒

A = 30°

and

B = 60°

[Q A + B = 90°]

∴ Ratio of angles, A : B : C = 30° : 60° : 90° = 1 : 2 : 3

5. For a ∆XYZ, it is given that

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ∴ Q ⇒ ⇒ Q

y

= tan

Z

y−z = y+ z

y2 − z 2 y+ z

X x = 2 y+ z

{Q x2 + z 2 = y2}

a b c = = sin A sin B sin C 3 +1 b = sin (105° ) sin 30°

6. By sine rule, ⇒

A c B

⇒

105°

30°

b 45°

a = ( 3+1)

b=

( 3 + 1 )2 1 1 1 ⋅ ⋅ ⋅ 2 (sin 45° cos 60° + cos 45° sin 60° ) 2 2 (3 + 1 + 2 3 ) (4 + 2 3 ) 1 ⋅2 2 =  4 2  1 1 1 4 2 (1 + 3 ) 3 ⋅ + ⋅   2 2  2 2 (1 + 3 )2 1 + 3 sq cm = 2 2(1 + 3 )

7. Since, a 2 + 2a , 2a + 3 and a 2 + 3a + 8 form sides of a

X

x

=

=

∆ = (s − x)(s − z ) s(s − x)(s − y)(s − z ) = (s − x)2 (s − z )2 s2 − sy = s2 − (x + z )s + xz s(x + z − y) = xz (x + z )2 − y2 = 2xz x2 + z 2 = y 2 π y= 2 X +Y + Z = π π X + Z = =Y 2 X + Z =Y z 1− X 1 − cos x y tan = = z 2 1 + cos x 1+ y

Y

1 1 ( 3 + 1)sin 30° sin 45° ab sin 45° = ( 3 + 1) 2 2 sin 105°

=

X Z 2y tan + tan = 2 2 x+ y+ z ∆ ∆ y + = s(s − x) s(s − z ) s (s − z + s − x) ∆ =y (s − x) (s − z )

z

=

C

( 3 + 1) sin 30° sin 105°

triangle. Now, a 2 + 3a + 8 < (a 2 + 2a ) + (2a + 3) ⇒

a 2 + 3a + 8 < a 2 + 4a + 3

⇒

a >5

...(i)

Also, (a 2 + 3a + 8) + (2a + 3) > a 2 + 2a ⇒ ⇒

3a > −11 11 a>− 3

Again, (a 2 + 3a + 8) + (a 2 + 2a ) > 2a + 3 ⇒

2a 2 + 3a + 5 > 0

which is always true. ∴ Triangle is formed, if a > 5 1 8. Given, ∆≤ (a + b + c) abc 4 1 ⇒ (a + b + c) abc ≥ 1 4∆ (a + b + c) abc ≥1 ⇒ 16∆2 2s abc ≥1 ⇒ 16∆2 sabc ⇒ ≥1 8 ⋅ s (s − a ) (s − b) (s − c) abc ⇒ ≥1 8 (s − a ) (s − b) (s − c) abc ≥ (s − a ) (s − b) (s − c) ⇒ 8 Now, put s − a = x ≥ 0, s − b = y ≥ 0, s − c = z ≥ 0 s−a + s−b=x+ y 2s − a − b = x + y c=x+ y Similarly, a = y + z, b = x + z (x + y) ( y + z ) (x + z ) ⇒ ⋅ ⋅ ≥ xyz 2 2 2 which it true. Now, equality will hold, if x = y = z ⇒

a=b=c

⇒ Triangle is equilateral.

...(ii)

Properties of Triangles 539 9. If the triangle is equilateral, then A = B = C = 60° ⇒ tan A + tan B + tan C = 3 tan 60° = 3 3

⇒

cos (B − C ) 1 + p = cos (B + C ) 1 − p

⇒

cos (B − C ) = −

Conversely assume that,

Since, B or C can vary from 0 to 3π / 4

But in ∆ABC, A + B = 180° − C

∴

0 ≤ B − C < 3π / 4 1 ⇒ − < cos (B − C ) ≤ 1 2 1 1+ p From Eqs. (ii) and (iii), − < ≤1 2 2 ( p − 1)

Taking tan on both sides, we get tan ( A + B) = tan (180° − C ) tan A + tan B = − tan C 1 − tan A tan B

⇒ tan A + tan B = − tan C+ tan A tan B tan C ⇒ tan A + tan B + tan C = tan A tan B tan C = 3 3

⇒ −

⇒ None of the tan A, tan B, tan C can be negative So, ∆ABC cannot be obtuse angle triangle.

⇒

Also, AM ≥ GM 1 ⇒ [tan A + tan B + tan C ] ≥ [tan A tan B tan C ]1/3 3 1 3 ≥ 3. (3 3 ) ≥ (3 3 )1/3 ⇒ ⇒ 3 So, equality can hold if and only if

⇒

 1 + 2 (1 − 2 )  p −  1 − 2  ≤0 2 ( p − 1)

⇒

2p > 0 and p−1

( p − ( 2 + 1 )2 ) ≥0 ( p − 1)

+

c 1)

and

( p < 1 or p > ( 2 + 1)2)

2

i.e.

p ∈ (−∞ , 0) ∪ [( 2 + 1)2, ∞ )

or

p ∈ (−∞ , 0) ∪ [3 + 2 2 , ∞ )

12. It is given that a , b, c and area of triangle are rational. B = 2

(s − c) (s − a ) s (s − b)

s (s − a )(s − b)(s − c) s (s − b) a+ b+ c are Again, a , b, c are rational given, s = 2 rational, Also, (s − b) is rational, since triangle is rational, therefore we get ∆  B is rational. tan   =  2  s (s − b) =

⇒ a 2 + b2 + c2 + 2ab + 2bc + 2ca ≥ 3(ab + bc + ca )

11. Since,

1

⇒

We have, tan

⇒ a 2 + b2 + c2 ≥ ab + bc + ca ⇒ (a + b + c)2 ≥ 3(ab + bc + ca )

1

+

–

p < 0 or p ≥ ( 2 + 1)2

a 2 + b2 + c2 < 2ab + 2bc + 2ca (a + b + c)2 < 4(ab + bc + ca )

+

On combining above expressions, we get

Similarly, a 2 < ab + ac and b2 < bc + ab

⇒

+

– 0

10. By using triangular inequality,

1+ p ≤1 2 ( p − 1)

2p ≥ 0 and p−1

or A = B = C or when the triangle is equilateral.

∴

1 1+ p and < 2 2 ( p − 1)

…(iii)

1+ p 1 + p − 2 p+ 2 + 1 ≥ 0 and ≤0 p−1 2 ( p − 1)

tan A = tan B = tan C

⇒

…(ii)

[Q B + C = 3π / 4]

tan A + tan B + tan C = 3 3

⇒

(1 + p) 2 (1 − p)

C ∆ is rational. = 2 s (s − c)

Similarly,

tan

Therefore

a , tan

B C , tan are rational. 2 2

which shows that, (i) ⇒ (ii). Again, it is given that, B C a , tan , tan are rational, then 2 2

540 Properties of Triangles tan

A  π B + C = tan  −  2 2 2  1  B + C = cot   =  2  B C  tan  +  2 2

=

 a b c  using sine rule, sin A = sin B = sin C   

 B C 1 − tan   ⋅ tan    2  2 =  B C tan   + tan    2  2 Since, tan (B / 2) and tan (C / 2) are rational, hence tan ( A / 2) is a rational. 2 tan A / 2 Now, sin A = as tan ( A / 2) is a rational 1 + tan 2 A / 2 number, sin A is a rational . Similarly, sin B and sin C are. Thus, a, sin A, sin B, sin C are rational, therefore (ii) ⇒ (iii). Again, a, sin A, sin B, sin C are rational. By the sine rule, a b c = = sin A sin B sin C a sin B a sin C and c = ⇒ b= sin A sin A Since a, sin A, sin B and sin C are rational, Hence, b and c are also rational. 1 Also, ∆ = bc sin A 2

Therefore, (iii) ⇒ (i).

abc sin 2 A ⋅ sin (B − C ) abc sin (B − C ) = b2 − c2 (b2 − c2) ⋅ sin 2 A

=

ab2r sin (B − C ) ar sin (B − C ) = 1 − r2 b2 − b2r 2

p≤

⇒

ar 1 − r2

[Q sin (B − C ) ≤ 1]

14. Given, cos A cos B + sin A sin B sin C = 1 sin C =

⇒

13. Let ABC be a triangle with base BC = a and altitude AD = p, then A

⇒

...(i) [Q sin C ≤ 1]

1 − cos A cos B ≤ sin A sin B

⇒

1 ≤ cos ( A − B)

⇒

cos ( A − B) ≥ 1

⇒

cos ( A − B) = 1

⇒

A − B =0

[Q as cos (θ ) ≤ 1]

On putting A = B in Eq. (i), we get

⇒

⇒

⇒

1 − cos 2 A sin 2 A

sin C = 1

⇒

∴ b

1 − cos A cos B sin A sin B

1 − cos A cos B ≤1 sin A sin B

⇒

C = π /2

Now,

p

=

sin C =

As b, c and sin A are rational, so triangle is rational number. Therefore, a , b, c and triangle are rational.

c

abc sin A ⋅ sin (B + C ) sin (B − C ) (b2 sin 2 A − c2 sin 2 A )

A+ B+C=π π π π  A+ B= Q A = B and C =  ⇒ A=  2 2 4  π π π sin A : sin B : sin C = sin : sin : sin 4 4 2 1 1 a :b:c= : :1 2 2 = 1 :1 : 2

B

D

C a

1 Area of ∆ABC = bc sin A 2 1 Also, area of ∆ABC = ap 2 1 1 ∴ ap = bc sin A 2 2 bc sin A p= ⇒ a abc sin A ⇒ p= a2 ⇒

abc sin A ⋅ (sin 2 B − sin 2 C ) p= a 2(sin 2 B − sin 2 C )

15. Let a , b, c are the sides of a ∆ABC. Given,

cos A + cos B + cos C =

3 2

⇒

b2 + c2 − a 2 a 2 + c2 − b2 a 2 + b2 − c2 3 + + = 2bc 2ac 2ab 2

⇒

ab2 + ac2 − a3 + ba 2 + bc2 − b3 + ca 2 + cb2 − c3 = 3abc

⇒ a (b − c)2 + b (c − a )2 + c (a − b)2 (a + b + c) = [(a − b)2 + (b − c)2 + (c − a )2] 2 ⇒ (a + b − c) (a − b)2 + (b + c − a ) (b − c)2

+ ( c + a − b) ( c − a )2 = 0 [as we know, a + b − c > 0, b + c − a > 0, c + a − b > 0]

Properties of Triangles 541 ∴ Each term on the left of equation has positive coefficient multiplied by perfect square, each term must be separately zero. ⇒

a=b=c

∴ Triangle is an equilateral. 1 1 a 16. Since, ∆ = ap1 ⇒ = 2 p1 2∆

Area of ∆ABC′ = =

= 2 ( 3 − 1) sq. units =|2 ( 3 + 1) − 2 ( 3 − 1)|= 4 sq units

1 b 1 c = , = p2 2∆ p3 2∆

∴

1 1 1 1 + − = (a + b − c) p1 p2 p3 2∆

Alternate Solution C′ D

2(s − c) s − c s(s − c) ab = = = ⋅ 2∆ ∆ ab s∆ ab 2C = ⋅ cos 2  a + b + c ∆     2

1 2 2∆ p1 = a 2∆ p2 = b

Similarly,

2∆ c

1. Let ABC be a given triangle with vertices B(0, 2), C (4, 3) and let third vertex be A (a , b) A (a , b )

a 2 2 4 18. In ∆ ABC , by sine rule, = = sin A sin 30° sin C

E F

C = 45° , C′ = 135° C = 45°

(0,0)

⇒ A = 180° − (45°+30° ) = 105°

When, C′ = 135° ⇒ A = 180° − (135° + 30° ) = 15° A

2

15° 2

2 C

4

2

135°

45°

C′

Area of ∆ABC = =

B

Topic 3 Circumcircle, Incircle, Escribed, Orthocentre and Centroid of a Triangle

8 (abc)3 (abc)2 ⋅ p1 p2 p3 = = abc 64R3 8R3

⇒

30°

A

= Area of ∆ACC′ 1 1 = AD × CC′ = × 2 × 4 = 4 sq units 2 2

8∆3 abc abc ∆= 4R

Since,

When,

C

Difference of areas of ∆ABC and ∆ABC′

p1 p2 p3 =

∴

2 2

2

Here, AD = 2 , DC = 2

and p3 =

Now,

2

2 2

2ab C cos 2 2 (a + b + c)∆

17. We know that, ∆ = ap1 ⇒

1 × 4 × 2 2 sin (15° ) 2

Difference of areas of triangle

Similarly,

=

1 AB × AC sin A 2

30°

1 AB × AC sin A 2 1 × 4 × 2 2 sin (105° ) 2

=4 2 ×

3 +1 2 2

= 2 ( 3 + 1) sq. units

B

(0, 2) B

D

C (4,3)

Also, let D , E and F are the foot of perpendiculars drawn from A , B and C respectively. b −0 3 −2 Then, AD⊥ BC ⇒ × = −1 a −0 4 −0 [if two lines having slopes m1 and m2, are perpendicular then m1m2 = −1] …(i) ⇒ b + 4a = 0 and CF⊥ AB b −2 3 −0 ⇒ × = −1 a −0 4 −0 ⇒ 3 b − 6 = −4 a …(ii) ⇒ 4a + 3b = 6 From Eqs. (i) and (ii), we get −b + 3b = 6 ⇒ 2b = 6

542 Properties of Triangles ⇒

b =3

∴ Slope of BC is

3 and [from Eq. (i)] a=− 4 So, the third vertex  3  (a , b) ≡  − , 3 , which lies in II quadrant.  4 

Now, equation of line BC is given by y − y1 = m(x − x1 ), where (x1 , y1 ) are coordinates of B. 13  35 ∴ y − (− 10) = x −   61 2 13 ⇒ y + 10 = (2x − 35) 61 × 2

2. A E

H

⇒ ⇒

F

(1, 1) C 3x–2y+6=0

3. B 4x+5y–20=0

Let equation of AB be 4x + 5 y − 20 = 0 and AC be 3x − 2 y + 6 = 0 3 Clearly, slope of AC = 2 a [Q slope of ax + by + c = 0 is − ] b ∴Slope of altitude BH, which is perpendicular to  2 1  . Q mBH = − AC = −  mAC  3  Equation of BH is given by y − y1 = m(x − x1 ) 2 Here, m = − , x1 = 1 and y1 = 1 3 2 y − 1 = − (x − 1) ∴ 3 ⇒ 2x + 3 y − 5 = 0 Now, equation of AB is 4x + 5 y − 20 = 0 and

35 On substituting y = − 10 in 2x + 3 y − 5 = 0, we get x = 2  35  B , − 10 ∴ 2  Solving 4x + 5 y − 20 = 0 and 3x − 2 y + 6 = 0, we get coordinate of A. 12x + 15 y − 60 = 0  ⇒ 23 y = 84 12x − 8 y + 24 = 0  84 10  10 84 ,  ⇒x= ⇒A  23 23 23 23   84  y2 − y1   23 − 1 61 Now, slope of AH =  . =  = 10  x2 − x1   − 1 − 13   23 ⇒

y=

Q BC is perpendicular to AH.

122 y + 1220 = 26x − 455 26x − 122 y − 1675 = 0

PLAN (i) cos C =

a2 + b 2 − c 2 2 ab

(ii) R =

∆ abc ,r= 4∆ s

where, R, r , ∆ denote the circumradius, inradius and area of triangle, respectively. Let the sides of triangle be a , b and c. Given,

x=a + b

⇒

y = ab x2 − c2 = y (a + b)2 − c2 = y

⇒

a 2 + b2 + 2ab − c2 = ab

⇒ ⇒ ⇒ Q

equation of BH is 2x + 3 y − 5 = 0 Solving these, we get point of intersection (i.e. coordinates of B). 4x + 5 y − 20 = 0   ⇒ y = − 10 4x + 6 y − 10 = 0

 1   Q mBC = − m  AH 

13 61

∴

a 2 + b2 − c2 = − ab a 2 + b2 − c2 1 = − = cos 120° 2ab 2 2π ∠C = 3 ∆ abc r 4∆2 = ,r = ⇒ R= R s (abc) 4∆ s 1  2π   4  ab sin     3  2 =  x+ c ⋅ y⋅ c 2 r 3y = R 2c (x + c)

2

4. First solve each option separately. (a) If a, sin A, sin B are given, then we can determine a a b= sin B, c = sin C. So, all the three sin A sin A sides are unique. So, option (a) is incorrect. (b) The three sides can uniquely make an acute angled triangle. So, option (b) is incorrect. (c) If a, sin B, R are given, then we can determine a sin B . So, sin C can be b = 2R sin B, sin A = b determined. Hence, side c can also be uniquely determined. (d) If a, sin A, R are given, then b c = = 2R sin B sin C But this could not determine the exact values of b and c.

Properties of Triangles 543 1 4

5. Here, R2 = MC 2 = (a 2 + b2)

[by distance from origin]

1 2 c 4

[by Pythagoras theorem]

=

Area of ∆ PQS =

1 ⋅ 4 2 ⋅ 2 = 4 2 sq units3 2 Y

P

Y A

X'

C

⇒

a

R=

B

c 2

Ratio of areas of ∆PQS and ∆PQR is 1 : 4.

9. Let a non-right angled ∆PQR. Now, by sine rule P

= a + b + c−c

Since,

∆ = 16 2

Now,

s=

6 2 +6 2 +4 2 2 =8 2

16 2 r= 8 2 =2

7. Equation of circumcircle of ∆PRS is (x + 1) (x − 9) + y2 + λy = 0 It will pass through (1, 2 2 ), then − 16 + 8 + λ ⋅ 2 2 = 0 8 λ= =2 2 2 2

∴ Equation of circumcircle is x2 + y2 − 8 x + 2 2 y − 9 = 0 Hence, its radius is 3 3 . Alternate Solution 2 2 2 3 PR sin θ = 2R

Let ∠ PSR = θ ⇒ sin θ = ∴ ⇒ ⇒

PR = 6 2 = 2R ⋅ sin θ R=3 3

8. Coordinates of P and Q are (1, 2 2 ) and (1, − 2 2 ). Now, PQ = (4 2 )2 + 02 = 4 2 Area of ∆ PQR =

O

∆ s

6. Radius of incircle is, r =

1 ⋅ 4 2 ⋅ 8 = 16 2 sq units 2

q=1

rS

=a+b

⇒

X

(9, 0)

Y'

X

2(r + R) = 2r + 2R = 2s − 2c + c

∴

R

Q (1, -2 2)

Next, r = (s − c) tan (C / 2) = (s − c) tan π / 4 = s − c ∴

(1,0) O

S (-3, 0) (-1, 0)

M (a /2, b /2)

b

(1, 2 2)

Q

E

¾

p=Ö 3

R

P q r = = = 2 × circumradius sin P sin Q sin R ⇒

r 3 1 = = = 2 ×1 sin P sin Q sin R

[circumradius = 1 unit]

1 3 and sin Q = 2 2 ⇒ P = 120º and Q = 30º (Q ∆PQR is non-right angled triangle) So, R = 30º ⇒ r = 1, so ∆PQR is an isosceles triangle. And, since RS and PE are the median of ∆PQR, so ‘O’ is centroid of the ∆PQR. Now, Option (a), From Apollonius theorem, 2(PE 2 + QE 2) = PQ 2 + PR2 3  ⇒ 2  PE 2 +  = 1 + 1  4 3 1 1 PE 2 = 1 − ⇒ PE 2 = ⇒ PE = units ⇒ 4 4 2 1 1 and OE = PE = units [QO divides PE is 2 : 1] 3 6 Option (b), Again from Apollonius theorem, 2(PS 2 + RS 2) = PR2 + QR2 1  2 + RS 2 = 1 + 3 ⇒ 4  ⇒

sin P =

⇒ RS 2 = 2 −

1 7 7 units ⇒ RS 2 = ⇒ RS = 4 4 2

544 Properties of Triangles 11. Given a ∆XYZ, where 2s = x + y + z

Option (c), 1 (OE ) (ST ) 2 1 1 = × [(PS )sin 60º ] 2 6 1 1 3 = × × 12 2 2 3 square units = 48

Area of ∆SOE =

X

Y

10. We have,

⇒

s = 9λ, s = 4λ + x, s = 3λ + y

and

s = 2λ + z s = 9λ, x = 5λ, y = 6λ, z = 7λ

Now,

Q

∆ = s(s − x)(s − y)(s − z ) [Heron’s formula]

10 3

10

PQ 2 + QR2 − PR2 2PQ ⋅ QR

3 300 + 100 − PR2 = 2 200 3

⇒ 300 = 300 + 100 − PR2 ⇒ PR = 10 Since, PR = QR = 10 ∴ ∠QPR = 30° and ∠QRP = 120° 1 Area of ∆ PQR = PQ ⋅ QR ⋅ sin 30° 2 1 1 = × 10 3 × 10 × = 25 3 2 2 Radius of incircle of Area of ∆ PQR ∆ PQR = Semi - perimetre of ∆ PQR 25 3 25 3 ∆ = = s 10 3 + 10 + 10 5( 3 + 2) 2 ⇒ r = 5 3 (2 − 3) = 10 3 − 15 abc and radius of circumcircle (R) = 4∆ 10 3 × 10 × 10 = = 10 4 × 25 3

i.e.

⇒ ∴

30°

cos 30° =

s−x s− y s−z s = = = = λ (let) 4 3 2 9

or

P

In ∆ PQR

Z

x

s−x s− y s−z = = 4 3 2 3s − (x + y + z ) s = = 4+3+2 9

∴

1 1 1 pq sin R ( 3 )(1) ∆ 2 = 2 Q Inradius of ∆PQR = = 2 s 1 ( p + q + r ) 1 ( 3 + 1 + 1) 2 2 3 = (2 − 3 ) units 2 Hence, options (a), (b) and (d) are correct.

By cosine rule

y

z

Option (d),

∠ PQR = 30° PQ = 10 3 QR = 10

s−x s− y s−z = = 4 3 2

and

r=

∴Area of circumcircle of ∆ PQR = πR2 = 100 π Hence, option (b), (c) and (d) are correct answer.

R

Also, ⇒ and Now, ⇒ ⇒

= 9λ ⋅ 4λ ⋅ 3λ ⋅ 2λ = 6 6λ2 8π πr 2 = 3 8 r2 = 3 xyz (5λ )(6λ )(7λ ) 35λ R= = = 4∆ 4 6 4 ⋅ 6 6λ2 r2 =

8 ∆2 216λ4 = = 3 S2 81λ2 8 8 2 = λ 3 3

…(i)

…(ii) …(iii)

[from Eq. (ii)]

λ =1

(a) ∆XYZ = 6 6λ2 = 6 6 ∴ Option (a) is correct. (b) Radius of circumcircle (R) =

35 35 λ= 4 6 4 6

∴ Option (b) is incorrect. X Y Z (c) Since, r = 4R sin ⋅ sin ⋅ sin 2 2 2 ⇒ ⇒

2 2 35 X Y Z = 4⋅ sin ⋅ sin ⋅ sin 2 2 2 3 4 6 4 X Y Z = sin ⋅ sin ⋅ sin 35 2 2 2

∴ Option (c) is correct. X +Y 2 Z  (d) sin 2  = cos    2   2 X +Y Z s(s − z ) 9 × 2 3 as = 90° − = = = 2 2 xy 5 ×6 5 ∴ Option (d) is correct.

Properties of Triangles 545 12. Let a straight line through the vertex P of a given

∆ PQR intersects the side QR at the point S and the circumcircle of ∆ PQR at the point T.

14. Since, sides of a triangle subtends α , β, γ at the centre. A

Points P , Q , R, T are concyclic, then PS ⋅ ST = QS ⋅ SR PS + ST Now, > PS ⋅ ST [Q AM > GM] 2

γ α

P

β

C

B

∴ Q

R

S T

and Also, ⇒ ⇒ ⇒ ∴

1 1 + > PS ST

α + β + γ = 2π

...(i)

Now, arithmetic mean π  π  π  cos  + α  + cos  + β + cos  + γ 2  2  2  = 3

2 2 = PS ⋅ ST QS ⋅ SR

As we know that, AM ≥ GM, i.e. π π π AM is minimum, when + α = + β = + γ 2 2 2 or α = β = γ = 120°

SQ + QR > SQ ⋅ SR 2 QR > SQ ⋅ SR 2 1 2 > SQ ⋅ SR QR

∴ Minimum value of arithmetic mean π  = cos  + α  2 

2 4 > SQ ⋅ SR QR

= cos (210° ) = −

1 1 2 4 + > > PS ST QS ⋅ SR QR

15. Here, central angle =

abc ∆ and r = 4∆ s R abc s abc ⋅ s = ⋅ = 4∆ ∆ r 4∆2 abc = 4 (s − a ) (s − b) (s − c)

3 2

360° = 40° 9

13. We have, R =

But a : b : c = 4 : 5 : 6 a b c = = =k ⇒ 4 5 6 ⇒

a = 4k, b = 5k, c = 6k 1 Now, s = (a + b + c) 2 1 15k = (4k + 5k + 6x) = 2 2 (4k) (5k) (6k) R = ∴ r  15k   15k   15k  4 − 4k  − 5k  − 6k   2   2  2  30k3 =  15 − 8  15 − 10  15 − 12 k3       2   2  2  30 ⋅ 8 16 = = 7 ⋅5 ⋅3 7

C 20° r 1

[given] [let]

1

A1

In ∆ACM ,

A2

M

1 = sin 20° r

⇒ r = cosec 20° ∴ Radius of circle = cosec 20°

16. Since, the circles with radii 3, 4 and 5 touch each other externally and P is the point of intersection of tangents. A

4

C2

5 P

4 3

5 3

C3

C1

546 Properties of Triangles Now, in right angled ∆IHJ,

⇒ P is incentre of ∆C1 C 2 C3 .

∠ JIH = π / 2 − A / 2

Thus, distance of point P from the points of contact

[Q ∠ IEA = 90°, ∠ IAE = A /2 and ∠ JIH = ∠ AIE]

= inradius (r) of ∆C1C 2 C3 i.e.

r=

∆ = s

s (s − a )(s − b)(s − c) s

In ∆JIH, r A r  π A tan  −  = 1 ⇒ cot = 1  2 2  r − r1 2 r − r1

where, 2s = 7 + 8 + 9 ⇒ s = 12 (12 − 7)(12 − 8)(12 − 9) 5 ⋅4 ⋅3 = = 5 12 12

Hence, r =

Similarly,

2π n 2 r sin 2 n [since, I n is area of regular polygon] 2I n 2π [Q r = 1] …(i) = sin n n π On = nr 2 tan 2

17. We know that, I n =

⇒ and

On adding above results, we get cot A / 2 + cot B / 2 + cot C / 2 = cot A / 2 cot B / 2 cot C / 2 r1 r2 r3 r1r2 r3 + + = ⇒ r − r1 r − r2 r − r3 (r − r1) (r − r2) (r − r3 )

19. It is clear from the figure that, OA = R A

[since, On is area of circumscribing polygon] On π …(ii) = tan n n

⇒

In =

[from Eq. (i)]

(1 − (2I n / n )2 )

AI = 4R sin (B / 2) sin (C / 2)

Again,

∠GOA = B ⇒ OAG = 90° − B

Therefore, ∠IAO = ∠IAC − ∠OAC 1 ( A + 2B − 180° ) 2 1 1 = ( A + 2B − A − B − C ) = (B − C ) 2 2 = A / 2 − (90° − B) =

A/2 A/2

In ∆OAI,OI 2 = OA 2 + AI 2 − 2(OA )( AI ) cos (∠IAO ) = R2 + [4R sin (B / 2) sin (C / 2)]2

K

r1

 B − C − 2R ⋅ [4R sin (B / 2) sin (C / 2)] cos    2 

90° r1 90°

E

= [R2 + 16R2 sin 2(B / 2) sin 2(C / 2)

r1

H I

r sin ( A / 2)

r = 4R sin ( A / 2) sin (B / 2) sin (C / 2)

But ∴

A

r2

IF sin ( A / 2)

Q ∆AIF is right angled triangle, so =

angles are right angles and JK = JH .

F

O

AI =

18. The quadrilateral HEKJ is a square, because all four

J

I

C

2π n

2

On (1 + 2

G

B

2 I n 1 + 1 − (2I n / n ) = On 2

∴

R

90° r

B

°

⇒

1 + cos

90

A/2

F

On dividing Eq. (i) by Eq. (ii), we get 2π sin 2I n n = π On tan n In π = cos 2 = On n

B r2 C r and cot = 3 = 2 r − r2 2 r − r3

cot

 B − C  − 8R2 sin (B / 2) sin (C / 2) cos    2  

r3

= R2[1 + 16 sin 2(B / 2) sin 2(C / 2) B

Therefore, ⇒

D

HE = JK = r1

C

= R2[1 + 8 sin (B / 2) sin (C / 2)

90°

and

IH = r − r1

 B − C  − 8 sin (B / 2) sin (C / 2) cos    2  

IE = r

[given]

  B − C    2 sin (B / 2) sin (C / 2) − cos   2   

Properties of Triangles 547 = R2[1 + 8 sin (B / 2) sin (C / 2)   B − C    B + C  B − C   − cos   + cos  cos   2       2 2    B + C  = R21 − 8 sin (B / 2) sin (C / 2) cos    2      π A  = R21 − 8 sin (B / 2) sin (C / 2) cos  −   2 2   A B C π Q 2 + 2 + 2 = 2  = R2[1 − 8 sin ( A / 2) sin (B / 2) sin (C / 2)]   r  = R2 1 − 8    = R2 − 2Rr  4 R   Now, in right angled ∆BIO, ⇒

OB2 = BI 2 + IO 2 R2 = BI 2 + R2 − 2Rr

⇒ ⇒

2Rr = BI 2 2 Rr = r 2 /sin 2(B / 2)

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

2R = r /sin 2(B / 2) 2 2R sin B / 2 = r R (1 − cos B) = r ∆ abc (1 − cos B) = 4∆ s 4∆2 abc (1 − cos B) = s 2 2 2  a +c −b 4∆2 abc 1 − = 2ac s   2ac − a 2 − c2 + b2  4∆2 abc  = 2ac s  

4∆2 s

⇒

b [b2 − (a − c)2] =

⇒

b [b2 − (a − c)2] = 8(s − a )(s − b)(s − c)

⇒

b [{ b − (a − c)}{ b + (a − c)}] = 8(s − a )(s − b)(s − c)

⇒

b [(b + c − a )(b + a − c)] = 8(s − a )(s − b)(s − c)

⇒ ⇒ ⇒ ⇒

b [(2s − 2a )(2s − 2c)] = 8(s − a )(s − b)(s − c) b [2 ⋅ 2 (s − a )(s − c)] = 8(s − a )(s − b)(s − c) b = 2s − 2b 2b = a + c

which shows that b is arithmetic mean between a and c.

20. Since, r1 , r2 and r3 are exradii of ∆ABC are in HP. ∴ ⇒ ⇒ ⇒ ⇒

1 1 1 , , are in AP. r1 r2 r3 s−a s−b s−c are in AP. , , ∆ ∆ ∆ s − a , s − b, s − c are in AP. − a , − b, − c are in AP. a , b, c are in AP.

3 and C is given to be obtuse. 2 2π = a 2 + b2 − 2ab cos C ⇒ C= 3 2π = 62 + 102 − 2 × 6 × 10 × cos = 14 3 ∆ 225 × 3 ∴ r= =3 ⇒ r2 = 2 s  6 + 10 + 14     2

21. sin C =

24 Vectors Topic 1 Scalar Product of Two Vectors Objective Questions I (Only one correct option)

7. Two adjacent sides of a parallelogram ABCD are

1. Let A( 3, 0, −1), B ( 2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the mid-point of AC. If G divides BM in the ratio 2 : 1, then cos ( ∠GOA) (O being the origin) is equal to (2019 Main, 10 April I) (a)

1 15

(b)

1 2 15

(c)

1 30

(d)

1 6 10

π π with $i, with $j and 3 4 $ θ ∈( 0, π ) with k, then a value of θ is (2019 Main, 9 April II)

2. If a unit vector a makes angles

(a)

5π 6

(b)

π 4

(c)

5π 12

(d)

$ , b = b $i + b $j + 2 a = i$ + $j + 2 k 1 2 $ be three vectors such c = 5 $i + $j + 2 k projection vector of b on a is a. If perpendicular to c, then|b | is equal to

3. Let

2π 3

$ k

that the a + b is

(b) 4 (d) 32

4. If lines x = ay + b, z = cy + d and x = a ′ z + b′,

y = c′ z + d ′ are perpendicular, then (2019 Main, 9 Jan II)

(a) ab′+ bc′+1 = 0 (c) aa ′+ c + c′ = 0

(b) bb′+ cc′+1 = 0 (d) cc′+ a + a ′ = 0

5. Let O be the origin and let PQR be an arbitrary triangle. The point S is such that OP ⋅OQ + OR ⋅ OS = OR ⋅ OP + OQ ⋅ OS = OQ ⋅ OR + OP ⋅ OS Then the triangle PQR has S as its (a) centroid (c) incentre

(a)

8 9

(b)

17 9

(c)

1 9

(d)

4 5 9

8. Let P, Q, R and S be the points on the plane with

position vectors − 2$i − $j , 4$i , 3$i + 3$j and − 3$i + 2$j, respectively. The quadrilateral PQRS must be a (2010)

and

(2019 Main, 9 Jan II)

(a) 6 (c) 22

→ → given by AB = 2$i + 10$j + 11k$ and AD = − $i + 2$j + 2k$ . The side AD is rotated by an acute angle α in the plane of the parallelogram so that AD becomes AD’. If AD’ makes a right angle with the side AB, then the cosine of the angle α is given by (2010)

(2017 Adv.)

(b) orthocentre (d) circumcentre

(a) parallelogram, which is neither a rhombus nor a rectangle (b) square (c) rectangle, but not a square (d) rhombus, but not a square

9. Let two non-collinear unit vectors a$ and b$ form an acute angle. A point → P moves, so that at any time t the position vector OP (where, O is the origin) is P is farthest from given by a$ cos t + b$ sin t. When→ $ be the unit origin O, let M be the length of and u OP → (2008, 3M) vector along OP. Then, a$ + (a) u$ = | a$ + a$ − (b) u$ = | a$ − a$ + (c) u$ = | a$ + a$ − (d) u$ = | a$ −

$b and M = (1 + $b| $b and M = (1 + $b| $ b and M = (1 + $| b $b and M = (1 + $b|

$ )1/ 2 a$ ⋅ b $ )1/ 2 a$ ⋅ b $ )1/ 2 2 a$ ⋅ b $ )1/ 2 2 a$ ⋅ b

→ $ , →b = i$ − $j + k $ and → $ be 6. Let a = $i + $j + k c = $i − $j − k

→ → $ →c = $i + $j − k$ .A vector 10. Let, a = $i + 2$j + k$ , b = $i − $j + k,

three vectors. A vector v in the plane of a and b, → 1 , is given by whose projection on c is (2011) 3

coplanar to a and b has a projection along c of 1 magnitude , then the vector is (2006, 3M) 3

→

$ (a) $i − 3$j + 3 k $ (c) 3 $i − $j + 3 k

$ (b) − 3 $i − 3$j − k $ (d) $i + 3 $j − 3 k

→

→

→

(a) 4 $i − (c) 2 $i +

$j + 4 k$ $j + k$

→

→

(b) 4 $i + $j − 4 k$ (d) None of these

Vectors 549 → → →

11. If a , b , c are three non-zero, non-coplanar vectors →

→ →

→

and b1 = b − → c1

→ →

b⋅a→ → → b⋅a→ a , b2 = b + a, →2 →2 | a| | a|

→ →

→ → → 18. Let a = 2$i − $j + k$ , b = $i + 2$j − k$ and c = $i + $j − 2 k$ be

→ →

→ →

→

→

→

c ⋅ a → c ⋅ b → → → c ⋅ a → c ⋅ b1 → b , c2 = c − b1 , a − a− = c− → → → → | b|2 | a |2 | b |2 | a|2 →

→

→ →

→

→ →

→

c ⋅ a → c ⋅ b2 → → → c ⋅ a → b 2 , c4 = a − a− a. → → 2 → 2 | a |2 | a| | b2 |

→

c3 = c −

(b) (d)

→ → → {a , b1 , c2 } → → → {a , b2 , c4 }

→

→

→

→

12. If a and b1 are two unit vectors such that a + 2 b and →

→

5 a − 4 b , are perpendicular to each other, then the →

→

(2002, 1M)

angle between a and b is (a) 45°

→ →

13. If →

→

→

→

→

c

are

→

→

unit

vectors,

(2001, 2M)

→

→

→ →

→ →

→ →

(1995, 2M)

(b) – 25 (d) 25 →

to vectors a = (1, 1, 0) and b = ( 0, 1, 1) is (a) one (c) three

→

(1987, 2M)

(b) two (d) infinite

→

→

→

→

→

→

→

→

( a − d) ⋅ ( b − c ) = ( b − d) ⋅ ( c − a ) = 0 The point D, then, is the… of the ∆ ABC.

(1984, 2M)

rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new → system, a has components p + 1 and 1, then (1986, 2M) 1 3

(a) p = 0

(b) p = 1or p = −

1 (c) p = − 1or p = 3

(d) p = 1or p = − 1

→ →

→

→

→

→

→

→

→

→

A + B. Then, the length of vector A + B + C is ... . (1981, 2M)

True/False →

→

→

23. The points with position vectors a + b, a − b and →

→

a + k b are collinear for all real values of k.

(1984, 1M)

60i$ + 3$j, 40i$ − 8$j, a$i − 52$j are collinear, if

→

→

→

24. Find 3-dimensional vectors v1 , v 2 , v3 satisfying →

→

→ →

→

→ →

→ →

→

→

v1 ⋅ v1 = 4, v1⋅ v2 = − 2, v1 ⋅ v3 = 6, →

v2⋅ v2 = 2, v2⋅ v3 = − 5, v3 ⋅ v3 = 29

(2001, 5M)

25. Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices. (2001, 5M)

17. The points with position vectors (b) a = 40 (d) None of these

→

Let A be perpendicular to B + C, B to C + A and C to

Analytical & Descriptive Questions

16. A vector a has components 2 p and 1 with respect to a

(a) a = − 40 (c) a = 20

→

→

15. The number of vectors of unit length perpendicular →

→

to a non-zero vector b are ……and……respectively.

→

→

|u|= 3,|v|= 4 and|w|= 5, then u⋅ v + v⋅ w + w⋅ u is (a) 47 (c) 0

→

20. The components of a vector a along and perpendicular

22. Let A, B, C be vectors of length 3, 4, 5 respectively. →

→

Fill in the Blanks

→ → →

→

→

some x , y ∈ R, let c = xa + yb + (a × b ). If|c| = 2 and the vector c is inclined at the same angle α to both a and b, then the value of 8 cos2 α is ......... . (2018 Adv.)

then

14. Let u, v and w be vectors such that u + v + w = 0. If →

19. Let a and b be two unit vectors such that a ⋅ b = 0. For

→ → →

(b) 9 (d) 6

→ →

(1993, 2M)

(b) 2 $i + 3$j + 3 k$ (d) 2 i$ + $j + 5 k$

vectors a , b , c and d respectively such that

|a − b|2 +|b − c|2 +| c − a|2 does not exceed (a) 4 (c) 8

(a) 2 $i + 3$j − 3 k$ (c) − 2 i$ − $j + 5 k$

21. A, B, C and D, are four points in a plane with position

2 (d) cos−1    7

and

a,b

→

projection on a is of magnitude 2 / 3 , is

(1988, 2M)

(b) 60°

1 (c) cos−1    3

→

three vectors. A vector in the plane of b and c, whose

Integer & Numerical Answer Type Question

Then, which of the following is a set of mutually orthogonal vectors? (2005, 1M) → → → (a) {a , b1 , c1 } → → → (c) {a , b2 , a3 }

Objective Question II (One or more than one correct option)

(1983, 1M)

26. In a ∆ ABC, D and E are points on BC and AC respectively, such that BD = 2DC and AE = 3EC.Let P be the point of intersection of AD and BE. Find (1993, 5M) BP / PE using vector methods.

550 Vectors →

→ A (1) = 6i$ + 2$j, B ( 0) = 3$i + 2$j

27. Determine the value of c, so that for all real x, the vector cx $i − 6$j − 3k$ and x $i + 2$j + 2cx k$ make an (1991, 4M) obtuse angle with each other.

28. In a ∆ OAB, E is the mid-point of BO and D is a point on AB such that AD : DB = 2 : 1. If OD and AE intersect at P, determine the ratio OP : PD using methods. (1989, 4M)

→

31. Suppose that p , q and r are three non-coplanar

vectors in R3 . Let the components of a vector s along p , q and r be 4, 3 and 5, respectively. If the components of this vector s along ( − p + q + r ), ( p – q + r ) and ( − p – q + r ) are x , y and z respectively, then the value of 2x + y + z is (2015 Adv.)

(1988, 3M)

30. Let A ( t ) = f1( t )i$ + f2( t )$j and B ( t ) = g( t )i$ + g2( t )$j, →

t ∈ [0, 1], f1 , f2 , g1g2 are continuous functions. If A ( t ) →

→

and B ( t ) are non-zero vectors for all t and A ( 0) = 2$i,

B (1) = 2$j + 6$j.

→

Integer & Numerical Answer Type Questions

and OC a diagonal. Let D be the mid-point of OA. Using vector methods prove that BD and CO intersect in the same ratio. Determine this ratio. →

→

Then, show that A ( t ) and B ( t ) are parallel for some t.

29. Let OACB be a parallelogram with O at the origin

→

and

→

32. If a, →

→

b and

→ 2

→

→

c

→ 2

are →

unit

vectors

satisfying

→ 2

| a − b| +| b − c| +| c − a| = 9, then →

→

→

| 2 a + 5 b + 5 c| is equal to

(2012)

Topic 2 Vector Product of Two Vectors Objective Questions I (Only one correct option) 1. Let a = i$ − 2 $j + k$ and b = i$ − $j + k$ be two vectors. If c is a vector such that b × c = b × a and c ⋅ a = 0, then (2020 Main, 8 Jan II) c ⋅ b is equal to (a)

1 2

(c) −

(b) − 1 2

3 2

(d) − 1

If a vector perpendicular to both the vectors a + b and a − b has the magnitude 12, then one such vector is (2019 Main, 12 April II)

$) 2$j + k $) 2$j − k

$) (b) ± 4 (2i$ − 2$j − k $) (d) ± 4 (− 2$i − 2$j + k

3. Let α ∈ R and the three vectors $ , b = 2i$ + $j − αk $ a = αi$ + $j + 3k $ and c = αi$ − 2$j + 3k. Then, the set S = { α : a, b and c are coplanar}

(2019 Main, 12 April II)

4. If the length of the perpendicular from the point

x y−1 z +1 3 is , (β, 0, β ) (β ≠ 0) to the line, = = −1 1 0 2 then β is equal to (2019 Main, 10 April +I) (b) − 2 (d) 1

r

r r r r β1 is parallel to α and β 2 is perpendicular to α, then r r β1 × β 2 is equal to (2019 Main, 9 April I) 1 $ $) (3i − 9$j + 5k 2 $ (c) −3$i + 9$j + 5k (a)

1 $) (−3$i + 9$j + 5k 2 $) (d) 3i$ − 9$j − 5k (b)

Then|a × b| = r is possible if 3 2

(a) 0 < r ≤ (c) 3

3 3 0, then 2a = 2 a 2 + b2

→ → (a + b) × ( 2i$ + 3$j + 4k$ ) = 0

⇒

→

| u | + | v | =| w | | a + b| | a − b| + = 2 2 2 a +b a 2 + b2

Q

→

a || ( b − c) or b − c = λ a

⇒

→

and, similarly, | v | = | projection vector of w = $i + $j along PS| |(i$ + $j ) ⋅ (ai$ + b$j )| | a − b| = = a 2 + b2 a 2 + b2

→

9. Plan If a × b = a × c →

→

a × b= c × a

=

4 + 9 − 2c ⋅ a = 9

⇒

→

sides of a parallelogram PQRS, so area of parallelogram PQRS = | PQ × PS | (given) = 2ab = 8 … (i) ∴ ab = 4 According to the question, | u | = |projection vector of w = $i + $j along PQ|

(2)2 + (3)2 − 2c ⋅ a = 9 ⇒

→

12. Given vectors PQ = ai$ + b$j and PS = ai$ − b$j are adjacent

=2

4+4+1

→

∴

k$

6

|c| =

→

a × a+ a × b+ a × c=a

Similarly,

But a × b = 2 1 − 2 = 2$i − 2$j + k$ 1 1 0 ∴

→

⇒

2

Again,

C

→

a

(c − a ) ⋅ (c − a ) = 9 2

b

B

|c − a |2 = 9

⇒

→

c

8. We have, a = 2i$ + $j − 2k$ ⇒

→ →

→

11. By triangle law, a + b + c = 0

⇒ 4a 2 = 2a 2 + 2b2 ⇒ a 2 = b2 ⇒ a=b From Eqs. (i) and (ii), we get a =2 = b ∴ a + b =4 and the length of diagonal PR = | a$i + b$j + a$i − b$j| = |2ai$ |

… (ii)

= 2a = 4 And, the angle bisector of vector PQ and PS is along the vector λ ± ((ai$ + b$j ) ± (ai$ − b$j )) 2 a + b2

Vectors 565 13. Given, |a| = 12 ,| b| = 4 3

R

a + b+ c= 0 ⇒

a = − ( b + c)

b

We have,

|a|2 = | b + c|2

⇒

|a|2 = | b|2 + |c|2 + 2 b ⋅ c

⇒

144 = 48 + |c|2 + 48

⇒

|c| = 48

a

c

P

Q

→

|c| = 4 3 |c|2 = |a|2 + | b|2 + 2 a ⋅ b

Also, ⇒

48 = 144 + 48 + 2 a ⋅ b

⇒

⇒

|u|2 =| a|2 + cos 2 θ|b|2 − 2 cos θ (a⋅ b)

⇒

|u|2 = 1 + cos 2 θ − 2 cos 2 θ

⇒

|u|2 = 1 − cos 2 θ ⇒ |u|2 = sin 2 θ

→

→

→

⇒

→ →

→

→

→

→

→

→ →

→

∴ |u|2 =|v|2

→ → →

→

→ →

→ →

→ →

u ⋅ a = [a − (a ⋅ b) b] ⋅ a = a ⋅ a − (a ⋅ b) (b ⋅ a )

→ →

→

→

→ → →

→

u ⋅ b = [a − (a ⋅ b) b] ⋅ b → →

→ →

→ →

→ →

→ → →

= 2 (144) (48) − (− 72)2

= a ⋅ b − (a ⋅ b) (b ⋅ b) = a ⋅ b − a ⋅ b|b|2

= 2 (12) 48 − 36 = 48 3

= a⋅ b − a⋅ b =0

→ →

→

P1 and P2 through origin. Normal to plane p1 is →

n1 = [(2$j + 3k$ ) × (4$j − 3k$ )] = −18i$ Normal to plane p2 is → → → OA is parallel to ± (n1 × n 2) = 54 $j − 54k$ . ∴ Angle between 54 ($j − k$ ) and (2i$ + $j − 2k$ ) is So,

1  54 + 108 cos θ = ±   =±  3 ⋅ 54 ⋅ 2  2 π 3π θ= , 4 4 →

→

→

→

= cos θ

[Q|a|= 1,|b|= 1, given]

→

→

→

→ → →

u = a − (a ⋅ b) b ⇒ |u|=|a − (a⋅ b) b| →

→ →

→ →

→

→

→

→

→ →

→ →

→ →

→

PQ = c , QR = a and RP = b, so a + b + c =0 ∴ |a|2 + a ⋅ b + a ⋅ c = 0 and a ⋅ b + | b |2 + c ⋅ b = 0 and, also given that|a | = 3,| b | = 4 a ⋅c − a ⋅ b |a | 3 and = = c ⋅ a − c ⋅ b |a | + | b | 7

… (i) … (ii)

→

T

→

→

S

Q TR is resultant of RS and ST →

→

→

→

vectors. ⇒ PQ × (RS + ST) ≠ 0. But for Statement II, we have

We have, a ⋅ b =|a||a| cos θ

→

→

→

non-collinear vectors, then θ ≠ 0 and θ ≠ π .

|u|2 =|a − (a ⋅ b) b|2

→

u ⋅ (a + b) = u ⋅ a + u ⋅ b = u ⋅ a

→

15. Let θ be the angle between a and b. Since, a and a are

→ → →

→

17. Since, PQ is not parallel to TR.

Hence, (b) and (d) are correct answers.

→

→

⇒ 7 [(a ⋅ c ) − (a ⋅ b )] = 3 [(c ⋅ a ) − (c ⋅ b )] from Eqs. (i) and (ii), on putting the values of a ⋅ c and c ⋅ b, we get ⇒ 7 [− 9 − (a ⋅ b ) − (a ⋅ b )] = 3 [ − 9 − (a ⋅ b ) + 16 + (a ⋅ b )] ⇒ 7 [− 9 − 2 (a ⋅ b )] = 3 × 7 ⇒ a⋅b = −6 Q |a × b |2 = (|a || b |)2 − (a ⋅ b )2 = 144 − 36 = 108

→ ^ n 2 = ($j − k$ ) × (3 i + 3$j) = 3i$ − 3$j − 3k$

→

→

16. It is given that, in a ∆PQR

14. Let vector AO be parallel to line of intersection of planes

→

→

…(i)

⇒ |u|+ u ⋅ (a + b) =|u|+ u ⋅ a ≠|v|

→

→

→ →

→ →

|u|+ |u⋅ b|=|u|+ 0 =|u|=|v|

Also,

∴ Option (b) is not correct.

→ →

→

∴

|c|2 + |a| = 24 + 12 = 36 2

→

→

|v|2 = sin 2 θ

→ →

∴ Option (a) is correct.

⇒

→

⇒

→

|c|2 − |a| = 24 − 12 = 12 2

→

→

|v|2 =|a × b|2 ⇒ |v|2 =|a|2|b|2 ⋅ sin 2 θ

∴ |u|+ |u ⋅ a|= sin θ + sin 2 θ ≠ |v|

∴ Option (c) is correct.

Now,

[given]

⇒

⇒ |a × b + c × a| = 2|a × b| = 2 |a|2| b|2 − (a ⋅ b)2

∴

→

→

a × b= c× a

and

→

v=a × b

Also,

= (a )2 − cos 2 θ = 1 − cos 2 θ = sin 2 θ

a × b + c × a = 2a × b

Also,

→ →

→

∴Option (d) is correct. Also,

→

→

Now,

a ⋅ b = − 72

→

|u|2 =|a − cos θ b|2

2

⇒

→

⇒

→

→

→

U

R P

Q

PQ × RS = 0 →

→

which is not possible as PQ not parallel to RS. Hence, Statement I is true and Statement II is false.

566 Vectors 1 → 2

18. cos(P + Q ) + cos(Q + R) + cos(R + P ) = − (cos R + cos P + cos Q ) 3 Max. of cos P + cos Q + cos R = 2

→ → AB = $i + 2$j − 3k$ and AC = 2$i + 2$j − 3k$

Min. of cos(P + Q ) + cos(Q + R) + cos(R + P ) is = −

3 2

19. sin R = sin(P + Q ) 20. Let $i be a unit vector in the direction of b, $j in the →

→

direction of c . Note that c = $j → → → → (b × c ) =|b||c|sin α k$ = sin α k$

and

→ → where, k$ is a unit vector perpendicular to b and c. →

→

→

|b × c|= sin α

b× c ⇒ k$ = → → |b × c|

→ →

→

b× c → $ a⋅ = a ⋅ k = a3 → → |b × c|

→ → →

→ → →

→

→

→

(b × c )

|b × c|2 →

→

= ±

∴

Unit vector = ±

where, and

→

→

→

→

(b × c ) |b × c|

→ = a1$i + a 2$j + a3 k$ = a

→

→

→ →

1 1 ⋅ sin 30°

|λ |=

⇒

→

→

→

⇒ |λ |= 2

⇒ λ=±2

→

A = ± 2 (B × C) →

→

→

→

⇒

⇒

→ → (PQ) × (PR)

∴

→ → | PQ × PR|

⇒

→ → |PQ × PR|= 4 4 + 1 + 1 = 4 6 4(2i$ + $j + k$ ) (2$i + $j + k$ ) =± =± 4 6 6 → → |PQ × PR|

→

→

→

→

and a × c = b × d

→

→

→

→

→

→

→

a × b−a × c= c ×d −b ×d →

⇒

→ → | PQ× PR|

→ PR = − $i + 3$j − k$ $i $j k$ → → PQ × PR = 1 1 −3 −1 3 −1

→

24. Given, a × b = c × d

⇒

= $i ( − 1 + 9) − $j(− 1 − 3) + k$ (3 + 1) $ = 8i + 4j + 4k

∴

→

|A|=| λ ||B × C|, where A , B, C are unit vectors.

→ → (PQ) × (PR)

→ PQ = $i + $j − 3k$

→ → PQ × PR

→

A = λ (B × C)

⇒

21. A unit vector perpendicular to the plane determined by P ,Q, R

→

Hence, given statement is true. →

a⋅ (b × c )

= a1 b + a 2 c + a3

→

∴

→ → →

→

→

→ a ⋅ c = a ⋅ $j = $j ⋅ (a1$i + a 2$j + a3 k$ ) = a 2

→

→ →

⇒

→

∴ (a ⋅ b ) b + (a ⋅ c ) c +

⇒

→ →

23. Given, A⋅ B = A⋅ C = 0 →

→ → →

0

1 → → ⇒ Area of triangle = |AB × AC| 2 1 = ⋅ 2 ⋅ 9 + 4 = 13 sq units 2

a ⋅ b = a ⋅ $i = $i ⋅ (a1i$ + a 2$j + a3 k$ ) = a1

and

∴

∴

⇒ A is perpendicular to both B and C.

a = a1$i + a 2$j + a3 k$

Now,

and

→

→

Let

$i $j k$ → → AB × AC = 1 2 −3 = 2(−3$j − 2k$ ) 2 0

→

⇒

→

22. Area of ∆ABC = |AB × AC|

→

→

→

→

→

a × (b − c ) = ( c − b) × d

→

→

→

→

→

→

→

→

→

a × (b − c ) − ( c − b) × d = 0 →

→

→

a × (b − c ) − d × (b − c ) = 0 →

→

→

→

→

→

→

→

→

→

(a − d ) × (b − c ) = 0 ⇒ (a − d )||(b − c ) →

→

(a − d ) ⋅ (b − c ) ≠ 0 → →

→ →

→ →

→ →

a⋅ b + d⋅ c ≠ d ⋅ b + a ⋅ c

25. (i) Since,

→ →

→

→

u ⋅ v =|u||v|cos θ

→

→

→

→

u × v =|u||v|sin θ n$

and

→

→

where, θ is the angle between u and v and n$ is unit →

→

vector perpendicular to the plane of u and v. → →

→

→

→

|u ⋅ v|2 =|u|2|v|2 cos 2 θ and

Again,

→

→

→

→

→

→

→

→

→

→

→

|u × v|2 =|u|2|v|2 sin 2 θ =|u|2 |v|2 sin 2 θ → →

∴ |u ⋅ v|2 + |u × v|2 =|u|2 |v|2 (cos 2 θ + sin 2 θ ) =|u|2 |v|2 →

→

→

→

…(i)

(ii)|u + v + ( u × v)| →

→

→

2 →

→

→

→

→

=|u + v|2 + |u × v|2 + 2 (u + v) ⋅ (u × v)

Vectors 567 →

→

→ →

→

→

=|u|2 + |v|2 + 2 u ⋅ v + |u × v|2 + 0 →

→

→

→

[Q u × v is perpendicular to the plane of u and v] →

→

→

→

→ → 2

∴ |u + v + (u × v)| + |1 − u ⋅ v| 2

→

→

→ →

→

→

→

→

Topic 3 Scalar Triple Product/Dot Product/Mixed Product 1. Given vectors, µ$i + $j + k$ , $i + µ$j + k$ , $i + $j + µk$ will be coplanar, if

→

→ →

µ 1 1 1 µ 1 =0 1 1 µ

→ →

=|u|2 + |v|2 + 2 u ⋅ v + |u × v|2 + 1 − 2 u ⋅ v + |u ⋅ v|2 →

=|u|2 + |v|2 + 1 + |u|2 |v|2 → 2

→ 2

[from Eq. (i)]

→ 2

→ 2

→

=|u| (1 + |v| ) + (1 + |v| ) = (1 + |v| ) (1 + |u|2 ) → → →

26. Let the position vectors of points A, B, C, D be a , b, c and →

d, respectively. →

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

AB = b − a , BC = c − b, AD = d − a,

Then,

µ = 1 − 2 = − 1.

BD = d − b, CA = a − c , CD = d − c → → → → → → Now,|AB × CD + BC × AD + CA × BD| →

→

→

→

→

→

→

→

2. We know that, if a , b , c are coplanar vectors, then [a b c ] = 0 →

→

→

→

=|(b − a ) × (d − c ) + ( c − b) × (d − a ) + (a − c ) × (d − b)| → → → → → → → → → → → → → → =|b × d − a × d − b × c + a × c + c × d − c × a − b × d →

→

→

→

→

→

→

→

→

→

+ b × a + a × d − a × b − c × d + c × b| →

→

→

→

→

⇒ µ (µ 2 − 1) − 1 (µ − 1) + 1 (1 − µ ) = 0 ⇒ (µ − 1) [ µ (µ + 1) − 1 − 1] = 0 ⇒ (µ − 1) [ µ 2 + µ − 2] = 0 ⇒ (µ − 1) [(µ + 2) (µ − 1)] = 0 ⇒ µ = 1 or − 2 So, sum of the distinct real values of

→

= 2| a × b + b × c + c × a )

…(i)

Also, area of ∆ ABC → → → 1 → → 1 → = |AB × AC| = |(b − a ) × ( c − a )| 2 2 1 → → → → → → → → = |b × c − b × a − a × c + a × a| 2 1 → → → → → → …(ii) = |a × a + b × c + c × a| 2 From Eqs. (i) and (ii),

∴

→ → → 27. Since, OA1 , OA2,... , OAn are all vectors of same magnitude and angle between any two consecutive vectors is same i.e. (2π / n ). 2π → → $ OA1 × OA2 = a 2 ⋅ sin ⋅p n

∴

…(i)

where, $p is perpendicular to plane of polygon. n −1

Now,

→ → ∑ (OA i × OA i + 1 ) =

i =1

n −1

2π ∑ a 2 ⋅ sin n ⋅ $p i =1

= (n − 1) ⋅ a 2 ⋅ sin

2π $ ⋅p n

4

1 λ 2 4

4

=0

λ2 − 1

⇒ 1 { λ (λ2 − 1) − 16} − 2((λ2 − 1) − 8) + 4 (4 − 2λ ) = 0 ⇒ λ3 − λ − 16 − 2λ2 + 18 + 16 − 8λ = 0 ⇒ λ3 − 2λ2 − 9λ + 18 = 0 ⇒ λ2(λ − 2) − 9 (λ − 2) = 0

⇒ (λ − 2)(λ2 − 9) = 0 ⇒ (λ − 2) (λ + 3) (λ − 3) = 0 ∴ λ = 2, 3 or − 3 If λ = 2 , then $i $j k$ a × c = 1 2 4 = $i (6 − 16) − $j(3 − 8) + k$ (4 − 4) 2 4 3 = − 10$i + 5$j $i $j k$

→ → → → → → |AB × CD + BC × AD + CA × BD| 2 (2 area of ∆ ABC ) = 4 (area of ∆ ABC)

1 2

If λ = ± 3, then a × c = 1 2 4 = 0 2 4 8 (because last two rows are proportional). →

→

→

→

→

→

3. Let angle between a and b be θ1 , c and d be θ 2 and a × b →

→

and b × d be θ. →

⇒

→

→

→

(a × b) ⋅ ( c × d ) = 1

Since,

sin θ1 ⋅ sinθ 2 ⋅ cosθ = 1

⇒

θ1 = 90° , θ 2 = 90° , θ = 0° →

→ →

→

→

→

→

→

⇒ a ⊥ b, c ⊥ d , (a × b)||( c × d )

→ → = (n − 1) [ OA1 × OA2]

So,

→ → = (1 − n ) [OA2 × OA1 ] = RHS

⇒

→

→

→

→

→

→

→

→

a × b = k ( c × d ) and a × b = k ( c × d )

→

→

→

→

→

→

(a × b) ⋅ c = k ( c × d ) ⋅ c

568 Vectors →

→

→

→

→

→

6. We know that, volume of parallelopiped whose edges

and (a × b) ⋅ d = k ( c × d ) ⋅ d →→→

⇒

[a b c ] = 0 → → →

and

→ → →

[a b d ] = 0

∴

→ → →

⇒ a , b, c and a , b, d are coplanar vectors, so options (a) and (b) are incorrect. →

→

→

→

Let b||d ⇒ b = ± d →

→

→

⇒ ⇒

→

→

→

(a × b) ⋅ ( c × d ) = 1 ⇒

As

→

→

→

(a × b) ⋅ ( c × b) = ± 1

→ →→

→→→

→

→

→ → →

[ a × b c b] = ± 1 ⇒ [ c b a × b] = ± 1 →

→

→

→

c ⋅ [b × (a × b)] = ± 1 ⇒

→

c ⋅ [a − (b ⋅ a ) b] = ± 1

→ →

⇒

→ →

c⋅a = ± 1

[Q a ⋅ b = 0] ←

which is a contradiction, so option (c) is correct.

b

→

→

→

α

←

a

60º

→

c =± b

→

→

→

→

→

→

→

→

←

←

d

As

(a × b) ⋅ ( c × d ) = 1

⇒

(a × b) ⋅ (b × a ) = ± 1

a 1

1 a = 1 + a3 − a

a

0

1

→ → = U ⋅ (3i$ − 7$j − k$ ) =| U||3$i − 7$j − k$ | cos θ

d=±a

and

1 →→→ [a b c ] = 0

Let f (a ) = a3 − a + 1 ⇒ f ′ (a ) = 3a 2 − 1 ⇒ f ′ ′ (a ) = 6a For maximum or minimum, put f ′ (a ) = 0 1 1 , which shows f (a ) is minimum at a = ⇒a =± 3 3 1 and maximum at a = − . 3 → → 7. Given, V = 2i$ + $j − k$ and W = $i + 3k$ →→ → → [ U V W ] = U ⋅ [( 2$i + $j − k$ ) × ( $i + 3k$ )]

Let option (d) is correct. ⇒

→→→

are a , b , c = [a b c ].

→→→

c

→ which is maximum, if angle between U and 3$i − 7$j − k$ is 0 and maximum value =|3$i − 7$j − k$ | = 59 −1 1−x y x 1+ x− y

1 0

→→→

8. [a b c ] = x 1

which is a contradiction, so option (d) is incorrect. Alternatively, options (c) and (d) may be observed from the given figure. 4. The volume of the parallelopiped with coterminus edges $ , c$ is given by [a$ $b $c] = a$ ⋅ (b $ × c$ ) as a$ , b

Applying C3 → C1 + C3 , 1 0

0

x 1 1 =1 y x 1+ x

Z

Therefore, it neither depends on x nor y. → → →

→

c

b

Y

←

a

$ c$ ]2 = 1 1 − 1 − 1  1 − 1 + 1 [a$ b  4 2  2 4 2

 1 1 1  − =  4 2 2

Thus, the required volume of the parallelopiped 1 cu unit = 2

i.e.

→

→

→

[2 a − b

→

2b− c

→ → →

→

→

2 c − a] = 0

→ →→

→→ →

Therefore, (c) is the answer. → → →

11. Since, a , b, c are linearly dependent vectors.

− λ2 1 1 − λ2 1

1

→

⇒

4 3 4 =0 1 α β

Applying C 2 → C 2 − C1 , C3 → C3 − C1, 1 0 0 −1

= 0 ⇒ − (β − 1) = 0 ⇒ β = 1

0

1 α −1 β −1

1 1

=0

−λ

2

2

→

[a b c ] = 0 1 1 1

4

⇒ λ − 3λ − 2 = 0 ⇒ (1 + λ ) (λ − 2) = 0 ⇒ λ = ± 2 2 2

→

⇒

5. Since, given vectors are coplanar

2

→

→

→→ →

$ a$ ⋅ c$ a$ ⋅ a$ a$ ⋅ b 1 1 /2 1 /2 2 $ $ $ $ $ $ $ $ $ Now, [a b c] = b ⋅ a b ⋅ b b ⋅ c = 1 / 2 1 1 / 2 $ c$ ⋅ c$ c$ ⋅ a$ c$ ⋅ b 1 /2 1 /2 1

6

→

10. [u v w] =|v w u|= [w u v] = − [ v u w]

X

∴

→

2 c − a are also coplanar vectors.

←

⇒

→

9. If a , b, c are coplanar vectors, then 2 a − b, 2 b − c and

←

→

|c|= 3

Also, ⇒ ⇒

1 + α + β =3 2

2

[given] $ [given, c = i + α$j + βk$ ]

1 + α2 + 1 = 3 ⇒ α2 = 1⇒ α = ± 1

Vectors 569 →

→

→

→

→

→

→

12. ( a + b + c ) ⋅ [(a + b) × (a + c )] →

→

→

→

→

→

→ →

→

→

→

→

→

→

→

→

→ →

→

→ →

→

→

→

→

= { a⋅ (a × c ) + a⋅ (b × a ) + a⋅ (b × c )} + { b⋅ (|a × c ) → →

→

→ →

→

→ →

→

⇒

→ →

→

+ b⋅ (b × a ) + b⋅ (b × c )} + { c⋅ (a × c ) + c⋅ (b × a ) + c⋅ (b × c )} →→→

→→→

→→→

→→→

= [a b c ] + [b a c ] + [ c b a ] = [a b c ]

…(i)

→

→ →

a⋅ d = 0

⇒

⇒

→

→

→

→

α =0

and

→

→

→

→

→

→

→

→

→

→

A → →

→

→

→

→

→

→

B

→

= A⋅ (B × A + B × B →

→

→

→

→

C →

→

+ B × C + C × A + C × B + C × C) → → → → → → = A⋅ (B × A) + A⋅ (B × C) + A⋅ (C × B)

x+ y+ z =0

→

→

→→→

→→→

= [ A B A] − [ A B C] = 0 →

[from Eq. (ii)] …(iii)

19. Let θ be the angle between u$ and v. →

∴ 1 6

→

| u × v| = 1 ⇒ | u | v|sin θ = 1 →

∴

| v|sin θ = 1

[Q| u | = 1] …(i) P

1 $ $ (i + j − 2k$ ) d=± 6 v

14. Since, three vectors are coplanar. c

1

θ O

Q

u →

→

Clearly, there may be infinite vectors OP = v, such that $ P is always 1 unit distance from u.

0 a c Applying C1 → C1 − C 2 , 1 0 1 = 0 0 c b

∴ Option (b) is correct.

⇒ − 1 (ab − c ) = 0

Again, let φ be the angle between ww and u × v.

ab = c → → → → π (a × b) =|a||b|sin ⋅ n$ 6 → → → 1 → → → $ ⋅c (a × b) ⋅ c = |a||b|⋅ n 2 →→→ 1 → → [a b c ] = |a||b| ⋅ cos 0° 2 2

15. Since,

⇒

→

2

cos φ = 1 ⇒ φ = 0 →

w=u×v

Thus, →

Now, if u$ lies in XY -plane, then $j k i →

→

u × v = u1 v1

vector perpendicular to both a and b. 2

a 2 a3  →→→ 2 1 → 2 → 2 b2 b3   = [a b c ] = 4 ⋅|a| |b| c2 c3  1 = (a12 + a 22 + a32 ) (b12 + b22 + b32 ) 4

 2 −3 0   16. The volume of parallelopiped = [a b c ] =   1 1 −1   3 0 −1  = 2 (−1) + 3 (−1 + 3) = − 2 + 6 = 4

→

w ⋅ (u × v) = 1 ⇒ | w|| u × v|cos φ = 1

⇒

Q n$ is perpendicular to both a, b and c is also a unit →

→

∴

→ →

 a1 ∴   b1  c1

O

[Q it is a scalar triple product of three

→

0 1 =0 c b

→

→

→

x2 + x2 + 4x2 = 1 ⇒ x = ±

1 c

→

18. A ⋅ {(B + C) × (A + B + C)}

…(ii)

From Eqs. (i), (ii) and (iii),

a

→

→

→

→

2x + z = 0

a

→

→ →

 0 1 −1   −1 0 1 = 0 ⇒    x y z 

∴

→

$ a ⊥ b and c||n

[b c d ] = 0

⇒

→

→

→→→

⇒

→

vectors of the form A, B + C, A + B + C]

x− y=0 ⇒ x= y

Also,

→

i.e. a ⊥ b and c perpendicular to both a and b.

[Q d being unit vector] Since,

→

|a||b||c||sin θ ⋅ cos α | = | a||b||c| π |sin θ ||cos ⋅ α |= 1 ⇒ θ = 2 →

d = x$i + y$j + zk$

where, x2 + y2 + z 2 = 1

→

∴

→

13. Let

→

⇒ ||a||b|sin θ n$ ⋅ c|=|a||b||c|

= (a + b + c ) ⋅ [a × a + a × c + b × a + b × c ] → →

→

|(a × b ) ⋅ c|=|a||b||c|

17. Given,

∴

∴

→→→

⇒

u2 0 v2 v3

or

u = ui$ + u2$j

$ w = (u2v3 ) $i − (u1v3 )$j + (u1v2 − v1u2)k 1 $ $ $) = (i + j + 2k 6 1 −1 , u1v3 = u2v3 = 6 6 u2v3 = − 1 or| u1 | = | u2| u1v3

∴ Option (c) is correct.

570 Vectors $ Now, if u$ lies in XZ-plane, then u = u1i$ + u3 k $i u × v = u1

$j 0

$ k u3

v1

v2

v3

→

∴

⇒ ⇒

$ ⇒ w = (− v2u3 ) $i − (u1v3 − u3 v1 ) $j + (u1v2) k

→

$ = 1 ($i + $j + 2k $) ⇒ k 6 1 2 and u1v2 = ⇒ − v2u3 = 6 6 ∴

23.

→

→

→

→

→

+

(C × A ) ⋅ B

→

→

[B A C] → → →

Therefore, (a) is the answer. (b) u⋅ ( v ⋅ w) is not meaningful, since v ⋅ w is a scalar quantity and for dot product both quantities should be vector. Therefore, (b) is not the answer.

⇒

a a 2 a3 a2 1 2 b b 1 + b b2 b3 = 0 c c2 c3 c c2 1

⇒

1 a a2 (1 + abc) 1 b b2 = 0 1 c c2

→ →

(d) u × ( v ⋅ w) is not meaningful, since v ⋅ w is a scalar quantity and for cross product, both quantities should be vector. Therefore, (d) is not the answer. →

But (1, a , a 2), (1, b, b2), (1, c, c2) are non-coplanar.

→ → → → adjacent sides =|OA × OC|=| a × b|

→

→

→

→

p = 6q ⇒

∴ →

→

→

→

→

→

→

→

→

→

→

→

→ →

→

→ →

→

→→→

→→→

= a⋅ (b × c ) − b⋅ ( c × a ) = [a b c ] + [a b c ] →→→

→ →

→

= 2 [a b c ] = 2 a⋅ (b × c )

k =6

Hence, it is a true statement.

a 1 1 1 b 1 =0 1 1 c

→ →

→ →

→ →

26. Since, X⋅ A = X⋅ B = X⋅ C = 0 →

→ → →

→→→

⇒ X is perpendicular to A , B, C, therefore [A B C] = 0

a / (1 − a ) 1 / (1 − b) 1 / (1 − c) 1 −1 0 =0 1

⇒

→

25. (a − b) ⋅ {(b − c ) × ( c − a )} = (a − b) ⋅ (b × c − b × a + c × a )

→

Applying R2 → R2 − R1, R3 → R3 − R1, 1 1 a 1 − a b −1 0 =0 1−a 0 c−1 ⇒

abc = − 1

→

22. Since, vectors are coplanar. ∴

1 a a2 1 b b2 ≠ 0 1 c c2

⇒

=|a × b|+ 5|a × b|= 6|a × b| ∴

1 a a2 1 b b2 = 0 1 c c2

⇒ Either (1 + abc) = 0 or

21. Since, q = area of parallelogram with OA and OC as

→

=0

1 + c3

→ → →

and p = area of quadrilateral OABC 1 → → 1 → → = |OA × OB|+ |OB × OC| 2 2 → → → → → 1 → 1 = |a × (10 a + 2 b) + |(10 a + 2 b) × b| 2 2

→ → →

[C A B]

a

(c) (u⋅ v) w is meaningful, since it is a simple multiplication of vector and scalar quantity. Therefore, (c) is the answer.

→

→→→

[A B C] − [A B C]

[C A B]

c2

c

→ →

→ →

→→→

=

a a 2 1 + a3 24. Since, b b2 1 + b3 = 0

→

20. (a) u ⋅ ( v × w) is a meaningful operation.

→

→

→ → →

+

[C A B]

→ → →

→

C⋅ (A × B)

[A B C] → → →

→

B ⋅ (A × C)

→ → →

=

∴Option (d) is wrong. →

→

A⋅ (B × C)

| u2| = 2| u3 | →

a 1 1 + + =0 1−a 1−b 1−c 1 1 1 −1 + + + =0 1 − a 1 − b 1 −c 1 1 1 + + =1 1−a 1−b 1−c

⇒

0

−1

a 1 1 (1) − (−1) + (1) = 0 1−a 1−b 1−c

Hence, given statement is true. 27.

 1 α → → = sec (u + v)  → → 2 2  | u + v|  → → → w+ v 1 β → →  = sec ( v + w) y= → → 2 2  | v + w|  → → → w+ u 1 γ → →  z= = sec (w + u )  → → 2 2 | w + u|  →

x=

→

→

u+ v

…(i)

Vectors 571 →

=

→ →

→ →

→

→→→

[x × y y × z z × x ] = [x y z ]2

Since,

[from Eq. (i)]

1 α β γ → →→ → → → sec2 ⋅ sec2 ⋅ sec2 [u + v v + w w + u ]2 …(ii) 64 2 2 2 →

→ →

→ →

→

→→ →

and [u + v v + w w + u ] = 2 [u v w] →

→ →

→ →

…(iii)

→

…(i)

Similarly, (b1 + b2 + b3 ) ≥ b12 + b22 + b32 (c1 + c2 + c3 ) ≥ c12 + c22 + c32 L3 ≥ [(a 12 + a 22 + a32 ) (b12 + b22 + b32 ) (c12 + c22 + c32 )]1/ 2

⇒

L3 ≥ V →

→

→

[from Eq. (i)] →

29. Given equation is w + (w × u ) = v →

→

→

→

→

→

→

→

→

→ → →

→

→

→

u × w + u × (w × u ) = u × v → → →

⇒ u × w + (u ⋅ u ) w − (u ⋅ w) u = u × v → → → → → → → → u × w + w − (u ⋅ w) u = u × v

…(ii) →

Now, taking dot product of Eq. (i) with u, we get → →

→

→

→

→ →

u ⋅ w + u⋅ (w × u ) = u ⋅ v → →

⇒

→ →

→

→

→

u⋅ w = u ⋅ v [Q u ⋅ (w × u ) = 0]

…(iii)

→

Now, taking dot product of Eq. (i) with u, we get → →

→

→

→

→ →

v ⋅ w + v⋅ (w × u ) = v ⋅ v

⇒

→ →

→→ →

→ →

→

→

→

→

→

…(v)

→

→

→

→

→ →

(u × v) ⋅ w = |w|2 − (u ⋅ w)2

→

→

→

→

→ →

→

→ →

|w|2 + 0 = v ⋅ w → 2

…(vi)

→

→

→

→

→

|w| = 1 − (u × v) ⋅ w

[from Eq. (iv)]

→

→

→

→

→ →

→

→ →

⇒ ⇒

→

→

→

→ →

2 (u × v) ⋅ w = 1 − (u ⋅ v)2 →

→

→

(u × v) ⋅ w =

[from Eq. (iii)]

→ → 2

1 1 1 − (u ⋅ v) ≤ 2 2

→ →

[Q (u ⋅ v)2 ≥ 0]

→ →

→

→ →

→

→

perpendicular to v. → 30. Here, AB = − $i + 5$j − 3k$ → → AC = − 4i$ + 3$j + 3k$ and AD = i$ + 7$j + (λ + 1) k$ → → → We know that, A , B, C , D lie in a plane if AB, AC, AD are coplanar i.e. −1 5 −3 → → → [AB AC AD] = 0 ⇒ −4 3 3 =0

→

1

7 λ+1

→

1 − v ⋅ w = (u × v) ⋅ w

−17λ + 146 = 0 146 ∴ λ= 17 $ $ $ 31. Given vectors a = 2i + j − k and b = $i + 2$j + k$ So,

a + b = 3$i + 3$j

⇒

|a + b| = 3 2

Since, it is given that projection of c = α a + β b on the vector (a + b ) is 3 2, then (a + b ). c =3 2 |a + b| (a + b ) . (αa + βb ) = 18

⇒ α (a. a ) + β (a. b ) + α (b. a ) + β (a . b ) = 18

→ →

⇒ − (u × v) ⋅ w − v ⋅ w + 1 = 0 ⇒

→

(u × v) ⋅ w + |w|2 − (u ⋅ w)2 = 1 − (u × v) ⋅ w − (u ⋅ w)2

⇒

→→ →

v ⋅ w + [ v w u ] = 1 ⇒ v⋅ w + [ v w u ] − 1 = 0 →

→

⇒

u × [w + (w × u )] = u × v

⇒

→

⇒ −1 (3λ + 3 − 21) − 5 (−4λ − 4 − 3) − 3 (−28 − 3) = 0

Taking cross product with u, we get

→

→ → 2

…(i)

→

→

→

The equality holds if and only if u ⋅ v = 0 or iff u is

(a1 + a 2 + a3 ) ≥ a12 + a 22 + a32

→

→

…(ii)

= a12 + a 22 + a32 + 2a1a 2 + 2a1a3 + 2a 2a3 ≥ a12 + a 22 + a32

→

→

Again, from Eq. (v), we get

c12 + c22 + c32

Now, (a1 + a 2 + a3 )2

⇒

→

0 + |w| − (u ⋅ w) = (u × v) ⋅ w

⇒

→

→

⇒

⇒

⇒ L3 ≥ [(a1 + a 2 + a3 ) (b1 + b2 + b3 ) (c1 + c2 + c3 )]

∴

→

→ 2

→

[using AM ≥ GM]

and

→

w ⋅ w + (w × u ) ⋅ w = v ⋅ w

(a + a 2 + a3 ) + (b1 + b2 + b3 ) + (c1 + c2 + c3 ) Now, L = 1 3 ≥ [(a1 + a 2 + a3 ) (b1 + b2 + b3 ) (c1 + c2 + c3 )]1/3

⇒

→

→

V = |a ⋅ (b × c )|≤ a12 + a 22 + a32 b12 + b22 + b32

→

Taking dot product of Eq. (i) with w, we get

→→ → 1 α β γ = sec2 ⋅ sec2 ⋅ sec2 ⋅ 4 [u v w]2 64 2 2 2 β γ 1 →→ → 2 2α = ⋅ [u v w] sec ⋅ sec2 ⋅ sec2 16 2 2 2 →

→

(u × w) ⋅ w + w ⋅ w − (u ⋅ w) (u ⋅ w) = (u × v) ⋅ w

⇒

→

∴[x × y y × z z × x ]

28.

→

Taking dot product of Eq. (ii) with w, we get

…(iv)

⇒

6α + 3β + 3α + 6β = 18

⇒

9α + 9β = 18

⇒

(α + β) = 2

…(i)

572 Vectors Now, for minimum value of (c − (a × b )) . c

$ − (a$ ⋅ b $ ) c$ = ⇒ (a$ ⋅ c$ ) b

= (α a + β b − (a × b )) . (αa + βb ) = α 2 (a. a ) + αβ (a. b ) + αβ (a . b ) + β 2(b . b ) [Q (a × b ) . a = 0 = (a × b ) . b ] = 6 α 2 + 6 αβ + 6β 2 = 6 (α 2 + β 2 + αβ ) = 6 [(α + β )2 − αβ ] = 6 [4 − αβ ] = 6 [4 − α (2 − α )] = 6 [4 − 2α + α 2] The minimum value of 6 (4 − 2α + α 2) = 6(3) = 18 D , if a > 0] [As minimum value of ax2 + bx + c = − 4a

1. We have, (a × c ) + b = 0

4. Given,

⇒

⇒ a × (a × c ) + a × b = 0 (taking cross product with a on both sides) $i $j k$

⇒

⇒ (a ⋅ c )a − (a ⋅ a ) c + 1 −1 0 = 0 1

On comparing, we get $ = − 3 ⇒ |a$ ||b $ |cos θ = − 3 a$ ⋅ b 2 2 3 $ |= 1 ] [Q |a$ | = |b ⇒ cos θ = − 2 π 5π  ⇒ cos θ = cos  π −  ⇒ θ =   6 6

⇒

Topic 4 Vector Triple Product

1

1

[Q a × (b × c ) = (a . c )b − (a ⋅ b )c] ⇒ 4($i − $j) − 2c + (− $i − $j + 2k$ ) = 0 [Q a ⋅ a = ($i − $j)($i − $j) = 1 + 1 = 2 and a ⋅ c = 4] ⇒ 2c = 4i − 4$j − $i − $j + 2k$ 3$i − 5$j + 2k$ 9 + 25 + 4 19 |c|2 = = ⇒ ⇒ c= 2 4 2 1 1 2. Given, a × (b × c) = b ⇒ (a ⋅ c)b − (a ⋅ b)c = b 2 2 [Q a × (b × c) = (a ⋅ c)b − (a ⋅ b)c] On comparing both sides, we get 1 …(i) a⋅c = 2 and …(ii) a⋅b = 0 Q a , b and c are unit vectors, and angle between a and b is α and angle between a and c is β, so 1 [from Eq. (i)] |a||c|cos β = 2 1 [Q|a|= 1 =|c|] cos β = ⇒ 2 π π 1  …(iii) Q cos = β= ⇒  3 2  3 and|a||b|cos α = 0 [from Eq. (ii)] π ⇒ α= 2 From Eqs. (iii) and (iv), we get π π π |α − β| = − = = 30º 2 3 6 $ 3. Given,|a| = |b$ | = |c$ | = 1 $ × c$) = 3 ( b $ + c$) and a$ × ( b 2 $ × c$ ) = 3 ( b $ + c$ ) Now, consider a$ × ( b 2

→ → → 1 → → → (a × b) × c = b c  a 3 → → → 1 → → → − c × (a × b) = b c  a 3 → → → → → → 1 → → → − ( c⋅ b) ⋅ a + ( c⋅ a ) b = bc a 3 → →  → → → 1 → →  3 bc+ ( c⋅ b) a = ( c⋅ a ) b

Since, a and b are not collinear. → → → → 1 → → ∴ c⋅ b + bc= 0 and c⋅ a = 0 3 → →  → → 1 → → 1 ⇒ bc cosθ + bc= 0 ⇒ bc  cos θ +  = 0  3 3 1 [Q |b|≠ 0,|c|≠ 0] ⇒ cos θ + = 0 3 1 3

cos θ = −

⇒

⇒ sin θ =

8 2 2 = 3 3 → →

5. As we know that, a vector coplanar to a , b and →

→

→

→

orthogonal to c is λ {( a × b) × c} . ∴ A vector coplanar to (2i$ + $j + k$ ), (i$ − $j + k$ ) and orthogonal to 3$i + 2$j + 6k$ = λ [{(2i$ + $j + k$ ) × (i$ − $j + k$ )} × (3i$ + 2$j + 6k$ )] = λ [(2$i − $j − 3k$ ) × (3$i + 2$j + 6k$ )] = λ (21$j − 7k$ ) ∴ Unit vector = + →

(21$j − 7k$ ) (21) + (7) 2

→

→

2

=+

(3$j − k$ ) 10

→ → →

→ → →

6. We know that, a × (a × b) = (a ⋅ b) a − (a⋅ a ) b ∴

→ (i$ + $j + k$ ) × ($j − k$ ) = (i$ + $j + k$ ) − ( 3 )2 b

⇒

→ → − 2$i + $j + k$ = $i + $j + k$ − 3 b ⇒ 3 b = + 3$i →

b = $i

∴ …(iv)

3 $ 3 b+ c$ 2 2

7. If θ is the angle between P1 and P2 , then normal to the planes are N1 N2 ∴

→ → = a × b,  → → = c × d 

N1 × N 2 = 0

Then,|N 1| × |N 2|sin θ = 0 ⇒

sin θ = 0 ⇒ θ = 0

Vectors 573 8.

→

11. There are three vectors given a , b and c , such that

In this question, vector c is not given, therefore, we → → → cannot apply the formulae of a × b × c (vector triple product).

NOTE

→

→

→

→

→

→

Now,|(a × b) × c|=|a × b||c|sin 30°

(Q it is given angle between b and c is

i$ $j k$ → → |a × b|= 2 1 −2 = 2i$ − 2$j + k$

Again,

1 1 →

…(i)

→

→

→

|c − a|= 2 2 →

|c − a|2 = 8

⇒

(c − a ) ⋅ (c − a ) = 8

⇒

→

→ →

→

→

→ →

→ →

→

→

[given]

→

⇒

→

→ →

c⋅ c − c⋅ a − a⋅ c + a⋅ a =8

⇒

Now, as a is perpendicular to the vector b × c, so π |a × (b × c)|= |a||b × c|sin  2  π = |a||b × c|=|a||b||c|sin 3 3 = 30 = ( 3 )(5)(4) 2 Hence answer is 30.

12. From the given information, it is clear that

→ →

|c|2 + |a|2 − 2 a ⋅ c = 8 → 2

→

a=

→

⇒

|c| + 9 − 2|c|= 8

⇒

|c|2 − 2|c|+ 1 = 0

→

→

→

⇒

→

→

⇒

→

→

→

→

→

→

→

→

→

[as a ⋅ b = 0]

2 2 2



O

P

X (1, 0, 0)

R (0, 0, 1) Z

Now, →

→ → →

= λ (a × b) × c = λ {(a ⋅ c ) v − (b ⋅ c ) a } = λ {(1 + 1 + 4) ($i + 2$j + k$ ) − (1 + 2 + 1) ($i + $j + 2k$ )} = λ {6 $i + 12 $j + 6 k$ − 6$i − 6$j − 12k$ } = λ {6 $j − 6 k$ } = 6λ { $j − k$ } 1 ⇒ Option (a) is correct. 6 1 and for λ = − ⇒ Option (d) is correct. 6

→

T (1, 1, 1)

∴ A vector coplanar to a and b, and perpendicular to c

For λ =

→ →

1 1 1 S , , 

and c = i$ + $j + k$ → → →

→

(0, 1, 0) Q

→

→

→

Y

10. Let a = $i + $j + 2k$ , b = $i + 2$j + k$

→

→ →

 1 1 1 S= , , .  2 2 2

→

→

→

→

→

→

→

13. Here, P(1, 0, 0), Q (0, 1, 0), R (0, 0, 1), T = (1, 1, 1) and

On equating the coefficient of c, we get → → 1 a⋅ b = − 2 → → 1 |a||b|cos θ = − ⇒ 2 1 ∴ cos θ = − 2 3π θ= ⇒ 4

→

→

= 4 ⋅1 + 1 = 5

→

a × (b × c ) =

→

→

= [2 a + b] ⋅ [b + 2 a ] = 4 a 2+ b2

b+ c 2 → → → → → → 1 → 1 → (a⋅ c ) b − (a⋅ b) c = b+ c 2 2 →

→

= (2 a + b) ⋅ [a 2 b − (a ⋅ b) ⋅ a + 2 b2 ⋅ a −2 (b⋅ a ) ⋅ b]

→ → →  1 3 From Eq. (i),|(a × b) × c| = (3) (1) ⋅   =  2 2 →

$i − 2$j → → → → ⇒ |a|= 1,|b|= 1, a ⋅ b = 0 5

Now, (2 a + b) ⋅ [(a × b) × (a − 2 b)]

(|c|− 1)2 = 0 ⇒|c|= 1

9. Since,

π ) 3

 1 ⇒ 5|c|  = 10 ⇒ |c| = 4  2

0

⇒ |a × b|= 22 + (−2)2 + 1 = 4 + 4 + 1 = 9 = 3 Since,

|a| = 3 ,|b|= 5 and b ⋅ c = 10 π So,|b||c|⋅ cos = 10 3

and

→

→

→

→

p = SP = OP − OS

1 1  1 1 =  i$ − $j − k$  = (i$ − $j − k$ ) 2 2 2  2 → → 1 q = SQ = (− i$ + $j − k$ ) 2 → → 1 r = SR = (− i$ − $j + k$ ) 2 → → 1 $ $ $ t = ST = (i + j + k) 2

574 Vectors →

i$

→

$j

→

k$

→

→

→

→

→

→

→

→ → →

→

→→→ →

Similarly, →

→

→

→

→

→

→

→→→ →

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→ →

→

→

→ →

→

→ → →

→ → →

→→→ →

→→→

→ → →

→

= [A C B] (B − C)

→→→

→

→

→

→

→

Now, [(A + B) × (A + C)] × (B × C) ⋅ (B + C) →

= ([A C B]{ B − C}) ⋅ (B + C) →

→→→ →

→→→ →

…(iii)

→→→ →

→→→ →

→→→ → →→→ →

→ →

→

…(i)

→ →

→ →

→

→

→

→

→

→ → →

…(ii)

→

[X A A × B] = X⋅ { A × (A × B)} → → →

→ →

→ →

→ →

→ →

→

→

= [A C B]{|B|2 −|C|2 } = 0

→

→

→

and A × B. Let us consider, →

→

→

→

→

A⋅X =c

Since, →

→

X = l A + m (A × B)

→

→

→

→

∴ A⋅ { l A + m (A × B)} = c ⇒ l|A|2 + 0 = c c l= ⇒ → 2 |A | Also,

→

→

→

→

→

→

→

→

→

→

→

A × X = B ⇒ A × { l A + m (A × B)} = B →

→

→

⇒

l (A × A ) + m { A × (A × B)} = B

⇒

0 − m|A|2 B = B

⇒

m=−

→

→

→

1 → 2

|A|

→

= [A C B]{(B − C) ⋅ (B + C)} →

→

→

→

[Q [a b c ] = 0, if any two of a , b, c are equal]

→

→→→ →

→

= [B A C] B − [A C B] C

→

→→→ →

So, X can be represented as a linear combination of A

[Q (a × b) × c = (a ⋅ c ) b − (b ⋅ c ) a ]

→→→

→

= ( A⋅ B)( X⋅ A ) − ( A⋅ A )( X⋅ B) = 0

→

→

→

→

→

→

→

→→→ →

→→→

→ → →

+ {(A × C) ⋅ C} B − {(A × C) ⋅ B} C

→

…(ii)

→

⇒ X , A , A × B are coplanar.

→ →

→

→

= X⋅ {(A⋅ B) A − (A⋅ A ) B}

→ → [Q A × A = 0]

→

→

→

→

and A × X = B ⇒ A⋅ B = 0 and X⋅ B = 0

= {(B × A ) ⋅ C} B − {(B × A ) ⋅ B } C

→→→

→ →→ →

17. (i) Given, A ⊥ B ⇒ A⋅ B = 0

= (B × A ) × (B × C) + (A × C) × (B × C)

→

→

→

= [B × A + A × C + B × C] × (B × C)

→

→

∴ Parallel to a.

∴ [(A + B) × (A + C)] × (B × C) →

→

→

= B × A + A × C+ B × C

→

→

→

3 2

= A × A + B × A + A × C+ B × C

→

→→→

→→→ →

Now,

→

→

→→→ →

− [a c d ] b = − 2 [b c d ] a

→

→→→

→

(a × c ) × (d × b) = [a d b] c − [ c d b] a

15. Now, (A + B) × (A + C)

→→→ →

…(i)

= [a c d ] b − [b c d ] a + [a d b] c − [b c d ] a − [a d b] c

π 3 ⇒θ= 6 2

cos θ =

→

→→→ →

= [a c d ] b − [b c d ] a

→ → → → → → → → → → → → ( a × b ) × ( c × d ) + (a × c ) × (d × b) + (a × d ) × (b × c )

→

cos θ = ±

→

→

From Eqs. (i), (ii) and (iii),

→

⇒ For θ to be acute,

→ → →

Also, (a × d ) × (b × c ) = − (b × c ) × (a × d )

→

4 cos 2 θ = 3

→

→

= [a d b] c − [a c d ] b

⇒

→

→

→

4 cos 2 θ − 4 − 8 cos 2 θ = 1

→

→

→

→

⇒

→

→

= (b × c ) × (d × a ) = [b d a ] c − [ c d a ] b

→ →

→

→

→

→

→

→

→

4 cos 2 θ ⋅|a|2 + |c|2 − 2 ⋅ 2 cos θ a⋅ c =|b|2

→

→ → →

→ → →

→

→

→

= { a ⋅ ( c × d )} b− { b ⋅ ( c × d )} d

( 2 cos θ a − c )2 = (− b)2

→

→

→

→

→

→ → →

→

( 2 cos θ ) a − c + b = 0

→

→ → →

→

⇒

→

→

→→→ →

(a⋅ c ) a − (a⋅ a ) c + b = 0

⇒

→

= [a d b] c − [b c d ] a

⇒

⇒

→

→

a × (a × c ) + b = 0

14. Given,

→

[Q (a × b) × c = (a ⋅ c ) b − (b ⋅ c ) a ]

1 $ 1 k = = 0.5 2 2

∴|( p × q) × ( r × t )| =

→

c ×d=e

→

$j k$ 1 1 (8k$ ) = k$ 2 0 = 16 2 −2 2 0

1 Now, ( p × q ) × ( r × t ) = 16

→

(a × b) × e = (a ⋅ e ) b − (b ⋅ e ) a

i$ 2

→

→

Let

$j k$ i$ → → 1 1 r× t = −1 −1 1 = (−2i$ + 2 $j ) 4 4 1 1 1

and

→

16. Considering first part (a × b) × ( c × d )

1 1 −1 −1 = (2i$ + 2 $j ) 4 −1 1 −1

1 p×q= 4

→

→

[Q |B|=|C|, given]

    → c  →  1  → → ∴ X = (A × B) A− →   → 2 2 |A|  |A| 

Vectors 575 →

(ii) Since, vector A has components A1 , A2, A3 in the coordinate system OXYZ. → ∴ A = A i$ + A $j + A k$ 1

2

3

When the given system is rotated about an angle of π / 2 , the new X-axis is along old Y-axis and new Y-axis is along the old negative X-axis , whereas z remains same. Hence, the components of A in the new system are ( A2, − A1 , A3 ). → ∴ A becomes ( A $i − A $j + A k$ ) . 2

1

→

→

→

→

⇒ 3λ − 6 = 4λ − 2 ⇒ λ = −4 2. We have, a = 2i$ + λ 1$j + 3k$ ; b = 4$i + (3 − λ 2)$j + 6k$ and

$, c = 3$i + 6$j + (λ 3 − 1)k

such that b = 2a Now, b = 2a $ = 2 (2$i + λ $j + 3k $) ⇒4$i + (3 − λ 2)$j + 6k 1 $ = 4i$ + 2λ $j + 6k $ ⇒ 4i$ + (3 − λ )$j + 6k 2

⇒

1

→

→

→ →

c ⊥ a ⇒ c⋅ a =0

Since,

→

Taking dot product of a in Eq. (i), we get → →

→ →

→ →

⇒6 + 6λ 1 + 3(λ 3 − 1) = 0 ⇒6λ 1 + 3λ 3 + 3 = 0 … (ii) ⇒ 2λ 1 + λ 3 = − 1 Now, from Eq. (i), λ 2 = 3 − 2λ 1 and from Eq. (ii) λ 3 = − 2λ 1 − 1 ∴ (λ 1 , λ 2, λ 3 ) ≡ (λ 1 , 3 − 2λ 1 , − 2λ 1 − 1) 1 If λ 1 = − , then λ 2 = 4, and λ 3 = 0 2  1  Thus, a possible value of (λ 1 , λ 2, λ 3 ) =  − , 4, 0  2 

→

→ →

c ⋅ a = p a ⋅ a + q b ⋅ a ⇒ 0 = p|a|2 + q|b ⋅ a| →   $ $ $ Q →a = 2i + j + k   |a|= 22 + 1 + 1 = 6.   →→  $ $ $ $ $ $  a⋅ b = (2i + j + k ) ⋅ (i + 2 j − k ) =2 + 2 −1 =3  

⇒

0 = p ⋅ 6 + q ⋅ 3 ⇒ q = − 2p

On putting in Eq. (i), we get →

→

→

→

→

c = p a + b (−2 p)

⇒

→

→

→

→

c = p a − 2 p b ⇒ c = p (a − 2 b)

→

c = p[(2i$ + $j + k$ ) − 2 (i$ + 2$j − k$ )]

⇒

→ c = p (−3$j + 3k$ ) ⇒ |c|= p (−3)2 + 32

→

⇒ →

→

|c|2 = p2 ( 18 )2 ⇒

⇒ ⇒

|c|2 = p2 ⋅ 18 →

1 = p2 ⋅ 18

⇒

p2 =

1 18

→

c=±

∴

[Q|c|= 1] p=±

⇒

1 3 2

1 (− $j + k$ ) 2

→ → →

4. Since, p, q , r are mutually perpendicular vectors of same magnitude, so let us consider → → →  |p|=|q|=|r|= λ and   → → → → → →  p⋅ q = q⋅ r = r⋅ p = 0 →

→

→

→

→

→

→

→

→

→

→

…(i)

Given, p × {(x − q ) × p} + q × {(x − r ) × q }

(3 − 2λ 1 − λ 2)$j = 0

⇒ 3 − 2λ 1 − λ 2 = 0 … (i) ⇒ 2λ 1 + λ 2 = 3 Also, as a is perpendicular to c, therefore a . c = 0 $ ) ⋅ (3i$ + 6$j + (λ − 1)k $)=0 ⇒ (2i$ + λ 1$j + 3k 3

…(i)

where, p and q are scalars.

Topic 5 Solving Equations and Reciprocal of Vectors write c = λb for some non-zero scalar λ. Let the vectors α = (λ − 2) a + b and β = (4λ − 2)a + 3 b are collinear, where a and b are non-collinear. ∴ We can write α = kβ, for some k ∈ R − {0} ⇒ (λ − 2)a + b = k[(4λ − 2)a + 3 b] ⇒ [(λ − 2) − k (4λ − 2)] a+ (1 − 3k)b = 0 Now, as a and b are non-collinear, therefore they are linearly independent and hence (λ − 2) − k (4λ − 2) = 0 and 1 − 3k= 0 ⇒ λ − 2 = k(4λ − 2) and 3k = 1 1 1  Q 3k = 1 ⇒ k = ⇒ λ − 2 = (4λ − 2)  3  3

→

c = pa+ qb

3

1. Two vectors c and d are said to be collinear if we can

→

3. It is given that c is coplanar with a and b, we take

→

→

+ r × {(x − p) × r } = 0 ⇒

→ →

→

→

→ →

→

→ →

→

→

→ →

→

→

→ →

→

(p⋅ p) (x − q ) − { p⋅ (x − q ) } p + (q⋅ q ) (x − r ) →

→ →

→

→

→

− {q⋅ (x − r )} q + ( r⋅ r ) (x − p) − { r⋅ (x − p)} r = 0 → →

→

→ →

→ →

→ → →

⇒ x{(p ⋅ p) + (q⋅ q ) + ( r⋅ r )} − (p ⋅ p) q → →

→ → →

→ → →

→ → →

→ → →

− (q⋅ q )r − ( r⋅ r ) p = (x⋅ p) p + (x⋅ q ) q + (x⋅ r ) r ⇒

→ → → → →→ → 3 x |λ |2 − ( p + q + r )|λ |2 = ( x ⋅ p ) p →→ → →→ → + (x⋅ q) q + (x⋅ r ) r →

Taking dot of Eq. (ii) with p, we get → → → → → → 1 3 (x⋅ p)|λ |2 −| λ |4 = (x⋅ p)|λ |2 ⇒ x⋅ p = |λ |2 2

…(ii)

576 Vectors →

⇒

→

Similarly, taking dot of Eq. (ii) with q and r respectively, we get → →

x⋅ q =

|λ |2 → → = x⋅ r 2

→ ( r − $i + $j) ⋅ [($i − $j) × ($i + k$ )] = 0 ⇒ {(x$i + y$j + zk$ ) − (i$ − $j)} ⋅ [i$ × i$ + i$ × k$ − $j × $i − $j × k$ ] = 0 ⇒ {(x − 1) $i + ( y + 1) $j + zk$ )} ⋅ [ − $j + k$ − i$ ] = 0

− (x − 1) − ( y + 1) + z = 0

…(ii)

→

Let a = a1$i + a 2$j + a3 k$ →

Since, a is parallel to Eqs. (i) and (ii), we obtain a3 = 0

→ If θ is the angle between a and $i − 2$j + 2k$ , then

(1) (1) + (− 1) (− 2) 3 =± cos θ = ± 1+1 1+4+4 2 ⋅3 cos θ = ±

^

^

^

(4 i + 3 j) ⋅ (a i + b j) = 0

4a + 3b = 0 ⇒ a : b = 3 : − 4 →

^

^

∴ c = λ (3 i − 4 j), where λ is constant of ratio. → Let the required vectors be a = p $i + q$j → →

a⋅ b

→

→

1=

4 p + 3q 5

⇒

4 p + 3q = 5 →

... (i)

→ →

a⋅ c

. → |c| 3 λp − 4 λq 2= ⇒ 3 p − 4q = 10 ⇒ 5λ On solving above equations, we get p = 2, q = − 1 → ∴ c = 2 $i − $j →

B = x$i + y$j + zk$

8. Let

→ $ → C = $j − k$ A = i$ + $j + k,

Given, →

→

→

Also, given A × B = C ⇒

^

^

^

^

^

(z − y) i − (z − x) j + ( y − x) k = j − k

⇒

z − y = 0, x − z = 1, y − x = − 1

Also,

A⋅ B = 3 ⇒ x + y + z = 3

→ →

On solving above equations, we get 5 2 x= ,y=z= 3 3

→ Thus, a vector in the direction a is $i − $j .

⇒

⇒

^

⇒

→

⇒

a1 = − a 2, a3 = 0

→

b ⋅c =0

Also, projection of a on c is

→ $ ]=0 [( r − (i$ − $j) ) (i$ − $j) (i$ + k)

⇒

→

∴

…(i) z =0 Equation of the plane containing i$ − $j and i$ + k$ is

a1 + a 2 − a3 = 0

→

| b|

⇒

and

→

Since, b and c are perpendiculars to each other. Then,

→

(x − 1) i$ ⋅ k$ + y$j ⋅ k$ + zk$ ⋅ k$ = 0

⇒

7. Let c = a ^i + b ^j

Projection of a on b is

→ ( r − $i ) ⋅ [ i$ × (i$ + $j) ] = 0 $ − $i } ⋅ [$i × $i + $i × $j] = 0 {(x $i + y $j + zk) {(x − 1) $i + y$j + zk$ } ⋅ [k$ ] = 0

⇒

1 ⋅ (− $j + k$ ) 2

→

→ [( r − $i) $i ($i + $j) ] = 0

⇒

−x = y

Hence, the required unit vector is a$ = ±

|λ |2 → → → → → → → (p+ q + r) 3 x |λ |2 − ( p + q + r )|λ |2 = 2 → → → 1 → → → ⇒ 3→ x = (p+ q + r ) + (p + q + r ) 2 → 1 → → → x = (p + q + r ) ⇒ 2 5. Equation of the plane containing $i and $i + $j is

⇒

⇒

→ a = − x$j + x k$ = x (− $j + k$ ) ⇒ |a| = 2|x|

∴

∴ Eq. (ii) becomes

⇒

4x + 4 y = 0 →

1 2

⇒ θ=

3π π or 4 4

∴

5 2 2 $ B =  i$ , $j, k  3 3 3  →

$ is the 9. Since, v$ is unit vector along the incident ray and w unit vector along the reflected ray. Hence, a$ is a unit vector along the external bisector of $ and w. $ v $ −v $ = λa$ ∴ w

6. Any vector coplanar with $i + $j + 2k$ and $i + 2$j + k$ is ^ 90° − θ a

given by →

a = x ($i + $j + 2k$ ) + y ($i + 2$j + k$ ) = (x + y) $i + (x + 2 y)$j + (2x + y) k$

This vector is perpendicular to $i + $j + k$ , if (x + y) 1 + (x + 2 y)1 + (2x + y) 1 = 0

^ w

^ v θ θ

Mirror

Vectors 577 On squaring both sides, we get $ ⋅v $ = λ2 ⇒ 2 − 2 cos 2θ = λ2 ⇒1+1−w

Thus, lines (i), (ii), and (iii) are concurrent is equivalent to say that there exist scalars t1 , t2 and t3 such that

⇒

(t2 − t3 ) a + (t3 − t1 ) b + (t1 − t2) c + t1 (t2 − t3 ) p

→

λ = 2 sin θ

$ and w. $ where, 2 θ is the angle between v $ −v $ = 2 sin θ ⋅ a$ = 2 cos (90° − θ ) a$ = − (2 a$ ⋅ v $ ) a$ Hence, w $ $ $ $ $ ⇒ w = v − 2 (a ⋅ v) a →

→

→

10. Let the position vectors of A , B, C be a, b and c → →

→

respectively and that of P , Q , R be p, q and r, → respectively. Let h be the position vector of the orthocentre H of the ∆ PQR. We have, HP ⊥ QR. Equation of straight line passing through A and → → → perpendicular to QR i.e. parallel to HP = P − h is →

→

→

→

r = a + t1 (p − h )

…(i)

where, t1is a parameter. Similarly, equation of straight line through B and →

→

→

→

perpendicular to RP is r = b + t2 (q − h )

…(ii)

→

→

→

perpendicular to PQ is r = c + t3 ( r − h )

→

1 → 1 → → → d= c+ r −h t3 t3

…(vi)

From Eqs. (iv) and (v), 1 1 → 1 → 1 → → →  −  d = a − b + p −q  t1 t2  t1 t2

…(vii)

→

…(viii)

→ Eliminating d from Eqs. (vii) and (viii), we get 1 1   1 → 1 → → →  −   a − b + p −q  t2 t3   t1 t2   1 1   1 → 1 → → → = −   b− c + q − r  t1 t2  t2 t3 

So, this is the condition that the lines from P perpendicular to BC, from Q, perpendicular to CA and from R perpendicular to AB are concurrent (by changing ABC and PQR simultaneously). $i + 3$i 11. F is mid-point of BC i.e. F = = 2$i and AE ⊥ DE 2 [given]

→

→

^

B(i )

^

C(3 i )

Now, volume of the tetrahedron 1 = (area of the base) (height) 3 2 2 1 = (area of the ∆ ABC) (DE) 3 3

→ 1 → But area of the ∆ ABC = |BC × BA | 2 1 = |2$i × ($j + k$ )| = | $i × $j + $i × k$ | =|k$ − $j| = 2 2 2 2 1 ∴ = ( 2 ) (DE ) ⇒ DE = 2 3 3

But

→

(4)2 = AE 2 + (2)2 AE 2 = 12 → 2λ + 1 $ 1 $ 1 $ AE = i+ j+ k − ($i + $j + k$ ) λ+1 λ+1 λ+1 =

→

+ t2 (t2 − t1 ) c + t1 t2 (t3 − t2) p →

^

F(2 i )

Let E divides AF in λ :1 . The position vector of E is given by 1 $ 1 $ 2 λ$i + 1 ($i + $j + k$ )  2λ + 1 $ j+ k =  i+  λ +1 λ+1 λ+1 λ+1

⇒ t2 (t3 − t2) a + t2 (t1 − t3 ) b +

D

1

⇒

→

[multiplying both sides by t1 t2 t3 ]

→

^

E

⇒

= (t2 − t1 ) [t3 b − t2 c + t2t3 (q − r )]

t22

^

AD 2 = AE 2 + DE 2

→

→

→

Since, ∆ ADE is a right angle triangle, then

⇒ (t3 − t2) [t2 a − t1 b + t1t2 (p − q )]

→

→

+ λ 1 (λ 2 − λ 3 ) a + λ 2 (λ 3 − λ 1 ) b + λ 3 (λ 1 − λ 2) c = 0 1 where, λ i = for i = 1, 2, 3 ti

⇒

and from Eqs. (v) and (vi), 1 1 → 1 → 1 → → → c+q −r  − d= b−  t2 t3  t2 t3

→

→

…(iii)

…(v)

→

→

(λ 2 − λ 3 ) p + (λ 3 − λ 1 ) q + (λ 1 − λ 2) r

^

1→ 1→ → → d = b+q −h t2 t2

→

→

On dividing by t1 t2 t3 , we get

A(i + j + k)

If the lines (i), (ii) and (iii) are concurrent, then there → exists a point D with position vector d which lies on all of them, that is for some values of t1 , t2 and t3 , which implies that 1→ 1→ → → …(iv) d = a + p−h t1 t1

→

→

→ → → + t2 (t3 − t1 ) q + t3 (t1 − t2) r = 0

Again, equation of straight line through C and →

→

→

(t1 − t3 ) q + t2 t3 (t2 − t1 ) r = 0

⇒

→ |AE|2 =

λ $ λ $ λ $ i− j− k λ+1 λ+1 λ+1 1 3λ2 [λ2 + λ2 + λ2] = 2 (λ + 1) (λ + 1)2

578 Vectors Therefore, 12 = ⇒

3λ2 (λ + 1)2

→

→

4 (λ + 1)2 = λ2

⇒

Equation of MN is r =

4λ + 4 + 8λ = λ 2

2

⇒

3λ2 + 8λ + 4 = 0

⇒

3λ + 6λ + 2λ + 4 = 0

→

⇒ 3λ (λ + 2) + 2 (λ + 2) = 0 (3λ + 2) (λ + 2) = 0

⇒

λ = − 2 / 3, λ = − 2

When λ = − 2 / 3, position vector of E is given by  2λ + 1 $ 1 $ 1 $ j+ k  i+  λ +1 λ+1 λ+1 =

2 ⋅ (−2 / 3) + 1 $ 1 1 $j + i+ k$ − 2 /3 + 1 −2 / 3 + 1 −2 / 3 + 1

−4 / 3 + 1 $ 1 $ 1 i+ j+ k$ −2 + 3 −2 + 3 −2 + 3 3 3 3 −4 + 3 1 $ 1 $ i$ + j+ k = − i$ + 3$j + 3k$ = 3 1 /3 1 /3 1 /3 =

and when λ = − 2 , position vector of E is given by, 2 × (−2 ) + 1 $ 1 $ 1 $ −4 + 1 $ $ $ i − j−k i+ j+ k= −1 −2 + 1 −2 + 1 −2 + 1 = 3$i − $j − k$ Therefore, − i$ + 3$j + 3k$ and +3i$ − $j − k$ are the answer.

12. Let O be the origin of reference. Let the position vectors → → of A and B be a and b respectively.

…(i) →

1 → 1 1 → 1→ b + (α − 1) ⋅ a+ a 2 (α + 1) α+1 2

=

→ 1 → 1 b+ (α − 1 + α + 1) a (α + 1) 2 (α + 1)

=

→ → → 1 (b + α a ) = r1 (α + 1)

⇒ P lies on MN. → → →

13. Since, a , b , c are non-coplanar vectors. →→→

⇒

[a b c ] ≠ 0 → → → → → → a×b+ b × c = pa+ qb+ r c

→

Also,

→ →

→

Taking dot product with a , b and c respectively both sides, we get →→→

p + q cos θ + r cos θ = [a b c ] p cos θ + q + r cos θ = 0

… (i) …(ii)

p cos θ + q cos θ + r = [a b c ]

…(iii)

→→→

and

On adding above equations, p+ q+ r=

↓

C(b + αa)

N

=

→→→

↓

B(b)

→ 1→ → 1 a + k b + (α − 1) a  2 2  

For k = 1 / (α + 1) {which is the coefficient of b in r1}, we get → 1→ → 1 → 1 r= a+ b + (α − 1) a  2 α + 1  2 

2

⇒

→ 1 → (b + α a ) α+1

Thus, position vector of P is r1 =

2 [a b c ] 2 cos θ + 1

…(iv)

On multiplying Eq. (iv) by cos θ and subtracting Eq. (i), we get →→→

p (cos θ − 1) =

2 [a b c ] cos θ → → → − [a b c ] 2 cos θ + 1

⇒

p=

[a b c ] (1 − cos θ ) (2 cos θ + 1)

Similarly,

q=

−2 [a b c ] cos θ (1 + 2 cos θ ) (1 − cos θ )

and

r=

[a b c ] (1 + 2 cos θ ) (1 − cos θ )

P

→→→

O

A(a)

M(a/2)

→ → → Since, BC||OA , BC = α OA = α a for some constant α. →

→

→→→

→

Equation of OC is r = t (b + α a ) and →

→

→→→

→

→

equation of AB is r = a + λ (b − a ) Let P be the point of intersection of OC and AB. Then, →

→

→

→

→

at point P, t (b + α a ) = a + λ ( b − a ) for some values of t and λ. →

⇒

→

(t α − 1 + λ ) a = (λ − t ) bs →

→

Since, a and b are non-parallel vectors, we must have tα − 1 + λ = 0 and λ = t ⇒ t = 1 / (α + 1)

Now,

→ →

→ →

→ →

→ →

→ →

→ →

a⋅ a a⋅ b a⋅ c

1 cos θ cos θ 1 cos θ [a b c ] = b⋅ a b⋅ b b⋅ c = cos θ → → → → → → 1 cos θ cos θ c⋅ a c⋅ b c⋅ c →→→ 2

Applying R1 → R1 + R2 + R3

1

1

1

= (1 + 2 cos θ ) cos θ 1 cos θ cos θ cos θ 1

Vectors 579 Applying C 2 → C 2 − C1 , C3 → C3 − C1 1 0 = (1 + 2 cos θ ) ⋅ cos θ 1 − cos θ cos θ

0

→

∴ and

1 − cos θ

→→→

[a b c ] = ( 1 + 2 cos θ ) ⋅ (1 − cos θ )

→ →

→ →

→ →

…(ii)

x (a⋅ b) + y (b⋅ b) + z ( c⋅ b) = 0

and

→

→

→

→ →

→ →

→ →

→ →

→ →

→ →

a

∴

1 r= 1 + 2 cos θ

…(iii)

∆ =0 ⇒

a⋅ a

b

a⋅ b

c

→

a⋅ c = 0

b⋅b b⋅b b⋅c

16. Since, ($i + $j + 3k$ ) x + (3$i − 3$j + k$ ) y + (−4$i + 5$j )z →

→

→

R × B= C × B $i $j k$ i$ $j

= λ ($i x + $j y + k$ z ) k$

⇒ x + 3 y − 4z = λx, x − 3 y + 5z = λy,3x + y + 0 z = λz ⇒ (1 − λ ) x + 3 y − 4z = 0, x − (3 + λ ) y + 5z = 0,

⇒

x y z = 4 −3 7 1 1 1 1 1 1 ⇒ ( y − z )i$ − (x − z ) $j + (x − y) k$ = − 10i$ − 3$j + 7k$ ⇒ y − z = − 10, z − x = − 3, x − y = 7

3x + y − λ z = 0 Since,

→ →

→→→

15. Given that, a b c are coplanar vectors. ∴ There exists scalars x, y, z not all zero, such that →

→

x a + y b + z c =0

(x, y, z ) ≠ (0, 0, 0)

∴ Non-trivial solution.

and R⋅ A = 0 ⇒ 2x + z = 0 On solving above equations, x = − 1, y = − 8 and z = 2 → ∴ R = − $i − 8$j + 2k$

→

→ →

⇒ Non-trivial solutions

1 −2 cos θ , q= p= 1 + 2 cos θ 1 + 2 cos θ

→

→ →

Since, Eqs. (i), (ii) and (iii) represent homogeneous equations with (x, y, z) ≠ (0, 0, 0) .

→ 14. Let R = x$i + y$j + z k$

∴

→ →

x (a⋅ a ) + y (a⋅ b) + z (a⋅ c ) = 0

= (1 + 2 cos θ ) ⋅ (1 − cos θ )2 ⇒

→

Taking dot with a and b respectively, we get 0 0

…(i)

⇒ ⇒

∆ =0 1−λ 1 3

3 −4 − (3 + λ ) 5 = 0 1

−λ

⇒ (1 − λ ) (3λ + λ2 − 5) − 3 (− λ − 15) − 4 (1 + 9 + 3λ ) = 0 ⇒ − λ3 − 2λ2 − λ = 0 ⇒ λ (λ + 1)2 = 0 ∴

λ = 0, − 1

25 3D Geometry Topic 1 Direction Cosines and Direction Ratios of a Line Objective Question I (Only one correct option) 1. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0 and l2 = m2 + n 2, is π (a) 3

π (b) 4

π (c) 6

(2014 Main)

π (d) 2

Topic 2 Straight Line in Space and Shortest Distance Objective Questions I (Only one correct option) 1. The distance of the point having position vector

$ from the straight line passing through the − $i + 2$j + 6k $ is point (2, 3, − 4) and parallel to the vector, 6i$ + 3$j − 4k (2019 Main, 10 April II)

(a) 2 13

2.

(b) 4 3

(c) 6

(d) 7

The vertices B and C of a ∆ABC lie on the line, x + 2 y −1 z = = such that BC = 5 units. Then, the area 3 0 4 (in sq units) of this triangle, given that the point (2019 Main, 9 April II) A(1, − 1, 2) is (a) 34 (c) 5 17

(b) 2 34 (d) 6

position vectors of the points A , B and C with respect to the origin O. If the distance of C from the bisector of the 3 , then the sum of acute angle between OA and OB is 2 all possible values of β is (2019 Main, 11 Jan II) (b) 3

(c) 4

(d) 2

x−3 y+ 2 z + 4 lies in the plane, 4. If the line, = = −1 2 3 (2016 Main) lx + my − z = 9, then l2 + m2 is equal to (a) 26 (c) 5

(b) 18 (d) 2

x−2 y−3 z −4 x−1 y−4 z −5 5. If the lines and = = = = 1 1 −k k 2 1

are coplanar, then k can have (a) (b) (c) (d)

any value exactly one value exactly two values exactly three values

(a)

3 2

(c) −

9 2 3 (d) − 2

(b) 2 9

Objective Questions II (One or more than one correct option) 7. Let L1 and L 2 be the following straight lines. x−1 y z −1 x−1 y z −1 and L 2 : = = . = = 1 −1 3 1 −3 −1 Suppose the straight line x − α y −1 z − γ L: = = l m −2 lies in the plane containing L1 and L 2 and passes through the point of intersection of L1 and L 2. If the line L bisects the acute angle between the lines L1 and L 2, then which of the following statements is/are TRUE? L1 :

3. Let 3 $i + $j, $i + 3 $j and βi$ + (1 − β)$j respectively be the

(a) 1

x−1 y+ 1 z −1 x−3 y− k z and = = = = 2 3 4 1 2 1 intersect, then the value of k is (2004, 1M)

6. If the lines

(2012 Main)

(a) α − γ = 3 (c) α − γ = 1

(b) l + m = 2 (d) l + m = 0

(2020 Adv.)

8. Three lines L1 : r = λ$i, λ ∈ R, L : r = k$ + µ$j, µ ∈ R 2

and L3 : r = $i + $j + vk$ , v ∈ R are given. For which point(s) Q on L 2 can we find a point P on L1 and a point R on L3 so that P , Q and R are (2019 Adv.) collinear? $ (a) k $ + 1 $j (c) k 2

$ + $j (b) k $ − 1 $j (d) k 2

3D Geometry 581 3. The distance of the point (1, − 5, 9) from the plane

9. Let L1 and L 2 denote the lines

x − y + z = 5 measured along the line x = y = z is

r = $i + λ (− i$ + 2$j + 2k$ ), λ ∈ R r = µ(2i$ − $j + 2k$ ), µ ∈ R

and

(2016 Main)

(a) 3 10

respectively. If L3 is a line which is perpendicular to both L1 and L 2 and cuts both of them, then which of the (2019 Adv.) following options describe(s) L3 ? 2 $ ) + t (2$i + 2$j − k $ ), t ∈ R (a) r = (2$i − $j + 2k 9 1 $ ) + t (2$i + 2$j − k $ ), t ∈ R (b) r = (2$i + k 3 2 $ ) + t(2$i + 2$j − k $ ), t ∈ R (c) r = (4$i + $j + k 9 $ ), t ∈ R (d) r = t(2$i + 2$j − k

(b) 1

(c) −1

(d) − 2

(2013 Adv.)

(c) 3

(d) 4

12. A line l passing through the origin is perpendicular to the lines

(2013 Adv.)

$,− ∞ < t< ∞ l1 : ( 3 + t ) i$ + ( −1 + 2 t )$j + ( 4 + 2 t )k $,−∞ < s< ∞ l2 : (3 + 2s)i$ + (3 + 2s)$j + (2 + s) k Then, the coordinate(s) of the point(s) on l2 at a distance of 17 from the point of intersection of l and l1 is (are) 7 7 5 (a)  , ,   3 3 3

(b) (−1, − 1, 0)

(c) (1, 1, 1)

7 7 8 (d)  , ,   9 9 9

Topic 3 Equation of a Plane Objective Questions I (Only one correct option) 1. The mirror image of the point (1, 2, 3) in a plane is 4 1  7  − , − , −  . Which of the following points lies on  3 3 3 [JEE Main 2020, 8 Jan, Shift II] this plane? (b) (− 1, − 1, 1) (d) (− 1, − 1, − 1)

2. Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is[JEE Main 2020, 7 Jan, Shift I] (a) (6, 5, 2) (c) (6, 5, −2)

(b) (4, 3, 2) (d) (3, 4, −2)

1 3

(d)

(c) 3

(a) (1, − 4, 1) (c) (2, 4, 1)

1 3

(b) (1, 4, − 1) (d) (2, − 4, 1)

x−2 y+ 1 z −1 intersects the plane = = 3 2 −1 2x + 3 y − z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to

6. If the line

(b)

14

(c) 2 7

(d) 2 14

7. A perpendicular is drawn from a point on the line

x−1 y+ 1 z = = to the plane x + y + z = 3 such that the −1 2 1 foot of the perpendicular Q also lies on the plane x − y + z = 3. Then, the coordinates of Q are (2019 Main, 10 April II)

(a) (− 1, 0, 4) (c) (2, 0, 1)

(b) (4, 0, − 1) (d) (1, 0, 2)

1 2 3 3 units from the planes 4x − 2 y + 4z + λ = 0 and 2x − y + 2z + µ = 0, respectively, then the maximum (2019 Main, 10 April II) value of λ + µ is equal to

8. If the plane 2x − y + 2z + 3 = 0 has the distances and

(a) 13

(b) 15

(c) 5

(d) 9

9. If Q(0, − 1, − 3) is the image of the point P in the plane 3x − y + 4z = 2 and R is the point (3, − 1, − 2) , then the (2019 Main, 10 April I) area (in sq units) of ∆PQR is (a)

(a) (1, − 1, 1) (c) (1, 1, 1)

(b)

(2019 Main, 12 April I)

value(s) (b) 2

20 3

(2, 1, 4) to the plane containing the lines $ ) and r = (i$ + $j) + µ (− i$ + $j − 2k $) r = (i$ + $j) + λ( i$ + 2$j − k (2019 Main, 12 April II) is

(a) 14

y z are coplanar. Then, α can take = −1 2 − α

(a) 1

(d)

planes 2x − y + 2z − 4 = 0 and x + 2 y + 2z − 2 = 0, passes (2019 Main, 12 April II) through the point

y z and 11. Two lines L1: x = 5, = 3 − α −2 L2 : x = α ,

10 3

5. A plane which bisects the angle between the two given

drawn respectively on the lines y = x, z = 1 and y = − x, z = −1. If P is such that ∠QPR is a right angle, then the possible value(s) of λ is (are) (2014 Adv.) 2

(c)

4. The length of the perpendicular drawn from the point

(a) 3

10. From a point P(λ , λ , λ ), perpendiculars PQ and PR are

(a)

(b) 10 3

91 2

(b) 2 13

(c)

91 4

(d)

65 2

10. Let P be the plane, which contains the line of

intersection of the planes, x + y + z − 6 = 0 and 2x + 3 y + z + 5 = 0 and it is perpendicular to the XY -plane. Then, the distance of the point (0, 0, 256) from P is equal to (2019 Main, 9 April II) (a) 63 5

(b) 205 5

(c)

11 5

(d)

17 5

11. A plane passing through the points (0, − 1, 0) and (0, 0, 1) π with the plane y − z + 5 = 0, also 4 passes through the points (2019 Main, 9 April I)

and making an angle

(a) ( 2 , 1, 4) (c) (− 2 , − 1, − 4)

(b) (− 2 , 1, − 4) (d) ( 2 , − 1, 4)

582 3D Geometry x−1 y+ 1 z −2 meets the plane, = = 2 3 4 x + 2 y + 3z = 15 at a point P, then the distance of P from (2019 Main, 9 April I) the origin is

12. If the line

(a) 7 / 2

(b) 9 / 2

(c)

(d) 2 5

5 /2

13. The vector equation of the plane through the line of

intersection of the planes x + y + z = 1 and 2x + 3 y + 4z = 5, which is perpendicular to the plane (2019 Main, 8 April II) x − y + z = 0 is $)+ 2= 0 (b) r × ($i + k $ $)+ 2= 0 (d) r ⋅ (i − k

$)− 2= 0 (a) r ⋅ (i$ − k $ $)+ 2= 0 (c) r × (i − k

14. The length of the perpendicular from the point (2, − 1, 4) on the straight line, (a) (b) (c) (d)

x+3 y−2 z = = is −7 10 1 (2019 Main, 8 April I)

greater than 3 but less than 4 less than 2 greater than 2 but less than 3 greater than 4

on the vector perpendicular to the plane containing the $ and $i + 2$j + 3k $ , is vectors $i + $j + k (2019 Main, 8 April I) (a)

(b)

6

(c) 3 6

(d)

3 2

16. The equation of a plane containing the line of

intersection of the planes 2x − y − 4 = 0 and y + 2z − 4 = 0 and passing through the point (1, 1, 0) is (2019 Main, 8 April I)

(a) x − 3 y − 2z = − 2 (c) x − y − z = 0

(b) 2x − z = 2 (d) x + 3 y + z = 4

17. If an angle between the line,

(2019 Main, 12 Jan II)

 2 2 x − 2 y − kz = 3 is cos− 1   , then value of k is  3 3 5 3 5

(a)

3

(b)

5

(c) −

(d) −

5

3

18. Let S be the set of all real values of λ such that a plane passing through the points (− λ2, 1, 1), (1, − λ2, 1) and (1, 1, − λ2) also passes through the point (− 1, − 1, 1). (2019 Main, 12 Jan II) Then, S is equal to

(a) { 3 , −

3 } (b) {3, − 3}

(c) {1, − 1}

(d) { 3 }

19. The perpendicular distance from the origin to the plane containing the two lines,

x+ 2 y−2 z + 5 = = 3 5 7

x −1 y − 4 z + 4 and , is = = 1 4 7 (a) 11 6

11 (b) 6

(2019 Main, 12 Jan I)

(c) 11

(d) 6 11

x−3 y+ 1 z −6 x+ 5 y−2 z −3 20. Two lines and = = = = 1 3 −1 −6 7 4 intersect at the point R. The reflection of R in the (2019 Main, 11 Jan II) xy-plane has coordinates (a) (2, − 4, − 7) (c) (− 2, 4, 7)

(a) 17

(b) 7

(c) 5

(d) 12

x −3 y + 2 z − 1 22. The plane containing the line and = = −1 2 3 also containing its projection on the plane 2x + 3 y − z = 5, contains which one of the following points? (2019 Main, 11 Jan I) (a) (−2, 2, 2) (b) (2, 2, 0)

(c) (2, 0, − 2)

(d) (0, − 2, 2)

23. The direction ratios of normal to the plane through the

points (0, − 1, 0) and (0, 0, 1) and making an angle π /4 with the plane y − z + 5 = 0 are (2019 Main, 11 Jan I)

(a) 2, − 1, 1 (c) 2, 2 , − 2

(b) 2 , 1, − 1 (d) 2 3 , 1, − 1

(b) (2, − 4, 7) (d) (2, 4, 7)

points (−3, − 3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ? (2019 Main, 10 Jan II)

(a) (4, − 1, 7) (b) (2, 1, 3)

(c) (−2, 3, 5)

(d) (4, 1, − 2)

25. On which of the following lines lies the point of intersection of the line, plane, x + y + z = 2 ?

x−4 y−5 z−5 = = 1 1 −1 x−2 y−3 z+ 3 (c) = = 2 2 3

(a)

x −4 y −5 z −3 = = 2 2 1

and the

(2019 Main, 10 Jan II)

x+ 3 4− y z + 1 = = 3 3 −2 x−1 y− 3 z + 4 (d) = = 1 2 −5

(b)

26. The plane passing through the point (4, − 1, 2) and

x+ 1 y−2 z −3 and = = 2 1 −2

the plane,

through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x − 5 y = 15, then 2α − 3β is (2019 Main, 11 Jan II) equal to

24. The plane which bisects the line segment joining the

15. The magnitude of the projection of the vector 2$i + 3$j + k$

3 2

21. If the point (2, α , β) lies on the plane which passes

x+ 2 y−2 z + 1 = = −1 3 2 x−2 y−3 z −4 also passes through the point and = = 1 2 3 parallel to the lines

(2019 Main, 10 Jan I)

(a) (−1, − 1, − 1) (c) (1, 1, 1)

(b) (1, 1, − 1) (d) (−1, − 1, 1)

27. Let

be a point on the line A $ and B(3, 2, 6) be a r = (1 − 3µ )i$ + (µ − 1)$j + (2 + 5µ )k point in the space. Then, the value of µ for which the vector AB is parallel to the plane x − 4 y + 3z = 1 is (2019 Main, 10 Jan I)

(a)

1 4

(b) −

1 4

(c)

1 8

(d)

1 2

28. The equation of the plane containing the straight line x y z = = and perpendicular to the plane containing the 2 3 4 x y z x y z straight lines = = and = = is 3 4 2 4 2 3 (2019 Main, 9 Jan II)

(a) 5x + 2 y − 4z = 0 (c) 3x + 2 y − 3z = 0

(b) x + 2 y − 2z = 0 (d) x − 2 y + z = 0

29. The plane through the intersection of the planes x + y + z = 1 and 2x + 3 y − z + 4 = 0 and parallel to Y -axis also passes through the point (2019 Main, 9 Jan I) (a) (3, 3, −1) (c) (3, 2, 1)

(b) (−3, 1, 1) (d) (−3, 0, −1)

3D Geometry 583 30. The equation of the line passing through

(−4, 3, 1), parallel to the plane x + 2 y − z − 5 = 0 and intersecting the line (2019 Main, 9 Jan I) x+1 y−3 z −2 is = = −3 2 −1

x+4 y− 3 z −1 = = 3 1 −1 x+4 y− 3 z −1 (c) = = 1 1 3 (a)

x+4 y− 3 z −1 = = 1 1 −1 x −4 y + 3 z + 1 (d) = = 2 1 4

(b)

31. If L1 is the line of intersection of the planes 2x − 2 y + 3z − 2 = 0, x − y + z + 1 = 0 and L 2 is the line of intersection of the planes x + 2 y − z − 3 = 0, 3x − y + 2z − 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L 2 is (2018 Main) (a)

1

4 2

(b)

1

3 2

(c)

1

2 2

(d)

1 2

32. The length of the projection of the line segment joining

the points (5, −1, 4) and (4, − 1, 3) on the plane, (2018 Main) x + y + z = 7 is

2 3 1 (c) 3

(a)

(b) (d)

2 3 2 3

$ If u is perpendicular to a and u ⋅ b = 24, then and b = $j + k. 2 (2018 Main) |u| is equal to (b) 315 (d) 84

34. If the image of the point P(1, − 2, 3) in the plane 2x + 3 y − 4z + 22 = 0 measured parallel to the line x y z = = is Q, then PQ is equal to 1 4 5 (2017 Main) (a) 3 5 (c) 42

(b) 2 42 (d) 6 5

plane passing through the point (1, − 1, − 1) having normal perpendicular to both the lines x−1 y+ 2 z −4 x−2 y+ 1 z + 7 , is and = = = = −2 −1 −1 1 3 2 20 units 74 5 units (c) 83

10 units 83 10 (d) units 74

(2017 Main)

(1, 1, 1) and perpendicular to the planes 2x + y − 2z = 5 and 3x − 6 y − 2z = 7 is (2017 Adv.) (b) − 14x + 2 y + 15z = 3 (d) 14x + 2 y + 15z = 31

37. Let P be the image of the point (3, 1, 7) with respect to the

plane x − y + z = 3. Then, the equation of the plane passing through P and containing the straight line x y z = = is 1 2 1 (2016 Adv.)

(a) x + y − 3z = 0 (c) x − 4 y + 7z = 0

(b) 3x + z = 0 (d) 2x − y = 0

(b) x + 3 y + 6 z = − 7 (d) 2 x + 6 y + 12 x = − 13

39. The distance of the point (1, 0, 2) from the point of intersection of the line plane x − y + z = 16 is (a) 2 14

x−2 y+ 1 z −2 and the = = 3 4 12

(b) 8

(2015 Main)

(c) 3 21

(d) 13

x−1 y−3 z −4 in the plane = = 3 1 −5 (2014 Main) 2x − y + z + 3 = 0 is the line

40. The image of the line

x+ 3 y−5 z−2 = = 3 1 −5 x−3 y+ 5 z−2 (c) = = 3 1 −5 (a)

x+ 3 y−5 z+ 2 = = 5 −1 −3 x−3 y+ 5 z−2 (d) = = 5 −1 −3 (b)

41. Perpendicular are drawn from points on the line x+2 y+1 z = = to the plane x + y + z = 3. The feet of −1 2 3 (2013 Adv.) perpendiculars lie on the line

x y−1 z − 2 = = −13 5 8 x y−1 z − 2 (c) = = 4 3 −7

x y−1 z − 2 = = −5 2 3 x y−1 z − 2 (d) = = 2 5 −7

(b)

42. Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2 y + 4z + 5 = 0 is 3 (a) 2

5 (b) 2

(2013 Main)

7 (c) 2

(d)

9 2

43. The equation of a plane passing through the line of

intersection of the planes x + 2 y + 3z = 2 and x − y + z = 3 and at a distance 2 / 3 from the point (2012) (3, 1, − 1) is (b) 2 x + y = 3 2 − 1 (d) x − 2 y = 1 − 2

44. The point P is the intersection of the straight line joining the points Q(2, 3, 5) and R (1, − 1, 4) with the plane 5x − 4 y − z = 1. If S is the foot of the perpendicular drawn from the point T (2, 1, 4) to QR, then the length of the line segment PS is (2012)

(a)

(b)

36. The equation of the plane passing through the point (a) 14x + 2 y − 15z = 1 (c) 14x − 2 y + 15z = 27

(a) 2 x + 6 y + 12 z = 13 (c) x + 3 y + 6 z = 7

(a) 5 x − 11y + z = 17 (c) x + y + z = 3

35. The distance of the point (1, 3, −7) from the

(a)

2x − 5 y + z = 3, x + y + 4z = 5 and parallel to the plane (2015 Main) x + 3 y + 6z = 1 is

(a)

33. Let u be a vector coplanar with the vectors a = 2$i + 3$j − k$

(a) 336 (c) 256

38. The equation of the plane containing the lines

1 2

(b)

2

(c) 2

(d) 2 2

45. If the distance of the point P (1, − 2, 1) from the plane x + 2 y − 2z = α, where α > 0, is 5, then the foot of the perpendicular form P to the plane is (2010) 8 4 7 (a)  , , −  3 3 3 1 2 10   (c)  , ,  3 3 3 

4 (b)  , 3 2 (d)  , 3

4 , 3 1 , 3

1  3 5  2

46. Equation of the plane containing the straight line x y z = = and perpendicular to the plane containing 2 3 4 x y z x y z the staight lines = = and = = is 3 4 2 4 2 3 (2010)

(a) x + 2 y − 2 z = 0 (c) x − 2 y + z = 0

(b) 3 x + 2 y − 2 z = 0 (d) 5 x + 2 y − 4 z = 0

584 3D Geometry 47. A line with positive direction cosines passes through the

54. In R3 , consider the planes P1 : y = 0 and P2 : x + z = 1. Let

point P (2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals

P3 be a plane, different from P1 and P2, which passes through the intersection of P1 and P2. If the distance of the point (0, 1, 0) from P3 is 1 and the distance of a point (α , β , γ ) from P3 is 2, then which of the following relation(s) is/are true? (2015 Adv.)

(2009)

(a) 1

(b)

2

(c)

(d) 2

3

48. If P is (3, 2, 6) is a point in space and Q be a point on the →

^

^

^

^

^

^

line r = ( i − j + 2 k ) + µ (− 3 i + j + 5 k ). Then, the value of µ for which the vector PQ is parallel to the plane (2009) x − 4 y + 3z = 1, is 1 (a) 4

1 (b) − 4

1 (c) 8

1 (d) − 8

49. A plane passes through (1, − 2, 1) and is perpendicular to two planes 2x − 2 y + z = 0 and x − y + 2z = 4, then the distance of the plane from the point (1, 2, 2) is (2006, 3M)

(a) 0

(b) 1

(c) 2

(d) 2 2

x y z 50. A variable plane + + = 1 at a unit distance from a b c origin cuts the coordinate axes at A , B and C. Centroid 1 1 1 (x, y, z ) satisfies the equation 2 + 2 + 2 = K . The x y z (2005, 2M) value of K is (a) 9 (c) 1/9

51. The value of k such that plane 2x − 4 y + z = 7, is (a) 7 (c) No real value

(b) 3 (d) 1/3

x−4 y−2 z − k lies in the = = 1 1 2 (2003, 1M)

(b) – 7 (d) 4

Objective Questions II (One or more than one correct option) 52. Let α , β, γ , δ be real numbers such that α 2 + β 2 + γ 2 ≠ 0 and α + γ = 1. Suppose the point (3, 2, − 1) is the mirror image of the point (1, 0, − 1) with respect to the plane αx + βy + γz = δ. Then which of the following statements is/are TRUE? (2020 Adv.) (a) α + β = 2 (c) δ + β = 4

(b) δ − γ = 3 (d) α + β + γ = δ

53. Consider a pyramid OPQRS located in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) with O as origin, and OP and OR along the X-axis and the Y-axis, respectively. The base OPQR of the pyramid is a square with OP = 3. The point S is directly above the mid-point T of diagonal OQ such that (2016 Adv.) TS = 3. Then, π 3 (b) the equation of the plane containing the ∆OQS is x− y=0 (c) the length of the perpendicular from P to the plane 3 containing the ∆OQS is 2 (d) the perpendicular distance from O to the straight line 15 containing RS is 2 (a) the acute angle between OQ and OS is

(a) 2α + β + 2γ + 2 = 0 (c) 2α + β − 2γ − 10 = 0

(b) 2α − β + 2γ + 4 = 0 (d) 2α – β + 2γ − 8 = 0

55. In R3 , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P1 : x + 2 y − z + 1 = 0 and P2 : 2x − y + z − 1 = 0. Let M be the locus of the foot of the perpendiculars drawn from the points on L to the plane P1. Which of the following (2015 Adv.) point(s) lie(s) on M ? 5 2 (a)  0 , − , −   6 3 1 5  (c)  − , 0 ,   6 6

1 1 1 (b)  − , − ,   6 3 6 2 1  (d)  − , 0 ,   3 3

56. Let P1 : 2x + y − z = 3 and P2 : x + 2 y + z = 2 be two planes. Then, which of the following statement(s) is (are) TRUE? (2018 Adv.) (a) The line of intersection of P1 and P2 has direction ratios 1, 2, −1 3x − 4 1 − 3 y z (b) The line = = is perpendicular to the line 9 9 3 of intersection of P1 and P2 (c) The acute angle between P1 and P2 is 60° (d) If P3 is the plane passing through the point (4, 2, − 2) and perpendicular to the line of intersection of P1 and P2, then the distance of the point (2, 1, 1) from the 2 plane P3 is 3

57. If the straight lines

x−1 y+ 1 z = = and 2 K 2

x+1 y+1 z are coplanar, then the plane(s) = = 5 2 k (2012) containing these two lines is/are (a) y + 2 z = − 1 (c) y − z = − 1

(b) y + z = − 1 (d) y − 2 z = − 1

Integer & Numerical Answer Type Questions 58. Three lines are given by

and

r = λ$i , λ ∈ R r = µ (i$ + $j), µ ∈ R r = ν (i$ + $j + k$ ), ν ∈ R

Let the lines cut the plane x + y + z = 1 at the points A , B and C respectively. If the area of the triangle ABC is ∆ then the value of (6∆ )2 equals .......... (2019 Adv.)

59. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the Z-axis. Let the distance of P from the X-axis be 5. If R is the image of P in the XY -plane, then the length of PR is ......... . (2018 Adv.)

3D Geometry 585 Passage Based Problems

Match the Columns

Read the following passage and answer the questions. Consider the lines x+1 y+2 z+1 x−2 y+ 2 z −3 , L2 : L1 : = = = = 3 1 2 1 2 3 (2008, 12M)

60. The distance of the point (1, 1, 1) from the plane passing through the point (−1, − 2, − 1) and whose normal is perpendicular to both the lines L1 and L 2 , is

(a) 2 / 75 unit (c) 13 / 75 units

(b) 7 / 75 unit (d) 23 / 75 units

Match the conditions/expressions in Column I with values statements in Column II.

65. Consider the lines

x−1 y z+3 x−4 y+ 3 z + 3 and the = = , L2 : = = 2 −1 1 1 1 2 planes P1 : 7x + y + 2z = 3, P2 : 3x + 5 y − 6z = 4. Let ax + by + cz = d the equation of the plane passing through the point of intersection of lines L1 and L 2 and perpendicular to planes P1 and P2. Match List I with List II and select the correct answer using the code given below the lists. (2013 Adv.)

L1 :

61. The shortest distance between L1 and L 2 is (a) 0 unit (c) 41/5 3 unit

List I

(b) 17 / 3 units (d) 17 /5 3 units

62. The unit vector perpendicular to both L1 and L 2 is $ − $i + 7$j + 7k (a) 99 − $i + 7$j + 5k$ (c) 5 3

− $i − 7$j + 5k$ (b) 5 3 7 $i − 7$j − k$ (d) 99

Assertion and Reason

a=

1.

13

Q.

b=

2.

–3

R.

c=

3.

1

S.

d =

4.

–2

Codes P (a) 3 (c) 3

Q 2 2

R 4 1

S 1 4

P (b) 1 (d) 2

Q 3 4

R 4 1

S 2 3

66. Consider the following linear equations

For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.

ax + by + cz = 0,

(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I (c) Statement I is true; Statement II is false

P2 : x + y − z = − 1

A.

Statement I Atleast two of the lines L1 , L 2 and L3 are non-parallel. Statement II The three planes do not have a common point. (2008, 3M)

64. Consider the planes 3x − 6 y − 2z = 15 and 2x + y − 2z = 5. Statement I The parametric equations of the line of intersection of the given planes are x = 3 + 14t , y = 1 + 2t , z = 15 t. ^

^

^

Statement II The vectors 14 i + 2 j + 15 k is parallel to (2007, 3M) the line of intersection of the given planes.

a + b + c ≠ 0 and

Column II p.

The equations represent planes meeting only at a single point

a2 + b 2 + c 2 = ab + bc + ca B.

a + b + c = 0 and a2 + b 2 + c 2 ≠ ab + bc + ca

q.

The equations represent the line x = y = z

C.

a + b + c ≠ 0 and

r.

The equations represent identical planes

s.

The equations represent the whole of the three-dimensional space

a2 + b 2 + c 2 ≠ ab + bc + ca

P3 : x − 3 y + 3z = 2

Let L1 , L 2, L3 be the lines of intersection of the planes P2 and P3 , P3 and P1 , P1 and P2, respectively.

cx + ay + bz = 0

Column I

(d) Statement I is false; Statement II is true

63. Consider three planes P1 : x − y + z = 1

bx + cy + az = 0,

(IIT 2007, 6M)

(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I

and

List II

P.

D.

a + b + c = 0 and a2 + b 2 + c 2 = ab + bc + ca

Analytical & Descriptive Questions 67. Find the equations of the plane containing the line 2x − y + z − 3 = 0, 3x + y + z = 5 and at a distance of from the point ( 2, 1, − 1).

1 6

(2005, 2M)

68. A plane is parallel to two lines whose direction ratios are (1, 0, –1) and (–1, 1, 0 ) and it contains the point (1, 1, 1). If it cuts coordinate axes at A,B,C. Then find the volume of the tetrahedron OABC. (2004, 2M)

586 3D Geometry 69. T is a parallelopiped in which A,B,C and D are vertices

70. (i) Find the equation of the plane passing through the

of one face and the face just above it has corresponding vertices A′ , B ′ , C ′ , D ′ , T is now compressed to S with face ABCD remaining same and A′ , B ′ , C ′ , D ′ shifted to A′′ , B ′′ , C′′ , D′′ in S. The volume of parallelopiped S is reduced to 90 % of T. Prove that (2004, 2M) locus of A′ ′ is a plane.

points (2,1,0), (5,0,1) and (4,1,1). (ii) If P is the point (2, 1, 6), then the point Q such that PQ is perpendicular to the plane in (a) and the mid-point of PQ lies on it. (2003, 4M)

Answers Topic 1 1. (a) Topic 2 1. (d) 5. (c) 9. (a,b,c)

2. (a) 6. (b) 10. (c)

Topic 3 1. (a) 3. (b) 7. (c) 11. (a) 15. (d) 19. (b) 23. (b, c)

2. 4. 8. 12. 16. 20. 24.

3. (a) 7. (a,b) 11. (a, d)

4. (d) 8. (c,d) 12. (b, d)

(c)

5. 9. 13. 17. 21. 25.

(c) (a) (b) (c) (a) (d)

6. 10. 14. 18. 22. 26.

(d) (a) (d) (a) (b) (d)

(d) (c) (a) (a) (c)

27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 66.

(a) (b) (b) (d) (a) (c) (a) (a, b) (8) (d)

28. 32. 36. 40. 44. 48. 52. 56. 60. 64.

29. 33. 37. 41. 45. 49. 53. 57. 61. 65.

(d) (d) (d) (a) (a) (a) (a,c) (c, d) (c) (d)

(c) (a) (c) (d) (a) (d) (b, c, d) (b, c) (d)

30. 34. 38. 42. 46. 50. 54. 58. 62.

(a) (b) (c) (c) (c) (a) (b, d) (0.75) (b)

(a)

A → r , B → q; C → p; D → s

67. 2x − y + z − 3 = 0; 62x + 29y + 19z – 105 = 0 68.

(c)

9 cu units 2

70. (i) x + y − 2z = 3 (ii) Q(6, 5, – 2)

Hints & Solutions Topic 1 Direction Cosines and Direction Ratios of a Line

1. Let point P whose position vector is (− $i + 2$j + 6k$ ) and a

1. We know that, angle between two lines is a1a 2 + b1b2 + c1c2

cos θ =

a12

+

b12

+

c12

a 22

+

Topic 2 Straight Line in Space and Shortest Distance

b22

+

c22

straight line passing through Q(2, 3, − 4) parallel to the $. vector n = 6i$ + 3$j − 4k P(–1,2,6)

l + m + n =0 ⇒

l = − ( m + n)

⇒

(m + n ) = l

⇒

m + n 2 + 2mn = m2 + n 2

2

2

⇒

d

2

[Q l 2 = m2 + n 2, given]

2 mn = 0 m=0 ⇒ l = −n

When

Hence, (l , m, n ) is (1, 0, − 1). n = 0, then l = − m

When

Hence, (l , m, n ) is (1, 0, –1). 1+0+0 1 = ∴ cos θ = 2× 2 2 ⇒

θ=

π 3

Q(2,3,–4)

n=6^ i+3^ j–4^ k

Q Required distance d = Projection of line segment PQ perpendicular to vector n. |PQ × n | = |n| $ $ , so Now, PQ = 3i + $j − 10k $i $j $ k $ PQ × n = 3 1 − 10 = 26$i − 48$j + 3k 6 3

−4

3D Geometry 587 d=

So,

Since, AD ⊥ BC, a1a 2 + b1b2 + c1c2 = 0 3 × (3λ − 3) + 0(2) + 4(4λ − 2) = 0 ⇒ 9λ − 9 + 0 + 16λ − 8 = 0 ⇒ 25λ − 17 = 0 17 ⇒ λ= 25 68 1 ,1, . ∴Coordinates of D =   25 25

(26)2 + (48)2 + (3)2 (6)2 + (3)2 + (4)2 A(1, –1, 2)

B

C

D

x+2 y–1 z = = 0 4 3

2

1 68   AD = 1 −  + (−1 − 1)2 + 2 −    25 25

Now,

676 + 2304 + 9 2989 = 49 = 7 units = 36 + 9 + 16 61

=

x + 2 y −1 z = = 3 0 4 $ Vector along line is, a = 3$i + 4k

2. Given line is

2

 −18  24 =   + (−2)2 +    25   25

576 324 2 +4+ = 34 625 625 5 1 1 2 34 ∴Area of ∆ABC = BC × AD = × 5 × 2 2 5 = 34 sq units

3. According to given information, we have the following

and|BC| = 5 units Now, area of required ∆ABC 1 = |BC || b ||sin θ| 2

figure. A(Ö3, 1)

…(ii)

+1 Ö3 +1 , 2 Ö3 2 M

[where θ is angle between vectors a and b] |a × b| , Q |b|sin θ = |a| $ i$ $j k

⇒

$ |a × b|= 3 0 4 = 8$i + 6$j − 6k 3 –2 2 | a × b| = 64 + 36 + 36

C (b, (1–b))

2

B(1, Ö3)

Clearly, angle bisector divides the sides AB in OA : OB, i.e., 2 : 2 = 1 : 1 [using angle bisector theorem]

|a| = 9 + 16 = 5

So, D is the mid-point of AB and hence coordinates of D  3 + 1 3 + 1 are  ,  2   2

2 34 ∴ |b|sin θ = 5 On substituting these values in Eq. (i), we get Required area =

D

2

O

= 136 = 2 34 and

2

=

and vector joining the points (1, − 1, 2) to (−2, 1, 0) is $ b = (1 + 2)$i + (−1 − 1)$j + (2 − 0)k $ = 3$j − 2$j + 2k

Q

2

1 2 34 ×5 × = 34 sq units 2 5

Alternate Method x+ 2 y−1 z Given line is = = = λ (let) 3 0 4 A(1, –1, 2)

…(i)

Now, equation of bisector OD is   3+1 − 0   (x − 0) ⇒ y = x ( y − 0) =  2   3+1  − 0   2 ⇒ x− y =0 According to the problem, 3 β − (1 − β ) = CM = 2 2

B

D

C

x+2 y–1 z = = 0 4 3

Since, point D lies on the line BC. ∴Coordinates of D = (3λ − 2, 1, 4λ ) Now, DR of BC ⇒ a1 = 3, b1 = 0, c1 = 4 and DR of AD ⇒ a 2 = 3λ − 3, b2 = 2, c2 = 4λ − 2

[Distance of a point P ( x1 , y1 ) from the line ax + by1 + c  ax + by + c = 0 is 1  a 2 + b2  ⇒ |2 β − 1 | = 3 ⇒ 2 β = ± 3 + 1 ⇒ 2 β = 4, − 2 ⇒ β = 2, − 1 Sum of 2 and −1 is 1.

588 3D Geometry x−3 y+ 2 z + 4 lies in the plane = = −1 2 3 lx + my − z = 9, therefore we have 2l − m − 3 = 0 [Q normal will be perpendicular to the line] ...(i) ⇒ 2l − m = 3 and 3l − 2m + 4 = 9 [Q point (3, − 2, − 4) lies on the plane] ...(ii) ⇒ 3l − 2m = 5 On solving Eqs. (i) and (ii), we get l = 1and m = − 1 ∴ l 2 + m2 = 2

4. Since, the line

7. Equation of given straight lines

x−1 y z −1 = = 1 −1 3 x−1 y z −1 and L2 : = = 1 −3 −1 having vector form respectively, are r = ($i + k$ ) + λ( $i − $j + 3k$ ) and r = ($i + k$ ) + v( − 3$i − $j + k$ ) L1 :

Q ($i − $j + 3k$ ) . ( −3$i − $j + k$ ) = − 3 + 1 + 3 = 1 is positive, ∴Angle between supporting line vectors of lines L1 and L 2 is acute, and point of intersection of given lines L1 and L 2 is (1, 0, 1). Now, vector along the acute angle bisector of vectors (i$ − $j + 3k$ ) and (−3i$ − $j + k$ ) is (− i$ − $j + 2k$ ) or (i$ + $j − 2k$ ).

5. Condition for two lines are coplanar. x1 − x2 l1

y1 − y2 z1 − z2 m1 n1 =0

l2

m2

n2

It is given that line x−α y−1 z −γ is the bisector of the acute angle L: = = l m −2 between the lines L1 and L 2, so l = 1 and m = 1 1 −α 0 −1 1− γ and = = 1 1 −2 ⇒ α = 2 , γ = −1 ∴ α − γ = 3, l + m = 2

where, (x1 , y1 , z1 ) and (x2, y2, z2) are the points lie on lines (i) and (ii) respectively and < l1 , m1 , n1 > and < l2, m2, n2 > are the direction cosines of the lines (i) and (ii), respectively. 2 −1 3 −4 4 −5 1 −1 −1 ∴ 1 1 − k =0 ⇒ 1 1 − k =0 k

2

1

k

2

1

⇒ 1(1 + 2k) + (1 + k ) − (2 − k) = 0 2

⇒

k2 + 2k + k = 0

⇒

k + 3k = 0 2

⇒

k = 0, − 3

If 0 appears in the denominator, then the correct way of representing the equation of straight line is x−2 y−3 ; z =4 = 1 1

6. Since, the lines intersect, therefore they must have a point in common, i.e. x−1 y+ 1 z −1 = = =λ 2 3 4 x−3 y− k z and = = =µ 1 2 1 ⇒

x = 2λ + 1, y = 3λ − 1 z = 4λ + 1

and ⇒

x = µ + 3, y = 2 µ + k, z = µ are same. 2λ + 1 = µ + 3 3λ − 1 = 2 µ + k ⇒ 4λ + 1 = µ

On solving Ist and IIIrd terms, we get, 3 λ = − and µ = − 5 2 ∴ k = 3λ − 2 µ − 1 9  3 k = 3  −  − 2 (−5) − 1 = ⇒  2 2 9 ∴ k= 2

8.

Key Idea Points, A, B, C are collinear ⇒ AB , BC are collinear vectors ⇒ AB = λBC for some non-zero scalar λ.

Given lines, L1 : r = λ$i , λ ∈ R L 2 : r = µ $j + k$ , µ ∈ R L : r = $i + $j + vk$ , v ∈ R

and

3

…(i) …(ii) …(iii)

Now, let the point P on L1 = (λ , 0, 0) the point Q on L 2 = (0, µ , 1), and the point R on L3 = (1, 1, v) For collinearity of points P , Q and R,there should be a non-zero scalar ‘m’, such that PQ = m PR ⇒ (− λi$ + µ$j + k$ ) = m [(1 − λ )i$ + $j + νk$ ] ⇒ ⇒v=

λ µ 1 = = λ −1 1 ν 1 µ and λ = where, µ ≠ 0 and µ ≠ 1 µ µ −1

⇒ Q ≠ k$ and Q ≠ k$ + $j Hence, Q can not have coordinater (0, 0, 1) and (0, 1, 1) Hence, options (c) and (d) are correct.

9. Given lines L1: r = $i + λ (− $i + 2$j + 2k$ ), λ ∈ R and L 2 : r = µ (2i$ − $j + 2k$ ), µ ∈ R and since line L3 is perpendicular to both lines L1 and L 2.

3D Geometry 589 Now, PR ⊥ L 2

Then a vector along L3 will be, $i $j k$ –1

2

2 = $i (4 + 2 ) − $j(−2 − 4) + k$ (1 − 4)

2

–1

2

∴ 1(λ − β ) + (−1)(λ + β ) + 0(λ + 1) = 0 λ −β − λ −β = 0 ⇒

= 6$i + 6$j − 3k$ = 3(2$i + 2$j − k$ )

…(i)

Now, let a general point on line L1. P(1 − λ , 2λ , 2λ ) and on line L 2. as Q(2 µ , − µ , 2 µ ) and let P and Q are point of intersection of lines L1, L3 and L 2, L3 , so direction ratio’s of L3 …(ii) (2µ + λ − 1, − µ − 2λ , 2 µ − 2λ ) 2 µ + λ − 1 −µ − 2λ 2 µ − 2λ Now, [from Eqs.(i) and (ii)] = = 2 2 −1 1 2 ⇒ λ = and µ = 9 3 2 4  8 2 2 4 So, P  , ,  and Q  , − ,   9 9 9 9 9 9 Now, we can take equation of line L3 as r = a + t(2i$ + 2$j − k$ ), where a is position vector of any point on line L3 and possible vector of a are 2 4  1  8 $ 2 $ 2 $  4 2  i + j + k or  $i − $j + k$  or  $i + k$  9 9 3 9 9  9 9  3 

β =0

Hence, R (0, 0, − 1) Now, as ∠QPR = 90° [as a1a 2 + b1b2 + c1c2 = 0, if two lines with DR’s a1 , b1 , c1 ; a 2 , b2, c2 are perpendicular] ∴ (λ − λ )(λ − 0) + (λ − λ )(λ − 0) + (λ − 1)(λ + 1) = 0 ⇒

(λ − 1) (λ + 1) = 0

⇒

λ =1 λ = −1

or

λ = 1, rejected as P and Q are different points. ⇒ λ = −1

11.

Key Idea If two straight lines are coplanar, x − x1 y − y1 z − z1 i.e. = = a1 b1 c1 and

x − x2 y − y2 z − z2 are coplanar = = a2 b2 c2 A (x1, y1, z1) (a1,b1,c1) DR’s

Hence, options (a), (b) and (c) are correct.

10.

Key Idea (i) Direction ratios of a line joining two points ( x 1, y1, z1 ) and ( x 2, y2, z2 ) are x 2 − x 1, y2 − y1, z2 − z1. (ii) If the two lines with direction ratios a1, b1,c1; a2, b 2,c 2 are perpendicular, then a1 a2 + b1 b 2 + c1 c 2 = 0

Line L1 is given by y = x ; z = 1 can be expressed x y z −1 =α L 1: = = 1 1 0 ⇒

[say]

x = α, y = α,z = 1

Let the coordinates of Q on L 1 be (α , α ,1). Line L 2 given by y = − x, z = −1 can be expressed as x y z+1 [say] = =β L 2: = 1 −1 0 ⇒

B (x ,y 2 2 ,z 2)

(a2,b2,c2)

Then, ( x 2 − x 1, y2 − y1, z2 − z1 ), ( a1, b1, c1 ) and ( a2, b 2, c 2 ) are coplanar, x 2 − x 1 y2 − y1 z2 − z1 i.e. a1 b1 c1 =0 a2

x = α,

and

⇒

y z = 3 − α −2

5 −α 0 0 3 −α 0

−1

0 −2

=0

2 −α

⇒

(5 − α ) [(3 − α ) (2 − α ) − 2] = 0

⇒

(5 − α ) [α 2 − 5 α + 4] = 0

∴ 1(λ − α ) + 1 ⋅ (λ − α ) + 0 ⋅ (λ − 1) = 0 ⇒ λ = α

⇒

(5 − α ) (α − 1) (α − 4) = 0

Hence, Q (λ , λ , 1)

∴

P (λ, λ, λ)

R (0, 0, –1)

Direction ratios of PR are λ − β , λ + β , λ + 1.

…(i)

y z = −1 2 − α

x−α y−0 z −0 = = −1 0 2 −α

⇒ Q (λ, λ, 1)

c2

x−5 y−0 z −0 = = − (α − 3) 0 −2

⇒

Direction ratios of PQ are λ − α , λ − α , λ − 1. Now, PQ ⊥ L1

b2

x = 5,

Here,

x = β , y = − β , z = −1

Let the coordinates of R on L 2 be ( β , − β , − 1).

DR’s

α = 1, 4, 5

…(ii)

590 3D Geometry 12.

Key Idea Equation of straight line is l :

∴ Equation of the plane is 2 1 4    1 x +  + 1  y −  + 1 z −  = 0    3 3 3

x − x1 y − y1 z − z1 = = a b c

Since, l is perpendicular to l1 and l 2 . So, its DR’s are cross-product of l1 and l 2 . Now, to find a point on l 2 whose distance is given, assume a point and find its distance to obtain point.

Let l :

…(i) ⇒ x+ y+ z =1 From the option the point (1, − 1, 1) satisfy the plane x+ y+ z =1. Hence, option (a) is correct.

x−0 y−0 z −0 = = a b c

2. Equation of plane passing through the points (2, 1, 0),

which is perpendicular to $ ) + t (i$ + 2$j + 2k $) l : (3i$ − $j + 4k

(4, 1, 1) and (5, 0, 1) is x − 2 y − 1 z − 0 x − 2 y − 1 z  4 − 2 1 − 1 1 − 0 = 0 ⇒  2 0 1   = 0  −1 1  5 − 2 0 − 1 1 − 0  3 ⇒ (x − 2)(0 + 1) − ( y − 1)(2 − 3) + z (−2 − 0) = 0 ⇒ x − 2 + y − 1 − 2z = 0 …(i) ⇒ x + y − 2z − 3 = 0 Now, let the image of point R(2, 1, 6) w.r.t. plane (1) is (x1 , y1 , z1 ), then x1 − 2 y1 − 1 z1 − 6 = = 1 1 −2  2 + 1 − 12 − 3  = − 2 2   1 + 12 + (−2)2

1

$ ) + s(2$i + 2$j + k $) l2 : (3$i + 3$j + 2k $i $j k $ $ ∴ DR’s of l is 1 2 2 = −2$i + 3$j − 2k 2 2 1 x y z = = = k1 , k2 −2 3 −2 Now, A (−2k1 , 3k1 , − 2k1 ) and B(−2k2, 3k2, − 2k2). l:

Since, A lies on l1. $ = (3 + t )i$ + (−1 + 2 t )$j ∴ (−2k1 )i$ + (3k1 )$j − (2k1 )k $ + (4 + 2 t )k ⇒ ∴

3 + t = − 2k1, −1 + 2 t = 3k1 , 4 + 2 t = − 2k1

x1 − 2 y1 − 1 z1 − 6 = = −2 1 1 2 × 12 = =4 6 ⇒ x1 = 6, y1 = 5, z1 = − 2 ∴Image of point R(2, 1, 6) w.r.t, plane (1) is (6, 5, −2).

k1 = − 1 ⇒ A(2, − 3, 2)

⇒

Let any point on l2 (3 + 2s, 3 + 2s, 2 + s) (2 − 3 − 2s)2 + (−3 − 3 − 2s)2 + (2 − 2 − s)2 = 17 ⇒

9s2 + 28s + 37 = 17

⇒

9s2 + 28s + 20 = 0

⇒

9s2 + 18s + 10s + 20 = 0

⇒

( 9s + 10) (s + 2) = 0

3. Equation of line passing through the point (1, − 5, 9) and parallel to x = y = z is x−1 y+ 5 z −9 = = =λ 1 1 1 Thus, any point on this line is of the form (λ + 1, λ − 5, λ + 9).

−10 s = − 2, ⋅ ∴ 9  7 7 8 Hence, (−1, − 1, 0) and  , ,  are required points.  9 9 9

Now, if P(λ + 1, λ − 5, λ + 9) is the point of intersection of line and plane, then (λ + 1) − (λ − 5) + λ + 9 = 5 ⇒ λ + 15 = 5 ⇒ λ = − 10 ∴ Coordinates of point P are (− 9, − 15, − 1). Hence, required distance

Topic 3 Equation of a Plane 1. Since, the mirror image of the point A(1, 2, 3) in a plane 4 1  7 is B − , − , −  , then mid-point of AB lies in the  3 3 3 plane and dr’s of line AB is the dr’s, of normal to the plane. 7 4 1  2− 3−  1 − 3 3 3 Now, mid-point of AB is P  , , 2 2   2    2 1 4 P − , ,   3 3 3 7 4 1  and dr’s of line AB is 1 + , 2 + , 3 +   3 3 3  10 10 10 ~ , ,  ~ (1, 1, 1) 3 3 3

(say)

= (1 + 9)2 + (− 5 + 15)2 + (9 + 1)2 = 102 + 102 + 102 = 10 3

4.

Key Idea QLength of the perpendicular drawn from point (x 1, y 1, z 1) to the plane ax + by + cz + d = 0 is | ax 1 + by 1 + cz 1 + d | d1 = a 2 + b2 + c2

Given line vectors $ ) and r = (i$ + $j) + λ (i$ + 2$j − k

…(i)

$) r = (i$ + $j) + µ (− i$ + $j − 2k

…(ii)

3D Geometry 591 Now, a vector perpendicular to the given vectors (i) and (ii) is $ i$ $j k

And, similarly for point ‘Q’, we get 3(3r2 + 2) + (2r2 − 1) + 4(− r2 + 1) = 16 ⇒

n = 1 2 −1 −1 1 − 2

So, point is Q (5, 1, 0)

= 36 + 16 + 4

∴The equation of plane containing given vectors (i) and (ii) is − 3 (x − 1) + 3( y − 1) + 3 (z − 0) = 0 ⇒ −3x +3 y + 3z = 0 …(iii) ⇒ x− y− z =0 Now, the length of perpendicular drawn from the point (2, 1, 4) to the plane x − y − z = 0, is |2 − 1 − 4| 3 d1 = = = 3 3 1+1+1 Key Idea Equation of planes bisecting the angles between the planes a 1x + b1y + c1z + d 1 = 0 and a 2 x + b2 y + c2z + d 2 = 0, are a 1x + b1y + c1z + d 1 a x + b2 y + c2z + d 2 =± 2 a 12 + b12 + c12 a 22 + b22 + c22

Equation of given planes are and

2x − y + 2z − 4 = 0

…(i)

x + 2 y + 2z − 2 = 0

…(ii)

Now, equation of planes bisecting the angles between the planes (i) and (ii) are 2x − y + 2z − 4 x + 2 y + 2z − 2 =± 4+1+4 1+4+4 ⇒

2x − y + 2z − 4 = ± (x + 2 y + 2z − 2)

On taking (+ ve) sign, we get a plane x − 3y = 2

…(iii)

On taking (− ve) sign, we get a plane …(iv) 3x + y + 4z = 6 Now from the given options, the point (2, − 4, 1) satisfy the plane of angle bisector 3x + y + 4z = 6 6. Equation of given line is x−2 y+ 1 z −1 = = =r −1 3 2 line is R(3r + 2, 2r − 1, − r + 1) Let the coordinates of point P are (3r1 + 2, 2r1 − 1, r1 + 1) and Q are (3r2 + 2, 2r2 − 1, − r2 + 1). Since, P is the point of intersection of line (i) and the plane 2x + 3 y − z + 13 = 0, so 2(3r1 + 2) + 3(2r1 − 1) − (− r1 + 1) + 13 = 0 6r1 + 4 + 6r1 − 3 + r1 − 1 + 13 = 0

⇒

13r1 + 13 = 0 ⇒ r1 = − 1

So, point P(− 1, − 3, 2)

= 56 = 2 14 Key Idea Use the foot of perpendicular Q( x 2 , y 2 , z 2) drawn from point P( x 1, y 1, z 1) to the plane ax + by + cz + d = 0, is P(x1,y1,z1 ) given by

7.

x 2 − x 1 y 2 − y 1 z2 − z1 = = a b c =−

ax 1 + by 1 + cz 1 + d a2 + b2 + c 2

Q(x2,y2,z2) ax+by+z+d=0

Let a general point on the given line x−1 y+ 1 z = = =r −1 2 1

(say)

is P (2r + 1, − 1 − r , r ). Now, let foot of perpendicular Q (x1 , y1 , z1 ) to be drawn from point P (2r + 1, − 1 − r , r ) to the plane x + y + z = 3, then x1 − 2r − 1 y + 1 + r z − r = = 1 1 1  2r + 1 − 1 − r + r − 3 =−    1+1+1 ⇒

x1 − 2r − 1 = y + r + 1 1 2 = z − r = (3 − 2r ) = 1 − r 3 3 4r 5r r x1 = 2 + , y = − and z = 1 + ⇒ 3 3 3 4r 5r r  So, point Q ≡ 2 + ,− , 1 +  , lies on the plane  3 3 3 x − y + z = 3 also. 4r 5r r So, 2+ + + 1 + =3 ⇒ r =0 3 3 3 Therefore, the coordinates of point Q are (2, 0, 1).

8. Equation of given planes are

(let) …(i)

Now, coordinates of a general point over given

⇒

PQ = (5 + 1)2 + (1 + 3)2 + 22

Now,

$ (1 + 2) = i$ (−4 + 1) − $j(− 2 − 1) + k $ $ $ = − 3i + 3 j + 3k

5.

7r2 = 7 ⇒ r2 = 1

and

2x − y + 2z + 3 = 0 4x − 2 y + 4z + λ = 0 2x − y + 2z + µ = 0

…(i) …(ii) …(iii)

Q Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is | d1 − d2| distance = a 2 + b2 + c2 ∴Distance between planes (i) and (ii) is 1 | λ − 2(3)| = 16 + 4 + 16 3 ⇒ | λ − 6| = 2 ⇒ λ − 6 = ± 2 ⇒ λ = 8 or 4

[given]

592 3D Geometry and distance between planes (i) and (iii) is 2 |µ − 3| = 4+1+4 3

11. Let the equation of plane is [given]

⇒ |µ − 3| = 2 ⇒ µ − 3 = ± 2 ⇒ µ = 5 or 1 So, maximum value of (λ + µ ) at λ = 8 and µ = 5 and it is equal to 13.

9. Given, equation of plane 3x − y + 4z = 2 ...(i) and the point Q(0, − 1, − 3) is the image of point P in the plane (i), so point P is also image of point Q w.r.t. plane (i). Let the coordinates of point P is (x1 , y1 , z1 ), then x1 − 0 y1 + 1 z1 + 3 = = 3 −1 4 [3(0) − 1(−1) + 4(−3) − 2] = −2 32 + (−1)2 + 42 [∴ image of the point (x1 , y1 , z1 ) in the plane ax + by + cz + d = 0 is (x, y, z ), where x − x1 y − y1 z − z1 − 2 (ax1 + by1 + cz1 + d ) ] = = = a b c a 2 + b2 + c2 x − 0 y1 + 1 z1 + 3 (1 − 12 − 2) 26 = = ⇒ 1 =2 = =1 3 −1 4 26 26 ⇒ P (x1 , y1 , z1 ) = (3, − 2, 1) Now, area of ∆PQR, where point R(3, − 1, − 2) 1 → → 1 = PQ × PR = (−3i$ + 2 2 i$ $j 1 = −3 1 2 0 1 =

$j − 4k$ ) × (0i$ + $j − 3k$ ) k$

1 −4 = i$ − 9 $j − 3k$ 2 −3

1 91 sq units 1 + 81 + 9 = 2 2

10. Equation of plane, which contains the line of intersection of the planes x + y + z − 6 = 0 and 2x + 3 y + z + 5 = 0, is (x + y + z − 6) + λ (2x + 3 y + z + 5) = 0 …(i) ⇒ (1 + 2λ )x + (1 + 3λ ) y + (1 + λ )z + (5λ − 6) = 0 Q The plane (i) is perpendicular to XY -plane (as DR’s of normal to XY -plane is (0, 0, 1)). ∴ 0(1 + 2λ ) + 0(1 + 3λ ) + 1(1 + λ ) = 0 ⇒ λ = −1 On substituting λ = − 1 in Eq. (i), we get − x − 2 y − 11 = 0 …(ii) ⇒ x + 2 y + 11 = 0 which is the required equation of the plane. Now, the distance of the point (0, 0, 256) from plane P is 0 + 0 + 11 11 = 1+4 5 [Q distance of (x1 , y1 , z1 ) from the plane ax + by + cz − d = 0, is

ax1 + by1 + cz1 − d    a 2 + b2 + c2

…(i) ax + by + cz = d Since plane (i) passes through the points (0,−1, 0) and (0, 0, 1), then − b = d and c = d …(ii) ∴ Equation of plane becomes ax − dy + dz = d π with the plane Q The plane (ii) makes an angle of 4 y − z + 5 = 0. π −d − d cos = 2 4 a + d2 + d2 1 + 1 [Q The angle between the two planes a1x + b1 y + c1z + d = 0 and a 2x + b2y + c2z + d = 0 is  a1 a 2 + b1 b2 + c1 c2  cos θ = a12 + b12 + c12 a 22 + b22 + c22  1 |−2d| = ⇒ a 2 + 2d 2 = |− d| ⇒ 2 2 2 a + 2d 2 ⇒ a 2 + 2d 2 = 4d 2 ⇒

[squaring both sides]

a = 2d ⇒ a = ± 2d 2

2

So, the Eq. (ii) becomes ± 2x − y + z = 1

…(i)

Now, from options ( 2 , 1, 4) satisfy the plane − 2x − y + z = 1

12. Equation of given plane is …(i) x + 2 y + 3z = 15 x −1 y + 1 z −2 …(ii) and line is, = = = r (let) 2 3 4 So, the coordinates of any point on line (ii) is P (1 + 2r , − 1 + 3r , 2 + 4r ). Q Point P is intersecting point of plane (i) and line (ii) ∴(1 + 2r ) + 2 (–1 + 3r ) + 3 (2 + 4r ) = 15 ⇒

1 + 2r − 2 + 6r + 6 + 12r = 15 ⇒ 20r = 10 1 ⇒ r= 2 3    1  ∴Coordinates of P = 1 + 1, − 1 + , 2 + 2 = 2, , 4    2  2 Now, distance of the point P from the origin = 4+

1 1 81 9 = units + 16 = 20 + = 4 4 4 2

13. Since, equation of planes passes through the line of intersection of the planes x+ y+ z =1 and 2x + 3 y + 4z = 5, is (x + y + z − 1) + λ (2x + 3 y + 4z − 5) = 0 ⇒ (1 + 2λ )x + (1 + 3λ ) y + (1 + 4λ )z − (1 + 5λ ) = 0 …(i) Q The plane (i) is perpendicular to the plane x − y + z = 0. ∴ (1 + 2λ ) − (1 + 3λ ) + (1 + 4λ ) = 0 [Q if plane a1x + b1 y + c1z + d1 = 0 is perpendicular to plane a 2x + b2y + c2z + d2 = 0, thena1a 2 + b1b2 + c1c2 = 0]

3D Geometry 593 ⇒

3λ + 1 = 0

1 λ=− ⇒ 3 So, the equation of required plane, is 2 3 4 5     1 −  x + 1 −  y + 1 −  z − 1 −  = 0     3 3 3 3

…(ii)

1 1 2 x− z + =0 ⇒ x− z + 2 =0 3 3 3 $)+ 2 =0 Now, vector form, is r ⋅ ($i − k ⇒

14. Equation of given line is

x+ 3 y−2 z = = = r (let) −7 10 1 Coordinates of a point on line (i) is A (10r − 3, − 7r + 2, r )

…(i)

Now, let the line joining the points P(2, − 1, 4) and A (10r − 3, − 7r + 2, r ) is perpendicular to line (i). Then, PA ⋅ (10i$ − 7$j + k$ ) = 0 [Q vector along line (i) is (10$i − 7$j + k$ )] ⇒ [(10r − 5)$i + (−7r + 3)$j + (r − 4)k$ ] ⋅[10i$ − 7$j + k$ ] = 0 ⇒ 10(10r − 5) − 7(3 − 7r ) + (r − 4) = 0 ⇒ 100r − 50 − 21 + 49r + r − 4 = 0 1 ⇒ 150r = 75 ⇒ r = 2 3 1  So, the foot of perpendicular is A 2, − ,   2 2 1 [put r = in the coordinates of point A] 2 Now, perpendicular distance of point P(2, − 1, 4) from the line (i) is 2

 1   3 PA = (2 − 2) +  − + 1 +  − 4  2   2 1 49 50 5 = + = = 4 4 4 2

Now, equation of family of planes passes through the line of intersection of given planes (i) and (ii) is …(iii) (2x − y − 4) + λ ( y + 2z − 4) = 0 According to the question, Plane (iii) passes through the point (1, 1, 0), so (2 − 1 − 4) + λ (1 + 0 − 4) = 0 ⇒ −3 − 3 λ = 0 ⇒ λ = −1 Now, equation of required plane can be obtained by putting λ = − 1 in the equation of plane (iii). ⇒ (2x − y − 4) − 1( y + 2z − 4) = 0 ⇒ 2x − y − 4 − y − 2z + 4 = 0 ⇒ 2x − 2 y − 2z = 0 ⇒ x− y− z =0

17. Clearly, direction ratios of given line are 2, 1, –2 and direction ratios of normal to the given plane are 1, –2, –k. As we know angle ‘θ’ between line and plane can be obtained by |a1 a 2 + b1 b2 + c1c2| sin θ = a12 + b12 + c12 a 22 + b22 + c22 |2 − 2 + 2k|

So, sin θ =

4 + 1 + 4 1 + 4 + k2

 |2k| 2 2 ⇒ sin  cos −1  = 3   3 5 + k2 ⇒

2

2 |k| 3 5+ k

=

2

1 3

2

 1 2 2 −1 1 Q cos θ = 3 ⇒ sin θ = 3 ⇒ θ = sin 3  2 2 1 = sin −1  ⇒ cos −1 3 3

3

which lies in (3, 4).

15. The normal vector to the plane containing the vectors (i$ + $j + k$ ) and (i$ + 2$j + 3k$ ) is n = ($i + $j + k$ ) × ($i + 2$j + 3k$ )  i$ $j k$  = 1 1 1    1 2 3  = $i (3 − 2) − $j (3 − 1) + k$ (2 − 1) = $i − 2$j + k$ Now, magnitude of the projection of vector 2$i + 3$j + k$ on normal vector n is |(2i$ + 3$j + k$ ) ⋅ n| |(2$i + 3$j + k$ ) ⋅ ($i − 2 $j + k$ )| = |n| 1+4+1 =

2√2

⇒ ⇒

and

4k = 5 + k2 3k2 = 5 5 k=± 3 2

⇒

18. According to the question points (−λ2, 1, 1), (1, − λ2, 1) and (1, 1, − λ2) are coplanar with the point (−1, − 1, 1), so 0 2 1 − λ2 =0 1 − λ2 0 2 2

2

− λ2 − 1 Q condition of coplanarity is  y2 − y1 z2 − z1  x2 − x1   x −x  − = 0 − z z y y 3 1 3 1 1  3   x4 − x1 y4 − y1 z4 − z1 

16. Equations of given planes are …(i) …(ii)

√32–(2√2)2 =1

θ

|2 − 6 + 1| 3 3 units = = 2 6 6

2x − y − 4 = 0 y + 2z − 4 = 0

 −1 2 2  given θ = cos 3  

⇒

(−1 − λ2) [(1 − λ2)2 − 4] = 0

594 3D Geometry

So, equation of plane will be 1(x + 2) − 2( y − 2) + 1(z + 5) = 0 ... (iv) ⇒ x − 2 y + z + 11 = 0 Now, perpendicular distance from origin to plane is 11 11 = = 1+4+1 6 [Q perpendicular distance from origin to the plane |d| ] ax + by + cz + d = 0, is 2 a + b2 + c2 x−3 y+ 1 z −6 20. Let = = = a (say). Then, any point on −1 1 3 this line is of the form P (a + 3, 3a − 1, − a + 6) x+ 5 y−2 z −3 Similarly, any point on the line. = = = −6 7 4 b(say), is of the form Q (7b − 5, − 6b + 2, 4b + 3) Now, if the lines are intersect, then P = Q for some a and b. ⇒ a + 3 = 7b − 5 3a − 1 = − 6b + 2 and − a + 6 = 4b + 3 ⇒ a − 7b = − 8, a + 2b = 1 and a + 4b = 3 On solving a − 7b = − 8 and a + 2b = 1 , we get b = 1 and a = − 1 , which also satisfy a + 4b = 3 ∴ P = Q ≡ (2, − 4, 7) for a = − 1 and b = 1 Thus, coordinates of point R are (2, − 4, 7) and reflection of R in xy-plane is (2, − 4, − 7)

figure. 2, –5, 0 n

2x–5y=15

p

A (3, 4, 2)

B (7, 0, 6) C (2, a, b)

From figure, it is clear that (AB × BC) = p and n = 2$i − 5$j + 0k$ $i $j k$ ∴

p= 4 −5

−4

4

α

β −6

[Q AB = (7 − 3)$i + (0 − 4)$j + (6 − 2)k$ = 4$i − 4$j + 4k$ and BC = (2 − 7)$i + (α − 0)$j + (β − 6)k$ = − 5$i + α$j + (β − 6)k$ ]

= $i( −4β + 24 − 4α ) − $j( 4β − 24 + 20) + k$ ( 4α − 20) ⇒ p = (24 − 4α − 4β )$i + $j(4 − 4β ) + k$ (4α − 20) Now,as the planes are perpendicular, therefore p ⋅ n = 0 ⇒ ((24 − 4α − 4β )$i + (4 − 4β )$j + (4α − 20)k$ ) ⋅ (2$i − 5$j + 0k$ )) = 0 ⇒ 2(24 − 4α − 4 β ) − 5(4 − 4 β ) + 0 = 0 ⇒ 8(6 − α − β ) − 4(5 − 5 β ) = 0 ⇒ 12 − 2α − 2 β − 5 + 5 β = 0 ⇒ 2 α − 3 β = 7

22. Let the direction vector of the line x−3 y+ 2 z −1 is b = 2$i − $j + 3k$ . = = −1 2 3 Since, the required plane contains this line and its projection along the plane 2x + 3 y − z = 5, it will also contain the normal of the plane 2x + 3 y − z = 5. n b×n

B

A

L

z–1 y+2 x–3 = = –1 3 2

required plane (ABCD) b P

3 y– z= 5

19. Let the equation of plane, containing the two lines x + 2 y −2 z + 5 = = 3 5 7 x −1 y −4 z + 4 is and = = 1 4 7 … (i) a (x + 2) + b( y − 2) + c(z + 5) = 0 Q Plane (i) contain lines, so ... (ii) 3a + 5b + 7c = 0 and ... (iii) a + 4b + 7c = 0 From Eqs. (ii) and (iii), we get a b c = = 35 − 28 7 − 21 12 − 5 a b c = = ⇒ 7 − 14 7 a b c = = ⇒ 1 −2 1

21. According to given information, we have the following

2x+

⇒ (1 + λ2) [(1 − λ2 − 2) (1 − λ2 + 2)] = 0 ⇒ (1 + λ2)2 (3 − λ2) = 0 [Q1 + λ2 ≠ 0 ∀ λ ∈ R] ⇒ λ2 = 3 λ=± 3 ⇒

C

D Projection of line L

Normal vector of the plane 2x + 3 y − z = 5 is n = 2$i + 3$j − k$ . Now, the required plane contains b = 2$i − $j + 3k$ and n = 2$i + 3$j − k$ . ∴ Normal of the required plane is b × n.

3D Geometry 595 Since, the plane contains the line x − 3 y + 2 z −1 , therefore it also contains the point = = −1 2 3 a = 3$i − 2$j + k$ .

6(x − 0) + 10( y − 2) + 2(z − 5) = 0 [Q P(0, 2, 5) line on the plane] ⇒ 3x + 5 y − 10 + z − 5 = 0 … (i) ⇒ 3x + 5 y + z = 15

Now, the equation of required plane is(r − a ) ⋅ (b × n ) = 0 x−3 y+ 2 z −1 2 −1 3 =0

On checking all the options, the option (4, 1, − 2) satisfy the equation of plane (i).

2

3

−1

⇒ (x − 3) [1 − 9] − ( y + 2) [ − 2 − 6] + (z − 1) [6 + 2] = 0 ⇒ − 8x + 8 y + 8z + 32 = 0 ⇒ x − y − z = 4 Note that (2, 0, − 2) is the only point which satisfy above equation.

23. Let the equation of plane be a (x − 0) + b( y + 1) + c (z − 0) = 0 [Q Equation of plane passing through a point (x1 , y1 , z1 ) is given by a (x − x1 ) + b( y − y1 ) + c(z − z1 ) = 0] ... (i) ⇒ax + by + cz + b = 0 Since, it also passes through (0, 0, 1), therefore, we get ...(ii) c+ b = 0 Now, as angle between the planes ax + by + cz + b = 0 π and y − z + 5 = 0 is . 4  π  |n1 ⋅ n 2| ; where n1 = a$i + b$j + c k$ ∴cos   =  4  |n1||n 2| and n = 0$i + $j − k$ 2

⇒

1 |(a$i + b$j + ck$ ) ⋅ (0$i + $j − k$ )| = 2 a 2 + b2 + c2 0 + 1 + 1 =

|b − c| a + b2 + c2 2 2

⇒ a 2 + b2 + c2 = |b − c|2 = (b − c)2 = b2 + c2 − 2bc ⇒ a 2 = −2bc [Using Eq. (ii)] ⇒ a 2 = 2b2 a = ± 2b ⇒ ⇒ Direction ratios (a , b, c) = (± 2 , 1, − 1) So, options (b) and (c) are correct because 2, 2 , − 2 and 2 , 1, − 1. are multiple of each other.

25. Given equation of line is

x −4 y −5 z −3 …(i) = = = r (let) 2 2 1 ⇒ x = 2r + 4; y = 2r + 5 and z = r + 3 ∴ General point on the line (i) is P (2r + 4, 2r + 5, r + 3) So, the point of intersection of line (i) and plane x + y + z = 2 will be of the form P (2r + 4, 2r + 5, r + 3) for some r ∈ R. ⇒ (2r + 4) + (2r + 5) + (r + 3) = 2 [Q the point will lie on the plane] ⇒ 5r = −10 ⇒ r = −2 So, the point of intersection is P(0, 1, 1) [putting r = −2 in (2r + 4, 2r + 5, r + 3)]

Now, on checking the options, we get x −1 y −3 z + 4 contain the point (0, 1, 1) = = 1 2 −5

26 Let a be the position vector of the given point (4, − 1, 2). $ Then, a = 4$i − $j + 2k The direction vector of the lines x+ 2 y−2 z + 1 and = = −1 3 2 are respectively $ b1 = 3$i − $j + 2k $ and b = i$ + 2$j + 3k 2

Now, as the plane is parallel to both b1 and b2 [Q plane is parallel to the given lines] So, normal vector (n ) of the plane is perpendicular to both b1 and b2. ⇒ n = b1 × b2 and Required equation of plane is

24. Let the given points be A(− 3, − 3, 4) and B (3, 7 6).

(r − a ) ⋅ n = 0 ⇒ (r − a ) ⋅ (b1 × b2) = 0

Then, mid-point of line joining A , B is  − 3 + 3 − 3 + 7 4 + 6 , , P  = P (0, 2, 5)  2 2 2  Q The required plane is perpendicular bisector of line joining A, B, so direction ratios of normal to the plane is proportional to the direction ratios of line joining A, B. So, direction ratios of normal to the plane are 6, 10, 2. [Q DR’s of AB are 3 + 3, 7 + 3, 6 − 4, i.e. 6, 10, 2] Now, equation of plane is given by a (x − x1 ) + b ( y − y1 ) + c (z − z1 ) = 0

x − 2 y − 3 z −4 = = 1 2 3

⇒

x−4 3 1

y+ 1 z −2 −1 2 =0 2

3

$ ) − (4i$ − $j + 2k $)  Q r − a = (xi$ + y$j + zk   $ $ $ = (x − 4) i + ( y + 1) j + (z − 2) k   and we know that, [a b c] = a ⋅ (b × c)   a1 a 2 a3   = b1 b2 b3     c1 c2 c3   ⇒ (x − 4) (− 3 − 4) − ( y + 1) (9 − 2) + (z − 2) (6 + 1) = 0

596 3D Geometry ⇒

− 7 (x − 4) − 7 ( y + 1) + 7 (z − 2) = 0

⇒

(x − 4) + ( y + 1) − (z − 2) = 0

⇒

x+ y− z −1 =0

⇒ x(− 4 + 30) − y(32 + 20) + z (24 + 2) = 0 ⇒ 26x − 52 y + 26z = 0 ⇒ x − 2y + z = 0

(1, 1, 1) is the only point that satisfies. $ r = (1 − 3 µ )i$ + (µ − 1) $j + ( 2 + 5 µ ) k Clearly, any point on the above line is of the form (1 − 3 µ , µ − 1, 2 + 5µ) Let A be (− 3 µ + 1, µ − 1, 5 µ + 2) for some µ ∈ R. Then, AB = (3 − (− 3 µ + 1)) i$ + (2 − (µ − 1)) $j $ [Q AB = OB − OA ] + (6 − (5µ + 2))k $ = (3 µ + 2) $i + (3 − µ )$j + (4 − 5 µ ) k Normal vector (n ) of the plane x − 4 y + 3z = 1 is $ n = i$ − 4$j + 3k

… (i) …(ii)

Q AB is parallel to the plane. ∴n is perpendicular to the AB. ⇒ AB ⋅ n = 0 $ ] ⋅ [$i − 4$j + 3k $]=0 ⇒ [(3 µ + 2)i$ + (3 − µ )$j + (4 − 5 µ )k [From Eqs. (i) and (ii)] ⇒ (3 µ + 2) − 4(3 − µ ) + 3 (4 − 5 µ ) = 0 ⇒ − 8 µ + 2 = 0 1 ⇒ µ= 4

28. Let P1 be the plane containing the lines x y z x y z and = = = = . 3 4 2 4 2 3 For these two lines, direction vectors are b = 3$i + 4$j + 2k$ and b = 4$i + 2$j + 3k$ . 1

2

A vector along the normal to the plane P1 is given by $i $j k$ n1 = b1 × b 2 = 3 4 2 4 2 3 = $i (12 − 4) − $j(9 − 8) + k$ (6 − 16) = 8$i − $j − 10k$ x y z Let P2 be the plane containing the line = = and 2 3 4 perpendicular to plane P1. x y z For the line = = , the direction vector is 2 3 4 b = 2$i + 3$j + 4k$ and it passes through the point with position vector a = 0$i + 0$j + 0k$ . Q P2 is perpendicular to P1, therefore n1 and b lies along the plane. Also, P2 also passes through the point with position vector a. ∴ Equation of plane P2 is given by x−0 y−0 z −0 (r − a ) ⋅ (n1 × b ) = 0 ⇒

Key Idea Equation of plane through the intersection of two planes P1 and P2 is given by P1 + λP2 = 0

29.

27. Given equation of line is

8 2

−1 3

− 10 = 0 4

The plane through the intersection of the planes x + y + z − 1 = 0 and 2x + 3 y − z + 4 = 0 is given by (x + y + z − 1) + λ (2x + 3 y − z + 4) = 0, where λ ∈ R ⇒ (1 + 2λ )x + (1 + 3λ ) y + (1 − λ )z + (4λ − 1) = 0, where …(i) λ ∈R Since, this plane is parallel to Y -axis, therefore its normal is perpendicular to Y -axis. ⇒ {(1 + 2λ )$i + (1 + 3λ )$j + (1 − λ )k$ } ⋅ $j = 0 ⇒

1 + 3λ = 0

1 3 Now, required equation of plane is 2 3 1   4    1 −  x + 1 −  y + 1 +  z +  − − 1 = 0   3    3 3 3 −1 [substituting λ = in Eq. (i)] 3 ⇒ x + 4z − 7 = 0 Here, only (3, 2, 1) satisfy the above equation. x + 1 y −3 z − 2 is of the form 30. Any point on the line = = −3 2 −1 (−3λ − 1, 2λ + 3, − λ + 2) x + 1 y −3 z − 2 [take = = = λ ⇒ x = − 3λ − 1, −3 −1 2 ⇒

λ=−

y = 2λ + 3 and z = − λ + 2] So, the coordinates of point of intersection of two lines will be (− 3λ − 1, 2λ + 3, − λ + 2) for some λ ∈ R. Let the point A ≡ (− 3λ − 1, 2λ + 3, − λ + 2) and B ≡ (−4, 3, 1) Then, AB = OB − OA $ = (−4$i + 3$j + k$ ) − {(−3λ − 1)$i + (2λ + 3)$j + (− λ + 2) k} = (3λ − 3)$i − 2λ$j + (λ − 1)k$ Now, as the line is parallel to the given plane, therefore AB will be parallel to the given plane and so AB will be perpendicular to the normal of plane. $ is normal to the plane. ⇒ AB⋅ λ = 0, where n = i$ + 2$j − k $ ⋅ ($i + 2$j − k$ ) = 0 ⇒ ((3λ − 3)$i − 2λ$j + (λ − 1)k) ⇒ 3(λ − 1) − 4λ + (−1)(λ − 1) = 0 [Q If a = a1$i + a 2$j + a3 k$ and b = b1$i + b2$j + b3 k$ , then a ⋅ b = a1b1 + a 2b2 + a3 b3 ] ⇒ 3λ − 3 − 4λ − λ + 1 = 0 ⇒ − 2λ = 2 ⇒ λ = −1

3D Geometry 597 Now, the required equation is the equation of line joining A(2, 1, 3) and B(−4, 3, 1), which is x − (− 4) y − 3 z − 1 = = 2 − (−4) 1 − 3 3 − 1

Alternative Method Clearly, DR’s of AB are 4 − 5, − 1 + 1, 3 − 4, i.e. −1, 0, − 1 and DR’s of normal to plane are 1, 1, 1. Now, let θ be the angle between the line and plane, −1 + 0 −1 then θ is given by sin θ = (−1)2 + (−1)2 12 + 12 + 12 2 2 = = 3 2 3

[Q Equation of line joining (x1 , y1 , z1 ) and (x2, y2, z2) is x − x1 y − y1 z − z1  = = x2 − x1 y2 − y1 z2 − z1  ⇒ or

x+ 4 y−3 z −1 = = −2 6 2 x+ 4 y−3 z −1 = = −1 3 1

B (4, –1, 3)

[multiplying by 2]

θ

A (5, –1, 4)

C

31. L1 is the line of intersection of the plane 2x − 2 y + 3z − 2 = 0 and x − y + z + 1 = 0 and L 2 is the line of intersection of the plane x + 2 y − z − 3 = 0 and 3x − y + 2z − 1 = 0 $  $i $j k  Since Li is parallel to 2 −2 3  = $i + $j   1 −1 1  $   $i $j k $ L 2 is parallel to1 2 −1 = 3$i − 5$j − 7k   3 −1 2  5 8  Also, L 2 passes through  , , 0 . 7 7  [put z = 0 in last two planes] So, equation of plane is x − 5 y − 8 z    7 7 1 0  = 0 ⇒ 7x − 7 y + 8z + 3 = 0  1  3 −5 −7    Now, perpendicular distance from origin is   3 3 1 =  2 = 2 2 162 3 2 7 + 7 + 8   32.

Key Idea length of projection of the line segment joining a 1 and (a − a1 ) × n a 2 on the plane r ⋅ n = d is 2 |n|

Length of projection the line segment joining the points (5, −1, 4) and (4, −1, 3) on the plane x + y + z = 7 is B (4, –1, 3) (a2) ^

^

θ

Plane : x + y + z = 7

2 2 1 ⇒ cosθ = 1 − sin2 θ = 1 − = 3 3 3 1 Clearly, length of projection = AB cosθ = 2 3 2 = 3 ⇒ sinθ =

33.

[Q AB =

2]

Key Idea If any vector x is coplanar with the vector y and z, then x = λy + µz

Here, u is coplanar with a and b. ∴ u = λa + µb Dot product with a, we get u ⋅ a = λ (a ⋅ a) + µ (b ⋅ a) ⇒ 0 = 14λ + 2µ ^

^

^

^

…(i) ^

[Q a = 2 i + 3 j − k, b = j + k, u ⋅ a = 0] Dot product with b, we get u ⋅ b = λ (a ⋅ b ) + µ (b ⋅ b ) …(ii) [Qu ⋅ b = 24] 24 = 2 λ + 2 µ Solving Eqs. (i) and (ii), we get λ = − 2, µ = 14 Dot product with u, we get |u |2 = λ (u ⋅ a) + µ (u ⋅ b ) |u |2 = − 2(0) + 14(24) ⇒ |u |2 = 336 x y z 34. Any line parallel to = = and passing through 1 4 5 P(1, − 2, 3) is P (1, –2, 3)

x y z = = 1 4 5

^

n=i+j+k

R

2x + 3y – 4z + 22 = 0

(5, –1, 4) C

A(a1)

Q

(a − a1 ) × n = |(− i − k ) × ( i + j + k )| AC =  2 | n | ^ ^ ^   |i + j + k| ^

AC =

^

^

^

^

2 2 |i − k| ⇒ AC = = 3 3 3 ^

^

x−1 y+ 2 z −3 = = =λ 1 4 5 Any point on above line can be written as (λ + 1, 4λ − 2, 5λ + 3).

(say)

598 3D Geometry ∴Coordinates of R are (λ + 1, 4λ − 2, 5λ + 3). Since, point R lies on the above plane. ∴

2(λ + 1) + 3(4λ − 2) − 4(5λ + 3) + 22 = 0

⇒ λ =1

Hence, the image of Q(3, 1, 7) is P(− 1, 5, 3). To find equation of plane passing through P (− 1, 5, 3) x y z and containing = = 1 2 1 DR’s (1, 2,1)

So, point R is (2, 2, 8). Now, PR = (2 − 1)2 + (2 + 2)2 + (8 − 3)2 = 42 ∴

(0, 0, 0)

35. Given, equations of lines are

P (–1, 5, 3)

x−1 y+ 2 z −4 = = −2 1 3

⇒

x−2 y+ 1 z + 7 = = −1 2 −1

and

∴Any vector n perpendicular to both n1 , n 2 is given by n = n1 × n 2 $j i$ k$ n = 1 − 2 3 = 5$i + 7$j + 3k$ 2 −1 −1

∴Equation of a plane passing through (1, − 1, − 1) and perpendicular to n is given by 5x + 7 y + 3z + 5 = 0

∴Required distance =

5 + 21 − 21 + 5 5 +7 +3 2

2

2

=

10 units 83

36. Let the equation of plane be ax + by + cz = 1. Then a + b + c=1 2a + b − 2c = 0 3a − 6b − 2c = 0 ⇒ a = 7b, c =

15b 2

2 14 15 ,a = ,c= 31 31 31 ∴ 14x + 2 y + 15z = 31 b=

⇒ x (6 − 5) − y (3 + 1) + z (5 + 2) = 0 ∴ x − 4 y + 7z = 0

38. Let equation of plane containing the lines 2x − 5 y + z = 3 and x + y + 4z = 5 be (2x − 5 y + z − 3) + λ (x + y + 4z − 5) = 0 ⇒ (2 + λ ) x + (λ − 5) y + (4λ + 1) z − 3 − 5λ = 0

⇒ ∴

α − 3 β − 1 γ − 7 − 2 (3 − 1 + 7 − 3) = = = 2 −1 1 1 1 + (− 1)2 + (1)2 α −3 =1 −β = γ − 7 = −4 α = − 1, β = 5, γ = 3 Q (3,1,7)

…(i)

This plane is parallel to the plane x + 3 y + 6z = 1. 2 + λ λ − 5 4λ + 1 = = ∴ 1 3 6

On taking last two equations, we get 6λ − 30 = 3 + 12λ ⇒ − 6λ = 33 ⇒ λ = −

11 2

So, the equation of required plane is 11 11    −11   44 − 5 y +  − + 1 z − 3 + 5 × =0 2 −  x +        2 2 2 2 7 21 42 49 − x− =0 y− z+ ⇒ 2 2 2 2 ⇒

x + 3 y + 6z − 7 = 0

39. Given equation of line is x−2 y+ 1 z −2 = = =λ 3 4 12

37. Let image of Q(3, 1, 7) w.r.t. x − y + z = 3 be P (α, β, γ). ∴

y−0 z −0 2 −0 1 −0 =0

On taking first two equations, we get 11 6 + 3λ = λ − 5 ⇒ 2λ = − 11 ⇒ λ = − 2

5(x − 1) + 7( y + 1) + 3(z + 1) = 0 ⇒

x−0 1 −0

−1 −0 5 −0 3 −0

Let n1 = i$ − 2$j + 3k$ and n 2 = 2i$ − $j − k$

⇒

(x, y, z)

PQ = 2PR = 2 42

[say]…(i)

and equation of plane is …(ii) x − y + z = 16 Any point on the line (i) is, ( 3λ + 2, 4λ − 1, 12λ + 2) Let this point of intersection of the line and plane. ∴ (3λ + 2) − (4λ − 1) + (12λ + 2) = 16 ⇒ ⇒

11λ + 5 = 16 11λ = 11 ⇒ λ = 1

So, the point of intersection is (5, 3, 14). x–y+z=3

Now, distance between the points (1, 0, 2) and (5,3,14) = (5 − 1)2 + (3 − 0)2 + (14 − 2)2

P(α, β, γ)

= 16 + 9 + 144 = 169 = 13

3D Geometry 599 40. Here, plane, line and its image are parallel to each

43.

other. So, find any point on the normal to the plane from which the image line will be passed and then find equation of image line.

i.e. ( a1x + b1 y + c1 z + d 1 ) + λ ( a2 x + b 2 y + c 2 z + d 2 ) = 0 (ii) Distance of a point ( x 1, y1, z1 ) from

Here, plane and line are parallel to each other. Equation of normal to the plane through the point (1, 3, 4) is x−1 y−3 z −4 [say] = = =k −1 2 1

ax + by + cz + d = 0 | ax 1 + by1 + cz1 + d| = a2 + b 2 + c 2

Equation of plane passing through intersection of two planes x + 2 y + 3z = 2 and x − y + z = 3 is (x + 2 y + 3z − 2) + λ (x − y + z − 3) = 0 ⇒ (1 + λ ) x + (2 − λ ) y + (3 + λ ) z − (2 + 3λ ) = 0 2 whose distance from (3, 1, − 1) is . 3 |3 (1 + λ ) + 1 ⋅ (2 − λ ) − 1 (3 + λ ) − (2 + 3λ )| 2 = ⇒ 3 (1 + λ )2 + (2 − λ )2 + (3 + λ )2

Any point in this normal is (2k + 1 , − k + 3, 4 + k).  2k + 1 + 1 3 − k + 3 4 + k + 4 Then,  , ,  lies on plane.   2 2 2  6 − k  8 + k ⇒ 2 (k + 1) −   + 3 = 0 ⇒ k = −2  +  2   2  Hence, point through which this image pass is ( 2k + 1 , 3 − k, 4 + k) i.e. [ 2(−2) + 1 , 3 + 2, 4 − 2] = (−3, 5 , 2) x+ 3 y−5 z −2 Hence, equation of image line is . = = 3 1 −5

41.

x+2 y+1 z = = =λ −1 2 3

⇒

Let foot of perpendicular from ( 2λ − 2, − λ − 1, 3λ ) to x + y + z = 3 be (x2, y2, z2). x2 − (2λ − 2) y2 − (− λ − 1) z2 − ( 3λ ) = ∴ = 1 1 1 ( 2 λ − 2 − λ − 1 + 3 λ − 3) =− 1+1+1 ⇒

4λ 3 2λ 7λ 5λ x2 = , y2 = 1 − , z2 = 2 + 3 3 3 x2 − 0 y2 − 1 z2 − 2 λ= = = −7 / 3 2 /3 5 /3

3λ2 + 4λ + 14

=

2 3

⇒

3λ2 = 3λ2 + 4λ + 14 7 λ=− ⇒ 2 7 21 7 7     ∴ 1 −  x + 2 +  y + 3 −  z − 2 −  = 0     2 2 2 2 5x 11 1 17 − + =0 y− z+ ⇒ 2 2 2 2 or 5x − 11 y + z − 17 = 0

44.

x = 2λ − 2, y = − λ − 1, z = 3λ

|− 2λ |

⇒

Key Idea To find the foot of perpendiculars and find its locus. Formula used Footof perpendicular from ( x 1, y1, z1 )to ax + by + cz + d = 0 be ( x 2, y2, z2 ), then − ( ax 1 + by1 + cz1 + d ) x 2 − x1 y − y1 z − z1 = = 2 = 2 a b c a2 + b 2 + c 2

Any point on

Key Idea (i) Equation of plane through intersection of two planes,

PLAN It is based on two concepts one is intersection of straight line and plane and other is the foot of perpendicular from a point to the straight line.

Description of Situation (i) If the straight line x − x1 y − y1 z − z1 = = =λ a b c

x2 − 2λ + 2 = y2 + λ + 1 = z2 − 3λ = 2 −

∴ ⇒

Hence, foot of perpendicular lie on x y−1 z −2 x y−1 z −2 = = = ⇒ = 2 / 3 −7 / 3 5 / 3 2 5 −7

intersects the plane A x + B y + C z + d = 0. Then, (a λ + x1 , b λ + y1 , c λ + z1 ) would satisfy A x + B y + C z + d =0 (ii) If A is the foot of perpendicular from P to l. Then, (DR’s of PA) is perpendicular to DR’s of l. P

42. Given planes are 2x + y + 2z − 8 = 0 and 2x + y + 2z +

5 =0 2

∴ Distance between two planes =

| c1 − c2| a 2 + b2 + c2

=

5 −8 − 21 / 2 7 2 = = 2 2 2 3 2 2 +1 +2

A

⇒

l

PA ⋅ l = 0

Equation of straight line QR, is x −2 y −3 z −5 = = 1 −2 −1 −3 4 −5

600 3D Geometry ⇒

x −2 y −3 z −5 = = −1 −4 −1

⇒

x−2 y−3 z −5 = = =λ 1 4 1

Also, equation of plane containing …(i)

∴ P (λ + 2 , 4λ + 3, λ + 5) must lie on 5x − 4 y − z = 1. ⇒

5 (λ + 2) − 4 (4λ + 3) − (λ + 5) = 1

⇒

5λ + 10 − 16 λ − 12 − λ − 5 = 1

⇒

− 7 − 12λ = 1 −2 λ= 3  4 1 13 P , ,  3 3 3 

∴ or

Again, we can assume S from Eq.(i), as S ( µ + 2, 4 µ + 3, µ + 5) ∴ DR’s of TS = < µ + 2 − 2, 4 µ + 3 − 1, µ + 5 − 4 > = < µ, 4 µ + 2, µ + 1 > and DR’s of QR = < 1, 4, 1 > Since, perpendicular 1 ( µ ) + 4 (4 µ + 2) + 1 ( µ + 1) = 0 1 9 3 and S  , 1,  ⇒ µ=− 2 2 2

⇒ ax + by + cz = 0 where, n1 : n 2 = 0 ⇒

…(i)

8a − b − 10c = 0

…(ii)

and n 2 ⊥ (2$i + 3$j + 4k$ ) ⇒

2a + 3b + 4c = 0

…(iii)

From Eqs (ii) and (iii), a b c = = − 4 + 30 − 20 − 32 24 + 2 ⇒

a b c = = 26 − 52 26 a b c = = 1 −2 1

⇒

…(iv)

From Eqs. (i) and (iv), required equation of plane is x − 2y + z = 0 1 47. Since, l=m = n = 3

(2

,−1

,2)

∴

normal to be n 1 = a$i + b$j + ck$

x y z = = and DR’s of 2 3 4

2

2

P

2

1 1  9 13   3 4 ∴ Length of PS =  −  + 1 −  +  −  = 2 3    2 3 3 2

45. Distance of point P from plane = 5 1 −4 −2 −α 3

∴

5=

⇒

α = 10

∴ Equations of line are A (x, y, z) x + 2 y – 2z = a

Foot of perpendicular x−1 y+ 2 z −1 5 = = = −2 1 2 3 x=

⇒

x y z = = . 4 z 3

x −2 = y + 1 = z −2 = r

∴ ⇒

x y z = = 3 4 2

⇒

2 (r + 2) + (r − 1) + (r + 2) = 9 4r + 5 = 9 ⇒ r = 1 Q (3 , 0 , 3)

∴ PQ = (3 − 2) + (0 + 1)2 + (3 − 2)2 = 3 2

48. Given, OQ = (1 − 3 µ ) i$ + (µ − 1) $j + (5 µ + 2) k$ and OP = 3 $i + 2$j + 6 k$

[where, O is origin]

PQ

$i $j k$ n 1 = 3 4 2 = ( 8 $i − $j − 10k$ ) 4 2 3

i – 4j+3k x – 4y+3z=1

n (a, b, c)

DR’s (0,0,0)

(2,3,4)

[say]

∴ Any point on the line is Q ≡ (r + 2 , r − 1 , r + 2)

8 4 7 , y= ,z=− 3 3 3

46. The DR’s of normal to the plane containing

⇒

x −2 y+1 z −2 = = 1/ 3 1/ 3 1/ 3

Q Q lies on the plane 2x + y + z = 9

7 8 4 Thus, the foot of the perpendicular is A  , , −  . 3 3 3

and

Q 2x + y + z = 9

P (1,– 2,1)

Now, PQ = (1 − 3 µ − 3) $i + (µ − 1 − 2) $j + (5 µ + 2 − 6) k$ = (−2 − 3 µ ) $i + ( µ − 3) $j + (5 µ − 4) k$ Q PQ is parallel to the plane x − 4 y + 3z = 1. ∴ − 2 − 3 µ − 4 µ + 12 + 15 µ − 12 = 0

3D Geometry 601 ⇒

8µ = 2 ⇒ µ =

1 4

52. Since, the point A(3, 2, − 1) is the mirror image of the

49. Let the equation of plane be a (x − 1) + b ( y + 2) + c (z − 1) = 0 which is perpendicular to 2x − 2 y + z = 0 and x − y + 2z = 4. ⇒ ⇒ ⇒

2a − 2b + c = 0 and a − b + 2c = 0 a b c = = −3 − 3 0 a b c = = ⋅ 1 1 0

So, the equation of plane is x − 1 + y + 2 = 0 x+ y+ 1 =0 |1 + 2 + 1| Its distance from the point (1, 2, 2) is =2 2 2 x y z 50. Since, + + = 1 cuts the coordinate axes at a b c A (a , 0, 0), B (0, b, 0), C (0, 0, c). And its distance from origin = 1 or

point B(1, 0, − 1) with respect to the plane αx + βy + γ z = δ, then α β γ = = 3 − 1 2 − 0 (− 1) − (− 1) α β γ ⇒ = = 1 1 0 ⇒ γ =0 it is given that α + γ = 1 ⇒ α = 1, so β = 1. And, the mid-point of AB, M (2, 1, − 1) lies on the given plane, so 2α + β − γ = δ ⇒ 2 (1) + (1) − 0 = δ ⇒ δ =3 ∴ α + β = 2 , δ − γ = 0, δ + β = 4.

53. Given, square base OP = OR = 3 ∴

P (3, 0, 0), R = (0, 3, 0) S

Y B (0,b,0) P

A

O Z

∴

or

O

X P (3,3,0)

C (0,0,c)

x=

a b c , y= ,z= 3 3 3

From Eqs. (i) and (ii), 1 1 1 + + =1 9 x 2 9 y2 9 z 2 1 1 1 or + + =9 = K x2 y2 z 2 ∴

K =9

51. Given equation of straight line x−4 y−2 z − k = = 1 1 2 Since, the line lies in the plane 2x − 4 y + z = 7. Hence, point (4, 2, k) must satisfy the plane. ⇒

8 −8 + k = 7 ⇒ k = 7

Q(3,3,0)

X

3 3  Also, mid-point of OQ is T  , , 0 . 2 2  …(i)

where, P is centroid of triangle.  a + 0 + 0 0 + b + 0 0 + 0 + c ∴ P (x, y, z ) =  , ,    3 3 3 ⇒

Y

T

(a,0,0)

1 =1 1 1 1 + + a 2 b2 c2 1 1 1 + 2 + 2 =1 2 a b c

R(0,3,0)

θ

…(ii)

Since, S is directly above the mid-point T of diagonal OQ and ST = 3. 3 3  i.e. S  , , 3 2 2  3 3  Here, DR’s of OQ (3, 3, 0) and DR’s of OS  , , 3 . 2 2  9 9 + 9 1 2 2 = = ∴ cos θ = 27 3 9 9 18 ⋅ + +9 9+9+0 2 4 4 ∴ Option (a) is incorrect. Now, equation of the plane containing the ∆OQS is x 3

y 3

z 0 =0⇒

3 /2 3 /2 3

x y z 1 1 0 =0 1 1 2

⇒ x(2 − 0) − y(2 − 0) + z (1 − 1) = 0 ⇒ 2x − 2 y = 0 or x − y = 0 ∴ Option (b) is correct. Now, length of the perpendicular from P(3, 0, 0) to the |3 − 0| 3 plane containing ∆OQS is = 1+1 2 ∴ Option (c) is correct.

602 3D Geometry If (α , β , γ ) is foot of perpendicular from P on P1, then α − λ β + 3λ γ + 5λ [say] = = =k 1 2 −1 ⇒ α = λ + k, β = 2k − 3λ, γ = − k − 5λ which satisfy P1 : x + 2 y − z + 1 = 0 ⇒ (λ + k) + 2 (2k − 3λ ) − (− k − 5λ ) + 1 = 0 1 k=− ⇒ 6 1 1 1 ∴ x = − + λ , y = − − 3 λ , z = − 5λ 6 3 6 which satisfy options (a) and (b).

Here, equation of RS is x−0 y−3 z −0 = = =λ −3 / 2 3 /2 3 3 3 x = λ , y = − λ + 3, z = 3λ ⇒ 2 2 To find the distance from O(0, 0, 0) to RS. Let M be the foot of perpendicular. O (0, 0, 0)

56. We have,

R M S (0,3,0)  3λ 3λ   3, 3 3  , ,3 λ 3 −  2   2 2  2  

Q ⇒

OM ⊥ RS ⇒

OM ⋅ RS = 0

1 9λ 3  3λ  − 3 −  + 3(3λ ) = 0 ⇒ λ = 3 4 2 2

and

P1 : 2x + y − z = 3 P2 : x + 2 y + z = 2

Here,

→ n1 = 2i$ + $j − k$

and

→ n2 = i$ + 2 $j + k$

(a) Direction ratio of the line of intersection of P1

1 25 15 30 1 5  ∴ M  , , 1 ⇒ OM = + +1= = 2 2  4 4 2 4

→

∴ Option (d) is correct.

54. Here, i.e.

i$

P3 : (x + z − 1) + λy = 0 P3 : x + λy + z − 1 = 0

…(i)

whose distance from (0, 1, 0) is 1. |0 + λ + 0 − 1| =1 ∴ 1 + λ2 + 1 ⇒ ⇒

∴ Equation of P3 is 2x − y + 2z − 2 = 0. Q Distance from (α , β , γ ) is 2. ∴ ⇒

$j

k$

i.e. 2 1 −1 = (1 + 2) i$ − (2 + 1) $j + (4 − 1)k$ 1 2 1 = 3 (i$ − $j + k$ ) Hence, statement a is false.

|λ − 1| = λ2 + 2 λ 2 − 2λ + 1 = λ2 + 2 ⇒ λ = −

→

and P2 is θ n1 × n2

1 2

|2α − β + 2 γ − 2| =2 4+1+4 2 α − β + 2γ − 2 = ± 6

⇒ 2 α − β + 2γ = 8 and 2 α − β + 2γ = − 4

55. Since, L is at constant distance from two planes P1 and P2. Therefore, L is parallel to the line through intersection of P1 and P2. $i $j k$ DR’s of L = 1 2 –1 2 –1 1 $ = i (2 − 1) − $j (1 + 2 ) + k$ (− 1 − 4) = $i − 3$j − 5k$ ∴ DR’s of L are (1, –3, –5) passing through (0, 0, 0). Now, equation of L is x−0 y−0 z −0 = = −3 1 −5 x y z For any point on L, = [say] = =λ 1 −3 −5 i.e. P (λ , − 3λ , − 5λ )

4  y − 1   x− 3x − 4 1 − 3 y z 3 z 3 = (b) We have, = = = ⇒ 3 −3 3 9 9 3 This line is parallel to the line of intersection of P1 and P2. Hence, statement (b) is false. (c) Let acute angle between P1 and P2 be θ. We know that, → → $ ⋅ (i$ + 2j$ + k) $ n1 ⋅ n 2 (2i$ + $j − k) cos θ = = $ $ $ $ $ $ → → | n | | n | | 2i + j − k | | i + 2j + k | 1

2

=

2 + 2 −1 1 = 6× 6 2

θ = 60° Hence, statement (c) is true. (d) Equation of plane passing through the point (4, 2, − 2 ) and perpendicular to the line of intersection of P1 and P2 is 3 (x − 4) − 3 ( y − 2) + 3 (z + 2) = 0 ⇒ 3x − 3 y + 3z − 12 + 6 + 6 = 0 ⇒ x− y+ z =0 Now, distance of the point (2, 1, 1) from the plane x − y + z = 0 is D=

2 −1 + 1 2 = 1+1+1 3

Hence, statement (d) is true.

3D Geometry 603 57.

PLAN If the straight lines are coplanar. They the should lie in same plane.

Description of Situation If straight lines are coplanar. x2 − x1 y2 − y1 z2 − z1 ⇒

b1

c1

a2

b2

c2

=0

0

x−1 y+ 1 z = = 2 K 2 x+1 y+1 z = = are coplanar. 5 2 k 0

2 K 5 2

2 =0 ⇒ K2 =4 ⇒ K = ± 2 K

Since, and 2 ⇒

a1

∴

n1 = b1 × d 1 = 6j − 6k, for k = 2

∴

n 2 = b 2 × d 2 = 14j + 14k, for k = − 2

59. Let P(α, β, γ ) and R is image of P in the xy-plane. ∴ R(α, β , − γ ) Also, Q is the image of P in the plane x + y = 3 x − α y − β z − γ −2 (α + β − 3) = = ∴ = 1 1 0 2 x = 3 − β, y = 3 − α, z = γ Since, Q is lies on Z-axis ∴ β = 3, α = 3, z = γ ∴ P(3, 3, γ ) Given, distance of P from X-axis be 5. 5 = 32 + γ 2

∴

25 − 9 = γ 2 ⇒ γ = ± 4 Then,

PR = |2γ| = |2 × 4| = 8

60. The equation of the plane passing through the point

(−1, − 2, − 1) and whose normal is perpendicular to both the given lines L1 and L 2 may be written as

So, equation of planes are (r − a ) ⋅ n1 = 0

(x + 1) + 7 ( y + 2) − 5 (z + 1) = 0 ⇒ x + 7 y − 5z + 10 = 0

⇒

y − z = − 1 and (r − a ) ⋅ n 2 = 0

The distance of the point (1, 1, 1) from the plane

⇒

y+ z = −1

=

58. Given three lines

and

r = λ$i , λ ∈ R, r = µ (i$ + $j), µ ∈ R r = v (i$ + $j + k$ ), v ∈ R

cuts the plane x + y + z = 1 at the points A , B and C, respectively. So, for point A, put (λ , 0, 0) in the plane, we get λ + 0 + 0 = 1 ⇒ λ = 1 ⇒ A ≡ (1,0, 0). Similarly, for point B, put (µ , µ , 0) in the plane, we get µ + µ + 0 = 1 ⇒ 1 1 1  µ = ⇒ B ≡  , , 0 . 2 2  2 and for point C, put (v, v, v) in the plane we get v+ v+ v = 1 1  1 1 1 ⇒ v= ⇒ C≡ , ,   3 3 3 3 1 → → Now, area of ∆ABC = | AB× AC|= ∆ 2 1$ 1$ AB = − i + j, and Q 2 2 2$ 1$ 1 $ AC = − i + j + k 3 3 3 $i $j k$ ∴ AB × AC = −1 2 1 2 0 −2 3

1 3

61. The shortest distance between L1 and L 2 is {(2 − (−1)) $i + (2 − 2) $j + (3 − (−1)) k$ } ⋅ (− $i − 7$j + 5k$ ) 5 3 =

(3$i + 4k$ ) ⋅ (− $i − 7$j + 5k$ ) 17 units = 5 3 5 3

62. The equations of given lines in vector form may be → written as L1 : r = (− $i − 2$j − k$ ) + λ (3$i + $j + 2k$ )

and

→ L 2 : r = (2i$ − 2$j + 3k$ ) + µ (i$ + 2$j + 3k$ )

Since, the vector is perpendicular to both L1 and L 2 . $i $j k$ 3 1 2 = − $i − 7$j + 5k$ 1 2 3 ∴ Required unit vector (− i$ − 7$j + 5k$ ) 1 = = (− i$ − 7$j + 5k$ ) 2 2 2 5 3 (−1) + (−7) + (5)

63. Given three planes are

1 3

 1  1  1 2 1 = $i   − $j –  + k$  – +  = ($i + $j + k$ )  6  6  6 6 6 1 1 ⇒|AB × AC| = 3= 6 2 3 1 ∆= ⇒ 4 3 1 3 ⇒ (6∆ )2 = 36 = = 0.75 16 × 3 4

1 + 7 − 5 + 10 13 units = 1 + 49 + 25 75

P1 : x − y + z = 1 P2 : x + y − z = − 1 and

P3 : x − 3 y + 3z = 2

On solving Eqs. (i) and (ii), we get x = 0, z = 1 + y which does not satisfy Eq. (iii). As

x − 3 y + 3z = 0 − 3 y + 3 (1 + y) = 3 (≠ 2)

So, Statement II is true.

...(i) ...(ii) ...(iii)

604 3D Geometry Next, since we know that direction ratios of line of intersection of planes a1x + b1 y + c1z + d1 = 0

B. a + b + c = 0 and a 2 + b2 + c2 ≠ ab + bc + ca ⇒ ∆ =0

a 2 x + b2y + c2z + d2 = 0 is

and

⇒ The equations have infinitely many solutions.

b1c2 − b2c1 , c1a 2 − a1c2, a1b2 − a 2b1

ax + by = (a + b) z, bx + cy = (b + c) z

Using above result,

⇒

Direction ratios of lines L1 , L 2 and L3 are

⇒

0, 2, 2 ; 0, − 4, − 4 ; 0, − 2, − 2 Since, all the three lines L1 , L 2 and L3 are parallel pairwise.

⇒

The equations represent planes meeting at D. a + b + c = 0 and a 2 + b2 + c2 = ab + bc + ca

Since, direction ratios of planes are

⇒

< 3, − 6, − 2 > and < 2, 1, − 2 >

[say]

Hence, Statement I is false. But Statement II is true. x − 1 y − 0 z − (−3) 65. L1 : = = 2 1 −1 $i

$j k $ Normal of plane P : n = 7 1 2 3 5 −6 $ (32) = − 16i$ + 48$j + 32 k $ = $i (−16) − $j(−42 − 6) + k $ DR’s of normal n = $i − 3$j − 2k

67. Equation of plane containing the lines 2x − y + z − 3 = 0 and 3x + y + z = 5 is (2x − y + z − 3) + λ (3x + y + z − 5) = 0 ⇒ (2 + 3λ ) x + (λ – 1) y + (λ + 1) z − 3 − 5λ = 0 Since, distance of plane from ( 2, 1, − 1) to above plane is 1 / 6. 6 λ + 4 + λ − 1 − λ − 1 − 3 − 5 λ 1 = ∴  2 2 2 6  (3λ + 2) + (λ − 1) + (λ + 1) ⇒

2K 1 + 1 = K 2 + 4 − k1 = k2 − 3 ⇒ k1 = 2 and k2 = 1

∴ Point of intersection (5, − 2, − 1)

6 (λ − 1)2 = 11λ2 + 12λ + 6 24 λ = 0, − 5

⇒

∴ Equations of planes are 2x − y + z − 3 = 0 and 62x + 29 y + 19z − 105 = 0

Point of intersection of L1 and L 2.

68. Let the equation of plane through (1, 1, 1 ) having a, b, c as DR’s of normal to plane, a (x − 1) + b ( y − 1) + c (z − 1) = 0 and plane is parallel to straight line having DR’s.

Now equation of plane, 1 ⋅ (x − 5) − 3( y + 2) − 2(z + 1) = 0 ⇒

x − 3 y − 2z − 13 = 0

⇒

x − 3 y − 2z = 13

∴

a → 1, b → − 3, c → − 2, d → 13 a

b

c =−

a

⇒

⇒

1 (a + b + c) [(a − b)2 + (b − c)2 + (c − a )2] 2 ∆ =0 a = b = c≠0

⇒ The equations represent identical planes.

a − c=0 − a + b =0

and ⇒

a=b=c

∴ Equation of plane is or

b

A. If a + b + c ≠ 0 and a 2 + b2 + c2 = ab + bc + ca and

(1, 0, − 1) and (− 1, 1, 0)

c

66. Let ∆ = b c a

a = b = c=0

⇒ The equations represent whole of the three-dimensional space.

x = 14 λ + 3, y = 2 λ − 1, z = 15 λ

and

∆ ≠0

only one point.

For z = 0, we get x = 3, y = − 1

⇒

ax = ay ⇒ x = y = z

C. a + b + c ≠ 0 and a 2 + b2 + c2 ≠ ab + bc + ca

64. Given planes are 3x − 6 y − 2z = 15 and 2x + y − 2z = 5.

⇒

ax + by + cy = 0

⇒

Hence, Statement I is false.

Then the DR’s of line of intersection of planes is < 14, 2, 15 > and line is x −3 y + 1 z −0 = = =λ 14 2 15

(b2 − ac) y = (b2 − ac) z ⇒ y = z

x−1 + y−1 + z −1 =0 x y z + + = 1. 3 3 3

Its intercept on coordinate axes are A (3, 0, 0), B (0, 3, 0), C (0, 0, 3). Hence, the volume of tetrahedron OABC 3 0 0 1 1 27 9 = [a b c] = 0 3 0 = = cu units 6 6 6 2 0 0 3

3D Geometry 605 69. Let

the equation of the plane ABCD be ax + by + cz + d = 0, the point A′′ be (α , β , γ ) and the height of the parallelopiped ABCD be h. | aα + bβ + c γ + d | = 90 % h ⇒ a 2 + b2 + c2

⇒

⇒

x + y − 2z = 3

(ii) Let the coordinates of Q be (α , β , γ ). Equation of line PQ ⇒

Since, mid-point of P and Q

aα + bβ + c γ + d = ± 0.9 h a 2 + b2 + c2

 α + 2 β + 1 γ + 6 , ,  ,  2 2 2 

∴ Locus is ax + by + cz + d = ± 0. 9h a + b + c 2

2

2

Hence, locus of A ′′ is a plane parallel to the plane ABCD.

70. (i) Equation of plane passing through (2, 1, 0) is a (x − 2) + b ( y − 1) + c (z − 0) = 0 It also passes through (5, 0, 1) and (4, 1, 1). ⇒ 3a − b + c = 0 and 2a − 0b + c = 0 a b c On solving, we get = = –1 –1 2 ∴ Equation of plane is − (x − 2) − ( y − 1) + 2(z − 0) = 0 − (x − 2) − y + 1 + 2z = 0

x−2 y−1 z−6 = = 1 1 −2

which lies in line PQ. ⇒

γ+6 α −2 β+ 1 −6 −1 −2 2 2 = 2 = −2 2 1

 γ + 6  β + 1  α + 2 1 − 6 − 1 − 2  − 2 + 1    2   2   2 = =2 1 ⋅ 1 + 1 ⋅ 1 + (−2) (−2)   α + 2   γ + 6  β + 1 Q  2  − 1  2  − 2  2  = 3   ⇒ α = 6, β = 5, γ = − 2 ⇒ Q (6, 5, − 2)

26 Miscellaneous Objective Questions I (Only one correct option) 1. A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x % of the people read both the newspapers, then a possible value of x can be (2020 Main , 4 Sep I) (a) 55

(b) 29

(c) 65

(d) 37

2. Let S be the set of all real roots of the equation, 3x ( 3x − 1) + 2 =| 3x − 1| +| 3x − 2|. Then S

(2020 Main, 8 Jan II)

(a) (b) (c) (d)

is a singleton. is an empty set. contains at least four elements. contains exactly two elements.

(b) P ∧ (P ∨ Q ) (d) Q → (P ∧ (P → Q ) (2020 Main, 3 Sep I)

(b) (~ p ) ∨ q (d) (~ p ) ∨ (~ q)

5. The boolean expression ~ ( p ⇒ (~ q )) is equivalent to (2019 Main, 12 April II)

(b) q ⇒ ~ p (c) p ∨ q

(d) (~ p ) ⇒ q

6. If the data x1 , x2 , … , x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2000, then the standard deviation of this data is (2019 Main, 12 April I) (a) 2 2

(b) 2

(c) 4

(d)

2

7. If the truth value of the statement p → (~ q ∨ r ) is false (F), then the truth values of the statements p, q and r are respectively (2019 Main, 12 April I) (a) T, T and F (c) T, F and T

8. The negation of the boolean expression ~ s ∨ (~ r ∧ s) (a) s ∧ r

(b) ~ s ∧ ~ r

(2019 Main, 10 April II)

(c) s ∨ r

(d) r

9. If both the mean and the standard deviation of 50 observations x1 , x2 , K , x50 are equal to 16, then the mean of ( x1 − 4)2,( x2 − 4)2 , K , ( x50 − 4)2 is (2019 Main, 10 April II)

(a) 480

(b) 400

(c) 380

(b) ( p ∧ q) ∨ ( p ∧ ~ q) (d) ( p ∨ q) ∧ (~ p ∨ ~ q)

11. If for some x ∈ R, the frequency distribution of the marks obtained by 20 students in a test is Marks Frequency

2 (x + 1)

2

3

5

7

2x − 5

x − 3x

x

2

(2019 Main, 10 April I)

(b) 2.8 (d) 3.2

known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then, the percentage of the population who look into advertisements is (2019 Main, 9 April II) (a) 13.5 (c) 12.8

(b) 13 (d) 13.9

13. The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, y 67, 70, y are 42 and 35 respectively, then is equal to x (2019 Main, 9 April II)

7 (a) 3

7 (b) 2

(d) 525

8 (c) 3

(d)

9 4

14. If p ⇒ ( q ∨ r ) is false, then the truth values of p, q , r are respectively (a) T, T, F (c) F, F, F

(b) T, F and F (d) F, T and T

is equivalent to

(a) ( p ∨ q) ∨ ( p ∨ ~ q) (c) ( p ∨ q) ∧ ( p ∨ ~ q)

12. Two newspapers A and B are published in a city. It is

4. The proposition p → ~ ( p ∧ ~ q ) is equivalent to

(a) p ∧ q

(2019 Main, 10 April I)

(a) 3.0 (c) 2.5

(2020 Main, 8 Jan I)

(a) q (c) (~ p ) ∧ q

tautology ?

Then, the mean of the marks is

3. Which one of the following is a tautology? (a) (P ∧ (P → Q )) → Q (c) P ∨ (P ∧Q )

10. Which one of the following Boolean expressions is a

(2019 Main, 9 April II)

(b) T, F, F (d) F, T, T

15. For any two statements p and q, the negation of the expression p ∨ (~ p ∧ q ) is

(a) ~ p ∧ ~ q (c) p ∧ q

(2019 Main, 9 April I)

(b) ~ p ∨ ~ q (d) p ↔ q

16. If the standard deviation of the numbers −1, 0, 1, k is 5 where k > 0, then k is equal to

10 (a) 2 3

(b) 2 6

5 (c) 4 3

(2019 Main, 9 April I)

(d)

6

Miscellaneous 607 17. A student scores the following marks in five tests 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is (2019 Main, 8 April II)

(a)

10 3

(b)

10 3

(c)

100 3

(d)

100 3

18. Which one of the following statements is not a tautology?

(2019 Main, 8 April II)

(a) ( p ∧ q) → (~ p ) ∨ q (c) p → ( p ∨ q)

(b) ( p ∧ q) → p (d) ( p ∨ q) → ( p ∨ (~ q))

26. The expression ~ (~ p → q ) is logically equivalent to (2019 Main, 12 Jan II)

(a) p ∧ ~ q (c) ~ p ∧ q

(b) p ∧ q (d) ~ p ∧ ~ q

27. The Boolean expression

(( p ∧ q ) ∨ ( p ∨ ~ q )) ∧ (~ p∧ ~ q ) is equivalent to (2019 Main, 12 Jan I)

(a) p ∧ q (c) p ∧ (~ q)

(b) p ∨ (~ q) (d) (~ p ) ∧ (~ q)

28. If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is

19. The sum of all natural numbers ‘n’ such that 100 < n < 200 and HCF (91, n) > 1 is (a) 3203 (c) 3221

(2019 Main, 8 April I)

and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is (2019 Main, 8 April I) (b) 49

(c) 48

(d) 40

21. All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is (2019 Main, 8 April I) (a) 180 (c) 160

(b) 175 (d) 162

in India, then you are a citizen of India’’, is (2019 Main, 8 April I)

(a) If you are not a citizen of India, then you are not born in India. (b) If you are a citizen of India, then you are born in India. (c) If you are born in India, then you are not a citizen of India. (d) If you are not born in India, then you are not a citizen of India.

23. The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4, then the absolute value of the difference of the other two observations, is (2019 Main, 12 Jan II)

(b) 7

(c) 5

(d) 3

24. In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is (2019 Main, 12 Jan II) (a)

1 6

(b)

1 3

(c)

2 3

(d)

5 6

25. Let Z be the set of integers. If 2

A = { x ∈ Z : 2( x + 2) ( x − 5 x + 6) = 1} and B = { x ∈ Z : − 3 < 2x − 1 < 9}, then the number of subsets of the set A × B, is (2019 Main, 12 Jan II) (a) 212

(b) 218

(c) 215

(c) 51

(d) 31

subsets A of S such that the product of elements in A is even, is (2019 Main, 12 Jan I) (a) 250 (250 − 1) (c) 250 + 1

(d) 210

(b) 250 − 1 (d) 2100 − 1

30. A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn,   mean of X then   is equal to  standard deviation of X  (2019 Main, 11 Jan II) 4 3 3 (c) 3 2 (a)

22. The contrapositive of the statement ‘‘If you are born

(a) 1

(b) 30

29. Let S = { 1, 2, 3, ... , 100}. The number of non-empty

(b) 3303 (d) 3121

20. The mean and variance of seven observations are 8

(a) 45

(2019 Main, 12 Jan I)

(a) 50

(b) 4 (d) 4 3

31. Contrapositive of the statement “If two numbers are not equal, then their squares are not equal” is (2019 Main, 11 Jan II)

(a) If the squares of two numbers are not equal, then the numbers are not equal. (b) If the squares of two numbers are equal, then the numbers are equal. (c) If the squares of two numbers are not equal, then the numbers are equal. (d) If the squares of two numbers are equal, then the numbers are not equal.

32. If q is false and p ∧ q ←→ r is true, then which one of the following statements is a tautology? (2019 Main, 11 Jan I)

(a) p ∨ r (c) ( p ∨ r ) → ( p ∧ r )

(b) ( p ∧ r ) → ( p ∨ r ) (d) p ∧ r

33. The outcome of each of 30 items was observed; 10 items gave an outcome

1 − d each, 10 items gave 2

1 each and the remaining 10 items gave 2 1 outcome + d each. If the variance of this outcome 2 4 data is , then|d| equals 3 (2019 Main, 11 Jan I)

outcome

2 3 (c) 2 (a)

5 2 (d) 2 (b)

608 Miscellaneous 34. Consider the following three statements : P : 5 is a prime number. Q : 7 is a factor of 192. R : LCM of 5 and 7 is 35. Then, the truth value of which one of the following statements is true ? (2019 Main, 10 Jan II)

(b) 6 : 7 (d) 5 : 8

numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then, the number of students who did not opt for any of the three courses is (2019 Main, 10 Jan I)

(b) 102 (d) 1

37. Consider the statement : ‘‘P ( n ) : n 2 − n + 41 is prime.’’ Then, which one of the following is true? (2019 Main, 10 Jan I)

Both P (3) and P (5) are true. P (3) is false but P (5) is true. Both P (3) and P (5) are false. P (5) is false but P (3) is true.

then the

i =1

(b) 4

(c) 2

(d) 3

44. The boolean expression ~ ( p ∨ q ) ∨ (~ p ∧ q ) is equivalent to (a) ~ p (c) q

(2018 Main)

(b) p (d) ~ q

following system of linear equations x + y + z = 1, x + ay + z = 1 and ax + by + z = 0 has no solution, then S is

(2017 Main)

(a) an infinite set (b) a finite set containing two or more elements (c) singleton set (d) an empty set

46. The statement ( p → q ) → [(~ p → q ) → q ] is (2017 Main) (a) a tautology (c) equivalent to p → ~ q

(b) equivalent to ~ p → q (d) a fallacy

 1  x S = { x ∈ R : f ( x ) = f ( − x )}; then S

[~ (~ p ∨ q) ∨ ( p ∧ r )] ∧ (~ q ∧ r ) is equivalent to (2019 Main, 9 Jan II) (a) ~ p ∨ r (b) ( p ∧ ~ q) ∨ r (c) ( p ∧ r ) ∧ ~ q (d) (~ p ∧ ~ q) ∧ r

39. In a group of data, there are n observations, n

Σ( xi + 1)2 = 9n i =1

(a) 9

47. If f ( x ) + 2 f   = 3x, x ≠ 0 and

38. The logical statement

n

and

Σ ( xi − 1)2 = 5n, i =1

the standard deviation of the data is (2019 Main, 9 Jan II) (b) 7 (d) 5

48. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true? (2016 Main)

(a) 3a 2 − 26a + 55 = 0 (c) 3a 2 − 34a + 91 = 0

(b) 3a 2 − 32a + 84 = 0 (d) 3a 2 − 23a + 44 = 0

equivalent to

( p ⊕ q ) ∧ (~ p ⋅ q ) is equivalent to p ∧ q, where ⊕, ⋅ ∈{ ∧,∨}, then the ordered pair (⊕, ⋅) is (b) (∧, ∧) (d) (∨, ∨)

(2016 Main)

(a) is an empty set (b) contains exactly one element (c) contains exactly two elements (d) contains more than two elements

49. The Boolean expression ( p ∧ ~ q ) ∨ q ∨ (~ p ∧ q ) is

40. If the Boolean expression

(a) (∧, ∨) (c) (∨, ∧)

∑ (xi − 5)2 = 45,

45. If S is the set of distinct values of b for which the

36. In a class of 140 students numbered 1 to 140, all even

(a) 2 (c) 5

9

and

(2018 Main)

(2019 Main, 10 Jan I)

x , x2 , .... , xn . If

9

∑ (xi − 5) = 9

standard deviation of the 9 items x1 , x2 , … , x9 is

is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is

(a) (b) (c) (d)

(2018 Main)

(a) B ⊂ A (b) A ⊂ B (c) A ∩ B = φ (an empty set)(d) neither A ⊂ B nor B ⊂ A

i =1

35. The mean of five observations is 5 and their variance

(a) 42 (c) 38

A = {( a , b) ∈ R × R :|a − 5|< 1and|b − 5|< 1}; B = {( a , b) ∈ R × R : 4( a − 6)2 + 9( b − 5)2 ≤ 36}. Then,

43. If

(a) (P ∧ Q ) ∨ (~ R ) (b) P ∨ (~ Q ∧ R ) (c) (~ P ) ∨ (Q ∧ R ) (d) (~ P ) ∧ (~ Q ∧ R )

(a) 4 : 9 (c) 10 : 3

42. Two sets A and B are as under

(2019 Main, 9 Jan I)

41. 5 students of a class have an average height 150 cm

(a) ~ p ∧ q (c) p ∨ q

(2016 Main)

(b) p ∧ q (d) p ∨ ~ q

50. The negation of ~ s ∨ (~ r ∧ s) is equivalent to (a) s∧ ~ r (c) s ∧ (r ∨ ~ s)

(b) s ∧ (r ∧ ~ s) (d) s ∧ r

(2015 Main)

51. The mean of the data set comprising of 16

and variance 18 cm 2. A new student, whose height is 156 cm, joined them. The variance (in cm 2 ) of the height of these six students is (2019 Main, 9 Jan I)

observations is 16. If one of the observations valued 16 is deleted and three new observations valued 3,4 and 5 added to the data, then the mean of the resultant data is (2015 Main)

(a) 16

(a) 16.8

(b) 22

(c) 20

(d) 18

(b) 16.0

(c) 15.8

(d) 14.0

Miscellaneous 609 52. If A and B are two sets containing four and two elements, respectively. Then, the number of subsets of the set A × B each having atleast three elements are (2015 Main) (a) 219

(b) 256

(c) 275

three collinear points. A, B and C on a line leading to the foot of the tower are 30°, 45° and 60° respectively, then the ratio AB : BC is (2015 Main) (b) 3 : 2 (d) 2: 3

54. The statement ~ ( p ↔ ~ q ) is (a) equivalent to p ↔ q (c) a tautology

(2014 Main)

(b) equivalent to ~ p ↔ q (d) a fallacy

55. A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After 1 s, the elevation of the bird from O is reduced to 30°. Then, the speed (in m/s) of the bird is (a) 40 ( 2 − 1) (c) 20 2

(b) 40 ( 3 − 2 ) (d) 20 ( 3 − 1)

(2014 Main)

Y = { 9 ( n − 1) : n ∈ N } , where N is the set of natural numbers, then X ∪ Y is equal to (2014 Main) (b) Y − X

(c) X

(d) Y

57. The variance of first 50 even natural numbers is (2014 Main)

(a)

833 4

(b) 833

(c) 437

(d)

437 4

58. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given? (2013 Main) (a) Mean (c) Mode

(b) Median (d) Variance

59. If A and B two sets containing 2 elements and 4 elements, respectively. Then, the number of subsets of A × B having 3 or more elements, is (2013 Main) (a) 256

(b) 220

(c) 219

(d) 211

60. The number log2 7 is (a) an integer (c) an irrational number

(b) a rational number (d) a prime number

61. If log0.3 ( x – 1) < log0. 09( x – 1), then x lies in the interval (a) (2, ∞ ) (c) (–2, – 1)

(b) (1, 2) (d) None of these

(1985, 2M)

62. Consider any set of 201 observations x1 , x2 ,... , x200 ,

x201.It is given that x1 < x2 < K < x200 < x201.Then, the mean deviation of this set of observations about a point k is minimum, when k equals (1981, 2M)

(a) (x1 + x2 + K + x200 + x201 ) / 201 (b) x1 (c) x101 (d) x201

(1980. 2M)

(b) 2 (d) None of these

Objective Questions II (One or more than one correct option) 64. Let f : R → ( 0, 1) be a continuous function. Then, which of the following function(s) has (have) the value zero at some point in the interval (0, 1) ? (a) e − x

∫

x

f (t ) sin t dt

0 π −x 2 f 0

(c) x − ∫

(t ) cos t dt

(b) f (x) +

π 2 f (t ) 0

∫

sin t dt

(d) x9 − f (x)

65. If 3x = 4x −1 , then x is equal to 2 log3 2 (a) 2 log3 2 − 1 1 (c) 1 − log 4 3

(2017 Adv.)

(2013 Adv.)

2 (b) 2 − log 2 3 2 log 2 3 (d) 2 log 2 3 − 1

Integer & Numerical Answer Type Questions

56. If X = { 4n − 3n − 1 : n ∈ N } and

(a) N

2 log10 x – logx ( 0.01), for x > 1, is

(a) 10 (c) – 0.01

(d) 510

53. If the angles of elevation of the top of a tower from

(a) 3 :1 (c) 1: 3

63. The least value of the expression

66. Let|X|denote the number of elements in a set X. Let

S = { 1, 2, 3, 4, 5, 6} be a sample space, where each element is equally likely to occur. If A and B are independent events associated with S, then the number of ordered pairs ( A, B) such that 1 ≤|B| 2 ) be the vertices of a regular polygon of n sides with its centre at the origin. Let ak be the position vector of the point A k, k = 1,2,..., n. If

n −1

∑

k=1

(2015)

( a k ⋅ a k +1 ) =

n −1

∑

k=1

(i)

1

(ii)

2

(iii)

8

(iv)

9

( a k ⋅ a k +1 ) , then the minimum

value of n is R.

If the normal from the point P( h,1) on the ellipse

x 2 y2 + = 1 is perpendicular to the line x + y = 8, then 6 3

the value of h is S.

1  1  −1  −1 2 Number of positive solutions satisfying the equation tan−1   + tan   = tan (2 / x ) is  2 x + 1  4x + 1

612 Miscellaneous Codes P (a) (iv) (c) (iv)

Q (iii) (iii)

R (ii) (i)

S (i) (ii)

P (b) (ii) (d) (ii)

Q (iv) (iv)

R (iii) (i)

S (i) (iii)

79. Match the statements given in Column I with the intervals/union of intervals given in Column II. Column I

Column II

A.

 2 iz   The set Re   : z is a complex number,| z| = 1, z ≠ ± 1} 2 1 − z  

p.

( −∞, − 1) ∪ (1, ∞ )

B.

 8 ( 3)x − 2  The domain of the function f( x ) = sin−1   is 2 ( x − 1) 1 − 3 

q.

( −∞, 0) ∪ ( 0, ∞ )

C.

 1 If f(θ) = − tanθ  −1

r.

[2 , ∞ )

D.

If f( x ) = x 3/ 2( 3x − 10), x ≥ 0, then f( x ) is increasing in

s.

( −∞, − 1] ∪ [1, ∞ )

t.

( −∞, 0] ∪ [2 , ∞ )

tanθ 1 − tanθ

1  π tanθ, then the set { f(θ): 0 ≤ θ < } is 2 1 

80. Match the statements given in Column I with the values given in Column II. Column I A.

→

If a = $j +

→ 3 k$ , b = − $j +

(2011)

Column II

→ 3 k$ and c = 2 3 k$ form a triangle, then the →

(2011)

p.

π 6

q.

2π 3

r.

π 3

s.

π

t.

π 2

→

internal angle of the triangle between a and b is B.

b π If ∫ [f( x ) − 3x ]dx = a 2 − b 2, then the value of f   is a  6

C.

The value of

D.

 for| z| = 1, z ≠ 1 is given by  arg  1   The maximum value of 1− z  

π2 log 3

5/ 6

∫ 7/ 6 sec ( πx )dx is

81. Match the statements of Column I with values of Column II.

(2010)

Column I

Column II

A.

x −2 y−1 z+ 1 and A line from the origin meets the lines = = 1 1 –2 x − 8/ 3 y+ 3 z−1 at P and Q respectively. If length PQ = d , then d 2 is = = −1 2 1

B.

The value of x satisfying tan− 1( x + 3) − tan− 1( x − 3) = sin− 1( 3 / 5) is

C.

Non-zero vectors a , b and c satisfy a ⋅ b = 0,(b − a ) ⋅ (b + c ) = 0 and 2|b + c| = |b − a|.

→ →

→

→

→

→ →

→

→

→

→

→

→

→

→

p.

–4

q.

0

r.

4

s.

5

t.

6

→

If a = µ b + 4 c , then the possible value of µ is D.

9x  x Let f be the function on [− π, π ] given by f ( 0) = 9 and f ( x ) = sin   sin   for x ≠ 0.  2  2 The value of

2 π

π

∫ f (x )dx is

−π

Miscellaneous 613 82. Match the statements/expressions given in Column I with the values given in Column II.

(2009)

Column I A. B.

Column II

Root(s) of the equation 2 sin2 θ + sin2 2 θ = 2

p.

π /6

6x 3x Points of discontinuity of the function f ( x ) =   cos   , where [y] denotes the largest integer less  π   π 

q.

π /4

r.

π /3

s.

π /2

t.

π

than or equal to y C.

Volume of the parallelopiped with its edges represented by the vectors $i + $j , $i + 2 $j and $i + $j + π k$

D.

Angle between vectors a and b, where a , b and c are unit vectors satisfying a + b +

→

→

→ →

→

→

→

→

→

3c = 0

83. Match the statements/expressions given in Column I with the values given in Column II. Column I

Column II

π − cos x = 0 in the interval  0,  is  2

(A)

The number of solutions of the equation xesin x

(B) (C)

Value(s) of k for which the planes kx + 4 y + z = 0, 4x + k y + 2 z = 0 and 2 x + 2 y + z = 0 intersect in a straight line is Value(s) of k for which| x − 1| + | x − 2| + | x + 1| + | x + 2| = 4k has integer solution(s) is

(D)

If y ′ = y + 1 and y( 0) = 1, then value(s) of y(ln 2 ) is

(p)

1

(q)

2

(r)

3

(s)

4

(t)

5 (2009)

84. Match the conics in Column I with the statements/expressions in Column II. Column I

Column II

A.

Circle

p.

The locus of the point (h, k) for which the line hx + ky = 1 touches the circle x 2 + y 2 = 4

B.

Parabola

q.

Points z is the complex plane satisfying| z + 2| − | z − 2| = ± 3

C.

Ellipse

r.

Points of the conic have parametric representation x =

D.

Hyperbola

 1− t 2  2t 3 , y = 1+ t 2 1+ t 2

s.

The eccentricity of the conic lies in the interval 1 ≤ x < ∞

t.

Points in the complex plane satisfying Re ( z + 1)2 = | z|2 + 1 (2009)

85. Match the statements/expressions in Column I with the open intervals in Column II. Column I

Column II

A.

Interval contained in the domain of definition of non-zero solutions of the differential equation ( x − 3)2 y ′ + y = 0

B.

Interval containing the value of the integral

C.

Interval in which atleast one of the points of local maximum ofcos x + sin x lies

D.

− π, π    2 2  0, π     2  π , 5π    8 4   0, π     8

p.

5

∫1 (x − 1)(x − 2 )(x − 3)(x − 4)(x − 5)dx

q.

2

Interval in which tan

−1

r.

(sin x + cos x ) is increasing

s.

(− π , π )

t.

(2009)

86. Match the statements/expressions given in Column I with the values given in Column II. Column I x + 2x + 4 is x +2 2

A.

The minimum value of

B.

Let A and B be 3 × 3 matrices of real numbers, where A is symmetric, B is skew-symmetric and ( A + B) ( A − B) = ( A − B)( A + B). If ( AB)t = ( −1)k AB, where ( AB)t is the transpose of the matrix AB, then the possible values of k are

Column II p.

0

q.

1

614 Miscellaneous Column I C.

Let a = log 3 log 3 2. An integer k satisfying 1 < 2

D.

If sin θ = cos φ, then the possible values of

1 π

( − k + 3− a )

Column II

< 2 , must be less than

 θ ± φ − π  are    2

r.

2

s.

3 (2008 6M)

87. Match the statements/expressions given in Column I with the values given in Column II. Column I ∞

A.

∑ tan

i =1

−1 

1   2  = t , then tan t is  2i 

Sides a, b, c of a ∆ABC are in AP and cos θ1 =

B.

Column II

a b c , , cos θ 2 = , cos θ 3 = b+c a+c a+ b

p.

2 3

q.

1

θ θ then tan2  1  + tan2  3  is 2 2 C.

A line is perpendicular to x + 2 y + 2 z = 0 and passes through (0, 1, 0). The perpendicular distance of this line from the origin is

5 3

r.

(2006 6M)

88. Match the statements/expressions given in Column I with the values given in Column II. Column II

Column I A.

Two rays in I quadrant x + y = | a | and ax − y = 1 intersect each other in the interval 2a a ∈( a0, ∞ ), then value of  0  is  3 

p.

2

B.

→ → Point (α, β, γ ) lies on the plane x + y + z = 2. If a = α $i + β$j + γ k$ and k$ × (k$ × a$ ) = 0 ,

q.

2 3

then γ is C.

 1 (1 − y 2 ) dy  +  0 ( y 2 − 1) dy ∫ 0  ∫ 1 

r.

D.

If sin A sin B sin C + cos A cos B = 1, then the value of sin C is

s.

 1 1 − x dx  +  0 1 + x dx ∫ 0  ∫ −1  1 (2006, 6M)

Fill in the Blanks 1 x 1 x = and ( x + y ) = − , then 2 y 2 y (1990, 2M) x = ……… and y = ……… .

89. If x > 0, y < 0, x + y +

90. The

solution of log7 log5 ( x + 5 + x ) = 0 is … .

the

equation

The mean square deviations about –1 and +1 of a set of observations are 7 and 3, respectively. Find the standard deviation of this set of observations. (1981, 2M)

Integer & Numerical Answer Type Questions 93. The total number of distincts x ∈ [0, 1] for which x

t2

∫0 1 + t4 dt = 2x − 1 is

Analytical & Descriptive Questions 91. The marks obtained by 40 students are grouped in a frequency table in class intervals of 10 marks each. The mean and the variance obtained from this distribution are found to be 40 and 49, respectively. It was later discovered that two observations belonging to the class interval ( 21 - 30) were included in the class interval ( 31 - 40) by mistake. Find the mean and the variance after correcting the error. (1982, 3M)

92. The mean square deviations of a set of observations x1 , x2 , ... , xn about a point c is defined to be 1 n ∑ ( xi − c)2. n i =1

94. Let

F (x) =

x2 +

∫x

π 6

2 cos2t dt

(2016 Adv.)

for

all

x ∈R

and

 1 f : 0,  → [0, ∞ ) be a continuous function. For  2  1 a ∈ 0, , if F ′ ( a ) + 2 is the area of the region  2 bounded by x = 0 , y = 0 , y = f ( x ) and x = a, then f( 0) is (2015 Adv.)

95. The value of

  1 1 1 1 …  is 4− 4− 4− 6 + log3/ 2  3 2 3 2  3 2 3 2  (2012)

Miscellaneous 615

Answers 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 41. 45. 49. 53. 57. 61. 65. 69. 73.

(a) (a) (b) (a) (b) (a) (c) (a) (c) (a) (c) (d) (c) (a) (b) (a) (c) 119 (a)

2. 6. 10. 14. 18. 22. 26. 30. 34. 38. 42. 46. 50. 54. 58. 62. 66. 70. 74.

(a) (b) (a) (b) (d) (a) (d) (d) (b) (c) (b) (a) (d) (a) (d) (c) 1523 (b) (b)

3. 7. 11. 15. 19. 23. 27. 31. 35. 39. 43. 47. 51. 55. 59. 63. 67. 71. 75.

4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 61. 64. 68. 72.

(a) (a) (b) (a) (d) (b) (d) (b) (a) (d) (c) (c) (d) (d) (c) (d) (8) (b) (a)

76. A → p, q, s; B → p, t; C → p, q, r, t; D → s

(b) (a) (d) (b) (c) (a) (d) (b) (c) (a) (a) (b) (a) (d) (c) (c, d) (3748) (c)

77. A → p, r,s; B → p; C → p, q; D → s, t 78. (a) 79. A → s, B → t, C → r, D → r 80. A → q; B → p; C → s; D → s 81. A → t, B → p and r, C → q, D → r 82. A → q, s; B → p, r, s t; C → t; D → r 83. A → p, B → q, s; C → q, r, s, t; D → r 84. A → p, q, s; B → p, t; C → r, D → q, s 85. A → p, q, s , B → p, t; C → q, r, s, t; D → s 86. A → r, B → q, s; C → r, s; D → p, r 87. A → q, B → p, C → r 88. A → q, B → p, C → r, D → s 89. – 1 and 2

90. (4)

91. 49.25

93. (1)

94. (3)

95. (4)

92.

3

Hints & Solutions 1. Let n ( A ) = Number of people read newspaper A = 63%

Case-II If|t − 1| + |t − 2| = 1, for 1 < t ≤ 2 then t2 − t + 2 = 1 ⇒ t 2 − t + 1 = 0 have no real solution. Case-III If|t − 1| + |t − 2| = 2t − 3, for t > 2 then t 2 − t + 2 = 2t − 3 ⇒ t 2 − 3t + 5 = 0 have no real solution. Since, S be the set of all real roots of the given equation, then S is a singleton set. Hence, option (a) is correct.

n (B) = Number of people read newspaper B = 76% and n ( A ∩ B) = Number of people read both = x% Q n ( A ∪ B) = n ( A ) + n (B) − n ( A ∩ B) and 76 ≤ n ( A ∪ B) ≤ 100 ⇒ 76 ≤ 63 + 76 − x ≤ 100 ⇒ 0 ≤ 63 − x ≤ 24 ⇒ 39 ≤ x ≤ 63 Hence, option (a) is correct.

2. Given equation 3x (3x − 1) + 2 = |3x − 1| + |3x − 2| Let 3x = t > 0, ∀ x ∈ R, so equation is t 2 − t + 2 = |t − 1| + |t − 2| 1 − t + 2 − t , 0 < t ≤ 1 Q |t − 1| + |t − 2|= t − 1 + 2 − t , 1 < t ≤ 2  t >2 t − 1 + t − 2 , 3 − 2t , 0 < t ≤ 1 , 1 < t ≤2 = 1  t >2 2t − 3, Now, Case-I If|t − 1| + |t − 2| = 3 − 2t, for 0 < t ≤ 1 then t 2 − t + 2 = 3 − 2t ⇒ t 2 + t − 1 = 0 −1 ± 5 ± 5 − 1 ⇒ t= = 2 2 5 −1 ∈ (0, 1] Q 2 5 −1 is one of the solution. ∴ t= 2

…(i)

3. From the truth table P ∨ Q P ∧ Q P → Q P ∧ (P → Q ) P ∧ (P ∨ Q ) P ∨ (P ∧ Q )

P

Q

T

T

T

T

T

T

T

T

T

F

T

F

F

F

T

T

F

T

T

F

T

F

F

F

F

F

F

F

T

F

F

F

P ∧ (P → Q ) → Q

Q → ( P ∧ ( P → Q ))

T

T

T

T

T

F

T

T

∴ P ∧ (P → Q ) → Q is a tautology. Hence, option (a) is correct.

616 Miscellaneous 4. Given proposition p ⇒ ~ ( p ∧ (~ q)) = p ⇒ (~ p ∨ q) = (~ p) ∨ ((~ p) ∨ q) = (~ p) ∨ q Hence, option (b) is correct.

[by De Morgan Law]

6.

9. It is given that both mean and standard deviation of 50 [Q p ⇒ q ≡ ~ p ∨ q]

observations x1, x2, x3 , K , x50 are equal to 16, Σx So, mean = i = 16 50

Σx i2  Σx i  −   n  n

2

2

Σxi2 − (16)2 = (16)2 50

⇒

Σxi2 = 2 × 256 = 512 50 Now, mean of (x1 − 4)2, (x2 − 4)2, K , (x50 − 4)2

⇒

⇒

x1 + x2 + x3 + x4 = 44 x5 + x6 + x7 + x8 + x9 + x10 and = 16 6 ⇒ x5 + x6 + x7 + x8 + x9 + x10 = 96

… (i)

… (ii) 44 + 96 140 So, mean of given 10 observations = = = 14 10 10 Since, the sum of squares of all the observations = 2000 2 x12 + x22 + x32 + K + x10 = 2000

Σx2  Σx  Now, σ = (standard deviation ) = i −  i  10  10  2000 = − (14)2 = 200 − 196 = 4 10 2

…(i)

Σxi2  Σxi  −  = 16 50  50 

and standard deviation =

Given 10 observations are x1 , x2, x3 , K , x10 x1 + x2 + x3 + x4 ∴ = 11 4

… (iii) 2

2

So, σ = 2

7.

[Q p∧ ~ p ≡ F]

Key Idea Formula of standard deviation ( σ ), for n observations =

[Q ~ ( p ∧ q) = ~ p∨ ~ q] [Q ~ ( p ∨ q) = ~ p∧ ~ q] [Q p ∧ (q ∨ r ) ≡ ( p ∧ q) ∨ ( p ∧ r )]

= (s ∧ r )

5. Given boolean expression is ~ ( p ⇒ (~ q)) ≡ ~ ((~ p) ∨ (~ q)) ≡ p∧q

= s ∧ (~ ((~ r ) ∧ s)) = s ∧ (r ∨ (~ s)) = (s ∧ r ) ∨ (s ∧ (~ s))

Key Idea Use formula : p → q = ~p ∨ q

Given statement is p → (~ q ∨ r ) = ~ p ∨ (~ q ∨ r )

=

Σ (xi − 4)2 Σ (xi2 − 8xi + 16) = 50 50

=

Σxi2  Σx  16 Σ1 − 8 i  +  50  50 50

  16 = 512 − (8 × 16) +  × 50   50

…(ii)

[from Eqs. (i) and (ii)]

= 512 − 128 + 16 = 400

10. Option (a) ( p ∨ q) ∨ ( p ∨ (~ q)) ≡ p ∨ (q∨ ~ q) is tautology, [Q q ∨ (~ q) ≡ T and p ∨ T ≡ T] Option (b) ( p ∧ q) ∨ ( p ∧ (~ q)) ≡ p ∧ (q ∨ ~ q) not a tautology, [Q q ∨ ~ q ≡ T and p ∧ T ≡ p] Option (c) ( p ∨ q) ∧ ( p ∨ (~ q)) ≡ p ∨ (q ∧ ~ q) not a tautology [Q q ∧ ~ q ≡ F and p ∨ F ≡ p] Option (d) ( p ∨ q) ∧ ((~ p) ∨ (~ q)) ≡ ( p ∨ q) ∧ (~ ( p ∧ q)) not a tautology.

Now, from the options (a) When p = T, q = T and r = F then ~ p ∨ (~ q ∨ r ) = F ∨ (F ∨ F) = F (b) When p = T, q = F and r = F then ~ p ∨ (~ q ∨ r ) = F ∨ (T ∨ F) = T (c) When p = T, q = F and r = T then ~ p ∨ (~ q ∨ r ) = F ∨ (T ∨ T) = T (d) When p = F, q = T and r = T then ~ p ∨ (~ q ∨ r ) = T ∨ (F ∨ T) = T 8.

Key Idea Use De-morgan’s law, Distributive law and Identity law as p ∨ F ≡ p

The given boolean expression is ~ s ∨ ((~ r ) ∧ s) Now, the negation of given boolean expression is ~ (~ s ∨ ((~ r ) ∧ s))

11. Key Idea Use n ∑ fi

= Number of students;

i =1

and Mean ( x ) =

Σ fi x i Σ fi

The given frequency distribution, for some x ∈ R, of marks obtained by 20 students is Marks

2

Frequency

(x + 1)

2

3

5

7

2x − 5

x − 3x

x

2

Q Number of students = 20 = Σ fi ⇒ (x + 1)2 + (2x − 5) + (x2 − 3x) + x = 20 ⇒ (x2 + 2x + 1) + (2x − 5) + (x2 − 3x) + x = 20 ⇒ 2x2 + 2x − 24 = 0 ⇒ x2 + x − 12 = 0 [as x > 0] ⇒ (x + 4)(x − 3) = 0 ⇒ x = 3

Miscellaneous 617 Now, mean (x ) =

Σ fi xi Σ fi

15. Q p ∨ ( (~ p) ∧ q)

2(x + 1)2 + 3(2x − 5) + 5(x2 − 3x) + 7x = 20 2(4)2 + 3(1) + 5(0) + 7(3) 32 + 3 + 21 56 = = = = 2 .8 20 20 20

16. Given observations are −1, 0, 1 and k.

Hence, option (b) is correct.

Also, standard deviation of these four observations = 5

12. Let the population of city is 100. Then, n ( A ) = 25, n (B) = 20 and n ( A ∩ B) = 8 A

17

B

8

12

[by Distributive law] = ( p ∨ (~ p)) ∧ ( p ∨ q ) [Q p ∨ (~ p) is tautology] = p∨ q So negation of p ∨ ((~ p) ∧ q) = ~ [ p ∨ (~ p) ∧ q] = ~ ( p ∨ q) [by Demorgan’s law] = (~ p) ∧ (~ q)

U

n(U)=100

Venn diagram

So, n ( A ∩ B ) = 17 and n ( A ∩ B) = 12 According to the question, Percentage of the population who look into advertisement is  30   40  × n( A ∩ B ) + × n ( A ∩ B) 100  100   50  + × n ( A ∩ B) 100   30   40   50  = × 17 +  × 12 +  × 8  100   100   100 

=

= 5.1 + 4.8 + 4 = 13.9

13. Given ten numbers are 10, 22, 26, 29, 34, x, 42, 67, 70, y and their mean = 42 10 + 22 + 26 + 29 + 34 + x + 42 + 67 + 70 + y = 42 ∴ 10 300 + x + y ⇒ = 42 10 …(i) ⇒ x + y = 120 and their median (arranged numbers are in increasing order) = 35 34 + x = 35 ⇒ 2 ⇒ 34 + x = 70 ⇒ x = 36 On substituting x = 36 in Eq. (i), we get 36 + y = 120 ⇒ y = 84 y 84 7 ∴ = = x 36 3

14. Given statement p ⇒ (q ∨ r ) is false. Q p → (q ∨ r ) = (~ p) ∨ (q ∨ r ) Now, by trial and error method, if truth value of p is T, q is F and r is F, then truth value of (q ∨ r ) is F. So, truth value of [(~ p) ∨ (q ∨ r )] is false. Thus, if truth value of p, q, r are T, F, F, then the statement p → (q ∨ r ) is false.

2

(−1)2 + (0)2 + (1)2 + k2  −1 + 0 + 1 + k −  = 5   4 4 [Q if x1 , x2.... xn are n observation, then standard 2 1 n 2 1 n  deviation = Σ xi −  Σ xi   n i=1   n i=1  2 + k2 k2 [squaring both sides] ⇒ − =5 4 16 8 + 4k2 − k2 8 + 3k2 =5 ⇒ =5 ⇒ 16 16 ⇒ 8 + 3k2 = 80 ⇒ 3k2 = 72 ⇒ k2 = 24 ⇒ k = 2 6 or −2 6 [Q k > 0] k =2 6 ⇒

∴

17. Let the marks in sixth tests is ‘x’, so

41 + 45 + 43 + 54 + 57 + x = 48 (given) 6 x 240 + x ⇒ = 48 ⇒ 40 + = 48 6 6 x ⇒ = 8 ⇒ x = 48 6 Now, standard deviation of these marks mean =

=

412 + 452 + 432 + 542 + 572 + 482 − 482 6 [Q standard deviation (SD) =

=

Σxi2 − ( x )2 ] 6

(412 − 482) + (452 − 482) + (432 − 482) + (542 − 482) + (572 − 482) 6

=

(−7 × 89) + (−3 × 93) + (−5 × 91) + (6 × 102) + (9 × 105) 6

=

945 + 612 − 455 − 279 − 623 6

=

1557 − 1357 200 100 10 = = = 6 6 3 3

18. ( p ∧ q) → (~ p) ∨ q ≡ (~ ( p ∧ q)) ∨ ((~ p ) ∨ q) ≡ ((~ p ) ∨ (~ q)) ∨ ((~ p ) ∨ q) ≡ (~ p ) ∨ (~ q) ∨ q [Q (~ p ) ∨ (~ p ) ≡ ~ p ] ≡ (~ p ) ∨ T [Q ~ q ∨ q ≡ T] ≡T So, it is a tautology [Q ((~ q) ∨ q) is tautology]

618 Miscellaneous (b) ( p ∧ q) → p ≡ ( p ∧ q) ∨ p ≡ ((~ p ) ∨ (~ q)) ∨ p [Q~ ( p ∧ q) ≡ (~ p ) ∨ (~ q)] ≡ (~ p ∨ p ) ∨ (~ q) is tautology. [Q~ p ∨ p is a tautology and (~ q) ∨ T ≡ T] (c) Q p → ( p ∨ q) ≡ (~ p ) ∨ ( p ∨ q) [Q p → q is equivalent to (~ p ∨ q)] ≡ (~ p ∨ p ) ∨ q is tautology. [Q (~ p ∨ p ) is tautology and q ∨ T ≡ T] (d) ( p ∨ q) → ( p ∨ (~ q)) ≡ (~ ( p ∨ q)) ∨ ( p ∨ (~ q)) ≡ ((~ p ) ∧ (~ q)) ∨ ( p ∨ (~ q)) ≡ ( p ∨ (~ q) ∨ ((~ p ) ∧ (~ q)) ≡ ( p ∨ (~ q) ∨ (~ p )) ∧ ( p ∨ (~ q) ∨ (~ q)) ≡ (T ∨ (~ q)) ∧ ( p ∧ (~ q)) ≡ T ∧ ( p ∧ (~ q))

≡ p ∧ (~ q), which is not a tautology.

19. The natural numbers between 100 and 200 are 101, 102, 103, …, 199. Since, 91 = 13 × 7, so the natural numbers between 100 and 200 whose HCF with 91 is more than 1 are the numbers which are either divisible by 7 or 13. So, the required sum of numbers between 100 and 200 = (sum of numbers divisible by 7) + (sum of numbers divisible by 13) − (sum of numbers divisible by 91) =

14

8

r =1

r =1

∑ (98 + 7r ) + ∑ (91 + 13r ) − (182)

 14 × 15  8 × 9 = (98 × 14) + 7  + (91 × 8) + 13   − (182)  2   2  = 1372 + 735 + 728 + 468 − 182 = 3303 − 182 = 3121

20. Let the remaining two observations are a and b. According to the question, 2 + 4 + 10 + 12 + 14 + a + b Mean = =8 7 ⇒ 42 + a + b = 56 …(i) ⇒ a + b = 14 and variance a 2 + b2 + 4 + 16 + 100 + 144 + 196 = − 82 = 16 7 a 2 + b2 + 460 ⇒ − 64 = 16 7 a 2 + b2 + 460 ⇒ = 80 7 ⇒ a 2 + b2 + 460 = 560 …(ii) ⇒ a 2 + b2 = 100 We know that, (a + b)2 = (a 2 + b2) + 2ab [from Eqs. (i) and (ii)] ⇒ (14)2 = 100 + 2ab ⇒ 196 = 100 + 2ab ⇒ 2ab = 96 ⇒ ab = 48 So, product of remaining two observations is 48.

21. Given digits are 1, 1, 2, 2, 2, 2, 3, 4, 4. According to the question, odd numbers 1, 1, 3 should occur at even places only.

Even places

∴ The number of ways to arrange odd numbers at even 3! places are 4C3 × 2! and the number of ways to arrange remaining even 6! numbers are . 4 !2 ! So, total number of 9-digit numbers, that can be formed using the given digits are 3! 6! 4 C3 × × = 4 × 3 × 15 = 180 2 ! 4 !2 !

22. Given statement is ‘‘If you are born in India, then you are a citizen of India’’. Now, let statement p : you are born in India and q : you are citizen of India. Then, given statement, ‘‘If you are born in India then you are a citizen of India’’ is equivalent to p ⇒ q. Q The contrapositive of statement p ⇒ q is ~ q ⇒ ~ p. ∴ The contrapositive of the given statement is ‘‘If you are not a citizen of India, then you are not born in India.

23. Given mean x = 4 variance σ 2 = 5.20 and numbers of observation n = 5 Let x1 = 3, x2 = 4, x3 = 4 and x4 , x5 be the five observations 5

So, ⇒ ⇒ ⇒

∑ xi = 5 ⋅ x = 5 × 4 = 20

i =1

x1 + x2 + x3 + x4 + x5 = 20 3 + 4 + 4 + x4 + x5 = 20 x4 + x5 = 9

…(i)

5

Now, variance σ 2 = ⇒ ⇒ ⇒ ⇒ ⇒ Q ∴ ⇒ ⇒

∑ xi2

i =1

− ( x )2 5 x12 + x22 + x32 + x42 + x52 − (4)2 = 5.20 5 9 + 16 + 16 + x42 + x52 = 16 + 5.20 5 41 + x42 + x52 = 5 × 21 .20 x42 + x52 = 106 − 41 x42 + x52 = 65

…(ii)

(x4 + x5 )2 = x42 + x52 + 2x4x5 [from Eqs. (i) and (ii)] 81 = 65 + 2x4x5 16 = 2x4x5 …(iii) x4x5 = 8

Miscellaneous 619 Now, (|x4 − x5|)2 = x42 + x52 − 2x4x5 [from Eqs. (ii) and (iii)] = 65 − 16 = 49 |x4 − x5| = 7

⇒

⇒

i =1

⇒

24. Let C and S represent the set of students who opted for NCC and NSS respectively. Then, n (C ) = 40, n (S ) = 30, n (C ∩ S ) = 20 and n (U ) = 60 Now, n (C ∪ S ) = n (C ∪ S )

50

∑ xi

i =1

50

= 31

Total number of non-empty subsets of ‘ S ’ = 2 100 − 1 Now, numbers of non-empty subsets of ‘ S ’ in which only odd numbers {1, 3, 5, … , 99} occurs = 2 50 − 1 So, the required number of non-empty subsets of ‘ S ’ such that product of elements is even. = (2 100 − 1) − (2 50 − 1) = 2 100 − 2 50 = 2 50(2 50 − 1).

= 60 − [40 + 30 − 20] = 10 10 1 So, required probability = = 60 6 − 5 x + 6)

i =1

29. Given, set S = {1, 2, 3, ... , 100 }.

= 60 − [n (C ) + n (S ) − n (C ∩ S )]

2

50

∑ xi = 50(30 + 1) = 50 × 31

∴ Mean of 50 observations = (x ) =

= n (∪ ) − n (C ∪ S )

25. Given, set A = { x ∈ Z : 2( x + 2)( x

50

∑ xi − (30 × 50) = 50

= 1}

2

( x + 2) ( x −5 x + 6 )

Consider, 2 = 1 = 2º ⇒ (x + 2) (x − 3) (x − 2) = 0

30. Number of white balls = 30

⇒

x = −2, 2, 3

⇒

A = { −2 , 2 , 3 }

Also, we have set B = { x ∈ Z : − 3 < 2x − 1 < 9} Consider, −3 < 2x − 1 < 9, x ∈ Z ⇒ −2 < 2x < 10, x ∈ Z ⇒ −1 < x < 5, x ∈ Z ⇒ B = {0, 1, 2, 3, 4} So, A × B has 15 elements. ∴ Number of subsets of A × B = 215 . [Q If n ( A ) = m, the number of possible subsets = 2m]

26. Since, the expression p→ q ≡ ~ p ∨ q So, ~ p→ q ≡ p ∨ q and therefore ~ (~ p → q) ≡ ~ ( p ∨ q) [by De Morgan’s law] ≡ (~ p) ∧ (~ q)

27. Let the given Boolean expression

and number of red balls = 10 Let p = probability of success in a trial = probability of 30 3 getting a white ball in a trial = = . 40 4 and q = probability of failure in a trial 3 1 =1 − p=1 − = 4 4 Here, n = number of trials = 16. Clearly, X follows binomial distribution with 3 parameter n = 16 and p = . 4 3 ∴ Mean of X = np, = 16. = 12 4 3 1 and variance of X = npq = 16. . = 3 4 4 mean of X 12 Now, = =4 3 standard deviation of X 3 [Q SD = variance]

31. We know that, contrapositive of p → q is ~ q → ~ p

(( p ∧ q) ∨ ( p∨ ~ q)) ∧ (~ p∧ ~ q) ≡ r Now, let us construct the following truth table p

q

~p

~q

p∧ q

p∨ ~ q

~ p∧ ~ q

( p ∧ q) ∨ ( p∨ ~ q)

r

Therefore, the contrapositive of the given statement is ‘‘If the squares of two numbers are equal, then the numbers are equal’’.

T

T

F

F

T

T

F

T

F

32. Given, ( p ∧ q) ↔ r is true. This is possible under two

T

F

F

T

F

T

F

T

F

F

T

T

F

F

F

F

F

F

F

F

T

T

F

T

T

T

T

r ≡ ~ p∧ ~ q

Clearly,

28. Let the 50 observations are x1 , x2, x3 , ... , x50. Now, deviations of these observations from 30 are (x1 − 30), (x2 − 30), (x3 − 30), ... , (x50 − 30) 50

Q

∑ (xi − 30) = 50

i =1

(given)

cases Case I When both p ∧ q and r are true, which is not possible because q is false. Case II When both ( p ∧ q) and r are false. ⇒ p ≡ T or F; q ≡ F, r ≡ F In this case, (a) p ∨ r is T or F (b) ( p ∧ r ) → ( p ∨ r ) is F→ (T or F) , which always result in T. (c) ( p ∨ r ) → ( p ∧ r ) is (T or F) → F, which may be T or F. (d) p ∧ r is F.

620 Miscellaneous Putting y = xt in Eq. (i), we get

33. We know, Σx − µ2 n 2 2 1  1  1 10 – d + 10 × + 10 + d  2  2 4 = 30

σ2 =

2

x(1 + t ) = 13

2

1  1 1    10 − d  + 10 × + 10  + d   2  2  2 −   30     Qµ = Σxi   n  

=

∴

1 20  + d 2 + 5 / 2 4  30

 1 −   4

⇒

97(t 2 + 2t + 1) = 169 (1 + t 2)

⇒

(169 − 97)t 2 − 194t + (169 − 97) = 0

⇒

36t 2 − 97t + 36 = 0

⇒

(4t − 9) (9t − 4) = 0 9 t= 4 4 t= 9

36.

P

Q

R

~P

~Q

~R

P ∧Q

T

F

T

F

T

F

F

A

Q ∧ R ~Q ∧ R F

F

T

F

Clearly, the truth value of P ∨ (~ Q ∧ R) is T.

35. Let 1, 3, 8, x and y be the five observations. Σx i n 1+3+8+ x+ y x= =5 ⇒ 5 ⇒ x + y = 25 − 12 = 13 ⇒ x + y = 13 Σ(xi. − x )2 and variance = σ 2 = n

Then, mean x =

(given)

…(i)

Let A be the set of even numbered students then 140  n( A) = = 70 ([.] denotes greatest integer function)  2  Let B be the set of those students whose number is divisible by 3, 140  then n (B) = = 46  3  ([.] denotes greatest integer function) Let C be the set of those students whose number is divisible by 5, 140  then n (C ) = = 28  5  Now,

([.] denotes greatest integer function) 140  n ( A ∩ B) = = 23  6  (numbers divisible by both 2 and 3)

(1 − 5)2 + (3 − 5)2 + (8 − 5)2   + (x − 5)2 + ( y − 5)2 = 9.2 (given) = 5 ⇒ 16 + 4 + 9 + (x2 − 10x + 25) + ( y2 − 10 y + 25) = 46 2 ⇒ x + y2 − 10(x + y) = 46 − 79 [Qx + y = 13] ⇒ x2 + y2 − 10 × 13 = − 33 x2 + y2 = 97 y = t ⇒ y = xt x

B

C

(P ∧ Q ) ∨ (~ R) P ∨ (~ Q ∧ R) (~ P ) ∨ (Q ∧ R) (~ P ) ∧ (~ Q ∧ R)

Let

… (iv)

or

P : 5 is a prime number, is true statement. Q : 7 is a factor of 192, is false statement and R : LCM of 5 and 7 is 35, is true statement. So, truth value of P is T, Q is F, R is T Now let us check all the options.

⇒

x2(1 + t 2) = 97 Dividing Eq. (iii) by Eq. (iv), we get

⇒

34. Since, the statements

T

…(iii)

x2(1 + t )2 169 = x2(1 + t 2) 97

15 + 20 d 2 1 1 2d 2 1 2 2 2 = − = + − = d 30 4 4 3 4 3 2 2 4 2 d = ⇒ d = 2 ⇒|d| = 2 3 3

F

⇒ x (1 + t )2 = 169 Putting y = xt in Eq. (ii), we get 2

…(ii)

n (B ∩ C ) =

140  =9  15  (numbers divisible by both 3 and 5)

140  n (C ∩ A ) = = 14  10  (numbers divisible by both 2 and 5) n (A ∩ B ∩ C ) =

140  =4  30  (numbers divisible by 2, 3 and 5)

Miscellaneous 621 and n ( A ∪ B ∪ C ) = Σn ( A ) − Σn ( A ∩ B ) + n ( A ∩ B ∩ C ) = (70 + 46 + 28 ) − (23 + 9 + 14) + 4 = 102 ∴ Number of students who did not opt any of the three courses = Total students − n ( A ∪ B ∪ C ) = 140 − 102 = 38

37. Given statement is “P (n ) : n 2 − n + 41 is prime”. Clearly P(3) : 32 − 3 + 41 = 9 − 3 + 41 = 47 which is a prime number. and P(5) : 52 − 5 + 41 = 25 − 5 + 41 = 61, which is also a prime number. ∴ Both P(3) and P(5) are true.

( p ∧ q)

≡ [( p ∧ ~ q) ∨ ( p ∧ r )] ∧ (~ q ∧ r ) (Q ~ (~ p ∨ q) ≡ ~ (~ p) ∧ ~ q ≡ p ∧ ~ q by De Morgan’s law) (distributive law) ≡ [p ∧ (~ q ∨ r )] ∧ (~ q ∧ r )] (associative law) ≡ p ∧ [(~ q ∨ r ) ∧ (~ q ∧ r )] (commutative law) ≡ p ∧ [(~ q ∧ r ) ∧ (~ q ∨ r )] ≡ p ∧ [{(~ q ∧ r ) ∧ (~ q )} ∧ {(~ q ∧ r ) ∧ r ] (distributive law) (idempotent law) ≡ p ∧ [(~ q ∧ r ) ∨ (~ q ∧ r )] (idempotent law) ≡ p ∧ [~ q ∧ r ] (associative law) ≡ p ∧ ~ q ∧ r ≡ ( p ∧ r ) ∧ (~ q)

( p ∧ q)

and

[take ∨ ≈ ∪ and ∧ ≈ ∩] (B) Consider, ⊕ = ∧ and ⋅ = ∨. In that case, we get ( p ∧ q) ∧ (∼ p ∧ q)

≡ null set

(C) Consider, ⊕ = ∨ and ⋅ = ∧ . In that case, we get ( p ∨ q) ∧ (∼ p ∧ q)

≡

(∼ p ∧ q)

and

(D) Consider, ⊕ = ∨ and u = ∨. In that case, we get ( p ∨ q) ∧ (∼ p ∨ q)

≡

q

and

we have

n

∑ (xi + 1)

= 9n

2

5

...(i)

i =1

Mean, x =

n

∑ (xi − 1)

and

2

= 5n

Σ xi

i =1

5

On subtracting Eq. (ii) from Eq. (i) is, we get i =1 n

variance =

and

∑ xi = n

i =1

i =1

i =1

⇒

i =1

n

i =1

n

− ( x )2

5

5

i =1

6

Now, new mean =

Σ xi

i =1

6

5

=

= 1 ⇒ mean (x ) = 1 n

Now, standard deviation =

∑ (xi − x )2 n

∑ (xi − 1)2

i =1

n

40. Let us check all the options (A) Consider, ⊕ = ∧ and ⋅ = ∨. In that case, we get

Σ xi + 156

i =1

6

⇒ xnew = 151 and new variance

i =1

n

=

Σ xi2

− (150)2 = 18 ⇒ Σ xi2 = 112590 …(ii)

n

∑ xi

…(i)

5

Σ xi2

n

∑ 4xi = 4n ⇒

i =1

5

n

∑ {(xi + 1)2 − (xi − 1)2} = 4n

5

= 150 ⇒ Σ xi = 750

...(ii)

i =1

⇒

≡

41. Let x1 , x2, x3 , x4 , x5 be the heights of five students. Then,

39. We have,

⇒

(∼ p ∨ q)

and

38. Clearly, [~ (~ p ∨ q) ∨ ( p ∧ r )] ∧ (~ q ∧ r )

⇒

∧

=

750 + 156 [using Eq. (i)] 6

6

Σ xi2

=

i =1

=

i =1

6

− (xnew )2

5

=

5n = 5 n

Σ xi2 + (156)2 6

− (151)2

112590 + (156)2 − (151)2 6 = 22821 − 22801 = 20

=

[using Eq. (ii)]

622 Miscellaneous = − a (a − 1) a =1 ∆ = ∆1 = ∆ 2 = ∆3 = 0 ∆ for b = 1 only x + y + z = 1, x+ y+ z =1 and x+ y+ z =0 i.e. no solution Hence, for no solution b = 1 only

42. We have, |a − 5| < 1 and|b − 5| < 1 ∴ −1 < a − 5 < 1 and −1 < b − 5 < 1 ⇒ 4 < a < 6 and 4 < b < 6 Now, 4(a − 6)2 + 9(b − 5)2 ≤ 36 (a − 6)2 (b − 5)2 ⇒ + ≤1 9 4 Taking axes as a-axis and b-axis (6, 7)

b P

p

The set A represents square PQRS inside set B representing ellipse and hence A ⊂ B. Key idea Standard deviation is remain unchanged, if observations are added or subtracted by a fixed number

We have, 9

9

∑ (x1 − 5) = 9 and ∑ (x1 − 5)2 = 45 i =1

i =1

9

∑ (x1 − 5)2

i =1

SD =

9

⇒

44.

SD =

  9  ∑ (x1 − 5)  i =1 −  9      

45  9 −  9  9

q

(9, 5)

a

43.

2

2

⇒ SD = 5 − 1 = 4 = 2

x ≡p→ q

~p ~p → q

[By distributive law] [~ q ∨ q = t]

45. Q ∆ = 1 a 1 = 1 (a − b) − 1 (1 − a ) + 1 (b − a 2) a

b 1

= − (a − 1) 1 1 1

2

∆1 = 1 a 1 = 1 (a − b) − 1 (1) + 1 (b) = − (a − 1) 0 b 1 1 1 1 ∆ 2 = 1 1 1 = 1 (1) − 1 (1 − a ) + 1 (0 − a ) = 0 a 0 1 1 and ∆3 = 1 a

T

T

F

T

T

T

F

F

F

T

F

T

F

T

T

T

T

T

T

F

F

T

T

F

T

T

Hence, it is a tautology. 1 ...(i) 47. We have, f (x) + 2 f   = 3x, x ≠ 0  x 1 On replacing x by in the above equation, we get x 3  1 f   + 2 f (x) =  x x  1 3 ...(ii) ⇒ 2 f (x) + f   =  x x On multiplying Eq. (ii) by 2 and subtracting Eq. (i) from Eq. (ii), we get  1 6 4 f (x) + 2 f   =  x x  1 f (x) + 2 f   = 3x  x − − − 6 3 f (x) = − 3x x 2 f (x) = − x x

⇒ Now, consider ⇒ ⇒

f (x) = f (− x) 2 2 4 = 2x −x=− + x ⇒ x x x 2x2 = 4 ⇒ x2 = 2 x=± 2

⇒

Hence, S contains exactly two elements.

48. We know that, if x1, x2, ..., xn are n observations, then their standard deviation is given by 1  Σx  Σxi2 −  i   n n We have, (3.5)2 =

1 1 a 1 = 1 ( − b) − 1 (− a ) + 1 (b − a 2) b 0

x→ y

T

We have, ~ ( p ∨ q) ∨ (~ p ∧ q) ≡ (~ p ∧ ~ q) ∨ (~ p ∧ q) [Q By De-Morgan’s law ~ ( p ∨ q) = (~ p ∧ ~ q)]

y ≡ (~ p → q) → q

T

Key idea Use De-Morgan’s and distributive law.

≡ ~ p ∧ (~ q ∨ q) ≡ ~p∧t ≡ ~p 1 1 1

(Q RHS is not equal)

46. The truth table of the given expression is given below :

Q (6, 6)

(6, 5) (0, 5) (3, 5) (4, 5) S R (6, 4) (6, 3)

For

⇒

2

(22 + 32 + a 2 + 112)  2 + 3 + a + 11 −    4 4

49 4 + 9 + a 2 + 121  16 + a  − =   4  4 4

2

2

Miscellaneous 623 ⇒ ⇒ ⇒

49 134 + a 2 256 + a 2 + 32a = − 4 4 16 49 4a 2 + 536 − 256 − a 2 − 32a = 4 16 2 49 × 4 = 3a − 32a + 280 ⇒ 3a 2 − 32a + 84 = 0

In ∆EDB, ⇒

BD = h

≡ (s ∧ r ) ∨ (s ∧ ~ s) ≡ (s ∧ r ) ∨ F

[Q s∧ ~ s is false]

≡ s∧ r x1 + x2 + x3 + K + x16 51. Given, = 16 16

h h tan 60° = ⇒ CD = CD 3

⇒

⇒

54.

30° 45° 60° B C D

A

AB AD – BD AB h 3 − h = ⇒ = h BC BD – CD BC h− 3 3 −1 AB h ( 3 – 1) AB × 3 = ⇒ = BC h ( 3 – 1) BC ( 3 − 1) 3 AB 3 = ⇒ AB : BC = 3 : 1 BC 1

Now,

p

q

~p

~q

T F T F

F T T F

F T F T

T F F T

p↔q p↔ ~ q ~ p↔q F F T T

T T F F

~ ( p↔ ~ q)

T T F F

F F T T

Hence, ~ ( p ↔ ~ q) is equivalent to ( p ↔ q).

16

55. In ∆OA 1B1,

i= l

tan 45° =

∑ xi = 16 × 16

⇒

h

In ∆EDC ,

49. Consider, ( p ∧ ~ q) ∨ q ∨ (~ p ∧ q) ≡ [( p ∧ ~ q) ∨ q] ∨ (~ p ∧ q) ≡ [( p ∨ q) ∧ (~ q ∨ q)] ∨ (~ p ∧ q) ≡ [( p ∨ q) ∧ t ] ∨ (~ p ∧ q) ≡ ( p ∨ q) ∨ (~ p ∧ q) ≡ ( p ∨ q ∨ ~ p) ∧ ( p ∨ q ∨ q ) ≡ (q ∨ t ) ∧ ( p ∨ q) ≡ t ∧ ( p ∨ q) ≡ p ∨ q 50. ~ (~ s ∨ (~ r ∧ s)) ≡ ~ (~ s)∧ ~ (~ r ∧ s) ≡ s ∧ (~ (~ r ) ∨ ~ s) ≡ s ∧ (r ∨ ~ s)

E

h BD

tan 45° =

Sum of new observations

A1B1 OB1

⇒

20 = 1 ⇒ OB1 = 20 OB1 A1

18

= ∑ yl = (16 × 16 − 16) + (3 + 4 + 5) = 252

A2

i= l

Number of observations = 18

20 m 45°

18

∴

New mean =

∑ yi

i= l

18

=

52. Given, n ( A ) = 4, n (B) = 2 ⇒ n ( A × B) = 8 Total number of subsets of set ( A × B) = 28

30°

O

252 = 14 18

20

In ∆OA2 B2, tan 30° =

B1

20 OB2

B2

⇒ OB2 = 20 3

Number of subsets of set A × B having no element (i. e. φ) = 1 .

⇒

B1B2 + OB1 = 20 3 ⇒ B1B2 = 20 3 − 20

⇒

Number of subsets of set A × B having one element = 8C1

Now,

B1B2 = 20 ( 3 − 1) m Distance Speed = = 20 ( 3 − 1) m/s Time

Number of subsets of set A × B having two elements = 8C 2 ∴ Number of subsets having alteast three elements = 28 − (1 + 8C1 + 8 C 2) = 28 − 1 − 8 − 28 = 28 − 37 = 256 − 37 = 219

53. According to the given information, the figure should be as follows: Let the height of tower = h ED In ∆EDA, tan 30° = AD 1 ED h = = 3 AD AD ⇒

AD = h 3

56. We have, X = {4n − 3n − 1 : n ∈ N } X = {0, 9, 54, 243, . K } Y = { 9, n − 1 : n ∈ N } Y = {0, 9, 18, 27, K } It is clear that, X ⊂ Y ∴ X ∪Y =Y

57. Here, X =

[put n = 1, 2, 3, K] [put n = 1, 2, 3, K]

∑ Xi = 2 + 4 + 6 + 8 + K + 100 = 50 × 51 = 51

50 50 [Q ∑ 2n = n (n + 1), here n = 50]

n

Variance, σ 2 = =

1 n

∑ Xi2 − (X )2

1 2 (2 + 42 + K + 1002) − (51)2 = 833 50

624 Miscellaneous π 2

58. If initially all marks were xi , then σ12 =

∑ (xi − x )2

0

N

∴ Option (b) is incorrect.

Now, each is increased by 10. σ 22 =

∴

∑ [(xi + 10) − (x + 10)]2 N

(c) Let h (x) = x −

= σ12

59. Given, n ( A ) = 2, n (B) = 4 ∴ n ( A × B) = 8 The number of subsets of A × B having 3 or more elements = 8C3 + 8C 4 +…+ 8C 8 = (8 C 0 + 8 C 1 + 8 C 2 + 8 C 3 + . . .+ 8 C 8 ) − (8 C 0 + 8 C 1 + 8 C 2 ) 2n = nC 0 + nC 1 + . . . + nC n = 28 − 8 C 0 − 8 C 1 − 8 C 2

π −1 2

0

0

h (0) = − ∫ f (t ) cos t dt < 0 ⇒h (1) = 1 −

61. Given, log 0.3 (x − 1) < log 0. 9 (x − 1)

∴ Option (c) is correct. (d) Let g (x) = x9 − f (x) ⇒ g (0) = − f (0) < 0 g (1) = 1 − f (1) > 0 ∴ Option (d) is correct.

65. 3x = 4x − 1

⇒ x=

Here, x − 1 > 0 and log( 0.3 ) (x − 1) < log( 0.3 )2 (x − 1) 1 ⇒ x > 1 and log 0.3 (x − 1) < log 0.3 (x − 1) 2 ⇒ x > 1 and log( 0.3 ) (x − 1) < 0 ⇒ x > 1 and x > 2 ∴ x ∈ (2, ∞ )  201 + 1 ∴Median of the given observation =   th item = x101  2 

Now, deviations will be minimum, if we taken from the median. ∴ Mean deviation will be minimum, if k = x101.

1 1− 2 log3 2

=

1 1 1− log3 4

=

i elements in set A and j elements in set B. Now, according to information 1 ≤ j < i ≤ 6. So, total number of ways of choosing sets A and B=Σ

Σ

6

1 ≤ j 0

x log3 3 = (x − 1) log34 ⇒ x = 2 log3 2 ⋅ x − log3 4 2 log3 2 ⇒ x (1 − 2 log3 2) = −2 log3 2 ⇒ x = 2 log3 2 − 1

which is only possible for irrational number.

or

π 2

⇒

60. Let x = log 2 7 ⇒ 2x = 7

Using AM ≥ GM, we get 1 log10 x + 1/ 2 1  log10 x  ≥  log10 x   2 log10 x 1 ⇒ ≥2 log10 x + log10 x

∫ f (t ) cos t dt,

Taking log3 on both sides, we get

= 256 − 1 − 8 − 28 = 219

 1 1  = 2 log10 x + 2 = 2  log10 x +   log10 x log10 x

π −x 2 0

So, variance will not change whereas mean, median and mode will increase by 10.

∴

∫ f (t )sin t dt always positive

(b) f (x) +

and

  1 Q log b = logb a    a [Q alog a b = b] [Q a n = a + (n − 1)d ]

Y = {9, 16, 23, … , 14128}

X ∩ Y = {16, 51, 86,… } tn of X ∩ Y is less than or equal to 10086 ∴ tn = 16 + (n − 1) 35 ≤ 10086 ⇒ n ≤ 288.7 ⇒ n = 288 Q n (X ∩ Y ) = n (X ) + n (Y ) − n (X ∩ Y ) ∴ n (X ∩ Y ) = 2018 + 2018 − 288 = 3748

Miscellaneous 625 69. Given, X has exactly 5 elements and Y has exactly 7 elements. a1 a2

b1 b2

a7

b5

∴ n (X ) = 5 and n (Y ) = 7 Now, number of one-one functions from X to Y is α = 7P5 = 7C5 × 5 ! Number of onto functions from Y to X is β 1, 1, 1, 1, 3 or 1, 1, 1, 2, 2 7! 7! ∴ β= × 5! + × 5! 3!4! (2 !)3 3 !

From Eqs. (i) and (iii), we get 16h 2 h2 + = 9 ⇒ 25h 2 = 81 9 9 12 and k = + h=+ ⇒ 5 5 18 12 So, 2h + k = + =6 5 5 Now, equation common chord XY of circles C1 and C 2 is C1 − C 2 = 0 ⇒ 6x + 8 y = 18 ⇒ 3x + 4 y = 9 … (iv) Now, N

Y C3

∴

7

7

C1P =

≡ p∧ ~q∧ ~ p∧ q Hence, it is a fallacy statement. So, Statement I is true. Statement II ( p → q) ↔ (~ q → ~ p) ≡ ( p → q) ↔ ( p → q) which is always true, so Statement II is true. Alternate Solution Statement I ( p ∧ ~ q) ∧ (~ p ∧ q) q T

~ p ~q F

p ∧ ~ q ~ p ∧ q ( p ∧ ~ q ) ∧ (~ p ∧ q )

F

F

F

F F

T

F

F

T

T

F

F

T

T

F

F

T

F

F

F

T

T

F

F

F

Hence, it is a fallacy. Statement II ( p → q) ↔ (~ q → ~ p) ~ q → ~ p is contrapositive of p → q. Hence, ( p → q) ↔ ( p → q) will be a tautology.

71. It is given that, the centres of circles C1, C 2 and C3 are co-linear, ∴

0 0 1 3 4 1 = 0 ⇒ 4h = 3k

…(i)

h k 1 and MN is the length of diameter of circle C3 , so

Z 3x+4y=9

9 5

81 144 12 = ⇒ PY = 25 25 5 12 24 Q XY = 2PY = 2 × = 5 5 9 12     3  + 4   −9 5  5 6 Now, C3 P = = 5 5 36 864 2 2 2 So, PW = C3W − C3 P = 36 − = 25 25 {Q C3W = r = 6} 12 6 24 6 ⇒ ⇒ ZW = 2PW = PW = 5 5 length of ZW = 6 ∴ length of XY 1 Now, area of triangle MZN = (MN ) (PZ ) 2 1 1  = × (12 )  WZ  {Q MN = 12} 2  2 24 6 72 6 = 3WZ = 3 = 5 5 1 and area of triangle ZMW = (ZW )(MP ) 2 1  24 6  12 6  9 =   (MG + GP ) = 3 +  2 5  5  5 9  Q MG = 3 and GP =  5  12 6  24 288 6 =   = 5 5 25

MN = 3 + (3 − 0)2 + (4 − 0)2 + 4 = 3 + 5 + 4 = 12 So, radius of circle C3 , r = 6 … (ii) Since, the circle C3 touches C1 at M and C 2 at N , so |C1 C3| = |r − 3| ⇒

X C1

Now, PY 2 = GY 2 − GP 2 = 9 −

≡ p∧ ~ p∧ ~q∧ q≡ f ∧ f ≡ f

p

(0,0) C1

M

70. Statement I ( p ∧ ~ q) ∧ (~ p ∧ q)

T

C2

P

3

= ( C3 + 3 C3 ) 5 ! = 4 × C3 × 5 ! β − α (4 × 7C3 − 7C5 ) 5 ! = 5! 5! = 4 × 35 − 21 = 140 − 21 = 119 7

(3,4) C2

W

h 2 + k2 = 3 ⇒ h 2 + k2 = 9 …(iii)

∴

72 6 Area of triangle MZN 5 5 = = Area of triangle ZMW 288 6 4 25

626 Miscellaneous Q Common tangent of circles C1 and C3 is C1 − C3 = 0 2 2   9 12  ⇒ (x2 + y2 − 9) −  x −  +  y −  − 36 = 0   5 5   18 24 … (v) x+ y + 18 = 0 ⇒ 3x + 4 y + 15 = 0 ⇒ 5 5 Q Tangent (v) is also touches the parabola x2 = 8 αy, 2

∴

15 10  3 ⇒α= − 2 α −  = −  4 4 3

So combination (iv), (S) is only incorrect. Hence, option (b) is correct. length of ZW 72. Q = 6 length of XY

⇒ Now,

satisfy x2 + y2 = a 2. So, option (a) is correct.

...(i)

2a (15 − 2a ) = 4(a + 15 − 2a )

⇒

73. Either option (a) or (c) is correct. But a = 2 and (−1, 1)

15a − 2a 2 = 30 − 2a

⇒ 2a 2 − 17a + 30 = 0 ⇒ a = 5 / 2, 6 Hence, q − 2a = 10 − 2a = 5 or 2

77. A → P, R, S; B → P; C → P, Q; D → S, T A. Since, 2(a 2 − b2) = c2 ∴

74. Either option (b) or (c) is correct. 1  Satisfying the point  3 ,  in (II), we get  2 a 2 = 4. ∴ The conic is x + 4 y2 = 4 2

1 Now, equation of tangent at ( 3 , ) is 3x + 2 y = 4, which 2 is the given equation. So, option (b) is correct.

75. Either option (a), (b) or (c) is correct. Satisfying the point (8, 16) in (III), we get ∴ The conic is y2 = 32x = 16(2x)4 Now, equation of tangent at (8, 16) is y ⋅ 16 = 16(x + 8) ⇒ y = x + 8, which is the given in equation. So, option (a) is correct.

76. A → P, Q; B → P, Q; C → P, Q, S, T; D → Q, T 3α + β = ± 2 3

[using sine law]

⇒ 2 sin (x − y) ⋅ sin(x + y) = sin 2 z ⇒ 2 sin (x − y) ⋅ sin z = sin 2 z [Q x + y + z = π] sin (x − y) 1 ⇒ = =λ sin z 2  nπ  Also, cos (nπλ) = 0 ⇒ cos   = 0 ⇒ n =1, 3, 5  2  B. 1 + cos 2X − 2 cos 2Y = 2 sin X ⋅ sin Y ⇒ 1 + 1 − 2 sin 2 X − 2(1 − 2 sin 2 Y ) = 2 sin X ⋅ sin Y ⇒

− 2a 2 + 4b2 = 2ab 2

 a  a a 2 + ab − 2b2 = 0 ⇒   +   − 2 = 0  b  b a   a a = 1, − 2 ⇒  + 2  − 1 = 0 ⇒   b b b C. OX = 3 $i + $j,OY = $i + 3 $j , OZ = βi$ + (1 − β)$j ⇒

β − (1 − β)

...(i)

and ...(ii) α − 3β = 2 On solving Eqs. (i) and (ii), we get α = 2 , − 1 ⇒ α = 2 ,1 B. Since, if any function is differentiable, then it is continuous for all x. By continuity at x = 1, ...(i) − 3a − 2 = b + a 2 − 6ax, x < 1 By differentiability, f ′ (x) =  x>1  b, ∴ − 6a = b From Eqs. (i) and (ii), we get − 3a − 2 = − 6a + a 2 2 ⇒ a − 3a + 2 = 0 ⇒ a = 1, 2 C. (3 − 3ω + 2ω 2)4n + 3 + (2 + 3ω − 3ω 2 )4n + 3

2(sin 2 x − sin 2 y) = sin 2 z

Angle bisector of OX and OY is along the line y = x and its distance from (β, 1 − β) is

A. Projection of (α$i + β$j) on ( 3 i$ + $j) = 3 3α + β = 3 ⇒ 2

b = 15 − 2a 2ab = 4 (a + b)

⇒

So, combination (ii), Q is only correct. Hence, option (c) is correct.

⇒

+ (− 3 + 2ω + 3ω 2)4n + 3 = 0 ⇒ {(− 3 + 2ω + 3ω )ω } + { − 3 + 2ω + 3ω 2)ω 2 }4n + 3 + {(− 3 + 2ω + 3ω 2)}4n + 3 = 0 ⇒ (− 3 + 2ω + 3ω 2)4n + 3 {ω 4n + 3 + (ω 2)4n + 3 + 1} = 0 ⇒ ω n + ω 2n + 1 = 0 which is true only when n is not a multiple of 3. ∴ n = 1, 2, 4, 5 2ab D. Here, = 4 and 2(5 − a ) = b − 5 a+b 4n + 3

2

2

=

3 ⇒ 2β − 1 = ± 3 2

∴ β = 2 , − 1 or β = 1, 2 D. Area bounded by x = 0, x = 2, y2 = 4x Y (0, 3)

X′

O

y=3 (2, 2√2)

(2, 0)

... (ii) Y′

x=2

When α = 0, then 2 Area bounded, F (α ) = 2 × 3 − ∫ 4 x dx 0

X

Miscellaneous 627 2

 x3/ 2  4 3/ 2 8 =6 −2   = 6 − 3 ⋅ (2 − 0) = 6 − 3 2 / 3 2  0 8 2 ∴ F (α) + =6 3 When α = 1, then x≥2 3x − 3,  y = x − 1 + x − 2 + x =  x + 1, 1 < x < 2  3 − x, x≤1 

...(i)

Y 3

⇒

=1 ⇒

⇒ ⇒

π , t ∈Z 4 8 4t + 1 , t ∈Z = ⇒ n= 4t + 1 4 = tπ +

∴ The minimum value of n = 8 (R) PLAN

y= 3

Equation of normal at the point y2 x2 ( a cos θ, b sin θ) of ellipse 2 + 2 = 1 is given by b a

X

O 1 2

2π n 2π π tan = tan n 4

= r 2(n − 1) cos

(2, 2√2) X′

2π n 2π tan n 2π n 2 n

⇒ r 2(n − 1) sin

ax sec θ − by cosec θ = a 2 − b 2

Equation of normal is Y′

∴ Area bounded,

6 x sec θ − 3 y cosec θ = 3

x=2

8 F (α ) = (6 − 1) − ∫ 2 x dx ⇒ F (α ) = 5 − 2 0 3 8 ...(ii) ∴ F (α ) + 2 =5 3 8 Hence, values of F (α ) + 2 are 5 or 6. 3 [from Eqs. (i) and (ii)] 2

78. (P) y = cos (3 cos −1 x) ⇒ y′ =

3 sin (3 cos −1 x) 1 − x2

1 − x2 y′ = 3 sin (3 cos −1 x) ⇒ ⇒

−x 1−x

2

y′ + 1 − x2 y′ ′ = 3 cos (3 cos −1 x) ⋅

− xy′ + (1 − x2) y′ ′ = − 9 y ⇒

−3 1 − x2

1 [(x2 − 1) y′ ′ + xy′ ] = 9 y

Its slope is

Q Slope of normal = Slope of line perpendicular to 1 x + y = 8 ⇒ tan θ = 2 So, normal is 6 x

Consider a polygon(S) of n sides with centre at origin O

An

A3

2π/ n

A1

A2

Let | OA1 | = | OA 2| = K| OA n |= r [say] 2π 2π 2 2 | a k × a k + 1 | = r sin ⇒ | a k ⋅ a k + 1 | = r cos n n   n −1 n −1 ⇒  ∑ a k × a k + 1  = ∑ a k ⋅ a k + 1    k =1 k =1

3 − 3 × 3 y=3 2

3x − 3 y = 3 ⇒ x − y = 1 As it passes through (h , 1). ∴ h −1 =1 ⇒ h =2  x+ y (S) Plan tan −1 x + tan −1 y = tan −1    1 − x y Given equation  1  1  −1  −1 tan −1   + tan   = tan  2x + 1  4x + 1

(Q) PLAN (i) Angle subtended by a side of n sided regular polygon 2π at the centre = . n (ii) | a × b| = | a || b| sin θ (iii) | a ⋅ b| = | a || b| cos θ (iv) tan θ = tan α ⇒ θ = nπ + α, n ∈ Z

6 sec θ =1 3 cosec θ

⇒

tan −1

 2  2 x 

1 1   +  2x + 1 4x + 1  −1  2    = tan  2 1 x  1 −  (2x + 1) (4x + 1)  

  6x + 2 −1  2  tan −1   = tan  x2 2 x 1 4 x 1 1 ( ) ( ) + + −   3x + 1 2 3 2 ⇒ = ⇒ 3x + x = 8x2 + 6x 4 x2 + 3 x x 2 ⇒ x (3x2 − 7x − 6) = 0 −2 ,3 ⇒ x (x − 3) (3x + 2) = 0 ⇒ x = 0, 3 So, only positive solution is x = 3. ⇒

(P) → (iv); (Q) → (iii); (R) → (ii); (S) → (i)

79. (A) Given,|z| = 1 ⇒ Let

z⋅ z = 1 ⇒

2 iz 2 iz 2i = = 2 2 z −z z⋅ z − z 1−z

z = x + iy

628 Miscellaneous ∴

z − z = 2 iy =

2 i 1 =− − 2 iy y

Using cosine law,

…(i)

→

cos C =

y= 1 − x ⇒ −1 ≤ y≤1 1 and ⇒ − 1 ≤ y and y ≤ 1 ⇒ − 1 ≥ y 2

where

→

→

⇒ (B)

 8 (3x − 2)  (B) f (x) = sin −1   , for domain  1 − 32( x − 1) 

⇒

3

b

∫ a f (x) dx − 2 (b

⇒

∫ a f (x) dx = (a

∴

−1 ≤

3x − 3x − 2 ≤1 1 − 3x ⋅ 3( x − 2)

⇒

∫ a f (x) dx =

⇒

(3x + 1) (3x − 2 − 1) ≥0 (3x − 1 + 1) (3x − 1 − 1)

3x − 3x − 2 ≥ −1 1 − 3x ⋅ 3x − 2

b

 x2  f ( x ) dx − 3   = (a 2 − b2) ∫a  2a b

9 ⋅ (3x − 2) − (3x − 2) 8 (3x − 2) ≤1 ≤1 ⇒−1 ≤ 2( x − 1) 1 − 32( x − 1) 1 −3

b

b

2

2

− a 2) = (a 2 − b2)

3 2 b2 − a 2 (b − a 2) = 2 2  π π ⇒ f (x) = x ⇒ f   =  6 6

− b2) +

b2 − a 2 2

1

z

z

z –1 z

2

3x − 3x − 2 ⇒ x ∈ (− ∞ , 1] ∪ (2, ∞ ) and ≤1 1 − 3x ⋅ 3x − 2

1 z –1

(3x − 2 − 1) (3x + 1) ≥0 (3x − 1 + 1) (3x − 1 − 1)

⇒

4 + 4 − 12 − 4 − 1 = = 2 ×2 ×2 8 2

=

2π ] 3

∠ C = 120° =

−1 ≤

1

→

2|a||b|

1 ≥1 y

 2 iz  ⇒ Re   ∈ (− ∞ , − 1] ∪ [1, ∞ )  1 − z 2

⇒

→

|a|2 + |b|2 − |c|2

π log e 3 2

(C)

5/ 6

∫ 7/ 6 sec (πx) dx 5/ 6

0

and x ∈ (− ∞ , 0) ∪ [1, ∞ ) ⇒ x ∈ (− ∞ , 0] ∪ [2, ∞ )

⇒

1 tan θ 1 (C) f (θ ) = − tan θ 1 tan θ 1 −1 − tan θ Applying

R1 → R1 + R3

0

0

2

1 f (θ ) = − tan θ tan θ 1 −1 − tan θ = 2 (tan 2 θ + 1) = 2 sec2 θ ≥ 2 f (θ ) ∈ [2 , ∞ ) (D) f (x) = x3/ 2 (3x − 10); x ≥ 0 3 1   f ′ (x) = x3/ 2 ⋅ 3 + ⋅ x1/ 2 (3x − 10) = 3x1/ 2 x + (3x − 10) 2 2   3 1/ 2 = x {2x + 3x − 10} 2 2

⇒

π loge 3

(D)

→

| − arg (1 − z )| | arg (1 − z )|

From figure, arg (z − 1) is maximum ⇒ π

81. (A) Equation of the line passing through origin is

A

c =2 √ 3k

B

1 c

a +3b+5c =0

Also,

8 3 2 a

∴

=0

a (− 1) − b (3) + c (− 5) = 0 ⇒ − a − 3 b − 5 c = 0

⇒

a = j + √3 k

→

−1

2 1 x y z = = ⇒ 1 −2 a b c a b

→

b = – j +√3k

for |z| = 1 ⇒ | arg (1 − z )− 1|

⇒

→

→

1 (1 − z )

⇒

80. (A) |a| = 1 + 3 = 2 ⇒ | b| = 1 + 3 = 2, | c| = 12 = 2 3 C

π 1  { loge 3} = π  ⇒ loge 3 3 

 loge | 3|− loge 

arg

⇒

15 1/ 2 = x (x − 2) ⇒ x ≥ 2 2 →

π 2  log e |sec πx + tan πx|   log e 3  π 7/ 6 π  5π 5π + tan log e sec log e 3  6 6 7π 7π − log e sec + tan  6 6 

⇒

1

...(i)

−3 1 −1 1 = 0 b

c

2 b 10 c  10  2 a (− 2) − b   + c   = 0 ⇒ 2a + − =0 3  3 3 3

Miscellaneous 629 ⇒

...(ii) 3a + b − 5 c = 0 a b c a b c From Eqs. (i) and (ii), = = ⇒ = = − 20 20 − 8 5 −5 4 Equation of line is x y z = = =λ 5 −5 4 x −2 y −1 z + 1 Also = = = k1 −2 1 1 8 x− 3 = y + 3 = z −1 = k Now, 2 2 −1 1

[say ]

...(iii)

[say ]...(iv)

[say ] ...(v)

Point on Eq. (iii) is (5λ , − 5λ , + 4λ ). Point on Eq. (iv) is (2 + k1 , 1 − 2k1 , − 1 + k1 ).  8 Point on Eq. (v) is  + 2k2, − 3 − k2, 1 + k2 .  3 On solving, 2 + k1 + 1 − 2k1 = 0 ⇒ − k1 + 3 = 0 ⇒

k1 = 3 ⇒ P ≡ (5, − 5, 2) 8 Again, for Q, + 2k2 − 3 − k2 = 0 3 1  10 − 10 4 , ,  ⇒ k2 − = 0 ⇒ k2 = 1 / 3 ⇒ Q ≡  3 3 3 3 2

2

= 9 − 6 (1 − cos x) − 6 (1 − cos 3x) + 4 (1 − cos x) (1 − cos 3x) = 1 + 6 cos x + 6 cos 3x − 4 cos x −4 cos 3x + 4 cos x cos 3x 9x π π sin 2 2 dx = 4 1 + 2 cos x + 2 cos 3x Let I = ∫ ∫ π 0 π − π sin x 2 4 + 2 (cos 4x + cos x) dx = × π = 4 π 82. (A) 2 sin 2 θ + sin 2 2 θ = 2 ⇒ sin 2 2θ = 2 cos 2 θ ⇒ 4 sin 2 θ cos 2 θ = 2 cos 2 θ 1 ⇒ cos 2 θ = 0 or sin 2 θ = 2 1 π π or ⇒ cos θ = 0 or sin θ = ± ⇒θ = ± 4 2 2 6x  3x  (B) f (x) = cos  π   π  6x  Possible points of discontinuity of are  π  6x nπ π π π = n, n ∈I ⇒ x = ⇒ x= , , , π π 6 3 2 6 ⇒ lim f (x) = 0 cos 0 = 0 ⇒ lim

2

54  2  5  5 PQ =   +   +   =  3  3  3 3 54 PQ 2 = d 2 = ⇒ =6 9  x + 3 − x + 3 6 3 − 1  3 (B) tan − 1  =  = tan   ⇒ 2  4  1 + (x2 − 9)  x −8 4

x→

Now,

⇒

Similarly, discontinuous at x = 1 1

→

→

→

→

→

→

→

...(i)

→ → a− µ b = |b – a| 4

2 b+

⇒

→ → (4 − µ ) b + a 2 = |b − a| 4 →

(4 − µ )2 b 2 a 2 → 2 → 2 3 a 2 (4 − µ )2 − 4 → 2 ⇒ + =b +a ⇒ = ⋅b 4 4 4 4 ⇒

→2

→2

→2

3 a = b (4 − µ ) − 4 b 2

0 0 = π cu unit

→

→

→

→

→

→

⇒ a+b=− 3 c

→

→

⇒|a + b|2 = | 3 c|2

→ →

83. (A) Let f (x) = xesin x − cos x

→

→

π π , ,π 3 2

⇒ a 2 + b2 + 2 a ⋅ b = 3c2 ⇒ 2 + 2 cos θ = 3 1 π ⇒ cos θ = ⇒ θ= 3 2

→

Also,

→

f (x) = 1 cos 0 = 1

(D) Given, a + b + 3 c = 0 →

→

+

1 1 π

⇒ (b − a ) ⋅ (4 b + a − µ b) = 0 ⇒ (4 − µ ) b 2− a 2 = 0 →

V = 1 2

(C) Here,

3x = 48 ⇒ x = ± 4 → →  → → → a − µ b (C) (b − a ) ⋅  b + =0  4    →

π x→   6

∴ Discontinuous at x = π / 6 .

2

→

π− 6

...(ii)

From Eqs. (i) and (ii), 3 (4 − µ ) = (4 − µ ) − 4 (4 − µ )2 − 3 (4 − µ ) − 4 = 0 ⇒ µ = 0, 5 µ = 5 is not admissible.  9 x sin    2  x  3 x (D) f (0 ) = 9, f (x) = =  3 − 4 sin 2   3 − 4 sin 2  x   2 2 sin 2 3x 3x x 2x = 9 − 12 sin − 12 sin 2 +16 sin 2 ⋅ sin 2 2 2 2 2

f ′ (x) = esin x + xesin x cos x + sin x ≥ 0  π For interval x ∈ 0 ,  , f is strictly increasing.  2  π π ∴ f (0) = − 1 and f   = e ⇒ One solution  2 2 k 4 1

2

(B) Since, ⇒ ⇒

4 k 2 =0 2 2 1

k (k − 4) − 4 (0) + 1 (8 − 2k) = 0 k2 − 6k + 8 = 0 ⇒ k = 2 , 4

(C) Let y = |x − 1| + |x − 2| + |x + 1| + |x + 2| 3 For solutions, 4k ≥ 6 ⇒ k ≥ 2

630 Miscellaneous y = − 4x

y=

−4

y = 4x

x+

4

y=

y =6

−1

−2

1

2x

+

5

On adding Eqs. (i) and (ii), we get, 2I = 0 ⇒ I = 0 f ′ (x) = − 2 cos x sin x + cos x = cos x(1 − 2 sin x) = 0 f ′ (x)

– ve +ve −π - p 2

Points of local maxima are

2

5  1 − sin x −   4 2 2 1  For y to be maximum. sin x −  = 0  2 1 π ⇒ sin x = ⇒ x = nπ + (− 1)n , n ∈ I 6 2 (D) Let y = tan −1 (sin x + cos x) cos x − sin x dy = dx 1 + (sin x + cos x)2

x2 + y2 = 1 3

π −π 2

Hence, locus is an ellipse. (s) Eccentricity x = 1 ⇒ Parabola 1 < x < ∞ ⇒ Hyperbola (t) Let z = x + iy

ππ 4 2

(x + 1)2 − y2 = x2 + y2 + 1 ⇒ 2x = 2 y2 ⇒ x = y2

Hence, locus is parabola.

86. (A) Let y =

x2 + 2 x + 4 x+2

⇒ x2 + (2 − y) x + (4 − 2 y) = 0

85. (A) Given, (x − 3)2 ⋅ y′ + y = 0 dy y =− dx (x − 3)2

⇒

1 + ln C (x − 3)

⇒

∫

⇒

dy dx = −∫ y (x − 3)2

(2 − y)2 − 4 (4 − 2 y) ≥ 0

⇒

y + 4 y − 12 ≥ 0 ⇒

1

y = Cex − 3 , C ≠ 0

⇒

A 2 − AB + BA − B2 = A 2 + AB − BA − B2

Alternate Solution Given differential equation is homogeneous linear differential equation and has x = 3 as a singular point, hence x = 3 cannot be in domain of solution. 5 (B) Let I = ∫ (x − 1) (x − 2) (x − 3) (x − 4) (x − 5) dx

⇒

B A t = (−1)k AB

⇒

− B ⋅ A = (−1)k AB

1

L et x − 3 = t ⇒ dx = dt (t + 2) (t + 1) t (t − 1) (t − 2) dt

I = ∫ (x − 1) (x − 2) (x − 3) (x − 4)(x − 5) dx 1

...(i)

AB = BA and ( AB)t = (−1)k AB t

B ⋅ A = (−1)

k+1

[Q Bt = − B, A t = A]

AB ⇒ (−1)k + 1 = 1

∴ k + 1 is even or k is odd. −a (C) 1 < 2( − k + 3 ) < 2 ⇒ 0 < − k + 3− a < 1 Given, a = log3 log3 2 ∴

Q Integrand is an odd function. ⇒ I = 0 Let

⇒

⇒

Alternate Solution 5

y ≤ − 6, y ≥ 2

∴ Minimum value of y is 2. (B) Since, ( A + B)( A − B) = ( A − B)( A + B) ⇒

−2

[Q D ≥ 0]

2

∴ Domain of y is x ∈ R − {3}

2

π

Clearly, by graph cos x > sin x is true for option (s).

Since, Re (z + 1)2 = |z|2 + 1

I=∫

π 5π . , 6 6

y=

Let tan θ = t

∴

–ve +ve – ve p p 5p p 6 2 6

Alternate Solution y = cos 2 x + sin x

Hence, locus is a hyperbola.  1 − t2 2t (r) Let x = 3   , y= 1 + t2  1 + t 2

ln y =

[say]

Sign scheme for first derivative

(q) ||z + 2|− |z − 2|| = 3 and 2 − (−2) = 4 > 3

⇒

f (x) = cos 2 x + sin x

2

Hence, locus is a circle.

⇒

...(ii)

1

4

(C) Let

Then, x = 3 cos 2 θ , y = sin 2 θ ⇒

b

I = ∫ (5 − x) (4 − x) (3 − x) (2 − x) (1 − x) dx

Integer values of k are 2, 3, 4, 5. dy (D) Given, = y + 1 ⇒ ln|( y + 1)| = x + C dx ⇒ ln 2 = C ⇒ ln| y + 1| = x + ln 2 Put x = ln 2 ∴ ln ( y + 1) = ln 2 + ln 2 = ln 4 ⇒ y+1 =4 ⇒ y =3 1 1 84. (p) = 2 ⇒ h 2 + k2 = 2 2 4 h +k

⇒

b

∫ a f (x) dx = ∫ a f (a + b − x) dx

Using,

3a = log3 2 ⇒ 3− a = log 2 3

…(i)

k < log 2 3 < 2

…(ii)

and 1 + k > log 2 3 > 1 ⇒ k > 0

…(iii)

From Eqs. (ii) and (iii), 0 < k 0 and y = >0 a+1 a +1 Q Rays intersect each other in Ist quadrant. ∴ x > 0, y > 0 ⇒ a + 1 > 0 2a 0 2 and|a|− 1 > 0 ⇒ a > 1 ⇒ a 0 = 1 ⇒ = 3 3

α + β + γ =2 ⇒ γ =2

Also, (C)

1

∫ 0 (1 − y ) dy

Also,

 (2i + 1) − (2i − 1)    1 + (2i − 1) (2i + 1)  −1

→ $ ⇒→ $ =γ (B) Given, a = αi$ + β$j + γ k a⋅ k

2

1

∫0

+

1

∫ 0 (y

1 − x dx +

2

− 1) dy = 2∫

0

∫ −1 =2 ∫

1 0

(1 − y2) dy =

4 3

1 + x dx 1 0

1 − x dx =

4 3

(D) sin A sin B sin C + cos A cos B ≤ sin A sin B + cos A cos B = cos ( A − B) ⇒

cos ( A − B) ≥ 1 ⇒ cos ( A − B) = 1 ⇒ sin C = 1 x 1 89. Since, x > 0, y < 0 and x + y + = y 2 x y x 1 1 and (x + y) ⋅ = − + = ⇒ − y 2x y 2 2 1 1 x Let +t= =t ⇒ − 2t 2 y ⇒ 2t 2 − 1 = t ⇒ 2t 2 − t − 1 = 0 x   1 but x > 0 and y < 0 ∴ < 0  t=−  ∴ y 2 neglecting 1    x 1 x 1 ⇒ =− ⇒ y = − 2x ⇒ x + y + = y 2 y 2 1 1 ⇒ x − 2x − = ⇒ x = − 1 and y = 2 2 2

90. Given equation is log7 log5 ( x + 5 + x ) = 0 ⇒ log5 ( x + 5 + x ) = 1 ⇒ x+5 =5– x ⇒ x+5 + x =5 ⇒ x + 5 = 25 + x – 10 x ∴

⇒ 10 x = 20 ⇒ x = 2

x=4

91. Given, n = 40, x = 40, var (x) = 49

Σfixi = 40 ⇒ Σfixi = 1600 40 1 Also, var (x) = 49 ⇒ Σ fi (xi − 40)2 = 49 40 1 49 = (Σxi2 fi ) − 2Σxi fi + 40 Σfi ∴ 40 1 ⇒ 49 = (Σxi2 fi ) − 2 (1600) + 40 × 40 40 ⇒

x=

∴ Σxi2fi = 1649 × 40 Let (21-30) and (31-40) denote the kth and (k + 1) th class intervals, respectively. Then, if before correction fk and fk + 1 are frequencies of those intervals, then after correction (2 observations are shifted from (31-40) to (21-30)), frequency of kth interval becomes fk + 2 and frequency of (k + 1)th interval becomes fk + 1 − 2  40  1  ∑ fi xi + ( fk + 2)xk + ( fk + 1 − 2)xk + 1  ∴ xnew =   40 i = 1  i ≠ k k + , 1  

632 Miscellaneous 1 40

⇒ xnew =

 40  2 {(xk − xk + 1 )}  ∑ fi xi  + 40 i = 1 

(1, 1) y = 2x −1

40

1 1 = (− 10) = 39 . 5 fi xi + 40 i∑ 20 =1

1/2

 40  2 2 1  ∑ fi (xi − 39. 5) + fk (xk − 39. 5)  = i 1  40  = 2   ( . ) where i ≠ k , k + 1 f x 39 5 + − k+1 k+1   1 2 2 = [Σ ( fi xi − 79 fi xi + (39. 5) fi )] 40 40 40 40 1 1 1 fi ⋅ xi2 − 79  ∑ fixi + (39 . 5)2 ⋅ fi = ∑ ∑  40 40 40 i =1

i =1

i =1

O

From the graph, the total number of distinct solutions for x ∈ (0, 1] = 1. [as they intersect only at one point]

94. Since, F ′ (a ) + 2 is the area bounded by x = 0, y = 0, y = f (x) and x = a.

∑ (xi − c) , about c. 2

Using Newton-Leibnitz formula, f (a ) = F ′ ′ (a ) and f (0) = F ′ ′ (0)

i =1

⇒

∑ xi2 + 2

i =1

∑

∑ xi + n = 7n and

i =1

n

n

⇒

n

xi2

i =1

∑ xi2 − 2

i =1 n

+ 2 ∑ xi = 6 n and i =1

n

n

⇒

n

∑ xi

−n ⇒ x=

∑ xi

i =1

n

i =1

∑

xi2

a

∫0 f (x) dx = F ′ (a ) + 2

∴

n

Also, given that mean square deviation about −1 and +1 are 7 and 3, respectively. n n 1 1 ⇒ (xi + 1)2 = 7 and (xi − 1)2 = 3 ∑ ∑ n i =1 n i =1 n

Given, F (x) = ∫

i =1 n

−2

∑ xi

1 n

Again differentiating, π π π      F ′′ (x) = 4 cos 2 x2 +  − 2x cos  x2 +  sin  x2 +  2x      6 6 6   + {4 cos x ⋅ sin x} π   2 2 π   2 π  2 = 4 cos  x +  − 4x cos  x +  sin  x2 +      6 6 6  

= 2n

i =1

=1

− x )2 =

i =1

1 n

n

∑ (xi − 1)2 =

i =1

x

for all x ∈ [0, 1]. f (x) is increasing. At x = 0, f (0) = 0 and at x = 1, f (1) = ∫

1

0

Because, 0
0, 4 1 + x4 01+ t

93. Let f (x) = ∫

0 < f (1)
0 - tan- 1 çç 2 è 10 x ø

(b) lim cot(S n ( x)) = x, for all x > 0 n® ¥

(a) f is decreasing in the interval (- 2, - 1) (b) f is increasing in the interval (1, 2) (c) f is onto 3 (d) Range of f is éê - , 2 ùú ë 2 û

(c) The equation S 3 ( x) =

p has a root in (0, ¥) 4

1 (d) tan(S n ( x)) £ , for all n ³ 1and x > 0 2

1 8

13. Let E , F and G be three events having probabilities P ( E ) = , 1 1 1 and P (G ) = , and let P ( E Ç F Ç G ) = . For any 6 4 10 event H, if H C denotes its complement, then which of the following statements is (are) TRUE? P (F ) =

1 (b) P(E C Ç F Ç G ) £ 15 5 (d) P(E Ç F Ç G ) £ 12 C

(b) (I - FE ) (I + FGE ) = I (d) (I - FE ) (I - FGE ) = I

15. For any positive integer n, let S n : ( 0, ¥ ) ® R be

(a) S10 ( x) =

Then, which of the following statements is(are) TRUE?

1 (a) P(E Ç F Ç GC ) £ 40 13 (c) P(E È F È G ) £ 24

(a)| FE | = | I - FE || FGE | (c) EFG = GEF

defined by S n ( x ) =

(c) | (EF )3 | > | EF |2

12. Let f : R ® R be defined by f ( x ) =

14. For any 3 ´ 3 matrix M , let | M | denote the determinant

C

C

16. For any complex number w = c + id , let arg( w ) Î ( - p , p ], where i = - 1. Let a and b be real numbers such that for all complex numbers z = x + iy æz +aö p satisfying arg ç ÷ = , the ordered pair ( x, y ) lies è z +bø 4 on the circle x2 + y2 + 5x - 3 y + 4 = 0 Then, which of the following statements is (are) TRUE? (a) a = - 1 (c) ab = - 4

(b) ab = 4 (d) b = 4

4 JEE Advanced

Solved Paper 2021

SECTION 4 l

l

l

l

This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE INTEGER. For each question, enter the correct integer corresponding to the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If ONLY the correct integer is entered; Zero Marks : 0 In all other cases.

17. For x Î R, the number of real roots of the equation 2

19. Let u , v and w be vectors in three-dimensional space, where u and v are unit vectors which are not perpendicular to each other and u × w = 1, v × w = 1, w× w = 4

2

3x - 4 | x - 1 | + x - 1 = 0 is

18. In a D ABC, let AB = 23, BC = 3 and CA = 4. Then, the value of

cot A + cot C is cot B

If the volume of the parallelopiped, whose adjacent sides are represented by the vectors u , v and w, is 2, then the value of | 3u + 5 v | is

Paper 2 SECTION 1 l

l

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This section contains SIX (06) questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is (are) correct answer(s). For each question, choose the option(s) corresponding to (all) the correct answer(s). Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is(are) chosen; Partial Marks : +3 If all the four options are correct but ONLY three options are chosen; Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of which are correct; Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct option; Zero Marks : 0 If unanswered; Negative Marks : - 2 In all other cases. For example, in a question, if (a), (b) and (d) are the ONLY three options corresponding to correct answers, then choosing ONLY (a), (b) and (d) will get +4 marks; choosing ONLY (a) and (b) will get +2 marks; choosing ONLY (a) and (d) will get +2marks; choosing ONLY (b) and (d) will get +2 marks; choosing ONLY (a) will get +1 mark; choosing ONLY (b) will get +1 mark; choosing ONLY (d) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get -2 marks.

1. Let S 1 = {( $i , $j, k$ ) : i$ , $j, k$ Î {1, 2, ¼ , 10}},

2. Consider a D PQR having sides of lengths p , q and r

S 2 = {( $i , $j ) : 1 £ $i < $j + 2 £ 10, i$ , $j Î {1, 2, ¼ , 10}}

opposite to the angles P , Q and R, respectively. Then, which of the following statements is (are) TRUE?

S 3 = {( $i , $j, k$ , l ) : 1 £ $i < $j < k$ < l, $i , $j, k$ , l Î {1, 2, ¼ , 10}}

p2 2qr æq - r ö æ p- r ö (b) cos R ³ ç ÷ cos P + ç ÷ cos Q èp+ qø èp+ qø sinQ sin R q+ r (c) and cos R > . r q

and S 4 = {( $i , $j, k$ , l ) : i$ , $j, k$ and l are distinct elements in {1, 2, … , 10}}. If the total number of elements in the set S r is nr , r = 1, 2, 3, 4, then which of the following statements is (are) TRUE? (a) n1 = 1000 (c) n3 = 220

(b) n2 = 44 n4 = 420 12

(d)

(a) cos P ³ 1 -

JEE Advanced Solved Paper 2021 5 é

1ö æ ex æ e2 ö - x ÷e ç x - ÷ + ççe 2 è 2ø è 4 ÷ø e x æ1 e2 ö - x ö æ ÷e (d) f( x) = ç - x÷ + ççe + è ø 2 2 4 ÷ø è

p pù

3. Let f : ê - , ú ® R be a continuous function such that ë 2 2û

(c) f( x) =

p

f ( 0 ) = 1 and ò 3 f ( t ) dt = 0 0

Then, which of the following statements is (are) TRUE? (a) The equation f( x) - 3cos 3 x = 0 has at least one solution in æ 0, p ö. ç ÷ è 3ø 6 (b) The equation f( x) - 3sin 3 x = - has at least one solution in p æ 0, p ö. ç ÷ è 3ø

5. Let O be the origin and OA = 2i$ + 2$j + k$ , 1 OB = i$ - 2$j + 2k$ and OC = (OB - l OA ), for some 2 9 l > 0. If | OB ´ OC | = , then which of the following 2 statements is (are) TRUE? 3 (a) Projection of OC on OA is - . 2 9 (b) Area of the DOAB is . 2 9 (c) Area of the D ABC is . 2 (d) The acute angle between the diagonals of the p parallelogram with adjacent sides OA and OC is . 3

x

(c) lim

x® 0

xò f(t ) dt 0

1- ex

2

= - 1.

x

(d) lim

x® 0

sin xò f(t ) dt 0

x2

= - 1.

4. For any real numbers a and b, let ya , b ( x ), x Î R, be the solution of the differential equation

dy + ay = xebx , y(1) = 1. dx

6. Let E denote the parabola y2 = 8x. Let P = ( - 2, 4 ), and let Q and Q¢ be two distinct points on E such that the lines PQ and PQ¢ are tangents to E. Let F be the focus of E. Then, which of the following statements is (are) TRUE ?

Let S = { ya , b ( x ) : a , b Î R}. Then, which of the following functions belong(s) to the set S ? x2 - x e + 2 x2 - x (b) f( x) = e 2 (a) f( x) =

æe - 1 ö e - x ç ÷ è 2ø 1 + æçe + ö÷ e è 2ø

(a) The D PFQ is a right-angled triangle. (b) The DQPQ¢ is a right-angled triangle. (c) The distance between P and F is 5 2. (d) F lies on the line joining Q and Q¢.

x

SECTION 2 l

l

l

l

l

l

This section contains THREE (03) question stems. There are TWO (02) questions corresponding to each question stem. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value corresponding to the answer in the designated place using the mouse and the on-screen virtual numeric keypad. If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +2 If ONLY the correct numerical value is entered at the designated place; Zero Marks : 0 In all other cases.

Question Stem for Question Nos. 7 and 8

Question Stem for Question Nos. 9 and 10

Question Stem

Question Stem

Consider the region R = {( x, y ) Î R ´ R : x ³ 0 and y2 £ 4 - x}. Let F be the family of all circles that are contained in R and have centers on the X -axis. Let C be the circle that has largest radius among the circles in F. Let (a , b) be a point, where the circle C meets the curve y2 = 4 - x.

Let f1 : ( 0, ¥ ) ® R and f2 : ( 0, ¥ ) ® R be defined by

7. The radius of the circle C is _____ . 8. The value of a is _____ .

f1 ( x ) =

x 21

(t ò0 Õ j =1

j ) j dt , x > 0 and f2 ( x ) = 98( x - 1)50

- 600( x - 1)49 + 2450, x > 0, where for any positive integer n and real numbers a1 , a2 , ¼ , an , Õ ni = 1 ai denotes the product of a1 , a2 , ¼ , an . Let mi and ni , respectively, denote the number of points of local minima and the number of points of local maxima of function fi , i = 1, 2 in the interval ( 0, ¥ ).

6 JEE Advanced

Solved Paper 2021 é p 3p ù f ( x ) = sin 2 x, for all x Î ê , ú. Define ë8 8 û

9. The value of 2m1 + 3n1 + m1 n1 is _____. 10. The value of 6m2 + 4 n2 + 8m2 n2 is _____. Question Stem for Question Nos. 11 and 12

Si =

3p 8

ò f ( x ) × g i ( x ) dx, i = 1, 2 p 8

Question Stem é p 3p ù é p 3p ù Let g i : ê , ú ® R, i = 1, 2, and f : ê , ú ® R be ë8 8 û ë8 8 û functions such that g1 ( x ) = 1, g 2 ( x ) = | 4 x - p | and

11. The value of

16S 1 is _____. p

12. The value of

48S 2 p2

is ______ .

SECTION 3 l

l

l

l

This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is the correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e, the question is unanswered); Negative Marks : - 1 In all other cases.

y1 ( x ) = e- x + x, x ³ 0,

Paragraph Let M = {( x, y ) Î R ´ R : x2 + y2 £ r2 },

y 2 ( x ) = x2 - 2x - 2e- x + 2, x ³ 0,

where r > 0. Consider the geometric progression an =

1

, 2n - 1 n = 1, 2, 3, ¼ . Let S 0 = 0 and, for n ³ 1, let S n denote the sum of the first n terms of this progression. For n ³ 1, let C n denote the circle with center ( S n - 1 , 0 ) and radius an , and Dn denote the circle with center ( S n - 1 , S n - 1 ) and radius an . 1025 13. Consider M with r = . Let k be the number of all those 513 circles C n that are inside M . Let l be the maximum possible number of circles among these k circles such that no two circles intersect. Then, (a) k + 2 l = 22 (c) 2 k + 3l = 34

(b) 2 k + l = 26 (d) 3k + 2 l = 40

14. Consider M with r =

( 2199 - 1) 2

2198 circles Dn that are inside M is (a) 198

(b) 199

. The number of all those

(c) 200

(d) 201

Paragraph Let y1 : [ 0, ¥ ) ® R, y 2 : [ 0, ¥ ) ® R, f : [ 0, ¥ ) ® R and g : [ 0, ¥ ) ® R be functions such that f ( 0 ) = g ( 0 ) = 0,

f (x ) =

x

ò- x (| t | - t

and g ( x ) =

x2

ò0

2

2

) e- t dt , x > 0

t e- t dt , x > 0.

15. Which of the following statements is TRUE ? 1 (a) f( ln 3 ) + g ( ln 3 ) = . 3 (b) For every x > 1, there exists an a Î(1, x) such that y1( x) = 1 + ax (c) For every x > 0, there exists a b Î(0, x) such that y2 ( x) = 2 x(y1(b ) - 1). 3 (d) f is an increasing function on the interval éê 0, ùú. ë 2û

16. Which of the following statements is TRUE? (a) y1( x) £ 1, for all x > 0 (b) y2 ( x) £ 0, for all x > 0 2 1 2 2 (c) f( x) ³ 1 - e - x - x3 + x5 , for all x Î æç 0, ö÷ è 2ø 3 5 1ö 2 3 2 5 1 7 æ (d) g ( x) £ x - x + x , for all x Î ç 0, ÷. è 2ø 3 5 7

JEE Advanced Solved Paper 2021 7

SECTION 4 l

l

l

l

This section contains THREE (03) questions. The answer to each question is a NON-NEGATIVE INTEGER. For each question, enter the correct integer corresponding to the answer using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If ONLY the correct integer is entered; Zero Marks : 0 In all other cases.

17. A number is chosen at random from the set {1,2,3, … ,2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then, the value of 500 p is ___ . x2 y2 18. Let E be the ellipse + = 1. For any three distinct points 16 9 P , Q and Q¢ on E, let M ( P , Q ) be the mid-point of the line segment joining P and Q, and M ( P , Q ¢ ) be the mid-point of the line segment joining P and Q¢. Then, the maximum

possible value of the distance between M ( P , Q ) and M ( P , Q ¢ ), as P , Q and Q¢ vary on E, is ___ .

19. For any real number x, let [x] denote the largest integer less than or equal to x. If 10 é 10x ù I = òê ú dx, then the value of 9I is ___ . x + 1û 0ë

Answer with Explanations Paper 1 1. (b) x 2 + y2 + x + 3y = 0. Method (1) As we know that, mirror image of orthocenter lie, on circumcircle. Image of (1, 1), in X-axis is (1, - 1). and Image of (1, 1) in x + y + 1 = 0 is (- 2, - 2). \ The required circle will be passing through both (1, - 1) and (- 2, - 2). \ Only x 2 + y 2 + x + 3y = 0 satisfy both. Method (2) One of the vertex is the point of intersection of X-axis and the line x + y + 1 = 0, say A, which is A º (- 1, 0). Y

C (µ, 0) )

A (–1, 0) O H

(1

,1

X′

1 2 ´ = -1 Þ m = 0 2 m -1

Þ

X

- 2ö Now, centroid of DABC is æç 0, ÷. è 3 ø As we know that, centroid (G) divides the line joining circumcentre (O) and orthocentre (H) in the ratio 1 : 2. 1 O (α, β)

2 G (0, –2/3)

H (1, 1)

2a + 1 = 0 and 2b + 1 = - 2 Þ a = - 1 / 2 and b = - 3 / 2 - 1 - 3ö , Þ Circumcentre = æç ÷ è 2 2 ø \ Equation of circumcircle which is passing through C(0, 0) is x 2 + y 2 + x + 3y = 0 Method (3) Equation of circle passing through C(0, 0) is x 2 + y 2 + 2gx + 2 fy = 0

… (i)

Since Eq. (i), also passes through (- 1, 0) and (1, - 2). x+y+1=0

–1 B (λ, –λ –1)

B (1, –2)

Y′ x+y+1=0

Let vertex B(l , - l - 1). Since, AC ^ BH \ l = 1 Þ B º (1, - 2) Let vertex C be (m , 0). Q AH ^ BC \ (Slope of AH) ´ (Slope of BC) = - 1

H(1, 1)

A (–1, 0)

C (0, 0)

y=0

8 JEE Advanced

Solved Paper 2021

1 − 2g = 0 ⇒

Then,

On solving y = 0 and x + y = 2, we get

g =1/ 2

B ≡ (2, 0)

5+ 1 − 4f = 0 ⇒

f = 3/ 2 1 3 2 2 ∴Equation of circumcircle is x + y + 2 × x + 2 × y = 0 2 2 and

On solving x = 9 / 4 and x = 3y, we get D ≡ (9 / 4, 3 / 4)

x 2 + y 2 + x + 3y = 0

i.e.

and C ≡ (9 / 4, 0) → (obviously)

2. (a) 11

So, required area = shaded region 3/ 2 1 / 2 1 1 1 = 2 0 1 + × 2 2 9 / 4 3/ 4 1

32 Method (1) Y x+y=2

(3, 1)

y=1

D

(3/2, 1/2)

 − 2 1 − 3  −  9 − 9   + 1 × 1 × 3        2 4  8 8  2 4 4 3 1 2 3 = − × + 4 2 2 32 8 ×1 3 8 + 3 11 sq units + = = = 8 × 4 32 32 32 =

(9/4, 3/4)

A X′

O

y=0

C X (9/4, 0)

B (2, 0)

Required area = Area of ∆OCD − Area of ∆OBA 1 3 1 1 9 = ×  − 0 × − × (2 − 0) ×  4 2 2 2 4 27 1 27 16 11 = − = − = 32 2 32 32 32

x=9/4

Required area = Shaded region On solving x + y = 2and x = 3y, we get 3 1 A ≡  ,   2 2

3. (a)

1 × 2

Method (2)

x=3y Y′

 9 − 2 × 3 / 4   4 

1 5 E1={1, 2, 3} S1={1, 2} E2={3} F2={1, 2, 3, 4} S2={1, 2}{1, 3} {1, 4} {2, 3}{2, 4} {3, 4} G2={1, 2, 3, 4, 5}

{2, 3, 4, 5}

S1={1, 3} E2={2} F2={1, 3, 4}

S1={2, 3} E2={1} F2={1, 2, 3, 4} {1, 2}{1, 3} {1, 4} {2, 3}{2, 4} {3, 4} {1, 2, 3, 4, 5}

{2, 3, 4, 5}

{2, 3}

{1, 3} {1, 4} {1, 2, 3, 4, 5}

{3, 4} {2, 3, 4, 5}

{2, 3} {1, 3}

S3= {1, 2} E3={1, 2, 3}

P(S1 ∩ (E1 = E 3)) P (A1 , 2) To find : Probability P = = P (E1 = E 3) P (A) where P(A) = P (A1 , 2) + P (A1 , 3) + P (A2, 3) Also, A1 , 2 represents 1, 2 chosen at start and similarly others. Now, P (A1 , 2) =

1 1 1 × 3C1 1 3 1 1 1 1 × 4 × 5 = × × = × × 3 C2 C2 3 6 10 3 2 10

1 1 × 2C 1 1 2 1 P (A1 , 3) = × 4 1 × 5 = × × 3 C2 C2 3 3 10 P (A2, 3) = =

1 × 3C 1  3 C2 × 1 1 1  × 4 × 4 + 4 1 × 5  3  C2 C2 C2 C2  1  3 1 3 1  1 1 1 1 1  × × + × = × + × 3  6 6 6 10  3  2 6 2 10 

P(A) =

1 1 1 1 2 1 1 1 1 1 1 × + ×0+ × + × + × + × 2 10 2 6 3 10 3 3  2 10 2 =

0 

1 1  1 = 3  4  12

1 1 1   ×    ∴Required probability = 3 2 10 1 12 1 12 1 = × = 60 1 5

P (A1 , 2)   Q P = P(A)   

4. (c) Both P and Q are true. QLength of direct distance ≤ length of arc i.e.| z2 − z1 | = length of line AB ≤ length of arc AB.

JEE Advanced Solved Paper 2021 9 Y C (z3)

(zK–1) (zK)

θ3

θK

X′

Now, determinant of α 2 γ M = | M |= β 1 0 = α − 2β + γ = 1 −1 0 1

B (z2)

θ2 θ1

A (z1)

O

[from Eq. (iv)]

Equation of plane P is given by

X

x − 2y + z = 1 Hence, perpendicular distance of the point (0, 1, 0) from the |0 − 2 × 1 + 0 − 1| | 3| plane = P= 6 12 + (− 2)2 + 12

Y′

2

| z3 − z2 | = length of line BC ≤ length of arc BC. ∴Sum of length of these 10 lines ≤ sum of length of arcs (i.e. 2π) (because θ1 + θ2 + θ3 + … + θ10 = 2π given)

9. (9) 10. (77.14)

∴| z2 − z1 | + | z3 − z2| + … + | z1 − z10|≤ 2π → P is true.

According to the question,

And | zk2 − zk2 − 1 |= | zk − zk − 1|| zk + zk − 1| | zk + zk − 1 |≤ | zk| + | zk − 1 |≤ 2

⇒

2 |≤ 2 ∴| z22 − z12 | + | z32 − z22 |+ … + | z12 − z10

(| z2 − z1 | + | z3 − z2|+ … + | z1 − z10|) ≤ 2 (2π)

5. (76.25) p1 = 1 − p(all 3 numbers are ≤ 80) 3

p2 = 1 − p (all three numbers are > 40) 3

60  27 98 = = 1 −   =1 −  100  125 125 125p2 125 98 98 = × = = 24.50 4 4 125 4

8. (1.50) 7 x + 8 y + 9z − (γ − 1) = A(4 x + 5y + 6z − β) + B(x + 2y + 3z − α)

R ≡ (x1 , y1) and S(x 2 , y2)

Q

C cuts y − 1 = 2x at R and S.

So,

| 2x 2 − 4 x 2 |= 3λ2 x=±

3 |λ | 2

∴

| x1 − x 2 | =

and

| y1 − y2 | = 2| x1 − x 2 |= 2 6 | λ |

6 |λ |

Q

RS 2 = 270

⇒

(x1 − x 2) + (y1 − y2) = 270

⇒

( 6λ)2 + (2 6 | λ |)2 = 270

⇒

30λ2 = 270 ⇒ λ2 = 9

⇒

On equating the coefficients, 4A + B = 7

… (i)

5A + 2B = 8

… (ii)

− (γ − 1) = − Aβ − αB

… (iii)

2

(given) 2

x + 2y − 2 = 0

On solving x + 2y − 2 = 0 with C, we get x2 = and| y1 − y2 | =

On solving Eqs. (i) and (ii), we get A = 2 and B = − 1

12 12 2 λ ⇒ | x1 − x 2 | = 2 |λ | 7 7 12 1 |λ | | x1 − x 2 |= 7 2

Hence, D ≡ (R′ S ′)2 = (x1 − x 2)2 + (y1 − y2)2

From Eq. (iii), we get − γ + 1 = − 2β − α(− 1) α − 2β + γ = 1

|(x 2)2 − (y − 1)2 | = λ2 3× 3

x + x 2 y1 + y2  Now, mid-point of RS is  1 ,  ≡ (0, 1)  2 2  −1 and slope of RS = 2and slope of R′ S ′ = 2 1 ∴ Equation of R′ S ′ : y − 1 = − x i.e. 2y − 2 = − x 2

7. (1)

⇒

x 2− y+1 = λ2 3

C :| 2x 2 − (y − 1)2 | = 3 λ2

⇒

125 − 64 61 80  = = 1 −   =  100  125 125 625p1 625 61 5 = × = × 61 = 76.25 4 4 125 4

6. (24.50)

and

C:

⇒ Let

≤ 4π → Q is true.

So,

x 2 + y −1 3

C:

As we know that,

So,

| 3| 9 D =   = = 1.5  6 6

⇒

= … (iv)

12 × 45 12 . × 9 × 5= ≈ 7714 7 7

10 JEE Advanced

Solved Paper 2021

11. (a, b and d)

– –4

For option (a)

1 2 3  1 0 0 =  8 13 18  0 0 1  =     2 3 4   0 1 0

1 3 2   8 18 13 = F    2 4 3 

1 0 0 1 0 0 1 0 0 2 P =  0 0 1   0 0 1  =  0 1 0       0 1 0  0 1 0  0 0 1 

and lim f (x) = 1 x→ ± ∞

∴

− 3 11  Range =  ,  2 6 

Hence, f (x) is into. Y 11/6 1

For option (b) | EQ + PFQ− 1 | = | EQ | + | PFQ− 1 | ∴ and

… (i) –4

| E | = 0 and| F | = 0 and| Q |≠ 0 | EQ | = | E || Q | = 0 | P || F | =0 | PFQ− 1 | = | Q|

f (x) has local maxima at x = − 4 …(ii)

⇒ RQ = EQ2 + PF = EQ2 + P 2EP = EQ2 + EP

[Q P 2 = I ]

= E(Q2 + P) ⇒

| R || Q | = | E || Q + P | = 0

⇒

| R | = 0 (as| Q |≠ 0)

and local minima at x = 0.

13. (a, b and c) For option (a)

| RQ | = | E(Q2 + P) | 2

X

O –3/2

Let R = EQ + PFQ− 1

⇒

+∞

Here, f is decreasing in the interval (− 2, − 1) and f is increasing in the interval (1, 2). −3 11 [from Eq. (i)] Now, f(− 4) = , f(0) = 6 2

Hence, option (a) is correct.

Q

+ 0

Sign scheme for f ′(x)

1 0 0 1 2 3  1 0 0 PEP =  0 0 1   2 3 4   0 0 1       0 1 0  8 13 18  0 1 0

and

+

–∞

P(E ∩ F ∩ GC) = P(E ∩ F) − P(E ∩ F ∩ G) ≤ P(E) − P(E ∩ F ∩ G) [Q| E | = O]

1 1 1 − = 8 10 40

=

… (iii)

From Eqs. (ii) and (iii), we get Eq. (i) is true.

For option (b)

Hence, option (b) is correct.

P(E C ∩ F ∩ G) = P(F ∩ G) − P(E ∩ F ∩ G) ≤ P(F) − P(E ∩ F ∩ G)

For option (c)

=

|(EF) |> | EF | 3

2

For option (c)

i.e. 0 > 0 which is false.

P(E ∪ F ∪ G) ≤ P(E) + P(F) + P(G) 1 1 1 3 + 4 + 6 13 = + + = = 8 6 4 24 24

For option (d) P2 = I

Q

1 1 1 − = 6 10 15

⇒ P− 1 = P

∴

P − 1 FP = PFP = PPEPP = E

For option (d)

So,

E + P − 1 FP = E + E = 2E

P(E C ∩ F C ∩ GC ) = 1 − P(E ∩ F ∩ G) ≥ 1 −

⇒ and

Tr(E + P P

−1

−1

FP) = Tr(2E) = 2Tr(E)

… (iv) P (E C ∩ F C ∩ GC)≥

EP + F

⇒

PEP + F = 2PEP

∴

Tr(2PEP) = 2Tr(PEP) = 2Tr(F) = 2Tr(E)

[Q F = PEP] … (v)

From Eqs. (iv) and (v), option (d) is also correct.

12. (a and b) Given, f (x) =

x 2 − 3x − 6 x 2 + 2x + 4

13 24

11 5 > 24 12

14. (a, b and c) Q

I − EF = G− 1

and

G − EFG = I

⇒ G − GEF = I

… (i) … (ii)

Clearly, GEF = EFG → option (c) is correct. … (i)

⇒

(x 2 + 2x + 4) (2x − 3) − (x 2 − 3x − 6) (2x + 2) f ′(x) = (x 2 + 2x + 4)2

⇒

5x(x + 4) f ′(x) = 2 (x + 2x + 4)2

Also, (I − FE) (I + FGE) = I − FE + FGE − FEFGE = I − FE + FGE − F(G − I) E = I − FE + FGE − FGE + FE = I→ option (b) is correct but option (d) is incorrect. Q (I − FE)(I − FGE) = I − FE − FGE + F(G − I) E = I − 2FE

JEE Advanced Solved Paper 2021 11 Now,

(I − FE) (− FGE) = − FE

So,

α = 1 and β = 4

⇒

| I − FE || FGE | = | FE |

Hence,

αβ = 1 × 4 = 4 and β = 4

→ option (a) is correct. n

∑ cot

15. (a and b) For option (a) Sn (x) =

− 1 1

k =1

 

+ k(k + 1) x 2   x 

17. (4) Given, 3x 2 − 4|x 2 − 1 | + x − 1 = 0 –∞

can be written as Sn (x) =

n

∑ tan

−1

(k + 1) x − kx  1 + kx ⋅ (k + 1) x   

k =1

=

n

k =1

  nx = tan− 1   2  1 + (n + 1) x  Now, S10 (x) = tan

10 x    2  1 + 11 x 

−1

+∞ –1

1

For − 1 ≤ x ≤ 1 i.e., x ∈ [− 1, 1] From Eq. (i) , we get

−1 −1 ∑ [tan (k + 1) x − tan (kx)]

= tan− 1 (n + 1) x − tan− 1 x

… (i)

3x 2 − 4(− x 2 + 1) + x − 1 = 0 ⇒

3x 2 + 4 x 2 − 4 + x − 1 = 0

⇒

7x2 + x − 5 = 0

⇒

x=

− 1 ± 1 + 140 (2 × 7)

Here, both values of x are acceptable. For| x |> |i.e. x ∈ (− ∞ , − 1) ∪ (1, ∞)

 10 x  π  1 + 11 x 2  π = − cot −1   = − tan− 1   2 2  1 + 11 x  2  10 x 

From Eq. (i), we get 3x 2 − 4(x 2 − 1) + x − 1 = 0 ⇒

Option (a) is correct. For option (b)

x2 − x − 3 = 0 x=

⇒

x   lim cot(Sn (x)) = cot  tan− 1  2   n→ ∞  x   1   = cot  tan− 1    = cot(cot − 1 x) = x, x > 0  x  Option (b) is correct.

1 ± 1 + 12 2

Again here, both values of x are acceptable. Hence, total number of solutions is 4.

18. (2) Given, AB = 23 = c BC = 3 = a

For option (c) S3(x) =

π 4

⇒

3x =1 1 + 4x2

CA = 4 = b A

⇒ 4 x 2 − 3x + 1 = 0 has no real root.

c

Option (c) is incorrect. For option (d) For x = 1, tan(Sn (x)) =

1 n which is greater than for n ≥ 3. 2 n+ 2

Option (d) is incorrect.

16. (b and d) Circle x 2 + y2 + 5x − 3y + 4 = 0 cuts the real axis ( X-axis) at (− 4, 0), (− 1, 0). π 4 π 4

–β

–α

 z + α π implies z is on arc and (−α, 0) and (− β , 0) arg   = 4  z + β π subtend on z. 4

B

b

a

C

cos A cos C + cot A + cot C sin A sin C Now, = cos B cot B sin B b2 + c 2 − a2 a2 + b2 − c 2 + 2bc(sin A) 2ab(sin C) = c 2 + a2 − b2 2ac(sin B) b2 + c 2 − a2 a2 + b2 − c 2 + 4∆ 4∆ = 2 c + a2 − b2 4∆ =

b2 + c 2 − a2 + a2 + b2 − c 2 c 2 + a2 − b2

=

2 × 16 2b 2 32 = = =2 a + c 2 − b 2 9 + 23 − 16 16 2

12 JEE Advanced

Solved Paper 2021

19. (7) Given, volume of the parallelopiped = 2 i.e. [u v w] =

2,

… (i)

where u, v and w are adjacent sides of parallelopiped. u⋅ u u⋅ v u⋅ w Also, [u v w]2 = v ⋅ u v ⋅ v v ⋅ w = 2 (from Eq. (i))

⇒ ⇒ or ∴

4λ2 − 2λ = 0 u ⋅ v = 0 (Rejected) 1 u⋅ v = 2 1 u⋅ v = 2

∴| 3u + 5v|2 = 9 | u |2 + 25| v|2 + 3 × 5 × 2 × u ⋅ v

w⋅u w⋅v w⋅w

= 9 + 25 + 30 ×

Let u ⋅ v = λ and substitute rest values, we get 1 λ 1 [Qgiven u ⋅ w = 1, v ⋅ w = 1, w ⋅ w = 4] λ 1 1 =2 1 1 4

1 2

= 49

(Q|u| = |v| = 1, given)

∴ | 3u + 5v | = 7

Paper 2 1. (a, b and d)

=

r − r(cos P + cos Q) r − r cos P − r cos Q = p+ q p+ q

=

r − (q − p cos R) − (p − q cos R) p+ q

=

(r − p − q) + (p + q) cos R p+ q

n1 = number of elements in S1 = 10 × 10 × 10 = 1000 n2 = number of elements in S2 = 8 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 Q1 ≤ i < j + 2 ≤ 10 ⇒ j ≤ 8 and i ≥ 1     8×9 =8+ = 44 ∴(i = 1, j = 1, 2, 3, …8), (i = 2, j = 1, 2, 3, … 8), 2   (i = 3, j = 2, 3, 4, … 8) and so on   n3 = number of elements in S3 =

10

C4

10

P4 = 10 × 9 × 8 × 7

For option (d)

2 2 2 2. (a and b) For option (a) cos P = q + r − p

2qr

P

r

For Option (c) q + r sin Q + sin R 2 sin Q ⋅ sin R = ≥ p sin P sin P → option (c) is incorrect.

n4 = 420 12

⇒

(Q r < p + q)

Hence, option (b) is correct.

(selecting 4 numbers and arranging in increasing order) 10 × 9 × 8 × 7 = = 210 24 n4 = number of elements in S4 =

r − (p + q) ≤ cos R p+ q

= cos R +

q

…(i)

If p < q and p < r, then p is the smallest side and hence one of Q or R can be obtuse. So, one of cos Q or cos R can be negative. p p Therefore, cos Q > and cos R > cannot hold always. r q Option (d) is incorrect.

3. (a, b and c) Q

q + r ≥ 2 2

Q ⇒

p

R

2

q 2r 2 (AM ≥ GM)

q 2 + r 2 ≥ 2qr

Given, f(0) = 1 and ∫

⇒

2qr − p 2 2qr

cos P ≥ 1 −

p2 → option (a) is correct. 2qr

For option (b) (q − r) cos P + (p − r) cos Q p+ q =

(q cos P + p cos Q) − r (cos P + cos Q) p+ q

f () t dt = 0

For option (a) Consider a function g(x) =

From Eq. (i), we get cos P ≥

π /3

0

x

∫0 f ()t dt − sin 3x

g(x) is continuous g(0) = g(π / 3) = 0

and

differentiable

function

and

∴ By Rolle’s theorem, g′(x) = 0 has atleast one solution in (0, π / 3). π i.e. g′(x) = f (x) × 1 − 3cos 3x = 0 for some x ∈  0,  .  3 For option (b) Consider a function φ(x) =

x

∫0 f ()t dt + cos 3x +

6 x π

JEE Advanced Solved Paper 2021 13 φ(x) is continuous and differentiable function as well as φ(0) = φ(π / 3) = 1

For α = 1 ye x =

x2 − 1 + e 2

1 x2 − x  e +  e −  e − x  2 2

⇒ y=

Hence, by Rolle’s theorem, φ′(x) = 0 has atleast one solution in (0, π / 3). π 6 i.e. φ′(x) = f (x) × 1 − 3sin 3x + = 0 for some x ∈  0,  .  3 π

Option (a) is correct.

For option (c)

⇒

ye αx = x ⋅

e( α + β) x e( α + β) x dx − ∫1 × (α + β) (α + β)

 form 0     0

⇒

ye αx = x ⋅

e( α + β) x e( α + β) x − + c1 (α + β) (α + β)2

(Using L-Hospital Rule)

⇒

y=

Case (II) If α + β ≠ 0

x

x ∫ f () t dt

L = lim

Let

0

x→ 0

1−e

x f (x) +

L = lim

⇒

x2

x→ 0

x

∫0 f ()t dt

− 2 x ex

2

∴ form 0     0

Again, using L′-Hospital rule L = lim

x→ 0

xf ′(x) + f (x) + f (x) 2

− 4 x 2 e x − 2e x

2

0 + 2 f (0) = = −1 −0−2

(Q f(0) = 1)

(Cancelling e αx from both sides) y=

⇒

y= x

P = lim

Let

 form  

0 2

x→ 0

x

0  0

Applying L-Hospital Rule, t dt sin x ⋅ f (x) + cos x ⋅ ∫ f () 0

x→ 0

2x

 form 0     0

Again using L-Hospital Rule,

∴ So, y =

t dt] [cos x ⋅ f (x) + sin x ⋅ f ′(x) + cos x ⋅ f (x) − sin x ∫ f () 0

Given,

x→ 0

and Also, Now,

Integrating factor (IF) = e ∫ α dx = e α x

⇒

y×e

αx

=

∫ xe

1=

1  ex  e2  − x  x −  +  e −  e → option (c) is correct.   2 2 4 

( α + β) x

dx

OA = 2$i + 2$j + k$

1 (OB − λOA) 2 | OB × OC | = 9 / 2 1 OB × OC = OB × (OB − λOA) 2 −λ λ OB × OA = (OA × OB) = 2 2

OC =

We need to fond OA × OB for this,

βx αx ∫ x e ⋅ e dx

… (i)

Case (I) If α + β = 0

(Qa × b = − b × a and a × a = 0) $i $j k$ OA × OB = 2 2 1 1 −2 2

From Eq. (i), we get ⇒

y e αx =

0 ⋅x ∫ x e dx =

∫ x dx =

x2 + C 2

= 6$i − 3$j − 6k$ = 3(2$i − $j − 2k$) … (ii) So,

OB × OC =

Given, y() 1 = 1 i.e. when x = 1, then y = 1 From Eq. (ii), we get 1⋅ eα =

e 1 c1 e2 × + ⇒ c1 = e − e 2 2 4

OB = $i − 2$j + 2k$

2 1 × f (0) + 0 × f ′(0) + 1 × f (0) − 0 × 0 ⇒ P= 2 1+ 0+1− 0 ⇒ P= =1 2 4. (a and c) Given, dy + αy = x ⋅ e βx which is a linear dx differential equation. So, the solution is y × e αx =

1 ex  −x  x −  + c1 e 2  2

5. (a, b and c) x

P = lim

From Eq. (i), 1 1 + C ⇒ C = eα − 2 2

⇒

From Eq. (ii), we get y e αx =

… (iii)

Given, y() 1 =1

x

P = lim

e βx  1  − αx x −  + c1 e α+β  α + β

Putting α = β = 1 in Eq. (iii), we get

For option (d) t dt sin x ⋅ ∫ f ()

x ⋅ e βx e βx − + c1 e − αx α + β (α + β)2

x2 − 1 + eα 2

⇒

3λ $ $ (2i − j − 2k$) 2

3λ 4 + 1 + 4 = 9/ 2 2

9λ 9 9 = 9/ 2 ⇒ |λ |= 2 2 2 |λ |= 1 ⇒ λ = ± 1

But

λ> 0

∴

λ =1

… (i)

14 JEE Advanced So, OC = ⇒

Solved Paper 2021 So, point P(−2 , 4) lies on the directrix of parabola y 2 = 8 x.

1 (OB − OA) (By putting λ = 1) 2 1 1 1 OC = (− $i − 4$j + k$) = − $i − 2$j + k$ 2 2 2

Hence, ∠QPQ′ =

and chord QQ′ is a focal chord and segment PQ subtends a right angle at the focus.

Option (a) OC ⋅ OA | OA | 1 (− 2 − 8 + 1) −3 = 2 = 3 2

Slope of PF = − 1

Projection of OC and OA =

32 = 4 2

7. (1.50)

1 1 | OA × OB |= | OA | OB |sin 90° 2 2 9 1 = | 3 × 3| = 2 2

8. (2.00) x ≥ 0, y 2 ≤ 4 − x

Given,

Let equation of circle be

Area of the ∆ ABC =

=

PF = (2 + 2)2 + (0 − 4)2 =

∴

Option (c)

=

(Q PF ⊥ QQ′)

2 Now, slope of QQ′ = =1 t1 + t2

Option (b) Area of ∆OAB =

π (by the definition of director circle) 2

1 2

(x − h)2 + y 2 = h2

1 | AB × AC | 2

$i

$j

−1

−4

… (i)

Y y2=4–x

k$

1 −1 −5 / 2 − 4 2 (0, 0)

9 1 $ 1 | 6 i − 3$j − 6k$ |= × 3 × 3 = 2 2 2

O

(h, 0)

X (4, 0)

Option (d) The acute angle between the diagonals of the parallelogram with adjacent sides OA and OC = θ (OA + OC) ⋅ (OA − OC) ⇒ = cosθ | OA + OC || OA − OC |  3 $i + 3 k$  ⋅  5 $i + 4$j + 1 k$       2  2 2  cosθ = 2 3 90 2× 4 2 18 1 1 = = ≠ 3 2 × 90 5 2

⇒

6. (a, b and d) Given,

E : y2 = 8 x

Solving Eq. (i) with y 2 = 4 − x, we get x 2 − 2hx + 4 − x = 0 ⇒

x 2 − x(2h + 1) + 4 = 0

…(ii)

For touching/tangency, Discriminant (D) = 0 (2h + 1)2 = 16 ⇒ 2h + 1 = ± 4

i.e. [∴ θ ≠ π / 3]

… (i)

and P ≡ (− 2, 4) Now, directrix of Eq. (i) is x = − 2

⇒

2h = ± 4 − 1 −5 3 (Rejected) because part of circle lies outside R. ⇒h = , h = 2 2 3 So, h = = radius of circle (c). 2 Putting h = 3 / 2 in Eq. (ii), x 2 − 4 x + 4 = 0 ⇒ (x − 2)2 = 0 ⇒ x = 2

Y

α=2

So, Q

y2=8x

9. (57) 10. (6) f1 (x) =

P (–2, 4) F (2, 0)

O

X′

X

x

∫0 (t − 1)

1

(t − 2)2(t − 3)3(t − 4)4 … (t − 21)21 dt

⇒ f1 ′(x) = (x − 1) (x − 2)2(x − 3)3(x − 4)4 … (x − 21)21 Sign Scheme for f1 ′(x )

Q′

x=–2

Y′

From sign scheme of f1 ′(x), we observe that f (x) has local minima at x = 4k + 1, k ∈ W i.e. f1 ′(x) changes sign from − ve to + ve which are x = 1, 5, 9, 13, 17, 21 and f (x) has local maxima at x = 4k + 3, k ∈ W i.e. f1 ′(x) changes sign from +ve to –ve, which are x = 3, 7, 11, 15, 19.

JEE Advanced Solved Paper 2021 15 So, m1 = number of local minima points = 6

⇒

and n1 = number of local maxima points = 5 Also, f2(x) = 98(x − 1)

− 600(x − 1)

49

⇒

2n < 1026 ⇒

n ≤ 10

∴Number of circles inside be10 = k. Clearly, alternate circle do not intersect each other i.e. C1 , C3 , C5 , C7 , C9 do not intersect each other as well as C2, C4 , C6, C8 and C10 do not intersect each other.

Hence, 2m1 + 3n1 + m1 n1 = 2 × 6 + 3 × 5 + 6 × 5 = 57 50

1 1 > 2n 1026

+ 2450

⇒

f2 ′(x) = 98 × 50(x − 1)

⇒

f2 ′(x) = 98 × 50(x − 1)48 (x − 1 − 6)

Hence, maximum 5 set of circles do not intersect each other.

⇒

f2 ′(x) = 98 × 50(x − 1) (x − 7)

∴

49

− 600 × 49(x − 1)48

48

–

–∞

–

+

1

So, +∞

7

 2199 − 1  Now, 2 s n − 1 + a n <  198    2

Clearly, m2 = 1 and n2 = 0 So, 6m2 + 4n2 + 8m2n2 = 6 + 0 + 0 = 6 3π / 8 3π / 8  1 − cos 2x  11. (2) sin2 x ⋅1 dx = ∫ S1 = ∫   dx π /8 π /8   2 3π / 8

3π / 8

∫ sin

2

∴

3π / 8

∫π / 8

2 n−2

2

1 −  2n − 2  2 …(i)

π /8

S2 =

2 2−

π 3π 3π π sin2  + − x  4  + − x  − π dx  8   8  8 8

+

1 n −1

2

2⋅ 2n − 2 2198

So, number of circles = 199

15. (c) ψ1 (x) = e − x + x, x ≥ 0

y=|4x–π|

ψ2(x) = x 2 − 2x − 2e − x + 2, x ≥ 0 A1 π/8

O

A2 3π/8

π/4 b

[∫ f (x) dx = a

⇒

S2 =

3π / 8

∫π / 8

f (x) = π/2

g(x) =

… (ii) Put

Adding Eqs. (i) and (ii), we get 2S2 =

x

3π / 8

∫π / 8| 4 x − π | dx

∴

13. (d) Qan = n1− 1 and Sn = 2 1 − 1n   

⇒ ⇒

t ⋅ e − t dt, x > 0

t = z 2 ⇒ dt = 2z dz g(x) =

x

∫0 z ⋅ e

− z2

⋅ 2z dz x

2

1025 1025 ⇒ Sn < 513 513 1  1025 1 1025  21 − n  < ⇒ 1− n <  2  513 2 1026 −1 −1 1 1 1 − n 0

2

x 2

From figure,

2

− t2

0

∫a f (a + b − x) dx]

cos x | π − 4 x | dx

2

= 2 ∫ (t − t 2) e − t dt

b

2

x

∫− x (| t | − t )e

⇒ ∴

f (x) + g(x) = 1 − e − x

2

2

f ( ln 3) + g( ln 3) = 1 − e −(

ln 3) 2

= 1 − e − ln 3 = 1 −

1 2 = 3 3

Hence, option (a) is incorrect. Q ⇒

ψ1 (x) = e − x + x ψ′1 (x) = 1 − e

−x

… (iii) < 1 for x > 1

16 JEE Advanced

Solved Paper 2021

Then, for α ∈(1, x), ψ1 (x) = 1 + αx does not true for α > 1.

⇒

17. (214) Given, set = {1, 2, 3, … , 2000}

So, option (b) is incorrect. Now, ψ2(x) = x 2 − 2x − 2e − x + 2

Let E1 = Event that it is a multiple of 3 = {3, 6, 9, … , 1998}

ψ′2 (x) = 2x − 2 + 2e − x = 2(x + e − x ) − 2

∴

= ψ′2 (x) = 2ψ1 (x) − 2

and E 2 = Event that it is a multiple of 7

[from Eq. (iii)]

n(E1) = 666 = {7, 14, … , 1995}

By LMVT, ψ (x) − ψ2(0) , for β ∈(0, x) ψ′2 (β) = 2 x−0 ⇒

∴

ψ2(x) − 0 = ψ′2 (β) ⇒ ψ2(x) = x ⋅ ψ′2 (β) = 2x (ψ1 (β) − 1) x−0

Hence, option (c) is correct.

16. (d) For option (a) ψ1 (x) = e − x + x ψ′1 (x) = 1 − e − x ⇒ ψ′1 (x) = 0 at x = 0

∴

n(E 2) = 285

E1 ∩ E 2 = multiple of 21 = {21, 42, … , 1995}

Here, ψ1 (x) is always increasing.

n(E1 ∩ E 2) = 95 ∴

P(E1 ∪ E 2) = P(E1) + P(E 2) − P (E1 ∩ E 2) 666 + 285 − 95 856 P(E1 ∪ E 2) = = = p (given) 2000 2000 856 856 Hence, 500p = 500 × = = 214 2000 4

18. (4) As we know that, in a triangle, sides joining the mid-points of two sides is half and parallel to the third side. 1 M1 M 2 = QQ′ ∴ 2

Now, ψ1 (x) > 1 (Q ψ1 (0) = 1)

Y

Thus, option (a) is incorrect. For option (b)

P

ψ2(x) = x 2 − 2x − 2e − x + 2 A'

ψ′2 (x) = 2x − 2 + 2e − x

⇒

ψ′′2(x) = 2 − 2e

⇒

−x

Q

Maximum value of QQ′ is AA′

For Option (c) x

2

x

2

f (x) = 2∫ (t − t 2)e − t dt = (− e − t )0x − 2∫ t 2 ⋅ e − t

2

dt

Hence, maximum value of M1 M 2 =

2 x   t4 = 1 − e − x − 2∫ t 2 1 − t 2 + + K dt 0 2!  

=1 − e

− x2

19. (182) I=

Given,

2x 3 2x 5 2x 7 2x 9 − + − + … 3 5 2⋅ 7 6 ⋅ 9 2x 3 2x 5 1 − < 0 in  0,   2 3 5

2

f (x) − 1 + e − x +

x2

∫0

t ⋅ e − t dt

From Eq. (i), we get

t = u ⇒ dt = 2udu g(x) =

2 −u

∫x 2⋅ u e 3

g(x) =

2

5

du =

x

∫0 7

 u4  2u2 1 − u2 + … du 2!  

I=

1/ 9

∫0

10 x    dx + x + 1 

2/ 3

10 x   dx + + 1 x 

∫1 / 9 

⇒

2x 3 2x 5 2x 7 1 g(x) − + − ≤ 0, ∀x ∈  0,   2 3 5 2⋅ 7

⇒

g(x) ≤

2x 3 2x 5 1 7 1 − + x , ∀ x ∈  0,   2 3 5 7

9

 10 x   dx  x + 1

∫2/ 3 

+

9

2x 2x 2x 2x − + − … 3 5 2⋅ 7 9 ⋅ 6

Hence, option (d) is correct.

… (i)

10 x = 9 ⇒ 10 x = 9 x + 9 ⇒ x = 9 x +1

2

x

10 x    dx  x + 1

2 10 x = 4 ⇒ 10 x = 4 x + 4 ⇒ x = 3 x +1

For option (d) g(x) =

10 

∫0

1 10 x = 1 ⇒ 10 x = x + 1 ⇒ x = 9 x +1

Hence, option (c) is also incorrect.

∴

1 AA′ = 4 2

0

0

Put

Q′

ψ′2 (x) > 0, ∀x > 0

Thus, option (b) is incorrect.

⇒

A X

≥ 0, ∀x > 0

⇒ ψ′2 (x) is increasing ⇒ ψ′2 (x) > ψ′2 (0) ⇒

M2

M1

=

1/ 9

∫0

0 ⋅ dx +

2/ 3

9

10 

∫9

10 x    dx  x + 1

10

∫1 / 91 ⋅ dx + ∫2/ 3 2⋅ dx + ∫9 3⋅ dx

= 0 + [x]12// 39 + [2x]92/ 3 + [3x]10 9 =

2 1 4 2 1 182 − + 18 − + 30 − 27 = 21 − − = 3 9 3 3 9 9

Hence, 9I = 182