Objective mathematics: for engineering entrance examinations [3rd ed] 9788131723630, 8131723631

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The Pearson Guide to

OBJECTIVE MATHEMATICS For Engineering Entrance Examinations

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The Pearson Guide to

OBJECTIVE MATHEMATICS For Engineering Entrance Examinations Third Edition

IIT (Screening Test), AIEEE (CBSE), CEE (Delhi), UPSEAT (UP), CEET (Haryana), PET (MP), GGS Indraprastha University, Jamia Millia Islamia University, AMU, PET (Rajasthan), EAMCET (Andhra Pradesh), BCA/BSc (Hons) Computer Science and other Common Engineering Entrance Examinations in Orissa, Bihar, Tamil Nadu (TNPCEE), Karnataka (CET), Assam and West Bengal (WBJEE), MBA and MCA

J K Sharma

Anita Khattar

Dinesh Khattar

Professor Faculty of Management Studies University of Delhi Delhi

Department of Mathematics Sarvodaya Vidyalaya Delhi

Head Department of Mathematics Kirori Mal College University of Delhi Delhi

Chandigarh • Delhi • Chennai

The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book. Copyright © 2009 Dorling Kindersley (India) Pvt. Ltd. This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise), without the prior written permission of both the copyright owner and the above-mentioned publisher of this book. ISBN: 978-81-317-2363-0 First Impression Published by 2009 Dorling Kindersley (India) Pvt. Ltd., licensees of Pearson Education in South Asia. Head Office: Knowledge Boulevard, a Floor 7th-8(A) Sector 62, Noida, India. Registered Office: 14 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India. Laser typeset by Tantla Compositions Services Pvt. Ltd., Chandigarh Printed in India by Anand Sons

PREFACE to the Third edition I take great pleasure in presenting to the readers, the third revised edition of the book. The third edition of objective Mathematics includes all the basic features of the earlier editions. However, some typical problems have been added in each chapter. These are designed to test the mental, academic and creative skills of the students. Questions from recent examination papers of various engineering entrance examinations have been included in every chapter. Chapters on Mathematical Reasoning and Mathematical Induction have been included in this new edition. In preparing this third edition, I am greatly indebted to many teachers as well as students throughout the country who made constructive criticism and extended valuable suggestions for the improvement of the book. Any suggestions to ensure further improvement of the book will be greatly acknowledged.

Dinesh Khattar

PREFACE The last twenty years of preparing/guiding students to success in different engineering entrance examinations has made us realize that what students really need is a book that imparts the necessary skills to solve questions in the shortest possible time without compromising on the theoretical aspect of the subject. This book introduces the concept briefly and then goes straight into solving real-life problems. The graded problems, techniques to solve them, and huge number of exercises and test papers with solutions will help the student gain the necessary skills and confidence to take the examinations successfully. The book is noteworthy in the following aspects: •

Each chapter contains concise definitions and explanations of basic/fundamental principles and illustrative examples to enable the students to recall the subject matter of the chapter before attempting to answer the questions.

•

Worked out solutions to a large number of problems have been provided. However, we urge students to attempt the same on their own and only if they fail should they refer to the solutions provided.

•

Problems have been categorised into various types, and working rules to help the students in solving them have also been provided.

•

Large number of problems that have been asked in the competitive examinations in the recent past have been included in the text.

•

To enable the students make a self assessment, practice exercises covering all the topics in each chapter have been provided at the end.

•

In order to help the students decide how much emphasis they should give to various topics in different competitive examinations, a smart table has been given in the beginning. This table provides the information about how many questions have been asked from various topics during 2001–04 in IIT(Screening test), AIEEE (CBSE), CEE (Delhi) and UPSEAT.

•

Nine Model Test Papers based on different competitive examinations have been provided at the end to facilitate students understand the pattern and the type of questions asked in different examinations.

The answers to almost all unsolved problems have been thoroughly checked. While every effort has been made to weed out typos, it is possible that a few might have crept in. We will be grateful to the readers for bringing these errors to our notice. It is earnestly hoped that this book will build a strong foundation for success in any competitive examination. We wish to place on record our sincere thanks to our friends and colleagues for their help and suggestions in planning and preparing the manuscript of this book. We would like to thank Mr. Dinesh Kaushik and Mr. Pawan Tyagi for their cooperation in typesetting the book. Suggestions and comments from our esteemed reader to improve the book in content and style are always welcome and will be greatly appreciated and acknowledged. Thank you for choosing our book. May you find it stimulating and rewarding!

Authors

CONTENTS





Preface to Third Edition

v

Preface

vi

  1. Functions   2. Limits   3. Continuity and Differentiability

1 43 75

  4. Differentiation

107

  5. Applications of Derivatives

137



  6. Indefinite Integration

189



  7. Definite Integral and Area

231



  8. Differential Equations

305



  9. Coordinates and Straight Lines

325



10. Pair of Straight Lines

371



11. Circles

388



12. Conic Sections

443



13. Complex Numbers

515



14. Sequences and Series

565



15. Quadratic Equations and Inequations

621



16. Permutations and Combinations

665



17. Binomial Theorem

695



18. Exponential and Logarithmic Series

729



19. Matrices

739



20. Determinants

763



21. Trigonometric Ratios and Identities

793



22. Trigonometric Equations

829



23. Inverse Trigonometric Functions

847



24. Properties and Solutions Triangles

866



25. Heights and Distances

892



26. Probability

904



27. Statistics

949



28. Vectors

969



29. Three Dimensional Geometry

1021



30. Statics

1047



31. Dynamics

1086



32. Set Theory

1113



33. Numerical Methods

1128



34. Linear Programming

1139



35. Hyperbolic Functions

1146





Model Test Papers

1151





Solutions to Model Test Papers

1187

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1

Functions

CHAPTER

Summary of concepts Function or Mapping Let X and Y be any two non-empty sets and there be correspondence or association between the elements of X and Y such that for every element x ∈ X, there exists a unique element y ∈ Y, written as y = f (x). Then we say that f is a mapping or function from X to Y, and is written as



= set of all image points in Y under the map f.



= f (X) = { f (x) : f (x) ∈ Y; x ∈ X}

The set Y is also called the co-domain of f. Clearly f (X) ⊆ Y.

f : X → Y such that y = f (x), x ∈ X, y ∈ Y.

Intervals in R Real Function If f : X → Y be a function from a non-empty set X to another non-empty set Y, where X, Y ⊆ R (set of all real numbers), then we say that f is a real valued function or in short a real function.

Features of a Mapping f : X → Y (i) For each element x ∈ X, there exist is unique element y ∈ Y. (ii) The element y ∈ Y is called the image of x under the mapping f. (iii) If there is an element in X which has more than one image in Y, then f : X ∈ Y is not a function. But distinct elements of X may be associated to the same element of Y. (iv) If there is an element in X which does not have an image in Y, then f : X → Y is not a function. Throughout this chapter a ‘function’ will mean a ‘real function’.

Value of a Function The value of a function y = f (x) at x = a is denoted by f (a). It is obtained by putting x = a in f (x).

Domain and Range of a Function If f : X → Y be a function, then the set X is said to be the domain of f and range of f

The set of all real numbers lying between two given real numbers is called an interval in R. Let a and b be any two real numbers such that a < b, then we define the following types of intervals. Closed Interval  The set of all real numbers x such that a ≤ x ≤ b is called a closed interval and is denoted by [a, b]. That is [a, b] = {x : x ∈ R; a ≤ x ≤ b} Open Interval  The set of all real numbers x such that a < x < b is called an open interval and is denoted by (a, b) or ]a, b[. That is (a, b) = {x : x ∈ R; a < x < b} Semi-Closed or Semi-Open Interval  The intervals [a, b) = {x : x ∈ R; a ≤ x < b} and (a, b] = {x : x ∈  R; a < x ≤ b} are called semi-closed or semi-open intervals. They are also denoted by [a, b[ and ]a, b] respectively. Note that the set R can be thought of as the open interval (– ∞, ∞), so that R = (– ∞, ∞) = {x : x ∈ R; – ∞ < x < ∞}. Also, the infinite intervals in R can be given by (– ∞, a), (a, ∞), (– ∞, a], [a, ∞).

2

types of functions

Objective Mathematics

one-one or Injective function A function f : X → Y is said to be one-one or injective if distinct elements of X have distinct images in Y.

A function f : X → Y is an into function if it is not an onto function.

many-one function A function f : X → Y is said to be many-one if there exists atleast two distinct elements in X whose images are same

1. If X and Y are any two finite sets having m and n elements respectively, where 1 ≤ n ≤ m, then the number of onto functions from X to Y is given by n

∑ (−1)

n−r n

r =1

Rule for checking whether the function f : X → Y is one-one or many-one 1. (a) Consider any two points x, y ∈ X.

Cr r m

2. Any polynomial function f is onto if degree is odd and into if degree of f is even. Rule for checking whether the function f : X → Y is onto or into

(b) Put f (x) = f (y) and solve the equation.

(i) Find the range of the function f.

(c) If we get x = y only, then f is one-one, otherwise it is many-one.

(ii) If range of f = Y, then f is onto, otherwise it is into.

2. If a function is either strictly increasing or strictly decreasing in the whole domain (or equivalently, f ′ (x) > 0 or f ′ (x) < 0, ∨ x ∈ X), then it is one-one, otherwise it is many-one. 3. If any straight line parallel to x-axis intersects the graph of the function atmost at one point, then the function is one-one, otherwise it is many-one (i.e. it intersects the graph of the function in atleast two points).

Bijective function A function f : X → Y is said to be bijective, if f is both one-one and onto.

4. Any continuous function f (x) which has atleast one local maxima or local minima is many-one. 5. All even functions are many-one. 6. All polynomials of even degree defined on R have atleast one local maxima or minima and hence are many one on the domain R. Polynomials of odd degree can be one-one or many-one. Note: If X and Y are any two finite sets having m and n elements respectively, then the number of one-one functions from X to Y would be

 n Pm , if n ≥ m    0, if n < m

= 

onto or Surjective function A function f : X → Y is said to be onto or surjective if every element of Y is the image of some element of X under the map f.

If X and Y are any two finite sets having the same number of elements, say n, then the number of bijective functions from X to Y is n!.

Some Important functions constant function A function f : R → R defined as f (x) = c, ∨ x ∈ R, where c is a constant, is called a constant function. Its domain is R and range is singleton set {c}. The graph of a constant function is a straight line parallel to x-axis when x is the independent variable.

[ x] + [ y ] , if {x} + { y} < 1 (b) [x + y] =  [ x] + [ y ] + 1, if {x} + { y} ≥ 1

Identity Function  The function f : R → R defined as f (x) = x, ∨ x ∈ R, is called the identity function. Its domain is R and range is also R. The graph of the identity function is a straight line passing through origin and inclined at an angle of 45º with x-axis. Modulus Function or Absolute Value Function  The function f : R → R, defined as

where {x} denotes the fractional part of x. (c) n ≤ x < n + 1 ⇔ [x] = n (d) n1 ≤ [x] ≤ n2 ⇒ n1 ≤ x < n2 + 1 (e) x – 1 < [x] ≤ x ( f ) [[x]] = [x] (g) [n + x] = n + [x], where n is any integer

 0 (h) [x] + [– x] =  −1

if x ∈ I . if x ∉ I

Fractional-Part Function  The function f : R →  R defined as f (x) = x – [x] or f (x) = {x}, where {x} denotes the fractional part of x, is called the fractional-part function. Its domain is R and range is [0, 1). The graph of the fractional part function is as shown below:

 x, if x > 0  f (x) = | x | =  0, if x = 0 − x, if x < 0  is called the absolute value function or modulus function. Its domain is R and its range is [0, ∞). The graph of the modulus function is as shown in the adjoining figure. Greatest Integer (Step or Integral) Function  The function f : R → R defined as f (x) = [x] is called the greatest integer function, where  [x] = integral part of x or greatest integer not greater than x or greatest integer less than or equal to x. i.e.   f (x) = n, where n ≤ x < n + 1, n ∈ I (the set of integers).

Note: (a) If x is an integer, then x = [x] ⇒ {x} = 0 ⇒ {[x]} = 0 (b) [{x}] = 0 (c) 0 ≤ {x} < 1

0, if x ∈ integer (d) {x} + {– x} =  1, if x ∉ integer Signum  Function  The function f : R → R defined as,

| x | for x ≠ 0  f (x) =  x  0 for x = 0

Its domain is R and range is I. The graph of the greatest is called the signum function. integer function is as shown below: Its domain is R and range is the set {– 1, 0, 1}. The graph of the signum function is as shown below:

Reciprocal Function  The function f : R\{0} → R defined 1 by f (x) = , is called the reciprocal function. Its domain as x well as range is R\{0}. The graph of the reciprocal function is as shown below:

3

(a)  [x] ≤ x < [x] + 1

Functions



4

Polynomial Function  A function f : R → R, defined by f (x) = a0 + a1 x + a2 x2 + ... + anxn, where n ∈ N and a0, a1, a2, ..., an ∈ R, is called a polynomial function. If an ≠ 0, then n is called the degree of the polynomial. The domain of a polynomial function is R.

Objective Mathematics

p( x) , q( x) where p (x) and q (x) are polynomials over the set of real numbers and q (x) ≠ 0, is called a rational function. Its domain is R\ {x | q (x) = 0}.

Rational  Function  A function of the form f (x) =

Trigonometric Functions Exponential Function  Let a (≠ 1) be a positive real number. Then the function f : R → R, defined by f (x) = ax, is called the exponential function. Its domain is R and range is (0, ∞). The graph of the exponential function is as shown below:



Function • y = sin x

Domain R

Range [– 1, 1]

• y = cos x

R

[– 1, 1]

• y = tan x

  π R\ (2n + 1) n ∈ I 2  

R

• y = cot x

R\{nπ | n ∈ I}

R

• y = sec x • y = cosec x

  π R\ (2n + 1) n ∈ I (– ∞, – 1] ∪ [1, ∞) 2   R\{nπ | n ∈ I}

(– ∞, – 1] ∪ [1, ∞)

Inverse Trigonometric Functions x log a (i)  a = e e , (a > 0). (ii)  a loga x = x, (a > 0, a ≠ 1). x

(iii)  log ab =

log bc , c > 0 and c ≠ 1. log ca

1 (iv)  log b = , provided a ≠ 1 and b ≠ 1. log ba a

Logarithmic Function  Let a (≠ 1) be a positive real number. Then the function f : (0, ∞) → R, defined by f (x) = loga x, is called the logarithmic function. Its domain is (0, ∞) and range is R. The graph of the logarithmic function is as shown below:

Function

Domain

Range  π π  − 2 , 2  [0, π]

• y = sin–1x

– 1 ≤ x ≤ 1

• y = cos–1x

– 1 ≤ x ≤ 1

• y = tan–1x

– ∞ < x < ∞

 π π  − ,  2 2

• y = cot–1x

– ∞ < x < ∞

(0, π)

• y = cosec–1x

(– ∞, – 1] ∪ [1, ∞)

 π   π  − 2 , 0  ∪  0,   2 

• y = sec–1x

(– ∞, – 1] ∪ [1, ∞)

 π π  0, 2  ∪  , π  2  

Operations on Functions Let f and g be two real functions with domain D1 and D2 respectively. Then, (i) The sum function (f + g) is defined by ( f + g) (x) = f (x) + g (x), ∨ x ∈ D1 ∩ D2 The domain of f + g is D1 ∩ D2 (ii) The difference function ( f – g) is defined by (i)  loga a = 1, loga 1 = 0

− ∞, if a > 1 (ii)  loga 0 =  + ∞, if 0 < a < 1 (iii)  loge x is also denoted as: ln x.

( f – g) (x) = f (x) – g (x), ∨ x ∈ D1 ∩ D2 The domain of f – g is D1 ∩ D2 (iii) The product function fg is defined by ( fg) (x) = f (x) ⋅ g (x), ∨ x ∈ D1 ∩ D2 The domain of fg is D1 ∩ D2

f The domain of is D1 ∩ D2 \ {x : g (x) = 0} g (v) The scalar multiple function cf is defined by (cf ) (x) = c ⋅ f (x), ∨ x ∈ D1 The domain of cf is D1

composition of functions Let f and g be two real functions with domain D1 and D2 respectively. If range of f ⊆ domain of g, then composite function is defined (gof ) by (gof ) (x) = g ( f (x)), ∨ x ∈ D1 Also, if range of g ⊆ domain of f, then composite function is defined ( fog) by ( fog) (x) = f (g (x)), ∨ x ∈ D2.

odd function A function f (x) is said to be odd if f (– x) = – f (x) for every real number x in the domain of f. even function A function f (x) is said to be even if f (– x) = f (x) for every real number x in the domain of f. Some Important Results 1. The graph of an odd function is symmetric about origin and it is placed either in the first and third quadrant or in the second and fourth quadrant. 2. The graph of an even function is symmetric about the y-axis. 3. To express a given function f (x) as the sum of an even and odd function, we write f (x) = where

1 1 [ f (x) + f (– x)] + [ f (x) – f (– x)]. 2 2 1 [ f (x) + f (– x)] is an even function and 2

1 [ f (x) – f (– x)] is an odd function. 2 4. f (x) = 0 is the only function which is both even and odd.

Some useful results

5. If f (x) is an odd function, then f ‘ (x) is an even function provided f (x) is differentiable on R.

Let f : X → Y and g : Y → Z.

6. If f (x) is an even function, then f ‘ (x) is an odd function provided f (x) is differentiable on R.

(a) If both f and g are one-one, then so is gof. (b) If both f and g are onto, then is also onto gof. (c) If gof is one-one, then f is one-one but g may not be oneone. (d) If gof is onto, then g is onto but f may not be onto. (e) If f and g are bijective, then is also bijective gof. (f) It may happen that gof may exist and fog may not exist. Moreover, even if both gof and fog exist, they may not be equal.

7. If f and g are even functions, then fog is also an even function, provided fog is defined. 8. If f and g are odd functions, then fog is also an odd function, provided fog is defined. 9. If f is an even function and g is an odd function, then fog is an even function. 10. If f is an odd function and g is an even function, then fog is an even function. 11. For a real domain, even functions are not one-one.

Inverse functions If the function f : X → Y is both one-one and onto, then we define inverse function f –1 : Y → X by the rule y = f (x) ⇔ f –1 (y) = x, ∨ x ∈ X, ∨ y ∈ Y as shown in the figure below:

even and odd extensions A function f (x) defined on the interval [0, a] can be extended to [– a, a], so that f (x) becomes an even or odd function on the interval [– a, a]. If this extension is an even function, it is called even extension and if this extension is an odd function, it is called odd extension. Let g be the extension. Then for even extension, we define

 f ( x) , if x ∈[0, a ] g (x) =   f (− x), if x ∈ [− a, 0] and for odd extension, we define Rule to find the inverse of a function Let f : X → Y be a bijective function. • Put f (x) = y. • Solve the equation y = f (x) to obtain x in terms of y. Interchange x and y to obtain the inverse of f.

 f ( x) , if x ∈[0, a ] g (x) =  − f (− x), if x ∈ [− a, 0].

periodic function A function f (x) is said to be a periodic function of x, provided there exists a real number T > 0 such that f (x + T) = f (x), ∨ x ∈ R.

5

odd and even functions

Functions

f (iv) The quotient function   is defined by g f f ( x)  g  (x) = g ( x) , ∨ x ∈ D1 ∩ D2 \ {x : g (x) = 0}

6

The smallest positive real number T, satisfying the above condition is known as the period or the fundamental period of f (x).

Trigonometric Functions (i) sin x and cos x are defined for all real values of x.

Objective Mathematics

(ii) tan x and sec x are defined for all real values of x except

Rule for testing the periodicity of a function 1. Put f (T + x) = f (x) (x) and solve this equation to find the positive (x values of T independent of x. 2. If no positive value of T independent of x is obtained, then f (x ((x) x) is a non-periodic function. 3. If positive values of T independent of x are obtained, then f (x) (x) is (x a periodic function and the least positive value of T is the period of the function f (x). (x). (x

x = (2n + 1)

π , where n ∈ I. 2

(iii) cot x and cosec x are defined for all real values of x except x = nπ, where n ∈ I. Inverse Trigonometric Functions (i) sin–1x and cos–1x are defined for – 1 ≤ x ≤ 1. (ii) tan–1x and cot–1x are defined for all real values of x. (iii) sec–1x and cosec–1x are defined for x ≤ – 1 or x ≥ 1. Logarithmic Functions logb a is defined when a > 0, b > 0 and b ≠ 1.

Hints for Solving Problems on Periodic Functions 1. Constant function is periodic with no fundamental period. 2. If f (x ((x)) is periodic with period T, then also periodic with same period T.

1 and f ( x)

ax is defined for all real values of x, where a > 0.

f ( x) are

3. If f (x) (x) is periodic with period T1 and g (x (x ((x)) is periodic with period T2, then f (x ((x) x) + g (x ((x)) is periodic with period equal to l.c.m of T1 and T2, provided there is no positive k such that f ((kk + x) = g (x) ( ) (x and g (kk + x) = f (x ((x). x). 4. If f (x ((x) x) is periodic with period T, then kf ((ax ax + b), is also periodic with period

Exponential Functions

T , where a, b, k ∈ R and a, k ≠ 0. | a|

5. sin x, cos x, sec x and cosec x are periodic functions with period 2π.

Rules for solving problems on domain of a function 1. (x ( – a)) ((x – b) > 0 ⇒ x < a or x > b, for a < b 2. (x ( – a)) ((x – b) < 0 ⇒ a < x < b, for a < b 3. | x | < a ⇒ – a < x < a 4. | x | > a ⇒ x < – a or x > a

a > b k , if b > 1 k a < b , if b < 1

5. logb a > k ⇒ 

6. tan x and cot x are periodic functions with period π.

6.

7. | sin x |, | cos x |, | tan x |, | cot x |, | sec x | and | cosec x | are periodic functions with period π.

7.

8. sinn x, cosn x, secn x and cosecn x are periodic functions with period 2π when n is odd or π when n is even. 9. tann x and cotn x are periodic functions with period π. 10. If f (x) (x) is a periodic function with period T and g (x) (x ( ) is any func(x tion such that domain of f ⊂ domain g, then gof is also periodic with period T.

Rules for Finding the Domain of a Function Algebraic Functions (i) Denominator should be non-zero. (ii) Expression under the square root should be non-negative.

x2 = | x | n

x n = | x |, if n is even and

n

x n = x, if n is odd.

Rules for finding the range of a function y = f (x) 1. Find the domain of the function y = f (x ((x). x). 2. If the domain is an infinite interval, solve the equation y = f (x ((x) x) and find x in terms of y to get x = g (y ((y). ). Find the real values of y for which x is real. The set of values of y so obtained constitutes the range of ff. Note that if finite number of values of x are excluded from the domain, find the values of y for these values of x and exclude these values of y from the range of f found earlier. 3. If the domain is a finite interval, find the least and greatest value of y for values of x in the domain. If a is the least value and b the greatest value of y, then range ( f ) = [[a, a, b].

muLtIpLe-cHoIce QueStIonS choose the correct alternative in each of the following problems: 1. Domain of f (x) = 1 (a)  , 4 1 (c)  , 4

1 2  1 3 

sin −1 ( 2 x ) + π/6 is  −1 1  (b)  ,   4 2  1 1 (d)  − ,   4 3

2. The domain of definition of the function y=

 5x − x2  is log10   4 

(a) [1, 4] (c) [1, 4)

(b) (1, 4) (d) (1, 4]

(a) [– 2, – 1) ∪ [1, 2] (c) [– 2, – 1] ∪ [1, 2]

x − 1− x

is

(a) [1, ∞) (c) (1, 5)

9 − x2 is sin (3 − x)

1− x

(b) [2, 3) (d) None of these

15. If f (x) =

1− x

(d) None of these

7. The domain of the function f (x) = (a) (– ∞, – 2) ∪ [4, ∞) (c) (– ∞, – 2) ∪ (4, ∞)

1 [ x]2 − [ x] − 6

(b) (– ∞, – 2] ∪ [4, ∞) (d) None of these

(a) [– 1, 3] (c) [– 1, 3)

 3 − 2x  3 − x + cos   5  −1

(b) (– 1, 3] (d) None of these

x2 + 1 , ([.] denotes the greatest integer func[ x] tion), 1 ≤ x < 4, then

10. If  f (x) =

 17  (a) range of f is  2,   3 (b) f is monotonically increasing in [1, 4] 17 (c) the maximum value of f (x) is 3 17 (d) the maximum value of f (x) is 4 11. The domain of the function 1 is f (x) = 12 9 x − x + x4 − x + 1

7

π  (a) 2nπ +  2 n ∈1  

∪  2nπ +

7π 11π  , 2 nπ +  6 6 

7π   (c) 2nπ +  6 n ∈1  7π 11 π   , 2 nπ + (d)  2nπ +  ∪  6 6  n,m ∈I

π   2mπ +  2

16. The domain of the function x−5 f (x) = log10 2 − 3 x + 5 is x − 10 x + 24

(b) [– 2, 1[ (d) None of these

9. The domain of the function f (x) = is

3| x | − x − 2 and g (x) = sin x, then domain of

n ∈1

is

(b) R – {1, – 1} (d) None of these

definition of f o g (x) is

(b)

8. The domain of definition of the function 1 y = log (1 − x) + x + 2 is 10 (a) [– 2, 1] (c) [– 2, 0[ ∪ ]0, 1[

(b) (– ∞, 5) (d) [1, 5]

(a) {1} (c) x > 3, x ∉ 1

−1

6. Let f : (–∞, 1] → (–∞, 1] such that f (x) = x (2 – x). Then f –1(x) is

(c)

x − 1 + 5 − x is

 1 + x3  2 f (x) = sin − 1  3 / 2  + sin(sin x) + log (3 {x} + 1)(x + 1), 2 x   where { } represents fractional part function, is

5. The domain of the function f (x) =

(b) 1 –

(b) (– 1, 2) ∪ [3, ∞) (d) None of these

14. The domain of the function,

 1  , 1 (d)   2 

1− x

(a) [– 1, 2) ∪ [3, ∞) (c) [– 1, 2] ∪ [3, ∞)

13. The domain of the function f (x) =

 1  1  , + ∞ (c)  −∞, −  ∪   2  2

(a) 1 +

( x + 1)( x − 3) is ( x − 2)

given by 2

1   1   ,1 ∪ (a)  −1, − 2   2   (b) [– 1, 1]

(a) (2, 3) (c) (2, 3]

(b) (1, ∞) (d) (– ∞, ∞)

12. The domain of the function f (x) =

(b) (– 2, – 1] ∪ [1, 2] (d) (– 2, – 1) ∪ (1, 2)

4. The domain of the function f (x) =

(a) (– ∞, – 1) (c) (– 1, 1)

(a) (4, 5) (c) (4, 5) ∪ (6, ∞)

(b) (6, ∞) (d) (4, 5] ∪ (6, ∞)

17. The domain of the function f (x) = (a) (0, ∞) (c) (– ∞, ∞)

1 | x | − x is

(b) (– ∞, 0) (d) None of these

18. The domain of the function f (x) =  3 (a)  0,  2

(b) (0, 3)

 3 (c)  −∞,  2

 3 (d)  0,  2

log10

3− x is x

19. The domain of the function f (x) = log10 | 4 – x2 | is (a) (– ∞, ∞)\{– 2, 2} (b) (0, ∞) (c) (– ∞, 0) (d) None of these

Functions

  1  3. The domain of the function f (x) = sin –1 log 2  x 2    2   is

8

20. The domain of the function f (x) = is

Objective Mathematics

(a) (– ∞, 1) (c) [0, 1]

1 − 1 − 1 − x2

(a) (4, 6) (c) [4, 6)

(b) (– 1, ∞) (d) [– 1, 1]

1 −| x| is 2 −| x|

21. The domain of definition of f (x) = (a) (– ∞, ∞)\[– 1, 1] (b) (– ∞, ∞)\[– 2, 2] (c) [– 1, 1] ∪ (– ∞, – 2) ∪ (2, ∞) (d) None of these

(b) (2, ∞) (d) φ

2 −| x|  f (x) = cos  + [log (3 − x)]−1 is  4  (b) [– 6, 2) ∪ (2, 3] (d) [– 6, 3)

24. The domain of the function f (x) = log 2 log3 log4 x is (b) (4, ∞) (d) None of these

25. The domain of the function f (x) = sin

–1

 x − 3   – log10 (4 – x) is 2 

(a) (1, 4) (c) [1, 4)

(b) [1, 4] (d) (1, 4]

26. The domain of the derivative of the function f (x) = tan–1x, |x| ≤ 1 1 (|x| – 1), |x| > 1 is 2 (a) R – {0} (c) R –­{–1}

(b) R – {1} (d) R – {–1, 1}

27. The domain of the function f (x) = (a) (– ∞, – 3] ∪ (2, 5) (c) (– ∞, – 3] ∪ [2, 5]

x+3 is (2 − x)( x − 5)

(b) (– ∞, – 3) ∪ (2, 5) (d) None of these

3   28. The domain of the function f (x) = cos –1   4 + 2 sin x  is π  π  (a)  − + 2nπ, + 2nπ  6  6  π  π  (b)  − + 2nπ, + 2nπ  6 6

 π π  (c)  − + 2nπ, + 2nπ  6 6   π π  (d)  − + 2nπ, + 2nπ  6  6

f (x) = log10 [1 – log10 (x2 – 5x + 16)] is (a) (2, 3) (c) (2, 3]

1 log [1 − | x |]

(c) x ! {x}

−1

(a) [4, ∞) (c) (– ∞, 4)

30. The domain of the function

(a)

23. The domain of the function

(a) [– 6, 3)\{2} (c) [– 6, 3]

(b) [4, 6] (d) None of these

(b) [2, 3] (d) [2, 3)

31. Which of the following is a function ([.] denotes the greatest integer function, {.} denotes the fractional part function]?

22. The domain of the function −1 1 1 + 2sin x + f (x) = is 1− x x−2 (a) (– ∞, ∞)\{1} (c) [– 1, 1)

29. The domain of the function f (x) = log10 ( x − 4 + 6 − x ) is

x! {x}

(b)

log ( x − 1)

(d)

1 − x2

32. The domain of the function f (x) = log10 sin (x – 3) + (a) (3, 4) (c) (3, π + 3)

16 − x 2 is

(b) (– 4, 4) (d) None of these

33. The domain of the function f (x) = log x cos x is  π π (a)  − ,  \{1} 2 2

 π π (b)  − ,  \{1}  2 2

 π π (c)  − ,  2 2

(d) None of these

4   34. The domain of the function f (x) = sin –1    3 + 2 cos x  is

π π ≤ x ≤ 2nπ + , n ∈ I 6 6 π (b) 0 ≤ x ≤ 2nπ + , n ∈ I 6 π π < x < 2nπ + , n ∈ I (c) 2nπ – 6 6 π ≤ x ≤ 0, n ∈ I (d) 2nπ – 6

(a) 2nπ –

 1   (where 35. Range of the function f defined by f (x) =   sin{x}  [.] and {.} respectively denote the greatest integer and the fractional part functions) is (a) I, the set of integers (b) N, the set of natural numbers (c) W, the set of whole numbers (d) {2, 3, 4, ...} 36. The domain of the function f (x) = tan–1 x ( x + 1) + sin −1 x 2 + x + 1 is (a) [– 1, 0] (b) {– 1, 0} (c) (– ∞, – 1] ∪ [0, ∞) (d) (– ∞, ∞)

3

46. The domain of the function f (x) = cos –1 (x + [x] ), where [⋅] denotes the greatest integer function, is

38. The domain of the function

(a) [– 1, 1] (c) (– 1, 0)

1 f (x) = log 1  x −  + log 2 4 x 2 − 4 x + 5 is 2  2

n times

(d) None of these

39. The domain of the function f (x) = x − [ x] , where [x] denotes the greatest integer less than or equal to x, is 2

(a) (0, ∞) (c) (– ∞, ∞)

2

(b) (– ∞, 0) (d) None of these

40. The domain of the function f (x) =

1 is | sin x | + sin x

(b) (2nπ, (2n + 1) π) π π (c)  (4n − 1) , (4n + 1)   2 2 (d) None of these 2 f ( n) + 1 , n = 1, 2, .... and f (1) = 2, then 2

(a) 52 (c) 48

(b) 49 (d) 51

f (x) =

C3x – 1 +

40 – 6x

(a) {2, 3} (c) {1, 2, 3, 4}

C8x – 10 is,

(a) (– ∞, – 3) ∪ (3, ∞) (c) (– ∞, – 3] ∪ [3, ∞)

1 −| x |  is cos −1   2 

(b) [– 3, 3] (d) φ

44. The domain of the function f (x) =

e

sin −1 (log16 x 2 )

1 (a)  , 4  4 

1 1   (b)  −4, −  ∪  , 4 4 4  

1  (c)  −4, −  4 

(d) None of these

45. The domain of the function f (x) = log 

1 x+ 2  

x 2 − 5 x + 6 is

 (a)  3 , 2  ∪ (2, 3) ∪ (3, ∞) 2  3  (b)  , ∞  2 

48. The domain of the function

x − 3 − 2 x − 4 − x − 3 + 2 x − 4 is

f (x) =

(b) (– ∞, 4] (d) (– ∞, 4)

49. The number of solutions of the equation a f (x) + g (x) = 1 0, a > 0, g (x) ≠ 0 and has minimum value is 2 (a) One (b) Two (c) Zero (d) Intinitely many

  16 − x 2 50. The domain of the function f (x) = cos log    3− x is (a) (– 4, 4)

(b) (– 4, 3)

(c) (– ∞, – 4) ∪ (3, ∞)

(d) None of these

f (x) =

(b) {1, 2, 3} (d) None of these

43. The domain of the function f (x) =

(b) [2n, ∞) (d) None of these

   

51. The domain of the function

42. The domain of the function 24 – x

(a) (2n – 1, ∞) (c) (2n – 2, ∞)

(a) [4, ∞) (c) (4, ∞)

(a) (– 2nπ, 2nπ)

41. If f (n + 1) = f (101) equals

(b) [0, 1) (d) None of these

47. The domain of the function log 2 log 2 log 2 ...log 2 x  is f (x) = 

1  (b)  , ∞  2 

1  (a)  , ∞  2  (c) (– ∞, ∞)

9

1  (c)  , ∞  2  (d) None of these

 2 x − 1  3 tan x is +e 1 − 3 x + 3 cos −1   3  (a) [– 1, 2] (b) (– 1, 2) (c) (– ∞, ∞) (d) None of these

f (x) =

3 + 2 ( x + 25) − 0.5 + (x – 5)0.5 + 5 (x – 3)0.5 is 5 − ( x + 25)0.5

(a) [5, ∞) (c) (– 25, 5)

(b) [3, ∞) (d) None of these

52. Let f : (4, 6) → (6, 8) be a function defined by f (x) =  x x +   (where [ . ] denotes the greatest integer func2 tion), then f– 1 (x) is equal to

is

 x (a) x −   2

(b) – x – 2

(c) x – 2

(d)

1  x x+  2

53. The domain of the function f (x) = (a) (– ∞, ∞) (b) (– ∞, ∞)\{nπ | n ∈ I}

  π (c) (– ∞, ∞)\ (2n + 1) n ∈ I 2   (d) None of these

4

 1  log 3   is  cos x 

Functions

37. The domain of the function

10

54. The domain of the function

63. The range of the function f (x) = loge (3x2 – 4x + 5) is

Objective Mathematics

 1    f (x) = log3   − log 1 1 + 1 5  − 1 is    2 x 

 11 (a)  −∞, log e   3

 11  (b) log e , ∞   3 

11 11 (c)  − log e , log e  (d) None of these  3 3  π2 55. If f : R → R, g : R → R be two given functions then − x 2 is 64. The range of the function y = 3 sin f (x) = 2 min { f (x) – g (x), 0} equals 16 (a) (– ∞, 1) (c) (1, ∞)

(b) (0, 1) (d) None of these

(a) f (x) + g (x) – | g (x) – f (x) | (b) f (x) + g (x) + | g (x) – f (x) | (c) f (x) – g (x) + |g (x) – f (x) | (d) f (x) – g (x) – | g (x) – f (x) | 56. Let f (x) be a function defined on [0, 1] such that

x f (x) =  1 − x,

(b) 1 + x (d) None of these

57. The domain of the function

58. The domain of the function f (x) = x

1 log x

is

(b) (0, ∞) (d) [0, ∞)\{1}

59. The domain of the function f (x) =

4 − x2 [ x] + 2 , where [x]

denotes the greatest integer less than or equal to x, is (a) [– 1, 2] (b) (– ∞, – 2) (c) (– ∞, – 2) ∪ [– 1, 2] (d) None of these 60. The range of the function y =  (a) 0, 

1 2 

 1  (c)  − , 0  2 

x is 1 + x2

 1 (b)  − ,  2

1 2 

(d) None of these

x2 is 1 + x2 (b) [0, 1] (d) None of these

61. The range of the function y = (a) [0, 1[ (c) ]0, 1[

1 is 62. The range of the function y = 2 − sin 3 x 1  (a)  , 1 3 1  (c)  , 1 3 

(d) None of these

1  (b)  , 1 3

 11  (a)  −∞, 3   

 11  (b)  −∞, 3  

 11  (c)  , ∞   3

 11  (d)  , ∞  3 

1  (a) (– ∞, ∞)\  , 1 5 

(b) (– ∞, ∞)

(c) (– ∞, ∞)\{1}

(d) None of these

 x2  67. The range of the function y = sin    is  1 + x 2  –1

 π (a)  0,  2

 π (b) 0,   2

 π (c) 0,   2

(d) None of these

 π 68. If A =  x : ≤ x ≤  6 f (A) is equal to

π  and f (x) = cos x – x (1 + x), then 3

 1 π π2 3 π π2  (a)  2 − 3 − 9 , 2 − 6 − 36     1 π π2 3 π π2  (b)  2 + 3 − 9 , 2 + 6 − 36     1 π π2 3 π π2  (c)  2 − 3 − 9 , 2 − 6 − 36  (d) None of these 69. The range of the function f (x) = tan (a) [0, (c) [0,

3 ] 3 )

(b) (0, (d) (0,

π2 − x 2 is 9

3)

3] 70. Period of the function log (sin (x – [x]) ([ . ] denotes the greatest integer function) is –1 

(d) None of these

3 x 2 − 4 x + 5 is

x 2 − 3x + 2 66. The value of the function f (x) = 2 lies in the x + x−6 interval

f (x) = log3 [– (log3 x)2 + 5 log3 x – 6] is (a) (0, 9) ∪ (27, ∞) (b) [9, 27] (c) (9, 27) (d) None of these

(a) (0, ∞)\{1} (c) [0, ∞)

3 3  (b)  − , 2   2

65. The range of the function f (x) =

x ∈Q x ∉Q

Then for all x ∈ [0, 1], fof (x) is (a) a constant (c) x

3  (a) 0, 2   3  (c)  − , 0  2 

71. The range of the function f (x) = sin x – cos x is

2 , 2 ) (c) [0, 2 ] (a) (–

(b) [–

2, 2] (d) None of these

72. The range of the function f (x) = (a) R\{0} (c) {– 1, 1}

x is | x|

(b) R\{– 1, 1} (d) None of these

81. The period of the function f (x) = sin4 2x + cos 4 2x is π 2 π (c) 4

(a)

(b)

π 8

(d) None of these

82. The period of the function f (x) =

tan x is

(a) π π (c) 2

(b) 2π

(a) 1 ≤ n < 2 (c) 1 ≤ n ≤ 2

(b) 1 < n < 2 (d) None of these

(d) None of these

 8x + 5  is 73. If f : R → R is a function satisfying the property 83. The period of the function f (x) = cos   4 π  f (2x + 3) + f (2x + 7) = 2, x ∈ R, then the period (a) 2π (b) π of f (x) is (d) None of these (c) π2 (a) 2 (b) 4 84. The period of the function f (x) = x – [x], where [x] de(c) 8 (d) 12 notes the greatest integer less than or equal to x, is   4 − x2  (a) 2 (b) 1 74. The range of the function f (x) = sin log   (c) 4 (d) None of these 1 − x     is   85. If T1 is the period of the function y = e 3(x – [x]) and T2 is (a) [0, 1] (b) (– 1, 0) the period of the function y = e 3x – [3x] ([.] denotes the (c) [– 1, 1] (d) (– 1, 1) greatest integer function), then 75. Suppose f (x) = (x + 1)2 for x ≥ –1, If g(x) is the function T whose graph is the reflection of the graph of f (x) with (b) T1 = 2 (a) T1 = T2 3 respect to the line y = x, then g(x) equals (c) T = 3T (d) None of these 1 2 1 x f (x) (b) x + 1 2 , x > −1 (a) − x − 1 , x ≥ 0 86. If e + e = e, then range of the function of f is ( ) (a) (– ∞, 1] (b) (– ∞, 1) (c) x + 1, x ≥ –1 (d) x − 1 , x ≥ 0 (c) (1, ∞) (d) [1, ∞) 76. Let the function f : R → R be defined by 87. If the period of the function f (x) = sin ( [n] x), where f (x) = 2x + sin x, [n] denotes the greatest integer less than or equal to x ∈ R. Then f  is n, is 2π, then (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 77. The period of the function f (x) = a sin kx + b cos kx is

2π (a) k π (c) | k |

2π (b) | k | (d) None of these

88. Let f (x) = (– 1)[x] (where [ . ] denotes the greatest integer function), then (a) Range of f is {– 1, 1} (b) f is an even function (c) f is an odd function (d) lim f (x) exists, for every integer n x→n

78. The period of the function f (x) = cos x2 is

89. The period of the function

(a) 2π (b) π π (c) (d) None of these 2 79. The period of the function f (x) = | sin 4x | + | cos 4x | is π π (b) (a) 8 2 π (c) (d) None of these 4 80. The period of the function f (x) = sin x is

 πx   πx  is f (x) = cos   − sin   n!   (n + 1)! 

(a) π π (c) 2

(b) 2π (d) None of these

(a) 2 (n + 1)! (c) (n + 1)

(b) 2 (n!) (d) not periodic

90. The function f (x) = k | cos x | + k2 | sin x | + φ (k) has period π if k is equal to 2 (a) 1 (c) 3

(b) 2 (d) None of these

91. The period of the function 3(sin

2 πx + x − [ x ] + sin 4 πx )

denotes the greatest integer function, is

, where [⋅]

11

(b) 2π (d) None of these

Functions

(a) 1 (c) π/2

12

Objective Mathematics

1 2 (c) 2

(a)

101. The period of the function f (x) = sin4 x + cos 4 x is π (a) π (b) 2 (c) 2π (d) None of these

(b) 1 (d) non periodic.

92. The period sin θ is 2

(a) π2 (c) 2π

102. Which of the following functions has period π ?

(b) π (c) π/2

|sin x | − |cos x | 93. The period of the function f (x) = |sin x + cos x | is π 2 (c) π

(a)

(b) 2π (d) None of these

94. The domain of the function f (x) = log 

1 x+ 2  

2 πx   πx  (a) 2 cos  + 3 sin    3   3  (b) | tan x | + cos 2x π π (c) 4 cos  2 πx +  + 2 sin  πx +    2 4 (d) None of these

| x – x – 6 |, 2

103. The function f (x) =

where [⋅] denotes the greatest integer function, is 3  (a)  , 3  ∪ (3, ∞) 2

3  (b)  , 3  ∪ (3, ∞) 2

3  (c)  , ∞  2

(d) None of these

95. If f (x) is defined on (0, 1), then the domain of definition of f (e x) + f (ln | x | ) is (a) (– e, – 1) (c) (– ∞, – 1) ∪ (1, ∞)

(b) (– e, – 1) ∪ (1, e) (d) (– e, e)

(a) [1, 4] (c) [0, 5]

 5x − x2  exists for log10   4  (b) [1, 0] (d) [5, 0]

104. If f (x) is defined on (0, 1), then the domain of definition of f (sin x) is (a) (2nπ, (2n + 1) π), n ∈ I π π  (b)  (2n + 1) , (2n + 3)  , n ∈ I 2 2 (c) ((n – 1) π, (n + 1) π), n ∈ I

(d) None of these 96. The period of the function f (x) = 2 sin x + 3 cos 2x 105. The period of the function f (x) = x [x] is is (a) 1 (b) 2 (a) π (b) 2π (c) non periodic (d) None of these. π (c) (d) None of these 106. If the period of the function f (x) = tan ( [k ] x), where 2 [⋅] denotes the greatest integer function, is π, then 97. The period of the function

1, when x is a rational f (x) = 0, when x is irrational is   (a) 1 (b) 2 (c) non-periodic (d) None of these 98. The domain of sin –1 [log3 (x/3)] is (a) [1, 9] (c) [– 9, 1]

(b) [– 1, 9] (d) [– 9, – 1]

99. The period of the function 2 πx πx + sin f (x) = cos is 5 4 (a) 5 (b) 8 (c) 12 (d) 40

(a) 1 < k < 2 (c) k = 1, 2

(b) 1 ≤ k < 2 (d) None of these

107. The period of the function f (x) = sin 5x + cos 3 x is (a) 3 π (c) non periodic

(b) π (d) None of these

108. The period of the function

πx , 2 where [x] denotes the greatest integer ≤ x, is (a) 4 (b) 1 (c) 2 (d) non-periodic f (x) = 3x + 3 – [3x + 3] + sin

109. The value of n ∈ I for which the function sin nx f (x) = has 4π as its period is x f (x + k) = 1 + [2 – 5 f (x) + 10 { f (x)}2 – 10 { f (x)}3 sin   n + 5 { f (x)}4 – { f (x)}5]1/5 (a) 2 (b) 3 for all real x and some positive constant k, then the period (c) 4 (d) 5 of the function f (x) is 110. π is the period of the function (a) k (a) | sin x | + | sin x | (b) sin4x + cos4x (b) 2k 1 + 2 cos x (c) non periodic (c) sin (sin x) + sin (cos x) (d) sin x (2 + sec x) (d) None of these 100. Let f be a real valued function with domain R satisfying

cos 2 x sin 2 x − 1 + tan x 1 + cot x 112. The period of the function (d) f (x) = 1 –

f (x) =

π 2 (d) None of these

sin x + sin 2 x + sin 4 x + sin 5 x is cos x + cos 2 x + cos 4 x + cos 5 x

π 3 (c) π

(a)

π 4 (d) None of these

115. A function f from the set of natural numbers to integers defined by n − 1  2 , where n is odd is f (n) =  − n , where n is even  2 (a) one-one but not onto (b) onto but not one-one (c) one-one and onto both (d) neither one-one nor onto

(a) an even function (b) an odd function (c) a periodic function (d) neither an even nor an odd function 122. A function whose graph is symmetrical about the y-axis is given by (a) f (x) = sin [log (x +

sec 4 x + cosec 4 x x 3 + x 4 cot x (c) f (x + y) = f (x) + f (y) ∨ x, y ∈ R (d) None of these (b) f (x) =

(a) f (x) = (3x + 3– x) (b) f (x) = cos [log (x + 1 + x 2 )] (c) f (x + y) = f (x) + f (y) ∨ x, y ∈ R (d) None of these

n

∑ f (r )

is

r =1

7n (a) 2

7(n + 1) (b) 2 7 n(n + 1) (c) 7n (n + 1) (d) 2 117. If f is an even function defined on the interval [– 5, 5], then the real values of x satisfying the equation  x +1  f (x) = f   are x + 2

124. Let f (x) = sin x + cos x, g(x) = x2 – 1. Then g( f (x)) is invertible in the domain  π  (a)  − , 0  2 

 π  (b)  − , π   2 

 π π (c)  − ,   4 4

 π (d) 0,   2

125. If f : R → S, defined by f (x) = sin x – onto, then the interval of S is (a) [0, 1] (c) [0, 3]

(b)

x 2 + 1 )]

123. A function whose graph is symmetrical about the origin is given by

116. If f : R → R satisfies f (x + y) = f (x) + f ( y),

−1 ± 5 2 −2 ± 5 (c) 2

3 + log10 ( x 3 − x) , is 4 − x2

121. The function f (x) = log( x + x 2 + 1) , is

114. Let f be a real valued function with domain R satisfy1 ing 0 ≤ f (x) ≤ and for some fixed a > 0, 2 1 f (x + a) = − f ( x) − ( f ( x)) 2 ∨ x ∈ R, 2 then the period of the function f (x) is (a) a (b) 2a (c) non-periodic. (d) None of these

(a)

119. Domain of definition of the function:

(a) 3x2 + 4x + 8 log (1 + | x | ) (b) 3x2 – 4x + 8 log (1 + | x | ) (c) 3x2 + 4x – 8 log (1 + | x | ) (d) None of these

(b)

for all x, y, ∈ R and f (1) = 7, then

(b) odd (d) None of these

120. Let the function f (x) = 3x2 – 4x + 8 log (1 + | x | ) be defined on the interval [0, 1]. The even extension of f (x) to the interval [– 1, 1] is

113. The period of the function f (x) =

1 + x 2 )] is

(a) (1, 2) (b) (– 1, 0) ∪ (1, 2) (c) (1, 2) ∪ (2, ∞) (d) (– 1, 0) ∪ (1, 2) ∪ (2, ∞)

(b)

(c) 2π

(a) even (c) constant

f (x) =

sin 8 x cos x − sin 6 x cos 3 x is cos 2 x cos x − sin 3 x sin 4 x

(a) π

118. The function f (x) = sec [log (x +

−3 ± 5 2

(d) None of these

3 cos x + 1 , is

(b) [– 1, 1] (d) [– 1, 3]

126. The graph of the function y = f (x) is symmetrical about the line x = 2, then (a) f (x) = f (–x) (c) f (x + 2) = f (x – 2)

(b) f (2 + x) = f (2 – x) (d) f (x) = – f (– x)

13

(a) f (x) = tan (3x – 2) (b) f (x) = {x}, the fractional part of the number x (c) f (x) = x + cos x

Functions

111. Which of the following functions is non-periodic ?

14

127. The function f : R → R, defined by f (x) = [x], ∨ x ∈ R, is

Objective Mathematics

(a) one-one (b) onto (c) Both one-one and onto (d) neither one-one nor onto

138. The function f : R → R, defined by

128. Let f : R → R be a function defined by f (x) = sin (2x – 3), then f is (a) injective (c) bijective

(b) surjective (d) None of these

129. Which of the following functions from I to itself are bijections? (a) f (x) = x + 3 (c) f (x) = 3x + 2

(b) f (x) = x5 (d) f (x) = x2 + x

130. Which of the following functions (is) are injective map(s)? (a) f (x) = x2 + 2, x ∈ (– ∞, ∞) (b) f (x) = | x + 2 |, x ∈ [– 2, ∞) (c) f (x) = (x – 4) (x – 5), x ∈ (– ∞, ∞) 4 x 2 + 3x − 5 , x ∈ (– ∞, ∞) (d) f (x) = 4 + 3x − 5 x 2 131. Let A = {x ∈ R | – 1 ≤ x ≤ 1} = B. Then the mapping f : A → B given by f (x) = x | x | is (a) Injective but not surjective (b) Surjective but not injective (c) Bijective (d) None of these 132. If the function f : (– ∞, ∞) → B defined by f (x) = – x2 + 6x – 8 is bijective, then B = (a) [1, ∞) (b) (– ∞, 1] (c) (– ∞, ∞) (d) None of these 133. Let f : R → R be defined by f (x) = Then f is

134. The number of bijective functions from a set A to itself when A contains 106 elements is (b) (106)2 (d) 2106

135. The number of surjections from A = {1, 2, ..., n}, n ≥ 2 onto B = {a, b} is (a) nP2 (c) 2n – 1

(b) 2n – 2 (d) None of these

136. Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A to B is (a) 144 (c) 24

f (x) = x – [x], ∨ x ∈ R is (a) one-one (b) onto (c) Both one-one and onto (d) neither one-one nor onto 139. Let f (x) = [x]2 + [x + 1] – 3, where [⋅] denotes the greatest integer function, then (a) f (x) ≠ 0 for all real x (b) f (x) = 0 for only two real values (c) f (x) = 0 for infinite number of values of x (d) None of these 140. The function f : R → R defined by f (x) = 4x + 4| x | is (a) one-one and into (c) one-one and onto

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

141. Let f : R → R be a function defined by x2 − 8 f (x) = 2 , then f is x +2 (a) one-one but not onto (b) one-one and onto (c) onto but not one-one (d) neither one-one nor onto 142. Let f : R → R be a function defined by f (x) = x + x 2 , then f is (a) injective (c) bijective

(b) surjective (d) None of these

x−a , where a ≠ b. 143. Let f : (– ∞, 2] → (– ∞, 2] be a function defined by x−b f (x) = 4x – x2. Then f –1 (x) is

(a) Injective but not surjective (b) Surjective but not injective (c) Bijective (d) None of these

(a) 106 (c) (106)!

(a) one-one but not onto (b) onto but not one-one (c) Both one-one and onto (d) neither one-one nor onto

(b) 12 (d) 64

2 π − 2 137. If A =  x : − ≤ x ≤  , B = {y : – 1 ≤ y ≤ 1} and 5 5   f (x) = cos (5x + 2), then the mapping f : A → B is

(a) 2 – (c)

4− x

4− x

(b) 2 +

4− x

(d) None of these

144. Let f : [4, ∞) → [4, ∞) be a function defined by f (x) = 5x (x – 4), then f –1 (x) is (a) 2 –

1 (c)   5

4 + log 5 x

(b) 2 +

4 + log 5 x

x ( x − 4)



(d) None of these

145. If f : R → R is given by f (x) = 3x – 5, then f –1 (x)

1 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 (a) is given by

146. Let f : R → R be given by f (x) = (x + 1)2 – 1, x ≥ – 1. Then f –1 (x)

x +1 (c) does not exist because f is not one-one . (d) does not exist because f is not onto (b) = – 1 –

147. Let f : R → R be given by f (x) = (x + 1)2 – 1, x ≥ – 1. Then, the set of values of x for which f (x) = f –1 (x) is given by (a) {0} (c) {– 1}

(b) {0, – 1} (d) None of these

148. Let f be an injective map with domain {x, y, z} and range {1, 2, 3} such that exactly one of the following statements is correct and the remaining are false : f (x) = 1, f (y) ≠ 1, f (z) ≠ 2. The value of f –1 (1) is (a) y (b) x (c) z (d) None of these 149. The value of the parameter α, for which the function f (x) = 1 + αx, α ≠ 0 is the inverse of itself, is (a) – 2 (c) 1

(b) – 1 (d) 2 [Based on IIT 1992]

150. The inverse of the function f (x) = (a)

1 1 − x  log a   1 + x  2

1+ x  (c) log a   1 − x 

(b)

a x − a− x is a x + a− x

1 1 + x  log a   1 − x  2

(d) None of these

151. If the function f : [1, ∞) → [1, ∞) is defined by f (x) = 2x (x – 1), then f –1 (x) is

1 (a)   2

x ( x −1)

1 (c)   1 − 1 + 4 log 2 x   2 (d) not defined 152. The inverse of the function y = [1 – (x – 3) 4]1/7 is (b) 3 – (1 – x7)1/4 (d) None of these

153. If f (x) = 1 – x, x ∈ [– 3, 3], then the domain of f [ f (x)] is (a) [– 2, 3] (c) [– 2, 3)

x < −1 | x |,  (b) 2 − | x |, − 1 ≤ x ≤ 1 | 2 − x |, x > 1 

| 2 − x |, x < −1  (c) | x |, − 1 ≤ x ≤ 1 2 − | x |, x > 1 

(d) None of these

156. If f (x) = 2 − x and g (x) = of f [g (x)] is

(b) (– 2, 3) (d) (– 2, 3]

1− x   1  ; x ≠ 0 then f [ f (x)] + f  f    154. If f (x) = 1+ x   x  (a) < 2 (b) ≥ 2 (c) = 2 (d) None of these

| x |, x ≤ 1 , then f [ f (x)] is equal to 155. If f (x) =   2 − x, x > 1

1 − 2 x , then the domain

 1 (a)  −∞,   2

1  (b)  , ∞  2 

 3 (c)  −∞, −   2

(d) None of these

157. Let f be a function with domain [– 3, 5] and let g (x) = | 3x + 4 |. Then the domain of ( fog) (x) is 1  (a)  −3,   3

1  (b)  −3,  3 

 (c)  −3, 1  3 

(d) None of these

1 , x ≠ 0, 1, then the graph of the function 1− x y = f [ f { f (x)}], x > 1 is

158. If f (x) =

(a) a straight line (c) an ellipse

(b) a circle (d) a pair of straight lines

159. Let f : R → R be a function defined by f (x) = x – [x], where [⋅] denotes the greatest integer function, then f –1 (x) is (a) [x] – x

1 (b)   1 + 1 + 4 log 2 x  2

(a) 3 + (1 – x7)1/4 (c) 3 – (1 + x7)1/4

2 − | x |, x < −1 (a) | x |, − 1 ≤ x ≤ 1 | 2 − x |, x > 1 

(c) not defined

1 x − [ x] (d) None of these (b)

160. If f (x + y, x – y) = xy, then the arithmetic mean of f (x, y) and f ( y, x) is (a) y (c) 0 161. If f (x) = 64x3 +

(b) x (d) None of these

1 1 3 and a, b are the roots of 4x + x x

= 3, then (a) f (a) = 12 (c) f (a) = f (b)

(b) f (b) = 11 (d) None of these

162. If f, g, h are functions from R to R such that f (x) = x2 – 1, 0, x ≤ 0 g (x) = x 2 + 1 ∨ x ∈ R and h (x) =   x, x ≥ 0 then the composition function ho ( fog) is given by (a) x2 (c) x

(b) 0 (d) None of these

15

x +1

Functions

(a) = – 1 +

16

163. If f (x) = cos (log x), then f (x) ⋅ f (y) –

Objective Mathematics

 1   x  + f ( xy ) is equal to f 2   y  

(a) 2 (c) 0

(b) 1 (d) None of these

171. If f (x) = sin [π2] x + sin [– π2] x, where [⋅] denotes the greatest integer function, then π (a) f    = 1 2

(b) f (π) = 2

π (c) f    = – 1 4

(d) None of these 164. If R denotes the set of all real numbers, then the function f : R → R defined f (x) = | x | is 172. Let f (x) = max. {(1 – x), (1 + x), 2}, ∨ x ∈ R. Then

1 + x, x ≤ −1  (a) f (x) = 2, − 1 < x < 1 1 − x, x ≥ 1 

(a) one-one only (b) onto only (c) both one-one and onto (d) neither one-one nor onto

1 1 – 3, ∨ x (≠ 0) ∈ R, then f (x) 165. If 3 f (x) + 5 f    = x x = (a)

1 3   + 5 x − 6  14  x

1  3  (c)  − + 5 x + 6  14  x

(b)

1  3   − + 5 x − 6  14  x

(d) None of these

1 − x, x ≤ −1  (b) f (x) = 1, − 1 < x < 1 1 + x, x ≥ 1  1 − x, x ≤ −1  (c) f (x) = 2, − 1 < x < 1 1 + x, x ≥ 1 

166. If f (x + y) = f (x) + f (y) – xy – 1 for all x, y ∈ R and (d) None of these f (1) = 1, then the number of solutions of f (n) = n, n 173. The distinct linear function (s) which map (s) [– 1, 1] ∈ N is onto [0, 2] is (are) (a) one (b) two (a) x + 1, – x + 1 (c) no solution (d) None of these (b) x – 1, x + 1 1 (c) – x + 1 , g (x) = f [ f (x)] and h (x) = f [ f { f (x)}], 167. If f (x) = 1− x (d) None of these then the value of f (x) ⋅ g (x) ⋅ h (x) is 174. Let f (x) be a function defined on [– 1, 1]. If the area (a) 1 (b) – 1 of the equilateral triangle with two of its vertices at (c) 0 (d) None of these 3 (0, 0) and (x, f (x)) is , then the function f (x) is 168. If g [ f (x)] = | sin x | and f [g (x)] = (sin )2, then 4 (a) f (x) = sin2x, g (x) = (a) x 2 − 1 (b) – 1 + x 2 (b) f (x) = sin x, g (x) = | x | 2 (c) f (x) = x , g (x) = sin (d) 1 + x 2 = (c) ± 1 − x 2 (d) f and g cannot be determined

4x , then 4x + 2

169. Let f be a function defined on [– 2, 2] and is given by

175. If f (x) =

−1, − 2 ≤ x ≤ 0 f (x) =   x − 1, 0 < x ≤ 2

 1   2   1996  f   + f   + ... + f   is equal to 1997 1997 1997 

and g (x) = f (| x |) + | f (x) |. Then g (x) is equal to

 − x, − 2 ≤ x < 0  (a) 0, 0 ≤ x 20, x 2 x3 g′ (x) = + + ... is 189. Domain of e x = 1 + x + 2! 3! (a) – 1 (b) 1 (a) (1, ∞) (b) (– ∞, ∞) (c) 2 (d) None of these (c) (0, ∞) (d) None of these 182. If S is the set of all real x and such that 190. The range of the function f (x) = | x – 1 | is 2x − 1 is positive, then S contains (a) (– ∞, 0) (b) [0, ∞) 2 x3 + 3x 2 + x (c) (0, ∞) (d) (– ∞, ∞) 3 1   3 2 (b)  − , −  (a)  −∞, −  191. Domain of 4 x − x is   2 2 4 (a) R\[0, 4] (b) R\(0, 4) (c) (0, 4) (d) [0, 4] 1  1 1   (c)  − ,  (d)  , 3   4 2 2  192. The domain of the function f (x) = 2 − 2 x − x 2 is 183. The range of the real function

x+2 is x2 − 8x − 4

1  1   (a)  −∞, −  ∪  − , ∞   4   20  1 1 (b)  −∞, −  ∪  − , ∞   4   20 

 1  1  (c)  −∞, −  ∪  − , ∞   4   20  (d) None of these 184. If the function f, g, h are defined from the set of real numbers R to R such that f (x) = x2 – 1, g (x) =

0, if x ≤ 0 x 2 + 1 , h (x) =   x, if x ≥ 0

then the composite function (hofog) (x) = x=0 0,  (a)  x 2 , x > 0 − x 2 , x < 0 

0, x = 0 (b)  2 x , x ≠ 0

0, x ≤ 0 (c)  2 x , x > 0

(d) None of these

(a) – 2 – 3 ≤ x ≤ – 2 + (b) – 2 ≤ x ≤ 2

3

3 ≤x≤–1+ 3 (d) – 3 ≤ x ≤ 3 193. If f (x) = 2x3 + mx2 – 13x + n and 2, 3 are roots of the equation f (x) = 0, then the values of m and n are (c) – 1 –

(a) 5, 30 (c) – 5, 30 194. If x ∈ R and P =

(b) – 5, – 30 (d) None of these

x2 , then P lies in the inx − 2x2 + 4 4

terval 1 (a) 0,   2 

3 4 (b)  ,  4 5

(c) 0, 

(d) 0, 

1 3 

1 4 

1 + x  3x + x3 195. If f (x) = log   , then f [g (x)]  and g (x) = 1− x 1 + 3x 2 is equal to (a) f (3x) (c) 3 f (x)

(b) [ f (x)]3 (d) – f (x)

17

x2 − x is x2 + 2x

Functions

178. The domain of definition of the function y (x) given 185. The range of the function f (x) = by the equation 2 x + 2y = 2 is

18

2x − 1 is 2 x + 3x 2 + x

196. If S is the set of all real x such that

3

Objective Mathematics

positive, then S contains 1 (a)  − ,  4

1  2

1 (b)  , 3  2 

3 1 (c)  − ,   2 4

206. If e x = y +

3 1 (d)  − ,   2 2

x

(b) 8 (d) 3

is equal to

201. The domain of the function f (x) = log (1 – x) + is

x2 − 1

(b) (– ∞, – 1] (d) [– 1, 1]

(a) R\{nπ : n ∈ I} (c) R\{2nπ : n ∈ I}

log

1 is | sin x |

(b) one-one but not onto (d) many-one onto

1  1 210. Let f   x +  = x2 + 2 (x ≠ 0) then f (x) is equal to x x (a) x2 – 1 (b) x2 – 2 (d) None of these (c) x2

(a) {– 3, 0} (c) [0, 3]

(b) R \ (– π, π) (d) (– ∞, ∞)

(b) [– 3, 0] (d) (– 3, 0)

212. Let f : R → R be a mapping defined by, f (x) = x3 + 5, then f (a) (5 – x) (c) 5 – x 1/3

(b) (sin x) (d) sin x/x2 2

204. Let f : R → R be defined by f (x) = 3x – 4, then f  (x) is

1 (x + 4) 3 (c) 3x + 4

x defined as f : (0, ∞) to (0, ∞), f (x) 209. Let f (x) = 1 + x is

is defined on the set S, where S is equal to

–1

(a)

, n ∈ N}

f (x) = cot–1 [ ( x + 3) x ] + cos–1 ( x 2 + 3 x + 1 )

f (x) = sin x, g : R → R, g (x) = x2 is (a) sin x + x (c) sin x2

(b) x < 0 (d) – ∞ < x < ∞

211. The function

203. The composite mapping fog of the maps f : R → R, 2

2x + 1 is given by x − 10 x − 11 2

(a) one-one onto (c) many-one into

5 2 (d) None of these

(b)

202. The domain of the function y =

3

, n ≥ 0, n ∈ I} (b) R – { (c) R (d) R – {0}

1 200. If 2 f (x) – 3 f    = x2, x is not equal to zero, then f (2) x

(a) (0, 1) (c) (1, ∞)

e x + e− x 2

207. The domain of definition of the function

(a) R – {±

(b) (1, 7/3) (d) (1, 11/7)

7 4 (c) – 1

(d)

  x f (x) = cot–1   , x ∈ R is 2 2  x − [x ] 

x2 + x + 2 is x2 + x + 1

(a) –



208. The domain of the function

(b) {1, cos 1} (d) {0}

(a) (1, ∞) (c) (1, 7/5)

e −e 2

(b) ex – e–x

−x

(a) x > 0 (c) x ≠ – 1, x ≠ 11

π π 1} (d) {y : y < 1}

229. If the function

220. R is the set of real numbers and f : R → R and g : R → R are defined by f (x) = 3x2 + 2 and g (x) = 3x–1 for all x ∈ R. Then (a) ( fog) (x) = 27x2 – 18x + 5 (b) ( fog) (x) = 27x2 + 18x – 5 (c) (gof ) (x) = 9x2 – 5 (d) (gof ) (x) = 9x2 + 5. 221. If x is real, then the expression

x 2 + 34 x − 71 x2 + 2x − 7

4 3

(b) 1

(c) 0

(d)

(a) 0 < x ≤ 1 (c) – ∞ < x ≤ 0

(b) 0 ≤ x ≤ 1 (d) – ∞ < x < 1

1+ x  231. The function f ( x) = log   satisfies the equation 1− x 

x − x +1 x −1 2

(a) cannot lie between – 1 and 3 (b) always lies between – 1 and 3

(a)

3 4 230. The domain of definition of the function y (x) given by the equation 2 x + 2y = 2 is

(a) cannot lie between 5 and 9 (b) always lies between 5 and 9 (c) is not real (d) None of these. 222. If x is real, then the expression

π  π  f (x) = cos2x + cos2   + x  – cos x . cos   + x  3  3  is constant (independent of x), then the value of this constant is

(a)  f(x1) f(x2) = f(x1 + x2) (b)  f(x + 2) – 2f(x + 1) + f(x) = 0 (c)  f(x) + f(x + 1) = f(x2 + x)  x +x  (d)   f ( x1 ) + f ( x2 ) = f  1 2   1 + x1 x2 

Functions

−2

 x − 1 (a) loge    x + 1 

20

Objective Mathematics

232. The largest interval lying in  − π , π  for which the  2 2   function x  – x2 –1  f (x ) = 4 + cos  − 1 + log (cos x) is defined, is 2   π (b)  [0, π] (a)   0,   2  π π (c)    − ,   4 2

 π π (d)    − ,   2 2

233. If A = {1, 2, 3, 4}, B = {1, 2, 3}, then number of mappings from A to B is (a)  34 (c)  43

(b)  12 (d)  27

x π 234. If f : 0,  → [0, ∞) be a function defined by y = sin   , 2  2 then f is (a)  injective (c)  bijective

(b)  surjective (d)  none of these

235. Let f(x) = |x| has an inverse, its domain is (a)  R (c)  {0, 1}

(b)  {–1, 1} (d)  none of these

238. If f(x) = e x and g(x) = loge x, then which of the following is true? (a)  f{g(x)} ≠ g {f(x)} (b)  f{g(x)} = g {f(x)} (c)  f{g(x)} + g {f(x)} = 0 (d)  f{g(x)} – g {f(x)} = 1   x  239.   The domain of sin −1 log 3    is  3   (a)  [1, 9] (c)  [– 9, 1]

240. Let f : N → N defined by f(x) = x2 + x + 1, x ∈ N, then f is (a)  one-one onto (b)  many-one onto (c)  one-one but not onto (d)  None of the above 241. The domain of the function f(x) = log 2x – 1 (x – 1) is

(b)  one-one into (d)  many-one into

237. Let the functions f, g, h be defined from the set of real numbers R to R such that f(x) = x2 – 1, g(x) =

( x 2 + 1)

0, if x < 0 and h( x) =  ,  x, if x ≥ 0 then ho (fog) (x) is defined by (b)  x2 (d)  none of these

(a)  x (c)  0

1  (b)    , ∞  2  (d)  None of these

(a)  (1, ∞) (c)  (0, ∞) 242. Let f ( x) =

236. The function f : R → R defined by f(x) = |(x – 1) (x – 2)|, is (a)  one-one onto (c)  many-one onto

(b)  [–1, 9] (d)  [– 9, – 1]

1  πx  − tan   , 1 < x < 1 and 2  2 

g ( x) = 3 + 4 x − 4 x 2 , then dom (f + g) is given by 1  (a)    , 1 2 

1  (b)    , − 1 2  

 1  (c)    − , 1  2 

 1  (d)    − , − 1  2 

243. Let f : N → Y be a function defined as f(x) = 4x + 3, where Y = {y ∈ N : y = 4x + 3 for some x ∈ N}. Then, the inverse of f(x) is 3y + 4 3 y+3 (c)   g ( y ) = 4

(a)   g ( y ) =

y+3 4 y −3 (d)   g ( y ) = 4

(b)   g ( y ) = 4 +

solutions 1. (b) For f (x) to exist, we must have π π sin –1(2x) + ≥ 0  or,  sin –1 2x ≥ –  6 6 π π –1 Since –  ≤ sin (2x) ≤ 2 2 π π So, – ≤ sin –1 (2x) ≤ 6 2  π π or, sin   −  ≤ 2x ≤ sin   2 6

⇒ –

1 ≤ 2x ≤ 1 2

1 1 ≤ x≤ 4 2



⇒ −



 1 1 ∴  Domain =  − ,   4 2

2. (a) For y to be defined, we must have

 5x − x2  5x − x2 ≥ 0 ⇒ ≥ 100 (a) log10     4  4 ⇒ 5x – x2 ≥ 4 ⇒ x2 – 5x + 4 ≤ 0 ⇒ (x – 1) (x – 4) ≤ 0 ⇒ 1 ≤ x ≤ 4.



1 2 x ≤ 21 ⇒2 ≤ 2 –1

⇒ [x] < – 2 or [x] > 3 But [x] < – 2 ⇒ [x] = – 3, – 4, – 5, ...∴x < – 2. Also, [x] > 3 ⇒ [x] = 4, 5, 6, ... ∴ x≥4 Domain of f = (– ∞, – 2) ∪ [4, ∞). 8. (c) log10 (1 – x) is defined for 1 – x > 0 ⇒ x < 1 ...(1) log10 (1 – x) ≠ 0 ⇒ 1 – x ≠ 100 ⇒ 1 – x ≠ 1 ⇒ x ≠ 0  ...(2)

x + 2 is defined for x + 2 ≥ 0 ⇒ x ≥ – 2 From (1), (2) and (3), we get domain of y = ([– 2, ∞) ∩ (– ∞, 1))\{0} = [– 2, 1) \ {0} i.e. [– 2, 0[ ∪ ] 0, 1[.

...(3)

[∵ the base = 2 > 1] ...(1) 9. (a) 3 − x is defined for 3 – x ≥ 0 ⇒ 1 ≤ x2 ≤ 4 Now, 1 ≤ x2 ⇒ x2 – 1 ≥ 0 i.e. (x – 1) (x + 1) ≥ 0 ⇒ x ≤ 3 ...(1) ⇒ x ≤ – 1 or x ≥ 1  ...(2)  3 − 2x  cos–1    is defined for Also, x2 ≤ 4 ⇒ x2 – 4 ≤ 0 i.e. (x – 2) (x + 2) ≤ 0 5  ⇒ – 2 ≤ x ≤ 2  ...(3) 3 − 2x From (2) and (3), we get the domain of f ≤1 –1 ≤ 5 = ((– ∞, – 1] ∪ [1, ∞)) ∩ [– 2, 2] = [– 2, – 1] ∪ [1, 2]. ⇒ – 5 ≤ 3 – 2x ≤ 5 ⇒ – 8 ≤ – 2x ≤ 2 4. (d) For f (x) to be defined, we must have ⇒ – 1 ≤ x ≤ 4 ...(2) From (1) and (2), we get domain of f x – 1 − x 2 ≥ 0  or  x ≥ 1 − x 2 > 0 (– ∞, 3] ∩ [– 1, 4] = [– 1, 3]. 1 ∴ x2 ≥ 1 – x2 or x2 ≥ . x2 + 1 2 . 10. (a) We have, f (x) = [ x] Also, 1 – x2 ≥ 0 or x2 ≤ 1. When x ∈ [1, 2) then f (x) = x2 + 1 1  1  1  Now, x2 ≥ ⇒ x − ≥0 x+    ⇒ Rf = [2, 5).  2 2  2 When x ∈ [2, 3) then 1 1 5  x2 + 1 or x ≥ . ⇒ x ≤ – ⇒ Rf =  , 5  . f (x) = 2 2 2 2 Also, x2 ≤ 1 ⇒ (x – 1) (x + 1) ≤ 0 When x ∈ [3, 4) then ⇒ – 1 ≤ x ≤ 1 10 17  x2 + 1  1  1 f (x) = ⇒ Rf =  ,  . 2 2 , 1 and x ≤ 1 ⇒ x ∈  . Thus, x > 0, x ≥  3 3 3  2  2 ∴ Rf = [2, 17/3). 5. (b) 9 − x 2 is defined for 11. (d) f (x) is defined for x12 – x9 + x4 – x + 1 > 0 2 9 – x ≥ 0 ⇒ (3 – x) (3 + x) ≥ 0 ⇒ x4 (x8 + 1) – x (x8 + 1) + 1 > 0 ⇒ (x – 3) (x + 3) ≤ 0 ⇒ (x8 + 1) x (x3 – 1) + 1 > 0. ⇒ – 3 ≤ x ≤ 3 ...(1) If x ≥ 1 or x ≤ – 1, then the above expression is sin–1 (3 – x) is defined for positive. – 1 ≤ 3 – x ≤ 1 ⇒ – 4 ≤ – x ≤ – 2 If – 1 < x ≤ 0, the above inequality still holds. ⇒ 2 ≤ x ≤ 4  ...(2) Also, sin–1 (3 – x) = 0 ⇒ 3 – x = 0 or  If 0 < x < 1, then x = 3 ...(3) x12 – xp (x8 + 1) + (x4 + 1) > 0 From (1), (2) and (3), we get the domain of f : [∵ x4 + 1 > x8 + 1 and so x4 + 1 > x (x8 + 1)]. ([– 3, 3] ∩ [2, 4]) \ {3} = [2, 3). The domain of f = (– ∞, ∞). 6. (b) We have, y = x (2 – x) 2 12. (a) We have, ⇒ x – 2x + y = 0



2 ± 2 1− y = 1 ± 1− y 2 But x ≤ 1. So, x = 1 – 1 − y



Therefore, f –1(x) = 1 –



⇒ x =

7. (a) f (x) is defined for [x]2 – [x] – 6 > 0

1− x .

⇒ ([x] – 3) ([x] + 2) > 0

( x + 1)( x − 2)( x − 3) ( x + 1)( x − 3) . = ( x − 2) ( x − 2) f (x) is defined for (x + 1) (x – 2) (x – 3) ≥ 0 or [x – (– 1)] (x – 2) (x – 3) ≥ 0 ⇒ – 1 ≤ x ≤ 2 or x ≥ 3. Also, x – 2 ≠ 0 i.e. x ≠ 2. Domain of f = ([– 1, 2] ∪ [3, ∞))\{2} = [– 1, 2) ∪ [3, ∞).

f (x) =

21

(b)

Functions

5x − x2 > 0 ⇒ 5x – x2 > 0 ⇒ x (x – 5) < 0 4 ⇒ 0 < x < 5 From (a) and (b), we get the domain of f = [1, 4] ∩ (0, 5) = [1, 4]. 3. (c) For f (x) to be defined, we must have 1 – 1 ≤ log2  x 2  ≤ 1 2 

22

13. (d)  x − 1 is defined for x – 1 ≥ 0 i.e. x ≥ 1   5 − x is defined for 5 – x ≥ 0 or x ≤ 5 

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

Objective Mathematics

From (1) and (2), we get domain of f (– ∞, 5] ∩ [1, ∞) = [1, 5]. 14. (d) We have, 2  1 + x3  f (x) = sin − 1  3 / 2  + sin (sin x) + log (( 3x{ x+} 1+)1) .  2x 

2

sin (sin x) to be defined sin (sin x) ≥

and for



0, 0 ≤ sin x < 1, x ∈ [2nπ, (2n + 1) π ], n ∈ I



 x − 3/ 2 + x3/ 2  For sin − 1   to be defined 2  −1 ≤



f (x) =

∴ fog (x) =

⇒ ⇒



i.e. x (x – 3) < 0 ⇒ 0 < x < 3



 From (1) and (2), we get domain of f =  0,

3| x | − x − 2 and g (x) = sin x 3|sin x | − sin x − 2

which is defined if 3 | sin x | – sin x – 2 ≥ 0 If sin x > 0 then 2 sin x – 2 ≥ 0 ⇒ sin x ≥ 1 sin x = 1 ⇒ x = 2nπ + π/2. If sin x < 0 then – 4 sin x – 2 ≥ 0 – 1 ≤ sin x ≤ – 1/2

x−5 is defined if x 2 − 10 x + 24 ( x − 4)( x − 5)( x − 6) x−5 > 0 i.e. >0 2 ( x − 4) 2 ( x − 6) 2 x − 10 x + 24

16. (c) log10



⇒ ⇒

(x – 4) (x – 5) (x – 6) > 0 4 < x < 5 or x > 6 

...(2) 3 . 2 

1 – 1 − 1 − x 2 ≥ 0, 1 – 1 − x ≥ 0 and 1 – x2 ≥ 0 1 – x2 ≥ 0 ⇒ (x + 1) (x – 1) ≤ 0 ⇒ – 1 ≤ x ≤ 1. Clearly for these values, the other two inequalities hold. Thus domain of f = [– 1, 1]. 1 −| x| 21. (c) f (x) is defined if 2 − | x | ≥ 0 and 2 – | x | ≠ 0 (1 − | x |)(2 − | x |) ⇒ ≥ 0 and x ≠ – 2, 2 (2 − | x |) 2 ⇒ (| x | – 1) (| x | – 2) ≥ 0 and x ≠ – 2, 2 ⇒ | x | ≤ 1 or | x | > 2 ⇒ – 1 ≤ x ≤ 1 or (x < – 2 or x > 2) Domain of f = [– 1, 1] ∪ (– ∞, – 2) ∪ (2, ∞). 22. (d)

−1 x

is defined if – 1 ≤ x ≤ 1 

1 is defined if x – 2 > 0 i.e. x > 2 x−2 From (1), (2) and (3), domain of f = φ (empty set).

...(1) ...(2) ...(3)

 2 −| x|  23. (a) cos–1   is defined for 4 

...(1)

3 ...(2) x + 5 is defined for all real x. From (1) and (2), we get domain of f = (4, 5) ∪ (6, ∞). 17. (b) f (x) is defined for | x | – x > 0 i.e. | x | > x, which is true for x < 0 only. Thus domain of f = (– ∞, 0).

1 is defined if 1 – x ≠ 0 i.e. x ≠ 1 1− x 2sin

7π 11π   π   ∴ x ∈  2nπ + , 2 nπ +  ∪ 2nπ +  , n,m ∈ I.  6 6   2



x (3 − x) >0 x2

2

x − 3/ 2 + x3/ 2 x − 3/ 2 + x3/ 2 ≤ 1 . But ≥1, 2 2 x − 3/ 2 + x3/ 2 =1 ∴ 2 ⇒ x–3/2 = x3/2 = 1 ⇒ x3 = 1 ⇒ x = 1. Thus Domain = φ. 15. (d) We have,

3− x > 0 or x

19. (a) log10 | 4 – x2 | is defined for all x except when 4 – x2 = 0 i.e., x = ± 2. Hence domain of f = (– ∞, ∞)\{– 2, 2}. 20. (d) f (x) is defined if

For log (( 3x{ x+} 1+)1) to be defined {x} ≠ 0 ⇒ x ∉ I







2 −| x | ≤1 4 ⇒ – 4 ≤ 2 – | x | ≤ 4 ⇒ – 6 ≤ – | x | ≤ 2 ⇒ – 2 ≤ | x | ≤ 6 ⇒ | x | ≤ 6 ⇒ – 6 ≤ x ≤ 6  ...(1) –1 ≤

1 log (3 − x) is defined if log (3 – x) ≠ 0 and

3 – x > 0 ⇒ 3 – x ≠ e0 = 1 and x < 3 ⇒ x ≠ 2 and x < 3  From (1) and (2), we get domain of f 3− x 3− x is defined for log10  ≥0 18. (d) log10  = [– 6, 6] ∩ ((– ∞, 3) \ {2})  x   x  = [– 6, 3) \ {2}. 3− x ≥ 100 = 1 ⇒ 3 – x ≥ x ⇒ 24. (b) f (x) is defined if log3 log4 x > 0, log4 x > 0 x and x > 0 ⇒ 2x ≤ 3 ⇒ x ≤  ...(1) ⇒ log4 x > 3º = 1, x > 4º and x > 0 ⇒ x > 41, x > 1 and x > 0 ⇒ x > 4 3− x Also, log10  Domain of f = (4, ∞).  is defined for x 

...(2)



26. (c) f (x) = tan–1x, –1 ≤ x ≤ 1

1 (– x – 1), x < –1 2 1 (x + 1), x > 1. 2 1 , −1 < x < 1 ∴ f ′(x) = 1 + x2 1 1 ,x>1 –  , x < –1 2 2 1 1 , f ′(–1 + 0) = 1 + −1 + 0 2 = ( ) 2 1 f ′(–1 + 0) = – 2 1 1 1 f ′(1 – 0) = 1 + 1 − 0 2 = , f ′(1 + 0) = ( ) 2 2

∴ f ′(–1) does not exist. ∴ domain of f ′(x) = R – {–1}. 27. (a) We have, f (x) =

=

x+3 (2 − x)( x − 5)

( x + 3)(2 − x)( x − 5) (2 − x)( x − 5)



f (x) is defined if (x + 3) (2 – x) (x – 5) ≥ 0 and (2 – x) (x – 5) ≠ 0 ⇒ (x + 3) (x – 2) (x – 5) ≤ 0 and x ≠ 2, 5 ⇒ (x ≤ – 3 or 2 ≤ x ≤ 5) and x ≠ 2, 5 ⇒ x ≤ – 3 or 2 < x < 5 Domain of  f = (– ∞, – 3] ∪ (2, 5). 3 28. (a) f (x) is defined if – 1 ≤ ≤1 4 + 2 sin x Since 4 + 2 sin x > 0 for all real x, therefore 3 ≤ 1 ⇒ 3 ≤ 4 + 2 sin x 4 + 2 sin x 1 ⇒ sin x ≥ – 2 π π ⇒ – + 2nπ ≤ x ≤ + 2nπ, n ∈ I 6 6 π  π  Domain of f =  − + 2nπ, + 2nπ  . 6  6  29. (b) f (x) is defined if x – 4 ≥ 0 and 6 – x ≥ 0 ⇒ x ≥ 4 and x ≤ 6 Domain of  f = [4, ∞) ∩ (– ∞, 6] = [4, 6].



23

30. (a) f (x) is defined if 1 – log10 (x2 – 5x + 16) > 0 and x2 – 5x + 16 > 0 ⇒ log10 (x2 – 5x + 16) < 1 and 2 5  39  x −  + >0  2 4

Functions

 x − 3 x−3 25. (c) sin–1  ≤1  is defined if – 1 ≤ 2  2 ⇒–2 ≤x–3 ≤2⇒1 ≤x≤5  ...(1) log10 (4 – x) is defined if 4 – x > 0 i.e. x < 4  ...(2) From (1) and (2), we get domain f = [1, 5] ∩ (– ∞, 4) = [1, 4).

⇒ x2 – 5x + 16 < 101 = 10 2    5  39 ∵ x − + > 0 for all real x     2 4   

⇒ x2 – 5x + 6 < 0 ⇒ (x – 3) (x – 2) < 0 ⇒ 2 < x < 3. Domain of f = (2, 3). 31. (c) Since only x ! {x} is defined so x! {x} represents a function. 32. (d) log10 sin (x – 3) is defined if sin (x – 3) > 0

⇒ 2nπ < x – 3 < 2nπ + π, n ∈ I



⇒ 2nπ + 3 < x < 2nπ + π + 3, n ∈ I

...(1)

16 − x is defined for 16 – x ≥ 0 ⇒ (4 – x) (4 + x) ≥ 0 ⇒ (x – 4) (x + 4) ≤ 0 ⇒–4 ≤x≤4  ...(2) The domain of f is the common part of (1) and (2) i.e. 3 < x ≤ 4.  Thus Domain of f = (3, 4]. 33. (d) f (x) is defined if cos x > 0, x > 0 and x ≠ 1 π π 0 and x ≠ 1 ⇒ – 2 2 π  ⇒  0 < x <  \{1} 2 2



 π Domain ( f ) =  0,  \{1}. 2

34. (a) f (x) is defined if – 1 ≤

2

4 ≤1 3 + 2 cos x

Since 3 + 2 cos x > 0 for 4 ≤ 1 ⇒ cos x ≥ 3 + 2 cos x π ≤ x ≤ 2nπ + ⇒ 2nπ – 6

all real x, this gives 1 2 π , n ∈ I. 6

35. (b) ∵  {x} ∈ [0, 1) ∴  sin {x} ∈ [0, sin 1) but f (x) is defined if sin {x} ≠ 0

 1   1  1 ∈ , ∞ . ∴   ∈{1, 2, 3.....} . sin{x}  sin 1   sin{x} 

∴

36. (b) tan–1 x ( x + 1) is defined if



x (x + 1) ≥ 0  or  x ≤ – 1 or x ≥ 0  sin

–1

...(1)

x + x + 1 is defined if 0 ≤ x2 + x + 1 ≤ 1. 2

2

1 3  Now x 2 + x + 1 =  x +  + ≥ 0  2 4 for all real x  Also, x2 + x + 1 ≤ 1 ⇒ x (x + 1) ≤ 0

...(2)

24

⇒–1 ≤x≤0  From (1), (2) and (3), we get domain of  f = {– 1, 0}.

...(3)

Objective Mathematics



| x | ≤ 3 ⇒ – 3 ≤ x ≤ 3.



Also



1 − | x|  cos–1   ≥0 2 

37. (a) 3 1 − 3 x and e3 tan x is defined for all real x.  2x − 1 2x − 1 ≤1   is defined if – 1 ≤ 3  3



3 cos–1



⇒ – 3 ≤ 2x – 1 ≤ 3 ⇒ – 1 ≤ x ≤ 2 Domain of f  = [– 1, 2].

1  1 1 > 0 i.e. x > 38. (b) log1/ 2  x −  is defined if x – 2 2 2

log 2 4 x 2 − 4 x + 5 is defined if 4x2 – 4x + 5 > 0 2   1 x − + 1 > 0    ⇒ 4   2   which is true for all real x.

40. (b) f (x) is defined if | sin x | + sin x > 0 ⇒ sin x > 0 ⇒ 2nπ < x < 2nπ + π Domain of f = (2nπ, (2n + 1) π). 2 f ( n) + 1 and f (1) = 2 2 ∴  f (101) = f (1) + 100 × 1/2 = 2 + 50 ∴ f (101) = 52.

41. (a) We have, f (n + 1) =

42. (a) 24 – xC3x – 1 is defined if, 24 – x > 0, 3x – 1 ≥ 0 and 24 – x ≥ 3x – 1 1 25 and x ≤ ⇒ x < 24, x ≥ 3 4 1 25 ⇒ ≤x≤  ...(1) 3 4 40 – 6x C8x – 10 is defined if 40 – 6x > 0, 8x – 10 ≥ 0 and 40 – 6x ≥ 8x – 10



20 25 5 ,x≥ and x ≤ 3 7 4 25 5 ⇒ ≤x≤  7 4 25 5 ≤x≤ From (1) and (2) we get 7 4 But 24 – x ∈ N, ∴ x must be an integer,



∴ x = 2, 3.



Hence domain ( f ) = {2, 3}.



⇒

x
0,  x +  ≠ 1 2 2  

x2 – 5x + 6 ≠ 0 ⇒ (x – 2) (x – 3) ≠ 0 ⇒ x ≠ 2, 3 ...(1)



1   x + 2  > 0 ⇒ x ≥



1 3  1   x + 2  ≠ 1 ⇒ x ∉  2 , 2   



From (1), (2) and (3), we get domain of f



3  =  , 2  ∪ (2, 3) ∪ (3, ∞). 2



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

46. (b) f (x) is defined if – 1 ≤ x + [x] ≤ 1 Case I. x is an integer. Let x = n, n ∈ I. Then x + [x] = n + n = 2n. 1 1 So, we get, – 1 ≤ 2n ≤ 1 ⇒ – ≤n≤ 2 2 ...(2) ⇒ n = 0 ...(1) Case II. x is not an integer. Let x = n + k, n ∈ I, 0 < k < 1 Then, x + [x] = n + k + [n + k] = 2n + k So, we get, – 1 ≤ 2n + k ≤ 1 ⇒ n = 0 ∴ x = k, 0 < k < 1 ...(2) From (1) and (2), we get Domain of ( f ) = [0, 1). 47. (d) f (x) is defined if (1) x > 0 (2) log2 x > 0 ⇒ x > 20 = 1. (3) log2 log2 x > 0 ⇒ log2 x > 20 = 1 ⇒ x > 21 = 2.



⇒ log2 x > 22 ⇒ x >



Continuing like this, we get x > 22



∴ Domain of f = ( 22





2...( n −1) times

2...( n −1) times

, ∞).

48. (a) We have,

f (x) =



=



f (x) is defined if x – 4 ≥ 0  or  x ≥ 4. Domain of  f = [4, ∞).

x −3− 2 x − 4 − x −3+ 2 x − 4 x − 4 +1

49. (c) We have, a f (x) + g (x) = 0, a > 0 Since minimum values of g (x) is 1/2 ∴ g (x) > 0 and a f (x) > 0 x. ∴ a f (x) + g (x) > 0, Hence number of solutions is zero.

16 − x 2 >0 50. (b) f (x) is defined if 3− x ⇒ 16 – x2 > 0 and 3 – x > 0 ⇒ (x – 4) (x + 4) < 0 and x < 3 ⇒ – 4 < x < 4 and x < 3  or  – 4 < x < 3 Domain of f = (– 4, 3). 51. (a) f (x) is defined if x + 25 ≥ 0, x ≠ 0, x – 5 ≥ 0, x – 3 ≥ 0 ⇒ x ≥ – 25, x ≠ 0, x ≥ 5, x ≥ 3 ⇒ x ≥ 5. Domain of f = [5, ∞). 52. (c) Since f : (4, 6) → (6, 8) ⇒ f (x) = x + 2 ∴ f – 1 (x) = x – 2. 53. (c)   f (x) is defined if  1  1 log3  ≥ 0, | cos x | > 0 and | cos x | ≠ 0  | cos x |  1 0 | cos x | ≥ 3 = 1 and | cos x | ≠ 0



⇒



  1 ∵ | cos x | > 0 for all real x 



  1 ≥ 1 for all real x  ⇒ | cos x | ≠ 0 ∵  | cos x | 



π ,n∈I 2 π   Domain of f = R\ (2n + 1) : n ∈ I . 2   ⇒ x ≠ (2n + 1)

54. (b) f (x) is defined if 1   1 – log 1 2 1 + 1 5  – 1 > 0, 1 + 1 5 > 0, x ≠ 0 x x

1   ⇒ log 1 2 1 + 1 5  < – 1, x1/5 + 1 > 0, x ≠ 0 x −1





x − 4 −1 −



25

(4) log2 log2 log2 x > 0 ⇒ log2 log2 x > 20 = 1 ⇒ log2 x > 21 = 2 ⇒ x > 22 (5) log2 log2 log2 log2 x > 0 ⇒ log2 log2 log2 x > 20 = 1 ⇒ log2 log2 x > 21



1 1 ⇒ 1 + 1 5 >   , x > (– 1)5, x ≠ 0 2 x 1 > 1, x > – 1 and x ≠ 0 ⇒ x1 5 ⇒ 0 < x < 1 and x > – 1 ∴ Domain ( f ) = (0, 1).

Functions



⇒ 0 < x < 1.

55. (d) We have, f (x) = 2 min {| f (x) – g (x)|, 0}

f ( x) > g ( x) 0 = 2( f ( x) − g ( x)), f ( x) ≤ g ( x)   f ( x) − g ( x) − | f ( x) − g ( x) |, =  f ( x) − g ( x) − | f ( x) − g ( x) |, 

f ( x) > g ( x) f ( x) ≤ g ( x)

∴ f (x) = f (x) – g (x) – | g (x) – f (x) |. 56. (c) We have, x f (x) = 1 − x, 

x ∈Q x ∉Q

 f ( x), ∴  f 0 f (x) = 1 − f ( x)   x,  1 − x,  = 1 − x,  1 − (1 − x),

x ∈ Q, x ∉ Q, 1 − x ∈ Q, 1 − x ∉ Q,

f ( x) ∈ Q , f ( x) ∉ Q x ∈Q x ∈Q x ∉Q x ∉Q

 x, x ∈ Q =  x, x ∉ Q  ∴  f (x) = x, x ∈ [0, 1]. 57. (c) f (x) is defined if – (log3 x)2 + 5 log3 x – 6 > 0 and x > 0 ⇒ (log3 x – 3) (2 – log3 x) > 0 and x > 0 ⇒ (log3 x – 2) (log3 x – 3) < 0 and x > 0 ⇒ 2 < log3 x < 3 and x > 0 ⇒ 32 < x < 33 ⇒ 9 < x < 27 Domain of f = (9, 27). 58. (a) f (x) is defined if log x ≠ 0 and x > 0 ⇒ x ≠ 100 = 1 and x > 0 Domain of f  = (0, 1) ∪ (1, ∞) = (0, ∞) \ {1}. 59. (c) f (x) is defined if

4 − x2 ≥ 0 and [x] + 2 ≠ 0 [ x] + 2

Case I. 4 – x2 ≥ 0 and [x] + 2 > 0 ⇒ (x – 2) (x + 2) ≤ 0 and [x] > – 2 ⇒ – 2 ≤ x ≤ 2 and x ≥ – 1 ⇒ – 1 ≤ x ≤ 2  Case II. 4 – x2 ≤ 0, [x] + 2 < 0 ⇒ (x – 2) (x + 2) ≥ 0 and [x] < – 2 ⇒ (x ≤ – 2 or x ≥ 2) and x < – 2

...(1)

26

⇒ x < – 2 

Objective Mathematics



...(2)

From (1) and (2), we get domain ( f ) = (– ∞, – 2) ∪ [– 1, 2].

60. (b) Clearly, y is defined for all real x. ∴ Domain of y = (– ∞, ∞). x ⇒ y + x 2y = x We have, y = 1 + x2 or x2y – x + y = 0 For x to be real, 1 – 4y2 ≥ 0, y ≠ 0⇒ (1 – 2y) (1 + 2y) ≥ 0, y ≠ 0 ⇒ (2y – 1) (2y + 1) ≤ 0, y ≠ 0 1 1 ,y≠0 ⇒ – ≤ y ≤ 2 2 Also, if y = 0, we get x = 0 so that y = 0 is also in the range.  1 1 ∴ Range of y =  − ,  .  2 2 61. (a) Clearly, y is defined for all real x. ∴ Domain of y = (– ∞, ∞) x2 We have, y = ⇒ x 2y + y = x 2 1 + x2 y y i.e. x = 1 − y = 1− y

y (1 − y ) (1 − y )



⇒ x2 =



For x to be real, 1 – y ≠ 0 i.e. y ≠ 1 and y (1 – y) ≥ 0 ⇒ y (y – 1) ≤ 0 and y ≠ 1 ⇒ 0 ≤ y < 1 Range of y = [0, 1).

62. (c) We have, y =

1  2 − sin 3 x



Domain of y = (– ∞, ∞)



From (1), 2 – sin 3x =



...(1)

1 ⇒ sin 3x = y 2y −1 1   2 y − 1 y ⇒x= sin–1  y  3 1   ∵ y = 2 − sin 3 x ∴ y > 0     as − 1 ≤ sin 3 x ≤ 1



For x to be real, – 1 ≤



⇒ – y ≤ 2y – 1 ≤ y





 11  Range of f = log e , ∞  . 3  π2 – x2 ≥ 0 64. (a) For y to be defined, 16



2  π  π π π − x 2 ∈ 0,  . Clearly, for x ∈  − ,  ,  4  4 4  16



 π Since sin x is an increasing function on 0,  .  4



Therefore, sin 0 ≤ sin



⇒ 0 ≤ 3 sin



⇒0 ≤y≤



3   ∴ Range of y = 0, . 2  

⇒ 2y – 1 ≥ – y and 2y – 1 ≤ y 1 ⇒ y ≥ and y ≤ 1 3 1  ∴ Range of y =  , 1 . 3 

2

3 2

which is true for all real x ∴ Domain ( f ) = (– ∞, ∞)



Let y = 3 x 2 − 4 x + 5 ⇒ y2 = 3x2 – 4x + 5 i.e. 3x2 – 4x + (5 – y2) = 0 For x to be real, 16 – 12 (5 – y2) ≥ 0



⇒y≥

11 3

 11  , ∞ ∴ Range of y =  3 .  2 66. (b) f (x) is defined if x + x – 6 ≠ 0 i.e. (x + 3) (x – 2) ≠ 0 i.e. x ≠ – 3, 2 ∴ Domain ( f ) = (– ∞, ∞) \ {– 3, 2}. x 2 − 3x + 2 Let y = x2 + x − 6 2 ⇒ x y + xy – 6y = x2 – 3x + 2

 2  11 4 5 ⇒ 3   x 2 − x +  > 0 ⇒ 3    x −  +  > 0,    3 9  3 3 

π2 − x2 16

π2 − x 2 ≤ sin π 16 4 3 ≤ 2

65. (c) f (x) is defined if 3x2 – 4x + 5 ≥ 0 2  2  11 5  2 4 x − +  ≥0    x x − + ≥ 0 ⇒ 3  ⇒ 3 3 9  3 3   



2y −1 ≤1 y ( ∵ y > 0)

π π  π  π  ⇒  − x   + x  ≥ 0 ⇒  x −   x +  ≤ 0 4 4 4 4 π π ≤x≤ ⇒– 4 4  π π ∴ Domain of y =  − ,   4 4





63. (b) f (x) is defined if 3x2 – 4x + 5 > 0



which is true for all real x. ∴ Domain ( f ) = (– ∞, ∞) Let y = loge (3x2 – 4x + 5) ⇒ ey = 3x2 – 4x + 5. ⇒ 3x2 – 4x + (5 – ey ) = 0 For x to be real, 11 16 – 12 (5 – ey) ≥ 0 ⇒ 12 ey ≥ 44 ⇒ ey ≥ 3 11 ⇒ y ≥ loge 3

 x2  67. (b) Clearly, for y to be defined,  ≤1  1 + x 2  which is true for all x ∈ R. So, the domain = (– ∞, ∞).



 x2  x 2 = sin y Now, y = sin–1  2  ⇒ 1 + x  1 + x2 sin y ⇒x= 1 − sin y For x to be real, sin y ≥ 0 and 1 – sin y > 0

 π ⇒ 0 ≤ sin y < 1 ⇒ y ∈ 0,   2  π ∴ Range = 0,  .  2 68. (a) We have f (x) = cos x – x (1 + x)

π π Since, in the interval  ,  , cos x decreases and 6 3 x (1 + x) increases,

π π ∴ f (x) decreases in  ,  . 6 3



π π π π ∴ f   ≤ f (x) ≤ f   , x ∈  ,  3 6 6 3



 1 π  π  3 π  π  Hence f (A) =  − 1 +  , − 1 +    2 3  3  2 6  6 



 1 π π2 3 π π2  =  − − , − − .  2 3 9 2 6 36 



 π π ∴ Domain of f =  − ,  .  3 3



The greatest value of f (x) = tan



y y π + sin −1 ⇒x= . 4 2 2 y ≤1 For x to be real, – 1 ≤ 2 x –

π = sin–1 4



⇒ – 2≤ y ≤ 2.



∴ Range of y = [– 2 , 2 ].

72. (c) Clearly, f (x) is defined for all real x except at x = 0. ∴ Domain of f = (– ∞, ∞) \ {0} x Now, for x > 0, f (x) = = 1 and for x < 0, f (x) x x = – 1. = −x 0 At x = 0, f (x) = = not defined. 0 Thus, f (x) has only two values, namely 1 and – 1. ∴ Range of f = {– 1, 1}. 73. (b) We have, f (2x + 3) + f (2x + 7) = 2 Replace x by x + 1, f (2x + 5) + f (2x + 9) = 2 Now replace x by x + 2, f (2x + 7) + f (2x + 11) = 2  From (1) – (3),

Since

...(1) ...(2) ...(3)

4 − x2

0, ∴ we have

1 – x > 0 and 4 – x2 > 0

π2 − 0 , when 9 π2 π2 − 9 9

π . 3

∴ The greatest value of f (x) = 3 and the least value of f (x) = 0 ∴ Range of f = [0, 3 ]. 70. (a) We have,  f (x) = log (sin {x}) ⇒   f (x + 1) = f (x) ⇒ period of f = 1. –1

⇒ x < 1 and (x – 2) (x + 2) < 0 ⇒ x < 1 and – 2 < x < 2 ⇒ – 2 < x < 1 ∴ Domain of f = (– 2, 1).  4 − x2  Since – ∞ < log  < ∞  1− x  ⇒

  4 − x2 – 1 ≤ sin log    1− x

  ≤ 1  

∴ Range of f = [– 1, 1]. 75. (d) The graph of f (x) has the equation y = (x + 1)2, x ≥ –1. The reflection of the graph of f (x) is obtained by

27

⇒

4 − x2 > 0, 4 – x2 > 0 and 1 – x ≠ 0 1− x

x = 0 and the least value of f (x) = tan , when x =



π  2 sin  x −   4

74. (c) For f (x) to be defined,

π π ≤x≤ . – 3 3

⇒

Let y =

we get f (2x + 3) – f (2x + 11) = 0

π – x2 ≥ 0 9





i.e.,  f (2x + 3) = f (2x + 11) ⇒ T = 4.

2

69. (a) For f (x) to be defined,



π  2 sin  x −  .  4 Clearly, f (x) is defined for all real x. ∴ Domain of f = (– ∞, ∞).

71. (b) We have, f (x) = sin x – cos x =

Functions

⇒ x2 (y – 1) + x (y + 3) – (6y + 2) = 0 For x to be real, (y + 3)2 + 4 (y – 1)  (6y + 2) ≥ 0 ⇒ 25y2 – 10y + 1 ≥ 0 i.e. (5y – 1)2 ≥ 0 which is true for all real y. Range of f = (– ∞, ∞).

28

interchanging x, y. So, the graph of g(x) has the equation x = (y + 1)2, y ≥ –1

Objective Mathematics

∴ f (x) = sin4 2x + cos4 2x is a periodic function 1 π with period ⋅ i.e. π . 2 2 4 82. (a) Let f (T + x) = f (x) ⇒ tan (T + x) = tan x ⇒ tan (T + x) = tan x ⇒ T + x = nπ + x, n ∈ I Clearly, from here, the least positive value of T independent of x is π. Therefore, f (x) is a periodic function of period π. 83. (c) cos x is a periodic function with period 2π, therefore  8x + 5  f (x) = cos    will be a periodic function with 4π 

∴ y = x − 1 because y ≥ –1 ∴ φ(x) = x − 1 , x ≥ 0. 76. (a) f ′(x) = 2 + cos x > 0. So, f (x) is strictly m.i. so, f (x) is one-to-one and only. 77. (b) We have, f (x) = a sin kx + b cos kx

2π = π 2. 8 4π 84. (b) Let  f (T + x) = f (x) ⇒  (T + x) – [T + x] = x – [x] ⇒  T = [T + x] – [x] = an integer. Clearly, from here, the least positive value of T independent of x is 1. Hence, f (x) is a periodic function of period 1. period

=

  a b sin kx + cos kx  a 2 + b2  2 2 2 2  a +b  a +b

85. (c) Let g (x) = e3{x} ⇒ T1 = 1

=

a 2 + b 2 (cos θ sin kx + sin θ cos kx),



and f (x) = e{3x} ⇒ T2 = 1/3



∴  T1 = 3T2.

where cos θ = =

a +b 2

2

a a 2 + b2

86. (b) We have, ex + ef (x) = e ⇒ ef (x) = e – ex

sin (kx + θ),

⇒ f (x) = log (e – ex). For f (x) to be defined, e – ex > 0 ⇒ e1 > ex ⇒x 0 ⇒ e1 > ey ⇒ y < 1 ∴ Range of f = (– ∞, 1).

2π which is a periodic function of period | k | . 78. (d) Let f (T + x) = f (x) ⇒ cos (T + x)2 = cos x2 ⇒ (T + x)2 = 2nπ ± x2 Clearly, from here, no positive value of T independent of x is possible because x2 on R.H.S. can be cancelled only when T = 0. ∴ f (x) is a non periodic function.

87. (a) sin x is a periodic function with period 2π, therefore

79. (b) Since, | sin x | + | cos x | is a periodic function with π , ∴ f (x) = | sin 4x | + | cos 4x | is a periodic 2 π 1 π i.e. . function with period 8 4 2

period

80. (d) Let f (T + x) = f (x) ⇒ sin

T + x = sin x

sin ( [n] x) is a periodic function with period

2π . [ n]

But the period of f (x) is 2π (given).

2π = 2π ⇒ [n] = 1 [ n] ⇒ [n] = 1 ⇒ 1 ≤ n < 2.



∴

88. (a) f (x) = (– 1)[x] = {– 1, 1}, since [x] ∈ I. ⇒ T + x = nπ + (– 1)n x Clearly, from here, no positive value of T independ- 89. (a) Since sin x and cos x are periodic functions with period 2π. ent of x is possible because x on RHS can be 2π cancelled only when T = 0.  πx  = 2 (n!) ∴ Period of cos   = π   n! ∴ f (x) is a non-periodic function. n! and period of sin 81. (c) Since sin4 x + cos4 x is a periodic function with π  πx  2π period , = 2 (n + 1)!  (n + 1)!  = π 2 (n + 1)!



2 (n + 1)!} = 2 (n + 1)!.

90. (a) Since | sin x | + | cos x | is a periodic function with π π period , therefore period of f (x) will be when 2 2 k = 1. 91. (b) sin2 πx =

1 − cos 2 πx . 2

96. (b) 2 sin x is a periodic function of period 2π and 3 cos 2x is a periodic function of period π.

∴  f (x) is a periodic function of period = L.C.M. Since cos 2πx is a periodic function will pe{2π, π} = 2π. 2π riod = 1, therefore sin 2 πx is periodic with 97. (d) For every rational number T, we have 2π period 1. ...(1) 1, when x is a rational f (T + x) = 0, when x is irrational , x – [x] is a periodic function with period 1. ...(2)   1 sin4 πx = (sin2 πx)2 = (1 – cos 2πx)2 but there is no least positive value of T for which 4 f (T + x) = f (x) because there are infinite number of 1 = (1 + cos2 2πx – 2 cos 2πx) rational numbers between any two rational numbers. 4 Therefore, f (x) is a periodic function having no 1 fundamental period. = (3 + cos 4πx – 4 cos 2πx) 8  x  −1  S ince cos 4πx is a periodic function with period 98. (a) f (x) = sin  log 3    exists if  3  2π 1 = and cos 2πx is a periodic function with period 4π 2 2π = 1, therefore, period of sin4 πx is equal to 2π  1  L.C.M. (1, 1) 1 ...(3) L.C.M. 1,  = H.C.F. (1, 2) = 1 = 1 2

x x −1≤ log 3   ≤ 1 ⇔ 3−1 ≤ ≤ 31 ⇔ 1 ≤ x ≤ 9. 3 3 99. (d) Since cos

From (1), (2) and (3), we get Period of 3(sin

2 πx + x − [ x ] + sin 4 πx )

= 1.

1 − cos 2θ 2π , ∴  period = = π. 2 2 93. (c) We have, |sin( π + x)| − |cos ( π + x)| f (π + x) = |sin( π + x) + cos( π + x)|

92. (b) sin2θ =





f (x + k) = 1 + [1 + {1 – f (x)}5]1/5 ⇒ f (x + k) – 1 = [1 – ( f (x) – 1)5]1/5 ⇒ φ  ( x + k ) = [ 1 – { φ  ( x ) } 5 ] 1 / 5 ,  where φ (x) = f (x) – 1 ⇒ φ (x + 2k) = [1 – {φ (x + k)}5]1/5 ⇒ φ (x + 2k) = [1 – {1 – (φ (x)) 5 }] 1/5  = φ (x), ∨ x ∈ R ⇒ f (x + 2k) – 1 = f (x) – 1 ⇒ f (x + 2k) = f (x), ∨ x ∈ R ∴ f (x) is periodic with period 2k.

∴ f (x) is periodic with period π. 94. (b) For f (x) to be defined,

⇒  (x – 3) (x + 2) ≠ 0, x + ⇒ x ≠ 3, – 2, x ≥

1 1 ≥ 1, x + ∉ [1, 2) 2 2

1 3  1 , x ∉  ,  2 2 2

3  ⇒ x ∈  , 3  ∪ (3, ∞). 2 95. (a) Since the domain of f is (0, 1), ∴ 0 < ex < 1 and 0 < ln | x | < 1

2π πx 2 π = 5 and sin 4 is a periodic function with 5 2π period π = 8. 4 ∴ f (x) is periodic with period = L.C.M. (5, 8) = 40.

100. (b) We have,

|sin x | − |cos x | = |sin x + cos x | = f (x) for all x.

1 1   x2 – x – 6 ≠ 0,  x +  > 0,  x +  ≠ 1 2 2  

2 πx is a periodic function with period 5

101. (b) f (x) = sin4x + cos4x = (sin2x + cos2x)2 – 2 sin2x cos2x

1 1 1 − cos 4 x  (sin 2 x) 2 = 1 −   2 2 2 3 1 = + cos 4 x 4 4 Since, cos x is periodic with period 2π = 1−

∴  period of f (x) =

2π π = . 4 2

29

⇒ log 0 < x < log 1 and e0 < | x | < e1 ⇒ – ∞ < x < 0 and 1 < | x | < e ⇒ x ∈ (– ∞, 0) and x ∈ ((– ∞, – 1) ∪ (1, ∞)) ∩ (– e, e) ⇒ x ∈ (– ∞, 0) and x ∈ (– e, – 1) ∪ (1, e) ⇒ x ∈ (– e, – 1).

∴ Period of f (x) = L.C.M. of {2 (n!),

Functions



30

2 πx   πx  102. (b) Period of 2 cos  + 3 sin    3   3 

Objective Mathematics



 2π 2π  = L.C.M.  ,  = 6.  2 π π  3 3



π π   Period of 4 cos   2 πx +  + 2 sin  πx +    2 4



2π 2π  = L.C.M.  , =2  2 π π 

 2π  Period of | tan x | + cos 2x = L.C.M.  π,  = π.  2  103. (a) The function f (x) =

110. (d) The period of | sin x | + | cos x | and sin4x + cos4x π . sin (sin x) + sin (cos x) has period 2π. The 2 1 + 2 cos x can be written in a simplifunction sin x (2 + sec x) cos x = cot x, so it has period π. fied form as sin x π 111. (c) tan (3x – 2) is a periodic function with period . 3 The function f (x) = {x} is periodic with period 1. The function in (d) can be written as is

f (x) = 1 –

 5x − x2  exists for log10   4 



5x − x2 ≥ 1 ⇒ 5x – x2 – 4 ≥ 0 ⇒ x ∈ [1, 4]. 4

=

104. (a) Since the domain of f is (0, 1), ∴ 0 < sin x < 1 ⇒ 2nπ < x < (2n + 1) π, n ∈ I. 105. (c) Let n ≤ x < n + 1 Then, f (x) = x ⋅ n, where n changes with x. Clearly no constant k > 0 is possible for which f (x) = f (x + k) corresponding to all x. ∴ f (x) is a non periodic function. 106. (b) The period of f (x) =

π [k ] .

But the period of f (x) = π (Given)

π [k ] = π =

=

cos3 x sin 3 x − sin x + cos x sin x + cos x =1–

sin 3 x + cos3 x sin x + cos x

2 2 1 – (sin x + cos x)(sin x + cos x − sin x cos x) (sin x + cos x) 1  1  1 – 1 − sin 2 x  = sin 2x  2  2

which is periodic with period π. The function x + cos x is non periodic as x is non periodic. 112. (b) We have, 2 sin 8 x cos x − 2 sin 6 x cos 3x f (x) = 2 cos 2 x cos x − 2 sin 3x sin 4 x

=

(sin 9 x + sin 7 x) − (sin 9 x + sin 3x) (cos 3 x + cos x) + (cos 7 x − cos x)

sin 7 x − sin 3 x 2 cos 5 x sin 2 x = = tan 2x, cos 7 x + cos 3 x 2 cos 5 x cos 2 x π ⇒ [k] = 1 ⇒ 1 ≤ k < 2. . which is periodic with period 2 2π 107. (c) The period of sin 5x is and that of cos 3x 113. (a) We have 5 (sin 5 x + sin x) + (sin 4 x + sin 2 x) 2π 2π f (x) = 2π (cos 5 x + cos x) + (cos 4 x + cos 2 x) is . As and do not have a common 3 3 5 2 sin 3x cos 2 x + 2 sin 3x cos x = multiple, f (x) is non periodic. 2 cos 3x cos 2 x + 2 cos 3x cos x πx has 108. (a) 3x + 3 – [3x + 3] has the period 1 and sin 2 sin 3 x (cos 2 x + cos x) 2 = 2 cos 3 x (cos 2 x + cos x) = tan 3x, 2π i.e., 4. Therefore, the period of f (x) the period π π 2 . which is periodic with period 3 is L.C.M. (1, 4) = 4. 114. (b) We have, f (x + 2a) = f ((x + a) + a) 109. (a) For n = 2, we have 1 2 = – f ( x + a ) − ( f ( x + a )) 4 sin x 2 cos x 2 cos x sin 2 x 2 sin x cos x 2 = = f (x) = x x x 1 sin   sin   sin   = 2 2 2 2 x 2 = 4 cos cos x. 1 1  2 − f ( x) − ( f ( x)) 2 −  − f ( x) − ( f ( x)) 2  – x 2  2 The period of cos x is 2π and that of cos is 2 1 1 1 1 2π − f ( x) + ( f ( x)) 2 = −  − f ( x)  = f (x). = – = 4π. 4  2 2 2 1/ 2 ∴

[k ] = 1

Hence the period of f (x) is 4π.



=

Hence, f (x) is periodic with period 2a.

r =1

= f (1) + 2f (1) + 3f (1) + ... + n f (1) [since f (x + y) = f (x) + f ( y)] = (1 + 2 + 3 + ... + n) f (1) =

n (n + 1) 7 n (n + 1) . ⋅7 = 2 2

−1 ± 5  ...(1) 2 Since f (x) is an even function defined on [– 5, 5], ⇒x=

∴

f (– x) = f (x), ∨ x ∈ [– 5, 5]

 x +1  ⇒ x = –   x + 2

therefore sin (log (x +

x 2 + 1 )) must be odd.

123. (c) A function whose graph is symmetrical about the origin must be odd. (3x + 3– x) is an even function. Since cos x is an even function and log (x +

1 + x2 ) 1 + x 2 )) is

is an odd function, ∴ cos (log (x + an even function. If f (x + y) = f (x) + f (y) ∨ x, y ∈ R, then f (x) must be odd (See Q. 122).

⇒ x2 + 3x + 1 = 0

−3 ± 5 . 2 From (1) and (2), the values of x are

⇒ x=

x 2 + 1 ) are odd functions,

4 4 Also, sec x + cosec x is an odd function. 3 4 x + x cot x Now, let f (x + y) = f (x) + f (y) ∨ x, y ∈ R ∴ f (0 + 0) = f (0) + f (0) ∴ f (0) = 0. f (x – x) = f (x) + f (– x) or 0 = f (x) + f (– x) i.e. f (– x) = – f (x) ∴ f (x) is odd.

 x +1  x +1 117. (a), (b) Since f (x) = f    ⇒ x = x + 2 x+2 ⇒ x2 + x – 1 = 0

Since sin x and log (x +

..(2)

124. (c) g( f (x)) = g(sin x + cos x)  π π It will be invertible when x ∈  − ,  .  4 4 125. (d) –2 ≤ sin x – 3 cos x ≤ 2

−1 ± 5 −3 ± 5 . and 2 2 ⇒ –1 ≤ sin x – 3 cos x +1 ≤ 3 118. (a) Since the function sec x is an even function and 2 ∴ range of f = [–1, 3] log (x + 1 + x ) is an odd function, therefore For f to be onto S = [–1, 3] 2 the function sec [log (x + 1 + x )] is an even function. 126. (b) Graph is symmetrical about the line x = 2 ⇒ f (2 – x) = f (2 + x) 3 + log10 ( x 3 − x) 119. (d) f (x) = 2 [Graph is symmetrical about y-axis 4− x For domain of f (x) i.e., about x = 0 iff f (–x) = f (x) x3 – x > 0 i.e., f (0 – x) = f (0 + x)]. ⇒ x (x – 1) (x + 1) > 0. x1 = x2 127. (d) Let f (x1) = f (x2) ⇒ [x1] = [x2] ⇒  the region is (– 1, 0) ∪ (1, ∞) 2 Also,  4 – x ≠ 0 ⇒ x ≠ ± 2 [For example, if x1 = 1.4 and x2 = 1.5, then [1.4] = [1.5] = 1] ⇒  the region is (– ∞, 2) ∪ (2, ∞) ∴ f is not one-one. ∴  Common region is (– 1, 0) ∪ (1, 2) ∪ (2, ∞).

31

Functions

120. (a) To make f (x) an even function in the interval n − 1 n , where is odd [– 1, 1], we re-define f (x) as follows:  2 115. (c) f (n) =   f ( x), 0 ≤ x ≤ 1   f (x) =  − n , where n is even  f (− x), − 1 ≤ x < 0  2 3 x 2 − 4 x + 8 log (1 + | x |), 0 ≤ x ≤ 1  and f : N → I, where N is the set of natural numbers =  2  and I is the set of integers. 3 x + 4 x + 8 log (1 + | x |), − 1 ≤ x < 0 Let x, y, ∈ N and both are even, then f (x) = f ( y) Thus, the even extension of f (x) to the interval x y = –  ⇒ x = y. Again x, y ∈ N and both ⇒– [– 1, 1] is 3x2 + 4x + 8 log (1 + | x | ). 2 2 121. (b) f (x) = log( x + x 2 + 1) are odd, then f (x) = f ( y) 2 ⇒ f (– x) = log(− x + x + 1) x −1 y −1 ⇒ ⇒x=y = ⇒ f (x) + f (– x) = 2 2 i.e., mapping is one-one. log( x + x 2 + 1) + log(− x + x 2 + 1) Since each negative integer is an image of even = log (1) = 0 natural number and positive integer is an image of Hence f (x) is an odd function. odd natural number. So mapping is onto. Hence, mapping is one-one onto. 122. (d) A function whose graph is symmetrical about the n y-axis must be even. 116. (d) ∑ f (r ) = f (1) + f (2) + f (3) + ... + f (n)

32

Objective Mathematics

Also, f is not onto as its range I (set of integers) is a proper subset of its co-domain R.

134. (c) The total number of bijective functions from a set A containing 106 elements to itself is (106)!.

128. (d) Since sin (2x – 3) is a periodic function with period π, therefore f is not injective. Also, f is not surjective as its range [– 1, 1] is a proper subset of its co-domain R.

135. (b) We know that if X and Y are any two finite sets having m and n elements respectively, where 1 ≤ n ≤ m, then the number of onto functions from X to Y is given by

129. (a) The function f (x) = x5 is not a surjective map from I to itself as 3 ∈ I does not have any preimage in I. The function f (x) = 3x + 2 is not a surjection from I to itself as 4 ∈ I does not have any pre-image in I. The function f (x) = x2 + x is not one-one as f (– 4) = f (3) but – 4 ≠ 3. The function f (x) = x + 3 is a bijection from I to itself. 130. (b) The function f (x) = x2 + 2, x ∈ (– ∞, ∞) is not injective as f (1) = f (– 1) but 1 ≠ – 1. The function f (x) = (x – 4) (x – 5), x ∈ (– ∞, ∞) is not one-one as f (4) = f (5) but 4 ≠ 5.



n

∑ (−1)

n−r n

r =1

Cr r m .

Thus, the number of surjective mappings is 2



∑ (−1) r =1

2− r 2

C r r n = (2n – 2).

136. (c) We know that if X and Y are any two finite sets having m and n elements respectively, then the number of one-one functions from X to Y is  n Pm if n ≥ m  given by =   . Thus, the number of  0 if n < m 4 injective mappings is P3 = 24.

4 x 2 + 3x − 5 , x ∈ (– ∞, ∞) is 137. (c) Let t = 5x + 2, then A = {t : 0 ≤ t ≤ π} 4 + 3x − 5 x 2 ∴  f (t) = cos t which is bijective in [0, π]. also not injective as f (1) = f (– 1) but 1 ≠ – 1. Hence, f (x) is bijective. For the function, f (x) = | x + 2 |, x ∈ [– 2, ∞). Let f (x) = f (y), x, y ∈ [– 2, ∞) ⇒ | x + 2 | = 138. (b) Let f (x1) = f (x2) ⇒ x1 – [x1] = x2 – [x2] x1 = x2 for if x1 = 2.7 and x2 = 3.7 | y + 2 | then f (x1) = f (x2) = 0.7 but x1 ≠ x2 ⇒x+2= y+2 ∴ f is not one-one. Clearly, f is onto. ⇒ x = y. So, f is an injection. 139. (c) We have, f (x) = [x]2 + [x + 1] – 3 = [x]2 + [x] + 1 – 3 131. (c) Let x, y be any two elements in A such that = [x]2 + [x] – 2 = ([x] + 2) ([x] – 1) f (x) = f (y) ⇒ x | x | = y | y | ⇒ x = y ∴ f is injective. f (x) = 0 ⇒ [x] + 2 = 0 or [x] – 1 = 0 i.e., [x] = – 2 or [x] = 1 Also, range of f = f (A) = B, ∴ f is surjective. Hence, f is a bijection. [x] = – 2 ⇒ – 2 ≤ x < – 1 [x] = 1 ⇒ 1 ≤ x < 2. 132. (b) Since the function f is bijective, therefore f is onto. Therefore range of f = B. 140. (a) Since, for different x, 4x and 4| x | are different Let y = – x2 + 6x – 8 ⇒ x2 – 6x + (8 + y) = 0 positive numbers, ∴ f is one-one. Also, f is not onto as its range 6 ± 36 − 4 (8 + y ) 6 ± 4 − 4y (0, ∞) is a proper subset of its co-domain R. ⇒ x = = 2 2 141. (d) Since f (x) = f (– x) ∴ f is not one-one. For x to be real, 4 – 4y ≥ 0 ⇒ y ≤ 1 Let y ∈ R. Then f (x) = y ∴ B = range of f = (– ∞, 1]. x2 − 8 ⇒y = 2 133. (a) Let x, y ∈ R be such that x +2 x−a y−a = f (x) = f (y) ⇒ 8 + 2y x−b y−b ⇒ x2 = . 1− y ⇒ (x – a) (y – b) = (x – b) (y – a) For x to be real, (8 + 2y) (1 – y) ≥ 0 ⇒ xy – bx – ay + ab = xy – ax – by + ab and 1 – y ≠ 0 ⇒ (y + 4) (y – 1) ≤ 0 and y ≠ 1 ⇒ (a – b) x = (a – b) y ⇒ x = y ⇒–4 ≤y 1 

Objective Mathematics

161. (c) We have,

x ≤ 1 and | x | ≤ 1 | x |, f (| x |) = 2 − | x |, x ≤ 1 and | x |> 1 =  | x |, − 1 ≤ x ≤ 1  2 − | x |, x < −1



= | 2 – x |, x > 1 ∴

⇒ –

3 1 ≤x≤ 2 2

 3 1 Hence, the domain of f [g (x)] is  − ,  .  2 2 157. (b) ( fog) (x) = f [g (x)] = f ( | 3x + 4 | ). Since the domain of f is [– 3, 5], ∴ – 3 ≤ | 3x + 4 | ≤ 5 ⇒ | 3x + 4 | ≤ 5 ⇒ – 5 ≤ 3x + 4 ≤ 5 ⇒ – 9 ≤ 3x ≤ 1 1 ⇒–3 ≤x≤ . 3 1  ∴ Domain of fog is  −3,  . 3 

=

1− x x −1 = . −x x

=



= (3)3 – 12 ⋅ 3 = 27 – 36 = – 9.



156. (d) f [g (x)] = f ( 1 − 2 x ) = 2 − 1 − 2 x For f [g (x)] to be defined, 1 – 2x ≥ 0 and 2 –

1 158. (a) We have f [ f (x)] = = 1 − f ( x)





2 − | x |, x < −1  f [ f (x)] = | x |, − 1 ≤ x ≤ 1 . | 2 − x |, x > 1 

1 − 2x ≥ 0 1 3 ⇒ x≤ and x ≥ –  2 2

f (a) = 64 a3 + 3

x > 1 and 2 − x ≤ 1 | 2 − x |, f (2 – x) = 2 − (2 − x), x > 1 and 2 − x > 1 (impossible) 

1 1 3 3 = (4a) + a a3



1 1 1   4a +  − 3 ⋅ 4a ⋅  4a +  a a a

1   Since a, b are roots of 4 x + x = 3    ∴ 4a + 1 = 3   a Similarly, f (b) = – 9 ∴ f (a) = f (b) = – 9.

162. (a) Since f (x) = x2 – 1 and g (x) =

1−

1 1− x

2

163. (c) We have f (x) ⋅ f (y) –

 1   x   f   + f ( xy ) 2   y  

= cos (log x) ⋅ cos (log y) –

 1    x   cos log    + cos {log ( xy )} 2  y     

= cos (log x) ⋅ cos (log y) – + cos (log x + log y)}

1

x2 + 1 ∨ x ∈ R

∴ f [g (x)] = [g (x)] – 1 = x + 1 – 1 = x 2, ∨ x ∈ R ∴ [ho ( fog)] (x) = h [( fog) (x)] = h [ f {g (x)}] = h (x2) = x2 [∵ x2 ≥ 0] 2

1 {cos (log x – log y) 2

1 = cos (log x) ⋅ cos (log y) – ⋅ 2 cos (log x) 2 cos (log y) = cos (log x) ⋅ cos (log y) – cos (log x) ⋅ cos (log y) = 0.

164. (d) 1 = x. 165. (b) We have, x −1 1− 1 1 x 3 f (x) + 5 f    = – 3, ∨ x (≠ 0) ∈ R  ...(1)   x x Hence, the graph of the function y = f [ f { f (x)}], x > 1 is a straight line passing through origin. 1 ⇒ 3 f    + 5 f (x) = x – 3  ...(2) 159. (c) Since x – [x] = 0 for all integral values of x, therex –1 fore, the function is not one-one. Hence f  (x) does 1  not exist.  Replacing x by x  160. (c) Let x + y = a and x – y = b Multiplying (1) by 3 and (2) by 5 and subtracting, a+b a−b and y = ⇒ x = we get 2 2 3  9 f (x) – 25 f (x) =  − 9  – (5x – 15) ∴ f (x + y, x – y) = xy x  a+b a−b a 2 − b2 3 ⇒ f (a, b) = = ⋅ ⇒ – 14 f (x) = – 5x + 6 2 2 4 x 2 2 2 2 x −y y −x 1  3  ∴ f (x, y) = and f ( y, x) = ⇒ f (x) =    − + 5 x − 6  , ∨ x (≠ 0) ∈ R. 4 4  14  x ∴

1 f [f {f (x)}] = 1 − f ( f ( x)) =



35

∴ f   π  = sin 9π – sin 5π = 1 – 0 = 1 2 2 f (π) = sin 9π – sin 10π = 0

9π 1 π f    = sin – sin 10π = – 1. 4 2 4 4 172. (c) For x ≤ – 1, 1 – x ≥ 2 and 1 – x ≥ 1 + x ∴ max {(1 – x), 2, (1 + x)} = 1 – x For – 1 < x < 1, 0 < 1 – x < 2 and 0 < 1 + x < 2. ∴ max {(1 + x), 2, (1 + x)} = 2. For x ≥ 1, 1 + x ≥ 2, 1 + x > 1 – x ∴ max {(1 – x), 2, (1 + x)} = 1 + x

167. (b) We have,

 1  = g (x) = f [ f (x)] = f   1 − x 



Also,



 x − 1 = h (x) = f [ f { f (x)}] = f   x 



∴

1 1 1− 1− x

=

x −1 . x

1 = x. x −1 1− x 1 x −1 ⋅ ⋅ x = – 1. f (x) ⋅ g (x) ⋅ h (x) = 1− x x

168. (a) When f (x) = sin2x and g (x) =

x,

( fog) (x) = f [g (x)] = f ( x ) = (sin x )2 and (gof ) (x) = g [ f (x)] = g (sin2x) = | sin x | When f (x) = sin x and g (x) = | x | ( fog) (x) = f (g (x)) = f (| x |) = sin | x | ≠ (sin x )2 When f (x) = x2 and g (x) = sin x ( fog) (x) = f [g (x)] = f (sin x ) = (sin x )2 and (gof ) (x) = g [ f (x)] = g (x2) = sin x 2 = sin | x | ≠ | sin x |. 169. (b) We have,

171. (a) We have, f (x) = sin [π2] x + sin [– π2] x = sin 9x + sin (– 10) x = sin 9x – sin 10x

−1, − 2 ≤ x ≤ 0 f (x) =   x − 1, 0 < x ≤ 2 Since x ∈ [– 2, 2], therefore | x | ∈ [0, 2]. Therefore, f (| x | ) = | x | – 1, ∨ x ∈ [– 2, 2] − x − 1, x ∈[−2, 0]  =   x − 1, x ∈[0, 2] Also, −2≤ x 0, then f (– 1) = 0 and f (1) = 2. ⇒ – a + b = 0 and a + b = 2 ⇒ a = 1 and b = 1. If a < 0, then f (– 1) = 2 and f (1) = 0 ⇒ – a + b = 2 and a + b = 0 ⇒ a = – 1 and b = 1. Hence, f (x) = x + 1 or f (x) = – x + 1. 174. (c) The length of the side of the equilateral triangle is

( x − 0) 2 + ( f ( x) − 0) 2 i.e.

x 2 + ( f ( x)) 2 .

∴ Area of the equilateral triangle =

3 4

[x2 + { f (x)}2] But the area = ...(1)

3 (given) 4

3 2 [x + { f (x)}2] = 4 ⇒ x2 + { f (x)}2 = 1

∴

...(2)

− x − 1 + 1, − 2 ≤ x < 0  g (x) = f (| x |) + | f (x) | =  x − 1 + 1 − x, 0 ≤ x < 1  x − 1 + x − 1, 1 ≤ x ≤ 2   − x, − 2 ≤ x < 0  0 ≤ x 0 =

1 1 ⇒ x ∈ (– ∞, – 1) ∪  − , 0  ∪  , ∞  . 2    2 2  x2 + 1 Thus (a) and (d) are the correct answers. is maximum, i.e. when x2 + 1 is minimum i.e. at x = 0. x+2 183. (a) Let f (x) = 2 ∴ minimum value of f (x) is f (0) = – 1. x − 8x − 4 177. (b) The maximum value of f (x) = cos x + cos ( 2 For f (x) to be defined, x2 – 8x – 4 ≠ 0 ⇒ x ≠ 4 x) is 2 which occurs at x = 0. Also, there is no ±2 value of x for which this value will be attained ∴ Domain of f = R\{4 + 2 ,4–2 } again. x+2 For range, put y = 2 178. (d) We have, 2x + 2y = 2 ⇒ 2y = 2 – 2x x − 8x − 4 log (2 − 2 x ) ⇒ x2y – (8y + 1) x – (4y + 2) = 0 ⇒ y = log 2 For x to be real, (8y + 1)2 + 4y (4y + 2) ≥ 0 and x x y ≠0 For y to be defined, 2 – 2 > 0 ⇒ 2 < 2 x ⇒ 80y2 + 24y + 1 ≥ 0 and y ≠ 0 Since, 2 is an increasing function, therefore we ⇒ (20y + 1) (4y + 1) ≥ 0 and y ≠ 0 get x < 1. 1 ⇒ y ≤ – 179. (a) Since f (x + y) = f (x) ⋅ f ( y) for all natural numbers 4 x and y and f (1) = 2, 1 so, f (2) = f (1 + 1) = [ f (1)]2 = 22  or y ≥ – and y ≠ 0 20 and f (3) = f (2 + 1) = f (2) ⋅ f (1) = 23. Also, if y = 0, we get x + 2 = 0 Continuing this process, we find that f (n) = 2n, ⇒x=–2 k since if f (k) = 2 , then so that y = 0 is also in the range. k k + 1 f (k + 1) = f (k) ⋅ f (1) = 2 ⋅ 2 = 2 . n

n

∑ f (a + k ) = ∑ f (a) f (k )

Now,

k =1

= f (a)

n

∑2

= f (a)

k =1

k

n

∑ f (k ) k =1

n

= 2 f (a) (2 – 1)

...(1)

k =1

n

It is given that

∑ f (a + k ) = 16 (2

n

– 1)

...(2)

k =1

From (1) and (2), we get f (a) = 8. So, f (a) = 23 ⇒ 2a = 23. ∴ a = 3. 180. (a), (c) [π ] = 9 and [– π ] = – 10 Thus, f (x) = cos [π2] x + cos [– π2] x = cos 9x + cos (– 10) x = cos 9x + cos 10x 2

2



π 9π So, f   = cos + cos 5π = – 1 2 2



f (π) = c os 9π + cos 10π = – 1 + 1 = 0 f (– π) = cos (– 9π) + cos (– 10π) = – 1 + 1 = 0



π 9π 10 π f   = cos + cos = 4 4 4

181. (b) g (x) = f [f (x)] = f [ | x – 2 | ] = f (x – 2) as x > 20

1 1 − = 0. 2 2

1  1   ∴ Range of f =  −∞, −  ∪  − , ∞  .  4   20 

184. (b) We have, ( fog) (x) = f [g (x)] =f

(

) (

x2 + 1 =

x2 + 1

)

2

– 1 = x2

∴ (hofog) (x) = h [( fog) (x)] = h (x2) 0 if x = 0 =  2 .  x if x ≠ 0  2 185. (a) For f (x) to be defined, x + 2x ≠ 0 ⇒ x (x + 2) ≠ 0 i.e. x ≠ 0, – 2. ∴  Domain of f = R\{0, – 2} x2 − x , x ≠ 0, – 2 For range, put y = 2 x + 2x 2y +1 x −1 ⇒ yx + 2y = x – 1 or x = , ⇒ y = 1− y x+2 which is not defined for y = 1. 1 and x ≠ – 2 for any real Also, x ≠ 0 ⇒ y ≠ – 2 value of y. 

1 ∴  Range of f = R\ 1, −  . 2 

 x + 2 x > −2   x + 2 = 1,  186. (a) Since, f (x) =  . 2 x +  = −1, x < −2   − ( x + 2)

196. (b) Let f (x) =

∴ Range of f = {1, – 1}. ( x 2 − 3)3 + 27

187. (a) 2 is minimum when (x2 – 3)3 + 27 is minimum. Since (x2 – 3)3 + 27 = x6 – 9x4 + 27x2



1  which clearly contains  , 3  . 2  197. (b) The number of functions = 23 = 8. π 0 i.e. x > 1 ∴ Domain of f = (1, ∞).

= x ⋅

x2 − x + 1 x −1 xy – y = x2 – x + 1 x2 – (y + 1) x + (y + 1) = 0 x to be real, (y + 1)2 – 4 (y + 1) ≥ 0 (y + 1) (y – 3) ≥ 0 ⇒ y ≤ – 1 or y ≥ 3 y cannot lie between – 1 and 3.

222. (a) Let y =

216. (d) Since f (x) is an odd periodic function with period  2 ∴ f (– x) = – f (x) and f (x + 2) = f (x) ∴ f (2) = f (0 + 2) = f (0) and f (– 2) = f (– 2 + 2) = f (0) Now, f (0) = f (– 2) = – f (2) = – f (0) ⇒ 2 f (0) = 0 i.e. f (0) = 0 ∴ f (4) = f (2 + 2) = f (2) = f (0) = 0 Thus, f (4) = 0.



y2 + 289 – 34y ≥ – 7y2 – 71 + 78y 8y2 – 112y + 360 ≥ 0 y2 – 14y + 45 ≥ 0 (y – 9) (y – 5) ≥ 0 ⇒ y ≤ 5 or y ≥ 9 y cannot lie between 5 and 9.

39

⇒ ⇒ ⇒ ⇒ ∴

 y +1 1  log    3 − y  2

Functions

∴ x =

[∵ Discriminent ≥ 0]



2x   1 + x2 − 2x  1 − 1 + x 2   2x  f   = log    = log  1 + x 2 + 2 x   1 + x 2  1 + 2 x  2  1 + x 



= log 

1 − x  1 − x  = 2 log   = 2 f (x).  1 + x   1 + x  2

224. (d) Let x1, x2 ∈ R, then f (x1) = f (x2) ⇒ x12 + x1 = x22 + x2 ⇒ (x1 – x2) (x1 + x2 + 1) = 0. ∴ x2 = x1 or x2 = – (x1 + 1). ∴ f is many one. Again, let y ∈ R and x be such that 1  [– 1 ± 1 − 4 y ]. f (x) = y ⇒ x2 + x = y ⇒ x = 2 1 x ∉ R, if y > , ∴ f is into mapping. 4 Hence, f is many one into mapping. 225. (a) We have, f (x) =

2 x + 2− x 2

∴ f (x + y) ⋅ f (x – y)

1 x + y 1  (2 + 2– x – y) ⋅   (2x – y + 2– x + y) 2 2 1 [(22x + 2– 2x) + (22y + 2–2y)] = 4 1  [ f (2x) + f (2y)].  = 2 226. (b) We have, g (x) = 1 + x =

and f [g (x)] = 3 + 2 x + x

...(1)

Also, f [g (x)] = f (1 +

...(2)

x ) From (1) and (2), we get

f (1 +

x ) = 3 + 2 x + x.

x = y or x = ( y – 1)2. ∴ f ( y) = 3 + 2 ( y – 1) + ( y – 1)2 = 3 + 2y – 2 + y2 – 2y + 1 = 2 + y2 ∴ f (x) = 2 + x2.

Let 1 +

227. (d) We have, f (x) = (a – xn)1/n = y ∴ f ( y) = (a – yn)1/n = [a – {(a – xn)1/n}n]1/n

= [a – (a – xn)]1/n = (xn)1/n = x.

40

Objective Mathematics

228. (d) For y to be defined, we must have x2 – 1 ≥ 0 and x – 1 > 0 ⇒ | x | ≥ 1 and x > 1 ⇒ x > 1. ∴ Domain of y is (1, ∞).

237. (b) We have, ho(fog) (x) = hof{g(x)} = hof { ( x 2 + 1)}

229. (d) Since the function is constant ∨ x ∈ R, therefore, the value of the constant is obtained by assigning any suitable value to x. Choose x = 0, the desired value of the constant π π = cos20 + cos2 – cos 0 ⋅ cos  3 3 1 1 5 1 3 =1+ − = − = . 4 2 4 2 4 230. (d) We have, 2x > 0, ∨ x ∈ R Thus, 2x = 2 – 2y < 2, ∨ x ∈ R. As 2x is increasing function, we get – ∞ < x < 1. 231. (d) We have,



 1 + x1   1 + x2  f ( x1 ) + f ( x2 ) = log   + log   1 x − 1    1 − x2   1 + x1   1 + x2   = log      1 − x1   1 − x2  

x1 + x2  1 + 1 + x x 1 2 = log   1 − x1 + x2  1 + x1 x2

   x1 + x2   = f   1 + x1 x2   

= h{( x 2 + 1) 2 − 1} = h{x2 + 1 – 1} = x2 238. (b) We have, f {g ( x)} = eloge x = x g{f(x)} = loge ex = x f{g(x)} = g{f(x)}

and ∴ 239. (a) Clearly,    ⇒  ⇒  ⇒  ∴ 

x −1 ≤ log 3   ≤ 1 3 1 x ≤ ≤3 3 3 1≤x≤9 x ∈ [1, 9]   x  Domain of sin −1 log 3    is [1, 9].  3  

240. (a) Let x, y ∈ N be such that f(x) = f(y) ⇒  ⇒  ⇒  ∴ 

x2 + x + 1 = y2 + y + 1 (x – y) (x + y + 1) = 0 x = y or x = (– y – 1) ∉ N f is one-one

  Also, f is onto. 241. (a) We have,

232. (a) Clearly, the domain of f is π π = R ∩ [0, 4] ∩  x : − < x < and cos x > 0  2 2   π π π = R ∩ [0, 4] ∩  ,  = 0,  2 2  2 233. (a) Since every element of set A can be associated to any element in set B, therefore, total number of mappings from A to B is 34. 234. (a) Since, 0 ≤ x ≤ π ⇒ 0 ≤ x ≤ π 2 2 4 x 1 ∴  0 ≤ sin ≤ 2 2  1  ⇒   0,  ⊂ [0, ∞) 2  Hence, function is injective.

x – 1 > 0 and 2x – 1 > 0 and 2x – 1 ≠ 1 1 ⇒  x > 1, x > and x ≠ 1 2 ⇒  x > 1 ∴Domain of f(x) is (1, ∞) 242. (c) We have, domain of f is D1 = (–1, 1). For the function g(x) – (2x – 3) (2x + 1) ≥ 0 ⇒  (2x – 3) (2x + 1) ≤ 0 1 3 ⇒  − ≤ x ≤ 2 2  1 ∴  Domain of g(x) is D2 =  − ,  2

3 2 

235. (c) For (a)   and (b)   f is not 1–1, but for (c) f is 1–1.

Therefore, domain of (f + g) = D1 ∩ D2

236. (d) Clearly, from the graph, f is many-one into function.

1 =  − , 1  2 

y

243. (d) We have, y = 4x + 3 ⇒ 

O

x

x=

y −3 4

∴  The inverse of f(x) is g ( x) =

y −3 . 4

1. Let A be a set of n distinct elements. Then the total number of distinct functions from A to A is (a) 2n (b) n2 (c) nn (d) None of these 2. Given f (x) = 4x8, then 1  1  1 (a) f'   = f'  −  (b) f′(x) = f'  −  2  2  2  1 1 (c)  f  −  = f    2 2 3. Let f (x) =

1  1 (d) f   = f'  −  2  2

sin x . If D is the domain of f, then D 1 + sin x

contains. (a) (– 2π, – π) (b) (4π, 6π) (c) (0, π) (d) (2π, 4π) 4. Let the function f : R → R be defined by f (x) = 2x + sin x, x ∈ R. Then f is (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 5. Suppose f (x) = (x + 1)2 for x ≥ – 1, If g (x) is the function whose graph is the reflection of the graph of f (x) with respect to the line y = x, then g (x) equals (a) – x – 1, x ≥ 0

(b)

x + 1 , x ≥ – 1

(d)

(c) 6. 7. 8.

1 ,x>–1 ( x + 1) 2

x, x ≥ 0

(a) f (– a)

1 (b) f    a

(c) f (a2)

 −a  (d) f    a − 1 

12. Let f (x) = x, g (x) = = 1, if

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

is is is is

1 and h (x) = f (x) ⋅ g (x). Then, h (x) x

an irrational number a real number ≠ 0 a real number a rational number

13. If the real-valued function f (x) = n equals (b)

1 (c) –  3

(d) 2

1 + 2 x − x 2 is [ x]

(a) (1, 2] (b) (0, 2) (c) (0, 1) (d) [1, 2] 15. The range of the function π −π 0

41

Functions

EXERCISES FOR SELF-PRACTICE

42

Objective Mathematics

 x, when x is an even integer 19. If f (x) =  , then f (x) is 0, when x is an odd integer (a) even (b) odd (c) both even and odd (d) neither even nor odd 20. If f (x) = 3x + 10, g (x) = x2 – 1, then ( fog)– 1 is equal to: 1/ 2 1/ 2  x + 7  x − 7 (a)  (b)     3   3 

x − 3 (c)   7 

x + 3 (d)   7 

1/ 2



1/ 2

23. The value of 24.

2 + 2 + 2 + ..... ∞ is:

(a) 5 (c) 2 If 3x – 3x – 1 = 6, then xx (a) 2 (c) 9

(b) 3 (d) None of these is equal to: (b) 4 (d) None of these

25. Inverse of the function y = 2x – 3 is: x+3 2 1 (c) 2x − 3

(a)

(b)

x−3 2

(d) None of these

26. If x is real, then value of the expression

21. Domain of the function

lies between:

1/ 2

  5x − x2  is: f (x) = log10  4     (a) – ∞ ≤ x ≤ ∞ (b) 1 ≤ x ≤ 4 (c) 4 ≤ x ≤ 16 (d) – 1 ≤ x ≤ 1

(a) 5 and 4 (c) – 5 and 4

x 2 + 14 x + 9 x2 + 2x + 3

(b) 5 and – 4 (d) None of these

27. Which of the following is one-one function: (a) ex (c) sin x

22. Let f : R → R be a function defined by: x−m f (x) = , where m ≠ n. Then: x−n (a) f is one-one onto (b) f is one-one into (c) f is many one onto (d) f is many one into

2

(b) ex (d) None of these

28. The value of ex – x is always: (a) greater than 1 for all real values (b) less than 1 for all real values (c) greater than 1 for some real values (d) None of these

Answers

1. (c) 11. (c) 21. (b)

2. (c) 12. (b) 22. (b)

3. (c) 13. (c) 23. (c)

4. (a) 14. (a) 24. (b)

5. (d) 15. (b) 25. (a)

6. (a) 16. (b) 26. (c)

7. (b) 17. (a) 27. (a)

8. (a) 18. (d) 28. (c)

9. (a) 19. (b)

10. (d) 20. (a)

2

Limits

CHAPTER

Summary of conceptS Limit defined Let a function f be defined at every point in the neighbourhood of a (an open interval about a) except possibly at a. If as x approaches closer and closer to a, but not equal to a, then the value of the function f (x) approaches a real number l. The number l is referred to as the limit of f (x) as x tends to a and we write it as

1.

lim [ f ( x) + g ( x)] = lim f ( x) + lim g ( x) = l + m. x→a x→a x→a

2.

lim [ f ( x) − g ( x)] = lim f ( x) – lim g ( x) = l – m. x→a x→a

3.

lim k ⋅ f ( x) = k ⋅ lim g ( x) = kl, x→a x→a

x→a

where k is a constant.

lim f ( x) = l

4.

f ( x) ⋅ lim g ( x) = lm. lim [ f ( x) ⋅ g ( x)] = lim x→a

Note that f (x) approaches l means the absolute difference between f (x) and l, i.e., | f (x) – l | can be made as small as we please. When the values of f (x) do not approach a single finite value as x approaches a, we say that the limit does not exist.

5.

lim f ( x) l x→a  f ( x)  = = lim   lim g ( x ) m x → a g ( x)   x→a

6.

f [ g ( x)] = f lim g ( x) = f (m). lim ( fog ) ( x) = lim x→a x→a

x→a

righthand Limit We say that right hand limit of f (x) as x tends to ‘a’ exists and is equal to l1 if as x approaches ‘a’ through values greater than ‘a’, the values of f (x) approach a definite unique real number l1 and we write

x→a

x→a

(provided m ≠ 0).

(

x→a

In particular,

(

)

)

log g ( x) = log lim g ( x) = log m. (i) lim x→a x→a (ii) lim e g ( x ) = e

lim f ( x) = l1 or f (a + 0) = l1.

lim g ( x ) x→a

x→a

x → a+

= em. n

7.

Working Rule To evaluate lim f ( x)

8.

x → a+

1. Put x = a + h in f (x) to get lim f ( a + h) 2. Take the limit as h → 0.

h→0

lim[ f ( x)]n = lim f ( x)  = ln, for all n ∈ N.  x→a  x → a Sandwich Theorem (or Squeeze Principle). If f, g and h are functions such that f (x) ≤ g (x) ≤ h (x) for all x in some neighbourhood of the point a (except possibly at h ( x) , then lim g ( x) = l. x = a) and if lim f ( x) = l = lim x→a x→a

Lefthand Limit We say that left hand limit of f (x) as x tends to ‘a’ exists and is equal to l2 if as x approaches ‘a’ through values less than ‘a’, the values of f (x) approach a definite unique real number l2 and we write lim f ( x) = l2 or f (a – 0) = l2.

x → a−

x→a

evaLuation of LimitS The problems on limits can be divided into the following categories:

Working Rule To evaluate lim f ( x) . x → a−

1. Put x = a – h in f (x) to get lim f ( a − h) . h→0 2. Take the limit as h → 0.

aLgebra of LimitS If lim f ( x) = l and lim g ( x) = m, then following results are x→a

true:

x→a

algebraic Limits The following methods are useful for evaluating limits of algebraic functions:

44

method of factorisation and g (a) ≠ 0, then we have

If f (x) and g (x) are polynomials

Objective Mathematics

f ( x) f (a) f ( x) lim lim = x→a = . x → a g ( x) lim g ( x) g (a ) x→a

Now, if f (a) = 0 = g (a), then (x – a) is a factor of both f (x) and g (x). We cancel this common factor (x – a) from both the numerator and denominator and again put x = a in the given expression. If we get a meaningful number then that number is the limit of the given expression, otherwise we repeat this process till we get a meaningful number.

Some Useful Summations n (n + 1) 2 n (n + 1) (2n + 1) (ii) Σ n2 = 12 + 22 + 32 + ... + n2 = 6 (i) Σ n = 1 + 2 + 3 + ... + n =

2

 n (n + 1)    2  a (1 − r n ) (iv) Σ arn – 1 = a + ar + ar2 + ... + arn – 1 = ; 1− r

(iii) Σ n3 = 13 + 23 + 33 + ... + n3 = 

provided r < 1.

method of rationalisation This method is useful where radical signs (i.e., expressions of the form a ± b ) are in- trigonometric Limits volved either in the numerator or in the denominator or both. The numerator or (and) the denominator (as required) is For finding the limits of trigonometric functions, we use trigo(are) rationalised and limit taken after cancelling out the com- nometric transformations and simplify. The following results are quite useful. mon factors. sin x 1. (i) lim =1 (ii) lim cos x = 1 xn − an x →0 x →0 x n – 1 Standard formula lim = na , where n ∈ Q, the x→a x − a tan x sin −1 x (iii) lim =1 (iv) lim =1 set of rational numbers. x →0 x → 0 x x π tan −1 x sin x 0 =1 (vi) lim = . (v) lim Limit of an algebraic function when x → ∞ x→0 x →0 180 x x f ( x) as x → ∞, g ( x) where f (x) and g (x) are algebraic functions of x, it is convenient to divide all the terms of f (x) and g (x) by the highest power of x in numerator and denominator both and use the following standard limits:

In order to find the limit of a function of the type

(i) (ii)

lim

x →∞

(iv)

if a > 1 if a = 1 if − 1 < a < 1 if a ≤ −1

a0 x p + a1 x p −1 + ... ... + a p −1 x + a p b0 x q + b1 x q −1 + ... ... + bq −1 x + bq  a0 b ,  0 = 0, ∞,  

if p = q if p < q if p > q

x3 x5 + − ... to ∞ 3! 5! 2 4 x x + − ... to ∞ cos x = 1 – 2! 4! 3 2 5 x x + ... to ∞ + tan x = x + 3 15 2 2 2 12 ⋅ x 3 12 ⋅ 32 5 1 ⋅ 3 ⋅ 5 7 x + x + ... to ∞ sin–1x = x + + 7 ! 3! 5! 2 2 2 ⋅ 22 4 2 ⋅ 22 ⋅ 42 6 x + x + x + ... to ∞ (sin–1 x)2 = 2! 4! 6!

(i) sin x = x –

(v)

∞,   1, (i) lim a n =   n →∞ 0 ,  does not exist,

x →∞

Some Useful Expansions

(iii)

1 = 0, if p > 0. lim x →∞ x p

lim

x→a

(ii)

1 =0 x

Some Useful Limits

(ii)

f (a + h) , where a ≠ 0, on taking x = a + h. 2. lim f ( x) = lim h→0

(vi) tan–1 x = x –

x3 x5 + − ... to ∞ . 3 5

exponential and Logarithmic Limits For finding the limits of exponential and logarithmic functions, the following results are useful: (i) lim x→0

ex − 1 =1 x

(ii) lim

ax − 1 = loge a, a > 0 x

(iii) lim x→0

ax − bx a = log e   ; a, b > 0 x b

(iv) lim x→0

(1 + x) n − 1 =n x

x→0

h→0

log x = 0, (m > 0) xm log a (1 + x) = loga e, (a > 0, a ≠ 1) (viii) lim x→0 x x a  (ix) lim 1 +  = ea x →∞ x  (vii) lim

x →∞

f ( x)

(x) (xi)

 1  = e, where f (x) → ∞ as x → ∞. lim 1 +  x →∞ f ( x)   lim (1 + f ( x)1/ f ( x ) = e.

x→a

Some Useful Expansions (i) ex = 1 +

x x 2 x3 + + + ... to ∞ 1! 2! 3!

(ii) e–x = 1 –

x x 2 x3 + − + ... to ∞ 1! 2! 3!

x 2 x3 (iv) loge (1 – x) = – x – − − ... to ∞, – 1 ≤ x < 1 2 3 ( x log a ) 2 + ... to ∞ 2!

n (n − 1) 2 x + ... to ∞, – 1 < x < 1, 2!

(vi) (1 + x)n = 1 + nx +

n being any negative integer or fraction.

The expansion formulae mentioned above can be used with advantage in simplification and evaluation of limits. For example, lim

cos x − 1 +

x→0

x

4

x2 2

4

exists. ∞ (ii)   form : If lim f ( x) = ∞ and lim g ( x) = ∞, then x→a x→a ∞ ′ f ( x) f ( x) lim = lim , provided the limit on the R.H.S. x → a g ( x) x → a g ′ ( x) exists.

Working Rule (i) L’Hospital’s Rule is applicable only when

6

f ( x) becomes of the g ( x)

0 ∞ or . 0 ∞ 0 ∞ (ii) If the form is not or , simplify the given expression till it 0 ∞ ∞ 0 reduces to the form or and then use L’Hospital’s rule. ∞ 0 form is

(iii) For applying L’Hospital’s rule differentiate the numerator and denominator separately.

Note: L’ Hospital’s rule cannot be applied in every problem. Consider the example, x→0

3 x + sin 2 x  ∞  form  . 3 x − sin 2 x  ∞

Here, if we apply L’ Hospital’s rule, we get 3 x + sin 2 x 3 + 2 cos 2 x = lim 3 x − sin 2 x x → ∞ 3 − 2 cos 2 x . Now, both the numerator and denominator are undefined because lim cos 2 x does not exist. lim

x→∞

x→∞

We can find the above limit as:

Some Useful Limits (i) If lim f ( x) = A > 0 and lim g ( x) = B, then x→a

x→a

lim [ f ( x)]g ( x ) = AB. x→a

(ii) If lim f ( x) = 1 and lim g ( x) = ∞, then x→a

0 (i)   form : If lim f ( x) = 0 and lim g ( x) = 0, then x→a x→a 0 ′ f ( x) f ( x) = lim , provided the limit on the R.H.S. lim x → a g ′ ( x) x → a g ( x)

lim

  x x x x2 1 − 2! + 4! − 6! + ... − 1 + 2 = lim x→0 x4 1 1  1 = lim  + terms containing x and its powers  = = . x → 0 4!   4! 24 2

Besides the methods given above to evaluate limits, there is yet another method for finding limits, usually known as L’Hospital’s Rule as given below for in deter minant forms:

Note that sometimes we have to repeat the process if the form is 0 ∞ or again. 0 ∞

x 2 x3 (iii) loge (1 + x) = x – + − ... to ∞, – 1 < x ≤ 1 2 3

(v) ax = ex log a = 1 + x log a +

evaLuation of LimitS uSing L’HoSpitaL’S ruLe

45

n

Limits

1  (v) lim (1 + x)1/ x = lim 1 +  = e x→0 n →∞ n  1/ h a lim ( 1 + a h ) (vi) =e

x→a

lim [ f ( x)]g ( x ) = e x→a

lim g ( x ) [ f ( x ) − 1] x→a

 sin 2 x  3 + 2  2 x  3 + 2(0) 3 x + sin 2 x = lim = lim x→∞ 2 x sin   3 − 2(0) x → ∞ 3 x − sin 2 x 3 − 2  2 x  = 1, since lim

x→∞

.

sin 2 x = 0. 2x

46

multiple-choice questions

Objective Mathematics

Choose the correct alternative in each of the following problems: 1. The integer n for which lim

( cos x − 1) ( cos x − e x )

x →0

x

n

finite nonzero number is (a) 1 (c) 3 2. lim

(b) 2 (d) 4 is equal to

n 2 + 1 + 4n 2 − 1

1 3 1 (c) –  5

(b) – 

(a)

(1 + x)1/ 3 − (1 − x)1/ 3 is x

x→0

2 (a) 3 (c) 1

1 (b) 3 (d) None of these

1 − cos 2 x 2x

x→0

1 3

(d) None of these

3. The value of lim

4. lim

11. In a circle of radius r, an isosceles triangle ABC is inscribed with AB = AC. If the ∆ ABC has perimeter

1 − 2 + 3 − 4 + 5 − 6 + ... − 2n

n →∞

is (b) – 1 (d) Does not exist

 x 4 sin(1/x) + x 2  5. The value of lim   is x→∞ 1+ | x |3   (a) 1 (b) – 1 (c) 0 (d) ∞  3x 2 + 2 x + 1  6. lim  2  x →∞  x +x+2  (a) 3 (c) 1

is equal to (b) 9 (d) None of these

  7. The value of lim  x + x + x − x  is x →∞   1 (a) 2 (b) 1 (c) 0 (d) None of these x

 x + 5x + 3  8. lim  2  equals x→0  x +x+2  2

P = 2  2 hr − h 2 + 2 hr  and area A = h 2 hr − h 2 ,   A where h is the altitude from A to BC, then lim 3 is equal h → 0+ P to 1 (a) 128 r (b) 128r 1 (c) (d) None of these 64 r

2 x + 23 − x − 6 is equal to 2− x / 2 − 21− x (a) 8 (b) 4 (c) 2 (d) None of these

12. lim x→2

n k sin 2 (n!) , 0 < k < 1, is equal to n+2 (a) ∞ (b) 1 (c) 0 (d) None of these

13. lim

(a) λ (c) zero

6x +1 3x + 2

x + cos x is is a 10. The value of xlim →∞ x + sin x (a) – 1 (b) 0 (c) 1 (d) None of these

(a) e4 (c) e3

(b) e2 (d) e

n →∞

14. Let f (2) = 4 and f' (2) = xf (2) − 2 f ( x) Then lim x→2 x−2 (a) 2 (c) – 4

is given by (b) – 2 (d) 3

2 x +3 3 x +5 5 x is x →∞ 3x − 2 + 3 2 x − 3

15. The value of lim

2 3 1 (c) 3 (a)

16. lim x→0

(a)

(

(b)

x 3 z 2 − ( z − x) 2 3

8 xz − 4 x 2 + 3 8 xz z

11/ 3

3

(d) None of these



)

4

is equal to

1 223/ 3 ⋅ z (d) None of these

(b)

2 (c) 221/3 z

n! is equal to (n + 1)!− n! (a) 0 (b) ∞ (c) 1 (d) None of these

17. lim

n →∞

x

 x−3 9. For x ∈ R, lim   x→∞ x + 2   (a) e (b) e–1 (c) e–5 (d) e5

4.

4

n5 + 2 − 3 n 2 + 1

n →∞ 5

n 4 + 2 − 2 n3 + 1 (b) 0 (d) ∞

18. The value of lim (a) 1 (c) – 1

is

x →∞ 4

(a) 1 (c) 0

x2 + 1 − 3 x2 + 1 4

5

4

x +1 − x −1

is equal to (b) – 1 (d) None of these

28. lim x →0

x2 x4 x2 x2  8  1 − cos − cos + cos cos  is equal to 8  2 4 2 4 x 

1 16 1 (c) 32

1 16 1 (d) – 32 (b) –

(a)

1 − cos 2( x − 1) x −1 (a) exists and it equals 2 ln(1 + 2h) − 2 ln(1 + h) 29. The value of lim is h →0 (b) exists and it equals – 2 h2 (c) does not exist because (x – 1) → 0 (a) 1 (b) – 1 (d) does not exist because left hand limit is not equal to (c) 0 (d) None of these right hand limit

20. lim x →1

1 1 1 1 + + 2 + ... + n 2 2 2 is equal to 21. lim n →∞ 1 1 1 1 + + 2 + ... + n 3 3 3 4 3 1 (c) 3

(b)

(a)

2 3

22. The value of lim (a) 1 (c) 0

(b) – 1 (d) None of these

  x  1 − tan  2   (1 − sin x) is 23. lim π  x   x→ 3 2 1 + tan    ( π − 2 x)  2  1 8 1 (c) 32

(a)

x→0

(c)

1 3 2 (d) –  3

(b) – 

−1

x →0

2 2 − (cos x + sin x)3 is π 1 − sin 2 x x→

27. The value of lim

3 (a) 2 1 (c) 2

4

2 (b) 3 (d)

(b) – 1 (d) None of these

(b) n (d) None of these 1/ x 2

tan x − sin x is equal to x3 1 1 (a) (b) – 2 2 (c) 1 (d) – 1

26. lim

x →0

(a) 1 (c) 0

is equal to (a) ∞ (c) k

log (3 + x) − log (3 − x) = k, the value of k is x

−1

32. The value of lim log e (sin x) tan x is

34. lim log n −1 (n) ⋅ log n (n + 1) ⋅ log n +1 (n + 2)...log k (n k )  n −1  n →∞ 

(d) ∞

2 3

x − sin x is equal to x + cos 2 x (a) – 1 (b) 1 (c) 0 (d) None of these x →∞

1− x

(b) 0

(a) 0

(b) etan a (d) ecos a

 1 + 3 x  1− x is 33. The value of lim   x →∞ 2 + 3 x   (a) 0 (b) – 1 (c) e (d) 1

3n + 2n 24. The value of lim n is n →∞ 3 − 2 n (a) – 1 (b) 1 (c) 0 (d) ∞ 25. If lim

(a) ecot a (c) esin a 31. lim

(d) None of these

x5 is x →∞ 5 x

1

 sin x  x − a 30. lim   , a ≠ nπ, n is an integer, equals x→a  sin a 

2

 1 + 5x2  35. The value of lim   x →0 1 + 3 x 2   (a) e2 (b) e (c) e–1 (d) None of these

is

1/sin x

 1 + tan x  36. lim   x →0 1 + sin x   (a) 0 (c) – 1

is equal to (b) 1 (d) None of these

 1 e1/ n e 2 / n e( n −1)/ n  37. The value of lim  + + + ... +  is n →∞ n n n n   (a) 1 (c) e – 1

 π  38. lim  tan  + x   x →0 4    (a) e (c) e3

(b) 0 (d) e + 1 1/ x

is equal to (b) e2 (d) e–1

47

2

Limits

19. lim

48

39. The value of lim x →1

x n + x n −1 + x n − 2 + ... + x 2 + x − n is x −1

Objective Mathematics

n (n +1) 2 (c) 1

(b) 0

(a)

(d) n

40. Let f (a) = g (a) = k and their nth derivatives f n (a), gn (a) exist and are not equal for some n. Further if f (a) g ( x) − f (a) − g (a) f ( x) + g (a) lim = 4, then the x→a g ( x) − f ( x) value of k is (a) 4 (b) 2 (c) 1 (d) 0 x2

41. The value of lim

2 ∫ cos t dt

0

x →0

(a) 3 2 (c) – 1 42. If lim

x→0

x sin x

is

(b) 1 (d) None of these

x − sin x is nonzero finite, then n may be x − sin n x n

n

equal to (a) 1 (c) 3

(b) 2 (d) None of these

43. lim(sin x + 1 − sin x ) is equal to x →∞

(a) 1 (c) 0

(b) – 1 (d) None of these

 x  44. lim sec −1   = x→∞  x + 1 (a) 0 (c) π/2

(b) π (d) does not exist

xn 45. lim x = 0, (n integer), for x →∞ e (a) no value of n (b) all values of n (c) only negative values of n (d) only positive values of n x sin ( x − [ x]) , where [⋅] denotes the greatest integer x −1 function, is equal to

46. lim x →1

(a) 1 (c) ∞

(b) – 1 (d) does not exist

47. lim (1 + x) (1 + x2) (1 + x4) ... (1 + x2n), | x | < 1, is n→∞

equal to

1 (a) x −1 (c) 1 – x 48. lim

x →∞

x →0

2 3 3 (c) 2

(a) – 1 (c) 0

1 + sin x − 3 1 − sin x is x 2 3 3 (d) – 2 (b) –

(a)

50. If f (x) =

3

x2n − 1 , then lim f(x) is n→∞ x2n + 1 (b) 1 (d) ∞

(1 + x)(1 − x 2 )(1 + x 3 )(1 − x 4 )...(1 − x 4 n ) is equal to x →−1 [(1 + x )(1 − x 2 )(1 + x 3 ) (1 − x 4 )...(1 − x 2 n )]2

51. lim

(a) 4nC2n (c) 2 ⋅ 4nC2n

(b) 2nCn (d) 2 ⋅ 2nCn.

1 is equal to 1 + n sin 2 nx (a) – 1 (b) 0 (c) 1 (d) ∞

52. lim

n →∞

53. lim log tan x (tan 2 x) is equal to x → 0+

(a) 1 (c) 0 54. If f (x) = (a) 0 (c) – 1

(b) – 1 (d) None of these

∫

2 sin x − sin 2 x dx , x ≠ 0, then lim f ′ ( x) is x→0 x3 (b) ∞ (d) 1

11/ x + 21/ x + 31/ x + ... + n1/ x  55. The value of lim   x →∞ n   (a) n! (b) n (c) (n – 1)! (d) 0 56. lim

π x→ 2

nx

is

sin x − (sin x)sin x equals 1 − sin x + ln sin x

(a) 1 (c) 3

(b) 2 (d) 4

x  2  57. lim (where [ . ] denotes the greatest integer x → π / 2 ln(sin x ) function) (a) does not exist (b) equals 1 (c) equals 0 (d) equals – 1 58. lim lim(1 + cos 2 m n!πx) is equal to m →∞ n →∞

1 (b) 1− x (d) x – 1

x + sin x = x − cos x

(a) 0 (c) – 1

49. The value of lim

(a) 2 (c) 0

59. If f (a) = 2, f ' (a) = 1, g (a) = – 1, g′ (a) = 2, then

g ( x) f (a) − g (a) f ( x) is equal to x−a (a) 3 (b) 5 (c) – 3 (d) 0 lim x→a

(b) 1 (d) None of these

(b) 1 (d) None of these

(a) 0 (c) does not exist

(b) 1 (d) sin 1

1 (1 − cos 2 x) 61. The value of lim 2 is x →0 x (a) 1 (b) – 1 (c) 0 (d) None of these

e x − e− x − 2 x is equal to x →0 x − sin x (a) 1 (b) – 1 (c) 2 (d) 0

62. lim

63. lim

3

x →1

x2 − 2 3 x + 1 is equal to ( x − 1) 2

1 (a) 9 (c)

1 (b) 6

1 3

1 + ax + bx + c) lim( x→ α

is

(a + h) 2 sin( a + h) − a 2 sin a is h →0 h (b) 2a sin a – a2 cos a (a) 2a sin a + a2cos a 2 (c) 2a cos a + a sin a (d) None of these

65. The value of lim

66. lim tan x log sin x is equal to x →0

(a) 1 (c) 0

(b) – 1 (d) None of these

67. If g (x) = – (a) (c) – 

g ( x) − g (1) is equal to 25 − x 2 , then lim x →1 x −1

3 24

1 24

(b)

1 24

(d) None of these

68. The values of constants a and b so that

 x2 + 1  − ax − b  = 0 are lim  x →∞ x + 1   (a) a = 1, b = – 1 (c) a = 0, b = 0 69. The value of lim x →0

(b) a = – 1, b = 1 (d) a = 2, b = – 1

27 − 9 − 3 + 1 is 2 − 1 + cos x x

x

πx 2a

is equal to (b) e2/π (d) e– π/2

72. The value of lim 11 cos π x→ 

2x

2

+ 21 cos

2x

cos 2 x

2 + ... + n1 cos x  

(a) 0

(b) n

(c) ∞

(d)

is

n (n +1) 2

x →0

(b) ea (α – β) (d) None of these

(a) log | a (α – β) | (c) ea (β – α)

tan

a

64. If α and β be the roots of ax2 + bx + c = 0, then 1 ( x −α )

71. lim  2 − x  x→a a  (a) eπ/2 (c) e– 2/π

73. lim (cos x + a sin bx) x is equal to

(d) None of these

2

x

(a) 4 2 (log 3)2

(b) 8 2 (log 3)2

(c) 2 2 (log 3)2

(d) None of these

2

(a) e − a b 2

(c) e a b

(b) e ab

2

(d) e − b

2a

sin x n , m, n ∈ N, is equal to x →0 (sin x ) m (a) 1 (b) – 1 (c) 0 (d) ∞

74. lim

sin x

 sin x  x −sin x 75. The value of lim  is  x →0  x  (a) 1 (b) – 1 (c) 0 (d) None of these 76. The value of lim  3 n 2 − n3 + n  is n →∞   (a)

1 3

(b) –

1 3

(c)

2 3

(d) –

2 3

 n2 + 1 + n  77. lim   is equal to n →∞ 4 3  n + n − 4 n  (a) 0 (b) 1 (c) – 1 (d) ∞ 78. The value of lim  3 (n + 1) 2 − 3 (n − 1) 2  is n →∞   (a) 1 (c) 0

(b) – 1 (d) ∞

 1 1 1 1  is equal to 79. lim  + + + ... + n →∞ 1 ⋅ 2 ( 2 ⋅ 3 3 ⋅ 4 n n + 1)   (a) 1 (c) 0

(b) – 1 (d) None of these

49

π − cos −1 x is equal to x →−1 x +1 1 1 (b) (a) 2π π 1 (c) (d) None of these 2

70. lim

Limits

 sin([ x − 3])  60. lim   , where [ . ] represents greatest integer x→0  [ x − 3)  function, is

50

Objective Mathematics

 2n n +1 n n (−1) n  80. The value of lim  2 cos − ⋅ 2  is n →∞ 2 n − 1 2n − 1 1 − 2n n + 1   (a) 1 (b) – 1 (c) 0 (d) None of these 8 x2 +3

 2x2 + 3  81. The value of lim  2 is  x →∞ 2 x + 5   (a) e8 (b) e– 8 4 (c) e (d) e– 4

90. lim sin( a + 3h) − 3 sin(a + 23h) + 3 sin( a + h) − sin a is equal h →0 h to (a) sin a (b) – sin a (c) cos a (d) – cos a x

 x2 − 2x + 2  91. The value of lim  2  is x →∞ x − 4 x + 1   (a) e–2 (c) 1

(b) e2 (d) 0

sin 2 x + a sin x be finite, then the value of a and 92. If f (9) = 9 and f ' (9) = 1, then lim 3 − f ( x) is equal x3 x →9 3− x to the limit are given by (a) 0 (b) 1 (a) – 2, 1 (b) – 2, – 1 (c) – 1 (d) None of these (c) 2, 1 (d) 2, – 1

82. If lim x →0

93. Let f (x) be a twice differentiable function and f ′′ (0) 3 f ( x) − 4 f (3 x) + f (9 x) = 5, then lim is equal to x →0 x2 (a) 30 (b) 120 (c) 40 (d) None of these

1/sin x

 1 + tan x  83. The value of lim  is  x →0 1 + sin x   (a) 1 (b) – 1 (c) 0 (d) ∞ 84. The values of a, b and c such that

x3

 2 1+ x 94. The value of lim  3 x 2 + 1  is x →∞ 4 x − 1  

−x

a e − b cos x + ce = 2 are x sin x (a) a = 1, b = – 2, c = 1 (b) a = 1, b = 2, c = – 1 (c) a = 1, b = 2, c = 1 (d) a = – 1, b = 2, c = 1 lim

x

x →0

 x+6 85. lim   x →∞ x + 1  

(a) 0 (c) 1

log (1 + x + x 2 ) + log (1 − x + x 2 ) is equal to x →0 sec x − cos x (a) 1 (b) – 1 (c) 0 (d) ∞

95. lim

x+4

is equal to

(a) e– 5 (c) 0

(b) e5 (d) None of these

86. If f (5) = 7 and f ' (5) = 7, then lim x →5

given by

96. The value of lim(sin x) tan x is x→

x f (5) − 5 f ( x) is x−5

(b) 1 (d) ∞

x →0

(b) 28 (d) – 35

1   87. If a, b, c, d are positive, then lim 1 +  x →∞  a + bx  d/b c/a (a) e (b) e (c) e(c + d)/a + b (d) e 88. If a = min {x2 + 4x + 5, x ∈ R} and b = lim θ→0

n

∑a

r

⋅ b n − r is

r =0

n +1

(a) 0 (c) – 1

π 2

97. lim x x is equal to

(a) – 28 (c) 35

then the value of

(b) ∞ (d) – 1

(a)

2 −1 4 ⋅ 2n

(b) 2n + 1 – 1

(c)

2n +1 − 1 3 ⋅ 2n

(d) None of these

3 1x 89. If lim (1 + a ) +3 8e1 x = 2, then x →0 1 + (1 − b ) e (a) a = 1, b = (– 3)1/3 (b) a = 1, b = 31/3 1/3 (c) a = – 1, b = – (3) (d) None of these

(a) 0 (c) – 1

c + dx

=

1 − cos 2θ , θ2

(b) 1 (d) None of these

98. The value of lim a 2 − x 2 cot x→a

2a π 4a (c) π

(a)

 99. The value of lim log a x →3  (a) loga 6 (c) loga 2 100. lim x→2

3

2a π 4a (d) – π (b) –

x−3  is x + 6 − 3  (b) loga 3 (d) None of these

10 − x − 2 is equal to x−2

1 12 1 (c) 6

(a)

π a−x is 2 a+x

1 12 1 (d) – 6 (b) –

3 2 (c) 3

3 2 (d) does not exist

(a)

to (a) 4 (c) 2

(b) –

(a) eb (c) e

(a) 4 (c) 0

x →−∞

x →0

(a) e (c) e–1

(b) a = 1, b =

1 2

x → 0+

(a) 1 (c) 0

(d) None of these

x

is equal to

(a) 1 (c) ∞

(a) 1 (c) 0

n

∑ (−1) r =1

r

⋅ tr , then

x →∞

(a) – 1 (c) – 1

n →∞

2 3 1 (d) – 3 (b) –

107. If f (x), g (x) be differentiable functions and f (1) = g (1) = 2 then f (1) g ( x) − f ( x) g (1) − f (1) + g (1) lim is equal to x →1 g ( x) − f ( x) (a) 0 (b) 1 (c) 2 (d) None of these

a0 x n + a1 x n −1 + a2 x n − 2 + ... + an ; m, n > 0, is equal to x →∞ b x m + b x m −1 + b x m − 2 + ... + b 0 1 2 m

108. lim

(c)

(d) None of these

109. lim (1 + cos π x)cot 2 π x is equal to

x →0

is

(a) a (c)

a 2 − ax + x 2 − a 2 + ax + x 2 a+x − a−x (b) – a

a

(d) –

ln x − 1 is equal to | x−e| 1 (a) e (c) e

a

118. lim x →e

1 e (d) Does not exist

(b) –

119. The value of lim [ x 2 ] is x→ 2

(a) 1 (c) 0

(b) 2 (d) Does not exist

sin (2k − 3) x ,x 0 is x →∞  x 

105. The value of lim 1 − cos (14− cos x) is x →0 x 1 1 (b) (a) 4 8

(a)

(b) – 4 (d) Does not exist

113. lim log tan x (sin x) is equal to

(b) e2 (d) 1

1 (c) 16

(b) eb/2 (d) None of these

x →3

and b are given by

104. lim (cos x + sin x)

is

integer function, is equal to

103. If lim ( x − x + 1 − ax − b) = 0, then the values of a

1 x

n

112. lim ([ x − 3] + [3 − x] − x) , where [⋅] denotes the greatest

2

1 2 1 (c) a = 1, b = – 2

(b) – 4 (d) – 2

b  111. The value of lim 1 + tan  n →∞ n 

 ln cos x  102. lim   is equal to 2 x →0 4  1 + x − 1 (a) 2 (b) – 2 (c) 1 (d) – 1

(a) a = – 1, b =

tan (3k − 4) x , x > 0 2x and lim f ( x) exists, then the value of k is given by

(b) – 1

=

(d) None of these



x →0

51

2 x 2 − 4 f ( x) is equal x→2 x−2

110. If f (2) = 2 and f ′ (2) = 1 then lim

Limits

101. If f (x) is differentiable and f ' (2) = 6, f '(1) = 4, then f ( 2 + 2h + h 2 ) − f ( 2 ) lim is equal to h →0 f (1 + h + h 2 ) − f (1)

52

Objective Mathematics

5 4 4 (c) 5

5 4 4 (d) – 5

(a)

(b) –

1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + ... + n (n + 1) is equal to n3 (a) 1 (b) – 1 1 (d) None of these (c) 3 x +1 x +1 3 −5 is 122. The value of lim x x →∞ 3 − 5 x 1 (a) 5 (b) 5 (c) – 5 (d) None of these 121. lim

n →∞

n −1 1 2  1 123. lim 1 + e n + e n + ... + e n  is equal to n →∞ n   (a) e (b) – e (c) e – 1 (d) 1 – e

130. lim π x→ 4

(a) 5 2 (c) 2

(b)

1 2

(c) –

(a) 0 (c) – 1

x →1

(a) 0 (c) – 1

x

x →0

4− x − x (a) 0 (c) – 1

x7 − 2 x5 + 1 is x3 − 3x 2 + 2 (b) 1 (d) None of these

(a)

e x + e − x + 2 cos x − 4 is equal to x →0 x4 (a) 0 (b) 1 1 1 (d) – (c) 6 6

133. lim

3 24 3 3 (c) 3 4 2 (a)

3 24 3 3 (d) – 3 4 2 (b) –

y3 as (x, y) → (1, 0) along the line x →1 x − y 2 − 1 y →0

129. lim

3

y = x – 1 is given by (a) 1 (c) 0

(b) ∞ (d) None of these

(b) –

1 2

(d) – 1

x − 2a + x − 2a x 2 − 4a 2

x→2 a

1 a 1 (c) – a

(a)

is equal to

1 2 a 1 (d) – 2 a (b)

134. lim (tan x cot β)1/( x −β ) is equal to x →β

1 sin β cos β 1 (c) – sin β cos β

(a)

(b) sin β cos β (d) None of these

135. lim(−1)[ x ] , where [x] denotes the greatest integer less x→n

than or equal to x, is equal to (a) (– 1)n (c) 0

(b) (– 1)n – 1 (d) Does not exist

log e [ x] , where [x] denotes the greatest integer less x than or equal to x, is

136. lim

x →∞

127. lim

2− 2+ x 128. lim 3 is equal to x→2 2 − 3 4− x

1 2

(c) 1

is equal to (b) 1 (d) Does not exist

(b) 1 (d) None of these

x +1 π   is 132. The value of lim x  tan −1 − x →∞ + 2 4  x 

(d) None of these

125. The value of lim

126. lim

1 2

(b) 3 2 (d) None of these

131. lim cos  π n 2 + n  , n ∈ Z is equal to n →∞  

 1  1 1 1 124. lim  + + + ... +  is equal to n →∞ 1 ⋅ 3 ( )( ) 3 ⋅ 5 5 ⋅ 7 2 n + 1 2 n + 3   (a) 1

4 2 − (cos x + sin x)5 is equal to 1 − sin 2 x

(a) 1 (c) 0

(b) – 1 (d) Does not exist

137. If α is a repeated root of ax2 + bx + c = 0, then tan (ax 2 + bx + c) lim is x →α ( x − α) 2 (a) a (b) b (c) c (d) 0 138. Let f (x)  = sin x,  x ≠ nπ, where n ∈ Z and = 2,    g (x)

x  = nπ = x2 + 1, x ≠ 2, then lim g [ f ( x)] is



x = 2.

= 3, (a) 1 (c) 3

x →0

(b) 0 (d) Does not exist

139. If the rth term, tr , of a series is given by n r . Then lim ∑ tr is tr = 4 2 n →∞ r =1 r + r +1

(c)

1 3

1 2

x→2

(d) None of these

 x4 + x2 + x + 1  140. lim   2 x →−1  x − x +1 

1− cos ( x +1) ( x +1)2

is equal to 1/ 2

2 (b)   3

(a) 1 1/ 2

3 (c)   2



(d) e1/2

2

(log x) , n > 0 is equal to xn (a) 1 (b) 0 (c) – 1 (d) ∞

141. lim

x →∞

4 142. The value of lim 3x sin  x  is x →∞ 3  (a) 4 log 3 (b) 3 log 4 (c) 4 (d) None of these

  tan[ x] , [ x] ≠ 0    where 143. If f (x) =  [ x]  0, 0 [ ] x =  

[x] denotes the greatest integer less than or equal to x, then lim f ( x) equals (a) 1 (c) 0

x →0

(b) – 1 (d) Does not exist

2x − x2 is equal to x→2 x x − 22

144. lim

log 2 − 1 log 2 + 1 (c) 1 (a)

log 2 + 1 log 2 − 1 (d) – 1

(b)

n

x  145. lim  cos  is equal to n →∞ n  (a) e1 (c) 1

(b) e–1 (d) None of these

 tan −1 ([ x] + x)  , [ x] ≠ 0   2 [ ] − x x 146. If f (x) =   0 , [ x] = 0 

1+ 2 + x − 3 is equal to x−2 1 1 (a) (b) 8 3 3

148. lim

53

(b)

(c) 8 3

(d)

3

149. Let f (x) = x – [x] where [x] denotes the greatest integer { f ( x)}2 n − 1 , then g (x) = ≤ x and g (x) = lim n →∞ { f ( x )}2 n + 1 (a) 0 (b) 1 (c) – 1 (d) None of these 150. lim  1 2 + 2 2 + ... + n 2  is equal to n →∞ 1 − n 1− n 1 − n   1 (a) 0 (b) – 2 1 (d) None of these (c) 2 151. lim (1 − x) tan  π x  is equal to x →1  2  π 2 (a) (b) 2 π π 2 (c) – (d) – 2 π 152. Let f : R → R be such that f (1) = 3 and f ′ (1) = 6. 1

 f (1 + x )  x Then lim   equals x →0  f (1)  (a) 1 (b) e1/2 2 (c) e (d) e3  π  π  2  3 sin  + h  − cos  + h   6 6      is equal to 153. lim  h →0 3h ( 3 cos h − sin h) 4 4 (b) – (a) 3 3 2 3 (c) (d) 3 4 154. lim x →0

(1 − cos 2 x)sin 5 x is equal to x 2 sin 3 x

6 5 10 (c) 3 (a)

(b)

3 10

(d) –

3 10

where [x] denotes the greatest integer less than or equal to  e x − esin x  155. lim  x, then lim f ( x) is equal to  is equal to x → 0 x − sin x x →0   (a) – 1 (b) 0 1 (b) 1 (a) – (c) 1 (d) None of these 2 x2 π e − cos x (d) Does not exist (c) is 156. lim x →0 4 x2 x 1 2  x−3 (b) (a) 147. For x ∈ R, lim  =  x →∞ x + 2 2 3   3 (a) e (b) e–1 (c) (d) 2 (c) e–5 (d) e5 2

Limits

(a) 1

54

157. If lim

( sin nx ) ( a − n ) nx − tan x  x2

x →0

= 0, then find the

Objective Mathematics

value of a (a)

1 n

(c) n +

(b) n −

1 n

1 n

(d) None of these

1   x sin , x ≠ 0 x  158. If f (x) = , then lim f ( x) equals x →0 0 ,x = 0 (a) 1 (c) – 1

(b) 0 (d) None of these

159. If f ( x) = ∫

x + sin x f ′ ( x) is equal to dx, then xlim →∞ x + cos x

(a)  0 (c)  ∞

(b)  1 (d)  –1

1 + sin x − cos x + log(1 − x) is 160. The value of lim x→0 x3 (a)  –1 (b)  1/2 (c)  –1/2 (d)  1

161. If S1 = Σ n, S2 = Σ n2, S3 = Σ n3, then the value of S   S1 1 + 3  8  is equal to  lim n→∞ S 22 (a)  3/32 (c)  9/32 162. lim

x→∞

(b)  0 (d)  

3 10

 x + bx + 4  163. The value of lim  2  is x → ∞ x + ax + 5   2

(a)  

b a

(c)  1

(b)  3 (d)  None of the above

x+2 165. lim   x →∞  x +1  (a)  1 (c)  e2

(b)  0 (d)  

4 5

x +3

is (b)  e (d)  e3

166. lim x loge(sin x) is equal to x→ 0

(a)  –1 (c)  1

(b)  loge 1 (d)  None of these

1 + tan x    1 + sin x 

cos ec x

167. lim  x →0 (a)  

is equal to

1 e

(b)  1

(c)  e

(d)  e2

168. lim (–1)[x] is equal to x→n

(a)  (–1)n (c)  (–1)n – 1

(b)  0 (d)  does not exist.

ax 2 + bx + c = 2 then (a, b, c) is ( x − 1) 2

x →1

(2 x − 3) (3 x − 4) is equal to (4 x − 5) (5 x − 6)

1 10 1 (c)   5

(a)  9 (c)  1

169. If lim

(b)  3/64 (d)  9/64

(a)  

164. If f be a function such that f(9) = 9 and f ′ (9) = 3, f ( x) − 3 then lim is equal to x →9 x −3

(a)  (2, – 4, 2) (c)  (2, 4, – 2) sec2 x

170. lim x→

∫

x2 −

8 π (c)   2 π

(a)  

f (t ) dt

2

π 4

(b)  (2, 4, 2) (d)  (2, – 4, – 2)

π2 16

equals 2 f (2) π

f (2)

(b)  

1 f   2

(d)  4f(2)

( x − 1) n ; 0 < x < 2, m and n are intelog cos m ( x − 1) gers, m ≠ 0, n > 0, and let p be the left hand derivative of |x – 1| at x = 1. If lim g(x) = p, then x →1

171. Let g(x) =

(a)  n = 1, m = 1 (c)  n = 2, m = 2

(b)  n = 1, m = – 1 (d)  n > 2, m = n

SOLUTIONS 1. (c) n cannot be negative integer for then the limit = 0 x 2 sin 2 x x 2 . e − cos x = 1 lim e − cos x , Limit = lim 2 n − 2 x →0 x 2 x →0 x n − 2  x 22.  2 (n ≠ 1 for then the limit = 0)

1 e x + sin x = lim . x → 0 2 ( n − 2 ) x n −3 So, if n = 3, the limit is

1 which is finite. 2 ( n − 2)

If n = 4, the limit is infinite.

2

2

n + 1 + 4n − 1

[1 + 3 + 5 + 7 + ... + (2n − 1)] − (2 + 4 + 6 + ... + 2n) 1 1 n 1+ 2 + n 4 − 2 n n n n [2 ⋅1 + (n − 1) ⋅ 2] − [2 ⋅ 2 + (n − 1) ⋅ 2] 2 2 = lim n →∞  1 1  n  1+ 2 + 4 − 2  n n   n n ⋅ 2n − 2 (n + 1) 2 = lim 2 n →∞  1 1  n  1+ 2 + 4 − 2  n n   = lim

n →∞

n2 − n2 − n

= nlim →∞

 1 n  1+ 2 + 4 − n  −n = lim n →∞  1 n  1+ 2 + 4 − n  −1 = nlim →∞ 1 1 1+ 2 + 4 − 2 n n

1   n2 

−1 −1 = . 1+ 2 3

 a 3 − b3   Using a − b = 2  a + ab + b 2   2 = lim x → 0 (1 + x ) 2 / 3 + (1 + x )1/ 3 (1 − x )1/ 3 + (1 − x ) 2 / 3 2 2 = . = 1+1+1 3 

x → −∞

= lim

x→0

2x

1 − (1 − 2 sin 2 x)

|sin x | 2 sin 2 x = lim . x→0 x 2x

Let f (x) =

|sin x | x

hh |sin (0 + h) | limsinsin =lim =1 h → 0 h→0 h→0 0+h h− h

Then, f (0 + 0) = lim

sin h |sin (0 − h) | = hlim = – 1. →0 −h −h ∵  f (0 + 0) ≠ f (0 – 0)

and f (0 – 0) = lim

h→0

∴  the limit does not exist.

 x 4hsin(1/x) + x 2  sin 5. (b) hlim   =  lim 3 → 0 − h 1+ | x | y →∞  

 3 =   1

6/3

= 9.

7. (a) lim  x + x + x − x   x →∞   

x+ x+ x −x x+ x+ x + x

x+ x

= xlim →∞

1   x 1 +   x

 4x + 1  = xlim 1 + 2  →∞ x + x + 2 

x+ x+ x + x

1/ 2

1 − y 4 sin + y 2 y 1+ y 3

[Putting x = – y; as x → – ∞, y → ∞]

1 1 = 1+1 = 2 .

x

[(1 + α)1/ α ]αx = xlim →∞

1 4+ 4x + 1 x →0 where α = 2 = 1 2 x + x +2  x x + + 2  x x    as x → ∞ 4+

1 x

→ 4 as x → ∞ 1 2 + 2 x x ∴  Given limit = e4. and αx =

1+

x

2x

x→0

    

1/ 2   1 1  1+ x  1 + + 1  x x    8. (a) Given limit

− (1 − x)1/ 3 x→0 x 1 (1 + x) − (1 − x) ⋅ = lim x → 0 (1 + x ) 2 / 3 + (1 + x )1/ 3 (1 − x )1/ 3 + (1 − x ) 2 / 3 x

= lim

2 1   3 + x + x2  = xlim →∞  1 + 1 + 22 x x 

= xlim →∞

1/ 3

1 − cos 2 x

1 x 2 3+ x

6+

x →∞

3. (a) lim (1 + x)

4. (d) lim

6x +1

 2 3x + 2 6. (b) lim  3 x2 + 2 x + 1  x →∞ x + x + 2  

= lim

1   n2  =

1  sin  y 1 + − 1  y    y  −1 + 0 = = – 1. = ylim →∞ 1 1+ 0 1+ 3 y

55

1 − 2 + 3 − 4 + 5 − 6 + ... − 2n

n →∞

Limits

2. (b) lim

x

    9. (c) lim  x − 3  = lim  1 − (3 / x)  x→∞ x + 2 x → ∞ 1 + (2 / x)     x 3  1 −  e −3 x = lim = 2 = e–5. x x→∞ e 2  1+    x 1 1 + cos   h x + cos x h lim 10. (c) xlim →∞ x + sin x = h → 0 1 1 + sin   h h

1    Putting x = h ; as x → ∞, h → 0    1 1 + h cos h = 1+ 0 = lim h→0 1 1+ 0 1 + h sin h

56

Objective Mathematics

1 1  , ∵−1 ≤ sin h ≤ 1 and −1 ≤ cos h ≤ 1 



∴ when h → 0, h cos 1 → 0 and h sin 1 → 0.  h h  = 1. 11. (b) lim

h → 0+

h 2 hr − h 2 A = lim P 3 h → 0+ 8  2 hr − h 2 + 2 hr  3  

= lim h→0 = lim h→0

=

(

h ⋅ h 2r − h

8 2 2r

12. (a) lim x→2

)

3

= lim x→2

(2

1 = . 128 r

x/2

+ 2) (2 − 2) (2 − 2) (2 x / 2 − 2)

x→2

n k sin 2 (n!) n k sin 2 (n!) = nlim →∞ 2  n →∞ n+2 n 1 +  n  

13. (c) lim

z= nlim →∞

sin 2 (n!) a finite quantity = 2   ∞ n1− k 1 +  n 

[ ∵ sin2 (n !) always lies between 0 and 1. Also, since 1 – k > 0, ∴ n1 – k → ∞ as n → ∞] = 0. 14. (c) lim xf (2) − 2 f ( x) x→2 x−2 xf (2) − 2 f (2) + 2 f (2) − 2 f ( x) = lim x→2 x−2 = xlim →2

f ( x) − f (2) ( x − 2) f (2) − 2 lim x→2 x−2 x−2

= f (2) – 2f ′ (2) = 4 – 2 × 4 = – 4. 3 5 15. (a) lim 2 x + 3 x + 5 x 3 x →∞ 3x − 2 + 2 x − 3

= xlim →∞

2 x +3 x +5 x 2 3 x 3− + 3 x 3 2− x x 3

3

z 2 − ( z − x) 2

( 3 8 xz − 4 x 2 + 3 8 xz ) 4

= lim x →0

x 3 2 xz − x 2 ( x 8z − 4x + 3 8z 3

3

x4 3

3

3

x )4

4

=

2z − x

3

2z

 3 8z − 4x + 3 8z  2 3 8z      n! n! = lim 17. (a) lim n →∞ ( n + 1)!− n ! n →∞ ( n + 1) n !− n ! = lim n →∞

x

43

=

1 . 223 3 ⋅ z

1 1 = lim = 0. n + 1 − 1 n→∞ n

4

n5 + 2 − 3 n 2 + 1

n →∞ 5

n 4 + 2 − 2 n3 + 1

18. (b) lim

4

x

= lim(2 x / 2 + 2) (2 x − 2) = (2 + 2) ⋅ (4 – 2) = 8.



x

= lim x →0

22 x − 6 ⋅ 2 x + 8 ( 2 x − 4) ( 2 x − 2) = lim x/2 x→2 2 −2 ( 2 x / 2 − 2) x/2

2 . 3

3

( 2 2 x + 23 − 6 ⋅ 2 x ) / 2 x 2 x + 23 − x − 6 lim − x/2 1− x = x → 2 1 2 2 −2 − 2x/2 2x

= lim x→2

16. (b) lim

3

2r − h

2r

=

x →0

8 ⋅ h ⋅ h  2r − h + 2r  8  2r − h + 2r 

3 5 + 3/10 1/ 6 x x = xlim →∞ 2 1 3 3 − + 1/ 6 3 2 − x x x [Dividing the numerator and denominator by the highest power x1/2] 2+

5

2 1 − n2 3 3 1 + 2 5 n n = lim n →∞ 2 1 n 4 5 5 1 + 4 − n3 2 2 1 + 3 n n n5 4 4 1 +

n5 4 4 2 n2 3 1 1+ 5 − 3 2 3 1+ 2 32 n n n n = lim n →∞ n 4 5 2 n3 2 2 1 5 1+ − 1+ 3 32 n n 4 n3 2 n [Dividing the numerator and denominator by the highest power n3/2]

1 4 2 1 1 1+ 5 − 5 6 3 1+ 2 14 0−0 n n n n = = 0. = lim n →∞ 0 −1 1 5 2 2 1 + − + 1 1 n 7 10 n4 n3 2

x2 + 1 − 3 x2 + 1

x →∞ 4

x4 + 1 − 5 x4 − 1

19. (a) lim

1 1 − x2 3 ⋅ 3 1 + 2 2 x x = lim x →∞ 1 1 x 4 1 + 4 − x4 5 5 1 − 4 x x x 1+

1 1 1 − 1 3 3 1+ 2 2 x x x = lim x →∞ 1 1 5 1 4 1+ − 1− 4 x 4 x1 5 x 1+

h = lim tan h→0 2

2

x →1

2

=–

2 ⋅1=–

2.

RHL = lim+ x →1

h  h  tan   sin  1  1 2 2 ⋅ = hlim .  h  = → 0 32 h 32     2  2 

2 sin( x − 1) 2 sin(1 − h − 1) = lim − h →0 ( x − 1) (1 − h − 1)

2 − sin h −h

= lim h →0 =–

tan

2 sin( x − 1) ( x − 1)

LHL = lim−

8h 3

h 2

h  h  sin 2  1 1 2 ⋅ × × = hlim →0 4 h h  4 ×2  2  2 

1 − cos 2( x − 1) 2 sin 2 ( x − 1) = lim 20. (d) lim x →1 x →1 x −1 x −1 = lim x →1

2 sin 2

57

1− 0 = = 1. 1− 0

Limits

[Dividing the numerator and denominator by x]

2 lim h →0

sin h h

24. (b) lim

n→ ∞

2 sin (1 + h − 1) 2 sin( x − 1) = lim h →0 (1 + h − 1) ( x − 1)

2 sin h sin h = 2 lim = h →0 h h Since LHL ≠ RHL,

= lim h →0

2⋅ 1 =

3 +2 = lim 3n − 2n n → ∞ n

n

2 1+   3

n

2 1−   3

n

[Dividing by 3n]

2 1+ 0 2 [ ∵ < 1, ∴   → 0 as n → ∞] 1− 0 3 3 = 1. n

=

2.

log (3 + x) − log (3 − x) = k x Applying L’Hospital rule

25. (c) lim

x →∞

∴ lim x →1

1 − cos 2( x − 1) does not exist. x −1

 1 1  3 + x + 3 − x    ⇒ xlim =k →∞ 1

1 1 1 1 1 1 − n +1 1− 1 + + 2 + ... + n 3 2 2 2 2 = lim × 21. (a) lim n →∞ 1 1 n →∞ 1 1 1 1− 1 − n +1 1 + + 2 + ... + n 2 3 3 3 3

1 1 + =k 3 3 2 . ⇒k= 3

⇒

n +1

1 1−   4 1− 0 4 2 ⋅  n +1 = ⋅ = lim n →∞ 3 3 1+ 0 1 1−   3

[ ∵ xn + 1 → 0 as n → ∞ if 0 < x < 1] =

22. (c) lim

x →∞

tan −1 x − sin −1 x  x →0 x3

4 . 3

1 1 − 1 + x2 − x2 1 = lim 2 x →0 3x

x5 x5 x5 = lim x log 5 = lim kx , where k = log 5 x x x →∞ →∞ 5 e e

= lim x →∞



= lim x →0



= lim x →0

1  1 1 k2 1 k3 1 k4 1   5 + k ⋅ 4 + ⋅ 3 + ⋅ 2 + ⋅  2! x 3! x 4! x  x  x



= lim x →0

 k5  k6 +  x + ...   5!  6! 



5

x   k 2 x 2 k 3 x3 k 4 x 4 x5k 5 k 6 x6 + + + + + ...  1 + kx + ! ! ! ! ! 2 3 4 5 6  

= lim x →∞

+

0   form  0 

26. (b) lim

1 = = 0. ∞ π π 23. (c) Put x = − h as x → , h → 0 2 2 π h 1 − tan  −   4 2  (1 − cos h) ∴  Given limit = lim ⋅ h→0 ( 2h)3 π h 1 + tan  +   4 2

[Using L’Hospital’s Rule]

1 − x 2 − (1 + x 2 ) 3 x 2 (1 + x 2 ) 1 − x 2 (1 − x 2 ) − (1 + x 2 ) 2 3 x 2 (1 + x 2 ) 1 − x 2  1 − x 2 + (1 + x 2 )   

−3 − x 2 3 (1 + x ) 1 − x 2  1 − x 2 + (1 + x 2 )    3 1 = – =− . 6 2

27. (a) lim π x→ 4

2

 2 2 − (cos x + sin x)3  0  form  1 − sin 2 x 0 

= limπ x→

4

−3 (cos x + sin x) 2 (− sin x + cos x) −2 cos 2 x [Using L’Hospital’s Rule]

58

= limπ x→

4

Objective Mathematics

= limπ x→

4

−3 (cos x + sin x)(cos 2 x − sin 2 x) −2 cos 2 x



= lim x→ a

−3 (cos x + sin x)cos 2 x −2 cos 2 x



x+a cos  = lim  2  ex→ a



= e sin a = ecot a.

3 (cos x + sin x) 3  1 1  3 + = ⋅ . = 2 2  2 2 2 x→ 4

= limπ 28. (c) lim x →0

x2 x2 x2 x2  8  1 − cos − cos + cos cos  8  2 4 2 4 x 



= lim x →0

x2  x4  x2  8  1 − cos  − cos 1 − cos   8  2 4 2  x 



= lim x →0

x2   x2  8 1 − cos  1 − cos  8  2  4 x 



2 2 8 2 x 2 x 2 2 ⋅ sin ⋅ sin = lim x →0 x 8 4 8





 x2 sin 32   24 = lim x →0 x 8  x   4 =

2

  x2   x 2  2  sin    ⋅  28   4  x     8

sin x x = 1− 0 cos 2 x 1+ 0 1+ x 1−

 sin x  cos 2 x ∵ x → 0, x → 0 as x → ∞   



2

   x 2 2  ⋅    8   



x→0

log e sin x ∞    form  ∞ cot x cot x = lim  [Using L’Hospital’s Rule] x → 0 − cosec 2 x − cos x ⋅ sin x) = 0. = lim( x→0

= lim x →0

1

1 + 3x  33. (d) lim   x →∞  2 + 3 x 

ln(1 + h) 2 − ln(1 + 2h) h →0 h2

3 =   3

= – lim



  (1 + h) 2  h2  ln  ln 1 +   1 + 2h   1 + 2h  = – lim = – lim 2 h→0  h 2 h→0 h   1 + 2h  (1 + 2h)



 h2  ln 1 +  1 + 2h  1 ⋅ = – lim =–1 h→0 1 + 2h  h2   1 + 2h 

1

  x 1 +3   = lim  x  x →∞ 2  + 3  x

−

1  1    −1 x x 

0 −1

= 10 = 1.

34. (c) lim log n −1 (n) ⋅ log n (n + 1) ⋅ log n +1 (n + 2)...log k (n k )  n −1 n →∞   log n log(n + 1) log(n + 2) log(n k )  ⋅ ⋅ ... = nlim  →∞ log( n − 1) log n log(n + 1) log(n k − 1)  

= nlim →∞

1

 sin x  x − a  sin x − sin a  x − a 30. (a) lim   = lim 1 +  x → a  sin a  x→ a  sin a sin x − sin a

sin a  ( x − a )sin a  sin x − sin a   x a sin − sin     1 lim + = x→ a        sin a   

sin x − sin a ( x − a )sin a

1− x 1− x

 log m   Using log n m = log n   

log(1 + x)   = 1  Using lim x→0 x 

e

x − sin x 31. (b) lim = lim x →∞ x + cos 2 x x →∞

x →0

1 . 32

= lim x→ a

cos a

32. (c) lim log e (sin x) tan x = lim tan x ⋅ log e sin x (0 ⋅ ∞ form)





  x − a   x − a  1 sin  2   2   sin a  

= 1.

ln(1 + 2h) − 2 ln(1 + h) 29. (b) lim h →0 h2



2 x+a x−a 1 ⋅ cos  sin  2   2  sin a x−a

e

log n k log n  ∞ = k lim  form  n → ∞ log( n − 1) ∞  log(n − 1)

= k lim

n →∞

1 1

n

[Using L’Hospital’s Rule]

n −1

 1 = k lim 1 −  = k. n →∞  n 1 + 5x2  35. (a) lim  x→0  1 + 3x 2  

1

x2

= e

1 1+ 5 x 2   lim −1 x → 0 x 2 1+ 3 x 2 

lim g ( x )[ f ( x ) −1]   [ f ( x ) ]g ( x ) = e x → a   Using lim x→ a  provided f ( x) → 1 and g ( x) → ∞ as x → a   

36. (b) Let f (x) =

2

1 + tan x 1 + sin x

⇒ lim

= e 2.

x→a

and g (x) =

1 sin x

 1 + tan x  ∴ lim  x →0 1 + sin x   

=e

⇒ k = 4.

1 . sin x

Clearly f (x) → 1 and g (x) → ∞ as x → 0.

x2

41. (b) lim x →0

1  1+ tan x  −1 lim  x →0 sin x  1+ sin x 

lim g ( x )  f ( x ) −1 g ( x)    f ( x) = e x →a  Using lim  x→a    lim

1  tan x − sin x    1+ sin x 

= e x →0 sin x 

lim

1− cos x

∫ cos t dt 0

2

= lim x →0

2 x cos x 4 x cos x + sin x

= lim x →0

2 cos x 4 − 8 x 4 sin x 4 2 cos x − x sin x

2−0 = 1. = 2−0



1



x

 1 + tan x  = lim  x →0 1 − tan x    = e

1

x

= e



=

39. (a) lim x →1

2 tan x lim x →0 x (1− tan x ) 1  tan x  lim 2  ⋅  x  1− tan x

e x →0

= e 2.

x n + x n −1 + x n − 2 + ... + x 2 + x − n x −1

0   0 form   

nx n −1 + (n − 1) x n − 2 + ... + 2 x + 1 1 [Using L’Hospital’s Rule] n (n +1) = n + (n – 1) + ... + 2 + 1 = . 2

= lim x →1



40. (a) lim

x→a

f (a ) g ( x) − f (a) − g (a) f ( x) + g (a) =4 g ( x) − f ( x)

Applying L’ Hospital rule ⇒ lim

f (a ) g' ( x) − g (a ) f ' ( x) =4 g' ( x) − f ' ( x)

⇒ lim

kg' ( x) − kf ' ( x) =4 g' ( x) − f ' ( x)

x→a

x→a

x − sin x = 1. x − sin x

x →∞



2 cos = lim x →∞

x +1 + x x +1 − x ⋅ sin 2 2



2 cos = lim x →∞

x +1 + x 1 ⋅ sin 2 2( x +1 + x )



= 2 × (some number between – 1 and 1) × sin 0 = 0.

x 44. (d) For x > 0,  ∵ x < x + 1,  ∴  x + 1 < 1

1  1+ tan x  lim  −1 x →0 x  1− tan x 

1  1 + tan x  ∵ 1 − tan x → 1 and x → ∞ as x → 0   

[Using L’Hospital’s Rule again]

43. (c) lim(sin x + 1 − sin x )

1 ⋅ (e ) − 1 1 = (e − 1) lim 1 n = lim n →∞ n →∞  e n (e1 n − 1) −1    n 1   = (e – 1) × 1 = (e – 1).

  π 38. (b) lim  tan  + x   x →0   4

[Using L’Hospital’s Rule]

42. (a) Clearly, for n = 1, lim x→0

 1 + e1 n + (e1 n ) 2 + ... + (e1 n ) n −1  = lim   n →∞ n   1n n



0   0 form   



x sin x

= e x →0 cos x (1+sin x ) = e0 = 1.

 1 e1 n e 2 n e( n −1) n  37. (c) lim  + + + ... +  n →∞ n n n n  

k[ g' ( x) − f ' ( x)] =4 g' ( x) − f ' ( x)

59

2

lim

= e x→ 0 1+3 x

Limits

= e

2 x2 1 lim ⋅ x → 01+ 3 x 2 x 2

x  −1  ∴  sec  x + 1  is not defined    x  sec −1   does not exist. Hence xLt →∞  x +1 45. (b) Case I. n is a positive integer xn nx n −1 lim x = lim x x →∞ e x →∞ e = lim x →∞

n (n − 1) x n − 2 n! = ... = lim x x →∞ e ex [Using L’Hospital’s Rule repeatedly]

= 0. Case II. n is a negative integer. xn x−m lim x = lim x x →∞ e x →∞ e [Putting n = – m, where m is a positive integer] 1 1 = = lim = 0. x →∞ x m e x ∞ Case III. n = 0 1 1 xn lim x = lim x = = 0. x →∞ e x →∞ e ∞ Hence lim

x →∞

xn = 0 for all values of n. ex

60

(1 + h)sin (1 + h − [1 + h]) 1 + h −1 (1 + h)sin (1 + h − 1) =  lim h →0 h sin h 1 + h) =  lim(  = 1. h →0 h (1 − h)sin (1 − h − [1 − h]) LHL = lim h →0 1− h −1 (1 − h)sin (1 − h) = lim = – ∞. h →0 −h Since LHL ≠ RHL,

Case III. | x | = 1

46. (d) RHL = lim h →0

Objective Mathematics

x sin ( x − [ x]) does not exist. x −1 47. (b) lim (1 + x) (1 + x2) (1 + x4) ... (1 + x2n)

lim f ( x) = lim n →∞

(1 + x)(1 − x 2 )(1 + x 3 )(1 − x 4 )...(1 − x 4 n ) x →−1 [(1 + x )(1 − x 2 )(1 + x 3 ) (1 − x 4 )...(1 − x 2 n )]2



(1 − x 2 )(1 + x 2 )(1 + x 4 )...(1 + x 2 n ) 1− x . . . . . . . . . 1 − x4n + 2 1 for | x | < 1. = lim = n →∞ 1− x 1− x = lim

n →∞

sin x 1+ x + sin x x = 1 + 0 = 1. 48. (b) lim = lim x →∞ x − cos x x →∞ 1 − cos x 1− 0 x   1 sin x = lim y sin   = O × (a finite quantity)  ∵lim x →∞ y → 0 x  y     cos x   =0. Similarly lim = 0. x →∞ x   49. (a) lim x →0

 = lim x →0

3

1 + sin x − 3 1 − sin x x

1 (1 + sin x) − (1 − sin x) × (1 + sin x) 2 / 3 + (1 + sin x)1/ 3 (1 − sin x) 2 / 3 + (1 − sin x) 2 / 3 x

2⋅ = lim x →0 ×

sin x x

(1 + sin x)

2/3

1 + (1 + sin x) (1 − sin x)1/ 3 + (1 − sin x) 2 / 3

= xlim →−1

1 + x 2 n +1 1 − x 2 n + 2 1 − x4n ... × × × 1+ x 1 − x2 1 − x2n

= xlim →−1

x 2 n +1 − (−1) 2 n +1 x 2 n + 2 − (−1) 2 n + 2 × x − (−1) x 2 − (−1) 2 ×...×

x →1

(1 − x)(1 + x)(1 + x 2 )(1 + x 4 )...(1 + x 2 n ) = lim n →∞ 1− x

1 −1 = 0. 1+1

51. (a) lim

∴ lim n→∞

n →∞

=

2n + 1 2n + 2 2n + 3 2n + 4 4n ⋅ ⋅ ⋅ ... = 4nC2n. 1 2 3 4 2n

52. (b), (c) Case I. x ≠ mπ (m is an integer) 1 1 lim = = 0. n →∞ 1 + n sin 2 nx ∞

Case II. x = mπ (m is an integer) lim

n →∞

1 1 = = 1. 1 + n sin 2 nx 1

log (tan 2 x)  ∞  log tan x (tan 2 x) = lim  form  53. (a) xlim x →0+ log (tan x )  ∞ → 0+   log m   Using log n m =  log n   1 (2 sec 2 2 x) x 2 tan = lim [Using L’Hospital’s Rule] x →0 1 (sec 2 x) tan x

50. (a), (b), (c) Case I. | x | < 1 x2n − 1 0 − 1 lim f ( x) = lim 2 n = = – 1. n →∞ n →∞ x +1 0 +1 Case II. | x | > 1 1 1 − 2n x = 1 − 0 = 1. lim f ( x) = lim n →∞ n →∞ 1 1 + 2n 1 + 0 x

0   0 form   

= lim x →0

2 sin x cos x 2 sin 2 x = lim  sin 2 x cos 2 x x →0 sin 4 x

= lim x →0

4 cos 2 x [Using L’Hospital’s Rule] 4 cos 4 x

= 1. 2 sin x − sin 2 x dx x3 d 2 sin x − sin 2 x 2 sin x − sin 2 x dx = ⇒ f ′ (x) = dx ∫ x3 x3

54. (d) f (x) =

∫

1/ 3

1 2 =2×1× = . 3 3

x 4 n − (−1) 4 n x 2 n − (−1) 2 n

∴ lim f' ( x) = lim x →0

= lim x →0

x →0

2 sin x − sin 2 x  x3

2 cos x − 2 cos 2 x  3x 2

0   0 form    0   0 form   

[Using L’Hospital’s Rule] = lim x →0

−2 sin x + 4 sin 2 x  6x

0   0 form    [Using L’Hospital’s Rule]

[Using L’Hospital’s Rule] 6 = = 1. 6  11 x + 2 1 x + 3 1 x + ... + n 1 x  55. (a) lim   x →∞  n    1 + 2 + 3 + ... + n    = lim y →0 n   y





= e = e

y

y

y

nx

 sin([ x − 3])   sin( − 4)  60. (c) L.H.L. = lim−   =   x→0  [ x − 3]   −4 

 sin 4  =   =–1  4 



∵ π 20.

13. (a) When t ≥ 0, | t | = t ∴ x = 2t – t = t and y = t2 + t.t = 2t2 ⇒ y = 2x2, when x ≥ 0 [As t ≥ 0 ⇒ x ≥ 0]

87

⇒ (a + 1) + 1 = xlim →0

Continuity and Differentiability



88



Also, for t ≤ 0, | t | = – t



∴ x = 2t + t = 3t and y = t2 + t (– t) = 0

Objective Mathematics



sin ( x − 1), if x ≥ 1 =  sin ( 1 − x ), if x < 1



⇒ y = 0 when x ≤ 0 [As t ≤ 0 ⇒ x ≤ 0]



Hence the function is defined as :



Now, L (gof ) ' (1) = lim



2 x 2 y = f (x) =  0



= lim



∴ (gof ) (x) is not differentiable at x = 1.

,x≥0 ,x≤0



Now, L f ′ (0) = hlim →0

f (0 − h) − f (0) 0 = lim =0 −h h→0 −h



and R f ′ (0) = hlim →0

2h 2 f (0 + h) − f (0) = hlim =0 →0 h h



f (1 − h) − f (1) 31− h − 3 = hlim →0 −h −h

 3− h − 1    = 3 log 3 = 3 hlim →0  −h  f (1 + h) − f (1) R f ′ (1) = hlim →0 h 4 − (1 + h) − 3 = lim h→0 h −h = hlim = – 1. →0 h

Since L f ′ (1) ≠ R f ′ (1), therefore, f (x) is not differentiable at x = 1 but f (x) is continuous at x = 1 (as L f ′ (1) and R f ′ (1) are finite). 15. (c) The function f (x) can be rewritten as

 x3 , − 1 < x < 1 f (x) =   x, x ≤ − 1 or x ≥ 1 f (1 − h) − f (1) Now, L f ′ (1) = hlim →0 −h

( gof ) (1 − h) − ( gof ) (1) −h

sin  1 − (1 − h)  − sin 0

h→0

−h

= lim

h→0

sin h →– ∞ −h

17. (c) We have,

Since L f ′ (0) = R f ′ (0), therefore, f (x) is differentiable and hence continuous at x = 0. 14. (b) L f ′ (1) = hlim →0

h→0



 x 1 − x , x ≤ − 1   x 1 + x , − 1 < x < 0 f (x) =  .  x , 0 ≤ x 0 f (x) =  =  xe , x > 0 If x = – 1, then x = x . So, f (x) = x.  0 0 , x = 0 ,x=0 If – 1 < x < 0, then x < x3. So, f (x) = x3.   3 3  If x = 0, then x = x . So, f (x) = x .  If 0 < x < 1, then x > x3. So, f (x) = x. f (x) is differentiable as well as continuous 3 If x = 1, then x = x . So, f (x) = x. everywhere except possibly at x = 0. If x > 1, then x < x3. So, f (x) = x3. Now, Thus, f (x) = x, x ≤ – 1 f ( 0 + h ) − f ( 0) R f ′ (0) = hlim x3, – 1 < x ≤ 0   x, 0 < x ≤ 1   x3, x > 1. →0 h Clearly, f (x) is not differentiable at x = – 1, 0, 1. − 2/ h he − 0 = hlim = hlim e– 2/h = 0 →0 →0 1  2 h x≠0  x sin , x 24. (a), (b)  We have, g (x) =  f (0 − h) − f (0) lim − h − 0 L f ′ (0) = lim = h→0 = 1. 0, x = 0. h→0 −h −h For x ≠ 0, 1  1 1 1 Since L f ' (0) ≠ R f '(0), therefore f (x) is not dif g′ (x) = x2 cos     − 2  + 2x sin  = – cos  +   x x x x ferentiable at x = 0 but f (x) is continuous at x = 0 1 2x sin  . [as L f ′ (0) and R f ′ (0) are finite]. x For x = 0

27. 1 x sin − 0 g ( x ) − g ( 0) x lim = g′ (0) = lim x→0 x→0 x−0 x 1 28. = lim x sin = 0. x→0 x 1 1  x≠0 2 x sin − cos , ∴ g′ (x) =  x x 0, x=0 29. 1 g′ is not continuous at x = 0 as cos  is not continux ous at x = 0. Also, f is not differentiable at x = 0.

2

(d) Since | x – 1 |, | x – 2 | and cos x are continuous functions and the sum of continuous functions is also continuous, so the given function is continuous every where, i.e.,, there is no point of discontinuity. (c) Since logax is continuous where x > 0 and a > 0, a ≠ 1, therefore log (1 + x) is continuous when 1 + x > 0 i.e., x > – 1. ∴ interval = (– 1, + ∞). 1 (a) Let f (x) = log | x | . The points of discontinuity of f (x) are those points where f (x) is undefined or infinite. It is undefined

89



For f to be continuous at x = 0, we must have

Continuity and Differentiability



90

Objective Mathematics

when x = 0 and is infinite when log | x | = 0, | x | RHL = lim {x – [x]} = lim {x – n} = p – n x→ p x→ p = 1, i.e., x = ± 1. x> p ∴ Set of points of discontinuity = {– 1, 0, 1}. f ( p) = p – [ p] = p – n 1 30. (b) We have, f (x) = 1 − x . ∴ f (x) is continuous at all non-integral points p. 35. (c) Let x = n, n ∈ Z As at x = 1, f (x) is not defined, x = 1 is a point of discontinuity of f (x). Then, LHL = xlim (x) = n; →n If x ≠ 1, f [ f (x)] xn

Since, LHL ≠ RHL, therefore f (x) is discontinuous at all integers n. Now, let x = p, n < p < n + 1, where n is an integer. (x) = n + 1, Then, LHL = xlim →p x< p

x> p

36.

 e[ x ] + | x | − 2 , x≠0  32. (d) f (x) =  [ x] + | x | − 1 , x=0 

37.

e[ x ] + | x| − 2 e−1 − 2 lim f ( x) = lim− = − x → 0 [ x] + | x | x→0 −1 e[ x ] + | x| − 2 ex − 2 lim lim f ( x) = lim+ →−∞. = + + x → 0 [ x] + | x | x→0 x→0 x 33. (b) We have,

38.

39.



f (x) = lim (sin x)



π π  0, if |sin x | < 1 i.e. − 2 + nπ < x < 2 + nπ =  1, if |sin x | = 1 i.e. x = nπ + π , n ∈ I  2

f ( p) = ( p) = n + 1. Since LHL = RHL = f ( p), therefore f (x) is continuous at all non-integral points p. (a) As the function log | x | is not defined at x = 0, therefore the set of points of discontinuity is {0}. (c) As the function | sin x | is defined for all real x, therefore the function f (x) is continuous at all real x. Hence, the required set is φ (empty set).  |sin x | (b) The function is not defined for x = nπ, n ∈ I.  sin x Hence the set of points of discontinuity is {nπ : n ∈ I}. (b) f is continuous at x = π/4, if lim f ( x) = f (π/4).

2n

x → π/ 4

x → π/ 4



lim

{x – [x]} = lim {x – (n – 1)} x→n

= n – (n – 1) = 1. RHL = xlim {x – [x]} = lim {x – n} = n – n = 0. →n x→n

xn

x → π/ 4

Now, L = lim (sin 2 x)

n→∞

Then LHL =

(x) = n + 1 RHL = xlim →p



∴

f (π/4) = e–1/2 = 1/ e .

x+ y f ( x) + f ( y ) f , f (0) = 0 and f ′ (0) = 3 = 3 3   f ( x + h) − f ( x ) f ′ (x) = hlim →0 h  3 x + 3h  f  − f ( x)  3  = hlim →0 h f (3 x) + f (3h) f (3 x) + f (0) − 3 3 = hlim →0 h f (3h) − f (0) = hlim =3 →0 3h

RHL = lim

h→0



= lim [6 – 5 (1 + h)] = 6 – 5 = 1 h→0



 ∴ f (x) continuous at x = 1. At x = 3 : LHL = lim (3 – h) = hlim [6 – 5 (3 – h)] = – 9, →0



f (3) = 3 – 3 = 0.



0   0 form   

lim  f (x) = lim 1 − tan x  x → π/ 4 4x − π

x → π/ 4

lim

x → π/ 4

2 1 −sec 2 x = − = − 4 4 2

∴  For f (x) to be continuous at x =

π 4

π 1 f  =– 2 4 43. (a) Differentiability at x = 0 :



f (0 − h) − f (0) −h − h (e − 1/ h − e1/ h ) − h (e − 1/ h + e1/ h )

L f ′ (0) = hlim →0 =  lim

h→0

e− 2 / h − 1  = – 1. e− 2 / h + 1 f ( 0 + h ) − f ( 0) R f ′ (0) = hlim →0 h

 =  hlim →0

1/ h

− 1/ h

h (e − e ) = hlim →0 h (e1/ h + e − 1/ h )

f (x) =

xn

∞

∑ n!

(log a)n =

n=0

= ex log a =



( x log a ) n n! n=0 ∞

∑

= a x. f (0 − h) − f (0) a− h − 1 lim = −h −h h→0

= logea

f ( 0 + h ) − f ( 0) h

R f ′ (0) = lim

h→0



= lim

h→0

ah − 1 = logea h

Since L f ′ (0) = R f ′ (0), ∴ f (x) is differentiable at x = 0. Since every differentiable function is continuous, therefore, f (x) is continuous at x = 0. 46. (b) We have, for x ≠ 0,





h→0

L  f ′ (0) = hlim →0

Therefore, the only point of discontinuity is x = 3. 42. (a) We have,

= lim (1 + h)2 [(1 + h)1/h ]– 2

= 1 × e– 2 = e– 2. Since LHL ≠ RHL,  ∴ f (x) is not continuous at x = 0. 45. (d) We have

h→0

Since LHL ≠ f (3),  ∴ f (x) is discontinuous at x = 3.

1 1 2− +  h h

h→0

2 2− h

h→0

1 (2 + 3) = 1. and f (1) = 5 Since LHL = RHL = f (1),



f (0 + h) = lim (h + 1)

= lim (1 + h)

RHL = lim f (1 + h) h→0

h→0

h→0

1 [2 (1 – h)2 + 3] = 1 = lim h→0 5



1 1 2− −  h h

= lim (1 – h)2 = 1

LHL = lim f (1 – h) h→0

f (0 – h) = lim (− h + 1)

LHL = hlim →0

1 − e− 2 / h = lim 1 + e − 2 / h = 1. h→0 Since L f ′ (0) ≠ R f  ‘ (0), ∴ f (x) is not differentiable at x = 0. But since L f ′ (0) and R f ′ (0) are finite, therefore f (x) is continuous at x = 0. Hence, f (x) is continuous every where but not differentiable at x = 0.

f (x) = 

x x x + + 1 + x ( x + 1) (2 x + 1) (2 x + 1) (3 x + 1) + ... to ∞

=

x ∑ [( − 1 ) + 1] [nx + 1] n x n =1

=

∑  (n − 1) x + 1 − nx + 1

∞

∞



1

1

  1  = nlim 1 −  = 1. →∞ nx + 1  n =1

 

1, x ≠ 0  For x = 0, f (x) = 0. Thus, f (x) = 0, x = 0    Clearly, lim f (x) = lim f (x) = 1 ≠ f (0). x → 0−

x → 0+

So, f (x) is discontinuous at x = 0. 47. (b) For x ≠ 0, we have x /1+ x x /1+ x f (x) = x + = x + x / 1 + x = x + 1. 1 1− 1+ x For x = 0, f (x) = 0  x + 1, x ≠ 0 Thus, f (x) =  0 , x = 0  Clearly, lim f (x) = lim+ f (x) = 1 ≠ f (0). x → 0−

x→0

91

  f (0) = 0 ⇒ c = 0 ∴  f (x) = 3x. 41. (b) At x = 1 :

Continuity and Differentiability

44. (a) The only doubtful point is x = 0.

∴  f (x) = 3x + c,

92

So, f (x) is discontinuous and hence not differentiable at x = 0.

Objective Mathematics

48. (b) Continuity at x = 2 : LHL = lim f (2 – h) h→0

h→0

RHL = lim f (2 + h) h→0

= lim [(2 + h)2] + [– 2 – h]2 h→0

= lim {4 + (– 3)2} = 13. h→0

Since LHL ≠ RHL,  ∴ f (x) is not continuous at x = 2. Differentiability at x = 2 : f ( 2 − h ) − f ( 2) LHD = hlim →0 ( 2 − h) − 2 7−8 → ∞. = lim h→0 −h So, f (x) is not differentiable at x = 2. Hence, the function f (x) is neither continuous nor derivable at x = 2. 49. (b) For x ≠ – 1, we have

f (x) = 1 – 2x + 3x2 – 4x3 + ...∞ 1 = (1 + x)– 1 = 1 + x . 1 lim f (– 1 – h) = lim 1 − 1 − h → – ∞ h→0 h→0 So, f (x) is not continuous at x = – 1. Also, lim

h→0

f (− 1 − h) − f (−1) = hlim (−1 − h) − (−1) →0

−1 −1 h −h

1+ h → ∞. = lim h→0 h2 So, f (x) is not derivable at x = – 1. Hence, f (x) is neither continuous nor derivable at x = – 1. 1 50. (a) Since f (x) = , ∴  sin x ∉ [0, 1] [sin x] ⇒  x ∉ [2nπ, (2n + 1) π] – (4n + 1) π/2, n ∈ I. ∴ f (x) is not continuous if x ∈ (2nπ, 2nπ + π), n ∈ I. 51. (b) Since f (x) is continuous at x =

Thus the required value is f ′ (0). f (h) − f (0) = k (say) h f (0) − f (0 − h) ∴  f ′ (0 – ) = hlim →0 h f (0) − f (h) = hlim =– k →0 h + – f ′ (0 ) ≠ f ′ (0 ), but both are finite so f (x) is continuous at x = 0 but not differentiable at x = 0. 54. (c) Since f (x) is continuous at x = 0, ∴ lim f (x) = f (0). 53. (b) Let f ′ (0 + ) = hlim →0

x→0



⇒

⇒

b (1 − sin x) (π − 2 x) 2 = a

lim

 π  b 1 − sin  + h   cos x 2   = lim 2 = a h→0 3 cos 2 x  π  π − 2  2 + h     2

x → π/ 2

lim

h→0



= f (a) + lim f (h) = f (a) + f (0)

h→0

h→0

= f (a + 0) = f (a). ∴  f (x) is continuous at x = a. Since x = a is any arbitrary point, therefore f (x) is continuous for all x. 55. (c) We have,

f (x) = nlim →∞



= nlim →∞

(2 sin x) 2 n 3 − (2 cos x) 2 n n

(2 sin x) 2 n ( 3 ) 2 n − (2 cos x) 2 n

f (x) is discontinuous when ( 3 ) 2 n − (2 cos x) 2 n = 0 i.e.,, cos x =

1 − sin x = 3 cos 2 x

x → π/ 2 x > π/ 2

lim f (x) = lim f (a + h)

x→a

= lim [ f (a) + f (h)] [ ∵ f (x + y) = f (x) + f ( y)]

π , 2

lim

x → π/ 2 x < π/ 2

Take any point x = a, then at x = a



π ∴ LHL = RHL = f   2 2



b  sin h / 2  1 = hlim   =a → 0 8  h/2  3 1 b = = a. ⇒ 3 8 ∴ a = 1/3 and b = 8/3. f ( x) 52. (d) We have, f (0) = 0, g (x) = , x f ( x) = lim f' ( x) = f ′ (0) lim g ( x) = lim x→0 x→0 x→0 x 2

{3 + (– 2)2} = 7. = hlim →0



1 b (1 − cos h) = hlim =a →0 3 4h 2

⇒

= lim [(2 – h)2] + [– 2 + h]2



⇒

⇒x =

π . 6

3 2

56. (c) Since f (x) is continuous in [0, 1], therefore

  n  n   lim f  2 n + 1  = f  nlim →∞ 2 n + 1  n→∞   



1 = f   = 2. 2

L f ′ (0) = hlim →0

f (h) − f (0) h2 = hlim =0 →0 h h



log cos h = hlim → 0 log (1 + h 2 ) 

f (− h) − f (0) −h



− tan h = hlim → 0 2 h / (1 + h 2 ) = 1/2.

− ( − h) − 0 = – 1. −h Since R f ′ (0) ≠ L f ′ (0), ∴ f ′ (0) does not exist.

= hlim →0

At x = 1: f (1 + h) − f (1) h 2 [(1 + h) − (1 + h) + 1] − 1 lim h→0 h

R f ′ (1) = hlim →0

= =

h2 + h lim = lim (h + 1) = 1 h→0 h h→0

L f ′ (1) = hlim →0 =

lim

h→0

(1 − h) 2 − 1 f (1 − h) − f (1) = hlim →0 −h (1 − h) − 1

h 2 − 2h = lim (2 – h) = 2. −h h→0

Since R f ′ (1) ≠ L f ′ (1), ∴ f ′ (1) does not exist. ∴ f (x) is not differentiable at x = 0 and x = 1. x x 58. (b) f (x) = 1 + x , x > 0; f (x) = 1 − x , x < 0;

f (x) = 0, x = 0.

If x > 0 or x < 0, f (x) is a rational function of x, having non-zero denominator. Hence f (x) is continuous and differentiable for x > 0 or x < 0. So the only doubtful point is x = 0. At x = 0 : h −0 f (h) − f (0) 1+ h lim lim = h→0 =1 R f ′ (0) = h → 0 h h −h −0 f (− h) − f (0) 1 − ( − h) lim lim L f ′ (0) = h → 0 = h→0 =1 −h −h

0   form  0 

Since L f ′ (0) = R f ′ (0), therefore f (x) is differentiable at x = 0. Since differentiability ⇒ continuity, therefore f (x) is continuous at x = 0. 60. (b) We have, f (0 − h) − f (0) L f ′ (0) = lim −h h→0 − h (1 + h sin 1 / h) −h 1 1  = hlim + h sin   →0 h  h

=  hlim →0

= ∞ + 0 × a finite quantity = ∞ ∴ f (x) is not differentiable at x = 0. Clearly, f (x) is continuous at x = 0. 61. (c) We have f (2 ⋅ 5 − h) − f (2 ⋅ 5) L f ′ (2 ⋅ 5) = lim −h h→0



[2 ⋅ 5 − h − 2] − [2 ⋅ 5 − 2] = hlim →0 −h



= hlim →0



and R f ′ (2 ⋅ 5) = hlim →0



= lim



∴  f ′ (2 ⋅ 5) = 0.



Also, L f ′ (5) = lim



2−3 → ∞. = lim [5 − h − 2] − [5 − 2] = hlim →0 − h h→0 −h

h→0

0 =0 −h

f (2 ⋅ 5 + h) − f (2 ⋅ 5) h

[ 2 ⋅ 5 + h − 2 ] − [ 2 ⋅ 5 − 2] 0 = lim =0 h→0 h h

h→0

f (5 − h) − f (5) −h

Since R  f ′ (0) = L f ′ (0) = 1, ∴ f ′ (0) = 1 i.e., f (x) is differentiable at x = 0. Since differentiability ⇒ continuity,  Hence, f ′ (2 ⋅ 5) = 0 while f ′ (5) does not exist. ∴ f (x) is also continuous at x = 0. Hence f (x) is continuous as well as differentiable π  π for all real x. f  − h − f   π 4  4 59. (a) We have, 62. (c) L f ′   = lim  h→0 −h 4 f (0 − h) − f (0) lim L f ′ (0)= hlim = h → 0 − h log cos h  π π   →0 −h  tan  4 − h   −  tan 4  − h log (1 + h 2 )     = lim 0 − 1 → ∞   = lim h→0 h→0 −h −h log cos h 0  = lim  π  form  log (1 + h 2 ) h→0 0  ∴ f ′    does not exist. 4 − tan h f (0 − h) − f (0) = lim 2h / (1 + h 2 ) = – 1/2. h→0 63. (c) L f ′ (0) = lim   −h h→0 h log cos h f (0 + h) − f (0) lim R f ′ (0) = hlim = h → 0 h log (1 + h 2 ) 1 − 1 − h2 →0 h = hlim   →0 −h

93



R f ′ (0) = hlim →0

Continuity and Differentiability

57. (d) At x = 0 :

94



Objective Mathematics





= lim   h→0

= lim

h→0

1 − (1 − h 2 ) x −h −1 1 + 1 − h2 h→0

= lim



= lim

1− 1− h

h→0

h→0

h→0

1 + 1 − h2

h p cos

−1 = . 2

= lim

h→0

1+ 1− h



2

=

1 . 2



Therefore, f (x) is not differentiable at x = 0. Since L f ′ (0) and R f ′ (0) are finite, therefore, f (x) is continuous at x = 0. Hence, f (x) is continuous but not differentiable at x = 0. f (0 − h) − f (0) 64. (b) L f ′ (0) = lim h→0 −h



= lim

h→0

= hlim →0

1− e −h

 h2  h ⋅ 1 − + ... 2!   = lim = – 1. h→0 −h f ( 0 + h ) − f ( 0) R f ′ (0) = lim h→0 h 2

1 − e− h = lim = 1. h→0 h Hence, f (x) is not differentiable at x = 0. Since L f ′ (0) and R f ′ (0) are finite, therefore, f (x) is continuous at x = 0. Hence, f (x) is continuous but not differentiable at x = 0. 65. (a) Continuity at x = 0: 1 LHL = lim f (0 – h) = lim (– h) p cos = 0 if p h→0 h→0 h >0 RHL = lim f (0 + h) = lim hp cos h→0

h→0



=  lim

(− h) p cos

h→0

= lim (– h) h→0



−h p – 1

cos

= lim h p – 1 cos h→0

1 = 0 if p > 1 h

1 −0 h 1 = 0 if p – 1 > 0, i.e., p > 1; h

(| 2 − h − 1| + | 2 − h − 3|) − (| 2 − 1| + | 2 − 3|) = hlim →0 −h | 1 − h | + | 1 + h | − 2 1− h +1+ h − 2 = hlim = hlim →0 →0 −h −h 0 = hlim = 0. →0 −h f ( 2 + h ) − f ( 2) R f ′  (2) = hlim →0 h (| 2 + h − 1 | + | 2 + h − 3|) − 2 = lim h→0 h



= lim |1 + h | + | −1 + h | − 2 h→0 h



= lim 1 + h + 1 − h − 2 = lim 0 = 0. h→0 h→0 h h

Since L f ′ (2) = R f ′ (2), therefore f (x) is differentiable at x = 2 and f ′ (x) = 0 at x = 2. 67. (b) We have,  0, 0 ≤ x ≤ b  2b3  −8 f ′ (x) =  x + 2 , b < x ≤ a 9x 9  −2 3 3  2 (a − b ), x > a 9x

∴ f ′ (a+) = lim f ′ (a + h)



 − 2 (a 3 − b3 )  ⋅ = lim  2  h→0  9 ( a + h) 



=

1 = 0 if p > 0 h

and f (0) = 0. ∴ f (x) is continuous at x = 0 if p > 0 Differentiability at x = 0 : f (0 − h) − f (0) L f ′ (0) = lim −h h→0



1/ 2

1/ 2





− h2

   h4 2 − ...   1 − 1 − h + ! 2    −h

h

1 −0 h

∴ f (x) is differentiable at x = 0 if p > 1. f (2 − h) − f (2) 66. (b) L f ′ (2) = lim −h h→0

2

h 1

f ( 0 + h ) − f ( 0) h

R f ′ (0) = lim

f ( 0 + h ) − f ( 0) h

R f ′ (0) = lim



1

h→0

− 2 a 3 − b3 ⋅ . 9 a2

and f ′ (a– ) = lim f ′ (a – h) h→0



 −8 2b3  ( a − h) + = lim   h→0 9 ( a − h) 2   9



=

− 8a 2b3 − 8a 3 + 2b3 + 2 = 9 9a 9a 2

∴ f ′ (a+ ) ≠ f ′ (a– ). 68. (a) We have, f (1 − h) − f (1) L f ′ (1) = lim −h h→0



0 −1 = hlim →∞ →0 −h ∴ f ′ (1) does not exist. f (2 − h) − f (2) Also, L f ′ (2) = lim −h h→0



(2 − h) [2 − h] − (2 − 1) [2] −h 2−h−2 = lim =1 −h h→0

= lim

h→0

f (2 + h) − f (2) h (1 + h) [2 + h] − (2 − 1) [2] = lim h→0 h 2 + 2 h − 2 = lim = 2. h→0 h ∴ f ′ (2) also does not exist. 69. (c) We have, and R  f ′ (2) = lim

h→0

f (0 − h) − f (0) −h

L f ′ (0) = lim

h→0

= lim 3 − 1 → – ∞ h→0 −h ∴ f (x) is not differentiable at x = 0 Also, if x < 0 or x ≥ 0 then | x | ≥ 0 ∴ f ( | x | ) = 2 | x | + 1 for all x. f ( 0 + h ) − f ( 0) ∴ R f ′ (0) = lim h→0 h 2h + 1 − 1 =2 = lim h→0 h f (0 − h) − f (0) and L f ′ (0) = lim −h h→0 = lim

h→0

2 (− h) + 1 − 1  = 2 −h

h 2 + 5h = hlim = 5. →0 h ∴ f ( | x | ) is not differentiable at x = 2 but f ( | x | ) is continuous at x = 2 (as L f ′ (2) and R f ′ (2) are finite). Hence, f ( | x | ) is continuous but not differentiable in (– 3, 3). f ( 4 + h ) − f ( 4) 71. (b) f ′ (4) = hlim →0 h f ( 4 + h ) − f ( 4 + 0) = hlim →0 h 2 f (4) ⋅ f (h) − 2 f (4) f (0) = hlim →0 h



[Using f (x + y) = 2f  (x) ⋅ f ( y) for all x, y] f (h) − f (0) = 2f (4) hlim = 2f (4) × f ′ (0) →0 h = 2 × 2 × 3 = 12. f ( 4 + h ) − f ( 4) h f ( x ) f ( h) − f ( x ) = hlim →0 h 1 + h φ ( h) − 1 = f (x) hlim = f (x) hlim φ (h) →0 →0 h = f (x) ⋅ 1 = f (x) f ( x − h) − f ( x ) and L f ′ (x) = hlim →0 −h

72. (c) R f ′ (x) = hlim →0

= hlim →0



1 − h φ ( − h) − 1 = f (x) ⋅ 1 = f (x). −h Hence f ′ (x) exists and is equal to f (x). = f (x) hlim →0

73. (a) Let f (x) be a periodic function with period T. i.e., f (x + T) = f (x). Now, f ′ (x) = lim

h→0

∴ f ( | x | ) is differentiable at x = 0. 70. (b) If – 3 < x < – 1 then 1 < | x | < 3

The nature of f ( | x | ) changes only at x = 2. f (2 − h) − f (2) L f ′ (2) = lim −h h→0

= lim 5 − (2 − h) − (4 + 2 − 3) = – 1 h→0 −h f (2 + h) − f (2) R f ′ (2) = hlim →0 h [(2 + h) + (2 + h) − 3] − (4 + 2 − 3) = hlim →0 h 2



f ( x + h) − f ( x ) h

f ( x + T + h) − f ( x + T ) h f ( x + h) − f ( x ) = f ′ (x). h

∴ f ′ (x + T) = lim

h→0

and if – 1 ≤ x < 0 then 0 < | x | ≤ 1. 5 − | x |, 0 < | x | ≤ 1  ∴ f ( | x | ) = 5 − | x |, 1 < | x | < 2  2 | x | + | x | − 3, 2 ≤ | x | < 3

f ( x ) f ( − h) − f ( x ) f ( − h) − 1 = f (x) hlim →0 −h −h





= hlim →0

Similarly, f “ (x + T) = f “ (x). ∴ f ′ and f “ are also periodic functions with the same period. 74. (d) We have,

  | x|  sgn   sgn [sgn (x)] =   x  sgn (0) 



 | xx |  | x| =  x   0

,x≠0 ,x=0

,x≠0 ,x=0

95

(1 − h) [1 − h] − 1 [1] = hlim →0 −h

Continuity and Differentiability



96

x  ,x≠0 ∴ f (x) =  | x | 0 , x = 0 

Objective Mathematics



Continuity at x = 0 :

LHL = hlim f (0 – h) = lim →0

h→0

−h = – 1 and f (0) = 0 |− h|

∴ f (x) is not continuous and hence not differentiable at x = 0. 75. (a) When t < 0, x = 2t – (– t) = 3t < 0 and y = t2 + t (– t) = 0 ∴ y = 0 when x < 0. When t ≥ 0, x = 2t – t = t ≥ 0 and y = t2 + t (t) = 2t2 = 2x2 ∴ y = 2x2 when x ≥ 0. Thus, the function f is given by : 0 , x < 0 f (x) =  2 2 x , x ≥ 0

Since, L f ′ (0) = R f ′ (0), therefore f (x) is differentiable at x = 0. Since differentiability ⇒ continuity, therefore f (x) is continuous at x = 0. Hence, f (x) is continuous as well as differentiable in the interval [– 1, 1]. 76. (b) If | 2sin x | < 1, f (x) x x = nlim = = x. →∞ (2 sin x) 2 n + 1 0 +1 If | 2 sin x | = 1, f (x) = nlim →∞

x (2 sin x) 2 n + 1

x 1 = = x. 1+1 2 If | 2 sin x | > 1, f (x) = nlim →∞





1 or 2

−1 2

sin x


1 2

Clearly, f (x) is not continuous at x = nπ ±

The only doubtful point is x = 0. f (0 − h) − f (0) 0−0 L f ′ (0) = lim = hlim = 0. →0 −h −h h→0 R f ′ (0) = hlim →0

⇒ | sin x | >

and | 2 sin x | > 1

78. (a) We have, f (x) = 2x + | x – x2 |, – 1 ≤ x ≤ 1

= 2x + | x | | 1 – x |, ­– 1 ≤ x ≤ 1



2 x + (− x) (1 − x), − 1 ≤ x < 0 =  2 x + x (1 − x), 0 ≤ x ≤ 1



2  x + x , − 1 ≤ x < 0  = 2 3 x − x , 0 ≤ x ≤ 1

Since the polynomial functions are continuous as well as differentiable everywhere, so the only doubtful point is the point x = 0.

= hlim →0

 1  (cos x + sin x)  83. (b) We have, f (x) =   2 

(− h + h 2 ) − 0 −h



= hlim (1 – h) = 1. →0



R f ′ (0) = hlim →0



= hlim (3 – h) = 3. →0



f ( 0 + h ) − f ( 0) (3h − h 2 ) − 0 = hlim →0 h h

Since L f ′ (0) ≠ R  f ′ (0), therefore f (x) is not differentiable at x = 0 but f (x) is continuous at x = 0 (as L  f ′ (0) and R  f ′ (0) are finite). Hence f (x) is continuous but not differentiable in the interval [– 1, 1]. 79. (c) Since [x] and e| x | are not differentiable at x = 0, therefore, for f (x) to be differentiable at x = 0, we must have a = 0, b = 0 and c can be any real number. 80. (a) Since f (x) is continuous at x = 0, therefore lim f (0 – h) = lim f (0 + h) = f (0) h→0



h→0

log (1 + 3h) − log (1 − 2h) h log (1 + 3h) log (1 − 2h) +2 = hlim 3 →0 3h − 2h

⇒ a = hlim →0

= 3 × 1 + 2 × 1 = 5.   81. (b) Since f (x) is differentiable at x = 0, therefore L f ′ (0) = R f ′ (0) ⇒ hlim →0 ⇒ hlim →0

f (0 − h) − f (0) f ( 0 + h ) − f ( 0) = hlim →0 −h h (a + b | − h | + c | − h |4 ) − a −h = hlim →0



f (1 + h) = hlim →0



 [1 + cos (π + π h)]n + 1   nlim  n = hlim →0  → ∞ [1 + cos (π + π h)] − 1 



 (1 − cos π h) n + 1  lim   = hlim n →0  n → ∞ (1 − cos π h) − 1 





 0 + 1   = – 1. = hlim →0  0 −1 f (1 – 0) = hlim f (1 – h) →0  [1 + cos (π − π h)]n + 1  lim lim  = h → 0    n → ∞ [1 + cos (π − π h)]n − 1   (1 − cos π h) n + 1  lim   = hlim   n →0  n → ∞ (1 − cos π h) − 1   0 + 1 = hlim  =–1 → 0    0 −1 0 +1 (1 + cos π) n + 1 Also, f (1) = nlim   = = – 1. → 0 (1 + cos π) n − 1 0 −1 Since, f (1 + 0) = f (1 – 0) = f (1), therefore f (x) is continuous at x = 1. x

1

∫ t cos t

dt

0

3

⇒ – b = b i.e., b = 0. , 0 < x 0  sgn x = 0, x = 0 −1, x < 0 

⇒ k = hlim →0



= hlim →0



= hlim →0



=



= hlim →0



= hlim →0

( fog ) (− 2 − h) − ( fog ) (− 2) −h |[− 2 − h]| − |[ − 2]| −h

sin 2 (π − h) log [1 + π − 2π (π − h) + (π − h) 2 ] 2

1 − cos h sin 2 h ⋅ 2 h log (1 + h 2 ) 1  sin h / 2  h2  sin h  ⋅  ⋅  ⋅  2  h / 2  log (1 + h 2 )  h  2

2

1 . ∴ k = 1/2. 2

94. (a) Since f (x) is continuous at x = π/2 ∴ x lim f (x) = f (π/2) → π/ 2  π  π  sin cos  − h   − cos  − h  2  2   ⇒ k = hlim 2 →0  π  π − 2  2 − h    

Clearly, f (x) is continuous as well as differentiable at x = 0. 90. (b) ( fog) (x) = f (g (x)) = f ([x]) = | [x] |. Now, L ( fog) ‘ (– 2)

1 + cos (π − h) ( π − π + h) 2 ·

 x5 ,x>0  5 0 ,x=0 Therefore, f (x) = x sgn x =   5 − x , x < 0



= hlim →0

 sin (sin h) − sin h  0 = hlim  form  →0 4h 2 0 

= hlim →0



= hlim →0



=

cos (sin h) cos h − cos h 8h

0   form  0  2 − sin (sin h) cos h − sin h cos (sin h) + sin h 8

0 = 0.     ∴ k = 0. 8



1 −3 ⋅ 4 ⋅ (− 8) 8 1 = = . = 4 −4 −2 64 8 ⋅ 4 ⋅ (3) 3



lim f (1 – h) = lim a (1 – h)2 + b = a + b, h→0

h→0

lim f (1 + h) = lim (1 + h) + 3 = 4 and h→0

h→0

f (1) = 4.

∴ f (x) will not be continuous at x = 1 if a + b ≠ 4.

97. (b) We have, 4x + 3  | y | = 5y

⇒



Taking limit as h → 0, we get



96. (d) We have,

⇒ 4x + 3y = 5y if y ≥ 0

 2 x, x ≥ 0  and 4x – 3y = 5y if y < 0 ⇒ y =  1  2 x, x < 0

Clearly, y is continuous at x = 0 but not differentiable at x = 0. 2, x ≥ 0  dy = 1 . Also, dx  2 , x < 0 98. (d) We have hlim f (3 – h) →0 = hlim | 3 – (3 – h) | + (3 + 3 – h) →0 = hlim (h + 6) = 6, hlim f (3 + h) →0 →0

f ( x + h) − f ( x ) = – 10x + 3x2h h



lim

h→0

f ( x + h) − f ( x ) = – 10x ⇒ f ′ (x) = – 10x. h

101. (c) Since | x – 1 |, | x – 1 | 2, etc, are continuous at x = 1 ∴ f (x) is continuous at x =1 for all ak ∈ R. Also, | x – 1 |2, | x – 1 |4, etc, are all differentiable at x = 1, whereas | x – 1 |, | x – 1 |3, etc, are not differentiable at x = 1. Therefore, f (x) is differentiable at x = 1 for all a2k + 1 = 0. 102. (b) We have,



f ( x + h) − f ( x ) h f ( x ) ⋅ f ( h ) − f ( x) = hlim →0 h [Using f (x + y) = f (x) ⋅ f ( y)]



= f (x) hlim →0



f ′  (x) = hlim →0

f ( h) − 1 h



1 + h φ ( h) + h 2 φ ( h) ψ ( h) − 1 = f (x) hlim →0 h lim = f (x)  [ φ (h) + h φ (h) ψ (h)]



= f (x) (a + 0 ⋅ a ⋅ b) = a ⋅ f (x).



h→0

103. (c) For f (x) to be continuous at x = 0, we must have  (1 + x)cot x f (0) = lim x→0

= hlim | 3 – (3 + h) | + (3 + 3 + h) →0

⇒ log  f (0) = lim  log (1 + x)cot x x→0

= hlim → 0 (h + 7) = 7



= lim  cot x log (1 + x) x→0

Since hlim f (3 – h) ≠ hlim f (3 + h), therefore f (x) →0 →0



log (1 + x)   = lim x→0 tan x



= lim   x→0

is not continuous and hence not differentiable at x = 3.

log (1 + x) x ⋅ = 1 ⋅ 1 = 1. x tan x ∴ f (0) = e1 = e.

99. (a) Clearly π [x – π] is an integral multiple of π for all x as [x – π] is an integer for all x. Therefore, sin (π [x – π]) = 0, for all x 104. (a) Since f (x) is continuous for all x, therefore, it is Also, 4 + [x]2 ≠ 0 for all x. continuous at x = 1 also. Therefore, f (x) = 0 for all x. ∴ f (1) = hlim f (1 – h) →0 Hence, f (x) is continuous as well as differentiable lim for all x. ⇒ 1 = h → 0 [a (1 – h)2 + b]

99

Continuity and Differentiability

95. (b) For f (x) to be continuous everywhere, we must 100. (b) We have, have f (s + t) = f (s) + as t + 3s 2 t2 for all s, t ∈ R  ...(1) f (0) = lim f (x) Putting s = 3 and t = 2 in (1), we get x→0 f (5) = f (3) + 6a + 108 (256 − 8 x)1/ 4 − 4 0  ⇒ 52 = 4 + 6a + 108 form = lim   x → 0 16 − 4 (64 + 3 x )1/ 3 0  [Putting f (3) = 4 and f (5) = 52] ⇒ 6a = – 60 i.e., a = – 10. 1 (256 − 8 x) − 3 / 4 ⋅ (− 8) Therefore, from (1), we have 4 = lim x→0 f (s + t) = f (s) – 10st + 3s2 t2 −4 − 2/3 (64 + 3 x) ⋅ (3) Putting s = x and t = h, we get 3 f (x + h) = f (x) – 10xh + 3x2 h2 [Using L' Hospital’s Rule]

100

⇒ a + b = 1

...(1)

Also, f (x) is differentiable at x = 1

Objective Mathematics

⇒ hlim →0

f (1 − h) − f (1) f (1 + h) − f (1) = hlim →0 −h h

108. (b), (c) Clearly, x = 1 is a point of discontinuity of the 1 function f (x) = . 1− x  1   = If x ≠ 1, then ( fof ) (x) = f [ f (x)] = f  1 − x  x −1 , which is discontinuous at x = 0. x If x ≠ 0 and x ≠ 1, then

1 −1 a (1 − h) 2 + b − 1 | 1 + h| lim ⇒ hlim = →0 h→0 −h h ⇒ hlim →0

(a + b − 1) + (h 2 − 2h) a 1−1− h = hlim →0 −h h (1 + h)

 x − 1 ( fo fof ) (x) = f [( fof ) (x)] = f   = x,  x  which is continuous everywhere.

⇒ 2a = – 1 (Using a + b = 1) −1 ∴ a = . 2 3 . Hence, a + b = 1 ⇒ b = 1 – a = 2

Hence f 3n (x) = ( fofof )n(x) = x, which is continuous everywhere.

105. (c) Since the function f (x) is continuous at x = 0, therefore, f (0 + h) lim f (0 – h) = f (0) = hlim →0 h→0

⇒ lim   (1 − |tan h |)

a | tan h |

h→0

e = b = hlim →0

sin 3 h sin 2 h

e ⇒ hlim [(1 – | tan h | )– 1/|tan h |]– a = b = hlim →0 →0

So, the only points of discontinuity are x = 0 and x = 1. e[ h ] + h − 1 f (0 + h) = hlim 109. (a), (d) lim+   f (x) = lim x→0 →0 x→0 [ h] + h



and t = 3 ⇒

1 = 3 ⇒ x = 7/3. x−2

1+ x − 4 1+ x x

(1 /3)(−2 /3) 2 (1 /4) (−3 /4) 2  1   1  x + ... − 1 + x + x + ... 1 + 3 x + 2! 4 2!      = lim x→0 x



 1  −1 3  + + x + terms containing x 2 and x 12  9 32    higher powers lim = x→0  x 1 1 = . ∴ f (0) = . 12 12

1 . e Since lim+ f (x) ≠ lim− f (x), therefore, f (x) is not

=1–

x→0

continuous at x = 0.

x→a

107. (a) For f (x) to be continuous at x = 0, we must have 3

e−1− h − 1 e−1 − 1 = ( − 1 − h) −1

110. (c) Let a be any real number. lim f (x) = lim x = a

Therefore, the values of x which make the function y discontinuous are x = 2, 3 and 7 . 2 3

f (0) = lim x→0

= hlim →0

x→0

1 is discontinuous t2 − t − 6

at the points where t2 – t – 6 = 0 i.e., (t + 2) (t – 3) = 0 ⇒ t = – 2, 3. 1 3 =–2 ⇒x= But t = – 2 ⇒ x−2 2

eh − 1 =1 h

e[ − h ] − h − 1 lim f (0 – h) =0 hlim f (x) = and xlim − →0 →0 h→0 [ − h] − h

sin 3 h / 3 h 3 ⋅ sin 2 h / 2 h 2

−3 ⇒ e– a = b = e3/2 ⇒ a = and b = e3/2. 2 1 106. (b) Clearly, the function t = is discontinuous at x x−2 = 2 and the function y =

= hlim →0

   

x→a



(when x → a through rational values) lim f (x) = lim (1 – x) = 1 – a x→a x→a



(when x → a through irrational values) lim f (x) will exist only when a = 1 – a i.e., when

x→a

a =

1 . 2

1 , then xlim f (x) will not exist and →a 2 hence f (x) will be discontinuous at x = a where a 1 . We have shown that ≠ 2 lim f (x) = 1 and f  1  = 1 1   x→ 2 2 2 2 Thus if x ≠

1 . 2 −π π  111. (a) 3 ≤ 3 + 2 cos x ≤ 5 for x ∈  ,   2 2 Hence f (x) is continuous at x =

f (x) = [3 + 2 cos x] is discontinuous at those points where 3 + 2 cos x is an integer.



 2x  1−  2  1 + x 

(not possible) 1 . 2



3 + 2 cos x = 4 if cos x =



So x has two values π and − π . 3 3 3 + 2 cos x = 5 if cos x = 1. So, x = 0. ∴ The number of values of x = 2 + 1 = 3. 

h 3 φ ( h) = hlim →0 h



 h2 φ (h) = 0 × φ (0) = hlim →0 h→0



∴ f (x) = (– 1)n = ± 1



Hence f (x) is discontinuous for x = n1/3, n ∈ I.

114. (a), (c) The function f (x) is defined when – 1 ≤ sin x < 0 or sin x = 1. ⇒ x ∈ ((2n + 1) π, 

(2n + 2) π) ∪ 2nπ + π  , n ∈ I. 2 



When x ∈ (2nπ + π, 2nπ + 2π), f (x) = – 1 ∴ f (x) is a constant function. Hence f (x) is continuous when x ∈ (2nπ + π, 2nπ + 2π).

π  π f  − h − f   2   2 Now, L f ′ (π/2) = hlim →0 −h 1 −1  π  sin − h   2    = hlim →0 −h 1 −1 [cos h] lim = h→0 = does not exist. −h

Hence, f (x) is not differentiable at x = π/2.

115. (c) Let g (x) = x3 – 3, then g (x) is an increasing function on the interval (1, 2). Since g (1) = – 2 and g (2) = 5, therefore between – 2 and 5 there are 6 points where f (x) is discontinuous (as [x3 – 3] is discontinuous at the points where x3 – 3 is an integer).

d  2x    dx  1 + x 2 

2 (1 − x 2 ) (1 + x 2 ) 2



 −2 1 + x 2 , if | x | < 1 − 2 1 − x2  ⋅ = =  1 + x 2 |1 − x 2 |  2 , if | x | > 1 1 + x 2

(1 + x 2 ) 2 − 4 x 2

x

Clearly, f (x) is differentiable everywhere except at the points where | x | = 1 i.e., x = ± 1. Hence, f (x) is differentiable on  (– ∞, ∞) \ { – 1, 1}.



= 0. 113. (a) Let x3 = n, n ∈ I ⇒ x = n1/3

x

=

[ f (x) = x3 φ (x)] 117. (a) f ′ (x) = lim h→0

[ φ is continuous at x = 0, ∴ lim φ (h) = φ (0)]



− (1 + x 2 )

2



f ( x + h) − f ( x ) 112. (c) f ′ (x) = hlim →0 h f ( x ) + f ( h) − f ( x ) = hlim →0 h [ f (x + y) = f (x) + f ( y)]

−1

116. (c) f ′ (x) =

101

−π π , 2 2

= hlim →0

f ( x + h) − f ( x ) h

f ( x ) ⋅ f ( h) − f ( x ) h



[ f (x + y) = f (x) ⋅ f ( y)] 

= f (x) lim  f (h) − 1   h→0 h  



= f (x) lim 1 + h φ (h) log 2 − 1 h→0 h [ f (x) = 1 + x φ (x) log 2]



= f (x) log 2 hlim φ (h) →0



= f (x) ⋅ log 2 ⋅ 1



= log 2



f (x)

∵ lim φ (h) = 1  h → 0 

.

118. (b) We have, f (x + y2n + 1) = f (x) + { f ( y)}2n + 1 ...(1) ⇒ f (0) = f (0) + { f (0)}2n + 1 [Putting x = y = 0] ⇒ f (0) = 0. Also, from (1) f (1) = f (0) + { f (1)}2n + 1 [Putting x = 0, y = 1] = 0 + { f (1)}2n + 1 [ f (0) = 0] ⇒ f (1) [{ f (1)}2n – 1] = 0 ⇒ f (1) = 0 or 1 or – 1. Now, f ′ (0) = lim x→0

f ( x ) − f ( 0) f ( x) = lim x→0 x x

[ f (0) = 0]

Since f ′ (0) ≥ 0 [Given] f ( x) ≥0 x ⇒ f (x) ≥ 0 for x > 0. ∴ f (1) ≠ – 1. Also, f (1) = 0 ⇒ f (x + 1) = f (x) + { f (1)}2n + 1 [Putting y = 1 in (1)] = f (x)  ⇒ f (2) = f (3) = ... = 0, ⇒ lim

x→0

Continuity and Differentiability

Now, 3 + 2 cos x = 3 if cos x = 0. So, x =

102



which is not true. Thus f (1) = 1.

Objective Mathematics



∴ f (x + 1) = f (x ) + 1



⇒  f (x) is a function whose value is increased by 1 when x is increased by 1, ∴ it is a linear function whose graph is a straight line passing through origin ( f (0) = 0) having slope 1. Hence f (x) = x. ∴ f ′ (6) = 1.

f  ′ (– 1+) = 0. Similarly, f ′ (1–) = 0 and f ′ (1+) = 1, so f is differentiable everywhere except at x = – 1, 1.

119. (c) Let x0 be any arbitrary real number. Case I. x0 is rational Then f (x0) = 1. 122. (c) The function f (x) is continuous at all points where In any vicinity of a rational point there are irrational points, where f (x) = 0. Hence, in any vicinity of x0 there are points x for which 1 – cos2x ≥ 0 ⇒ | cos x | ≤ 1 | ∆ y | = | f (x0) – f (x) | = 1. 2 2 Case II. x0 is irrational ⇒ π + 2nπ ≤ x ≤ 3π + 2nπ Then f (x0) = 0. 4 4 In any vicinity of an irrational point there are rational or 5π + 2nπ ≤ x ≤ 7 π + 2nπ, n ∈ I. points at which f (x) = 1. Hence, it is possible to 4 4 find the values of x for which 1 | ∆ y | = | f (x0) – f (x) | = 1. suffers a discontinuity at 123. (b) The function u = x −1 Thus, in both cases, the difference ∆ y does not tend the point x = 1. to zero as ∆ x → 0. Therefore, x0 is a point of discontinuity. Since x0 is an arbitrary point, the Dirichlet 1 function f (x) is discontinuous at each point. The function f (x) = 2 suffers a discontinuity u +u−2 120. (a) We have, at the points where u2 + u – 2 = 0 i.e., u = – 2 and u = 1. Using these values of u, the corresponding 1 values of x are obtained by solving the equations lim f (5 – h) = lim tan– 1 h→0 h→0 (5 − h) − 5 1 1  −1  –2= and 1 = i.e., x = 1/2 and x = 2. = hlim tan– 1   →0 x −1 x − 1  h 

= tan– 1 (– ∞) = − π 2

Hence, the composite function is discontinuous at three points x = 1/2, x = 1 and x = 2.

and lim f (5 + h) = hlim tan– 1 →0 h→0



1 (5 + h) − 5

π 1 = hlim tan– 1   = tan– 1 (∞) = . →0 2 h f (5 – h) ≠ hlim f (5 + h), therefore, f (x) Since hlim →0 →0

has discontinuity of the first kind at x = 5. 1 − x, 121. (a), (c)  f (x) = 2, 1 + x, 

x ≤ −1 −1 < x ≤ 1 x >1

lim f ( x) = lim (1 − x) −

x → −1−

x → −1

= 2 = lim+ f ( x)

124. (a), (b), (d)  We have  x, if x is an integer  f (x) = n, if x is not an integer i.e.  n < x < n +1  0, if x is an integer and  g (x) =  2  x , otherwise Now, ( gof ) (x) = g[ f (x)] = g (integer)  [ f (x) is an integer ∨ x ∈ R] = 0. ∴  (gof ) (x) is a constant function, so it is continuous and differentiable at all x ∈ R. Also, ( fog) (x) = f [g (x)]

x → −1

a nd 

lim f ( x) = 2, so f is continuous at all x →1

points. f ′ (– 1–) = lim− h→0



= lim− h→0

f (−1 + h) − f (−1) h

1+1− h − 2 = – 1. h

0, if x ∈ I  = n, if n < x < n + 1   or − n + 1 < x < − n , n = 0, 1, 2, ... Thus ( fog) (x) is continuous and differentiable at all x except when x = ± n , n = 1, 2, 3,... .

h→0

f (0 + h) − f (0) f (0 − h) − f (0) = lim h → 0 h −h f (− h) − f (0) = – hlim →0 h ( ) f h − f (0) = – lim h→0 h [ f (– h) = f (h)]

⇒ lim

= lim

h→0



h→0

f ( h) − 1  = f (x) ⋅ lim    h→0 h   = f (x) lim

h→0

= f (x) hlim g (h) lim G (h) = ab f (x). ∴ k = ab →0 h→0

126. (b) L f ′ (1) = lim f (1 − h) − f (1) h→0 −h

129. (c) f (x) will be a continuous function at x = π f   = limπ f (x) x→ 2 2 1 − sin x ⇒ λ = limπ π − 2 x) 2 ( x→

 (1 − h) 2 3 13  − (1 − h) +  − 2  4 2 4  = lim h→0 −h



h→0

= lim

h→0

sin x 1 8 = 8. ∴ λ = 1/8.

|1 + h − 3| − 2 f (1 + h) − f (1) = lim h→0 h h

2−h−2 = ­– 1. h

Since L f ′ (1) = R f ′ (1), therefore f (x) is derivable at x = 1. Hence f (x) is continuous at x = 1. π , we must have 127. (b) For f to be continuous at x = 4 lim f (x) = f  π  π   x→ 4 4 π  tan  − x  4   lim But π f (x) = lim x→ π cot 2 x x→ 4

− cos x = xlim π → 2 ( π − 2 x ) ( − 2) 2

h+4 h 2 + 4h = lim = hlim = – 1. →0 −4 h→0 − 4h

π if 2

0   form  0  0   form  0 

2

(1 + h 2 − 2h) − 6 (1 − h) + 13 − 8 − 4h

R f ′ (1) = lim

1 + h g (h) G (h) − 1 h

= f (x) hlim g (h) G (h) →0

⇒ f ′ (0) = 0.

h→0

f ( x ) ⋅ f ( h) − f ( x ) h [ f (x + y) = f (x) ⋅ f (x)]

f (h) − f (0) = 0 ⇒ 2 f ′ (0) = 0 ⇒ 2 lim h→0 h [ f ′ (0) exists]

= lim

f ( x + h) − f ( x ) h

103

128. (d) f ′ (x) = lim

=

lim

π x→ 2

130. (c) Since, x is not defined for negative values of x, therefore, f is neither continuous nor differentiable at x = 0. 131. (d)

y

3 2 x'

1 -3

-2 -1 O

1

2

3

4

x

4



 1 − tan x  = xlim π tan 2x   → 4  1 + tan x 



=



=



2 (1) 2 1 = = = . (1 + 1) 2 4 2

lim x→

π 4

lim x→

π 4

2 tan x 1 − tan x ⋅ 1 − tan 2 x 1 + tan x 2 tan x (1 + tan x) 2

π 1 ∴ f   = . 4 2

y' Clearly, from the figure, f is continuous and derivable for all x ∈ (R – I).

132. (a) We have, h(x) = min {x, x2}  x, for x ≤ 0  =  x 2 , for 0 < x < 1  x, for x ≥ 1  Clearly from the figure, h(x) is continuous everywhere, but not derivable at 0 and 1.

Continuity and Differentiability

125. (a) Since f ′ (0) exists, ∴ R f ′ (0) = L f ′ (0)

104

y

Objective Mathematics

1   ⇒ k = lim (2 + h) 2 + e 2 − (2 + h )  h→0  



−1



−1

−1

⇒ k = lim  4 + h 2 + 4h + e h  h→0 x



x

O



⇒ k = [4 + 0 + 0 + ε–∞]–1 ⇒ k = 1

4

138. (d) Given that, f(x) = ex sin x, x ∈ [0, π] y

133. (a) Since |x| and sin x are continuous for all x, therefore f(x) = sin |x| is continuous for all x ∈ R. 134. (c) f(x) being a polynomial function, is continuous for all real values of x except at x = 2. At x = 2, LHL = lim− x – 1 = lim 2 – h – 1 = 1 h→0

x→2

RHL = lim+ 2x – 3 = lim 2(2 + h) – 3 = 1 h→0

x→2

f(2) = 2(2) – 3 = 1

and

∴ LHL = RHL = f(2) Thus, f(x) is continuous for all real values of x. 135. (a)

y

At x = 0, f(0) = 0 and at x = π, f (π) = 0 Also, f(x) is continuous and differentiable in the interval. [0, π]. ∴ f(x) satisfies conditions of Rolle’s theorem in [0, π]. Hence, option (d) is the required answer. 139. (c) We have, at x = a 3 3 3 3 LHL = lim x − a = lim ( a − h) − a − x→a

= lim h→0

x−a

h→0

a−h−a

[(a − h) + a + a (a − h)] 1 2

2

= (a2 + a2 + a2) = 3a2 Since, f(x) is continuous at x = a. ∴  LHL = f(a) ⇒  3a2 = b y=−x+1

y = x +1 (0,1)

140. (c) Continuity at x = 0: LHL = lim−

tan x − tan h = lim− =1 h→0 −h x

RHL = lim

tan x tan h = lim =1 h → 0 x h

x →0

x

O

x →0

Clearly, from the figure, f(x) ≥ 1, for every x ∈ R.

Since LHL = RHL = f(0) = 1, therefore f(x) is continuous at x = 0. Differentiability at x = 0:

136. (d) We have,

2  1 e2 x − 1 − 2 x lim  − 2 x  = lim x →0 x → 0 x (e 2 x − 1)  x e −1 2x

= lim

2e − 2  (e 2 x − 1) + 2 x e 2 x

= lim

4e 2 x =1 4e 2 x + 4 xe 2 x

x →0

x →0

(using L’Hospital’s rule)

137. (b) f(x) is continuous from right at x = 2, x→2



1



−1

⇒ lim  x 2 + e 2 − x  = k x → 2+





f (0 − h) − f (0) −h

tan (−h) h 2 2h 4 −1 + + + ... 15 = lim 3 =0 = lim −h h→0 h→0 −h −h h→0

x→ 0

if lim+ f(x) = f(2) = k

h→0

RHD = lim

f(x) is continuous at x = 0, if lim f(x) = f(0) i.e.,, 1 = f(0)

LHD = lim

f (0 + h) − f (0) h

tan h h 2 2h 4 −1 + + ... 15 = lim h = lim 3 =0 h→0 h→0 h h  ince LHD = RHD, therefore f(x) is differentiable at S x=0 Hence, Option (c) is the required answer. 141. (c) At no point, function is continuous.

x →0

h→ 0

and lim f(x) = lim 51/x = lim 5–1/h = 0 − − h→ 0 x →0

∴ f is differentiable at x = 0.

x →0

Also, f(0) = λ[0] = 0 ∴ f(x) is continuous for x = 0, whatever λ may be. f (1 − h) − f (1) 143. (c) Now, f ′ (1 ) = lim h →0 −h –

 1  (1 − h − 1) sin  −0  1− h −1  = lim h →0 −h

h →0

x →0

x →0

x →0

x →0

Since L.H.L ≠ R.H.L ∴ f (x) is discontinuous at x = 0 log e (1 + x 2 tan x) sin x 3 This function is continuous at x = 0, then

145. (a) Given, f (x) =

f (1 + h) − f (1) h

log e (1 + x 2 tan x) = f (0) x→0 sin x 3

lim

 1  (1 + h − 1) sin  −0 1  1+ h −1  = lim = lim sin h →0 h →0 h h Since f ′ (1–) ≠ f´ (1+) ∴ f is not differentiable at x = 1. Again,   1   − sin 1 (0 − h − 1) sin    0 − h −1  , f ′ (0− ) = lim  h →0 −h   1   1    1  − (h + 1) cos   ×  (h + 1) 2   + sin  h + 1  + ! h         = lim  h →0 1 = cos 1 – sin 1 1   (0 + h − 1) sin  − sin 1 0 + h − 1   + and f ′ (0 ) = lim h →0 h   1   −1   1  (h − 1) cos    (h − 1) 2  + sin  h − 1   − 1 h        = lim  h →0 1 

2 x, x < 0 f (x) =   2 x + 1, x ≥ 0 L.H.L = lim f ( x) = lim 2 x = 0 − − R.H.L = lim f ( x) = lim 2 x + 1 = 1 + +

1  1 = lim sin  −  = − lim sin h →0 h → 0 h  h and f′ (1+) = lim

144. (c) We have,

(using L’ Hospital’s rule)

   x3 log e 1 + x 2  x + + ...  3    ⇒ lim = f (0) 9 15 x→0 x x − ... x3 − + 3! 5! log e (1 + x 3 ) ⇒ lim = f (0) x→0 x 9 x15 x3 − + − ... 3! 5! [neglecting higher power of x in x2 tan x] x6 x9 x3 − + − 2 3 ⇒ lim = f (0) ⇒ 1 = f (0) x→0 x 9 x15 3 − ... x − + 3! 5! 146. (a) We have lim t→x

t 2 f ( x) − x 2 f (t ) =1 t−x

⇒ x2f´(x) – 2xf(x) + 1 = 0 ⇒ f(x) = cx2 + 1 also f(1) = 1 3x ⇒c= 2 3 2 1 Hence, f(x) = x 2 + 3 3x

EXERCISEs FOR SELF-PRACTICE 1. If f (x) = x3Sgn, then 

(a) f is continuous but not derivable at x = 0 (b) f ′ (0 +) = 2 (c) f ′ (0 – ) = 1 (d) f is derivable at x = 0.

2. In order that the function f (x) = (x + 1)cot x is continuous at x = 0, f (0) must be defined as

1 e (c) f (0) = e

(a) f (0) =

(b) f (0) = 0 (d) None of these

3. The domain of the derivative of the function

 tan −1 x, | x | ≤ 1  is f (x) =  1  (| x | − 1), | x | > 1 2

Continuity and Differentiability

x →0

105

= cos 1 – sin 1 Since, f ′ (0–) = f′ (0+),

142. (c) lim+ f(x) = lim+ λ[x] = lim λ[h] = 0

106

(a) R – [0] (c) R – {– 1}

Objective Mathematics

4. If f (x) = x

(

(b) R – {1} (d) R – {– 1, 1}

(a) 3

(c) 9 (d) 12 9. The value of the constants a, b and c for which the function

)

x − x + 1 , then

(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 these. 2 x − sin −1 x 5. If the function f (x) = is continuous at each 2 x + tan −1 x point of its domain, then the value of f (0) is

(a) 2

1 (c) –  3

(b) (d)

(b) 6

2 3



 (1 + ax)1/ x ,x0  ( x + 1)1/ 2 − 1

may be continuous at x = 0, are −2 2 (a) a = log , b= , c = 1 3 3

2 2 , b= , c = – 1 3 3 2 2 (c) a = log , b = , c = 1 3 3

(b) a = log

2 3

1  x≠0  x sin , x 6. If f (x) =  , then, at x = 0, the function 0, x=0 f (x) is: (a) Differentiable (b) Not differentiable (c) Continuous but not differentiable (d) None of these

(d) None of these

10.

7. If f (x) = | x – 3 |, then f ′ (3) is: (a) – 1 (b) 1 (c) 0 (d) Does not exist



sin 3 x , x≠0 x 8. The function f (x) = k , x = 0, 2 is continuous at x = 0, then k =



 −1/ x2 1 sin   , x ≠ 0 e x then graph of the function If f (x) =   0 ,x=0  f (x) (a) is broken at the point x = 0 (b) is continuous at the point x = 0 (c) has a definite tangent at the point x = 0 (d) does not have a definite tangent at the point x = 0.

11. The function f (x) = a [x + 1] + b [x – 1], where [x] is the greatest integer function is continuous at x = 1, if: (a) a + b = 0 (b) a = b (c) 2a – b = 0 (d) None of these

Answers

1. (d) 11. (a)

2. (c)

3. (c)

4. (b)

5. (b)

6. (c)

7. (d)

8. (b)

9. (c)

10. (b)

4

Differentiation

CHAPTER

Summary of concepts Derivative of a Function Let y = f (x) be a function defined on the interval [a, b]. Let for a small increment δx in x, the corresponding increment in the value of y be δy. Then y = f (x) and y + δy = f (x + δx) On subtraction, we get δy = f (x + δx) – f (x) or

δy f ( x + δx ) − f ( x ) = δx δx

Taking limit on both sides when δx → 0 we have, δy f ( x + δx ) − f ( x ) = lim . δ x δx → 0 δx if this limit exists, is called the derivative or differential coeffidy or f ' (x). Thus cient of y with respect to x and is written as dx lim

δx → 0

∴

dy δy f ( x + δx ) − f ( x ) = lim = δlim . x → 0 δx δx → 0 dx δx

Derivative at a Point The value of f′ (x) obtained by putting x = a, is called the deriva dy  tive of f (x) at x = a and it is denoted by f ' (a) or   .  dx  x = a

Standard Derivatives The following formulae can be applied directly for finding the derivative of a function: 1.

d (sin x) = cos x dx

2.

d (cos x) = – sin x dx

3.

d (tan x) = sec2x dx

4.

d (cot x) = – cosec2x dx

d 5.   (sec x) = sec x tan x dx

d dx d 7. dx d 8. dx d 9. dx 6.

(cosec x) = – cosec x cot x (ex ) = ex (ax ) = ax logea, a > 1 (logex) =

1 , x>0 x

d (xn) = nxn – 1 dx d 1 11. (sin–1x) = ,–1< x 0 and x2 + y2 ≤ 1 14. cos–1x ± cos–1y = π – cos–1 ( xy  1 − x 2 1 − y 2 ) ,  2x  15. sin–1   = 2 tan–1x 1 + x2  1 − x2  16. cos–1  = 2 tan–1x  1 + x 2  17. tan–1

1 1− x = cos–1x 2 1+ x

SucceSSive Differentiation dy is again a function of x dx and is called the first derivative of y w.r.t. x. If the first derivative is differentiable, its derivative is called second derivative of d2y the original function and is denoted by or y2. If the second dx 2 derivative is differentiable, its derivative is called the third ded3y or y3 and rivative of the original function and is denoted by dx 3 so on. This process of differentiating a function more than once is called successive differentiation. Let y = f (x) be a function of x, then

Differentiation of a function given in the form of a Determinant

π 1 − tan x 6. tan  − x  = 4  1 + tan x –1

–1

 x+ y   1 − xy  ,

provided x, y > 0 and xy > 1 9. tan–1x – tan–1y = tan–1  x − y  ; if x, y > 0  1 + xy  π = tan–1x + cot–1x = sec–1x + cosec–1x 10. sin–1x + cos–1x = 2

f ( x ) g ( x ) h( x ) If y = p ( x) q ( x) r ( x) , then u ( x) v( x) w( x) f' ( x) g' ( x) h' ( x) dy = p( x) q( x) r ( x) dx u ( x) v( x) w( x) f ( x ) g ( x ) h( x ) f ( x ) g ( x ) h( x ) + p' ( x) q' ( x) r' ( x) + p ( x) q ( x) r ( x) u' ( x) v' ( x) w' ( x) u ( x) v( x) w( x) Note that differentiation of a determinant can be done in columns also.

109

provided x, y ≥ 0 and x2 + y2 ≤ 1

if x, y > 0 and x2 + y2 > 1

a+x a−x

Some useful trigonometric and inverse trigonometric transformations

–1

11. sin–1x ± sin–1y = sin–1 ( x 1 − y 2 ± y 1 − x 2 ) ,

Differentiation

Important Substitutions to Reduce the Function to a Simpler Form

110

multiple-choice questions

Objective Mathematics

Choose the correct alternative in each of the following problems:  xa  1. If f (x) =  b  x  to

a +b

 xb  ⋅ c  x 

(a) 1 (c) xa + b + c  sin x  2. If f (x) =  n   sin x  is equal to

m+n

 sin x  ⋅ p   sin x 

(a) 0 (c) cosm + n + p x

then

 xc  ⋅ a  x 

n

n+ p

(a)

dy 1 = dx a + b cos x

(c)

dy 1 = dx a − b cos x

c+a

, then f ′ (x) is equal

(b) 0 (d) None of these m

3. If y =

b+c

 sin x  ⋅ m   sin x  p

b sin x d2y = (a + b cos x) 2 dx 2 −b sin x d2y (d) = (a − b cos x) 2 dx 2 (b)

9. If sin y = a sin (x + y), then

p+m

, then f ′ (x)

(b) 1 (d) None of these

1 1 1 + + 1 + xβ − α + x γ − α 1 + x α − β + x γ − β 1 + x α − γ + xβ − γ

dy = dx

(a)

sin a sin a sin 2 (a + y )

(c) sin a sin2 (a + y) 10. If f (x) =

4. If y = ( x + 1 + x 2 ) n , then (1 + x 2 ) (a) n2y (c) – y

(b) – n2y (d) 2x2y

d2y dy +x is dx 2 dx

5. If φ (x) = log5 log3 x, then φ′ (e) is equal to (a) e log 5 1 (c) elog 5

(b) – e log 5

1+ x 1 + log x

dy is dx

x − 1 − x2

12. If y = tan –1

(d) None of these

(d)

sin 2 (a − y ) sin a

1 − log x 1 + log x log x (d) (1 + log x) 2

(b)

(c) not defined

(a)

sin 2 (a + y ) sin a

(b) 0 (d) Does not exist

11. If x y = e x – y then (a)

(b)

1 − cos x π , then f ′   is equal to 1 − sin x 2

(a) 1 (c) ∞

(a) 0 (b) 1 (c) (α + β + γ ) xα + β + γ – 1 (d) None of these

dy is dx

−1 1− x −x

2

x + 1− x

2

dy is equal to dx

, then



(b)

1 1 − x2



(c) (d) None of these 1 − x2  2x − 1 dy 2 and f ′ (x) = sin x , then is equal 6. If y = f  2 1  x + 1  , then g′ (x) is 13. If g is the inverse of f and f ′ (x) = dx 1 + x3 to equal to 2x − 1   2 + 2x + x2  (a) sin  2 ⋅  x + 1   ( x 2 + 1) 2  2

2x − 1   2 + 2 x − 2 x2  (b) sin  2  x + 1   ( x 2 + 1) 2  2

2x − 1  2 + 2x − x  (c) sin  2 ⋅  x + 1   ( x 2 + 1) 2  2

7. If f (x) = log x (log x), then f ′(x) at x = e is (a) e (c)

(b)

2 e

8. If y =

1 e

(d) 0  a−b x tan −1  tan  , a > b > 0, then 2  a+b a −b 2

2

2

(b)

(c) [g (x)]3 14. If y = cos –1

2

(d) None of these

1 1 + [ g ( x)]3 (d) None of these

(a) 1 + [g (x)]3

1 1 + x2 1 (c) 2 (1 + x 2 ) (a)

1 + x2 + 1 2 1 + x2

, then (b)

dy = dx

−1 2 (1 + x 2 )

(d) None of these

15. Let f (x) be a polynomial function of second degree. If f (1) = f (– 1) and a, b, c are in AP, then f ′ (a), f ′ (b) and f ′ (c) are in (a) AP (b) GP (c) HP (d) Arithmetico-Geometric progression

1 1 + tan −1 2 x + x +1 x + 3x + 3 1 dy −1 + tan 2 + ... to n terms, then = x + 5x + 7 dx

17. If y = tan –1

2

1 1 1 1 + − (b) 2 2 2 1 + ( x + n) 1 + x 1 + ( x + n) 1 + x 2 2 1 − (c) (d) None of these 1 + ( x + n) 2 1 + x 2

(a)

18. If y = etan x , then cos2 x dy (a) (1 – sin2x) dx dy (c) (1 + sin2x) dx 19. If y = x (log x ) (a)

log log x

, then

d y = dx 2 2

dy (b) – (1 + sin2x) dx (d) None of these dy is equal to dx

y log y (2 log log x + 1) x log x

x2

x

∑ tan r =1

(a)

−1

(c) 0 23. If y = eax sin bx, then (a) – (a2 + b2) y (c) – y

26. The derivative of tan –1

1 + x2 − 1 w.r.t. x

cos–1

1 + 1 + x2 2 1 + x2

is

(a) 1 1 (c) 2

(b) – 1 (d) None of these

2u 1 1− u 1 2u  + sin −1 27. If y = tan  cos −1 and x = , 2 2  1 − u2 2 1 + 2 1 + u u   dy = then dx (b) 0 (d) None of these π , where 4

1 2 (c) 1 (a)

(b)

2

(d) None of these

d2y is equal to dx 2 dy (b) x –2 dx (d) None of these

30. If x = t t and y = t t , then

(b) y (d) None of these

dy 1 then is equal to dx 1+ r + r2

1 1 + x2

(b) P (x) ⋅ P''' (x) (d) None of these.

29. If x = a (cos θ + θ sin θ) and y = a (sin θ – θ cos θ), π d2y π where 0 < θ < , then at x = is equal to 2 2 dx 4 4 2 8 2 (a) (b) aπ aπ 4 (c) (d) None of these aπ 2

21. If y = (sin –1 x)2, then (1 – x2)

22. If y =

(a) – P (x) ⋅ P''' (x) (c) P (x) ⋅ P′′ (x)

f ′ (1) = 2 and g′ ( 2 ) = 4, is

20. If y = a cos (log x) + b sin (log x), then

dy + 2 dx dy (c) – x + 2 dx

2 d  3d y y 2  = dx  dx 

28. The derivative of f (tan x) w.r.t. g (sec x) at x =

(d) None of these

(a) x

2

(a) – 1 (c) 1

2 y log y (log log x + 1) x log x

d2y dy +x = dx 2 dx (a) 0 (c) – y

25. If y 2 = P (x), a polynomial of degree n ≥ 3, then

2

x log x (b) (2 log log x + 1) y log y (c)

(b)

1 1 + (1 + x) 2

(d) None of these d2y dy − 2a is equal to dx 2 dx (b) (a2 + b2) y (d) None of these

t

t  1 t t ⋅ (1 + log t )log t +  t  (a) (1 + log t ) t  1 t t 1 + log t +  t  (b) (1 + log t ) t  1 t t (1 + log t )log t +  t  (c) 2 (1 + log t )

(d) None of these

dy is equal to dx

111

2x 1 − x2 24. The differential coefficient of tan –1 w.r.t. 1 − 2x2 1 1 sec–1 at x = is equal to 2 2x2 − 1 1 1 (a) (b) – 2 2 (c) – 1 (d) None of these.

Differentiation

16. If S n denotes the sum of n terms of a G.P. whose comdSn mon ratio is r, then (r – 1) is equal to dr (a) (n – 1) Sn + n Sn – 1 (b) (n – 1) Sn – n Sn – 1 (c) (n – 1) Sn (d) None of these

112

Objective Mathematics

dy 31. Let y = x3 – 8x + 7 and x = f (t). If = 2 and x = 3 dt dx at t = 0, then at t = 0 is given by dt 19 (a) 1 (b) 2 2 (c) (d) None of these 19 32. If ax2 + 2hxy + by 2 = 1, then

d2y is equal to dx 2

ab − h 2 (a) (hx + by ) 2

h 2 − ab (b) (hx + by ) 2

h 2 + ab (c) (hx + by ) 2

(d) None of these

33. If y = sin x, then

d2 (cos7x) is equal to dy 2

sin x x x x ⋅ cos 2 cos 3 ... to ∞ = , then x 2 2 2 x 1 x x 1 1 1 tan + 2 tan 2 + 3 tan 3 + ... to ∞ = – cot x +   2 2 2 2 2 2 x x 1 x x 1 1 1 tan + 2 tan 2 + 3 tan 3 + ... to ∞ = cot x – 2 2 2 2 2 2 x 1 1 1 2 x 2 x 2 sec + 4 sec 2 + ... + ∞ = cosec x – 2 x 22 2 2 2 1 x x 1 1 sec 2 + 4 sec 2 2 + ... + ∞ = – cosec2x + 2 2 x 2 2 2 2

34. If cos

(b) (c) (d)

 3π  35. If f (x) = | cos x |, then f ′   is equal to  4  1 1 (a) – (b) 2 2 (c) 1 (d) None of these. a+ x 36. If f (x) =   b+x

(a) m2y (c) – m2y

(b) my (d) None of these

39. Let f be a twice differentiable function such that f ′′(x) = – f (x) and f ′ (x) = g (x). If h (x) = [ f (x)]2 + [g (x)]2 and h (5) = 11 then h (10) = (a) 11 (c) – 1

(b) 0 (d) None of these

40. A function f (x) is so defined that for all x, [ f (x)]n = f (nx). If f ′ (x) denotes derivative of f (x) w.r.t. x, then f ′ (x) ⋅ f (nx) = (a) f (x)

(b) 0

(c) f (x) ⋅ f ′ (nx)

(d) None of these

41. If y = x – x2, then the derivative of y 2 w.r.t. x2 is (a) 2x2 + 3x – 1 (c) 2x2 + 3x + 1

(b) 2x2 – 3x + 1 (d) None of these

ax + b , where a, b, c are constants then x2 + c (2xy′ + y) y′′′ is equal to

42. If y =

(a) 35 cos3x – 42 cos5x (b) 35 cos3x + 42 cos5x (c) 42 cos3x – 35 cos5x (d) None of these.

(a)

38. If y = sin (m sin –1 x), then (1 – x2) y′′– xy′ is equal to

a +b+ 2 x

, then f ′ (0) is equal to

 a a 2 − b2   a  (a)  2 log +   b ab   b  

a +b

 a b2 − a 2   a  (b)  2 log +   b ab   b  

a +b

 a a 2 + b2   a  (c)  2 log +   b ab   b   (d) None of these

a +b

(a) 3 (xy′′+ y′) y′′

(b) 3 (xy′ + y′′ ) y′′

(c) 3 (xy′′+ y′) y’

(d) None of these

43. If f (x) = (ax + b) sin x + (cx + d) cos x, then the values of a, b, c and d such that f ′ (x) = x cos x for all x are (a) b = c = 0, a = d = 1 (c) c = d = 0, a = b = 1

(b) b = d = 0, a = c = 1 (d) None of these d 2P is equal to d θ2 (b) 2a2 + 2b2 – 3P (d) None of these

44. If P = a2 cos2θ + b 2 sin 2θ, then P + (a) 2a2 + 2b2 + 3P (c) 2a2 – 2b2 – 3P 45. If y = then (a)

ax 2 bx c + + +1 , ( x − a )( x − b)( x − c) ( x − b)( x − c) x − c

y' = y 1 a b c  + +   x a− x b− x c− x

a b c  (b)  + +  a − x b − x c − x  (c)

1 a b c  + +   x x−a x−b x−c

(d) None of these

π 37. If f (x) = | cos x – sin x |, then f ′    is equal to 2 (a) 1 (b) – 1 (c) 0 (d) None of these

46. A triangle has two of its vertices at P (a, 0), Q (0, b) and the third vertex R (x, y) is moving along the straight line dA y = x. If A be the area of the triangle, then = dx a−b a−b (a) (b) 2 4 a+b a+b (c) (d) 2 4

(a)

f1φ2 − φ1 f 2 f12

f φ −φ f (b) 1 2 3 1 2 f1

(c)

φ1 f 2 − f1φ2 f13

(d) None of these,

where the suffixes denote diff. w.r.t. t.  5 x + 12 1 − x 2 48. If y = sin –1   13  1

(a)

1 − x2 3

(c)

1 − x2

 dy  , then is equal to  dx  1



(b) −



(d) None of these

54. If f, g, h are differentiable functions of x and f ( xf )' ( x 2 f )"

f' (a) f ( x 3 f" )'

g' g ( x 3 g" )'

h' h ( x 3h" )'

(b)

f f' ( x 2 f" )'

g g' ( x 2 g" )'

h h' ( x 2 h" )'

(c)

f f' ( x 3 f" )'

g g' ( x 3 g" )'

h h' ( x 3h" )'

1 − x2

(d) None of these

 dx   dy  then   +   is equal to  dt   dt  2

(a) f (t) – f ′′(t)

(b) [ f (t) – f ′′(t)]2

(c) [ f (t) + f ′′(t)]2

(d) None of these

50. If x = sec θ – cos θ, y = sec nθ – cos nθ, then dy (x2 + 4)   is equal to  dx  2

(b) n (4 – y ) (d) None of these

(a) n (y – 4) (c) n2 (y2 + 4) 2

2

2

f1 ( x) f 2 ( x) f 3 ( x) F (x) = g1 ( x) g 2 ( x) g 3 ( x) , then F′ (a) is equal to h1 ( x) h2 ( x) h3 ( x) (b) – a (d) None of these

52. If y = k sin px, then the value of the determinant y y3 y6

y1 y4 y7

55. If 2x3 – 3x2y 2 + 4x – y + 7 = 0 and y (1) = 1 then the value of y′′(1) is equal to (a) – (c) 56. If

343 474

(b) –

474 343

(d)

474 343

343 474

1 − x 6 + 1 − y 6 = a3 (x3 – y 3), then

2

51. If f r (x), g r (x), hr (x), r = 1, 2, 3 are polynomials in x such that f r (a) = g r (a) = hr (a), r = 1, 2, 3 and

(a) a (c) 0

h ( xh)' ( x 2 h)"

then ∆′ (the derivative of ∆ w.r.t. x) is given by

49. If x = f (t) cos t – f ′ (t) sin t, y = f (t) sin t + f ′ (t) cos t 2

g ( xg )' ( x 2 g )"

∆=

y2 y5 is equal to y8

dy is equal to dx

(a)

x2 1 − y6 y 2 1 − x6

(b)

y2 1 − y6 x2 1 − x6

(c)

x2 1 − x6 y2 1 − y6

(d) None of these

f ( x1 ) + f ( x2 ) + ... + f ( xn )  x + x + ... + xn  57. Let f  1 2 ,  = n n   where all xi ∈ R are independent to each other and n ∈ N. If f (x) is differentiable and f ′ (0) = a, f (0) = b, then f ′ (x) is equal to (a) a (c) b

(b) 0 (d) None of these

58. If f (x) = (1 – x) n , then the value of

(a) 1

(b) 0

(c) – 1

(d) None of these,

where yn denotes nth derivative of y w.r.t. x.

f (0) + f ′ (0) +

f"(0) f n (0) + ... + is equal to 2! n!

(a) 2n (c) 2n – 1

(b) 0 (d) None of these

53. If f (x), g (x), h (x) are polynomials in x of degree 2 and 59. If y = f (x3), z = g (x5), f ′ (x) = tan x and g′ (x) = sec x, dy f g h then the value of is dz F (x) = f' g' h' , then F′ (x) is equal to f" (a) 1 (c) – 1

g"

h" (b) 0 (d) None of these

(a)

3 tan x 3 ⋅ 5 x 2 sec x 5

(b)

(c)

3 x 2 tan x 3 ⋅ 5 sec x 5

(d) None of these

5 x 2 sec x 5 ⋅ 3 tan x 3

113

d2y is equal to dx 2

Differentiation

47. If x = f (t), y = φ (t), then

114

60. If f ′ (x) = φ (x) and φ′ (x) = f (x) for all x. Also, f (3) = 5 and f ′ (3) = 4. Then the value of [ f (10)]2 – [ φ (10)]2 is

Objective Mathematics

(a) 0 (b) 9 (c) 41 (d) None of these

(a) 0 (c) – 1

1 dy 61. If 8 f (x) + 6 f   = x + 5 and y = x2 f (x), then at dx  x x = – 1 is equal to (a) 0 (c) –

1 14

1 (b) 14 (d) None of these

62. If cos y = x cos (a + y) and the value of k is (a) sin a (c) 1

k dy = then 2 1 + x − 2 x cos a dx

(b) cos a (d) – sin a

63. If e x + e y = e x + y, then the value of (a) 0 (c) 1

dy at (1, 1) is dx

(b) – 1 (d) None of these.

(b) 45 (d) None of these

(b) 1 (d) None of these xn

n! nπ 70. If f (x) = cos x cos 2 nπ sin x sin 2 n d [f (x)]x = 0 is dx n (a) 0 (c) – 1 71. If

64. Let f be a function defined for all x ∈ R. If f is differentiable and f (x3) = x5 for all x ∈ R (x ≠ 0), then the value of f ′ (27) is (a) 15 (c) 0

 m ( x)  69. If f (x) = log    , m (1) = n (1) = 1 and m′ (1) = n′ (1)  n ( x)  = 2, then f ′ (1) is equal to

2 4

then the value of

8

(b) 1 (d) None of these

d  1 + x 4 + x8  3   = ax + bx then dx  1 + x 2 + x 4 

(a) a = 4, b = 2 (c) a = – 2, b = 4

(b) a = 4, b = – 2 (d) None of these

72. Let f (x) = α (x) β (x) γ (x) for all real x, where α (x), β (x) and γ (x) are differentiable functions of x. If f ′ (2) = 18 f (2),  α′ (2) = 3α (2), β' (2) = – 4β (2) and γ′ (2) = k γ (2), then the value of k is (a) 14 (c) 19

(b) 16 (d) None of these

π π  2 2  1 dy 1 and x 4 + y4 = t2 + 2 , then is 73. If f (x) = cos x + cos  x + 3  + sin x sin  x + 3  and   t dx t 5 equal to g    = 3 then (gof ) (x) is equal to 4 y y (a) 1 (b) 2 (a) (b) – x x (c) 3 (d) None of these 3 30 n x x 74. Let f (x) = (x + 2) . If f  (x) is a polynomial of degree (c) (d) – y y 20, where f n (x) denotes the nth derivative of f (x) w.r.t. x, then the value of n is 66. If f (x) = x 2 − 10 x + 25 , then the derivative of f (x) on (a) 60 (b) 40 the interval [0, 7] is (c) 70 (d) None of these.

65. If x2 + y 2 = t +

(a) 1 (c) 0

(b) – 1 (d) Does not exist

67. Let f (x) = 22x – 1 and φ (x) = – 2 x + 2x log 2. If f ′ (x) > φ′ (x), then (a) 0 < x < 1 (c) x > 0

(b) 0 ≤ x < 1 (d) x ≥ 0

1 68. Let f (x) be a polynomial function satisfying f (x) ⋅ f     x 1 = f (x) + f    . If f (4) = 65 and l1, l2, l3 are in G.P.  x then f ′ (l1), f ′ (l2), f ′ (l3) are in (a) A.P. (b) G.P. (c) H.P. (d) None of these

75. Let f (x) = 3x2 + 4x g′ (1) + g′′ (2) and g (x) = 2x2 + 3x f ′ (2) + f ′′(3) then (a) f ′ (1) = 22 + 12 f ′ (2) (b) g′ (2) = 44 + 12 g′ (1) (c) f ′′(3) + g′ ′ (2) = 10 (d) None of these 76. If y = xn – 1 log x, then x2y2 + (3 – 2n) xy1 is equal to (a) – (n – 1)2 y (b) (n – 1)2y (c) – n2y (d) n2y 77. If 2x = y1/5 + y–1/5, then (x2 – 1) y2 + xy1 = ky where k is equal to (a) – 25 (b) 25 (c) 16 (d) – 16 dy π at x = is equal to dx 4 (b) 2 (d) None of these

78. If y = [(tan x)tan x]tan x , then (a) 1 (c) 0

(a) – 1 (c) 0

(b) 1 (d) None of these

π 80. If f (x) = cos x cos 2x cos 4x cos 8x, then f ′    is 4 (a) – 1 (b) 2 (d) None of these. (c) 2 π  81. If f (x) = sin  [ x] − x 5  , 1 < x < 2 and [x] denotes the 2   π greatest integer less than or equal to x, then f ′  5   2 is equal to π (a) 5   2 (c) 0

4/5

π (b) – 5   2 (d) None of these 4/5



1 x 1 (c) |x|

(a)

(b) –

1 x

(d) None of these

89. If the parametric equation of a curve is given by θ x = cos θ + log tan and y = sin θ, 2 d2y = 0 are given by then the points for which dx 2 π ,n∈I 2 (c) θ = (2n + 1) π, n ∈ I (d) θ = 2nπ, n ∈ I.

(a) θ = nπ, n ∈ I

(b) θ = (2n + 1)

90. If f (x) = | x – 3 | and φ (x) = ( fof ) (x), then for x > 10,  φ′ (x) is equal to (a) 1 (c) – 1

(b) 0 (d) None of these

82. If f (x) + f (y) + f (z) + f (x) ⋅ f (y) ⋅ f (z) = 14 for all x, y, z dy  2x  ∈ R, then 91. If y = sin–1  is equal to  , then 1 + x2  dx  (a) f (0) = 2 2 (b) f ′ (x) = 0, for all x ∈ R (a) , when – 1 < x < 1 (c) f ′ (x) > 0, for all x ∈ R 1 + x2 (d) None of these 2 , when x < – 1 or x > 1 (b) + x2 1 83. Let f (x) be a polynomial of degree 3 such that f (3) = 1, 2 f ′ (3) = – 1, f ′′(3) = 0 and f ′′′ (3) = 12. Then the value (c) ­–  , when – 1 < x < 1 of f ′ (1) is 1 + x2 (a) 12 (c) – 13

(b) 23 (d) None of these

sin x cos x sin x dy 84. If y = cos x − sin x cos x , then is equal to dx x 1 1 (a) 1 (c) 0

(b) – 1 (d) None of these

85. If f (x) = xn; then the value of f' (1) f" (1) f"' (1) (−1) n f n (1) + − + ... + is 1! 2! 3! n! (a) 2n (b) 0 (c) 2n – 1 (d) None of these f (1) –

86. If f (x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ... (cos nx + i sin nx) and f (1) = 1 then f ′′ (1) is equal to (a)

n (n + 1) 2

 n (n + 1)  (b)   2   2

 n (n + 1)  (c) –   2   2

(d) None of these

87. Let f (x) = sin x, g (x) = 2x and h (x) = cos x. If φ (x) = π [go ( f h)] (x), then φ′′   is equal to 4 (a) 4 (b) 0 (c) – 4 (d) None of these

(d) None of these 92. If f (x) =

x 2 + 6 x + 9 , then f ′ (x) is equal to (a) 1 for x < – 3 (b) – 1 for x < – 3 (c) 1 for all x ∈ R (d) None of these

93. If f (x) = | (x – 4) (x – 5) |, then f ′ (x) is equal to (a) – 2x + 9, for all x ∈ R (b) 2x – 9 if 4 < x < 5 (c) – 2x + 9 if 4 < x < 5 (d) None of these 94. If f (x) = | x – 1 | and g (x) = f [ f { f (x)}], then for x > 2, g′ (x) is equal to (a) – 1 if 2 ≤ x < 3 (c) 1 for all x > 2

(b) 1 if 2 ≤ x < 3 (d) None of these

95. If x2 + y2 = a2, then a is equal to (a)

(1 + y' 2 )3 | y" |

(b)

(c)

(1 + y' 2 )3 2 | y" |

(d) None of these

96. If y = tan

–1

( )  + tan

 log e  x3  log (ex3 )  

  

| y" | (1 + y' 2 )3

−1

  log (e 4 x3 )   log e 12 x 

( )

  d2y  , then dx 2  

is equal to (a) 1 (c) – 1

(b) 0 (d) None of these

115

88. If f (x) = log | 2x |, x ≠ 0, then f ′ (x) is equal to

Differentiation

dy 79. If y = (1 + x) (1 + x2) (1 + x 4) ... (1 + x2n), then at dx x = 0 is

116

97. If f (x) = xm, m being a non-negative integer, then the value of m for which f ′ (α + β) = f ′ (α) + f ′ (β), for all α, β > 0, is

Objective Mathematics

(a) 1 (c) 0

(b) 2 (d) None of these

98. If xexy = y + sin2x, then at x = 0, (a) – 1 (c) 0

dy is equal to dx

(b) 1 (d) None of these

99. If x2 + y2 = 1, then (a) yy′′– 2 (y′)2 + 1 = 0 (c) yy′′+ (y′)2 – 1 = 0 100. Let f (x) =

(b) yy′′+ (y′)2 + 1 = 0 (d) yy′′+ 2 (y′)2 + 1 = 0

x 3 sin x cos x 6 −1 0 , where p is a constant. p p2 p3

d3 [ f (x)] at x = 0 is dx 3 (a) p (b) p + p2 3 (d) independent of p (c) p + p Then

101. Let F (x) = f (x) g (x) h (x) for all real x, where f (x), g (x) and h (x) are differentiable functions. At some point x0, if F′ (x0) = 21 F (x0), f ′ (x0) = 4 f (x0), g′ (x0) = – 7 g (x0) and h′ (x0) = kh (x0) then k is equal to (a) 24 (c) – 12

(b) 12 (d) – 24

  dy  (a > 0); then 1 +     dx   is given by a 3 2 (c) 3a

2

  

2

d2y dx 2

at t =

cos 2t π 6

(b) a 2

(a)

(d)

2a 3

103. If f (x) is a polynomial of degree n (> 2) and f (x) = f (k – x), (where k is a fixed real number), then degree of f ′ (x) is (a) n (c) n – 2 104. If x + y y′′(0) = 2

(b) n – 1 (d) None of these 2

2 −π (a) e 2 a 2 −π (c) – e 2 a

= ae

( x ) , a > 0, then, assuming y > 0,

tan −1 y

2 π (b) – e 2 a (d) None of these

x sin x cos x f' ( x) 105. If f (x) = x 2 tan x − x 3  , then lim    is equal to x→0 x 2 x sin 2 x 5 x (a) – 4 (c) 0

(a) A (x) (c) C (x)

(b) B (x) (d) f (x)

107. The function y defined by the equation xy – log y = 1 satisfies x ( yy′′+ y′ 2) – y′′+ kyy′ = 0. The value k is (a) – 3 (c) 1

(b) 3 (d) None of these

108. If the function y (x) represented by x = sin t, y = aet

2

+ bet

2

 π π , t ∈  − ,  satisfies the equation  2 2

(1 – x2) y′′– xy′ = ky, then k is equal to (a) 1 (c) 2

(b) – 2 (d) None of these

109. The derivative of sec–1

1

w.r.t.

2x − 1 2

1 − x 2 at x =

1 2

is (a) – 4 (c) 2

(b) 4 (d) – 2

110. If f (x) = logx (lnx) then f ′ (x) at x = e is

102. Given : x = a cos t cos 2t and y = a sin t 3

106. If α is a repeated root of a quadratic equation f (x) = 0 and A (x), B (x), C (x) be polynomials of degree > 2, then the determinant A( x) B( x) C( x) A(α) B(α) C(α) is divisible by A' (α) B' (α) C' (α)

(b) 4 (d) None of these

1 e 1 (c) –  e

(a)

(b) e

111. If f (x) =

x −x df −1 ( x) , then is equal to 2 x + 2x dx

(d) 0 2

3 (1 − x) 2 1 (c) (1 − x) 2 (a) – 

112. If y = x + ex, then (a)

1 (1 + e x ) 2

(c) –

ex (1 + e x )3

113. If x = e y + e 1− x x x (c) 1+ x

(a)

y + ... to ∝

(b)

3 (1 − x) 2

(d) None of these d 2x is dy 2 (b) –

ex (1 + e x ) 2

(d) ex. , x > 0 then

dy is dx

(b)

1 x

(d)

1+ x x

114. If y = f (x) is an odd differentiable function defined on (– ∞, ∞) such that f ′ (3) = – 2, then f ′ (– 3) equals (a) 4 (b) 2 (c) – 2 (d) 0

(a) tan x

(b) cot x

(c) f (cos x)

(d)

116. If y =

sin x + y , then

1 x

dy equals dx

(a)

cos x 2y −1

(b)

cos x 1− 2y

(c)

sin x 1− 2y

(d)

sin x 2y −1

 1 + cos d  −1  tan 125. dx  1 − cos  

x 2 x 2

   is equal to   

(a)  –1/4 (c)  –1/2

 1 − x2  dy 117. If y = log  , then = 2  dx 1+ x  4 x3 (a) 1 − x4 1 (c) 4 − x4

(b)  1/4 (d)  1/2

126. If f ′′ (x) = – f(x), where f(x) is a continuous double differentiable function and g(x) = f ‘ (x). If F(x) =

4x (b) – 1 − x4 4 x3 (d) – 1 − x4

2

2

  x    x   f    +  g    and F(5) = 5, then F(10) is   2    2 

(a)  0 (b)  5 (c)  10 (d)  25 118. Let f and g be differentiable functions satisfying g′ (a) = 2, g (a) = b and fog = I (identity function). x −1 −1 x + 1 + sin −1 , then dy is 127. If y = sec Then f ′ (b) is equal to x −1 x +1 dx 2 (a)  1 (b)  0 (a) 2 (b) 3 1 (c)   x − 1 (d)   x + 1 (c) (d) None x +1 x −1 2 dx 128. Let φ(x) be the inverse of the function f(x) and 119. If x + y = 4, then at y = 1, is dy 1 d φ(x) is f ' ( x) = , then (a) – 1 (b) – 3 1 + x5 dx (c) 3 (d) None of these 1 1 (b)  (a)   2 5 d y dy 1 + [φ( x)] 1 + [ f ( x)]5 120. If y = log ( x + 1 + x 2 ), then is equal to 2 dx dx (c)  1 + [φ(x)]5 (d)  1 + f(x) 2x ( x 2 + 1)3/ 2 x (c)   2 ( x + 1)3/ 2

(a)   −

129. If f(x) is differentiable function and f ′′ (0) = a, then 2 f ( x) − 3 f (2 x) + f (4 x) is equal to lim x→0 x2 (a)  3a (b)  2a (c)  5a (d)  4a

1

(b)  

x2 + 1 x (d)   − 2 ( x + 1)3/ 2

121. Derivative of log | x | w.r.t. | x | is (a)   1 x (c)   ± 1 x

(b)  

 dy  1+    dx 

2

is equal

(b)  sec2 θ (d)  | sec θ |

π 123. The derivative of the function cot −1 ( cos 2 x ) at x = 6 is 2 1 (b)   (a)   3 3

(c)   3

(d)   6

sec x is

1

sec x sin x 4 x 1 (b)   (sec x )3/ 2 ⋅ sin x 4 x (c)   1 x sec x sin x 2 (d)   1 x (sec x )3/ 2 ⋅ sin x 2 (a)  

(d)  none of these

122. If x = a cos3 θ, y = a sin 3 θ, then to (a)  tan2 θ (c)  sec θ

130. Differential coefficient of

1 |x|

131. If x = y 1 − y 2 , then dy is equal to dx 1 − y2 (a)  x (b)   1 + 2 y2 2 1 − y (c)   (d)  0 1 − 2 y2

117

d  −1 1 + x 2 − 1   is equal to 124. dx  tan  x   x2 (a)   1 (b)   2 1 + x 2 ( 1 + x 2 − 1) 1 + x2 1 (c)   2 (d)   2(1 + x 2 ) 1 + x2

Differentiation

115. The differential coefficient of f (sin x) with respect to x where f (x) = log x is

118

132.

Objective Mathematics

d 2x is equal to dy 2 1 (a)   2 dy    dx   

2 (b)    d y   dy    2  dx   dx 

(c)   1 (–1 + loge 2) 3e (d)   − 1 (1 + loge 2) 3e

2

 d y   dy  (d)    − 2     dx   dx  2

2 (c)   d y dx 2

2

133. If f(x) = mx2 + nx + p, then f ' (1) + f ' (4) – f ' (5) is equal to (a)  m (c)  n 134. If y =

(b)  –m (d)  –n dy 1− x + y is equal to , then (1 – x2) dx 1+ x

(a)  1 (c)  2

(b)  – 1 (d)  0

2 135. y = eax sin bx, d y − 2a dy + a 2 y + is dx 2 dx

(a)  –a2y. (c)  –ay

d 2x equals dy 2

137.

−1

−1

 d2y  (a)    2   dx 

 d2y  dy (b)   −  2     dx   dx  −2

 d 2 y  dy (c)    2     dx   dx 

 d 2 y  dy (d)   −  2     dx   dx 

−3

−3

138. 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). Then, for N = 1, 2, 3,…,  1 1  1 + ... + (a)   −4 1 + +  (2 N − 1) 2   9 25

….

 1 1  1 (b)   4 1 + + + ... + 2 9 25 (2 N − 1)  

(b)  –b2y (d)  –by

136. If f(x) = log x3 (loge x2), then f´ (x) at x = e is (a)   1 (1 – loge 2) 3e (b)   1 (1 + loge 2) 3e

 1 1  1 (c)   −4 1 + + + ... + 2 9 25 (2 N + 1)    1 1  1 (d)   4 1 + + + ... + 2 9 25 (2 N + 1)  

SOLUTIONS 1. (b) We have



a = x

2 − b2



a = x

2 − b2 + b2 − c2 + c2 − a 2



4. (a) We have, y = ( x + 1 + x 2 ) n 

f (x) = (xa – b)a + b ⋅ (xb – c)b + c ⋅ (xc – a)c + a ⋅ xb

2 − c2

⋅ xc



2 − a2



= x0 = 1.

∴ f ′ (x) = 0.

2. (a) We have, f (x) = (sinm – nx)m + n ⋅ (sinn – px)n + p ⋅ (sinp – mx)p + m 2 − n2



m = sin



m = (sin x)



x ⋅ sin n

2 − p2

x ⋅ sin p

2 − n2 + n2 − p 2 + p 2 − m2

2 − m2

x



= (sin x)0 = 1

∴ f ′ (x) = 0.

3. (a) We have,

1

y=



γ

+

1 α

γ

+

1 α

x α + xβ + x γ = α = 1 ∴ x + xβ + x γ

dy = 0. dx

= or

n[ x + 1 + x2 ] 1 + x2 dy = dx

ny 1 + x2

⇒ y12 (1 + x2) = n2 y2.



Again differentiating, we get



2y1 y2 (1 + x2) + 2xy12 = 2n2yy1



Dividing by 2y1, we get

y2 (1 + x2) + xy1 = n2y d2y dy or (1 + x 2 ) + x = n2y. dx 2 dx 5. (c) We have,

β

x x x x x x + 1+ β + β 1+ γ + γ xα xα x x x x xα xβ xγ + + = α x + xβ + x γ x α + xβ + x γ x α + xβ + x γ 1+



β



Differentiating Eq. (1), we get  dy x  = n [ x + 1 + x 2 ]n − 1 1 +  2 dx  x + 1  



 log x  φ (x) = log5 log3 x = log5    log 3 



= log5 (log x) – log5 (log 3)

...(1)



φ′ (x) =



∴ φ′ (e) =

1 1 1 ⋅ ⋅ –0 log 5 log x x

=

1 1 1 ⋅ ⋅ log 5 log e e



=

1 . a + b cos x



∴

−1 b sin x d2y (−b sin x) = . 2 2 = (a + b cos x) (a + b cos x) 2 dx

9. (b) sin y = x sin (a + y) sin y x = sin (a + y )

1 . elog 5



 2x − 1  6. (b) We have, y = f  2   x +1



dy  2 x − 1   ( x + 1) 2 − (2 x − 1) ⋅ 2 x  = f ′  2  ⋅ dx ( x 2 + 1) 2  x +1  

dx sin (a + y ) cos y − sin y cos (a + y ) = dy sin 2 (a + y ) sin (a + y − y ) = sin 2 (a + y ) ⇒

2



⇒



 2x − 1  = sin  2   x +1



2

 2 + 2x − 2x2  ⋅  2 2  ( x + 1) 

2   2x − 1   2x − 1   2 sin =  f' ( x) = sin x , ∴ f'  2   2    x +1  x + 1   

7. (b) Given f (x) = logx (log x)

log e (log e x) = loge (loge x) × logx e = log e x log e x ×



∴ f ′ (x) =



1 [1 − log e (log e x)] = x [log e x]2



1 1 1 × − log e (log e x) × log e x x x (log e x) 2

[   loge e = 1 and loge 1 = 0]

dy = dx





a−b x 1 ⋅ × sec 2 ⋅ 2 2 2 2 2 a + b a −b  a−b x 1+  tan  2  a+b 2

1 ⋅ a−b a+b

=

=

1

1 x a − b sin 2 1+ ⋅ a + b cos 2 x 2 2

⋅

a−b x sec 2 2 a+b

x ( a + b)cos 2 x 1 2 ⋅ ⋅ sec 2 a + b ( a + b)cos 2 x + ( a − b)sin 2 x 2 2 2 1 x x x x    a  cos 2 + sin 2  + b  cos 2 − sin 2     2 2 2 2

sin 2 (a + y ) dy = . sin a dx

∴

10. (d) f ′ (x) =

=

(1 − sin x)(sin x) − (1 − cos x)(− cos x) (1 − sin x) 2

sin x − sin 2 x + cos x − cos 2 x sin x + cos x − 1 = (1 − sin x) 2 (1 − sin x) 2



1+ 0 −1 π 0 ∴ f ′   = = (1 − 1) 2 0 2



π Hence, f ′   does not exist. 2

11. (d) xy = ex– y

Taking log on both sides, we get



y log x = x – y x ⇒ y = 1 + log x



8. (a), (b)

=



1 [1 − log e (log e e)] 1 ∴ f ′ (x)at x = e = e = [log e]2 e



119

=

dy ∴ = dx

y (log x + 1) = x

(1 + log x) − x ⋅ (1 + log x)

2

1 x =

log x . (1 + log x) 2

12. (b) Put x = cos θ ⇒ θ = cos–1 x

 cos θ − sin θ   1 − tan θ  –1 ∴ y = tan–1   = tan   cos θ + sin θ   1 + tan θ  



π  π π = tan–1 tan  − θ  = – θ= – cos–1 x 4 4 4 



∴

 1  dy  = = 0 – − 2  1− x  dx

13. (a) We have, g = inverse of f = f

1 1 − x2 –1



⇒ g (x) = f –1 (x) ⇒ f [g (x)] = x.



Differentiating w.r.t. x, we get



f ′ [g (x)] ⋅ g′ (x) = 1



1 ∴ g′ (x) = f' g ( x) = 1 + [g (x)]3 [ ]



.

  1 1 , ∴ f'[ g ( x)] =  f' ( x) = 3 3 + x + [ g ( x )] 1 1  

Differentiation

log (log x) – log5 (log 3) log 5



120

14. (c) Put x = tan θ ⇒ θ = tan–1x 1 + tan 2 θ + 1

Objective Mathematics



∴ y = cos–1



= cos–1



θ  θ 1 = cos–1  cos  = = tan–1x 2 2 2 

2 1 + tan 2 θ

1 + cos θ = cos–1 2

sec θ + 1 2 sec θ

= cos–1 2 cos 2 θ

18. (c) We have, y = etan x ⇒ log y = tan x Differentiating w.r.t. x, we get 1 dy dy = sec2x or = y sec2x y dx dx dy =y ⇒ cos2x dx Differentiating again w.r.t. x, we get d2y dy dy cos2x – 2 cos x sin x = dx 2 dx dx

2

2

1 dy = . ( + x2 ) 2 1 dx 15. (a) Let f (x) = ax2 + bx + c





∴

Then,  f (1) = a + b + c and f (– 1) = a – b + c Since f (1) = f (– 1) ⇒ a + b + c = a – b + c ⇒ 2b = 0  or  b = 0 i.e.,, f (x) = ax2 + c ∴  f ′ (x) = 2ax. f ′ (a) = 2a2, f ′ (b) = 2ab, f ′ (c) = 2ac Now 2f ′ (b) = f ′ (a) + f ′ (c) if 2 ⋅ 2ab = 2a + 2ac if 2b = a + c if  a, b, c are in AP, which is given. ∴  f ′ (a), f ′ (b), f ′ (c) are in AP. 2

a (r n − 1) ⇒ (r – 1) Sn = arn – a r −1 Differentiating both sides w.r.t. r, we get

16. (b) We have, Sn =

dS (r – 1) n   + Sn = narn – 1 – 0 dr dSn ⇒ (r – 1) = narn – 1 – Sn dr



= n [nth term of G.P.] – Sn = n (Sn – Sn – 1) – Sn



= (n – 1) Sn – n Sn – 1.

17. (b) We have,

y = tan–1







⇒ log log y = log log x ⋅ log log x + log log x



Differentiating w.r.t. x, we get y log y dy = [2 log log x + 1]. x log x dx



20. (c) We have,



y = a cos (log x) + b sin (log x)  ...(1) dy 1 1 ⇒ = – a sin (log x) ⋅   + b cos (log x) ⋅   dx x x  [Differentiating w.r.t. x] dy ⇒ x  = – a sin (log x) + b cos (log x) dx d2y dy + dx 2 dx



⇒ x



= – a cos (log x) ⋅



⇒ x2 



= – y. 

1 1 – b sin (log x) ⋅ x x

d2y dy +x = – [a cos (log x) + b sin (log x)] dx 2 dx [From (1)]

21. (a) We have, y = (sin x)  –1

2

dy = 2 sin–1x ⋅ dx

...(1) 1

dy dx

1 − x2

1 + ... to n terms 1 + ( x + 5 x + 6)



= 2 sin–1x



 dy  ⇒ (1 – x2)   = 4 (sin–1x)2 = 4y  [Using (1)]  dx 

+ [tan–1 (x + 3) – tan–1 (x + 2)] + ... + [tan  (x + n) – tan (x + n – 1)] –1 

= tan–1 (x + n) – tan–1x 1 1 dy − ∴ = . 1 + ( x + n) 2 1 + x 2 dx

1 − x2



i.e.,

2

⇒ (1 – x2) ⋅ 2 ⋅

−1  ( x + 3) − ( x + 2)  + tan   + ... to n terms  1 + ( x + 3)( x + 2) 





⇒ log y = (log x)log log x ⋅ log x

2

= [tan–1 (x + 1) – tan–1x] + [tan–1 (x + 2) – tan–1 (x + 1)]





⇒ 

−1 + tan

–1

log log x





(log x ) 19. (a) We have, y = x

1 1 + tan −1 2 2 1 + ( x + x) 1 + ( x + 3 x + 2)

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



d2y dy = (1 + sin 2x) . dx 2 dx

i.e., cos2x

=

dy dx

4

⇒ (1 – x2) 

 dy  dy d2y ⋅ + (– 2x) ⋅   dx dx 2  dx 

d2y dy =x +2 2 dx dx [Dividing both sides by 2

22. (b) We have,

y=

x

∑ tan r =1

−1

1 = 1+ r + r2

x

∑ tan r =1

−1

 (r + 1) − r     1 + (r + 1) r 

2

dy ] dx

(r + 1) − tan −1 r 





= [tan–12 – tan–11 + tan–13 – tan–12 + ...

r =1



+ tan–1x – tan–1 (x – 1) + tan–1 (x + 1) – tan–1x]

= [tan–1 (x + 1) – tan–11] 1 dy ∴ = . 1 + ( x + 1) 2 dx



23. (a) We have,

y = eax sin bx

...(1)



dy = eax ⋅ b cos bx + a eax sin bx ⇒ dx



= b eax cos bx + ay



d2y dy ⇒ = ba eax cos bx + beax ⋅ b (– sin bx) + a   2 dx dx

[From (1)]

...(2)



 dy  dy = a  − ay    – b2y + a [From (1) and (2)] dx dx  



dy = 2a – (a2 + b2) y dx



d2y dy ∴ – 2a = – (a2 + b2) y. dx 2 dx

24. (c) Let y = tan–1

2x 1 − x 1 − 2x2

2

and t = sec–1

dy = – 2. dθ



⇒



1   Also, t = sec–1  2 cos 2 θ − 1    = sec–1 (sec 2θ) = 2θ



⇒



∴





= P (x) P′′′ (x). 

 1 + x2 − 1   26. (a) Let y = tan–1    x  

dt dy  = – 1.  ∴ = – 1. dθ dt  x = 1



2 1 + x2

Put x = tan θ  sec θ − 1   1 − cos θ  –1 ∴ y = tan–1   = tan    tan θ   sin θ  = tan–1 tan

θ θ = 2 2

1 + sec θ 1 + cos θ = cos–1 2 sec θ 2 θ θ = cos–1 cos = 2 2 θ dy ∴ y = = t; ∴ = 1. 2 dt 1 1  −1 −1 27. (c) y = tan  (2 tan u ) + (2 tan u )  2 2  and  t = cos–1

 −1 2u  2u = tan  tan = x.  = 1 − u2  1 − u2 

y2 = P (x) dy = P′ (x) ⇒ 2y  dx

28. (a) Let y = f (tan x) and u = g (sec x)

⇒

dy = f ′ (tan x) sec2x dx



and

du = g′ (sec x) ⋅ sec x tan x dx



∴

dy dy = dx du



∴

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

dy dy d2y ⋅ + 2y ⋅ = P′′ (x) dx dx dx 2



d2y  dy  = y2 P′′ (x) ⇒ 2y2   + 2y3 ⋅ dx 2  dx  2

 dy  d2y ⇒ 2y3 = y2 P′′ (x) – 2y2   dx dx 2 1 = y2 P′′ (x) – [P′′ (x)]2 2

dy = 1. dx

∴

2

⇒2



1 + 1 + x2

and t = cos–1







[ y2 = P (x)]

= tan (tan–1u + tan–1u) = tan (2 tan–1u)

25. (b) We have,

= y2 P′′′ (x)



1 2x2 − 1

dy 1 P′′ (x) + y2 P′′′ (x) – 2P′ (x) ⋅ P′′ (x) dx 2 = P′ (x) P′′ (x) + y2 P′′ (x) – P′ (x) P′′ (x) dy    2 y dx = P' (x) = 2y





Put x = cos θ Then,  2 cos θ sin θ  –1 y = tan–1   = tan (– tan 2θ) = – 2θ 2  1 − 2 cos θ 

dt =2 dθ dy dy = dθ dt



2 d  y3 d y   dx 2  dx

⇒2

121

−1

=

dy  du  x = π 4

2

= [from 2]

f ′(tan x) sec 2 x du = g ′(sec x) sec x tan x dx π  f'  tan  4  = π π  g'  sec  sin 4 4 

f' (1) 1 g' ( 2 ) ⋅ 2

=

2×2 = 4

1 . 2

Differentiation

x

∑  tan



122

29. (b)

Objective Mathematics





dx = a (– sin θ + 1 ⋅ sin θ + θ ⋅ cos θ) = aθ cos θ dθ dy = a (cos θ – 1 ⋅ cos θ + θ sin θ) = aθ sin θ dθ a θ sin θ dy dy dx = = = tan θ. ∴ a θ cos θ dθ dθ dx ⇒

∴

1 sec3 θ d2y dθ 2 2 = sec θ ⋅ = sec θ ⋅ = a θ cos θ aθ dx 2 dx sec3 π d2y 4 = 8 2 . =  dx 2  θ = π aπ π a  4 4  

30. (a) We have,

x = t = e dx ⇒ = et log t (1 + log t) = tt (1 + log t) dt t

t log t

t

Also,  y = t t ⇒ log y = tt log t = et log t ⋅ log t 1 dy 1 ⇒ = et log t (1 + log t) log t + et log t ⋅ y dt t 1 dy tt t  ⇒ = t ⋅ t (1 + log t )log t +  dt t 

∴

dy dt dy = = dx dt dx

t  1 t t (1 + log t )log t +  t  . (1 + log t )

dy 31. (c) We have, y = x3 – 8x + 7 ⇒ = 3x2 – 8 dx It is given that when t = 0, x = 3. dy ∴ When t = 0, = 3 ⋅ 32 – 8 = 19. dx dy dy dt =  ...(1) Also, dx dx dt dy dy = 19 and = 2, Since, when t = 0, dx dt 2 dx 2 ∴  from (1), 19 = ⇒ = . dx dt dt 19 32. (b) Differentiating the given equation w.r.t. x, we get dy dy + 2hy + 2by =0 2ax + 2hx dx dx dy ax + hy =– ⇒ dx hx + by d2y ⇒  dx 2  dy  dy      (hx + by )  a + h dx  − (ax + hy )  h + b dx        =–  (hx + by ) 2     dy 2   2  y (ab − h ) + dx (h x − abx)  =– (hx + by ) 2

dy   (h 2 − ab)  y − x  dx   = 2 (hx + by ) (h 2 − ab)  ax + hy  h 2 − ab ⋅ y + x ⋅ = .   (hx + by ) 2  hx + by  (hx + by ) 2 dy 33. (a) We have, y = sin x ⇒ = cos x. Now, dx d2 d  d 7  7  cos x  2  (cos x) = dy dy  dy  =



d dy d = dy

=

 dx  6  7 cos x ⋅ (− sin x)  dy   (– 7sin x cos5x)



= – 7 [cos x ⋅ cos5x – 5cos4x sin2x] 



= – 42 cos5x + 35 cos3x

dx dy

34. (a), (c) Taking log on both sides of the given equation, we get, x x x log cos + log cos 2 + log cos 3 + ...∞ 2 2 2  = log sin x – log x Differentiating w.r.t. x 1 x 1 x 1 − tan − 2 tan 2 − ...∞ = cot x – 2 2 2 2 x ⇒

x 1 x 1 1 tan + 2 tan 2 + ...∞ = – cot x + 2 2 2 2 x

Differentiating again w.r.t. x, we get x 1 x 1 1 2 sec 2 + 4 sec 2 2 + ...∞ = cosec2x – 2 . x 2 2 2 2 π 35. (b) When < x < π, cos x < 0, ∴ | cos x | = – cos x 2 ∴ f (x) = – cos x ⇒ f ′ (x) = sin x  3π  3π ∴ f ′   = sin = 4 4  

1 . 2

36. (b) We have, log f (x) = (a + b + 2x) [log (a + x) – log (b + x)] Differentiating both sides w.r.t. x, we get f' ( x) = 2 [log (a + x) – log (b + x)] f ( x) 1   1 − + (a + b + 2x)   a + x b + x ⇒ f ′ (0) = f (0)   1 1   2 (log a − log b) + (a + b)  a − b     



=

a   b

a +b

 a b2 − a 2   2 log +  b ab  

π , cos x > sin x ∴ cos x – sin x > 0. 4 π π – 2x + 2

⇒

22x + 2x – 2 > 0

⇒ ⇒

(2x – 1) (2x + 2) > 0 2x – 1 > 0 [  2x + 2 > 0 for all x]

⇒

2x > 1 ∴ x > 0.

68. (b) Since f (x) is a polynomial function satisfying 1 1 f (x) ⋅ f   = f (x) + f    ,  x  x

⇒ cot y (1 – x cos a) = – x sin a  x sin a  ⇒ y = cot–1    x cos a − 1 

– e– y

f (x3) = x5

= x4 + y4 + 2x2y2 – 2 ⇒ x2y2 = 1

1 8 f (x) + 6 f   = x + 5 for all x  x Therefore, 1 1 8 f   + 6 f (x) = + 5 x x  

dy  ∴ = dx



∴ f ′ (27) = f ′ (33) =

∴ [ f (10)]2 – [φ (10]2 = [ f (3)]2 – [φ (3)]2 = [ f (3)]2 – [ f ′ (3)]2

64. (a) We have, for all x (x ≠ 0)

–x

=1

∴ f (x) = xn + 1 or f (x) = – xn + 1 If f (x) = – xn + 1, then f (4) = – 4n + 1 ≠ 65 So, f (x) = xn + 1. Since f (4) = 65 ∴ 4n + 1 = 65 ⇒ n = 3 ∴ f (x) = x3 + 1 ⇒ f ′ (x) = 3x2 ∴ f ′ (l1) = 3l12 , f ′ (l2) = 3l22 , f ′ (l3) = 3l32 Since l1, l2, l3 are in G.P., ∴ f ′ (l1), f ′ (l2), f ′ (l3) are also in G.P. 69. (a) f ′ (x) =

n( x)  n( x)m' ( x) − m( x)n' ( x)    m( x )  [n( x)]2 

⇒ f ′ (1) =

n(1)  n(1)m' (1) − m(1)n' (1)    = 0. m(1)  [n(1)]2 

∴

n

d [ f (x)]x = 0 dx n

n! n! nπ nπ cos = cos 2 2 nπ nπ sin sin 2 2



Thus,

2 4 =0 8

[ C1 and C2 are identical]

71. (b) We have, d  (1 + x 2 + x 4 )(1 − x 2 + x 4 )    = ax3 + bx dx  (1 + x 2 + x 4 ) 

∴

75. (a), (b), (c)  We have,

f ′ (x) = 6x + 4g′ (1)

...(1)



f ′′(x) = 6

...(2)



g′ (x) = 4x + 3 f ′ (2)

...(3)



g′′ (x) = 4

...(4)

From (1), f ′ (1) = 6 + 4g′ (1) = 6 + 4 [4 + 3 f ′ (2)] [  g′ (1) = 4 + 3 f ′ (2)] = 22 + 12 f ′ (2)

⇒ – 2x + 4x3 = ax3 + bx ⇒ a = 4 and b = – 2.

From (3), g′ (2) = 8 + 3 f ′ (2) = 8 + 3 [12 + 4g′ (1)]

72. (c) We have, f (x) = α (x) β (x) γ (x), for all real x

[ f ′ (2) = 12 + 4g′ (1)] = 44 + 12 g′ (1)

⇒ f ′ (x) = α′ (x) β (x) γ (x) + α (x) β′ (x) γ (x) + α (x) β (x) γ′ (x)

Also, from (2) and (4), f ′′(3) + g′′ (2) = 6 + 4 = 10. 76. (a) We have, y1 = (n – 1)xn – 2 log x + xn – 1 ⋅

⇒ 18 f (2) = 3α (2) β (2) γ (2) – 4α (2) β (2) γ (2)

1 x



+ kα (2) β (2) γ (2)

⇒ xy1 = (n – 1) xn – 1 log x + xn – 1



[ f ′ (2) = 18 f (2), α′ (2) = 3α (2), β′ (2) = – 4β (2)

Differentiating again w.r.t. x, we get



= (n – 1) y + xn – 1 

⇒ x2y2 + xy1 = (n – 1) xy1 + (n – 1) xn – 1

⇒ 18 f (2) = (– 1 + k) α (2) β (2) γ (2) = (k – 1) f (2) ⇒  k – 1 = 18  ∴ k = 19.

π π   73. (c) f ′ (x) = – 2 cos x sin x – 2 cos   x +   sin   x +  3 3   π π   + cos x sin  x +    + sin x cos   x +  3 3   2π  π   = – sin 2x – sin   2 x +  + sin   x + x +  3  3   π π   π = – 2 sin  2 x +  cos  + sin   2 x +  3 3 3   π π   = – sin   2 x +  + sin   2 x +  = 0. 3 3   ⇒

= (n – 1) xy1 + (n – 1) (xy1 – (n – 1) y)

[ f (2) = α (2) β (2) γ (2)]



f (x) = constant for all x.

π π 5 But, f (0) = cos2 0 + cos2 + sin 0 ⋅ sin = 3 3 4

...(1)

y1 + xy2 = (n – 1) y1 + (n – 1) xn – 2

and γ′ (2) = kγ (2)] 

n = 70.

d (1 – x2 + x4) = ax3 + bx ⇒ dx

⇒ f ′ (2) = α′ (2) β (2) γ (2) + α (2) β′ (2) γ (2) + α (2) β (2) γ′ (2)

127

74. (c) f (x) is a polynomial of degree 90. f ′ (x) reduces the degree of f (x) by one. Thus, in order to get a polynomial of degree 20, we must reduce the degree of f (x) by 70. Hence, f (x) should be differentiated 70 times to get a polynomial of degree 20.



[using (1)]

⇒ x2y2 + (3 – 2n) xy1 + (n – 1)2 y = 0. 77. (b) We have,

4x2 = ( y1/5 + y–1/5)2 = y2/5 + y–2/5 + 2 = ( y1/5 – y–1/5)2 + 4 ⇒ y1/5 – y–1/5 = 2 x 2 − 1 . Adding with the given relation, we get 2y1/5 = 2  x + x 2 − 1    ⇒ y =  x + x 2 − 1      2 4 ⇒ y1 = 5 ( x + x − 1) 1 + 

⇒ y1 =

5 ( x + x 2 − 1) 5 x −1 2

⇒ (x2 – 1) y12 = 25y2

=

x   x2 − 1  5y x2 − 1

5

Differentiation

2 n! n! dn nπ  nπ  70. (a) n [ f (x)] = cos  x + 4  cos dx 2  2  nπ  nπ  8 sin  x +  sin 2  2 

5 for all x. 4 5 (gof ) (x) = g [ f (x)] = g    = 3. 4

f (x) =

∴

128

Differentiating again w.r.t. x, we get

⇒ 3 f ′ (x) [1 + { f (x)}2] = 0

(x – 1) 2y1 y2 + y1 ⋅ 2x = 50 yy1

⇒ f ′ (x) = 0, for all x.

2

2

Objective Mathematics



[Dividing by 2y1] 83. (b) Let f (x) = a (x – 3)3 + b (x – 3)2 + c (x – 3) + d 2 ⇒ (x – 1) y2 + xy1 = 25y. ∴ k = 25. Then, f (3) = 1 = d ⇒ d = 1

78. (b) We have, log y = tan2x ⋅ log tan x ⇒ ⇒ ∴

1 dy 1 = 2 tan x sec2x ⋅ log tan x + tan2x ⋅ sec2x y dx tan x

∴ f (x) = 2 (x – 3)3 – x + 4 ⇒ f ′ (x) = 6 (x – 3)2 – 1

dy  = 1 ⋅ 1 ⋅ 2 (2 ⋅ 0 + 1) = 2. dx  x = π

∴ f ′ (1) = 6 (4) – 1 = 23. 84. (a)

79. (b) We have, 2n

y = (1 + x) (1 + x2) (1 + x4) ... (1 + x ) n +1

(1 − x)(1 + x)(1 + x 2 )(1 + x 4 )...(1 + x 2 ) 1 − x2 = 1− x 1− x n +1

(1 − x) ⋅ −2n +1 ⋅ x 2 −1 + (1 − x 2 dy ⇒ = (1 − x) 2 dx

n +1

sin x cos x sin x cos x sin x = 0 – sin x cos x sin x + 1 − sin x cos x x 1 1

dy  = 1. dx  x =0

1  sin x ⋅ cos16 x ⋅ 16 − sin 16 x ⋅ cos x  ⇒ f ′ (x) = a 16  sin 2 x   1  1 ⋅ 1 ⋅ 16 − ⋅0  π 1  2 2  = ∴ f ′   = 2  16  4  1         2

= 0 + (cos2x + sin2x) = 1. 85. (b) We have, f (x) = xn ⇒ f (1) = 1 = nC0

82. (a), (b) We have, f (x) + f (y) + f (z) + f (x) ⋅ f (y) ⋅ f (z) = 14 ...(1) Putting x = y = z = 0, we get 3 f (0) + [ f (0)]3 = 14 ⇒ [ f (0)]3 + 3 f (0) – 14 = 0

3 f ′ (x) + 3 [ f (x)]2 ⋅ f ′ (x) = 0

f"(1) n (n −1) = = nC 2 2! 2! f'"(1) n (n − 1)(n − 2) n = = C3 3! 3! f n (1) n! = = nC n n! n! f' (1) f" (1) f"' (1) (−1) n f n (1) + − + ... + 1! 2! 3! n! = nC0 – nC1 + nC2 – nC3 + ... + (– 1)n nCn ∴ f (1) –

= (1 – 1)n = 0. 86. (c) f (x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x)

for all x, y, z ∈ R

Differentiating w.r.t. x, we get





4/5 4/5  π π π π ⇒ f ′  5  = – 5   ⋅ sin   = – 5   . 2 2  2 2

3 f (x) + [ f (x)]3 = 14

f'(1) n = = nC 1 1! 1!



π 5 ∴ f (x) = sin  − x  = cos x5 2  ⇒ f ′ (x) = – sin x5 ⋅ 5x4

⇒ f (0) = 2.



2.

81. (b) Since, 1 < x < 2. ∴ [x] = 1

Now, putting y = z = x in (1), we get

sin x cos x sin x + cos x − sin x cos x 1 0 0



)

sin 16 x 80. (c) f (x) = 2 sin x cos x cos 2 x cos 4 x cos 8 x = 4 2 sin x 2 sin x



cos x − sin x cos x sin x cos x sin x dy = cos x − sin x cos x + − sin x − cos x − sin x dx 1 1 x x 1 1



n

∴

f ′′(3) = 0 = 2b ⇒ b = 0 f ′′′ (3) = 12 = 6a ⇒ a = 2.

dy = [(tan x)tan x]tan x ⋅ tan x sec2x (2 log tan x + 1) dx

4

=

f ′ (3) = – 1 = c ⇒ c = – 1



... (cos nx + i sin nx) = cos (x + 2x + 3x + ... + nx) + i sin (x + 2x + 3x + ... + nx) n (n + 1) n (n + 1) = cos x + i sin x 2 2 ⇒ f ′ (x) = ⇒  f ′′(x)

n (n + 1)  n (n + 1) n (n + 1)  − sin x + i cos x 2  2 2 

 n (n + 1)  = –   f ( x)   2   2

 n (n + 1)   n (n + 1)  ∴ f ′′ (1) = –   f (1) = –   . 2  2    2

2

92. (b) f (x) =

∴  [go ( f h)] (x) = g [( f h) (x)] = g [ f (x) ⋅ h (x)]

1 if x ≥ −3  . ∴ f ′ (x) =  −1 if x < −3

= g (sin x cos x) = 2sin x cos x = sin 2x i.e., φ (x) = sin 2x

93. (c) f (x) = | (x – 4) (x – 5) | ( x − 4)( x − 5) if ( x − 4)( x − 5) ≥ 0   =  −( x − 4)( x − 5) if ( x − 4)( x − 5) < 0 

π π ∴ φ′′    = – 4 sin = – 4. 4 2   log 2 x, 2 x > 0 i.e. x > 0   88. (a) f (x) = log | 2x | =  log(−2 x), 2 x < 0 i.e. x < 0  1  1   2 x × 2, x > 0   x , x > 0   =   ⇒ f ′ (x) =   1 × −2, x < 0   1 , x < 0  −2 x   x 

∴ f ′ (x) =

2 if x ≤ 4 or x ≥ 5  x − 9 x + 20  =  2 − ( x − 9 x + 20 ) if 4< x 2)

1 , x ≠ 0. x



89. (a) We have, dx 1 sec θ / 2 = – sin θ + = – sin θ + θ θ dθ 2 tan θ / 2 2 sin cos 2 2 1 1 − sin 2 θ cos 2 θ = – sin θ + = = sin θ sin θ sin θ dy = cos θ dθ dy d θ dy cos θ sin θ = = = tan θ ∴ dx d θ dx cos 2 θ

and

∴

d2y dθ sin θ = sec2θ ⋅  = dx 2 dx cos 4 θ

d2y = 0 ⇒ sin θ = 0 dx 2



90. (a) Since x > 10, therefore, f (x) = x – 3 ∴ φ (x) = ( fof ) (x) = f [ f (x)] = f (x – 3) = x – 3 – 3 = x – 6

 ⇒ φ′ (x) = 1.



dy = dx = 2⋅

1 1−

4x2 (1 + x 2 ) 2

1 + x2 (1 − x 2 ) 2

⋅

 (1 + x 2 ) ⋅ 2 − 2 x ⋅ 2 x  ×  (1 + x 2 ) 2  

1 − x2 1 1 − x2 ⋅ = 2 ⋅   2 2 2 | 1 − x | 1 + x2 (1 + x )

 x − 3, if x ≥ 3  = | (x – 2) – 1 | = | x – 3 | =   − x + 3, if 2 ≤ x < 3 1, if x ≥ 3  ∴ g′ (x) =  . −1, if 2 ≤ x < 3

95. (a) We have, x2 + y2 = a2 ⇒ 2x + 2yy’ = 0 x ⇒ y’ = – . y Differentiating again w.r.t. x, we get 1 + y' 2 y′2 + yy′′+ 1 = 0 ⇒ y = – y" 2   x ∴ a = x2 + y 2 = y 1 +    y

⇒ θ = nπ, n ∈ I

91. (a)

f [ f (x)] = f (x – 1) = | (x – 1) – 1 | = | x – 2 | = (x – 2) (  x > 2) ∴ g (x) = f [ f { f (x)}] = f (x – 2)

2

Also,

( x + 3) 2 = | x + 3 |

 x + 3, x + 3 ≥ 0 i.e. x ≥ −3  =  − x − 3, x + 3 < 0 i.e. x < −3 

87. (c) We have, ( f h) (x) = f (x) ⋅ h (x) = sin x cos x

⇒ φ′ (x) = 2 cos 2x and φ′′ (x) = – 4 sin 2x

x2 + 6x + 9 =



= −

(1 + y' 2 ) ⋅ 1 + y' 2 y"

=

(1 + y' 2 )3 . | y" |

96. (b) We have,

−1  1 − 3 log x  −1  4 + 3 log x   + tan   y = tan   1 + 3 log x   1 − 12 log x 

= tan–11 – tan–1 (3 log x) + tan–14 + tan–1 (3 log x)  −1 −1 −1  A − B    Using tan A − tan B = tan  1+AB       A + B   −1  −1 −1  and tan A + tan B = tan     1 − AB   

129

2

Differentiation

 2  1 + x 2 , − 1 < x < 1  dy  =  −2 ∴   dx , x or x < − > 1 1 1 + x 2 

n (n + 1) n (n + 1)   n (n + 1)   x + i sin x = –   cos 2   2 2  

130



Objective Mathematics

= tan–11 + tan–1 (4) dy d2y ∴ = 0 and hence = 0. dx dx 2

102. (d) We have, −a sin 3t cos t sin 2t  dx  = a  − sin t cos 2t − =  cos 2t dt cos 2t  

97. (b), (c) We have, f ′ (α + β) = f ′ (α) + f ′ (β) ⇒ (α + β)m – 1 = αm – 1 + βm – 1

and

Since, for m > 2, the above equality is not valid ∴ we must have m = 2. Also, for m = 0, f ′ (x) = 0 for all x. So the equality is trivially true. 98. (b) When x = 0, we get y = 0. Differentiating both sides of the given equation w.r.t. x, we get  dy  dy exy + xexy  x + y  = + 2 sin x cos x dx dx

⇒

dy dy dt = = – cot 3t. dx dx dt

⇒

dt −3cosec 2 3t ⋅ cos 2t d2y 2 = 3 cosec 3t ⋅ = dx a sin 3t dx 2

2   ∴ 1 +  dy     dx  



99. (b) Differentiating x2 + y2 = 1 w.r.t. x, we get 2x + 2yy’ = 0 or x + yy’ = 0. Again differentiating w.r.t. x, we get 1 + y′y’ + yy′′= 0 or 1 + (y′)2 + yy′′= 0.

∴

3 = –   cosec33t cos 2t a

Putting x = 0, y = 0, we get  dy   dy  e0 + 0 e0  0 ⋅  + 2 sin 0 cos 0 + 0 =   dx  ( 0, 0 )  dx  ( 0, 0 ) dy  = 1. dx  ( 0, 0 )

sin t sin 2t  a cos 3t dy  = a cos t cos 2t −  = cos 2t dt cos 2t  



= (1 + cot 2 3t )

⇒

3

3

d2y dx 2

2

2

  dy  2  1 +      dx  

 3 3  −  cosec 3t cos 2t  a 3

2

d2y dx 2

at

t =



π is 6

100. (d) We have,





3x 2 f ′ (x) = 6 p

cos x − sin x −1 0 p2 p3

6 x − sin x − cos x f ′′(x) = 6 −1 0 p p2 p3 and f ′′′ (x) =

6 ∴ f ′′′ (0) = 6 p

6 − cos x sin x 6 −1 0 p p2 p3 −1 −1 p2



[ R1 and R2 are identical]

101. (a) We have,

3 cos



+ ... + an – 1 [x + (k – x)] + an.

It may be noted that n must be even for otherwise f (x) will become a polynomial of degree n – 1. Clearly, f ′ (x) is a polynomial of degree n – 1.

104. (c) When x = 0, y > 0 ⇒ y = aeπ/2.  aking log on both sides of the given equation, T we get  y 1 log (x2 + y2) = log a + tan–1   . 2 x

Differentiating both the sides w.r.t. x, we get 1 2 x + 2 yy' ⋅ = 2 x2 + y 2

1

( x)

1+ y

2

⋅

xy' − y x2

⇒ x + yy′ = xy′ – y Differentiating again w.r.t. x, we get



⇒ log F (x) = log f (x) + log g (x) + log h (x) Differentiating both the sides w.r.t. x, we get



F' ( x) f' ( x) g' ( x) h' ( x) = + + F(x) f (x) g (x) h(x)



⇒



⇒ 21 = 4 – 7 + k ⇒ k = 24.

F' ( x0 ) f' ( x0 ) g' ( x0 ) h' ( x0 ) = + + F(x0 ) f (x0 ) g (x0 ) h(x0 )

2a . 3

f (x) = a0 [xn + (k – x)n] + a1 [xn – 1 + (k – x)n – 1]

F (x) = f (x) g (x) h (x)



π 3

=

103. (b) Clearly, f (x) must be of the form

0 0 =0 p3

a

...(1)

1 + (y′)2 + yy′′= xy′′+ y’ – y’ ⇒ 1 + (y′)2 = (x – y) y′′ 

⇒ y′′=

When x = 0, we get, from (1), y’ = – 1 2 2 −π ∴ y′′(0) = = – e 2. π/ 2 −ae a

1 + ( y' ) 2 . x− y



x sin x − sin x + x 2 tan x −3 x 2 . 2 x sin 2 x 5



∴ For x ≠ 0, 1 sin x cos x 1 cos x cos x f' ( x) 2 2 tan x − x + x sec x − x3 = 2 x x 2 2 cos 2 x 5 x 2 sin 2 x 5x

⇒ f ′ (x) =

= – 2 – 2 = – 4.

∆′ (x) =

and ⇒

dz =– dx

=

1 − x2

dy  = 4. dz  x = 1 2

log (log x) log x

1 − log (log x)  x (log x) 2

∴ f ′ (e) =

1 − log (log e) 1 = . e (log e) 2 e

x2 − x x2 + 2x



Clearly, f (0) and f (– 2) are not defined.



So domain of f = R\{0, – 2}.



Then, in this domain, we have



y = f (x) =



A ′ ( x) B′ ( x) C′ ( x) A(α) B(α) C(α) A' (α) B' (α) C' (α)

x

1 1 1 ⋅ − log (log x) ⋅ log x x x (log x) 2

111. (b) We have, f (x) =

A( x) B( x) C( x) 106. (d) Let ∆ (x) = A(α) B(α) C(α) , then A' (α) B' (α) C' (α)



dy dy dx 2 = = dz dz dx x

∴

log x ⋅

1 0 1 1 1 1 1 0 0 f' ( x) ⇒ lim = 2 1 0 + 0 1 0 + 0 0 0 x→0 x 2 0 0 2 2 0 2 0 5



and z = 1 − x 2 dy 2 ⇒ =– dx 1 − x2

110. (a) We have, f (x) = logx (lnx) =

1 sin x − sin x + x tan x −3 x 2 2 sin 2 x 5





1 = cos–1 (2x2 – 1) = 2cos–1x 2x2 − 1



or x =

x −1  x+2

⇒ yx + 2y = x – 1

2y +1 2x + 1 , i.e., f –1 (x) = 1− y 1− x

df −1 ( x) 2 (1 − x) + 2 x + 1 3 = = . (1 − x) 2 dx (1 − x) 2

∴

dy Since, ∆ (α) = 0 = ∆’ (α), therefore α is a repeated root of  = 1 + ex 112. (c) y = x + ex ⇒ ∆ (x) and α is a repeated root of the quadratic equation dx f (x) = 0, so ∆ is divisible by f (x). dx dx d 2x −1 1 ex ⋅ ∴ =  ⇒ = (1 + e x ) 2 dy dy dy 2 107. (b) Differentiating the given equation w.r.t. x, we get 1 + ex



xy′ + y ⋅ 1 –

1 y′ = 0 y





⇒ xyy′ – y′ + y2 = 0 i.e., (xy – 1) y′ + y2 = 0



Differentiating again w.r.t. x, we get



⇒ x (yy′′+ y′ ) – y′′+ 3yy’ = 0

108. (c) We have, ∴

113. (a) x = ey + x ⇒ log x = x + y

(xy – 1) y′′+ y′ (xy′ + y ⋅ 1) + 2yy’ = 0 dx dy = cos t and = dt dt

dy dt dy = = dx dt dx

2 ( a + b) et cos t

2 (a + b) et 2

=

2y 1 − x2

 dy  ⇒ (1 – x2)   = 2y2  dx  Differentiating w.r.t. x, we get (1 – x2) 2y′ y′′– 2x (y′ )2 = 4yy’ ⇒ (1 – x2) y′′– xy′ = 2y [dividing by 2y′] 2

∴ k = 2.

⇒

∴ k = 3.

2

1 −e x ex ⋅ . x 2 x = – (1 + e ) 1 + e (1 + e x )3

=

2

dy 1 =1+  dx x

114. (c) Since f (x) is odd

∴ 

dy 1− x = . dx x

∴ f (– x) = – f (x)

⇒ f ′ (– x) (– 1) = – f ′ (x) i.e., f ′ (– x) = f ′ (x) ∴ f ′ (– 3) = f ′ (3) = – 2. 115. (b) We have,

y = f (sin x)  or  y = log (sin x)

∴

1 dy x cos x = cot x. = sin x dx

116. (a) y =

sin x + y

⇒ 2y

⇒ y2 = sin x + y

dy dy = cos x + dx dx

131

109. (b) Let y = sec–1

1 sin x cos x cos x cos x x f ′ (x) = 2 x tan x − x 3 + x 2 sec 2 x − x3 2 sin 2 x 5 x 2 x 2 cos 2 x 5 x

Differentiation

105. (a) We have,

132

Objective Mathematics

dy = cos x dx cos x dy = . ( 2 y − 1) dx

⇒ (2y – 1) ∴

π 3 sin   3 2 =   π   π  1 + 1  1 + 1 cos cos   3  3  2  2      

117. (b) We have, y = log (1 – x2) – log (1 + x2) dy −2 x 2x (1 + x + 1 − x ) = = – 2x ⋅   − 2 2 dx 1− x 1+ x 1 − x4 4x . =– 1 − x4 118. (c) We have, 2

⇒



fog = I ⇒ ( fog) (x) = x for all x



⇒ f ′ [g (x)] ⋅ g′ (x) = 1, for all x



⇒ f ′ [g (a)] ⋅ g′ (a) = 1



⇒ f ′ [g (a)] =

1 1 = g' (a ) 2

( g′ (a) = 2) 1 ⇒ f ′ (b) =  (  g (a) = b) 2 119. (b) Differentiating the given equation w.r.t. y, we get 1 dx 1 + =0 2 x dy 2 y dx =– dy

x = y

y −4 y

 ∴

dx  1− 4 = – 3.  = dy  y =1 1

d  −1 1 + tan 2 θ − 1   = dx  tan  tan θ  

1 dy d = (1 + x 2 ) −1/ 2 = log ( x + 1 + x 2 ) = 1 + x2 dx dx d2y 1 x = − (1 + x 2 ) −3/ 2 ⋅ 2 x = − dx 2 2 (1 + x 2 )3/ 2

∴

121. (b) Let |x| = t, 

∴ 

d 1 1 (log t ) = = dt t |x|

122. (d) We have, dy = 3a sin2 θ cos θ dθ  and dx = –3a cos2 θ sin θ dθ 3 a sin 2 θ cos θ ⇒  dy = dy dx  = = − tan θ 3a cos 2 θ(− sin θ) dx d θ d θ 2

∴ 

 dy  2 1 +   = 1 + tan 2 θ = sec θ =| sec θ |  dx 

123. (a) We have, d 1 1 (− sin 2 x) [cot −1 ( cos 2 x )] = − ⋅ ⋅2 dx 1 + cos 2 x 2 cos 2 x =

sin 2 x (1 + cos 2 x) cos 2 x

Therefore, the value of the derivative at x = π is 6

(put x = tan θ)

=

 1 − cos θ  d d θ d θ tan −1  tan −1 tan =  = θ dx sin dx 2 dx  2   

=

1 1 d tan −1 x = 2(1 + x 2 ) 2 dx

 1 + cos d  −1  tan 125. (a) We have, dx  1 − cos    x  2 cos 2  d  −1 4 = d  tan = dx  dx 2 x  2sin  4   =

120. (d) We have,

2 3

2   124. (d) We have, d  tan −1 1 + x − 1   dx  x 



⇒

3 2= 3

=

2

x 2 x 2

     

x  −1  tan cot  4 

1 − d  −1  π x   −d  π x  + =−  tan tan  +   =  dx  4  2 4   dx  2 4 

126. (b) We have, f ′′ (x) = – f(x) ⇒ d (f ′ (x)) = – f(x) ⇒ g ′ (x) = – f(x) dx 2 2   x    x  Now, F ( x) =  f    +  g      2    2    x  x 1 ⇒ F '( x) = 2  f    ⋅ f '   ⋅ 2   2 2     x  x 1 x x  +2  g    ⋅ g '   ⋅ = f   ⋅ f '   − 2 2 2 2   2 

x x f   f '  =0 2 2 ∴ F(x) is constant  Since F(5) = 5, ∴ F (10) = 5.

 x +1  x −1  127. (b) We have, y = sec −1  + sin −1     x −1   x +1  x −1   x −1  = cos −1  + sin −1     x +1  x +1 ⇒ 

y=

π 2

∴ 

dy =0 dx

128. (c) We have, φ(x) = f –1(x) ⇒  x = f[φ (x)] On differentiating both sides w.r.t. x, we get 1 = f ′ [φ(x)]. φ ' (x)

f ' [φ( x)] =

∴ 

1  1 + [φ( x)]5

=−

∴  φ ' (x) = 1 + [φ(x)]5

129. (a) lim 2 f ( x) − 3 f (2 x) + f (4 x) x→0 x2 2 f ' ( x) − 6 f ' (2 x) + 4 f ' (4 x) = lim x→0 2x  (by L’ Hospital’s rule) = lim

x→0

 =

2 f '' ( x) − 12 f '' (2 x) + 16 f '' (4 x) 2 (by L’ Hospital’s rule)

2 f '' (0) − 12 f '' (0) + 16 f '' (0) 6a = = 3a 2 2

130. (b) Let y =

sec x

⇒ dy = 1 (sec x ) −1/ 2 ⋅ d (sec x ) dx 2 dx 1 1 = ⋅ sec x ⋅ tan x ⋅ 2 x 2 sec x =

dy = aeax sinbx + beax cosbx dx = ay + beax cosbx d2y dy =a + abe ax cos bx − b 2e ax sin bx of dx 2 dx

….(ii)

dy + abeax cosbx – b2y [using (i)] dx log (2 log x) 136. (a) Since, f (x) = log x3 (log x2) = 3 log x On differentiating w.r.t. x, we get 1 2 1 log x ⋅ ⋅ − log (2 log x) 2 log x x x f ′ (x) = 3 (log x) 2

= a

1 1 − log 2  1   = (1 − log 2) 3e  (1) 2  3e

dx 1  dy  = = dy dy / dx  dx 

−1

−1

 dx  d  dy  dx  =    dy  dx  dx  dy −2 −3  d 2 y   dy   dx   d 2 y   dy  d 2x = − 2      = −  2    ⇒ 2 dy  dx   dx   dy   dx   dx  138. (a) Given: g(x) = log f(x)

131. (c) We have, x = y 1 − y 2 dy y (− y ) dy 1 − y2 + dx 1 − y 2 dx  dy  1 − y 2 − y 2   = 1 or dx  1 − y 2  

….(i)

⇒ 

d ⇒ dy

1

dy  y2  1 − y2 − dx  1 − y2 

1− x 1− x 1 =− ⋅ 1− x 1 + x 1 − x2

dy + y=0 dx 135. (b) We have, y = eax sin bx

137. (d) Since,

(sec x )1/ 2 ⋅ sin x ⋅ sec x 4 x 1 = (sec x )3/ 2 ⋅ sin x 4 x

1=

×

(1 – x2)

⇒ f ′ (e) =

1 sin x (sec x )1/ 2 4 x cos x

=

1 (1 + x) 1 − x 2

133

1 1 Since f ' ( x) = 1 + x5 f ' [φ( x)]

Differentiation

⇒  φ ' (x) = =

  =1  

1− y dy = dx 1 − 2 y 2

2

−d 2 y d 2 x d  dx  d  1  dx 2 132. (d) We have, =  =  = dy 2 dy  dy  dy  dy / dx   dy  2  dx    133. (c) We have, f(x) = mx2 + nx + p

⇒ g (x + N) = log f (x + N) f (x + 1) = x f(x) ∴ f (x + N) = (x + N – 1) f (x + N – 1) = (x + N – 1)(x + N – 2) f (x + N – 2)… = (x + N – 1)(x + N – 2)… … …(x + 1) xf (x) ∴ log f(x + N) = log(x + N – 1) + log(x + N – 2)  + … log (x + 1)+log x log f(x) ⇒ log f(x + N) – log f(x) = logx + log(x + 1)  + …log(x + N – 1) ⇒ g(x + N) – g(x) = log x+ log(x + 1)  + … + log (x + N – 1) 1 and g ′ (x + N) – g ′ (x) = 1 + 1 + ... + x x −1 x + N −1

∴ g ′ (x + N) – g′ (x) = 1  1 1 1 puting x = we get − 2 + + ... 2 2 x ( x + 1) ( x + N − 1) 2   1− x   134. (d) We have, y =   1+ x 1 1 1 1  1  = −4 1 + 1 + ... g ′′  N +  = g ''   = −  ... + +  2   1 2  3 2  (−1) 1 2 2   2 N 1 −    9 − 1− x × 1+ x ×         dy − + x x 2 1 2 1 2 2 2        =   dx ( 1 + x )2    1 1 1  1  1 + 1 + ...  = −4 1 + 1 + ... 1 1 + x + 1 − x    N +  = g ''   = −  2 2 2 2  −  2 (2n − 1)   2  1 3  2N − 1   9 2  1 − x 2         2 2 2   =      (1 + x) ⇒  f ' (x) = 2mx + n  ∴f ' (1) + f ' (4) – f ' (5) = (2m + n) + (8m + n) – (10m + n) = n

134

EXERCISEs FOR SELF-PRACTICE

Objective Mathematics

1. Differential co-efficient of sec (tan – 1 x) is x 1 (a) (b) 2 1+ x 1 + x2 x (c) x 1 + x 2 (d) 1 + x2 2. If f (0) = 1, f ′ (0) = – 1, f (x) > 0 for all x then there exists a function f (x) such that (a) f ′ (x) < 0 for all x (b) – 1 < f ′′(x) < 0 for all x (c) – 2 ≤ f ′′(x) ≤ – 1 for all x (d) f ′′(x) ≤ – 2 for all x. 3. If 2x + 2y = 2x + y, then the value of (a) 0 (c) 1

dy at x = y = 1 is dx

(b) – 1 (d) 2.

1 − x2   2x  –1 4. The derivative of sin–1   is 2  w.r.t. sin    1 + x   1 + x 2  (a) – 1

(b) 1 (c) 2 (d) 1 2 5. Let f be a polynomial. Then, the second derivative of f (ex) is

(a) f ′′(ex) e2x + f ′ (ex) ex (b) f ′′(ex) ex + f ′ (ex) (c) f ′′(ex) e2x + f ′′(ex) ex (d) f ′′(ex).

6. If y = log (sin x), then

 x x − x− x  9. If f (x) = cot–1   , then f ′ (1) is equal to 2   (a) – 1 (c) log 2

(b) 1 (d) – log 2

2t 1− t2 dy , then = 2 , y = 1+ t 1+ t2 dx

10. If x = (a)

2t t2 −1

(b)

2t t2 +1

(c)

2t 1− t2

(d) None of these

 1 + x2 − 1   w.r.t. tan–1x is 11. Derivative of tan–1  x   (a) 1 1 (c) 2

(b) 2 (d) None of these

 3a 2 x − x 3  dy = 12. If y = tan–1   2 2  , then  a (a − 3x )  dx 3a 2 a + x2 a (c) 2 a + x2

(a)

2

3a a2 + x2 3 (d) 2 a + x2 (b)

13. Differential coefficient of esin d y equals dx 2 2

(a) sec x tan x (b) – cosec x cot x (c) sec2x (d) – cosec2x 7. If y = cos 2x sin 3x, then yn is equal to nπ  nπ    (a) 6n sin   2 x +  cos  3 x +   2  2  nπ  nπ    (b) 6  cos   2 x +  cos  3 x +   2 2 

(c) e

w.r.t. sin–1x is −1

(a) sin–1x cos −1 x

−1 x

(b) esin x (d) cos–1x



dy  cos x  14. If y = tan–1   , then is equal to dx  1 + sin x  1 2 (c) 1

1 2 (d) None of these

(b) – 

(a)

15. Differential coefficient of sin–1x w.r.t. cos–1 1 − x 2 is

n

 n nπ  nπ     5 sin  5 x + 2  + sin  x + 2     (d) None of these dy If y = (sin x)tan x, then is equal to dx (a) (sin x)tan x ⋅ (1 + sec2x ⋅ log sin x) (b) tan x ⋅ (sin x)tan x – 1 ⋅ cos x (c) (sin x)tan x ⋅ sec2x ⋅ log sin x (d) tan x ⋅ (sin x)tan x ­– 1

(c) 8.

1 2

2

(a)

1− x

2



1

(c) – 

1 + x2

(b)

1 1 − x2

(d) None of these

16. Differential coefficient of tan–1  (a) –  (c)

1 2

1 2

1 − x2 w.r.t. cos–1x2 is 1 + x2

(b) 1 (d) None of these

1 (1 + x 2 ) 1 (c) 1 + x2 (a)

(d) None of these

(a) 2 x (loge x − 1) ⋅ log e x (b) x (loge x − 1) 2 (c) log e x x (d) x

dy = dx

(a) 1 (b) – 1 (c) 2 (d) None of these

28.

sec x is

(sec x )3 / 2 sin x

2 1− x

29. If y = (a) – (c)

(b) sec xº tan xº (d)

180 sec x º tan x º π

22. Differential coefficient of cos–1 x w.r.t. (a)

x

(b) –  x

(c)

1 x

(d) – 

1 x

(b)

1 − x is

π 2

(d) None of these

2

(c) cos 2x

dy = dx

π sec x º tan x º 180



d (sin–12x 1 − x 2 ) is equal to dx

(a) – 

4 x 1 (b) sec x sin x 4 x 1 (c) x (sec x )3 / 2 sin x 2 1 (d) x sec x sin x 2

(a) sec x tan x

1 − x2

(c) 0

(d) e y

21. If y = sec xº, then

2

(a)

1 (b) x

20. Differential coefficient of

(c)



dP at V = 9 equals dV

(a) 0 (b) e (c) 2e (d) 1/e 27. d (cos–1x + sin–1x) is dx

⋅ log e x

1 (a) x log x 1 (c) log x

1

3 2

26. If f (x) = logx (ln x) then f ′ (x) at x = e is

19. If y = log log x, then ey 

(a)

(c) ± 



2

(b)

1 − x2

(d) None of these.

x + x + x + ...∞ , then 1 2y −1

1 xy

(b)

dy is equal to dx

1 2y −1

(d) 1

30. The differential coefficient of x6 w.r.t. x3 is (a) 6x5 (b) 3x2 (d) x3 (c) 2x3  sin x + cos x  dy 31. If y = tan– 1   is  , then  cos x − sin x  dx (a) 1/2 (b) π/4 (c) 0 (d) 1. 32. If xp yq = (x + y)p + q, then dy/dx is equal to

 1+ x   + xx 23. The first derivative of the function cos −1  sin 2   (a) y/x (b) py/qx w.r.t. x at x = 1 is (c) x/y (d) qy/px 3 (a) (b) 0 4 x .... ∞ dy , then is: 33. If e x + e x + e 1 1 dx (c) (d) –  2 2 x y (a) (b) 24. The values of x, at which the first derivative of the func1− x 1+ y 1   tion  x +   x

2

w.r.t. x is

3 are 4

(c)

y 1− y

(d)

1− y y

135

(b) ± 

25. If PV = 81, then

18. The differential coefficient of the function x loge x w.r.t. x is

(loge x − 1)

1 2 2 (d) ±  3

(a) ± 2

Differentiation

dy is equal to dx 1 (b) –  (1 + x 2 )

17. If x 1 + y + y 1 + x = 0, then

136

34. If x = a sin θ and y = b cos θ, then

Objective Mathematics

(a)

a sec 2 θ b2 −b (c) 2 sec3 θ a

(a) 4 (c) 1/2

d2y is  dx 2

(b) 1 (d) 6.

(b)

−b sec 2 θ a

36. If y = sin x + ex, then

(d)

−b sec3 θ a2

(a) (– sin x + ex)– 1

35. Let f (x + y) = f (x) f ( y) 3, then f ′ (5) equals:

x, y ∈ R, f (5) = 2, f ′ (0) = (c)

sin x − e x (cos x + e x )3

d 2x equals dy 2 (b)

sin x + e x (cos x + e x ) 2

(d)

sin x + e x (cos x + e x )3

Answers

1. 11. 21. 31.

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

2. 12. 22. 32.

(a) (a) (c) (a)

3. 13. 23. 33.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

(d) (c) (d) (c)

7. (c) 17. (b) 27. (c)

8. (a) 18. (a) 28. (b)

9. (a) 19. (b) 29. (b)

10. (a) 20. (a) 30. (c)

5

Applications of Derivatives

CHAPTER

Summary of conceptS tangentS and normalS geometrical meaning of derivative at a point The derivative of a function f (x) at a point x = a is the slope of the tangent to the curve y = f (x) at the point [a, f (a)]. Consider a curve y = f (x) and a point P(x, y) on this curve. If tangent to the curve at P(x, y) makes an angle θ with the positive dy direction of x-axis, then, at the point P(x, y): = tan θ = m = dx gradient or slope of tangent to the curve at P (x, y).

[ the normal to the curve at the point P(x1, y1) is a line perpendicular to the tangent at the point P (x1, y1) and passing through it. Therefore, slope of normal at P (x1, y1) =

−1 −1 . = Slope of tangent at P ( x1 , y1 )  dy    dx ( x1 , y1 )

Important Results

equation of tangent The equation of a tangent to a curve y = f (x) at a given point P (x1, y1) is given by  dy  (x – x1) y – y1 =    dx  ( x , y ) 1

1. If

dy > 0, the tangent makes an acute angle with the x-axis. dx

2. If

dy < 0, the tangent makes an obtuse angle with the x-axis. dx

3. If

dy = 0, the tangent is parallel to x-axis. dx

1

[Using point slope form of equation of the straight line]

4. If the tangent is perpendicular to x-axis, then

equation of normal The equation of normal to a curve y = f (x) at a given point P (x1, y1) is given by −1 (x – x1) y – y1 =  dy    dx ( x , y ) 1

dy dy = ∞, i.e. = 0. dx dx 5. If the tangent is equally inclined to the axes, then

dy = tan 45º or tan 135º = ± 1. dx

1

angle of Intersection of two curves Let y = f (x) and y = g (x) be two curves intersecting at a point P (x1, y1). Then, the angle of intersection of these two curves is defined as the angle between the tangents to the two curves at their point of intersection.

138

(iii) Subtangent = TM = | y cot θ | = Objective Mathematics

y  dy   dx   

 dy  (iv) Subnormal = MN = | y tan θ | = y   .  dx  If θ is the required angle of intersection, then, θ = ± (θ1 – θ2), where θ1 and θ2 are the inclination of tangent to the curves y = f (x) and y = g (x) respectively at the point P.

IncreaSIng and decreaSIng functIonS (monotonIcIty) Increasing function A function f(x) is said to be an increasing function on an interval I, if x1 < x2 ⇒ f (x1) ≤ f (x2), ∨ x1, x2 ∈ I.

Working rule for finding the angle of intersection 1. Find f ′ (x) and g′ (x). 2. If f ′ (x) × g′(x) = – 1, then the two curves are said to cut each other orthogonally, wherever they cut. 3. If the product is not – 1, solve the equation of the two curves to get their point of intersection. If (α, β) be their point of intersection, then find f ′ (α) and g′ (α). Let m1 = f ′ (α) and m2 = g′ (α). 4. If θ is the angle between the tangents, then tan θ = ±

f ' (α) − g' (α) m1 − m2 =± . 1 + f ' (α) g' (α) 1 + m1m2

Increasing Function Strictly Increasing function A function f (x) is said to be a strictly increasing function on an interval I, if

Repeat this process for other points of intersection. Note: The two curves are said to touch each other at their point of intersection (α, β), if the slope of their tangents at (α, β) are equal.

x1 < x2 ⇒ f (x1) < f (x2), ∨ x1, x2 ∈ I.

length of tangent, length of normal, Sub-tangent and Subnormal Let the tangent and normal at the point P (x, y) on the curve meet the axis of x at the points T and N respectively. Let M be the foot of the ordinate at P. Then, Strictly Increasing Function decreasing function A function f (x) is said to be a decreasing function on an interval I, if x1 < x2 ⇒ f (x1) ≥ f (x2), ∨ x1, x2 ∈ I Decreasing Function Strictly decreasing function A function f (x) is said to be a strictly decreasing function on an interval I, if

(i) Length of the tangent = PT = | y cosec θ |

=

 dy  y 1+    dx  y 1 + cot θ = dy dx

2

x1 < x2 ⇒ f (x1) > f (x2) ∨ x1, x2 ∈ I.

2

(ii) Length of the normal = PN = | y sec θ | =

 dy  y 1 + tan 2 θ = y 1 +    dx 

2

Strictly Decreasing Function

test for monotonicity of functions (i) (ii) (iii) (iv)

f (x) is increasing in [a, b] if f ′ (x) ≥ 0, ∨ x ∈ [a, b]. f (x) is strictly increasing in [a, b] if f ′ (x) > 0, ∨ x ∈ [a, b]. f (x) is decreasing in [a, b] if f ′ (x) ≤ 0, ∨ x ∈ [a, b]. f (x) is strictly decreasing in [a, b] if f ′ (x) < 0, ∨ x ∈ [a, b].

Remarks 1. If a function f (x) is strictly increasing (strictly decreasing) on an interval I, then f –1 exists and is also strictly increasing (strictly decreasing). 2. If f (x) is monotonic on an interval I, then f (x) has at the most one zero in the interval I. 3. If the functions f (x) and g (x), both are increasing or decreasing on an interval I, then the composite function (gof ) (x) is an increasing function on I. 4. If the function f (x) is increasing and g (x) decreasing on an interval I, then the composite function (gof ) (x) is decreasing on the interval I.

(i) The points at which a function attains either the local maximum value or local minimum value are called the extreme points and both local maximum and local minimum values are called the extreme values of the function f (x). (ii) The local maximum and local minimum values are also known as relative maximum and relative minimum values respectively.

Working rule to determine the points of local maxima and local minima method I (first derivative test) 1. For the function y = f (x), find f ′ (x). 2. Put f ′ (x) = 0 and solve this equation for x. Let its roots be a, b, c etc. These points are called stationary points or critical points. 3. At x = a, determine the sign of f ′ (x) for values of x slightly less than a and that for values of x slightly greater than a. (i) If f ′ (x) changes sign from positive to negative as x increases through a, then x = a is a point of maximum.

5. A function may be increasing in some interval I1 and decreasing in some other interval I2.

maxIma and mInIma of functIonS local maximum A function y = f (x) is said to have a local maximum value at a point x = a, if f (x) ≤ f (a), ∨ x ∈ (a – h, a + h), for small h > 0, i.e., f (a) is the greatest of all the values of f (x) in the interval (a – h, a + h).

(ii) If f ′ (x) changes sign from negative to positive as x increases through a, then x = a is a point of minimum.

The point x = a is called a point of local maximum of the function f (x). local minimum A function y = f (x) is said to have a local minimum value at a point x = a, if f (x) ≥ f (a), ∨ x ∈ (a – h, a + h), for small h > 0, i.e., f (a) is the smallest of all the values of f (x) in the interval (a – h, a + h).

(iii)f ′ (x) does not change sign as x increases through a, then x = a is neither a point of maximum nor a point of minimum. Such a point is called a point of inflexion. We repeat this process for other values of x and examine them for maxima or minima.

Method II (Second Derivative Test) 1. For the function y = f (x), find f ′ (x) and f ′′ (x). 2. Put f ′ (x) = 0 and solve this equation for x. Let its roots be a, b, c etc.

3. At x = a (i) if f ′′ (a) < 0, then x = a is a point of local maxima;

The point x = a is called a point of local minimum of the function f (x).

(ii) if f ′′ (a) > 0, then x = a is a point of local minima; (iii) if f ′′ (a) = 0, we cannot say any thing.

139

Remarks

Applications of Derivatives

monotonic function A function f (x) is said to be monotonic on an interval I if it is either increasing or decreasing on I.

140

Objective Mathematics

General Test  In the cases where the second derivative vanishes, the method discussed above fails to give any result. In those cases, we make use of still higher derivatives, and the following working rule proves very useful. If f (x) has a derivative at x = a such that (i) f ′ (a) = f ′′ (a) = f ′′′ (a) = ... = f  n – 1 (a) = 0, and (ii) f n (a) exists and is not zero, then for n odd, f (a) is not an extreme value, while for n even f (a) The theorem states that between two points with equal oris a maximum or minimum value according as f n (a) is negative dinates on the graph of f, there exists atleast one point where the or positive. tangent is parallel to x-axis.

Greatest and Least Values of a Function in a Closed Interval (Absolute Maximum and Absolute Minimum)

Algebraic Interpretation Between two zeros a and b of f (x) (i.e., between two roots a and b of f (x) = 0) there exists atleast one zero of f′ (x).

If f (x) is continuous in an interval [a, b], then greatest or absolute maximum value of f (x) = max. { f (a), f (b), values of f (x) at all Lagrange’s Mean Value Theorem critical points in (a, b)}. Also, least or absolute minimum value of f (x) = min.  If a function f defined on the closed interval [a, b], is {f (a), f (b), values of f (x) at all critical points in (a, b)} If a function is defined and continuous on an interval which (i) continuous on [a, b] and is not a closed interval, then it cannot have any greatest or least (ii) derivable on (a, b), then there exists atleast one real number c between a and b (a < c < b) such that value other than local maximum or local minimum values. f (b) − f ( a ) f ′ (c) = b−a Rolle’s and Lagrange’s Mean Geometrical Interpretation  The theorem states that Value Theorem between two points A and B on the graph of f there exists atleast Rolle’s Theorem one point where the tangent is parallel to the chord AB. If a function f defined on the closed interval [a, b], is (i) continuous on [a, b], (ii) derivable on (a, b) and (iii) f (a) = f (b), then there exists atleast one real number c between a and b (a < c < b) such that f ′ (c) = 0. Geometrical Interpretation  Let the curve y = f (x), which is continuous on [a, b] and derivable on (a, b), be drawn.

MULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. The tangent to the curve y = x3 – 2x2 + x – 2 is parallel to x-axis at the point (a) (1, – 2)  1 50  (c)  , −   3 27  2. The tangent to the curve

(b) (– 1, 2)  1 50  (d)  ,   3 27  x + y = 4 is equally inclined

to the axes at the point (a) (1, – 2) (c) (4, – 4)

(b) (4, 4) (d) (– 4, 4)

3. The tangent to the curve x2 + y2 = 25 is parallel to the line 3x – 4y = 7 at the point

(a) (– 3, – 4) (c) (– 3, 4)

(b) (3, – 4) (d) (3, 4)

4. The minimum value of a tan2x + b cot2x equals the maximum value of a sin2 θ + b cos2 θ where a > b > 0, when (a) a = b (c) a = 3b

(b) a = 2b (d) a = 4b

x  5. The points on the curve y2 = 4a  x + a sin  at which a  the tangent is parallel to x-axis, lie on (a) a straight line (c) a circle

(b) a parabola (d) an ellipse

(a) m < 1 (c) | m | > 1

(b) | m | ≤ 1 (d) None of these

ax 7. If the slope of the curve y = b − x at the point (1, 1) is 2, then the values of a and b are (a) 1, – 2 (c) 1, 2

(b) – 1, 2 (d) None of these

8. If the tangent at each point of the curve 2 3 x – 2ax2 + 2x + 5 y= 3 makes an acute angle with the positive direction of x-axis, then (a) a ≥ 1 (c) a ≤ – 1

(b) – 1 ≤ a ≤ 1 (d) None of these

9. The angle between the tangents to the curve y2 = 2ax at a the points where x = is 2 π π (b) (a) 6 4 π π (c) (d) 3 2 10. The equation of the tangent to the curve y = 9 − 2 x 2 at the point where the ordinate and the abscissa are equal, is (a) 2x + y – 3 3 = 0

(b) 2x + y + 3 3 = 0 (d) None of these

(c) 2x – y – 3 3 = 0 11. The angle between the tangents at those points on the curve y = (x + 1) (x – 3) where it meets x-axis is (a) ± tan–1

15 8

8 (b) ± tan–1    15 

π (d) None of these 4 12. The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point t = 2 is (c) ±

7 1 (b) 6 2 6 (c) (d) 1 7 13. The equation of the tangent to the curve y = 2x2 + 5x at the point where the line y = 3 cuts the curve in the first quadrant is (a)

(a) 14x – 2y – 1 = 0 (c) 14x + 2y – 1 = 0

(b) 14x – 2y + 13 = 0 (d) None of these

2 2 14. The angle between the tangents to the curve x + y 2 2 a b = 1 at the points (a, 0) and (0, b) is

π 4 π (c) 3

(a)

(b)

π 2

(d) None of these

(a) at an angle π 3 (c) orthogonally

π 4 (d) None of these

(b) at an angle

16. The angle of intersection of the parabolas y2 = 4ax and x2 = 4ay at origin is π π (b) 3 2 π (c) (d) None of these 4 17. If a < 0, f (x) = eax + e–ax and S = {x : f (x) is monotonically decreasing}, then S equals (a)

(a) {x : x > 0} (c) {x : x > 1 }

(b) {x : x < 0} (d) {x : x < 1}

18. The two curves y2 = 4x and x2 + y2 – 6x + 1 = 0 at the point (1, 2) (a) intersect orthogonally π (b) intersect at an angle 3 (c) touch each other (d) None of these 19. The critical points of the function f (x) = (x – 2)2/3 (2x + 1) are (a) –1 and 2 (c) 1 and (–1/2)

(b) 1 (d) 1 and 2

20. The angle of intersection of the curves y = 2 sin2x and y π = cos 2x at x = is 6 π (a) (b) π 4 3 (c) π (d) None of these 2 21. An extreme value of the function f (x) = (sin–1 x)3 + (cos–1 x)3, (–1 < x < 1) is (a)

7 π3 8

3 (c) π 32

(b)

π3 8

(d)

π3 16

22. The angle at which the curves y = sin x and y = cos x intersect in [0, π] is (a) ± tan–1 2 2

(b) ± tan–1 2

 1  (c) ± tan–1    2

(d) None of these

α  x log x, x > 0 23. If f (x) =  , find the value of α so that the x=0  0, function obeys the Rolle’s theorem in [0, 1]

(a) – 2 (c) 0

(b) – 1 1 (d) 2

141

15. The curves x3 – 3xy2 = a and 3x2y – y3 = b, where a and b are constants, cut each other

Applications of Derivatives

6. If m be the slope of a tangent to the curve e2y = 1 + 4x2, then

142

n

n

Objective Mathematics

x  y x y 24. The line a + b = 2 touches the curve  a  +  b  = 2 at the point (a, b) for (a) n = 2 only (b) n = – 3 only (c) n is any real number (d) None of these

a 2

a (c) (d) None of these 3 26. Let f (x) and g(x) be defined and differentiable for x ≥ x0 and f (x0) = g(x0), f ′(x) > g′(x) for x > x0, then (b) f (x) = g(x), x > x0 (d) None of these

27. A function f such that f ′(a) = f ′′(a) = f ′′′ (a) = ... = f (2n) (a) = 0 and f has a local maximum value b at x = a, if f (x) is (a) (x – a)2n+2 (c) b – (x – a)2n+2

(b) b – 1 – (x + 1 – a)2n–1 (d) (x – a)2n+2 – b

28. f (x) = x3 + ax2 + bx + 5 sin2 x is an increasing function in the set of real numbers if a and b satisfy the condition (a) a – 3b – 15 > 0 (c) a2 – 3b + 15 < 0 2

(b) a – 3b + 15 > 0 (d) a > 0 b > 0 2

29. The curve y – exy + x = 0 has a vertical tangent at the point (a) (1, 1) (c) (0, 1)

(b) at no point (d) (1, 0)

30. If f (x) satisfies the conditions for Rolle’s theorem in [3,5] then

∫

5

3

f ( x)dx equals to

(a) 2 (c) 0

(b) –1 (d) – 4/3

31. Let f (x) = x3 + bx2 + cx + d, 0 < b2 < c. Then, f (a) is bounded (c) has a local minima

(b) has a local maxima (d) is strictly increasing

32. Tangents are drawn from the origin to the curve y = sin x. Their points of contact lie on the curve (a) x2y2 = x2 + y2 (c) x2y2 = y2 – x2

π 4

35. The points where the normal to the curve makes equal intercepts on the axes are

(b)

(a) f (x) < g(x), x > x0 (c) f (x) > g(x), x > x0

(c) inclined at an angle (d) None of these

25. The length of the perpendicular from the origin to the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ) at any point θ is (a) a

(a) parallel (b) perpendicular

(b) x2y2 = x2 – y2 (d) None of these

33. If the line y = 2x touches the curve y = ax2 + bx + c at the point where x = 1 and the curve passes through the point (– 1, 0), then the values of a, b and c are 1 1 1 1 (a) a = 2 , b = 1, c = 2 (b) a = 1, b = 2 , c = 2 1 1 (c) a = 2 , c = 2 , b = 1 (d) None of these 34. The two tangents to the curve ax2 + 2hxy + by2 = 1, a > 0 at the points where it crosses x-axis, are

 (a) ±    (b) ±  

xy = a + x

a a  , 2a − ± 2a  2 2  a a  , − 2a + ± 2a  2 2 

a  a  , 2a + ± 2a  (c) ±  2  2  (d) None of these 36. The maximum value of (x – p)2 + (x – q)2 + (x – r)2 will be at x equal to p+q+r 3 (c) qpr (a)

(b) 3 qpr 2 2 2 (d) p + q + r

37. The normals to the curve x = a (θ + sin θ), y = a (1 – cos θ) at the points θ = (2n + 1) π, n ∈ I are all (a) parallel to x-axis (b) parallel to y-axis (c) parallel to the line y = x

(d) None of these 38. The value of θ, θ ∈ [0, π/2] for which the sum of intercepts on co-ordinate axes by tangent at point ( 3 3 cos θ, sin θ) x2 + y 2 = 1 is minimum, is: of ellipse 27 (a) π/6 (b) π/4 (c) π/3 (d) π/2 39. The equation of the tangent to the curve 1  2 x≠0  x sin , x at the origin is y=  0, x=0 (a) x = 0 (c) y = 0

(b) x = y (d) None of these

40. Number of possible tangents to the curve y = cos (x + y), –3π ≤ x ≤ 3π, that are parallel to the line x + 2y = 0, is (a) 1 (c) 3

(b) 2 (d) 4

41. The condition that the line lx + my = 1 may be normal to the curve y2 = 4ax is (a) al3 – 2alm2 = m2 (b) al2 + 2alm3 = m2 (c) al3 + 2alm2 = m3 (d) al3 + 2alm2 = m2. 42. The point in the interval [0, 2π], where f (x) = ex sin x has maximum slope, is (a) π/4 (c) π

(b) π/2 (d) 3π/2

44. If the normal at the point “t1” on the curve xy = c2 meets the curve again at “t2”, then 3 (a) t1 t2 = 1

3 (b) t1 t2 = – 1

(c) t t = – 1

(d) t t = 1.

3 1 2

3 1 2

45. The tangent to the curve 3xy2 – 2x2y = 1 at (1, 1) meets the curve again at 16 1 (a)  − , −  20   5

16 1 (b)  , −  20   5

 16 1  (c)  − ,   5 20 

(d) None of these

3  1  2 (1 − x ) 2 − 8 3   3   1  2 2 (b) f (x) = (1 − x ) + 8 3   1 (c) f (x) = – 3 {sin x + 7}

(a) f (x) = –

x+ y

(b) a (d) None of these 2/3

 y +  b

2/3

= 1 meets

the coordinate axes in A and B, where p = OA and q = OB, O being the origin, then the locus of ( p, q) is p q p q = 1 (b) 2 + 2 = 2 + a b b2 a 2 p2 q2 (c) 2 + 2 = 1 (d) None of these a b 49. The part of the tangent to the curve xy = c2 included between the coordinate axes, is divided by the point of tengency in the ratio 2

2

(a)

(a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) None of these 50. The portion of the tangent to the curve a − a2 − y2 a x = a − y + log 2 a + a2 − y2 intercepted between the curve and x-axis, is of length. 2

(b) 2y + x = 0 (d) None of these

52. The equation of the normal to the curve y = e–2 | x | at the 1 point where the curve cuts the line x = is 2 (a) 2e (ex + 2y) = e2 – 4 (b) 2e (ex – 2y) = e2 – 4 (c) 2e (ey – 2x) = e2 – 4 (d) None of these

(a) 9 5 (c)

9 5 4

9 5 2 (d) None of these

(b)

(b) a > 0, b < 0 (d) a < 0, b < 0

55. If y = f (x) be the equation of an ellipse to which the line y = 2x + 3 is a tangent at the point where x = 2, then

x 48. If the tangent to the curve   a

2

(a) 2y – x log 2 = 0 (c) 2y + x log 2 = 0

(a) a > 0, b > 0 (c) a < 0, b > 0

47. The sum of intercepts of the tangent to the curve = a upon the coordinates axes is

2

51. The equation of the normal to the curve y = 1 – 2x/2 at the point of intersection with the y-axis is

54. If the line ax + by + c = 0 is a tangent to the curve xy = 4, then

(d) None of these

2

(d) None of these

53. The area of the triangle formed by the positive x-axis, the tangent and the normal to the circle x2 + y2 = 9 at (2, 5 ) is

46. The function f whose graph passes through the point  7  and whose derivative is x 1 − x 2 , is given by 0 ,   3

(a) 2a (c) 2 2 a

(b) a

143

(a) 3 (b) a (c) no maximum value (d) None of these

a 2 (c) 2a

(a)

(a) f ′ (2) = 2 (b) f (2) = 2 f ′ (2) (c) f (2) + f ′ (2) + f ′′ (2) = 2 (d) None of these 56. The length of the portion of the tangent, at any point on the curve x = a cos3θ, y = a sin3θ, intercepted between the coordinate axes, is (a) a (c) 3a

(b) 2a (d) None of these

57. If the curve y = px2 + qx + r passes through the point (1, 2) and the line y = x touches it at the origin, then the values of p, q and r are (a) p = 1, q = – 1, r = 0 (b) p = 1, q = 1, r = 0 (c) p = – 1, q = 1, r = 0 (d) None of these 58. The length of the perpendicular from the origin to the tangent to the curve y = e4x + 2 drawn at the point x = 0 is (a)

4 17

(b)

(c)

2 17

(d) None of these

3 17

59. The area of the triangle formed by a tangent to the curve 2xy = a2 and the coordinate axes is

Applications of Derivatives

43. If f (x) = a – (x – 3)89, then greatest value of f (x) is

144

(a) 2a2 (c) 3a2

(b) a2 (d) None of these

Objective Mathematics

60. Tangents are drawn to the parabola x2 = 4y at its points of intersection with another parabola y2 = 4x. The point of intersection of the tangents drawn, is given by (a) (2, 0) (c) (– 2, 0)

(b) (0, 2) (d) None of these

π 4 with the straight line y = 3x + 5, then the point of contact is

61. If a tangent to the parabola y2 = 8x makes an angle

1  (a)  , 2  2  1 (c)  , − 2  2 

(b) (0, 0)

62. The set of critical points of (a) {0} (c) {2, –2}

x

π t + 2 2

(d) None of these

(a) decreases on (0, 1) ∪ (2, ∞) (b) increase on (– ∞, 0) ∪ (1, 2) (c) has a local maximum value 0 (d) has a local maximum value 1. (b) k2 = 4 (d) None of these

71. For the curve, x = a cos3θ, y = a sin3θ, at any point θ, is

(b) {0, 2, – 2} (d) {2}

(b) is – 2 3 (d) None of these

64. The area of the triangle, formed by the x-axis and the tangent and the normal to the curve y = 6x – x2 at the point (5, – 5), is 85 (a) 425 (b) 8 8 5 (c) (d) None of these 8 65. If the curve y = x2 + bx + c touches the line y = x at the point (1, 1), then the values of x for which the curve has a negative gradient are 1 (a) x < 2 1 (c) x < – 2

(b)

69. The function f (x) = x2(x – 2)2

(a) k2 = 8 (c) k2 = 2

1 63. If P = x3 – 1 and Q = x – , x ∈ (0, x) then minimum 3 x x value of P/Q2 (a) is 2 3 (c) does not exist

(a) π 4 π (c) + t 4 2

70. The curves y2 = 2x and 2xy = k cut at right angles if

1 (d)  − , 2   2  x2 − 4

68. The angle made by the tangent at any point on the curve x = a (t + sin t cos t), y = a (1 + sin t)2, with x-axis, is

1 (b) x > 2

(d) x > – 1 2 66. The angle subtended by x2 + y2 = 25 at the point (10, 0) is

(a) the length of the tangent is asin2θ (b) the length of the normal is a tan θ sin2θ (c) the length of sub-tangent = a sin2θ cos θ (d) satisfies all the three conditions. 72. The sub-normal at any point of the curve x2y2 = a2 (x2 – a2) varies as (a) (abscissa)– 3 (c) (ordinate)– 3

(b) (abscissa)3 (d) None of these

73. The sub-tangent at any point of the curve xmyn = am+n varies as (a) (abscissa)2 (c) abscissa

(b) (ordinate)2 (d) ordinate

74. The sum of tangent and sub-tangent at any point of the curve y = a log (x2 – a2) varies as (a) abscissa (c) ordinate

(b) product of the coordinates (d) None of these

75. For the curve xm + n = am – n y2n, where a is a positive constant and m, n are positive integers, (a) (sub-tangent)m ∝ (sub-normal)n (b) (sub-normal)m ∝ (sub-tangent)n (c) the ratio of subtangent and subnormal is constant (d) None of these.

76. The length of the normal at any point on the ellipse x2 y 2 + a 2 b 2 = 1 varies as (a) π/6 (b) π/4 (a) (abscissa)3 (c) π/2 (d) π/3 (b) (ordinate)3 (c) the ⊥ r from origin on tangent 67. The intercepts made on the coordinate axes by the tangent (d) None of these m m –1 m at any point on the curve y = ax + x are 2 2 ay 77. The angle between the ellipse x + y = 1 and the circle − ax 2 2 (a) , a b (m − 1) a + mx m (a + y ) x2 + y2 = ab at their point of intersection is ay − ax a−b a+b (b) , (a) tan–1  (b) tan–1    m ( a + x) (m − 1) a + mx  ab   ab  ay ax (c) , (c) tan–1  a − b  (d) None of these (m − 1) a + mx m (a + x)    2 ab  (d) None of these

79. The points where the curves x2 + y2 = 2a2 and 2y2 – x2 = a2 cut orthogonally are  2  4 4  2  a  a (b)  ± a, ± (a)  ± a, ± 3 3 3 3      a 2  ,± a (c)  ± 3  3 

(d) None of these

80. The two curves y = 3x and y = 5x intersect at an angle  log 3 − log 5  (a) tan–1    1 + log 3 ⋅ log 5   log 3 + log 5  (b) tan–1    1 − log 3 ⋅ log 5 

(d) None of these

2 ordinate

(b) ordinate

(c)

2 ordinate

(d) None of these

82. For the parabola y2 = 4ax, the ratio of the sub-tangent to the abscissa is (b) 2 : 1 (d) x2 : y.

83. Let f (x) = cos 2πx + x – [x] ([ . ] denotes the greatest integer function). Then number of points in [0, 10] at which f (x) assumes its local maximum value, is (b) 10 (d) infinite

84. The subtangent, ordinate and subnormal to the parabola y2 = 4ax at a point (different from the origin) are in (b) AP (d) None of these

85. The curve y = ax3 + bx2 + cx + 5 touches the x-axis at A (– 2, 0) and cuts the y-axis at a point B where its slope is 3. The values of a, b and c are 1 3 (a) a = , b = – , c = 3 2 4 1 3 (b) a = – , b = – , c = 3 2 4 1 3 (c) a = , b = , c = 3 2 4 (d) None of these 86. The normal to the curve 5x5 – 10x3 + x + 2y + 6 = 0 at P (0, – 3) meets the curve again at the point (a) (– 1, 1) (c) (– 1, – 5)

88. The tangent to the graph of the function y = f (x) at the π point with abscissa x = 1 forms an angle of and at 6 π and at the point x = 3 the point x = 2 an angle of 3 π an angle of with the positive direction of x-axis. The 4 value of

∫

3

1

f ' ( x) f '' ( x) dx +

continuous) 4 3 −1 3 3

∫

3

2

f '' ( x) dx is f ′′(x) being

(b) 3 3 − 1 2

4− 3 (d) None of these 3 89. The abscissa of the point on the curve xy = (c + x)2, the normal at which cuts off numerically equal intercepts from the axes of coordinates, is (a)

(a)

(a) GP (c) HP

(b) b = {1, 2} (d) None of these

(c)

81. If at any point on a curve the sub-tangent and sub-normal are equal, then the length of the normal is equal to

(a) 0 (c) 9

(a) 1 ≤ b ≤ 2 (c) b ∈ (–∞, –1)

(a)

 log 3 + log 5  (c) tan–1    1 + log 3 ⋅ log 5 

(a) 1 : 1 (c) x : y

 x 3 − x 2 + 10 x − 5, x ≤ 1 87. Let f (x) =  2 −2 x + log 2 (b − 2) x > 1 The set of values of b for which f (x) has greatest value at x = 1, is given by

(b) (1, – 1) (d) (– 1, 5)

c 2

(b) c

2 c

(d) –

c 2 90. The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 2 = 0 (c)

(a) cut at right angles (b) touch each other π (c) cut at an angle 3 π (d) cut at an angle 4 91. The intercept made by the tangent to the curve y =

x

∫ | t | dt ,

which is parallel to the line y = 2x, on

0

x-axis is equal to (a) 1 (c) 2

(b) – 1 (d) – 2.

x2 y 2 = 1 where the tangent is + 4 16 equally inclined to the axes is

92. A point on the curve

 2 −8  , (a)    5 5

 2 8  , (b)    5 5

 −2 8  (c)  (d) all the above ,   5 5 93. The area bounded by the tangent to the curve y = loge x at the point (2, 0) and the coordinate axes is 1 sq. unit 2 (c) 1 sq. unit

(a)

(b) 2 sq. unit (d) None of these

145

(a) set of critical points is {–1/2, 0, 1/2} (b) f (x) is increasing in (–∞, –1/2] ∪ (0, 1/2] (c) f (x) is decreasing in [–1/2, 0) ∪ [1/2, ∞) (d) None of these

Applications of Derivatives

78. For the function f (x) = 2x2 – ln | x |

146

94. Any tangent to the curve y = 3x7 + 5x + 3

Objective Mathematics

(a) is parallel to x-axis (b) is parallel to y-axis (c) makes an acute angle with the x-axis (d) makes an obtuse angle with the x-axis 95. The function f (x) = cot–1 x + x increases in the interval (a) (1, ∞) (c) (–∞, ∞)

(b) (–1, ∞) (d) (0, ∞)

8 (c)  4, −   3

(d) None of these

104. The line

8  3

x y + = 1 touches the curve y = be–x/a at a b

(a) (– a, ba)

f (x) = (x + 1)1/3 – (x – 1)1/3 on [0, 1] is

b (c)  a,   a

(b) 2 (d) 1/3

97. The equation of the normal to the curve y = (1 + x) y + sin–1 (sin2x) at x = 0 is (a) x + y = 2 (c) x – y = 1

(b)  −4, 

the point

96. The greatest value of (a) 1 (c) 3

(a)  4, 8   3

(b) x + y = 1 (d) None of these

105. For the curve x = t2 – 1, y = t2 – t, the tangent is parallel to x-axis where (a) t =

1 3

(c) t = 0

98. If the function

 a (b)  a,   b (d) None of these

(b) t = –

1 3

(d) t = 1

2 f (x) = 2x3 – 9ax2 + 12a2x + 1, 106. For the curve x = 3 cos θ, y = 3 sin θ, 0 ≤ θ ≤ π, the where a > 0, attains its maximum and minimum at p and q tangent is parallel to the x-axis when respectively such that p2 = q, then a equals (a) θ = π (b) θ = 0 (a) 3 (b) 1 π π (c) θ = (d) θ = 1 (c) 2 (d) 3 2 2 107. The length of the subtangent to the curve x2 + xy + y2 99. On the ellipse 4x2 + 9y2 = 1, the points at which the = 7 at (1, 3) is tangents are parallel to the line 8x = 9y are (a) 3 (b) 5 2 1 (c) 15 (d) 3/5 (a)  2 , 1  (b)  − ,   5 5 5 5 108. The curve y – exy + x = 0 has a vertical tangent at the point 2 1 2 1 (c)  − , −  (d)  , −  .  5 5 5 (a) (1, 1) (b) at no point 5 (c) (1, 0) (d) (0, 0) 100. If the normal to the curve y = f (x) at the point (3, 4) 3 2 109. The curve y = ax + bx + cx is inclined at 45º to x-axis makes an angle 3π with the positive x-axis, then f ′(3) = at (0, 0) but it touches x-axis at (1, 0), then the values 4 of a, b, c are given by 3 (a) – 1 (b) – (a) a = 1, b = – 2, c = 1 4 4 (b) a = 1, b = 1, c = – 2 (c) 3 (d) 1. (c) a = – 2, b = 1, c = 1 101. A curve y = f (x) passes through the point P (1, 1). The normal to the curve at P is a ( y – 1) + (x – 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is (a) y = ea (x–1) (c) y = e

a ( x −1) 2

(b) y = ea (1–x)

(d) y = e

a ( x +1) 2

.

102. If the line ax + by + c = 0 is a normal to the curve xy = 1, then (a) a > 0, b > 0 (b) a > 0, b < 0 (c) a < 0, b > 0 (d) a < 0, b < 0 103. The point on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes is

(d) a = – 1, b = 2, c = 1 π   x sin , x > 0 x 110. If f (x) =  , then in the interval (0, 1), f ′ (x) 0, x=0 vanishes at (a) exactly one point (b) exactly two points (c) at no point (d) infinite number of points 111. If f (x) and g (x) are differentiable functions for 0 ≤ x ≤ 1 such that f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2, then in the interval (0, 1), (a) f ′ (x) = 0 for all x (b) f ′ (x) = 2g′ (x) for atleast one x (c) f ′ (x) = 2g′ (x) for atmost one x (d) None of these

n – 1

(a) exactly one root (c) atleast one root

2

(b) atmost one root (d) no root

113. If a + b + c = 0, then the equation 3ax2 + 2bx + c = 0 has, in the interval (0, 1) (a) atleast one root (c) no root

(b) atmost one root (d) None of these

114. If α and β (α < β) be two different real roots of the equation ax2 + bx + c = 0, then (a) α > –

b 2a

(b) β < –

b 2a

b < β 2a

(d) β < – b < α 2a n n – 1 + ... + a1x = 0 has a 115. If the equation anx + an – 1x positive root x = α, then the equation (c) α < –

nanxn – 1 + (n – 1) an – 1xn – 2 + ... + a1 = 0 has a positive root, which is (a) smaller than α (c) equal to α

(b) greater than α (d) greater than or equal to α.

sin x sin a sin b π 116. If f (x) = cos x cos a cos b , where 0 < a < b < , 2 tan x tan a tan b then the equation f ′ (x) = 0 has, in the interval (a, b) (a) atleast one root (c) no root

(b) atmost one root (d) None of these

121. If the function f (x) and g (x) are continuous in [a, b] and differentiable in (a, b), then the equation f (a ) f (b) = (b – a) g (a ) g (b) interval [a, b],

f (a) g (a)

f ' ( x) g' ( x)

has, in the

(a) atleast one root (b) exactly one root (c) atmost one root (d) no root 1 for all x and f (0) = 0, then 1 + x2

122. If f ′ (x) =

(a) f (2) < 0.4 (c) 0.4 < f (2) < 2

(b) f (2) > 2 (d) f (2) = 2

123. If a, b, c be non-zero real numbers such that 1

∫ (1 + cos 0

=

8

x)(ax 2 + bx + c) dx

2

∫ (1 + cos

8

x)(ax 2 + bx + c) dx = 0,

0

then the equation ax2 + bx + c = 0 will have (a) one root between 0 and 1 and other root between 1 and 2 (b) both the roots between 0 and 1 (c) both the roots between 1 and 2 (d) None of these 124. The value of c in Lagrange’s theorem for the function f (x) = log sin x in the interval  π , 5π  is 6 6    (a)

π 4

(b)

π 2

2π (c) (d) None of these a0 a a a + 1 + 2 + ... + n −1 + an = 0, then the equation 3 2 n +1 n n −1 a0xn + a1xn – 1 + ... + an – 1x + an = 0 has, in the interval 125. The value of c in Lagrange’s theorem for the function  1 (0, 1),  x cos   , x ≠ 0 f (x) = in the interval [– 1, 1] is  x  (a) exactly one root (b) atleast one root  x=0 (c) atmost one root (d) no root 0, (a) 0 (b) 1 118. The equation x log x = 3 – x has, in the interval (1, 3), 2 (a) exactly one root (b) atmost one root 1 (c) – (d) non existent in the interval (c) atleast one root (d) no root. 2 119. Between any two real roots of the equation ex sin x = 1, 126. If 27a + 9b + 3c + d = 0, then the equation the equation ex cos x = – 1 has 4ax3 + 3bx2 + 2cx + d = 0 has atleast one real root lying (a) atleast one root (b) exactly one root between (c) atmost one root (d) no root (a) 0 and 1 (b) 1 and 3 120. If f (x) is differentiable in the interval [2, 5], where (c) 0 and 3 (d) None of these 1 1 f (2) = and f (5) = , then there exists a number c, 127. Let f be a function which is continuous and differenti5 2 able for all real x. If f (2) = – 4 and f ′ (x) ≥ 6 for all 2 < c < 5 for which f ′ (c) = x ∈ [2, 4], then 1 1 (a) f (4) < 8 (b) f (4) ≥ 8 (b) (a) (c) f (4) ≥ 12 (d) None of these 2 5 2 1 1 28. The function f (x) = 2x – log | x |, x ≠ 0 is increasing (c) (d) None of these in the interval 10

117. If

147

anx + an – 1x + ... + a2x + a1x + a0 = 0, n positive integer, has two different real real roots α and β, then between α and β, the equation nanxn – 1 + (n – 1) an – 1xn – 2 + ... + a1 = 0 has n

Applications of Derivatives

112. If the polynomial equation

148

1   (a)  − 1 , 0  ∪  1 , ∞  (b)  −∞, −  ∪  0,  2   2  2 

Objective Mathematics

1 1 (c)  − ,   2 2

1  2

(d) None of these

129. The set of values of x for which log (1 + x) < x, is (a) x < 0 (c) 0 < x < 1

(b) x > 0 (d) None of these

130. The values of x for which 1 + x ln (x + (a) x ≤ 0 (c) x ≥ 0

x2 + 1 ) ≥

1 + x 2 , are (b) 0 ≤ x ≤ 1 (d) None of these

131. If 0 < x < π , then 2 x2 (a) cos x < 1 – 2

x2 (b) cos x > 1 – 2

x 2

x 2

2

(c) cos x = 1 –

(d) cos x ≥ 1 –

2

132. The function f (x) = x – log (1 + x), x > – 1 is increasing in the interval (a) (0, ∞) (c) (– ∞, 0)

(b) (– 1, 0) (d) None of these

133. The range of values of x for which the function x , x > 0 and x ≠ 1, may be decreasing, is log x (a) (0, e) (b) (e, ∞) (c) (0, e)\{1} (d) None of these f (x) =

134. The function f (x) = x1/x is increasing in the interval (a) (e, ∞) (b) (– ∞, e) (c) (– e, e) (d) None of these 135. log x – tan–1x increases in the interval (a) (– ∞, 0) (c) (– ∞, ∞)

(b) (0, ∞) (d) None of these

136. The real number x when added to its inverse gives the minimum value of the sum at x equal to (a) 2 (c) –1 137. The function f (x) = val π (a)  − , 0   2  (c) (0, π) π , then 2 2 (a) > sin x π x sin x (c) < 1 x

138. If 0 < x
cos x (b) cos (sin x) < cos x (c) cos (sin x) > sin (cos x) (d) cos (sin x) < sin (cos x) 140. (1 + x)p ≤ 1 + xp, where (a) p > 1 (c) x > 0

(b) 0 ≤ p ≤ 1 (d) x < 0

141. If ax2 + b ≥ c for all positive x, where a, b > 0, x then (a) 27ab2 ≥ 4c3 (c) 4ab2 ≥ 27c3

(b) 27ab2 < 4c3 (d) None of these

b ≥ c for all positive x, where a, b > 0, x

142. If ax + then

c2 c2 (b) ab ≥ 4 4 c (d) None of these (c) ab ≥ 4 2x 143. The function f (x) = log x – is increasing in the 2+ x interval (a) ab
2 5 145. The function f (x) = – 2x3 + 21x2 – 60x + 41, in the interval (– ∞, 1), is (b) ≤ 0 (d) ≥ 0

(a) < 0 (c) > 0 146. If 0 < α < β < α (a) tan β < β tan α tan α α (c) < tan β β

π , then 2 tan β α > tan α β tan α α (d) > tan β β (b)

sin x is decreasing in the interx 147. A point on the parabola y 2 = 18x at which the ordinate increases at twice the rate of the abscissa is  π 0 , (b)   (a)  −9 , 9  (b) (2, – 4)  2  8 2 (d) None of these 9 9 (c) (2, 4) (d)  ,  8 2 2 –x 148. Let y = x e , then the interval in which y increases 2 sin x (b) < w.r.t. x, is : π x (d) sin x > 1 x

(a) (– ∞, ∞) (c) (2, ∞)

(b) (– 2, 0) (d) (0, 2)

(a) x > 0 (c) x > 1

(b) x < 0 (d) x < 1

150. If y = 2x + arc cot x + ln [ 1 + x 2 – x], then y (a) increases in [0, ∞[ (b) decreases in [0, ∞[ (c) neither increases nor decreases in [0, ∞[ (d) increases in ] – ∞, 0] 151. The function f defined by f (x) = (x + 2) e–x is (a) decreasing for all x (b) decreasing in (– ∞, – 1) and increasing (– 1, ∞) (c) increasing for all x (d) decreasing in (– 1, ∞) and increasing in (– ∞, – 1) 152. The function f (x) =

ln ( π + x) is ln (e + x)

(a) increasing on (0, ∞) (b) decreasing on (0, ∞) (c) increasing on (0, π/e), decreasing on (π/e, ∞) (d) decreasing on (0, π/e), increasing on (π/e, ∞) 153. The function f (x) = tan x – x (a) somtimes increases and sometimes decreases (b) never increases (c) never decreases (d) can’t say. 154. Let the function f (x) = sin x + cos x, be defined in [0, 2π], then f (x) π π (a) increases in  ,  4 2  π 5π  (b) decreases in  ,  4 4 

1 (b) g (x) increases in  0,   2 1 (c) g (x) decreases in  , 1 2  1 (d) g (x) decreases in  , ∞  2  158. The function f (x) =

4 sin x − 2 x − x cos x , 0 < x < 2π 2 + cos x

 π  3π  (a) increases in  0,  ∪  , 2π   2  2   π 3π  (b) increases in  ,  2 2   3π  π (c) decreases in  0,  ∪  , 2π  2 2     π 3π (d) decreases in  ,  2 2  159. A function y = f (x) has a second order derivative f ′′(x) = 6(x – 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x – 5, then the function is (b) (x – 1)3 (d) (x + 1)2

| x −1| x2 (a) increases in (– ∞, 0) ∪ (1, 2) (b) increases in (0, 1) ∪ (2, ∞) (c) decreases in (0, 1) ∪ (2, ∞) (d) decreases in (– ∞, ∞) ∪ (1, 2)

160. The function f (x) =

 π π  (d) decreases in 0,  ∪  , 2π  4 2     –1

(sin x + cos x), be defined

 π  5π  (a) increases in 0,  ∪  , 2π    4  4 π 5 π   (b) decreases in  ,  4 4   5π  π (c) increases in  0,  ∪  , 2π  4 4      π 5π  (d) decreases in  4 , 4  156. The function f (x) = | x + 2 | + | x – 1 | is (a) increasing in (1, ∞) (b) increasing in [1, ∞) (c) decreasing in (– ∞, – 2] (d) decreasing in (– ∞, – 2)

1 (a) g (x) increases in  −∞,  2 

(a) (x + 1)3 (c) (x – 1)2

 5π 7 π   π (c) increases in 0,  ∪  ,  4  4 4  

155. Let the function f (x) = tan in [0, 2π], then f (x)

161. The function f (x) = 3cos4x + 10cos3x + 6cos2x – 3, 0 ≤ x ≤ π is (a) increasing in  π , 2π  2 3  (b) increasing in  0, π  ∪  2 π , π      2  3 (c) decreasing in  0, π  ∪  2 π , π   2  3  π 2π  (d) decreasing in  ,  2 3  162. The values of k for which the function f (x) = kx3 – 9x2 + 9x + 3 may be increasing on R are (a) k > 3 (b) k < 3 (c) k ≤ 3 (d) None of these 163. The least possible value of k for which the function f (x) = x2 + kx + 1 may be increasing on [1, 2] is

149

157. If g (x) = f (x) + f (1 – x) and f ′′ (x) < 0 for 0 ≤ x ≤ 1, then

Applications of Derivatives

149. If a < 0, the function (eax + e–ax) is a monotonic decreasing function for all values of x, where

150

(a) 2 (c) 0

(b) – 2 (d) None of these

Objective Mathematics

164. If f (x) = 2x3 + 9x2 + λx + 20 is a decreasing function of x in the largest possible interval (– 2, – 1) then λ is equal to (a) 12 (c) 6

(b) – 12 (d) None of these

165. The function f (x) = sin4x + cos4x increases in the interval π 3π  (b)  ,  4 8 

(a)  0, π   8

 5π 3π  3π 5π  (c)  , (d)  ,  8 4   8 8  166. Let f (x) = cot–1 [g (x)], where g (x) is an increasing function for 0 < x < π. Then f (x) is (a) increasing in (0, π) (b) decreasing in (0, π) (c) increasing in  0, π  and decreasing in  π , π   2 2  (d) None of these 167. Let f ′ (x) > 0 and g′ (x) < 0 for all x ∈ R. Then, (a) f  [ g (x)] > f  [g (x + 1)] (b) f [g (x)] > f [g (x – 1)] (c) g [  f (x)] > g [ f (x + 1)] (d) g [  f (x)] > g [  f (x – 1)] 168. If the function f (x) = 3 cos | x | – 6ax + b increases for all x ∈ R, then the range of values of a is given by (a) a > – 1 2 (c) a ≤ b

(b) a < –

1 2

(d) a ≥ b

169. The function f (x) = 2 log (x – 1) – x2 + 2x + 3 increases in the interval (a) (– ∞, 0) ∪ (1, 2) (c) (0, 1) ∪ (2, ∞)

(b) (– ∞, 0) ∪ (2, ∞) (d) None of these

170. The equation x + ex = 0 has (a) only one real root (c) no real root

(b) only two real roots (d) None of these

171. The interval in which the function 2x3 + 15 increases less rapidly than the function 9x2 – 12x, is (a) (– ∞, 1) (c) (2, ∞)

(b) (1, 2) (d) None of these

172. Let h (x) = f (x) – [ f (x)] + [ f (x)] for every real number x. Then 2

3

(a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general 173. Consider the following statements S and R:

Which of the following is true? (a) Both S and R are wrong (b) Both S and R are correct, but R is not the correct explanation for S (c) S is correct and R is the correct explanation for S (d) S is correct and R is wrong 174. Let f (x) = interval

∫ e ( x − 1)( x − 2) dx . Then f decreases in the x

(a) (– ∞, – 2) (c) (1, 2)

(b) (– 2, – 1) (d) (2, + ∞)

175. The value of a in order that f (x) = sin x – cos x – ax + b decreases for all real values is given by (a) a ≥ 2 (c) a ≥ 1

(b) a < 2 (d) a < 1

3 x 2 + 12 x − 1, − 1 ≤ x ≤ 2 176. If f (x) =  , then 2< x≤3 37 − x, (a) f (x) is increasing on [– 1, 2] (b) f (x) is continuous on [– 1, 3] (c) f ′ (2) does not exist (d) all of these 177. Let f and g be increasing and decreasing functions respectively from [0, ∞) to [0, ∞). Let h (x) = f (g (x)). If h (0) = 0, then h (x) is (a) always zero (c) always positive

(b) always negative (d) strictly increasing

178. The normal to the curve x = a(1 + cosθ), y = a sin θ at θ always passes through the fixed point (a) (0, 0) (c) (a, 0)

(b) (0, a) (d) (a, a)

179. Given that f ′ (x) > g′ (x) for all real x and f (0) = g (0), then (a) f (x) > g (x) ∨ x ∈ (0, ∞) (b) f (x) < g (x) ∨ x ∈ (– ∞, 0) (c) f (x) < g (x) ∨ x ∈ (0, ∞) (d) f (x) > g (x) ∨ x ∈ (– ∞, 0) 180. If f ′′ (x) < 0 ∨ x ∈ (a, b), then f ′ (x) = 0 (a) exactly once in (a, b) (b) atmost once in (a, b) (c) atleast once in (a, b) (d) None of these 181. The maximum value of x1/x, x > 0 is (a) e1/e (c) 1 182. The minimum value of (a) e

e

(b)  1  e (d) None of these x is log x 1 e (d) None of these

(b)

S: Both sin x and cos x are decreasing functions in the (c) 1 interval  π , π    2  183. The maximum value of sin x + 1 sin 2x + 1 sin 3x, 2 3 R: If a differentiable function decreases in an interval π 0≤x≤ , is (a, b), then its derivative also decreases in (a, b). 2

(c) 1 – 2 2 (d) None of these 2 4 184. If P (x) = a0 + a1x + a2x + ... + anx2n be a polynomial in x ∈ R with 0 < a1 < a2 ... < an , then P (x) has (a) no point of minimum (b) only one point of minimum (c) only two points of minimum (d) None of these 185. If y = a loge x + bx2 + x has its extreme values (i.e., maximum or minimum value) at x = 1 and x = 2, then the values of a and b are

1 4 ,b= 6 3 4 1 (c) a = , b = – 3 6

(a) a = –

(b) a = –

4 ,b= 1 3 6

(d) None of these

P (2, – 1), then the values of a and b are (b) a = 0, b = – 1 (d) a = – 1, b = 0

sin ( x + a ) 187. If y = ; a ≠ b, then y has sin ( x + b) (a) maximum at x = 0 (b) minimum at x = 0 (c) neither maximum nor minimum (d) None of these 1 , x (a) x = 1 is a point of maximum (b) x = – 1 is a point of minimum (c) maximum value > minimum value (d) maximum value < minimum value.

188. For the function y = x +

(a) y must be maximum at x = a (b) y must be minimum at x = a (c) y may not have a maximum or minimum at x = a (d) it is a constant function 195. The fraction exceeding its pth power by the greatest number possible, where p ≥ 2, is 1 (a)    p

1

p −1



(c) p1/p – 1

ax + b 186. If the function y = has an extremum at ( x − 4)( x − 1) (a) a = 0, b = 1 (c) a = 1, b = 0

(a) maximum at x =1 7 (b) minimum at x = 5 (c) neither maximum nor minimum at x = 2 (d) all the above d2y dy 194. For a function y = f (x), if = 0 and = 0 at a dx 2 dx point x = a, then

1 (b)    p

p −1



(d) None of these

196. If y = x2 + ax + b has a minimum at x = 3 and the minimum value is 5, then the values of a and b are (a) a = 6, b = – 14 (c) a = 14, b = – 6

(b) a = – 6, b = 14 (d) a = – 14, b = 6

197. The greatest value of the function f (x) = xe–x in [0, ∞) is (a) 0 (b) 1 e (c) – e (d) None of these 198. If 2a + 3b + 6c = 0, then at least one root of the equation ax2 + bx + c = 0 lies in the interval (a) (2, 3) (c) (0, 1)

(b) (1, 2) (d) (1, 3)

199. The greatest value of the function 1  1  f (x) = tan–1x – log x in  , 3  is 2  3  π 1 (a) π + 1 log 3 (b) − log 3 3 4 6 4 189. If xz = 1, where x > 0, then the least value of x + z π 1 π 1 is (c) − log 3 (d) + log 3 3 4 6 4 (a) 1 (b) 2 200. The maximum value of the function (c) – 2 (d) None of these y = x (x – 1)2, 0 ≤ x ≤ 2 is 190. The minimum value of loga x + logx a, 0 < x < a, is 4 (a) 0 (b) (a) 1 (b) 2 27 (c) – 2 (d) None of these (c) – 4 (d) None of these 191. The minimum value of 2 log10 x – logx .01, x > 1, is n2 (a) 1 (b) – 1 201. The largest term in the sequence xn = n3 + 200 , n ∈ N, (c) 2 (d) None of these is 49 8 192. For a differentiable curve y = f (x) having atleast two (a) (b) extremum in the interval [a, b], 543 89 1 (a) two of its maximum values occurs successively (c) (d) None of these 52 (b) two of its minimum values occur successively 202. The shortest distance of the point (0, 0) from the curve (c) maximum and minimum values occur alternatively 1 (d) None of the above (ex + e–x) is y = 2 193. The function x (a) 2 (b) 1 f (x) = ∫  2 (t − 1)(t − 2)3 + 3 (t − 1) 2 (t − 2) 2  dt has (c) 3 (d) None of these 1

151

(b) 2 2 – 1

Applications of Derivatives

(a) 1 + 2 2

152

203. The greatest height of the graph of the curve y = 6 cos x – 8 sin x above the x-axis is

Objective Mathematics

(a) 4 (c) 10

(b) 8 (d) None of these

204. The minimum value of a2sec2x + b2cosec2x, 0 < a < b, is (a) a + b (c) (a + b)4

(b) (a + b)2 (d) None of these

205. The points on the curve xy2 = 1 which are nearest to the origin are 1/ 3 −1/ 6  1 1/ 3    (a)  1  , ±  1   (b)   , 2−1/ 6   2    2    2   1/ 3  1  (c)  2 , ±   2 

−1/ 6

   

(d) None of these.

206. N characters of information are held on magnetic tape, in batches of x characters each; the batch processing time is α + βx2 seconds; α, β are constants. The optimum value of x for fast processing is β α (a) (b) α β (c)

α β

(d)

β α

207. AB is a diameter of a circle and C is any point on the circumference of the circle, then (a) area of ∆ABC is maximum when it is an isosceles (b) area of ∆ABC is minimum when it is an isosceles (c) the perimeter of ∆ABC is minimum when it is isosceles (d) the perimeter of ∆ABC is maximum when it is isosceles. 208. If xy = k and z = lx + my, where k, l, m are positive constants, then the minimum value of z is (a) mlk (c) 2mlk

(b) 2 mlk (d) None of these

209. The number of values of x where the function f (x) = 2 (cos 3x + cos 3 x) attains its maximum is (a) 1 (c) 0

(b) 2 (d) infinite

210. If the slope of the tangent to the curve y = excos x is minimum at x = a, 0 ≤ a ≤ 2π, then the value of a is (a) 0 (c) 2π

(b) π (d) None of these

π , then cos α ⋅ cos β has a maximum value 2 at β equal to π π (a) 4 (b) 2 π (c) 6 (d) None of these

211. If α + β =

212. Let f (x) = 1 + 3x2 + 32x4 + ... + 330 ⋅ x60. Then f (x) has

(a) atleast one maximum (b) exactly one maximum (c) atleast one minimum (d) exactly one minimum. 213. Let the function f (x) be defined as: −1 2  tan α − 3 x , 0 < x < 1 f (x) =  x ≥1 − 6 x, f (x) can have a maximum at x = 1 if the value of α is (a) 0 (b) 2 (c) 1 (d) None of these

214. If the roots of the equation x3 – ax2 + 4x – 8 = 0 are real and positive, then the minimum value of a is (a) 2 (b) 6 (d) None of these (c) 3 3 4 215. Let f (x) = (x – 2)2 xn, n ∈ N. Then f (x) has a (a) minimum at x = 2, ∨ n ∈ N (b) minimum at x = 2 if n is even (c) minimum at x = 0 if n is even (d) minimum at x = 0 if n is odd. 216. For the function f (x) =

x

sin t dt , where x > 0, t 0

∫

(a) maximum occurs at x = nπ, n even (b) minimum occurs at x = nπ, n odd (c) maximum occurs at x = nπ, n odd (d) minimum occurs at x = nπ, n even. 217. For the function f (x) =

x

∫e

−t 4

4

(4 − t 2 ) dt ,

2

(a) maximum occurs at x = 2 (b) minimum occurs at x = – 2 (c) maximum occurs at x = – 2 (d) minimum occurs at x = 2 218. A function f is such that f ′ (4) = f ′′ (4) = 0 and f has minimum value 10 at x = 4. Then f (x) = (a) 4 + (x – 4)4 (c) (x – 4)4

(b) 10 + (x – 4)4 (d) None of these

219. The difference between the greatest and the least value of the function f (x) =

x

∫ (6t

2

− 24) dt on [1, 3] is

0

(a) 14 (c) 4

(b) 10 (d) None of these

220. The range of values of k for which the function f (x) = (k2 – 7k + 12) cos x + 2 (k – 4) x + log 2 does not possess critical points, is (a) (1, 5) (c) (1, 4)

(b) (1, 5) – {4} (d) None of these

221. The range of values of k for which the function f (x) = (2k – 3) (x + tan 2) + (k – 1) (sin4x + cos4x) does not possess critical points, is 4 (a)  −∞,  ∪ (2, ∞) 3 

4 (b)  , 2  3 

4 (c)  , ∞  3 

(d) (2, ∞)

223. The minimum value of the function f (x) = 2 | x – 2 | + 5 | x – 3 |, ∨ x ∈ R is (a) 3 (c) 5

(b) 2 (d) 7

224. The maximum value of x2/3 + (x – 2)2/3 is (a) 0 (c) 22/3

(b) 2 (d) None of these

225. The minimum value of the function x p x−q f (x) = , where 1 + 1 = 1, p > 1 is + p q p q (a) 1 (c) 2

(b) 0 (d) None of these

226. If (x – a)2n (x – b)2m + 1, where m and n are positive integers and a > b, is the derivative of a function f, then (a) x = a gives neither a maximum nor a minimum (b) x = a gives a maximum (c) x = b gives a minimum (d) x = b gives neither a maximum nor a minimum 227. On the interval [0, 1], the function x25 ⋅ (1 – x)75 takes its maximum value at the point

(b)  –6 (d)  0

232. If the function f(x) = kx3 – 9x2 + 9x + 3 is monotonically increasing in every interval, then (a)  k < 3 (c)  k > 3

(b)  k ≤ 3 (d)  k ≥ 3

233. The function f(x) = 2 + 4x2 + 6x4 + 8x6 has (a)  only one maxima (b)  only one minima (c)  no maxima and minima (d)  many maxima and minima 234. The length of the largest interval in which the function 3 sin x – 4 sin3 x is increasing, is (b)   π (a)   π 3 2 3 π (c)   (d)  π 2 235. The maximum value of x1/x is (a)  1/ee (c)  e1/e

(b)  e (d)  1/e

236. The function f defined by f(x) = 4x4 – 2x + 1 is increasing for (a)  x < 1 (c)  x < 1/2

(b)  x > 0 (d)  x > 1/2

237. The function f ( x) = x + 2 has a local minimum at 2 x (a)  x = – 2 (b)  x = 0 (c)  x = 1 (d)  x = 2

1 4 1 (c) 1 (d) 238. Angle between the tangents to the curve y = x2 – 5x + 3 2 6 at the points (2, 0) and (3, 0) is 228. The value of k so that the sum of the cubes of the (a)  π/2 (b)  π/6 roots of the equation x2 – kx + (2k – 3) = 0 assumes (c)  π/4 (d)  π/3 the minimum value, is x 239. The function x is increasing, when (a) k = 1 (b) k = 3 (b)   x < 1 (a)   x > 1 (c) k = 0 (d) None of these e e 1 (c)  x < 0 (d)  for all real x 229. Let f (x) = sin x + cos 2x. Then 2 240. For what value of a, f(x) = – x3 + 4ax2 + 2x – 5 is 4 3 decreasing ∀ x. (a) min f (x)  π  π 3 4 x∈ 0 ,  x∈ 0 ,  (a)  (1, 2) (b)  (3, 4)  2  2 (c)  R (d)  no value of a 2 x2 + 1 (c) min f (x) >   (d) min f (x) < 3 −t 2  π  π 241. If f ( x) = ∫ 2 e dt , then f(x) increases in 3 x∈ 0 ,  x∈ 0 ,  2 x  2  2 (a) 0

(b)

230. A private telephone company serving a small community makes a profit of Rs. 12 per subscriber, if it has 725 subscribers. It decides to reduce the rate by a fixed sum for each subscriber over 725, thereby reducing the profit by 1 paise per subscriber. Thus, there will be profit of Rs. 11.99 on each of the 726 subscribers, Rs. 11.98 on each 727 subscribers etc. The number of subscribers which will give the company the maximum profit, is (a) 961 (c) 963

(b) 962 (d) None of these

231. If the curres ay + x2 = 7 and x3 = y cut orthogonally at (1,1), then a is equal to

(a)  (–2, 2) (c)  (0, ∞)

(b)  no value of x (d)  (–∞, 0)

242. The function x5 – 5x4 + 5x3 – 1 is (a)  neither maximum nor minimum at x = 0 (b)  maximum at x = 0 (c)  maximum at x = 1 and minimum at x = 3 (d)  minimum at x = 0 x 243. The maximum value of f ( x) = on [–1, 1] is 4 + x + x2 (a)   − 1 (b)   − 1 3 4 (c)   1 (d)   1 5 6

153

(a)  1 (c)  6

Applications of Derivatives

222. The minimum value of e( x4 − x3 + x2 ) is (a) e (b) e2 (c) 1 (d) None of these

154

Objective Mathematics

244. A value of c for which the conclusion of mean value theorem holds for the function f(x) = loge x on the interval [1, 3] is (b)   1 log e 3 (a)  2log3 e 2 (c)  log3 e (d)  loge 3 245. The function f(x) = tan–1 (sin x + cos x) is an increasing function in (a)  (π/4, π/2) (c)  (0, π/2)

(b)  (–π/2, π/4) (d)  (–π/2, π/2)

246. The tangent to the curve y = e drawn at the point (c, ec) intersects the line joining the points (c –1, ec –1) and (c + 1, ec + 1) x

(a)  on the left of x = c (b)  on the right of x = c (c)  at no point (d)  at all points 247. If the Mean Value theorem is f(b) – f(a)  Then for the function x2 – 2x + 3 in 1,  c is (a)  6/5 (b)  5/4 (c)  4/3 (d)  7/6

= (b – a) f ′ (c). 3 , the value of 2 

p and maxima at – 3

p 3

p and maxima at 3 p p (c)  The cubic has minima at both and – 3 3 p p (d)  The cubic has maxima at both and – 3 3

p 3

(a)  The cubic has minima at (b)  The cubic has minima at –

256. The minimum value of 2log10 x – logx 0.01, x > 1 is…. (a)  2 (c)  6

(b)  4 (d)  8

257. In between any two real roots of an ex sinx = 1 there exists how many roots satisfying equation ex cosx = – 1 (a)  at least one root (c)  at most one root

(b)  no root (d)  none of these.

258. If the curves x2 = 9A(9 – y) and x2 = A(y + 1) intersect orthogonally, then the value of A is

(a)  3 (b)  4 (c)  5 (d)  7 248. The equation of the curve which passes through the point (0, 1) and has the slope 3x2 + 2x + 5 at any point 259. The total number of local maxima and local minima of (x, y) is (2 + x)3 , −3 < x ≤ −1 the function f(x) =  2/3 is (a)  y = 3x3 + 2x2 + 5x + 1 (b)  y = 2x3 + 3x2 + 5x + 1 −1 < x < 2  x , 3 2 3 2 (c)  y = x + x + 5x + 1 (d)  y = x + x + 5x – 1 (a)  0 (b)  1 249. The abscissa of the point on the curve y = a (ex/a + (c)  2 (d)  3 –x/a e ), where the tangent is parallel to the x-axis, is 260. Let f(x) be a non-constant twice differentiable function (a)  0 (b)  a 1 defined on (– ∞, ∞) such that f(x) = f(1 – x) and f '   (c)  2a (d)  – 2a 4  = 0.  e2 x − 1  This section contains 4 multiple correct answer(s) type 250. Function f ( x) =  2 x  is questions. Each question has 4 choices (A), (B), (C) and (D),  e +1 out of which ONE OR MORE is/are correct. Then (a)  increasing (b)  decreasing (a)  f” (x) vanishes at least twice on [0, 1] (c)  even (d)  None of the above 1 −  x

251. The function f ( x) = x( x + 3) e  2  satisfies all the conditions of Rolle’s theorem in [–3, 0]. The value of c is (a)  0 (c)  –2

(b)  –1 (d)  –3

x −x 252. If f ( x) = cot  2  (a)  –1 (c)  log 2 −1

x

−x

  , then f ′ (1) is equal to  (b)  1 (d)  – log 2

253. If a + b = 8, then ab is greatest when (a)  a = 4, b = 4 (c)  a = 6, b = 2

(b)  a = 3, b = 5 (d)  None of these

254. How many real solutions does the equation x7 + 14x5 + 16x3 + 30x – 560 = 0 have? (a)  7 (c)  3

(b)  1 (d)  5

255. Suppose the cubic x3 – px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?

1 (b)   f '   = 0 2 1/ 2

(c)  

∫

−1/ 2 1/ 2

(d)  

∫ 0

1  f  x +  sin x dx = 0 2 

f (t )esin πt dt =

1

∫

f (1 − t )esin πt dt

1/ 2

261. A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then 1 1 2 + < PS ST QS × SR 1 1 2 + > (b)   PS ST QS × SR 1 1 4 (c)   + < PS ST QR 1 1 4 (d)   + > PS ST QR

(a)  

1. (a), (c) We have, x  ...(1) y = x3 – 2x2 + x – 2  ...(1) 5. (b) We have, y2 = 4a  x + a sin   a  dy = 3x2 – 4x + 1 dy x 1 x dx   dy =4a 1 + a ⋅ cos ⋅  = 4a 1 + cos  ⇒ 2y Since the tangent is parallel to x-axis, = 0. dx a a a   dx That is 2 dy 3x – 4x + 1 = 0   or  (x – 1) (3x – 1) = 0 Since the tangent is parallel to x-axis, =0 1 dx ⇒  x = 1, . 3 x x ⇒ 4a 1 + cos  = 0 ⇒ cos = – 1 = cos π 50 1 a a  and if x ∴ If x = , from (1) we get, y = – 27 3 x ∴ = π i.e., x = aπ. a = 1, from (1) we get, y = – 2. Putting this value of x in equation (1), we get y2 = 4a (x + a sin π) = 4a (x + 0) = 4ax.  1 50  The points are (1, – 2) and  , −  . ∴ y2 = 4ax, which is a parabola.  3 27  2. (b) We have,  1

⇒

2 x

x+ y = 4  +

...(1)

1 dy = 0,  or  2 y dx

dy =– dx

y x

.

Since the tangent is equally inclined to the axes, dy = tan 45º or tan 135º i.e., 1 or – 1. Thus dx –

y x

= ± 1. This gives y = x.

From (1), x + x = 4 ⇒ x = 4. Also, y = x = 4. The point is (4, 4). 3. (b), (c)   We have, x2 + y2 = 25  ⇒

2x + 2y

dy = 0,  or  dx

...(1)

x dy =– . y dx

Now, slope of the line 3x – 4y = 7 is m =

3 . 4

Since the tangent is parallel to the given line, x dy 3 3 4 = ⇒– = i.e., y = – x. ...(2) y dx 4 4 3 From (1) : x2 +

16 2 x = 25 ⇒ x = ± 3. 9

If x = 3, from (2), y = –

4 (3) = – 4. 3

If x = – 3, from (2), y = –

4 (– 3) = 4. 3

The points are (3, – 4) and (– 3, 4). 4. (d) Minimum value of a tan2 x + b cot2 x is 2 ab and maximum value of a sin2 θ + b cos2 θ is a ( a sin2 θ + b cos2 θ = (a – b) sin2 θ + b) Given: a = 2 ab .  ∴ a = 4b.

6. (b) We have, e2y = 1 + 4x2 ⇒

dy 4x 4x = 2y = . dx e 1 + 4x2

∴  Slope of tangent = m = ⇒

| m | =

4| x | ≤1 1 + 4 | x |2

 (1 − 2 | x |) 2 ≥ 0    7. (c) We have,

⇒ e2y ⋅ 2

dy = 8x dx

4x 1 + 4x2

⇒ 1 + 4 | x |2 −4 | x |≥ 0   4| x |  ⇒ ≤1 2  1 + 4| x |

ax y = b−x

⇒

(b − x) a − ax ⋅ (−1) ab dy = = . 2 (b − x) (b − x) 2 dx

∴

dy  ab dx  (1, 1) = (b −1) 2 = 2 (given) 

...(1)

Since the curve passes through the point (1, 1), therefore, a 1= i.e., a = b – 1 ...(2) b −1 Putting a = b – 1 in (1), we get (b − 1) b = 2 ⇒ b = 2. ∴ a = 2 – 1 = 1. (b − 1) 2 Hence, a = 1, b = 2. 8. (b) We have, y = ⇒

2 3 x – 2ax2 + 2x + 5 3

dy = 2x2 – 4ax + 2. dx

Since, the tangent makes an acute angle with the positive direction of x-axis, therefore dy ≥ 0 ⇒ 2x2 – 4ax + 2 ≥ 0 for all x dx

Applications of Derivatives

155

Solutions

156

Objective Mathematics

⇒ 16a2 – 16 ≤ 0 ( Disc. = (4a)2 – 4 (2) (2) ≤ 0) ⇒ a2 – 1 ≤ 0 i.e., (a – 1) (a + 1) ≤ 0 ⇒ – 1 ≤ a ≤ 1.

dx dy = 2t + 3 and = 4t – 2 dt dt dy dt dy 4t − 2 ∴ = = . dx dt dx 2t + 3

12. (c) We have,

9. (d) We have, y2 = 2ax ...(1) a a 2 Put x = ; y = 2a     ⇒ y = ± a 2 2

Thus, slope of the tangent to the curve at the point t = 2 is

a  a  ∴ The points are  , − a  and  , − a  2   2  Differentiating (1) w.r.t x, we get a dy dy 2y = 2a ⇒ = y dx dx a dy a At  a , a  : = = = 1 = m1 (say).   y dx a 2  a a  dy a At  , − a  : = = = – 1 = m2 (say). y −a 2  dx Since m1m2 = – 1, the two tangents are at right angles. 10. (a) Putting y = x in y = 9 − 2 x 2 , we get x = 9 − 2 x 2 ⇒ x2 = 9 – 2x2 ⇒ x = 3 , – 3 . Since y > 0, therefore, the point is ( 3 , 3 ). Now, we have, y2 = 9 – 2x2 Differentiating w.r.t. x, we get dy 2y = – 4x dx ⇒

2x dy =– y dx dy  dx  (

∴

3)

3 ) = – 2 (x –

3 ) is

3) i.e., 2x + y – 3 3 = 0.



11. (b) We have, y = (x + 1) (x – 3) = x2 – 2x – 3 ...(1) For points on x-axis, y = 0. This gives (x + 1) (x – 3) = 0 ⇒ x = – 1, 3. Therefore, the curve meets x-axis at (– 1, 0) and (3, 0). Differentiating the given equation w.r.t. x, we get dy = 2x – 2. dx dy  ∴ dx  = – 4 = tan θ1 i.e., θ1 = tan–1 (– 4)  ( −1, 0 ) and

dy  = 4 = tan θ2 i.e., θ2 = tan–14 dx  ( 3, 0 )

Hence, the angle between these tangents = ± (θ1 – θ2) = ± [tan–1 (– 4) – tan–1 (4)] =

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

Therefore, the point of intersection in the first 1  quadrant is  , 3  . 2  Now, differentiating equation (1) w.r.t. x, we get 1 dy dy  = 4x + 5 ⇒ = 4   + 5 = 7. dx dx  ( 1 , 3) 2 2 1  Therefore, the equation of tangent at  , 3  is 2  1  ( y – 3) = 7  x −  i.e., 14x – 2y – 1 = 0. 2  x2 y 2 + =1 a 2 b2 −b 2 x 2 x 2 y dy dy =0 ⇒ = ⇒ 2 + 2 a2 y a b dx dx

2 3 = – 2. 3

So, the equation of tangent at ( 3 , ( y –

13. (a) y = 2x2 + 5x  y =3  Solving equations (1) and (2), we get 2x2 + 5x – 3 = 0 i.e., (2x – 1) (x + 3) = 0 1 ⇒ x = , x = – 3. 2

14. (b) We have, =–

3,

4 (2) − 2 dy  6 = = . 2 (2) + 3 dx  t =2 7

 −4 − 4  8  = ± tan–1 ± tan–1   . 15  1 + (−4)(4) 

∴ tan θ1 =

dy  π = ∞ ⇒ θ1 = dx  ( a , 0 ) 2

and  tan θ2 =

dy  = 0 ⇒ θ2 = 0 dx  ( 0, b )

Hence, the angle between the two tangents is π π –0= . θ = θ1 – θ2 = 2 2 15. (c) The two curves are x3 – 3xy2 = a  ...(1) and 3x2y – y3 = b  ...(2) Differentiating (1) w.r.t. x, we get x2 − y 2 dy dy 3x2 – 3y2 – 6xy =0 ⇒ = 2 xy dx dx Differentiating (2) w.r.t. x, we get 6xy + 3x2

2 xy dy dy dy – 3y2 =0 ⇒ =– 2 x − y2 dx dx dx

The product of

dy for the two curves dx

∴ The curves cut each other orthogonally.

⇒ 16. (b) Differentiating y = 4ax w.r.t. x, we get 2a dy dy 2y = 4a ⇒ = . y dx dx 2

∴

dy  2a π dx  ( 0, 0 ) = 0 = ∞ = tan 2 = tan θ1 (say)

∴ θ1 =

π . 2

dy  0 = = 0 = tan 0 = tan θ2 (say) dx  ( 0, 0 ) 2a

∴ θ2 = 0. ∴ Angle between the two curves = (θ1 – θ2) =

π . 2

17. (b) Since f (x) is monotonically decreasing ∴ f ′ (x) = a eax – a e–ax < 0 ⇒ eax > e–ax ( a < 0) ⇒ e2ax > 1 ⇒ 2ax > 0 ⇒ 2x < 0 ( a < 0) ∴ x < 0. 18. (c) We have, y2 = 4x and x2 + y2 – 6x + 1 = 0  Differentiating (1) w.r.t. x, we get 2 dy dy 2y =4 ⇒ = y dx dx ∴

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

157

Differentiating (2) w.r.t. x, we get dy  dy π = – 2 sin 2x ⇒  π = – 2 sin dx  dx 3 x= 6 = – 3 = m2 (say). Hence, angle between the two curves is

=

3π 3π2 π3 (sin −1 x) 2 − sin −1 x + 2 4 8

=

3π  π π2  3π3 π3 −1 2 −1 + (sin x) − 2 ⋅ ⋅ sin x +  − 2  3 16  32 8

=

3π  −1 π  π3  π3 7 π3  ⋅  sin x −  + ∈  ,  2  4  32  32 32 

22. (a) We have, y = sin x  and y = cos x  The two curves intersect at x =

π in [0, π]. 4

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

Differentiating (1) w.r.t. x, we get dy  1 dy = cos x. ∴ = = m1 (say). dx  x = π 2 dx

dy  2 = = 1 = m1 (say). dx  (1, 2 ) 2

4

Differentiating (2), w.r.t. x, we get dy  1 dy = – sin x. ∴ = m2 (say)  π =– dx  2 dx x= 4

dy  3 −1 = = 1 = m2 (say) dx  (1, 2 ) 2

Hence, the angle between the two curves is

Since m1 = m2, therefore the two curves touch each other at (1, 2). 19. (b) We have, f (x) = (x – 2)2/3 (2x + 1) ⇒ f ′(x) = 2/3 (x – 2)–1/3 (2x + 1) + (x – 2)2/3.2 10 ( x − 1) = 3 ( x − 2)1/ 3 Now, f ′(x) = 0  ⇒ x = 1 and f ′(x) is not defined at x = 2. 20. (b) We have, y = 2sin2x  and y = cos 2x

3 = m1 (say)

2

Differentiating (2) w.r.t. x, we get 3− x dy dy 2x + 2y –6=0 ⇒ = y dx dx ∴

3 = 2

 m − m2  π 2π θ = ± tan–1  1 or .  = ± tan–1 3 = 1 + m m 3 3 1 2   21. (c) f (x) = (sin –1 x + cos –1 x) 3  – 3sin –1 x cos –1 x(sin –1  x + cos–1 x) f (x) = (sin– 1x + cos– 1x)3 – 3 sin– 1 ⋅ cos– 1  (sin– 1x + cos– 1x)  3 π 3π  π  −  − sin −1 x  sin −1 x = 8 2 2 

Differentiating x2 = 4ay w.r.t. x, we get dy dy x 2x = 4a ⇒ = dx dx 2a ∴

dy  1 =4 ⋅   ⋅ dx  x = π 2 6

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

 m1 − m2  θ = ± tan–1  1 + m m  = ± tan–12 2 . 1 2   23. (d) f (x) = xα log x For Rolle’s theorem to be applicable, f (x) is continuous in [0, 1] f (x) is differentiable in (0, 1); f ′(x) = 0 ∴ f (0) = f (1) given f ′(x) = xα–1[1 + α log x] = 0 As xα–1 ≠ 0 −1 ∴ 1 + α log x = 0  or  α = log x log x < 0 for x ∈ [0, 1] 1 . ∴ α must be + ve ∴ α = 2

Applications of Derivatives

=

Differentiating (1) w.r.t. x, we get dy = 4 sin x cos x dx

 x 2 − y 2   − 2 xy   × 2 2  = – 1.  2 xy   x − y 

158

n

n

Objective Mathematics

x  y 24. (c) We have,   +   = 2 a b nx n −1 ny n −1 dy =0 ⇒ + n an b dx n

Now,

dx xe xy − 1 = = 0 for vertical tangents dy 1 − ye xy

∴ xexy – 1 = 0 ⇒ exy = 1/x, y = 0 ⇒ x = 1. 30. (d) Let f (x) = (x – 3) (x – 5) = x2 – 8x + 15

n −1

dy  b ⋅a b = − n n −1 = − . ⇒ dx  a b a (a, b)

5

∴ ∫ f ( x)dx = 3

∴  The equation of tangent at (a, b) is b x y y–b =– (x – a) ⇒ + = 2. a a b

x y = 2 touches the curve at + a b (a, b), for all n.

5

∫ (x 3

2

− 8 x + 15)dx

3

 x3 8 x 2  =  3 − 2 + 15 x   5 125  − 100 + 75 − [9 − 36 + 45] =   3 

∴ The line

50 − 54 4 =– . 3 3 31. (d) f (x) = x3 + bx2 + cx + d = [50/3 – 18] =

25. (a) We have, x = a (cos θ + θ sin θ) and y = a (sin θ – θ cos θ) dx = a (– sin θ + θ cos θ + sin θ) = aθ cos θ ⇒ dθ dy = a (cos θ + θ sin θ – cos θ) = aθ sin θ and dθ dy d θ a θ sin θ dy = = = tan θ. ∴ dx d θ a θ cos θ dx

32. (b) The given curve is

 ence, the equation of the normal at any point θ H on the curve is given by [ y – a (sin θ – θ cos θ)]

y = sin x dy = cos x. ⇒ dx

cos θ [x – a (cos θ + θ sin θ)] sin θ ⇒ x cos θ + y sin θ = a (cos2θ + sin2θ) = a. Therefore, the length of the perpendicular from the origin to the normal

 o, the equation of tangent through the origin (0, S 0) is dy y (x – 0) = x cos x, ∴ = cos x ...(2) y–0= dx x



f ′(x) = 3x2 + 2bx + c; D = 4 (b2 – 3c) As, 0 < b2 < c ∴ b2 – 3c < 0 ⇒ f ′(x) > 0 Hence, function is strictly increasing.

=–



=

0 ⋅ cos θ + 0 ⋅ sin θ − a cos 2 θ + sin 2 θ

= a.

26. (c) Since f (x0) = g(x0)  and  f ′(x) > g′(x), ∀x > x0

Squaring and adding (1) and (2), we get y2 y2 + 2 = 1, ∴ x2y2 = x2 – y2. x 33. (a) The given curve is y = ax2 + bx + c

⇒  f (x) – g(x) is an increasing function ∀x > x0. ∴ f (x) – g(x) > f (x0) – g(x0) = 0 ∴ f (x) > g(x), ∀x > x0. 27. (c) Since f 2n + 1 (a) = 0  f 2n + 2(a) = –ve and f (a) = b, ∴  f (x) = – (x – a)2n + 2 + b = b – (x – a)2n + 2. 28. (c) We have, f (x) = x3 + ax2 + bx + 5 sin2x ⇒  f ′(x) = 3x2 + 2ax + b + 5 sin 2x Since f (x) is an increasing function ∴ 3x2 + 2ax + b – 5 > 0, ∀x ∈ R ⇒ 4a2 – 4 · 3 · (b – 5) < 0 ∴ a2 – 3b + 15 < 0. 29. (d) We have, y – exy + x = 0 ⇒

 dx  dx + = 0. 1 – exy  x + y dy   dy

...(1)

Since the point (– 1, 0) lie on it ∴ a – b + c = 0 ...(2) Also, y = 2x is a tangent to (1) at x = 1, so that y = 2. Since the point (1, 2) lies on (1), ∴ a + b + c = 2  ...(3)

∴  f ′(x) – g′(x) > 0



...(1)

dy  = (2ax + b)](1, 2 ) dx  (1, 2 ) = 2, ∴ 2a + b = 2 ...(4) 1 1 , b = 1, c = . Solving (2), (3) and (4) : a = 2 2

Also 

34. (a) When the curve meets x-axis, then 1 y = 0 ⇒ ax2 = 1 ⇒ x = ± a  1   1  , 0 . , 0  and  − Hence the points are  a a     Differentiating the given equation w.r.t. x, we get



dy   dy 2ax + 2h  y + x  + 2by =0 dx  dx 

The point (a cos θ, b sin θ) is (3 3 cos θ, sin θ). Tangent at the above point is

159

⇒

ax + hy dy =– . hx + by dx

⇒

dy  dx  

x3 3 cos θ y sin θ =1 + 9 1 ∴  Sum of intercepts =

Applications of Derivatives

1

 a

 , 0 

dy  a =– and dx    − h 

1 a

 , 0 

a =– . h

Hence the tangents at these points are parallel. 35. (c) We have, xy = (a + x)2 ⇒

a2 dy a2 + 2a + x ⇒ =1– 2 . y= x dx x

I f the normal makes equal intercepts then its slope is ± 1. x2 = ± 1 ⇒ x2 = x2 – a2 or ⇒ – 2 x − a2 x2 = – (x2 – a2) a a2 i.e., x = ± . ⇒ a2 = 0 or x2 = 2 2 a a2 a + 2a ± = ± 2a ± Then, y = + 2a . a 2 2 ± 2 a   a , 2a + ± 2a  . Hence the points are ±  2   2 36. (a) We have,

9 1 + 3 cos θ sin θ

or s = 3 3 sec θ + cosec θ ds = 3 3 sec θ tan θ – cosec θ cot ⇒ dθ 1 1   ∴  tan θ = ⇒θ ⇒ tan3 θ = 3 3 3 d 2s π is positive at θ = . d θ2 6 π Therefore, sum is minimum at θ = . 6 39. (c) We have, 1 h 2 sin − 0 dy  h lim = h→0 = lim h sin h→0 dx  ( 0, 0 ) h

θ=0

π 6

=

1 = 0. h

∴ Slope of the tangent at origin is 0. Hence, the equation of the tangent at (0, 0) is y – 0 = 0 (x – 0) ⇒ y = 0.

40. (c) We have, y = cos (x + y) dy dy  = sin (x + y) 1 + ⇒  ...(1) z = (x – p)2 + (x – q)2 + (x – r)2 dx dx   Differentiating equation (1) with respect to x, we get Since, the tangents are parallel to the line x + 2y = 0 dz = 2(x – p) + 2(x – q) + 2(x – r) ...(2) 1 1 = sin (x + y) 1 −  ⇒ sin (x + y) = –1 ∴ – dx 2 2 dz  For minima or maxima, put =0 dx ⇒ x + y = –π/2, 3π/2, –5π/2 Form equation (2), we get

2(x – p) + 2(x – q) + 2(x – r) = 0 ⇒ 3x – ( p + q + r) = 0 1 or, x = ( p + q + r) 3 ⇒ the given equation is maximum at 1 x = ( p + q + r). 3 37. (a) We have, x = a (θ + sin θ) and y = a (1 – cos θ) dx dy = a (1 + cos θ) and = a sin θ ⇒ dθ dθ 2 sin θ cos θ dy dy d θ sin θ 2 2 = tan θ = = = 2 cos 2 θ dx dx d θ 1 + cos θ 2 2 dy  π = tan (2n + 1) = ∞. ∴ dx   θ = ( 2 n +1) π 2

∴

∴ Slope of the normal is 0. Hence, the normal is parallel to x-axis at θ = (2n + 1) π. 38. (a) Here a2 = 27, b2 = 1,

a=3 3,b=1



–1 ≤ y ≤ 1.

41. (d) Let P (x1, y1) be a point on the curve y2 = 4ax

...(1)

Differentiating w.r.t. x, we get 2a dy dy    = 4a ; ∴ = . 2y  y1 dx dx  ( x1 , y1 ) Hence, the equation of normal at (x1, y1) is y y – y1 = – 1 (x – x1) 2a ⇒ xy1 + 2ay = y1 (x1 + 2a) But lx + my = 1 is also normal; hence coefficients must be proportional. y 2a y1 ( x1 + 2a ) ∴ 1 = = l m 1 2al 1 , x1 = – 2a. m l Hence, from (1),  1 − 2al  4 a 2l 2 = 4a   2 m  l  ⇒ y1 =

⇒ al3 + 2alm2 = m2.

160

42. (b) We have, f ′(x) = ex cos x + sin x.ex and

f ′′(x) = –sin x ex + cos x ex + cos x ex

Objective Mathematics



+ sin x ex.

Now, f ′′(x) = 2 cos x e = 0 ⇒ cos x = 0 x

⇒ x = π/2.



7 1 8 = –­ +c ⇒c = 3 3 3 / 3 2 1 ∴ f (x) = – (1 − x 2 ) − 8 . 3 47. (b) Let P (x1, y1) be a point on the curve ∴

{

Also, f ′′’(x) = –2 sin x e + 2 cos x e = –ve x

x

x + y = a  ...(1) Differentiating w.r.t. x, we get

∴  slope is maximum at x = π/2.

1 1 dy dy  + =0 ⇒ =– 2 x 2 y dx dx  ( x1 , y1 )

43. (b) f (x) = a – (x – 3)89 ≤ a (equality holds at x = 3). 44. (b) We have, xy = c2 dy dy y ⇒ x +y⋅1=0 ⇒ =– ; dx dx x ∴

3 ⇒ t1 t2 = – 1 (as t1 – t2 ≠ 0).

⇒

a [from (1)]



a ⋅ a = a.

=

48. (c) Let P (x1, y1) be a point on the curve

x   a

2/3

 y +  b

2/3

= 1 ...(1)

Differentiating w.r.t. x, we get 2

2

−1

−1



2  x  3 1 2  y  3 1 dy   ⋅ +   ⋅ 3 a  a 3 b  b dx

⇒

2/3 dy   b   y1  = –     dx  ( x1 , y1 )  a   x1 

=0

1/ 3

The equation of the tangent at (1, 1) is 1 ( y – 1) = (x – 1) 4 ⇒ x = 4y – 3.

b y – y1 = –   a ⇒

x x a

13 23 1

∴

46. (a) We have, f′(x) = x 1 − x 2

1/ 3

 y1     x1 

y y b

13 23 1

(x – x1)

x  =  1 a

2/3

y  + 1 b

2/3

=1

p2 q2 x 2 / 3a 4 / 3 y 2 / 3b 4 / 3 + 2 = 1 2 + 1 2 2 a b a b



2 3/2

+c

+

2/3

 [from (1)] Hence, p = x11 3a 2 3 and q = y11 3b 2 3

 ence, the tangent meets the curve again at H 1   16 − , −  . 20   5 1 (1 − x ) 1 − x 2 dx = – 2 3/2

.

The equation of the tangent at (x1, y1) is

When this tangent meets the curve again, then 3 ⋅ (4y – 3) y2 – 2 (4y – 3)2 y = 1 ⇒ 20y3 – 39y2 + 18y + 1 = 0 1 16 ⇒ x = 1, 1, – . ⇒ y = 1, 1, – 20 5

7 Since it passes through  0 ,   3

.

Hence, the intercepts on the axes are a x1 and a y1 . ∴  Sum of the intercepts = a ( x1 + y1 )

Differentiating w.r.t. x, we get dy dy – 4xy – 2x2 =0 3y2 + 6xy dx dx 1 dy  = . ⇒  4 dx  (1, 1)

3/2 1 1 − x2 ) + c ( 3

x y + = x1 + y1 = x1 y1 x y + = 1. a x1 a y1

⇒

45. (a) We have, 3xy2 – 2x2y = 1

⇒ f (x) = –

x1

y1 (x – x1) x1

y – y1 = –

 c The equation of the normal at  ct1 ,  is t 1  c 2 y – = t1 (x – ct1). t1  c Since, this normal passes through  ct2 ,  , theret2   fore, c c − = t12 (ct2 – ct1) t2 t1

∫x

y1

Equation of tangent at (x1, y1) is

1 dy  =– 2 .  t dx  ( ct1 , c t1 ) 1

⇒ f (x) =

}

x  =  1 a

2/3

y  + 1 b

2/3

= 1.

[from (1)]

49. (a) Let P (x1, y1) be a point on the curve xy = c2 ...(1)

 Differentiating w.r.t. x, we get

x

dy  y1 dy + y ⋅ 1 = 0 ⇒ dx  =– . x1  ( x1 , y1 ) dx

 herefore, the tangent at P(x1, y1) meets x-axis at T A (2x 1 , 0) and y-axis at B (0, 2y 1 ). Clearly P (x1, y1) is the mid point of AB.

53. (c) We have, dy =0 x2 + y2 = 9 ⇒ 2x + 2y dx dy  2 =– . ∴ dx  ( 2, 5 ) 5 The equation of tangent is 2 (x – 2) ( y – 5 ) = – 5 and the equation of normal is

a − a2 − y2 a a 2 − y 2 + log  ...(1) 2 a + a2 − y2 Differentiating w.r.t. x, we get



x=

dx = dy

a2 − y2 dy  ⇒ = y dx  ( x1 , y1 )

y1 a 2 − y12

i.e., 2e (ex – 2y) = e2 – 4.



50. (b) Let P(x1, y1) be a point on the curves

1 1  e  y−  = x−  e 2 2  

161



.

(y –

Equation of the tangent at (x1, y1) is y1 (x – x1). y – y1 = 2 a − y12

5)=

5 (x – 2) 2

...(1)

...(2)

The tangent and the normal intersect x-axis at the 9  points A  , 0  and O (0, 0) respectively. 2  Therefore, the area of the triangle OAP 1 9 9 5 = × × 5 = . 2 2 4

This passes through A. ∴ α = x1 – a 2 − y12 . ∴ AP2 = (x1 – x1 + a 2 − y12 )2 + (y1 – 0)2 = a 2 – y 12 + y 12 = a 2. Hence, the length of the portion of the tangent to the given curve, intercepted between the curve and x-axis is a. 51. (d) We have, y = 1 – 2 x/2

54. (a), (d)  We have, xy = 4

...(1)

For y-axis, x = 0 ∴ y = 1 – 20 = 1 – 1 = 0. Differentiating (1) w.r.t. x, we get dy  x dy 1 1 = − 2 2 ⋅ log 2 ⇒ =– log 2. dx  ( 0, 0 ) dx 2 2 Therefore, equation of normal is 2 ( y – 0) =  (x – 0) ⇒ 2x – y log 2 = 0. log 2 1 , y = e–1 52. (b) At the point x = 2 1 Since, x = > 0, ∴ y = e–2x 2 Differentiating w.r.t. x, we get dy  dy −2 = – 2e–2x ⇒ = – 2e–1 = . dx   1 , 1  dx e 2 e

Thus, the equation of normal is



⇒ x ⋅ 

dy + y ⋅ 1 = 0 dx

dy y 4 =– = – 2  ( xy = 4) x dx x 4 ∴ Slope of tangent = –  2 . x a Slope of the line ax + by + c = 0 is = –  . b Since the given line is a tangent to the curve 4 a a ∴ – 2 = – ⇒ >0 x b b It is possible only when a > 0, b > 0 or a < 0, b < 0. i.e.,

dy = f ′ (x). dx ∴ Slope of the tangent to the ellipse at the point where x = 2, is = f ′ (2) = 2 [ y = 2x + 3 is a tangent to the ellipse at x = 2] Hence, f ′ (2) = 2.

55. (a) We have, y = f (x) ⇒

Applications of Derivatives

Equation of tangent at (x1, y1) is y – y1 = – y1 (x – x1) ⇒ xy1 + yx1 = 2x1 y1 x x y 1 + ⇒ = 1. 2 x1 2 y1

162

56. (a) We have,

Objective Mathematics

x = a cos θ and y = a sin θ dy dy = – 3acos2θ sin θ and = 3a sin2θ cos θ ⇒ dθ dθ 3

3

dy d θ dy 3a sin 2 θ cos θ sin θ = = =– . dx d θ dx −3a cos 2 θ sin θ cos θ The equation of tangent, at any point (a cos3θ, a sin3θ) on the curve, is sin θ (x – a cos3θ) (y – a sin3θ) = – cos θ This meets the coordinates axes at A (a cos θ, 0) and B (0, a sin θ). ∴ AB = (a cos θ − 0) 2 + (0 − a sin θ) 2 = a. ∴

57. (b) Since the curve passes through the point (1, 2) ∴ 2 = p + q + r Also, the curve passes through the origin, ∴ r = 0. The equation of the tangent at (0, 0) is dy  (x – 0) ⇒ y = qx. y – 0 = dx   ( 0, 0)

...(1)

61. (b), (c)  Let P (x1, y1) be a point on the parabola y2 = 8x ...(1) Differentiating equation (1) w.r.t. x, we get 4 dy  dy =8 ⇒ = = m1 (say). 2y  y dx  ( x1 , y1 ) dx 1

But y = x is the tangent at origin, ∴ q = 1. ∴ (1) ⇒ 2 = p + 1 + 0 i.e., p = 1. Hence, p = 1, q = 1, r = 0. 58. (b) We have, y = e4x + 2  ...(1) Putting x = 0, we get y = 3. So, the given point is (0, 3). Differentiating equation (1) w.r.t. x, we get dy  dy = 4e4x ⇒ = 4. dx  ( 0, 3) dx The equation of the tangent at (0, 3) is

(y – 3) = 4 (x – 0) i.e., 4x – y + 3 = 0.

∴ Length of the ⊥ r from origin to the tangent

| 4(0) − 0 + 3| = = 16 + 1

3 . 17

59. (b) Let P (x1, y1) be a point on the curve

2xy = a2 

...(1)

Differentiating equation (1) w.r.t. x, we get

∴  The area of the required triangle is 1 (2x1) (2y1) = 2 (x1 y1) = a2. = 2 60. (a) We have, ...(1) x2 = 4y  ...(2) and y2 = 4x Solving the equation (1) and (2), we get x = 0 and x = 4. Thus, the two parabolas intersect at the points (0, 0) and (4, 4). Now, differentiating equation (1) w.r.t x, we get dy dy x ⇒ = . 2x = 4 dx dx 2 dy  dy  = 0 and = 2. ⇒ dx  ( 4, 4 ) dx  ( 0, 0 ) Thus, the two tangents to the parabola x2 = 4y are (y – 0) = 0 (x – 0) and (y – 4) = 2 (x – 4) ⇒ y = 0 and 2x – y – 4 = 0 Clearly, the two tangents intersect at (2, 0).

dy  y dy 2x + 2y ⋅ 1 = 0 ⇒ =– 1.  dx x dx  ( x1 , y1 ) 1

Equation of the tangent at P (x1, y1) is y y – y1 = – 1 (x – x1) x1 ⇒ xy1 + yx1 = 2x1 y1 x y + ⇒ = 1. 2 x1 2 y1 Hence, intercepts on x-axis and y-axis are 2x1 and 2y1 respectively.

Also, the slope of the given line is = 3 = m2 (say) 4 −3 m1 − m2 y1 π = ⇒1= ∴ tan 12 1 + m1m2 4 1+ y1 ⇒ y1 = 0 or y1 = – 2. 1 The corresponding values of x1 are 0, . 2 1  Hence, the points of contact are (0, 0) and  , − 2  . 2  62. (c) f (x) =

x2 − 4 x

, f (x) = 0 for x = ±2.

4  ∴ f (x) = ±  x −  = f ′(x) x  4  ⇒ ±  x + 2  ≠ 0. x   63. (a) P = x3 –

P ∴ Q2

1 ,Q x3   x − =

= x –

1 x

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

1   x −  + 3 x 1 3  = = x −  + 1  1 x    x −  x −  x x  Clearly, the minimum does not exist.

y = 6x – x2 ⇒

∴

dy  = – 4. dx  ( 5, 5)

dy = 6 – 2x dx

Y –y =



x-intercept, put Y = 0 mx m −1 (a + x) − ax ∴ X = x – m − 2 = x (am − a + mx) (m − 1) a + mx ay Similarly, intercept on y-axis = . m (a + x)

Hence, the equation of the tangent at (5, 5) is y – 5 = – 4 (x – 5) ⇒ 4x + y = 25. 25 ∴ Co-ordinates of A are  , 0  .  4  Also, the equation of the normal at (5, 5) is

x m − 2 (am − a + mx) (X – x) my m −1

68. (c) We have, x = a (t + sin t cos t) and y = a (1 + sin t)2 dx ⇒ = a (1 + cos 2t) dt dy and = 2a cos t (1 + sin t) dt = a (2 cos t + sin 2t) dy dt dy 2 cos t + sin 2t 1 + sin t ∴ = = = dx dt dx 1 + cos 2t cos t

1 y–5= (x – 5) ⇒ x – 4y = – 15. 4 ∴ Co-ordinates of B are (– 15, 0). 1 85 425 ×5= . Thus, the area of the ∆APB = × 2 4 8 2 65. (a) We have, y = x + bx + c dy = 2x + b. ⇒ dx Since the curve touches the line y = x at the point (1, 1) ∴ (2 x + b)](1, 1) = 1 i.e., 2 + b = 1 ⇒ b = – 1. Also, the curve passes through the point (1, 1) ∴ 1 = 1 + b + c i.e., c = – b = 1. dy = 2x – 1. ∴ y = x2 – x + 1 ⇒ dx dy 1 < 0 ⇒ 2x – 1 < 0 ⇒ x < . Now, dx 2 66. (d) sin θ =

5 1 π = , ∴θ = 10 2 6

∴ 2θ =

π 3



t t t cos + sin 1 + tan π t  2 2 2 + = t t = t = tan  4 2  . cos − sin 1 − tan 2 2 2

π t Therefore, the tangent makes an angle + with 4 2 x-axis. 69. (d)

f (x) = x2 (x – 2)2 f ′(x) = 2x (x – 2)2 + 2x2(x – 2) = 4x (x – 2) (x – 1)

∴

f ′(x) ≥ 0 for x ∈ [0, 1] ∪ [2, ∞)

and

f ′(x) ≤ 0 for x ∈ (–∞, 0] ∪ [1, 2]

70. (a) Let P (x1, y1) be the point of intersection of the two curves. We have,

y2 – 2x ⇒ 2y

dy =2 dx 1 = . y1

 dy   ⇒ m1 =     dx   ( x1 , y1 )

and 2xy = k ⇒ x 67. (b) We have, ym = axm – 1 + xm ⇒ mym – 1

dy = a (m – 1) xm – 2 + mxm – 1 dx



dy +y =0 dx

y1  dy   ⇒ m2 =    =– . x 1  dx   ( x1 , y1 )

163



x m − 2 (am − a + mx) dy = . my m −1 dx Equation of tangent: ⇒

Applications of Derivatives

64. (a) We have,

164

Objective Mathematics

Since the two curves intersect at right angles,  1  y  ∴ m1m2 = – 1 ⇒    −  = – 1  y1   x1 

⇒ x1 = 1 2 and hence from y12 = 2x1, we get y1 = 2. Since (x1, y1) also lies on 2xy = k ∴ k2 = 4 x12 y12 = 4 × 1 × 2 = 8.

71. (d) We have, x = acos3θ and y = asin3θ dx dy = – 3acos2θ sin θ and = 3a sin2θ cos θ ⇒ dθ dθ dy d θ dy = = – tan θ. ∴ dx d θ dx Length of the tangent =



2

dy dx

= a sin 3 θ 1 + tan 2 θ tan θ = asin2θ.

Length of the normal =

 dy  y 1+    dx 

 dy  y 1+    dx 

2

= asin3θ 1 + tan 2 θ = atanθ sin2θ dy Length of sub-tangent = y dx =

a sin 3 θ = asin2θ cos θ. tan θ

72. (a) We have, x2y2 = a2 (x2 – a2) ...(1) dy + y2 ⋅ 2x = a2 ⋅ 2x ⇒ x2 ⋅ 2y dx

75. (a) We have xm + n = am – n ⋅ y2n ⇒ (m + n) log x = (m – n) log a + 2n log y Differentiating w.r.t. x, we get 2n dy m+n dy y m+n = ⇒ = ⋅ . y dx x dx x 2n y 2n Sub-tangent = dy =x⋅ m +n dx m

 2n  m ∴ (sub-tangent)m =   ⋅x  m+n  dy  m + n y2 ⋅ Sub-normal =  y  = 2n x  dx  n

2n m+n y ∴ (Sub-normal)n =   ⋅ n  2n  x n

xm+n m+n =   ⋅ m−n n  2n  a ⋅ x = constant xm  ...(2) From (1) and (2), we have (Sub-tangent)m ∝ (sub-normal)n.

2 2 76. (c) We have, x + y = 1 ⇒ 2 x + 2 y dy = 0, 2 2 a b a 2 b 2 dx 2 dy b x =– 2 ∴ dx a y



p = length of the ⊥r from (0, 0) on the tangent dy x −y dx =



a2 − y2 dy ⇒ = . xy dx dy a 2 − y 2 x 2 (a 2 − y 2 ) a 4 = = = 3 dx x x3 x [ from (1) x2 (a2 – y2) = a4] ∴ Sub-normal = y

⇒ The sub-normal varies inversely as the cube of its abscissa.

 dy  1+    dx 

∴ Sub-tangent =

y

dx dy

=

nx ∝ x. m

74. (b) We have, y = a log (x2 – a2) dx dy 2x x2 − a2 =a⋅ 2 ⇒ = . ⇒ 2 dy 2ax dx x −a ∴ Sum of tangent and sub-tangent dx  dy  dx x2 + a2 x2 − a2 +y =y⋅ = y 1 +   ⋅ + y dy  dx  dy 2ax 2ax 2

=

xy ∝ product of the co-ordinates a

2

p′ = length of normal = y

 dy  1+    dx 

2

 b2 x2  dy   − y = y − 2 − y ∴ p × p′ = y  x − a y dx    



73. (c) We have, xmyn = am + n ⇒ m log x + n log y = (m + n) log a m n dy dx + ∴ =0 ⇒ = – nx x y dx dy my

...(1)

− (b 2 x 2 + a 2 y 2 ) b2a 2 =– 2 a a2



=



= – b2 = constant.

∴ p′ ∝

1 . p

77. (a) Let (x1, y1) be the point of intersection. Solving the two equations, we get 2 2 2 2 x12 = a (ab − b ) and y12 = b (a − ab)  2 2 2 a − b2 (a − b )

...(1)

Also, x2b2 + y2a2 = a2b2 ⇒

b 2 x1 dy  = – = m1 (say) a 2 y1 dx  ( x1 , y1 )

and x2 + y2 = ab ⇒

x dy  = – 1 = m2 (say)  y1 dx  ( x1 , y1 )

= tan

–1

 (a 2 − b 2 ) x1 y1   2 2 2 2   a y1 + b x1 

a 2b 2 (ab − b 2 )(a 2 − ab) = tan–1  a − b  .    ab  a 2b 2 78. (d) f (x) = 2x2 – ln | x |

165

 m − m2  θ = tan–1   1   1 + m1m2   log 3 − log 5  = tan–1    .  1 + log 3 log 5 



81. (a) We are given, sub-tangent = sub-normal ⇒

y dy dx

=

y

dy dx

⇒

dy = ± 1. dx

= tan–1

1 =0 x ⇒ 4x | x | = 1 ⇒ x = 1/2. f ′(x) = 4x –

79. (b) Let (x1, y1) be the point of intersection where the curves cut orthogonally. Now, x2 + y2 = 2a2 x dy   ⇒ = – 1 = m1 (say)  y1 dx  ( x1 , y1 ) and 2y2 – x2 = a2 

x1 dy  ⇒ = = m2 (say)  2 y1 dx  ( x1 , y1 )

 dy  1+    dx 

So, length of the normal = y

=y ⋅

2 =

2

2

(ordinate).

82. (b) We have, y = 4ax 2

dy dy = 4a i.e., = dx dx y y Sub-tangent = dy = 2a y dx ∴ Sub-tangent : Abscissa = ⇒ 2y

2a . y =

y2 4ax = = 2x. 2a 2a

2x : x = 2 : 1.

83. (b) f (x) = cos (2πx) + {x} f (x) is periodic with period 1 and f (x) has one local maximum in x ∈ [0, 1].

Now, the product of the slopes at (x1, y1) is 84. (a) We have, y2 = 4ax x12 2a dy dy = – 1 ⇒ x12 = 2 y12  ...(1) = m1 × m2 = – = 4a i.e., = . ⇒ 2y 2 y12 y dx dx Also, (x1, y1) lies on the curve, x2 + y2 = 2a2 y y y2 ∴ x12 + y12 = 2a2  ...(2) Sub-tangent = dy = 2a = 2a y dx From (1) and (2) 2a dy Sub-normal = y =y × = 2a. 2 2 2 y ⇒ 3 y1 = 2a ⇒ y1 = ± a and hence dx 3 y2 4 , y, 2a are in G.P. Clearly, x1 = ± a. 2a 3 ...(1) 85. (b) We have, y = ax3 + bx2 + cx + 5  Thus, the points of intersection of the two curves dy are = 3ax2 + 2bx + c  ...(2) ⇒ dx  4 2  Since the curve touches x-axis at A (– 2, 0) a  .  ± a, ± 3   3 dy  ∴ = 0 ⇒ 12a – 4b + c = 0 ...(3) 80. (a) We have, dx  ( −2, 0 ) ...(1) y = 3x Also, the curve passes through the point A (– 2, 0) x and y = 5 ...(2) ∴ – 8a + 4b – 2c + 5 = 0  ...(4) Clearly, the two curves intersect at the point (0, 1). dy  =3 Also, it is given that Differentiating (1) w.r.t. x, we get dx  x =0 dy  dy ⇒ 3a (0)2 + 2b (0) + c = 3 i.e., c = 3. = 3x log 3, ∴ dx  = log 3 = m1 (say)  ( 0, 1) dx Solving (3) and (4), we get Differentiating (2) w.r.t. x, we get dy  dy = 5x log 5, ∴ dx  = log 5 = m2 (say).  ( 0, 1) dx ∴ The angle between the two curves is given by

a=–

1 3 and b = – . 2 4

Therefore, a = –

1 3 , b=– and c = 3. 2 4

Applications of Derivatives

Hence, the angle between the two curves is  x1 b 2 x1   − 2  y a y θ = tan–1 m1 − m2 = tan–1  1 2 2 1   b x1  1 + m1m2  1+ 2 2  a y1  

166

86. (b), (c)  We have,

Objective Mathematics

5x – 10x + x + 2y + 6 = 0 dy =0 ⇒ 25x4 – 30x2 + 1 + 2 dx dy 1 ⇒ =– (25x4 – 30x2 + 1). dx 2 dy  1 ∴ dx  =– .  ( 0 , − 3) 2 5

3

...(1)

Equation of the normal to the curve at P (0, – 3) is (y + 3) = 2 (x – 0) i.e., y = 2x – 3 ...(2) Substituting the value of y from (2) in (1), we get 5x5 – 10x3 + 5x = 0 ⇒ 5x (x2 – 1)2 = 0 ⇒ x = 0, 1, – 1. When x = 0, y = – 3; when x = 1, y = – 1 and when x = – 1, y = – 5. ∴ The normal at P (0, – 3) meets the curve again at the points (1, – 1) and (– 1, – 5). 87. (d) For x £ 1 2  1  29  f ′ (x) = 3x2 – + 10 = 3  x −  +    > 0 3 3  

=

( f' (3) )

2

− ( f' (1) )

2

2  1 − 1/3  =  + 1 − 3  2 

(

+ f' (3) − f' (2)

)

=

1 4 +1− 3 = − 3 3 3

4−3 3 . 3 89. (a), (d)  We have, xy = (c + x)2 dy dy 2(c + x) − y ⇒ x + y = 2 (c + x) ⇒ = dx dx x =

⇒

dy = dx

(c + x ) 2 ( x2 − c2 ) x = . x2 x

2 (c + x ) −

At the point, the normal cuts off numerically, equal  dx  intercepts on axes, slope of normal = –   = ± 1  dy  − x2 ⇒ 2 =±1 x − c2 ⇒ – x2 = x2 – c2 or – x2 = – (x2 – c2) c2 or c2 = 0 2 c ,  c2 = 0 is not possible. ⇒ x = ± 2

⇒ x2 =

 o f (x) is an increasing function for x £ 1, for x > 1, S f′ (x) = – 1 So, f (x) is decreasing functon for x > 1. 90. (a) We have  x3 – 3xy2 + 2 = 0 Now f (x) will have greatest value at x = 1. If and  3x2y – y3 – 2 = 0

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

Diff. (1) and (2) w.r.t. x, we obtain  dy   dy  x2 − y 2 −2 xy   = and  dx  = 2 dx 2 xy x − y2  C1  C2 Since m1 × m2 = –1, therefore the two curves cut at right angles. 91. (a), (b)  We have, y =

x

∫ | t | dt .

dy = | x |. dx Since the tangent is parallel to the line y = 2x ∴ | x | = 2 ⇒ x = ± 2. 0

lim f ( x) ≤ f (1) ⇒ lim f (1 + h) ≤ 5



x → 1+

h→0

− 1(1 + h) + log 2 (b 2 − 2) ≤ 5 ⇒ lim h→0 ⇒ ⇒ ⇒ ⇒

– 2 + log2 (b2 – 2) ≤ 5 ⇒ log2(b2 – 2) ≤ 7 b2 – 2 ≤ 128 b2 ≤ 130 but b2 > 2 2 < b2 ≤ 130

∴ b ∈ [− 130 , − 2 ) ∪ ( 2 , 130 ] . 88. (d) We have, f ′(1) = tan π/6 =

3 and f ′(3) = tan π/4 = 1.

f ′ (2) = tan π/3 =

∫

3

1

f' ( x) f'' ( x) dx + 3



1 , 3

∫

3

2

f'' ( x) dx

 ( f' ( x) )2  3  + [ f' ( x)]2 =  2  1

Differentiating w.r.t. x, we get

When x = 2, y =

2

∫ | t | dt 0

When x = – 2, y =

−2

2

=

∫ | t | dt 0

∫ t dt

= 2.

0

0

=–

∫ −t dt

=–2

−2

The tangent at (2, 2) is y – 2 = 2 (x – 2) i.e., 2x – y – 2 = 0 ∴ x-intercept = 1. (Putting y = 0) Also, the tangent at (– 2, – 2) is y + 2 = 2 (x + 2) i.e., 2x – y + 2 = 0 ∴ x-intercept = – 1 (Putting y = 0) 92. (d) We have, ⇒

x2 y 2 + = 1 4 16

4x 2 x 2 y dy dy =0 ⇒ =– . + y 4 16 dx dx

...(1)

 2 −8   2 8   − 2 8   − 2 −8  , , , ,  ,  ,  ,  .  5 5  5 5  5 5  5 5 93. (c) We have, y = loge x

⇒

dy   y (0) dy = 1 ⋅  + dx  x =0 dx  1

 log1 + 0 = y (0) = 1. x =0 

∴ The equation of the normal at (0, 1) is ( y – 1) = – 1 (x – 0) i.e., x + y = 1. 98. (c) We have, f (x) = 2x3 – 9ax2 + 12a2x + 1 ∴ f ′(x) = 6x2 – 18ax + 12a2 = 0 ⇒ 6[x2 – 3ax + 2a2] = 0 ⇒ x2 – 3ax + 2a2 = 0 ⇒ x2 – 2ax – ax + 2a2 = 0 ⇒ x (x – 2a) – a (x – 2a) = 0 ⇒ (x – a) (x – 2a) = 0 ⇒ x = a,  x = 2a Now, f ′′(x) = 12x – 18a

dy 1 1 dy  ⇒ = ⇒ = .  dx x 2 dx  ( 2, 0 ) The equation of the tangent at (2, 0) is 1 (x – 2) i.e., 2y = x – 2. (y – 0) = 2 1 ∴ Area of the ∆OAB = (2) (1) = 1 square 2 unit. dy = 21x6 + 5, dx which is positive ∨ x ∈ R. Therefore, the tangent makes an acute angle with x-axis.

94. (c) We have,

95. (c) We have, f (x) = cot–1 x + x ⇒ f ′(x) = – 

1 +1 1 + x2

x2 . Clearly, f ′(x) > 0 for all x 1 + x2 therefore f (x) increases in (–∞, ∞).

=

96. (b) We have, f (x) = (x + 1)1/3 – (x – 1)1/3  1 1 1 − ∴ f ′(x) =  . 3  ( x + 1) 2/3 ( x − 1) 2/3  Clearly, f ′ (x) does not exists at x = ± 1. Now, f ′ (x) = 0 ⇒ (x – 1)2/3 = (x + 1)2/3 ⇒ x = 0 Clearly f ′ (x) ≠ 0 for any other value of x ∈ [0, 1]. The value of f (x) at x = 0 is 2. Hence, the greatest value of f (x) is 2.

∴ f ′′(a) = 12a – 18a = –6a < 0 ∴  f (x) will be maximum at x = a i.e.,,

p =a

Also, f ′′(2a) = 24a – 18a = 6a ∴  f (x) will be minimum at x = 2a i.e.,,



q = 2a

Given: p2 = q ⇒ a2 = 2a ⇒ a = 2. 99. (b), (d)  We have, 4x2 + 9y2 = 1

...(1)

Differentiating w.r.t. x, we get

4x dy dy =0 ⇒ =– . 9y dx dx The tangent at point (x, y) will be parallel to the the line 4x 8 = i.e., x = – 2y. 8x = 9y if – 9y 9 Subsituting x = – 2y in (1), we get 8x + 18y

4 (– 2y)2 + 9y2 = 1 or 25y2 = 1 ⇒ y = ±

1 . 5

Thus, the points where the tangents are parallel to  2 1 2 1 8x = 9y are  − ,  and  , −  .  5 5 5 5 100. (d) Slope of normal to y = f (x) at (3, 4) is Thus,

−1  3π  = tan    4  f'(3)

 ⇒ f ′ (3) = 1.

−1 . f'(3)

π π π = – 1. = tan  +  = – cot 4 2 4

167

4x dy =– = ± ⇒ y = 4x, – 4x. y dx 2 4 Subtituting in (1), we get x2 = i.e., x = ± 5 5 ∴ The points on the curve, where the tangent is equally inclined to the axes, are ∴

97. (b) We have, y = (1 + x)y + sin–1 (sin2x) dy ⇒ dx dy  y  2 sin x cos x + log (1 + x)  +  = (1 + x)y  1 1 + x dx   (1 − sin 4 x) 2

Applications of Derivatives

Since the tangent is equally inclined to the axes,

168

1 . a ⇒ Slope of the tangent at (1, 1) is a

101. (a) Slope of the normal at (1, 1) is –

Objective Mathematics

i.e.,

dy  = a dx  (1, 1)

⇒ 1 = e–x/a ⇒ –

...(1)

dy ∝y We are given that dx dy = ky, where k is some constant dx

dy = k dx y log | y | = kx + c, where c is a constant | y | = ekx + c y = ± ec ekx = Aekx, where A is a constant. Since the curve passes through (1, 1), therefore 1 = Aek ⇒ A = e– k Therefore, y = e– k ⋅ ekx = ek (x – 1) dy dy  = kek (x – 1) ⇒ =k dx dx  (1, 1) ⇒ a = k [Using (1)] Thus, the required curve is y = ea (x – 1). ⇒

...(1)

6y = ± 1 x2

...(2)

( any line making equal intercepts on axes will have its slope as 1 or – 1) Now from (2), we have x2 x2 or y = y =– 6 6 Solving these with the equation (1), we get the 8 8 points  4,  ,  4, −  .  3  3 104. (d) We have, y = be–x/a dy b –x/a =– e dx a x y Since the line + = 1 touches (1) a b b −1 a b –x/a b –x/a ∴ =– e ⇒– a =– e 1b a a ⇒

105. (d) We have, x = t2 – 1, y = t2 – t dy dy = 2t and = 2t – 1 ⇒ dt dx dy dt dy 2t − 1 ∴ = = . dx dt dx 2t Since the tangent is parallel to x-axis, dy 1 ∴ =0 ⇒t= . dx 2 106. (d) We have, x = 3 cos θ and y = 3 sin θ dx dy = – 3 sin θ and = 3 cos θ ⇒ dθ dθ dy d θ dy 3 cos θ cos θ = = =– . ∴ dx d θ dx −3 sin θ sin θ

On differentiating (1) w.r.t. ‘x’, we get 2x + x

(2 x + y ) dy dy dy =0⇒ =– + y + 2y ( x + 2 y) dx dx dx

y − y ( x + 2 y) Length of subtangent = dy = 2x + y dx ∴  Length of subtangent at (1, –3) = 15. 108. (c) We have, y – ex y + x = 0. Taking log on both sides, we get log (x + y) = log ex y = xy log e = xy 1  dy  dy ⇒ +y  + 1 = x ( x + y )  dx  dx

Differentiating w.r.t. x, we get x2 dy dy 18y = 3x2 ⇒ = . 6y dx dx Slope of the normal = –

∴ y = be0 = b. Hence, the required point is (0, b).

Since the tangent is parallel to x-axis, dy π = 0 ⇒ cos θ = 0 ⇒ θ = . ∴ dx 2 107. (c) We have, x2 + xy + y2 = 7 ...(1)

1 102. (b), (c) We have xy = 1 ⇒ y = x dy 1 ∴ =– 2 . x dx ∴  The slope of the normal = x2. If ax + by + c = 0 is normal to the curve xy = 1 a a     ∴  – >0 then x2 = – b b ⇒ a > 0, b < 0 or a < 0, b > 0. 103. (a), (c) We have, 9y2 = x3 

x = 0 i.e., x = 0. a

...(1)

⇒

 dy  1 1 − x = y –  dx  y + x y+x 

⇒

 dy  1 1 − x = y –  dx  y + x y +x 

y ( y + x) − 1 dy = 1 − x( y + x) dx Since the curve has a vertical tangent, dy = ∞ ⇒ 1 – x(y + x) = 0 ∴ dx which is satisfied by the point (1, 0). ⇒

109. (a) We have, y = ax3 + bx2 + cx dy = 3ax2 + 2bx + c dx dy  = c = tan 45º = 1 (Given) ⇒ c = 1. ∴ dx  ( 0, 0 ) ⇒

dy  = 3a + 2b + c = 0 dx  (1, 0 )

 ( x-axis is tangent at (1, 0)) ⇒ 3a + 2b + 1 = 0 which is true if a = 1, b = – 2. Hence, a = 1, b = – 2, c = 1. π π π – cos , x > 0, which x x x remains finite and unique for (0, 1]. Hence f (x) is differentiable and continuous in (0, 1]. π f (0 + h) = lim h sin = 0 = f (0), Also lim h→0 h→0 h ∴ f (x) is continuous at x = 0 also. Also, f (0) = 0 = f (1). Hence, Rolle’s theorem is applicable in [0, 1]. Consider the intervals  1 1   r + 1 , r  , (r = 1, 2, 3, ...).  

110. (d) We have, f ′ (x) = sin

 learly, Rolle’s theorem is applicable in each these C intervals. Hence, by Rolle’s theorem f ′ (cr) = 0,  1 1 , . where cr ∈   r +1 r   1 , As   r +1

1  ⊂ [0, 1], cr ∈ [0, 1]. r

Hence, f ′ (x) = 0 at cr ; r = 1, 2, 3, ... Thus, f ′ (x) = 0 at infinite number of points. 111. (b) Let h (x) = f (x) – 2g (x), x ∈ [0, 1]. ⇒ h′ (x) = f ′ (x) – 2g′ (x) Also, h (0) = f (0) – 2g (0) = 2 – 2 ⋅ 0 = 2. h (1) = f (1) – 2g (1) = 6 – 2 ⋅ 2 = 2. ∴ h (0) = h (1). Since f (x) and g (x) are differentiable in [0, 1], h (x) is also differentiable in [0, 1]. Hence h (x) is also continuous in [0, 1]. So, all the conditions of Rolle’s theorem are satisfied. Hence, there exists a point c, 0 < c < 1 for which h′ (c) = 0. ∴ f ′ (c) – 2g′ (c) = 0 i.e., f ′ (c) = 2g′ (c). 112. (c) Let f (x) = anxn + an – 1xn–1 + ... + a2x2 + a1x + a0, which is a polynomail function in x of degree n. Hence f (x) is continuous and differentiable for all x. Let α < β. We are given, f (α) = 0 = f (β). By Rolle’s theorem, f ′ (c) = 0 for some value c, α < c < β. Hence the equation f ′ (x) = nanxn–1 + (n – 1) an – 1xn–2 + ... + a1 = 0 has atleast one root between α and β. 113. (a) Let f (x) = ax3 + bx2 + cx, x ∈ [0, 1]. ∴ f ′ (x) = 3ax2 + 2bx + c.

 ince f (x) is a polynomial function of x, it is conS tinuous and differentiable for all x ∈ [0, 1]. Also, f (0) = 0; f (1) = a + b + c = 0. ∴ f (0) = f (1). Applying Rolle’s theorem, f ′ (k) = 0 for atleast one value k, 0 < k < 1. Hence k is a root of the equation 3ax2 + 2bx + c = 0, where 0 < k < 1. 114. (c) Let f (x) = ax2 + bx + c. Then f (α) = 0 = f (β). Also, f (x) is continuous and differentiable in [α, β] as it is a polynomial function of x. Hence, by Rolle’s theorem, there exists a k in (α, β), such that b f ′ (k) = 0 ⇒ 2ak + b = 0 ⇒ k = – . 2a b ∴ α < – < β. 2a n 115. (a) Let f (x) = anx + an – 1xn – 1 + ... + a1x. Then f (α) = 0 (Given). Also f (0) = 0. Moreover, f (x) is continuous and differentiable in [0, α] as it is a polynomial function of x. Hence, by Rolle’s theorem, there exists a ‘c’ in (0, α) such that f ′ (x) = 0 for x = c i.e., nanxn – 1 + (n – 1) an – 1xn – 2 + ... + 2a2x + a1 = 0. sin a sin a sin b 116. (a) Here f (a) = cos a cos a cos b = 0. tan a tan a tan b Also f (b) = 0. Moreover, as sin x, cos x and tan x are continuous and π differentiable in (a, b) for 0 < a < b < 2 , therefore f (x) is also continuous and differentiable in [a, b]. Hence, by Rolle’s theorem, there exists a real number c in (a, b) such that f ′ (c) = 0. 117. (b) Let

x n −1 x2 x n +1 xn + a1 + a2 + ... + an −1 + an x . 2 n +1 n n −1 Then f (x) is continuous and differentiable in [0, 1], as it is a polynomial function of x. Also, f (0) = 0 a a a a and f (1) = 0 + 1 + 2 + ... + n −1 + an = 0. 2 n +1 n n −1 (Given) Hence, by Rolle’s theorem, there exists atleast one real number c ∈ (0, 1) such that f ′ (c) = 0 i.e., c is a root of the equation a0xn + a1xn – 1 + ... + an – 1x + an = 0.

f (x) = a0

118. (c) Let f (x) = (x – 3) log x Then, f (1) = – 2 log 1 = 0 and f (3) = (3 – 3) log 3 = 0. As, (x – 3) and log x are continuous and differentiable in [1, 3], therefore (x – 3) log x = f (x) is also continuous and differentiable in [1, 3]. Hence, by Rolle’s theorem, there exists a value of x in (1, 3) such that

169

Also,

Applications of Derivatives



170



f ′ (x) = 0 ⇒ log x + (x – 3)

Objective Mathematics

⇒ x log x = 3 – x.

Now, 0 < c < 2,

1 =0 x

119. (a) Let α, β (α < β) be any two real roots of f (x) = e – sin x. Then, f (α) = 0 = f (β) Moreover, f (x) is continuous and differentiable for x ∈ [α, β]. Hence, from Rolle’s theorem, there exists atleast one x in (α, β) such that f ′ (x) = 0 ⇒ – e– x – cos x = 0 ⇒ – e–x (1 + ex cos x) = 0 ⇒ ex cos x = – 1.

and

–x

120. (c) As f (x) is differentiable in [2, 5], therefore, it is also continuous in [2, 5]. Hence, by mean value theorem, there exists a real number c in (2, 5) such that 1 1 − f (5) − f (2) 1 ⇒ f ′ (c) = 2 5 = . f ′ (c) = 5−2 10 3 f (a) g (a)

121. (a) Let h (x) =

f ( x) = f (a) g (x) – g (a) f (x). g ( x)

Then, h' (x) = f (a) g′ (x) – g (a) f ′ (x) =

f (a) g (a)

f ' ( x) . g' ( x)

Since f (x) and g (x) are continuous in [a, b] and differentiable in (a, b), therefore, h (x) is also continuous in [a, b] and differentiable in (a, b). So, by Mean Value Theorem, there exists atleast one real number c, a < c < b for which h (b) − h (a ) , h′ (c) = b−a ∴ h (b) – h (a) = (b – a) h′ (c) Here h (a) =

f (a) g (a)

∴ From (1) :

= (b – a)



f (a) f (a) = 0, h (b) = g (a) g (a)

f (a) g (a)

...(1) f (b) . g (b)

f (b) = (b – a) h′ (c) g (b)

1 for all x.  1 + x2

 rom (1), it follows that f (x) is differentiable at all F x, therefore f (x) is also continuous at all x. ∴ By mean value theorem in [0, 2] f (2) − f (0) 1 = f ′ (c) = where 0 < c < 2 2−0 1 + c2 f ( 2) − 0 1 2 = or f (2) =  2 1 + c2 1 + c2

2 > 0.4 1 + c2

...(4)

From (2), (3) and (4) it follows that 0.4 < f (2) < 2. 123. (a) Let f ( y) =

y

∫ (1 + cos

8

x)(ax 2 + bx + c) dx

0

⇒ f ′ (y) = (1 + cos8y) (ay2 + by + c)

...(1)

1

∫ (1 + cos

Now, f (1) =

8

x)(ax 2 + bx + c) dx = 0

0

and  f (2) =

2

∫ (1 + cos

8

x)(ax 2 + bx + c) dx = 0

0

Also, f (0) = 0. ∴ f (0) = f (1) = f (2). Now by Rolle’s theorem for f (x) in [0, 1]. f ′ (α) = 0, for atleast one α, 0 < α < 1 and by Rolle’s theorem for f (x) in [1, 2], f ′ (β) = 0, for atleast one β, 1 < β < 2. From (1),  f ′ (α) = 0 ⇒ (1 + cos8α) (aα2 + bα + 2) = 0. But 1 + cos8α ≠ 0, ∴ aα2 + bα + c = 0, i.e., α is a root of the equation ax2 + bx + c = 0. Similarly f ′ (β) = 0 ⇒ aβ2 + bβ + c = 0, i.e., β is a root of the equation ax2 + bx + c = 0. But the equation ax2 + bx + c = 0, being a quadratic equation, cannot have more than two roots. ∴ The equation ax2 + bx + c = 0 has one root α between 0 and 1 and other root β between 1 and 2. 124. (b) We have, f (x) = log sin x 1 cos x = cot x ⇒ f ′ (x) = sin x

applicable. ...(1)

⇒ f ′ (x) > 0, for all x ( 1 + x2 > 0)

⇒

2 2 2 > = = 0.4 1 + c2 1 + 22 5

Clearly f (x) is continuous and differentiable  π 5π  in  ,  . Hence mean value theorem is 6 6 

f (a ) f' (c) . g (a ) g' (c)

122. (c) We have, f ′ (x) =

or

2 2 2 < or < 2 ...(3) 1 + c2 1 + 02 1 + c2

∴



...(2)

 π 5π  ∴ There exists a real number c in  ,  such 6 6  that



π  5π  f  − f   6  6  f ′ (c) = 5π π − 6 6 log sin

⇒ cot c =

5π π − log sin 6 6 2 π 3

of the polynomial f ′ (x) = 4ax3 + 3bx2 + 2cx + d lying between 0 and 3.

127. (b) By mean value theorem, there exists a real number c ∈ (2, 4) such that f (4) − f (2) f (4) + 4 f ′ (c) = ⇒ f ′ (c) = 4−2 2 Since f ′ (x) ≥ 6 ∨ x ∈ [2, 4] f (4) + 4 ≥6 ∴ f ′ (c) ≥ 6 ⇒ 2 ⇒ f (4) + 4 ≥ 12

⇒ f (4) ≥ 8.

128. (a) We have, f (x) = 2x – log | x |, x ≠ 0 Case I. When x < 0, f (x) = 2x – log | x | = 2x – log (– x) 1 1 4x2 − 1 ∴ f ′ (x) = 4x – (– 1) = 4x – = . (−x) x x Case II. When  x > 0, f (x) = 2x2 – log | x | = 2x2 – log x 1 ∴ f ′ (x) = 4x – x Thus, when 1 4x2 − 1  x < 0 or x > 0, f ′ (x) = 4x – = x x 4x2 − 1  : Sign scheme for x 

2

Thus, f (x) is an increasing function in the interval

 1  1   − , 0  ∪  , ∞  . 2 2

129. (b) Let f (x) = log (1 + x) – x 1 x –1=– ⇒ f ′ (x) = 1+ x 1+ x ⇒ f ′ (x) < 0, for x > 0 ⇒ f (x) is decreasing for x > 0 ⇒ f (x) < f (0), for x > 0

⇒ 1 + x log (x +

x2 + 1 ) ≥

131. (b) Let f (x) = cos x – 1 +

⇒ f ′ (x) = – sin x + x Let Then

1 + x2

≥ 0,

1 + x 2 , for x ≥ 0.

x2 2

g (x) = x – sin x. g′ (x) = 1 – cos x.

Clearly, g′ (x) > 0, for 0 < x
0, for 0 < x < 2

π 2

g (x) > g (0), for 0 < x
f (0), for 0 < x
0, for 0 < x < 2 2 x2 π i.e., cos x > 1 – , for 0 < x < . 2 2 132. (a) We have, f (x) = x – log (1 + x), x > – 1 x (1 + x) 1 x = = ⇒ f ′ (x) = 1 – (1 + x) 2 1+ x 1+ x cos x – 1 +

Sign scheme for

x (1 + x) : (1 + x) 2

f ′ (x) > 0 if x < – 1 or x > 0. But x > – 1. ∴ f ′ (x) > 0 if x > 0. Thus, f (x) is increasing in the interval (0, ∞).

171

Applications of Derivatives

⇒ log (1 + x) – x < 0, for x > 0 1 1 log − log i.e., log (1 + x) < x, for x > 0. 2 2 ⇒ cot c = =0 2π 130. (c) Let f (x) = 1 + x log (x + x 2 + 1 ) – 1 + x 2 3 1  π 5π  π ⇒ f ′ (x) = 1 ⋅ log (x + x 2 + 1 ) + x ⋅ ⇒c= ∈  , . x + x2 + 1 2 6 6  125. (d) f (x) is continuous in the interval [– 1, 1], but f (x)  x  x  × 1 + − is not differentiable at x = 0. Hence mean value  x2 + 1  1 + x2 theorem is not applicable. So, no c can be found. x x − 126. (c) Let f (x) = ax4 + bx3 + cx2 + dx. = log (x + x 2 + 1 ) + x2 + 1 x2 + 1 Then, f (0) = 0 and f (3) = 81a + 27b + 9c + 3d = log (x + x 2 + 1 ). = 3 (27a + 9b + 3c + d) Clearly, f ′ (x) ≥ 0, for x ≥ 0 = 0 ( 27a + 9b + 3c + d = 0) ⇒ f (x) is increasing for x ≥ 0 Therefore 0 and 3 are roots of the polynomial f (x). ⇒ f (x) ≥ f (0), for x ≥ 0 So, by Rolle’s theorem, there exists atleast one root

172

133. (c) We have, f (x) =

Objective Mathematics

⇒ f ′ (x) =

x , x > 0, x ≠ 1. log x

log x −1 (log x) 2

134. (b) We have, f (x) = x 1 ⇒ f ′ (x) = 2 (1 – log x) x1/x. x f ′ (x) > 0 if 1 – log x > 0, i.e., log x < 1 ⇒ x < e.  ∴  f (x) is increasing in the interval (–∞, e). 1/x

135. (b) Let f (x) = log x – tan–1x 1 + x2 − x 1 1 − = x (1 + x 2 ) x 1 + x2 2



1 3   x −  + 2 4 > 0 for all x > 0. = x (1 + x 2 )

 ence, f (x) is an increasing function in the interval H (0, ∞). 136. (a) We have, f (x) = x +

1 x

∴ f ′(x) = 1 –

1 =0 x2

Now, f ′′(x) =

2 .  ∴  f ′′ (1) = 2 > 0 x3

⇒ x=±1

 herefore, f (x) will be minimum at x = 1 and the T minimum value is 1 = 1 + 1 = 2. f (1) = 1 + 1 137. (b) We have, f (x) = sin x x x cos x − sin x cos x ( x − tan x) = 2 x x2 π But tan x > x and cos x > 0, for 0 < x < 2 π  ∴ f ′ (x) < 0 in the interval  0,  .  2 π Thus, f (x) is decreasing in  0,  .  2 ⇒ f ′ (x) =

 sin x ,  138. (b), (c)  Let f (x) =  x 1, Then f ′ (x) =

cos x ( x − tan x) π < 0 if x ∈  0,  . x2  2

π   tan x > x and cos x > 0 when 0 < x <  2 

f ′ (x) < 0 if log x – 1 < 0, i.e., log x < 1 ⇒ x < e. But x > 0 and x ≠ 1 ∴ f ′ (x) < 0 if x > 0, x ≠ 1, x < e. Thus, f (x) is decreasing for (0, e) \ {1}.

⇒ f ′ (x) =

=

x≠0 x=0

x cos x − sin x x2

 π ∴ f (x) is decreasing in  0,  . Since  2 π ,  0< x < 2 π 2 sin x < < 1. ∴ f   < f (x) < f (0) ⇒ 2 π x   139. (a), (c)  We know that π  ...(1) 2  π Since cos x is decreasing in  0,  , cos (sin x) >  2 cos x. π π , ∴ 0 < cos x < 1 < Also, since 0 < x < 2 2 sin x < x if 0 < x
cos x > sin (cos x) if 0 < x < 140. (b), (c)  Let f (x) = 1 + xp – (1 + x)p ⇒ f ′ (x) = pxp – 1 – p (1 + x)p – 1

 1 = p  1 − 1− p 1− p  ( ) x x + 1  

≥ 0 if p ≥ 0, 1 – p ≥ 0, x > 0 ∴ f (x) increases when x > 0 and 0 ≤ p ≤ 1. Since x > 0, f (x) > f (0) ∴ 1 + xp – (1 + x)p > 0    ( f (0) = 0) i.e., 1 + xp > (1 + x)p if 0 ≤ p ≤ 1 and x > 0. 141. (a) Let f (x) = ax2 +

b – c; x > 0; a, b > 0 x

⇒ f ′ (x) = 2ax –

b 2a  b  = 2  x3 −  . 2  x x 2a 

 b  b ,∴x=    2a  2a

f ′ (x) = 0 gives x3 = Since ax2 +

1/ 3

b ≥ c, ∴ f (x) ≥ 0 for all x > 0. x

  b 1 / 3  ∴ f     ≥ 0   2a   ⇒ a  b   2a   

2/3

1/ 3

 2a  + b   b 

– c≥0

⇒ a ⋅  b  + b − c  b      2a 2a

1/ 3

3b  b  ≥ c⋅    2a  2 2 ∴ 27ab ≥ 4c3.

27b3 b ≥ c3 ⋅ 8 2a

⇒

> 0.

1/ 3

⇒

≥0

π . 2

b ax − b = . x2 x2 b b =0 ⇒x =   2 x a

⇒ y = 9/2 ∴ x =

1/ 2



f ′ (x) = 0 ⇒ a –

.

b ≥ c; ∴ f (x) ≥ 0 for all x > 0 But ax + x

1/ 2

dy = x2e–x (– 1) + 2xe–x = e–x x (2 – x) dx y increases in the interval where ⇒

a + b    b

1/ 2

c2 . 4 2x 143. (b) We have, f (x) = log x – 2+ x



=



– c≥0

⇒ 2 ab ≥ c ⇒ ab ≥

⇒ f ′ (x) =

x2 + 4 ( x 2 + 4) x = > 0, for x > 0 x (2 + x) 2 x 2 ( x + 2) 2

∴ f (x) is increasing for x > 0. 144. (c) We have, y = x3 (x – 2)2 ⇒

dy = 3x2 (x – 2)2 + 2x3 (x – 2) dx



= x2 (x – 2) (5x – 6)

dy 6 > 0 if (x – 2) (5x – 6) > 0 i.e., x < or x > 2. dx 5 6 or x > 2. 5 145. (c) We have, f (x) = – 2x3 + 21x2 – 60x + 41 ∴ f (x) is increasing for x
f (1) as x < 1. Since f (1) = 0. ∴ f (x) > 0 for x ∈ (– ∞, 1).

dy >0 dx

⇒ e–x ⋅ x (2 – x) > 0

⇒ x (2 – x) > 0 ⇒ x (x – 2) < 0 ⇒ 0 < x < 2. 149. (a) Let f (x) = eax + e–ax a (e 2 ax − 1) . e ax f (x) is a decreasing function if f ′ (x) < 0 a (e 2 ax − 1) ⇒ < 0 ⇒ e2ax – 1 < 0 ⇒ e2ax < 1 e ax ⇒ 2ax < 0 ⇒ x > 0 ( a < 0) Thus, f (x) is a decreasing function for x > 0. ⇒ f ′ (x) = a (eax – e–ax) =

1  2 (2 + x) − 2 x  ( x + 2) 2 − 4 x −  = 2 x  (2 + x) x (2 + x) 2 

146. (b) Let f (x) = x tan x

y2 81 9 = = 18 4 × 18 8

9 9 ∴ Required point is  ,  8 2 148. (d) We have, y = x2e–x

 b 1/ 2  ∴ f  a   ≥ 0   b ⇒ a    a

dx   dy dy dx = 18 ⇒ 2y⋅ 2 = 18  = 2  dt  dt dt  dt

150. (a), (d) We have, y = 2x + cot–1x + log [ 1 + x 2 – x]

 ⇒

dy  x  1 1 =2– + × − 1 2 2 2 dx 1+ x 1+ x − x  1+ x 



=



Now,

2x2 + 1 1 (2 x 2 + 1) − 1 + x 2 − = . 2 2 1+ x 1+ x 1 + x2 dy ≥ 0 dx

⇒ (2x2 + 1) – 1 + x 2 ≥ 0 ⇒ (2x2 + 1)2 ≥ 1 + x2 ⇒ 4x4 + 3x2 ≥ 0, which is true for all real values of x. ∴ y increases for all real values of x. 151. (d) We have, f (x) = (x + 2) e–x ⇒ f ′ (x) = – (x + 2) e–x + e–x = – e–x (1 + x).

f (x) is increasing if f ′ (x) > 0.

⇒ – e–x (1 + x) > 0 ⇒ 1 + x < 0 i.e., x < – 1.

Also, f (x) is decreasing if f ′ (x) < 0 π   ⇒ – e–x (1 + x) < 0 ⇒ 1 + x > 0 i.e., x > – 1. ⇒ f ′ (x) = tan x ⋅ 1 + x sec2x > 0, for x ∈  0,  .  2 Hence, f (x) decreases in (– 1, ∞) and increases in  π (– ∞, – 1). 0 , So, f (x) is increasing for x ∈  .  2 ln ( π + x) 152. (b) We have, f (x) = π . ∴ f (β) > f (α) Since, 0 < α < β < ln (e + x) 2 1 1 α tan β ln (e + x) ⋅ − ln ( π + x) ⇒ β tan β > α tan α i.e., > . π+ x e+ x ⇒ f ′ (x) = β tan α 2 [ln (e + x)]

173

∴ 2y

2

Applications of Derivatives

⇒ f ′ (x) = a –

147. (d) Given curve is y2 = 18x

b – c; x > 0; a, b > 0 x

142. (b) Let f (x) = ax +

174

Objective Mathematics

(e + x)ln (e + x) − ( π + x)ln ( π + x)

=

 < 0 on (0, ∞) since 1 < e < π ∴ f (x) decreases on (0, ∞).

( π + x)(e + x) [ ln (e + x)]

2

153. (c) We have, f (x) = tan x – x ⇒ f ′ (x) = sec2x – 1 = tan2x ≥ 0, ∨ x ∈ R ⇒ f (x) never decreases. 154. (b), (c)  We have, f (x) = sin x + cos x ⇒ f ′ (x) = cos x – sin x = f ′ (x) > 0 if 0 ≤ x + i.e., –

π 2 cos   x +  4 

π π 3π π < or 1 2,   hus, f (x) is increasing in (1, ∞) and decreasing in T (– ∞, – 2). 157. (b), (c)  We have, f ′′ (x) < 0, for 0 ≤ x ≤ 1 ⇒ f ′ (x) is a decreasing function in [0, 1]. Now, g′ (x) = f ′ (x) – f ′ (1 – x). g (x) is increasing if g′ (x) > 0 ⇒ f ′ (x) > f ′ (1 – x) ⇒ x < 1 – x ( f ′ (x) is decreasing) 1 . i.e., x < 2 1 ∴ g (x) is an increasing function for 0 < x < . 2 Also, g (x) is decreasing if g′ (x) < 0 ⇒ f ′ (x) < f ′ (1 – x) ⇒ x > 1 – x  ( f ′ (x) is decreasing) 1 . i.e., x > 2 1 < x < 1. ∴ g (x) is a decreasing function for 2

π π 3π 0 for all x, therefore π 3π 0 if cos x > sin x 2 2 π 5π π ⇒ 0 ≤ x < or < x ≤ 2π Hence, f (x) is an increasing function in  0,  ∪ 4 4  2 π 5π and f ′ (x) < 0 if cos x < sin x ⇒ < x < .  π 3π   3π  4 4  , 2π  and a decreasing function in  2 , 2  .    2   5π  π  , 2 π Thus, f (x) is increasing in 0,  ∪    4  159. (b) f ′′(x) = 6(x – 1)  4  π 5π  and decreasing in  , ⇒ f ′(x) = 3(x – 1)2 + c ...(1) . 4 4  But at (2, 1), y = 3x – 5 is tangent to y = f (x) 156. (a), (d)  We have, ∴ f ′ (2) = 3 ∴ from (1), 3 = 3 + c ⇒ c = 0 −2 x − 1, x < −2  ∴ f ′(x) = 3(x – 1)2 ⇒ f (x) = (x – 1)3 + k   − 2 ≤ x ≤ 1 f (x) = | x + 2 | + | x – 1 | = 3, But the curve passes through (2, 1) 2 x + 1, x > 1    ∴ 1 = (2 – 1)3 + k ⇒ k = 0 ∴ f (x) = (x – 1)3. Also, f ′ (x) < 0 if

x − 2 x < 1, x ≠ 0  x3 ,  ∴ f ′ (x) = does not exist, x = 1 2 − x  3 , x >1  x Clearly, f ′ (x) > 0 for x < 0 or 1 < x < 2 and f ′ (x) < 0 for 0 < x < 1 or x > 2. Thus, f (x) is increasing for (– ∞, 0) ∪ (1, 2) and decreasing for (0, 1) ∪ (2, ∞). 161. (a), (c) f (x) = 3 cos4x + 10 cos3x + 6 cos2x – 3, 0 ≤x≤π ⇒ f ′ (x) = – 12 cos3x sin x – 30 cos2x sin x – 12 cos x sin x = – 6 sin x cos x (cos x + 2) (2 cos x + 1). π 2π , , π. 2 3 2π π 0, for 3 2 2π π and f ′ (x) < 0, for 0 < x < or < x < π. 3 2 Thus, f (x) is increasing in the interval  π , 2 π  2 3 

f ′ (x) = 0, for x = 0,

π 2π and decreasing in the interval  0,  ∪  , π  .  2  3  162. (a) We have, f (x) = kx3 – 9x2 + 9x + 3 ⇒ f ′ (x) = 3kx2 – 18x + 9. Since f (x) is increasing on R, therefore, f ′ (x) > 0 ∨ x ∈ R ⇒ 3kx2 – 18x + 9 > 0 ∨ x ∈ R ⇒ kx2 – 6x + 3 > 0 ∨ x ∈ R ⇒ k > 0 and 36 – 12k < 0 [ ax2 + bx + c > 0 ∨ x ∈ R ⇒ a > 0 and discrininant < 0] ⇒ k > 3. Hence, f (x) is increasing on R, if k > 3. 163. (b) We have, f (x) = x + kx + 1 2

⇒ f ′ (x) = 2x + k. Also, f ′′ (x) = 2. Now, f ′′ (x) = 2, ∨ x ∈ [1, 2] ⇒ f ′′ (x) > 0, ∨ x ∈ [1, 2] ⇒ f ′ (x) is an increasing function in the interval [1, 2]. ⇒ f ′ (1) is the least value of f ′ (x) on [1, 2] But f ′ (x) > 0 ∨ x ∈ [1, 2]  [ f (x) is increasing on [1, 2]] ∴ f ′ (1) > 0, ∨ x ∈ [1, 2] ⇒ k > – 2. Thus, the least value of k is – 2.

f ′ (x) < 0 ⇒ 6x2 + 18x + λ < 0. The value of λ should be such that the equation 6x2 + 18x + λ = 0 has roots – 2 and – 1. λ ⇒ λ = 12. 6 165. (b) We have, f (x) = sin4x + cos4x Therefore, (– 2) (– 1) =

⇒ f ′ (x) = 4 sin3x cos x – 4 cos3x sin x = – 4 sin x cos x (cos2x – sin2x) = – 2 (2 sin x cos x) cos 2x = – 2 sin 2x cos 2x = – sin 4x. f (x) is an increasing function if f ′ (x) > 0 ⇒ – sin 4x > 0 ⇒ sin 4x < 0 π π 0] Thus, f (x) is decreasing in (0, π). 167. (a), (c)  f ′ (x) > 0 ∨ x ∈ R ⇒ f (x) is increasing ∨ x ∈ R ∴ f (x) < f (x + 1) and f (x) > f (x – 1) ∨ x ∈ R ⇒ g ( f (x)) > g ( f (x + 1))  and g ( f (x)) < g ( f (x – 1)) as g (x) is decreasing ∨ x ∈ R. Also, g′ (x) < 0 ∨ x ∈ R ⇒ g (x) is decreasing ∨ x ∈ R ⇒ g (x) > g (x + 1) and  g (x) < g (x – 1) ∨ x ∈ R ∴ f (g (x)) > f (g (x + 1)) and f (g (x)) < f (g (x – 1)) ∨ x ∈ R, as f (x) is increasing ∨ x ∈ R. 168. (b) We have, f (x)  = 3 cos | x | – 6ax + b   = 3cos x – 6ax + b ( cos (– x) = cos x) ⇒ f ′ (x) = – 3sin x – 6a. Since f (x) is an increasing function ∨ x ∈ R ⇒ f ′ (x) > 0 ∨ x ∈ R ⇒ – 3sin x – 6a > 0 ∨ x ∈ R Inparticular, at x =

π 2

1 . 2 169. (a) We have, f (x) = 2 log (x – 1) – x2 + 2x + 3 – 3 – 6a > 0 ⇒ a < –

⇒ f ′ (x) =

1 − ( x − 1) 2  2 − 2x + 2 = 2   x −1  x −1 

175

164. (a) Since f (x) is decreasing in the interval (– 2, – 1), therefore,

Applications of Derivatives

 x −1  x 2 , x ≥ 1 | x −1| 160. (a), (c)  f (x) = = 1 − x  x2 , x < 1, x ≠ 0  x 2

176

Objective Mathematics

− 2 x ( x − 1)( x − 2) − 2 x ( x − 2) = ( x − 1) 2 x −1 − 2 x ( x − 1)( x − 2) Sign scheme for : ( x − 1) 2 =

f ′ (x) > 0 if x ∈ (– ∞, 0) or x ∈ (1, 2) ∴ f (x) is increasing in the interval (– ∞, 0) ∪ (1, 2). 170. (a) Let f (x) = x + ex = 0. Since f (– ∞) = – ∞ and f (+ ∞) = ∞, ∴ f (x) = 0 has a real root. Let the real root be α. Then f (α) = 0. Now, f ′ (x) = 1 + ex > 0, ∨ x ∈ R ∴ f (x) is an increasing function ∨ x ∈ R. ∴ for any other real number β, f (β) > f (α) or f (β) < f (α). But f (α) = 0; so, f (β) ≠ 0. ∴ f (x) = 0 has no other real root. Hence, the equation has only one real root. 171. (b) Let f (x) = 2x3 + 15 and g (x) = 9x2 – 12x. Then, f ′ (x) = 6x2 > 0 ∨ x ∈ R ∴ f (x) is an increasing function ∨ x ∈ R. 2 . Also, g′ (x) > 0 ⇒ 18x – 12 > 0 i.e., x > 3 2 Thus, f (x) and g (x) both increase for x > . 3 Let F (x) = f (x) – g (x). Since f (x) increases less rapidly than the function g (x), therefore, F (x) is a decreasing function. ⇒ F′ (x) < 0 ⇒ 6x2 – 18x + 12 < 0 i.e., 6 (x2 – 3x + 2) < 0 ⇒ (x – 1) (x – 2) < 0 ⇒ 1 < x < 2. Hence, the function 2x3 + 15 increases less rapidly than the function 9x2 – 12x on the interval (1, 2). 172. (a), (c) We have, h′ (x) = f ′ (x) [1 – 2 f (x) + 3 ( f (x))2] 2 1  = 3 f ′ (x) ( f ( x)) 2 − f ( x) +  3 3 

174. (c) We have, f ′ (x) = ex (x – 1) (x – 2). As ex > 0, f ′ (x) < 0 if and only if (x – 1) (x – 2) < 0 i.e., if 1 < x < 2. 175. (a) We have, f (x) = sin x – cos x – ax + b ⇒ f ′ (x) = cos x + sin x – a. f (x) is a decreasing function for all real values of x, if f ′ (x) < 0 ∨ x ∈ R ⇒ cos x + sin x < a ∨ x ∈ R.  s the maximum value of cos x + sin x is 2 , the A above is possible when a ≥ 2 . ∴ f (x) decreases for each x ∈ R if a ≥ 2 . 176. (d) In the interval [– 1, 2], f ′ (x) = 6x + 12 > 0. Hence f (x) is increasing in [– 1, 2]. Now, f (x) being a polynomial in x is continuous in – 1 ≤ x < 2 and in 2 < x ≤ 3. We check at x = 2.

lim f (2 – h)= lim 3 (2 – h)2 + 12 (2 – h) – 1 h→0 h→0 = 12 + 24 – 1 = 35

lim f (2 + h) = lim 37 – (2 + h) = 35. h→0 h→0

Also, f (2) = 3 (2)2 + 12 (2) – 1 = 35. ∴ f (x) is continuous at x = 2 and hence in the interval [– 1, 3]. f (2 − h) − f (2) Now, L f ′ (2) = lim h→0 −h 2 3 ( 2 − h ) + 12 ( 2 − h ) − 1 − 35 = lim h→0 −h 2 = lim 3h − 24h = 24. h→0 −h f (2 + h) − f (2) R f ′ (2) = lim h→0 h 37 − ( 2 + h ) − 35 = – 1. = lim h→0 h  ince, L f ′ (2) ≠ R f ′ (2), therefore, f ′ (2) does not S exist. 177. (a) Since g (x) is decreasing, ∴ g (x2) ≤ g (x1) when x2 ≥ x1. Since f (x) is increasing, ∴ f [g (x1)] ≥ f [g (x2)] ⇒ h (x1) ≥ h (x2) when x2 ≥ x1. ⇒ h (x) is a decreasing function of x and h (0) = 0. Also, domain of h = [0, ∞) and range of h = [0, ∞). ∴ h (x) = 0, ∨ x ∈ [0, ∞).

2  1  2 = 3 f ′ (x)   f ( x) −  +  3  9    Note that h′ (x) < 0 whenever f ′ (x) < 0 and h′ (x) dx dx/d θ −a sin θ > 0 whenever f ′ (x) > 0. Thus, h (x) increases (de- 178. (c) = = = – tan θ dy dy/d θ a cos θ creases) whenever f (x) increases (decreases). Equation of normal at θ is 173. (d) S is clearly correct. R is incorrect, can be seen by sin θ 3π  (y – a sinθ) = ( x − a (1 + cos θ )) taking f (x) = sin x, x ∈  π, . cos θ   2 

I n this interval f (x) decreases from 0 to – 1 but f ′ (x) = cos x increases from – 1 to 0.

Clearly this line passes through (a, 0).

180. (b) Suppose, there are two points x1 and x2 in (a, b) such that f ′ (x1) = f ′ (x2) = 0. By Rolle’s theorem applied to f ′ on [x1, x2], there must then be a c ∈ (x1, x2) such that f ′′ (c) = 0. This contradicts the given condition f ′′ (x) < 0 ∨ x ∈ (a, b). Hence our assumption is wrong. Therefore, there can be atmost one point in (a, b) at which f ′ (x) is zero. 181. (a) Let y = x1/x, x > 0  ...(1) 1 log x ⇒ log y = x log x Let z = log y = x 1 x ⋅ − log x ⋅ 1 dz 1 − log x x  ⇒ = = 2 x dx x2 dz 1 − log x =0 ⇒ = 0 ⇒ log x = 1 ⇒ x = e. dx x2  1 x 2 ⋅  −  − (1 − log x) ⋅ 2 x d 2z − 3 + 2 log x  x = = x4 dx 2 x3 ∴

d 2z  − 3 + 2 log e −3 + 2 1 = = – 3 < 0.  = dx 2  x = e e3 e e3

∴ x = e gives maximum value of z and hence, for y also  ( z = log y). Also, from (1), Max. (y) = e1/e. x log x −1 ⇒ f ′ (x) = 182. (a) Let f (x) = log x (log x) 2 For maximum or minimum, f ′ (x) = 0 ⇒ log x = 1 ⇒ x = e. 1 2 log x ⋅ (log x) 2 − (log x − 1) ⋅ x x Now, f ′′ (x) = (log x) 4 1 −0 1 ⇒ f ′′ (x)]x = e = e = > 0. 1 e ∴ f (x) is minimum at x = e. ∴ Minimum value of f (x) = f (e) = 1 1 183. (d) Let y = sin x + sin 2x + sin 3x 2 3 dy = cos x + cos 2x + cos 3x ⇒ dx

e = e. 1

For maximum or minimum,

177

Since, f ′ (x) > g′ (x) ∨ x ∈ R, therefore, h′ (x) = f ′ (x) – g′ (x) > 0 ∨ x ∈ R. ⇒ h (x) is an increasing function ∨ x ∈ R But h (0) = f (0) – g (0) = 0, So, for x > 0, we must have h (x) > h (0) = 0 and for x < 0, we have h (x) < h (0) = 0. ⇒ f (x) > g (x) ∨ x ∈ (0, ∞) and f (x) < g (x) ∨ x ∈ (– ∞, 0).

= cos 2x + 2 cos 2x cos x = cos 2x (1 + 2 cos x). dy =0 dx

⇒ cos 2x (1 + 2 cos x) = 0 ⇒ cos 2x = 0 or cos x = –

1 2

π 2π or x = 2 3 π 2π ∴ x = as x = is not required. 4 3 ⇒ 2x =

Now, ⇒

d2y = – sin x – 2 sin 2x – 3 sin 3x dx 2

d2y =–  dx 2  x = π

1 1 − 2 − 3⋅ < 0. 2 2

4

π  y is maximum, when x = 4 and maximum ∴ value is 4  1 1 1 1 1 8+3 2 y ]x = π = + + ⋅ = =  + . 2 3 2 4 2 2 3 2 6 2 184. (b) We have, P (x) = a0 + a1x2 + a2x4 + ... + anx2n ⇒ P′ (x) = 2a1x + 4a2x3 + ... + 2nan x2n – 1. For maximum or minimum, P′ (x) = 0 ⇒ x (2a1 + 4a2x2 + ... + 2nanx2n – 2) = 0 ⇒ x = 0  [ each ai > 0 and powers of x are even] Now, P′′(x) = 2a1 + 12a2x2 + ... + 2n (2n – 1) anx2n – 2 ∴ P′′ (x)]x = 0 = 2a1 > 0 i.e., P (x)  has a minimum at x = 0 only. 185. (d) We have, y = a loge x + bx2 + x dy a = + 2bx + 1 ⇒ dx x Since y has its extreme values at x = 1 and x = 2. dy  dy  = 0 and =0 ∴ dx  x =1 dx  x = 2 ⇒ a + 2b + 1 = 0 ...(1) and a + 8b + 2 = 0 ...(2) Solving (1) and (2), we get −2 1 a= and b = – . 3 6 ax + b  ...(1) 186. (c) We have, y = ( x − 4)( x − 1) dy a ( x − 4)( x − 1) − (ax + b)(2 x − 5) = dx ( x − 4) 2 ( x − 1) 2 For an extremum (maximum or minimum), dy =0 dx

⇒

Applications of Derivatives

179. (a), (b)   Let h (x) = f (x) – g (x)

178

Objective Mathematics

⇒ a (x – 4) (x – 1) – (ax + b) (2x – 5) = 0...(2) Since, y has an extremum at P (2, – 1), therefore, x = 2 satisfies the equation (2) and y = – 1 when x = 2. ∴ a (– 2) (1) – (2a + b) (4 – 5) = 0 i.e., b = 0 2a + b and – 1 = (2 − 4)(2 − 1) [Putting x = 2, y = – 1 in (1)]



⇒ 2 = 2a + b ⇒ 2 = 2a i.e., a = 1. sin ( x + a ) sin ( x + b)

dy dx sin ( x + b) ⋅ cos ( x + a ) ⋅ 1 − sin ( x + a )cos ( x + b) ⋅ 1  = sin 2 ( x + b) sin (b − a ) = ≠ 0 for any x as a ≠ b. sin 2 ( x + b) ⇒

Hence, y has neither maximum nor minimum.

1 dy 1 ⇒ =1– 2 . x x dx dy =0 For maximum or minimum, dx 1 ⇒ 1 – 2 = 0 ⇒ x2 = 1, i.e., x = ± 1. x

188. (d) We have, y = x +

Now,

d2y =2>0  dx 2  x =1

2 d2y = 3. ∴ x dx 2

x+ 1  x  ≥ 2  x ⋅ 1  ( A.M. ≥ G.M.) = 2    x  2     ∴ y ≥ 2 ∴ Min. (y) = 2. 2

2

2 ≥ 2.

 ∴ y ≥ 2, ∴ Min. (y) = 2.

1 1 = = logx a z log a x 2



1  z− 1  =  + 2, z z 

whose minimum value is 2.

∴ 2 log10 x – logx .01 = z +

4  z− 2  =  + 4, z z 

2

whose minimum value is 4.

193. (d) We have, f (x) =

x

∫ 2 (t − 1)(t − 2) 1

3

+ 3 (t − 1) 2 (t − 2) 2  dt

⇒ f ′ (x) = 2 (x – 1) (x – 2)3 + 3 (x – 1)2 (x – 2)2 = (x – 1) (x – 2)2 [2 (x – 2) + 3 (x – 1)] = (x – 1) (x – 2)2 (5x – 7). For maximum or minimum, f ′ (x) = 0 ⇒ (x – 1) (x – 2)2 (5x – 7) = 0 7 . ⇒ x = 1, 2, 5 Sign scheme for f ′ (x):

195. (a) Let y = x – xp, where x is the fraction ⇒

dy = 1 – pxp – 1. dx

For maximum or minimum,

1  1   1  = x−  liter: y = x + A = ( x )2 +   +   x  x x

∴ loga x + logx a = z +

−2 4 =– . log10 x z

d2y = 0 at x = a, then we have to obtain dx 2 d3y d4y , , ... and so on to ascertain the existence dx 3 dx 4 of points of extremum.

1 1    xz = 1; ∴ z = , x > 0 x x  

⇒

= logx 10–2 = – 2 logx 10 =

194. (c) When

⇒ y is maximum when x = – 1. Also, Max. (y) = – 2 and Min. (y) = 2 ∴ Max. (y) < Min. (y).

190. (b) Let z = loga x

1 100

Clearly, f (x) has maximum value at x = 1, minimum 7 and neither maximum nor minimum value at x = 5 at x = 2.

⇒ y is minimum when x = 1. d2y =–2 f (8).

6k = k ∴ f (0) = f (1) ⇒ f ′(x) = 0 6 i.e., equation ax2 + bx + c = 0 has at least one root between 0 and 1. Also, f (0) =

199. (b) We have, 1 log x 2 1 1 − ( x − 1) 2 = . ⇒ f ′ (x) = − 2 1+ x 2x 2 x (1 + x 2 )

f (x) = tan–1x –

Now, f (1) = tan–1 1 –

1 π log 1 = 2 4

 1   1  1 π 1 f   log 3 = + log 3  = tan–1  + 3    3 4 6 4 1 π 1 and f ( 3 ) = tan–1 ( 3 ) – log 3 = − log 3 . 4 3 4 π 1 Hence, the least value of f (x) is − log 3 . 3 4 2 200. (b) We have, y = x (x – 1)

dy ⇒ = (x – 1)2 ⋅ 1 + 2x (x – 1) dx

= 3x2 – 4x + 1 = (x – 1) (3x – 1). dy For maximum or minimum, =0 dx

3

3

f (x) =

∴ f ′ (x) = 0 ⇒ x = 1.

d2y  dx 2  x = 1 = – 2 < 0.

1 ∴ y is maximum when x = and maximum value 3 is 4 . y ]x = 1 = 27 3 d2y Also, = 2 > 0.  dx 2  x =1  ∴ y is minimum when x = 1.

⇒ 6 + a = 0 ∴ a = – 6. Also, y]x = 3 = 5 ⇒ 9 + (– 6) (3) + b = 5



d2y = 6x – 4. ∴ dx 2

1 , 1. 3

179

dy = 2x + a dx

49 8 and f (8) = . 543 89

49 . 543 202. (b) Let P (x, y) be the point on the curve which is nearest to O (0, 0). 1 (e2x + e–2x + 2). Let z = OP2 = x2 + y2 = x2 + 4 dz 1 ⇒ = 2x + (e2x – e–2x) dx 2 dz =0 For maximum or minimum, dx 1 ⇒ 2x + (e2x – e–2x ) = 0 2 e −2 x − e 2 x ⇒ = 2x 2 ⇒ x = 0 is a solution and then 1 (e0 + e0) = 1.  y = 2 d2y Also, = 2 + e2x + e–2x > 0, 2 dx  hence z is minimum. Hence the largest value is

∴ The shortest distance OP =

02 + 12 = 1.

203. (c) We have, y = 6 cos x – 8 sin x ⇒

dy = – 6 sin x – 8 cos x. dx

⇒ – 6 sin x = 8 cos x ⇒ sin x = ±

∴

dy =0 dx

⇒ tan x = –

4 3 , cos x =  . 5 5

4 3

Applications of Derivatives

196. (b) We have, y = x2 + ax + b ⇒

180

When sin x =

4 3 , cos x = – 5 5

∴

Objective Mathematics

 3 4 y = 6   −  – 8   = – 10 and  5 5 4 3 , cos x = when sin x = – 5 5 then

 4 3 y = 6   – 8  −  = 10.  5 5 Hence, the greatest height above the x-axis is 10.

then

204. (b) Let f (x) = a2sec2x + b2cosec2x ⇒ f ′ (x) = 2a2sec2x tan x – 2b2cosec2x cot x 2a 2 sin x 2b 2 cos x =0 − ∴ f ′ (x) = 0 ⇒ cos3 x sin 3 x 2 b ⇒ tan4x = 2 a

α dT = 0 ⇒ x2 = ⇒x= β dx

α . β

α  2α  d 2T = N  3  > 0 at x = . 2 β dx x  α ∴ T is minimum for x = . β 207. (a) Let the equation of the circle be x2 + y2 = a2 Also,

Let A (a, 0), B (– a, 0) be the ends of the diameter and C (x, y) be any point on the circle. Then, area of ∆ABC 1 × AB × y =A = 2 2 = ay = a a − x 2 .

b a i.e., cot2x = . a b b a and cosec2x = 1 + . Hence sec2x = 1 + a b ⇒ tan2x =

Also, f ′′ (x) = 2a2 (sec4x + 2sec2x tan2x)  + 2b2 (cosec4x + 2cosec2x cot2x) > 0,  for all x b2 . a2 Hence the minimum value of f (x) is a+b b+a + b2 ⋅ = (a + b)2. = a2 ⋅ a b 205. (a) Let P (x, y) be such a point then OP2 = x2 + y2. ⇒ f (x) is minimum at tan4x =

1 dS 1 ⇒ = 2x – 2 . x x dx dS 1 1 = 0 ⇒ 2x – 2 = 0 ⇒ x3 = . ∴ x dx 2

Let S = OP2 = x2 +

1/ 3

1 ∴ x =   2



d 2S 2 Also, =2+ 3 x dx 2

1/ 3

1 ∴ S is minimum at x =   2

z

= lx + my = lx + dz mk ⇒ =l– 2 . dx x ∴

( xy = k)

dz mk =0 ⇒l– 2 =0 dx x

⇒ x = ± Now,

mk . l

d 2z 2mk = dx 2 x3 ⇒

d 2z   dx 2  x =

mk l

=

2mk  mk     l 

3

2

mk . l

Also,

1/ 3

α  N (α + βx2) = N  + βx  x x 

mk  x

∴ z is minimum at x =

.

1 Thus, for nearest point x =   2 −1/ 6 1 y =±   . 2

 α  dT = N  − 2 + β . ⇒ x dx  

208. (b) We have,



d 2S 2 ⇒ =2+ >0  1 dx 2  x =  1 1/ 3   2 2  

206. (c) T =

∴  A is maximum if x = 0 i .e., C lies on y-axis and then CAB is an isosceles triangle.

and then

min. z = l ⋅

mk   + m ⋅ l

= 2 lmk .

k  = mk l

lmk + lmk

209. (a) We have, f (x) = 2 (cos 3x + cos 3 x) 3− 3  3+ 3  = 4 cos  ≤4  2  x ⋅ cos  2  x    

>0

This is possible only when x = 0.

dS = ex (– sin x + cos x – cos x – sin x) = – 2ex dx sin x

Now, 

dS = 0 ⇒ – 2ex sin x = 0 ⇒ x = 0, π, 2π dx ( 0 ≤ x ≤ 2π)

d 2S Also,  2 = – 2ex (cos x + sin x) > 0 at x = π dx only ∴ S is minimum at x = π. Thus, a = π. 1 (2 cos α cos β) 211. (a) Let y = cos α ⋅ cos β = 2 1 [cos (α + β) + cos (α – β)] = 2 1 cos π + cos  π − 2β   1 sin 2β.   =  2 2  2  2 π π i.e., β = Clearly, y is maximum when 2β = 2 4

=

f (x) = 1 + 3x2 + 32 ⋅ x4 + ... + 330 ⋅ x60 ⇒ f ′ (x) = x (6 + 4 ⋅ 32 ⋅ x2 + ... + 60 ⋅ 330 ⋅ x58) f ′ (x) = 0 ⇒ x = 0. Also, f ′′ (x) = 1 ⋅ (6 + 4 ⋅ 32 x2 + ... + 60 ⋅ 330 ⋅ x58)  + x (4 ⋅ 32 ⋅ 2x + ... + 60 ⋅ 330 ⋅ 58x57) ⇒ f ′′ (0) = 6 > 0.  ∴ f (x) has minimum at x = 0 only. 213. (d) We have,  − 6 x, 0 < x < 1 f ′ (x) =  − 6, x ≥ 1 ∴ f ′ (1 – h) = – 6 (1 – h) < 0

⇒ f ′ (x) = 2 (x – 2) ⋅ xn + (x – 2)2 ⋅ nxn – 1

= xn – 1 ⋅ (x – 2) [(n + 2) x – 2n] 2n ∴ f ′ (x) = 0 ⇒ x = 0, 2, . n+2 Now,



f ′ (0 – h) = (– h)n – 1 (– h – 2) ((n + 2) (– h) – 2n)

= (– ve)n – 1 ⋅ (– ve) ⋅ (– ve) = (– ve)n – 1

and

f ′ (0 + h) = hn – 1 (h – 2) ((n + 2) h – 2n)

= (+ ve) (– ve) (– ve) = +ve

∴ f (x) has a minimum at x = 0 if n is even. Also, f ′ (2 – h) = (2 – h)n – 1 ⋅ (– h) ⋅ (2 (2 – h) – nh) = (2 – h)n – 1 ⋅ (– h) ⋅ (4 – 2h – nh) = (+ ve)n – 1 ⋅ (– ve) ⋅ (+ ve) = – ve and f ′ (2 + h) = (2 + h)n – 1 ⋅ h ⋅ (2 (2 + h) + nh) = +ve. ∴ f (x) has a minimum at x = 2, ∨ n ∈ N. 216. (c), (d)  We have, sin x x cos x − sin x and f ′′ (x) = f ′ (x) = x x2 For maximum or minimum, f ′ (x) = 0 sin x = 0 ⇒ sin x = 0; x ≠ 0. x ∴ x = nπ; n = 1, 2, 3, ... ( x > 0). nπ cos nπ − sin nπ (nπ) 2 cos nπ (−1) n = .  = nπ nπ ∴ Extreme points are x = nπ, n = 1, 2, 3, ..., where the maximum occurs at x = π, 3π, 5π, ... and the minimum occurs at x = 2π, 4π, 6π, ... At x = nπ, f ′′ (x) =

− x4

4

(4 – x2).

For maximum or minimum, f ′ (x) = 0

f ′ (1 + h) = – 6 < 0.

⇒ e

= 4, αβγ = 8.

− x4

4

(4 – x2) = 0 ⇒ x = – 2, 2.

Also, − x4



f ′′ (x) = e



< 0 if x = 2  and > 0 if x = −2 

214. (b) Let α, β, γ be the roots of the given equation. Then, α + β + γ = a, αβ + αγ + βγ

= (x – 2) ⋅ xn – 1 [2x + n (x – 2)]

217. (a), (b)  We have, f ′ (x) = e

 ince f ′ (x) does not change sign as x passes through S 1, therefore, f (x) does not have a maximum or minimum at x = 1, whatever be the value of α.



4

 (– 2x) + e

− x4

4

 ⋅ (4 – x2) ⋅ (– x3)

181

αβγ

⇒

212. (d) We have,

and

3

215. (a), (c)  We have, f (x) = (x – 2)2 ⋅ xn

210. (b) The slope of the tangent to the curve y = ex cos x is dy = ex (– sin x + cos x) S= dx

.

α+β+ γ Since A.M. ≥ G.M. ⇒ ≥ 3 a 3 ⇒ ≥ 8 ⇒ a ≥ 6. 3 ∴ The minimum value of a = 6.

Applications of Derivatives

3+ 3  and it is equal to 4 when both cos  x and  2    3− 3  cos  x are equal to 1 for a value of x.  2   

182

∴ f (x) has a maximum at x

=2

Objective Mathematics

= (2k – 3) – (k – 1) 2 sin 2x cos 2x = (2k – 3) – (k – 1) sin 4x. Since f (x) does not have critical points, therefore, f ′ (x) = 0 does not have any solution in R. ⇒ (2k – 3) – (k – 1) sin 4x = 0 has no solution in R 2k − 3 is not solvable in R ⇒ sin 4x = k −1 2k − 3 2k − 3 2k − 3 >1 ⇒ < – 1 or >1 ⇒ k −1 k −1 k −1

and minimum at x = – 2. 218. (b) Since f ′ (4) = f ′′ (4) = 0, therefore, f (x) = (x – 4)n + k, where n ≥ 3 But f has minimum at x = 4, so n = 4. ∴ f (x) = (x – 4)4 + k. Since f (4) = 10, therefore, k = 10. Thus, f (x) = (x – 4)4 + 10. 219. (a) We have, f (x) =

x

∫ (6t

2

− 24) dt

⇒ 2k – 3 < – k + 1 or 2k – 3 > k – 1

0

⇒ f ′ (x) = (6x2 – 24) ⋅ 1

4  4 or k > 2 ⇒ k ∈  −∞,  ∪ (2, ∞). 3 3 

⇒ k
3 f (x) = (k – 7k + 12) cos x + 2 (k – 4) x + log 2 Thus, we find that f (x) has a minimum value = (k – 3) (k – 4) cos x + 2 (k – 4) x + log 2 2 at x = 3. ⇒ f ′ (x) = – (k – 3) (k – 4) sin x + 2 (k – 4) = (k – 4) [– (k – 3) sin x + 2] 224. (b) Let f (x) = x2/3 + (x – 2)2/3 Since f (x) does not have critical points, therefore 2 –1/3 2 ⇒ f ′ (x) = x + (x – 2)–1/3 f ′ (x) = 0 does not have any solution in R. 3 3 –1/3 Now, f ′ (x) = 0 ⇒ x + (x – 2)–1/3 = 0 ⇒ k ≠ 4 and 2 – (k – 3) sin x = 0 is not solvable ⇒ (x – 2)–1/3 = – x–1/3 ⇒ x – 2 = – x in R 2 ⇒ x = 1 is the critical point. is not solvable in R ⇒ k ≠ 4 and sin x = 2  1 −4 / 3 1 k −3  − x − ( x − 2) −4 / 3  Also, f ′′ (x) = 2 3  3 3  >1 ⇒ k ≠ 4 and k −3 2 −4 / 3  x + ( x − 2) −4 / 3  =– ⇒ k ≠ 4 and | k – 3 | < 2 9 2 −4 / 3 4 ⇒ k ≠ 4 and – 2 < k – 3 < 2 1 + (−1) −4 / 3  = – ∴ f ′′ (1) = – b, (x – b)2m + 1 is positive ( 2m + 1 is odd). Thus, f ′ changes sign from negative to positive as x passes through b and so, f has a minimum at x = b. Since 2n is an even integer, (x – a)2n does not change sign as x passes through a i.e., f ′ (x) does not change sign as x passes through a. Hence f has neither a maximum nor a minimum at x = a. 227. (b) Let f (x) = x25 ⋅ (1 – x)75

=

475 = 0 i.e., x = 237.5. 2 But x is a positive integer, so x can be taken 237 or 238. P is maximum when x –

 ince P (237) = P(238), for maximum profit, total S number of subscribers should be 725 + 237 = 962 or 725 + 238 = 963. Since there is no profit from the 963rd subscriber, therefore number of subscribers should be 962.

231. (c) ⇒ f ′ (x) = 25x24 (1 – x)75 – 75x25 (1 – x)74 = 25x24 (1 – x)74 (1 – 4x) 1 are the critical points. f ′ (x) = 0 ⇒ x = 0, 1, 4 25 75 1 1 3 Now, f (0) = 0, f (1) = 0 and f   =   ⋅   . 4 4 4 1 . ∴ f (x) takes its maximum value at x = 4 228. (b) Let α and β be the roots of the given equation. Then α + β = k and αβ = (2k – 3) Let z = α3 + β3 = (α + β)3 – 3αβ (α + β) = k3 – 3k (2k – 3) = k3 – 6k2 + 9k dz = 3k2 – 12k + 9 = 3 (k2 – 4k + 3) ∴ dk  = 3 (k – 1) (k –3) dz = 0 ⇒ 3 (k – 1) (k – 3) = 0 Now, dk  ⇒ k = 1, 3. d 2z  Also,  = (6k − 12) k = 3 = 6 (3) – 12 > 0. dk 2  k = 3 Hence, z is minimum when k = 3.

π  π π or x = 0 ≤ x ≤  2 2 6 

⇒ x =



Now, f (0) =

On differentiating w.r.t. x, respectively, we get and 3 x 2 = dy dx

2  dy  m1 =   = − a  dx (1,1)

 dy  and  m2 =   = 3(1)  dx (1,1) Since, the two curves cut each other orthogonally

∴  m1m2 = – 1 ⇒  − 2 × 3 = −1 a 232. (d) We have,

or

a=6

f ′ (x) = 3kx2 – 18x + 9 f(x) is increasing on every interval, if  f ′ (x) ≥ 0, ∀ x ∈ R ∴  3kx2 – 18x + 9 ≥ 0, ∀ x ∈ R ⇒  If k > 0 and (–18)2 – 4(3k) (9) ≤ 0. ⇒  If k > 0 and 324 – 108k ≤ 0 ⇒  If k > 0 and k ≥ 3 ∴ k ≥ 3.

and f ′′ (x) = 8 + 72x2 + 240x4 ≥ 8, ∀ x ∈ R For maxima and minima, put 1 2

3 π π 1 1 , f    = and f    = . 4 2 2 6 2

Of these values, the minimum value is

ay + x2 = 7 and x3 = y

⇒ f ′ (x) = 8x + 24x3 + 48x5, ∀ x ∈ R

⇒ f ′ (x) = cos x – sin 2x = cos x (1 – 2 sin x). ∴ f ′ (x) = 0



We have,

233. (b) We have f(x) = 2 + 4x2 + 6x4 + 8x6

1 cos 2x 229. (a), (d)  We have, f (x) = sin x + 2

⇒ cos x (1 – 2 sin x) = 0 ⇒ cos x = 0 or sin x =

2 2 1  475    475   − x − 870000 −  .    2   100  2  

1 4 3 < < . 2 3 2



f ′ (x) = 0

⇒ x(8 + 24x2 + 48x4) = 0 ⇒ x = 0 Since, f ′′ (x) > 0, therefore, f has a local minima at x = 0. Moreover, f has no other critical point and f ′ (x) exists at all x ∈ R. Therefore, there can be only one minima.

183

230. (b) Let the additional number of subscribes be x, so the number of subscribers becomes 725 + x and then x   the profit per subscriber is Rs. 12 − .  100 

Applications of Derivatives

f ′ (x) = 0 gives xp + q = 1 ⇒ x = 1. When x < 1, f ′ (x) < 0 and when x > 1, f ′ (x) > 0

184

234. (a) Let f(x) = 3 sin x – 4 sin3 x = sin 3x

Objective Mathematics

 π π Since, sin x is increasing in the interval  − ,   2 2 π π − ≤ 3x ≤ 2 2 π π − ≤x≤ 6 6 Therefore, the length of interval = 235. (c) Let y = x1/x

π  π π − − = 6  6  3

⇒ log y = 1 log x x log x Let f(x) = x f ′ (x) = 1 − log x x2 Put f ′ (x) = 0 for maxima and minima ⇒ 1 − log x = 0 x2 ⇒ 1 – log x = 0 ⇒ x = e Now, f '' ( x) =

− x − 2(1 − log x) x < 0 at x = e ( x 2 )2

∴ at x = e, f(x) is maximum and maximum value of x1/x = e1/e. 236. (d) We have, f(x) = 4x4 – 2x + 1 ⇒ f ′ (x) = 16x3 – 2 Put f ′ (x) = 0 for maxima and minima ⇒ 16x3 – 2 = 0 ⇒ x = 1/2 Since for x > 1 , f ′ (x) > 0 2 ∴ f(x) is increasing for x > 1 . 2 237. (d) We have, f ( x) =

x 2 + 2 x

1 2 − 2 x2 Put f ′ (x) = 0 for maxima or minima, we get

⇒

f ' ( x) =

1 2 − =0 2 x2 ⇒ x2 = 4 ⇒ x = ± 2

f ′′ (x) = 4 ⇒ f ′′ (2) = 4 = 1 > 0 x3 8 2 4 1 and f ′′ (–2) = − = − < 0 8 2 ∴ f(x) is minimum at x = 2. Now, 

238. (a) We have, y = x – 5x + 6 2

⇒ dy = 2 x − 5 dx

 dy  = 4 − 5 = −1 and ∴ m1 =    dx (2, 0)  dy  = 6 −5 =1  m2 =    dx (3, 0) Since, m1m2 = – 1 ∴ Angle between the tangents is π . 2 239. (a) Let y = xx ⇒ dy = x x (1 + log x) dx For y to be increasing, dy > 0 dx ⇒ xx (1 + log x) > 0 or 1 + log x > 0

1 e

x>

or

240. (d) We have, f(x) = – x3 + 4ax2 + 2x – 5 ⇒f ′ (x) = –3x2 + 8ax + 2

Since, f(x) is decreasing ∀x, therefore f ′ (x) < 0 ⇒ –3x2 + 8ax + 2 < 0  rom above, it is clear that for no value of a, f(x) is F decreasing. x2 + 1

241. (d) We have, f ( x) = ∫x2 ⇒ f ′ (x) = 2xe –( x = 2x(e − ( x

2

+1) 2

+1) 2

2

− 2 xe − x – 2xe–x4 4

− e− x )

 ince, e − x > e − ( x S be x < 0. 4

2

e − t dt

4

2

+ 1) 2

, therefore for f ′ (x) > 0, it should

242. (c) Let f(x) = x5 – 5x4 + 5x3 – 1 ⇒ f ′ (x) = 5x4 – 20x3 + 15x2 For maximum or minimum, put f ′ (x) = 0 ⇒ 5x4 – 20x3 + 15x2 = 0 ⇒ x2(5x2 – 20x + 15) = 0 ⇒ 5x2(x – 1) (x – 3) = 0 ⇒ x = 0, 1, 3 Now,  f ′′ (x) = 5(4x3 – 12x2 + 6x) At x = 1, f ′′ (1) = 5(4 – 12 + 6) = – 10 < 0, maximum. At x = 3, f ′′ (3) = 5(4 × 27 – 12 × 9 + 6 × 3)  = 90 > 0, minimum. At x = 0, f ′′ (0) = 5(0 – 12 × 0 + 6 × 0) = 0, f ′′′ (x) = 5(12x2 – 24x + 6) ⇒ f ′′′ (0) = 30 ≠ 0 ∴ f (x) is maximum at x = 1, minimum at x = 3 and x = 0 is point of inflection. 243. (d) We have, x 4 − x2 ⇒ f '( x) = 2 4+x+x (4 + x + x 2 ) 2 On the interval [–1, 1], f ′ (x) > 0 f ( x) =

Therefore, f(x) is an increasing function on [–1, 1]

∴

244. (a) Using mean value theorem, we have f (3) − f (1) ⇒ 1 log 3 − log 1 = 3 −1 c 2 2 ⇒ c = = 2log 3 e log e 3

e2 x − 1 e −2 x − 1 1 − e 2 x ⇒ f (− x) = −2 x = 2x e +1 e + 1 1 + e2 x e2 x − 1 = − 2x = − f ( x) ∴  f(x) is an odd function. e +1 f ( x) =

e2 x − 1 e2 x + 1 4e 2 x ⇒ f ' ( x) = > 0, ∀ x ∈ R (1 + e 2 x ) 2 ⇒ f(x) is an increasing function.

1 (cos x − sin x) 1 + (sin x + cos x) 2

Again, f ( x) =

π  2 cos  x +  4  = 1 + (sin x + cos x) 2 3π π 1

190

metHoDS of InteGratIon

Thus,

Objective Mathematics

method of transformation When the integrand is a trigonometric function, we transform the given function into standard integrals or their algebraic sum by using trigonometric formulae:

∫ [ f ( x)] f ′ (x) dx =

n ∫ t dt =

=

[ f ( x)]n + 1 n +1

n

(iv) When the integrand is of the form

1 − cos 2mx 2 1 + cos 2mx cos2mx = 2 sin mx mx cos sin mx = 2 sin 2 2 3 sin mx − sin 3 mx sin3mx = 4 3 cos mx + cos 3 mx cos3mx = 4

(ii) (iii) (iv) (v)

Thus,

(i) Choose the first and second function in such a way that the derivative of the first function and the integral of the second function can be easily found. (ii) In case of integrals of the form

By suitable substitution, the variable x in

∫

f ( x ) dx is changed

into another variable t so that the integrand f (x) is changed into F (t) which is some standard integral or algebraic sum of standard integrals. There is no general rule for finding a proper substitution and the best guide in this matter is experience. However, the following suggestions will prove useful. (i) If the integrand is of the form f ′ (ax + b), then we put ax + b = t and dx =

1 dt. a

f ' (t )

dt 1 = a a

dt

=

∫

1 n

∫ f ' (t ) dt

1 1 = f (t) = f (xn). n n (iii) When the integrand is of the form [ f (x)] n. f ′ (x), we put f (x) = t and f ′ (x) dx = dt.

n

function and f (x ((x) x) as the second function. (iii) In case of integrals of the form

log x ) ∫ (log

n

· dx, take 1 as the

second function and (log (logx)) as the first function. (logx n

(iv) Rule of integration by parts may be used repeatedly, if required. (v) If the two functions are of different type, we can choose the first function as the one whose initial comes first in the word ‘‘ILATE’’, where I

— Inverse Trigonometric function

L — Logarithmic function

E — Exponential function.

f ' (t ) dt

xn = t and nxn – 1 dx = dt.

∫ f ' (t ) n

n

T — Trigonometric function

f (t ) f ( ax + b ) ⋅ = a a (ii) When the integrand is of the form xn – 1 f ′ (xn), we put

f ' ( x n ) dx =

∫ f ( x) . x dx, take x as the first

A — Algebraic function

=

n −1

dx.

Working Hints

method of Substitution

∫x



In words, integral of the product of two functions = first function × integral of the second – integral of (differential of first × integral of the second function).

(x) 2 sin A sin B = cos (A – B) – cos (A + B).

Thus,

dt = log t = log f (x). t

 du

(ix) 2 sin A cos B = sin (A + B) + sin (A – B)

∫

∫

∫ (uv) dx = u. ∫ vdx − ∫  dx ⋅ ∫ vdx 

(viii) 2 cos A cos B = cos (A + B) + cos (A – B)

∫

dx =

The process of integration of the product of two functions is known as integration by parts. For example, if u and v are two functions of x, then

(vii) cot2mx = cosec2mx – 1

f ' ( ax + b ) dx =

f ' ( x)

∫ f ( x)

method of Integration by parts

(vi) tan2mx = sec2mx – 1

Thus,

f' ( x ) , we put f ( x)

f (x) = t and f ′ (x) dx = dt.

The following trigonometric identities are important: (i) sin2mx =

tn +1 n +1

(vi) In case, both the functions are trigonometric, take that function as second function whose integral is simpler. If both the functions are algebraic, take that function as first function whose derivative is simpler. (vii) If the integral consists of an inverse trigonometric function of an algebraic expression in x, first simplify the integrand by a suitable trigonometric substitution and then integrate the new integrand.

method of partial fractions for rational functions Integrals of the type

p ( x)

∫ g ( x)

dx can be integrated by resolving

the integrand into partial fractions. We proceed as follows:

caSe 1: When the denominator contains non-repeated linear factors. That is g (x) = (x – α1) (x – α2) ... (x – αn).

In such a case write f (x) and g (x) as:

f ( x) A1 A2 An + + ... + = x − α1 x − α 2 x − αn g ( x) where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M. caSe 2: When the denominator contains repeated as well as non-repeated linear factors. That is

Some SpecIaL InteGraLS 1.

∫

dx x 1 = tan − 1 + C x + a2 a a

2.

∫

dx 1 x−a log = +C x − a2 2a x+a

3.

∫a

4.

∫

a −x

5.

∫

x + a2

6.

∫

x − a2

∫

a 2 − x 2 dx =

∫

x 2 + a 2 dx =

g (x) = (x – α1)2 (x – α3) ... (x – αn). In such a case express f (x) and g (x) as:

f ( x) A1 A2 A3 An + + ... + + = g ( x) x − α1 ( x − α1 )2 x − α 3 x ( − αn ) where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M. caSe 3: When the denominator contains a non repeated quadratic factor which cannot be factorised further:

7.

8.

g (x) = (ax2 + bx + c) (x – α3) (x – α4) ... (x – αn).

2

2

2

dx 1 a+x log = +C − x2 2a a−x

dx 2

2

dx 2

dx 2

= sin– 1

= log x + x 2 + a 2 + C = log x + x 2 − a 2 + C

x x a2 a2 − x2 + sin– 1 +C a 2 2

x 2

where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M.

9.

∫

x 2 − a 2 dx =

(a)

In such a case write f (x) and g (x) as:

Note:

where A1, A2, ... An are constants to be determined by comparing the coefficients of various powers of x on both sides after taking L.C.M. Corresponding to repeated linear factor (x – a)r in the denominator, a sum of r partial fractions of the type

A1 A2 Ar + ... + + is taken. x − a ( x − a )2 ( x − a )r

x 2

x2 − a2 −

a2 2

Integrals of the form

g (x) = (ax2 + bx + c)2 (x – α5) (x – α6) ... (x – αn)

An + ... + x − α ( n)

a2 2

log x + x 2 − a 2 +C

caSe 4: When the denominator contains a repeated quadratic factor which cannot be factorised further. That is

f ( x) A1 x + A 2 A3 x + A 4 A5 = + + g ( x) ax 2 + bx + c ( ax 2 + bx + c )2 x − α 5

x2 + a2 +

2 2 log x + x + a + C

In such a case express f (x) and g (x) as:

f ( x) A1 x + A 2 A3 An + + ... + = 2 ax + bx + c x − α 3 x − αn g ( x)

x +C a

(c)

− x 2 ) dx ,

(b)

∫ f (a

f ( x 2 − a 2 ) dx ,

(d)

∫

∫ f (a ∫

2

2

+ x 2 ) dx ,

a − x f  dx,  a + x 

Working Rule Integral

∫ f (a ∫ f (x

Substitution

) dx + a ) dx dx ∫ f (x − a ) ∫

2

x = a sin θ or x = a cos θ

2

2

x = a tan θ or x = a cot θ

2

2

x = a sec θ or x = a cosec θ

2

−x

a − x f dx or  a + x 

∫

a + x dx f  a − x 

x = a cos 2θ

191

Indefinite Integration

• Check degree of p (x) and g (x). caSe 5: If the integrand contains only even powers of x • If degree of p (x) > degree of g (x), then divide p (x) by g (x) (i) Put x2 = z in the integrand. till its degree is less, i.e., put in the form (ii) Resolve the resulting rational expression in z into partial p ( x) f ( x) fractions = r (x) + (iii) Put z = x2 again in the partial fractions and then integrate g ( x) g ( x) both sides. where degree of f (x) < degree of g (x).

192

∫ sin

Integrals of the form

m

x cos n x dx

Working Rule (i) Divide the numerator and denominator by cos2x.

Objective Mathematics

Working Rule

(ii) In the denominator, replace sec2x, if any, by 1 + tan2x.

(i) If the power of sin x is an odd positive integer, put cos x = z.

(iii) Put tan x = z ⇒ sec2x dx = dz.

(ii) If the power of cos x is an odd positive integer, put sin x = z.

(iv) Integrate the resulting rational algebraic function of z.

(iii) If the power of sin x and cos x are both odd positive integers, put sin x = z or cos x = z.

(iv) In the answer, put z = tan x.

(iv) If the power of sin x and cos x are both even positive integers, use De’ Moivre’s theorem as follows :

a cos x + b sin x

∫ c cos x + d sin x dx

Integrals of the form:

Let cos x + i sin x = z. Then cos x – isin x = z–1 Adding these, we get z+



Working Rule

1 1 = 2 cos x and z – = 2i sin x z z

(i) Put Numerator = λ (denominator) + µ (derivative of denominator)

By De’ Moivre’s theorem, we have

1 1 zn + n = 2cos nx and zn – n = 2i sinnx ...(1) z z n m 1 1  1  1 ⋅ n z +  z −  m ∴ sinmx cosnx = (2 i ) 2  z   z  1 =

2m + n

1 ⋅ m i

n

m

1  1   z +   z −  . z z

Now expand each of the factors on the R.H.S. using Binomial theorem. Then group the terms equidistant from the beginning and the end. Thus express all such pairs as the sines or cosines of multiple angles. Further integrate term by term. (v) If the sum of powers of sin x and cos x is an even negative integer, put tan x = z.

a cos x + b sin x = λ (c cos x + d sin x) + µ (– c sin x + d cos x). (ii) Equate coefficients of sin x and cos x on both sides and find the values of λ and µ. (iii) Split the given integral into two integrals and evaluate each integral separately, i.e.,

a cos x + b sin x

∫ c cos x + d sin x dx = λ

∫ 1 dx + µ ∫ = λx + µ log | a cos x + b sin x |.

(iv) Substitute the values of λ and µ found in step 2.

Integral of the form dx , (a) ∫ a + b cosx dx (c) ∫ a + b cos x + c sin x

dx (b) ∫ , a + b sinx

− c sin x + d cos x dx a cos x + b sin x

Integrals of the form

a + b cos x + c sin x

∫ e + f cos x + g sin x dx

Working Rule (i) Put Numerator = l (denominator) + m (derivative of denominator) + n

Working Rule

x x 1 − tan 2 tan 2 and sin x = 2 so that the given (i) Put cos x = x x 1 + tan 2 1 + tan 2 2 2 x integrand becomes a function of tan . 2 x 1 x (ii) Put tan =z⇒ sec2 dx = dz 2 2 2 2

(iii) Integrate the resulting rational algebraic function of z

x (iv) In the answer, put z = tan ⋅ 2

dx ∫ a + b cos2 x ,

(c)

∫ a cos

2

(b)

+ m (– f sin x + g cos x) + n (ii) Equate coefficients of sin x, cos x and constant term on both sides and find the values of l, m, n. (iii) Split the given integral into three integrals and evaluate each integral separately, i.e.,

a + b cos x + c sin x

∫ e + f cos x + g sin x dx ∫

= l 1dx + m

− f sin x + g cos x

∫ e + f cos x + g sin x dx + n dx

∫ e + f cos x + g sin x

Integrals of the form (a)

a + b cos x + c sin x = l (e + f cos x + g sin x)

dx

∫ a + b sin

dx x + b sin x cos x + c sin 2 x

2

x

,

= lx + m log | e + f cos x + g sin x | +n

dx

∫ e + f cos x + g sin x dx

(iv) Substitute the values of l, m, n found in Step (ii).

∫

(b)

∫

ax + bx + c

(c)

∫

ax + bx + c dx a

dx

Working Rule (i) Make the coefficient of x2 unity by taking the coefficient of x2 outside the quadratic. (ii) Complete the square in the terms involving x, i.e., write ax2 + bx + c in the form a [x ± α)2 ± β2].

,

2

(iii) The integrand is converted to one of the nine special integrals. (iv) Integrate the function.

2

px + q

Integrals of the form (a) ∫ ax 2 + bx + c dx, (b) Integral

∫ ax

193

(a)

dx , ax 2 + bx + c

∫

px + q

dx,

ax 2 + bx + c

(c)

∫ ( px + q )

ax 2 + bx + c dx

Working rule

px + q dx + bx + c

Put px + q = λ (2ax + b) + µ or px + q = λ (derivative of quadratic) + µ. Comparing

2

the coefficient of x and constant term on both sides, we get p = 2aλ and q = bλ + µ p bp   ⇒λ= and µ =  q −  Then the integral becomes 2a  2a 

∫ ax

px + q dx + bx + c

2

px + q

∫

ax 2 + bx + c

=

p 2a

=

bp  dx p  ⋅ log | ax2 + bx + c | +  q −  ∫ 2  2a 2a ax + bx + c

In this case the integral becomes

dx

px + q

∫ =

∫ ( px + q )

2ax + b dx dx +  q − bp  2   + bx + c 2a  ∫ ax 2 + bx + c

∫ ax

ax 2 + bx + c dx

ax 2 + bx + c p a

dx =

p 2a

∫

bp   dx +  q − ∫ 2a  ax 2 + bx + c

dx

2ax + b

bp   ax 2 + bx + c +  q −   2a  ∫

ax + bx + c 2

dx ax 2 + bx + c

The integral in this case is converted to p 2 ∫ ( px + q ) ax + bx + c dx = 2a ∫ (2ax + b) ax 2 + bx + c dx

bp   +  q −  2a  ∫ =

ax 2 + bx + c dx

p (ax2 + bx + c)3/2 + 3a

bp    q −  2a  ∫

ax 2 + bx + c dx

P ( x)

x2 + 1 4 Integrals of the form: dx, where P (x) is a ∫ x + kx 2 + 1 dx or ax 2 + bx + c x2 − 1 polynomial in x of degree n. ∫ x 4 + kx 2 + 1 dx, where k is a constant positive, negative

Integrals of the form:

∫

or zero. Working Rule: Write

∫

Working Rule

P( x)

2 n–1 ax + bx + c dx = (a0 + a1x + a2x + ... + an– 1 x )

2

ax + bx + c 2

+k

∫

dx

ax + bx + c 2

where k, a0, a1, ... an – 1 are constants to be determined by differentiating the above relation and equating the coefficients of various powers of x on both sides.

(i) Divide the numerator and denominator by x2. (ii) Put x –

1 1 = z or x + =z x x

whichever substitution, on differentiation gives, the numerator of the resulting integrand. (iii) Evaluate the resulting integral in z (iv) Express the result in terms of x.

Indefinite Integration

Integrals of the form

194

∫P

Integrals of the form: quadratic functions of x.

dx , where P, Q are linear or Q

Objective Mathematics

Integral

Substitution

1

∫ (ax + b)

∫ (ax

2

dx + bx + c )

px + q

ax 2 + bx + c

dx

2

∫

dx

x m ( a + bx )

p

Substitution Put a + bx = zx

,

cx + d = z2

where either (m and p are positive integers) or (m and p are fractions, but m + p = integers > 1)

px + q = z2

∫ x (a + bx )

1 px + q = z

(i) p is a + ve integer

Apply Binomial theorem to (a + bxn)p.

(ii) p is a – ve integer

Put x = zk where k = common denominator of m and n.

n p

m

dx,

where m, n, p are rationals.

dx

∫ ( px + q )

∫ (ax

dx

cx + d

Integral

x=

+ b ) cx 2 + d

∫

Integrals of the form: n are integers.

1 ⋅ z

(

f x, ( ax + b )

α/n

) dx, where α and

Working Rule

(iii)

m +1 is an integer n

(iv)

m +1 + P is an integer n

Put ax + b = zn.

α/n

, ( ax + b )

β/ m

) where α, β, m, n are integers.

(ii) Integrate only the first integral by parts, i.e.,

Put ax + b = zk, where k = l.c.m. (n, m).

x ∫ e  f ( x ) + f ' ( x ) dx =

Integrals of the form (a) (b) (c)

x =  f ( x ). e −



m

∫ (a + bx)

p

p

m

p

Integral

∫ (a + bx)

m

n p

dx.

Substitution Put a + bx = z p

dx,

m is a + ve integer

∫ f ' ( x).e

x

x

dx  +

∫e

x

x

f ' ( x ) dx

f ' ( x ) dx

Integrals of the form

dx

m

∫ e f ( x) dx + ∫ e

= ex f (x) + C.

dx

∫ x (a + bx) dx ∫ x (a + bx) a ∫ x (a + bx ) xm

Working Rule (i) Split the integral into two integrals.

Working Rule

x

Put a + bxn = xn zk where k = denominator of fraction p.

x Integrals of the form: ∫ e  f ( x ) + f ' ( x ) dx

Integrals of the form

∫ f ( x, (ax + b)

Put a + bx n = z k , where k = denominator of p.

where the initial integrand reappears after integrating by parts. Working Rule (i) Apply the method of integration by parts twice. (ii) On integrating by parts second time, we will obtain the given integrand again. Put it equal to I. (iii) Transpose and collect terms involving I on one side and evaluate I.

Choose the correct alternative in each of the following problems: 1.

1 + x4

∫ (1 − x )

4 3/2

(a)

6.

dx =

1

+c

(b)

1 x − 2 x 1 +c 1 2 +x x2 2

(c)

2. If

1 +c 1 2 − x x2

1 sin (2x – c) + a, then 2

π , a = arbitrary constant 2 π (b) c = , a = arbitrary constant 6 π (c) c = , a = arbitrary constant 4 (d) None of these

(b) cosec α log

sin ( x + α ) +C sin x

(c) cosec α log

sec ( x + α ) +C sec x

(d) cosec α log

sec x + C. sec ( x + α )

(a) c =

3. If

∫

4. If

dx

∫ 1 + sin x

1 1 (b) A = , B = 9 5

∫x

dx

 a+x − a−x

∫ 

(a) – (b)

a−x dx is equal to a + x 

a2 − x2 + C a2 − x2 + C

(c) –

x2 − a2 + C

(d) None of these 9.

∫ [1 + tan x. tan ( x + α)

dx is equal to

sin x (a) cot α . log sin ( x + α ) + C sin x (b) tan α. log sin ( x + α ) + C sin ( x + α) + C sin x (d) None of these (c) cot α. log

=

(a)

1 log 3

1 − x3 − 1

(b)

1 log 3

1 − x2 + 1

1 log 3

8.

x  = tan  + a  + b, then 2 

1 − x3

(c)

cos3 x + cos5 x dx is 2 x + sin 4 x

∫ sin

(a) sin x – 6 tan–1 (sin x) + c (b) sin x – 2 (sin x)–1 + c (c) sin x – 2 (sin x)–1 – 6 tan–1 (sin x) + c (d) None of these

(d) None of these

π (a) a = – , b = arbitrary constant 4 π , b = arbitrary constant (b) a = 4 π (c) a = , b = arbitrary constant 2 (d) None of these 5.

7. The value of

−9 −5   cos3 x dx = − 2  A tan 2 x + B tan 2 x  + C, then 11 sin x  

−1 1 (a) A = ,B= 5 9 1 1 (c) A = –  , B = 9 5

is equal to

sin x (a) cosec α log sin ( x + α ) + C

(d) None of these

∫ (sin 2 x + cos 2 x) dx =

dx

∫ sin x. sin ( x + α)

1 − x3 + 1 1 − x2 − 1 1 1− x

3

+c

1 (d) log 1 − x 3 + c 3

+c +c

10.

∫

ex − 1 dx is equal to ex + 1

( (b) log (e (c) log (e

) − 1) + sec − 1) – sec

(a) log e x + e 2 x − 1 – sec– 1 (ex) + C x

2x

+ e

x

− e2 x

(d) None of these

– 1

(ex) + C

– 1

(ex) + C

195

Indefinite Integration

mULTIPLE-CHOICE QUESTIONS

196

11.

∫ tan ( x − α) tan ( x + α)

tan 2x dx is equal to

17. If

Objective Mathematics

4 4 8 , B = –  , C = 15 35 25 4 4 8 , B = –  , C = –  (b) A = 15 35 25 4 4 8 (c) A = , B = –  , C = 15 35 25 (d) None of these

sec 2x . sec ( x − α ) +C sec ( x + α )

(c) log

(d) None of these 12. If

∫

dx A B x = 2 + + log + C, then x 4 + x3 x x x +1

1 , B = 1 2 1 (c) A = –  , B = 1 2

(a) A =

13.

∫

dx

x 2 (1 + x 4 )

3/ 4

(a) –

(1 + x )

(c) –

(1 + x )

4 1/ 4

x

4 3/ 4

14.

x

∫ ( x + 1)

∫ sin

(b)

1 2

19.

(b) (1 + x x

)

4 1/ 4

(d) None of these

4

x dx is equal to x+3x

∫

(a) x –

6 5/ 6 3 2/3 x − x – 2x1/2 + 3x1/3 + 6x1/6 5 2 + 6 log (1 + x1/6) + C

(b) x –

6 5/ 6 3 2/3 x + x – 2x1/2 + 3x1/3 – 6x1/6 5 2 + 6 log (1 + x1/6) + C

6 5/ 6 3 2/3 x + x – 2x1/2 + 3x1/3 – 6x1/6 5 2 + 6 log (1 + x 1/6) + C (d) None of these (c) x +

 +C

20. If

dx is equal to x + cos 4 x

(a) (– sin α, cos α) (c) (sin α, cos α) 21.

 1  2 tan– 1  tan 2 x  + C  2 

 1  1 cot 2 x  + C tan– 1    2 2 (d) None of these

then f (x) =

cos x dx =

1 log f (x) + c, 2 (b − a 2 )

= Ax + B log sin(x – a) + C, then value

of (A, B) is

1  1  tan– 1  tan 2 x  + C   2 2

∫ f ( x) sin x

sin x

∫ sin( x − α) dx

∫ cos x

(b) (cos α, sin α) (d) (– cos α, sin α)

dx is equal to cos 2 x

(a)

5   1 2  tan x + tan 2 x  + C 5  

(b)

5   1 2  cot x + tan 2 x  + C 5  

(c)

5   1 2  tan x − tan 2 x  + C 5  

(c)

16. If

3 3  = −  f ( x ) + g ( x )  + C, then   8 2 sin x cos x 11



is equal to

( x + 1)2

dx

3



+C

x2 + 2x + 1

∫

(a) f (x) = tan–8/3 x, g (x) = tan–2/3x (b) f (x) = tan8/3x, g (x) = tan–2/3x (c) f (x) = tan–8/3x, g (x) = tan2/3x (d) None of these

x2 + 2x + 2 + C (d) None of these x +1

(c) –

(a)

x + 2x + 2 2

18. If

(d) None of these

x 2 + 2 x + 2 + C (b) – x +1

(a)

15.

+ C

dx 2

(b) A = 1, B = – 

is equal to

+ C

⋅ (1 + x 5/ 2 )1/ 2 dx

(a) A = – 

sec 2x +C sec ( x − α ). sec ( x + α )

(b) log

13/ 2

= A (1 + x5/2)7/2 + B (1 + x5/2)5/2 + C (1 + x5/2)3/2, then

sec 2x . sec ( x + α ) +C sec ( x − α )

(a) log

∫x

(d) None of these

2

22.

dx

∫ cos x − sin x

is equal to

(a)

1 1 (b) 2 2 a 2 sin 2 x + b 2 cos 2 x a sin x − b 2 cos 2 x

(a)

1  x 3π  log tan  −  + C 2 8  2

(c)

1 1 (b) 2 a 2 cos 2 x + b 2 sin 2 x a cos 2 x − b 2 sin 2 x

(b)

1 x log cot   + C 2 2

1  x π log tan  −  + C 2 8 2

(d)

1  x 3π  log tan  +  + C 2 8  2

(a) log

x  + C x + cos x

(b) log

dx is equal to x sin 2 x

(c) log

1  + C x + cos x

(d) log| x + cos x | + C

∫ cos

3

1  5/ 2  2    cot x + tan x  + C   5

(a)

30.

∫

x2 − 2 x3

x2

(a) (c) 25.

∫

(b) –

x2 − 1 x2

(d) –

x2 x2 − 1 x2 − 1 x2

(a) 2

1 − x + cos − 1

x + x − x2 + C

(b) 2

1 − x − cos − 1

x + x − x2 + C

1 − x + cos − 1

x + x − x2 + C

(d) None of these 26.

∫

d (cos θ)

32.

1 − cos θ

dx

∫ sec x + cos ec x

(b) θ + c (d) sin–1 (cos θ) + c =

1 1 g (x), then f ( x) − 2 2 2

 x π (a) f (x) = sin x – cos x, g (x) = log cot  +  2 8  x π (b) f (x) = sin x – cos x, g (x) = log tan  +  2 8  x π (c) f (x) = cos x – sin x, g (x) = log tan  +  2 8  x π (d) f (x) = cos x – sin x, g (x) = log cot  +  . 2 8 28. If

∫

2 1 + sin x dx = – 4 cos (ax + b) + c, then the value

of (a, b) is equal to 1 π (a) , 2 4 (c) 1, 1

x2 x2

2

2

2

2

2

2

2

2

2

x x ⋅ tan 2 2 dx is equal to 31. ∫ cos x x x 1 − 2 sin 1 + sin 1 2 2 − log log +C (a) x x 2 1 + 2 sin 1 − sin 2 2 x x 1 + 2 sin 1 + sin 1 2 − log 2 +C log (b) x x 2 1 − 2 sin 1 − sin 2 2 x x 1 + 2 sin 1 − sin 1 2 2 log − log +C (c) x x 2 1 − 2 sin 1 + sin 2 2 (d) None of these

2

(a) cos–1 θ + c (c) sin–1 θ + c 27. If

) dx is equal to +a )+ x +a +C +a )−2 x +a +C +a )− x +a +C

x2 + a2

(d) None of these

1− x dx is equal to 1+ x

(c) – 2

x + cos x +C x

sin



x2 − 1

dx is equal to

(a) x log x + x 2

dx is equal to

x2 − 1

∫ log ( x +

( (b) x log ( x + (c) x log ( x +

1   2  tan x + tan 5 / 2 x  + C   5 1 (c) 2  tan x − tan 5 / 2 x  + C   5 (d) None of these (b)

24.

∫ x ( x + cos x)

197

29.

π (b) 1, 2 (d) none of these

∫ sin

−1

x dx is equal to a+x

(a) (x + a) tan– 1

x − ax + C a

(b) (x + a) tan– 1

x + ax + C a

(c) (x + a) cot– 1

x − ax + C a

(d) None of these 33.

∫

tan x a + b tan 2 x

dx, a > b > 0, is equal to

(a)

 1 b  +C log  cos x + cos 2 x + a − b  a−b 

(b)

 1 b  +C log  sin x + sin 2 x + a − b  a−b 

(c) –

 1 b  +C log  cos x + cos 2 x + a − b  a−b 

(d) None of these

Indefinite Integration

23.

cos x + x sin x

(c)

198

34. For the function f (x) = 1 + 3x log 3, the antiderivative F assumes the value 7 for x = 2. At what value of x does the curve y = F(x) cut the abscissa?

Objective Mathematics

(a) x = 3 (c) x = 0 35.

(b) x = 1 (d) None of these

tan x

∫

41.

(

)

(a) log tan 2 x + 1 + tan 4 x + C

( 1 (c) log ( tan x + 4

42.

) 1 + tan x ) + C

1 log tan 2 x + 1 + tan 4 x + C 2

(b)

2

∫

4

43.

then x 1+ x 2x

(c) f (x) = 37.

2

1 + x2

−x



(b) f (x) =



(d) None of these

dx

∫

1 + x2 44.

sin x sin ( x + α )

cos α + sin α cot x + C

(a) 2 cosec α

cos α + sin α cot x + C

(b) – 2 cosec α

45.

x

∫ log (1 + cos x) − x tan 2 

dx is equal to

x 2 (b) log (1 + cos x) (c) x log (1 + cos x) (d) None of these

(a) x tan

39. Let f (x) =

40.

(a)

−1 2

(c)

2

∫

∫

(c)

2 3/ 2

and f (0) = 0, then f (1) = (b)

)

x −1 1− x

2

+ C

+ C

sin 3 x dx 1 + cos 2 x + cos 4 x

2

x

∫ cos x. log tan 2

(a) 1 –

1 2 47.

is equal to

(b)

(

2 1− x 1− x

π 4

(c) tan 1 –

(d) None of these.

1− x

(

∫ (1 + cos x)

is equal to

dx is equal to

∫

(x

2

+ sin 2 x ) sec 2 x dx and f (0) = 0, 1 + x2

then f  (1) =

(1 + x )

(1 + x ) x − x 2 ( x − 1)

(a)

1 − sin x − x / 2 ⋅e dx is equal to 1 + cos x x x (a) sec ⋅ e − x / 2 + C (b) – sec ⋅ e − x / 2 + C 2 2 x (c) tan ⋅ e − x / 2 + C (d) None of these 2

∫

46. If f (x) =

dx

dx

− sin 2 x – sin x + C 2 − sin 2 x (d) + sin x + C. 2 (b)

(a) sin x. log tan

(d) None of these  38.  

cos 5 x + cos 4 x dx is equal to 1 − 2 cos 3 x

x + x + C 2 x (b) sin x log tan –x+C 2 x (c) – sin x log tan – x + C 2 (d) None of these

cos α + sin α cot x + C

(c) cosec α

(d) None of these

(a) sec–1(sec x + cos x) + C (b) sec–1(sec x – cos x) + C (c) sec–1(cos x – tan x) + C (d) None of these

is equal to

3

∫

 x −1  (b) ex  +C  1 + x 2 

sin 2 x  + sinx + C 2 sin 2 x – sinx + C (c) 2

1 1 − tan −1 x arc tan x − 2 − log f (x) + C, dx = 4 3x3 6x 3 x

(a) f (x) =

dx is equal to

(a)

(d) None of these 36. If

2

 1− x  (a) ex  + C  1 + x 2  1 + C (c) ex. 1 + x2

dx is equal to

sin 4 x + cos 4 x

∫

1 − x  e   1 + x  x

)

∫

(d) None of these

π 4

(

log x + 1 + x 2

(a) +C

(b)

1+ x

2

π – 1 4

(d) None of these

)

dx is equal to

(

)

2 1 log x + 1 + x 2  + C   2

(

)

2

(b) log x + 1 + x 2  + C   1 (c) log x + 1 + x 2  + C  2  (d) None of these

(

)

49.

3

x − sin x )

cos 2 x (a) esin x (sec x – x) + C (c) esin x (tan x – x) + C

dx is equal to (b) esin x (x – sec x) + C (d) None of these

∫

x 2 + 1  log ( x 2 + 1) − 2 log x  dx is equal to x4

(a)

( x 2 + 1)3 / 2 x3

2  x2 + 1   − log  2   + C  x  3

(b)

( x + 1) 3x3

  x + 1 log  2  −  x  

2

3/ 2

2

 x2 + 1  ( x 2 + 1)3 / 2  2  − log  2   + C 3 3x  x   3 (d) None of these

50. If

∫





54. If

2  +C 3

(c)

55.

1 log x 56.

(d) None of these 51.

3x + 1

∫ ( x − 1) ( x + 1) dx is equal to 1 1 1 + − log (a) ( x − 1) 2 2 ( x − 1) 4

x +1 x −1 + C

−1 1 1 − + (b) log ( x − 1) 2 2 ( x − 1) 4

x +1 x −1 + C

−1 1 1 + + log ( x − 1) 2 2 ( x − 1) 4

x +1 x −1 + C

57.

∫

 x + 2 ex   x + 4 

x2

∫ ( x sin x + cos x)

52. If

2

dx =

x cos x −x (c) f (x) = cos x

f ( x) + tan x + C, x sin x + cos x

∫

(b) f (x) =

cos x x

(d) None of these

x4 dx is equal to ( x − 1) ( x 2 + 1)

(a)

x2 1 1 + x +  log (x – 1) –  log (x2 + 1) 2 2 4 1  tan– 1x + C – 2

dx is equal to

(a)

xe x + C x+4

 x + 2 (b) ex  +C  x + 4 

(c)

ex + C x+4

(d) None of these

dx

∫ sin x + sin 2 x

is equal to

1 1 log (1 – cos x) – log (1 + cos x) 6 2 2 + log (2 cos x + 1) + C 3 1 1 (b) log (1 – cos x) + log (1 + cos x) 6 2 2 log (2 cos x + 1) + C – 3 1 1 (c) log (1 – cos x) + log (1 + cos x) 6 2 2  log (2 cos x + 1) + C + 3 (d) None of these

then (a) f (x) =

2

(a)

(d) None of these



x3 − 1 dx is equal to x3 + x

3

(c)

53.

∫

(b) 2 (d) –2

1 log (x2 + 1) – tan– 1x + C 2 1 log (x2 + 1) – tan– 1x + C (b) x – log x + 2 1 log (x2 + 1) – tan– 1x + C (c) x – log x – 2 (d) None of these

1 log x

1 , g (x) = log (log x) (c) f (x) = log x

cot x + Q, then P equals

(a) x + log x +

then

(b) f (x) = log x, g (x) =

cot x

∫ sin x cos x dx = P

(a) 1 (c) –1

 1   log (log x ) +  dx = x [ f (x) – g (x)] + C, (log x )2  

(a) f (x) = log (log x), g (x) =

x2 1 1 +x+ log (x – 1) +  log (x2 + 1) 2 2 4 1  tan–1 x + C – 2 x2 1 1 – x +  log (x – 1) +  log (x2 + 1) (c) 2 2 4 1 +  tan–1x + C 2 (d) None of these (b)

−1

58.

∫

e tan x (1 + x + x2) dx is equal to (1 + x 2 ) −1

(a)

e tan x 1 + x2

(c) xe tan

−1 x

(b) e tan

−1 x

 · (1 + x2)

(d) None of these

199

∫e

( x cos .

Indefinite Integration

48.

sin x

200

59.

∫ sin x.

(c) f (x) = tan

1 1 + 2 sin x 1 1 + sin x log +C − log 2 2 1 − sin x 1 − 2 sin x 4

(a)

Objective Mathematics

α π x  tan   −   2 4 2 (d) None of these

cosec 4x dx is equal to

1 + 2 sin x 1 1 + sin x log +C + log (b) 2 2 1 − sin x 1 − 2 sin x 4 1

64. If

(a) f (x) =

1 − 2 sin x 1 1 + sin x − log log +C (c) 1 − sin x 2 2 1 + 2 sin x 4 65.

 cos x + sin x  1 sin 2 x log  − log sec 2x + C 2  cos x − sin x  2

x



∫ cos  b log a  (a) –

66.

x  x x    cos  b log  + b sin  b log   + C   1 + b 2  a a 

x  x x    cos  b log  − b sin  b log   + C   1 + b 2  a a 

(c)

x  x x    cos  b log  + b sin  b log   + C   1 + b 2  a a 

∫ (x (a)

2

67.

2

2

1 − a2 )

x  −1 x − a tan − 1  + C  b tan a b

(b)

x 1  −1 x + a tan − 1   + C  b tan b2 − a 2  b a

(c)

1 b2 − a 2

x  −1 x + a tan − 1   + C  − b tan b a

(d) None of these. dx π 63. If ∫ (0 < α < and α is a constant) 1 + sin α. cos x 2 = 2 sec α tan– 1 [f (x)] + C, then (a) f (x) = cot

x π α tan  −  4 2 2

(b) f (x) = tan

x π α tan  −  4 2 2

1 2

1+ x dx is equal to 3 x

∫ 1+

sin 3 x dx ∫ (cos4 x + 3 cos2 x + 1) tan −1 (sec x + cos x) =

∫x

x1/ 2 dx is equal to − x1/ 3

1/ 2

5 2 1 1   1 1 6 3 3 2 x x x x x (a) 6  + + + + + x 6 + log( x 6 − 1) + C 6 5  4 3 2   5 2 1 1   1 1 x x6 x3 x2 x3 (b) 6  − + − + + x 6 − log( x 6 − 1) + C 6 5  4 3 2   5 2 1 1   1 1 x x6 x3 x2 x3 (c) 6  + − + − + x 6 + log( x 6 − 1) + C 6 5  4 3 2  

x2 dx is equal to + a ) ( x2 + b2 )

(b

(d) L =

(d) None of these

(d) None of these 62.

(b) g(x) = log x

(a) tan–1 (sec x + cos x) + c (b) log tan–1 (sec x + cos x) + c 1 +c (c) (sec x + cos x )2

dx is equal to

(b)

+ Lx + c,

3 5/3 3 4/3 x − x + x + C 5 4 3 5/3 3 4/3 (b) x + x + x + C 5 4 3 5/3 3 4/3 (c) x + x – x + C 5 4 (d) None of these

(c)

61.

2

(a)

 cos x + sin x  1 sin 2 x (a) log  + log sec 2x + C 2  cos x − sin x  2

 cos x − sin x  1 sin 2 x log  − log sec 2x + C 2  cos x + sin x  2 (d) None of these

x2 2

(c) L = 1

(d) None of these

(b)

1

then

1

 cos x + sin x  dx is equal to 60. ∫ cos 2x log   cos x − sin x 



∫ x log 1 + x  dx = f (x) log (x + 1) + g(x) log x

(d) None of these 68.

dx

∫ ( x − α) (β − x) , where β > α, is equal to  2x + α + β  (a) sin– 1  + C  β − α   x − α − β (b) sin– 1  +C  β − α   2x − α − β  (c) sin– 1  +C  β − α  (d) None of these

dx 5

(a)

3 4

4

x −1 + C x+2

(b)

(c)

4 3

4

x+2 + C x −1

(d) None of these

70. If

dx

∫ (1 + x )

4 3

x −1 +C x+2

4

(a) f (x) =

1− x 1 + x2

(b) f (x) =

(c) f (x) =

x2 − 1 x2 + 1

(d) None of these

2

71.

∫

 3x − 4  = x + 2, then 74. If   3 x + 4  (a) ex – 2 ln

1 sin– 1 [  f (x)] + C, then 2

=–

1 − x2

2

4  1 + x 3 / 4 − log (1 + x 3 / 4 ) + C 3 4 (c)  1 − x 3 / 4 + log (1 + x 3 / 4 ) + C 3 (d) None of these (b)

is equal to

( x − 1) ( x + 2) 3

4

1+ x 1 − x2

75.

−

5

 +C 

4

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 (b) 6  − + 9 8 7  −



76.

6

5

−

4

 

+

+

5

(d) None of these 72.

(x

∫x

2

− 1)

x 4 + 3x 2 + 1

(a) log x + (b) log x −

(c)

(c) log x + x + 3

+C

2

∫ 1+ (a)

x 4

x3

dx is equal to

4  1 + x 3 / 4 + log (1 + x 3 / 4 ) + C 3

dx is equal to

)

tan x + cot x

)

dx is equal to

2 cos– 1 (sin x – cos x) + C

4

+ C 

 3  x x + x +1  e ∫  1 + x 2 3/2  dx is equal to )  (

(c)

(d) None of these 73.

−

77.

dx is equal to

1 1 + x2 + 2 − 3 + C x x 2

∫(

(a)

1 1 + x2 + 2 + 3 + C x x

(

x 1+ 3 x

(d) None of these

(1 + x ) (1 + x )5 / 6 (1 + x )2 / 3  6

x + 3 x2 + 6 x

(b) 2 sin– 1 (sin x – cos x) + C

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 (c) 6  + − 9 8 7  

∫

(a) sin– 1 (sin x – cos x) + C

(1 + x ) (1 + x )5 / 6 (1 + x )2 / 3  + C +

3x − 4 +c 3x + 4

(a) 3 x2/3 + 6 tan– 1 x1/6 + C 2 (b) 3 x2/3 – 6 tan– 1 x1/6 + C 2 − 3 2/3 x + 6 tan– 1 x1/6 + C (c) 2 (d) None of these

(1 + x ) (1 + x )5 / 6 (1 + x )2 / 3  +

is equal to

(d) None of these

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 + + (a) 6  9 8 7 

6

∫ f ( x) dx

(b) – 8 ln x − 1 + 2 x + c 3 3 (c) 8 ln x − 1 + x + c 3 3

2

x dx is equal to 3 1+ x + 2 1+ x

−

201

∫

x 2e x

+c

(b)

+c

(d)

(1 + x )

2 1/ 2

ex

(1 + x )

2 1/2

78. The value of

∫

1+ x −1 1+ x +1

 1 + x − 1 (b) ln  +c  1 + x + 1 (c) 2 1 + x + c 1+ x −1 1+ x +1

x (1 + x 2 )

1/2

xe x

(1 + x )

2 1/2

1− x dx is x

(a) 2 1 + x + ln

(d)

ex

+c

+c

+c

+c

Indefinite Integration

69.

202

79.

2

∫ (2 − x )

2

3

2− x dx is equal to 2+ x

Objective Mathematics

4 2 + x 3  2 − x 

2/3

3 2 − x (c) 4  2 + x 

2/3

(a)

80. The value of

∫x

(b)

 + C

(d) None of these

dx

n

(1 + x n )

1− n

1 n

1−

1 n

1−

1  1 1 − n  x  1+ n 

1  1 1 − n  x  1− n 

81.

1 n

1 1− n

1  1 1 + n  n −1  x 

(d) –

1− x

∫1+x

2

2

⋅

dx 1 + x4

85.

+c 86.

is equal to

(b)

x 2 1 sin − 1  2  +C 2  x + 1

87.

(a) f (θ) = sin 2θ + 1

(b) f (θ) = 1 – sin 2θ

(c) f (θ) = sin 2θ – 1

(d) None of these

∫

x3

(1 + x )

2 1/ 3

dθ

88.

∫

dx

( 1+ x

)

(n ≠ ± 1) =

n

1  zn +1 zn −1  + + C, 2  n + 1 n − 1 

1 + x2

(b) z =

(c) z = x +

1 + x2

(d) None of these

(1 + x ) 4

∫ (1 − x )

4 3/ 2

(a)

 xn  1 log  n +c n  x + 1 

(c)

(d) None of these

−x

2

(a) z = x –

dx dx is equal to n + 1)

∫ x (x

 xn  +c (c) log  n  x + 1 

dx is equal to

where

= cosec– 1 [ f (θ)] + C, then

 xn + 1  1 (b) log  n  + c n  x 

x 2

(a) 20 (1 + x2)2/3 (2x2 – 3) + C 3 (b) 3 (1 + x2)2/3 (2x2 – 3) + C 20 (c) 3 (1 + x2)2/3 (2x2 + 3) + C 20 (d) None of these

x 2 1 sin − 1  2   + C 2  x + 1

(sin θ − cos θ) ∫ (sin θ + cos θ) sin θ cos θ + sin 2 θ cos2 θ

(a)

dx is equal to

(1 + xe )

 1 + xe x  1 (c) log  +C +  xe x  1 + xe x (d) None of these

(d) None of these

83.

( x + 1)

 x  1 (b) log  +C +  1 + xe x  1 + xe x

+c

x 2 2 sin − 1  2 +C  x + 1

82. If

∫x

 xe x  1 +C (a) log  +  1 + xe x  1 + xe x

(a)

(c)

, n ∈ N, is

+c

1−

(c) –

(a) tan– 1x +

+C

+c

x4 + 1 ∫ x6 + 1 dx is equal to 1 tan– 1 x3 + C 3 1 (b) tan– 1x – tan– 1 x3 + C 3 1 (c) – tan x – tan– 1 x3 + C 3 (d) None of these

2/3

 + C

1  1 (a) 1 + n  x  1− n  (b)

3 2 + x 4  2 − x 

84.

89.

∫

x 1− x 2x

1 − x4

(x x

(a) (c)

4

2

4

1 + x2 – x

dx is equal to + C

(b)

+ C

(d)

− 1)

x + x2 + 1 4

x x4 + x2 + 1 2x x4 + x2 + 1

−x 1 − x4 − 2x 1 − x4

+C +C

dx is equal to

 + C

(b)

x4 + x2 + 1 +C x

 + C

(d)

x4 + x2 + 1 +C 2x

x −1 x +x +x

(a) tan– 1

3

2

dx is equal to

x + x + 1 + C x

(b) 2 tan– 1

x + x +1 + C x

(c) 3 tan– 1

x 2 + x + 1 + C x

(1 + x 2 )

 (a) – log   x −    (b) log   x −  

98.

1   +  x − x

1   +  x − x

2  1  − 2  + C x  

2  1  + 2  + C x  

2   1 1  (c) log   x −  +  x −  − 2  + C  x x    

dx

∫ x (a + bx)

2

=

a + bx a a + bx (c) z = x

(b) z =

(a)

2sec α (tan x + tan α) + C

(b)

2sec α (tan x − tan α) + C

(c)

2sec α (tan α − tan x) + C

∫

3

x2

1 12 (2 + x2/3)5/4 + C (2 + x2/3)9/4 – 3 5

dx is equal to

(a) (1 + x1/3)3/2 + C (c) 2 (1 + x1/3)3/2 + C

(b) – (1 + x1/3)3/2 + C (d) None of these

dx

∫x

1 − x3

1 − x3 − 1

= a log

1 − x3 + 1

(a) 1/3

(b) 2/3

(c) – 1/3

(d) – 2/3

+ C, then a =

1 + x dx is equal to x

∫

(a) 2

 1 + x − 1  +C 1 + x – 2 log   1 + x + 1 

(b) 4

 1 + x + 1  +C 1 + x + 2 log   1 + x − 1 

4e x + 6e − x dx = Ax + Bln (9e2x – 4) + C, then x − 4e − x

∫ 9e

35 3  , B = , C = a constant 36 2 3 − 35 , C = a constant (b) A = , B = 2 36 3 − 35 , C = A constant (c) A = – , B =   2 36 (d) None of these

(d) None of these

(d) None of these 95.

(c)

(a) A = –

a + bx b

sec x dx is equal to sin (2 x + α) + sin α

1+ 3 x

2 12 (2 + x2/3)9/4 – (2 + x2/3)5/4 + C 3 5

99. If

− 1  z2 b3  − 3bz + 3b 2 log z +  + C, 4  a 2 z

(a) z =

∫

(b)

 1 + x − 1 1 + x + 2 log   +C  1 + x + 1  (d) None of these

where

94.

2 12 (2 + x2/3)9/4 + (2 + x2/3)5/4 + C 3 5

(c) 4

(d) None of these 3

(a)

97. If

dx is equal to

1 + x4

(2 + x 2 / 3 )1/ 4 dx is equal to

(d) None of these

(a) tan– 1 (tan x + cot x) + C (b) tan– 1 (cot x – tan x) + C (c) tan– 1 (tan x – cot x) + C (d) None of these

93.

1/ 3

2

dx 91. ∫ 6 is equal to sin x + cos 6 x

∫x

∫x

2

(d) None of these

92.

96.

203

∫ ( x + 1)

100.

∫

3

1+ 4 x x

dx is equal to

(

 1+ 4 x (a) 12    7 

(

 1+ 4 x (b) 12    7 

(

 1+ 4 x (c) 6    7 

)

7/3

(1 + x ) + 4

4/3

  +C  

4/3

  +C  

4

)

7/3

)

7/3

(d) None of these

(1 + x ) − 4

4

(1 + x ) − 4

4

4/3

  +C  

Indefinite Integration

90.

204

∫ f ( x) sin x cos x

101. If

dx =

Objective Mathematics

then f (x) is equal to

1 log [ f (x)] + C, 2 (b − a 2 ) 2

106.

(c)

2

(d) None of these 102.

∫

3

x

7

107.

1 + 3 x 4 dx is equal to

21 1 + 3 x4 32 3

4

3

4

8/ 7

∫

(x − x )

5 1/ 5

8/ 7

+C

8/ 7

+C

5 1   − 1 24  x 4

(b)

5  1 1 − 4  x 24

(c) –

∫e

104.

∫x

(1 + x ) 3

2

109.

2/3

+C

+C

1/ 2

(a)

+C

(c)

110.

is equal to

∫

2 5/ 4

1+ x

1 + x2

1+ x

(a) z =

 1  x + (a) 3  log +C 1 + x1/ 3 1 + 3 x  

2

− x 4

dx 4

(d) A = 1

(1 + x )

−2 x 4

6/5

4

4

=

is equal to

+ C

(b)

+ C

(d)

111.

 1+ x 1  − (c) 3  log 3  +C 1+ 3 x  x  3

∫ (a)

(d) None of these dx = a cos 8x + C, then

(b)

2 x 4

1 + x2

4

1 + x2

x

1 + x4 x

(b) z =

1 + x4

(d) None of these

(

− 1 + x2 + log x + 1 + x 2 x

( + log ( x +

) +C

) 1+ x ) + C

1 + x2 + log x + 1 + x 2 + C x

(a) a =

−1 16

(b) a =

1 8

(c)

(c) a =

1 16

(d) a =

−1 ⋅ 8

(d) None of these

x

+C

x 4

x 2 + 1 dx is equal to x2

1 + x2

+C

 1 1 1+ z log − tan −1 z  + C, where 2  2 1− z 

4 4 (c) z = − 1 + x x

 1+ 3 x 1  + (b) 3  log 3  +C 1+ 3 x  x 

∫ tan 2 x − cot 2 x

. cos ec3 x (cos2x + sin x + cos x + sin x cos x) dx

dx

∫x

6/5

cos 8 x + 1

+C

(c) A = – 1

1/ 3

105. If

1  x2  2a  a + bx 2 

cos ec x

(d) None of these dx

3/ 2

sin 8 x − cos8 x dx = A sin 2x + B, then 2 x cos 2 x 1 1 (a) A = –  (b) A = 2 2

+C

5 1   − 1 24  x 4

+ C

∫ 1 − 2 sin

108. If

6/5

(a)

3/ 2

(a) e cosec x (cosec x + cot x) + C (b) e cosec x + C (c) – e cosec x (cosec x + cot x) + C (d) None of these

dx is equal to

x6

1  x2  3a  a + bx 2 

is equal to

+C

(d) None of these 103.

dx is equal to

(d) None of these

( ) 32 (d) (1 + x ) 21 7 (c) (1 + x ) 32

(a)

2 5/ 2

1  x2  (b) 3a  a + bx 2 

1 a cos x − b 2 sin 2 x 2

(a + bx )

(a) –

1 (a) 2 2 a sin x − b 2 cos 2 x 1 (b) 2 2 a sin x + b 2 cos 2 x (c)

∫

x2

2

(a) tan

−1

(log x) + C x

120.

x 2 + 4 + C

(c) 2 114.

(b)

∫ x. (x ) . x x

1 x2 + 4

(a) x + C (c) xx. log x + C

(b) (xx)x + C (d) None of these

(d) θ = 2 sin– 1 (x + 1)

∫ u v" dx = uv′ – vu′ + a, then a = (a) ∫ u" .v dx (b) ∫ u' .v dx (c) ∫ uv' dx (d) None of these

121. If

122. If

115. The equation of a curve passing through origin is given by y =

written in the form x = g ( y), then (a) g (y) = (b) g (y) = (c) g (y) =

sin sin

4

−1

(4 y )

−1

(4 y )

123.

(a) 3 sin x – (3x + 4) cos x (b) 3 sin x + (3x + 4) cos x (c) – 3 sin x + (3x + 4) cos x (d) None of these

[ f ( x) g' ( x) + g ( x) f ( x).g ( x)

124.

f' ( x)] [log f (x) + log g (x)] dx is

(a) f (x) g (x) log (f (x) g (x)) + C

118. If f ′′ (x) = sec2x and f (0) = f ′ (0) = 0 then

dx

(a) A =

2 3

5 (c) B = 3

(d) None of these

∫ sin x d (cos x)

is equal to

(b) f (x) = sec2x (d) None of these

x 4  = A tan– 1  B tan +  + C, then  2 3 (b) A =

d2

∫ dx

2

(b)

1  sin 2 x  − x + C   2  2

(d) None of these

(tan–1x) dx is equal to

1 +C 1 + x2 (b) tan– 1 x + C 1 log (1 + x2) + C (c) x tan–1 x – 2 (d) None of these

∫

dx 2ax − x 2

= ( f o g) (x) + C, then

x+a a x−a –1 (b) f (x) = sin x, g (x) = a x−a –1 (c) f (x) = cos x, g (x) = a x−a –1 (d) f (x) = tan x, g (x) = a sec 2x 126. If ∫ dx = f [g (x)] + C, then (a) f (x) = sin– 1x, g (x) =

(d) log f (x).g (x) + C

∫ 5 + 4 sin x

(b) ag (ax + b) + C

1 (c) [g(ax + b) + C] a

125. If

(b) 1 [ log f ( x) g ( x)]2 + C 2 (c) [ log f ( x) g ( x)]2 + C

119. If

dx is equal

(a)

equal to

(a) f (x) = log sec x (c) f (x) = log secx + x

∫ f (ax + b)

(a) g (ax + b) + C

(a)

116. The anti-derivative of the function (3x + 4) | sin x |, when 0 < x < π, is given by

∫

dx = g (x) + C, then

sin 2 x – x + C 2 1  sin 2 x  + x   + C (c)   2 2

sin − 1 ( 4 y )

(d) None of these

117.

∫ f ( x)

to

3 4 ∫ x cos x dx. If the equation of the curve is

3

sin 4 θ  −1   3θ − 2 sin 2θ +  + C, 8 4 

 1  (c) θ = 2 sin– 1    1 + x 

(2 log x + 1) dx is equal to

( xx )

=

 1  (b) θ = sin– 1    1 + x 

+C

(d) None of these

x + 2x 2

(a) θ = sin–1(x + 1)

(d) None of these (c) tan x + C x 1 w.r.t (x2 + 3) is equal to 113. Integral of x2 + 4 x 2 + 4 + C

dx

5

where

(b) tan– 1 (log x) + C

−1

(a)

∫ (1 + x)

1 3

5 (d) B = –  3

(a) dom ·  f (x) = R – {0} (c) f ′ (x) =

205

1 w.r.t. log x is 1 + (log x) 2

Indefinite Integration

112. Integral of

(b) range of g (x) = R

1 π , ∨ x ∈ R+ (d) g ′ (x) = – cosec2   − x  . 2x 4 

206

x +1 ⋅ If 127. Let f (x) = x+2

Objective Mathematics

= 

 f ( x)  ∫  x 2 

 2 f ( x) − 1   1 g   −h 2  2 f ( x) + 1  

1/ 2

dx

132.

1+ x dx is equal to 1− x

(a)   − sin

f ( x) − 1   , then f ( x) + 1 

(a) g (x) = log | x |, h (x) = log | x | (b) g (x) = log | x |, h (x) = tan– 1x (c) g (x) = tan– 1x, h (x) = log | x | (d) None of these

∫

−1

x − 1 − x2 + c

(b)   sin

−1

x + 1 − x2 + c

(c)   sin

−1

x − 1 − x2 + c

(d)   − sin

−1

x − x2 − 1 + c

133. ∫ 32x3(log x)2 dx is equal to

(a)  8x4(log x)2 + c 128. Let f (x) be a polynomial of degree three satisfying (b)  x4 {8(log x)2 – 4(log x) + 1} + c f (0) = – 1 and f (1) = 0. Also, 0 is a stationary point (c)  x4 {8(log x)2 – 4 log x} + c of f (x). If f (x) does not have an extremum at x = 0, (d)  x3 {(log x)2 – 2 log x} + c f ( x) dx is equal to then ∫ 3 cos x − 1 x x −1 e dx is equal to 134. ∫ sin x + 1 2 x e x sin x (a) + C (b) x + C. (b)   c − (a)  ex cos x + c 2 1 + sin x x3 + C (d) None of these (c) ex e x cos x (c)   c − (d)   c + 6 1 + sin x 1 + sin x 129. If In = ∫ tan n x dx , then I0 + I1 + 2 (I2 + ... + I8) + I9 + −1 e m tan x I10 is equal to 135. ∫ dx equals 1 + x2 2 9  tan x tan x tan x  m tan –1 x + + ... + −1 1 (a)  + c (b)   e (a)   e tan x + c 2 9   1 m −1 1  tan x tan 2 x tan 9 x  (c)   e m tan x + c (d)  none of these + + ... + (b) –  m 2 9   1 dx is 136. The value of ∫ 2 sin x cos 2 x  cot x cot 2 x cot 9 x  + + ... + (c)  (a)  tan x – cot x + c (b)  tan x + cot x + c 2 9   1 (c)  sec x – tan x + c (d)  none of these  cot x cot 2 x cot 9 x  + + ... + (d) –  2 9   1 130.

∫x

( x + 1)

dx is eqaul to

(1 + x e )

x 2

1 xe x +C (a) log − 1 + xe x 1 + xe x 1 xe x (b) log +C + x 1 + xe 1 + xe x (c) log

137.

∫x

2

dx is equal to + 4 x + 13

(a)  log (x2 + 4x + 13) + c (b)  

1  x+2 tan −1  +c 3  3 

(c)  log (2x + 4) + c

2x + 4 +c ( x 2 + 4 x + 13) 2

(d)   1

∫ [( x − 1) ( x + 2) ]

138. The value of

3

5 1/ 4

1/ 4

x

1 + xe 1 +C + xe x 1 + xe x

(a)  

4  x −1  3  x + 2 

(c)  

4  x +1  3  x − 2 

1/ 4

+c

(b)  

4  x +1  3  x + 2 

+c

(d)  

4  x −1  3  x − 2 

1/ 4

(d) None of these ex 131. ∫ dx is equal to x (2 + e ) (e x +1)  e x +1  (a)   log  x  + c  e +2 

 ex + 2  (b)   log  x +c  e +1 

 e x +1  (c)    x  + c  e +2 

 ex + 2  (d)    x +c  e +1 

139.

∫

ax/2 a− x − a x

(a)  

dx is

+c

1/ 4

+c

dx is equal to

1 1 sin −1 (a x ) + c (b)   tan −1 (a x ) + c log a log a

(c)   2 a − x − a x + c

(d)  log (ax – 1) + c

ex ∫ (2 + e x ) (e x + 1) dx is equal to

π (c)   x + log sin  x −  + c 4 

207

 ex + 1  (a)   log  x +c e +2  ex + 1  (c)    x +c e +2

π  (d)   x − log cos  x −  + c 4 

Indefinite Integration

140.

 ex + 2  (b)   log  x +c  e +1   ex + 2  (d)    x +c  e +1 

141. ∫ [sin (log x) + cos (log x)] dx is equal to (a)  x cos (log x) + c (c)  cos (log x) + c

142.

(b)  sin (log x) + c (d)  x sin (log x) + c

1 + x + x + x2 ∫ x + 1 + x dx is equal to 1 1+ x + c 2 2 (b)   (1 + x)3/ 2 + c 3

(a)  

(c)   1 + x + c (d)  2 (1 + x)3/2 + c 143.

dx

∫ cos x +

equals

3 sin x

1 x π (a)   log tan  +  + c 2  2 12  1 x π (b)   log tan  −  + c 2  2 12  x π (c)   log tan  +  + c  2 12  x π (d)   log tan  −  + c  2 12  x f o f o .. o f . 144. Let f ( x) = for n ≥ 2 and g ( x) =    (1 + x n )1/ n f occurs n times Then ∫ xn – 2g(x) dx equals

(a)  

1 1− 1 (1 + nx n ) n + c n(n − 1) 1−

1 n

(b)  

1 (1 + nx n ) n −1

(c)  

1 1+ 1 (1 + nx n ) n + c n(n + 1)

(d)  

1 1+ 1 (1 + nx n ) n + c n +1

145. The value of

2∫

+c

sin x dx is π  sin  x −  4 

146.

∫ x( x )

x x

(2log x + 1) dx = ….

(a)  (xx)x + c

(b)  log (x)x + c

(c)  xx + c

(d)  none of these.

147. Let f(x) = f(x). Then

x fofo ... of ) for n ≥ 2 and g(x) = (  (1 + x n )1/ n f occurs n times

∫x

n– 2

g(x)dx equals

(a)  

1 1− 1 (1 + nx n ) n + K n(n − 1)

(b)  

1 1− 1 (1 + nx n ) n + K n −1

(c)  

1 1+ 1 (1 + nx n ) n + K n(n + 1)

(d)  

1 1+ 1 (1 + nx n ) n + K n +1

148. Let l =

∫e

4x

ex e− x dx, J = ∫ −4 x dx. 2x + e +1 e + e −2 x + 1

The, for an arbitrary constant C, the value of J – l equals  e4 x − e2 x + 1  1 (a)   log  4 x +C 2x 2  e + e +1  e2 x + e x + 1  1 (b)   log  2 x +C x 2  e − e +1  e2 x − e x + 1  1 (c)   log  2 x +C x 2  e − e +1  e4 x + e2 x + 1  1 (d)   log  4 x +C 2x 2  e − e +1 1 + sin x 149. The area of the region between the curves y = cos x 1 − sin x y = and bounded by the lines x = 0 and cos x π x= is 4 2 −1

(a)  

∫ 0

t (1 + t 2 ) 1 − t 2

2 −1

(b)  

∫ 0

π (a)   x + log cos  x −  + c 4 

(c)  

π (b)   x − log sin  x −  + c 4 

(d)  

4t (1 + t 2 ) 1 − t 2

2 +1

∫ 0

4t (1 + t ) 1 − t 2 2

2 +1

∫ 0

t (1 + t ) 1 − t 2 2

dt

dt dt dt

208

solutions

Objective Mathematics

1. (b) I =



=

Let

∫

1 + x4

∫

dx =

(1 − x ) ( x + 1 x ) dx 4 3/2

∫1

x 3 ( x + 1/ x 3 ) dx

4. (a)

(1 − x )

4 3/2

dx

=

∫



=

1 2



π x = – tan  −  + k, 4 2

3

2  2 − x  x

3/2

 −2  ⇒  3 − 2 x  dx = dt x

1 – x2 = t x2

5. (a) I =

∫ x (1 − x )

3 −1/2

−1

dt 2 3 ∫ 1 − t2 = 2 dt = 2 ⋅ 1 log t − 1 + c 3 ∫ t2 −1 3 2 t +1

π  2 ∫ cos  2 x −  dx  4

∴

1 π  = 2 . sin  2 x −  + K,  2 4 where K is arbitrary constant

−1

1  ln  3 

=

1 π  sin  2 x −  + K.  4 2 π and a = arbitrary constant. ∴ C = 4 3. (b)  Since, the sum of powers of

6. (a)

3 11 sin x and cos x = = – 4 = negative even − 2 2 integer, ∴ put tan x = z ⇒ sec2x dx = dz





=

=

∫ ∫

∫

dx tan x. cos x 11

(1 + z ) dz

4

∫

2

tan

11/ 2

x

dx

2

x

=

∫ (z

− 11/ 2

+z

− 7/2

)

1 1 and B = 5 9

=

1 sin α

sin (( x + α ) − x ) ∫ sin x. sin ( x + α)

∫ cot x − cot ( x + α)

sin x. sin ( x + α ) dx

= cosec α [log sin x – log sin (x + α)] + C

7. (c) I = dx

=

∫

∫

(cos

2

sin x + C. sin ( x + α )

x + cos 4 x ) cos xdx

sin 2 x + sin 4 x

1 − t 2 + (1 − t 2 ) t2 + t4

2

dt,  where sin x = t

(1 − t ) (2 − t ) dt t (1 + t ) (2 − t ) dt 2−t 2 dt − 2

∫

2

2

∫t

2

2

2

∫ t2 + t4 dt 3dt = 2⋅ ∫ 2 − 4 ∫ 2 dt + ∫ dt 2 t t (1 + t ) =

2

dx

(sin ( x + α) cos x − cos ( x + α) sin x)

= cosec α

= –

∴ A =

=−

∫

dz

2 − 9/ 2 2 − 5/ 2 z − z +C 9 5 1 1  − 9/ 2 x + tan − 5 / 2 x  + C. = – 2  tan  9 5

2

= cosec α

=

2

z11/ 2

=

(1 + tan x) sec

tdt

1 − x3 − 1  +c. 1 − x 3 + 1 

∫ sin x. sin ( x + α)

= cosec α log

cos3 x dx tan11 x. cos11 x

−2

  ∫ (t )  3  1 − t

I =

=

∫

2 tdt 2 tdt = − 3 x3 3 1 − t2

⇒ x–1dx = –

cos3 x dx = sin11 x

dx

Let 1 – x3 = t2 ⇒ –­ 3x2dx = 2t dt



∴

π x  −  dx 4 2

π and b = arbitrary constant. 4

∴ a = –

2. (c) We have, dx =

2

 x π = tan  −  + k. 2 4



3 − +1

∫ (sin 2 x + cos 2 x)

∫ sec

where k is any arbitrary constant.

1 1  ⇒  x + 3  dx = – dt  x  2 1 dt 1 t 2 1 ∴ I = – = ⋅ +c = +c 3/2 ∫ 2 t 2 −3/2 + 1 t 1 +c. = 1 2 −x x2

dx π  1 + cos x  − x  2 

∫ 1 + sin x

dx

−2 –6 tan–1 (t) + t + c t = sin x – 2(sin x)–1 – 6 tan–1 (sin x) + c.



= tan 2x – tan (x – α) – tan (x + α).

∴

∫ tan ( x − α) tan ( x + α)



=

∫ [ tan 2x − tan ( x − α) − tan ( x + α)]  dx



=

1 log | sec 2x |  – log  | sec (x – α) |  2

=

8. (a)

 a+x ∫  a − x −

∫

dx

z dz dx = − ∫ 2 2 z a −x 2 [Put a – x2 = z2 ⇒ – 2x dx = 2 zdz]

12. (c)

cos α

∫ cos ( x + α) cos x

sin [ ( x + α) − x ] = cot α. ∫ cos ( x + α) cos x dx.

∫ [ tan ( x + α) − tan x]

13. (a)

∫

∫

θ 2 θ cos 2 sin

(1 − cos θ) ∫ cos θ dθ

=

∫

=

∫ (sec θ − 1) d θ = log (sec θ + tan θ) – θ + C

(

)

x 2x = log e + e − 1 – sec– 1 (ex ) + C.

=

tan ( x − α) + tan ( x + α) 1 − tan ( x − α). tan ( x + α)

 1  dx x ( x + 1) 

−

 1 1 +  x2 x ( x + 1)  dx

−

1 1 1  + − x 2 x x + 1  dx

2

∫  x

=

−1 1 + + log | x | – log | x + 1 | + C 2x2 x

3

3

1 and B = 1. 2

∫x

−1 4

dt

∫t

3/ 4

1 dx −1    Putting 1 + x 4 = t ⇒ x 5 = 4 dt 



1/ 4

+C

dx x2 + 2x + 2

2

∫ ( x +1)

dx 2

( x +1) 2 + 1

=

sec 2 θ d θ ∫ tan 2 θ sec θ

[Putting x + 1 = tan θ ⇒ dx = sec2θ dθ]

 =

dx 1  x 5 1 + 4   x 

(1 + x 4 )1/ 4 + C. x

∫ ( x +1) =

3/ 4

∫

1  −1 t1/ 4 ⋅ + C = – 1 + 4  x 4 1/ 4

= –  14. (c)

dx = (1 + x 4 )3 / 4

2

−1 cos θ dθ = +C = − 2 sin θ θ

∫ sin

( x + 1) 2 + 1 +C x +1 x + 1  ∵ tan θ = 1 

11. (b) tan 2x = tan [(x – α) + (x + α)]

−

=

=

sec θ − 1 ⋅ tan θ d θ   sec θ + 1

θ θ  2 sin cos  2 2  dθ =   cos θ  

3

∫  x

=

[Put ex = sec θ ⇒ dx tan θ d θ]

1

∫  x

=

dx

sin ( x + α ) + C. sin x

ex − 1 dx = ex + 1

( x + 1) − x dx = x 3 ( x + 1)

x −1 1 = 2 x 2 + x + log x + 1 + C.

= cot α. log sin ( x + α) − log| sin x | + C  

10. (a)

∫

∴ A = –

sin ( x + α) cos x − cos ( x + α) sin x dx = cot α. ∫ cos ( x + α) cos x

= cot α. log

dx = + x3

dx

sin α = cot α. ∫ dx cos ( x + α) cos x

= cot α.

4

sec 2x + C. sec ( x − α). sec ( x + α)

1

dx



∫x

= log

1

a 2 − x 2 + C.

∫ 1 + tan x. tan ( x + α) =



2x

=–z+c=–

9. (c)

(a + x ) − (a − x )

=

– log | sec (x + α) | + C

a − x  dx a + x 

∫ (a − x) (a + x)

=

tan 2x dx



=–

x2 + 2x + 2 + C. x +1

Indefinite Integration

dt dt −6∫ + dt 1 + t2 ∫ t2

= 2∫

209

⇒ tan (x – α). tan (x + α) tan 2x

1 dt 1  = 6 ∫ 2 − dt − 4 ∫ 2 + ∫ dt t 1 + t 2  t

210

15. (a)

∫ sin

4

dx dx = x + cos 4 x ∫ (sin 2 x + cos 2 x) 2 − 2 sin 2 x cos 2 x

Objective Mathematics

2dx = ∫ 2 − sin 2 2 x =

sec 2 2 x dx = 2 2x

∫

∴ put tan x = z ⇒ sec2x dx = dz ∴

dt

( 2)

2

+t2

∫

dx 3

[Putting tan 2x = t ⇒ 2sec2 2x dx = dt] 1  t  tan − 1   2  + C 2

=



=

16. (c) f (x) sin x cos x = f' ( x)

∫ ( f ( x))

⇒

⇒ –

2

1  1  tan 2 x  + C. tan– 1   2  2

1 1 ⋅ ⋅ f' ( x) 2 (b 2 − a 2 ) f ( x )

dx =

=

(b

2

2 − a 2 ) (cos 2 x − sin 2 x )

2 b 2 ( 2 cos 2 x − 1) + a 2 ( 2 sin 2 x − 1)

17. (c)

f (x) =

∫x

13 / 2

1/ 2

dx

20. (b)

5 3/ 2 5 / 2 1/ 2 = ∫ x . x . (1 + x )

5 4 dx = ∫ x . z. z dz 5

5 3/ 2  5/ 2 2  Putting 1 + x = z ⇒ 2 x dx = 2 z dz  4  i.e. x 3 / 2 dx = z dz  5

 = 4 5

a 2 − b2 2

−1 . b 2 cos 2 x + a 2 sin 2 x + c

. (1 + x 5 / 2 )

2 2 ∫ z ⋅ (z – 1) dz = 2

4 5

    

2 4 2 ∫ z ( z − 2 z + 1) dz

4  z7 2z5 z3  = +  +C  − 5 7 5 3 4 8 = (1 + x5/2)7/2 – (1 + x5/2)5/2 35 25  ∴ A =

4 8 , B = –  35 25

dx

∫ 3

(1 + z ) 2



=

∫



=

∫ (z



3 3  − 8/3 x + tan − 2 / 3 x  + C = –  tan 8  2

3

tan11 x − 11/ 3

=

∫

sin x ⋅ cos12 x cos11 x 11

z

11/ 3

dz

+ z − 5 / 3 ) dz

x dx = x+3x

∴

∫

=6

∫ 

=6

∫

∫

z 3 . 6 z 5 dz z 6 dz =6∫ 3 2 z +z z +1

 z6 − 1 1  dz + z +1 z + 1  ( z 3 − 1) ( z +1) ( z 2 − z +1) dz dz + 6 ∫ z +1 z +1

5 4 3 2 = 6  ∫ ( z − z + z − z + z − 1)  dz + 6 log (z + 1)

1 b 2 cos 2 x + a 2 sin 2 x −

sec 4 xdx

=

19. (b) Put x = z6 ⇒ dx = 6z5 dz.

2 2 ∫ (b − a ) sin 2x.dx

=

sin x cos x 11

∴ f (x) = tan–8/3x and g (x) = tan–2/3x.

1  − cos 2 x  = (b 2 − a 2 )    f ( x) 2

⇒ f (x) =

−11 1 − = – 4 = – ve even 3 3

integer,

2 sec 2 2 x ∫ 2 sec2 2 x − tan 2 2 x dx

∫ 2 + tan

=2

18. (a) ∵ Sum of powers =

4 +  (1 + x5/2)3/2 + C. 15 4 ⋅ and C = 15

= z6 –

6 5 3 4 z + z – 2z3 + 3z2 – 6z + 6 log (z + 1) + C. z 2

=x–

6 5 / 6 3 2/3 x + x – 2x1/2 + 3x1/3 5 2 – 6x1/6 + 6 log (1 + x1/6) + C.

sin x

∫ sin( x − α) dx = Ax + B log sin (x – α) + c sin x B = A+ cos( x − α) sin( x − α) sin( x − α) ⇒ sin x = A sin (x – α) + B cos (x – α)

⇒

= (A cos α + B sin α)sin x



+ (– A sin α + B cos α) cos x

∴ A cos α + B sin α = 1 and – A sin α + B cos α = 0 ∴ B = sin α, A = cos α. 21. (d)   ∫

=

dx = cos x cos 2 x

∫

sec 2 x 1 − tan x 2

∫

dx =

dx cos x

∫

1 − tan 2 x 1 + tan 2 x dx

1 − z2

[Putting tan x = z ⇒ sec2x dx = dz] = sin– 1 z + C = sin– 1 (tan x) + C.

1 π x π log tan  + +  + c 4 2 8 2

=

1  x 3π  log tan  +  + c 2 8  2 dx dx 23. (b) ∫ = ∫ 3 cos x sin 2 x 2 tan x cos3 x 1 + tan 2 x =

=

∫

sec 4 x 1 dx = 2 tan x 2

∫

24. (d)

∫x

dx

x −1

= sec– 1x – 2

∫x

x −1 2

−2 ∫

∫ sec x + cos ec x

x

x2 − 1

sec θ tan θ 3 θ tan θ dθ

∫ sec

= sec– 1 x – 2



= sec x –



sin 2θ  = sec– 1 x –  θ +  +C  2 



= sec– 1 x – sec– 1x –

∫ (1 + cos 2θ)

∴ 

∫

∫

∫

=

1 2sin x cos x dx 2 ∫ sin x + cos x

=

1 2

∫

=

1 2

∫ (sin x + cos x) dx

=

1 1 (– cos x + sin x) – 2 2 2

=

1 1 (sin x – cos x) – log tan  x + π  + C. 2 2 8 2 2

∫ 

dθ

(sin x + cos x )2 − 1 sin x + cos x

dx −

1 dx ∫ 2 sin x + cos x



= – 4  ∫ cos 2θ. 2 sin 2 θ d θ



= – 4  ∫ cos 2θ (1 − cos 2θ) d θ



= – 4  ∫ cos 2θ d θ + 2 ∫ (1 + cos 4θ) d θ

dx π  sin  x +   4

 x π f (x) = sin x – cos x and g (x) = log tan  +  . 2 8 2 1 + sin x dx = – 4 cos (ax + b) + c

∫

2 (sin x / 2 + cos x / 2) dx = – 4 cos (ax + b) + c = – 4 cos (ax + b) + c



π x ⇒ − 4 cos  +  = – 4 cos (ax + b) 4 2 ∴ a = 1/2, b = π/4. 29. (a)

1 − cos 2θ 1 + cos 2θ (– 4 sin 2θ cos 2θ) dθ

= – 4  ∫ tan θ. cos 2θ. sin 2θ d θ

∫

⇒ 2 2 (sin x / 2 − cos x / 2) + c

x2 − 1 +C x2



211

dx 1 1 + cos x sin x

=

⇒

x2 − 1 = – + C. x2 25. (c) Put x = cos2 2θ ⇒ dx = – 4 cos 2θ sin 2θ dθ 1− x dx = 1+ x

dx

∴ 28. (a)

2 ∫ cos θ  dθ



+ C.

= sin– 1 (cos θ) + c

1 − cos 2 θ

dx 3

[Putting x = sec θ ⇒ dx = sec θ tan θ dθ]

–1

1  sin 4θ + C 2

1 − x + cos– 1 x + x . 1 − x

d (cos θ)

z

x2 − 2

dx =

=–2

27. (b)

4

1   2  tan x + tan 5 / 2 x  + C.   5

2



∫

(1 + z ). 2 z dz

 z5  2 z + +C = 5 

3

= – 2sin 2θ + 2θ +

26. (d)

[Putting tan x = z2 ⇒ sec2x dx = 2z dz] =



30. (c)

cos x + x sin x

∫ x ( x + cos x) 1

∫

dx =

( x + cos x ) − x (1 − sin x ) x ( x + cos x )

1 − sin x 

∫  x − x + cos x 

dx



=



= log | x | – log | x + cos x | + C



= log

∫ log ( x +

dx

x + C. x + cos x x2 + a2

(

)

dx

= x  ∫ log x + x 2 + a 2 –

)

∫ x⋅ x+

1 x2 + a2

 1 + 

  dx x2 + a2  x

Indefinite Integration

dx 1 π 1  = sec  x +  dx ∫ ∫  π 4 2 2 cos  + x    4

22. (d) I =

Objective Mathematics

)

212

(

= x log x + x 2 + a 2 − ∫ = x log x + x + a

(

31. (b)

∫

2

x x ⋅ tan 2 2 dx = cos x

sin

2

)− sin

∫

x

dx

=–

x + a + C.

=–

x +a 2

2

x  ⋅ 2 

2

x sin 2 cos

1 − 2 sin 2

x x cos dx 2 2 = ∫  2 x  2 x 1 − sin  1 − 2 sin  2 2

2

x 2

x 2

dz

∫ (a − b) z 1 a−b

  dx

z dz

=–

 1 b  + C. log  cos x + cos 2 x + a − b  a−b 

2

=

dz −2∫ 1 − z2

2

1  2   −z 2 1 +z 1 1+ z 1 log 2 − 2 ⋅ log = +C 1 1 2 1− z −z 2. 2 2 x 1 + 2 sin 1 + sin 1 2 = ⋅ log − log x 2 1 − 2 sin 1 − sin 2

33. (c)

x = a+x

∫ θ.

x 2 + C. x 2

∫ sin



 tan 2 θ tan 2 θ  = 2a θ. −∫ d θ 2 2  



= a θ tan2θ – a



= a θ tan2θ – a (tan θ – θ) + C



= a θ (1 + tan2θ) – a tan θ + C



= (x + a) tan– 1

∫ =–

a + b tan 2 x

∫

∫ (sec

dx =

dz

az 2 + b (1 − z 2

x log 3) dx = x + 3 + c

2

)

a cos 2 x + b sin 2 x

1 2

∫

dx =

∫

tan x. sec 2 x 1 + tan 4 x

dx

dz 1 + z2

(

)

=

1 log z + 1 + z 2 2

=

1 log tan 2 x + 1 + tan 4 x 2

+C

(

∫

=

tan − 1 x dx = x4

∫ θ cot

=–

2

) + C.

=–  dx

dz

[Putting cos x = z ⇒ sin x dx = – dz]

=–

θ

∫ tan

4

θ

sec2θ d θ

θ cos ec 2θ d θ

θ cot 3 θ 1 + 3 3

3 = – θ cot θ + 1 3 3

θ − 1) d θ

sin x

sin 4 x + cos 4 x

∴ 

x − ax + C. a

∫

tan x

36. (a) Put x = tan θ ⇒ dx = sec2θ d θ

2a tan θ sec2θ dθ

∴

tan x

x

1   2 2  Putting tan x = z ⇒ tan x. sec x dx = dz  2

32. (a) Put x = a tan2θ ⇒ dx = 2a tan θ sec2θ dθ −1

∫ (1 + 3

∴ F(x) = x + 3x – 4 = 0 ⇒ x = 1. 35. (b)   ∫

 1 1  = 2 ∫ dz − 2 1 − z 2  1 − 2z

∫

+C

Since F (2) = 2 + 9 + c = 7 ⇒ c = –4

x 1 x    Putting sin 2 = z ⇒ 2 cos 2 dx = dz 

=

2

 1 b  log  z + z 2 + a − b  a−b 

∴ F(x) =

∫ (1 − z ) (1 − 2 z )

dz

 b  z2 +    a−b

34. (b) f (x) = 1 + 3x log 3.

2

2

dz

=–

sin 2

=2

∫

+b

2

∫ cot

3

θ dθ⋅

∫ cot θ

1 1 θ cot3θ + 3 3

(cosec2θ – 1) dθ

∫ cot θ

cosec2θ dθ –

∫ cot θ 1 1 cot 2 θ 1 θ cot3θ – ⋅ − log sin θ + C 3 3 2 3

 x  1 1 tan − 1 x − 2 − log   + C. 3 2 3x 6x 3  1+ x  x ⋅ ∴ f (x) = 1 + x2 =–

1 3 dθ

=

=

3

dx x sin ( x + α )

42. (b) dx

∫ ∫

z

∫ z dz

[Putting cos α + sin α cot x = z2 ⇒ – sin α cosec2x dx = 2 zdz] = – 2 cosec α .  cos α + sin α cot x  + C. 38. (c)

=

x

∫ log (1 + cos x) dx − ∫ x tan 2

dx

x sin x x ∫ 1 + cos x dx − ∫ x tan 2 dx

= x log (1 + cos x) +

∫ x tan 2 dx − ∫ x tan 2

x

 sin 2 x  + sin x  + C. =–   2 

∫

x

=–

dx

3/ 2

=

sec 2 θ d θ ∫ sec3 θ

44. (a)

=

Thus, f (x) =

∫

x

dθ = sin θ + C =

∫ cos θ

Since f (0) = 0,

∫ (1 + cos x) 2

x + C. 2 sin 3 x

1+ x therefore C = 0. x . Hence, f (1) = 1 + x2

dx

(1 + x )

x−x

2

=

2

+ C.

1 2

2 sin θ cos θ d θ

∫ (1 + sin θ)

sin 3 x

∫ cos x (sec x + cos x) cos x

=

∫

=

∫z

sin θ − sin θ 2

(

∫e

x

 1− x   1 + x 2  dx =

)

2 x −1  + C. +C = x 1− x

2

∫e . x

1 + x − 2x 2

(1 + x )

2 2

sec 2 x + 1 + cos 2 x

 dx

sin 3 x dx cos 2 x (sec x + cos x )

(sec x + cos x )2 − 1

dz z2 − 1

4

dθ 1 − sin θ = 2∫ = 2∫ d θ = 2 (tan θ – sec θ) cos 2 θ 1 + sin θ  x 1 − =2  1 − x 1 − 

dx

1 + cos 2 x + cos 4 x

=

  sin 3 x dx = dz   Putting sec x + cos x = z ⇒ cos 2 x  

[Putting x = sin2θ ⇒ dx = 2 sin θ cos θ d θ]

41. (c)

 ez dz = – ez sec z + C

[Putting x = tan θ ⇒ dx = sec2θ dθ]



40. (a)

d



∫ sec z + dz (sec z )

= – e– x/2 · sec

(1 + x 2 )

cos

−x    Put 2 = z ⇒ dx = − 2 dz 

= x log (1 + cos x). dx

sin 3 x (cos 5 x + cos 4 x ) dx sin 3 x − sin 6 x

x x − sin 2 2 43. (b) ∫ e– x/2 dx ∫ 2 x 2 cos 2 z = – ∫ (sec z + sec z tan z ) e dz

= x log (1 + cos x) +

39. (b) We have, f (x) =

∫

1 − sin x ⋅ e– x/2 dx = 1 + cos x

x  dx



∫ log (1 + cos x) − x tan 2 

cos 5 x + cos 4 x dx = 1 − 2 cos 3 x

3x 3x   9x x  cos   2 cos cos   2 sin    dx 2 2 2 2 = ∫ 9x 3x − 2 cos sin 2 2 3x x = – ∫ 2 cos cos dx = − ∫ (cos 2 x + cos x ) dx 2 2

sin x. (sin x cos α + cos x sin α ) 3

cos ec 2 x −2 dx = sin α cos α + sin α cot x

∫

dx

 1 2 x  x −  dx  =  ∫ e  2 2 1 + x (1 + x 2 )    1 d  1  ex =  ∫ e ⋅  + dx = + C. 2 2   dx  1 + x   1 + x2 1 + x

= sec– 1 z + c = sec– 1 (sec x + cos x) + C. 45. (b) ∫ cos x. log tan = log tan

x dx 2

x. sin x – 2

∫ sin x.

x – ∫ 1. dx 2 x – x + C. = sin x. log tan 2

= sin x ⋅ log tan

46. (c) We have, f (x) =

x

=

∫

∫

(x

x 2 + (1 − cos 2 x ) 1 + x2

2

x 1 ⋅ 2 2 dx x tan 2

sec 2

+ sin 2 x )

1 + x2

⋅ sec2x dx

sec2x dx

213

∫ sin

Indefinite Integration

37. (b)

214

Objective Mathematics

 2 1  ∫  sec x − 1 + x 2  dx

=

= tan x – tan– 1 x + C. ∵ f (0) = 0, ∴ C = 0.

 x2 + 1  ( x 2 +1)3 / 2  2  − log  2   + C. 3 3x  x  3 50. (a) Put log x = z ⇒ x = ez ⇒ dx = ez dz =

Thus, f (x) = tan x – tan– 1x. Hence, f (1) = tan 1 – tan– 1 1 = tan 1 –

π . 4

 1  1 z  ∴ ∫  log (log x ) +  dx = ∫  log z + 2  e dz z (log x )2    1   1 1  = ∫   log z −  +  + 2   ez dz z   z z  

47. (a) Put x = tan θ ⇒ dx = sec2θ dθ ∴

∫

(

log x + 1 + x 2 1+ x

)

dx

2

∫

=

∫ sec θ.

log (sec θ + tan θ) dθ d

=

∫ log (sec θ + tan θ). d θ

=

2 1 log (sec θ + tan θ) + C 2 

=

2 1  2  log x + + x 1  + C. 2 

(

48. (b) ∫ esin x

3

x − sin x )

∫e

=

∫ x (e

sin x

cos x ) dx –

d = ∫x (e sin x) dx – dx = xesin x –

∫e

sin x

∫e

=

=

∫e

sin x

sin x

∴ ∴

x + 1 log ( x 2 + 1) − 2 log x  dx 49. (c) ∫ x4 1 1 1  = ∫ 1 + 2 ⋅ 3 ⋅ log 1 + 2  dx  x x x 

( x − 1)

 −1  2z 1 2 − ∫ ⋅ z 3 / 2 dz  log z. z 3 2  3  −1 3/2 2 3/2 = z log z + z +C 3 9

−

2

=

53. (a)

1 2 ( x − 1)

2

−

 1 1 +  dx 4 ( x − 1 ) 4 ( x + 1)  x +1 x −1

1 1 + log 2 ( x − 1) 4

+ C.

1 =z x sin x + cos x −1

(sinx + x cosx – sinx) dx = dz

( x sin x + cos x )2 x cos x

 dx = – dz,

( x sin x + cos x )2 x cos x

∫ ( x sin x + cos x) x2

∫ ( x sin x + cos x)

2

2

 dx = – z.

dx =

x

x cos x

∫ cos x ⋅ ( x sin x + cos x)

 cos x. 1 + x sin x  ∫ (− z )  cos x 

x ⋅ (– z) – cos x

=

−x + sec 2 x  dx cos x ( x sin x + cos x ) ∫

=

−x + tan x + C. cos x ( x sin x + cos x )

x4

∫ ( x − 1) ( x

2

−x ⋅ cos x

∴ f (x) =

3/ 2

+

=

z log z dz 1 1 −1    Putting 1 + x 2 = z ⇒ x 3 dx = 2 dz 

2

−1

⇒

.1 dx

3

3

⇒

d (sec x) dx dx

2

∫



(sec x tan x) dx

= esin x (x – sec x) + C.

−1 = 2

3x + 1

∫  ( x − 1)

52. (c) Put

sin x sin x – e . sec x − ∫ e . sec x. cos x dx 



1   dz z  

∫ ( x − 1) ( x + 1)

we have dx

(x cos x – sec x tan x) dx

sin x

1 d  +  log z − z  dz 

 1 1   ez  log z −  + C = x log (log x ) − + C  z log x   1 ∴ f (x) = log (log x) and g (x) = ⋅ log x 51. (b) Using partial fractions,

dx

cos 2 x

=

(log (sec θ + tan θ)) dθ

   log z − 

x

=

)

( x cos .

∫e

=

log ( tan θ + sec θ)  · sec2θ d θ sec θ

=

z 3/ 2  2   − log z  + C 3 3

=

2

+ 1)

dx =

(x

4

− 1) + 1

∫ ( x − 1) ( x

2

+ 1)

dx

dx

2

dx

=

∫ ( x + 1)  dx

  dx +1) 

=

1

1

∫  2 ⋅ x − 1 − 2 ⋅ x

+

2

x 1 1  − ⋅  + 1 2 x 2 + 1 dx

[Using partial fractions] 1 x2 + x + log (x – 1) 2 2 1 1  – log ( x 2 + 1) − tan–1x + C. 4 2

=

54. (d)

=–

cosec xdx , let cotx = t, cosec2 x dx = –dt cot x

∫

dt = – 2 t

x −1 3 +x

∫x =

t + c = – 2 cot x + c

x + x − ( x + 1) dx = ∫ x3 + x

=

∫ e ( tan z + sec z ) dz

=

∫e



( x + 1)



2

  dx = + 1) 



z

z

= x – log x +

dx

1 − x   dx 2 + 1  

1

∫ 1 −  x + x

1 log (x2 + 1) – tan– 1 x + C. 2

 2  56. (a) e x  x + 2  dx = e x  x + 4 x + 4  dx ∫  x + 4  ∫  ( x + 4 )2  4  x x  ∫ e  x + 4 + ( x + 4)2  dx

 x d  x  +   x + 4   dx + 4 x dx   x = e x⋅ + C. x+4

57. (b)

dz ∫ (1 − z 2 ) (1 + 2 z )

[Putting cos x = z ⇒ sin x dx = – dz]

dx

1 dx 4 ∫ cos x (1 − 2 sin 2 x )



=

1 4

∫ (1 − sin x) (1 − 2 sin x)



=

1 4

∫ (1 − z ) (1 −2 z )

=

cos x

2

2

dx

dz

2

2

[Putting sin x = z ⇒ cos x dx = dz]

=



=



=

= 

1 2

dz

∫ 1 − 2z 1

2 2 1 2 2

2

−

1 4

dz

∫ (1 − z ) 2

log

1+ 2 z 1 1+ z +C − log 1− z 1− 2 z 4

log

1 + 2 sin x 1 1 + sin x log + C. − 4 1 − sin x 1 − 2 sin x

 cos x + sin x  sin 2 x log  2  cos x − sin x  –

∫

sin 2 x cos x − sin x 2 dx ⋅ ⋅ 2 cos x + sin x (cos x − sin x )2

=

 cos x + sin x  sin 2 x sin 2 x log  cos x − sin x  − ∫ cos 2 x dx 2

=

 cos x + sin x  1 sin 2 x log  − log sec2x + C. 2  cos x − sin x  2

dz

∫ ( z + 1) ( z − 1) (2 z + 1)

sin x

∫ sin 4 x

 cos x + sin x  60. (b) ∫ cos 2x log  dx  cos x − sin x 

dx ∫ sin x (1 + 2 cos x)

sin x dx = − = ∫ 2 − 1 cos x ) (1 + 2 cos x ) (

=

dz

+ C.

∫ 4 cos x cos 2 x

x

dx ∫ sin x + sin 2 x =

−1 x

=

2

∫e



d  tan z + ( tan z )  dz  dz



1 dx x + dx = ∫ 1 dx − ∫ dx − ∫ x 1 + x2 ∫ x2 + 1

=

1

2

= ez tan z + C = xe tan

dx

[Using partial fractions]

=

4

e tan x z 2 2 ∫ 1 + x 2 (1 + x + x ) dx = ∫ e (1 + tan z + tan z ) dz

59. (a) ∫ sin x. cosec 4x dx =

3

∫ 1 − x ( x

1

1 1 2 log (1 – z) + log ( z + 1) − log (2z + 1) + C 6 2 3 1 1 log (1 + cos x) = log (1 − cos x ) + 6 2 2 log (2 cos x + 1) + C. – 3 58. (c) Put tan– 1 x = z i.e., x = tan z

2

∴ P = – 2 55. (b)

1

=

∴

∫

3

1

−1

cot x ∫ 2 cos x dx = P cot x + Q sin x sin x I=

1

∫  6 ⋅ z − 1 + 2 ⋅ z + 1 − 3 ⋅ 2 z + 1

[Using partial fractions] 1



2

215

1

∫  x − 1 + ( x −1)( x

Indefinite Integration

 x2 − 1

=

216

x   dx =



∫ cos  b log a

61. (c) Let I =

z  ∫ cosbz ⋅ ae dz

64. (d) x log 1 + ∫ 

Objective Mathematics

x   z z  Put log a = z ⇒ x = ae ⇒ dx = ae dz 

∫e

= a

z

where L =

z ∫ e cos bz dz

∫e

z

sin bz dz



= ez cos bz + b e sin bz − b ∫ e cos bz dz 



= ez cos bz + bez sin bz – b2 L

z

z



62. (a)

63. (b)

=

∫ (x

2

=

x  x x    cos  b log  + b sin  b log   + C. 2     1+ b  a a  x2

+ a 2 ) ( x2 + b2 )

dx

b2 ( x2 + a 2 ) − a 2 ( x2 + b2 )

=

1 b − a2

∫



=

1 b − a2

∫  x



x x 1  b tan − 1 − a tan − 1  + C. = 2 2   b −a b a

(x



2

dx ∫ 1 + sin α. cos x =

2

∫

2

=

+ a 2 ) ( x2 + b2 )

− 1)

log (1 + x) −

2

∴  f (x) =

x2 x log x 2 + + c 4 2

x2 − 1 x2 1 ,L= . , g ( x) = – 4 2 2

1+ x (1 + z 3 ) 3 z 2 2 2 dx = ∫ 1 + z dz = 3 ∫ z ( z − z + 1)dz 1+ 3 x  z5 z 4 z3  = 3 ∫ ( z 4 − z 3 + z 2 ) dz = 3 ∫  − +  +C 4 3 5 3 5/3 3 4/3 = x − x +x+C 5 4

dx

b2 a2  − 2 dx 2 +b x + a 2 

dx

66. (b) I =

∫ (cos

4

sin 3 x dx x + 3 cos x + 1) tan −1 (sec x + cos x ) 2

Let tan–1 (sec x + cos x) = t ⇒

1 − tan 2

1 1 + (sec x + cos x )

2

(sec x tan x – sin x) dx = dt

sin 3 xdx = dt cos 4 x + 3 cos 2 x + I dt ∴  I = ∫ = log | t | + c = log | tan–1 t  (sec x + cos x) | + c. ⇒

67. (a) Put x = z6 ⇒ dx = 6z5 dz

dz 2 ⋅ 1 − sin α ∫  1 + sin α  2 2  1 − sin α  + z  

2 ⋅ 1 − sin α

2

∴∫

x 2 1 + sin α. x 1 + tan 2 2 x sec 2 2 dx = ∫ x (1 + sin α ) + (1 − sin α ) tan 2 2 =

(x

1 dx x2 x2 − + +c log x 2 ∫1+ x 2 4

65. (a) Put x = z3 ⇒ dx = 3z2 dz



2

∫ ( x − 1) dx −

ae z (cos bz + b sin bz) ∴ I = 1 + b2

dx

x2 1 x2 x2 x − ∫ log x + ∫ dx dx − 2 2 1+ x 2 2

1 x2 log (1 + x) − 2 2

=

ez (cos bz + sin bz) 1 + b2

∴ L=

∫ ( x log (1 + x) − x log x)

= log (1 + x)

= ez cos bz + b



= f (x) log (x + 1) + g(x) log x2 + Lx + c I=

⋅ cos bz dz = a L,

1  dx x

∴

x1/ 2 z6 ( z 6 − 1) + 1 dz dx = 6 dz = 6 ∫ x1/ 2 − x1/ 3 ∫ z −1 ∫ ( z − 1)

x   2 x  Putting tan 2 = z ⇒ sec 2 dx = 2dz 

=6

 1 − sin α 1 − sin α tan– 1  1 + sin α  1 + sin α

 z6 z5 z 4 z3 z 2  + + + z + log ( z − 1) + C =6  + + 5 4 3 2 6 

 x  π α  = 2 sec α tan– 1  tan tan  −   + C  4 2  2  ∴ f (x) = tan

x π α ⋅ tan  −  . 4 2 2

 z + C 



∫  z

5

+ z 4 + z3 + z 2 z + 1 +

1  dz z − 1 

 x x 5/ 6 x 2 / 3 x1/ 2 =6  + + + 5 4 3 6 1/ 3  x 1/ 6 1/ 6 + 2 + x + log ( x − 1)  + C 

α + β  β − α     −  x − 2  2 

∫

69. (b)



=

dx

=

( x − 1) ( x + 2) 3

4

∫  x −1

5

dx 3/ 4

 x + 2 

( x + 2 )2

− 1) ( z 3 + 1)

3

z 2 ( z + 1)

12 z 3

(1 − z )

4 2

(1 − z ) ×

4 2

9

⋅

1 dz z3 72. (a)

∫x =

∫

(x

− 1)

2

x + 3x + 1 4

2

x −1 4 4 dz = ⋅ 4 + C. x+2 3∫ 3

1 − x2 1 − x2 = t  ⇒ = t2 2 1+ x 1 + x2 1− t2 ⇒ x2 = 1+ t2 −4t ⇒ 2x dx = dt. (1 + t 2 ) 2 dx

∫ (1 + x ) 2

=

∫

=–

= ∴

=

1 − x2

−2t

(1 + t )

2 2

1 2

∫

⋅

−2t

∫ x 1+ t

(

)

2 2

2

1   x +  + 1 x

dt 1− t2

=

1 − x2 . 1 + x2

dx =

(x

x2

∫

2

− 1)

x2 + 3 +

1 x2

217

dx

dz z2 + 1

  1 1   Putting x + x = z ⇒ 1 − x 2  dx = dz   

73. (b) Put x = z4 ⇒ dx = 4z3 dz. ∴ 1

 1− t2  1− t2 + − 1 1 2 1+ t2  1 + t 

−1 sin– 1 t + C 2

∫

= log x + 1 + x 2 + 1 + 3 + C. x x2

1+ t2 1+ t2 1+ t2 ⋅ ⋅ dt 2 1− t2 2t

−1 1 − x2 sin −1 + C. 1 + x2 2 f (x) =

⋅

z 2 + z3

= log z + z 2 + 1 + C

70. (a) Put

 ∴

− 1) 6 z 5 dz

dz

dx =

1  1 − 2  x





6

 (1 + x )3 / 2 (1 + x )4 / 3 (1 + x )7 / 6 − + = 6   9 8 7  (1 + x ) (1 + x )5 / 6 (1 + x )2 / 3   − + −  + C. 6 5 4 

5/ 4

  x −1 4 1+2z 4 = z ⇒ x=   Putting 4 x 2 1 z + −   3 12z dz   and dx =  (1 − z 4 ) 2      3 2 and = x +  1 − z 4  

=

(z

 z9 z8 z 7 z 6 z5 z 4  = 6   − + − + −  + C 6 5 4 9 8 7

dx

3/ 4

∫

(z

∫

8 7 6 5 4 3 = 6  ∫ ( z − z + z − z + z − z ) dz

∫ ( x − 1) ( x + 2) =

x dx = 1+ x + 2 1+ x

= 6  ∫ z 3 ( z 3 − 1) ( z 2 − z + 1) dz

2

 2x − α − β  = sin–1  + C.  β − α 



3

= 6  ∫

dx 2

∫

∴

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

∫

=

− x 2 + x (α + β) − αβ dx

∫

=

71. (b) Put (1 + x) = z6 ⇒ dx = 6z5 dz

dx

∫

=

Indefinite Integration

dx

∫ ( x − α) (β − x)

68. (c)

 dt

2

∫1+

x 4

x3

dx =

z 2 .4 z 3 ∫ 1 + z 3 dz = 4

z 3 .z 2 ∫ z 3 + 1 dz

1   3 2 4 ( y − 1) dy  Putting z + 1=y ⇒ z dz = dy  ∫ 3   3 y 4 = (y – log y) + C 3 =

=

4 [1 + x3/4 – log (1 + x3/4)] + C. 3

 3x − 4  74. (b) We have f  =x +2  3 x + 4  Let

3x − 4 = t ⇒ 3x – 4 = 3xt + 4t 3x + 4

⇒ x =

4t + 4 3 (1 − t )

4t + 4 +2 3 (1 − t )

218

∴ f (t) =

Objective Mathematics

∴ f (x) =

 1 1   t − 1 − dt = 2t + ln  +c = 2 ∫ dt + 2 ∫   t − 1 t + 1   t + 1 

4x + 4 + 2 = 4 ( x − 1) + 8 + 2 3 (1 − x ) 3 (1 − x ) 4 8 − 3 3 ( x − 1)



=2–

∴

∫ f ( x) dx

 1 + x − 1 2 1 + x + ln   +c.  1 + x + 1 2− x 2 − 2z3 = z3 ⇒ x = 2+ x 1 + z3 2 −12 z dz 2 ⇒ dx = ∴ 2 . ∫ ( 2 − x )2 (1 + z 3 )

79. (b) Put

2 8 ln | x – 1 | + c. x− 3 3

=

75. (a) Put x = z6 ⇒ dx = 6z5 dz

76. (b)

x + 3 x2 + 6 x

∫

∴

∫



= 6  ∫



3 4 =  z + 6 tan– 1 z + C 2



=

∫(

tan x + cot x  dx =

=

(

x 1+ 3 x

)

dx =

= =

2

+ z 4 + z ) 6 z 5 dz

∫ ∫

80. (d) I =

 sin x cos x  +  dx cos x sin x 

∫

sin x + cos x

sin x + cos x 1 − (sin x − cos x )

2

sin x + cos x dx 2 sin x cos x

dx =

2

∫

dz 1 − z2

[Putting sin x – cos x = z ⇒ (sin x + cos x) dx = dz] = =

2 sin– 1 z + C 2 sin– 1 (sin x – cos x) + C.

 x  1 x  e + / 3 2 ∫  1 + x 2 1 + x 2  dx ( )   x Let f (x) = 1 + x2

77. (d) I =

⇒ f ′ (x) = =

1 + x2 −

x2

1 + x2 2 1 + x ( ) 1

(1 + x )

2 3/2

∴  I = ex f (x) + c =

ex x 1 + x2

∫

⋅ z⋅

3 2

∫x

2/3

+ C.

dx

n

dz

(1 + x n )

1/n

=

∫

dx 1  x n + 1 1 + n   x 

1/n

1 −n = t ⇒ n + 1 dx = dt xn x

Let 1 +

  1 dt 1  t − 1/n + 1  +c ∴ I = – =– n ∫ t1/n n  − 1 + 1     n 1−

1

1 n  1 + n  x = – +c n −1 1 − x2 dx 81. (b) ∫ ⋅ 1 + x2 1 + x4 1  1 − 2  dx x =–

∫



z

=

∫

+c.

= 

2

1  1 1   x +   x +  − 2.x. x x x dz

= –  ∫

=

1− x dx . Let 1 + x = t2, dx = 2t dt x t2 −1 + 1 t ⋅ 2t dt ∴ I = ∫ 2 =2 ∫ 2 dt t −1 t −1

78. (a) I =



dx

1 − (1 − 2 sin x cos x )

2

3 2 + x = 4  2 − x 

∫  2

∫

(− 12 z ) (1 + z )

2 − x dx 2+ x

2

2

2 − 2z3   2 − 1 + z 3  −3 dz 3 + C. = = 2 ∫ z3 4z2

z 6 (1 + z 2 )

3 2/3  x + 6 tan– 1 x1/6 + C. 2

sin x + cos x  dx = sin x cos x

2

6

z5 + z3 + 1 1  dz = 6  ∫  z 3 + 2  dz  z + 1 z2 + 1

)

∫

(z

=

3

  1 1   Putting x + x = z ⇒ 1 − x 2  dx = dz    z −2 2

dt 1 − 2t 2



1 −1    Putting z = t ⇒ dz = t 2 dt 

1 dt = 2 ∫  1 2 2   −t 2  2 1 sin– 1  z  + C 2 =

 t  1 sin– 1  +C  1 / 2  2

 2x 1 sin– 1  x 2 + 1  + C. 2

∫

=

(sin

2

θ − cos 2 θ) d θ   sin θ cos θ +

(sin θ + cos θ)2

(sin 2θ + 1) (sin 2θ + 1)2 − 1

= −∫

dt t t2 − 1

f (θ) = sin 2θ + 1. dx n + 1)

1 dx 1  1 1 I = n ∫ t (t − 1) = n ∫  t − 1 − t  dt

=

∫x

∫ (t

(x

∫ (x

2

4

− x 2 + 1) + x 2

+ 1) ( x − x + 1) 4

3 2

∫ (t

3

− 1) tdt

3  t5 t2  +C − 2  5 2 

− t ) dt =

=

3 2  t (2t3 – 5) + C 20

=

3  (1 + x2)2/3 (2x2 – 3) + C. 20

4

 x  ⇒ 1 +  dx = dz  1 + x2 

1 + x 2 = z z dx. z−x

2

z2 − 1 ⋅ 2z

⇒ x =

 x  1 1  t − 1 = n log  t  + c = n log  x n + 1  + c . x4 + 1 dx = 6 +1

2 1/ 3

Also, 1 + x2 = (z – x)2 = z2 – 2 xz + x2

n

∫x

(1 + x )

3 2

⇒ dz =

Put xn + 1 = t,  nxn–1dx = dt

84. (a)

∫

x2 ⋅ x

= 

87. (c) Put x +

∫ x (x



dx =

dx =

= cosec– 1 t + C = cosec– 1 (sin 2θ + 1) + C.

83. (a) I =

(1 + x )

2 1/ 3



[Putting sin 2θ + 1 = t ⇒ 2 cos 2θ dθ = dt] ∴

x3

2

1 1  − 2 4

2 cos 2θ d θ

= –  ∫

∫

∴

219

86. (b) Put 1 + x2 = t3 ⇒ 2x dx = 3t2 dt

(sin θ − cos θ) d θ ∫ (sin θ + cos θ) sin θ cos θ + sin 2 θ cos2 θ

∴

∫

dx

dx x dx + + 1 ∫ ( x 3 )2 + 1

1 + x2 − x n

=



= 1 2

1 d (x ) dx ∫ x 2 + 1 + 3 ∫ x3 2 + 1 dx ( )

)

n

=

∫  1+ x (

z − x  dz = z 

∫ z . 



2

2

(

( 1 + x + x) 2

dx

∫ (z

n

∫z . n

2

n

) − x  2

n

dx

2 z 2 − ( z 2 − 1) 2z2

dz

1  z n +1 z n −1  + + C,  2  n + 1 n − 1 

+ z n − 2 ) dz =

3

=

= tan– 1x +

where z = x +

1 tan– 1 x3 + C. 3

88. (a)

85. (a) Put 1 + xex = t

`

∫

1 + x4

(1 − x )

4 3/ 2

1   3 + x  x

⇒ (x + 1) exdx = dt ∴

∫x

( x + 1)

(1 + xe )

x 2

dx =

( x + 1) e x dx

∫ xe

x

(1 + xe )

 = =

x 2

=



1

∫  − t

2

1 – log t + log (t – 1) + C t

1  xe x  + log  + C. 1+ xe x  1 + xe x 

∫ 1

2  2 − x  x

dt

∫ (t − 1) t

3/ 2

1  x  2 − x2  x 

3/ 2

dx

3

dx =

−1 dz 2 ∫ z 3/ 2

 1 − dz  1  2  Put x 2 − x = z ⇒  x 3 + x  dx = 2   

2

1 1  − +  dt (by partial fractions) t t − 1

 t −1 1 = + log   +C t  t  =

=

∫

dx =

1  x2  2 + x2  x 

1 + x2 .

= z–1/2 + C =

89. (b)

∫x

=

∫

1 − x4

+ C.

1  x2  x2 − 2    x dx = dx ∫ 2 2 x4 + x2 + 1 1 x .x x + 1 + 2 x 1   x − 3  x 1 dz dx = ∫ 1 2 z +1 x2 + 1 + 2 x x4 − 1

2

x

Indefinite Integration

82. (a)

220

 dz  1 1  2  Putting x + x 2 = z ⇒  x − x 3  dx = 2   

Objective Mathematics

z +1 + C =

=

x 4 + x 2 + 1 + C. x x −1

= 90. (b)

∫ ( x + 1) =

x2 + 1 +

∫

x3 + x 2 + x

(x

1 + C. x2

2 = log  z + z + 2  + C

 1 = log   x −  +  x  

2  1   x −  + 2  + C. x   a 93. (c) Put a + bx = zx ⇒ x = z−b

dx

− 1)

2

( x 2 + 2 x + 1).x x + 1x + 1

dx



=

=

∫

2

1 + tan 6 x

dx =

∫

(1 + t )

94. (a)

2 2

∫

dt

1 + t6

=

)

∫ (1 + t ) (1 − t 2

2

1  1 + 2  dt t = = ∫ 1 2 − + t 1 t2

+t

4

)

dt =

a

∫ ( z − b)

2

1

⋅

a3

⋅

dz

a2 z 2

( z − b)3 ( z − b)2

1  3b 2 b3  z − 3b + − 2  dz 4 ∫ a  z z 

b3  − 1  z2 2  + C, bz b log z − + + 3 3 a 4  2 z 

sec x

sin ( 2 x + α ) + sin α

=

∫

=

∫

[Putting tan x = t ⇒ sec x dx = dt]

(1 + t

= −

dz

−1 ( z − b ) −1  z 3 − 3 z 2b + 3 zb 2 − b3  dz = 4 ∫  4 ∫ 2  dz a z a  z2

sec x dx

2 sin ( x + α ) cos x

2

2 2

2

( z − b )2

where z =

sec6 x dx ∫ 1 + tan 6 x

(1 + tan x) . sec x 2

=

=–

 x2 + x + 1   + C. – 1  = 2 tan    x 

2

3

−a

3



dz = 2 tan– 1 z + C 1 + z2

dx 91. (c) ∫ sin 6 x + cos6 x =

dx

∫ x (a + bx)

∴

2 z dz ∫ ( z 2 + 1).z

  1 1  2  Putting x + x + 1 = z ⇒ 1 − x 2  dx = 2 z dz   

= 2∫

⇒ dx =



1  1 − 2  x dx = = ∫ 1 1   + + + + x x 2 1   x x 

  1 1   Putting x − x = z ⇒ 1 + x 2  dx = dz   



1+ t dt 2 + t4

∫1− t

1  1 + 2  dt t ∫  12 =  t −  + 1 t

2

=

∫z

dz +1

 dx

2 sin x ( x + α ) / cos x 1 2

∫

 dx

sec 2 x dx = tan x cos α + sin α

1 2 cos α

∫

dt t

⇒ sec2x dx = dt/cos α]



  1 1   Putting t − t = z ⇒ 1 + t 2  dt = dz     1 = tan – 1 z + C = tan– 1   t −  t = tan– 1 (tan x – cot x) + C.

( )

dx

[Putting tan x cos α + sin α = t

2

1  1 + 2  dx 1 + x2 ) ( x dx = 92. (b) ∫ ∫ 2 1 x 1 + x4 x + 2 x 1  1 + 2  dz x = ∫ dx = ∫ 2 2 z + 2 1   x −  + 2 x

sec 2 x

a + bx ⋅ x

95. (c)

=

1 ⋅2 t + C 2 cos α

=

2 sec α tan x cos α + sin α + C

=

2sec α ( tan x + tan α ) + C.

∫

1+ 3 x 3

x2

Put 1 + x

dx =

1/3

∫x

=z 2

2

⇒ x–2/3 dx = 6z dz

−2 3

1

1 2   3 1 + x  dx

1 1 −2   Here m = 3 , n = 3 , p = 2    ∴ m + 1 = 1 (an integer )    n



= 2 (1 + x1/3)3/2 + C.

x2

=–

35 −3 , C = a constant. , B= 36 2 100. (b) Put 1 + x1/4 = z3 ⇒ dx = 12 z2 x3/4 dz ⇒ A =

96. (b) Put 2 + x2/3 = z4 ⇒ dx = 6z3. x1/3 dz

∫ x (2 + x )

∴

2 / 3 1/ 4

1/ 3

dx =

∫x

1/ 3

3 1/3 . z (6z . x ) dz

 z9 2z5  4 4 = 6 ∫ z ( z − 2) dz = 6 ∫  − +C 5  9 2 12 = (2 + x2/3)9/4 – (2 + x2/3)5/4 + C. 3 5 97. (a) Put 1 – x3 = t2 ⇒ – 3x2 dx = 2t dt dx

∴

∫x

=

1 log 3

1 − x3 t −1 t +1

∫x

=

x2

dx =

1 − x3

3

+C=

dt ∫ t2 − 1

2 3

1 − x3 − 1

1 log 3

1 − x3 + 1

∴

= 

∫ 1 + z

99. (a)

−x

2

z ∫ x 4z

(

x

)

dz = 4

2x

4e + 6 dx = 2x −4

∫ 9e

 2x  Put 9e − 4=t       





 z7 z4  = 12  −  + C 4 7



 (1 + x1/ 4 )7 / 3 (1 + x1/ 4 ) 4 / 3  − = 12   + C. 7 4 

∫z .x 3

1/ 4

1/ 2

⋅ (12 z2 x3/4) dz

∫ z (z

dz = 12

3

3

− 1) dz

1 f ( x) sin x cos x dx = 2 (b 2 − a 2 log [ f (x)] + C )

∫z

z2 − 1 dz

⇒

2

102. (a) Put 1 + x4/3 = z7 ⇒ x1/3 dx = ∴

=

35 3  log (9e2x – 4) –  (log 9 + 2x) + C 36 4

3  log (9e2x) + C 4





2

dx

∫

3

x

7

1 + 3 x 4 dx =

21 6 z dz 4

 21

∫ z  4

 z 6  dz 

 =

 35 3  = ∫ −  dt (by partial fractions)  36t 4 (t + 4) 

35  log (9e2x – 4) – 36

f ' ( x)

∫  f ( x)

=

1 ⇒ f (x) = a 2 sin 2 x + b 2 cos 2 x ⋅

 1 (4 + t ) 9  35 1  2x ⇒ 2e dx = dt  dt 36 9  dt  ⇒ dx = 2 (t + 4) 

=

sin x cos x − 2a 2 sin x cos x )  dx

2

−1 ⇒ – b2 cos2x – a2 sin2x = f ( x )

⇒ e2 x =

35 3  log t –  log (t + 4) + C = 36 4

∫ (2b



35 + 2t

∫ 9t (t + 4) dt

1 1 ⋅ ⋅ f ′ (x) 2 (b 2 − a 2 ) f ( x )

f ' ( x) ⇒ 2 (b2 – a2) sin x cos x =  f ( x ) 2  

1   1 z − 1 dz = 4  z + log +C − 1  2 z + 1  

4e + 6e dx = x − 4e − x

∫ 9e

= 12

x

f (x) sin x cos x =

 1 + x − 1 = 4  1 + x + 2 log    + C.  1 + x + 1  x



dx =

∫x

z

∫

∫

x = z2 ⇒ dx = 4 z x dz

 =4

1+ 4 x

Differentiating both sides w.r.t x, we get

1 + x dx x

∫

3

∴

101. (b) We have, + C.

1 ∴ a = . 3 98. (c) Put 1 +

3   3 35 x+  log (9e2x – 4) +  C − log 3  2 2 36

221

∫

3

∫ z. 6 z dz = 2z3 + C.

dx =

Indefinite Integration

1+ 3 x

∴

103. (c)

∫

=

(

21 1 + 3 x4 32

(x − x )

5 1/ 5

x6

)

8/ 7

21 z 8 ⋅ +C 4 8

+ C.

1   4 − 1 x dx = ∫ x5

1/ 5

dx =

−1 1/ 5 dt t 4 ∫

dx −1  1   Putting x 4 − 1 = t ⇒ x 5 = 4 dt  1  = −5 t6/5 + C = −5  4 − 1  24 24  x

6/5

+ C.

222

104. (a) Put x = z3 ⇒ dx = 3z2 dz

∫x

Objective Mathematics

∴

=3

(

dx

1+ x 3

1

)

2

3 z dz

∫ z (1 + z )

=

3

1

1

∫  z − 1 + z − (1 + z ) 

2

= 3∫

2

dz

z ( z + 1)

  dz 

 1  x = 3  log 1 + x1/ 3 + 1 + x1/ 3  + C.

=

sin 8 x − cos8 x dx 2 x cos 2 x

=

x2

(a + bx 2 )

=– a

5/ 2

dz

(z

2

∴

107. (c)

∫e =

x

∫ (x z)

dx =

− b)

2

5

 − az    dz 2  x ( z 2 − b ) 

(1 − 2 sin

∫

(1 − 2 sin

−1 sin 2x + B. 2

x ( z 4 − 1)

1 z −1

+ C.

dx

(1 + x )

2 5/ 4





=–2



=

∫x

(z

4

2 x

dz 4

− 1) z 2

cosec x cot x (cot x + cosec x) dx

dx

= – (cosec x + cot x) e

1 



=–



=

∫ (1 + z ) (1 − z )



=

1 2

4

∫ xz  x

=

1 + x4

∫x

z2

4

(z

4

− 1)

)

= −2

dz

∫z

2

=

2

1

∫  1 − z

1 z4 − 1 − z3

x 3 ( z 4 − 1)

2

2

−

− z3

2

 dz = –

2

z2

2



z

 − 2z3    dz 4  x ( z − 1) 

  dz 3 4  ( z − 1) 

∫

(cosec2x + cosec x cot x) dx

cosec x

2

2

⇒ x4 =

∴

+ (cosec x cot x + cosec x)] dx

cosec x

(x

1/ 2

4 5/ 4

2 + C. z

+ C.

1 + x2

2

∫e

1

∫x

=

⇒ dx =

cosec x [cosec x cot x (cot x + cosec x) = ∫e

+

4

dz

2

110. (a) Put 1 + x4 = x4z4

3/ 2

dx

x cos 2 x )

2

−2 z 3

1/ 2

4

cosec x ∫ e (cot2x cosec x + cosec2x + cot x cosec2x

∫e

x cos 2 x )

−1 . 2

∫x

+ cosec x cot x) dx

=

2

cos 2 x. (1 − 2 sin 2 x cos 2 x )

cosec3x (cos2x + sin x + cos x + sin x cos x) dx

cosec x

dx

x cos 2 x )

2

x − cos 2 x ) (sin 2 x + cos 2 x )

2

2

dz 1 ∫ a 2 z 4 = 3az 3  + C

1  x2  = 3a  a + bx 2  cosec x

(sin

⇒ dx =

⇒ dx = x ( z 2 − b )2 dz

4 4

(1 − 2 sin

109. (b) Put 1 + x2 = x2z4 ⇒ x2 =

− az

∫x z

x − cos 4 x ) (sin 4 x + cos 4 x )

4

∴ A =

a 106. (b) Put a + bx2 = x2z2 ⇒ x2 = z 2 − b

=– a

∫

(sin

= –  cos 2x dx = ∫

−1 sin 8 x dx 2 ∫

1 ∴ a = . 16

∫

∫

=–

1 cos8x + C. 16

∴

(cosec2x + cosec x cot x) dx + C

cosec x

× [(sin2x + cos2x)2 – 2 sin2x cos2x] dx

cos 8 x + 1 ∫ tan 2 x − cot 2 x dx 2 cos 2 4 x = ∫ sin 2 2 x − cos 2 2 x ⋅ sin 2x cos 2x dx

∫ sin 4 x cos 4 x dx =

 (cosec2x + cosec xcot x) dx

∫ 1 − 2 sin

=

1/ 3

∫e

+

cosec x

– ecosec x (cosec x + cot x) + C. 108. (a)

 1  = 3  log z − log (1 + z ) + 1 + z  + C.

=–



2

(by partial fractions)

105. (c)

∫e

– 2

∫z

dz

1  dz 1 + z 2 

z2 dz −1

4

dz

where z =

=

Since the curve passes through (0, 0), ∴ C = 0.

1 4 1 + x4 . x

∴  y =

111. (a) Put x = tan θ ⇒ dx = sec2θ dθ x2 + 1  dx = x2

∫

∴

∫

=

∫

sec θ. sec 2 θ dθ tan 2 θ

sec θ (1 + tan 2 θ)

tan 2 θ sec θ = ∫ dθ + tan 2 θ

dθ

=

cos θ dθ + log (sec θ + tan θ) + C 2 θ



(

)

= − 1 + x + log x + 1 + x + C x  x x 1  and cos θ = ∵ tan θ = , ∴ sin θ =  2 1 1+ x 1 + x2   1

∫ 1 + (log x)

2

d (log x)

2

dt

∫1+ t

=

[Putting log x = t ⇒ d (log x) = dt]

= tan– 1 t + C = tan– 1 (log x) + C.

∫

d ( x 2 + 3) x +4 2

∫ (3 x + 4 )

sin x dx = – (3x + 4) cos x  +  ∫ 3 cos x dx

= – (3x + 4) cos x + 3 sin x⋅ 117. (b)

∫



 f ( x ) g' ( x ) + g ( x ) f' ( x ) f ( x) ⋅ g ( x)

=

d ( x 2 + 3)

∫

(x

2

+ 3) + 1

=

∫

dt t +1

t +1 + C = 2

x 2 + 4 + C.

2 2 d  x x 114. (b) ( x )  = dxd  x x  = dxd e x . log x  dx  2 = e x . log x . d (x2. log x) dx 1  2 = e x . log x .  log x. 2 x + x 2 .   x

=

∫

 f ( x ) g' ( x ) + g ( x ) f' ( x ) log f (x) g (x) dx f ( x) ⋅ g ( x)

=

∫

log t dt t ⇒ [f (x) g ′ (x) + g (x) f ′ (x)] dx = dt]



1 1 (log t)2 + C = [log f (x)⋅ g (x)]2 + C. 2 2 118. (a) We have, f ′′ (x) = sec2x ⇒ f ′ (x) =

∫ sec

x x



=

∫ tan x

dx + C2 = log sec x + C2

Since f (0) = 0, ∴ C2 = 0 ∴ f (x) = log sec x. 119. (a), (c) 

115. (c) We have, y =



⋅ (2 log x + 1) dx = ( x x )x + C⋅

∫x

3

cos x 4

1 dx = cost dt 4∫

1   4 3  Putting x = t ⇒ x dx = 4 dt 

x dx + C1 = tan x + C1.

∴ f ′ (x) = tan x ⇒ f (x)

dx

∫ 5 + 4 sin x

x

∫ x ⋅ (x )

2

Since f ′ (0) = 0, ∴ C1 = 0

= ( x x ) ⋅ x (2 log x + 1)⋅ ∴

[log f (x) + log g (x)] dx

=

[Putting x2 + 3 = t ⇒ d (x2 + 3) = dt] =2

sin − 1 ( 4 y ) ⋅

[Putting f (x). g (x) = t

2



113. (c)

4

∴ The antiderivative of (3x + 4) | sin x | is

−1 = + log (sec θ + tan θ) + C sin θ

112. (b)

⇒x=

116. (a) In the interval (0, π), sin x is positive, therefore, (3x + 4) | sin x | = (3x + 4) sin x.

∫ secθ d θ

2

1 sin x4 ⇒ x4 = sin–1 (4 y) 4



∫ sin

=

1 1 sin t + C = sin x4 + C. 4 4

223

 1 1 1+ z log − tan −1 z  + C,  2 2 1− z 

=

=

∫

sec 2

x 2

=

∫

dx x   2 tan   2 5+4  2 x   1 + tan 2

x x 5 tan 2 + 8 tan + 5 2 2

=2

∫ 5z

2

dz + 8z + 5

x   2 x  Putting tan 2 = z ⇒ sec 2 dx = 2dz 

Indefinite Integration



224

=

2 5

dz

∫

Objective Mathematics

8 z +1+ z 5 2

=

2 dz 2 2 ∫ 5  4 3 + z +     5 5



2 5 ⋅ tan–1 5 3

=

4  z+  5 + C  3     5 

x   5 tan + 4 2   2 –1 tan  =  +C 3 3  



2 5 ∴ A = and B = ⋅ 3 3

1 ⋅  z2

z 5 dz

(1 − z )2 z

=–

z

∫

1 − z2

dz = –

∫ sin

4

2

+

=–

1 4 1 8



∫ 1 − 2 cos 2θ +

θ dθ

1 + cos 4θ  dθ  2

sin 4θ    3θ − 2 sin 2θ +  + C, 4 

 1  ⋅ where θ = sin– 1z = sin– 1    1 + x 

= uv′ – vu′ +



∴a=



∫ u'' v dx ⋅

∫ u'' v dx ⋅ 1 dz a

122. (c) Put ax + b = z ⇒ dx =



∴

∫ f (ax + b)

dx =

1 a

∫ f (z)

dz =

1 [g (ax + b) + c]. a 123. (b) ∫ sin xd (cos x ) = ∫ sin x ( − sin x ) dx





=

1 − cos 2 x dx ∫ 2 = −1  x − sin 2 x  + C   2  2 

=–

1  sin 2 x  − x  + C. =   2  2



=



∴



d d

∫ dx  dx tan

1 [g (z) + c] a

−1

dx 2ax − x 2

=

dx

∫

a − ( x − 2ax + a 2 ) 2

dx

∫

a − ( x − a) 2

2

f (x) = sin–1x

2

x − a = sin– 1  + C.  a 

1 d 2 2 sec 2 x (sec 2 x + tan 2 x ) dx ∫ (sec 2 x + tan 2 x)

∫ sec 2x dx =

=

1 log | sec 2x + tan 2x | + C. 2



=

1 log 1 + sin 2 x + C 2 cos 2 x



1 log = 2









x−a ⋅ a

and g (x) =





 x  dx 

1 d (tan–1x) = + C. 1 + x2 dx

125. (b)   ∫



121. (a) ∫ uv'' dx = uv′ – ∫ u' v' dx = uv′ – vu' − ∫ vu'' dx   

=

2 (1 − z ) z

[Putting z = sin θ ⇒ dz = cos θ dθ] =–



(tan–1x) dx=

2



= –∫ 4

∫ dx

126. (a), (b), (c) 

1 120. (b) Put 1 + x = z 1 ⇒ dx = –  2 dz. z dx ∴ ∫ 5 (1 + x ) x 2 + 2 x 

d2

124. (a)

π  1 + cos  − 2 x  2  +C π  sin  − 2 x  2 

π 1 log cot  − 4 2 1 log ∴ f (x) = 2 =

 x  + C.  π  | x | and g (x) = cot  − x  ⋅ 4 

∴ dom⋅ f (x) = (– ∞, ∞) – {0} and range g (x) = (– ∞, ∞)⋅ Also, f ′ (x) =

π  1 , ∨ x ∈ R+ and g ′ (x) = cosec2  − x  ⋅ 4  2x

127. (a) Put f (x) = y2 =



⇒x=





∴

2 y2 − 1 1 − y2

 f ( x)  ∫  x 2 

x +1 x+2 and  dx =

1/ 2

dx =

∫ y⋅

2y

(1 − y )

2 2

dy.

1 − y2 2y ⋅ dy 2 2 y − 1 (1 − y 2 )2

y2  1 1  − 2 2  dy = 2 ∫    dy 2 2 − 1 1 − y y ( )( )  2 y − 1 y − 1



=2∫



=

1  log 2



=

1  log 2



Thus, g (x) = log | x | and h (x) = log | x |.

2

2 y −1 y −1 − log y +1 2 y +1 2 f ( x) − 1 2 f ( x) + 1

− log

f ( x) − 1 f ( x) + 1





Since f (0) = – 1 and f (1) = 0 ⇒ d = – 1 and a + b + c + d = 0 ⇒ d = – 1 and a + b + c = 1. ...(1) Since 0 is a stationary point of f (x), ∴ f ‘ (0) = 0 ⇒ 3a (0)2 + 2b (0) + c = 0 ⇒ c = 0. Since f (x) does not have an extremum at x = 0, ∴ f ′′ (0) = 0 ⇒ b = 0. and ∴ from (1), a = 1. So, f (x) = x3 – 1. f ( x) ∴ ∫ 3 dx = ∫ 1 dx = x + C. x −1

129. (a) We have,

I n



=



=









∫ tan ∫ tan

n

∫ tan

x dx =

n−2

= 8x4(log x)2 – 16 ∫ x3 log x dx  x4 1 x 4  = 8 x 4 (log x) 2 − 16 log x ⋅ − ∫ ⋅ dx  4 x 4   4 2 4 3 = 8x (log x) – 4x log x + 4 ∫ x dx = 8x4 (log x)2 – 4x4 log x +x4 + c = x4 [8(log x)2 – 4 log x + 1] + c

n−2

x (sec2x – 1) dx

n−2

x sec2 x dx –

∫ tan

n−2

= (I2 + I0) + (I3 + I1) + (I4 + I2) + (I5 + I3)



=  tan x + tan x + ... + tan x  ⋅  1 2 9 

130. (b)

∫x



=

∫ (t − 1) t



=

∫  t − 1 − t − t



1 t −1 1 = log | t – 1 | – log | t | + + C = log + +C t t t



= log



( x + 1)

(1 + xe ) dt

 1

9

dx =

x 2

x

(1 + xe )

x 2

dx

[Putting 1 + xex = t ⇒ (x + 1) ex dx = dt]

2

1

1 dt 2  

1 xe x + C. + 1 + xe x 1 + xe x dt

 1

1 

∫ (2 + t ) (1 + t ) = ∫ 1 + t − 2 + t  dt

= log (1 + t) – log (2 + t) + c  1 + ex = log  x 2+e 132. (c)

 + c 

1+ x 1+ x dx = ∫ dx 1− x 1 − x2 dx x =∫ +∫ dx = sin −1 x − 1 − x 2 + c 1 − x2 1 − x2

∫

(cos x)e x −(1 + sin x) sin x − cos 2 x x −∫ e dx 1 + sin x (1 + sin x) 2 ex dx 1 + sin x

e x cos x ex ex +∫ dx − ∫ dx + c 1 +sin x 1 + sin x 1 + sin x e x cos x = +c 1 + sin x

=

−1

135. (c) Let I = ∫

e m tan x dx 1 + x2

Put m tan–1 x = t dx 1 = dt 2 1+ x m −1 1 1 1 ∴ I = ∫ et dt = et + c = e m tan x + c m m m sin 2 x + cos 2 x dx 136. (a) Let I = ∫ sin 2 x cos 2 x = ∫ (sec2x + cosec2x)dx = tan x – cot x + c

(by partial fractions)

131. (a) Put ex = t ⇒ ex dx = dt ∴

=

⇒

( x + 1) e x

∫ xe

(cos x − 1)e x e x cos x ex dx = ∫ dx − ∫ dx sin x + 1 1 + sin x 1 + sin x

−∫

x dx

+ (I6 + I4)+ (I7 + I5) + (I8 + I6) + (I9 + I7) + (I10 + I8) 2

∫

x ⋅ tan2x dx

n −1 = tan x – In – 2 n −1 n −1 ∴ In + In – 2 = tan x , n ≥ 2 n −1 ∴ I0 + I1 + 2 (I2 + I3 + ... + I8) + I9 + I10



 x4 1 x 4  = 32 (log x) 2 − ∫ 2log x ⋅ ⋅ dx  4 x 4  

134. (d)

∫ tan

=

133. (b) ∫ 32x3(log x)2 dx

137. (b)

∫x

2

dx + 4 x + 13

dx 1  x+2 = tan −1  +c ( x + 2) 2 + 32 3  3  x −1 1 1 =t ⇒ dx = dt 138. (a) Let x+2 ( x + 2) 2 3 =∫

∫ [( x − 1)

∴

=

4 3

1 1 dx = ∫ t −3/ 4 dt 5 1/ 4 ( x + 2) ] 3

1/ 4

 x −1   x+2  

139. (a) Let I = ∫ =∫

3

+c ax/2

a

−x

− ax

ax/2 a− x/ 2 1 − a2x

dx

dx = ∫

ax 1 − a2x

Let ax = t ⇒ ax log a dx = dt

dx

225



Indefinite Integration

128. (b) Let f (x) = ax3 + bx2 + cx + d.

226

∴ I =

∴

Objective Mathematics

=

d (1 + nx n ) n 2 x n −1 1 1 dx dx = 2 ∫ dx = 2∫ n (1 + nx n )1/ n n (1 + nx n )1/ n

1 1 dt log a ∫ 1 − t 2

1 sin −1 t + c log a

1 sin −1 (a x ) + c log a ex dx 140. (a) Let I = ∫ x (2 + e ) (e x + 1) Put ex = t =

⇒ ex dx = dt ∴ I = ∫

∴

dt (2 + t ) (t + 1)

 1 1  = ∫ −  dt (1 + t ) (2 + t)   = log (1 + t) – log(2 + t) + c = log (1 + ex) – log (2 + ex) + c  1+ e  = log  x  + c  2+e  141. (d) ∫ [sin (log x) + cos (log x)] dx x

= ∫ sin (log x) dx + ∫ cos (log x) dx x cos (log x) dx + ∫ cos (log x) dx + c x = x sin (log x) + c

= x sin (log x) − ∫

142. (b)

∫

1 + x + x + x2 dx x + 1+ x

=∫ =∫

(1 + x) 2 + x x + 1 x + 1+ x

dx

1+ x( 1+ x + x) dx ( x + 1+ x)

2 (1 + x)3/ 2 + c 3 dx 1 π  = sec  x −  dx 143. (a) ∫ 3 cos x + 3 sin x 2 ∫ 

1  x π π log tan  − +  + c 2 2 6 4

1 x π = log tan  +  + c 2  2 12  f ( x) x = [1 + f ( x) n ]1/ n (1 + 2 x n )1/ n x and f ( f ( f ( x))) = (1 + 3 x n )1/ n x g ( x) = ( f o f o . . . o f )( x) = n 1/ n  (1 ) + nx n times

144. (a) We have, f ( f ( x)) =

∴

∴

∫x

n−2

g ( x) dx = ∫

 1 1  = 2∫  cot t +  dt 2  2 = t + log | sin t | + c π  = x + log sin  x −  + c 4  146. (a) ∫ x( x x ) x (2log x + 1) dx = ∫1. dt Let (xx)x = t = t + c ⇒ x log xx = log t = (xx)x + c ⇒ x2 log x = log t 1 dt ⇒ 2x log x + x = t dx ⇒ (2log x + 1) x · (xx)x dx = dt

= ∫ 1 + x dx =

=

1 1− 1 (1 + nx n ) n + c n(n − 1) sin x 145. (c) Let I = 2 ∫ dx π  sin  x −  4  π Put x − = t 4 ⇒ dx = dt π  sin  + t  dt 4  ∴ I = 2∫ sin t

=

x n −1 dx (1 + nx n )1/ n

147. (a) We have f ( x) x = [1 + f ( x) n ]1/ n (1 + 2 x n )1/ x x Also, fofof(x) = (1 + 3 x n )1/ n

fof(x) =

∴ g(x) = ( fofo...of ) ( x) = n terms

Thus,

∫x

n−2

g ( x)dx = ∫

n 2 x n −1dx 1 1 = 2∫ = n (1 + nx n)1/ n n 2 =

x n −1dx (1 + nx n )1/ n

∫

d (1 + nx n ) dx dx (1 + nx n )1/ n

1 1− 1 (1 + nx n ) n + k . n(n − 1)

148. (c) J – l =



∫  e

−4 x

 e− x ex − 4x dx −2 x 2x + e +1 e + e +1

 e3 x − e x  = ∫  4x dx 2x  e + e +1 =

x (1 + nx n )1/ n

e 2 x (e x − e − x ) dx 4x + e 2 x + 1)

∫ (e

π/4

0

dt dt Put e t ⇒ dx = x or e t (t − 1 / t ) dt ⋅ ∴J–1= ∫ 2 (t + 1 + 1 / t ) t x

π/4

=

0

1  1 − 2  t   =∫ dt  2 1  t + + 1   t2  

π/4

=

 | (tan x / 2 + 1) |

1 − tan 2 x / 2 0 put tan x/2 = t ⇒ dx =

∫

1 as u = t + t

2t 1− t

4t

∫

1+ t2 1− t2

0

1 t + −1 1 t = log 1 2 t + +1 t

2

0

2 −1

1 u −1 log 2 u +1

[as 0 < × < π/4]

(1 + t ) 2 −1

=

dx

2dt

∴ Area =

1 du 1 du − 2 ∫ u −1 2 ∫ u +1

| (1 − tan x / 2) |  dx 1 − tan 2 x / 2 

2

2 tan x / 2

∫

2 tan x / 2 1 + tan 2 x / 2 1 − tan 2 x / 2 1 + tan 2 x / 2

1−

−

 1 − tan x / 2

π/4

=

1 1  Put t + = u ⇒ 1 − 2  dt = du t t   du ∴J–l= ∫ 2 u −1

=

 2 tan x / 2  1+ 1 + tan 2 x / 2   1 − tan 2 x / 2  1 + tan 2 x / 2 

∫  0

1  1 − 2  t   = ∫ dt 2  1 t 1  +  −  t

=

∫

 1 + sin x 1 − sin x − cos x cos x 

∫ 

149. (b) Required Area =

2

⋅

2dt (1 + t 2 )

dt

=

 t2 − t +1  1 log  2  2  t + t +1

=

 e2 x − e x + 1  1 log  2 x  as t = ex x 2  e + e −1 

Exercises for self-practice 1. ∫ x log x dx is equal to 2

(b) x log x − 2 2 x (c) log x − 2 x2 (d) log x + 2 2

2.

tan x

∫ log sec x dx

x log x + 4 x log x − 4 x log x − 4

∫

(

(d) None of these

x +c 4 2 x +c 4 x2 +c 4 2

4.

1

∫ sin x + cos x

dx =

1 π  log tan  x +  + c  4 2 1 x π  (b) log tan  +  + c 2 8 2  x π (c) log tan  +  + c 2 8

(a)

is equal to

5.

sin x − cos x dx is equal to 2 + sin 2 x

(

)

)

(c) log sin x + cos x − 2 + sin 2 x + C

(a) log (log (log x)) + c (b) log (log tan x) + c (c) log (log sec x) + c (d) None of these 3.

(

(b) – log sin x + cos x + 2 + sin 2 x + C

x x (a) log x − +c 2 4 2

)

(a) log sin x + cos x + 2 + sin 2 x + C

(d)

1  x π log tan  +  + c 2 8 2

∫e

(1 + tan x + tan 2 x) dx =

x

(a) ex cos x + c (b) ex tan x + c (c) ex sec x + c (d) ex sin x + c

 dx 

  dx   

227

(e x − e − x ) ∫ e2 x + 1 + e−2 x )dx

Indefinite Integration

=

228

6. If

∫ g ( x) dx

= g (x), then

∫ g ( x)[ f ( x) + f '( x)] dx is equal

to Objective Mathematics

5 3 −5 (d) 4

5 4 5 (c) –  6

(b) – 

(a)

(a) g (x) f (x) (b) g (x) f ′ (x) (c) g (x) f (x) – g (x) f ′ (x) (d) g (x) f 2(x)

1 dx = f (x) + A, where A is any arbitrary 4 2 15. If ∫ 3 sec x cos ec x dx = K tan x + L tan x + M cot  ( 1 + ) x x ∫ constant, then the function f (x) is x + const., then (b) 2 tan– 1 x (a) 2 tan–1x 1 (a) K = – 1, L = 0, M = 1 (b) K = , L = 1, M = 2 (c) 2 cot–1 x (d) loge (1 + x) 3 1 16. ∫ e x dx is equal to: (c) K = , L = 2, M = – 1 (d) None of these 3

7. If

8.

∫e

x

(a) e

(1 − cot x + cot 2 x) dx =

(a) e cot x + c (c) ex cosec x + c x

(b) – e cot x + c (d) – ex cosec x + c

1 (b) e 2

x

9. If ∫ x sin x dx = – x cos x + α, then α = (a) sin x + C (c) C 10.

(b) cos x + C (d) None of these

x2 + 1 dx is equal to 2 − 1)

∫ x (x

x2 − 1 + C x x + C (c) log 2 x +1 (a) log

11.

17.

∫e

18.

1 x e (sin x – cos x) + C 2 1 (d) (tan– 1x2)3 + C 2

t 3t 2

1 (c) –  (tan–1x3)2 + C 2 13.

∫ x cos x

2

dx is equal to :

1 sin2x + C 2 1 (c) – sin x2 + C 2

(a) –

14. If

2x + 3

∫ ( x − 1) ( x

2

1 = loge ( x − 1)5/ 2 ( x 2 + 1) a  − tan– 1x + A, 2 where A is any arbitrary constant, then the value of ‘a’ is

+ A , A is an arbitrary constant.

dt =

(c)

1 −3t 2 e + C 6

1 2 (d) –  e −3t + C 6

xe x

∫ ( x + 1)

2

dx =

(a)

ex  + C 1+ x

(b)

xe x  + C (1 + x)

(c)

ex (1 + x)3

(d)

ex +C (1 + x) 2

1 dx = x − x3

1 (1 − x 2 ) log +C 2 x2

(c) log x (1 – x2) + C

∫ x (1 + log x) dx x

(1 + log x) 2 (c) x2x 21.

∫ (a

2

(d)

(1 − x) +C x(1 + x)

1 x2 log +C 2 1 − x2

(b) xx log x (d) xx

dx = + x 2 )3 / 2

x ( a 2 + x 2 )3/ 2 x (c) 2 2 a (a + x 2 )1/ 2

(a)

(b) log

=

2

(a)

(b)

+ 1) dx

x

1 3t 2 (b) –  e + C 6

(a)

20. 1 sin2x + C 2 1 (d) sin x2 + C 2

+A

1 3t 2 e + C 6

19. ∫

1 (b) (tan– 1x3)2 + C 6 1 (d) (tan– 1x2)3 + C 2

x

(a)

(c)

(a) tan x + C

+A

(d) 2 ( x + 1) e

(a) ex (sin x – cos x) + C (b) ex (sin x + cos x) + C

–1 3

x



x2 − 1 +C x x (d) – log 2 +C x +1

x 2 tan − 1 x 3 dx is equal to 12. ∫ 1 + x6

+A

(c) 2 ( x − 1) e

(b) – log

x ∫ e sin x dx is equal to

x

(b)

1 a 2 (a 2 + x 2 )1/ 2

(d) None of these

e2 x − 1

(a) sin  (e ) + C (c) tan–1 (ex) + C 23. The value of

1 (a) sin −1 2

(

∫

3

3

3

30.

)

x  (b) – 2 sin–1   2 sin  + C 2 (c) 2 sin–1 

(

)

1  [sin (log x) + cos (log x)] + C 2 x (b)  [sin (log x) + cos (log x)] + C 2 x (c)  [sin (log x) – cos (log x)] + C 2 1 (d)  [sin (log x) – cos (log x)] + C 2

33.

(1 + sin x) dx is 25. ∫ e (1 + cos x) (a)

1 x x  e  sec  + C 2 2

(b) ex sec 

x +C 2

(c)

1 x x  e  tan  + C 2 2

(d) ex tan 

x +C 2

dx 26. ∫ is equal to x( x 4 − 1)

27.

∫ sin

4

(a) a

1 x4 − 1 log 4 + C 4 x

(d) log

x4  + C x −1 4

sin 2 x dx is equal to x + cos 4 x

28. The value of x

∫

a

(b) tan–1 (x tan2x) + C (d) None of these

x

x

loge a + C

(b) 2a

x

loge a + C

(c) 2a

x

log10a + C

(d) 2a

x

loga e + C

log x

dx is equal to

x2 2 (d) ex (b)

dx

∫

x2 − 1

is

(a) sin–1x

(b) log  x + x 2 − 1  

(c) log  x − x 2 − 1  

(d) – cos–1x

∫ sin

−1

x dx is equal to

(a) x sin– 1x –

1 − x2 + C

(b) x sin– 1x +

1 − x2 + C

(c) – x sin– 1x +

1 − x2 + C

(d) None of these 34.

∫ x log (1 + x)

dx is equal to

1 [2 (x2 – 1) log (1 + x) – x2 + 2x] + C 4 1 (b) [2 (x2 – 1) log (1 + x) + x2 + 2x] + C 4 1 [2 (x2 – 1) log (1 + x) + x2 – 2x] + C (c) 4

(a) (b)

(a) tan–1 (tan2x) + C (c) 2 tan–1 (tan2x) + C

∫e

1 log (log x) + C 2

(d) (log x)2 + C

2

32. The value of

x

x4 − 1  + C x4

(b)

x + log x 2 (c) a log x

(a)

(c) log 

1 (log x)2 + C 2

(a)

3 1 sin (e x ) 3

is equal to

31. One value of

∫ cos (log x) dx is

1 x4 log 4 +C 4 x −1

(d)

(c) log (log x) + C

x  (d) 2 sin–1   2  + C 2

(a)

dx

∫ x log x (a)

2 sin x + C

24. The value of

3

(b) sin (e x )

(c) 3sin (e x )

1 + sec x dx is

2 sin x + C

3

(a) ex sin (e x )

(b) cos–1 (ex) + C (d) sec–1 (ex) + C

x

–1

29. ∫ x 2 ⋅ e x ⋅ cos (e x ) dx is equal to

is:

229

dx

∫

dx is

(d) None of these 35. If

4e x + 6e − x dx = Ax + B log (9e2x – 4) + C, then x − 4e − x

∫ 9e

(a) A =

3 2

(c) 4

(b) B =

35 36

(d) None of these

x dx 36. ∫ 1 + x4 (a) log (1 + x2) + k (c)

1 tan–1 x2 + k 2

(b) tan–1 x2 + k (d) None of these

Indefinite Integration

22. The value of

230

37.

xe x

∫ (1 + x)

Objective Mathematics

(a)

ex +c x +1

(b) ex (x + 1) + c

−e x

(c) x + 1 2 + c ( ) 38. If

∫xe

2x

(a) (3x – 1)/4 (b) (2x + 1)/2 (c) (2x – 1)/4 (d) (x – 4)/6

dx is equal to

2

39. If

ex +c (d) 1 + x2

∫

2 1 + sin x dx = 2 cos (ax + b) + c then the value

of (a, b)

π 2 (d) None of these

1 π , 2 4 (c) 1, 1

(b) 1,

(a)

dx is equal to e f (x) + c, where c is constant 2x

of integration, then f (x) is

Answers

1. 11. 21. 31.

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

2. 12. 22. 32.

(c) (b) (b) (b)

3. 13. 23. 33.

(b) (d) (d) (b)

4. 14. 24. 34.

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

5. 15. 25. 35.

(b) (b) (b) (d)

6. 16. 26. 36.

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

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. 19. 29. 39.

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

10. (a) 20. (d) 30. (c)

7

Definite Integral and Area

CHAPTER

Summary of concepts The Definite Integral Let F (x) be any antiderivative of f (x), then for any two values of the independent variable x, say a and b, the difference F (b) – F (a) is called the definite integral of f (x) from a to b and is b

denoted by

∫ f ( x) dx . Thus

6.

2a

∫ o

7.

2a

∫ o

a

b

∫ f ( x) dx

8.

= F (b) – F (a),

a

a

The reason for using the term ‘‘definite integral’’ follows from the fact that the value of the definite integral in independent of the particular choice of the antiderivative of f (x). For, if F (x) + C is any other antiderivative of f (x), then b

∫

∫

f ( x) dx = F (x) + C

b



= [F (b) + C] – [F (a) + C]



= F (b) – F (a),

∫

a

f ( x) dx + ∫ f (2a − x) dx o

0  a dx = f ( x)  2 ∫ f ( x) dx  o

if f (2a − x) = − f ( x)   if f (2a − x) = f ( x)  

 a  2 ∫ f ( x)dx if f (− x)=f ( x) i.e. f ( x) is even  f ( x) dx =  o  0 if f (− x) = f ( x) i.e. f ( x) iss odd   b

∫ f ( x) dx

∫ f (a + b − x)

=

a

10.

a

o

b

9.

b

dx

a

f ( x)

∫ f ( x) + f (a + b − x)

b−a 2

=

a

11. If f (x) ≥ 0 on the interval [a, b], then

12. If f(x) ≤ g(x) on the interval [a, b], then

a

13.

b

∫ f ( x) dx a

b

∫ f ( x) dx , it is sufficient to find an antiderivaa

tive of f (x), say F (x) and subtract the value of F (x) at the lower limit a from its value at the upper limit b.

b

∫

1.

∫ f ( x) dx

a

=–

a

∫ f ( x) dx a

4. 5.

c

=

∫ a

a

a

b

a

14. If f (x) is continuous on [a, b], m is the least and M is the greatest value of f (x) on [a, b], then m (b – a) ≤

b

∫ f ( x) dx

≤ M (b – a)

15. If f (x) is a periodic function of period a, i.e.,

o

o

∫ o

∫ f ( x) dx a

b

=

∫ f ( y) dy

na

(a)

a

na

(b)

c

dx

f ( x) a dx = 2 f ( x) + f (a − x)

∫ f ( x)

dx = n

o

b

f ( x) dx + ∫ f ( x) dx, where a < c < b

∫ f ( x) dx = ∫ f (a − x) a

2. 

b

b

3.

∫ f ( x)

b

∫ g ( x)

≤ ∫ | f ( x)| dx

f (a + x) = f (x), then b

b

a

a

Properties of Definite Integrals

≥ 0.

f ( x) dx ≤

a

dx

which is same as before.

b

∫ f ( x) dx a

a

Thus to find

a

−a

where F (x) is any antiderivative of f (x). The numbers a and b are called the limits of integration; a is the lower limit and b is the upper limit. Usually F (b) – F (a) is abbreviated by writing b F (x) .

f ( x) dx =

∫

a

∫ f ( x) dx o

f ( x) dx = (n – 1)

o

∫ f ( x) dx o

b + na

(c)

a

∫

na

f ( x) dx =

b

∫ f ( x) dx, where b ∈ R

+

0

b+a

(d)

∫ b

f ( x) dx independent of b.

232

b + na

(e)

∫ b

b

a

∫ f ( x) dx

f ( x) dx = n ∫ f ( x) dx, where n ∈ I

Objective Mathematics

where nh = b – a.

In particlular, (i) if b = 0, (ii) if n = 1,

∫

b+a

a

Put a = 0 and b = 1 ⇒ nh = 1 ⇒ h =

o

Substitute a = 0, b = 1 and h =

∫

h→0

∫ f ( x) dx

f ( x) dx =

b

∫ a

a

= f (c) (b – a), where a < c < b.

(i) Find the rth term of the given series and express it in the form 1  r  Then the given series can be written as f  ⋅ n n

lim 1 n

n→∞

In the functions g (x) and h (x) are defined on [a, b] and differentiable at all points x ∈ (a, b) and f (t) is continuous on [a, b], then d dx

b

∫ | f ( x )| dx Working Rule

1

lim

n→∞

Note: Here each term tends to zero when n → ∞, so addition or omission of finite number of terms does not affect the required limit.

∫

π/2

a

β

2.

with proper sign of f (x) in each (a, α), (α, β), (β, b).

Summation of SerieS with the help of Definite integral aS the limit of a Sum If f (x) is a continuous and single valued function defined on the

follows:

∫ cos

n

x dx

0

∫ sin

n

x cosmx dx

0

dx

interval [a, b], then the definite integral

π/2

2  n −1 n − 3 n − 5   n ⋅ n − 2 ⋅ n − 4 ⋅ ⋅ ⋅ 3 , when n is odd  =  1 π  n −1 n − 3 n − 5 , when n is even ⋅ ⋅ ⋅ ⋅ ⋅ ⋅   n n−2 n−4 2 2  

b

α

sin n x dx =

0

∫ f ( x) dx

a

If n is a positive integer, then π/2

(iii) Check the sign of f (x) by taking any point in each of these subintervals.

∫ f ( x) dx + ∫ f ( x) dx + ∫ f ( x)

r1 n

r2 , n where r1 and r2 are the least and greatest values of r respectively.

(ii) Divide the interval (a, b) into subintervals (a, α), (α, β) and (β, b).

=

∫.

and upper limit of integration = nlim →∞

Let the roots be α and β such that a < α < β < b.

b

by

n→∞

(i) Solve the equation f (x) = 0.

β

∑

Also, lower limit of integration = lim

1.

α

r =1

Some uSeful reDuction formulae

a

(iv) Use the property

r

n

∑ f  n 

(ii) Write the above series equal to a definite integral by replacing n r by x and by dx, n

 h ( x)   ∫ f (t ) dt  = d [h (x)]⋅ f [h (x)] – d [g (x)]⋅ f [g (x)]  g ( x)  dx dx

integration of moD functionS

1 . n

Working Rule

a

Differentiation unDer integral Sign

...(1)

1 in (1), we get n 1 1 n r lim f   = ∫ f ( x) dx. ∑ n→∞ n n 0 r =1

b

f 2 ( x) dx ⋅ ∫ g 2 ( x) dx

If a function f (x) is continuous on the interval [a, b], then there exists a point c ∈ (a, b) such that

∫ f ( x) dx

a

o

f ( x) ⋅ g ( x) dx ≤

b

r =1

o

h or nlim →∞

∫ f ( x) dx = n ∫ f ( x) dx

a

17.

b

a

For any two functions f (x) and g (x), integrable on the interval [a, b], the Schwarz – Bunyakovsky inequality holds b

n

∑ f (a + rh) = ∫ f ( x) dx

na

b

16.

= hlim h [ f (a + h) + f (a + 2h) + ⋅ ⋅ ⋅ + f (a + nh)] →0

a

o

b

∫ f ( x) dx a

is defined as

m−3 2 1   m −1   m + n ⋅ m + n − 2 ⋅ ⋅ ⋅ 3 + n ⋅ 1 + n ; if m is odd   and n may be even or odd     m−3 n −1 n − 3 m −1 1 2  ⋅ ⋅⋅⋅ ⋅ ⋅ ⋅⋅⋅ =   m+n m+n−2 2+n n n−2 3   m n if is even and is odd   m −1 m−3 1 n −1 n − 3 1 π  ⋅ ⋅⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ,  2+ n n n − 2 2 2  m + n m + n − 2  if m is even and n is evenn 

∫ sin

m

x ⋅ cos n x dx

0

=

[(m − 1) (m − 3) ...][(n − 1) (n − 3) ⋅ ⋅⋅] ( m + n ) ( m + n − 2) ⋅ ⋅ ⋅

to be multiplied by

4. If we have two curve y = f (x) and y = g (x), such that y = f (x) lies above the curve y = g (x) and both are above the axis of x then the area bounded between them and the ordinates x = a and x = b (b > a), is given by

π when m and n are both even integers. 2

Area of Plane Regions 1. The are bounded by the curve y = f (x), x-axis and the ordinates x = a. and x = b (where b > a) is given by

b

A=

∫ a

b

f ( x) dx − ∫ g ( x) dx a

i.e., upper curve area – lower curve area. 5. The area bounded by the curves y = f (x) and y = g (x) between the ordinates x = a and x = b is given by b

A=

b

∫ y dx =

∫ f ( x) dx

a

a

2. The area bounded by the curve x = g (y), y – axis and the abscissae y = c and y = d (where d > c) is given by

c

A= d

A=

∫ x dy = c

∫ a

d

∫ g ( y) dy .

b

f ( x) dx + ∫ g ( x) dx , c

where x = c is the point of intersection of the two curves.

c

3. The area bounded by the curve y = h (x), x-axis and the two ordinates x = a and x = b is given by

Curve Tracing In order to find the area bounded by several curves, it is important to have rough sketch of the required portion. The following steps are very useful in tracing a cartesian curve f (x, y) = 0. Step 1: Symmetry



A=

c

b

a

c

∫ y dx + ∫ y dx

where c is a point in between a and b.

(i) The curve is symmetrical about x-axis if all powers of y in the equation of the given curve are even. (ii) The curve is symmetrical about y – axis if all powers of x in the equation of the given curve are even. (iii) The curve is symmetrical about the line y = x, if the equation of the given curve remains unchanged on interchanging x and y. (iv) The curve is symmetrical in opposite quadrants, if the equation of the given curve remains unchanged when x and y are replaced by – x and – y respectively.

233

π/2

Definite Integral and Area

These formulae can be expressed as a single formula :

234

Step 2: Origin

Objective Mathematics

If there is no constant term in the equation of the given curve, then the curve passes through the origin. In that case, the tangents at the origin are given by equating to zero the lowest degree terms in the equation of the given curve. For example, the curve y3 = x3 + axy passes through the origin and the tangents at the origin are given by axy = 0 i.e., x = 0 and y = 0.

the coefficient of the highest power of y in the equation of the given curve. (ii) The horizontal asymptotes or the asymptotes parallel to x-axis of the given curve are obtained by equating to zero the coefficient of the highest power of x in the equation of the given curve. Step 5: Region

Find out the regions of the plane in which no part of the curve lies. To determine such regions we solve the given equation for (i) To find the points of intersection of the curve with y in terms of x or vice-versa. Suppose that y becomes imaginary X-axis, put y = 0 in the equation of the given curve and get for x > a, the curve does not lie in the region x > a. the corresponding values of x. Step 6: Critical Points (ii) To find the points of intersection of the curve with dy Y-axis, put x = 0 in the equation of the given curve and get Find out the values of x at which = 0. dx the corresponding values of y. At such points y generally changes its character from an increasing function of x to a decreasing function of x or viceStep 4: Asymptotes versa. Find out the asymptotes of the curve. Step 7: (i) The vertical asymptotes or the asymptotes parallel to y-axis of the given curve are obtained by equating to zero Trace the curve with the help of the above points. Step 3: Intersection with the Co-ordinate Axes

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. The value of the integral

π

∫e

cos 2 x

⋅ cos3 (2n + 1) x dx, n

2

5.

2.

π/ 2

∫ 0

3.

6.

 x cos   ⋅ sin x dx is equal to 2 5

−4  1  (b) 1 −  7  8 2 

4 1  (c) 1 −  7 8 2 

(d) None of these

∞

2 ∫0  e x  dx (where [ . ] denotes the greatest integer function) equals (b) e2 (d) 2 e

(c) 0 −4

4. If

∫

f ( x) dx = 4 and

−1

∫ f ( − x ) dx

1 6 1 (c) 4

(a)

(b)

(d) None of these π

∫ x (sin

7. The value of the integral 3 π2 (a) 64 2 3π (c) 256

1 12

0

(b)

4

x cos 4 x) dx is

3 π2 128

(d) None of these r

0

= 7, then the value

is

(a)

(b)

r3 3

r3 (d) None of these 2 9. The area of the figure bounded by f (x) = sin x, g(x) = cos x in the first quadrant is (c)

(b) –3 (d) None of these

r 6 3

2

−2

(a) 2 (c) 5

sin 2 x ⋅ cos 2 x dx is equal to 3 x + cos3 x) 2

∫ (sin

−4

∫ ( 3 − f ( x) ) dx

(b) 1 (d) 101/2

8. The value of the integral ∫ xy dx, where y = r 2 − x 2 , is

1

of

π/ 4

0

2 1  (a) 1 −  7 8 2 

1   x −  dx = x 

(a) 1/4 (c) 0

(b) π (d) None of these

(a) loge 2

101

1/2

0

integer, is (a) 0 (c) 2π

1

∫ x cos ec

(b)

3 +1

(a) –

(d) None of these π

∫e

10. The value of the integral

x

π/ 2

(a) e– π/2 (c) eπ/2

⋅

1 − sin x dx is 1 − cos x

a

19.

2π

∫e

12. The value of the integral

x

0

2 (e2π + 1) 5 (c) – 2 (e2π – 1) 5 (a)

(b)

π/6

π 4 π (c) 12

(d) None of these

π/ 2

∫ (tan x + cot x)

dx is equal to

(a)

π log 2 2

1

(c) π log 2

(d) None of these ∞

x

∫ (1 + x ) (1 + x )

21. The value of the integral

dx is

2

0

π 2 π (c) 6

(b)

π 4

(d) None of these π

∫ log (1 + cos x )

22. The value of the integral

dx is

0

(d) None of these

∫ e ( x − α ) dx x2

π 6

π log 2 2

(b) –

(d) None of these

(b)

(a)

π –1 2

0

sin x dx is equal to sin x + cos x

∫

14. If

is equal to

a2 − x2

(a)

π/3

13.

dx

π + 1 2 π (c) 1 – 2

20.

π2 log 2 2

(d) None of these

(a)

π x sin  +  dx is  4 2

(b) – 2 (e2π + 1) 5

∫a+ 0

11. The area bounded by the curves y = ln, y = ln | x |, y = | ln x | and y = | ln | x | | is (b) 2 sq. units (d) None of these

(b)

(c) π2 log 2

(b) eπ/4 (d) None of these

(a) 5 sq. units (c) 4 sq. units

π2 log 2 2

235

) 3 − 1)

2 −1

Definite Integral and Area

( (c) 2 ( (a) 2

(a)

π log 2 2

(b) π log 2

(c) – π log 2

= 0, then

(d) None of these

0

(a) 1 < α < 2 (c) 0 < α < 1 15.

π/ 2

∫

(b) α < 0 (d) α = 0

23. The value of

(b) –

(c) π log 2 16. If

∫

r2

0

xf ( x) dx =

17.

∫

1

0

(a)

π log 2 2

(d) None of these

24. The value of c for which the area of the figure bounded by the curve y = 8x2 – x5, the straight lines x = 1 and x = c and the x-axis is equal to 16/3, is

2 5 , t > 0 then f  4  = t  25  5

2 (a) 5 2 (c) – 5

5 (b) 2

(c) π 18. The value of the integral

25.

(b)

(c) 3

(d) –1

1

∫

log (1 + x )

(d) 1

(b)

π –1 2

(d) 1 π

∫ x log sin x dx is 0

8 − 17

(a) 2

0

1− x dx = 1+ x π + 1 2

is

(a) –π (b) –2π (c) –­3π (d) –4π [ . ] representing the greatest integer function.

0

π log 2 2

∫ [ 2 sin x] dx 0

log sin x dx is equal to

(a)

2π

1 + x2

dx is equal to

(a)

π log 2 8

(b)

(c)

π log 2 4

(d) None of these

26. The value of the integral π 2 (c) 0

(a)

π 1 log 8 2

π

sin 2kx  dx, where k ∈I, is 0 sin x

∫

(b) π (d) None of these

236

27.

π/ 2

∫ 0

cos 2 θ d θ is equal to cos 2 θ + 4 sin 2 θ

Objective Mathematics

π 2 π (c) 6

π log 2 2

29.

π

∞

∞

1  dx  ∫0 log  x + x  1 + x 2 is

37.

x 3 cos 4 x sin 2 x dx is equal to 2 − 3πx + 3 x 2

38. If

π2 8 π2 (d) . 32

π2 16 π2 (c) 4

π

π2 4

x 2 sin x

∫ ( 2 x − π ) (1 + cos x ) 2

(c)

π 6

(b)

n

dx is

n n2 − 1 −n (d) 2 n +1 (b)

et dt ∫0 t + 1 = a, then 1

b

e − t dt is equal to ∫ b −1 t − b − 1 (b) ­–a e–b (d) None of these

π2 2

dx = sin x +1 e − π /2

∫

(a) 0 (c) –π/2

(b) 1 (d) π/2

40. The value of the integral

3

∫ | ( x − 1) ( x − 2) ( x − 3) | 1

(d) None of these

(a) 1 3 (c) 9 4

 x2  31. Area bounded by the curves y =  + 2  ([ . ] denotes 64   the greatest integer function), y = x – 1 and x = 0 above the x-axis is (a) 2 (c) 4

π/3

41.

π

x 2 cos x

∫ (1 + sin x )

2

dx is

(b) – π (2 + π) (d) π (2 – π)

33. The area bounded by y = x e y = 0 is (a) 4 (c) 1

| x |

a

and the lines | x | = 1,

0

0

34. If ∫ f (2a − x)dx = µ and ∫ f ( x) dx = λ, then

∫

2a

0

f ( x)dx

equals to: (a) 2λ – µ (c) µ – λ 35. The value of the integral

(b) λ + µ (d) λ – 2µ π/ 4

sin x ⋅ cos x dx is 2 x + sin 4 x

∫ cos 0

(b)

π 6 3

(d) None of these

(d) None of these

sin x + cos x dx is equal to sin 2 x (b) sin − 1 1 ( 3 − 1) 2 (d) None of these

3/ 2

42.

(b) 6 (d) 2

a

(b) 9 2

(a) 2 sin − 1 1 ( 3 − 1) 2 (c) 2 sin − 1 1 ( 3 + 1) 2

0

(a) π (2 + π) (c) π (π – 2)

∫

π/6

(b) 3 (d) None of these

32. The value of the integral

π 3 π (c) 4 3

ln ( x) dx is

where n is an integer > 1, is

2

(a)

∫

π /2

39.

0

(a)

)

(a) a e–b (c) b e–b

(b)

30. The value of the integral

1 + x2

n 1 − n2 n (c) 2 n +1

(d) π log 2.

(a)

(x +

dx

e4 e

(a)

∫π 0

∫ 0

π log 2 2

(b) –

(c) – π log 2

dx = a, the value of

(a) e4 – e (b) e4 – a (c) 2e4 – a (d) 2e4 – e – a

(d) None of these

28. The value of the integral

∫e

x2

1

(b) 0

(a)

(a)

36. Given

2

∫ | x sin πx |

dx is equal to  

−1

(a)

1 1 1 +  π π

(b)

1 1 1 −  π π

(c)

1 1   − 1 ππ 

(d) None of these

43. For x ∈ R and a continuous function f, 1 + cos 2 t

let  I1 =

∫

x f { x (2 − x)} dx

sin 2 t 1 + cos 2 t

and  I2 =

∫

f { x (2 − x)} dx . Then I1/I2 is

sin 2 t

(a) 0 (c) 2

(b) 1 (d) 3

dx is

0

2 (a) π −2 π

(c)

0

0

∫ log (1 + tan x ) dx

(b) I3 > I1 > I2

(c) I1 > I2 > I3

(d) I3 > I2 > I1

(b) k 4 (d) k 8

1 2 1 (c) 16

2

∫| x

(c) − 3 x 2 + 2 x | dx is

3

(b) n – 1

1 n −1

(b) 4 log 4 3

56.

n +1

∫

f ( x) dx = n2. The value

∫

10π

0

(b) 8 (d) 18

(b) 14 (d) None of these

1 − x   −1 50. If I = ∫ cos 2 cot  dx , then 1 + x  0  1

51. If A =

∫ 1

A A B of e e 1

(b) 1 = –1/2 (d) None of these

2

A B2 2 A + B2

cosec θ

B −1 is −1

∫ 1

dt , then the value t (1 + t 2 )

∫ cos

p

x cos qx dx, then

(a) f (p, q) =

q f ( p – 1, q – 1) p+q

(b) f (p, q) =

p f ( p – 1, q – 1) p+q

(c) f (p, q) =

−p f ( p – 1, q – 1) p+q

(d) f (p, q) =

−q f ( p – 1, q – 1) p+q

−2

t dt and B = 1 + t2

π/ 2

0

n

sin θ

dx is

|sin x | dx is

57. If f (p, q) =

f ( x) dx is

(a) I > 1/2 (c) 0 < 1 < 1/2

2

(b)

(a) 20 (c) 10

(d) None of these

49. Suppose for every integer n,

1 + 2 cos x

∫ ( 2 + cos x )

1 2 −1 (d) 4

1 4 −1 (c) 2

2

(a) 16 (c) 19

(d) None of these

(a)

 x + 1  x − 1   +  − 2 dx is  x − 1  x + 1

∫

θ dθ, where n is a positive integer, then

0

48. The value of the integral

4

n

55. The value of the integral

(b) 1 (d) 0

(a) 2 log 4 3 4 (c) log 3

∫ tan

π/ 2

−1

∫

π/ 4

(d) None of these

(a) 1

2−x

− 1/ 2

1 n −1

n (In – 1 + In + 1) =

1 4

(c)

2

x dx, then Un + Un – 2 = (b) n – 1

54. If In =

(d) None of these

(a) – 1 (c) 2

n

(a) 1

∫ log 2 + x dx is equal to

1/ 2

∫ tan

0

(b)

(a)

1

π/ 4

0

0

of

(a) I1 > I3 > I2

53. If Un =

is

46. The value of the integral

0

0

π/4

k 4 k (c) – 8

∫ sin ( cos x ) dx

=

∫ cos x dx , then

and  I3 =

π

(a) –

π/2

2

π/2

−1 π

(d)

∫ cos ( sin x ) dx ; I 0

45. If ∫ log sin x dx = k, then the value of

47.

π/2

52. If I1 =

1 (b) π

(b) cosec θ (d) 1

237

∫ | x cos πx | dx is

58.

π

∫ x sin

6

x cos 4 x dx is equal to

0

(a)

3π2 512

(b)

3π2 256

(c)

3π2 1024

(d) None of these

Definite Integral and Area

(a) sin θ (c) 0

1

44. The value of the integral

238

59. In =

Objective Mathematics

(a)

∫

π/4

0

tan n x dx , then nlim n [In + In + 2] equals →∞

1 2

60.

π/3

∫ cos

4

68.

(d) zero 3φ sin2 6φ dφ is equal to

π

2 x (1 + sin x )

−π

1 + cos 2 x

∫

π 32 5π (c) 96

61.

(b)

7π 96 69.

9/ 2

− 1/ 2

dx is equal to

(a) 63π  a5 4

(b) 63π  a5 2 63 π a5 (d) 8

(c) 63π a5 2

∫x

3

1

∫x

7π (b) 4 7π (d) 16

5

∞

(1 + x )

6 7

0

72.

73.

dx is equal to 8 (b) 45

74.

(a) 0 (c) – 1

67.

(d) None of these

π [ x] dx is equal to 2

2

∫ sin

2π

∫

(b) – 1 (d) None of these

1 − cos 2 x dx is equal to 2 (b) – 2 (d) – 4

∫ [2 x]

dx is equal to

(a)

1 2

(b)

−1 2

(c)

3 2

(d)

−3 2

π

∫ | cos θ − sin θ | (a)

(b) 1 (d) None of these

dθ is equal to

2

(c) 3 3 2

[ x 2 ] dx is

75.

∫ [x ] 2

dx is equal to

−2

(a) 2 – (c)

π2 2 3

0

−1

2

(c)

(d) None of these

is equal to

0

(b)

−1

∫ [ x] dx, where [.] denotes the greatest integer function,

∫

π2 6 3

(a) 2 (c) 4

(d) None of these

π2 4 3

(a)

0

1

66.

x+π/4 dx is equal to 2 − cos 2 x − π/ 4

∫

(a) 1 (c) 0

π+2 (b) 8

4 (a) 45 2 (c) 45 65.

(b) 2 (d) – 2

2

x2

∫

π 2

x  dx is equal to

0

1+ x dx is equal to 1 − x2

3π + 8 (a) 24 3π + 8 (c) 16 64.

71.

2

0

(d)

π/ 4

70.

0

63.

∫ 

2 x − x 2 dx is equal to

7π (a) 2 7π (c) 8

(b) π2

(a) 1 (c) – 1

0

62.

3

0

∫ x ( 2a − x )

dx is

(c) zero

(d) None of these

2a

(d) None of these

π2 4

(a)

0

(a)

(b) 2 − 2

(c) 2 + 2

(b) 1

(c) ∞

2 −2

(a)

2

2 – 1

1.5

(b) 2 + (d) –

2 2 –

3 +5

2 ∫ [ x ] dx, where [.] denotes the greatest in integer func0

tion, is equal to

(a) 10 − 2 3 − 2 2 (b) 10 + 2 3 − 2 2 (c) 10 − 2 3 + 2 2 (d) None of these

(b) 2 2 (d) None of these

dx is equal to

x

∫ f (t ) dt

85. If f (x) is an odd function, then

0

(a)

2

(d) None of these

1

(a) odd (b) even (c) neither even nor odd (d) None of these 86. If f (x) is an even function, then

77. ∫ [ x + 1] dx is equal to −2

(a) 0 (c) – 1 78.

100 π

∫

(b) 1 (d) None of these

∫ [ x − 1]

(d) 400 2

| cos x | dx is equal to (b) 20 (d) None of these

n2

0

(a)

n (n + 1) (4n + 1) 6

(b)

(c)

n (n − 1) (4n − 1) 6

(d) None of these

∫ |sin x |

n (n − 1) (4n + 1) 6

(b) 20 (d) None of these

] dt, where n is any positive integer, is equal to

( ) (b) n n − (1 + 2 + 3 + ⋅ ⋅ ⋅ + n ) (c) − n n + (1 + 2 + 3 + ⋅ ⋅ ⋅ + n )

(a) n n + 1 + 2 + 3 + ⋅ ⋅ ⋅ + n

(d) None of these 84.

∫ ( x − [ x ])

−1

dx is equal to

t

∫ n

2

sin z 2 z

dz, then dy is dx

equal to tan t 2t 2 tan t (c) 2 t

(a)

(b)

2t 2 tan t

(d) None of these

∫

sin x

π/2

sin x + cos x

0

π 4

(b)

dx π 2

(d) 1

x2

∫ sin

t dt is equal to

0

1 3 −1 (c) 3

2 3 −2 (d) 3 (b)

91. If f (x) be a periodic function of period a, then

a

(a)

∫

a

(b) n  ∫ f ( x)  dx

f ( x)  dx

0

0

a

(c) (n – 1) ∫ f ( x)  dx

(d) None of these

0

92. lim

h→0

1 h

x+h

∫ x

dz z + z2 + 1

(a) 0 (c)

is equal to

(b) 1 x +1 2



na

∫ f ( x) a

dx is equal to

0

(a) 50 (b) 100 (c) 200 (d) None of these

(d) None of these

z dz and y =

(a)

0

100

∫ sin

1 90. lim 3 x→0 x

dx is equal to

(a) 18 (c) 40 2

0

(c) zero

π

n

sin t

`88. If x =

0

10 π

(b) n ∫ f ( x) dx

a

89. Evaluate

∫  x  dx is equal to

∫ [t

dx

a

∫ f ( x)  dx

(a)

− 20 π

83.

∫ f ( x)

0

(b) 3 (d) 2

(a) 40 (c) 60

a

(a)

(c) (n – 1) ∫ f ( x)   dx

dx is equal to

(a) – 3 (c) –2

82.

na

is equal to

−1

81.

(a) odd (b) even (c) neither even nor odd (d) None of these

(b) 50 2

2

∫

dt is

0

(c) 200 2

20 π

∫ f (t )

87. If f (x) is a periodic function of period a, then

1 − cos 2 x dx is equal to

(a) 100 2

80.

x

0

0

79.

is

a

(b) 2 2

(c) 3 2

239

π

∫ |sin x + cos x |

1 x + x2 + 1



(d) None of these

Definite Integral and Area

76.

240

x3

dt 93. If f (x) = ∫ , then f ′′ (x) is equal to 1 + t4 1

Objective Mathematics

(a)

6 x (1 − 5 x ) (1 + x12 ) 2

(b)

(c)

− 6 x (1 − 5 x12 ) (1 + x12 ) 2

(d) None of these

12

−1 2 ∫ (tan t ) dt 0

x4

∫ sin

2 π x = 0 and x = a is a + a sin a + π cos a, then f   = 2 2 2 2 1 (a) 1 (b) 2 (c) 1 (d) None of these 3

6 x (1 + 5 x ) (1 + x12 ) 2 12

x2

94. lim x→0

101. Let f (x) be a continuous function such that the area bounded by the curve y = f (x), x-axis and the lines

is equal to

1 102. If 2 f (x) + 3f   = 1 − 2 x ≠ 0, then x  x equal to

t dt

0

(a) 1

−1 1 (d) 2 2 95. The intervals of increase of the function f (x) defined by x

∫ (t

2

−1

(b)

(c)

2 1 log 2 + 5 2

(d) None of these

103. If an =

π/ 2

∫ (|sin x | + |cos x |)

0

(a) G.P. (c) H.P. 104. If an =

(b) A.P. (d) None of these

π/ 4

∫ cot

(b) 40 (d) None of these

∫ ( sin

− 100 π

4

dx. then

x + cos 4 x ) dx is equal to

(a) 100 π (c) 200 π

(b) 150 π (d) None of these

99. If f (x) and φ (x) are continuous functions on the interval [0, 4] satisfying f (x) = f (4 – x), φ (x) + φ (4 – x) = 3 and 4

∫ f ( x)

dx = 2, then

0

4

∫ f ( x)

φ (x) dx =

0

(a) 3 (c) 2

(b) 6 (d) None of these

100. If f (y) = e , g(y) = y; y > 0 and y

F(t) =

∫

t

0

f (t − y ) g ( y ) dy , then

(a) F(t) = 1 – e–t (1 + t) (c) F(t) = tet

(b) F(t) = et – (1 + t) (d) F(t) = te–t

(b) A.P. (d) None of these log ( sin x )

3

3 log ( sin x 3 )

∫1 x x = φ (k) – φ (1) then the possible value of k is

105. If φ ′ (x) =

, x ≠ n π, n ∈ I and

(a) 27 (c) 9

−2

(a) f (x) is continuous in [–1, 1] (b) f (x) is differentiable in [– 1, 1] (c) f ′ (x) is continuous in [– 1, 1] (d) f ′ (x) is differentiable in [– 1, 1].

x dx, then a2 + a4, a3 + a5, a4 + a6 are

(a) G.P. (c) H.P.

x

∫ | x + 1|

n

0

in

dx is equal to

(a) 20 (c) 10

98.

−2 1 log 2 − 5 2

cos 2 nx dx, then a2 – a1, a3 – a2, a4 – a3 are sin x

∫

0

100

dx is

1

in

+ 2t ) ( t − 1) dt are

10π

97. Let f (x) =

∫ f ( x)

2

(a) (– ∞, – 2) ∪ (2, – ∞) (b) (– ∞, – 2) ∪ (– 1, 0) ∪ (1, – ∞) (c) (– 1, 0) ∪ (1, ∞) (d) None of these 96.

−2 1 log 2 + 5 2

(b) – 1

(c)

f (x) =

(a)

2

106.

16

∫ log x

 dx

(b) 18 (d) None of these dz, where z = x4, is equal to

1

(a) 16 log 2 + 15 4

(b) 16 log 2 − 15 4

15 (c) − 16 log 2 +  4 

(d) None of these

107. Let f (x) be a polynomial of degree 2 satisfying f (0) = 1, 2

f ′ (0) = – 2 and f ′′ (0) = 6, then to (a) 6 (c) 9 108. Let In =

∫ f ( x)

dx is equal

−1

(b) 0 (d) None of these π/ 2

∫ sin

n

x dx, n is a positive integer. Then

0

(a) In : In – 2 = n : (n – 1) (b) In > In – 2

(c) n (In – 2 – In) = In – 2 (d) None of these

∫ log

(a) 1 (c) 0

3e x cos (3e ) dx is x

117. The value of the integral

3π / 2

∫ [sin x] dx, where [.] denotes

π/ 2

π 6

the greatest integer function, is π −π (a) (b) 2 2

(b) – 1 (d) None of these

(c) 0 (d) π 110. If (– 1, 2) and (2, 4) are two points on the curve y = f (x) and if g (x) is the gradient of the curve at the 14 + 24 + 34 + ... + n 4 13 + 23 + 33 + ... + n3 118. lim is − lim point 5 n→∞ n → ∞ n n5 2

∫ g ( x) dx

(x, y), then

1 30 1 (c) 4

is

(a)

−1

(a) 2 (c) 0

(b) – 2 (d) 1

111. If f (a + b – x) = f (x), then

112.

(a)

a+b 2

∫

b

(c)

b−a 2

∫

b

5

∫e

f [φ ( x )]

a

a

∫

b

a

x f ( x) dx is equal to

a+b 2

∫

b

f ( x) dx

a+b 2

∫

b

(d)

a

a

f ( x) dx f (a + b − x) dx

x→0

∫

0

l

(a) 3 (c) 1

(b) 2 (d) 1

(b)

a log 2 π

(d)

2a log 2 π

e6

 log x  ∫1  3  dx, where [.] denotes the greatest integer function, is

115. The value of the integral

(a) 0 (c) e6 + e3

(b) e6 – e3 (d) e3 – e6

116. The value of the integral I = (a)

1 n +1

1 1 (c) − n +1 n + 2

(b)

∫

1

0

dx, then

1− l

2π

dx

∫ 1+ e

sin x

(b) 2 I1 = I2 (d) None of these

is equal to

(a) π π (c) 2

where a, b, c are constants, then c =

−a log 2 π

∫ sin ( x (1 − x))

0

a  b tan x + c  dx = 0, 114. If ∫  |tan x | + 1 + sec x  − π/3  3

(c)

dx and

1− l

(a) I1 = 2 I2 (c) I1 = I2

is

π/3

(a) a log 2

∫ x sin ( x (1 − x))

l

I2 =

121.

2

x sin x

dx is

(b) – 8 (d) – 9

120. If I1 =

(b) 0 (d) None of these

113. The value of lim

∫ max {(1 − x), (1 + x), 2}

(a) 8 (c) 9

to

sec t dt

1 5

2

3

x2

(d)

119. The value of

⋅ f ' [φ ( x)] ⋅ φ ′ (x) dx, where φ (3) = φ (5), is equal

(a) 1 (c) 3

(b) zero

−2

f (b − x) dx (b)

x (1 − x) n dx is

1 n+2

1 1 (d) + n +1 n + 2

241

109. The value of the integral

π 3

122.

π/6

∫ sec

2

(b) 2π (d) None of these x d ( x − [ x]) is equal to

0

(a)

3

1 3

(b)

(c) 1

(d) None of these

123. Let f (x) = minimum {x + | x |, x – [x]}, where [.] de1

notes the greatest integer function. Then

∫ f ( x)

dx is

−1

equal to 1 (a) 2

(b)

(c) 1

−1 2

(d) – 1

124. If φ and f are two continuous functions, then the value π/ 4

of the integral

∫

{ f ( x) + f (− x)} . {g (x) – g (– x)} dx

− π/ 4

is π 4 −π (c) 4

(a)

(b) 0 (d) None of these

Definite Integral and Area

log

242

125. If f (a – x) = f (x) and

a/2

∫

19

f ( x) dx = p, then

133.

a

Objective Mathematics

∫ f ( x)

dx is equal to

(a) 2p (c) p

(b) 0 (d) None of these

(a) b – a (c) b + a

3 sin x3 dx = F(k) – F(1), e x then one of the possible values of k, is 4

log 3

∫

  f (0) + 4  1 (b)  f ( 0 ) + 2 6 1 2

(a)

(

log x + 1 + x 2

log 1/ 3

)

dx is

(b) 2 log 3 (d) None of these

a

128. If I =

∫ (α sin

5

x + β tan 3 x + γ cos x) dx, where α,β,γ

−a

b

∫ x f ( x) dx equals

b

∫ f ( x)

dx

a

(c) 0 2π

∫

1 + sin

0

(a) 0 (c) 8 131.

∫e

− ax

b

∫ f ( x) a

(d) None of these

130. The value of

∞

a−b (b) 2

x dx is 2

dx

cos bx dx equals

b (a) 2 a + b2 a (c) 2 a + b2

b (b) 2 a − b2

∫ |(1 − x)|

137. If

x dx is 132. The value of ∫ 2 −4 x 3

15 (b) 2 + log e   7 (c) 2 + 4 loge 3 – 4 loge 7 + 4 loge5 (d) 2 – tan– 1 (15/7)

dx = p log (qe – 1) – r, then x −1

1

(b) p = 1, q = 2, r = 1 (d) None of these

138. The smallest interval [a, b] such that 1 dx ∫0 1 + x 4 ∈ [a, b] is given by  1  (a)  , 1  2 

(b) [0, 1]

1 (c)  , 1 2 

3 (d)  , 1 4 

139. If f (x + y) = f (x). f (y) for all x, y where f ′ (0) = k ≠ 0, then f (x) can be expressed as (a) aekx (c) kx 140. If

2

15 (a) 2 − log e   7

(b) 0 (d) 4

∫ 2e 0

(d) None of these 5

dx equals

(a) p = 1, q = 1, r = –1 (c) p = 1, q = 2, r = –1

(b) 2 (d) 4

0

1 6

(a) – 2 (c) 2

a

a+b (a) 2

 1 f   + f (1)  2   1   f   + f (1)  2 

−1

(b) α, β, γ, a (d) None of these

129. If f (a + b – x) = f (x), then

dx equals

1

136.

are constants, then the value of I depends on (a) γ, a (c) α, β, a

1

∫ f ( x)

  1  f ( 0 ) + 4 f  2  + f (1)      (d) None of these (c)

(a) log 3 (c) 0

| x| dx, a < b is a x (b) a – b (d) | b | – | a |

0

(b) 16 (d) 64

127. The value of the integral

b

∫

135. If f (x) = a + bx + cx2, then

1

(a) 15 (c) 63

(b) 10– 11 (d) 10– 9

134. The value of

 esin x  126. Let d F(x) =  , x > 0 dx  x 

∫

sin x dx is less then 1 + x8

(a) 10– 10 (c) 10– 7

0

If

∫

10

0

1

∫e

x2

(b) a cos kx + b sin kx (d) None of these

( x − α) dx = 0, then

0

(a) 1 < α < 2 (c) 0 < α < 1 141.

(b) α < 0 (d) α = 0

tan − 1 x dx equals x 0

1

∫

(a)

π/ 2

x

∫ sin x

dx

0

π/ 2

(c)

∫ 0

sin x dx x

(b)

1 2

π/ 2

x

∫ sin x

dx

0

(d) None of these

0

1

(a)

(b)

143. The value of

π/ 2

∫

151.

− x2

0

0

(a) – 1 (c) 1 + e–1

(b) 2 (d) None of these

144. If f (x) = ae + be + cx satisfies the conditions x

2x

log 4

∫ [ f ( x) − cx]

f (0) = – 1,  f ′ (log 2) = 31 and

0

dx = 39 , 2

152.

dx equals π π   sin  x −  ⋅ sin  x −  3 6  

(a) 4 log

3

(b) – 4 log

(c) 2 log

3

(d) None of these

∫ ( cox px − sin qx )

2

(a) – π (c) π 1

2

× (ax2 + bx + c) dx =

∫ (1 + cos

∫ (1 + cos 0

8

8

x)

3

dx, where p and q are integers,

x) (ax2 + bx + c) dx,

1

0

(a) 0 (c) 1/π

0

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

146. The value of

∫

cos x log

− 1/ 2

(a) 0 −1 (c) 2

k

(b) – 1/π (d) 2/π

π

∫ x f ( sin

154. If

0 π/ 2

∫ f ( sin

3

3

x + cos 2 x ) dx =

x + cos 2 x ) dx, then k =

0

1+ x dx is 1− x

(b)

(b) 0 (d) 2π

153. ∫ |sin 2π x | dx is equal to

then the quadractic equation ax2 + bx + c = 0 has

(a) π 2 (c) 2 π

1 2

(d) None of these

(b) π (d) None of these

155. The value of

π/2

∫ |sin x − cos x |

(a) π (c) 25

(b) 2 ( 2 − 1)

(c) 2 2 (d) 2 ( 2 + 1) x 156. If f (x) = A. 2 + B, where f ′ (1) = 2 and

(b) 0 (d) None of these

3

∫ f ( x)

π

148. The value of

∫ (1 − x ) 2

(c) 2π – π3

sin x cos2x dx is (b) π – (d)

x < 1  x, 149. If f (x) =  , then x − 1 , x ≥1   (a) 1 5 3

dx = 7, then

0

(a) A =

−π

(a) 0

dx is

0

(a) 0 sin10 x (6x9 – 25x7 + 4x3 – 2x) dx equals

− π/ 2

(c)

10 8 6 4 2 π ⋅ ⋅ ⋅ ⋅ ⋅ 11 9 7 5 3 2

−π

(b) b = – 6 (d) a = 3

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

∫

(b)

is equal to

(a) a = 5 (c) c = 2

π/ 2

x dx is

(d) 0

π

then

147.

10 8 6 4 2 ⋅ ⋅ ⋅ ⋅ 11 9 7 5 3

(c) 1

∫ (1 + e ) dx is 1

11

−1

1 3 1 (d) 8

1 2 1 (c) 4

(a)

∫ sin

150. The value of the integral

π3 3

(b) B =

7 – 2π3 2 2

∫ x f ( x) 2

dx equals 157.

1 log 2 7 3 ( log )

2

(log 2) 2 − 1

(c) A =

7 (log 2) 2 − 1 3 (log 2) 2 

(d) B =

1 log 2

0

4 (b) 3 5 (d) 2

243

∫

3 + 2 3  cos x dx = k log   , then k is 3   3 + 4 sin x

4

∫ log [ x]

dx equals

1

(a) log 6 (c) log 2

(b) log 3 (d) None of these

Definite Integral and Area

142. If

π / /3

244

x

∫ cos t

158. If f (x) =

2

dt, x > 0, then f ′ (x) is equal to

(a)

π 1 + log 2 4 2

(b)

π 1 − log 2 4 2

1/ x

Objective Mathematics

π 1  (c) –  + log 2  4 2 

1 1 1 sin x (a) 2 cos 2 + x x 2 x 1 1 1 cos x (b) 2 cos 2 − x x 2 2 1 1 1 cos x (c) 2 cos 2 + x x 2 x (d) None of these 159. The area of the region bounded by the curves y = | x – 1 | and y = 3 – | x | is (a) 2 sq. units (b) 3 sq. units (c) 4 sq. units (d) 6 sq. units

 1  + 166. nlim →∞  2n − 12 equal to π (a) 4 π (c) 6 167. The value of

(d) None of these 1

4n − 2

2

1 + ⋅ ⋅ ⋅ +  is n 6n − 3  1

+

2

π 2 π (d) 3 (b)

 n 1 n n lim  2 2 + 2 + 2 + ⋅ ⋅ ⋅ +  is 2 2 + 1 + 2 + 3 2 n n n n 

n→∞

160. Let f (x) be a function satisfying f ′(x) = f (x) with f (0) = 1 and g(x) be a function that satisfies f (x) + g(x) = x2. Then the value of the integral (a) e – e − 5 2 2 2

∫

1

0

f ( x) g ( x) dx , is

(b) e + e − 3 2 2 2

2 2 (c) e – e − 3 (d) e + e + 5 2 2 2 2  1 1 1  161. nlim + + ⋅⋅⋅   is equal to →∞ n + 1 n + 2 n + n 

(a) 3 log 2 (c) 2 log 2

(b) log 2 (d) None of these

1 n n 1 + + ⋅ ⋅ ⋅+  is equal to 162. lim  + 3 3 n→∞ n ( n + ) ( n + ) n 1 2 8  3 1 (b) (a) 8 4 1 (c) (d) None of these 8 163. The value of the nlim →∞  1  1 1 n2   is + + + ⋅⋅⋅ +  n 2 n2 − 1 n 2 − 22 n 2 − (n − 1) 2  2

π 4 (c) π

π π (b) 4 3 π (c) (d) None of these 2 164. The area bounded by the curve y = 2x – x2 and the straight line y = – x is given by (a) 9 (b) 43 2 6 35 (c) (d) None of these 6  n +1 n+2 n+3 1 + 2 + + ⋅ ⋅ ⋅ +  is equal to 165.  nlim →∞  2 2 n + 22 n 2 + 32 n n +1

(b)

2 r sec2 r is equal to ⋅ 2 n2 n

n

∑

168. lim

n→∞

r =1

(a) tan 1 (c)

(b)

1 tan 1 2 1 n

169. nlim →∞

n

∑ sin

1 tan 1 3

(d) None of these 2k

r =1

rπ is equal to 2n 2k ! 2 (k !)

(a)

2k ! 2 (k !) 2

(b)

(c)

2k ! 2k (k !) 2

(d) None of these

2

(a)

π 2 (d) None of these

(a)

170. lim

n →∞

2k

k

(n !)1/ n is equal to n

(a) e

(b) 1 e

(c) e – 1

(d) None of these

1 2   n   171. nlim  1 + n   1 + n  ⋅ ⋅ ⋅  1 + n   →∞     

1/ n

(b) e 2 (d) 4 e

(a) 2 e e (c) 4

 12   22   n2  172. nlim  1 + 2   1 + 2  ⋅ ⋅ ⋅  1 + 2   →∞ n  n   n   (a) 2e (c) 2e

π+4 2 π−4 2

is equal to



(b) e



(d) e

π+4 2 π−4 2

1/ n

is equal to

(a)

1 p +1

(b)

1 (c) p+2

1 p

1

(d) None of these 181.

4n 2 − 1

+

1 4n 2 − 4

+ ⋅⋅⋅ +

 , 3n 2 + 2n − 1  1

Sn is equal to then nlim →∞ π 4 π (c) 3

π/2

∫ 0

(a)

1 2

dx is

−1 2

176. If f and g are continuous functions on [0, π] satisfying f (x) + f (π – x) = g (x) + g (π – x) = 1, then π

dx is equal to

0

(a) π π (c) 2

(b) 2π 3π (d) 2

2

∫

f ( x) dx is equal

−2

n →∞

179.

(b) 5 +

9 log 2 4

(d) None of these

 12 22 1 + 3 +⋅ ⋅ ⋅ +  is equal to  3 3 3 2n  1 + n 2 + n

(a) 1 log 2 3

(b) 1 log 2 2

(c) log 2

(d) None of these

π/ 2

 ax + bx + c  log  2 ⋅ ( a + b ) |sin x | dx is equal to  ax − bx + c  −π/ 2

∫

(c)

π log (a + b) 2

dx is

a +b (b) π log    2  (d)

1

∫e

x2

dx is

(a) greater than e (b) less than e (c) greater than 1 (d) less than 1 184. The value of α which satisfies

α

∫ cos x dx

= cos 2α,

0

α ∈ [0, 2π]  is π (a) 6 π (c) 2

(b)

π 3

(d) None of these 100

∫2

x −[ x]

dx, where [.] denotes

the greatest integer function, is (a) log 2 (c) 100 log 2

π a +b log   2  2 

(b) 50 log 2 (d) None of these

186. The value of the integral 3



−2



∫ cot

−1

  x − 1 −1  x + 1    + cot    is  x +1   x − 1 

(a)

5π 2

(b)

(c)

π 2

(d) None of these

187. Let I1 =

2 − tan 2 z

∫

3π 2

x f ( x (3 −x)) dx and

sec2 z

2

(a) π log (a + b)

∫ |log 2 x |

0

to

178. lim

e/2

(b) 1 – e– 1 (d) None of these

185. The value of the integral

177. If f (x) = | 2x – 1 | + | x – 1 |, then

9 (a) 5 − log 2 4 9   (c) −  5 + log 2  4  

(b) 2 log (a + 1) (d) 2 log a

0

(d) 0

∫ [ f ( x) + g ( x)]

182. The value of the integral

183. The value of the integral

sin 8 x log (cot x) cos 2 x

(b)

(c) 1

(a) log (a + 1) (c) 3 log a

(a) 1 + e– 1 (c) e– 1 – 1

(b)

175. The value of the integral

xa − 1 ∫0 log x dx, where a > 0, is equal to 1

1/ 2 e

π 6 π (d) 2

(a)

(b) B = x – [x] (d) B = x + [x]

(a) A = [x] – 1 (c) A = [x] + 1

174. If Sn = 1  +  2n

x

1 180. If x > 0 and ∫ [ x] dx = [x]  A + B  , where [.] denotes 2  0 the greatest integer function, then

2 − tan 2 z

I2 =

∫

f ( x (3 −x)) dx,

sec2 z

where f is a continuous function and z is any real number, then I1/I2 = (a) 3/2 (c) 1

(b) 1/2 (d) None of these

245

n

n→∞

1p + 2 p +⋅ ⋅ ⋅ + n p  , p > – 1, is equal to

p +1

Definite Integral and Area

1

173. lim

246

π/ 2

∫ log sin x

188. If

4

dx = a, then the value of

196. The value of the integral

0

3

Objective Mathematics

∞

log (1 + x 2 ) ∫0 1 + x 2

dx in terms of a is

(a) 2a (c) a/2

(b) – 2a (d) a log 2

b + 2T

197. If I1 =

f ( x) dx is equal to

∫

(a)

b

f ( x) dx

∫

(b)

a+T

∫

f ( x) dx

b

f ( x) dx

∫ f ( x)

(d)

a

dx

198. The value of the integral

(c) 3 [x] 199. The value of the integral

48

(b)

(c)

73

(d) None of these

(a) ae (c) a (1 – e)

66

192. If the variables x and y are cannected by the relation y 2 dz x= ∫ , then d y is proportional to 3 1 1 + 6z dx 2 (a) y (c) y3

2 x − 9 x + 12 x + 4 3

1

2

(a) 1 < I < 1 2 3 1 (c) < I < 1 4 194.

∫{ x} 3

tan θ

201.

, then

α

(d) None of these

(c) π

0

(d) None of these π 2

∫ (|sin x | + |cos x |) (b) 2a π (d) independent of a

(b) sin −1

2+ x dx is equal to 2−x

(a) π + 1 2 (c) π + 1

(b) 7 2

a

(a) aπ aπ (c) 2

∫ 0

is equal to

195. The value of the integral

2

α β α (d) cos −1 β

β−α

(a)

202.

(b) 2 tan θ (d) None of these

dx , β > α > 0, is equal to ( x − α) (β − x)

∫

(b) 1 < I < 1 4 3

a+

dx, a ∈ Z+, is

(b) a (e + 1) (d) a (e – 1)

(a) tan2θ (c) 0

dx, where {x} denotes the fractional part of x,

(a) 5 2 3 (c) 2

[ x]

is

β

dx

∫

ex

1 200. If f (x) is function satisfying f     + x2 f (x) = 0 for cot θ  x all x (x ≠ 0), then the value of the integral ∫ f ( x) dx

(b) y2 (d) None of these 2

193. If I =

a

∫e 0

1

(a)

dx is

(b) 1 [x] 2 (d) 2 [x]

(a) [x]

(b) 0, 4) (d) None of these

(2 x + 3) (3 x 2 + 4) dx cannot exceed

∫ ( x − [ x ]) 0

191. The value of the interval

∫

where [x] and {x}

0

2[ x]

k

 5 3/ 2 1/ 2   x + 2 x − 3 x  dx < 0, lie in the interval ∫ k k 12 

2

∫{x} dx,

(d) I2 = a I1

a

(a) (0, 1) (c) (1, 4)

a

and I2 =

(a) I2 = (a – 1) I1 (b) I1 = (a – 1) I2 (c) I1 = a I2

190. All values of k for which the integral 1

 x 2  dx, − 14 x + 49  +  x 2 

denote, respectively, the integral and fractional parts of x and a is a positive integer, then

a+T

b+T

(c)

a

∫ [ x] dx 0

a + 2T b+T



2

where [.] denotes the greatest integer function, is 3 (a) 1 (b) 2 1 (c) (d) None of these 2

189. If f (x) is a periodic function with period T, then

∫

∫ x

e

203. If dx is

 1

(b) π + 3 2 (d) None of these 1

∫  log x − ( log x ) 2



(a) a = e, b = – 2 (b) a = e, b = 2 (c) a = – e, b = 2 (d) None of these

2

  dx = a + b , then  log 2

(b)

b a dx

1

∫ (1 + x )

205. If

0

1 − x2

(a) a = π, b = (c) a =

= a , then b

π , b = – 2 2 (d) a = b 2 2

206. The value of

∫ ( ax

−2

3

(b) value of c (d) value of a and b

207. If d g (x) = f (x), where f (x) is continuous in [a, b], dx then b

∫

dx = a + b log 2 , then 3

(a) a =

3 3 ,b= 4 2

(b) a =

3 3 ,b= 2 4

(c) a =

−3 3 ,b= 2 2

(d) a =

3 −3 ,b= 2 2

dx is equal to tan x − cot x

∫ 0

π 2

(a)

+ bx + c ) dx depends on the

(a) value of b (c) value of a

∫ x log 1 + 2 

π/ 2

213.

(b) a = π, b = 2 2

2

x



1

0

a b 1 (d) ab

(a) ab (c)

212. If

f ( x) ⋅  g (x) dx equals

(b)

(c) 0

4

214. The value of the integral

∫ |log

(a)

 e9  1 log 4  15  4 4 

(b)

(c)

 e9  1 log 4  15  2 4 

(d) None of these

215. The value of the integral

 415  1 log 4  9  4 e 

1

∫ 0

1 f   dx  x

1

(b)

1

209. If f (x) =

∫ x f ( x) 1

(d)

∫ f ( x)

1 2 log   n 3

2 (b) n log   3

(c)

1 3 log   n 2

3 (d) n log   2

216. The value of the integral 4

(a)

4 3 log 3

dx

+

1 3 12 log 4



42 x + 1 − 32 x − 1 dx is 12 x 0

1

∫

(b)

0

x

sin t f ′ (x) = ∫1 t dt, then xlim →∞

(a) 1 (c) – 1

(a)

dx

0

1 1 (c) ∫ f   dx  x 0 x

dx is n + 1)

∫ x (x 1

1  1 2  n  f   + f   + ⋅ ⋅ ⋅ + f    is equal to n   n  n    n 

(b) 0 (d) ∞

x | dx is

4

1/ 4

31/ n

208. nlim →∞ (a)

π 4

(d) None of these

a

1 (a) [ f (b)]2 − [ f (a )]2  2 (b) f (b) – f (a) 1 (c) [ g (b)]2 − [ g (a )]2  2 (d) g (b) – g (a)

247

dx ∫0 [ax + (1 − x) b]2 is equal to 1

Definite Integral and Area

204.

4

(c)

4 3 log 3

4

+

3 log

3 4



4 4 3 log 3

3

+

4 log

3 4

(d) None of these 41 π / 4

217. The value of the integral

∫

|cos x | dx is

0

1 (e1/n + e2/n + e3/n + ··· + en/n) is equal to 210. nlim →∞ n (a) e (b) 1 – e 211.

π/ 2

∫ sin θ cos θ

(b) e – 1 (d) None of these a 2 sin 2 θ + b 2 cos 2 θ dθ is equal to

0

(a) (c)

(a) 20 −

1 2

(b) 20 +

1 2

(c) 19 +

1 2

(d) 19 −

1 2

2

218. If

∫ f ( x)

dx = 2 and

−3

a 2 + ab + b 2 2 ( a + b)

(b)

a 2 + ab + b 2 ( a + b)

a 2 + ab + b 2 3 ( a + b)

(d) None of these

−3

of the integral

5

∫ [5 + f ( x ) ]

∫ f ( x)

2

dx is

5

(a) 2 (c) 4

(b) 3 (d) 5

dx = 9, then the value

248

y

219. If

∫e

− t2

0

x2

dt + ∫ sin t dt = 0, then dy dx 0 2

Objective Mathematics

(a) 2 x sin 2 x 2 e y

2

∫ cos

2n

x dx is equal to

π/ 2

∫ cos

2n

0 π/ 2

∫ cos

(c) 2k

π

(b) 2k ∫ cos 2 n x dx

x dx

0

2n

x  dx

(d) None of these

0

f (m, n) =

1k + 2k + 3k + ⋅ ⋅ ⋅ + n k  221. lim   is equal to nk +1 n→ ∞   1 (b) (a) 1 k +1 k 1 (c) (d) None of these k+2 3

x−4

∫x

n −1

(log x)m dx, then f (m, n) =

0

(a) m  f (m – 1, n) n (c) n f (m – 1, n) m 230.

∞

∫x

n

(b) − m f (m – 1, n) n (d) None of these

e– x dx (n is a + ve integer) is equal to

(a) n ! (c) (n – 2)!

dx is

−3

(a) 0 (c) – 6

1

0

∫ | x − 4|

222. The value of the integral

dx is equal to

x

2 2 (a) π (b) π 4 8 2 π (c) (d) None of these 12 229. If m and n are positive integers and

0

(a) k

x

∫1 + e

2 2 (a) π (b) π 6 8 π2 (c) (d) None of these 12 1 1 1+ x 228. ∫ log dx is equal to 1− x 0 x

2

kπ

∞

0

2

(d) − x sin 2 x 2 e y 220. The integral

is equal to

2

2 2 y (b) − 2 x sin x e

(c) x sin 2 x 2 e y

227.

(b) 6 (d) None of these

231. If In =

π/ 2

∫x

n

(b) (n – 1)! (d) None of these sin x dx, where n > 1, then

0

223. The value of the integral is π ( 2n) !

(a)

22 n ⋅ ( n!)

2



 n   2n + 1  !  2  

2



x dx (n even integer),

π (a) In + n (n – 1) In – 2 = n   2

n

π ⋅ n!

π (b) In + n (n – 1) In – 2 = n   2

n −1

 n   2n ⋅  !  2  

2



π (c) In – n (n – 1) In – 2 = n   2 (d) None of these

(d) None of these

π

3

224. The value of

n

0

(b)

π ⋅ n!

(c)

π

∫ cos

∫ |1 − x

2

232.

| dx is

225. If

(b) 14/3 (d) 1/3 π/ 2

π

∫ x f (sin x) dx 0

(a) π/4 (c) 0

I2 =

∫

f (− a)

(a) –1 (c) 2

∫

f (sin x) dx , then A is

0

233.

(b) π (d) 2π

226. If f (x) = f (a)

= A

ex , I1 = 1 + ex

4

2

dθ is equal to

0

−2

(a) 7/3 (c) 28/3

1 − cos θ

∫ sin θ (1 + cos θ )

n

∫ 0

f (a)

∫

(a) 8 2 15 32 2 (c) 15 a xn

xg {x (1 − x)} dx

and

(b) –3 (d) 1

(d) None of these dx is equal to

(a)

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ (2n − 1) (– π an) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ 2n

(b)

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ (2n − 1) (π an) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ 2n

(c)

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ (2n − 1) (π an – 1) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅ 2n

f (− a)

I g{x(1 − x)} dx , then the value of 2 is I1

ax − x 2

(b) 64 2 15

(d) None of these

(a) 2 (c) 8

(b) 4 (d) None of these

236. The area bounded by the semi-circle y = its diameter y = 0 is

4 − x 2 and

(a) 2π (b) π π (d) None of these (c) 2 237. The area bounded by y = logex, x – axis and the ordinate x = e is given by (a) 4 (c) 1

(b) 1 2 (d) None of these

238. The area bounded by the curve | x | + y = 1 and axis of x is given by (a) 2 (c) 4

(b) 1 (d) None of these

239. The area of the bounded by y = cos x, y = 0, | x | = 1 is given by (a) sin 1 (c) 4 sin 1

(b) 2 sin 1 (d) None of these

240. The area of the region bounded by x-axis and the curves defined by y = tan x, − π ≤ x ≤ π and y = cot x, π ≤ x ≤ 3π is 3 3 6 2 (a) log 2 (c) 3 log 3

(b) 2 log 2 (d) None of these

241. If the ordinate x = a divides the area bounded by x-axis, part of the curve y = 1 + 8 and the ordinates x = 2, x2 x = 4, into two equal parts, then a is equal to (a)

(b) 2 2 2 (c) 3 2 (d) None of these 242. The area bounded by y = | x – 1|, y = 0 and | x | = 2 is  

246. The area bounded by y = cos x, y = 1 + x and x – axis is (a) 1

(b) 1 2

(c) 3 (d) None of these 2 247. The area bounded by the lines | x |+ | y | = 1, is (a) 1 (b) 2 (c) 4 (d) None of these 248. The area of the portion of the circle x2 + y2 = 64 which is exterior to the parabola y2 = 12x is (b) 8 (8π + 3 ) (a) 16 (8π + 3 ) 3 3 16 (c) (d) None of these (8π − 3 ) 3 249. The area bounded by the curve y = sin– 1x and the lines x = 0 , | y | = π is 2 (a) 2 (b) 4 (c) 8 (d) 16 250. The area of one the curvilinear triangles formed by the curves y = sin x, y = cos x and x-axis is (a) 2 + 2 (c) 2 + 2 2

(b) 2 − 2 (d) None of these

251. The total area enclosed by the lines y = | x |, y = 0 and | x | = 1 is (a) 2 (c) 1

(b) 4 (d) None of these

252. The area bounded by y = tan x, y = cot x, x-axis in 0 ≤ x ≤ π is 2 (a) 3 log 2 (b) log 2 (c) 2 log 2 (d) None of these

(a) 4 (b) 5 (c) 3 (d) None of these 243. The area bounded by the y = | sin x |, x-axis and the lines | x | = π is (a) 2 (c) 4

(b) f (x) = sin (3x + 4) + 3 (x – 1) cos (3x + 4) (c) f (x) = sin (3x + 4) – 3 (x – 1) cos (3x + 4) (d) None of these

(b) 1 (d) None of these

2 2 253. The area of the region bounded by the ellipse x + y a 2 b2 = 1 and the ordinates x = ± ae where b2 = a2 (1 – e2)

249

Definite Integral and Area

234. The area of the smaller part of the circle x2 + y2 = a2, 244. The smaller area enclosed by the circle x2 + y2 = a2 and the line x + y = a is a , is given by cut off by the line x = 2 2 2 (b) a (π + 2) (a) a (π – 2) 4 4 a2  π a2  π   (b) (a) 2  + 1  − 1 a 2 2 2 2   (2 – π) (d) None of these (c) 4 π (c) a 2  − 1 (d) None of these 245. If the area bounded by the curve y = f (x), x-axis and 2  the ordinates x = 1 and x = b is (b – 1) sin (3b + 4), then 235. The area bounded by the curve y = 2cos x and the x-axis from x = 0 to x = 2π is (a) f (x) = cos (3x + 4) + 3 (x – 1) sin (3x + 4)

250

Objective Mathematics

and e < 1, is given by

the coordinate axes in the first quadrant is

( ) (b) 2 ab ( e 1 − e − sin e ) (c) ab e 1 − e − sin e ( )

(a) 1 (c) 3

2 −1 (a) 2 ab e 1 − e + sin e 2

2

263. The area bounded by the lines y = 2, x = 1, x = a and the curve y = f (x), which cuts the last two lines above the first line for all a ≥ 1, is equal to

−1

−1

(d) None of these 254. The area bounded by y = e , y = e and x = 1 is 1 1 (a) e + + 2 (b) e + − 2 e e 1 (d) None of these (c) e − + 2 e 255. The area bounded by the curve x = 2 – y – y2 and y-axis is (b) 7 (a) 9 2 2 5 (c) (d) None of these 2 256. The area of the smaller part bounded by the semi-circle y = 4 − x 2 , y = x 3 and x-axis is x

π 3 4π (c) 3

(a)

(b)

(b) 2 (d) None of these

– x

2π 3

(d) None of these

257. The area of the region bounded by the curve y = and y = | x – 1 | is

5 − x2

(a) 5π + 1 (b) 3π + 1 4 2 4 2 5 π 1 3 (d) π − 1 (c) − 4 2 4 2 258. The area lying above x-axis, bounded by the circle x2 + y2 = 2ax and the parabola y2 = ax is π 2 π 2 (b)  −  a 2 (a)  +  a 2  4 3  4 3 π 1 2  (d) None of these (c)  +  a  4 3 259. The area bounded by the curves x2 + y2 = 25, 4y = | 4 – x2 | and x = 0, above x-axis is (a) 2 + 25 sin − 1 4 (b) 2 + 25 sin − 1 4 2 5 4 5 25 1 − 1 (c) 2 + (d) None of these sin 2 5 260. The area bounded by y = x3 – 4x and x – axis is

2 3/ 2 ( 2a ) − 3a + 3 − 2 2  ⋅ Then f (x) = 3 (a) 2  2x x ≥ 1 (c) 2

x , x ≥ 1

(b)

2x , x ≥ 1

(d) None of these

264. The area of the region in xy – plane enclosed by the curve a2y2 = x2 (a2 – x2) is 4 4 (a)  a (b)  a2 3 3 2 2 (c)  a (d)  a2 3 3 265. The area bounded by the normal at (1, 2) to the parabola y2 = 4x (y ≥ 0), the curve itself and the axis of the parabola is 2 4 (a) (b) 3 3 8 10 (c) (d) 3 3 266. The area bounded by the curve xy2 = 1 and the lines x = 1, x = 2 is (a) 4 ( 2 – 1)

(b) 4 ( 2 + 1)

(d) 2 ( 2 + 1). (c) 2 ( 2 – 1) 267. The area bounded by the curve y = 2x4 – x2, x-axis and the two ordinates corresponding to the minima of the function, is 3 5 (a) (b) 120 120 1 7 (c) (d) 20 120 268. The area contained between the curve y = x-axis is (a) πa2 (c) 3πa2

a2 and x + a2 2

(b) 2πa2 (d) None of these

269. The area bounded by y = [x], x-axis and the two ordinates x = 1 and x = 1.7 is 17 (a) (b) 1 10 17 7 (d) (c) (a) 4 (b) 8 5 10 (c) 16 (d) None of these 270. The area of the loop of the curve y2 = x (1 – x)2 is 261. The area bounded by the curve y = x (3 – x)2, the 7 (a) 8 (b) x-axis and the ordinates of the maximum and minimum 15 15 points of the curve is given by 4 (c) (d) None of these (a) 1 (b) 2 15 (c) 4 (d) None of these et + e − t and y = et − e − t is a point 262. The area bounded by the curve y = sin x + cos x and 271. For any real t, x = 2 2

(a) 1 t1 2 (c) 2 t1

(b) t1 (d) None of these

272. The area bounded by the curve y2 (a2 + x2) = x2 (a2 – x2) is (a) a2 (π – 2) (c) a2 (π – 1)

(b) a2 (π + 2) (d) a2 (π + 1)

273. The area of the region bounded by x = 1 , x = 2, y = 2 logex and y = 2x is (a)

4− 2 5 3 + log 2 + log 2 2 2

(b)

4− 2 5 3 − log 2 − log 2 2 2

(c)

4− 2 5 3 − log 2 + log 2 2 2

(d) None of these 2 274. The area included between the parabola y = x (a > 0) 4a 8a 3 is and the witch of Agnessi y = 2 2 x + 4a

4 (a) a 2  2 π −   3

4 (b) a 2  2 π +   3

4 (c) a 2  π +  (d) None of these  3 275. If the ordinate x = a divides the area bounded by x-axis, part of the curve y = 1 + 8 and the ordinates x = 2, x2 x = 4 into two equal parts, then a = (a)

2

(b) 2 2

(a) 2 (c) 4

(b) 1 (d) None of these

280. The area bounded by the curves x + y = 1 and x + y = 1 is 1 1 (a) (b) 3 6 1 (c) (d) None of these 2 281. The area of the region bounded by the curves y = x2 2 is and y = 1 + x2 1 2 (a) π + (b) π + 3 3 2 (c) π − (d) None of these 3 282. The area above x – axis, bounded by the line x = 4 and the curve y = f (x), where f (x) = x2, 0 ≤ x ≤ 1 and f (x) = x , x ≥ 1, is (a) 1 (b) 2 (c) 4 (d) 5 283. The area of the region bounded by the parabolas x = – 2y2 and x = 1 – 3y2 is 4 2 (a) (b) 3 3 1 (c) (d) None of these 3 284. The area common to the curves y=

9 − x 2 , x2 + y2 = 6x in y ≥ 0 is

(a) 3 π +

9 3 4

(b) 3 π −

9 3 4

(c) 3 2 (d) None of these 9 3 9 3 (c) 3 π − (d) 3 π + 276. The area bounded by the curves x2 + y2 = 4; x2 = – 2 y 2 2 and x = y is 2 285. The area of the loop the curve ay = x2 (a – x) is 1 1 (a) 2 π + (b) π − 2 2 3 3 (b) 4 a (a) 8 a 1 15 15 (c) − π + (d) None of these 2 3 2 a (c) (d) None of these 15 277. The area of the region bounded by the curves y = ex 286. The area of the region bounded by x2 + y2 – 2x ≤ 0, logex and y = log e x is x + y ≤ 1; y ≥ 0 is ex 2 2 π 1 π 1 (a) − (b) + (a) e − 5 (b) e + 5 8 2 8 2 2e 2e 2 2 π 1 π 1 (c) − (d) + (c) e + 5 (d) e − 5 4 2 4 2 4e 4e 278. The ratio of the areas into which the circle x2 + y2 = 287. The curve y = a  x + bx passes through the point (1, 2) and the area enclosed by the curve, the x-axis 64 is divided by the parabola y2 = 12x, is and the line x = 4 is 8 square units. The values of a 4π − 3 4π + 3 and b are (b) (a) 8π + 3 8π − 3 (a) a = 3, b = 1 (b) a = 3, b = – 1 4π − 3 (c) a = – 3, b = 1 (d) a = – 3, b = – 1 (d) None of these (c) 8π − 3

251

279. The area enclosed between the curves y = loge (x + e), 1 and the x – axis is x = loge y

Definite Integral and Area

on the hyperbola x2 – y2 = 1. The area bounded by the hyperbola and the lines joining its centre to the points corresponding to t1 and – t1 is

252

288. The area of the portion of the circle x2 + y2 = 1, which lies inside the parabola y2 = 1 – x, is

Objective Mathematics

(b) π + 2 (a) π − 2 2 3 2 3 π 4 π (c) + (d) − 4 2 3 2 3 289. The area bounded by the parabolas y2 = 5x + 6 and x2 = y is (a) 19 (b) 21 5 5 23 27 (c) (d) 5 5 290. The area bounded by the parabolas y2 = 4a (x + a) and y2 = – 4a (x – a) is 16 2 8 (a) a (b) a 2 3 3 4 2 (c) a (d) None of these 3 291. The area inside the parabola 5x2 – y = 0 but outside the parabola 2x2 – y + 9 = 0 is (a) 8 3 (c) 4  3

(b) 12 3 (d) None of these

2 292. The area bounded by the parabolas y = 4x2, y = x 9 and the line y = 2 is

(a) 20 2 3

(b) 10 2 3

(c) 40 2 3

(d) None of these

293. The area bounded by the circle x2 + y2 = 8, the parabola x2 = 2y and the line y = x in y ≥ 0 is

296. The area bounded by the curves y = sin x and y = cos x between two consecutive points of their intersection is (a) 2 (c) 3 2

(b) 2 2 (d) None of these

297. The area of the region formed by x2 + y2 – 6x – 4y + 12 ≤ 0, y ≤ x and x ≤ 5/2 is (a)

π 3 +1 − 6 8

(b)

π 3 −1 + 6 8

π 3 −1 (d) None of these − 6 8 298. If An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = π/4, then, for n > 2, (c)

(a) An + An – 2 = (c) An >

1 n −1

1 2n + 2

(b) An + An – 2 < (d) An
0 be a fixed number. Suppose f is a continuous function such that for all x ∈ R, f (x + T) = f (x). If

∫ f ( x) dx T

I=

0

3 I 2 (c) 3I

(a)

316.

2π

∫e

x 2

0

3 + 3T

3

f ( 2 x ) dx , is

(b) 2I (d) 6I

 x π sin  +  dx = 2 4 (b) eπ (d) 2 2

(a) 2π (c) 0

(1 – t2) dt are

∫

then the value of

317. Suppose that f ′′ (x) is continuous for all x and 1 (b) x = – 2 f (0) = f ′ (1) = 1. If ∫ t f '' (t ) dt = 0, then the value 1 (c) x = 0 (d) x = 0 2 of f (1) is 309. The area of the region bounded by the curve y = x – x2, (a) 3 (b) 2 x-axis between x = 0 and x = 1 is 1 (c) 4 (d) None of these 5 1 2 (a) (b) 6 2 a dx 1 1 = π , then a equals 318. If ∫ (c) (d) 1 + 4x2 8 0 3 6 310. The area between the curve y = 1 – | x | and x-axis is (b) 1 (a) π 1 2 2 (a) (b) 2 3 π (c) (d) 1 1 4 (d) 1 (c) e2 2 log x 319. The value of ∫e −1 xe dx is 1 1 1 311. If ∫ f ( x) dx = 1, ∫ x f ( x) dx = a, ∫ x 2 f ( x) dx = a2, 5 3 0 0 0 (b) (a) 1 2 2 2 then ∫ ( a − x ) f ( x) dx equals (c) 3 (d) 5 1

(a) x = 1, – 1

0

π

320.

(a) 4a2 (b) 0 (c) 2a2 (d) None of the above 312. The integral

∫

1/2

−1/2

∫ (1 − x ) 2

π − 2 π3 2 π3 (c) π − 3

321. If

b

∫ (b − 4x )

(b) 0

(c) 1

(d) 2log 1 2

(a) 3 (c) 2

esin x dx is

322. If I1 =

π/ 2

∫ cos x ⋅ 0

(a) 1 (c) 0

(b) 2π – π3

(a)

1 (a) – 2

313. The value of

sin x cos2x dx is equal to

−π

  1 + x  [ x ] + log   dx equals  1 − x   

(d) 0 dx ≥ 6 – 5b, b > 1, then b equals

1

(b) 4 (d) 1 3π

∫ f ( cos x ) 2

0

(b) e – 1 (d) – 1

253

2

314. The value of

(a) I1 = 5I2 (c) I1 = 3I2

dx and I2 =

π

∫ f ( cos x ) 2

0

(b) I1 = I2 (d) None of these

dx, then

Definite Integral and Area

305. The area of the region bounded by y = | [x – 2] |, the x-axis and the lines x = – 1 and x = 2 is

254

323. Area of the region bounded by y = x-axis lying in Ist quadrant is

Objective Mathematics

(a) 18 sq. units (c) 36 sq. units

x , x = 2y + 3 &

log 5

332.

ex

0

(b) 9 sq. units (d) 27/4 sq. units

324. The equation of the common tangent to the curves y2 = 8x and xy = –1 is (b) y = 2x + 1 (d) y = x + 2

(a) 3y = 9x + 2 (c) 2y = x + 8

∫

e x −1 dx = e +3 x

(a) 4 + π (c) 4 – π 333. If u10 =

(b) 2 + π (d) 3 + 2π

π/ 2

∫x

10

sin x dx, then the value of u10 + 90u8 is

0

π (a) 9   2

π (b) 10   2

π (c)   2

π (d) 9   2

9

 n n n n  + 2 + 2 + ⋅⋅⋅ + 2 325. lim  2  2 2 2 n→∞ + + + + 1 n 2 n 3 n n n2   is equal to

9

9

(a) log 2

8

(b) 0 334. Let f(x) be a continuous function in R such that f(x) (c) 1 (d) π 3 3 4 + f(y) = f(x + y). If ∫ f ( x) dx = k , then ∫−3 f ( x) dx is 0 3 326. The area enclosed between the curves y = x and equal to y = x is, (in square units) (a)  2k (b)  0 k (c)   (d)  –2k (a) 5 (b) 5 2 3 4 (c) 5 12

335.

(d) 12 5 x3

dt 327. The derivative of F (x) = ∫ (x > 0) is log t 2 x (a)

1 3log x

(b)



1 1 − 3 log x 2 log x 3x log x

(d)

ecos x ⋅ sin x, | x | ≤ 2 328. If f (x) =  , then 2, otherwise 3

f ( x) dx is equal to

−2

(a) 0 (c) 2

(b) 1 (d) 3

329. nlim   1 + 2 2 + 3 3 +⋅ ⋅ ⋅ + n n is equal to →∞ n5 / 2 1

(a)

∫x

(b) 5 2 (d) 1

x dx

0

(c) 0 π

330. The value of

3 2 ∫ sin x cos x dx is

−π 4 (a) π 2 (c) 0

4 (b) π 4 (d) None of these x

1 − x  331. The function F (x) = ∫ log   dx is 1 + x  0 (a) an odd function (c) an even function

π/ 2

−π / 2

sin 2 x cos 2 x (sin x + cos x) dx is equal to

(a)   2 15 (c)   6 15 336. The value of

(b)   4 15 (d)   8 15

∫

π/ 2 0

log | tan x + cot x | dx is

2

(c) (log x)– 1· x (x – 1)

∫

∫

(b) a periodic function (d) None of these

(a)  π log 2 (b)  –π log 2 π (c)   log 2 (d)   − π log 2 2 2 337. The ratio of the areas between the curves y = cos x and y = cos 2x and x-axis from x = 0 to x = π is 3 (a)  1 : 2 (b)  2 : 1 (d)  1 : 4 (c)   3 :1 338. The value of the integral (a)  π (c)   π 2

∫

b

a

x dx is x + a+b− x

(b)   1 (b − a ) 2 (d)  b – a

339. The area bounded by y = log x, x-axis and ordinates x = 1, x = 2 is (a)   1 (log 2)2 2 (c)  log 4/e 340. The value of the integral (a)   3 2 (c)  1

(b)  log 2/e (d)  log 4

∫

6

3

x dx is 9− x + x

(b)  2 (d)   1 2

0

350.

| sin 3 θ | d θ is (b)   3 8

∫

1000

0

e x −[ x ] dx is

255

π

∫

1000 (b)   e − 1 e −1 (c)   4 (d)  π e −1 (c)  1000 (e – 1) (d)   3 1000 342. The area bounded by the curves y = log x, y = log | x |, 1 sin x 1 cos x 351. Let I = ∫ y = |log x | and y = | log | x || is dx and J = ∫ dx 0 0 x x (a)  4 sq unit (b)  6 sq unit Then, which one of the following is true ? (c)  10 sq unit (d)  none of these 10 2 2 (b)   I < and J < 2 (a)   I > and J > 2 343. ∫ | x( x − 1) ( x − 2) | dx is equal to 0 3 3 (a)  160.05 (b)  1600.5 2 2 (c)   I < and J > 2 (d)   I > and J < 2 (c)  16.005 (d)  none of these 3 3 344. 30 |x3 + x2 + 3x| dx is equal to 3 52. The area of the plane region bounded by the curves ∫ 2 2 x + 2y = 0 and x + 3y = 1 is equal to (a)   171 (b)   171 2 4 (b)   1 sq unit (a)   5 sq unit 3 3 170 170 (c)   (d)   2 4 3 (d)   4 sq unit (c)   sq unit 3 3 3 x +1 345. The value of ∫ 2 dx is 353. The area bounded by the curve y = 2x – x2 and the line 2 x ( x − 1) y = –x is given by (b)   log 16 − 1 (a)   log 16 + 1 (a)  1/2 (b)  9/3 9 6 9 6 (c)  9/2 (d)  none of these.

(a)  0

(c)   2log 2 − 1 6 346.

x) dx + ∫

5π / 4

π/ 4

∫

π/ 4

(sin x − cos x) dx + ∫

(cos x − sin x) dx + ∫

π/ 4

2π

5π / 4

347. The value of

(sin x − cos x) dx + ∫

π/ 4

2π

∫

1

348. The value of the integral

(b)   1 (3π + 4) 6 1 (d)   (3π − 4) 6

∫

a

0

1 π 1 − 16a 3  4 3 

(b)  

1 3 π 1 a − 16  4 3 

(d)  

x 4 dx is (a + x 2 )4 2

1  π 1 + 16a 3  4 3  1 3 π 1 a + 16  4 3 

349. The area bounded by the curves y2 = 4a2(x – 1) and lines x = 1 and y = 4a is (a)  4a2 sq unit (c)   16a 3

(b)   16a sq unit 3

2

sq unit

(d)  None of these

is ..

(cos x − sin x) dx (a)  log 2

x4 + 1 dx is x2 + 1

(a)   1 (3 − 4π) 6 1 (c)   (3 + 4π) 6

∫ log [ x] dx 2

(b)   2 2 − 2 (d)   4 2 − 2 0

(c)  

354.

(cos x − sin x) dxis equal to

(a)   2 − 2 (c)   3 2 − 2

(a)  

4

(d)   log 4 − 1 3 6 π/ 4

0

(a)  e1000 – 1

(b)  log 3 (d)  none of these.

(c)  log 5 355. If I10 =

π/ 2

∫x

10

sin x dx. Then the value of I10 + 90 I8 is

0

(a)  10(π/2)6

(b)  10(π/2)9

(c)  10(π/2)7

(d)  none of these.

356. If I = π/ 2

∫

π/ 2

∫

cos(sin x)dx, j =

0

π/ 2

∫

sin(cos x) dx, K =

0

cos x dx then

0

(a)  I > J > K (c)  K > J > I 357. If g(x) =

∫

x

0

(b)  J > K > I (d)  I > K > J

cos 4 t dt, then g(x + π) is equal to

(a)  g(x) + g(π)

(b)  g(x) – g(π)

(c)  g(x) · g(π)

(d)  

358. Let Sn =

n

∑ k =1

g ( x) g (π)

n and Tn = n + kn + k 2 2

n −1

∑n k =0

n = 1, 2, 3,…  . Then, π 3 3 π (c)   Tn < 3 3

(a)   S n
3 3 (b)   S n >

2

n for + kn + k 2

Definite Integral and Area

341. The value of

256

solutions

Objective Mathematics

π

∫e

1. (a) Let I=

cos 2 x

 Put 1 + tan 3 x = z    1  ⇒  tan 2 x sec 2 x dx = dz   3  2  −1  1 1  1 =    = –  − 1 = ⋅ 32  3 z 1 6 

⋅ cos3  (2n + 1) x dx

0

π

=

∫e

cos 2 ( π − x )

∫e

cos 2 x

0 π

=

⋅ cos3 (2n + 1) (π – x) dx

⋅ cos3 ( 2n + 1) π − ( 2n +1) x  dx

7. (c) Let I =

0

π

∫e

=–

cos 2 x

=

0

π/ 2

2. (c) Let I =

∫ 0

π/ 2

∫

=2

0

0

 x cos5    · sin x dx 2

x  x cos   sin dx = 2 2 2

∴ I =

1/ 2

∫

z (−2) dz

0

 − 4  1  4 1  − 1 = 1 −  =  ·   7  2  7 8 2    7

z  = – 4.    7 1 ∞

ln 2

2 3. (a) ∫  x  dx = 0 e 

∫ 1dx + 0

4. (b)

∫ f ( x ) dx

−1

2

∫ ( 3 − f ( x ) ) dx −4

2

−2

= 4 + 25 = 29.

−1

−1

1 1 101  ∴ I = –  ∫ cos ec  t −  dt t  1/2 t

∫ 0

π/ 4

=

∫ 0

3

tan 2 x ⋅ sec 2 x

(1 + tan x ) 3

2

x + cos3 x )

2

2

 dx. =

1

dx

2

dz

∫ 3z

0

2

−1 z dz 2 r2

∫

= 1 (r3 ) ⋅ 3

π/2

∫ sin xdx + ∫ cos xdx = ( − cos x ) π/4

π

∫e

x

⋅

π/4 0

2

1 − sin x dx 1 − cos x

x

=

∫e

x

π/2 π

∫e

π/2

sin x ⋅ cos x

(sin

0

π/2

=

⇒ I = – I ⇒ 2I = 0 ⇒ I = 0. 6. (a) Let I =

∫

π

2

x cos 4 x dx

0

x x  1 − 2 sin 2 cos 2  = ∫e   dx x π/2   2 sin 2   2

Let 1 = t, – 1 dx = dt x2 x

4

0

r2

10. (c) Let I =

π

1 1 101  5. (c) I = ∫ cos ec  x −  dx   x x  1/2 2

2

∫ sin

2 2 ∫ x r − x dx =

 z 3/ 2  z dz =    3 0

r2

2

−4

π/ 2

r

1  1  =  =2– + 1 + 1 − 2  2 

2

∫ f ( − x ) dx = ∫ f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx

π/ 4

1 2

0

= 7,

π 2

−1   2 2  Put r − x = z ⇒ x dx = 2 dz   

π/4 −4

−4 1

I=

∫ xy dx =



9. (d) A =

f ( x ) dx = 7 + 18 = 25

∫

⇒

r

0

= ln 2

= 4 and

2 3 ⋅ 1⋅ 3 ⋅ 1 π = 3π ⋅ ⋅ 8⋅ 6⋅ 4⋅ 2 2 256

8. (b) Let I =

=

0

−4

π

π sin 4 x cos 4 x dx 2 ∫0 = 2·

= π · 

ln 2

=

x cos 4 x dx – I

∞

2 2 dx + ∫  x  dx x   ln 2  e 

∫  e 0

4



6

x x    Putting cos 2 = z ⇒ sin 2 dx = − 2 dz    1/ 2

π

∫ sin

=π

6

7

∫ (π − x) sin4 (π – x)cos4 (π – x) dx 0

∴ 2 I = 0 ⇒ I = 0.

x cos 4 x ) dx

4

0

π

cos3 (2n + 1) x dx = – I.

π

∫ x ( sin

x

x 1 2 x − cot  dx  cos ec 2 2 2  x d  x   − cot 2 + dx  − cot 2   dx   π

x  =  − e x cot  2 π/2  π π =  − e π cot + e π / 2 cot  = eπ/2. 2 4 

+ ( sin x )π/4 π/2

∫ ln xdx = 4 ·  εlim →0

Required area = 4

0

 x ( ln x − 1)  ε = 4 ·  εlim →0 1

1

∫ ln xdx

1

∫ e ( x − α ) dx

( x − α)

must be +ve

257

1

and –ve both for x ∈ (0, 1) i.e., ex (x – α) = 0 for one x∈ (0, 1)

Definite Integral and Area

11. (c) y = ln x, y = ln | x |, y = | ln x |, y = | ln | x | |

14. (a) 

x2

= 0, ∴ e x

2

0

ε

= 4 sq. units

∴ α ∈ (0, 1). 15. (b) Let I =

π/ 2

∫ log sin x

dx =

0

=

π/ 2

∫ log cos x

π/ 2

π



∫ log sin  2 − x 

dx

0

dx.

0

π/ 2

∫ ( log cos x + log cos x )

⇒ 2 I =

dx

0

π/ 2

∫ log ( sin x cos x )

=

dx

0

π/ 2

0

π/ 2

∫ ( log sin 2 x − log 2 )

= 12. (b) Let I =

2π

∫e 0

x

 sin 2 x   dx 2 

∫ log 

=

dx

0

π π sin  +  dx 4 2

=

π/ 2

∫ log sin 2 x

π/ 2

∫ dx

dx – log 2

0

2π

 π x  = sin  +  e x  −   4 2  0

2π

π

1

x

∫ 2 cos  4 + 2  ⋅ e

x

dx

0

2π

∫ 0

 1 π x sin  +  e x dx    2 4 2 

∫

π/6

π/3

=

∫

π/6

π/3

=

∫

π/6

cos x dx sin x + cos x π/3

∫

π/6

∴ I =

sin x dx sin x + cos x

π π  sin  + − x  6 3  dx π π  π π  sin  + − x  + cos  + − x  6 3  6 3 

∴ 2I =

π ⋅ 12

π/ 2 0

0

sin x + cos x π π/3 dx = [ x ]π / 6 = ⋅ 6 sin x + cos x

π/ 2

π

∫ log sin z dz − 2 log 2 0

∵log sin z = log sin ( π − z )   π  π/ 2   log sin z dz = log sin z dz ∴ 2 ∫  ∫  0 0   π/ 2

π

∫ log sin x dx − 2 log 2

=

⇒ I = − 2 (e 2 π +1) ⋅ 5 13. (c) Let I =

1 2 2

=

1  −1 2 π 1  −1 2 π 1  1 =  I e − e −  –  −  2 2 2 2 2 4

π/3

∫ log sin z dz − log 2 ⋅ [ x]

[Putting 2x = z in the first integral, ∴ 2 dx = dz]

  π π = sin  π +  e 2 π − sin  4 4    2π 1   π x x  cos  +  e  + 2  4 2 0

0

π

1 = 2

0

∴ I = 16. (a)

∫

=I–

π log 2 2

−π log 2 . 2

t2

xf ( x) dx = 2 t 5 By Leibnitz’s rule, 5 2 2t [t2 f (t2)] = 5t 4 or f (t2) = t 5 2 2. 4 Putting t = , f   = 5 5 25   0

17. (b) l =

∫

1

0

1− x dx 1+ x

Put x = cos 2θ, dx = – 2 sin 2θ dθ 0

∴ l = − 2 ∫π sin 2θ tan θ d θ = 2 ∫

π 4

0

4

2 tan 2 θ dθ 1 + tan 2 θ

258

π  π  = 4  ∫ 4 θ d θ − ∫ 4 cos 2 θ d θ 0 0  

=

Objective Mathematics

sin 2θ  π  = 4   − 2 θ + 2  0 4  π 1 = π − 2  +   4 2

∫ x log sin x

∞

21. (b) Let I =

dx

=

∫ ( π − x ) log sin ( π − x ) ∫ ( π − x ) log sin x

dx

=

1 2x 1 dx 1 dx dx + − ∫ 2 2 ∫ ∫ 4 0 x +1 2 0 x +1 2 0 1 + x

=π

dx

∞

π

∫ log sin x

dx = 2π

0

π/ 2

∫ log sin x

dx

0

−π = 2π· log 2 [see Q. No. 15] 2 − π2 log 2 ⋅ ∴ I = 2 a

dx

∫a+

=

π/ 2

=

a2 − x2

0

∫ 0

a cos θ d θ a + a cos θ

[Putting x = a sin θ = ⇒ dx = a cos θ dθ] π/ 2

∫ 0

π/ 2

cos θ dθ = 1 + cos θ

π/ 2

π/ 2

1 = ∫ 1 dθ − 2 0

∫ 0

∫ 0

  1 1 −  dθ  1 + cos θ 

θ sec dθ 2

 1   1+ 2    1 1 1 + x2  1 π x  − log = log   + ⋅ 2 4 4  (1 + x) 2  x =0 2 2 1     + 1   x  x = ∞  = 0 – 0 + π = π. 4 8 22. (c) Let I = =

0

⇒ 2 I =

0 π

=

π/ 2

∫

dx

0

 sin x cos x  log  +  dx  cos x sin x 

π/ 2

=



2



∫ log  sin 2 x 

dx

π/ 2

π/ 2

0

0

∫ 1 dx −

∫

∫ log sin

x) dx =

0

dx = 2· 2

log sin 2 x dx

π π log 2 – 1 ⋅ log sin z dz = ∫ 2 2 0 1     Putting 2 x = z ⇒ dx = 2 dz   

2

x dx

π/ 2

∫ log sin x

dx

0

[see Q. No. 15]

2π

∫ [ 2 sin x] dx  

23. (a) I =

0

–2 ≤ 2 sin x ≤ 2

2 sin x = 1 ⇒ sin x = 1 ⇒ x = π , 5π 2 6 6



2 sin x = –1 ⇒ sin x = – 1 ⇒ x = 7 π , 11π 2 6 6

0

= log 2

π

∫ log sin x

2

−π  = 4·  log 2    2  ∴ I = – π log 2.

0

=

∫ log (1 − cos

dx

π

0

π  π π =  − 1 · − tan 2 4 2 

dx

0

π

= 2

∫ log ( tan x + cot x )

π

∫ log (1 + cos x) (1 − cos x) 0

π/ 2

dx

∫ log [1 + cos (π − x)] dx =  ∫ log (1 − cos x )

2

θ  = θ − tan  2 0 

20. (c) Let I =

π

∫ log (1 + cos x) 0

π

π/ 2

=

∞

1  1 + x2  ∞ 1 =  log    +  tan − 1 x  0  (1 + x )2   2  4 0 

∫ log sin x dx – I

⇒ 2 I = π



∞

∞

π

0

19. (b) Let I =

(by partial fraction)

1 1 1  =  log (1 + x 2 ) − log (1 + x ) + tan − 1 x  4 2 2  0

0



dx

∞ 1  x +1 1  −   dx 2 ∫0  x 2 +1 1 + x  ∞

0 π



2

=

0

=

x

∫ (1 + x ) (1 + x ) 0

π



0

= π log 2.

π π = π − − 1 = − 1 . 2 2 18. (a) Let I =

∫ log sin z dz

π π = log 2 + log 2 [see Q. No. 15] 2 2

π 4

π

π/ 2

π 1 log 2 – ⋅ 2 2 2

=

+

π/6

π/2

5 π/6

0

π/6

π/2

∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx π

7 π/6

11π/6

5 π/6

π

7 π/6

∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx + ∫ [ 2 sin x] dx

11π / 6 π/6

=

π/2

5 π/6

π/6

π/2

7 π/6



∫

–

11π/6

dx − 2

π

∫

dx −

7 π/6

∫

dx

=

11π/6



2π π 4π π = – π 2π = –π. = − − − − 3 6 3 6 3 3

=

24. (d) y′ = 16x – 5x4 = 0 ⇒ x = 0, x3 = 16 5 c

2

1

− x 5 ) dx = 16 3

∫ 8 x dx − ∫ x dx 2

5

−1

= 16 3

[Putting x = tan θ ⇒ dx = sec2θ dθ] π/ 4

=

∫ log (1 + tan θ )

=



=

  dθ

=

∫ 0

0

2   log   dθ  1 + tan θ 

π/ 4

∫ 1 dθ

= log 2

0

–I

0

π/ 2

∫ log ( tan θ + cot θ )

=

π

sin 2 kx ∫0 sin x dx =

sin 2k (π − x) ∫0 sin (π − x) dx

π

sin (2k π − 2kx) dx = – sin x 0

∫

π

sin 2kx dx = – I 0 sin x

∫

∴ 2 I = 0 ⇒ I = 0.

= π log 2 [see Q. No 20] π

=

x 3 cos 4 x sin 2 x dx 2 − 3πx + 3 x 2

∫π ∫

(π − x)

cos 4 ( π − x ) sin 2 ( π − x )

π − 3π ( π − x ) + 3 ( π − x ) 2

0

=  ∫

3

(π

π/ 2

∫ cos 0

π/ 2

=

∫ 0

2

cos 2 θ dθ θ + 4 sin 2 θ

cos θ dθ = 1 + 3 sin 2 θ

= −1 3

2

π/ 2

∫ 0

π/ 2

∫ 0

1 + cos 2θ dθ 2 + 3 (1 − cos 2θ )

( 5 − 3 cos 2θ ) − 8 5 − 3 cos 2θ

dx

π2 − 3πx + 3 x 2

0

π

∫

(π

− 3π2 x + 3πx 2 ) cos 4 x sin 2 x

3

π2 − 3πx + 3 x 2 π

⇒ 2 I = π

∫

(π

2

(π

= π·2

π/ 2

∫ cos

4

2

− 3πx + 3 x 2 )

x sin 2 x dx = 2π·

0

∴ I = π /32. π

x 2 sin x

∫ (2 x − π) (1 + cos

2

π

=

x)

dx

(π − x) 2 sin (π − x) dx 2 (π − x))

∫ (2π − 2 x − π) (1 + cos 0

π

(π2 − 2πx + x 2 ) sin x dx 2 x) 0

=

∫ (π − 2 x) (1 + cos

=

∫ (π − 2 x) (1 + cos 0

(π2 − 2π x) sin x dx – I 2 x)

dx

3 ⋅ 1⋅ 1 π ⋅ 6⋅ 4⋅ 2 2

2

30. (a) Let I =

dx

dx – I

− 3πx + 3 x 2 ) cos 4 x sin 2 x

0

π

dθ

2

− 3π2 x + 3πx 2 − x 3 ) cos 4 x sin 2 x

3

0

27. (c) Let I =

dθ

0

0

π

2

sec 2 θ dθ 2 θ [Putting x = tan θ ⇒ dx = sec2θ dθ]



=

⇒ 2I = π log 2   ∴  I = π log 2. 4 8

=

1  dx



0

26. (c) Let  I =

∞ −π 2 +  tan − 1 ( 2 z )  0 6 3

=

2

∫ log ( tan θ + cot θ ) sec

π

∫

1 z2 +   2

π/ 2

=

0

π/ 4

dz

∫

∫ log  x + x  1 + x

0

 1 − tan θ  1 +  dθ =  1 + tan θ 

∞

∞

π

∫ log 1+ tan  4 − θ 

π/ 4

0

sec 2 θ d θ θ) − 3 (1 − tan 2 θ)

2

π −π 2 π + ⋅ = . 6 6 3 2

29. (d) Let I =

dθ

π

∫ 5 (1 + tan

−π 1 + 6 3

0

π/ 4

π/ 2

0

16 16    ∴   c = –1. = 3 3 1 π/ 4 log (1 + tan θ ) log (1 + x ) dx = ∫ sec2θ dθ 25. (a) Let I = ∫ 2 + 1 x (1 + tan 2 θ) 0 0 

0

[Putting tan θ = z ⇒ sec2θ dθ = dz]

1

−1

dθ

∫ 5 − 3 cos 2θ

−π 8 dz + ∫ 2 6 3 0 8z + 2

28. (d) Let I =

1

If c = –1,

0

π/ 2

8 3

∞

2π

= (5π/6 – π/6) – (7π/6 – π) – 2(11π/6 – 7π/6) – (2π – 11π/6)

∫ (8 x

∫ 1⋅ dθ +

= −π + 8 6 3

∫ 0 ⋅ dx + ∫ 1⋅ dx + ∫ 1⋅ dx + 0 0

π/ 2

259

= −1 3

∫ [ 2 sin x] dx

Definite Integral and Area

2π

–

260

⇒ 2 I = π

Objective Mathematics



π

sin x ∫0 1 + cos2 x dx = – π

= 2 (e – e + 1) = 2

−1

dt ∫1 1 + t 2

[Putting cos x = t ⇒ sin x dx = – dt]

2 1 π π = π  tan − 1 t  = π  +  = π −1 2 4 4 π2 ∴ I = . 4 31. (b) x ∈ (–8, 8)   y = 2

x ∈ (–8 2 , –8] ∪ [8, 8 2 )   y = 3



34. (b)

1 ×2×2 2 = 6 – 2 = 4 sq. units Required area = 2 × 3 –

2a

∫ f ( x) dx 0

0

∫

=λ–

=

a

2a

0

a

∫ f ( x) dx + ∫ f ( x) dx

f (2a − t ) dt

a

[Put x = 2a – t ⇒ dx = –dt]

 0

∫ f (2a − x) dt

=l+

a

= λ + µ. π/ 4

sin x ⋅ cos x dx 2 x + sin 4 x

∫ cos

35. (b) Let I =

0

1 2 32. (d)

π

x 2 cos x

∫ (1 + sin x)

π

=2

 −1   x   + 2 ∫ dx sin sin x x 1 + 1 +  0 0

= – π2 + 2

π

π

x

∫ 1 + sin x x

∫ 1 + sin x

Let I =

dx =

0

(π − x) ∫0 1 + sin x dx = π

⇒ 2 I = π

π

(π − x)

∫ 1 + sin (π − x) 1

∫ 1 + sin x

dx

dx – I.

0

1 − sin x dx = π 2 0 cos x

∫

π

∫ ( sec

= [ π (tan x − sec) ]0

2

x − sec x tan x ) dx

0

x

− ex )

1 0

∫z 1

dz +3

2

0

1 π  1  tan − 1  .  = 3 6 3  3 2

e4

x ∫ e dx = a, let I =

36. (d) Given

∫

2

Put ln (x) = t2 ⇒ 1 dx = 2tdt x

∴  I =

∫e

t2

( ) − ∫ e dt = 2e

⋅ 2t 2 dt = t et

∫ 0

(x +

2

1 + x 2 dz; dx 1 + x2

∞

)

n

2

t2

4

– e – a.

1

 x  dx = dz ⇒ 1 +  1 + x2  

37. (b) Put x +  1 + x 2 = z ⇒ z dx =

2

1

1

∴

ln ( x) dx

e

1

∞

( xe

0

 −1  z  tan − 1  =    3  1  3

33. (d) y = x e| x | = 1, y = 0 x ∫ xe dx = 2

dx

2

−1    Putting cos 2 x = z ⇒ sin 2 x dx = 2 dz   

2

= π [(tan π – sec π) – (tan 0 – sec 0)] = 2π ⇒ I = π. π x 2 cos x dx So, by (1) ∫ 2 0 (1 + sin x )  = – π2 + 2π = π (2 – π).

1

sin 2 x dx = – cos 2 2 x + 3



0

π

Required area = 2

dx

2

sin 2 x



=

π



∫ 0

...(1)

0

π

=

 1 − cos 2 x  +  2  

∫ 2 (1 + cos 2 x) + (1 − cos 2 x)

π/ 4

=2

dx  π

sin 2 x

0

0

π

0

π/ 4

0

 2 = x 

∫ 1 + cos 2 x 2

dx

2

π/ 4

=∫ 1

1 1 z+  1 + x 2 dz ∞ 2  z  dz =∫ n +1 zn ⋅ z z 1





1 2

=

∞

∫ 1

  1 1 2 ,∴2 1 + x = z +  z z 2

z2 + 1 1 dz = zn + 2 2

∞

∫ (z

3 −1  3 − 1 = sin − 1 z  1 −2 3 = 2 sin– 1  .    2    2

42. (a) x sin πx = 0, gives x = 0 or πx = nπ, n integer.

+ z − n − 2 ) dz

−n

∴ x = 0; x = n, n = 0, ± 1, ± 2, ··· ; out of

1

which 0, 1 ∈  −1, 

∞



1  z1 − n z − n −1  n − ⋅   = 2 2 1 − n (n + 1) 1 n −1

=

et dt ∫0 t + 1 = a,  Let I = 1

38. (b) Given

3/ 2

b

e − t dt ∫b − 1 t − b − 1

dt = dZ =

1

∫

e

0

1

=

dZ = Z +1

eZ − b

∫ Z + 1 ⋅ dZ 0

0

1

∫

e

− (1 − Z ) − b + 1

0

dZ 1− Z − 2

=

e Z dZ = –e–b · a 0 Z +1

=

−1

0

1

1

3/ 2

−1

1

∫ x sin πx dx − ∫ x sin πx

1 1 1  1 −1 1 =  −  −  2 −  = 1 +  . π π  π π π π

dx = π ⇒ I = π . 2

1 + cos 2 t

3

1

1 + cos 2 t

3

+  ∫ |( x − 1) ( x − 2) ( x − 3)| dx

∫ ( 2 − x ) f ( x ( 2 − x ) ) dx

=

sin 2 t

2

2

∫(x

3

xf ( x ( 2 − x ) ) dx

sin 2 t

= ∫ |( x − 1) ( x − 2) ( x − 3)| dx 

=

∫

43. (d) I1 =

2



− 6 x 2 + 11 x − 6 ) dx

⇒ 2I1 = 2I2 ⇒

1

3



–

∫(x

3

− 6 x 2 + 11x − 6 ) dx

2

2

 x4  11 2 3 =   − 2 x + x − 6 x  4 2  1

3

x  11 2 3 –  − 2x + x − 6x 4 2  2 4



 11 1  = (4 − 16 + 22 − 12) −  − 2 + − 6   2 4    81  99  –  − 54 + − 18  − (4 − 16 + 22 − 12)  = 9 . 4 2     2 π/3

π/3 sin x + cos x sin x + cos x dx = ∫ dx 41. (a) Let I = ∫ sin 2 x − (1 − sin 2 x) 1 π/6 π/6

π/3

=

∫

π/6

sin x + cos x 1 − (sin x − cos x)

dx

1  −x  –  cos πx + 2 sin πx  π π 1



40. (b) ∫ |( x − 1) ( x − 2) ( x − 3)| dx 1

dx

3/ 2

b b  esin x dx  = ∫ sin x  ∫ f ( x) dx = ∫ f (a + b − x) dx  + 1 e a π /2 a 

∫

3/ 2

1

π /2

π /2

1

1

1  −x  =   cos πx + 2 sin πx    π π  −1

dx 39. (d) I = ∫ sin x +1 π /2 e

2I =

0

0

∫ x sin πx dx + ∫ x sin πx dx − ∫ x sin πx

1

= –e–b ·  ∫

3/ 2

1

∫ | x sin πx | dx + ∫ | x sin πx | dx + ∫ | x sin πx | dx

=

π /2

π /2

dx

−1

−Z − b + 1

3 . 2

−1

Let t – b + 1 = Z



∫ | x sin πx |

∴ 

2

dx =

3 −1 2

∫

1− 3 2

1 − z2

I1 = 1. I2

1

1/ 2

0

0

∫ | x cos πx | dx =

1/ 2

=

∫

x cos πx dx −

0

∫ | x cos πx | dx +

1

∫ x cos πx

1

∫ | x cos πx |

dx

1/ 2

dx

1/ 2

1 1 x  x  =  sin πx + 2 cos πx  −  sin πx + 2 cos πx  π π π 0  π 1 / 2 1/ 2

1

1 1   −1 1  1 =  − 2 − 2 −  = . 2π  π  2π π   π 45. (c) I =

dz

= 2 · I2 – I1

44. (b) x cos πx = 0 gives x = 0, πx = π , 3π etc. 2 2 ∴ x = 0, 1 ∈ [0, 1] 2 ∴

π/4

π/4

0

0

∫ log (1 + tan x ) dx = ∫ log (1 + tan ( π/4 − x ) ) dx

 sin ( π/4 − x ) + cos ( π/4 − x )   dx = ∫ log  cos ( π/4 − x ) 0   π/4

261

[Putting sin x – cos x = z ⇒ (cos x + sin x) dx = dz]

) ( 1 + x + x) = 1

Definite Integral and Area

(

∵ 1+ x 2 − x   ∴ 1 + x 2 − x = 

262

π/4





2

∫ log  1 + tan x  dx

=

= (log 2) π/4 – I

0

Objective Mathematics

  k = –π log 2  ∴  I = –  k 8

⇒ I = π ⋅log 2 .   8

0

1

2

0

1

3 2 3 2 ∫ | x − 3x + 2 x | dx + ∫ | x − 3x + 2 x | dx 1

∫(x

=

2

− 3 x 2 + 2 x ) dx − ∫ ( x 3 − 3 x 2 + 2 x ) dx

3

0

−1

1 − x   dx . 1 + x 

0   et cos–1 x = 2θ ⇒ x = cos 2θ ⇒ dx = –2 L sin 2θdθ

1 − cos 2θ cot–1 1 + cos 2θ = cot–1 (tan θ)

2

46. (a) ∫ | x 3 − 3 x 2 + 2 x | dx =



1

∫ cos 2 cot

50. (b) I =

1

1

π = cot–1cot  − θ  = π − θ 2 2  π   cos 2  − θ  = cos(π –­ 2θ) = –cos 2θ = –x 2 



2

 x4  x4 3 3 2 2 =  − x +x  −  −x +x  4 0  4 1

sin θ

1 16 1 =  − 1 + 1 −  − 8 + 4  +  − 1 + 1 = 1 . 2 4   4  4 

1

 x2  1 ∫0 − xdx =  − 2  = – 2 1

∴  I = 51. (c) A =

t dt

∫ 1+ t

2

1

cosec θ

∫

=

2 −x 2 +x , then f (– x) = log  47. (d) Let f (x) = log   2 + x   2 − x 

1

. Let t = 1 , dt = – 1 dx x2 x

cosec θ 1 −1 dx dx =– ∫ = –B ⋅ 2⋅ 2 x x x +1 x (1 + x 2 ) 1 2 x

2 −x = – log  = – f (x)·  2 + x  So, the function f (x) is odd.



2−x

1

∫ log 2 + x

∴

dx = 0.

−1

2

∫

− 1/ 2

 x+1 x − 1  −    x − 1 x + 1

1/ 2

∫

=

− 1/ 2

1/ 2

=

2

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

1/ 2

48. (b)

∫

− 1/ 2

2

1/ 2

dx =

∫

− 1/ 2

x + 1 x −1 − dx x − 1 x +1

1/ 2 4x 4x dx = 2 ∫ 2 dx x −1 x −1 0

 4x  ∫0  x 2 − 1  dx

3 4 = – 4  log  = 4 log   . 4  3

4

∫

f ( x)dx =

−2

−1

∫

f ( x) dx +

−2 2

∫

0

∫

−1 3

0

I2 = I3 =

∫ cos ( cos x ) dx 0

∫ sin ( cos x ) dx ∫ cos x dx 0

f1 (x) = cos (sin x) Area under curve f1 is more than f2 where f2 (x) = sin (cos x), x ∈ [0, π/2] ∴  I1 > I2    area under f3 is more than others ∴  I3 > I1 > I2. 53. (c) We have, Un = π/ 4

4

f ( x) dx + ∫ f ( x) dx + ∫ f ( x) dx

+



= 4 + 1 + 0 + 1 + 4 + 9 = 19.

2

π/2

π/2

1

f ( x) dx + ∫ f ( x) dx



1

= 0.

0

= – 4 log 1 − x 2 1/ 2 ( ) 0 

49. (c)

∫ cos ( sin x ) dx =

−A −1 −1

π/2

 4x  1  < 0 in the integral  0,   ∵ 2  2   x −1



1

0

1/ 2



A+B

π/2

(  integrand is even)

= –2

∆ = e

A A2 B 2 −1 = 1 A 1 2A 2 −1

A2 B2 A 2 + B2

A

52. (b) I1 =

2



∴  A + B = 0.

= =

∫ tan

n

x dx =

0

n−2

x ⋅ (sec 2 x − 1) dx

∫ tan

n−2

x ⋅ sec 2 x dx −

0

π/ 4

∫ tan

n−2

x ⋅ tan 2 x dx

0

∫ tan

0 π/ 4

3

π/ 4

π/ 4

∫ tan 0

n−2

x dx

π/ 4

π/ 4

∫ tan

θ dθ

0

∫ tan

Now, In + 1 =

n

n +1

θ dq

0

58. (a) Let I =

π/ 4



=

∫ tan

n −1



=

∫ tan

n −1

θ ⋅ tan 2 θ dθ.



=

∫ tan

n −1

θ sec 2 θ d θ −

0



n −1

θ dθ

i.e., ∫ 0

10π

0

59. (b) In + In

=

∫ t dt 1

∴ In + In

6

x cos 4 x dx

π/3

∫ cos 0

π/2 π = 10  ∫ sin x dx + ∫ sin x dx  π/2  0 





π/2 π = 10 [ − cos x ]0 + [ − cos x ]π/2   

=



= 10 [1 + 1] = 20.  x cos qx dx

=

π/ 2

 61. (d)

tan n x (1 + tan 2 x ) dx

tan n x sec 2 x dx

n

+2

=

where t = tan x 1 n +1

⇒ nlim n [In + In + 2] →∞ n n 1 = = = 1. = lim n ⋅ n→∞ 1  n +1 n +1 n 1 +  n 

=

π/ 2  sin qx  p = cos p x ⋅ + cos p − 1 x sin x sin qx dx  ∫ q 0  0 q π/ 2 p =0+ cos p − 1 x [cos (q – 1) x – cos qx cos x] dx q ∫0 [Since cos (q – 1) x = cos qx cos x + sin qx sin x ∴ cos (q – 1) x – cos qx cos x = sin qx sin x]

π/4

∫

0

π/4

0

=



0

∫

=

+2

0

π/ 2

∫ cos

∫ sin

2 ∴ I = 3π . 512

|sin x | dx

57. (b) f (p, q) =

π/ 2

2 5 ⋅ 3 ⋅ 1⋅ 3 ⋅ 1 π ⋅ = 3π 10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 2 256

= 2π·

 sin x  2 cos x + 1 dx =   2 (2 + cos x)  2 + cos x  0

p

x cos 4 x dx – I

0

2 cos x + 1 . (2 + cos x) 2

π/ 2

6

⇒ 2 I = π· 2·

1 1 =  − 0  = . 60. (c) 2 2 

∫

x cos 4 x dx

0

– In – 1

 56. (a)

π

∫ sin

= π

I ntegrating both sides w.r.t. x between the limits 0 and π , 2 π/ 2 2 cos x + 1 we get [P]0π / 2 = ∫ dx 2 0 ( 2 + cos x ) π/ 2

6

0

π/ 4

(2 + cos x) ⋅ cos x − sin x ⋅ (− sin x) ⇒ dP = (2 + cos x) 2 dx =

∫ (π − x) sin

=

⇒ In – 1 + In + 1 = 1 or n (In – 1 + In + 1) = 1. n sin x 55. (b) Let P = 2 + cos x



(π − x) cos 4 (π − x) dx

π

0

 tan θ  =    n 0 n

∫ tan

6

0

θ ⋅ (sec 2 θ − 1) dθ π/ 4

x cos 4 x dx

6

∫ (π − x) sin

=

0

π/ 4

π

∫ x sin 0 π

0

π/ 4

=

 p p ⇒ 1 +  f ( p, q) = f ( p – 1, q – 1) q q  p f ( p – 1, q – 1). ⇒ f ( p, q) = p+q

1 . n −1

∴ Un + Un – 2 = 54. (a) We have, In =



– Un – 2

2a

3φ ⋅ sin2 6 φ d φ

π

1 cos 4 θ ⋅ sin2 2 θ dθ 3 ∫0 1    Putting 3φ = θ ⇒ d φ = 3 d θ    π 1 4 2 cos θ ⋅ (2 sin θ cosθ) dθ 3 ∫0 π 4 = ∫ cos 6 θ ⋅ sin2θ dθ 30 π/ 2 4 6 2 x 2 ∫ cos θ ⋅ sin θ dθ 3 0 8 5 ⋅ 3⋅1 π = ⋅ ⋅ = 5π . 3 8⋅ 6⋅ 4⋅ 2 2 96

∫x

9/ 2

0



4

(2a − x) − 1/ 2 dx π/ 2

=

∫ 0

(2a )9 / 2 sin 9 θ ⋅ 4a sin θ cos θ d θ 2a − 2a sin 2 θ

263



Definite Integral and Area

p p f ( p − 1, q − 1) − f ( p, q ) q q

π/ 4

 tan n − 1 x  =    n − 1 0

[Putting x = 2a sin2θ ⇒ dx = 4a sin θ cos θ dθ]

264



π/ 2

∫ sin

= 64a5

10

66. (d)

θ dθ

Objective Mathematics

0

9 ⋅ 7 ⋅ 5 ⋅ 3⋅1 π = 64a5 · ⋅   = 63π a5. 10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 2 8 62. (c)

2

∫x

2 x − x 2 dx

3

0

π/ 2

=

∫ 0

8 sin 6 θ 4 sin 2 θ − 4 sin 4 θ (4 sin θ cos θ) dθ [Putting x = 2 sin θ ⇒ dx = 4 sin θ cos θ dθ] 2



π/ 2

∫ sin

= 64

8



0

1 2

=–

∫ 0

π/ 2

1 = 2 =

1 2

∫ 0

0

∫ cos

2

θ

π/ 2

1+ cos θ sin θ dθ 1 − cos θ

0

  = cos θ  = − sin θ d θ  − sin θ  = d θ 2 

64. (b)

∫ 0

=

=

x

1 3

1 3

∫ 0

68. (b)

) (

=5–

3− 2. 1

3

1

5

θ dθ =

0

1

dx =

−1

π

2 x (1 + sin x ) 1 + cos 2 x

=

∫

π

−π

1 4⋅ 2 = 8 . ⋅ 3 5⋅3 45

0

1

−1

0

= – ( x) −1 = – 1.

∵[ x] = − 1 if − 1 ≤ x < 0   0 if 0 ≤ x < 1 

2

3/ 2 2

2.

dx

π 2x x sin x dx + 2 ∫ dx 2 − π 1 + cos x 1 + cos 2 x

∫

π

0

x sin x dx 1 + cos 2 x

( π − x ) sin ( π − x ) 1 + cos 2 ( π − x ) π ( π − x ) sin x π

0

1 + cos 2 x

0

⇒ I = 4π

∫

π

2I = 4π

∫

π

0

0

π sin x x sin x dx − 4 ∫ dx 0 1 + cos 2 x 1 + cos 2 x

sin x dx 1 + cos 2 x

 Put cos x = t ⇒ – sin x dx = dt ⇒ sin x dx = –dt 1 1 − dt dt = 2π ∫ = 2π  tan −1 t  −1 −1 1 + t 2 1 + t2 π π = 2π [tan–1 (1) – tan–1 (–1)] = 2π  +  4 4 = 2π × π = π2. 2

∫

−1

1

3

69. (b) ∫  x  dx = 0

∫ (−1) dx + ∫ 0 dx

 0

 Putting x 3 = tan θ    1  ⇒ x 2 dx = sec 2 θ d θ    3

dx = ( x) 1 + 2 ( x)

2

−π

∴ I = 2π

sec θ d θ  sec7 θ

dx

1.5

2

θ dθ

dx

2

2

3  2 −1+ 2  − 2  = 2 – 2 

∫

0

2

∫ cos

1

∫ 0 dx + ∫ 1 ⋅ dx + ∫ 2

∫

2

2

∵[ x 2 ] = 0, 0 ≤ x < 1    1, 1 ≤ x < 2     2, 2 ≤ x < 1.5 

I = 4

∫ cos

)

1.5

2

0



π/ 2

) (

3− 2 +3 2− 3

2

2

∫

1 cos 2 θ d θ + 2

3

∫ [ x ] dx = ∫ [ x ] dx + ∫ [ x ] dx + ∫ [ x ]

I = 4

θ (1 + cos θ) dθ

3

2

2

2 −1 + 2

1.5

2

1⋅ dx + ∫ 2dx + ∫ 3dx 3

(



2

3

2

=0+1

cos θ / 2 θ θ cos 2 θ ⋅ ⋅ 2 sin cos dθ sin θ / 2 2 2

π/ 2

∫ [ x]

65. (c)

1

I = 0 + 4

(1 + x 2 )7 π/ 2

0

2



1 1 π 1 2 = π + 1 = 3π + 8 . = ⋅ ⋅ + ⋅ 2 2 2 2 3⋅1 8 3 24 ∞

1

=

0

∫

∫ 0 dx + ∫

2

1

2 cos 2 θ / 2 cos θ ⋅ ⋅ sin θ dθ 2 sin 2 θ / 2

∫ cos

π/ 2

[ x 2 ] dx + ∫ [ x 2 ] dx + ∫ [ x 2 ] dx + ∫ [ x 2 ] dx

0

2

π/ 2

1 = 2

=

1

0

1

 π/ 2

∫

=

  Putting x 2  ⇒ 2 x dx  ⇒ x dx 

1 = 2

=



1 + x2 dx 1 − x2

5 ∫x

[ x 2 ] dx =

0

7 ⋅ 5 ⋅ 3 ⋅ 1⋅ 1 π ⋅ = 7π . 10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 2 8

= 64· 1

2

0

θ ⋅ cos2θ dθ

0

63. (a)

67. (b)

∫

1

3

0

1

∫  x  dx + ∫  x  dx  0 if 0 ≤ x < 1  ∵  x  =   1 if 1 ≤ x < 3 

 1

=

3

∫ 0 dx + ∫1 dx = x 0

1

3 1

= 2.

x π dx + = ∫ − cos 2 2 x 4 − π/ 4



π ⋅2 4

π/ 4

∫ 0

π/ 4

dx dx ∫ − cos 2 2x − π/ 4

dx 2 − cos 2 x

π/ 4

∫

=

π2 . 6 3

2

71. (a) ∫ sin 0

π[ x]  dx = 2

1

=

∫ sin 0

∫ sin 0

π/ 4

=

2

− 1/ 2

72. (c)

∫ 0

∫

∫ (cos θ − sin θ)

= (sin θ + cos θ)

75. (a)

π



=

2π

∫ |sin x | dx + ∫ |sin x | dx π

0

∵sin x = 0 gives x = nπ, n = 0, ±1, ± 2, ...; out of which only x = π ∈ (0, 2π) ⋅   

 =

π

2π

0

π

∫ sin x dx − ∫ sin x

dx = – [cos x]0π + [cos x]2ππ

= – (– 1 – 1) + (1 + 1) = 4. −1/ 2

2

∫ [2 x]

73. (c)

−1





 ∵[2 x] =      

dx =

1/ 2

0

∫ [2 x] dx + ∫ [2 x] dx + ∫ [2 x]

−1

−1/ 2

+

0

1

3/ 2

1/ 2

1

∫ [2 x] dx +

−2 if −1 ≤ x < 1 0 if 0 ≤ x < ; 2 3 2 if 1 ≤ x < ; 2

−1 ; 2

dx

∫ [2 x] dx +

2

∫ [2 x] dx

3/ 2

−1  ≤ x < 0; 2  1 1 if ≤ x < 1;   2  3 3 if ≤ x < 2   2

+ 3 ( x)

2 3/ 2

=

3 . 2

π

∫ |cos θ − sin θ|

dθ

dθ –

π

∫ (cos θ − sin θ)

dθ

π/ 4

π/ 4

π

− (sin θ + cos θ)

0

π/ 4

 1 1    1   1 + + =    − 1 − (−1) −  2   2   2  2

0 if 0 ≤ x < 1 1 if 1 ≤ x < 2 

∫ |sin x | dx

3/ 2 1

π/ 4

2.

=2

0

1/ 2

0

2π

3/ 2

+ 2 ( x)

1

|cos θ − sin θ | d θ +

π/ 4

= 2 – 1 = 1.

1 − cos 2 x dx = 2

+ ( x)

0

2

2π

1

∵cos θ − sin θ = 0 gives cosθ = sin θ    π   ∴ tan θ = 1 = tan   4   π  ∴ θ = n π + , n = 0, ± 1, ± 2, ⋅ ⋅ ⋅ ⋅,  4     π out of which only θ = ∈ (0, π) ⋅  4  

=

2

1

1/ 2 −1

0

π⋅ 0 π ⋅1 dx + ∫ sin dx 2 2 1

=0+ x

2

0

π [ x] π [ x] dx + ∫ sin dx 2 2 1



3/ 2

π

2

∵[ x] =  

1

∫ 1dx + ∫ 2dx + ∫ 3dx

74. (b) ∫ |cos θ − sin θ | dθ

 1

0

− 1/ 2

– ( x)



1

π 2

∫ 0 ⋅ dx +

 tan − 1 3 z  π π π tan − 1 3 = ⋅   = 3 2 3 2 3 3   0

=

1/ 2

− 1/ 2

= – 2 ( x)

π/ 4 dx π sec 2 x = dx 2 ∫ 1 − tan x 2 0 1 + 3 tan 2 x 0 2− 2 1 + tan x 1 π dz =  [Put tan x = z ⇒ sec2x dx = dz] 2 ∫0 1 + 3 z 2

π 2

∫

(−1)dx +



  x 1 is an odd function while is even  ∵ 2 − cos 2 x 2 − cos 2 x   =

0

(−2)dx +

−1

π/ 4

=0+

∫

=

265

x+π/4 dx ∫ − π / 4 2 − cos 2 x

∫ [x ] 2

−2

2

dx = 2

∫[x ] 2

dx 

[ integrand is even]

0

2 3 1 = 2  ∫ [ x 2 ] dx + ∫ [ x 2 ] dx + ∫ [ x 2 ] dx +  0 1 2

2



∫ [ x ] dx  2

3



∵[ x 2 ] = 0 if 0 ≤ x < 1; 1 if 1 ≤ x < 2 ;    2 if 2 ≤ x < 3 ; 3 if 3 ≤ x < 2  



1 = 2  ∫ 0 dx +  0 = 2 ( x)

2 1

2

3

2



2

3



∫ 1 dx + ∫ 2 dx + ∫ 3 dx  1

+ 4 ( x)

(

3 2

+ 6 ( x)

2 3

)

= 10 − 2 3 − 2 2 . π

76. (b) |sin x + cos x | dx ∫ 0

−1 if

3π / 4

=

∫ 0

|sin x + cos x | dx +

π

∫

3π / 4

|sin x + cos x | dx

∵sin x + cos x = 0 ⇒ cos x = − sin x    ⇒ tan x − 1 ⇒ x = 3π ∈ (0, π)    4

Definite Integral and Area

− 1/ 2

π/ 4

70. (a)

266

3π/ 4

=

∫

π

∫

(sin x + cos x) dx −

3π/ 4

0

Objective Mathematics

∵ x  = 0 if 0 ≤ x < 1; 1 if 1 ≤ x < 4;       2 if 4 ≤ x < 9 ⋅ ⋅ ⋅ ; (n −1) if (n − 1) 2 ≤ x < n 2  

(sin x + cos x) dx

= (− cos x + sin x)

3π / 4 0

π

− (− cos x + sin x)

3π / 4

 1 1     1 1  + + =   = 2 2 .  +1 − 1 −  2 2 2 2       1

−1

0

1

−2

−2

−1

0

∫ [ x + 1] dx = ∫ [ x + 1] dx + ∫ [ x +1] dx + ∫ [ x + 1]

77. (a)

−1

0

1

∫ (−1) dx + ∫ 0 ⋅ dx + ∫1

−2

−1

dx = – ( x)

dx

∫

+ ( x)

−2

100 π

1 − cos 2 x dx =

1 0

= 0.

1

4

( n − 1)2

2

∫ |sin x |

π

10 π

∫ |sin x |

dx

0

[ sin x ≥ 0 in [0, π]]

n

∫ [t

83. (b)

1

2

−1

0

1



=

− 2 if −1 ≤ x < 0; −1 if 0 ≤ x < 1  0 if 1 ≤ x < 2 

0

1

2



80. (d) Let I =

0

20 π

∫

1

= – 2 ( x)

|cos x | dx = 2

∫

− 20 π

∫ |cos x |

∴  I = 2⋅ 20

0 −1

− ( x) 0 = – 3.

|cos x | dx



= 40 π/ 2

= 40

∫ 0

dx π/ 2

π

0 π

π/ 2

∫ |cos x | dx + 40 ∫ |cos x |

= 40[sin x]



= 40 (1 – 0) – 40 (0 – 1) = 80.

∫  x  dx = 0

n

] dt + ⋅ ⋅ ⋅

∫

2

∫ 0 dt +

∫ 1 ⋅ dt +

(

) (

n

3

∫ 2 dt + ⋅ ⋅ ⋅

1

2

2 −1 + 2

3− 2

∫

(n − 1) dt

)

+ ⋅ ⋅ ⋅ (n − 1)

(

∫ ( x − [ x ])

).

1

∫ ( x − [ x ])

dx = 100

0

0

1

= 100

1

∫ x dx − 100 ∫ [ x] dx 0

4

9

0

1

4

x

∫ f (t )

85. (b) Let φ (x) =

∫  x  dx + ∫  x  dx + ∫  x  dx ∫

( n − 1)2

 x   dx

n − n −1

84. (a) Since x – [x] is a periodic function of period 1 100

dt, then

a

−x

φ (– x) =

∫ a



f (t ) dt = –

x

∫

    

n −1

1  x2  = 100   − 100 ∫ 0 dx = 50.  20 0

1

[t 2 ] dt

n −1

1

− 40 [sin x]

+ ⋅ ⋅ ⋅ +

2

2

0

n2



3

∫ [t

(

dx

π π/ 2



] dt +

= n n − 1 + 2 + 3 + ⋅⋅⋅ + n

∴

π/ 2

2



cos x dx − 40 ∫ cos x dx π/ 2 0

2

∫ [t 1

0

1

0

n2

] dt +

1

=

 [ integrand is even] Since | cos x | is a periodic function of period π,



2

0

π

sin x > 0 in (0, π)]

∵[t 2 ] = 0 if 0 ≤ t < 1; 1 if 1 ≤ t < 2  2 if 2 ≤ t < 3 ; 3 if 3 ≤ t < 2 ⋅ ⋅⋅;   (n −1) if n − 1 ≤ t < n 

= 1 ⋅ 20π

dx

0

] dt

∫ [t



2

∫ (− 2) dx + ∫ (−1) dx + ∫ (0) dx

−1

∫ sin x

0

=

0

∵[ x − 1] =  

π

= – 9 [cos x] = – 9 (– 1 – 1) = 18.

∫ [ x − 1] dx = ∫ [ x − 1] dx + ∫ [ x − 1] dx + ∫ [ x − 1] dx



0

[

0

−1

∫ |sin x | dx = 9

π 0

1

2 [− cos x]0π = 200 2 .

2

π



2 ∫ sin x  dx

0

79. (a)

dx = 9

π

2 ∫ |sin x | dx = 100



(n − 1) dx

∫

n (n + 1) (2n + 1) n (n − 1) (4n + 1) = . 6 6

π

Since | sin x | is a periodic function of the period π

81. (b)

0

82. (a) Since | sin x | is a periodic function of period π,

0

= 100

n2

∴

0

∴   I = 100

9

= n3 –

0

100 π

78. (c) Let I =

−1

4

= 1 (4 – 1) + 2 (9 – 4) + ⋅ ⋅ ⋅ + (n – 1) [n2 – (n – 1)2] = – 1 – 4 – 9 ⋅ ⋅ ⋅ – (n – 1)2 + (n – 1)⋅ n2 = – (12 + 22 + 32 + ⋅⋅⋅ + n2) + n3

∵[ x + 1] = −1 if − 2 ≤ x < − 1; 0 if −1 ≤ x < 0;   1 if 0 ≤ x 0



π/ 2 π/ 2  π = 400  ∫ sin 4 x + ∫ cos 4  − x  dx  2   0 0 



= 800 ⋅ 

⇒ (x + 2) (x + 1) x (x – 1) > 0 ⇒ x ∈ (– ∞, – 2) ∪ (– 1, 0) ∪ (1, ∞). 96. (b)

−1

− 100 π

2

x

100 π

98. (b)

 tan − 1 x 2    1 x2  lim  =  sin x 2  2 x→0  2   x 



x

Clearly, f (x) is continuous as well as differentiable in the interval [– 1, 1]. Also, f ′ (x) = | x + 1 | ⇒ f ′ (x) is continuous in [– 1, 1] but differentiable everywhere except at x = – 1.

[Using L’ Hospital’s Rule)



− x2 −x ; 2

∫ | x + 1| dx + ∫ | x + 1| dx

 x2  = −  + x   2 

d 2 (tan − 1 t ) 2  2 ⋅ (x ) t=x dx = lim    x→0 sin t  4 ⋅ d ( x 4 )   t = x dx

−1

−1

−2

t ) 2 dt

0

x4

=

−1

0



−2

=

∫ (tan

x→0

∫ − ( x + 1) dx

When – 1 ≤ x ≤ 1

6 x (1 + x12 − 6 x12 ) 6 x (1 − 5 x12 ) = . 12 2 (1 + x ) (1 + x12 ) 2 x2

= 20 (1 + 1) = 40.

97. (a), (b), (c)  When – 2 ≤ x ≤ – 1

 (1 + x12 ) ⋅ 2 x − x 2 ⋅ 12 x11  ∴ f ′′ (x) = 3   (1 + x12 ) 2  

94. (d) lim

   π ∵ in  0, 2  both sin x and cos x are positive     



 1  d 3 1 ⋅ 3x2 ⇒ f ′ (x) = 1 + t 4  3 ⋅ dx ( x ) = 1 + x12  t = x

=

dx

∫ (sin x + cos x)

π/ 2

10π

∫ (|sin x | + |cos x |) dx

=



99. (a)

∫

x dx = 800 ⋅

3⋅1 π ⋅ = 150 π. 4⋅ 2 2

4

∫ f (4 − x) φ (4 − x)

dx

0

4

=

0

∵|sin x | + |cos x | is a periodic    π   function of period   2

f ( x) φ ( x) dx =

0

π/ 2

∫ (|sin x | + |cos x |) dx = 20 ∫ (|sin x | + |cos x |) dx 0

4

0

4

0

 π 20   2

∫ sin

   

∫ f ( x) ⋅ (3 − φ ( x))

dx

0



∵ f ( x) = f (4 − x)  and φ (x) + φ (4 − x) = 3  

0

⇒ 2 I = 3 ⋅ 2 100. (b) F(t) =

∫

t

0



t

∴ a2 – a1, a3 – a2, a4 – a3 are in H⋅P.

f (t − y ) g ( y ) dy

∫e

=

∴ I = 3.

t−y

0

t

⋅ y dy = e  ·  ∫ e t

0

−y

⋅ y dy



= et ( −te − t − 0 ) − ( e 

) 

−y t

0

π a a + sin a + cos a 101. (b) We have, ∫ f ( x) dx = 2 2 2 0 Differentiating w.r.t a, we get

f (a) = a +

1 π (sin a + a cos a ) − sin a 2 2

π π π 1 π 1 + − Put a = ;f   = = . 2 2 2 2 2 2 1 1 102. (a) We have, 2 f (x) + 3 f    = –2 x  x 1 ⇒ 2 f   + 3 f (x) = x – 2  x Solving the above two equations, we get − 2 3x 2 + − 5x 5 5

f (x) = ∴

2

∫ f ( x)

2

 −2

∫  5 x +

dx =

1

1

 −2 3x 2  =  5 log x + 10 − 5 x    2

3x 2  −  dx 5 5 2

6 4  3 2  −2 log 2 + −  −  −  =  5 5   10 5   5 1  −2 log 2+  . =  2  5

103. (c) an – an – 1=

π/ 2

∫ 0

= −1 2 =–

π/ 2

∫ 0

π/ 2

∫ 0

π/ 2

=–

∫ 0

cos 2 nx − cos 2 (n − 1) x dx, n ≥ 2 sin x

cos 2 (n − 1) x − cos 2 nx dx sin x

sin (2n − 1) x sin x dx sin x

x + cot n + 2 x) dx, n ≥ 2

∫ cot

n

x cos ec 2 x dx π/ 4

 − cot n + 1 x  −1 =  =  1 n + n +1  0 ∴ a2 + a4, a3 + a5, a4 + a6 are in H⋅P. 105. (a)

3 log (sin x 3 ) dx = x 1

27

3

∫

∫

1

dz   3  Putting x = z ⇒ 3dx = x 2   

 27

=

∫ φ ( z)

dz = [φ ( z )]1 = φ (27) – φ (1). 27

1

∴ k = 27. 16

106. (b) ∫ log x  dz = 1

2

2

∫ log x d ( x ) 4

=4

1

∫x

3

log x dx

1

2

4 4 = 4 log x ⋅ x − 1 ⋅ x dx  x ∫x x  1

2

2  4 x4   x4 x4  = 4  log x −  =  x log x −  4 1 16 1  4

 = (16 log 2 – 4) –  0 − 

1  4

15   = 16 log 2 −  . 4 

⇒ f ′ (x) = 2ax + b and f ′′ (x) = 2a. We are given f (0) = c = 1, f ′ (0) = b = – 2 and f ′′ (0) = 2a = 6 ⇒ a = 3, b = – 2 and c = 1. ∴ f (x) = 3x2 – 2x + 1. 2

∴

∫ f ( x)

dx =

−1

2

∫ (3x

2

− 2 x + 1) dx

−1

= [ x 3 − x 2 + x]2− 1



= (8 – 4 + 2) – (– 1 – 1 – 1) = 6 + 3 = 9. π/ 2

108. (c) In =

∫ sin 0

n

x dx =

π/ 2

∫ sin

n −1

x sin x dx

0

π/ 2

sin (2n − 1) x dx

log (sin z ) dz z

107. (c) Let f (x) = ax2 + bx + c

1



n

0

= et [–te–t – (e–t – e0)]

2

∫ (cot 0

π/ 4

=

= et [–te–t – e–t + 1] = [1 – e–t (1 + t)] et = et – (1 + t). a

π/ 4

104. (c) an + an + 2 =

t −y t −y   = et ( − ye )0 − ∫0 1 ( −e ) dy   



1 (2n − 1)

=–

269

∫

π/ 2

=  cos (2n − 1) x   (2n − 1)  0

f ( x) dx – I

n −1 =  − sin x cos x  0 +

π/ 2

∫ (n − 1) sin 0

n−2

x cos 2 x dx

Definite Integral and Area

4

=3

270

π/ 2

∫

= (n – 1)

sin n − 2 x (1 − sin 2 x) dx

113. (c)

0

Objective Mathematics

π/ 2

∫ sin

= (n – 1)

n−2

x dx − (n − 1)

0

π/ 2

∫ sin

n

x dx

⇒ n In = (n – 1) In – 2 ⇒ In =

n −1 In – 2 n

⇒ n (In – 2 – In) = In – 2. Also, In : In – 2 = (n – 1) : n and In – 2 > In.

∴

∫ log

x

π 3

3e x cos (3e x ) dx =

π/3

∫ 3 cos 3t  dt

π/6

π 6

= [sin 3t ]π / 6 = sin π – sin π/3

π 2

∫ g ( x)

∫ f '( x) dx

dx =

= lim x→0



Again applying L’Hospital rule

∫

111. (b) I =

a

0 + 2 sec 2 0 = 1. 0 + 2 cos 0

114. (c) We have π/3

a  b tan x + c  dx = 0  |tan x | + + sec 3 1 x  − π/3  π/3

a   |tan x | + c  dx = 0 3   − π/3

∫

−1

[ f ( x)]−1 2

= f (2) – f (– 1)

⇒ ⇒

x f ( x) dx b

= ∫ ( a + b − x ) f ( a + b − x ) dx a

b

= ∫ ( a + b − x ) f ( x) dx a

[Since given f (a + b – x) = f (x)]



2 x ⋅ 2 sec 2 x 2 ⋅ tan x 2 ⋅ 2 x+ 2 sec 2 x 2 − x sin x + cos x + cos x

= lim x→0

= 4 – 2 [As f (– 1) = 2 and f (2) = 4] = 2. b

= (a + b)

∫

b

= (a + b)

∫

b

a

a

⇒

f ( x) dx – I

∫

b

a

f ( x) dx

a +b b ∴ I =   ∫ f ( x) dx .  2  a 5

112. (b) e f [ φ ( x )] ⋅ f ' [φ ( x)] ⋅ φ ' ( x) dx ∫

π/3

a 3

∫

− π/3

2a 3

π/3

2a 3

π/3

∫

e f ( t ) ⋅ f ' (t ) dt [Putting φ (x) = t ⇒ φ′ (x) dx = dt]

= 0. [ φ (3) = φ (5)].

∫ tan x dx + 0

2π c=0 3

2a 2π −a (log 2 − 0) + log 2 . c=0 ⇒c= 3 3 π

 log x  115. (b) When 1 < x < e3,   =0  3   log x  and when e3 < x < e6,   = 1.  3  ∴

e6

 log x  ∫1  3  dx =

= 116. (c) I =

∫

1

0

φ (5)

f (t ) = e  φ ( 3) = e f[φ (5)] – e f [φ (3)]

0

2π c = 0 [ | tan x | is even] 3

⇒

φ ( 3)



∫ |tan x | dx +

2π =0 3

2a 2π π/3 c=0 [log | sec x |]0 + 3 3

3

=

|tan x | dx + c ⋅

⇒

b

f ( x) dx − ∫ x f ( x) dx a

∴    2I = (a + b)

φ(5)

sec 2 x 2 ⋅ 2 x x cos x + sin x

π/3   tan x tan x is an odd function ∴ ∫ dx = 0  ∵  1 + sec x  − π / 3 1 + sec x

2

−1

=

Applying L’Hospital rule

⇒

110. (a) Since g (x) = f ′ (x) ∴

x→0

∫

= – 1.

2

sec 2 t dt  0   form  x sin x  0 

0

=

109. (b) Put e = t ⇒ e dx = dt log

x2

0

∴ In = (n – 1) In – 2 – (n – 1) In

x

∫ lim



e3

e6

1

e3

e3

e6

 log x   log x  ∫1  3  dx + ∫3  3  dx e

6 3 ∫ 0 dx + ∫ 1 dx = (e – e ).

x (1 − x) n dx

Let 1 – x = z ⇒ – dx = dz

∴ I =

∫

0

1

(1 − z ) z n (− dz ) =

∫

1

0

(1 − z ) z n dz

n

2π

dx 121. (a) Let I = ∫ = sin x 0 1+ e

− zn + 1) 1

1 1 z  z − =   = n +1 − n + 2 .  n + 1 n + 2 0 n +1

117. (b)

n+2

π

∫ [sin x]  dx = ∫ [sin x] dx + ∫ [sin x]

π/ 2

π/ 2

π

3π / 2

π/ 2

π

∫ 0 dx + ∫

π

( − 1) dx = – [ x]

π/6

14 + 24 + 34 + ... + n 4 13 + 23 + 33 + ... + n3 − lim 5 n→∞ n n5

∫

1 r 1 1 ∑   − lim x lim n r = 1 n  n → ∞ n n → ∞ n

∑  n 

3

π/6

x  1 dx =   – 0 = . 5  5 0

2

∫ max {(1 − x), (1 + x), 2}

∴

1

l

=

1− l

 l

=

dx

∫ x sin [ x (1 − x)]

⇒ 2 I1 = I2.

1− l

0

0

1

1

0

0

∫ x dx − ∫ x dx + ∫ x dx − ∫ 0 dx −1

2 0

x 2

=

−

−1

0

2 1

x  x  + 2  −1 2 2

= 0

1 . 2

124. (b) S ince f (x) + f (– x) is an even function and g (x) – g (– x) is an odd function, therefore, { f (x) + f (– x)}⋅ {g (x) – g (– x)} is an odd function, therefore π/ 4

∫

{ f ( x) + f (− x)} ⋅ {g ( x) − g (− x)} dx = 0.

∫

f ( x) dx =

a/2

∫

f ( x) dx +

0

∫

a

∫

f ( x) dx

a/2

0

=p–

f (a − z ) dz

a/2

[ Putting x = a – z in the second integral so that dx = – dz]

dx

=p+

dx

∫ sin [(1 − x) x] dx − ∫ x sin [(1 − x) x]

1− l

1

0

a/2

∫ 0

b  b  ∵ ∫ f ( x) dx = ∫ f (a + b − x) dx  a  a  l

( x) dx

∫ ( x + | x |) dx + ∫ ( x − [ x]) dx

−1

a

1− l

∫ (1 −x) sin [(1 − x) x]

0

−1

125. (a)

2 2 =  x − x  + [2 x]1− 1 +  x + x  = 9. 2 − 2 2 1  

120. (b) We have, I1 =

−1

0

dx

2

l

−1

=

1

−1

1

0

2

−1

0

− π/ 4

∫ (1 − x) dx + ∫ 2 dx + ∫ (1 + x)

−2

1

=

−2 −1

1 . 3

=

∫ f ( x) dx = ∫ f ( x) dx + ∫ f

1

and 1 – x > 1 + x ∴ max {(1 – x), (1 + x), 2} = 1 – x. For – 1 < x < 1, we have 0 < 1 – x < 2 and 0 < 1 + x  1– x ∴ max {(1 – x), (1 + x), 2} = 1 + x.

t dt

123. (a) Since | x | ≥ – [x] for all position x and | x | ≤ –­ [x] for all negative x, therefore

r =1 5

1 1 x 4 dx − lim ⋅ × ∫ x 3 0 n→∞ n

1

0

r

n

2

0

= tan t 0

119. (c) For – 2 ≤ x ≤ – 1, we have 1 – x ≥ 2

=

π/6

0

= nlim →∞

= 2π

0

2

n→∞

=

∫ 1 dx

∫ sec x d ( x − [ x]) = ∫ sec

∴

3



 π 122. (b) Let x – [x] = t. In the interval 0,  , [x] = 0  6 ∴ x = t

lim

n

− sinx

0

∴ I = π .

3π / 2 π

−π  3π  − π = . =–  2 2  

118. (d)

dx

∫ 1+ e

2π

1 + esinx ∫o 1 + esin x dx =

∴ 2 I =

dx

=

sin x

0

3π / 2

sin ( 2 π − x )

0

2π

3π / 2

=

∫e

∫ 1+ e

2π

dx

esinx dx +1

2π

=

2π

dx = I2 – I1

a/2

=p+

∫

 f (a − z ) dz  ∵  f (a − x) dx 

0

=p+

a/2

∫

f ( x) dx

0

[

f (a – x) = f (x)]

= p + p = 2p.

b

∫ a

 ∵ 

271

1

0

a  f ( x) dx = − ∫ f ( x) dx  b  b

∫f a

b  ( x) dx = ∫ f ( y ) dy  a 

Definite Integral and Area

∫ (z

=

272

d esin x F(x) = ,x>0 dx x

126. (d)

131. (c) Let I =

Objective Mathematics

∫

64

1

64

Using eq. (1), we get

132. (b)

(

)

2 Then, f (– x) = log − x + 1 + x



(x +

= – log

1 + x2

)

5

)

∴

∫

(

log x + 1 + x 2

log 1/ 3

)

= – f (x)

 15  = 2 + loge   . 7 sin x dx 8 10 1 + x 19

19

≤

dx = 0.

5 3 128. (a) We have, I = ∫ (α sin x + β tan x + γ cos x)  dx −a

∫ sin

=α



5

a

∫ tan

x dx + β

−a

3

x dx + γ

−a

=

−a

=α×0+ β×0+ γ×2

∫ cos x

dx

b

∫ (a + b − x) f (a + b − x)

a

=

b

a

a

∫ [(a + b) − x] f ( x)  dx = (a + b) ∫ f ( x) dx

∴ I = 130. (c)

2π

∫

a+b 2

1 + sin

0

2π

=

a

b

∫ 0

sin 2

b

∫ f ( x)

∴ I = dx

∫

b

−x dx = a – b = – b – (– a) = | b | – | a |. a x

∫

(iii) If a < 0 < b, then 0



I =

∫

= ­–

a

x dx 2

= 4 [(1 + 1)] = 8.

b

| x| | x| dx = dx + ∫ x 0 x 0

b

a

0

∫ dx + ∫ dx

b

−x x dx + ∫ x a 0 x 0

∫

dx

= a + b = b – (– a) = | b |– | a |.

Hence (d) is correct answer. 1 135. (c) f (0) = a, f (1) = a + b + c, f   = a + b + c . 2 4 2

dx

x x  x x  = ∫  sin + cos  dx = 4  − cos + sin  4 4   4 4 0 

−1 − 7 19 1 1 1  x  = 10− 7 − 19− 7 < 10− 7 1 and y = – x + 1,  x < 1

f (sin 3 x + cos 2 x) dx

0

f (sin 3 x + cos 2 x)  dx

275

π ⋅2 2

Definite Integral and Area

=

∴ k = π.

0

155. (b)

π/2

∫ |sin x − cos x |

dx

0

π/4

π/2

0

π/4

∫ |sin x − cos x | dx + ∫ |sin x − cos x |

=

π/4

π/2

0

π/4

dx

∫ (sin x − cos x) dx + ∫ (sin x − cos x)

=–

dx

= [ cos x + sin x ]0 + [ − cos x − sin x ]π / 4 π/4

π/2

Also, y = 3 – | x | ⇒ y = 3 – x,  x > 0

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

(

)

and y = 3 + x,  x < 0 Solving y = x – 1 and y = 3 – x

156. (a), (b)  We have f (x) = A ⋅ 2x + B

⇒ x – 1 = 3 – x ⇒ x = 2

⇒ f ′ (x) = A ⋅ 2x log 2

and y = 3 – 2   i.e., y = 1 1 . log 2

f ′ (1) = 2 ⇒ 2 = 2 A log 2 ⇒ A = 3

∫



0

AB2 = (2 – 1)2 + (1 – 0)2 = 1 + 1 = 2 ∴ AB = 2 BC2= (2 – 0)2 + (1 – 3)2 = 4 + 4

 1  ⋅ 2 x + B  dx = 7 f ( x) dx = 7 ⇒ ∫  log 2  0 3

∴ BC = 2 2 Area of rectangle

3

 1  ⋅ 2 x + Bx  = 7 ⇒  2  (log 2) 0

ABCD = AB × BC =

160. (c) Since f ′(x) = f (x) and f (0) = 1

8 1 + 3B − =7 2 (log 2) (log) 2

⇒

therefore let f (x) = ex Also, f (x) + g(x) = x2 Again,

7 (log 2) 2 −1 ⇒ B = 3 (log 2) 2 

∫

4

2

4

2

3

=

1

2

3

4

1

2

3

∫ log 1 dx + ∫ log 2 dx + ∫ log 3 dx

= log 2 + log 3 = log 6. 158. (c) f (x) =

x

∫ cos t

2

dt =

1/ x

1/ x 1/ x

=–

2 ∫ cos t dt + 0



0

∫ cos t

2

dt +

x

∫ cos t

2

dt

0

x

∫ cos t dt . 2

0

d 1 2 ∴  f′(x) = – cos t 2  1   t = dx  x  + cot t  t =   x 1 1 1 = 2 cos 2 + cos x. x x 2 x

x

d dx

( x)

f ( x) g ( x) dx =

∫

1

0

∫ e (x 1

0

x

∴ g(x) = x2 – ex 2

− e x ) dx =

∫ (x e 1

0

2 x

− e2 x )

x 2e x dx − ∫ e 2 x dx 1

0

1 1 2x 1 2 x x =  x e − ∫ 2 xe dx  − e  0 0 2 1 1 2 2 x x x 0 =  x e − 2 ( xe − e )  0 − (e − e ) 2 1 1 =  x 2e x − 2 xe x + 2e x  − (e 2 − e0 ) 0 2 1 2 1 2 x 1   =  ( x − 2 x + 2) e  − e + 0 2 2

∫ log [ x] dx + ∫ log [ x] dx + ∫ log [ x] dx

1

=

3

1

0

1 7 (log 2) 2 − 1 . ∴ A = and B = log 2 3 (log 2) 2  157. (a) ∫ log [ x] dx =

2 × 2 2 = 4 sq. units

= [(1 – 2 + 2) e1 – (0 – 0 + 2) e0] – 1 e 2 + 1 2 2 = [e – 2] – 1 e 2 + 1 = e – 1 e 2 − 3 . 2 2 2 2 n  1 1 1  1 = nlim + +⋅ ⋅ ⋅ + 161. (b) lim  ∑  → ∞ n→ ∞ +r n + 1 n + 2 n + n n r = 1   

= lim

n→∞

n

∑

r =1

1 1 1 1 1 ⋅ =∫ dx = [log (1 + x)]0 = log 2. n 1+ r 01+x n

276

1 n2 n2 1 162. (a) nlim + + ⋅⋅⋅ +  →∞  + 3 3 8n   n (n + 1) (n + 2)

Objective Mathematics



 n2 n2 n2 n2  = nlim + + + ⋅⋅⋅ +   3 3 3 →∞ (n + 1) (n + 2) ( n + n)3   (n + 0) n

n

= nlim →∞

n

2

∑ (n + r )

r=0

= nlim →∞

3

∑

r=0

1 1 ⋅ n  r 3 1 +   n

1

 −1  dx −1  1  3 ∫0 (1 + x)3 =  2 (1 + x)2  0 = 2  4 − 1 = 8 . 163. (c) lim 1

1 1 =  tan − 1 x  + 1 log (1 + x 2 )   0 2  0 = π + 1 log 2 . 4 2  1 1 1  + + + ⋅⋅⋅ + 166. (b) nlim →∞ 2 2  2n − 1 4n − 2 6n − 32

 1  + = nlim →∞  2n − 12

=

= lim

n→∞

 1  1 1 1   + + +⋅ ⋅ ⋅ +  n 2 n2 − 1 n 2 − 22 n 2 − (n −1) 2 

=

 1 1 1 = nlim  + + + ⋅⋅⋅ →∞  n 2 − 02 n 2 − 12 n 2 − 22

=

  2 2 n − (n − 1) 



+ n −1

= nlim →∞

=

1

∑

n −r 2

r=0

dx

1

∫

1 − x2

0

n −1

2

= nlim →∞

∑

r=0

1 1 ⋅ n 1 − r 2 / n2

1 = sin − 1 x  = π . 0 2

164. (a) Required area =

∫

3

0

4n − 2

n

n→∞

1

1

1

2x − x

0

0

2

dx =

n→∞

1 n

∑

r =1

  2n ⋅ n − n  1

2

1 2r r 2 − n n2

[1 − ( x − 2 x + 1)] 2

1

−1 = sin ( x − 1)  0

1 − ( x − 1) 2

 −π  π =0–  .  = 2 2  



 n 1 n n + 2 + ⋅⋅⋅ +  167. (a) nlim  2 2 + 2 2 2 →∞ 2n  n +2 n +3  n +1  n n n n  + 2 + ⋅⋅⋅ + 2 = nlim    2 2 2 2 →∞  2 n +3 n + n2  n +1 n + 2 n n 2 = lim ∑ n→∞ r =1 n + r r =1 n

= nlim ∑ →∞

( y1 − y2 ) dx

+ ⋅⋅⋅ +

dx

∫ 0

dx

1

= lim

2

1

1

∫ ∫

2r n − r

r =1

6n − 3

2

n

1

∑

1

+

2

1  n 

2

1 π −1 =  tan x  0 = . 4

168. (c) lim

n→∞

r

n

∑n r =1

2

r

n

= lim

n→∞

⋅ sec 2

r 1+ 2 n 2

⋅

1 = n

1

1

∫1 + x

2

dx

0

r2 n2

∑  n ⋅ sec r =1

1

2



r2  1 =  n2  n

1

∫ x sec

2

x 2 dx

0

1

= =

∫

3

0

(2 x − x 2 ) − (− x) dx =

∫

3

0

(3 x − x 2 ) dx

1 sec 2 t dt [Putting x2 = t ⇒ 2x dx = dt] 2 ∫0

= 1 [tan t ]10 = 1 tan 1. 2 2

3

 3x 2 x3  −  = 27 – 9 = 9 =  3 0 2 2  2  n +1 n +2 n+3 165. (a) lim  2 2 + 2 + 2 +⋅ ⋅ ⋅ + 2 2 n→∞  n +1 n + 2 n + 3

169. (a) lim

n→∞

1  n

 n +1 n+2 n+3 n+n  = nlim + 2 + ⋅⋅⋅ + 2 →∞  2 2 + 2  2 2 n +2 n +3 n + n2   n +1 n

= nlim ∑ →∞

r =1

1+ x ∫0 1 + x 2 dx = 1

=

1 1+ r / n 1 n+r lim ⋅ 2 2 2 2 = n→∞ ∑ n n +r r = 11 + r / n 1

dx

∫1 + x 0

1 2x dx 2 ∫0 1 + x 2 1

2

+

1

=

∫ sin 0



1 n

n

∑ sin r =1

2k

2k

rπ 2n

πx 2 dx = 2 π

π/ 2

∫ sin

2k

t dt

0

πx 2    Putting 2 = ⇒ dx = π dt   

=

2 (2k − 1) (2k − 3) ⋅ ⋅ ⋅ ⋅ 1 π ⋅ ⋅ π 2k (2k − 2) ⋅ ⋅ ⋅ ⋅ 2 2

=

[(2k − 1) (2k − 3) (2k − 5) ⋅ ⋅ ⋅ 1] [2k ⋅ (2k − 2) ⋅ ⋅ ⋅ ⋅ 2] 2k [k (k − 1) (k − 2) ⋅ ⋅ ⋅ 1] [2k ⋅ (2k − 2) ⋅ ⋅ ⋅ 2]

=

(2k ) ! . 22k ⋅ (k !) 2

170. (b) Let y = lim

n→∞



(n !)1/ n n! = lim  n  n→∞ n n  

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ n  = nlim  →∞  nn  



n→∞

= [ x log x ]0 − ∫ x ⋅ 1

0

∴ y = e– 1 =

n→∞

1 n



∴ y = e

= lim

n→∞

2

  dx 

log 2 +

π −2 2

= elog 2 ⋅ e

π−4 2

= 2e

π−4 2

.

n

1

∑r

n p +1

p

r =1

1 r ⋅  n n

n

= lim ∑ n→∞

r =1

1

1

 1  + 174. (b) lim Sn= nlim →∞ n→∞  4n 2 − 0 1

+

 n −1

= lim

n→∞

dx

p

p +1 x p + 1  1 p  x x a dx =  ∫0  p + 1p + 1 = p + 1 . 0 0

1/ n

∑

r=0

4n − 2 2

2

1 4n 2 − 12 +⋅ ⋅ ⋅ +

  4n − (n − 1)  1

2

2

1 ( 2n) 2 − r 2

0

x dx = [ x log (1 + x)]0 − ∫ 01+ x 1

1

(1 + x) − 1 dx 1+ x 0

=

n −1 1 lim, ∑ 2 n→∞ r=0

=

1 2

1

∫

 1  = log 2 – ∫ 1 −  dx x 1 + 0 1

= log 2 – [ x − log (1 + x)]10 = log 2 – [(1 – log 2) – 0] = 2 log 2 – log e = log 4 ⋅ e ∴ y = 4 . e

1

∫ 0

1  r  1−    2n 

=

⋅

1 n

1 π dx= 1 ⋅  2 sin − 1 x  = .   6 x 2 2    0 1−   2 2

175. (d) Let f (x) =

1/ n

2

1

sin 8 x ⋅ log (cot x) cos 2 x

π  ⇒ f  − x  = 2 

2 2 2 172. (c) Let y = lim 1 + 1  1 + 2  ⋅ ⋅ ⋅ 1 + n      n→∞ 2  2  n  n   n2  

⇒ log y = lim 1 n→∞ n

dx

1 p p p 173. (a) nlim → ∞ n p + 1 [1 + 2  + ⋅ ⋅ ⋅ + n  ]

2   1   n   1 + n   1 + n  ⋅ ⋅ ⋅  1 + n         

= log 2 –

2

0

1 −1 π = log 2 – 2  x − tan x  0 = log 2 + – 2. 2



1 ⇒ log y = lim × log  n→∞ n

∫ log (1 + x)

1

0

1

r  log 1 +  = ∑ n  r =1



∫ 1 − 1 + x

= log 2 – 2

=

1

∫ log (1 + x )

0

1 dx = 0 – 1 = – 1. x

n

1

1 2x =  x log (1 + x 2 )  − ∫ x ⋅ dx 0 1 + x2 1

1 . e



 r2  log 1 + 2  = ∑ n  r =1  n

1/ n

1 2   n  171. (d) Let y = nlim  1 + n   1 + n  ⋅ ⋅ ⋅  1 + n   →∞       

1 = lim n→∞ n

= lim

1

1 n

1







n 1 2  log n + log n + ⋅ ⋅ ⋅ + log n    1 n r 1 log = ∫ log x dx = nlim ∑ →∞ n r =1 n 0

= lim

    12  22  n2  log 1 + 2  + log 1 + 2  + ⋅ ⋅ ⋅ log 1 + 2   n  n  n     

1/ n

n 1 1 2 3 ⋅ log  ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅  ⇒ log y = nlim →∞ n n n n n



 π  π  sin 8  − x  ⋅ log cot  − x     2  2  π  cos 2  − x  2 

sin (4 π − 8 x) log tan x − sin 8 x ⋅ log cot x = =– f (x) cos 2 x cos ( π − 2 x) π/2

∴

∫ 0

sin 8 x log cot x dx = 0. cos 2 x

277

2k (2k − 1) (2k − 2) (2k − 3)  2 ⋅ 1 2k ⋅ [(k ( k − 1) (k − 2)1] ⋅ 2k [k ⋅ (k − 1)(k − 2) 1]

Definite Integral and Area

=

278

176. (a)

π

∫

f ( x) ⋅ g ( x) dx =

0

π

∫

π/ 2

f (π − x) g (π − x)

Objective Mathematics

π

∫ [1 − f ( x)] ⋅

=

 −π  = π log (a + b) + 2 ⋅  log 2    2  

[1 – g (x)] dx (given)

0

a +b = π log (a + b) – π log 2 ⋅ = π log  .  2 

π



∫ [1 − f ( x) − g ( x) + f ( x) ⋅ g ( x)]

=

dx

0



π

π

π

0

0

0

∫ [ f ( x) + g ( x)]

∫1 ⋅ dx

dx =

0

2

∫

−2

x

= π. =

0

0

1

−2

0

=

2

x + ∫ (2 − 1 + x − 1) dx



1

0

1

   x x x2  x =  2 x − 2 − 2 log 2  +  2 log 2 − 2   − 2  0 2



0

1

2

l

l+k

0

1

l −1

l

∫ [ x] dx + ∫ [ x] dx + ⋅ ⋅ ⋅ + ∫ [ x] dx + ∫ [ x] dx

= 0 + 1 + 2 + ⋅ ⋅ + (l – 1) + l ⋅ k

x x ∫ (1 − 2 + 1 − x) dx + ∫ (2 − 1 + 1 − x) dx

=

∫ [ x] dx

=

0

f ( x) dx



l+k

∫ [ x] dx

∴

π

π

⇒

180. (a), (b)  Let [x] = l then x = l + k, 0 ≤ k < 1.

∫1 ⋅ dx − ∫ [ f ( x) + g ( x)] dx + ∫ f ( x) ⋅ g ( x) dx

=

dx

0

0



177. (b)

∫ log sin x

= π log (a + b) + 2

1 1 (l – 1) l + lk = [x] ([ x] − 1) + [ x] ( x − [ x]) 2 2

1  = [x]  ([ x] − 1) + ( x − [ x])  2  ∴ A = [x] – 1 1

181. (a) Let f (a) =

and B = x – [x].

a

x −1

∫ log x

dx

0

2

  x2 +  2 x log 2 + − 2 x  2  1



9 log2. 4

=5+

n→∞

1

=

∫x

r 1 n  n   r2  = nlim ∑  3 3 3  →∞ n r =1  r  r +n    +1 n

e/2

∴

x2 1 dx = 1 log ( x 3 + 1)    0 +1 3

π/ 2

 ax 2 + bx + c  179. (b) Let I = ∫ log  2 ⋅ (a + b) |sin x | dx  ax − bx + c  − π/ 2 =

∫

− π/ 2 π/ 2

+

∫

−π/ 2

 ax 2 + bx + c  log  2  dx  ax − bx + c  log (a + b) dx +

π/ 2

∫

= 0 + log (a + b) [ x]

+2

π/ 2

∫ log |sin x |

|log 2 x | dx =

dx

  ax 2 + bx + c  ∵log  2  is an odd function and  ax − bx + c   log | sin x | is an even function 

    

e

1 | log t | dt 2 1∫/ e

e 1  1  ∫ |log t | dt + ∫ |log t | dt  2 1/ e 1 



=



e 1  1 − log t dt + log t dt   = ∫ ∫ 2  1/ e 1 



=

1 (− t log t ) 2



=

1 1 1 1  log + 1 − + e − e + 1 2  e e e 



=

1 1  −2   + 2  = 1 −  .=  2 e e  

log | sin x | dx

0



∫

−π/ 2 π/ 2 −π/ 2

= log (a + 1) + C.

1 dt 2

1/ 2 e

1 1 (log 2 – log 1) = log 2. 3 3

π/ 2

1

∫ a + 1 da + C

182. (b) Put 2x = t ⇒ dx =

3

0

=

r =1

1

1 xa + 1 x a log x dx = ∫ x a dx = = a +1 0 log x 0

If a = 0, then f (a) = 0, ∴ C = 0. Hence, f (a) = log (a + 1).

2

n

0

∴ f (a) =

 12 22 1 + 3 +⋅ ⋅ ⋅ ⋅ +  178. (a) lim  3 3 3 n→∞ 1 + n + n n 2 2  = lim ∑

1

⇒ f ′ (a) = ∫

1 1/ e

+t

1 1/ e

e e + (t log t ) 1 − t 1  

183. (b), (c)  We have, 0 < x2 < 1, ⇒ e0 < e x 1

1

0

0

1

1

0

0

2

< e1

x x ⇒ ∫1 dx < ∫ e dx < ∫ e dx ⇒ 1 < ∫ e dx < e . 2

2

∫

=–2

0 π/ 2

0

⇒ sin α = 1 – 2 sin2α

∫ log sin θ

=–2

i.e., 2 sin2α + sin α – 1 = 0

b + 2T

∫

189. (a), (d) 

100

∫

2 x − [ x ] dx = k

100

∑

∫

k =1

∫

k −1

∑2

(1 − k )

∫ f ( z)

100

∑ 2− ( k − 1)

k =1

k

∫

k −1

= 2 x dx

∫ f ( x)

= 100 log 2. 

=

 −1 x − 1 −1 x + 1  ∫− 2  cot x + 1 + cot x − 1  dx 

∫  tan

−1

−2

dx. Hence, (a) and (d) are correct.

k

1 k k

190. (a)

 ⋅ (2k – 2k – 1) log 2

3

=

dz

a



k =1

3

f ( y + T) dy

b

100

186. (a)

∫

a+T

[Again, putting z = y + T so that dz = dy]



2 x − [ x ] dx

2 x − ( k − 1) dx =

∑ (2 − 1)

b+T

b

k −1

= log 2

∴

5

∫  2 x

⇒

π π 5π = ∫ dx = [3 − (2)] = . −2 2 2 2

3/ 2

1

1 k k

 + 2 x − 3 x1/ 2  dx  k

( x5 / 2 + x 2 − 2 x3/ 2 ) k

1

5

∫  2 x

k k

3/ 2

1

⇒ k +

x +1 x + 1 + cot − 1  dx x −1 x −1

1

(

= k+ k −2

(

k –20

2 ∫ (2 x + 3) dx ⋅ ∫ (3x + 4) dx

≤

2 − tan 2 z

=

0 be a fixed number. Suppose f is a continuous 2 2 function such that for all x ∈ R, f (x + T) = f (x). If I (a)

=

∫

(a)

T

0

f ( x) dx then the value of

3 I 2

3 + 3T

3

f (2 x) dx , is

∫

1/2

−1/2

 1 + x    [ x] + log   dx equals  1 − x   

1 2

10. The value of

(d) 2log π

−π

(a) π π (c) 2

cos 2 x

∫ 1+ a

x

∫ log (1 + tan x)

dx =

π log 2 4 (c) π log 2 tan x 8

17.

π/2

∫ log (tan x)

π log 2 8 π (d) log 3 8

(b)

dx =

0

(b) 0

(c) 1

π/4

(a)

(d) 6I

9. The integral

16.

0

(b) 2I

(c) 3I

(a) –

∫

1 2

dx, a > 0 is (b) aπ

(a) π 4 π (c) 2 18.

1

∫ sin 0

(d) 2π.

m m m   11. lim 1   1  +  2  +  n − 1   is equal to n→∞ n n  n   n 

−1

(d) 2

 2 x  dx =  1 + x 2 

π + log 2 4 (c) π + log 2 2

(a)

(b) 0

π − log 2 4 π (d) − log 2 2 (b)

∫ 0

(a) π ab π 2ab

(c) 20.

∫ 0

1

∫ |sin 2π x |

(a)

(b) π 4 (d) (a + b) π

0

∫x

−1

2

(b)

24.

1 e 2 e

(d) 1 –

dx = + 2x + 2 (b) π/4 (d) – π/4

(a) 0 (c) π/2 =

(a) 0 (c) 2

x4 + 1 ∫0 x 2 + 1 dx is 1

(c) 1 (3π + 4) 6

(d) 1 (3 + 4π) 6

26. If f (x) is an odd function of x then

∫

f (cos x) dx is equal to

− π/2 π/2

(a) 0

(b)

∫

f (cos x) dx

0

π/2

(c) 2

∫ 0

=

∞

(b)

π 24

(d) 0



1

dx

∫ log  x + x  1 + x

π log 2 2 (c) – π log 2 (a) –

2

= π log 2 2 (d) π log 2 (b)

  1 1 1 1  + + + ... 31. lim  2 2 2 2 n→∞  n n +n n + 2n n + (n − 1) n  (a) 2  2

(b) 2

(c) 2 – 2  2

(d) 2

1

x dx

∫ (1 − x)

5/ 4

2 –2

=

15 16 3 (c) – 16

(b) 1 (3 – 4π) 6

π/2

2

(a)

(b) 1 (d) 4

(a) 1 (3π – 4) 6



∫ 4 + 9x

3 (d) ( π − 8) 2

0

−2

25. The value of

dx

2/3

π3 4

0

32.

2

∫ | x | dx

29.

30.

(d) 0

22. The area bounded by the the curves y = xex, y = xe– x and line x = 1 is

23.

(b)

(a) π 12 (c) π 4

(b) 1 π

1 (a) 1 –  e 2 (c) e

( π3 − 8) 4

0

0

2 π −1 (c) π

(d) 2 : 1

3 (c) ( π − 16) 4

dx is equal to

(a)

(b) 1 : 1

28. The area in the first quadrant between x2 + y2 = π2 and y = sin x is

a sin x + b cos x dx = sin x + cos x

π 4 (c) (a + b) π 2

2 : 1

(c) 1 : 2

(d) πab

(a) (a + b)

21.

(a)

π (b) ab

2

π/2

27. The ratio of the areas bounded by the curves y = cos x and y = cos 2x between x = 0, x = π/3 and x-axis, is

dx = a 2 cos 2 x + b 2 sin 2 x

π/2

f (sin x) dx

(d)

∫

− π/2

f (cos x) dx

3 16 −16 (d) 3

(b)

33. Let f : (0, ∞) → R and F (x2) =

x2

∫

f (t ) dt .

0

If F (x2) = x2 (1 + x), then f (4) equals (b) 7 (a) 5 4 (c) 4 (d) 2.  1 1 1  34. lim  + + ... +  = n → ∞ 2n + 1 2n + 2 2n + n   1 (a) log e   3

2 (b) log e   3

3 (c) log e   2

4 (d) log e   3

303

π/2

Definite Integral and Area

19.

304

Objective Mathematics

35. The area of the region bounded by y = | x – 1 | and y = 1 is (a) 2 (b) 1 (c) 1/2 (d) None of these

42.

∫

π/2

0

sin 2θ d θ is equal to: sin 2θ + cos 2θ

(a) π/2 (c) π/4 36. The value of ∫ [2 sin x] dx, where [.] represents the great-

(b) 2π/3 (d) π/6 

2π

π

43. If g(x) =

est integer function, is 5π 3 5π (c) 3 (a) –

(d) – 2π

(b)

(c) 0 and – 4 π

(d) 4 and 0 π

2

x

1

cos 4 t dt , then g(x + π) is equal to:

(d) g(x)/g(π)

π (a) π and – 2 2

∫e

x

0

(a) g(x) + g(π) (b) g(x) – g(π) (c) g(x) g(π)

(b) – π

1 1 πx 37. If f (x) = A sin   + B, f '   = 2 and f ( x) dx ∫   2  2  0 2A then the constants A and B are, respectively = π

38.

∫

π 3 and 2 π

(b) e (e – 1)

(c) 0

(d) None of these

∫

π/2

0

|sin x − cos x | dx is equal to:

(a) 2 2 – 2

(b)

(c) 1/ 2

(d) 0

45. If I =

 1 1  dx is equal to  − 2  x x

e  (a) e  − 1 2 

44.

∫

cos x

π/2

0

(a)

π 2

(b)

(c)

π 6

(d) None of these

1

π 4 (c) 1

40.

2π

∫ 0

41. If un =

∫

0

(d)

π

0

x tan x dx is: sec x + cos x

3π 2 4

(b)

1 2n + 1

π2 3

2 (d) π 2

48. The value of the integral b are integers), is:

1 (b) n +1

1 2n − 1

∫

2 (c) π 4

tan n x dx , then un + un – 1:

1 (a) n −1 (c)

(a)

(b) 2 (d) 4 π/4

 x + 2 dx where [ ] is the greatest   (b) 22 (d) None of these

47. The value of

x dx is equal to 2

(a) zero (c) 8

9

0

(a) 31 (c) 23

π2 32

(d) None of these

1 + sin

∫

π 4

integer function: (b)

(a)

then the value of I is:

sin x + cos x

46. Find the value of

tan − 1 x 39. ∫ dx is equal to 1 + x2 0

2

∫

π

−π

(cos ax − sin bx) 2 dx (a and

(a) –π

(b) 0

(c) π

(d) 2π 

Answers

1. 11. 21. 31. 41.

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

2. 12. 22. 32. 42.

(c) (c) (c) (d) (c)

3. 13. 23. 33. 43.

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

4. 14. 24. 34. 44.

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

5. 15. 25. 35. 45.

(a) (b) (a) (b) (b)

6. 16. 26. 36. 46.

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

7. 17. 27. 37. 47.

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

8. 18. 28. 38. 48.

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

9. 19. 29. 39.

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

10. 20. 30. 40.

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

8

Differential Equations

CHAPTER

Summary of conceptS Differential equation an equation involving an independent variable, a dependent variable and the derivatives of the dependent variable, is called a differential equation. a few examples of differential equations are as follows: dy 1. x + y = x3, dx d3y d2y dy + 2 +6 + 7 y = 0, 2. 3 2 dx dx dx 2

 d2y   dy  3.  2  + 4   + 3 y = 0, dx  dx      dy  2  4. 1 +      dx  

3

3/ 2

=

d3y . dx 3

orDer of a Differential equation The order of the highest derivative appearing in a differential equation is called the order of the differential equation.

Degree of a Differential equation The power of the highest order derivative appearing in a differential equation, after it is made free from radicals and fractions, is called the degree of the differential equation. For example in the above examples of differential equations the

• • • •

are also solutions of this differential equation. The solution y = sin x + c, where c is an arbitrary constant, is called the general solution of (1), since every solution of (1) can be obtained from y = sin x + c for a suitable choice of c.

general anD particular SolutionS The solution of a differential equation which contains a number of arbitrary constants equal to the order of the differential equation is called the general solution. Thus, the general solution of a differential equation of the nth order has n arbitrary constants. a solution obtained by giving particular values to arbitrary constants in the general solution is called a particular solution.

formation of a Differential equation Let f (x, y, c1, c2, ... cn) = 0 be the solution of a differential equation, where c1, c2, ..., cn are n arbitrary constants. If we eliminate these n constants, we obtain the differential equation of the nth order satisfied by the given solution values. Any equation taken together with n relations obtained by differentiating it n times helps us to eliminate the n constants. Working Rule 1. Write the given equation. 2. Differentiate the given equation w.r.t. independent variable x as many times as the number of arbitrary constants.

differential equation (1) is of first order and first degree, differential equation (2) is of order 3 and degree 1, differential equation (3) is of order 2 and degree 2, and differential equation (4) is of order 3 and degree 2.

Solution of a Differential equation any relation between the dependent and independent variables (not involving the derivatives) which, when substituted in the differential equation reduces it to an identity is called a solution of the differential equation. For example, the solution to the differential equation: dy = cos x ...(1) dx is y = sin x. However y = sin x + 1, y = sin x – 3

3. Eliminate the arbitrary constants with the help of the given equation and the equations obtained by differentiation to get the required differential equation.

Solution of firSt orDer anD firSt Degree Differential equationS The following methods may be used to solve first order and first degree differential equations. 1. Variable Separable Differentiable equations a differential equation of the form dy =0 dx f (x) dx + g (y) dy = 0, f (x) + g (y)

or

is said to have separated variables.

...(2)

306

Objective Mathematics

We substitute x = X + h and y = Y + k in eqn. (5), where h, k are constants to be determined suitably. dx dy We have = 1 and = 1, so that dX dY

Integrating eqn. (2), we obtain dy

∫ f ( x) dx + ∫ g ( y) dx dx

= c,

where c is an arbitrary constant. (2).

dy dy d Y d X d Y = ⋅ ⋅ = . dx d Y d X dx d X

Hence ∫ f ( x) dx + ∫ g ( y ) dy = c is the solution of eqn.

2. equations reducible to Variable Separable form Sometimes in a given differential equation, the variables are not separable. But, some suitable substitution reduces it to a form in which the variables are separable. For example, the differential dy equations of the type = f (ax + by + c) can be reduced to varidx able separable form by substituting ax + by + c = t. The reduced variable separable form is : dt = dx. b f (t ) + a

3. Homogeneous equations a differential equation of the form ...(3)

...(4)

To solve this equation, we put y = Vx, where V is a function of x. Differentiating y = Vx w.r.t x dy dV . =V+x dx dx Substituting in eqn. (2), we obtain V + x d V . = φ (V) dx ⇒

...(6)

ah + bk + c = 0, ah + Bk + C = 0. These equations give bC −Bc ac − aC ,k= ⋅ aB − a b a B − ab

...(7)

Now equation (6) becomes d Y aX + bY = , d X aX + BY which being a homogeneous equation can be solved by means of the substitution Y = VX. type ii: Consider a differential equation of the form

where f (x, y) and g (x, y) are homogeneous functions of x and y of the same degree, i.e., degree of each term in f (x, y) and g (x, y) is same, is called a homogeneous differential equation. If f (x, y) and g (x, y) are homogeneous functions of degree n each, then we can write f (x, y) = xn f1 (y/x) and g (x, y) = xn g1 (y/x).  y dy Now eqn. (3) takes the form =φ   dx x

a X + b Y + ( ah + bk + c ) dY = d X a X + BY + (a h + B k + C) Choose h and k so that

h=

Integrating both sides to obtain the solution of this differential equation.

dy f ( x, y ) = dx g ( x, y )

Now eqn. (5) becomes

dx dV = , x φ(V) − V

which being a differentiable equation with separated variables, can be solved. 4. equations reducible to the Homogeneous form type i: Consider a differential equation of the form : dy ax + by + c a b , where ...(5) = ≠ dx ax + By + C a B This is clearly non-homogeneous. In order to make it homogeneous, we proceed as follows :

a b dy ax + by + c , where = k (say) = = a B d x ax + By +C Since a B – a b = 0, the above method fails in view of eqn. (7). We have

dy k (a x + B y ) + c = dx ax + By +C

...(8)

Substitute ax + By = z so that a+B Now eqn. (8) becomes

dy dz = ⋅ dx dx

dz kz + c = B⋅ + a, dx z+C which is an equation with variables separable. 5. linear Differential equations a differential equation of the form dy + Py = Q, dx

...(9)

where P and Q are functions of x (or constants), is called a linear differential equation of the first order. Working Rule For Solving

dy + Py = Q. dx

1. Find integrating factor (I. F.) = e ∫

Pdx

.

2. The solution of the differential equation is y (I.F) =

∫ Q (I.F) dx + c,

where c is constant of integration.

where R and S are functions of y alone (or constants). The integrating factor in this case is given by

 x  y dx − x dy (iii) d   = y2  y

I.F. = e ∫ . The solution of this equation is

y x dy − y dx (iv) d   = x x2  

R dy

x ⋅ (I.F.) =

∫ (S x (I.F.) ) dy + C ,

(v) d (log xy) =

where C is the constant of integration. 2. The fact elogt = t will be frequently used in the solution of linear equations. 6. Equations Reducible to the Linear Form Consider a differential equation of the form:

y dx + x dy xy

y x dy − y dx (vi) d  log  = x xy  (vii) d (xm yn) = xm – 1 yn – 1 (my dx + nx dy)

dy n + Py = Q y ,  dx

...(10)

1 x dx + y dy (viii) d  log ( x 2 + y 2 )  = x2 + y 2 2 

where P and Q are functions of x. This equation can be reduced (ix) d  1 log x + y  = x dy − y dx   to the linear form as follows: x− y x2 − y 2 2 n Dividing both sides of eqn (10) by y , we get y x dy − y dx (x) d  tan −1  = dy Py –n+1 = Q  ...(11) y– n x x2 + y 2  dx d [ f ( x, y ) ] 1-n dy dz f ¢ ( x, y ) Put y– n + 1 = z ⇒ (– n + 1) y– n = ⋅ = (xi) n dx dx 1- n f [ ( x, y ) ] Substituting in eqn. (11), we obtain x dx + y dy (xii) d x 2 + y 2 = dz x2 + y 2 + (1 − n) Pz = (1 – n) Q, dx which is now a linear equation with z as the dependent variable.

(

)

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. If φ(x) is a differentiable function then the solution of dy + (yφ′ (x) – φ(x) φ′(x)) dx = 0 is (a) y = (φ(x) – 1) + ce (b) yφ(x) = (φ(x))2 + c (c) yeφ(x) = φ(x) eφ(x) + c (d) (y – φ(x)) = φ(x) e–φ(x)

–φ(x)

(b) 2 (d) None of these

dy + 4x2 tan y = e x sec y satisfying 3. The solution of x dx y(1) = 0, is 3

x

(a)  tan y = (x–2)e logx (c) tan y = (x – 1) ex x–3

y2 = 4a (x + a) is a solution, is (a) 1 (c) 3

2. The order of the differential equation, of which xy = ce x + be – x + x2 is a solution, is (a) 1 (c) 3

4. The degree of the differential equation, of which

(b) sin y = e (x – 1)x (d) sin y = ex(x – 1)x–3 x

–4

(b) 2 (d) None of these

5. The differential equation of the family of curves y = e x (A cos x + B sin x), where A and B are arbitrary constants, is 2 (a) d y + 2 dy + 2 y = 0 dx 2 dx 2 (b) d y − 2 dy − 2 y = 0 dx 2 dx 2 (c) d y − 2 dy + 2 y = 0 dx 2 dx

(d) None of these

307

Solution by Inspection The following derivatives must be remembered as they 1. Sometimes a first order differential equation which is not are very useful in solving some differential equations expressible as eqn (9) becomes a linear equation of the directly. form (i) d (x + y) = dx + dy dx + Ry = S, (ii) d (xy) = ydx + xdy dy

Differential Equations

Remarks

308

6. The differential equation of all circles passing through the origin and having their centres on the x-axis is

Objective Mathematics

(a) y2 = x2 + 2xy (c) x2 = y2 + xy

dy dx

(b) y2 = x2 – 2xy

dy dx

dy dx

(d) None of these

7. The differential equation of family of parabolas with foci at the origin and axis along the x-axis is dy  dy  − y = 0 (a) y   + 2 x  dx  dx 2

(c) ex + 2 = 3

dy  dy  (c) y   + 2 x + y  = 0  dx  dx 2

8. The differential equation that represents all parabolas each of which has a latus rectum 4a and whose axes are parallel to x-axis, is d 2 y  dy  d 2 y  dy  +   = 0 (b) 2a 2 +   = 0 2  dx   dx  dx dx 3

3

d 2 y  dy  −   = 0 (d) None of these dx 2  dx  3

9. The differential equation of all parabolas whose axes are parallel to the axis of y, is 3 (b) d y = – 1 dx 3

(c) d y = 0 dx 3 3

(d) None of these

10. The degree of differential equation dy 1 dy 1 dy x = 1 +   +   +   + ............ is  dx    2! dx 3!  dx  2

(a) three (c) not defined

3

(b) one (d) None of these

x2+ y 2+ 1 11. The solution of dy = , satisfying y(1) = 0, is 2 xy dx given by (a) a hyperbola (b) a circle (c) y2 = x(1 + x) – 10 (d) (x – 2)2 + (y – 3)2 = 5 12. The order and degree of the differential equation dy   1 + 3  dx

2/3

2 cos y (d) None of these

2 15. The solution of the equation d y = e –2x is y = dx 2

(d) None of these

3 (a) d y = 1 dx 3

(a) (x + a) (1 + ay) = – 4a2y (b) (x + a) (1 – ay) = 4a2y (c) (x + a) (1 – ay) = – 4a2y (d) None of these cos y dx + (1 + 2e– x) sin y dy = 0, when x = 0, y = π is 2 (a) ex – 2 = 3 2 cos y (b) ex + 2 = 2 cos y

2

(c) 2a

dy y – x dy = a  y 2 +  is  dx  dx

14. The particular solution of

dy  dy  − y = 0 (b) x   + 2 y  dx  dx

(a) a

13. The equation of the curve passing through the point 1   a, −  and satisfying the differential equation a

3 = 4 d y are dx 3

2 (a) 1,   3

(b) (3, 1)

(c) (3, 3)

(d) (1, 2)

−2 x −2 x (a) e (b) e + cx + d 4 4 (c) 1 e–2x + cx2 + d (d) None of these 4 16. The equation of the family of curves for which subnormal is constant, is

(a) y2 = cx + k (c) x2 = 2cy + k

(b) y2 = 2cx + k (d) None of these

17. The equation of the curve for which the cartesian subtangent varies as the reciprocal of the square of the abscissa, is (a) x = c e y (c) y = c e

3 / 3k

x3 / 3 k



(b) x = c e y



(d) None of these

2 / 3k



18. A curve is such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2). The equation of the curve is (a) xy = 1 (c) xy = 3

(b) xy = 2 (d) None of these

19. The equation of the curve through the point (1, 1) and 2ay , is whose slope is x ( y − a) (a) ya . x2a = ey – 1 (b) ya. x2a = ey 2a a y – 1 (c) y . x = e (d) None of these 20. Solution of the differential equation (x + 2y 3) dy = y is dx (a) x = y2 (c + y2) (b) x = y (c – y2) (c) x = 2y (c – y2) (d) x = y (c + y2) 21. The differential equation of all non-vertical lines in a plane is 2 (a) d y = 0 dx 2 (c) dy = 0 dx

d 2x =0 dy 2 dx (d) =0 dy (b)

23. The degree and order of the differential equation of the family of all parabolas whose axis is x-axis, are respectively (a) 2, 1 (c) 3, 2

(b) 1, 2 (d) 2, 3

π 24. The equation of a curve passing through 1,  and  4 sin 2 y at (x, y) is having slope x + tan y (a) x = tan y (c) x = 2 tan y

(b) y = tan x (d) y = 2 tan x

25. The equation of the curve satisfying the equation −1 y

(1 + y2) dx + ( x − e − tan ) dy = 0 and passing through origin is −1 y

= cot– 1y

−1 y

= tan– 1y

(a) x ⋅ e tan (b) x ⋅ ecot (c) y ⋅ e

tan −1 x

= tan x

−1 y

= tan– 1y

(d) x ⋅ e tan

26. The solution of the differential equation (1 + y2) + (x – e

dy ) = 0, is dx

(a) (x – 2) = k (b) 2x e tan (c) x e tan

−1 y

−1 y

(d) x e 2 tan

−1 y

= e 2 tan

−1 y

+k

= tan–1 y + k tan −1 y

x (b) log y = cos   + C  y x (c) log x = cos   + C  y (d) None of these 32. The general solution of the differential equation  2 xy − x  dy + ydx = 0 is   x =C y

(a) log x +

y = C x

(b) log y −

(c) log y +

x = C y

(d) None of these

33. The general solution of the differential equation

– 1

− tan −1 y

y (a) log x = cos   + C x

+k

dy + y g′ (x) = g (x). g′ (x), dx where g (x) is a given function of x, is (a) g (x) + log [1 + y + g (x)] = C (b) g (x) + log [1 + y – g (x)] = C (c) g (x) – log [1 + y – g (x)] = C (d) None of these 34. Solution of the equation (x – y) (2dy – dx) = 3dx – 5dy is (a) 2y – x = log (x – y + 2) + C (b) 2x – y = log (y – x + 2) + C (c) 2y + x = log (x – y + 2) + C (d) None of these

= e 27. The equation of the curve which passes through the point (2a, a) and for which the sum of the cartesian sub tangent and the abscissa is equal to the constant xdy − ydx 35. Solution of the equation xdx + ydy + 2 = 0 is a, is x + y2 (a) y (x – a) = a2 (b) y (x + a) = a2  c + x2 + y 2  (c) x (y – a) = a2 (d) x (y + a) = a2. (a) y = x tan   2  28. The general solution of the differential equation dy = y dx  c + x2 + y 2  tan x – y 2 sec x is (b) x = y tan   2  (a) tan x = (c + sec x) y (b) sec y = (c + tan y) x (c) sec x = (c + tan x) y (d) None of these  c − x2 − y 2  (c) y = x tan   2  29. The equation of the curve satisfying the equation (xy – x2) (d) None of these dy = y 2 and passing through the point (–1, 1) is 36. The equation of the curve satisfying the differential dx equation y (x + y 3) dx = x (y 3 – x) dy and passing through (a) y = (log y – 1) x (b) y = (log y + 1) x the point (1, 1) is (c) x = (log x – 1) y (d) x = (log x + 1) y

309

Differential Equations

22. The equation of a curve passing through (0, 1) and hav- 30. The general solution of the differential equation y (x2y + e x) dx – e x dy = 0 is − ( y + y3 ) at (x, y) is ing gradient 1 + x + xy 2 (a) x3y – 3ex = cy (b) x3y + 3ex = cy (c) y3x – 3ey = cx (d) y3x + 3ey = cx (a) xy + tan– 1y = π (b) xy + tan– 1y = π 31. Solution of the differential equation 2 4 y y   (c) xy – tan– 1y = π (d) xy – tan– 1y = π x sin dy =  y sin − x  dx is   x x 2 4

310

(a) y3 – 2x + 3x2y = 0 (c) y3 + 2x – 3x2y = 0

(b) y3 + 2x + 3x2y = 0 (d) None of these

Objective Mathematics

37. A curve passes through the point (5, 3) and at any point (x, y) on it, the product of its slope and the ordinate is equal to its abscissa. The curve is a (a) parabola (c) hyperbola

(b) ellipse (d) circle

38. Differental equation of y = sec (tan – 1 x) is (a) (1 + x 2 ) dy = y + x + c dx dy 2 =y–x+c (b) (1 + x ) dx (c) (1 + x 2 ) dy = xy + c dx x dy = +c (d) (1 + x 2 ) y dx 1  2 2  xy − e ⋅ 3  dx –x y dy = 0, given y = 0 when x = 1 is x e (a) y = 2 (c) y =

1 2  4 − x  x

e1 2  + x  3  x4

(a) x log (xy) = 2 (x – y) (b) y log (xy) = 2 ( y – x) (c) x log (xy) = 2 (x + y) (d) None of these 45. If y+ d (xy) = x (sin x + log x), then dx (a) y = cos x + 2 sin x + 2 cos x + x log x − x + c x x2 3 9 x2 (b) y = − cos x − 2 sin x + 2 cos x + x log x − x + c x x2 3 9 x2 (c) y = − cos x + 2 sin x + 2 cos x − x log x − x + c x x2 3 9 x2 (d) None of these

39. The solution of the equation

2

y (2 y − x) , 44. The solution of the differential equation dy = x (2 y + x) dx which satisfies y = 1 at x = 1, is

e 1 (b) y =  4 − x 2   3 x 2

(d) None of these

40. The solution of the equation  y2  y sin x dy = cos x  sin x −  , 2  dx given y = 1 when x = π is 2 (a) y2 = sin x (b) y2 = 2 sin x 2 (c) x = sin y (d) x2 = 2 sin y  f ( y / x)  41. Solution of the equation x dy =  y + x dx is f ' ( y / x)   x (a) f   = cy  y

y (b) f   = cx x

y (c) f   = cxy x

(d) None of these

46. The general solution of the differential equation (1 + tan y) (dx – dy) + 2xdy = 0 is (a) x (sin y + cos y) = sin y + cey (b) x (sin y + cos y) = sin y + ce– y (c) y (sin x + cos x) = sin x + cex (d) None of these 47. Solution of the equation dy = e x – y (e x – e y) is dx x

(a) ey = e x − 1 + ce − e y

(c) ex = e y − 1 + ce − e

(b) ey = e x − 1 + cee

x

(d) None of these

dy dy 48. Solution of the equation x   + ( y − x ) − y = 0 is  dx  dx 2

(a) (x – y + c) (xy – c) = 0 (b) (x + y + c) (xy – c) = 0 (c) (x – y + c) (2xy – c) = 0 (d) ( y – x + c) (xy – c) = 0 49. Solution of equation (xy4 + y) dx – xdy = 0 is (a) 4x4y3 + 3x3 = cy3 (c) 3x4y3 + 4x3 = cy3

(b) 3x3y4 + 4y3 = cx3 (d) None of these

50. A curve C has the property that if the tangent drawn 42. The solution of the equation log dy = 9x – xy + 6, given at any point P on C meets the coordinate axis at A and dx B, then P is the mid- point of AB. If the curve passes that y = 1 when x = 0, is through the point (1, 1), then the equation of the cure is (a) 3e6y = 2e9x– 6 + 6e6 (a) xy = 2 (b) xy = 3 (b) 3e6y = 2e9x + 6 – 6e6 (c) xy = 1 (d) None of these (c) 3e6y = 2e9x + 6 + e6 (d) None of these 43. The differential equation of all straight lines passing through origin is (a) y = (d)

x

dy dx

dy = y – x dx

(b) dy = y + x dx (d) None of these

51. A normal is drawn at a point P (x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k, then the differential equation describing such a curve is 2 2 (a) y dy = ± k 2 − y 2 (b) x dy = ± k − x dx dx

(c) y dy = ± dx

y 2 − k 2 (d) x dy = ± x 2 − k 2 dx

(b) x3 = 3y3 sin x (d) None of these

(a) y3 = 3x3 sin x (c) x3 = y3 sin x

(a) 5 (c) 3

53. The differential equation which is satisfied by all the curves, y = Ae 2x + Be –x/2, where A and B are non-zero constants, is

2 (b) 2 d y + 3 dy − 2 y = 0 dx 2 dx

54. The degree of the differential equation of all tangent lines to the parabola y 2 = 4ax is (b) 2 (d) None of these

55. A solution of the differential equation dy  dy  + y = 0 is   − x dx dx 2

(b) y = 2x (d) y = 2x2 – 4

)

ves y 2 = 2c x + c , where c is a positive parameter, is of (a) order 1 (c) degree 3

(b) order 2 (d) degree 4

57. The order of the differential equation whose general solution is given by y = c1 cos (2x + c2) – (c3 + c4)  a x + c5 + c6 sin (x – c7) is (a) 3 (c) 5

(b) 4 (d) 2.

58. The degree of the differential equation d4y   dx 4 

3/ 5

−5

d3y d2y dy +6 2 −8 + 5 = 0 is 3 dx dx dx

(a) 2 (c) 4

(b) 3 (d) 5

59. The order of the differential equation satisfying 1 − x 4 + 1 − y 4 = a (x2 – y2). is (a) 1 (c) 3

(b) 2 (d) None of these

60. The degree of the differential equation d3y  dy  d 4 y is + x   = 4 log 3  dx  dx dx 4 (a) 1 (b) 3 (c) 4 (d) None of these 4

311

63. The solution of the equation, x 2 d y = log x , when x = dx 2 1, y = 0 and dy = −1 is dx (a)   y = 1 (log x) 2 + log x 2 (b)   y = 1 (log x) 2 − log x 2 (c)   y = − 1 (log x) 2 + log x 2 (d)   y = − 1 (log x) 2 + log x 2 64. Which of the following is a solution of the differential 2

56. The differential equation representing the family of cur-

(

62. The equation of the curve whose tangent at any point (x, y) makes an angle tan – 1 (2x + 3y) with x-axis and which passes through (1, 2) is

2

2 (c) 2 d y − 3 dy + 2 y = 0 dx 2 dx (d) None of these

(a) y = 2 (c) y = 2x – 4

(b) 4 (d) 2

(a) 6x + 9y + 2 = 26e3 (x – 1) (b) 6x – 9y + 2 = 26e3 (x – 1) (c) 6x + 9y – 2 = 26e3 (x – 1) (d) None of these

2 (a) 2 d y − 3 dy − 2 y = 0 dx 2 dx

(a) 1 (c) 3

61. The order of the differential equation whose general solution is given by y = (c1 + c 2) cos (x + c 3) – c4 e x + c5 where c1, c 2, c3, c4, c5 are arbitrary constants, is

Differential Equations

52. Solution of the equation x2y – x3 dy = y4 cos x, when dx y (0) = 1 is

dy  dy  equation   − x + y=0 dx dx   (a)  y = 2x – 4 (c)  y = 2

(b)  y = 2x2 – 4 (d)  y = 2x

65. Given that, dy = ye x such that x = 0, y = e. The value dx of y(y > 0), when x = 1 will be (a)  e (c)  ee

(b)   1 e (d)  none of these

dy 2 yx 2 + = 66. The solution of the differential equation dx 1 + x 2 (1 + x 2 ) dy 2 yx 2 + = is dx 1 + x 2 (1 + x 2 ) (a)  y(1 + x2) = c + x y = c + tan −1 x 1 + x2 (c)  y log (1 + x2) = c + tan–1 x

(b)  

(d)  y(1 + x2) = c + sin–1 x 67. The solution of the differential equation x dy − y dx = x 2 + y 2 dx is 2 2 2 (a)   x + x + y = cx

2 2 (b)   y − x + y = cx

(c)   x − x 2 + y 2 = cx

2 2 2 (d)   y + x + y = cx

68. The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants, is of

312

Objective Mathematics

(a)  first order and second degree (b)  first order and first degree (c)  second order and first degree (d)  second order and second degree

76. The differential equation

(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

(b)  x + y3 = cy (d)  none of these

70. The equation of family of a curve is y 2 = 4a(x + a), then differential equation of the family is (a)  y = y’ + x (c)  y = 2y’ x + yy’2

(b)  y = y” + x (d)  y” + y’ + y2 = 0

71. The solution of the differential equation dy 2x 1 is + y= dx 1 + x 2 (1 + x 2 ) 2 (a)  y(1 – x ) = tan x + c (b)  y(1 + x2) = tan–1 x + c (c)  y(1 + x2)2 = tan–1 x + c (d)  y(1 – x2)2 = tan–1 x + c 2

–1

(c)   log ( y + x 2 + y 2 ) + log y + c = 0 x (d)   sinh −1   + log y + c = 0  y 2 73. The solution of the equation d y = e −2 x is dx 2

(a)   y = 1 e −2 x + cx + d (b)   y = 1 e −2 x + cx + d 4 2 4 (c)   y = 1 e −2 x + cx 2 + d (d)   y = 1 e −2 x + cx 3 + d 4 4

77. The differential equation of all coaxial parabolas y 2 = 4a (x – b), where a and b are arbitrary constants, is 2

2 d 2 y  dy  (b)   y 2 +   = 1 (a)   y d y + dy = 1 dx  dx  dx 2 dx 2 2 d 2 y  dy  (c)   y 2 +   = 0 (d)   y d y + dy = 0 dx  dx  dx 2 dx

78. The solution of the differential equation ( x 2 − yx 2 ) dy + y 2 + xy 2 = 0 dx dy ( x 2 − yx 2 ) + y 2 + xy 2 = 0 is dx x 1 1 x 1 1 (a)   log   = + + c (b)   log   = + + c y x y    y x y

72. The general solution of y 2 dx + (x2 – xy + y 2) dy = 0 is −1  x  (a)   tan   + log y + c = 0  y −1  x  (b)   2 tan   + log x + c = 0  y

determines a

family of circles with

69. Solution of (x + 2y 3) dy = y dx is (a)  x = y3 + cy (c)  y2 – x = cy

1 − y2 dy = dx y

(c)   log( xy ) =

1 1 1 1 + + c (d)   log( xy ) + + = c x y x y

79. The differential equation of all circles which pass through the origin and whose centre lies on y-axis is (a)   ( x 2 − y 2 ) dy − 2 xy = 0 dx (b)   ( x 2 − y 2 ) dy + 2 xy = 0 dx dy (c)   ( x 2 − y 2 ) − xy = 0 dx (d)   ( x 2 − y 2 ) dy + xy = 0 dx 80. The differential equation of the family of circles with fixed radius 5 unit and centre on the line y = 2 is (a)  (x – 2)y´2 = 25 – (y – 2)2 (b)  (y – 2)y´2 = 25 – (y – 2)2 (c)  (y – 2) 2y´2 = 25 – (y – 2)2 (d)  (x – 2)2y´2 = 25 – (y – 2)2

dy = y tan xx –− 2 sin 81. x y = 2e 2x – e –x is a solution of the differential equation dx sin x, is (a)  y2 + y1 + 2y = 0 (b)  y2 – y1 + 2y = 0 1 (c)  y2 + y1 = 0 (a)  y sin x = c + sin 2x (b)   y cos x = c + sin 2 x 2 (d)  y2 – y1 – 2y = 0 1 (c)  y cos x = c – sin 2x (d)   y cos x = c + cos 2 x 1 − y2 dy 2 82. The differential equation determines a fam= dx y 75. The differential equation of all circles passing through ily of circles with the origin and having their centres on the x-axis is (a)  variable radii and a fixed centre at (0, 1) (b)   x 2 = y 2 + 3 xy dy (a)   x 2 = y 2 + xy dy (b)  variable radii and a fixed centre at (0, –1) dx dx (c)  fixed radius 1 and variable centres along the x-axis (c)   y 2 = x 2 + 2 xy dy (d)   y 2 = x 2 − 2 xy dy (d)  fixed radius 1 and variable centres along the y-axis dx dx

74. The solution of the differential equation

5. (c) We have, y = ex (A cos x + B sin x) ...(1) Differentiating w.r.t. x, we get dy = ex (A cos x + B sin x) + ex (– A sin x + B cos x) dx

∴ The solution is y.e

φ(x)

=

∫e

φ( x)

= y + ex (– A sin x + B cos x)

.φ( x.).φ' ( x) dx = φ(x)e

φ(x)

– e

φ(x)

+c

y = (φ(x) – 1) + c.e–φ(x). 2. (b) We have, xy = ce x + be – x + x2 Differentiating w.r.t. x, we get dy x + y = cex – be– x + 2x dx Differentiating again,

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

d y dy = cex + be– x + 2 = xy – x2 + 2 +2 dx 2 dx [Using (1)] Hence, the required differential equation is x

x

dy    x + y  . dx

dy   dy . x2 + y2 – 2x  x + y  = 0, i.e., y2 = x2 + 2xy  dx  dx 7. (a) Let the directrix be x = – 2a and latus rectum be 4a. Then, the equation of the parabola is (distance from focus = distance from directrix), ...(1) x2 + y2 = (2a + x)2 or y2 = 4a (a + x)  Differentiating w.r.t. x, we get dy 1 dy = 2a or a = y y ⋅ dx 2 dx

sin y  x4 = xex – ex + c x = 1, y = 0  ∴ c = 0 ∴ sin y = ex (x – 1) x–4.

...(1)

dy y dy = 4a i.e., a = ⋅ dx 2 dx

Substituting the value of a in (1), we get y dy    x + 2 dx  dy  dy  + y   dx  dx

...(1)

Putting this value of g in (1), we get

4 log x 4 = x4 dx = e e∫ x The solution is ex ∴ tx4 = ∫ x 4 . 3 dx = xex – ex + c x

or y = 2 x

[from (1) and (2)]

d2y dy + 2y = 0, −2 dx 2 dx

where g is an arbitrary constant. Differentiating w.r.t. x, we get dy x +y + g = 0. i.e., g = – dx

I.F. =

dy dx

 dy  dy – y +  − y   dx  dx

x2 + y2 + 2gx = 0,

dy 4 ex 3. (b) We have, cos y  +  sin y = 3 dx x x dy dt Let sin y = t  ⇒  cos y  = dx dx dt 4 ex ∴ +   t= 3 dx x x

y2 = 2 y

=

6. (a) The equation of circles passing through the origin and having their centres on the x-axis is

The order of this differential equation is 2.

2y

d2y dy =  – ex (A cos x + B sin x) dx 2 dx  + ex (– A sin x + B cos x)

which is the required differential equation.

d2y dy +2 − xy + x 2 − 2 = 0. 2 dx dx

4. (b) We have y2 = 4a (x + a) Differentiating w.r.t. x, we get

...(2)

Differentiating again,

or

2

[from (1)] 

2

  dy  2  dy , or y 1 −  dx   = 2x   dx which is the required differential equation. The degree of the differential equation is 2.

Putting this value of a in (1), the differential equation is dy  y dy  + x y2 = 2 y   dx  2 dx  dy   dy  or y   + 2 x   − y = 0.  dx   dx  2

8. (b) Equation of the family of such parabolas is ( y – k)2 = 4a (x – h) where h and k are arbitrary constants. Differentiating w.r.t. x, we get dy = 2a  dx Differentiating again,

( y – k)



( y– k)

...(1)

...(2)

d 2 y  dy  +   = 0 dx 2  dx  2

...(3)

Differential Equations

1. (a) dy + (yφ′(x) – φ(x) φ′(x)) dx = 0 dy + yφ′(x) = φ(x) φ′(x) dx φ' ( x ) dx I.F = e ∫ = eφ(x)

313

solutions

314

Putting value of y – k from (2) in (3), we get

2a

14. (c) We have, cos y dx + (1 + 2e– x) sin y dy = 0

Objective Mathematics

d y  dy  +   = 0, dx 2  dx 

which is the required differential equation. 9. (c) Such parabolas are given by

d y = 0. dx 3 This is the required differential equation.

∴ The solution is ex + 2 = 3 15. (b)

 dy   

dy = ln x dx ∴  order = 1  and degree = 1.

10. (b) x = e  dx  ⇒

x + y +1 dy = ⇒ 2xydy = (x2 + 1) dx + y2dx 2 xy dx 2



2

=



1 dx 2  

∫ 1 + x

Integrating ,

⇒ y dx – x dy = ay2 dx + a dy ⇒ y (1 – ay) dx = (x + a) dy dx dy − =0 ⇒ x + a y (1 − ay )

dy = c (a constant) dx

y2 k = cx + i.e., y2 = 2cx + k. 2 2

17. (c) We have, Cartesian sub-tangent α i.e.,

y k = 2 dy / dx x

Integrating, log y =

3

 2 dy  dy = a y +  13. (c) We have, y – x  dx  dx

e −2 x + cx + d. 4

or y dy = c dx.

x = 1, y = 0,  ∴  0 = 1 – 1 + c ⇒ c = 0 y2 = x2 – 1  ⇒ x2 – y2 = 1. 2

dy e −2 x = +c dx −2

16. (b) Cartesian sub-normal = y

y2 1 =x– +c ⇒ y2 = (x2 – 1 + cx) x x

 d3y  dy   12. (d) 1 + 3  =  4 3  . dx  dx 

2 cos y.

d y = e–2x dx 2

⇒ y =

 x2+ 1  xd ( y 2 ) − y 2 dx =  2  dx 2 x  x 

2.

2

⇒

2

∫ d ( y / x)

1 2 i.e., c = 3



Differentiating once again,

⇒

− sin y ex ∫ e x + 2 dx − ∫ cos y dy = log c

⇒ ex + 2 = c cos y, where 1 + 2 = c.

2

3

⇒

⇒

 ex + 2  = log c ⇒ log   cos y 

d y d y 1 i.e. 2 = ⋅ dx 2 dx 2a 2

Differentiating again, 1 = 2a

⇒

dx sin y + dy = 0 1 + 2e − x cos y

x ⇒ log (e + 2) − log cos y = log c

(x – h)2 = 4a (y – k), where h, k, a are three arbitrary constants. dy . Differentiating w.r.t. x, (x – h) = 2a dx

11. (a)

⇒

3

2

or

1 square of abscissa dy x2 = dx y k

x3 + log c or y = . 3 3k ce x / 3k

18. (b) L et P (x, y) be any point on the curve, PM the perpendicular to x-axis PT the tangent at P meeting the axis of x at T. As given OT = 2. OM = 2x. Equation of the tangent at P (x, y) is

Integrating, we get

log (x + a) – log y + log (1 – ay) = log C

or log 

(a + x ) (1 − ay ) y = log C i.e., (x + a) (1 – ay) = Cy.

1  Since the curve passes through  a, −  ,  a C i.e., C = – 4a2. ∴ 2a × (1 + 1) = − a So, (x + a) (1 – ay) = – 4a2y.

dy (X − x ) dx It intersects the axis of x where Y = 0 Y– y=

i.e., – y = OT.

dx dy (X – x) or X = x – y = dy dx

2ay dy = x ( y − a) dx y−a 2a dy = dx y x

19. (a) We have, slope = ⇒

Integrating both sides, we get a log | y | – y = – 2a log | x | + log c ⇒ ya . x2a = cey.

or xy + tan– 1 y = C. This passes through (0,1), therefore, tan– 11 = C π . i.e., C = 4 Thus, the equation of the curve is π . ny + tan– 1 y = 4 23. (b) Parabola, whose axis is x-axis, is given by y2 = 4ax Differentiating, we get dy dy = 4a ⇒ y = 2a 2y dx dx Again differentiating, we get

d 2 y  dy  y 2 +  = 0  his passes through (1, 1), therefore 1 = ce i.e., c = T  dx  dx 1 . So, the equation of the curve is ya . x2a = ey – 1. ∴  degree = 1, order = 2. e 20. (d) We have, 24. (a) We have, sin 2 y dy x tan y dx dy dy − = = y ⇒ y = x + 2y3 (x + 2y3) = ⇒ x + tan y dx sin 2 y sin 2y dy dx dx dy dx 1 I.F. = − ∫ sin 2 y − x = 2y2, ⇒ e = elog cot y = cot y . dy y Hence the solution is which is a linear equation, if we take x as the detan y pendent variable. x cot y = ∫ ⋅ cot y dy + C. sin 2 y 1 log   2 1 1  y 1 sec y − ∫ dy = e– log y = e = ⋅ I.F. = ∫ pdy = e y e = ∫2 dy + C. = tan y + C y tan y 1 2 1 ∴ The solution is x. = ∫ 2 y ⋅ dy + c  π S ince the curve passes through 1,  , therey y  4 1 2 2 fore, 1 = 1 + C i.e., C = 0. = y + c or x = y (c + y ). ⇒ x ⋅  y Thus, the equation of curve is x = tan y. 21. (a) The general equation of all non vertical lines in a 25. (d) We have (1 + y2) dx + (x – e– tan  – 1y) dy = 0 plane is ax + hy = 1, where h ≠ 0 dx 1 1 − tan −1y e + x = ⇒ 2 2 dy d2y d2y dy 1 + y 1+ y ⇒ a + h =0 ⇒h 2 =0 ⇒ = 0. dx dx dx 2 1 I.F. = ∫ 1 + y 2 dy = tan–1y. e − ( y + y3 ) dy 22. (b) We have, = 1 + x 1 + y2 ⋅ Hence the solution is ( ) dx −1y − 1y −1y 1 x ⋅ e tan = ∫ ⋅ e − tan ⋅ e tan dy + C. 2 2 1 + x (1 + y ) 1+ y dx or =– dy y + y3 dy + C = tan– 1y + C = ∫  x (1 + y 2 )  1 1 + y2 +    = –  y ( y 2 + 1) y (1 + y 2 )  This passes through the origin,  ∴ 0 = 0 + C i.e.,   −1 dx x + ⇒ = y 1 + y2 ⋅ ( ) dy y dy = elog y = y. ∫ e y Hence the solution is, 1 x ⋅ y = –  ∫ y 1 + y 2 ⋅ y dy + C ( ) dy + C = – tan– 1 y + C =– ∫ 1 + y2

I.F. =

2

C=0  ence, the equation of the curve is x.e tan H tan–1y. 26. (b) (1 + y 2) + (x – e tan

−1 y

)

dy =0 dx

dx −1 + x = e tan y dy



(1 + y2)



dx 1 e tan y + 2  x = 1 + y2 dy 1 + y

−1

−1y

=

315

dx dy + =0 x y log C i.e., xy = C. ∴  C = 2. xy = 2.

Differential Equations

dx = 2x or dy Integrating, log x + log y = This passes through (1, 2),  Hence the required curve is Hence x – y

316



I.F. = e

∫ p dy

= e

∫

1 dy 1 + y2

Objective Mathematics

The solution is −1 y



x . e tan



x e tan

−1 y



x e tan

−1 y



2x e tan

tan ∫e

=

−1 y

tan = e

−1 y

−1

.

e tan y . dy + k1 1 + y2

−1

−1 y

=

e 2 tan y ∫ 1 + y 2 . dy + k1

=

1 2 tan −1 y e + k1 2

2 tan = e

−1 y

30. (b) We have y (x2y + ex) dx – ex dy = 0 dy = x2y2 + yex ⇒ ex dx 1 dy 1 − Dividing by y2ex, we get 2 = x2e– x y dx y

+k

27. (a) We have, Cartesian subtangent + abscissa = constant dx y +x = a +x = a ⇒ y ⇒ dy dy / dx dy dx =  ⇒ y a−x Integrating, we get

y = log y + C x or y = x (log y + C). This passes through the point (– 1, 1), ∴ 1 = – 1 (log 1 + C)  i.e.,  C = – 1. Thus, the equation of the curve is  y = x (log  y – 1). or

log y + log (x – a) = log C

dV 1 −1 dy = ⋅ = V so that 2 y dx dx y dV We thus have + V = –x2e– x, which is linear dx Put

1dx ∴ I. F. = e ∫ = e x. Hence the solution is

∴ y (x – a) = C.

V⋅ ex = –

As the curve passes through the point (2a, a), we have

or



C = a 2.

Hence the required curve is y (x – a) = a2. dy = y tan x – y2 sec x dx

28. (c) We have ⇒

1 dy 1 − tan x = – sec x. y 2 dx y

Putting

dV 1 −1 dy = =V⇒ 2 , y dx dx y

we obtain

dV + tan x ⋅ V = sec x which is linear. dx

I. F. = e ∫ = elog sec x = sec x. Hence the solution is 1 2 sec x = tan x + c V sec x = ∫ sec x dx + c or y or sec x = y (c + tan x). tan x dx

29. (a) We have, (xy – x2) ⇒ y2 Putting

dy = y2 dx

−1 1 dx 1 1 dx = xy – x2 ⇒ 2 − ⋅ = 2 x dy x y y dy dV −1 dx 1 = = V so that 2 , x dy dy x

we obtain

dV V 1 + = 2 , which is linear. dy y y



I. F. = e

2 −x

. e x dx +

x3 C 1 x e = − + y 3 3

C 3 or

x3y + 3ex = Cy.

y −x x ⋅ y x sin x dy dV Put y = Vx so that =V+ x . dx dx

dy = 31. (a) We have, dx

Hence V + x

y sin

1 dV V sin V − 1 = = V− sin V dx sin V

1 dV = − ⇒ sin V dx

dx + sin V dV = C. x ∫ y ⇒ log x – cos V = C ⇒ log x = cos  +C x y dy = which is homogeneous. 32. (c) We have, x − 2 xy dx ⇒ x

∫

dV dy = x +V dx dx V dy Hence x +V= 1− 2 V dx Put y = Vx so that

⇒ x ⇒

dV V 2V 3 / 2 = −V = dx 1− 2 V 1− 2 V

dx 1− 2 V 1 1 = dV =  3 / 2 −  dV x 2 V3/ 2  2V V

Integrating, we get – C + log x = – V1/2 – log V = –

1

∫ y dy

∫x e

= elog y = y.

Hence the solution is Vy =

1

∫y

2

⋅ y dy + C

or

log y +

x = C. y

x – log y + log x y

⇒

dV

∫1− V

=

∫ g' ( x) dx

 ⇒ – log (1 – V) = g (x) – C ⇒ g (x) + log (1 – V) = C ∴ g (x) + log [1 + y – g (x)] = C. 34. (a) We have, (2x – 2y + 5) dy = (x – y + 3) dx x− y+3 dy ⋅ ⇒ = 2 x − y) + 5 ( dx Put x – y = V so that dy dV dy dV = 1− 1− = or dx dx dx dx



∴ The equation becomes V+3 dV dV V+3 V+2 or = =1– 1− = ⋅ 2 V + 5 dx dx 2V + 5 2V + 5 ⇒ dx =

 1  2V + 5 dV. dV =  2 + V + 2  V+2 

Integrating, x = 2V + log (V + 2) + C, ⇒ x = 2 (x – y) + log (x – y + 2) + C as V = x – y. or 2y – x = log (x – y + 2) + C is the required solution. 35. (c) We have, x dx + y dy + ⇒

x dy − y dx =0 x2 + y 2

 y 1 d (x2 + y2) + d tan– 1   = 0 x 2

y 1 c (x2 + y2) + tan– 1 x = 2 2 y ⇒ x2 + y2 + 2 tan– 1 x = c. Integrating,

 c − x2 − y 2  ∴ y = x tan   is the required solution. 2  36. (c) We have, y (x + y3) dx = x (y3 – x) dy ⇒ y3 ( y dx – x dy) + x ( y dx + x dy) = 0  y dx − x dy  ⇒ x2y3.   + xd (xy) = 0  x2 ⇒ –

 y  d ( xy ) y d  + 2 2 = 0 x x y x

Integrating, we get

 y −  x − 2 1 y2 + ⇒ 2 x 2 xy ⇒ y3 + 2x +

1 xy

317

2

=C

+C=0 2cx2y = 0.

Since it passes through the point (1, 1), therefore 1 + 2 + 2C = 0 −3 or C = ⋅ 2  ∴  The curve is y3 + 2x – 3x2y = 0. 37. (c) We have,

dy ⋅ y = x (Given) dx

⇒ y dy = x dx Integrating,

y2 x2 C i.e., y2 = x2 + C. = + 2 2 2

Since the curve passes through the point (5, 3), ∴ 9 = 25 + C ⇒ C = – 16. So the curve is y2 = x2 – 16 i.e., x2 – y2 = 16 which is clearly a rectangular hyperbola. 38. (c) y = sec (tan– 1x) Let, tan–1x = t 1 dx = dt  or  ⇒ 1 + x2 ∴ y = sec t

dt 1 = dx 1 + x2

dy = sec t tan t dt dy sec t tan t xy = = ∴ 2 dx 1+ x 1 + x2 dy ⇒ (1 + x 2 ) = xy + c. dx ⇒

e dx x3 e −1 2 2  (x . 2y dy – y · 2x dx) = 3  dx ⇒ x 2 x2 d ( y 2 ) − y 2 d ( x2 ) − 2e 1 ⇒ dx = ⋅ x4 x3 x 4  y2   y2  −2e − 2e ⇒ d  2  = 7 dx ⇒ ∫ d  2  = ∫ 7 dx x x x  x 

39. (b) We have, xy 2 dx – x2y dy =

x −6 y2 = – 2e x −7 dx = – 2e. +C 2 −6 ∫ x e 1 ⇒ y2 = ⋅ 4 + Cx2. 3 x e Putting y = 0, x = 1 we get 0 = +C 3 −e ⇒C= ⋅ 3 e1 2 ∴ The solution is y2 =  4 − x  .  3 x 40. (a) Put y2 sin x = V. dy dV sin x + y2 cos x = . Then 2y dx dx ⇒

Differential Equations

dy = (g (x) – y). g’ (x) dx dy dV Put g (x) – y = V ⇒ g’ (x) – = dx dx dV Hence g′(x) – = V · g ′ (x) dx dV dV = (1 – V) g’ (x) ⇒ = g ′ (x) dx ⇒ 1 −V dx

33. (b) We have,

318

So, the given equation becomes

1 dV = sin x cos x 2 dx

⇒ 2 

Objective Mathematics

⇒ d V = 2 sin x cos x dx Integrating, we get V = sin2x + C. i.e., y2 sin x = sin2x + C π Putting y = 1, x = , we get 1 = 1 + C 2  i.e., C = 0 2 2 2 So, the solution is y sin x = sin x i.e., y = sin x.

Integrating, we get 2log x + 2V + log V = C ⇒ log (x2V) = – 2V + C

y f ( y / x) dy + = which is homogeneous. x f' ( y / x ) dx dy dV =V+ x , Put y = Vx so that dx dx we obtain



∴ C = log 1 + 2 = 2. So the solution is log (xy) = 2 – i.e., x log (xy) = 2 (x – y). 45. (d) We have, y + ⇒ x ⇒

f (V) dV dx =V+ dV = V+x f' ( V) dx x

∴ The solution is

e e = 6 9

+

e6 . 18

I.F. = e

e 18 6

y = mx  ...(i) H ence, the required differential equation of all such lines is  dy  [   m = dy/dx]. y =  x  dx  dy dV =V+ x , we get 44. (a) Putting y = Vx so that dx dx

⇒ x

dV − 2V = dx 2V + 1

x2 x2 +c log x − 3 9

⇒ (1 + tan y) dx = (1 + tan y – 2x) dy dx 2 + ⇒ x = 1, which is linear in x. dy 1 + tan y 2

2 cos y

1

∫ 1 + tan y dy 

= e

∫ sin y + cos y dy

cos y − sin y 

∫ 1 + cos y + sin y  dy = e = ey + log (cos y + sin y) = (cos y + sin y) ey. So, the solution is

43. (d) The equation of all the straight lines passing through origin (0, 0) is



(sin x + log x) dx + c

46. (b) We have, (1 + tan y) (dx – dy) + 2x dy = 0

⇒ 3e6y = 2e9x + 6 + e6

V ( 2V − 1) dV = V+x dx (2V + 1)

2

2 2 x i.e., y = – cos x + sin x + 2 cos x + log x – x x 3 x c + 9 x2

e6 y e9 x + 6 +C = 6 9 e6 e6 +C Putting x = 0, y = 1, we get = 6 9

9x + 6

∫x

= – x2cos x + 2x sin x + 2 cos x +

Integrating, we get

6y

d + 2y = x (sin x + log x) dx

dy 2 + · y = sin x + log x, which is linear in y. dx x

y ⋅ x2 =

 y ⇒ log f (V) = log cx ⇒ f   = cx. x dy = 9x – 6y + 6 42. (c) We have, log dx dy = e9x – 6y + 6 = e9x + 6 · e– 6y ⇒ dx ⇒ e6y dy = e9x + 6 dx

i.e., C =

d (xy) = x (sin x + log x) dx

1

log f (V) = log x + log c



2y x

2 ∫ dx I.F. = e x = e2log x = x2. So, the solution is

Integrating, we get

2y . x When x = 1, y = 1,

⇒ log (xy) = C –

 xf ( y / x )  41. (b) We have, x dy =  y + f' y / x  dx ( )  ⇒

dx  1 +  2 +  d V = 0  x V

y xey (sin y + cos y) = ∫ e (sin y + cos y ) dy + c i.e., xey (sin y + cos y) = eysin y + c.

i.e.,

x (sin y + cos y) = sin y + ce– y.

47. (a) We have, 

dy = e x – y (e x – e y) dx dy + e x ⋅ e y = e 2x. ⇒ ey dx

Putting ey = V so that e y

dy dV = , we get dx dx

dV + ex ⋅ V = e2x, which is linear in V. dx



x

∫e

ex

x

∫e

z

V ⋅ ee =

e ⇒ V ⋅e =

.e

2x

dx + c

Since P is the mid point of AB, ∴ 2x = x – y

⋅ z dz + c [Putting ex = z ⇒ ex dx = dz]

 ⇒ V ⋅ee

x

x = (z – 1) ez + c = (e − 1) e

ex

+c

 dy  dy 48. (a) We have, x   + (y – x) – y=0  dx  dx  dy   dy  + y = 0 ⇒  − 1  x  dx   dx  2

dy dy = 1 or x =– y dx dx

dy and solution of x =– y dx 

dy dx + = 0 is y x

i.e., 

log (xy) = log c  i.e.,  xy = c. Hence general solution is (x – y + c) (xy – c) = 0. 49. (c) We have, xy4 dx + y dx – x dy = 0 ⇒ x dx +

y dx − x dy =0 y4 2

 x  y dx − x dy =0 ⇒ x3 dx +   ⋅ y2  y x x ⇒ x3 dx +   d   = 0  y  y 1 x x + 4 3  y  4

∴  Equation of curve is xy = 1. 51. (a) We have, k = PQ = length of normal  dy  ⇒ k = y 1 +    dx 

3

= c′

or 3x4y3 + 4x3 = c y3, which is the required solution. 50. (c) Equation of tangent at P (x, y) is dy (X – x). Y– y= dx

dy = ± dx

2

⇒

k2  dy  = 1+   2  dx  y

2

k 2 − y2 ,

which is the required differential equation. dy = y4cos x dx dy – x2y = y4cos x i.e., x3 dx Dividing by, – y4x3, we get

52. (b) We have, x2y – x3

−1 dy 1 1 1 + ⋅ = 3 cos x y 4 dx y 3 x x 1 −1 dy 1 d V = , we get Putting 3 = V so that 4 y dx 3 dx y

1 dV 1 1 dV 3 3 + V = 3 cos x ⇒ + V = 3 cos x, 3 dx x x dx x x which is linear in V. 3



2

Integrating, we get

log y + log x = log C ⇒ log (xy) = log C or xy = C.

∴ y

dy = 1 is y = x + c dx

The solution of

Integrating, we get Since the curve passes through (1, 1), ∴ C = 1.

x

⇒ ey = ex – 1 + ce − e .

⇒

dx dy dy dx ⇒ + and 2y = y − x =0 dy dx y x

∫ I.F. = e x

dx

= e3log x = x3.

So the solution is 3 x3 V = ∫ x 3 . 3 cos x dx + C = 3 sin x + C. x x3 ⇒ = 3 sin x + C. y3 Putting x = 0, y = 1, we get C = 0. Hence the solution is x3 = 3y3 sin x. 53. (a) We have, y = Ae2x + Be– x/2 

...(1)

dy 1 = 2ae 2 x − Be − x / 2  dx 2

...(2)

⇒

d2y 1 = 4ae 2 x + Be − x / 2  dx 2 4 Eliminating A, between (1) and (2), we get 2 dy  Be– x/2 =  2 y −   5 dx

and

It meets the coordinate axes in A and B

∴ from (1), Ae2x = y – Be– x/2 = y −

...(3)

...(4)

dy  2  2 y −  dx 5

319

 dx  dy   ∴ A ≡  x − y , 0  and B ≡  0, y − x  . dy    dx 

Differential Equations

x

x e dx I.F. = e ∫ = ee . So, the solution is

320

⇒ Ae2x =

58. (b) We have,

1 dy   y + 2   5 dx

...(5)

Objective Mathematics

So, from (3), (4) and (5), we get 4 dy  1 2  dy  d2y =  y + 2  + ⋅  2 y −  5 dx 4 5 dx dx 2 d2y dy or 2 2 − 3 – 2y = 0. dx dx

which is a differential equation of degree 2. 55. (c) D  irect substitution of y = 2x – 4 in the equation shows that it is a solution of the given differential equation. ...(1)

Differentiating w.r.t. x, we get dy = 2c 2y dx dy . ⇒ c = y dx

⇒ y – 2 x

y

dy  dy  =2 y   dx  dx  2

dy dx

5

59. (a) Put x2 = sin α, y 2 = sin β. ∴  Given equation reduces to   cos α + cos β = a (sin α – sin β) α + β α − β α + β α − β = 2a cos  cos  . sin  ⇒ 2 cos   2   2   2   2  α + β α − β α + β α − β = 2a cos  2 cos  cos  . sin   2   2   2   2   α − β ⇒ cot  = a ⇒ α – β = 2 cot– 1 a.  2  ⇒ sin– 1x2 – sin– 1y2 = 2 cot– 1 a. Differentiating w.r.t. x, we get 1 1 dy ⋅ 2x − ⋅ 2y 4 4 dx = 0 1− x 1− y ⇒

x 1 − y4 dy = , y 1 − x4 dx

which is a differential equation of first order and first degree. 60. (d) Since, the given differential equation is not a polynomial in differential coefficients, so its degree is not defined.

Putting this value of c in (1), dy  we get y2 = 2y dx  x + 

d3y d2y dy − 6 +8 −5 3 2 dx dx dx

 hich is a differential equation of order 4 and w degree 3.

2

)

= 5 3

dy  dy  ⇒ x   − y + a = 0,  dx  dx

(

3/ 5

d4y   d3y  d2y dy − 5 ⇒  4  =  5 3 − 6 2 + 8 dx dx  dx   dx 

54. (b) The equation of any tangent to the parabola a y2 = 4ax is y = mx + , m where m is any arbitrary constant. dy = m. Differentiating w.r.t x, we get dx Substituting the value of m in (1), we get dy a y = x + dx dy dx

56. (a), (c)  We have, y 2 = 2c x + c .

d4y   dx 4 

  

3/ 2

61. (c) We can write y = A cos (x + B) – ce x where A = c1 + c2, B = c3 and C = c4 eC5. dy = – A sin (x + B) – Cex dx

3

dy    dy  ⇒  y − 2 x  = 4 y  dx  , dx      hich is a differential equation of order 1 and w degree 3. 57. (c) We have, y = c1 cos (2x + c2) – (c3 + c4)  ac + c5 + c6 sin (x – c7) c5 = c1 cos (2x + c2) – c8. a . ax + c6 sin (x – c7) where c3 + c4 = c8. ⇒ y = c1 cos (2x + c2) – c9 ax + c6 sin (x – c7), where c9 = c8. ac 5. S ince the above relation contains five arbitrary constants, so the order of the differential equation satisfying it is 5.

d2y = – A cos (x + B) – Cex dx 2

d2y = – 2cex dx 2 d 3 y dy d2y ⇒ = – 2cex = + +y 3 dx dx dx 2 ⇒

d 3 y d 2 y dy – y=0 − + dx 3 dx 2 dx which is a differential equation of order 3.

⇒

dy = tan θ = 2x + 3y, where θ is the dx angle that the tangent at any point (x, y) on the curve makes with x-axis. dy ⇒ − 3 y = 2x, which is a linear dx − 3 dx = e– 3x. ∴ I. F. = e ∫

62. (a) We have

⇒ y.e– 3x =

− 3x

∴ Solution of differential equation is 1 y (1 + x 2 ) = ∫ (1 + x 2 ) dx + c (1 + x 2 ) ⇒ y(1 + x2) = x + c

dx + C

−2 −3 x 2 −3 x xe − e + C 3 9

Since it passes through (1, 2)  ∴  C = −2 − 3 x 2 − 3 x 26 −3 xe − e + e 3 9 9 or 2 + 6x + 9y = 26 e3(x – 1).

26 −3 e 9

∴ ye– 3x =

63. (d) We have, d y log x = 2 dx 2 x ⇒ dy = − log x + 1 dx + c ∫ x2 dx x log x 1 =− − +c x x At x = 1, y = 0 and dy = −1 ⇒ c = 0 dx dy (log x + 1)  log x 1  ∴ =− ⇒ ∫ dy = − ∫  +  dx dx x x  x 2

⇒ y = − 1 (log x) 2 − log x + c1 2 At x = 1, y = 0 ⇒ c1 = 0 ∴ y = − 1 (log x) 2 − log x 2 64. (a) We have, 2

dy  dy   dx  − x dx + y = 0  

…(i)

From option (a) y = 2x – 4 ⇒ dy = 2 ∴ From Eq. (i), dx (2)2 – x(2) + 2x – 4 = 0 ⇒ 0 = 0 Hence, option (a) is correct answer. 65. (c) We have, dy = ye x dx dy ⇒ ⇒ log y = ex + c = e x dx y When x = 0, y = e ∴ log e = e0 + c ⇒ 1 = 1 + c ⇒ c=0 ∴ log y = ex When x = 1, log y = e1 = e ∴ y = ee

66. (a) We have, dy 2 yx 1 , which is a linear differential =  + 2 dx (1 + x ) (1 + x 2 ) equation. 2 x / (1+ x 2 ) dx IF = e∫ p dx = e ∫ = elog(1+ x ) = (1 + x2) 2

321

∫ 2x. e

Differential Equations

∴  Solution is y.e– 3x =

67. (d) We have, x dy − y dx = x 2 + y 2 dx ⇒ x dy = x 2 + y 2 dx + y dx 2 2 ⇒ x dy = ( x + y + y ) dx

x2 + y 2 + y x Now, put y = vx ⇒ dy = v + x dv dx dx ∴ dy = dx

∴ v + x dv = dx ⇒

∫

dv 1 + v2

x 2 + v 2 x 2 + vx x

=∫

dx x

⇒ log (v + 1 + v 2 ) = log x + log c ⇒ y + x 2 + y 2 = cx 2 68. (c) We have, Ax2 + By2 = 1 ⇒ 2 Ax + 2 By dy = 0  dx

…(i)

 dy  2 d 2 y  ⇒ 2 A + 2 B   + y 2  = 0  dx   dx 

…(ii)

From Eqs. (i) and (ii), 2

y

d 2 y  dy  y  dy  + −  =0 dx 2  dx  x  dx 

 hich is the required differential equation whose orW der is 2 and degree is 1. 69. (a) Given equation can be rewritten as dx x = + 2 y2 dy y ⇒

dx x − = 2 y2 dy y

∴ IF = e ∫ Pdx = e ∫ ∴ Solution is,

1 − dy y

=

1 y

x = 2 y dy + c = y 2 + c y ∫

⇒ x = y3 + cy 70. (c) The given equation is y2 = 4ax + 4a2 On differentiating w.r.t. x, we get 2yy′ = 4a On putting the value of 4a in Eq. (i), we get y 2 y '2 y 2 = 2 yy ' x + 4 4

…(i)

322

⇒ y2 = 2yy’ x + y2y′ 2 ⇒ y = 2y’ x + yy’2

Objective Mathematics

71. (b) The given equation is dy 2 1  + , which is a linear differeny= dx 1 + x 2 (1 + x 2 ) 2 tial equation. IF = e

∫ Pdx

=e

On differentiating w.r.t. x, we get dy 2x + 2 y + 2g = 0 dx dy   ⇒ 2 g = −  2 x + 2 y  dx   From Eq. (i),

log(1+ x 2 )

=1+ x

2

dy   x 2 + y 2 +  −2 x − 2 y  x = 0 dx   ⇒ x 2 + y 2 − 2 x 2 − 2 xy dy = 0 dx ⇒ y 2 = x 2 + 2 xy dy dx 76. (c) We have,

∴ The complete solution is 1 y (1 + x 2 ) = ∫ (1 + x 2 ) ⋅ dx + c (1 + x 2 )2 ⇒ y(1 + x2) = tan–1 x + c 72. (a) We have, y2 dx + (x2 – xy + y2) dy = 0 x 2 − xy + y 2 dy = 0 y2 2 dx  x   x  +   −   +1 = 0 ⇒ dy  y   y 

1 − y2 dy = dx y y dy = ∫ dx ⇒ ∫ 1 − y2

⇒ dx +

x ⇒ x = vy y dx dv ⇒ =v+ y dy dy dv ⇒ y = −(v 2 + 1) dy dv dy ⇒ ∫ 2 +∫ =0 v +1 y Let v =

⇒ − 1 − y 2 = x + c ⇒ (x + c)2 + y2 = 1, which represents a family of circles with centre (–c, 0) and radius = 1. 77. (c) We have,

y2 = 4a (x – b)

⇒ 2 y dy = 4a dx

⇒ tan–1 v + log y + c = 0 −1  x  ⇒ tan   + log y + c = 0  y 2 73. (b) Given equation is d y = e −2 x dx 2

Again differentiating, we get 2

2y

d2y  dy  + 2  = 0 dx 2  dx  2

⇒ y

On integrating both sides, we get

d 2 y  dy  + =0 dx 2  dx 

dy e −2 x = +c 78. (a) We have −2 dx dy ( x 2 − yx 2 ) + y 2 + xy 2 = 0 Again integrating, we get −2 x dx e + cx + d y= ⇒ x2 (1 – y) dy = – (1 + x)y2 dx 4 dy 74. (d) We have, 1− y = y tan x − 2 sin x 1+ x  ⇒ dy = −  2  dx dx 2 y  x  ⇒  dy − y tan x = −2 sin x , which is a linear differdx  1 1  1 1 ⇒  2 −  dy = −  2 +  dx ential equation y y x x    ∫ p dx –∫ tan x dx IF = e =e = cos x On integrating, we get ∴ The solution 1  1  − − log y = −  − + log x  + c y(cos x) = ∫ – 2 sin x cos x dx + c y x   = – ∫ sin 2x dx + c 1 1 ⇒ log x − log y = + + c cos 2 x x y ⇒ y cos x = +c x 1 1 2 ⇒ log   = + + c 75. (c) General equation of all such circles which pass  y x y through the origin and whose centre lie on x-axis, 79. (a) The equation of all circles which passes through is the origin and whose centre lies on y-axis is x2 + y2 + 2gx = 0

…(i)



x2 + y2 – 2ky = 0

…(i)

⇒ 2 x + 2 y dy − 2k dy = 0 dx dx

⇒ y′2 (y – 2)2 = 25 – (y – 2)2 81. (d) Given that, y = 2e2x – e–x On differentiating w.r.t. x, we get y1 = 4e2x + e–x Again differentiating w.r.t x, we get y2 = 8e2x – e–x Now, y2 – y1 – 2y = 8e2x – e–x – 4e2x – e–x – 4e2x + 2e–x =0

On putting this value of k in Eq. (i), we get    x  x + y − 2 + y y = 0  dy   dx  2

323

x +y  dy   dx   

Differential Equations

⇒ k =

2

 dy  ⇒   ( y − 2) 2 = 25 − ( y − 2) 2  dx 

2

80. (c) The equation of family of circles with centre on y = 2 and of radius 5 is (x – α)2 + (y – 2)2 = 52

…(i) 82. (c)

⇒ x2 + α2 – 2αx + y2 + 4 – 4y = 25 ⇒ 2 x − 2α + 2 y dy − 4 dy = 0 dx dx

1 − y2 dy = dx y y dy = ⇒ ∫ 1 − y2

∫ dx

⇒ − 1 − y 2 = x + c

⇒ α = x + dy ( y − 2) dx On putting the value of α in Eq. (i), we get

⇒ (x + c)2 + y2 = 1.

2

dy   2 2  x − x − dx ( y − 2)  + ( y − 2) = 5  

centre (–c, 0) and, which determines a family of circles with radius

c 2 − c 2 + 1 = 1.

EXERCISES FOR SELF-PRACTICE 5. dy = e–2y and y = 0 when x = 5, then the value of x dx for y = 3 is e6 + 9 (b) (a) loge 6 2 (c) e6 + 1 (d) e5 6. 1. If

The solution of the differential equation dy = e y + x + e y – x is dx

 d2y  d3y  2  + 4 = x2 – 1, then + dx 3  d3y   dx   dx 3 

(a) ex (x + 1) + 1 = y (b) ex (x – 1) + 1 = y (c) ex (x + 1) = y (d) None of these The curve for which the slope of the tangent at any point equals the ratio of the obscissa to the ordinate of the point is (a) a circle (b) an ellipse (c) a rectangular hyperbola (d) None of these 7. Solution of the differential equation dy y = sin x is + dx x (a) x ( y + cos x) = cos x + C (b) x ( y – cos x) = sin x + C (c) x ( y + cos x) = sin x + C (d) None of these 8. The second order differential equation is: (b) y′ y′′ + y = sin x (a) y′ 2 + x = y2 (c) y′′′ + y′′ + y = (d) y′ = y x− y 9. If y′′z = , then its solution is: x+ y

(a) m = 3, n = 1 (c) m = 3, n = 2

(a) y2 + 2xy – x2 = C (c) y2 – 2xy – x2 = C

2. If

dI = 3cos y . sin y, then I is equal to dx

3cos y + c log 3 (c) sin y + c

(b) 3cos y + c

(a) –

3. 4.

(d) None of these dy y – x = 2 is The solution of dx (b) 2x – 2 · 2 y = k (a) 2x + 2 y = k 1 1 1 1 (c) x − y = k (d) 2 + y = k 2 2 2 2 If m and n are the order and degree of the differential equation 5

 d2y   dx 2 

3

(b) m = 3, n = 3 (d) m = 3, n = 3

(b) y2 + 2xy + x2 = C (d) y2 – 2xy + x2 = C

324

Objective Mathematics

10. The solution of y′ = 1 + x + y2 + xy2, y (0) = 0 is:  x2  (a) y2 = exp  x +  − 1 2    x2  (b) y2 = 1 + c exp  x +  2  

2 3 (a) y = ae (2a – x)

(b) y = ae

−2 3

(a – x)

x+a x+a

2 3

(c) y = ae (2a + x) x + a −

2

(d) y = ae 3 (2a − x) x + a

(c) y = tan (c + x + x2)  x2  (d) y = tan   x +  2   11. 12. 13. 14. 15. 16. 17.

where A is an arbitrary constant. dy = 4 is 18. The solutuion of x2 + y2 dx Family y = Ax + A3 of curves is represented by the dif (a) x2 + y2 = 12x + C (d) x2 + y2 = 3x + C ferential equation of degree: (c) x3 + y3 = 3x + C (d) x3 + y3 = 12x + C (a) 3 (b) 2 19. The solution of x dx + y dx = x2y dy – xy2 dx is (c) 1 (d) None of these (a) x2 – 1 = c (1 + y2) (b) x2 + 1 = c (1 – y2) Family of curves y = ex (A cos x + B sin x), represents the 3 3 (c) x – 1 = c (1 + y ) (d) x3 + 1 = c (1 – y3) differential equation : 20. The family of curves in which the subtangent at any point 2 d2y dy (b) – 2y (a) d y = 2 dy – y =2 of a curve is double the abscissa, is given by 2 2 dx dx dx dx (a) x = cy2 (b) y = cx2 2 2 2 2 d y dy d y dy (d) x = cy (d) y = cx – 2y (d) +y (c) = =2 dx 2 dx dx 2 dx 21. Consider the differential equation Solution of y dx – x dy = x2y dx is: 3/2   dy  2   d2y  x2 − x2 2 2 ye ye = cx (b) = cx (a) 1 +    = k  2    dx    dx  2 2 − x2 (c) y 2e x = cx2 (d) y e = cx2 Find the degree and order of the equation. dy For solving = (4x + y + 1), suitable substitution (a) 2, 2 (b) 3, 2 dx is: (c) 2, 3 (d) None of these (a) y = Vxm (b) y = 4x + V 22. Find the integral factor of expansion (c) y =4x (d) y + 4x + 1 = V dy If integrating factor of + 2xy = x2 – 1. (x2 + 1) dx x (1 – x2) dy + (2x2y – y – ax3) dx = 0 is ∫ p dx then p = e 2x (b) 2 (a) x2 + 1 2 x 2 − ax 3 x +1 (b) 2x2 – 1 (a) 2 x (1 − x 2 ) x −1 (c) 2 (d) None of these 2x2 − 1 x +1 2x2 − 1 (a) (d) x (1 − x 2 ) ax 3 23. Differential equation of y = sec (tan–1 x) is The order and degree of the differential equation: dy 1/ 3 =y +x+c (a) (1 + x2) d 2 y  dy  dx +   + x1/4 = 0 are respectively: 2 dx  dx  dy (b) (1 + x2) =y – x +c dx (a) 2, 3 (b) 3, 3 (c) 2, 6 (d) 2, 4 dy = xy + c (c) (1 + x2) dx Solution of the differential equation dy dy x a+x + xy = 0 is: = (d) (1 + x2) +c dx dx y

Answers

1. (b) 11. (a) 21. (a)

2. (a) 12. (b) 22. (b)

3. (c) 13. (c) 23. (c)

4. (c) 14. (d)

5. (d) 15. (d)

6. (c) 16. (a)

7. (c) 17. (a)

8. (b) 18. (d)

9. (a) 19. (a)

10. (d) 20. (a)

9

Coordinates and Straight Lines

CHAPTER

Summary of conceptS DiStance formula

 x1 + x2 y1 + y2   2 , 2 .  

The distance between two points A (x1, y1) and B (x2, y2) is given by

AB =

( x2 − x1 ) 2 + ( y2 − y1 ) 2

Note 1. If the distance between two points is given, then use ± sign. 2. The distance of any point A(x, y) from origin is x2 + y 2 .

Note: The coordinates of any point on a line joining the two  λx2 + x1 λy2 + y1  points A and B are given by  λ + 1 , λ + 1  . Such a point   divides the given line in the ratio λ : 1. If λ is positive, then the point divides internally and if λ is negative, then the point divides externally.

Section formula

Important Tips In order to prove that a given figure is a

1. The coordinates of the point P (x, y) dividing the line segment joining the two points A (x1, y1) and B (x2, y2) internally in the ratio m1 : m2, are given by x=

(i) Square: Prove that the four sides are equal and the diagonals are also equal. (ii) Rhombus (but not a square): Prove that the four sides are equal but the diagonals are not equal.

m1 x2 + m2 x1 m y + m2 y1 ,y= 1 2 m1 + m2 m1 + m2

(iii) Rectangle: Prove that the opposite sides are equal and the diagonals are also equal. (iv) Parallelogram (but not a ractangle): Prove that the opposite sides are equal but diagonals are not equal. Note that in each of these cases diagonals bisect each other.

2. The coordinates of the point P (x, y), dividing the line segment joining the two points A (x1, y1) and B (x2, y2) externally in the ratio m1 : m2, are given by x=

m1 x2 − m2 x1 m y − m2 y1 ,y= 1 2 m1 − m2 m1 − m2

area of a triangle The area of ∆ABC with vertices A (x1, y1), B (x2, y2) and C (x3, y3) is given by:

∆=

=

1 [x ( y – y ) + x2 ( y3 – y1) + x3 ( y1 – y2)] 2 1 2 3

1 [(x y + x2 y3 + x3 y1) – (x2 y1 + x3 y2 + x1y2)] 2 1 2

= 3. The coordinates of the mid point of the line segment joining the two points A (x1, y1) and B (x2, y2) are given by

x1 1 x2 2 x3

y1 1 y2 1 . y3 1

326

tranSlation of axeS

Objective Mathematics

Sometimes a problem with a given set of axes can be solved more easily by translation of axes. The translation of axes involves the shifting of the origin to a new point, the new axes remaining parallel to the original axes.

Notes: (i) Area of a triangle is always taken as positive. (ii) If area of a triangle is given, then use ± sign. (iii) If the three points A, B, C are collinear, then area of ∆ABC is zero. (iv) The area of a quadrilateral, whose vertices are A (x1, y1), B (x2, y2), C (x3, y3) and D (x4, y4), is

 x1   x2

x y1 + 2 x3 y2

y2 x + 3 y3 x4

y3 x + 4 y4 x1

y4 y1

  

Let OX, OY be the original axes and O′  be the new origin. Let coodinates of O′  referred to original axes, i.e., OX, OY be (h, k). Let O′ X′  and O′ Y′  be drawn parallel to and in the same If the area of a quadrileteral is zero, then its four vertices direction as OX and OY respectively. Let P be any point in the lie on a stright line i.e., points are collinear. (v) The area of a polygon of n sides with vertices A1 (x1, y1), plane having coordinates (x, y) referred to old axes and (X, Y) referred to new axes. Then, A2 (x2, y2), ..., An (xn, yn) is x = OM = ON + NM = ON + O′ M′  xn −1 yn −1 x2 y2 xn yn  1  x1 y1 = h + X = X + h or X = x – h + + + .... + =  . xn yn x3 y3 x1 y1  2  x2 y2 and y = MP = MM′  + M′ P = NO′  + M′ P =

1 2

= k + Y = Y + k. or Y = y – k (vi) If a1x + b1 y + c1 = 0, a2x + b2 y + c2 = 0 and a3x + b3 y + c3 = 0 are the equations of the sides of a triangle, then the area Thus, the point whose coordinates were (x, y) has now the coorof the triangle is dinates (x – h, y – k).

1 = 2C1C 2C3

a1 a2 a3

b1 b2 b3

2

c1 c2 , c3

rotation of axeS

rotation of axes without changing the origin Let OX, OY be the original axes and OX′ , OY′  be the new axes where C1, C2, C3 are the cofactors of c1, c2, c3 in the de- obtained by rotating OX and OY through an angle θ in the anterminant i.e., C1 = a2b3 – a3b2, C2 = a3b1 – a1b3 and ticlockwise sense. Let P be any point in the plane having coorC3 = a1b2 – a2b1. dinates (x, y) w.r.t. axes OX and OY and (x′ , y′ ) w.r.t. axes OX′  and OY′ . Then,

locuS

The locus of a moving point is the path traced by it under certain geometrical condition or conditions. For example, if a point moves in a plane under the geometrical condition that its distance from a fixed point O in the plane is always equal to a constant quantity a, then the curve traced by the moving point will be a circle with centre O and radius a. Thus, locus of the point is a circle with centre O and radius a. Working rule to find the locus of a point (a) Let the coordinates of the moving point P be (h, k). (b) Using the given geometrical conditions, find the relation between h and k. This relation must contain only h, k and known quantities. (c) Express the given relation in h and k in the simplest form and then put x for h and y for k. The relation, thus obtained, will be the required equation of the locus of (h, k).

x = x′   cos θ – y′  sin θ; y = x′   sin θ + y′  cos θ and  

x′   = x cos θ + y sin θ; y′  = – x sin θ + y cos θ.

Note: 1. The above transformation of coordinates may be displayed by a table

x

cos θ

– sin θ

y

sin θ

cos θ

2. If f (x, y) = 0 is the equation of a curve then its transformed equation is f (x′  cos θ – y′  sin θ, x′ sin θ + y′ cos θ) = 0.



3. If f (x′ , y′ ) = 0 is the transformed equation, then the equation w.r.t. original axes is f (x cos θ + y sin θ, – x sin θ + y cos θ) = 0. 4. The angle of rotation of coordinate axes in order to remove the xy term in the equation

 2h  1 ax + 2hxy + by + 2gx + 2fy + c = 0 is tan −1  . 2  a−b 2

2

5. If the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is translated to the form in which first degree terms are missing, then

 hf − bg gh − af  , the coordinates of the new origin are  2 2 ,  ab − h ab − h  where ab ≠ h2. Change of Origin and Rotation of Axes  If origin is changed to O′  (h, k) and axes are rotated about the new origin O′  by angle θ in the anticlockwise sense such that the new coordinates of P (x, y) become (x′ , y′ ), then the equations of transformation will be x = h + x′  cos θ – y′  sin θ and y = β + x′  sin θ + y′  cos θ.

(b) If three points A, B, C are collinear, then slope of AB = slope of BC = slope of AC. (c) If a line is equally inclined to the axes, then it will make an angle of 45º or 135º with x-axis (i.e., positive direction of x-axis) and hence its slope will be tan 45º or tan 135º, = ± 1. (d) Slope of the line joining two points (x1, y1) and (x2, y2) is given as

y1 − y2 y2 − y1 Difference of ordinates m = x − x = x − x = Difference of abscissae . 1 2 2 1

Intercept of a Line on the Axes (i) Intercept of a line on x-axis  If a line cuts x-axis at a point (a, 0), then a is called the intercept of the line on x-axis. | a | is called the length of the intercept of the line on x-axis. Intercept of a line on x-axis may be positive or negative. (ii) Intercept of a line on y-axis  If a line cuts y-axis at a point (0, b), then b is called the intercept of the line on y-axis and | b | is called the length of the intercept of the line on y-axis. Intercept of a line on y-axis may be positive or negative.

Equations of Lines Parallel to Axes Equation of x-axis  The equation of x-axis is y = 0. Equation of y-axis  The equation of y-axis is x = 0. Equation of a line parallel to y-axis  The equation of the straight line parallel to y-axis at a distance ‘a′  from it on the positive side of x-axis is x = a. If a line is parallel to y-axis, at a distance a from it and is on the negative side of x-axis, then its equation is x = – a. Equation of a line parallel to x-axis  The equation of the straight line parallel to x-axis at a distance b from it on the positive side of y-axis is y = b. If a line is parallel to x-axis, at a distance b from it and is on the negative side of y-axis, then its equation is y = – b.

equation of a Straight line in various forms General Equation of a straight Line An equation of the form: ax + by + c = 0, where a, b, c are any real numbers not all zero, always represents a straight line. Equation of a straight line is always of first degree in x and y. π  Slope of a Line  If a line makes an angle θ  θ ≠  with the 2  positive direction of x-axis, the slope or gradient of that line is usually denoted by m, i.e., tan θ = m. Note:

Slope-Intercept Form The equation of a straight line whose slope is m and which cuts an intercept c on the y-axis is given by y = mx + c.

If the line passes through the origin, then c = 0 and hence (a) The slope of a line parallel to x-axis = 0 and perpendicular the equation of the line will become y = mx. to x-axis is undefined.

327

y′

Coordinates and Straight Lines

x′

328

Point-Slope Form

Objective Mathematics

The equation of a straight line passing through the point (x1, y1) and having slope m is given by ( y – y1) = m (x – x1).

Two-Point Form The equation of a straight line passing through two points (x1, y1) and (x2, y2) is given by (y – y1) =

y2 − y1 (x – x1). x2 − x1

where r is the distance of the point (x, y) from the point (x1, y1). The coordinates (x, y) of any point P on the line at a distance r from the point A (x1, y1) can be taken as (x1 + r cos θ, y1 + r + sin θ) or (x1 – r cos θ, y1 – r sin θ).

Intercept Form The equation of a straight line which cuts off intercepts a and b on x-axis and y-axis respectively is given by

x y + = 1. a b

Reduction of the general equation to different standard forms Slope-Intercept Form: The general form, Ax + By + C = 0, of the straight line can be reduced to the form y = mx + c by expressing y as:

A C x− = mx + c B B A C and c = – . where   m = – B B Thus, slope of the line Ax + By + C = 0 is y=–

m=–

Normal Form (or Perpendicular Form) The equation of a straight line upon which the length of the perpendicular from the origin is p and the perpendicular makes an angle α with the positive direction of x-axis is given by x cos α + y sin α = p.

In normal form of equation of a straight line p is always taken as positive and α is measured from positive direction of x-axis in anticlockwise direction between 0 and 2π.

Parametric Form (or Symmetric Form)

coefficient of x A =− . coefficient of y B

Intercept Form: The equation Ax + By + C = 0 can be reduced to the form

x y + = 1 by exa b

pressing it as: Ax + By = – C A B or –  x − y = 1, where C ≠ 0 C C x y x y or − C/A + − C/B = 1, which is of the form + = 1, a b C C and b = – are intercepts on x-axis and where a = – A B y-axis, respectively. Intercept of a straight line on x-axis can be found by putting y = 0 in the equation of the line and then finding the value of x. Similarly intercept on y-axis can be found by putting x = 0 in the equation of the line and then finding the value of y.

Normal Form To reduce the equation Ax + By + C = 0 to the form x cos α + y sin α = p, first express it as:

Ax + By = – C ...(1) The equation of a straight line passing through the point (x1, y1) and making an angle θ with the positive direction of x-axis is Case 1. If C < 0 or – C > 0, then divide both sides of eqn. (1) by A 2 + B2 , we get given by

x − x1 y − y1 = =r cos θ sin θ

A A +B 2

2

x+

B A +B 2

2

y=−

which is of the form x cos α + y sin α = p,

C A + B2 2

and

p=–

A

, sin α

A 2 + B2

Thus to write the equation of any line parallel to a given line, do not change the coefficient of x and y and change the constant term only.

B A 2 + B2

C

equation of a line perpenDicular to a given line

A + B2 2

caSe 2. If C > 0 or – C < 0, then divide Eqn. (1) by

−A

– A + B , we get 2

x−

2

B

A 2 + B2 A 2 + B2 which is of the form x cos α + y sin α = p, where and

p=

cos α = –

A A 2 + B2

, sin α = –

y=

C A 2 + B2 B

A 2 + B2

The equation of a line perpendicular to a given line ax + by + c = 0 is bx – ay + k = 0, where k is a constant.

point of interSection of two given lineS Let the two given lines be

C A 2 + B2

a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0.

.

angle between two interSecting lineS

Solving these two equations, the point of intersection of the given two lines is given by

 b1c2 − b2c1 c1a2 − c2 a1  ,  .  a1b2 − a2b1 a1b2 − a2b1 

The angle θ between two lines y = m1x + c1 and y = m2x + c2 is given by concurrent lineS m1 − m2 The three given lines are concurrent if they meet in a point. tan θ = ± 1 + m m , 1 2 provided no line is perpendicular to x-axis and the acute angle θ is given by tan θ = Note:

m1 − m2 1 + m1m2 .

(a) If both the lines are perpendicular to x-axis, then the angle between them is 0º. (b) If slope of one line is not defined (one of the lines is perpendicular to x-axis and other makes an angle θ with the positive direction of x-axis), then angle between them = π – θ. (c) The two lines are parallel if and only if m1 = m2. (d) The two lines are perpendicular if and only if m1 × m2 = – 1.

conDition for two lineS to be coinciDent, parallel, perpenDicular or interSecting Two lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 are a b c (i) Coincident, if 1 = 1 = 1 ; a2 b2 c2 (ii) Parallel, if

a1 b1 c1 = ≠ ; a2 b2 c2

(iii) Perpendicular, if a1a2 + b1b2 = 0;

Working rule: Following three methods can be used to prove that the three lines are concurrent 1. Find the point of intersection of any two lines by solving them simultaneously. If this point satisfies the third equation also, then the given lines are concurrent. 2. The three lines P ≡ a1x + b1 y + c1 = 0, Q ≡ a2x + b2 y + c2 = 0, R ≡ a3x + b3 y + c3 = 0 are concurrent if

a1 a2 a3

b1 b2 b3

c1 c2 = 0 c3

3. The three lines P = 0, Q = 0 and R = 0 are concurrent if there exist constants l, m and n, not all zero at the same time, such that lP + mQ + nR = 0. This method is particularly useful in theoretical results.

poSition of two pointS relative to a line Two points (x1, y1) and (x2, y2) are on the same side or on opposite sides of the line ax + by + c = 0 according as the expressions: ax1 + by1 + c and ax2 + by2 + c have same sign or opposite signs.

length of perpenDicular from a

point on a line a b (iv) Intersecting, if 1 ≠ 1 i.e., if they are neither coincident a2 b2 The length of the perpendicular from the point (x1, y1) to the line nor parallel. ax + by + c = 0 is given by equation of a line parallel to a given line The equation of a line parallel to a given line ax + by + c = 0 is ax + by + k = 0, where k is a constant.

p=

a x1 + b x2 + c

. a 2 + b2 The length of perpendicular from origin (0, 0) to the given line is

a 2 + b2 .

329

cos α =

Coordinates and Straight Lines

where,

330

Distance between Two parallel lines

Objective Mathematics

The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is given by d=

| c1 − c2 | a +b 2

2

.

(a) The distance between two parallel lines can also be obtained by taking a suitable point (take y = 0 and find x or take x = 0 and find y) on one straight line and then finding the length of the perpendicular from this point to the second line. (b) Area of a parallelogram or a rhombus, equations of whose sides are given, can be obtained by using the following formula pp Area = 1 2 , sin θ where p1 = DL = distance between lines AB and CD, p2 = BM = distance between lines AD and BC, θ = angle between adjacent sides AB and AD.

Equation of the Bisector of the Acute and Obtuse Angle between Two Lines  Let the equations of the two lines be

a1x + b1 y + c1 = 0 

...(1)

and

a2x + b2 y + c2 = 0 

...(2)

where c1 > 0 and c2 > 0. Then the equation

a1 x + b1 y + c1 a +b 2 1

2 1

=+

a2 x + b2 y + c2 a22 + b22

is the bisector of the acute or obtuse angle between the lines (1) and (2) according as a1a2 + b1b2 < 0 or > 0. Similarly, the equation

a1 x + b1 y + c1 a +b 2 1

2 1

=−

a2 x + b2 y + c2 a22 + b22

is the bisector of the acute or obtuse angle between the lines (1) and (2) according as a1a2 + b1b2 > 0 or < 0. Note: If a1a2 + b1b2 > 0, then the origin lies in obtuse angle and if a1a2 + b1b2 < 0, then the origin lies in acute angle.

Equations of Lines Passing through the point of intersection of two given lines The equation of any line passing through the point of intersection of the lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 is

In the case of a rhombus, p1 = p2. Thus,

(a1x + b1 y + c1) + k (a2x + b2 y + c2) = 0, p12 where k is a parameter. The value of k can be obtained by using Area of rhombus = . one more conditions which the required line satisfies. sin θ 1 Also, area of rhombus = dd 2 1 2 Standard Points of a Triangle where d1 and d2 are the lengths of two perpendicular diagoCentroid of a Triangle  The point of intersection of nals of a rhombus. the medians of the triangle is called the centroid of the triangle. The centroid divides the medians in the ratio 2 : 1 EquationS of straight lines (2 from the vertex and 1 from the opposite side).

passing through a given point and making a given angle with a given line

The equations of the striaght lines which pass through a given point (x1, y1) and make a given angle α with the given straight line y = mx + c are y – y1 =

m ± tan α (x – x1) 1 m tan α

EquationS of the bisectors of the Angles between two lines The equations of the bisectors of the angles between the lines a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0 are given by

a1 x + b1 y + c1 a12 + b12

=±

a2 x + b2 y + c2 a22 + b22

.

Any point on a bisector is equidistant from the given lines.

The coordinates of the centroid of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) are

 x1 + x2 + x3 y1 + y2 + y3  ,  . 3 3  

Incentre of a Triangle The point of intersection of the internal bisectors of the angles of a triangle is called the incentre of the triangle.

The coordinates of the incentre of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) are

 ax1 + bx2 + cx3 ay1 + by2 + cy3   a+b+c , a + b + c  .  Note: The incentre of he triangle formed by (0, 0), (a, 0) and   ab ab (0, b) is  , . 2 2 2 2  a +b + a +b a+b+ a +b  ex-centres of a triangle A circle touches one side outside the triangle and the other two extended sides then circle is known as excircle.

Solving any two of the above equations, we get the circumcentre ((x,, y). (x 3. (a) If the equations of the three sides of the triangle are given, first of all find the coordinates of the vertices of the triangle by solving the equations of the sides of the triangle taken two at a time. (b) Find the coordinates of the middle points of two sides of the triangle. (c) Find the equations of the perpendicular bisectors of these two sides and solve them. This will give the coordinates of the circumcentre of the triangle. 4. If angles A, B, C and vertices A(x A (x ( 1, y1), B (x ( 2, y2) and C (x ( 3, y3) of a ∆ABC are given, then its circumcentre is given by

sinn 2A+x2 sin sin 2B+x B+x3 sin 2C  x1 si  sinn 2A+ ssin , 2 B+ in 2B+ sin 2C 

Let ABC be a triangle then there are three excircles, with three excentres I1, I2, I3 opposite to vertices A, B and C respectively. If the vertices of triangle are A (x1, y1), B (x2, y2) and C (x3, y3) then

 − ax1 + bx2 + cx3 − ay1 + by2 + cy3   I1 =  − a + b + c , −a + b + c    ax1 − bx2 + cx3 ay1 − by2 + cy3  I2 =  a − b + c , a − b + c    ax1 + bx2 − cx3 ay1 + by2 − cy3  I3 =  a + b − c , a + b − c  .  circumcentre The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. It is the centre of the circle which passes through the vertices of the triangle and so its distance from the vertices of the triangle is same and this distance is known as the circum-radius of the triangle. Working Rules 1. Let A (x1, y1), B (x2, y2) and C (x3, y3) be the vertices of the ∆ABC and let circumcentre be P (x, y). Then (x, y) can be found by solving (OA)2 = (OB)2 = (OC)2 or

(x – x1)2 + ( y – y1)2 = (x – x2)2 + ( y – y2)2 = (x – x3)2 + ( y – y3)2

y1 sin sin 2A+y2 sin sin 2B+ B+yy3 sin 2C  sinn 2A+ ssin B+ sin 2C  in 22B+ Note: 1. The circumcentre of a right angled triangle is the mid point of its hypotenuse. 2. The circumcentre of the triangle formed by (0, 0), (x1, y1) and (x2, y2) is

 y2 ( x12 + y12 ) − y1 ( x22 + y22 ) ,  2 ( x1 y2 − x2 y1 )  x2 ( x12 + y12 ) − x1 ( x22 + y22 )  . 2 ( x2 y1 − x1 y2 ) 

orthocentre The orthocentre of a triangle is the point of intersection of altitudes. Working Rules 1. Let O be the orthocentre. Since AD ⊥ BC, BE ⊥ CA and CF ⊥ AB, then OA ⊥ BC

and

OB ⊥ CA OC ⊥ AB.

331

Coordinates and Straight Lines

2. Let D, E and F be the mid points of the sides BC, CA and AB of the ∆ABC respectively. Then, OD ⊥ BC, OE ⊥ AC, OF ⊥ AB. slope of OD × slope of BC = – 1 slope of OE × slope of AC = – 1 slope of OF × slope of AB = – 1

332

Objective Mathematics

 x1 tan A+x2 tan B+x3 tan C  tan A+ tan B+ tan C ,  y1 tan A+y2 tan B+y3 tan C  tan A+ tan B+ tan C  Note: Solving any two of these, we can get coordinates of O. 2. (a) Write down the equations of any two sides of the triangle. (b) Find the equations of the lines perpendicular to these two sides and passing through the opposite vertices. (c) Solve these equations to get the coordinates of the orthocentre. 3. If angles A, B and C and vertices A ((x1, y1), B ((x2, y2) and C ((x3, y3) of a ∆ABC are given, then orthocentre of ∆ABC is given by

1. If any two lines out of three lines, i.e., AB, BC and CA are perpendicular, then orthocentre is the point of intersection of two perpendicular lines. 2. The orthocentre of the triangle with vertices (0, 0), (x1, y1) and (x2, y2) is

 x1 x2 + y1 y2    x1 x2 − y1 y2   ( y1 − y2 )   .  , ( x1 − x2 )  x y − x y  x1 y2 − x2 y1    2 1 1 2 

multiple-choice queStionS choose the correct alternative in each of the following problems: 1. The intercepts on the straight line y = mx by the line y = 2 and y = 6 is less than 5, then m belongs to  4 4 (a)  − ,   3 3

 4 3 (b)  ,   3 8

4  4   (c)  −∞,  ∪  , , ∞  3  3  

4  (d)  , ∞  3 

2. Let A (2, – 3) and B (–2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line (a) 3x + 2y = 5 (c) 2x + 3y = 9

(b) 2x – 3y = 7 (d) 3x – 2y = 3

3. The ratio in which the line y – x + 2 = 0 divides the line joining (3, – 1) and (8, 9) is (a) 2:3 (c) – 2:3

(b) 3:2 (d) – 3:2

4. The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is (a)

x y x y + = 1 and + = 1 2 3 2 1

(b)

x y x y − = – 1 and + =1 2 3 −2 1

x y x y (c) + = –1 and + = –1 2 3 −2 1 (d)

x y x y − = 1 and + =1 2 3 −2 1

5. The extremities of the diagonal of a parallelogram are the points (3, – 4) and (– 6, 5). If the third vertex is the point (– 2, 1), then the coordinates of the fourth vertex are (a) (1, 0) (c) (0, 1)

(b) (– 1, 0) (d) (0, – 1)

6. The length of the median through A of a triangle whose vertices are A (– 1, 3), B (1, – 1) and C (5, 1), is (a) 5 (c) 1

(b) 4 (d) None of these

7. The coordinates of the circumcentre of the triangle having vertices (– 2, – 3), (– 1, 0) and (7, – 6) are (a) (– 3, 3) (c) (3, – 3)

(b) (3, 3) (d) (– 3, – 3)

8. The coordinates of A, B, C are (6, 3), (– 3, 5) and (4, – 2) respectively and P is any point (x, y). The ratio of the areas of triangles PBC and ABC is x+ y−2 | x + y − 2| (b) 7 7 x+ y+2 (c) (d) None of these 7 9. The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio 1 : 2 is (a)

(a)

x2 y 2 l 2 + = 1 4 9

(b)

x2 y 2 l2 + = 4 1 9

(c)

x2 y 2 l2 − = 1 4 9

(d)

x2 y 2 l 2 − = 4 1 9

2 ⋅ a2 + a + 1 3 1 a2 + a + 1 (c) 3 (a)

2 a2 − a + 1 3 1 a2 − a + 1 (d) 3

(b)

11. What is the equation of the locus of a point which moves such that 4 times its distance from the x-axis is the square or its distance from the origin? (a) x2 + y2 – 4y = 0 (c) x2 + y2 – 4x = 0

(b) x2 + y2 – 4 | y | = 0 (d) x2 + y2 –­4 | x | = 0

12. x-coordinates of two points B and C are the roots of equation x2 + 4x + 3 = 0 and their y-coordinates are the roots of equation x2 – x – 6 = 0. If x-coordinate of B is less than x-coordinate of C and y-coordinate of B is greater than the y-coordinate of C and coordinates of a third point A be (3, – 5), then the length of the bisector of the interior angle at A is (a)

7 2 3

(b)

14 2 3

5 2 (d) None of these 3 13. The point A divides the join of P (– 5, 1) and Q (3, 5) in the ratio k : 1. The two values of k for which the area of ∆ABC, where B ≡ (1, 5) and C ≡ (7, – 2), is equal to 2 units, are 31 (a) 7, 14 (b) – 7, 9 31 31 (c) 7, (d) 17, – 9 9 14. Equation of the straight line making equal intercepts on the axes and passing through the point (2, 4) is (c)

(a) 4x – y – 4 = 0 (c) x + y – 6 = 0

(b) 2x + y – 8 = 0 (d) x + 2y – 10 = 0

15. If A, B, C, D are points whose coordinates are (– 2, 3), (8, 9), (0, 4) and (3, 0) respectively, then the ratio in which AB is divided by CD, is (a) 11 : 23 (c) 23 : 47

(b) 11 : 47 (d) None of these

16. The straight line ax + by + c = 0 divides the join of points A (x1, y1) and B (x2, y2) in the ratio (a)

ax1 + by1 + c ax2 + by2 + c

(b) –

ax1 + by1 + c ax2 + by2 + c

(c)

ax2 + by2 + c ax1 + by1 + c

(d) –

ax2 + by2 + c ax1 + by1 + c

17. The value of k for which the points (k, 2 – 2k), (– k + 1, 2k) and (– 4 – k, 6 – 2k) are collinear, is 1 2 (c) k = 1

(a) k =

1 2 (d) k = – 1

(b) k = –

19. If α, β, γ are the real roots of the equation x3 – 3px2 + 3qx – 1 = 0, then the centroid of the triangle  1  1   1 having vertices  α,  ,  β,  and  γ ,  are  α  β  γ (a) ( p, q) (c) (– p, q)

(b) ( p, – q) (d) (– p, – q)

20. The area of the quadrilateral whose vertices are (– 3, 2), (7, – 6), (– 5, – 4) and (5, 4) is (a) 68 (c) 80

(b) 70 (d) None of these

21. The locus of the point whose distance from x-axis is twice that from y-axis, is (a) y = x (c) x = y

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

22. The locus of the moving point P such that 2PA = 3PB, where A is (0, 0) and B is (4, – 3), is (a) 5x2 + 5y2 + 72x + 54y + 225 = 0 (b) 5x2 + 5y2 – 72x – 54y + 225 = 0 (c) 5x2 + 5y2 – 72x + 54y + 225 = 0 (d) None of these 23. The points (3, 3), (h, 0) and (0, k) are collinear if 1 1 1 + = 3 h k 1 1 1 (c) − = 3 k h

(a)

(b)

1 1 1 − = 3 h k

(d) None of these

24. If (1, 4) be the C.G. of a triangle and the coordiantes of its any two vertices be (4, – 8) and (– 9, 7), then the area of the triangle is(a) 84

(b) 132

333 (d) None of these (c) 2 25. The coordinates of incentre of the triangle whose sides are 3x – 4y = 0, 5x + 12y = 0 and y – 15 = 0, are (a) (1, 8) (c) (– 1, 8)

(b) (1, – 8) (d) (– 1, – 8)

26. The coordiantes of the orthocentre of the triangle, having vertices (0, 0), (2, – 1) and (– 1, 3), are (a) (4, – 3) (c) (4, 3)

(b) (– 4, 3) (d) (– 4, – 3)

27. The coordiantes of the orthocentre of the triangle, formed by lines xy = 0 and x + y = 1, are (a) (0, 0) (c) (– 2, 1)

(b) (2, – 1) (d) None of these

28. The coordiantes of the orthocentre of the triangle, having sides 3x – 2y = 6, 3x + 4y + 12 = 0 and 3x – 8y + 12 = 0, are

333

x defined as f : (0, ∞) to (0, ∞), f (x) is 1+ x (a) one-one onto (b) one-one but not onto (c) many-one into (d) many-one onto

18. Let f (x) =

Coordinates and Straight Lines

10. The distance between two parallel lines is unity. A point P lies between the lines at a distance a from one of them. The length of a side of an equilateral triangle PQR, vertex Q of which lies on one of the parallel lines and vertex R lies on the other line, is

334

Objective Mathematics

 1 23  (a)  − ,  6 9 

23   1 (b)  − , −   6 9

 1 23  (c)  ,  6 9 

(d) None of these

45   36 (c)  , −  7 7 

 36 45  (d)  ,  7 7 

36. If the point A is symmetric to the point B (4, – 1) with respect to the bisector of the first quadrant, then the length of AB is

(a) 5 (b) 5 2 (c) 3 2 (d) 3 29. Two vertices of a triangle are (3, – 1) and (– 2, 3) and 37. The equation of the straight line which makes an angle its orthocentre is origin, the coordinates of the third of 15º with the positive direction of x-axis and cuts an vertex are intercept of length 4 on the negative direction of y-axis, 45  is  36 45   36 (b)  − , −  (a)  − ,   7 7   7  7 (a) (2 – 3 ) x – y – 4 = 0

30. If a triangle has its orthocentre at (1, 1) and circumcentre 3 3 at  ,  , then the coordinates of the centroid of the 2 4 triangle are 5 4 (a)  , −  3 6

4 5 (b)  ,  3 6

 4 5 (c)  − ,   3 6

5  4 (d)  − , −   3 6

31. If G be the centroid and I be the incentre of the triangle with vertices A (– 36, 7), B (20, 7) and C (0, – 8) and 25 205 λ, then the value of λ is GI = 3 1 (a) 25 (b) 25 4 25 32. If a vertex of a triangle of two sides through it centroid of the triangle (c)

(d) None of these be (1, 1) and the middle points be (– 2, 3) and (5, 2) then the is

5  (a)  , 3  3 

5  (b)  , − 3  3 

5 (c)  − , 3   3 

5 (d)  − , − 3   3 

33. The equation of the locus of a point such that the sum of its distances from (0, 2) and (0, – 2) is 6, is (a)

x2 y 2 − = 1 5 9

(b)

x2 y 2 − =1 9 5

(c)

x2 y 2 + = 1 5 9

(d)

x2 y 2 + =1 9 5

34. A ladder of length ‘a’ rests against the floor and a wall of a room. If the ladder begins to slide on the floor, then the locus of its middle point is (a) x2 + y2 = a2 (c) x2 + y2 = 2a2

(b) 2 (x2 + y2) = a2 (d) 4 (x2 + y2) = a2

35. A and B are two fixed points. The locus of a point P such that ∠APB is a right angle, is (a) x2 + y2 = a2 (c) 2x2 + y2 = a2

(b) x2 – y2 = a2 (d) None of these

3 ) x + y – 4 = 0 (c) (2 + 3 ) x – y – 4 = 0 (d) None of these (b) (2 –

38. The equation of the straight line cutting off an intercept 8 from x-axis and making an angle of 60º with the positive direction of y-axis is (a) x + (c) y –

3 y = 8 3 x = 8

(b) x – 3 y = 8 (d) None of these

39. A line through the point A (2, 0), which makes an angle of 30º with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15º. The equation of the straight line in the new position is 3)x–y–4+2 3 =0 (b) (2 – 3 ) x + y – 4 + 2 3 = 0 (c) (2 – 3 ) x – y + 4 + 2 3 = 0 (d) None of these

(a) (2 –

40. The equation of the straight line, the portion of which intercepted between the coordinate axes being divided by the point (– 5, 4) in the ratio 1 : 2, is (a) 8x + 5y = 60 (c) – 8x + 5y = 60

(b) 8x – 5y = 60 (d) None of these

41. The coordinates of a point on the line x + y = 4 that lies at a unit distance from the line 4x + 3y – 10 = 0 are (a) (3, 1) (c) (3, – 1)

(b) (– 7, 11) (d) (7, – 11)

42. The equation of the internal bisector of ∠BAC of ∆ABC with vertices A (5, 2), B (2, 3) and C (6, 5), is (a) 2x + y + 12 = 0 (c) 2x + y – 12 = 0

(b) x + 2y – 12 = 0 (d) None of these

43. The range of values of θ in the interval (0, π) such that the points (3, 5) and (sin θ, cos θ) lie on the same side of the line x + y – 1 = 0, is π (b)  0,  (a)  0, π   4  2 π π (c)  ,  (d) None of these 4 2 44. A rectangle has two opposite vertices at the points (1, 2) and (5, 5). If the other vertices lie on the line x = 3, then the coordinates of the other vertices are (a) (3, – 1), (3, – 6) (c) (3, 2), (3, 6)

(b) (3, 1), (3, 5) (d) (3, 1), (3, 6)

(a) 9x + 20y + 96 = 0 (c) 9x + 20y – 96 = 0

(b) 9x – 20y + 96 = 0 (d) None of these

46. The equation of the straight line which passes through the point (3, 4) and whose intercept on y-axis is twice that on x-axis, is (a) 2x – y = 10 (c) 2x + y = 10

(b) x + 2y = 10 (d) None of these

47. The equation of the straight line whose intercepts on x-axis and y-axis are respectively twice and thrice of those by the line 3x + 4y = 12, is (a) 9x + 8y = 72 (c) 8x + 9y = 72

(b) 9x – 8y = 72 (d) None of these

48. The equation of the straight line passing through the origin and the middle point of the intercept of the line ax + by + c = 0 between the axes, is (a) ax + by = 0 (c) bx + ay = 0

(b) ax – by = 0 (d) bx – ay = 0

 (a)  2, 

3 2 

 3 (b)  2, − 2  

 1 3 (c)  2 + (d) None of these , 2  2  55. The acute angle between the line x + y = 3 and the line joining the points (1, 1) and (– 3, 4) is 3 (a) tan–1   7

3 (b) π – tan–1   7

1 (c) tan–1   7

1 (d) π – tan–1   7

(

56. Let P = (–­1, 0), Q = (0, 0) and R = 3, 3 3

) be three

points. Then the equation of the bisector of the angle PQR is (a)

3 x+ y =0 2

(b) x + 3 y = 0

3 49. If m1 and m2 are the roots of the equation x2 + ( 3 + 2) x y =0 (d) x + (c) 3x + y = 0 2 + ( 3 – 1) = 0, then the area of the triangle formed 57. The equation of the straight line, passing through by the lines y = m1x, y = m2x and y = 2 is the point (2, – 4) and perpendicular to the line (a) 33 − 11 (b) 33 + 11 8x – 4y + 7 = 0, is (c) 33 + 7 (d) None of these (a) x + 2y + 6 = 0 (b) x – 2y + 6 = 0 50. Through the point P (α, β), where αβ > 0 the straight (c) 2x + y + 6 = 0 (d) 2x – y + 6 = 0 x y = 1 is drawn so as to form with coordinate 58. The equation of the straight line, having x-intercept equal line + a b 4 to – and is perpendicular to the line 2x – 5y + 8 = 0, axes a triangle of area S. If ab > 0, then the least value 5 of S is is (a) αβ (c) 4αβ

(b) 2αβ (d) None of these

(a) 2x + 5y + 4 = 0 (c) 2x – 5y + 4 = 0

(b) 5x – 2y + 4 = 0 (d) 5x + 2y + 4 = 0.

51. The equation of the straight line upon which the length 59. The equation of the perpendicular bisector of the line of perpendicular from origin is 3 2 units and this segment joining the points (1, 4) and (3, 6) is perpendicular makes an angle of 75º with the positive (a) x – y – 7 = 0 (b) x + y – 7 = 0 direction of x-axis, is (c) x + y + 7 = 0 (d) None of these (a) ( 3 – 1) x + ( 3 + 1) y – 12 = 0 60. A line 4x + y = 1 through the point A (2, – 7) meets the (b) ( 3 – 1) x + ( 3 + 1) y + 12 = 0 (c) ( 3 + 1) x + ( 3 – 1) y – 12 = 0 (d) None of these

52. If the straight line drawn through the point P ( 3 , 2) π with the x-axis meets the line and making an angle 6 3 x – 4y + 8 = 0 at Q, then the length of PQ is (a) 4 (c) 6

(b) 5 (d) None of these

53. The coordinates of the points at a distance 4 2 units from the point (– 2, 3) in the direction making an angle of 45º with the positive direction of x-axis, are (a) (2, 7), (– 6, – 1) (c) (2, – 7) , (– 6, – 1)

(b) (2, 7), (6, – 1) (d) None of these

line BC whose equation is 3x – 4y + 1 = 0 at the point B. If AB = AC, then the equation of the line AC is (a) 52x – 89y + 519 = 0 (b) 52x + 89y – 519 = 0 (c) 52x + 89y + 519 = 0 (d) None of these 61. The coordinates of the foot of the perpendicular drawn from the point (2, 3) to the line y = 3x + 4 are  1 37  (a)  − ,   10 10 

 1 37  (b)  , −   10 10 

1 37 (c)  − , −  (d) None of these  10 10  62. The image of the point (– 8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is (a) (16, – 2) (c) (16, 2)

(b) (– 16, 2) (d) (– 16, – 2)

335

54. A line joining two points A (2, 0) and B (3, 1) is rotated about A in anticlockwise direction through an angle 15º. If B goes to C in the new position, then the coordinates of C are

Coordinates and Straight Lines

45. A straight line passes through the point (– 4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point. The equation of the line is

336

63. The image of the point (3, – 8) under the transformation (x, y) → (2x + y, 3x – y) is

Objective Mathematics

(a) (– 2, 17) (c) (– 2, – 17)

(b) (2, 17) (d) (2, – 17)

64. The image of the point P (3, 5) with respect to the line y = x is the point Q and the image of Q with respect to the line y = 0 is the point R (a, b), then (a, b) = (a) (5, 3) (c) (– 5, 3)

(b) (5, – 3) (d) (– 5, – 3)

65. A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q, respectively. Then the point O divides the segment PQ in the ratio (a) 1:2 (c) 2:1

(b) 3:4 (d) 4:3

72. If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0 (a, b and c being distinct and different from 1) are con1 1 1 = current, then + + 1− a 1− b 1− c (a) 1 (c) 0

73. The distance between the lines 5x – 12y + 2 = 0 and 5x – 12y – 3 = 0 is 3 5 (b) 13 13 7 (c) (d) None of these 13 74. The sum of the abscissas of all the points on the line x + y = 4 that lie at a unit distance from the line 4x + 3y – 10 = 0, is (a)

66. The equation of the straight line passing through the point of intersection of the lines x – y = 1 and 2x – 3y + 1 = 0 and parallel to the line 3x + 4y = 14 is 75. (a) 3x + 4y + 24 = 0 (b) 3x + 4y – 24 = 0 (c) 4x + 3y + 24 = 0 (d) 4x + 3y – 24 = 0 67. The equation of the straight line passing through the point of intersection of lines 3x – 4y – 7 = 0 and 12x – 5y – 13 = 0 and perpendicular to the line 2x – 3y + 5 = 0 is

68. The equation of the straight line passing through the point of intersection of the lines x + 3y + 4 = 0 and 3x + y + 4 = 0 and equally inclined to the axes is (a) x – y = 0 (c) x – y + 2 = 0

(b) x + y + 2 = 0 (d) None of these

69. The distance of the point (1, 2) from the straight line with slope 5 and passing through the point of intersection of lines x + 2y = 5 and x – 3y = 7 is (a)

64 650

(b)

(c)

132 650

(d) None of these

16 650

70. If a and b are the intercepts of a straight line on the xaxis and y-axis respectively and p be its perpendicular distance from the origin, then 1 1 1 (a) 2 = 2 + 2 p a b

1 1 1 (b) 2 = 2 − 2 p a b

(c) p2 = a2 + b2

(d) None of these

71. If the family of lines x (a + 2b) + y (a + 3b) = a + b passes through the point for all values of a and b, then the coordinates of the point are (a) (2, 1) (c) (– 2, 1)

(a) 3 (c) 4

(b) (2, – 1) (d) None of these

(b) – 3 (d) – 4

If p1 and p2 are the lengths of the perpendiculars from the origin to the straight lines x sec θ + y cosec θ = a and x cos θ – y sin θ = a cos 2θ respectively, then the value of 4 p12 + p22 is (a) 4a2 (c) a2

(b) 2a2 (d) None of these

76. The equation of the straight line that can be drawn through the point (4, – 5) so that its distance from the point (– 2, 3) is equal to 12, is (a) x – 2y = 3 (c) 2x – y = 3

(a) 33x + 22y + 13 = 0 (b) 33x + 22y – 13 = 0 (c) 33x – 22y + 13 = 0 (d) None of these

(b) – 1 (d) None of these

(b) 2x + y = 3 (d) no such line is possible

77. The equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin, is (a) x – 2y + (c) x + 2y +

(b) x + 2y – 5 = 0 5 = 0 (d) x – 2y – 5 = 0 5 = 0 78. A line L has intercepts a and b on the coordinate axes. When the axes are rotated through an angle, keeping the origin fixed, the same line L has intercepts p and q. Then, (a)

1 1 1 1 + = + a 2 b2 p 2 q 2

(b)

1 1 1 1 − = − a 2 b2 p 2 q 2

 1 1 1 1  (c) a 2 + b 2 = 2  p 2 + q 2    (d) None of these 79. A straight road passes through two towns, one 5 km. east 1 and other 2 km. north from a tower. A rest house is 2 to be constructed by the side of the road. The nearest position of the rest house from the tower is (a) 1 km. east and 2 km. north (b) 2 km. east and 1 km. north (c) 1 km. east and 1 km. north (d) None of these

(a) 9x – 7y – 41 = 0 (c) 7x + 9y – 3 = 0

(b) 7x + 9y – 3 = 0 (d) None of these

81. The equation of the bisector of the acute angle between the lines 3x + 4y – 11 = 0 and 12x – 5y – 2 = 0 is (a) 11x + 3y + 17 = 0 (b) 11x + 3y – 17 = 0 (c) 3x + 11y + 17 = 0 (d) None of these

(b) 7x – 9y + 3 = 0 (d) None of these

83. The value of k so that the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent, is (a) 7 (c) 5

(a) all real values of t

(b) some real values of t

−4 ± 7 (c) t = 8

(d) None of these

91. P (3, 1), Q (6, 5) and R (x, y) are three points such that the angle RPQ is a right angle and the area of ∆RPQ = 7, then the number of such points R is (a) 0 (c) 2

82. For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 =  0, the equation of the bisector of the angle which contains the origin is (a) 7x + 9y + 3 = 0 (c) 7x + 9y – 3 = 0

(b) – 7 (d) – 5

(b) 1 (d) 4

92. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is (a) square (b) circle (c) straight line (d) two intersecting lines 93. If P (1, 0), Q (– 1, 0), R (2, 0) are three given points, then the locus of point S satisfying the relation SQ2 + SR2 = 2SP2 is (a) a straight line | | to x-axis (b) a straight line | | to y-axis (c) circle through the origin (d) circle with centre at the origin.

84. If the equal sides AB and AC (each equal to a) of a right angled isosceles triangle ABC be produced to P and Q so that BP ⋅ CQ = AB2, then the line PQ always passes 94. A line passes through the point (2, 2) and is perpendicular through the fixed point to the line 3x + y = 3. Its y intercept is (a) (a, 0) (b) (0, a) 1 2 (c) (a, a) (d) None of these (b) (a) 3 3 85. If a straight line cuts intercepts from the axes of coordi4 nates the sum of the reciprocals of which is a constant k, (c) 1 (d) 3 then the line passes through the fixed point (a) (k, k)

1 1 (b)  ,  k k 

(c) (k, – k)

(d) (– k, k)

86. The three lines 3x + 4y + 6 = 0, and 4x + 7y + 8 = 0 are (a) sides of a triangle (c) parallel

2x+

3y+2 2=0

(b) concurrent (d) None of these

87. The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x – 2y = 7. Then PQRS must be  a (a) rectangle (c) cyclic quadrilateral

(b) square (d) rhombus

88. The equation of the straight line equally inclined to the axes and equidistant from the points (1, – 2) and (3, 4)  is (a) x + y + 1 = 0 (c) x – y – 2 = 0

(b) x + y + 2 = 0 (d) x – y – 1 = 0

(a) (– 1, 2) (c) (1, 2)

(b) (1, – 2) (d) (– 1, – 2)

95. A line L passes through the points (1, 1) and (2, 0). Another line M which is perpendicular to L passes through 1 the point  , 0  . Then the area of the triangle formed 2  by the lines L, M and y-axis is 25 16 25 25 (c) (d) 32 4 96. The coordinates of the foot of the perpendicular from the point (2, 4) on the line x + y = 1 are (a)

25 8

(b)

1 3 (a)  2 , 2   

 1 3 (b)  − 2 , 2   

4 1 (c)  3 , 2   

1 3 (d)  4 , − 2   

97. Let PQR be a right angled isosceles triangle, right angled at P (2, 1). If the equation of the line QR is 89. Without changing the direction of coordinates axes, 2x + y = 3, then the equation representing the pair of origin is transferred to (α, β) so that the linear terms in lines PQ and PR is the equation x2 + y2 + 2x – 4y + 6 = 0 are eliminated. (a) 3x2 – 3y2 + 8xy + 20x + 10y + 25 = 0 The point (α, β) is (b) 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0 (c) 3x2 – 3y2 + 8xy + 10x + 15y + 20 = 0 (d) 3x2 – 3y2 – 8xy – 10x – 15y – 20 = 0

337

90. The point (2t2 + 2t + 4, t2 + t + 1) lies on the line x + 2y = 1 for

Coordinates and Straight Lines

80. For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, the equation of the bisector of the obtuse angle between them is

338

98. The orthocentre of the triangle formed by (0, 0), (8, 0) and (4, 6) is

Objective Mathematics

8 (a)  4,   3 (c) (4, 3)

(b) (3, 4) (d) (– 3, 4)

104. Let the algebraic sum of the perpendicular distances from the points A (2, 0), B (0, 2), C (1, 1) to a variable line be zero. Then all such lines (a) are concurrent (b) pass through the fixed point (1, 1) (c) touch some fixed circle (d) pass through the centroid of ∆ ABC.

99. P is a point on either of the two lines y – 3 | x | = 2 at a distance of 5 units from their point of intersection. 105. If α + β + γ = 0, the line 3αx + β y + 2γ = 0 passes The coordinates of the foot of the perpendicular from through the fixed point P on the bisector of the angle between them are  2 2  (b)  , 2  (a)  2,   1   1   3 3  (a) 0, 2 (4 + 5 3 )  or 0, 2 (4 − 5 3 )  depending on     2 (c)  −2,  (d) None of these which line the point P is taken  3  1  106. The area of the region enclosed by 4 | x | + 5 | y | ≤ 20 (b) 0, 2 (4 + 5 3 )    is 1   (a) 10 (b) 20 (c) 0, 2 (4 − 5 3 )  (c) 40 (d) None of these  

5 5 3  (d)  2 , 2  .   100. Let P be the image of the point (– 3, 2) with respect to x-axis. Keeping the origin as same, the coordinate axes are rotated through an angle 60º in the clockwise sense. The coordinates of point P with respect to the new axes are 2 3 −3 (3 3 + 2)  (a)  , −  2  2  2 3 − 3 3 3 + 2 (b)  , 2   2  (2 3 − 3) 3 3 + 2  (c)  − , 2 2   (d) None of these

x y 1 + + a b c = 0 always passes through a fixed point, that point is

107. If a, b, c are in H.P. then the straight line

(a) (– 1, – 2) (c) (1, – 2)

(b) (– 1, 2) 1 (d) 1, −   2

108. The distance of the middle point of the line joining the point (a sin θ, 0) and (0, a cos θ) from the origin is a 2 (c) a (sin θ + cos θ)

(a)

a (sin θ + cos θ) 2 (d) a

(b)

109. Two vertices of a triangle are (2, – 1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 units then third vertex is: (a) (0, 5) or (4, 1) (c) (5, 0) or (4, 1)

(b) (5, 0) or (1, 4) (d) (0, 5) or (1, 4)

101. A square is constructed on the portion of the line 110. The figure which is designed by the line ax ± by ± c = 0 is: x + y = 5 which is intercepted between the axes, on the side of the line away from origin. The equations to the (a) Rectangle (b) Square diagonals of the square are (c) Rhombus (d) None of these (a) x = 5, y = – 5 (b) x = 5, y = 5 111. The locus of the mid point of the intercept of the line (c) x = – 5, y = 5 (d) x – y = 5, x – y = – 5 x cos α + y sin α = p between coordinate axes is : 102. If one of the diagonals of a square is along the line (b) x–2 + y–2 = p–2 (a) x–2 + y–2 = 4 p–2 x = 2y and one of its vertices is (3, 0), then its sides (c) x2 + y2 = 4 p–2 (d) x2 + y2 = p2 through this vertex are given by the equations 112. If the points (x, y), (x′, y′ ) and (x – x′, y – y′ ) are col(a) y – 3x + 9 = 0, 3y + x – 3 = 0 linear, then, (b) y + 3x + 9 = 0, 3y + x – 3 = 0 (a) xy = x′ y′ (b) xx′ = yy′ (c) y – 3x + 9 = 0, 3y – x + 3 = 0 (c) xy′ = x′ y (d) None of these (d) y – 3x + 3 = 0, 3y + x + 9 = 0 x y 103. The line (p + 2q) x + ( p – 3q) y = p – q for different 113. If for a variable line + = 1, the condition a–2 + b–2 a b values of p and q passes through the fixed point = c–2 (c is a constant) is satisfied, then the locus of foot 2 2 3 5 of the perpendicular drawn from origin to this is: (b)  ,  (a)  ,  5 5 2 2 c2 (b) x2 + y2 = 2c2 (a) x2 + y2 = 2 3 3 2 3   (c)  ,  (d)  ,  5 5 5 5 (c) x2 + y2 = c2 (d) x2 – y2 = c2

(a) 2 (c) 4

(b) 0 (d) 1

115. If the centroid and circumcentre of a triangle are (3, 3) and (6, 2) respectively, then the orthocentre is (a) (– 3, 5) (c) (3, – 1)

(b) (– 3, 1) (d) (9, 5)

116. A stick of length l rests against the floor and a wall of a room. If the stick begins to slide on the floor, then the locus of its middle point is (a) an ellipse (c) a circle

(b) a parabola (d) a straight line

3 and which is 2 concurrent with lines 4x + 3y – 7 = 0 and 8x + 5y – 1 = 0 is

117. The equation of the line with slope –

(a) centroid (c) circumcentre

(b) incentre (d) orthocentre

(a rational point is a point both of whose coordinates are rational numbers) 125. Let PS be the median of the triangle with vertices P (2, 2), Q (6, – 1) and R (7, 3). The equation of the line passing through (1, – 1) and parallel to PS is (a) 2x – 9y – 7 = 0 (c) 2x + 9y – 11 = 0

(b) 2x – 9y – 11 = 0 (d) 2x + 9y + 7 = 0

126. The incentre of the triangle with vertices (1, and (2, 0) is

 3 (a) 1,   2  

3 ), (0, 0)

2 1  (b)  ,  3 3

2 3  1  (d) 1, (c)  ,   3 2  3    127. If the lines ax + by + c = 0, bx + cy + a = 0 and 118. The equations ax + by + c = 0 and dx + ey + f = 0 cx + ay + b = 0 are concurrent (a + b + c ≠ 0) then represent the same straight line if and only if (a) a3 + b3 + c3 – 3a bc = 0 a b (b) a + b = 0 (b) c = f (a) = d e (c) a + b + c = 0 a b c (d) a2 + b2 + c2 + ab + bc + ca = 0 (d) a = d, b = e, c = f = = (c) d e f 128. If the lines x – 2y – 6 = 0, 3x + y – 4 = 0 and λ x + 119. The distance between the lines 4x + 3y = 11 and 4y + λ2 = 0 are concurrent, then 8x + 6y = 15 is (a) λ = 2 (b) λ = – 3 (a) 2y – 3x – 2 = 0 (c) 3x + 2y – 63 = 0

(b) 3x + 2y – 2 = 0 (d) None of these

7 10 (c) 4

7 2 (d) None of these

(a)

(b)

120. The point (3, 2) is reflected in the y-axis and then moved a distance 5 units towards the negative side of y-axis. The coordinates of the point thus obtained are (a) (3, – 3) (c) (3, 3)

(b) (– 3, 3) (d) (– 3, – 3)

121. If a, b, c are in A.P., then the straight line ax + by + c = 0 will always pass through a fixed point whose coordinates are (a) (– 1, – 2) (c) (– 1, 2)

(b) (1, 2) (d) (1, – 2)

122. If x1, x2, x3 as well as y1, y2, y3 are in G.P. with the same common ratio, then the points (x1, y1), (x2, y2) and (x3, y3) (a) lie on a straight line (b) lie on an ellipse (c) lie on a circle (d) are vertices of a triangle 123. 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

(b) a = 3, b = 4 (d) a = 3, b = 5

(c) λ = 4

(d) None of these

129. If A and B are two variable points on axes of x and y respectively, such that OA + OB = c, the locus of the foot of perpendicular from origin to AB is (a) x2 + y2 = cxy (b) x + y = cxy (c) (x + y) (x2 + y2) = cxy (d) x2 + y2 = c2 130. Area of the triangle formed by the points [(a + 3) (a + 4), a + 3], [(a + 2) (a + 3), a + 2] and [(a + 1) (a + 2), a + 1] is (a) 25a2 (c) 24a2

(b) 5a2 (d) None of these

131. If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in (a) H.P. (c) A.P.

(b) G.P. (d) None of these

132. Equation of a straight line passing through the point of intersection of x – y + 1 = 0 and 3x + y – 5 = 0 and perpendicular to one of them is (a) x + y + 3 = 0 (c) x – 3y – 5 = 0

(b) x + y – 3 = 0 (d) x + 3y + 5 = 0

133. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

339

Coordinates and Straight Lines

114. The number of integer values of m, for which the 124. If the vertices P, Q, R of a ∆PQR are rational points, x-coordinate of the point of intersection of the lines which of the following points of the ∆PQR is (are) 3x + 4y = 9 and y = mx + 1 is also an integer, is always rational point (s) ?

340

(a) square (c) straight line

(b) circle (d) two intersecting lines

Objective Mathematics

134. The vertices of a triangle are (0, 0), (3, 0) and (0, 4). Its orthocentre is at 3  (a)  , 2  2 

(b) (0, 0)

 4 (d) None of these (c) 1,   3 135. The vertices of a triangle are (0, 3), (– 3, 0) and (3, 0). The coordinates of its orthocentre are (a) (0, 2) (c) (0, 3)

(b) (0, – 3) (d) (0, – 2)

136. The lines x cos α + y sin α = p1 and x cos β + y sin β = p2 will be perpendicular if π π (b) α = (a) α ± β = 2 2 π (d) α = β (c) | α – β | = 2 137. The vertices of a ∆OBC are O (0, 0), B (– 3, – 1) and C (– 1, – 3). The equation of a line parallel to BC and intersecting sides OB and OC whose distance from the 1 , is origin is 2 1 1 = 0 (b) x + y – =0 2 2 1 1 = 0 (d) x + y + =0 (c) x + y – 2 2 138. The medians AD and BE of a triangle with vertices A (0, b), B (0, 0) and C (a, 0) are perpendicular to each other if (a) x + y +

(a) a =

b 2

(c) ab = 1

(b) b =

a 2

(d) a = ±

2b

139. The equation of a straight line passing through the point (– 5, 4) and which cuts off an intercept of 2 units between the lines x + y + 1 = 0 and x + y – 1 = 0 is (a) x – y + 10 = 0 (b) x – y + 9 = 0 (c) 2x – y + 14 = 0 (d) x – 2y + 13 = 0 140. The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1 (a) lies in the IIIrd quadrant (b) lies in the IInd quadrant (c) lies in the Ist quadrant (d) cannot be found   141. The point (– 4, 5) is the vertex of a square and one of its diagonals is 7x – y + 8 = 0. The equation of the other diagonal is (a) 7x – y + 23 = 0 (c) x – 7y = 31

(b) x + 7y = 31 (d) None of these

 8 142. The points  0, 3  , (1, 3) and (82, 30) are vertices of   (a) an obtuse angled triangle (b) an acute angled triangle (c) a right angled triangle (d) None of these 143. In a ∆ABC, the equation of the perpendicular bisector of AC is 3x – 2y + 8 = 0. If the coordinates of the points A and B are (1, – 1) and (3, 1) respectively, then the circumcentre of the ∆ABC is

 4 14  (a)  5 , 5   

 4 14  (b)  − 5 , 5   

 4 14  (c)  5 , − 5   

(d) None of these

(e) All of these 144. Given vertices A (1, 1), B (4, – 2) and C (5, 5) of a triangle, the equation of the perpendicular dropped from C to the interior bisector of the angle A is (a) y – 5 = 0 (c) x – 5 = 0

(b) y + 5 = 0 (d) x + 5 = 0

145. The sides AB, BC, CD and DA of a qaudrilateral have the equations x + 2y = 3, x = 1, x – 3y = 4, 5x + y + 12 = 0 respectively. The angle between the diagonals AC and BD is

π π (b) 3 6 π (c) (d) None of these 2 146. The condition to be imposed on β so that (0, β) lies on or inside the triangle having sides y + 3x + 2 = 0, 3y – 2x – 5 = 0 and 4y + x – 14 = 0 is (a)

(a) 0 < β < (c)

5 3

5 7 ≤β≤ 3 2

(b) 0 < β
0 and y = 3x, x > 0, then a belongs to 2 1  (a)  (3, ∞) (b)    , 3  2   1 (d)    0,   2

172. The triangle joining the points P(2, 7), Q(4, –1), R(–2, 6) is (a)  scalene triangle (b)  isosceles triangle (c)  right angled triangle (d)  equilateral triangle

(a)  x + y + 3 = 0 (c)  x – y + 3 = 0

(b)  x – y – 3 = 0 (d)  3x + y – 7 = 0

179. The equation of the line bisecting perpendicularly the segment joining the points (–4, 6) and (8, 8) is (a)  6x + y – 19 = 0 (c)  6x + 2y – 19 = 0

(b)  y = 7 (d)  x + 2y – 7 = 0

180. The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept – 4. Then, a possible value of k is (a)  1 (c)  –2

(b)  2 (d)  –4

181. The equation of straight line equally inclined to the axis and equidistant from the points (1, – 2) and (3, 4) is ax + by + a = 0 where

173. Let P = (–1, 0), Q = (0, 0) and R = (3,3 3) be three points. The equation of the bisector of the angle PQR is (a)   3 x + y = 0

(b)   x + 3 y = 0 2

(c)   3 x + y = 0 2

(d)   x + 3 y = 0

174. Let O (0, 0), P (3, 4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are 4  (a)    , 3  3 

(b)  x + y = 5 (d)  x – y = – 5

178. The bisector of the acute angle formed between the lines 4x – 3y + 7 = 0 and 3x – 4y + 14 = 0 has the equation

(a)  right angled (b)  isosceles (c)  equilateral (d)  none of these

1  (c)    −3, −  2 

(a)  x – y = 5 (c)  x + y = – 5

 2 (b)    3,   3

(a)  a = 1, b = 1, c = 1 (b)  a = 1, b = – 1, c = – 1 (c)  a = 1, b = 1, c = 2 (d)  none of these. 182. The x-axis, y-axis and a line passing through the point A(6, 0) form a triangle ABC. If ∠A = 30º, then the area of the triangle, in sq unit is (a)   6 3

(b)  12 3

(c)   4 3

(d)   8 3

183. The straight line 3x + 4y – 5 = 0 and 4x = 3y + 15 interesect at the point P. On these lines the points Q and R are chosen so that PQ = PR. The slopes of the lines QR passing through (1, 2) are (a)  – 7, 1/7 (c)  7, – 1/7

(b)  7, 1/7 (d)  3, – 1/3

 4 4 2 (c)    3,  (d)    ,  3   3 3 184. The tangent to the curve y = ex drawn at the point (c, ec) intersects the line joining the points (c – 1, ec – 1) 175. The circumcentre of a triangle formed by the line xy + and (c + 1, ec + 1) 2x + 2y + 4 = 0 and x + y + 2 = 0 is (a)  (–1, –1) (c)  (1, 1)

(b)  (0, –1) (d)  (–1, 0)

176. A pole stands vertically inside a triangular park ABC. If the angle of elevation of the top of the pole from each corner of the park is same, then in park the foot of the pole is at the (a)  centroid (c)  incentre

(b)  circumcentre (d)  orthocentre

177. The point P(a, b) lies on the straight line 3x + 2y = 13 and the point Q(b, a) lies on the straight line 4x – y = 5, then equation of line PQ is

(a)  on the left of x = c (b)  on the right of x = c (c)  at no point (d)  at all points 185. Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD and AB = 2CD. 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 (a)  3 (c)  3/2

(b)  2 (d)  1

4 (a)    , 3  3 

2 (b)    3,   3

4 (c)    3,   3

4 2 (d)    ,  3 3

solutions 1. (c) We have y = mx, y = 2, y = 6  et the point of intersection of line y = mx by lines L y = 2 and y = 6 are P (2/m, 2) and Q (6/m, 6) ∴ Intercepts on y = mx is 2

2 6 2  −  + (2 − 6) m m

PQ = PQ < 5 ∴

∴ Required lines are

y x – =1 3 2



and

x y + = 1. −2 1

5. (b) Since the diagonals of a parallelogram bisect each other, therefore the coordinates of the mid point E of diagonal AC are

(given) 2

2 6  −  + 16 < 5  m m

2

2 6 ⇒  −  + 16 < 25 m m

2 6 4 4 − m> m m 3 3 4  4   ∴ m ∈  − ∞, − ∪ , ∞  3  3   ⇒–3
0, ∴ sin θ + cos θ – 1 > 0 π  ⇒ sin θ + cos θ > 1 ⇒ sin  + θ  > 4

1 2

π  π π π 3π ⇒ sin  + θ  > sin ⇒ 0) 

...(2)

Since the line (1) passes through the point P (α, β) ∴

α β α aβ + + = 1 or = 1 a b a 2S

[Using (2)]

or a2β – 2aS + 2αS = 0. Since a is real, ∴ 4S2 – 8αβS ≥ 0



or S ≥ 2αβ

1    S = ab > 0 as ab > 0  2

Hence the least value of S = 2αβ. 51. (a) Let AB be the required line and OL be perpendicular to it.

∴ m1 + m2 = – ( 3 + 2) and m1m2 = ( 3 – 1) ∴ m1 – m2 =

11 ).

or 4S2 ≥ 8αβS

49. (b) Since m1 and m2 are the roots of the equation



0 2 1 = |  m 2 1 2 m2

Coordinates and Straight Lines

47. (a) The equation of the given line is 3x + 4y = 12 which can be written in the intercept form as

(m1 + m2 ) 2 − 4m1m2

= 3 + 4 + 4 3 − 4 3 + 4 = 11 . The coordinates of the vertices of the given triangle are  2  2   (0, 0),  m , 2  and  m , 2  . 1 2

Given, OL = 3 2 and ∠LOA = 75º. ∴ Equation of line AB is

352

x cos 75º + y sin 75º = 3 2 ...(1) (Normal form)

Objective Mathematics

Now cos 75º = cos (30º + 45º) 3 1 1 1 × − × = = 2 2 2 2

3 −1 2 2

and sin 75º = sin (30º + 45º) 1 1 3 1 3 +1 − × = × = 2 2 2 2 2 2 ∴  From (1), equation of line AB is

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

or ( 3 – 1) x + ( 3 + 1) y – 12 = 0. 52. (c) The given line is

0 −1 = 1 = tan 45º 2−3 ∴ ∠BAX = 45º. Given ∠CAB = 15º. ∴ ∠CAX = 60º.



AB =

∴ Slope of line AC = tan 60º = 3 .  ow, line AC makes an angle of 60º with positive N direction of x-axis and AC = AB = = 2. ∴ Coordinates of C are 

(2 +

 i.e 2 + 1 ,  2 

3. 2 

(3 − 2) 2 + (1 − 0) 2

2 cos 60º, 0 +

2 sin 60º)

55. (c) Let A ≡ (1, 1) and B ≡ (– 3, 4). ∴  Slope of AB, m1 =

1− 4 3 =– . 1+ 3 4

Given line is x + y = 3. 3 x – 4y + 8 = 0 ...(1) Let PQ = r. Then, the coordinates of Q are  π π  3 r r, 2 +  .  3 + r cos , 2 + r sin  or  3 + 6 6 2 2   Since the point Q lies on the given line,

1 = – 1. 1 If θ is the acute angle between the two lines, then ∴  Slope of given line, m2 = –

3 − +1 − 1 m m 1 2 4 tan θ = = = . 3 1 + m1m2 7 1+ 4 1 ∴ θ = tan–1   . 7

 r  3  3  3+ r  – 4 2 +  + 8 = 0   2 2   56. (c) The equation of the bisector of ∠PQR is y = tan 120o.x  or  y = − 3x . ⇒ 6 + 3r – 16 – 4r + 16 = 0 or r = 6. Hence, PQ = 6. 53. (a) We know that the coordinates of points on the line making an angle θ with positive direction of x-axis at a distance r from the point (x1, y1) are ∴

57. (a) Equation of any line perpendicular to the line 8x – 4y + 7 = 0 is 4x + 8y + K = 0. (x1 + r cos θ, y1 + r sin θ) and (x1 – r cos θ, y1 – r sin θ). ∴ Coordinates of the required points will be (– 2 ± 4 2 cos 45º, 3 ± 4 2 sin 45º) or (– 2 ± 4, 3 ± 4) or (2, 7) and (– 6, – 1). 54. (c) Slope of line

Since it passes through the point (2, – 4) ∴ 4 (2) + 8 (– 4) + K = 0 or K = 24. ∴ Equation of the line is 4x + 8y + 24 = 0 or x + 2y + 6 = 0. 58. (d) Equation of any line perpendicular to the line 2x – 5y + 8 = 0 is 5x + 2y + k = 0  Intercept of the line (1) on x-axis

...(1)

k  [obtained by putting y = 0 in (1)] 5 k 4 =– or  k = 4. Given : – 5 5 ∴ Equation of the line is 5x + 2y + 4 = 0.



=–

or x + 3y – 11 = 0 ...(2) Solving (1) and (2), we get  1 37  1 37 , y = ∴ A ≡  − , x=– . 10 10  10 10 62. (d) Equation of the given line is Now, Slope of AB =

6−4 = 1, 3 −1

∴ Slope of CD = – 1 ( CD ⊥ AB) Since C is the mid point of AB, 1 + 3 4 + 6  , ∴ Coordinates of C are    or (2, 5). 2 2  ∴ Equation of CD is (y – 5) = – 1 (x – 2) or x + y – 7 = 0. 60. (c) Since AB = AC ∴ ∠ABC = ∠ACB = θ (say)

4x + 7y + 13 = 0 ...(1) Let Q (α, β) be the image of the point P (– 8, 12) w.r.t. line (1). Then, PQ ⊥ line (1) and PC = CQ. Equation of the line PC is 7 (x + 8) 4 [PC is ⊥ to the line (1) and passes through  (– 8, 12)] or 7x – 4y + 104 = 0 ...(2) Solving (1) and (2), we get x = – 12 and y = 5. ∴ C ≡ (– 12, 5). Since C is mid point of PQ,

3 and slope of AB = – 4. 4 Let slope of AC be m. Equating the two values of tan θ, we get 3 3 −4 − −m 19 3 − 4m 4 = ⇒ = 4 8 4 + 3m 3 3 1− 4 × 1+ m 4 4 52 ⇒ 76 + 57m = 24 – 32m ⇒ m = – 89 52 ∴ Equation of AC is (y + 7) = – (x – 2) 89 Now, slope of BC =

(y – 12) =

∴ – 12 =

α−8 β + 12 and 5 = 2 2

⇒ α = – 16 and β = – 2. ∴ Q ≡ (– 16, – 2). 63. (a) Let (x1, y1) be the image of the point (x, y) under the given transformation.

or 52x + 89y + 519 = 0. 61. (a) The equation of given line is

Then, x1 = 2x + y = 2 (3) – 8 = – 2 and y1 = 3x – y = 3 (3) – (– 8) = 17. Hence, the image is (– 2, 17). 64. (b) Let (x1, y1) be the image of the point P (3, 5) w.r.t. the line y = x. Then, x1 = 5, y1 = 3.

3x – y + 4 = 0 The slope of the line (1) is = 3 ∴  slope of PA (⊥ to given line) = –

...(1) 1 . 3

∴ Q = (5, 3). Since the image of the point Q (5, 3) w.r.t. the line y = 0 is (a, b). ∴ a = 5 and b = – 3. ∴ (a, b) = (5, – 3).

353

∴  Equation of PA is 1 (x – 2) or 3y – 9 = – x + 2 (y – 3) = – 3

Coordinates and Straight Lines

59. (b) Let A ≡ (1, 4) and B ≡ (3, 6). Let CD be the perpendicular bisector of the line segment AB.

354

67. (a) Given lines are

Objective Mathematics

65. (b) OP:OQ = OR:OS.

1 The equation of SR is y = x 2  −12 −6  9 9  ∴ R =  5 , 10  and S =  5 , 5     

9 −6 9   −12  5 λ + 5 5 λ + 10  ∴ O = (0, 0) =  ,  λ +1   λ + 1    −12λ 9 −6λ 9 + = 0, + ∴ = 0 ⇒ 12 λ = 9 5 5 5 10 ∴ λ =

3 . So, OP:OQ = 3:4. 4

3x – 4y – 7 = 0 ...(1) 12x – 5y – 13 = 0 ...(2) and 2x – 3y + 5 = 0 ...(3) Equation of any line through the point of intersection of lines (1) and (2) is (3x – 4y – 7) + k (12x – 5y – 13) = 0 or (3 + 12k) x – (4 + 5k) y – (7 + 13k) = 0  ...(4) 3 + 12k . Slope of line (4) = 4 + 5k 2 . Slope of line (3) = 3 Since line (4) is perpendicular to line (3), 3 + 12k 2 × = – 1 or 6 + 24k = – 12 – 15k ∴ 4 + 5k 3 6 . or 39k = – 18. ∴ k = – 13 Putting the value of k in equation (4), the equation of the required line is 72  30     3 −  x −  4 −  y – (7 – 6) = 0 13 13 or – 33x – 22y – 13 = 0 or 33x + 22y + 13 = 0. 68. (a), (b)  Equation of any line through the point of intersection of given lines is (x + 3y + 4) + k (3x + y + 4) = 0 or (1 + 3k) x + (3 + k) y + (4 + 4k) = 0

66. (b) Given lines are x – y – 1 = 0 ...(1) 2x – 3y + 1 = 0 ...(2) and 3x + 4y – 14 = 0 ...(3) Equation of any line through the point of intersection of lines (1) and (2) is (x – y – 1) + k (2x – 3y + 1) = 0 or (1 + 2k) x – (1 + 3k) y + (k – 1) = 0...(4) 1 + 2k . 1 + 3k 3 . Slope of line (3) is = – 4 Since line (4) is parallel to line (3), or 4 + 8k = – 3 – 9k, ∴k=–

7 . 17

 utting the value of k in (4), the equation of reP quired line is 3 4 24 x+ y− =0 17 17 17 or 3x + 4y – 24 = 0.

(1 + 3k ) . (3 + k )

Since the line (1) is equally inclined to the axes, ∴ –

(1 + 3k ) = ± 1 ⇒ k = – 1, 1. (3 + k )

 o, the required lines are x – y = 0 and x + y + S 2 = 0. 69. (c) The two lines x + 2y = 5 and x – 3y = 7 intersect 2  29 at the point P  , −  . 5 5 2  29 Equation of the line through the point P  , −  5 5

Slope of line (4) is =

1 + 2k 3 =– ∴ 1 + 3k 4 

Slope of line (1) is = –

...(1)

and having slope 5 is 29   2 = 5  x −  y + 5 5 or 5y + 2 = 5 (5x – 29) or 25x – 5y – 147 = 0. ...(1) Now, length of the perpendicular from the point (1, 2) to the line (1) is

p =

| 25 (1) − 5 (2) − 147 | (25) 2 + 52

=

132 . 650

x y + = 1. ...(1) a b  ince the length of the perpendicular drawn from S the origin (0, 0) upon line (1) is p, ∴ p =

∴

1 = p

0 0 + −1 a b 1 1 + 2 2 a b

=

1 1 1 + a 2 b2

1 1 1 1 1 + 2 or = 2+ 2. 2 a b p a b

71. (b) The given equation can be written in the form

42 + 32

=1 ⇒

4α + 3 (4 − α) − 10 16 + 9

=1

[Using (1)] ⇒ | α + 2 | = 5 ⇒ α + 2 = ± 5 ⇒ α = 3 or – 7. ∴ required sum = 3 – 7 = – 4. 75. (c) We have, |− a|

p1 =

sec θ + cosec θ 2

2

⇒ p12 =

a2 sec θ + cosec 2θ 2

2 2 2 = a sin θ cos θ ⇒ 4 p1 = a2 sin2 2θ ...(1) 1 | − a cos 2θ | Also, p2 = = | – a cos 2θ | cos 2 θ + sin 2 θ 2

⇒ p2 = a2 cos2 2θ Adding (1) and (2), we get 2 2 4 p1 + p2 = a2. 2

...(2)

76. (d) Equation of any line through the point (4, – 5) is a (x + y – 1) + b (2x + 3y – 1) = 0 which is the equation of a straight line passing through y + 5 = m (x – 4) the point of intersection of the lines x + y – 1 = 0 and ⇒ mx – y – 4m – 5 = 0. 2x + 3y – 1 = 0. The point of intersection of these lines is Since its distance from the point (– 2, 3) is equal (2, – 1). Hence the given family of lines passes through to 12, the point (2, – 1) for all values of a and b. 72. (a) Since the given lines are concurrent,

∴ ⇒

a 1− a 1− a a 1 1 = 0 ⇒ =0 b −1 1 0 1 b 1 c −1 1 0 1 1 c [Applying C2 → C2 – C1, C3 → C3 – C1] a (b – 1) (c – 1) – (1 – a) (c – 1) – (1 – a) (b – 1) = 0

[Expanding along R1] ⇒ a (1 – b) (1 – c) + (1 – a) (1 – c) + (1 – a) (1 – b) = 0 a 1 1 + + =0 ⇒ 1− a 1− b 1− c

[Dividing by (1 – a) (1 – b) (1 – c)]

1 1 1 + + =0 ⇒ – 1 + 1− a 1− b 1− c 1 1 1 + + = 1. 1− a 1− b 1− c 73. (b) The distance between the given parallel lines is | 2 − (−3)| 5 = p = 52 + (−12) 2 13 . ⇒

74. (d) Let (α, β) be any point on the line x + y = 4. Then, α + β = 4 ...(1) Since the point (α, β) is at a unit distance from the line 4x + 3y – 10 = 0, we get

∴

| m (−2) − 3 − 4m − 5| m2 + 1

= 12

⇒ | – 6m – 8 | = 12 m 2 + 1 ⇒ (6m + 8)2 = 144 (m2 + 1) ⇒ (3m + 4)2 = 36 (m2 + 1) ...(1) ⇒ 27m2 – 24m + 20 = 0 Since the discriminant of quadratic equation (1) is (– 24)2 – 4 · 27 · 20 = – 1584 which is negative, so there is no real value of m. Hence no such line is possible. 77. (b), (c)  Let a and b be the intercepts cut off by the line on x-axis and y-axis respectively. Given a = 2b. The equation of the line is x y x y + + = 1 or =1 a b 2b b or x + 2y – 2b = 0 ...(1)  ince the length of the perpendicular from the origin S upon the line (1) is 1 ∴

| 0 + 2 (0) − 2b | 12 + 22

=1 ⇒

| 2b | =1 5

2b 5 =±1 ⇒b=± . 5 2 So, the equations of the required lines are x + 2y – 5 = 0 and x + 2y + 5 = 0. ⇒

78. (a) Since the line L has intercepts a and b on the coordinate axes, therefore its equation is

355

4α + 3β − 10

Coordinates and Straight Lines

70. (a) Since a and b are the intercepts of the given line on x-axis and y-axis respectively, therefore its equation is

356

x y + =1 a b

⇒ 13 (– 4x – 3y + 6) = – 5 (5x + 12y + 9) ⇒ 27x – 21y – 123 = 0 or 9x – 7y – 41 = 0.

...(1)

Objective Mathematics

When the axes are rotated, its equation with respect x y + to the new axes and same origin will become p q = 1 ...(2)

81. (b) Re-writing the given equations so that the constant terms are positive, we have – 3x – 4y + 11 = 0  ...(1) and – 12x + 5y + 2 = 0  ...(2) Now, a1a2 + b1b2 = (– 3) (– 12) + (– 4) (5) = 16 > 0. ∴ The equation of the bisector of the acute angle between the lines (1) and (2) is

I n both the cases, the length of the perpendicular from the origin to the line will be same. ∴

1 1 1 + a 2 b2

1

=

1 1 + p2 q2

or

1 1 1 1 + = + a 2 b2 p 2 q 2



which is the required relation. 79. (a) Let O be the foot of the tower and A and B be two towns. We take O as origin and OA and OB as x-axis and y-axis respectivley. 5 km. Given : OA = 5 km and OB = 2 Let OC ⊥ AB. Then, C will be the position of the rest house. x y Equation of AB is + 5 = 1 5 2 or

x + 2y = 5

− 3 x − 4 y + 11 −12 x + 5 y + 2 = − 9 + 16 (−12) 2 + 52

⇒ 13 (– 3x – 4y + 11) = – 5 (– 12x + 5y + 2) ⇒ 11x + 3y – 17 = 0. 82. (c) Re-writing the given equations so that constant terms are positive, we have – 4x – 3y + 6 = 0 ...(1) and 5x + 12y + 9 = 0 ...(2) Now, a1a2 + b1b2 = (– 4) (5) + (– 3) (12)  = – 20 – 36 = – 56 < 0. So, the origin lies in acute angle. ∴ The equation of the bisector of the acute angle between the lines (1) and (2) is

...(1)

− 4x − 3y + 6

(− 4) 2 + (− 3) 2

= +

5 x + 12 y + 9 52 + (12) 2

⇒ – 52x – 39y + 78 = 25x + 60y + 45 ⇒ 7x + 9y – 3 = 0. 83. (b) Since the given lines are concurrent, therefore,



 ince OC is ⊥ to AB and passes through O (0, 0) S therefore its equation is 2x – y = 0 ...(2) Solving (1) and (2), we get C ≡ (1, 2). Thus, the rest house should be 1 km. east and 2 km. north of the tower.



2 −3 k 3 −4 −13 = 0 8 −11 −33

⇒ 2 (132 – 143) + 3 (– 99 + 104) + k (– 33 + 32) = 0 ⇒ – 22 + 15 – k = 0 or k = – 7.

84. (c) We take A as the origin and AB and AC as x-axis and y-axis respectively.

80. (a) The given lines are 4x + 3y – 6 = 0 or – 4x – 3y + 6 = 0  and 5x + 12y + 9 = 0 

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

Here a1 = – 4, a2 = 5, b1 = – 3, b2 = 12. Now, a1a2 + b1b2 = – 20 – 36 = – 56 < 0. ∴ The equation of the bisector of the obtuse angle between the lines (1) and (2) is − 4x − 3y + 6

(− 4) + (− 3) 2

2

= −

5 x + 12 y + 9 52 + (12) 2

Let AP = h, AQ = k. Equation of the line PQ is

...(1)

Given, BP ⋅ CQ = AB2 ⇒ (h – a) (k – a) = a2 ⇒ hk – ak – ah + a2 = a2 a a + =1 or h k

x + 2y = 1 if (2t2 + 2t + 4) + 2 (t2 + t + 1) = 1 i.e., 4t2 + 4t + 5 = 0. Here, discriminent = 16 – 4 ⋅ 4 ⋅ 5 = – 64 < 0. ∴  No real value of t is possible. Hence, the given point cannot lie on the line.

or ak + ha = hk ...(2)

91. (c) Since the angle RPQ is a right angle, ∴ Slope of RP × slope of PQ = – 1

From (2), it follows that line (1) i.e., PQ passes through the fixed point (a, a).

1− y 5 −1 × = – 1 ⇒ 3x + 4y = 13 3− x 6−3 ...(1) Also, area of ∆RPQ = 7 ⇒

85. (b) Let the equation of the line be x y + =1 ...(1) a b Its intercepts on x-axis and y-axis are a and b respectively. Given : ⇒

1 1 + =k a b

1 1 + =1 ak bk

or

1/k 1/k + =1 a b

From (2) it follows that the line (1) passes through 1 1 the fixed point  ,  . k k



3x + 4y + 6 = 0

...(1),

...(2) 2x + 3y + 2 2 = 0 and 4x + 7y + 8 = 0 ...(3) Lines (1) and (2) intersect at the point (– 2, 0). Since the point (– 2, 0) also satisfies the equation (3), therefore, the three lines are concurrent. 87. (d) Since the product of slopes of the diagonals is – 1, therefore, the diagonals are at right angles. Hence, PQRS is a rhombus. 88. (d) The coordinates of the middle point of the line joining  1 + 3 −2 + 4  the points (1, – 2) and (3, 4) are  2 , 2   i.e., (2, 1).  learly, the line x – y – 1 = 0 passes through the C point (2, 1) and is equally inclined to the axes. ∴ x – y – 1 = 0 is the equation of the required line. 89. (a) The given equation is ...(1) x2 + y2 + 2x – 4y + 6 = 0 Putting x = x′ + α, y = y′ + β in (1), we get x′ 2 + y′ 2 + x′ (2α + 2) + y′ (2β – 4) + (α2 + β2 + 2α – 4β + 6) = 0 To eliminate linear terms, we should have 2α + 2 = 0 and 2β – 4 = 0 ⇒ α = – 1 and β = 2 ∴ (α, β) ≡ (– 1, 2).

x y 1 1 |  3 1 1  | = 7 2 6 5 1

⇒

1 [x (1 – 5) – y (3 – 6) + 1 (15 – 6)] = ± 7 2

⇒ – 4x + 3y + 9 = ± 14 ⇒ – 4x + 3y = 5 ...(2) and – 4x + 3y = – 23 ...(3) Solving equations (1) and (2) and (1) and (3), we get two different coordinates of the point R. So, there are two such points R.

...(2)

86. (b) Given lines are

⇒

92. (a) Let the two perpendicular lines be the coordinate axes and the point be P (x, y). Then, the sum of the distances of P (x, y) from these lines is | x | + | y | = 1 (Given). This gives four lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 1, which form a square. 93. (b) Let S ≡ (x, y).



Given : SQ2 + SR2 = 2SP2 ⇒ [(x + 1)2 + y2] + [(x – 2)2 + y2] = 2 [(x – 1)2 + y2] ⇒ (x2 + y2 + 2x + 1) + (x2 + y2 – 4x + 4) = 2 (x2 + y2 – 2x + 1) ⇒ x = –

3 , which is a straight line parallel to 2

y-axis. 94. (d) The given line is 3x + y = 3

...(1)

The slope of line (1) is = – 3 ∴ Slope of the line which is ⊥ to line (1) is 1 = 3 Now, equation of the line through the point (2, 2) 1 (x – 2) and ⊥ to line (1) is (y – 2) = 3 ⇒ 3y – 6 = x – 2 ⇒ x – 3y + 4 = 0 ...(2) 4 . [Putting x = 0 in (2)] Its y-intercept = 3

357

90. (d) The point (2t2 + 2t + 4, t2 + t + 1) lies on the line

x y + = 1 h k

Coordinates and Straight Lines



358

Objective Mathematics

95. (b) Equation of the line L is 0 −1 (x – 2) ⇒ y = – x + 2 (y – 0) = 2 −1 Equation of the line M is 1  1 (y – 0) = 1  x −  ⇒ y = x – 2 2 5 3 Solving (1) and (2), we get A ≡  ,  . 4 4

...(1)

...(2)

m − (−2) 1 ∴ 1 + (−2) m = ± tan 45º ⇒ m = 3, – . 3 ∴ Equations of lines PQ or PR are

1 (x – 2) 3 ⇒ 3x – y – 5 = 0 and x + 3y – 5 = 0. Their combined equation is (3x – y – 5) (x + 3y – 5) = 0 i.e., 3x2 – 3y2 + 8xy – 20x – 10y + 25 = 0. (y – 1) = 3 (x – 2)

and (y – 1) = –

98. (a) Let A ≡ (0, 0), B ≡ (8, 0) and C ≡ (4, 6).

Now, area of ∆ABC 5 3 1 4 4 1 1 = |  0 − 1 1  | = 2 2 2 0 2 1

5  1   4  − 2 − 2     6−0 3 =– . 4−8 2 Equation of the line through A (0, 0) and ⊥ to BC is 2 (x – 0) i.e., 2x – 3y = 0 ...(1) (y – 0) = 3 6−0 3 = . Slope of CA = 4−0 2 Slope of BC =



=

1 25 25 × . = 2 8 16

96. (b) The given line is x + y = 1 

...(1)

Slope of line (1) is = – 1. ∴ Slope of line PQ [⊥ to line (1)] is = 1. Equation of PQ is (y – 4) = 1 (x – 2)

 quation of the line through B (8, 0) and ⊥ to E CA is 2 (x – 8) i.e., 2x + 3y = 16 ...(2) (y – 0) = – 3 8 Solving (1) and (2), the orthocentre is  4,  .  3 99. (b) Equation of two lines are

⇒ x – y + 2 = 0

...(2)

 1 3 Solving (1) and (2), we get Q ≡  − 2 , 2    Thus, the coordinates of the foot of the perpendicular  1 3 are  − 2 , 2  .   97. (b) Given line is 2x + y = 3    ...(1) Slope of line (1) is = – 2. Let m be the slope of PQ or PR. Since ∠PQR = ∠PRQ = 45º.

y

=

3 x + 2, if x ≥ 0

and y = – 3 x + 2, if x ≤ 0. Clearly y ≥ 2.

105. (b) The given line is 3αx + β y + 2γ = 0

102. (a) The equation of one of the diagonals of the square is

⇒ 3αx + β y + 2 (– α – β) = 0  ( α + β + γ = 0) ⇒ α (3x – 2) + β (y – 2) = 0 ⇒ the given line passes through the point of intersection of the lines 3x – 2 = 0 and y – 2 = 2  0 i.e.,  , 2  , for all values of α and β. 3 106. (c) The four lines enclosing the given region are 4x + 5y = 20, 4x – 5y = 20, – 4x + 5y = 20 and – 4x – 5y = 20. Clearly, the four lines form a rhombus having diagonals of length 10 and 8. 1 × 10 × 8 = 40. ∴  Required area = 2

x = 2y ...(1) Let the equation of one of the sides of the square through the vertex (3, 0) be (y – 0) = m (x – 3) ...(2) Since the angle between (1) and (2) is 45º, 1    m − 2  2m − 1 =±1 ∴ tan 45º = ±  ⇒ 1 2+m  1 + m ⋅  2  1 ⇒ m = 3, – . 3 Putting these values of m in (2), we get y – 3x + 9 = 0 and 3y + x – 3 = 0 as the required two sides of the square through the vertex (3, 0). 103. (d) The equation of the given line can be re-written as p (x + y – 1) + q (2x – 3y + 1) = 0

107. (c) Since a, b, c are in H.P., ∴

 ⇒

1 1 2 + = a c b

1 1 1 , , are in A.P. a b c ...(1)

359

Coordinates and Straight Lines

which, clearly, passes through the point of intersecAlso, y-axis is the bisector of the angle between the tion of the lines two lines. P1, P2 are two points on these lines, at a distance 5 units from A. Q is the foot of the ⊥ x + y –1=0 ...(1) from P1 and P2 on the bisector ( y-axis). and 2x – 3y + 1 = 0 ...(2) Then, the coordinates of Q are (0, 2 + 5 cos 30º) for different values of p and q.  5 3  1  Solving (1) and (2), we get the coordinates of the =  0, 2 + 2  =  0, 2 (4 + 5 3 )  .   2 3 point of intersection as  ,  . 100. (a) Since P is the image of the point (– 3, 2) w.r.t. x5 5 axis, therefore, the coordinates of P are (– 3, – 2). 104. (a), (b), (d)  Let the variable line be Let (x′, y′ ) be the coordinates of P w.r.t. new axes. ax + by + c = 0  ...(1) Then, x′ = x cos α + y sin α = – 3 cos (– 60º) – 2 sin (– 60º) (2a + c) + (2b + c) + (a + b + c) Given :  =0 3 2 3−3 a 2 + b2 + c2 . = – + 3 = ⇒ 3a + 3b + 3c = 0 or a + b + c = 0. 2 2 So, the equation of the line becomes y′ = – x sin α + y cos α = 3 sin (– 60º) – 2 cos (– 60º) 3 3 + 2 ax + by – a – b = 0 = –  2  or a (x – 1) + b (y – 1) = 0. ⇒ the line passes through the point of intersection    of lines x – 1 = 0 and y – 1 = 0 i.e., the fixed ∴ Coordinates of P are  2 3 − 3 , − 3 3 + 2  .  2     2 point (1, 1). S  o, all such lines are concurrent. Also, (1, 1) is the 101. (b) Clearly, the equations of the two diagonals are centroid of the ∆ABC. x = 5 and y = 5.

360

The given line is

x y 1 + + =0 a b c

Objective Mathematics

⇒

2 1 x y + +  −  = 0 b a a b

⇒

1 1 (x – 1) + (y + 2) = 0 a b

 p  , 0 Then A ≡  cos α  [Using (1)]

⇒ The given line passes through the point of intersection of x – 1 = 0 and y + 2 = 0 i.e., (1, – 2) which is a fixed point. 108. (a) Let P ≡ (a sin θ, 0) and Q ≡ (0, a cos θ) Let R be the middle point of PQ. Then  a sin θ + 0 0 + a cos θ   a sin θ a cos θ  , ,  R ≡   =   2 2 2 2   he distance of the middle point R from the origin T is  a sin θ  a  − 0  +  cos θ − 0    2  2 2



= =

1 2

a 2 sin 2 θ + a 2 cos 2 θ =

[obtained by putting y = 0 in the given equation] p   and B ≡  0,  sin α  [obtained by putting x = 0 in the given equation] Since C (x1, y1) is the mid point of AB, p p and 2y1 = ∴ 2x1 = cos α sin α p p ⇒ cos α = 2 x and sin α = 2 y 1 1 p2 p2 −2 −2 ⇒ 4 x 2 + 4 y 2 = 1 or x1 + y1 = 4 p–2 1 1 ∴ Required locus is x–2 + y–2 = 4 p–2. 112. (c) Since the points (x, y), (x′, y′ ) and (x – x′, y – y′ ) are collinear, therefore

2



a . 2

113. (c) The given line is

...(1)

x y − =0 b a

...(2)

Let the third vertex be C (α, β).

Area of ∆ABC =

1 2

y 1 y' 1 = 0 ⇒ xy′ = x′ y. y − y' 1

x y + =1  a b Equation of the line ⊥ to (1) from (0, 0) is

109. (b) Let A ≡ (2, – 1) and B ≡ (3, 2). Then, α + β = 5 (given)

x x' x − x'

...(1)

2 −1 1 3 2 1 = ± 4 (given) α β 1

⇒ β – 3α = 1

...(2)

or β – 3α = – 15

...(3)

Solving (1) and (2), we get, α = 1, β = 4 Solving (1) and (3), we get, α = 5, β = 0. Thus, the third vertex is either (5, 0) or (1, 4). 110. (c) 111. (a) Let C (x1, y1) be the mid point of the intercept AB of the line x cos α + y sin α = p between the coordinate axes.



 et (α, β) be the point of intersection of (1) and L (2). α β + =1 ...(3) Then, a b α β − =0 ...(4) and b a Squaring and adding (3) and (4), we get 1 1 1 1 α2  2 + 2  + β2  2 + 2  = 1 a b b a 1 1 ⇒ (α2 + β2)  2 + 2  = 1 a b 1 1 1  1 ⇒ (α2 + β2)  2  = 1  2 + 2 = 2  c a b c ⇒ α2 + β2 = c2. Thus, locus of (α, β) is x2 + y2 = c2. 114. (a) The given lines are 3x + 4y = 9 ...(1) and y = mx + 1 ...(2) Solving (1) and (2), we get the x-coordinate of the 5 . point of intersection as x = 4m + 3 Since x-coordinate is an integer, ∴ 4m + 3 = ± 5 or 4m + 3 = ± 1.

115. (a) Let orthocentre be (α, β). 2 (6) + 1 × α 2 ( 2) + 1 × β and 3 = 2 +1 2 +1 ⇒ α = – 3 and β = 5. ∴ Orthocentre is (– 3, 5). Then, 3 =

116. (c) 117. (b) The equation of any line through the point of intersection of the given lines is (4x + 3y – 7) + k (8x + 5y – 1) = 0 ⇒ (4 + 8k) x + (3 + 5k) y – 7 – k = 0.

x2 x3 y2 y3 122. (a) Let x = x = r and y = y = r 1 2 1 2 ⇒ x2 = x1r, x3 = x1r2, y2 = y1r and y3 = y1r2. We have, x1 ∆ = x2 x3

=

x1 0 0

y1 1 x1 y2 1 = x1r y3 1 x1r 2 y1 1 0 1− r 0 1− r

y1 1 y1r 1 y1r 2 1



[Applying R3 → R3 – rR2 and R2 → R2 – rR1] ( 4 + 8k ) 3 = 0   ( R2 and R3 are identical) =– (given) 3 + 5k  2 Thus, (x1, y1), (x2, y2), (x3, y3) lie on a straight ⇒ 8 + 16k = 9 + 15k or k = 1. line. ∴ Required line is 12x + 8y – 8 = 0 or 3x + 2y 123. (c) PQRS will represent a parallelogram if and only if – 2 = 0. the mid point of PR is same as that of QS. That is, 118. (c) The equations ax + by + c = 0 and dx + ey + f = 0 if and only if represent the same line if their slopes are equal and 1+ 5 4 + a 2+7 6+b y-intercepts are equal. = = and 2 2 2 2 a d c f = − = − and – ∴ – ⇒ a = 2 and b = 3. b e b e b c a b c a b 124. (a) Let P ≡ (x1, y1), Q ≡ (x2, y2); R ≡ (x3, y3), = = = = ⇒ and ⇒ . e f d e f d e where xi, yi (i = 1, 2, 3) are rational numbers. Now, the centroid of ∆PQR is  11  119. (a) Any point on the line 4x + 3y = 11 is  0,  3  x1 + x2 + x3 y1 + y2 + y3  ,   3 3   11 w hich is rational point. Incentre, circumcentre 0 × 8 + 6 × − 15 7 . 3 Required distance = and orthocentre depend on sides of the trian= 10 64 + 36 gle which may not be rational even if vertices Its slope = –

120. (d) Reflection in the y-axis gives the new position as (– 3, 2).

are so. For example, for P (0, 1) and Q (1, 0); PQ = 2 . 125. (d) Slope of median PS is

 hen it moves towards the negative side of y-axis W through 5 units, then the new position is (– 3, 2 – 5) i.e., (– 3, – 3). 121. (d) Since a, b, c are in A.P.,

∴ 2b = a + c or

a – 2b + c = 0

1− 2 2 =− . m = 13 9 −2 2 The equation of the line passing through (1, – 1) and parallel to PS is

y+1=–

2 (x – 1) or 9

2x + 9y + 7 = 0.

361

⇒ The line ax + by + c = 0 passes through the point (1, – 2).

Coordinates and Straight Lines

 olving these, only integer values of m are – 1 S and – 2. ∴ m = – 1, – 2.

362

126. (d) Let A ≡ (1,

3 ), B ≡ (0, 0) and C ≡ (2, 0).

Objective Mathematics

Then, AB =

1 + 3 = 2, BC = 2 and

2 2 CA = (1 − 2) + ( 3 − 0) = 2. Thus, AB = BC = CA. As ∆ABC is an equilateral triangle, the incentre coincides with the centroid of the triangle which is



 0 +1+ 2 0 + 0 + 3   1  ,  i.e., I 1, . I  3 3 3   

Thus, OA =

127. (a) Since the lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent a b c ∴ b c a =0 c a b

Since OA + OB = c (given), we get 2 2 h2 + k 2 + h + k = c k h  ⇒ (h2 + k2) (h + k) = chk. Therefore, required locus is (x2 + y2) (x + y) = cxy.

=0



1 ∆ = 2

[Applying R1 → R1 + R2 + R3 and taking (a + b + c) common from R1]



=

1 0 0 ⇒ (a + b + c) b c − b a − b c a−c b−c

=0

[Applying C2 → C2 – C1, C3 → C3 – C1] ⇒ (a + b + c) [(c – b) (b – c) – (a – b) (a – c)] =0 [Expanding along R1] ⇒ (a + b + c) (ab + bc + ca – a2 – b2 – c2) = 0 ⇒ a3 + b3 + c3 – 3abc = 0 ( a + b + c ≠ 0).

128. (a) Since the lines x – 2y – 6 = 0, 3x + y – 4 = 0 and λ x + 4y + λ2 = 0 are concurrent 1 −2 −6 ∴ 3 1 − 4 = 0 ⇒ 7λ2 + 14λ – 56 = 0 λ 4 λ2 ⇒ λ2 + 2λ – 8 = 0 ⇒ λ = 2, – 4. 129. (c) Let P (h, k) be any point on the locus, then

OP ⊥ AB.

2 2 h2 + k 2 , OB = h + k h k

130. (d) Area of the triangle formed by the given points is 1 1 1 ⇒ (a + b + c) b c a c a b





h . k So, equation of AB is h (x – h) ⇒ ky + hx = h2 + k2. y– k=– k x y + 2 ⇒ =1 (h 2 + k 2 ) (h + k 2 ) h k ∴ Slope of AB = –

1 2

(a + 3)(a + 4) a + 3 1 (a + 2)(a + 3) a + 2 1 (a + 1)(a + 2) a + 1 1

4a + 10 2 0 2a + 4 1 0 (a + 1)(a + 2) a + 1 1

[Applying R1 → R1 – R3, R2 → R3 – R3] 1 [(4a + 10) – 2 (2a + 4)] =  2 [Expanding along C3] 1 (2) = 1. = 2

131. (c) Since the given lines are concurrent, ∴

a 12 1 a 12 1 = 0 ⇒ b a − 1 0 =0 b 13 1 c−b 1 0 c 14 1

[Applying R3 → R3 – R2, R2 → R2 – R1] ⇒ b – a – c + b = 0 or 2b = a + c ⇒ a, b, c are in A.P. 132. (b) Given lines are 3x + y – 5 = 0 ...(1) and x – y + 1 = 0 ...(2) Slope of line (1) is = – 3 and slope of line (2) is = 1. Equation of any line through the point of intersection of the given lines is (3x + y – 5) + k (x – y + 1) = 0. Since this line is ⊥ to one of the given lines, 3+ k 1 = – 1 or ⇒ k = – 1 or – 5. k −1 3 Therefore, the required straight line is x + y – 3 = 0 or x – 3y + 5 = 0.

∴

134. (b) Let O ≡ (0, 0), A ≡ (3, 0) and B ≡ (0, 4).  he triangle OAB is a right angled triangle, right T angled at O. Therefore, O (0, 0) is the orthocentre of the triangle. 135. (c) Let A ≡ (0, 3), B ≡ (– 3, 0) and C ≡ (3, 0) be the vertices of ∆ABC with O, the mid point of BC as origin and OA as y-axis and BC as x-axis.

Slope of median AD is

m1 = –

Slope of median BE is

m2 =

Since AD ⊥ BE, 

⇒–

b a

2b a

363

| α | + | β | = 1 (given), whose graph is a square.

∴ m1 × m2 = – 1 2b b × =–1 ⇒a=± a a

2 b.

139. (b) Since the ⊥ distance between the given lines is 2 , ∴ the required line is a straight line ⊥ to the given parallel lines and passing through (– 5, 4). The equation of any line ⊥ to the given lines is x – y + k = 0. This passes through (– 5, 4), ∴ – 5 – 4 + k = 0 ⇒ k = 9. Hence, the required line is x – y + 9 = 0.

Then, OA ⊥ BC. Slope of AB =

3−0 = 1. 0+3

⇒ ∠OBA = 45º,   ∴  ∠BAC = 90º i.e., CA ⊥ AB. Hence, orthocentre is (0, 3). π π or α – β = 136. (c) The lines are ⊥ if β – α = 2 2 π i.e., if | α – β | = . 2 137. (a) The equation of line BC is x + y + 4 = 0. ∴ Equation of a line parallel to BC is x + y + k = 0. k 1 from the origin. ∴ This is at a distance 2 2 1 . = 2 1 . ∴ k = ± 2

140. (d) External division in the ratio 1 : 1 is not defined. 141. (b) Diagonals of a square bisect each other at right angles. Thus the diagonal BD is ⊥ to the diagonal AC.

 ince BC and the required line are on the same S side of the origin 1 . ∴ k = 2 1 = 0. Hence, the required line is x + y + 2 138. (d) The coordinates of the middle points D and E are Let the equation of BD be x + 7y + k = 0 ...(1) The line (1) passes through the point B (– 4, 5) ∴ – 4 + 35 + k = 0 ⇒ k = – 31. ∴  The equation of diagonal BD is x + 7y – 31 = 0.

 8 142. (d) Let A ≡  0, 3  , B ≡ (1, 3) and C ≡ (82, 30).   a  a b D ≡  , 0  , E ≡  ,  2 2 2

82 10 10 , BC = 27 . 10 , AC = 3 3 Since AB + BC = AC, So A, B, C are collinear.

Then, AB =

Coordinates and Straight Lines

133. (a) If α and β are the lengths of perpendiculars, then

364

143. (b) Let the coordinates of C be (h, k).

Objective Mathematics

∴ Slope of line CD (⊥ to AO) = ∞. Also, the line CD passes through (5, 5). ∴ Equation of line CD is x – 5 = 0. 145. (c) Solving the given equations in pairs, we get the coordinates of the vertices as

 h +1 k −1  Then mid point of AC i.e.,  2 , 2  lies on   3x – 2y + 8 = 0 and AC is ⊥ to this line. 2 (k – 1) + 8 = 0 ∴ 3 (h + 1) – 2 2 and

k +1 3 × =–1 h −1 2

or 3h – 2k = – 21 and 2h + 3k = – 1. Solving these equations, we get h = – 5, k = 3. ∴ C ≡ (– 5, 3). The equation of the ⊥ bisector of AB is y – 0 = (– 1) (x – 2) or x + y = 2. Circumcentre is the point of intersection of these right bisectors of AB and AC. Solving these right  4 14  bisectors we get circumcentre as  − 5 , 5  .   144. (c) AO divides BC in the ratio of arms of the angle. Equation of AB is y = – x + 2.

A ≡ (– 3, 3), B ≡ (1, 1), C ≡ (1, – 1), D  ≡ (– 2, – 2) −1 − 3 4 =− = – 1 m1 = slope of AC = 1+ 3 4 −2 − 1 −3 =− = 1. m2 = slope of BD = −2 − 1 −3

m1m2 = (– 1) (1) = – 1.

So, diagonals AC and BD are perpendicular. 146. (c) Clearly point (0, β) lies on y-axis. Drawing the graph of the three straight lines, we  7  5 see that Q ≡  0, 2    and  P ≡  0, 3  .    

Therefore, the point (0, β) lies on or inside ∆ABC, 5 7 ≤β≤ . when 3 2 147. (c) Let A ≡ (2, 4), B ≡ (2, 6) and C ≡ (2 + 3 , 5). Then, AB =

(2 − 2) 2 + (6 − 4) 2 =

0+4 = 2



BC =

(2 + 3 − 2) 2 + (5 − 6) 2 =

3 +1 = 2



CA =

(2 + 3 − 2) 2 + (5 − 4) 2 =

3 + 1 = 2.

∴ AB = BC = CA, so the triangle is equilateral. Equation of AC is y = x. ∴ Angle between AB and AC = 90º. ∴ The bisected angle = 45º. Length of AB = 3 2 . Length of AC = 4 2 . BO AB 3 = = . Also, OC AC 4  31  ∴ Coordinates of O are  7 , 1 .  

1 −1 Slope of AO = 31 = O. 1− 7

148. (d) Since the centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1, ∴ if circumcentre is (α, β) then 2 × α + 1 × (−3) ⇒ 2α = 9 + 3 3= 2 +1 2 × β +1× 5 ⇒ 2β = 9 – 5 and 3 = 2 +1 ∴ α = 6, β = 2 ∴ Circumcentre is (6, 2). 149. (c) If O ≡ (0, 0), A ≡ (2, 0) and B ≡ (0, 3), then π (since A lies on x-axis and B lies on ∠AOB = 2 y-axis). So, the ∆AOB is a right angled triangle.

m +1=–

 ∴ m =

3 (1 – m)

3 +1 ( 3 + 1)( 3 + 1) = =2+ 3 −1 ( 3 − 1)( 3 + 1)

3.

 ubstituting these values of m in (1), the two sides S of the equilateral triangle are y – 3 = (2 ± 3 ) (x – 2). 154. (b) Let the vertices of the equilateral triangle be (x1, y1), (x2, y2) and (x3, y3). If none of xi and yi (i = 1, 2, 3) are irrational, then 1 2

x1 x2 x3

y1 1 y2 1 = rational. y3 1

Since the point R lies on the line 5x + y + 6 = 0

area of ∆ =

 α + 4   β − 13  ∴ 5    +6=0 + 2   2 

But the area of an equilateral triangle =

⇒ 5α + β + 19 = 0

...(1)

β + 13 Also, slope of PQ is = α−4 and slope of given line = – 5. Since PQ is ⊥ to given line

3 (side)2 4

= irrational. Thus, the two statements are contradictory. Therefore, both the coordinates of the third vertex cannot be rational. 155. (b) Coordinates of the point P, dividing the join of A (6, 6) and B (1, – 3) in the ratio 2 : 3 are

 β + 13  ∴   (– 5) = – 1 α−4 ⇒ 5β + 65 = α – 4 or α – 5β – 69 = 0  ...(2) Solving (1) and (2), we get α = – 1 and β = – 14 ∴ The image point is (– 1, – 14). 151. (c) Let A ≡ (– a, – b), B ≡ (0, 0), C ≡ (a, b) and D ≡ (a2, ab) b b , slope of BC = Then, slope of AB = a a

α =

Thus, the points A, B, C and D are collinear.

and  β =

152. (c) According to the question, x + 1 = 4

2 (1) + 3 (6) 2+3

ab − b b (a − 1) b and sloep of CD = a 2 − a = a (a − 1) = a .



⇒x=3 and y – 2 = 5 ⇒ y = 7 So, the point is (3, 7).

153. (a) Equation of the side through the vertex (2, 3) is y – 3 = m (x – 2) ...(1) The side is inclined at an angle of 60º to the third side x+y=2 ...(2) Since slopes of (1) and (2) are m and – 1, m − (−1) ∴ tan 60º = ± 1 + m (−1) 

or m + 1 = ±

Taking +ve sign, we get m + 1 = ∴ m =

3 (1 – m).

3 –

3 −1 ( 3 − 1)( 3 − 1) =2– = 3 +1 ( 3 + 1)( 3 − 1)

3m 3.

=

20 =4 5

2 (−3) + 3 (6) − 6 + 18 12 = = 2+3 5 5

 12  ∴ (α, β) ≡  4,  5

Since the point P lies on the line x + λ y = 2  12  5 ∴ 4 + λ    = 2 or λ = – . 5 6 156. (a) We have

(x – 2) (x – 3) + (y – 1) (y + 4) = 0

 y + 4   y −1 ⇒  ×  =–1 x − 2   x − 3 π . ⇒ RP ⊥ RQ or ∠PRQ = 2 ∴ The point R lies on the circle whose diameter is PQ. 13 Now, area of ∆PQR = 2

365

Now, taking –ve sign, we get

Coordinates and Straight Lines

150. (c) Let Q (α, β) be the image of the point P (4, – 13). Then, the coordinates of the mid point R are  α + 4 β − 13  ,  . 2 2 

366

159. (d) The given lines are

Objective Mathematics

3x + 4y = 1 and (1 + c) x + 3c2y = 2 Solving (1) and (2), we get 1 ⇒ × 2

13 26 × (altitude) = 2

⇒ altitude =

Now, and

26 = radius 2

⇒ there are two possible positions of R. 157. (a) Let DL be the required line.

x=

Let BL = x

1  ow, Area of trap. ABLD = N area of trap. ABCD 2 (Given) 1 1 1 (x + 2) × 4 =  ⋅  (4 + 2) × 4 ⇒ 2 2 2  ⇒ x = 1. ∴ Coordiantes of the point L are (– 1, 4). So, equation of the line DL is

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

3c 2 − 8 5−c , y= 9c − 4c − 4 9c 2 − 4c − 4 2

3c 2 − 8 −5 = =–5 c →1 9c − 4c − 4 9−4−4

lim lim c →1

2

5−c 4 = = 4. 9c − 4c − 4 9−4−4 2

∴ The limiting position of the point of intersection of the given lines is (– 5, 4). 160. (b) The point (1, β) lies on the line x = 1, for all real β. Clearly, from the figure, it will lie on or inside the triangle formed by the given lines if 0 ≤ β ≤ 1.

161. (a) Coordinates of U are (3 – 3 cos 60º, 3 sin 60º)

3 3 3 i.e.,  2 , 2  .

Coordinates of S are (6, 2 ⋅ 3 sin 60º) i.e., (6, 3 3 ).



8−4 (x – 1) or 2x – y + 6 = 0. 1+1 158. (c) Coordinates of any point on the line through (2, 3) and parallel to the line x – y = 4, at a distance r, are (2 + r cos θ, 3 + r sin θ), where tan θ = 1.

( y – 8) =

∴  The equation of US is 3 3 2 (x – 6) 3 6− 2 1 (x – 6)  or x – 3 y + 3 = 0. ⇒ y – 3 3 = 3

y –3 3 =

If this point lies on the line 3x + 2y = 17, then 3 (2 + r cos θ) + 2 (3 + r sin θ) = 17 1 1 + 6 + 2r ⋅ = 17 ⇒ 6 + 3r ⋅  2 2 5 ⇒ r=5 ⇒ r = 2. 2

3 3−

162. (c) The refracted ray passes through the point (5, 0) and makes an angle 120º with positive direction of x-axis ∴ The equation of the refracted ray is (y – 0) = tan 120º (x – 5)

or

3 x + y – 5 3 = 0.

168. (b) Since, the given lines are concurrent. ∴

3 4 1 5 λ 3 =0 2 1 −1

⇒ 3(–λ – 3) – 4(–5 – 6) + 1 (5 – 2λ) = 0 ⇒ –3λ – 9 + 20 + 24 + 5 – 2λ = 0 ⇒ –5λ + 40 = 0 ⇒λ=8 163. (a) G being the centroid, divides AD in the ratio 2 : 1.

169. (b) L et three points be A(–2, –5), B(2, – 2) and C(8, a). If three points are collinear, then

slope of AB = slope of BC

5 ⇒ −2 + 5 = a + 2 ⇒ a = 2 2+2 8−2

170. (c) Let P(1, 1), Q(–1, – 1) and R (− 3, 3) be the vertices of ∆PQR. Since AG = 2, ∴  GD = 1, ∴ Coordinates of D, using section formula, are D (1, 3). Now AD = 1 + 2 = 3, 3  ⇒ BD = 3 . ∴ tan 60º = BD ∴ B ≡ (1, 3 + 3 ) and C ≡ (1, 3 – 3 ). 1 , it’s angle with positive 3

164. (c) Slope of given line is x-axis is 30º.

 ow, lines making an angle 30º from it are either xN axis (ie, y = 0) or at an angle 60º with positive x-axis (i.e., y = 3 x + λ ). 165. (c) For the point (a, a) to fall between the lines x + y + 2 = 0 and x + y – 2 = 0. (2a + 2) (2a – 2) < 0



QR = (− 3 + 1) 2 + ( 3 + 1) 2 = 2 2

and PR = (− 3 − 1) 2 + ( 3 − 1) 2 = 2 2 PQ = QR = PR Which shows, triangle PQR is an equilateral triangle.

∴

171. (b) The graph of equations x – 2y = 0 and 3x – y = 0 is as shown in the figure. Since, given point (a, a2) lies in the shaded region, therefore a – 2a2 < 0 and 3a – a2 > 0 1 ⇒ a ∈ (−∞, 0) ∪  , ∞  2  1  and a ∈ (0, 3) ⇒ a ∈  , 3  2 

⇒ –1 < a < 1 ⇒ |a| < 1

y

166. (a) The point of intersection of the lines 3x + y + 1 =  4 7 0 and 2x – y + 3 = 0 is  − 5 , 5  . The equation of  



167. (b) We have, d1 =

and d 2 =

9 32 + 42

=

3 = 0 i.e., 5x + 5y – 3 = 0 5

9 5

15 15 15 2 = = . 2 2 2 ⋅ 5 10 3 +4

∴ Distance between L1 and L2 is d = d1 – d2 =

y=

4 7 3 + =a ⇒ a= 5 5 5

∴ Equation of line is x + y −



3x −

−

0

line which makes equal intercepts with axes is x + y = a. ∴

9 15 18 − 15 3 − = = . 5 10 10 10

PQ = (1 + 1) 2 + (1 + 1) 2 = 2 2

Then,

x

y −2

=0

O

x

172. (c) We have, PQ = (4 − 2) 2 + (−1 − 7) 2 = 68 QR = (−2 − 4) 2 + (6 + 1) 2 = 85 2 2 and RP = (−2 − 2) + (6 − 7) = 17

Since QR2 = RP2 + PQ2 therefore Since, triangle PQR is a right angled.

367

3 (x – 5)

Coordinates and Straight Lines

⇒ y = –

Slope of the line QM =

Objective Mathematics

2π =− 3 3

y M

R (3,3√3)

2π/3

P (−1, 0)

π/3 Q(0, 0)

x

∴ 3a + 2b = 13…(i) and point Q(b, a) is on 4x – y = 5 ∴ 4b – a = 5 …(ii) On solving Eqs. (i) and (ii), we get a = 3, b = 2 ∴ P(a, b) → (3, 2) and Q(b, a) → (2, 3) ∴ equation of PQ is 3−2 (x – 3) or x + y = 5 y–2= 2−3 178. (c) The equation of bisector of acute angle formed between the lines 4x – 3y + 7 = 0 and 3x – 4y + 14 = 0 is 4x − 3y + 7 3 x − 4 y + 14 =− 16 + 9 16 + 9 or x – y + 3 = 0

⇒ 7x – 7y + 21 = 0 Hence, equation of line QM is y = – or

3x

3x + y = 0

174. (c) Since, triangle is an isosceles triangles. Hence, centroid is the desired point.  4 ∴ Coordinates of R are  3, 3  .   y

P(3, 4)

R O (0, 0)

x

Q(6, 0)

179. (a) E  quation of the line passing through (–4, 6) and (8, 8) is 8−6 ( y − 6) =   ( x + 4) 8+4 ⇒ 6y – x – 40 = 0 …(i) Now, equation of any line perpendicular to Eq. (i) is 6x + y + λ = 0 …(ii) This line passes through the mid point of (–4, 6) and (8, 8), which is (2, 7) ∴ 6 × 2 + 7 + λ = 0 ⇒ 19 + λ = 0 ⇒ λ = – 19 On putting λ = – 19 in Eq. (ii), we get the required line 6x + y – 19 = 0. 180. (d) Since, slope of PQ =

4−3 1 = 1− k 1− k

A

175. (a) y

x′

(−2, 0)

(0, 0) O

x

P

M

1

( K +2 1 , 72

(0, − 2)

Q

(

368

173. (a) QM is the bisector of the angle PQR.

∴ Slope of AM = (k – 1) y′

The Vertices of the triangle are (–2, 0), (0, 0), (0, – 2) Since, In a right angled triangle circumcentre is mid point of hypotenuse, therefore the point (–1, –1) is the circumcentre. 176. (a) The foot of the pole is at the centroid. 177. (b) Since, point P(a, b) is on 3x + 2y = 13

y−

∴ Equation of AM is

  k + 1  7 = (k − 1)  x −   2   2 

For y-intercept, x = 0, y = – 4 ∴ −4 − ⇒

7  k +1 = −(k − 1)   2  2 

15 k 2 − 1 = 2 2

⇒ k2 – 1 = 15 or k = ± 4

∴ −

a = ±1 ⇒ a = ± b … b

…(i)

distance of line ax + by + c = 0 from (1, –2) | a − 2b + c | = a 2 + b2 distance of line ax + by + c = 0 from (3, 4) | 3a + 4b + c |

=



a +b 2

Given:

a +b 2

184. (a) The equation of the tangent to the curve y = ex at (c, ec) is y – ec = ec(x – c) …(1) Equation of the line joining the given points is y – ec – 1 =

2

| a − 2b + c |

4 −m ∴ tan 45º = 3 4 1+ m 3 1 ⇒ m = , – 7 7

=

| 3a + 4b + c |

e c (e − e −1 ) [x – (c – 1)] …(2) 2

Eliminating y from (1) and (2), we get

a 2 + b2

2

369

a b

[x – (c – 1)] [2 – (e – e–1)] = 2e–1

⇒ 3a + 4b + c = ± (a – 2b + c) e + e −1 − 2 or x – c = 2 − (e − e −1 ) < 0 ⇒ x < c. ⇒ a + 3b = 0 …(ii) if taking +ive 2a + b + c = 0 … . (iii) if taking –ive ⇒ the line (1) and (2) meet on the left of the line From (i) and (ii) we get a = b = 0 which is not possible x = c. so taking (i) and (iii), we get (taking a = – b from (i)) 1 a+c=0⇒c=–a 185. (b) Given: 18 = (3α) (2r) ⇒ αr = 6 2 a : b : c = a : –a : – a = 1 : –1 : –1 or a = 1, b = – 1, c = –1 2r (x – 2α) is a tangent to the circle The line y = − OB α 182. (a) In ∆AOB, tan 30º = (x – r)2 + (y – r)2 = r2 OA ∴ 2α = 3r and αr = 6

y A ( 6, 0 )

C ( α, 2r )

( 0, 2r ) D

30º

∴ r = 2.

(r,r) B O

x (0, 0 )

⇒

1 OB = 6 3

⇒ OB =

A

B ( 2α , 0 )

186. (c) Since, ∆ is isosceles, hence centroid is the desired point.

6 3 =2 3 3

1 ∴Area of ∆ AOB = × OA × OB 2 1 = × 6 × 2 3 = 6 3 sq unit 2 183. (a) The given equations are

3x + 4y – 5 = 0 …(i) and 4x – 3y – 15 = 0 …(ii)  ince, there lines are perpendicular to each other, so S ∠QPR is right angle and PQ = PR. Hence, ∆PQR is a right angle isosceles triangle. ∴ ∠PQR = ∠PRQ = 45º 3 4 Slope of PQ = − and slope of PR = 4 3 Let slope of QR = m

( 3, 4 )

( 0, 0 )

( 6, 0 )

187. (d) Let α = β = θ = π/6   3  3  1 ,O   , O = 1,  R =  Then, P =  O, −  2   2   2 

which are non-collinear therefore option (D) is correct

Coordinates and Straight Lines

181. (b) Slope of given line ax + by + c = 0 is −

370

EXERCISES FOR SELF-PRACTICE

Objective Mathematics

1. Let P = (– 1 0), Q = (0, 0) and R = (3, 3 3 ) be three points. Then the equation of the bisector of the angle PQR is (a)

3 x + y = 0 2

(b) x + 3 y = 0

(c)

3x + y = 0

(d) x +

3 y 2

(b) 3 : 4 (d) 4 : 3

3. The line passing thro′  (0, 1) and perpendicular to the line x – 2y + 11 = 0 is (a) 2x + 2y + 11 = 0 (c) 2x – y + 3 = 0

(b) 2x + y – 1 = 0 (c) 2x – y + 1 = 0

4. Two points (a, 0) and (0, b) are joined by a straight line. Another point on this line is (a) (– 3a, 2b) (c) (3a, – 2b)

(b) (a, b) (d) (a2 · ab)

5. The straight line 5x + 4y = 0 passes through the point of intersection of the lines (a) x + y – 2 = 0, 3x + 4y – 7 = 0 (b) x – y = 0, x + y = 0 (c) x + 2y – 10 = 0, 2x + y + 5 = 0 (d) none of these

(a) 120º (c) 180º

(b) right angled (d) None of these

(b) 135º (d) 90º

9. The graph of the function cos x ⋅ cos (x + 2) – cos2(x + 1) is a (a) straight line passing through the point (0, – sin21) with slope 2 (b) straight line passing through the origin (c) parabola with vertex (1, – sin21) π 2  (d) straight line passing through the point  , − sin 1 2  and parallel to the x-axis. 10. In a right angled isosceles triangle the ratio of the circum radius and in radius is (a) 2 ( 2 + 1) : 1

(b) ( 2 + 1) : 1

(c) 2 : 1

(d)

2 :1

11. The equation of straight line passing through point of intersection of the straight lines 3x – y + 2 = 0 and 5x – 2y + 7 = 0 and having infinite slope is: (a) x = 2 (c) x = 3

6. The orthocentre of the triangle formed by the lines x + y = 1, 2x + 3y = 6 and 4x – y + 4 = 0 lies in quadrant number (a) I (c) III

(a) isosceles (c) equilateral

8. The angle between the lines 3x + y – 7 = 0 and x + 2y + 9 = 0 will be

2. A straight line throught the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. Then the point O divides the segment PQ in the ratio (a) 1 : 2 (c) 2 : 1

7. The straight lines x + y – 4 = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is

(b) x + y = 3 (d) x = 4

12. The straight lines x + y – 4 = 0, 3x + y – 4 = 0, x = 3y – 4 = 0 form a triangle, which is: (a) isosceles (c) equilateral

(b) right angled (d) None of these

(b) II (d) IV

Answers

1. (c) 11. (c)

2. (b) 12. (a)

3. (b)

4. (c)

5. (b)

6. (a)

7. (a)

8. (b)

9. (d)

10. (b)

10

Pair of Straight Lines

CHAPTER

Summary of conceptS General HomoGeneouS equation of equation of tHe BiSectorS of tHe Second deGree anGleS Between tHe pair of lineS An equation of the form ax2 + 2hxy + by2 = 0 is called a homogeneous equation of second degree. It represents a pair of straight lines both passing through the origin. (a) The lines are real and distinct if h2 >ab. (b) The lines are real and coincident if h2 = ab. (c) The lines are imaginary if h2 < ab. (d) If the two lines represented by ax2 + 2hxy + by2 = 0 are y = m1x and y = m2x, then ( y – m1x) (y – m2x) = y2 +

∴

m1 + m2 = –

2h a xy + x 2 = 0 b b

2h a and m1m2 = . b b

The equations of the bisectors of the angles between the pair of lines represented by x 2 − y 2 xy ...(1) ax2 + 2hxy + by2 = 0 is = a−b h Remarks 1. If a = b, the bisectors are x2 – y2 = 0 i.e., x – y = 0, x + y = 0. 2. If h = 0, the bisectors are xy = 0, i.e., x = 0, y = 0. 3. Since in eqn. (1), coefficient of x2 + coefficient of y2 = 0, the two bisectors are always perpendicular to each other.

anGle Between tHe pair of StraiGHt lineS

General equation of Second deGree

The angle θ between the pair of straight lines represented by ax2 + 2hxy + by2 = 0 is given by

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

tan θ = ±

2 h 2 − ab a+b

and the acute angle is given by tan θ =

2 h 2 − ab 2 h 2 − ab = . a+b |a +b|

Remarks 1. The two lines are coincident if and only if h2 – ab = 0 2. The two lines are perpendicular if and only if a + b = 0, i.e., coefficient of x2 + coefficient of y2 = 0. 3. The equation to the pair of straight lines passing through the origin and perpendicular to the pair of lines is given by bx2 – 2hxy + ay2 = 0. Thus, to obtain the equation to the pair of lines passing through the origin and perpendicular to the pair of lines ax2 + 2hxy + by2 = 0 We should interchange the coefficients of x2 and y2 and change the sign of the term containing xy.

Angle θ between the pair of lines ax2 + 2hxy + by2 = 0 can also be obtained from a+b cos θ = ( a − b) 2 + 4h 2

The general equation of second in degree x and y ...(1)

will represent a pair of straight lines if and only if

a (i) ∆ ≡ h g

h b f

g f c

=0

or abc + 2fgh – af 2 – bg2 – ch2 = 0 and (ii) h2 – ab ≥ 0. Remarks 1. Angle between the lines: The angle θ between the lines represented by eqn. (1) is given by θ = tan– 1

2 h 2 − ab . a+b

Note that the angle between the pair of lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is same as the angle between the lines ax2 + 2hxy + by2 = 0. 2. The two lines represented by eqn. (1) will be parallel if h2 = ab and bg2 = af 2. 3. point of intersection of the lines: The point of intersection of the two lines represented by eqn. (1) is

 bg − hf af − gh  , 2  2 .  h − ab h − ab 

372

The point of intersection can also be determined with the help of partial differentiation as follows.

Objective Mathematics

Let φ ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Then

∂φ = 2ax + 2hy + 2g ∂x

13. The lines ax2 + 2hxy + by2 = 0 and lx + my + n = 0 form an isos-

[Differentiating φ w.r.t. x, treating y as constant]

∂φ = 2hx + 2by + 2f ∂y

and

∂φ

∂φ = 0. Thus, we ∂y

ax + hy + g = 0 and hx + by + f = 0

x y 1 = = fh − bg gh − af ab − h 2

 bg − fh af − gh  , 2 . 2  h − ab h − ab 

4. Bisectors: The equations of the bisectors of the angles between the lines represented by eqn. (1) are given by

( x − x′) 2 − ( y − y′) 2 ( x − x′) ( y − y′) , = a−b h where (x′, y′) is the point of intersection of the lines represented by (1). 5. Distance between the parallel lines : If the two lines represented by eqn. (1) are parallel, then the distance between the two parallel lines is given by

g − ac . a ( a + b) 2

6. The equation to the pair of straight lines through the origin and parallel to the pair of lines represented by eqn. (1) is ax2 + 2hxy + by2 = 0. 7. The equation to the pair of lines through (α, β) and perpendicular to the pair of lines ax2 + 2hxy + by2 = 0 is b (x – α)2 – 2h (x – α) (y – β) + a (y – β)2 = 0. 8. The area of the triangle formed by ax + 2hxy + by = 0 and lx + 2

2

n 2 h 2 − ab my + n = 0 is given by . am 2 − 2hlm + bl 2

a α + 2hαβ + bβ ( a − b) + 4h 2

2

method to find the separate equations of the lines when their joint equation is given Working Rules Method I: (a) Write down the equation of pair of lines (1) as a quadratic equation in x (or y). (b) Write down the value of x in terms of y (or value of y in terms of x). (c) Do cross multiplication and bring all the terms on LHS making RHS equal to zero. The two equations thus obtained will be the equations of the two lines. Method II: (a) Factorize the homogeneous part ax2 + 2hxy + by2 into two linear factors, say a1x + b1 y and a2x + b2 y. (b) Add constants c1 and c2 in the linear factors obtained in (a) to get a1x + b1 y + c1 and a2x + b2 y + c2. (c) Write ax2 + by2 + 2hxy + 2gx + 22fy + c

a1c2 + a2c1 = 2gg and b1c2 + b2c1 = 22f. Solving the above two equations, we obtain the values of c1 and c2.

2

.

a α 2 + 2hαβ + bβ2 + 2 g α + 2 f β + c

.  

= (a1x + b1 y + c1) × (a2x + b2 y + c2).

10. The product of the perpendiculars from (α, β) to the pair of lines ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 is

( a − b) 2 + 4h 2

 . 2 (bl − hm) 2 (am − hl ) ,  2 2 2 2   3(bl − 2hlm + am ) 3 (bl − 2hlm + am ) 

Comparing coefficients of x and y on both sides, we get

9. The product of the perpendiculars from (α, β) to the pair of lines 2

h

− ln (a + b) − mn (a + b)   am 2 − 2hlm + bl 2 , am 2 − 2hlm + bl 2 

∴ (x, y) = 

.

11. The equation to the pair of lines through the origin and forming an equilateral triangle with the line ax + by + c = 0 is given by (ax + by)2 – 3 (ay – bx)2 = 0. Also, the area of the equilateral triangle is

2

15. The orthocentre of the triangle formed by the lines ax2 + 2hxy + by2 = 0 and lx + my + n = 0 is

Solving the two equations, we get

ax2 + 2hxy + by2 = 0 is

2

lm

= 0 and For point of intersection, ∂x have

d= 2

celes triangle if l − m = a − b . 14. The centroid of the triangle formed by the lines ax2 + 2hxy + by2 = 0 and lx + my = 1 is

[Differentiating φ w.r.t.y, treating x as constant]



(a) square if (a – b) fg + h (f 2 – g2) = 0, a + b = 0 (b) rectangle if (a – b) fg + h (f 2 – g2) ≠ 0, a + b = 0 (c) rhombus if (a – b) fg + h (f 2 – g2) = 0, a + b ≠ 0 (d) parallelogram if (a – b) fg + h (f 2 – g2) ≠ 0, a + b ≠ 0.

c2 . 3 (a 2 + b 2 )

12. The two pairs of straight lines ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 and ax2 + by2 + 2hxy = 0 form a

(d) Substitute the values of c1 and c2 in a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0, we get the required equations.

remark: A pair of lines is equally inclined to another pair of lines if the joint equation of bisectors of angles between both pair of lines is same.

equations of the lines joining the origin to the points of intersection of a given line and a given curve y − mx = 1 and the c 2 2 curve be ax + 2hxy + by + 2gx + 2fy + k = 0

Let the straight line be y = mx + c or

373

Let the line cuts the curve at the points P and Q, then the joint equation of OP and OQ is

This is done by making the equation of the curve homogeneous with the help of the equation of the line.

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. The separate equations of the straight lines whose joint equatiion is x2 – 5xy + 6y2 = 0, are (a) x + 2y = 0, x – 3y = 0 (b) x – 2y = 0, x – 3y = 0 (c) x – 2y = 0, x + 3y = 0 (d) x – 2y = 0, x – 3y = 0

(a) ab (a + b) + 6abh + 8h3 = 0 (b) ab (a + b) – 6abh + 8h3 = 0 (c) ab (a + b) + 3abh + 4h3 = 0 (d) None of these

2. The area bounded by the angle bisector of the lines x2 – y2 + 2y = 1 and the line x + y = 3 is (a) 2 (c) 4

(b) 3 (d) 6

3. If the equation 3x2 – 8xy + λ y2 = 0 repersents two perpendiculat lines, then the value of λ is (a) 3 (c) 2

(b) – 3 (d) None of these

4. If the lines represented by 2x2 + 8xy + ky2 = 0 are coincident, then the values of k is (a) 8 (c) 4

(b) – 8 (d) None of these

5. The angle between the lines represented by (b) 2θ (d) None of these

6. The gradient of one of the lines x + hxy + 2y = 0 is twice that of the other, then h = 2

(a) ± 3 (c) ± 2

2

3 2 (d) ± 1

(b) ±

7. If λx2 – 10xy + 12y2 + 5x – 16y – 3 = 0 represents a pair of straight lines, then the value of λ is (a) 4 (c) 2

10. If one of the straight lines given by the equation ax2 + 2hxy + by2 = 0 coincides with one of those given by a′x2 + 2h′xy + b′y2 = 0, then (a) (ab′ – a′ b)2 + 4 (ah′ – a′ h) (bh′ – b′ h) = 0 (b) (ab′ – a′ b)2 + 4 (a′ h – ah′ ) (bh′ – b′ h) = 0 (c) (ab′ + a′ b)2 + 4 (a′ h – ah′ ) (bh′ – b′ h) = 0 (d) None of these 11. If two of the straight lines represented by ax3 + bx2y + cxy2 + dy3 = 0 are at right angles, then (a) a2 + ac + bd – d2 = 0 (b) a2 + ac – bd + d2 = 0 (c) a2 – ac + bd + d2 = 0 (d) a2 + ac + bd + d2 = 0 12. If the distance of a point (x1, y1) from each of two straight lines , which pass through the origin of coordinates, is δ, then the two lines are given by

x2 + 2xy sec θ + y2 = 0 is (a) 4θ (c) θ

9. I f t h e s l o p e o f o n e o f t h e l i n e s g i v e n b y ax2 + 2hxy + by2 = 0 be the square of the other, then

(b) 3 (d) 1

8. The two straight lines

(a) (x1 y – xy1)2 = δ2 (x2 + y2) (b) (x1 y + xy1)2 = δ2 (x2 + y2) (c) (x1 y – xy1)2 = δ2 (x2 + y2) (d) None of these 13. The equation m (x3 – 3xy2) + y3 – 3x2y = 0 represents three straight lines (a) which are equally inclined to one another (b) two of which are at right angles (c) two of which are coincident (d) which pass through origin. 14. The coordinates of the centroid of the triangle whose sides are 12x2 – 20xy + 7y2 = 0 and 2x – 3y + 4 = 0 are

x2 (tan2 θ + cos2 θ) – 2xy tan θ + y2 sin2θ = 0 make with the axis of x angles such that the difference of their tangents is

8 −8 (a)  ,  3 3 

8 8 (b)  − ,  3 3 

(a) 4 (c) 2

8 8 (c)  ,  3 3

8 8 (d)  − , −   3 3

(b) 3 (d) None of these

Pair of Straight Lines

2

ax2 + 2hxy + by2 + (2gx + 2fy)  y − mx  + k  y − mx  = 0.  c   c     

374

Objective Mathematics

15. The equation of two straight lines through the point (x1, y1) and ⊥ to the lines given by ax2 + 2hxy + by2 = 0 is 25. (a) b (x – x1)2 – 2h (x – x1) (y – y1) + a (y – y1)2 = 0 (b) b (x – x1)2 + 2h (x – x1) (y – y1) + a (y – y1)2 = 0 (c) a (x – x1)2 – 2h (x – x1) (y – y1) + b (y – y1)2 = 0 (d) None of these 16. The equation of two straight lines through the point (x1, y1) and parallel to the lines given by ax2 + 2hxy + by2 = 0 is (a) a ( y – y1)2 + 2h (x – x1) ( y – y1) + b (x – x1)2 = 0 (b) a ( y – y1)2 – 2h (x – x1) ( y – y1) + b (x – x1)2 = 0 (c) b ( y – y1)2 + 2h (x – x1) ( y – y1) + a (x – x1)2 = 0 (d) None of these 17. The equation x4 + bx3y + cx2y2 + dxy3 + ey4 = 0 represents two pairs of perpendicular straight lines if (a) b + d = 1 and e = – 1 (b) b + d = 0 and e = – 1 (c) b + d = 0 and e = 1 (d) None of these

(b) pq = – 1 (d) None of these

19. If the pair of straight lines ax2 + 2hxy – ay2 = 0 and bx2 + 2gxy – by2 = 0 be such that each bisects the angle between the other, then (a) hg = – ab (c) hb = ag

(b) hg = ab (d) hb = – ag

20. The lines bisecting the angle between the bisectors of the angles between the lines ax2 + 2hxy + by2 = 0 are given by (a) (a – b) (x2 – y2) – 4hxy = 0 (b) (a – b) (x2 + y2) + 4hxy = 0 (c) (a – b) (x2 – y2) + 4hxy = 0 (d) None of these 21. The equations a2x2 + 2h (a + b) xy + b2y2 = 0 and ax2 + 2hxy + by2 = 0 represent (a) two pairs of perpendicular straight lines (b) two pairs of parallel straight lines (c) two pairs of straight lines which are equally inclined to each other (d) None of these 22. The lines ax2 + 2hxy + by2 = 0 are equally inclined to the lines ax2 + 2hxy + by2 + λ (x2 + y2) = 0 for (a) λ = 1 only (c) for any value of λ

(b) λ = 2 only (d) None of these

23. If ax2 – y2 + 4x – y = 0 represents a pair of lines, then a is equal to (a) – 16 (c) 4

(b) (a + b)2 = 2h2 (d) None of these

The point of intersection of the pair of straight lines given by the equation 6x2 + 5xy – 4y2 + 7x + 13y – 3 = 0, is (a) (1, 1) (c) (– 1, 1)

(b) (1, – 1) (d) (– 1, – 1)

26. A pair of straight lines x2 – 8x + 12 = 0 and y2 – 14y + 45 = 0 are forming a square. The co-ordinates of its centroid are (a) (4, 6) (c) (4, 8)

(b) (4, 7) (d) (4, 9)

27. The distance between the two lines represented by the equation 9x2 – 24xy + 16y2 – 12x + 16y – 12 = 0 is (a)

8 5

(b)

6 5

11 (d) None of these 5 28. If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents two parallel straight lines, then (c)

18. If pairs of straight lines x 2 – 2pxy – y 2 = 0 and x2 – 2qxy – y2 = 0 be such that each pair bisects the angle between the other pair, then (a) pq = 1 (c) pq = 2

(a) (a + b)2 = 4h2 (c) (a + b)2 = h2

(b) 16 (d) – 4

24. One of the lines ax2 + 2hxy + by2 = 0 will bisect an angle between the coordinate axes if

(a) h2 = ab (c) bg2 = af 2

(b) g2 = ac (d) ag2 = bf 2

29. The angle between the lines joining the origin to the points of intersection of the line y = 3x + 2 and the curve x2 + 2xy + 3y2 + 4x + 8y – 11 = 0 is   (a) tan­ –1 2 2  3   

 2 (b) tan­ –1    3 

  (c) tan ­–1 3 2  3   

(d) None of these

30. All chords of the curve 3x2 – y2 – 2x + 4y = 0 which subtend a right angle at the origin pass through the point (a) (1, 2) (c) (1, – 2)

(b) (– 1, 2) (d) (– 1, – 2)

31. The straight lines joining the origin to the points of intersection of the two curves ax2 + 2hxy + by2 + 2gx = 0 and a′x2 + 2h′xy + b′y2 + 2g′x = 0 are at right angles, if (a) g (a′ + b′ ) = g′ (a + b) (b) g (a + b) = g′ (a′ + b′ ) (c) g (a′ – b′ ) = g′ (a – b) (d) None of these 32. The angle between the straight lines joining the origin to the points of intersection of 3x2 + 5xy – 3y2 + 2x + 3y = 0 and 3x – 2y = 1 is (a)

π 3

(b)

π 4

π π (d) 6 2 33. If the curve x2 + y2 + 2gx + 2fy + c = 0 intercepts on the line lx + my = 1, a length which subtends a right angle at the origin, then (c)

34. The equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the straight line x – y = 2 with the curve 5x2 + 11xy – 8y2 + 8x – 4y + 12 = 0 is (a) x2 + 30xy + y2 = 0 (c) x2 + 30xy – y2 = 0

(b) x2 – 30xy + y2 = 0 (d) None of these

35. The pair of lines which join the origin to the points of intersection of the line y = mx + c with the curve x2 + y2 = a2 are at right angles, if (a) c2 = a2 (1 + m2) (c) 2c2 = a2 (1 – m2)

(b) 2c2 = a2 (1 + m2) (d) None of these

36. If the sum of the slopes of the lines given by x2 – 2cxy – 7y2 = 0 is four times their product, then c has the value (a) 2 (c) 1

(b) –1 (d) –3  

37. The straight lines represented by the equation 135x2 – 136xy + 33y2 = 0 are equally inclined to the line (a) x – 2y = 7 (c) x – 2y = 4

(b) x + 2y = 7 (d) 3x + 2y = 4

(a) a + b = 2 | h  | (c) a – b = 2 | h |

(b) a + b = – 2h (d) (a – b)2 = 4h2

44. The angle between the lines joining the origin to the points of intersection of the line x 3 + y = 2 and the curve x2 + y2 = 4 is (a) (c) 45. The is a θ is

π 6

π 4 π π (d) 2 3 2 2 equation x – 3xy + λy + 3x – 5y + 2 = 0, where λ real number, represents a pair of straight lines. If the angle between the lines, then cosec2 θ = (b)

(a) 3 (c) 10

(b) 9 (d) 100

46. Which of the following pair of straight lines intersect at right angles ? (a) 2x2 = y (x + 2y) (c) 2y (x + y) = xy 

(b) (x + y)2 = x (y + 3x) (d) y =  2x

47. The equation x2y2 – 9y2 + 6x2y + 54y = 0 represents (a) a pair of straight lines and a circle (b) a pair of straight lines and a parabola (c) a set of four straight lines forming a square (d) None of these

38. The pair of straight lines joining the origin to the x2 y2 intersection of the curve 2 + 2 = 1 by the line a b 48. If the equation hxy + gx + fy + c = 0, (h ≠ 0) represents two straight lines, then lx + my + n = 0 are coincident if (a) a2n2 + b2l2 = m2 (c) a2l2 + b2m2 = n2

(b) a2m2 + b2l2 = n2 (d) None of these

39. The condition that the slope of one of the lines represented by ax2 + 2hxy + by2 = 0 should be λ times the slope of the other is (a) ab (1 + λ)2 = h2λ (c) ab (1 + λ)2 = 3h2λ

(b) ab (1 + λ)2 = 2h2λ (d) ab (1 + λ)2 = 4h2λ

40. If the distance of a given point (α, β) from each of two straight lines through the origin is d, then (a) (αx – βy)2 = d2 (x2 + y2) (b) (αy – βx)2 = d2 (x2 + y2) (c) (αx + βy)2 = d2 (x2 + y2) (d) None of these 41. If the lines joining the origin to the points of intersection of y = mx + 1 with x2 + y2 = 1 are ⊥, then m is equal to (a) 2 (c) – 1

(b) 1 (d) – 2

42. The equations of lines joining the origin to the point of intersection of circle x2 + y2 = 3 and the line x + y = 2 are (a) y – (3 + 2 2 ) x = 0 (b) x – (3 + 2 2 ) y = 0 (c) y – (3 – 2 2 ) x = 0 (d) x – (3 – 2 2 ) y = 0

(a) 2fgh = c2 (c) fgh = c2

49. The equation represents (a) pair of lines (c) a hyperbola

(b) 2fg = ch (d) fg = ch

( x − 2) 2 + y 2 +

( x + 2) 2 + y 2 = 4

(b) a parabola (d) None of these

50. If the ratio of gradients of the lines, represented by ax2 + 2hxy + by2 = 0 is 1 : 3, then the value of the ratio h2 : ab is (a)

1 3

(b)

3 4

4 (d) 1 3 51. The equation 2x2 + Kxy + 2y2 = 0 represents a pair of real and distinct lines if (c)

(a) K ∈ (– 4, 4) (b) K ∈ R (c) K ∈ (– ∞, – 4) ∪ (4, ∞) (d) None of these 52. If the equation 3x2 + 6xy + my2 = 0 represents a pair of straight lines inclined at an angle π, then m is equal to (a) 3 (c) 9

(b) 6 (d) any real number.

375

43. If one of the lines of the pair ax2 + 2hxy + by2 = 0 bisects the angle between positive direction of the axes, then a, b and h satisfy the relation

Pair of Straight Lines

(a) c (l2 + m2) + 2 (gl + fm + 1) = 0 (b) c (l2 + m2) – 2 (gl + fm + 1) = 0 (c) c (l2 + m2) + 2 (gl + fm – 1) = 0 (d) None of these

376

53. The equation x3 + xy2 – 2y3 = 0 represents

Objective Mathematics

(a) three straight lines passing through origin (b) three real straight lines (c) three distinct points (d) None of these 54. The three lines whose combined equation is (3x2 + 2xy – 3y2)  (x – y + 2) = 0 form a triangle which is (a) equilateral (c) obtuse angled

(b) right angled (d) None of these

55. The straight lines represented by (y – mx)2 = a2 (1 + m2) and ( y – nx)2 = a2 (1 + n2) form a (a) rectangle (c) rhombus

(b) trapezium (d) None of these

56. If two of the lines ax3 + bx2y + cxy2 + dy3 = 0 (a ≠ 0) makecomplementary angles with x-axis in anticlockwise sense then (a) a (a – c) – d (b – d) = 0 (b) d (a – c) + a (b – d) = 0 (c) a (a – c) + d (b – d) = 0 (d) None of these 57. The centroid of the triangle formed by the pair of lines 2x2 – 27y2 – 3xy + 4x – 3y + 2 = 0 and the line 4x – 3y – 26 = 0 is (a) (3, – 2) (c) (4, 0)

(b) (4, 2) (d) None of these

58. The orthocentre of the triangle whose three sides are given by the combined equation 2

(15x + 16xy – 48y ) ( y – 15) = 0, is (b) (0, 33) (d) (– 33, 0)

59. The pair of lines

3 x2 – 4xy + 3 y2 = 0 are rotated π in the anticlockwise sense. The about the origin by 6 equation of the pair in the new position is (a) x2 – (c)

3 xy = 0

3 x2 – xy = 0

(b) xy –

3 y2 = 0

(d) None of these

60. The equation of the image of the lines y = | x | by the line x = 2 is (a) y = | x – 4 | (c) | y | + 4 = x

(b) | y | = x + 4 (d) None of these

61. If the lines represented by the equation 3y2 – x2 + 2 3 x – 3 = 0 are rotated about the point ( 3 , 0) through an angle 15º, one clockwise direction and other in anticlockwise direction, then the equation of the pair of lines in the new position is (a) y2 – x2 + 2 3 x + 3 = 0 (b) y2 – x2 + 2 3 x – 3 = 0 (c) y2 – x2 – 2 3 x + 3 = 0 (d) None of these 62. Two pairs of straight lines have the equations y2 + xy – 12x2 = 0 and ax2 + 2hxy + by2 = 0. One line will be common among them if

(b) a – 8h + 16b = 0 (d) a + 6h + 9b = 0

63. The image of the pair of lines represented by 3x2 + 4xy + 5y2 = 0 by the line mirror x = 0 is (a) 3x2 – 4xy + 5y2 = 0 (b) 3x2 – 4xy – 5y2 = 0 (c) 5y2 – 4xy – 3x2 = 0 (d) None of these 64. Mixed term xy is to be removed from the general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, one should rotate the axes through an angle θ, given by tan 2θ equal to (a)

a−b 2h

(b)

2h a+b

a+b 2h (d) 2h a−b 65. If the angle between the lines represented by equation y2 + kxy – x2 tan2 A = 0 is 2 A, then k is equal to (c)

(a)  2 (c)  0

(b)  –2 (d)  4

66. If one of the lines of my2 + (1 – m2)xy – mx2 = 0 is a bisector of the angle between the lines xy = 0, then m is (a)   −

1 2

(b)  –2

(c)  ±1

(d)  2

67. If ax + by + hx + hy = 0, h ≠ 0 represents a pair of straight lines, then 2

2

(a) (0, – 33) (c) (33, 0)

(a) a + 8h – 16b = 0 (c) a – 6h + 9b = 0

2

(a)  a + b = 0 (c)  b + h = 0

(b)  a + h = 0 (d)  ab – h2 = 0

68. If pairs of straight lines x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 be such that each pair bisects the angle between the other pair, then (a)  pq = 1 (c)  pq = 2

(b)  pq = – 1 (d)  pq = – 2

69. The equation y2 – x2 + 2x – 1 = 0 represents (a)  a hyperbola (b)  an ellipse (c)  a pair of straight lines (d)  a rectangular hyperbola 70. The slopes of the lines represented by x2 + 2hxy + 2y2 = 0 are in the ratio 1 : 2, then h equals (a)   ± 1 2 (c)  + 1

3 2 (d)  ± 3

(b)  ±

71. The distance between the pair of lines represented by the equation x2 – 6xy + 9y2 + 3x – 9y – 4 = 0 is (a)  

15 10

(b)  

(c)  

5 2

(d)  

1 2

1 10

(a)  3 and (c)  –3 and

1 3

(b)  –3 and − 1 3

1 3

(d)  3 and −

1 3

73. The lines ax2 + 2hxy + by2 = 0 are equally inclined to the lines ax2 + 2hxy + by2 + k(x2 + y2) = 0 for

(b)  k = 1 (d)  none of these.

74. The acute angle between the lines joining the origin to the points of intersection of the line 3x + y = 2 and the circle is x2 + y2 = 4, is (a)  π/2 (c)  π/4

(b)  π/3 (d)  π/6

Solutions 1. (b) Comparing with

ax2 + 2hxy + by2 = 0, we get



5 and b = 6. 2 The straight lines represented by



ax2 + 2hxy + by2 = 0



‘a’ = 1, 2h = – 5, i.e.,, h = –



are ax + (h +



and ax + (h – h 2 − ab ) y = 0. ∴ The required equations are



 −5 1 · x +  +  2

h 2 − ab ) y = 0

 25 − 6   y = 0 4 

 −5 − and 1 · x +   2

 25 − 6   y = 0 4 

are perpendicular if a + b = 0 ⇒ 3 + λ = 0 ⇒ λ = – 3.

4. (a) The lines represented by the given equation will be coincident if h2 – ab = 0. Here, a = 2, b = k, h = 4. Substituting the values, we get (4)2 – 2k = 0 ⇒ k = 8. 5. (c) Let φ be the angle between the lines represented by ...(1) x2 + 2xy sec θ + y2 = 0. ∴ tan φ =

2 (h 2 − ab) . a+b

Here a = 1, b = 1, h = sec θ

 −5 1  + y=0 i.e., x +   2 2

Hence tan φ =

 −5 1  −   y = 0 and  x +   2 2 i.e.,, x – 2y = 0 and x – 3y = 0. Hence the given equation represents the lines x – 2y = 0 and x – 3y = 0.

⇒ tan φ =

2. (a) The triangle is bounded by the lines



x + y = 3, x = 0 and y = 1

Hence, required area = 1 × 2 × 2 = 2. 2 3. (b) The lines represented by 3x2 – 8xy + λ y2 = 0

2 (sec 2 θ − 1×1) 1+1

2 (sec 2 θ − 1) = tan θ 2

∴ φ = θ.  ence the angle between the lines represented by H (1) is θ. 6. (a) If m1, m2 are gradients of lines represented by ax2 + 2hxy + by2 = 0, then 2h a , m 1m 2 = . m1 + m2 = – b b (Given) If m1 = m, then m2 = 2m 2 h h or 9m2 = ...(1) ∴ m + 2m = – 4 2 1 1 or m2 = ...(2) and m · 2m = 2 4 Eliminating m2 from (1) and (2), we get h2 = 9 or h = ± 3. 7. (c) The given equation will represent a pair of straight lines if abc + 2fgh – af 2 – bg2 – ch2 = 0 and h2 – ab ≥ 0 5 ⇒ λ (12) (– 3) + 2 (– 8)    (– 5) – λ (– 8)2 2

377

(a)  k = 2 (c)  any value of k

Pair of Straight Lines

72. The slopes of the lines given by the equation 3x2 + 10xy + 3y2 – 15x – 21y + 18 = 0 are

378

Objective Mathematics

Then, a + 2hm + bm2 = 0 and a′ + 2h′m + b′m2 = 0 m m2 1 = = ⇒ (a ′b − ab′ ) 2 (b′h − bh′ ) 2 (ah′ − a ′h)

2

5 – 12    – (– 3) (– 5)2 = 0 2 ⇒ – 100 λ + 200 – 75 + 75 = 0 ⇒ λ – 2 = 0 ⇒ λ = 2. Also, for λ = 2, h2 – ab = 25 – 2 × 12 = 1 > 0. ∴ λ = 2 is valid. 8. (c) We have a = tan 2 θ + cos 2 θ, h = – tan θ and b = sin2 θ.

⇒ m =



(ah′ − a ′h) a ′b − ab′ , m2 = (b′h − bh′ ) 2(b′h − bh′ )

Eliminating m, we obtain (a ′b − ab′ ) 2 (ah′ − a ′h) = 4 (b′h − bh′ ) 2 (b′h − bh′ )

Hence, (ab′ – a′b)2 + 4 (ah′ – a′h) (bh′ – b′h) = 0.  et y = m1x and y = m2x be the two lines represented L by the given equation so that ...(1) 11. (d) ax3 + bx2y + cxy2 + dy3 = 0 2h a and m1m2 = m1 + m2 = − i s a homogeneous equation of third degree in x and b b y. It, therefore, represents three straight lines through 2 tan θ tan 2 θ + cos 2 θ the origin. Let the slopes of the lines be m1, m2, m3. i.e., m1 + m2 = and m1m2 = . sin 2 θ sin 2 θ Then, m1, m2 and m3 are the roots of Now, (m1 – m2)2 = (m1 + m2)2 – 4m1 m2 ...(2) dm3 + cm2 + bm + a = 0 a 2 2 2 Product of the roots = m1m2m3 = − 4 tan θ 4 (tan θ + cos θ) = − d ...(3) 4 2 sin θ sin θ Since the two lines represented by (1) are at right 4 [tan 2 θ − (tan 2 θ + cos 2 θ) sin 2 θ] = angles, let the lines with slopes m1, m2 be perpensin 4 θ dicular. 4 Then, m1m2 = – 1. [tan 2 θ (1 − sin 2 θ) − cos 2 θ sin 2 θ] = sin 4 θ a a or m3 =  ...(4) ∴ from (3), (– 1) m3 = − 4 2 2 2 2 d d [tan θ cos θ − cos θ sin θ ] = 4 But m3 is a root of (2) sin θ =

∴ dm33 + cm32 + bm3 + a = 0 Substituting the value of m3 from (4),

4 4 sin 4 θ [sin 2 θ − sin 2 θ cos 2 θ] = = 4. 4 sin θ sin 4 θ

Hence, | m1 – m2 | = 2. 9. (b) Let y = m1x and y = m2x be the two lines represented by ax2 + 2hxy + by2 = 0 so that 2h a and m1m2 =  b b Given : m2 = m12 2h ∴, from (1), m1 + m12 = −  b a a i.e., m13 =  and m1 · m12 = b b m1 + m2 = −

...(1)

or a3 + a2c + abd + ad2 = 0 or a2 + ac + bd + d2 = 0, which is the required condition. 12. (a) Any line through the origin is

...(2)

y = mx or mx – y = 0.  δ = distance of (x1, y1) from the line mx – y = 0

...(3)

∴ δ = ±

 he required condition is obtained by eliminating T m1 between (2) and (3). 3  2h  Cubing (2), we get (m1 + m12 )3 =  −  b  ⇒ m13 + m16 + 3m13 (m1 + m12 ) = –

 a3   a2  a we get d  3  + c  2  + b   + a = 0 d  d  d 

8h 3 b3

a a2 a  2h  8h 3 + 2 + 3 −  = – 3   b b b b b  [Using (2) and (3)] ⇒ ab2 + a2b – 6abh = – 8h3 or ab (a + b) – 6abh + 8h3 = 0, which is the required condition. ⇒

10. (a) Let y = mx be the common line of ax2 + 2hxy + by2 = 0 and a′x2 + 2h′xy + b′y2 = 0

mx1 − y1

1 + m2 or δ  (1 + m ) = (mx1 – y1)2 y But y = mx or m = x ∴ The required equation is 2



2

 y2  y  δ2   2 + 1 =  ⋅ x1 − y1  x  x 

...(1)

2

or δ2 (x2 + y2) = (x1 y – xy1)2. 13. (a), (d)  Dividing by x3, the given equation reduces to  3 y 2  y3 3 y m  1 − 2  + 3 − =0 x  x x  ⇒ m (1 – 3 tan2 θ) + tan3 θ – 3 tan θ = 0,



where

y  = tan θ. x

⇒ tan 3θ = tan φ , where m = tan φ ⇒ 3θ = φ, φ + π , φ + 2π φ φ π φ 2π ⇒ θ = , + , + . 3 3 3 3 3  hus, the given equation represents three lines T through the origin which are equally inclined to one another. 14. (c) We have, 12x2 – 20xy + 7y2 = 0 ⇒ (2x – y) (6x – 7y) = 0 So, the equations of three sides of the triangle are 2x – y = 0, 6x – 7y = 0, 2x – 3y + 4 = 0. The coordinates of vertices of the triangle are (0, 0), (7, 6) and (1, 2). [Solving the above three equations, pairwise] ∴ Centroid of the triangle is

 0 + 7 + 1 0 + 6 + 2  i.e.,  8 8  . ,    3, 3 3 3     15. (a) Let y = mx

...(1)

x2 + pxy – y2 = 0 and x2 + qxy – y2 = 0, then x4 + bx3y + cx2y2 + dxy3 + ey4 ≡ (x2 + pxy – y2) (x2 + qxy – y2) ...(1) Equating coefficients of y4; e = 1 3 ...(2) Equating coefficients of x y; b = p + q ...(3) Equating coefficients of xy3; d = – p – q Adding (2) and (3), we get b + d = 0 ∴ b + d = 0 and e = 1. 18. (b) The equation of the bisectors of the angles between the lines x2 – 2pxy – y2 = 0 is

x2 − y 2 xy = 1 − (− 1) −p

or

x2 − y 2 xy = − 2 p

2 ...(1) xy – y2 = 0 p Also, x2 – 2qxy – y2 = 0 ...(2) is the equation of the bisectors of the angles between the same lines (given). From (1) and (2), by comparing coefficients, we get

i.e.,

x2 +

1 2/ p −1 1 = =   i.e.,  1 = − or pq = – 1. b e one of the lines represented by ax + 2hxy + 1 − 2q −1 pq 2 by = 0 19. (a) The equation of the bisectors of the angles between Then, m is a root of the equation the lines ax2 + 2hxy – ay2 = 0 2 bm + 2hm + a = 0 ...(2) x2 − y 2 xy The equation of the line through (x1, y1) and ⊥ to is = a − (− a) h (1) is x − x1 or hx2 – 2axy – hy2 = 0 ...(1) 1 y – y1 = − (x – x1) ⇒ m = −  ...(3) y − y1 m ...(2) Also, bx2 + 2gxy – by2 = 0 i  s the equation of the bisectors of the angles between Substituting this value of m in (2), we obtain 2 the same lines (given).  x − x1   x − x1  b  2 h a = 0 + − + From (1) and (2), by comparing coefficients  y − y   y − y1  1 h − 2a − h i.e., hg = – ab. = = or b (x – x1)2 – 2h (x – x1) (y – y1) + a (y – y1)2 = 0, b 2g −b which is the equation of the required line 20. (c) The equation of the bisectors of the angles between pair. x 2 − y 2 xy 16. (c) Let y = mx ...(1) the lines ax2 + 2hxy + by2 = 0 is = a−b h 2 be one of the lines represented by ax + 2hxy + a − b i.e., x2 – y2 = xy by2 = 0 h Then, m is a root of the equation a−b bm2 + 2hm + a = 0 ...(2) or x 2 − xy − y 2 = 0 ... (1) h The equation of the line through (x1, y1) and parallel to (1) is The equation of the bisectors of the angles between y − y1 the lines (1) is y – y1 = m (x – x1) ⇒ m =  ...(3) x − x1 x2 − y 2 xy a−b   =  Here a = 1, b = 1, h = − 2h  − a b ( ) 1 − − 1 Substituting this value of m in (2), we obtain − 2h 2  y − y1   y − y1  b + 2h  +a = 0  2h  or x2 – y2 = 2xy  −  x − x1   x − x1   a − b  or b (y – y1)2 + 2h ( y – y1) (x – x1) + a (x – x1)2 = 0. or (a – b) (x2 – y2) + 4hxy = 0 2

379

17. (c) Let the two pairs of perpendicular lines be

3 tan θ − tan 3 θ = tan 3θ 1 − 3 tan 2 θ

Pair of Straight Lines

⇒ m =

380

21. (c) The given equations are



Objective Mathematics

a x + 2h (a + b) xy + b y = 0 ... (1) 2 2 ... (2) and ax + 2hxy + by = 0 The equation of the bisectors of the angles between the line pair (1) is x2 − y 2 xy x 2 − y 2 xy or = = 2 2 a −b h ( a + b) a−b h 2 2

2 2

 hich is same as equation of the bisectors of angles w between the line pair (2). Thus, the two line pairs are equally inclined to each other. 22. (c) Equation of the bisectors of the angle between the lines ax2 + 2hxy + by2 + λ (x2 + y2) = 0 is

x2 − y 2 xy = (a + λ ) − (b + λ ) h

or

x 2 − y 2 xy = a−b h

 hich is same as the equation of the bisectors of w angles between the lines ax2 + 2hxy + by2 = 0. Thus, the two line pairs are equally inclined to each other for any value of λ. 23. (b) We have, ax2 – y2 + 4x – y = 0 Here, a = a, h = 0, b = – 1, f = – a For the straight lines, h g ⇒

1 , g = 2, c = 0 2

h b f

g f =0 c

a 0 2 0 −1 −1 / 2 = 0 2 1/ 2 0

⇒ −

a + 4 = 0  or  a = 16. 4

24. (a) Slopes of the given lines are roots of the equation ...(1) bm2 + 2hm + a = 0 But one of the lines has slope + 1 or – 1, ∴ putting m = ± 1 in (1), we get a + b = ± 2h. Squaring both sides, we obtain (a + b)2 = 4h2. 25. (c) The given equation can be written as 2

2

6x + (5y + 7) x – (4y – 13y + 3) = 0. Solving it as a quadratic in x, we get x

=



=



− (5 y + 7) ± (5 y + 7) 2 + 24 (4 y 2 − 13 y + 3) 12

− (5 y + 7) ± 121 y 2 − 242 y + 121 12 − (5 y + 7) ± 11( y − 1) = 12 y − 3 − 4y + 1 6 y − 18 − 16 y + 4 = or , , , 2 3 12 12

...(1)

 hese are the two lines represented by the given T equation. Solving the equations (1) simultaneously, we get x y 1 . = = 1 − 12 9 + 2 8 + 3 ∴ x = – 1, y = 1. ∴  Point of intersection is (– 1, 1). 26. (b) x2 – 8x + 12 = 0 (x – 6) (x – 2) = 0 x = 2, 6 y2 – 14y + 45 = 0 (y – 9) (y – 5) = 0 y = 5, 9 2+6 5+9 = = 4, = =7 2 2 ∴ Centroid = (4, 7). 27. (a) Comparing the given equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we have a = 9, b = 16, h = – 12, g = – 6, f = 8, c = – 12. Now,  h2 – ab = (– 12)2 – (9) (16) = 144 – 144 = 0. Also, bg2 – af 2 = (16) (36) – (9) (64) = 576 – 576 = 0. Since h2 = ab and bg2 = af 2, so the two lines represented by the given equation are parallel. Thus, the distance between the two lines = 2

  1  ⇒ a 0 +  −   − 0 + 2[2]  4  

or 2x – y + 3 = 0; 3x + 4y – 1 = 0

36 − (9) (− 12) 24 8 g 2 − ac = or . =2 9 (9 + 16) 15 5 a ( a + b)

28. (a), (c)  If one of the two lines represented by the given equation is lx + my + n = 0, the second being parallel to it, will be lx + my + n′ = 0. ∴ (lx + my + n) (lx + my + n′)  ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c Equating coefficients of like terms, we get a = l2 c = nn′ 2f = m (n + n′)

...(1) b = m2 ...(3) h = lm ...(5) 2g = l (n + n′)

...(2) ...(4) ...(6)

[  of (1), (2) and (4)] Now, ab = l2m2 = h2 ∴ h2 = ab. l2 l 2m2 Also, bg2 = m2 · (n + n′)2 = (n + n′)2 = af 2 4 4 [ of (2), (6), (1) and (5)] ∴ bg2 = af 2. Clearly, g2 ≠ ac and ag2 ≠ bf 2. 29. (a) The given line is y = 3x + 2 or

y − 3x = 1... (1) 2

Making x2 + 2xy + 3y2 + 4x + 8y – 11 = 0 homogeneous with the help of (1), we obtain

 y − 3x   y − 3x  + 8y  –11  = 0.  2   2  On simplification, we get 7x2 – 2xy – y2 = 0 ⇒ a = 7, h = – 1, b = – 1. This equation represents two lines joining the origin to the points of intersection of the given line and the curve. The angle between these two lines is given by 2



tan θ =

2 h 2 − ab 2 1 + 7 = = a+b 7 −1

8 2 2 . = 3 3

2 2  Hence, θ = tan– 1    is the required angle.  3  30. (c) Let lx + my = 1

...(1)

be any chord of the curve 3x2 – y2 – 2x + 4y = 0, ...(2) which subtends a right angle at the origin. Making (2) homogeneous with the help of (1), we obtain 3x2 – y2 – 2x (lx + my) + 4y (lx + my) = 0 or (3 – 2l) x2 + (4m – 1) y2 + 2 (2l – m) xy = 0. These angles are right angles, if (3 – 2l) + (4m – 1) = 0 or l – 2m = 1, which shows that (1) passes through the fixed point (1, – 2). 31. (a) The given curves are

33. (a) Equation of the curve is

x2 + y2 + 2gx + 2fy + c = 0 Equation of the line is lx + my = 1

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

 quation of the lines joining the origin to the points E of intersection of (1) and (2) is obtained by making (1) homogeneous with the help of (2). ∴ x2 + y2 + 2 (gx + fy) (lx + my) + c  (lx + my)2 = 0 or (1 + 2gl + cl2) x2 + 2 (gm + f l + clm) xy + (1 + 2fm + cm2) y2 = 0 These lines will be at right angles if coefficient of x2 + coefficient of y2 = 0 i.e., if 1 + 2gl + cl2 + 1 + 2fm + cm2 = 0 or if c (l2 + m2) + 2 (gl + fm + 1) = 0 34. (c) The equation of lines joining the origin to the points of intersection of the given line and curve is 5x2 + 11xy – 8y2 + (8x – 4y) x−  2



x − y y  + 12   2  = 0 2

or 12x2 – xy – 3y2 = 0 x2 − y 2 xy = 12 − (− 3) − 1 2 or x2 + 30xy – y2 = 0. Equation of bisectors is

...(1) ax2 + 2hxy + by2 + 2gx = 0, ...(2) and a′x2 + 2h′xy + b′y2 + 2g′x = 0. From (1), we obtain 1 (ax2 + 2hxy + by2) ...(3) 2x = − g Making (2) homogeneous with the help of (3), we obtain g' (ax2 + 2hxy + by2) = 0 a′x2 + 2h′xy + b′y2 – g

35. (b) The equation of the pair of lines joining the origin to the points of intersection of given line and curve are  y − mx  x 2 + y 2 – a 2   =0  c  2

or (c2 – a2m2) x2 + 2a2 mxy + (c2 – a2) y2 = 0 They are ⊥ if (c2 – a2m2) + (c2 – a2) = 0 i.e., 2c2 = a2 (1 + m2).

or g (a′x2 + 2h′xy + b′y2) – g′ (ax2 + 2hxy + by2) = 0 ...(4) 36. (a) Given pair of lines is x2 – 2cxy – 7y2  = 0 (4) represents two straight lines joining the origin to the points of intersection of (1) and (2). The lines will be ⊥ if, coefficients of x2 + coefficients of y2 = 0 i.e., (ga′ – g′a) + (gb′ –­ g′b) = 0 Hence, g′ (a + b) = g (a′ + b′). 32. (d) Equation of the line is 3x – 2y = 1

...(1)

and the equation of the curve is ...(2) 3x2 + 5xy – 3y2 + 2x + 3y = 0 M aking (2) homogeneous with the help of (1), we get

2c −4 ⇒c=2 = 7 7 a 2h 2c 1   m1 + m2 = − b = − 7 ; m1m2 = b = − 7 

Given: m1 + m2 = 4m1m2 ⇒ – 

381

3x2 + 5xy – 3y2 + (2x + 3y) (3x – 2y) = 0 ...(3) or 9x2 + 10xy – 9y2 = 0 which is the equation of the lines joining the origin to the points of intersection of (1) and (2). Here a = 9, b = – 9. Since a + b = 0, ∴ the two lines given by (3) are at right angles.

Pair of Straight Lines

 y − 3x  x2 + 2xy + 3y2 + 4x     2 

382

37. (b) Given pair of lines is 2

2

Objective Mathematics

135x – 136xy + 33y = 0 ...(1)  he equation of bisectors of angles between pair T of lines (1) is

x 2 − y 2 xy = a−b h

or

x2 − y 2 xy = 135 − 33 − 68

∴ 

or

2

 β − mα  2 = d or   =d 2  1+ m 

β − mα 1 + m2 y    β − α  x

2

 y2  = d2 1 + 2  x  

y   From (1), y = mx ∴ m = x  or 2x2 + 3xy – 2y2 = 0 ...(2)  or (x + 2y) (2x – y) = 0 or (αy – βx)2 = d2 (x2 + y2). One of the bisectors is x + 2y = 0 which is parallel to the line x + 2y = 7. 41. (b), (c)  Making the equation x2 + y2 = 1 homogeneous Hence, the line x + 2y = 7 is equally inclined to the with the help of y – mx = 1, we get given lines. x2 + y2 – ( y – mx)2 = 0. x2 y 2 The above equation represents a pair of straight lines pass...(1) 38. (c) Given curve is 2 + 2 = 1 ing through origin. The two lines are ⊥ if coefficient of a b x2 + cofficient of y2 = 0 and given line is lx + my + n = 0 ...(2) y2 = 0 i.e., 1 + 1 – 1 – m2 = 0 or m = 1, – 1. In order to find the equation of the lines joining origin and the point of intersection of the line (2) 42. (a), (b), (c) (d)  Making the equation of circle x2 + y2 = 3 homogeneous using the equation of line x + y = 2, and curve (1), we make the equation (1) homogenewe get ous with the help of (2). From (2),

lx + my =1 −n

∴ From (1),

x 2 y 2  lx + my  + = a 2 b 2  − n 



2

x2 + y2 = 3   x + y  or x2 + y2 – 6xy = 0  2   

2

1  1 m2  l 2  2ml or x2  2 − 2  − 2 xy + y2  2 − 2  = 0 ...(3) n  n n  a b This is of the form Ax2 + 2Hxy + By2 = 0. The lines given by (3) will be coincident if l 2m2  1 l 2   1 m2  = 2 − 2 2 − 2  4 n n  b n  a l2 m2 1 or 0 = 2 2 − 2 2 − 2 2 or a2l2 + b2m2 = n2. ab nb an

H2 – AB = 0 or

39. (d) Given equation is ax2 + 2hxy + by2 = 0

...(1)

 et one of the lines represented by (1) be y – mx L = 0. Then the equation of the other line represented by (1) will be y – λ mx = 0. ∴ m + λ m = – 2h/b ...(2) and m (λ m) = a/b ...(3) Squaring equation (2) and dividing it by (3), we get

2 2 ∴ y = 6 x ± (36 x − 4 x ) = (3 ± 2 2 ) x. 2 One of the lines is y = (3 ± 2 2 ) x which is given in (a). This can also be written as 1 y = (3 – 2 2 ) y, x= (3 + 2 2 )

which is given in (d).  he other line is y = (3 – 2 2 ) x, which is given T in (c) This can also be written as 1 y = (3 + 2 2 ) y, x= 3−2 2 which is given in (b). 43. (b) Bisector of the angle between the positive directions of the axes is y = x.  ince it is one of the lines of the given pair of S lines ax2 + 2hxy + by2 = 0, we have x2 (a + 2h + b) = 0 or a + b = – 2h.

(1 + λ ) 2 4h 2 / b 2 4h 2 = = λ a/b ab 44. (c) Making the equation of the curve x2 + y2 = 4 hoor ab (1 + λ )2 = 4h2 λ, which is the required condimogeneous, using the equation of the line tion. x 3 + y = 2, we get 40. (b) Let the equation of any line through the origin be 1 x 2 + y 2 − 4 ⋅ ( x 3 + y ) 2 = 0 or x2 + 3 xy = 0 y – mx = 0  ...(1) 4 I t is given that the length of the ⊥ from (α, β) to line (1) is d.

∴ a = 1, b = 0, h =

3 /2. So, the angle between the two lines is

45. (c) Comparing the given equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we have a = 1, h = – 3/2, b = λ, g = 3/2, f = – 5/2, c = 2. The given equation will represent a pair of straight lines if abc + 2fgh – af 2 – bg2 – ch2 = 0  − 5  − 5  3  − 3 ⇒ (1) (λ) (2) + 2   – 1           2   2  2  2  2

3 –λ   2

⇒ λ = 2.

2

 − 3 – 2   =0  2  2

If θ is the angle between the lines, then

tan θ =

2 h 2 − ab = a+b

2

( x − 2) 2 + ( x + 2) 2 = 4



9 −2 1 4 = + 1 2 3

which is satisfied only if x = 2 or x = – 2. ∴ Only (2, 0) and (­– 2, 0) on x-axis satisfy the given equation and not the line y = 0. 50. (c) Let m, 3m be the gradients of the lines represented by ax2 + 2hxy + by2 = 0. 2h ...(1) b a and m · 3m = 3m2 = ...(2) b Substituting the value of m from (1) in (2), we get ∴ m + 3m = 4m = –

 h  a h2 4 ⇒ = . 3   −  =  2b  b ab 3 2



51. (c) The given equation will represent a pair of real and distinct lines if h2 > ab k  i.e.,    > (2) (2) or k2 > 16 or (k – 4) (k + 4) > 0 2 2

∴ cosec θ = 1 + cot θ = 10. 2

2

46. (a) The equation 2x2 = y (x + 2y) or 2x2 – xy – 2y2 = 0 i.e., k ∈ (– ∞, – 4) ∪ (4, ∞). represents a pair of straight lines passing through 52. (a) Since the angle between the pair of straight lines origin represented by the given equation is π 2 2 ( coefficients of x + coefficient of y = 0) 47. (c) We have, x2y2 – 9y2 – 6x2y + 54y = 0 ⇒ y2 (x2 – 9) – 6y (x2 – 9) = 0 ⇒ y ( y – 6) (x – 3) (x + 3) = 0 ⇒ y = 0, y = 6, x = 3, x = – 3. So, the given equation represents four straight lines which form a square. 48. (d) Comparing the given equation with the equation

6 ∴ h2 = ab i.e.,   2

2

= 3m i.e., m = 3.

53. (d) The given equation can be rewritten as (x – y) (x2 + xy + 2y2) = 0. x – y = 0 represents a straight line passing through origin. But the equation x2 + xy + 2h2 = 0. represents two imaginary straight lines as

1 −7 ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get – 1.2 = < 0. h2 – ab = 4 4 a = 0, b = 0, h = h/2, g = g/2, f = f /2, c = c. Since the given equation represents a pair of straight 54. (b) The equation 3x2 + 2xy – 3y2 = 0 represents a pair lines, therefore of straight lines which are mutually perpendicular . abc + 2fgh – af 2 – bg2 – ch2 = 0 Therefore, the given three lines form a right angled 2 triangle. h  f   g  h ⇒ 0 + 2       – 0 – 0 – c   = 0 2  2   2  2 55. (c) The straight lines represented by 2 ( y – mx)2 = a2 (1 + m2) fgh ch − ⇒ = 0 or fg = ch.  are y – mx = ± a  1 + m 2 4 4 49. (d) We have, or

( x − 2) 2 + y 2 + ( x + 2) 2 + y 2 = 4

( x − 2) 2 + y 2 = 4 –

( x + 2) 2 + y 2

Squaring both the sides, we get (x – 2)2 + y2  = 16 + (x + 2)2 y2 – 8 ⇒

i.e., y – mx = a  1 + m 2 

...(1)

y – mx = – a  1 + m 2 

...(2)

and

Also, the straight lines represented by ( y – nx)2 = a2 (1 + n2) are y – nx = ± a  1 + n 2 ( x + 2) 2 + y 2

( x + 2) 2 + y 2 = x + 2.

 quaring again, we get y2 = 0, which represents S two coincident lines y = 0.

i.e.,

y – nx = a  1 + n 2

...(3)

and

y – nx = – a  1 + n

...(4)

2

 ince the lines (1) and (2) are parallel, so the disS tance between (1) and (2) is

383

Pair of Straight Lines

θ = tan

 ut, if we substitute y = 0 in the given equation, B we get

2 h 2 − ab π = tan– 1 3 = . a+b 3

– 1

384

a 1 + m2 + a 1 + m2



1 + m2

= | 2a |.

Objective Mathematics

 imilarly the lines (3) and (4) are parallel lines S and the distance between them is | 2a |. Since the distance between parallel lines are same, hence the four lines form a rhombus. 56. (c) Let y = mx be one of the lines represented by the given equation, then

ax3 + bmx3 + cm2x3 + dm3x3 = 0.

or dm3 + cm2 + bm + a = 0. Let its roots be m1, m2, m3. a ∴ m1 m2 m3 = – d If m1 = tan α, then m2 = tan (90º – α)

...(1) (Given)

∴ m2 = cot α. ∴ m1 m2 = 1. a . From (1), m3 = – d Since m3 is the root of the above cubic equation, we have  a2   a3   a d   − 3  + c   2  + b   −  + a = 0  d d   d  a 3 ca 2 ab + − +a = 0 d2 d2 d ⇒ – a3 + ca2 – abd + ad2 = 0

⇒ −

⇒ – a2 + ca – bd + d2 = 0 or a (a – c) + d (b – d) = 0.

59. (c) The given equation of pair of straight lines can be rewritten as

57. (c) The straight lines represented by the equation

3x – 4y = 0, 5x + 12y = 0 and y – 15 = 0  et ABC be the triangle whose sides BC, CA and L AB have the equations y – 15 = 0, 3x – 4y = 0 and 5x + 12y = 0 respectively. Solving these equations pairwise, we can obtain the coordinates of the vertices A, B, C as A (0, 0); B (– 36, 15) and C (20, 15) respectively. AD is a line passing through A (0, 0) and ⊥ to y – 15 = 0. So, the equation of AD is x = 0. The equation of any line ⊥ to 3x – 4y = 0 is 4x + 3y + λ = 0. This passes through (– 36, 15) if – 144 + 45 + λ = 0 ⇒ λ = 99. So, the equation of BE is 4x + 3y + 99 = 0. Solving the equations of AD and BE, we get x = 0, y = – 33. Thus, the coordinates of the orthocentre are (0, – 33).

2x2 – 27y2 – 3xy + 4x – 3y + 2 = 0 are x + 3y + 1 = 0 and 2x – 9y + 2 = 0.

( 3 x – y) (x – 3 y) = 0 Their separate equations are y=



3 x and y =

1 x 3

or y = tan 60º x and y = tan 30º x. After rotation, the separate equations are y = tan 90º x and y = tan 60º x or x = 0 and y = 3  x. ∴ The combined equation in the new position is The two lines intersect at the point A (– 1, 0) Let AB : x + 3y + 1 = 0 ...(1) and

AC : 2x – 9y + 2 = 0

Then, BC : 4x – 3y – 26 = 0 Solving (1) and (3), we get B ≡ (5, – 2). Solving (2) and (3), we get C ≡ (8, 2). ∴ Coordinates of centroid of ∆ ABC are

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

 − 1 + 5 + 8 0 − 2 + 2  i.e., (4, 0) ,   3 3

58. (a) The straight lines represented by the given equation are

x ( 3 x – y) = 0 or

3 x2 – xy = 0.

60. (a) The image of the lines y=x and y = – x by the line x = 2 are given by the lines y = x – 4 and y = – (x – 4). Their combined equation is y = | x – 4 |.

Their separate equations are 3 y – x + 3 = 0 and 3 y + x – or

y=

F (X, Y) = (a cos2 θ + 2h cos θ sin θ + b sin 2 θ) X2 

+ 2 [(b – a) cos θ sin θ + h (cos2 θ – sin2 θ)] XY + (a sin2 θ – 2h cos θ sin θ + b cos2 θ) Y2.

3 =0

1 1  x – 1 and y = –  x + 1. 3 3

or y = (tan 30º) x – 1 and y = (tan 150º) x + 1.

Now, coefficient of XY = 0, then we get 2h a−b or tan 2θ = . cot 2θ = a−b 2h 65. (c) We have, tan 2 A = ⇒

2 k 2 + tan 2 A − tan 2 A + 1

2 k 2 + tan 2 A 2 tan A = 2 1 − tan A 1 − tan 2 A

⇒ tan A = k 2 + tan 2 A ⇒ k = 0 66. (c) Equations of bisectors of lines xy = 0 are y = ± x. Put y = ± x in my2 + (1 – m2)xy – mx2 = 0, we get After rotation through an angle of 15º, the lines mx2 + (1 – m2)x2 – mx2 = 0 are ⇒ (1 – m2)x2 = 0 ( y – 0) = tan 45º (x – 3 ) ⇒ m = ± 1 and ( y – 0) = tan 135º (x – 3 ) 67. (a) Given equation is or y = x – 3 and y = – x + 3 . ax2 + by2 + hx + hy = 0 Their combined equation is On comparing with the standard equation ( y – x + 3 ) (y + x – 3 ) = 0 Ax2 + By2 + 2Hxy + 2Gx + 2Fy + C = 0, we get. or

y2 – x2 + 2 3 x – 3 = 0.

62. (b), (d)  The equation y2 + xy – 12x2 = 0 can be rewritten as

y = – 4, 3. x The two pairs will have a line common if – 4 or 3 is a root of 2  y  y b   + 2h   + a = 0. x x ( y + 4x) ( y – 3x) = 0

⇒

∴ 9b + 6h + a = 0 or 16b – 8h + a = 0. 63. (a) Replacing x by – x, the image of the pair of lines 3x2 + 4xy + 5y2 = 0 is 3x2 + 4y (– x) + 5y2 = 0 or 3x2 – 4xy + 5y2 = 0

A = a, B = b, H = 0, G = h , F = h and C = 0 2 2 Condition for pair of straight lines is ABC + 2FGH – AF2 – BG2 – CH2 = 0 ∴ a × b × 0 + 2 × h × h × 0 2 2 2

2

h h −a   − b   − 0 × 0 = 0 2 2 2 2 ⇒ − ah − bh = 0 4 4 ⇒ a + b = 0

68. (b) Equations of bisectors between the lines x2 – 2pxy – y2 = 0 is x2 − y 2 xy = 1 − (−1) − p 2 xy ⇒ x 2 + − y2 = 0 p This will be same as the x2 – 2qxy – y2 = 0 1 ⇒ pq + 1 = 0. =1 ∴ − pq 69. (c) Given equation is y2 – x2 + 2x – 1 = 0

64. (d) Rotating the axes through an angle θ, we have x = X cos θ – Y sin θ and y = X sin θ + Y cos θ.  f (x, y) = ax2 + 2hxy + by2.

On comparing the given equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 we get, a = – 1, h = 0, b = 1, g = 1, f = 0, c = – 1

385

After rotation, new equation is

Pair of Straight Lines

61. (b) The given equation of pair of straight lines can be rewritten as ( 3 y – x + 3 ) ( 3 y + x – 3 ) =0

386

∴ abc + 2fgh – af 2 – bg2 – ch2 =1+0+0–1–0=0 Hence, the given equation represents a pair of straight lines.

2

⇒ 3  y  + 9  y  +  y  + 3 = 0 x x x      

Objective Mathematics

  y y  ⇒ 3   + 1  + 3 = 0 x x     

70. (b) The given equation is x2 + 2hxy + 2y2 = 0

⇒

Here, a = 1, 2h = 2h and b = 2 Let the slopes of lines are m1 and m2.

Therefore, the slopes of the two lines represented by the given equation are –3 and − 1 . 3 73. (c) Given equations are

∴ m1 : m2 = 1 : 2 Let m1 = m and m2 = 2m. ∴ m1 + m2 = − 2h 2 ⇒ m + 2m = – h ⇒ h = – 3m ⇒ m · 2m = 1 and m1m2 = a b 2 ⇒ m = ± 1  2

…(i)

ax2 + 2hxy + by2 = 0

….(i)

and ax2 + 2hxy + by2 + k(x2 + y2) = 0

….(ii)

or (a + k) x2 + 2hxy + (b + k) y2 = 0 Equation of bisector of equation (i) is x 2 − y 2 xy = a−b h

…(ii)

Equation of bisector of (ii) is

From Eqs. (i) and (ii), we get h = ± 3 2 71. (c) The given equation is x2 – 6xy + 9y2 + 3x – 9y – 4 = 0 On comparing this equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get a = 1, h = – 3, b = 9, g = 3 , f = − 9 and c = – 4 2 2 ∴ Required distance = 2

y y 1 = −3 and = − x x 3

x2 − y 2 xy = or (a + b) − (b + k ) h

x 2 − y 2 xy = a−b h

Since equations of bisector of (i) and (ii) are same. ∴ lines ax2 + 2hxy + by2 = 0 are equally inclined to lines 2 ax + 2hxy + by2 + k(x2 + y2) = 0 for any value of k. 74. (b) Given equation of line is 2 2 3x + y = 2 and equation of circle is x + y = 4

f 2 − bc b ( a + b)

Now, we makes a given equation homogeneous  3x + ∴ x2 + y2 = 4   2  2 2 2 2 ⇒ x + y = 3x + y +

2

 9  − 2  + (9) (4)   = 2 9 (9 + 1)

y  

2

2 3xy

⇒2x2 + 2 3 xy = 0 81 + 144 225 = 2 = 4 × 90 4 × 90

= 2

Here a = 2, h =

5 2

3 and b = 0

 2 (3) 2 − 0 ∴ θ = tan–1   2+0  π = tan–1 ( 3) = 3

72. (b) The given equation can be written as 2

 y  y 3   + 10   + 3 = 0 x   x

   

EXERCISEs FOR SELF-PRACTICE 1. The area of the triangle by the lines x2 + 4xy + y2 = 0, x + y = 1 is (a) (c) 1

3

(b) 2

3 (d) 2

2. 2x2 + 7xy – 3y2 + 8x + 14y + λ = 0 will represent a pair of straight lines when λ = (a) 2 (c) 6

(b) 4 (d) 8

(a) c2 – 4 = 0 (c) c2 – 9 = 0

11. For what value of ‘p’, y2 + xy + px2 – x – 2y + p = 0 represent 2 straight lines: 1 (a) 2 (b) 3 1 1 (c) (d) . 4 2

(d) c2 – 8 = 0 (d) c2 – 10 = 0.

4. The angle between the lines given by x2 – y2 = 0 is (a) 15º (b) 45º (c) 75º (d) 90º 5. The equation x2 + kxy + y2 – 5x – 7y + 6 = 0 represents a pair of straight lines, then k is: (a) 5/3 (b) 10/3 (c) 3/2 (d) 3/10. 6. The gradient of one of the lines ax2 + 2hxy + by2 = 0 is twice that of the other, then: (a) h2 = ab (b) h = a + b (c) 8h2 = 9ab (d) 9h2 = 8ab 7. The sum of the slopes of the lines represented by 4x2 + 2hxy – 7y2 = 0 is equal to the product of the lopses then h is (a) – 2 (b) – 4 (c) 4 (d) – 6 8. If the equation x2 + y2 + 2gx + 2fy + 1 = 0 represents a pair of lines, then 1 (b) f 2 – g2 = 1 (a) f 2 + g2 = 2 (d) g2 – f 2 = 0 (c) f 2 + g2 = 1 9. The angle between the pair of lines given by equation x2 + 2xy – y2 = 0 is π π (b) (a) 6 3 π (c) (d) 0 2

12. 13.

Equation 3x2 + 7xy + 2y2 + 5x + 5y + 2 = 0 represents: (a) a pair of straight lines (b) an ellipse (c) a hyperbola (d) None of these The equation 8x2 + 8xy + 2y2 + 26x + 13y + 15 = 0 represents a pair of straight lines. The distance between them is:

(a) 7 / 5

(b) 7 /2 5

(c)

(d) None of these

7 /5

14. If the two pairs of lines x2 – 2mxy – y2 = 0 and x2 – 2nxy – y2 = 0 are such that one of them represents the bisectors of the angle between the other, then: (a) mn + 1 = 0 (b) mn – 1 = 0 (c) 1/m + 1/n = 0 (d) 1/m – 1/n = 0

Answers

1. (d) 11. (c)

2. (d) 12. (a)

3. (c) 13. (b)

4. (d) 14. (a)

5. (b)

6. (a)

7. (a)

8. (c)

9. (c)

10. (a)

387

10. The angle between the straight lines x2 – y2 – 2y – 1= 0 is (a) 90º (b) 60º (c) 75º (d) 36º

Pair of Straight Lines

3. The pair of straight lines joinign the origin to the points of interesection of the line y = 2 2 x + c and the circle x2 + y2 = 2, are at right angles, if

11

Circles

CHAPTER

Summary of conceptS circle

nature of the circle

A circle is the locus of a point which moves in a plane such that its distance from a fixed point always remains constant. The fixed point is called the centre and the constant distance is called the radius of the circle.

Standard equation of a circle The equation of a circle with the centre at (h, k) and radius a, is (x – h)2 + ( y – k)2 = a2 If the centre of the circle is at the origin and radius is a, then the equation of circle is x2 + y2 = a2.

General equation of a circle The general equation of a circle is of the form ...(1) x2 + y2 + 2gx + 2 fy + c = 0, where g, f and c are constants. The coordinates of its centre are (– g, – f ) and radius is g2 + f 2 − c .

1. If g2 + f 2 – c > 0, then the general eqn. (1) represents real circle with centre (– g, – f ). 2. If g2 + f 2 – c = 0, then the general eqn. (1) represents a circle whose centre is (– g, – f ) and radius is zero i.e., the circle coincides with the centre represented by a point (– g, – f ). It is, therefore called a point circle. 3. If g2 + f 2 – c < 0, the radius of the circle is imaginary but the centre is real. Such a circle is called a virtual circle or imaginary circle as it is not possible to draw such a circle.

different forms of the equation of a circle 1. Circle with centre at the point (h, k) and which touches the axis of x Since the circle touches the x-axis, the radius of the circle = k.

conditionS for an equation to repreSent a circle A general equation of second degree ax2 + 2hxy + by2 + 2gx + 2 fy + c = 0 in x, y represents a circle if (i) coefficient of x2 = coefficient of y2, i.e., a = b, (ii) coefficient of xy is zero, i.e., h = 0.

to find the centre and radius of a circle whose equation is Given Working Rule 1. Make the coefficients of x2 and y2 equal to 1 and right hand side equal to zero. 2. The coordinates of centre will be (h, k), where

h=–

1 (coefficient of x) and 2

3. Radius =

k = –

h 2 + k 2 − constant term

1 (coefficient of y) 2

Therefore, the equation of the circle is: (x – h)2 + (y – x)2 = k2 or

x2 + y2 – 2hx – 2ky + h2 = 0.

2. Circle with centre at the point (h, k) and which touches the axis of y Since the circle touches the y-axis, the radius of the circle = h.

(x – h) – (y – k) =  h

or

x2 + y2 – 2hx – 2ky – k2 = 0.

2

2

(d) (e)

If the circle touches y-axis, then | CD | = 0 Thus, c = f 2. If the circle touches both the axes, then | AB | = 0 = | CD | Thus, c = g2 = f 2.

3. Circle with radius a and which touches both the coordinate axes Since the centre of the circle may be in any of the four Parametric Equations of a Circle quadrants, therefore it will be any one of the four points (±a, ±a). Thus, there are four circles of radius a touching (a) The parametric equations of a circle x2 + y2 = a2 are both the coordinate axes, and their equations are x = a cos θ, y = a sin θ, 0 ≤ θ < 2π. θ is called parameter and the point P (a cos θ, a sin θ) is called the point ‘θ’ on the circle x2 + y2 = a2. Thus, the coordinates of any point on the circle x2 + y2 = a2 may be taken as (acos θ, a sin θ).



(x ± a)2 + ( y ± a)2 = a2

or

x2 + y2 ± 2ax ± 2ay + a2 = 0.

4. Equation of the Circle on the Line Joining two Point as as Diameter The equation of the circle drawn on the line segment joining two given points A (x1, y1) and B (x2, y2) as diameter is

(b) The parametric equations of a circle (x – h)2 + ( y – k)2 = a2 are x = h + a cos θ, y = k + a sin θ, 0 ≤ θ < 2π. is called the point ‘θ’ on this circle. Thus the coordinates of any point on this circle may be taken as (h + a cos θ, k + a sin θ).

(x – x1) (x – x2) + (y – y1) (y – y2) = 0.  x + x y + y2  Its Centre =  1 2 , 1  and  2 2   y1 − y2   x1 − x2   .   +  2  2  2

Radius =

2

Intercepts made by a Circle on the axes (a) The Length of the intercept made by the circle x2 + y2 + 2gx + 2fc + c = 0 on

Position of a Point with respect to a circle Let S ≡ x2 + y2 + 2gx + 2 f y + c = 0, be a circle and P (x1, y1) be a point in the plane of S, then S1 ≡ x12 + y12 + 2 gx1 + 2 fy1 + c . The point P (x1, y1) lies outside, on or inside the circle S according as S1 > , = or < 0, respectively.

x-axis = AB = 2 g 2 − c y­-axis = CD = 2 f 2 − c (b) Intercepts are always positive. (c) If the circle touches x-axis then | AB | = 0 Thus, c = g2.

Note:  Let S be a circle and P (x1, y1) and Q (x2, y2) be two points in the plane of S, then they lie (i) on the same side of S iff S1 and S2 have same sign, (ii) on the opposite sides of S iff S1 and S2 have opposite signs,

389



2

Circles

Therefore, the equation of the circle is

390

Circle through three points

Contact of Two Circles

Objective Mathematics

The equation of the circle through three non-collinear points The two circles having centres at A (x1, y1) and B (x2, y2) and radii (x1, y1), (x2, y2) and (x3, y3) is r1 and r2 respectively will x2 + x12 + x22 + x32 +

y2 y12 y22 y32

x x1 x2 x3

y y1 y2 y3

1 1 1 =0 1

(a) intersect in two real distinct points if and only if | r1 – r2 | < AB < r1 + r2,

Note: 1. If the two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 meet the coordinate axes in four distinct points, then those points are concyclic if a1a2 = b1b2. (b) touch each other externally if and only if Also, the equation of the circle passing through those concyclic points is (a1x + b1y + c1) (a2x + b2y + c2) – (a1b2 + a2b1) xy = 0.

AB = r1 + r2 and their point of contact C is given by

2. The equation of the circumcircle of the triangle formed by the line ax + by + c = 0 with the coordinate axes is ab (x2 + y2) + c (bx + ay) = 0.

Intersection of a Line and a Circle

rx +r x r y +r y  C≡  1 2 2 1, 1 2 2 1 . r1 + r2   r1 + r2 (c) touch each other internally if and only if AB = | r1 – r2 |, and their point of contact C is given by

Let S be a circle with centre C and radius a. Let l be any line in the plane of the circle and d be the perpendicular distance from C to the line l, then rx −r x r y −r y  (a) l intersects S in two distinct points iff d < a C≡  1 2 2 1, 1 2 2 1 . r1 − r2   r1 − r2 (b) l intersects S in one and only point iff d = a, i.e., the line l touches the circle iff perpendicular distance from the cen- (d) One circle lies outside the other if AB > r + r . 1 2 tre to the line l must be equal to radius of the circle. (c) l does not intersect S iff d > a.

Length of intercept made by a circle on a line If a line l meets a circle S, with centre C and radius a, in two (e) One circle is contained in the other if AB < |r1 – r2|. distinct points and if d is the perpendicular distance of centre C from the line l, then the length of the intercept made by the circle on the line | AB | = 2 a 2 − d 2 .

Notations If S ≡ x2 + y2 –a2, then S1 ≡ x12 + y12 − a 2 and T ≡ xx1 + yy1 – a2. Note:  If the points of intersection of a line l and a circle S are If S ≡ x2 + y2 +2gx + 2f y + c, then known, then the distance between these points is the required 2 2 S1 ≡ x1 + y1 + 2 gx1 + 2 fy1 + c length of intercept and there is no need of using the above and T ≡ xx1 + yy1 + g (x + x1) + f ( y + y1) + c. formula.

a Given point

PT =

1. Equation of the tangent to the circle x + y = a at the point (x1, y1) on it is: xx1 + yy1 = a2. 2. Equation of the tangent to the circle 2

2

x12 + y12 + 2 gx1 + 2 fy1 + c

2

x2 + y2 + 2gx + 2f y + c = 0 at the point (x1, y1) on it is

xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0.

power of a point with respect to a circle

3. Equation of the tangent to the circle x2 + y2 = a2 at the point The power of a point P (x1, y1) with respect to a given circle (a cos θ, a sin θ) on it is: x2 + y2 + 2gx + 2f y + c = 0 is defined as: x cos θ + y sin θ = a Power = PA × PB [Parametric form of equation of tangent] where PAB is any line through P intersecting the circle in Note: The equation of the tangent at the point (x1, y1) on the A and B. If PT is the length of the tangent from a point P to a given circle S = 0 is T = 0. circle, the PT2 is also the power of the point P with respect to the given circle. Thus, power of the point P (x1, y1) w.r.t. the circle equation of the tangent in Slope form S = 0 is S1. The equation of a tangent of slope m to the circle x2 + y2 = a2 is y = mx ± a 1 + m 2 . The coordinates of the point of contact are   ±

am 1+ m

2

,∓

a 1+ m

2

  .

Note: 1. The power of the point outside the circle is positive. 2. The power of the point on the circle is zero. 3. The power of the point inside the circle is negative.

condition of tangency The straight line y = mx + c will be a tangent to the circle x + y = a if c = ± a 2

2

2

pair of tangents

1+ m . 2

Note: A line will touch a circle if and only if the length of the perpendicular from the centre of the circle to the line is equal to the radius of the circle.

The equation of the pair of tangents drawn from the point P (x1, y1) to the circle S = 0 is SS1 = T2, where

tangents from a point outside the circle Working Rule 1. Let the point be (x1, y1). Write the equation of a straight line passing through the point (x1, y1) and having slope m i.e., ( y – y1) = m (x – x1)

...(1)

2. Find the length of the perpendicular from the centre of the circle to the line (1) and equate it to the radius of the circle. Call this equation as (2). 3. Obtain the value of m from the eqn. (2). 4. Substitute this value of m in equation (1) to obtain the required equation of tangent.

and

S : x2 + y2 + 2gx + 2fy + c, 2 2 S1 : x1 + y1 + 2 gx1 + 2 fy1 + c T : xx1 + yy1 + g (x + x1) + f (y + y1) + c.

Note: The pair of tangents from (0, 0) to the circle x2 + y2 + 2gx + 2fy + c = 0 are at right angles if g2 + f 2 = 2c.

chord of contact of tanGentS Let P be a point outside a circle. If two tangents PA and PB be drawn to the circle, then the chord AB is called the chord of contact of tangents drawn from the point P to the circle.

391

Circles

S1 is obtained from S by putting x1 and y1 in place of length of the tangent from a point to a x and y respectively and T is obtained by putting xx1 in place circle x + x1 y + y1 in place of x and in The length of the tangent that can be drawn from the point of x2, yy1, in place of y2, 2 2 P (x1, y1) to the circle S = 0 is S1 , where the coefficients of x2 place of y. and y2 in the equation of the circle are unity. Thus, the length of the tangent from the point P (x1, y1) to circle x2 + y2 + 2gx + 2f y tanGent to a circle at + c = 0 is given by

392

The equation of the chord of contact of tangents drawn from the point P (x1, y1) is given by

at the point (x1, y1) on it is

Objective Mathematics

x − x1 y − y1 . = x1 + g y1 + f Alternate Method  Equation of the line joining (x1, y1) and centre of the circle gives the equation of normal to the circle.

T = 0,  i.e., xx1 + yy1 = a2 for the circle x2 + y2 = a2 and xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0 for the circle x2 + y2 + 2gx + 2fy + c = 0.

Pole and Polar If a straight line is drawn from a point P to meet the circle at Q and R and the tangents to the circle at Q and R meet at a point T1, then the locus of T1 is called the polar of P with respect to the circle.

Conjugate Points and lines Two points are said to be conjugate points with respect to a circle if the polar of either passes through the other. Two straight lines are said to be conjugate lines if the pole of either lies on the other. Note: 1. P (x1, y1) and Q (x2, y2) are conjugate points w.r.t. the circle x2 + y2 + 2gx + 2fy + c = 0 if x1x2 + y1y2 + g (x1 + x2) + f (y1 + y2) + c = 0 2. If P and Q are conjugate points w.r.t. a circle with centre at O and radius r, then PQ2 = OP2 + OQ2 – 2r2.

Chord with a given middle point The point P is called the pole of its polar. The polar of the point P (x1, y1) w.r.t. the circle S = 0 is given by T = 0, i.e.,

xx1 + yy1 + = a2, for the circle x2 + y2 = a2 and xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0

The equation of the chord of the circle S = 0, whose middle point is (x1, y1), is given by T = S1, i.e.,   xx1 + yy1 + g (x + x1) + f (y + y1) + c = x12 + y12 + 2gx1 + 2fy1 + c, for the circle x2 + y2 + 2gx + 2f y + c = 0 and



2 2 2   xx1 + yy1 – a2= x1 + y1 − a for the circle x2 + y2 = a2.

for the circle x2 + y2 + 2gx + 2f y + c = 0.

Note:

Note:

1. If (x1, y1) is the mid point of the chord PQ of the circle

S = 0, then the length PQ = 2 −S1 , where S1 < 0. 1. If the point P lies outside the circle, then the polar and the 2. The coordinates of middle point of the chord lx + my + n chord of contact of this point P are same straight line. 2. If the point P lies on the circle, then the polar and the tan= 0 of the circle x2 + y2 = a2 are  − ln , − mn  .  l 2 + m2 l 2 + m2  gent to the circle at P are same straight line. 3. The coordinates of the pole of the line lx + my + n = 0 with 3. Let (x , y ) be the mid point of the chord PQ of the circle 1 1 2 2 S = 0. If the tangents at P and Q meet at R, then the area of respect to the circle x2 + y2 = a2 are  − a l , − a m  .  n n  (− S1 )3 / 2 , where a is the radius of the circle. ∆PQR = 4. If (x1, y1) is the pole of the line lx + my + n = 0 S1 + a 2 w.r.t. the circle x2 + y2 + 2gx + 2fy + c = 0, then r2 x1 + g y + f , where r is the radius of Common Chord of Two circles = = 1 lg + mf − n l m The chord joining the point of intersection of two circles is called the circle. their common chord.

Normal to A Circle at a Given Point

If S = 0 and S′ = 0 be two interesting circles, then the equation of their common chord is S – S′ = 0.

The normal to a circle, at any point on the circle is a straight line which is perpendicular to the tangent to the circle at that point and always passes through the centre of the circle. 1. Equation of the normal to the circle x2 + y2 = a2 at the point y x = . (x1, y1) on it is: y x1 1 Note:  In order to find the equation of the common chord of two 2. Equation of the normal to the circle circles, first of all make the coefficients of x2 and y2 in the equa x2 + y2 + 2gx + 2f y + c = 0 tion of the two circles equal to unity.

393

Length of the Common Chord  The length of the common chord PQ is given by

Circles

PQ = 2 PR = 2 (AP) 2 − (AR) 2 , where AP is the radius of the circle S = 0 and AR is the length of perpendicular from A to the common chord PQ. Note: 1. The length of the common chord of two circles becomes maximum when it is a diameter of the smaller one between Note: them. 1. If the two circles cut each other at two different points, 2. If the length of common chord is zero, then the two circles then their radical axis will be their common chord. touch each other and the common chord becomes the common tangent to the two circles at the common point of contact.

Angle of intersection of two circles The angle between the two circles is the angle between their tan 2. If the two circles touch each other (internally or externalgents at their point of intersection. ly), then their radical axis will be the common tangent at The angle of intersection θ of two circles the point of contact. S ≡ x2 + y2 + 2g1x + 2f1 y + c1 = 0 and S′ ≡ x2 + y2 + 2g2x + 2f2 y + c2 = 0

Properties of Radical Axis 1. The radical axis of two circles is perpendicular to the line joining the centres. 2 g12 + f12 − c1 . g 22 + f 22 − c2 2. Radical  Centre  The radical axis of three circles taken in pairs meet at a point, called the radical centre of the circles. Coordinates of radical centre can be found by solving the Orthogonal Intersection of two Circles equations S1 = S2 = S3 = 0. Two circles are said to intersect orthogonally when they intersect 3. The radical centre of three circles described on the sides of at right angles. a triangle as diameters is the orthocentre of the triangle. The condition for the circles x2 + y2 + 2g1x + 2f1 y + c1 4. If two circles cut a third circle orthogonally, then the radi= 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 to intersect orthogonally is cal axis of the two circles will pass through the centre of given by the third circle. 2g1g2 + 2 f1 f2 = c1 + c2. is given by cos θ = ±

2 g1 g 2 + 2 f1 f 2 − c1 − c2

Note:  If P and Q are conjugate points w.r.t. the circle S = 0, then the circle on PQ as diameter is orthogonal to S = 0.

Equation of a Circle through the intersection of a line and a Circle

Radical Axis

Equation of the family of circles passing through the points of intersection of the circle S = 0 and the line L = 0 is

The radical axis of two circles is the locus of a point which moves in such a way that the lengths of the tangents drawn from it to the two circles are equal. The equation of radical axis of two circles S ≡ x2 + y2 + 2g1x + 2f1 y + c1 = 0 and

S′ ≡ x2 + y2 + 2g2x + 2f2 y + c2 = 0

is 2 (g1 – g2) x + 2 ( f1 – f2) y + c1 – c2 = 0 or S – S′ = 0.

S + kL = 0, where k is a parameter, k ≠ – 1.

Equation of a CIrcle through the intersection of two Circles Equation of the family of circles passing thought the points of intersection of circles S = 0 and S′ = 0 is S + kS′ = 0, where k is a parameter, k ≠ – 1.

394

equation of a circle throuGh two Given pointS

Objective Mathematics

Equation of the family of circles passing through two given points (x1, y1) and (x2, y2) is x (x – x1) (x – x2) + ( y – y1) ( y – y2) + k x1 x2 = 0, k ∈ R.

y 1 y1 1 y2 1

k can be found from the third condition.

Working Rule

coaxal SyStem of circleS A system of circles is said to be coaxal if each pair of circles of the system has the same radical axis.

equation of a System of coaxal circles in the Simplest form The equation of a system of coaxal circles is x2 + y2 + 2gx + c = 0, where g is a constant and c is a parameter and the line of centre of the circles is taken as x-axis and the radical axis as the y-axis. Note:

1. Find the coordinates of centres C1, C2 and radii r1, r2 of two given circles. 2. Find the coordinates of the point P dividing C1C2 in the ratio r1 : r2 externally. Let P ≡ (h, k). 3. Write the equation of any line through the point P(h, k), i.e., ( y – k) = m (x – h)

...(1)

4. Find the two values of m, using the fact that the length of perpendicular on line (1) from the centre C1 of one circle is equal to its radius r1. 5. Substituting these values of m in eqn. (1), the equations of two direct common tangents are obtained.

transverse common tangents

1. The equation of a system of coaxal circles, when the equa- The transverse common tangents to the two circles intersect at a tion of the radical axis and one circle of the system are point (say θ) which lies on the line joining the centres C1 and C2 P ≡ lx + my + n = 0 of the two circles and divide C1C2 internally in the ratio of their and S ≡ x2 + y2 + 2gx + 2fy + c = 0 radii r1 and r2. respectively, is S + kP = 0 (k is an arbitrary constant) 2. The equation of a coaxal system of circles when the equation of any two circles of the system are S ≡ x2 + y2 + 2g1x + 2f1 y + c1 = 0 and S′ ≡ x2 + y2 + 2g2x + 2f2 y + c2 = 0, is S + k (S – S′) = 0 or S′ + k (S – S′) = 0, i.e., S + kS′ = 0 (k ≠ –1).

limitinG pointS of a coaxal SyStem Limiting points of coaxal system of circles, are point circles of the system i.e., circles of radius zero. The limiting points of the system of circles x2 + y2 + 2gx + c = 0 are (±

c , 0).

common tanGentS to two circleS direct common tangents The direct Common tangents to the two circles meet at a point (say P) which lies on the line joining the centres C1 and C2 of the two circles and divide C1C2 externally in the ratio of their radii say (r1 and r2)

Working Rule 1. Find the coordinates of centres C1, C2 and radii r1, r2 of two given circles. 2. Find the coordinates of the point θ dividing C1C2 ,in the ratio r1 : r2 internally. Let P ≡ (h, k). 3. Write the equation of any line through the point P (h, k) ( y – k) = m (x – h)

...(1)

4. Find the two values of m, using the fact that the length of perpendicular on (1) from the centre C1 of one circle is equal to its radius r1. 5. Substituting these values of m in eqn. (1), the equations of two transverse common tangents are obtained.

Note: 1. If two circles do not intersect (c1c2 > r1 + r2), then they have two transverse and two direct common tangents. 2. If two circles intersect (c1c2 < r1 + r2), then they have two direct tangents only. 3. If two circles touch externally (c1c2 = r1 + r2), then they have one transverse and two direct common tangents. 4. If two circles touch internally (c1c2 = r1 – r2), then they have only one common tangent.

Let the circle be

S ≡ x2 + y2 + 2gx + 2fy + c = 0

and the line be

L ≡ lx + my + n = 0.

Slope of C1C2 × Slope of the line L = – 1 and mid point of C1 (– g, – f ) and C2 (x1, y1) lies on the line lx + my + n = 0 x −g y − f  i.e., l  1 + n = 0 + m 1  2   2 

...(1)

...(2)

Solving eqns. (1) and (2), to get value of (x1, y1). Then the required image circle is where

(x – x1)2 + (y – y1)2 = r2. r=

g2 + f 2 − c .

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: (a) intersecting circles (b) non-intersecting circles (c) touching circles (d) touching or non-intersecting circles

1. The radius of the circle, having centre at (2, 1) whose one of the chords is a diameter of the circle x2 + y2 – 2x – 6y + 6 = 0 is (a) 1

(b) 2

(c) 3

(d)

7. The locus of the foot of the perpendicular from the origin to the line which always passes through a fixed point 2. The area of an equilateral triangle inscribed in the (h, k) is circle (a) parabola (b) ellipse 3

x2 + y2 + 2gx + 2f y + c = 0 is

(c) hyperbola

(a)

3 3 2 ( g + f 2 − c) 2

(b)

3 3 2 ( g + f 2 − c) 4

(c)

3 3 2 ( g + f 2 + c) 4

(d) None of these

(a) x2 + y2 + 16x + 4y – 32 = 0 (b) x2 + y2 – 16x + 4y + 32 = 0 (c) x2 + y2 – 16x + 4y – 32 = 0 (d) None of these

3. The circle x2 + y 2 – 8x + 4y + 4 = 0 touches (a) x-axis (b) y-axis (c) both axis (d) neither x-axis nor y-axis

9. A circle has radius 3 and its centre lies on the line y = x – 1. The equation of the circle, if it passes through (7, 3), is

4. If sin [cot– 1 (x + 1)] = cos (tan – 1 x), then x = 1 2 10. 9 (c) 0 (d) 4 5. The radius of the circle passing through the point (6, 2) and two of whose diameters are x + y = 6 and x + 2y = 4 is (a) –

1 2

(a) 4 (c) 20

(b)

(b) 6 (d) 20 6. The co-axial system of circles given by x2 + y 2 + 2gx + c = 0 for c < 0 represents

(d) circle.

8. If the equations of the two diameters of a circle are x + y = 6 and x + 2y = 4 and the radius of the circle is 10, then the equation of the circle is

(a) x2 + y2 + 8x – 6y + 16 = 0 (b) x2 + y2 – 8x + 6y + 16 = 0 (c) x2 + y2 – 8x – 6y – 16 = 0 (d) x2 + y2 – 8x – 6y + 16 = 0 The equation of a circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of the lines 3x + 5y = 1 and (2 + c) x + 5c 2y = 1 as c tends to 1, is (a) 25 (x2 + y2) + 20x + 2y – 60 = 0 (b) 25 (x2 + y2) – 20x + 2y + 60 = 0 (c) 25 (x2 + y2) – 20x + 2y – 60 = 0 (d) None of these

11. The equation of the circle which touches the line 5x + 12y = 1 and which has its centre at (3, 4) is

395

The radius of the image circle remains unchanged but centre changes. Let the centre of image circle be (x1, y1). Then,

Circles

Image of the Circle by the Line Mirror

396

 62  (a) (x – 3)2 + (y – 4)2 =    11 

2

Objective Mathematics

 62  (b) (x – 3)2 + (y – 4)2 =    17 

2

 62  (c) (x – 3)2 + (y – 4)2 =    13  (d) None of these

2

19. If (2, – 1) lies on the circle x2 + y 2 + 2gx + 2f y + c = 0 which is concentric with the circle x2 + y 2 + 4x – 6y + 3 = 0, then c is (a) 19 (c) 21

12. The equation of the circle which touches the axis of y at a distance + 4 from the origin and cuts off an intercept 6 from the axis of x is (a) x2 + y2 – 10x – 8y + 16 = 0 (b) x2 + y2 + 10x – 8y + 16 = 0 (c) x2 + y2 – 10x + 8y + 16 = 0 (d) None of these 13. The equation of a circle passing through the origin and making intercepts 4, 5 on the coordinate axes is (a) x2 + y2 – 4x + 5y = 0 (b) x2 + y2 – 4x – 5y = 0 (c) x2 + y2 + 4x + 5y = 0 (d) None of these 14. A circle of radius 2 lies in the first quadrant and touches both the axes of coordinates. The equation of the circle with centre at (6, 5) and touching the above circle externally is (a) x2 + y2 + 12x – 10y + 52 = 0 (b) x2 + y2 – 12x + 10y + 52 = 0 (c) x2 + y2 – 12x – 10y + 52 = 0 (d) None of these 15. The equation of the circle whose centre is (3, –1) and which cuts off an intercept of length 6 from the line 2x – 5y + 18 = 0, is (a) x2 + y2 – 6x + 2y + 28 = 0 (b) x2 + y2 + 6x + 2y – 28 = 0 (c) x2 + y2 – 6x – 2y + 28 = 0 (d) x2 + y2 – 6x + 2y – 28 = 0. 16. The equation of the circle, the end points of whose diameter are the centres of the circles x + y + 6x – 14y = 1 and x + y – 4x + 10y = 2, is 2

2

2

2

(a) x2 + y2 + x – 2y – 41 = 0 (b) x2 + y2 + x + 2y – 41 = 0 (c) x2 + y2 + x + 2y + 41 = 0 (d) None of these 17. The sides of a square are x = 2, x = 3, y = 1 and y = 2. The equation of the circle drawn on the diagonals of the square as its diameter, is (a) x2 + y2 – 5x – 3y + 8 = 0 (b) x2 + y2 + 5x – 3y + 8 = 0 (c) x2 + y2 + 5x + 3y – 8 = 0 (d) None of these 18. The equation to the circle, touching y-axis at (0, 3) and making intercept of 8 units on the x-axis, is (a) x2 + y2 – 10x + 6y + 9 = 0 (b) x2 + y2 – 10x – 6y + 9 = 0 (c) x2 + y2 + 10x + 6y + 9 = 0 (d) x2 + y2 + 10x – 6y + 9 = 0

(b) – 19 (d) none of these .

20. A circle of radius 5 touches the coordinate axes in the first quadrant. If the circle makes one complete roll on x-axis along the positive direction of x-axis, then its equation in the new position is (a) x2 + y2 – 10 (2π + 1)x – 10y + 100π2 + 100π + 25 = 0 (b) x2 + y2 + 10 (2π + 1)x – 10y + 100π2 + 100π + 25 = 0 (c) x2 + y2 – 10 (2π + 1)x + 10y + 100π2 + 100π + 25 = 0 (d) None of these 21. The intercept on the line y = x by the circle x2 + y 2 – 2x = 0 is AB. Equation of the circle on AB as a diameter is (a) x2 + y2 + x + y = 0 (c) x2 + y2 – x – y = 0

(b) x2 + y2 – x + y = 0 (d) x2 + y2 + x – y = 0

22. The equation of the circle which has two normals (x – 1) (y – 2) = 0 and a tangent 3x + 4y = 6 is (a) x2 + y2 – 2x – 4y + 4 = 0 (b) x2 + y2 + 2x – 4y + 5 = 0 (c) x2 + y2 = 5 (d) (x + 3)2 + (y – 4)2 = 5 23. The line joining (5, 0) and (10 cos θ, 10 sin θ) is divided internally in the ratio 2 : 3 at P. If θ varies, then the locus of P is (a) a pair of straight lines (b) a circle (c) a straight line (d) None of these 24. The abscissae of two points A and B are the roots of the equation x2 + 2ax – b 2 = 0 and their ordinates are the roots of the equation x2 + 2px – q2 = 0. The equation of the circle with AB as diameter is (a) x2 + y2 + 2ax + 2py + b2 + q2 = 0 (b) x2 + y2 – 2ax – 2py – b2 – q2 = 0 (c) x2 + y2 + 2ax + 2py – b2 – q2 = 0 (d) None of these 25. The length of the intercept made by the circle whose diameter is the line joining the end points (– ­4, 3) and (12, – 1) on y-axis, is (a) 2 13

(b) 4 13

(c) 8 13

(d) None of these

26. The lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154. Then the equation of this circle is (a) x2 + y2 + 2x – 2y = 62 (b) x2 + y2 + 2x – 2y = 47 (c) x2 + y2 – 2x + 2y = 47 (d) x2 + y2 – 2x + 2y = 62 27. The equation of the circle passing through the point (0, 0) and the points where the straight line 3x + 4y = 12, meets the axes of coordinates, is

28. The centres of the three circles x2 + y2 = 1, x2 + y2 + 6x – 2y = 1 and x2 + y2 – 12x + 4y = 1 (a) form a right angled triangle (b) form an isosceles triangle (c) are collinear (d) None of these 29. The equation of the circle, if its centre is (4, 5) and its circumference passes through the center of the circle x2 + y2 + 4x – 6y = 12, is (a) x2 + y2 – 8x – 10y + 1 = 0 (b) x2 + y2 + 8x – 10y + 1 = 0 (c) x2 + y2 – 8x + 10y + 1 = 0 (d) None of these 30. The equation of that diameter of the circle x2 + y2 – 6x + 2y – 8 = 0, which passes through the origin, is (a) x – 3y = 0 (c) 3x – y = 0

(b) x + 3y = 0 (d) None of these

31. The equation of the circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and of double its radius is (a) x2 + y2 – 6x + 12y + 75 = 0 (b) x2 + y2 – 6x – 12y + 75 = 0 (c) x2 + y2 – 6x + 12y – 75 = 0 (d) None of these 32. The equation of the circle of radius 5 with centre on x-axis and passing through the point (2, 3), is (a) x2 + y2 – 12x + 11= 0 (b) x2 + y2 + 4x – 21 = 0 (c) x2 + y2 + 12x + 11= 0 (d) x2 + y2 – 4x + 21 = 0 33. The locus of a point, which moves in such a way that the ratio of its distances from two given points is constant (≠ 1), is (a) circle (c) ellipse

(b) parabola (d) hyperbola.

(b) (1 − 2 , −2)

(c) (1, − 2 + 2 )

(d) None of these

37. The equation of circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is (a) x2 + y2 = 9a2 (c) x2 + y2 = 4a2

(b) x2 + y2 = 16a2 (d) x2 + y2 = a2

38. The equation of the circle described on the common chord of the circles x2 + y 2 – 4x – 5 = 0 and x2 + y 2 + 8y + 7 = 0 as diameter is (a) x2 + y2 – 2x + 4y + 1 = 0 (b) x2 + y2 – 2x – 4y + 1 = 0 (c) x2 + y2 + 2x + 4y + 1 = 0 (d) None of these 39. Equation of tangent to the circle x2 + y 2 = 10, at the point whose abscissa is 1, is (a) x + 3y = 10 (c) x – 3y = 10

(b) 3x + y = 10 (d) 3x – y = 10

40. If the lines a1 x + b1 y + c1 = 0 and a2 x + b 2y + c 2 = 0 cut the coordinate axes in concyclic points, then (a) a1a2 = b1b2 (c) a1b2 = a2b1

(b) a1b1 = a2b2 (d) None of these

41. The equation of the circle on the chord x cos α + y sin α – p = 0 of the circle x2 + y2 = a2 as diameter is (a) x2 + y2 – a2 – p (x cos α + y sin α – p) = 0 (b) x2 + y2 – a2 – 2p (x cos α + y sin α – p) = 0 (c) x2 + y2 – a2 + 2p (x cos α + y sin α – p) = 0 (d) None of these 42. Two rods of lengths a and b slide along the axes which are rectangular is such a manner that their ends are concyclic. The locus of the centre of the circle passing through these points is (a) 4 (x2 + y2) = a2 + b2 (b) x2 – y2 = a2 – b2 (c) 4 (x2 – y2) = a2 – b2 (d) x2 + y2 = a2 + b2 43. The equation of that normal to the circle x2 + y 2 – 2x = 0 which is parallel to the line x + 2y = 3 is

(a) x – 2y – 1 = 0 (b) x + 2y + 1 = 0 34. ABCD is a square with side whose length is l, if AB and (c) 2x + y – 1 = 0 (d) x + 2y – 1 = 0 AD be taken as axes, then the circle circumscribing 44. A circle of radius 2 lies in the first quadrant and touches the square will have the equation both the axes of coordinates. The equation of the cir(a) x2 + y2 = l (x + y) (b) x2 + y2 = l (x – y) cle with centre at (6, 5) and touching the above circle (c) x2 + y2 = l (x + 2y) (d) x2 + y2 = l (2x + y) externally is 35. From the origin, chords are drawn to the circle (a) x2 + y2 – 12x – 10y + 42 = 0 x2 + y 2 – 2y = 0. The locus of the middle points of (b) x2 + y2 – 12x – 10y – 52 = 0 these chords is (c) x2 + y2 – 12x – 10y + 52 = 0 2 2 2 2 (d) None of these (a) x + y – y = 0 (b) x + y – x = 0 2 2 2 2 (c) x + y – 2x = 0 (d) x + y – x – y = 0 45. The circles x2 + y 2 – 10x + 16 = 0 and x2 + y 2 = r 2 intersect each other in two distinct points if 36. A square is inscribed in the circle x2 + y 2 – 2x + 4y + 3 = 0. Its sides are parallel to the coordinate axes. Then (a) r < 2 (b) r > 8 one vertex of the square is (c) 2 < r < 8 (d) 2 ≤ r ≤ 8

397

(a) (1 + 2 , −2)

Circles

(a) x2 + y2 + 4x – 3y = 0 (b) x2 + y2 – 4x – 3y = 0 (c) x2 + y2 – 4x + 3y = 0 (d) None of these

398

46. The coordinates of the middle point of the chord which the circle x2 + y 2 + 4x – 2y – 3 = 0 cuts off on the line y = x + 2, are

Objective Mathematics

 −3 1  (a)  ,   2 2

3 1 (b)  ,  2 2

 −3 −1  (c)  ,   2 2

 3 −1  (d)  ,  2 2 

47. Circles are drawn through the point (2, 0) to cut intercept of length 5 units on the x-axis. If their centres lie in the first quadrant, then their equation is (a) x2 + y2 – 9x + 2ky + 14 = 0 (b) 3x2 + 3y2 + 27x – 2ky + 42 = 0 (c) x2 + y2 – 9x – 2ky + 14 = 0 (d) x2 + y2 – 2kx – 9y + 14 = 0

55. The equation of the circle, which touches the line x = y at the origin and passes through the point (2, 1), is (a) x2 + y2 – 5x + 5y = 0 (b) x2 + y2 + 5x – 5y = 0 (c) x2 + y2 + 5x + 5y = 0 (d) None of these 56. The equation of the tangent, from the point (0, 1) to the circle x2 + y 2 – 2x – 6y + 6 = 0, is (a) y – 1 = 0 (c) 4x + 3y – 3 = 0

(b) 4x + 3y + 3 = 0 (d) y + 1 = 0

57. The line y = mx + c intersects the circle x2 + y 2 = r 2 at the two real distinct points if (a) –  r  1 + m 2 < c < r

1 + m2

(b) –  c  1 − m 2 < r < c

1 + m2

(c) –  r 1 − m 2 < c < r 1 + m 2 48. The equation of the circles, whose diameter is the com(d) None of these mon chord of the circles x2 + y 2 + 2x + 3y + 1 = 0 and 58. If the tangents are drawn to the circle x2 + y 2 = a2 at the x2 + y 2 + 4x + 3y + 2 = 0 is points where x cos α + y sin α = p meets the circle, (a) 2x2 + 2y2 – 2x + 6y + 1 = 0 then their point of intersection is 2 2 (b) 2x + 2y + 2x – 6y + 1 = 0  a 2 sin α a 2 cos α   a 2 cos α a 2 sin α  (c) 2x2 + 2y2 + 2x – 6y – 1 = 0 (b)  (a)  , , 2 2  (d) 2x + 2y + 2x + 6y + 1 = 0 p p  p p    2 2 49. The two circles x + y + 2ax + c = 0 and  a cos α a sin α  (c)  (d) None of these , x2 + y2 + 2by + c = 0 touch each other if p 2   p2 1 1 1 1 1 1 (b) 2 − 2 = 2 (a) 2 − 2 = 59. The points of contact of the two tangents drawn from a b c a b c the point (4, – 2) to the circle x2 + y 2 = 10, are 1 1 1 1 1 1 (c) 2 + 2 = (d) 2 + 2 = 2 (a) (1, 3), (3, 1) (b) (1, – 3), (3, – 1) a b c a b c (c) (1, – 3), (3, 1) (d) None of these 50. The value of k for which the line 4x + 3y + k = 0 touches 60. A line meets the coordinate axes in A and B. A circle the circle 2x2 + 2y 2 = 5x is is circumscribed about the ∆AOB. If m, n are the dis5 −5 (b) 4 4 45 −45 (c) (d) 4 4 51. The equation of tangent to the circle x2 + y 2 = 25, which is inclined at an angle of 30º to the axis of x, is (a)

(a)

3 x + y + 10 = 0

(c) x ­–

3 y + 10 = 0

(b)

3 x – y + 10 = 0

(d) x –­

3 y – 10 = 0

52. The equation of the tangent to the circle x2 + y2 – 2x – 4y – 4 = 0, which is parallel to the line 3x – 4y – 1 = 0, is (a) 3x – 4y + 20 = 0 (c) 3x – 4y – 10 = 0

(b) 3x + 4y + 20 = 0 (d) 3x + 4y – 10 = 0

53. The equation of the tangent to the circle x2 + y2 – 4x – 6y – 12 = 0, which is perpendicular to the line 4x + 3y = 7, is (a) 3x – 4y + 31 = 0 (c) 3x + 4y + 31 = 0

(b) 3x – 4y – 19 = 0 (d) 3x + 4y – 19 = 0

54. The equation of the circle, touching the axis of x at the origin and the line 3y = 4x + 24, is (a) x2 + y2 + 24y = 0 (c) x2 + y2 – 24y = 0

(b) x2 + y2 – 6y = 0 (d) x2 + y2 + 6y = 0

tances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is (a) m (m + n) (c) n (m + n)

(b) m + n (d) None of these

61. If the chord of contact of tangents from a point on the circle x2 + y 2 = a2 to the circle x2 + y 2 = b 2 touches the circle x2 + y 2 = c 2, then a, b, c are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

62. The equation of the circle which touches both the axes and the straight line 4x + 3y = 6 in the first quadrant and lies below it is (a) 4x2 + 4y2 – 4x – 4y + 1 = 0 (b) x2 + y2 – 6x – 6y + 9 = 0 (c) x2 + y2 – 6x – y + 9 = 0 (d) 4 (x2 + y2 – x – 6y) + 1 = 0 63. If the polar of P with respect to the circle x2 + y 2 = a2 touches the circle (x – f )2 + (y – g)2 = b 2, then its locus is given by the equation (a) ( f x + gy – a2)2 = a2 (x2 + y2) (b) ( f x + gy – a2)2 = b2 (x2 + y2) (c) ( f x – g y – a2)2 = a2 (x2 + y2) (d) None of these

2

2

3   3   (a)  x − 3 −  =9  + y −  2  2 2

2

3   3   + y − (b)  x −  =9   2  2 (c) x2 + (y – 3)2 = 9 (d) (x – 3)2 + y2 = 9 65. If the equation of a circle is (4a – 3) x2 + ay2 + 6x – 2y + 2 = 0, then its centre is (a) (3, –1) (c) (– 3, 1)

(b) (3, 1) (d) None of these

66. If the equation of a circle is 3x2 + 3y2 + kxy + 9x + (k – 6) y + 3 = 0, then its radius is 3 17 (b) (a) 2 2 2 (c) (d) None of these 3 67. The equation x2 + y 2 – 8x + 6y + 25 = 0 represents (a) a circle (c) a point

(b) a pair of straight lines (d) None of these

68. If from any point on the circle x2 + y 2 = a2, tangents are drawn to the circle x2 + y 2 = b 2 (a > b), then the angle between the tangents is b a (b) 2 sin −1 a b b (c) 2 sin −1 (d) None of these a 69. The equation of the circle passing through the point (2, – 1) and having two diameters along the pair of lines (a) sin −1

(a) 1

(b) – 1

1 −1 (d) 2 2 73. If the distances from the origin of the centres of the three circles x2 + y 2 + 2αix = a2 (i = 1, 2, 3) are in G.P., then the lengths of the tangents drawn to them from any point on the circles x2 + y 2 = a2 are in (c)

(a) A.P. (c) H.P.

(b) G.P. (d) None of these

74. The coordinates of the mid point of the chord intercepted by the circles x2 + y 2 = a2 from the line ax + by = c are ac   bc , (a)  2  a + b 2 a 2 + b 2  bc   ac , (b)  2  a + b 2 a 2 + b 2  −bc   − ac (c)  2 ,  a + b 2 a 2 + b 2  (d) None of these 75. If the tangents PQ and PR are drawn to the circle x2 + y 2 = a2 from the point P(x1, y1), then the equation of the circumcircle of ∆PQR is (a) x2 + y2 – xx1 – yy1 = 0 (b) x2 + y2 + xx1 + yy1 = 0 (c) x2 + y2 – 2xx1 – 2yy1 = 0 (d) None of these 76. An equilateral triangle has two vertices (–2, 0) and (2, 0), and its third vertex lies below the x-axis. The equation of the circumcircle of the triangle is (a)

3 (x2 + y2) – 4y + 4 3 = 0

(b)

3 (x2 + y2) – 4y – 4 3 = 0

2x2 + 6y2 – x + y – 7xy – 1 = 0 is

(c)

3 (x2 + y2) + 4y + 4 3 = 0

(a) x + y + 10x + 6y – 19 = 0 (b) x2 + y2 + 10x – 6y + 19 = 0 (c) x2 + y2 + 10x + 6y + 19 = 0 (d) None of these

(d)

3 (x2 + y2) + 4y – 4 3 = 0

2

2

70. The slope of the tangent at the point (h, h) of the circles x2 + y 2 = a2 is (a) 0 (c) –  1

(b) 1 (d) depends on h

71. The pole of the chord of the circle x2 + y 2 = 16 which is bisected at the point (– 2, 3), with respect to the circle is  −32 48  , (a)   13 13 

 32 48  (b)  ,   13 13 

 −32 −48  (c)  ,  13 13 

(d) None of these

77. If the circle x2 + y 2 + 2gx + 2f y + c = 0 bisects the circumference of the circle x2 + y 2 + 2g′x + 2f ′y + c′ = 0, then, (a) 2g′ (g – g′ ) + 2f ′ ( f – f ′ ) = c – c′ (b) 2g (g – g′ ) + 2f  ( f – f ′) = c – c′ (c) g′ (g – g′) + f ′ ( f – f ′) = c – c′ (d) None of these 78. If the centroid of an equilateral triangle is (3, 2) and its one vertex is (2, –3) then the equation of its circumcircle is (a) x2 + y2 – 6x – 4y + 13 = 0 (b) x2 + y2 – 6x – 4y – 13 = 0 (c) x2 + y2 – 6x – 4y – 39 = 0 (d) None of these

399

72. If a circle passes through the points where the lines 3kx– 2y – 1 = 0 and 4x –3y + 2 = 0 meet the coordinate axes then k =

Circles

64. To which of the following circles, the line y – x + 3 = 0 3 3   , is normal at the point  3 +  ?  2 2

400

79. The equation of chord of contact of the tangents drawn from the point (1, 2) to the circle x2 + y 2 + x – 2y – 5 = 0 is

Objective Mathematics

(a) 3x + 2y – 13 = 0 (b) 3x + 2y + 13 = 0 (c) no chord of contact exists (d) None of these 80. If the chord of contact of the tangents drawn from the point (2, 2) to the circle x2 + y 2 = 16 meets the coordinate axes in A and B, then the circumcentre of ∆OAB is 8 8  16 8  (b)  ,  (a)  ,  3 3  3 3  8 16  (c)  ,  3 3 

(d) None of these

81. If the lines 12x + 5y + 16 = 0 and 12x + 5y – 10 = 0 are tangents to the same circle, then the radius of this circle is (a) 1 (c) 4

(b) 2 (d) none of these

82. The equation of the circle on the join of the points A and B as diameter, where the ordinates of A, B are the roots of the equation y 2 – 7y + 12 = 0 and the abscissae are the roots of the equation x2 – 3x + 2 = 0, is (a) x2 + y2 – 3x + 7y (b) x2 + y2 + 3x – 7y (c) x2 + y2 + 3x + 7y (d) x2 + y2 – 3x – 7y

+ 14 = 0 + 14 = 0 + 14 = 0 + 14 = 0

83. If the equation of the common tangent at the point (1, – 1) to the two circles, each of radius 13, is 12x + 5y – 7 = 0, then the centres of the two circles are (a) (13, 4), (– 11, 6) (b) (13, 4), (– 11, – 6) (c) (13, –­4), (– 11, – 6) (d) (– 13, 4), (– 11, – 6)

87. A point moves so that the sum of the squares of its distances from the four sides of a square is constant. The locus of the point is (a) a circle (c) a hyperbola

88. The coordinates of a point P on the circle x2 + y 2 – 4x – 6y + 9 = 0 such that ∠POX is minimum, where O is the origin and OX is the x-axis, are

(a) (3, 4), (9, 12) (c) (– 3, 4), (9, 12)

(b) (3, 2), (9, 12) (d) (3, 4), (9, – 12)

(a) x2 + y2 – 4 3 y – 4 = 0 (b) x2 + y2 + 4 3 y – 4 = 0 (c) x2 + y2 – 4 3 x – 4 = 0 (d) x2 + y2 + 4 3 x – 4 = 0

 −36 15  (b)  ,  13 13 

 14 12  (c)  ,   27 27 

(d) None of these

x2 + y2 – 8y – 4 = 0 (a) touch externally (c) intersect

(b) touch internally (d) do not touch

90. The locus of the centre of a circle which passes through the point (0, 0) and cuts off a length 2b from the line x = c, is (a) y2 + 2cx = b2 + c2 (c) y2 + 2cy = b2 + c2

(b) x2 + cx = b2 + c2 (d) None of these

91. The equation of the circle inscribed in the triangle, formed by the coordinate axes and the line 12x + 5y = 60, is given by (a) x2 + y2 + 4x + 4y + 4 = 0 (b) x2 + y2 – 4x – 4y + 4 = 0 (c) x2 + y2 – 4x – 4y – 4 = 0 (d) None of these 92. If the equation of the in circle of an equilateral triangle is x2 + y 2 + 4x ­– 6y + 4 = 0, then the equation of the circumcircle of the triangle is (a) x2 + y2 + 4x ­+ 6y – 23 = 0 (b) x2 + y2 + 4x ­– 6y – 23 = 0 (c) x2 + y2 – 4x ­– 6y – 23 = 0 (d) None of these 93. If the circle x2 + y 2 – 4x ­+ 6y + k = 0 does not touch or intersect the axes and the point (2, 2) lies inside the circle, then

85. A variable chord is drawn through the origin to the circle x2 + y 2 – 4y = 0. Locus of the centre of the circle drawn on this chord as diameter is 94. (a) x2 + y2 + 2y = 0 (b) x2 + y2 – 2y = 0 (c) x2 + y2 + 2x = 0 (d) x2 + y2 – 2x = 0 86. If the coordinates of two consecutive vertices of a regular hexagon which lies completely above the x-axis, are (–2, 0) and (2, 0), then the equation of the circle, circumscribing the hexagon, is

 36 15  (a)  ,   13 13 

89. The two circles x2 + y 2 – 2x – 4y = 0 and

84. The coordinates of two points on the circle x2 + y2 – 12x – 16y + 75 = 0, one nearest to the origin and the other farthest from it, are

(b) an ellipse (d) None of these

(a) 4 < k < 9 (c) 9 < k < 12

(b) 4 < k < 12 (d) None of these

If the tangents PA and PB are drawn from the point P (– 1, 2) to the circle x2 + y 2 + x ­– 2y – 3 = 0 and C is the centre of the circle, then the area of the quadrilateral PACB is (a) 4 (c) does not exist

(b) 16 (d) None of these

95. The coordinates of the poles of the common chord of the circles x2 + y 2 = 12 and x2 + y 2 – 5x + 2y – 2 = 0 with respect to the circle x2 + y 2 = 12 are  −12  (a)  6,   5 

12   (b)  −6,   5

12 (c)  6,   5

(d) None of these

π 3 π (d) 4

π 2 π (c) 6

(b)

(a)

(a) 4 (c) 2

97. If a > 2b > 0 then the positive value of m for which y = mx – b 1 + m 2 is a common tangent to x2 + y 2 = b 2 and (x – a)2 + y 2 = b 2 is (a)

2b



(b)

a − 4b 2b 2

2

a − 4b 2b b (c) (d) a − 2b a − 2b 98. The equation of the circle which cuts orthogonally the circle x2 + y 2 – 6x + 4y – 3 = 0, passes through (3, 0) and touches y–axis is 2

2

(a) x2 + y2 + 6x + 6y + 9 = 0 (b) x2 + y2 – 6x + 6y + 9 = 0 (c) x2 + y2 + 6x – 6y + 9 = 0 (d) x2 + y2 – 6x – 6y + 9 = 0 99. The radius of the circle which passes through the origin and cuts orthogonally each of the circles x2 + y2 – 4x + 4 = 0 and x2 + y2 – 6x + 8y – 2 = 0 is (a) 2 (b) 2 (c) 3 (d) None of these 100. The equation of the circle which cuts the circles x2 + y2 – 4x – 6y + 11 = 0 and x2 + y2 – 10x – 4y + 21 = 0 orthogonally and has 2x + 3y = 7 as diameter, is (a) x2 + y2 – 4x + 2y + 3 = 0 (b) x2 + y2 + 4x + 2y + 3 = 0 (c) x2 + y2 – 4x – 2y + 3 = 0 (d) None of these 101. The equation of the circle which intersects the circles x2 + y 2 – 6y + 1 = 0, and x2 + y 2 – 4y + 1 = 0 orthogonally and touches the line 3x + 4y + 5 = 0 is (a) x2 + y2 = 1 (b) 4x2 + 4y2 – 15x – 4 = 0 2 2 (c) 4x + 4y – 15x + 4 = 0 (d) None of these 102. If P, Q is a pair of conjugate points with respect to a circle S, then the circle on PQ as diameter (a) touches the circle S

π 4 (c) cuts the circle S orthogonally (d) None of these (b) cuts the circle S at an angle

103. If the equations to two circles whose radii are a, a′ reS S' =0 spectively be S = 0, S′ = 0, then the circles ± a a' (a) touch each other (b) cut each other orthogonally (c) do not cut each other (d) None of these

(b) 3 (d) None of these

105. The point of contact and the common tangent of the two circles x2 + y 2 – 2x – 4y – 20 = 0 and x2 + y2 + 6x + 2y – 90 = 0 are (a) (5, 5), 4x + 3y + 35 = 0 (b) (5, 5) 4x – 3y + 35 = 0 (c) (– 5, 5), 4x + 3y – 35 = 0 (d) (5, 5) 4x + 3y – 35 = 0 106. The equation of the circle which passes through the intersection of the circles x2 + y 2 = 4 and x2 + y2 – 2x –4y + 4 = 0 and has radius 2 2 is (a) x2 + y2 – 4x – 8y + 12 = 0 (b) 5x2 + 5y2 + 4x + 8y + 36 = 0 (c) 5x2 + 5y2 + 4x + 8y – 36 = 0 (d) None of these 107. The equation of the circle through the points of intersection of the circles x2 + y 2 + 2x + 3y – 7 = 0 and x2 + y2 + 3x – 2y – 1 = 0 and the point (1, 2) is (a) x2 + y2 + 4x + 7y + 5 = 0 (b) x2 + y2 + 4x + 7y – 5 = 0 (c) x2 + y2 + 4x – 7y + 5 = 0 (d) None of these 108. The equation of the circle through the points of intersection of the circles x2 + y 2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 4y – 12 = 0 and cutting the circle x2 + y2 – 2x + 3 = 0 orthogonally is (a) x2 + y2 + 9x – 7y – 12 = 0 (b) x2 + y2 + 9x + 7y – 12 = 0 (c) x2 + y2 + 9x + 7y + 12 = 0 (d) None of these 109. If from any point P, tangents PT, PT′ are drawn to two given cirles with centres A and B respectively; and if PN is the perpendicular from P on their radical axis, then PT2 – PT′ 2 = (a) PN.AB (c) 4PN.AB

(b) 2PN.AB (d) None of these

110. If A, B, C be the centres and r1, r 2 , r 3 the radii of three coaxal circles, then r12 .BC + r22 .CA + r23 .AB = (a) BC.CA.AB (c) 2BC.CA.AB

(b) – BC.CA.AB (d) None of these

111. The limiting points of the coaxal system determined by the circles x2 + y 2 – 2x – 6y + 9 = 0 and x2 + y2 + 6x – 2y + 1 = 0 are 3 −14  (a) (– 1, 2),  ,  5 5 

3 14 (b) (– 1, 2),  ,  5 5 

 −3 14  (c) (– 1, 2),  ,   5 5

(d) None of these

401

104. The length of the of tangent from the radical centre of the three circles x2 + y 2 + 4x + 7 = 0, 2x2 + 2y 2 + 3x + 5x + 9 = 0 and x2 + y 2 + y = 0 to the second circle is

Circles

96. The angle of intersection of the circles x2 + y 2 = 4 and x2 + y 2 = 2x + 2y is

402

112. The equation of the circle which passes through the origin and belongs to the coaxal system whose limiting points are (1, 2) and (4, 3), is

Objective Mathematics

(a) 2x2 + 2y2 – x – 7y = 0 (b) 2x2 + 2y2 + x – 7y = 0 (c) 2x2 + 2y2 + x + 7y = 0 (d) None of these 113. Two distinct chords drawn from the point ( p, q) on the circle x2 + y 2 = px + q y, where pq ≠ 0, are bisected by the x-axis. Then (b) p2 = 8q2 (d) p2 > 8q2

(a) | p | = | q | (c) p2 < 8q2

114. The area of the triangle formed by joining the origin to the points of intersection of the line x 5 + 2 y = 3 5 and circle x2 + y 2 = 10. is (a) 3 3

(b) 4 3

(c) 5 3

(d) 6 3

115. If the line (y – 2) = m (x + 1) intersects the circle x2 + y2 + 2x – 4y – 3 = 0 at two real distinct points, then the number of possible values of m is (a) 2 (c) any real value of m

(b) 1 (d) None of these

116. The number of points on the circle x2 + y 2 – 4x – 10y + 13 = 0 which are at a distance 1 from the point (– 3, 2) is (a) 1 (c) 3

(b) 2 (d) None of these

117. The centre of a circle of radius 4 5 lies on the line y = x and satisfies the inequality 3x + 6y > 10. If the line x + 2y = 3 is a tangent to the circle, then the equation of the circle is 2

2

23   23   (a)  x −  +  y −  = 80    3 3 17   17   (b)  x +  +  y +  = 80  3  3 2

2

2

2

23   23   (c)  x +  +  y −   3  3

= 80

(d) None of these 118. The number of common tangents to the circles x2 + y 2 = 4 and x2 + y 2 – 8x + 12 = 0 is (a) 1 (c) 3

(b) 2 (d) 4

119. If (α, β) is a point on the chord PQ of the circle x2 + y 2 = 19, where the coordinates of P and Q are (3, –4) and (4, 3) respectively, then (a) α ∈ [3, 4], β ∈ [– 4, 3] (b) α ∈ [– 4, 3], β ∈ [3, 4] (c) α ∈ [3, 3], β ∈ [– 4, 4] (d) None of these

120. The equation of the circle which has a tangent 2x – y – 1 = 0 at (3, 5) on it and with the centre on x + y = 5, is (a) x2 + y2 + 6x – 16y + 28 = 0 (b) x2 + y2 – 6x + 16y – 28 = 0 (c) x2 + y2 + 6x + 6y – 28 = 0 (d) x2 + y2 – 6x – 6y – 28 = 0 121. From any point on the circle x2 + y 2 = a 2 tangents are drawn to the circle x2 + y 2 = a2 sin 2 α. The angle between them is (a) α/2 (c) 2α

(b) α (d) None of these

122. If the point (2, k) lies outside the circles x2 + y2 + x – 2y – 14 = 0 and x2 + y2 = 13, then (a) k ∈ (– 3, – 2) ∪ (3, 4) (b) k∈ (–3, 4) (c) k∈ (– ∞, – 3) ∪ (4, ∞) (d) k∈ (– ∞, – 2) ∪ (3, ∞) 123. If the point (k + 1, k) lies inside the region bounded by the curve and y-axis, then k belongs to the interval x=

25 − y 2 .

(a) (– 1, 3) (c) (– ∞, – 4) ∪ (3, ∞)

(b) (– 4, 3) (d) None of these

124. If the two circles x2 + y 2 = 4 and x2 + y 2 – 24x – 10y + a2 = 0, a ∈ Ι, have exactly two common tangents, then the number of possible values of a is (a) 11 (c) 0

(b) 13 (d) 2

125. Polar of origin (0, 0) w.r.t. the circle x2 + y2 + 2λx + 2µy + c = 0 touches the circle x2 + y2 = r2 if (a) c = r ( λ2 + µ2) (c) c2 = r2 ( λ2 + µ2)

(b) r = c ( λ2 + µ2) (d) r2 = c2 ( λ2 + µ2)

126. The number of common tangents to the circles x2 + y2 – 6x – 2y + 9 = 0 and x2 + y2 – 14x – 8y + 61 = 0 is (a) 2 (c) 1

(b) 3 (d) 4

127. If the line 3x + ay – 20 = 0 cuts the circle x2 + y 2 = 25 at real distinct or coincident points, then a belongs to the interval (a) [− 7 , 7 ] (b) (− 7 , 7 ) (c) (− ∞ − 7 ] ∪ [ 7 , ∞) (d) None of these 128. If the circles x2 + y 2 – 9 = 0 and x2 + y 2 + 2ax + 2y + 1 = 0 touch each other then a is (a) – 4/3 (c) 1

(b) 0 (d) 4/3

(a) x2 + y2 + 1 = x (c) x2 + y2 + 1 = – x

(b) x2 + y2 + 1 = y (d) x2 + y2 – 1 = – x

(a) (– 2, 0), (0, 2) (c) (2, 0), (0, – 2)

130. The equation of any tangent to the circle (a) y = m (x + 3) + 6 1 + m 2 + 4

(a) 4 ( 2 + 1)

(b) y = m (x + 3) – 6 1 + m 2 + 4

(b) 4 ( 2 − 1) (d) None of these

(c) 2 ( 2 − 1) 139. With respect to the circle x2 + y 2 + 6x – 8y – 10 = 0,

(c) y = m (x – 3) + 6 1 + m 2 − 4 (d) y = m (x – 3) – 6 1 + m 2 − 4 131. If the tangents are drawn from any point on the line x + y = 3 to the circle x2 + y 2 = 9, then the chord of contact passes through the point (b) (3, 3) (d) None of these

132. A foot of the normal from the point (3, – 2) to a circle is (4, – 1). If the line 3x – y = 2 is a diameter of the circle, then the equation of the circle is

(a) The chord of contact of tangents from (2, 1) is 5x – 3y – 8 = 0 (b) the pole of the line 5x – 3y – 8 = 0 is (2, 1) (c) the polar of the point (2, 1) is 5x – 3y – 8 = 0 (d) all of these 140. The locus of the centres of circles passing through the origin and cutting the circle x2 + y 2 + 6x – 4y + 2 = 0 orthogonally is (a) 2x – 3y + 1 = 0 (c) 3x – 2y + 1 = 0

(a) x2 + y2 + 3x – 13y – 16 = 0 (b) x2 + y2 + 3x + 13y + 16 = 0 (c) x2 + y2 + 3x + 13y – 16 = 0 (d) None of these 133. The common chord of the circle x2 + y2 + 8x + 4y – 5 = 0 and a circle passing through the origin and touching the line y = x, passes through the fixed point 5 5 (a)  ,   12 12 

5 −5 (b)  ,   12 12 

−5 5 (c)  ,   12 12 

(d) None of these

(b) 2x + 3y +1 = 0 (d) None of these

141. For the circles S1 ≡ x2 + y2 – 4x – 6y – 12 = 0 and S2 ≡ x2 + y2 + 6x + 4y – 12 = 0 and the line L ≡ x + y = 0 (a) L is the common tangent of S1 and S2 (b) L is the common chord of S1 and S2 (c) L is radical axis of S1 and S2 (d) L is perpendicular to the line joining the centres of S1 and S2 142. Locus of the centre of a circle of radius 4 which touches the circle x2 + y 2 – 4x + 2y – 4 = 0 externally is

134. The equation of a circle of radius 2 touching the circles x2 + y2 – 4 | x | = 0 is (a) x 2 + y 2 + 2 3 y + 2 = 0

(a) x2 + y2 – 4x + 2y – 44 = 0 (b) x2 + y2 + 4x + 2y – 44 = 0 (c) x2 + y2 – 4x – 2y – 44 = 0 (d) None of these 143. A variable circle passes through the point P (1, 2) and touches the x-axis. The locus of the other end of the diameter through P is

(b) x 2 + y 2 + 4 3 y + 8 = 0 (c) x 2 + y 2 − 4 3 y + 8 = 0 (d) None of these 135. If the equation of a circle is x + y = 4 and x + y − 4 < 0 , then the line xx1 + yy1 = 4 is the 2

(b) (2, 0), (0, 2) (d) None of these

138. If the equations of four circles are (x ± 4)2 + ( y ± 4)2 = 42, then the radius of the smallest circle touching all the four circles is

x2 + y2 + 6x – 8y – 11 = 0 is

(a) (3, 5) (c) (5, 3)

2

2 1

2 1

(a) tangent to the circle at the point (x1, y1) (b) chord of contact of tangents drawn from the point (x1, y1) to the circle (c) polar of the point (x1, y1) w.r.t. the circle (d) all of these 136. The coordinates of a point on the line y = 2 from which the tangents drawn to the circle x2 + y 2 = 25 are perpendicular, are (a) ( 46 , 2)

(b) (− 46 , 2)

(c) ( 37 , 2)

(d) (− 37 , 2)

(a) (x – 1)2 = 8y (b) (x + 1)2 = 8y (c) ( y – 1)2 = 8x (d) None of these 144. The point A (1, 5) is joined to any point P of the circle x2 + y 2 = 4. The locus of the middle point of AP as P moves on the circle, is (a) 2 (x2 + y2) – 2x – 10y – 11 = 0 (b) 2 (x2 + y2) –2x – 10y + 11 = 0 (c) 2 (x2 + y2) +2x + 10y – 11 = 0 (d) None of these 145. The chords of contact of the pair of tangents drawn from each point on the line 2x + y = 4 to the circle x2 + y 2 = 1 pass through the point

403

137. If the tangents are drawn from the point (2, –2) to the circle x2 + y 2 = 4, then the coordinates of the points of contact are

Circles

129. A circle C and the circle x2 + y 2 = 1 are orthogonal and have radical axis parallel to y-axis, then C can be

404

Objective Mathematics

 −1 1  (a)  ,   2 4

 1 −1  (b)  ,  2 4 

 −1 −1  (c)  ,   2 4

1 1 (d)  ,  2 4

146. Let C be any circle with centre (0, circle C, there can be

(a) π 2 (c) π

2 ). Then, on the

147. If the tangent at any point P on the circle x2 + y 2 = 2 cuts the axes in L and M, then the locus of the middle point of LM is (b) x– 2 + y– 2 = 1 (d) None of these

148. If two distinct chords, drawn from the point ( p, q) on the circle x2 + y 2 = px + qy (where pq ≠ 0) are bisected by the x-axis, then (a) p2 = q2 (c) p2 < 8q2

(b) p2 = 8q2 (d) p2 > 8q2

149. Let L1 be a straight line passing through the origin and L2 be the straight line x + y = 1. If the interecpts made by the circle x2 + y 2 – x + 3y = 0 on L1 and L2 are equal, then which of the following equations can represent L1? (a) x + y = 0 (c) 7y + x = 0

(b) x – y = 0 (d) x – 7y = 0

150. The locus of the centres of the circles which cut the circles x2 + y 2 + 4x – 6y + 9 = 0 and x2 + y 2 – 4x + 6y + 4 = 0 orthogonally is (a) 8x – 12y + 5 = 0 (c) 12x – 8y + 5 = 0

(b) 8x + 12y – 5 = 0 (d) None of these

151. The number of common tangents to the circles x2 + y 2 = 4 and x2 + y 2 – 6x – 8y – 24 = 0 is (a) 0 (c) 3

(b) 1 (d) 4



3 π (d) 6

155. If the circles x2 + y 2 + 2x + 2ky + 6 = 0 and

(a) at the most one rational point (b) at the most two rational points (c) at the most three rational points (d) None of these

(a) x– 2 + y– 2 = 2 (c) x– 2 + y– 2 = 4

4

(b) π

x2 + y2 + 2ky + k = 0 intersect orthogonally, then k is −3 −3 (b) – 2 or 2 2 3 3 (c) 2 or (d) – 2 or 2 2 156. A point moves such that the sum of the squares of its distances from the sides of a square of side unity is equal to 9. The locus of the point is a circle such that (a) 2 or

(a) Centre of the circle coincides with that of square. 1 1 (b) Centre of the circle is  ,  2 2 (c) radius of the circle is 2 (d) all the above are true 157. The angle between a pair of tangents drawn from a point P to the circle x2 + y 2 + 4x – 6y + 9sin 2α + 13cos2 α = 0 is 2α. The equation of the locus of the point P is (a) x2 + y2 + 4x – 6y + 4 = 0 (b) x2 + y2 + 4x – 6y – 9 = 0 (c) x2 + y2 + 4x – 6y – 4 = 0 (d) x2 + y2 + 4x – 6y + 9 = 0 158. The intercept on the line y = x by the circle x2 + y 2 – 2x = 0 is AB. Equation of the circle with AB as a diameter is (a) x2 + y2 + x + y = 0 (c) x2 + y2 + x – y = 0

(b) x2 + y2 – x – y = 0 (d) none of these

159. The locus of the centres of the circles which touch the two circles x2 + y 2 = a2 and x2 + y 2 = 4ax externally is (a) 12x2 – 4y2 – 24ax + 9a2 = 0 (b) 12x2 + 4y2 – 24ax + 9a2 = 0 (c) 12x2 – 4y2 + 24ax + 9a2 = 0 (d) None of these

152. Let A0 A1 A 2 A 3 A4 A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths 160. The locus of the centre of a circle which touches exof the line segments A0 A1, A0 A 2 and A0 A4 is ternally the circle x2 + y 2 – 6x – 6y + 14 = 0 and also touches the y-axis is given by the equation. 3 (b) 3 3 (a) (a) x2 – 6x – 10y + 14 = 0 4 (b) x2 – 10x – 6y + 14 = 0 3 3 (c) 3 (d) (c) y2 – 6x – 10y + 14 = 0 2 (d) y2 – 10x – 6y + 14 = 0 153. A triangle has two of its sides along the axes. If the third side touches the circle x2 + y 2 – 2ax – 2ay + a2 161. The coordinates of the point at which the circles = 0, then the equation of the locus of the circumcenter x2 + y2 – 4x – 2y – 4 = 0 and x2 + y2 – 12x – 8y –36 = 0 touch of the triangle is each other, are (a) (3, –2) (a) 2a (x + y) = 2xy + a2 (b) 2a (x – y) = 2xy + a2 (b) (– 2, 3) 2 (c) 2a (x + y) = 2xy – a (c) (3, 2) (d) None of these (b) None of these 2 2 154. The ∆PQR is inscribed in the circle x + y = 25. If Q and R have coordinates (3, 4) and (–4, 3) respectively, 162. The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle x2 + y 2 = 9 is then ∠QPR is equal to

1 1 (c)  ,  2 2

1  (d)  , −21/ 2  2 

170. The equations of the tangents drawn from the origin to the circle x2 + y 2 – 2rx ­– 2hy + h2 = 0, are (a) x = 0 (b) y = 0 (c) (h2 – r2)x – 2rhy = 0 (d) (h2 – r2)x + 2rhy = 0

163. Three circles touch one another externally. If the tangents at their points of contact meet at a point whose 171. The area of the triangle formed by the tangents from distance from a point of contact is 4, then the ratio of the point (4, 3) to the circle x2 + y 2 = 9 and the line the product of the radii to the sum of the radii of the joining their point of contact is circles, is (a) 12 (b) 6 (a) 4 : 1 (b) 8 : 1 (c) 4 (d) None of these (c) 16 : 1 (d) None of these 172. A variable circle passes through the fixed point A( p, q) 164. The equations of three circles are given: and touches x-axis. The locus of the other end of the diameter through A is x2 + y2 = 1, x2 + y2 – 8x + 15 = 0, x2 + y2 + 10y + 24 = 0. The coordinates of the point such that the tangents drawn from it to three circles are equal in length, are 5 (a)  2,   2

 −5  (b)  2,   2 

5 (c)  −2,   2

−5 (d)  −2,   2 

165. If a circle passes through the points of intersection of the coordinate axes with the lines λx – y + 1 = 0 and x – 2y + 3 = 0, then the value of λ is (a) 2 (c) – 1

(b) 1 (d) – 2

166. The number of circles which pass through the origin and cut off equal chords of 2 units from the straight lines y = x and y = –x, is (a) 1 (c) 4

(b) 2 (d) None of these

167. If the two circles (x – 1)2 + ( y – 3) = r 2 and x2 + y2 – 8x + 2y + 8 = 0 intersect in two distinct points, then (a) 2 < r < 8 (c) r = 2

(b) r < 2 (d) r > 2

168. If a circle passes through the point (a, b) and cuts the circle x2 + y 2 = 4 orthogonally, then the locus of its centre is (a) 2ax – 2by + (a2 + b2 + 4) = 0 (b) 2ax + 2by – (a2 + b2 + 4) = 0 (c) 2ax + 2by + (a2 + b2 + 4) = 0 (d) 2ax – 2ab – (a2 + b2 + 4) = 0 169. If a circle passes through the point (a, b) and cuts the circle x2 + y 2 = k2 orthogonally, then the equation of the locus of its centre is (a) 2ax + 2by – (a2 + b2 + k2) = 0 (b) 2ax + 2by – (a2 – b2 + k2) = 0 (c) x2 + y2 – 3ax ­– 4by + (a2 + b2 – k2) = 0 (d) x2 + y2 – 2ax ­– 3by + (a2 – b2 – k2) = 0

(a) ( y – p)2 = 4qx (c) (x – p)2 = 4qy

(b) (x – q)2 = 4py (d) ( y – q)2 = 4px

173. The equation of the circle whose diameters lies on 2x + 3y = 3 and 16x – y = 4 and which passes through (4, 6) is (a) x2 + y2 = 40 (b) 5(x2 + y2) – 4x – 8y = 200 (c) x2 + y2 + 4x + 8y = 200 (d) 5 (x2 + y2) – 3x – 8y = 200 174. The number of tangents to the circle x2 + y2 – 8x – 6y + 9 = 0 which pass through the point (3, –2) is (a) 2 (c) 0

(b) 1 (d) None of these

175. A stick of length l rests against the floor and a wall of a room. If the stick begins to slide on the floor, then the locus of its middle point is (a) an ellipse (c) a circle

(b) a parabola (d) a straight line.

176. If the radical axis of the circles x2 + y 2 + 2gx + 2fy + c = 0 and 2x2 + 2y 2 + 3x + 8y + 2c = 0 touches the circle x2 + y 2 + 2x + 2y + 1 = 0, then (a) g = 3/4 or f = 2 (c) g = 3/4 and f ≠ 2

405

1 3 (b)  ,  2 2

Circles

3 1 (a)  ,  2 2

(b) g ≠ 3/4 and f = 2 (d) None of these

177. Two circles x2 + y 2 = 6 and x2 + y 2 – 6x + 8 = 0 are given. Then the equation of the circle through their points of intersection and the point (1, 1) is (a) x2 + y2 – 4y + 2 = 0 (b) x2 + y2 – 6x + 4 = 0 (c) x2 + y2 – 3x + 1 = 0 (d) None of these 178. If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10π, then the equation of the circle is (a) x2 + y2 + 2x + 2y – 23 = 0 (b) x2 + y2 – 2x – 2y – 23 = 0 (c) x2 + y2 – 2x + 2y – 23 = 0 (d) x2 + y2 + 2x – 2y – 23 = 0

406

179. The tangent to the circle x2 + y 2 = 5 at the point (1, –2) also touches the circle x2 + y 2 – 8x + 6y + 20 = 0. Then its point of contact is

Objective Mathematics

(a) (3, – 1) (c) (– 1, – 1)

(b) (– 3, 0) (d) (– 2, 1)

189. If the tangent at the point P on the circle x2 + y 2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point Q on the y-axis then the length of PQ is (a) 4 (c) 5

(b) 2 5 (d) 3 5

180. The extremities of the diameter of a circle have coordi- 190. A circle passes through (0, 0), (a, nates (– 4, 3) and (12, – 1). The length of the intercept coordinates of its centre are which the circle makes on the y-axis is b a a b (a) 2 13 (b) 4 13 (a)  ,  (b)  , 2 2 2 2 (c) 2562 (d) None of these (c) (b, a) (d) (a, b)

181. The distance from the centre of the circle x2 + y 2 = 2x to straight line passing though the points of intersection of the two circles x2 + y 2 + 5x – 8y + 1 = 0 and x2 + y2 – 3x – 7y – 25 = 0 is 1 (b) 2 (a) 3 (c) 3 (d) 1 182. The equation of the circle passing through (1, 0) and (0, 1) and having smallest possible radius is (a) x2 + y2 – x – y = 0 (c) x2 + y2 – 2x – y = 0

(b) x2 + y2 + x + y = 0 (d) x2 + y2 – x – 2y = 0

183. The equation of the tangents drawn from the origin to the circle x2 + y 2 – 2px – 2qy + q2 = 0 are prependiculat if (a) p = q/2 (c) p2 = q2

(b) q = p/2 (d) p2 + q2 = 1

184. The radical axis of the circles, belonging to the coaxal system of circles whose limiting points are (1, 3) and (2, 6), is (a) x – 3y – 15 = 0 (c) x – 3y + 15 = 0

(b) x + 3y – 15 = 0 (d) 2x + 3y – 15 = 0

185. The area of a circle centred at (1, 2) and passing through (4, 6) is (a) 30 π sq. units (c) 15 π sq. units

(b) 5 π sq. units (d) 25 π sq. units

186. The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangent to the same circle. The radius of this circle is 3 2 7 (c) 10 (a)

3 4 4 (d) 5

(b)

0) and (0, b). The   

191. Given the circles x2 + y 2 – 4x – 5 = 0 and x2 + y 2 + 6x – 2y + 6 = 0. Let P be a point (α, β) such that the tangents from P to both the circles are equal. Then (a) 10α + 2β + 11 = 0 (c) 2α ­– 10β + 11 = 0

(b) 10α – 2β + 11 = 0 (d) 2α + 10β + 11 = 0.

192. The equation of the chord of the circle x2 + y 2 – 4x = 0, whose mid point is (1, 0), is (a) x = 1 (c) y = 1

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

193. The greatest distance of the point P (10, 7) from the circle x2 + y 2 – 4x – 2y – 20 = 0 is (a) 10 (c) 5

(b) 15 (d) None of these

194. If 2x – 3y = 0 is the equation of the common chord of the circles x2 + y 2 + 4x = 0 and x2 + y 2 + 2λy = 0, then the value of λ is (a) 0 (c) 2

(b) 1 (d) 3

195. The equation of the line parallel to tangent to the circle x2 + y 2 = r 2 at the point (x1, y1) and passing through origin is (a) xy1 + x1 y = 0 (c) xx1 + yy1 = 0

(b) xx1 – yy1= 0 (d) xy1 – x1 y = 0

196. If the line 2x – y + k = 0 is a diameter of the circle x2 + y2 + 6x – 6y + 5 = 0, then k is equal to (a) 12 (c) 6

(b) 9 (d) 3

197. The tangent to the circle x2 + y 2 = 9, which is parallel to y-axis and does not lie in third quadrant, touches the circle at the point (a) (– 3, 0) (c) (0, 3)

(b) (3, 0) (d) (0, – 3)

187. The equation of the circle described on the common chord of the circles x2 + y 2 + 2x = 0 and x2 + y 2 + 2y 198. The angle between the tangents from the origin to the circle (x – 7)2 + ( y + 1)2 = 25 is = 0 as diameter is (a) x2 + y2 + x + y = 0 (c) x2 + y2 – x – y = 0

(b) x2 + y2 – x + y = 0 (d) x2 + y2 + x – y = 0

188. The equations of tangents to the circle x2 + y 2 = 25 which are inclined at angle of 30º to the x-axis are (a) y = x 3 ± 5 (c) ± 3 y = x + 10

3 y = x ± 10 (d) None of these

(b)

π π (b) 3 6 π π (c) (d) 8 2 199. Through a fixed point (h, k) secants are drawn to the circle x2 + y 2 = a2. The locus of the mid points of the secants intercepted by the given circle is (a)

(b) x2 + y2 + x + y – 3 = 0 (c) x2 + y2 – x – y – 3 = 0 (d) None of these

201. The locus of the mid point of the chord of the circle x2 + y 2 – 2x – 2y – 2 = 0 which makes an angle of 120º at the centre is: (a) x2 + y2 – 2x – 2y + 1 = 0 (b) x2 + y2 + x + y – 1 = 0 (c) x2 + y2 – 2x – 2y – 1 = 0 (d) None of these 202. The equation of the system of coaxal circles that are tangent at ( 2 , 4) to the locus of the point of intersection of mutually ⊥ tangents to the circle x2 + y 2 = 9, is (a) (x2 + y2 – 18) + λ ( 2 x + 4y – 18) = 0 (b) (x + y – 18) + λ (4x + 2 y – 18) = 0 2

2

(c) (x2 + y2 – 16) + λ ( 2 x + 4y – 16) = 0 (d) None of these 203. The equation of the circle, which touches the circle x2 + y 2 – 6x + 6y + 17 = 0 externally and to which the lines x2 – 3xy – 3x + 9y = 0 are normal, is (a) x2 + y2 – 6x – 2y – 1 = 0 (b) x2 + y2 + 6x – 2y + 1 = 0 (c) x2 + y2 – 6x – 2y + 1 = 0 (d) None of these 204. A tangent drawn from the point (4, 0) to the circle x2 + y 2 = 8 touches it at a point A in the first quadrant. The coordinates of another point B on the circle such that AB = 4, are

(a)  x2 + y2 – 6x + 7 = 0 (c)  x2 + y2 – 3x + 4 = 0

(b)  x2 + y2 – 2x – 2y + 1 = 0 (d)  x2 + y2 + 2x – 4y + 4 = 0

209. The equation of the circumcircle of the triangle formed by the lines x = 0, y = 0, 2x + 3y = 5 is (a)  6(x2 + y2) + 5 (3x – 2y) = 0 (b)  x2 + y2 – 2x – 3y + 5 = 0 (c)  x2 + y2 + 2x – 3y – 5 = 0 (d)  6(x2 + y2) – 5 (3x + 2y) = 0 210. The value of λ, for which the circle x2 + y 2 + 2λx + 6y + 1 = 0 intersects the circle x2 + y 2 + 4x + 2y = 0 orthogonally, is (a)  11/8 (c)  –5/4

(b)  –1 (d)  5/2

211. If the lines 3x – 4y – 7 = 0 and 2x – 3y – 5 = 0 are two diameters of a circle of area 49π sq unit, the equation of the circle is (a)  x2 + y2 + 2x – 2y – 62 = 0 (b)  x2 + y2 – 2x + 2y – 62 = 0 (c)  x2 + y2 – 2x + 2y – 47 = 0 (d)  x2 + y2 + 2x – 2y – 47 = 0 212. If the area of the circle 4x2 + 4y 2 – 8x + 16y + k = 0 is 9 π sq unit, then the value of k is (a)  4 (c)  –16

(b)  16 (d)  ± 16

213. The value of k so that x2 + y 2 + kx + 4y + 2 = 0 and 2(x2 + y 2) – 4x – 3y + k = 0 cut orthogonally, is

(b)   − 8 (a)   10 3 3 10 8 (c)   − (d)   (a) (2, – 2) (b) (– 2, 2) 3 3 (c) (2, 2) (d) (– 2, – 2) 214. The centre of the circle, which cuts orthogonally each 205. If a circle passes through the point (a, b) and cuts the of the three circles circle x2 + y 2 – k2 = 0 orthogonally, then the equation x2 + y2 + 2x + 17y + 4 = 0, x2 + y2 + 7x + 6y + 11 = 0 and of the locus of its centre is x2 + y2 – x + 22y + 3 = 0, is (a)  2ax + 2by – (a2 + b2 + k2) = 0 (a)  (1, 2) (b)  (3, 2) (b)  2ax + 2by – (a2 – b2 + k2) = 0 (c)  (2, 3) (d)  (0, 2) 2 2 2 2 2 (c)  x + y – 2ax – 3by + a – b – k = 0 215. The radical centre of the circles x2 + y 2 – 16x + 60 = (d)  x2 + y2 – 3ax – 4by + (a2 + b2 – k2) = 0 0, x2 + y 2 – 12x + 27 = 0 and x2 + y 2 – 12y + 8 = 0, 206. The equation of the circle which touches both the axes is and the line x + y = 1 and lies in the first quadrant is 33   33   3 4 (b)   13,  (a)    , − 13  (x – c)2 + (y – c)2 = c 2, where c is 4    4  (a)  1 or 6 (b)  2 or 3  33  (c)    , 13  (d)  none of these (c)  1 or 4 (d)  none of these  4  207. The line y = mx + c is a normal to the circle x2 + y 2 216. From the point P(16, 7) tangents PQ and PR are drawn + 2gx + 2fy + c = 0, if to the circle x2 + y 2 – 2x – 4y – 20 = 0. If c be the (a)  c = 0 (b)  mg = c + f centre of the circle, then area of quadrilateral PQCR (c)  mg ≠ c + f (d)  None of these is

407

Circles

208. The equation of the circle having centre on the line x + 2y – 3 = 0 and passing through the point of intersection of the circles 200. Equation of the circle, whose diameter is the chord x x2 + y2 – 2x – 4y + 1 = 0 + y = 1 of the circles x2 + y 2 = 4, is 2 and x + y2 – 4x – 2y + 4 = 0, is 2 2 (a) x + y – x – y + 3 = 0 (a) 2(x2 + y2) = hx + ky (b) x2 + y2 = hx + ky (c) x2 + y2 + hx + ky = 0 (d) None of these

408

(a)  450 sq unit (c)  50 sq unit

(b)  15 sq unit (d)  75 sq unit

(c)  Two circles interesect in two points (d)  They touch each other externally

Objective Mathematics

217. Consider a family of circles which are passing through 222. The length of the common chord of the circles x2 + y 2 the point (–1, 1) and are tangent to x-axis. If (h, k) are + 2x + 3y + 1 = 0 and x2 + y 2 + 4x + 3y + 2 = 0 the coordinates of the centre of the circles, then the is set of values of k is given by the interval (b)   2 2 (a)   9 (a)  0 < k < ½ (b)  k ≥ 1/2 2 (c)  – 1/2 ≤ k ≤ 1/2 (d)  k ≤ 1/2 3 (c)   3 2 (d)   2 2 2 2 2 218. Two circles x + y – 2x – 3 = 0 and x + y – 4x – 6y – 8 = 0 are such that 223. The point diametrically opposite to the point P (1, 0) (a)  they touch internally on the circle x2 + y 2 + 2x + 4y – 3 = 0 is (b)  they touch externally (a)  (3, – 4) (b)  (– 3, 4) (c)  they intersect in two points (c)  (– 3, – 4) (d)  (3, 4) (d)  they are non-intersecting 224. The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are 219. The limiting points of coaxial-system determined by tangents to the same circle then radius of this circle the circles x2 + y 2 + 5x + y + 4 = 0 and x2 + y 2 + 10x is … – 4y – 1 = 0 are (a)  1/2 (b)  3/4 (a)  (0, 3) and (2, 1) (b)  (0, – 3) and (–2, – 1) (c)  4/3 (d)  5/4 (c)  (0, 3) and (1, 2) (d)  (0, – 3) and (2, 1) 225. Let a and b be non-zero real numbers. Then, the equa220. If the line y cos α = x sin α + a cos α be a tangent tion (ax2 + by 2 + c) (x2 – 5xy + 6y 2) = 0 represents to the circle x2 + y 2 = a2, then (a)  four straight lines, when c = 0 and a, b are of the same (a)  sin2 α = 1 (b)  cos2 α = 1 sign 2 2 2 2 (c)  sin α = a (d)  cos α = a (b)  two straight lines and a circle, when a = b, and c is of 221. For the given circles x2 + y 2 – 6x – 2y + 1 = 0 and sign opposite to that of a x2 + y 2 + 2x – 8y + 13 = 0 which of the following is (c)  two straight lines and a hyperbola, when a and b are of true? 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 (a)  One circle lies inside the other sign and c is of sign opposite to that of a (b)  One circle lies completely outside the other

Solutions 1. (c) Centre of circle ≡ (1, 3) Centre of circle C ≡ (1, 2) Equation of the circle C is (x – 2)2 + (y – 1)2 = r2 x2 + y2 – 4x – 2y + 5 – r2 = 0 Equation of common chord is 2x – 4y + 1 + r2 = 0 Since, it passes through (1, 3), r2 = 9 ⇒ r = 3. 2. (b) Given Circle is x2 + y 2 + 2gx + 2f y + c = 0

...(1)

 et C be its centre and PQR be an equilateral trianL gle inscribed in the circle, then C ≡ (– g, – f ) and radius of the circle CQ =

g2 + f 2 − c .

From ∆ QLC, QL = CQ sin 60º = ∴ QR = 2QL =

3 g2 + f 2 − c 2

3 ⋅ g2 + f 2 − c

Now, area of ∆PQR =

3 ⋅ QR 2 4



=

3 ⋅ 3 ( g 2 + f 2 − c) 4



=

3 3 2 ( g + f 2 − c) . 4

3. (b) We have, x2 + y2 – 8x + 4y + 4 = 0 On comparing with standard equation of circle x2 + y2 + 2gx + 2fy + c = 0, we get

=

(− 4) 2 + (2) 2 − 4

= 16 + 4 − 4 = 4 Here, radius of circle is equal to x-coordinate of the centre, ∴ circle touches y-axis. 4. (a) sin[cot–1(x + 1)] = cos(tan –1 x) 1

⇒



(1 + x) 2 + 1

=

1

1 x= − . 2

∴  r = CP =

4 + 16 = 20 = 2 5 .

6. (a) ∵ family of circles passing through the point of intersection of two circles is given by S + λS′ = 0 ∴ option ‘a’ is correct.

1 . Slope of a line ⊥ to line (1) is = – m

...(1)

 he equation of the line ⊥ to (1) and through T (0, 0) is 1 ( y – 0) = (x – 0) m 1 x ...(2) m To find the locus of the foot of the ⊥ i.e., the locus of the point of intersection of the line (1) and (2), we have to eliminate m from (1) and (2).  x Putting the value of m   = −  from (2) in (1), y 

or y = −

we get

x y – k = − ( x − h) y

or or

Since the centre lies on the line y = x – 1 ∴ k = h – 1 ...(1) Since the circle passes through the point (7, 3), therefore, the distance of the centre from this point is the radius of the circle. ∴ 3 =

(h − 7) 2 + (k − 3) 2

(h − 7) 2 + (h − 1 − 3) 2 ⇒ 9 = (h – 7)2 + (h – 4)2

[using (I)]

⇒ 3 =

⇒ h2 – 11h + 28 = 0 or (h – 7) (h – 4) = 0 ⇒ h = 7 or h = 4.

7. (d) The equation of a line through (h, k) is y – k = m (x – h)

So, the equation of the required circle is (x – 8)2 + (y + 2)2 = (10)2 or x2 + y2 – 16x + 4y – 32 = 0 9. (d) Let the centre of the circle be (h, k).

1 + x2

5. (d) Centre is point of intersection of two diameters, i.e., the point is C (8, – 2)



Circles

g 2 + f 2 − c2

∴ Radius of the circle =

409

g = – 4, f = 2 and c = 4 ∴ Co-ordinates of the centre are  (­– g, – f ), i.e., (4, – 2)

y2 + ky = – x2 + hx x2 + y2 – hx – ky = 0, which is a circle.

8. (c) The two diameters of the circle are x + y = 6 ...(1) and x + 2y = 4 ...(2) Solving (1) and (2), we get x = 8 and y = –2. Hence, the centre of the circle is (8, –2). The radius of the circle = 10 (Given).

For h = 7, we get k = 6 from (1) and for h = 4 , we get k = 3 from (1). Hence, the two circles which satisfy the given conditions are (x – 7)2 + (y – 6)2 = 9 or x2 + y2 – 14x + 12y + 76 = 0 and (x – 4)2 + (y – 3)2 = 9 or x2 + y2 – 8x – 6y + 16 = 0. 10. (c) Let A ≡ (2, 0). Given lines are 3x + 5y = 1  and (2 + c) x + 5c2y = 1

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

Multiplying equation (1) by c2 and subtracting (2) from it, we get c2 − 1 . (3c2 – c – 2) x = c2 – 1 or x = 2 3c − c − 2 (c − 1) (c + 1) c +1 2 Now, lim x = lim = lim = . c →1 c →1 (c − 1) (3c + 2) c →1 3c + 2 5 2 . ∴ x coordinate of centre = 5 1 2 , y= − . From (1), when x = 25 5 2 1 Hence, the centre of the circle is  , −  . 5 25  Also, the circle passes through the point A (2, 0). 2 1     2 −  +  0 +  5 25 2

∴ radius of the circle =

2

410

Thus, equation of the required circle is

Objective Mathematics

2 1     x −  +  y +  5 25 2



2

=

64 1 + 25 625

or 25 (x2 + y2) – 20x + 2y – 60 = 0. 11. (c) Since the circle touches the line 5x + 12y = 1, ∴ length of the ⊥ from the centre (3, 4) to the given line is equal to the radius.

(x – 4) (x – 0) + ( y – 0) ( y – 5) = 0 or x2 + y2 – 4x – 5y = 0. 14. (c) Given AC = 2. ∴ A ≡ (2, 2). Let B ≡ (6, 5). ∴ AB =

(2 − 6) 2 + (2 − 5) 2 = 5.

∴ BC = AB – AC = 5 – 2 = 3.

∴ radius =

| 5 × 3 + 12 × 4 − 1 | 62 = . 13 25 + 144

Thus, equation of the circle with centre (3, 4) and 62  is 2 13  62  (x – 3)2 + ( y – 4)2 =   .  13 

radius 

12. (a) Let the centre of the circle be C. Let the circle touches y-axis at A and cuts off intercept DE from x-axis. Let B be the mid point of DE.

[∵ the two circles touch each other externally] ∴ The equation of the required circle is (x – 6)2 + ( y – 5)2 = 32 or x2 + y2 – 12x – 10y + 52 = 0. 15. (d) Let C be the centre of the circle, then C ≡ (3, – 1).

Now, OA = 4 = BC and BD = 3. ∴ Radius of the circle = = 5.

BD 2 + BC 2 = 9 + 16

Also, OB = AC = radius = 5. ∴ Centre of the circle is (5, 4). Thus, equation of the required circle is

(x – 5)2 + ( y – 4)2 = (5)2

or

x2 + y2 – 10x – 8y + 16 = 0.

13. (b) Let the intercepts on the coordinate axes be OA and OB. Then OA = 4 and OB = 5. ∴ the coordinates of A and B are (4, 0) and (0, 5) respectively π Since ∠AOB = , AB is a diameter. 2 ∴ The required equation of the circle is

Equation of line AB is 2x – 5y + 18 = 0 and AB = 6. ∴ AL = 3. CL = length of the  ⊥ from C on AB

=

| 2 × 3 − 5( − 1) + 18 | (2) 2 + (−5) 2

= 29

∴ radius of the circle AC =

=

AL2 + CL2

32 + 19 = 38 .

Thus, equation of the required circle is (x – 3)2 + ( y + 1)2 = 38 or x2 + y2 – 6x + 2y – 28 = 0.

Hence, equation of circle is (x + 3) (x – 2) + (y – 7) (y + 5) = 0 ⇒ x2 + y2 + x – 2y – 41 = 0. 17. (a) Let ABCD be a square. Let the equations of its sides AB, BC, CD and DA be y = 1, x = 3, y = 2 and x = 2 respectively.

is concentric with the circle

x2 + y2 + 4x – 6y + 3 = 0

...(2)

∴ (1) and (2) will have the same centre i.e., – g = – 2 and – f = 3 or g = 2 and f = – 3. Substituting the values of g and f in (1), we get

x2 + y2 + 4x – 6y + c = 0

Since this circle passes through the point (2, – 1), ∴ 4 + 1 + 8 + 6 + c = 0 or c = – 19. 20. (a) Let C be the centre of the circle in its initial position and D be its centre in the new position.

Then, A ≡ (2, 1), B ≡ (3, 1), C ≡ (3, 2), D ≡ (2, 2).  ince diagonals of square are the diameters of the S circle, ∴ equation of the circle is (x – 2) (x – 3) + ( y – 1) ( y – 2) = 0 ⇒ x2 + y2 – 5x – 3y + 8 = 0. 18. (b), (d)  Let r be the radius of the circle. It is clear from the figure that two circles satisfying the given conditions are possible. Let C and D be the centres of the circles in the first and second quadrants respectively.

Then C ≡ (5, 5) and D ≡ (5 + 10π, 5)  ow centre of the N (5 + 10π, 5) and its tion will be (x – 5 – 10π)2 + ( y or x2 + 25 + 100π2 – 10y = 25

circle in the new position is radius is 5, therefore its equa– 5)2 = 52 – 10x – 20πx + 100π + y2 + 25

or x2 + y2 – 10 (2π + 1) x – 10y + 100π2 + 100π + 25 = 0. 21. (c) Given circle is x2 + y 2 – 2x = 0 given line is y = x Putting y = x in (1), we get 2x2 – 2x = 0 ⇒ x = 0, 1

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

Given AB = 8, ∴ AL = 4 and since P ≡ (0, 3) ∴ OP = 3. ∴ CL = 3, where L is the mid point of AB. Now AC =

AL2 + CL2 = 42 + 32 = 5.

∴ C ≡ (5, 3) and D ≡ (– 5, 3).  hus, equation of the required circle having centre T at  C (5, 3) is ( x – 5)2 + ( y – 3)2 = 52 or x2 + y2 – 10x – 6y + 9 = 0. a nd equation of the required circle having cente at D (– 5, 3) is (x + 5)2 + ( y – 3)2 = 52 or x2 + y2 + 10x – 6y + 9 = 0.

or

From (2), y = 0, 1 Let A ≡ (0, 0), B ≡ (1, 1) Equation of required circle is (x – 0) (x – 1) + ( y – 0) (y – 1) = 0 x2 + y2 – x – y = 0.

22. (a) The two normals are x – 1 = 0 and y – 2 = 0. Their point of intersection (1, 2) is the centre and radius = length of the ⊥ from centre (1, 2) on tangent 3x | 3.1 + 4.2 − 6 | + 4y = 6 = = 1. 9 + 16

411

19. (b) Since the circle x2 + y 2 + 2gx + 2f y + c = 0 ...(1)

Circles

16. (a) The centres of the given circles are (– 3, 7) and (2, – 5). It is given that the points (– 3, 7) and (2, – 5) are the extremities of the diameter of required circle.

412

∴  Equation of the circle is (x – 1)2 + (y – 2) = 12. i.e., x2 + y2 – 2x – 4y + 4 = 0.

Objective Mathematics

23. (b) Let P ≡ (α, β), A ≡ (5, 0) and B  ≡ (10 cos θ, 10 sin θ).  ince the point P divides AB internally in the ratio S 2 : 3, therefore 20 cos θ + 15 = 4 cos θ + 3 5 20 sin θ + 0 and β = = 4 sin θ. 5 ⇒ α – 3 = 4 cos θ and β = 4 sin θ. Squaring and adding, we get (α – 3)2 + β2 = 16. ∴ Locus of P (α , β) is (x – 3)2 + y2 = 16, which represents a circle.

α=

24. (c) Given equations are

x2 + 2ax – b2 = 0

...(1)

and x2 + 2px – q2 = 0

...(2)

 et the roots of (1) be α and β and that of (2) be γ L and δ. Then, α + β = – 2a and αδ = – b2. and γ + δ = – 2p, γδ = – q2. Let A ≡ (α, γ) and B = (β, δ). The equation of the circle with AB as diameter is (x – α) (x – β) + (y – γ) (y – δ) = 0 ⇒ x2 + y2 – (α + β)x – (α + δ)y + αβ + γδ = 0 ⇒ x2 + y2 + 2ax + 2py – b2 – q2 = 0. 25. (b) Let A ≡ (– ­4, 3) and B ≡ (12, – 1).  he equation of the circle having A and B as the T ends of a diameter is (x + 4) (x – 12) + ( y – 3) (y + 1) = 0 ...(1) or x2 + y2 – 8x – 2y – 51 = 0 ∴ Length of the intercept made by the circle (1) on y-axis =2

f 2 − c = 2 (−1) 2 + 51 = 2 52 = 4 13 .

 ince it passes through the points (0, 0), (4, 0) and S (0, 3), therefore, c = 0 ...(2), 16 + 8g + c = 0 ...(3) and 9 + 6f + c = 0 ...(4) 3 . Putting 2 the values of g, f and c in (1), the equation of the required circle becomes

Solving (3) and (4), we get g = – 2 and f = –



x2 + y2 – 4x – 3y = 0.

28. (c) Centres of the three given circles are A (0, 0),

B (– 3, 1) and C (6, –2) respectively.

0 0 1 1 −3 1 1 Area of ∆ ABC = 2 6 −2 1

= 0.

Hence, the centres A, B, C are collinear. 29. (a) Centre of the required circle is C (4, 5). Centre of the circle x2 + y2 + 4x – 6y – 12 = 0 is P (– 2, 3). ∵ The required circle passes through the point (–2, 3), ∴ radius of required circle,

CP =

(4 + 2) 2 + (5 − 3) 2 = 40 .

Thus, equation of the required circle is

(

)



(x – 4)2 + ( y – 5)2 =

or

x2 + y2 – 8x – 10y + 1 = 0.

40

2

30. (b) The equation of the given circle is x2 + y2 – 6x + 2y – 8 = 0. Its centre is C (3, – 1). Diameter through the origin is the line joining the origin O (0, 0) and the centre C (3, –1) of the circle. ∴ Its equation is −1 − 0 (x – 0) or 3y = –x y–0= 3−0 or x + 3y = 0.

26. (c) The centre of the circle is the point of intersection of the diameters 2x – 3y = 5 and 3x – 4y = 7 i.e. the point (1, – 1). 31. (c) The equation of given circle is I f r is the radius of the circle, then its area πr2 = 154 (Given) 22 × r 2 = 154 ⇒ r = 7. ⇒ 7 ∴ Equation of the circle is (x – 1)2 + ( y + 1)2 = 72 or x2 + y2 – 2x + 2y = 47. 27. (b) Given line is 3x + 4y = 12. It cuts x-axis at (4, 0) and y-axis at (0, 3). Let the equation of the required circle be x2 + y2 + 2gx + 2f y + c = 0



Its radius =

...(1)

9 + 36 − 15 = 30 .

Equation of any circle concentric with (1) is

x2 + y2 – 6x + 12y + λ = 0

Its radius =

9 + 36 − λ = 45 − λ

But radius of (2) = 2 × radius of (1) (Given) ∴

...(1)

x2 + y2 – 6x + 12y + 15 = 0

45 − λ = 2 30

or 45 – λ = 4 × 30

∴ λ = – 75. Hence the equation of the required circle is x2 + y2 – 6x + 12y – 75 = 0.

...(2)

413

32. (a), (b)  Since the centre of the circle lies on x–axis, ∴  ordinate of the centre is zero.

Circles

Let the coordinates of the centre be (h, 0). Radius of the circle = 5

(Given)

∴ Equation of the circle is

(x – h)2 + ( y – 0)2 = 52 x2 + y2 – 2hx + h2 – 25 = 0.

or

...(1)

Since it passes through (2, 3), ∴ 4 + 9 – 4h + h2 – 25 = 0

35. (a) Let P (x1, y1) be the middle point of the chord drawn from the origin to the given circle.

or h2 – 4h – 12 = 0 or (h – 6) (h + 2) = 0 or h = 6, – 2.

Then, the equation of the chord is T = S1

When h = 6, (1) becomes

x2 + y2 – 12x + 11 = 0

...(A)

i.e., xx1 + yy1 – ( y + y1) = x12 + y12 − 2 y1

...(B)

∴ – y1 = x12 + y12 − 2 y1 i.e.,

When h = –2, (1) becomes

x + y + 4x – 21 = 0 2

2

which passes though (0, 0) x12 + y12 − y1 = 0.

( A) and (B) are the equations of the required circles.

∴ Locus of the middle point (x1, y1) is x2 + y2 – y =0

33. (a) Take the join of two given points A and B as x-axis, their mid point O as origin and a line through O⊥r to AB as y-axis.

36. (d) T he centre of the given circle is (1, – 2). Since the sides of the square inscribed in the circle are parallel to the coordinate axes, so the x-coordinate of any vertex cannot be equal to 1 and its y-coordinate cannot be equal to – 2. Hence none of the points given in (a), (b) and (c) can be the vertex of the square. 37. (c) The centroid of an equilateral triangle is the centre of its circum centre and the radius of the circle is the distance of any vertex from the centroid i.e., radius of the circle

I f AB = 2c, then coordinates of A and B are (–  c, 0) and (c, 0). Let P (x, y) be any point on the locus. Then,

PA = k (given) PB (x +c) 2 + y 2

= k. ( x − c) 2 + y 2 Squaring both sides and cross-multiplying, we get (x + c)2 + y2 = k2 [(x – c)2 + y2] or (1 – k2) x2 + (1 – k2) y2 + 2c (1 + k2) x + c2 (1 – k2) = 0, which clearly represents a circle. Hence, the locus of P is the circle. ∴

34. (a) Each side of the square is l and AB, AD are the coordinate axes. ∴ C  oordinates of B and D are (l, 0) and (0, l) respectively. Since∠ BAD = 90º, BD is a diameter of the circumcircle of square ABCD. ∴  Equation of circumcircle is (x – l) (x – 0) + ( y – 0) ( y – l) = 0 or x2 + y2 = l (x + y).

= distance of centroid from any vertex 2 2 = (Median) = (3a) = 2a. 3 3 Hence, equation of circle whose centre is (0, 0) and radius 2a is (x – 0)2 + ( y – 0)2 = (2a)2   or  x2 + y2 = 4a2. 38. (a) Given circles are S1 ≡ x2 + y2 – 4x – 5 = 0 ...(1) ...(2) and S2 ≡ x2 + y2 + 8y + 7 = 0 Equation of common chord say AB of circles (1) and (2) is S1 – S2 = 0 or – 4x – 8y – 12 = 0 or x + 2y + 3 = 0 ...(3) Equation of any circle having AB as a chord is x2 + y2 – 4x – 5 + k (x + 2y + 3) = 0 or x2 + y2 – (4 – k) x + 2ky + 3k – 5 = 0 ...(4) 4−k  Its centre is C  , −k  .  2  I f line AB is the diameter of circle (4), then 4− k  C , − k  will lie on line (3)  2 

414

∴

4−k − 2k + 3 = 0 or k = 2. 2

Objective Mathematics

 utting the value of k in (4), we obtain the equaP tion of the required circle as x2 + y2 – 2x + 4y + 1 = 0. 39. (a), (c)  Equation of the circle is x2 + y 2 = 10

...(1)

Let the point be (1, h). As it lies on (1) ∴

1 + h2 = 10 ⇒ h = ± 3.

∴ There are two points (1, 3) and (1, –3) on the circle whose abscissa is 1. Equation of tangent at (1, 3) is x + 3y = 10 Equation of tangent at (1, – 3) is x – 3y = 10. 40. (a) The line a1 x + b1y + c1 = 0 cuts the coordinate axes at A (– c1/a1, 0) and B (0, – c1/b1) and the line a2 x + b 2y + c 2 = 0 cuts the axes at C (– c 2/a2, 0) and D (0, – c 2/b 2).

∴ Locus of (h, k) is 4 (x2 – y2) = a2 – b2. 43. (d) Equation of circle is x2 + y 2 – 2x = 0

...(1)

Its centre is C (1, 0). Equation of any line parallel to x + 2y = 3 ...(2) is x + 2y + k = 0. If it is a normal to the circle, it must pass through centre C, ∴ 1 + 0 + k = 0 or k = – 1. ∴ Equation of normal to (1) parallel to the line (2) is x + 2y – 1 = 0.

 o, AC and BD are chords along x-axis and y-axis S respectively, intersecting at origin O. Since A, B, C, D are concyclic, therefore OA.OC = OB.OD 44. (c) Let A and B be the centres of the two circles.  −c   −c   −c   −c  ⇒  1  .  2  =  1  .  2   a1   a2   b1   b2 

or a1a2 = b1b2.

41. (b) Given circle is x2 + y 2 – a2 = 0

...(1)

and equation of chord of circle (1) is x cos α + y sin α – p = 0

...(2)

 ow equation of any circle passing through the point N of intersection of circle (1) and line (2) is x2 + y2 – a2 + k (x cos α + y sin α – p) = 0

...(3)

or x2 + y2 + kx cos α + ky sin α – (a2 + kp) = 0 ...(4)

Then, A ≡ (2, 2) and B ≡ (6, 5).

 − k cos α − k sin α  Centre of circle (4) is C   ,  .  2 2

Also, AC = 2 (radius of given circle) Let CB = r (radius of required circle)

I f line (2) is the diameter of circle (4), then centre C will lie on line (2).

Since the two circles touch each other externally. ∴ AC + CB = AB

k cos α k sin α cos α − ⋅ sin α – p = 0 2 2 ⇒ k = – 2p. Putting the value of k in (3), we obtain the equation of the required circle as x2 + y2 – a2 – 2p (x cos α + y sin α – p) = 0.

⇒ 2 + r =

∴ −

42. (c) Let AB and DE be the two rods of lengths a and b respectively so that the points A, B, D and E are concyclic. Let the centre of the circle be C (h, k). Then k2 +

a2 = h2 + 4

b2 = (radius)2 4

⇒ 4 (h2 – k2) = a2 – b2.

(5 − 2) 2 + (6 − 2) 2

⇒ 2 + r = 5. ∴ r = 3. Thus, the equation of required circle with centre at (6, 5) and radius 3 is (x – 6)2 + ( y – 5)2 = 32 or x2 + y2 – 12x – 10y + 52 = 0. 45. (c) Centres of the given circles are C1 ≡ (5, 0) and C2 ≡ (0, 0). Also, their radii are r1 = 3 and r2 = r. Since the two circles cut each other in two distinct points, therefore | r1 – r2 | < C1C2 < r1 + r2 ⇒ r – 3 < 5 < r + 3 ⇒ 2 < r < 8.

 ince 2x + 1 = 0 is a diameter of (1), ∴ C lies S on it

y=x+2 or x – y + 2 = 0

Circles

...(1)

⇒ – 2 (λ + 2) + 1 = 0 ⇒ λ = – Centre of circle is C (– 2, 1). Draw CM ⊥ PQ, then M is the mid point of PQ. Equation of any line ⊥ to PQ is x + y + k = 0 If it passes through C (– 2, 1) then – 2 + 1 + k = 0 or k = 1. Equation of CM is x + y + 1 = 0. ...(2) 1 3 Solving (1) and (2), we obtain x = – and y = . 2 2  −3 1  ∴ Coordinates of M are  ,  .  2 2 47. (c) It is clear from the figure that the coordinates of 9  centre of such circles are  , k  . 2  9   x −  2

2

49. (c) Clearly the two circles will touch if the distance between their centres = sum of their radii. Now, the centre and radius of x2 + y 2 + 2ax + c = 0 are (– a, 0) and

a 2 − c respectively.

The centre and radius of x2 + y2 + 2bx + c = 0 are (0, – b) and

b 2 − c respectively.

Hence, the two circles will touch if

(− a − 0) 2 + (0 + b) 2 = a 2 − c + b 2 − c

Squaring both sides, we get

or c =

+ ( y – K)2

3 . 2

Substituting in (1), we obtain 2x2 + 2y2 + 2x + 6y + 1 = 0 which is the required equation of the circle.

a2 + b2 = a2 – c + b2 – c + 2

∴  Equation of such circles is

(a 2 − c) (b 2 − c)

(a 2 − c) (b 2 − c)

Squaring again, we get c2 = (a2 – c) (b2 – c) = a2b2 – c (a2 + b2) + c2 or c (a2 + b2) = a2b2 1 1 1 Dividing by a2b2c, we obtain 2 + 2 = . a b c 50. (a), (d)  Equation of line is 4x + 3y + k = 0 Equation of circle is 2x + 2y – 5x = 0 5 or x2 + y2 – x = 0 2 Since the line touches the circle, 2

2



25 9  =  − 2  + ( k – 0)2 = 2  4

+ k 2.

or x2 + y2 – 9x – 2ky + 14 = 0. 48. (d) T he common chord of the given circles is given by

415

46. (a) Equation of chord PQ is

...(1)

2

...(2)

5  ∴ length of ⊥ from centre  , 0  on (1) is nu4  25 5 +0−0 = . 16 4

merically equal to radius

(x2 + y2 + 4x + 3y + 2) – (x2 + y2 + 2x + 3y + 1) = 0 or 2x + 1 = 0. The equation of any circle through the points of intersection of x2 + y2 + 4x + 3y + 2 = 0 and 2x + 1 = 0 is (x2 + y2 + 4x + 3y + 2) + λ (2x + 1) = 0 ...(1) or x2 + y2 + 2 (λ + 2) x + 3y + (2 + λ) = 0

⇒

−3   Its centre is C  − ( λ + 2),  .  2

⇒ k = –5 ±

5 4   + 3 (0) + k 4 16 + 9

⇒ 5 + k = ±

=±

5 4

5 25 ×5=± 4 4 25 5 = 4 4

or

−45 . 4

416

51. (c), (d)  Equation of circle is x2 + y 2 = 25 = 52.

Objective Mathematics

If it touches the line 3y = 4x + 24 i.e., 4x – 3y + 24 = 0, then the length of ⊥ from centre (0, – f ) on the line is numerically = radius f.

Equation of tangents to it having slope m are y = mx ± 5

1 + m2 . 1 . Here, m = tan 30º = 3 ∴ Equations of tangents are 1 1 x ± 5 1+ = y = 3 3

10 1 x± 3 3

3 y = x ± 10 or x –

or

x–

and

3 y + 10 = 0

3 y – 10 = 0.

Equation of any line parallel to it is 3x – 4y + k = 0 Equation of circle is x2 – y2 – 2x – 4y – 4 = 0 Its centre is (1, 2) and radius =

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

1 + 4 + 4 = 3.

 he line (1) touches the circle (2) if the length of ⊥ T from (1, 2) on (1) is numerically equal to 3. 3(1) − 4(2) + k =±3 9 + 16

or if – 5 + k = ± 15

or if k = 20 and k = – 10.

 utting these values in (1), the required tangents P are 3x – 4y + 20 = 0 and 3x – 4y – 10 = 0.

∴ g = –

...(1)

x2 + y2 – 4x – 6y – 12 = 0

...(2) 4 + 9 + 12 = 5.

 he line (1) touches the circle (2) if the length of T ⊥ from (2, 3) on (1) is numerically = 5. 3(2) − 4(3) + k =±5 9 + 16 or if – 6 + k = ± 25 or if k = 31 and k = – 19. Putting these values in (1), the required tangents are 3x – 4y + 31 = 0 and 3x – 4y – 19 = 0. i.e., if

54. (a), (b)  Let the equation of the circle be x + y + 2gx + 2f y + c = 0

5 2

and f = – g =

5 . 2

Hence, the equation of circle is x2 + y2 – 5x + 5y = 0.

Equation of any line ⊥ to it is 3x – 4y + k = 0 Equation of circle is, Its centre is (2, 3) and radius =

...(1) x2 + y2 + 2gx + 2f y + c = 0 Equation of tangent to (1) at origin (0, 0) is x ⋅ 0 + y ⋅ 0 + g (x + 0) + f ( y + 0) + c = 0 or gx + f y + c = 0. But it is given to be x – y = 0. g f ⇒ f = – g. = ∴ c = 0, 1 −1 Since (1) passes through the point (2, 1) ∴ 4 + 1 + 4g + 2f + c = 0 or 4g – 2g + 5 = 0

56. (a), (c)  Equation of circle is

53. (a), (b)  The given line is 4x + 3y = 7.



⇒ 3f + 24 = ± 5f ⇒ f = 12, – 3.  utting these values of f in (2), the equations of P circles are x2 + y2 + 24y = 0 and x2 + y2 – 6y = 0. 55. (a) Let the equation of the circle be

52. (a), (c)  The given line is 3x – 4y – 1 = 0.

i.e., if

4(0) − 3 (− f ) + 24 =± f 16 + 9

∴

x2 + y2 – 2x – 6y + 6 = 0.

...(1) 1 + 9 − 6 = 2.

Its centre is (1, 3) and radius =

Equation of any line through (0, 1) is y – 1 = m (x – 0) or mx – y + 1 = 0 ...(2) If it touches the circle (1), then the length of ⊥ from centre (1, 3) on (2) is numerically equal to radius 2 i.e., m − 3 +1



m2 + 1

= ±  2

4 . 3 Substituting these values of m in (2), the required tangents are y – 1 = 0 and 4x + 3y – 3 = 0. or

(m – 2)2 = 4 (m2 + 1).

∴ m = 0, –

57. (a) Given line is y = mx + c

...(1)

Equation of tangent to (1) at origin (0, 0) is

and the given circle is x + y = r Solving (1) and (2), we get







2

2

...(1)

x ⋅ 0 + y ⋅ 0 + g (x + 0) + f ( y + 0) + c = 0

or gx + f y + c = 0. But it is given to be axis of x i.e., y = 0 ∴ g = 0, c = 0. ∴ Equation (1) becomes x2 + y2 + 2f y = 0 ...(2)

2

2

2

(1 + m2) x2 + 2mcx + c2 – r2 = 0

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

 or two real distinct points of intersection, both the F roots of (3) must be real and distinct. ∴ 4m2c2 – 4 (1 + m2) (c2 – r2) > 0 ⇒ c2 < r2 (1 + m2) ⇒ – r

1 + m2 < c < r 1 + m2 .

b2 a 2 + b2

.

x cos α + y sin α = p ...(1) ∴ Diameter = a 2 + b 2 = m + n. ...(2) meets the circle x2 + y2 = a2 2 2 2. E quation of chord of contact of tangents from 61. (b) Let (x1, y1) be any point on the circle x + y = a 2 2 2 then x1 + y1 = a ...(1) P (x1, y1) 2 ...(3) to (2) is xx1 + yy1 = a Since (1) and (3) represent the same line, comparing coefficients x1 y a2 = 1 = , cos α sin α p  a 2 cos α a 2 sin α  , . ∴ the required point is (x1, y1) =  p p   59. (c) Equation of circle is x2 + y 2 = 10

...(1)

 quation of the chord of contact of tangents drawn E from the point (4, – 2) to (1) is 4x – 2y = 10 or 2x – y = 5 ...(2) The required points of contact are obtained by solving (1) and (2) simultaneously. From (2), y = 2x – 5. Putting this value of y in (1), we get x2 + (2x – 5)2 = 10 or x2 – 4x + 3 = 0 ⇒ x = 1, 3. When x = 1, y = 2(1) – 5 = ­–3 and when x = 3, y = 2(3) – 5 = 1. Hence, the points of contact are (1, –3) and (3, 1). 60. (b) Let A ≡ (a, 0) and B ≡ (0, b). Since ∠AOB = 90º, ∴ AB is the diameter.

 quation of chord of contact of tangents from E (x1, y1) to ...(2) the circle x2 + y2 = b2 is xx1 + yy1 = b2 ∵ (2) touches the circle x2 + y2 = c2, ∴ the length of ⊥ from centre (0, 0) on (2) is = radius c. b2 b2 ⇒ = c  or   = c [Using (1)] x12 + y12 a2 or

62. (a) L et a be the radius of the circle which touches both the axes and lies in the first quadrant, then its centre is C (a, a). Since the circle also touches the line 4x + 3y = 6, ∴ length of ⊥ from centre (a, a) on the line = radius of the circle. i.e.,

a b ∴ Centre of the circle is  ,  and radius 2 2 1 2 a + b2 . = 2 ∴ Equation of the circle is a  b   x −  +  y −  2 2 2



2

=

1 (a2 + b2) 4

or x + y – ax – by = 0. Equation of tangent to the circle at O (0, 0) is ax + by = 0 ...(1) ∴ m = length of ⊥ from a2  A (a, 0) on (1) = a 2 + b2 2

2

b2 = ac. Hence, a, b, c are in G.P.

4a + 3a − 6 1 =± a⇒a= , 3. 2 16 + 9

Since the circle lies below the given line, 1 ∴ a = . 2 ∴ Equation of the required circle is 1  1   x −  +  y −  2 2 2



2

1 =   2

2

or 4x2 + 4y2 – 4x – 4y + 1 = 0. 63. (b) Let P be the point (x1, y1). Polar of P w.r.t. the circle x2 + y2 = a2 is xx1 + yy1 = a2

...(1)

417

and n = length of ⊥ from (0, b) on (1) =

Circles

58. (b) Let P (x1, y1) be the point of intersection of tangents at the points where the line

418

∵ (1) touches the circle (x – f )2 + ( y – g)2 = b2 ∴ length of ⊥ from centre ( f, g) on (1) = radius b

Objective Mathematics

⇒ ±

fx1 + gy1 − a x12 + y12

2

Also, OT = b and OP = a. From right angled triangle OPT, sin θ =

b a

b . a b . ∴ Angle between the tangents = 2θ = 2 sin– 1 a

=b

⇒ θ = sin– 1

or ( fx1 + g y1 – a2)2 = b2 ( x12 + y12 ). ∴ Locus of P (x1, y1) is ( fx + g y – a2)2 = b2 69. (a) The two diameters of the circle are given by the (x2 + y2). equation 3 3  64. (d) T he point  3 + ,  does not satisfy circles  2 2 given in (a) and (c). ∴ (a) and (c) cannot be the correct choices. The  3 3  which , centre of circle given in (b) is   2 2  does not lie on the line y – x + 3 = 0. ∴ The circle given in (b) cannot be the correct choice. The centre (3, 0) of circle given in (d) lie on the line y – x + 3 = 0. Thus, the line is normal at the given point on the circle given in (d). 65. (c) Since the given equation represents a circle, therefore, 4a – 3 = a i.e., a = 1 ( ∵ coefficients of x2 and y2 must be equal). So, the circle becomes x2 + y2 + 6x – 2y + 2 = 0. ∴ The coordinates of centre are (– 3, 1). 66. (a) Since the given equation represents a circle, therefore, k = 0 ( ∵ coefficient of xy must be zero)  o, the circle becomes 3x2 + 3y2 + 9x – 6y + 3 = 0 S or x2 + y2 + 3x – 2y + 1 = 0.  3  Its centre is  − , 1 and radius =  2 

9 3 +1−1 = 4 2

67. (c) The given equation can be written as (x – 4)2 + ( y + 3)2 = 0 ⇒ x = 4, y = –3, which gives a point. 68. (c) Let P be any point on x2 + y 2 = a2.

2x2 + 6y2 – x + y – 7xy – 1 = 0. or (x – 2y – 1) (2x – 3y + 1) = 0 or x – 2y – 1 = 0  and 2x – 3y + 1 = 0  Solving (1) and (2), we get the coordinates centre as (– 5, – 3). ∴ Equation of the circle is (x + 5)2 + ( y + 3)2 = (2 + 5)2 + (– 1 + ⇒ x2 + y2 + 10x + 6y – 19 = 0.

...(1) ...(2) of the 3)2

70. (c) The equation of tangent to the circle at the point (h, h) is hx + hy = a 2. h = – 1. h 71. (a) Equation of the circle is x2 + y 2 = 16 ∴  Slope = –

...(1)

 quation of the chord having (–2, 3) as its mid E point is T = S1 i.e., x (– 2) + y (3) – 16 = (– 2)2 + (3)2 – 16 or – 2x + 3y = 4 + 9 or 2x – 3y + 13 = 0 ...(2) Let (x1, y1) be the pole of (2) w.r.t. (1). Polar of (x1, y1) w.r.t. the circle (1) is xx1 + yy1 = 16. ...(3) S ince (2) and (3) both represent the polar of (x1, y1) 16 32 x y 48 ∴ 1 = 1 = − or x1 = − , y1 = . 2 −3 13 13 13  32 48  Hence, pole of (2) is  − ,  .  13 13  72. (c) The line 3kx – 2y –1 = 0 meets x-axis and y-axis 1  1 at A  , 0  and B  0, −  respectively and the  2  3k  line 4x – 3y + 2 = 0 cuts x-axis and y-axis at C  1   2  − , 0  and D  0,  respectively. 3 2

 et PT, PT′ be the tangents to the concentric L circle x 2 + y 2 = b 2. Join OP, OT and OT′. Then ∠OPT = ∠OPT′ = θ (say) and OT ⊥ PT.

i.e., if x12 + y12 – a2 + k ( x12 + y12 – a2) = 0 ⇒ k = – 1. Hence, from (3), equation of required circle is x2 + y2 – a2 – (xx1 + yy1 – a2) = 0 or x2 + y2 – xx1 – yy1 = 0.

1 1 1 2  ⋅  =  ⋅  3| k | 2 2 3

⇒ | k | =

1 . 2

 ccording to the given geometrical position (see A figure), k must be positive, 1 ∴  k = . 2

Circles

 ⇒

76. (d) Let ABC be the equilateral triangle with vertices B and C as (– 2, 0) and (2, 0) respectively.

73. (b) T he centre of the three given circles are (– α1, 0), (– α2, 0) and (­– α3, 0).  he distances of the three points from the origin are T α1, α2 and α3. Given: α1, α2 and α3 are in G.P. i.e.,

α 22 = α1α3.

...(1)

 ow, coordinates of any point on the circle x + y2 N = a2 are (a cos θ, a sin θ). ∴ The lengths of the tangents drawn from the point (a cos θ, a sin θ) to the three given circles are 2



2α1 a cos θ ,

2α 2 a cos θ and

2α 3 a cos θ

which, in view of (1), are in G.P. 74. (b) Equation of the circle is x2 + y 2 = a2 ...(1) Let (x1, y1) be the mid point of the chord ...(2) ax + by = c Equation of chord of (1) having (x1, y1) as its mid point is T = S1 i.e., xx1 + yy1 – a2 = x12 + y12 – a2 or

xx1 + yy1 = x12 + y12

 ...(3)

Since (2) and (3) represent the same line, ∴ Comparing coefficients, we get x1 y1 x12 + y12 = = = λ (say) a b c ⇒ x1 = aλ, y1 = bλ and x12 + y12 = cλ λ2 (a2 + b2) = cλ c ∴ λ = 2 . a + b2 ac bc , y1 = 2 ∴ x1 = 2 a + b2 a + b2

 − 12   − 2   or G   0, G  0, .  3  3 

 ince for an equilateral triangle, the circumcentre S and the centroid G coincide, 4 4  − 2 ∴ D ≡  0, .  . Also, circumradius = 4 + 3 =  3 3 Thus, the circumcircle is 2



2   4   = (x – 0)2 +  y +   3   3

2

3 (x2 + y2) + 4y – 4 3 = 0.

or

77. (a) Given circles are S1 ≡ x2 + y2 + 2gx + 2f y + c = 0

and S2 ≡ x + y + 2g ′y + 2 f ′y + c′ = 0 2

2

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

 quation of common chord of circles (1) and (2) E is

or



S1 – S2 = 0

i.e., 2 (g – g′) x + 2 ( f – f ′) y + c – c′ = 0 ...(3)

.

 ince circle (1) bisects the circumference of circle S (2), therefore, common chord will be the diameter of circle (2) and hence centre (– g′, – f ′) of circle (2) will lie on line (3).

 ac bc  , Hence the mid point is  2 .  a + b 2 a 2 + b 2  75. (a) Given circle is x2 + y 2 – a2 = 0

Then, the coordinates of the vertex A are (0, – 12 ) [Vertex A is below x-axis]. ∴ Coordinates of the centroid of ∆ABC are

...(1)

 ince PQ and PR are tangents to the circle (1), S therefore QR is chord of contact of point P (x1, y1) and hence equation of QR is xx1 + yy1 – a2 = 0 ...(2) Now, equation of any circle through the point of intersection Q and R of circle (1) and line (2) is ...(3) x2 + y2 – a2 + k (xx1 + yy1 – a2) = 0

∴ – 2 (g – g′) g′ – 2 ( f – f ′) f ′ + c – c′ = 0 or

2g′ (g – g′ ) + 2f ′ ( f – f ′) = c – c′.

78. (b) Since for an equilateral triangle, the circumcentre and the centroid coincide, therefore coordinates of the circumcentre are (3, 2). Also, circumradius =

(3 − 2) 2 + (2 + 3) 2 =

∴ Equation of the circumcircle is

419

Circle (3) will be circumcircle of ∆PQR if it passes through the point P (x1, y1).

Since the four points are concyclic, therefore OA ⋅ OC = OB ⋅ OD

26 .

420

(

)

2

Objective Mathematics

(x – 3)2 + ( y – 2)2 = 26 or x2 + y2 – 6x – 4y + 9 + 4 = 26 or x2 + y2 – 6x – 4y – 13 = 0. 79. (c) Let S ≡ x2 + y 2 + x – 2y – 5 = 0. Since S](1, 2) = 1 + 4 + 1 – 4 – 5 = – 3 < 0, ∴ the point (1, 2) lies inside the circle. So, no chord of contact exists. 80. (d) Let S : x2 + y 2 – 16 = 0

or i.e.,

(1 ± 12, – 1 ± 5) (13, 4) and (­– 11, – 6).

84. (a) The equation of the circle is x2 + y2 – 12x – 16y + 75 = 0. Join its centre C with O. Let OC meets the circle in A and B. Centre of the circle is C ≡ (6, 8) and radius

AC =

36 + 64 − 75 = 5.

Since S](2, 2) = 4 + 4 – 16 < 0, ∴ the point (2, 2) lies inside the given circle. So, no chord of contact exists and hence ∆OAB does not exist. 81. (a) Since the given two lines are parallel and distance between two parallel tangents is equal to the diameter of the circle. ∴ Radius =

1 | 16 − (−10) | 13 ⋅ = = 1. 2 13 122 + 52

82. (d) T he ordinates of the points A, B are the roots of the equation y 2 – 7y + 12 = 0 i.e., ( y – 3) ( y – 4) =0 i.e., y = 3, 4.  lso, the abscissae of the two points are the roots A of the equation x2 – 3x + 2 = 0 i.e., (x – 1) (x – 2) = 0 i.e., x = 1, 2. ∴ A ≡ (1, 3) and B ≡ (2, 4).  hus, the equation of the circle on the join of the T points A (1, 3) and B (2, 4) as diameter is

(x – 1) (x – 2) + ( y – 3) (y – 4) = 0

or x2 + y2 – 3x – 7y + 14 = 0. 83. (b) Let A, B, be the centres of the two circles. Slope 12 of the common tangent = − 5

Also, OC = 36 + 64 = 10. ∴ OA = OC – AC = 5. Now, the point on the circle nearest to the origin is A and farthest from it is B. ∵ OA : AC = 5 : 5 = 1 : 1 ⇒ A is the mid point of OC. 0 + 6 0 + 8 , ∴ the coordinates of A are    2 2  i.e., (3, 4). Let the coordinate of B be (h, k). Since C is the mid point of AB 3+ h 4+ k = 6, =8 ∴ 2 2 ⇒ h = 9, k = 12. ∴ The coordinates of B are (9, 12). 85. (b) Let x = my be any chord through the origin. Then, (1 + m2) y2 – 4y = 0 4 . ⇒ y = 0, 1 + m2 ∴ Points of intersection of the chord and the given  4m 4  , circle are O (0, 0) and A  .  1 + m 2 1 + m 2 

∴ Slope of AB is 1 5 . = tan θ = – −12 / 5 12 The point (1, – 1) lies on the line AB and the points A and B are at a distance 13 from the point (1, – 1). ∴ Coordinates of A and B are 5 (1 ± 13 cos θ, – 1 ± 13 sin θ), where tan θ = . 12 5 12 i.e., 1 ± 13⋅ , − 1 ± 13⋅   13 13 

 et the centre of the circle drawn on OA as diameter L be (h, k), then (h, k) is the mid point of OA. 2m 2 ∴ h = , k= 1 + m2 1 + m2 h ⇒ h = km or m = . k  h2  ∴ 2 = k 1 + 2  or 2k = h2 + k2 k   Hence, locus of (h, k) is x2 + y2 – 2y = 0. 86. (a) It is clear from the figure that coordinates of the circumcentre are (0, 2 tan 60º) or (0, 2 3 )

(0 − 2) 2 + (2 3 − 0) 2

= 4 + 12 = 4. ∴ Equation of the circumcircle is

(x – 0)2 + ( y – 2 3 )2 = (4)2

i.e., x2 + y2 – 4 3 y – 4 = 0. 87. (a) Take the centre of the square as origin and axes parallel to its sides.

421

OC 2 − CP 2 = 13 − 4 = 3.

∴ OP =

Also, circumradius =

22 + 32 = 13 .

∵ C ≡ (2, 3), ∴ OL = 2. From the figure, OM = OL + LM = OL + HP ∴ OP cos θ = 2 + 2 sin θ or 3 cos θ = 2 + 2 sin θ or 3 = 2 sec θ + 2 tan θ or 3 – 2 tan θ = 2 sec θ or 9 + 4 tan2θ – 12 tan θ = 4 (1 + tan2θ) 5 or 5 = 12 tan θ ∴ tan θ = 12 12 5 ∴ cos θ = and sin θ = . 13 13  36 15  ∴ P ≡ (OP cos θ, OP sin θ) i.e., P ≡  ,  .  13 13  89. (b) Given circles are x2 + y 2 – 2x – 4y = 0

...(1)

and x + y – 8y – 4 = 0 ...(2) Let A and B be the centres and r1 and r2 the radii of circles (1) and (2) respectively, then 2



2

A ≡ (1, 2), B ≡ (0, 4) r1 =

Now AB =

(1 − 0) 2 + (2 − 4) 2 =

5 , r2 = 2 5 . 5.

and r1 + r2 = 3 5 , | r1 – r2 | = |  5 – 2 5  | = 5 . Thus AB = | r1 – r2 | , hence the two circles touch each other internally.

Let side of square be 2a. The equations of sides are AD : x = a, BC : x = – a AB : y = a, CD : y = – a Let P (x, y) be any point on locus. Distances of P from the sides of square are x – a, x + a, y – a and y + a

90. (a) Let the centre of the circle be C (x1, y1).

By the given condition, ( x – a)2 + (x + a)2 + ( y – a)2 + ( y + a)2 = constant = 4c2 (say) ∴ 2x2 + 2y2 = 4c2 – 4a2 or x2 + y2 = 2 (c2 – a2), which is a circle. 88. (a) Given circle is x + y – 4x – 6y + 9 = 0 2

2

...(1)

Its centre is C (2, 3) and radius is 2. Let OP and ON be the two tangents from 0 to circle (1), then ∠POX will be minimum when OP is tangent to the circle at P. Let ∠POX = θ, then ∠LCP = θ.

As it passes through (0, 0), its radius = OC =

x12 + y12 .

Let AB be the line x = c meeting the circle in A and B. Draw CM ⊥ AB. Join CB. CB = radius =

x12 + y12 .

CM = length of ⊥ from C on AB = x1 – c. Now AB = 2b (given). CB2 − CM 2 = b

∴ 2BM =2b or or CB – CM = b 2

or

2

2

or x12 + y12 – (x1 – c)2 = b2

y12 + 2cx1 – c2 = b2.

∴ Locus of (x1, y1) is y2 + 2cx = b2 + c2.

Circles

Now CP = 2, OC =

422

do not exist. Thus, the quadrilateral PACB cannot be formed.

91. (b) Let the radius of the circle be a.

Objective Mathematics

95. (a) Equations of the circles are

Then the centre is C ≡ (a, a).  lso, the distance of C (a, a) from the line 12x + 5y A = 60 is a. | 12a + 5a − 60 | ∴ =a 122 + 52 | 17 a − 60 | = a. 13 or 17a – 60 = ± 13a or a = 15, 2.

or

 −12  Hence pole of (3) w.r.t. (1) is  6, .  5 

It is clear from the figure that a ≠ 15. ∴ a = 2. ∴ The euqation of the incircle is (x – 2)2 + (y – 2)2 = 22 or x2 + y2 – 4x – 4y + 4 = 0.

96. (d) Equations of the circles are x2 + y 2 = 4

x2 + y2 + 4x – 6y + 4 = 0.

I ts incentre is (– 2, 3) and inradius = = 3.

...(1)

and x2 + y2 = 2x + 2y ...(2) Centre of (1) is C1 ≡ (0, 0); Radius of (1) = r1 = 2; Centre of (2) is C2 ≡ (1, 1);

92. (b) Given equation of the incircle is

...(1) S1 : x2 + y2 – 12 = 0 ...(2) and S2 : x2 + y2 – 5x + 2y – 2 = 0 Common chord of (1) and (2) is 5x – 2y – 10 = 0 ...(3) (S1 – S2 = 0) Let (x1, y1) be its pole w.r.t. (1) Polar of (x1, y1) w.r.t. (1) is xx1 + yy1 = 12 ...(4) As (3) and (4) both represent the polar of (x1, y1) w.r.t. (1), ∴ comparing coefficients x1 y1 −12 12 = = ⇒ x1 = 6, y1 = –  . 5 −2 −10 5

4+9−4

 ince in an equilateral triangle, the incentre and the S circumcentre coincide, ∴ Circumcentre ≡ (– 2, 3).  lso, in an equilateral triangle, circumradius = 2 A (inradius) ∴ Circumradius = 2 ⋅ 3 = 6. ∴ The equation of the circumcircle is (x + 2)2 + ( y – 3)2 = (6)2 or x2 + y2 + 4x – 6y – 23 = 0. 93. (c) The centre of the given circle is (2, 3) and the radius = 4 + 9 − k i.e., 13 − k .  ince the given circle does not touch or intersect S the coordinate axes and the point (2, 2) lies inside the circle ∴ x-coordinate of centre > radius i.e., 2 > 13 − k , y-coordinate of centre > radius i.e., 3 > 13 − k and 4 + 4 – 8 – 12 + k < 0 ⇒ 4 > 13 – k, 9 > 13 – k and – 12 + k < 0 ⇒ k > 9, k > 4 and k < 12 ⇒ 9 < k < 12 94. (c) The given circle is S : x2 + y 2 + x – 2y – 3 = 0.  ince S]P (– 1, 2) = 1 + 4 – 1 – 4 – 3 = –3 < 0, the S point P (– 1, 2) lies inside the circle. Consequently, the tangents from the point P (– 1, 2) to the circle

Radius of (2) = r2 = 2 . d = distance between centres = C 1C 2 = 1 + 1 = 2 . If θ is the angle of intersection of two circles, then

( ) ( 2) 2

2 2 2 (2) 2 + 2 − cos θ = r1 + r2 − d = 2r1r2 2⋅ 2⋅ 2 π ∴ θ = . 4

2

=

1 . 2

97. (a) Any tangent to x2 + y 2 = b 2 is y = mx – b 1 + m 2 . It touches (x – a)2 + y2 = b2 if or

ma − b 1 + m 2 m2 + 1

= b

ma = 2b 1 + m 2 or m2a2 = 4b2 + 4b2m2

∴ m = ±

2b a − 4b 2 2

.

98. (d) Let the equation of the circle be x2 + y2 + 2gx + 2f y + c = 0 ...(1) ∵ it cuts the circle x2 + y2 – 6x + 4y – 3 = 0 orthogonally ∴ 2[g (– 3) + f (2)] = c – 3 or – 6g + 4 f = c – 3 or 6g – 4 f + c – 3 = 0 ...(2) ∴ (1) passes through (3, 0) ∴ 6g + c + 9 = 0 ...(3)

⇒ ±

15 8 ∴ the equations of the two circles are x2 + y2 = 1

and

...(1) x2 + y2 + 2g x + 2f y + c = 0 ∵ It passes through the origin, ∴ c = 0. ∴ (1) cuts x2 + y2 – 4x + 4 = 0 orthogonally. ∴ 2 [g (–2) + f (0)] = c + 4 or –4g = 4, ∴ g = 1. ∵ (1) cuts x2 + y2 – 6x + 8y – 2 = 0 orthogonally ∴ 2 [g (– 3) + f (4)] = c – 2 or – 6g + 8f = –2 or 6 + 8f = – 2, ∴ f = – 1. Putting these values of g, f and c in (1), the required equation is x2 + y2 – 2x – 2y = 0. 2.

100. (c) Let the equation of the circle be

x2 + y2 + 2gx + 2f y + c = 0

...(1)

∵ (1) cuts two given circles orthogonally. ∴ 2 [g (–2) + f (–3)] = c + 11 or

4g + 6f + c + 11 = 0

and

2 [g (– 5) + f (– 2)] = c + 21

or

10g + 4f + c + 21 = 0

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

∵ (1) has 2x + 3y – 7 = 0 as diameter ∴ centre of (1)

i.e., (– g, – f ) lies on it.

– 2g – 3f ­– 7 = 0

∴

or 2g + 3f + 7 = 0 ...(4)

3g – f + 5 = 0

Solving (4) and (5), g = –2, f = –1. Substituting these values of g, f  and c in (1), the equation of required circle is x2 + y2 – 4x – 2y + 3 = 0. 101. (a), (b)  Let the equation of the circle be

x2 + y2 + 2gx + 2fy + c = 0

∵ It cuts the two given circles orthogonally



(x – x1) (x – x2) + (y – y1) ( y – y2) = 0

x2 + y2 – (x1 + x2) x – ( y1 + y2) y + a2 = 0 ...(3) [Using (2)] Clearly, circles (1) and (3) cut each other orthogonally. 103. (b) For the sake of convenience, choose the line of centres of the given circles S = 0 and S′ = 0 as x-axis. Let their centres be (k, 0) and (– k, 0) respectively. ∴ The equations of the circles are

S ≡ (x – k)2 + y2 – a2 = 0,



S′ ≡ (x + k)2 + y2 – a′ 2 = 0.

Now the circles

S S' ± = 0 are a a'

x 2 + y 2 − 2kx + k 2 − a 2 a x 2 + y 2 − 2kx + k 2 − a' 2 =0 a' 2k (a − a' ) x + (k2 – aa′) = 0 ...(1) or x2 + y2 + (a + a' )



=±

2k (a + a ') x + (k2 + aa′) = 0 ...(2) (a − a ')

Since for the circles (1) and (2), 2g1g2 + 2f1f2 = c1 + c2 is satisfied, hence the circles

and 2 [g (0) + f (– 2)] = c + 1 ⇒ – 4 f = c + 1

each other orthogonally.

∴ (1) becomes x2 + y2 + 2gx – 1 = 0

...(2)

∵ (2) touches the line 3x + 4y + 5 = 0

...(3)

...(2)

or x2 + y2 – (x1 + x2) x – ( y1 + y2) y + (x1x2 + y1 y2) = 0

∴ 2 [g (0) + f (– 3)] = c + 1 ⇒ – 6 f = c + 1 ∴ f = 0, c = – 1.

...(1)

Equation of circle on PQ as diameter is

and x2 + y2 + ...(1)

15 x – 1 = 0. 4

Let the given circle be S ≡ x2 + y2 – a2 = 0 Polar of P w.r.t. (1) is xx1 + yy1 = a2. ∵ P, Q are conjugate points, ∴ polar of P passes through Q ⇒ x1x2 + y1 y2 = a2

...(5)

∴ From (2), – 8 – 6 + c + 11 = 0 and c = 3.

x2 + y2 –

102. (c) Let the coordinates of P and Q be (x1, y1) and (x2, y 2) respectively.

Subtracting (2) from (3), 6g – 2f + 10 = 0 or

g2 +1 ,

∴ g = 0 or –

99. (b) Let the equation of the circle be

Its centre is (1, 1) and radius =

−3 g + 5 = 9 + 16

Circles

∴ length of ⊥ from centre (–g, 0) on (3) = radius g2 +1

S S' ± = 0 cut a a'

104. (c) Equations of the circles are

x2 + y2 + 4x + 7 = 0 2x2 + 2y2 + 3x + 5y + 9 = 0

423

Subtracting (2) from (3), we get 4 f + 12 = 0 ∴ f = –3. ∵ (1) touches y-axis i.e., x = 0 ∴ y2 + 2f y + c = 0 has equal roots ⇒ 4 f 2 – 4c = 0 or c = f 2 = 9. From (3), 6g + 9 + 9 = 0 ∴ g = – 3. Hence, the required equation of the circle is x2 + y2 – 6x – 6y + 9 = 0.

...(1)

424

x2 + y 2 +

i.e.,

3 5 9 x+ y+ = 0 2 2 2

...(2)

Objective Mathematics

and x2 + y2 + y = 0 ...(3) Radical axis of (1) and (3) is 4x – y + 7 = 0 ...(4) Radical axis of (2) and (3) is 3 3 9 x+ y+ =0 2 2 2 i.e., x + y + 3 = 0. ...(5) Radical centre is the point of intersection of (4) and (5). From (4) and (5), by cross-multiplication x y 1 = = −3 − 7 7 − 12 4 + 1



or x = –2, y = – 1.

∴ Radical centre is (– 2, – 1). Length of tangent from radical centre (– 2, – 1) to (2)

=

3 5 9 (−2) 2 + (−1) 2 + (−2) + (−1) + 2 2 2



=

4 +1− 3 −

5 9 + = 2 + 2 = 4 = 2. 2 2

105. (d) Equations of the given circles are S1 : x2 + y2 – 2x – 4y – 20 = 0

...(1)

and S2 : x + y + 6x + 2y – 90 = 0 Centre of (1) is C1 (1, 2);

...(2)



2

2

radius of (1) is r1 =

1 + 4 + 20 = 5.

Centre of (2) is C2 (– 3, – 1); radius of (2) is

r2 =



d = C 1C 2 =

9 + 1 + 90 = 10. 2

∵ r2 – r1 = 10 – 5 = 5 = d, ∴ the two circles touch internally. The point of contact divides C1C2 externally in the ratio r1 : r2 = 5 : 10 = 1 : 2. ∴ The coordinates of the point of contact are 1(− 3) − 2(1) 1(−1) − 2 (2)  , ∵   i.e., (5, 5). 1− 2  1− 2 The equation of the common tangent is the equation of the radical axis i.e., S1 – S2 = 0 ⇒ – 8x – 6y + 70 = 0 or 4x + 3y – 35 = 0. 106. (a), (c)  Equations of circles are

λ 2 + ( 2 λ ) 2 + 4 + 8λ =

5λ 2 + 8λ + 4 .

But it is given to be 2 2 . ∴

5λ 2 + 8λ + 4 = 2 2

Squaring, we have 5λ2 + 8λ + 4 = 8 2 . 5 When λ = –2, from (3), x2 + y2 – 4x – 8y + 12 = 0. which gives λ = – 2,

When λ =

2 , from (3), 5x2 + 5y2 + 4x + 8y – 36 5

= 0. which are required equations of the circles. 107. (c) The given circles are

...(1) S1 ≡ x2 + y2 + 2x + 3y – 7 = 0 2 2 and S2 ≡ x + y + 3x – 2y – 1 = 0 ...(2) Their common chord is S1 – S2 = 0 or – x + 5y – 6 = 0 or L ≡ x – 5y + 6 = 0 ...(3) The equation of any circle through the points of intersection of (1) and (2) is S1 + kL = 0 or (x2 + y2 + 2x + 3y – 7) + k (x – 5y + 6) = 0 ...(4) If it passes through the point (1, 2), then (1 + 4 + 2 + 6 – 7) + k (1 – 10 + 6) = 0 or 6 – 3k = 0. ∴ k = 2. ∴ From (4), the required equation of the circle is (x2 + y2 + 2x + 3y – 7) + 2 (x – 5y + 6) = 0 or x2 + y2 + 4x – 7y + 5 = 0. 108. (b) The given circles are

(1 + 3) + (2 + 1) = 16 + 9 = 5. 2

Its radius =

S1 ≡ x2 + y2 – 4 = 0

and S2 ≡ x2 + y2 – 2x – 4y + 4 = 0

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

Their common chord is L ≡ 2x + 4y – 8 = 0. S1 + λL = 0

i.e., or

(x2 + y2 – 4) + λ (2x + 4y – 8) = 0 x2 + y2 + 2λx + 4λy – 4 – 8λ = 0 

 4 − k   6−k  2    −  ( −1) +  −  (0) = –12 + 3  2 2    ⇒ k = 13. Substituting for k in (4), the required circle is x2 + y2 + 9x + 7y – 12 = 0. 109. (b) Let the two given circles be

Any circle through the intersection of (1) and (2) is

...(1) S1 ≡ x2 + y2 – 4x – 6y – 12 = 0. 2 2 and S2 ≡ x + y + 6x + 4y – 12 = 0. ...(2) Their common chord is S1 – S2 = 0 or – 10x – 10y = 0 or L ≡ x + y = 0 ...(3) Equation of any circle through the points of intersection of (1) and (2) is S1 + kL = 0 or (x2 + y2 – 4x – 6y – 12) + k (x + y) = 0 ...(4) or x2 + y2 – (4 – k)x – (6 – k) y – 12 = 0 If it cuts the circle x2 + y2 – 2x + 3 = 0 orthogonally, then

...(3)

x2 + y2 + 2g1x + c = 0 ...(1) ...(2) and x2 + y2 + 2g2x + c = 0 Their centres are A (– g1, 0) and B (– g2, 0). ∴ AB = g1 – g2.

PT =

x + y + 2 g1 x1 + c ;



PT′ =

x12 + y12 + 2 g 2 x1 + c

 ubstituting these values in (5), the limiting points S are 3 14 (– 1, 2) and  ,  . 5 5 

2 1

 adical axis of (1) and (2) is 2 (g1 – g2) x = 0 or 112. (a) Since the limiting points are point circles belonging R to the coaxal system, x = 0. ∴ their equations are (x – 1)2 + (y – 2)2 = 0 PN = length of ⊥ from P on radical axis = x1. and (x – 4)2 + ( y – 3)2 = 0 ∴ PT2 – PT′2 ...(1) i.e., x2 + y2 – 2x – 4y + 5 = 0 = ( x12 + y12 + 2 g1 x1 + c) − ( x12 + y12 + 2 g 2 x1 + c) 2 2 + y – 8x – 6y + 25 = 0 ...(2) and x = 2x1 (g1 – g2) = 2 PN.AB. Equation of any circle coaxal with the circles (1) 110. (b) Let the three coaxial circles be and (2) is x2 + y2 – 2x – 4y + 5 + λ  (x2 + y2 – 8x – 6y + 25) = 0 ...(3) x2 + y2 + 2grx + λ = 0 (r = 1, 2, 3). (3) will pass through the origin, if Coordinates of A, B, C, the centres, are 1 (– g1, 0), (– g2, 0), (– g3, 0). 5 + 25λ = 0 i.e., λ = – 5 Also, r1 = g12 − λ , r2 = g 22 − λ , r3 = g 32 − λ . ∴ (3) becomes, x2 + y2 – 2x – 4y 1 Now, BC . CA . AB = (g2 – g3) . (g3 – g1) . (g1 – g2). + 5 – (x2 + y2 – 8x – 6y + 25) = 0 5 Also, r12 .BC + r22 .CA + r32 .AB or 2x2 + 2y2 – x – 7y = 0 = ( g12 − λ) ( g 2 − g 3 ) + ( g 22 − λ) ( g 3 − g1 ) which is the required equation.  + ( g 32 − λ ).( g1 − g 2 ) 113. (d) Given circle is x2 + y2 = px + qy. = g12 (g2 – g3) + g22 (g3 – g1) + g 32 (g1 – g2)  p q Since the centre of the circle is  ,  , so (p, q) and – λ (g2 – g3 + g3 – g1 + g1 – g2)  2 2 = – (g1 – g2) (g2 – g3) (g3 – g1) (0, 0) are the end points of a diameter. As the two chords are bisected by x-axis, the chords will cut the  [ Σa2 (b – c) = – (a – b) (b – c) (c – a)] circle at the points (x1, – q) and (x2, – q), where x1 , = – BC . CA . AB. x are real. 2

111. (b) The equations of two circles are

x2 + y2 – 2x – 6y + 9 = 0 ...(1) ...(2) and x2 + y2 + 6x – 2y + 1 = 0 Their radical axis is 8x + 4y – 8 = 0 or 2x + y – 2 = 0 ...(3) The equation of any circle coaxal with the given circles is x2 + y2 – 2x – 6y + 9 + λ (2x + y – 2) = 0 or ...(4) x2 + y2 +(2λ – 2) x + (λ – 6) y + (9 – 2λ) = 0 λ  The centre of this circle is [(1 – λ),  3 −  ]  2 Its radius =



=

λ (1 − λ ) 2 + (3 − ) 2 − (9 − 2 λ ) 2

5λ 2 − 3λ + 1 . 4

For limiting points, its radius = 0 5λ 2 i.e., − 3λ + 1 = 0 or 5λ2 – 12λ + 4 = 0 4 or 5λ2 – 10λ – 2λ + 4 = 0 or (λ – 2) (5λ – 2) = 0 ∴ λ = 2, 2 . 5

...(5)

The equation of the line joining these points is y = –q. Solving y = – q and x2 + y2 = px + qy, we get x2 – px + 2q2 = 0. The roots of this equation are x1 and x2. Since the roots are real and distinct, ∴ discriminent > 0 i.e., p2 – 8q2 > 0 or p2 > 8q2. 114. (c) Given circle is x2 + y 2 = 10 ...(1) and the straight line is x 5 + 2 y = 3 5 

...(2)

425



2 1

Circles

Let P be the point (x1, y1). Then,

426

Objective Mathematics

Solving (1) and (2), we get the two points of intersection 1 1  of the line and the circle as A  5 − 2 5 , 5 + 2 5  3 3  1 1  and B.  5 + 2 5 , −5 + 2 5  3 3 

(

(

) (

) (

∴ Area of ∆OAB =

)

)

1 × OD × AB2



=

1 | 0( 5 ) + 2 (0) − 3 5 | 180 × × 2 3 5+4



=

1 3 5 2 5 ×3 × × = 5 3. 2 3 3

115. (c) The given line passes through the point (– 1, 2). Given circle is S ≡ x2 + y2 + 2x – 4y – 3 = 0. Since S](– 1, 2) = 1 + 4 – 2 – 8 – 3 < 0, ∴ (– 1, 2) is an interior point of the circle. Thus, m can have any real value. 116. (d) Given circle is

So, the equation of the circle is 2 2 23   23   x − + y −     = 80. 3  3 118. (c) The equations of the circles are x2 + y 2 = 4 ...(1) and x2 + y2 – 8x + 12 = 0 ...(2) Centre of (1) is C1 ≡ (0, 0) and radius r1 = 2 Centre of (2) is C2 ≡ (4, 0) and radius r2 = 2 d = distance between centres = C1C2 = 4. Since C1C2 = r1 + r2, ∴ the two circles touch each other externally. Hence 3 common tangents can be drawn to the two circles. 119. (a) Clearly, the point (α, β) is either an internal point or one of the end points of the line segment joining P (3, – 4) and Q (4, 3). ∴ 3 ≤ α ≤ 4

and – 4 ≤ β ≤ 3.

120. (a) Clearly, the centre of the circle lies on the line through the point (3, 5) and ⊥ to the tangent 2x – y – 1 = 0. The equation of such line is −1 ...(1) ( x − 3) i.e., x + 2y = 13 2 Also, it is given that centre lies on the line x + y = 5 ...(2) Solving (1) and (2), we obtain the coordinates of the centre of the circle as C ≡ (– 3, 8).

( y – 5) =

x2 + y2 – 4x – 10y + 13 = 0 Its centre is C ≡ (2, 5) and radius = 4 (2 − 5) + (−3 − 2) 2

Also, AC =

2

= 9 + 25 = 34 = 5.83. ∴ AB = AC – BC = 5.83 – 4 = 1.83 > 1. ∴ There is no point on the circle at a distance 1 from the point (– 3, 2). 117. (a) Let the centre of the circle be C = (a, a) (∵ it lies on the line y = x). S ince the line x + 2y = 3 is a tangent to the circle, ∴ length of the ⊥ from (a, a) to the line = radius

Also, radius of the circle =

36 + 9 = 45 .

∴ Equation of the circle is

(x + 3)2 + (y – 8)2 = ( 45 )2

i.e., x2 + y2 + 6x – 16y + 28 = 0 121. (c) Let the angle between the tangents be 2θ. From the figure, a sin α sin θ = = sin α a ⇒ θ = α.

4 5 i.e.,

| a + 2(a ) − 3 | =±4 5 4 +1



⇒a=

⇒ (3a – 3) = ± 20

23 −17 , . 3 3

 23 23  So, the coordinates of the centre are  ,  or  3 3  −17 −17  ,   . But the centre satisfies the inequality 3 3  3x + 6y > 10.  23 23  Thus, C ≡  ,   3 3



Thus, the required angle = 2θ = 2α.

122. (c) Since the point (2, k) lies outside the circle ∴ or or or

x2 + y2 + x – 2y – 14 = 0 4 + k2 + 2 – 2k – 14 > 0 k2 – 2k – 8 > 0 (k + 2) (k – 4) > 0 k∈ (– ∞, – 2) ∪ (4, ∞)

...(1)

So, r1 + r2 < d, ∴ the two circles do not cut each other and hence the number of common tangents is 4. 127. (c) The length of the ⊥ from the centre (0, 0) of the given circle to the line 3x + ay – 20 = 0 is ⇒ ⇒ ⇒ ⇒ ⇒

x + y = 4

...(1)

and x + y – 24x – 10y + a = 0 2

2

Centre of (1) is C1 ≡ (0, 0) and radius r1 = 2 Centre of (2) is C2 ≡ (12, 5) and radius

...(2)

20

7 ) (a –

7)≥0

7 ] ∪ [ 7 , ∞).

128. (a), (d)  The equations of the circles are x2 + y 2 – 9 = 0

...(1)

x + y + 2ax + 2y + 1 = 0 2

...(2)

2

Centre of (1) is C1 ≡ (0, 0) and radius r1 = 3

r2 = 169 − a d = distance between centres

Centre of (2) is C2 ≡ (– a, – 1) and radius

144 + 25 = 13.



I f the two circles have exactly two common tangents, then

=

≤ 5 ⇒ a2 + 9 ≥ 16 ⇒ a2 – 7 ≥ 0

⇒ a∈ (– ∞, –

and

2

= C 1C 2 =

9 + a2

⇒ (a –

2

2

2

20

∴

124. (b) The equations of the circles are

| 3(0) + a (0) − 20 |

9+a 9 + a2 Radius of the given circle = 5. Since the line cuts the circle at real distinct or coincident point,

2k2 + 2k – 24 < 0 and k > – 1 k2 + k – 12 < 0 and k > – 1 (k + 4) (k – 3) < 0 and k > – 1 – 4 < k < 3 and k > – 1 – 1 < k < 3.

2

=

169 – a2 > 0 and r1 + r2 > d

⇒ (a – 13) (a + 13) < 0 and 2 +

d = distance between centres = C1C2 =

169 − a

> 13

⇒ – 13 < a < 13 and 169 – a > 121

a2 + 1 .

The two circles will touch each other, if

2

a 2 + 1 − 1 = a.

r2 =

r1 ± r2 = d

i.e., a 2 + 1 = 3 ± a ⇒ a2 + 1 = (3 ± a)2

2

⇒ a =

⇒ – 13 < a < 13 and a2 – 48 < 0 ⇒ – 13 < a < 13 and – ⇒ –

48 < a
0 ∴ length of ⊥ from (0, 0) to (1) = radius r or (k – 3) (k + 3) > 0 0+0+c ⇒ = ± r ⇒ c2 = r2 (λ2 + µ2). or k∈ (– ∞, – 3) ∪ (3, ∞) ...(2) λ 2 + µ2 The common solution of (1) and (2) is given by 126. (d) The equations of the circles are k∈ (– ∞, – 3) ∪ (4, ∞). x2 + y2 – 6x – 2y + 9 = 0 ...(1) 123. (a) Since the point (k + 1, k) lies inside the region and x2 + y2 – 14x – 8y + 61 = 0 ...(2) Centre of (1) is C1 ≡ (3, 1) and radius r1 = 1 bounded by x = 25 − y 2 and y-axis, Centre of (2) is C2 ≡ (7, 4) and radius r2 = 2 ∴ (k + 1)2 + k2 – 25 < 0 d = distance between centres and k + 1 > 0  = C1C2 = 16 + 9 = 5.

428

x2 + y2 = a2 is given by y = mx ± a

1 + m2 .

Objective Mathematics

∴ Equation of any tangent to the given circle will be

1+ m

y – 4 = m (x + 3) ± 6

or y = m (x + 3) ± 6 1 + m

2

2

or (– 8x – 4y + 5) + g (2x – 2y) = 0, which passes through the point of intersection of 5 5 8x + 4y – 5 = 0 and x = y i.e., the point  ,  .  12 12  134. (b), (c)  The given circles are x2 + y 2 – 4x = 0, x > 0

+ 4.

131. (b) T he coordinates of any point on the line x + y = 3 are (k, 3 – k). The equation of chord of contact of tangents drawn from (k, 3 – k) to the circle x2 + y 2 = 9 is

i.e., (x – 2)2 + y2 = 22, x > 0. and x2 + y2 + 4x = 0, x < 0 i.e., (x + 2)2 + y2 = 22, x < 0.

k · x + (3 – k). y = 9  or  (3y – 9) + k (x – y) = 0 which clearly passes through the intersection of 3y – 9 = 0 and x – y = 0 i.e., (3, 3). 132. (c) Clearly, the line joining the points (3, – 2) and (4, – 1) is also a diameter of the circle.  o, the two diameters of the circle are S 3x – y = 2

...(1)

−1 + 2 ( x − 4) i.e., x – y = 5...(2) 4−3 The centre of the circle is intersection of (1) and (2).  −3 −13  . So, the coordinates of centre are C ≡  ,  2 2 

and ( y + 1) =

 lso, radius of the circle = distance between A  −3 −13   ,  and (4, –1) 2 2  2

2



2

=

121 . 2

133. (a) Given circle is x2 + y 2 + 8x + 4y – 5 = 0

...(1)

Let the equation of the second circle be x2 + y2 + 2g x + 2f y + c = 0.

Since it passes through origin, ∴

c = 0.

x2 + y2 + 2g x + 2f y + c = 0

...(2)

The equation of common chord of (1) and (2) is

2 (g – 4) x + 2 ( f – 2) y + 5 = 0

...(3)

Since the line y = x touches the circle (2) ∴ x2 + x2 + 2g x + 2f x = 0 has equal roots i.e., f + g = 0. ∴ From (3), the equation of common chord is

∴ Equations of the required circles are

(x – 0)2 + ( y ∓

12 )2 = 22

i.e., x2 + y2 + 2 12 y + 8 = 0

2 (g – 4) x + 2 (– g – 2) y + 5 = 0

Since the tangents drawn from (x1, 2) to the given circle are at right angles, so the point (x1, 2) must also lie on the director circle whose equation is

x2 + y2 = 2.25

∴

x

2 1

i.e., x2 + y2 = 50.

+ 4 = 50 ⇒ x’ = ± 46 .

So, the points are ( 46 , 2) and (­– 46 , 2).

So, the equation becomes

12 ).

136. (a), (b)  Let the point on the given line be (x1, 2).

i.e., x2 + y2 + 3x + 13y – 16 = 0.



12 ) and (0, –

135. (c) Since x12 + y12 – 4 < 0, so the point (x1, y1) lies inside the given circle. Thus, the given line cannot be a tangent to the given circle. Also, it cannot be chord of contact of tangents.

∴ Equation of the circle is 3  13    x +  +  y +  2 2

circles are at (0,

and x2 + y2 – 2 12 y + 8 = 0.

121  −3   −13  + 1 = =  − 4 +  .  2   2  2 2



Clearly, from the figure, the centres of the required

137. (c) Given circle is x2 + y 2 = 4. ...(1)  et P and Q be the points of contact. Then the chord L PQ is the chord of contact of the tangents from (2, – 2) whose equation is 2x – 2y = 4 or x – y = 2

...(2)

 olving (1) and (2), we get the points of contact as S (2, 0) and (0, – 2). 138. (b) Clearly, from the figure, the radius of the smallest circle touching the given circles is

∴ d = r1 + r2 ⇒

( x1 − 2) 2 + ( y1 + 1) 2 = 4 + 3

2 2 ⇒ x1 + y1 – 4x1 + 2y1 + 5 = 49

∴ locus of (x1, y1) is



=

42 + 42 − 4

i.e., 4 2 − 4 .

139. (b), (c)  Given circle is S ≡ x2 + y 2 + 6x – 8y – 9 = 0 Since S](2, 1) = 4 + 1 + 12 – 8 – 10 = – 1 < 0, so the point (2, 1) lies inside the given circle. ∴ Chord of contact of tangents from (2, 1) does not exist. Also, polar of (2, 1) w.r.t. the circle is 2x + y + 3 (x + 2) – 4 ( y + 1) – 10 = 0 i.e., 5x – 3y – 8 = 0. 140. (c) Let the equation of one of the circles be x2 + y2 + 2g x + 2f y + c = 0. Since it passes through origin, ∴ c = 0. So, the equation becomes x2 + y2 + 2g x + 2f y = 0. Since it cuts the circle x2 + y2 + 6x – 4y + 2 = 0 orthogonally, ∴ 2g (3) + 2f (– 2) = 0 + 2 ⇒ – 6 (– g) + 4 (– f ) = 2. Thus, the locus of the centre (– g, – f ) is – 6x + 4y = 2 or 3x – 2y + 1 = 0. 141. (b), ( c), (d)  Centre of S1 is C1 ≡ (2, 3) and radius of S1 is r1 = 5. Centre of S2 is C2 ≡ (– 3, – 2) and radius of S2 is r 2 = 5. Also, d = distance between centres

= C 1C 2 =

25 + 25 =

50

∴ | r1 – r2 | = 0 and r1 + r2 = 10. Since | r1 – r2 | < C1C2 < r1 + r2, therefore circles S1 and S2 intersect.

Equation of the common chord of S1 and S2 is S1 – S2 = 0 i.e., x + y = 0, which is also the equation of the radical axis and hence it is ⊥ to the line joining the centres C1 and C2 of the two circles.

142. (a) Let the centre of the circle S1 be C1 (x1, y1)

Its radius = r1 = 4.

Given circle is S2 ≡ x2 + y2 – 4x + 2y – 4 = 0 Its centre is C2 (2, –1) and radius = r2 =

4 + 1 + 4 = 3.

x2 + y2 – 4x + 2y – 44 = 0.

143. (a) The equation of any circle touching x-axis is of the form (x – h)2 + ( y – k)2 = k2. Let the coordinates of the other end of the diameter through P be (α, β) α +1 β+2 = h and =k Then, 2 2 i.e., α = 2h – 1 and β = 2k – 2 Also, (h – 1)2 + (k – 2)2 = (radius)2 = k2  α +1  − 1 ⇒   2 

2

β + 2 β +1  − 2 =  +   2   2  2

...(1) 2

⇒ (α – 1)2 + (β – 2)2 = (β + 2)2. ⇒ (α – 1)2 = 8β. ∴ Locus of (α, β) is (x – 1)2 = 8y. 144. (b) Let the coordinates of the point P be (x1, y1). Let Q (α, β) be the middle point of AP. Then, α =

x1 + 1 y1 + 5 and β = 2 2

⇒ x1 = 2α – 1 and y1 = 2β – 5. Since the point P (x1, y1) lies on the circle x2 + y2 = 4, ∴

...(1)

x12 + y12 = 4

 ubstituting the values of x1 and y1 from (1), we S get (2α – 1)2 + (2β – 5)2 = 4 ⇒ 4 (α2 + β2) – 4α – 20β + 22 = 0. Thus, locus of (α, β) is 4 (x2 + y2) – 4x – 20y + 22 = 0 or 2 (x2 + y2) – 2x – 10y + 11 = 0. 145. (d) A point on the line 2x + y = 4 is of the form (α, 4 – 2α). Equation of the chord of contact is T = 0 i.e., αx + (4 – 2α) y = 1 or (4y – 1) + α (x – 2y) = 0. T his line clearly passes through the point of intersection of 4y – 1 = 0 and x – 2y = 0, i.e., 1 1 through the point  ,  . 2 4 146. (b) Let the equation of the circle be

x 2 + ( y − 2 ) 2 = a2

⇒ x 2 + y 2 − 2 2 y = c,

429

Circles

Also, d = distance between the centres. = ( x1 − 2) 2 + ( y1 + 1) 2 . Since the two circles touch each other externally,

430

Objective Mathematics

where c = a2 – 2 → Rational number. Let (x1, y1), (x2, y2), (x3, y3) be three distinct rational points on the circle, then x12 + y12 – 2 2 y1 = c ...(1)

x22 + y22 – 2 2 y2 = c

...(2)



x

...(3)

2 3

+ y – 2 2 y3 = c 2 3

 omparing the irrational parts of the equations, C we get y 1 = y 2 = y 3 ...(4) Comparing the rational parts of the equations, we get x12 + y12 = x22 + y22 = x32 + y32 ∴ y1 = y2 = y3, ∴ x12 = x22 = x32 . ∴ The only possible values of x are ± x1, ± x2, ± x 3. ∴ There can be at the most two rational points on the circle C. 147. (a) Let the coordinates of the point P be (x1, y1).

As B lies on the circle x2 + y2 = px + qy, we have

(– p + 2h)2 + (– q)2 = p (– p + 2h) + q (– q)

⇒ 2p2 + 2q2 – 6ph + 4h2 = 0 ⇒ 2h2 – 3ph + p2 + q2 = 0

 s there are two distinct chords from A (p, q) which A are bisected on x-axis, there must be two distinct values of h satisfying (1). ⇒ D = 9p2 – (4) (2) (p2 + q2) > 0 ⇒ p2 > 8q2. 149. (b), ( c)  Let the equation of line L1 be y = mx. Intercpts made by L1 and L2 on the circle will be equal if L1 and L2 are at the same distance from the centre of 1 −3 the circle. Centre of the given circle is  ,  .  2 2 Therefore,



1 3 − −1 2 2 = 1+1

m 3 + 2 2 ⇒ 1 + m2

The equation of the tangent at (x1, y1) is xx1 + yy1 = 2.

⇒ 7m2 – 6m – 1 = 0 ⇒ (7m + 1) (m – 1) = 0

 2 2  Then, L ≡  , 0  and M ≡  0,  .  y1   x1 



Let (α, β) be the middle point of LM, then

...(1)

⇒ m = 1,

4 = 2

m+3 1 + m2

−1 . 7

Thus, two chords are y = x and 7y + x = 0.

150. (a) Let the equation of one of the circles be 2 2 + 0 and 2β = 0 + x2 + y2 + 2g x + 2f y + c = 0. x1 y1 1 Since it cuts the given circles orthogonally, 1 ⇒ x1 = and y1 = . ...(1) β ∴ 2g (2) + 2f (– 3) = c + 9 α Since the point (x1, y1) lies on the circle and 2g (– 2) + 2f (3) = c + 4  x2 + y2 = 2 i.e., 4g – 6f = c + 9 and – 4g + 6f = c + 4 2 2 ∴ x1 + y1 = 2 On subtracting, we get, 8g – 12f = 5



2α =

i.e., – 8 (– g) + 12 (– f ) = 5.  ubstituting the values of x1 and y1 from (1), we S get So the locus of (– g, – f ) is – 8x + 12y = 5. 1 1 151. (b) Given circles are x 2 + y 2 – 4 = 0 ...(1) + = 2. α 2 β2 ...(2) and x2 + y2 – 6x – 8y – 24 = 0 1 1 ∴ The locus of (α, β) is 2 + 2 = 2 ≡ (0, 0) and radius C  entre of circle (1) is C 1 x y r1 = 2 or x– 2 + y– 2 = 2. Centre of circle (2) is C2 ≡ (3, 4) and radius 148. (d) Suppose AB is a chord of the circle through A (p, q) havr2 = 7. ing M (h, 0) as its mid point. Then coordinates of B are Also d = distance between the centres (– p + 2h, – q). = C1C2 = 5.

 ince d = r2 – r1, therefore the given circles touch S internally, as such they can have just one common tangent at the point of contact. 152. (c) We have (A0 A1)2 = ⇒ A0 A1 = 1

1 3 + = 1. 4 4

2 3  3  (A0 A2)2 =   +   2  2 

2

=

9+3 = 3. 4

157. (d) Centre of the circle

⇒ A0A2 = 3 . Similarly, A0 A4 = 3 . Thus, (A0 A1) (A0 A2) (A0 A4) = 3.

x2 + y2 + 4x – 6y + 9 sin2α + 13 cos2α = 0 is C (– 2, 3) and its radius is

x y + = 1. α β For the circle, centre ≡ (a, a) and radius = a. Since the third side touches the circle, a a + −1 α β ∴ a= ...(1) 1 1 + α 2 β2

153. (a) Let the third side be

Vertices of the triangle are (0, 0), (α, 0) and (0, β), ∴ if the circumcentre is (γ, δ) then α β and δ = . γ= 2 2  a  a 1   1 − 1 ∴ From (1), a 2  2 + 2  =  +   γ δ 2 2  4r 4δ 

22 + (−3) 2 − 9 sin 2 α − 13 cos 2 α



=

4 + 9 − 9 sin 2 α − 13 cos 2 α = 2 sin α.

 et P (h, k) be any point on the locus. The L ∠APC = α. π Also, ∠PAC = , i.e., ∆APC is a right triangle. 2

2

⇒ 2a (γ + δ) – a2 = 2γδ So, the locus of (γ, δ) is 2a (x + y) = 2xy + a2.

Thus, sin α =

AC = PC

2 sin α (h + 2) 2 + (k − 3) 2

⇒ (h + 2) 2 + (k − 3) 2 = 2 ⇒ (h + 2)2 + (k – 3)2 = 4 or h2 + k2 + 4h – 6k + 9 = 0. Thus, required equation of the locus is x2 + y2 + 4x – 6y + 9 = 0.

4 and m2 = slope of OR 3 −3 . As m1m2 = – 1, ∠QOR = π 2 . = 4 158. (b) E quation of any circle passing through the point π of intersection of x2 + y 2 – 2x = 0 and y = x is Thus, ∠QPR = 4 x2 + y2 – 2x + λ (y – x) = 0 or x2 + y2 – (2 + λ) x + λy = 0.

154. (c) Let m1 = slope of OQ =

 2 + λ −λ  Its centre is  , .  2 2   or AB to be the diameter of the required circle, F the centre must lie on AB, i.e., ( ∵ angle subtended at the centre of a circle is double the angle subtended in the alternate segment). 155. (a) Using the condition, 2gg1 + 2ff1 = c + c1, we get

2 (1) (0) + 2 (k) (k) = 6 + k

or 2k2 – k – 6 ⇒ k = −3 , 2. 2

= 0 or (2k + 3) (k – 2) = 0

2+λ −λ = ⇒ λ = – 1. 2 2

Thus, equation of required circle is x2 + y2 – x – y = 0. 159. (a) Let x2 + y 2 + 2g x + 2f y + c = 0 be the variable circle. Since it touches the given circles externally ∴

(− g − 0) 2 + (− f − 0) 2 = g 2 + f 2 − c + a ..(1)

431

y = 1, x = 0 and x = 1. Let the moving point be (x, y). Then, y2 + (y – 1)2 + x2 + (x – 1)2 = 9 is the equation of the locus. ⇒ 2x2 + 2y2 – 2x – 2y – 7 = 0, 1 1 which represents a circle having centre  ,  (which 2 2 is also the centre of the square) and radius 2.

Circles

156. (d) Let the sides of the square be y = 0,

432

and

(− g − 2a ) 2 + (− f − 0) 2 = g 2 + f 2 − c + 2a



...(2)

Objective Mathematics

Subtracting (1) from (2), we get

( g + 2a ) 2 + f 2 =

g2 + f 2 + a .

Squaring both sides, we get (g + 2a)2 + f 2 = a2 + g2 + f 2 +2a

163. (c) Let r1, r 2 , r 3 be the radii of circles with centres A, B, C respectively. These circles touch one another externally as shown in figure. Let O be their radical centre. Also, O is the incentre of ∆ABC. Since OL = 4, ∴ the radius of the incircle is 4. Now AB = r1 + r 2 = c

g2 + f 2

⇒ 4ag + 4a2 = a2 + 2a g 2 + f 2 ⇒ (4g + 3a)2 = 4 (g2 + f 2) or (– 4 (– g) + 3a)2 = 4 [(– g)2 + (– f)2]. ∴ Locus of centre (– g, – f ) is (– 4x + 3a)2 = 4 (x2 + y2) or 12x2 – 4y2 – 24ax + 9a2 = 0. 160. (d) Let (h, k) be the centre of the circle whose locus is to be obtained. It touches y-axis, hence its radius is h. The circle x2 + y 2 – 6x – 6y + 14 = 0 has centre (3, 3) and radius = 3 + 3 − 14 = 2 and touches the former circle, hence, the distance between their centres = sum of their radii. 2

or

2

(h − 3) 2 + (k − 3) 2 = h + 2

⇒ h2 – 6h + 9 + k2 – 6k + 9 = h2 + 4h + 4 ⇒ k2 – 10h – 6k + 14 = 0

x2 + y2 – 4x – 2y – 4 = 0 and x2 + y2 – 12x – 8y – 36 = 0 Centre of circle (1) is C1 ≡ (2, 1) and

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

and radius = r1 = 4 + 1 + 4 = 3. Centre of circle (2) is C2 ≡ (6, 4) 88

Also, d = distance between C1 and C2 = C1 C2 = 16 + 9 = 5. Since d ≠ r1 ± r2, ∴ the two circles do not touch each other. 162. (d) Let the equation of the circle be x2 + y2 + 2g x + 2f y + c = 0. As this circle passes through (0, 0) and (1, 0), we get 1 c = 0 and g = − . 2 By the given condition, circle can touch the given circle internally only. ∴

g2 + f 2 = 3 –

g2 + f 2

9 4 1 9 ⇒ ⇒ f 2 = 2 ⇒ f = ± 2 . + f2 = 4 4 1  1  Thus, the centre is  , 2  or  , − 2  . 2  2  ⇒ 2 g 2 + f 2 = 3 ⇒ g2 + f 2 =

a+b+c = r 1 + r 2 + r 3. 2 So, s – a = r1, s – b = r2 and s – c = r3.

∴ s =

Thus, area of ∆ABC =

s (s − a) ( s − b) ( s − c)



(r1 + r2 + r3 )r1r2 r3

=

area of ∆ABC = radius semiperimeter of ∆ABC

of incircle

161. (d) Given circles are

36 + 16 + 36 =

BC = r2 + r3 = a CA = r1 + r3 = b

W e know that

∴ Desired locus is y2 – 10x – 6y + 14 = 0.

and radius = r2 =



∴

(r1 + r2 + r3 )r1r2 r3 r1 + r2 + r3

=4 ⇒

r1r2 r3 = 16. r1 + r2 + r3

Therefore, the required ratio = 16:1. 164. (b) L et (x1, y 1) be the point. As the tangents from (x1, y1) to the first two circles are equal, (x1, y1) is on the radical axis of the circles, its equation being S1 – S2 ≡ (x2 + y2 – 1) – (x2 + y2 – 8x + 15) = 0 or 8x – 16 = 0 or x – 2 = 0 ...(1)  imilarly (x1, y1) is on the radical axis of the second S and third circle whose equation is S2 – S3 ≡ x2 + y2 – 8x + 15 – (x2 + y2 + 10y + 24) = 0 or 8x + 10y + 9 = 0 ...(2) 5 Solving (1) and (2), we get x = 2 and y = − . 2 5  ∴ The required point is  2, −  .  2 165. (a) Let the lines cuts the x-axis at A and B, then 1 and OB = – 3. OA = − λ Also, if the lines cut the y-axis at C and D, then 3 . OC = 1 and OB = 2 Now if the circle passes through A, B, C and D then 3 1 OA × OB = OC × OD ⇒  −  (– 3) = 1 ×  λ 2 ⇒ λ = 2.

169. (a) Let (α, β) be the centre of the circle passing through (a, b). ∴ Its equation is (x – α)2 + (y – β)2 = (α – a)2 + (β – b)2 ⇒ x2 + y2 – 2αx – 2βy = a2 + b2 – 2aα – 2bβ  ...(1) Since (1) cuts the circle x2 + y2 = k2 orthogonally, O + O = a2 + b2 – 2aα – 2bβ + k2

α2 + β2 .

Then radius =

2

 2  β − α  ∴ α2 + β2 =    +  2   2 

...(1)

and

2

...(2)

 ∵OC 2 = CQ 2 + OQ 2   2 2 2 and OC = CP + OP ,     2   where OP = OQ =   2  β−α β + α  and CQ = , CP = 2 2   Solving (1) and (2), we get (α, β) = (0, 1), (0, – 1), (1, 0) and (– 1, 0). ∴ There are 4 such circles. 167. (d) C entre of the circle (x – 1)2 + (y – 3)2 = r 2 is (1, 3) and radius is r.  entre of the circle x + y – 8x + 2y + 8 = 0 is C (4, – 1) and its radius = 16 + 1 − 8 = 3. 2

⇒ 2aα + 2bβ – (a2 + b2 + k2) = 0. Thus, the locus of (α, β) is

2

 2   β + α 2 α +β =    +  2   2  2

[∵ 2g1g2 + 2f1 f2 = c1 + c2]

2



2ax + 2by – (a2 + b2 + k2) = 0.

170. (a), ( c)  The equation of the given circle can be written as (x – r)2 + ( y – h)2 = r2, which shows that the circle touches the y-axis at y = h. ∴ y-axis or x = 0 is a tangent to the circle. Let the general tangent from the origin to the given circle be y = mx ...(1). ∴ The length of ⊥ from centre (r, h) on (1) must be equal to the radius r of the circle. Thus,

2

∴ Distance between their centrres is (1 − 4) 2 + (3 + 1) 2 = 5 and sum of the radii = r + 3. Since the two circles intersect, ∴ 5 < r + 3 ⇒ r > 2.

1 + m2

⇒ (mr – h)2 = r2 (1 + m2) ⇒ – 2mrh + h2 = r2 ⇒ m =

h2 − r 2 . 2rh

y= ...(1)

h2 − r 2 x ⇒ (h2 – r2)x – 2rhy = 0. 2rh

171. (a) Length of the tangents from the point P (3, 4) to the circle x2 + y 2 = 9 is

c ircle (1) cuts the circle x2 + y2 – 4 = 0 orthogonally ∴ 2g . 0 + 2f. 0 = c – 4 ⇒ c = 4.

=r

Substituting the value of m in (1), the tangent is

168. (b) Let the variable circle be x2 + y2 + 2gx + 2fy + c = 0

mr − h



PA = PB =

(3) 2 + (4) 2 − 9 = 4

433

Circles

Since circle (1) passes through (a, b), ∴ a2 + b2 + 2ga + 2fb + c = 0 ∴ Locus of centre (–g, –f ) is 2ax + 2by – (a2 + b2 + 4) = 0.

166. (c) Let the centre of one such circle be (α, β).

434

Objective Mathematics

∴ Area of ∆PAB = 2 (area of ∆PAC) 1 × 3 × 4 = 12. = 2 × 2 172. (c) Equation of the circle having AB as diameter is (x – p) (x – α) + ( y – q) (y – β) = 0

x y l2 By (1),  1  +  1  = . 2 2 4 2



2

∴ Locus of mid point of AB is

x2 + y2 =

l2 , which is a circle. 4

or x + y – (p + α) x – (q + β) y + pα + qβ = 0 ...(1) putting  y = 0, we get x2 – (p + α) x + pα + qβ = 0  ...(2) 176. (a) The given circles are S1 : x2 + y2 + 2g x + 2f y + c = 0 Since circle (1) touches x-axis 3 ∴ discriminant of equation (2) = 0 and S2 : x2 + y2 + x + 4y + c = 0. 2 2 2 ⇒ (p + α) = 4 (pα + qβ) ⇒ (p – α) = 4qβ The equation of the radical axis of the two circles ∴ Locus of B(α, β) is (p – x)2 = 4qy is 2 or (x – p) = 4qy 3 ) x + (2f – 4) y = 0 S1 – S2 = 0 i.e., (2g – 173. (d) Solving the given equations 2 i.e., (4g – 3) x + (4f – 8) y = 0. 2x + 3y = 3 and 16x – y = 4, we obtain the centre 3 8 Since it touches the circle of the circle as  ,  .  10 10  x2 + y2 + 2x + 2y + 1 = 0 2

2

2

3  8   4 −  +  6 −  10 10

Radius of the circle =

=

1369 2704 = + 100 100

2

4073 . 100

∴ Equation of the circle is 2



3  8   x −  +  y −  10 10

2

=

4073 100

⇒ (10x – 3) + (10y – 8) = 4073 ⇒ 100 (x2 + y2) – 60x – 160y = 4000 ⇒ 5 (x2 + y2) – 3x – 8y = 200. 2

2

174. (a) Let S ≡ x2 + y 2 – 8x – 6y + 9 = 0. Now S for (3, – 2) = 9 + 4 – 24 + 12 + 9 > 0, ∴ the point (3, ­– 2) lies outside the circle. ∴ Two tangents can be drawn to the circle from the point (3, – 2). 175. (c) Let OX denote the floor and OY denote wall. Let AB be the stick such that AB = l.

∴

(4 g − 3) (−1) + (4 f − 8) (−1) (4 g − 3) 2 + (4 f − 8) 2

= 1.

⇒ (4g – 3)2 + (4f – 8)2 + 2 (4g – 3) (4f – 8) = (4g – 3)3 + (4f – 8)2 3 or f = 2. ⇒ (4g – 3) (4f – 8) = 0 ⇒ g = 4 177. (c) Any circle through the given circles is x2 + y2 – 6 + k (x2 + y2 – 6x + 8) = 0. Since it passes through (1, 1), ∴ 1 + 1 – 6 + k (1 + 1 – 6 + 8) = 0 ⇒ 4k – 4 = 0 ⇒ k = 1 ∴ The circle is x2 + y2 – 6 + x2 + y2 – 6x + 8 = 0 or 2x2 + 2y2 – 6x + 2 = 0 or x2 + y2 – 3x + 1 = 0. 178. (c) According to question two diameters of the circle are 2x + 3y + 1 = 0 and 3x – y – 4 = 0

Solving, we get x = 1, y = –1

∴ Centre of the circle is (1, –1)

Given 2πr = 10π ⇒ r = 5

∴ required circle is (x –1)2 + (y + 1)2 = 52 or x2 + y2 – 2x + 2y – 23 = 0. 179. (a) Equation of tangent to the circle

Let OA = x1, OB = y1. ∴ x12 + y12 = l2 ...(1) Again A ≡ (x1, 0) and B ≡ (0, y1). x y ∴ Mid point of AB is  1 , 1  . 2 2

x2 + y2 = 5 at (1, – 2) is x – 2y – 5 = 0 ...(1) Let this line touches the circle x2 + y2 – 8x + 6y + 20 = 0 at (x1, y1) ∴ Equation of tangent at (x1, y1) is xx1 + yy1 – 4 (x + x1) + 3 ( y + y1) + 20 = 0 or x (x1 – 4) + y ( y1 + 3) – 4x1 + 3y1 + 20 = 0 ...(2) Now (1) and (2) represent the same line

⇒ q2(x2 + y2 – 2px – 2qy + q2) – (–px – qy + q2)2 = 0

 nly the point (3, – 1) satisfies it. Hence the point O of contact is (3, – 1). (x + 4) (x – 12) + (y – 3) (y + 1) = 0. This meets the y-axis where x = 0 ∴ – 48 + y2 – 2y – 3 = 0 or y2 – 2y – 51 = 0 ∴ y1 + y2 = 2 and y1 y2 = –51.



=

( y1 + y2 ) 2 − 4 y1 y2

4 + 204 = 208 =

16 × 13 = 4 13

∴ Required intercept = 4 13 . 181. (b) T he equation of the straight line passing through the points of intersection of given circles is (x2 + y2 + 5x – 8y + 1) – (x2 + y2 – 3x + 7y – 25) = 0 i.e.,

The two tangents are ⊥ if

q2 + q2 – p2 – q2 = 0 (sum of coefficients of x2 and y2 = 0)

180. (b) Equation of the circle is

∴ | y1 – y2 | =

= (x ⋅ 0 + y ⋅ 0 – p(x + 0) – q (y + 0) + q2)2

8x – 15y + 26 = 0

...(1)

⇒ q = p . 2

2

184. (b) T he limiting points are circles of the coaxal system with zero radius, so the equations of these point circles are and ∴ ⇒ ⇒ or

S1 ≡ (x – 1)2 + ( y – 3)2 = 02 ...(1) S2 ≡ (x – 2)2 + (y – 6)2 = 02 ...(2) The equation of their radical axis is S1 – S2 = 0 1 – 2x – 6y – (4 – 4x + 36 – 12y) = 0 2x + 6y – 30 = 0 x + 3y – 15 = 0.

185. (d) Radius of the circle is

Also, centre of the circle x2 + y2 – 2x = 0 is (1, 0). ∴ Distance of the point (1, 0) from the straight line (1) is

=

8 (1) − 15 (0) + 26 64 + 225

=

34 = 2. 17

182. (a) Let the equation of the circle be

x2 + y2 + 2g x + 2f y + c = 0

...(1)

Radius of the circle is R = g + f − c Since (1) passes through (1, 0) and (0, 1) ∴ 1 + 2g + c = 0 and 1 + 2f + c = 0 ⇒ g = f and c = – (1 + 2g) 2

∴ R =

2

g 2 + g 2 + (1 + 2 g )

For R to be minimum or maximum, d (2g2 + 2g + 1) = 0 ⇒ 4g + 2 = 0. dg ∴ g = –

1 . 2

d2 1 , 2 (4g + 2) = 4 > 0. 2 dg 1 Hence, R is minimum at g = ­– 2 1 −1 Also, when g = – ,f= and c = 0. 2 2 At g = –

Hence, the equation of required circle is x2 + y2 – x – y = 0. 183. (c) The equations of pair of tangents drawn from the origin to the given circle, are

SS1 = T2

r

=

(4 − 1) 2 + (6 − 2) 2



=

9 + 16 = 5.

∴ Area of circle = πr2 = 25π sq. units. 186. (b) The given lines are

3x – 4y + 4 = 0 and 3x – 4y –

7 =0 2

Distance between the given parallel lines 7 4+ 2 = 15 = 3 . = 2×5 9 + 16 2  ince the given || lines are tangent to the same S circle, 3 ∴ diameter of the circle = . 2 3 . ∴ radius of the circle = 4 187. (a) Given circles are S1 ≡ x2 + y 2 + 2x = 0 and S2 ≡ x + y + 2y = 0 2

2

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

 quation of common chord say AB of circles (1) E and (2) is

S1 – S2 = 0 ⇒ 2x – 2y = 0

...(3)

Equation of any circle having AB as a chord is

x2 + y2 + 2x + k (2x – 2y) = 0

435



Circles

i.e., (x2 + y2 – 2px – 2qy + q2) (0 + 0 – 0 – 0 + q2)

x1 − 4 y1 + 3 −4 x1 + 3 y1 + 20 = = 1 −2 −5 ⇒ – 2x1 + 8 = y1 + 3 or 2x1 + y1 – 5 = 0. ∴

436

Objective Mathematics

or x2 + y2 + (2 + 2k) x – 2ky = 0 ...(4) Its centre is C (– (1 + k), k). If line AB is the diameter of circle (4), then C (– (1 + k), k) will lie on line (3) 1 ∴ – 2 (1 + k) – 2k = 0 or k = –  . 2  utting the value of k in (4), we get the equation P of the required circle as 1 x2 + y2 + 2x – (2x– 2y) = 0 2 or x2 + y2 + x + y = 0. 188. (b) Equations of tangents are y = mx ± a

1 + m 2 ...(1) 1 and a = 5. Here m = tan 30º = 3 ∴ From (1), the equations of tangents are

y=

1 x±5 3

1+

1 3

3 y = x ± 10.

or

189. (c) Let P = (x1, y1).

Since AD ⊥ BC, ∴ Slope of BC =

−1 . 0

−1 ∴ Equation of BC is ( y – 0) = (x – 1) 0 or x = 1. 193. (b) Since S1 = 102 + 72 – 4 × 10 – 2 × 7 – 20 > 0 s o, P lies outside the circle. Join P with the centre C (2, 1) of the given circle. Suppose PC cuts the circle at A and B then, PB is the greatest distance of P from the circle

PC =

(10 − 1) 2 + (7 − 1) 2 = 10



BC =

4 + 1 + 20 = 5



PB = PC + CB = 10 + 5 = 15.

194. (d) The given circles are S1 ≡ x2 + y2 + 4x = 0 and S2 ≡ x2 + y2 + 2λy = 0 The equation of their common chord is S1 – S2 = 0 i.e., 4x – 2λy = 0 or 2x – λy = 0. Also, 2x – 3y = 0 is the common chord. ∴ λ = 3.

The tangent at P is ...(1) 195. (c) The equation of tangent to the circle x2 + y 2 = r 2 xx1 + yy1 + 3(x + x1) + 3(y + y1) – 2 = 0 Coordinates of Q satisfy (1), at the point (x1, y1) is xx1 + yy1 = r 2.  5x – 2y + 6 = 0, x = 0 Any line parallel to this line is xx1 + yy1 = k. So, 3x1 + 6y1 + 7 = 0 and Q = (0, 3) It passes through origin (0, 0), ∴ k = 0. ∴ The equation of the line is xx1 + yy1 = 0. ∴ PQ2 = x 2 + (y – 3)2 = x 2 + y 2 − 6 y + 9 1

1

1

1

1

= 11 – 6x1 – 12y1 (∵ x12 + y12 + 6 x1 + 6 y1 − 2 = 0) = 11 – 2(3x1 + 6y1) = 11 – 2(–7) = 25 So, PQ = 5.

196. (b) The centre of the given circle is (– 3, 3). Since the centre lies on the diameter 2x – y + k = 0 ∴ – 6 – 3 + k = 0 ⇒ k = 9.

190. (b) Circle through the points (0, 0), (a, 0) and (0, b) 197. (b) Any line parallel to y-axis is x = k. is

x2 + y2 – ax – by = 0.

a b Its centre is  ,  . 2 2 191. (b) Lengths of tangents from the point P (α, β) to the given circles are equal ∴

α 2 + β 2 − 4α − 5 =

α 2 + β 2 + 6α − 2β + 6

⇒ 10α – 2β + 11 = 0. 192. (a) Let BC be the chord having mid point D (1, 0). The centre of the given circle is A (2, 0).

I f it touches the circle x2 + y2 = 9, then ⊥ distance from the centre (0, 0) of the circle to the line x = k, must be equal to radius 3. |0− k | i.e., =3 ⇒k=±3 1 ∴ k = 3.\  (∵ line does not lie in the IIIrd quadrant) ∴ The equation of the tangent line is x = 3. This meets the circle when 9 + y2 = 9 ⇒ y = 0. ∴ Point of contact is (3, 0). 198. (c) Given equation of circle is 

∴ Slope of AD =

0−0 = 0. 1− 2

(x – 7)2 + (y + 1)2 = 25

⇒ x2 + 49 – 14x + y2 + 2y + 1 = 25 ⇒ x2 + y2 – 14x + 2y + 25 = 0 Equation of pair of tangents drawn from the origin to the given circle is SS1 = T2 ⇒ (x2 + y2 – 14x + 2y + 25) (0 + 0 – 14 × 0 + 2 × 0 + 25)

2

2

202. (a) Centre of the circle x2 + y 2 = 9 is (0, 0) and any tangent to the circle is

⇒ 25x2 + 25y2 – 350x + 50y + 625

= 49x2 + y2 + 625 – 14xy – 350x + 50y

⇒ – 24x2 + 24y2 + 14xy = 0 Since, coefficient of x2 + coefficients of y2

= 24 – 24 = 0, so, the required angle is

199. (b) Equation of circle is x2 + y 2 = a2

∴ Locus of mid point of chord is x2 + y2 – 2x – 2y + 1 = 0.

437

⇒ (x + y – 14x + 2y + 25) 25 = (– 7x + y + 25) 2

π . 2 ...(1)

 et the middle point of secant be (α, β). Then, the L equation of secant is α2 + β2 = αx + βy. Since it passes through (h, k) α2 + β2 = αh + βk. ∴ Locus of the mid point (α, β) is  x2 + y2 = hx + ky. 200. (c) Given equations of the circle and the line are ...(1) x2 + y2 – 4 = 0 and x + y – 1 = 0 ...(2) Equation of the circle through the intersection of (1) and (2) is ...(3) (x2 + y2 – 4) + k (x + y – 1) = 0

x cos α + y sin y = 3 ...(1) Its distance from centre (0, 0) is equal to radius 3. Any tangent to x2 + y2 = 9 but ⊥ to (1) is obtained by replacing α by (α – 90º) and its equation is x cos (α – 90º) + y sin (α – 90º) = 3 or x cos (90º – α) – y sin (90º – α) = 3 or x sin α – y cos α = 3 ...(2) Squaring and adding (1) and (2) we get x2 + y2 = 18 which is a circle concentric with the given circle. ...(3) ∴ Locus is S ≡ x2 + y2 – 18 = 0 Equation of tangent to (3) at ( 2 , 4) is P ≡ 2 x + 4y – 18 = 0. ∴ System of coaxal circles is S + λP = 0. 203. (c) Centre of given circle is (3, –3) and radius = 9 + 9 − 17 = 1.

k k Centre of the circle is  − , −  .  2 2  ince the line (2) is the diameter of the circle, S therefore centre of the circle must lie on it k k – 1 = 0 or k = – 1. − 2 2 Substituting the value of k in (3), we obtain the equation of the required circle as x2 + y2 – 4 – 1 (x + y – 1) = 0 or x2 + y2 – x – y – 3 = 0. i.e.,

−

201. (a) Given equation of circle is

x2 + y2 – 2x – 2y – 2 = 0.

Its centre is (1, 1) and radius =

1 + 1 + 2 = 2. = OB.

Equation of pair of lines is x2 – 3xy – 3x + 9y = 0 or (x – 3) (x – 3y) = 0. Now, x = 3 passes through the centre of the required circle, ∴ (3, b) is the centre of the required circle. Also, x = 3y passes through the centre of the required circle ∴ 3 = 3b or b = 1. ∴ Centre of the circle is (3, 1). Also, radius is given by

∴ sin 30º = OP/2 or OP = 1.

∴ Equation of the required circle is (x – 3)2 + ( y – 1)2 = 9 or x2 + y2 – 6x – 2y + 1 = 0.

Let mid point of chord AB be (h, k) Since, OP = 1 ⇒ (h – 1)2 + (k – 1)2 = 1 or h2 + k2 – 2h – 2k + 1 = 0

r =

(3 − 3) 2 + (1 − (−3)) 2 – 1 = 4 – 1 = 3.

In ∆OPB, ∠OBP = 30º.

204. (a), (b)  Equation of given circle is x2 + y 2 = 8

...(1)

Circles

= (x × 0 + y × 0 – 7 (x + 0) + 1 (y + 0) + 25)2

438

 et A (h, k) be the point of contact, in the first quadL rant, of tangent from P (4, 0) to the circle (1). Equation of tangent at A (h, k) is hx + ky = 8. It passes through P (4, 0), ∴ 4h = 8 or h = 2.

Objective Mathematics

Since, A (h, k) lies on the circle, we get 

h2 + k2 = 8

or

4 + k2 = 8 or k = 2 (∴ k > 0)

⇒ x2 – 2(1 – k)x – 2(2 + k)y + 1 – 3k = 0 Since, its centre (1 – k, 2 + k) lies on the line x + 2y – 3 = 0. ∴ 1 – k + 2(2 + k) – 3 = 0 ⇒ k = – 2 Hence, the equation of required circle is x2 + y2 – 2x – 4y + 1 – 2(2x – 2y – 3) = 0 ⇒ x2 + y2 – 6x + 7 = 0 209. (d) O n solving the given equations, the required points are



∴ A ≡ (2, 2). Let the coordinates of point B on circle (1), be (a, b) such that AB = 4. ∴ a2 + b2 = 8 ...(2) and AB2 = (a – 2)2 + (b – 2)2 = 16 ...(3) Solving (2) and (3), we get a = 2, b = – 2 or a = –2, b = 2 Hence, the coordinates of B are (2, – 2) or (– 2, 2).

A(0, 0), (0, 5/3), (5/2, 0) Let the equation of circumcircle be …(i) x2 + y2 + 2gx + 2fy + c = 0 Eq. (i) passes through (0, 0), we get c = 0 Similarly, Eq. (i) passes through (0, 5/3) and (5/2, 0), we get 2f = – 5/3 and 2g = – 5/2 ∴ Required equation of circle is 5 5 x2 + y 2 − x − y = 0 2 3 or 6x2 + 6y2 – 15x – 10y = 0 210. (c) Since the given two circles intersect orthogonally

∴ 2gg′ + 2ff ′= c + c′ ⇒ 2λ × 2 + 6 × 1 = 1 + 0 ⇒ 4λ + 6 = 1 2 2 of that circle is ( x1 − a ) + ( y1 − b) . ⇒ λ = − 5 Since, the two circles cut orthogonally, therefore, 4 (distance between the centres)2 = sum of square of the 211. (c) The point of intersection of two diameters are x = radii 1 and y = – 1. 2 2 2 2 2 ⇒ (x1 – 0) + (y1 – 0) = k + (x1 – a) + (y1 – b) ∴ Centre of circle is (1, – 1). ⇒ 2ax1 + 2by1 – (a2 + b2 + k2) = 0 Let r be the radius of circle, then Hence, locus of centre is πr2 = 49π ⇒ r = 7 unit 2ax + 2by – (a2 + b2 + k2) = 0 ∴ Equation of required circle is 206. (a) A circle which touches both the axes and radius (x – 1)2 + (y + 1)2 = 49 c > 0 (centre in the first quadrant) has its centre at ⇒ x2 + y2 – 2x + 2y – 47 = 0 (c, c) and its equation is

205. (a) Let (x1, y1) be the centre of the circle, then radius

(x – c)2 + (y – c)2 = c2 Since, circle touches the line 4x + 3y – 12 = 0. | 4c + 3c − 12 | ∴ c = 42 + 32 ⇒ 5c = |7 c – 12| ⇒ 7c – 12 = ± 5c ⇒ c = 1 or 6 207. (b) For the line y = mx + c to be a normal to the circle, the centre of the circle must lie on the line. ⇒ –f = m(–g) + c or mg = c + f 208. (a) Equation of any circle through the intersection of given two circles is x2 + y2 – 2x – 4y + 1 + k (2x – 2y – 3) = 0

212. (c) The given circle is

x2 + y2 – 2x + 4y + k = 0 4

Radius of circle =

1+ 4 −

k k = 5− 4 4

Area of circle = 9π (given) k  ⇒ π  5 −  = 9π 4  ⇒ k = – 16 213. (c) Since the given circles cut orthogonally ∴ 2 ×

k k  3 × (−1) + 2 × 2 ×  −  = 2 + 2 4 2  

⇒ − k − 3 = 2 + k ⇒ k = − 10 2 3

215. (d) The radical axis of circles Ist and IInd is S1 – S2 = 0 ⇒ –4x + 33 = 0 and the radical axis of circles IInd and IIIrd is S2 – S3 = 0 ⇒ –12x + 12y + 19 = 0 On solving Eqs. (i) and (ii), we get 33 and 20 x= y= 4 3  33 ∴ The coordinates of radical centre are  ,  4 216. (d) Given equation of circle is

…(i)

…(ii)

20  . 3 

P(16,7)

r = 12 + 22 + 20 = 5 PC = (16 − 1) 2 + (7 − 2) 2

= 225 + 25 = 250

In Δ PCQ,

PQ = PC 2 − QC 2



= ( 250 ) 2 − (5) 2



= 225 = 15

21 = 6.58

∴ Two circles intersect each other at two points. 219. (b) We have, S1 ≡ x2 + y2 + 5x + y + 4 = 0

and S2 ≡ x2 + y2 + 10x – 4y – 1 = 0 ∴ Equation of the coaxial system of circles is

S1 + λS2 = 0

⇒ (x2 + y2 + 5x + y + 4) + λ (x2 + y2 + 10x – 4y – 1) = 0 ⇒ ( x 2 + y 2 ) + 5(1 + 2λ ) x + (1 − 4λ ) y + 4 − λ = 0 (1 + λ ) (1 + λ ) 1+ λ The centre of the circle is

∴ Centre is (1, 2) and radius,



= 1 + 32 = 10 = 3 ⋅ 16

⇒ r2 – r1 = 2.58



R

Now,



(r2 – r1) < C1C2 < (r1 + r2)

Q



Now, C1C2 = (2 − 1) 2 + (3 − 0) 2

and r1 + r2 = 2 +

x2 + y2 – 2x – 4y – 20 = 0 5 C (1,2)

are C1(1, 0), r1 = 2 and C2(2, 3), r2 = 21

∴ Area of quadrilateral PQCR = 2 · area of ΔDPCQ 1 = 2 ⋅ PQ ⋅ QC 2 217. (b) E quation of circle which touches x-axis and having coordinates of centre as (h, k), is (x – h)2 + (y – k)2 = k2 It is passing through (–1, 1), therefore (–1 – h)2 + (1 – k)2 = k2

 5(1 + 2λ ) (1 − 4λ )  ,− −  2(1 + λ )   2(1 + λ )

…(i)

For limiting, points, radius = 0 ⇒

25(1 + 2λ ) 2 (1 − 4λ ) 2 (4 − λ ) + − =0 4(1 + λ ) 2 4(1 + λ ) 2 (1 + λ )

⇒ 25(1 + 2λ)2 + (1 – 4λ)2 – 4 (4 – λ) (1 + λ) = 0 ⇒ 25(4λ2 + 4λ + 1) + (16λ2 – 8λ + 1) 

– 4(–λ2 + 3λ + 4) = 0

⇒ 120λ2 + 80λ + 10 = 0 ⇒ (6λ + 1) (2λ + 1) = 0 ⇒ λ = − 220. (b) The line

1 1 and − 6 2

y = x tan α + a is a tangent to the circle x2 + y2 = a2, if a2 = a2(1 + tan2 α) ⇒ sec2 α = 1 ⇒ cos2 α = 1 221. (d) The centres of the given circles are C1 ≡ (3, 1) and C2 ≡ (– 1, 4) and corresponding radii are r1 = 32 + 12 − 1 = 3 and r2 = (−1) 2 + 42 − 13 = 2

439

x2 + y2 + 2gx + 2f y + c = 0 Since, this circle cuts orthogonally each of the given three circles. ∴ 2g + 17 f = c + 4 …(i) 7g + 6f = c + 11 …(ii) and –g + 22f = c + 3 …(iii) On solving Eqs. (i), (ii) and (iii) we get g = – 3, f = – 2 Therefore, the centre of the circle is (3, 2).

⇒ h2 + 2h – 2k + 2 = 0 ∵ D ≥ 0 ⇒ 2k – 1 ≥ 0 ⇒ k ≥ 1 2 218. (c) The centres and radii of circles x2 + y2 – 2x – 3 = 0 and x2 + y2 – 4x – 6y – 8 = 0

Circles

214. (b) Let the equation of the circle be

440

Objective Mathematics

α +1 β+0 = −1 and = −2 2 2 ⇒ α = – 3 ad β = – 4

2 2 Now, C1C2 = (−1 − 3) + (4 − 1)



∴

= 42 + 32 = 25 = 5

∴ C1C2 = r1 + r2 Hence, two circles touch externally. 222. (b) Let S1 ≡ x2 + y2 + 2x + 3y + 1 = 0 and S2 ≡ x2 + y2 + 4x + 3y + 2 = 0, then equation of common chord is S2 – S1 = 0 P

C2

∴ Required point is (– 3, –4). 224. (b) Lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are parallel because slope of 1st line = 3 and 4 slope of 2nd line = 6 = 3 . 8 4 Also they are tangents to the circle so distance between lines = diameter of circle. Let one point on the line

C1

M

L1

Q ⇒ 2x + 1 = 0 3 3  Here, C1  −1, −  , r1 = = C1P 2 2  3 17  and C2  −2, −  , r2 = 2 2  

L2

C1M = perpendicular distance from C1 to the common chord 2x + 1 = 0 ⇒ C1M = Now,

| −2 + 1| 22

=

| 6 × 0 − 8 ×1 − 7 |

1 2

62 + (−8) 2

PQ = 2 PM = 2 (C1P) 2 − (C1M ) 2 2

3 1 = 2   −  2 2

3x – 4y + 4 = 0 be (0, 1) So distance of (0, 1) from the line 6x – 8y – 7 = 0 is

2

=2

9 1 − = 2 2. 4 4

=

15 3 = 10 2

∴ diameter = 3 unit 2 225. (b) We have,

∴ radius = 3 unit 4

(ax2 + by2 + c) (x2 – 5xy + 6y2) = 0 ⇒ ax2 + by2 + c = 0 or x2 – 5xy + 6y2 = 0

223. (c) Given equation of circle can be rewritten as

c ⇒ x2 + y2 =  −  iff a = b, x – 2y = 0 and x – 3y = 0  a Hence, the given equation represents two straight lines and a circle, when a = b and c is of sign opposite to that of a.



(x + 1)2 + (y + 2)2 = (2 2 ) 2 Let required point be Q (α , β). Then, mid point of P(1, 0) and Q (α , β) is the centre of the circle.

EXERCISEs FOR SELF-PRACTICE 1. The equation 3 (x2 + y 2) + 5x – 7y – 2 = 0 represents (a) a pair of straight lines (b) an empty set (c) a circle (d) a degenerate circle 2. The equation of the circle having its centre on the line x + 2y – 3 = 0 and passing through the points of intersection of the circles

x2 + y2 – 2x – 4y + 1 = 0 and x2 + y2 – 4x – 2y + 4 = 0 is (a) x2 + y2 + 2x – 4y + 4 = 0 (b) x2 + y2 – 2x – 2y + 1 = 0 (c) x2 + y2 – 6x + 7 = 0 (d) x2 + y2 – 3x + 4 = 0 3. The circles x2 + y 2 – 12x – 12y = 0 and x2 + y 2 + 6x + 6y = 0

4. If the tangent at the point P on the circle x2 + y 2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point Q on the y-axis then the length of PQ is (a) 4

(b) 2 5

(c) 5

(d) 3 5

5. If a > 2b > 0, then the positive value of m for which y = mx − b 1 + m 2 is a common tangent to x2 + y 2 = b 2 and (x – a)2 + y 2 = b 2, is 2b

(a)

a 2 − 4b 2 2b (c) a − 2b



a 2 − 4b 2 (b) 2b b (d) a − 2b

6. The length of the common chord of the circles x2 + y2 + 2x + 3y + 1 = 0 and  x2 + y2 + 4x + 3y + 2 = 0, is: a (b) 2 2 (a) 2 (c) 3 2

(d) 3 2

7. The gradient of the radical axis of the circles x2 + y2 – 3x – 4y + 5 = 0 and  3x2 + 3y2 – 7x + 8y + 11 = 0 is: (a) 1 (b) – 1 3 10 1 (c) – (d) – 2 3 2 8. If a circle whose centre is (– 1, 1) touches the straight line x + 2y + 12 = 0, then the coordinates of the point of contact are 7 (a)  − , − 4   2 

18 21 (b)  − , −   5 5

(c) (2, – 7)

(d) (– 2, – 5)

9. The equation of the circle in the first quadrant which touches each axis at a distance 5 from the origin is (a) x + y + 5x + 5y + 25 = 0 (b) x2 + y2 – 10x – 10y + 25 = 0 (c) x2 + y2 – 5x – 5y + 25 = 0 (d) x2 + y2 + 10x + 10y + 25 = 0 2

2

10. The equation of radical axis of the circles x2 + y2 + x – y + 2 = 0 and  3x2 + 3y2 – 4x – 12 = 0 will be (a) 2x2 + 2y2 – 5x + y – 14 = 0 (b) 7x – 3y + 18 = 0 (c) 5x – y + 14 = 0 (d) None of these

(a) a circle (b) a pair of two distinct straight lines (c) a pair of coincident straight lines (d) a point 12. The equations of tangents to the circle x2 + y2 – 6x – 6y + 9 = 0 drawn from the origin are (a) x = 0 (c) y = 0

(b) x = y (d) x + y = 0

13. From the origin, chords are drawn to the circle (x – 1)2 + y 2 = 1. The equation of the locus of the mid points of these chords is (a) x2 + y2 – x = 0 (c) x2 + y2 – 2x = 0

(b) x2 + y2 + x = 0 (d) None of these

14. Let x2 + y 2 – 4x –2y – 11 = 0 be a circle. A pair of tangents from the point (4, 5) with a pair of radii form a quadrilateral of area (a) 8 sq. units (c) 4 sq. units

(b) 16 sq. units (d) None of these

15. The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to the same circle. The radius of the circle (a) 4 3 (c) 4

(b) 3 4 (d) None of these

16. In the coaxal system of circles x2 + y 2 + 2gx + c = 0, where g is a parameter, if c > 0 then the circles are of (a) orthogonal type (c) touching type

(b) non-intersecting type (d) intersecting type.

17. The circle x2 + y 2 – 8x + 4y + 4 = 0 touches (a) neither of the two axes (b) x-axis (c) y-axis (d) both x and y-axis 18. The locus of the centre of a circle which touches externally the given two circles is (a) hyperbola (c) parabola

(b) circle (d) ellipse

19. The radical centre of the three circles described on the three sides of a triangle as diameter is (a) the circumcentre (c) the centroid

(b) the incentre (d) the orthocentre

20. The cut points of circle x2 + y 2 = 25 and x2 + y2 – 8x + 7 = 0 are (a) (4, 3) and (4, – 3) (c) (– 4, 3) and (4, 3)

(b) (4, – 3) and (3, 4) (d) (4, 3) and (3, 4)

21. A circle passing through the origin has its centre (α, β), then the equation of tangent at the origin is (a) αx – βy = 0 (c) y = 0

(b) αx + βy = 0 (d) None of these

441

11. The equation x2 + y 2 + 4x + 6y + 13 = 0 represents

Circles

(a) intersect in two points (b) touch each other externally (c) touch each other internally (d) None of these

442

Objective Mathematics

22. The equation of the circle passing through (2a, 0) and 25. The equation of normal to the circle whose radical axis w. r. t. the circle x2 + y 2 = a2 is x x2 + y2 + 6x + 4y – 3 = 0 at (1, 1) is a , is = (a) y + 1 = 0 (b) y + 4 = 0 2 (c) y + 2 = 0 (d) y = 0 (a) x2 + y2 + 2ay = 0

(b) x2 + y2 + 2ax = 0

(c) x2 + y2 – 2ay = 0

(d) x2 + y2 – 2ax = 0

23. The locus of the middle point of the chord of length 2l to the circle x2 + y 2 = a2 is (a) x2 + y2 = l2 + a2 (b) 2x2 + 2y2 = l2 + a2 (c) x2 + y2 = a2 – l2 (d) 2x2 + 2y2 = a2 – l2

(a) 12x + 8y + 5 = 0 (c) 5x + 8y + 12 = 0

(b) 8x + 12y + 5 = 0 (d) None of these

27. The locus of the centre of a circle which cuts orthogonally the circles x2 – 20x + 4 = 0 and which touches x = 2 is:

24. If 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to the same circle, then the radius of that circle is (a) 3 2 (c) 11 10

26. The locus of the centre of circle which cuts the circles x2 + y 2 + 4x – 6y + 9 = 0 and x2 + y 2 – 4x + 6y + 4 = 0 orthogonally is:

(b) 3 4 (d) 1 10

(a) y2 = 16x + 4 (c) x2 = 16y + 4

(b) x2 = 16y (d) y2 = 16x

28. Two tangents are drawn from a pt. P on radical axis to the two circles touching at Q and R respectively then the triangle formed by joining P,Q,R is: (a) isosceles (c) right angled

(b) equilateral (d) None of these

Answers

1. (c) 11. (a) 21. (b)

2. (c) 3. (b) 12. (a), (c) 13. (a) 22. (c) 23. (a)

4. (c) 14. (c) 24. (b)

5. (a) 15. (b) 25. (c)

6. (b) 16. (b) 26. (c)

7. (b) 17. (c) 27. (d)

8. (b) 18. (a) 28. (a)

9. (b) 19. (a)

10. (b) 20. (a)

Conic Sections (Parabola,  Ellipse and Hyperbola)

12 CHAPTER

SUMMARY OF CONCEPTS CONIC SECTION A conic section or conic is the locus of a point which moves in a plane in such a way that its distance from a fixed point bears a constant ratio to its distance from a fixed straight line. The fixed point is called the focus and the fixed line is called the directrix of the conic. The constant ratio is called the eccentricity of the conic and is denoted by e. If

• • • • •

e = 1, the conic is called Parabola. e < 1, the conic is called Ellipse. e > 1, the conic is called Hyperbola. e = 0, the conic is called Circle. e = ∞, the conic is called pair of straight lines.

2. Section of a right circular cone by a plane parallel to its base is a circle as shown in the figure.

Important Terms 1. Axis  The straight line passing through the focus and perpendicular to the directrix of the conic is known as its axis. 2. Vertex  A point of intersection of a conic with its axis is known as a vertex of the conic. 3. Centre  The point which bisects every chord of the con 3. Section of a right circular cone by a plane parallel to a ic passing through it, is called the centre of the conic. generator of the cone is a parabola as shown in the 4. Focal Chord  A chord passing through the focus is figure. known as focal chord of the conic. 5. Latus Rectum  The focal chord which is perpendicular to the axis is known as latus rectum of the conic. Double Ordinate  A chord of the conic which is perpendicular to the axis is called the double ordinate of the conic.

Remark The curves defined above are called conic sections, because these are obtained when a right circular cone is cut by a plane in various ways.

Section of a right circular cone by different planes 1. Section of a right circular cone by a plane passing through its vertex is a pair of straight lines passing through the vertex as shown in the figure.

4. Section of a right circular cone by a plane not parallel to any generator of the cone and not perpendicular or parallel to the axis of the cone is an ellipse as shown in the figure.

444

5. Section of a right circular cone by a plane parallel to the axis of the cone is a hyperbola as shown in the figure.

CENTRE OF CONIC

Objective Mathematics

Working Rule Let S ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 be the given conic. 1. Differentiate S partially w.r.t. x treating y as constant, we get

∂S = 2ax + 2hy + 2g. ∂x 2. Differentiate S partially w.r.t. y treating x as constant, we get

EQUATION OF CONIC Let S (h, k) be the focus and QN be the directrix whose equation is Ax + By + C = 0

∂S = 2hx + 2by + 2f. ∂y

3. Equating these two equations to zero and solving for x and y, we get the coordinates at centre as,  hf − bg gh − af  , (x, y) =  2 2  .  ab − h ab − h 

PARAbOLA

Let P (x, y) be any point on the conic. From P, draw PN ⊥ QN. If e is the eccentricity of the conic, then by definition

PS = e ⇒ PS2 = e2 PN2 PN or (x – α)2 + (y – β)2 = e2 This is the cartesian equation of the conic which, on simplification, takes the form ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 where a, b, c, f, g and h are constants. Thus the equation of a conic is an equation of second degree in x and y.

gENERAL EQUATION The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents • a pair of straight lines if ∆ = 0 where ∆ = abc + 2fgh – af 2 – bg2 – ch2 or

• • • •

a ∆ = h g

h b f

g f , c

a circle if ∆ ≠ 0, a = b and h = 0, a parabola if ∆ ≠ 0 and h2 = ab, an ellipse if ∆ ≠ 0 and h2 < ab, and a hyperbola if ∆ ≠ 0 and h2 > ab.

A parabola is the locus of a point which moves in a plane such that its distance from a fixed point (called the focus) is equal to its distance from a fixed straight line (called the directrix). Let S be the focus, QN be the directrix and P be any point on the parabola. Then, by definition, PS = PN where PN is the length of the perpendicular from P on the directrix QN.

Latus Rectum of the Parabola Let the given parabola be y2 = 4ax. In the figure LSS′ (a line through focus ⊥ to axis) is the latus rectum. Also by definition, LSL′ = 2 ( 4a ⋅ a ) = 4a = double ordinate (Any chord of the parabola y2 = 4ax which is ⊥ to its axis is called the double ordinate) through the focus S. Note: Two parabolas are said to be equal when their latus

recta are equal.

Parametric Equations of a Parabola Clearly x = at2, y = 2at satisfy the equation y2 = 4ax for all real values of t. Hence the parametric equations of the parabola y2 = 4ax are x = at2, y = 2at, where t is the parameter. Also, (at2, 2at) is a point on the parabola y2 = 4ax for all real values of t. This point is also described as the point ‘t’ on the parabola.

Focal Chord Any chord to the parabola which passes through the focus is called a focal chord of the parabola. Four standard forms of the parabola are shown in the table below:

Standard Equation

445

y = 4ax (a > 0)

y2 = – 4ax (a > 0)

x2 = 4ax (a > 0)

x2 = – 4ax (a > 0)

Vertex

A (0, 0)

A (0, 0)

A (0, 0)

A (0, 0)

Focus

S (a, 0)

S (– a, 0)

S (0, a)

S (0, – a)

Equation of directrix

x=–a

x=a

y=–a

y=a

Equation of axis

y=0

y=0

x=0

x= 0

Length of latus rectum

4a

4a

4a

4a

Extermities of latus rectum

(a, ± 2a)

(– a, ± 2a)

(± 2a, a)

(± 2a, – a)

Equation of latus rectum

x=a

x=–a

y=a

y=–a

Equation of tangents at vertex

x=0

x=0

y=0

y=0

Focal distance of a point P (x, y)

x+a

x–a

y+a

y–a

Parametric coordinates

(at2, 2at)

(– at2, 2at)

(2at, at2)

(2at, – at2)

Eccentricity (e)

1

1

1

1

Shape of the parabola

EQUATION OF A ChORD

EQUATIONS OF TANgENT IN DIFFERENT FORMS (a) The equation of chord joining the points (x1, y1) and (x2, y2) on the parabola y2 = 4ax is: y (y1 + y2) = 4ax + y1y2

(b) The equation of chord joining the points (a (at22, 2at2) is: y (t1 + t2) = 2 (x + at1t2)

1. Point Form The equation of the tangent to the parabola y2 = 4ax at the point (x1, y1) is

, 2at1) and

(c) Length of the chord y = mx + c to the parabola y2 = 4ax is 4 given by 2 1 + m 2 a (a − mc) . m

yy1 = 2a (x + x1).

Note: The equation of tangent at (x1, y1) can also be obtained x + x1 y + y1 , y by and by replacing x2 by xx1, y2 by yy1, x by 2 2 xy1 + x1 y . This method is used only when the equation of xy by 2 parabola is polynomial of second degree in x and y.

Condition for the Chord to be a Focal Chord The 2. Parametric Form The equation of the tangent to the parabola y2 = 4ax at the point (at2, 2at) is chord joining the points (a t12 , 2at1) and (a t22 , 2at2) passes through focus provided t1t2 = – 1. t y = x + a t 2. Length of Focal Chord The length of a focal chord joining 3. Slope Form The equation of tangent to the parabola y2 = 4ax in terms of slope ‘m’ is the points (a t12 , 2at1) and (a t22 , 2at2) is (t2 – t1)2. a y = mx + . Note: The length of the focal chord through the point ‘t’ m 2 2 on the parabola y = 4ax is a (t + 1/t) .  a 2a  The coordinates of the point of contact are  2 ,  m m  Condition for Tangency and Point of Contact The a line y = mx + c touches the parabola y2 = 4ax if c = and POINT OF INTERSECTION OF TANgENTS m The point of intersection of tangents drawn at two different  a 2a  the coordinates of the point of contact are  2 ,  . points of contact P (a t12 , 2at1) and Q (a , 2at2) on the parabola m m  y2 = 4ax is

Conic Sections (Parabola, Ellipse and Hyperbola)

Four Standard Forms of the Parabola 2

446

Objective Mathematics

R ≡ [2a + a ( t12 + t22 + t1t2), – at1t2 (t1 + t2)]. R ≡ (at1t2, a (t1 + t2)). Notes: 1. Angle between tangents at two points P (at12 , 2at1) and Q (at22 , 2at2) on the parabola y2 = 4ax is:



θ = tan –1

t2 − t1 1 + t1t2

Remarks 1. If the normal at the point P (a t12 , 2at1) meets the parabola y2 = 4ax again at Q (a t22 , 2at2), then t2 = – t1 – 2 .

t1 2. The G. M. of the x-coordinates of P and Q (i.e. at12 x at22 Note that PQ is normal = at1t2) is the x-coordinate of the point of intersection of to the parabola at P and tangents at P and Q on the parabola. not at Q. 3. The A.M. of the y-coordinates of P and Q

2. If the normals at the points (a t12 , 2at 1 ) and (a t22 , 2at2) meet on the parabola y2 = 4ax, then  2at1 + 2at2  = a (t1 + t2 )  is the y-coordinate of the  i.e.  t1t2 = 2. 2

point of intersection of tangents at P and Q on the parabola.

4. The orthocentre of the triangle formed by three tangents to Co-normal Points the parabola lies on the directrix. Any three points on a parabola normals at which pass through a common point are called co-normal points

Equations of Normal in Different Forms

Remarks

This implies that if three normals are drawn through a point 1. Point Form  The equation of the normal to the parabola (h, k), then their slopes are the roots of the cubic : 2 y = 4ax at a point (x1, y1) is k = mh – 2am – am3 y (i) The sum of the slopes of the normals at co-normal points y – y1 = – 1 (x – x1). 2a is zero, i.e. m + m + m = 0. 1

2

3

(ii) The sum of the ordinates of the co-normal points is zero (i.e. – 2am1 – 2am2 – 2am3 = – 2a (m1 + m2 + m3) = 0). (iii) The centroid of the triangle formed by the co-normal 3 y + tx = 2at + at . points lies on the axis of the parabola [the vertices of the triangle formed by the co-normal points are (am 21, –2am1), 3. Slope Form  The equation of normal to the parabola 2 (a m22 , – 2am2) and (am 32, – 2am3). Thus, y-coordinate of y = 4ax in terms of slope ‘m’ is the centroid becomes 3 y = mx – 2am – am . − 2a (m1 + m2 + m3 ) − 2a Note: The coordinates of the point of contact are (am2, =  × 0 = 0. 3 3 – 2am). Hence, the centroid lies on the x-axis, i.e. axis of the Condition for Normality  The line y = mx + c is a normal parabola.] to the parabola (iv) If three normals drawn to any parabola y2 = 4ax from a y2 = 4ax if c = – 2am – am3. given point (h, k) be real, then h > 2a.

2. Parametric Form  The equation of the normal to the parabola y2 = 4ax at the point (at 2, 2at) is

Point of Intersection of Normals

Position of a point with respect to a parabola

The point of intersection of normals drawn at two different points of contact P (a t12 , 2at1) and Q (a t22 , 2at2) on the parabola The point (x1, y1) lies2 outside, on or inside the parabola y2 = 4ax according as y1 – 4ax1 >, = or < 0, respectively. y2 = 4ax is

Two tangents can be drawn from a point to a parabola. The two tangents are real and distinct or coincident or imaginary according as the given point lies outside, on or inside the parabola.

The polar of a point P (x1, y1) with respect to the parabola y2 = 4ax is T = 0 where T ≡ yy1 – 2a (x + x1).

Remarks

1. Polar of the focus is the directrix. 2. Any tangent is the polar of its point of contact. The equation of the pair of tangents drawn from a point 3. Pole of a given line lx + my + n = 0 with respect to the P (x1, y1) to the parabola y2 = 4ax is SS1 = T2, parabola y2 = 4ax is  n − 2am  2 2 where S ≡ y – 4ax, S1 ≡ y1 – 4ax1 and T ≡ yy1 – 2a (x + x1)  , . l l  4. Pole of the chord joining (x1, y1) and (x2, y2) is

Equation of the Pair of Tangents

 y1 y2 y1 + y2  . ,   2   4a

5. The point of intersection of the polars of two points Q and R is the pole of QR. 6. If the polar of P (x1, y1) passes through Q (x2, y2), then the Chord of Contact polar of Q will pass through P and such points are said to be conjugate points. The equation of chord of contact of tangents drawn from a point P (x1, y1) to the parabola y2 = 4ax is T = 0 where T ≡ yy1 – 2a  7. If the pole of a line ax + by + c = 0 lies on the another line a1x + b1 y + c1 = 0 then the pole of the second line will lie on (x + x1). the first and such lines are said to be conjugate lines. 8. The locus of poles of focal chord of the parabola y2 = 4ax is the directrix.

Diameter of a Parabola Chord with a Given Mid Point The equation of the chord of the parabola y2 = 4ax with P (x1, y1) as its middle point is given by T = S1

2 where T ≡ yy1 – 2a (x + x1) and S1 ≡ y1 – 4ax.

Diameter of a conic is the locus of middle points of a series of its parallel chords.

Equation of Diameter of a Parabola The equation of the diameter bisecting chords of slope m of the 2a . parabola y2 = 4ax is y = m

Pole and Polar Let P be a given point. Let a line through P intersect the parabola at two points A and B. Let the tangents at A and B intersect at Q. The locus of point Q is a straight line called the polar of point P with respect to the parabola and the point P is called the pole of the polar.

Prepositions on the Parabola

1. The tangent at any point P on the parabola bisects the angle between the focal chord through P and the perpendicular from P on the directrix. 2. The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus. 3. Tangents at the extremities of any focal chord intersect at right angles on the directrix. 4. Any tangent to a parabola and the perpendicular on it from the focus meet on the tangent at the vertex.

447

Equation of Polar of a Point

Conic Sections (Parabola, Ellipse and Hyperbola)

Number of tangents drawn from a point to a parabola

448

Objective Mathematics

1. Symmetry (a) On replacing y by – y, the above equation remains unAn ellipse is the locus of a point which moves in a plane so that changed. So, the curve is symmetrical about x-axis the ratio of its distance from a fixed point (called focus) and a (b) On replacing x by – x, the above equation remains unfixed line (called directrix) is a constant which is less than one. changed. So, the curve is symmetrical about y-axis This ratio is called eccentricity and is denoted by e. For an el2. F  oci  If S and S′ are the two foci of the ellipse and their colipse, e < 1. ordinates are (ae, o) and (– ae, 0) respectively, then distance between foci is given by SS ′ = 2ae.

ELLIPSE

3. D  irectrices If ZM and Z ′ M′ are the two directrices of the a a ellipse and their equations are x = and x = − respece e tively, then the distance between directrices is given by 2a . ZZ ′ = e 4. A  xes  The lines AA′ and BB′ are called the major axis and minor axis respectively of the ellipse. Let S be the focus, QN be the directrix and P be any point The length of major axis = AA′ = 2a PS The length of minor axis = BB′ = 2b = e  or  PS = e PN, on the ellipse. Then, by definition, PN 5. C  entre  The point of intersection C of the axes of the ele < 1, where PN is the length of the perpendicular from P on the lipse is called the centre of the ellipse. All chords, passing directrix QN. through C are bisected at C. An Alternate Definition  An ellipse is the locus of a point 6. V  ertices  The end points A and A′ of the major axis are that moves in such a way that the sum of its distances from two known as the vertices of the ellipse fixed points (called foci) is constant. A ≡ (a, 0) and A′ ≡ (– a, 0)

EQUATION OF AN ELLIPSE IN STANDARD 7. F ocal Chord  A chord of the ellipse passing through its focus is called a focal chord. FORM The standard form of the equation of an ellipse is:

x2 y 2 = 1 (a > b), + a 2 b2 where a and b are constants.

8. O  rdinate and Double Ordinate  Let P be a point on the ellipse. From P, draw PN ⊥ AA′ (major axis of the ellipse) and produce PN to meet the ellipse at P ′. Then PN is called an ordinate and PNP ′ is called the double ordinate of the point P. 9. L  atus  Rectum  If LL ′ and NN ′ are the latus rectum of the ellipse, then these lines are ⊥ to the major axis AA ′, passing through the foci S and S ′ respectively.

 b2  L ≡  ae,  ,  a 

 − b2  L′ ≡  ae, , a  

  b2  − b2  N ≡  − ae,  , N′ ≡  − ae, a a    Length of latus rectum = LL ′ =

2b 2 = NN ′. a

a

10. By definition, SP  =  ePM = e   − x  = a – ex e 



a  S′ P = e   + x  = a + ex. e Thus implies that distances of any point P (x, y) lying on the ellipse from foci are : (a – ex) and (a + ex). In other words

   and

Some Terms and Properties Related to an Ellipse

SP + S ′ P = 2a i.e., sum of distances of any point P (x, y) lying on the ellipse from foci is constant.

A sketch of the locus of a moving point satisfying the equax2 y 2  ccentricity of the Ellipse  Since, SP = ePM, theretion 2 + 2 = 1 (a > b), has been shown in the figure given 11. E a b fore above. SP2  =  e2 PM2

(x – ae)2 + y2 = x + a e – 2aex + y 2

2 2

2

=

2

x2 y 2 =1 + a 2 b2

(a – ex)2 a2 – 2aex + e2x2

x (1 – e ) + y = a (1 – e ) x2 y2 = 1. + 2 2 a a (1 − e 2 ) 2

2

2

2

2

449

 a (x – ae)2 + (y – 0)2 = e2  − x  e 

for all real values of θ. Thus, x = a cos θ, y = b sin θ are the parametric equations of the ellipse

x2 y 2 = 1, where the + a 2 b2

parameter 0 ≤ θ < 2π. Hence the coordinates of any point on the ellipse x2 y 2 x2 y 2 =1 + On comparing with 2 + 2 = 1, we get a 2 b2 a b may be taken as (a cosθ, b sin θ). This point is also called b2 the point ‘θ’. b2 = a2 (1 – e2) or e = 1 − 2 . a The angle θ is called the eccentric angle of the point (a cosθ, 12. Auxiliary Circle The circle drawn on major axis AA′ bsin θ) on the ellipse. as diameter is known as the Auxiliary circle. 14. Equation of Chord The equation of the chord joining the points P ≡ (a cosθ1, bsin θ1) and Q ≡ (acos θ2 , bsin θ2) is

x  θ + θ2  y  θ + θ2  cos  θ1 − θ2  + sin  1 cos  1 =  .  2  b  2   2  a

POSITION OF A POINT wITh RESPECT TO AN ELLIPSE The point P (x1, y1) lies outside, on or inside the ellipse

Let the equation of the ellipse be equation of its auxiliary circle is:

x2 y 2 2 2 = 1. Then the = 1 according as x1 + y1 − 1 > 0, = 0 or < 0. + a 2 b2 2 2 a b

x2 y 2 + a 2 b2

CONDITION FOR TANgENCY AND POINTS OF CONTACT

x2 + y2 = a2.

Let Q be a point on auxiliary circle so that QM, perpendicular to major axis meets the ellipse at P. The points P and Q are The condition for the line y = mx + c to be a tangent to the ellipse called as corresponding point on the ellipse and auxiliary circle x 2 y 2 = 1 is that c2 = a2m2 + b2 and the coordinates of the + respectively. a 2 b2 The angle θ is known as eccentric angle of the point P on points of contact are the ellipse.   a 2m b2 It may be noted that the CQ and not CP is inclined at θ with ,∓ ±  2 2 2 2 2 2 a m +b a m +b   x-axis. 13. Parametric Equation of the Ellipse The coordiTwo standard forms of the ellipse are shown below along nates x = a cos θ and y = b sin θ satisfy the equation with their properties: Two Standard Forms of the Ellipse Standard equation

x2 y 2 = 1 (a > b) + a 2 b2

(Horizontal Form of an Ellipse)

x2 y 2 = 1 (a, b) + b2 a 2

(Vertical Form of an ellipse)

Shape of the Ellipse

(Continued on following page)

Conic Sections (Parabola, Ellipse and Hyperbola)

or

450

(Continued)

Two Standard Forms of the Ellipse

Objective Mathematics

Centre

(0, 0)

(0, 0)

Equation of major axis

y=0

x=0

Equation of minor axis

x=0

y=0

Length of major axis

2a

2a

Length of minor axis

2b

2b

Foci

(± ae, 0)

(0, ± ae)

Vertices

(± a, 0)

(0, ± a)

Equation of directrices

Eccentricity

x=±

a e

y=±

a e

e=

a 2 − b2 a2

e=

a 2 − b2 a2

Length of latus rectum

2b 2 a

2b 2 a

Ends of latra-recta

 b2  ± ae ± ,  a 

 b2   ± a , ± ae 

Parametric coordinates

(a cos θ, b sin θ)

(a cos θ, b sin θ)

Focal raddis

P = a – ex1 and S′ P = a + ex1

SP = a – ey1 and S′ P = a + ey1.

Sum of focal radii SP + S′ P =

2a

2a

Distance between foci

2ae

2ae

2a e

2a e

Distance between directrices Tangents at the vertices

x=±a

EQUATION OF TANgENT IN DIFFERENT FORMS

y=±a

The coordinates of the points of contact are

  2 2 2 1. Point Form The equation of the tangent to the ellipse a m +b  2 2 x y = 1. + Note: a 2 b2 xx1 yy1 1. Number of Tangents Drawn From a Point Two tangents at the point (x1, y1) is 2 + 2 = 1 a b can be drawn from a point to an ellipse. The two tangents are real and distinct or coincident or imaginary according as the Note: The equation of tangent at (x1, y1) can also be obtained given point lies outside, on or inside the ellipse. x + x1 y + y1 by replacing x2 by xx1, y2 by yy1, x by , y by , and 2. Director Circle It is the locus of points from which perpen2 2 dicular tangents are drawn to the ellipse. xy + x y xy by 1 1 . This method is used only when the equation The equation of Director Circle of the ellipse 2 x2 y 2 of ellipse is a polynomial of second degree in x and y. = 1 is x2 + y2 = a2 + b2. + a 2 b2 2. Parametric Form The equation of the tangent to the el- 3. The product of perpendiculars from the foci on any tangent to x2 y 2 lipse 2 + 2 = 1 at the point (a cos θ, b sin θ) is x2 y 2 the ellipse 2 + 2 = 1 is equal to b2. a b x y a b cos θ + sin θ = 1. a b EQUATION OF NORMAL IN DIFFERENT 3. Slope Form The equation of tangent to the ellipse x2 y 2 = 1 in terms of slope ‘m’ is + a 2 b2 y = mx ± a 2 m 2 + b 2

 a 2m ,∓ ± a 2m2 + b2 

b2

FORMS

1. Point Form The equation of the normal to the ellipse a 2 x b2 y x2 y 2 − = 1 at the point (x , y ) is = a2 – b2. + 1 1 x1 y1 a 2 b2

3. Slope Form  The equation of normal to the ellipse

x2 y 2 = 1 in terms of slope ‘m’ is + a 2 b2 m (a 2 − b 2 ) y = mx ± a 2 + b2m2 Notes:

where

1. The coordinates of the points of contact are

  a2 mb 2 ,± ±  2 2 2 2 2 2 a +b m a +b m  

T≡

xx1 yy1 x2 y 2 + 2 − 1 and S1 ≡ 12 + 12 –1. 2 a b a b

CHORD OF CONTACT

The equation of chord of contact of tangents drawn from a point x2 y 2 P (x1, y1) to the ellipse 2 + 2 = 1 is T = 0, where a b 2. Condition for Normality  The line y = mx + c is a norxx1 yy1 T ≡ 2 + 2 −1. mal to the ellipse a b 2 2 2 2 2 2 m ( a − b ) x y . 2 + 2 = 1 if c2 = (a 2 + b 2 m 2 ) a b

3. Number of Normals  In general, four normals can be drawn to an ellipse from a point in its plane, i.e. there are four points on the ellipse, the normals at which it will pass through a given point. These four points are called the conormal points.

POLE AND POLAR

4. If α, β, γ, δ are the eccentric angles of the four points on Let P be a given point. Let a line through P intersect the ellipse the ellipse such that the normals at these points are concur- at two points A and B. Let the tangents at A and B intersect at Q. rent, then (α + β + γ + δ) is an odd multiple of π. The locus of point Q is a straight line called the polar of point P 5. If α, β, γ are the eccentric angles of three points on the w.r.t. the ellipse and the point P is called the pole of the polar. ellipse

x2 y 2 = 1, the normals at which are concurrent, + a 2 b2

then sin (α + β) + sin (β + γ) + sin (γ + α) = 0.

EQUATION OF THE PAIR OF TANGENTS The equation of the pair of tangents drawn from a point

Equation of Polar of a Point x2 y 2 The polar of a point P (x1, y1) w.r.t. the ellipse 2 + 2 = 1 is a b xx yy T = 0, where T ≡ 21 + 21 − 1 . a b Notes:

P (x1, y1) to the ellipse SS1 = T2

x2 y 2 x2 y 2 – 1, S1 ≡ 12 + 12 – 1 + a 2 b2 a b xx1 yy1 T ≡ 2 + 2 −1 a b

where S ≡ and

x2 y 2 = 1 is + a 2 b2

1. Polar of the focus is the directrix 2. Any tangent is the polar of its point of contact. 3. Pole of a given line lx + my + n = 0 w.r.t. the ellipse

 − a 2l − b 2 m  x2 y 2 ,  . = 1 is + n  a 2 b2  n

4. If the polar of P (x1, y1) passes through Q (x2, y2), then the

polar of Q will pass through P and such points are said to be conjugate points

451

Conic Sections (Parabola, Ellipse and Hyperbola)

2.  Parametric Form  The equation of the normal to the CHORD WITH A GIVEN MID-POINT x2 y 2 ellipse 2 + 2 = 1 at the point (a cosθ, b sin θ) is x2 y 2 a b The equation of the chord of the ellipse 2 + 2 = 1 with a b ax secθ – by cosec θ = a2 – b2. P (x1, y1) as its middle point is given by T = S1 ax by − or = a2 – b2. cos θ sin θ

452

Objective Mathematics

5. If the pole of a line lx + my + n = 0 lies on the anoth- a fixed line (called directrix) is a constant which is greater then er line l ′ x + m ′ y + n ′ = 0, then the pole of the second one. This ratio is called eccentricity and is denoted by e. For a line will lie on the first and such lines are said to be hyperbola,e > 1. conjugate lines. 6. The point of intersection of any two lines is the pole of the line joining the pole of the two lines.

DIAMETER OF AN ELLIPSE The locus of the middle points of a system of a parallel chords of an ellipse is called a diameter of the ellipse. Equation of a Diameter The equation of the diameter bisecting chords of slope m of the ellipse

is

x2 y 2 = 1 + a 2 b2 b2 y = – 2 x. am

Note:  Diameter of an ellipse always passes through its centre. Thus a diameter of an ellipse is its chord passing through its centre. CONJUGATE DIAMETERS Two diameters of an ellipse are said to be conjugate diameters if each bisects the chords parallel to the other.

Let S be the focus, QN be the directrix and P be any point on the hyperbola. Then, by definition,

PS = e  or  PS = e PN, e > 1, PN

where PN is the length of the perpendicular from P on the directrix QN. An Alternate Definition  A hyperbola is the locus of a point which moves in such a way that the difference of its distances from two fixed points (called foci) is constant.

EQUATION OF A HYPERBOLA IN STANDARD FORM The general form of standard hyperbola is:

x2 y 2 − = 1, a 2 b2 1. Major and minor axes of an ellipse is also a pair of conju- where a and b are constants. gate diameters. 2. If m1 and m2 be the slopes of the conjugate diameters 2 2 2 = 1, then m1m2 = −b . of an ellipse x + y 2 2 2 a b a 3. The eccentric angles of the ends of a pair of conjugate diameters differ by a right angle. Notes:



i.e. if PCP ′ and QCQ ′ is a pair of conjugate diameters and if eccentric angle of P is θ, then eccentric angles of Q, P ′, π Q ′ (proceeding in anticlockwise direction) will be θ + , 2 3π respectively. θ + π and θ + 2

Some Terms and Properties Related to a Hyperbola A sketch of the locus of a moving point satisfying the equation x2 y 2 − = 1, has been shown in the figure given above. a 2 b2 1.  Symmetry  Since only even powers of x and y occur in the above equation , so the curve is symmetrical about both the axes. Hyperbola 2.  Foci  If S and S′ are the two foci of the hyperbola and A hyperbola is the locus of a points which moves in a plane so their coordinates are (ae, 0) and (­– ae, 0) respectively, then that the ratio of its distances from a fixed point (called focus) and distance between foci is given by SS′ = 2ae.

the curve meets the line joining the foci S and S ′, are called the vertices of the hyperbola.

12.  Equation  of  Chord  The equation of the chord joining the points

7. Focal Chord  A chord of the hyperbola passing through its focus is called a focal chord.



x2 y 2 =1 − a 2 b2

y = b tan θ are the 4.  Axes  The lines AA ′ and BB ′ are called the transverse for all real values of θ therefore, x = a sec θ, x2 y 2 axis and conjugate axis respectively of the hyperbola. parametric equations of the hyperbola 2 − 2 = 1, where the a b parameter 0 ≤ θ < 2π. The length of transverse axis = AA ′ = 2a Hence, the coordinates of any point on the hyperbola The length of conjugate axis = BB ′ = 2b. x2 y 2 − 2 = 1 may be taken as (a sec θ, b tan θ). This point is also 2 b 5.  Centre  The point of intersection C of the axes of the a hyperbola is called the centre of the hyperbola. All chords, called the point ‘θ’. The angle θ is called the eccentric angle of the point passing through C, are bisected at C. (a secθ, b tan θ) on the hyperbola. 6.  Vertices  The points A ≡ (a, 0) and A ′ ≡ (– a, 0), where

8. Focal Distances of a Point  The difference of the focal distances of any point on the hyperbola is constant and equal to the length of the transverse axis of the hyperbola. If P is any point on the hyperbola, then

S′ P – SP = 2a = Transverse axis.



 b  L ≡  ae,  ,  a 

 −b  L′ ≡  ae, , a  



 b2  N ≡  − ae,  , a 

 − b2  N′ ≡  − ae,  a  

2

Length of latus rectum = LL′ =

2b 2 = NN′. a

10. Eccentricity of the Hyperbola  We know that

SP = e PM  or  SP2 = e2 PM2

or

(x – ae)2 + (y – 0)2 = e2   x − a    e 



2

(x – ae)2 + y2 = (ex – a)2





x2 + a2e2 – 2aex + y2 = e2x2 – 2aex + a2





x2 (e2 – 1) – y2 = a2 (e2 – 1)





x2 y2 − 2 2 = 1. 2 a a (e − 1)

On comparing with

x2 y 2 = 1 , we get − a 2 b2

   b2 = a2  (e2 – 1)  or  e =

1+

b2 a2

x cos  θ1 − θ2  – y sin  θ1 + θ2  = cos  θ1 + θ2  .       a b  2   2   2  or x a sec θ1 a sec θ2



9.  Latus Rectum  If LL′ and NN ′ are the latus rectum of the hyperbola then these lines are perpendicular to the transverse axis AA′, passing through the foci S and S ′ respectively. 2



P ≡ (a sec θ1, ,b tanθ1) and Q ≡ (a sec θ2, b tan θ2) is

y 1 b tan θ1 1 = 0 b tan θ2 1

CONJUGATE  HYPERBOLA The hyperbola whose transverse and conjugate axes are respectively the conjugate and transverse axes of a given hyperbola is called the conjugate hyperbola of the given hyperbola. The conjugate hyperbola of the hyperbola x2 y 2 =1 − a 2 b2 is

–

 x2 y 2  x2 y 2 + 2 = 1   i.e. 2 − 2 = − 1 2 a b a b  

453

11.  Parametric Equations of the Hyperbola  Since coordinates x = a sec θ and y = b tan θ satisfy the equation

Conic Sections (Parabola, Ellipse and Hyperbola)

3. Directrices  ZM and Z′ M′ are the two directrices of the a a hyperbola and their equations are x = and x = – respectively, e e 2a then the distance between directrices is given by zz ′ = . e

454

Properties of hyperbola and conjugate are given below in the table:

Objective Mathematics

Properties of Hyperbola and its Conjugate hyperbola Standard equation

Conjugate hyperbola

x2 y 2 − =1 a 2 b2

2 2 − x2 y 2 + 2 = 1 or x − y = – 1 2 a b a 2 b2

centre

(0, 0)

(0, 0)

Equation of transverse axis

y=0

x=0

Equation of conjugate axis

x=0

y=0

Length of Transverse axis

2a

2b

Length of Conjugate axis

2b

2a

Foci

(± ae, 0)

(0, ± be)

Equation of directrices

x=±

Vertices

(± a, 0)

Eccentricity

e=

Length of latus rectum

a e

y=±

b e

(0, ± b)

a 2 + b2 a2

a 2 + b2 b2

e=

2b 2 a

2a 2 b

Parametrc Coordinates

(a sec θ, b tan θ)

(bsec θ, a tan θ)

Focal radii

SP = ex1 – a and S′ P = ex1 + a

SP = ey1 – b and S′ P = ey1 + b

Difference of focal radii (S′ P – SP)

2a

2b

Tangents at the vertices

x=±a

y=±b

Note:

The equation of tangent at (x1, y1) can also be obtained x + x1 y + y1 , y by and by replacing x2 by xx1, y2 by yy1, x by 2 2 The point P (x1, y1) lies outside, on or inside the hyperbola xy by xy1 + x1 y . This method is used only when the equation of 2 x2 y 2 x2 y 2 − 2 = 1 according as 12 − 12 – 1 > 0, = 0 or < 0. 2 hyperbola is a polynomial of second degree in x and y. a b a b

POSITION OF A POINT wITh RESPECT TO A hYPERbOLA

2. Parametric Form The equation of the tangent to the hyx2 y 2 perbola 2 − 2 = 1 at the point (a sec θ, b tan θ) is a b x sec θ – y tan θ = 1. The condition for the line y = mx + c to be a tangent to the hya b 2 2 perbola x − y = 1 is that c2 = a2m2 – b2 and the coordinates of 3. Slope Form The equation of tangent to the hyperbola a 2 b2 x2 y 2 the points of contact are − = 1 in terms of slope ‘m’ is a 2 b2   a 2m b2 y = mx ± a 2 m 2 − b 2 . ±  ,± 2 2 2 2 2 2   − − a m b a m b The coordinates of the points of contact are  

CONDITION FOR TANgENCY AND POINTS OF CONTACT

 ±  

Equation of Tangent in Different Forms 1. Point Form The equation of the tangent to the hyperbola x 2 y 2 = 1 at the point (x , y ) is − 1 1 a 2 b2 xx1 yy1 − 2 = 1. a2 b

a 2m a 2m2 − b2

,±

  a 2 m 2 − b 2  b2

Notes: 1. Number of Tangents From a Point Two tangents can be drawn from a point to a hyperbola. The two tangents are real and distinct or coincident or imaginary according as the given point lies outside, on or inside the hyperbola.

x2 y 2 − = 1 is x2 + y2 = a2 – b2. a 2 b2

xx1 yy1 x2 y 2 − 2 − 1 and S1 ≡ 12 − 12 –1. 2 a b a b

CHORD OF CONTACT

The equation of chord of contact of tangents drawn from a 2 2 Equations of Normal in Different Forms point P (x1, y1) to the hyperbola x − y = 1 is T = 0, where 2 2 a b 1.  Point Form  The equation of the normal to the hyperbola T ≡ xx1 − yy1 − 1 . a2 b2 x2 y 2 − = 1 at the point (x , y ) is 1 1 a 2 b2 a 2 x b2 y + = a2 + b2. x1 y1

POLE AND POLAR

2 2 The polar of a point P (x1, y1)w.r.t. the hyperbola x − y = 1 is 2 2 a b 2.  Parametric Form  The equation of the normal to the hyxx yy T = 0, where T ≡ 21 − 21 − 1 2 2 x y a b perbola 2 − 2 = 1 at the point (a sec θ, b tan θ) is a b Notes: ax by 2 2 + =a +b. 1. Pole of a given line lx + my + n = 0 w.r.t. the hyperbola sec θ tan θ 2 2 x 2 y 2 = 1 is  − a l , − b m  . 3.  Slope Form  The equation of normal to the hyperbola   − n  a 2 b2  n x2 y 2 − = 1 in terms of slope ‘m’ is 2. Polar of the focus is the directrix a 2 b2 3. Any tangent is the polar of its point of contact m (a 2 + b 2 ) 4. If the polar of P (x1, y1) passes through Q (x2, y2) then the . y = mx ± a 2 − b2m2 polar of Q will pass through P and such points are said to Notes: be conjugate points. 5. If the pole of a line lx + my + n = 0 lies on the another line 1. The coordinates of the points of contact are l ′ x + m ′ y + n ′ = 0, then the pole of the second line will lie   on the first and such lines are said to be conjugate lines. a2 mb 2 ±  ,∓ 2 2 2 2 2 2   a −b m a −b m   EQUATION OF A DIAMETER OF A

2. Number of Normals  In general, four normals can be HYPERBOLA drawn to a hyperbola from a point in its plane, i.e. there The equation of the diameter bisecting chords of slope m of the are four points on the hyperbola, the normals at which will x2 y 2 b2 pass through a given point. These four points are called the hyperbola 2 − 2 = 1 is y = 2 x . a b am co-normal points. 3. Tangent drawn at any point bisects the angle between the lines joining the point to the foci, whereas normal bisects CONJUGATE DIAMETERS the supplementary angle between the lines. Two diameters of a hyperbola are said to be conjugate diameters if each bisects the chords parallel to the other. If m1 and m2 be the EQUATION OF THE PAIR OF TANGENTS x2 y 2 slopes of the conjugate diameters of a hyperbola 2 − 2 = 1, a b The equation of the pair of tangents drawn from a point b2 then m1m2 = 2 . 2 2 x y a P (x1, y1) to the hyperbola 2 − 2 = 1 is a b ASYMPTOTES OF HYPERBOLA SS1 = T2 2 2 2 2 2 2 x y x y The lines x − y = 0 i.e. y = ± bx are called the asymptotes where S ≡ 2 − 2 – 1, S1 ≡ 12 − 12 – 1 2 2 a b a a b a b of the hyperbola. xx yy The curve comes close to these lines as x → ∞ or x → and T ≡ 21 − 21 − 1 . – ∞ but never meets them. In other words, asymptote to a curve a b touches the curve at infinity.

CHORD WITH A GIVEN MID POINT

Notes: x2 y 2 x2 y 2 The equation of the chord of the hyperbola 2 − 2 = 1 with 1. The angle between the asymptotes of 2 − 2 = 1 is a b a b b 2tan– 1   . P (x1, y1) as its middle point is given by T = S1 where a

455

T≡

Conic Sections (Parabola, Ellipse and Hyperbola)

2. Director Circle  It is the locus of points from which ⊥ tangents are drawn to the hyperbola. The equation of director circle of the hyperbola

456

Objective Mathematics

2. Asymptotes are the diagonals of the rectangle passing Properties of Rectangular Hyperbola, through A, B, A ′, B ′ with sides parallel to axes. x2 – y2 = a2 3. A hyperbola and its conjugate hyperbola have the same 1. The equations of asymptotes of the rectangular hyperbola asymptotes. are y = ± x. 4. The asymptotes pass through the centre of the hyperbola. 2.  The transverse and conjugate axes of a rectangular hyper 5. The bisector of the angle between the asymptotes are the bola are equal in length. coordinate axes. 6. The product of the perpendiculars from any point on the 3. Eccentricity, e = b2 1+ 2 = 2 . 2 2 x y a hyperbola 2 − 2 = 1 to its asymptotes is a constant a b

Properties of Rectangular Hyperbola xy = c2

a 2b 2 equal to 2 . a + b2

1. Equation of the chord joining ‘t1’ and ‘t2’ is:

x + yt1t2 – c (t1 + t2) = 0 7. Any line drawn parallel to the asymptote of the hyperbola 2. Equation of tangent at (x1, y1) is: would meet the curve only at one point. 8. A hyperbola and its conjugate hyperbola have the same x y xy1 + x1 y = 2c2 or  + = 2. asymptotes. x y 1

RECTANGULAR HYPERBOLA If asymptotes of the standard hyperbola are perpendicular to each other, then it is known as Rectangular Hyperbola. Thus, –1

2 tan

π b = 2 a



⇒ b = a  or  x2 – y2 = a2

1

3. Equation of tangent at ‘t’ is: x + yt = 2c. t 4. Point of intersection of tangents at ‘t1’ and ‘t2’ is:

is general form of the equation of the rectangular hyperbola. If we take the coordinate axes along the axsymptotes of a rectangular hyperbola, then equation of rectangular hyperbola becomes : xy = c2, where c is any constant. In parametric form , the equation of rectangular hyperbola is x = ct, y = c/t, where t is the parameter. The point (ct, c/t) on the hyperbola xy = c2 is generally re- ferred as the point ‘t’.

 2ct1t2 2c  ,    t1 + t2 t1t2  5. Equation of normal at (x1, y1) is: xx1 – yy1 =x12 – y12. 6. Equation of normal at ‘t’ is: xt3 – yt – ct4 + c = 0 7. The equation of the chord of the hyperbola xy = c2 whose middle point is (x1, y1) is: T = S1 i.e. xy1 + x1 y = 2x1 y1. 8. The slope of the tangent at the point (ct, c/t) is – 1/t2, which is always negative. Hence tangents drawn at any point to xy = c2 would always make an obtuse angle with the xaxis. 9. The slope of the normal at the point (ct, c/t) is t2, which is always positive. Hence normal drawn to xy = c2 at any point would always make an acute angle with the x-axis.

MULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. The focal distance of a point on a parabola y 2 = 8x is 4, its coordinates are (a) (2, 4) (c) (– 2, 4)

(b) (2, – 4) (d) (– 2, – 4)

4. The normal to the parabola y 2 = 8x at (2, 4) meets the parabola again at (a) (18, 12) (c) (– 18, 12)

2. The equation of the parabola, whose focus is (3, – 4) and directrix is the line x + y – 2 = 0, is (a) x2 + 2xy + y2 – 8x + 20y + 46 = 0 (b) x2 – 2xy + y2 – 8x + 20y + 46 = 0 (c) x2 – 2xy + y2 + 8x – 20y + 46 = 0 (d) None of these

(b) (18, – 12) (d) None of these

5. The equation of the tangent to the parabola y 2 = 16x inclined at an angle of 60º to x-axis, is (a) 3x – (c)

3 y + 4 = 0

3 x – y + 4 = 0

(b) 3x +

3y+4=0

(d) None of these

6. The equation of the tangent to the parabola y 2 = 16x, perpendicular to the line 2x – y + 5 = 0 is 3. The equation of the tangent to the parabola y = 6x at the point whose ordinate is 6, is (a) 2x – y + 2 = 0 (b) 2x + y + 2 = 0 (a) x + 2y + 6 = 0 (b) 2x – y + 6 = 0 (c) x + 2y – 16 = 0 (d) x + 2y + 16 = 0 (c) x – 2y + 6 = 0 (d) None of these 2

(a) 2x + y – 12 = 0 (c) x + 2y – 12 = 0

(b) 2x – y – 12 = 0 (d) None of these

8. The condition that the line the parabola y 2 = 4ax is (a) p3 = 2ap2 + aq2 (c) q3 = 2ap2 + aq2

x y = 1 be a normal to + p q

(b) p3 = 2aq2 + ap2 (d) None of these

9. The value of k for which the line x + y + 1 = 0 touches the parabola y 2 = kx is (a) – 4 (c) 2

(b) 4 (d) – 2

1 (a)  0,   2

1 (b)  , 0  2 

−1 (c)  0,   2 

(d) None of these

19. The locus of the poles of normal chords of the parabola y 2 = 4ax is (a) (x + 2a) y2 – 4a3 = 0 (b) (x + 2a) y2 + 4a2 = 0 (c) (x + 2a) y2 + 4a3 = 0 (d) None of these 20. The locus of the poles of tangents to the parabola y 2 = 4ax w.r.t. the parabola y 2 = 4bx is the parabola

2 2 (b) y2 = 4b x (a) y2 = 4a x b a 2 2 4 a 4 2 2 a a (c) x = (d) x = b y (a) c = am – (b) c = m – y b a m m a 21. If the polar of any point w.r.t. the parabola y 2 = 4ax (c) c = am + (d) None of these m touches the circle x2 + y 2 = 4a2, then the locus of the point is the curve 11. A straight line touches the circle x2 + y 2 = 2a2 and the parabola y 2 = 8ax. The equation of the line is (a) x2 – y2 = 2a2 (b) x2 – y2 = a2 (c) x2 – y2 = 3a2 (d) x2 – y2 = 4a2 (a) y = x + 2a (b) y = – x – 2a (c) y = x – 2a (d) None of these 22. Chords of the parabola y 2 = 4ax are drawn at fixed

10. The straight line y = mx + c touches the parabola y 2 = 4a (x + a) if

12. The equation of the common tangent to the parabolas y 2 = 4ax and x2 = 4by is given by (a) xa1/3 + yb1/3 + a2/3 b2/3 = 0 (b) xb1/3 + ya1/3 + a2/3 b2/3 = 0 (c) xa1/3 + yb1/3 – a2/3 b2/3 = 0 (d) None of these

13. The equation of the normal to the parabola y 2 = 4x, which passes through the point (3, 0), is (a) y = 0 (c) y = – x + 3

(b) y = x – 3 (d) All of these

14. The point of intersection of two perpendicular tangents to a parabola lies on the (a) axis (c) directrix

(b) tangent at the vertex (d) None of these

15. The angle between the two tangents drawn from the point (1, 4) to the parabola y 2 = 12x is 1 1 (b) tan– 1   (a) tan– 1   3 2 (c) tan– 1 (2)

(d) None of these

16. If two tangents to the parabola y 2 = 4ax make angles θ1 and θ2 with x-axis where tan 2 θ1 + tan 2 θ2 = c, then the locus of their point of intersection is (a) y2 – ax = 2cx2 (c) y2 + 2ax = cx2

(b) y2 – 2ax = cx2 (d) None of these

distance ‘a’ from the focus. The locus of their poles w.r.t. the parabola is (a) y2 = 4x (2a + x) (c) y2 = 4x (2a – x)

(b) y2 = 2x (2a + x) (d) None of these

23. The locus of the mid points of chords of the parabola y 2 = 4ax, which pass through a fixed point (h, k), is given by (a) y2 – 2ax + ky + 2ah = 0 (b) y2 – 2ax – ky – 2ah = 0 (c) y2 – 2ax – ky + 2ah = 0 (d) None of these 24. The locus of the mid points of the focal chords of the parabola y 2 = 4ax is another parabola whose vertex is given by (a) (a, 0) (c) (– a, 0)

(b) (0, a) (d) None of these

25. The locus of mid points of chords of the parabola y 2 = 4ax which touch the circle x2 + y 2 = a2, is given by (a) ( y2 – 2ax)2 = a2 ( y2 – 4a2) (b) ( y2 + 2ax)2 = a2 ( y2 + 4a2) (c) ( y2 – 2ax)2 = a2 ( y2 + 4a2) (d) None of these

26. If the point (at2, 2at) be one extremity of a focal chord of the parabola y 2 = 4ax, then the length of the chord is 17. The locus of the poles of the focal chords of a parabola is the 2 2  1  1 (b) a  (a) a  t + t −     (a) axis t t   (b) directrix 2 1 (c) tangent at the vertex (c) 2a   t −  (d) None of these t (d) None of these 

457

18. The pole of the line 2x = y with respect to the parabola y 2 = 2x is

Conic Sections (Parabola, Ellipse and Hyperbola)

7. The equation of the normal to the parabola y 2 = 4x which is parallel to the line y – 2x + 5 = 0 is

458

27. The normal at the point (bt12, 2bt1) on a parabola meets the parabola again in the point (b 22, 2bt2), then

Objective Mathematics

2 t1 2 (c) t2 = t1 – t1

(a) t2 = –t1 –

2   t1 2 (d) t2 = t1 + t1 (b) t2 = –t1 +

28. If the focal distance of a point on a parabola y 2 = 4x is 6, then the coordinates of the point are (a) (5, 2 5 ) (c) (5, 5 )

(b) (5, – 2 5 ) (d) None of these

29. If QQ' is a double ordinate of a parabola y 2 = 9x, then the locus of its point of trisection is (a) y2 = x (c) y2 = 6x

(b) y2 = 3x (d) None of these

30. If three points E, F, G are taken on the parabola y2 = 4ax so that their ordinates are in G.P. , then the tangents at E and G intersect on the (a) directrix (c) ordinate of F

(b) axis (d) None of these

31. The length of the side of an equilateral triangle, inscribed in the parabola y 2 = 8x so that one angular point is at the vertex, is (a) 16 3

(b) 8 3

(c) 4 3 (d) None of these 2 32. For all parabola x + 4x + 4y + 16 = 0, the equation of the axis and the directrix are given by (a) x + 2 = 0, y – 2 = 0 (c) x + 2 = 0, y + 2 = 0

(b) x – 2 = 0, y + 2 = 0 (d) None of these

33. The equation of the parabola having axis parallel to yaxis and which passes through the points (0, 4), (1, 9) and (4, 5), is (a) y = −19 x 2 + 79 x + 4 12 12 − 19 79 (b) y = x2 + x−4 12 12 (c) y = 19 x 2 + 79 x + 4 12 12 (d) None of these 34. The normal chord to a parabola y 2 = 4ax at the point whose ordinate is equal to abscissa subtends an angle θ at the focus where θ = (a) π (b) π 6 4 π π (c) (d) 3 2 35. The locus of the point through which pass three normals to the parabola y 2 = 4ax such that two of them make angles α and β respectively with the axis and tanα tan β = 2, is (a) y2 = 2ax (c) y2 = 4ax

(b) y2 = – 2ax (d) y2 = ax

36. If the ordinates of points P and Q on the parabola y 2 = 12x are in the ratio 1 : 2, then the locus of the point of intersection of the normals to the parabola at P and Q is (a) 343y2 = 12 (x – 6)3 (b) 343y2 = 12 (x + 6)3 (c) 343y2 = – 12 (x – 6)3 (d) None of these 37. If the tangents are drawn from the point (x1, y1) to the parabola y 2 = 4ax, then the length of their chord of contact is 1 ( y12 − 2ax1 ) ( y12 + 4a 2 ) |a| 1 ( y12 − 4ax1 ) ( y12 + 4a 2 ) (b) |a| 1 ( y12 − 4ax1 ) ( y12 + 2a 2 ) (c) |a| (d) None of these

(a)

38. Through the vertex O of a parabola y 2 = 4x, chords OP and OQ are drawn at right angles to one another. The locus of the middle point of PQ is (a) y2 = 2x + 8 (c) y2 = 2x – 8

(b) y2 = x + 8 (d) None of these

39. The curve described parametrically by x = t2 + t + 1, y = t2 – t + 1 represents (a) a pair of straight lines (b) an ellipse (c) a parabola (d) a hyperbola. 40. If (4, 0) is the vertex and y-axis, the directrix of a parabola, then it focus is (a) (8, 0) (c) (0, 8)

(b) (4, 0) (d) (0, 4)

41. If tangents are drawn to the parabola y 2 = 4ax at points whose abscissaes are in the ratio m2 : 1, then the locus of their point of intersection is the curve (a) y2 = (m1/2 – m– 1/2)2 ax (b) y2 = (m1/2 + m– 1/2)2 ax (c) y2 = (m1/2 + m– 1/2)2x (d) None of these 42. If (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then its equation is (a) x2 + 8y = 32 (c) x2 – 8y = 32

(b) y2 + 8x = 32 (d) y2 – 8x = 32.

43. If the tangent to the parabola y 2 = 4ax meets the axis in T and tangent at the vertex A in Y and the rectangle TAYG is completed, then the locus of G is (a) y2 + 2ax = 0 (c) x2 + ay = 0

(b) y2 + ax = 0 (d) None of these

44. The focus of the parabola x2 – 2x – y + 2 = 0 is (a) (1, 0) 5 (c) 1,   4

5 (b)  0,   4 5 (d)  , 1 4 

(a) m = 1 (c) m = 4

(b) m = 2 (d) m = 3

53. The line x – y + 2 = 0 touches the parabola y 2 = 8x at the point (a) (2, – 4) (c) (4, – 4

(b) (1, 2 2 ) (d) (2, 4)

2 ) 54. Two tangents are drawn from the point (– 2, – 1) to the parabola y 2 = 4x. If α is the angle between them, then tan α = (a) 3 (c) 2

(b) 1/3 (d) 1/2

64. The centroid of the triangle formed by joining the feet of the normals drawn from any point to the parabola y 2 = 4ax, lies on (a) axis (c) latus rectum

(b) directrix (d) tangent at vertex

65. The pole of the line lx + my + n = 0 with respect to the parabola y 2 = 4ax is : n 2am  (a)  ,  l  l

n − 2am  (b)  ,  l  l

− n 2am  (c)  ,  l   l

− n − 2am  (d)  ,  l   l

459

Conic Sections (Parabola, Ellipse and Hyperbola)

45. If P1Q1 and P2Q2 are two focal chords of the parabola 55. The slope of the normal at the point (at2, 2at) of the y 2 = 4ax, then the chords P1P2 and Q1Q2 intersect on parabola y 2 = 4ax is the (a) 1/t (b) t (a) directrix (c) – t (d) – 1/t (b) axis 2 56. If the points (au , 2au) and (av 2, 2av) are the extremities (c) tangent at the vertex of a focal chord of the parabola y 2 = 4ax, then (d) None of these (a) uv – 1 (b) uv + 1 = 0 46. Equation of the parabola whose vertex is (– 3, – 2), axis (c) u + v = 0 (d) u ­– v = 0 is horizontal and which passes through the point (1, 2) 57. The equation of the tangent to the parabola y 2 = 16x is which is perpendicular to the line y = 3x + 7 is (a) y2 + 4y + 4x – 8 = 0 (b) y2 + 4y – 4x + 8 = 0 (a) y – 3x + 4 = 0 (b) 3y – x + 36 = 0 (c) y2 + 4y – 4x – 8 = 0 (d) None of these (c) 3y + x – 36 = 0 (d) 3y + x + 36 = 0 47. If tangents are drawn from any point on the line x + 58. The normals to the parabola y 2 = 4ax from the point 2 4a = 0 to the parabola y = 4ax, then their chord of (5a, 2a) are contact subtends an angle θ at the vertex, where θ is equal to (a) y = x – 3a (b) y = – 2x + 12a (c) y = – 3x + 33a (d) y = x + 3a π π (a) (b) 59. Tangents at the extremities of any focal chord of a pa4 3 rabola y 2 = 4ax intersect π (c) (d) None of these (a) at right angles 2 (b) on the directrix 2 48. Axis of the parabola x – 3y – 6x + 6 = 0 is (c) on the tangent at vertex (a) x = ­– 3 (b) y = – 1 (d) None of these (c) x = 3 (d) y = 1 60. The portion of a tangent to a parabola y 2 = 4ax cut off 49. The angle subtended by the double ordinate of length between the directrix and the curve subtends an angle 2a of the parabola y 2 = ax, at the vertex is θ at the focus, where θ = π π (a) (b) (a) π (b) π 4 3 4 3 π (c) π (d) None of these (c) (d) None of these 2 2 50. If the two parabolas y 2 = 4a (x – k1) and x2 = 4a ( y – k2) 61. If the parabola y2 = 4ax passes through the point (1, – 2), always touch each other, k1 and k2 being variable paramthen the tangent at this point is eters, then their point of contact lies on the curve (a) x + y – 1 = 0 (b) x – y – 1 = 0 (a) xy = a2 (b) xy = 2a2 (c) x + y + 1 = 0 (d) x – y – 1 = 0 (c) xy = 4a2 (d) None of these 62. The length of latus rectum of the parabola 51. If a variable circle is described to pass through (a, 0) and 4y2 + 2x – 20y + 17 = 0 is touch the line x + y = 0, then the locus of the centre (a) 3 (b) 6 of the circle represents (c) 1/2 (d) 9 (a) a parabola (b) an ellipse 63. The line y = mx + c touches the parabola x2 = 4ay if (c) pair of straight lines (d) None of these (a) c = –­ am (b) c = – a/m 52. The line y = mx + 1 is a tangent to the parabola y 2 = (c) c = – am2 (d) c = a/m2 4x, if

460

66. If b, k are intercepts of a focal chord of the parabola y 2 = 4ax then k is equal to

Objective Mathematics

ab b−a a (c) b−a

(a)

b b−a ab (d) a−b

(a) P and Q both lie inside the parabola (b) P lies inside whereas Q lies outside the parabola (c) P lies outside whereas Q lies inside the parabola (d) P and Q both lie outside the parabola

(b)

67. The length of the normal chord to the parabola y 2 = 4x which subtends a right angle at the vertex is : (a) 6 3 (c) 2

(b) 3 3 (d) 1

68. If lx + my + n = 0 is a tangent to the parabola x = y then condition of tangency is: 2

(a) l2 = 2mn (c) m2 = 4/n

(b) l = 4m2n2 (d) l2 = 4mn

69. A line PQ meets the parabola y 2 = 4ax in R such that PQ is bisected at R. If the coordinates of P are (x1, y1), then the locus of Q is the parabola (a) ( y + y1)2 = 8a (x + x1) (b) ( y – y1)2 = 8a (x + x1) (c) ( y + y1)2 = 8a (x – x1) (d) None of these 70. If a chord which is normal to the parabola at one end and subtends a right angle at the vertex, then slope of the chord is (a) 1

(b) – 2

(c)

(d) 1/ 2 2 71. Coordinates of any point on the parabola, whose focus is  −3   , − 3  and the directrix is 2x + 5 = 0 is given by  2  (a) (2t2 + 2, 2t – 3) (c) (2t2 – 2, 2t + 3)

(b) (2t2 – 2, 2t – 3) (d) None of these

72. The conic represented by the equation is (a) ellipse (c) parabola

ax + by = 1

(b) Hyperbola (d) None of these

(a) 1 (c) – 1

78. If the tangents to the parabola y 2 = 4ax at (x1, y1) and (x2, y 2) intersect at (x3, y 3), then (a) x1, x2, x3 are in G. P (c) y1, y2, y3 are in G.P

(b) x1, x2, x3 are in A. P (d) y1, y2, y3 are in A. P

79. The locus of the middle points of parallel chords of a parabola y 2 = 4ax is a (a) straight line parallel to x-axis (b) straight line parallel to y-axis (c) straight line parallel to a bisector of the angles between the axes (d) None of these 80. Tangents are drawn from a point P to the parabola y 2 = 8x such that the slope of one tangent is twice the slope of other. The locus of P is (a) a circle (c) a parabola

(b) a straight line (d) an ellipse

81. A circle has its centre at the vertex of the parabola x2 = 4y and the circle cuts the parabola at the ends of its latus rectum. The equation of the circle is (a) x2 + y2 = 5 (c) x2 + y2 = 1

(b) x2 + y2 = 4 (d) None of these

82. The arithmetic mean of the ordinates of the feet of the normals from (2, 6) to the parabola y 2= 4x is (a) 5 (c) 0

(a) focus at (­– 3, – 2) (c) directrix x = – 5

(d) None of these 40

(b) – 2 (d) 2

(b) (0, 1) (d) None of these

(b) 0 < a < 1 (d) None of these

(b) 6 (d) None of these

(b) m1m2 = ­– 1 (d) None of these

84. The parametric representation (3 + t2, 3t – 2) represents a parabola with

(b) 8 5

75. A ray of light is coming along the line which is parallel to y-axis and strikes a concave mirror whose intersection with the xy-plane is a parabola (x – 4)2 = 4 (y + 2). After reflection, the ray must pass through the point (a) (4, – 1) (c) (– 4, 1)

(a) 0 ≤ a ≤ 1 (c) 0 < a ≤ 1

(a) m1 + m2 = 0 (c) m1m2 = 1

25 [(x– 2)2 + ( y – 4)2] = (4x – 3y + 12)2 is

74. If the parabola x2 = ay makes an intercept of length on the line y – 2x = 1, then a is equal to

77. The point (2k, K) lies inside the region bounded by the parabola x2 = 4y and its latus rectum. Then,

83. If y + 3 = m1 (x + 2) and y + 3 = m2 (x + 2) are two tangents to the parabola y 2 = 8x, then

73. The length of the latus rectum of the parabola

(a) 16 5 12 (c) 5

76. With respect to the parabola y 2 = 2x, the points P (4, 2) and Q (1, 4) are such that

(b) vertex at (3, – 2) (d) All of these

85. The graph represented by the equation x = sin 2 t, y = 2cos t is (a) a portion of a parabola (b) a parabola (c) a part of a sine graph (c) a part of a hyperbola. 86. At what point on the parabola y 2= 4x, the normal makes equal angles with the axes ? (a) (4, 4) (c) (4, – 4)

(b) (9, 6) (d) (1, – 2)

(a) C = 1/4 (c) C > 1/2

(b) C = 1/2 (d) None of these

88. If PSQ is the focal chord of the parabola y 2 = 8x such that SP = 6, then the length SQ is (a) 6 (c) 3

(b) 4 (d) None of these

89. The circle on focal radii of a parabola as diameter touches the (a) axis (b) directrix (c) tangent at the vertex (d) None of these 90. If the two parabolas y 2 = 4a  (x – 2) and x2 = 4a ( y – 3) touch each other, then their point of contact lies on a (a) circle (c) ellipse

(b) parabola (d) hyperbola

91. If ASB is a focal chord of a parabola such that AS = 2 and SB = 4, then the latus rectum of the parabola is (a) 8 (b) 16 3 3 25 (c) (d) None of these 3 92. If the line x – 1 = 0 is the directrix of the parabola y 2 – kx + 8 = 0, then one of the values of k is (a) 1 8 (c) 4

(b) 8

(a) 3 (c) – 9

(b) 9 (d) – 3

(d) 1 4 93. If x + y = k is normal to y 2 = 12x, then k is

(a) x = a (c) x + 4a = 0

(b) x + a = 0 (d) x + 2a = 0

98. The angle between the tangents drawn from the point (1, 4) to the parabola y 2 = 4x is (a) π/2 (c) π/4

(b) π/3 (d) π/6

99. The parabola y 2 = 4ax passes through the point (2, – 6), then the length of its latus rectum is (a) 9 (c) 18

(b) 16 (d) 6

100. If b and c are the lengths of the segments of any focal chord of a parabola y 2 = 4ax, then the length of the semi latus rectum is bc (b) bc b+c 2bc (c) b + c (d) b +c 2 101. If t is the parameter for one end of a focal chord of the parabola y 2 = 4ax, then its length is (a)

1 (a) a  t −  t  2

1 (c) a  t −  t 

1 (b) a  t +  t  (d) a  t + 

1  t

2

102. Consider the equation of a parabola y 2 + 4ax = 0, where a > 0. Which of the following is false? (a) Tangent at the vertex is x = 0 (b) Directrix of the parabola is x = 0 (c) Vertex of the parabola is at the origin (d) Focus of the parabola is at (a, 0)

94. A line bisecting the ordinate PN of a point P (at2, 2at), 2 2 t > 0, on the parabola y2 = 4ax is drawn parallel to the 103. The two parabolas y = 4x and x = 4y intersect at a point P, whose abscissae is not zero, such that axis to meet the curve at Q. If NQ meets the tangent at the vertex at the point T, then the coordinates of T are (a) the tangents to each curve at P make complementary angles with the x-axis 4 (b) ( 0, 2at ) (a)  0, at  (b) they both touch each other at P  3  (c) they cut at right angles at P (d) None of these 1 2   (c)  at , at  (d) (0, at) 4 104. If the line 3x – 4y + 5 = 0 is a tangent to the parabola   y 2 = 4ax, then a is equal to 95. The difference of the squares of the perpendiculars drawn from the points (a ± k, 0) on any tangent to a parabola (a) − 5 (b) − 4 4 3 y 2 = 4ax is 5 15 (c) (d) (a) 4 (b) 4a 4 16 (c) 4k (d) 4ak 105. The locus of a point whose sum of the distances from 96. If the normal drawn from the point on the axis of the the origin and the line x = 2 is 4 units, is 2 parabola y = 8ax whose distance from the focus is 8a and which is not parallel to either axis, makes an angle θ with the axis of x, then θ is equal to

(a) π 6 π (c) 3

(b) π 4 (d) None of these

(a) y2 = – 12 (x – 3) (c) x2 = 12 ( y – 3)

(b) y2 = 12 (x – 3) (d) x2 = – 12 ( y – 3)

106. The angle between the tangents drawn from the origin to the parabola y 2 = 4a (x – a) is (a) 90º (c) tan–11/2

(b) 30º (d) 45º

461

97. Two perpendicular tangents to y 2 = 4ax always intersect on the line

Conic Sections (Parabola, Ellipse and Hyperbola)

87. Three normals to the parabola y 2 = x are drawn through a point (C, 0), then

462

107. The axis of the parabola 9y 2 – 16x – 12y – 57 = 0 is

Objective Mathematics

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

(b) x + 3y = 3 (d) y = 3

108. If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y 2 = 4ax and x2 = 4ay, then (a) d2 + (2b – 3c)2 = 0 (c) d2 + (2b + 3c)2 = 0

(b) d2 + (3b + 2c)2 = 0 (d) d2 + (3b – 2c)2 = 0

109. The eccentricity of an ellipse with its centre at the origin is 1/2. If one of the directrices is x = 4, the the equation of the ellipse is (a) 4x2 + 3y2 = 12 (c) 3x2 + 4y2 = 1

(b) 3x2 + 4y2 = 12 (d) 4x2 + 3y2 = 1

2 2 116. The eccentricity of the ellipse x + y = 1 if its latus a 2 b2 rectum is equal to one-half of its minor axis, is

1 (b) 3 2 2 (c) 1 (d) None of these 2 117. The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum of length 4 and distance between foci 4 2 is (a) x2 + 2y2 = 24 (b) 2x2 + y2 = 24 2 2 (c) x + 2y = 16 (d) 2x2 + y2 = 16 (a)

118. 110. The locus of the midpoint of the line segment joining the focus to a moving point on the parabola y 2 = 4ax is another parabola with the directrix a (a) x = –a (b) x = – 2 a (c) x = 0 (d) x = 119. 2 111. The coordinates of the focus of the parabola y2 – 4y – 6x + 13 = 0 are 3 (a)  , 2  2  (c)  2,  112. For the

3  2

(b) (3, 2) (d) None of these

ellipse 12x2 + 4y 2 + 24x – 16y + 25 = 0,

(a) centre is (– 1, 2) (b) length of axes are 3 and 1 2 (c) eccentricity is 3 (d) All of these 113. The radius of the circle passing through the foci of x2 y2 + = 1 and having its centre at the ellipse 16 9 (0, 3) is (a) 4 (b) 3 (c) 12 (d) 7/2 114. The equation to the ellipse whose foci are (± 2, 0) and eccentricity 1 is 2 (a)

x2 y2 = 1 + 12 16

2 2 (b) x + y = 1 16 12

2 x2 + y = 1 (d) None of these 16 8 115. The equation of the ellipse referred to its axes as the axes of coordinates with latus rectum 8 and eccentricity 1 , is 2

(c)

(a) x2 + 2y2 = 64 (c) x2 + 4y2 = 16

(b) 2x2 + y2 = 64 (d) None of these

The equation of the ellipse, referred to its axes as the axes of coordinates which passes through the point (– 3, 1) and has eccentricity 2 , is 5 (a) 5x2 + 3y2 = 64 (c) 5x2 + 3y2 = 32

(b) 3x2 + 5y2 = 64 (d) 3x2 + 5y2 = 32

The equation of the ellipse referred to its axes as the axes of coordinates with minor axis 4 and the distance between the foci 2, is

2 2 2 2 (b) x + y = 1 (a) x + y = 1 5 4 4 5 2 2 2 2 x y x y = 1 (d) = 1 (c) + + 6 4 4 6 120. The equation of the ellipse, referred to its axes as the axes of coordinates, which passes through the points (2, 2) and (1, 4), is 2 2 2 2 (b) x + y = 1 (a) x + y = 1 20 5 5 20 2 2 2 x y x y2 = 1 = 1 (d) (c) + + 25 16 16 25 121. The locus of the middle point of the intercept of the tangetns drawn from an external point to the ellipse x2 + 2y 2 = 2, between the coordinate axes is

1 1 + = 1 x2 2 y 2 1 1 + = 1 (c) 2x2 4 y 2 (a)

1 =1 4x2 + 2 y 2 1 1 + (d) =1 2x2 y 2 (b)

122. The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latus rectum, is 5 −1 (b) 5 + 1 2 2 1 − 5 (c) (d) None of these 2 123. The eccentricity of the ellipse, if the minor axis is equal to the distance between the foci, is (a)

3 2 1 (c) 2

(a)

(b)

2 3

(d)

2

125. The equation of the tangent to the ellipse x2 + 4y 2 = 25 at the point whose ordinate is 2, is (a) 3x + 8y – 25 = 0 (c) 3x – 8y + 25 = 0

(b) 8x + 3y – 25 = 0 (d) 8x – 3y + 25 = 0

2 2 126. The equation of the normal to the ellipse x + y = a 2 b2 1 at the end of the latus rectum in the first quadrant, is

(a) x + ey – ae3 = 0 (c) x – ey – ae3 = 0

(b) x – ey + ae3 = 0 (d) None of these

127. The coordinates of the point where the line 2 2 lx + my + n = 0 touches the ellipse x + y = 1, are 2 2 a b

 a 2l b 2 m  , (a)   n   n

 − a 2l − b 2 m  (b)  ,  n   n

 a 2l − b 2 m  (c)  , (d) None of these  n   n 128. The condition that the line x cosα + y sinα = p may be 2 2 a tangent to the ellipse x + y = 1, is 2 2 a b (a) a2 cos2 α + b2 sin2 α = p2 (b) a2 cos2 α + b2 sin2 α = 2p2 (c) a2 sin2 α + b2 cos2 α = p2 (d) None of these 129. If the normal at the end of a latus rectum of an ellipse passes through one extremity of the minor axis, then (a) e4 + e2 – 1 = 0 (c) e4 – e2 – 1 = 0

(b) e4 – e2 + 1 = 0 (d) None of these

130. The equation of the tangent to the ellipse x2 + 3y 2 = 3 which is perpendicular to the straight line 4y = x – 5, is (a) 4x + y + 7 = 0 (c) x + 4y + 7 = 0

(b) 4x + y – 7 = 0 (d) x + 4y – 7 = 0

131. The equation of the tangent to the ellipse 4x2 + 3y 2 = 5 which is parallel to the straight line y = 3x + 7 is (a) 2 3 x − 6 3 y + 155 = 0

(a) (x2 + y2)2 = a2x2 + b2y2 (b) (x2 + y2)2 = b2x2 + a2y2 (c) (x2 – y2)2 = a2x2 + b2y2 (d) None of these 134. The locus of the point of intersection of two tangents 2 2 to the ellipse x + y = 1 which are inclined at angles 2 2 a b θ1 and θ2 with major axis such that θ1 + θ2 is constant, is (a) 2xy cot α = x2 – y2 + b2 – a2 (b) 2xy cot α = x2 – y2 + b2 – a2 (c) 2xy cot α = x2 + y2 + b2 – a2 (d) None of these 135. The locus of the point of intersection of two tangents 2 2 to the ellipse x + y = 1 which are inclined at angles 2 2 a b θ1 and θ2 with the major axis such that tan 2θ1 + tan 2θ2 is constant, is (a) 4x2y2 + 2 (x2 – a2) (y2 – b2) = k (x2 – a2)2 (b) 4x2y2 – 2 (x2 – a2) (y2 – b2) = k (x2 – a2)2 (c) 4x2y2 – 2 (x2 – a2) (y2 – b2) = k (x2 – a2)2 (d) None of these 136. The condition that the line xcos α + ysin α = p may be 2 2 a normal to the ellipse x + y = 1, is 2 2 a b (a) (a2 – b2)2 = p2 (a2sec2 α + b2 cosec2α) (b) (a2 – b2)2 = p2 (a2cosec2 α + b2 sec2α) (c) (a2 + b2)2 = p2 (a2sec2 α + b2 cosec2α) (d) None of these 137. If θ and φ are the eccentric angles of the ends of a 2 2 focal chord of the ellipse x + y = 1, then 2 2 a b (a) cos θ − φ = e cos θ + φ 2 2 θ + φ θ−φ (b) (a2 – b2) cos2 = a2cos2 2 2 (c) cos θ + φ  = e cos θ − φ 2 2 (d) None of these

(c) 6 3 x − 2 3 y + 155 = 0 (d) 6 3 x − 2 3 y − 155 = 0 x y = 1 intercepts + a 2 b2 lengths h and k on the axes, then 2

2

(a) π 3 π (c) 6

(b) π 4 π (d) 2

463

2 2 (b) a + b = 1 h2 k 2 2 2 (d) a + b = 1 2 k h2 the perpendicular from C, the y2 + 2 = 1 on any tangent is b

2 2 138. If the distance of a point on the ellipse x + y = 1 6 2 from the centre is 2, then the eccentric angle is

(b) 2 3 x − 6 3 y − 155 = 0

132. If any tangent to the ellipse

2 2 (a) a − b = 1 h2 k 2 2 2 (b) a − b = 1 2 k h2 133. The locus of the foot of 2 centre of the ellipse x a2

Conic Sections (Parabola, Ellipse and Hyperbola)

124. The equation of the ellipse with focus (– 1, 1), directrix x – y + 3 = 0 and eccentricity 1 , is 2 2 2 (a) 7x + 2xy + 7y + 10x + 10y + 7 = 0 (b) 7x2 + 2xy + 7y2 + 10x – 10y + 7 = 0 (c) 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0 (d) None of these

464

Objective Mathematics

139. The point of intersection of tangents to the ellipse x 2 y 2 = 1 at the points where the line + a 2 b2 x cos α + y sinα = p meets it, is  a 2 sin α b 2 cos α   a 2 cos α b 2 sin α  , , (a)   (b)   2 p p  p2    p  a 2 cos α b 2 sin α  (c)  ,  (d) None of these p p  

(a) b2l2 + a2m2 = 4 (c) a2l2 + b2m2 = 4

(b) b2l2 + a2m2 = 2 (d) None of these

147. The locus of the poles of the tangents to the ellipse x 2 y 2 = 1 w.r.t. the circle x2 + y 2 = a2 is + a 2 b2 (a) parabola (b) ellipse (c) hyperbola (d) circle 2 2 148. The locus of the poles of tangents to the ellipse x + y 2 2 a b = 1 w.r.t. the concentric ellipse c 2 x2 + d2y 2 = 1 is

140. If the chord of contact of tangents from the point (a) a2c4x2 + b2d4y2 = 1 (b) a4c2x2 + b4d2y2 = 1 2 2 (h, k) to the ellipse x + y = 1 touches the circle 2 4 2 2 4 2 (c) a c x – b d y = 1 (d) None of these a 2 b2 149. The locus of poles of tangents to x2 + y 2 = d2 with 2 2 2 x + y = c , then the point (h, k) lies on the ellipse 2 2 respect to the ellipse x + y = 1 is 2 2 2 2 (b) x + y = 1 (a) x + y = 1 a 2 b2 4 4 2 2 2 2 c c a b a b 2 2 2 2 (b) x + y = 1 (a) x + y = 2 2 2 4 4 4 4 2 d d2 a b a b (c) x + y = 1 (d) None of these 4 4 4 c a b 2 2 (c) x + y = 1 (d) None of these 141. From a point P, two tangents are drawn to the ellipse d2 b4 a 4 2 2 x y = 1 and the line joining the points of contact + 150. The locus of the mid-points of a system of parallel a 2 b2 chords of an ellipse is a is at a given distance ‘d’ from the centre. The locus of (a) straight line (b) parabola P is (c) pair of straight lines (d) None of these 2 2 2 2 (b) x + y = 1 (a) x + y = 1 2 2 4 4 4 d d2 a b a 4 b4 151. If chords of the ellipse x + y = 1 pass through a 2 2 a b 2 2 (c) x + y = 2 (d) None of these fixed point (h, k), then the locus of their middle points d4 a 4 b4 is a 142. Tangents are drawn from any point on the ellipse (a) parabola (b) ellipse (c) hyperbola (d) None of these x 2 y 2  = 1 to the circle x2 + y 2 = r 2. The chord of + a 2 b2 152. The locus of the middle points of the chords of the 2 2 contact touches the ellipse ellipse x + y = 1, which pass through the positive a 2 b2 (a) a2x2 + b2y2 = r2 (b) b2x2 + a2y2 = r2 end of the major axis, is 2 2 2 2 4 (c) a x + b y = r (d) None of these 2 2 2 2 143. The pole of the line 8x – 15y = 20 with respect to the (b) x + y = y (a) x + y = x 2 2 2 2 ellipse 4x2 + 5y 2 = 20 is a b a a b b 2 2 2 2 (a) (2, – 3) (b) (2, 3) (c) x + y  = x + y (d) x + y = x − y (c) (– 2, 3) (d) (­– 2, – 3) a 2 b2 a b a 2 b2 a b 144. The equation of a straight line through the point (1, 4) and conjugate to straight line 9x + 2y = 1 w.r.t. the ellipse 3x2 + 2y 2 = 1, is (a) 2x + 3y = 11 (c) 3x – 2y = 11

(b) 2x – 3y = 11 (d) 3x + 2y = 11

145. The locus of a point whose polar with respect to the 2 2 ellipse x + y = 1 touches the parabola y 2 = 4kx is 2 2 a b (a) parabola (c) hyperbola

(b) ellipse (d) circle

146. The pole of lx + my = 1 with respect to the ellipse x 2 y 2 = 1 lies on the ellipse x 2 y2 + 2 = 1 if + 2 2 2 4a 4b a b

153. The locus of the middle points of the chords of the 2 2 ellipse x + y = 1 which are at a constant distance 2 2 a b ‘d’ from the centre is 2  x2 y 2   x2 y 2  (a)  2 + 2  = d2  a 2 − b 2    b  a 2

 x2 y 2   4 + 4 b  a

2

 x2 y 2   4 + 4 b  a

 x2 y 2  (b)  2 + 2  = d2 b  a  x2 y 2  (c)  2 − 2  = d2 b  a (d) None of these

2 2 161. Let P be a variable point on the ellipse x + y = 1 2 2 a b with foci F1 and F2. If A is the area of the triangle PF1F2, then the maximum value of A is

(a) 2abe (c) 1 abe 2

(b) abe

170. If the focal distance of an end of the minor axis of any ellipse (referred to its axes as the axes of x and y respectively) is k and the distance between the foci is 2h, then its equation is (a)

(d) None of these

x2 y2 + 2  = 1 2 k k + h2 x2 y2 + 2  = 1 2 k k − h2

(b)

x2 y2 + 2 =1 2 k h − k2

2 2 (d) x + y =1 2 2 162. The foci of the ellipse 25 (x + 1) + 9 (y + 2) = 225 k h are 16 sin θ) to the 171. If the tangent at the point (4 cos θ, (a) (– 1, 2), (6, 1) (b) (– 1, – 2), (1, 6) 11 ellipse 16x2 + 11y 2 = 256 is also a tangent to the circle (c) (1, – 2), (1, – 6) (d) (– 1, 2), (– 1, – 6) x2 + y 2 – 2x = 15, then the value of θ is 163. The equation of the tangents drawn at the ends of the major axis of the ellipse 9x2 + 5y 2 – 30y = 0 are (a) ± π (b) ± π 4 3 (a) y = ± 3 (b) x = ± 5 π (c) (d) ± (c) y = 0, y = 6 (d) None of these 2

2

2

(c)

465

Conic Sections (Parabola, Ellipse and Hyperbola)

154. The eccentricity of the curve represented by the equa- 164. The eccentricity of the ellipse 9x 2 + 5y 2 – 30y = 0 is tion x2 + 2y 2 – 2x + 3y + 2 = 0 is (a) 1 (b) 2 (a) 0 (b) 1/2 3 3 (c) 1 / 2 (d) 2 3 (c) (d) None of these 4 155. The sum of the focal distances from any point on the ellipse 9x2 + 16y 2 = 144 is 165. If eccentricity of ellipse becomes zero, then it takes the form of (a) 32 (b) 18 (c) 16 (d) 8 (a) a parabola (b) a straight line (c) a circle (d) None of these 156. If P ≡ (x, y), F1 ≡ (3, 0), F2 ≡ (– 3, 0) and 16x2 + 25y 2 2 2 = 400, then PF1 + PF2 equals = 1 166. If S' and S are the foci of the ellipse x + y (a) 8 (b) 6 a 2 b2 (c) 10 (d) 12 and P (x, y) a point on it, the value of SP + S' P is 157. The number of values of c such that the straight line (a) 2b (b) 2a 2 (c) a – b (d) a + b x 2 y = 4x + c touches the curve + y = 1 is 4 167. The angle between the pair of tangents drawn from the point (1, 2) to the ellipse 3x2 + 2y 2 = 5 is (a) 0 (b) 1 (c) 2 (d) infinite  12 5  6 5 –1  2 2  (a) tan (b) tan–1   5  158. Eccentric angle of a point on the ellipse x + 3y = 6    5  at a distance 2 units from the centre of the ellipse is 3 5 (a) π (b) π (c) tan–1  5    (d) None of these   4 3 (c) 3π (d) 2π 168. The eccentricity of the ellipse which meets the straight 4 3 line x + y = 1 on the axis of x and the straight line 159. The radius of the circle passing through the foci of the 7 2 2 2 x y = 1 on the axis of y and whose axes lie along x y = 1, having its centre at (0, 3) is ellipse − + 3 5 16 9 the axes of coordinates is (a) 4 (b) 3 (a) 2 6 (b) 3 2 (c) 12 (d) 7/2 7 7 160. An ellipse has OB as a semi minor axis, F, F ' are its (c) 6 (d) None of these foci and the angle FBF ' is a right angle. Then the ec7 centricity of the ellipse is 169. The area of quadrilateral formed by tangents at the 1 2 2 (b) 1 (a) = 1 is ends of latus rectum of ellipse x + y 2 2 9 5 (c) 3 (d) None of these (a) 9 sq. units (b) 27 sq. units 2 (c) 27/2 sq. units (d) 27/4 sq. units

466

172. The line 2x + y = 3 cuts the ellipse 4x2 + y 2 = 5 at P and Q. If θ be the angle between the normals at these points, then tan θ =

 2 91 3 105  ,  (a)  14   7

Objective Mathematics

(a) 3 (b) 3 4 5 (c) 1 (d) 5 2 173. The equation of the chord joining two points (x1, y1) and (x2, y 2) on the rectangular hyperbola xy = c 2 is

 2 91 − 3 105  ,  (b)  14   7

x y (a) + =1 x1 + x2 y1 + y2 (b)

x y + =1 x1 − x2 y1 − y2

(c)

x y + =1 y1 + y2 x1 + x2

(d)

x y + =1 y1 − y2 x1 − x2

174. For the ellipse 3x2 + 4y 2 – 6x + 8y – 5 = 0, (a) centre is (1, – 1) (b) eccentricity is 1 2 (c) foci are (3, – 1) and (– 1, – 1) (d) All of these are true 175. For the ellipse 25x2 + 45y 2 = 9, (a) eccentricity is 2 (b) latus rectum is 2 3 3 2   (c) foci are  ± , 0  (d) All of these are true  5  176. The latus rectum of the conic 3x2 + 4y 2 – 6x + 8y – 5 = 0 is (a) 3 (c)

2 3

(b) 3 2 (d) None of these

2 2 177. Let E be the ellipse x + y = 1 and C be the circle 9 4 x2 + y 2 = 9. Let P and Q be the points (1, 2) and (2, 1) respectively. Then

(a) Q lies inside C but outside E (b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but outside E

 2 105 − 3 91  ,  (c)  − 7 14    2 105 3 91  ,  (d)  − 7 14   180. S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is (a) 1/4 (c) 1/2

(b) 1/3 (d) 2/3

181. If the polar with respect to the parabola y 2 = 4x touches x2 y 2 the ellipse 2 + 2 = 1, then the locus of its pole is α β (a)

x2 y2 x 2 β2 y 2 − + = 1 (b) =1 α 2 4a 2 α 2 / β 2 α 2 4a 2

(c) α2x2 + β2y2 = 1

(d) None of these

182. The locus of mid-points of focal chords of the ellipse x 2 y 2 = 1 is + a 2 b2 ex x2 y 2 + 2 = 2 a a b (c) x2 + y2 = a2 + b2

(a)

x2 y 2 ex − 2 = 2 a b a (d) None of these (b)

2 2 183. If x + y = 2 touches the ellipse x + y 2 2 a b a b its eccentric angle θ is equal to

(a) 0 (c) 45º

= 1, then

(b) 90º (d) 60º

2 2 = 1 and the straight line y = mx 184. The ellipse x + y 2 2 a b + c intersect in real points only if

(a) a2m2 < c2 – b2 (c) a2m2 ≥ c2 – b2

(b) a2m2 > c2 – b2 (d) c ≥ b

185. An ellipse slides between two lines at right angles to one another. The locus of its centre is

(a) a parabola (b) an ellipse 178. The number of real tangents that can be drawn to the (c) a circle (d) None of these ellipse 3x2 + 5y 2 = 32 and 25x2 + 9y 2 = 450 passing 186. If the angle between the straight lines joining foci and through (3, 5) is 2 2 = 1 is the ends of minor axis of the ellipse x + y (a) 0 (b) 2 a 2 b2 90º, then the eccentricity is (c) 3 (d) 4 2 2 179. A point on the ellipse x + y = 1 at a distance equal 16 9 to the mean of the lengths of the semi-major axis and semi-minor axis from the centre is

(a) 1 2 1 (c) 2

(b)

3 2

(d) None of these

193. If θ and φ are the eccentric angles of the extremities of a 2 2 pair of conjugate diameters of the ellipse x + y = 1, a 2 b2 then θ – φ = (a) π 4 (c) π 2

(b) π 3 (d) None of these

2 2 (a) ( x − 1) − ( y − 4) 25 / 4 75 / 4

=1

2 2 (b) ( x + 1) − ( y + 4) 25 / 4 75 / 4

=1

2 2 (c) ( x − 1) − ( y − 4) 75 / 4 25 / 4 (d) None of these

=1

467

Conic Sections (Parabola, Ellipse and Hyperbola)

187. If a chord joining two points whose eccentric angles are 194. If CP and CD is a pair of semi-conjugate diameters of 2 2 − a2 = 1, then CP2 +CD2 = α and β so that tan α ⋅ tan β = 2 , subtend an angle θ the ellipse x + y b a 2 b2 x2 y 2 (a) a2 + b2 (b) a2 – b2 at the centre of the ellipse 2 + 2 = 1, then θ = a b 1 2 2 (c) (a + b ) (d) None of these 2 (a) π (b) π 2 3 195. The area of the parallelogram formed by tangents at the extremities of two conjugate diameters of the ellipse π (d) None of these (c) x 2 y 2 = 1 is equal to 4 + a 2 b2 188. If the extremities of a line segment of length l move in two fixed perpendicular straight lines, then the locus of (a) ab (b) 2ab that point which divides this line segment in the ratio (c) 4ab (d) None of these 1 : 2, is 196. If P and D are the extremities of a pair of conjugate 2 2 (a) a parabola (b) an ellipse = 1, then the locus diameters of the ellipse x + y (c) a hyperbola (d) None of these 2 2 a b 189. If a circle of radius r is concentric with the ellipse of the middle point of PD is x 2 y 2  = 1, then the common tangent is inclined to 2 2 2 2 1 + (b) x + y (a) x + y = 2 a 2 b2 2 2 2 2 a b a b 2 the major axis at an angle 2 2 x y 1 (d) None of these (c) +  r 2 − b2   r 2 − b2  a 2 b2 4 (a) tan–1  2 (b) tan–1  2 2  2  a −r  r −a  197. If CP and CD be any two semi-conjugate diameters 2 2 2 2 a −r  = 1 and the circles with CP of the ellipse x + y 2 2 (c) tan–1  2 (d) None of these 2  a b r − b   and CD as diameters intersect in R, then R lies on the curve 2 2 = 1 intercepts 190. If any tangent to the ellipse x + y (a) 2 (x2 + y2)2 = a2x2 + b2y2 16 9 (b)  (x2 + y2)2 = a2x2 + b2y2 equal lengths l on the axes, then l = (c) (x2 + y2)2 = 2 (a2x2 + b2y2) (a) 3 (b) 5 (d) None of these (c) 5 (d) None of these 198. The orbit of the earth is an ellipse with eccentricity 191. If the chords of contact of tangents from two points 1/60 with the sum at one focus, the major axis being 2 2 approximately 186 × 106 miles in length. The shortest = 1 are per(α, β) and (γ, δ) to the ellipse x + y 5 2 and longest distance of the earth from the sun is αγ = pendicular, the βδ (a) 9145 × 104 miles, 9455 × 104 miles (b) 9147 × 104 miles, 9457 × 104 miles (a) 4 (b) − 4 (c) 9145 × 106 miles, 9455 × 106 miles 25 25 (d) None of these (d) − 25 (c) 25 199. The line 2 x + 6 y = 2 is a tangent to the curve 4 4 x2 – 2y 2 = 4. The point of contact is 2 2 x y 2 192. The line y = 2t meets the ellipse = 1 in real + (a) ( 6 , 1) (b) (7, − 2 6 ) 9 4 points if (c) (2, 3) (d) (4, − 6 ) (a) | t | ≤ 1 (b) | t | > 1 200. The equation of the hyperbola whose foci are (6, 4) (c) | t | < 3 (d) None of these and (­– 4, 4) and eccentricity 2, is

468

Objective Mathematics

201. The equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity 3 , is (a) 7x2 – 2y2 + 12xy + 2x + 14y – 22 = 0 (b) 2x2 – 7y2 + 12xy – 2x + 14y – 22 = 0 (c) 7x2 – 2y2 + 12xy – 2x + 14y – 22 = 0 (d) None of these 202. The eccentricity of the hyperbola 3x2 – y 2 = 4 is (a)

1 2

(b)

2

(d) 1 2 203. The equation of the hyperbola referred to its axes as axes of coordinates whose distance between the foci is 16 and eccentricity 2 , is (a) x2 – y2 = 16 (b) x2 – y2 = 32 2 2 (d) None of these (c) x – y = 64 (c) 2

204. The equation of the hyperbola referred to its axes as axes of coordinates whose latus rectum is 4 and eccentricity is 3, is (a) 2x2 – 16y2 = 1 (c) 16x2 – 2y2 = 1

(b) 4x2 – 16y2 = 1 (d) None of these

2 2 205. The equation of the tangent to the hyperbola x − y  = 1, 4 3 parallel to the line y = x + 2, is

(a) y = – x + 1 (c) y = – x – 1

(b) y = x + 1 (d) y = x – 1

206. The equation of the tangent to the hyperbola 3x2 – y 2 = 3 which is perpendicular to the line x + 3y – 2 = 0 is (a) y = 3x + (c) y = 3x –

6 6

(b) y = – 3x +

6 (d) None of these

207. The equation of the tangent to the hyperbola 3x2 – 4y 2 = 12 which makes equal intercepts on the axes, is (a) y = x + 1 (b) y = – x + 1

(b) y = x – 1 (d) y = – x – 1

208. The locus of the foot of the perpendicular from the 2 2 = 1 on a variable centre of the hyperbola x − y 2 2 a b tangent is (a) (x2 + y2)2 = a2x2 + b2y2 (b) (x2 + y2)2 = a2x2 – b2y2 (c) (x2 + y2)2 = b2x2 – a2y2 (d) None of these 209. The length of the straight line x – 3y = 1 intercepted by the hyperbola x2 –­ 4y 2 = 1 is (a) 6 5 (b) 3 10 5 5 6 (c) (d) None of these 10 5 210. Tangents are drawn to the hyperbola 5x2 – y 2 = 5 from the point (0, 2). Their equations are

(a) y = ± 3x + 2 (c) y = ± 2x + 3

(b) y = ± 3x – 2 (d) y = ± 2x – 3.

2 2 211. Two tangents are drawn to the hyperbola x − y = 1 2 2 a b such that the product of their slopes is c 2. The locus of their point of intersection is

(a) x2 – a2 = c2 (y2 + b2) (c) y2 + b2 = c2 (x2 – a2)

(b) x2 + a2 = c2 (y2 – b2) (d) None of these

212. The line lx + my + n = 0 touches the hyperbola x 2 y 2  = 1 if − a 2 b2 (a) a2l2 – b2m2 = n2 (c) b2l2 – a2m2 = n2

(b) a2l2 + b2m2 = n2 (d) None of these

213. The line x cos α + y sin α = p will be a tangent to the 2 2 = 1 if hyperbola x − y 2 2 a b (a) a2cos2α + b2sin2α = p2 (b) a2cos2α – b2sin2α = p2 (c) b2cos2α – a2sin2α = p2 (d) None of these 214. The line lx + my = 1 will be a normal to the hyperbola x 2 y 2 = 1 if − a 2 b2 (a)

b2 a 2 − = (a2 + b2)2 l 2 m2

2 2 (b) a − b = (a2 – b2)2 2 l m2 2 2 (c) a − b = (a2 + b2)2 2 l m2 (d) None of these

215. The coordinates of the point at which the line 3x + 4y 2 2 = 1, are = 7 is a normal to the hyperbola x − y 2 2 a b   7a 2 7b 2 (a)  , 2 2 2 2   3 (a + b ) 4 (a + b )    7b 2 7a 2 (b)  , 2 2 2 2   3 (a + b ) 4 (a + b )    7a 2 7b 2 (c)  , 2 2 2 2   3 (a − b ) 4 (a − b )  (d) None of these 216. If the polars of (x1, y1) and (x2, y 2) w.r.t. the hyperbola x 2 y 2 = 1 are at right angles, then x1 x2 = − y1 y2 a 2 b2 2 (a) − a 2 b

2 (b) a 2 b

4 (c) a 4 b

4 (d) − a 4 b

(a) x2 + y2 = 16 (c) x2 + y2 = 4

(b) x2 + y2 = 8 (d) None of these

2 2 = 1 219. If the polar of a point with respect to x + y 2 2 a b 2 2 touches the hyperbola x − y = 1, then the locus of a 2 b2 the point is

(a) given hyperbola (c) circle

(b) ellipse (d) None of these

220. The locus of the poles of normal chords of the hyperbola x 2 y 2 = 1 is the curve − a 2 b2 (a)

a 6 b6 a 6 b6 + 2 = (a2 + b2)2 (b) 2 + 2 = (a2 – b2)2 2 x y x y

(c)

a 6 b6 a 6 b6 − 2 = (a2 – b2)2 (d) 2 − 2 = (a2 + b2)2 2 x y x y

221. The locus of the poles w.r.t. the parabola y = 4ax of the tangents to the rectangular hyperbola x2 – y 2 = a2 is the ellipse

2

(a) 4x2 + y2 = a2 (c) 4x2 + y2 = 4a2

(b) 4x2 + y2 = 2a2 (d) None of these

222. The equation of the chord of the hyperbola 4x2 – y 2 = 4, which is bisected at the point (2, – 3) is (a) 8x + 3y = 7 (c) 3x + 8y = 7 223. The locus 2 bola x − 9 (1, 2) is a

(b) 8x – 3y = 7 (d) 3x – 8y = 7

2

 x2 y 2   4 − 4 b  a

2

 x2 y 2   2 + 2 b  a

 x2 y 2  (b)  2 + 2  = c2 b  a  x2 y 2  (c)  4 − 4  = c2 b  a (d) None of these

226. If e, e' be the eccentricities of a hyperbola and its conjugate, then (a) e2 + e' 2 = 1

(b) 1 − 1 e 2 e′2

= 1

(c) 1 + 1 = 1 (d) None of these e 2 e′2 227. The equation of the diameter which is conjugate to y 2 2 = 1, is = 3x w.r.t. the hyperbola x − y 4 9 (a) y = 3 x (b) y = −3 x 4 4 4 − (c) y = x (d) y = 4 x 3 3 228. The condition for two diameters of the hyperbola x 2 y 2 = 1 represented by − a 2 b2 Ax2 + 2Hxy + By2 = 0 to be conjugate is (a) Ab2 = Ba2 (c) Aa2 = Bb2

(b) Aa2 = – Bb2 (d) None of these

of the middle points of the chords of hyper- 229. The area of the triangle formed by the asymptotes and any tangent to the hyperbola x2 – y 2 = a2 is y 2 = 1 which pass through the fixed point 4 (a) 4a2 (b) 3a2 2 hyperbola whose centre is (c) 2a (d) a2

1 1 (a)  ,  2 2

1 (b) 1,   2

1 (c)  , 1 2 

(d) (1, 1)

2 2 224. P is a variable point on the hyperbola x − y = 1 a 2 b2 whose vertex is A (a, 0). The locus of the middle point of AP is 2 2 (a) (2 x − a ) − 2 y = 1 2 2 a b 2 2 (b) (2 x − a ) − 4 y = 1 2 2 a b 2 2 (c) (2 x − a ) − 8 y = 1 a2 b2 (d) None of these

x2 y 2 = 1, are − a 2 b2 tangents to the circle described on the line joining the foci as diameter. The locus of their poles w.r.t. the hyperbola is

230. A series of chords of the hyperbola

2 2 2 2 1 1 (b) x + y = 2 (a) x + y = 2 2 4 4 2 2 a + b a + b2 a b a b 2 2 1 (d) None of these (c) x − y = 2 4 4 a + b2 a b 231. If the chord of contact of tangents from a point P to 2 2 = 1 subtends a right angle at the hyperbola x − y a 2 b2 the centre, then the locus of P is

(a) a parabola (c) a hyperbola

(b) an ellipse (d) None of these

469

Conic Sections (Parabola, Ellipse and Hyperbola)

217. The line 3x + 2y + 1 = 0 meets the hyperbola 4x2 – y 2 x2 y 2 = 4a2 in the points P and Q. The coordinates of the 225. A variable chord of the hyperbola 2 − 2 = 1 is a b point of intersection of the tangents at P and Q are a tangent to the circle x2 + y 2 = c 2. The locus of its (b) ( 3a2, 8a2) (a) (– 3a2, 8a2) middle point is (c) ( 3a2, – 8a2) (d) None of these 2  x2 y 2   x2 y 2  218. The locus of pole of any tangent to the circle x2 + y 2 (a)  2 − 2  = c2  a 4 − b 4    b  a = 4 w.r.t. the hyperbola x2 – y 2 = 4 is the circle

470

Objective Mathematics

232. The normal to the hyperbola 4x2 – 9y 2 = 36 meets the 241. The length of the common chord of the ellipse axes in M and N and the lines MP, NP are drawn at ( x − 1) 2 ( y − 2) 2 2 2 + = 1 and the circle (x – 1) + (y – 2) right angles to the axes. The locus of P is the hyper9 4 bola = 1, is (a) 9x2 – 4y2 = 169 (a)  2 (b)   2 (b) 4x2 – 9y2 = 169 (c)  0 (d)  5 (c) 3x2 – 4y2 = 169 (d) None of these 242. The value of m, for which the line y = mx + 25 3 is 2 2 2 3 233. The normal to the rectangular hyperbola xy = c at a normal to the conic x − y = 1 , is the point ‘t’ meets the curve again at a point ‘t’ such 16 9 that 2 (b)   3 (a)   − (a) t3t' = – 1 (b) t2 t' = – 1 3 (c) tt' = – 1 (d) None of these (c)   − 3 (d)  none of these 234. The eccentricity of the hyperbola whose latus rectum 2 is half of its transverse axis, is 243. The value of m, for which the line y = mx + 2 becomes 1 (b) 2 (a) a tangent to the conic 4x2 – 9y 2 = 36, are 2 3 2 2 2 (a)   ± (b)   ± (c) 3 (d) None of these 3 3 2 4 2 8 235. The equation of the hyperbola, referred to its axes as (c)   ± (d)   ± 3 axes of coordinates, given that the distances of one of 9 its vertices from the foci are 9 and 1 units, is 244. The locus of the vertices of the family of parabolas x2 y 2 x2 y 2 = 1 a3 x2 a 2 x = 1 (b) (a) − − y= + − 2a is 16 9 9 16 3 2 35 2 2 3 (b)   xy = (a)   xy = (d) None of these (c) x − y = – 1 16 4 16 9 64 105 236. If the line y = 2x + λ touches the hyperbola 16x2 – 9y 2 (c)   xy = (d)   xy = 105 64 = 144, then λ is equal to (a) ± 2 5

(b)

(c) ± 3 5

(d) None of these

5

245. In an ellipse the distance between its foci is 6 and its minor axis is 8, the eccentricity of the ellipse is

(b)   3 5 9x2 – 16y2 + 72x – 32y – 16 = 0 is (d)   1 2 (a) 5/4 (b) 4/5 (c) 9/16 (d) 16/9 246. Axis of a parabola is y = x and vertex and focus are at a distance 2 and 2 2 respectively from the origin. 238. If the tangent at the point (2 sec θ, 3 tan θ) of the hy2 2 Then equation of the parabola is = 1 is parallel to 3x – y + 4 = 0, perbola x − y (a)  (x – y)2 = 8(x + y – 2) (b)  (x + y)2 = 2(x + y – 2) 4 9 (c)  (x – y)2 = 4(x + y – 2) (d)  (x + y)2 = 2(x – y + 2) then the value of θ is

237. The eccentricity of the hyperbola

(a) 45º (c) 30º

(b) 60º (d) 75º

2 2 239. The number of tangents to the hyperbola x − y = 1 4 3 through (4, 1) is

(a) 1 (c) 0

(b) 2 (d) 3

240. Three normals to the parabola y 2 = x are drawn through a point (c, 0), then (a)   c = 1 4 1 (c)   c > 2

(b)   c = 1 2 (d)  none of these

(a)   4 5 1 (c)   52

2 2 247. If e1 is the eccentricity of the ellipse x + y = 1 and 16 25 e 2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e 2 = 1, then equation of the hyperbola is 2 2 (a)   x − y = 1 9 16

2 2 (b)   x − y = −1 16 9

2 2 (c)   x − y = 1 (d)  none of these 9 25 248. The equation of the common tangent to the curves y 2 = 8x and xy = – 1 is

(a)  3y = 9x + 2 (c)  2y = x + 8

(b)  y = 2x + 1 (d)  y = x + 2

(a)  y = x ± 1 (c)  y = x + 1/2

(b)  y = x – 1/2 (d)  y = 1 – x

250. The foci of the conic section 25x2 + 16y 2 – 150x = 175 are (a)  (0, ± 3) (c)  (3, ± 3)

(b)  (0, ± 2) (d)  (0, ± 1)

251. The product of perpendiculars drawn from any point of a hyperbola to its asymptotes is 2 2 (a)   a b 2 a + b2 ab (c)   a+ b

2 2 (b)   a + b 2 2 ab (d)   ab a 2 + b2

2 2 252. Equation of tangents to the ellipse x + y = 1 which 9 4 are perpendicular to the line 3x + 4y = 7, are

(a)   4 x − 3 y = ± 20

(b)   4 x − 3 y = ± 12

(c)   4 x − 3 y = ± 2

(d)  4x – 3y = ± 1

253. The focus of the parabola y 2 – x – 2y + 2 = 0 is 1  (a)    , 0  4 

(b)  (1, 2)

5  (c)    , 1 4 

3 5 (d)    ,  4 2

254. A hyperbola, having the transverse axis of length 2 sin θ, is confocal with the ellipse 3x2 + 4y 2 = 12. Then its equation is (a)  x2 cosec2θ – y2 sec2θ = 1 (b)  x2 sec2θ – y2 cosec2θ = 1 (c)  x2 sin2θ – y2 cos2θ = 1 (d)  x2 cos2θ – y2 sin2θ = 1 255. The value of c for which the line y = 2x + c is a tangent to the parabola y2 = 4a (x + a), is (b)   3a 2 5 (c)  2a (d)   a 2 256. The radius of the circle passing through the foci of the 2 2 ellipse x + y = 1 and having its centre (0, 3) is 16 9 (a)  4 (b)  3 (a)  a

(c)   12

(d)   7 2

257. The equation of the director circle of the hyperbola 9x2 – 16y 2 = 144 is (a)  x2 + y2 = 7 (c)  x2 + y2 = 16

(b)  x2 + y2 = 9 (d)  x2 + y2 = 25

(a)  π/6 (c)  π/3

(b)  π/4 (d)  2π/3

259. The point of intersection of the curves whose parametric equations are x = t2 + 1, y = 2t and x = 2S, y = 2 is S given by (a)  (1, – 3) (c)  (–2, 4)

(b)  (2, 2) (d)  (1, 2)

260. If the line y = 2x + λ be a tangent to the hyperbola 36x2 – 25y 2 = 3600, then λ is equal to (a)  16 (c)  ± 16

(b)  – 16 (d)  None of these

261. The eccentricity of the hyperbola can never be equal to 9 1 (b)   2 5 9 1 (c)   3 (d)  2 8 262. The number of values of c such that the straight line 2 y = 4x + c touches the curve x + y 2 = 1, is 4 (a)  0 (b)  2 (c)  1 (d)  ∞ (a)  

263. The locus of the equation x2 – y 2 = 0 is (a)  a circle (b)  a hyperbola (c)  a pair of lines (d)  a pair of lines at right angles 264. Locus of mid point of any focal chord of y 2 = 4ax is (a)  y2 = a(x – 2a) (c)  y2 = 2a(x – a)

(b)  y2 = 2a(x – 2a) (d)  None of the above

265. The product of perpendiculars drawn from any point if, a hyperbola to its asymptotes, is a 2b 2 a 2 + b2 ab (c)   a+ b

a 2 + b2 a 2b 2 ab (d)   2 a + b2 266. If the tangent at the point (2 sec θ, 3 tan θ) of the 2 2 hyperbola x − y = 1 is parallel to the line 3x – y + 4 9 4 = 0. Then value of θ is (a)  

(b)  

(a)  30º (c)  90º

(b)  60º (d)  45ºº

267. Length of the straight line x – 3y = 1 intercepted by the hyperbola x2 – 4y 2 = 1 is … (a)   6 10 5 (c)   10

(b)   3 10 2 5 (d)   6

471

258. The line 3x + 5y = 15 2 is a tangent to the ellipse x2 y 2 + = 1 , at a point whose eccentric angle is 25 9

Conic Sections (Parabola, Ellipse and Hyperbola)

2 2 249. The equation of tangent to the curve x − y = 1 which 3 2 is parallel to y = x, is

472

268. Three normals are drawn to the parabola y 2 = x through point (a, 0). Then

Objective Mathematics

(a)  a = 1/2 (c)  a > 1/2

(b)  a = 1/4 (d)  none of these.

269. Through the vertex of the parabola y = 4x chords OP and OQ are drawn at right angles to one another. The locus of middle point of PQ is 2

(a)  y = x + 8 (c)  y2 = 2x – 8

(b)  y = – 2x + 8 (d)  y2 = x – 8.

2

2

(c)  C1 and C2 intersect (but do not touch) at exactly two points (d)  C1 and C2 neither intersect nor touch each other 272. Let P (x1, y1) and Q (x2, y 2), y1 < 0, y 2 < 0, be the end points of the latus rectum of the ellipse x2 + 4y 2 = 4. The equations of parabolas with latus rectum PQ are (a)  x2 + 2 3 y = 3 +

3

(b)  x – 2 3 y = 3 +

3

2

(c)  x + 2 3 y = 3 – 3 270. A hyperbola, having the transverse axis of length 2 (d)  x2 – 2 3 y = 3 – 3 sinθ, is confocal with the ellipse 3x2 + 4y 2 = 12. Then its equation is 273. Consider a branch of the hyperbola x 2 − 2 y 2 − 2 2 x − 4 2 y − 6 = 0 2 2 x − 2 y − 2 2 x − 4 2 y − 6 = 0 with vertex at the point A. Let B be one (a)  x2 cosec2θ – y2 sec2θ = 1 of the end points of its latus rectum. If C is the focus 2 2 2 2 (b)  x sec θ – y cosec θ = 1 of the hyperbola nearest to the point A, then the area (c)  x2 sin2θ – y2 cos2θ = 1 of the triangle ABC is (d)  x2 cos2θ – y2 sin2θ = 1 2 3 (b)   (a)   1 − −1 271. Consider the two curves C1 : y 2 = 4x, C2 : x2 + y 2 – 6x 3 2 + 1 = 0. Then, 2

(a)  C1 and C2 touch each other only at one point (b)  C1 and C2 touch each other exactly at two points

2 3

(c)   1 +

(d)  

3 +1 2

SOLUTIONS 1. (a), (b) The given parabola is y = 8x ...(1) Here 4a = 8 ∴ a = 2. Let the required point be (x1, y1). Its focal distance = a + x1 = 4 (given) ⇒ 2 + x1 = 4 ∴ x1 = 2. The point lies on (1) 2 ∴ y1 = 8 × 2 = 16 ⇒ y1 = ± 4. Hence we get two points (2, 4) and (2, – 4).

and

PM = length of ⊥ from P (x, y) on (1)



| x + y − 2| | x + y − 2| = . 1+1 2

2

2. (b) Let S (3, – 4) be the focus and ZM, the directrix

x + y – 2 = 0

...(1)

=

2 2 | x + y − 2| ∴ From (2), ( x − 3) + ( y + 4) = . 2 Squaring and cross multiplying, we get 2 (x2 – 6x + 9 + y2 + 8y + 16) = x2 + y2 + 4 + 2xy – 4y – 4x or x2 – 2xy + y2 – 8x + 20y + 46 = 0 which is the required equation of the parabola.

3. (c) Let x1 be the abscissa of the point whose ordinate is 6. Since the point (x1, 6) lies on y2 = 6x, ∴ 36 = 6x1 or x1 = 6. ∴ The point is (6, 6). Equation of parabola is y2 = 6x (Here 4a = 6, ∴ 3 ). a= 2 Equation of tangent at (6, 6) is 3 (x + 6) 2 or x – 2y + 6 = 0.

y⋅6=2 ⋅ Let P (x, y) be any point on the parabola. Joint SP and draw PM ⊥ ZM. Then, by definition, SP = PM Now SP =

( x − 3) 2 + ( y + 4) 2

...(2)

[ yy1 = 2a (x + x1)]

4. (b) Equation of parabola is y2 = 8x Here 4a = 8, ∴ a = 2. ∴ Equation of normal at (2, 4) is

...(1)

⇒ y – 4 = – (x – 2) or x + y – 6 = 0 ...(2) From (2), y = 6 – x. Putting this value of y in (1), (6 – x)2 = 8x or x2 – 20x + 36 = 0 or (x – 2) (x – 18) = 0. ∴ x = 2, 18. When x = 2, y = 6 – x = 6 – 2 = 4 and when x = 18, y = 6 – x = 6 – 18 = – 12. ∴ Normal (2) meets the parabola (1) in (2, 4) and (18, – 12). Hence, the normal at (2, 4) meets the parabola again in (18, – 12). 5. (a) Since the tangent is inclined at 60º to x– axis, ∴ m = slope of the tangent = tan 60º = Equation of parabola is y2 = 16x. Here 4a = 16, ∴ a = 4. ∴  Equation of the tangent is

y=

or 3x –

3 ⋅x+

4 3

3.

a   y = mx +  m 

3 ⋅ y + 4 = 0.

6. (d) Equation of parabola is y2 = 16x Here 4a = 16, ∴ a = 4. Equation of given line is 2x – y + 5 = 0 Its slope = 2. Since the tangent is ⊥ to (2), ∴ slope of tangent = −1 . 2 ∴ m = −1 . 2 ∴ Equation of tangent is

4 y = −1 x + − 1 /2 2

or

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

...(3) ...(4)

From the first two members of (4), mp = – q , ∴ m = – q/p. From the first and third members of (4) mp = m (2a + a m 2) or or

 −q  p = 2a + am2 = 2a + a     p  2 aq p = 2a + p2

2

or p3 = 2ap2 + aq2, which is the required condition. 9. (b) Equation of the line is x + y + 1 = 0 or y = – x – 1 ...(1)  omparing it with y = mx + c, we get, m = – 1 and C c = – 1. ...(2) Equation of the parabola is y2 = kx Comparing it with y2 = 4ax, we get 4a = k,  ∴  a = k . 4 (1) touches (2), if c = a/m i.e., k /4 i.e. if k = 4. if – 1 = −1 10. (c) The given line is y = mx + c

...(1)

and the parabola is y = 4a (x + a) 2

...(2)

Putting y = mx + c in (2), we get (mx + c)2 = 4a  (x + a)



y=– 1 x–8 2

or

m x + 2  (mc – 2a) x + (c2 – 4a2) = 0 2 2

...(3)

(1) touches (2) if the roots of (3) are equal i.e. if discriminant of (3) = 0

or x + 2y + 16 = 0.

⇒ 4 (mc – 2a)2 – 4m2 (c2 – 4a2) = 0

7. (b) Equation of parabola is y2 = 4x.

⇒ m2 c2 – 4amc + 4a2 – m2c2 + 4a2m2 = 0

Here 4a = 4, ∴ a = 1.

⇒ – 4amc + 4a2 + 4a2m2 = 0 ⇒ – mc + a + am2 = 0 ⇒ mc = am2 + a ⇒ c = am + a . m 11. (a), (b)  Equation of the parabola is

Equation of given line is y – 2x + 5 = 0. Its slope = 2. Since the normal is parallel to the given line, ∴ slope of normal = 2 ⇒ m = 2. ∴ Equation of the normal is y = 2x – 2 ⋅ 1 ⋅ 2 – 23 i.e. 2x – y – 12 = 0. x y = 1 + 8. (a) The line p q will be a normal to the parabola y2 = 4ax if, for some value of m, it is identical with y = mx – 2am – am3

i.e. mx – y = (2am + am3) Comparing coefficients, we get 3 m −1 = = 2am + am 1/ p 1/ q 1

y2 = 8ax ...(1) ...(2)

Any tangent to it is given by y = mx + 2a  m [Here ‘4a’ = 8a ⇒ ‘a’ = 2a] Since it touches the circle x2 + y2 = 2a2,

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

∴ length of ⊥ from centre (0, 0) on (2) is equal to the radius 2 a

473

4 y (x – 2) ( y − y1 ) = − 1 ( x − x1 )  2×2 2a  

Conic Sections (Parabola, Ellipse and Hyperbola)

y – 4 =−

474

Objective Mathematics

2a / m

⇒

⇒ 2a = ± m

1 + m2

=± 2a

 quation of any tangent to (1), in the slope form, E is y = mx + a . m

2a 1 + m2 .

If it passes through (x1, y1), then y1 = mx1 +

4 = 2 (1 + m2) or 2 = m2 (1 + m2) m2 or m4 + m2 – 2 = 0 or (m2 + 2) (m2 – 1) = 0 ∴ m2 = – 2, 1. But m2 ≠ – 2 as it gives imaginary values of m. ∴ m2 = 1 or m = ± 1. Hence from (2) , y = ± x ± 2a i.e. y = ± (x + 2a), which are the required equations of common tangents. Squaring,

12. (a) Any tangent to the parabola y2 = 4ax is

y = mx + a m

...(1)

a It meets x2 = 4by where x2 = 4b  mx +  m   or mx2 – 4bm2x – 4ab = 0 ...(2) If the line (1) be a tangent to the second parabola, then roots of (2) must be equal. The condition for this is (“b2 – 4ac” = 0) 16b2m4 + 16abm = 0 1/ 3 a or m3 = − a . ∴ m = –   = – a . 1/ 3 b b b Substituting this value of m in (1), we get 1/ 3

1/ 3  b1/ 3  y=– a x + a   − 1/ 3  b1/ 3  a  1/ 3 x – a2/3 b1/3 or y = – a b1/ 3



or m2x1 – my1 + a = 0 ...(2) This is a quadratic in m and has two roots, say m1 and m2 which are the slopes of the tangents from (x1, y1) to the parabola. a  ...(3) ∴ m1m2 = product of roots of (2) = x1  ut the tangents are at right angles to each other B ∴ m1m2 = – 1 ...(4) a From (3) and (4), = – 1 or x1 = – a. x1 ∴ Locus of (x1, y1) is x = – a, which is the directrix of the parabola. 15. (a) The given parabola is y2 = 12x

I f it passes through the point (1, 4), then 4 =m+ 3 m or m2 – 4m + 3 = 0 ⇒ m = 3, 1. These are the slopes of tangents from (1, 4) to (1). If θ be the angle between the tangents, then 3 −1 = 1 1 + 3 ⋅1 2

1 ∴  θ = tan– 1   . i.e. xa1/3 + yb1/3 + a2/3 b2/3 = 0, 2 w hich is the required equation of the common tangent. 16. (b) The given parabola is y2 = 4ax ...(1)

Here 4a = 4, ∴ a = 1. ∴ Equation of any normal to (1) is y = mx – 2m – m3 

...(2) [y = mx – 2am – am3]

If it passes through (3, 0), then 0 = 3m – 2m – m3 or m3 – m = 0 or m (m2 – 1) = 0 

∴ m = 0, ± 1.

 utting these values of m in (2), the equations of the P normals are y = 0, y = x – 3 and y = – x + 3. 14. (c) Let the equation of the parabola be y2 = 4ax ...(1) Let (x1, y1) be the point of intersection of two ⊥ tangents to the parabola (1).

...(1)

Here 4a = 12 ∴ a = 3.  quation of any tangent to (1) , in slope form, is E y = mx + 3 . m

tan θ =

13. (d) The given parabola is y2 = 4x

a m

...(1)

 quation of any tangent to (1), in slope form , E is y = mx + a ...(2) m Let P (x1, y1) be the point of intersection of the tangents to (1). If (2) passes through P (x1, y1), then y1 = mx1 + a m or x1m2 – y1m + a = 0 ...(3) This is a quadratic in m and has two roots, say m1 and m2, which are the slopes of the tangents from P to (1). ∴ m1 + m2 =

y1 a , m 1m 2 = . x1 x1

 ince the inclinations of tangents are given to be θ1 S and θ2,

or (m1 + m2)2 – 2m1m2 = c or or

y 2a =c − x x1 2 1 2 1

Substituting for m from (5) in (6), we have

y12 – 2ax1 = cx12 .

a⋅

∴  Locus of P(x1, y1) is y2 – 2ax = cx2. y2 = 4ax

...(1)

Let (x1, y1) be the pole of any focal chord. Polar of (x1, y1) w.r.t. (1) is  yy1 = 2a (x + x1). Since this is a focal chord, it passes through the focus (a, 0) ∴ 0 = 2a (a + x1) or x1 + a = 0. ∴ Locus of (x1, y1) is x + a = 0 i.e. x = – a, which is the directrix of the parabola (1). 18. (a) The parabola is y2 = 2x

4a 2 + 2a + x1 = 0 or (x1 + 2a) y12 + 4a3 = 0. y12

Hence, locus of (x1, y1) is (x + 2a) y2 + 4a3 = 0.

17. (b) Let the equation of the parabola be

...(1) 1   Here 4a = 2, ∴ a = 2   

20. (b) Equation of any tangent to the parabola y2 = 4ax is ...(1) y = mx + a m L et (x 1 , y 1 ) be its pole w.r.t. the parabola y2 = 4bx ...(2)



Polar of (x1, y1) w.r.t. (2) is yy1 = 2b (x + x1)

...(3)

 ince (1) and (3) both represent the same line i.e. polar S of (x1, y1) w.r.t. (2), ∴ comparing coefficients y1 2b 2bx1 2b 2bmx1 = = or y1 = =  1 m a/m m a

...(4)

2b From the first two members of (4), m = y  ...(5) 1

Let (x1, y1) be the pole of the line a 2x – y = 0 ...(2) From the last two members of (4), m2 = x  ...(6) 1 w.r.t. (1). Polar of (x1, y1) w.r.t. (1) is Eliminating m between (5) and (6), we have 4b 2 a yy1 = 2 ·  1 (x + x1) [ yy1 = 2a (x + x1)] 4b 2 x1 . y 2 = x or y 2 = 2 a 1 1 1 4b 2 ...(3) or x – y1 y + x1 = 0 x. ∴ Locus of (x1, y1) is y2 = a Since (2) and (3) both represent the same line i.e. polar of (x1, y1) w.r.t. (1), therefore, comparing 21. (d) Let (x1, y1) be any point. coefficients Polar of (x1, y1) w.r.t. the parabola y2 = 4ax 1 y x 1 1 1 = 2a (x + x1) or 2ax – y1y + 2ax1 = 0 ...(1) is yy = = ⇒ x1 = 0, y1 = . 1 1 0 2 2 ...(2) Since it touches the circle x2 + y2 = 4a2 whose centre is (0, 0) and radius 2a, 1 Hence the pole of (2) is  0,  . ∴ length of ⊥ from centre (0, 0) on (1) is equal to  2 the radius 2a. 2 ...(1) 19. (c) The given parabola is y = 4ax 2ax1 ⇒ ± 4a 2 + y 2 = 2a or x12 = 4a2 + y12 Any normal to (1) is given by y = mx – 2am – am3

...(2)

Let the pole of (2) w.r.t. (1) be (x1, y1). Polar of (x1, y1) w.r.t (1) is yy1 = 2a (x + x1)

...(3)

 ince (2) and (3) represent the same line i.e. polar of S (x1, y1), ∴  comparing coefficients, we get

2ax1 y1 = 2a = − 2am − am3 1 m

...(4)

1

or x12 – y12 = 4a2. ∴ Locus of (x1, y1) is x2 – y2 = 4a2. 22. (a) The given parabola is y2 = 4ax

From the first two members of (4), 2a  m = y1

...(5)

...(1)

Let (x1, y1) be the pole of any chord of (1). ∴ Equation of chord [i.e. the polar of (x1, y1)] is yy1 = 2a (x + x1) or 2ax – y1 y + 2ax1 = 0 ...(2) Focus of the parabola (1) is (a, 0) Since the distance of (2) from focus (a, 0) is ‘a’ ∴ ±

2a ⋅ a − 0 + 2ax1 4a 2 + y12

=a

475

...(6)

Conic Sections (Parabola, Ellipse and Hyperbola)

From the last two members of (4) x1 or am2 + 2a + x1 = 0 1= − 2a − am 2

let m1 = tan θ1 and m2 = tan θ2. Then, tan2θ1 + tan2θ2 = c ⇒ m12 + m22 = c

476

⇒ ± 2 (a + x1) =

4a 2 + y12

2a   2 a   at − 2  +  2at +  t   t   2

Objective Mathematics

⇒ 4 (a + x1)2 = 4a2 + y12 or 8ax1 + 4 x12 = y 2 1



=



1   1 = a t2 − 2  + 4 t +  t  t  



2 2   1   1  = a  t +   t −  + 4  t   t  

2

2 ⇒ y1 = 4x1 (2a + x1). ∴ Locus of (x1, y1) is y2 = 4x (2a + x).

23. (c) The given parabola is y2 = 4ax.

2

2

 et (x1, y1) be the midpoint of the chord passing L through the point (h, k). 2 2 2 Equation of the chord having (x1, y1) as its mid  1  1 = a t + 1 . = a t + t +   point is     t t  t   2 T = S1 i.e. yy1 – 2a (x + x1) = y1 – 4ax1 2 27. (a) Equation of the normal at point (bt12, 2bt1) on the or yy1 – 2ax = y1 – 2ax1. parabola is Since it passes through (h, k), 2 y = – t1x + 2bt1 + bt13 ∴ ky1 – 2ah = y1 – 2ax1. The locus of the mid point (x1, y1) is ky – 2ah = y2 – 2ax,  or  y2 ­– 2ax – ky + 2ah = 0. 24. (a) The equation of the parabola is y2 = 4ax

...(1)

Let (x1, y1) be the mid point of a focal chord of the parabola. Equation of chord having (x1, y1) as its mid point is T = S1 i.e. yy1 – 2a (x + x1) = y12 – 4ax1 or yy1 – 2ax = y12 – 2ax1. Since it passes through the focus (a, 0) ∴ 0 – 2a2 = y12 – 2ax1 or y12 = 2a (x1 – a). Thus, the locus of the mid point (x1, y1) is y2 = 2a (x – a) , which is a parabola whose vertex is (a, 0). 25. (c) The given parabola is y2 = 4ax

It also passes through (b t22 , 2 bt2) 2bt2 = – t1. b t22 + 2bt1 + b t 3 1 ⇒ 2t2 – 2t1 = – t1 ( t22 – t12 ) = t1 (t2 + t1) (t2 – t1)

⇒ 2 = – t1(t2 + t1) 2 t1 2 ⇒ t2 = ­– t1 – t1

⇒ t2 + t1 =

28. (a), (b)  Given parabola is y2 = 4x Here 4a = 4 ⇒ a = 1.

...(1)

Let (x1, y1) be the mid point of any chord of (1), then its equation is T = S1 i.e. yy1 – 2a (x + x1) = y12 – 4ax1 or yy1 – 2ax = y12 – 2ax1 or 2ax – yy1 + ( y12 – 2ax1) = 0

...(2)

Since it touches the circle x2 + y2 = a2, ∴ length of ⊥ from the centre (0, 0) on (2) = radius a ⇒ ±

0 − 0 + y12 − 2ax1 4a 2 + y12

=a

or ( y12 – 2ax1)2 = a2 (4a2 + y12 ) ∴ Locus of (x1, y1) is (y2 – 2ax)2 = a2 (y2 + 4a2). 26. (a) Let the other extremity of the focal chord be (at 12, 2at1). −1 . ∴ tt1 = – 1 or t1 = t ∴ T he coordinates of the other extremity are  a − 2a   2, . t  t Length of the chord = distance between the points  a − 2a  (at2, 2at) and  2 ,  t  t

∴ Equation of directrix is x = – 1 Let P (x1, y1) be the point on the parabola having focal distance 6. ∴ y12 = 4x1 ...(1) Also, SP = PM (by definition of parabola) ∴ PM = 6 ⇒ x1 + 1 = 6 ⇒ x1 = 5. From (1), y12 = 4 × 5. ∴ y1 = ± 20 i.e. ± 2 5 ∴ The required points are (5, 2 5 ) and (5, – 2 5 ). 29. (a) Let R and T be the points of trisection of double ordinate QQ '. Let (h, k) be the coordinates of R. Then, AL = h and RL = k. ∴ RT = RL + LT = k + k = 2k. Since RQ = TR = Q' T = 2k ∴ LQ = LR + RQ = k + 2k = 3k. Thus, the coordinates of Q are (h, 3k). Since (h, 3k) lies on the parabola y2 = 9x

This is of the form X2 = – 4aY, where Y = y + 3, X = x + 2, 4a = 4. ∴ E  quation of the axis is given by X = 0 i.e. x + 2 = 0. Also, Equation of the directrix is Y =a i.e. y + 3 = 1 or y + 2 = 0. ∴ 9k2 = 9h or k2 = h.

33. (a) The equation of parabola parallel to y-axis is

Hence, the locus of (h, k) is y2 = x. 30. (c) Given parabola is y = 4ax 2

...(1)

The tangents at E and G are 2 t1 y = x + a t1

∴ 4 = 0 + 0 + c ⇒ c = 4

...(2)

9 = a + b + c ⇒ a + b = 5

and 5 = 16a + 4b + c ⇒ 16a + 4b = 1 

...(3)

2

31. (a) Let ABC be the equilateral triangle inscribed in the parabola

...(3)

(∵c = 4) (∵c = 4)

...(2)

...(4) and t3 y = x + a t3 Solving (3) and (4), we get 2 x = at1t3 = a t2  [from (2)]. Since the x-coordinate of the point of intersection 2 2 is a t2 , the point lies on the line x = a t2 i.e. on the 2 ordinate of F (a t2 , 2at2).



...(1)

 ince it passes through the points (0, 4), (1, 9) and S (4, 5),

Let the coordinates of E, F and G be respectively 2 2 3 (a t1 , 2at1), (a t2 , 2at2) and (a t2 , 2at3). Since ordinates of E, F and G are in G.P. 2 ∴ (2at2)2 = (2at1) (2at3) or t2 = t1t3

y = ax2 + bx + c

...(4)   (∵c = 4) .

Solving (3) and (4), we get 19 79 and b = ...(5) 12 12 Substituting the values of a, b and c from (2) and (5) in (1), we obtain the equation of parabola as −19 2 79 x + x+4. y= 12 12 34. (d) Let the normal at P (a t12 , 2at1) meets the parabola again at Q (a t22 , 2at2).

a =−

y2 = 8x  ...(1)

 et l be the length of the side of equilateral triL angle.

∴ PQ is a normal chord and 2 t2 = − t1 − t1

∴ AB = BC = CA = l.

Given : 2at1 = a t12

∴ The coordinates of B are (l cos 30º, l sin30º)

⇒ t1 = 2

l 3 l  i.e.  .  2 , 2    Since the point B lies on (1),

∴ from (1), t2 = – 3.

l ∴   2  

2

l 3 = 8    2 

Thus, P ≡ (4a, 4a) and Q ≡ (9a, – 6a). ∴ Slope of SP =

or l = 16 3 . slope of SQ =

32. (c) Equation of parabola is x2 + 4x + 4y + 16 = 0 ⇒ (x2 + 4x + 4) + 4y + 12 = 0

...(1)

...(1)

4 4a − 0 = and 3 4a − a

− 6a − 0 −3 = . 9a − a 4

Since slope of SP × slope of SQ = – 1 ∴ ∠ PSQ = π/2 i.e. PQ subtends a right angle at the focus S.

Conic Sections (Parabola, Ellipse and Hyperbola)

477

⇒ (x + 2)2 = – 4 ( y + 3).

478

35. (c) Let (h, k) be the point of intersection of three normals to the parabola y 2 = 4ax.

Objective Mathematics

The equation of any normal to y2 = 4ax is

∴ x1 = at1t2 and y1 = a (t1 + t2)

y = mx – 2am – am3.

or t1t2 = x1/a and t1 + t2 = y1/a

If it passes through (h, k) then

k = mh – 2am – am3

or

am3 + m (2a – h) + k = 0

...(1)

Let roots of (1) be m1, m2, m3, then from (1), k mm1m2m3 = − ...(2) a Also, m1 = tan α , m2 = tan β and tan α ⋅ tan β = 2 ∴ m1m2 = 2

...(3)

From (2) and (3), we get m3 =

 et the tangents from P touch the parabola at L Q (a t12 , 2at1) and R (a t22 , 2at2), then P is the point of intersection of tangents.

−k a. 2a

Since m3 is a root of (1), ∴ a m33 + m3 (2a – h) + k = 0 k  −k  ( 2a − h) + k = 0 ⇒ a   −  2a  2a

...(2)

(at12 − at22 ) 2 + (2at1 − 2at2 ) 2

Now QR =

2 2 2 = a (t1 − t2 ) [(t1 + t2 ) + 4]



2 = | a | | t1 − t2 | (t1 + t2 ) + 4



= | a | (t1 + t2 ) 2 − 4t1t2 ⋅ (t1 + t2 ) 2 + 4



= |a|

y12 4 x1 y12 − ⋅ +4 a2 a a2



= |a|

y12 − 4ax1 ⋅ |a|



1 2 2 2 = | a | ( y1 − 4ax1 ) ⋅ ( y1 + 4a ) .

[Using (2)]

y12 + 4a 2 |a|

3

⇒

−k3 kh −k + + k = 0 ⇒ k2 – 4ah = 0. 8a 2 2a

38. (c) Given parabola is y 2 = 4x

∴ Required locus of (h, k) is y2 – 4ax = 0. 36. (a) Given parabola is y 2 = 12x

...(1)

Here 4a = 12, ∴ a = 3.

Slope of OP =

Let P ≡ (3 t12 , 6t1) and Q ≡ (3 t2 , 6t2). 2

...(2)

 et R (α, β) be the point of intersection of the L normals to parabola (1) at Q and P, then 2 2 2 α = 2a + a  t1 + t2 + t1t2) = 6 + 21 t1  ...(3) (∵t2 = 2t1 )

 and β = – at1t2 (t1 + t2) = – 18 t 3 1

...(4)

α −6 β2 6 =  and from (4), t1 = 324  21  3

From (3), t

6 1

∴ β2 =

324 (α − 6)3 (21)3

or 343β2 = 12 (α – 6)3.

Hence locus of R (α, β) is 343y2 = 12 (x – 6)3. 37. (b) Given parabola is y 2 = 4ax Let P ≡ (x1, y1).

2t1 2 2 = and slope of OQ = . t12 t1 t2

Since OP ⊥ OQ, ∴

6t1 1 Given : 6t = or t2 = 2t1 2 2

...(1)

...(1)

Here 4a = 4, ∴ a = 1. Let P ≡ t12 , 2t1) and Q ≡ t22 , 2t2).

4 = – 1 or t1t2 = – 4 ...(2) t1t2

Let R (α, β) be the middle point of PQ, then t 2 + t22 α= 1 ...(3) and β = t1 + t2 ...(4) 2 From (4), β2 = t12 + t2 + 2t1t2 = 2α – 8 2



[From (2) and (3)]

Hence locus of R (α, β) is y2 = 2x – 8. 39. (c) Given, x = t2 + t + 1 and y = t2 – t + 1 ⇒ x + y = 2 (t2 + 1) and x – y = 2t. Eliminating t, we get

x+ y=

or

1 ( x − y)2 + 2 2

x2 – 2xy + y2 – 2x – 2y + 4 = 0

Here a = 1, h = – 1, b = 1, 

g = – 1, f = –1, c = 4.

∴ ∆ = abc + 2fgh – af 2 – bg2 – ch2 = 1 ⋅ 1 ⋅ 4 + 2 (– 1) (– 1) (– 1) – 1 ⋅ 1 – 1 ⋅ 1 – 4 ⋅ 1 = 4 – 2 – 1 – 1 – 4 = ­– 4 ≠ 0. Also, h2 – ab = (­– 1)2 – 1 ⋅ 1 = 0. Hence, the given curve represents a parabola.

∴ distance between the vertex and the directrix = 4. ∴ Distance of focus from the directrix = 2 × 4 = 8. ∴ Coordinates of the focus, which is on x-axis, are (8, 0). 41. (b) Consider two points P (a t12 , 2at1) and Q (a t22 , 2at2) on the parabola y 2 = 4ax. at12 m2 Given : at 2 = or t1 = mt2 1 2

...(1)

 et R (h, k) be the point of intersection of tangents  L at P and Q. Then, h = at1t2 and k = a (t1 + t2) ⇒ h = am t22 and k = a (mt2 + t2)

∴ midpoint of diagonals TY and GA is same ⇒

x1 + 0 − at 2 + 0 y +0 0 + at = and 1 = 2 2 2 2

⇒ x1 = – at2

...(1)

and y1 = at

[Using (1)]

...(2)

Eliminating t from (1) and (2), we get y  x1 = − a  1   a 2



2 ⇒ y1 + ax1 = 0.

∴ The locus of G (x1, y1) is y2 + ax = 0.

k h and t2 = a (m + 1) am Equating the two values of t2, we get k2 h 2 a (m + 1) 2 = am ⇒ t22 =

44. (c) The equation of the given parabola can be written in the form (x – 1)2 = y – 1 which is of the form X2 = Y, 2

1   (m + 1) 2  . ⇒ k2 = ah ⇒ k2 = ah  m + m  m ∴ Required locus is y2 = ax (m1/2 + m– 1/2)2. 42. (a) It is clear from the figure that equation of directrix is y = 6. Let P (x, y) be any point on the parabola. Then, by definition, PS = PM ⇒ PS2 = PM2 ⇒ (x – 0)2 + ( y – 2)2 = (6 – y)2 ⇒ x2 + y2 – 4y + 4 = 36 + y2 – 12y ⇒ x2 + 8y = 32, which is the required equation of parabola.

43. (b) L et P (at 2 , 2at) be any point on the parabola y 2 = 4a x. T he equation of t angent at P is ty= x + at2.  ince the tangent meets the axis of parabola in T S and tangent at the vertex A in Y, ∴ coordinates of T and Y are (– at 2, 0) and (0, at) respectively. Let the coordinates of G be (x1, y1).

Since TAYG is a rectangle,

where X = x – 1 and Y = y – 1. Here 4a = 1, ∴ a = 1/4. ∴ The coordinates of focus are X = 0, Y = a 1 i.e. x – 1 = 0, y – 1 = 4 or x = 1, y =

 5 i.e. 1, 4 

5 . 4

45. (a) Let the coordinates of P1 and P2 be (a t1 , 2at1) and (a t22 , 2at2) respectively. 2

Since P1Q1 and P2Q2 are focal chords, ∴ coordinates of Q1 and Q2 are  a − 2a   a − 2a   respectively.  2,  and  2 ,  t2 t2   t1 t1 

∴ Equation of P1P2 is

y (t1 + t2) = 2 (x + at1t2)

...(1)

and equation of Q1Q2 is   −1  −1   − t   1 y  , −  = 2  x + a     t t 2   1  t1   t2    ⇒ – y (t1 + t2) = 2 (xt1t2 + a) Adding (1) and (2), we get

...(2)

Conic Sections (Parabola, Ellipse and Hyperbola)

479

40. (a) Since y-axis is the directrix and (4, 0) is the vertex of the parabola,

480

0 = 2 (1 + t1t2) (x + a)



⇒ x + a = 0

t1t2 ≠ – 1),

Objective Mathematics

50. (c) Given parabolas are y 2 = 4a (x – k1)

which is the directrix of the parabola. 46. (c) Since the axis is horizontal and vertex is (­– 3, – 2), ∴  the equation of the parabola must be of the form

( y + 2)2 = 4a (x + 3).

It passes through (1, 2), so 16 = 16a i.e. a = 1. Hence, the equation of the required parabola is ( y + 2)2 = 4 (x + 3) or y2 + 4y – 4x – 8 = 0.



Since m1 × m2 = – 1, ∴ ∠PAQ ∴ π/2 i.e. the double ordinate subtends a right angle at the vertex.

and

x = 4a ( y – k2) 2

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

Let (α, β) be their point of contact. Equation of tangent to (1) at (α, β) is βy = 2a (x – k1 + α) or 2ax – βy = 2a (k1 – α)

...(3)

Equation of tangent to (2) at (α, β) is

αx = 2a ( y – k2 + β)

or αx – 2ay = 2a (β – k2)

...(4)

 ince (3) and (4) are identical, comparing coefficients S of x and y in (3) and (4), we get 2a β = α 2a ⇒ αβ = 4a2 . i.e. the point of contact (α, β) lies on the curve xy = 4a2. 51. (a)  Let the centre of the circle be (α, β). 47. (c) Let R (– 4a, k) be any point on the line x = – 4a. The equation of chord of contact PQ w.r.t. P (– 4a, k) is

y ⋅ k = 2a (x – 4a) 

...(1)

 aking equation of parabola y = 4ax homogeneous M using (1), we get 2

 2ax − yk  y = 4ax   ⇒ 8a2x2 – 8a2y2 – 4akxy = 0. 2  8a  This represents the pair of straight lines AP and AQ. Since coefficient of x2 + coefficient of y2 = 0 ∴ ∠ PAQ = 90º i.e. chord of contact PQ subtends a right angle at the vertex. 2

48. (c) The given equation of parabola can be written in the from (x – 3)2 = 3 ( y + 1). ∴ E  quation of axis of the parabola is x – 3 = 0 i.e. x = 3.

Since the circle passes through (a, 0), ∴ its radius =

(α − a ) 2 + β 2 .

Also, the circle touches the line x + y = 0, | α + β| 2 2 = (α − a ) + β 2 ⇒ (α + β)2 = 2 [(α – a)2 + β2]

so

⇒ α2 + β2 + 2αβ = 2α2 + 2β2 – 4aα + 2a2 ⇒ α2 + β2 – 2αβ – 4αa + 2a2 = 0. Hence locus of (α, β) is

x2 + y2 – 2xy – 4ax + 2a2 = 0.

Here, h2 – ab = 0 but

∆ = abc + 2fgh – 2f 2 – bg2 – ch2 ≠ 0

so it represents a parabola. 52. (a) The equation of parabola is y 2 = 4x. Here 4a = 4, ∴ a = 1.

The line y = mx + 1 will be a tangent to the given 49. (c) Let PQ be a double ordiante of length 2a of the parabola, if c = a/m i.e. 1 = 1/m or m = 1. parabola y2 = ax. Then, the coordinates of P and Q are (a, a) and (a, – a) respectively. 53. (d) Given parabola is y 2 = 8x. Here 4a = 8, ∴ a = 2. Let the point of contact be (x1, y1). Equation of tangent to the parabola at (x1, y1) is

a−0 Now, slope of AP = m1 = = 1 and a−0 slope of AQ = m2 =

−a − 0 = – 1. a−0

yy1 = 2 ⋅ 2 (x + x1) i.e. 4x – y1y + 4x1 = 0

...(1)

But the line x – y + 2 = 0

...(2)

is also a tangent at (x1, y1), so comparing coefficients in (1) and (2), we get − y1 4 4 x1 = −1 = ⇒ x1 = 2 and y1 = 4. 1 2 ∴ The point of contact is (2, 4).

a = 1.

 quation of pair of tangents drawn from the point E (­– 2, – 1) to the parabola is SS1 = T2 i.e. ( y2 – 4x) [(– 1)2 – 4 (–2)] = [y (– 1) – 2 ⋅ 1(x – 2)]2



⇒ 9 (y2 – 4x) = (– 2x – y + 4)2 Since α is the angle between the tangents, 2 h 2 − ab 2 4 − 4 (− 8) = = | – 3 | = 3. a+b 4−8

55. (c) The equation of the normal to the parabola y 2 = 4ax at the point (at2, 2at) is

...(1)

2

and

t2 y = x + a t

y = – tx + 2at + at3

∴ Slope of normal = – t.

 he point of intersection of (1) and (2) is given by T x = at1t2 and y = a (t1 + t2). ∴ x = – a

( t1t2 = – 1).

Thus (1) and (2) intersect on the directrix.

 ence the two tangents (1) and (2) intersect at right H angles on the directrix. 60. (c) The equation of the tangent at P (at2, 2at) to

y2 = 4ax is ty = x + at2

...(1)

It meets the directrix x = – a. ∴ ty = – a + at2 ⇒ y =

56. (b) E quation of the line joining (au 2 , 2au) and (av 2, 2av) is

...(2)

2 2

1 1  he product of the slopes of (1) and (2) = t ⋅ t T 1 2 = – 1.

⇒ 4x2 – 8y2 + 4xy + 20x – 8y + 16 = 0

∴  tan α =

t1 y = x + a t1

a (t 2 − 1) . t

2 a (v − u ) ( x − au 2 ) a (v 2 − u 2 )

y – 2au =

2 ( x − au 2 ) . u+v

or y – 2au =

 ince it is a focal chord, therefore focus (a, 0) must S lie on it 2 2 ∴ 0 – 2au = u + v (a − au ) ⇒ uv = – 1. 57. (d) Given parabola is y 2 = 16x.

 a (t 2 − 1)  . Thus, (1) meets the directrix at Q  − a, t   2at − 0 Now, slope of PS is m1 = at 2 − a

Here 4a = 16, ∴ a = 4. Slope of given line is 3, ∴ slope of tangent (⊥ to given line) is m = ∴ Equation of tangent to the parabola is −1 4 a y = mx + i.e. y = 3 x + (−1 / 3) m i.e. 3y + x + 36 = 0.

−1 . 3

slope of QS is m2 =

2t = t 2 − 1 and

a (t 2 − 1) / t − 0 (t 2 − 1) − = . −a − a 2t

 ince m1m2 = ­– 1, therefore PQ subtends a right S angle at the focus.

61. (c) Since the parabola y 2 = 4ax passes through the point (1, – 2), ∴ (­– 2)2 = 4a (1) ⇒ a = 1. 58. (a), ( b)  Equation of normal to the parabola y 2 = 4ax 2 at the point (at , 2at) is Equation of tangent to the parabola at (1, – 2) is y + tx = 2at + at3 ...(1) yy1 = 2a (x + x1) or y (­– 2) = 2 (1) (x + 1) Since it passes through the point (5a, 2a), ∴ 2a + 5at = 2at + at3 ⇒ t3 – 3t – 2 = 0 or (t – 2) (t + 1)2 = 0 ∴ t = – 1, – 1, 2.

62. (c) The given parabola can be written in the form

Putting t = – 1 and t = 2 in (1), the normala from (5a, 2a) are y = x – 3a and y = – 2x + 12a. 59. (a), (b)  Let P (a t , 2at1) and Q (a t , 2at2) be the extremities of a focal chord of the parabola so that t1t2 = – 1. 2 1

or x + y + 1 = 0.

2 1 5  y −   = 2 ( x − 4) . 2 

∴ Length of latus rectum = 4a =

2 2

The equations of the tangents at P and Q are

1 . 2

63. (c) Given parabola is x2 = 4ay

...(1)

and the line is y = mx + c

...(2)

481

Here 4a = 4, ∴



Conic Sections (Parabola, Ellipse and Hyperbola)

54. (a) Given parabola is y 2 = 4x.

482

Eliminating y between (1) and (2), we get

x2 – 4amx – 4ac = 0

...(3)

Objective Mathematics

I f line (2) touches the parabola (1), then the roots of (3) must be real and equal. ∴ (­­– 4am)2 = 4 ⋅ 1⋅ (­– 4ac) ⇒ c = – am2.

1  = a  2 + 1 = k t 

(given)

Dividing (1) by (2), we get t2 =

...(2)

b . k

64. (a) Let (h, k) be the point of intersection of three normalsto the parabola y 2 = 4ax. The equation of any normal to y 2 = 4ax is

y = – tx + 2at + at3.

If it passes through (h, k), then

k = – th + 2at + at3

or at3 + (2a – h) t – k = 0

...(1)

If the normals to the parabola from A (h, k) meet it at the points

P (a t , 2at1), Q (a t , 2at2) and R (a t , 2at3), 2 1

2 2

2 3

then roots of (1) will be t1,t2,t3.

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

Let (x1, y1) be the pole of (2) w.r.t. (1). The equation of the polar of (x1, y1) w.r.t. (1) is yy1 = 2a (x + x1)

i.e. 2ax – y1 y + 2ax1 = 0



y2 = 4x

...(1)

2 t2 = − t1 − t

...(3)

I f PQ subtends a right angle at the vertex (0, 0), then the slope of OP × slope of OQ = ­– 1 ⇒

2t1 2t2 −4 x 2 =–1 ⇒t = 2 t1 t12 t2

2a − y1 2ax1 = = l m n

∴ Coordinates of P and Q are (2, ∓ 2 2 ) and (8, ∓ 4 2 ) ∴ PQ =

36 + 72 = 6 3 .



we get mx2 + lx + n = 0

2am n and y1 = − . l l

69. (a) Let the coordinates of Q be (h, k). Since the point R lies on the parabola, let its coordinates be (at2, 2at).

and

(at 2 − a ) 2 + (2at − 0) 2

= a (t2 + 1) = b (given) SQ=

a   − 2a  − 0  2 − a +  t   t  2

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

∴ l2 = 4mn.

66. (a) If coordinates of one end of focal chord are P (at2, 2at) then the coordinates of other end will be  a − 2a  Q  2, . t  t



...(1)

I f (1) is a tangent to (2), roots of (3) are real and equal.

 n − 2am  ∴ The required pole is  , . l  l

∴ SP =

...(3)

From (2) and (3), t1 = ± 2 , t2 = ∓ 2 2 .

 ince (2) and (3) both represent the polar of (x1, y1) 68. (d) Eliminating y between S w.r.t. (1), lx + my + n = 0 ∴ comparing coefficients, we get and x2 = y

∴ x1 =

...(2)

1

 ence the C.G. of ∆ PQR lie on x = 0 i.e. on the H axis of the parabola.



2



1 2a (2at1 + 2at2 + 2at3 ) = Σt1 = 0. 3 3

and the line is lx + my + n = 0

67. (a) Normal at the point P  t1 , 2t1) on the parabola meets it again at the point Q  t , 2t2), where

If G ( x , y ) is the C.G of the ∆ PQR, then

65. (b) Given parabola is y 2 = 4ax

ab b b +1 = ⇒ k = b−a . k a

2 2

∴ Σt1= 0, Σt1t2 = 2a – h, t1t2t3 = k/a.

x =

Putting in (1),

...(1) 2

Since R is mid point of PQ, x +h y +k ∴ at2 = 1 and 2at = 1 2 2

 y1 + k  x1 + h ⇒ ( y1 + k)2 = 8a (x1 + h)   = 2a  4a  2

Hence, locus of Q (h, k) is (y + y1)2 = 8a (x + x1). 70. (c) Chord PQ is normal to the parabola at Q and its equation is

73. (a) The given equation of the parabola can be written as  4 x − 3 y + 12 (x – 2)2 + (y – 4)2 =   (4) 2 + (− 3) 2 

2

  .  

∴ The coordinates of focus are (2, 4) and the equation of directrix is 4x – 3y + 12 = 0. The distance of the focus from the directrix

y = mx – 2am – am3 ...(1)

The slope of normal PQ is m = tan θ. The joint equation of OP and OQ is obtained by making the equation of parabola y2 = 4ax homogeneous with the help of (1). Thus joint equation of lines OP, OQ is

y2 = 4ax (mx – y)/(2am + am3)

or m (2 + m2) y2 + 4xy – 4mx2 = 0 ...(2) π Since ∠POQ = , the sum of the coefficients of 2 x2 and y2 in (2) must be zero, i.e. m (2 + m2) – 4m = 0 or m (m2 – 2) = 0  s m ≠ 0, we get m2 = 2 or m = ± 2 . Thus, m A = 2. 71. (b) The equation of the parabola is 3  2x + 5   2  x +  + ( y + 3) =   2  2   2

2



=

4 (2) − 3 (4) + 12 42 + (− 3) 2

=

8 . 5

8 16 = . 5 5 74. (a), ( b)  Solving the two equation x2 = ay and y –2x = 1, we get ∴ The length of latus rectum = 2 ×



x2 = a (2x + 1) or x2 – 2ax – a = 0.

∴ x1 + x2 = 2a and x1x2 = – a.  o, the given line cuts the parabola at two points S (x1, y1) and (x2, y2). 2 Now, ( 40 ) = (x1 – x2)2 + (y1 – y2)2

 x2 x2  ⇒ 40 = ( x1 − x2 ) 2 +  1 − 2  a   a

[Given]

2

 ( x + x )2  = ( x1 − x2 ) 2 1 + 1 2 2  a  

 4a 2  = [( x1 + x2 ) 2 − 4 x1 x2 ]  2 + 1 = 5(4a2 + 4a).  a  ∴ a2 + a – 2 = 0 or (a + 2) (a – 1) = 0 ∴ a = 1, – 2.

 2 9  ⇒ 4  x + + 3 x  + 4 [y2 + 9 + 6y] 4    = (4x2 + 25 + 20x) 2 ⇒ ( y + 6y + 9) – 2 (x + 2) = 0 75. (a) The equation of axis of the parabola is x – 4 = 0 which is parallel to y-axis. So the ray of light is or ( y + 3)2 = 2 (x + 2). parallel to the axis of the parabola. We know that Clearly, x = 2t2 – 2 and any ray parallel to the axis of a parabola passes y = 2t – 3 satisfy it for all t. through the focus after reflection. 72. (c) The given conic is

ax + by = 1

Squaring both sides,



ax + by + 2

abxy = 1

or ax + by – 1 = – 2 abxy . Squaring again, (ax + by – 1)2 = 4abxy or a2x2 – 2abxy + b2y2 – 2ax – 2by + 1 = 0...(1) Comparing the equation (1) with the equation Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0 ∴ A = a2, H = – ab, B = b2, G = – a,  F = –b, C = 1.

∴ The ray must pass through the point (4, – 1). 76. (b) Let S ≡ y 2 – 2x. Then S ]P (4, 2)= (2)2 – 2 (4) = 4 – 8 < 0 ∴ Point P lies inside the parabola. Also, S]Q (1, 4) = (4)2 – 2 (1) = 16 – 2 = 14 > 0 ∴ Point Q lies outside the parabola. 77. (b) Let S ≡ x2 – 4y Since the point (2a, a) lies inside the parabola, ∴ S](2a, a) = 4a2 – 4a < 0

483

Then, ∆ = ABC + 2FGH – AF2 – BG2 – CH2 = a2b2 – 2a2b2 – a2b2 – a2b2 – a2b2 = – 4a2b2 ≠ 0 and H2 = a2b2 = AB. So we have ∆ ≠ 0 and H2 – AB = 0. Hence the given equation represents a parabola.

Conic Sections (Parabola, Ellipse and Hyperbola)

x1 + h y +k and t = 1 Equating the two 2a 4a values of t, we get

⇒ t2 =

484

i.e. 4a (a – 1) < 0 or a (a – 1) < 0

...(1)

Objective Mathematics

 lso, the vertex A (0, 0) and the point (2a, a) are A on the same of the line y = 1 (the equation of latus rectum) So, a – 1 < 0 i.e. a < 1

...(2)

∴ Equation of circle is x2 + y2 = 5. 82. (c) Let the feet of the normals from the point (2, 6) be (x1, y1), (x2, y 2) and (x3, y 3). Then, arithmetic mean of the ordinates of the feet y + y2 + y3 of the normals = 1 = 0. 3 [ the sum of the ordinates of the feet of three normals is zero.] 83. (b) Clearly the two tangents, having slopes m1 and m2, meet on the line x = – 2, which is the directrix of the parabola y 2 = 8x, therefore the two tangents must be at right angles, i.e. m1m2 = – 1.

From (1) and (2), we have a (a – 1) < 0 or

0 < a < 1.

78. (a), (d)  Let (x1, y1) ≡ (a t12 , 2at1) and (x2, y 2) ≡ (a t22 , 2at2).



Then, (x3, y3) = [at1t2, a (t1 + t2)] ∴ x1x2 = a t12 ⋅ a t22 = (at1t2)2 = x32 and

1 y3 = a (t1 + t2) = ( y1 + y2 ) 2

∴ x1, x2, x3 are in G. P and y1, y2, y3 are in A.P. 79. (a) The locus of the middle point of the parallel chords of a parabola is a straight line parallel to the axis of the parabola and here the axis is y = 0 i.e. xaxis. 80. (c) Let P ≡ (h, k). Equation of any tangent to the parabola is 2 y = mx + (Here a = 2). m 2 Since it passes through (h, k), ∴ k = mh + m ⇒ m2h – km + 2 = 0. Its roots are m1 and 2m1 (given). So, m1 + 2m1 = ⇒ 3m1 =

k 2 and m1 ⋅ 2m1 = . h h

k 1 2 and m1 = h h

1 k2 ⇒ 9   = 2 or k2 = 9h. h h ∴ The locus of (h, k) is y2 = 9x, which is a parabola.

84. (b) We have, x = 3 + t2 and y = 3t – 2 ⇒ x – 3 = t2 and y + 2 = 3t ⇒ ( y + 2)2 = 9 (x – 3) which is a parabola with vertex at (3, – 2), focus  21  3 at  , − 2  and directrix x = . 4  4  85. (b) We have, x = sin 2 t and y = 2 cos t ⇒ sin2t = x and cos2t = ⇒ x+

y2 4

y2 = 1 or y2 = – 4 (x – 1) 4

which represents a parabola. 86. (d) T he equation of a normal to y 2 = 4x at (m2, – 2m) is y = mx – 2m – m3. If the normal makes equal angles π = 1. with the coordinates axes, then m = tan 4 Thus, the required point is (1, – 2). 87. (c) Given parabola is y 2 = x. Here 4a = 1, ∴ a = 1/4. ∴ Equation of the normal to the parabola is y = mx −

1 1 m − m3 . 2 4

[ y = mx – 2am – am3]

If it passes through (c, 0), then 1 1 3 0 = mc − m − m 2 4 ⇒ m = 0 or m = ± 2 c −

1 . 2

For three normals, values of m should be real. 1 ∴ c − > 0 i.e. c > 1/2. 2

81. (a) Coordinates of the vertex of the parabola x2 = 4y 88. (c) Since the semi latus rectum of a parabola is the are (0, 0) and the ends of latus rectum are (2, 1) harmonic mean between the segment of any focal and (­– 2, 1). chord of a parabola, therefore SP, 4, SQ are in ∴ Centre of the circle is (0, 0) and radius of the H.P. circle is 6 ⋅ SQ SP ⋅ SQ ⇒ 4 = 2 SP + SQ ⇒ 4 = 2 ⋅ 6 + SQ ⇒ SQ = 3. = (2) 2 + (1) 2 = 5 .

485

In the present case, y2 = 12x

Let P (at , 2at) be any point on the parabola.

⇒ 4a = 12 or a = 3

Then SP is focal radii of the parabola.

and x + y = k ⇒ y = (– 1) x + k . or m = – 1.

The equation of a circle with SP as diameter is

∴ k = – 2 (3) (–1) – 3 (­– 1)3 = 9.

2



(x – a) (x – at2) + ( y – 0) ( y – 2at) = 0.

It meets y-axis at x = 0 ∴ y2 – 2aty + a2t2 = 0 i.e. (y – at)2 = 0 ⇒ y-axis meets the circle only at one point.  herefore, the circle touches the tangent at the T vertex.

94. (a) Equation of the line parallel to the axis and bisecting the ordinate PN of the point P (at2, 2at) is y = at which meets the parabola y 2 = 4ax at the point 1  Q  at 2 , at  . 4 

90. (d) Let P (x1, y1) be the point of contact of the two given parabolas

y2 = 4a (x – 2)

and

...(1)

x = 4a (y – 3)

...(2)

2

Equation of tangent at P to (1) is

yy1 = 2a (x + x1) – 8a

or 2ax – y1 y + (2ax1 – 8a) = 0

Equation of tangent at P to (2) is



xx1 = 2a ( y + y1) – 12a

or x1x – 2ay – (2ay1 – 12a) = 0

...(3)

...(4)

Since (3) and (4) represent the same line, ∴

2a − y1 = ⇒ x1 y1 = 4a2. x1 − 2a

∴ point of contact (x 1, y 1) lies on the curve xy = 4a2.

which is a hyperbola.

91. (b) We know that semi-latus rectum of a parabola is the harmonic mean of segments of a focal chord. AS ⋅ SB 16 ∴  Semi latus rectum = 2 AS + SB = . 6 ∴  Latus rectum =

32 16 i.e. . 6 3

8  92. (c) The given parabola is y 2 = kx – 8 = k  x −  . k   hifting the origin to (8/k, 0), the parabola becomes S Y2 = kX where X = x – 8/k and Y = y. −k Directrix of this parabola is X = 4 8 −k = . or x − k 4 8 k − This will coincide with x = 1 if =1 k 4 ⇒ 32 – k = 4k ⇒ k + 4k – 32 = 0 2

2

or (k + 8) (k – 4) = 0 ⇒ k = ­– 8, 4. Thus, k = 4. 93. (b) y = mx + k is a normal to y 2 = 4ax if

k = – 2am – am3.

Coordinates of N are (at2, 0). 0 − at Equation of NQ is y = ( x − at 2 ) , 1 2 2 at − at 4 which meets the tangent at the vertex, x = 0, at the 4 point y = at . 3 95. (d) E quation of any tangent to the parabola y 2 = 4ax is

ty = x + at2

...(1)

The difference of the squares of the perpendiculars (a + k + at 2 ) 2 (a − k + at 2 ) 2 from (a ± k, 0) to (1) is 1 + t2 =

4a (1 + t 2 ) k = 4ak. 1 + t2

96. (c) The focus of the parabola y 2 = 8ax is (2a, 0).  o, the coordinates of the point on the axis of the S parabola at a distance 8a from the focus is (10a, 0). Equation of a normal to the parabola y2 = 8ax is

y = mx – 4am – 2am3.

Since it passes through (10a, 0), ∴ 0 = 10am – 4am –2am3 ⇒ 2am (3 – m2) = 0 ⇒ m2 = 3  π ⇒ m = ± 3 = tan  ±  .  3

m ≠ 0)

97. (b) E quations of two tangents to the parabola y 2 = 4ax are y = mx +

a m

...(1)

and y = m′x +

a m'

...(2)



Conic Sections (Parabola, Ellipse and Hyperbola)

89. (c) Let the parabola be y 2 = 4ax.

486

Since (1) and (2) are ⊥,

103. (a) Solving x2 = 4y and y 2 = 4x, we get x = 0, y = 0

Objective Mathematics

∴ mm' = – 1 ⇒ m ' = – 1/m. −1 x − am So, from (2), y = m

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

1  (1) – (3) ⇒ 0 =  m +  ( x + a ) ⇒ x + a = 0. m  ∴ The two tangents intersects on the line x + a = 0. 98. (b) Equation of tangent is

y = mx +

1 1   or  4 = m + m m

⇒  m2 – 4 m + 1 = 0 ⇒ m1 + m2 = 4, m1m2 = 1 m1 − m2 ∴ tan θ = 1 + m m = 1 2

=

12 = 2

(m1 + m2 ) 2 − 4m1m2 1 + m1m2 π . 3

3 ⇒θ=

99. (c) Given parabola is y 2 = 4ax. Since it passes through (2, – 6), ∴ 36 = 4a (2) ⇒ a =

9 36 9 = .   ∴  4a = 4   = 18. 8 2 2

∴ length of latus rectum = 18.

P is (4, 4).

∴

 he equations of the tangents to the two parabolas at T (4, 4) are

2x – y – 4 = 0

...(1)

and x – 2y + 4 = 0

...(2)

m1 = slope of (1) = 2 and m2 = slope of (2) = Since m1m2 = 1 i.e. tan θ1 ⋅ tan θ2 = 1 ∴ θ1 and θ2 are such that θ1 +θ2 = i.e. θ2 =

π 2

π − θ1 . 2

104. (d)  The given line is y = Here m =

1 . 2

3 5 x+ . 4 4

3 5 and c = . 4 4

We know that y = mx + c touches the parabola a y2 = 4ax if c = . m ∴ The given line touches 5 a 5 3 15 y2 = 4ax if = ⇒a= × = . 4 3/ 4 4 4 16

100. (d) Since the semi latus rectum of a parabola is the 105. (a) Let the coordinates of the point be (h, k). harmonic mean between the segments of any focal Distance of the point from origin chord of the parabola. = (h − 0) 2 + (k − 0) 2 = h 2 + k 2 . ∴ l is the harmonic mean between b and c. Hence, l =

 istance of the point from the line x –­ 2 = 0 is D h – 2.

2bc . b+c

101. (d) T he value of the parameter for the other end of the 1 focal chord is = − . t ∴ The coordinates of the other end of the focal  a − 2a  chord are  2 , . t  t ∴ Length of the focal chord 2a   2 a   at − 2  +  2at +  t   t   2

=

2

4 = a t +

1 4 − 2 + 4t 2 + 2 + 8 4 t t

4 = a t +

1 4 + 4t 2 + 2 + 6 t4 t 2

2  2 1   1  2 1  = a t + t2 + 2 = a t + 2 + 2 = a t +  . t t     

102. (a), (c)  Equation of parabola is y 2 = – 4ax. Its focus is at (–a, 0).

According to the given condition , h2 + k 2 + h – 2 = 4 ⇒

h 2 + k 2 = 6 – h.

Squaring both sides, we have,

h2 + k2 = 36 + h2 – 12h or k2 = – 12 (h – 3).

∴ The path of the point is y2 = – 12 (x – 3). 106. (a) Equations of tangents drawn from the origin to the parabola y 2 = 4a (x – a) is SS1 = T2 i.e. [ y2 – 4a (x – a)] [0 – 4a (0 – a)] = [ y × 0 – 2a (x – 0)]2



⇒ ( y2 – 4ax + 4a2)4a2 = 4a2x2 

or y2 – 4ax + 4a2 = x2

or x2 – y2 + 4ax – 4a2 = 0.  ince the sum of coefficients of x2 and y2 is zero, S ∴ the angle between the tangents is 90º. 107. (a) The given equation of the parabola can be written as

61   (3y – 2)2 = 16  x +  . 16  

∴ The axis of the parabola is 3y – 2 = 0.

and x2 = 4ay

...(2)

3  ⇒ ( y – 2)2 = 6  x −  2 

Putting the value of y from (2) in (1), we get

x4 16a 2

⇒ ( y – 2)2 = 4 ⋅

= 4ax ⇒ x (x3 – 64a3) = 0 ⇒ x = 0, 4a From (2), y = 0, 4a

 ince line 2bx + 3cy + 4d = 0 passes through A and S B ∴ d = 0 and 8 ab + 12ac = 0 ( a ≠ 0)

⇒ 2b + 3c = 0.

109. (b) Since directrix is parallel to y-axis hence axes of the ellipse are parallel to the coordinate axes. Let the equation of the ellipse be x2 y2 = 1 (a > b) 2 + a b2 2

b b2 1 3 = e = 1 – 2 ⇒ 2 = 1 – e2 = 1 – a a 4 4 Also, one directrix is x = 4 2



a 1 = 4 ⇒ a = 4e = 4. =2 ⇒ e 2



b 2 =

3 2 3 a = .4=3 4 4 x2 y2 + =1 4 3

or 3x + 4y = 12. 2

110. (c) α =

3 . 2

∴ Coordinates of focus are x −

Let A ≡ (0, 0), B ≡ (4a, 4a)

∴ Required ellipse is

3 x − 2

487

...(1)

i.e. (3, 2).

3 3 = ,y–2=0 2 2

112. (d) The given equation can be written in the form

12 (x + 1)2 + 4 (y – 2)2 = 3

or

( x + 1) 2 ( y − 2) 2 + = 1 1/ 4 3/ 4

...(1)

Coordinates of centre of the ellipse are given by x + 1 = 0 and y – 2 = 0 i.e. (­ – 1, 2). I f a and b are the lengths of the semi major and 3 2 1 semi axes, then a2 = , b = . 4 4 ∴ Length of major axis = 2a =

3 and

length of minor axis = 2b = 1. 3 1 − 2 2 a − b 2 Eccentricity is given by e2 = =4 4= . 2 3 a 3 4 2 ∴ e = . 3 113. (a) We have a2 = 16 and 9 = b2 = a2 (1 – e2) = 16 ( 1– e2)

2

at 2 + a 2at + 0 ,β= ⇒ 2α = at2 + a, at = β 2 2

7 4

So, e =

Thus, the foci are (±

7 , 0) The radius of required circle is

=

( 7 − 0) 2 + 32 =

114. (b) Let the ellipse be It is given that e = β2 + a or 2aα = β2 + a2 a2 4a  a ∴ the locus is y2 = x−  2  2 ∴ 2α = a

a  = 4b(x – b),  b =  2 



∴ directrix is (x – b) + b = 0 or x = 0. 111. (b) Given equation of parabola is

y – 4y – 6x – 13 = 0 2

⇒ y2 – 4y + 4 – 6x + 9 = 0 ⇒ ( y – 2)2 – 6 (x – 3/2) = 0

7 + 9 = 4.

y2 x2 + 2 =1 2 b a 1 and ae = 2 2

Therefore, a = 4 Now, b2 = a2 (1 – e2) ⇒ b2 = 12 Thus, the required ellipse is 115. (a) Latus rectum = Also e =

x2 y2 + = 1. 16 12

2b 2 = 8, ∴ b 2 = 4a a

...(1)

1 , ∴ b2 = a2 (1 – e2) 2

 ⇒ 4a = a2 1 − 

1  2

[Using (1)]

Conic Sections (Parabola, Ellipse and Hyperbola)

108. (c) Given parabolas are y 2 = 4ax

488

Objective Mathematics

∴ From (1), the equation of the ellipse is x2 y2 + = 1 or 3x2 + 5y2 = 32. 32 / 3 32 / 5

1  a, ∴ a = 8 and b2 = 4 × 8 = 32. 2 x2 y 2 + =1 ∴ Equation of ellipse is 64 32

⇒ 4 =

119. (a) Let the equation of the ellipse be

or x + 2y = 64. 2

2

x2 y 2 + = 1. a b2

x2 y 2 116. (b) Equation of ellipse is 2 + 2 = 1. a b

Given : minor axis = 2b = 4

2b 2 Its latus rectum = and minor axis = 2b. a 1 . minor axis (Given) rectum = 2 ∴

2b a

⇒ b2 = 4, Latus

∴ 4 = a2 – 1 or a2 = 5. ∴ From (1), the equation of the ellipse is x2 y 2 +  5 4

= b or a = 2b.

a 2 − b2 4b 2 − b 2 3 ∴ e2 = = = . 2 a 4b 2 4

Latus rectum =

x2 y 2 + = 1. a 2 b2

2b 2 = 4 or b2 = 2a a

...(1)

Distance between the foci = 2ae = 4 2 ⇒  ae = 2 2 

...(2)

  b2 = a2 (1 – e2), ∴ 2a = a2 – a2e2 [Using (1) and (2)]

x2 y 2 + = 1 a 2 b2 Since it passes through (2, 2) and (1, 4), 4 4 + ∴ =1  a 2 b2 1 16 and 2 + 2 = 1 a b

1 a2

But a cannot be negative, ∴ a = 4. So, b2 = 2a = 8.



9 5 + 2 =1 2 a 3a 32 or 3a2 = 32 ∴ a2 . 3

1−

16 20

1 5

121. (c) x2 + 2y2 = 2  or

...(1)

x2 y 2 + 5 20

...(2)  2 . ∵e = 5  

∴ From (2),

9×3 1 + 2 Substituting the value of a2 in (2), we get 32 b 1 5 32 or b2 . =1 ⇒ 2 b 32 5

= 1.

x2 y 2 +   =1 2 1

∴ a = 2 ,  b = 1 So tangent at ( 2 cos θ, sin θ) is x cos θ + y sin θ = 1 2



2 3 2   a 5 5

...(3)

⇒ a2 = 5

Equation of ellipse is

=1

118. (d) Let the equation of the ellipse be

 Also, b2 = a2 (1 – e2) = a2 1 − 

...(2)

Multiplying equation (3) by 4 and subtracting (2) 60 from it, we get 2 = 3 ⇒ b2 = 20. b

(a – 4) (a + 2) = 0     ∴  a = 4, –2.

x2 y 2 = 1 + a 2 b2 Since (1) passes through 9 1 + (– 3, 1), = 1 a 2 b2

...(1)

Substituting this value of b2 in (3), we get

⇒  2a = a2 – 8 ⇒ a2 – 2a – 8 = 0

x2 y 2 + ∴ Equation of ellipse is 16 8 or x2 + 2y2 = 16.

= 1.

120. (b) Let the equation of the ellipse be

3 . 2

117. (c) Let the equation of the ellipse be

or

and distance between foci = 2ae = 2 ⇒ ae = 1. Since b2 = a2 (1 – e2) = a2 – a2e2,

2

Hence e =

...(1)

∴ Coordinates of intercept at x and y axis are

  2 1  , 0   0,    cos θ   sin θ 

Let mid-point be (h, k). ∴ 2h = ⇒

2 , 2k = cos θ

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

1 sin θ

⇒ 2ae =

2b 2 a

⇒ ae =

Equation of tangent to the ellipse at (3, 2) is

a 2 (1 − e 2 ) a



∴ e =

2



−1 ± 1 + 4 −1 ± 5 = . 2 2

For an ellipse, 0 < e < 1, ∴ rejecting e =

−1 − 5 2

have

x (– 3) + 4y (2) = 25 i.e. – 3x + 8y = 25

or 3x – 8y = ­– 25.

126. (c) The end of the latus rectum in the first quadrant is  b2   ae,  . a  which is negative, we  b2  a2 x b2 y − 2 Equation of normal at  ae, a  is ae b / a  

5 −1 . 2 123. (c) Given : Minor axis = distance between the foci

x (3) + 4y (2) = 25 i.e. 3x + 8y = 25.

Equation of tangent to the ellipse at (– 3, 2) is

or e = 1 – e or e + e – 1 = 0 2

e=

=a – b 2

⇒ 2b = 2ae or b2 = a2e2 ⇒ a2 (1 – e2) = a2e2 ⇒ 1 – e2 = e2 1 ⇒ 2e2 = 1, ∴ e = . 2 124. (b) Let P (x, y) be any point on the ellipse whose focus is S (– 1, 1). Let ZM be the directrix x – y + 3 = 0

...(1)

or

 a 2 x b2 y  − = a 2 − b2   y1  x1 

2

a x − ay e

 a 2 − b2  = a2e2 ∵e 2 =  a2  

or x – ey – ae3 = 0. 127. (b) Let (x1, y1) be the point of contact of the line

lx + my + n = 0 x2 y 2 with the ellipse 2 + 2 = 1 a b Equation of tangent at (x1, y1) to (2) is xx1 yy1 + 2 = 1 a2 b

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

...(3)

 ince (1) and (3) represent the same line i.e. tangent S at (x1, y1), ∴ comparing coefficients Join SP and draw PM ⊥ on ZM. Then, by definition,

SP = e PM

Now, SP =

...(2)

( x + 1) + ( y − 1) PM = length of ⊥ from (x, y) on (1)



2

2

x− y+3 x− y+3 = . = 1+1 2

Also, e =

128. (a) Let (x1, y1) be the point of contact of the line

1 x− y+3 ∴ From (2), ( x + 1) + ( y − 1) = ⋅ . 2 2 Squaring and cross-multiplying 2

8 (x2 + y2 + 2x – 2y + 2) = x2 + y2 + 9 – 2xy – 6y + 6x or

2

2

7x + 7y + 2xy + 10x – 10y + 7 = 0,

which is the required equation. 125. (a), (c)  The given ellipse is x2 + 4y2 = 25

 − a 2l − b 2 m  ,   n  .  n



1 . 2

2

x1 / a 2 y / b2 −1 = 1 = l m n 2 −a l −b 2 m , y1 = . ⇒ x1 = n n ∴ Coordinates of the point of contact are

...(1)

 et the abscissa of the point be x1. Then (x1, 2) lies L on (1). ⇒ x12 + 4 (4) = 25 or x12 = 9 ⇒ x1 = ± 3.



x cos α + ysin α = p x2 y 2 with ellipse 2 + 2 = 1 a b Equation of tangent at (x1, y1) to (2) is xx1 yy1 + 2 a2 b

= 1

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

...(3)

 ince (1) and (3) represent the same line i.e. tangent S at (x1, y1) ∴ comparing coefficients 1 x1 / a 2 y / b2 = 1 = p cos α sin α ⇒ x1 =

a 2 cos α b 2 sin α , y1 = . p p

489

∴ There are two points (3, 2) and (– 3, 2) on (1) whose ordiante is 2.

Conic Sections (Parabola, Ellipse and Hyperbola)

122. (a) Given : Distance between the foci = length of latus rectum

490

131. (c), (d) 

Since (x1, y1) lies on the ellipse,

Objective Mathematics

∴

x y + a b

⇒

a 4 cos 2 α b 4 sin 2 α = 1 or a2cos2α + b2sin2α = p2, + a2 p2 b2 p 2

Equation of the ellipse is 4x2 + 3y2 = 5

x y2 + = 1 5/4 5/3 5 5 so that a2 = and b2 = 4 . 3 Equation of line is y = 3x + 7 2

2 1 2

2 1 2

...(1)

or

=1

which is the required condition.

...(2)

Its slope = 3. ∴ Slope of tangent [which is || to (2)] = 3.

x2 y 2 129. (a) Let the equation of the ellipse be 2 + 2 = 1. a b Let the normal at the extremity L of the latus rectum passes through the extremity B ' of the minor axi.  b  Coordinates of L are  ae,  and coordinates of a  B' are (0, – b). 2

∴ m = 3. So, equations of tangents to (1) with slope 3 are y = 3 x ±





= 3x ±

5 5 + (9) 3 4 155 12

 y = mx ± a 2 + b 2 m 2   

or 6 3 x − 2 3 y ± 155 = 0 ⇒ 6 3 x − 2 3 y + 155 = 0 and 6 3 x − 2 3 y − 155 = 0 . 132. (b) Equation of ellipse is

a 2 ⋅ x b2 ⋅ y − 2  quation of the normal at L is E ae b /a

x2 y 2 + = 1 a 2 b2

...(1)

Equation of tangent to (1) at any point P (x1, y1) is xx1 yy1 + 2 = 1 ...(2) a2 b The tangent at P meets x-axis i.e. y = 0 where

 a 2 x b2 y  − = a 2 − b2  = a2 – b2  x y 1  1  ax − ay = a2 – b2. or e

a2 xx1 = 1 or x = x . a2 1

⇒ a2b2 = (a2 – b2)2

But the intercept on x-axis is given to be h a2 a2 ∴ = h or x1 = x1 h

But b2 = a2 (1 – e2).

The tangent at P meets y-axis i.e. x = 0 where

If it passes through B' (0, – b), then 0 + ab = a2 – b2

∴ a2 × a2 (1 – e2) = [a2 – a2 (1 – e2)]2

yy1 b2 = 1 or . 2 b y1

⇒ a (1 – e ) = a (1 – 1 + e ) 4

2

4

2 2

⇒ 1 – e2 = e4 or e4 + e2 – 1 = 0.

But the intercept on y-axis is given to be k.

130. (a), (b)  Equation of ellipse is x2 + 3y2 = 3 x y + or 3 1 2

∴

2

= 1

...(1)

so that a = 3 and b = 1 2

2

Equation of line is 4y = x – 5 Its slope =

...(2)

1 . 4

∴ Slope of tangent [which is ⊥ to (2)] = – 4. ∴ m = – 4.  o, equation of tangents to (1) S 4 are

...(3)

with slope –

2 2 2 y = − 4 x ± 3 × 16 + 1 ­  y = mx ± a m + b  ⇒ 4x + y = ± 7

⇒ 4x + y – 7 = 0 and 4x + y + 7 = 0.

b2 b2 = k or y1 k y1

...(4)

2 2 Since P lies on (1), ∴ x1 + y1 = 1 2 2 a b

or

a 4 / h2 b4 / k 2 = 1 + a2 b2

or

a 2 b2 + = 1. h2 k 2

133. (a) Any tangent to the ellipse is y = mx + a 2 m 2 + b 2

[Using (3) and (4)]

x2 y 2 + = 1 in the slope form a 2 b2 ...(1)

slope of any line through C (0, 0) ⊥ to (1) is −1 −x ( x − 0) or y = y–0= m m

...(2)

 he required locus of the foot of ⊥ from C on (1) is T obtained by eliminating m between (1) and (2). From (1), y – mx =

a 2m2 + b2

or cos θ ⋅

p cos α sin α = − sin θ ⋅ = 2 − a b2 a b

ap − bp ⇒ cos θ = (a 2 − b 2 ) cos α , sin θ = (a 2 − b 2 )sin α Squaring and adding,

⇒ ( y – mx)2 = a2m2 + b2.



Substituting for m from (2), we get 2

 x2  a2 x2 + b2 y+  = y y2 

1=

or

or (x2 + y2)2 = a2x2 + b2y2, which is the required locus.

 a2 p2 b2  +  2  2 (a − b )  cos α sin 2 α  2

(a2 – b2)2 = p2 (a2sec2 α + b2 cosec2 α).

137. (a), (b)  Let P (θ) and Q (φ) be the ends of a focal chord through S.

134. (a) Proceeding as in Q 186, we get

m1 + m2 =

2 x1 y1 x12 − a 2

y12 − b 2 x12 − a 2

and m1m2 =

Given : θ1 + θ2 = constant = α (say) ∴ tan (θ1 + θ2) = tan α ⇒

tan θ1 + tan θ2 m + m2 = tan α or 1 = tan α 1 − tan θ1 ⋅ tan θ2 1 − m1m2

⇒

2 x1 y1 / ( x12 − a 2 ) tan α 1 − ( y12 − b 2 ) / ( x12 − a 2 )

or

+ b2 – a2

Hence, locus of P is 2xy cot α = x2 – y2 + b2 – a2. 135. (b) Proceeding as in Q. No. 182, we get

m1 + m2 =

2 x1 y1 x12 − a 2

⇒

θ−φ ae θ+φ cos + 0 = cos 2 a 2 θ−φ θ+φ Hence, cos = ecos 2 2 2 2 θ+ φ 2 θ−φ ⇒ e cos = cos 2 2 ⇒

y12 − b 2 x12 − a 2

and m1m2 =

Given : tan2 θ1 + tan2 θ2 = constant = k (say) ⇒ m12 m22

Since it passes through focus S (ae, 0), ∴

2 x1 y1 = tan α x12 − y12 + b 2 − a 2

2 ⇒ 2x1 y1 cot α = x1 – y12

Equation of chord PQ is x cos θ + φ + y sin θ + φ a 2 b 2 θ−φ = cos . 2

= k or (m1 + m2)2 – 2m1m2 = k

4 x12 y12 2 ( y12 − b 2 ) − =k 2 2 2 ( x1 − a ) ( x12 − a 2 )

θ+φ a 2 − b2 2 θ−φ cos 2 = cos 2 2 2 a

 2 a 2 − b2  ∵e =   a2   θ+φ θ−φ = a 2 cos 2 . or (a 2 − b 2 ) cos 2 2 2 138. (b) Equation of the ellipse is

or 4 x12 y12 – 2( x12 – a2) y12   – b2) = k( x1 – a2)2. Hence locus of P is

x2 y 2 + 6 2

= 1.

2



4x2y2 – 2 (x2 – a2) (y2 – b2) = k(x2 – a2)2.

2 2 136. (a) The normal to the ellipse x + y 2 2 a b

ax by at the point ‘θ’ is − cos θ sin θ

= 1

= a – b 2

2

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

If it is the same as the line x cos α + y sin α = p then comparing coefficients, we get cos α p sin α = = 2 a / cos θ a − b2 −b / sin θ

...(3),

Here a2 = 6 and b2 = 2 ⇒ a =

6 and b =

2 . Let ‘θ’ be the eccentric angle of the point so that the coordinates of the point are ( 6 cos θ, 2 sin θ) .

Since distance of this point form the centre C (0, 0) is 2. ∴

6 cos 2 θ + 2 sin 2 θ = 2

⇒ 6cos2 θ + 2sin2 θ = 4 ⇒ 6cos2 θ + 2 (1 – cos2 θ) = 4 or 4cos2 θ = 2 1 ⇒ cos θ = ± . 2 ∴ θ =

π 3π 5π 7 π , , , 4 4 4 4

[∴ 0 ≤ θ < 2π]

491

−x y 

Conic Sections (Parabola, Ellipse and Hyperbola)

or m=

492

139. (c) Let the line xcos α + y sinα = p

Objective Mathematics

x y meets the ellipse 2 + 2 a b in P and Q. 2

...(1)

2

=1 

...(2)

Let the tangents at P and Q meet in T (x1, y1). Then PQ is the chord of contact of tangents from T (x1, y1) to the ellipse.  quation of the chord of contact of tangents from E T (x1, y1) to the ellipse is

 he line joining the points of contact of tangents to T the ellipse from P is the chord of contact of P. Equation of chord of contact of tangents from P is xx1 yy1 + 2 = 1 ...(1) a2 b Centre of ellipse is C (0, 0). Since distance of (1) from C = d (given) 1

∴ ±

x y + a b 2 1 4

2 1 4

=d⇒

x12 y12 1 + = 2 . d a 4 b4

Hence, locus of P (x1, y1) is 142. (c) Any point on the ellipse

x2 y 2 1 + = 2 . d a 4 b4

x2 y 2 + =1 a 2 b2

is P (acos θ, b sin θ). xx1 yy1 + 2 = 1 a2 b

T = 0 i.e

...(3)

 ince (1) and (3) represent the same line i.e. chord S of contact of tangents from T,

 he equation of the chord of contact of tangents T from P to the circle x2 + y2 = r2 is

xacos θ + y ⋅ bsinθ = r2

or

x y cos θ + 2 sin θ = 1 r2 / a r /b

∴ comparing coefficients, we get

which clearly touches the ellipse

x1 / a 2 y / b2 1 = 1 = cos α sin α p



2 2 .⇒ x1 = a cos α , y1 b sin α p p

or a2x2 + b2y2 = r4.

 a 2 cos α b 2 sin α  Hence, the required point is  ,  . p p   140. (a) The chord of contact of tangents from P (h, k) to the ellipse

x2 y 2 + a 2 b2

= 1 is

hx ky + a 2 b2

1 h k + a 4 b4 2

2

=c⇒

143. (a) The given line is 8x – 15y = 20 and the given ellipse is 4x2 + 5y2 = 20

...(1)

h2 k 2 1 + = 2 . c a 4 b4

x2 y 2 1 Hence P (h, k) lies on the ellipse 4 + 4 = 2 . c a b 141. (b) Let the coordinates of P be (x1, y1). Equation of the x2 y 2 ellipse is 2 + 2 = 1. a b

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

Let (x1, y1) be the pole of (1) w.r.t. (2) Equation of polar of (x1, y1) w.r.t. (2) is

= 1

I f it touches the circle x2 + y2 = c2, then length of ⊥ from the centre (0, 0) of the circle on (1) is equal to the radius c of the circle. ⇒ ±

x2 y2 =1 + 2 2 (r / a) ( r / b) 2 2

4xx1 + 5yy1 = 20

...(3)

 ince (1) and (3) both represent the same line i.e. S polar of (x1, y1), ∴ comparing coefficients, we get 4x1 5 y1 20 = = 8 −15 20 ⇒ x1 = 2, y1 = – 3.



∴ Pole of (1) is (2, – 3). 144. (d) Equation of the ellipse is 3x2 + 2y2 = 1 Equation of the line is 9x + 2y = 1

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

Let P (x1, y1) be the pole of (2) w.r.t. (1). Equation of polar of (x1, y1) w.r.t. (1) is

3xx1 + 2yy1 = 1

...(3)

 ince (2) and (3) both represent the same line i.e. S polar of P, ∴ comparing coefficients, we get

3 x1 2 y1 1 = = ⇒ x1 = 3, y1 = 1. 9 2 1

∴ P ≡ (3, 1).

∴ Equation of the required line PQ is

4 −1 y – 4 = 1 − 3 ( x − 1)

or y – 4 = −

3 ( x − 1) 2

or 2y – 8 = – 3x + 3 or 3x + 2y = 11.



T = 0 i.e.

or

y= −

x2 y 2 + = 1 is a 2 b2

b x1 b x+  a 2 y1 y1 2

or

...(1)

a b2 k − a 2 ky1 b2 ⇒ = or = 2 m − b x1 y1 b 2 x1 y1 2 a y1

−b 4 y12 = 2 x1 . ak

∴ Locus of P is y2 = –

if

a 4l 2 b 4 m 2 + =1 4a 2 4b 2

b4 x , which is a parabola. a 2k

x2 y 2 + = 1 at any point ‘θ’ is a 2 b2



x y cos θ + sin θ = 1 a b

493

...(1)

c2x2 + d2y2 = 1

...(2)

Polar of (x1, y1) w.r.t. the ellipse (2) is

c2xx1 + d2yy1 = 1

...(3)

 s (1) and (3) represent the same line, i.e. polar of A (x1, y1) w.r.t. (2), ∴ comparing coefficients, cos θ sin θ 1 2 2 ac 2 x1 = bd 2 y1 = 1 ⇒ cos θ = ac x1, sin θ = bd y1. Squaring and adding, we get 1 = a2c4 x12 + b2d4 y12 . Hence, locus of (x1, y1) is a c x + b2d4y2 = 1. 2 4 2

149. (b) Equation of the tangent to the circle x2 + y2 = d2 at any point ‘θ’ is x cos θ + y sinθ = d Let (x1, y1) be its pole w.r.t. the ellipse x2 y 2 = 1 + a 2 b2

or if a2l2 + b2m2 = 4,

x2 y 2 147. (b) Equation of the tangent to the ellipse 2 + 2 = 1 a b at any point ‘θ’ is x y cos θ + sin θ = 1 a b

...(1)

x2 + y2 = a2 

...(2)

Polar of (x1, y1) w.r.t. (2) is xx1 + yy1 = a2

...(3)

 s (1) and (3) represent the same line i.e. polar of A (x1, y1), ∴ comparing coefficients, x1 y1 a2 = = cos θ / a sin θ / b 1

ax1 by ⇒ cos θ = sin θ = a2. x by ∴ cos θ = 1 and sin θ = 21 . a a Squaring and adding, we get

...(1)

...(2)

Polar of (x1, y1) w.r.t. (2) is

xx1 yy1 + 2 = 1 a2 b

...(3)

As (1) and (3) represent the same line, i.e. polar of (x1, y1) w.r.t. (2), ∴ comparing coefficients,

cos θ sin θ x1 / a 2 = y1 / b 2 = d

⇒ cos θ =

dx1 dy , sin θ = 21 a2 b

Squaring and adding, we get

Let (x1, y1) be its pole w.r.t. the circle



y1 a

x2 y2 + 2 =1 2 4a 4b

which is the required condition.



148. (a) The equation of the tangent to the ellipse



x2 y 2 146. (c) Pole of lx + my = 1 w.r.t. the ellipse 2 + 2 = 1 a b is (a2l, b2m). It will lie on the ellipse

or a2 x12 + b2 y12 = a4.

 ence, locus of (x1, y1) is a2x2 + b2y2 = a4, which H is an ellipse.



Since (1) touches the parabola y2 = 4kx ∴ c =

x12 b 2 y12 + 4 a2 a

Let (x1, y1) be its pole w.r.t. the ellipse

xx1 yy1 + 2 =1 a2 b 2

1=

2

145. (a) Let P (x1, y1) be any point on locus. Polar of P (x1, y1) w.r.t. the ellipse



d 2 x12 d 2 y 2 + 4 a4 b



1=

or

x12 y12 1 + = 2 . d a 4 b4

Hence, locus of (x1, y1) is

x2 y 2 1 + = 2 . d a 4 b4

150. (a) Let m be the slope of the system of parallel chords x2 y 2 of the ellipse 2 + 2 = 1. Let M (x1, y1) be the a b mid point of any one chord of the system. Then its equation is

Conic Sections (Parabola, Ellipse and Hyperbola)

 he line through Q (1, 4), conjugate to (2), passes T through the pole of (2) i.e. through P (3, 1).

494



Objective Mathematics

Its slope is =

T = S1 i.e.,

which represent an ellipse.

xx1 yy1 x12 y12 + − 1 = + – 1. a2 b2 a 2 b2

Here, a2 = 1/8 and b2 = 1/16. 1 1 − a 2 − b2 8 16 = 1  i.e. e = 1 . 2 ∴ e = = 2 a2 2 1/ 8 155. (d) The given equation of ellipse can be written in the x2 y 2 + form = 1. 16 9

−b 2 x1 . a 2 y1

But the slope is given to be m. −b 2 −b 2 x1 = m ⇒ y 1 = 2 x 1. 2 am a y1 −b 2 Hence locus of (x1, y1) is y = 2  x, am ∴

Here a2 = 16 and b2 = 9.

 hich is a straight line through the centre (0, 0) w of the ellipse. Equation of the chord having (x1, y1) as its mid point is

T = S1 i.e.,



= 2a = 8.

156. (c) PF1 + PF2=

151. (b) Let (x1, y1) be the mid point of any chord

xx1 yy1 x2 y 2 + 2 − 1 = 12 + 12 − 1 . 2 a b a b

a = 4.

∴

 um of the focal distances from any point on the S ellipse

= ( x − 3) 2 +

( x − 3) 2 + y 2 + ( x + 3) 2 + y 2 (400 − 16 x 2 ) 25



+

(400 − 16 x 2 ) 25

( x + 3) 2 +

If it passes through the fixed point (h, k), then

=

hx1 ky1 x12 y12 + + = . a2 b2 a 2 b2

∴ Locus of (x1, y1) is

x 2 y 2 hx ky + − − = 0, a 2 b2 a 2 b2

which is another ellipse. 152. (a) Let (x1, y1) be the mid point of any chord. Then, equation of the chord is

T = S1 i.e.,

[ (25 – 3x) + (25 + 3x)] = 10.

157. (c) The line y = 4x + c touches the ellipse if

2 2 c = ± 4 ⋅ (4) + (1)

i.e., c = ±

c = ± a 2 m 2 + b 2   

65 .

158. (a) (c)  The equation of ellipse can be written in the form

xx1 yy1 x2 y 2 + 2 − 1 = 12 + 12 – 1. 2 a b a b

x2 y2 + = 1. 2 ( 6) ( 2 )2

Positive end of the major axis is A′(a, 0).



If the chord passes through A’ (a, 0), then ax1 x2 y 2 x12 y12 x + = 12 + 12 or = 1. 2 a a b a 2 b2 a x2 y 2 x Hence, locus of (x1, y1) is 2 + 2 = . a b a

coordinates are

Let the eccentric angle of the point be θ, then its

(

∴ ( 6 cos θ − 0) 2 + ( 2 sin θ − 0) 2 = 4

Then, equation of the chord is

⇒ 6 cos2 θ + 2 (1 – cos2 θ) = 4

xx1 yy1 x12 y12 + − 1 + = –1 a2 b2 a 2 b2 If its distance from the centre C(0, 0) is ‘d’, then

⇒ 4cos2 θ = 2



x12 y12 + 2 2 2  x2 y 2   x12 y12  b ± a = d ⇒ + = d2  14 + 14   2 2 2 2  b  x1 y b  a a + 1 a 4 b4 x y  x y  ∴ Locus of (x1, y1) is  2 + 2  = d2  4 + 4  . a b a b     2

2

2

154. (c) The given equation can be written in the form

( x − 1) 2 ( y + 3 / 4) 2 + = 1, 1/ 8 1 / 16

)

6 cos θ, 2 sin θ .

 ince the distance of the point from the centre is S 2 units

153. (b) Let (x1, y1) be the mid point of any chord.

2

x2 + y2 = 1 4

2

⇒ cos θ = ±

1 π 3π . ∴ θ = or . 4 4 2

159. (a) The given equation of the ellipse is

x2 y 2 + = 1. 16 9

Here, a2 = 16 and b2 = 9. ⇒ a = 4 and b = 3. Eccentricity of the ellipse is given by a 2 − b2 16 − 9 7 e2 = = = . a2 16 16 ∴ e =

7 . 4

i.e.,

( 7 , 0) and (–

 − 7 ⋅ 4, 0   4  

7 , 0).



=

( 7 − 0) + (0 − 3)

160. (a) Since ∠ FBF ′ =

2

=

The equation of tangents at (0, 6) and (0, 0) are 9x ⋅ 0 + 5y ⋅ 6 – 15 ( y + 6) = 0

and 9x ⋅ 0 + 5y ⋅ 0 – 15 ( y + 0) = 0 respectively. 7 + 9 = 4.

π 2

∴ ∠ FBC = ∠ F′ BC =

(0, 3 + a) and (0, 3 – a)

i.e. (0, 3 + 3) and (0, 3 – 3) or (0, 6) and (0, 0).

Thus, radius of required circle 2



π 4

So the tangents are y = 6 and y = 0.  uppose the tangents at P and Q meet in A (h, k). S Equation of the chord of contact of the tangents through A (h, k) is hx ky + = 1 6 3 Since (4) and (2) represent the same line

...(4)



h/6 1 k /3 cos θ = sin θ = 1 2 ⇒ h = 3 cos θ and k = 3sin θ. ∴

∴ CB = CF

⇒ b = ae



⇒ b = a e ⇒ a2 (1 – e2) = a2e2



⇒ 2e2 = 1 ⇒ e = 1/ 2 .

2

Thus, coordinates of A are (3cos θ, 3sin θ).

2 2

 he joint equation of the tangents at A is given by T T2 = SS1

161. (b) Let P ≡ (a cos θ, b sin θ)

hx ky  i.e.  + − 1 6 3  





= abe sin θ .

 x2 y 2   h2 k 2  − 1  + − 1  =  + 3 3  6   6 

= 1 ⋅ 2ae ⋅ bsin θ 2

Clearly, A is maximum when sin θ = 1. ∴ Maximum value of A = abe. 162. (d) Given equation of ellipse can be written in the form ( x + 1) 2 ( y + 2) 2 + = 1. 9 25 Eccentricity of the ellipse is a 2 − b2 25 − 9 16 = = . ∴ a2 25 25

e = 4/5.

∴ coordinates of foci are (– 1 + 0, – 2 ± ae) i.e. (– 1, – 2 + 4) and (– 1, – 2 – 4) i.e. (– 1, 2) and (– 1, – 6). 163. (c) The given equation of the ellipse can be written in x 2 ( y − 3) 2 + the form = 1. 5 9  ere a2 = 9, b2 = 5 and centre of the ellipse is H (0, 3).

 h2 1  h2 k 2 −k2 1 −  + − 1 = + 36 6  6 3 18 6 

and b = coefficient of y2 in (5) =

 − k 2 1  h2 k 2 − h2 1 −  + − 1 = + . 9 3 6 3 18 3 

We have, a + b =

−1 2 1 1 (h + k 2 ) + + 18 6 3



=

−1 1 (9 cos 2 θ + 9 sin 2 θ) + 18 2



=

−1 1 (9) + = 0. 18 2

Here, a2 = 25 and b2 = 9 ⇒ a = 5 and b = 3.

e2 =

...(5)

Let a = coefficient of x2 in (5)

Then, A = area of ∆ PF1F2 a cos θ b sin θ 1 1 0 1 = ae = 2 0 1 − ae

2

 hus, (5) represents two lines which are at right T angles to each other. 164. (b) The given equation of the ellipse can be written in x 2 ( y − 3) 2 + =1 the form 5 9 Here, a2 = 9 and b2 = 5. Eccentricity of the ellipse is given by e2 =

a 2 − b2 9−5 4 = = . ∴ a2 9 9

e =

2 . 3

495

  7 ⋅ 4, 0  and  4  



Conic Sections (Parabola, Ellipse and Hyperbola)

∴ Coordinates of the ends of major axis are

∴ Coordinates of foci are

496

165. (c) Equation of ellipse is

Objective Mathematics

Now, b2 = a2 (1 – e2).

x2 y 2 + = 1 a 2 b2

...(1)

When e = 0, b = a.  o, equation (1) becomes x2 + y2 = a2, which is a S circle. 166. (b) SP = a – ex and S’ P = a + ex 167. (a) Equations of pair of tangents drawn from the point (1, 2) to the ellipse 3x2 + 2y2 = 5 is SS1 = T2 i.e.  (3x2 + 2y2 – 5) (3 · 12 + 2 · 22 – 5) = (3x ⋅ 1 + 2y ⋅ 2 – 5)2

...(1)

 ocal distance of one end of minor axis say F (0, b) is k.



Here a = 9, h = – 12, b = – 4.



∴ 2ae = 2h ⇒ ae = h

...(2)

From (1) and (2)

⇒ 9x2 – 24xy – 4y2 + 30x + 40y ­– 55 = 0 2 ∴ tan θ = 2 h − ab a+b

Since distance between foci = 2h

∴ a + e (0) = k ⇒ a = k

⇒ 6 (3x2 + 2y2 – 5) = (3x + 4y – 5)2

x2 y 2 + = 1. a 2 b2

170. (c) Let the equation of the ellipse be

Let e be the eccentricity of the ellipse.

∴ SP + S’ P = 2a.



or 2x + 3y = 9 9 x-intercept = , y-intercept = 3 2 9 1 Required area = 4 × shaded area = 4 × × 3 × = 2 2 27 sq. units

b2 = a2 (1 – e2) = k2 – h2.

∴ The equation of the ellipse is

2 144 + 36 = 9−4

x2 y2 + 2 = 1. 2 k k − h2

171. (b) The equation of the tangent at

2 180 12 5 = = 5 5

16   sin θ  to the ellipse  4 cos θ, 11  

 12 5  ∴ θ = tan– 1   .  5 



168. (a) Let the equation of the ellipse be

x2 y 2 + = 1. a 2 b2 It is given that it passes through (7, 0) and (0, – 5)



∴ a2 = 49 and b2 = 25.

or 4x cos θ +

Since b2 = a2 (1 – e2), ∴ 25 = 49 (1 – e2) 25 25 24 ⇒ 1 – e2 = ⇒ e2 = 1 − = 49 49 49

This touches the circle (x – 1)2 + y2 = 42 if

2 6 ⇒e= . 7

∴ e2 =

 16  sin θ  = 256 16x (4 cos θ) + 11y   11  11 y sin θ = 16.

4 cos θ − 16 16 cos 2 θ + 11sin 2 θ

=4

⇒ (cos θ – 4)2 = 16cos2 θ + 11sin2 θ

169. (b) Here  a = 9, b = 5 2

16x2 + 11y2 = 256 is

2

∴

b = a (1 – e ) 2

2

2

⇒ 15cos2 θ+ 11sin2 θ + 8cos θ – 16 = 0 ⇒ 4cos2 θ + 8cos θ – 5 = 0 ⇒ (2cos θ – 1) (2cos θ + 5) = 0

4 2 ,e= 9 3

⇒ cos θ = ⇒ θ = ±

1 2

π . 3

−5   ∵cos θ = 2 is not possible   

172. (b) Solving 2x + y = 3 i.e. y = 3 – 2x and

 Tangent at L  2, 

5  is 3

x.2 y.5/3 + = 1  ∴  9 5

2x y + =1 9 3



4x2 + y2 = 5, we get



4x2 + (3 – 2x)2 = 5 ⇒ 8x2 – 12x + 4 = 0

⇒ 2x2 – 3x + 1 = 0 ⇒ (2x – 1) (x – 1) = 0. 1 ∴ x = , 1. 2 1 When x = 1, y = 3 – 2 = 1 and when x = , 2  y = 3 – 1 = 2.

1  Equation of tangent at P   , 2  is 2  1 4 x   + y (2) = 5 i.e., 2x + 2y = 5. 2

a2 =

Latus rectum =

2b 2 = a

2 2 = . 3 3 5  5

Its slope = – 1. ∴ Slope of the normal = 1. Again, equation of tangent at Q (1, 1) is 4x (1) + y (1) = 5 i.e., 4x + y = 5. Its slope = – 4. 1 ∴ Slope of the normal = . 4 1 − (1/4) 3/ 4 3 ∴ tan θ = 1 + (1) (1/4) = = . 5/4 5 173. (a) The mid-point of the chord is

 x1 + x2 y1 + y2  ,  . 2   2

The equation of the chord in terms of its mid-point is T = S1  y1 + y2   x1 + x2  or x   +y    2   2 

⇒ x (y1 + y2) + y (x1 + x2) = (x1 + x2) (y1 + y2) x y ⇒ x +x + y +y = 1 1 2 1 2 174. (d) The given equation can be written in the form

( x − 1) 2 ( y + 1) 2 + =1 4 3

⇒

X2 Y2 + = 1 where x – 1 = X and y + 1 = Y. 4 3

⇒ x – 1 = 0 and y + 1 = 0 i.e. (1, – 1). 1−

b2 = a2

1−

3 1 = . 4 2

Foci are X = ± ae, Y = 0 1  ⇒ x – 1 = ±  4 ×  = ± 2, y + 1 = 0 2  Hence, foci are (3, – 1) and (­– 1, – 1). 175. (d) The given equation can be written in the form

176. (a) The given equation can be written in the form ( x − 1) 2 ( y + 1) 2 + = 1. 4 3



Comparing with

x2 y 2 + = 1, a 2 b2

we get a = 2, b =

3.

Hence, latus rectum =

x2 y2 + = 1. 9 / 25 5 / 25

x2 y2 Comparing with a 2 + a 2 (1 − e 2 ) = 1, we get

2b 2 2 ( 3 )2 = = 3. a 2

177. (d) Since 12 + 22 = 5 < 9 and 22 + 12 = 5 < 9, ∴ both P and Q lie inside C.

and

12 22 1 + = +1>1 9 4 9

22 1 16 + 9 25 + = = 0, the point (3, 5) lies outside the ellipes 3x2 + 5y2 = 32. Also, 25 × 32 + 9 × 52 – 450 = 0, ∴ the point (3, 5) lies on the ellipse 25x2 + 9y2 = 450. So the required number of tangents is 3. 179. (a), (b)  Given ellipse is

Centre is X = 0, Y = 0

Eccentricity, e =

 2  Foci ≡ (± ae, 0) =  ± , 0  .  5 

Also,

 x1 + x2   y1 + y2  =2      2   2 



5 25 1 . 5

x2 y 2 x2 y 2 + = 1 i.e. 2 + 2 = 1 16 9 4 3

∴ Lengths of semi-major and semi-minor axes are 4 and 3 respectively. So, the mean of these lengths is 7/2. Let P (4cos θ, 3sin θ) be a point on the ellipse at a distance 7/2 from the centre (0, 0). 49 ∴ 16cos2 θ + 9sin2 θ = 4 49 ⇒ 16cos2 θ + 9 (1 – cos2 θ) = ⇒ 28cos2 θ = 13 4 13 91 105 = ± and sin θ = ± 28 14 14 So, the coordinates of the required point are

⇒ cos θ = ±



 4 91 3 105   2 91 3 105  ,  i.e.  ±  ± , . 14 14  7 14   

497

9 and a2 (1 – e2) = 25 3 2 ⇒ a = , e = and b = 5 3

Conic Sections (Parabola, Ellipse and Hyperbola)

1  ∴ Points of intersection are P   , 2  and Q (1, 1). 2 

498

∴ θ = 45º.

180. (c) We have, S ≡ (ae, 0),

T ≡ (­– ae, 0) and B ≡ (0, b).

184. (c) Equation of line is y = mx + c

Objective Mathematics

Since STB is an equilateral triangle

...(1)

x y + = 1 ...(2) a 2 b2 To find the points of intersection, we have to solve (1) and (2) simultaneously. 2

2

Equation of ellipse is

Substituting for y from (1) in (2), we have ∴ ST2 = TB2 ⇒ 4a2e2 = a2e2 + b2



⇒ 3a2e2 = b2 = a2 (1 – e2) ⇒ 3e2 = 1 – e2 ⇒ 4e2 = 1 ⇒ e =

1 . 2

181. (a) Let P (h, k) be the pole.  hen, the equation of the polar of P (h, k) w.r.t. y2 T = 4x is 2a 2ah x+ k y = 2a (x + h) i.e. y = . k k Since it touches the ellipse

x2 y 2 + =1 α 2 β2



x 2 (mx + c) 2 + =1 a2 b2

⇒ (b2 + a2m2) x2 + 2mca2x + a2 (c2 – b2) = 0 ...(3) ( 1) intersects (2) in real points if discriminant of (3) ≥ 0 ⇒ 4m2c2a4 – 4a2 (b2 + a2m2) (c2 – b2) ≥ 0 ⇒ m2c2a2 – b2c2 + b4 – m2c2a2 + a2b2m2 ≥ 0 ⇒ – b2c2 + b4 + a2b2m2 ≥ 0 ⇒ a2m2 ≥ c2 – b2.

185. (c) Let the two given lines be taken as the coordinate axes.

 2ah  2  2a  2 2 2 2 2 ∴   = α   + β [c = a m + b ]  k   k  2

2

⇒ 4a2h2 = 4a2α2 + k2β2. So, locus of (h, k) is 4a2x2 = 4a2α2 + β2y2 i.e.,

x2 y2 − = 1. α 2 4a 2 α 2 / β 2

182. (a) Let (h, k) be the mid point of a focal chord. Then its equation is T = S1 xh ky h2 k 2 i.e., 2 + 2 − 1 = 2 + 2 − 1 . a b a b Since it passes through (ae, 0) ∴

h2 k 2 hae = + . a 2 b2 a2

∴ Locus of (h, k) is

Let C (α, β) be the centre of the ellipse in any position. Here the position of centre C changes as the ellipse slides. Let a and b be the semi major and semiminor axes of the ellipse. Equation of the director circle of the ellipse is

x2 y 2 xe + = a 2 b2 a

183. (c) Equation of any tangent to the ellipse x2 y 2 + = 1 is a 2 b2 x y cos θ + sin θ = 1 ...(1) a b x y + = 2 touches the given ellipse. Also, a b



(x ­– α)2 + (y – β)2 = a2 + b2

∴ α2 + β2 = a2 + b2. Hence, locus of C (α, β) is x2 + y2 = a2 + b2. b−0 b 186. (c) Slope of BS is m1 = 0 − ae = − ae and slope of BS’ is

Comparing coefficients in (1) and (2), we get 1 cos θ / a sin θ / b = = 2 1/ a 1/ b 1 ⇒ cos θ = = sin θ 2

...(1)

 ince OX and OY are mutually perpendicular tanS gents to sliding ellipse for all its positions, therefore, O (0, 0) will lie on circle (1).



b−0 b m2 = 0 + ae = . ae

−b b x =–1 ae ae

2 9k 2 + 9h = l2 or 9h2 + 36k2 = 4l2 4

⇒ b2 = a2e2

∴ locus of P (h, k) is 9x2 + 36y2 = 4l2, which is an ellipse.

⇒ a2 (1 – e2) = a2e2

189. (b) Equation of circle of radius r concentric with the 2 2 ellipse x + y = 1 is 2 2 a b

⇒ 1 – e2 = e2 ⇒ 2e2 = 1. 1 ∴ e = . 2



187. (a) Let P ≡ (a cos α, b sin α)

x 2 + y 2 = r 2

...(1)

Equation of any tangent to the ellipse

and Q ≡ (a cos β, b sin β)



x y = 1 is + a 2 b2 2

2

y = mx + a 2 m 2 + b 2 (where m = tan θ).

I f it is a tangent to the circle (1), then ⊥ distance from the centre (0, 0) of the circle to it must be equal to the radius r Slope of CP = m1 = Slope of CQ = m2 = ∴ m1m2 =



b sin α − 0 b = tan α. and a cos α − 0 a b sin β − 0 b = tan β. a cos β − 0 a

m2 + 1

=r

⇒ a2m2 + b2 = m2r2 + r2 ⇒ (a2 – r2) m2 = r2 – b2.

b2 b2 − a 2 × 2 2  · tan α tan β = a a2 b

∴ m =

 − a2  ∵ tan α ⋅ tan β = 2  b  

= – 1.

a 2m2 + b2

i.e. ±

∴ θ = ∠PCQ = 90º.

r 2 − b2 a2 − r 2 r 2 − b2 a2 − r 2

i.e. tan θ =

∴ θ = tan– 1

r 2 − b2 . a2 − r 2

188. (b) Let the two fixed ⊥ straight lines be the coordinates axes. 190. (b) The equation of any tangent to the given ellipse Let P (h, k) be the point whose locus is required. is

x y cos θ + sin θ = 1. 4 3

The tangent meets x-axis at A   4 , 0  and y-axis  cos θ   3  at  0, .  sin θ  Given :

⇒ cos θ =

Let PA : PB = 1 : 2 Then PA = k=

and  h = or

2l l and PB = . 3 3

l sin θ or 3k = lsin θ 3

3 4 and sin θ = l l

⇒ cos2 θ + sin2 θ = ...(1)

2l cos θ 3

3h = lcos θ 2

3 4 =l= sin θ cos θ

...(2)

16 9 + ⇒ l2 = 25. ∴ l = 5. l2 l2

191. (d) The equations of chord of contact of tangents from two points (α, β) and (r, δ) to the given ellipse are xα yβ + = 1 ...(1) 5 2

Conic Sections (Parabola, Ellipse and Hyperbola)

∴ m1m2 = – 1 ⇒

499

Squaring and adding (1) and (2), we get

Since ∠ SBS ′ = 90º,

500

and

x γ yδ = 1 + 5 2

...(2)

195. (c) Let PCP ' and DCD ' be a pair of conjugate diameters of the given ellipse.

Objective Mathematics

Since (1) and (2) are ⊥, ∴

− 2α − 2 γ x =–1 5β 5δ

⇒

25 αγ = − . 4 βδ

We have, P ≡ (acos φ, bsin φ),

192. (a) Putting y = 2t2 in the equation of the given ellipse

x2 y 2 + 9 4 = 1, we get

D ≡ (– asin φ, bcos φ)

P' ≡ (– acos φ, – b sin φ)

and D ' ≡ (asin φ, – b cos φ). Now, the area of ||gm EFGH

x 2 4t 4 =1 + 9 4

⇒ x2 = 9 (1 – t4) = 9 (1 – t2) (1 + t2).  his will give real values of x if 1 – t2 ≥ 0 i.e. T | t | ≤ 1. 193. (c) Let PCP ' and DCD ' be a pair of conjugate diameters. Let P ≡ (acos φ, bsin φ) and D ≡ (acos θ, bsin θ).



= 4 × area of ||gm CPED



= 4 (2 × area of ∆ CPD)



=8×



= 4 (abcos2 φ + absin2 φ) = 4ab.

1 2

0 0 1 a cos φ b sin φ 1 − a sin φ b cos φ 1

196. (b) We have P ≡ (acos φ, bsin φ) and D ≡ (– asin φ, bcos φ) Let the middle point of PD be (x1, y1).

The slope of CP is

m =

b sin φ b sin φ − 0 = a cos φ a cos φ − 0

and the slope of CD is m ' =

b sin θ . a cos θ

For conjugate diameters, mm ' = − ∴

b2 . a2

2 b sin φ b sin θ ⋅ = −b a cos φ a cos θ a2

or sin φ sin θ + cos φ cos θ = 0 ⇒ cos (θ – φ) = 0 ⇒ θ – φ = π . 2 194. (a) Let P ≡ (a cos θ, bsin θ)

∴ x1 =

a (cos φ − sin φ) , 2



y1 =

b (sin φ + cos φ) . 2

∴

1 x12 y12 1 + = ⋅2 = . 2 a 2 b2 4

2 2 1 Hence the locus of (x1, y1) is x + y = 2 2 2 a b

197. (a) We have C ≡ (0, 0) ≡ (acos φ, bsin φ) and

D ≡ (– asin φ, b cos φ).

The equation of the circle with CP as diameter is (x – 0) (x – acos φ) + ( y – 0) (y – bsin φ) = 0 or x2 + y2 = axcos φ + bysin φ ...(2) Squaring and adding (1) and (2), we see that the point of intersection R of (1) and (2) lies on the curve

2 (x2 + y2)2 = a2x2 + b2y2.  ince the eccentric angles of the extremities of a pair S of conjugate diameter differ by a right angle, 198. (a) Let the orbit of the earth be the ellipse x2 y 2 + = 1 a 2 b2

 π  π  ∴ D ≡  a cos  + θ  , b sin  + θ   2 2     



i.e. (– asin θ, bcos θ).

 ength of major axis = 2a = 186 × 106 miles L (given)

∴ CP2 = a2cos2 θ + b2sin2 θ and CD2 = a2sin2 θ + b2cos2 θ ⇒ CP2 + CD2 = a2 + b2.

⇒ a = 93 × 106 miles. 1 Also, eccentricity e = (given). 60

...(1)

= SA, where S is (ae, 0) and A is (a, 0)

1 = a – ae = (93 × 10 ) 1 −  = 9145 × 104 miles 60   6

and longest distance of the earth from the sun = SA ', where S is (ae, 0) and A ' is (– a, 0) 1 = a + ae = (93 × 106) 1 +  = 9455 × 104 miles. 60   199. (d) Equation of tangent at any point (x1, y1) is xx1 yy1 = 1 − 4 2 Given equation is 2 x + 6 y = 2

203. (b) Distance between foci = 16 ⇒ 2ae = 16 or ae = 8

...(1)

2,

...(2)

Since b2 = a2 (e2 – 1),

8 = 4 2. 2

∴ b2 = 32 (2 – 1) = 32. Thus, equation of the hyperbola is

= (6 + 4) 2 + (4 − 4) 2 = 10. But SS ' = 2ae, ∴ 2ae = 10 or ae = 5. ⇒ a = 5/2. ∴ b = a (e – 1) = 25 (4 − 1) = 75/4. 4 2

 , the centre of the hyperbola is the mid point of C SS'. 6 − 4 4 + 4 ∴ Coordinates of C are  ,  i.e. (1, 4). 2   2

x2 y 2 = 1 or x2 – y2 = 32. − 32 32

204. (c) Here, e = 3 and latus rectum = 4. ∴

e = 2 ∴ 2a = 5

2 

4 +4 2 2 a + b = 3 = 4. ∴ e2 = 2 4 a 3 ⇒ e = 2. ∴ Eccentricity = 2.

∴ from (1) a =

200. (a) SS' = distance between foci

2

4 and b2 = 4. 3

...(1)

x1 = 4,  y1 = − 6 .



x2 y2 − =1 4/3 4

Since e =

Comparing (1) and (2),



∴ a2 =

∴ Shortest distance of the earth from the sun

2b 2 = 4 or b2 = 2a. a

Now, b2 = a2 (e2 – 1) ⇒ 2a = a2 (9 – 1) or 4a = 1 1 . ∴ a = 4 1 1 ∴ b2 = 2 × = . 4 2

Hence, the equation of the hyperbola is

Thus, equation of hyperbola is

 ( x − h) 2 ( y − k ) 2  ( x − 1) 2 ( y − 4) 2 − − = 1 . =1  2 2 25 / 4 75 / 4 a b  

or 16x2 – 2y2 = 1.

201. (c) Let P (x, y) be any point on the hyperbola and S (1, 2) be the focus. Draw PM ⊥ on the corresponding directrix. Then, by definition, SP = e PM ⇒

( x − 1) 2 + ( y − 2) 2 =

3.

2x + y − 1 4 +1

⇒ 5 [(x – 1)2 + (y – 2)2] = 3 (2x + y – 1)2 ⇒ 5x2 + 5y2 – 10x – 20y + 25

= 12x2 + 3y2 + 3 + 12xy – 12x – 6y

205. (b), (d)  Equation of the hyperbola is

Equation of line is y = x + 2

...(1)

Slope of line = 1. ∴ Slope of tangents which are parallel to (1) is 1 i.e.,

m = 1.

∴ Equations of tangents are y = 1  ⋅ x ± 4 ⋅ 1 − 3  y = mx ± a 2 m 2 − b 2   

2

which is the required equation of the hyperbola.

x2 y 2 = 1. − 4 3

Here a2 = 4 and b2 = 3.

⇒ 7x – 2y + 12xy – 2x + 14y – 22 = 0, 2

x2 y2 =1 − 1 / 16 1 / 2

i.e. y = x ± 1.

501

202. (c) The given equation of the hyperbola can be written in the form

Conic Sections (Parabola, Ellipse and Hyperbola)

 et the sun be at the focus S (ae, 0). Then the earth L will be at shortest and longest distance from the sun when the earth is at the extremities of the major axis which are respectively nearest and farthest from this focus S.

502

209. (c) The equation of the line is x – 3y = 1

206. (a), (c)  Equation of hyperbola is 3x2 – y2 = 3.

and that of the hyperbola is x – 4y = 1 2

x2 y 2 − = 1, so that a2 = 1, b2 = 3. 1 3

or Objective Mathematics

slope of tangents = 3. ∴ m = 3.

∴ From (2), (3y + 1)2 – 4y2 = 1 ⇒ 9y2 + 6y + 1 – 4y2 = 1 ⇒ 5y2 + 6y = 0 ∴ y = 0, − 6 . 5

Thus, equations of tangents are y = 3x ± 1 ⋅ 9 − 3

When y = 0, from (3), x = 0 + 1 = 1.

i.e. y = 3 x ± 6 .

When y = −

207. (a), (b), (c), (d)  The given hyperbola is

x2 y 2 − = 1. 4 3

Here a2 = 4 and b2 = 3. If the intercepts made by the tangents are equal in magnitude and sign, then slope of tangents = ­– 1 ⇒ m = – 1. ∴ Equations of tangents are y = −1 ⋅ x ± 4 (−1) 2 − 3

I f the intercepts made by the tangents are equal in magnitude but opposite in sign, then slope of tangents = 1 ⇒ m = 1. ∴ Equations of tangents are y = 1 ⋅ x ± 4 (1) 2 − 3 or y – x = ± 1. x2 y 2 = 1 − a 2 b2

in the slope is y = mx + a 2 m 2 − b 2 

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

slope of tangent = m. ∴ slope of any line ⊥ to it = −

1 m

2

∴ PQ =

 13   1 +  +  0 + 5  



=

324 36 + 25 25



=

360 = 6 10 . 25 5



y= −

x2 x2 + a 2 ⋅ 2 − b2 y y a 2 x2 − b2 y 2

or

x2 + y2 =

or

(x2 + y2)2 = a2x2 – b2y2.

2

210. (a) Equation of the hyperbola is 5x2 – y2 = 5 x 2 y 2 = 1 ...(1) − 1 5 Here, a2 = 1 and b2 = 5. Equation of any tangent to (1) in the slope form is or

2 y = mx + 1 ⋅ m − 5

...(2)



2=

m2 − 5

⇒ 4 = m2 – 5 ⇒ m = ± 3. ∴ From (2), the equations of the tangents are

...(3)

 he required locus is obtained by eliminating the T parameter m between (2) and (3). Substituting for m from (3) in (2), we get

6  5

If it passes through (0, 2), then

Equation of ⊥ from centre (0, 0) of (1) on (2) is y – 0 = − 1 ( x − 0) m x or m = −  y

6 18 13 , from (3), x = − +1 = − . 5 5 5

 13 − 6  ∴ Coordinates of P and Q are (1, 0) and  − , .  5 5 

or x + y = ± 1.

208. (b) Equation of any tangent to

...(2)

From (1), x = 3y + 1 ...(3)

Since the tangents are ⊥ to the line, therefore,



...(1)

 et (1) meets (2) in P and Q. The coordinates L of P and Q are obtained by solving (1) and (2) simultaneously.

Equation of line is x + 3y – 2 = 0. 1 Slope of line = − . 3



2



y = ± 3x +

9 − 5 or y = ± 3x + 2.

2 2 ...(1) 211. (c) Equation of hyperbola is x − y = 1 a 2 b2 Let the tangents be drawn from (x1, y1). Equation of any tangent to (1) in the slope form is



2 2 2 y = mx + a m − b .

As it passes through (x1, y1) ∴ y1 – mx1 =

a 2m2 − b2

2 2 ⇒ y1 – 2mx1y1 + m2 x1 = a2m2 – b2

 his is a quadratic in m. Let its roots be m1 and T m 2 which are the slopes of the tangents from (x1, y1) to (1) y 2 + b2 . m1m2 = 12 x1 − a 2

If it is the same as the given line lx + my = 1 then comparing coefficients, we get

l m 1 = = 2 a + b2 a / secθ b / tan θ

⇒

1 m tan θ l secθ = = 2 a + b2 b a

But m1m2 = c2 (given). ∴

y12 + b 2 = c2 x12 − a 2

2 or ( y1 + b2) = c2 ( x12 – a2).

Hence, locus of (x1, y1) is (y2 + b2) = c2 (x2 – a2). 212. (a) Let lx + my + n = 0

...(1)

be a tangent to x + y = 1 at ‘θ’ a 2 b2 Equation of tangent to (2) at ‘θ’ is 2



2

x y sec θ − tan θ = 1 a b

...(2)

...(1)

be a tangent to x − y = 1 a 2 b2 at ‘θ’. Equation of tangent to (2) at ‘θ’ is

...(2)

x  sec θ – y tan θ = 1 a b It is the same as the given line (1), ∴ comparing coefficients, we get sec θ − tan θ 1 = = a cos α b sin α p

...(3)



a cos α −b sin α , tan θ = . p p

2

2 

2

ax by = a 2 + b 2 + sec θ tan θ

...(1) 2

...(2)

...(3)

3 4 7 = = 2 a + b2 a / secθ b / tan θ



7a 7b and tan θ = 3 (a 2 + b 2 ) 4 (a 2 + b 2 )

 ence, the coordinates of the foot of the normal H are   7a 2 7b 2 (asec θ, btan θ) i.e.  . , 2 2 2 2   3 (a + b ) 4 (a + b ) 

xx1 yy1   =  1 − 2 a2 b

Its slope =

2 2 214. (c) Any normal to the hyperbola x − y = 1 is a 2 b2



which is the required condition.



or a cos α – b sin α = p , which is the required condition. 2 

a 2 b 2 = (a2 + b2)2, − l 2 m2

216. (d) Polar of (x1, y1) w.r.t. the given hyperbola is  

a 2 cos 2 α b 2 sin 2 α − =1 p2 p2 2

or

2

⇒ sec θ =

Also, sec2 θ – tan2 θ = 1 ∴

a2 b2 − 2 2 =1 2 2 l (a + b ) m (a + b 2 )2 2

As (1) and (3) both represent the normal at ‘θ’, ∴  comparing coefficients, we get

which is the required condition.

⇒ sec θ =

∴

be a normal to the hyperbola x − y = 1 a 2 b2 at (asec θ, btan θ). Equation of normal at (asec θ, btan θ) is ax by = a 2 + b 2 + sec θ tan θ

Also, sec2 θ – tan2 θ = 1 2 2 2 2 ∴ a l − b m = 1 or a2l2 – b2m2 = n2, 2 n n2

2

b m (a 2 + b 2 )

Also, sec2 θ – tan2 θ = 1

2

− al bm ⇒ sec θ = and tan θ = n n

2

tan θ =

215. (a) Let 3x + 4y = 7

− tan θ −1 secθ = = bm n al

213. (b) Let xcos α + ysin α = p

a , l (a + b 2 ) 2



...(3)

 ince it is the same as the given line (1), comparS ing coefficients

⇒ sec θ =

...(2)

x1 / a 2 b 2 x1 = . y1 / b 2 a 2 y1

Similarly, slope of polar of (x2, y2) w.r.t the given b2 x hyperbola is 2 2 . a y2 Polars are ⊥, ∴

...(1)

or

4 x1 x2 = −a . 4 y1 y2 b

b 2 x1 b 2 x2 ⋅ = – 1. a 2 y1 a 2 y2

503

+ b2) = 0 ...(2)

Conic Sections (Parabola, Ellipse and Hyperbola)

2 2 ⇒ ( x1 – a2) m2 – 2x1y1m + ( y1

504

217. (a) The required point is clearly the pole of the given line.

Objective Mathematics

Let (x1, y1) be the pole of 3x + 2y + 1 = 0 ...(1) w.r.t. the hyperbola 4x2 – y2 = 4a2

...(2)

Polar of (x1, y1) w.r.t. (2) is 4xx1 – yy1 = 4a  ...(3) 2

⇒ sec θ = x1 , tan θ = − y1 b a 2 2 2 2 Since sec θ – tan θ = 1, ∴ x1 − y1 = 1. 2 2 a b 2 2 ∴ Locus of (x1, y1) is x − y = 1, 2 2 a b

 s (1) and (3) represent the same line i.e. polar of A which is the given hyperbola. (x1, y1), ∴ comparing coefficients, we get 220. (d) Equation of any normal to the hyperbola

4 x1 − y1 4a 2 = = ⇒ x1 – 3a2, y1 = 8a2. 2 3 −1

 ence the required point i.e. the pole of (1) is H (– 3a2, 8a2). xcos θ + ysin θ = 2 ...(1) Let (x1, y1) be its pole w.r.t. the hyperbola ...(2) x2 – y2 = 4 ...(3) Polar of (x1, y1) w.r.t. (2) is xx1 – yy1 = 4 As (1) and (3) both represent the polar of (x1, y1), ∴ comparing coefficients, we get − y1 4 x1 = = sin θ cos θ 2

⇒ cos θ =

⇒ cos2 θ + sin2 θ = or

x

2 1

x y + 4 4

2 1

+ y = 4. 2 1

219. (a) Equation of any tangent to the hyperbola



...(1)

Let (x1, y1) be its pole w.r.t.

x 2 y 2 = 1 + a 2 b2

x1 / a 2 − y1 / b 2 1 = = 2 a + b2 a / secθ b / tan θ

xx1 yy1 = 1 + 2 a2 b

x1 / a 2 y / b2 = 1 =1 sec θ / a − tan θ / b

or

x1 − y1 = =1 a secθ b tan θ

∴

a6 b6 − 2 2 =1 2 2 x (a + b ) y1 (a + b 2 ) 2

or

a 6 b6 − = (a2 + b2)2. x12 y12

2 1

2

a 6 b6 − = (a2 + b2)2. x2 y 2

2 2 221. (c) Equation of hyperbola is x − y = 1. 2 2 a a x y Any tangent to it is sec θ − tan θ = 1. a a

i.e. x sec θ – y tan θ = a i.e. y tan θ = xsec θ – a

...(1)

y2 = 4ax

...(2)

Polar of (x1, y1) w.r.t. (2) is yy1 = 2a (x + x1) or ...(3)

 ince (1) and (3) represent the same line i.e. polar of S (x1, y1) w.r.t. (2), ∴ comparing coefficients

a3 b3 , tan θ = 2 2 x1 (a + b ) y1 (a 2 + b 2 )

Let (x1, y1) be its pole w.r.t. the parabola ...(2)

Polar of (x1, y1) w.r.t. (2) is



Hence, locus of (x1, y1) is

which is the circle itself. x2 y 2 − = 1 is a 2 b2 x y sec θ − tan θ = 1 a b

...(2)

Also, sec2 θ – tan2 θ = 1

∴ Locus of (x1, y1) is x2 + y2 = 4,



...(1)

Polar of (x1, y1) w.r.t. (1) is xx1 − yy1 = 1 ...(3) a2 b2 As (2) and (3) both represent the polar of (x1, y1), ∴ comparing coefficients

⇒ sec θ =

x1 − y1 and sin θ = . 2 2 2 1

is

x 2 y 2 = 1 − a 2 b2 ax by = a 2 + b 2 + sec θ tan θ

Let (x1, y1) be its pole w.r.t. (1).

218. (c) Any tangent to the circle x2 + y2 = 4 is





yy1 = 2ax + 2ax1

...(3)

 s (1) and (3) both represent the polar of (x1, y1) A w.r.t. (2), ∴ comparing coefficients, tan θ −a = secθ = y1 2 ax1 2a a y ⇒ sec θ = − , tan θ = − . x1 2 x1

Also, sec2 θ – tan2 θ = 1, ∴

a2 y2 − 12 = 1 2 x1 4 x1

2

⇒

Hence, locus of (x1, y1) is 4x2 + y2 = 4a2.

2 2 2 ⇒ 4 x1 − 4ax1 + a − 4 y1 = 1 2 2 a b

222. (a) Equation of hyperbola is 4x2 – y2 = 4 or

x 2 y 2 = 1 − 1 4

...(1)

 quation of the chord of (1) having (2, – 3) as its E mid point is

x (2) y (− 3) (2) 2 (− 3) 2 − − 1 [T = S1] − −1 = 1 4 1 4

or 2 x + 3 y = 4 − 9 4 4

or 8x + 3y = 7.

223. (c) Let (x1, y1) be the middle point of a chord of the 2 2 hyperbola x − y = 1. 9 4 Equation of the chord having (x1, y1) as its middle xx1 yy1 x2 y 2 = 1 − 1 point is − 9 4 9 4 If it passes through (1, 2), then

x1 2 y1 x2 y 2 − = 1 − 1 9 4 9 4

∴ Locus of (x1, y1) is

x2 y 2 x 2y = − − 9 4 9 4

or

 x2 x   y 2 2 y  −  − −  =0 4   9 9  4

or

x2 − x y 2 − 2 y =0 − 9 4 2

or

1  x −  ( y − 1) 2 2 1 1  − = − 9 4 9⋅4 4

which is a hyperbola a whose centre is  1 , 1 . 2  224. (b) Let (x1, y1) be the mid point of the variable chord AP (where A is fixed while P varies) of the hyperbola

x2 y 2 = 1. − a 2 b2

4 x12 4 x1 4 y12 =1 − + 1 − a2 a b2

505

4a2 – y12 = 4 x12 or 4 x1 + y12 = 4a2.

⇒

(2 x1 − a ) 2 4 y12 − 2 = 1. a2 b

2 2 Hence locus of (x1, y1) is (2 x − a ) − 4 y = 1. a2 b2

225. (a) Let (x1, y1) be the middle point of a variable chord x2 y 2 of the hyperbola 2 − 2 = 1 a b Equation of the chord is

xx1 yy1 x2 y 2 − 2 = 12 − 12 2 a b a b

...(1)

[T = S1]

I f this chord touches the circle x2 + y2 = c2, then length of ⊥ from the centre (0, 0) on (1) is equal to the radius c. x12 y12 − 2 2 ⇒ ± a 2 b 2 = c x1 y + 1 a 4 b4 2 2  x2 y 2   2  ⇒  x1 − y1  = c2   14 − 14  2 b  b2  a a

 x2 y 2   x2 y 2  Hence locus of (x1, y1) is  2 − 2  = c2  4 + 4  . b  b  a a 2

x2 y 2 − = 1. a 2 b2 y 2 x2 Then, the conjugate hyperbola is 2 − 2 = 1. b a

226. (c) Let the equation of the hyperbola be

Their eccentricities e and e' are given by b2 = a2 (e2 –­ 1) and a2 = b2 (e' 2 – 1) Multiplying the corresponding sides, we have a2b2 = a2b2 (e2 – 1) (e' 2 – 1) ⇒ 1 = (e 2 – 1) (e' 2 – 1) ⇒ 1 = e2e ' 2 – e' 2 – e2 + 1 1 1 ⇒ e2 + e' 2 = e2e' 2 ⇒ 2 + 2 = 1. e e′

2 2 Equation of the chord having (x1, y1) as its mid 227. (a) Equation of the hyperbola is x − y = 1. points is 4 9 2 2 xx1 yy1 x12 y12 Here a = 4 and b = 9 − 2 = 2 − 2 (T = S1) a2 b a b Equation of given diameter is y = 3x = m1x, where m1 = 3 Let the equation of diameter conjugate to y = 3x be As it passes through the fixed point A (a, 0) y = m2x. 2 2 2 2 ax1 x y 4 x 4 x 4 y 1 1 1 1 1 b2 = or ∴ − − − 2 =0 Then, m1m2 = 2 a2 a 2 b2 a2 a b a

Conic Sections (Parabola, Ellipse and Hyperbola)

i.e.

506

Objective Mathematics

⇒ 3m2 =

9 4

or m2 =

3 . 4

230. (a) Equation of the hyperbola is



(x – ae) (x + ae) + (y – 0) (y – 0) = 0 x 2 + y 2 = a 2e 2

or

...(2)

Equation of chord i.e. polar of (x1, y1) w.r.t. (1) is

Ax2 + 2Hxy + by2 = 0 be y = mx and y = m 'x. The two diameters are conjugate, b2 a2

 quation of the circle on the join of foci as diE ameter is

Let (x1, y1) be the pole of a chord of (1).

228. (c) Let the two diameters represented by

...(1)

Also m, m' are the roots of Bm2 + 2Hm + A = 0 A ...(2) ∴ mm' = B 2 A From (1) and (2) , b = 2 B a or Aa2 = Bb2 which is the required condition.

xx1 yy1 − 2 = 1 a2 b

...(3)

 ince it touches the circle (2), the length of ⊥ from the S centre (0, 0) of (2) on (3) is equal to radius ae. ⇒ ±

or

1 x12 y12 + a 4 b4

= ae

1 1 x12 y12 . + 4 = 2 2 = 2 4 a + b2 ae a b  2 a 2 + b2  ∵e =  a2  



229. (d) Equation of any tangent to x2 – y2 = a2 i.e.

x2 y 2 − =1 a2 a2

∴ Locus of (x1, y1) is

x y sec θ − tan θ = 1 a a or x sec θ – y tan θ = a

is

...(1)

Equation of other two sides of the triangle are x – y = 0

...(2)

x + y = 0

and

...(3)

[The two asymptotes of the hyperbola x2 – y2 = a2 are x – y = 0 and x + y = 0]  olving (1) (2) and (3) in pairs, the coordinates of S the vertices of the triangle are (0, 0);

...(1)

Its foci are (ae, 0) and (­– ae, 0).

Hence the equation of the conjugate diameter is 3 y=  x. 4

∴ mm' =

x2 y 2 − = 1 a 2 b2

  a a ,−   sec θ + tan θ sec θ + tan θ  

. a a and  ,   sec θ − tan θ sec θ − tan θ 

231. (b) The equation of the chord of contact of tangents to the given hyperbola from P (x1, y1) is T = 0 xx1 yy1 ...(1) − 2 = 1 a2 b The joint equation of the two lines joining the centre to the points of intersection of (1) and the given hyperbola is

i.e.,



=

 1 a2 a2 +  2 2 2 2  2  sec θ − tan θ sec θ − tan θ 



=

1 2 (a + a 2 ) 2



= a 2.

[

sec2 θ – tan2 θ = 1]

x2 y 2 yy   xx − =  21 − 21  a 2 b2 b  a 2



...(2)

 he two lines given by (2) will be at right angles if T the sum of the coefficients of x2 and y2 is zero.  1 x12   −1 y12  ∴  2 − 4  +  2 − 4  = 0 a  b b  a or

∴ Area of triangle

1 x2 y 2 = 2 . + a + b2 a 4 b4

1 1 x12 y12 + = 2 − 2 . a b a 4 b4

2 2 1 1 Hence the locus of P (x1, y1) is x + y = 2 − 2 , 4 4 a b a b which is an ellipse.

232. (a) The given hyperbola can be written in the form x2 y 2 − = 1. 9 4

 13  It meets x-axis in M   sec θ, 0  and 3 

 13  y-axis in N   0, tan θ  .  2 

⇒ 1 + e = 9e – 9 ⇒ e =

Let the coordinates of P be (x1, y1). 13 Since MP ⊥ NP, x1 = NP = secθ , 3 13 y1 = MP = tan θ 3 ∴ sec θ = ⇒

2 1

 25  − 1 = 9. Also, b2 = a2 (e2 – 1) = 16    16  Thus, from (1), equation of hyperbola is

( sec2 θ – tan2 θ = 1)

Hence the locus of (x1, y1) is 9x2 – 4y2 = 169. 233. (a) The equation of the normal to xy = c2 at the point ‘t’ is ty = t3x + c – ct4 or t3x – ty + c – ct4 = 0

...(1)

 ow, the equation of the line joining ‘t’ and ‘t' ’ N is c c / t′ − c / t y– = (x – Ct) t ct ′ − ct 1 yt − c = − ( x − ct ) tt ′ t

⇒

⇒ ytt' – ct' = – x + ct ...(2)

Since (1) and (2) represent the same line, ∴, on comparing the coefficients of x and y, we get tt ′ 1 = − ⇒ t3t ' = – 1. t3 t

2b 2 1 = ( 2a ) a 2



⇒ 2e2 – 2 = 1 ⇒ e2 =

∴ e =

3 . 2

3 . 2

x2 y 2 − = 1 a 2 b2

( x + 4) 2 ( y + 1) 2 − = 1. 16 9 Here a2 = 16 and b2 = 9.

b2 9 25 1+ = ⇒ e = 5/4. 2 = a 16 16

x ⋅ 2 sec θ y ⋅ 3 tan θ = 1. − 4 9

Slope of the tangent =

235. (a) Let the equation of hyperbola be

x2 y 2 − =1 9 16 Here, a2 = 9 and b2 = 16. The given line is y = 2x + λ. Here, m = 2 and c = λ. If the line y = 2x + λ touches the hyperbola 16x2 – 9y2 = 144 then c2 = a2m2 – b2 ⇒ λ2 = 9 (4) – 16 = 20.  ∴  λ = ± 2 5 .



⇒ 2b2 = a2 ⇒ 2a2 (e2 – 1) = a2

236. (a) The given equation of hyperbola can be written as

238. (c) Equation of tangent at the point (2sec θ, 3tan θ) of x2 y 2 − the hyperbola = 1 is 4 9

x2 y 2 − = 1. a 2 b2

Given :

x2 y 2 − = 1. 16 9

∴ e2 = 1 +

234. (c) Let the equation of the hyperbola be



237. (a) The given hyperbola can be written in the form

⇒ x + ytt' – c (t + t ' ) = 0



5 . 4

5  a (1 + e) = 9, ∴ a  1 +  = 9 ⇒ a = 4. 4 

3 x1 2 y1 and tan θ = 13 13

9x 4y − = 1 169 169 2 1

...(1)

3 3 sec θ = . 2sin θ 2 tan θ

Slope of given line = 3. Since the tangent is parallel to the given line, 3 ∴ 2sin θ = 3 1 ⇒ sin θ = 2 ⇒ θ = 30º.

507

Its vertices are A (a, 0) and A ' (­– a, 0) and foci are S (ae, 0) and S ' (– ae, 0). Given : S ' A = 9 and SA = 1 ⇒ a + ae = 9 and ae – a = 1 ⇒ a (1 + e) = 9 and a (e – 1) = 1 9 a (1 + e) ⇒ = 1 a (e − 1)

Conic Sections (Parabola, Ellipse and Hyperbola)

The equation of any normal to the given hyperbola 3x 2y is sec θ + tan θ = 13.

508

239. (b) Since

16 1  x2 y 2 − −1> 0 , − − 1 = 4 3 4 3  ( 4, 1)

⇒ y + 2a =

Objective Mathematics

∴ the point (4, 1) lies outside the hyperbola, hence the number of tangents through (4, 1) is two. 1 4 Any normal to this parabola with slope m is 1  1 y = mx − 2  m  − m3 , 4  4 The point (c, 0) lies on this,

240. (c) Here, a =

a3 3

2

3  9 a3   x + 4a  − 16a 2 × 3   2

35a  a 3  3   ⇒  y + x+ = 16  3  4a   35a Let h = − 3 and k =− 4a 16 ⇒ hk = 105 64 Hence, the locus of vertices of a parabola is 105 xy = . 64

m 1 3 − m 2 4 ⇒  m = 0 or c − 1 = 1 m 2 2 4

if 0 = mc −

245. (b) Here, 2ae = 6 ⇒ ae = 3

1 ⇒  m = 0 or m = ± 2 c − . 2  or three normals through (c, 0) all the three values of F m should be real and distinct.

…(i)

and 2b = 8 ⇒ b = 4 ⇒ b2 = 16 ⇒ a2(1 – e2) = 16 ⇒ a2 – (ae)2 = 16 ⇒ a2 – 32 = 16 ⇒ a2 = 25 ⇒ a = 5 From Eqs. (i) and (ii), we get

…(ii)

∴  c > 1 . 2 3 3 241. (c) Clearly, from the figure, the two curves do not e= = a 5 intersect each other. y 246. (a) As distance of vertex from origin is is 2 2 .

2 and focus

∴ V (1, 1) and F (2, 2) (i.e., lying on y = x), where, length of latusrectum = 4a = 4 2 (where a = ∴ By definition of parabola

(1, 2) x′

x

O

2)

y y=x

y′ 242. (a) The given line is a normal to the given conic if N

(a 2 + b 2 )2 a 2 b 2 = 2 − c2 m 1

x +y − 2 = 0

O

16 2 = 12 ⇒ m = ± m2 3

243. (b) T he line y = mx + 2

…(i)

will be a tangent to the conic 4x2 – 9y 2 = 36

…(ii)

if 2 = ± 9m 2 − 4 ⇒

m=±

2 2 3

244. (d) Given equation can be rewritten as

y + 2a =

a3 3

M

(1,1)V

2 ⇒ (16 + 9) × 9 = 16 − 9 625 × 3 m2 1

⇒

F(2,2)

P

 2 3   x + 2a x   

x

PM2 = (4a) (PN), where PN is length of perpendicular upon x + y – 2 = 0 (ie, tangent at vertex).  x+ y−2 ( x − y)2 = 4 2  2 2   ⇒ (x – y)2 = 8 (x + y – 2) ⇒

247. (b) We have e1 = 1 −

16 3 = , and foci of ellipse is (0, ± 3). 25 5

∴

⇒ b2x12 – a2y12 = a2b2

∴ Equation of hyperbola is x − y = −1 . 16 9 2

2

y = mx +



2 ,m≠0 m

This intersect xy = – 1.

2  ∴ x  mx +  = −1 m 

2 2 249. (a) We have, x − y = 1  3 2

…(i)

2x 3y  =1 ⇒ x = 3y 2

…(ii)

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

⇒ y = 4

y=±2

⇒

250. (c) The given equation of conic is 25x2 – 150x + 16y2 = 175

⇒ 25(x2 – 6x + 9 – 9) + 16y2 = 175 ⇒ 25(x – 3) + 16y = 400 2

b 2 x12 − a 2 y12 (a 2 + b 2 )2

=

a 2b 2  a 2 + b2

[using Eq. (i)]

252. (a) Let the equation of tangent, which is perpendicular to the line

Here, a2 = 9, b2 = 4 and m = 4 . 3 2 4   ∴ λ2 = 9 ×   + 4 ⇒ λ2 = 20 3 ⇒ λ = ±

20 20

253. (c) The given equation of parabola is

∴ Points of contact are (3, 2) and (– 3, – 2). ∴ Equations of tangents are ⇒ y – 2 = (x – 3) and x – y – 1 = 0 ⇒ y + 2 = x + 3





x2 y 2 − = 0 is a 2 b2

∴ Equation is 4x – 3y = ±

3y2 y2 − =1 4 2 2



3x + 4y = 7, be 4x – 3y = λ. Since, it is a tangent to the ellipse. ∴ λ2 = a2m2 + b2

dy 2 x ⇒ = dx 3 y

Now, slope of the line y = x is 1. Since, tangent is parallel to given line, then

x2 y 2 − =0 a 2 b2

∴ Product of perpendicular from (x1, y1) to pair of lines

⇒ m2x2 + 2x + m = 0 ∴ D = 0 ⇒ 22 – 4m2 · m = 0 ⇒ m = 1 Hence, the required common tangent is y = x + 2.

⇒ 2 x − y dy = 0 3 dx

…(i)

The asymptotes of given hyperbola are

248. (d) Equation of any tangent to the parabola is

x12 y12 − =1 a 2 b2

509

⇒ e2 =

2

2 2 ⇒ ( x − 3) + y = 1 16 25

Here, a2 = 16 and b2 = 25 a2 3 = b2 5 Hence, the foci of conic section are (3, ± 3).

∴ e = 1 −

2 2 251. (a) Let the equation of hyperbola be x − y = 1. a 2 b2

Let (x1, y1) be any point on the hyperbola.

y2 – x – 2y + 2 = 0 ⇒ (y – 1)2 = x – 1 ⇒ Y2 = X, where Y = y – 1, X = x – 1 Here, a = 1 4 1  ∴ Focus is (a, 0) ie,  , 0  4  ⇒ X = 1 , Y = 0 4 ⇒ x – 1 = 1 , y – 1 = 0 4

⇒ x = 5 , y =1 4

5  ∴ Required focus is  , 1 . . 4  2 2 254. (a) The equation of given ellipse is x + y = 1 4 3

Here, a = 2, b =

3

∴ 3 = 4 (1 – e2)

1 ∴ ae = 1 2 Hence, the eccentricity e1, of the hyperbola is given by 1 = e1 sin θ

⇒ e =

Conic Sections (Parabola, Ellipse and Hyperbola)

5 3 Since, hyperbola is passing through foci (0, ± 3).

Since, e1e2 = 1

510

Objective Mathematics

⇒ e1 = cosec θ ⇒ b2 = sin2 θ (cosec2θ – 1) = cos2 θ Hence, the hyperbola is

x2 y2 − =1 sin 2 θ cos 2 θ



which represent the same line. ∴

or x2cosec2θ – y2 sec2θ = 1 255. (d) T he intersection of given line y = 2x + c and given parabola y2 = 4ax + 4a2 is (2x + c)2 = 4ax + 4a2 ⇒ 4x2 + 4 (c – a) x + (c2 – 4a2) = 0 Since, it is a tangent line. ∴ D = 0 ⇒ 16(c – a)2 – 4 × 4 (c2 – 4a2) = 0 ⇒ c2 + a2 – 2ac – c2 + 4a2 = 0 ⇒ 2ac = 5a2 ⇒ c = 5a 2



x y + =1 16 9 2

∴ e = 1 −

(0 ± 7 ) 2 + (3 − 0) 2

= 7+9 =4

257. (a) The given equation of hyperbola can be rewritten as x2 y 2 − =1 16 9

∴ Required equation of director circle is x2 + y2 = 16 – 9 or x2 + y2 = 7 258. (b) The parameric equation of the given ellipse

259. (b) Eliminating t from x = t2 + 1, y = 2t, we obtain y2 = 4x – 4 Similarly, eliminating S from x = 2S, y = 2 , S we get, xy = 4 Hence, point of intersection is (2, 2). 260. (c) If y = 2x + l is a tangent to given hyperbola

⇒ λ = ±

a 2 m2 − b2 (100) (4) − 144 = ±

256 = ±16

⇒ c2 = 4 × 42 + 1 = 65

(± 7 , 0).

x y + = 1 is 25 9 2

⇒ θ= π 4

262. For ellipse, condition of tangency is c 2 = a2 m2 + b 2

9 7 = 16 4

∴ Radius of circle =

1 1 and sin θ = 2 2

But in option (b) 2 < 1 . 3

∴ Coordinates of foci are (± 7 , 0)  ince, centre of circle is (0, 3) and passing through S foci



⇒ cos θ =

261. (b) Since, e > 1 always for hyperbola.

Here, a2 = 16, b2 = 9



cos θ 3 sin θ 5 and = = 5 3 15 2 15 2

then, λ = ±

256. (a) Given equation of ellipse is 2

3x 5y =1 + 15 2 15 2

2

x = 5 cos θ, y = 3 sin θ The equation of tangent to the ellipse at the point (5 cos θ, 3 sin θ) is x y cos θ + sin θ = 1 5 3 Since, the given equation of tangent is

⇒ c = ±

65

263. (b) The equation x2 – y2 = 0 is an equation of rectangular hyperbola. Therefore, the locus of the equation x2 – y2 = 0 is a hyperbola. 264. (d) Let the mid point be P(h, k). Equation of this chord is T = S1 ie., ky – 2a (x + h) = k2 – 4ah It must passes through (a, 0). (–2a) (a + h) = k2 – 4ah Hence, the locus is y2 = 2ax – 2a2 265. (a) Let the equation of hyperbola be

x2 y 2 − = 1. a 2 b2

Let (x1, y1) be any point on the hyperbola.

x12 y12 − =1 a 2 b2

⇒ b2x12 – a2y12 = a2b2  The asymptotes of given hyperbola are

…(i)

x2 y 2 − =0 a 2 b2

∴ Product of perpendicular from(x1, y1) to pair of lines

( A − B)2 + 4 H 2

=

b 2 x12 − a 2 y12 (b 2 + a 2 ) 2

2 2 = ab  [from Eq. (i)] 2 a + b2 266. (a) Equation of tangent at (x´ y´) on hyperbola

x2 y 2 xx ' yy ' − = 1 is 2 − 2 = 1 a 2 b2 a b ∴

⇒ 4a – 2 > 0 ⇒ a > 1 2 269. (c) Let the coordinates of P and Q are (t12 , 2t1), (t22, 2t2)

O

x'

⇒ y = 3 sec θ x − 3 2 tan θ tan θ

y'

slope of tangent = 3 sec θ 2 tan θ ∵ tangent is parallel to the line 3x – y + 4 = 0 ∴  3 sec θ = 3 2 tan θ

267. (a) Finding point of intersection of x – 3y = 1 and x2 – 4y2 = 1 ∴ (1 + 3y)2 – 4y2 = 1 2 2 ⇒ 1 + 6y + 9y – 4y = 1 ⇒ 5y2 + 6y = 0 ⇒ y (5y + 6) = 0 ⇒ y = 0 or –6/5 If y = 0 then x = 1

slope of OP =

2t1 − 0 2 = t12 − 0 t1

slope of OQ =

2t2 − 0 2 = t2 2 − 0 t2

6  13   1 + 5  +  0 + 5     

∴ Slope of OP × slope of OQ = – 1 2 2 ⇒ ⋅ = –1 ⇒ t1 t2 = –4 …..(i) t1 t2



t1 + t2 = k

… .(iii)

from (i), (ii) and (iii)

 6  5 − 18 −13 − 5  = 5 = 5  

2

2

324 36 360 6 + = = 10 25 25 25 5

268. (c) Equation of normal in slope form to the parabola y 2 = x is

Q(t22 ,2t2 )

Let the coordinates of mid point of PQ be (h, k) ∴ t12 + t2 2 = 2h … .(ii)

 −13 6  so points of intersection are (1, 0),  ,−  5  5

∴ =

x

Since OP ⊥ OQ,

⇒ sinθ = 1 ⇒ θ = 30 2

so length of intercept =

R(h,k)

90º

(0,0)

(2sec θ) x (3tan θ) y − =1 4 9

If y = − 6 then x = 1 + 3 5

P (t12 ‚2t1)

y

1 1  y = mx – 2   m −  m3  4 4 

⇒ 4mx – 4y – m3 – 2m = 0 ∵ (a, 0) lies on the normal So 4m × a – 4 × 0 – m3 – 2m = 0 ⇒ m(m2 + 2 – 4a) = 0 ⇒ m = 0 or m2 + 2 – 4a = 0 If m = 0, then from (i) y = 0 i.e. x - axis is one normal

(t1 + t2)2 = t12 + t22 + 2t1t2 ⇒ k2 = 2h + 2 × –4 ⇒ k2 = 2(h – 4) ⇒ y 2 = 2(x – 4) which is required locus. 2 2 270. (a) The equation of given ellipse is x + y = 1 4 3

⇒ a = 2, b =

3

⇒ 3 = 4 (1 – e2) ⇒ e = 1 2 so that ae = 1 Hence the eccentricity e1, of the hyperbola is given by 1 = e1sinθ ⇒ e1 = cosecθ ⇒ b2 = sin2θ(cosec2θ – 1) = cos2θ Therefore, the equation of hyperbola is x2 y2 2 2 2 2 − = 1 or x cosec θ – y sec θ = 1 2 sin θ cos 2 θ 271. (b) T he circle and the parabola touch each other at x = 1 i.e. at the points (1, 2) and (1, – 2) as shown in the figure.

Conic Sections (Parabola, Ellipse and Hyperbola)

| Ax12 + 2 Hx1 y1 + By12 |

511

If m2 + 2 – 4a = 0 ⇒ m2 = 4a – 2 ∵m2 > 0

x2 y 2 − = 0 is a 2 b2



(given y1 and y2 less than 0). Co-ordinates of mid-point of PQ are

512

(1,2)

Objective Mathematics

1 R ≡  0, −  . 2 

(3, 0 ) ( 1 , 0)

PQ = 2 3 = length of latus rectum.

2√2

⇒ two parabola are possible whose vertices are   3 1 3 1 −  and  0, −   0, − 2 2 2 2   

( 1 , –2 )

272. (b, c)  The given ellipse can be written as

Hence the equations of the parabolas are x2 – 2 3 y = 3 + 3 and x2 + 2 3 y = 3 – 3 . 273. (b) We have, x 2 − 2 y 2 − 2 2 x − 4 2 y − 6 = 0 2 2 ⇒ ( x − 2 ) − ( y − 2 ) = 1 4 2 2 2 x y ⇒ − =1 22 2

R Q (x2, y2 )

P (x2, y2 )

e=

x y =1 + 4 1 2

1+

2

2 = 4

3 2

2 Hence required Area = 1 (ae − a ) b 2 a

Now, b2 = a2(1 – e2) ⇒ e = 3 2

=

1 1 ⇒ P  3, −  and Q  − 3, −  2 2  

=

 2_ 1 _ 3 × 2 − 1 ×  2  _ 2   2 3 − 1 Hence option (B) is correct 2

Exercises for self-practice 1. The locus of the mid point of the line segment joining the focus to a moving point on the parabola y2 = 4ax is another parabola with the directrix (a) x = – a (b) x = –  a 2 a (c) x = 0 (d) x = 2 2 2. The tangents at the points (a t1 , 2at1) and (at 22, 2at2) on the parabola y2 = 4ax are at right angles if (a) t1t2 = ­– 1 (c) t1t2 = 2

(b) t1t2 = 1 (d) t1t2 = ­– 2

3. The line y = mx + 1 is a tangent to the parabola y2 = 4x if (a) m = 1 (c) m = 4

(b) m = 2 (d) m = 3

(a) 1 t

(b) t

(d) −1 t 5. The equation of normal at the point (0, 3) of the ellipse 9x2 + 5y2 = 45 is (c) – t

(a) y – 3 = 0 (c) x-axis

(b) y + 3 = 0 (d) y-axis

6. The eccentricity of ellipse 5x2 + 4y2 = 1 is (a) 1 2 1 (c) 5

(b)

1 2

(d) 1 3

7. An ellipse has its centre at (1, – 1) and semi major axis = 8, which passes through the point (1, 3). Then the 4. The slope of the normal at the point (at , 2at) of pa  2 equation of the ellipse is rabola y = 4ax is, 2

(b) ( x + 1) + ( y + 1) = 1 64 16 2

2

2 2 (d) ( x − 1) + ( y + 1) = 1. 16 64 2 2 8. The number of normals to the hyperbola x − y = 1 2 2 a b from an external point is

(a) 2 (c) 6

(b) 4 (d) 5

2 2 9. Equation of the normal to the hyperbola x − y = 1 at 2 2 a b the point (a sec θ , b tan θ) is

(a)

ax by =a–b − sec θ tan θ

(b)

ax by = a2 – b2 − sec θ tan θ

(c)

ax by = a2 + b2 + sec θ tan θ

(d)

ax by = a2 – b2 + sec θ tan θ

2 2 10. The parametric equations of the hyperbola x − y 2 2 a b = 1 are

(a) x = a tan θ, y = b sec θ (b) x = a sec θ , y = b tan θ (c) x = 2 a, y = b (d) None of these 11. The eccentricity of hyperbola is (a) 0 (c) more than 1

(b) 1 (d) less than 1

12. The equation of the hyperbola referred to its axes as axes of coordinates and whose distance between the foci is 16 and eccentricity is 2 , is (a) x2 – y2 = 16 (c) x2 – 2y2 = 16

(b) x2 – y2 = 32 (d) y2 – x2 = 16

13. The equation of the line touching both the parabolas y2 = 4x and x2 = – 32y is (a) x + 2y + 4 = 0 (c) x – 2y – 4 = 0

(b) 2x + y + 4 = 0 (d) x – 2y + 4 = 0.

1 2

(b)

(c)

2

(d) > 2

2 2 17. The equation of the normal to the hyperbola x − y = 16 9 1 at (–4, 0) is:

(a) y = 0 (c) x = 0

(b) y = x (d) x = –y

18. The length of the latusrectum of the parabola whose focus is (3, 3) and directrix is 3x – 4y – 2 = 0 is: (a) 2 (c) 4

(b) 1 (d) None of these

19. The angle between the pair of tangents drawn from the point (1, 2) to the ellipse 3x2 + 2y2 = 5 is: (a) tan–1(12/5)

(b) tan–1 6/ 5

(c) tan–1(12 / 5 )

(d) tan–1(12 5 )

2 2 20. The foci of the ellipse x + y = 1 and the hyperbola 2 x2 y 2 coincide, then16the bvalue of b2 is: − 144 81

(a) 1 (c) 7

(b) 6 (d) 9

21. The equation x2 – 2xy + y2 + 3x + 2 = 0 represents: (a) a parabola (c) an ellipse

(b) a hyperbola (d) a circle

2 2 22. If the eccentricity of the two ellipse x + y = 1 and 169 25 x 2 y 2 = 1 are equal, then the value of a/b is: + 2 2 a b (a) 5/13(c) (b) 6/13 (c) 13/5 (d) 13/6

23. Equation of locus of a point whose distance from point (a, 0) is equal to its distance from y-axis is: (a) y2 + 2ax + a2 = 0 (b) y2 + 2ax + a2x2 = 0 (c) y2 + 2axy + a2x2 = 0 (d) y2 – 2ax + a2 = 0 24. The locus of the poles of normal chords of an ellipse is given by: (a)

a 6 b6 a 3 b3 + 2 = (a2 – b2)2 (b) 2 + 2 = (a2 – b2)2 2 x y x y

(c)

a 6 b6 a 3 b3 2 2 2 + = (a + b ) (d) + = (a2 + b2)2 x2 y 2 x2 y 2

14. Number of integral points lying inside the parabola y2 = 25. Eccentricity of the rectangular hyperbola is: 8x and circle x2 + y2 = 16 are (a) 2 (b) 2 (a) 15 (b) 17 1 (c) 19 (d) 12 (c) 1 (d) 2 2 2 15. If the normal at (ap , 2ap) on y = 4ax makes 90º at the vertex, then 26. Eccentricity of the curve x2 – y2 = a2 is: (a) p = – 1 (b) c = 1/2 (a) 2 (b) 2 (c) c > 1/2 (d) None of these (c) 4 (d) None of these

513

−1 2

(a)

2

(c) ( x − 1) + ( y + 1) = 1 64 16 2

16. Eccentricity of rectangular hyperbola is:

Conic Sections (Parabola, Ellipse and Hyperbola)

2 2 (a) ( x + 1) + ( y − 1) = 1 64 16

514

Answers

Objective Mathematics



1. (c) 11. (c) 21. (a)

2. (a) 12. (b) 22. (c)

3. (a) 13. (d) 23. (d)

4. (c) 14. (b) 24. (a)

5. (d) 15. (d) 25. (b)

6. (c) 16. (c) 26. (b)

7. (c) 17. (a)

8. (b) 18. (a)

9. (c) 19. (b)

10. (b) 20. (c)

13

Complex Numbers

CHAPTER

Summary of conceptS complex number An expression of the form x + iy, where x and y are real numbers and i = −1 is called a complex number. It is usually denoted by z, i.e., z = x + iy where x is called the real part and y the imaginary part of the complex number z and may be denoted by Re (z) and Im (z) respectively. A complex number may also be defined as an ordered pair z = (x, y) of real numbers.

Integral powerS of i Integral powers of i are defined as i0 = 1, i1 = i, i2 = – 1, i3 = i2 ⋅ i = (– 1) i = – i; i4 = (i2)2 = (– 1)2 = 1, i5 = i4 ⋅ i = i, i6 = i4 ⋅ i2 = – 1 and so on. If n is a positive integer such that n > 4, then to find in, we first divide n by 4. Let m be the quotient and r be the remainder. Then n = 4m + r, where 0 ≤ r < 4. Thus in = i(4m + r) = i4m ⋅ ir = (i4)m ⋅ ir = ir ( ∵ i4 = 1) For any two real numbers a and b, a × b = ab is true only when atleast one of a and b is either 0 or positive. In fact, 2 −a × −b = (i a ) (i b ) = i ab = – ab , where a and b are positive real numbers.

3. multiplication Multiplication of two complex numbers z1 = a + ib and z2 = c + id is defined as z1 ⋅ z2 = (a + ib) ⋅ (c + id) = (ac – bd) + i (ad + bc). 4.

Division

Division of two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, where x2 + iy2 ≠ 0, is defined as z1 x + iy1 ( x + iy1 )( x2 − iy2 ) = 1 = 1 z2 x2 + iy2 ( x2 + iy2 )( x2 − iy2 )

z–1 =

Two complex numbers are said to be equal if and only if their real parts and imaginary parts are separately equal, i.e.,

Note: Inequality relation does not hold good in case of complex numbers having non-zero imaginary parts. For example, the statement 8 + 5i > 4 + 2i makes no sense.

algebra of complex numberS 1. addition For two complex numbers z1 = a1 + ib1 and z2 = a2 + ib2, their sum is defined as z = z1 + z2 = (a1 + a2) + i (b1 + b2) 2. Subtraction For two complex numbers z1 = a1 + ib1 and z2 = a2 + ib2, the subtraction of z2 from z1 is defined as z1 – z2 = z1 + (– z2) = (a1 – a2) + i (b1 – b2).

x1 x2 + y1 y2 + i ( x2 y1 − x1 y2 ) x22 + y22

=

x1 x2 + y1 y2 i ( x2 y1 − x1 y2 ) . + x22 + y22 x22 + y22

5. multiplicative Inverse of a non-zero complex number Multiplicative inverse of a non-zero complex number z = a + ib is defined as

equalIty of complex numberS

a + ib = c + id ⇔ a = c and b = d.

=

1 1 a − ib a − ib 1 = = × = 2 a + ib a + ib a − ib a + b2 z a b = 2 −i 2 a + b2 a + b2

i.e.z– 1 =

Re (z ) [ − Im (z )] . +i | z |2 | z |2

conjugate of a complex number Conjugate of a complex number z = a + ib is defined as = a – ib. For example, 4 + 5i = 4 – 5i and 4 − 5i = 4 + 5i. Properties of Conjugate (i) ( z ) = z if and only if z is purely real

(ii) z =

(iii) z = –

if and only if z is purely imaginary

(iv) z + z = 2 Re (z) and z – z = 2i Im (z)

516

(v) z + z = z + z 1 2 1 2

y=±

Objective Mathematics

 a 2 + b2 − a    2  

(vi) z1 − z2 = z1 − z2

and

(vii) z1 z2 = z1 ⋅ z2

From (2), we can determine the sign of xy. If xy > 0, then x and y will have same sign. Thus

z1  z1  (viii) = ,z ≠0  z  z2 2 2 (ix) If z = f (z1), then z = f (

( ) = (z )

(x) z

n

  2  a 2 + b2 − a   a + b2 + a  a + ib = ±    +i     2 2      If xy < 0, then

)

n

(xi) z1 z2 + z1 z2 = 2 Re ( z1 z2) = 2 Re (z1 z2 )

  2  a 2 + b2 − a   a + b2 + a  a + ib = ±    −i     2 2      Thus, square roots of z = a + ib are :  | z | +a | z| − a  +i ±   for b > 0 and 2 2  

moDuluS of a complex number Modulus of a complex number z = a + ib, denoted as mod (z) or | z |, is defined as |z| =

a 2 + b 2 , where a = Re (z), b = Im (z).

Sometimes, | z | is called absolute value of z. Note that | z | ≥ 0. For example, if z = 3 + 2i, then | z | =

32 + 22 =

Properties of Modulus

arganD plane anD geometrIcal repreSentatIon of complex numberS

(i) | z | ≥ 0 and | z | = 0 if and only if z = 0, i.e., x = 0, y = 0 (ii) | z | = | z | = | – z | = | − z | . (iii) z z = | z |2 (iv) – | z | ≤ Re (z) ≤ | z | and – | z | ≤ Im (z) ≤ | z | (v) | zn | = | z |n (vi) | z1z2 | = | z1 | | z2 | (vii)

z1 |z | = 1 z2 | z2 |

(viii) | z1 ± z2 | ≤ | z1 | + | z2 | (ix) | z1 – z2 | ≥ | z1 | – | z2 | (x) | z1 + z2 |2 + | z1 – z2 |2 = 2 ( | z1 |2 + | z2 |2 ) (xi) | z1 + z2 |2 = | z1 |2 + | z2 |2 + 2 Re (z1 z2 ) (xii) | z1 – z2 |2 = | z1 |2 + | z2 |2 – 2 Re (z1 z2 )

Square rootS of a complex number Let z = a + ib and let the square root of z be the complex number x + iy. Then (a + ib) = (x + iy)2 = (x2 – y2) + (2xy) i Equating real and imaginary part, we get a = x2 – y2 and

b = 2xy Now, x2 + y2 =

( x 2 − y 2 )2 + 4 x 2 y 2 =

Solving the equation (1) and (3), we get x=±

 a 2 + b2 + a    2  

Let O be the origin and OX and OY be the x-axis and y-axis respectively. Then, any complex number z = x + iy = (x, y) may be represented by a unique point P whose coordinates are (x, y). The representation of complex numbers as points in a plane forms an Argand diagram. The plane on which complex numbers are represented is known as the complex plane or Argand’s plane or Gaussian plane. The x-axis is called the real axis and y-axis the imaginary axis. The complex number z = x + iy is known as the affix of the point (x, y) which it represents.

polar form of a complex number

Let O be the origin and OX and OY be the x-axis and y-axis respectively. Let z = x + iy be a complex number represented by the point P (x, y). Draw PM ⊥ OX. Then, ...(1) OM = x and PM = y. Join OP. Let ...(2) OP = r and ∠XOP = θ. Then

a + ib = x + iy or

 | z | +a | z | −a  −i ±   for b < 0 2 2  

13 .

a 2 + b2

z = x + iy = r (cos θ + i sin θ)

...(3)

This form of z is called polar or trigonometric form. Comparing real and imaginary parts, we get and

x = r cos θ y = r sin θ

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

2

2

x +y 2

2

= |z|

Thus, r is known and is equal to the modulus of the complex number z. Substituting the value of r in (1) and (2), we get x y cos θ = and sin θ = ...(3) 2 2 2 x +y x + y2

 π  π (iv) – i = 0 + i (– 1) = cos  −  + i sin  −   2  2

  π  π  2 cos  −  + i sin  −      4  4    3π   3π   (vi) – 1 – i = 2 cos  −  + i sin  −   . 4 4   (v) 1 – i =

y logarithm of complex number . x 1 2 2 −1 β ( α ≠ 0) The form z = r (cos θ + i sin θ) = reiθ of the complex number log (α + iβ) = log (α + β ) + i tan α 2 z is called exponential form. iπ log (iβ) = log β + Any value of θ  satisfying (3) is known as amplitude or 2 argument of z and written as θ = arg (z) or θ = amp z. Dividing (2) by (1), we get tan θ =

The unique value of θ such that – π < θ ≤  π for which x = r cos θ and y = r sin θ, is known as the principal value of the argument.

eular’s formula eiθ = cos θ + i sin θ π π + i sin = eiπ/2 2 2 iπ iπ   iπ log i = log e 2 = , log (log i) = log   2 2 π iπ = log i + log   = + log (π/2). 2 2 e– iθ = cos θ – i sin θ, i = cos

The general value of the argument is (2nπ + θ), where n is an integer and θ is the principal value of arg (z). While reducing a complex number to polar form, we always take the principal value. If x > 0, y > 0 (i.e., z is in first quadrant), then  y arg z = θ = tan– 1   . x If x < 0, y > 0 (i.e., z is in second quadrant), then arg z = θ = π – tan– 1

 y  .  | x | 

If x < 0, y < 0 (i.e., z is in third quadrant), then y arg z = θ = – π + tan– 1   . x If x > 0, y < 0 (i.e., z is in fourth quadrant), then | y| arg z = θ = – tan– 1   .  x  Note: Argument of the complex number 0 is not defined. Properties of Argument (i) arg (z1z2) = arg (z1) + arg (z2) (ii) arg

 z1   z  = arg z1 – arg z2  z  = 2 arg z  z

(iii) arg  

(iv) arg (zn) = n arg z

z 

(v) If arg  2  = θ, then arg  1  = 2kπ – θ where k ∈ I.  z1   z2  (vi) arg

If z and z′ are two complex z numbers then argument of z' is the angle through which Oz′ must be turned in order that it may lie along Oz. z | z | i(θ – θ′ ) | z | ei θ = = e iθ' z' | z' | | z' | e =

| z | iα e . | z' |

In general, let z1, z2 z3 be three vertices of a ∆ABC described in the counter-clockwise sense. Draw OP and OQ parallel and equal to AB and AC respectively. Then the point P is z2 – z1 and Q is z3 – z1 and z3 − z1 OQ CA iα | z3 − z1 | eiα = (cos α + i sin α) = e = |z − z | . z2 − z1 OP 2 1 BA

2

z 

concept of rotation

= – arg z

particular cases of polar form (i) 1 = 1 + i0 = cos 0 + i sin 0 (ii) – 1 = – 1 + i0 = cos π + i sin π π π + i sin (iii) i = 0 + i1 = cos 2 2

517

r = x + y or r = 2

Complex Numbers

Squaring (1) and (2) and adding, we get

518

Objective Mathematics

Note that arg (z3 – z1) – arg (z2 – z1) = α is the angle through which OP must be rotated in the anticlockwise direction so that it may lie along OQ. z −z |z − z | Note: We can also write 3 1 = 3 1 e − i ( 2 π − α ) . z2 − z1 | z2 − z1 | In this case, we are rotating OP in clockwise direction by an angle (2π – α). Since the rotation is in clockwise direction, we are taking negative sign with angle (2π – α).

Some useful relations

De moIvre’S theorem

x2 + y2 = (x + iy) (x – iy) x3 + y3 = (x + y) (x + yω) (x + yω2) x3 – y3 = (x – y) (x – yω) (x – yω2) x2 + xy + y2 = (x – yω) (x – yω2), in particular, x2 + x + 1 = (x – ω) (x – ω2) (v) x2 – xy + y2 = (x + yω) (x + yω2), in particular, x2 – x + 1 = (x + ω) (x + ω2) (vi) x2 + y2 + z2 – xy – xz – yz = (x + yω + zω2) (x + yω2 + zω) (vii) x3 + y3 + z3 – 3xyz = (x + y + z) (x + ω y + ω2z) (x + ω2y + ω z)

If n is any integer, then (cos θ + i sin θ)n = cos nθ + i sin nθ.

nth rootS of unIty

Important Results (i) If n is any rational number, then cos nθ + i sin nθ is one of the values of (cos θ + i sin θ)n (ii) (cos θ + i sin θ)– n = cos nθ – i sin nθ (iii) (cos θ – i sin θ)n = cos nθ – i sin nθ (iv)

1 = (cos θ + i sin θ)–1 = cos θ – i sin θ cos θ + i sin θ

(v) (sin θ + i cos θ)n ≠ sin nθ + i cos nθ.

 nπ  nπ − nθ  + i sin  − nθ   2  2 

In fact, (sin θ + i cos θ)n = cos  

(vi) (cos θ + i sin φ)n ≠ (cos nθ + i sin nφ) (vii) (cos θ1 + i sin θ1) (cos θ2 + i sin θ2) ... (cos θn + i sin θn) = cos (θ1 + θ2 + ... + θn ) + i sin (θ1 + θ2 + ... + θn)

(i) (ii) (iii) (iv)

Since 1 = cos 0 + i sin 0, therefore (1)1/n = (cos 0 + i sin 0)1/n 2π r + 0 2π r + 0 = cos + i sin ; r = 0, 1, 2, ..., (n – 1) n n 2π r 2π r + i sin ; r = 0, 1, 2, ..., (n – 1) = cos n n 2rπ

i ; r = 0, 1, 2, ..., (n – 1) e n = 1, e(i2π/n), e(i4π/n), ..., e[i2(n – 1)π/n] = 1, α, α2, α3, ..., αn – 1, where α = e(i2π/n)

=

Properties of nth Roots of Unity (i) 1 + α + α2 + ... + αn – 1 = 0 (ii) α ⋅ α ⋅ α2 ⋅ ... αn – 1 = (– 1)n – 1

rootS of a complex number

(iii) The n, nth roots of unity lie on the unit circle | z | = 1 and form the vertices of a regular polygon of n sides.

If z = r (cos θ + i sin θ) and n is a positive integer, then

(iv) nth roots of unity form a G.P. with common ratio e(i2π/n).

1   2k π + θ  + i sin = r n cos  z  n   where k = 0, 1, 2, 3, ... (n – 1). Cube roots of unity Let z = 11/3 or z3 – 1 = 0 ⇒ (z – 1) (z2 + z + 1) = 0. 1 n

i.e.,

z = 1,

 2k π + θ     , n  

−1 + i 3 −1 − i 3 , . 2 2

−1 + i 3 −1 − i 3 , then ω2 = . Thus 2 2 Cube roots of unity are 1, ω, ω2.

Put ω =

(i) 1 + ω + ω2 = 0

Distance between two points z1 and z2 | z 1 – z2 |

Section formula If the line segment joining A (z1) and B (z2) is divided by the point P (z) in the ratio m1 : m2, then z=

m1 z2 + m2 z1 m1 + m2

But if P divides AB externally in the ratio m1 : m2, then m1 z2 − m2 z1 m1 − m2

If P is mid point of AB, then z =

(ii) ω3 = 1 (iii) ω3n = 1, ω3n + 1 = ω, ω3n + 2 = ω2 (iv) ω = ω2 and (ω)2 = ω, 2 πi 3

, ω2 = e

2 πi − 3

(v) The cube roots of unity lie on the unit circle and divide the circumference into three equal parts. (vi) If a + bω + cω = 0, then       a = b = c provided a, b, c are real. 2

Distance formula is given by

z=

Properties of Cube Roots of Unity

ωω = ω3, ω = e

geometry of complex numberS

z1 + z2 . 2

condition for collinearity Three points z1, z2 and z3 will be collinear if there exists a relation az1 + bz2 + cz3 = 0 (a, b and c are real), such that a + b + c = 0. In other words

z1 Three points z1, z2 and z3 are collinear if z2 z3

z1 1 z2 1 = 0. z3 1

Equation of straight line through z1 and z2 is given by z − z1 z − z1 = z2 − z1 z2 − z1 or

z z1 z2

z 1 z1 1 = 0 z2 1

Equation of Ellipse If | z – z1 | + | z – z2 | = 2a, where 2a > | z1 – z2 |, then the point z describes on ellipse having foci at z1 and z2 and a ∈ R+.

The general equation of straight line is = 0, where b is a real number.



Equation of the perpendicular bisector

Equation of Hyperbola If | z – z1 | – | z – z2 | = 2a, where 2a < | z1 – z2 |, then the point z describes a hyperbola having foci at z1 and z2 and a ∈ R+.

The equation of the perpendicular bisector of the line segment Some Properties of Triangle joining points A (z1) and B (z2) is 1. If z1, z2, z3 are the vertices of triangle then centroid z0 may be given as 2 2 z ( z1 − z2 ) + z ( z1 − z2 ) = | z1 | − | z2 | z + z + z3 z0 = 1 2 3 2. If z1, z2, z3 are the vertices of an equilateral triangle then the Equation of a Circle circumcentre z0 may be given as The equation of a circle with centre z0 and radius r is | z – z0 | = r. z12 + z22 + z32 = 3  z02 The general equation of a circle is z z + a z + a z + b = 0, where b is a real number. The centre of this circle is ‘– a’ and its 3. If z1, z2, z3 are the vertices of an equilateral triangle then radius is a a − b . z 2 + z 2 + z 2 = z  z + z  z + z  z 1

Important Results (i)



z − z1 = k is a circle if k ≠ 1 and is a line if k = 1. z − z2

(ii) The equation | z – z1 |2 + | z – z2 |2 = k represents a circle 1 if k ≥ | z1 – z2 |2. 2

2

3

1 2

2 3

3 1

1 1 1 + + = 0. z1 − z2 z2 − z3 z3 − z1

or

4. If z1, z2, z3 are the vertices of an isosceles triangle, right angled at z2, then z12 + z22 + z32 = 2 z2 (z1 + z3). 5. If z1, z2, z3 are the vertices of right angled isosceles triangle then (z1 – z2)2 = 2 (z1 – z3) (z3 – z2).

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. ω is an imaginary cube root of unity. If (1 + ω2) m = (1 + ω4) m , then least positive integral value of m is (a) 6 (c) 4

(b) 5 (d) 3

2. Complex numbers z1, z 2 and z3 in A.P. (a) lie on a line (c) lie on ellipse

(b) lie on a parabola (d) lie on circle

3. 1 + i2 + i4 + i6 + ... + i2n is (a) positive (c) 0

(b) negative (d) cannot be determined.

4. The product of cube roots of –1 is equal to (a) –1 (c) –2

(b) 0 (d) 4

1 + i  1 − i  5.  +  1 − i   1 + i  2

(a) 2i (c) – 2

2

is equal to (b) – 2i (d) 2

6. If n is any positive integer, then the value of equals (a) 1 (c) i

(b) – 1 (d) – i

7. The smallest positive integer n for which (1 + i)2n = (1 – i)2n is (a) 1 (c) 3

(b) 2 (d) 4

i 4 n +1 − i 4 n −1 2

519

 ( z − z3 )( z1 − z4 )  (iii) If arg  2  = ± π, 0, then the points z1,  ( z1 − z3 )( z2 − z4 )  z2, z3, z4 are concyclic. (iv) | z – z0 | < r represents interior of the circle | z – z0 | = r and | z – z0 | > r represents exterior of the circle | z – z0 | = r.

Complex Numbers

Equation of Straight Line

520

8. If | z2 – 1 | = | z | + 1, then z lies on

Objective Mathematics

(a) a circle (b) the imaginary axis (c) the rea axis (d) an ellipse 9. Let z, w be complex numbers such that z + i w = 0 and zw = π Then arg z equals (b) π/2 (d) 5π/4

(a) 3π/4 (c) π/4

10. If (x + iy)1/3 = a + ib, then (a) 2 (a2 – b2) (c) 8 (a2 – b2) 11.

x y = + a b

(b) 4 (a2 – b2) (d) None of these

24 10 + i 13 13

(b)

24 10 − i 13 13

(c)

10 24 + i 13 13

(d)

10 24 − i 13 13

12. Let z1 and z2be two roots of the equation z2 + az + b = 0, z being complex. Further, assume that the origin, z1 and z2 form an equilateral triangle. Then (b) a2 = b (d) a2 = 4b

13. If iz4 + 1 = 0, then z can take the value 1+ i 2 1 (c) 4i

(b) cos

π π + i sin 8 8

(d) i

14. If |z| = 1, w = z − 1 and z ≠ –1, then real part of w =? z +1 (a)

−1 z −1

2



(c) 0

z 1 . (b) z +1 z −12 (d)

2 z −1

2

15. 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

16. If z = x + iy, x, y real, then | x | + | y | ≤ k | z |, where k is equal to (a) 1 (c) 3

(b) 2 (d) None of these

17. The value of i1/3 is (a) (c)

3+i 2 1+ i 3 2

(a)

(a 2 + 1) 2 2a 2 + 1

(b)

(a 2 + 1) 2 2a 2 − 1

(c)

(a 2 + 1) 2 4a 2 − 1

(d) None of these a + ib , then (x2 + y2)2 is equal to c + id

a 2 + b2 c2 + d 2 a 2 − b2 (c) 2 c + d2 (a)

(b)

c2 + d 2 a 2 + b2

(d) None of these

20. The square roots of 7 + 24i are

(a)

(a)

(a + i)2 = p + iq, then p2 + q2 = 2a − 1

19. If x + iy =

1 − 2i 4 − i = + 2 + i 3 + 2i

(a) a2 = b (c) a2 = 3b

18. If

(b)

3−i 2

1− i 3 (d) 2

(a) ± (3 + 4i) (c) ± (4 + 3i)

(b) ± (3 – 4i) (d) ± (4 – 3i)

21. The square roots of – 2 + 2 3 i are (a) ± (1 + (c) ± (– 1 +

3 i) 3 i)

(b) ± (1 –

3 i)

(d) None of these

22. The smallest positive integer n for which (1 + i)2n = (1 – i)2n is (a) 4 (c) 2

(b) 8 (d) 12 x

1 + i  23. If  = 1, then  1 − i  (a) x = 4n, where n is any positive integer (b) x = 2n, where n is any positive integer (c) x = 4n + 1, where n is any positive integer (d) x = 2n + 1, where n is any positive integer

24. If z = x – iy and z1/3

 x y  p + q  = p + iq, then is equal ( p2 + q2 )

to (a) 2 (c) 1

(b) –1 (d) –2 n

1 + i  25. The smallest integer n for which  = 1, is  1 − i  (a) 2 (c) 8

(b) 4 (d) 12

26. On the Argand plane the complex number lies in the (a) first quadrant (c) 3rd quadrant 27. If

(1 + 2i ) 1− i

(b) 2nd quadrant (d) 4th quadrant

(cos x + i sin x)(cos y + i sin y ) = A + iB, then (cot u + i )(1 + i tan v)

(a) A = sin u cos v cos (x + y – u – v) (b) B = sin u cos v sin (x + y – u – v) (c) A = cos u sin v cos (x + y – u – v) (d) B = cos u sin v sin (x + y – u – v)

3

(a) (0, 2) (c) (0, – 2)

(b) (– 2, 0) (d) None of these

29. The conjugate complex number of

2−i is (1 − 2i ) 2

 2   11  (a)   +   i  25   25 

 2   11  (b)   −   i  25   25 

 2   11  (c)  −  +   i  25   25 

 2   11  (d)  −  −   i  25   25 

30. The value of ii is (a) ω π (c) 2

(b) – ω2 (d) None of these

31. The inequality | z – 4 | < | z – 2 | represents the region given by (a) Re (z) > 0 (c) Re (z) > 3

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

32. If (1 + i) (1 + 2i) (1 + 3i) ... (1 + ni) = α + iβ then 2 ⋅ 5 ⋅ 10 ... (1 + n2) =

1 ∆ = ωn ω 2n

ωn ω2n 1

ω 2n 1 is equal to ωn

(a) 0 (c) ω

(b) 1 (d) ω2

2z + 1 is – 2, then the locus of iz + 1 the point representing z in the complex plane is

40. The imaginary part of

(a) circle (c) a parabola

(b) a straight line (d) None of these

41. If α, β are the complex cube roots of unity, then α3 + β3 + α–2 β–2 = (a) 0 (c) – 3

(b) 3 (d) None of these

(a) 0 (c) 2

3n

=

(b) 1 (d) 3

43. If ω is a complex cube root of unity, then (1 – ω + ω2) (1 – ω2 + ω4) (1 – ω4 + ω8) (1 – ω8 + ω16) = (b) 14 (d) None of these

1 + i  1 − i  + 44. The value of  is equal to  2   2  8

= 1, then locus of z is (b) y-axis (d) y = 1

35. If z = x + iy and ‘a’ is a real number such that | z – ai | = | z + ai |, then locus of z is (a) x-axis (c) x = y

39. If 1, ω, ω2 are the cube roots of untiy, then

(a) 12 (c) 16

(a) the x-axis (b) the line y = 5 (c) a circle through the origin (d) None of these

(a) x-axis (c) x = 1

3 – 2i 2 3 (d) 2 +    i 2 (b)

3n

33. The complex number z = x + iy which satisfy the equaz − 5i = 1 lie on tion z + 5i

z−i z+i

3 + 2i 2 3 (c) 2 –   i 2

(a)

 −1 + i 3   −1 − i 3  42.  +   2 2    

(a) α – iβ (b) α2 – β2 (c) α2 + β2 (d) None of these

34. If

38. If z satisfies the equation | z | – z = 1 + 2i, then z =

(b) y-axis (b) x2 + y2 = 1

36. The locus represented by | z – 1 | = | z + i | is (a) a circle of radius 1 (b) an ellipse with foci at 1 and – i (c) a line through the origin (d) a circle on the join of 1 and – i as diameter

(a) 4 (c) 8

8

(b) 6 (d) 2

45. The continued product of the four values of  cos 

π + i sin  3 

 π   3  

3/ 4

(a) – 1 (c) 2

is (b) 1 (d) – 2

 1 + cos θ + i sin θ  = cos (nθ) + i sin (nθ), then n = 46. If   sin θ + i + i cos θ  4

(a) 2 (c) 4

(b) 3 (d) None of these

37. For any complex number z, the minimum value of 47. If z = cos  π  + i sin  π  , r = 1, 2, 3, ..., then r  3r   3r  | z | + | z – 1 | is z1 z2 z3 ... ∞ = (a) 0 (b) 1 (c)

1 2

(d)

3 2

(a) i (c) 1

521

3

(b) – i (d) – 1

Complex Numbers

1 + i  1 − i  28. If   −  = x + iy, then (x, y) = 1 − i  1 + i 

522

48. If a = cos α + i sin α, b = cos β + i sin β,

Objective Mathematics

b c a + + = 1, then c a b cos (β – γ) + cos (γ – α) + cos (α – β) =

c = cos γ + i sin γ and

3 2 (c) 0

(b) –

(a)

3 2

(d) 1

49. If x = ω – ω – 2, then the value of 2

x4 + 3x3 + 2x2 – 11x – 6 is (a) 1 (c) 2

50. If 1, ω, ω2 be the three cube roots of unity, then (1 + ω) (1 + ω2) (1 + ω4) (1 + ω8) ... to 2n factors =

51. The value of

(b) – 1 (d) None of these 10



∑  sin k =1

(a) 1 (c) i

2 πk 2 πk  − i cos  is 11 11  (b) – 1 (d) – i

1 3   3 + 4i  is 52. The value of  +  1 − 2i 1 + i   2 − 4i  1 9 − i 4 4 1 9 (c) − + i 4 4

1 9 + i 4 4 1 9 (d) − − i 4 4 (b)

(a)

53. The value of

57. The value of

1 is 1 − cos θ + 2i sin θ

(b) z1 + z3 = z2 + z4 (d) None of these

4 (cos 75º +i sin 75º ) is 0.4 (cos 30º +i sin 30º )

10 (1 + i) 2 5 (1 + i) (c) 2 (a)

58. The value of (b) – 1 (d) None of these

(a) 1 (c) 0

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

(b)

10 (1 – i) 2

(d) None of these

(cos 2θ − i sin 2θ)7 (cos 3θ + i sin 3θ) −5 is (cos 4θ + i sin 4θ)12 (cos 5θ + i sin 5θ) −6

(a) cos 33θ + i sin 33θ (c) cos 47θ + i sin 47θ

(b) cos 33θ – i sin 33θ (d) cos 47θ – i sin 47θ

59. (cos 2θ + i sin 2θ) –5 (cos 3θ – i sin 3θ)6 (sin θ – i cos θ)3 = (a) cos 25θ + i sin 25θ (c) sin 25θ + i cos 25θ 60. Given z =

(b) cos 25θ – i sin 25θ (d) sin 25θ – i cos 25θ

q + ir p + iq 1 + iz , then if = 1+ p 1 + r 1 − iz

(a) p2 + q2 + r2 = 1 (c) p2 + q2 – r2 = 1

(b) p2 + q2 + r2 = 2 (d) None of these

3 = a + ib, then a2 + b2 = 2 + cos θ + i sin θ (a) 2a – 3 (b) 2a + 3 (c) 4a – 3 (d) 4a + 3

61. If

62. If

3

a − ib = x – iy, then

(a) x + iy (c) y + ix

3

a + ib =

(b) x – iy (d) y – ix a b + = x y (b) 4 (x2 – y2) (d) None of these

θ 2 tan 1 2 (a) −i 5 + 3 cos θ 5 + 3 cos θ

63. If

θ 2 tan 1 2 (b) +i 5 + 3 cos θ 5 + 3 cos θ

64. If 1, α, α2, ..., αn – 1 are the n nth roots of unity, then n −1

∑ i =1

(d) None of these

(b) 1 (d) n2

55. The complex numbers z1, z2, z3 are the vertices A, B, C of a parallelogram ABCD, then the fourth vertex D is 1 (z1 + z2) 2 1 (z1 + z2 + z3) (c) 2

(b)

1 is equal to 2 − αi

(n − 2) 2n −1 + 1 2n − 1 (n − 2) ⋅ 2n −1 (c) 2n − 1 (a)

54. If 1, ω, ω2, ..., ωn – 1 are the n nth roots of unity, then (1 – ω) (1 – ω2)...(1 – ωn – 1) is equal to

(a)

a + ib = x + iy, then

(a) 4 (x2 + y2) (c) 2 (x2 – y2)

θ 2 cot 1 2 (c) −i 5 + 3 cos θ 5 + 3 cos θ

(a) 0 (c) n

3

1 (z1 + z2 + z3 + z4) 4

(d) z1 + z3 – z2

56. The points z1, z2, z3, z4 in the complex plane are the vertices of a parallelogram taken in order if and only if

(b) (n – 2) ⋅ 2n (d) None of these

65. All the roots of the equation a1 z3 + a2 z2 + a3 z + a4 = 3, where | ai | ≤ 1, i = 1, 2, 3, 4, lie outside the circle with centre origin and radius (a) 1 2 (c) 3

1 3 (d) None of these (b)

66. If z1 and z2 are 1 – i and – 2 + 4i respectively, then z z  Im  1 2  is equal to  z1  (a) 1 (c) 2

(b) – 1 (d) – 2

(a) zero (c) ω

(b) 1 (d) ω2

68. If ω is an imaginary cube root of unity, then (1 – ω) (1 – ω2) (1 – ω4) (1 – ω5) is equal to (a) 3 (c) 18

(b) 9 (d) None of these

76. If z = (a)

3π 3π   2  cos + i sin   4 4 

(b)

π π  2  cos + i sin   4 4

(c)

7π 7π   2  cos + i sin   4 4 

69. If n is a positive integer, then n

n

 −1 + −3   −1 − −3  =   +  is equal to 2 2     (a) – 1 if n is a multiple of 3 (b) 2 if n is not a multiple of 3 (c) – 1 if n is not a multiple of 3 (d) 2 if n is a multiple of 3 70. In a geometrical progression, first term and common ratio 1 ( 3 + i). Then the absolute value of the are both 2 nth term of the progression is (a) 2n (c) 1

(b) 4n (d) None of these

71. If 1, ω, ω2 are the three cube roots of unity, then (1 – ω + ω2) (1 – ω2 + ω4) (1 – ω4 + ω8) ... to 2n factors = n

2n

(a) 2 (c) 24n

(b) 2 (d) None of these

1 + 7i , then the polar form of z is (2 − i)2

(d) None of these 77. If z is a point on the Argand plane such that | z – 1 | = 1, z−2 is equal to then z (a) tan (arg z) (c) i tan (arg z)

(b) cot (arg z) (d) None of these

78. If z is a non-real complex number lying on the circle | z | = 1, then z is equal to  arg z  1 − i tan   2  (a)  arg z  1 + i tan   2  (c)

 arg z  1 + i tan   2  (b)  arg z  1 − i tan   2 

1 − i tan (arg z ) 1 + i tan (arg z )

(d) None of these

79. If x = a + b, y = aα + bβ and z = aβ + bα, where α and β are complex cube roots of unity, then xyz = (a) 2 (a3 + b3) (c) a3 + b3

(b) 2 (a3 – b3) (d) a3 – b3

(a) a3 + b3 (c) a3 – b3

(b) 3 (a3 + b3) (d) 3 (a3 – b3)

72. If α and β are the complex cube roots of unity, then  80. If x = a + b, y = aω + βω2, z = aω2 + bω, then x3 + y3 α4 + β4 + α–1 β–1 = + z3 = (a) 0 (b) 1 (c) – 1 73. If z =

(d) None of these ( z1 + z2 ) z1 , where z1 = 1 + 2i and z2 = 1 – i, z2 z1

then (a) modulus of z is (b) arg z = tan–1

1 2

26

19 17

(c) modulus of z is

1 2

(d) arg z = – π + tan–1

3α 2 α (c) 2

75.

65 19 17

(a) – 2 (c) – 1

(b) 2 (d) 1

(a) – 1, ω (c) ω, ω2

(b) – 1, ω2 (d) None of these

83. The region of Argand diagram defined by

(b) α (d) None of these

−1 − −1 − −1 − ... to ∞ = (a) 1 (c) ω

6

 3 + i i − 3  81.   +  =  2   2 

82. The common roots of the equations z3 + 2 z2 + 2 z + 1 = 0 and z1985 + z100 + 1 = 0 are

74. If z = cos α + i sin α, then the amplitude of z2 + z is (a)

6

(b) – 1 (d) ω2

| z – 1 | + | z + 1 | ≤ 4 is (a) interior of an ellipse (b) exterior of a circle (c) interior and boundary of an ellipse (d) None of these 84. The solution of the equation | z | – z = 1 + 2i is 3 – 2i 2 3 (c) 2 –  i 2

(a)

(b)

523

(1 + ω2 + 2ω)3n – (1 + ω + 2ω2)3n is equal to

Complex Numbers

67. The value of

3 + 2i 2

(d) None of these

524

85. The complex number z satisfying the equations | z – i | = | z + 1 | = 1 is

Objective Mathematics

(a) 0 (c) – 1 + i

(b) 1 + i (d) 1 – i

86. If z = a + ib where a > 0, b > 0, then 1 (a – b) 2 1 (a + b) (c) | z | < 2 (a) | z | ≥

(b) | z | ≥

1 (a + b) 2

(d) None of these

87. If for the complex numbers z1 and z2, | z1 + z2 | = | z1 – z2 |, then amp z1 ~ amp z2 = π 2

(a) π

(b)

π (c) 4

(d) None of these

|1 − z1 z2 |2 − | z1 − z2 |2 = k (1 – | z1 |2 ) (1 – | z2 |2 ), then k is equal to

89. The locus of the complex number z in an argand plane satisfying the inequality

(a) a circle (c) exterior of a circle

2  is 3

(b) interior of a circle (d) None of these

90. Let z be a complex number satisfying z2 + z + 1 = 0. If n is not a multiple of 3, then the value of zn + z2n = (a) 2 (c) 0

(b) – 2 (d) – 1

91. If z1, z2 and z3 are the vertices of an equilateral triangle, then z12 + z22 + z32 = (a) 1 (c) z1 z2 + z1 z3 + z2 z3

(a) Re (z) > 0 (c) Re (z) > 2

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

97. For any complex number z, the minimum value of | z | + | z – 1 | is (a) 1 1 (c) 2

(b) 0 3 (d) 2

a0 + a3 + a6 + ... = (a) 3n (c) 3n – 2

(b) 3n – 1 (d) None of these

99. If 1, a1, a2, ..., an – 1 are the n nth roots of unity, then (1 – a1) (1 – a2) (1 – a3) ... (1 – an – 1) = (a) n + 1 (c) n – 1

(b) – 1 (d) 4

 | z − 1| + 4   log1/ 2  > 1  where | z − 1| ≠   3 | z − 1| − 2 

96. The inequality | z – 4 | < | z – 2 | represents the region given by

98. If (1 + x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n, then

88. If for the complex numbers z1 and z2,

(a) 1 (c) 2

(a) a straight line (b) an ellipse (c) a circle with centre origin and radius 2 (d) a circle with centre origin and radius unity

(b) – 1 (d) None of these

92. The locus of the complex number z in the Argand plane z−3 = 2, is if z+3 (a) a straight line (b) a single point (c) a circle (d) None of these 93. The locus of the complex number z in the Argand plane 1 − iz = 1, is if z−i (a) a circle (b) x-axis (c) y-axis (d) None of these 94. The complex number z satisfying the equations | z | – 4 = | z – i | – | z + 5i | = 0, is (a) 3 – i (b) 2 3 – 2i (c) – 2 3 – 2i (d) 0 95. The equation | z – 1 |2 + | z + 1 |2 = 4 represents on the Argand plane

(b) n (d) None of these

100. The closest distance of the origin from a curve given as a z + a z + a a = 0 (a is a complex number) is |a| 2 Im a (d) |a|

(a) 1 (c)

(b)

Re a |a|

101. For any integer n, the argument of z = (a)

π 6

π 3 2π (d) 3

( 3 + i ) 4 n +1 is (1 − i 3 ) 4 n

(b)

π 2 (e) all of the above (c)

102. The principal value of the arg z and | z | of the complex number z=

(1 + cos θ + i sin θ)5 is (cos θ + i sin θ)3

(a) –

θ θ , 32 cos5 2 2

(b)

(c) –

θ θ , 16 cos4 2 2

(d) None of these

103. If z = cos θ + i sin θ, then 1 zn 1 n (b) z + n z 1 n (c) z − n z 1 n (d) z − n z

n (a) z +

= 2 cos nθ = 2n cos nθ = 2i sin nθ = (2i)n sin nθ

θ θ , 32 cos5 2 2

(a) i cot nθ (c) tan nθ (n is an integer)

111. If n is a positive integer, then ( 3 + i)n + ( 3 – i)n is equal to nπ 6 nπ (c) 2n – 1 cos 6

(b) i tan nθ (d) cot nθ

(a) 2n cos

105. If n is an integer, then (1 + cos θ + i sin θ)n + (1 + cos θ – i sin θ)n = θ nθ cos 2 2 θ nθ sin (b) 2n cosn 2 2 θ nθ (c) 2n + 1 cosn sin 2 2 θ nθ cos (d) 2n + 1 cosn 2 2 106. If m = cos α + i sin α and n = cos β + i sin β, then m−n = m+n α −β α −β (a) i tan   (b) i cot      2 2 

107. If xr = cos (a) 1 (c) 0

nπ 3 nπ (c) 2n + 1 cos 3

(a) 2n – 1 cos

(b) – 1 (d) None of these

108. If a = cos α + i sin α, b = cos β + i sin β, a b c + + = – 1, then b c a cos (β – γ) + cos (γ – α) + cos (α – β) =

n

(a) 2 2  cos  n

(c) 2 2

(b) 1 (d) None of these

1 m n (a) z w + m n = 2 cos 2 (mθ + nφ) z w 1 m n (b) z w + m n = 2 cos 2 (mθ – nφ) z w z m wn (c) n + m = 2 cos 2 (mθ – nφ) w z z m wn (d) n + m = 2 cos 2 (mθ + nφ) w z (m and n being integers)

(b) 2n cos

nπ 3

(d) None of these

 cos 

nπ 4

n

(b) 2 2

+1

cos 

nπ 4

(d) None of these

(b) ± 4, ± 4 i (d) None of these

115. If ω is an imaginary cube root of unity then (1 + ω – ω 2)7 equals (a) 128ω (c) 128ω2

(b) –128ω (d) – 128ω2

116. The roots of the equation z4 + 1 = 0 are (a) (± 1 ± i) 1 (± 1 ± i) (c) 2

(b) (± 2 ± 2i) (d) None of these

117. If 1, ω, ω2, ... ωn – 1 are the n nth roots of unity and z1 and z2 are any two complex numbers, then n −1

∑ |z

1

k =0

c = cos 2γ + i sin 2γ and d = cos 2δ + i sin 2δ, then 1 abcd + = abcd

110. If z = cos 2θ + i sin 2θ and w = cos 2φ + i sin 2φ, then

−1

nπ 4

(a) ± 2, ± 2 i (c) ± 1, ± i

109. If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β,

(a) 2  cos (α + β + γ + δ) (b) 2 cos (α + β + γ + δ) (c) cos (α + β + γ + δ) (d) None of these

(d) None of these

114. The values of (16)1/4 are

c = cos γ + i sin γ and

(a) 0 (c) – 1

nπ 6

113. If α, β are the roots of the equation x2 – 2x + 2 = 0, then αn + βn is equal to

α −β (d) cot    2 

π π + i sin r , then x1 x2 x3 .... ∞ = r 2 2

(b) 2n + 1 cos

112. If n is a positive integer, then (1 + i 3 )n + (1 – i 3 )n =

(a) 2n cosn

α −β (c) tan    2 

525

z 2n − 1 = z 2n + 1

+ wk z2 |2 =

(a) n [ | z1 |2 + | z2 |2] (b) (n – 1) [ | z1 |2 + | z2 |2] (c) (n + 1) [ | z1 |2 + | z2 |2] (d) None of these 118.

10

∑i

n

is equal to

n=1

(a) 0 (c) i + 1

(b) i – 1 (d) – 1

119. If p, q, r are positive integers and ω be an imaginary cube root of unity and f (x) = x3p + x3q + 1 + x3r + 2, then f (ω) = (a) ω (c) 0

(b) – ω2 (d) None of these

120. If α, β, γ are the cube roots of a real number p, then xα + yβ + z γ = for non-zero x, y, z xβ + y γ + zα

Complex Numbers

104. If z = cos θ + i sin θ, then

526

Objective Mathematics

(a)

−1 + 3 i 2

(b)

−1 − 3 i 2

(c)

1+ 3i 2

(d)

1− 3i 2

(a) a : 1 (c) a : 10 131. If

121. The quadratic equation whose one root is a square root of – 47 + 8 −3 is (a) x2 – 2x + 49 = 0 (c) x2 – 2x – 49 = 0

(b) x2 + 2x + 49 = 0 (d) None of these

122. The locus of the point z satisfying the condition z −1 π = is arg z +1 3 (b) circle (d) None of these

2z + 1 is – 2, then the locus iz + 1 of the point representing z in the complex plane is

123. If the imaginary part of

(a) a circle (c) a parabola

(b) a straight line (d) None of these

124. The integral solution of the equation (1 – i)n = 2n is (a) n = 0 (c) n = – 1

(b) n = 1 (d) None of these

125. The greatest value of the moduli of complex numbres 4 = 2 is z satisfying the equation z − z (a) (c)

5 5 + 1

z −1 is purely imaginary, then z +1

(a) | z | > 1 (c) | z | = 1

(b) 5 – 1 (d) None of these

(b) | z | < 1 (d) None of these

132. If | z – i | < 1, then | z + 12 – 6i | (a) < 14 (c) > 14 133.

(b) < 16 (d) = 14

i − −i is equal to (a) i 2

(a) a straight line (c) a parabola

(b) 2a : 1 (d) None of these

(c) 0

1 i 2 (d) – i 2

(b)

134. If i z3 + z2 – z + i = 0, then (a) | z | < 1 (c) | z | = 1

(b) | z | > 1 (d) | z | = 0

135. If | z1 | = | z2 | = ... = | zn | = 1, then (a) = | z1 + z2 + ... + zn | (c) > | z1 + z2 + ... + zn |

1 1 1 + + ... + z1 z2 zn

(b) < | z1 + z2 + ... + zn | (d) = 1

136. The locus of the complex number z in an argand plane satisfying the equation Arg (z + i) – Arg (z – i) =

π is 2

(a) boundary of a circle (b) interior of a circle (c) exterior of a circle (d) None of these

126. The range of real number α for which the equation 137. The greatest value of | z + 1 | if | z + 4 | ≤ 3 is z + α | z – 1 | + 2i = 0 has a solution is (a) 4 (b) 5 (c) 6 (d) None of these  5 5  3 3 , ,  (a)  − (b)  −  1000  1  2 2  3   2 2  i 138.  − + =  2 2      5 5  5 (c) 0, (d)  −∞, −  ∪  , ∞  (a) 1 (b) ω 2     2 2   (c) ω2 (d) None of these 127. The argument of (a) 60º (c) 210º 128. The argument of

is (b) 120º (d) 240º

1− i 3 is 1+ i 3

139. If the cube roots of unity are 1, ω, ω2, then the roots of the equation (x – 1)3 + 8 = 0 are (a) – 1, 1 + 2ω, 1 + 2ω2 (b) – 1, 1 – 2ω, 1 – 2ω2 (c) – 1, – 1, – 1 (d) None of these 140. arg bi (b > 0) is

π (a) π (b) π 2π 2 (b) 3 3 π (c) – (d) 0 4π 2π (c) (d) – 2 3 3 z1 z2 + 129. The smallest positive integer n for which (1 + i)2n 141. Let z1 and z2 be two complex numbers such that z z1 2 = (1 – i)2n is = 1, then (a) 4 (b) 8 (a) z1, z2 are collinear (c) 2 (d) 12 (b) z1, z2 and the origin from a right angled triangle 130. If | z | = 2 and locus of 5z – 1 is the circle having radius (c) z1, z2 and the origin form an equilateral triangle a and z12 + z22 − 2 z1 z2 cos θ = 0, then | z1 | : | z2 | = (d) None of these (a)

(a) only even n (c) only positive n

(b) only odd n (d) all n

143. The equation z  z + a  z + a  z + b = 0, b ∈ R represents a circle (not point circle) if (a) | a |2 > b (c) | a | > b

(b) | a |2 < b (d) | a | < b

144. If z4 = (z – 1)4, then the roots are represented in the argand plane by the points that are (a) collinear (b) concyclic (c) vertices of a parallelogram (d) None of these rπ rπ + i sin then z1 z2 z3 z4 = 145. If zr = cos 10 10 (a) 0 (c) – 1

(b) 1 (d) None of these

146. If f (x) = x4 – 8x3 + 4x2 + 4x + 39 and f (3 + 2i) = a + ib, then a : b = (a)

1 8

(c)

1 8

1 4 1 (d) – 8 (b) –

4 3 4 (c) – π + tan– 1 3

(b) tan– 1

4 3

(d) None of these

148. If | z – 4 + 3i | ≤ 2, then the least and the greatest values of | z | are (a) 3, 7 (c) 3, 9

(b) 4, 7 (d) None of these

149. If z1, z2, z3 are non-zero, non collinear complex numbers 2 1 1 = + , then the points z1, z2, z3 lie such that z1 z 2 z3 (a) in the interior of a circle (b) on a circle passing through origin (c) in the exterior of a circle (d) None of these 150. If z2 – 2z cos θ + 1 = 0, then z2 + z–2 is equal to (a) 2 cos 2θ (c) 2 cos θ 151. e 2 mi cot

−1 p

(a) 0 (c) – 1

 pi + 1  ⋅  pi − 1 

to (a) 0 (c) (1 + (– 1)n) ⋅ in

(b) 2 (d) None of these

153. If S (n) = in + i – n, where i = −1 and n is a positive integer, then the total number of distinct values of S (n) is (a) 1 (c) 3

(b) 2 (d) 4

154. The maximum value of | z | when z satisfies the condition 2 z+ = 2 is z (a)

3 – 1

(b)

3 + 1

(c)

3

(d)

2 +

(b) 2 sin 2θ (d) 2 sin θ

1 2 (c) | z + z  | = 1 (a) | z + z  | =

(b) z + z = 1 (d) None of these

156. If 1, ω, ω are the cube roots of unity, then the value of (1 + ω)3 – (1 + ω2)3 is 2

(b) 2 (d) 0

157. Let p be a complex number such that | a | < 1 and z1, z2, ..., zn be the vertices of a polygon such that zk = 1 + a + a2 + ... a k, then the vertices of the polygon lie within the circle (a) | z – a | = (c) z − 158. (i +

1 |1 − a |

(b) | z – 1 | =

3 )100 + (i –

3 )100 + 2100 =

(a) 1 (c) 0

(b) – 1 (d) None of these

159. If z1, z2, z3, z4 are represented by the vertices of a rhombus taken in the anticlockwise order, then (a) z1 + z2 = z3 + z4 (b) z1 – z2 + z3 – z4 = 0 z2 − z4 π = (c) amp z1 − z3 2 z1 − z2 π (d) amp = z3 − z 4 2 160. Let zk (k = 0, 1, 2, ..., 6) be the roots of the equation (z + 1)7 + z7 = 0, then

6

∑ Re ( z ) k

k =0

(a) 0 (b) 1 (d) None of these

1 |1 − a |

1 1 = (d) None of these 1− a |1 − a |

m

=

3

155. If | z | = Max. {| z – 1 |, | z + 1 |}, then

(c) –

3 2 7 (d) 2

(b) 7 2

is equal to

527

(1 + i ) 2 n 2n , n ∈ I, is equal + 2n (1 − i ) 2n

(a) 2ω (c) – 2

147. If | z – 25 i | ≤ 15, then the least positive value of arg z is (a) π – tan– 1

152. The complex number

Complex Numbers

n

 z−i  142. If ω =  , n integral, then ω lies on the unit  1 + iz  circle for

528

161. If z1 ≠ – z2 and | z1 + z2 | =

1 1 + , then z1 z2

Objective Mathematics

(a) at least one of z1, z2 is unimodular (b) z1 ⋅ z2 is unimodular (c) both z1, z2 are unimodular (d) None of these 162. If z = x + iy satisfies amp (z – 1) = amp (z + 3i) then the value of (x – 1) : y is equal to (a) 2 : 1 (c) 1 : 3

(b) – 1 : 3 (d) None of these

163. The common roots of the equations z3 + 2 z2 + 2 z + 1 = 0 and z149 + z100 + 1 = 0 are (a) – 1, ω, ω2 (c) ω, ω2

(b) 1, ω, ω2 (d) None of these

164. If a complex number lies in the third quadrant then its conjugate lies in the quadrant (a) first (c) third 165. The equation z2 + z a (a) straight line (c) hyperbola

(b) second (d) None of these 2

– 2 | z |2 + z + z = 0 represents (b) circle (d) parabola

166. If z1, z2, z3, z4 are the four complex numbers represented by the vertices of a quadrilateral taken in order such z4 − z1 π = then the that z1 – z4 = z2 – z3 and amp z2 − z1 2 quadrilateral is a (a) square (c) rectangle

(b) rhombus (d) a cyclic quadrilateral

167. The equation | z – i | + | z + i | = k represents an ellipse if k = (a) 1 (c) 4

(b) 2 (d) None of these

168. Let z1 and z2 be two non real complex cube roots of unity and | z – z1 |2 + | z – z2 |2 = λ be the equation of a circle with z1, z2 as ends of a diameter, then the value of λ is (a) 4 (c) 2

(b) 3 (d) 2

169. The equation | z + i | + | z – i | = k represents a hyperbola if (a) k > 2 (c) – 2 < k < 2

(b) 0 < k < 2 (d) None of these

170. The region in the Argand diagram defined by | z – 3 | + | z + 3 | < 6 is the interior of the ellipse with major axis along (a) real axis (c) y = x

(b) imaginary axis (d) y = – x

171. If the area of the triangle on the argand plane formed by the complex numbers – z, iz, z – iz is 600 square units, then | z | is equal to

(a) 10 (c) 30

(b) 20 (d) None of these

172. If | z + z  | + | z – z  | = 8, then z lies on (a) a circle (c) a square

(b) a straight line (d) None of these

173. The complex numbers given by 1 – 3i, 4 + 3i and 3 + i represent the vertices of (a) a right angled triangle (b) an isosceles triangle (c) an equilateral triangle (d) None of these  z + 2i  = 0, then z lies on a circle with 174. If Re   z + 4  centre (a) (– 2, – 1) (c) (2, – 1)

(b) (– 2, 1) (d) (2, 1)

 z + 2i  = 0, then z lies on the curve 175. If Im   z + 2  (a) x2 + y2 + 2x + 2y = 0 (b) x2 + y2 – 2x = 0 (c) x + y + 2 = 0 (d) None of these 176. Let z be a complex number with modulus 2 and argu2π , then z is equal to ment 3 (a) – 1 + i 3

(b) 1 – i 3

1 i 3 (c) − + 2 2

(d) None of these

177. If

z−2 (z ≠ – 2) is purely imaginary then | z | is equal z+2

to (a) 1 (c) 3

(b) 2 (d) 4

178. The cube roots of unity (a) lie on the circle | z | = 1 (b) are collinear (c) form an equilateral triangle (d) None of these 179. If z1 and z2 are two complex numbers such that Re (z1 + z2) = 0 and Im (z1 z2) = 0, then (a) z1 = z2

(b) z1 = z2

(c) z1 = – z2

(d) None of these

180. If the complex numbers z1, z2, z3, z4 taken in that order in the argand plane represent the vertices of a parallelogram, then (a) z1 + z2 = z3 + z4 (c) z1 + z4 = z2 + z3

(b) z1 + z3 = z2 + z4 (d) None of these

181. The locus of z satisfying Im (z2) = 4 is (a) a circle (b) a rectangular hyperbola (c) a pair of straight lines (d) None of these

(b) an ellipse (d) a rectangular hyperbola

183. If P (x) and Q (x) are two polynomials such that f (x) = P (x3) + x Q (x3) is divisible by x2 + x + 1, then (a) P (x) is divisible by (x – 1) but Q (x) is not divisible by x–1 (b) Q (x) is divisible by (x – 1) but P (x) is not divisible by x–1 (c) Both P (x) and Q (x) are divisible by x – 1 (d) f (x) is divisible by x – 1

(b) y = – x (d) y = – x + 1

185. If | z – 1 | + | z + 3 | ≤ 8, then the range of values of | z – 4 | is (b) [0, 8] (d) [5, 9]

1 2

(b) –

(c) 0 187. If i =

1 2

 1 i 3 + 3 − + 2   2

is equal to

(b) – 1 + i 3

(c) i 3

(d) – i 3

188. For positive integers n1, n2 the value of the expression (1 + i ) n1 + (1 + i 3 ) n1 + (1 + i 5 ) n2 + (1 + i 7 ) n2 , where i=

−1 is a real number if and only if (a) n1 = n2 + 1 (b) n1 = n2 – 1 (c) n1 = n2 (d) n1 > 0, n2 > 0 189. The value of the expression 1 ⋅ (2 – ω) (2 – ω2) + 2 ⋅ (3 – ω) (3 – ω2) + ...  + (n – 1) ⋅ (n – ω) (n – ω2), where ω is an imaginary cube root of unity, is

1 (n – 1) n (n2 + 3n + 4) 2 (d) None of these

(c)

13

∑ (i

n

+ i n +1 ) , where i =

−1 ,

(b) i – 1 (d) 0

193. If arg (z) < 0, then arg (– z) – arg (z) = (a) π π (c) – 2

1 1 1 + + z1 z2 z3

(b) – π π (d) 2

= 1, then | z1 + z2 + z3 | is (b) less than 1 (d) equal to 3

195. If i z3 + z2 – z + i = 0, then | z | = (a) 1 (c) – 1

365

(a) 1 – i 3

1 (a) (n – 1) n (n2 + 3n + 4) 4 1 (n – 1) (n + 1) (n2 + 3n + 4) (b) 4

(a) i (c) – i

(a) equal to 1 (c) greater than 3

−1 , then

 1 i 3 4 + 5 − + 2   2

(b) x = 1, y = 3 (d) x = 0, y = 0

192. The value of the sum

| z3 | =

(d) None of these 334

(a) x = 3, y = 1 (c) x = 0, y = 3

194. If z1, z2, z3 are complex numbers such that | z1 | = | z2 | =

 8π   8π  186. If α = cos   + i sin   , then  11   11  Re (α + α2 + α3 + α4 + α5) is equal to (a)

6i −3i 1 4 3i −1 = x + iy, then 20 3 i

equals

| z – i Re (z) | = | z – Im (z) | then z lies on

(a) (0, 8) (c) [1, 9]

191. If

(b) – 128ω (d) – 128ω2

n =1

184. If z is a complex number satisfying (a) y = x (c) y = x + 1

(a) 128ω (c) 128ω2

(b) i (d) – i

196. Suppose z1, z2, z3 are the vertices of an equilateral triangle inscribed in the circle | z | = 2. If z1 = 1 + i 3 then z2 and z3 are equal to (a) – 2, 1 – i 3

(b) 2, 1 – i 3

(c) – 2, 1 + i 3

(d) None of these

197. If α, β, γ are the cube roots of p, p < 0 then for any xα + yβ + z γ = x, y and z, xβ + y γ + zα (a) ω2 (c) 1

(b) ω (d) None of these

198. The complex numbers sin x + i cos 2x and cos x – i sin 2x are conjugate to each other, for (a) x = nπ

(b) x = 0

1 (c) x =  n +  π  2

(d) no value of x

199. For any two complex numbers z1, z2 and any real numbers a and b, | az1 – bz2 |2 + | bz1 + az2 |2 = (a) (a2 + b2) [ | z1 |2 + | z2 |2] (b) 2 (a2 + b2) [ | z1 |2 + | z2 |2] (c) 4 (a2 + b2) [ | z1 |2 + | z2 |2] (d) None of these

529

(a) a parabola (c) a circle

190. If ω is an imaginary cube root of unity, then (1 + ω – ω2)7 equals

Complex Numbers

182. The curve represented by Re (z2) = 4 is

530

200. If z1 and z2 are two non-zero complex numbers such that | z1 + z2 | = | z1 | + | z2 |, then arg z1 – arg z2 is equal to

Objective Mathematics

(a) – π

(b) –

(c) π

(d)

π 2

208. The number of solutions of the equation z2 + | z |2 = 0, where z ∈ C is (a) 2 (b) 3 (c) 1 (d) infinitely many 209. (1 + i)6 + (1 – i)3 =

π 2

(a) – 2 – 10i (c) 2 – 10i

(b) – 2 + i (d) 2 + i

1 1 x x   = 2 cos θ, then xn + n is equal to 210. If x + sin + cos − i tan x x x   2 2 201. If the expression   is real, then the x (a) 2 sin nθ (b) cos nθ 1 + 2i sin   2 (c) sin nθ (d) 2 cos nθ (n is a positive integer) set of all possible values of x is (a) 2nπ, n ∈ I π (c) (2n + 1) , n ∈ I 2

(b) nπ, n ∈ I

211. If

(d) None of these

(a) 0 (b) real and positive (c) real and negative (d) purely imaginary

(b)

1 | z |2 2

(d) None of these

204. If the complex numbers z1, z2, z3 are the vertices A, B, C respectively of an isosceles right angled triangle with right angle at C, then (z1 – z2)2 = k (z1 – z3) (z3 – z2), where k = (a) 1 (c) 4

θ 2 θ (c) i cot 2

(b) cot θ

(d) 1

(d) i tan

θ 2

(b) 0 (d) – 1

207. If ω is an imaginary cube root of unity, then the value 1 1 1 is + − of 1 + 2ω 2 + ω 1 + ω (b) 1 (d) – 2

(b) z12 + z22 = z1 z2

(c) z22 − z12 = z1 z2

(d) z12 − z22 = z1 z2

213. If α + β = tan–1 z, z = x + iy and α is constant then, the locus of z is (a) x2 + y2 + 2x sin α = 1 (b) x2 + y2 + 2y tan 2α = 1 (c) cot 2α (x2 + y2) = 1 + x (d) x2 + y2 + 2x cot 2α = 1  | z |2 − | z | +1  < 2, then the locus of z is 214. If log 3   2 + | z | 

(a) a2 – b2 = 1 (c) a2 – b2 = 0

206. If ω is the nth root of unity, then 1 + ω + ω + ... + ωn –1 is

(a) 0 (c) – 1

3 4

(b) | z | = 5 (d) None of these

215. If a, b, x are real numbers such that  1 − ix    = a – ib, then 1 + ix 

1+ a = 1− a

2

(a) 2 (c) 1

(b)

(a) z12 + z22 + z1 z2 = 0

(a) | z | < 5 (c) | z | > 5

(b) 2 (d) None of these

205. If a = cos θ + i sin θ, then (a) cot

1 3 2 (c) 3 (a)

212. The three vertices of a triangle are represented by the complex numbers 0, z1 and z2. If the triangle is equilateral, then

203. The area of the triangle on the Argand plane formed by the complex numbers z, iz and z + iz is

1 (c) | z |2 4

= 2 represents a circle, then its radius is

equal to

202. Let z1 and z2 be complex numbers such that z1 ≠ z2 and | z1 | = | z2 |. If z1 has positive real part and z2 has z +z negative imaginary part, then 1 2 may be z1 − z2

(a) | z |2

z−2 z−3

(b) a2 + b2 = 1 (d) a2 + b2 = 0

 z − 1 is 216. If | z | = 1, then the value of   z + 1  (a) 0 (c) purely imaginary 217. If z = x + iy and ω =

(b) purely real (d) complex number 1 − iz , then | ω | = 1 implies that, z−i

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

228. If ( 3 + i)100 = 299 (a + ib), then b =

(b) 1 (d) i

219. If z1 and z2 are complex numbers, such that z1 + z2 is a real number, then

(b) 2 (d) None of these

229. The equation z2 = z has (a) an infinite number of solutions (b) four solutions (c) two solutions (d) no soution

(a) z1 = – z2 (b) z2 = z1 (c) z1 and z2 are any two complex numbers (d) z1 = z1 , z2 = z2 220. If | z – 25i | ≤ 15, then | maximum amp (z) – minimum amp (z) | is equal to 3 3 (a) sin–1   – cos–1   5 5 (b)

(a) 3 (c) 1

230. The centre of a square ABCD is at z = 0. The affix of the vertex A is z1. Then the affix of the centroid of the triangle ABC is (a)

π 3 + cos −1   5 2

z1 3

π π   cos ± i sin  2 2

π π  (b) z1  cos ± i sin   2 2 z1 (c) (cos π ± i sin π) 3

3 (c) π – 2 cos–1   5 3 (d) cos–1   5

(d) z1 (cos π ± i sin π)

231. If z1, z2, z3 are three complex numbers in A.P., then 221. For x1, x2, y1, y2 ∈ R, if 0 < x1 < x2, y1 = y2 and z1 = x1 + iy1, they lie on 1 (a) a circle (b) an ellipse (z1 + z2), then z1, z2 and z3 z2 = x2 + iy2 and z3 = 2 (c) a straight line (d) a parabola satisfy 2 2 (a + 1) (a) | z1 | < | z3 | < | z2 | (b) | z1 | > | z2 | > | z3 | = x + iy, then x2 + y2 is equal to 232. If 2a − i (c) | z  | < | z  | < | z  | (d) | z  | = | z  | = | z  |. 1

2

3

1

2

3

222. The complex number which satisfies the equation z+

| z + 1 | + i = 0 is

(a) 2 – i (c) 2 + i

(b) – 2 – i (d) – 2 + i

223. If α is a complex number such that α2 + α + 1 = 0, then α31 is (a) 1 (c) α2

(b) 0 (d) α

224. If z is complex number, then (a) | z2 | < | z |2 (c) | z2 | = | z |2

(b) | z2 | ≥ | z |2 (d) | z2 | > | z |2

225. The origin and the roots of the equation z2 + pz + q = 0 form an equilateral triangle if (a) q = p (c) p2 = 3q 2

(b) q = 3p (d) p2 = q 2

226. The principal value of the amplitude of (1 + i) is π (a) 2 3π (c) 4

π (b) 12 (d) π

a − ib   227. tan i log is equal to a + ib   2ab a + b2 2ab (c) 2 a − b2

(a)

2

(b)

(a)

(a 2 + 1) 4 4a 2 + 1

(b)

(c)

(a 2 − 1) 2 (4a 2 − 1) 2

(d) None of these

233. The solution set of the equation | 5x – 3 | = – 1 is (a) φ  3 (c)   5

(d) ab

(b) {0} (d) None of these

234. The equation not representing a circle is given by (a)

z −1 =1 z +1

 z − 1 π = (b) Arg   z + 1  2 (c) z  z + i z – i  z + 1 = 0 1 + z  (d) Re  =0  1 − z  235. If atleast one value of the complex number z = x + iy satisfies the condition | z + 2 | = a2 – 3a + 2 and the inequality | z + i 2 | < a2, then (a) a < 2 (c) a = 2

a 2 − b2 2ab

(a + i)2 4a 2 + 1

(b) a > 2 (d) None of these

236. If z2 + (p + iq) z + r + is = 0 where p, q, r, s are nonzero, has real roots, then (a) pqs = s2 + q2r (c) prs = q2 + r2p

(b) pqr = r2 + p2s (d) qrs = p2 + s2q

531

(a) – 1 (c) – i

Complex Numbers

218. i2 + i4 + i6 + ... (2n + 1) terms =

532

237. The reciprocal of 3 +

Objective Mathematics

(a)

3 7 − i 16 16

(c)

7 + 3i

7 i is (b)

6

3 7 − i 4 4

(d) 3 –

(a) 0 (c) 2

7i

(a) ω = ω (c) ω4 = 1

(b) ω = 1 (d) ω14 = ω2

z1 + z12 − z22 + z1 − z12 − z22

5

239. The complex number z = x + iy which satisfies the z − 3i = 1 lies on equation z + 3i

240. If n = 4m + 3, m integral, then in is equal to (a) – i (c) 1

3 + 2i 2 3 (d) – 2i 2 (b)

3 i 2

2

 2i  243. The least positive integer n such that   1 + i  positive integer, is (a) 16 (c) 4

(c)   | z | (d)  none of these 3 252. The maximum value of |z| when z satisfies the condition 2 = 2, is z+ z 2

n

is a

(b) 8 (d) 2

244. The real value of α for which the expression 1 − i sin α is purely real is 1 + 2i sin α π 2

π 2 (d) none of these (b) (n + 1)

(c) nπ (n is an integer)

(a + ib) (a − ib) − a − ib a + ib

− 2b3 (a + b 2 )2 (c) 0 (a)

2

(a)   3 − 1

(b)   3 + 1

(d)   2 + 3 253. If z = x + iy is a variable complex number such that  z −1  π arg   = , then  z +1 4 (c)   3

(a)  x2 – y2 – 2x = 1 (c)  x2 + y2 – 2y = 1 254. The value of

10

= x + iy, then x = 6 a 2b (a + b 2 )2 (d) None of these

(b)

2

(b) – 1 (d) 2



∑  sin k =1

(a)  1 (c)  –i

2

246. If α, β are non-real cube roots of unity, then αβ + α5 + β5 is equal to (a) 3 (c) 0

(b)   | z | 2

(a)  |z|2

(a) z1, z2 and the origin form an equilateral triangle (b) z1, z2 are collinear (c) z1, z2 and the origin form a right angled triangle (d) None of these

245. If

(b)   3 4 (d)  no unique value

251. The area of the triangle on the Argand diagram formed by the complex numbers z, iz and z + iz, is

z1 z2 + = 1. Then z2 z1

2

(b)  ω2 (d)  none of these

(a)   1 2 (c)  1

242. Let z1 and z2 be two complex numbers such that

(a) (2n + 1)

249. If 1, ω, ω2 are the cube roots of unity, then

250. If z be complex number satisfying z4 + z3 + 2z2 + z + 1 = 0, then | z | is equal to

241. The solution of the equation | z | = z + 1 + 2i is

(c) 2 –

(b) | z1 | (d) None of these

(a)  1 (c)  ω

(b) i (d) – 1

(a) 3 – 2i

(a) | z1 + z2 | (c) | z2 |

is equal to

a + bω + cω2 + d ω3 is equal to c + d ω + aω + bω2

(b) the line y = 3 (d) the x-axis

(a) a circle (c) the y-axis

(b) 1 (d) – 2

248. If z1 and z2 are any two complex numbers, then

238. If ω is a complex cube root of unity, then 6

6

1 + i 3  1 − i 3  247. The value of   +  is 1 − i 3  1 + i 3 

(b)  x2 + y2 – 2x = 1 (d)  x2 + y2 + 2x = 1

2k π 2k π  is + i cos 11 11  (b)  –1 (d)  i

255. If w = α + iβ, where β ≠ 0 and z ≠ 1, satisfies the  w−w z   is purely real, then the set condition that    1− z  of values of z is (a)  |z| = 1, z ≠ 2

(b)  |z| = 1 and z ≠ 1

(c)  z = z

(d)  none of these

256. If z ∈ C and |z + 4 | ≤ 3, then the greatest value of |z + 1| is

257. If

266. The amplitude of sin

z+i = 3 , then radius of the circle is z −i

(a)  

2 21

(c)   3

(b)  

258. If arg (z) = θ, then arg ( ) is equal to (a)  θ – π z (b)  π – θ (c)  θ (d)  –θ 259. If z is a complex number such that z − 1 is purely z +1 imaginary, then |z| is equal to (a)  0 (c)   2

(b)  1 (d)  none of these

(b)   π 15 (d)   π 5

(a)   2π 5 π (c)   10

1 21

(d)   21

π π  + i 1 − cos  is 5 5 

533

(b)  5 (d)  3

2 2 267. The value of a + bω + cω + a + bω + cω will be 2 b + cω + aω c + aω + bω2

(a)  1 (c)  2 268.

(b)  – 1 (d)  – 2

1 1 ( z1 + z2 ) + z1 z2 + ( z1 + z2 ) − z1 z2 is equal to 2 2 (a)  |z1 + z2| (c)  |z1| + |z2|

(b)  |z1 – z2| (d)  |z1| – |z2|

260. In which quadrant of the complex plane, the point 269. The equation 1 + 2i lies? z z + (2 – 3i)z + (2 + 3i) z + 4 = 0 represents a circle of 1− i radius (a)  Fourth (b)  First (a)  2 (b)  3 (c)  Second (d)  Third (c)  4 (d)  6 z lie 270. If z1 = a + ib and z2 = c + ib are complex numbers 261. If |z| = 1 and z ≠ ± 1, then all the values of 1 − z2 such that | z1| = | z2| = 1 and Re ( z1 z2 ) = 0, then the on (a)  a line not passing through the origin (b)  |z| = 2 (c)  the x-axis (d)  the y-axis 262. For three complex numbers 1 – i, i, 1 + i which of the following is true? (a)  They form a right triangle (b)  They are collinear (c)  They form an equilateral triangle (d)  They form an isosceles triangle. 263. If ω is an imaginary cube root of unity and x = a + b, y = aω + bω2, z = aω2 + bω, then x2 + y 2 + z2 is equal to (a)  6ab (c)  6a2b2

(b)  3ab (d)  3a2b2

264. The points representing the complex numbers z, for which |z – a|2 + |z + a|2 = b 2 lie on (a)  a straight line (b)  a circle (c)  a parabola (d)  a hyperbola

(a) (c)

| w1 | = 1

| w1w2 | = 0

(b)

| w2 | = 1

(d) None of these

271. The conjugate of a complex number is that complex number is (a)   − 1 i −1 (c)   − 1 i +1

1 . Then, i −1

(b)   1 i +1 (d)   1 i −1

272. A man walks a distance of 3 units from the origin towards the north-east (N 45º E) direction. From there, he walks a distance of 4 units towards the north-west (N 45º W) direction to reach a point P. Then the position of P in the Argand plane is (a)  3eiπ/4 + 4i (b)  (3 – 4i) eiπ/4 (c)  (4 + 3i) eiπ/4 (d)  (3 + 4i) eiπ/4

265. The value of (cos 20° + i sin 20°) (cos 75° + i sin 75°)    (cos 10° + i sin 10°   is sin 15° − i cos 15° (a)  0 (c)  i

pair of complex numbers ω1 = a + ic and ω2 = b + id satisfies

(b)  – 1 (d)  1

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

z lie 1 − z2

Complex Numbers

(a)  6 (c)  4

534

solutions

Objective Mathematics

9. (a) Arg zw = π

1. (d) (1 + ω2)m = (1 + ω4)m

∴ (– ω) = (– ω )   or  ω = ω which is only satisfied for m = 3, as ω3 = ω6 = 1. m

2 m

m

2. (a) Given that z1, z2, z3 are in A.P.

...(1)

z + iw = 0 ⇒ z = – iw

2m

⇒ z = iw ⇒ w = – iz From (i), arg (–iz2) = π

⇒ arg (–i) + 2arg z = π ∴ 2z2 = z1 + z3 π 3π If z1 = 1, z2 = 3 and z3 = 5, then + 2argz = π ⇒ argz = . ⇒ 2 4 arg (z1) = π/3, arg (z2) = π/3, arg (z3) = π/3 ⇒ complex numbers z1, z2 and z3 lie on a straight 10. (b) We have, (x + iy)1/3 = a + ib line. Cubing both sides, we get 3. (d) Given expression x + iy = (a + ib)3 = a3 + 3a2 (ib) + 3a (ib)2 + (ib)3 2 4 6 2n = 1 + i + i + i + ... + i = a3 + 3a2 bi + 3ab2 i2 + i3 b3 = 1 – 1 + 1 – 1 + ... + (– 1)n, = a3 + 3a2 bi – 3ab2 – ib3 [∵ i3 = i2 ⋅ i = – i] which cannot be determined unless n is known. 2 2 2 = a (a – 3b ) + ib (3a – b2). ∴ x = a (a2 – 3b2) and y = b (3a2 – b2) 4. (a) Let product of cube roots of (–1) = z x y ∴  z = [cos π + i sin π]1/3 ⇒ = a2 – 3b2 and = 3a2 – b2 a b 1/3 or,  z = [cos (2n + 1)π + i sin (2n + 1) π] x y where n = 0, 1, 2 + ∴ = 4 (a2 – b2). a b ∴  Roots with respect to the values of 1 − 2i 4 − i 11. (d) We have, + n = 0, 1, 2 2 + i 3 + 2i i π/3 i π 5π i/3 are  e , e , e (1 − 2i )(3 + 2i ) + (4 − i )(2 + i ) ∴ Product of roots ei π/3, ei π, e5π i/3 = 9i π/3 3πi (2 + i )(3 + 2i ) =e =e i.e., product of roots = cos 3π + i sin 3π = – 1 3 + 2i − 6i − 4i 2 + 8 + 4i − 2i − i 2 = 6 + 4i + 3i + 2i 2 5. (c) We have,

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



2

[

(1 + i)2 = 1 + i2 + 2i = 2i



=

16 − 2i 16 − 2i 4 − 7i = × 4 + 7i 4 + 7i 4 − 7i



=

64 − 112i − 8i + 14i 2 50 − 120i = ( 4) 2 − ( 7 i ) 2 16 + 49



=

50 − 120i 10 24 = − i. 65 13 13

(1– i)2 = 1 + i2 – 2i = – 2i]

= – 1 – 1 = – 2.

1 i− 4 n +1 4 n −1 i 4 n ⋅ i − i 4 n ⋅ i −1 i − i i = 6. (c) We have, = 2 2 2

12. (c) If z1,z2 and z3 are the vertices of an equilateral triangle. Then

[∵ i4n = (i4)n = (1)n = 1]

=

i 2 − 1 −1 − 1 −2 1 = = = = i. 2i 2i 2i −i

7. (b) Given that, (1 + i)2n = (1 – i)2n

⇒  (1 – 1 + 2i) n = (1 – 1 – 2i) n



2 n in = 2 n (– 1) n in



1 = (– 1) n ∴   The smallest value of n is 2.

8. (b)

z 12 + z 22 + z 32 = z 1z 2 + z 2z 3 + z 3z 1 But z3 = 0 ⇒ z12 + z22 = z1z2 ⇒ (z1 + z2)2 = 3z1z2

...(1)

z + az + b = 0 then z1 + z2 = –a and z1z2 = b 

18. (c) We have, p + iq =

2

Putting in (1), we have

(–a) = 3b ⇒ a = 3b 2

2

1 i = =i i i π π Let z4 = cos + i sin 2 2 2

π π  ∴ z = cos + i sin  2 2  By using De-Moivre’s theorem, we get π π z = cos + i sin . 8 8 14. (c) Let z = cos θ + i sin θ; | z | = 1 cos θ − 1 + i sin θ z −1 = w= cos θ + 1 + i sin θ z +1 1/ 4

−2 sin 2 θ/2 + i 2 sin θ/2.cos θ/2 w= 2 cos 2 θ/2 + i 2 sin θ/2.cos θ/2 = i

sin θ/2 cos θ/2

...(2)

Multiplying (1) and (2), we get (a + i)2 (a − i)2 ( p + iq) ( p – iq) = × 2a − 1 2a + 1 2 [( a + i )( a − i )] ⇒ p2 – (iq)2 = (2a − 1)(2a + 1)

13. (b) Given than, iz4 + 1 = 0 ⇒ z4 = −

(a + i)2 ...(1) 2a − 1 Changing i to – i, we get (a − i)2 p – iq = 2a + 1

535

Again z1, z2 are the roots of

⇒ p2 + q2 =

19. (a) We have, x + iy =

Re (w) = 0. 15. (b) min |z1 – z 2| = minimum distance between the circles

a + ib  c + id

...(1)

Changing i to – i, we get

a − ib  c − id

x – iy =

...(2)

Multiplying (1) and (2), we get

 cos θ/2 + i sin θ/2  θ   = i tan cos θ / 2 + i sin θ / 2 2  

(a 2 + 1) 2 . 4a 2 − 1

(x + iy) (x – iy) =

⇒ x2 – (iy)2 = ⇒ x2 + y2 =

a + ib × c + id

a − ib c − id

(a + ib)(a − ib) (c + id )(c − id )

a 2 + b2 . c2 + d 2

Squaring both sides, we get

= 12 – 5 – 5 = 2.

(x2 + y2)2 =

a 2 + b2 . c2 + d 2

20. (c) Let a + bi be a square root of 7 + 24i ∴ (a + bi)2 = 7 + 24i ⇒ a2 + 2abi + i2  b2 = 7 + 24i ⇒ (a2 – b2) + 2abi = 7 + 24i ⇒ a2 – b2 = 7 and 2ab = 24 ...(1) ...(2) Now (a2 + b2)2 = (a2 – b2)2 + 4a2 b2 = (7)2 + (24)2 = 49 + 576 = 625 16. (b) For every a ∈ R, | a | =

a

2

∴ | a | = a Now (| x | – | y |)2 ≥ 0 ⇒ | x |2 + | y |2 – 2 | x | | y | ≥ 0 ⇒ 2 | x | | y | ≤ | x |2 + | y |2 ⇒ | x |2 + | y |2 + 2 | x | | y | ≤ 2 | x |2 + 2 | y |2 ⇒ (| x | + | y |)2 ≤ 2 (x2 + y2) ⇒ (| x | + | y |)2 ≤ 2 | z |2 2

2

∴ | x | + | y | ≤

21. (a) Let a + bi be a square root of – 2 + 2 3 i

2 | z |.

π π  17. (a) i1/3 =  cos + i sin   2 2 π π = cos + i sin = 6 6

∴ a2 + b2 = 625 = 25 ...(3) Solving (1) and (3), we get 2a2 = 32 or a2 = 16, ∴ a = ± 4 2b2 = 18 or b2 = 9, ∴ b = ± 3. From (2), ab = 12 which is positive, therefore either a = 4, b = 3 or a = – 4, b = – 3. Hence the two square roots are 4 + 3i and – 4 – 3i i.e., ± (4 + 3i).

∴ (a + bi)2 = – 2 + 2 3 i

1/ 3

3 i = + 2 2

3+i . 2

a2 + 2abi + i2b2 = – 2 + 2 3 i

⇒ (a2 – b2) + 2abi = – 2 + 2 3 i

Complex Numbers



536

Objective Mathematics

∴ a2 – b2 = – 2

...(1)

2ab = 2 3 Now (a2 + b2)2 = (a2 – b2)2 + 4 a2b2

...(2)



= (– 2)2 + (2 3 )2 = 4 + 12 = 16



∴ a2 + b2 = ± 16 = 4 Solving (1) and (3), we get

...(3)



2a2 = 2, or a2 = 1, ∴ a = ± 1



2b2 = 6, or b2 = 3, ∴ b = ±

 rom (2), ab = F either

a = 1, b =

= 3

3 or a = – 1, b = –

3 i i.e., ± (1 +

and – 1 –

27. (a), (b)  We have,

3 which is positive, therefore,

Hence the two square roots are 1 +

3.

3 i).

2n

 (1 + i ) 2  =1 ⇒    2 

 (1 + i )(1 + i )  1 + i  23. (a)  =    1 − i   (1 − i )(1 + i ) 

x

⇒ – i – i = x + iy; x + iy = 0 – 2i

24. (d) Given z1/3 = p + iq ⇒ x – iy = ( p+ iq)3

= p + 3p iq + 3pi q + i q



= p3 – 3pq2 + i (3p2q – q3) x = p(p2 – 3q2), y = (q2 – 3p2) x y + = – 2(p2 + q2) p q

1+ i 1+ i 1+ i = 25. (b) Consider × 1− i 1− i 1+ i =

Hence (x, y) = (0, – 2). 29. (c)

2−i 2−i (2 − i ) (−3 + 4i ) = = × (1 − 2i ) 2 −3 − 4i −3 − 4i −3 + 4i

=

3 3

Equating real and imaginary parts, we get



3

⇒ i3 – (– i)3 = x + iy ⇒ – i + i3 = x + iy

⇒ (i)x = (i)4n, where n is any positive integer. ⇒ x = 4n.

∴

[cos (u + v) − i sin (u + v)] [cos (u + v) − i sin (u + v)]

1+ i 1− i = i and =– i 1− i 1+ i 3

x



×

1 + i  1 − i  − = x + iy ∴   1 − i   1 + i 

1 + i  ⇒  = (i)x = 1 (given)  1 − i 

2 2

sin u cos v [cos ( x + y ) + i sin ( x + y )] [cos (u + v) + i sin (u + v)]

28. (c) Since,

e

2

=

sin u cos v [cos ( x + y − u − v) + i sin ( x + y − u − v)] cos 2 (u + v) + sin 2 (u + v) = sin u cos v cos (x + y – u – v)  + i sin u cos v sin (x + y – u – v) ∴ A = sin u cos v cos (x + y – u – v) and B = sin u cos v sin (x + y – u – v).

x  (1 + i ) 2  1 − 1 + 2i  =  =  2   2    1− i 

3

sin u cos v (cos x + i sin x)(cos y + i sin y ) (cos u + i sin u )(cos v + i sin v)

=

=1

2n

x

(cos x + i sin x)(cos y + i sin y ) sin v   cos u   + i  1 + i   sin u   cos v 



2n

1 − 1 + 2i  ⇒  =1 2   ⇒ i2n = 1 ⇒ n = 2 is the required smallest positive integer.

(cos x + i sin x)(cos y + i sin y ) (cot u + i )(1 + i tan v)

=

3i

22. (c) We have, (1 + i)2n = (1 – i)2n

1 + i  ⇒   1 − i 

(1 + 2i ) (1 + 2i )(1 + i ) 1 = = (– 1 + 3i) 1− i 1 − i2 2 which lies in the 2nd quadrant.

26. (b) We have,

(1 + i ) 2 1 + 2i + i 2 1 + 2i − 1 = = =i 2 1− i 1 − i2 1+1 n

1 + i  = (i)n = 1. ∴   1 − i  Now (i)n = 1 ⇒ n = 4, 8, 12, 16, ... Hence the smallest integer, n = 4.

 2   11  11i − 2 = −  +   i .  25   25  9 + 16

 2−i   −2  11 =  − i. ∴   25  25  (1 − 2i ) 2  30. (d) Let A = ii ⇒ log A = i log i ⇒ log A = i log (0 + i) = i [log 1 + i tan–1 ∞] π π  ⇒ log A = i  0 + i  = –  2 2 ⇒ A = e– π/2. 31. (c) Given | z – 4 |2 < | z – 2 |2 ⇒ | (x – 4) + iy |2 < | (x – 2) + iy |2 ⇒ (x – 4)2 + y2 < (x – 2)2 + y2 ⇒ – 4x < – 12 ⇒ 4x > 12; x > 3 ⇒ Re (z) > 3.

⇒ (1 + 1) ⋅ (1 + 4) ⋅ (1 + 9) ... (1 + n2) = α2 + β2 ⇒ 2 ⋅ 5 ⋅ 10 ... (1 + n2) = α2 + β2. 33. (a)

z − 5i z + 5i

x + i ( y − 5) x + i ( y + 5)

=1 ⇒



=1

=

⇒ x2 + ( y – 5)2 = x2 + (y + 5)2

⇒

z−i z+i

=1 ⇒

x + i ( y − 1)

2

x + i ( y + 1)

2

x + i ( y − 1) x + i ( y + 1)

(2 x + 1) + 2iy  1 − y − ix  ⋅ (1 − y ) + ix  1 − y − ix 

(2 x + 1)(1 − y ) + 2 xy + i [− x (2 x + 1) + 2 y (1 − y )] (1 − y ) 2 + x 2

 2z + 1 ∵ Imaginary part of  iz + 1  = – 2.

⇒ 20y = 0; y = 0, which is x-axis. 34. (a) We have,

=

537

= | α + iβ |2



40. (b) Let z = x + iy 2 z + 1 2 ( x + iy ) + 1 ∴ = iz + 1 i ( x + iy ) + 1

=1

∴ We have,

− x (2 x + 1) + 2 y (1 − y ) = – 2 (1 − y ) 2 + x 2

or – 2x2 – 2y2 – x + 2y = – 2 (1 + y2 – 2y) – 2x2

=1

i.e., x + 2y – 2 = 0, which is a straight line.

⇒ x2 + (y – 1)2 = x2 + (y + 1)2 ⇒ 4y = 0; y = 0, which is x-axis.

41. (b) Let α = ω and β = ω2

35. (a) We have, | z – ai | = | z + ai |

1 1 ⋅ ω2 ω4 1 = 1 + (1)2 + =3 (1) 2

Now α3 + β3 + α–2 β–2 = ω3 + ω6 +

⇒ | x + i ( y – a) | = | x + i (y + a) | ⇒ x2 + ( y – a)2 = x2 + ( y + a)2 ⇒ 4ay = 0; y = 0, which is x-axis. 2

2

= ω3 + (ω3)2 +

1 (ω 3 ) 2

[∵ ω3 = 1]

36. (c) We have, | z – 1 | = | z + i | ⇒ | (x – 1) + iy | = | x + i (y + 1) | ⇒ (x – 1)2 + y2 = x2 + (y + 1)2 ⇒ x + y = 0, which is a line through the origin. 37. (b) We have, | z1 + z2 | ≤ | z1 | + | z2 | ⇒ | z1 | + | z2 | ≥ | z1 + z2 | ⇒ | z | + | z – 1 | = | z | + | 1 – z | ≥ | z + (1 – z) | = 1 [∵ | z – 1 | = | 1 – z |] ∴ | z | + | z – 1 | ≥ 1 ⇒ Minimum value of | z | + | z – 1 | is 1. 38. (b) Let z = x + iy ∴ | z | – z = 1 + 2i ⇒

x 2 + y 2 – (x + iy) = 1 + 2i

⇒ ( x + y 2

⇒

2

1 39. (a) ∆ = ω n ω 2n

ωn ω2n 1

8

= cos 2π + i sin 2π + cos 2π – i sin 2π = 2 cos 2π = 2 (1) = 2  [By De-Moivre’s theorem]

3 2 3 – 2i. 2

 π  π  45. (b) We have, cos   + i sin    3  3   = (cos π + i sin π)1/4

ω 2n 1 ωn n

8

8

= 1(ω – 1) – ω (ω – ω ) + ω (ω – ω ) = 1(1 – 1) – 0 + ω 2n (ωn – ω n) = 0. 3n

(1 – ω + ω2) (1 – ω2 + ω4) (1 – ω4 + ω8) (1 – ω8 + ω16) = (1 + ω2 – ω) (1 – ω2 + ω) (1 – ω + ω2) (1 – ω2 + ω) [∵ ω4 = ω3 ⋅ ω = ω; ω8 = (ω3)2 ⋅ ω2 = ω2; ω16 = (ω3)5 ⋅ ω = ω and ω3 = 1] = (– ω – ω) ( – ω2 – ω2) (– ω – ω) (– ω2 – ω2) = (– 2ω) (– 2ω2) (– 2ω) (– 2ω2) = 16 ⋅ ω6 = 16 (ω3)2 = 16 (1)2 = 16.

π π  π π  = cos + i sin  + cos − i sin  4 4  4 4 

If y = – 2, x 2 + 4 – x = 1 ⇒ x2 + 4 = (1 + x)2

∴ z = x + iy =

43. (c) We have,

8

– x) + i (– y) = 1 + 2i

3n

= ω3n + (ω2)3n = (ω3)n + (ω6)n = 1n + 1n = 2.

1 + i  1 − i  + 44. (d) We have,   2   2 

x 2 + y 2 – x = 1 and y = – 2.

⇒ 2x = 3; x =

3n

 −1 + i 3   −1 − i 3  42. (c) We have,  +   2 2    

2n

2n

2n

n

4n

3

4



= [cos (2kπ + π) + i sin (2kπ + π)]1/4



(2k + 1) π (2k + 1) π   = cos + i sin  4 4 

where k = 0, 1, 2, 3

Complex Numbers

32. (c) We have, | 1 + i |2 ⋅ | 1 + 2i |2 ⋅ | 1 + 3i |2 ... | 1 + ni |2

538

Objective Mathematics

(2k + 1) π  = cis   where k = 0, 1, 2, 3 4  Continued product of the four values   π   3π   5π   7 π   = cis    +   +   +    = cis (4π) = 1.  4   4   4   4   1 + cos θ + i sin θ  46. (c) We have,   sin θ + i + i cos θ 







θ  θ 24 cos 4   cos   + i sin 2  2 = θ  θ 24 cos 4   sin   + i cos 2  2

 θ     2 

4

= – 1 – 2 = – 3.

⇒ x2 + 4x + 7 = 0.  ividing x4 + 3x3 + 2x2 – 11x – 6 by x2 + 4x + D 7, we get

 θ     2 

4

50. (a) We have, (1 + ω) (1 + ω2) (1 + ω4) (1 + ω8) ... to 2n factors = (1 + ω) (1 + ω2) (1 + ω3 ⋅ ω) (1 + ω6 ⋅ ω2) ... to 2n factors 

to 2n factors [∵ ω3 = ω6 = ... = 1]

= [(1 + ω) (1 + ω) ... to n factors] [(1 + ω2) [∵ i4 = 1]

(1 + ω2) ... to n factors]



= (1 + ω)n (1 + ω2)n = [(1 + ω) (1 + ω2)]n = (1 + ω + ω2 + ω3)n = (0 + 1)n = 1

π π 47. (a) Since z r = cos  r  + i sin  r  , 3  3  r = 1, 2, 3, ... we have, z1 · z2 · z3 ...∞ π π π π  π π   =  cos + i sin   cos 2 + i sin 2   cos 3 + i sin 3  ...∞  3 3 3 3  3 3  π π π   + 2 + 3 + ... 3 3 3

 π   3   1 1 −   3

[∵ 1 + ω + ω2 = 0, ω3 = 1]. 51. (c) We have, 

10

∑  sin k =1

2 πk 2 πk  − i cos  11 11  10



=

so [cos (α – β) + cos (β – γ) + cos (γ – α)]  + i [sin (α – β) + sin (β – γ) + sin (γ – α)] = 1 + i0 On equating real parts, we get cos (α – β) + cos (β – γ) + cos (γ – α) = 1.

49. (a) We have, x = ω – ω2 – 2 or x + 2 = ω – ω2 Squaring, x + 4x + 4 = ω + ω – 2ω 4

3



∑  −i

2

sin

k =1

10



=– i



∑  cos k =1

2 πk 2 πk  − i cos  11 11 

2 πk 2 πk  + i sin  =– i 11 11 

10

∑e

 = – i ∑ e  k =1



= – i (sum of 11th roots of unity – 1)



= – i (0 – 1) = i.

2 πk i 11

i

k =1



10

π π = cos + i sin = 0 + i ⋅ 1 = i. 2 2 48. (d) We have, a = cos (α – β) + i sin (α – β) b b = cos (β – γ) + i sin (β – γ) c c = cos (γ – α) + i sin (γ – α) and a b c a As + + = 1, c a b

2



[∵ ω3 = 1]

= (1 + ω) (1 + ω2) (1 + ω) (1 + ω2) ...

cis 2θ = 4 = cis 4θ i cis (−2θ)

2

= ω2 + ω – 2

= 0 + 1 = 1.

4

Now cis 4θ = cis nθ ⇒ n = 4.





= (0) (x2 – x – 1) + 1

 θ θ 2 θ  2 cos  2  + 2i sin  2  cos  2    =   2 sin  θ  cos  θ  + 2i cos 2  θ           2  2 2

 π    = cos  3  + i sin 1 1 −   3

= ω2 + ω3 ⋅ ω – 2ω3

x4 + 3x3 + 2x2 – 11x – 6 = (x2 + 4x + 7) (x2 – x – 1) + 1

4

π π π  = cos  + 2 + 3 + ... + i sin 3 3 3 



 − 1 

3   1 52. (b) We have,  +  1 − 2i 1 + i 

 3 + 4i    2 − 4i 

=

1 + i + 3 (1 − 2i )  3 + 4i  (4 − 5i )(3 + 4i )   = (1 − i − 2i 2 )(2 − 4i ) (1 − 2i )(1 + i )  2 − 4i 

=

12 + i − 20i 2 32 + i = (3 − i )(2 − 4i ) 6 − 14i + 4i 2

 32 + i   2 + 14i  64 + 450i + 14i 2 =  =     2 − 14i   2 + 14i  22 + 142 =

50 + 450i 1 9 = + i. 200 4 4

53. (c) We have, =

1 1 − cos θ + 2i sin θ

1 (1 − cos θ − 2i sin θ) ⋅ 1 − cos θ + 2i sin θ (1 − cos θ − 2i sin θ)

2 πk 11

10[cos (75º −30º ) + i sin (75º −30º )] cos 2 30º + sin 2 30º 10 = 10 [cos 45º + i sin 45º] = (1 + i). 2 58. (d) We have, using De-Moivre’s Theorem, the given expression

θ θ θ − 2i ⋅ 2 sin cos 2 2 2 = 2 2 θ θ   2 θ  2 sin  + 4  2 sin cos  2 2 2 2 sin 2

(cos θ + i sin θ) −14 (cos θ + i sin θ) −15 (cos θ + i sin θ) 48 (cos θ + i sin θ) −300 = (cos θ + i sin θ)– 47 = cos 47θ – i sin 47θ.

θ θ 2 sin 2 1 − 2i cot  2 2 = θ θ θ  2 sin 2  2 sin 2 + 8 cos 2  2 2 2

=

59. (c) We have, sin θ – i cos θ = – i2  sin θ – i cos θ

θ θ 1 − 2i cot 2 2 = = 2 + 3 (1 + cos θ)  2θ 2 θ 2 θ 2  sin + cos  + 6 cos  2 2 2 1 − 2i cot

θ θ 2 cot 1 2 2 . = = −i 5 + 3 cos θ 5 + 3 cos θ 5 + 3 cos θ 1 − 2i cot

54. (c) Let an = 1 (an – 1) = (a – 1) (a – ω) (a – ω2) ... (a – ωn – 1) ⇒ (a – ω) (a – ω ) ... (a – ω 2

n –1

an − 1 )= a −1

⇒ lim (a – ω) (a – ω2) ... (a – ωn – 1) a→1 = lim a→1

=

an − 1 a −1

⇒ (1 – ω) (1 – ω2) ... (1 – ωn – 1) = n. 55. (d) Let the fourth vertex D be z4. Since the points z1, z2, z3, z4 are the vertices of a parallelogram, therefore, equating the complex numbers corresponding to the mid points of the two diagonals, we have z1 + z3 z2 + z4 = 2 2 or z1 + z3 = z2 + z4

= – i (cos θ + i sin θ). ∴ The given expression, using De-Moivre’s theorem, is = (– i)3 [cos (– 25θ) + i sin (– 25θ)] = i [cos 25θ – i sin 25θ] = sin 25θ + i cos 25θ. 60. (a) We have,

z =

q + ir − r + iq ; ∴ iz = 1+ p 1+ p

∴ By componendo and dividendo,

1 + p − r + iq 1 + iz = 1 + p + r − iq 1 − iz

∴

1 + p − r + iq 1 + iz p + iq p + iq = if = 1 + p + r − iq 1 − iz 1+ r 1+ r

or p (1 + p + r) = ⇒ p (1 + p + r) and q (1 + p + r)

+ q2 + i {q (1 + p + r) – pq} (1 + r) (1 + p – r) + iq (1 + r) + q2 = (1 + r) (1 + p – r) – pq = q (1 + r), {this is obviously true}. ∴ The condition is p (1 + p + r) + q2 = (1 + r) (1 + p – r) or p + p2 + pr + q2 = 1 + p – r + r + pr – r2 or p2 + q2 + r2 = 1.



∴ z4 = z1 + z3 – z2. 56. (b) Since the points z1, z2, z3, z4 are the vertices of a parallelogram, therefore, the mid points of the two diagonals must be same. Equating the complex numbers corresponding to the mid-points of the two diagonals, we have

z1 + z3 z2 + z4 = 2 2

or z1 + z3 = z2 + z4. 4 (cos 75º +i sin 75º ) 57. (a) We have, 0.4 (cos 30º +i sin 30º )

10 (cos 75º +i sin 75º )(cos 30º −i sin 30º ) = cos 2 30º −i 2 sin 2 30º

61. (c) We have, ⇒

3 = a + ib 2 + cos θ + i sin θ

3{(2 + cos θ) − i sin θ} = a + ib (2 + cos θ) 2 + sin 2 θ = a + ib

or ∴ a =

3 (2 + cos θ) −3 sin θ and b = 5 + 4 cos θ 5 + 4 cos θ

∴ a2 + b2 =

9 (2 + cos θ) 2 + 9 sin 2 θ (5 + 4 cos θ) 2

9 {5 + 4 cos θ} 9 = (5 + 4 cos θ) 2 5 + 4 cos θ



=

and

4a – 3 =

12 (2 + cos θ) 9 –3= 5 + 4 cos θ 5 + 4 cos θ

∴ a2 + b2 = 4a – 3.

539

1 − cos θ − 2i sin θ (1 − cos θ) 2 + 4 sin 2 θ

Complex Numbers

=

540

62. (a) We have,

Objective Mathematics

z1 z2 a − ib = x – iy (1 − i )(−2 + 4i ) 66. (c) We have, z = 3 ⇒ a – ib = (x – iy) 1+ i 1  = x3 – 3x2 ⋅ iy + 3x (iy)2 – (iy)3 −2 + 2i + 4i + 4 (2 + 6i )(1 − i ) 8 + 4i = = = 4 + 2i. = = (x3 – 3xy2) – i (3x2y – y3) 1+ i 1 − i2 2 ∴ a + ib = (x3 – 3xy2) + i (3x2y – y3) z z  ∴ Im  1 2  = 2. = x3 + 3x2 ⋅ (iy) + 3x (iy)2 + (iy)3 = (x + iy)3  z1  ∴ 3 a + ib = x + iy. 67. (a) (1 + ω2 + 2ω)3n – (1 + ω + 2ω2)3n 63. (b) We have, 3 a + ib = x + iy = (ω)3n – (ω2)3n = 1 – 1 = 0 a + ib = (x + iy)3 = (x3 – 3xy2) + i (3x2y – y3) [∵ 1 + ω + ω2 = 0 and ω3 = 1] ∴ a = x3 – 3xy2 and b = 3x2y – y3. 68. (b) ∵  ω4 = ω3 ⋅ ω = ω and ω5 = ω3 ⋅ ω2 = ω2 b a 2 2 2 2 = x – 3y and = 3x – y ∴ ∴ (1 – ω) (1 – ω2) (1 – ω4) (1 – ω5) y x = (1 – ω) (1 – ω2) (1 – ω) (1 – ω2) a b 2 2 2 2 2 2 + ∴ = x – 3y + 3x – y = 4 (x – y ). = (1 – ω)2 (1 – ω2)2 = [(1 – ω) (1 – ω2)]2 x y = [1 – (ω + ω2) + ω3]2 2 n – 1 64. (a) Since 1, α, α , ..., α are the n nth roots of = [1 – (– 1) + 1]2 unity, [∵ 1 + ω + ω2 = 0, ∴ ω + ω2 = – 1] 2 ∴ xn – 1 = (x – 1) (x – α) (x – α2) ... (x – αn – 1) = (3) = 9. ⇒ log (xn – 1) = log (x – 1) + log (x – α) 69. (c), (d)  Clearly n = 3m or 3m + 1 or 3m + 2. + log (x – α2 ) + ... + log (x – αn – 1) −1 + −3 −1 − −3 Differentiating both sides w.r.t. ‘x’, we get = ω, then = ω2 Let 2 2 n −1 n 1 1 1 1 + + + ... + = Now given expression x − 1 x − α x − α2 x − α n −1 xn − 1 3

n

Putting x = 2, we get

∴



n 2 n −1 1 1 1 1 = + + + ... + 2n − 1 1 2 − α 2 − α2 2 − α n −1 n −1

n ⋅ 2 n −1 –1= 2n − 1 n −1

Hence

∑ i =1

∑ i =1

1 2 − αi

1 n ⋅ 2n −1 − 2n + 1 (n − 2) 2n −1 + 1 = . i = 2−α 2n − 1 2n − 1

65. (c) We have, a1 z3 + a2 z2 + a3 z + a4 = 3 ⇒ | 3 | = | a1 z3 + a2 z2 + a3 z + a4 | ⇒ 3 ≤ | a1 z3 | + | a2 z2 | + | a3 z | + | a4 | ⇒ 3 ≤ | a1 | | z3 | + | a2 | | z2 | + | a3 | | z | + | a4 | ⇒ 3 ≤ | z |3 + | z |2 + | z | + 1

(∵ | ai | ≤ 1)

⇒ 3 ≤ 1 + | z | + | z | + | z | < 1 + | z | + | z | + | z |3 + ... ∞ 2

3

⇒ 3 < 1 + | z | + | z |2 + | z |3 + ... ∞ 1 ⇒ 3 < 1 −| z | ⇒ 1 – | z |
. 3

⇒

2

n

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

Case I. When n = 3m Given expression = ωn + ω2n = ω3m + ω6m = (ω3)m + (ω6)m = 1 + 1 = 2. Case II. When n = 3m + 1 ωn + ω2n = ω3m + 1 + ω6m + 2 = ω3m ⋅ ω + ω6m ⋅ ω2 = 1 ⋅ ω + 1 ⋅ ω2 = ω + ω2 = – 1 [∵ 1 + ω + ω2 = 0] Case III. When n = 3m + 2 ωn + ω2n = ω3m + 2 + ω6m + 4 = ω3m ⋅ ω2 + ω6m ⋅ ω4 = ω2 + ω4 = ω2 + ω3 ⋅ ω = ω2 + ω = – 1. [∵ 1 + ω + ω2 = 0] 70. (c) Here a =

1 1 ( 3 + i), r = ( 3 + i) 2 2

1  ∴ Tn = arn –1 =  ( 3 + i ) 2  Put

n

3 1 = r cos θ, = r sin θ. 2 2

∴ arg (z2 +

∴ Tn = [r (cos θ + i sin θ)]n

π π  =  cos + i sin   6 6



| Tn | =

z

n

= cos 

= tan–1

nπ nπ + i sin  6 6

3α α ⋅ sin 2 2 = tan– 1  tan α  = α .   3α α 2 2 2 cos ⋅ cos 2 2

75. (c), (d)  Let x = Then x =

71. (b) We have, ( 1 – ω + ω2) (1 – ω2 + ω4) (1 – ω4 + ω8)  (1 – ω8 + ω16)... to 2n factors = (1 – ω + ω2) (1 – ω2 + ω) (1 – ω + ω2) (1 – ω2 + ω) ... to 2n factors. [∵ ω4 = ω, ω8 = ω2, ω16 = ω etc.] 2

2

= (22 ω3) (22 ω3) ... to n factors [∵ (– 2ω) (– 2ω ) = 2 ω = 2 ] 2



2

3

2

−1 ± − 3 −1 ± 1 − 4 ⋅ 1 ⋅ 1 = 2 2 ⋅1 −1 ± 3i = ω or ω2. = 2



∴ x =

76. (a) We have, z =

α Hence

= ω and β = ω2. α 4 + β4 + α–1 β–1 = ω4 + ω8 + ω–1 ω–2 = ω3 ⋅ ω + ω6 ⋅ ω2 + (ω3)–1 = ω + ω2 + 1 = 0. [∵ ω3 = 1]

73. (a), (d)  We have,

z =

(1 + 2i + 1 + i )(1 + 2i ) (2 + 3i )(1 + 2i ) = (1 − i )(1 − 2i ) (1 − i )(1 − 2i )



=

(− 4 + 7i )(−1 + 3i ) −4 + 7i = (−1) 2 + (− 3) 2 −1 − 3i



=

−17 −19i −17 19 − i. = 10 10 10 2

∴ | z | =



=

 17   19    +   10 10

2

=

1 17 2 + 192 10

1 1 650 = 26 . 10 2

arg z = tan– 1

  ∵ plotting of 

19 − 19 10 –1 17 = – π + tan 17 . − 10

  17 19   − , −  is in the 3rd quadrant  . 10 10 

74. (c) We have, z2 + z = (cos α + i sin α)2  + (cos α – i sin α) = (cos2α – sin2α + i 2 sin α cos α + cos α – i sin α) = (cos 2α + cos α) + i (sin 2α – sin α)

1 + 7i (2 − i)2

=

(1 + 7i )(3 + 4i ) 1 + 7i 1 + 7i = = (3 − 4i )(3 + 4i ) 2 4 + i − 4i 3 − 4i

=

− 25 + 25i =–1+ i 32 + 42

= (22)n = 22n.

72. (a) Since α, β are the complex cube roots of unity, we may write

−1 − −1 − −1 − ... to ∞ −1 − x or x2 = – 1 – x or x2 + x + 1 = 0

= (– 2ω) (– 2ω ) (– 2ω) (– 2ω ) ... to 2n factors

sin 2α − sin α cos 2α + cos α

2 cos

nπ   2 nπ + sin 2  cos  = 1. 6 6 



) = tan– 1

541

π , 6

∴ r = | z | =

(−1) 2 + 12 =

2

Again, z = – 1 + i, ∴ x = – 1, y = 1. 1 y π Also, tan θ = = −1 = 1; ∴ θ = . x 4 ∵ x = – 1 < 0 and y = 1 > 0, hence z lies in the 2nd quadrant π 3π = . 4 4 3π 3π   2  cos + i sin .  4 4 

Hence arg z = π – θ = π – ∴ Polar form of z = 77. (c) Since | z – 1 | = 1,

∴ let z – 1 = cos θ + i sin θ Then, z – 2 = cos θ + i sin θ – 1

θ θ θ = – 2 sin2 2 + 2i sin  2  cos  2



= 2i sin 

and

θ θ θ   cos + i sin  2 2 2  z = 1 + cos θ + i sin θ

= 2 cos2 = 2 cos 

...(1)

θ θ θ + 2i sin cos 2 2 2 θ    cos θ + i sin θ  . 2  2 2

...(2)

From (1) and (2), we get θ z−2 = i tan  2 = i tan (arg z) z θ   ∵ arg z = from (2)  2

Complex Numbers

These give r = 1 and θ =

542

83. (c) We have, | z – 1 | + | z + 1 | ≤ 4

78. (b) Since | z | = 1, ∴ let z = cos α + i sin α

Objective Mathematics



∴ z =

α 2 α 1 + tan 2 2 1 − tan 2

α α α 1 − tan 2 + 2i tan 2i tan 2 2 2 = + α α 1 + tan 2 1 + tan 2 2 2



α  1 + i tan 1 + i tan  2 = = α  α  1 − i tan 1 + i tan  1 − i tan  2 2



 arg z  1 + i tan   2  . =  arg z  1 − i tan    2 

2

α 2 α 2

( ∵ arg z = α)

( x − 1) 2 + y 2 + ( x + 1) 2 + y 2 ≤ 4

⇒

where z = x + iy

 ⇒ A + B ≤ 4 where A =

...(1)

( x − 1) 2 + y 2 and B =

( x + 1) 2 + y 2

But [(x – 1)2 + y2] – [(x + 1)2 + y2] = – 4x A2 – B2 = – 4x

i.e.,

...(2)

Dividing (2) by (1), we get ( x − 1) 2 + y 2 –



( x + 1) 2 + y 2 ≤ – x

( x − 1) 2 + y 2 ≤ 4 – x

⇒ 2

⇒ 3x2 + 4 y2 ≤ 12 [Squaring and simplifying]

79. (c) We have, α = ω and β = ω2 Then, xyz = (a + b) (aω + bω2) (aω2 + bω) = (a2  ω + abω2 + abω + b2 ω2) (aω2 + bω) = a3 + b3 + a2 b (1 + ω + ω2) + ab2 (1 + ω + ω2) ( ∵ 1 + ω + ω2 = 0) = a3 + b3. 80. (b) We have,

x2 y 2 + ≤1 4 3 which represents the interior and boundary of the ellipse.

or

84. (a) We have, | z | – z = 1 + 2i

x + y + z = (a + b) + (aω + bω2) + (aω2 + bω) = a (1 + ω + ω2) + b (1 + ω + ω2) = a × 0 + b × 0 = 0

...(1)

⇒

x 2 + y 2 – (x + iy) = 1 + 2i, where z = x + iy

⇒

x 2 + y 2 – x = 1 and y = – 2

Hence from the relation

[Comparing real and imaginary parts]

x3 + y3 + z3 – 3xyz

3 and y = – 2. 2

= ( x + y + z) (x2 + y2 + z2 – yz – zx – xy) = 0 by (1),

⇒ x =

we obtain x3 + y3 + z3 = 3xyz

∴ The solution of the given equation is

but

xyz = a3 + b3

[See previous problem]

Hence x + y + z = 3 (a + b3). 3

3

3

3

3 + i = i 3 + i 2 = – i  −1 + 3 i  = – iω   2  2i 2  and

85. (a), (c)  Let z = x + iy. Then,

81. (a) We have,

2 i− 3 = i −i 3 2i 2

 −1 − 3 i  2 = – i    = – iω 2   6

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

 1  ∵ i = − i 

6

= (– iω)6 + (– iω2)6 = i6 (ω6 + ω12) = – 1 (1 + 1) = – 2. 82. (c) We have, z3 + 2 z2 + 2 z + 1 = 0 ⇒ (z + 1) (z2 + z + 1) = 0. Its roots are – 1, ω and ω2. The root z = – 1 does not satisfy the equation z1985 + z100 + 1 = 0 but z = ω and z = ω2 satisfy it. Hence ω and ω2 are the common roots.

3 – 2i. 2

| (x + iy) – i | = | (x + iy) + 1 | = 1

or

x 2 + ( y − 1) 2 =

( x + 1) + y 2 = 1

∴ x2 + y2 – 2y + 1 = x2 + y2 + 2x + 1 i.e., x = – y and x2 + y2 – 2y + 1 = 1 From (i) and (ii), x2 + x2 + 2x = 0; or x (x + 1) = 0 ∴ x = 0, – 1; ∴ y = 0, 1 ∴ z = x + iy = 0, – 1 + i. 86. (b) As (a – b)2 ≥ 0, a2 + b 2 ≥ 2ab But | z | =

...(i) ...(ii)

...(i)

a 2 + b 2 ; so from (i), | z |2 ≥ 2ab

∴ | z |2 + a2 + b2 ≥ a2 + b2 + 2ab | z |2 + | z |2 ≥ (a + b)2; ∴ 2 | z |2 ≥ (a + b)2

or ∴

2 | z | ≥ a + b as | z | is positive.

∴ | z | ≥

1 (a + b). 2

– 1

y1 y2 − x y − y2 x1 x1 x2 = tan– 1 2 1 y1 y2 x1 x2 + y1 y2 1+ ⋅ x1 x2



= tan



= tan–1 ∞, by (i)

∴ | amp z1 – amp z2 | =

Since ∆ABC is equilateral ∴ AB = BC = CA ∴ | z1 – z2 | = | z2 – z3 | = | z3 – z1 |

π . 2

{∵ zz = | z |2 } = (1 − z1 z2 )(1 − z1 z2 ) − ( z1 − z2 )( z1 − z2 ) ,

= (1 − z1 z2 )(1 − z1 z2 ) − ( z1 − z2 )( z1 − z2 ) {∵ z1 = z1} = 1 − z1 z2 − z1 z2 + z1 z1 z2 z2 − z1 z1 + z1 z2 + z1 z2 − z2 z2

 | z − 1| + 4  1  3 | z − 1| − 2  > 1 = log1/ 2   2

| z − 1| + 4 1 ⇒ 3 | z − 1| − 2 < 6 ⇒ | z – 1 | > 3 which is an exterior of a circle. 90. (d) We have, z2 + z + 1 = 0

92. (c) Let z = x + iy. Then

If n is not a multiple of 3, then we can write n = 3m + r, where m ∈ I and r = 1 or 2, then 2n = 6m + 2r If r = 1, then 2r = 2 ∴ zn + z2n = (z3)m ⋅ zr + (z3)2m ⋅ z2r = zr + z2r = z + z2 = – 1 [Using (1)] If r = 2, then 2r = 4 ∴ 2n = 3 (m + 1) + 1 ∴ zn + z2n = (z3)m ⋅ zr + (z3)m + 1 ⋅ z1 = z2 + z = – 1

z1 − z2 z 2 − z3 = z3 − z 2 z1 − z3

z−3 =2 z+3

| x − 3 + iy | = 2 ⇒ | x + 3 + iy | = 2

⇒ | x – 3 + iy |2 = 22 ⋅ | x + 3 + iy |2 ⇒ (x – 3)2 + y2 = 4 [(x + 3)2 + y2] ⇒ 3x2 + 3 y2 + 30x + 27 = 0 ⇒ x2 + y2 + 10x + 9 = 0, which represents a circle. 93. (b) Let z = x + iy 1 − iz z−i

=1

⇒

1 − i ( x + iy ) x + iy − i

⇒

1 + y − ix x + i ( y − 1)

⇒ (z – 1) (z2 + z + 1) = 0, ∴ z3 = 1.

Hence z + z = – 1.

x − 3 + iy x + 3 + iy

Given,

...(1)

...(ii)

z12 + z22 + z32 – z1 z2 – z2 z3 – z3 z1 = 0.

or

⇒

∴ k = 1.

2n

...(i)

z12 − z1 z2 − z1 z3 + z2 z3 = z3 z2 − z22 − z32 + z2 z3

or

2 = 1 + | z1 |2 | z2 |2 − | z1 |2 − | z2 |2 = (1 – | z1 | ) (1 – | z2 |2 ).

n

[Each ratio = 1]

z −z  z −z  ∴ arg  1 2  = arg   2 3   z3 − z 2   z1 − z3  From (i) and (ii), we get

{∵ z1 − z2 = z1 − z2 and 1 = 1}

89. (c) We have, log1/ 2

z 2 − z3 z1 − z3

=

Also ∠CBA = ∠ACB

= (1 − z1 z2 )(1 − z1 z2 ) − ( z1 − z2 )( z1 − z2 )



z1 − z2 z3 − z 2

∴

2 2 88. (a) We have, |1 − z1 z2 | − | z1 − z2 |

543

Now z1 + z2 = (x1 + x2) + i ( y1 + y2) and z1 – z2 = (x1 – x2) + i ( y1 – y2) As | z1 + z2 | = | z1 – z2 |, we get (x1 + x2)2 + ( y1 + y2)2 = (x1 – x2)2 + ( y1 – y2)2 or, x1 x2 + y1 y2 = 0 ...(i) y1 y2 Now amp z1 – amp z2 = tan– 1 x – tan– 1 x 1 2

91. (c) Let A, B, C be the vertices of the equilateral triangle represented by the complex numbers z1, z2 and z3 respectively.

Complex Numbers

87. (b) Let z1 = x1 + iy1 and z2 = x2 + iy 2

⇒ ⇒ ⇒ ⇒

=1

=1 ⇒

(1 + y ) 2 + x 2 x 2 + ( y − 1) 2 = 1

(1 + y)2 + x2 = x2 + (y – 1)2 1 + y2 + 2 y + x2 = x2 + y2 – 2 y + 1 4 y = 0 y = 0, which is the equation of x-axis.

94. (b), (c)  We have two equations | z | – 4 = 0 and | z – i | – | z + 5i | = 0  Putting z = x + iy, these equations become

544

Objective Mathematics

| x + iy | = 4 i.e., x2 + y2 = 16 and | x + iy – i | = | x + iy + 5i | or x2 + (y – 1)2 = x2 + (y + 5)2 i.e., y = – 2 Putting y = – 2 in (1), x2 + 4 = 16

...(1)

...(2)

or x = ± 2 3 . Hence the complex numbers z satisfying the given equations are z1 = (2 3 , – 2) and z2 = (– 2 3 , – 2)

[Putting z = x + iy]

⇒ 2 (x + y + 1) = 4, ∴ x + y2 = 1 2

2

2

or | z |2 = 1 ⇒ | z | = 1 (since | z | cannot be –ve) Thus the equation (1) represents all points z on the circle with centre origin and radius unity. 96. (d) We have, | z – 4 | < | z – 2 | ⇒ | z – 4 |2 < | z – 2 |2 ⇒ | x + iy – 4 |2 < | x + iy – 2 |2  [Putting z = x + iy] 2 2 2 ⇒ (x – 4) + y < (x – 2) + y2 ⇒ x2 – 8x + 16 + y2 < x2 – 4x + 4 + y2 ⇒ – 4x < – 12 ⇒ x > 3 ⇒ Re (z) > 3.

=

| a |2 a (0) + a | 0 | + a a |a| = = . 2 | a | 2| a | 2

 i π6   2 e 

= 2⋅e

99. (b) Let

1 = x; ∴ xn = 1; ∴ xn – 1 = 0 ∴ xn – 1 = (x – 1) (x – a1) (x – a2) ... (x – an – 1) ∴ (x – a1) (x – a2) (x – a3) ... (x – an – 1)



n

=

xn − 1 1 − xn = = 1 + x + x2 + ... + xn – 1. x −1 1− x

 Putting x = 1, we get (1 – a1) (1 – a2) (1 – a3) ... (1 – an – 1) = n.

=

4n

2 4 n +1 e

i ( 4 n +1)

24 n e

i (12 n +1)

− i 4n

π 6

π 3

πi

π 6

2 nπ i ⋅e 6 = 2⋅e

= 2 ⋅ eπ i/6 

(e2nπ i = 1)

∴ arg z =

π . 6

102. (a) We have, z =

(1 + cos θ + i sin θ)5 (cos θ + i sin θ)3

θ θ  2 θ 1 + 2 cos − 1 + 2i sin cos  2 2 2 =  cos 3θ + i sin 3θ

5

θ θ θ cos + i sin  2 2 2 cos 3θ + i sin 3θ

6

32 cos5 =

= 32 cos5

θ  5θ 5θ  + i sin  {cos 3θ – i sin 3θ} cos 2  2 2

= 32 cos5

θ cos  5θ − 3θ + i sin  5θ − 3θ      2  2  2 

= 32 cos5

θ cos  − θ  + i sin    2 2 

98. (b) Putting x = 1, ω, ω2 respectively, 3n = a0 + a1 + a2 + ... + a2n (1 + ω + ω2)n = a0 + a1 ω + a2 ω2 + ... + a2n ω2n (1 + ω2 + ω4)n = a0 + a1 ω2 + a2 ω4 + ... + a2n ω4n Adding these, 3n + (1 + ω + ω2)n + (1 + ω2 + ω4)n = 3a0 + a1 (1 + ω + ω2) + a2 (1 + ω2 + ω4) + a3 (1 + ω3 + ω6) + ... n n n ∴ 3 + 0 + 0 = 3a0 + 3a3 + 3a6 + ... ∴ 3n – 1 = a0 + a3 + a6 + ....

4 n +1

 − i π3   2 e 

97. (a) ∵ 1 = 1 – z + z, we have 1 = | 1 – z + z | ≤ | 1 – z | + | z | = | z – 1 | + | z | Thus, | z | + | z – 1 | ≥ 1 and so the minimum value of | z | + | z – 1 | is 1.

( 3 + i ) 4 n +1 (1 − i 3 ) 4 n

101. (a) We have, z =

...(1)

⇒ (x – 1)2 + y2 + (x + 1)2 + y2 = 4 



=

that is, z1 = 2 3 – 2i, z2 = – 2 3 – 2i. 95. (d) We have, | z – 1 |2 + | z + 1 |2 = 4

100. (b) T he closest distance = length of the perpendicular from the origin on the line a z + a z + a a = 0

 θ   −   2 

 ence, the modulus and argument of z is 32 H and  − θ  respectively.   2

 5 θ cos 2 

1 1 = cos θ + i sin θ = cos θ – i sin θ. z ∴ zn = (cos θ + i sin θ)n = cos nθ + i sin nθ,

103. (a), (c) We have,

1 = (cos θ – i sin θ)n = cos nθ – i sin nθ zn 1 1 Hence z n + n = 2 cos nθ and z n − n = 2i sin nθ. z z

and

104. (b) We have, z 2n − 1 (cos θ + i sin θ) 2 n − 1 = 2n z +1 (cos θ + i sin θ) 2 n + 1 cos 2nθ + i sin 2nθ − 1  = cos 2nθ + i sin 2nθ + 1 (Using De Moivre’s Theorem)



i sin nθ cos nθ + i 2 sin 2 nθ = cos 2 nθ + i sin nθ cos nθ ( ∵ i2 = – 1)



i sin nθ (cos nθ + i sin nθ) = cos nθ (cos nθ + i sin nθ) = i tan nθ.



(1 + cos θ + i sin θ)n + (1 + cos θ – i sin θ)n θ θ  2 θ = 1 + 2 cos − 1 + 2i sin cos  2 2 2  θ = 2 cos 2

θ θ  cos 2 + i sin 2 

= 2n cos n

θ 2

n

n

+ 2n cos n

 

n

θ θ θ  + 1 + 2 cos 2 − 1 − 2i sin cos  2 2 2  n

θ 2

θ θ  cos 2 − i sin 2 

n

[By De Moivre’s Theorem] θ θ nθ nθ = 2n + 1 cosn   cos  . ⋅ 2 cos 2 2 2 2

106. (a) We have, m – n = (cos α – cos β) + i (sin α – sin β) = – 2 sin

α+β α−β sin 2 2



+ 2i cos

∴ m – n = 2i sin

α+β α−β sin 2 2

α−β 2

α+β α + β  cos 2 + i sin 2  

...(1)

Now, m + n = (cos α + cos β) + i (sin α + sin β) α+β α−β α+β α−β = 2 cos cos + 2i sin cos 2 2 2 2 ∴ m + n = 2 cos

α−β 2

α+β α + β  cos 2 + i sin 2  

...(1) π π π + + + ... is a G.P., whose We note that 2 2 2 23 1 and the first term is a = common ratio is r = 2 π . Thus, 2 π π π π a 2 = π. + 2 + 3 + ... = = 1 2 2 2 1− r 1− 2 ∴ From (1), x1 x2 x3 ... ∞ = cos π + i sin π = – 1.

 nθ nθ   nθ nθ     cos 2 + i sin 2  +  cos 2 − i sin 2    

= 2n cos n

π π   cos 3 + i sin 3  ... 2 2 π π π  π π π  = cos  + 2 + 3 + ... + i sin  + 2 + 3 + ... 2 2 2 2 2 2

105. (d) We have,

n

π π  π π  x1 x2 x3 .... ∞ =  cos + i sin   cos 2 + i sin 2  2 2 2 2

108. (c) We have,

a = (cos α + i sin α) (cos β – i sin β) b a = cos (α – β) + i sin (α – β). or b b = cos (β – γ) + i sin (β – γ) Similarly, c c = cos (γ – α) + i sin (γ – α). and a a b c Putting these values in + + = – 1, we get b c a Now

[cos (α – β) + cos (β – γ) + cos (γ – α)] + i [sin (α – β) + sin (β – γ) + sin (γ – α)] = – 1 = – 1 + 0 i. Comparing real and imaginary parts, we get  cos (α – β) + cos (β – γ) + cos (γ – α) = – 1. 109. (b) We have, abcd = cos (2α + 2β + 2γ + 2δ)  ∴



+ i sin (2α + 2β + 2γ + 2δ) abcd = [cos (2α + 2β + 2γ + 2δ) + i sin (2α + 2β + 2γ + 2δ)]1/2

or ...(2)



abcd = cos (α + β + γ + δ) + i sin (α + β + γ + δ)

...(1)

[De Moivre’s Theorem]

From (1) and (2), we obtain α −β m−n = i tan  . 2  m+n

1 1 = cos α – i sin α, = cos β – i sin β  a b

∴ 

1 = cos (α + β + γ + δ) abcd – i sin (α + β + γ + δ)

...(2)

545



Complex Numbers

107. (b) We have,

(1 − 2 sin 2 nθ) + 2i sin nθ cos nθ − 1 = (2 cos 2 nθ − 1) + 2i sin nθ cos nθ + 1

546

Objective Mathematics

Adding (1) and (2), we obtain 1 abcd + = 2 cos (α + β + γ + δ). abcd 110. (a), (c)  We have, z m

= (cos 2θ + i sin 2θ)

m

1 = cos 2nφ – i sin 2nφ. ωn 1 z mω n + m n z ω

Similarly,





= {cos 2 (mθ + nφ) + i sin 2 (mθ + nφ)} + {cos 2 (mθ + nφ) – i sin 2 (mθ + mφ)} = 2 cos 2 (mθ + nφ). zm ωn + ωn zm

111. (b) Let

nπ  nπ n = 2  2 cos .  = 2n + 1 cos 3  3

= (cos 2mθ + i sin 2mθ) (cos 2nφ – i sin 2nφ) + (cos 2nφ + i sin 2nφ) (cos 2mθ – i sin 2mθ)

= 2 cos 2 (mθ – nφ).

⇒ r2 = 2 and tan θ = 1. ⇒ r =

1 π ⇒ r = 2, θ = 3 6

π π  ∴ ( 3 + i)n = 2n   cos + i sin  6 6  or ( 3 + i)n = 2n cos 

...(1)

n

n

 nπ       ...(1) 6 

+1

nπ cos   .  4 

114. (a) We have (16)1/4 = (24)1/4 = 2 (1)1/4 = 2 (cos 0 + i sin 0)1/4

...(2)

Adding (1) and (2), we obtain

 nπ  ( 3 + i)n + ( 3 – i)n = 2 ⋅ 2n cos   6



nπ = 2n + 1 cos   .  6 

1 1   = 2 cos (2k π + 0) + i sin (2k π + 0) , k = 0, 1, 2, 3 4 4   = 2 × 1, 2 × i, 2 × – 1, 2 × – i = ± 2, ± 2i. 115. (d) We have (1 + ω – ω2)7 = (– ω2 – ω2)7 = (–2)7 (ω2)7 = – 128 w2. 116. (c) We have, z4 + 1 = 0 ⇒ z4 = – 1

112. (c) Let r cos θ = 1 and r sin θ = ⇒ r2 = 4 and tan θ = π ⇒ r = 2, θ = 3

n  nπ nπ  2 + i sin = 2  cos  4 4 

= 22

  nπ   nπ   ( 3 – i) = 2 cos   − i sin        6  6 

n

 nπ  2 αn + βn = (1 + i)n + (1 – i)n = 2 × 2 cos   4

Similarly, n

π , 4

Adding (1) and (2), we obtain

n

 nπ    + i sin 6

π π 2  cos + i sin   4 4

n  π π 2 or (1 + i)n= 2  cos + i sin  4 4



2 and  θ =

n  nπ nπ  2 − i sin Similarly, (1 – i)n = 2  cos   ...(2) 4 4 

3 = r cos θ and 1 = r sin θ so that

n

2± 4−8 = 1 ± i. 2

Thus α = 1 + i and β = 1 – i and so αn + βn = (1 + i)n + (1 – i)n Now, put 1 = r cos θ and 1 = r sin θ

= [cos 2 (mθ – nφ) + i sin 2 (mθ – nφ)]

r2 = 4 and tan θ =

x=

and so 1 + i =

+ [cos 2 (mθ – nφ) – i sin 2 (mθ – nφ)]

n

nπ nπ  nπ nπ  n n + i sin − i sin = 2  cos  + 2  cos  3 3 3 3 



+ {cos 2mθ – i sin 2mθ} {cos 2nφ – i sin 2nφ}

Also,

π π π π n n = 2  cos + i sin  + 2  cos − i sin  3 3 3 3

113. (b) Solving x2 – 2x + 2 = 0, we obtain

= (cos 2mθ + i sin 2mθ) (cos 2nφ + i sin 2nφ)



Now (1 + i 3 )n + (1 – i 3 )n n

= cos 2mθ + i sin 2mθ, and ωn = (cos 2φ + i sin 2φ)n = cos 2nφ + i sin 2nφ. 1 Since = cos 2θ – i sin 2θ, z 1 = cos 2mθ – i sin 2mθ  zm

∴

π π  ∴ 1 ± i 3 = 2  cos ± i sin  .  3 3

3

3

⇒ z = (cos π + i sin π)1/4 ⇒ z = cos

1 1 (2kπ + π) + i sin (2kπ + π), 4 4 k = 0, 1, 2, 3.

Hence the four roots of z4 + 1 = 0 are 1 (± 1 ± i). 2

 117. (a) Since 1, ω, ω2, ... ωn unity n −1

∑ω

∴ 

n −1

∑ |z k =0

1

=

∑zz k =0

1 1

n −1

118. (b)

n −1

k =0

n −1



= n | z1 | + 0 + 0 + n | z2 |



= n ( | z1 | + | z2 |  ).

k =0

2

n

119. (c) We have, f (ω) = ω + ω

3q + 1

+w

3r + 2

=ω +ω ⋅ω+ω ⋅ω = (ω3)p + (ω3)q ⋅ ω + (ω3)r ⋅ ω2 = 1 + ω + ω2 = 0. 3q

3r

 = 

−1

⇒

⇒ x2 + y2 –

[∵ ω3 = 1]

xα + yβ + z γ xα + yαω + zαω 2 = xβ + y γ + zα xαω + yαω 2 + zα

3

2 y – 1 = 0, which is a circle. 3

2 ( x + iy ) + 1 2 x + 1 + 2iy 2z + 1 = i ( x + iy ) + 1 = (1 − y ) + ix iz + 1

=

[(2 x + 1) + 2iy ] [(1 − y ) − ix] (1 − y ) 2 + x 2

=

(2 x + 1)(1 − y ) + 2 xy + i [− x (2 x + 1) + 2 y (1 − y )] . (1 − y ) 2 + x 2

 2z + 1 Since, Im    =–2 iz + 1 

zαω 2 + yαω 4 + xαω 3 xαω + yαω 2 + zα [ ∵ xα = xαω3 and yαω = yαω4]

⇒

− x (2 x + 1) + 2 y (1 − y ) =–2 (1 − y ) 2 + x 2

⇒ – 2x2 – 2y2 – x + 2y = – 2 (1 + y2 – 2y) – 2x2 ⇒ x + 2y – 2 = 0, which is a straight line.

ω 2 ( zα + yαω 2 + xαω ) −1 − 3 i = ω2 = 2 xαω + yαω + zα 2 or

y y − x −1 x +1 = π 3 y2 1+ 2 x −1

2y π = tan = x2 + y 2 − 1 3

123. (b) We have,

∴ β = αω and γ = αω2.

=

y y π − tan −1 = x −1 x +1 3   z1 ∵ Arg z = Arg z1 − Arg z2  2

2

120. (a), (b)  Since α, β, γ are the cube roots of the same number p

x + iy − 1 π π z −1 = ⇒ arg = x + iy + 1 3 3 z +1 [Putting z = x + iy]

⇒ tan

= i – 1 – i + 1 + i – 1 – i + 1 + i – 1 = i – 1.

Now,

x2 + 2x + 49 = 0.

2

3p



+ (– 1 – 4 3 i) (– 1 + 4 3 i) = 0



k =0

2

= i + i2 + i3 + i4 + i5 + i6 + i7 + i8 + i9 + i10

3p

x2 – (– 1 – 4 3 i – 1 + 4 3 i) x



2

n=1



– 1 – 4 3 i is

n −1

k

k =0

2

10

i.e., x2 – 2x + 49 = 0. Also, the quadratic equation having one root

| + ∑ z1 z2 (ω ) + ∑ z1 z2ω + ∑ | z2 | k

=

∑i

x + (1 + 4 3 i) (1 – 4 3 i) = 0

−1 ⇒ tan

+ z1 z2 (ω ) k + z1 z2 ω k + z2 z2 (ω k )(ω ) k

2

1

x – (1 + 4 3 i + 1 – 4 3 i) 2

+ ω k z2 )( z1 + (ω ) k z2 )

1



k =0

1 + 4 3 i is



122. (b) We have, arg

=0

∑ (z

n −1

∑ |z

k

k =0

+ ω k z2 |2 =

n −1



∑ (ω )

= 0 and

−1 + 3 i 2

 −1 ± 3 i −1 ∓ 3 i  then ω 2 = ∵ If ω =  2 2  

124. (a) Put 1 = r cos θ and – 1 = r sin θ π . 4 Then given equation takes the form

⇒ r =

2 and θ = –

547

– 47 + 8 −3 are ± (1 + 4 3 i) The quadratic equation, having one root

i.e.,

n −1

k

k =0

Now,

are the n, nth roots of

– 1

121. (a), (b)  The square roots of the complex number

Complex Numbers

π π 3π 3π + i sin  , cos  + i sin  , 4 4 4 4 5π 5π 7π 7π + i sin  , cos  + i sin  cos  4 4 4 4 1 1 = (1 + i), (– 1 + i), 2 2 1 1  (– 1 – i), (1 – i). 2 2 ⇒ z = cos 

548



 ( 2 ) cos  n

 π  −  + i sin 4

 π   −   4 

n



= 2n

Objective Mathematics

nπ nπ   − i sin 2n/2 cos = 2 n. 4 4  

or

3 2 ∴ tan θ = = 1 2

Equating real and imaginary parts, we get

These are satisfied only for n = 0 Hence, n = 0 is the only solution. z−

125. (c) We have, 2 = ⇒ | z | –

4 z

4 z

≥ | z | –

128. (d)

4 z

1− i 3 1 3 (1 − i 3 ) 2 −2 − 2 3 i = = = − − i. 1+ i 3 2 2 4 4  1 3  2π ∴ arg  − 2 − 2 i  = – (π – tan–1 3 ) = – . 3

≤2

129. (c) We have, (1 + i)2n = (1 – i)2n

⇒ | z |2 – 2 | z | – 4 ≤ 0 or ( | z | – 1)2 – 5 ≤ 0 ⇒ ( | z | – 1)2 ≤ 5 or | z | – 1 ≤ ⇒ | z | ≤

5

5 + 1.

Hence the greatest value of | z | is

5 + 1.

126. (a) Let z = x + iy.

1 + i  ⇒   1 − i 

Equating real and imaginary parts y + 2 = 0 ∴ y = – 2  and x + α ( x − 1) 2 + 4 = 0

1 − 1 + 2i  ⇒    2

2n

 (1 + i ) 2   2 

2n

=1

= 1 ⇒ (i)2n = 1

Let α = 5z – 1 ⇒ | α + 1 | = 5 | z | = 5 (2) = 10 ⇒ | α + 1 | = 10 ∴ locus of α i.e., 5z – 1 is the circle having centre at – 1 and radius 10. ∴ a = 10. 2 2 Again z1 + z2 − 2 z1 z2 cos θ = 0

∴ x = α2 (x2 – 2x + 5) 2

2

or  (1 – α2) x2 + 2α2 x  – 5α2 = 0 Since x is real, ∴ D = B2 – 4AC ≥ 0 ⇒ 4α4 + 20α2 (1 – α2) ≥ 0 ⇒ – 4α4 + 5α2 ≥ 0  2 5 ⇒ 4α2  α −  ≤ 0. 4 Now α being real implies α is +ve and hence we  5  5 5 conclude that α2 – is –ve or α + 2  α − 2     4 is –ve.

z  z ⇒  1  − 2 1 cos θ + 1 = 0 z2  z2  ⇒ ⇒

2

127. (a) Let z =

=1 ⇒

130. (c) Given | z | = 2.

⇒ x + i ( y + 2) + α ( x − 1) 2 + y 2 = 0.

5 ≤α≤ 2

2n

⇒ n = 2 is the smallest positive integer.

We have, z + α | z – 1 | + 2i = 0

⇒ –

y  ∵ tan θ = x 

3

−1 ∴ Argument of z = θ = tan ( 3 ) = 60º.

nπ nπ = 2n/2 and – sin = 0. 4 4

cos

1 3 −1 − i 3 = − −i 2 2 2

=

5 . 2

1− i 3 1+ i 3

(1 − i 3 ) x (1 − i 3 ) 1 + 3i − 2 3 i = 1 − 3i 2 (1 + i 3 )(1 − i 3 ) 2



=



1− 3 − 2 3i =  1+ 3

2 z1 = 2 cos θ ± 4 cos θ − 4 = cos θ ± i sin θ z2 2

z1 z2

= | cos θ ± i sin θ | = 1 =

a 10

∴ | z1 | : | z2 | = a : 10. 131. (c) Let

z −1 = iy, where y is real z +1

⇒

1 z +1 = iy z −1

⇒

2z 1 + iy = 2 1 − iy

[by componendo and dividendo]

1 + iy ⇒ z = ⇒ | z | = 1 − iy

1 + y2 1 + y 2 = 1.

132. (a) Given, | z – i | < 1 [∵ i2 = – 1]

Now, | z + 12 – 6i | = | (z – i) + (12 – 5i) |

133. (a), (d)  We have, i = 0 + i ⋅ 1 =

1 (0 + 2i) 2

[(1 + i) – (1 – i)] = ±

2 i.



=



= | z1 + z2 + ... + zn |



= | z1 + z2 + ... + zn |. [ ∵ |  z  | = | z | ]

140. (b) Since b > 0, bi represents a point on the positive side of the imaginary axis on which the argument π . of every point is 2

⇒ z12 + z22 + z32 = z1 z2 + z2 z3 + z3 z1, where z3 = 0. ⇒ z 1 , z 2 and the origin form an equilateral triangle. n

n n  z−i  1  z−i  n = = 142. (d) We have, ω =       i  = (– i)  i ( z − i)   1 + iz 

from (1)

136. (a) We have, Arg (z + i) – Arg (z – i) =

π 2

∴ | ω | = | (– i)n | = | – i |n = 1 for all n. ∴ ω lies on unit circle for all n. 143. (a) We have, z  z + a  z + a  z + b = 0

⇒ z  z + a  z + a  z + a a = a a – b

⇒ (z + a) ( z + a ) = a a – b ⇒ | z + a |2 = | a |2 – b.

z +i ∴ Re   =0 z −i

 his represents a circle (not point circle) if T | a |2 > b.

z +i z +i  +  z −i z −i ⇒ =0 2 z +i z + i  ⇒  z − i  +  z − i  = 0

144. (a) We have, z4 = (z – 1) 4  z − 1 ⇒   = 11/4 = 2 n4πi , z  e

⇒ (z + i) ( z + i) + (z – i) ( z – i) = 0 ⇒ z  z + i (z + z ) – 1 + z  z – i (z + z ) – 1 = 0 ⇒ 2 (z  z ) = 2 ⇒ z  z = 1 or ⇒ | z | = 1.

∴ x – 1 = (– 8)1/3 = – 2, – 2ω, – 2ω2. Hence, x = – 1, 1 – 2ω, 1 – 2ω2.

2 2 ⇒ z1 + z2 = z1 z2

z1 z z + 2 + ... + n 2 | z1 |2 | z2 |2 | zn |

z +i π ⇒ Arg  =  z − i  2

= ω1000 = ω999 ⋅ ω = ω.

z1 z2 141. (c) We have, z + z = 1 2 1

z1 z z + 2 + ... + n z1 z1 z2 z2 zn zn

1 1 1 + + ... + z1 z2 zn =

1000

⇒ (x – 1)3 = – 8

...(1)

Now,

= ω,

139. (b) We have, (x – 1)3 + 8 = 0

1 = – i. i

135. (a) Given, | z1 | = | z2 | = ... = | z n | = 1 

≤ | z + 4 | + | – 3 | = | z + 4 | + 3

 1 3  ∴  − 2 + 2 i 

⇒ i z2 (z – i) – (z – i) = 0

Now z = – i ⇒ | z | = | i | = 1 and z2 = – i ⇒ | z2 | = | – i | ⇒ | z |2 = 1 ⇒ | z | = 1 Thus, in both cases | z | = 1.



1 3 138. (b) Since − + i 2 2

134. (c) Given, i z3 + z2 – z + i = 0 ⇒ (z – i) (i z2 – 1) = 0 ⇒ z = i or z2 =

137. (c) We have, | z + 1 | = | z + 4 – 3 | = | (z + 4) + (– 3) | ≤ 3 + 3 = 6. ( ∵ | z + 4 | ≤ 3) Hence the greatest value of | z + 1 | is 6.

1 1 (1 + i2 + 2 ⋅ 1 ⋅ i) = (1 + i)2 = 2 2 1 ∴ i =± (1 + i) 2 1 ∴ −i = ± (1 – i). 2 1 Hence i − −i = ± 2 

| z |2 = 1

n = 0, 1, 2, 3. Since for all these values of z, 

z −1 = 1 so they lie on the line bisecting perpenz

dicularly the join of z = 1 and z = 0.

549

 he equation represents a circle centred at origin T and radius 1 unit.

Complex Numbers

≤ | z – i | + | 12 – 5i | [ ∵ | z1 + z2 | ≤ | z1 | + | z2 | ] < 1 + 13 = 14. Hence | z + 12 – 6i | < 14.



550

145. (c) We have,

148. (a) Let z = x + iy

Objective Mathematics

π π  2π 2π   + i sin   cos + i sin z1 z2 z3 z4 =  cos  × 10 10   10 10  3π 3π   4π 4π   cos + i sin + i sin   cos    10 10 10 10  

 π 2 π 3π 4π   π 2π 3π 4π  + + + + = cos  +  + i sin  +  10 10 10 10  10 10 10 10  = cos π + i sin π = – 1.

146. (d) Let x = 3 + 2i

⇒ (x – 3)2 = 4i2 = – 4

⇒ x2 – 6x + 13 = 0

...(1)

)

x 2 − 6 x + 13 x 4 − 8 x 3 + 4 x 2 + 4 x + 39 ( x 2 − 2 x − 21 x 4 − 6 x 3 + 13 x 2 − + − – 2x3 – 2x3 + – 21x2 – 21x2 + – – 96x

–   9x2 + + 12x2 – – + 30x + 126x + +

or or

( x − 4) 2 + ( y + 3) 2 ≤ 2 (x – 4)2 + (y + 3)2 ≤ 22

 hus z lies in the interior or on the boundary of the T circle whose centre is (4, – 3) and radius is 2.  4x + 26x + + –

39

Least value of | z | = OA = OC – AC = 5 – 2 = 3. Greatest value of

39 273

 | z | = OB = OC + CB = 5 + 2 = 7. Thus, 3 ≤ | z | ≤ 7. 149. (b) We have, 2 1 1 z +z + = = 3 2 z1 z 2 z3 z 2 z3

312

Thus, x4 – 8x3 + 4x2 + 4x + 39

Given, | z – 4 + 3i | ≤ 2. ∴ | x + iy – 4 + 3i | ≤ 2

= (x2 – 6x + 13) (x2 – 2x – 21) – 96x + 312

∴ f (3 + 2i) = 0 (x2 – 2x – 21) – 96 (3 + 2i) + 312 = 24 – 192i = a + ib ∴ a = 24 and b = – 192. a 1 24 ∴ = =– . b 8 −192 147. (b) Since | z – 25i | ≤ 15, therefore distance between z and 25i is less than or equal to 15.

2 z 2 z3 ⇒ z1 = z + z . 2 3  z2 − z4   z1 − z3  Now   z1 − z4   z2 − z3 

 hus point z will lie in the interior and boundary T of the circle whose centre is (0, 25) and radius is 15. Let OP be tangent to the circle at point P. Let ∠POX = θ. Then, ∠OCP = θ Now, OC = 25, CP = 15 ∴ Now, tan θ = OP = 20 = 4 CP 15 3

OP = 20.

4 ∴ Least positive value of arg z = θ = tan–1   . 3



   z2 − z4  =    2 z 2 z3 − z 4    z + z 2 3

  2 z 2 z3 − z3   z 2 + z3   − z z 2 3    

z2 z3 ( z 2 − z3 ) 2 z2 z3 ( z2 + z3 )( z2 − z3 ) =  z 2 + z3

[taking z4 = 0]

1 (a real number). 2 Hence points z1, z2, z3 and origin are concyclic and therefore z1, z2, z3 lie on a circle passing through the origin.



=

 = cos θ ±

That is, the maximum value of | z | is 1 + cos 2 θ − 1 i sin θ

− sin θ = cos θ ± 2

2

2

= cos θ ± i sin θ. When z = cos θ + i sin θ. z2 + z–2 = cos 2θ + i sin 2θ + (cos 2θ – i sin 2θ) = 2 cos 2θ and when z = cos θ – i sin θ, z2 + z–2 = cos 2θ – i sin 2θ + cos 2θ + i sin 2θ = 2 cos 2θ. 151. (b) Let cot –1 p = θ, then p = cot θ ∴ e 2 mi cot





−1 p

 pi + 1  ⋅  pi − 1 

m

 i cot θ + 1  = e 2 miθ ⋅   i cot θ − 1 

2 miθ  i (cot θ − i )  ⋅ = e  i (cot θ + i ) 

m

2 miθ  cos θ − i sin θ  ⋅ = e  cos θ + i sin θ 

−i θ  2 miθ  e ⋅  iθ  = e e 

m

= [(1 + i)2] n = (1 + i2 + 2i) n

(1 + i ) 2 n 2n 2n 2n i n + + n = n n 2n n (1 − i ) (−2) i 2 2

1 1 + (−1) n i 2 n 1 + (−1) n (i 2 ) n + in =   = = n n n n (−1) i (−1) i (−1) n i n   =

1 + (−1) n (−1) n 1 + (−1) 2 n 1+1 2 = = = . (−1) n i n (−1) n i n (−1) n i n (−1) n i n 2n

153. (c) We have, S (n) = in + i–n = in +

1 i +1 = in in

(−1) n + 1 = , n = 1, 2, 3, 4, ... in ∴ values of S (n) are 0, – 2, 2, 0, – 2, 2, ... ∴ Total number of distinct values of S (n) is 3. 154. (b) We have, | z | = z +

2 2 − z z

⇒ z  z = (z + 1) ( z + 1) = z  z + z + z + 1 ⇒ z + z = – 1, ∴ | z + z  | = 1. 156. (d) We have, (1 + ω)3 – (1 + ω2)3

= (– ω2)3 – (– ω)3 = – ω6 + ω3 = 0.

157. (c) We have, zk = 1 + a + a2 + ... + ak = 1 − a k +1 = 1− a 1− a

zk −

m

= (1 – 1 + 2i)n = 2n in. (1 – i)2n = [(1 – i)2]n = (1 + i2 – 2i)n = (1 – 1 – 2i)n = (– 2)n in ∴ 

⇒ z  z = z  z – z – z + 1 ⇒ z + z = 1 Also, | z | = | z + 1 | ⇒ | z |2 = | z + 1 |2

⇒

= e2miθ (e–2iθ)m = e2miθ ⋅ e–2miθ = e0 = 1.

152. (d) We have, (1 + i)2n

1 1− a

| a |k +1 1 = |1 − a | < |1 − a |



z−

1 1− a

3 ≤ | z | – 1 ≤

3

( ∵ | a | < 1)

158. (c) We have, i + and

i –

∴ (i +  2ω  =  i 

1 = |1 − a | . 3 =

3 =

−1 + 3 i 2 2ω ⋅ = 2 i i

−1 − 3 i 2 2ω 2 ⋅ = 2 i i

3 )100 + (i – 100

 2ω 2  +  i 

3 )100 + 2100

100

+ 2100

2100 (ω100 + ω200) + 2100 = 2100 (ω + ω2) + 2100 i100 = – 2100 + 2100 = 0. =

159. (b), (c)  Since diagonals other z +z z + z4 ∴ 1 3 = 2 2 2 ⇒ z1 – z2 + z3 – z4 = Also, since diagonals angles

2 ⇒ | z | ≤ 2 + | z | ⇒ | z |2 ≤ 2 | z | + 2

⇒ –

1 − ak + 1 1− a

∴ Vertices of the polygon z1, z2, ..., zn lie within the circle

2 2 ≤ 0 z + z + | z| .

⇒ | z |2 – 2 | z | + 1 ≤ 1 + 2 ⇒ ( | z | – 1)2 ≤ 3

3.

⇒ | z |2 = | z – 1 |2 ⇒ z  z = (z – 1) ( z – 1)

⇒ z k –

m

2 miθ  cot θ − i  ⋅ = e  cot θ + i  m

3

155. (c) We have, | z | = | z – 1 |

= cos θ ±



3 ≤ | z | ≤ 1 +

551

⇒1–

2 cos θ ± 4 cos 2 θ − 4 ⇒ z = 2

z 2 − z0 π ∴ amp z − z = 1 0 2

of a rhombus bisect each = z0 (say) 0. of a rhombus are at right

Complex Numbers

150. (a) We have, z2 – 2z cos θ + 1 = 0

552

Objective Mathematics

⇒ – 4 y2 + 2x = 0 1 or y2 = x, which is a parabola. 2

z2 + z4 π 2 ⇒ amp = z1 + z3 2 z1 − 2 z2 −

166. (c), (d)  We have, z1 – z4 = z2 – z3

z2 − z4 π ⇒ amp z − z = . 1 3 2 7 160. (c) Let zk = xk + iyk , we have (zk + 1)7 + zk = 0 7 ⇒ (zk + 1)7 = – zk

⇒ | zk + 1 | = | zk | ⇒ | zk + 1 | = | zk | 7

7

⇒ (xk + 1)2 + yk2 = xk2 + yk2

167. (c) We have, | z – i | + | z + i | = k

1 ⇒ 2xk + 1 = 0 or xk = – . 2 6

k =0

∑x

=

k

k =0

161. (b) We have, | z1 + z2 | =

k

1 1 + z1 z2

=

⇒ x + ( y – 1) – x – ( y + 1)2 2

z1 + z2 z1 z2

2

=k

 ⇒

2

{

x 2 + ( y − 1) 2 − x 2 + ( y + 1) 2

x 2 + ( y − 1) 2 − x 2 + ( y + 1) 2 =

−4y  k

From (1) and (2), 2 x 2 + ( y − 1) 2 = k – ( ∵ z 1 ≠ – z 2)

⇒ | z1 z2 | = 1.

...(1)

[Putting z = x + iy]

7 . 2

=–

 1  ⇒ | z1 + z2 | 1 − =0  | z1 z2 |  162. (c) We have, amp (z – 1) = amp (z + 3i) ⇒ tan–1

x 2 + ( y − 1) 2 + x 2 + ( y + 1) 2 = k

⇒

6

∑ Re ( z )

z4 − z1 π Also, amp z − z = 2 1 2 ⇒ angle at z1 is a right angle. ∴ It is a rectangle and hence a cyclic quadrilateral.

⇒ | xk + iyk + 1 |2 = | xk + iyk |2

Thus,

z1 + z3 z + z4 = 2 2 2 i.e., the diagonals bisect each other. ∴ It is a parallelogram. or

y y+3 = tan– 1 x −1 x

⇒ k = 4.

⇒ 3 (x – 1) = y

...(2)

4y k

16  2  ⇒ 4x2 +  4 − 2  y = k2 – 4. k For an ellipse, 4 –

⇒ xy = (x – 1) (y + 3)

}

16 > 0 and k2 – 4 = 0 k2

168. (b) We have, | z – ω |2 + | z – ω2 |2 = λ

∴ (x – 1) : y = 1 : 3.

⇒ λ = | ω – ω2 |2 = | ω2 + ω4 – 2ω3 | = | ω2 + ω – 2 | = | – 1 – 2 | = 3.

163. (c) We have, z3 + 2 z2 + 2 z + 1 = (z + 1) (z2 – z + 1) + 2 z (z + 1)

169. (c) Proceed as in Q. 167.

= (z + 1) (z + z + 1)

170. (a) The equation | z – (3 + 0i) | + | z – (– 3 + 0i) | < 6 represents the interior of ellipse with foci at (3, 0) and (– 3, 0). So, major axis is along real axis.

2

∴ Roots of the equation z3 + 2 z2 + 2 z + 1 = 0 1 (– 1 ± i 3 ) i.e., – 1, ω, ω2. are – 1, 2 Now, (– 1)149 + (– 1)100 + 1 = – 1 + 1 + 1 ≠ 0, but,

(ω)149 + (ω)100 + 1 = ω2 + ω + 1 = 0

and

(ω2)149 + (ω2)100 + 1 = ω + ω2 + 1 = 0,

∴ only ω and ω2 satisfy both the equations. 164. (b) If z = – a – ib, a > 0, b > 0, then z = – a + ib i.e., second quadrant. 165. (d) We have, z2 + z

2

– 2 | z |2 + z + z = 0

⇒ (x + iy) + (x – iy) – 2 (x + y ) + x + iy + x – iy = 0 2

2

2

2

[Putting z = x + iy] ⇒ 2x2 + 2 (iy)2 – 2x2 – 2y2 + 2x = 0

171. (b) A rea of the triangle on the argand plane formed 3 | z |2. by the complex numbers – z, iz, z – iz is 2 3 ∴ | z |2 = 600 ⇒ | z | = 20. 2 172. (c) We have, | z + z  | + | z – z  | = 8 ⇒ 2 | x | + 2 | y | = 8 or | x | + | y | = 4 which represents a square. 173. (d) Let z1 = 1 – 3i, z2 = 4 + 3i and z3 = 3 + i. Then z3 divides the line segment joining z1 and z2 in the ratio 2 : 1 internally. So, the points z1, z2, z3 are collinear.

x + iy + 2i x + i ( y + 2) z + 2i = = Then, x + iy + 4 ( x + 4) + iy z+4



=

[ x + i ( y + 2)][( x + 4) − iy ] ( x + 4) 2 + y 2

 z + 2i  Since Re   = 0 ⇒ x2 + y2 + 4x + 2y = 0, z+4  which represents a circle with centre (– 2, – 1). 175. (c) Let z = x + iy x + iy + 2i x + ( y + 2) i z + 2i = = Then, x + iy + 2 ( x + 2) + iy z+2 =



=

3

 1 i 3 and | 1 – ω2 | = 1 −  − − 2   2 

( x 2 + 4 x + y 2 + 2 y ) + i (2 x + 4 y + 8) = ( x + 4) 2 + y 2



| ω – ω2 | = |  3 i | =

[ x + ( y + 2) i ] [( x + 2) − iy ] ( x + 2) 2 + y 2

553



=

3 i 3 + 2 2

=

3.

Therefore, 1, ω, ω2 form an equilateral triangle. 179. (c) Let z1 = x1 + iy1 and z2 = x2 + iy 2 Since Re (z1 + z2) = 0 ⇒ x1 + x2 = 0 or x2 = – x1. Also, Im (z1 z2) = 0 ⇒ x1 y2 + x2 y1 = 0 ⇒ x1 y2 – x1 y1 = 0 ⇒ y2 = y1 ∴ z1 = x1 + iy1 = – x2 + iy2 = – (x2 – iy2) = – z2 . 180. (b) L et the complex numbers z1, z2, z3, z 4 represent the vertices A, B, C and D of a parallelogram respectively.

( x 2 + y 2 + 2 x + 2 y ) + i (2 x + 2 y + 4) ( x + 2) 2 + y 2

 z + 2i  Since Im   =0 ⇒x+y+2=0 z+4  which represents a straight line. 176. (a) Given : | z | = 2 and arg (z) =

2π . 3

∴ If z = r (cos θ + i sin θ), then 2π r = 2 and θ = 3

181. (b) We have, Im (z2) = 4

2π 2π   + i sin ∴ z = 2   cos  3 3 



 1 3 = 2   − 2 + i 2  = (– 1 + i 3 ).

177. (b) Let z = x + iy Then,

x + iy − 2 ( x − 2) + iy z−2 = = x + iy + 2 ( x + 2) + iy z+2

[( x − 2) + iy ][( x + 2) − iy ] ( x 2 + y 2 − 4) + i (4 y) = = ( x + 2) 2 + y 2 ( x + 2) 2 + y 2 z−2 is purely imaginary, z+2 ∴ x2 + y2 – 4 = 0 ⇒ x2 + y2 = 4 ⇒ | z |2 = 4 ⇒ | z | = 2. Since

178. (a), (c)  Clearly, cube roots of unity 1, ω, ω2 satisfy | z | = 1. 2  3 3 Also, | 1 – ω | =   +   2   2 

2

⇒ | 1 – ω | =

3

Since AB = DC ∴ z2 – z1 = z3 – z4 ⇒ z1 + z3 = z2 + z4.

2

=3

⇒ Im [(x2 – y2) + 2ixy] = 4 [Putting z = x + iy] ⇒ 2xy = 4 or xy = 2, which is a rectangular hyperbola. 182. (d) We have, Re (z2) = 4 ⇒ Re [(x2 – y2) + 2ixy] = 4 [Putting z = x + iy] ⇒ x2 – y2 = 4, which is a rectangular hyperbola. 183. (d) We have, x2 + x + 1 = (x – ω) (x – ω2) Since f (x) is divisible by x2 + x + 1, f (ω) = 0, f (ω2) = 0 ∴ P (ω3) + ω Q (ω3) = 0 or P (1) + ω Q (1) = 0 ...(1) and P (ω6) + ω2 Q (ω6) = 0 or P (1) + ω2 Q (1) = 0 ...(2) From (1) and (2) we obtain

P (1) = 0 and Q (1) = 0.

∴ Both P (x) and Q (x) are divisible by x – 1. Since  f (x) = P (x) + x Q (x), we get f (x) is divisible by x – 1.

Complex Numbers

174. (a) Let z = x + iy

554

184. (a), (b)  Let z = x + iy, then Re (z) = x, Im (z) = y Now, | z – i Re (z) | = | z – Im (z) |

Objective Mathematics

189. (a) r th term of the given series

⇒ | x + iy – ix | = | x + iy – y | ⇒ x + (y – x) = (x – y) + y 2

2

2

 his is a real number irrespective of the values of T n1 and n2. = r [(r + 1) – ω] [(r + 1) – ω2] = r [(r + 1)2 – (ω + ω2) (r + 1) + ω3] = r [(r + 1)2 – (– 1) (r + 1) + 1] = r [r2 + 3r + 3] = r3 + 3r2 + 3r. Thus, sum of the given series

2

⇒ y2 = x2 or y = ± x. Thus, z lies on y = ± x. 185. (c) Given | z – 1 | + | z + 3 | ≤ 8 ∴ z lies inside or on the ellipse whose foci are (1, 0) and (– 3, 0) and vertices are (– 5, 0) and (3, 0).

( n −1)

=

∑ (r

3

+ 3r 2 + 3r )

r =1

= 1 (n − 1) 2 n 2 + 3 ⋅ 1 (n − 1)(n)(2n − 1) + 3 ⋅ 1 (n − 1) n 4 6 2 1 1 3 = (n – 1) (n)  (n − 1) n + (2n − 1) +  2 2 4 = 1 (n – 1) n [n2 – n + 4n – 2 + 6] 4 Now, | z – 4 | is distance of z from (4, 0). Minimum distance is 1 and maximum is 9. 8π 8π 186. (b) We have, α = cos   + i sin   = 8 π i  11   11  e 11 2 3 4 5 Re (α + α + α + α + α )

190. (d) We have, (1 + ω – ω2)7 = (– ω2 – ω2)7 = (– 2)7 (ω2)7 = – 128ω14 = – 128ω2. 6i −3i 1 191. (d) We have, x + iy = 4 3i −1 20 3 i

2 3 4 5 2 3 4 5 = α+α +α +α +α +α+α +α +α +α 2



= 1 (n – 1) n (n2 + 3n + 4). 4

2 3 4 5 2 3 4 5 = −1 + (1 + α + α + α + α + α + α + α + α + α + α ) 2

= −1 + 0  2 =–

= (– 3i) sum of 11, 11th roots of unity)

[C2 and C3 are proportional] ∴ x + iy = 0 ⇒ x = 0, y = 0.

1 . 2

187. (c) We know that ω = − 1 + 3 i 2 2  1 i 3 Thus, 4 + 5  − + 2   2 = = = =

4 4 4 4

+ + + +

6i 1 1 4 −1 −1 = 0 20 i i

334

192. (b)

n =1

 1 i 3 + 3 − + 2   2

3i=

n

+ i n +1 ) =

13

∑i

n

(1 + i )

n =1

 i (1 − i13 )   i (1 − i )  = (1 + i)   = (1 + i)   1 − i  1− i   

365

= (1 + i) i = – 1 + i.

5ω334 + 3ω365 5 (ω3)111 ω + 3 (ω3)123 ω2 5 (1)111 ω + 3 (1)123 ω2 5ω + 3ω2 = 1 + 2ω + 3 (1 + ω + ω2)

= 1 + 2ω + 3 (0) = 1 + 2ω = 1 – 1 +

13

∑ (i

193. (a) As – θ = arg (z) < 0, we take z = r [cos (– θ) + i sin (– θ)] = r (cos θ – i sin θ) 3 i.

188. (d) Using i3 = – i, i5 = i and i7 = – i, we can write the given expression as (1 + i ) n1 + (1 − i ) n1 + (1 + i ) n2 + (1 − i ) n2 n n n 2 4 = 2 [ 1 C0 + 1 C 2i + 1 C 4i + ...]

+ 2 [ n2 C0 + n2 C 2i 2 + n2 C 4i 4 + ...] n1 n1 n1 = 2 [ C0 − C 2 + C 4 + ...]

+ 2 [ n2 C0 − n2 C 2 + n2 C 4 − ...]

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

194. (a) Since | z1 | = | z2 | = | z3 | = 1,

=

we get z1  z = z2  z3 = z3  z3 = 1. Now, 1 =

z1 + z2 + z3

=

555

ω3 = ω 2. ω

198. (d) sin x + i cos 2x and cos x – i sin 2x are conjugate of each other

1

1 1 1 + + z1 z2 z3

 x + yω + zω2  1  2 = ω  x + yω + z ω 

1 = ω

= |   z1 + z2 + z3  |

if ⇒ ⇒ ⇒

= | z1 + z2 + z3 |.

195. (a) Given that i z3 + z2 – z + i = 0 Dividing both sides by i we get z3 + 1 z2 – z + 1 = 0. i i

sin x + i cos 2x = cos x − i sin 2 x sin x + i cos 2x = cos x + i sin 2x sin x = cos x and cos 2x = sin 2x tan x = 1 and tan 2x = 1, which is not possible for any values of x.

199. (a) We have, | az1 – bz2 |2 + | bz1 + az2 |2

But 1 = i = – i i i2

= (az1 – bz2) (a z1 – b z2 ) + (bz1 + az2) (b z1 + a z2 )

⇒ z3 – i z2 + i z + 1 = 0 ⇒ (z – i) (z2 + i) = 0 ⇒ either z – i = 0 or z2 + i = 0 ⇒ z = i or z2 = – i ⇒ | z | = | i | = 1 or | z2 | = | – i | = 1 i.e., | z |2 = 1 or | z | = 1.

[ ∵ | z |2 = z z ]



= a2 z1  z1 – ab z1  z2 – ba z2  z1 + b2 z2  z2 + b2 z1  z1 + ab z1  z2 + ab z2  z1 + a2 z2  z2 = a2 | z1 |2 + b2 | z2 |2 + b2 | z1 |2 + a2 | z2 |2

= (a2 + b2) [ | z1 |2 + | z2 |2]. 196. (a) A (z1), B (z2), C (z3) lie on | Z | = 2 whose centre is 200. (a), (c)  Since | z1 + z2 | = | z1 | + | z2 |, ∴ P, O, Q where P at O (0, 0) and radius 2. is the affix of z1, O is the affix of origin and Q the π affix of z2 must lie in the same straight line. Thus, z1 = 1 + i 3 hence | z1 | = 2 and Arg (z1) = 3 arg z – arg z = ± π. 1

2

x x   sin + cos − i tan x   2 2 201. (a)   is real x 1 + 2i sin   2

I n turn | z2 | = | z3 | = 2 and Arg (z2) = Arg (z1) + 120º = 180º ∴ z2 = – 2 Further Arg (z3) = Arg (z2) + 120º = 300º

 Hence z3 = 2 cos 

π   2π −  + i sin 3

π    2π −   3 

1 i 3 π π = 2 cos − i sin  = 2  − =1– 2  2 3 3 

3i

Thus, z2 = – 2 and z3 = 1 – i 3 . 197. (a) Cube roots of p are – p1/3, – p1/3 ω, – p – 1/3 ω2.

x x x   sin 2 + cos 2 − i tan x  1 − 2i sin 2  is real if x 1 + 4 sin 2 2 or if

2 sin2

or if

sin

∴

− p ( x + yω + z ω ) xα + yβ + z γ = − p1/ 3 ( xω + yω2 + z ) xβ + y γ + zα x + yω + z ω  xω + yω2 + zω3

2

where x1 ≠ x2, y1 ≠ y2 and x12 + y12 = x22 + y22 . z1 + z2 ( x1 + x2 ) + i ( y1 + y2 ) Now z − z = ( x − x ) + i ( y − y ) 1 2 1 2 1 2 =

=

[ ∵ ω3 = 1]

[( x

1

+ x2 ) + i ( y1 + y2 )] [ ( x1 − x2 ) − i ( y1 − y2 )] ( x1 − x2 ) 2 + ( y1 − y2 ) 2

[( x12 − x22 ) + ( y12 − y22 )] + i [ x1 y1 − y1 x2 + y2 x1

2



x x =0 ⇒ = nπ ⇒ x = 2nπ. 2 2

202. (d) Let z1 = x1 + iy1 and z2 = x2 + iy 2

Let α = – p1/3, β = – p1/3 ω, γ = – p1/3 ω2 1/ 3

x + sin x + tan x = 0 2

=

− y2 x2 − x1 y1 + x1 y2 − x2 y1 + x2 y2 ] ( x1 − x2 ) 2 + ( y1 − y2 ) 2

Complex Numbers

∴  arg (– z) = π – θ Thus, arg (– z) – arg (z) = π – θ + (θ) = π.

556

2 i ( x1 y2 − y1 x2 ) = ( x − x )2 + ( y − y )2 1 2 1 2

207. (a) We have,

1 1 1 + − 1 + 2ω 2 + ω 1 + ω

Objective Mathematics

x1 y1 = a purely imaginary or 0 if x = y . 2 2

=

1 1+ ω − 2 − ω 1 1 + − = 1 + 2ω (2 + ω )(1 + ω ) 1 + 2ω (2 + ω )(1 + ω )

x1 y1 If x = y then x1 + iy1 = k (x2 + iy2). 2 2

=

2 + ω + 2ω + ω 2 − 1 − 2ω 1 + ω + ω2 = (1 + 2ω )(2 + ω )(1 + ω ) (1 + 2ω )(2 + ω )(1 + ω )

If k = 1, z1 = z2, which is not true and if k ≠ 1, | z1 | ≠ | z2 |.



z1 + z2 z1 − z2 is purely imaginary.

∴

208. (d) Putting z = x + iy in z2 + | z |2 = 0, we get

203. (b) If z = x + iy, then the area of triangle ABC

1 2

=

x −y x− y

1 y 1 x x+ y 1

⇒ and is a

[ ∵ area is always positive]

= 1 + 6i + 15i2 + 20i3 + 15i4 + 6i5 + i6 + 1 – i3 – 3i + 3i2 = 1 + 6i – 15 – 20i + 15 + 6i – 1 + 1 + i – 3i – 3 = – 2 – 10i. 210. (d) We have, x+

Now

∴ (z2 – z3) = ± i (z1 – z3) ∴

⇒ (z2 – z3)2 = – (z1 – z3)2 2 2 ⇒ z22 + z32 − 2 z2 z3 = − z1 − z3 + 2 z1 z3

z + z − 2 z1 z2 = 2 ( z1 z3 − z1 z2 + z2 z3 − z ) 2 3

⇒ (z1 – z2)2 = 2 [z1 – z3] [z3 – z2]. 205. (c) Since a = cos θ + i sin θ, 1 + cos θ + i sin θ 1+ a ∴ = 1 − cos θ − i sin θ 1− a





=

1 = 2 cos θ ⇒ x2 – 2x cos θ + 1 = 0 x

⇒ x = cos θ ± i sin θ ⇒ xn = cos nθ ± i sin nθ

1 | z |2 . 2 204. (b) Since | CA | = | CB | and ∠ACB = 90º =

2 2

(x + iy)2 + x2 + y2 = 0 2x2 + 2ixy = 0 ⇒ x2 = 0 and xy = 0 ⇒ x = 0 y may have any value ⇒ z = iy for all y ∈ R, solution.

209. (a) (1 + i)6 + (1 – i)3

[∵ Vertices have (x, y), (– y, x), (x – y, x + y) as coordinates] x y 1 1 1 (x2 + y2) = −y x 1 = 2 2 0 0 −1

2 1

0 = (1 + 2ω )(2 + ω )(1 + ω ) = 0.

[(1 + cos θ) + i sin θ] [(1 − cos θ) + i sin θ] [(1 − cos θ) − i sin θ] [(1 − cos θ) + i sin θ]

θ θ i ⋅ 2 sin ⋅ cos 2i sin θ 2 2 = = (1 − cos θ) 2 + sin 2 θ 2 θ 2 sin 2 θ . = i cot 2

206. (b) As ω is the nth root of unity so ωn – 1 = 0 ⇒ (ω – 1) (1 + ω + ω2 + ... + ωn – 1) = 0 Hence, 1 + ω + ω2 + ... + ωn – 1 = 0 or ω – 1 = 0 i.e., ω = 1.

1 1 = cos θ ± i sin θ = cos θ  ∓ i sin θ x 1 = cos nθ ∓ i sin nθ xn

∴ xn +

1 = 2 cos nθ. xn

211. (c) Put z = x + iy, we get

( x − 2) + iy ( x − 3) + iy

=2

⇒ (x – 2)2 + y2 = 4 [(x – 3)2 + y2] ⇒ x2 + y2 – 4x + 4 = 4 [x2 + y2 – 6x + 9] ⇒ 3 (x2 + y2) – 20x + 32 = 0 20 32 x + = 0. ⇒ x2 + y2 – 3 3  10  Its centre is  , 0  3 and radius =

100 32 − = 9 3

100 − 96 2 = . 9 3

212. (b) Since the triangle is equilateral ∴ | z1 – 0 | = | z2 – z1 | = | 0 – z2 | ⇒ | z1 |2 = | z2 – z1 |2 = | z2 |2 ⇒ z1  z1 = (z2 – z1) ⋅ ( z2 – z1 ) = z2  z2

Also, z2  z2 = (z2 – z1) ( z2 – z1 ) = (z2 – z1)

⇒ ⇒ ⇒ ∴

( z1 − z2 ) z2 z1

∴ z1 z2 = (z2 – z1) (z1 – z2)

557

(1 + y)2 + x2 = x2 + (y – 1)2 1 + y2 + 2y + x2 = x2 + y2 – 2y + 1 4y = 0 ⇒ y = 0 which is x-axis. z lies on the real axis.

= – 1 + 1 – 1 + ... to (2n + 1) terms = – 1. ( ∵ no. of terms is odd)

⇒ z12 + z22 = z1 z2. 213. (d) Since α + iβ = tan – 1 z = tan – 1 (x + iy)

219. (b) Let z1 = x1 + iy1

∴ α – iβ = tan– 1 (x – iy) ∴ 2α = tan– 1 (x + iy) + tan– 1 (x – iy)

∴

x + iy + x − iy 2x = tan– 1 1 − ( x + iy )( x − iy ) 1 − ( x2 + y 2 )

∴ tan 2α =

=1

218. (a) We have, i2 + i4 + i6 + ... (2n + 1) terms

= z2 z1 – z12 − z22 + z1 z2

= tan– 1

1 + y − ix x + ( y − 1) i

⇒

z1 = x1 – iy1

Now z2 = z1 ⇒ z1 + z2 = z1 + z1 ⇒ x1 + iy1 + x1 – iy1 = 2x1, which is real. Hence result holds if z2 = z1 .

2x 1 − ( x2 + y 2 )

220. (c) We have, max. amp (z) = amp (z2),

∴ 1 – x2 – y2 = 2x cot 2α ⇒ x2 + y2 + 2x cot 2α = 1.  | z |2 − | z | +1  0 ⇒ | z | < 5. 1 − ix = a – ib ...(1) 1 + ix Changing i to – i, we get 1 + ix = a + ib ...(2) 1 − ix Multiplying corresponding sides of (1) and (2), we get 1 = a2 + b2 i.e., a2 + b2 = 1.

215. (b) We have,

216. (c) We have, | z | = 1, or

x2 + y 2 = 1

or x2 + y2 – 1 = 0.  z − 1   ( x − 1) + iy  ×  ( x + 1) − iy     Now  =  z + 1   ( x + 1) + iy   ( x + 1) − iy  =

[ ∵ x + y – 1 = 0] 2

217. (c) Let z = x + iy 1 − i ( x + iy ) 1 − iz =  ω= x + iy − i z−i

Now, amp (z1) = θ1  15  3 = cos–1   = cos–1   25 5 and amp (z2) =



=

π + θ2 2

 15  π + sin–1   25 2

3 π + sin–1   . 5 2

=

∴  | max. amp (z) – min. amp (z) | 

=

π 3 3 + sin −1 − cos −1 2 5 5

π π 3 3 + − − cos −1 − cos −1 2 2 5 5



=



3 = π – 2 cos–1   . 5

221. (a) Let y1 = y 2 = y

( x 2 + y 2 − 1) + 2iy 2iy = . 2 2 ( x + 1) + y ( x + 1) 2 + y 2 2

min. amp (z) = amp (z1).

Since x1 < x2 ⇒ x1 < ⇒

∴ | ω | = 1

x12 + y 2
2 or a < – . 2 But a > 0 from (1), ∴ a > 2. 236. (a) Putting z = x + iy, we get

⇒

=1

=1 ⇒

x + ( y − 3) i x + ( y + 3) i

=1

x 2 + ( y − 3) 2 =1 x 2 + ( y + 3) 2

⇒ x2 + y2 – 6y + 9 = x2 + y2 + 6y + 9 ⇒ 12y = 0 ⇒ y = 0 which is x-axis. ∴ z lies on x-axis. 240. (a) We have, in = i4m + 3 = i4m ⋅ i3

= (i4) m ⋅ (– i)

= (1)m ⋅ (– i) = – i.

241. (d) We have, | z | = z + 1 + 2i

i.e., if 2 (− 2 − 0) 2 + (0 + 2 ) 2 < a 2 − 3a + 2 ± a

2

z − 3i z + 3i

x + iy − 3i ⇒ x + iy + 3i



 ince | z + 2 | = a2 – 3a + 2 represents a circle S with centre at A (– 2 , 0) and radius a 2 − 3a + 2 and | z + 2 i | < a2 represents the interior of the circle with centre at B (0, 2 ) and radius a, ∴ there will be a complex number satisfying the given condition and the given inequality if the distance AB is less than the sum or difference of the radii of the two circles.

⇒ 2 ± a
0

s 2 ps − +r = 0 q2 q

or s2 – pqs + rq2 = 0 ⇒ pqs = s2 + rq2.

(a 2 + 1) 2 (a 2 + 1) 2 (a 2 + 1) 4 ⋅ = . ∴ x2 + y2 = 2a − i 2a + i 4a 2 + 1

234. (a)

Putting in (iii), we get

...(iii) ...(iv)

⇒

x 2 + y 2 = x + iy + 1 + 2i = x + 1 + (2 + y) i

⇒

x 2 + y 2 = x + 1 and 0 = 2 + y ⇒ y = – 2

∴

x2 + 4 = x + 1

⇒ x2 + 4 = x2 + 2x + 1

⇒ 2x = 3 ⇒ x =

∴ Solution = x + iy =

3 . 2

3 – 2i. 2

z1 z2 242. (a) We have, z + z = 1 2 1 ⇒ z12 + z22 = z1 z2 ⇒ z12 + z22 + z32 = z1 z2 + z1 z3 + z2 z3, where z3 = 0 ⇒ z1, z2 and the origin (since z3 = 0) form an equilateral triangle.

(x + iy)2 + (p + iq) (x + iy) + r + is = 0 243. (a) We have, ⇒ (x2 – y2 + px – q y + r) 2 2 −4  + i (2xy + py + qx + s) = 0  2i  = 4i = 2 2   i 1 + + 2i (1 + i ) 1 + i  ...(i) ⇒ x2 – y2 + px – q y + r = 0 and 2xy + py + qx + s = 0 ...(ii) −4 −2 = = = 2i If the roots are real, then y = 0 2i i

559

∴ z1 + z3 = 2 z2 or z2 =

∴ (i) gives x2 + px + r = 0 and (ii) gives pqx + s = 0 s From (iv), x = – q

Complex Numbers

231. (c) ∵ z1, z2, z3 are in A.P.

560

 2i  ∴   = 4 i2 = – 4 1+ i 

2 2 = 2 | z1 |2 + 2 |  z1 − z2  | + 2 |  z2  | [By (i)]

Objective Mathematics

 2i  ∴   = (– 4)2 = 16. 1+ i 

= | z1 + z2 |2 + | z1 – z2 |2 + 2 | z1 + z2 | | z1 – z2 |

2

4

= 2 | z1 |2 + 2 | z2 |2 + 2 |  z12 − z22  |

8

= ( | z1 + z2 | + | z1 – z2 | )2 Taking square root of both sides, we get

244. (c) We have,

z1 + z12 − z22 + z1 − z12 − z22 = | z1 + z2 | + | z1 – z2 |.

1 − i sin α (1 − i sin α) (1 − 2i sin α) 1 + 2i sin α = (1 + 2i sin α) (1 − 2i sin α)

2 3 2 3 249. (b) a + bω + cω + d ω = 1 ⋅ ω(a + bω + cω + d ω ) 2 2 c + d ω + aω + bω ω c + d ω + aω + bω

1 − 3i sin α − 2 sin α 1 − 4i 2 sin 2 α 2



=



=

(1 − 2 sin 2 α) − 3i sin α , 1 + 42 sin 2 α

=

= 1 ⋅ 1 = 1 ⋅ ω3 = ω2 ω ω

which is purely real if sin α = 0 i.e., if α = nπ, where n is an integer.

250. (c) We have,

(a + ib) 2 (a − ib) 2 − 245. (c) Since = x + iy a − ib a + ib (a + ib) − (a − ib) a 2 + b2 3

⇒ 





⇒ (z2 + z + 1) (z2 + 1) = 0

= x + iy

∴ z = i, –i, ω, ω2

= x + iy ⇒ x = 0.

and z + iz = (x – y) + i(x + y)

246. (c) Take α = ω, β = ω

∴ αβ + α + β = ω ω + ω + ω = ω3 + ω2 ω3 + ω9 ω = 1 + ω2 + ω = 0 [Since ω9 = 1 and ω3 = 1] 5

2

5

10

247. (c) The given expression 6

3  1 3  i  − i 2  +2 2  3  1 3  i i + 2   2 2 

1  + = 2 1  2 −  − =   − 

6

252. (b) We have,

6



= 1 + 1 = 2 (since ω = 1).

248. (d) We know that | z1 + z2 |2 + | z1 – z2 |2 = 2 [ | z1 | + | z2 | ]

...(i)

Now  z1 + z12 − z22 + z1 − z12 − z22    + z1 − z12 − z22

2

⇒ 1 –

3 ≤ |z| – 1 ≤

3 ≤ |z| ≤ 1 +

3

3

Hence, the maximum value of |z| is 1 +

2

2

2 ⇒ |z|2 ≤ 2 |z| + 2 |z|

⇒ (|z| – 1)2 ≤ 3 ⇒ 3

z1 + z12 − z22

2 2 2 2 − ≤ z+ + z z z |z|

⇒ |z|2 – 2|z| + 1 ≤ 1 + 2

6

=

|z|= z+

⇒ |z| ≤ 2 +

6 1  ω2  ω =   +  2  = ω6 + 6 ω  ω  ω 



1 y 1 x x+ y 1

1 1 = [ x 2 + y 2 ] = | z |2 2 2

6

2

x 1 −y So, required area = 2 x− y

= 1 |[x (x – x – y) – y (–y – x + y) 2  + 1 (– yx – y2 – x2 + xy)]|

 −1 − 3 i     −1 + 3 i      −  2 2      +  −1 + 3 i     −1 − 3 i       −    2 2      



For each value of z, | z | = 1 251. (b) Let z = x + iy ∴ iz = –y + ix,

2

5

z4 + z3 + 2z2 + z + 1 = 0

⇒ z2(z2 + z + 1) + 1(z2 + z + 1) = 0

3

[a 3 + i 3b3 + 3a 2ib − 3ab 2 ] − [a 3 − i 3b3 − 3ia 2b − 3ab 2 ] a 2 + b2

⇒

1 (aω + bω2 + c + d ω) ⋅ ω aω + bω2 + d ω + c

2

+ 2 | z12 − ( z12 − z22 )|

253. (c) Given: z = x + iy ∴

z − 1 x + iy − 1 = z + 1 x + iy + 1



=

( x − 1) + iy ( x + 1) − iy × ( x + 1) + iy ( x + 1) − iy

3.

∴

 2y  z −1  −1  ∴ arg   = tan  2 + 2 −  x y 1  z +1    π 2y −1  ⇒ tan  2 =  2  x + y −1  4 ⇒ 254. (c)

10

∑ k =1

2y π = tan = 1  x2 + y 2 − 1 4

(given) ⇒ x2 + y2 – 2y = 1

=i∑ k =1

2k π 2k π    cos 11 − i sin 11   

 = i ∑  e k =1  10

2kπ − 11

 10  = i ∑   k = 0 

 − 2 k π    e 11  − 1 = −i   

z1 = z1 ⇒

w−w z w−w z = 1− z 1− z

=

( x − 1) + iy ( x + 1) − iy × ( x + 1) + iy ( x − 1) − iy

=

x2 − 1 + y 2 2y + i ( x + 1) 2 + y 2 ( x + 1) 2 + y 2

x2 − 1 + y 2 =0 ( x + 1) 2 + y 2 ⇒ x2 + y2 – 1 = 0 260. (c) Let z = 1 + 2 i 1− i (1 + 2 i ) (1 + i ) 1 3 = =− + i (1 − i ) (1 + i ) 2 2  n comparing with z = x + iy, we have x is negative O and y is positive, therefore it lies in the second quadrant. 261. (d) Let z = cosθ + i sin θ

⇒ w − wz − wz + wz ⋅ z = w − zw − wz + wz ⋅ z

∴

z cos θ + i sin θ = 1 − z 2 1 − (cos 2 θ + i sin 2 θ )



=

⇒ ( w − w) + ( w − w) | z |2 = 0 ⇒ ( w − w) (1− | z |2 ) = 0 ⇒ |z|2 = 1 {as, w –

|z| = 1 and z ≠ 1

w

≠ 0, since β ≠ 0}

256. (a) |z + 1| =| (z + 4) – 3| ≤|z+4|+|–3|≤3+3=6

Hence,

⇒ x2 + (y + 1)2 = 3 {x2 + (y – 1)2}

258. (d) Let z = a + ib b ∴ arg (z) = θ = tan–1   a ∵ z = a – ib  b ∴ arg ( z ) = φ = tan–1  −   a

–1

= – tan

b a =–θ  

12 + 12

⇒ |z1| =

2 , |z2| = 1 and |z3| =

2

263. (a) Given: x = a + b, y = aω + bω2, z = aω2 + bω

∴ Radius of circle = g 2 + f 2 − c = (−2) 2 − 1 = 3

and |z3| =

 ence, given complex numbers form an isosceles triH angle.

⇒ x2 + y2 + 2y + 1 = 3x2 + 3y2 – 6y + 3 Here, g = 0, f = – 2, c = 1

z lies on the imaginary axis. ie x = 0 1 − z2

2 2 2 ∴ | z1 |= 1 + 1 , | z2 |= 1

z+i | x + ( y + 1) i | =3⇒ =3 z −i | x + ( y − 1) i |

⇒ x2 + y2 – 4y + 1 = 0

cos θ + i sin θ i = 2sin 2 θ − 2i sin θ cos θ 2 sin θ

262. (d) Let z1 = 1 – i, z2 = i and z3 = 1 + i

257. (c) Let z = x + iy ∴

[∵ |z| = x2 + y2]

⇒ |z| = 1

255. (b) Let z1 = w − w z be purely real 1− z ∴

z − 1 x + iy − 1 = z + 1 x + iy + 1

Since, it is purely imaginary, therefore

2k π 2k π    sin 11 + i cos 11    10

561

259. (b) Let z = x + iy

x 2 + y 2 + 2iy − 1 x2 + 1 + 2 x + y 2

Complex Numbers

=



∴ x2 + y2 + z2 = (a + b)2 + (aω + bω2)2 + (aω2 + bω)2 

= a2 + b2 + 2ab + a2ω2 + b2ω4 + 2abω3 + a2ω4 + b2ω2 + 2abω3 = a2 + b2 + 2ab + a2ω2 + b2ω4 + 2ab + a2ω + b2ω2 + 2ab = a2 (1 + ω + ω2) + b2 (1 + ω + ω2) + 6ab = a2 (0) + b2 (0) + 6ab

(∵ 1 + ω + ω2 = 0)

= 6ab 264. (b) Let z = x + iy ∴ |z – a|2 + |z + a|2 = b2 ⇒ |(x – a) + iy|2 + |(x + a) + iy|2 = b2

270. (c) (a, b, c,)

562

⇒ (x – a)2 + y2 + (x + a)2 + y2 = b2 2x + 2y + 2a = b

⇒

2

2

2

2

Objective Mathematics

It is given that | z1| = 1, | z2| = 1 and Re ( z1 z2 ) = 0 ∴ a2 + b2 = 1 ------ (1) c2 + d2 = 1 ------ (2) and , ac + bd = 0 ------ (3)

2 2 ⇒ x + y = b − 2a 2 which represents a circle. 2

2

265. (b) 

(cos 20° + i sin 20°) (cos 75° + i sin 75°) (cos 10° + i sin 10° sin 15° − i cos 15° =

ei 20° ⋅ ei 75° ⋅ ei 10° −i (cos 15° + i sin 15°)

Putting b = –c in (1), we get a2 + c2 = 1 ⇒ | w1 | = 1 Putting a = d in (1) , we get b2 + d2 = 1 ⇒ | w2 | = 1 Now, a = d and b = –c gives ab + cd = 0 ⇒ Re = 0 ⇒ ( w1 ω2 )

π π  266. (c) We have, sin + i 1 − cos  5 5  π π π cos + i 2sin 2 10 10 10

∴

π π π cos + i sin  10  10 10  π sin π 10 = tan ∴ Amplitude, tan θ = π 10 cos 10 π ⇒ θ = 10 2 2 267. (b) a + bω + cω + a + bω + cω b + cω + aω2 c + aω + bω2 2

1 1  1  z = =− = i +1  i − 1  −i − 1

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

P

ω(a + bω + cω ) ω (a + bω + cω ) + bω + cω2 + a cω2 + a + bω 2 = ω + ω = – 1 [∵ 1 + ω + ω2 = 0] =

2

1 i −1

271. (c) Let z =

= 2 sin



From (1) λ 2 (c 2 + a 2 ) = 1 Thus, a = b, b = –c or a = –d and b = c

i 90° (0 + i ) =−e =− = −1 i i

= 2 sin

−b a = = λ, say ⇒ a = λd , b = −cλ c d

⇒ λ 2 = 1 using (2) ⇒ λ = ±1

ei 105° = − i 15° ie



From (3),

2

4 3ei�/4 A

1 1 ( z1 + z2 ) + z1 z2 + ( z1 + z2 ) − z1 z2 2 2

268. (c)

1 1 | z1 + z2 |2 + | z1 − z2 |2 2 2 1 2 = ⋅ 2[| z1 | + | z2 |2 ] =| z1 | + | z2 |  2

3

=

π/4 O

{∵ |z2| = |z|2}

269. (b) Let z = x + iy, then the given equation becomes (x + iy) (x – iy) + (2 – 3i) (x + iy)

273. (d) Let z = cosθ + sinθ, so that z cos θ + sin θ = 1 − z 2 1 − (cos 2θ + i sin 2θ)

+ (2 + 3i) (x – iy) + 4 = 0



x

=

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

⇒ x2 + y2 + 4x + 6y + 4 = 0

=

 herefore, given equation represents a circle with raT dius

i 2sin θ

Hence

⇒ x2 + y2 + 2x + 3y – 3ix + 2iy 

= 22 + 32 − 4 = 3

+ 2x – 2iy + 3ix + 3y + 4 = 0

z lies on the imaginary axis i.e., x = 0. 1 − z2

1. If n is any positive integer, then the value of equals (a) –i (c) 1

i 4 n +1 − i 4 n −1 2

(b) i (d) –1

(a) | z1 | = | z2 | (b) | z1 – z2 | = | z1 | (c) | z1 + z2 | = | z1 | + | z2 |

π 3 The area of the triangle on the argand plane formed by the complex numbers – z, iz, z – iz is 1 | z |2 (b) | z |2 (a) 2 3 (c) | z |2 (d) None of these 2 If z = 1 + i, then the multiplicative inverse of z2 is

(d) | arg z1 – arg z2 | =

2. If α, β are two different complex numebrs such that 11. β−α | α | = 1, | β | = 1, then the expression 1 − αβ equals 1 (a) 1 (b) 2 (c) 2 (d) None of these (1 + i ) n is 3. The smallest positive integer n, for which 12. (1 − i ) n − 2 a real number is (a) 2 (b) 1 (a) 2i (b) 1 – i (c) 3 (d) 4 i i (c) (d) – 4. The value of (i)i is 2 2 (a) 2 2 (b) e–π/2 (1 − i )3 13. The number is equal to 1 − i3 (c) ω2 (d) ω (a) – 2 (b) 1 1 3 5. Let ω = – + i . Then the value of the determinant (c) – 1 (d) i 2 2 1 1 1 14. Which of the following is not applicable for a complex 2 2 number ? 1 −1 − ω ω is (a) Inequality (b) Division 2 4 ω ω 1 (c) Subtraction (d) Addition. (a) 3ω (b) 3ω(ω – 1) 15. If z = – 1, then the principal value of the arg (z2/3) is (c) 3ω2 (d) 3ω(1 – ω) 10π π (b) 6. For all complex numbers z1, z2 satisfying | z1 | = 12 and (a) 3 3 | z2 – 3 – 4i | = 5, the minimum value of | z1 – z2 | is 2 π (a) 0 (b) 2 (c) or 2π (d) π 3 (c) 7 (d) 17 16. 17.

7. The region of the z-plane for which z−a = 1 (Re a ≠ 0) is z+a (a) x-axis (b) y-axis (c) the straight line x = | a | (d) None of these 8. 9. 10.

If n is any integer, then (a) 1, – 1, i, – i (c) 1, – 1 The square roots of 3 –

(a) ± ( 3 – 2i) (c) ± (2 + i)

in is (b) i, – i (d) i 4i are

(b) ± ( 3 + 2i) (d) ± (2 – i)

18. If ω is one of the cube roots of unity other than 1, then the expression (1 – ω) (1 – ω2) (1 + ω4) (1 + ω8) is equal to  i − 1 The least positive integer n which reduce  to a (a) 0 (b) 3  i + 1  real number is (c) 1 (d) 2 (a) 2 (b) 3 19. If ω is cube root of unity, then the value of (1 + ω – ω2) (c) 4 (d) 5 (1 – ω + ω2) is (a) 1 (b) 0 Which of the following are correct for any two complex (c) 2 (d) 4 number z1 and z2? 20. If z, iz and z + iz are the vertices of a triangle whose (a) | z1 z2 | = | z1 | | z2 | area is 2 units then the value of | z | is (b) Arg (z1 z2) = (Arg z1) (Arg z2) (a) 4 (b) 2 (c) | z1 + z2 | = | z1 | + | z2 | (c) – 2 (d) None of these (d) | z1 – z2 | ≥ | z1 | – | z2 |. 2π 2π + i sin , then the quadratic equation 21. If a = cos 7 7 If the vertices of an equilateral triangle are situated at z = 0, 2 whose roots are α = a + a + a4 and β = a3 + a5 + a6 is z = z and z = z , then which of the following are true ? 1

2

n

563

Complex Numbers

EXERCISEs FOR SELF-PRACTICE

564

Objective Mathematics



(a) x2 (b) x2 (c) x2 (d) x2

– + – +

x x x x

30. The complex roots of (– 1)1/3 are

+2=0 –2=0 –2=0 +2=0

22. If xn = cos (a) 1 (c) 0

(a) 1 ± 3 i 2

π π + i sin n , then x1 x2 x3 x 4 ... xn = n 2 2 (b) – 1 (d) None of these π 2 (d) None of these (b)

π 2 24. | (1 – i) (1 + 2i) (2 – 3i) | = (c) –

(a) 13 (c)

(c) ± i (a) i (c) 1

(a) cosα + i sinα (c) eiα/2 34. If

(b) – 20 + 16i 3 (d) – 20

(b)

1 (c) 3

2 (d) 3

(b) 2sin nθ (d) sin nθ n

is an integer, then n is equal to

(a) 1 (c) 3

(b) 2 (d) 4

36. Find the value of (1 + 2ω + ω2)3n – (1 + ω + 2ω2)3n is: (a) 0 (b) 1 (c) ω (d) ω2 37. The value of i1/3 is:

1+ i is 1− i

(a) 1

(b) 1 2

(b) 2

(d)

29. Argument of 1 –

1 1 + x = 2cos θ, then xn + n is equal to: x x

1 + i 3  35. If   1 − i 3 

1 2

(a) 0

(b) cosα/2 + i sinα/2 (d) eiα

(a) 2cos nθ (c) cos nθ

3 i then x4 + 4x2 – 8x + 39 =

27. The minimum value of | 2z – 1 | + | 3z – 2 | is

28. Modulus of

n

r α sin r α where r = 1, 2, 3, ... n +i n2 n2 lim n → ∞ z1z2z3 ... zn is equal to:

25. (a + b) (aω + bω2) (aω2 + bω) = (a) 0 (b) a3 + b3 3 3 (c) 3 (a + b ) (d) 6 (a3 + b3)

(a) 20 + 16i 3 (c) – 52

(b) –i (d) –1

33. If z r = cos 

(d) 130

26. If x = 2 +

1± 3i 4 + z –100 is equal to: (d)

31. If z + z –1 = 1 then z100

(b) 130

13

−1 ± 3 i 2

1 + i  32. The smallest positive integer n for which  =1  1 − i  is: (a) 3 (b) 2 (c) 4 (d) None of these

23. The amplitude of 0 is (a) 0

(b)



2

38.

3 i is

π 3 π (c) – 3

2π 3 2π (d) – 3



(b)

(a)

3+i 3−i (b) 2 2 1+ i 3 1− i 3 (c) (d) 2 2 If z = x + iy then the area of a triangle with vertices z, iz and z + iz is: 3 (a) | z |2 (b) | z |2 2 1 1 (c) | z |2 (d) | z |2 2 4

(a)



Answers

1. 11. 21. 31.

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

2. 12. 22. 32.

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

3. 13. 23. 33.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

(b) (c) (b) (c)

6. 16. 26. 36.

(b) (a) (b) (a)

7. 17. 27. 37.

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

8. 18. 28. 38.

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

9. (a), (d) 10. (a) (b) (d) 19. (d) 20. (b) 29. (c) 30. (a)

14

Sequences and Series

CHAPTER

Summary of concepts Sequence

Arithmetic Progression (A.P.)

A succession of numbers a1, a2, ..., an formed according to some definite rule is called a sequence. A sequence is a function whose domain is the set N of natural numbers and range a subset of real numbers or complex numbers. A sequence whose range is a subset of real numbers is called a real sequence. Since we shall be dealing with real sequences only, we shall use the term sequence to denote a real sequence.

A sequence whose terms increase or decrease by a fixed number is called an arithmetic progression. The fixed number is called the common difference of the A.P. In an A.P., the first term is usually denoted by a, the common difference by d and the nth term by tn. Obviously

Notation  The different terms of a sequence are usually denoted by a1, a2, a3, ... or by t1, t2, t3, ... The subscript (always a natural number) denotes the position of the term in the sequence. The term at the nth place of a sequence, i.e., tn is called the general term of the sequence. A sequence is said to be finite or infinite according as it has finite or infinite number of terms.

a, a + d, a + 2d, ..., a + (n – 1) d, ...

Illustrations

d = tn – tn – 1. Thus, an A.P. can be written as For example, (i) 1, 3, 5, 7, 9, ... Since, 2nd term – Ist term = 3rd term – 2nd term = 4th term – 3rd term = ... = 2, the sequence 1, 3, 5, 7, ... are in A.P. whose first term is 1 and common difference is 2. (ii) 5, 3, 1, – 1, – 3, – 5, – 7, ... are in A.P. whose first term is 5 and common difference is – 2.

1. 1, 4, 7, 10, ... 19. In this sequence each term is obtained by The nth term of an Arithmetic Progression  If a is the first term and d is the common difference of an A.P., then its nth adding 3 to the previous term. 2. 2, – 4, 8, – 16, ... In this sequence each term is obtained by term tn is given by multiplying the preceding term by – 2. tn = a + (n – 1) d 3. 2, 3, 5, 7, 11, 13, ... This is the sequence of prime num- Notes: bers. (i) If an A.P. has n terms, then the nth term is called the last Note that sequence (1) is a finite sequence whereas others term of A.P. and it is denoted by l. That is are infinite sequences. l = a + (n – 1) d. (ii) Three numbers a, b, c are in A.P. if and only if Series b – a = c – b, i.e., if and only if a + c = 2b. A series is obtained by adding or subtracting the terms of a se- (iii) If a is the first term and d the common difference of an A.P. having m terms, then nth term from the end is quence. (m – n + 1)th term from the beginning. Thus A series is finite or infinite according as the number of nth term from the end = a + (m – n) d. terms in the corresponding sequence is finite or infinite. (iv) Any three numbers in an A.P. can be taken as a – d, a, a + d. Any four numbers in an A.P. can be taken Progressions as a – 3d, a – d, a + d, a + 3d. Similarly 5 numbers in A.P. can be taken as a – 2d, a – d, a, a + d, a + 2d. If the terms of a sequence follow certain pattern, then the sequence is called a progression. Following are the three special Sum of n terms of an A.P.  The sum of n terms of an A.P. types of progressions: with first term ‘a’ and common difference ‘d’ is given by (i) Arithmetic Progression (A.P.) n Sn = 2 [2a + (n – 1) d] (ii) Geometric Progression (G.P.) (iii) Harmonic Progression (H.P.)

566

Objective Mathematics

Notes: (i) If Sn is the sum of n terms of an A.P. whose first term is ‘a’ and last term is l, then n Sn = (a + l). 2 (ii) If common difference d, number of terms n and the last term l, are given then n Sn = [2l – (n – 1) d] 2

Now, b or and

(i) (ii)

If a1, a2, a3, ..., an are in A.P., then (a) a1 + k, a2 + k, ..., an + k are also in A.P. (b) a1 – k, a2 – k, ..., an – k are also in A.P. (c) ka1, ka2, ..., kan are also in A.P. a a a (d) 1 , 2 , ..., n , k ≠ 0 are also in A.P. k k k If a1, a2, a3, ... and b1, b2, b3, ... are two A.P.s, then (a) a1 + b1, a2 + b2, a3 + b3, ... are also in A.P. (b) a1 – b1, a2 – b2, a3 – b3, ... are also in A.P. (c) a1b1, a2b2, a3b3, ... are also in A.P. a1 a2 a3 (d) b , b , b , ... may not be in A.P. 1 2 3 If a1, a2, a3, ..., an are in A.P., then (a) a1 + an = a2 + an – 1 = a3 + an – 2 = ...

= a + (n + 2 – 1) d = a + (n + 1) d b−a , where d is common difference of A.P. n +1 b−a A1 = a + d = a +  ,  n +1  d=

b−a A2 = a + 2d = a + 2  ,  n +1 

(iii) tn = Sn – Sn – 1.

Properties  of  A.P.

= (n + 2)th term of A.P.







 b−a An = a + nd = a + n  .  n +1 

Note:  The sum of n arithmetic means between two given numbers is n times the single A.M. between them, i.e., if a and b are two given numbers and A1, A2, ..., An are n arithmetic means between them, then a+b A1 + A2 + ... + An = n  .  2 

Geometric Progression (G.P.)

A sequence (finite or infinite) of non-zero numbers in which every term, except the first one, bears a constant ratio with its preceding term, is called a geometric progression. (iii) The constant ratio, also called the common ratio of the G.P., is usually denoted by r. For example, in the sequence, 1, 2, 4, 8, ..., a + ar + k (b) ar = r − k , 0 ≤ k ≤ n – r. 2 4 8 2 = = = ... = 2, which is a constant. (iv) If nth term of a sequence is a linear expression is n then the 1 2 4 sequence is an A.P. Thus, the sequence is a G.P. whose first term is 1 and the com (v) If the sum of first n terms of a sequence is a quadratic ex- mon ratio is 2. pression in n, then the sequence is an A.P.

Arithmetic Mean (A.M.) Single Arithmetic Mean  A number ‘A’ is said to be the single A.M. between two given numbers a and b provided a, A, b are in A.P. For example, since 2, 4, 6 are in A.P., therefore, 4 is the single A.M. between 2 and 6. n-Arithmetic Means  The numbers A1, A2, ..., An are said to be the n arithmetic means between two given numbers a and b provided

a, A1, A2, ..., An , b are in A.P.

For example, since 2, 4, 6, 8, 10, 12 are in A.P., therefore, 4, 6, 8, 10 are the four arithmetic means between 2 and 12. Inserting Single A.M. between Two given Numbers  Let a and b be two given numbers and A be the A.M. between them. Then, a, A, b are in A.P. Thus a+b A – a = b – A or 2A = a + b,  or  A = . 2 Inserting n-Arithmetic Means between Two given Numbers  Let A1, A2, ..., An be the n arithmetic means between two given numbers a and b. Then a, A1, A2, ..., An, b are in A.P.

nth Term of a G.P.

If a is the first term and r is the common ratio of a G.P., then its nth term tn is given by tn = ar n – 1. Notes: (i) If a is the first term and r is the common ratio of a G.P., then the G.P. can be written as a, ar, ar 2, ..., arn – 1, ... (a ≠ 0). (ii) If a is the first term and r is the common ratio of a finite G.P. consisting of m terms, then the nth term from the end is given by ar m – n. (iii) The nth term from the end of a G.P. with last term l and l common ratio r is ( n −1) . r a (iv) Three numbers in G.P. can be taken as , a, ar; r Four numbers in G.P. can be taken as a a 3 , , ar, ar 3; r r Five numbers in G.P. can be taken as a a 2 , , a, ar, ar2. r r

G b or G2 = ab, or G = = a G



ab

[∴ G > 0]

Inserting n-Geometric Means between Two given Numbers  Let G1, G2, G3, ..., Gn be the n geometric means Sum of n terms of a G.P. between two given numbers a and b. Then, a, G1, G2, G3, ..., Gn, The sum of first n terms of a G.P. with first term a and common b are in G.P. ratio r is given by a (r n − 1) Now, b = (n + 2)th term of G.P. Sn = r − 1 , r ≠ 1. = ar n + 1, where r is the common ratio Notes: (i) When r = 1, Sn = a + a + a + ... upto n terms = na. (ii) If l is the last term of the G.P., then lr − a , r ≠ 1. Sn = r −1

Sum of an Infinite G.P. when | r | < 1 The sum of an infinite G.P. with first term a and common ratio r is a ; when | r | < 1  i.e.,  – 1 < r < 1. S∞ = 1− r

Properties of G.P. (i) If a1, a2, a3, ... are in G.P., then • a1k, a2k, a3k, ... are also in G.P. a a a • 1 , 2 , 3 , ... are also in G.P. k k k 1 1 1 • a , a , a , ... are also in G.P. 1 2 3 • ak1, ak2, ak3 ... are also in G.P. (ii) If a1, a2, a3, ... and b1, b2, b3, ... are two G.P.s, then • a1b1, a2b2, a3b3, ... are also in G.P. a1 a2 a3 • b , b , b , ... are also in G.P. 1 2 3 • a1 ± b1, a2 ± b2, a3 ± b3, ... may not be in G.P. (iii) If a1, a2, a3, ..., an are in G.P., then • a1 an = a2 an – 1 = a3 an – 2 = . . . ar − k ar + k , 0 ≤ k ≤ n – r. (iv) If a1, a2, a3, ..., is a G.P. of positive terms, then log a1, log a2, log a3, ..., is also an A.P. and vice-versa.

• ar =

Geometric Mean (G.M.) Single Geometric Mean  A number G is said to be the single geometric mean between two given numbers a and b if a, G, b are in G.P. For example, since 2, 4, 8 are in G.P., therefore 4 is the G.M. between 2 and 8.

1

or

r

n+1

b  b  n +1 = or r =   a a 1

and



 b  n +1 G1 = ar = a   a 2  b  n +1 G2 = ar 2 = a   a 



 2

 b  n +1 Gn = ar = a   a n

Notes: (i) The product of n geometric means between two given numbers is nth power of the single G.M. between them i.e., if a and b are two given numbers and G1, G2, ..., Gn are n geometric means between them, then G1G2G3 ... Gn = ( ab )n. (ii) If A and G are respectively arithmetic and geometric means between two positive numbers a and b then (a) A > G (b) the quadratic equation having a, b as its roots is x 2 – 2Ax + G2 = 0 (c) the two positive numbers are A ± A 2 − G 2 .

Harmonic Progression (H.P.) A sequence of non-zero numbers a1, a2, a3, ... is said to be a harmonic progression if the sequence 1 1 1 , ... is an A.P. , , a1 a2 a3 1 1 1 , ... is a H.P., since For example, the sequence 1, , , 4 7 10 the sequence obtained by taking reciprocals of its corresponding terms i.e., 1, 4, 7, 10, ... is an A.P. 1 1 1 , , ... A general H.P. is , a a + d a + 2d

nth Term of an H.P.

1 nth term of H.P. = nth term of the corresponding A.P. n-Geometric Means  The numbers G1, G2, ..., Gn are said to be the n geometric means between two given positive numbers a Note: 1 1 1 and b if a, G1, G2, ..., Gn , b are in G.P. (i) Three numbers a, b, c are in H.P. if and only if , , For example, since 1, 2, 4, 8, 16 are in G.P., therefore 2, 4, 1 1 1 2ac a b c 8 are three geometric means between 1 and 16. + = 2 ⋅  ,  i.e.,  b = . are in A.P. i.e., a c b a+c Inserting Single G.M. between Two given Numbers (ii) No term of H.P. can be zero Let a and b be two given positive numbers and G be the G.M. (iii) There is no general formula for finding the sum of n terms of H.P. between them. Then a, G, b are in G.P. Thus

567

b c = i.e., if and only if b2 = ac. a b

Sequences and Series

(v) Three numbers a, b, c are in G.P. if and only if

568

(iv) Reciprocals of terms of H.P. are in A.P. and then properties of A.P. can be used.

Objective Mathematics

Harmonic Mean (H.M.) Single Harmonic Mean  A number H is said to be the single harmonic mean between two given numbers a and b if a, H, b are in H.P. 1 1 1 1 For example, since , , are in H.P., therefore, is 3 2 3 4 1 1 and . the H.M. between 2 4 n-Harmonic Means  The numbers H1, H2, ..., Hn are said to be the n harmonic means between two given numbers a and b if a, H1, H2, ..., Hn, b are in H.P. That is .

1 1 1 1 1 , , are in A.P. , , ... a H1 H 2 Hn b

a+b ,G= 2 2 (ii) G = AH (iii) A ≥ G ≥ H. (i) A =

ab and H =

2ab a+b

Some Special Sequences 1. The sum of first n natural numbers n (n + 1) . Σn = 1 + 2 + 3 + ... + n = 2 2. The sum of squares of first n natural numbers n (n + 1)(2n + 1) . Σn2 = 12 + 22 + 32 + ... + n2 = 6 3. The sum of cubes of the first n natural numbers 2  n (n + 1)  3 3 3 3 3 Σn = 1 + 2 + 3 + ... + n =   .  2  Notes:  If nth term of a sequence is

Inserting Single H.M. between Two given Numbers Tn = an3 + bn2 + cn + d, Let a and b be two given numbers and H be the H.M. between then the sum of n terms is given by them. Then, a, H, b are in H.P. That is Sn = ΣTn = aΣn3 + bΣn2 + cΣn + Σd, 2ab . H= which can be evaluated using the above results. a+b Inserting n-Harmonic Means between Two given Numbers  Let H1, H2, ..., Hn be the n harmonic means between two given numbers a and b. Then, a, H1, H2, ..., Hn, b are in H.P. Thus 1 1 , 1 , ... 1 1 , H n , b are in A.P. a H1 H 2 1 Now, = (n + 2)th term of A.P. b 1 + (n + 2 – 1) d, = a where d is common difference of A.P. Thus, we get 1 1 − a−b b a = ab (n + 1) . d= n +1 1 a−b bn + a 1 1 Now,  H = +d= + ab (n + 1) = ab (n + 1) a a 1 ab (n + 1) or H1 = bn + a 1 1 1 2 ( a − b) 2a + (n − 1) b H 2 = a + 2d = a + ab (n + 1) = ab (n + 1) ab (n + 1) or H2 = 2a + (n − 1) b





In general, we have Hn =

ab (n + 1) . na + b

Relation between A.M., G.M. and H.M.

Arithmetico-Geometric Sequence (A.G.S.) A sequence is said to be an arithmetico geometric sequence if its each term is the product of the corresponding terms of an A.P. and a G.P. In other words, if a1, a2, a3, ... is an A.P. and b1, b2, b3, ... is a G.P., then the sequence a1b1, a2b2, a3b3, ... is an arithmetico-geometric sequence. Notes: (i) The general form of an arithmetico-geometric sequence is a, (a + d) r, (a + 2d) r 2, ... The nth term of this sequence is: [a + (n – 1) d] rn – 1. (ii) The sum of n terms of an A.G. sequence is given by a dr dr n [ a + (n − 1) d ] r + − Sn = (1 − r ) (1 − r ) 2 (1 − r ) 2 − , r ≠ 1. (1 − r ) n

(iii) The sum of an infinite A.G. sequence is given by a dr S∞ = (1 − r ) + (1 − r ) 2 , where | r | < 1.

Method of Differences Suppose a1, a2, a3, ... is a sequence such that the sequence a2 – a1, a3 – a2, ... is either an A.P. or a G.P. The nth term ‘a’n of this sequence is obtained as follows: S = a1 + a2 + a3 + ... + an–1 + an S = a1 + a2 + ... + an–2 + an–1 + an ⇒ an = a1 + [a2 – a1) + (a3 – a2) + ... + (an – an–1)] Since the terms within the brackets are either in an A.P. or in a G.P., we can find the value of an, the nth term, We can now

Let A, G, and H be arithmetic, geometric and harmonic means between two numbers a and b. Then, find the sum of the n terms of the sequence as S =

n

∑a k =1

k

.

Choose the correct alternative in each of the following problems: 1. If the pth term of an A.P. is q and the qth term is p, then its (p + q)th term is (a) 0 (c) p + q

(b) p – q (d) None of these

2. The number of odd numbers between 60 and 360 is (a) 148 (c) 153

(b) 150 (d) None of these

3. The number of numbers lying between 100 and 500 that are divisible by 7 but not by 21 is (a) 57 (c) 38

(b) 19 (d) None of these

4. The middle term in the following arithmetic progression 20, 16, 12, ..., – 176 is (a) – 46 (c) – 80

(b) – 76 (d) None of these

5. In the following two A.P.s, 2, 5, 8, 11, ... to 60 terms

and 3, 5, 7, 9, ... to 50 terms, the number of terms that are identical is (a) 17 (b) 15 (c) 19 (d) None of these

(a) 25 (c) 31

(b) 36 (d) None of these

12. If the nth term of a series is 3 + n , then the sum of 105 4 terms is (a) 1470 (c) 1530

(b) 1360 (d) None of these

13. The sum of first 24 terms of the A.P. a1, a2, a3, ... if it is known that a1 + a5 + a10 + a15 + a20 + a24 = 225, is (a) 865 (c) 930

(b) 900 (d) None of these

14. If a, b, c be respectively the sum of p, q and r terms of an A.P., then a b (q – r) + (r – p) + c (p – q) is equal to p q r (a) 1 (c) 0

(b) – 1 (d) None of these

15. The maximum sum of the series 20 + 19 1 + 18 2 + 18 + ... is 3 3

(a) 310 (b) 290 6. In the series 3, 7, 11, 15, ... and 2, 5, 8, ... each continued (c) 320 (d) None of these to 100 terms, the number of terms that are identical 16. The sum of all natural numbers less than 200, that are is divisible neither by 3 nor by 5, is (a) 21 (b) 27 (a) 10730 (b) 10732 (c) 25 (d) None of these (c) 15375 (d) None of these

7. If m times the mth term of an A.P. is equal to n times 17. The four numbers in A.P., such that their sum is 50 and the nth term, then its (m + n)th term is greatest of them is 4 times the least, are (a) 1 (b) – 1 (a) 7, 11, 15, 19 (b) 6, 11, 16, 21 (c) 0 (d) None of these (c) 5, 10, 15, 20 (d) None of these 8. If a, b, c be the pth, qth and rth terms respectively of an A.P., then a (q – r) + b (r – p) + c (p – q) = (a) a (c) b

(b) 0 (d) c

18. If the sum of four integers in A.P. is 24 and their product is 945, then the numbers are (a) 3, 5, 7, 9 (c) 4, 8, 12, 16

(b) 5, 8, 11, 14 (d) None of these

9. If the first, second and the last terms of an A.P. are a, 19. If the first and the last terms of an A.P. are – 4 and 146 b, c respectively, then the sum is and the sum of the A.P. is 7171, then the number of (a + b)(a + c − 2b) (b + c)(a + b − 2c) terms in the A.P. and the common difference is (b) (a) 2 (b − a ) 2 (b − a ) (a) 101 and 2 (b) 98 and 3 (a + c)(b + c − 2a ) 2 (c) (d) None of these 2 (b − a ) 3 (c) 101 and (d) None of these 2 10. Four different integers form an increasing A.P. If one of these numbers is equal to the sum of the squares of the 20. The sum to n terms of the sequence other three numbers, then the numbers are log a, log ar, log ar2, ... is (a) – 2, – 1, 0, 1 (b) 0, 1, 2, 3 (b) n log a2 rn – 1 (a) n log a2 rn – 1 (c) – 1, 0, 1, 2 (d) None of these 2 11. The number of terms in the series 20, 19 1 , 18 2 , ... (c) 3n log a2 rn – 1 (d) None of these 3 3 2 of which the sum is 300, is

569

Sequences and Series

 MULTIPLE-CHOICE QUESTIONS

570

21. The sum of 11 terms of an A.P. whose middle term is 30, is

Objective Mathematics

(a) 320 (c) 340

(b) 330 (d) 350

22. The sum of an A.P. is 525. If its first term is 3 and last term is 39, then the common difference is (a) 3 (b) 1 2 (c) 1 (d) None of these 2 23. The common difference of an A.P., whose first term is 100 and the sum of whose first six terms is five times the sum of the next six terms, is (a) 10 (c) 5

(b) – 10 (d) – 5

24. The interior angles of a polygon are in A.P. If the smallest angle is 120º and the common difference is 5, then the number of sides of the polygon is (a) 16 (c) 9

(b) 8 (d) None of these

25. The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the l 2 − a2 , then k is equal to common difference is k − (l + a ) (a) S (c) 3S

(b) 2S (d) None of these

26. If x18 = y21 = z28, then 3, 3 logy x, 3 logz y, 7 logx z are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

27. If the sum of any number of terms in a sequence is always three times the squared number of these terms, then the sequence is (a) an A.P. (c) an H.P.

(b) a G.P. (d) None of these

28. The sum of all two digit numbers which when divided by 4, yield unity as remainder, is (a) 1100 (c) 1210

(b) 1200 (d) None of these

29. The sum of numbers of three digits which are divisible by 7, is (a) 70336 (c) 70330

(b) 70331 (d) None of these

30. The sum of all odd numbers of four digits which are divisible by 9, is (a) 2754,000 (c) 2752,000

(b) 2753,000 (d) None of these

31. Let Sn denotes the sum of n terms of an A.P. whose first term is a. If the common difference d = Sn – k Sn – 1 + Sn – 2 then k = (a) 1 (c) 3

(b) 2 (d) None of these

32. If the sum of the first n terms of a sequence is of the form An2 + Bn where A, B are constants independent of n, then the sequence is (a) an A.P. (c) an H.P.

(b) a G.P. (d) None of these

33. If the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k is equal to (a) 1 (b) n −1 n n n + 1 n (c) (d) + 1 2n n 34. The sum of positive terms of the series 10 + 9 4 + 9 1 + ... is 7 7 (b) 437 (a) 352 7 7 852 (c) (d) None of these 7 35. The minimum number of terms from the beginning of the series 20 + 22 2 + 25 1 + ..., so that the sum may 3 exceed 1568, is 3 (a) 25 (c) 28

(b) 27 (d) 29

36. The maximum sum of the A.P. 40, 38, 36, 34, ... is (a) 390 (c) 460

(b) 420 (d) None of these

37. If S1, S2, S3 are the sum of n, 2n, 3n terms of an A.P., then 3 (S2 – S1) = (a) S3 (c) 4S3

(b) 2S3 (d) None of these

38. If the first, second and last terms of an A.P. are a, b and 2a respectively, then its sum is (a)

ab 2(b − a )

(b)

(c)

3ab 2 (b − a )

(d) None of these

ab b−a

39. If 51 + x + 51 – x, a and 25x + 25–x are three consecutive 2 terms of an A.P., then the values of a are given by (a) a ≥ 12 (c) a < 12

(b) a > 12 (d) a ≤ 12

40. If log10 2, log10 (2x – 1) and log10 (2x + 3) be three consecutive terms of an A.P., then (a) x = 0 (c) x = log2 5

(b) x = 1 (d) x = log10 2.

41. The sum of two numbers is 2 1 . If an even number of 6 arithmetic means are inserted between them and their sum exceeds their number by 1, then number of means inserted is

42. If m arithmetic means are inserted between 1 and 31 so that the ratio of the 7th and (m – 1)th means is 5 : 9, then the value of m is (a) 9 (c) 13

(b) 11 (d) 14

43. In an A.P. of even number of terms, the sum of the odd terms is 24 while that of even terms is 30. If the last term exceeds the first by 10.5, then the number of terms is (a) 8 (c) 10

(b) 6 (d) None of these

44. If there are (2n + 1) terms in A.P., then the ratio of the sum of odd terms and the sum of even terms is (a) n : (n + 1) (c) (n – 1) : n

(b) (n + 1) : n (d) None of these

45. If S1 is the sum of an arithmetic series of ‘n’ odd number of terms and S2, the sum of the terms of the series in S odd places, then 1 = S2

51. A polygon has 25 sides, the lengths of which starting from the smallest side are in A.P. If the perimeter of the polygon is 2100 cm and the length of the largest side 20 times that of the smallest, then the length of the smallest side and the common difference of the A.P. is (a) 8 cm, 6 1 cm (b) 6cm, 6 1 cm 3 3 (c) 8cm, 5 1 cm (d) None of these 3 52. A club consists of members whose ages are in A.P., the common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250 years, then the number of members in the club are (a) 15 (c) 20

(b) 25 (d) 30

53. If four numbers in A.P. are such that their sum is 28 and the ratio of the product of first and third with the product of second and fourth is 8 : 15, then the numbers are (a) 3, 5, 7, 9 (c) 3, 6, 9, 12

(b) 4, 6, 8, 10 (d) None of these

54. If four numbers in A.P. are such that their sum is 20 and (a) 2n (b) n n +1 n +1 sum of their squares is 120, then the numbers are n + 1 n + 1 (a) 1, 4, 7, 10 (b) 3, 5, 7, 9 (c) (d) 2n n (c) 2, 4, 6, 8 (d) None of these 46. If the ratio of the sum of n terms of two A.P.s is 55. If the sum of three numbers in A.P. is 12 and the sum (3n – 13) : (5n + 21), then the ratio of 24th terms of of their cubes is 288, then the numbers are the two progressions is (a) 2, 4, 6 (b) 1, 4, 7 (a) 2 : 3 (b) 2 : 1 (d) 1, 3, 5 (d) None of these (c) 1 : 2 (d) None of these 56. If the series of natural numbers is divided into groups 47. The sum of each of two sets of three terms in A.P. is 15. (1); (2, 3, 4) : (5, 6, 7, 8, 9); ... and so on, then the The common difference of the first set is greater than sum of the numbers in the nth group is that of the second by 1 and the ratio of the products (b) n2 + (n + 1)2 (a) n3 + (n + 1)3 of the terms in the first set and that of the second set (c) (n – 1)2 + n2 (d) (n – 1)3 + n3 is 7 : 8. The two sets of numbers are 57. If a is the first term, d the common difference and Sk the (a) 3, 5, 7 and 4, 5, 6 (b) 3, 5, 7 and 7, 8, 9 S (c) 2, 4, 6 and 4, 5, 6 (d) 21, 5, – 11 and 22, 5, – 12. sum to k terms of an A.P., then for kx to be independSx 48. The least value of n such that ent of x 1 + 3 + 5 + 7 + ... to n terms ≥ 500, is (a) a = 2d (b) a = d (a) 18 (b) 22 (c) 2a = d (d) None of these (c) 23 (d) 19 58. If d is the common difference of an A.P. and Sn is the 49. A man arranges to pay off a debt of Rs. 3600 by 40 sum to its first n terms, then annual instalments which are in A.P. When 30 of the (a) d = Sn – Sn – 1 + Sn – 2 instalments are paid he dies leaving one third of the (b) d = Sn – 2Sn – 1 – Sn – 2 debt unpaid. The value of the 8th instalment is (c) d = Sn – 2Sn – 1 + Sn – 2 (a) Rs. 35 (b) Rs.50 (d) None of these (c) Rs. 65 (d) None of these 50. If The ratio of the sum of m terms and n terms of an 59. If θ1, θ2, θ3, ..., θn are in A.P. whose common difference is d, then A.P. is m2 : n2, then the ratio of its mth and nth terms is sin d [sec α1 sec α2 + sec α2 sec α3 + ... + sec αn – 1 sec αn] = (a) (2m + 1) : (2n + 1) (b) (m – 1) : (n – 1) (a) sec αn – sec α1 (b) sin αn – sin α1 (c) (2m – 1) : (2n – 1) (d) None of these (c) cos αn – cos α1 (d) tan αn – tan α1

571

(b) 8 (d) None of these

Sequences and Series

(a) 12 (c) 6

572

60. Let a1, a2, a3, ..., an be in A.P.

Objective Mathematics

If

1 1 1 + + ... + a1an a2 an −1 an a1

=

k 1 1 1  + + ... +  , then k is equal to a1 + an  a1 a2 an 

(a) 1 (c) 3

(b) 2 (d) None of these

61. If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are (a) 5, 10, 15, 20 (c) 3, 7, 11, 15

(b) 4, 10, 16, 22 (d) None of these

62. If the lengths of sides of a right angled triangle are in A.P., then their ratio is (a) 2 : 3 : 4 (c) 4 : 5 : 6

(b) 3 : 4 : 5 (d) None of these

(a) 352 (c) 852

(b) 652 (d) None of these

70. The sum of n terms of m A.P.s are S1, S2, S3, ..., Sm. If the first term and common difference are 1, 2, 3, ..., m respectively, then S1 + S2 + S3 + ... + Sm = (a) 1 mn (m + 1) (n + 1) 4 1 mn (m + 1) (n + 1) (b) 2 (c) mn (m + 1) (n + 1) (d) None of these 71. If 5th and 8th terms of a G.P. are 32 and 256 respectively, then the 4th term of the G.P. is (a) 8 (c) 16

(b) 12 (d) 20

72. If 5th, 8th and 11th terms of a G.P. are p, q and s respectively, then

(a) p2 = qs (b) q2 = ps 63. If n arithmetic means are inserted between 1 and 31 such (c) s2 = pq (d) None of these that the ratio of the first mean and nth mean is 3 : 29, 73. The 3rd term of a G.P. is the square of the first term. If then the value of n is the second term is 8, then the 6th term is (a) 10 (b) 12 (a) 128 (b) 64 (c) 13 (d) 14 (c) 32 (d) None of these 64. Between two numbers whose sum is 2 1 , an even number 74. If the four numbers forming a G.P. are such that the third 6 term is greater than the first by 9 and the second term of arithmetic means are inserted. If the sum of these is greater than the fourth by 18, then the numbers are means exceeds their number by unity, then the number (a) 3, – 6, 12, – 24 of means are (b) 2, – 8, 32, – 128 (a) 12 (b) 10 (c) 1, – 3, 9, – 27 (c) 8 (d) None of these (d) None of these 65. The sum of n arithmetic means between two numbers 75. If a, b, c are three consecutive terms of an A.P., then is 20. If the last mean is double of the first and one ka, kb, kc are three consecutive terms of number is three times the other, then the numbers are (a) an A.P. (b) a G.P. (a) 150, 100 (b) 50, 100 (c) an H.P. (d) None of these (c) 150, 50 (d) None of these 76. If pth, qth and rth terms of a G.P. are themselves in G.P., 66. If x, y, z are in A.P., then then p, q, r are in (x + 2y – z) (2y + z – x) (z + x – y) = (a) A.P. (b) G.P. (a) 4xyz (b) 2xyz (c) H.P. (d) None of these (c) xyz (d) None of these 77. If x, y, z are in G.P., then log x, log y, log z are in 67. If a, b, c are in A.P. and p is the A.M. between a and (a) A.P. (b) G.P. b and q is the A.M. between b and c, then (c) H.P. (d) None of these (a) a is the A.M. between p and q 1 1 1 (b) b is the A.M. between p and q , , are the three consecutive terms 78. If x + y 2y y + z (c) c is the A.M. between p and q of an A.P., then x, y, z are the three consecutive terms (d) None of these of 68. If 11 A.M.s are inserted between 28 and 10, then the middle term in the series is (a) 15 (c) 21

(b) 19 (d) None of these

69. The digits of a positive integer having three digits are in A.P. and their sum is 15. If the number obtained by reversing the digits is 594 less than the original number, then the number is

(a) an A.P. (c) an H.P.

(b) a G.P. (d) None of these

1 1 1 79. If a, b, c are in G.P., then a2 b2 c2  3 + 3 + 3  = a b c  (a) a + b + c (c) a3 + b3 + c3

(b) ab + ac + bc (d) None of these

1

 p p  p−q (a)  q  q 

 qq  q− p (b)  p  p 

(1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + ... upto n terms is x 2 (1 − x n )  1  (a) n −  1− x  1− x 

(d) None of these

(b)

x 3 (1 − x n )  1  n −  1− x  1− x 

(c)

1  x(1 − x n )  n −  1− x  1− x 

1

1

 p p  p+q (c)  q  q 

81. If a, b, c are respectively the xth, yth and zth terms of a G.P., then ( y – z) log a + (z – x) log b + (x – y) log c = (a) 1 (c) 0

(d) None of these

(b) – 1 (d) None of these

90. Sum of the series :

82. The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 while the third is increased by 1, three consecutive terms of a G.P. result. The three numbers are (a) 3, 7, 11 (c) 1, 7, 11

(b) 12, 7, 2 (d) None of these

83. If p, q, r are in A.P. and x, y, z are in G.P., then xq – r ⋅ yr – p ⋅ zp – q = (a) 1 (c) – 1

91. Sum of the series :

(b) 2 (d) None of these

.5 + .55 + .555 + ... upto n terms is

84. If the common ratio, the last term and the sum of a G.P. are 3, 486 and 728 respectively, then the first term and the number of terms are (a) 2, 5 (c) 2, 6

(b) 3, 6 (d) None of these

85. The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an A.P. The three numbers are (a) 8, 16, 32 (b) 10, 18, 26 (c) 9, 16, 23 (d) None of these 86. If the sum of the first three terms of a G.P. is 21 and the sum of the next three terms is 168, then the first term and the common ratio is (a) 3, 4 (c) 3, 2

9 + 99 + 999 + ... upto n terms is (a) 1 (10n – 9n – 10) 9 1 (10n + 1 – 9n – 10) (b) 9 (c) 1 (10n + 1 + 9n – 10) 9 (d) None of these

(b) 2, 4 (d) None of these

87. If the sum of the n terms of a G.P. be S, their product n

S P and the sum of their reciprocals R, then   = R (a) P (b) P2 (c) 2P2 (d) None of these

(a)

5 9

 2 1   n − 9 1 − 10n     

(b)

3 9

 1 1   n − 9 1 − 10n     

(c)

7 2 1  n − 1 − n    9 9  10  

(d)

5 9

 1 1   n − 9 1 − 10n     

92. The sum of three numbers in A.P. is 15. If 1, 3, 9 are added to them respectively, the resulting numbers are in G.P. The numbers are (a) 3, 5, 7 (c) 15, 5, – 5

(b) 2, 5, 8 (d) None of these

93. The sum of an infinite G.P., whose first term is 28 and fourth term is 4 , is 49 (a) 98 3

(b) 49 3

(c) 78 (d) None of these 3 88. If the sum of four numbers in G.P. is 60 and the 94. The product 91/3 ⋅ 91/9 ⋅ 91/27 ... to infinity is A.M. of the first and the last is 18, then the numbers (a) 9 (b) 3 are (c) 81 (d) None of these (a) 8, 12, 16, 20 95. If the sum of three numbers in G.P. is 63 and the product (b) 32, 16, 8, 4 (c) 4, 8, 16, 32 of the first and the second term is 3 of the third term, 4 (d) None of these then the numbers are

573

89. Sum of the series:

Sequences and Series

80. If pth and qth terms of a G.P. are q and p respectively, then (p + q)th term is

574

(a) 3, 12, 48 (c) 2, 10, 50

(b) 4, 12, 36 (d) None of these

Objective Mathematics

96. If, in a G.P., the (p + q)th term is a and the (p – q)th term is b, then pth term is (a) (ab)1/3 (c) (ab)1/4

(b) (ab)1/2 (d) None of these

97. If a, b, c, d are in G.P., then (a2 + b2 + c2) (b2 + c2 + d2) = (a) (ab + ac + bc)2 (c) (ab + bc + cd)2

(b) (ac + cd + ad)2 (d) None of these

98. Let a be the first term and b be the nth term of a G.P. If P is the product of n terms, then P2 = (a) ab (c) (ab)n/2

(b) (ab)n (d) None of these

99. If the continued product of three numbers in G.P. is 216 and the sum of their products in pairs is also 216, then the numbers are 12 , 6, 3 (5 ± (a) 5 ± 21 12 , 6, 3 (4 ± (b) 4 ± 17 12 , 6, 3 (7 ± (c) 7 ± 19 (d) None of these

21 ) 17 ) 19 )

100. If every even term of a series is a times the term before it and every odd term is c times the term before it, the first term being unity, then the sum to 2n terms of the series is n n (a) (1 − a )(1 − c a ) 1 − ca

n −1 n −1 (b) (1 + a )(1 − c a ) 1 − ca

n−2 n−2 (c) (1 − a )(1 − c a ) (d) None of these 1 − ca 101. The minimum number of terms of the series

1 + 3 + 9 + 27 + ... so that the sum may exceed 1000, is (a) 7 (c) 3

(b) 5 (d) None of these

102. If Sn denotes the sum of n terms of a G.P. whose first term is a and the common ratio r, then the sum S1 + S3 + S5 + ... + S2n – 1 is (a)

2a  1 − r 2n  3a  1 − r 2n  (b) n − r ⋅ n − r ⋅  2  1− r  1− r  1− r  1− r2 

1 − r 2n  a  (c) n − r ⋅  (d) None of these 1− r  1− r2  103. In a set of four numbers the first three are in G.P. and the last three are in A.P. with a common difference 6. If the first number is same as the fourth, the four numbers are (a) 3, 9, 15, 21 (c) 8, – 4, 2, 8

(b) 1, 7, 13, 19 (d) None of these

104. A number consists of three digits in G.P. If the sum of the right hand and left hand digits exceeds twice the middle digit by 1 and the sum of the left hand and middle digits is two third of the sum of the middle and right hand digits, then the number is (a) 469 (c) 468

(b) 376 (d) None of these

105. If ax = by = cz = k and x, y, z are in G.P., then (a) logb a = logc b (b) logc a = loga c (c) logb a = logb c (d) None of these 106. If S1, S2, S3, ..., Sp are the sum of the infinite geometric series whose first terms are 1, 2, 3, ..., p and whose 1 respectively, common ratios are 1 , 1 , 1 , ..., p +1 2 3 4 then S1 + S2 + S3 + ... + Sp = (a) p ( p + 1) 2 (c) p ( p + 3) 2

(b) p ( p + 2) 2 (d) None of these

107. If the first term of a G.P. exceeds the second term by 2 and the sum to infinity is 50, then the common ratio is (a) 3 5 1 (c) 5

(b) 2 5 (d) 4 5

108. If r > 1 and x = a + a + a + ... to ∞, r r2 y = b – b + b – ... to ∞ and r r2 z = c + c + c + ... to ∞, then xy = r2 r4 z (a) ab c bc (c) a

(b) ac b (d) None of these

109. Sum to infinity of the series 3 5 3 5 3 5 − + − + − +... is 4 4 2 4 3 4 4 45 4 6 (a) 7 (b) 2 15 5 (c) 1 (d) None of these 3 110. If one geometric mean G and two arithmetic means p and q be inserted between two numbers, then G2 is equal to (a) (3p – q) (3q – p) (c) (4p – q) (4q – p)

(b) (2p – q) (2q – p) (d) None of these

(a) 2n – 1 (c) 2– n + n – 1

(b) 2n – n – 1 (d) 1 – 2– n

112. If the A.M. between two numbers is 34 and their G.M. is 16, then the two numbers are (a) 64 and 8 (c) 8 and 4

(b) 64 and 4 (d) None of these

113. The A.M. between two numbers b and c is a and the two G.M.s between them are g1 and g2. If g13 + g 23 = kabc, then k is equal to (a) 1 (c) 3

(b) 2 (d) 4

114. If two geometric means g1 and g2 and one arithmetic mean A be inserted between two numbers, then g12 g2 + 2 = g2 g1 (a) 4A (c) 2A

(b) 3A (d) A

115. If x, y and z are 4th, 10th and 16th terms of a G.P., then x, y and z are in

(a) 2, 5, 8 (c) 3, 5, 7

(b) 26, 5, – 16 (d) None of these

122. If a, b, c are in G.P., then log an, log bn, log cn are in (a) A.P. (c) H.P

(b) G.P. (d) None of these

123. There are four numbers of which the first three are in G.P. whose common ratio is 1 and the last three are 2 in A.P. If the last number is two less than the first, then the four members are (b) 2, 1, 1 , 0 (a) 3, 1, 1 , – 1 3 3 2 (c) 4, 1, 1 , – 1 (d) None of these 4 2 1

1

1

124. If a x = b y = c z and a, b, c are in G.P., then x, y, z are in (a) A. P. (c) H.P.

(b) G.P. (d) None of these

125. The three numbers a, b, c between 2 and 18 are such that their sum is 25; the numbers 2, a, b are consecutive terms of an A.P. and the numbers b, c, 18 are consecu116. If a, b, c are in A.P., x is the G.M. between a and b, tive terms of a G.P. The three numbers are y is the G.M. between b and c, then b2 is (a) 3, 8, 14 (b) 2, 9, 14 (a) A.M. between x2 and y2 (c) 5, 8, 12 (d) None of these (b) G.M. between x2 and y2 , A are the two A.M.s between two numbers a and 126. If A 1 2 (c) H.M. between x2 and y2 b and G1, G2 be two G.M.s between same two numbers, (d) None of these A + A2 = then 1 117. If S1, S2, S3 be sums of a G.P. of n, 2n, 3n terms reG1 ⋅ G 2 spectively, then S1 (S3 – S2) = (a) a + b (b) a + b (a) (S2 – S1)2 (b) (S3 – S1)2 ab 2ab (c) S1 (S2 + S3) (d) None of these 2ab (c) (d) ab 118. If, in a G.P. of 3n terms, S1 denotes the sum of the first a+b a+b n terms, S2 the sum of the second block of n terms and 127. If the product of three terms of a G.P. is 512. If 8 S3 the sum of the last n terms, then S1, S2, S3 are in added to first and 6 added to second term, so that the (a) A.P. (b) G.P. numbers may be in A.P., then the numbers are (c) H.P. (d) None of these (a) 2, 4, 8 (b) 4, 8, 16 119. If the first term of an infinite G.P. is 1 and each term is (c) 3, 6, 12 (d) None of these twice the sum of the succeeding terms, then the common 128. If H is the harmonic mean between a and b, then ratio is (a) A. P. (c) H. P.

(b) G. P. (d) None of these

(a) 1 (b) 2 3 3 3 (c) (d) None of these 4 120. There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6. If the first number and the last number are equal, then the numbers are (a) 2, 4, 8, 12 (c) 3, 6, 12, 18

(b) 8, – 4, 2, 8 (d) None of these

H+a + H+b = H−a H−b (a) 1 (c) – 1

(b) 2 (d) – 2

129. If a, b, c are in A.P., p, q, r are in H.P. and ap, bq, cr are in G.P., then p + r p (c) + r

(a)

r = c+a p a c r b = +c p c b

(b)

p q + = c+a q p a c

(d) None of these

575

1 3 7 15 + + + +... is given by 2 4 8 16

121. The sum of three numbers in A.P. is 15. If 1, 4, 19 are added to them respectively, the resulting series is in G.P., then the numbers are

Sequences and Series

111. The sum to n terms of the series

576

130. If a, b, c are in G.P. and ax = by = cz, then x, y, z are in

Objective Mathematics

(a) A.P. (c) H.P.

(b) G.P. (d) None of these

131. If a, b, c are in A.P., b, c, d are in G.P. and c, d, e are in H.P., then a, c, e are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

132. If a, b, c are in H.P., then (a) 1 + 1 a b 1 (c) + 1 b c

1 1 = + b−a b−c

(b) 1 + 1 a c (d) None of these

133. If the roots of the equation x3 – 12x2 + 39x – 28 = 0 are in A.P., then their common difference is (a) ± 1 (c) ± 3

(b) ± 2 (d) ± 4

134. If S1, S2 and S3 denote the sum of first n1, n2 and n3 terms respectively of an A.P., then S1 S S (n2 – n3) + 2 (n3 – n1) + 3 (n1 – n2) = n1 n2 n3 (a) 0 (c) S1S2S3

(b) 1 (d) n1 n2 n3

135. If a, b, c, d, e, f are in A.P., then e – c is equal to (a) 2 (c – a) (b) 2 (d – c) (c) 2 ( f – d ) (d) d – c (b) 2ad (d) None of these

137. If the pth term of an H.P. is qr and qth term is pr then the rth term is (a) pq (c) 3pq

(b) 2pq (d) None of these

138. If a, b, c are in H.P. then (a) A.P. (c) H.P.

,

b , c are in c+a a+b

(b) G.P. (d) None of these

139. Given two numbers a and b. Let A denote the single A.M. and S denote the sum of n A.M.s between a and b, then S depends on A (a) n, a, b (c) n, a

(b) n, b (d) n

140. Let an be the nth term of the G.P. of positive numbers. 100

n =1

2n

= α and

the common ratio is

α β

(b) β α (d)

β α

141. The third term of a G.P. is 4. The product of first five terms is (a) 43 (c) 44

(b) 45 (d) None of these

142. If three positive numbers a, b, c are in H.P., then an + cn (a) > 2bn (c) < 2bn

(b) = 2bn (d) > bn

143. The H.M. of two numbers is 4. If their A.M. A and G.M. G satisfy the relation 2A + G2 = 27, then the numbers are (a) 2, 6 (c) 1, 3

(b) 3, 6 (d) None of these

144. If the G.M. between a and b be twice the H.M., then a is equal to b 2+ 3 2− 3 (a) (b) 2− 3 2+ 3 (c)

4+ 3 4− 3

(d)

4− 3 4+ 3

145. If x, 2x + 2, 3x + 3, ... are in G.P., then the fourth term is (b) – 27 (d) – 13.5

146. If a, b, c are in A.P. as well as in G.P., then

(a) ad (c) 3ad

∑a

(c)

α β

(a) 27 (c) 13.5

136. If a, b, c and d are in H.P., then ab + bc + cd =

Let

(a)

100

∑a n =1

2 n −1

= β, such that α ≠ β, then

(a) a = b ≠ c (c) a ≠ b ≠ c

(b) a ≠ b = c (d) a = b = c

147. If x, y, z are in G.P. and ax = by = cz, then (a) logc b = loga c (c) loga b = logc b

(b) loga c = logb a (d) logb a = logc b

148. If the A.M. and G.M. of the roots of a quadratic equation in x are p and q respectively, then its equation is (a) x2 – 2px + q2 = 0 (c) x2 – px + q = 0

(b) x2 + 2px + q2 = 0 (d) x2 – 2px + q = 0

149. If the A.M. between a and b is m times their H.M. then a :b = (a)

m + m +1 m − m +1

(b)

m − m −1 m + m −1

(c)

m + m −1 m − m −1

(d) None of these

150. If one A.M. A and two G.M.s p and q are inserted p2 q2 between two given numbers, then + = q p (a) A (c) 3A

(b) 2A (d) 4A

1

161. If a, b, c, d are four unequal positive numbers in H.P. then

1

(a) 1

(b) 2

(d) 5 . (c) 3 2 2 152. The value of x in (– π, π) which satisfies the equation 8(1+ |cos x | +| cos

2 x | + | cos3 x | + ... to ∞ )

= 43 is

(a) π (b) – π 3 3 2 π (c) (d) all of these 3 153. log3 2, log6 2, log12 2 are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

154. If a, b, c are in A.P. and a , b , c are in H.P., then 2

(a) a = b = c (c) b2 =

ac 8

2

2

(b) 2b = 3a + c (d) None of these

155. The sum of first n terms of the series 1 3 7 15 + + + +... is 2 4 8 16 (a) 2n – 1 (c) 2–n – n + 1

157. If

(b) 1 – 2–n (d) 2–n + n – 1

(b) 4 + 3 ac b 2 (d) None of these

1 1 = 1 1 , then a, b, c are in + + a c b−a b−c

(a) A.P. (c) H.P.

(b) G.P. (d) None of these

158. If one G.M. G and two A.M.s p and q be inserted between two given numbers, then (2p – q) (2q – p) = (a) G (c) 2G2 2

(b) 4G (d) None of these

159. If the A.M. of two numbers exceeds their G.M. by 2 and the G.M. exceeds their H.M. by 1 3 , then the numbers 5 are (a) 12, 4 (c) 16, 4

(b) 9, 4 (d) None of these

160. If a, b, c are three positive unequal numbers in H.P. then a5 + c5 (a) > b5 (c) > 3b5

1 1 , 162. If x > 1, y > 1, z > 1 are in G.P., then , 1 + ln x 1 + ln y 1 are in 1 + ln z (a) A.P. (c) H.P.

(b) G.P. (d) None of these

163. If the H.M. of two numbers is to their G.M. as 12 : 13, then the numbers are in the ratio (a) 4 : 9 (c) 2 : 9

(b) 9 : 4 (d) 9 : 2

164. If ax = by = cz = du and a, b, c, d are in G.P., then x, y, z, u are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

165. If S1, S2, S3, ..., Sn are the sum of infinite geometric series whose first terms are 1, 2, 3, ..., n and whose 1 respectively, common ratios are 1 , 1 , 1 , ..., (n + 1) then the value of 2 3 4 S 21 + S 22 + S 23 ... + S 22n –1 is equal to

1 1 1 1 1 1 156. If a, b, c are in H.P, then  + −   + −  = a b c b c a (a) 4 − 3 ac b 2 (c) 2 − 3 ac b 2

(a) a + d > b + c (b) a + c > b + d (c) b + c > a + d (d) None of these

(b) > 2b5 (d) None of these

(a) 1 [n (2n + 1) (4n + 1) – 3] 3 (b) 1 [n (2n + 1) (4n + 1) + 3] 3 1 (c) [n (2n – 1) (4n + 1) – 3] 3 (d) None of these 166. If a, b, c are in A.P. and a2, b2, c2 are in H.P. then (a) a = b = c

(b) – a , b, c are in G.P. 2

(c) – c , b, a are in G.P. (d) all of these 2 167. If a, b, c are in A.P. and a, mb, c are in G.P, then a, m2b, c are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

168. If x, 1, z are in A.P. and x, 2, z are in G.P. then x, 4, z are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

169. The (m + 1)th, (n + 1)th and (r + 1)th terms of an A.P. are in G.P. and m, n, r are in H.P., then the ratio of the first term to the common difference of the A.P. is (a) n 2 n (c) 3

(b) – n 2 (d) None of these

577

1

Sequences and Series

1

151. 2 4 ⋅ 4 8 ⋅ 816 ⋅ 16 32... is equal to

578

170. The product of n positive integers is 1, then their sum is a positive integer, that is

Objective Mathematics

(a) equal to 1 (b) equal to n + n2 (c) divisible by n (d) never less than n

(a) 9, 1 (c) 8, 2

171. Sum to infinity of the series

(b) 1 (a) 4 9 3 2 (c) (d) None of these 9 172. Sum to infinity terms of the series

(b) 5 24

(c) 11 (d) None of these 48 173. The sum of (33 – 23) + (53 – 43) + (73 – 63) + ... to 10 brackets is (a) 4960 (c) 5060

(b) 4860 (d) None of these

174. Sum up to 16 terms of the series 13 13 + 23 13 + 23 + 33 + + +... is 1 1+ 3 1+ 3 + 5 (a) 450 (c) 446

(b) 456 (d) None of these

175. Sum to 20 terms of the series 1.32 + 2.52 + 3.72 + ... is (a) 178090 (c) 188090

(b) 168090 (d) None of these

176. Sum to n terms of the series 2 + 5 + 14 + 41 + ... is (a) n + 1 (3n − 1) 2 4 (c) n + 1 (3n − 1) 2 2

(b) n + 3 (3n − 1) 2 4 (d) None of these

177. If H is the H.M. between P and Q, then the value of H H is + P Q PQ (a) 2 (b) P+Q (c)

P+Q PQ

a (b – c) x2 + b (c – a) x + c (a – b) = 0 are equal, then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) None of these

181. If x, y, z are in A.P., ax, by, cz are in G.P. and a, b, c are in H.P., then x y (b) x + z = a + c (a) + = a + c y x c a z x c a x z b c (c) + = + (d) None of these z x c b 182. α, β are the roots of x2 – 3x + a = 0 and γ, δ the roots of x2 – 12x + b = 0 and numbers α, β, γ, δ (in order) form an increasing G.P., then (a) a = 3, b = 12 (c) a = 2, b = 32 183. The value of (0.2) (a) 1

(b) a = 12, b = 3 (d) a = 4, b = 16. 1 1 1  log 5  + + + ...∞   4 8 16 

is

(b) 2

(c) 1 (d) 4. 2 184. 1 + 2.2 + 3.22 + 4.23 + ... + n.2n – 1 = (a) 1 + (1 + n) 2n (c) 1 – (1 – n) 2n

(b) 1 – (1 + n) 2n (d) None of these

185. Sum to infinity of the series 1 + 22x + 32x2 + 42x3 + ... to ∞, | x | < 1 is 1+ x 1− x (b) (a) (1 − x)3 (1 + x)3 (c)

2+ x (1 − x)3

(d) None of these

186. i – 2 – 3i + 4 ... to 100 terms = (a) 50 (1 – i) (c) 25 (1 + i)

(b) 25i (d) 100 (1 – i)

187. 100th term of the series 1 + 3 + 7 + 15 + ... is (a) 250 – 1 (c) 2100 – 1

(b) 280 – 1 (d) None of these

(d) None of these

188. Number of terms in the sequence 1, 3, 6, 10, 15, ..., 5050 is

(b) 41 495

189. If b + c − 2a , c + a − 2b , a + b − 2c are in A.P., then a b c a, b, c are in

  is 178. The value of .1 23 (a) 61 495 (c) 51 495

(b) 7, 4 (d) None of these

180. If the roots of the equation

2 5 2 11 − + − +... is 3 6 3 24

1 1 1 1 + + + +... is 5 7 2 53 7 4 (a) 9 48

179. The value of x + y + z is 15 if a, x, y, z, b are in A.P. while the value of 1 + 1 + 1 is 5 if a, x, y, z, b x z z 3 are in H.P. The values of a and b are

(d) None of these

(a) 50 (c) 100

(a) A.P. (c) H.P.

(b) 75 (d) 125

(b) G.P. (d) None of these

(a) A.P. only (b) A.P. and G.P. only (c) A.P., G.P. and H.P. (d) A.P. and H.P. only 191. If the sides of a right angled triangle are in G.P., then the cosine of the greater acute angle is (a)

1 1+ 5

(b)

1 1− 5

(c) 1 + 5 (d) None of these 2 192. If x + y + z = 1 and x, y, z are positive numbers such that (1 – x) (1 – y) (1 – z) ≥ kxyz, then k = (a) 2 (c) 8

(b) 4 (d) 16

193. If a + b + c = 3 and a > 0, b > 0, c > 0, then the greatest value of a2 b3 c2 is (a) 3 ⋅ 2 77 8 4 (c) 3 ⋅ 2 77 10

4

(b) 3 ⋅ 2 77 9

4

(d) None of these

194. If Sn = 1 + 1 + 1 + ... + 1 , then the least value of n 2 22 2n −1 such that 2 – Sn < 1 is 100 (a) 6 (b) 8 (c) 10 (d) None of these 195. The sum of first n terms of the series 1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + 4 ⋅ 4! + ... is (a) (n + 1)! – 1 (b) n! – 1 (c) (n – 1)! – 1 (d) None of these 196. The largest term of the sequence 1 4 9 16 , ... is , , , 503 524 581 692 (a) 16 692 (c) 49 1529

(b)

4 524

(d) None of these

197. The consecutive numbers of a three digit number form a G.P. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order and if we increase the second digit of the required number by 2, the resulting number forms an A.P. The number is (a) 139 (c) 931

(b) 193 (d) None of these

198. If a, b, c are digits, then the rational number represented by 0 ⋅ cababab ... is (a) 99c + ab 990

(b) 99c + 10a + b 99

(c) 99c + 10a + b 990

(d) None of these

(a) 2870 (c) 2970

(b) 2160 (d) None of these

200. If a1 = 0 and a1, a2, a3, ..., an are real numbers such that | ai | = | ai – 1 + 1 | for all i then the A.M. of the numbers a1, a2, ..., an has value x where (a) x ≤ – 1 (b) x ≥ – 1 2 2 (c) x < – 1 (d) None of these 2 201. If a, a1, a2, a3, ..., a2n , b are in A.P. and a, g1, g2, ..., g2n , b are in G.P. and h is the H.M. of a and b then a1 + a2 n a2 + a2 n −1 a + an +1 is equal to + + + n g1 g 2 n g 2 g 2 n −1 g n g n +1 (b) n h (c) nh (d) 2n h 202. If 0.272727..., x and 0.727272... are in H.P., then x must be (a) 2nh

(a) rational (c) irrational 203.

n

i

(b) integer (d) None of these

j

∑∑∑ 1

=

i =1 j =1 k =1

(a) n (n + 1)(2n + 1) 6

(b) n (n + 1) 2

n (n + 1)  (c)   2  

(d) n (n + 1)(n + 2) 6

2

204. Let a = 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5. Then (a) 55 ≥ a (c) 55 ≥ 6a 205. If Tn = n × (n!), then (a) (21)! – 1 (c) (21)! + 1

(b) 35 ≥ 5! (d) None of these 20

∑T n=1

n

is equal to

(b) (20)! – 1 (d) None of these

206. The sum of the products of the 2n numbers ± 1, ± 2, ± 3, ..., ± 2n taking two at a time is (a) – n (n + 1) 2 n ( n + 1)(2n + 1) (c) – 6

(b) n (n + 1)(2n + 1) 6 (d) None of these

207. If a, b, c are positive then the minimum value of alog b – log c + blog c – log a + clog a – log b is (a) 3 (b) 1 (c) 9 (d) 16 208. The largest value of the positive integer k for which nk + 1 divides 1 + n + n2 + ... + n127 is divisible by (a) 8 (c) 32

(b) 16 (d) 64

579

199. If Σ n = 210, then Σ n2 =

Sequences and Series

190. Non zero equal numbers are in

580

lim (1 + 3–1) (1 + 3–2) (1 + 3–4) (1 + 3–8) ... ( 1 + 3−2n ) 209. n→∞ is equal to

Objective Mathematics

(a) log34 (c) 1 – log43

(b) 1 2

(a) 1

(c) 3 (d) None of these 2 210. If x1, x2, x3 as well as y1, y2, y3 are in G.P. with the same common ratio, then the points (x1, y1), (x2, y2) and (x3, y3) (a) lie on a straight line (b) lie on an ellipse (c) lie on a circle (d) are vertices of a triangle 211. The determinant

xp + y yp + z 0

(a) x, y, z are in A.P. (c) x, y, z are in H.P.

x y xp + y

218. If 1, log3 (31–x + 2), log3 (4.3x – 1) are in A.P. then x equal

y z = 0 if yp + z

(b) x, y, z are in G.P. (d) xy, yz, zx are in A.P

(b) 1 – log34 (d) log43

219. The value of 21/4, 41/8, 81/6, + ...∞ is (a) 1 (c) 3/2

(b) 2 (d) 4

220. Fifth term of an G.P. is 2, then the product of its 9 terms is (a) 256 (b) 512 (c) 1024 (d) None of these 221. Let T, be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m,n,m ≠ n, Tm = 1 and Tn = 1 , then a – d equals n m

(b) 1 (c) 0 (d) 1 + 1 212. Let x be the arithmetic mean and y, z be two geometm n ric means between any two positive numbers. Then 222. The third term of a G.P. is the square of first term. If y3 + z3 the second term is 8, then the 6th term is  = xyz (a) 120 (b) 124 (a) 2 (b) 1 (c) 128 (d) 132 (c) 3 (d) 4 223. The sum of first n terms of the series12 + 2. 22 + 213. Let Tr be the rth term of an A.P., for r = 1, 2, 3, .... If for some positive integers m, n we have Tm = 1 and n Tn = 1 . Then Tmn equals m (a) 1 mn

(b) 1 + 1 m n

(a) A.P. (c) H.P.

(b) G.P. (d) no particular order

(a) 1/mn

2 32 + 2. 42 + 52 + 2. 62 + ... is. n(n + 1) when n is 2 even. When n is odd the sum is 2 (a) n(n + 1) 4

2 (b) n(n + 1) 2

2 n(n + 1)  (c) 3n(n + 1) (d)   2  2  (c) 1 (d) 0 2 2 2 214. The Arithmetic mean, Geometric mean and Harmonic 224. Suppose a, b, c are in A.P. and a , b , c are in G.P. If mean between two positive real quantities are themselves a < b < c and a + b + c = 3 then the value of a is in 2

215. H.M. between the roots of the equation x2 – 10x + 11 = 0 is (a) 1 5 (c) 21 20

(b) 5 21 11 (d) 5

n +1 n +1 216. If A.M. of a and b is a + b then n = n a + bn

(a) – 1 (c) 1

(b) 2 (d) 0

217. If the ratio of the sums of m and n terms of an A.P. is m2 : n2 then the ratio of its mth and nth terms is (a) m – 1 : n – 1 (c) 2m – 1 : 2n – 1

(b) 2m + 1 : 2n + 1 (d) None of these

2

(a)

1 2 2

(b)

1 2 3

(c)

1 1 − 2 3

(d)

1 1 − 2 2

a b aα − b b c bα − c 225. If = 0 and α ≠ 1 , then 2 1 0 2 (a) a, b, c are in A.P. (c) a, b, c are in H.P.

(b) a, b, c are in G.P. (d) None of these

226. In a G.P. if the (p + q)th term is m and (p – q)th term is n, then its pth term is (b) mn

(a) 0 (c)

mn

(d) 1 (m + n) 2

2 2 ) x – (4 +

5 ) x + 8 + 2 5 = 0 is

(a) 2 (c) 6

(b) 4 (d) 8

228. Let the harmonic mean and the geometric mean of two positive numbers be in the ratio 4 : 5. The two numbers are in the ratio (a) 1 : 1 (c) 3 : 1

(b) 2 : 1 (d) 4 : 1

229. The product of n positive numbers is unity their sum is (a) a positive integer (b) equal to n + 1 n (c) divisible by n (d) never less than n

(c) 10 × 81

232. The A.M., G.M. and H.M. of two numbers are x, y and z respectively. Then, which of the following is true ? (a) z < x < y (c) y < x < z

(b) x < y < z (d) z < y < x

233. The arithmetic mean between two numbers is A and the geometric mean is G. Then these numbers are A2 − G 2 ± A

)

ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a comd e f , , are in mon root if a b c (b) A.P. (d) None of these

235. Every term of a G.P. is positive and also every term is the sum of two preceding terms. Then the common ratio of the G.P. is (a) 1 − 5 2 5 −1 (c) 2

237. The first term of an infinite geometric progression is x and its sum is 5. Then (a) 0 ≤ x ≤ 10 (c) – 10 < x < 0

(b) 0 < x < 10 (d) x > 0

(b) 1 2 (d) None of these

239. The sum of first n terms of the series 2 12 + 2.22 + 32 + 2.42 + 52 + 5.62 + ... is n (n + 1) when n is 2 even. When n is old the sum is 2 (a) n (n + 1) 2

2 (b) n (n + 1) 2

n (n + 1)  (c)    2  2

(d) n (n + 1) 2

240. The sum of first n terms of two A.P., are 3n + 8, 7n + 15, then the ratio of their twelfth term is (a) 7 16

(b) 8 15

(c) 4 9

(d) 3 7

(a) 840 (c) 1680

234. If a, b, c are in G.P.; then the equation

(a) G.P. (c) H.P.

(d) None of these

241. The 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ..., is

(d) None of these

 A − 1 (c) log B − 1   A  B 

3 2 (c) – 1

(b) 10 × 41 2 (d) 41 × 81

(a) 10 × 41

(

(b) log1 – B (1 – A)

(a)

(a) 232 (b) 232 9909 9900 232 (c) (d) 232 9990 990 231. In an A.P. of 81 terms, the 41th term is 10. Then the sum of series is

1 (b) A2 + G 2 + A 2 (c) A ± A 2 − G 2

(a) logB A

238. cos x = b. For what b do the roots of the equation form an A.P. ?

230. The value of 0.2 34 is

(a)

B = 1 + rb + r2b + r3b + ... ∞, then a is equal to b

(b) (d) 1

5 +1 2

(b) 420 (d) 1600

242. If a + b , b, b + c are in A.P., then a, 1 , c are in 1 − ab 1 − bc b (a) H.P. (c) G.P.

(b) A.P. (d) None of these

243. If a, b, c are in G.P., then 1 , 1 , 1 are in a b c (a) G.P. (c) H.P. 244. The sum of the series

(b) A.P. (d) None of these 2 + 6 + 18 +... is 121 3 −1

(a) 242 ( 3 – 1)

(b)

(c) 243 ( 3 + 1)

(d) 121 ( 6 +

2)

581

(5 +

236. If A = 1+ ra + r2a + r3a + ... ∞ and

Sequences and Series

227. The harmonic mean of the roots of the equation

582

Objective Mathematics

 15 5 3  245. If x2 + 9y2 + 25x2 = xyz  + +  , then x, y and z  x y z are in (a) A.P. (c) A.G.P.

(b) G.P. (d) H.P.

246. If a , b , c are in H.P. then b c a (a) a2b, c2a, b2c are in A.P. (b) a2b, b2c, c2a are in H.P. (c) a2b, b2c, c2a are in G.P. (d) None of these 247. The next term of the sequence 1, 3, 6, 10, ... is (a) 16 (c) 15

(b) 13 (d) 14

256. The value of 0 ⋅ 234 is equal to (b)   232 (a)   232 999 900 116 (c)   (d)   210 495 999 257. If arithmetic mean of two positive numbers is A, their geometric mean is G and harmonic mean is H, then H is equal to (a)  G2 / A (c)  A / G2

(b)  A2 / G2 (d)  G / A2

258. The sum of n terms of two arithmetic series are in the ratio 2n + 3 : 6n + 5, then the ratio of their 13th terms is (a)  53 : 155 (c)  29 : 83

(b)  27 : 87 (d)  31 : 89

248. If p, q, r are in A.P., a is G.M. between p and q and 259. If a , a , ….., a are in HP, then the expression a a + 1 2 n 1 2 b is G.M. between q and r, then a2, q2, b2 are in a2 a3 +….+ an – 1 an is equal to (a) G.P. (b) A.P. (a)  (n – 1) (a1 – an) (c) H.P. (d) None of these (b)  na1ar 249. A person purchases one kg of tomatoes from each of (c)  (n – 1) a1an the 4 places at the rate of 1 kg, 2 kg, 3 kg and 4 kg (d)  n (a1 – an) per rupee respectively. On the average he has purchased 260. If the numbers a, b, c are in AP; b, c, d, are in GP x kg of tomatoes per rupee, then the value of x is and c, d, e are in HP, then a, c, e are in (a) 2 (b) 2.5 (a)  AP (b)  GP (c) 1.92 (d) None of these (c)  HP (d)  AGP 250. The fourth, seventh and tenth terms of a G.P. are p, q, 261. If x, y, z are in AP, then (x + 2y – z) (x + z – y) (z + r respectively then 2y – x) is equal to 2 2 2 (a) p = q + r (a)  xyz (b)  2xyz (b) q2 = pr (c)  4xyz (d)  5xyz (c) p2 = qr (d) pqr + pq + 1 = 0 262. If AM of two numbers is twice their GM, then the ratio of greatest number to smallest number is 251. If the roots of the equation x3 – 12x2 + 39x – 28 = 0 are in A.P., then their common difference will be : (a)   7 − 4 3 (b)   7 + 4 3 (a) ± 1 (c) ± 3

(b) ± 2 (d) ± 4

252. If the sum of series 2, 5, 8, 11, ..., is 60100, then n is, (a) 100 (c) 150

(b) 200 (d) 250

253. The sum of the series 1 + 2⋅2 + 3⋅22 + 4⋅23 + 5⋅24 + ... + 100⋅299 is (a) 99⋅2100 + 1 (c) 99⋅2100

(b) 100⋅2100 (d) 99⋅2100 + 1

(c)  21

(d)  5

263. If b , a , c are in AP, then a + b, b + c, c + a, will be in 2

(a)  AP (c)  HP

2

2

(b)  GP (d)  none of these

264. If the 7th term of a HP is 1 and the 12th term is 1 , then the 20th term is 10 25

(b)   1 (a)   1 41 45 254. In a GP, the last term is l and the common ratio is r. 1 (c)   (d)   1 Its nth term from the end is 49 37 (a)  lrn – 1 (b)  lr–n 265. The value of 21/4 · 41/8 · 81/16 161/32… is (c)   l (d)  none of these (b)   5 (a)   3 r n −1 2 2 255. The 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + (c)  2 (d)  1 … is 266. In a geometric progression consisting of positive, each (a)  1600 (b)  1680 term equals the sum of the next two terms. Then the (c)  420 (d)  840 common ratio of this progression equals

(c)   5

275. Sum of n terms of the series 1 + 3 + 7 + 15 + ... is 2 4 8 16

267. If the (p + q) th term of a geometric series is m and the (p – q) th term is n, then the p th term is (a)  (mn)1/2 (c)  m + n

(b)  mn (d)  m – n

268. Three numbers whose sum is 15 are in AP. If they are added by 1, 4 and 19 respectively they are in GP. The numbers are

270. If a = b are in

(b)  60 (d)  49

1/y

= c

(a)  AP (c)  HP

1/z

276. If |x| < 1, then the sum of the series 1 + 2x + 3x2 + 4x3 + … ∞ will be (a)   1 1− x (c)   1 1 + x2

(b)   1 1+ x 1 (d)   (1 − x) 2

n

269. The difference between two numbers is 48 and the difference between their arithmetic mean and their geometric mean is 18. Then, the greater of two numbers is

1/x

(b)  2–n (n – 1) (d)  2–n + n – 1

277. If S be the sum, P be the product and R be the sum of the reciprocals of n terms of a GP, then P2 is equal to

(a)  2, 5, 8 (b)  26, 5, – 16 (c)  2, 5, 8 and 26, 5, – 16 (d)  None of these

(a)  96 (c)  54

(a)  2–n (c)  2n (n – 1) + 1

S (a)     R

(b)   S R

n

R (c)     S

(d)   R S

278. If the AM of two numbers be A and GM be G, then the numbers will be and a, b, c are in GP, then x, y, z (a)  A ± (A2 – G2) (b)  GP (b)   A ± A2 − G 2 (d)  None of these

271. The sum of the series 1 + 3x + 6x2 + 10x3 + … ∞ will be (a)  

1 (1 − x) 2

(b)   1 1− x

(c)  

1 (1 + x) 2

(d)  

272. The sum of the series

(c)   A ± ( A + G ) ( A − G ) (d)   A ± ( A + G ) ( A − G ) 2

1 (1 − x)3

279. If sum of the series

∞

∑r

n

= S , for |r| < 1, then sum of

n=0

1 1 1 + + + ...∞ , 1⋅ 2 ⋅ 3 3 ⋅ 4 ⋅ 5 5 ⋅ 6 ⋅ 7

the series

∞

∑r

2n

, is

n=0

is

(a)  S2

(b)  

(a)  loge 2 − 1 2 (b)  log2 2

S2 2S + 1

(c)   2 S S 2 −1

(d)  

S2 2S − 1

(c)  loge 2 + 1 2 (d)  loge 2 + 1 273. Let two numbers have airthmetic mean 9 and geometric mean 4. Then, these numbers are the roots of the quadratic equation (a)  x2 – 18x – 16 = 0 (b)  x2 – 18x + 16 = 0 (c)  x2 + 18x – 16 = 0

(a)   7

n+1 2 a

274. If a + 2b + 3c = 12; (a, b, c, ∈ R), then ab c is 2 3

(c)  ≤ 26

(a)  – 4 (c)  12

(b)  – 12 (d)  4

281. Geometric mean of 7, 72, 73, …  .  . 7n is

(d)  x2 + 18x + 16 = 0 (a)  ≥ 23

280. The first two terms of a GP add upto 12. The sum of the third and the fourth terms is 48. If the terms of the GP are alternately positive and negative, then the first term is

(b)  ≥ 26 (d)  None of these

(b)   7 2 (c)   7

n−1 2

(d)  None of these

583

(b)   1 5 2 1 (d)   ( 5 − 1) 2

Sequences and Series

(a)   1 (1 − 5) 2

584

SOLUTIONS

Objective Mathematics

1. (a) Let the A.P. be a, a + d, a + 2d ...

1 1 ≤ k ≤ 29 and – ≤ k ≤ 16. or – From question, tp = q = a + (p – 1) d, ...(1) 2 8 ...(2) tq = p = a + (q – 1) d Hence, we have k = 0, 1, 2, 3, ..., 16. Corresponding to each value of k we get one identical term. Hence From (1) and (2), q – p = (p – q) d; ∴ d = – 1 there are 17 identical terms. ∴ From (1), q = a + (p – 1) (– 1); 6. (c)  Let the nth term of the first series = the mth term  ∴ a = q + p – 1. of the second series. ∴ tp + q = a + (p + q – 1) d ∴ 3 + (n – 1) ⋅ 4 = 2 + (m – 1) ⋅ 3, = ( p + q – 1) – (p + q – 1) = 0.

2. (b) Let the number of odd numbers between 60 and 360 be n.  ere, the first term, a = 61; the common difference, H d = 2; the nth term = 359. Then 359 = 61 + (n – 1) × 2; or n = 150. Hence, the number of odd numbers between 60 and 360 is 150. 3. (c) The numbers between 100 and 500 that are divisible by 7 are 105, 112, 119, 126, 133, 140, 147, ..., 483, 490, 497. Let such numbers be n. Then, 497 = 105 + (n – 1) × 7; or n = 57.  he numbers between 100 and 500 that are divisible T by 21 are 105, 126, 147, ..., 483. Let such numbers be m. Then 483 = 105 + (m – 1) × 21; or m = 19. Hence, the required number = n – m = 57 – 19 = 38. 4. (b), (c) Here, the first term, a = 20; the common difference, d = – 4 and the nth term = – 176. Then, – 176 = 20 + (n – 1) (– 4); or n = 50. Hence, the middle terms are 25th and 26th.

25th term = 20 + 24 × – 4 = – 76



26th term = 20 + 25 × – 4 = – 80.

5. (a) If possible let the mth term of the first A.P. be identical with nth term of the second A.P. Now, mth term of the 1st A.P. = 2 + (m – 1) 3 nth term of the 2nd A.P. = 3 + (n – 1) 2. ∴ 2 + (m – 1) 3 = 3 + (n – 1) 2 or 3m – 1 = 2n + 1 or 3m – 6 = 2n – 4,  (subtracting 5 from each side) or 3 (m – 2) = 2 (n – 2); n−2 m−2  or = = k (say); 3 2 ∴ m = 2k + 2 n = 3k + 2. But 1 ≤ m ≤ 60 and 1 ≤ n ≤ 50 ∴ 1 ≤ 2k + 2 ≤ 60 and 1 ≤ 3k + 2 ≤ 50

or

4n = 3m  or 

n m = = k (say) 3 4

∴ n = 3k and m = 4k. As each series is continued to 100 terms, n = 3k ≤ 100 and m = 4k ≤ 100 ∴ Possible values of k are 1, 2, 3, ..., 25 and corresponding to each value of k we get one identical term. Hence there are 25 identical terms. 7. (c) Let a be the first term and d be the common difference of A.P. Given ntn = mtm ∴ n [a + (n – 1) d] = m [a + (m – 1) d] ⇒ (m – n) a = d [n (n – 1) – m (m – 1)] ⇒ (m – n) a = d [(m – n) – (m2 – n2)] ⇒ (m – n) a = d (m – n) [1 – (m + n)] ⇒ a = d [(1 – (m + n)] [ m ≠ n] ⇒ – a = d (m + n – 1) ...(1) Now (m + n)th term is tm + n = a + (m + n – 1) d = a – a = 0 [from (1)] 8. (b) Let x be the first term and d be the common difference of A.P. Given: a = pth term of A.P. = x + (p – 1) d

b = qth term of A.P. = x + (q – 1) d

and

c = rth term of A.P. = x + (r – 1) d

∴ a – c = ( p – r) d,

b – a = (q – p) d, c – b = (r – q) d

...(1)

Now a (q – r) + b (r – p) + c (p – q)

= q (a – c) + r (b – a) + p (c – b)

= q (p – r) d + r (q – p) d + p (r – q) d  [From (1)] = d {q ( p – r) + r (q – p) + p (r – q)} = d ⋅ 0 = 0. 9. (c) We have, first term = a,

∴ T1 = a

Second term = b,  ∴ T2 = b Then common difference d = T2 – T1 = b – a Also, last term = c. ⇒ c = a + (n – 1) d ⇒n =

c−a+d . d

∴ Sum of n terms Sn =

n (a + l) = (b + c − 2a )(a + c) . 2 2 (b − a )

r {2x + (r – 1) y}, 2 a 1 = {2x + (p – 1) y} ∴ We get p 2 b 1 = {2x + (q – 1) y} q 2 c 1 and = {2x + (r – 1) y} r 2 Thus, the given expression 1 = [(q – r) {2x + (p – 1) y} + (r – p) 2  {2x + (q – 1) y} + {p – q) {2x + (r – and

10. (c) Let the number be a – d, a, a + d, a + 2d where a, d ∈ Z and d > 0 Given : (a – d)2 + a2 + (a + d)2 = a + 2d ⇒ 2d2 – 2d + 3a2 – a = 0 1 1 ± (1 + 2a − 6a 2 )   2 Since d is positive integer, ∴ 1 + 2a – 6a2 > 0 ∴ d =

1− 7  1+ 7   < a <  ⇒   6  6     Since a is an integer, ∴ a = 0, 1 [1 ± 1] = 1 or 0. Since d > 0, then d = 2  ∴ d = 1. Hence, the numbers are – 1, 0, 1, 2. 11. (a), (b)  We have, 2 and Sn = 300. 3 n  2  2 ⋅ 20 + (n − 1)  −   = 300 2   3 

a = 20, d = – ∴

⇒ n2 – 61n + 900 = 0 ⇒ (n – 25) (n – 36) = 0 ∴ n = 25 or 36. 12. (a) We have, Tn =

3+ n . 4

We get the series

1,

5 3 1 , , ... ∴ a = 1, d = 4 2 4

∴ S105 =

105  1   105  2 ⋅ 1 + (104) ⋅  =   × 28 = 1470. 2  4  2 

13. (b) Since, in an A.P., the sum of terms equidistant from the beginning and end is constant and equal to the sum of first and last term, ∴ We have,  a5 + a20 = a1 + a24, a10 + a15 = a1 + a24. Hence, the given relation gives 3 (a1 + a24) = 225 ⇒ a1 + a24 = 75,  24  ∴ S24 =   (a1 + a24) = 12 × 75 = 900.  2  14. (c) Let the A.P. be x, x + y, x + 2y, ... Given a =

b=

p {2x + (p – 1) y}, 2

q {2x + (q – 1) y} 2

585

( d = b – a)

c =

...(1) ...(2) ...(3)

1) y}]

1 [2x (q – r + r – p + p – q) + y {(q – r) 2 (p – 1) + (r – p) (q – 1) + (p – q) (r – 1)}] 1 [2x × 0 + y × 0] = 0. = 2

= 

15. (a) The given series is arithmetic whose first term = 2 20, common difference = – . 3 As the common difference is negative, the terms will become negative after some stage. So the sum is maximum if only positive terms are added.  2 Now tn = 20 + (n – 1)  −  ≥ 0  3  if 60 – 2 (n – 1) ≥ 0; or 62 ≥ 2n  or  31 ≥ n. ∴ The first 31 terms are non-negative. ∴ Maximum sum 31 2 × 20 + (31 − 1)  − 2     = S31 =    3  2  31 = {40 – 20} = 310. 2 16. (b) The required sum = (1 + 2 + 3 + ... + 199) – (3 + 6 + 9 + ... + 198) – (5 + 10 + 15 + ... + 195) + (15 + 30 + 45 + ... + 195) 199 66 39 (1 + 199) – (3 + 198) – (5 + 195) 2 2 2 13 + (15 + 195) 2 = 199 × 100 – 33 × 201 – 39 × 100 + 13 × 105 = 10732. =

17. (c) Let the four numbers in A.P. be α – 3β, α – β, α + β, α + 3β Given, α – 3β + α – β + α + β + α + 3β = 50 25 ⇒ 4α = 50. ∴ α = 2 and α + 3β = 4 (α – 3β) or 3α = 15β α 25 5 ∴ β = = = 5 5× 2 2 Hence, the four numbers are 5, 10, 15, 20.

Sequences and Series

(b + c − 2a ) (b − a )

⇒ n =

586

18. (a) Let the four numbers be

Objective Mathematics

α 3β, α – β, α + β, α + 3β Given, α – 3β + α – β + α + β + α + 3β = 24 ⇒ 4α = 24 ∴ α = 6 and (α – 3β) (α – β) (α + β) (α + 3β) = 945 ⇒ (α2 – 9β2) (α2 – β2) = 945 ⇒ (36 – 9β2) (36 – β2) = 945 ⇒ β4 – 40β2 + 144 = 105 ⇒ β4 – 40β2 + 39 = 0 ⇒ β4 – β2 – 39β2 + 39 = 0 ⇒ (β2 – 1) (β2 – 39) = 0 Since numbers are integers, ∴ β2 ≠ 39 2 ∴ β – 1 = 0 ∴ β = ± 1. Hence, the four integers are 3, 5, 7, 9. 19. (c) Here a1 = 4, an = last term = 146

22. (a) If n be the number of terms, then an = a1 + (n – 1) d where a1 is the first term and d the common difference ∴ 39 = 3 + (n – 1) d or (n – 1) d = 36 ...(1) Also Sn =

n (a1 + an) 2

⇒ 525 =

n (3 + 39) 2

⇒ 1050 = n (42) or n = 1050 ÷ 42 = 25 Putting n = 25 in (1), we get

(25 – 1) d = 36 ⇒ d = 36 ÷ 24 =

3 . 2

23. (b) Here a = 100 Let d be the common difference. Now a1 + a2 + ... + a6 = 5 (a7 + a8 + ... + a12)

∴ a1 + (n – 1) d = 146  a1 + a6   a7 + a12  ⇒ – 4 + (n – 1) d = 146 ⇒ 6   =5×6   2   2   ⇒ (n – 1) d = 146 + 4 = 150 ...(1) ⇒ a1 + a3 = 5 (a7 + a12) n [2a1 + (n – 1) d] We have, Sn = ⇒ a + (a + 5d) = [(a + 6b) + (a + 11d)] 2 ⇒ 2a + 5d = 10a + 85d n n ⇒ 7171 = [2 × – 4 + 150] = × 142 = 71n −8a a 2 2 =– ⇒ 80d = – 8a, or d = 80 10 ∴ n = 7171 ÷ 71 = 101. 100 Now from (1), (101 – 1) d = 150 ∴ d = – = – 10. [∴ a = 10] 3 10 ⇒ 100d = 150; d = 150 ÷ 100 = 2 Hence the required number of terms = 101 and 24. (c) Sum of the interior angles of a polygon of n sides. 3 common difference = . π 2 = (2n – 4) = (n – 2) π = (n – 2) ⋅ 180º 2 20. (a) The given sequence can be expressed as Also a = 120º, d = 5º log a, (log a + log r), (log a + 2 log r) ... n which is clearly an A.P. whose first term is log a ∴ [2 ⋅ 120º + (n – 1) 5º] = (n – 2) × 180º 2 and common difference is log r. 2 ⇒ n – 25n + 144 = 0 ⇒ (n – 9) (n – 16) = 0 The nth term = log a + (n – 1) log r ∴ n = 9, 16. But when n (a1 + an) Since sum to n terms = n = 16, Tn = a + (16 – 1) d 2  = 120º + 15 × 5º = 195º. n [log a + log a + (n – 1) log r] ∴ Sn = This is not possible as interior angle cannot be 2 greater than 180º. ∴ n = 9. n = [2 log a + (n – 1) log r] 2 n 2S 25. (b) We have, S = (a + l) ⇒ = n ...(1) n 2 a+l = log a2 rn – 1. 2 l−a Also, l = a + (n – 1) d ⇒ d = 21. (b) Let ‘a’ be the first term and ‘d’ the common difn −1 ference of the given A.P. l−a = [Using (1)] Middle term i.e., 6th term 2S −1 = a + (6 – 1) d a+l = a + 5d = 30 (Given) l 2 − a2 = . ∴ k = 2S. 11 [2a + (11 – 1) d] Now, S11 = 2S − (l + a ) 2 26. (a) Let x18 = y21 = z28 = k (say) 11 (2a + 10d) = 11 (a + 5d) = Taking log, we get 2 = 11 × 30 = 330. 18 log x = 21 log y = 28 log z = log k ...(1)



= 3 ⋅

21 7 = =3 6 2 28 Similarly, c = 3 ⋅  21 18 9 and d = 7 ⋅  = 28 2

=

21   18

[Using (1)]

1 2

1 2

1 1 , 4, 4 which 2 2 1 . are clearly in A.P. with common difference 2 Hence the four numbers are 3, 3

27. (a) Given, Sn = 3n2 ∴ Tn = Sn – Sn – 1 = 3n2 – 3 (n – 1)2 = 6n – 3. Putting n = 1, 2, 3, ..., the sequence is 3, 9, 15, 21,... which is clearly an A.P. 28. (c) The first two digit number which when divided by 4 leaves remainder 1 is 4 ⋅ 3 + 1 = 13 and last is 4 ⋅ 24 + 1 = 97. Thus we have to find the 13 + 17 + 21 + ... + which is an A.P. ∴ 97 = ⇒ n = 22. n [a + l] = 11 and Sn = 2 

sum 97 13 + (n – 1) ⋅ 4 ⋅ [13 + 97] = 11 × 110 = 1210.

29. (a) The least and the greatest number of three digits divisible by 7 are 105 and 994 respectively. ∴ It is required to find the sum 105 + 112 + 119 + ... + 994 Here, a = 105, d = 7, an = 994 Now, an = a + (n – 1) d ⇒ 994 = 105 + (n – 1) × 7 ⇒ 994 – 105 = 7 (n – 1) 889 7 ∴ n – 1 = 127 ⇒ n = 127 + 1 = 128 n ∴ Sum = [2a + (n – 1) d] 2 ⇒ 889 = 7 (n – 1) or n – 1 =



128 [2 × 105 + (128 – 1) × 7] 2 = 64 (210 + 889) = 64 × 1099 = 70336. =

30. (a) The odd numbers of four digits which are divisible by 9 are 1017, 1035, ..., 9999. These are in A.P. with common difference 18. Hence nth term = an = a + (n – 1) d, Here, a = 1017, d = 18 and l = 9999

587

∴ Sn = 250 × 11016 = 2754,000 which is the required sum. 31. (b) We have, an = Sn – Sn – 1

=4 =4

∴ 9999 = 1017 + (n – 1) × 18 ⇒ 18n = 9999 – 999 = 9000 ⇒ n = 9000 ÷ 18 = 500 n 500 (a1 + an) = (1017 + 9999) ⇒ Sn = 2 2

and ∴ d

...(1)

a n – 1 = S n – 1 – S n – 2 ...(2) = an – an – 1 = (Sn – Sn – 1) – (Sn – 1 – Sn – 2) [From (1) and (2)] = Sn – 2Sn – 1 + Sn – 2.

32. (a) We have, Sn = An2 + Bn ∴ Sn – 1 = A (n – 1)2 + B (n – 1) = A (n2 – 2n + 1) + B (n – 1) = An2 – 2An + A + Bn – B ∴ an = Sn – Sn – 1

= An2 + Bn – (An2 – 2An + A + Bn – B)



= An2 + Bn – An2 + 2An – A – Bn + B



= 2An + B – A

⇒ an – 1 = 2A (n – 1) + B – A Now an – an – 1 = 2An + B – A – 2A (n – 1) + B – A = 2A (a constant) Hence the sequence is an A.P. 33. (d) Let S1 denotes the sum of the first n even natural numbers. ∴ S1 = 2 + 4 + 6 + ... to n terms n [2 × 2 + (n – 1) × 2] = 2 n (4 + 2n – 2) ∴ S1 = 2 n = (2n + 2) = n (n + 1) 2

[ a1 = 2, d = 2]

...(1)

Let S2 denotes the sum of the first n odd natural numbers ∴ S2 = 1 + 3 + 5 + ... to n terms n [2 × 1 + (n – 1) 2] = 2 [ a1 = 1, d = 2] ∴ S2 =

n n (2 + 2n – 2) = (2n) = n2 2 2

...(2)

Dividing (1) by (2), S1 n (n + 1) n +1 1 = = =1+ 2 S2 n n n  1 n +1 Hence S1 = 1 +  S2 ∴ k = . n  n

Sequences and Series

If the given terms are a, b, c, d then  log x   a = 3, b = 3 logy x = 3   log y 

588

34. (c) Here a = 10, d = –

Objective Mathematics

3n [2a1 + (3n – 1) d] = S3 2 ∴ 3 (S2 – S1) = S3.

3 . 7



 3 Then, tn = 10 + (n – 1)  −  .  7

 3 tn is positive if 10 + (n – 1)  −  ≥ 0;  7

or 70 – 3 (n – 1) ≥ 0 1 or 73 ≥ 3n; or 24 ≥n 3 ∴ First 24 terms are positive. ∴ Sum of the positive terms 24  −3  2 × 10 + 23 ×  = S24 = 2  7

Now, Sn > 1568 ⇒

38. (c) Here a1 = a and a2 = b ∴ Common difference d = a2 – a1 = b – a Let n be the number of terms in the series ∴ an = 2a = a + (n – 1) d or (n – 1) d = a or (n – 1) (b – a) = a ∴ n – 1 = or n =

69   852 = 12  20 −  = . 7 7 

35. (d) It is in A.P. for which a = 20, d = 2

=

8 n   40 + (n − 1) 3  > 1568 2  

n 112 + 8n ⋅ > 1568 2 3 6 ⇒ n2 + 14n > × 1568 = 1176 8 ⇒ n2 + 14n – 1176 > 0, or (n + 42) (n – 28) > 0 As n is positive, n – 28 > 0 i.e., n > 28 ∴ Minimum value of n = 29.

⇒

 39. (a) Since 51 + x we have

Then, nth term tn < 0 ∴ 40 + (n – 1) (– 2) < 0 or, 42 – 2n < 0 or, 2n > 42 or, n > 21 ∴ least value of n = 22  ence first 21 terms of the A.P. are non negative. Sum H will be maximum if no negative terms is taken. ∴ Maximum sum 21 21 [2⋅40 + (20) (– 2)} = ⋅ 40 = 420. 2 2

37. (a) Let a1 be the first term and d the common difference of the given A.P. n Then S1 = {2a1 + (n – 1) d} 2 2n S2 = {2a1 + (2n – 1) d} 2 3n [2a1 + (3n – 1) d] and S3 = 2 ∴ 3 (S2 – S1)

n  2n  = 3  {2a1 + (2n − 1) d } − {2a1 + (n − 1) d } 2 2  3n = {4a1 + (4n – 2) d – 2a1 – (n – 1) d} 2

b n (a1 + an) = (a + 2a) 2(b − a ) 2 3ab . = 2 (b − a ) a + 5 1 – x, , 25x + 25–x are in A.P., 2

a = 51 + x + 51 – x + 25x + 25–x 2 Now put 5x = t so that t > 0, we then have

2



a = 5t +

or

36. (b) Let nth term be the first negative term.

= S21 =

a a+b−a b +1= = b−a b−a b−a

∴ Sum = 2 8 = 3 3

a b−a

 2 1  1 5 2 1 + t + 2 = t + 2  + 5 t +  t  t t   t

2 2   1   1 t − t 2 5 + 2 + + −      a =  t  t   2

2

1 1   =  t −  + 5  t −  + 12 ≥ 12. t  t  Thus values of a are given by the inequality a ≥ 12.

40. (c) Since the three terms are in A.P. ∴ ⇒ ⇒ ⇒ ⇒ ∴

2 log10 (2x – 1) = log10 2 + log10 (2x + 3) (2x – 1)2 = 2 (2x + 3) (  y – 1)2 = 2 (y + 3), where y = 2x y2 – 4y – 5 = 0 ⇒ (y – 5) (y + 1) = 0 y = 5 ( 2x = y ≠ – 1) x 2 = 5 ⇒ x = log2 5.

41. (a) Let a and b be the two numbers so that 13 a+b= (given) ...(1) 6 Let 2n (even) means be inserted between them so that a, x1, x2, ... x2n, b is an A.P. of (2n + 2) terms whose first term is a and last term is b and whose sum is 2n + 2 [a + b] = (n + 1) (a + b) ...(2) = 2 ∴ Sum of the means = Sum of the series  – (a + b)

13 by (1) 6 ⇒ 12n + 6 = 13n, ∴ n = 6.

⇒ 2n + 1 = n ⋅

Hence, the number of means inserted = 2n = 12. 42. (d) Let the means be x1, x2, ..., xm so that

45. (a) Let the odd number of terms of an arithmetic series be a, a + d, a + 2d, a + 3d, a + 4d, ....., a + (n – 1) d

1, x1, x2, ..., xm, 31 is an A.P. of (m + 2) terms. Now, 31 = Tm + 2 = a + (m + 1) d = 1 + (m + 1) d 30 m +1 x7 5 = Given : xm −1 9

∴ d =

∴

Then, S1 =

...(1)

S2 = a + (a + 2d) + (a + 4d) + ... to

T8 a + 7d 5 = = . Tm a + (m − 1) d 9

⇒ 9a + 63d = 5a + (5m – 5) d 30 ⇒ 4⋅1 = (5m – 68) m +1



=

n +1 {2a + (n – 1) d}. 4

1022 = 14. 73

 iven 24 = a + (a + 2d) + (a + 4d) + ... to n G terms 30 = (a + d) + (a + 3d) + (a + 5d) + ... to n terms and {a + (2n – 1) d} – a = 10.5 From these we get, n {2a + (n – 1) 2d} ...(1) 24 = 2 n [2 (a + d) + (n – 1) 2d}  2 21 and (2n – 1) d =  2 n 6 From (2) – (1), 6 = [2d]; ∴ d = . 2 n 30 =

S1 2n = . S2 n +1

b, b + c, b + 2c, ... n [2a + (n − 1) d ] 3n − 13 = Given: 2 n 5n + 21 [2b + (n − 1) c] 2

43. (a) Let the A.P. be a, a + d, a + 2d, ... to 2n terms

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

6 21 2n − 1 7 Putting in (3), (2n – 1) = or = n 2 n 4 or 8n – 4 = 7n; ∴ n = 4. ∴ The number of terms = 2n = 2 × 4 = 8. 44. (b) Let the A.P. containing (2n + 1) terms be a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, ..., a + 2nd. The sum of odd terms of this A.P. = a + (a + 2d) + (a + 4d) + ... to (n + 1) terms n +1 {2a + (n + 1 – 1) ⋅ 2d} = (n + 1) (a + nd) 2 The sum of even terms of this A.P.

=

n + 1  2a +  n + 1 − 1 × 2d       2   2× 2 

and

⇒ 73m = 1022.



=

n +1 terms 2

46. (c) Let the two A.P. be a, a + d, a + 2d, ...

⇒ 2m + 2 = 75m – 1020

∴ m =



∴ [Using (1)]

n {2a + (n – 1) d} 2

or

2a + (n − 1) d 3n − 13 = 2b + (n − 1) c 5n + 21

...(1)

which holds for all positive integral values of n. 24 th term of the first progression Now, 24 th term of the second proogression  

a + 23d b + 23c In order to find this ratio, we put n = 47 in (1) =

∴

2a + 46d 3 × 47 − 13 = 2b + 46c 5 × 47 + 21

or

a + 23d 128 1 = = . b + 23c 256 2

∴ The required ratio is 1 : 2. 47. (a), (d)  Let the first set of number be a – d, a, a + d. ∴ From question, the second set will be of the form

b – (d – 1), b, b + (d – 1)

and a – d + a + a + d = 15, i.e., a = 5 b – (d – 1) + b + b + (d – 1) = 15, i.e., b = 5. ∴ The two sets become 5 – d, 5, 5 + d and 6 – d, 5, 4 + d Again from question,

589

Sequences and Series

= (a + d) + (a + 3d) + (a + 5d) + ... to n terms n = {2 (a + d) + (n – 1) ⋅ 2d} = n (a + nd) 2 n +1 Hence the required ratio = . n

⇒ (2n + 1) (given)  = (n + 1) (a + b) – (a + b) = n ⋅ (a + b)

590



(5 − d ) 5 (5 + d ) 7 25 − d 2 7 = ; or = (6 − d ) 5 ( 4 + d ) 8 24 + 2d − d 2 8

Objective Mathematics

⇒ 200 – 8d2 = 168 + 14d – 7d2 ⇒ d2 + 14d – 32 = 0 ⇒ (d + 16) (d – 2) = 0 ∴ d = 2, – 16. ∴ The two sets are 3, 5, 7 and 4, 5, 6 or

21, 5, – 11 and 22, 5, – 12.

48. (c) Given, 1 + 3 + 5 + 7 + ... to n terms ≥ 500 n [2⋅1 + (n – 1) 2} ≥ 500 ⇒ n2 ≥ 500 2 ∴ either, n ≥ 500 or, n ≤ – 500

⇒

But n is a positive integer ∴ n ≥ 500 = 10 5 or, n ≥ 22.36 ∴ least value of n = 23. 49. (c) Let the first instalment be a and common difference of A.P. be d. Given, 3600 = sum of 40 terms 40 [2a + (40 – 1) d]  = 2 or, 3600 = 20 {2a + 39d} or, 180 = 2a + 39d  ...(i) After 30 instalments one third of the debt is unpaid 3600 = 1200 is unpaid and 2400 is paid Hence, 3 Now 2400 = 

30 {2a + (30 – 1) d} or, 160 = 2a + 29d 2 ...(ii)

Subtracting (ii) from (i), we get 20 = 10d  ∴d =2 From (i), 180 = 2a + 39⋅2 or, 2a = 180 – 78 = 102 ∴ a = 51 Now, value of the 8th instalment = a + (8 – 1) d = 51 + 7 ⋅ 2 = Rs. 65. 50. (c) Let ‘a’ be the first term and ‘d’ the common difSm ference of A.P. Using the given information, Sn m2 = 2 n ∴

m [ 2a + (m − 1) d ] m2 2 = 2 n n [ 2a + (n − 1) d ] 2

2a + (m − 1) d m = 2a + (n − 1) d n ⇒ 2an + (mn – n) d = 2am + (mn – m) d ⇒ 2an – 2am = (mn – m – mn + n) d ⇒

⇒ 2a (n – m) = (n – m) d ⇒ d = 2a (n – m ≠ 0 as m ≠ n) Now

=

am a + (m − 1) d a + (m − 1) ⋅ 2a = = an a + (n − 1) d a + (n − 1) ⋅ 2a a (1 + 2m − 2) 2m − 1 = a (1 + 2n − 2) 2n − 1

Hence the required ratio = (2m – 1) : (2n – 1). 51. (a) Let ‘a’ be the length of the smallest side and d cm the common difference. n Now, Sn = [2aa + (n – 1) d] 2 Here n = 25, S25 = 2100 25 [2aa + (25 – 1) d] 2 ⇒ a + 12d = 84 ⇒ 2100 =

...(1)

 he largest side = 25th side = a + (25 – 1)  T d = a + 24 d ∴ a + 24d = 20a (given) Multiplying (1) by (2), we get 2a + 24d = 168 Subtracting (2) from (3), we get a = 168 – 20a ⇒ 21a = 168, ∴ a = 168 ÷ 21 = 8 Hence from (1), we get 8 + 12d = 84 ⇒ 12d = 84 – 8 = 76 19 1 ∴ d = 76 ÷ 12 = = 6  cm. 3 3

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

Hence the length of the smallest side = 8 cm and 1 the common difference = 6 cm. 3 52. (b) Sn =

n [2a + (n – 1) d] 2

Here a = 1st term = 7 years, d = 3 months = year, Sn = 250 n ∴ 250 = 2

1 4

1   2 × 7 + (n − 1) × 4   

 n + 55     4  ⇒ 2000 = n2 + 55n ⇒ n2 + 55n – 2000 = 0 ⇒ (n – 25) (n + 80) = 0 ⇒ n = 25. ∴ Number of members in the club = 25. ⇒ 250 =

n 2

53. (b) Let the four numbers in A.P. be a – 3d, a – d, a + d, a + 3d. Then, (a – 3d) + (a – d) + (a + d) + (a + 3d) = 28 ⇒ 4a = 28, ∴ a = 7. Also

(a − 3d )(a + d ) 8 = (a − d )(a + 3d ) 15

⇒ 7 (49 – 3d2) = 46 × 7 × d

= tan α2 – tan α1 Similarly, sin d sec α2 sec α3 = tan α3 – tan α2    sin d sec αn – 1 sec αn = tan αn – tan αn – 1 On adding, we get the given expression = tan αn – tan α1.

⇒ (d – 1) (3d + 49) = 0 ∴ d = 1. ∴ Required numbers are 4, 6, 8, 10. 54. (c) Let the four members in A.P. be

a – 3d, a – d, a + d, a + 3d

Sum = 4a = 20 ∴ a = 5

60. (b) We have,

Sum of their squares = 4a2 + 20d2 = 120. ∴ 20d2 = 120 – 4 × 25 = 20 ∴ d = 1 or d = ± 1. Hence the numbers are 2, 4, 6, 8. 2

55. (a) Let the three numbers in A.P. be



=

a + an  1  a1 + an a1 + an + + ... + 1   a1 + an  an an a2 an − 1 an a1 



=

a + a1  1  a1 + an a2 + an −1 + + ... + n   a1 + an  a1an a2 an −1 an a1 



=

1 a1 + an

a – d, a and a + d.

Given : (a – d) + a + (a + d) = 12 ⇒ 3a = 12 ∴ a = 4. Also (a – d)3 + a3 + (a + d)3 = 288. ⇒ 3a2 + 6ad2 = 288

1 1 1 + + ... + a1an a2 an −1 an a1

 a1 + an a2 + an −1 a + a1  + + ... + n   a2 an −1 an a1   a1an

[ a1 + an = (a1 + d) + (an – d) = a2 + an – 1]  1 1  1  1 1   1 1 = a + a  a + a  +  a + a  + ... +  a + a   1 n  1 2  n   n −1  1  n 

⇒ 24d2 = 288 – 3 × 64 = 96 ∴ d2 = 4 or d ± 2 Hence the numbers are 2, 4, 6. 56. (d) The various groups of natural numbers are



2 = a +a 1 n

1 1 1  + + ... +  . a a a 2 n  1

G1 = (1), G2 = (2, 3, 4), G3 = (5, 6, 7, 8, 9) ... 61. (a) Let the four numbers in A.P. be  lso, the terms in each group form an A.P. with A a – 3d, a – d, a + d, a + 3d, which are in ascendcommon difference 1. ing order. Hence the sum of terms which will be in A.P. in the Then a – 3d + a – d + a + d + a + 3d = 50, or nth group. 25 a = 2n − 1 2 2 = [2 {1 + (n – 1) } + (2n – 1 – 1) ⋅ 1] 2 Also, a + 3d = 4 (a – 3d); or 3a = 15d; = (2n – 1) [n2 – 2n + 2 + n – 1] = (2n – 1) (n2 – n + 1) 5 or d = = 2 n3 – 3n2 + 3n – 1 = n3 + (n3 – 3n2 + 3n – 1) = n3 2 + (n – 1)3. Hence the four numbers are 5, 10, 15, 20. kx 62. (b) L et ABC be a right angled triangle in which [ 2a + (kx − 1) d ] Skx ∠B = 90º. 57. (c) We have, = 2 Sx x + − 2 a x 1 d ( ) [ ] ∴ b2 = a2 + c2. 2 a+b Since a, c, b are in A.P. ∴ c = k [(2a − d ) + kxd ] 2 = 2 2 (2a − d ) + xd a + b + 2ab ∴ b2 = a2 + ; Skx 4 For to be independent of x, 2a – d = 0 Sx or 5a2 + 2ab – 3b2 = 0 or 2a = d. 2 a a 58. (c) Sn – 2Sn – 1 + Sn – 2 = (Sn – Sn – 1) – (Sn – 1 – Sn – 2) ⇒ 5   + 2   − 3 = 0 b  b = d. =t – t n

n–1

59. (d) We have,



a −2 ± 4 + 60 = b 2×5 a 3 ⇒ = , as the negative value is inadmissible; b 5 ⇒

sin (α 2 − α1 ) sin d sec α1 sec α2 = cos α1 ⋅ cos α 2 ( d = α 2 – α 1)

591



⇒ 49 – 3d2 = 46d or 3d2 + 46d – 49 = 0

=

sin α 2 cos α1 cos α 2 sin α1 − cos α1 ⋅ cos α 2 cos α1 ⋅ cos α 2

⇒ 7 (a2 – 3d2) = 46ad

Sequences and Series

⇒ 15 [a2 – 3d2 – 2ad] = 8 [a2 – 3d2 + 2ad]

592

66. (a)

a b = = k (say) 3 5

or

Objective Mathematics

∴ a = 3k, b = 5k, ∴ c =

x+z or 2y = x + z ...(1) 2 ∴ (x + 2y – z) (2y + z – x) (z + x – y) = (x + x + z – z) (x + z + z – x) (2y – y)  [From (1)] = (2x) (2z) ( y) = 4xyz.

∴ y =

b 2 − a 2 = 4k.

Hence a : c : b = 3 : 4 : 5. 63. (d) Let the n arithmetic means be A1, A2, A3, ... An. Then 1, A1, A2, A3, ..., An , 31 are in A.P. Let the common difference = d. ∴ tn + 2 = 31 = 1 + (n + 2 – 1) d; ∴ d =

30 . n +1

∴ A1 = t2 = 1 + d = 1 +

30 31 + n = n +1 n +1

and An = tn + 1 = 1 + nd = 1 +

30 31n + 1  ⋅ n = n +1 n +1

Given, A1 : An = 3 : 29 31 + n 3 31 + n 3 ⇒ n + 1 = or = ; 31n + 1 29 31n + 1 29 n +1 ⇒ 899 + 29n = 93n + 3; ∴ ∴ n =

64n = 896;

896 = 14. 64

x, y, z are in A.P.

67. (b)

a, b, c are in A.P., ∴ 2b = a + c p is the A.M. between a and b a+b ∴ p = 2

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

q is the A.M. between b and c b+c 2 Adding (2) and (3)

∴ q =

p+ q=

=

...(3)

a+b b+c a + c + 2b + = 2 2 2 2b + 2b = 2b 2

[Using (1)]

p+q 2 Hence b is the A.M. between p and q.

∴ 2b = p + q or b =

64. (a) Let 2n arithmetic means be A1, A2, A3, ..., A2n 68. (b) L et A 1, A 2, A 3, ..., A 11 be 11 A.M.s between 28 and 10 between a and b. ∴ 28, A1, A2, ..., A11, 10 are in A.P. a+b Then A1 + A2 + A3 + ... + A2n = × 2n 2 Let ‘d’ be the common difference of A.P. 13n 13 Also the number of terms = 13. = × 2n = 6 6 10 = a13 = a1 + 12d = 28 + 12d 2 10 − 28 18 3 Given, A1 + A2 + A3 + ... + A2n = 2n + 1; ∴ d = =– =– 12 12 2 13n Total number of terms = 13 (odd) ; ∴ 2n + 1 = 6  13 + 1  or 12n + 6 = 13n; ∴ n = 6. ∴ Middle term =   th = 7th = a7  2  ∴ The number of means = 2n = 2 × 6 = 12.  3 65. (c) Let the n arithmetic means between two numbers a = a1 + 6d = 28 + 6  −  = 28 – 9 = 19. and b be A1, A2, A3, ..., An.  2 a+b Hence middle term of the A.P. = 19. Then A1 + A2 + A3 + ... + An = × n. 2 69. (c) Let the digit in the unit’s place be a – d a+b ∴ By question, × n = 20 ...(1) Digit in the ten’s place = a 2 and the digit in the hundred’s place be a + d Again by question, a = 3b ...(2) ∴ From (1), bn = 10 ...(3) Sum of digits = (a – d) + a + (a + d) = 3a Again by question, An = 2A1 Also sum = 15 (Given) ⇒

a + nb an + b =2 n +1 n +1

⇒ a + 10 = 2an + 2b, from (3) ⇒ 3b + 10 = 6bn + 2b, from (2) ⇒ b + 10 = 60, from (3) ⇒ b = 50 ∴ a = 150, from (2).

∴ 3a = 15 ⇒ a = 15 ÷ 3 = 5 Original number = (a – d) + 10a + 100 (a + d) = 111a + 99d = 111 × 5 + 99d = 555 + 99d Number formed by reversing the digits = (a + d) + 10a + 100 (a – d) = 111a + 99d = 111 × 5 – 99d = 555 – 99d

 hus the digit in the unit’s place is 5 – 3 = 2, in T the ten’s place is 5 and in the hundred’s place is 5+3=8 Hence the number is 852. 70. (a) We have, S1 = (n/2) [2 ⋅ 1 + (n – 1) ⋅ 1]

S2 = (n/2) [2 ⋅ 2 + (n – 1) ⋅ 2]



Sm = (n/2) [2 ⋅ m + (n – 1) ⋅ m]

∴ S1 + S2 + ... + Sm = n (1 + 2 + 3 ... + m) +

n (n −1) × (1 + 2 + ... + m) 2



=

m (m + 1)  n2 − n  n +  2 2  

=

m (m + 1) n (n + 1) 1 ⋅ = mn (m + 1) (n + 1). 2 2 4

n–1

⇒ ar4 = 32

...(1)

and ar7 = 256

...(2)

Dividing (2) by (1), r3 =

256 = 8 = 23 32

= ar = 2 (2) = 2 × 8 = 16.

⇒ ar = p ⇒ ar7 = q ⇒ ar10 = s 4

∴ q2 = (ar7)2 = a2 r14 Also, ps = ar4 × ar10 = a2 r14 Hence q2 = ps.

We have, a3 = (a1)2 ⇒ ar2 = a2 ⇒ r2 = a

...(1)

Also a2 = 8 ⇒ ar = 8

...(2)

Multiplying (1) and (2), we get ar = 8a ⇒ r = 8 3

∴ r = (8)1/3 = 2 ∴ From (1), a = (2)2 = 4.

Since ap , aq , ar in the G.P. ⇒ (aq)2 = ap × ar

77. (a) Since x, y, z are in G.P., z y = ⇒ xz = y2 ∴ y x ⇒ ⇒ ⇒ ⇒ ⇒

log log log log log

78. (b) Since

73. (a) Let a be the first term and r, the common ratio of G.P.



aq = ARq – 1  and  ar = ARr – 1

⇒ 2q – 2 = p + r – 2 ⇒ 2q = p + r ⇒ p, q, r in A.P.

72. (b) Let a be the first term and r, the common ratio of G.P.

3



⇒ A2 R2q – 2 = A2 Rp + r – 2

3

Then, a5 = p a8 = q a11 = r

76. (a) Let A be the first term and R the common ratio of G.P.

⇒ (ARq – 1)2 = ARp – 1 × ARr – 1

∴ r = 2 From (1), a (2)4 = 32 ⇒ 16a = 32, ∴ a = 2 Hence the 4th term of the G.P.

⇒ b – a = c – b ⇒ kb – a = kc – b kb kc ⇒ a = b ⇒ ka, kb, kc are in G.P. k k

Then ap = ARp – 1

∴ a5 = ar4 and a8 = ar7

3

a, ar, ar2, ar3 i.e., 3, – 6, 12, – 24.

75. (b) Since a, b, c are in A.P.

71. (c) Let a be the first term and r the common ratio of the G.P. Then, an = ar

 et 1st term, a = 1 and r be the common ratio of L G.P. According to the question, we have a3 = a1 + 9 ⇒ ar2 = a + 9 ⇒ a (r2 – 1) = 9 ...(1) Also, a2 = a4 + 18 ⇒ ar = ar3 + 18 ⇒ ar (r2 – 1) = – 18 ...(2) Dividing (2) by (1), we have r = – 2 Substituting r = – 2 in (1), we get a (4 – 1) = 9 ⇒ 3a = 9, ∴ a = 3 Hence the four numbers are

[ a = r 2]

Hence, a6 = ar5 = 4 (2)5 = 4 × 32 = 128.

(xz) = log ( y2) ⇒ log x + log z = 2 log y x + log z = log y + log y z – log y = log y – log x y – log z = log x – log y x, log y, log z are in A.P.

1 1 1 , , are in A.P. x + y 2y y + z

∴

1 1 1 1 – = – 2y x+ y y+z 2y

⇒

x + y − 2y 2y − y − z x− y y−z = ⇒ = 2 y ( x + y) 2 y ( y + z) x+ y y+z

⇒ (x – y) ( y + z) = (x + y) (y – z) ⇒ xy + xz – y2 – yz = xy – xz + y2 – yz ⇒ 2xz = 2y2 or y2 = xz z y ⇒ = y x ⇒ x, y, z are in G.P.

593

⇒ 198d = 594 ⇒ d = 594 ÷ 198 = 3

Sequences and Series

74. (a) Let a be the first term and r be the common ratio of the G.P.

∴ (555 + 99d) – (555 – 99d) = 594

594

Objective Mathematics

79. (c) Since a, b, c are in G.P. b c ∴ = ⇒ b2 = ac ...(1) a b

⇒ (d + 5) (d – 4) = 0 ∴ d = – 5, 4 When d = – 5 The numbers are 12, 7, 2. When d = 4, the numbers are 3, 7, 11.

1 1  1 Now, a2 b2 c2  3 + 3 + 3  a b c 

b 2 c 2 a 2 c 2 a 2b 2 + + = a b c



=



= a 3 + b 3 + c 3.

83. (a) Let d be the common difference of A.P. and R (≠ 0), the common ratio of G.P., then

ac ⋅ c 2 (b 2 ) 2 a 2 ⋅ ac + + a b c

[Using (1)]

80. (a) Let a be the first term and r be the common ratio of G.P. Given, ap = arp – 1 = q and aq = arq – 1 = p q ar p −1 = q −1 p ar

Dividing (1) by (2),

⇒ rp – q =

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

q q ⇒r=   p  p

q



= q⋅

q

1+

=

p p−q

q

r = 3; 486 = last term = nth term  = arn – 1 = a (3)n – 1 ⇒ a (3)n – 1 = 486 ...(1)

1 p−q

q

p p−q

=

and 728 = sum of n terms =

q p−q

p

q p−q

⇒ 1

q p−q q

p p−q

 q p  p−q =  q . p 

81. (c) Let A be the first term and R, the common ratio of G.P. Then a = ax = ARx – 1, b = ay = ARy – 1 and

c = az = ARz – 1

∴ ( y – z) log a + (z – x) log b + (x – y) log c = log ay – z + log bz – x + log cx – y

⋅R

(x – 1) (y – z) + (y – 1) (z – x) + (z – 1) (x – y)

]

= log (A0 × R0) = log 1 = 0. 82. (a), (b)  Let the three numbers in A.P. be a – d, a, a + d.

 Given (a – d) + a + (a + d) = 21 ⇒ 3a = 21, ∴

a (3n − 1) = 728 ⇒ a ⋅ 3n – a = 1456 2

⇒ a ⋅ 3n – 1 ⋅ 3 – a = 1456 ⇒ 486 × 3 – a = 1456

a = 21 ÷ 3 = 7

∴ The numbers are 7 – d, 7, 7 + d. I f the second number is reduced by 1 while the third number is increased by 1, the resulting numbers are 7 – d, 6, 8 + d which are given to be in G.P. 6 8+d ∴ = 7−d 6 ⇒ 36 = (7 – d) (8 + d) ⇒ 36 = 56 – d – d2 ⇒ d2 + d – 20 = 0

[using (1)]

⇒ a = 1458 – 1456 = 2. Substituting a = 2 in (1), we get 2 ⋅ 3n – 1 = 486 ⇒ 3n – 1 = 243 = 35 ∴ n – 1 = 5 ⇒ n = 6. Hence the first term = 2 and the number of terms = 6. 85. (a) Let the three numbers in G.P. be a, ar, ar2 ...(1) (given)

Subtracting 1, 7, 21 from the numbers, we get

= log [(ARx – 1)y – z ⋅ (ARy – 1)z – x ⋅ (ARz – 1)x – y] = log [A

a (r n − 1) r −1

∴ a + ar + ar2 = 56

= log (ay – z ⋅ bz – x ⋅ cx – y) y–z+z–x+x–y

84. (c) Let a be the first term and n be number of terms in the G.P. If r be the common ratio, then

q Now ap + q = qrp + q – 1 = arp – 1 rq = q    p q p−q

q = p + d, r = p + 2d and y = xR, z = xR2 so that q – r = – d, r – p = 2d, p – q = – d ∴ xq – r ⋅ yr – p ⋅ zp – q = xd ⋅ (xR)2d ⋅ (xR2)– d = (x– d ⋅ x2d ⋅ x– d) (R2d ⋅ R– 2d) = (x– d + 2d – d ) ⋅ (R2d – 2d) = x0 ⋅ R0 = 1 × 1 = 1.



a – 1, ar – 7, ar2 – 21 which are given to be in A.P.

∴ (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7) ⇒ ar – a – 6 = ar2 – ar – 14 ⇒ a – 2ar + ar2 = 8 Subtracting (2) from (1), we get 16  3ar = 48 ⇒ a = r 16 in (1), we have Substituting a = r 16 + 16 + 16r = 56 r 2 ⇒ 16r – 40r + 16 = 0 ⇒ 2r2 – 5r + 2 = 0 1 ⇒ (r – 2) (2r – 1) = 0, ∴ r = 2, 2



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

86. (c) Let a be the first term and r, the common ratio of G.P. Given a1 + a2 + a3 = 21



⇒ a + ar + ar2 = 21

⇒ a (1 + r + r ) = 21

...(1)

2

Again a4 + a5 + a6 = 168 (Given) ⇒ ar3 + ar4 + ar5 = 168

89. (a) We have,

⇒ ar3 (1 + r + r2) = 168 ...(2)

(1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + ... upto n terms

Dividing (2) by (1), r3 = 8, ∴ r =

8 =2 ∴ From (1), a (1+ 2 + 4) = 21 3



=



=



=

∴ a = 21 ÷ 7 = 3. 87. (b) Let a be the first term and r the common ratio of the G.P. ∴ S = a + ar + ar2 + ... + arn

– 1

=



a (1 − r n ) 1− r ...(1)

and P = a × ar × ar2 × ... × arn – 1 = an r1 + 2 + 3 ... + (n – 1) = an n ( n − 1) r 2 2 2n n (n – 1) ⇒ P = a r 1 1 1 1 + + ... + n −1 Also, R = + a ar ar 2 ar

n

...(2)

= (10 – 1) + (102 – 1) + (103 – 1) + ... to n terms = (10 + 102 + 103 + ... to n terms)

=

10 (10n − 1) 10 (10n − 1) −n – n = 10 − 1 9

=

1 (10n + 1 – 9n – 10). 9

91. (d) We have,

...(3)

Sn = .5 + .55 + .555 + ... upto n terms 5 = [.9 + .99 + .999 + ... n terms] 9

=

1  1   1  5   1 −  + 1 −  + 1 −  + ... 9  10   100   1000  



=

5 (1 + 1 + 1 + ...) −  1 + 1 + 1 + ...      2 3  10 10 10  9 



= 5 9

n

S P2 =   . R

88. (b), (c)  Let the numbers be a, ar, ar2, ar3. ∴ a (1 + r + r + r ) = 60 2

a + ar 2

Also or

3

3

= 18 

a (1 + r3) = 36

Dividing (1) by (2), we get

1 + r + r2 + r3 5 = 1 + r3 3

 x 2 (1 − x n )  n − . 1− x  

– (1 + 1 + 1 + ... to n terms)

From (2) and (3), we get

1 1− x

9 + 99 + 999 + ... upto n terms

S (1 − r n ) r − 1 =a arn – 1 = a2 r(n – 1) ⋅ R 1− r rn −1

∴  S  = a2n rn (n – 1) R

1 [(1 + 1 + 1 + ... n terms) 1− x – (x2 + x3 + x4 + ... to n terms)]

90. (b) We have,

1  1− (r n − 1) 1 1  r n  ⋅ ⇒ R = ⋅ = (r − 1) ar n −1 a  1 − 1    r ∴

1 − x 2 1 − x3 1 − x 4 + + + ... to n terms 1− x 1− x 1− x

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

 1   1−  1  10n   n −   10 1 − 1   10 



=

5 1 10  1  n − × 1 − n    9  10 9  10  



=

5 9

 1 1   n − 9 1 − 10n   .   

595

(1 + r )(1 + r 2 ) 5 = (1 + r )(1 − r + r 2 ) 3 ⇒ 5 – 5r + 5r2 = 3 + 3r2 ⇒ 2r2 – 5r + 2 = 0 1 ⇒ r = 2 or , 2 Putting r = 2 in (2), we get a (1 + 8) = 36 or a = 4 ∴ The numbers are 4, 8, 16, 32 1 and r = gives the numbers 32, 16, 8, 4. 2 ⇒

Sequences and Series

 hen r = 2, then from (3), a = 8 and the numbers w are 8, 16, 32. 1 when r = , then from (3), a = 32 and the numbers 2 are 32, 16, 8.

596

92. (a), (c)  Let the numbers be a – d, a, a + d.

Objective Mathematics

Then, a – d + a + a + d = 15 ⇒ 3a = 15; a = 5. a+3 a+d +9 = Given: a − d +1 a+3 ⇒ (a + 3)2 = (a – d + 1) (a + d + 9) ⇒ 64 = (6 – d) (14 + d) [ a = 5] ⇒ d2 + 8d – 20 = 0 ⇒ (d + 10) (d – 2) = 0; d = – 10 or 2 Hence the numbers are 15, 5, – 5 or 3, 5, 7. 93. (a) Let a be the first term and r be the common ratio, 4 4 4 Given a4 = ⇒ ar3 = or 28r3 = 49 49 49 1 4 1 = =   28 × 49 7×7×7 7

⇒ r3 =

3

1  · | r | < 1 7 28 a 28 × 7 98 28 = = = = . Now, (S)∞ = 6 1 1− r 6 3 1− 7 7

∴ r =

94. (b) 91/3 ⋅ 91/9 ⋅ 91/27 ... to infinity = 9

1 1 1 + + + ... to ∞ 3 9 27

= 9

3 1− 1

3

= 9

2

3 3

= 9

1 2

Given a + ar + ar2 = 63 3 ⋅ ar2 and a ⋅ ar = 4 a=

= 3.

= a



= a 2b 2 =

1

3 r 4

...(1)

...(2)

−5 ± 25 − 84 2 ∴ Real value of r is 4. Putting this value in (2), 3 × 4 = 3. a = 4 ∴ The three numbers are, 3, 3 × 4, 3 × 42, i.e., 3, 12, 48. 96. (b) Let the G.P. be x, xy, xy2, ... Then tp + q = xyp + q – 1 = a tp – q = xyp – q – 1 = b

a ; Dividing (1) by (2), y2q = b 1

p −1

 a  2q   b

p + q −1 p −1 − 2q 2q

1

ab .

97. (c) As a, b, c, d are in G.P., therefore b c d = = = r (say) a b c



⇒ b = ar, c = br = ar ⋅ r = ar2, d = cr = ar2 ⋅ r = ar3. 2 (a + b2 + c2) (b2 + c2 + d2) = (a2 + a2 r2 + a2 r4) (a2 r2 + a2 r4 + a2 r6) = a4 r2 (1 + r2 + r4) (1 + r2 + r4) = (a2 r + a2 r3 + a2 r5)2 = (a ⋅ ar + ar ⋅ ar2 + ar2 ⋅ ar3)2 = (ab + bc + cd)2. 98. (b) If r be the common ratio of the given G.P., then

99. (a) Let the three numbers in G.P. be a ⋅ a ⋅ ar = 216; r

Then

3 3 3 r+ r⋅r+ r ⋅ r2 = 63 4 4 4 r3 + r2 + r – 84 = 0 (r – 4) (r2 + 5r + 21) = 0;

 a  2q ∴y=   b

⋅b

p + q −1 2q

n ( n −1) 2

∴ P2 = a2n rn (n – 1) = (a ⋅ arn – 1)n = (ab)n.

∴ r = 4,

and

p + q −1 p −1 + 2q 2q



Again

Putting in (1), or or

1−

b =a⋅   a

⇒ P = an ⋅ r1 + 2 + 3 + ... + (n – 1) = an ⋅ r

95. (a) Let the three numbers be a, ar, ar2

or

∴ tp = xy

p–1

p + q −1 2q

b = arn – 1; P = a × ar × ar2 × ar3 × ... × arn – 1 1

1

b From (1), x = a    a

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

a , a, ar. r

or a3 = 63; or a = 6.

a a ⋅ a + a ⋅ ar + . ar = 216; r r

a2 (1 + r2 + r) = 216; r

or

or 36 (1 + r + r2) = 216r, or r2 + r + 1 = 6r ; or r2 – 5r + 1 = 0 or r =

5 ± 25 − 4 5 ± 21 = 2 2

Hence the three numbers in G.P. are 12 ; 6, 3 (5 ± 21 ) 5 ± 21 100. (d) According to the question, the series is 1 + a + ca + a (ca) + c (aca) + ... to 2n terms i.e.,

1 + a + ca + ca2 + c2 a2 + ... to 2n terms

∴ Sum = (1 + ca + c2 a2 + ... to n terms) + (a + ca2 + c2 a3 + ... to n terms)

=

1{1 − (ca ) n } 1 − ca

+

a {1 − (ca ) n } 1 − ca

=

(1 + a )(1 − c n a n ) . 1 − ca

3 −1 1(1 − 3 ) > 1000; or > 1000 2 1− 3

Taking, logarithm, we get x log a = y log b = z log c = log k.

or 3n > 2001; but 36 = 729 and 37 = 2187; ∴ n > 6; but n is a positive integer; ∴ n = 7, 8, 9, ... ∴ The minimum number of terms = 7. 102. (c) We have, Sn =

∴ x =

a [1 – r2n – 1]. 1− r Putting 1, 2, 3, ..., n for n in it and summing up we get S1 + S3 + S5 + ... + S2n – 1 a = 1 − r [(1 + 1 + ... to n terms) – (r + r3 + r5 + ... to n terms)]

∴ S2n – 1 =

2n r {1 − (r 2 ) n }  a    n −  = a n − r ⋅ 1 − r 2  . = 2 1− r  1− r 1 − 1 − r r     

103. (c) Let the last three numbers in A.P. be b, b + 6, b + 12 and the first number be a. Hence the four numbers are a, b, b + 6, b + 12 and

a, b, b + 6 are in G.P. i.e., b2 = a (b + 6)

or

b2 = (b + 12) (b + 6)

or

18b = – 72 ∴ b = – 4,

[∴ a = b + 12]

 log a  = log b a  ∵  log b 

logb a = logc b.

106. (c) We have,



S1 =

S2 =

1 1−

1 2

=2 Given

2

=3 1 3 ... ... ... ... ... ... ... ... ... ... ... 1−

... ... ... ... ... ... ... ... ... ... ... p =p+1 Sp = 1 1− p +1

∴ S1 + S2 + ... + Sp = 2 + 3 + 4 + ... + ( p + 1) = p [2 ⋅ 2 + (p – 1) ⋅ 1] = p ( p + 3) . 2 2

107. (d) Let a be the first term and r, the common ratio of G.P. | r | < 1.

Hence the four numbers are 8, – 4, 2, and 8. 104. (a) Let the three digits be a, ar and ar2

Given, a1 – a2 = 2

Given, a + ar2 = 2ar + 1

⇒ a – ar = 2

⇒ a (1 – r) = 2 ...(1)

⇒ 3a (1 + r) = 2ra (1 + r) 3 ⇒ (1 + r) (3 – 2r) = 0, ∴ r = 2 , – 1 3 , from (1), when r = 2 1 1 a = = = 4, 2 (r − 1) 2 3   − 1 2  1 when r = – 1, a = , which is not possible, for 4 a is an integer. 9 3 Hence a = 4, ar = 4 ·  = 6, ar2 = 4 ·   = 9. 4 2 ∴ Required number is 469.

or



From (i), a = – 4 + 12 = 8.

log a log b = log b log c

or

..(i)

⇒ a (r2 – 2r + 1) = 1 ⇒ a (r – 1)2 = 1 3 Also given, a + ar = 2 (ar + ar2)

log k log k log k , y = and z = . log a log b log c

z Since x, y, z are in G.P. ∴ y = y x log k log a log k log b ∴ × = × log b log k log c log k

a (1 − r n ) 1− r

Given, a = b + 12

Sequences and Series

Then

n

n

597

105. (a) Given, ax = by = cz = k (say)

101. (a) Let, the sum of n terms exceeds 1000.

Also given, S = 50

...(1) a ⇒ = 50 1− r

...(2)

Multiplying (1) and (2), we have a = 2 × 50 1− r



a (1 – r) ×



a2 = 100 = (10)2

∴ a = 10 Putting a = 10 in (1), we get 10 (1 – r) = 2 or 10 – 10r = 2  ∴ r = −8 = 4 . −10 5 108. (a) Since r > 1, 1 < 1 r ar a = ∴ x = 1 r −1 1− r

⇒ – 10r = 2 – 10 = – 8

598

Similarly, y =

Objective Mathematics

z=

and



b br =  1 r +1 1− −   r c

1 1− 2 r

cr r2 −1

∴ a – b =

2

=

2 ∴ xy = ar × br = abr 2 r −1 r +1 r −1

...(1)

...(2)

Dividing (2) by (1), we get

ab xy abr 2 r2 −1 = 2 × = . c z r −1 cr 2

109. (a) We have,





=

3 16 5 16 4 1 12 − 5 7 × − × = − = = . 4 15 16 15 5 3 15 15

110. (b) Let the two numbers be a and b,then

G =

ab or G2 = ab

Also, p and q are two A.M.s between a and b. ∴ a, p, q, b are in A.P. ∴ p – a = q – p and q – p = b – q ∴ a = 2p – q and b = 2q – p ∴ G2 = ab = (2p – q) (2q – p). 111. (c) We have, 

1 3 7 15 + + + +... to n terms 2 4 8 16

1 1 1  = n –  + + + ... n terms  2 4 8 

Subtracting (3) from (1), we get 2b = 8, ∴ b = 4. ∴ Required numbers are 64 and 4. 113. (b) The A.M. between b and c is a b+c ∴ = a ⇒ b + c = 2a 2



c c = br ⇒ r =   b



c c g1 = br = b      and  g2 = br2 = b    b b

∴

 c   c  2  g13 + g 23 = b3   +     b   b  



= b3 ×



= bc (2a) = 2abc.

3

1/ 3

112. (b) Let the numbers be a and b. ...(1)

c b

and

[ b + c = 2a]

...(i)

g2 g2 g1 = ⇒ 1 =a g1 g2 a g2 b g2 = ⇒ 2 =b g1 g2 g1 g 22 g12 + g1 g2

...(ii)

∴ From (i) and (ii), we get g2 g2 2A = 1 + 2 g2 g1 115. (b) We have, x = ar3, y = ar9, z = ar15 z z y y ⇒ = r6 and = r6 ⇒ = y y x x ⇒ x, y, z are in G.P. 116. (a) a, b, c are in A.P.

...(2)

2/3

 c b+c 1 +  = b2 c × b  b

114. (c) Let the two numbers be a and b. a+b ∴ A = or 2A = a + b 2 Again, a, g1, g2, b are in G.P. g1 g b = 2 = ∴ a g1 g2

∴ a + b =

1   1− 1  2n  1 = n – = n – 1 + n = n – 1 + 2 – n. 2 1− 1 2 2

ab = 16, or ab = 256 ∴ ∴ (a – b)2 = (a + b)2 – 4ab

Adding (1) and (3), 2a = 128, ∴ a = 64

Now

1  1  1  1  = 1 −  + 1 −  + 1 −  + 1 −  +... to n terms  2   4   8   16 

Since A.M. between a and b is 34 a+b ∴ = 34 or a + b = 68 2 Since G.M. between a and b is 16.

...(3)

1/ 3

3 5 5 3 3   5  Sum =  + 3 + 5 + ... to ∞  –  2 + 4 + 6 + ... to ∞  4 4 4 4 4  4 



3600 = 60

Again g1 and g2 are two G.M.’s between b and c ∴ b, g1, g2, c are in G.P. If r be the common ratio, then

3 5 3 5 3 5 − + − + − +... 4 4 2 4 3 4 4 45 4 6

3 5 2 4 4 − = 2 2 1 1 1−   1−   4 4

= (68)2 – 4 × 256 = 4624 – 1024 = 3600

∴ 2b = a + c x is the G.M. between a and b, ∴ x2 = ab. y is the G.M. between b and c, ∴ y2 = bc.

∴ a = – 4.  utting the value of a in (1), the required numbers are P 8, – 4, 2, 8.

a (1 − r n ) ; Then S1 = 1− r S2 =

a a+6 = ; or a2 = (a + 12) (a + 6) a + 12 a

or a2 = a2 + 18a + 72; or 18a = – 72;

117. (a) Let the G.P. be a, ar, ar2, ar3 ...



∴

a (1 − r 2 n ) a (1 − r 3n ) ; S3 = 1− r 1− r

∴ S1 (S3 – S2)

121. (a), (b),  Let the three numbers in  A.P. be a – d, a, a + d. Then a – d + a + a + d = 15; or 3a = 15; or a = 5. Again a – d + 1; a + 4, a + d + 19 are in G.P.



a (1 − r n ) a = × (r2n – r2n) 1− r 1− r



 ar 2 n  a 2r 2n (1 − r n )  (1 – rn)2 =  = 2 (1 − r ) 1 − r  



= (S2 – S1)2.

That is, 6 – d, 9, 24 + d are in G.P. 2

∴ (9)2 = (6 – d) (24 + d); or 81 = 144 – 18d – d2 or d2 + 18d – 63 = 0; or (d + 21) (d – 3) = 0; ∴ d = 3 or – 21. Hence the required numbers are 2, 5, 8 or 26, 5, – 16.

118. (b) Let the 3n terms of G.P. are a, ar, ar2, ...,

arn – 1, arn, arn + 1, arn + 2, ... ar2n – 1, ar2n, ar2n + 1, ar2n + 2, ..., ar3n – 1 122. (a) Since a, b, c are in G.P., therefore b2 = ac Then S1 = a + ar + ar2 + ... + arn – 1 =

S2 = arn + arn + 1 + arn + 2 + ... + ar3n – 1 n



a (1 − r n ) 1− r

=

=

ar (1 − r ) 1− r

Hence log an, log bn, log cn are in A.P.

(1 − r n ) 2 (1 − r ) 2

=

1

Given: tn = 2 (tn + 1 + tn + 2 + tn + 3 + ... to ∞) tn = 2 ⋅

tn+1 , 1− r

( common ratio = r)

1

or a, ∴ b2 or

a = k x, b = k y, c = k z. b, c are in G.P. = ac; or k2y = kx ⋅ kz = kx + z 2y = x + z. Hence x, y, z are in A.P.

125. (c) We have, a + b + c = 25 ...(1)

t 1− r 1− r = n +1 ; or =r tn 2 2 1 or 1 – r = 2r ; ∴ r = 3 . ∴



2, a, b are in A.P., ∴ 2a = 2 + b

...(2)



b, c 18 are in G.P., ∴ c = 18b

...(3)

2

From (1) and (2), 3b + 2c = 48; or 3b = 48 – 2c.

120. (b) As the last three terms are in A.P. whose common difference is 6 and the first term is equal to the last, the four numbers can be taken as follows : a + 12, a, a + 6, a + 12 But the first three are in G.P.

1

y 124. (a) We have, a x = b = c z = k (say)

119. (a) Let the G.P. be 1, r, r2, r3, ...

or

a a , , a – 2. 2 4

By question,

(1 − r n ) a (1 − r n ) ⋅ ar2n = S 1S 3 1− r 1− r Hence S1, S2, S3 are in G.P.

123. (b) The numbers can be taken as a,

a a , , a – 2 are in A.P. 2 4 a a ∴ 2 ·  = + a – 2; or a = 2. 4 2 1 , 0. Hence the numbers are 2, 1, 2

ar 2 n (1 − r n ) 1− r

Now (S2)2 = a2 r2n

or log b2n = log an + log cn or 2 log bn = log an + log cn.

n

S3 = ar2n + ar2n + 1 + ar2n + 2 + ... + ar3n – 1



or b2n = (ac)n = an cn

...(1)

∴ From (3), c2 = 6 (48 – 2c) = 288 – 12c or c2 + 12c – 288 = 0; or c2 + 24c – 12c – 288 = 0 or (c + 24) (c – 12) = 0; ∴ c = 12, as c ≠ – 24. ∴ From (3), b = 8 and from (2), a = 5.

599

x2 + y 2 ∴ b2 = 2 ∴ b2 is the A.M. between x2 and y2.

Sequences and Series

Now, x2 + y2 = ab + bc = b (a + c) = b (2b) = 2b2

600

126. (a) Given that a, A1, A2, b are in A.P.

Objective Mathematics

a + A2 A +b Therefore, A1 = , A2 = 1 2 2 1 ⇒ A1 + A2 = (a + b + A1 + A2) 2 1 1 (A1 + A2) = (a + b) ⇒ 2 2 ⇒ A1 + A2 = a + b

a+c  2 2 pr Since p, q, r in H.P., ∴ q = p + r 

129. (a) Since a, b, c are in A.P. ∴ b =

∴ G ⇒ G

2 1

= a G2 , G G

2 2

2 2

...(i)

i.e., b2 q2 = acpr Putting the value of b and q in (3), we get

...(ii)

 a + c   2 pr  = acpr      2   p+r 2

= b G1

= ab G1 G2 ⇒ G1 G2 = ab

a+b A + A2 Hence 1 = . ab G1G 2 127. (b) Let the three terms of G.P. be Their product = 512 ⇒

a , a, ar r

a × a × ar = 512 r

⇒ a3 = (8)3 ⇒ a = {(8)3}1/3 = 8 a Also + 8, a + 6, ar are in A.P. r a ⇒ 2 (a + 6) = + 8 + ar r 8 8 ⇒ 2 (8 + 6) = + 8 + 8r ⇒ 28 = + 8 + 8r r r 1 7 1 5 ⇒ +r+1= ⇒ +r– =0 r 2 r 2 5 ⇒ r2 – r + 1 = 0 ⇒ 2r2 – 5r + 2 = 0 2 1 ,2 ⇒ (2r – 1) (r – 2) = 0 ⇒ r = 2 1 For r = , the numbers are 16, 8, 4 2 and for r = 2, the numbers are 4, 8, 16. 128. (b)

H is the H.M. between a and b 2ab H 2b H 2a ∴ = and = a+b a a+b b a+b N ow, applying componendo and dividendo, we get H+a 3b + a = ...(1) H−a b−a H+b 3a + b 3a + b and = =– ...(2) H−b a−b b−a ∴ H =

Adding (1) and (2), we have  3a + b  H+a H+b 3b + a + = + −  H−a H−b b−a  b−a  3b + a 3a + b – b−a b−a 1 = (3b + a – 3a – b) b−a 2(b − a ) = 2. = b−a

=

...(2)

Since ap, bq, cr are in G.P., ∴ (bq)2 = (ap) (cr)

and a, G1, G2, b are in G.P. 2 1

...(1)

∴

...(3)

2

(a + c) 2 ( p + r )2 (a + c) 2 pr = ac ⇒ = 2 ac pr ( p + r)

a 2 + c 2 + 2ac p 2 + r 2 + 2 pr = ac pr p r a c + +2, + +2 = ⇒ r p c a ⇒

Hence

a c p r + = + . c a r p

130. (c) Let ax = by = cz = k (say) ∴ a = (k)1/x, b = (k)1/y, c = (k)1/z

...(1)

 ubstituting these values of a, b, c in (1), we S have ∴ ∴

1 1 + z

(k)2/y = (k)1/x ⋅ (k)1/z = (k ) x 2 1 1 + y = x z 1 1 1 , , are in A.P. x y z

Hence x, y, z are in H.P. 131. (b) Since a, b, c are in A.P. a+c ∴ b =  ...(i) 2 b, c, d are in G.P. ...(ii) ∴ c2 = bd Also, c, d, e, are in H.P. 2ce ∴ d =  ...(iii) c+e Substituting the values of b and d from (i) and (iii) respectively in (ii), we get

c2 =

a+c 2ce e (a + c) ⋅ or c = 2 c+e c+e

⇒ c2 + ce = ae + ce ⇒ c2 = ae, which shows that a, c, e are in G.P. 132. (b) We have, 1 1 1 1 + = + 2ac 2ac b−a b−c −a −c a+c a+c



=

a c 1 1 + = + . ac ac c a

133. (c) Let the roots be a – d, a, a + d Then, the sum of three numbers = 3a = 12 ∴ a = 4 is a root ∴ (x – 4) (x2 – 8x + 7) = 0 ⇒ (x – 4) (x – 1) (x – 7) = 0 ⇒ x = 1, 4, 7 or 7, 4, 1 ∴ d = ± 3.

or a – d = 3kad; a – d = k (ab + bc + cd) Hence ab + bc + cd = 3ad. 137. (a) Since the pth and the qth term of H.P. are respectively qr and pr, therefore the pth and qth terms of 1 1 and . the corresponding A.P. are respectively qr pr  et a and d be respectively the first term and the L common difference of the corresponding A.P.

134. (a) We have, n S1 = 1 [2a + (n1 – 1) d] 2 2S1 ⇒ n = 2a + (n1 – 1) d 1

138. (c)



+ [2a + (n2 – 1) d] (n3 – n1)

 = 0.

+ [2a + (n3 – 1) d] (n1 – n2)

135. (b) Let x be the common difference of the A.P. a, b, c, d, e.

[ d – c = x]

136. (c) Since a, b, c and d are in H.P., therefore 1 1 1 1 , , , are in A.P. a b c d



– – – – –

1 a 1 a 1 b 1 c 1 a

1 1 1 1 = – = – = k (say) c b d c = k; a – b = kab = k; b – c = kbc = k; c – d = kcd = 3k; a – d = k (ab + bc + cd)

=

1 pqr = = pq. a + (r − 1) d 1 + (r − 1) ⋅1

a, b, c are in H.P.; 1 1 ∴ , and 1 are in A.P. a b c ⇒ a + b + c , a + b + c , a + b + c are in A.P. a b c b+c a+b c + a ⇒ 1 + ,1+ ,1+ are in A.P. a c b b+c c+a a+b ⇒ , , are in A.P. a b c ⇒

1 b 1 b 1 c 1 d 1 d

p−q 1 ; or d = pqr pqr

1 1 p −1 = − pqr qr pqr



2S1 2S 2S3 (n – n3) + 2 (n3 – n1) + (n1 – n2) n1 2 n2 n3

∴ e = a + 4x; c = a + 2x ⇒ e – c = (a + 4x) – (a + 2x) = 2x = 2 (d – c)

( p – q) d =

∴ The rth term of H.P.

= [2a + (n – 1) d] (n2 – n3)



...(2)

∴ a =

n3 [2a + (n3 – 1) d] 2 2S3 ⇒ n = 2a + (n3 – 1) d 3

∴

and



S3 =

∴

...(1)

Subtracting (2) from (1), we get

n S2 = 2 [2a + (n2 – 1) d] 2 2S2 ⇒ n = 2a + (n2 – 1) d 2

∴

1 qr 1 a + (q – 1) d = pr

Then a + (p – 1) d =

139. (d)

a b c , , are in H.P. b+c c+a a+b

A is the single A.M. between a and b. a+b ∴ A = . 2 Let A1, A2, ... An be n A.M.s between a and b. ∴ a, A1, A2, ... An , b is an A.P. with common difference

d =

b−a . n +1

n (A1 + An) 2 n n = (a + d + b – d) = (a + b) 2 2 a+b ∴ S = n   = nA  2  Now, S = A1 + A2 + ... + An =

⇒

S = n. A

601

a+c a+c a+cc−a = a (c − a ) + −c (c − a ) =   c − a  ac 

Sequences and Series



602

140. (a) Let a be the first term and r, the common ratio of the given G.P. Then 100

Objective Mathematics

∑a

α=

n =1

2n

β=

100

∑a n =1

2 n −1

...(1)



...(2)

Then, a1 = a, a2 = ar, a3 = ar2, a4 = ar3, a5 = ar4 ∴ a1 × a2 × a3 × a4 × a5 = a × ar × ar2 × ar3 × ar4 = a5 r10 = (ar2)5 = 45. [ a3 = ar2 = 4] 142. (a) For the positive numbers a, c we have harmonic mean H = b { a, b, c are in H.P.} and geometric mean G = ac . ac > b.

a nc n 

...(2)

x2 – 4x + 1 = 0; 4 ± 16 − 4 4±2 3 = =2± 2 2

3.

Taking the positive sign,

x2 = (2 +



=

3 )2 =

(2 + 3 )2 4−3

(2 + 3 )2 (2 + 3 )2 = 22 − ( 3 ) 2 (2 + 3 )(2 − 3 )

2+ 3 a = . 2− 3 b 2− 3 a = . 2+ 3 b

145. (d) Since, x, 2x + 2, 3x + 3 are in G.P. ∴ (2x + 2)2 = x (3x + 3) or x2 + 5x + 4 = 0 ⇒ (x + 1) (x + 4) = 0 ∴ x = – 1, – 4.

143. (b) Let the numbers be a and b. If H is the harmonic mean then ...(1)

(Given)

But we know, AH = G2.

a+b 9 = , 2 2

a + b  ∵ A =  2  

∴ a + b = 9

...(2)

2ab 2ab ;∴4= Again H = 4 = a+b 9 ∴ ab = 18. From (2) and (3), we get

 hen x = – 1, the given series is – 1, 0, 0, ... which W do not form a G.P. Hence x = – 1 is rejected.  o, we take x = – 4 and the series in this case is S – 4, – 6, – 9 of which the fourth term

∴ 4A = G2; ∴ from (1), 2A + 4A = 27; 27 9 ∴ A = = 6 2 ∴

a 1 = x2, then x + = 4 or x2 + 1 = 4x b x



an + cn > ( ac )n > bn, ∴ an + cn > 2bn. 2

H = 4, 2A + G2 = 27 

b = 4. a

Similarly, taking the negative sign, we get

an + cn arithmetic mean = ; 2 an + cn A.M. > G.M., ∴ > 2 From (1) and (2), we get

If

∴ x =

∴

a nc n

a + b

or

...(1)

For the positive numbers an, cn, we have geometric mean =

ab , H = 2ab and G = 2H. a+b 4 ab ab = or a + b = 4 ab a+b

G =

or

141. (b) Let a be the first term and r, the common ratio of G.P.

But G > H; ∴

∴

⇒ β = a1 + a3 + ... + a199

⇒ β = a + ar2 + ... + ar198 ⇒ β = a (1 + r2 + ... + r198) α = r. From (1) and (2), we get β



∴ The numbers are 3 and 6. 144. (a), (b)  We have,

⇒ α = a2 + a4 + ... + a200

⇒ α = ar + ar3 + ... + ar199 ⇒ α = ar (1 + r2 + r4 + ... + r198) and

∴ a = 6, 3 and b = 3, 6.

27 3 =– = – 13.5. 2 2

146. (d) Since a, b, c are in A.P. a+c ∴ b = 2 Also a, b, c are in G.P., ∴ b2 = ac a+c    2 

...(2)

2

= ac, ∴ (a + c)2 – 4ac = 0

⇒ (a – c)2 = 0, ∴ a = c a+a = a 2

a – b = (a + b) 2 − 4ab = 92 − 4 × 18 = ± 3 ...(4)

∴ b =

∴ From (2) and (4), 2a = 9 ± 3 and 2b = 9

Hence a = b = c.

3;

...(1)

From (1) and (2), we have

...(3)

= (– 9) ×

[From (1)]

Since x, y, z are in G.P. ∴ y2 = xz k2 k2 log a log b = ⇒ = 2 (log b) log a log c log b log c

⇒

∴ S = 1. Hence, the given expression = 21 = 2.

⇒ logb a = logc b. 148. (a) Let the roots be α and β. α+β Then, = p and αβ = q (Given) 2 Now

152. (d) Let S = 1 + | cos x | + | cos2 x | + | cos3 x | + ... to ∞ It is an infinite G.P., whose common ratio r = | cos x | < 1 a 1 = ∴ S = 1 − r 1 −| cos x |

α+β = p ⇒ α + β = 2p 2 αβ = q ⇒ αβ = q2.

and

Hence, the given equation reduces to

Hence required equation is x2 – (α + β) x + αβ = 0 ⇒ x2 – 2px + q2 = 0. 149. (c) We have, ⇒

a+b 2ab =m ⋅ (Given) 2 a+b

( a + b) 2 = m 4ab

...(1)

( a + b) 2 –1= m–1 4ab ( a − b) 2 ⇒ = m – 1 4ab Dividing (1) by (2), we get ⇒

...(2)

( a + b) 2 m a+b = or = ( a − b) 2 m −1 a−b a = b

∴

p2 q2 + = a + b = 2A. q p 1

1

2

3

1

6 = log2   = log2 2 = 1 3 Also,

1

1

1 2 3 4 + + + + ... 8 16 32

= 2 4 ⋅ 2 8 ⋅ 216 ⋅ 2 32... = 2 4 1 2 3 4 + + + +...  4 8 16 32

...(1)

1 1 2 3 S = + + +...  2 8 16 32

...(2)

Let S = ∴

4

Subtracting (2) from (1), we get

1 1 − = log2 12 – log2 6 log12 2 log 6 2

 12  = log2   = log2 2 = 1.  6 1 1 1 1 − − = ∴ log 6 2 log 3 2 log12 2 log 6 2

151. (b) We have, 2 4 ⋅ 4 8 ⋅ 816 ⋅ 16 32...

1 1 , i.e., cos x = – , 2 2

153. (c) We have, 1 1 − = log2 6 – log2 3 log 6 2 log 3 2

150. (b) Let the two numbers be a and b. a+b Then, A = 2 Also, a, p, q, b are in G.P. p2 p q b q2 = = ;∴ = a and =b q a p q p

If – cos x =

2π 2π ,– . 3 3 π 2π π 2π , ,– ,– . ∴ x = 3 3 3 3

m + m −1 . m − m −1

∴

3 1− | cos x |

then x =

m . m −1

Applying componendo and dividendo, we get

1 1− | cos x |

= 43 or 2 = 26 3 1 −| cos x | 1 ⇒ =6 ⇒ = 1 −|cos x | 3 6 1 ⇒ 1 – | cos x | = 2 1 ∴ | cos x | = . 2 π π 1 , then x = ,– If cos x = 3 3 2 8

⇒ log3 2, log6 2, log12 2 are in H.P. 154. (a) Since a, b, c are in A.P. ∴ 2b = a + c Also, a2, b2, c2 are in H.P. ∴ b2 =

2a 2c 2 a2 + c2

603

1 1 1 1 1 S = + + + +... 2 4 8 16 32 1 1 2 1 = 4 = × = 1 4 1 2 1− 2

∴ x log a = y log b = z log c = k (say)

Sequences and Series

147. (d) Since ax = by = cz

604

Objective Mathematics

a+c 2a 2c 2 ∴   = 2 a + c2  2  ⇒ (a + c)2 (a2 + c2) = 8 a2 c2 2

⇒ (a2 + c2 + 2ac) (a2 + c2) = 8 a2 c2 ⇒ (a2 + c2)2 + 2ac (a2 + c2) – 8 a2 c2 = 0 ⇒ (a2 – c2)2 + 2ac (a – c)2 = 0 ⇒ (a – c)2 [(a + c)2 + 2ac] = 0 ⇒ a – c = 0 ∴ a = c. Hence, a = b = c 155. (d) We have, 1 3 7 15 + + + +... to n terms Sn = 2 4 8 16 1  1  1  1  = 1 −  + 1 −  + 1 −  + 1 −  + ... to n terms  2   4   8   16    1  n  1 −      2   = n – 1 − 1  n–  n  1  2  1− 2 1 n – 1 + n = n – 1 + 2–n = 2–n + n – 1. 2 1 2

=

=

156. (a) Since a, b, c are in H.P.



=

 1  1 1   b −  a − c    

2 2 1  1 1  4 1 1 1 − +    −  = 2 −  −  = b2  a c  ac  b a c



1  4 4  − 2 −  2 b b ac 



=

  1 1 2  2 2 4  ∵ +  =   = 2    a c   b  b 

4 3 − . ac b 2

157. (c) We have, 1 1 1 1 + = + b−a b−c a c ⇒

1 1 1 1 − = − b−a c a b−c

c−b+a b−c−a = (b − a ) c a (b − c) 1 1 ⇒ =– (b − a ) c a (b − c) ⇒

⇒ ab – ac = – bc + ac ⇒ 2ac = ab + bc = b (a + c) 2ac a+c ⇒ a, b, c are in H.P. ⇒ b =

Also, a, p, q, b are in A.P. Then, p – a = q – p = b – q. ∴ 2p – q = a and 2q – p = b. ∴ (2p – q) (2q – p) = ab = G2. 159. (c) Let the two numbers be a and b, where a > b.  et their A.M., G.M. and H.M. be respectively A, G, L and H. 8 Then A = G + 2 and G = H + , 5 8 or H = G – . 5 8  AH = G2 ⇒ (G + 2)  G −  = G2 5  ⇒ 2G = 16, ⇒ G = 8 = ∴ A = 8 + 2 = 10 =

ab ,

a+b ⇒ a + b = 20 ...(1) 2

∴ (a – b)2 = (a + b)2 – 4ab = 400 – 256 = 144. ∴ a – b = ± 12 ...(2) Solving (1) and (2), we get a = 16, b = 4. 160. (b) For the positive numbers a, c, we have the harmonic mean, H = b ( a, b, c are in H.P.)

2 1 1 = + . b a c The given expression

∴

 1  1 1  =  +  −   b  a c 

158. (a) Let the two given numbers be a and b. Then G2 = ab.

and the geometric mean, G = But G > H; ∴

ac > b

ac ...(1)

 or the positive numbers a5, c5, we have the geometF ric mean = a 5c 5 a5 + c5 and the arithmetic mean = . 2 5 5 ∴ a + c > a 5c 5 ; 2 or a5 + c5 > 2 ( ac )5 > 2b5. Hence a5 + c5 > 2b5. 161. (a) Since the unequal and positive numbers, a, b, c are in H.P, therefore, the H.M. of a and c = b, a+c and the A.M. of a and c = . 2 Since, A.M. > H.M., a+c > b or a + c > 2b  ...(1) 2 Again, since the unequal and positive numbers b, c, d are in H.P., therefore, the H.M. of b and d = c, b+d and the A.M. of b and d = . 2 Since A.M. > H.M., b+d > c; or b + d > 2c  ...(2) ∴ 2 Adding (1) and (2), we get a + d > b + c. ∴

⇒ b2 (a2 + c2) = 2 a2 c2 ⇒ b2 {(a + c)2 – 2ac} = 2 a2 c2 ⇒ b2 {4b2 – 2ac} = 2 a2 c2,

1 1 1 , , are in H.P. ⇒ 1 + ln x 1 + ln y 1 + ln z

⇒ 2b4 – b2 ac – (ac)2 = 0 ⇒ (2b2 + ac) (b2 – ac) = 0. ∴ either 2b2 + ac = 0 i.e., b2 = –

⇒ 13 ab = 6 (a + b)

i.e., – a = p, we get b 2 3 or 3 2

a 4 9 = or . b 9 4

Hence

i.e., the numbers are in the ratio 4 : 9 or 9 : 4. 164. (c) Since a, b, c, d are in G.P., we have b c d = = a b c y

z

...(i)



k k

1

⇒ k ⇒ ∴

x

1 1 − y x

=

k

1

k

1

= k

z

=

y

1 1 − z y

k

1

u

k

1

z

= k

165. (a) We have, S n =

n 1 1− n +1

=

n (n + 1) =n +1 n +1−1

⇒ Sn = (n + 1)2 2

2 So, S1 + S 2 S 2n–1 = 22 + 32 + ... + (2n)2 2

2+

1 (2n) (2n + 1) (4n + 1) – 1 6 1 = [n (2n + 1) (4n + 1) – 3]. 3 =

168. (c) Since, x, 1, z are in A.P. x+z ∴ 1 = . 2 Also, x, 2, z are in G.P., ∴ 2 = ∴

Hence x, y, z, u are in H.P.

a , b, c are in G.P. 2

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

a+c 2ac m 2b ⋅   = ac; or m2b = 2 a+c   ∴ a, m2b, c are in H.P.

1 1 − u z

1 1 1 1 1 1 – = – = – y x z y u z 1 1 1 1 , , , are in A.P. x y z u

= ac



u

Putting these values of a, b, c, d in (i), we get y

2

a+c  2 2 2 As a, mb, c are in G.P., m b = ac From (1) and (2), we get

a = k1/x, b = k1/y, c = k1/z, d = k1/u. 1

a+c ac ∴   2  2 

167. (c) As a, b, c are in A.P., b =

Let a = b = c = d = k. Then x

or b2 = ac;

i.e., (a – c)2 = 0 i.e., a = c; c , b, a, are in G.P. or – 2 a+c a+c ∴ b = = =a 2 2 i.e., a = b = c.

6p2 – 13p + 6 = 0 ⇒ (3p – 2) (2p – 3) = 0 ⇒ p =

using (1)

⇒ b2 (2b2 – ac) = a2 c2;

163. (a), (b)  Let the numbers be a, b, then clearly ab > 0, and 2ab 12 = (a + b) ab 13

Dividing by b, and putting

...(1)

2xz 2⋅2 = x+z 2

xz .

2

= 4.

Hence x, 4, z are in H.P. 169. (b)

m, n, r are in H.P. ∴ n =

2mr  m+r

...(1)

I f a be the first term and d the common difference of an A.P. then the (m + 1)th, (n + 1)th and (r + 1)th terms of A.P. are respectively a + md, a + nd, a + rd, which are in G.P. ∴ (a + nd)2 = (a + md) (a + rd) ⇒ a2 + 2and + n2d2 = a2 + ad (m + r) + mrd2 ⇒ d [2an + n2d] = d [a (m + r) + mrd] ⇒ a [2n – (m + r)] = – d (n2 – mr) n (m + r )   ⇒ a [2n – (m + r)] = – d  n 2 −  2   a n ⇒ =– . d 2

605

a+c  2 2a 2c 2 As a2, b2, c2 are in H.P., b2 = 2 a + c2

166. (d) As a, b, c are in A.P., b =

∴ y2 = xz Taking log on both sides, we get ⇒ 2 lny = lnx + lnz ⇒ 2 + 2 lny = (1 + lnx) + (1 + lnz) ⇒ 2 (1 + lny) = (1 + lnx) + (1 + lnz) ∴ 1 + lnx, 1 + lny, 1 + lnz are in A.P.

Sequences and Series

162. (c) Since, x, y, z are in G.P.

606

170. (d) Given x1 · x2 ...xn = 1 Since A.M. ≥ G.M.



Objective Mathematics

 x1 + x2 + ... + xn  ∴   ≥ (x1 · x2 ...xn)1/2 n   = (1)1/n = 1 ⇒ x1 + x2 + ... + xn ≥ n. Hence x1 + x2 + ... + xn can never be less than n. 171. (c) Let S =

2 5 2 11 − + − +... to ∞ 3 6 3 24

...(1)

1 Multiplying both sides by – , the common ratio 2 of G.P. –

1 2 5 8 +... to ∞ S =– + − 2 6 12 24

...(2)

Subtracting (2) from (1), we have

=



2 = − 3

1 2 1 1 2 = − = .  1 3 3 3 1− −   2

=

1−

1 25

+

1−

1 49

5 1 11 + = . 24 48 48

=

173. (a) Sum = (33 + 53 + 73 + ... to 10 terms)

– (23 + 43 + 63 + ... to 10 terms)

= (23 + 33 + 43 + 53 + ... to 20 terms) – 2 (23 + 43 + 63 + ... to 10 terms)



= (13 + 23 + 33 + ... to 21 terms) – 13 – 2 ⋅ 23 (13 + 23 + 33 + ... to 10 terms)



 21 × (21 + 1)  10 ⋅ (10 + 1)  =  − 1 − 16 ⋅   2 2    



= 2312 – 2202 – 1 = (231 + 220) (231 – 220) – 1 = 451 × 11 – 1 = 4961 – 1 = 4960.

2

174. (c) We have, t n

=

13 + 23 + 33 + ... + n3 1 + 3 + 5 + ... upto n terms

16.17.33 16.17 16 + + = 446. 24 4 4

175. (c) We have, tn = [nth term of 1, 2, 3, ...] 

× [nth term of 3, 5, 7, ...]2 = n (2n + 1) = 4n3 + 4n2 + n. 2

2 + (2 + 3) + (2 + 3 + 9) + (2 + 3 + 9 + 27) + ...

1 1 1 1 1   1  =  + 3 + 5 + ... to ∞  +  2 + 4 + 6 + ... to ∞  7 7 5 5 5  7  1 72

∴ S16 =

176. (b) Here the series is

172. (c) The given series,

1 5

n {2 + 2(n − 1)} 2 2 n n 1 + + . = 4 2 4 1 1 1 Σ n2 + Σ n+ Σ1 ∴ Sn = Σ tn = 4 2 4 1 n (n + 1)(2n + 1) 1 n (n + 1) 1 + ⋅ + ⋅n = ⋅ 4 6 2 2 4

2

1 2 2 × = . 3 3 9

∴ S =

n 2 (n + 1) 2 (n + 1) 2 4 = = 4 n2

n (n + 1)(2n + 1) n (n + 1)  n (n + 1)  +  =4 ⋅   + 4⋅ 6 2  2  2 1  = n2 (n + 1)2 + n (n + 1) (2n + 1) + n (n + 1); 3 2 2 1 ∴ S20 = 202 ⋅ 212 +  ⋅ 20 ⋅ 21 ⋅ 41 +  ⋅ 20 ⋅ 21 3 2 S20 = 188090.

2 1 1 1  −  − + + ... to ∞  3 2 4 8 



2

∴ Sn = Σ tn = 4 Σ n3 + 4 Σ n2 + Σ n

3 2 3 3 3 +... to ∞ S = − + − 2 3 6 12 24



=

 n (n + 1)     2 

t he difference of the consecutive terms being 3, 9, 27, ... ∴ tn = 2 + 3 + 9 + 27 ... to n terms

= 2 + [3 + 9 + 27 + ... to (n – 1) terms]



=2+



=

∴ Sn =

=



=

3 (1 − 3n −1 ) 3 =2– (1 – 3n – 1) 1− 3 2 1 1 + ⋅ 3 n; 2 2 1 1 Σ tn = Σ1+ Σ 3n 2 2 1 1 ⋅n+ (3 + 32 + 33 + ... + 3n) 2 2 n 1 3 (1 − 3n ) n 3 + ⋅ = + (3n − 1) . 2 2 1− 3 2 4

177. (a) Since H is the H.M. between P and Q,

2

∴

1 1 H H 2 + + = or = 2. P Q P Q H

  = .123232323... 178. (a) .1 23 1 23 23 + + +... = 10 1000 100000 1 1 1  1  + 23  3 + 5 + 7 + ...∞  = 10 10 10 10 

=

Given, k (1 + r) = 3, kr2 (1 + r) = 12 ∴ 3r2 = 12 or r = 2 ∴ k = 1. Also, k2r = a, k2 r5 = b. Putting for r and k, we get a = 2, b = 32.

99 + 23 122 61 = = . 990 990 495

1 1 179. (a) We know that sum of n A.M.’s between two quanti- 183. (d) We have, 1 + 1 + 1 + ...∞ = 4 A 1 = –1 2 =2 . 1 2 ties is equal to n times their single mean. 4 8 16 1− 2 Now x, y, z are three A.M.’s between a and b

a+b x+y+z=3   = 15  2 

∴ (0.2)

−1

or a + b = 10 ...(1) a, x, y, z, b are in H.P. 1 1 1 1 1 , , , , are in A.P. ∴ a x y z b 1 3 1 1 a+b 1 1 + + = ∴  +  =3   y 2 a b x z  2ab  or

5 3 ⋅ 10 , by (1) ∴ ab = 9 = 3 2ab



– Sn = 1 + [(2 – 1) 2 + (3 – 2) 22 + ... to (n – 1)] – 2tn





=

1 ⋅ (2n − 1) – 2tn 2 −1

...(i)

185. (a) Let S = 1 + 22x + 32x2 + 42x3 + ... to ∞ then, xS = 1 ⋅ x + 22x2 + 32x3 + ... to ∞

...(2) ...(3) [ a, b, c are in H.P.]

 ubstituting for y and b from (1) and (3) in (2), S we get 4a 2c 2 ( x + z ) 2 ( x + z )2 (a + c) 2 ⋅ = axcz ⇒ = 2 (a + c) 4 xz ac 2 2 x 2 + z 2 + 2 xz ⇒ = a + c + 2ac xz ac

x z a c x z a c ⇒ z + x + 2 = c + a + 2 ⇒ z + x = c + a .

= (42 – 32) x3 + ... to ∞

or (1 – x) S = 1 + 3x + 5x2 + 7x3 + ... to ∞...(i)

...(1)

[ ax, by, cz are in G.P.] 2ac b=  a+c

= 1 + 2 + 22 + ... to n terms – 2 ⋅ tn

∴ (1 – x) S = 1 + (22 – 1) x + (32 – 22) x2

[ x, y, z are in A.P.]



Now tn = (nth term of A.P. 1, 2, 3, 4, ...) ⋅ (nth term of G.P. 1, 2, 22, ...) = {1 + (n – 1) ⋅ 1} {1 ⋅ 2n – 1} = n 2n – 1. From (i), Sn = – (2n – 1) + 2n 2n – 1 = 1 – 2n + n 2n = 1 – (1 – n) 2n.

2ac b = . a+c

b2 y2 = ax ⋅ cz

2

Subtracting on, we get

Hence a, b, c are in H.P.



log 5 2−1

then  2 ⋅ Sn = 1 ⋅ 2 + 2 ⋅ 22 + ... + 2 ⋅ tn – 1 + 2 ⋅ tn

180. (c) Since a (b – c) + b (c – a) + c (a – b) = 0, ∴ x = 1 is a root of the given equation. Since both the roots are equal, therefore the other root is also 1.

181. (b) We have, 2y = x + z

log5 2

1 =   5

= 52 log5 2 = 5log5 ( 2 ) = 22 = 4. 184. (b) Let Sn = 1 + 2.2 + 3.22 + ... + tn

...(2)

Their sum = 1 + 1 = 2 −b (c − a ) ⇒ = 2 ⇒ – bc + ab = 2ab – 2ac a (b − c)

−1 = (5 )1/ 2



Hence a and b are the roots of t2 – 10 t + 9 = 0  [Using (1) and (2)] ⇒ t = 9, 1 ∴ 9, 1 are the required values of a and b.

⇒ 2ac = b (a + c) ∴

1 1 1  log 5  + + + ...∞   4 8 16 



and x (1 – x) S = 1 ⋅ x + 3x2 + 5 ⋅ x3 + ... to ∞ ...(ii) (i) – (ii) ⇒ (1 – x) S (1 – x) = 1 + 2x + 2x2 + 2x3 + ... to ∞

= 1 + 2x (1 + x + x2 + ... + to ∞) 1 1+ x = = 1 + 2x ⋅ 1− x 1− x

∴ S =

1+ x . (1 − x)3

186. (a) We have, S = i + 2i2 + 3i3 + ... + 100i100

...(i)

⇒ S.i. = i2 + 2i3 + ... + 99i100 + 100i101

...(ii)

(i) – (ii) ⇒ (1 – i) S

607

Sequences and Series

182. (c) Since α, β, γ, δ are in G.P. they may be taken as k, kr, kr2, kr3

 1   3 1 10  = 1 + 23 . = + 23 ⋅  10 990 10  1  1−  2  10  

608

= [i + i2 + i3 + ... 100 terms] – 100i101



Objective Mathematics

i (1 − i ) − (100) (i 2 )50 .i = 0 – 100i 1− i 100

=

∴ S = –

=

100i −100 (i + i 2 ) 100 i (1 + i ) =– = 1− i 2 1 − i2

Hence a1, a2, ..., an are in A.P., G.P. and H.P. 191. (a) Let the sides of the right angled triangle be a, ar, ar2 out of which ar2 is the hypotenuse, then r > 1.

−100 (i − 1) = – 50 (i – 1) = 50 (1 – i). 2

187. (c) Let S = 1 + 3 + 7 + 15 + .. + tn Again S = 1 + 3 + 7 + ... + tn – 1 + tn

...(i) ...(ii)

(i) – (ii) ⇒ 0 = [1 + 2 + 4 + 8 + ... to n terms] – tn or

tn = 1 ⋅

(2n − 1) = 2n – 1 2 −1

Now, a2 r4 = a2 + a2 r2 or r4 – r2 – 1 = 0

∴ t100 = 2100 – 1. 188. (c) Let S = 1 + 3 + 6 + 10 + 15 + ... + tn then S = 1 + 3 + 6 + 10 + ... + tn – 1 + tn (i) – (ii)

...(i) ...(ii)

⇒ 0 = (1 + 2 + 3 + 4 + ... to n terms) – tn n (n + 1) ⇒ tn = . 2 Given, 5050 =

n (n + 1) ⇒ n2 + n – 10100 = 0 2

−1 ± 1 + 40400 −1 ± 40401 = ⇒ n = 2 2 −1 ± 201 = = – 101,100 2 ∴ n = 100.

[ n is a positive integer]

189. (c) We have, b + c − 2a c + a − 2b a + b − 2c  are in A. P. , , a b c b + c − 2a c + a − 2b +3, +3, a b

⇒

a + b − 2c + 3 , are in A.P. c b+c+a c+a+b a+b+c , , are in A.P. ⇒ a b c 1 1 1 ⇒ , , are in A.P. ⇒ a, b, c are in H.P. a b c



190. (c) If a1 = a2 = ... = an ≠ 0, then

a2 – a1 = a3 – a2 = ... = an – an – 1 = 0, a a a2 = 3 = ... = n = 1 a a2 a1 n−1

and

1 1 1 1 1 1 − − = = ... = = 0. − a a a a a2 a1 3 2 n n −1

∴ r2 =

1± 5 2

1+ 5 . 2 Angle C is the greater acute angle

r > 1 ∴ r2 > 1

Now, cos C =

∴ r2 =

1 a 1 = 2 = . 2 1+ 5 r ar

192. (c) Since A.M. ≥ G.M. y + z ≥ yz 2 z + x ≥ zx 2 x+ y ≥ xy 2 Multiplying (i), (ii) and (iii), we get

∴

( y + z )( z + x)( x + y ) ≥ xyz 8 or (1 – x) (1 – y) (1 – z) ≥ 8xyz.

193. (a) Taking A.M. and G.M. of 7 numbers

a a b b b c c , , , , , , , we get 2 2 3 3 3 2 2 2⋅



1 a b c 2 3 2 7 + 3⋅ + 2⋅ 2 3 2 ≥  a   b   c          7  2   3   2   1

3  a 2 b 3c 2  7 ⇒ ≥  2 3 2 7 2 32  a 2 b 3c 2 37 ≥ 22 ⋅ 33 ⋅ 22 77 310 ⋅ 24 ⇒ a2 b3 c2 ≤ 77 ⇒

∴ greatest value of a2 b3 c2 =

310 ⋅ 24 . 77

...(i) ...(ii) ...(iii)

  1 n  1 ⋅ 1 −     2    =  1 1 −   2





Sn = 2 –

∴ 2 – Sn =

1   ∵ 2 − Sn <  100  

1 1 < 2n−1 100

⇒ 2 > 100 > 2 ⇒ 2 >2 ∴ n – 1 > 6 ∴ n > 7. Hence least value of n is 8. n–1

6

⇒ Sn = (2 – 1) 1! + (3 – 1) 2! + (4 – 1) 3!



+ (5 – 1) 4! + ... + [(n + 1) – 1] n! = (2 ⋅ 1! – 1!) + (3 ⋅ 2! – 2!) + (4 ⋅ 3! – 3!) + (5 ⋅ 4! – 4!) + ... + [(n + 1) n! – n!] = (2! – 1!) + (3! – 2!) + (4! – 3!) + (5! – 4!) + ... + [(n + 1)! – n!]



198. (c) Let R = 0 ⋅ cababab...

caba − ca 9900

⇒ R =

Sn = 1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! + 4 ⋅ 4! + ... + n ⋅ n!



 hus, digits are 1, 3, 9 and so the required number T is 931. ⇒ 102R = ca ⋅ bababa ... and 104R = caba ⋅ baba ... ⇒ (104 – 102) R = caba – ca

6

195. (a) Let



(r + 1)(r − 1) r +1 =2 ⇒ =2 (r − 1) 2 r −1

∴ r = 3 ∴ from (1), a = 1.

1 2n−1

n–1



a (r 2 − 1) 8 = a (r 2 − 2r + 1) 4

then ⇒

= (n + 1)! – 1! = (n + 1)! – 1.

196. (c) The general term of the given sequence is

1000c + 100a + 10b + a − 10c − a 9900 99c + 10a + b = . 990 n (n + 1) 199. (a) We have, Σ n = 210 ⇒ = 210 2 

∴

=

⇒ n2 + n – 420 = 0 ⇒ (n – 20) (n + 21) = 0 ⇒ n = 20, – 21. = 20 ( n = – 21 is not possible)

n

20 ⋅ 21 ⋅ 41 n (n + 1)(2n + 1) = = 2870. 6 6

∴ Σ n2 =

n Tn = 500 + 3n3 2



then

200. (b) We have, | ai | = | ai – 1 + 1 | 2 ⇒ ai2 = ai −1 + 2 ai −1 + 1

d Tn n (1000 − 3n3 ) = dn (500 + 3n3 ) 2

1

 1000  Now 6 <    3 

Putting i = 1, 2, 3, ..., n + 1, we get



For maximum or minimum of Tn d Tn =0 dn 1000  ∴ n =    3 

...(2)



a

=0



a

2 = a1 + 2a1 + 1

2 1 2 2

a32 = a2 + 2a2 + 1   � 2

3



1



2 an2 = an−1 + 2an – 1 + 1



2 an+1 = an2 + 2an + 1

3

G.M.

n

(1 + 3– 8) ... (1 + 3−2 )



∴ x is rational. n

n

S = (1 + 3–1) (1 + 3–2) (1 + 3–4) (1 + 3–8) ... ( 1 + 3−2 ) ⇒ (1 – 3– 1) S = (1 – 3– 1) (1 + 3– 1) (1 + 3– 2) (1 + 3– 4)

3 8 2⋅ ⋅ 11 11 = 48 ⇒ x = 121 3 8 + 11 11

j

log a – log b

208. (d) We have, 1 + n + n2 + ... + n127

8 3 ⇒ , x, are in H.P. 11 11

i

log c – log a

+ (log a – log b) log c = 0 ⇒ x = 1. ∴ From (1), alog b – log c + blog c – log a + clog a – log b ≥ 3.

Since 0.272727..., x and 0.727272... are in H.P.

n

...(1)

⋅b ⋅c Let x = a ⇒ log x = (log b – log c) log a + (log c – log a) log b

2ab   ∵ h = a + b   

202. (a) Let R = 0.272727...

∴ R =

a log b −log c ⋅ b log c −log a ⋅ c log a −log b  log b – log c

2n n ( a + b) = . = h ab



3

a log b −log c + b log c −log a + c log a −log b 3

210. (a) Let

x2 x3 y y = = r and 2 = 3 = r x1 x2 y1 y2

⇒ x2 = x1r, We have, x1 y1 x ∆ = 2 y2 x3 y3

x3 = x1r2 and y2 = y1r and y3 = y1r2, 1 1 = 1

x1 x1r x1r 2

y1 1 y1r 1 y1r 2 1

Using R3 → R3 – rR2 and R2 → R2 – rR1, we get ∆=

x1 0 0

y1 1 0 1− r 0 1− r

= 0

 [ R2 and R3 are identical] Thus, (x1, y1), (x2, y2), (x3, y3) lie on a straight line. 211. (b) Let ∆ =

xp + y yp + z 0

x y xp + y

y z yp + z

Applying R3 → R3 – pR1 – R2, we get

216. (d) A.M. of a, b, is A =

= – (xp + 2yp + z) (xz – y )



2

2

I f x, y, z are in G.P. then xz – y2 = 0 and hence ∆ = 0. 212. (a) Let the two positive numbers be a and b. a+b Then x = 2 . Also, a, y, z, b are in G.P. Let the common ratio be r. 1/ 3 b b Then b = ar3 ⇒ r3 = ⇒r =   a a b ∴ y = ar = a   a

But given A.M. =

a n +1 + b n +1 a + b a+b a+b = = = 0 2 a n + bn 2 a + b0 On comparing n + 1 = 1 or n = 0. 217. (c) Let ‘a’ be the first term and d be the common difference. Given

= a2/3 b1/3

b z = ar2 = a   = a1/3 b2/3 a Now, y3 + z3 = a2b + ab2 = (a + b) ab

 Thus,

y3 + z xyz

1 (a + b) a2/3 b1/3 a1/3 b2/3 2 1 = (a + b) ab 2 3 = 2.

1 213. (c) Tm = a + (m – 1) d = n and Tn = a + (n – 1) d = ⇒ (m – n) d = ⇒ d =

1 m

1 1 m−n – = n m mn

1 mn

∴ Tmn = a + (mn – 1) d 

= a + (n – 1) d + (mn – n) d =

1 1 + n (m – 1) = 1. m mn

214. (b) Relation between arithmetic mean A, geometric mean G and harmonic mean H is given by G2 = AH or G = AH . Clearly, A, G, H are in G.P. 215. (d) Harmonic mean of roots (α, β) is

H =

2αβ . α+β

But α + β = – ∴ H =

b c = 10, αβ = = 11 a a

2 × 11 11 = . 10 5

Sm m2 = 2 , where Sn n

Sn =

n [2a + (n – 1) d] 2

∴

m [ 2a + (m − 1) d ] m2 2 = 2 n n [ 2a + (n − 1) d ] 2

or

2a + (m − 1) d m = 2a + (n − 1) d n

and

Also, xyz =

a n +1 + b n +1 a n + bn

∴

1/ 3

2/3

a+b 2

Or replacing m by 2m – 1, n by 2n – 1

2 a + ( 2 m − 2) d 2m − 1 = 2a + (2n − 2) d 2n − 1

or

m th term (Tm ) 2m − 1 = . n th term (Tn ) 2n − 1

218. (b) 1, log3 (31–x + 2), log3 (4.3x – 1) are in A.P.

2 log (31–x + 2) = 1 + log3 (4.3x – 1)



log3 (31–x + 2) = log33 + log3 (4.3x – 1)



log3 (31–x + 2) = log3 [3(4. 32 – 1)]

31–x + 2 = 3 (4.3x – 1) 3.3–x + 2 = 12.3x – 3 Let 3x = 1

3 + 2 = 122 – 3t t 12r2 – t – 3 = 0 1 3 t = , 3 4 3x =

3 4

3 ⇒ x = log3   = log33 – log3 4 4 ⇒ x = 1 – log34 1 1 , P (E2) = 2 3 1 and P (E2) = 4 P(E1 ∪ E2 ∪ E3) = 1 – P ( E 1) P ( E 2) P ( E 3)



P(E1) =

611

y z 0

Sequences and Series

xp + y x ∆= yp + z y − ( xp 2 + 2 yp + z ) 0

612

Objective Mathematics



 1  1  1 = 1 – 1 −  1 −  1 −   2  3  4



=1–

or (a + d)2 = ± a(a + 2d) 2 2 or a + d + 2ad = ± (a2 + 2ad) Taking + sign, d = 0 (not possible as a < b < c). Taking – sign, 2a2 + 4ad + d2 = 0. 3 1 Also a + b + c = ⇒a+d= 2 2

1 2 3 3 × × = . 2 3 4 4

219. (b) 21/4. 41/8. 81/16...∞

= 2 = 2

1 2 3 + + + ...∞ 4 8 16

= 2

2

1  1  ∴  2a2 + 4a  − a  +  − a  = 0 2 2    

4  1  2 3 +  1+ +  22  2 22 23 

1− (1/ 2 )  1  1   + 22 1− (1/ 2 ) (1−1/ 2 )2 

1/ 2 = 2

or 4a2 – 4a – 1 = 0

2 ( 2 + 2)

1

= 2 = 2.

220. (b) ar4 = 2 a × ar × ar2 × ar3 × ar4 × ar5 × ar6 × ar7 × ar8  ...(1) = a9 r36 = (ar4)9 = 29 = 512. 221. (c) Given

1 = a + (m – 1)d n

1 m −1 1 – = n mn mn ∴ a – d = 0 Second method from (1) and (2) an + (mn – n)d = am + (mn – m)d ⇒ a (n – m) – d (n – m) = 0 ⇒ a – d = 0. [∴ n ≠ m]. from (1), a =

or ar2 = a2

...(1)

and T2 = 8 or ar = 8 From (1) and (2) r = 2, a = 4. ∴ T6 = ar5 = 4 (2)5 T6 = 128.

...(2)

Hence a =

Now Sn = S2n+1 = S2m + (2m + 1)th term 2

1 1 − . 2 2

225. (b) Expanding along R3, we get 2 ⇒ ⇒ or or ∴ ∴

b aα − b a aα − b − +0=0 c bα − c b bα − c 2 (b2α – bc – acα + bc) – (abα – ac – abα + b2) = 0 2α (b2 – ac) – (b2 – ac) = 0 (b2 – ac) (2α – 1) = 0 [ (2α – 1) ≠ 0] b2 – ac = 0 b2 = ac a, b, c are in G.P.

226. (c) Given Tp + q ⇒ arp + q – 1 = m

...(1)

Tp – q ⇒ arp – q – 1 = n

...(2)



a2 r2p – 2 = mn  or  arp – 1 =

∴ pth term =

+ nth term   n(n + 1) 2 , where n is even  ∵ S n 2  



(n −1)n 2 = + n 2 2



=

224. (d) b = a + d, c = a + 2d where d > 0. Now, a2, (a + d)2, (a + 2d)2 are in G.P. ∴ (a + d)4 = a2(a + 2d)2

mn .

α+β=

4+ 5 8+2 5 , αβ = 5+ 2 5+ 2

Harmonic mean of α, β is given by 2αβ H = α+β



[ n = 2m + 1]

n2 n 2 (n + 1) (n – 1 + 2) = . 2 2

mn

227. (b) If α, β are the roots of the given equation then

223. (b) Let n = 2m + 1 2m(2m + 1) 2

1 . 2

Multiplying (1) and (2), we get

2 Given T3 = T1

=

So, a
0. 2 2 2

...(1)

1 = a + (n – 1)d ...(2) and m m−n 1 = (m – n) d ⇒ d = (1)–(2) ⇒ mn mn



∴ a =

8+2 5  2   5 + 2   H = = 4. 4+ 5 (5 + 2 )

228. (d) Harmonic mean of a, b is H = Geometric mean G =

Given

2ab a+b

ab .

H 4 2 ab 4 = , so = G 5 a+b 5

By componendo and dividendo ( a + b )2 9 =   or  ( a − b )2 1



3 a+ b = . 1 a− b

a a 4 = 2 or = . b b 1 ...(1)

Let numbers be x1, x2, x.3, ..., xn Their Geometric mean G.M. = (x1 ⋅ x2 ⋅ x3 ... xn)1/n As their product = 1 ∴ G.M. = 11/n = 1. x1 + x2 + ... + xn n

x1 + x2 + ... + xn ≥1 n or x1 + x2 + ... + xn ≥ n i.e., Sum ≥ n ∴ Sum will never be less than n.

By (1),

= 0.2 + 0.034 + 0.00034 + ...



34 2 34 100 2 + × = + 1000 = 10 1000 99 10 1 − 1 100



=

234. (b) Since a, b, c are in G.P. b2 = ac. The first equation can be written as ax2 + 2 ac x + c = 0

d 2e f − + =0 a c ac

or 

d f 2e + =  a c b

B=

232. (d) Since A.M. > G.M. > H.M. ∴ x > y > z.

1+ 5  2

Also,

ab = G

or

ab = G2



1 A −1 = . A A

1 1 − rb

⇒ 1 – rb =

1 1 B −1 ⇒ rb = 1 – = . B B B

 A −1 ∴  a log r = log    A  ...(i)

 B −1 and  b log r = log    B 

...(ii)

Now, (a – b) = (a + b) – 4ab 2

2

= 4A2 – 4G2

∴ a – b = ± 2 A 2 − G 2

ac = b)

  1− 5 < 0 ∵ terms are + ve, and 2  

1 1 ⇒ 1 – ra = 1− ra A

233. (c) Let the required numbers be a, b

a + b = 2A

(

d e f , , are in A.P. a b c

⇒ ra = 1 –

81 [2a + 80d] = 81 [a + 40d] = 81 × 10. 2

or

root, substituting in 0 c +f=0 a

or

236. (c) A =

81 [2a + (81 – 1) d] 2

a+b =A 2

c . a

c )2 = 0  or  x = –

This is the common dx2 + 2ex + f = c – 2e ⋅ we get, d a

r =

2 34 198 + 34 232 + = = . 10 990 990 990

Then,

A2 − G 2 .

⇒ ar2 = a + ar ⇒ r2 = 1 + r 1± 1+ 4 1± 5 = ⇒ r2 – r – 1 = 0 ⇒ r = 2 2

231. (c) We have, T41 = a + 40d = 10 ∴ S81 =

A2 − G 2

235. (b) T3 = T1 + T2



=

b =A ±

Hence

230. (d) 0.2 34 = 0.234 34 34 34 ...





or ( a x +

229. (d) For n numbers A.M. ≥ G.M.

Now, A.M. =

A2 − G 2

Hence, numbers are A ±

3 +1 2 a Again by componendo and dividendo = 3 −1 2 b

a=A ±

Sequences and Series





613

Solving (i) and (iii)

5 a+b = 4. 2 ab

or

[from (i) and (ii)] ...(iii)

∴

a = b

 A −1 log    A − 1 .  A  = log B −1   A   B −1 B  log    B 

614

237. (d) Let the given series be x, xr, xr2 ... By hypothesis,

Objective Mathematics

x x x   or  1 – r =   or  r = 1 − 1− r 5 5 For an infinite G.P. to sum up to 5, | r | < 1 x 1  r 

n

= (a2 rn – 1)n = a2n rn(n – 1) = {a r

1 =S 1− r

⇒ 1 + r + r2 + .. ∞ = S ⇒

(∵ r < 1)

n

∞

∑r

279. (d) Given,

S = a + ar + ar2 + … + arn – 1

   a (1 − r n )  n   S Now,   =  (1 − r )   R   1 (1 − r n )   a   r n −1 (1 − r ) 

…(iii)

 S −1  1−    S 

=

2

2

280. (b) Since, a + ar = a (1 + r) = 12 and ar2 + ar3 = ar2 (1 + r) = 48

…(i) …(ii)

From Eqs. (i) and (ii) n

( n −1) 2 2

} =P

r2 = 4 ⇒ r = – 2 On putting the value of r in Eq. (i) we get

2

a = – 12

278. (c) ∴ Let the two numbers be a and b, then AM = a + b = A  2

…(i)

and GM = ab ⇒ G2 = ab Now, (a – b)2 = (a + b)2 – 4ab = (2A)2 – 4G2

281. (a) Geometric mean of 7, 72, 73, ……., 7n 1

= (7 × 7 2 × 73 × .........7 n ) n = [(7)1 + 2 + 3 + 4 +………..n]1/n

…(ii)

 n ( n +1)  = (7) 2   

1/ n

n

= 72

( n +1) / n

=7

( n +1) 2

EXERCISES FOR SELF-PRACTICE 1. If the sum of the first n terms of a series be 5n2 + 2n, then its second term is 56 15 (c) 17

(a)

27 14 (d) 16 (b)

n(n −1) Q, where Sn denotes the sum of 2 first n term of an A.P., the common difference is

2. If Sn = nP +

(a) 2P + 3Q (c) Q

(b) P + Q (d) 2Q

3. If a1, a2, ... , an are positive real numbers whose product is a fixed number c, then the minimum value of a1 + a2 + ... + an–1 + 2an is (a) n(2c)1/n

(b) (n + 1)c1/n

(c) 2nc

(d) (n + 1)(2c)1/n

1/n

4. Suppose a, b, c are in AP and a2, b2, c2 are in GP. If 3 a < b < c and a + b + c = 2 then the value of a is

(c)

2 2



(b)

13. The value of 1 + 2 + 3 + ... + n is

1 2 3

1 1 (d) − 2 2

1 1 − 2 3

5. If the third term of an A.P. is 12 and the seventh term is 24, then the 10th term is (a) 33 (c) 39

(b) 30 (d) 36

6. The sum of the series 12 + 32 + 52 + ... to ∞ n(2n − 1)(2n + 1) (a) 3 (c)

(2n + 1)3 3

(n − 1) 2 (2n + 1) (b) 6 (d)

n (n + 1) + (2n + 1) 2

(a)

(c) n2 14. If

1 (10n + 1 + 9n – 28) 27 (d) None of these

(c)

9. The sum of the series 1 + 4 + 7 + 10 + ... upto 20 terms is (a) 590 (c) 290

(b) 490 (d) None of these

10. If G.M. is 4 and A.M. is 5, then H.M. will be (a)

5 16

(b)

16 5

(c)

17 8

(d)

25 15

11. Harmonic mean between 1 and

(c)

32 (a) 17

17 (b) 32

2 (c) 17

17 (d) 2

12. If n ∈ N, then

n

∑m

2

is equal to

m =1

n (n + 1) 2

 n (n − 1)  (d)    2 

2

1 1 1 , ,  are in A.P. p q r

(b) p2, q2, r2 are in A.P. (d) p, q, r are in G.P

15. In a G.P., second terms is 2 and sum of infinite terms is 8, then its first term will be (a) 8 (c) 4

(b) 6 (d) 2

If A.M. and H.M. of the numbers are 2 are 12 respectively. Then G.M. is equal to

(a) 36 (c) 18

(b) 24 (d) 9.

17. The value of 2. 357 is (a)

2355 997

(b)

(c)

2379 999

(d) None of these

2355 999

18. If in an A.P. sum of n terms is 3n2 – n and common difference is 6, then first term will be (a) 1

(b) 2

(c) 3

(d) 4

19. If in a infinite G.P. every term is 10 times the sum of next terms, then common ratio of the G.P. is 1 1 (a) (b) 10 2 (c)

1 is 16

(b)

1 1 1 ⋅ ⋅ are in A.P. then q+r r+ p p+q

(a) p, q, r are in A.P.

7. If x, y, z are in H.P. (z > y > x) log (x + z) + log (z – 2y + x) = 16. (a) 2 log (z – x) (b) 2 log (y – z) (c) 4 log (x – z) (d) log (y – z) 8. The sum of the series 3 + 33 + 333 + ... n terms is 1 (a) (10n + 1 – 9n – 10) 27 1 (b) (10n + 1 + 10n – 9) 27

n (n −1) 2

619

1

1 11

(d)

10 11

20. If a, b, c are in H.P., then (a) a2 + c2 = 2b2 (b) (c) a2 + c2 > 2b2 (d) a2 + c2 < b2

a2 + c2 < 2b2

21. If in an A.P., 5th and 31st terms are 1 and 77 respectively, then its 18th term will be (a) – 38 (c) – 37

(b) 38 (d) – 36

(a)

m (m − 1)(2m − 1) 6

(b)

n (n − 1)(2n − 1) 6

22. If in an A.P., sum of p terms is q and the sum of q terms is p, then sum of ( p + q) terms will be

(c)

n (n + 1)(2n + 1) 6

(d)

m (m + 1)(2m + 1) 6

(a) – (p + q) (c) p + q + 1

(b) 0 (d) p – q

Sequences and Series

(a)

620

23. If sum of the terms of any A.P. is 36, with first term 1 and last term 11, then number of terms are

Objective Mathematics

(a) 10 (b) 6 (c) 12 (d) 3 24. The second term of a G.P. is 2 and sum of infinite number of terms is 8, then its first term will be 25.

1 1 (b) (a) 4 2 (c) 2 (d) 4 a b c , , If are in H.P., then a, b, c are in b+c c+a a+b (a) G.P. (b) A.P. (c) H.P. (d) None of these

26. If a, b are in A.P., a2, b 2, c 2 are in H.P. then : (a) a ≠ b ≠ c

(b) a2 = b2 =

(c) a, b, c are in G.P.;

(d)

c2 2

−a , b, c are in G.P. 2

30.

33. If a, b, c are three unequal numbers such that a, b, c are in A.P. and b – a, c – b, a – c are in G.P. then a : b : c is: (a) 1 : 2 : 3 (c) 1 : 3 : 2

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

34. If the roots of the equation x3 – 12x2 + 39x – 28 = 0 are in A.P. then their common difference will be: (a) ± 1 (c) ± 3

(b) ± 2 (d) ± 4

1 1 1 1 + + + ........ is 2 2.3 3.4 n(n + 1)

(b) x > –2

1 (c) x > 2

(d) None of these



a b c , , are in H.P. then: b c a

(a) a2b, c2a, (b) a2b, b2c, (c) a2b, b2c, (d) None of

b2c are in A.P. c2a are in H.P. c2a are in G.P. these

37. If S is the sum of infinite terms of a G.P. whose first term is ‘a’, then the sum of first ‘n’ terms is: a (a) S 1 −   S

  a n  (b) S 1 − 1 −     S  

n  a  (c) S 1 − 1 +     S  

(d) None of these

n

n −1 n (a) (b) n +1 n +1 1 (c) (d) n(n + 1) n +1 1 1 1 , , If x > 1, y > 1, z > 1 are in G.P. then 1 + ln x 1 + ln y 1 + ln z are in: (a) A.P. (b) H.P. (c) G.P. (d) None of these

38. If ‘H’ is the harmonic mean between P and Q, then  H   H  is:  +   P  Q

31. If A and G are arithmetic and geometric means and x2 – 2Ax + G2 = 0, then:

(a) 2

(a) A = G (c) A < G

(c)

(b) A > G (d) A = – G

2 4 8 + + + ... ∞ is a finite x x 2 x3

(a) x < 2

36. If

28. If a1/x = b1/y = c1/z and a, b, c are in G.P., then x, y and z are in: (a) A.P. (b) G.P. (c) H.P. (d) None of these

equal to:

(b) 6 (d) 10

number, then:

(b) f (b) = g (b) (d) f (c) = g (a)

29. The value of the series

(a) 4 (c) 8

35. If the sum of the series 1 +

27. If f (x) = ax + b and g (x) = cx + d, then f (g (x)) is equivalent to: (a) f (a) = g (c) (c) f (d) = g (b)

32. Let Sn denotes the sum of first n terms of an A.P. If S S2n = 2Sn, then the ratio of 3n is equal to: Sn

(b)

P+Q PQ

PQ P+Q

(d) None of these

Answers

1. 11. 21. 31.

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

2. 12. 22. 32.

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

3. 13. 23. 33.

(a) (b) (b) (a)

4. 14. 24. 34.

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

5. 15. 25. 35.

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

6. 16. 26. 36.

(a) (c) (c) (a)

7. 17. 27. 37.

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

8. 18. 28. 38.

(a) (b) (a) (a)

9. (a) 19. (c) 29. (b)

10. (b) 20. (c) 30. (b)

15

Quadratic Equations and Inequations

CHAPTER

Summary of concepts QUADRATIC EQUATION An algebraic expression of the form: ax2 + bx + c, where a (≠ 0), b, c ∈ R is called a quadratic expression. An equation of the form: ax2 + bx + c = 0, where a (≠ 0), b, c ∈ R is called a quadratic equation. The numbers a, b, c are called the coefficients of the quadratic equation and the expression b2 – 4ac is called its discriminant. Discriminant of a quadratic equation is usually denoted by D or ∆.

(x) The quadratic equation whose roots are reciprocals of the roots of ax2 + bx + c = 0 is cx2 + bx + a = 0 (i.e., the coefficients are written in reverse order). (xi) If a = 1, b, c ∈ Z and the roots are rational numbers, then these roots must be integers.

Sum and Product of the Roots If α and β are roots of ax2 + bx + c = 0, then Sum of roots = α + β =

Root of the Quadratic Equation A quadratic equation ax + bx + c = 0 has two and only two roots. Let α, β be two distinct roots. Then

Product of roots = αβ =

2

−b + b − 4ac −b − b − 4ac   and  β = . 2a 2a 2

α=

2

Nature of Roots of the Quadratic Equation (a) If D < 0, then roots α, β are imaginary (b) If D > 0, then roots α, β are real and distinct (c) If D = 0, then roots α, β are real and equal.

−b Coefficient of x =– a Coefficient of x 2

Constant term c = Coefficient of x 2 a

Formation of Equation with Given Roots If α and β are roots of f (x) = ax2 + bx + c = 0, then



f (x) = (x – α) (x – β) = 0



= x2 – (α + β)x + αβ = 0

i.e., x – (sum of the roots) x + (product of the roots) = 0. 2

COMMON ROOTS

One Root Common  If α is a common root of the equations (i) Roots are rational ⇔ D is a perfect square a1x2 + b1x + c1 = 0 ...(1) (ii) Roots are irrational ⇔ D is positive but not a per- and ...(2) a2x2 + b2x + c2 = 0 fect square. then we have (iii) If a + b + c = 0, then 1 is a root of the equation a1α 2 + b1α + c1 = 0 and a2α 2 + b2α + c2 = 0 ax2 + bx + c = 0 α2 α = (iv) If a and c are of opposite sign, the roots must be of oppo- These give b1c2 − b2c1 c1a2 − c2 a1 site sign 1 (v) If the roots are equal in magnitude but opposite in sign, = ( a1b2 − a2b1 ≠ 0). then b = 0, ac > 0. a1b2 − a2b1 (vi) If the roots are reciprocal of each other, then c = a. Thus, the required condition for one common root is 2 (vii) If ax2 + bx + c = 0 is satisfied by more than two values, it is (a1b2 − a2b1 )  (b1c2 − b2c1 ) = (c1a2 − c2 a1 ) and the value of the an identity and a = b = c = 0 and vice-versa. c a − c2 a1 bc −b c (viii) If ax2 + bx + c = 0, where a, b, c ∈ R, has one root common root is α = 1 2 or 1 2 2 1 . a1b2 − a2b1 c1a2 − c2 a1 p + iq, then the other root will be p – iq. Hence the imagiNote:

nary roots occur in conjugate pair. (ix) If ax2 + bx + c = 0, where a, b, c are rational, has one root p −+ q then the other root will be p − q . Hence irrational roots occur in conjugate pair if the coefficients are rational.

Both Roots Common  If the equations (1) and (2) have both roots common, then these equations will be identical. Thus the required condition for both roots common is:

a1 b1 c1 = = a2 b2 c2 .

622

Notes:

Objective Mathematics

(i) To find the common root of two equations, make the coefficient of second degree terms in two equations equal and subtract. The value of x so obtained is the required common root. (ii) If two quadratic equations with real coefficients have an imaginary root common, then both roots will be common 6. For D = 0, parabola touches x-axis in one point. and the two equations will be identical. The required condition is:

a1 b1 c1 = = a2 b2 c2 . (iii) If two quadratic equations have an irrational root common, then both roots will be common and the two equations will 7. For D < 0, parabola does not cut x-axis. be identical. The required condition is:

a1 b1 c1 = = a2 b2 c2 . (iv) If α is a repeated root of the quadratic equation f (x) = ax2 + bx + c = 0, then α is also a root of the equation f ′ (x) = 0. (v) If α is repeated common root of two quadratic equations f (x) = 0 and φ (x) = 0, then α is also a common root of the equations f ′ (x) = 0 and φ′ (x) = 0.

SYMMETRIC FUNCTION of the roots

GREATEST AND  LEAST VALUES OF A QUADRATIC EXPRESSION 1. If a > 0, then the quadratic expression y = ax2 + bx + c has no greatest value but it has least value

A function of α and β is said to be a symmetric function if it b 4ac − b 2 remains unchanged when α and β are interchanged. at x = – . 2 2 2 a 4 a For example, α + β + 2αβ is a symmetric function of α and β whereas α2 − β2 + 3αβ is not a symmetric function of α 2. If a < 0, then the quadratic expression y = ax2 + bx + c has no least value but it has greatest value and β. In order to find the value of a symmetric function of α and b 4ac − b 2 β, express the given function in terms of α + β and αβ. The folat x = – . 2a 4a lowing results may be useful. 1. α2 + β2 = (α + β)2 – 2αβ 2. α3 + β3 = (α + β)3 – 3αβ (α + β) 3. α4 + β4 = (α3 + β3) (α + β) – αβ (α2 + β2) 4. α5 + β5 = (α3 + β3) (α2 + β2) – α2β2 (α + β) 5. | α – β | =

(α + β) 2 − 4αβ

6. α2 – β2 = (α + β) (α – β) 7. α3 – β3 = (α – β) [(α + β)2 – αβ] 8. α4 – β4 =  (α + β) (α – β) (α2 + β2).

GRAPH OF A QUADRATIC EXPRESSION

SIGN OF QUADRATIC EXPRESSION If α, β are roots of the quadratic equation ax2 + bx + c = 0, then for x = α and x = β, the value of the expression ax2 + bx + c is equal to zero. For other real values of x, the expression ax2 + bx + c > 0 or 0) roots of the quadratic equation, then the sign of the expression ax2 + bx + c , x ∈ R is determined as follows:

We have, y  or  f (x) = ax2 + bx + c  where a, b, c, ∈ R, a ≠ 0. 1. The shape of the curve y = f (x) is a parabola 2. The axis of the parabola is y-axis (incase b = 0) or parallel to y-axis. 3. If a > 0, then the parabola opens upwards. 4. If a < 0, then the parabola opens downwards 5. For D > 0, parabola cuts x-axis in two distinct points

2. If α, β are real and equal (i.e., D = 0) roots of the quadratic equation, then the sign of the expression ax2 + bx + c, x ∈ R is as follows: ax2 + bx + c ≥ 0 if a > 0 and ax2 + bx + c ≤ 0 if a < 0

and

ax2 + bx + c < 0 if a < 0.

nature of rootS of a QuaDratIc eQuatIon WIth reSpect to one or tWo reaL numBerS Let f (x) = ax2 + bx + c, where a, b, c ∈ R, a ≠ 0. Let α, β (α < β) be the roots of the corresponding quadratic equation. Let k, k1, k2 ∈ R and k1 < k2. 1. nature of roots with respect to one real number (a) If both the roots of f (x) = 0 are greater than k, then b D ≥ 0, a f (k) > 0 and k < – 2a

(b) If both the roots of f (x) = 0 are less than k, then b D ≥ 0, a f (k) > 0 and k > – 2a

For example, consider f (x) = 2x5 – 6x4 + 7x3 – 8x2 + 5x + 3 + – + – + + Then, f (– x) = – 2x5 – 6x4 – 7x3 – 8x2 – 5x – – – – –

+3 +

Clearly, f (x) has 4 changes of signs and f (– x) has only one change of sign, Therefore, the equation f (x) = 2x5 – 6x4 + 7x3 – 8x2 + 5x + 3 = 0 has atmost four positive roots and one negative root. Also, the equation has atmost (4 + 1) = 5 real roots. 3. (a) A polynomial equation f (x) = 0 has exactly one root equal to α if f (α) = 0 and f ′ (α) ≠ 0. (b) A polynomial equation f (x) = 0 has exactly two roots equal to α if f (α) = 0 , f ′ (α) = 0 and f ′′ (α) ≠ 0. (c) In general, a polynomial equation f (x) = 0 has exactly n roots equal to α if f (α) = f ' (α) = f ′′ (α) = ... = f n – 1 (α) = 0 and f n (α) ≠ 0.

reLatIon BetWeen rootS anD (c) If one roots is less than k and other is greater than k, then coeffIcIentS of a poLynomIaL eQuatIon D > 0 and a f (k) < 0 2. roots with respect to two real numbers (a) If exactly one root of f (x) = 0 lies in the interval (k1, k2), then D > 0 and f (k1) · f (k2) < 0

(b) If both roots of f (x) = 0 lie between k1 and k2, then α+β < k2 D ≥ 0, a f (k1) > 0, a f (k2) > 0 and k1 < 2 (c) If k1 and k2 lie between the roots of f (x) = 0, then D ≥ 0, a f (k1) < 0 and a f (k2) < 0.

Important Results 1. Let f (x) = 0 be a polynomial equation. Let p and q be two real numbers. (a) If f (p) · f (q) < 0, then the equation f (x) = 0 has odd number of real roots between p and q. (b) If f (p) · f (q) > 0, then the equation f (x) = 0 has either no real root or even number of real roots between p and q. (c) If f (p) = f (q), then the equation f ′ (x) = 0 has at least one real root between p and q (This is due to Rolle’s Theorem) 2. (a) If the coefficients of the polynomial equation f (x) = 0 have p changes of signs, then the equation f (x) = 0 will have atmost p, positive roots. (b) If the coefficients of the polynomial equation f (– x) = 0 have q changes of signs, then the equation f (x) = 0 will have atmost q, negative roots.

Let f (x) = a0xn + a1xn – 1 + a2xn – 2 + ... + an – 1x + an = 0, a0, a1, a2, ... , an ∈ R, a0 ≠ 0 be a polynomial equation of degree n, having n roots α1, α2, ... αn. Then, 1. Sum of all roots σ1 = α1 + α2 + ... + αn

a1 a1 = Σ α1 = – a = (– 1)1 a 0 0 2. Sum of the product of two roots σ2 = α1α2 + α1α3 + ...

a2 a2 = Σ α1α2 = a = (– 1)2 a 0 0 3. Sum of the product of three roots σ3 = α1α2α3 + α2α3α4 + ...

a3 a3 = Σ α1α2α3 = – a = (– 1)3 a and so on. 0 0 ar r In general, σr = Σ α1α2 ... αr = (– 1) a . 0 Particular Cases Quadratic equation ax2 + bx + c = 0, then

If α, β are roots of the quadratic equation

α+β=–

b b and αβ = . a a

cubic equation If α, β, γ are roots of the cubic equation ax3 + bx2 + cx + d = 0, then σ1 = α + β + γ = −

bc σ2 = αβ + αγ + βγ = a a

b a

σ3 = αβγ = −

d a

623

(c) The polynomial equation f (x) = 0 will have atmost p + q real roots where p and q are the changes of signs of coefficients in f (x) and f (– x). (This is due to Descarte’s Rule of signs)

Quadratic Equations and Inequations

3. If α and β are imaginary (i.e., D < 0), then the expression ax2 + bx + c > 0 if a > 0

624

Biquadratic equation If α, β, γ, δ are roots of the biquadratic equation ax4 + bx3 + cx2 + dx + e = 0, then

Objective Mathematics

σ1 = α + β + γ + δ = −

b a

bc σ2 = αβ + αγ + αδ + βγ + βδ + γδ = a a σ3 = αβγ + αβδ + αγδ + βγδ = −

d a

σ4 = αβγδ =

3  Thus, f (x) > 0 if x ∈ (– ∞, – 2) ∪  − 1, 2  ∪ ( 4, ∞)   e . a

formatIon of a poLynomIaL eQuatIon from GIVen rootS

and

3 f (x) < 0 if x ∈ (– 2, – 1) ∪  , 4  . 2 

ratIonaL aLGeBraIc expreSSIon

P (x) where P (x) and Q (x) are polynoQ (x) If α1, α2, α3, ... αn are the roots of a polynomial equation of de- mials and Q (x) ≠ 0, is known as a rational algebraic expression. gree n, then the equation is xn – σ1xn – 1 + σ2xn – 2 – σ3xn – 3 + ... + (– 1)nσn = 0 where σr = Σ α1α2 .... αr.

An expression of the form

Sign Scheme for a rational algebraic expression in x

Particular Cases

Step 1: Factorise the numerator and denominator of the given rational expression into linear factors.

Quadratic equation If α, β are the roots of a quadratic equation, then the equation is x2 – (α + β)x + αβ = 0.

Step 2: Find the real values of x by equating all the factors to zero.

cubic equation If α, β, γ are the roots of a cubic equation, then the equation is:

Step 3: If n distinct real values of x are obtained then the entire line will be divided into (n + 1) parts.

x3 – σ1x2 + σ2x − σ3 = 0 or

x3 – (α + β + γ)x2 + (αβ + αγ + βγ)x − αβγ = 0.

Biquadratic equation If α, β, γ, δ are the roots of a biquadratic equation, then the equation is

Step 4: Plot all these points on the number line in order. Step 5: Start with ‘+’ sign from extreme right and change the sign alternatively in other parts.

x4 – σ1x3 + σ2x2 − σ3x + σ4 = 0 or

x – (α + β + γ + δ)x3 + (αβ + αγ + αδ 4

+ βγ + βδ + γδ)x2 − (αβγ + αβδ + αγδ + βγδ)x + αβγδ = 0.

SIGn of a poLynomIaL expreSSIon Step 1: Factorise the given polynomial expression as. k k k f (x) = ( x − α1 ) 1 ( x − α 2 ) 2 ( x − α 3 ) 3 ... kn − 1 ( x − αn − 1) ( x − α n ) kn where k1, k2, ... kn ∈ N and α1, α2, α3, ... αn ∈ R (α1 < α2 < α3 ... < αn). Step 2: Plot the points α1, α2, α3, ..., αn on the real line. Step 3: Mark plus sign in the interval of the right of the largest of these numbers i.e., on the right of αn.

Note: If the rational expression in x occurs under modulus sign, then first of all remove the modulus sign and then proceed. In order to remove the modulus sign, the following results may be useful: (a) | x | = k ⇔ x = ± k (b) | x | < k ⇔ = – k < x < k (c) | x | > k ⇔ = x < – k or x > k.

to find the values of a rational expression in x, where x is real Working Rule 1. Put the given rational expression equal to y and form the quadratic equation in x. 2. Find the discriminant D of the quadratic equation obtained in step 1.

Step 4: If kn is even, put ‘+’ sign of the left of αn and if kn is odd, put ‘–’ sign on the left of αn. Step 5: Consider the next interval and put a sign in it using the above rule. Thus, consider all the intervals.

3. Since x is real, therefore, put D ≥ 0. We get an inequation in y. 4. Solve the above inequation for y. The values of y so obtained determine the set of values attained by the given rational expression.

Note: The general quadratic expression ax2 + 2hxy + by2 + 2gx Step 6: The solution of f (x) > 0 is the union of all the intervals + 2fy + c in x and y may be resolved into two linear rational facin which there is a ‘+’ sign and the solution of tors if f (x) < 0 is the union of all the intervals in which there is a abc + 2fgh – af 2 – bg2 – ch2 = 0. ‘–’ sign. For example, consider the polynomial expression a h g 58 or 3 h b f = 0. 40 31  37 f (x) = (x + 2) (x + 1)  x −  (x – 4) 2  g f c

Choose the correct alternative in each of the following problems: 1. If a + b + c = 0 and a, b, c are rational, then the roots of the equation (b + c – a) x2 + (c + a – b) x + (a + b– c) = 0 are (a) rational (c) imaginary

(b) irrational (d) equal

2. The value of m for which the equation (1 + m) x2 – 2 (1 + 3m) x + (1 + 8m) = 0 has equal roots, is (a) 0 (c) 2

(b) 1 (d) 3

3. If α is a root of 4x2 + 2x – 1 = 0, then the other root is (a) 3α3 – 4α (c) 3α3 + 4α

(b) 4α3 – 3α (d) 4α3 + 3α

11. If r be the ratio of the roots of the equation 2 ax2 + bx + c = 0, then (r + 1) = r a2 b2 (b) (a) bc ca c2 (c) (d) none of these ab 12. If the roots of the equation x2 + a2 = 8x + 6a are real, then (a) a ∈ (– ∞, – 2] (c) a ∈ [8, ∞)

13. If the roots of the equation 9x2 + 4ax + 4 = 0 are imaginary, then (a) a ∈ (– 3, 3) (b) a ∈ [– 3, 3] (c) a ∈ (– ∞, – 3] ∪ [3, ∞) (d) None of these

4. If a, b, c are rational then the roots of the equation abc2x2 + 3a2cx + b2cx – 6a2 – ab + 2b2 = 0 are (a) imaginary (c) rational

(b) equal (d) irrational

(c2 – ab) x2 – 2 (a2 – bc) x + (b2 – ac) = 0 are equal, then

6. If 0 ≤ x ≤ π and 81sin to π (a) 6 π (c) 4

(b) b = 0 (d) a3 + b3 + c3 = 3abc 2x

+ 81cos

2x

= 30, then x is equal

π 2 3π (d) 4 (b)  

7. The set of values of p for which the roots of the equation 3x2 + 2x + p ( p – 1) = 0 are of opposite sign is (a) (­– ∞, 0) (c) (­1, ∞)

14. The value of m for which the roots of the equation x2 + (m – 2) x + m + 2 = 0 are in the ratio 2 : 3, is (a) 1 (b) – 1 2 2 (c) 26 (d) – 26 3 3 15. If the ratio of the roots of lx2 + nx + n = 0 is p : q, then

5. If the roots of the equation (a) a = 0 (c) c = 0

(b) a ∈ [– 2, 8] (d) None of these

(b) (0, 1) (d) (­0, ∞)

(a)

q + p

p + q

l = 0 (b) n

p + q

q + p

n =0 l

(c)

q + p

p + q

l = 1 (d) n

p + q

q + p

n =1 l

16. The roots of the equation x4 – 2x3 + x = 380 are

1 ± 5 −3 1 ± 5 −3 (b) − 5, 4, 2 2 (a2 + b2) x2 + 2 (bc + ad) x + (c2 + d2)  = 0 are real, then a2, −1 ± 5 − 3 1 ± 5 −3 (c) 5, 4, (d) − 5, − 4, bd, c2 are in 2 2 (a) A.P. (b) G. P. 17. If α, β are irrational roots of ax2 + bx + c = 0 (a, b, c (c) H.P. (d) None of these ∈ Q), then 2 9. If a (b – c) x + b (c – a) x + c (a – b)  = 0 has equal roots, (a) α = β then a, b, c are in (b) αβ = 1 (c) α and β are conjugate roots (a) A.P. (b) G. P. (d) α2 + β2 = 1 (c) H.P. (d) None of these

8. If the roots of the equation

10. If the roots of the equation ax2 + bx + c = 0 are in the ratio 1:n, then (a) na2 = bc (1 + n)2 (b) nb2 = ca (1 + n)2 (c) nb2 = ca (1 – n)2 (d) None of these

(a) 5, − 4,

18. If the difference of the roots of x2 – px + q = 0 is unity, then (a) p2 + 4q = 1 (c) p2 + 4q2 = (1 + 2q)2

(b) p2 – 4q = 1 (d) q2 + 4p2 = (1 + 2p)2

19. If one root of the equation x2 + px + q = 0 is the square of the other, then

Quadratic Equations and Inequations

625

mULTIPLE-CHOICE QUESTIONS

626

29. If α and β are roots of the equation

Objective Mathematics

(a) p3 + q2 – q (3p + 1) = 0 (b) p2 + q2 + q (1 + 3p) = 0 (c) p3 + q2 + q(3p – 1) = 0 (d) p2 + q2 + q(1 – 3p) = 0

A (x2 + m2) + A m x + c  m2x2 = 0, then A (α2 + β2) + A αβ + c α2β2 =

20. The roots of the equation (x – a) (x ­– b) = abx2 are always (a) real (c) depend upon b

(b) depend upon a (d) depend upon a and b

(a) 0 (c) – 1

30. The value of p for which the quadratic equation x2 – px + p + 3 = 0 has reciprocal roots is (a) 1 (c) 2

21. The equation x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1 has (a) no solution (c) two solutions

(b) one solution (d) more than two solutions

22. If α and β are the roots of the equation x2 + px+ q = 0, α 2 β2 is then the value of + β α p (3q − p ) q p (3q + p 2 ) (c) q 2

(a)

q (3 p − q ) p 2

(b)

(a) A.P. (c) H.P.

(d) None of these

1 1 1 = are equal + r x+ p x+q in magnitude and opposite in sign, then (a) p + q = r (b) p + q = 2r (c) product of roots = – 1  (p2 + q2) 2 (d) None of these

2

43 (a) 16

(b) 3

(c) – 3

(d)

34. If f (x) = 2x3 + mx2 – 13x + n and 2, 3 are roots of the equation f (x) = 0, then the values of m and n are

− 43 16 25. In a quadratic equation with leading coefficient 1, a student 35. reads the coefficient 16 of x wrongly as 19 and obtain the roots as – 15 and – 4. The correct roots are (a) 6, 10 (c) – 7, – 9

(b) – 6, –10 (d) None of these

26. The roots of the equation c−a , 1 b−c b−c , 1 (c) a−b

a−b , 1 b−c c−a (d) ,1 a−b

(b) one solution (d) more than two solutions

If q ≠ 0 and the equation x3 + px2 + q = 0 has a root of multiplicity 2, then p and q are connected by (a) p2 + 2q = 0 (c) 4p3 + 27q + 1 = 0

(a) a (c) c

(b) p2 – 2q = 0 (d) 4p3 + 27q = 0

(b) b (d) None of these

37. If sin θ and cos θ are the roots of the equation

is double the other and if l is real, then the greatest value of m is 7 9 38. (b) (a) 8 8 8 (c) (d) None of these 9 28. The equation 125x + 45x = 2.27x has (a) no solution (c) two solutions

(b) – 5, 30 (d) None of these

 α 2 β2  α β  a + +b +  = β α α β

(b)

27. If one root of the equation (l – m) x2 + lx + 1 = 0

(a) ­– 5, – 30 (c) 5, 30

36. If α and β are roots of the equation ax2 + bx + c = 0, then

(b – c) x2 + (c – a) x + (a – b) = 0 are (a)

(b) 1 : 3 : 4 (d) None of these

33. If the roots of the equation

(b) p2 – 3q (d) q2 + 3p

24. The value of k so that the equations x – x – 12 = 0 and kx2 + 10x + 3 = 0 may have one root in common, is

(b) G.P. (d) None of these

32. If the equations ax2 + bx + c = 0 and x2 + 2x + 3 = 0 have a common root, then a : b : c =

2

(a) q2 – 3p (c) p2 + 3q

(b) – 1 (d) – 2

31. If the sum of the roots of the equation ax2 + bx + c = 0 is equal to the sum of the reciprocals of their squares, then bc2, ca2 and ab2 are in

(a) 2 : 4 : 5 (c) 1 : 2 : 3

23. If α and β are the roots of the equation x + px + q = 0, then the value of (ωα + ω2β) (ω2α + ωβ), where ω is an imaginary cube root of unity, is

(b) 1 (d) None of these

ax2 + bx + c = 0, then (a) (a – c)2 = b2 – c2 (c) (a + c)2 = b2 – c2

(b) (a – c)2 = b2 + c2 (d) (a + c)2 = b2 + c2

If a, b, c are in G.P., then the ratio between the roots of the equation ax2 + bx + c = 0, is (a)

−1+ 3i 2

(b)

−1− 3i 2

(c)

1+ 3i 2

(d)

1− 3i 2

q + 4q1/ 4

p+2 q

(a) 1 (c) 1/3

) , where k is equal to k

(b) x2 + 13x – 30 = 0 (d) None of these

(a) x2 + 13x + 26 = 0 (c) x2 + 13x + 30 = 0

(b) 1/2 (d) 1/4

2 40. The solutions of the equation (3 | x | – 3)2 = | x | + 7 49. In writing an equation of the form ax + bx + c = 0; the coefficient of x is written incorrectly and the roots are which belongs to the domain of definition of the function found to be equal. Again in writing the same equation, y = x ( x − 3) , are given by the constant term is written incorrectly and it is found that one root is equal to that of the previous wrong 1 −1 (a) , – 2 (b) ,2 equation while the other is double of it. If the square 9 9 of the difference of the roots of the correct equation is 1 1 k times their product, then k is equal to ± , ± 2 (d) – , ­– 2 (c) 9 9 (a) 2 (b) 4 3x (c) 5 (d) None of these 41. The roots of the equation 2 x + 2 ⋅ 3 x − 1 = 9 are given by 50. If α and β are the roots of the equation x2 – p (x + 1) – q

2 (a) log2    , – 2 3

(b) 3, – 3

(c) – 2, 1 − log 3 log 2

(d) 1 – log23, 2

42. In copying a quadratic equation of the form x2 + px + q = 0, a student wrote the coefficient of x incorrectly and the roots were found to be 3 and 10; another student wrote the same equation but he wrote the constant term incorrectly and thus he found the roots to be 4 and 7. The roots of the correct equation are (a) 5, 6 (c) 4, 5

(a) 1 (c) 3

(b) 2 (d) None of these

44. If the roots of the equation x2 + px + q = 0 differ from the roots of the equation x2 + qx + p = 0 by the same quantity, then (a) p + q + 1 = 0 (c) p + q + 4 = 0

(b) p + q + 2 = 0 (d) None of these

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

α β x + p x + q = 0 and if , are the roots of β α xn + 1 + (x + 1)n = 0, then n must be n n

n

(a) even integer (b) odd integer (c) rational but not integer (d) None of these 46. 7 log7 ( x

2 − 4 x + 5)

= x – 1, x may have values

(a) 2, 3 (c) – 2, – 3

(b) 7 (d) 2, ­– 3

47. For the equation | x2 | + | x | – 6 = 0, the roots are (a) real and equal (c) real with sum 1

(b) real with sum 0 (d) real with product 0

α 2 + 2α + 1 β 2 + 2β + 1 is + α 2 + 2α + q β 2 + 2β + q

(a) 1 (c) 3

(b) 2 (d) 0

51. The sum of the real roots of the equation | x – 2 |2 + | x – 2 | – 2 = 0 is (a) 2 (c) 4

(b) 6 (d) 8

52. If α, β are the roots of the equation ax2 + bx + c = 0 and

(b) 4, 6 (d) None of these

43. The number of real roots of | x2 + 4 | x | + 3 | + 2x – 11 = 0 is

2n

= 0, then the value of

Sn = αn + βn, then aSn + 1 + bSn + cSn – 1 is equal to (a) 0 (c) a + b + c

(b) abc (d) None of these

53. If α, β are the roots of x2 – 2px + q = 0 and γ, δ are roots of

x2 – 2rx + s = 0 and α, β, γ, δ are in A.P., then (a) p – q = r2 – s2 (c) r – s = p2 – q2

(b) s – q = r2 – p2 (d) None of these

54. If a ≤ 0, then the root of the equation

x2 – 2a | x – a | – 3a2 = 0 is (a) (1 − 2 )a

(b) (− 1 + 6 )a

(c) (1 + 2 )a

(d) − (1 + 6 )a

55. The values of a for which the equation 2x2 – 2 (2a + 1) x + a (a – 1) = 0 has roots α and β satisfying the condition α < a < β, are (a) (– 3, 0) (c) (– ∞, – 3) ∪ (0, ∞)

(b) (0, ∞) (d) None of these

56. Let a, b, c ∈ R and a ≠ 0. If α is a root of a2x2 + bx + c = 0, β is a root of a2x2 – bx – c = 0 and 0 < α < β, then the equation a2x2 + 2bx + 2c = 0 has a root γ that always satisfies (a) γ = β α+β (c) γ = 2

(b) γ = α (d) α < γ < β

627

(p + 6

48. In the equation x2 + px + q = 0, the coefficient of x was incorrectly written as 17 instead of 13, then the roots were found to be – 2 and – 15. The correct equation is

Quadratic Equations and Inequations

39. If α, β be the roots of the equation x2 – px + q = 0 and α > 0, β > 0, then the value of α1/4 + β1/4 is

628

57. If α, β are roots of x2 – px + q = 0; γ, δ are roots of x2 – rx + s = 0 and α, β, γ, δ,are in G.P., then

Objective Mathematics

(a) p2s = r2q (c) p2r = s2q

(b) p2q = r2s (d) None of these

58. If a, b, c ∈ R and the equations ax2 + bx + c = 0 and x3 + 3x2 + 3x + 2 = 0 have two roots in common, then (a) a = b ≠ c (c) a = b = c

(b) a = b = – c (d) None of these

59. The set of all real x for which x2 – |x + 2| + x > 0, is (a) (–∞, –2), ∪ (2, ∞)

(b) (−∞, − 2 ) ∪ ( 2 , ∞)

(c) (–∞, –1) ∪ (1, ∞)

(d) ( 2 , ∞)

60. If  f (x) = x – [x], x (≠ 0) ∈ R, where [x] is the greatest integer less than or equal to x, then the number of 1 solutions of f (x) + f   = 1 are x (a) 0 (c) infinite

(b) 1 (d) 2

61. The quadratic equation with rational coefficients, one 1 , is of whose roots is 2+ 5 (a) x2 + 4x – 1 = 0 (c) x2 – 4x – 1 = 0

(b) x2 – 4x + 1 = 0 (d) None of these

62. If (log5x) + log5x < 2, then x belongs to the interval 2

1  (a)  , 5  25 (c) (1, ∞)

1 1  (b)  ,  5 5  (d) None of these

63. The number of solutions of the equation sin (e x) = 5x + 5– x is (a) 0 (c) 2

(b) 1 (d) infinite

a b (c ≠ 0) has equal + 64. If the equation p = x + c x −c 2x roots, then p is equal to (a) ( a − b ) 2 (c) a + b

(b) ( a + b ) 2 (d) a – b

65. If x1, x2 are roots of the equation ax2 + bx + c = 0, then the value of (ax1 + b)– 2 + (ax2 + b)– 2 is 2 (a) b − 4ac a 2c 2

2 (b) b − ac a 2c 2

2 (c) b − 2ac a 2c 2

(d) None of these

66. Let a, b, c be real and ax2 + bx + c = 0 has two real roots c b α and β where α < – 1 and β > 1, then 1 + + a a (a) < 2 (c) < 0

(b) < 1 (d) None of these

67. Let f (x) = 1 + 2x + 3x2 + ... + (n + 1)xn, where n is even. Then the number of real roots of the equation f (x) = 0 is (a) 0 (c) n

(b) 1 (d) None of these

68. If α and β are the roots of the equation x2 + px + q = 0 and α4, β4 are the roots of x2 – rx + s = 0, then the equation x2 – 4qx + 2q2 – r = 0 has (a) has two real roots (b) two positive roots (c) two negative roots (d) one positive and one negative root 69. If c, d are the roots of the equation (x – a) (x – b) – k = 0, then a, b are the roots of the equation (a) (x – c) (x – d) – k = 0 (b) (x – c) (x – d) + k = 0 (c) (x – c) (x – d) + 2k = 0 (d) None of these 70. If α, β are the roots of the equation x2 + px + q = 0 then α is a root of the equation β (a) px2 + (2q – p2) x + p = 0 (b) qx2 + (p2 – 2q) x + q = 0 (c) qx2 + (2q – p2) x + q = 0 (d) None of these 71. If α, β are the roots of the equation x2 + px + q = 0 and γ, δ are roots of x2 + rx + s = 0 then the value of (α – γ) (α – δ) (β – γ) (β – δ) is (a) q2 + s2 – pr (q + s) + s (p2 – 2q) – qr2 (b) q2 + s2 – pr (q + s) + s (p2 – 2q) + qr2 (c) q2 + s2 + pr (q + s) + s (p2 – 2q) + qr2 (d) None of these 72. The greatest negative integer satisfying x2 – 4x – 77 < 0 and x2 > 4 is (a) – 4 (c) – 7

(b) – 6 (d) None of these

73. If the equations k (6x2 + 3) + rx + 2x2 – 1 = 0 and 6k (2x2 + 1) + px + 4x2 – 2 = 0 have both the roots common, then the value of 2r – p is (a) 0 (c) – 1

(b) 1 (d) None of these

74. If x = 2 + 22/3 + 21/3 then the value of x3 – 6x2 + 6x is (a) 3 (c) 1

(b) 2 (d) None of these

75. If x2 + px + 1 is a factor of ax3 + bx + c, then (a) a2 + c2 = – ab (c) a2 – c2 = ab

(b) a2 – c2 = – ab (d) None of these

76. If α, β are the roots of the equation x2 + px + q = 0 and γ, δ are roots of x2 + rx + s = 0, then the value of (α – γ)2 + (β – γ)2 + (α – δ)2 + (β – δ)2 is (a) 2 (p2 + r2 – pr + 2q – 2s) (b) 2 (p2 + r2 – pr + 2q + 2s) (c) 2 (p2 + r2 – pr – 2q – 2s) (d) None of these

(a) 1 (c) ­– 1

(b) 3 (d) – 3

78. Both the roots of the equation (x – b) (x – c) + (x – a) (x – c) + (x – a) (x – b) = 0 are always (a) positive (b) negative (c) real (d) None of these 79. If one root of x2 + px + q = 0 is 5 – 3i, then the real values of p and q are (a) p = – 10, q = – 34 (c) p = 10, q = – 34

(b) p = – 10, q = 34 (d) None of these

80. The real values of p and q such that the equation x2 + px + q = 0 has 2 + i 3 as a root, are (a) p = – 4, q = 7 (c) p = 4, q = – 7

(b) p = – 4, q = – 7 (d) p = 4, q = 7

81. If the equations x – ax + b = 0 and x + bx – a = 0 have a common root, then 2

(a) a + b = 1 (c) a – b = 2

2

(b) a = b (d) a + b = 0 or a – b = 1

82. If one root of the equation x2 + px + 12 = 0 is 4, while the equation x2 + px + q = 0 has equal roots, the value of q is (a) 49/4 (c) 4

(b) 4/49 (d) None of these

83. The roots of the equation 2

3 1 1   x +  = 4 +  x −  are x 2 x (a) ± 1, 2, 1 (b) ± 1, 2, – 1 2 2 1 (c) ± 1, – 2, –  (d) None of these 2 84. The roots of the equation 32x + 1 + 32 = 3x + 3 + 3x are (a) 1, ­– 2 (c) – 1, 2

(b) 1, 2 (d) – 1, ­– 2

85. If the expression y2 + 2xy + 2x + my – 3 can be resolved into two rational factors, then m must be (a) any positive real number (b) any negative real number (c) ­– 2 (d) 3 86. The quadratic equation with rational coefficients whose 2+ 3 , is one root is 2− 3 (a) x – 14x + 1 = 0 (c) x2 – 14x – 1 = 0 2

(b) x + 14x + 1 = 0 (d) None of these 2

87. The rational values of a and b in ax2 + bx + 1 = 0 if 1 is a root, are 4+ 3 (a) a = 13, b = – 8 (c) a = 13, b = 8

(b) a = – 13, b = 8 (d) a = – 13, b = ­– 8

(a) x2 + b (b2 – 3c) x – c3 = 0 (b) x2 + b (b2 – 3c) x + c3 = 0 (c) x2 – b (b2 – 3c) x + c3 = 0 (d) None of these 89. A car travels 25 km an hour faster than a bus for a journey of 500 km. If the bus takes 10 hours more than the car, then the speed of the car and the bus is (a) 25 km/hr, 40 km/hr (b) 25 km/hr, 50 km/hr (c) 25 km/hr, 60 km/hr (d) None of these 90. The values of a, for which the quadratic equation 3x2 + 2 (a2 + 1) x + (a2 – 3a + 2) = 0 possesses roots of opposite sign, are (a) 1 < a < 2 (c) 1 < a < 3

(b) a ∈ (2, ∞) (d) None of these

91. Solution of the equation x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1 is (a) x ∈ [2, 3] (c) x ∈ (5, 10)

(b) x ∈ [2, 5] (d) x ∈ [5, 10]

92. If tan A and tan B are the roots of x2 – px + q = 0, then the value of sin2 (A + B) is (a)

p2 p + q2

(b)

p2 p + (1 − q ) 2

(c)

q2 p + (1 − q ) 2

(d)

p2 ( p + q)2

2

2

2

93. If a and b are rational and α, β be the roots of x2 + 2ax + b = 0, then the equation with rational coefficients one of whose roots is α + β + α 2 + β 2 is (a) x2 + 4ax – 2b = 0 (c) x2 – 4ax + 2b = 0

(b) x2 + 4ax + 2b = 0 (d) x2 – 4ax – 2b = 0

94. The real roots of the equation x2/3 + x1/3 – 2 = 0 are (a) 1, 8 (c) – 1, 8

(b) – 1, – 8 (d) 1, – 8

95. If α and β are the roots of the equation x2 – px + q = 0, 2 α2 and β , is then the equation whose roots are β α (a) qx2 + ( p3 – 3pq) x + q2 = 0 (b) qx2 – ( p3 + 3pq) x + q2 = 0 (c) qx2 – (p3 – 3pq) x + q2 = 0 (d) None of these 96. If the sum of the roots of a quadratic equation is 3 and the sum of their cubes is 7, then the equation is (a) 9x2 + 27x + 20 = 0 (c) 9x2 – 27x – 20 = 0

(b) 9x2 + 27x – 20 = 0 (d) 9x2 – 27x + 20 = 0

97. If one root of the equation px2 – 14x + 8 = 0 is six times the other, then p is equal to (a) 2 (c) 1

(b) 3 (d) None of these

629

88. The quadratic equation whose roots are cubes of the roots of x2 + bx + c = 0 is

Quadratic Equations and Inequations

77. If (a2 – 1) x2 + (a – 1) x + a2 – 4a + 3 = 0 is an identity in x, then the value of a is

630

98. If α and β are the roots of the equation (a) –

ax2 – 2bx + c = 0, then α3β3 + α2β3 + α3β2 =

Objective Mathematics

(a)

c2 (c – 2b) a3

(b)

c2 (c + 2b) a3

(c)

bc 2 a3

(d) None of these

(c)

99. Root of the equation 3x – 1 + 31 – x = 2 is (a) 1 (c) 0

(b) 2 (d) – 1

100. If c, d are the roots of the equation (x – a) (x – b) – k = 0, then the roots of the equation (x – c) (x – d) + k = 0 are (a) c, d (c) b, d

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

101. Solution of the equation x−2 + 4− x =

6 − x is

(a) x = 4 –

4 5

(b) x = 4 +

4 5

(c) x = 4 –

2 5

(d) x = 4 +

2 5

1 and 7 7

1 and 3 3

(b)

1 and 7 7

(d) None of these

x 3 108. The coefficient of x4 in the expansion of  − 2  2 x  equal to 405 504 (a) (b) 256 259 450 (c) (d) none of these 263

10

is

109. If x is real, then the maximum value of 3 – 6x – 8x is 17 33 (a) (b) 8 8 21 (c) (d) None of these 8 110. If x2 + px + q = 0 and x2 + p ′ x + q ′ = 0 have one root in common, then its value is q − q′ p′ − p p′ − p (c) q − q′ (a)

pq ′ − p ′q q − q′ q − q′ (d) pq ′ − p ′q (b)

102. If α ≠ β, but α2 = 5α – 3, β2 = 5β – 3, then the equaα β 111. If α, β are the roots of the equation x2 + x + 1 = 0, tion whose roots are and is β α then the equation whose roots are α19 and β7 is (a) x2 – 5x – 3 = 0 (c) 3x2 + 12x + 3 = 0

(a) x2 – x + 1 = 0 (c) x2 + x + 3 = 0

(b) 3x2 – 19x + 3 = 0 (d) None of these

103. If α, β are the roots of the equation ax2 + bx + c = 0, then 1 1 and is the equation whose roots are aα + b aβ + b (a) cax2 – bx + 1 = 0 (c) cax2 + bx – 1 = 0

(b) cax2 + bx + 1 = 0 (d) None of these

104. The quadratic equation whose roots are the squares of the sum and difference of the roots of ax2 + bx + c = 0 is

x

(a) x ∈ [0, 64] (c) x ∈ [0, 64)

x

> 1, are

(b) x ∈ (0, 64) (d) None of these

(b) α = 1, β = – 2 (d) α = 2, β = – 2

107. For real values of x, the expression between

271/x + 121/x = 2 ⋅ 81/x is (a) one (b) two (c) infinite (d) zero 113. If α, β are roots of the equation x2 + px + p2 + q = 0, then the value of α2 + αβ + β2 + q is (a) 0 (c) q

1 2 1 (c) 2 (a)

106. If α, β are the roots of the equation x2 + x α + β = 0 then the values of α and β are (a) α = 1, β = – 1 (c) α = 2, β = 1

112. The number of real solutions of the equation

(b) 1 (d) 2q

114. The value of m so that the equations 3x2 – 2mx – 4 = 0 and x (x – 4m) + 2 = 0 may have a common root is

(a) a4x2 + 2a2 (b2 – 2ca) x + b2 (b2 – 4ca) = 0 (b) a4x2 – 2a2 (b2 – 2ca) x – b2 (b2 – 4ca) = 0 (c) a4x2 – 2a2 (b2 – 2ca) x + b2 (b2 – 4ca) = 0 (d) None of these 1 1 105. The real values of x for which 372     3 3

(b) x2 + x + 1 = 0 (d) x2 – x + 3 = 0

x 2 − 3x + 4 lies x 2 + 3x + 4

1 2 (d) – 1 2

(b) –

115. If the equations ax2 + bx + c = 0 and bx2 + cx + a = 0 have a common root, then (a) a + b + c = 1 (c) a = b = c

(b) a + b + c = 0 (d) None of these

116. If ax2 + 2bx + c = 0 and a1x2 + 2b1x + c1 = 0 have a a b c , , are in A.P., then a1, b1, c1 common root and a1 b1 c1 are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

(a) 6a = 4b = – 3c (c) 6a = – 4b = 3c

(b) 3a = – 4b = 3c (d) None of these

(a) ( pr – q)2 (c) ( pq – r)2

(b) (qr – p)2 (d) None of these

118. If a, b, c are rational and ax2 + bx + c = 0 and 3x2 + 128. The expression x2 + 2 (a + b + c) x + 3 (bc + ca + ab) x – 5 = 0 have a common root, then 3a + b + 2c = will be a perfect square if (a) 0 (b) 1 (a) a + b + c = 0 (b) ac + bc + ab = 0 (c) 2 (d) None of these (c) a = b = c (d) None of these 119. If the equations x2 + abx + c = 0 and x2 + acx + b = 0 129. The value of k for which the expression have a common root, then their other roots satisfy the 3 x2 + (2 – k) x + k – will be a perfect square, is equation 4 (a) x2 + a (b + c) x + a2bc = 0 (a) 1 (b) 6 (b) x2 – a (b + c) x + a2bc = 0 (c) 7 (d) None of these 2 2 (c) x – a (b + c) x – a bc = 0 (d) None of these 130. If (ax2 + bx + c) y + a′x2 + b′x + c′ = 0, then the condition that x may be a rational function of y is 120. If one root of x2 + px + q = 0 is also the root of the equation x2 + mx + n = 0, then the other root of the first equation will satisfy the equation (a) nx2 + mqx + q2 = 0 (c) mx2 – mqx + q2 = 0

(b) mx2 + nqx + q2 = 0 (d) None of these

(a) (ac′ – a′ c)2 = (ab′ – a′  b) (bc′ – b′c) (b) (ab′ – a ′ b)2 = (ac′ – a ′ c) (bc′ – b′c) (c) (bc′ – b ′ c)2 = (ab′ – a ′ b) (ac′ – a′c) (d) None of these

121. If the equations x2 – px + q = 0 and x2 – ax + b = 0 131. If a (b – c) x2 + b (c – a) xy + c (a – b) y2 is a perfect have a common root and the roots of the second equasquare, then a, b, c are in tion are equal, then (a) A.P. (b) G.P. (a) aq = 2 (b + p) (b) aq = b + p (c) H.P. (d) None of these (c) ap = 2 (b + q) (d) ap = b + q 2 132. If the expression ax + by2 + cz2 + 2ayz + 2bzx + 2cxy 2 2 122. If the equations x – px + q = 0 and x – ax + b = 0 can be resolved into rational factors, then have a common root and the other root of the second (a) a + b + c = 0 (b) a3 + b3 + c3 = 3abc equation is the reciprocal of the other root of the first, (c) ab + ac + bc = 0 (d) None of these then (q – b)2 = 12 x (a) aq ( p – b)2 (b) bq (p – a)2 133. For all real values of x, (c) bq ( p – b)2 (d) None of these 4x2 + 9 123. If the equation x2 + 9y2 – 4x + 3 = 0 is satisfied for real values of x and y, then (a) 1 ≤ x ≤ 3 1 1 (c) – ≤ y ≤ 3 3

(b) 2 ≤ x ≤ 3 1 (d) ≤x≤1 3

124. The values of a which make the expression x2 – ax + 1 – 2a2 always positive for real values of x, are 2 2 1

(b) ≤ 2 (d) > 2

134. If the roots of the equations x2 – bx + c = 0 and x2 – cx + b = 0 differ by the same quantity then b + c is equal to (a) 4 (c) 0

(b) 1 (d) – 4

135. The equation whose roots are (a) 7x2 – 6x + 1 = 0 (c) x2 – 6x + 7 = 0

1 1 and is 3+ 2 3− 2

(b) 6x2 – 7x + 1 = 0 (d) x2 – 7x + 6 = 0

136. If α, β are roots of the equation x2 – 4x + 1 = 0, then the value of α3 + β3 is (a) 76 (c) – 52

(b) 52 (d) – 76

137. If x2 – 3x + 2, be one of the factors of the expression x4 – px2 + q, then (a) p = 4, q = 5 (b) p = 5, q = 4 (c) p = – 5, q = – 4 (d) None of these

631

127. If the equations x3 + 3px2 + 3qx + r = 0 and x2 + 2px + q = 0 have a common root, then 4 (p2 – q) (q2 – pr) =

Quadratic Equations and Inequations

117. If the two equations ax2 + bx + c = 0 and 2x2 – 3x + 4 = 0 have a common root, then

632

138. If the roots of the equation (a + b ) t – 2 (ac + bd) t + (c + d ) = 0 are equal, then 2

2

2

Objective Mathematics

(a) ab = cd (c) ad + bc = 0

2

2

(b) ac = bd a c (d) = b d

x − bx λ −1 = are ax − c λ +1 such that α + β = 0 then the value of λ is

139. If the roots α, β of the equation a−b (a) a+b

2

(b) c

a+b (d) (c) 1 a−b c 140. The number of real solutions of the equation x2 – 3 | x | + 2 = 0 are (a) 2 (c) 4

(b) 3 (d) 1

141. For what values of k will the equation x2 – 2 (1 + 3k) x + 7 (3 + 2k) = 0 have equal roots ? (a) 1, −10 (b) 2,  −10 9 9 (c) 3, −10 (d) 4, −10 9 9 142. If α, β are the roots of the equation lx2 + mx + n = 0, then the equation whose roots are α3β and αβ3 is (a) l4x2 – n l (m2 – 2nl) x + n4 = 0 (b) l4x2 + n l (m2 – 2nl) x + n4 = 0 (c) l4x2 + n l (m2 – 2nl) x – n4 = 0 (d) l4x2 – n l (m2 + 2nl) x + n4 = 0 143. If (2 + i 3 ) is a root of the equation x2 + px + q = 0, where p and q are real, then (p, q) equals (a) (4, 7) (c) (– 4, 7)

(b) (– 4, – 7) (d) (4, – 7)

144. If the roots of the equation x2 + a2 = 8x + 6a are real, then (a) a ∈ [2, 8] (c) a ∈ (2, 8)

(b) a ∈ [– 2, 8] (d) a ∈ (– 2, 8)

145. Let α, β be the roots of the quadratic equation x2 + px + p3 = 0 (p ≠ 0). If (α, β) is a point on the parabola y2 = x, then the roots of the quadratic equation are (a) 4, – 2 (c) 4, 2

(b) – 4, – 2 (d) – 4, 2

146. Let α, β be the roots of the equation (x – a) (x – b) = c, c ≠ 0. The roots of the equation (x – α) (x – β) + c = 0 are (a) a, c (c) a, b

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

147. If the roots of the equation x2 – 2ax + a2 + a – 3 = 0 are real and less than 3, then (a) a < 2 (b) 2 ≤ a ≤ 3 (b) 3 ≤ a ≤ 4 (d) a > 4

148. For all real x, the minimum value of

1 − x + x2 is 1 + x + x2

(b) 1 3 (d) 3

(a) 0 (c) 1

149. The number of real roots of the following equation | x |2 – 7 | x | + 12 = 0 is (a) 1 (c) 3

(b) 2 (d) 4

150. The equation x +

2 2 =1+ , has 1− x 1− x

(a) no real root (c) two equal roots

(b) one real root (d) infinitely many roots

151. The roots of an equation x3 – 9x2 + 14x + 24 = 0 are in the ratio 3 : 2. The roots will be (a) 6, 4, – 1 (c) – 6, 4, 1

(b) 6, 4, 1 (d) – 6, – 4, 1

2 152. Given that, for all real x, the expression x − 2 x + 4 x2 + 2x + 4 lies between 1 and 3. The values between which the 3 9 ⋅ 32 x + 6 ⋅ 3x + 4 expression lies are 9 ⋅ 32 x − 6 ⋅ 3x + 4

(a) 0 and 2 (c) – 2 and 0

(b) – 1 and 1 (d) 1 and 3. 3

153. The numerical difference of the roots of x2 – 7x – 9 = 0 is (a)

85

(c) 2 85

(b) 9 7 (d) 5

154. The quadratic equation 8sec2 θ – 6sec θ + 1 = 0 has (a) exactly two roots (b) exactly four roots (c) infinitely many roots (d) no roots 155. The value of x2 + 2bx + c is positive if (a) b2 < c (c) b2 – 4c < 0

(b) c2 < b (d) b2 – 4c > 0.

156. If sin α and cos α are the roots of the equation px2 + qx + r = 0, then (a) p2 + q2 – 2pr = 0 (c) p2 – q2 + 2pr = 0

(b) (p – r)2 = q2 + r2 (d) (p + r)2 = q2 – r2

157. The roots of the equation x2 + ax + b = 0 where a =2 b and b > 3 are (a) imaginary (b) equal (c) unequal (d) None of these 158. The quadratic equation whose roots are three times the roots of the equation 3ax2 + 3bx + c = 0 is (a) ax2 + 3bx + 3c = 0 (c) ax2 + bx + c = 0

(b) ax2 + bx + 3c = 0 (d) ax2 + 3bx + c = 0

2

(a) – a ( 6 + 1) (b) a ( 6 + 1) (c) a (1 –

2 ), a (–1 +

(d) a (1 +

2)

168. If α, β are the roots of the equation ax2 + bx + c =0 1 1 + , then and α2 + β2 = α β a, b, c are in G.P. b c a a b c are in A.P. (b) , , b c a b c a are in A.P. (c) , , a b c (a)

6)

160. If α ≠ β and α2 = 5α – 3, β2 = 5β – 3, then the equation having α/β and β/α as its roots is (a) 3x2 + 19x + 3 = 0 (c) 3x2 – 19x – 3 = 0

(b) 3x2 – 19 + 3 = 0 (d) x2 – 16x + 1 = 0

a b = 1 has roots equal in + x−a x−b magnitude but opposite in sign, then value of a + b is

161. The equation

(a) 1 (c) 0

(b) – 1 (d) None of these

162. The value of k for which the number 3 lies between the roots of the equation x2 + (1 – 2k) x + (k2 – k – 2) =0 is given by (a) 2 < k < 5 (c) 2 < k < 3

(b) k < 2 (d) k > 5

163. The minimum value of | x | x +

1 5 + | x – 3 | x − 2 2

is (a) 2 (c) 6

(b) 4 (d) 0

164. The solutions of the inequation 2x2 + 3x – 9 ≤ 0 are given by 3 3 (a) – 3 ≤ x ≤ – (b) – 3 ≤ x ≤ 2 2 3 (c) ≤ x ≤ 3 (d) None of these 2

(d) None of these 169. If a < b, then the solution of x2 + (a + b) x + ab < 0, is given by (a) x < – b or x < – a (c) x < a or x > b

(b) a < x < c (d) – b < x < – a

170. If α, β are the roots of the equation 8x2 – 3x + 27 = 1

1

 α 2  3  β2  3 0, then the value of   +   is β  α (a)

1 3

(c) 4

7 2 1 (d) 4

(b)

171. If 2 – 3x – 2x2 ≥ 0, then (a) x ≤ – 2

(b) – 2 ≤ x ≤

1 2

1 2 2 172. In the equation ax + bx + c = 0, if a ≠ 0 and b = 0, then the roots are (c) x ≥ – 2

(d) x ≤

(a) reciprocal to each other (b) reciprocal to each other but have same sign (c) equal (d) equal in magnitude but opposite in sign

173. If the roots of x2 – bx + c = 0 are two consecutive integers, then b2 – 4ac is x + 2 (3a + 5) x + 2 (9a + 25) = 0 are real, when ‘a’ (a) 1 (b) 0 equals (c) 2 (d) None of these −5 3 (a) (b) 174. If α, β are the roots of the equation x2+ px + p2 + q = 3 5 0, then the value of α2 + αβ + β2 + q is equal to 5 −3 (c) (d) 3 5 (a) 2q (b) q (c) 1 (d) 0 166. The value of λ for which the quadratic equation 2 x2 − 7 x + 7 = 9 is 3x2 + 2 (λ2 + 1) x + (λ2 – 3λ + 2) = 0 has roots of opposite 175. The number of real roots of 3 165. The roots of the equation 2

2

signs, lies in the interval. (a) (1, 2)

3 (b)  , 2  2 

(c) (– ∞, 1)

(d) (– ∞, 0)

167. The greatest and the least values of the expression x2 + 2x + 1 for all real values of x, are respectively x2 + 2x + 7 (a) 1, 0 (c) 1, 2

(b) 2, 3 (d) 2, 0

(a) zero (c) 1

(b) 2 (d) 4

176. If ax2 + bx + c = a (x – α) (x –­ β), then a (αx + 1)  (βx + 1) is equal to (a) cx2 – bx – a (c) cx2 – bx + a

(b) cx2 + bx + a (d) cx2 + bx + c

177. If α, β are non-real roots of ax2 + bx + c = 0, (a, b, c ∈ R), then (a) αβ = 1 (b) α = β (c) αβ = 1

(d) α = β

633

x – 2a | x – a | – 3a = 0, are given by 2

Quadratic Equations and Inequations

159. For a ≤ 0, the roots of the equation

634

178. The conditions that the equation ax2 + bx + c = 0 has both the roots positive is that

Objective Mathematics

(a) a and b are of the same sign (b) a, b and c are of the same sign (c) a and c have of the same sign opposite to that of b (d) b and c have the same sign opposite to that of a 179. If f (x) = 2x3 + mx2 – 13x + n and 2, 3 are roots of the equation f (x) = 0, then the values of m and n are (a) 5, 30 (c) – 5, 30

(b) – 5, – 30 (d) None of these

180. If two equations x2 + a2 = 1 – 2ax and x2 + b2 = 1 – 2bx have only one common root then (a) a – b = 1 (c) a – b = 2

(b) a – b = – 1 (d) | a – b | = 1

181. The value of k (k > 0) for which the equations x2 + kx + 64 = 0 and x2 – 8x + k = 0, both will have real roots is (a) 16 (c) – 16

(b) – 64 (d) 8

x 2 − bx m −1 = has roots equal in ax − c m +1 magnitude but opposite in sign then m equals

182. If the equation

b−a (a) b+a (c)

a+b (b) a−b

a−b a+b

(d) None of these

183. The value or values of p for which the equation 2x2 –

2 px + p = 0 has equal roots is or are

(a) 0 (c) 0, 4

(b) 4 (d) None of these

184. If p (q – r) x + q (r – p) x + r (p – q) = 0 has equal 2 = roots, then q 1 1 (b) +r (a) p + r p 1 1 + (c) p + r (d) p r 2

185. If a, b, c are in G.P., then equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 d e f are in have a common root if , , a b c (a) H.P. (c) A.P.

(b) G.P. (d) None of these

186. Let a, b and c be real numbers such that 4a + 2b + c = 0 and ab > 0. Then the quadratic equation ax2 + bx + c = 0 has (a) purely imaginary roots (b) only one root (c) real roots (d) complex roots

187. The smallest value of x2 – 3x + 3 in the interval  − 3,  is (a) – 20

(b) – 15

(c) 5

(d) 3 4

3  2

188. If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a , b and c are in c a b (a) Arithmetic progression (b) Geometric progression (c) Harmonic progression (d) Arithmetico-Geometric progression 189. If the equation y = λx + a 1 + λ 2 , regarded as a quadratic in λ, will have equal roots, then x2 + y2 is equal to (a) – a2 (c) 0

(b) a2 (d) none

190. If a, b, c ∈Q, then roots of the equation (b + c – 2a) x2 + (c + a – 2b) x + (a + b – 2c) = 0 are (a) irrational (c) equal

(b) rational (d) non-real

191. If the roots of (b – c) x2 + (c – a) x + (a – b) = 0 are equal, then a + c = (a) b2 (c) 2b

(b) b (d) 3b

192. If (7 − 4 3 ) x − 4 x + 3 + (7 + 4 3 ) x2 − 4 x + 3 = 14, then the value of x is given by 2

(a) 2, 2 ± (c) 3 ±

2 , 2 2

(b) 2 ±

3 , 3 (d) None of these

193. The solution of 6 + x – x2 > 0 is (a) – 1 < x < 2 (b) – 2 < x < 3 (c) – 2 < x < – 1 (b) None of these 194. The value of m for which the equation x3 – mx2 + 3x – 2 = 0 has two roots equal in magnitude but opposite in sign, is (a) 4/5 (c) 2/3

(b) 3/4 (d) 1/2

195. The value of a for which one root of the quadratic equation (a2 – 5a + 3)x2 + (3a – 1)x + 2 = 0 is twice as large as the other is (a)

2 3

(c)

1 3

2 3 1 (d) – 3 (b) – 

196. If the roots of a1 x2 + b1 x + c1 = 0 and a2x2 + b2x + c2 = 0 are same, then

208. If the roots of the equation px2 + qx + r = 0 and qx2 + 2 pr x + q = 0 are simultaneously real, then

197. The roots of the equation (3 – x)4 + (2 – x)4 = (5 ­– 2x)4 are (a) two real and two imaginary (b) all imaginary (c) all real (d) none of the above 198. If 4 ≤ x ≤ 9, then (a) (x – 4) (x – 9) ≤ 0 (c) (x – 4) (x – 9) < 0

(b) (x – 4) (x – 9) ≥ 0 (d) (x – 4) (x – 9) > 0

199. For all x ∈ R if mx2 – 9mx + 5m + 1 > 0, then m lies in the interval − 61  (a)  , 0  4 

4 61 (b)  ,   61 4 

4 (c) 0,   61 

−4  (d)  ,0  61 

200. If a, b, c are + ve real numbers, then the number of real roots of the equation ax2 + b | x | + c = 0 is (a) 0 (c) 4

(b) 2 (d) None of these

201. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation (a) x2 + 18x – 16 = 0 (b) x2 – 18x + 16 = 0 (c) x2 + 18x + 16 = 0 (d) x2 – 18x – 16 = 0 202. The real roots of | x | – 3x + 3 | x | – 2 = 0 are 3

(b) – 1 (d) – 1, 1

2

(a) ± 1 (c) 1, 2

(b) ± 2 (d) 0, 2

(a) 0, –1 (c) 0.1

(b) –1, 1 (d) –1, 2

(a)

p q = q r

(c) 2q = ±

(b) q = r ≠ 0 pr

(d) p = q, r ≠ 0

209. The set of values of m for which both roots of the equation x2 – (m + 1) x + m + 4 = 0 are real and negative consists of all m such that (a) – 3 ≥ m or m ≥ 5 (c) – 4 < m ≤ – 3

(b) – 3 ≤ m ≤ 5 (d) – 3 < m ≤ – 1

210. If α, β are the roots of the equation x2 + x α + β = 0, then the values of α and β are (a) α = 2 and β = – 2 (b) α = 2 and β = 1 (c) α = 1 and β = – 2 (d) α = 1 and β = – 1 211. The number of negative integral solutions of x2 · 2x + 1 + 2| x – 3 | + 2 = x2 · 2(| x – 3 | + 4) + 2x – 1 is (a) 4 (c) 1

(b) 2 (d) 0

212. If α, β are the roots of quadratic equation x2 + bx – c = 0, then the equation whose roots are b and c is (a) x2 + (αβ + α + β) x – αβ (α + β) = 0 (b) x2 + [(α + β) + αβ] x + αβ (α + β) = 0 (c) x2 – [(α + β) + αβ] x – αβ (α + β) = 0 (d) x2 + αx – β = 0. 213. The values of p and q (p ≠ 0, q ≠ 0) for which p, q are the roots of the equation x2 + px + q = 0 are (a) p = 1, q = – 2 (c) p = – 1, q = 2

(b) p = – 1, q = – 2 (d) p = 1, q = 2

203. If (1 – p) is a root of quadratic equation x2 + px + (1 – p) 214. The roots of the equation (a + c – b) x2 – 2cx + (b + c – a) = 0 are = 0 then its roots are (a) 1,

a+c−b b+c−a

(b) 1, b + c − a 2c

(c) 1,

b+c−a a+c−b

(d) 1,

204. The roots of the equation 4x – 3 · 2x + 3 + 128 = 0 are (a) 4 and 5 (c) 2 and 3

(b) 3 and 4 (d) 1 and 2

205. Given that ax2 + bx + c = 0 has no real roots and a + b + c < 0, then (a) c ≠ 0 (c) c > 0

(b) c < 0 (d) c = 0

206. The number of roots of the equation ( x + 2)( x − 5) ( x − 2) = is ( x − 3)( x + 6) ( x + 4) (a) 0 (c) 2

(b) 1 (d) 3

2c a+c−b

215. If one root of the equation x2 + px + 12 = 0 is 4, while the equation x2 + px + q = 0 has equal roots, then the value of q is (a) 3 (c) 49/4

(b) 12 (d) 4

216. If the roots of the equation ax2 + bx + c = 0 are in the m n b = ratio m : n, then + + n m ac (a) 0 (c) – 1

(b) 1 (d) None of these

635

(a) 0 (c) 1

Quadratic Equations and Inequations

207. If x2 – x + 1 = 0, then value of x3n is

a1 b c = 1 = 1 a2 b2 c2 (b) a1 = b1 = c1; a2 = b2 = c2 (c) a1 = a2, b1 = b2, c1 = c2 (d) c1 = c2 = 0 (a)

636

217. The equation

Objective Mathematics

x2 – 2 2 kx + 2 × e2 log k – 1 = 0 has the product of roots equal to 31, then, for what value of k it has real roots? (a) 1 (c) 3

(b) 2 (d) 4

218. The values of k for which the quadratic equation kx2 + 1 = kx + 3x – 11x2 has real and equal roots are (a) {– 11, – 3} (c) {5, – 7}

(b) r (d) None of these

(b) A/a (d) (A/a)2

221. If the sum of squares of the roots of the equation x2 – (a – 2)x – (a + 1) = 0 is least then the value of a is (b) 2 (d) 1

222. If one root of the = n ax2 – bx + c = 0 is square of the other, then (a) b – 4ac = 0 (c) ac = b3 2

(b) ac (a + c + 3b) = b (d) None of these

3

223. If x2 + 2ax + 10 – 3a > 0 for all x ∈ R, then (a) – 5 < a < 2 (c) a > 5

(b) a < – 5 (d) 2 < a < 5

224. If the equation 5x2 + 13x + k = 0 has roots α and 1 , α then k is equal to (a) 1 (c) – 5 225. Roots of the equation are (a) ± 3 (c) ± 5

(a) p = 1 (c) p = – 2

(b) p = 1 or 0 (d) p = –­2 or 0

(a) 0 < α < β (c) α < β < 0

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

230. Let α, β be roots of x2 – x + p = 0 and γ, δ be the roots of x2 – 4x + q = 0. If α, β, γ, δ are in G. P. then the integral values of p and q respectively, are (a) – 2, – 32 (c) – 6, 3

(b) ­– 2, 3 (d) ­– 6, – 32

231. 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 1 + c + a b . a

Ax2 + Bx + C = 0 are α + k, β + k then (B2 – 4AC)/(b2 – 4ac) is equal to

(a) 0 (c) – 1

228. If p and q are the roots of the equation x3 + px + q = 0, then

x2 + bx + c = 0, where c < 0 < b, then

220. If roots of ax2 + bx + c = 0 are α, β and roots of

(a) a/A (c) (a/A)2

(c) (– 1, 2)

5π   (b)  − 1, 6   π  (d)  , 2  6 

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

(b) {5, 7} (d) None of these

1 1 1 + = are 219. If the roots of the equation x+ p x+q r equal in magnitude but opposite in sign, then (p + q) equals (a) 2r (c) – 2r

 π 5π  (a)  ,  6 6 

(b) 13 (d) 5 2 x + 31 x + 7 = 2 x + 47 + 2 9 x −7 9

(a) < 0 (c) ≤ 0

(b) > 0 (d) None of these

232. If α and β are the roots of x2 + px + q = 0 and α4 and β4 are the roots of x2 – rx + s = 0, then the equation x2 – 4qx + 2q2 – r = 0 has always (a) two real roots (b) ­two positive roots (c) two negative roots (d) ­one positive and one negative root 233. Let a, b, c be real numbers, a ≠ 0. If α is a root of a2x2 + bx + c = 0, β is a root of a2x2 – bx – c = 0 and 0 < α < β, then the equation a2x2 + 2bx + 2c = 0 has a root γ that always satisfies (a) γ = α + β 2 (c) γ = α

(b) γ = α + β 2 (d) α < γ < β.

234. Solution of | x2 + 4x + 3 | + 2x + 5 = 0 is

2

(b) ± (d) ±

3 5

226. If one root of a quadratic equation is − i + 3 , then 2 the sum of the roots is (a) – i

(b)

3 /2

(c) + i

(d)

3

227. Let 2sin2 x + 3sin x – 2 > 0 and x2 – x – 2 < 0 (x is measured in radians). Then x lies in the interval

(a) 4 (c) – 1 –

(b) – 4

3

(d) 1 +

3 235. If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2) p2 – 2 (ab + bc + cd) p + (b2 + c2 + d2) ≤ 0 then a, b, c and d. (a) are in A.P. (c) are in H.P.

(b) are in G.P. (d) ­satisfy ab = cd

236. If the quadratic equation x2 + ax + b = 0 and x2 + bx + a = 0 (a ≠ b) have a common root, then the numerical value of a + b is (a) 1 (c) 2

(b) – 1 (d) – 2.

(a) a < 2 (c) 3 < a ≤ 4 238. The equation

(b) 2 ≤ a ≤ 3 (d) a > 4

x + 1 − x − 1 = 4 x − 1 has

(a) no solution (b) one solution (c) two solutions (d) more than two solutions 239. If α, β are the roots of the equation ax2 + bx + c = 0, (a ≠ 0) and α + δ, β + δ are the roots of Ax2 + Bx + C = 0, (A ≠ 0) for some constant δ, then 2 2 (a) b − 4ac = B − 4AC 2 a A2 2 2 (b) b − 2ac = B − 2AC 2 a A2 2 2 (c) b − 8ac = B − 8AC a2 A2 (d) None of these

2

| x – 2 | + | x – 2 | – 2 = 0 is (b) 4 (d) None of these

241. Let p and q be 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 A.P., then values of A and. B are (a) – 3, 77 (c) 3, 77

(b) 3, – 77 (d) None of these

242. The equation formed by decreasing each root of ax2 + bx + c = 0 by 1 is 2x2 + 8x + 2 = 0, then (a) a = – b (b) b = – c (c) c = – a (d) b = a + c (b) 1, 2 (d) 1, ­– 2

244. If the roots of x2 + x + a = 0 exceed a, then (a) 2 < a < 3 (c) – 3 < a < 3

(b) a > 3 (d) a < – 2

245. If the ratio of the roots of x2 + bx + c = 0 and x2 + qx + r = 0 be the same, then (a) r2c = b2q

(b) r2b = c2q

(c) rb = cq

(d) rc2 = bq2

2

2

246. If log2x + logx2 = 10 = log2y + logy2 and x ≠ y, then 3 x+y = (a) 2 (c) 37 6

(b)  – n2 (d)  – n4

249. If x2 + 2ax + 10 – 3a > 0 for all x ∈ R, then (a)  – 5 < a < 2 (b)  a < – 5 (c)  a > 5 (d)  2 < a < 5

(a)  0 (c)  4

(b) 65 8 (d) None of these

(b)  2 (d)  6

251. If α, β are the roots of the equation ax2 + bx + c = 0, α β + is equal to then αβ + b aα + b (a)   2 (b)   2 a b 2 (c)   (d)   − 2 c a 252. If the roots of the equations px2 + 2qx + r = 0 and qx 2 − 2 pr x + q = 0 be real, then (a)  p = q

(b)  q2 = pr

(c)  p = qr

(d)  r2 = pr

2

243. {x ∈ R : | x – 2 | = x2} = (a) – 1, 2 (c) – 1, – 2

(a)  n2 (c)  n4

250. The value of ‘c’ for which |α2 – β2| = 7 , where α and 4 β are the roots of 2x2 + 7x + c = 0, is

240. The sum of all real roots of the equation (a) 2 (c) 1

(a)  – 3 (b)  – 5 1 (d)   − 3 (c)   − 5 5 248. If α, β are the roots of the equation x2 – (1 + n2)x + 1 (1 + n2 + n4) = 0, then α2 + β2 is 2

253. The quadratic equations x2 – 6x + a = 0 and x2 – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then, the common root is (a)  1 (c)  3

(b)  4 (d)  2

254. If α and β are the roots of the equation ax2 + bx + c = 0, αβ = 3 and a, b, c are in AP, then α + β is equal to (a)  – 4 (c)  4

(b)  1 (d)  – 2

255. Let α, β be the roots of the equation x2 – px + r = 0 and α , 2β be the roots of the equation x2 – qx + r = 2 0. Then the value of r is (a)   2 (p – q) (2q – p) 9 (c)   2 (q – 2p)(2q – p) 9

(b)   2 (q – p)(2p – q) 9 (d)   2 (2p – q)(2q – p) 9

637

Quadratic Equations and Inequations

237. If the roots of the equation x2 – 2ax + a2 + a – 3 = 0 247. If the roots of 4x2 + 5k = (5k + 1) x differ by unity, are real less than 3, then then the negative value of k is

638

SOLUTIONS

Objective Mathematics

1. (a) We have, D = (c + a – b)2 – 4 (b + c – a) (a + b – c) = (a + b + c – 2b)2 – 4 (a + b + c – 2a)  (a + b + c – 2c)



= (– 2b)2 – 4 (­­– 2a) (– 2c) = 4 (b2 – 4ac)



= 4 [(– a – c)2 – 4ac] = 4 (a – c)2



= {2 (a – c)}2 = perfect square.

(1 + m) x2 – 2 (1 + 3m) x + (1 + 8m) = 0

...(1)

 et D be the discriminant of equation (1). Roots of L equation (1) will be equal if D = 0. 0 = 0 ⇒ m (m – 3) = 0 ⇒ m2 – 3m = 4 ∴ m = 0, 3. 3. (b) Let the other root be β, 2 1 =– then, α + β = – 4 2 1 ⇒β=– – α ...(1) 2 and 4α2 + 2α – 1 = 0. Now 4α3 – 3α = α (4α2 – 3) = α (1 – 2α – 3) [ ∵ 4α2 + 2α – 1 = 0] 1 = – 2α2 – 2α = –  (4α2) – 2α 2 1  (1 – 2α) – 2α [ ∵ 4α2 = 1 – 2α] =– 2 1 – α= β {from (1)} =– 2



4. (c) The discriminant = c (9a + b + 10a b + 24a b – 8ab ) 4

=y 81 ∴ y + = 30 ⇒ y2 – 30y + 81 = 0 y

2x

We get,

81sin

⇒ sin2 x =

3 ⇒ sin x = 4

= 27  or  34 sin

2x

= 33

π 3 = sin 3 2

⇒ x = π/3 Now if,

sin 2 x

81

⇒ sin x =

y – 3 = 0  or  y = 3 4 sin = 3  or  3

2x

= 31 ⇒ 4 sin2 x = 1

π π 1 = sin ⇒x= . 6 6 2

7. (b) Since the roots are of opposite sign, product of roots is negative ∴ p (p – 1) / 3 < 0 or 0 < p < 1 Thus, the required set of values is (0, 1). 8. (b) S ince the roots of the given equation are real, therefore the discriminant ≥ 0 ⇒ 4 (bc + ad)2 – 4 (a2 + b2) (c2 + d2) ≥ 0 ⇒ b2c2 + a2d2 + 2abcd – a2c2 – a2d2 – b2c2 – b2d2 ≥ 0 ⇒ (ac – bd)2 ≤ 0.  ut (ac – bd)2 cannot be negative as it is a square B of real number Hence a2, bd, c2 are in G.P.

= c2 (3a2 + b2)2 – 4 · abc2 (– 6a2 – ab + 2b2) 4

2x

∴ ac – bd = 0; or b2d2 = a2c2.

Hence 4α3 – 3α is the other root.

2 

= 30

2x

if, y – 27 = 0  or  y = 27

2. (a), (d)  We have,



sin Let 81

81 81sin

⇒ (y – 27)(y – 3) = 0

∴ Roots are rational.



⇒ 81sin 2 x +

2 2

3

3

9. (c) Since the roots are equal, ∴ B2 – 4AC = 0

⇒ 4 (a2 – bc)2 – 4 (c2 – ab) (b2 – ac) = 0

⇒ b2 (c – a)2 – 4ac (b – c) (a – b) = 0 ⇒ b2 (c2 + a2 – 2ac) – 4ac [ab – ac – b2 + bc] = 0 ⇒ b2 (c2 + a2 – 2ac + 4ac) + 4a2c2 – 4abc  (c + a) = 0 ⇒ [b (c + a)]2 + (2ac)2 – 2 · 2ac · b (c + a) = 0 ⇒ [b (c + a) – 2ac]2 = 0 ⇒ b (c + a) = 2ac 2ac ⇒ b = a + c

⇒ [a4 –­ 2a2bc + b2c2] – [b2c2 – ab3 – ac3 + a2bc] = 0

∴ b is H.M. of a and c i.e., a, b, c are in H.P.

= [c (3a2 –­ b2 + 4ab)]2, which is a perfect square. Hence the roots of the given equation are rational. 5. (a), (d)  The given equation is (c2 – ab) x2 – 2 (a2 – bc) x + (b2 – ac) = 0 Since the roots are equal, B2 – 4AC = 0

⇒ a (a3 + b3 + c3 – 3abc) = 0 ∴ Either a = 0 or a3 + b3 + c3 – 3abc = 0. sin 2 x

6. (a) 81

cos 2 x

+ 81

⇒ 81sin

2x

= 30

+ 811 − sin

2x

= 30

10. (b) Let α be a root. Then the other root is nα. −b ,  a c  and   α · nα = a

∴ α + nα =

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

⇒ (x2 – x – 20)(x2 – x + 19) = 0

2

 −b  c ca 2 (1 + n) 2 n ·    = ; or nb = a a  a (1 + n) 

or

⇒ (x – 5)(x + 4)(x2 – x + 19) = 0

2

Hence, the required roots of the equation are

nb2 = ca (1 + n)2.



11. (b) Given equation is ax2 + bx + c = 0

...(1)

Let the root of equation (1) be α and rα, then −b α + rα =  ...(2) a c  ...(3) and rα2 = a b From (2), α = – a (r + 1)

...(4)

Putting the value of α in (3), we get

rb 2 c b 2 (r + 1) 2 = 2 2 = or, a (r + 1) a ac r

12. (b) The given equation is x – 8x + a – 6a = 0 2



x4 – 2x3 + x – 380 = 0



Putting this value in (2), we get

639

16. (a) Given that

b . a (1 + n)

2

5, –4,

1 ± 1 − 76 1 ± 5 −3 ,  i.e.,  5, − 4, . 2 2

17. (c) Since α and β are roots of the equation ax2 + bx + c = 0 ∴ α =

− b + b 2 − 4ac 2a

and β =

− b − b 2 − 4ac 2a [ ∵ roots are irrational]



∴ b – 4ac > 0 and not a perfect square 2

∴ α and β are conjugate roots. 18. (c) Let the roots of the given equation be α and β. Then α + β = p and αβ = q.

Since the roots are real, ∴ B2 –­ 4AC ≥ 0

We have α – β = 1; or (α – β)2 = (1)2;

⇒ 64 – 4 (a2 – 6a) ≥ 0

⇒ (α + β)2 – 4αβ = 1 ⇒ p2 – 4q = 1.

⇒ – 4 (a + 2) (a – 8) ≥ 0

Now, p2 = 1 + 4q ⇒ p2 + 4q2 = 1 + 4q + (2q)2

⇒ [a – (– 2)] (a – 8) ≤ 0 ⇒ – 2 ≤ a ≤ 8.



13. (a) The given equation is 9x2 + 4ax + 4 = 0 Since the roots are imaginary, ∴ B –­ 4AC < 0 2

⇒ p2 + 4q2 = (1 + 2q)2.

19. (a) Let roots are α, α2 ∴ α + α2 = – p

...(1)

α =q 

...(2)

⇒ 16a2 – 36 × 4 < 0 ⇒ 16 (a2 – 9) < 0



⇒ 16 (a + 3) (a – 3) < 0 ⇒ – 3 < a < 3.

⇒ α3(1 + α)3 = – p3 ⇒ α3[1 + α3 + 3(α + α2)] = – p3

14. (b), (c)  Let the roots be 2α and 3α. Then, 2α + 3α = 2 – m; ⇒ α =

1 (2 – m); 5

Also, 2α · 3α = m + 2; 1 ⇒ 6 ·   (2 – m)2 = m + 2 25 ⇒ 6m2 – 49m – 26 = 0; ⇒ 6m2 – 52m + 3m – 26 = 0; ⇒ (3m – 26) (2m + 1) = 0 26 1 ∴ m = – or m = . 3 2 15. (b) Let the roots be α and β. Then α + β =

Now,

=

p + q

α p −n n = . ; αβ = ; and β q l l q + p

3

n l

n n − + α β α + β + αβ l l = 0. + + αβ = = β α αβ n l

⇒ q[1 + q – 3p] = – p3 ⇒ p3 = 3pq + q + q2 = 0 ⇒ p3 – q(3p – 1) + q2 = 0. 20. (a) The given equation can be written as (ab – 1) x2 + (a + b) x – ab = 0



Here, D = (a + b)2 + 4ab (ab – 1) = (a + b)2 – 4ab + 4 (ab)2 = (a – b)2 +  (2ab)2 ≥ 0 for all a, b.



∴ Roots are always real. 21. (d) Put

x − 1 = t ⇒ x – 1 = t2 or x = t2 + 1, The

given equation reduces to t 2 + 1 + 3 − 4t + t 2 + 1 + 8 − 6t = 1 where t ≥ 0. ⇒ | t – 2 | + | t – 3 | = 1, where t ≥ 0. This equation will be satisfied if 2 ≤ t ≤ 3. Therefore, 2 ≤

x − 1 ≤ 3 or 5 ≤ x ≤ 10.

∴ The given equation is satisfied for all values of x lying in [5, 10].

Quadratic Equations and Inequations

From (1), α = –

640

Objective Mathematics

Since l is real, ∴ discriminant ≥ 0; 9 . ⇒ 81 – 72m ≥ 0; or m ≤ 8 9 . Hence the greatest value of m is 8

22. (a) Since α and β are the roots of the equation x2 + px + q = 0 ∴ α + β = – p and αβ = q α 2 β2 α 3 + β3 (α + β)3 − 3αβ (α + β) + = = β α αβ αβ

Now

=

− p 3 − 3q (− p ) p (3q − p 2 ) = . q q

28. (b) The given equation can be written as

Putting (5/3)x = t, the equation becomes

23. (b) Since α and β are the roots of the equation x2 + px + q = 0



⇒ (t – 1) (t2 + t + 2) = 0 ⇒ t = 1 or t2 + t + 2 = 0

Now, (ωα + ω2β) (ω2α + ωβ)

= ω3α2 + ω4αβ + ω2αβ + ω3β2



= α2 + ωαβ + ω2αβ + β2



= α + β + (ω + ω )αβ



= α + β – αβ



= (α + β) – 2αβ – αβ = p – 3q.

2

2

 ut t2 + t + 2 = 0 does not have real solutions. B Therefore, t = 1 ⇒ (5/3)x = 1 ⇒ x = 0. [ ∵ ω3 = 1] 29. (a) Since α and β are the roots of the equation

2

2



[∵ 1 + ω + ω = 0 ]

2

t3 + t – 2 = 0 ⇒ t3 – 1 + (t – 1) = 0

⇒ (t – 1) (t2 + t + 1) + (t – 1) = 0

∴ α + β = – p and αβ = q.

2

(5/3)3x + (5/3)x = 2

2

A (x2 + m2) + Amx + cm2x2 = 0

or, (A + cm2) x2 + Amx + Am2 = 0 ...(1) Am Am 2 ∴ α + β = – A + cm 2 and αβ = A + cm 2

2

24. (b), (d)  Let α be the common root. Then, α2 – α – 12 = 0 and kα2 + 10α + 3 = 0

Now, A (α2 + β2) + Aαβ + cα2β2

Solving the two equations;

= A [(α + β)2 – 2αβ] + Aαβ + cα2β2

We get, α 1 α2 = − 12k − 3 = 10 + k 117



⇒ (– 12k – 3)2 = 117 (10 + k)





⇒ 9 (4k + 1)2 = 117 (10 + k) ⇒ (4k + 1)2 = 13 (10 + k) ⇒ 16k2 + 8k + 1 = 130 + 13k ⇒ 16k2 – 5k – 129 = 0 ⇒ 16k2 – 48k + 43k – 129 = 0 − 43 . ∴ k = 3 or k = 16 25. (b) Since coefficient of x = 16, ∴ sum of roots = – 16

 A 2m2 2Am 2  A 2m2 cA 2 m 4 = A − + + 2 2 2 2 A + cm  A + cm (A + cm 2 ) 2  (A + cm ) =

A 3 m 2 − 2A 2 m 2 (A + cm 2 ) + A 2 m 2 (A + cm 2 ) + cA 2 m 4 (A + cm 2 ) 2 0 = (A + cm 2 ) 2 = 0

30. (d) Let the roots of the given equation be α and Then α · 

1 = p + 3; α

26. (b) Since b – c + c – a + a – b = 0, ∴ one root = 1. a−b . Also, product of the roots = b−c a−b ∴ other root = . b−c 27. (a) Let the roots of the given equation be α and 2α Then α + 2α = Also, α · 2α =

l l 1 ; or α =  ·  m−l 3 m−l

1 2 l2 1 ; or  ·  = l−m 9 (l − m) 2 l−m

But l ≠ m. ∴ 2l2 – 9l + 9m = 0

⇒ p = – 2.

31. (a) Let α and β be the roots of the given equation; then α + β = −

Since constant term = (– 15) (– 4) = 60, ∴ correct answer is – 6, – 10.

1 α

Given α + β =

b c and αβ = . a a

1 1 α 2 + β2 (α + β) 2 − 2αβ = , + 2 = 2 2 α β (αβ) (αβ) 2

b 2 2c − 2 b b 2 − 2ca = a 2 a = ⇒ − a c2 c 2 a ⇒ 2ca2 = bc2 + ab2  Hence, bc2, ca2 and ab2 are in A.P. 32. (c) For the equation x2 + 2x + 3 = 0, Discriminant = (2)2 – 4 ⋅ 1 ⋅ 3 < 0. ∴ roots of x2 + 2x + 3 = 0 are imaginary. Since the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0 are given to have a common root, therefore

⇒ r (x + q + x + p) = (x + p) (x + q)

Now (sin θ + cos θ)2 = 1 + 2 sin θ cos θ 2c b2 a + 2c =1 + = a a2 a

∴

or x2 + x (p + q – 2r) + pq – r (p + q) = 0.

⇒ b2 = a (a + 2c) = a2 + 2ac

Let its roots be α and – α

⇒ b2 + c2 = a2 + 2ac + c2 = (a + c)2

Then α + (– α) = – (p + q – 2r)

Hence (a + c)2 = b2 + c2.

or

p + q = 2r.

38. (a), (b) Let α, β be the roots of equation ax2 + bx + c=0 b c and αβ = . ∴ α + β = − a a b c c a ∵ a, b, c are in G.P., ∴ a = b = a ⋅ b 2 c ⇒  b  = or (α + β)2 = αβ a a

Also, product = α (– α) = pq – r (p + q)

= pq –

1 ( p + q)2 = − ( p2 + q2 ) . 2 2

34. (b) Since 2 and 3 are roots of f (x) = 0 ∴ f (2) = 0 and f (3) = 0 ⇒ 4m + n – 10 = 0 and 9m + n + 15 = 0.

2

Solving the two equations, we get m = – 5, n = 30.

 35. (d) Let f (x) = x + px + q = 0 3

2

...(i)

α α ⇒ α2 + β2 + αβ = 0; or   + + 1 = 0; β β α −1± 1− 4 − 1 ± 3i = . = β 2 2

∴

Since f (x) = 0 has a root of multiplicity 2 ∴ f (x) = 0 and f ′ (x) = 3x2 + 2px = 0 have a common root.  he roots of 3x2 + 2px = 0 are x = 0 T and x = – 2p/3.

∴ α + β = p and αβ = q. Now, (α1/4 + β1/4)4 = [(α1/4 + β1/4)2]2

But x = 0 is not a root of f (x) = 0 ( ∵ q ≠ 0)



∴ common root of f (x) = 0

and

39. (d) Since α, β are the roots of the equation x2 – px + q = 0

2

f ′ (x) = 0 is x = – 2p/3.



∴ (– 2p/3)3 + p (– 2p/3)2 + q = 0 ⇒ 4p + 27q = 0. 36. (b) α and β are roots of the equation ax2 + bx + c = 0 b c and αβ = a a

=  p + 2 q + 2 (q )1/ 4   



=p+6

∴ α + β = – 



 α 2 β2  α β  a Now  β + α  + b  β + α 



=

a (α 3 + β 3 ) b (α 2 + β 2 ) + αβ αβ



=

a[(α + β) − 3αβ (α + β)] b[(α + β) − 2αβ] + αβ αβ



 b 3 c a  −  − 3 ⋅   a a  =  c a



=

q + 4 q1/ 4

2

p+2 q

∴ α1/4 + β1/4 =  p + 6 q + 4q1/ 4 



−

1/ 4   =  α + β + 2 αβ + 2(αβ)   



3

3

= [α1/2 + β1/2 + 2 (αβ)1/4]2

40. (d) the function y = x (x – 3) ≥ 0 i.e.,

p + 2 q  

1/ 4

x ( x − 3) is defined for

x ≤ 0 or x ≥ 3

...(1)

The given equation can be written as 2

 b 2  b  − b  −  − 2   a    a  +  c a

b3 3bc b3 2bc bc + + 2 − a2 a a a = a = b. c c a a

c  a 



9 | x |2 – 18 | x | + 2 = 0

⇒ (9 | x | – 1) ( | x | – 2) = 0 ⇒ | x | = 2 or | x | = 1/9. ∴ Solutions of the given equation are ± 2, ± 1/9. In the domain (1), the required solutions are – 2, – 1/9. 41. (c) We have, 2x + 2 ⋅ 33x /(x – 1) = 9 = 32 3x ⇒ (x + 2) log 2 + log 3 = 2 log 3 x −1   ⇒ (x + 2) log 2 + 3 x − 2 log 3 = 0  x − 1 

641

37. (d) Since sin θ and cos θ are the roots of the equation ax2 + bx + c = 0 b c and sin θ cos θ = ∴ sin θ + cos θ = − a a

Quadratic Equations and Inequations

both roots will be common. Hence both the equations are identical. a b c = = i.e., a : b : c = 1 : 2 : 3. ∴ 1 2 3 1 1 1 33. (b), (c)  We have, x + p + x + q = r

642

Now α/β is a root of xn + 1 + (x + 1)n = 0

  1 ⇒ (x + 2)  log 2 + log 3  = 0 x −1  

Objective Mathematics

⇒ x = – 2 or x = 1 –

n

n ⇒ α + 1 +  α + 1 = 0 n  β  β

log 3 . log 2

⇒

42. (a) In case of the first student,

α n + βn (α + β) n + =0 n β βn

⇒ (αn + βn) + (α + β)n = 0

p roduct of the roots = 3 × 10 = q. So the correct value of q is 30.

⇒ – pn + (– p)n = 0 [from (1) and (2)]

In case of the second student,

This is true only if n is an even integer.

s um of the roots = 4 + 7 = – p. So the correct value of p is – 11. ∴ The correct equation is x4 – 11x + 30 = 0 or (x – 5) (x – 6) = 0; ∴ x = 5, 6.

⇒ x2 – 4x + 5 = x – 1

(∵ a loga x = x)

⇒ x2 – 5x + 6 = 0 or (x – 3) (x – 2) = 0,

x2 – x – 6 = 0 ⇒ x = – 2, 3 ∵ x < 0 ∴ x = – 2 is the solution

∴ | x2 + 4 | x | + 3 | = x2 + 4 | x | + 3. So, the given equation can be written as ...(1)

For, x ≥ 0, | x | = x, ∴ equation is

If x ≥ 0, then (1) can be written as

=x–1

47. (b) For, x < 0, | x | = – x ∴ equation is

43. (b) Since x2 + 4 | x | + 3 > 0 for all x ∈ R.

x2 + 4 | x | + 3 + 2x – 11 = 0

2 − 4 x + 5)

∴ x = 2, 3.

∴ Roots of the correct equation are 5, 6.



46. (a) We have, 7 log7 ( x

x2 + x – 6 = 0 ⇒ x = 2, – 3

∵ x ≥ 0 ∴ x = 2 is the solution.

x2 + 6x – 8 = 0

 ence x = 2, – 2 are the solutions and their sum H is zero.

− 6 ± 36 + 32 = – 3 ± 17 . 2 As x ≥ 0, we get x = – 3 + 17 . If x < 0, then (1) can be written as

48. (c) Given : (– 2) (– 15) = q or q = 30.



49. (c) Let α and β be the roots of the correct equation

⇒ x =

∴ The correct equation is x2 + 13x + 30 = 0.

x2 – 4x + 3 + 2x – 11 = 0

or x – 2x – 8 = 0 2

⇒ (x – 4) (x + 2) = 0

⇒ x = – 2, 4. As x < 0, we get x = – 2. Thus, the given equation has two real roots. 44. (c) Let α and β be the roots of the equation

x2 + px + q = 0

and

γ, δ be the roots of the equation



x2 + qx + p = 0 

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

then α + β = – p, αβ = q, and

γ + δ = – q, γδ = p

ax2 + bx + c = 0. −b  Then α + β = a c  and αβ = a

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

When b is written incorrect, then let the roots be γ and γ. c = αβ, by virture of (2) ∴ γ · γ = a ...(3) or γ2 = αβ 

⇒ (α + β )2 – 4αβ =  (γ + δ)2 – 4γδ

 hen c is written incorrect, then let the roots be γ W and 2γ. −b = α + β [Using (1)] ∴ γ + 2γ = a

⇒ p2 – 4q = q2 – 4p ⇒ p2 – q2 + 4 (p – q) = 0

⇒ 9γ2 = (α + β)2

⇒ (p – q) (p + q + 4) = 0

⇒ 9αβ = (α – β)2 + 4αβ,

Given : α – γ = β – δ ⇒ α – β = γ – δ ⇒ (α – β)2 = (γ – δ)2

 ut p  ≠ q otherwise the two given equations will be B the same ∴ p – q ≠ 0. Hence p + q + 4 = 0. ∴ α + β = – p, αβ = q

...(1)

Also α, β are the roots of x2n + pnxn + qn = 0 ∴ αn + βn = – pn and αn βn = qn

⇒ (α – β) = 5αβ. 50. (a) Since α and β are roots of the equation

45. (a) Since α, β are the roots of x2 + px + q = 0

...(2)

[Using (3)]

2

x2 – p (x + 1) – q = 0

∴ α + β = p and αβ = – (p + q) ∴ (α + 1) (β + 1) = α + β + αβ + 1

=p– p– q+1=1– q

...(1)



∴ 2 f (a) < 0 ⇒ f (a) < 0.

2 (α + 1) 2 (β + 1) 2 + (q − 1)[(α + 1) 2 + (β + 1) 2 ] = (α + 1) 2 (β + 1) 2 + (q − 1) 2 + (q − 1) [(α + 1) 2 + (β + 1) 2 ] 2 (1 − q ) + (q − 1)[(α + 1) + (β + 1) ] . 2 (1 − q ) 2 + (q − 1)[(α + 1) 2 + (β + 1) 2 ] 2

=

2

2



∴ f (a) = 2a2 – 2 (2a + 1) a + a (a – 1) < 0 ⇒ – a2 – 3a < 0 ⇒ a (a + 3) > 0 ⇒ a ∈ (– ∞, – 3) ∪ (0, ∞). 56. (d) Since α is a root of a2x2 + bx + c = 0

[Using (1)] = 1.

51. (c) Put | x – 2 | = t.

∴ a2α2 + bα + c = 0 Also, β is a root of a x – bx – c = 0 ∴ a2β2 – bβ – c = 0 Let f (x) = a x + 2bx + 2c.

t + t – 2 = 0 or (t + 2) (t – 1) = 0

Since t + 2 = | x – 2 | + 2 > 0

f (x) is continuous on [α, β]

∴ we get t – 1 = 0 ⇒ | x – 2 | = 1

Also, f (α) = a2α2 + 2bα + 2c

⇒ x – 2 = ± 1



⇒ x = 3, 1. Thus, the sum of roots is 4.

and

52. (a) Since α, β are the roots of the equation 2

Now, aSn + 1 + bSn + cSn – 1 + c (αn – 1 + βn – 1) (∵ Sn = αn + βn)



= α n – 1 (aα2 + bα + c) + β n – 1 (aβ2 + bβ + c)



= αn – 1 ⋅ 0 + β n – 1 ⋅ 0 = 0.

∵ α, β, γ, δ are in A.P. ∴ β – α = δ – γ ⇒ (β – α)2 = (δ – γ )2 ⇒ (β + α)2 – 4βα = (δ + γ )2 – 4δγ ⇒ 4p2 – 4q = 4r2 – 4s; or s – q = r2 – p2. 54. (a), (b)  If x – a < 0, | x – a | = – (x – a) ∴ equation becomes x2 + 2a (x – a)

– 3a2 = 0

⇒ x2 + 2ax – 5a2 = 0 ⇒ x = – (1 + 6 )  a, (− 1 + 6 ) a ∵ x < a ≤ 0 ∴ x = (− 1 + 6 ) a If x – a ≥ 0, | x – a | = x – a ∴ the equation becomes x2 – 2a (x – a) – 3a2 = 0 ⇒ x2 – 2ax – a2 = 0 ⇒ x = (1 + 2 )  a, (1 − 2 )  a

∴ x = (1 − 2 )  a.

[Using (2)]

57. (a) We have, α + β = p; αβ = q, γ + δ = r and γ δ = s. δ β = γ α

α+β γ+δ ⇒ α − β = γ − δ , 

αβ = q, γ + δ = 2r and γ δ = s.

∵ x ≥ a and a ≤ 0

= 3a2β2 > 0

∴ f (x) = 0 has a root γ such that α < γ < β.

53. (b) We have, α + β = 2p;

f (β) = a β + 2bβ + 2c

∵ α, β, γ, δ are in G.P. ; ∴

= a (αn + 1 + βn + 1) + b (αn + βn)



[Using (1)]

∴ f (α) ⋅ f (β) < 0

∴ aα + bα + c = 0 and aβ + bβ + c = 0. b c and αβ = . Also, α + β = − a a

= – a2α2 < 0 2 2



ax2 + bx + c = 0 2

...(2)

2 2

2



...(1)

2 2

The given equation becomes

Here f (x) = 2x2 – 2 (2a + 1) x + a (a – 1)

(by componendo and dividendo)

⇒

(α + β ) 2 ( γ + δ) 2 = ; 2 (α + β) − 4αβ ( γ + δ) 2 − 4 γδ

⇒

r p2 = 2 r − 4s p − 4q 2

2

⇒ p2r2 – 4sp2 = p2r2 – 4qr2 ⇒ – 4sp2 = – 4qr2; or p2s = r2q. 58. (c) We have, x3 + 3x2 + 3x + 2 = 0 ⇒ (x + 1)3 + 1 = 0 ⇒ (x + 1 + 1) {(x + 1)2 – (x + 1) + 1} = 0 ⇒ (x + 2) (x2 + x + 1) = 0 ⇒ x = – 2,

− 1 ± 3i ⇒ x = – 2, ω, ω2. 2

 ince a, b, c ∈ R, ax2 + bx + c = 0 cannot have one real S and one imaginary root. Therefore, two common roots of ax2 + bx + c = 0 and x3 + 3x2 + 3x + 2 = 0 are ω, ω2. b = ω + ω2 = – 1 Thus, − a c = ω ⋅ ω2 = 1 ⇒ c = a ⇒ a = b and a ⇒ a = b = c.

643

(α + 1) 2 (β + 1) 2 + 2 (α + 1) + (q − 1) (β + 1) 2 + (q − 1)

Quadratic Equations and Inequations

55. (c) Since ‘a′ lies between the roots of the given equation

The given expression can be written as

644

59. (b) x < –2 ⇒ x2 + x + 2 + x > 0 or x2 + 2x + 2 > 0

Objective Mathematics

or ∴ x≥ or

(x + 1) + 1 > 0 which is true for all x. x < –2 –2 ⇒ x2 – x – 2 + x > 0 or x2 – 2 > 0 x < – 2 or x > 2 2

⇒ s in (ex ) ≥ 2. which is not possible for any real value of x. Hence the given equation has no real solution.

p (a + b) x + c (b − a ) = 2x x2 − c2

∴ x ∈ [−2, − 2 ) ∪ ( 2 , + ∞)



∴ Solution is x ∈ (– ∞, –2) ∪ [−2, − 2 ) ∪ ( 2 , + ∞)

⇒ p (x2 – c2) = 2 (a + b) x2 – 2c (a – b) x

i.e., (−∞, − 2 ) ∪ ( 2 , + ∞) .

⇒ x – [x] + ⇒ x+

⇒

⇒ (2a + 2b – p) x2 – 2c (a – b) x + pc2 = 0  his equation will have equal roots, if Discriminant T =0

1 60. (c) We have, f (x) + f   = 1 x

⇒ 4c2 (a – b)2 – 4pc2 (2a + 2b – p) = 0 ⇒ (a – b)2 – 2p (a + b) + p2 = 0

1 1 − =1 x  x 

⇒ p = (a + b) ± (a + b) − (a − b) 2

1 1 − 1 = [x] +   x  x



2 = a + b ± 2 ab = ( a ± b ) .

2 Also, ax1 + bx1 + c = 0 or x1 (ax1 + b) = – c

Since x is real, so (k + 1)2 – 4 ≥ 0

and

⇒ k2 + 2k – 3 ≥ 0 ⇒ (k + 3) (k – 1) ≥ 0

So, the given expression

⇒ k ≤ – 3 or k ≥ 1. Therefore, number of solutions is infinite. 61. (a) Let α = =

1 2− 5 = 2+ 5 (2 + 5 ) (2 − 5 )

2− 5 2− 5 = =–2+ 4−5 −1

5.

 ince irrational roots of a quadratic equation with S rational coefficients always occur in conjugate pair, therefore, other root β of the quadratic equation will be – 2 – 5 . Now, α + β = – 2 + 5 – 2 – 5 = – 4 and

αβ = (– 2 +

5 ) (– 2 –

[4c2 ≠ 0] 2

c 65. (c) Here x1 + x2 = − b and x1x2 = . a a

x2 + 1 − x ⇒ = (integer) k (say) x ⇒ x2 – (k + 1) x + 1 = 0



p a b = + 2x x+c x−c

64. (a), (b)  The given equation is

5)

ax22 + bx2 + c = 0; or x2 (ax2 + b) = – c.



=

1 1 x12 x22 + + 2 2 = (ax1 + b) (ax2 + b) c2 c2



=

1 {(x1 + x2)2 – 2x1x2} c2



=

1 c2

 b 2 2c  b 2 − 2ac .  2 −  = a c2a 2 a

66. (c) Since α < – 1 and β > 1 ∴ α + λ = – 1 and β = 1 + µ where λ, µ > 0. Now, 1 +

c b + a a

= 1 + αβ + | α + β |



= 1 + (– 1 – λ) (1 + µ) + | – 1 – λ + 1 + µ |



= (– 2)2 – ( 5 )2 = 4 – 5 = – 1 ∴ required equation is x2 – (α + β) x + αβ = 0



= 1 – 1 – µ – λ – λµ + | µ – λ |

or,



− µ − λ − λµ + µ − λ, if µ > λ =  − µ − λ − λµ + λ − µ, if λ > µ

x2 – (– 4) x – 1 = 0 or x2 + 4x – 1 = 0.

62. (a) We have, (log5 x)2 + log5 x < 2 Put log5 x = a then a2 + a < 2

∴ 1 +

⇒ a2 + a – 2 < 0 ⇒ (a + 2) (a – 1) < 0 ⇒ – 2 < a < 1 or – 2 < log5 x < 1 1 < x < 5. ∴ 5 – 2 < x < 5 i.e., 25 63. (a) Put 5x = y. Then the given equation becomes 2

1  1   + 2 [ 5x > 0] sin (ex) = y + =  y − y  y 

c b + a a

In both cases 1 +

= – 2λ – λµ or – 2µ – λµ c b + a a

< 0

( ∵ λ, µ > 0)

67. (a) For x > 0, f (x) > 0 so there is no positive real root of f (x) = 0. Now f (0) = 1 ≠ 0 so x = 0 is not a root.

f (x) = 1 + 2x + 3x2 + ...+ (n + 1) xn,

x f (x)

= x + 2x2 + ... + nxn + (n + 1) xn + 1

⇒ f (x)

=

=

1 1 [q (α + β)2 – p2αβ] = 2 [qp2 – p2q] = 0. β2 β α is a root of the equation β

Thus,

1 − x n + 1 (n + 1) x n + 1 − (1 − x) 2 (1 − x)

qx2 + (2q – p2) x + q = 0.



71. (b) Since α, β are roots of the equation x2 + px + q = 0 and γ, δ are roots of the equation x2 + rx + s = 0

1 − (n + 2) x n + 1 + (n + 1) x n + 2 (1 − x) 2

∴ α + β = – p and αβ = q, γ + δ = – r and γ s = s.

∴ The equation f (x) = 0 is equivalent to

P (x) = (n + 1) xn + 2 – (n + 2) x n + 1 + 1 = 0.



P (– x) = (n + 1) xn + 2 + (n + 2) x n + 1 + 1 = 0,

 hich has no change of sign so has no negative w real root.

 ow, the given expression = (α – γ ) (α – δ) (β – γ) N (β – δ) = [α2 – α (γ + δ) + γ δ] [β2 – β (γ + δ) + γ δ] = (α2 + α r + s) (β2 + β r + s) = (αβ)2 + αβr (α + β ) + s {(α + β )2 – 2αβ} + rs (α + β ) + (αβ) r2 + s2

Hence f (x) = 0 has no real root.

= q + qr (– p) + s (p – 2q) + rs (– p) + qr2 + s2

68. (a) Since α, β are the roots of x2 + px + q = 0

2

2

= q2 + s2 – pr  (q + s) + s (p2 – 2q) + qr2.

∴ α + β = – p and αβ = q.

72. (c) We have, x2 – 4x – 77 < 0 and x2 – 4 > 0

Also, α4, β4 are roots of x2 – rx + s = 0 ∴ α4 + β4 = r and α4β4 = s.

⇒ (x + 7) (x – 11) < 0 and (x – 2) (x + 2) > 0

⇒ q4 = (αβ)4 = s.

⇒ – 7 < x < 11 and x < – 2 or x > 2.

Now Disc. for x2 – 4qx + 2q2 – r = 0 is

∴ – 7 < x < – 2.



16q2 – 4 (2q2 – r) = 8q2 + 4r.

Hence, the greatest negative integer is – 7.

Since, r = α + β = (α + β ) – 2α β 4

4

2

2 2

73. (a) The two equations can be written as

2 2



= [(α + β)2 – 2αβ]2 – 2(αβ)2



= [p2 – 2q]2 – 2q2



= p4 + 4q2 – 4p2q – 2q2 = p4 + 2q2 – 4p2q



= 16q2 – 16p2q + 4p4 = (4q – 2p2)2 > 0

∴ the given equation has two real roots. 69. (b) We have, (x – a) (x – b) – k = 0 or, x2 – (a + b) x + ab – k = 0

...(1)

Since c and d are the roots of the equation (1) ⇒ c + d = a + b,

...(2)

and

...(3)

cd = ab – k

and x  (12k + 4) + px + (6k – 2) = 0. Divide by 2, we get

x2 (6k + 2) +

p x + (3k – 1) = 0 2

74. (b) We have, x – 2 = 22/3 + 21/3. Cube both sides, we get

(x – 2)3 = 22 + 2 + 3 ⋅ 22/3 ⋅ 21/3 (x – 2)



= 6 + 6 (x – 2)



or

x3 – 6x2 + 12x – 8 = – 6 + 6x.

∴ x3 – 6x2 + 6x = 2.

x – (a + b) x + ab = 0 2

⇒ x2 – (c + d) x + cd + k = 0 [putting the values of a + b and ab]

⇒ x2 – cx – dx + cd + k = 0 or,

x (x – c) – d (x – c) + k = 0

or, (x – c) (x – d) + k = 0. 70. (c) Since α, β are roots of the equation x2 + px + q = 0 ∴ α + β = – p and αβ = q. 2

α α Now q   + (2q − p 2 ) + q β β

...(3)

Comparing (1) and (3), we get p ∴ 2r – p = 0. r = 2

 ow the equation whose roots are a and b will N be



...(1)

2

∴ Disc. = 8q2 + 4p4 + 8q2 – 16p2q

x2 (6k + 2) + rx + (3k – 1) = 0

75. (c) We have, ax3 + bx + c = (x2 + px + 1) (ax + λ), where λ is a constant. ∴ Equating the coefficients of x2, x and constant term, we get 0 = ap + λ, b = pλ + a, c = λ ∴ p = −

λ c = − a a

Putting for λ and p in the second relation  c Hence, b =  −  c + a or ab = a2 – c2. a

645

=

= (1 + x + x2 + ... + xn) – (n + 1) xn + 1 1 − xn + 1 − (n + 1) x n + 1 = 1− x

Quadratic Equations and Inequations

⇒ (1 – x) f (x)

646

76. (c) Since α, β are roots of the equation x2 + px + q = 0 and γ, δ are roots of the equation x2 + rx + s = 0

Objective Mathematics

∴ α + β = – p and αβ = q, γ + δ = – r and γ δ = s. The given expression = 2 (α2 + β2) + 2 (γ2 + δ2) – 2 (αγ + βγ + αδ + βδ) = 2 {(α + β) ­– 2αβ} + 2 (γ + δ) – 2γ δ} 2

2

– 2 (α + β) (γ + δ)



= 2 (p2 – 2q) + 2 (r2 – 2s) – 2 (– p) (– r) = 2 (p2 + r2 – pr – 2q – 2s). 77. (a) Equating the coefficients of similar powers of x, we get a2 − 1 = 0 ⇒ a = ± 1   a −1= 0 ⇒ a =1  2 a − 4a + 3 = 0 ⇒ a = 1, 3 ∴ common value of a = 1.



78. (c) The given equation can be written as

3x2 – 2x (a + b + c) + bc + ca + ab = 0

Discriminant = 4 (a + b + c)2 – 12 (bc + ca + ab)

= 4 (a2 + b2 + c2 – bc – ca – ab)



= 2 [(b – c)2 + (c – a)2 + (a – b)2] ≥ 0.

Hence the roots are real. 79. (b) As one root is 5 – 3i, the other is 5 + 3i. Hence the quadratic equation should be x2 – (5 – 3i + 5 + 3i) x + (5 – 3i) (5 + 3i) = 0 or

x2 – 10x + 25 – 9i2 = 0; x2 – 10x + 34 = 0.

 his equation should be identical with x2 + px + T q = 0. 1 − 10 34 = = ; ∴ p = – 10, q = 34. ∴ 1 p 8 80. (a) As one root is 2 + i 3 , the other is 2 – i 3 .

1 83. (b) We have,  x +   x 2

1 3 ⇒  x +  −  x 2

2

= 4+

3 1  x −  2 x

1   x −  − 4 = 0 x

2   1 1  3  1 ⇒   x −  + 4 x  −  x −  − 4 = 0 2 x x x  

⇒ a 2 − 3 ⋅ a = 0, where a = x − 1  x 2

⇒ 2a2 – 3a = 0 ⇒ a (2a – 3) = 0 ∴ a = 0, 3 . 2 1 2 =0 ⇒ x –1=0 Now a = 0 ⇒ x − x ⇒x=±1 and a = 3 ⇒ x − 1 = 3 x 2 2 2 ⇒ 2x – 3x – 2 = 0 ⇒ 2x2 – 4x + x – 2 = 0 ⇒ (x – 2) (2x + 1) = 0 ∴ x = 2, – 1 2 Hence x = 1, – 1, 2, – 1 . 2 2x + 1 2 84. (c) We have, 3 + 3 = 3x + 3 + 3x ⇒ 32x  ⋅  31 + 9 = 3x   ⋅ 33 + 3x ⇒ 3a2 + 9 = 27a + a, where a = 3x ⇒ 3a2 – 28a + 9 = 0 or, (3a – 1) (a – 9) = 0 ∴ a = 1, 9 3 Now a = 1 ⇒ 3x = 1 = 3– 1 ⇒ x = – 1 3 3 and a = 9 ⇒ 3x = 9 = 32 ⇒ x = 2 Hence x = – 1, 2. 85. (c) The given equation can be written as

∴ 2 + i 3  + 2 – i 3 = – p; or p = – 4.



Again (2 + i 3 ) (2 – i 3 ) = q; or q = 7.

Here, D = (2x + m)2 – 4 (2x – 3)

81. (d) Let α be a common root of the given equations. Then α2 – aα + b = 0 and α2 + bα – a = 0 ⇒ (a + b) α – (a + b) = 0 ⇒ (a + b) (α – 1) = 0 ⇒ a + b = 0 or α = 1 If α = 1, then 1 – a + b = 0 ⇒ a – b = 1. 82. (a) Since 4 is a root of x2 + px + 12 = 0, we have

y2 + y (2x + m) + (2x – 3) = 0

= 4x2 + 4xm – 8x + m2 + 12 = 4x2 + 4x (m – 2) + (m2 + 12) = [4x2 + 4x (m – 2) + (m – 2)2]  + (m2 + 12) – (m – 2)2 = [2x + (m – 2)]2 + [12 + 4m – 4] For rational factors, D should be a perfect square ∴ 8 + 4m = 0 or m = – 2. 86. (a) Given : one root

16 + 4p + 12 = 0 or p = – 7.

2+ 3 2+ 3 4+4 3+3 = = 7 + 4 3, × 4−3 2− 3 2+ 3

Also, since the roots of x2 + px + q = 0 are equal,

=

we have

∴ other root = 7 –­ 4 3 . Hence the required quadratic equation is

p2 – 4q = 0 or 49 – 4q = 0

∴ q = 49/4.

...(1)

or

x – 14x + 1 = 0. 2

87. (a) One root =

1 4− 3 = 4− 3 × 4+ 3 4− 3 13

4 + 3 4 − 3 4+ 3 4− 3 x2 −  x+ + ⋅ =0 13  13 13  13 or 13x2 – 8x + 1 = 0. This equation must be identical with ax2 + bx + 1 = 0; ∴ a = 13 and b = – 8. 88. (b) Let α, β be the roots of x2 + bx + c = 0. Then α + β = – b, αβ = c.  oots of the required equation are α3, β3. So the R equation is x2 – (α3 + β3) x + α3β3 = 0 ⇒ x2 – {(α + β)3 – 3αβ (α + β)} x + (αβ)3 = 0 ⇒ x – (– b + 3cb) x + c = 0 3

then x = t2 + 1, t ≥ 0. The given equation reduces to

4+ 3 ∴ other root = 13 ∴ The quadratic equation is

2

647

x + (7 + 4 3 ) (7 – 4 3 ) = 0



x −1 = t

91. (d) Let

3

⇒ | t – 2 | + | t – 3 | = 1 ∴ 2 ≤ t ≤ 3 ⇒ 4 ≤ t2 ≤ 9 ⇒ 4 ≤ x – 1 ≤ 9 ⇒ 5 ≤ x ≤ 10. ∴ Solution of the original equation is

500 Time taken by the bus to cover 500 kms = x hours and time taken by the car to cover 500 kms 500 hours. = x + 25 500 50 50 500 = +1 = + 10 ⇒ Given, x + 25 x x + 25 x 50 50 50 ( x + 25 − x) − ⇒ = 1⇒ =1 x x + 25 x ( x + 25) ⇒ 1250 = x2 + 25x ⇒ x2 + 25x – 1250 = 0 ⇒ x2 + 50x – 25x – 1250 = 0

x ∈ [5, 10].



92. (b) Since tan A and tan B are roots of the equation x2 – px + q = 0 ∴ tan A + tan B = p and tan A tan B = q tan A + tan B p Now, tan (A + B) = 1 − tan A tan B = 1 − q ∴ sin2 (A + B) = 1 2

 1 − tan 2 (A + B)  tan 2 (A + B) 1 −  = 2 1 + tan 2 (A + B)  1 + tan (A + B) 

=



p2 p2 (1 − q ) 2 = 2 . = 2 p p + (1 − q ) 2 1+ 2 (1 − q )

93. (b) Since α, β are roots of x2 + 2ax + b = 0

α + β = – 2a and αβ = b.

Let y = α + β + α 2 + β 2 ⇒ ( y + 2a)2 = α2 + β2 = (α + β)2 – 2αβ = 4a2 – 2b ⇒ y2 + 4ay + 2b = 0 So the required equation is x2 + 4ax + 2b = 0. 94. (d) The given equation is

x2/3 + x1/3 – 2 = 0

⇒ (x + 50) (x – 25) = 0.

Put x1/3 = y, then

∴ x = 25. ∵ x ≠ – 50, Hence the speed of the bus = 25 km/hour and speed of the car = (25 + 25) = 50 km/hour.



90. (a) Roots are of opposite sign if ( a) roots are real and distinct (b) product is negative. So D = 4 (a2 + 1)2 – 12 (a2 – 3a + 2) > 0 a 2 − 3a + 2 0. Hence 1 < a < 2.



 α 2 β2  α 2 β2 x2 −  + x+ ⋅ =0 α β α β

Quadratic Equations and Inequations

x2 – (7 + 4 3  + 7 – 4 3 ) 

648

Objective Mathematics

⇒ αβx2 – {(α + β)3 – 3αβ (α + β)}x + (αβ)2 = 0

⇒ x ≥ 2, x ≤ 4 and x ≤ 6

⇒ qx – (p – 3pq) x + q = 0.

∴ 2 ≤ x ≤ 4. Now, the given equation is

2

3

2

96. (d) Let α and β be the roots of the required equation. ⇒ (α + β)3 – 3αβ (α + β) = 7 ⇒ 27 – 9αβ = 7

x – 2 + 4 – x + 2 ( x − 2) (4 − x) = 6 – x

⇒ 2 ( x − 2) (6 − x) = (4 – x).

20 x – 3x + 9 = 0; or 9x2 – 27x + 20 = 0. 14  ...(1) 97. (b) Let the roots be α and 6α, then 7α = p 8 ...(2) and 6α2 = p Putting the value of α from equation (1) in (2), we get 2



4 8 = or p2 = 3p or p (p – 3) = 0 ⇒ p = 3 p2 p [∵ p ≠ 0].

Again squaring both sides, we get

4 (– x2 + 8x – 12) = 16 + x2 – 8x

⇒ 5x2 – 40x + 64 = 0 ⇒ x = 4 + But 2 ≤ x ≤ 4.

ax – 2bx + c = 0 2

2b c and αβ = . a a

∴ α + β =

3

i.e.,

x2 – 5x + 3 = 0 [∵ α2 = 5α – 3 and β2 = 5β – 3]



∴ α + β = 5 and αβ = 3. Now,

99. (a) We have, 3

 2b  c 3 + 2bc 2 c 2 (c + 2b) = .   = 3 a a a3

x – 1

+3

1 – x

α β α 2 + β2 5 (α + β) − 6 = = + β α αβ αβ =

5 (5) − 6 19 α β ⋅ = and 3 3 β α

∴ Required equation is x 2 − i.e.,

3 x −1 ⇒ 3 ⋅ 3 + x = 2 3

3x2 – 19x + 3 = 0.

b c and αβ = . a a The required equation is  1 1  1 1 x2 −  x+ + ⋅ =0 aα + b aβ + b  aα + b aβ + b 

⇒ y2 – 6y + 9 = 0 ⇒ (y – 3)2 = 0 ∴ y = 3 ⇒ 3x = 31 ⇒ x = 1.

⇒ {a2αβ + ab (α + β) + b2} x2

100. (d) We have, (x – a) (x – b) – k = 0 ...(1)

Since the roots of equation (1) are c and d ∴ c + d = a + b,

...(2)

and

...(3)

cd = ab – k

Now (x – c) (x – d) + k = 0 ⇒ x2 – (c + d) x + cd + k = 0 ⇒ x2 – (a + b) x + ab = 0

– {a (α + β) + 2b} x + 1 = 0 ⇒ (ca – b + b ) x – (2b – b) x + 1 = 0 2

2

2

⇒ cax2 – bx + 1 = 0. 104. (c) Let α and β be the roots of ax2 + bx + c = 0. Then α + β = −

b c and αβ = . a a

The roots of the required equation are  (α + β)2 and (α – β)2.

[ Putting the values of a + b and ab from (2) and (3)]

The required equation is

⇒ (x – a) (x – b) = 0 ⇒ x = a, b.



101. (a) The given equation is defined for x – 2 ≥ 0,

4 – x ≥ 0 and 6 – x ≥ 0

19 x +1 = 0 3

∴α +β= −

y 3 + = 2, where y = 3x 3 y

⇒ x2 – (a + b) x + ab – k = 0

=1

103. (a) Since α, β are roots of the equation ax2 + bx + c = 0

=2

1

⇒

4 . 5

102. (b) Since α, β are roots of the x2 = 5x – 3

Now, α3β3 + α2β3 + α3β2 = (αβ)3 + α2β2 (β + α) c2 c =   + 2 a a

4 4 , 4− 5 5

∴ Solution of the original equation is x = 4 −

98. (b) Since α, β are roots of the equation 

6− x

Squaring both sides, we obtain

20 . 9 Hence the required equation is

⇒ 9αβ = 20 ⇒ αβ =

6⋅

x−2 + 4− x =



Then α + β = 3 and α3 + β3 = 7.

x2 – [(α + β)2 + (α – β)2] x + (α + β)2 (α – β)2 = 0

 b2  ⇒ x 2 −  2 + (α + β) 2 − 4βα  x a 

 b2 b2 ⇒ x −  2 + 2 − 4 a a 2

c b 2  b 2 4c  x+ 2  2 −  = 0 a a a a

or a4x2 – 2a2 (b2 – 2ca) x + b2 (b2 – 4ac) = 0. 105. (c) The given inequation is valid only when x ≥ 0

...(1)

The given inequation can be written in the form 372 − x −



x

⇒ 72 – x – ⇒ x +

Since x is real, ∴ 62 – 4 ⋅ 8 (y – 3) ≥ 0, or 36 – 32y + 96 ≥ 0 or 32y ≤ 132 ∴ y ≤ 132 or y ≤ 33 . 32 8 33 . Hence maximum value of y = 8 110. (a), (b)  Let α be the common root. Then α2 + pα + q = 0 and α2 + p′ α + q′ = 0

>1 x > 0

(∵ 3 > 1)

x – 72 < 0

By cross-multiplication, we get α2 α 1 = = pq ′ − p ′q q − q′ p′ − p .

⇒ ( x + 9) ( x – 8) < 0

pq ′ − p ′q q − q′ . Also α = . q − q′ p′ − p

∴ α =

x + 9 > 0 for all x ≥ 0

But

x – 8 < 0 ⇒ x < 8 ∴ 0 < x ≤ 64 {from (1)}.

∴

∴ The common root is

106. (b) Since α, β are roots of the equation

∴ α+ β=–

⇒ (x – ω) (x – ω2) = 0 ⇒ x = ω, ω2

α and αβ = β

⇒ α = 1 ∴ β = – 1 –

α = – 1 – 1 = –­ 2.

Thus, α = 1 and β = – 2.



⇒ ( y – 1) x2 + 3 ( y + 1) x + 4 ( y – 1) = 0 Since x is real, ∴ discriminant ≥ 0 ⇒ 9 ( y + 1)2 – 16 ( y – 1)2 ≥ 0 ⇒ 9 ( y2 + 2y + 1) – 16 (y2 – 2y + 1) ≥ 0 ⇒ – 7y2 + 50y – 7 ≥ 0 ⇒ 7y2 – 50y + 7 ≤ 0 1  1 ⇒ ( y – 7)  y −  ≤ 0 ⇒ ≤ y ≤ 7. 7 7 x 3  108. (a) Tr+1th term in the expansion of  − 2  2 x x Cr   2

10 − r

 3  − 2  x



=



=

10

Cr



=

10

Cr ⋅ x10 − 3 r ⋅

10

3 +  2

1/ x

= 2.

1/ x

= t, then the equation becomes t3 + t – 2 = 0

But t2 + t + 2 = 0 has no real roots, ∴ t = 1 3 ⇒   2

1/ x

=1 ⇒

1 =0 x

which is not possible for any value of x. 113. (a) Since α, β are roots of x2 + px + p2 + q = 0 ∴ α + β = – p and αβ = p2 + q = p2 – ( p2 + q) + q = 0.

114. (a), (b)  Let α be the common root. Then 3α2 – 2mα – 4 = 0 and α2 – 4mα + 2 = 0. By cross-multiplication, we get

(−1) r ⋅ 3r 210 − r

⇒ 10 – 3r = 4 ⇒ r = 2

∴ T3 =

(−1) 2 32 28

Hence the coefficient of x4 =

3/ x

⇒ (t – 1) (t2 + t + 2) = 0.



x10 − r (−1) r ⋅ 3r × 210 − r x2 r

C2 x 4 ⋅

3 Put   2 

3   2

∴ α2 + αβ + β2 + q = (α + β)2 – αβ + q

r

∴ x10 – 3r = x4 10

∴ α19 = ω19 = ω and β7 = ω14 = ω2. 112. (d) The given equation can be written as

⇒ yx2 + 3xy + 4y = x2 – 3x + 4

10

∴ α = ω and β = ω2 ∴ Required equation is x2 + x + 1 = 0.

x 2 − 3x + 4 x 2 + 3x + 4

107. (b) Let y =

q − q′ pq' − p ′q or . p′ − p q − q′

111. (b) We have, x2 + x + 1 = 0

x2 + x α + β = 0

10

C2 ×

32 405 = . 8 2 256

649

109. (b) Let y = 3 – 6x – 8x2 then 8x2 + 6x + y – 3 = 0.

Quadratic Equations and Inequations

 + (α + β) 2 [(α + β) 2 − 4αβ] = 0



α2 α 1 = = − 4m − 16m −4 − 6 −12m + 2m

⇒

α2 = − 20m

⇒

1 α2 1 ; or 2m2 = 1; ∴ m = ± . = α = 2 2m m 1

α 1 = −10 −10m

650

115. (b), (c)  Let the common-root be α

Then aα + bα + c = 0

Objective Mathematics

and bα2 + cα + a =

0.

By cross-multiplication, we get α α 1 = = ab − c 2 ac − b 2 bc − a 2 2



⇒ (bc – a2)2 = (ab – c2) (ac – b2) ⇒ ⇒ ⇒ ⇒ ⇒

a b c = k   (say) = = 3 1 −5

∴

2

a3 + b3 + c3 – 3abc = 0 (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 0 (a + b + c) {(a –­ b)2 + (b – c)2 + (c – a)2} = 0 a + b + c = 0 or (a – b)2 + (b – c)2 + (c – a)2 = 0 a + b + c = 0 or a = b = c.

116. (b) Let α be the common root. Then aα2 + 2bα + c = 0 and

a1α2 + 2b1α + c1 = 0

By cross-multiplication, we get

α2 α 1 = = 2(bc1 − b1c) ca1 − ac1 2 (ab1 − ba1 )

⇒ (ca1 – ac1) = 4 (bc1 – b1c) (ab1 – ba1)  ...(1) 2

∵ ∴

a b c , , are in A.P. , a1 b1 c1 b a c b − − = b1 a1 c1 b1 = k (say)

⇒ ab1 – a1b = – ka1b1 and bc1 – b1c = – kb1c1. Also 2k = k + k = or

b a c b ca1 − ac1 − + − = b1 a1 c1 b1 a1c1

ca1 – ac1 = 2ka1c1.

∴ From (1), 4k 2 a12c12 = 4 (– ka1b1) (– kb1c1) or

a1c1 = b12 . Hence a1, b1, c1 are in G.P.

117. (c) The roots of 2x2 – 3x + 4 = 0 are imaginary, because disc. = (– 3)2 – 4 · 2 · 4 < 0. Hence the common root must be imaginary. But imaginary roots occur in pairs. Hence both the roots will be common, i.e., two equations will be identical. So their co-efficients will be proportioanl a b c i.e., = = , ∴ 6a = – 4b = 3c. 2 −3 4 118. (a) We have 3x2 + x – 5 = 0. I ts discriminant = 1 – 4 · 3 (– 5) = 61, which is positive but not a perfect square. Hence both the roots of 3x2 + x – 5 = 0 must be irrational as the irrational roots occur in conjugate pair. But one root of ax2 + bx + c = 0 and 3x2 + x – 5 = 0 is common. Hence both the roots of ax2 + bx + c = 0 must also be irrational, that is both the roots of the given equations are common. Thus both the equations are the same.

⇒ a = 3k; b = k, c = – 5k. ∴ 3a + b + 2c = 9k + k – 10k = 10k – 10k = 0. 119. (b) Let α, β be the roots of x2 + abx + c = 0 and α, γ be the roots of x2 + acx + b = 0, α being the common root. ∴ α + β = – ab 

... (1)



αβ = c 



α + γ = – ac 

... (3)



αγ = b

... (4)

... (2)

From (1) – (3), β – γ = a (c – b) From (2)– (4), α (β – γ) = c – b α (β − γ ) c−b 1 = ; or α = . β−γ a (c − b ) a

∴

∴ From (2) and (4), and

β = c, i.e., , β = ac a

γ = b, i.e., , γ = ab. a

∴ The quadratic equation whose roots are β, γ is

x2 – (β + γ) x + βγ = 0

or,

x2 – (ac + ab) x + ac · ab = 0;

or, x – a (b + c) x + a2bc = 0. 2

120. (a) Let α be the common root and β the other root of x2 + px + q = 0 q ∴ αβ = q; or α = . β Since α is the root of x2 + mx + n = 0 q2 q ∴ α2 + mα + n = 0 ⇒ 2 + m + n = 0 β β ⇒ nβ2 + mqβ + q2 = 0. Hence β satisfies the equation nx2 + mqx + q2 = 0. 121. (c) Let α be the common root and β the other root of the equation x2 – px + q = 0. Then α + β = p  and

...(1)

αβ = q. 

...(2)

The roots of x – ax + b = 0 are α and α 2

∴ a2 = 4b.

...(3)

a . Also α + α = a ⇒ α = 2 From (1), β = p − ∴ p−

a 2q = 2 a

⇒ 2ap – 4b = 4q ∴ ap = 2 (b + q).

a 2q and from (2), β = . 2 a ⇒ 2ap – a2 = 4q [using (3)]

Also a +

 α Now, (q – b)2 =  αβ −   β

=

 1 = α2 β −   β

  α 1  ⋅βα (α + β) −  α +    β β  

2

= bq (p – a)2.

α3 + 3pα2 + 3qα + r = 0





α=

(i)

(ii)

(iii)

2( pr − q 2 ) pq − r

⇒ 9y2 – 1 ≤ 0



α=

pq − r 2( q − p 2 )

...(4)

...(5)

From (4) and (5),

Equation (1) can also be written as

...(2)

From (i) and (ii),

From (ii) and (iii),

9y + 0y + x – 4x + 3 = 0 2

0

...(3) α2 α 1 = = From (2) and (3), 2( pr − q 2 ) pq − r 2(q − p 2 )

⇒ 16 – 4 (9y2 + 3) ≥ 0 ⇒ 4 – 9y2 – 3 ≥ 0

⇒ −1 ≤ y ≤ 1 . 3 3

...(1)

(1) – α (2) ⇒ pα2 + 2qα + r = 0

Since x is real, ∴ (– 4)2 – 4 (9y2 + 3) ≥ 0

⇒ (3y – 1) (3y + 1) ≤ 0

2

4 (q2 – pr) (p2 – q) = (pq – r)2.

128. (c) The expression will be a perfect square if the roots of the corresponding quadratic equation are equal.

Since y is real ∴ 02 – 4.9 (x2 – 4x + 3) ≥ 0

Condition for that is D = 0

⇒ x – 4x + 3 ≤ 0

⇒ 4 (a + b + c)2 – 4 ⋅ 1 ⋅ 3 (bc + ca + ab) = 0

⇒ (x – 1) (x –­ 3) ≤ 0 ⇒ 1 ≤ x ≤ 3.

⇒ a2 + b2 + c2 – ab – bc – ca = 0

2

124. (a) Since the coefficient of x2 is 1 which is positive,

127. (c) Let α be the common root, then and α2 + 2pα + q =

2

123. (a), (c)  Given equation is x2 + 9y2 – 4x + 3 = 0  ...(1) or x2 – 4x + 9y2 + 3 = 0.

∴ the given expression is positive for all real values of x if D < 0. ⇒ (– a)2 – 4 (1 – 2a2) < 0 ⇒ 9a2 – 4 < 0 ⇒ (3a + 2) (3a – 2) < 0 ⇒ –  2 < a < 2 . 3 3

125. (c) Let y = 5 + 4x – x2, then x2 – 4x + y – 5 = 0. Since x is real, ∴ 16 – 4 (y – 5) ≥ 0 ⇒ 4y ≤ 36 ⇒ y ≤ 9. ∴ Maximum value of y is 9.

2

n

n

D = 4 [( Cr) – Cr – 1 Cr + 1] = 4 (a – b), where a = (nCr)2, b = nCr – 1 · nCr + 1 Now



=

n a Cr ⋅ nCr = n b Cr − 1 ⋅ nCr + 1

=

3 ⇒ (2 – k)2 – 4 ⋅ 1 ⋅   k −  = 0  4 ⇒ k2 – 8k + 7 = 0; ∴ k = 7, 1. 130. (a) The given equation can be written as (ay + a′) x2 + (by + b′) x + (cy + c′) = 0.

The condition that x may be a rational function of y is, (by + b′)2 – 4 (ay + a′) (cy + c′) is a perfect square; t hat is, (b2 – 4ac) y2 + (2bb′ – 4ac′ – 4a′c) y + b′ 2 – 4a′c′ is a perfect square.

n! n! (r − 1)! ( n − r + 1)! ⋅ r ! (n − r )! r ! (n − r )! n! (r + 1)! ⋅ (n − r − 1)! ⋅ n!



129. (a), ( c)  For perfect square, the discriminant of the corresponding equation = 0



126. (a) The discriminant of the given equation is n

⇒ 1 {(a – b)2 + (b – c)2 + (c – a)2} = 0; 2 ∴ a = b = c.

1  r +1 n − r +1  1  = 1 +  1 + >1 ⋅  r  n − r  r n−r

 he corresponding quadratic equation has discriminant T =0 that is, 4 (bb′ – 2ac′ – 2a′c)2 – 4 (b2 – 4ac) (b′ 2 – 4a′c′) = 0; or (ac′ + a′c)2 – 4aa′cc′ = abb′c + a′bb′c – a′c′ b2 – acb′ 2 or (ac′ – a′c)2 = (ab′ – a′ b) (bc′ – b′c).

651

⇒ roots of given equation are real and distinct.



α 1 = a and = b. β β 2

∴ a>b⇒D >0

Quadratic Equations and Inequations

122. (b) Let α and β be the roots of x2 – px + q = 0 and 1 be the roots of x2 – ax + b = 0. α and β Then α + β = p and αβ = q.

652

131. (c) The corresponding quadratic equation of the given expression is

Objective Mathematics



a (b – c) x2 + b (c – a) xy + c (a – b) y2 = 0 2

x x or a (b − c)   + b (c − a ) + c (a − b) = 0. y  y Let

x =X y

 he given expression will be perfect square if the disT criminant of the corresponding equation of the given expression is zero. ∴ b2 (c – a)2 – 4a (b – c) ⋅ c (a – b) = 0 or (bc + ab – 2ac)2 = 0; or b (a + c) = 2ac 2ac . a+c

132. (b) The given expression 2

 x   y  + 2c       z   z  = z2 [aX2 + bY2 + c + 2aY + 2bX + 2cXY], ...(1) x y = X and = Y. where z z The given expression can be resolved into rational factors when the expression under the brackets given in (1) can be resolved into rational factors. The condition for this is abc + 2 ⋅ a ⋅ b ⋅ c – a ⋅ a2 – b ⋅ b2 – c ⋅ c2 = 0 (∵ here f = a, g = b, h = c)

or

a3 + b3 + c3 = 3abc.

12 x 133. (a) Let 4 x 2 + 9 = y,

⇒ y2 ≤ 1; ∴ | y | ≤ 1. Hence

x2 – (sum of the roots) x + (product of the roots) = 0 2 ⇒ x −

6 1 x+ = 0 or 7x2 – 6x + 1 = 0 7 7



= (α + β)3 – 3αβ (α + β)

= (4)3 – 3 (1) (4) = 64 – 12 = 52.

137. (b) Since x – 3x + 2 = 0 is one of the factors of the expression x4 – px2 + q, therefore, on dividing the expression by factor, remainder = 0 i.e., on dividing x4 – px2 + q by 2

x2 – 3x + 2, the remainder

(15 – 3p) x + (2p + q – 14) = 0.

On comparing both sides, we get

15 – 3p = 0 or p = 5 and 2p + q – 14 = 0 or q = 4.

138. (d) Since the roots of the given equation are equal, ∴  B2 = 4AC ⇒ 4 (ac + bd)2 = 4 (a2 + b2) (c2 + d2) ⇒ a2c2 + b2d2 + 2abcd = a2c2 + a2d2 + b2c2 + b2d2 a c = . ⇒ (ad – bc)2 = 0 or ad = bc i.e., b d

Since α, β are roots of the above equation,

12 x 4 x2 + 9

≤ 1.

134. (d) We know that if α, β are roots of the equation Ax2 + Bx + C = 0, B2 − 4AC . A

 quation the value of α – β from both the given E equations, we get b 2 − 4c =

∴ The required equation is

or (λ + 1) x2 – [b (λ + 1) + (λ – 1) a] x + c (λ – 1) = 0

⇒ 1 – y2  ≥  0



1 1 1 = . × 3+ 2 3− 2 7

(λ + 1) x2 – b (λ + 1) x = (λ – 1) ax – c (λ – 1)

As x is real, D = 144 – 4 ⋅ 4y ⋅ 9y ≥  0

then α – β =

1 1 6 = + 7 3+ 2 3− 2

139. (a) Given equation can be written as

Now, 4yx2 – 12x + 9y = 0.



1 1 and . 3+ 2 3− 2

and product of the roots =

Now, α3 + β3

 x  y  y x 2 = z  a   + b   + c + 2a   + 2b   z z z  z



135. (a) Given roots are

( ∵ b ≠ c)

136. (b) Since α, β are roots of the equation x2 – 4x + 1 = 0, ∴ α + β = 4 and αβ = 1.

Hence a, b, c are in H.P. 2

⇒ b + c = – 4

Sum of the roots =

Then a (b – c) X2 + b (c – a) X + c (a – b)  = 0.

or b =

⇒ b2 – c2 = – 4 (b – c) ⇒ (b – c) (b + c + 4) = 0

c 2 − 4b ⇒ b2 – 4c = c2 – 4b

∴ α+β=

b ( λ + 1) + ( λ − 1) a ( λ + 1)

But α + β = 0 (given), ∴ b (λ + 1) + (λ – 1) a = 0 ⇒ (a + b) λ = a – b or λ =

a−b . a+b

140. (c) The given equation can be written as

|x |2 – 3 | x | + 2 = 0 ⇒ ( | x | – 1) ( |x | – 2) = 0

⇒ | x | = 1 or |x | = 2 ⇒ x = ± 1 or x = ± 2 ∴ Four roots are possible.

The equation (x – α) (x – β) + c = 0 may be written as

∴ B2 = 4AC i.e., 4 (1 + 3k)2 = 4 ⋅ 7 ⋅ (3 + 2k) ⇒ 9k2 – 8k – 20 = 0



− 10 ⇒ (9k + 10) (k – 2) = 0 ⇒ k = 2, . 9

⇒ x2 – (a + b) x + ab = 0

147. (a) Since the roots of the given equation are real

−m n and αβ = . l l

∴ B2 – 4AC ≥ 0 ⇒ 4a2 – 4 (a2 + a – 3) ≥ 0 ⇒ – a + 3 ≥ 0 or a ≤ 3.

For the required equation,

=

⇒ 32 – 2a (3) + a2 + a – 3 > 0 ⇒ a2 – 5a + 6 > 0 or (a – 2) (a – 3) > 0

n  m 2 2n  n = 3 (m2 – 2nl) − l  l 2 l  l

⇒ a < 2 or a > 3 n4 . l4

and product of the roots = α3β ⋅ β3α = α4β4 = ∴ The required equation is

x2 – (sum of roots) x + (product of roots) = 0 x2 −

or

x2l4 – ln (m2 – 2nl) x + n4 = 0.

148. (b) Let z =

⇒ z + zx + zx2 = 1 – x + x2

⇒ (z + 1)2 – 4 (z – 1) (z – 1) ≥ 0 ⇒ z2 + 2z + 1 – 4z2 + 8z – 4 ≥ 0 ⇒ – 3z2 + 10z – 3 ≥ 0 ⇒ – 3z2 + 9z + z – 3 ≥ 0

∴ p = – 4 and q = 7.

⇒ – 3z (z – 3) + 1 (z – 3) ≥ 0

144. (b) The given equation can be written as

⇒ (z – 3) (– 3z + 1) ≥ 0

x2 – 8x + a2 – 6a = 0

Since the roots of the above equation are real, ∴ B2 – 4AC ≥ 0 ⇒ 64 – 4 (a2 – 6a) ≥ 0

⇒

⇒ a – 6a – 16 ≤ 0 ⇒ (a + 2) (a – 8) ≤ 0



⇒ a ∈ [– 2, 8].

(| x | – 3) (| x | – 4) = 0

∴ | x | = 3 or | x | = 4 ⇒ x = ± 3 or x = ± 4.

145. (a) Since α, β are roots of the equation

150. (a) Since x = 1 does not satisfy the given equation,

x2 + px + p3 = 0 ∴ α + β = – p

...(1)

αβ = p  3

...(2)

Since (α, β) lies on parabola y2 = x ∴ β2 = α

...(3)

From (2) and (3), we get β3 = p3 or β = p.

∴ given equation has no real root. 151. (a) Since sum of three roots 6, 4, – 1 is equal to 9, ∴ (a) is true. 152. (d) Given

From (1) and (3), we get β2 + β = – p ∴ p2 + p = – p

⇒ p (p + 2) = 0 ⇒ p = – 2

(∵ p ≠ 0)

∴ α = 4, β = – 2.

⇒

1 x2 + 2 x + 4 < < 3 for all x ∈ R. 3 x2 − 2 x + 4

Let 3x + 1 = y

∴

x2 – (a + b) x + ab – c = 0.

Since α, β are roots of the above equation ∴ α + β = a + b and αβ = ab – c

1 x2 − 2 x + 4 < < 3 for all x ∈ R. 3 x2 + 2 x + 4

Then y ∈ R for all x ∈ R.

146. (c) The given equation can be written as

1 1 ≤ z ≤ 3. ∴ minimum value of z = . 3 3

149. (d) The given equation can be factorized as

2



1 − x + x2 1 + x + x2

For real x, B2 – 4AC ≥ 0

∴ other root β = 2 − i 3 ∴ α + β = 4 = – p and αβ = 7 = q

and

From (1) and (2), we have a < 2.

⇒ x2 (z – 1) + x (z + 1) + (z – 1) = 0

143. (c) Given, one root α = 2 + i 3 ,



...(2)

⇒ zx2 – x2 + zx + z + z – 1 = 0

n n4 (m 2 − 2nl ) x + 4 = 0 3 l l

i.e.,

...(1)

Since the root is less than 3, so f (3) > 0

sum of the roots = α3β + αβ3 = αβ (α2 + β2) = αβ [(α + β)2 – 2αβ]

[Using (1)]

⇒ (x – a) (x – b) = 0 or x = a, b.

142. (a) Since α, β are roots of the given equation ∴ α+β=

x2 – (α + β) x + αβ + c = 0

...(1)



32 x + 2 + 2 ⋅ 3x + 1 + 4 9 ⋅ 32 x + 6 ⋅ 3x + 4 = 2x + 2 2x x 3 − 2 ⋅ 3x + 1 + 4 9 ⋅3 − 6 ⋅3 + 4 =

y2 + 2 y + 4 y2 − 2 y + 4

...(1)

653

or a + b = α + β and ab = αβ + c.

Quadratic Equations and Inequations

141. (b) Since the given equation has equal roots

654

From (1),

Objective Mathematics

∴

1 y2 + 2 y + 4 < 0 2

2

2

2

156. (c) Since sin α and cos α are the roots of given equation −q  p

∴ sin α + cos α =

r  p

sin α cos α =

50 + 26 1 = 19 and αβ = (25 – 13) = 3 4 4 α β as its roots is Thus, the equation having and β α α β  αβ x2 – x  +  + =0 β α αβ



⇒ 3x2 – 19x + 3 = 0.

161. (c) The given equation can be written as

...(ii)

 he roots are equal in magnitude but opposite in T sign, hence their sum,

x2 – 2 (a + b) x + 3ab = 0

2 (a + b) = 0 or a + b = 0.

162. (a) Let f (x) = x2 + (1 – 2k) x + k2 – k – 2

q2 p2

sin2α + cos2α + 2 sin α cos α =



...(i)

Squaring (i), we get

 he number 3 lies between the roots of the given T equation, if f (3) < 0

2r q2 ⇒ 1 + = 2 p p

Now, f (3) = 9 + (1 – 2k) 3 + k2 – k – 2

⇒ p2 + 2pr = q2 157. (b) D = B2 – 4AC = a2 – 4b = 0

[∵ a = 2 b ]

∴ Roots are equal. 158. (a) Required equation is 1

= 10 – 7k + k2 = k2 – 7k + 10

Hence f (3) < 0 ⇒ k2 – 7k + 10 < 0

⇒ p2 – q2 + 2pr = 0.

2

5 − 13 5 + 13 and β = 2 2

 α2 + β2  ⇒ x2 – x  +1=0  αβ 

if c – b > 0 i.e., if c > b or b < c. 2

0

5 + 13 5 − 13 and β = 2 2

α2 + β2 =



and

or α =

2

⇒ cos θ = 2 or 4. Both are not possible.



⇒ (k – 2) (k –­ 5) < 0 ⇒ 2 < k < 5. 163. (c) The minimum value occurs at x = 0, because minimum value of | x | = 0. ∴ Minimum value

2

3  ⋅ 3ax + 3  ⋅ 3bx + 3  ⋅ c = 0

= |0| 0 +



⇒ 3ax + 9bx + 9 c = 0 ⇒ ax + 3bx + 3c = 0. 2

2

159. (c) For x ≥ a, the given equation becomes



x2 – 2a (x – a) – 3a2 = 0 ⇒ x2 – 2ax – a2 = 0 ⇒ x = a (1 ±

2 ). As a ≤ 0 and x ≥ a, the possible root is a (1 – 2 ). For x < a, the given equation becomes x + 2a (x – a) – 3a = 0 ⇒ x + 2ax – 5a = 0 2

5 ± 13 2

2

2

⇒ x = a (– 1 ±

6) As a ≤ 0 and x < a ∴ possible root is a (– 1 +

2

1 5 + | 0 − 3| + 0 − 2 2

= 1 + 3 + 5 = 6. 2 3

164. (b) We have, 2x2 + 3x – 9 ≤ 0 ⇒ 2x2 + 6x – 3x – 9 ≤ 0 ⇒ (x + 3) (2x – 3) ≤ 0 ⇒ – 3 ≤ x ≤ 3 . 2 165. (c) Roots of the given equation are real,

6 ).

160. (b) We have α2 = 5α – 3 ⇒ α2 – 5α + 3 = 0 ⇒α = 5 ± 13 .2



if D = b2 – 4ac ≥ 0

i.e., if 4 (3a + 5)2 – 8 (9a2 + 25) ≥ 0 i.e., if (3a + 5)2 – 2 (9a2 + 25) ≥ 0 i.e., if 9a2 + 30a + 25 – 18a2 – 50 ≥ 0

i.e., if (3a – 5) ≤ 0 or if 3a – 5 = 0 or if

a = 5/3.

166. (a) Since the roots are of opposite signs, so product of the roots is negative.



2 ⇒ λ − 3λ + 2 < 0 3

x2 + 2x + 1 =k x2 + 2x + 7

(say)

⇒ (1 – k) x2 + 2 (1 – k) x + (1 – 7k) = 0 This equation has real roots if B2 – 4AC ≥ 0 ⇒ 4 (1 – k)2 – 4 (1 – k) (1 – 7k) ≥ 0 ⇒ (1 – k) (1 – k – 1 + 7k) ≥ 0 ⇒ (1 – k) (6k) ≥ 0 ⇒ k (k – 1) ≤ 0 ⇒ 0 ≤ k ≤ 1 ∴ Least value is 0 and greatest value is 1. 168. (c) Since α, β are roots of the given equation −b c , αβ = a a

∴ α + β =

Now α2 + β2 =

−b/a b 2 2c ⇒ 2 − = c/a a a ⇒

α+β αβ

⇒ b c – 2ac = – a b Dividing throughout by abc, we get 2

2



b 2c −a 2c b a = ⇒ = − + a b c b a c

∴

b c a are in A.P. , , a b c

169. (d) x2 + (a + b) x + ab < 0 ⇒ (x + a) (x + b) < 0 ⇒ x + a < 0, x + b > 0 or x + a > 0, x + b < 0 ⇒ x < – a, x > – b or x > ­– a, x < – b ⇒ – b < x < – a or – a < x < – b Since a < b ∴ – a > – b Hence – b < x < – a. 170. (d) Since α, β are roots of the given equation ∴ α+β=

173. (a) The roots of the equation are given by

x=

b ± b 2 − 4c 2

2 b − b 2 − 4c If α = b + b − 4c and β = 2 2

Then α – β = 1 ⇒

b 2 − 4c = 1

⇒ b – 4c = 1. 2

174. (d) Since α, β are the roots of the given equation α + β = – p and αβ = p2 + q

∴ α2 + αβ + β2 + q = (α + β)2 – αβ + q

b 2 − 2ac −b = 2 a a 2

⇒ 2x2 + 3x – 2 ≤ 0 ⇒ (x + 2) (2x – 1) ≤ 0 ⇒ – 2 ≤ x ≤ 1 . 2 172. (d) We have, ax2 + c = 0 (since b = 0). ∴ roots are equal in magnitude but opposite in sign.

∴

1 1 + α β

⇒ (α + β)2 – 2αβ =

3 3 α+β 1 8 8 = = = = . 1/ 3 3 (αβ)1/ 3 4  27    2 8

171. (b) We have, 2 – 3x – 2x2 ≥ 0

⇒ λ2 – 3λ + 2 < 0 ⇒ (λ – 1) (λ – 2) < 0 ⇒ 1 < λ < 2. 167. (a) Let

1

 α 2  3  β 2  3 (α 3 )1/ 3 + (β3 )1/ 3 Now   +   = (αβ)1/ 3 β  α

3 27 , αβ = 8 8



= p2 – (p2 + q) + q = 0.

175. (b) We have 32 x

2 −7 x+7

= 32



⇒ 2x2 – 7x + 7 = 2 ⇒ 2x2 – 7x + 5 = 0



D = b2 – 4ac = (–7)2 – 4 × 2 × 5 = 49 – 40 = 9

The discriminant of this equation is positive. Hence, it has two real roots. 176. (c) We have, α + β =

−b c and αβ = a a

∴ a (αx + 1) (βx + 1) = a [αβx2 + (α + β) x + 1]



c b = a   x 2 − x + 1 = cx2 – bx + a. a a  

177. (d) Here b2 – 4ac < 0 ∴ α=

− b + i 4ac − b 2 2a

and β =

− b − i 4ac − b 2 2a

∴ α = β

178. (c) Since both the roots are positive b c −b c 0. >0, >0 ⇒ a a a a ∴ a and c have same sign opposite to that of b.

∴

655

1

2

Quadratic Equations and Inequations

i.e., if 9a2 – 30a + 25 ≤ 0

656

183. (c) Roots are equal if discriminant = 0

179. (c) Since 2, 3 are roots of f (x) = 0 ∴ 16 + 4m – 26 + n = 0

i.e., if 2p2 – 8p = 0

4m + n – 10 = 0

Objective Mathematics

i.e.,

...(1)

54 + 9m – 39 + n = 0

and

9m + n + 15 = 0

i.e.,

184. (d) Since p (q – r) + q (r – p) + r (p – q) = 0 ...(2)

∴ m = – 5. ∴ From (1), – 20 + n – 10 = 0 ...(1)

Since roots are equal rp − rq ∴   pq − pr = 1 ⇒ rp – rq = pq – pr

...(2)

⇒ 2rp = q (p + r)

⇒ n = 30 ∴ m = – 5, n = 30. 180. (c) We have, x2 + a2 = 1 – 2ax and x2 + b2 = 1 – 2bx

Let ‘β’ be the common root of both the quadratic equations ∴ from equation (1), ...(3)

Similarly from equation (2), we get β2 + 2 b β + b2 – 1 = 0

...(4)

2 p+r 1 1 = + . = q pr p r



x=

− 2b ± 4b 2 − 4ac −b = 2a a (Since b2 = ac as a, b, c, are in G.P.)



∴ From equation (3) and equation (4), we get

This is root of dx2 + 2ex + f = 0

β β 1 = 2 = 2 2a (b − 1) − 2b(a − 1) (a − 1) − (b 2 − 1) 2b − 2a

− b − b + f =0 + 2e  ∴ d   a   a 

2



∴

185. (c) Roots of ax2 + 2bx + c = 0 are given by

β2 + a2 = 1 – 2 a β

or, β2 + 2 a β + a2 – 1 = 0

∴ one root is 1 r ( p − q) ∴ other root = p (q − r ) .

(2) – (1) gives 5m + 25 = 0



⇒ 2p (p – 4) = 0 ⇒ p = 0, 4.

2

⇒

β2 β 1 = = 2(b − a )(ab + 1) − (b − a )(a + b) 2(b − a )

On solving, we get

b=–

( a + b) and β2 = ab + 1 2

 ( a + b)  ∴ (ab + 1) =  − 2   ⇒ (ab + 1) =

⇒ dac – 2eba + a2 f = 0 (∵ b2 = ac) ⇒ 2eb = dc + a f 2e dc + af dc + af d f = + = = ⇒ b b2 ac a c ⇒

a 2 + b 2 + 2ab 4

( ∵ b2 = ac)

d e f , , are in A.P. a b c

186. (c) Since one real root of the given equation is 2 ∴ the other root must also be real.

2

187. (d) We have, x2 – 3x + 3

⇒ (a – b)2 = 4 ⇒ a – b = 2 181. (a) Since x + kx + 64 = 0 and x – 8x + k = 0 have real roots 2

⇒ db2 – 2eba + a2 f = 0

2

⇒ a + b – 2ab = 4 2

2

2

∴ k2 – 256 ≥ 0 and 64 – 4k ≥ 0

2

∴ smallest value =

⇒ k2 ≥ 256 and 16 ≥ k ∴ k = 16

2

3 9 3 3   =  x −  + 3 − =  x −  + 2 4 2 4

[∵ k > 0].

182. (c) Given equation can be written as (m + 1) x2 – (m + 1) bx = (m – 1) ax – (m – 1) c i.e., (m + 1) x2 – [mb + b + ma – a] x + (m + 1) c = 0

3 , which lies in the interval 4

3   − 3,  . 2 188. (c) Given equation is

 ince its roots are equal in magnitude and opposite S in sign



ax2 + bx + c = 0



∴ sum of the roots = 0



Let α, β be the roots of this equation. c then α + β = – b and αβ = a a

⇒ mb + b + ma – a = 0 a−b ⇒ m (a + b) = a – b ⇒ m = a + b .

Also

α+β=

1 1 a 2 + β2 + 2 = 2 α β α 2β 2

2

194. (c) Since sum of two roots = 0 ∴ if α is the third root, then α = m

b b 2a  =   – c a c 2

∴ m3 – m3 + 3m – 2 = 0

b 2a b 2a b b c  ⇒ =   + ⇒ =  +  c c a c a c a 2

2a b c c a b ⇒ = + ⇒ , , are in A.P. b c a a b c ⇒

a b c , , are in H.P. c a b

⇒ λ2 (x2 – a2) – 2λxy + y2 – a2 = 0 Discriminant = 4x2y2 – 4 (x2 – a2) (y2 – a2) = 0 [∵ roots are equal] ⇒ x y – (x y – x a – y a + a4) = 0 2 2

2 2

2 2

⇒ x 2 + y 2 – a 2 = 0 ⇒ x 2 + y 2 = a 2. 190. (b) Put A = b + c – 2a

B = c + a – 2b



C = a + b – 2c

∴ A + B + C = 0  o equation can be written as Ax2 + Bx + C = 0 S where A + B + C = 0 C Since A + B + C = 0 ∴ roots are 1, A ∴ Roots are rational. 191. (c) Since b – c + c – a + a – b = 0,  ∴ one root is 1. Since, the product of roots = ∴ other root is

a−b , b−c

a−b . b−c

Since the roots are equal, ∴ ⇒ a + c = 2b.

195. (a) (a2 – 5a + 3) x2 + (3a – 1) x + 2 = 0 Let α and 2α be the roots of this equation, then (3a − 1) (3a − 1) ⇒ 3α = − 2 α + 2α = – (a − 5a + 3) (a 2 − 5a + 3)

a−b =1 b−c

∴ the given equation becomes 1 2 y+ = 14 where y = (7 − 4 3 ) x − 4 x + 3 y

2 2 ⇒ 2α2 = 2 (a − 5a + 3) (a − 5a + 3) 2

2

2  − (3a − 1)  ⇒ 2  2  = (a 2 − 5a + 3)  3(a − 5a + 3)  1 (3a − 1) 2 ⇒ = (a 2 − 5a + 3) 9(a 2 − 5a + 3) 2 ⇒ (3a – 1)2 = 9 (a2 – 5a + 3) ⇒ 9a2 + 1 – 6a = 9a2 – 45a + 27 ⇒ 45a – 6a = 27 – 1 ⇒ 39a = 26 2 . ⇒ a = 3 196. (a) Since the roots of the given equation are same, ∴  sum of the roots in both the equations are same and product of roots is also the same. ∴ ∴

− b1 − b2 c c , 1 = 2 = a1 a2 a1 a2 a1 b c = 1 = 1 . a2 b2 c2

197. (a) We have, (3 – x)4 + (2 – x)4 = (5 – 2x)4 = (3 – x + 2 – x)4 ⇒ (3 – x)4 + (2 – x)4

192. (a) Since (7 + 4 3 ) (7 − 4 3 ) = 1,



∴ m = 2/3.

and α. 2α =

189. (b) We have, ( y – λx)2 = a2 (1 + λ2)

2 2

= (3 – x)4 + (2 – x)4 + 4 (3 – x) (2 – x)3 + 4 (3 – x)3 (2 – x) + 6 (3 – x)2 (2 – x)2

⇒ 2 (3 – x) (2 – x) [2 (3 – x)2 + 3 (3 – x) (2 – x) 

+ 2 (2 – x)2] = 0

⇒ (3 – x) (2 – x) (18 + 2x2 – 12x + 18 – 9x – 6x 

+ 3x2 + 8 + 2x2 – 8x) = 0

⇒ (3 – x) (2 – x) (7x2 – 35x + 44) = 0 ⇒ x = 2, 3 and 7x2 – 35x + 44 = 0.

⇒ y – 14y + 1 = 0 ⇒ y = 7 ± 3

Since discriminant of 7x2 – 35x + 44 = 0 is negative, ∴ it has no real roots.

Now y = 7 + 4 3 ⇒ x2 – 4x + 3 = – 1

 ence the given equation has two real roots and H two imaginary roots.

2

⇒ x = 2, 2 Also y = 7 − 4 3 ⇒ x2 – 4x + 3 = 1 ⇒ x = 2 ± 2 .

198. (a) Since 4 ≤ x ≤ 9 ⇒ x – 4 ≥ 0, x – 9 ≤ 0 ⇒ (x – 4) (x – 9) ≤ 0.

657

⇒ x2 – x – 6 < 0 ⇒ (x – 3) (x + 2) < 0 ⇒ – 2 < x < 3.

 −b / a   b 2 ⇒  −  =   – c/a  a c/a ⇒ –

193. (b) We have, 6 + x – x2 > 0

Quadratic Equations and Inequations

(α + β ) 2 − 2αβ α +β 2   ⇒α+ β=  − (αβ ) 2  αβ  αβ 2

=

658

2

9 61  199. (c) The given expression = m  x −  − m +1> 0  2 4

Objective Mathematics

if m ≥ 0 and 1 − Hence 0 ≤ m
0 i.e., if m < 61 4

4  4 ∴ m ∈ 0, . 61  61 

∴ ax2 + b | x | + c > 0 for all real x ∴ ax2 + b | x | + c ≠ 0 for any real x ∴ no real solution is possible. 201. (b) Let the two numbers be α and β, then α + β = 9 2 and αβ = 42 = 16 ∴ required quadratic equation is x2 – (α + β) x + αβ = 0  or  x2 – 18x + 16 = 0. 202. (b) We have | x |3 – 3x2 + 3 | x | – 2 = 0 ⇒ | x |3 – 3 | x |2 + 3 | x | – 2 = 0 ⇒ (| x | – 2) (| x |2 – | x | + 1) = 0 ⇒ | x | = 2 or | x |2 – | x | + 1 = 0 ⇒ x = ± 2  [ ∵ | x |2 – | x | + 1 = 0 has no real roots as discriminant = 1 – 4 = – ve.] 203. (a) Since (1 – p) is a root of given equation, ∴ (1 – p)2 + p(1 – p) + 1 – p = 0 ⇒ (1 – p) [1 – p + p + 1] = 0 ⇒ p = 1 ∴ required equation is x + x = 0 ⇒ x = 0, –1. 2

204. (b) We have, 4x – 3 ⋅ 2x + 3 + 128 = 0 ⇒ 22x – 3 ⋅ 2x ⋅ 23 + 128 = 0 ⇒ 22x – 24 ⋅ 2x + 128 = 0 ⇒ y2 – 24y + 128 = 0 where 2x = y ⇒ ( y – 16) ( y – 8) = 0 ⇒ y = 16, 8 ⇒ 2x = 16 or 2x = 8 ⇒ x = 4 or 3. 205. (b) Since the equation ax2 + bx + c = 0 has no real roots ∴ the curve y = ax2 + bx + c does not meet x-axis. ∴ f (x) = ax2 + bx + c has the same sign for all value of x. But a + b + c < 0 and f  (1) = a + b + c < 0 ∴ f (0) < 0 ⇒ c < 0.

206. (b) Simplifying the given equation, we shall get a linear equation, ∴ number of roots = 1. 207. (c) Since x2 – x + 1 = 0 ∴ (x – 1) (x – x + 1) = 0 2

⇒ x3 – 1 = 0 ⇒ x3 = 1, ∴ x3n = 1.

∴ 4q2 – 4pr ≥ 0  and  4pr – 4q2 ≥ 0 i.e., q2 – pr ≥ 0 and pr – q2 ≥ 0 ⇒ q2 = pr or

p q = . q r

209. (a) Since both roots are real ∴ (m + 1)2 – 4 (m + 4) ≥ 0 ⇒ (m – 1)2 ≥ 42

200. (a) Since a, b, c are all +ve

∴ f (x) < 0 for all x,

208. (a) Since the roots of the given equation are real

⇒ m – 1 ≥ 4 or m – 1 ≤ – 4 ⇒ m ≥ 5 or m ≤ – 3. 210. (c) From the given equation, α + β = − α , αβ = β. ∴ α=1 

[Since β = 0 does not satisfy given equation]

∴ 1+ β= − α, ∴ β=–1 − α =–1–1=–2 ∴ α = 1, β = – 2. 211. (d) The given equation can be written as | x − 3| 1  [4 x 2 − 1] = 4 · 2|x – 3| 2x + 1  x2 −  = 2 4 

 2 1  x − 4 

⇒ 2x – 1 = 2|x – 3| 1 [ ∵ x2 = does not give negative integral value] 4 ⇒ x – 1 = ± (x – 3) ⇒ either x – 1 = x – 3 or x – 1 = – x + 3 ⇒ 1 = 3 (not possible) or 2x = 4 i.e., x = 2. ∴ Given equation does not give any negative integral solution. 212. (b) Since α, β are the roots of given equation ∴ α + β = – b, αβ = – c ∴ b = – (α + β), c = – αβ ∴ Sum of roots = b + c = –[(α + β) + αβ] and product of roots 

= bc = – (α + β). – αβ = (α + β) αβ.

∴ New equation is 

x2 + [(α + β) + αβ] x + (α + β) αβ = 0.

213. (a) Sum of the roots = p + q = – p and product of roots = pq = q [ ∵ q ≠ 0]

∴ p = 1 ∴ 1 + q = – 1 ⇒ q = – 2 ∴ p = 1, q = – 2.

214. (c) Since a + c – b + (– 2c) + (b + c – a) = 0 ∴ roots are 1,

b+c−a . a+c−b

215. (c) 4 is a root of equation x2 + px + 12 = 0 ⇒ 16 + 4p + 12 = 0 ⇒ p = – 7

∴ (α – β)2 = (α + β)2 – 4αβ = (b2 – 4ac)/a2

216. (a) Let α, β be the roots of the equation .

and (α + k) + (β + k)

ax + bx + c = 0 ...(i) α m m ⇒α= ∴ = β β n n −b c But α + β = , αβ =  [being roots of (i)] a a 2

− b m  m  c ,  β β = ∴  + 1 β = n  n  a a Solving we get, β = ∴ ∴

− bn = a ( m + n)

m+n n −b = n m a m+n b + =0 mn ac m + n

∴

⇒ e2 log k2

(given)

⇒ k = 16 Let roots be real ∴ D ≥ 0

⇒ (– 2 2 k)2 – 4.1 (2 e2 log k – 1) ≥ 0 ⇒ 8k2 – 4(2k2 – 1) ≥ 0 ⇒ 4 > 0, which is true for all values of k. ∴ k = 4 218. (c) ∵ kx2 + 1 = kx + 3x – 11x2 ⇒ (11 + k) x2 – (3 + k) x + 1 = 0 For real and equal roots, [– (3 + k)]2 – 4 (11 + k) 1 = 0

⇒ 9 + k2 + 6k – 44 – 4k = 0 ⇒ k2 + 2k – 35 = 0 ⇒ k = =

− 2 ± 4 − 4 (− 35) (1) 2

− 2 ± 4 + 140 − 2 ± 144 − 2 ± 12 = = = – 7, 5. 2 2 2

1 1 1 219. (a) We have, x + p + x + q = r ⇒ x2 + (p + q – 2r) x + {pq – (p + q) r} = 0 Since roots are equal and of opposite sign, ∴ sum of roots = – (p + q – 2r) = 0 ⇒ p + q = 2r.

2 B2 − 4AC A A2 2 = =   . b − 4ac a a2

∴ Sum of roots = α + α2 =

b , a

and product of roots = α3 =

c a

3

2 log k

⇒ k = 4 (k > 0 because log k is defined)



⇒ (b2 – 4ac)/a2 = (B2 – 4AC)/A2

222. (b) Let first root be α and second root be α2

2



= (B2 – 4AC)/A2

= (a – 1)2 + 5 which is minimum when a = 1.

217. (d) We have, x2 – 2 2 kx + 2 × e2 log k – 1 = 0 −1 = 31 1 – 1 = 31 ⇒ 2k2 = 32



= (α + β)2 – 2αβ = (a – 2)2 + 2 (a + 1)

n b + = 0. m ac

2e

= {(α + k) + (β + k)}2 – 4 (α + k) (β + k)

221. (d) Sum of squares of roots = α2 + β2

a −b = c ac

∴ Product of roots =



⇒

cn am

= – B/A, (α + k) (β + k) = C/A



∴ (α – β)2 = {(α + k) – (β + k)}2

2/3 3  c 1/ 3  c b ∴    +    =    a   a   a 

⇒

c c2 c + +3  a a a2

2/3

c   a

2/3

c + 3  a

1/ 3

c   a

4/3

=

1/ 3 2/3 c c2 c  c  c  b3 + 2 + 3   +    ⇒ a a  a   = 3 a   a  a

⇒

c c 2 3c b b3 + 2 + × = 3 a a a a a

⇒ a2c + ac2 + 3abc – b3 = 0 ⇒ ac (a + c + 3b) = b3. 223. (a) If x2 + 2ax + 10 – 3a > 0 then D < 0 for all x, i.e., ∆ = 4 a2 – 4 (10 – 3a) < 0 or a2 + 3a – 10 < 0 or a2 + 5a – 2a – 10 < 0 or (a + 5)(a – 2) < 0 ∴ – 5 < a < 2. 1 k 224. (d) Product of roots = α    = ⇒ k = 5. α 5 225. (c) We have,

2x + 31 2 x + 47 x 2 + 7 =0 − + 2 9 9 x −7

x 2 + 7 16 ⇒ x 2 − 7 − 9 = 0 ⇒ 9x2 + 63 – 16x2 + 112 = 0 ⇒ 7x2 = 175 ⇒ x = ±

25 = ± 5.

b3 a3

Quadratic Equations and Inequations

⇒ p = 4q ⇒ q = 49/4. 2

659

220. (d) α + β = – b/a, αβ = c/a

Equation x2 + px + q = 0 has equal roots

660

Objective Mathematics

226. (d) Since one of the roots is 3+i . root will 2 ∴ Sum of the roots =

3−i , therefore, the other 2

 c   ∵  a + 1 < 0  

 or 1 +

3.

227. (d) We have 2 sin2x + 3 sin x – 2 > 0 ⇒ (2 sin x – 1) (sin x + 2) > 0 ⇒ 2 sin x – 1 > 0 [ ∵ sin x + 2 > 0 ∨ x ∈ R] 1 ⇒ x ∈ (π/6, 5π/6). 2 Also, x2 – x – 2 < 0 ⇒ sin x >

⇒ (x – 2) (x + 1) < 0 ⇒ – 1 < x < 2. As 2 < 5π/6, we obtain that x must lie in (π/6, 2). 228. (b) We have p + q = – p, pq = q.

c b + 0

∴ roots are real. 233. (d) Since α is a root of the equation.

a2x2 + bx + c = 0

∴ a2α2 + bα + c = 0  ...(1) Since β is a root of the equation

a2x2 – bx – c = 0

If q = 0 then p = 0.

∴ a2β2 – bβ – c = 0

If q ≠ 0 then p = 1 and q = – 2.

Now let f (x) = a x + 2bx + 2c

Thus, p = 1 or 0.

∴ f (α)

229. (b) We have α + β = – b, αβ = c As c < 0, b > 0, we get α < 0 < β.

= a2α2 + 2 (bα + c)



= a α2 + 2 (– a2α2)



= – a 2α 2 < 0

2



Thus, α < 0 < β < | α |.



= a2β2 + 2 (a2β2)



= 3a β > 0.

α + β = 1, αβ = p, r + δ = 4, rδ = q As α, β, r, δ are in G.P. So β = αr, γ = αr 2, δ = αr3 ∴ p = αβ = α2r, q = γ δ = αr2 ⋅ αr3 = α2r5 2 5 ∴ q = α 2r = r4. p αr Also α + αr = 1, αr2 + αr3 = 4

∴ α (1 + r) = 1, αr2 (1 + r) = 4 ⇒ r2 = 4 q = r4 = 42 = 16 or 16p = q So, (a) is true. ∴ p 231. (a) Given equation is ax2 + bx + c = 0 Since α, β are the roots of the equation c −b , αβ = . ∴ α+β= a a Also, since α < – 1, β > 1 ∴ αβ < – 1 ⇒

c < – 1 or c + 1 < 0 a a

Let f (x) = ax + bx + c 2

As f (1) f (– 1) > 0 ∴ (a + b + c) (a – b + c) > 0 or (a + c)2 – b2 > 0  or  (a + c)2 > b2 c  b 1 +  >   a a 2

or

2

c b  ⇒ 1 +  < −  a a

[from (1)]

f (β) = a2β2 + 2 (bβ + c)

Also, β = – b – α < – α = | α | 230. (a) We have,

...(2)

2 2

[from (2)]

2 2

 ince f (α) and f (β) are of opposite sign and r is a S root of the equation f (x) = 0, ∴ r must lie between α and β. Thus α < r < β. 234. (b), (c)  When x2 + 4x + 3 ≥ 0 i.e., x ≥ – 1 or x ≤ – 3 Then | x2 + 4x + 3 | + 2x + 5 = 0 ⇒ x2 + 4x + 3 + 2x + 5 = 0 ⇒ x2 + 6x + 8 = 0 ⇒ (x + 2) (x + 4) = 0 ⇒ x = – 2 or x = – 4. Thus x = – 4 as x ∈ {x : x ≥ – 1} ∪ {x : x ≤ – 3} When x2 + 4x + 3 < 0 i.e., – 3 < x < –­ 1 Then | x2 + 4x + 3 | + 2x + 5 = 0 ⇒ – (x2 + 4x + 3) + 2x + 5 = 0 ⇒ x2 + 4x + 3 – 2x – 5 = 0 ⇒ x2 + 2x – 2 = 0 ⇒ x =

−2± 4+8 = −1± 3 2

⇒ x = − 1 − 3 because − 1 + 3 does not lie between – 3 and – 1. Hence we have either x = – 4 or x = − 1 − 3 . 235. (b) We have, (a2 + b2 + c2) p2 – 2 (ab + bc + cd) p  + (b2 + c2 + d2) ≤ 0 ⇒ (ap – b)2 + (bp – c)2 + (cp – d)2 ≤ 0 ⇒ (ap – b)2 + (bp – c)2 + (cp – d)2 = 0 ( a, b, c, d, p ∈ R) ⇒ ap – b = 0, bp – c = 0, cp – d = 0

⇒ b = c = d = p a b c

=

⇒ a, b, c, d are in G.P.

From (1) and (2), we get ...(1)

α2 + bα + a = 0

...(2)

(1) – (2) ⇒ (a – b) α = a – b [ ∵ a ≠ b]

⇒ α = 1 Substituting α = 1 in (1), we get

a + b = – 1 ⇒ | a + b | = 1

240. (b) Put | x – 2 | = t, then the equation becomes

t2 + t – 2 = 0 ⇒ (t + 2) (t – 1) = 0



Since, t ≥ 0, we get t – 1 = 0 or t = 1

Thus the sum of the roots is 4.

237. (a) We can write the given equation as (x – a)2 = 3 – a.

 e are given p + q = 2, pq = A and r + s = 18, W rs = B.

This shows that a ≤ 3 and x = a ± 3 − a .

Now 2 = p + q = 2a – 4d

 oth the roots of the given equation will be less B than 3 if the larger of the two roots is less than 3, that is, if

and 18 = r + s = 2a + 4d.

a + 3−a 1.

But 3 − a > 1 ⇒ 3 – a > 1 or a < 2 Thus a < 3 and a < 2 ⇒ a < 2. 238. (a) The given equation is

x + 1 − x − 1 = 4x − 1  Squaring both the sides, we get

...(1)

2 x + 1 + x – 1 – 2 x − 1 = 4x – 1

L.H.S of (1) =

5 / 4 +1 − 5 / 4 −1 =

ax2 + bx + c = 0. −b c ; αβ = . ∴ α + β = a a New roots are α – 1, β – 1. −b −8 −2= = ­4 Their sum = α + β – 2 = a 2 Their product = (α – 1) (β – 1)

2 ⇒ 2 x − 1 = 1 – 2x. Squaring again, we get 4 (x2 – 1) = 1 – 4x + 4x2 ⇒ 4x = 5 or x = 5/4. When x = 5/4,

3 1 =1 − 2 2

5 4 ⋅ −1 = 4 = 2 4 ∴ x = 5/4 does not satisfy (1). R.H.S of (1) =

239. (a) As α, β are roots of ax2 + bx + c = 0, we have α + β = –­ b/a, αβ = c/a Now (α – β)2 = (α + β)2 – 4αβ 2

2 =  − b  − 4c = b − 4ac  ...(1) a a2  a



241. (a) Let p = a – 3d, q = a – d, r = a + d, s = a + 3d. As p < q < r < s, we have d > 0

b 2 − 4ac B2 − 4AC . = a2 A2



∴ | x – 2 | = 1 ⇒ x – 2 = ± 1 ⇒ x = 1, 3.

Thus numerical value of a + b is 1.



661

and

...(2)

Quadratic Equations and Inequations

236. (a) Let α be the common root Then, α2 + aα + b = 0

B2 4C B2 − 4AC  − = A2 A A2

Now, as α + δ, β + δ are the roots of Ax2 + Bx + C = 0, we have α + δ + β + δ = –B/A , (α + δ) (β + δ) = C/A. Now (α – β)2 = [(α + δ) – (β + δ)]2  = (α + δ + β + δ)2 – 4 (α + δ) (β + δ)

c b + +1 = 1 a a [ ∵ New equation is 2x2 + 8x + 2 = 0]

= αβ – (α + β) + 1 =



∴ b = 2 i.e., b = 2a a Also, c + b = 0 ∴ b = – c. 243. (d) | x – 2 | = x2 ⇒ x – 2 = x2 or 2 – x = x2 ⇒ x2 – x + 2 = 0 or x2 + x – 2 = 0 ⇒ x2 + x – 2 = 0  [ ∵ x2 – x + 2 = 0 does not give any real root] ⇒ (x + 2) (x – 1) = 0 ⇒ x = – 2, 1. 244. (d) Let f (x) = x2 + x + a. Since both the roots of

x2 + x + a = 0 exceed a

Therefore, (i) Discriminant > 0 (ii) f (a) > 0 (iii) − 1 > a 2 1 4 f (a) > 0 ⇒ a2 + 2a > 0 ⇒ a < – 2 or a > 0

Now, Discriminant > 0 ⇒ 1 – 4a > 0 ⇒ a <

662

⇒ a2 + 3a – 10 < 0

Objective Mathematics

−1 −1 >a⇒a< 2 2 ∴ a < – 2. and,

⇒ (a + 5) (a – 2) < 0 ⇒ – 5 < a < 2

245. (c) Let α, β be the roots of equation x2 + bx + c = 0 and α′, β′ be the roots of the x2 + qx + r = 0. Then, α + β = – b ; αβ = c, α′ + β′ = – q, α′ β′ = r. It is given that

α+β α ′ + β′ α α′ ⇒ = = α−β α ′ − β′ β β′

250. (d) Since, α and β are the roots of the equation 2x2 + 7x + c = 0. 7 and αβ = c 2 2 7 2 2 Given, |α – β | = 4

Then, α + β = −

⇒

(α + β) 2 (α ′ + β ′ ) 2 2 = (α − β) (α ′ − β ′ ) 2

⇒ (α + β) (α – β) = ±

⇒

b2 q2 = 2 b − 4c q − 4r

⇒

2

246. (d) We have, 1 1 1 = 3 + = log 2 y + log 2 x log 2 y 3 1  [ ∵ x ≠ y] 3 ⇒ x + y = 8 + 21/3.

⇒ log 2 x = 3, log 2 y = ⇒ x = 23, y = 21/3

247. (c) Let the roots of the equation 4x2 – (5k + 1) x + 5k = 0 be α and α + 1. ∴ α + α + 1 = ⇒ α =

49 − 8c = ∓ 1 ⇒ 49 – 8c = 1 ⇒ c = 6 251. (d) Since, α and β are the roots of the equation ax2 + bx + c = 0. Then, α + β = − b and αβ = c a a 2 Also, α2 + β2 = (α + β)2 – 2 αβ = b − 2ac a2

Now,

(5k + 1) 4

β α (aα + b) + β (aβ + b) α + = αβ + b aα + b (αβ + b) (aα + b)

5k − 3 8

and α (α + 1) =

49 7 − 2c = ± 4 4

⇒

⇒ b2r = q2c.

log 2 x +

−7 2

7 4

= 5k 4

 b 2 − 2ac   b a  + b −  2 a  a  =  c 2  b 2  a  a + ab  − a  + b    

⇒ (5k − 3) (5k − 3 + 8) = 5k 8×8 4 ⇒ (5k – 3) (5k + 5) = 80 k ⇒ (5k – 3) (k + 1) = 16k ⇒ 5k2 – 14k – 3 = 0 ⇒ (k – 3) (5k + 1) = 0 1 ⇒ k = 3, − 5 Hence, negative value of k is −

a ( α 2 + β 2 ) + b ( α + β) αβ a 2 + ab (α + β) + b 2

=

b 2 − 2ac − b 2 2 =− a 2c − ab 2 + ab 2 a

252. (b) Given equations are

1 . 5

248. (a) Since, α and β are the roots of the equation x2 – (1 + n2)x + 1 (1 + n2 + n4) = 0 2 1 + n2 + n4 ∴ α + β = 1 + n2 and αβ = 2 2 2 2 Now, α + β = (α + β) – 2 αβ = (1 + n2)2 – (1 + n2 + n4) = n2 249. (a) Given: x2 + 2ax + 10 – 3a > 0 ∴ 4a2 – 4 (10 – 3a) < 0

px2 + 2qx + r = 0 and qx 2 − 2 pr x + q = 0 Since, they have real roots. 4q2 – 4pr ≥ 0 ⇒ q2 ≥ pr and from second equation 4pr – 4q2 ≥ 0 ⇒ pr ≥ q2 From Eqs. (i) and (ii), we get

…(i)

…(ii)

253. (d) Let the roots of x2 – 6x + a = 0 be α, 4β and that of x2 – cx + 6 = 0 be α, 3β. ∴ α + 4β = 6 and 4 αβ = a and α + 3β = c and 3 αβ = 6 ⇒ a = 8

∴ b = a + c 2 ⇒ b = a + 3a = 2a 2 Hence, α + β = – 2a = – 2 a 255. (d) The equation x2 – px + r = 0 has roots (α, β) and the equation α  x2 – qx + r = 0 has roots  , 2β  2 

254. (d) Since, α and β are the roots of the equation

⇒ r = αβ and α + β = p and α + 2β = q 2

ax2 + bx + c = 0 ∴ α + β = − b and αβ = c a a But αβ = 3

⇒ β = 2q − p and α = 2(2 p − q ) 3 3

∴ 3 = c ⇒ c = 3a a

….(i)

⇒ αβ = r = 2 (2q – p) (2p – q). 9

EXERCISES FOR SELF-PRACTICE 1. The set of all real x for which x2 – | x + 2 | + x > 0, is (a) (–∞, –2) ∪ (2, ∞)

(b) (−∞, − 2 ) ∪ ( 2 , ∞)

(c) (–∞, –1) ∪ (1, ∞)

(d) ( 2 , ∞)

2. Both the roots of the equation (x – b)(x – c) + (x – c) + (x – a)(x – a)(x – b)(x – b) = 0 are always (a) negative (c) real

(b) positive (d) None of these

3. The real roots of the equation 7log (a) 4 and 5 (c) 1 and 2

7(x2–4x + 5)

= x – 1, are

(b) 3 and 4 (d) 2 and 3

4. The value of ‘a’ for which the equations x3 + ax + 1 = 0 and x4 + ax2 + 1 = 0 have a common root is (a) –2 (c) –1

(b) 1, –1 (d) None of these

5. If the two equations a1 x2 + b1 x + c1 = 0 and a2 x2 + b2 x + c2 = 0 have a common root then the value of (a1b2 – a2b1) (b1c2 – b2c1) is (a) – (a1c2 – a2c1)2 (c) (a1c1 – a2c2)2

(b) (a1a2 – c1c2)2 (d) (a1c2 – c1a2)2

6. If the roots of the equation

l b + x+l +k x+b+k =2 are equal in magnitude but in opposite in sign, then the value of K is (l + b) (a) 0 (b) − 4

(l + b) (c) 4

(l + b) (d) 2

2x − 1 7. If X be the root of real numbers x such that 2 x 3 + 3 x 2 + x is positive, then X contains

3  (a)  − ∞, 2   

1  3 (b)  − 2 , − 4   

 1 1 (c)  − 4 , 2   

1  (d)  2 , 3   

8. If the difference of the roots of the equation x2 + kx + 7 = 0 is 6, then the possible values of k are (a) 4 (c) 8 9. If x is real and y = (a) y ≥ 1 (c) y ≤ – 11

(b) – 4 (d) – 8.

x2 − x + 3 x + 2 , then (b) y ≥ 11 (d) – 11 ≤ y ≤ 1

10. The real roots of the equation x2 + 5 | x | + 4 = 0 are (a) {– 1, 4} (c) {– 4, 4}

(b) {1, 4} (d) None of these

11. If a, b, c are positive real numbers then both the roots of the equation ax2 + bx + c = 0 (a) are always real and +ve (b) have –ve real parts (c) are real and negative (d) None of these 12. The maximum number of real roots of the equation x2n – 1 = 0 is (a) 3 (c) 2

(b) n (d) 2n

Quadratic Equations and Inequations

663

Also a, b, c are in AP.

⇒ a = 4 ⇒ a = 8 ⇒ a=8 6 3 ⇒ x2 – 6x + 8 = 0 ⇒ (x – 4) (x – 2) = 0 ⇒ x = 2, 4 and x2 – 2c + 6 = 0 ⇒ 22 – 2c + 6 = 0 ⇒ c=5 ∴ x2 – 5x + 6 = 0 ⇒ x = 2, 3 Hence, common root is 2.

664

13. If α, β are the roots of ax2 + bx + c = 0, then

1 1 , are α β

the roots of

Objective Mathematics

(a) ax2 + cx + a = 0 (c) bx2 + ax + a = 0

(b) cx2 + bx + a = 0 (b) cx2 + ax + b = 0

14. Let Pn (x) = 1 + 2x + 3x2 + ... + (n + 1) xn be a polynomial such that n is even. Then the number of real roots of Pn (x) = 0 is (b) n (d) None of these

(a) 1 (c) 0

15. If α, β are the roots of x2 – 3x + 1 = 0 then the equation 1 1 whose roots are α − 2 , β − 2 is (a) x2 + x – 1 = 0 (c) x2 – x – 1 = 0

(b) x2 + x + 1 = 0 (d) None of these

(a) 33x2 + 4x + 1 = 0 (b) 33x2 – 4x – 1 = 0 (c) 33x2 – 4x + 1 = 0 (d) 33x2 + 4x – 1 = 0 17. If the equation 4x2 + 3x + 7 = 0 has roots α and β, then 1 1 + is equal to α β

3 7

2 (c) 7

(b) –

3 7

(c)

(b)

m + n +1= 0

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

20. If α, β are the root of 9x2 + 6x + 1 = 0, then the equation 1 1 , is with the roots α β (a) 2x2 + 3x + 18 = 0 (c) x2 + 6x + 9 = 0

(b) x2 + 6x – 9 = 0 (d) x2 – 6x + 9 = 0

21. The values of a for which 2x2 – 2(2a + 1)x + a(a + 1) = 0 may have one root less than a and other root greater than a are given by: (b) –1 < a < 0 (d) a > 0 or a < – 1

22. The value of k for which one of the roots of x2 – x + 3k = 0 is double of one of the roots of x2 – x + k = 0 is: (a) 1 (c) 2

(b) –2 (d) None of these

23. If A is the A.M. of the roots of the equation x2 – 2αx + b = 0 and G is the G.M. of the roots of the equation x2 – 2bx + a2 = 0, then: (a) A > G (c) A = G

(b) A ≠ G (d) None of these

24. If one root of the eqn. x2 + px + q = 0 is 2 +2, 3 then values of p and q are: (a) –4, 1 (c) 2, 3

7 (d) 3

18. If ratio of the roots of equation x2 + x + 1 = 0 are m, n then (a) m + n + 1 = 0

(a) a > b > c (c) a > c > b

(a) 1 > a > 0 (c) a ≥ 0

16. If α, β are the roots of the equation x2 – 5x – 3 = 0 then 1 1 the equation whose roots are 2α − 3 , 2β − 3 will be

(a)

19. For real x, the function values provided

(b) 4, –1 (d) −2, − 3

25. If x2 – 5x + 6 > 0, then x ∈: (a) (–∞, 2) ∪ (3, ∞) (c) (2, 3)

(b) [2, 3] (d) None of these

26. The values of ‘a’ for which (a2 – 1)x2 + 2(a – 1)x + 2 is positive for any x are :

m n + + 1 = 0 (d) (m + n)2 = mn n m

(a) a ≥ 1 (c) a > –3

(b) a ≤ 1 (d) a ≥ –3 or a ≥ 1

Answers

1. (b) 2. (c) 10. (d) 11. (b) 19. (a),(b),(c),(d)

3. (d) 12. (c) 20. (c)

4. (a) 13. (b) 21. (d)

5. (d) 14. (c) 22. (b)

6. (b) 15. (c) 23. (c)

7. (a), (d) 8. (c), (d) 9. (a), (c) 16. (d) 17. (b) 18. (d) 24. (a) 25. (a) 26. (d)

16

Permutations and Combinations

CHAPTER

SUMMARY OF CONCEPTS FACTORiAl NOTATiON The continued product of first n natural numbers is called n factorial or factorial n and is denoted by n or n! Thus, n or n! = 1 ⋅ 2 ⋅ 3 ⋅ 4....... (n – 1)n

= n (n – 1) (n – 2) ........3 ⋅ 2 ⋅ 1 (in reverse order)

Important Results (i) When n is a negative integer or a fraction, n! is not defined. Thus, n! is defined only for positive integers. (ii) According to the above definition, 0! makes no sense. However we define 0! = 1. (iii) n! = n (n – 1)! (iv) (2n)! = 2n ⋅ n! [1 ⋅ 3 ⋅ 5 ⋅ 7 .......... (2n – 1)].

FUNdAMENTAl PRiNCiPlE OF COUNTiNg Multiplication Principle If an operation can be performed in ‘m’ different ways; following which a second operation can be performed in ‘n’ different ways, then the two operations in succession can be performed in m × n different ways. Addition Principle If an operation can be performed in ‘m’ different ways and another operation, which is independent of the first operation, can be performed in ‘n’ different ways. Then either of the two operations can be performed in (m + n) ways. Note: The above two principles can be extended for any

finite number of operations.

PERMUTATiON Each of the different arrangements which can be made by taking some or all of given number of things or objects at a time is called a Permutation. Note: Permutation of things means arrangement of things. The word arrangement is used if order of things is taken into account. Thus, if order of different things changes, then their arrangement also changes. Notations Let r and n be positive integers such that 1 ≤ r ≤ n. Then, the number of permutations of n different things, taken r at a time, is denoted by the symbol nPr or P (n, r).

Important Results on Permutations 1.

n

Pr =

n! = n(n – 1) (n – 2) ... {n – (r + 1)}, 0 ≤ r ≤ n. (n − r )!

2. Number of Permutations of n different things taken all at a time is: nPn = n!. 3. The number of permutations of n things, taken all at a time, out of which p are alike and are of one type, q are alike and are of second type and rest are all different is n! . p! q! 4. The number of permutations of n different things taken r at a time when each thing may be repeated any number of times is nr. 5. Permutations under Restrictions (a) Number of Permutations of n different things, taken r at a time, when a particular thing is to be always included in each arrangement, is: r ⋅ n – 1P r – 1 . (b) Number of permutations of n different things, taken r at a time, when s particular things are to be always included in each arrangement, is s! (r – (s – 1)) ⋅ n – sPr – s. (c) Number of permutations of n different things, taken r at a time, when a particular thing is never taken in each arrangement, is n–1 P r. (d) Number of permutations of n different things, taken all at a time, when m specified things always come together, is m! × (n – m + 1)! (e) Number of permutations of n different things, taken all at a time, when m specified things never come together, is n! – m!× (n – m + 1)!.

6.

Circular Permutations (a) Number of circular arrangements (permutations) of n different things is: (n – 1)! (b) Number of circular arrangements (permutations) of n different things when clockwise and anticlockwise arrangements are not different, i.e., when observation can be made from both sides is: 1 (n − 1)! . 2 (c) Number of circular permutations of n different things, taken r at a time, when clockwise, and anticlockwise orders are n Pr taken as different, is = . r (d) Number of circular permutations of n different things, taken r at a time, when clockwise and anticlockwise orders are not n different, is = Pr . 2r

666

Objective Mathematics

16. (a) Number of selections of zero or more things out of n different things Each of the different groups or selections which can be made by n C0 + nC1 + nC2 + ... + nCn = 2n some taking some or all of a number of things (irrespective of (b) Number of combinations of n different things selectorder) is called a combination. ing at least one of them is n C1 + nC2 + ... + nCn = 2n – 1. Note:  Combination of things means selection of things. Obviously, in selection of things order of things has no importance. (c) Number of selections of r things (r ≤ n) out of n identical things is 1. Thus, with the change of order of things selection of things does (d) Number of selections of zero or more things out of n not change. identical things = n + 1. Notations  The number of combinations of n different things (e)  Number of selections of one or more things out of n taken r at a time is denoted by nCr or C (n, r). Thus, identical things = n. n n! P (f) If out of (p + q + r + t) things, p are alike of one kind, n Cr = = r (0 ≤ r ≤ n) r !(n − r )! r ! q are alike of second kind, r are alike of third kind and t are different, then the total number of selections is n (n − 1) (n − 2) .... (n − r + 1) (p + 1) (q + 1) (r + 1) 2t – 1. = r (r − 1) (r − 2) ... 3⋅ 2 ⋅1 (g) The number of ways of selecting some or all out of p + q + r items where p are alike of one kind, q are If r > n, then nCr = 0. alike of second kind and rest are alike of third kind is Important Results on Combination [(p + 1) (q + 1) (r + 1)] – 1.

Combination

1. nCr = nCn – r 2. nC0 = nCn = 1, nC1 = n 3. If nCx = nCy then either x = y  or  y = n – x, i.e., x + y = n. 4. nCr + nCr – 1 = n + 1Cr 5. r · nCr = n · n – 1Cr – 1 6. n · n – 1Cr – 1 = (n – r + 1) nCr – 1 n C n − r +1 7. n r = C r −1 r 8. If n is even then the greatest value of nCr is nCn/2. 9. If n is odd then the greatest value of nCr is

n

C n +1 or nC n −1 . 2

2

10. nCr =

r decreasing numbers starting with n r increasing numbers starrting with 1



n (n − 1) (n − 2) ....(n − r + 1) 1 ⋅ 2 ⋅ 3.....r.



=

11. nPr = r! nCr = n (n – 1) (n – 2).....(n – r + 1).

17. (a) Number of ways of dividing m + n different things in two groups containing m and n things respectively (m ≠ n): (m + n)! m+n Cm = . m! n! (b) Number of ways of dividing m + n + p different things in three groups containing m, n and p things respectively (m ≠ n ≠ p): (m + n + p)! . m! n! p ! (c) Number of ways of dividing 2m different things in two groups, each containing m things and the order of the (2m)! . groups is not important, is 2!(m!) 2 (d) Number of ways of dividing 2m different things in two groups, each containing m things and the order of the groups is important, is

(2m)! . (m!) 2

(e) Number of ways of dividing 3m different things in three groups, each containing m things and the order (3m)! 13. nC0 + nC2 + nC4 + ... = nC1 + nC3 + nC5 + ... = 2n – 1. . of the groups is not important, is 3!(m!)3 14. Number of combinations of n different things taken r at (f) Number of ways of dividing 3m diferent things in a time three groups, each containing m things and the order (a) when p particular things are always included (3m)! n – p = Cr – p . of the groups is important, is n – p (m!)3 (b) when p particular things are never included = Cr  (c) when p particular things are not together in any selec- 18. (a) Number of ways of dividing n identical things into tion = nCr – n – pCr – p. r groups, if blank groups are allowed is n + r – 1Cr – 1. (b)  Number of ways of dividing n identical things into 15. (a) Number of selections of r consecutive things out of r groups, if blank groups are not allowed is n – 1Cr – 1. n things in a row = n – r + 1. (b) Number of selections of r consecutive things out of (c) Number of ways of dividing n identical things into r groups such that no group contains less than m things n things along a circle and more than k (m < k) things is coefficient of xn in n, when r < n the expansion of   .  (xm + xm + 1 + ... + x k ) r. 1, when r = n 12. nC0 + nC1 + nC2 + ... + nCn = 2n.

[(1 + x + x2 + ... + x p – 1) (1 + x + x2 + ... + xq – 1) × (1 + x + x2 + ... + xs – 1) ...] 24. The number of ways in which r identical things can be distributed among n persons when each person can get zero or more things: = coefficient of x r in (1 + x + x2 + ... + x r)n = coefficient of xr in (1 – x)–n = n + r – 1Cr 25. The number of non-negative integral solutions of the equation x1 + x2 + ... + xr = n is n + r – 1Cr. 26. The number of terms in the expansion of (a1 + a2 + a3 + ... + a n)r is n + r – 1Cr.

derangement Rearrangement of objects such that no one goes to its original place is called derangement. If ‘n’ things are arranged in a row, the number of ways in which they can be deranged so that none of them occupies its original place is n 1  1 1 1 r 1 . n!1 − + − + ... + (−1) n  = n!∑ (−1) r! n!  r =0  1! 2! 3!

= 33 + 11 + 3 + 1 = 48. Some Useful Results 1. If n distinct points are given in the plane such that no three of which are collinear, then the number of line segments formed = nC2 If m of these points are collinear (m ≥ 3), then the number of line segments is (nC2 – mC2) + 1. 2. The number of diagonals in an n-sided closed polygon = nC2 – n. 3. If n distinct points are given in the plane such that no three of which are collinear, then the number of triangles formed = nC3. If m of these points are collinear (m ≥ 3), then the number of triangles formed = nC3 – mC3. 4. If n distinct points are given on the circumference of a circle, then (a) Number of st. lines = nC2 (b) Number of triangles = nC3 (c) Number of quadrilaterals = nC4 and so on 5. The sum of the digits in the unit place of all numbers formed with the help of a1, a2, ..., an taken all at a time is = (n – 1)! (a1 + a2 + .... + an) (repetition of digits not allowed) 6. The sum of all n digit numbers that can be formed using the digits a1, a2, .... , an is (10n − 1) = (n – 1)! (a1 + a2 + .... + an) . 9

667

Permutations and Combinations

For example, if 4 letters are taken out of 4 different enve19. If n things form an arrangement in a row, the number of ways in which they can be deranged (any change in the lopes, then the number of ways in which they can be reinserted given order of the things) so that no one of them occupies in the envelopes so that no letter goes in to its original envelope 1 1 1 1 1 1  its original place is   = 4!1 − 1 + − +  = 24 1 − 1 + − + = 9. 2 3 4 ! ! ! 2 6 24     1 1 1  n 1  n! 1 − + − + ... + (−1) ⋅  . n!   1! 2! 3! NUMbER OF RECTANglES ANd SqUARES 20. The number of ways of selecting r things out of n things of (a) Number of rectangles of any size in a square of size n × n which p are of one type, q of second type , s of third type n n and so on is is ∑ r 3 and number of squares of any size is ∑ r 2 . r = coefficient of x in r =1 r =1 (b) Number of rectangles of any size in a rectangle of size [(1 + x + x2 + ... + x p ) (1 + x + x2 + ... + x q ) × np (1 + x + x2 + ... + x s) ...]. (n + 1) (p + 1) and number of squares n × p (n < p) is 4 21. The number of ways of selecting r things out of n things of n which p are alike and are of one kind, q are alike and are of of any size is ∑ (n + 1 − r ) ( p + 1 − r ) . second, s are alike and are of third kind and so on, is r =1 = coefficient of xr in ExPONENT OF PRiME p iN n! [(1 + x + x2 + ... + x p) (1 + x + x2 + ... + xq) × (1 + x + x2 + ... + x s) ...]. Exponent of a prime p in n! is denoted by Ep (n!) and is given 22. The number of ways of selecting r things out of n things of by which p are like and are of one kind, q are alike and are of n  n   n   n  Ep (n!) =   +  2  +  3  + .... +  k  , second kind and rest (n – p – q) things are all different is: p p p       p  = coefficient of xr in n [(1 + x + x2 + ... + x p) (1 + x + x2 + ... + xq) × where pk < n < pk + 1 and   denotes the greatest integer less  p (1 + x)n – p – q] n than or equal to . 23. The number of ways of selecting r things out of n things of  p which p are alike and are of one kind, q are alike and are For example, exponent of 3 in (100)! is of second kind, s are alike and are of third kind when each  100   100   100   100  thing is taken at least once: E3 (100!) =  + 2 + 3 + 4   3  3   3  3  = coefficient of xr – 3 in

668

MULTIPLE-CHOICE QUESTIONS

Objective Mathematics

Choose the correct alternative in each of the following problems: 1. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is (a) 140 (c) 280

11.

(b) 196 (d) 346

2. The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by (a) 6! ×5! (c) 5! × 4!

(b) 30 (d) 7! × 5!

3. If nCr denotes the number of combinations of n things taken r at a time, then the expression

13.

C Cr

4.

n – 1

(b) n+2Cr+1 (d) n+1Cr+1

n+1

(b) n Pr (d) None of these

5. How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order? (a) 360 (c) 120

(b) 240 (d) 480

6. If P r = P r + 1 and C r = C r – 1, then the values of n and r are n

n

n

n

(a) 4, 3 (b) 3, 2 (c) 4, 2 (d) None of these 7. If

n – 1

C3 +

n – 1

(a) n > 5 (b) n > 6 (c) n > 7 (d) None of these

9.

n

∑

r =1

(1⋅ 3⋅ 5....(2n − 1)) 2 (d) None of these (1⋅ 3⋅ 5....(4n − 1))

(a)

n + 1

(c)

n + 1

n–1

Pn + 1

(b) n + 1Pn + 1 –­1

Pn + 1 – 2

(d) None of these

(b) (– ∞, –2) (d) [ 3 , 2]

(b) 4 (d) None of these

15. If 18Cr = 18Cr + 2, then rC5 is equal to (a) 23 (c) 56 16.

(b) 46 (d) None of these 5

47

C4 +

∑

52 − j

j =1

C3 =

(a) 52C4 (c) 52C3 15

(b) 51C4 (d) None of these

C8 + 15C9 – 15C6 – 15C7 =

(a) 8 (c) 6

(b) 0 (d) None of these

(a) r = 24 (c) r = 7

(b) 21 (d) 8C3

Pr = r! (b) 2n (d) 2n + 1

(b) 60 (d) 100

(b) r = 14 (d) None of these

19. If 3 ⋅ x + 1C2+ 2P2 ⋅ x = 4 ⋅  xP2, x ∈ N, then x is equal to (a) 2 (c) 4

(b) 3 (d) None of these

20. The number of positive terms in the sequence xn =

10. The number of arrangements of the letters of the word Banana in which two Ns do not appear adjacently is (a) 40 (c) 80

(1⋅ 3⋅ 5....(4n − 1)) 2 (1⋅ 3⋅ 5....(2n − 1))

Cr = (k2 – 3) ⋅ nCr+1, if k lies in

n

(a) 2n – 1 (c) 2n – 1

(b)

18. If 28C2r : 24C2r – 4 = 225 : 11, then

8. The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is (a) 38 (c) 5

(c)

(a) 3 (c) 5

17.

C4 > nC3, then

(1⋅ 3⋅ 5....(4n − 1)) (1⋅ 3⋅ 5....(2n − 1)) 2

14. The positive integer r, such that 15C3r = 15Cr + 3, is equal to

Pr + r ⋅ n – 1Pr – 1 =

(a) n + 1Pr (c) r ⋅  nPr

(a)

(a) [– 3 , 3 ] (c) (2, ∞)

Cr+1+ nCr–1 + 2 × nCr equals n+2

C2n : 2nCn =

12. 1P1 + 2 ⋅ 2P2 + 3 ⋅ 3P3 + ... + n ⋅ nPn =

n

(a) (c)

4n

195 − 4 n Pn

(a) 2 (c) 4

n+3 n +1

P3 , n ∈ N is Pn + 1 (b) 3 (d) None of these

21. The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places is (a) 24 (c) 12

(b) 18 (d) 30

(a) 360 (c) 72

(b) 114 (d) 54

23. A gentleman invites 13 guests to a dinner and places 8 of them at one table and remaining 5 at the other, the tables being round. The number of ways he can arrange the guests is 11! (a) (b) 9! 40 12! 13! (d) (c) 40 40 24. A man has got seven friends. The number of ways in which he can invite one or more of his friends to dinner, is (a) 116 (c) 127

(b) 128 (d) None of these

25. The chief minister of 11 states of India meet to discuss the language problem. The number of ways they can seat themselves at a round table so that the Punjab and Madras Chief Ministers sit together is (a) 725760 (c) 725778

(b) 725748 (d) None of these

26. The number of ways in which n things of which r are alike, can be arranged in a circular order, is n! (n − 1)! (a) (b) r! r! n! (c) (d) None of these (r − 1)! 27. The number of ways in which 6 different beads can be string into a necklace is (a) 60 (c) 72

(b) 48 (d) None of these

28. There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is (a) 1048 (c) 1072

(b) 1024 (d) None of these

29. The number of numbers greater than 50,000 that can be formed by using the digits 3, 5, 6, 6, 7, 9 is (a) 36 (c) 54

(b) 48 (d) None of these

30. The number of ways in which n books can be arranged on a shelf so that two particular books shall not be together is (a) (n – 2) (n – 1)! (c) (n – 2) n!

(b) (n – 1) n! (d) None of these

31. The number of ways in which the letters of the word “STRANGE” can be arranged so that the vowels may appear in the odd places, is (a) 1440 (c) 1370

(b) 1470 (d) None of these

(a) 3265920 (c) 3268620

(b) 3265860 (d) None of these

33. Ten 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 (a) 69760 (c) 99748

(b) 30240 (d) None of these

34. The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, 6 taken all at a time is (a) 432 (c) 36

(b) 108 (d) 18

35. If there are 12 persons in a party, and if each of them shakes hands with each other, then number of handshakes happen in the party is (a) 66 (c) 72

(b) 48 (d) None of these

36. The number of ways in which a committee of 5 can be chosen from 10 candidates so as to exclude the youngest if it includes the oldest, is (a) 196 (c) 202

(b) 178 (d) None of these

37. The number of ways in which 7 people can be arranged at a round table so that 2 particular persons may be together, is (a) 132 (c) 240

(b) 148 (d) None of these

38. Three boys and three girls are to be seated around a table, in a circle. Among them, the boy X does not want any girl neighbour and the girl Y does not want any boy neighbour. The number of such arrangements possible is (a) 4 (c) 8

(b) 6 (d) None of these

39. There are 5 gentlemen and 4 ladies to dine at a round table. The number of ways in which they can seat themselves so that no two ladies are together is (a) 1770 (c) 3990

(b) 2880 (d) None of these

40. The number of ways in which the letters of the word “VOWEL” can be arranged so that the letters O, E occupy only even places is (a) 12 (c) 24

(b) 18 (d) None of these

41. The number of ways in which the letters of the word ‘FRACTION’ be arranged so that no two vowels are together, is (a) 17330 (c) 16440

(b) 14400 (d) None of these

669

32. The total number of 8 digits numbers which have all different digits is

Permutations and Combinations

22. All the letters of the word ‘EAMCET’ are arranged in all possible ways. The number of such arrangements in which two vowels are not adjacent to each other is

670

42. The number of words that can be formed from the letters of the word DAUGHTER so that the vowels always come together, is

Objective Mathematics

(a) 4320 (c) 5230

(b) 3470 (d) None of these

43. In an examination there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all answers correct is (a) 11 (c) 27

(b) 12 (d) 63

44. There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is (a) 102 (c) 210

(b) 18 (d) 1023

45. 12 persons are to be arranged to a round table If two particular persons among them are not to be side by side, the total number of arrangements is

52. The number of ways in which two 10-paise, two 20-paise, three 25-paise and one 50-paise coins can be distributed among 8 children so that each child gets only one coin, is (a) 1720 (c) 1570

(b) 1680 (d) None of these

53. The number of ways in which the letters of the word BALLOON can be arranged so that two L’s do not come together, is (a) 700 (c) 900

(b) 800 (d) None of these

54. There are three copies each of 4 different books. The number of ways in which they be arranged on a shelf is (a)

12! (3!) 4

(b)

11! (3!) 2

(c)

9! (3!) 2

(d) None of these

55. When a group photograph is taken, all the seven teachers should be in the first row and all the twenty students should be in the second row. If the two corners of the 46. The number of ways in which a committee of 3 ladies and second row are reserved for the two tallest students, 4 gentlemen can be appointed from a meeting consisting interchangeable only between them, and if the middle of 8 ladies and 7 gentlemen, if Mrs X refuses to serve seat of the front row is reserved for the principal. The in a committee if Mr. Y is a member is no. of such possible ‘arrangements’ is (a) 1960 (b) 1540 (a) 720 × 18! (b) 1370 × 18! (c) 3240 (d) None of these (c) 1440 × 18! (d) None of these (a) 9 (10)! (c) 45 (8)!

(b) 2 (10)! (d) 10!

47. 4 letter lock consists of three rings each marked with 10 different letters, the number of ways in which it is possible to make an unsuccessful attempt to open the lock, is (a) 899 (c) 479

(b) 999 (d) None of these

48. A telegraph has 5 arms and each arm is capable of 4 distinct positions, including the position of rest. The total number of signals that can be made is (a) 473 (c) 1173

(b) 1023 (d) None of these

49. There are stalls for 10 animals in a ship. The number of ways the shipload can be made if there are cows, calves and horses to be transported, animals of each kind being not less than 10, is (a) 59049 (c) 69049

(b) 49049 (d) None of these

50. The number of arrangements which can be made out of the letters of the word ALGEBRA, without changing the relative order (positions) of vowels and consonants, is (a) 72 (c) 36

(b) 54 (d) None of these

51. The number of arrangements of the letters of the word “pencil” so that e always comes earlier than i, is (a) 290 (c) 360

(b) 340 (d) None of these

56. The number of divisors of 9600 including 1 and 9600 are (a) 60 (c) 48

(b) 58 (d) 46

57. The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is (a) 6 (c) 12

(b) 18 (d) 9

58. Out of 10 red and 9 white balls, 5 red and 4 white balls can be drawn in number of ways = (a) 10C5 × 10C4 (c) 18C9

(b) 10C5 × 8C4 (d) None of these

59. In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. The number of telephone numbers having all six digits distinct is (a) 8400 (c) 9200

(b) 7200 (d) None of these

60. The number of diagonals that can be drawn by joining the vertices of an octagon is (a) 28 (c) 20

(b) 48 (d) None of these

61. There are six teachers. Out of them two are primary teachers, two are middle teachers and two secondary teachers. They are to stand in a row, so as the primary

(a) (7!)2 (c) 6! × 7!

at least one vowel is

(b) (6!)2 (d) 7!

64. The number of ways in which a team of eleven players can be selected from 22 players including 2 of them and excluding 4 of them is (a) 16C11 (c) 16C9

(a) 96 (c) 106

74. The number of other ways in which the letters of the work SIMPLETON can be rearranged is

(b) 16C5 (d) 20C9

65. A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each containing 6 questions. He is not permitted to attempt more than 5 questions from each group. The number of ways in which he can choose the 7 questions is (a) 780 (c) 820

(b) 640 (d) None of these

66. In how many ways can 6 persons be selected from 4 officers and 8 constables, if at least one officer is to be included (a) 224 (c) 896

(b) 672 (d) None of these

67. If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent. The number of points in which they intersect each other is

(a) 8! (c) 9! – 1

(a) 6480 (c) 5862

(a) 864 (c) 766

(a) 11920 (c) 11760

(b) 11820 (d) None of these

(a) 230 (c) 340

68. For the post of 5 teachers there are 23 applicants, 2 posts 78. are reserved for S. C. candidates and there are 7 S.C. candidates among the applicants. The number of ways in which the selection can be made is

70. If m + nP2 = 90 and and n are (a) 8, 3 (c) 8, 2

m – n

P2 = 30, then the values of m (b) 7, 2 (d) None of these

(b) 932 (d) None of these

77. Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. The number of words which have at least one letter repeated is (a) 48370 (c) 69760

(b) 32 (d) None of these

(b) 7290 (d) None of these

76. The number of old numbers lying between 40000 and 70000 that can be made from the digits 0, 1, 2, 4, 5, 7 if digits can be repeated in the same number is

(b) 190 (d) None of these

(a) 41 (c) 51

(b) 9! (d) None of these

75. In a town the car plate numbers contain only there or four digits, not containing the digit 0. The maximum number of cars that can be numbered is

(a) 220 (c) 250

69. A boy has 3 Library tickets and 8 books of his interest in the library. Out of these 8, he does not want to borrow Chemistry part II, unless Chemistry part I is also borrowed. The number of ways in which he can choose the three books to be borrowed is

(b) 84 (d) None of these

(b) 58630 (d) None of these

In a network of railways, a small island has 15 stations. The number of different types of tickets to be printed for each class, if every station must have tickets for other station, is (b) 210 (d) None of these

79. There are n concurrent lines and another line parallel to one of them. The number of different triangles that will be formed by the (n + 1) lines, is (a) (n −1) n (b) (n − 1) (n − 2) 2 2 n ( n +1 ) ( n + 1 ) ( n + 2) (c) (d) 2 2 80. In a certain season 45 matches were played between the teams of the first division football league, each team playing once against each of other teams. The number of teams who were there in the first division, is

671

Permutations and Combinations

teachers, middle teachers and secondary teachers are 71. Eleven animals of a circus have to be placed in eleven always in a set. The number of ways in which they can cages one in each cage. If 4 of the cages are too small do so, is for 6 of the animals, then the number of ways of caging the animals is (a) 34 (b) 48 (c) 52 (d) None of these (a) 304800 (b) 504800 (c) 604800 (d) None of these 62. In a football championship, there were played 153 matches. Every two teams played one match with each other. 72. A person wishes to make up as many different parties as The number of teams, participating in the championship he can out of 20 friends, each party consisting of the are same number. The number of persons he should invite at a time is (a) 14 (b) 22 (c) 18 (d) None of these (a) 8 (b) 10 (d) None of these (c) 12 63. Seven women and seven men are to sit around a circular table such that there is a man on either side of every 73. The number of words that can be formed from the letters woman, the number of seating arrangements are a, b, c, d, e, f, taken 3 at a time, each word containing

672

(a) 8 (c) 10

(b) 12 (d) None of these

Objective Mathematics

81. The number of diagonals in a polygon of n sides is

90. The number of ways in which p positive and n negative signs may be placed in a row so that no two negative signs shall be together is

(b) pCn (a) p + 1Cn (a) n (n − 3) (b) n (n − 1) (c) p – 1Cn (d) None of these 2 2 91. The number of ways in which 10 boys and 10 girls can (c) (n − 1) (n − 2) (d) None of these be seated in a row so that boys and girls alternate is 2 82. In a plane there are two sets of parallel lines, one of m (a) 3628800 (b) 3628730 lines and the other of n lines. If the lines of one set (c) 3628860 (d) None of these cut those of the other, then the number of different 92. From 3 mangoes, 4 apples and 2 oranges. The number parallelograms that can be formed is of selections of fruits that can be made, taking at least one of each kind is m ( m − 1 ) ( n − 1 ) mn ( m − 1 ) ( n − 1 ) (a) (b) 6 2 (a) 24 (b) 36 nm ( m − 1 ) ( n − 1 ) (c) 42 (d) None of these (c) (d) None of these 4 93. The total number of selections of at least one red ball 83. If each of 10 points on a straight line be joined to each from a bag containing 4 red balls and 3 green balls, of 10 points on a parallel line then the total number of balls of the same colour being identical, is triangles that can be formed with the given points as (a) 32 (b) 24 vertices, is (c) 16 (d) None of these (a) 860 (b) 900 94. There are p copies each of n different books. The number (c) 920 (d) None of these of different ways in which a non-empty selection can 84. A father with 8 children takes them 3 at a time to the be made from them, is zoological garden, as often as he can without taking the (b) pn – 1 (a) ( p + 1)n – 1 same 3 children together more than once. Then n–1 (c) ( p + 1) – 1 (d) None of these (a) number of times he will go is 56 95. The number of proper divisors of 1260, is (b) number of times each child will go is 21 (b) 26 (a) 42 (c) number of times a particular child will not go is 35 (c) 34 (d) None of these (d) all of these. 85. In an examination a candidate has to pass in each of the 96. papers to be successful. It the total number of ways to fail is 63, how many papers are there in the examination? 97. (a) 6 (b) 8 (c) 14 (d) None of these 86. The total number of selections from 4 boys and 3 girls if each selection has to contain at least one boy is

The number of proper divisors of 2520 is (a) 46 (c) 64

(b) 52 (d) None of these

Four boys picked 30 apples. The number of ways in which they can divide them if all the apples are identical, is (a) 5630 (c) 5456

(b) 4260 (d) None of these

98. A family consists of grandfather, 5 sons and daughters and 8 grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the 4 seats at each end and the grandfather refuses to have a grand 87. From 17 consonants and 5 vowels. The number of words child on either side of him. The number of ways in containing 3 consonants and 2 vowels that can be made which the family can be made to sit is if all the letters are different, is (a) 106 (c) 240

(b) 120 (d) None of these

(a) 816000 (c) 926000

(b) 736000 (d) None of these

88. From 5 consonants, 4 vowels and 3 capital letters, the number of words beginning with a capital letter and containing 3 consonants and 2 vowels that can be made is

(a) 11360 (c) 21530

(b) 11520 (d) None of these

99. A teaparty is arranged for 16 people along two sides of a large table with 8 chairs on each side. Four men want to sit on one particular side and two on the other side. The number of ways in which they can be seated is

6! 8!10! 8! 8!10! (b) 4! 6! 4! 6! 8! 8! 6! 89. A boat is to be manned by eight men of whom 2 can only (d) None of these (c) row on bow side and 1 can only row on stroke side; the 6! 4! number of ways in which the crew can be arranged is 100. The number of ways in which a mixed doubles game in tennis can be arranged from 5 married couples, if no (a) 4360 (b) 5760 husband and wife play in the same game, is (c) 5930 (d) None of these (a) 18880 (c) 21680

(b) 21600 (d) None of these

(a)

101. The number of seven letter words that can be formed by using the letters of the word SUCCESS so that the two C are together but no two S are together is (a) 24 (c) 54

(b) 36 (d) None of these

102. The sum of all five digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 (repitition of digits not allowed) (a) 3999960 (c) 3999963

(b) 3999954 (d) None of these

112. In a class tournament where the participants were to play one game with another, two class players fell ill, having played 3 games each. If the total number of games played is 84, the number of participants at the beginning was (a) 22 (c) 17

(b) 15 (d) None of these

113. The number of ways in which 10 examination papers can be arranged so that the best and the worst papers never come together, is (a) 9.9! (c) 4.9!

(b) 8.9! (d) None of these

103. The number of rectangles excluding squares from a 114. In a class of 10 students, there are 3 girls. The number of different ways in which they can be arranged in a rectangle of size 9 × 6 is row such that no two of the three girls are consecutive, (a) 391 (b) 791 is (c) 842 (d) None of these (a) 42.8! (b) 36.8! 104. The number of zeros at the end of 100! is (c) 24.8! (d) None of these (a) 36 (b) 18 115. On a new year day every student of a class sends a (c) 24 (d) None of these card to every other student. The postman delivers 600 105. The largest integer n such that 33! is divisible by 2n is (a) 30 (c) 32

(b) 31 (d) None of these

106. Four boys picked up 30 mangoes. The number of ways in which they can divide them if all mangoes be identical, is (a) 5456 (c) 5462

(b) 3456 (d) None of these

107. The product of r consecutive positive integers is divisible by (a) r! (c) (r + 1)!

(b) (r – 1)! (d) None of these

108. If 7 points out of 12 are in the same straight line, then the number of triangles formed is (a) 19 (c) 201

(b) 185 (d) None of these

109. Ten 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 (a) 69760 (c) 99748

(b) 30240 (d) None of these

110. In a certain test, ai students gave wrong answers to at least i questions where i = 1, 2, 3, ..., k. No student gave more than k wrong answers. The total number of wrong answers given is (a) a1 + a2 + ... + ak (c) a1 + a2 + ... + ak + 1

(b) a1 + a2 + ... + ak – 1 (d) None of these

111. The number of ordered triplets of positive integers which are solutions of the equation x + y + z = 100 is (a) 5081 (c) 4851

(b) 6005 (d) None of these

cards. The number of students in the class are (a) 42 (c) 25

(b) 34 (d) None of these

116. In an examination a minimum is to be secured in each of 5 subjects for a pass. The number of ways in which a student can fail is (a) 31 (c) 42

(b) 36 (d) None of these

117. A parallelogiam is cut by two sets of m lines parallel to its sides. The number of parallelogram then formed is (a) (mC2)2 (b) (m + 1C2)2 (c) (m + 2C2)2 (d) None of these 118. Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. The number of straight lines that can be formed by joining them is (a) 143 (c) 153

(b) 144 (d) None of these

119. The number of ways in which r letters can be posted in n letterboxes in a town is (a) nr (c) nPr

(b) rn (d) nCr

120. The number of lines drawn through 6 points lying on a circle is (a) 12 (c) 24

(b) 15 (d) 30

121. The least positive integral value of n which satisfies the inequality 10Cn – 1 > 2 · 10Cn is (a) 7 (c) 9

(b) 8 (d) 10

673

(b) 54 (d) None of these

Permutations and Combinations

(a) 46 (c) 60

674

122. The expansion nC r + 4. nC r–1 + 6. nC r–2 + 4. nC r– 3 + n Cr–4=

Objective Mathematics

(a) n + 4Cr (c) 4.nCr

(b) 2.n+4Cr–1 (d) 11.nCr.

123. The maximum number of points into which 4 circles and 4 straight lines intersect is (a) 26 (c) 56

(b) 50 (d) 72

124. The sum of all numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digit repeated is (a) 16 × 4! (c) 16 × 1111 × 3!

(b) 1111 × 3! (d) 16 × 1111 × 4!

125. The number of permutations of n different things, taken r at a time, in which p (r ≤ n – p) particular things will never occur is (a) P (n – p, r) (c) P (n,r) × P (n,p)

(b) P (n,r) – P (n,p) (d) P (n – p, r) × P (n,n – p)

126. There are n different books and m copies of each in a college library. The number of ways in which a student can make a selection of one or more book is (a) (m + 1)n – 1 mn! (c) (m!) m

(b) (m + 1)n – 1 (d) mnCn × nCn

127. The number of permutations of n different objects taken r (≥ 3) at a time which include 3 particular objects is (a) nPr × 3! (c) n–3Pr–3 × 3!

(b) nPr–3 × 3! (d) rP3 × n–3Pr – 3

128. There are n seats round a table numbered 1, 2, 3, ...n. The number of ways in which m(≤n) persons can take seats is (a) nPm 1 (c) .n Pm 2

(b) nCm × (m – 1)! (d) n –1Pm

129. Out of 18 points in a plane no three are in the same straight line except five points which are collinear. The number of straight lines that can be formed joining them is (a) 143 (c) 153

(b) 144 (d) None of these

130. The straight lines I1, I2, I3 are parallel and lie in the same plane. A total numbers of m points are taken on I1, n points on I2, k points on I3. The maximum number of triangles formed with vertices at these points are (a) m+n+kC3 (c) mC3 + nC3 + kC3

(b) m+n+kC3 – mC3 – nC3 – kC3 (d) None of these

131. The number of divisors of 9600 including 1 and 9600 are (a) 60 (c) 48

(b) 58 (d) 46

132. The maximum number of points of intersection of 8 straight lines is

(a) 56 (c) 16

(b) 28 (d) 8

133. The number of ways in which 6 red roses and 3 white roses can form a garland so that no two white roses come together is (a) 7255 (c) 7215

(b) 7200 (d) 7235

134. Twelve students compete for a race. The number of ways in which first three places can be taken as (a) 12! – 3 (c) 12.11.10

12! 3! (d) 3! (b)

135. An examination paper contains 8 question of which 4 has 3 possible answers, 3 have 2 possible answers each and the remaining one question has 5 possible answers. The total number of possible answers to all the questions is (a) 2880 (c) 94

(b) 78 (d) 3240

136. The total number of numbers greater than 100 and divisible by 5, that can be formed from the digits 3, 4, 5, 6 if no digit is repeated is (a) 24 (c) 36

(b) 48 (d) 12

137. There are 10 true-false questions. The number of ways, in which they can be answered is (a) 102 (c) 10

(b) 210 (d) 10!

138. The maximum number of points of intersection of 8 circles is (a) 56 (c) 24

(b) 28 (d) 16

139. The number of ways in which 6 red roses and 3 white roses can form a garland so that all the white roses come together is (a) 2160 (c) 2165

(b) 2155 (d) 2170

140. If C (n, 10) = C (n, 12), then n is equal to (a) 22 (c) 10 × 12

(b) 2 (d) None of these

141. The value of P (n, 1) + P(n, 2) + P(n, 3) + ... + P(n, n) is 2! 3! n! equal to (a) 2n + 1 (c) 2n – 1

(b) 2n – 1 (d) 2n

142. The number of signals that can be sent by 5 flags of different colours taking one or more at a time is (a) 300 (c) 450

(b) 225 (d) 325

143. The number of different words that can be formed from the letters of the word ‘pencil’ so that no two vowels are together is (a) 120 (c) 240

(b) 260 (d) 480

(a) 28 – 2 (c) 28 + 1

(b) 28 – 1 (d) 28

145. A polygon has 44 diagonals. The number of its sides are (a) 13 (c) 11

(b) 12 (d) 10

146. A library has a copies of one book, b copies of two books, c copies of each of three books and single copy of d books The total number of ways in which these books can be distributed is (a + 2b + 3c + d )! (a) a ! b! c!

(a + b + c + d )! (b) a ! b! c!

(a + 2b + 3c + d )! (c) a !(b!) 2 (c !)3

(d) None of these

147. If there are n number of seats and m number of people have to be seated, then how many ways are possible?

9! 2!+ 3!+ 4!

(b)

9! 2! 3! 4!

(c) 2! + 3! + 4! (2! 3! 4!) (d) 2! 3! – 4 154. The number of straight lines that can be drawn out of 10 points of which 7 are collinear (a) 23 (c) 25

(b) 21 (d) 24

155. In a steamer there are stalls for 12 animals and there are cows, horses and calves (not less than 12 of each) ready to be shipped. The total number of ways in which the ship load can be made is (a) 12C3 (c) 312

(b) 12P3 (d) 123

156. If C (10, 4) + C (10, 5) = C (11, r), then r equals (a) 6 (c) 4

(b) 5 (d) 3

157. The number of ways in which n ties can be selected from a rack displaying 3n different ties is (b) nCm (a) nPm 3n! (a) 3 × n! (b) (c) nCn × (m – 1)! (d) n–1Pm–1 n ! 2n ! 148. On a railway route there are 15 stations. The number 3n! (c) (d) 3n! of tickets required in order that it may be possible to 2n ! book a passenger from every station to every other 158. The number of ways in which 5 boys and 5 girls can is sit in a row so that all girls sit together is 15! 15! (a) 12600 (b) 7200 (b) (a) 13! 2! 13! (c) 86400 (d) 14400 15! 2! 149. If eight persons are to address a meeting then the number of ways in which a specified speaker is to speak before another specified speaker, is (c) 15!

(a) 40320 (c) 20160

(d)

(b) 2520 (d) None of these

150. How many numbers of 6 digits can be formed from the digits of the number 112233? (a) 30 (c) 90

(b) 60 (d) 120

151. Among 15 players, 8 are batsmen and 7 are bolwers, the probability that a team is chosen of 6 batsmen and 5 bowlers is 8

(a)

C6 x 7C5 15 C11

28 (b) 15

15 (c) (d) None of these 28 152. Ashok, Usha, Rani, Sonu are to give speeches in a class. The teacher can arrange the order of their presentation in (a) 12 ways (c) 4 ways

(b) 24 ways (d) 256 ways

153. How many different arrangements can be made out of the letters in the expansion A2 B3 C4, when written in full?

159. In a cricket championship there are 36 matches. The number of teams, if each plays one match with other are (a) 9 (c) 8 160. n – 2Cr + 2. n+1

(a) Cr (c) nCr + 1

(b) 10 (d) None of these n–2

Cr – 1 +

n–2

Cr – 2 equals (b) nCr (d) n – 1Cr

161. Given five line segments of length 2, 3, 4, 5, 6 units. Then the number of triangles that can be formed by joining these lines is (a) 5C3 – 3 (c) 5C3

(b) 5C3 – 1 (d) 5C3 – 2

162. How many numbers greater than 24000 can be formed by using digits 1, 2, 3, 4, 5 when no digit is repeated, is (a) 36 (c) 84

(b) 60 (d) 120

163. The number of words which can be made out of the letters of the word “MOBILE” when consonants always occupy odd places, is (a) 20 (c) 30

(b) 36 (d) 720

164. Among 14 players, 5 are bowlers. In how many ways a team of 11 may be formed with at least 4 bowlers? (a) 265 (c) 264

(b) 263 (d) 275

675

(a)

Permutations and Combinations

144. The value of (7C0 + 7C1) + (7C1 + 7C2) + ... + (7C6 + 7 C7) is

676

165. How many different nine digit numbers can be formed from the number 22 33 55 8 88 by rearranging its digits so that the odd digits occupy even positions?

Objective Mathematics

(a) 16 (c) 60

(b) 36 (d) 180

166. 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 (a) 6 (c) 8

(b) 7 (d) 9

167. 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 (a) 9 (c) 8

(b) 13 (d) None of these

168. A five digit number divisible by 3 is to be formed using the numerals 0, 1, 3, 4 and 5 without repetition. The total number of ways this can be done is (a) 216 (c) 240

(b) 600 (d) 3125

169. The total number of ways in which six ′+′ and four ′–′ signs can be arranged in a line such that no two ‘–’ signs occur together is (a) 26 (c) 34

(b) 35 (d) None of these

170. A box contains two white balls, three black balls and four red balls. The number of ways in which three balls can be drawn from the box if atleast one black ball is to be included in the draw, is (a) 32 (c) 128

(b) 64 (d) None of these

171. The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (a) 66666 (c) 93324

(b) 84844 (d) None of these

172. Sita has 5 coins each of the different denomination. The number of different sums of money she can form is (a) 32 (c) 31

(b) 25 (d) None of these

173. The number of ways in which the six faces of a cube be painted with six different colours is (a) 6 (c) 6C2

(b) 6! (d) None of these

174. Number of ways in which the letters of the word ARRANGE can be arranged such that both R do not come together is (a) 360 (c) 1260

(b) 900 (d) 1620

175. Out of 6 boys and 4 girls, a group of 7 is to be formed. In how many ways can this be done, if the group is to have a majority of boys?

(a) 120 (c) 100

(b) 90 (d) 180

176. There are 5 roads leading to a town from a village. The number of different ways in which a villager can go to the town and return back, is (a) 25 (c) 10

(b) 20 (d) 5

177. Every body in a room shakes hands with every body else. The total number of hand shakes is 66. The total number of persons in the room is (a) 11 (c) 13 178. mCr + 1 +

(b) 12 (d) 14 n

∑

k

k =m

Cr =

(a) nCr + 1

(b) n + 1Cr + 1

(c) Cr

(d) None of these

n

179. From 7 gentlemen and 4 ladies, a committee of 5 is to be formed. The number of ways in which this can be done so as to include atleast one lady is (a) 805 (c) 754

(b) 403 (d) None of these

180. The total number of ways of selecting five letters from the letters of the word INDEPENDENT is (a) 72 (c) 56

(b) 64 (d) None of these

181. The least positive integral value of x which satisfies the inequality 10Cx – 1 > 2. 10Cx is (a) 7 (c) 9

(b) 8 (d) 10

182. Two straight lines intersect at a point O. Points A1, A2, ..., An are taken on one line and points B1, B2, ..., Bn on the other. If the point O is not to be used, the number of triangles that can be drawn using these points as vertices, is (a) n (n – 1) (c) n2 (n – 1)

(b) n (n – 1)2 (d) n2 ( n – 1)2

183. The number of positive integers satisfying the inequality Cn – 2 – n + 1Cn – 1 ≤ 100 is

n+1

(a) nine (c) five

(b) eight (d) None of these

184. We are required to form different words with the help of the letters of the word INTEGER. Let m1 be the number of words in which I and N are never together and m2 be the number of words which begin with I and end with R, then m1\m2 is given by (a) 42 (c) 6

(b) 30 (d) 1/30

185. If a denotes the number of permutations of x + 2 things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of x – 11 things taken all at a time such that a = 182bc, then the value of x in

(a) 144 (c) 288

(b) 72 (d) 720

195. If S = (1) (1!) + (2) (2!) + (3) (3!) + ... + n (n!) then

(a) n/(S + 1) (b) (n + 1)/(S + 1) 186. The number of divisors a number 38808 can have, (c) (n + 1)!/(s + 1) (d) None of these excluding 1 and the number itself is 196. The ten digits of 1! + 2! + 3! + ... + 49! is (a) 70 (b) 72 (a) 1 (b) 2 (c) 71 (d) None of these (c) 3 (d) 4 187. The number of six digit numbers that can be formed 197. Let A = {x | x is a prime number and x < 30}. The from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do number of different rational numbers whose numerator not repeat and the terminal digits are even, is and denominator belong to A is (a) 90 (c) 91

(b) 180 (d) None of these

188. In a certain test, these are n questions. In this test 2n – i 198. Let S be he set of all functions from the set A to the students gave wrong answers to at least i questions, set A. If n (A) = k then n (S) is where i = 1, 2, 3, ..., n. If the total number of wrong (a) k! (b) kk answers given is 2047, then n is equal to k (c) 2 – 1 (d) 2k (a) 10 (b) 11 199. The number of numbers of 9 different nonzero digits (c) 12 (d) 13 such that all the digits in the first four places are less 189. The number of different 7 digit numbers that can be than the digit in the middle and all the digits in the last written using only the three digits 1, 2 and 3 with four places are greater than that in the middle is the condition that the digit 2 occurs twice in each (a) 2 (4!) (b) (4!)2 number is (c) 8! (d) None of these (a) 7P2 25 (b) 7C2 25 200. ABCD is a quadrilateral. 3, 4, 5 and 6 points are marked (c) 7C2 52 (d) None of these on the sides AB, BC, CD and DA respectively. The 190. A set contains (2n + 1) elements. If the number of number of triangles with vertices on different sides is subsets of this set which contain at most n elements is (a) 270 (b) 220 4096, then the value of n is (c) 282 (d) None of these (a) 6 (b) 15 201. There are 10 points in a plane of which no three points (c) 21 (d) None of these are collinear and 4 points are concyclic. The number of 191. Let p be a prime number such that p ≥ 11. Let n = p! different circles that can be drawn through at least 3 of + 1. The number of primes in the list n + 1, n + 2, n these points is + 3, ..., n + p – 1, is (a) 116 (b) 120 (a) p – 1 (b) 2 (c) 117 (d) None of these (c) 1 (d) None of these 202. The number of ways in which n different prizes can be 192. Five distinct letters are to be transmitted through a comdistributed among m (< n) persons if each is entitled to munication channel. A total number of 15 blanks is to receive at most n – 1 prizes, is be inserted between the two letters with at least three (a) nm – n (b) mn between every two. The number of ways in which this (c) mn (d) None of these can be done is 203. Total number of 6-digit numbers in which all the odd (a) 1200 (b) 1800 digits and only odd digits appear, is (c) 2400 (d) 3000 5 (a) (6!) (b) 6! 193. If N is the number of positive integral solutions of x1 2 x2 x3 x4 = 770. Then 1 (d) None of these (c) (6!) (a) N is divisible by 4 distinct primes 2 (b) N is a perfect square 204. The number of numbers greater than a million that can (c) N is a perfect 4th power be formed with the digits 2, 3, 0, 3, 4, 2 and 3 is (d) N is a perfect 8th power. (a) 360 (b) 340 1 1 1 2 (c) 370 (d) None of these 194. Let E =  +  +  +  + ... + up to 50 terms,  3 50   3 50  205. Ten different letters of an alphabet are given, words then with five letters are formed from these given letters. The number of words which have at least one letter (a) E is divisible by exactly 2 primes repeated, is (b) E is prime (c) E ≥ 30 (b) 105 – 10P4 (a) 105 – 10P5 (d) E ≤ 35 (c) 10P5 (d) None of these

677

(b) 12 (d) 18

Permutations and Combinations

(a) 15 (c) 10

678

206. The number of ways in which 16 identical things can be distributed among 4 persons if each person gets at least 3 things, is

Objective Mathematics

(a) 33 (c) 38

(c)  

(b) 35 (d) None of these

207. The number of ways in which 30 marks can be alloted to 8 questions if each question carries at least 2 marks, is (a) 115280 (c) 116280

(b) 117280 (d) None of these

208. The number of permutations of the letters of the word ‘CONSEQUENCE’ in which all the three E’s are together, is

9! 2!2! 9! (d)   2!3!

(a)  9!3!

(b)  

9! 2!2!3!

209. 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 (a)  360 (c)  96

(b)  192 (d)  48

SOLUTIONS 1. (b) The number of choices available to him is

Also, nCr = nCr – 1 ⇒ r + r – 1 = n

= 5C 4 × 8C 6 + 5C 5 × 8C 5

⇒ 2r – n = 1 7. (c) We have,

C 4 > nC 3 n! n! ⇒ nC4 > nC3 ⇒ > (n − 4)! 4! (n − 3)! 3!

= 5 × 4 × 7 + 8 × 7 = 140 + 56 = 196. 2. (a) The number of ways to sit men = 5! and the number of ways to sit women = 6C5 × 5! Total number of ways = 5! × 6C5 × 5! = 5! × 6 × 5! = 6! 5!. 3. (b) nCr+1 + nCr–1 + 2 . nCr

=

n+2

n+1

Cr+1 +

n+1

(n − 1)! r (n − 1)! + (n − r + 1)! (n − r )!

=

(n − 1)!  r  1+ (n − r − 1)!  n − r 

C3 +

n – 1

1 1 > 4 ⋅ 3!(n − 4)! 3!(n − 3)(n − 4)!

1 ⇒ 1 > n−3 4 ⇒ n – 3 > 4 ⇒ n > 7.

⇒ and ∴

4. (b) We have, Pr + r n – 1Pr – 1 =

⇒

n – 1

8. (b) Let the number of balls in the 3 boxes be x, y, z respectively, then x + y + z = 8, x, y, z ≥ 1

Cr

Cr+1.

n – 1

...(2)

Solving (1) and (2), we get r = 2 and n = 3.

5! 8! 5! 8! × + × 4!1! 6! × 2! 5!0! 5!3! 8×7 8×7×6 =5× +1× 2 3× 2 =

= nCr+1 + nCr + nCr–1 + nCr=



a + b + c = 5, where a = x – 1, b = y – 1, c = z – 1 a, b, c, are non-negative integers. Required number =

n+r–1

9. (a) We have,

∑

Cr =

n

r =1

[ (n – r)! = (n – r) (n – r – 1)!] n (n − 1)! = (n − r ) ⋅ (n − r − 1)!

n

Cr = r! n



=

∑

r =1



3+3–1

n

C5 = 7C5 = 7C2 = 21. n

n!

∑ (n − r )! r ! r =1

Cr

= C1 + C2 + .... + Cn = 2n – 1.

5. (a)

10. (a) Required number of arrangements = (total number of arrangements) – (number of arrangements in which 6! 5! Ns are together) = = 60 – 20 = 40. − 2!× 3! 3!

6. (b) We have, nPr = nPr + 1

11. (a) We have,

=

n! = nP r. (n − r )!

⇒

n! n! 1 = ⇒ =1 (n − r − 1)! (n − r )! (n − r )

or n – r = 1

= ...(1)

4n

4n ! n! n! C2 n × = 2n ! 2n ! 2n ! Cn

2n

{1⋅ 2 ⋅ 3⋅ 4 ⋅ 5 ⋅ 6....(4n − 1) 4n} n! n! (1⋅ 2 ⋅ 3⋅ 4 ⋅ 5 ⋅ 6.... 2n) 2 2n!

=

{1⋅ 3⋅ 5...(4n − 1)}{2 ⋅ 4 ⋅ 6....4n} n! n! {1⋅ 3⋅ 5...(2n − 1)}2 {2 ⋅ 4 ⋅ 6....2n}2 2n!

=

{1⋅ 3⋅ 5...(4n − 1)} 22 n {1⋅ 2 ⋅ 3....2n} n! n! {1⋅ 3⋅ 5...(2n − 1)}2 22 n {1⋅ 2 ⋅ 3....n}2 2n!

⇒

=

{1⋅ 3⋅ 5...(4n − 1)} 22 n 2n! n! n! {1⋅ 3⋅ 5...(2n − 1)}2 22 n (n!) 2 2n!

(28)! (28 − 2r )!(2r − 4)! = 225 × (28 − 2r )! 2r ! (24)! 11

⇒

28 × 27 × 26 × 25 = 225 2r (2r − 1) (2r − 2) (2r − 3) 11

{1⋅ 3⋅ 5...(4n − 1)} {1⋅ 3⋅ 5...(2n − 1)}2 12. (b) rth term = r ⋅ rPr = r ⋅ r! = (r + 1) – 1) r!. =

= (r + 1)! – r! Putting r = 1, 2, 3, ... n and adding, we get 1 P1 + 2 ⋅ 2P2 + 3 ⋅ 3P3 + ... + n ⋅ nPn = (2! – 1!) + (3! – 2!) + (4! – 3!) +  ... + ((n +1)! – n!) = (n +1)! – 1! = n + 1Pn + 1 – 1. 13. (d) n–1Cr = (k2 – 3) ⋅ nCr+1 ⇒ Since

r +1 = k2 – 3 n

r +1 ≤ 1, k ∈ [– 2, 2] and k2 – 3 > 0 n

∴ k ∈ ( − ∞, − 3 )   ∪ ( 3 , ∞ ) . Hence k ∈ [ 3 , 2] . 14. (a) We have, 15C3r = 15Cr + 3 ⇒ 3r = r + 3, or 3r + (r + 3) = 15 ⇒ 2r = 3, or 4r = 12 ⇒ r = 3 , or r = 3. 2 But r must be a positive integer, so r = 3. 15. (c) We have, 18Cr = 18Cr + 2 ⇒ r = r + 2, or r + (r + 2) = 18

5



= = = = =

∑

52 − j

j =1

C3 = 47C4 + 51C3 + 50C3 + 49C3 + 48C3 + 47C3

(47C4 + 47C3) + 48C3 + 49C3 + 50C3 + 51C3 48 C4 + 48C3 + 49C3 + 50C3 + 51C3 49 C4 + 49C3 + 50C3 + 51C3 50 C4 + 50C3 + 51C3 51 C4 + 51C3 = 52C4.

17. (b) We have, 15

C8 + 15C9 – 15C6 – 15C7



= (15C8 + 15C9) – (15C6 + 15C7)



= 16C9 – 16C7



= 16C9 – 16C9 = 0.

(

n

Cr + nCr + 1 (

n

679

⇒ 2r (2r ­– 1) (2r ­– 2) (2r ­– 3) = 11 × 28 × 27 × 26 × 25 225 = 11 × 28 × 3 × 26 = 11 × 14 × 2 × 3 × 13 × 2 = 11 × 12 × 13 × 14 = 14 (14 – 1) (14 – 2) (14 –3) ∴ 2r = 14 or r = 7. 19. (b) We have, 3 ⋅ x + 1C2+ 2P2 ⋅ x = 4 ⋅  xP2 ⇒

3⋅ ( x + 1) x + 2! ⋅ x = 4 ⋅ x (x – 1) 1⋅ 2

⇒ 3x2 + 3x + 4x = 8x2 – 8x ⇒ 5x2 – 15x = 0 ⇒ 5x (x – 3) = 0 Since, x ≠ 0, ∴ x = 3. 20. (c) We have, xn =

145 − 4 ⋅ n Pn

(

x ∈ N)

n+3 n +1

P3 Pn +1

195 (n + 3) (n + 2) (n + 1) − 4 ⋅ n! (n + 1)!



=



=–

195 (n + 3) (n + 2) − 4 ⋅ n! n!



16. (a) We have, C4 +

C2 r 225 = C2 r − 4 11

2 = 195 − 4n − 20n − 24 4 ⋅ n! 171 − 4n 2 − 20n = 4 ⋅ n! xn is positive.

⇒ 0 = 2 (absurd) or r = 8 8! 8×7×6 ∴ r P5 = 8C5 = = = 56. 5! 3! 3× 2

47

28 24

∴

171 − 4n 2 − 20n > 0 ⇒ 4n2 + 20n – 171 < 0 4 ⋅ n!

which is true for n = 1, 2, 3, 4. Hence, the given sequence has 4 positive terms.

21. (b) The 4 odd digits 1, 3, 3, 1 can be arranged in the 4! 4 odd places in = 6 ways. 2! 2! and even digits 2, 4, 2 can be arranged in the three 3! even places in = 3 ways. 2! Hence the required number of ways = 6 × 3 = = n + 1Cr + 1) 18.

Cr = nCn – r)

22. (c) First, we arrange 3 consonants in 3! ways and then at four places (two places between them and two places on two sides) 3 vowels can be placed in

Permutations and Combinations

18. (c) We have,

680

Objective Mathematics

1 ways. 2! Hence the required number



P (4, 3) ×

1 4! 1 × =6× 2! 1! 2



= 3! × P (4, 3) ×



= 6 × 24 × 1 = 72. 2

23. (d) The number of ways in which 13 guests may be 13! divided into groups of 8 and 5 = 13C5 = . 5! 8!

= (n – 1)! Thus x · r! = (n – 1)! Hence, x =

(n − 1)! r!

27. (a) 6 different beads can be arranged in a circle in (6 – 1)! = 5! ways. But in case of circular arrangements of beads to form a necklace, clockwise and anticlockwise orders cannot be distinguished. Hence, the required number of ways

=

1 1 × 5! = × 120 = 60. 2 2

 ow, corresponding to one such group, the 8 guests N may be seated at one round table in (8 – 1)! i.e., 28. (b) Each man can vote for any one of the 4 candidates and this can be done in 4 ways. 7! ways and the five guests at the other table in (5 – 1)! i.e., 4! ways. Similar is the case with every other man. But each way of arranging the first group of 8 per[ Repetition is allowed] sons can be associated with each way of arranging Hence 5 men can vote in 45 i.e., 1024 ways, the second group of 5, therefore, the two processes can be performed together in 7! × 4! ways. 29. (b) Since any number greater than 50,000 contains 5 Hence required number of arrangements digits 13! 13! 13! ∴ We are to take all the 5 digits out of which 6 × 7! × 4! = × 7! × 4! = . = is repeated twice. 5!8! 5.4!8.7! 40 5! 24. (c) Number of friends = 7. = 60. ∴ Number of permutations = 2! The men can invite one friend, two friends ...., or But out of these arrangements, we have to reject seven friends. those which begin with 3 because 3 at the extreme ∴ Required number of ways left gives a number less than 50,000. 7 7 7 7 7 7 7 = C1 + C2 + C3 + C4 + C5 + C6 + C7 Fixing at the extreme left, we have 7 7x 6 7x 6x 5 7×6×5 7×6 7 3×××× + + + + +1 = + 1 1x 2 1x 2 x 3 1× 2 × 3 1× 2 1 Now the remaining 4 digits out of which 6 is repeated twice can be arranged among themselves in = 7 + 21 + 35 + 35 + 21 + 7 + 1 = 127. 4! i.e., 12 ways. 25. (a) Treat the Panjab and Madras Chief Ministers as one 2! (P, M) + 9 others. ∴ The required no. of numbers = 60 – 12 = 48. ∴ We have to arrange 10 persons round a table. 30. (a) Given number of books are n, which can be arranged This can be done in (10 – 1)! = 9! ways. Corin = n! ways. responding to each of these 9! ways, the Panjab Consider the two particular books as one book. and Madras Chief Ministers can interchange ∴ n – 1 books can be arranged in (n – 1)! ways. their places in 2! ways. Associating the two operations, total number of ways Also the two books can be arranged in themselves in 2! ways. = 9! × 2! ∴ No. of ways in which two particular books are = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) × (2 × 1) always together = 725760. (n – 1) 1 × 2! = 2(n – 1)! ...(1) 26. (b) Let the required number of arrangements = x ∴ No. of ways in which two particular books are Consider any one of these x arrangements. never together If, in this arrangement, r alike things are replaced = n! – 2 ⋅ (n – 1)! = n (n ­– 1)! – 2 (n – 1)! by r different things, then these r things can be = (n – 2) (n – 1)! arranged among themselves in r! ways. Since one arrangement gives rise to r! arrange- 31. (a) There are 5 consonants and 2 vowels in the word ments STRANGE. Out of 7 places for the 7 letters, 4 places are odd and 3 places are even. ∴ x arrangements will give rise to xr! arrangements. 2 vowels can be arranged in 4 odd places in P (4, 2) ways = 12 ways and then 5 consonants can be arranged But if all the n things are different, then the number in the remaining 5 places in P(5, 5) ways of circular arrangements

32. (a) The number is to be of 9 digits, the first place can be filled in 9 ways only (as 0 cannot be in the leftmost position). Having filled up the first place, the remaining 8 places can be filled up by the remaining 9 digits in 9! 9 P8 = ways = 9! ways. (9 − 8)! Hence the required no. of numbers = 9 × 9! = 9 × (9 × 8 × 7 × 6!) = 8! × 56 × 6! = 8! × 56 × 720 = 3265920. 33. (a) No. of words in which all the 5 letters are repeated = (10)5 = 100000. a nd the no. of words in which no letter is repeated = 10P5 = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 = 30240. Thus the no. of words which have at least one letter repeated = 100000 – 30240 = 69760.

38. (a) As shown in figure, 3, 1, 2 and X are the three boys and 3, 4 and Y are three girls, Boy X will have neighbours as boys 1 and 2 and the girl Y will have neighbors as girls 3 and 4.] 1 and 2 can be arranged in P (2, 2) ways

= 2! = 2 × 1 = 2 ways.

Also 3 and 4 can be arranged in P (2, 2) ways = 2! = 2 × 1 = 2 ways. Hence required no. of permutations = 2 × 2 = 4.

34. (b) The total number of numbers that can be formed with the digits 3, 4, 5, 6 taken all at a time = P (4, 4) = 4! = 24. Each of the digits 3, 4, 5, 6 occurs in 3! = 3 × 2 = 6 times in unit’s place. ∴ Sum of the digits in the unit’s place of all the numbers = (3 + 4 + 5 + 6) × 6 = 18 × 6 = 108 35. (a) Total number of handshakes  The number of ways of selecting 2 persons from = among 12 persons

= 12C2 =

12 × 11 = 66. 2 ×1

36. (a) There are two different ways of forming the committee (i) oldest may be included (ii) oldest may be excluded (i) If oldest is included, then youngest has to be excluded and we are to select 4 candidates out of 8. This can be done in

8

C4 =

8! 8×7×6×5 = = 70 ways. 4! 4! 4×3× 2

(ii) If oldest is excluded, then we are to select 5 candidates from 9 which can be done in

39. (b) Even number can be formed by using either 2 or 4 in the unit’s place. ∴ The unit’s place can be filled up in 2 ways. Corresponding to each of these 2 ways, the remaining 4 places can be filled with the remaining 4 digits (in which 5 appears ‘2’ times and each of the others one time) in 4! 4 × 3 × 2! = = 12 ways. 2!1!1! 2! Hence the required number of even numbers = 2 × 12 = 24. 40. (a) V

O

W E

L

1 2 3 4 5 × × Q  O, E occupy only even places ∴ O and E can be arranged in two × marked places in 2P2 ways = 2! = 2. Also the remaining 3 letters can be arranged in themselves in 3! ways = 3 × 2 × 1 = 6. ∴ Required no. of words = 2P2 × 3! = 2 × 6 = 12.

9! 9×8×7×6 = = 126 ways 5! 4! 4 × 3 × 2 ×1 Hence the total number of ways in which committee can be formed = 126 + 70 = 196.

41. (b) The word ‘FRACTION’ consists of 8 different letters out of which (A, I, O) are 3 vowels and (F, R, C, T, N) are 5 consonants.

37. (c) (b) The two particular persons can be arranged among themselves in 2P2 ways i.e., 2! ways.

 irst of all, we arrange the consonants among themF selves.



9

C5 =

681

 aking them as one person and keeping him fixed, T we can arrange the remaining 5 persons among themselves in 5! ways. Hence the required no. of ways in which 2 particular persons come together = 5! × 2! = 120 × 2 = 240.

Permutations and Combinations

= 5 × 4 × 5 × 3 × 2 × 1 = 120 ways. Hence the required number of ways = P (4, 2) × P (5, 5) = 12 × 120 = 1440.

682

Objective Mathematics

This can be done in P (5, 5) = 5! = 120 ways. Let one such way be : × F × R × C × T × N × Now no two vowels are together if they are put at places marked (×). The 3 vowels can fill up these 6 places in P (6, 3) ways. Hence the total number of words = 120 × P (6, 3) = 120 × 6 × 5 × 4 = 120 × 120 = 14400. 42. (a) There are 8 different letters in the word ‘DAUGHTER’. When the vowels A, E, U are always together, they can be supposed to be put in a bracket and treated as one letter (A, E, U). The remaining 5 and one letter (A, E, U) i.e., 6 letters can be arranged in 6! ways.

47. (b) Each ring can be set in 10 different ways. ∴ The total number of ways of attempts = 10 × 10 × 10 = 103 = 1000. There is only one successful attempt. ∴  Required number of unsuccessful attempts = 1000 – 1 = 999. 48. (b) Each arm can be set in 4 ways. ∴ Five arms can be set in 4 × 4 × 4 × 4 × 4 ways. But this includes the way when all the arms are in the position of rest, when no signal is sent. Hence required number of signals = 45 – 1 = 1024 – 1 = 1023.

 ow corresponding to each of these 6 arrangements, N the letters of the group (A, E, U) can be arranged 49. (a) Each stall can be filled in 3 ways as there are three types of animals (animals of each category being among themselves in 3! ways. not less than 10). ∴ Required number of words = 6! × 3! = 720 × Shipload, i.e., filling up of 10 stalls, can be made 6 = 4320. in 43. (d) Each question can be answered in 4 ways and all 3 × 3 × ... up to 10 times = 310 = 59049. questions can be answered correctly in only one way.

∴ The required number of ways = 43 – 1 = 64 – 1 = 63. 44. (d) E ach bulb has two choices, either switched on or off ∴  Required number of ways = 210 – 1 = 1024 – 1 = 1023 [Since in one way when all are switched off, the hall will not be illuminated]. 45. (a) 12 persons can be seated round table in (11)! ways  he total number of ways in which 2 particular T persons sit side by side is (10)! × 2!. Hence the required number of arrangements = (11)! – (10)! × (2)! = 11 × (10)! – (10)! × (2)! = (10)! (11 – 2) = 9 (10)!. 46. (b) 3 ladies out of 8 can be selected in 8C3 ways and 4 gentlemen out of 7 in 7C4 ways.

50. (a) We have to arrnage the 3 vowels in their places i.e., in the 1st, 4th, 7th places and the four consonants in their places.

 he number of ways to arrange 3 vowels in 1st, T 4th and 7th places.

=

3! 2!

( there are 2 A’s)

 he number of ways to arrange four consonants in T their places = 4!. ∴ Required number of arrangements

=

3! × 4! = 3 × 24 = 72. 2!

51. (c) As e always comes earlier than i the order of e and i cannot be changed in the arrangement. So in the arrangement, they are to be treated as identical things. ∴ Required number of arrangements =

6! = 360. 2!

 ow each way of selecting 3 ladies is associated N with each way of selecting 4 gentlemen. 52. (b) Total no. of coins = 2 + 2 + 3 + 1 = 8. Hence the required number of ways 2 coins are 10 paise, 2 are 20 paise, 3 are 25 paise = 8C3 × 7C4 = 56 × 35 = 1960. and 1 is of 50 paise. We now find the no. of committees of 3 ladies and 4 ∴ Required no. of ways gentlemen in which both Mrs. X and Mr. Y are mem8! bers. In this case, we can select 2 other ladies from the 8 × 7 × 6 × 5 × 4 × 3 × 2 ×1 = = remaining 7 in 7C2 ways and 3 other gentlemen from 2! × 2! × 3! × 1! 2 ×1× 2 ×1× 3 × 2 ×1×1 the remaining 6 in 6C3 ways. = 8 × 7 × 6 × 5 = 1680. ∴ The no. of ways in which both Mrs. X and Mr. 7 8 Y are always included = C2 × C3 = 21 × 20 53. (c) There are in all seven letters in the word BALLOON = 420. in which L occurs 2 times and O occurs 2 times. Hence the required no. of committes in which Mrs. ∴ The number of arrangements of the seven letters X and Mr. Y do not serve together = 1960 – 420 7! = 1540. = 1260. of the word = 2! × 2!

∴ The no. of arrangements in which the two L’s come together 6! = = 6 × 5 × 4 × 3 = 360. 2!  ence the required no. of ways in which the two L’s H do not come together = 1260 – 360 = 900. 54. (a) Total number of books = 3 × 4 = 12 in which each of 4 different books is repeated 3 times. Hence the required number of arrangements 12! 12! = = . (3!) 4 3! × 3! × 3! × 3! 55. (c) Since the middle seat of the first row is reserved for the principal, the remaining 6 teachers can be arranged in 6! ways.



= 8C2 = 28 ways.

 ut of these 28 lines, 8 are sides and remaining O diagonals. Hence the number of diagonals = 28 – 8 = 20. 61. (b) There are 2 primary teachers. They can stand in a row in P (2, 2) = 2! = 2 × 1 ways = 2 ways. There are 2 middle teachers. They can stand in a row in P (2, 2) = 2! = 2 × 1 ways = 2 ways. There are 2 secondary teachers. They can stand in a row in P (2, 2) = 2! = 2 × 1 ways = 2 ways.

= 6 × 5 × 4 × 3 × 2 × 1= 720 ways.

 gain there are twenty students. The two corner A seats are to be used by two tall students. The two tall students can be arranged in two ways.  he remaining 18 students can be arranged in 18! T ways. Total number of arrangements for the second row = 2 × 18! ways. ∴ The required number of arrangements = 2 × 18! × 720 = 1440 × (18!). 56. (c) Q 9600 = 27 × 3 × 52 ∴ No. of divisors = (7 + 1) × (1 + 1) × (2 + 1) = 8 × 2 × 3 = 48. 57. (b) Required number of parallelograms

= 4C 2 × 3C 2 =

4! 3! 4×3 3 × × = 18. = 2! 2! 2!1! 2 ×1 1

58. (a) Required number of ways

= 10C5 × 10C4 ways.

59. (a) Suppose the first two digits are 41, remaining four digits are to be chosen from 0, 2, 3, 5, 6, 7, 8, 9 as to make all the digits distinct.

These three sets can be arranged in themselves in = 3! = 3 × 2 × 1 = 6 ways. Hence the required number of ways = 2 × 2 × 2 × 6 = 48. 62. (c) If n is the number of teams then n! n C2 = 153 or = 153 2!(n − 2)! ⇒ ⇒ ⇒ But

n (n – 1) = 153 × 2 n2 – n – 306 = 0 ⇒ (n – 18) (n + 17) = 0 n = 18, – 17 n ≠ = – 17 ∴ n = 18.

63. (c) 7 women can sit on a round table in (7 – 1) = 6! ways. When the women have been seated in any one of the 6!, the 7 men can be seated at 7 places marked M in 7! ways  ssociating the two operations, the total no. of A ways = 6! × 7!.

∴ Remaining four can be chosen in P (8, 4) ways. We have 8 letters and they are to be arranged at 64. (c) Out of 22 players, 2 are to be included and 4 are to be excluded. We have to select a team of 11 players. 4 places. So the remaining 9 players are to be selected from First place can be arranged in 8 different ways the remaining 16 players. This can be done in 16C9 Second place can be arranged in 7 different ways ways Third place can be arranged in 6 different ways Fourth place can be arranged in 5 different ways 65. (a) A candidate can attempt 5 questions from group 1 Number of such telephone numbers and 2 from group II or 4 from group 1 and 3 from = 8 × 7 × 6 × 5 = 1680. group II or 3 from group I and 4 from group III or Now considering the all given Ist two digits (41 2 from group I and 5 from group II. This can be or 42 or 46 or 64), total number of the telephone done in numbers 6 C 5 × 6C 2 + 6C 4 × 6C 3 + 6C 3 × 6C 4 + 6C 2 × 6C 5. = 1680 × 5 = 8400.

683

60. (c) The total number of lines obtained by joining 8 vertices of octagon

Permutations and Combinations

I f two L’s always come together, taking them as one letter, we have to arrange 6 letters in which O occurs 2 times.

684



= 6 × 15 + 15 × 20 + 20 × 15 + 15 × 6 = 90 + 300 + 300 + 90 = 780.

Objective Mathematics

66. (c) Required number of ways = 4C 1 × 8C 5 + 4C 2 × 8C 4 + 4C 3 × 8C 3 + 4C 4 × 8C 2 = 4 × 56 + 6 × 70 + 4 × 56 + 1 × 28 = 224 + 420 + 224 + 28 = 896. 67. (b) As no two lines are parallel and no three are concurrent, only two lines can meet at one point and every two lines meet at a point. ∴ They will intersect in 20

C2 =

20

C2 points

20! 20 × 19 × 18! 20 × 19 = = = 190. 18!(2!) 18! × 2! 2

68. (c) Total no. of applicants = 23 No. of S.C. applicants = 7 No. of non S.C. applicants = 23 – 7 = 16 Total posts of teachers = 5. We can select 2 teachers from 7 S.C. applicants and 3 other teachers from the other 16 applicants. The required number of ways in which selection can be made = C (16, 3) × C (7, 2) 16! 7! 16! 7! × × = = 3!(16 − 3)! 2!(7 − 2)! 3!13! 2! 5!

=

16 × 15 × 14 7 × 6 × = 11760 ways. 3 × 2 ×1 2 ×1

69. (a) The following are the different possibilities in which three books can be borrowed;

71. (c) 6 large animals can be caged in 7 large cages in 7 P6 = 7! ways. 5 small animals can be caged in remaining 5 cages (4 small + 1 large) in 5! ways. Hence the number of ways is

7! × 5! = 5040 × 120 = 604800.

72. (b) Since n = 20 is even, 20Cr is greatest when r = 20 2 = 10. Hence the maximum number of parties = 20C10. Thus he should invite 10 friends at a time in order to form the max. number of parties. 73. (a) The total number of words

= (2C1 × 4C2 + 2C2 × 4C1) 3! = (12 + 4) × 6 = 96.

74. (c) Total number of words = 9! ∴ The number of other words = 9! – 1. 75. (b) The maximum number of cars is the total number of four digit that can be be formed by the nine digits 1, 2, 3, 4, 5, 6, 7, 8, 9 and the digits can be repeated in the same number. The number of numbers of three digits = 93, ( digits can be repeated) The number of numbers of four digits = 94 ∴ The required number of cars = 93 + 94 = 93(1 + 9) = 7290. 76. (a) The numbers are of five digits having 4 or 5 in ten thousand’s place and 1 or 5 in unit’s place and the remaining digits are any of the given six digits.

∴ number of ways to fill ten thousand’s place = (i) When Chemistry part II is selected, then Chemistry 2 P1 = 2 part I is also borrowed and the third book is Number of ways to fill unit’s place = 2P1 = 2 selected from the remaining 6 books. Number of ways to fill other three places = 63. (ii) When Chemistry part II is not selected, in this ∴ Required number of numbers = 2 × 2 × 63 = case he has to select the three books from the 864. remaining 7 books. First choice can be made in C (6, 1) = 6 ways. 77. (c) Number of words of five letters, when letters can Second choice can be made in be repeated = 105 = 100000. 7×6×5 = 35 ways. C (7, 3) = Number of words of five different letters 1× 2 × 3 = 10P5 = 10 × 9 × 8 × 7 × 6 = 30240. Total number of ways in which he can choose the ∴ Number of words which have at least one letter three books to be borrowed = 6 + 35 = 41. repeated 70. (c) Since m + nP2 = 90 = 100000 – 30240 = 69760. ∴ (m + n) (m + n – 1) = 90 78. (b)  F or each pair of stations, two different types of ⇒ (m + n)2 – (m + n) – 90 = 0 tickets are required, Now, the number of selections ⇒ m + n = 10 or m + n = –9 of 2 stations from 15 stations = 15C2. But m + n ≠ –9, ∴ m + n = 10. ∴ Required number of types of tickets Also, we have m – nP2 = 30 ...(1) 15! = 15 × 14 = 210. = 2 15C2 = 2 ⇒ (m – n) (m – n – 1) = 30 2 !13! 2 ⇒ (m – n) – (m – n) – 30 = 0 79. (b) The number of triangles = number of selections of ⇒ m – n = 6 or m – n = –5 2 lines from the (n – 1) lines which are cut by the But m – n ≠ –5, last line ∴ m – n = 6. ...(2) (n − 1)! (n − 1) (n − 2) Solving (1) and (2), we get m = 8 & n = 2 = . = n – 1C 2 = 2!(n − 3)! 2

81. (a) The number of diagonals + number of sides = number of selections of two vertices from n vretices ∴ the number of diagonals. = nC2 – n

=

n (n − 1) n 2 − n − 2n n (n − 3) −n = = . 2 2 2

 umber of selections of any number of girls from N 3 girls

= 3C 0 + 3C 1 + 3C 2 + 3C 3 = 2 3.

∴ Required number of selections of at least one boy from 4 boys and 3 girls = (24 – 1)23 = 15 × 8 = 120. 87. (a) The number of selections of 3 consonants and 2 vowels

= 17C3 × 5C2 =

17 × 16 × 15 5 × 4 = 6800 × 3× 2 2

 or each selection of 3 consonants and 2 vowels, F the number of words = 5P5 = 5! = 120 ∴ Required number of words = 6800 × 120 = 816000.

82. (c) The number of parallelograms = number of selections of 4 lines, two from each set ( a parallelogram has 88. (b) Each arrangement has to contain 1 capital letter, 3 two sets of parallel sides) consonants and 2 vowels. So we have to select them n ( n − 1) m ( m − 1) mn ( m − 1) ( n − 1) = nC2 × mC2 = = . first from the given things. Number of selections of × 2 2 4 1 capital, 3 consonants and 2 vowels 83. (b) A triangle is formed for each selection of 2 points = 3C 1 × 5C 3 × 4C 2. from one line and 1 point from the other line. For each selection there are 5! arrangements ( the ∴ The number of triangles capital letter will occupy the first place. ∴ Required number of words  = 10C2 × 10C1 + 10C1 × 10C2 = 3C1 × 5C3 × 4C2 × 5! = 3 × 5 × 4 × 4 × 3 5! 10 x 9 10 × 9 = × 10 + 10 × 2 2 2 2 = 21600. = 900. 89. (b) First we have to select 2 men for bow side and 3 84. (d) (a) Number of times he will go for stroke side. ∴ The number of selections of the crew for two 8x 7×6 = 56. sides 3x 2 = 5C 2 × 3C 3. (b) Number of items each child will go For each selection there are 4 persons each on both = number of selections of 3 in which a particular sides who can be arranged in 4! × 4! ways. child is present ∴ Required number of arrangements = 1C1 × 7C2 = 1 × 7 × 6 = 21. 5×4 2 × 1 × 24 × 24 = 5C2 × 3C3 × 4! × 4! = (c) Number of times a particular child will not 2 go = 5760. = number of selections of 3 in which a particular  90. (a) The p positive signs can be arranged in 1 way as child is absent these are all identical. There are p + 1 places for 7×6×5 = 35. = 7C 3 = n negative signs ( p – 1 in between the +ve signs 3+ 2 and two ends). 

= 8C 3 =

Number of selections for n places form p + 1 places for n negative signs = p + 1Cn and for each selection ∴ Total number of ways to fail or pass of n places for n negative signs there is only one n C0 + nC1 + nC2 + ... + nCn = 2n. arrangement ( –ve signs are identical) But there is only one way to pass, i.e., when he fails ∴ Required number of arrangements of +ve and –ve in none. signs n ∴ Total number of ways to fail = 2 – 1. = 1 × p + 1C n × 1 = p + 1C n. ∴ From question, 2n – 1 = 63; ∴ 2n = 64 = 26 91. (a) 10 boys can be seated in a row in 10! ways. For ∴ n = 6. each of these, there are 9 places in between the 10 86. (b) Number of selections of at least one boy from 4 boys. If the 10 girls sit in these 9 places and one of boys the 2 ends then the boys and girls will alternate.

85. (a) Let the number of papers be n.



= 4C1 + 4C2 + 4C3 + 4C4 = 24 – 1.

∴ The 10 girls can be seated in 2 ×

10

P10 ways.

685

Then the number of matches = C2 =45 (given). (n −1) = 45; ∴ n2 – n – 90 = 0; ∴ 2 or (n – 10) (n + 9) = 0. ∴ n = 10, – 9 (impossible); so the required number of teams = 10. n

Permutations and Combinations

80. (c) Let the number of teams = n.

686

Objective Mathematics

∴ The required number of ways= 10! × 2 × 10P10 = 10! × 2 × 10! = 3628800. 92. (a) Number of selections of at least one mango from 3 mangoes = 3  umber of selections of at least one apple from 4 N apples = 4 Number of selections of at least one orange from 2 oranges = 2. ∴ Required number of selections of fruits = 3 × 4 × 2 = 24. 93. (c) Number of selections of at least one red ball from 4 identical balls = 4  umber of selections of any number of green balls N from 3 identical green balls = 3 + 1. ∴ Required number of selections of balls = 4 × (3 + 1) = 16. 94. (a) Number of selections of any number of copies of a book = p + 1, (because copies of the same book are identical things). Similarly for each book. ∴ Total number of selections = (p + 1) (p + 1)... to n factors = (p + 1)n. But this includes a selections which is empty i.e., zero copy of each is selected. Excluding this, the required number of nonempty selections = (p + 1) n – 1. 95. (c) Here 1260 = 2 × 2 × 3 × 3 × 5 × 7.

98. (b)

The total number of seats  1 grand father + 5 sons and daughters + 8 grand = children = 14 The grand children with to occupy the 4 seats on either side of the table = 4! ways = 24 ways and grand father can occupy a seat in (5 – 1) ways = 4 ways (Since 4 gaps between 5 sons and daughters) a nd the remaining seat can be occupied in 5! ways = 120 ways (5 seats for sons and daughters) Hence required number of ways = 24 × 4 × 120 = 11520.

99. (b) There are 8 chairs on each side of the table. Let the sides be represented by A and B. Let four persons sit on side A, then number of ways of arranging 4 persons on 8 chairs on side A = 8P4 and then two persons sit on side B. The number of ways of arranging 2 persons on 8 chairs on side B = 8P2 and the remaining 10 persons can be arranged in remaining 10 chairs in 10! ways.  ence the total number of ways in which the persons H can be arranged 8! 8!10! . = 8P4 × 8P2 × 10! = 4! 6! 100. (c) Let the sides of the game be A and B. Given 5 married couples, i.e., 5 husbands and 5 wives. Now, 2 husbands for two sides A and B can be selected out of 5 = 5C2 = 10 ways.

A selection from 2, 2, 3, 3, 5, 7 gives a divisor. After choosing the two husbands their wives are to be excluded (since no husband and wife play in the Total number of selections of one or more from 2, 2, 3, same game). So we are to choose 2 wives out of 3, 5, 7 remaining 5 – 2 = 3 wives i.e., 3C2 = 3 ways. 2 = (2 + 1) (2 + 1) 2 – 1 = 36 – 1 = 35. Again two wives can interchange their sides A and But one of the selections is the selection of all the B in 2! = 2 ways. six prime factors which corresponds to the divisor B y the principle of multiplication, the required  1260. number of ways = 10 × 3 × 2 = 60. ∴ Required number of proper divisors 101. (a) Considering CC as single object, U, CC, E can be = 35 – 1 = 34. arranged in 3! ways × U × CC × E × 96. (a) 2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7. Now the three S are to be placed in the four availTotal number of non-empty selections from 2, 2, 2, able places. 3, 3, 5, 7 Hence required no. of ways = 3! · 4C3 = 24. = (3 + 1) (2 + 1) 22 – 1 = 47. 102. (a) Required sum = (5 – 1)! (1 + 2 + 3 + 4 + 5) But one of the selections is the selection of all the  105 − 1  six prime factors which correspond to the divisor  9  = 24.15.11111 = 3999960. 2520. ∴ Required number of proper divisors = 47 – 1 103. (b) Here n = 6 and p = 9. = 46. ∴ No. of rectangles excluding square 97. (c) We know, the number of ways to divide n identical 6 n+r –1 C r – 1. things among r persons = 6.9 (6 + 1) (9 + 1) − ∑ (7 − r ) (10 − r ) = 4 r =1 Here n = 30, r = 4. ∴ Required number of ways=

=

33 × 32 × 31 = 5456. 3× 2

30 + 4 – 1

C4 – 1 = 33C3.

6



= 945 − ∑ (70 − 17 r + r 2 ) = 945 – 154 = 791. r =1

Now E2 (100!) 100  100  100  100  100  100  + + =  + + + 2 3 4 5 6  2   2   2   2   2   2  = 50 + 25 + 12 + 6 + 3 + 1 = 97 and

100  100  + 2 E5 (100!) =  = 20 + 4 = 24  5   5 

100! = 297.3b.524.7d... = 273.3b.(2 × 5)24.7d... = 273.3b.(10)24.7d... Hence number of zeros at the end of 100! is 24.

111. (c) The no. of triplets of positive integers which are solutions of x + y + z = 100

= Coeff. of x100 in (x + x2 + x3 + ...)3 = Coeff. of x100 in x3 (1 – x)–3 = Coeff. of x100 in (n + 1) (n + 2) n   x 3 1 + 3 x + 6 x 2 + ... + x + ... 2  



=

(97 + 1) (97 + 2) = 49 × 99 = 4851. 2

105. (b) I n terms of prime factors 33! can be written 112. (b) Suppose the two players did not play at all so that as 2a.3b5c.7d... the remaining (n – 2) players played n – 2C2 matches. Since these two players played 3 matches each,  33   33   33   33   33  Now, E2 (33!) =   +  2  +  3  +  4  +  5  hence the total number of matches is  2  2  2  2  2  n–1 C2 + 3 + 3 = 84 (given) = 16 + 8 + 4 + 2 + 1 = 31. Hence, the exponent of 2 in 33! is 31. (n − 2)(n − 3) = 78  or  n2 – 5n + 6 = 156 or ∴ Largest value of n is 31. 1.2

106. (a) Clearly, 30 mangoes can be distributed among 4 boys such that each boy can receive any number of mangoes.

or n2 – 5n – 150 = 0  or  (n – 15) (n + 10) = 0 ∴ n = 15 (n ≠ – 10).

C4 – 1 = 33C3

113. (b) The number of permutations of 10 papers when there is no restriction = 10P10 = 10!.

m, m + 1, m + 2, ... (m + r – 1), where m ∈ N ∴ Product = m ( m + 1) (m + 2) .. (m + r – 1)

 hen the best and the worst papers come together, W regarding the two as one paper, we have only 9 papers. These 9 papers can be arranged in 9P9 = 9! ways. But these two papers can be arranged among themselves in 2! ways. ∴ number of arrangements when the best and the worst papers do not come together = 10! – 9!.2! = 9! (10 – 2) = 8.9!.

Hence total number of ways =

=

30 + 4 – 1

33.32.31 = 5456. 1.2.3

107. (a) Let r consecutive positive integers be



=

(m − 1)!m (m + 1) (m + 2)... (m + r − 1) (m − 1)!



=

(m + r − 1)! (m + r − 1)! = r !. (m − 1)! r !(m − 1)!

= r!. m + r – 1Cr which is divisible by r! ( number)

m + r – 1

108. (b) Number of triangles formed =

=

12

Cr is a natural

C 3 – 7C 3

12.11.10 7.6.5 − = 220 – 35 = 185. 1.2.3 1.2.3

109. (a) Number of words in which all the 5 letters are repeated

114. (a) Number of girls = 3, number of boys = 7. Since there is no restriction on boys, therefore, first of all arrange the 7 boys.  ow 7 boys, can be arranged in a row in 7! ways. N ×B×B×B×B×B×B×B× If the girls are arranged at the places (including the two ends) indicated by crosses, no two of three girls will be consecutive. Now there are 8 places for 3 girls ∴ 3 girls can be arranged in 8P3 ways. ∴ Required number

8! = 10 = 100000.  = 8P3 × 7! = × 7! = 42 · 8!. 5! and the number of words in which no letter is repeated 115. (c) Let n be the number of students. = 10P5 = 10.9.8.7.6 = 30240. Now number of ways in which two students can be Hence the number of words which have at least  one selected out of n students is nC2. letter repeated ∴ number of pairs of students = nC2. = 100000 – 30240 = 69760. But for each pair of students, number of cards sent 110. (a) Total no. of wrong answers is 2 (since if there are two students A and B, A will = 1 · (a1 – a2) + 2.(a2 – a3) + ... send a card to B and B will send a card to A). 5

687

+ (k – 1) (ak – 1 – ak) + kak  = a1 + a2 + a3 + ... + ak.

Permutations and Combinations

104. (c) In terms of prime factors 100! can be written as 2a.3b.5c.7d...

688

∴ For nC2 pairs, number of cards sent = 2.nC2.

=

According to the question, 2. C2 = 600 n

Objective Mathematics

or, 2.

=(

n(n − 1) = 600 or, n2 – n – 600 = 0 2!

=

116. (a) The student will fail if he fails in one or more subjects.  ow the student can fail in one or more subjects N out of 5 subjects in 5 C 1 + 5C 2 + 5C 3 + 5C 4 + 5C 5 = 5C 0 + 5C 1 + 5C 2 + 5C 3 + 5C 4 + 5C 5 – 5C 0 = 25 – 1 = 31 ways. ∴  Required number = 31. 117. (c) Each set is having m + 2 parallel lines and each parallelogram is formed by chossing two st. lines from the first set and two st. lines from the second set. Two st. lines from the first set can be chosen in m + 2C2 ways and two st. lines from the second set can be chosen in m + 2C2 ways. Hence the total number of parallelograms formed = ( m + 2C 2) ( m + 2C 2) = ( m + 2C 2) 2.



= 18 × 17 − 5 × 4 + 1 2 2



= 153 – 10 + 1 = 144.

18

=

10! 10! > 2. (n − 1)!(11 − n)1 n!(10 − n)!

n + 3

Cr–3)

Cr – 1 +

n + 2

Cr – 2)

Cr – 1) + (n + 2Cr – 1 +

Cr – 1 =

n + 4

n + 2

Cr – 2)

C r.

Each line cuts 4 circles into 8. ∴ 4 lines cut four circles into 32 points. ∴ Reqd. number = 6 + 12 + 32 = 50. 124. (c) Reqd. sum = 3! (1 + 3 + 5 + 7) × (1 + 10 + 100 + 1000) [ each digit occurs 3! times in unit’s place, ten’s place, hundred’s place and thousands place]. = 3! × 16 × 1111 = 16 × 1111 × 3! 125. (a) n–pPr [ p things are excluded, ∴ no. of things left = n – p] 126. (a) The no. of ways in which a student can make a selection of one or more books. = (m + 1) (m + 1) ... (m + 1) – 1 = (m + 1)n – 1

127. (d) First we arrange 3 particular things in r places. This can be done in r P3 ways. Then, remaining n – 3 things can be arranged taken r – 3 at a time in n–3 Pr – 3 ways. ∴ Total no. of ways = r P3.

n–3

P r – 3.

128. (a) Since the seats are numbered, ∴ the arrangement is not circular. ∴ the required no. of ways = the no. of arrangements of n things taken m at a time = nP m.

J oining 3 points on the same line gives no triangle, such ∆′ s are mC3 + nC3 + kC3. ∴ Required number = m + n + kC 3 – mC 3 – nC 3 – kC 3. 131. (c)

22 1 = 7 . 3 3 ∴ Least +ve integral value of n = 8.

9600 = 27 × 3 × 52. ∴ No. of divisors = (7 + 1) × (1 + 1) × (2 + 1) = 48.

122. (a) nCr + 4.nCr – 1 + 6. nCr – 2 + 4.nCr – 3 + nCr – 4. = ( Cr + Cr – 1) + 3 ( Cr – 1 + Cr – 2)  + 3 (nCr–2 + nCr–3) + (nCr–3

Cr +

n + 2

Cr – 2)

130. (b) Total no. of points are m + n + k. The ∆′s formed by these points = m + n + kC3.

⇒ 3n > 22 ⇒ n >

n

Cr + 2 · (

n + 2

Cr – 1 +

Cr – 3

n+1

123. (b) 4 lines intersect each other in 4C2 = 6 points and 4 circles intersect in 4P2 = 12 points.

⇒ n > 2 (11 – n) i.e., n > 22 – 2n

n

+ ( Cr–2 +

n+1

n + 1

= 18C2 – (5C2 – 1) = 144.

n(n − 1)! (11 − n) (10 − n)! > 2. (n − 1)! (10 − n )!

n

Cr – 1) + 2(

n+1

n + 1

129. (b) The number of st. lines

n! (11 − n)! ⇒ > 2 (n −1)! (10 − n)!

n

n + 3

n + 1

[ student can make selection of 0, 1, 2, 3, ...m books out of each type]

[ no of letter boxes = n] ∴ total number of ways for r letters n × n × ... × n = = n r. r factors 6×5 120. (b) Reqd. number of lines = 6C2 = = 15. 1× 2 121. (b) We have, 10Cn–1 > 2.10Cn

⇒

n + 2

Cr +

= (n + 2Cr +

C 2 – 5C 2 + 1

119. (a) Since every letter can be posted in n ways

⇒

Cr + 3.n+1Cr–1 + 3.n + 1Cr–2 +

n + 1



or, (n – 25) (n + 24) = 0 ∴ n = 25, –24 But n ≠ –24 ∴ n = 25.

118. (b) Reqd. number of st. lines =

n+1

132. (b) Maximum number of points of intersection 8×7 = 28. = 8C 2 = 1× 2 + nC r – 4)

=

(6 − 1)!.6 P3 5! × 6 × 5 × 4 = = 120 × 60 = 7200. 2 2

134. (c) Number of ways in which first three places be taken by 12 students = 12 × 11 × 10. 135. (d) Total no. of possible answers = 34 × 23 × 51 = 81 × 8 × 5 = 3240.

∴ first person can be seated in n ways, second person can be seated in (n – 1) ways Similarly, mth person can be seated in (n – m + 1) ways ∴ Total number of ways = n(n – 1) (n – 2) ... (n – m + 1) = nPm ways.

15! 136. (d) Fix the digit 5 at the unit place. Now there are 148. (a) Required number of tickets = 15C2 = 3 2!13! P2 = 6 three digit numbers and 3P3 = 6 four digit numbers. Total number of numbers = 6 + 6 = 12. 149. (c) Let A, B be the corresponding speakers. Without any restriction the eight persons can be arranged 137. (b) Required no. of ways among themselves in 8! ways, but the number of ways in which A speaks before B and the number = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 10 of ways in which B speaks before A make up 8!. [ each question can be answered in two ways] Also number of ways in which A speaks before B 138. (a) Maximum no. of points = 8P2 = 8 × 7 = 56. is exactly same as the number of ways in which B speaks before A. ∴ the reqd. number of ways = 1 6!⋅ 3! 139. (a) Required no. of ways = = 720 × 3 = 2160. 2 2 (8!) = 20160. 140. (a) We have, C(n, 10) = c (n, 12) 150. (c) From 112233, the number of 6 digits that can be ⇒ n = 10 + 12 = 22. ( 10 ≠ 12) formed n n n n n 720 6! 141. (c) The given value = C1 + C2 + C3 +...+ Cn = 2 –1. = = = 90. 8 2! 2! 2! 142. (d) Required number of signals are 151. (a) Total number of ways = 15C11 5 P1 + 5P2 + 5P3 + 5P4 + 5P5 Favourable cases = 8C6 × 7C5 = 5 + 20 + 60 + 120 + 120 = 325. 8 C6 × 7C5 143. (d) In the word PENCIL, there are total number of Required probability = . 15 C11 alphabet = 6 out of which 4 are consonants and can be arranged in 4 ! ways 152. (b) Required no. of ways = 4! = 24. ∴ remaining two places can be filled by two vowels 153. (b) A2 B3 C4 written in full is AA BBB CCCC in 5P2 ways 5 ∴ Total number of ways = 4 ! × P2 = 24 × 20 = 9! 480. ∴ Required no. of ways = 2! 3! 4! 144. (a) (7C0 + 7C1) + (7C1 + 7C2) + ... + (7C6 + 7C7) 154. (d) The required no. of lines = 10C2 – 7C2 = 45 – 21 = = 8C1 + 8C2 + ... + 8C7 24. = 8C0 + 8C1 + 8C2 + ... + 8C7 + 8C8 – (8C0 +8C8) 155. (c) Each stall can be occupied by a cow, a horse or a = 28 – 1 (1 + 1) = 28 – 2. calf i.e., in 3 ways. 145. (c) Let n be the number of sides. ∴ The required number of ways = 312. n ∴ Number of diagonals = C2 – n = 44 (Given) 156. (b) C (10, 4) + C (10, 5) = C (11, 5) = C (11, r) n(n − 1) ⇒ r = 5. = 44 ⇒ −n 2 3n! ⇒ n2 – n – 2n = 88 157. (b) The required no. of ways = 3nCn = n ! 2n ! 2 ⇒ n – 3n = 88 ⇒ (n – 11) (n + 8) = 0 ⇒ n = 11

( n ≠ –8)

146. (c) Total no. of books = a + 2b + 3c + d.  ince there are b copies each of two books, c copies S each of three books and single copy of d books, ∴ total number of arrangements

=

(a + 2b + 3c + d )! a !(b!) 2 (c !)3

158. (c) Required no. of ways = 5!. 6! = 120 × 720 = 86400. 159. (a) Let n be the number of teams ∴

n



C2 = 36 ⇒

n(n −1) = 36 1.2

⇒ n (n – 1) = 72 = 9.8 ⇒ n = 9.

689

147. (a) Number of seats = n, and number of people = m

Permutations and Combinations

133. (b) Required number of ways

690

160. (b) n – 2Cr + 2 · n – 2Cr – 1 + n–2

n–2

n–2

Objective Mathematics

= ( Cr + C r – 1) + ( n–1 n–1 Cr + Cr – 1 = = nC r

Cr – 2

n–2

Cr – 1 + (

n

n–2

C r – 2)

C r – 1 + nC r = n + 1C r)

161. (a) We know that in any trinagle the sum of two sides is always greater than the third side.

∴ All the five digit numbers formed by the digits 1, 2, 3, 4, 5 are divisible by 3 and their number = 5! = 120.  hen we include 0, the four other digits whose sum W is divisible by 3 are 1, 2, 4 and 5. ∴  The number of numbers in this case

= 4 × 4! = 4 × 24 = 96. ∴ The triangle will not be formed if we select Hence the required number of numbers = 120 + segments of lengths (2, 3, 5), (2, 3, 6) or 96 = 216. (2, 4, 6). Hence no. of triangles formed = 5C3 – 3. 169. (b) Six ‘+’ can be arranged in one way only. After placing + sign we shall have 7 places in which we 162. (c) In the required numbers, digits at the ten thousand can place ‘–’ which can be done in 7C4 ways place may be 2, 3, 4 or 5. Total of such numbers which begin with 2 is 2 × 3! = 12. Numbers which begin with 3 are 4! = 24 Similarly, numbers which begin with 4 and 5 each will also be equal to 24. Therefore required number of numbers = 12 + 24 + 24 + 24 = 84. 163. (b) There are three odd places in the words formed by the letters of MOBILE. At these 3 places 3 consonants can be placed in 3! ways and the remaining 3 letters at the remaining three places can be placed in 3! ways. ∴ Number of words formed = 3! 3! = 36. 164. (c) Possible ways of selecting the team  4 bowlers + 7 others, 5 bowlers + 6 others Since = there are 5 bowlers and 9 other players. Total number of ways in which team can be selected

= 5C4 × 9C7 + 5C5 × 9C6 = 180 + 84 = 264.

165. (c) The four digits 3, 3, 5, 5, can be arranged at four even places in 4! = 6 ways and the remaining 2! 2! digits viz. 2, 2, 8, 8, 8 can be arranged at the five 5! = 10 ways. Thus, the number 2! 3! of possible arrangements is (6) (10) = 60.

odd places in

166. (b) Distinct n numbers which can be formed using digits 2, 5 and 7 are 3n. We have to find n so that 3n ≥ 900 ⇒ 3n–2 ≥ 100 ⇒ n – 2 ≥ 5 ⇒ n ≥ 7.



=

7! = 7 × 6 × 5 = 35 ways. 4! 3! 3 × 2 ×1

Thus the required number of ways is 35. 170. (b) The number of ways of selecting 3 balls out of total 9 (2 white, 3 black, 4 red balls) is 9C3 i.e., 9 × 8 × 7 6 = 84.  he number of ways of selecting 3 balls out of T non-black six balls is 6C3 i.e.,

6×8×4 = 20. 3 × 2 ×1

 herefore the number of ways of selecting 3 balls T out of 9 balls so as to include atleast one black ball = 84 – 20 = 64. 171. (c) The total number of numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time = 4P4 = 4! = 24. Consider the digit in the unit’s place in all these numbers. Each of the digits 2, 3, 4, 5 occurs in 3! = 6 times in the unit’s place ∴ total for the digits in the unit’s place = (2 + 3 + 4 + 5) 6 = 84 Since each of the digits 2, 3, 4, 5 occurs 6 times in any one of the remaining places ∴ the reqd. total = 84 (1 + 10 + 102 + 103) = 84 (1111) = 93324. 172. (c) Required no. of ways = 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 25 – 1 = 31. 173. (d) No. of ways =

6! = 1 (All faces are alike). 6!

So, the least value of n is 7. 167. (a) The required no. of ways

1 1 1 = 4!  − +  = 12 – 4 + 1 = 9.  2! 3! 4!

168. (a) We know that a number is divisible by 3 if the sum of its digits is divisible by 3.  ow the sum of the digits 1, 2, 3, 4 and 5 is 15, N which is divisible by 3.

174. (b) Total arrangements =

7! (as A, R are twice) 2! 2!

6! × 2! No. of arrangements with both R together = 2! 2! 6! . = 2! No. of arrangements such that both R do not come 7! 6! 6!  7  − − 1 = 900. = together are = 2! 2! 2! 2!  2 



(6 Boys & 1 Girl) ⇒ 6C6 × 4C1 = 4

or (5 Boys & 2 Girls) ⇒ 6C5 × 4C2 = 36 or (4 Boys & 3 Girls) ⇒ 6C4 × 4C3 = 60 ∴ required no. of ways = 4 + 36 + 60 = 100. No. of ways to return back = 5. Total no. of ways in which villager can go to town & return back are = 5 × 5 = 25. 177. (b) Let total number of persons be n. Since two persons shake hands, so number of ways of shaking hands = nC2. According to question, we have nC2 = 66 n (n − 1)(n − 2)! n! = 66 or = 66 or (n − 2)! 2 x 1 (n − 2)! 2! or n2 – n = 132 or n2 – n – 132 = 0 or n2 – 12n + 11n – 132 = 0 or n (n – 12) + 11 (n – 12) = 0 ⇒ n = – 11 or 12 But, n = – 11 is not possible. ∴  n = 12. n

∑

k

k=m

Cr

= mCr + 1 + mCr + m + 1Cr + ... + n – 1Cr + nCr = m + 1Cr + 1 + m + 1Cr + ... + n – 1Cr + nCr = m + 2Cr + 1 + m + 2Cr + ... + n – 1Cr + nCr











= nCr + 1 + nCr =



n + 1

4

7

4

7

7

4



Choice Ways (i) all the distinct (ii) 3 distinct, 2 alike (iii) 2 distinct, 3 alike (iv) 2 alike, 2 alike, 1 distinct (v) 3 alike, 2 alike

C5 3 C1 2 C1 3 C2 2

184. (b) The no. of words in which I and N are together is 6! = × 2! = 6! = 720. 2! ∴ The no. of words in which I and N are never together

= m1 =

7! – 720 = 1800. 2!

 he no. of words which begin with I and end T with R m1 5! = m2 = = 60.   ∴  = 30. m 2! 2 185. (b) We have,

x + 2

Px + 2 = a ⇒ a = (x + 2)!

x – 11

x! ( x − 11)!

Px – 11 = c ⇒ c = (x – 11)!

∴ a = 182 bc (x + 2)! = 182

x!  · (x – 11)! ( x − 11)!

= × × ×

6 5 C3 = 30 5 C2 = 20 4 C1 = 12 2

186. (a) Factorizing the given number, we have

38808 = 23 ⋅ 32 ⋅ 72 ⋅ 11

Therefore the total number of divisors

Total = 72

 ut this includes the division by the number itB self. Hence, the required number of divisors = 71 – 1 = 70.

∴ x ≤ 10

10

Cx – 1 > 2 ⋅ 10Cx ⇒ 1 > 2 ·  10

∴ x + 1 = 13. ∴ x = 12.

C1 × C1 = 4

181. (b) 10 ≥ x – 1 ⇒ x ≤ 11 and 10 ≥ x Cx C x −1

22 10 − x + 1 ⇒ 1 > 2 ⋅  ⇒ x > 22 – 2x ; ⇒ x > 3 x

C2 ≤ 100

⇒ (x + 2) (x + 1) = 182 = 14 × 13 6



1 ⇒ x > 7 ∴ x = 8. 3

C3 –

Cn – 1 ≤ 100

n + 1

⇒ 1 n (n + 1) {n − 1 − 3} ≤ 100 6 ⇒ n (n + 1) (n – 4) ≤ 600 It is true for n = 2, 3, 4, 5, 6, 7, 8, 9



180. (a) The choice are ase follows.

10

n + 1

1 n (n + 1) (n + 1) n (n − 1) − ≤ 100 ⇒ 6 2

and 4

= C4 ⋅  C1 + C3 ⋅  C2 + C2 ⋅  C3 + C4 ⋅  C4 = 140 + 210 + 420 + 35 = 805.



n + 1

Also, x P11 = b ⇒ b =

Cr + 1

179. (a) The required no. of ways 7

183. (b) n + 1Cn – 2 – ⇒

176. (a) No. of ways to go to a town = 5.

178. (b) mCr + 1 +

= (3 +1) (2 + 1) (1 + 1) – 1 = 71.

187. (d) Terminal digits are the first and last digits. Terminal digits are even ∴ 1st place can be filled in 3 ways and last place can be filled in 2 ways and remaining places can be filled in 5P4 = 120 ways.

691

182. (c) No. of triangles = 2nC3 – nC3 – nC3 2n (2n − 1) (2n − 2) 2n (n − 1) (n − 2) − = 6 6 1 2 = n (n – 1) (3n) = n (n – 1). 3

Permutations and Combinations

175. (c) Various alternatives to form a group of 7 with majority of boys; from 6 boys & 4 girls are →

692

 ence the number of six digit numbers so that the H terminal digits are even, is = 3 × 120 × 2 = 720.

Objective Mathematics

188. (b) ( b) The number of students answering exactly i(1 ≤ i ≤ n – 1) questions wrongly is 2n – i – 2n – i – 1. The no. of students answering all n questions wrongly is 2º.

Hence the total number of wrong answer is n −1



∑ i (2

n− i

− 2n − i −1 ) + n (2 )º = 2047

i =1

190. (a) The number of subsets of the set which contain at most n elements is C1 +

2n + 1

C2 + ... +

2n + 1

Cn = K (say)

We have 2K = 2 (2n + 1C0 + 2n + 1C1 + 2n + 1C2 + ... + 2n + 1Cn)

= (2n + 1C0 + 2n + 1C2n + 1) + (2n + 1C1 + 2n + 1C2n) + ... Cn + 1) (∴ nCr = nCn – r)



+ (2n + 1Cn +

2n + 1



2n + 1

C1 +

=

C0 +

2n + 1

2n + 1

C2 + ... +

2n + 1

1 x 4 + < 3 50 3

1 x  ⇒  +  = 1 for 34 ≤ x ≤ 50  3 50 

195. (a), (b), (c)  We have,

∴ number of ways for five places = 2 × 2 × 2 × 2 × 2 = 25 For 2, selecting 2 places out of 7 = 7C2 ∴ Required no. of ways = 7C2 ⋅ 25.

2n + 1

For 34 ≤ x ≤ 50, 1
2) points in each of two parallel lines. Every point on one line is joined to every point on the other line 21. by a line segment drawn within the lines. The number of points (between the lines) in which these segments intersect is (b) 2nC2 – 2 nC2 (a) nC2 × nC2 (c) 2nC2 – 2 nC1 + 2

(b) 1520 (d) 5067

(b) 120 (d) 14.

There are 10 points in a plane, out of these 6 are collinear. The number of triangles formed by joining these points is (a) 100 (c) 150

(b) 120 (d) None of these

22. The straight lines l1, l2, l3 are parallel and lie in the same plane. A total number of m points are taken on l1, n points on l2, k points on l3. The maximum number of triangles formed with vertices at these points are

(a) m + n + kC3 (b) m + n + kC3 – mC3 – nC3 – kC3 (c) mC3 + nC3 + kC3 (d) None of these

23. In an examination here are three multiple choice questions and each question has four choices. Number of sequences in which a student can fail to get all answers correct is

15. The number of selection of n different things taken r at (a) 11 (b) 15 a time which include a particular thing is (c) 80 (d) 63 (a) n . C (n – 1, r – 1) (b) r. C (n – 1, r – 1) 24. In how many ways 7 men and 7 women can sit on a (c) C (n – 1, r) (d) C (n – 1, r – 1) round table such that no two women sit together 16. The total number of permutations of the letters of the word “BANANA” is (a) ( 7) 2 (b) 7 × 6 (a) 60 (c) 720

(b) 120 (d) 24

(c) ( 6) 2

(d) 7

Answers

1. (c) 11. (d) 21. (a)

2. (b) 12. (a) 22. (b)

3. (c) 13. (a) 23. (d)

4. (a) 14. (b) 24. (b)

5. (b) 15. (d)

6. (a) 16. (a)

7. (c) 17. (a)

8. (a) 18. (a)

9. (c) 19. (a)

10. (c) 20. (a)

17

Binomial Theorem

CHAPTER

SUMMARY OF CONCEPTS BiNOMiAl ExPRESSiON An algebraic expression consisting of only two terms is called a binomial expression. For example, expressions such as: 4 x + a, 4x + 3y, 2x – y are all binomial expressions.

(iii) In the expansion of (x + y)n, the power of x goes on decreasing by 1 and that of y goes on increasing by 1 so that the sum of powers of x and y in any term is n. (iv) The binomial coefficients of the terms equidistant from the beginning and the end are equal.

General Term in the Expansion of (x + y)n

BiNOMiAl ThEOREM

In the binomial expansion of (x + y)n, the (r + 1)th term from the This theorem gives a formula by which any power of a bino- beginning is usually called the general term and it is denoted by mial expression can be expanded. It was first given by Sir Isaac Tr + 1, i.e., Newton. Tr + 1 = nCr xn – r yr

Binomial Theorem for Positive integral index If x and y are real numbers, then for all n ∈ N, (x + y)n = nC0 xn y0 + nC1 xn – 1 y1 + nC2 xn – 2 y2 + ...+ nCn – 1 + x1 yn – 1 + nCn x0 yn ...(1) Here nC0, nC1, nC2, ..., nCn are called binomial coefficients. Special Cases

Greatest Term (Numerically) in the Expansion of (1 + x)n Method 1 (a) Let Tr (the rth term) be the greatest term. (b) Find Tr – 1, Tr , Tr + 1 from the given expansion.

Tr Tr ≥ 1 and ≥ 1. This will give an inequality from Tr +1 Tr −1

(c) Put

where value or values of r can be obtained. (d) Then, find the rth term Tr which is the greatest term. Method 2

(a) Replacing y by – y, in (1) we get (x – y)n = nC0 xn y0 – nC1 xn – 1 y1 + nC2 xn – 2 y2 – ... + (– 1)n nCn x0 yn ...(2)

(a) Find the value of k =

(n + 1)| x | 1 +| x|

(b) Replacing x by 1 and y by x, we get (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn

(b) If k is an integer, then Tk and Tk + 1 are equal and both are greatest term.

(c) Replacing x by 1 and y by – x, we get (1 – x)n = nC0 – nC1 x + nC2 x2 – ... + (– 1)n nCn xn.

(c) If k is not an integer, then T[ k ] + 1 is the greatest term, where [k] is the greatest integral part of k. Note: To find the greatest term in the expansion of (x + y)n, write (x +

(d) Adding (1) and (2), we get (x + y) + (x – y) = 2 [x + C2 x n

n

n

n

n–2

y + C4 x 2

n

n–4

y + ...] 4

= 2 (sum of terms at odd places). The last term is nCn yn or nCn – 1 xyn – 1 according as n is even or odd respectively. (e) Subtracting (2) from (1), we get (x + y)n – (x – y)n = 2 [nC1 xn – 1 y + nC3 xn – 3 y3 + ...] = 2 (sum of terms at even places) The last term is nCn – 1 xyn – 1 or nCn yn according as n is even or odd respectively.



y)n = xn 1+



n

y and then find the greatest term in x 

n

y  1+ x  .  

MiddlE TERM iN ThE BiNOMiAl ExPANSiON

The middle term in the binomial expansion of (x + y)n depends upon the value of n. Notes: n  (i) The positive integer n is called the index of the binomial. (i) If n is even, then there is only one middle term i.e.,  + 1 2  (ii) Number of terms in the expansion of (x + y)n is n + 1 i.e., th term. one more than the index n.

696

Objective Mathematics

Important Results on Multinomial  n +1  (ii) If n is odd, then there are two middle terms i.e.,  th Theorem   2  n+3 th terms. and   1. The general term in the multinomial expansion is  2  n! x1r ⋅ x2r ...xkrk r1 ! r2 !...rk ! 1

2

pth Term from the End in the Binomial Expansion  of  (x + y)n

2. The total number of terms in the multinomial expansion is = number of non-negative integral solutions of the equapth term from the end in the expansion of (x + y)n is (n – p + 2)th tion term from the beginning. r1 + r2 + ... + rk = n = n + k – 1Cn or n + k – 1Ck – 1.

Properties of Binomial Coefficients

In the binomial expansion of (1 + x) , the coefficients C0, C1, C2, ... nCn are denoted by C0, C1, C2, ... Cn respectively. n

n

n

3. Coefficient of x1r1 ⋅ x2r2 ...xkrk in the expansion of

n! a1r1 ⋅ a2r2 ...akrk r1 ! r2 !...rk !

n

(a1x1 + a2x2 + ... + akxk)n =

1. If n is even, then greatest coefficient = nCn/2

4. Greatest coefficient in the expansion of (x1 + x2 + ... + xk)n n! = r k −r (q !) ⋅ [(q + 1)!]

2. If n is odd, then greatest coefficient is nC(n –1)/2 or nC(n +1)/2. 3. C0 + C1 + C2 + ... + Cn = 2n 4. C0 + C2 + C4 + ... = C1 + C3 + C5 + ... = 2n – 1 5. C0 – C1 + C2 – C3 + C4 – ... + (– 1)n Cn = 0 2

2

2

2

6. C 0 + C 1 + C 2 + ... C n =

(2n)! 2n = Cn (n!) 2

0, if n is odd 2 2 2 7. C 0 – C 1 – C 2 + ... =  n/2 n ( −1) ⋅ cn / 2 , if n is even 8. C0C1 + C1C2 + C2C3 + ... + Cn – 1 Cn = 2nCn – 1 9. C0Cr + C1Cr + 1 + ... Cn – rCn = 2nCn – r or 2nCn + r

Properties of nCr If 0 < r < n, n, r ∈ N, then 1. r ⋅ nCr = n ⋅ n – 1Cr – 1 n

C r n +1 C r +1 = r +1 n +1 n n–1 3. nCr = Cr – 1 r 2.

n

4.

n

Cr n − r +1 = C r −1 r

n

Cr r +1 = 5. n C r +1 n − r 6. nCr – 1 + nCr = n + 1Cr 7. nCx = nCy ⇒ x = y or x + y = n 8. nCr = nCn – r .

Multinomial Theorem for a positive integral index If x1, x2, ..., xk are real numbers, then for all n ∈ N, (x1 + x2 + ... + xk)n n! x1r1 ⋅ x2r2 ...xkrk , where = ∑ r1 + r2 + ...+ rk = n r1 ! r2 !...rk !

r1, r2, ..., rk are all non-negative integers.

where q is the quotient and r the remainder when n is divided by k. 5. The number of terms in the expansion of (x + y + z)n, where 1 (n + 1) (n + 2). n is a positive integer, is 2 6. Sum of all the coefficients is obtained by putting all the variables xi equal to 1 and it is equal to nm.

Binomial Theorem for any index If n is a rational number and x is real number such that | x | 1, n > 2 and the coefficients of (3r)th term and (r + 2)th term in the binomial expansion of (1 + x)2n are equal, then r = n , n even 2 (c) n

(a)

n 2 (d) 1

(b)

43. If rth and (r + 1)th term in the expansion of (1 + x)n are equal, then n = (a)

(1 + x) r − x 4x

(b)

(1 + x) r − x 3x

(c)

(1 + x) r − x x

(d)

(1 + x) r − x r

36. The two consecutive terms in the expansion of 44. The coefficient of x4 in the expansion of (3x + 2)74, whose coefficients are equal, are (1 + x + x2 + x3)11 is (a) 20th and 21st (a) 990 (b) 30th and 31st (b) 605 (c) 40th and 41st (c) 810 (d) None of these (d) None of these

699

37. If the sum of the coefficients in the expansion of (αx2 – 2x + 1)35 is equal to the sum of the coefficients in the expansion of (x – αy)35, then α is equal to

8

Binomial Theorem

27. The term independent of x in the expansion of

700

45. If (1 + x – 2x2)6 = 1 + a1 x + a2 x2 + ... + a12 x12, then a2 + a4 + a6 + ... + a12 =

Objective Mathematics

(a) 21 (c) 31

(b) 11 (d) None of these

46. Which term in the expansion of the binomial

56. If the coefficients of second, third and fourth terms in the expansion of (1 + x)2n are in A.P., then (a) 2n2 – 9n + 7 = 0 (c) n2 – 9n + 7 = 0

(b) 2n2 + 5n + 7 = 0 (d) None of these

57. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then n

∑k

21

2

⋅ Ck =

  a   b  3   +  3   a     b 

(a) n (n – 1) 2n – 1

(b) n (n + 1) 2n – 2

contains a and b to one and the same power ?

(c) (n + 1) (n + 2) 2n – 1

(d) None of these

(a) 5th (c) 10th

k =1

(b) 8th (d) None of these

58. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then 1 – (1 + x) C1 + (1 + 2x) C2 – (1 + 3x) C3 + ... =

47. If 7103 is divided by 25, then the remainder is (a) 20 (c) 18

(b) 16 (d) 15

48. The numerically largest term in the binomial expansion of (4 – 3x)7 when x =

2 is 3

(a) 46016 (c) 86016

(a) 0 (c) 2

59. If C0, C1, C2, ..., Cn are the coefficients of the expansion of n C (1 + x)n, then the value of ∑ k is 0 k +1 (a) 0

(b) 66016 (d) None of these

49. The greatest coefficient in the expansion of (1 + x)2n is (a) 2nCn (c) 2nCn – 2

(b) 2nCn – 1 (d) None of these

50. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then

(c)

a – nC1 (a – 1) + nC2 (a – 2) – ... + (– 1)n (a – n) is equal to

52. In the expansion of (1 + x)n, coefficients of 2nd, 3rd and 4th terms are in A.P. Then n is equal to (a) 7 (c) 11

(b) 9 (d) None of these

53. The sum of rational terms in the expansion of ( 2 + 31/15)10 is (a) 31 (c) 51 54. The fractional part of

(b) 41 (d) None of these

3n +1 − 2n − 5 (n + 1)(n + 2)

(b)

3n + 2 − 2n − 5 (n + 1)(n + 2)

(c)

3n + 2 + 2n − 5 (n + 1)(n + 2)

(d) None of these

61. The value of 

(183 + 73 + 3 ⋅ 18 ⋅ 7 ⋅ 25) 36 + 6 ⋅ 243 ⋅ 2 + 15 ⋅ 81 ⋅ 4 + 20 ⋅ 27 ⋅ 8 + 15 ⋅ 9 ⋅ 16 + 6 ⋅ 3 ⋅ 32 + 64 is (a) 0 (b) 1 (c) 2 (d) None of these

62. The integral part of ( 2 + 1)6 is (a) 198 (b) 196 (c) 197 (d) 199 63. Larger of 9950 + 10050 and 10150 is

24n is 15

(a)

1 15

(b)

(c)

4 15

(d) None of these

(a) 10150 (b) 9950 + 10050 (c) both are equal (d) None of these

2 15

55. In the expansion of (x + a)n if the sum of odd terms be P and the sum of even terms be Q, then 4PQ = (a) (x + a)n – (x – a)n (c) (x + a)2n – (x – a)2n

(d) None of these

(a)

51. If n is a positive integer greater than 1, then (b) a (d) None of these

2 n +1 − 1 n +1

2n − 1 n

22 ⋅ C0 23 ⋅ C1 2n + 2 ⋅ Cn + + ... + is equal to (n + 1)(n + 2) 1⋅ 2 2⋅3

(b) n2n – 1 (d) None of these

(a) n (c) 0

(b)

60. If C0, C1, C2, ..., Cn be the coefficients in the expansion of (1 + x)n, then

C0 + 3 ⋅ C1 + 5 ⋅ C2 + ... + (2n + 1) ⋅ Cn is equal to (a) n2n (c) (n + 1) 2n

(b) 1 (d) None of these

(b) (x + a)n + (x – a)n (d) None of these

64.

n

C0 −

n

n C1 n C 2 Cn + − ... + (−1) n = 2 3 n +1

1 n −1 1 (c) n +1

(a)

(b)

1 n

(d) None of these

C02 − C12 + C 22 − ... + (−1) n C 2n is equal to

(b) 125 (d) None of these

66. When 599 is divided by 13, the remainder is (a) 8 (c) 10

(b) 9 (d) None of these

67. The last digit of the number (32) is 32

(a) 4 (c) 8

(b) 6 (d) None of these

(a)

n

−1 n +1

3

(b)

3n + 2 − 1 (c) n+2

3 −1 n

(d) None of these

(a)

(2n)! (n!) 2

(b)

(c)

(2n)! (n − 2)!(n + 2)!

(d) None of these n

C0C1 + C1C2 + C2C3 + ... + Cn – 1 Cn =

(c)

(2n)! (n − 2)!(n + 2)!

(b)

(2n)! (n − 1)!(n + 1)!

(b)

(2n)! (n − r )!(n + r )!

(2n)!

(c) (n − 1)! [ ] 2

(d) None of these

73. The value of the sum of the series C0 ⋅ 15C1 + 14C1 ⋅ 15C2 + 14C2 ⋅ 15C3 + ... + 14C14 ⋅ 15C15 is

14

(a) 29C12

(b) 29C10

29

(d) 29C16

(c) C14

is

(a)

m+n

Cr – 1

(b) m + nCr

(c)

m+n

Cr + 1

(d) None of these 2

(b)

n (n + 1)(n + 2) 2 12

(d) None of these

78. If (1 + x)n = C0 + C1 x + C2 x2 + ... Cn xn, then (C0 + C1) (C1 + C2) ... (Cn – 1 + Cn) = k ⋅ C1C2C3 ...Cn where k = nn (n −1)!

(a)

(n + 1) n n!

(b)

(c)

(n + 1) n (n − 1)!

(d) None of these

79. If (1 + x)15 = C0 + C1 x + C2 x2 + ... + C15 x15, then the value of C2 + 2 C3 + 3 C4 + ... + 14 C15 is

80.

C0Cr + C1Cr + 1 + C2Cr + 2 + ... + Cn – r Cn = (2n)! (n!) 2

 201 (d)    100 

n (n + 2)(n + 1) 2 12 n (n + 1)(n + 2) (c) 12

(d) None of these

72. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn , then

(a)

 200  (c)    101 

(d) None of these

C0C2 + C1C3 + C2C4 + ... + Cn – 2 Cn = (2n)! (n!) 2

 201 (b)    102 

(a)

71. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn , then

(a)

j

 C  k 3  k  is ∑ k =1  C k −1 

(2n)! (b) (n − 1)!(n + 1)!

(2n)! (c) (n − 2)!(n + 2)!

 200  (a)    100 

n

(2n)! (n − 1)!(n + 1)!

70. If (1 + x) = C0 + C1 x + C2 x + ... + Cn x , then (2n)! (a) (n!) 2

200

∑ (1 + x)

77. If n is a positive integer and Ck = nCk , then the value of

C02 + C12 + C 22 + ... + C n2 =

2

(d) None of these

76. mCr + mCr – 1 ⋅ nC1 + mCr – 2 ⋅ nC2 + ... + mC1 ⋅ nCr – 1 + nCr =

69. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn , then

n

(c) n C n / 2

j=0

C1 C Cn + 23 ⋅ 2 + ... + 2n + 1 = 2 3 n +1

n +1

n/2 n Cn / 2 (b) (−1)

75. The coefficient of x100 in the expansion of

68. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then 2C0 + 22 ⋅

(a) 0

(a) 219923

(b) 16789

(c) 219982

(d) None of these

C1 C C C + 2 2 + 3 3 + ...n n = C0 C1 C2 C n −1 n (n −1) 2 (n + 1)(n + 2) (c) 2 (a)

(b)

n (n + 1) 2

(d) None of these

81. If (1 + x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n, then a0 + a3 + a6 + ... = (a) 3n + 1 (c) 3n – 1

(b) 3n (d) None of these

82. nCn + n + 1Cn + n + 2Cn + ... + n + kCn = (a) (c)

n+k–1

Cn + 1 Cn + 1

n+k+1

(b) n + kCn + 1 (d) None of these

701

(a) 225 (c) 325

74. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then for n even,

Binomial Theorem

65. For all n ∈ N, 24n – 15n – 1 is divisible by

702

83. The sum of coefficients in the expansion of (7x – 8y)149 is

Objective Mathematics

(a) 1 (c) 0

(b) – 1 (d) None of these

84. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then

92. 1 −

(a)

ΣΣ (Ci + C j ) 2 =

1 1 1⋅ 3 1 1⋅ 3 ⋅ 5 1 1⋅ 3 ⋅ 5 ⋅ 7 1 ⋅ + ⋅ − ⋅ + ⋅ ... = 2 2 2 ⋅ 4 2 2 2 ⋅ 4 ⋅ 6 23 2 ⋅ 4 ⋅ 6 ⋅ 8 2 4 1 3

(b)

2 3

(d)

4 3

0≤i≤ j ≤n

(a) (n – 1) ⋅ 2nCn + 22n (c) (n + 1) ⋅ 2nCn + 22n

(c) 1

(b) n ⋅ 2nCn + 22n (d) None of these

93. The coefficient of xn in the expansion of

85. If n is an even positive integer, then

(1 – 2x + 3x2 – 4x3 + ... to ∞)– n is

1 1 1 1 + + + ... + = 1!(n − 1)! 3!(n − 3)! 5!(n − 5)! (n − 1)!1! 2n −1 (n − 1)!

(a)

n (c) 2 n!

86.

n

∑ r =0

n

(b)

2n −1 n!

(b)

(2n)! [(n − 1)!]2

(c)

(2n)! (n!) 2

(d) None of these 94. The values of a and b so that the coefficient of xn in the a + bx may be 2n + 1, are expansion of (1 − x) 2

(b) 2n sin nx (d) None of these

87. The total number of terms in the expansion of (a + b + c + d)n, n ∈ N is n (n + 1)(n + 2) (a) 6 n (n + 1)(n + 2)(n + 3) (b) 6 (n + 1)(n + 2)(n + 3) (c) 6 (d) None of these

(a) 1, 1 (c) 2, 3

(b) 1, 2 (d) None of these

95. In the binomial expansion (a + bx)– 3 = values of a and b are (a) a = 2, b = 3 (c) a = 3, b = 2

88. The greatest cofficient in the expansion of (x + y + z + w)15 is (a)

15! 3!(4!)3

(b)

(c)

15! 2! (4!) 2

(d) None of these

places is

3 x 2 5 (c) 1 – x 8

(a) 1 – (b) 13063600 (d) None of these

1 + x − 2 x3 is 90. The coefficient of xn in the expansion of (1 − x)3 (b) 3n – 1

the value of a and b are

(c) 4n – 1

(d) 5n – 1

(a) 2, 12 (c) 1, 12

2r ! r ⋅2 (b) (r !) 2

2r ! 2 r ⋅2 (c) (r !) 2

(d) None of these

(b) 0.2085 (d) None of these

−

1 2

is

(b) 1 +

5 x 8

(d) None of these

98. If the binomial expansion of (a + bx)–2 is

(a) n

2r ! (a) (r !) 2

1 correct to four decimal 23

97. If x is so small that its square and higher powers may be (8 + 3 x) 2 / 3 is neglected, then the value of (2 + 3 x)(4 − 5 x)1/ 2

(2x – 3y + 4z)9 is

91. The coefficient of xr in the expansion of (1 − 4 x)

1 9 + x + ... , the 8 8

(b) a = 2, b = – 6 (d) a = – 3, b = 2

96. The approximate value of (a) 0.2084 (c) 0.2086

15! (3!)3 4!

89. The coefficient of x3 y4 z2 in the expansion of (a) 12063600 (c) 11063600

(2n)! n!(n − 1)!

(d) None of these

C r sin r x cos (n − r ) x =

(a) 2n – 1 sin (n – 1) x (c) 2n – 1 sin nx

(a)

1 – 3x + ..., then 4

(b) 2, 10 (d) None of these

99. The two values of the rational exponent m such that in the binomial expansion of (1 – x)m, the coefficient of x2 is 3, are (a) – 3, 2 (c) 3, – 2

(b) 3, 2 (d) – 3, – 2

10

1 is (1 − x)(1 − 2 x)(1 − 3 x) 1 n+2 n+3 (2 – 3 + 1) 2 1 n+2 n+3 (3 – 2 + 1) (b) 2 1 n+3 n+2 (2 – 3 + 1) (c) 2 (d) None of these

405 256 450 (c) 263

(a)

35 24 35 (c) a = 1, b = 24

(b) a = –

35 ,b=1 24

(d) None of these

(b) x n (d) None of these

103. If x is nearly equal to one, then the approximate value of ax b − bx a is xb − x a (a)

1 x −1

(c) x – 1

(d) None of these

(1 + 3 2 x)9 + (1 − 3 2 x)9 is

102. If x is nearly equal to 1, then the approximate value of mx m − nx n is m−n (a) x m (c) x m + n

(b)

(a) 9 (c) 5

(b) 0 (d) 10

110. The expression [x + (x3 – 1)1/2]5 + [x – (x3 – 1)1/2]5 is a polynomial of degree (a) 5 (c) 7

(b) 6 (d) 8

111. 7C0 + 7C1 + 7C2 + 7C3 + 7C4 + 7C5 = (a) 128 (c) 120

(b) 121 (d) 129

112. In the expansion of (1 + x)50, the sum of the coefficient of odd powers of x is (b) 249 (d) 251

(a) 0 (c) 250

113. If the coefficient of x2 in the expansion of 1 1 + is 246, then a = (1 + x) 2 (a + x) 2 1 2 1 (c) 4

1 1− x

(b)

(a)

(d) 1 – x

105. The term independent  of  x in the expansion of (2 − x)(2 + x) (when expanded as a power of x for| x |

(b) | x | < (d) x
2 (c) | x | < 2

(b) | x | > 2 (d) x < 2

140. ( 3 + 1)4 + ( 3 – 1)4 is equal to

n

∑ r =1

n

Pr = r!

(a) 2n + 1 (c) 2n – 1

(b) 2n – 1 (d) 2n

142. The value of (7C0 + 7C1) + (7C1 + 7C2) + ... + (7C6 + 7C7) is (a) 28 – 2 (c) 28

(b) 28 – 1 (d) None of these

143. The coefficient of x7 in the expansion of (1 – x4) (1 + x)9 is (a) – 48 (c) 9

(b) 48 (d) 27

144. aC0 + (a + b) C1 + (a + 2b) C2 + ... + (a + nb) Cn is equal to (a) (na + 2b) 2n (c) (2a + nb) 2n

(b) (na + 2b) 2n – 1 (d) (2a + nb) 2n – 1

145. In the expansion of (3x + 2)4, the coefficient of middle term is (a) 36 (c) 54

(b) 216 (d) 81

146. The integral part of (8 + 3 7 ) n is (a) an even integer (c) zero

(a) 256 (c) 128

(b) 64 (d) 0

150. The sum of coefficient in the expansion of (1 – x)10 is (a) 1024 (c) 1

(b) 0 (d) 102

151. If n is a positive integer, then 2 ⋅ 42n + 1 + 33n + 1 is divisible by (a) 27 (c) 9

(b) 11 (d) 2

152. Given positive integers r > 1, n > 2 and the coefficients of (3r) and (r + 2)th terms in the expansion of (1 + x)2n are equal, then (a) n = 2r (c) n = 3r

(b) n = 2r + 1 (d) None of these

153. Coefficient of xr in the expansion of (1 – 2x)–1/2 is

(a) a rational number (b) a negative integer (c) an irrational number (d) None of these 141.

149. The sum of the binomial coefficient in the expansion of (1 + x)7 is

(b) an odd integer (d) nothing can be said

(a)

2r ! (r !) 2 ⋅ 22 r

(b)

2r ! 2r (r + 1)!(r − 1)!

(c)

2r ! (r !) 2

(d)

2r ! (r !) 2 ⋅ 2r

154. The value of

C1 C3 C5 + + +... is equal to 2 4 6

(a)

2n + 1 n +1

(b)

2n − 1 n +1

(c)

2n + 1 n −1

(d)

2n n +1

k   155. If the absolute term in the expansion of  x − 2  x   405, then k is equal to

10

(a) ± 2 (c) ± 1

is

(b) ± 3 (d) None of these

156. The sum to (n + 1) terms of the series C0 C1 C 2 C3 − + − +... is 2 3 4 5 1 1 (b) (a) n (n + 1) n+2 1 (d) None of these (c) n +1

1 147. If (1 + ax)n = 1 + 8x + 24x2 + ..., then the values of a and n 157. If = 1 + a1 x + a2 x2 + ... then the value of ar are equal to 1 − 2x + x2 is (a) 1, 2 (b) 3, 6 (c) 2, 3

(d) 2, 4

148. The power of x occurring in the 7th term in the expansion  4x 8  −  is of   5 5x  2

(a) – 5 (c) – 3

(a) 2r (c) r

(b) r + 1 (d) r – 1

158. If the sum of the coefficients in the expansion of (α2 x2 – 2αx + 1)51 vanishes, then α is equal to

(b) 5 (d) 3

(a) – 2 (c) 1

(b) 2 (d) – 1.

705

10

Binomial Theorem

x 3  137. The coefficient of x in  − 2  2 x  4

706

159. The value of 2 C0 +

22 23 211 C1 + C2 + ... + C is 2 3 11 10

Objective Mathematics

112 − 1 11 211 − 1 (c) 11

113 − 1 11 311 − 1 (d) 11 (b)

(a)

n

1 160. If sn = ∑ n and tn = r = 0 Cr

161. 1 +

5 24 1 (c) 24

r tn then is ∑ n sn r = 0 Cr 1 n–1 2 2n − 1 (d) 2

n

2n

171.

162. Sum of the infinite series 2 1 2 5 1 2 5 8 1 ⋅ + ⋅ ⋅ + ⋅ ⋅ ⋅ + ...∞ is 3 2 3 6 2 2 3 6 9 23

163. In the expansion of (1 + x)n, the sum of coefficients of odd powers of x is (b) 2n – 1 (d) 2n – 1 m

164. The sum

10   20 

∑  i   m − i  , where 1= 0







 p   = 0 if p < q is the q

maximum when m is (a) 5 (c) 15

(b) 10 (d) 20

(a) 12C6 + 2

(b) 12C5

(c) 12C6

(d) 12C7

1   166. The middle term in the expansion of  x +  2x   1 ⋅ 3 ⋅ 5...(2n − 1) n! 1 ⋅ 3 ⋅ 5...(2n + 1) (c) n!

(b)

n (n + 1)(2n + 1) 6

(b)

(c)

n (n − 1)(2n − 1) 6

(d) None of these

(a) 198 (c) 197

n (n + 1) 2

(b) 196 (d) 199

173. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then the value of C0 + 2 C1 + 3 C2 + ... + (n + 1) Cn is (a) (n + 2) 2n – 1

(b) (n + 1) 2n

n–1

(c) (n + 1) 2

(d) (n + 2) 2n



(a)  260 (b)  259 61 (c)  2 (d)  none of these 175. If (1 + x)n = C0 + C1x + C2x2 +…+ Cnxn then C0 +

2n

is

1 ⋅ 3 ⋅ 5...(2n − 1) n!

(d) None of these

C1 C2 C + + ... + n is equal to 2 3 n +1

(a)  2n + 1 (c)  

(b)  

2 n +1 n +1

176. The expansion of

2 n +1 − 1 n +1

(d)  2n + 1 – 1 1 by binomial theorem will be (4 − 3 x)1/ 2

valid, if

167. The number of terms in the expansion of ( 5 + 4 11 )124 which are integers, is equal to (a) 0 (c) 31

(a)

174. In the expansion of (1 + x)60, the sum of coefficients of odd powers of x is

165. Find coefficient of t32 in the expansion of (1 + t2)12 (1 + t12) (1 + t24) is

(a)

C2 + 2 [2C2 + 3C2 + 4C2 + ... + nC2] =

172. Let n ∈ N and n < ( 2 + 1)6. Then the greatest value of n is

(b) 41/3 (d) None of these

(a) 2n + 1 (c) 2n

(b) 5 (d) None of these

n+1

(d) None of these

(a) 21/3 (c) 81/3

(b) 30 (d) 32

is

(b) 12 (d) None of these

(a) 40 (c) 41

n

1+

n

170. The number of irrational terms in the expansion of (41/5 + 71/10)45 is

1+ x  (b)   1− x 

1− x  (c)   1+ x 

(d) None of these

(a) 8 (c) 4

2nx n (n + 1) + + ...∞ equals 1+ x 2!

1+ x  (a)   1− x 

9 24

 13 a   169. If the second term in the expansion  a + −1 a   n C3 14 a5/2, then the value of n is C2

(b)

1 n 2

(b)

(a)

n

(a) n – 1 (c)

 52n  168. The value of   , n ∈ N where {} denotes the frac 24  tional part of x, is

(a)  x < 1 (c)   −

2 2 q, is − i m i   i =0    m

maximum when m is (a)  5 (c)  20

(b)  15 (d)  10

181. In the binomial expansion of (a – b)n, n ≥ 5, the sum of 5th a and 6th terms is zero, then equals b 5 6 (b)   (a)   n−4 n−5 n−5 n−4 (c)   (d)   6 5 182. The sum of the series 20C0 – 20C1 + 20C2 – 20C3 + … – … + 20 C10 is 1 20C 10 2

(a)  – 20C10

(b)  

(c)  0

(d)  20C10

(c)   n

25840 9

C0 n C2 n C4 n C6 + + + + … is equal to 1 3 5 7

2n +1 n +1 2n (c)   n +1

(a)  2 (c)  214

(b)  

(a)  

2 n +1 − 1 n +1

(d)  None of these

187. The value of 1 10 102 − n 2 nC1 + n n 81 81 81

2n

C2 −

(a)  2 1 (c)   2

103 2 n 102 n ⋅ C3 + .... + n is n 81 81

(b)  0 (d)  1 6

1  188. In the expansion of  x −  , the constant term is x  (a)  20 (c)  30

(b)  – 20 (d)  – 30

189. If 8Cr – 7C3 = 7C2, then r is equal to (a)  3 (c)  8

(b)  4 (d)  6

190. Coefficient of x5 in (1 + 2x + 3x2 + ….)3/2 is (a)  19 (c)  21

183. The sum of the last eight coefficients in the expansion of 191. If I = (1 + x)15 is 16

(d)  None of the above

15

(b)  2 (d)  none of these

3π

∫ 0

(b)  20 (d)  22 π

f (cos 2 x)dx & J = ∫ f (cos 2 x)dx. Then 0

(b)  I = J (d)  none of these.

(a)  I = 5J (c)  I = 3J

solUtions 10

1 1. (b) There are 11 terms in the expansion of  3 x − 2  . x   ∴

5th term from the end = t11 – 4



 1  = t7 = 10C6 (3x)10 – 6 ⋅  − 2   x 



=

6

10! 1 17010 ⋅ 34 x4 ⋅ 12 = . 6! 4! x x8

12

2   2. (a) When  3 x + 2  is expanded, the power of x goes 3x   on decreasing as the term proceed. Hence, it is expanded 12

in descending powers of x. So  2 + 3 x  , when ex 2   3x  panded, will be in ascending powers of x. 12

 2  Now, t8 in  2 + 3 x   3x 

12 − 7

 2  = 12C7  2   3x 

⋅ (3x)7

707

Binomial Theorem

177. If the expansion in powers of x of the function 184. The greatest term in the expansion of (1 + 3x)54, where x = 1/3 is 1 is a0 + a1x + a2x2 + a3x3 + …., then an is (1 − ax) (1 − bx) (a)  T28 (b)  T25 (c)  T26 (d)  T24 n n n +1 n +1 a −b a −b (b)   (a)   20 1   b−a b−a 3 1 + 185. The greatest term in the expansion of   is 3  b n +1 − a n +1 bn − a n 26840 24840 (c)   (d)   (b)   (a)   b−a b−a 9 9

708

10

Objective Mathematics

=

5 12!  2  ⋅  2  ⋅ (3x)7 7! 5!  3 x 

=

12 × 11 × 10 × 9 × 8 25 ⋅ 32 228096 ⋅ = . 3 x3 5× 4 × 3× 2 x

3. (b) We have, (x + a)n = nC0 xn + nC1 xn – 1 a1 + nC2 xn – 2 a2  + nC3 xn – 3 a3 + ... + nCn xn = (nC0 xn + nC2 xn – 2 a2 + ...)  + (nC1 xn – 1 a1 + nC3 xn – 3 a3 + ...) =A+B (x – a)n = nC0 xn – nC1 xn – 1 a1 + nC2 xn – 2 a2  – nC3 xn – 3 a3 + ... + nCn (– 1)n an n n n n–2 = ( C0 x + C2 x a2 + ...) n – ( C1 xn – 1 a1 + nC3 xn – 3 a3 + ...) = A – B ∴ A2 – B2 = (A + B) (A – B) = (x + a)n (x – a)n = (x2 – a2)n. 4. (b) rth term of ( 3 + 5 ) i.e., Tr+1 = 256Cr(3)(256–r)/2 (5)r/8 256 − r r The terms are integral if and are both 2 8 positive integers. 256

8

Hence total number of terms are 33. 5. (a) The general term is 10 − r

 x Tr + 1 = 10Cr   3    5− r

 3  2  2x

  

r

r/2

5 − 3r 1 2 3 2 = Cr     ⋅x 3 2     For term independent of x, 3r =0 5– 2

10

10 , which is not a positive integer. 3 Hence, there is no term independent of x. ⇒ r =

6. (d) We have (1 + 2x + 3x2 + ...)–3/2 = [(1 – x)–2]–3/2

= (1 – x)3

So, coefficient of x5 in (1 + 2x + 3x2 + ...)–3/2 = c oefficient of x5 in (1 – x)3 = 0. 7. (b) We have (1 + x + x2 + x3 + ...)2 = [(1 – x)–1]2 = (1 – x)–2 ∴ Coefficient of xn in (1 + x + x2 + ...)2

= coefficient of xn in (1 – x)–2



=

n +2–1

C2–1 =

n+1

C1 = n +1.

 c3  tr = 10Cr – 1 y10 – r + 1  y 2   

r −1

the rth term is

c3r − 3 = 10Cr–1 y11–r y 2 r − 2

= 10Cr – 1 y11 – r – 2r + 2 c3r – 3 Thus, tr = 10Cr – 1 y13 – 3r c3r – 3 Let the rth term contains y

...(1)

–2

15 = 5. 3 Therefore, 5th term will contain y– 2

∴ 13 – 3r = – 2 ⇒ r =

From (1), t5 = 10C5 – 1 c15 – 3 y– 2 = 10C4 c12 y– 2 ∴ coefficient of y– 2 = 10C4 c12 =

10!c12 10!c12 10 ⋅ 9 ⋅ 8 ⋅ 7 = = ⋅ c12 = 210 c12. 4!(10 − 4)! 4!6! 4 ⋅ 3 ⋅ 2 ⋅1

9. (a) (n + 1)th term from the end = [3n – (n + 1) + 2]th term from the beginning 1 = T2n + 1 = 3nC2n (2x)3n – 2n  −   x

2n

3n! ⋅ 2n ⋅ xn (– 1)2n ⋅ 1 2 n !n ! x2n 3n! = 2 n ⋅ x – n. 2 n !n ! =

∴ r = 0,8,16,24, 32, ..., 256



 c3  8. (d) In the expansion of  y + 2  y   given by

10 − ( r −1)

x 10. (b) We have, Tr= 10Cr – 1   3

 2   − x2   

r −1

11 − r

1 = 10Cr – 1   ⋅ (– 2)r – 1 x13 – 3r 3 For coefficient of x4, 13 – 3r = 4 ⇒ r = 3. 10

1  11. (c) When  + y 2  y  

is expanded, the powers of y go on

increasing as the terms proceed. Hence it is expanded 10

 1 in ascending powers of y. So  y 2 +  , when exy  panded, will be in descending powers of y. Hence, 10 × 9 × 8 × 7 2 t7 = 10C6 (y2)4  1  = y = 210 y2. 4 × 3 × 2 × 1 y   12. (a) The (p + 2)th term from the end 6

 the {(2n + 2) – ( p + 2 – 1)}th term from the = beginning = the (2n – p + 1)th term from the beginning 2n − p 1 = 2n + 1C2n – p x(2n + 1) – (2n – p)  −   x ( 2 + 1 )! n = (−1) p x 2 p − 2 n +1 . (2n − p )!( p + 1)!

(1 + x)

 1  = (1 + x)m  x + 1   1 +   x  x  

m

14. (c) We have, 3x 9 (2 + 3x)9 = 29 1 +  = 29 1 +   2   4 

=

9

 ∵ x = 

3  2

x (n + 1) ( x + 1) 9   (9 + 1) 4 9   +1 4

= 2 9 ⋅ 9C 3  9    4 15. (a) We have,

=

6

= 29 ⋅

90 12 =6 ≠ Integer 13 13

6

11

11

 1 = 311 1 −   3

1  ∵ x =  5 

 | x | (n + 1) (| x | + 1)

 1   − < 0  3 

 1   −  (11 + 1) =  3  =3  1   − + 1  3  The greatest terms in the expansion are T3 and T4 ∴  Greatest term (when r = 2) = 311 | T2 + 1 |  1 C2  −   3

11

2

= 311

11 ⋅ 10 1 × 1⋅ 2 9

= 55 × 39

and greatest term (when r = 3) = 311 | T3 + 1 |  1 C3  −   3

11

6

1  = tn + 1 – 6 = tn – 5 = nCn – 6 ( 3 2 )6   3  3



3

= 311

 1  C6 ( 3 2 ) n − 6  3   3

 1  n Cn − 6 ( 3 2 )6  3   3

1 ⇒  2 3   

n −12

1 1 ⇒  2 3 ⋅ 33   

6

n−6

12 − n

 − 13  3   

n−6

=

1 6

1 6

=

n −12

= 6–1;  or  6

n−12 3

= 6–1

n −12 = – 1;   ∴  n = 12 – 3 = 9. 3 2  1  18. (a) We have, t3 = mC2 (9x)m – 2 ⋅  −  . 3x   ⇒

312 9⋅8⋅7 7 × 313 ⋅ 12 = . 2 1⋅ 2 ⋅ 3 2

 5x  (3 – 5x)11 = 311 1 −  3  

= 311

 1  3 17. (c) We have, t7 = nC6 ( 2 )n – 6  3   3

Given,

= 2 9 ⋅ T 6 + 1 = 2 9 ⋅ 9C 6  9    4

= 311

= (x – 3)100 + 100C1 (x – 3)99 ⋅ 21  + 100C2 (x – 3)98 ⋅ 22 + ... + 100C100 2100 = [(x – 3) + 2]100 = (x – 1)100 = (1 – x)100  ∴ coefficient of x54 = 100C53 (– 1)53 = – 100C53.

n

The greatest term in the expansion is  T[m] + 1 = T6 + 1 = T7 Hence the greatest term = 29 ⋅ T7

∴ m =

C m ( x − 3)100− m ⋅ 2m

7th term from the end 9



100

m=0

m+n = (1 + x) = x– n (1 + x)m + n n x ∴ Required term independent of x = co-efficient of x0 in x– n (1 + x)m + n = coefficient of xn in (1 + x)m + n = m + nCn.

∴ m =

100

∑

16. (c) We have,

n

709

n

11 ⋅ 10 ⋅ 9 1 = 55 × 39 ×− 1⋅ 2 ⋅ 3 27

 rom above we see that the values of both greatest F terms are equal.

Also from question, mC2 = 105 ∴ m (m – 1) = 210  or m2 – m – 210 = 0; or m = 15, – 14 (absurd). 2

 1  ∴ t3 = 15C2 ⋅ (9x)13 ⋅  −  3x   15 ⋅ 14 1 = ⋅ 913 ⋅ x13 ⋅ = 35 ⋅ 913 ⋅ x12. 2 3x 19. (c), (d)  The given expression n

2 2n   =  x + 1   =  x + 1  .   x   x  

The number of terms = 2n + 1, which is odd. t The middle term = ( 2 n +1) +1 = tn + 1 2 1 = 2nCn x2n – n ⋅    x

n

=

(2n)! (n)! (n)!

=

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6...(2n − 1) ⋅ 2n (n)! (n)!

=

[1 ⋅ 3 ⋅ 5...(2n − 1)][2 ⋅ 4 ⋅ 6...2n] (n)! (n)!

Binomial Theorem

13. (a), (b)  We have,

710

Objective Mathematics

=

[1 ⋅ 3 ⋅ 5...(2n − 1)] 2n [1 ⋅ 2 ⋅ 3...n] (n)! [1 ⋅ 2 ⋅ 3....n ]

=

1 ⋅ 3 ⋅ 5...(2n − 1) ⋅ 2n. (n)!

Since rth term is independent of x 15 − 5r = 0 ∴ r = 3 2 Putting r = 3 in (1), we get t3 = 10C2 k2 ∴

It is given that,

20. (d) Putting x = 1 in (1 + 5x – 7x ) , the required sum of coefficients = (1 + 5 – 7)3165 = (– 1)3165 = – 1. 3 3165

21. (a) Putting a = b = 1 in (6a – 5b) , the required sum of coefficients = (6 – 5)n = 1n = 1. n

22. (c) The given expression =

8

1 27. (b) In the expansion of  x1/ 3 + x − 1/ 5  , (r + 1)th term 2  is given by

2 [1 + 9C2 (3 2 x)2 + 9C4 (3 2 x)4

23. (b) We have, tr + 1 = Cr (x )

4 15 – r

15

r

r

1 Then tr + 1 = 2nCr (x2)2n – r   = 2nCr x4n – 3r  x 4n − p 3

 k  Now, tr = 10Cr – 1 ( x )10 – r + 1  2  x  r −1 k = 10Cr – 1 ( x )11 – r 2 r − 2 x = Cr – 1 x

∴ tr = 10Cr – 1 x

r −1

k r −1 × 2 r − 2 = 10Cr – 1 x x

15 − 5 r 2

k r – 1

40 − 8r = 0, ∴ 15

...(1)

r = 5.

=–2 ⋅

6! 6! = – 50. − 2! 4! 3! 3! 3

1 29. (c) We have, t4 = nC3 (px)n – 3 ⋅    x  = nC 3 ⋅ p n – 3 ⋅ x n – 6 This is independent of x, provided n – 6 = 0; that is, n = 6. ∴  The value of the term independent of x = 6C3 ⋅ p3 = 20 p3. ∴ greatest coefficient = coefficient of the middle term = 2nCn

C 4n − p .

26. (a), (b) Let the rth term in the expansion be independent of x.





30. (a), (b)  Since the middle term has greatest coefficient, 3

11− r 2

40 − 8 r 15

= 6C0 – 6C1 x (1 – x) + 6C2 x2 (1 – x)2 – 6C3 x3 (1 – x)3 + ... to 7 terms = 6C0 – 6C1 x (1 – x) + 6C2 x2 (1 – 2x + x2)  – 6C3 x3 (1 – 3x + 3x2 – x3) + ... to 7 terms ∴ Co-efficient of x3 = – 2 ⋅ 6C2 – 6C3, (collecting coefficients of x3 from each term)

25. (a) Let tr + 1 contains x p.

Hence, the coefficient of xp =

⋅x

r

5

3

∴  The coefficient of x30 = 45C30.

2n

8− r

−1

T6 = 8C5 ⋅  1  ⋅ x 0 = 7.   2 2 6 28. (b) (1 – x + x ) = {1 – x (1 – x)}6

24. (c) (1 + 3x + 3x2 + x3)15 = {(1 + x)3}15 = (1 + x)45

10

= 8C r ⋅  1    2

(x )



 ence the given expression is independent of the H term containing x–16. Thus, the coefficient of x–16 = 0.

or r =



8− r

 ence the 6th term is independent of x. Putting r H = 5 in (1), we get

If this term contains x32, then 60 – 7r = 32; or r = 4 Hence the co-efficient of x32 = 15C4 (– 1)4 15 × 14 × 13 × 12 = = 1365 4 × 3 × 2 ×1 Again, if the term tr + 1 contains x–16, then 60 – 7r = – 16; 76 6 = 10 which is not possible. or r = 7 7

∴ 4n – 3r = p;

1 1 Tr + 1 = 8Cr  x 3  2 

The index of x is

 1  = 15C x60 – 7r (– 1)r. r − 3   x 

C2 k 2 = 405

∴ 45k = 405 or k2 = 9 ∴ k = ± 3.



 + 9C6 (3 2 x)6 + 9C8 (3 2 x)8] ∴ The number of non-zero terms is 5.

10

2

11− r − 2r + 2 2

=

(2n)! = 2n (2n − 1)(2n − 2)(2n − 3)...4 ⋅ 3 ⋅ 2 ⋅ 1 n !n ! n !n !

=

[(2n − 1)(2n − 3)...3 ⋅ 1] [2n (2n − 2)(2n − 4)...4 ⋅ 2] n !n !

r–1

k



=

[1 ⋅ 3 ⋅ 5...(2n − 1)] 2n [n (n − 1)(n − 2)...2 ⋅ 1] n !n !

...(1)



=

1 ⋅ 3 ⋅ 5...(2n − 1) 2n 1 ⋅ 3 ⋅ 5...(2n − 1) 2n n! = . n! n !n !

2 1 1 + = r (n − r ) (n − r + 1)(n − r ) r (r + 1)

or

Tr + 1 = 10Cr (– x2)r

∴ coefficient of x10 will be in the 6th term = – 10C5.

711



or 2 (n – r + 1) (r + 1) 10

2  The general term in the expansion of  x −  is x  r  2 10 10 – r 10 r r 10 Tr + 1 = Cr (x)  −  = Cr (– 1) ⋅ 2 ⋅ x – 2r.  x ∴ The term independent of x in the expansion will be 6th term = – 25 ⋅ 10C5. Hence the ratio is 1 : 25 or 1 : 32. 32. (b), (c)  Put log10 x = z Then, given expression = (x + xz)5. Now, T3 = 5C2 x3 (xz)2 = 10x3 + 2z = 106 ∴ x3 + 2z = 105. Taking log, we get (3 + 2z) log10 x = 5 log10 10 ⇒ (3 + 2z) z = 5 or 2z2 + 3z – 5 = 0 5 ⇒ (z – 1) (2z + 5) = 0 ⇒ z = 1, – 2 5 ∴ log10 x = 1 or – ∴ x = 101 or 10–5/2. 2 33. (b) We have, Tr – 1 = nCr – 2 ⋅ xr – 2, ∴ Coeff. is nCr – 2, and T2r + 3 = nC2r + 2 ⋅ x2r + 2, ∴ Coeff. is nC2r + 2

= r (r + 1) + (n – r + 1) (n – r)

or n2 – n (4r + 1) + 4r 2 – 2 = 0. 36. (b) Let the coefficient of r th and (r + 1)th terms be equal. We have, tr = 74Cr – 1 (3x)74 – (r – 1) (2)r – 1 = 74Cr – 1 375 – r 2r – 1 x75 – r and

tr + 1 = 74Cr 374 – r 2r x74 – r

Then,

74

Cr – 1 375 – r 2r – 1 = 74Cr 374 – r ⋅ 2r

or

(74)! (74)! ⋅3= ⋅2 (r − 1)! (75 − r )! (r )! (74 − r )!

or

3 2 = (r − 1)! (75 − r ) (74 − r )! r (r − 1)! (74 − r )!

or 3r = 150 – 2r, or r = 30. Hence, the required consecutive terms are 30th and 31st. 37. (c) We have, sum of the coefficients in the expansion of (αx2 – 2x + 1)35. = Sum of the coefficients in the expansion of (x – αy)35 ⇒ (α – 1)35 = (1 – α)35 [Putting x = y = 1] ⇒ (α – 1)35 = – (α – 1)35 ⇒ 2 (α– 1)35 = 0 ⇒ α – 1 = 0.  ∴  α = 1. 2n

 2 1 38. (a) The general term in the expansion of  x +  is x  ⇒ (r – 2) + (2r + 2) = n = 15 r 1 2n 2 2n – r ⋅   Tr + 1 = Cr (x )  ( nCp = nCq ⇒ p + q = n)  x or 3r = 15 ∴ r = 5. Index of x is 4n – 2r – r = p (Given). 34. (a) We have, 4n − p ∴ r = T2r + 1 = 43C2r ⋅ x2r, ∴ Coeff. is 43C2r 3 and Tr + 2 = 43Cr + 1 ⋅ xr + 1, ∴ Coeff. is 43Cr + 1 which must be an integer. Given 43C2r = 43Cr + 1 ⇒ 2r + r + 1 = 43 ∴ Coefficient is ⇒ 3r = 42. ∴ r = 14. (2n)! 2n C 4n − p = . 1    1 r–1 r r+1 3 35. (c) The co-efficients of x  , x  , x  in the expansion of ( 2n + p )  ! ( 4n − p )  !  3  3  (1 + x)n are respectively nCr – 1, nCr , nCr + 1, which are in A.P. 39. (c)  xn occurs in (n + 1)th term and their coefficients are 2n ∴ 2 ⋅ nCr = nCr – 1 + nCr + 1 Cn and 2n – 1Cn. Their ratio is or 2n Cn (2n)! n!(n − 1)! ⋅ = n! n! n! 2 n −1 Cn n!n! (2n − 1)! + 2⋅ = r !(n − r )! (r − 1)!(n − r + 1)! (r + 1)!(n − r − 1)! 2n ⋅ (2n − 1)! (n − 1)! ⋅ 2  = n ⋅ (n − 1)! (2n − 1)! or r (r − 1)!(n − r )(n − r − 1)! = 2 : 1 1 Hence, first coefficient is double the other. = (r − 1)!(n − r + 1)(n − r )(n − r − 1)! 40. (d) We have, (1 + x2)5 (1 + x)4 1 = (1 + 5C1 x2 + 5C2 x4 + ...) (1 + 4C1 x + 4C2 x2  + (r + 1)r (r − 1)!(n − r − 1)! + 4C3 x3 + 4C4 x4) Given nCr – 2 = nC2r + 2

Binomial Theorem

31. (b) The general term in the expansion of (1 – x2)10 is

712

Objective Mathematics

Putting x = – 1, we get 64 = 1 – a1 + a2 – ... + a12 Adding (1) and (2), we get 64 = 2 (1 + a2 + a4 + ...) ∴ a2 + a4 + a6 + ... + a12 = 31.

= (1 + 5x2 + 10x4 + ...) (1 + 4x + 6x2 + 4x3 + x4) The term giving x5 in the above product is (5x2) (4x3) + (10x4) (4x) = (20 + 40) x5 = 60x5. Hence, the coefficient is 60. 41. (d) Coefficient of x r in (1 – x)2n – 1 = 2n – 1Cr (– 1)r. ∴ ar = 2n – 1Cr (– 1)r; ∴ ar – 1 = 2n – 1Cr – 1 (– 1)r – 1 and a2n – r = 2n – 1C2n – r ⋅ (– 1)2n – r = 2n – 1C(2n – 1) – (2n – r) ⋅ (– 1)2n ⋅ (– 1)– r,  { nC r = nC n – r} = 2n – 1Cr – 1 ⋅ (– 1)2r ⋅ (– 1)– r, { (– 1)2r = 1 = (– 1)2n} = 2n – 1Cr – 1 (– 1)2r – r = 2n – 1Cr – 1 (– 1)r ∴ a2n – r = – 2n – 1Cr – 1 (– 1)r – 1 = – ar – 1; ∴ ar – 1 + a2n – r = 0.

...(2)

46. (c) Let (r + 1)th term in the expansion of the given binomial contains a and b to one and the same power. We have, Tr + 1 = 21Cr a7 – r/2 ⋅ b2r/3 – 7/2  ince the powers of a and b are the same, we S have 1 7– r = 2 r − 7 or r = 9. 2 3 2  ence (9 + 1)th, that is, 10th term is required H term. 47. (c) We have, 7103 = 7 (49)51 = 7 (50 – 1)51

42. (a) We have, t3r = 2nC3r – 1 x3r – 1

= 7 (5051 – 51C1 5050 + 51C2 5049 – ... – 1) tr + 2 = 2nCr + 1 xr + 1. 2n = 7 (5051 – 51C1 5050 + 51C2 5049 – ...) – 7 + 18 – 18 C3r – 1 = 2nCr + 1 – 1 = r + 1; or (3r – 1) + (r + 1) = 2n = 7 (5051 – 51C1 5050 + 51C2 5049 – ...) – 25 + 18 = 2 ; or 4r = 2n k is divisible by 25, = k + 18 (say) n ∴ remainder is 18. ⇒ r = 1 (impossible); or r = . 2 7 7–r r But r is a positive integer greater than 1. So the value 48. (c) Here | tr + 1 | = Cr ⋅ 4 ⋅ (3x) n and | tr | = 7Cr – 1 48 – r (3x)r – 1 of r is provided n is an even integer (> 2), otherwise 2 7! r has no value. | tr +1 | 1 (r − 1)! (8 − r )! ∴ = × ⋅ ⋅ 3x r ! ( 7 − r )! | t | 4 7 ! r 43. (c) We have, tr = tr + 1 8−r 3 2 8−r or nCr – 1 xr – 1 = nCr xr = ⋅ ⋅ = n! n! r 4 3 2r or =  x 8 (r − 1)! (n − r + 1)! r ! (n − r )! ∴ | tr + 1 | ≥ | tr | if 8 – r ≥ 2r or ≥ r . 3 or r = (n – r + 1) x or r (1 + x) = (n + 1) x ∴ If r = 1, 2 then | tr + 1 | > | tr | or n = (1 + x)r − x If r = 3, 4 then | tr + 1 | < | tr | x Thus we get | t1 | < | t2 | < | t3 | > | t4 | > ... Here n is a positive integer. and Given, ⇒ 3r ⇒ 2r

∴

∴ Numerically largest term

(1 + x)r − x is a positive integer. x

44. (a) We have, (1 + x + x2 + x3)11 = (1 + x)11 (1 + x2)11 = (1 + 11C1 x + 11C2 x2 + 11C3 x3 + 11C4 x4 + ...)  × (1 + 11C1 x2 + 11C2 x4 + ...) The terms which give x4 are  1 ⋅ 11C2 x4 + 11C2 x2 ⋅ 11C1 x2 + 11C4 ⋅ x4 ⋅ 1 Hence the coefficient of x4 is 11 C2 + 11C2 × 11C1 + 11C4

11 × 10 11 × 10 11 11 × 10 × 9 × 8 + × + 1⋅ 2 1⋅ 2 1 4 ⋅ 3 ⋅ 2 ⋅1 = 55 + 605 + 330 = 990. =



=

2 7! ⋅ 45 ⋅  3 ⋅ 2  = 86016. 2! 5!  3

The coefficient of tr +1 = The coefficient of tr

49. (a)

=

2n 2n

Cr C r −1

(2n)! (r − 1)! (2n − r + 1)! 2n − r + 1 ⋅ = r ! ( 2n − r ) (2n)! r

 he coefficient of tr + 1 ≥ the coefficient of tr, proT vided 2n − r + 1 ≥ 1; or 2n + 1 ≥ 2r r 1 2n + 1 or r ≤ . or r ≤ n + 2 . 2

45. (c) Given (1 + x – 2x2)6 = 1 + a1x + a2x2 + ... + a12x12 Putting x = 1, we get 0 = 1 + a1 + a2 + ... + a12

= | t3 | = 7C2 ⋅ 45 ⋅ (3x)2



...(1)

Hence the greatest co-efficient

54. (a)

50. (c) We have, tr + 1 = (2r + 1) ⋅ nCr = 2 (r ⋅ nCr) + nCr = 2n ⋅ n – 1Cr – 1 + nCr,  { r ⋅ nC r = n ⋅ Putting r = 0, 1, 2, ..., n we get t1 = 2n ⋅ 0 + nC0 t2 = 2n ⋅ n – 1C0 + nC1 t3 = 2n ⋅ n – 1C1 + nC2 t4 = 2n ⋅ n – 1C2 + nC3 ........................

tn + 1 = 2n ⋅

n–1

n–1

C r – 1}

n–1

1 + n C115 + n C 2152 + ... + n C n15n 15 1 1 + 15k = , where k ∈ N, = 15 + k 15 =

 24n  1 1  . ∴   =  +k = 15  15   15  55. (c) We have,

C n – 1 + nC n

Adding, sum = 2n [n – 1C0 +

16n 24n (1 + 15) n = = 15 15 15

C1 +

n–1

C2 + ...

 + n – 1Cn – 1] + [nC0 + nC1 + nC2 + ... nCn] = 2n ⋅ 2n – 1 + 2n = (n + 1) 2n. 51. (c) We have, tr + 1 = (– 1)r ⋅ nCr ⋅ (a – r)

(x + a)n = xn + nC1 xn – 1 a + nC2 xn – 2 a2  + nC3 xn – 3 a3 + ... n n n–2 2 n n–1 = (x + C2 x a + ...) + ( C1 x a + nC3 xn – 3 a3 + ...) =P+Q ∴ (x – a)n = P – Q, as the terms are alternatively positive and negative. ∴ 4PQ = (P + Q)2 – (P – Q)2 = (x + a)2n  – (x – a)2n.

= (– 1)r {a ⋅ nCr – r ⋅ nCr} = (– 1)r [a ⋅ nCr – n ⋅ n – 1Cr – 1] { r ⋅ nC r = n ⋅ n – 1C r – 1} Putting r = 0, 1, 2, ..., n we get 56. (a) We have, t 1 = a ⋅ nC 0 – n ⋅ 0 T2 = 2nC1 ⋅ x, T3 = 2nC2 x2, T4 = 2nC3 x3. t2 = – (a ⋅ nC1 – n ⋅ n – 1C0) The coefficient are given to be in A.P. t3 = a ⋅ nC2 – n ⋅ n – 1C1 ∴ 2 ⋅ 2nC2 = 2nC1 + 2nC3 n n–1 t4 = – (a ⋅ C3 – n ⋅ C 2) 2n (2n − 1) 2n 2n (2n − 1)(2n − 2) ........................ ⇒ 2 ⋅ = + 1⋅ 2 n n n–1 1 1⋅ 2 ⋅ 3 tn + 1 = (– 1) (a ⋅ Cn – n ⋅ C n – 1) 2 ⇒ 2n – 9n + 7 = 0. Adding, sum = a [nC0 – nC1 + nC2 – nC3 + ... + (–1)n ⋅ nCn] 57. (b) We have,  + n [n – 1C0 – n – 1C1 + n – 1C2 – ... + (– 1)n – 1 ⋅ n – 1Cn – 1] tr + 1 = (r + 1)2 ⋅ nCr+1 = (r + 1) ⋅ {(r + 1) ⋅ nCr + 1} = a × 0 + n × 0 = 0. = (r + 1) ⋅ {n ⋅ n – 1Cr} = n [r ⋅ n–1Cr + n – 1Cr} n n n = n {(n – 1) ⋅ n – 2Cr – 1 + n – 1Cr} 52. (a) C1, C2 and C3 are in A.P. Putting r = 0, 1, 2, 3, ... n – 1 and adding we get ⇒ 2 ⋅ nC2 = nC1 + nC3 the required sum (2n )! n! n! ⇒ = = n [(n – 1) {n – 2C0 + n – 2C1 + n – 2C2 + ... + n – 2Cn – 2} + 2! (n − 2)! (n − 1)! 3!(n − 3)!  + {n – 1C0 + n – 1C1 + n – 1C2 + ... + n – 1Cn – 1}] = n [(n – 1) ⋅ 2n – 2 + 2n – 1] 1 1 1 = + = n ⋅ 2n – 2 [(n – 1) + 2] = n (n + 1) ⋅ 2n – 2. n−2 (n − 1) (n − 2) 6 On solving the equation, we get n = 7. 53. (b) (r + 1)th term in the given expansion is given by tr + 1 = 10Cr 2

10 −r 2

r

a35 ,

where r = 0, 1, 2, ..., 10 For rational terms r = a multiple of 5 = 0, 5, 10 ...(1) 10 – r = a multiple of 2 = 0, 2, 4, 6, 8, 10 ...(2) From (1) and (2) possible values of r are : 0 and 10 ∴ sum of rational terms

58. (a) We have, 1 – (1 + x) C1 + (1 + 2x) C2 – (1 + 3x) C3 + ... = (1 – C1 + C2 – C3 + ...) + (– x ⋅ C1 + 2x ⋅ C2 – 3x ⋅ C3 + ...) = 0 + (– x ⋅ C1 + 2x ⋅ C2 – 3x ⋅ C3 + ...) = – x ⋅ C1 + 2x ⋅ C2 – 3x ⋅ C3 + ... Now | tr + 1 | = (r + 1) x ⋅ nCr + 1 = x ⋅ n ⋅ n – 1Cr ∴  The given expression = – x ⋅ n ⋅ n – 1C0 + 2x ⋅ n ⋅ n – 1C1 – 3xn ⋅ n – 1C2 + ... = nx (– n – 1C0 + n – 1C1 – n – 1C2 + ...) = nx · 0 = 0.

713

= t1 + t11 = 10C0 ( 2 )10 (31/5)0 + 10C10 ( 2 )0 (31/5)10 = 25 + 32 = 32 + 9 = 41.

Binomial Theorem

= the co-efficient of (n + 1)th term (2n)! = 2nCn = . (n!) 2

714

Objective Mathematics



62. (c) We have,

n

Cr r +1

59. (c) Here, tr + 1 =

1 1 = r + 1 ⋅ nC r = n + 1 ⋅

n+1

Cr + 1

P utting r = 0, 1, 2, ... n and adding we get, n C ∑0 k +k 1 1 = {n + 1C1 + n + 1C2 + n + 1C3 + ... + n + 1Cn + 1} n +1 1 2 n +1 − 1 = {2n + 1 – n + 1C0} = . n +1 n +1 60. (b) We have,

(x + a)n + (x – a)n = 2 (xn + nC2 xn – 2 a2 + nC4 xn – 4 a4 + nC6 xn – 6 a6 + ...) Put n = 6, x = 2 and a = 1, we get ( 2 + 1)6 + ( 2 – 1)6



= 2 [( 2 )6 + 15 ( 2 )4 ⋅ 1 + 15 ( 2 )2 ⋅ 1 + 1 ⋅ 1]

= 2 [8 + 15.4 + 15.2 + 1] = 2 (99) = 198

∴ ( 2 + 1)6 = 198 – ( 2 – 1)6 Now

2 – 1 = 1 ⋅ 4 – 1 = .4 < 1

tr + 1 =

2r + 2 n Cr 2r + 2 1 n = Cr ⋅ (r + 1)(r + 2) r + 2 r +1

∴ ( 2 – 1)6 < 1, and it is certainly + ive

=

2r + 2 1 ⋅ r + 2 n +1

Hence from (1), using (2), we get





2 r + 2  1 n +1  ⋅ C r +1  = n +1  r + 2 



2r + 2 1 n+2 ⋅ Cr + 2 = n +1 n + 2

n+1

Cr + 1



+ 2n + 2 ⋅ n + 2Cn + 2} 1 {(1 + 2)n + 2 – n + 2C0 – 2 ⋅ n + 2C1} (n + 1)(n + 2)



=

3

n+2

3 − 2n − 5 − 2 ( n + 2) − 1 = . (n + 1)(n + 2) (n + 1)(n + 2)

n+2

61. (b) The numerator is of the form a + b + 3ab (a + b) = (a + b) . where a = 18 and b = 7 ∴ Numerator = (18 + 7)3 = 253. For denominator, 31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243 6 C1 = 6, 6C2 = 15, 6C3 = 20. 3



6

3

C4 = 6C2 = 15, 6C5 = 6C1 = 6, 6C6 = 1

 + 6C 3 3 3 ⋅ 2 3 + 6C 4 3 2 ⋅ 2 4 + 6C 5 3 ⋅ 2 5 + 6C 6 2 6. This is clearly the expansion of (3 + 2)6 = 56 = (25)3 Numerator (25)3 = = 1. (25)3 Denominator



50 ⋅ 49 ⋅ 10048 + ... 1⋅ 2

= 10050 – 50 ⋅ 10049 +

50 ⋅ 49 48 1 ⋅ 2 ⋅ 100 – ...

Subtracting, we get 10150 – 9950 = 2 [50 ⋅ 10049 +



= 10050 + 2 ⋅

50 ⋅ 49 ⋅ 48 × 10047 + ...] 1⋅ 2 ⋅ 3

50 ⋅ 49 ⋅ 48 ⋅ 10047 + ... > 10050 1⋅ 2 ⋅ 3

Hence 10150 > 9950 + 10050. n Cr 1 64. (c) We have, | tr + 1 | = r + 1 = n + 1 ⋅ n + 1Cr + 1

∴ The given expression

3

∴ denominator = 36 + 6C1 35 ⋅ 21 + 6C2 34 ⋅ 22

∴

= 10050 + 50 ⋅ 10049 +

and 9950 = (100 – 1)50

 =

( 2 + 1)6 = 198 – (something +ve but < 1)

63. (a) We have, 10150 = (100 + 1)50

Putting r = 0, 1, 2, ..., n and adding we get, The given expression 1 {22 ⋅ n + 2C2 + 23 ⋅ n + 2C3 + ... = (n + 1)(n + 2)



...(2)

Therefore the integral part of ( 2 + 1)6 is 197.

1 n 1 n +1   ∵ r + 1 C r = n + 1 C r +1   



∴ 0 < ( 2 – 1)6 < 1.



...(1)



1 [n + 1C1 – n + 1C2 + n + 1C3 – n + 1C4 n +1 + ... to (n + 1) terms] 1 = [ n + 1C 0 – ( n + 1C 0 – n + 1C 1 + n + 1C 2 n +1

=



– to (n + 2) terms)]

1 1 = n + 1 [1 – (1 – 1)n + 1] = n + 1 .

65. (a) We have, 24n = (24)n = (16)n = (1 + 15)n ∴ 24n = 1 + nC1 ⋅ 15 + nC2 152 + nC3 153 + ... ⇒ 24n – 1 – 15n = 152 [nC2 + nC3 ⋅ 15 + ...] = 225 K, where K is an integer. Hence 24n – 15n – 1 is divisible by 225.

3

96

(1 + x)n = C0 + C1 x + C2 x2 + C3 x3

24

+ C4 x4 + ... + Cn – 2 xn – 2 + Cn – 1 xn – 1 + Cn xn  and (x + 1) = C0 x + C1 x n

n+1

=

1 {(1 + 2)n + 1 – n +1

n+1

C 0} =

3n +1 − 1 . n +1

69. (a) We have



(1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn ...(1) and (x + 1)n = C0 xn + C1 xn – 1 + C2 xn – 2 + ... + Cn ...(2) Multiplying (1) and (2) and equating the co-efficients of xn, we get C02 + C12 + C 22 + ... + C n2

...(2)



C0C2 + C1C3 + C2C4 + ... + Cn – 2 Cn



= the coefficient of xn – 2 in (1 + x)2n



= 2nCn – 2 =

(2n)! . (n − 2)! (n + 2)!

(x + 1)n = nC0 x n + nC1 xn – 1 + nC2 xn – 2 + ... + nCn – r x r + ... + nCn. The given series is the coefficient of xn + r in the product of R.H.S. of the above two. ∴ Sum of the series = coefficient of x n + r in (1 + x)n ⋅ (x + 1)n = coefficient of xn + r in (1 + x)2n (2n)! = 2nCn + r = . (n − r )! (n + r )!

 utting r = 0, 1, 2, ..., n and adding, we get the reP quired sum 1 {2 ⋅ n + 1C1 + 22 ⋅ n + 1C2 + ... + 2n + 1 ⋅ n + 1Cn + 1} n +1

+ C3 xn – 3

(1 + x)n = nC0 + nC1 x + nC2 x 2 + ... + nCr xr + ... + nCn xn,

Cr + 1

=

...(1)

72. (b) We have,

n

Cr 1 = 2r + 1 ⋅ ⋅ r +1 n +1

+ C2 x

n–2

 ultiplying (1) and (2) and equating the coefficients M of xn – 2, we get

68. (a) We have, tr + 1 = 2r + 1

n–1

+ C4 xn – 4 + ... + Cn – 2 x2 + Cn – 1 x + Cn

67. (b) (32)32 = (2 + 3 × 10)32 = 232 + 10k, where k ∈ N Therefore, last digits in (32)32 = last digit in (2)32 But 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32 ∴ 232 = (25)6 ⋅ 22 = (32)6 ⋅ 4 = (2 + 30)6 ⋅ 4 = (26 + 10r) 4, r ∈ N Last digit in 232 = last digit in (2)6 ⋅ 4 = last digit in 4×4=6 ∴ Last digit in (32)32 = 6.

n

73. (c) We have, (1 + x)14 =

14

C0 +

14

C1 x +

14

C2 x2 + ... +



14

C14 x14 ...(1)

and (x + 1)15 = 15C0 x15 + 15C1 x14 + 15C2 x13

+ 15C3 x12 + ... +

15

C15

...(2)

 ultiplying (1) and (2) and equating the co-efficient M of x14, we get 14

C0 ⋅ 15C1 + 14C1 ⋅ 15C2 + 14C2 ⋅ 15C3 + ... + 14C14 ⋅ 15C15

= the coefficient of x14 in (1 + x)29 = 29C14. = the coefficient of xn in (1 + x)2n 74. (b) We have, (2n)! 2n = Cn = . (1 – x)n = nC0 – nC1 x + nC2 x2 – ... + (– 1)n nCn xn, (n!) 2 and 70. (b) We have, (x + 1)n = nC0 xn + nC1 xn – 1 + nC2 xn – 2 + ... + nCn. (1 + x)n = C0 + C1 x + C2 x2 + C3 x3 + ... The given series is the co-efficient of xn in the n–1 n product of R.H.S. of the above two. + Cn – 1 x + Cn x  ...(1) ∴ Sum of the series = coefficient of x n in (1 – x)n ⋅ and (x + 1)n (x + 1)n = C0 xn + C1 xn – 1 + C2 xn – 2 + C3 xn – 3 + ... n 2 n = coefficient of x in (1 – x ) ...(2) + C n – 1 x + C n = coefficient of xn in Multiplying (1) and (2) and equating the co-efficients [nC0 + nC1 (– x2) + nC2 (– x2)2 + ... + nCn (– x2)n] of xn – 1, we get Since n is even, let n = 2m. Then C0C1 + C1C2 + C2C3 + ... + Cn – 1 Cn sum = coefficient of x2m in n–1 2n [2mC0 + 2mC1 (– x2) + 2mC2 (– x2)2 + ... + 2mC2m in (1 + x) = the co-efficients of x  (– x2)2m] (2n)! = 2nCn – 1 = . 2m m n n/2 (n − 1)! (n + 1)! = Cm (– 1) = Cn/2 (– 1) . 

Binomial Theorem

5 = 5 ⋅ 5 = (125) (625) = [13 × 9 + 8] (1 + 48 × 13)24 = (13 × 9 + 8) [1 + 24C1 × (48 × 13) + 24C2 (48 × 13)2 +...+ (48 × 13)24] = 8 + terms containing powers of 13. Hence remainder = 8. 99

715

71. (c) We have,

66. (a) We have,

716

75. (a) We know that

(n + 1) n! r !(n − r + 1)(n − r )!

Objective Mathematics

(1 + x)j = 1 + jC1x + jC2x2 + jC3x3 + ....  + jC100  x 100 + jC101 x 101 + ... + jC200  x 200



=

∴ Coefficient of x100 in the expansion of

or

n +1

r = 200

∑ (1 + x)



r = 200

j

∑

will be

j=0



= [100C100 +



 200  =  .  100 

j

C100

j=0

101

C100 +

102

Cr =

n +1 n C r n − r +1

...(2)

Putting n = 1, 2, 3, ..., n in (2), we get

C100 + ... +

200

(C0 + C1) (C1 + C2) ... (Cn – 1 + Cn)

C100]

76. (b) The given series



= n + 1 C ⋅ n + 1 C ... n + 1 C 1 2 n n n −1 1



=

(n + 1) n C1C2C3 ... Cn. n!

= mCr ⋅ nC0 + mCr – 1 ⋅ nC1 + ... + mC0 ⋅ nCr, 79. (a) We have, (1 + x)15 = C0 + C1 x + C2 x2 + ... + C15 x15 n m C0 = 1 = C0). ( (1 + x)15 C0 Now, (1 + x)m = mC0 + mC1 ⋅ x + m C2 x2 + ... ⇒ = + C1 + C2 x + C3 x2 + ... + C15 x14 x x m r m m  + Cr x + ... + Cm x , Differentiating both sides w.r.t. x, we get and (1 + x)n = nC0 + nC1 x + nC2 x2 + x ⋅ 15 (1 + x)14 − 1 ⋅ (1 + x)15  ... + nCr xr + ... + nCn xn x2 The given series is the coefficient of xr in the product of R.H.S. of above two. C = – 20 + C2 + 2 C3 x + 3 C4 x2 + ... + 14 · C15 x13 ∴ Sum of the series= coefficient of x r in (1 + x)m ⋅ x (1 + x)n Putting x = 1 on both sides, we get = coefficient of x r in (1 + x)m + n 15 ⋅ 214 – 215 = – C0 + C2 + 2 C3 + 3 C4 + ... + 14 C15 = m + nC r. ⇒ 214 (15 – 2) + 1 = C2 + 2 C3 + 3 C4 2 2 n n   C 1 − + n k    + ... + 14 C15 77. (a) ∑ k 3  k  = ∑ k 3   k  k =1  C k −1  k =1  ∴ The given series = 214 ⋅ 13 + 1 = 219923.

 n Ck n − k +1 = ∵ n  k  C k −1 

 n

=

n

∑ k (n − k + 1) = ∑ k (n + 1) 2

k =1

2

k =1

⇒ r ⋅

− 2k (n + 1) + k 2 

n

n

n

k =1

k =1

k =1

n (n + 1) = (n + 1) ⋅ − 2 (n + 1) ⋅ 2 n (n + 1)(2n + 1) n 2 (n + 1) 2 + 6 4 2

n (n +1) 2 [6 (n + 1) – 4 (2n + 1) + 3n] 12

C 1, C 1 + C 2 =

n+1

C1n + 1 C2 ... n + 1Cn (n + 1)! Now n + 1Cr = r !(n − r + 1)!

= n + (n – 1) + (n – 2) + ... + 1 =

...(1)

n (n + 1) . 2

...(1)

3n = a0 + a1 + a2 + a3 + a4 + a5 + a6

∴ The given expression n +1



Putting x = 1, ω and ω , in (1), we get

= =

C1 C C C + 2 2 + 3 3 + ...n n C0 C1 C2 C n −1

2

C2, ... Cn – 1 + Cn





= a0 + a1 x + a2 x2 + ... + a2n x2n

78. (a) We have, n+1

Cr =n –r +1 C r −1

81. (c) Given, (1 + x + x2)n

n (n +1) 2 n (n + 2)(n + 1) 2 = ⋅ (n + 2) = . 12 12

C0 + C1 =

Tr +1 n − r +1 Cr = = Tr r C r −1

Put r = 1, 2, 3, ..., n and add, we get

= (n + 1) 2 ∑ k − 2 (n + 1)∑ k 2 + ∑ k 3

=

80. (b) We have,

n +1

Cn

+ ... + a2n

...(A)

0 = a0 + a1 ω + a2 ω + a3 + a4 ω + a5 ω + a6 + ... 2



+ a2n ω2n 

2

...(B)

0 = a0 + a1 ω2 + a2ω + a3 + a4 ω2 + a5 ω + a6 + ...

+ a2n ω2n

Adding (A), (B) and (C), we get

...(C)

+ ... + (nCn sin nx cos 0x + nC0 sin 0x cos nx)]

3 = 3 n – 1. 3

∴ a0 + a3 + a6 + ... =

=

1 n [ C0 sin nx + nC1 sin nx + ... + nCn sin nx] 2

=

1 n 2n sin nx [ C0 + nC1 + ... + nCn] sin nx = 2 2

82. (c) We have, n

Cn +

n+1

Cn +

n+2

Cn + ... +

n+ k

Cn

= coeff. of xn in (1 + x)n + (1 + x)n + 1 + ... + (1 + x)



n+ k

.

Now, (1 + x)n + (1 + x)n + 1 + (1 + x)n + 2 + ... + (1 + x)n + k



∴

n

∑

n

r =0

C r sin r x cos (n − r ) x = 2n – 1 sin nx.

87. (c) The number of terms in the expansion of

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

(a + b + c + d)n is

n

=

=

1 1 (1 + x)n + k + 1 –  (1 + x)n x x n



C0 +

n+1

n+k+1

=

Cn +

n+2

Cn + ... +

Cn + 1 – 0 =

n+ k +1

n+ k

Cn

+ a2 x147 ⋅ y2 + ... + a149 y149



=

9! ⋅ (2 x)α1 (−3 y )α2 (4 z )α3 α1 ! α 2 ! α 3 !



=

9! ⋅ 2α1 (−3)α2 4α3 ⋅ x α1 y α2 z α3 α1 ! α 2 ! α 3 !

∴  Coefficient of x3 y4 z2

a0 + a1 + a2 + ... + a149 = (7 – 8)

149

=–1

⇒ Sum of the coefficients = – 1. i = 0, 1, 2, ..., (n – 1)

2 ΣΣ (Ci + C j )

0≤i≤ j ≤n

and =n⋅

2n

2 1

9! ⋅ 23 (– 3)4 ⋅ 42 3! 4! 2!

=

j = 1, 2, 3, ..., n i < j

= 13063600.

2 n

Cn + [(C0 + C1 + ... + Cn)2

90. (d)

= n ⋅ 2nCn + (2n)2 – 2nCn = (n – 1) ⋅ 2nCn + 22n.



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

=

85. (b) We have, 1 1 1 1 + + + ... + 1!(n − 1)! 3!(n − 3)! 5!(n − 5)! (n − 1)!1! =

1  n! n! n!  + + ... +   n! 1!(n − 1)! 3!(n − 3)! (n − 1)!1!

=

1 n 1 ( C1 + nC3 + nC5 + ... + nCn – 1) = ⋅ 2 n – 1. n! n!

86. (c) We have,

n

∑ r =0

=

n

C r sin r x cos (n − r ) x

1 n [( C0 sin 0x cos nx + nCn sin nx cos 0x) 2

+ (nC1 sin x cos (n – 1) x + nCn – 1 sin (n – 1) x ⋅ cos x]

8 × 81 × 16 = 1260 × 10368

1 + x − 2 x3 1 − x + 2 x − 2 x3 = 3 (1 − x) (1 − x)3

– ( C02 + C12 + ... + C 2n )]



[taking α1 = 3, α2 = 4, α3 = 2]



9⋅8⋅7 ⋅6⋅5 = 3! 2!

= n ( C + C + ... + C ) + 2 Σ Σ Ci Cj 0 ≤ i < j ≤ n 2 0

(n + 1)(n + 2)(n + 3) . 6

89. (b) The general term in the expansion of (2x – 3y + 4z)9

Putting x = y = 1, we get

84. (a)

C3 =

[Here n = 15, q = 3, r = 3, k = 4]

83. (b) Let (7x – 8y)149 = a0 x149 + a1 x148 y



n+3



Cn + 1

1  n n There is no term containing x in x (1 + x)  .  



C4 – 1 =

88. (a) The greatest coefficient is n! = r (q !) k − r [(q + 1)!]

Equating the coefficient of xn, we get

n +4–1

= (1 + 2x + 2x2) (1 – x)–2 = (1 + 2x + 2x2) {1 + 2x + 3x2 + ... + (n + 1) xn + ...} ∴ Coefficient of xn = 1 × {coefficient of xn in the bracket} + 2 × (coefficient of xn – 1 in the bracket) + 2 (coefficient of xn – 2 in the bracket) = 1 ⋅ (n + 1) + 2 ⋅ n + 2 ⋅ (n – 1) = n + 1 + 2n + 2n – 2 = 5n – 1. 91. (a) We have, (1 − 4 x)

−

1 2

1 1  1 −  − − 1 2 2  2 = 1+ (−4 x) + (− 4 x) 2 + ... 1! 2! −

717

+ [nC2 sin 2x cos (n – 2) x + nCn – 2 sin (n – 2) x ⋅ cos 2x]

n

Binomial Theorem

3n = 3 (a0 + a3 + a6 + ...)



718

Objective Mathematics

1 1   1  −  − − 1 ...  − − r + 1  2 2   2 + (− 4 x) r + ... r!

⇒ a = 2, b = – 6.

∴ Coefficient of

96. (b) We have,

1 1   1  −  − − 1 ...  − − r + 1  2 2   2 r x = (− 4) r r!

=

1 1 1 1    + 1  + 2  ...  r −  2 2 2 2 r (−1) r 4r = (−1) r!



=

1 ⋅ 3 ⋅ 5 ⋅ 7...(2r − 1) r 4 2r (r )!

=

(2r )! (2r )! ⋅ 4r = . 2 r !(2 ⋅ 4 ⋅ 6...2r ) (r !) 2 1  1  1 ⋅ (1 + 2)  1   +   1!  22  2!  2 2  −



97. (c) We have,

1 ⋅ (1 + 2) ⋅ (1 + 2 ⋅ 2)  1   2 3! 2 

3



1 2

3 =   2

−

1 2

2 . 3

= [(1 + x)–2]– n = (1 + x)2n.

∴ Coefficient of xn = 2nCn = 94. (a) We have,

=

2/3

 3x  1 +  8

2/3

 5x   3  2 1 + x  2 1 −    2  4

1

 2 2 =   = 3

93. (c) We have, (1 – 2x + 3x2 – 4x3 + ... to ∞)– n

(8 + 3 x) 2 / 3 (2 + 3x) (4 − 5 x) 8



4

−

2   5 1 −   25 

2

1 ⋅ (1 + 2) ⋅ (1 + 2 ⋅ 2) ⋅ (1 + 3 ⋅ 2)  1   2  − ... 4! 2   1 = 1 +   2

1 (1 – 0.08)–1/2 5

1

= 0 .2 0848 = 0.2085 correct to four places of decimals.

92. (b) The given series

+

1 = (25 − 2)

1 = 23

   1 3  −   −   1   1 2 2 2 = 1 +  −  (−0.08) + (−0.08) + ... 5   2 1⋅ 2    1 = [1 + 0.04 + .0024 + ...] 5  = 0.2 + 0.008 + 0.00048

r

=1−

−3b 9 = a4 8

⇒  a3 = 8 and

(2n)! . (n!) 2

a + bx = (a + bx) (1 – x)– 2 (1 − x) 2

= (a + bx) [1 + 2x + 3x2 + ... + (n + 1) xn + ...]

−1

− 1/ 2



 3x  = 1 +  8  



 2 3x  3  5  = 1 + ⋅ + ...  1 − x + ...  1 + x + ...   3 8  2  8 



5 1 3 5 x. = 1+  − +  x = 1 – 8 4 2 8

2/3

 3   5  1 + x  1 − x   2   4 

98. (a) We have, (a + bx)–2 =

1 – 3x (Given) 4 −2

 2b  x = a–2 1 − a  

a + bx = a (n + 1) + bn ∴ Coefficient of x in (1 − x) 2

 b  Also, (a + bx)–2= a–2 1 + x   a 

 ut a (n + 1) + bn = 2n + 1. Equating coefficient B of n and constant term, we get a + b = 2 and a = 1. So a = 1, b = 1 ∴ When a = 1, b = 1, coefficient of xn = 2n + 1.

[Neglecting x2 and higher powers of x]

n

1 9 95. (b) (a + bx)– 3 = + x ... 8 8 b or a − 3 1 + x  a   ⇒ 

−3

=

1 9 + x + ... 8 8

1  3b  1− x = 1 + 9 x 3  a  a  8 8



=

1 2bx − a 2 a3

1 2bx = 1 – 3x − 4 a 2 a3 1 1 2b ⇒ 2 = and 3 = 3 4 a a ∴

⇒ a2 = 4, ∴ ⇒ b =

a = 2  and 

2b =3 a3

3 3 3 3 a = × (2)3 = × 8 = 12 2 2 2

Hence a = 2, b = 12.

Also it is given that the coefficient of x2 = 3. m (m −1) =3 ∴ 2 ⇒ m2 – m = 6 ⇒ m2 – m – 6 = 0 ⇒ (m – 3) (m + 2) = 0 ⇒ m = 3, – 2. Hence, the required values of m are 3 and – 2. 100. (b) We have,

=

m n m (1 + h) m − n (1 + h) n Now, mx − nx = m−n m−n





=

1 4 9 − + 2 (1 − x) 1 − 2 x 2 (1 − 3 x)

ax b − bx a a (1 + h)b − b (1 + h) a = xb − x a (1 + h)b − (1 + h) a

1 9 (1 – x)–1 – 4 (1 – 2x)–1 + (1 – 3x)–1 2 2

1 [1 + x + x2 + ... + xn + ...] 2 – 4 [1 + 2x + (2x)2 + ... + (2x)n + ...]

=



+

9 [1 + (3x) + (3x)2 + ... + (3x)n + ...] 2

∴ Coefficient of xn =

=

1 [1 – 8.2n + 9.3n] 2

1 [1 – 2n + 3 + 3n + 2]. 2

101. (a) Since the given expression is equal to a + bx, so we have to neglect the terms x2 and higher powers of x.  

(1 − 3 x)1/ 2 + (1 − x)5 / 3 4− x

= [(1 – 3x)1/2 + (1 – x)5/3] [4 – x]– 1/2 x  1 −   4

=

1 3 5x    1   x   2 − x −  1 +  −   −   2 2 3    2   4 

=

1 19   x  2 − x  1 +  2 6  8 

−1/ 2

35 x = a + bx 24

Hence a = 1 and b = –

a (1 + bh) − b (1 + ah) (1 + bh) − (1 + ah)

=

1 1 1 a−b =– =– = . h x −1 1− x − ( a − b) h

104. (d) We have, (1 + x)30 = 1 + 30C1 + 30C2 x2 + 30C3 x3  + ... + 30C29 x29 + 30C30 x30 ∴ Sum of coefficients of odd powers of x is  30 C1 + 30C3 + ... + 30C29 = 230–1 = 229 (2 − x)(2 + x) 3/2 3/2 105. (a) (1 + x)(1 − x) = 1 + + 1+ x 1− x

= 1 + 3 [(1 + x)–1 + (1 – x)–1] 2

3 [(1 – x + x2 + ...) + (1 + x + x2 + ...)] 2 ∴ Term which is independent of x = 1 + 3 (1 + 1) = 4. 2 r  1  2n 2n – r − 106. (c) Tr + 1 = Cr x  2  x  =1+

Since it is independent of x, ∴ x2n – r – 2r = x0 2n ⇒r = ⇒ n is a multiple of 3. 3 107. (a) The general term is

1 x 19 x  = 1  2 − 35 x  = 1 – 35 x   = 2 + −  2 12  24 2 4 6  ∴1–

=



5  1  = 1 + (− x) + 1 + (− x)  (4–1/2) 3  2 

(m − n)[1 + (m + n) h] m−n = 1 + (m + n) h = (1 + h)m + n = xm + n. =

103. (b) Since x is nearly equal to one, let x = 1 + h where h is so small that its squares and higher powers are neglected, then

[By resolving into partial fractions]

m (1 + mh) − n (1 + nh) m−n

[Neglecting h2 and higher powers]





1 (1 − x)(1 − 2 x)(1 − 3 x)

=



Tr+1

 x = Cr   3 10

10 − r 2

10 − r

35 . 24



⋅

3r / 2 x2r  10 − r



− 2r   1 2  × 3r / 2 × x  2 Tr+1 = Cr   3   ∴ for term independent of x, we have,

10

719

Binomial Theorem

102. (c) Since x is very nearly equal to 1, let x = 1 + h, where h is very nearly equal to zero, so that h2 and higher powers of h may be neglected.

m (m −1) 2 x + ... 2 m (m −1) Coefficient of x2 = 2

99. (c) (1 – x)m = 1 – mx +

720

10 − r − 2r = 0 ⇒ r = 2 2

1 3 ∴ Coefficient of x2 = 246 ⇒ 3 + a 2  a 2  = 246  

Objective Mathematics

∴ Term independent of x

⇒ a4 =

4

1 1× 3 45 5 = = . = 10C2   × 3 = 45 × 81 27 3 3



x 3  108. (a) Tr+1th term in the expansion of  − 2  2 x 

10

10 − r

 x = 10 Cr   2 =

10

Cr

 3 − 2   x 

x10 − r (−1) r ⋅ 3r × = 210 − r x2 r

10

Cr ⋅ x10 − 3 r ⋅

(−1) r ⋅ 3r 210 − r

∴ x10 – 3r = x4 ⇒ 10 – 3r = 4 ⇒ r = 2 ∴ T3 =

10

C2 x 4 ⋅

(−1) 2 32 28

79 + 97 = (1 + 8)7 – (1 – 8)9

10

C2 ×

(1 + 3 2 x) + (1 – 3 2 x) 9

– (1 – 9C1 ⋅ 81 + 9C2 ⋅ 82 – ... 9C9 ⋅ 89) = 16 × 8 + 64 [(7C2 + ... + 7C7 ⋅ 85) – (9C2 – ... – 9C9 ⋅ 87)] = 64 (an integer) Hence 79 + 97 is divisible by 64. 115. (c) We have, (1 + 2x + x2)n =

3 405 = . 28 256

116. (c) (1 + x)27/5 = 1 +

= 5x7 + 10x6 + x5 – 10x4 – 10x3 + 5x

C 0 + 7C 1 + 7C 2 + 7C 3 + 7C 4 + 7C 5 + 7C 6 + 7C 7  – ( 7C 6 + 7C 7) = 27 – (7C6 + 7C7) = 128 – (7 + 1) = 120.

r =0

50

Cr x r

+

+



1 1 = 2 (50C0 + 50C1 + ... + 50C50) = (250) = 249. 2

1 x 1 +  a2  a 

−2

3

= (1 – 2x + 3x – 4x + ...) +

 1  2 x 3x 2 4 x3 + 2 − 3 + ... 1 − a2  a a a 

Cr x r =

2n

∑a r =0

r

xr

5 27  27 − 1  27 − 2   27 − 3   27 − 4  x        5   5   5  5!  5  5

27  27   27 − 2   27 − 3  − 1      5  5  5   5 6  27   27  x − 4  − 5   5  5  6!

+

  27  27  27   27 − 2  − 3 − 1     5 5 5  5    

 7  27  x +...   ×  − 6  5  7!

 27   27  − 5 − 4   5    5 

27 – 6 is negative i.e., 8th term is negative 5 in the expansion of (1 + x)27/5.

117. (c), (d)  The given expression 2

2n

27  27  27 − 1 x+  5  5  5

Here

1 1 + 113. (b) (1 + x) 2 (a + x) 2 = (1 + x) −2 +

r =0





∴ Sum of the coefficients of odd powers of x = 50C1 + 50C3 + ... + 50C49

2n

∑

  27  x4 27  27   27 − 1  − 2  − 3  5  5   5   5  4!

7

50

⇒

+

which is polynomial of degree 7. 111. (c) We have, 7C0 + 7C1 + 7C2 + 7C4 + 7C5

r

r

 x3 27  27   27 − 1  − 2  5  5   5  3!

= 2 [5C0 x5 + 5C2 x3 (x3 – 1) + 5C4 x (x3 – 1)2] = 2 [x5 + 10x3 (x3 – 1) + 5x (x3 – 1)2]

r

r

+

2 [9C0 + 9C2 (3 2 )2 + ... + 9C8 (3 2 )8] is 5.

∑

2n

∑a x

⇒ ar = 2nCr.

9

110. (c) [x + (x3 – 1)1/2]5 + [x – (x3 – 1)1/2]5

2n

∑a x r =0

2

2nd, 4th, 6th, 8th and 10th terms get cancelled. ∴ Number of non-zero terms in

112. (b) (1 + x)50 =

= (1 + 7C1 ⋅ 81 + 7C2 ⋅ 82 + ... + 7C7 ⋅ 87)

⇒ (1 + x)2n =

109. (c) In the expansion of



1 where | x | < | a |. 3

r =0

Hence the coefficient of x4 =



a =

114. (c) We have,

r

1 ; 81

  1 =  9 x −1 + 7 + x −1 1/ 5  ( 3 + 1 )   Given, T6 = 84

7

⇒ C5 (9 7

⇒ ⇒ ⇒ ⇒

x–1

+ 7)

7 −5

  1  x −1 1/ 5  = 84 ( 3 + 1 )   1

+ 7) ⋅

(3x −1 + 1)

⇒ 6α2 = –20α3 ⇒ α = 0, – 3/10. 126. (d) We know that if n is even, then nCr is greatest for

= 84

9x – 1 + 7 = 4 (3x – 1 + 1) 32x – 12 ⋅ 3x + 27 = 0 (3x – 3) (3x – 9) = 0 3x = 3, 9 ⇒ x = 1, 2.

r =

∴

24

n . 2

Cr is greatest for r = 12. 1 − x3 (1 − x)(1 + x + x 2 ) = 1− x 1− x

127. (d) 1 + x + x2 =

118. (d) The general term in the given expansion is

−3

r

 3  Tr + 1 = 6Cr ( x 5 )6 – r ⋅  3  = 6Cr ⋅ x15 – 4r ⋅ 3r  x  Now, 15 – 4r = 3 ⇒ r = 3. ∴ Coefficient of x3 = 6C3 ⋅ 33 = 540. 119. (b) In the binomial expansion of (1 + x)21, coefficient of xr = coefficient of xr + 1 ⇒ 21Cr = 21Cr + 1 ...(1) 21 21 ...(2) But Cr = C21 – r Comparing equation (1) and equation (2), we get, r + 1 = 21 – r 21 − 1 ∴ r = = 10. 2

 1 − x3  ∴ (1 + x + x2)– 3 =   = (1 – x)3 (1 – x3)– 3  1− x  = (1 – 3x + 3x2 – x3) (1 + 3x3 + 6x6 + ...) ∴ coefficient of x6 = 6 – 3 = 3. 10

∑

128. (b) We have,

20

r =0

But

20

and

20



∴

C r = 20C0 + 20C1 + ... + 20C10

C0 + 20C1 + ...

1 1 3 1⋅ 3 ⋅ 5  1 1− + ⋅ − +... = 1 +  8 8 16 8 ⋅ 16 ⋅ 24  4

− 1/ 2

=

2 . 5

20

C18 = 20C2 ... and

∑

20

∴ 10 – 2r = 0 ⇒ r = 5 ∴ constant term = 10C5 (– 1)5

=

9×8× 7× 6 = – (3 × 2 × 7 × 6) = – 252. 3× 4

= 220 + 20C10 – (20C10 + 20C9 + ... +

⇒ 2 [ C0 + C1 + ... + 20

3/ 2

C0 + 20C1 + ... +

20

20

20

C20)

C 0)

20

C10] = 2 + C10

C10 = 219 +

1 2

20

C10.

< 1 i.e., | x |
0. Now 3rd term in the expansion = T2 + 1 5 = C2 x5 – 2 (xlog 10 x)2 = 10,00,000 (Given) 3 + 2 log x 10 = 105. or x Taking logarithm of both sides, we get (3 + 2 log10 x) log10 x = 5 or 2y2 + 3y – 5 = 0, where log10 x = y. or ( y – 1) (2y + 5) = 0 or y = 1 or – 5 2 5 or log10 x = 1 or – ∴ x = 101 = 10 2 or 10–5/2. 137. (b) Let x4 occurs in Tr + 1 10 − r

Tr + 1

 x = 10Cr   2

r

x10 − 3r (−3) r  3 10  − 2  = Cr (2)10 − r  x 

∴ 10 – 3r = 4 ⇒ 3r = 10 – 4 = 6 ∴ r = 2

3x < 1 i.e., if  | x | < 2. 6

140. (a) ( 3 + 1)4 + ( 3 – 1)4 = ( 3 )4 + 4C1 ( 3 )3 (1) + 4C2 ( 3 )2 + 4C3 ( 3 ) + 4C4 + ( 3 )4 – 4C1 ( 3 )3 (1) + 4C2 ( 3 )2  – 4C 3 3 + 4C 4 = 2 [9 + 6 (3) + 1] = 56 = a rational number. 

5 5 – 11 = 18034 < 1,

∴ if f ′ = (5 5 – 12)2n + 1, then 0 < f ′ < 1 Now,

10 × 9 × 9 405 = . 1 × 2 × 256 256

coeff of xn in (1 – x)n + coeff. of xn in x(1–x)n (–1)n. nCn + coeff. of xn–1 in (1 – x)n (–1)n + (–1)n–1 nCn–1 (–1)n–1 (–­1 + n) (n – 1) (–1)n–1 (1 – n) (–1)n.

= = = = = =

R = I + f = (5 5 + 11)2n + 1, and 0 < f < 1. Now

=

(−3) 2 28

138. (b) The coeff. of xn in (1 + x) ( 1 – x)n

135. (c) Here f = R – [R] is the fraction part of R. Thus if I is the integral part of R, then

∴

Objective Mathematics

= 1 + mx + m (m − 1) x 2 + ...   2  

∴

722

134. (c) We have, (1 + x)m (1 – x)n

141. (c)

n

∑ r =1

n

Pr = r!

n

n!

n

∑ (n − r )! r ! = ∑ r =1

r =1

n

Cr

= C1 + C2 + ... + Cn = 2n – 1. 142. (a) (7C0 + 7C1) + (7C1 + 7C2) + ... + (7C6 + 7C7) = 8C1 + 8C2 + ... + 8C7 = (8C0 + 8C1 + 8C2 + ... + 8C7 + 8C8) – (8C0 + 8C8) = 28 – 1 (1 + 1) = 28 – 2. 143. (a) We have, (1 – x4) (1 + x)9 = (1 – x4) 

(1 + 9C1 x + 9C2 x2 + 9C3 x3

+ 9C 4 x 4 + 9C 5 x 5 + 9C 6 x 6 + 9C 7 x 7 + 9C 8 x 8 + 9C 9 x 9) ∴ coefficient of x7 = 9C7 – 9C3 = 9C2 – 9C3 =

 7 4 9×8 9×8× 7 = 36 1 −  = 36 × – = – 48. − 3   3 1× 2 1× 2 × 3

144. (d) Let S = aC0 + (a + b) C1 + (a + 2b) C2 + ...  + (a + nb) Cn Also S = (a + nb) Cn + ... + aC0 ∴ 2S = (2a + nb) (C0 + C1 + ...+ Cn) = (2a + nb) 2n ∴ S = (2a + nb) 2n – 1. 145. (b) Middle term = T3 = T2 + 1 = 4C2 (3x)2 (2)2 ∴ Coefficient of middle term = 6 × 9 × 4 = 216. 146. (b) We have, x = (8 + 3 7 )n = [x] + f where [x] ≤ x and 0 ≤ f  < 1 Now, let f ′ = (8 – 3 7 )n , where 0 < f ′ < 1 Also, [x] + f + f ′ = (8 + 3 7 )n + (8 – 3 7 )n

 ...(ii)

∴ L.H.S. is also an integer. therefore, f + f ′ = 1

( 0 < f + f ′ < 2)

Hence [x] + 1 = 2k

[From (ii)]



(n + 1)(n)(n − 1)(n − 2)(n − 3)(n − 4)  + ... 6!  1 = [n + 1C2 + n + 1C4 + ... + ...] n +1 1 2n − 1 = n + 1 [2n – 1] = n +1 +

[x] = 2k – 1 = odd integer.

147. (d) We have, na = 8 ⇒ n2 a2 = 64,

n (n −1) 2 a = 24 2

n (n −1) 2 a = 24 2 2n 8 ⇒ = ⇒ 6n = 8n – 8 n −1 3 ⇒

[ nC0 + nC2 + nC4 + ... = 2n – 1, ∴ nC2 + nC4 + ... = 2n – 1 – 1] 10 − r – 2r = 0 2 ∴ 10 – r – 4r = 0, ∴ 5r = 10 ∴ r = 2

155. (b) For absolute term

⇒ 2n = 8 ⇒ n = 4, ∴ a = 2.  4x   8  148. (c) We have, T7 = T6 + 1 = 9C6    −   5   5x  3

C2 k2 = 405 405 10 × 9 2 k = 405 ⇒ k2 = =9 ⇒ 45 1× 2 ∴ Absolute term =

6

∴  Power of x = 3 – 6 = – 3. 149. (c) Sum of binomial coefficients = (1 + 1)7 = 27 = 128. 150. (b) We have,

= 10C0 – 10C1 + 10C2 – 10C3 + ... +

10

156. (d) Since (1 – x)n = C0 – C1 x + C2 x2 – C3 x3 + ... ∴ x (1 – x)n = C0 x – C1 x2 + C2 x3 – C3 x4 + ... 1

⇒

∴ or ∴ ∴ But

2n

C3r – 1 = 2nCr + 1

either 3r – 1 = r + 1 3r – 1 + r + 1 = 2n either 2r = 2 or 4r = 2n either r = 1 or 2r = n r ≠ 1 ∴ n = 2r.

153. (d) Coefficient of xr    1  1 1  1 − − 1 − − 2  ...  − − r + 1 −    2  2   2   2 = (−2) r r!

=

∫ (C x − C x 0

1

2

+ C 2 x 3 − C3 x 4 + ...) dx

0

For L.H.S. put 1 – x = t, ∴ dx = – dt 1 . ∴ L.H.S. = (n + 1)(n + 2)

152. (a) We have, T3r = 2nC3r – 1 x3r – 1 By the given condition

dx

C C C = 0 − 1 + 2 −... upto (n + 1) terms 2 3 4

= 209, which is divisible by 11. [but not by 2 or 9 or 27] Tr + 2 = 2nCr + 1 xr + 1

n

1

C10 = 0.

151. (b) For n = 1, 2 ⋅ 42n+1 + 33n+1 = 2 ⋅ 43 + 34 = 128 + 81

and

∫ x (1 − x) 0

[Putting x = 1 on both sides] 

10

⇒ k = ± 3.

(1 – x)10 = 10C0 + 10C1 (– x) + 10C2 (– x)2  + 10C3 (– x)3 + ... + 10C10 (– x)10. ∴ Sum of the coefficients

1  (n + 1)(n) (n + 1)(n)(n − 1)(n − 2) + n + 1  2! 4!

=

∴ f + f ′ is an integer

157. (b)

1 = (1 – x)– 2 = 1 + 2x + 3x2 + ... 1 − 2x + x2

= 1 + a1 x + a2 x2 + ... ⇒ ar = r + 1.

158. (c) (α  x – 2αx + 1)51 = [(1 – αx)2]51 2 2

= (1 – αx)102 = A0 + A1 x + A2 x2 + Putting x = 1, we get (1 – α)102 = A0 + A1 + A2 + ... = 0 ∴ α = 1.

...(say) (Given)

159. (d) We know that (1 + x)10 = 10C0 + 10C1 x + 10C2 x2 + ... +

10

C10 x10

=

1 ⋅ 3 ⋅ 5....(2r − 1) (−1) r (−1) r 2r ⋅ 2r r!

Integrating both sides with respect to x from 0 to 2,

=

2r ! 2r ! 1 ⋅ 3 ⋅ 5....(2r − 1) = = . r !r !2 r (r !) 2 ⋅ 2r r!

we get

(1 + x)11 11

723

C1 C3 C5 + + +... 2 4 6 n n (n − 1)(n − 2) n (n − 1)(n − 2)(n − 3)(n − 4) + + ... = + 2 3!⋅ 4 5!⋅ 6

154. (b) We have,

Binomial Theorem

= 2 {8n + nC2 (8)n – 2 (3 7 )2 + nC4 (8)n – 4 (3 7 )4 + ...} = 2 (Integer) ∴ [x] + f + f ′ = Even integer = 2k (say)

724

=

10

x2 2

C0 x 0 + 10C1 2

2

0

Objective Mathematics

3 −1 2 = 2 10C0 + 11 2 11

i.e.,

x11 11

+ ... + 10C10 2

2 1 3 1 × × =– 3 2 2 2 ∴ Sum of given series

2

⇒x =–

0

3

2 3

10

C1 +

10

C2

 1 = 1 −   2

11

2 + ... + 11

 160. (c) tn =

= 2 ⋅ C0 + n

∑ r =0





n

r = Cr

= n = n

r =0

n

∑ r =0

n

n

∑ r =0

C10

22 23 211 C1 + C2 + ... + C . 2 3 11 10

n

∑

10

n

n − (n − r )   n Cn − r

1 – Cn − r

n

∑ r =0

n−r Cn − r

r = n . Sn – ∑ n = nSn – tn r = 0 Cr



∴ 2tn = n . Sn ⇒



−n

−n

Multiply and equate coefficient of x20 – m on both sides, Sum = coefficient of x20–m in (1 + x)30. The middle coefficient of (1 + x)30 is the greatest coefficient, i.e., the coefficient of x15, so, 20 – m = 15 or m = 5.

n

1+ x  =   . 1− x 

165. (a) (1 + t2)12 (1 + t12) (1 + t24) 

and

2 1 ⋅  3 2

n (n − 1) 2 2 5 1 x = ⋅ ⋅ 2  2 ⋅1 3 6 2

Tr + 1 = 124Cr ( 5 )124 – r ( 4 11 )r = 124Cr ⋅ 5(124 – r)/2 ⋅ 11r/4

For integer terms r = 0, 4, 8, 12, ..., 124. ...(2)

5 5 n −1 n −1 = ⇒ = ⇒ 5n = 2n – 2 4 2 n 2n 2 ⇒ 3n = – 2 ⇒ n = – 3 putting value of n in (1), we get 2 2 1 x= × – 3 3 2

C1t2 + 12C2t4 + ...) (1 + t12 + t24 + t36)

167. (d) The general term in the expansion is ...(1)

2 5 1 n (n − 1) 2 × × x 3 6 4 2 ⋅1 = 2 1 2 1 2 2 n x × × × 3 2 3 2

⇒

12

166. (b)

Now divide (2) by square of (1) we get ⇒

= (1 +

∴ coefficient of t32 = 12C6 + 2166.

Comparing these two get nx =

...(2)

(x + 1)20 = 20C0x20 + 20C1x19 + 20C2x18 + ... + 20Cmx20–m  + ... + 20C20

162. (b) Given series is 2 1 2 5 1 2 5 8 1 S = 1 + ⋅ + ⋅ ⋅ 2 + ⋅ ⋅ ⋅ 3 + ...∞ 3 2 3 6 2 3 6 9 2 and we know that (1 + x)n n (n − 1) 2 n (n − 1)(n − 2) 3 x + x + ...∞ = 1 + nx + 2! 3!



...(1)

 + ... + 10Cm.20C0, (1 + x) 10 = 10C0 + 10C1x + 10C2x2 + ... + 10Cmxm + ...  + 10 C10x10

tn n = . Sn 2

1− x  =   1+ x 

= (2)2/3 = (4)1/3.

164. (a) Expression = 10C0.20Cm + 10C1.20Cm–1 + 10C2.20Cm–2

161. (a) Given expansion 2x   = 1 −   1+ x 

− 2/3

(1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn Put x = 1, 2n = C0 + C1 + C2 + C3 ... Cn and put x = – 1 0 = C0 – C1 + C2 – C3 ... Cn Subtracting (2) from (1), we get 2n = 2 [C1 + C3 + C5 ...] ∴ C1 + C3 + C5 ... = 2n – 1.

n

n

1 =   2

163. (d) By Binomial Theorem

[∴ nCr = nCn–r]

 n  n −1 1 1 −n + n + ... + n + 0  Cr  Cn Cn −1 C1 

− 2/3

∴ Number of terms which are integers = 32. 168. (c) We have,

1 52n 1 (1 + 24) n = (25) n = 24 24 24

1 [nC0 + nC1 ⋅ 24 + nC2 ⋅ 242 + nC3 ⋅ 243 24  + ... + nCn × 24n] 1 = + [nC1 + nC2 ⋅ 24 + ... + nCn ⋅ 24n – 1] 24  52n  1 1 = + an integer.  ∴    = . 24 24  24  =

169. (c) Given : T2 = 14 a5/2  a  ⇒ nC1 (a1/13)n – 1 ⋅  − 1/ 2  = 14 a5/2 a  1

= 60C1 + 60C3 + 60C5 + … + 60C59 = 260 – 1 = 259 175. (c) We have, (1 + x)n = C0 + C1x + C2x2 + …+ n Cn x n On integrating w.r.t. x, we get

170. (c) Total number of terms in the expansion of

(1 + x) n +1 C x 2 C x3 C x n +1 = C0 x + 1 + 2 + ... + n n +1 2 3 n +1

(41/5 + 71/10)45 is 45 + 1 i.e., 46. The general term in the expansion is

Tr + 1 = Cr ⋅ 4 45

45 −r 5

⋅ 7

2 n +1 C C C = C0 + 1 + 2 + ... + n  (n + 1) 2 3 n +1 176. (d) The given expression can be rewritten as

r 10

⇒

Tr + 1 is rational if r = 0, 10, 20, 30, 40. ∴ Number of rational terms = 5. ∴ Number of irrational terms = 46 – 5 = 41.

 3  4−1/ 2 1 − x   4 

171. (a) We have, n + 1C2 + 2 [2C2 + 3C2 + 4C2 + ... + nC2] =

n+1

C2 + 2 [3C3 + 3C2 + 4C2 + ... + nC2]

=

n+1

C2 + 2 [4C3 + 4C2 + ... + nC2]

= =

n+1

=

n+2



C2 + 2 [5C3 + ... + nC2] n+1 C 2 + 2 ⋅ n + 1C 3 = n + 1C 2 + C3 +

n+1

C3 =

n+1

C3 +

n+1

C3

n (n + 1)(n + 2) n (n + 1)(n − 1) + 6 6

∴ ( 2 + 1)6 = n + f = 6C0 ( 2 )6 + 6C1 ( 2 )5 + ... + 6C6

Let ( 2 – 1)6 = f ′ = 6C0 ( 2 )6 – 6C1 ( 2 )5 + ..., 0 < f ′ < 1

∴ n + f + f ′  = 2 [6C0 ⋅ 23 + 6C2 ⋅ 22 + 6C4 ⋅ 2 + 6C6] = 198 ∴ 197 < n + f < 198 ⇒ 197 < ( 2 + 1)6 < 198. ∴ n < ( 2 + 1)6 ⇒ t he greatest value of the natural number n = 197. 173. (a) We have, (1 + x)n = C0 + C1 x + C2 x2 + C3 x3 + C4 x4 + ... + Cn xn 

...(1)



n (1 + x)n – 1 = C1 + 2 C2 x + 3 C3 x2 + 4 C4 x3



+ ... + n Cn xn – 1

...(2)

and n 2

n–1

...(3)

= C1 + 2 C2 + 3 C3 + 4 C4

+ ... + n Cn 

  a  n +1     −1 b  = a 0b n     a  −1   b    b n ( a n +1 − b n +1 ) b a n +1 − b n +1 ⋅ n +1 = a −b b a −b 178. (c) ∵ (1 – y)m (1 + y)n =

= (mC0 – mC1 y + mC2 y2 – ….) (nC0 + nC1 y + nC2 y2 + … .) ∴ a1 = coefficient of y in (1 – y)m (1 + y)n = nC1 – mC1 = 10 ⇒ n = m + 10 …(i) and a2 = coefficient of y2 in (1 – y)m (1 + y)n = nC2 – mC1 · nC1 + mC2 ∴ nC2 – mC1 · nC1 + mC2 = 10 n(n − 1) m(m − 1) + − mn = 10 2 2

⇒ (10 + m) (9 + m) + m(m − 1) − m(10 + m) = 10 2 2  ⇒ 45 − 5m +

Putting x = 1 in (1) and (2), we get 2n = C0 + C1 + C2 + C3 + C4 + ... Cn

= (1 + ax + a2x2 + …) (1 + bx + b2x2 + …) Hence, an = coefficient of xn in (1 – ax)– 1 (1 – bx)–1 = a0bn + abn – 1 + …. + anb0

⇒

Differentiating w.r.t. x, we get



3 x < 1 or − 4 < x < 4 4 3 3

177. (c) ∵ (1 – ax)– 1 (1 – bx)– 1

172. (c) Let ( 2 + 1)6 = n + f where n ∈ N and 0 < f < 1.



−1/ 2

and it is valid only when

n (n + 1)(2n + 1) = . 6



(put x = 1)

...(4)

Adding (3) and (4), we get (n + 2) 2n – 1 = C0 + 2 C1 + 3 C2 + ... + (n + 1) Cn.

[using Eq. (i)] 9m m m m − + − = 10 2 2 2 2 2

2

⇒ 45 – m = 10 ⇒ m = 35 ∴ n = 45 ∴ (m, n) = (35, 45)

[using Eq. (i)]

179. (c) Coefficient of T5 = nC4, T6 = nC5 and T7 = nC6 ∴ 2nC5 = nC4 + nC6

725

174. (b) The sum of coefficients of odd powers of x.

Binomial Theorem

⇒ n ⋅ a(n – 1)/13 ⋅ a3/2 = 14 a5/2 ⇒ n ⋅ a(n – 1)/13 = 14 a ⇒ n ⋅ a(n – 14)/13 = 14 ⇒ n = 14 n 14 C C ∴ n 3 = 14 3 = 4. C2 C2

726

    n! n! n! ⇒ 2  + =   (n − 5)! 5!  (n − 4)! 4! (n − 6)! 6!

55 ≥ r ⇒ r = 27 2

⇒

Objective Mathematics

∴ Greatest term in the expansion of (1 + 3x)54 is T28.

  6   5⋅6 ⇒ 2  + 1 = ( n − 5) ( n − 4) ( n − 5)     ⇒ n2 – 21n + 98 = 0 ⇒ n = 7 or 14

185. (c) We have,  1  Tr +1 = 3 ⋅30 Cr    3

m 10    20  m 180. (b) ∑     is the coefficient of x in the expani =0  i   m − i 

20  1  and Tr = 3 Cr −1    3

sion of (1 + x)10 (x + 1)20. Now,

10   20   30  ∴ ∑    =  i =0  i   m − i  m  m

n Now,    r  max

n  r = 2 , = r= n ± 1 ,  2

181. (d) Given: C4a

n–4

n

n–5

⇒ r
1, then x is equal to (a) 10 2 (c) 104

expansion of (1 + ix)4n – 2, n ∈ N, x > 0, is

(a) n

(b) n + 1  + 12 x  is equal to 200 (c) n – 1  (d) 2n.  11. The coefficient of the term independent of x in the 6

(b) 10 (d) None of these

10

3 2  expansion of  x / 3 + x  2  

is:

727

1  sion of  x −  x 

Binomial Theorem

188. (b) Let (r + 1)th term be the constant term in the expan-

728

Objective Mathematics

5 4 9 (c) 4 (a)

(b)

(a) a = 3b (c) b = a3

7 4

(d) None of these

12. The coefficient of x5 in the expansion + (1 + x)22 + ... + (1 + x)30 is: (b) 9C5 (a) 51C5 31 21 (c) C6 – C6 (d) 30C5 + 20C5

18. In the binomial expansion (a + bx)–3 = (a) a = 2, b = 3 (c) a = 3, b = 2

n

(a) 56 (c) 45

values of a and b are:

of (1 + x)21

x  13. If the coefficient of x and x in  2 +  are equal, then 3   n is equal to: 7

8

(b) 55 (d) 15

n

2 +1 (d) None of these n 5 The coefficient of x in the expansion of (1 + x2)5 (1 + x)4 is: (a) 30 (b) 60 (c) 40 (d) None of these If the sum of the coefficients in the expansion of (1 – 3x + 10x2)n is a and if the sum of the coefficients in the expansion of (1 + x2)n is b, then:

(c)

17.

1 9 + x + ... , the 8 8

(b) a = 2, b = –3 (d) a = –3, b = 2

19. Given positive integer r > 1, n > 2 and the coefficients of (3r)th and (r + 2)th terms in the binomial expansion of (1 + x)2n are equal. Then: (a) n = 2r (b) n = 3r (c) n = 2r + 1 (d) None of these

14. In the expansion of (1 + x)50, the sum of the coefficient 20. of odd powers of x is: (a) 0 (b) 249 50 (c) 2 (d) 251 n 15. The coefficient of x in the expansion of loge (1 + 3x + 2x2) is: n n +1 n  2 +1 ( − 1 ) (a) (−1)  n  (b) [2n + 1] 21.   n

16.

(b) a = b3 (d) None of these

10

x 3  The coefficient of x4 in the expansion of  − 2  2 x  is: 405 405 (b) (a) 256 259 (c)

450 263

(d) None of these

In the expansion of (1 + x)n, coefficients of 2nd, 3rd and 4th terms are in A.P. then n is equal to:

(a) 7 (c) 11

(b) 9 (d) None of these 1 22. If the binomial expansion of (a + bx)–2 is − 3 x + ....... 4 then (a, b): (a) (2, 12) (c) (–2, –12)

(b) (2, 8) (d) None of these

Answers

1. (a) 11. (a) 21. (a)

2. (c) 12. (c) 22. (a)

3. (a) 13. (b)

4. (b) 14. (b)

5. (b) 15. (b)

6. (a) 16. (b)

7. (d) 17. (b)

8. (d) 18. (a)

9. (a) 19. (c)

10. (a) 20. (a)

18

Exponential and Logarithmic Series

CHAPTER

Summary of conceptS exponential SerieS

Particular Cases

If x is any real number, then

(a) e =

x x x + + ... ∞ . ex = 1 + + 1! 2! 3! 2

3

is known as exponential series and ex is called exponential function.

exponential theorem For a > 0, ax = e x loge a =

x log e a x 2 (log e a ) 2 x 3 (log e a)3 + + + ... ∞ , 1! 2! 3! for all real values of x. 1+

Deductions from exponential Series (i) e– x = 1 −

–x

 x x3 x5  (iii) ex – e– x = 2 1! + 3! + 5! + ... ∞    (iv) e = 1 +

1 1 1 1 + + + + ... ∞ 1! 2! 3! 4!

(v) e– 1 = 1 −

1 1 1 1 + − + + ... ∞ 1! 2! 3! 4!

1 1   (vi) e + e– 1 = 2 1 + + + ... ∞   2! 4!  1 1 1  (vii) e – e– 1 = 2  + + + ... ∞  1! 3! 5!  the value of e The value of e lies between 2 and 3 i.e., 2 < e < 3. The value of e, correct upto ten places of decimals, is 2.7182818284.

∞

∞



1

∑ (n − 1)!

∴

n=0

∞

=

1

∑ n!

n=0

[Since 0! = 1 and

1

1 when n < m = 0 (n − m)!

∑ (n − k )! = e]

n=0

∞ ∞ 1 1 1 1 1 + + + ... ∞ = ∑ = ∑ ( n + 1)! n ! n=0 1! 2! 3! n =1 1 1 1 (c) e – 2 = + + ... ∞ 2! 3! 4! ∞ ∞ ∞ 1 1 1 = ∑ = ∑ = ∑ . ( n + 1)! ( n + )! 2 n = 2 n! n = 1 n=0

(b) e – 1 =

(d)

x x 2 x3 x 4 + − + + ... ∞ 1! 2! 3! 4!

 x2 x4  (ii) e + e = 2 1 + 2! + 4! + ... ∞    x

1 1 1 1 + + + + ... ∞ = 0! 1! 2! 3!

(e)

1 1 1 e + e −1 + + + ... = 1+ 2! 4! 6! 2 ∞ ∞ 1 1 = ∑ = ∑ n = 0 ( 2n )! n = 1 ( 2n − 2)! 1 1 1 e − e −1 + + + ... = 1 ! 3! 5! 2 ∞

=

1

∞

1

∑ (2n − 1)! = ∑ (2n + 1)! n =1

n=0

logarithmic SerieS If | x | < 1, then

x 2 x3 x 4 + − + ... ∞ 2 3 4 is known as logarithmic series. log e (1 + x) = x −

Deductions from logarithmic Series (i) log e (1 – x) = − x −

x 2 x3 x 4 − − − ... ∞ 2 3 4

  1 + x  x3 x5 (ii) log e  1 − x  = 2  x + 3 + 5 + ... ∞      (iii) log e 2 = 1 –

1 1 1 + − + ... ∞ 2 3 4

730

MULTIPLE-CHOICE QUESTIONS

Objective Mathematics

Choose the correct alternative in each of the following problems: 1. The sum of the series

2 4 6 + + + ... ∞ is 1! 3! 5!

(a) e (c) 3e

(b) 2e (d) 4e

2. The sum of the series 1 + (a) e (c) 3e

3 5 7 + + + ... ∞ is 2! 4! 6!

(b) 2e (b) None of these 2

4.

∞

∑ n=0

2

(b) – 2 (d) – 1.

(log e x) n!

(−1) n [cn 2 − (b + c) n + a ] n!

(b)

(−1) n [cn 2 + (b + c) n + a ] n!

(c)

(−1) n [cn 2 − (b + c) n − a ] n!

(d) None of these

1 1 1 1     3. The value of 1 + + + ... − 1 + + + ... is    3! 5!  2! 4! (a) 2 (c) 1

(a)

10. The sum of the series (a) (c) 11. If

n

is equal to

(a) loge x (c) x

n

a − bx is ex

ex = B0 + B1x + B2x2 + ... + Bn – 1xn – 1 1− x

(a)

1 (n − 2)!

(b)

(c)

1 n!

(d) None of these

n

(a)

(−1) (a − bn) n!

(b)

(c)

(−1) n (b + an) n!

(d) None of these

(−1) (a + bn) n!

1 + x (1 + x) 2 (1 + x)3 + + + ... is 1! 2! 3! 2e n! e (c) n!

(b)

(a)

−2 n

(b)

−1 n

(c)

1 n

(d)

2 n

13. If y = x −

4e n!

7. The coefficient of xn in the expansion of  a + bx (a + bx) 2 (a + bx)3  + + + ... is 1 + 1! 2! 3!   ea ⋅bn (n − 1)!

(a)

e ab n n!

(b)

(c)

eb ⋅ a n n!

(d) None of these

C (n, 0) + C (n, 1) + ... + C (n, n) is equal to P ( n, n ) n =1 ∞

8.

∑

(a) e2 (c) e2 – 1

(b) e2 + 1 (d) None of these

9. The coefficient of xn in the expansion of is

x 2 x3 + − ... , then x is equal to 2 3

(a) log (1 + y)

(d) None of these

(a + bx + cx 2 ) ex

1 (n − 1)!

12. If n is a multiple of 3, then the coefficient of xn in the expansion of log (1 + x + x2) (| x | < 1), is

6. The coefficient of xn in the series

(a)

(b) e − 1 (d) None of these

+ Bnxn + ..., then the value of Bn – Bn – 1 is

(b) log x (d) None of these

5. The coefficient of xn in the expansion of

e e−2

1 1⋅ 3 1⋅ 3⋅ 5 + + + ... is 1⋅ 2 1⋅ 2 ⋅ 3⋅ 4 1⋅ 2 ⋅ 3⋅ 4 ⋅ 5 ⋅ 6

(c) y +

y 2 y3 + + ... 2! 3!

(b) e y (d) e­– y– 1

14. The sum of the series 1+ 3 1+ 3 + 5 1+ 3 + 5 + 7 + + + ... ∞ is 2! 3! 4! (a) e – 1 (b) e + 1 (c) e (d) 2e

1+

15. The sum of the series 1 + (a) e (c) e– 1/2

1 1⋅ 3 1⋅ 3⋅ 5 + + + ... is 2! 4! 6!

(b) e3/2 (d) e 2

16. The sum of the series 1 + (a) log 3 3 (c) log  2

4

11 1 1   +   + ...∞ is 32 5 2

(b) 2 log 3 (d) None of these

(a) log 3 (c) log 10

26. log

  5 x 3 9 x 5 13 x 7 (a)  x + + + + ... 2 ⋅ 3 4 ⋅ 5 6 ⋅ 7  

(b) log 5 (d) None of these

18. The sum of the series

  5 x 3 9 x 5 13 x 7 (b) 2  x + + + + ... 2 ⋅ 3 4 ⋅ 5 6 ⋅ 7  

 1  1 1 + + + ... is 2 3 5 (2n + 1)5  2n + 1 3 (2n + 1)  n (a) log n +1

n +1 (b) log n

(c) log n

(d) None of these

1 1 1 1 1  19. log 2 + 2  + ⋅ 2 + ⋅ 3 + ... ∞  = 5 5 5 3 5  (a) log 3 (c) 3 log 3

  5 x 3 9 x 5 13 x 7 (c)  x − + − + ... 2 ⋅ 3 4 ⋅ 5 6 ⋅ 7   (d) None of these 27. The value of 1 +

(b) 2 log 3 (d) None of these

20. If α, β are the roots of the equation x2 ­– px + q = 0, then (α + β ) x −

α 2 + β 2 2 α 3 + β3 3 x + x ... is equal to 2 3

(a) log (1 + px + qx2) (c) log (1 + px – qx 2)

(a) 2 loge 2

(b) loge 2 – 1

(c) loge 2

4 (d) loge   e

(a) 1 (c) e2

(a) e (e + 1) (c) e (e – 1)

+

(1 + 3 ) (log e 3) 2 2! 2

(1 + 33 ) (log e 3)3 + ... ∞ is 3!

(a) 18 (c) 36 25. 1+

3 7 15 + + + ... ∞ is 2! 3! 4!

(b) e (1 – e) (d) 3e

24. The value of (1 + 3) log e 3 +

(b) 28 (d) None of these

2 2 2 + + + ... ∞ = 1⋅ 2 ⋅ 3 3⋅ 4 ⋅ 5 5 ⋅ 6 ⋅ 7

(a) 2 log e4 (c) 2 log e3

e − ea a −1

(b)

(c)

ea −1 − e a −1

(d) None of these

29. If y = x −

(b) 2 log e2 (d) None of these

ea − e a −1

3 5 9 15 23 + + + + + ... ∞ is 1! 2! 3! 4! 5! (b) 4e – 3 (d) 3e + 4

x 2 x3 x 4 + − + ... , then x is 2 3 4 (b) log (1 + y) (d) e y

(a) e y – 1 (c) e y + 1

  x6 x9 30. If y = −  x 3 + + + ... , then 2 3   (a) x3 = 1 – e y (b) x3 = e y 31. The sum of (a) 5e (c) 7e

(b) e– 2 (d) – 1

23. The sum of the series 1 +

(a)

(a) 4e + 3 (c) 3e – 4

22. The product of the following series 1 1 1 1 1     1 + + + ...   1 − + − + ... is 1! 2! 1! 2! 3!

1 + a 1 + a + a2 + + ... ∞ is 2! 3!

28. The sum of the series

(b) log (1 – px + q x 2) (d) None of these

1 1 1 − + ... upto ∞ is equal 21. The sum of the series 1.2 2.3 3.4 to

(1 + x)(1 − x ) / 2 = (1 − x)(1 + x ) / 2

731

3

1 1 1 11 −   +   ... is equal to 4 2 4 34

32. The value of (a) 1.648 (c) 1.444 33.

(b) x = log (1 + y) (d) x = 1 + e y

12 ⋅ 2 22 ⋅ 3 32 ⋅ 4 42 ⋅ 5 + + + + ... is 1! 2! 3! 4! (b) 3e (d) 2e e will be (b) 1.547 (d) 1.348

1 1 1 + + + ... ∞ is n! 2!(n − 2)! 4!(n − 4)! 2n − 1 n! 2n (c) n!

(a)

(b)

2n (n + 1)!

(d)

2n − 2 (n − 1)!

34. Coefficient of x4 in the expansion of (a)

5 24

(b)

4 25

(c)

24 25

(d)

25 24

1 − 3x − x 2 is ex

Exponential and Logarithmic Series

2

17. The series 3 log 2 +

732

35. The sum of series

1 1 1 + ... is + + 2! 4! 6!

Objective Mathematics

(a)

(e − 1) 2e

(b)

(e − 1) 2e

(c)

(e 2 − 1) 2

(d)

(e 2 − 1) e

2

42. Sum of the series 1 4 9 16 + + + + ... is 1 ⋅ 2 ⋅ 3 ⋅ 4 3 ⋅ 4 ⋅ 5 ⋅ 6 5 ⋅ 6 ⋅ 7 ⋅ 8 7 ⋅ 8 ⋅ 9 ⋅ 10 equal to

2

(a)  

1 1 (c)   log e 2 − 6 24

1 2 3 4 36. x 2 + x 3 + x 4 + x 5 + ... is 2 3 4 5 x x + log (1 + x) (b) + log (1 + x) (a) − 1+ x 1+ x x + log (1 − x) (c) 1− x

43. The value of 7 log2

(d) None of these

37. If 1, log9 (31 – x + 2) and log3 (4 · 3x – 1) are in AP, then x is equal to (a)  log43 (c)  1 – log34

 ea − e  (b)     −1  a −1 

a

a −a (d)   e + e a −1

(c)   e − a a −1 39. If S =

11 1 1 1 1 1 1 1  +  −  +  +  +  + ... , then S 2  2 3  4  22 32  6  23 33 

is equal to (b)  log 3 2 (d)   1 log 2 2

(a)  log 3 (c)  log 2 3 40. If S =

2n

(log x) , then S is equal to (2n)! n=0 ∞

∑

(a)  x + x

(b)  x – x

(c)   1 (x + x– 1) 2

(d)  none of these

–1

–1

2 3 41. The value of 1 – log 2 + (log 2) − (log 2) + ... is 2! 3!

(a)  log 3

(b)  log 2

(c)   1 2

(d)  none of these

3 − log e 2 2

(d)  none of these 16 25 81 is + 5 log 2 + 3 log 2 15 24 80

(a)  1

(b)  log2(105)

9 (c)  log2   8

8 (d)  log2   9

1+

2 2 3 38. The sum of the series 1 + a + 1 + a + a + 1 + a + a + a + ... 2! 3! 4! 1 + a 1 + a + a 2 1 + a + a 2 + a3 + + + ... is 2! 3! 4! a

(b)  

44. The coefficient of xn in the series

(b)  log34 (d)  1 – log43

(a)   e a

5 − log e 2 2

a + bx (a + bx) 2 (a + bx)3 + + + .... ∞ is 1! 2! 3!

(a)  

(ab) n n!

(c)   e a ⋅

bn n!

45. The value of

(b)   eb ⋅

an n!

(d)   e a + b ⋅

(ab) n n!

2 4 6 8 + + + + .... ∞ is 3! 5! 7! 9!

(b)   1 e

(a)  e (c)   e +

1 e

(d)   e −

1 e

46. The sum of the series 1 − 1 + 1 − ... upto infinity is 2! 3! 4! (a)  e– 2 (c)   e

−

1 2

(b)  e–1 1



(d)   e 2

7x 3x 47. In the expansion of e + e , the constant term is 5x e

(a)  0 (c)  2

(b)  1 (d)  None of these

3 3 3 48. 1 + 2 + 3 + 4 + ...∞ equals 2! 3! 4! (a)  5e (b)  4e (c)  3e (d)  2e

= e1 + x – 1 = e. ex – 1   x x2 xn + ... + + ... − 1 . = e 1 + + 1 ! 2 ! n !  

2 4 6 + + + ... 1! 3! 5!

1+1 3 +1 5 +1 + + + ... 1! 3! 5! 1 3 1 5 1 =1+ + + + + + ... 1! 3! 3! 5! 5! 1 3 1 1 1 =1+ + + + + + ... = e1 = e. 1! 2! 3! 4! 5! =

∴ Coefficient of xn = 7. (a) We have, ex = 1 + x + e (a + bx) = 1 +

3 5 7 + + + ... ∞ 2. (a) We have, 1 + 2! 4! 6!

2 1 4 1 6 1 = 1 +  +  +  +  +  +  + ... ∞  2! 2!   4! 4!   6! 6!  1 1 1 1 1 1 =1+ + + + + + + ... ∞ = e. 1! 2! 3! 4! 5! 6! e+e 2

3. (c) We have and

−1

1 1 + + ... 2! 4!

=1+

2

e − e  e + e  − =   2   2   = 4. (c)

e2 + e− 2

−1

(log e x) n = eloge x = x. ∑ n! n=0

 (− 1) n x n xn −1 + + ... . (n − 1)! n!  n

n −1

(− 1) (− 1) −b n! (n − 1)!

a (− 1) b (− 1) + = n! n! (n − 1)!

6. (c) We have,

n

=

n =1

21 22 23 + + + ... ∞ 1! 2! 3!

  2 2 2 23 + + ... − 1 = e2 – 1. = 1 + +  1! 2! 3!  a + bx + cx 2 = (a + bx + cx2) e– x ex  (− x) 2 (− x) n − 2 = (a + bx + cx2) 1 + (− x) + + ... + 2! (n − 2)! 

9. (a) We have,

 (− x) n − 1 (− x) n + + ... . (n − 1)! n!  Hence, coefficient of xn in the expansion of 

=

a (−1) n b (−1) n n c (−1) n n (n − 1) − + n! n (n − 1)! n (n − 1) (n − 2)!

=

a (−1) n bn (−1) n cn (n − 1) (−1) n − + n! n! n!

=

(−1) n (a – bn + cn2 – cn) n!

=

(−1) n [cn2 – (b + c) n + a]. n!

(a + bn).

1 + x (1 + x) 2 (1 + x)3 + + + ... 1! 2! 3!

 1 + x (1 + x)  (1 + x) = 1 + + + + ... – 1 1 ! 2 ! 3 !   2

2n

∑ n!

a (−1) n b (−1) n − 1 c (−1) n − 2  a + bx + cx 2  + + is =   n! (n − 1)! ( n − 2)! ex

 x x 2 x3 = (a – bx)  1 − + − + ...  1! 2! 3!

= (− 1) n ⋅

=

+

a − bx = (a – 6x) e– x ex

n

C (n, 0) + C (n, 1) + ... + C (n, n) = 2n

∞

2

e2 + e− 2 + 2 e2 + e− 2 − 2 = − 4 4 2+2 4 + 2 − e2 − e− 2 + 2 = = 1. = 4 4 4

∴  Coefficient of xn = a ⋅

8. (c)

ea ⋅bn . n!

C (n, 0) + C (n, 1) + ... + C (n, n) P ( n, n ) n =1

2

+ (− 1) n − 1

The coefficient of xn in the expansion =

∞

∞

5. (b) We have,

 bx b 2 x 2  bn xn + + ... + + ... . = e a 1 + 1! 2! n!  

∴   ∑

1 1 1 1     ∴ 1 + + + ... − 1 + + + ...    3! 5!  2! 4! −1

(a + bx) (a + bx) 2 + + ... 1! 2!

and P (n, n) = n!

1 1 e − e −1 =1+ + + ... 3! 5! 2 2

x 2 x3 + + ... 2! 3!

(a + bx) (a + bx) 2 + + ... = e (a – bx) = ea · ebx 1! 2!

∴ 1 +

2 +1 4 +1 6 +1 =1+ + + + ... ∞ 2! 4! 6!

e . n!

10. (b) The general term of the given series is

3

Tn =

1⋅ 3⋅ 5 ... (2n − 1) 1⋅ 2 ⋅ 3⋅ 4 ⋅ 5 ... (2n − 1) 2n = (2n)! (2n)!{2 ⋅ 4 ⋅ 6 ... 2n}

Exponential and Logarithmic Series

1. (b) The given series =

733

solutions

734

(2n)! 1 (1 / 2) n = n = n n! (2n)! 2 n! 2 n!

=

Objective Mathematics

∑T n =1

n

(1 / 2) n n! n =1 ∞

∞

Hence sum of the series =

=

∑

 (1 / 2)1 (1 / 2) 2 (1 / 2)3  + + + ...  = e1/2 –1= e − 1 .  2! 3!  1! 

=

1 1 1 1  T4 = + + 1! 2! 2! 3! .................... .................... Adding vertically, we get 1 1 1 1     S = 1 + + + ... ∞  + 1 + + + ... ∞   1! 2!   1! 2!  T3 =

11. (c) We have ex = (B0 + B1x + B2x2 + ... + Bn – 1 xn – 1 + Bnxn + ...,) 1− x

= e + e = 2e. 15. (a) The nth term of the given series is Tn =

⇒ ex = (1 – x) (B0 + B1x + B2x2 + ... + Bn–1xn–1 + Bnxn + ...,)   x 2 x3 xn ⇒ 1 + x + + + ... + + ... 2! 3! n!  

=

1⋅ 3⋅ 5 ... (2n − 3) (2n − 2)!

1⋅ 2 ⋅ 3⋅ 4 ⋅ 5 ⋅ 6 ... (2n − 3) (2n − 2) , for n ≥ 2 (2n − 2)!⋅ 2 ⋅ 4 ⋅ 6 ... (2n − 2) n −1

Comparing the coefficient of xn on both sides, we get

1   (2n − 2)! 2 ⇒ Tn = =  ...(1) n −1 (n − 1)! (2n − 2)!⋅ 2 ⋅ (n − 1)!

1 1 = Bn – Bn – 1 Hence, Bn – Bn – 1 = . n! n!

From (1), we have

= (1 – x) (B0 + B1x + B2x2 + ...+ Bn–1xn–1 + Bnxn + ...,)

Clearly T1 = Ist term of the series = 1 2

12. (a) We have, log (1 + x + x2)= log {(1 – x3)/(1 – x)} = log (1 – x3) – log (1 – x) 1 1 1 1 = [–x3 –  (x3)2 –  (x3)3 ...] + [x +  x2 +  x3 ...] 3 3 2 2 If n is multiple of 3, say n = 3m, where m ∈ Z, the coefficient of x3m



= −

1 1 3 1 2 + = − + = − . m 3m n n n

x 2 x3 x 4 + − + ... 13. (c) Since y = x − 2 3 4 ⇒  y = log e (1 + x) y 2 y3 + + ... = 1 + x ∴ ey = 1 + x  or  1 + y + 2! 3! 2 3 y y + + ... ∞ . ∴ x = y + 2! 3! 14. (d) The general term of the given series is 1 + 3 + 5 ... n terms T n = n! Sum of n terms of A.P. with a = 1, d = 2 = n! n n [2a + (n − 1) d ] [2 + (n − 1) × 2] 2 = = 2 n! n! =

n2 n n2 = = ( − 1 )! n n ( n − 1)! n!

=

n −1+1 n −1 1 + = (n − 1)! ( n − 1)! (n − 1)!

∴ Tn =

2

3

1 1  2   2  + + ... = e1/2 = 2! 3!

e.

16. (a) Given series 2

=1+

4

11 1 1   +   + ... 3 2 5 2

 1 1  1 3 1  1 5  2 =  2 + 3  2  + 5  2  + ...   1 3 1 5   1 = 2  x + x + x + ... , where x = 3 5   2 1 1+ x 1+ x = 2 ⋅ log = log 2 1− x 1− x 1 1+ 3/ 2 2 = log = = log 3. 1 1 /2 1− 2 17. (c) The given series 2

3

1 1 1 11 −   +   ... 4 2 4 34

  x 2 x3 + − ...  = log 23 +  x − 2 3  

1   where x = 4 

1 = log 8 + log (1 + x) = log 8 + log  1 +   4

Putting n = 1, 2, 3, 4, ... 

1 1 + 2 + 1!

= 3 log 2 +

1 1 + . (n − 2)! (n − 1)!

T1 = 0 + 1

3

1 1 1     2 2 T2 = 2 , T3 = , T4 = , and so on 1! 2! 3! Thus, the given series become

T2 = 1 +

1 1!

= log 8 + log 

5 = log   8 ×  4

5  = log 10. 4

Tn =

(1 + 3) log e 3 +

19. (a) The given series

1  1+ 1+ x  5 = log 2 + log  1 1− x 1 −   5 3 3 6 5   log 2 + log  ×  = log 2 + log = log  2 ×  = 5 4  2 2 = log 3. = log 2 + log

20. (a) Since α, β are the roots of x2 –­ px + q = 0 ∴ α + β = p and αβ = q. Now, the given series

 2n

1

∑  n! − n! n =1





(1 + 32 ) (1 + 33 ) (log e 3) 2 + (log e 3)3 + ... 2! 3!

  (log e 3) 2 (log e 3)3 + + ... = 1 + log e 3 + 2 ! 3 !     (3 log e3) 2 (3 log e3)3 + 1 + 3 log e3 + + + ... – 2 2! 3!   = eloge 3 + e3 loge 3 − 2

2

3

3

    α x αx β x βx = αx − + − ... + βx − + − ... 2 3 2 3     = log (1 + αx) + log (1 + βx) 3 3

2 2

3 3

= log [(1 + αx) (1 + βx)] = log [1 + (α + β) x + αβx2] = log (1 + px + qx2) 1 1 1 21. (d) ... − + 1.2 2.3 3.4

[ α + β = p and αβ = q].

1 1 1 + 2. –2. + ... 3 2 4 1 1 1 = 2 1 − + − ... – 1 = 2 log (1 + 1) – 1  2 3 4  = 1 – 2.

22. (a) We have, 1 1    1 1 1  1 + + + ... 1 − + − + ... 1! 2!  1! 2! 3! 

25. (b) The given series is 2 2 2 + + + ... ∞ 1+ 1⋅ 2 ⋅ 3 3⋅ 4 ⋅ 5 5 ⋅ 6 ⋅ 7 2 1 1 − = ; 1⋅ 2 ⋅ 3 1⋅ 2 2 ⋅ 3 2 1 1 − = and so on. ∴ The given sum 3⋅ 4 ⋅ 5 3⋅ 4 4 ⋅ 5

Now

 1 1   1 1  − + − + ... =1+   1⋅ 2 2 ⋅ 3   3⋅ 4 4 ⋅ 5 

1 1 1   1 1 1 1  = 1 + 1 − + − + ... −  − + − + ...   2 3 4 5  2 3 4  1 1 1  = 2 1 − + − + ... = 2 loge 2.  2 3 4  26. (a) We have, log 

(1 + x)(1 − x ) / 2 (1 − x)(1 + x ) / 2

1 1 (1 − x) log (1 + x) − (1 + x) log (1 − x) 2 2 1 = [log (1 + x) − log (1 − x)] 2 1 − x[log (1 + x) + log (1 − x)] 2 =

1 1 1 1 1   –  −  +  −  ... 2 2 3 3 4

= 2 log 2 – log e = log 4 – log e = log

log 3 log 3 = e e + e e − 2 = 3 + 3 3 – 2 = 28.

 1   1  1 1 + + ... −  + + ... =1+   1⋅ 2 3⋅ 4   2 ⋅3 4 ⋅5 

α +β 2 α +β 3 x + x − ... 2 3 2

= e · e– 1 = e0 = 1.

n =1

=

3

  x3 x5 1 + + ... , where x = = log 2 + 2  x + 3 5 5  

 = 1 − 

n

24. (b) We have,

 2n + 1 + 1   2n + 1  2n + 2 n +1 = log = log = log . 2n n  2n + 1 − 1   2n + 1 

2 2

∞

∞

∑T

= (e2 – 1) – (e – 1) = e2 – e = e (e – 1).

1   x3 x5 = 2 x + + + ... , where x = 2n + 1 3 5   1 1+ 2n + 1 1 1+ x = 2 ⋅ log = log 1 2 1− x 1− 2n + 1

= (α + β) x −

2n − 1 ⇒ Sn = n!

=

 x5 1  1 ⋅ 2  x + x3 + + ... 2  3 5  1 1 1  − ⋅ x ⋅ (− 2)  x 2 + x 4 + ... 2 4 2 

4 . e 1 = x+ + 3 = x+

1  2 1  x +  + 2 5

1 5 1 1 7  x +  +  x + ... 4 7 6

5 x 3 9 x 5 13 x 7 + + + ... . 2 ⋅3 4 ⋅5 6 ⋅ 7

735

 1  1 1 + + + ... = 2 3 5 ( ) ( ) n + n + n + 2 1 3 2 1 5 2 1  

Exponential and Logarithmic Series

23. (c) The general term of the given series is

18. (b) The given series

736

27. (b) The general term of the given series is

Objective Mathematics

1 + a + a 2 + ... + a n − 1 1⋅ (a n − 1) / (a − 1) = n! n! (Series in the numerator is G.P with first term as 1 1  an 1  and common ratio a) =  − . (a − 1)  n! n! Tn =

 utting n = 1, 2, 3, ... and adding, we get the P given series  1   a a 2 a3 1 1 1  = (a − 1)   1! + 2! + 3! + ... −  + + + ...  1! 2! 3!   a

=

1 e −e [(e a − 1) − (e1 − 1)] = . (a − 1) a −1

28. (b) Let S =

3 5 9 15 23 + + + + + ... ∞ 1! 2! 3! 4! 5!

Let S1 = 3 + 5 + 9 + 15 + 23 + ... + tn or S1 = 3 + 5 + 9 + 15 + .......... + tn – – – – – – – 0 = 3 + 2 + 4 + 6 + 8 + ........ tn or tn = 3 + (2 + 4 + 6 + 8 + ... upto n – 1) n −1 = 3 + [2 × 2 + (n – 1 – 1) 2] 2 n −1 (2n) = 3 + n (n – 1) 2 nth term of given series is

= 3+

x 2 x3 + + ... 2! 3! 1 1 1 =1+ + + + ... 2 8 48

32. (a) We have, ex = 1 + x + e = e1/2

∴

= 1 + .5 + .125 + .0208 + ... = 1.648 nearly. 33. (a) We have,

1 1 1 + + + ... ∞ n! 2!(n − 2)! 4!(n − 4)!

2n − 1 1 . [nc0 + nc2 + nc4 + ... ∞] = n! n!

=

1 − 3x + x 2 = (1 – 3x + x2) e– x ex   x 2 x3 x 4 = (1 – 3x + x2) 1 − x + − + ... ∞  2! 3! 4!  

34. (d) We have,

Coefficient of x4 = =

1 1 1 1 1 1 = + + + + 4! 2 2! 24 2 2

1 + 12 + 12 25 . = 24 24 1 1 1 + + + ... to ∞ 1! 2! 3! 1 1 1 + – + ... to ∞ e–1 = 1 – 1! 2! 3!

35. (b) e = 1 +

1 1  ∴ e + e–1 = 2 + 2  + + ...  2! 4!  ∴

2 (e − 1) 2 1 1 1  e + 1 − 2 + ... = . +  e  = 2e 2! 4! 2

3 1 3 + n (n − 1) 3 + n (n − 1) = n! n (n − 1)! = n! + (n − 2)! n! 36. (c) The general term of the given series is 1 1 ∴ S∞ = 3 ∑ n! + ∑ (n − 2)! = 3 (e – 1) + e = 4e – 3. n n +1 n + 1 − 1 n +1 x x = T n = n +1 n +1 x 2 x3 x 4 29. (a) We have, y = x − + − + ... ∞ xn +1  1  n +1 n +1 2 3 4 x = 1 − = x −  n +1 n + 1  ∴   e y = 1 + x ⇒ x = e y – 1. = log e (1 + x) Putting n = 1, 2, 3, ... 30. (a) We have, y = log e (1 – x3)  x 2 x3 x 4  ... + ∴ S = (x2 + x3 + x4 + ...) −  + ∴ e y = 1 – x3 ⇒ x3 = 1 – e y. 3 4  2 31. (c) The general term of the given series is  x2 x 2 x3 x 4  ... = x+ + − x − − − n (n + 1) n 2 (n + 1) 1− x  2 3 4  Tn = = (n − 1)! n! x − x2 + x2 x 2 (n − 1 + 1) + (n − 1 + 1) (n − 1) 2 + 3 (n − 1) + 2 + log (1 − x) = + log (1 − x) . = =  = 1 − x 1 − x (n − 1)! (n − 1)! Tn =

=

n −1 3 2 + + (n − 2)! (n − 2)! ( n − 1)!

1 4 2 + + = . (n − 3)! (n − 2)! (n − 1)! 1 1 1 ∴ Sn = Σ Tn = Σ + 4Σ + 2Σ (n − 3)! (n − 2)! (n − 1)!

= e + 4e + 2e = 7e.

37. (c) 2log9 (31 – x + 2) = 1 + log3 (4·3x – 1) ⇒

2 log 3 (31− x + 2) = log3 3 + log3 (4·3x – 1) log 3 9

⇒ 31 – x + 2 = 3(4·3x – 1) 3 3 ⇒ x + 2 = 12·3x – 3 ⇒ t + 2 = 12t – 3 3 ⇒ 12t2 – 5t – 3 = 0

(where t = 3x)

38. (b) Tn =

737

16 25 81 + 5 log 2 + 3 log 2 15 24 80

= 7 [4 log2 2 – log2 3 – log2 5] + 5 [2 log2 5 – log2 3 – 3 log2 2]

1 + a + a 2 + ... + a n −1  a n − 1  1 =  n!  a − 1  n!

 1  a − 1 a 2 − 1 a3 − 1 ∴ Sn = + + + ... − 1  a − 1  1! 2! 3!  =

43. (a) 7 log2

+ 3 [4 log2 3 – 4 log2 2 – log2 5] = log2 2 = 1 x x 2 x3 + + + ... ∞ 1! 2! 3! Put x = a + bx, on both sides, we get

44. (c) We have, ex = 1 +

 1 1 1  1  a a 2 a 3 + + ...  −  + + + ...   − 1  + a − 1  1! 2 ! 3 ! 1! 2 ! 3 !   

ea + bx = 1 +

(a + bx) (a + bx) 2 (a + bx)3 + + + ... 1! 2! 3!

1 (b) n (e a − e) − 1 ∴ Coefficient of xn in ea + bx = ea ⋅ a −1 n! 1 1 1 1 1 1  1 1 1 1 1 1  2n 39. (d) S =  ⋅ − ⋅ 2 + ⋅ 3 − ...  +  ⋅ − ⋅ 2 + ⋅ 3 − 45. ...  (b) Let S = 2 + 4 + 6 + 8 + ...∞  ∴ T = n 6 2 2 2 4 2  2 3 4 3 6 3  (2n + 1)! 3! 5! 7! 9! 1 1 1 1 1 1  1 1 1 1 1 1  1 1  ⋅ − ⋅ 2 + ⋅ 3 − ...  +  ⋅ − ⋅ 2 + ⋅ 3 − ...  Now, Tn = − 6 2 2 2 4 2  2 3 4 3 6 3  (2n)! (2n + 1)! 1 1 1 1 1 1  11 1 1 1 1  1 1 1 1 =  − ⋅ 2 + ⋅ 3 − ...  +  − ⋅ 2 + ⋅ 3 − ...  T1 = − , T2 = − 2 2 2 2 3 2 2 3 2 3 3 3      2! 3! 4! 5! =

1  1 1  1 = log 1 +  + log 1 +  2  2 2  3

……………… ……………… ………………

1 3 4 1 = log  ×  = log 2 2 2 3 2 40. (c) S = 1 +

(log x) 2 (log x) 4 (log x)6 + + + ... 2! 4! 6!

1 1 = (elog x + e − log x ) = ( x + x −1 ) 2 2

41. (c) We have, 1 – log 2 + =e

– log 2

=e

log 2– 1

42. (c) We have, Tn = =

(log 2) 2 (log 2)3 − + ... 2! 3!

1 =2 = 2 –1

n2 (2n − 1)2n(2n + 1) (2n + 2)

 1 1 1 1 + −   2 12(2n − 1) 4(2n + 1) 3(2n + 2) 

1 = 24

 1 1 1   1 − +4 −   2 n − 1 (2 n + 1) 2 n + 1 2 n + 2   

Putting n = 1, 2, 3, 4, … ∴ S∞ =

1  1 1 1 1 1  1 − + − + − ...  24  3 3 5 5 7 

=

1 1 1 1  + 4  − + − + ...  3 4 5 6 

1 1 1 [(4 log e 2 − 1)] = log e 2 − 24 6 24

∴S=

∞

∑T n =1

46. (b)

47. (c)

n

=

1 1 1 1 − + − + ... = e–1 2! 3! 4! 5!

1 1 1 1 1 1 − + − ... 1 − 1 + − + − ... = e −1 2! 3! 4! 2! 3! 4!

e 7 x + e3 x = e2x + e–2x Since, e5 x 2 x (2 x) 2 (2 x)3 e2x = 1 + + + + ...∞  1! 2! 3! and e–2x = 1 −

2 x (2 x) 2 (2 x)3 + + + ....∞  1! 2! 3!

…(i) …(ii)

 (2 x) 2 (2 x) 4  + + .....∞  ∴ e2x + e–2x = 2 1 + 2! 4!   Hence, the constant term is 2. 48. (a) The given series is 1 + Tn = =

23 33 43 + + + ...∞ 2! 3! 4!

n3 n2 n2 − 1 1 = = + n! (n − 1)! (n − 1)! (n − 1)!

n +1 1 1 3 1 + = + + (n − 2)! (n − 1)! (n − 3)! (n − 2)! (n − 1)!

 1 3 1  ⇒ ΣTn =Σ  + +  ( n − 3)! ( n − 2)! ( n − 1)!   = e + 3e + e = 5e

Exponential and Logarithmic Series

5 ± 169 3 = (as t j. For example, the matrix 1  0 A= 0   0

2 5 0 0

3 2 1 0

4  3 is an upper triangular matrix. 4  5 

Lower Triangular Matrix  A square matrix A = [aij] is called a lower triangular matrix if aij = 0 for all i < j. For example, the matrix

740

Objective Mathematics

2  3 A=  1   2

0 4 3 4

0 0 5 6

0  0 is a lower triangular matrix. 0  7 

 2 −1    A =  5 4  , then – A =  −6 0 

 −2 1     −5 −4  .  6 0 

Trace of a Matrix  The sum of the diagonal elements of Properties of Addition of Matrices a square matrix A is called the trace of A and is denoted by (i) Matrix addition is commutative. If A and B are two matritr (A). For example, if ces of the same order, then A + B = B + A. (ii) Matrix addition is associat ive. If A, B and C are three ma 1 3 4 trices of the same order, then (A + B) + C = A + (B + C).   A =  2 −1 6  , then tr (A) = 1 – 1 + 4 = 4. (iii) Existence of additive identity. If O is the zero matrix of  3 1 4  the same order as that of the matrix A, then A + O = A = O + A. Sub Matrix  A matrix which is obtained from a given matrix (iv) Existence of additive inverse. If A is any matrix, then by deleting any number of rows or columns or both is called a A + (– A) = O = (– A) + A. sub-matrix of the given matrix. For example, (v) Cancellation laws hold good in case of addition of matrices. If A, B, C are matrices of the same order, then 5 3 1   2 −1    A + B = A + C ⇒ B = C (left cancellation law)     is a sub-matrix of the matrix  4 2 −1  . and  B + A = C + A ⇒ B = C (right cancellation law) 3 5   6 3 5  Note:  The zero matrix plays the same role in matrix addition as Equality of Matrices  Two matrices A and B are said to be the number zero does in addition of numbers. equal if they are of same order and all the corresponding elements are equal. It is written as A = B. For example, 1+ 2 1+ 3  2 3 4  2 A=   and B =   5 1 0 2 + 3 1 0     are equal matrices, whereas  2 5 1 3 4 C=   and  2 3 1 2 1     are not equal, because their orders are not same.

Algebra of Matrices Addition of Matrices  Let A and B be two matrices each of order m × n. Then the sum matrix A + B is defined only if matrices A and B are of same order. The new matrix, say C = A + B is of order m × n and is obtained by adding the corresponding elements of A and B. Thus, if A = [aij]m × n and B = [bij]m × n are two matrices of same order then the sum A + B is defined to be the matrix of order m × n such that A + B = C = [aij + bij] for all i and j. 2 3 For example, if A =  1 0 then C = A + B =  2 + 4 3 − 2   1+1 0 +1

4  4 −2 3   and B =   5  1 1 4 4 + 3  6 1 7  , whereas =  5 + 4  2 1 9 

the addition of  4 −2  2 3 4 A=   and B =  1 1    1 0 5

Subtraction of Matrices

Let A and B be two matrices of the same order. Then by A – B, we mean A + (– B). In other words, to find A – B we subtract each element of B from the corresponding element of A. 2 3    For example, if A =  6 1  and B =  7 −2 

 −3 4     2 5  6 3 

 2+3 3− 4   5 −1      then A – B =  6 − 2 1 − 5  =  4 −4  .  7 − 6 −2 − 3   1 −5 

Mulitiplication of a Matrix by a Scalar Let A = [aij] be an m × n matrix and k be any scalar. Then the matrix obtained by multiplying each element of A by k is called the scalar multiple of A by k and is denoted by kA. Thus, if A = [aij]m × n, then kA = [kaij]m × n. For example, if 1 2 3   A =  4 −6 8  then  0 2 5  1 2 6 9   3  1   3A = 12 −18 24 and A =  2   2   0 6 15  0 

1 −3 1

3 2 4 5 2

   .   

is not defined since the two matrices are not of same order. If A Properties of Scalar Multiplication is any matrix, the negative of A, denoted by – A, is the matrix (i) If A and B are two matrices of the same order and k be a obtained by replacing each entry in A by its negative. For exscalar, then ample, if k (A + B) = kA + kB.

Multiplication of Matrices Two matrices A and B can be multiplied only if the number of columns in A (pre-multiplier) is same as the number of rows in B (first multiplier). For example, if A = [aij]m × n and B = [bjk]n × p are two matrices of order m × n and n × p respectively, then their product AB is of order m × p and is defined as n



(AB)i, k =

∑a j =1

ij

b jk = ai1 b1k + ai2 b2k + ... + ain bnk  b1k  b ... ain]  2 k     bnk

   = [ai1 ai2    = (ith row of A) (kth column of B). (AB)ik = Sum of the product of elements of ith row of A with the corresponding elements of kth column of B. Remarks 1. If A and B are square matrices of the same order, say n, then both the products AB and BA are defined and each is a sqaure matrix of order n. 2. In the matrix product AB, the matrix A is called premultiplier (prefactor) and B is called post-multiplier (postfactor). 3. The rule of multiplication of matrices is row column wise (or → ↓ wise), viz. the first row of AB is obtained by multiplying the first row of A with first, second, third, ... columns of B respectively. Similarly second row of A with first, second, third, ... columns of B respectively and so on.



c21 c12 c13 c14 c22 c23 c24 c31 c32 c33 c34

= (5) (– 2) + (3) (2) = – = (1) (4) + (– 4) (7) = – = (1) (1) + (– 4) (3) = – = (1) (6) + (– 4) (8) = – = (5) (4) + (3) (7) = 41 = (5) (1) + (3) (3) = 14 = (5) (6) + (3) (8) = 54 = (0) (– 2) + (2) (2) = 4 = (0) (4) + (2) (7) = 14 = (0) (1) + (2) (3) = 6 = (0) (6) + (2) (8) = 16.

4. 24 11 26

 −10 −24 −11 −26    Thus, AB =  −4 41 14 54   4 14 6 16  The product BA is not defined since the number of columns of B is not equal to the number of rows of A. This shows that matrix multiplication is not commutative. That is, for any two matrices A and B, it is usually the case that AB ≠ BA (even if both products are defined).

Properties of Matrix Multiplication (i) Multiplication is distributive over matrix addition. If A, B, C are m × n, n × p and n × p matrices respectively, then A (B + C) = AB + AC. (ii) Multiplication is associative. If A, B, C are matrices of order m × n, n × p and p × r respectively, then (AB) C = A (BC) (iii) Multiplicative identity. If A is an m × n matrix and In the identity matrix of order n × n and Im the identity matrix of order m × m, then ImA = A and AIn = A. In particular if A is a square matrix of order n, then

AIn = InA = A.  1 −4  − 2 4 1 6   (iv)  AB = 0 (null matrix) does not necessarily imply that A = 0 Let A =  5 3  and B =      0 −1  2 7 3 8    0 2  or  B = 0 or both = 0. For example, if A =   ≠0 0 0  and be two matrices. 1 1 0 0 Since the number of columns in A are equal to the numB=   ≠ 0, then AB =  . ber of rows in B, the product AB is defined. As order of ma0 0 0 0 trix A is 3 × 2 and B is 2 × 4, the product AB will be of order (v) If A is a square matrix of order n, then A2 is defined as AA. 3 × 4. In general Am = AA .... A (m times), where m is any positive  c11 c12 c13 c14  integer. AB =  c21 c22 c23 c24  . (vi) If I be a unit matrix, then I = I2 = I3 = ... = In.    c31 c32 c33 c34 

The entry c11 is obtained by summing the products of each Transpose of a Matrix entry in row 1 of A by the corresponding entry in column 1 of Let A be an m × n matrix. Then, the n × m matrix obtained by B, i.e., interchanging the rows and columns of A is called the transpose c11 = (1) (– 2) + (– 4) (2) = – 10. of A, and is denoted by A′ or At. Thus, Similarly, for c21, we use the entries in row 2 of A and those (i) if order of A is m × n, then, the order of A' is n × m. (ii) (i, j)th element of A = ( j, i)th element of A'. in column 1 of B:

741

Also,

Matrices

(ii) If k1 and k2 are two scalars and A is a matrix, then (k1 + k2) A = k1A + k2A. (iii) If k1 and k2 are two scalars and A is a matrix, then (k1k2) A = k1 (k2A) = k2 (k1A). (iv) If A is any matrix, then 1A = A.

742

Objective Mathematics

 2 4    2 −3 −1  −3 2  . For example, if A =   , then A' =  4 2 3    −1 3  3×2

Properties of the Transpose of a Matrix

(vi) The matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric. (vii) All positive integral powers of a symmetric matrix are symmetric. (viii) All positive integral powers of a symmetric matrix are symmetric. (ix) All positive odd integral powers of a skew-symmetric matrix are skew-symmetric and positive even integral powers of a skew-symmetric matrix are symmetric. (x) If A and B are symmetric matrices of the same order, then (a) AB – BA is a skew-symmetric matrix. (b) AB + BA is a symmetric matrix.

(i) Let A and B be two matrices of order m × n, then (A ± B)' = A′ ± B′. (ii) Let A be a matrix of order m × n and k be a scalar, then (kA)′ = kA′. (iii) Let A and B be two matrices of order m × n and n × p respectively. Then, (AB)′ = B′ A′. (iv) The double transpose of any matrix is the original matrix. For exampe, if A is any marix, then (A′ )' = A.

Orthogonal Matrix

Symmetric Matrix

A square matrix of order n × n is said to be orthogonal if AA′ = In = A′ A. For example, if

A square matrix A is said to be symmetric if A′ = A. That is, the matrix A = [aij]n × n is said to be symmetric provided aij = aji for all i and j.



2 1 5  For example, A =  1 0 −3  is symmetric, since    5 −3 6  2 1 5    A′ =  1 0 −3  = A.  5 −3 6 

Skew Symmetric Matrix A square matrix A is said to be skew symmetric, if A′ = – A. That is, the matrix A = [aij]n × n is skew-symmetric if aij = – aji for all i and j.  0 5 7   For example, A =  −5 0 3  = – A.  −7 −3 0  Remark:  Elements of main diagonal of a skew-symmetric matrix are all zero, because by definition, aii = – aii ⇒ 2aii = 0 or aii = 0 for all values of i.

Properties of Symmetric and Skew-Symmetric Matrices

A =

 2 −2  1  2 2   , then A′ =   2 2 2 2  2 2  −2 2  1

Also, AA' =

1  2 −2   2 2  1  8 0    =   8  2 2   −2 2  8  0 8  1 0 =   = I. 0 1



Similarly, A′ A = I. Hence A is orthogonal.

Idempotent Matrix A square matrix A is said to be idempotent if A2 = A.  2 −2 −4  For example, if A =  −1 3 4  , then    1 −2 −3   2 −2 −4   2 −2 −4     A = AA =  −1 3 4   −1 3 4   1 −2 −3   1 −2 −3  2

 2 −2 −4    =  −1 3 4  = A.  1 −2 −3  Hence A is idempotent.

Involutory Matrix (i) If A is a square matrix, then (a) A + A′ is symmetric A square matrix A is said to be involutory matrix if A2 = I. (b) A – A′ is skew-symmetric.  −5 −8 0  (ii) If A and B are two symmetric (or skew-symmetric) matriFor example, if A =  3 5 0  , then ces of the same order, then so is A + B.    1 2 −1  (iii) If A is symmetric (or skew-symmetric) matrix and k is a scalar, then kA is also symmetric (or skew-symmetric).  −5 −8 0   −5 −8 0   1 0 0  (iv) If A and B are symmetric matrices of the same order, then      2 A =  3 5 0   3 5 0  =  0 1 0  = I. the product AB is symmetric if and only if AB = BA.  1 2 −1   1 2 −1   0 0 1  (v) Every square matrix can be expressed uniquely as the sum of a symmetric and a skew-symmetric matrix. Hence A is involutory.

Properties of Transpose Conjugate

A square matrix A is said to be nilpotent matrix if there exists a positive integer m such that Am = 0. If m is the least positive integer such that Am = 0, then m is called the index of the nilpotent matrix A.

(i) (ii) (iii) (iv)

 ab A = AA =  2  −a 2

b2   , then − ab 

b 2   ab  − ab   −a 2

b2   0 0  =  = 0. −ab   0 0 

Hence the matrix A is nilpotent of the index 2.

(Aθ)θ = A (A + B)θ = Aθ + Bθ (kA)θ = kAθ, k being any number. (AB)θ = BθAθ.

Matrices

 ab For example, if A =  2  −a

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Nilpotent Matrix

Hermitian Matrix A square matrix A = [aij] is said to be Hermitian matrix if aij = a ij, ∀ i, j  i.e.,  A = –Aθ. 3 + 2i   4 For example,   is a skew–Hermitian matrix. 3 − 2 6  i 

Singular Matrix

Skew-Hermitian Matrix

A square matrix A is said to be singular matrix if determinant of A denoted by det A or | A | is zero, i.e., | A | = 0, otherwise, it is a non-singular matrix.

A square matrix A = [aij] is said to be a Skew-Hermitian matrix

 0 1 −1    For example, the matrix A =  4 −3 4   4 −3 4  0 1 −1 is singular as 4 −3 4 = 0. 4 −3 4 Note: See next chapter for expansion of the determinant.

Conjugate of a Matrix

if aij = – aij, ∨ i, j i.e., Aθ = – A.  0 3 + i For example,   is a Skew-Hermitian matrix. 0  3 − i Note: 1. If A is a Hermitian matrix then aii = aii ⇒ aii is real ∀ i, thus every diagonal element of a Hermitian matrix must be real. 2. A Hermitian matrix over the set of real numbers is actually a real symmetric matrix. 3. If A is a skew– Hermitian matrix then aii = – aii⇒ – aii + aii = 0 i.e., aii must be purely imaginary or zero. 4. A Skew–Hermitian matrix over the set of real numbers is actually a real symmetric matrix.

The matrix obtained from any given matrix A containing complex numbers as its elements, on replacing its elements by the corresponding conjugate complex numbers is called conjugate Adjoint of a Square Matrix of A and is denoted by A. Let A = [aij] be a square matrix of order n and let Cij be the cofac2 3+i   1− i tor of aij in the determinant | A |. Then the adjoint of A, denoted   2  , then For example, if A = 7 − 3i 1 + i by adj A, is defined as the transpose of the cofactor matrix.  5 6 3 − 4i  The adjoint of a square matrix A is obtained on replacing each (i, j)th element of A by the cofactor of the 2 3−i   1+ i (j, i)th element in | A |.   2 . A = 7 + 3i 1 − i  1 2 3  5 6 3 + 4i    For example, if A =  −1 0 1  , then we have  4 3 2  Transpose Conjugate of a Matrix The transpose of the conjugate of a matrix A is called transpose conjugate of A and is denoted by Aθ. 2   1− i For example, if A =   , then 3 + 2 1 + i i  1 + i 3 − 2i  Aθ =  . 1− i   2

C11 = – 3 C21 = 5 C31 = 2

C12 = 6 C22 = –10 C32 = – 4

C13 = – 3 C23 = 5 C33 = 2

2   −3 5 Thus, adj A =  6 −10 −4  .    −3 5 2 

Note that the conjugate of the transpose of A is the same as Properties of the Adjoint of a Matrix the transpose of the conjugate of A i.e., (A′) = (A′) = Aθ. (i) If A is a square matrix of order n, then If A = [aij]m × n, then Aθ = [bji]n × m where bji = aij A (adj A) = | A | In = (adj A) A, θ i.e., the (j, i)th element of A = the conjugate of (i, j)th element where In is a square matrix of order n. of A.

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Objective Mathematics

(ii) If A is a square matrix of order n, then adj (A′ ) = (adj A)′. The determinant of a square submatrix of order p × p is called a (iii) If A and B are two square matrices of the same order, minor of A of order p. For example, in the matrix then adj (AB) = (adj B) (adj A). 1 2 0 4   (iv) adj (adj A) = | A |n – 2 A, where A is a non-singular matrix. 2 3 1 0 , A=  ( n−1)2 4 1 3 6 , where A is a non-singular matrix. (v) | adj (adj A) | = | A |   (vi) Adjoint of a diagonal matrix is a diagonal matrix.  8 1 2 0 

Inverse of a Square Matrix

(i) every element of the matrix is the minor of order 1. 1 2 3 6 2 3 0 4 (ii) etc. , , , 2 3 2 0 4 1 1 0

Let A be any n-rowed square matrix. Then a square matrix B, such that AB = BA = I, is called inverse of A. The inverse of A denoted by A–1 is determined by the formula: 1 (adj A). A–1 = |A| It may be noted that AA– 1 = A– 1A = I.

are the minors of order 3.

Properties of the Inverse of a Matrix

Rank of a Matrix

(i) A square matrix is invertible if and only if it is nonsingular. (ii) The inverse of the inverse is the original matrix itself, i.e., (A–1)–1 = A. (iii) The inverse of the transpose of a matrix is the transpose of its inverse, i.e., (A' )–1 = (A–1)′. (iv) If A and B are two invertieble matrices of the same order, then AB is also invertible and moreover (AB)–1 = B–1 A–1. (v) Let A, B, C be square matrices of the same order n. If A is a non-singular matrix, then (a)  AB = AC ⇒ B = C (Left cancellation law) (b)  BA = CA ⇒ B = C (Right cancellation law) Note that these cancellation laws hold only if the matrix A is non-singular. (vi) If A is a non-singular matrix such that A is symmetric then A–1 is also symmetric. (vii) If A is a non-singular matrix, then | A–1 | = | A |–1.

A positive integer r is said to be the rank of a non zero matrix A, if

Elementary Transformations Any one of the following operations on a matrix is called an elementary transformation.

are the minors of order 2 3 1 0 2 1 0 2 3 1 (iii) 1 3 6 , 4 3 6 , 4 1 3 etc. 1 2 0 8 2 0 8 1 2

(i) there exists atleast one minor in A of order which is not zero, (ii) every minor in A of order greater than r is zero. It is written as ρ (A) = r. The rank of a zero matrix is defined to be zero. 2 1 For example, rank of   is 2 because the minor 1 3 2 1   =5≠0 1 3 1 0 0 1 0 0     and rank of  0 1 0  is 3 as the minor   0 1 0  = 1 ≠ 0.  0 0 1   0 0 1  2 1 3   For the square matrix A =   4 2 6  we have | A | = 0,  3 1 4  since first two rows are identical.

4 2 2 1 = 0, but = – 2 ≠ 0. Since atleast 3 1 4 2 (a) Interchanging any two rows (or columns). This transforone minor of order 2 is not zero, therefore the rank of A is 2. mation is indicated by Ri ↔ Rj ⋅ (Ci ↔ Cj) (b) Mulitplication of the elements of any row (or column) by a non-zero scalar quantity and indicated as Ri ↔ kRi Properties of Rank of a Matrix (Ci ↔ kCi ) (i) Rank of a matrix remains unaltered by elementary transfo (c) Addition of constant multiple of the elements of any row mations. to the corresponding element of any other row, indicated as (ii) No skew-symmetric matrix can be of rank 1. Ri → Ri + kRj. (iii) Rank of matrix A = Rank of matrix A′ Two matrices are said to be equivalent if one is obtained (iv) AA′ has the same rank as A. from the other by elementary transformations. The symbol ≈ is used for equivalence. Echelon Form of a Matrix

Minor

The minor

A non-zero matrix A is said to be in Echelon form if A satisfies the following conditions:

If m – p rows and n – p columns from matrix Am × n, are removed, 1. All the non-zero rows of A, if any, precede the zero rows. the remaining square submatrix of p rows and p columns is left.

1  0 0   0

2 1 0 0

3 3 1 0

4 5 2 0

5  1 1  0 

criterion of consistency Let AX = B be a system of n linear equations in n variables.

is in the Echelon form because. (i) there are three non-zero rows which precede the fourth zero row. (ii) Number of zeros in R2, R3, R4 are 1, 2, 5 which are in ascending order. (iii) The first non-zero element is unity. Note that there are three non-zero rows, so the rank is 3.

a13 a23 a33  am 3

 a1n    a2 n   a3n      ammnn 

AX = O where O is a null matrix of order n × 1.

1. Make a11 = 1, using elementary row transformations. 2. Make a21, a31, ..., am1 all zeros using the transformations. R2 → R2 – a21R1, R3 → R3 – a31R1, ..., Rm → Rm – am1R1. 3. Make a22 = 1, using elementary row transformations. 4. Make a32, a42, ... am2 all zeros using the transformations R3 → R3 – a32R2, R4 → R4 – a42R2, ..., Rm → Rm – am2R2. 5. Repeat this process, till the matrix A is reduced to Echelon form. 6. Count the number of non-zero rows in it, which will be the rank of the matrix A.

SoLutIon of a SyStem of LInear equatIonS by matrIx methoD Consider a system of linear equations a11x1 + a12x2 + ... + a1nxn = b1 a21x1 + a22x2 + ... + a2nxn = b2     

 a11 a12  a1n   x1   b1      a21 a22  a2 n   x2  =  b2               an1 an 2  ann   xn   bn X

B

(i) If | A | ≠ 0, then its only solution X = 0, is called the trivial solution. (ii) If | A | = 0, then AX = O have a non-trivial solution. It will have infinitely many solutions. Solution of a System of Linear Equations by Rank Method Let AX = B be a system of n linear equations in n variables. 1. Write the augmented matrix [A B]. 2. Reduce the augmented matrix to echelon form using elementary row-operations. 3. Determine the rank of coefficient matrix A and augmented matrix [A B] by counting the number of non-zero rows in A and [A B]. 4. If ρ (A) ≠ ρ (A B), then the system of equations is inconsistent. 5. If ρ (A) = ρ (A B) = the number of unknowns, then the system of equations is consistent and has a unique solution. 6. If ρ (A) = ρ (A B) < the number of unknowns, then the system of equations is consistent and has infinitely many solutions.

an1x1 + an2x2 + ... + annxn = bn. We can express these equations as a single matrix equation

A

homogeneous equations The system of equations AX = B is said to be homogeneous if the constants b1, b2 ... bn are all zero. That is, if the matrix B is a zero matrix and the system is of the form

Working rule for finding the rank of a matrix  a11 a12   a21 a22 Let A =  a31 a32     a a  m1 m 2

(i) If | A | ≠ 0, then the system of equations in consistent and has a unique solution given by X = A–1B. (ii) If | A | = 0 and (adj A) B = 0, then the system of equations is consistent and has infinitely many solutions. (iii) If | A | = 0 and (adj A) B ≠ 0, then the system of equations is inconsistent, i.e., it has no solution.

     

remarks: Let AX = O be a homogeneous system of linear equations. 1. If ρ (A) = number of variable, then AX = 0 have a trivial solution X = 0. 2. If ρ (A) < number of variables, then AX = 0 have a nontrivial solution. It will have infinitely many solutions.

745

Matrices

Let | A | ≠ 0, so that A–1 exists uniquely. Pre-multiplying 2. The number of zeros preceding the first non-zero element in a row is less than the number of such zeros in the suc- both sides of AX = B by A–1, we get ceeding row. A–1 (AX) = A–1B or (A–1A) X = A–1B 3. The first non-zero element in a row is unity. or IX = A–1B or X = A–1B. The number of non-zero rows of a matrix given in the EchHence X = A–1B is the unique solution of AX = B, | A | ≠ 0. elon form is its rank. For example, the matrix

746

multiple-choice questions

Objective Mathematics

Choose the correct alternative in each of the following problems: 5 −7   0 1. The matrix  −5 0 11  is known as    7 −11 11  (a) skew-symmetric matrix (b) symmetric matrix (c) diagonal matrix (d) upper triangular matrix a 0 0 2. If A =  0 a 0  then the value of | adj A | is  0 0 a  (a) a27 (c) a6

(b) a9 (d) a2

3. Which of the following is correct ? (a) skew symmetric matrix of even order is always singular (b) skew symmetric matrix of odd order is non-singular (c) skew symmetric matrix of odd order is singular (d) none of the above 4. The equations x + 2y + 3z = 1, x – y + 4z = 0, 2x + y + 7z = 1 have (a) only two solutions (b) only one solutions (c) no solution (d) infinitely many solutions. 5. If A is a non-zero column matrix of order m × 1 and B is a non-zero row matrix of order 1 × n, then rank of AB is equal to (a) n (c) 1

(b) m (d) None of these

 1   2  6. The matrix product  −2  × [4 5 2] ×  −3  equals  3   5   3  (a)  −6     9 

3 (b)  6   9 

 3  (c)  6     −9 

(d) None of these

7. If A is a square matrix, then AA ′ is a (a) skew-symmetric matrix (b) symmetric matrix (c) diagonal matrix (d) None of these

8. The inverse of a symmetric matrix is (a) diagonal matrix (b) symmetric matrix (c) skew-symmetric matrix (d) None of these 9. If the system of linear equations x + 2ay + az = 0 x + 3by + bz = 0 x + 4cy + cz = 0 has a non-zero solution, then a,b,c (a) are in A.P. (b) are in G.P. (c) are in H.P. (d) satisfy a + 2b + 3c = 0  i 0 10. If A =   , n ∈ N, then A4n equals 0 i  0 i  (a)    i 0

0 0 (b)   0 0

1 0 (c)   0 1

0 i  (d)    i 0

3 4  −2 −2  11. If A =   , B =  0 −2  , then (A + B)–1 = 2 4     (a) is a skew symmetric matrix (b) A–1 + B–1 (c) does not exist (d) none of these a b  α β  12. If A =   and A2 =  β α  , then b a    (a) α = a2 + b2, β = ab (b) α = a2 + b2, β = 2ab (c) α = a2 + b2, β = a2 – b2 (d) α = 2ab, β = a2 + b2 13. If AB = A and BA = B, then B2 is equal to (a) B (c) 1

(b) A (d) 0

1 2 14. If A =  , then adj A is equal to  2 1   −1 2  (a)   2 −1

 1 −2  (b)   −2 1 

 2 1 (c)   1 1

 1 −2  (d)   −2 1 

(a) α = 2ab, β = α2 + β2 (b) α = a2 + b2, β = a2 – b2 (c) α = a2 + b2, β = 2ab (d) α = a2 + b2, β = ab 16. If the matrix AB is zero, then (a) It is not necessary that either A = 0 or B = 0 (b) A = 0 or B = 0 (c) A = 0 and B = 0 (d) All the above statements are wrong 17. Let A be an invertible matrix, which of the following is not true ? (a) (A′ )–1 = (A–1)′ (c) (A2)–1 = (A–1)2

(b) A–1 = | A |–1 (d) None of these

18. If A = [aij] is a scalar matrix of order n × n such that aii = k for all i, then trace of A is equal to (a) kn (c) nk

n k (d) none of these (b)

 an  (a)  0  0 

0 an 0

0  0 a 

 an  (b)  0  0 

 an  (c)  0  0 

0 an 0

0   0  n  a 

 na 0 0  (d)  0 na 0     0 0 na 

24. If A, B are two square matrices such that AB = A and BA = B, then (a) only B is idempotent (b) A, B are idempotent (c) only A is idempotent (d) None of these 1 3 2  x  25. If [1 x 1]  0 5 1   1  = 0, then x is  0 3 2   −2 

19. If AB = O for the matrices 2  cos θ sin θ  and A =  cos θ  sin 2 θ   cos θ sin θ

 cos 2 φ cos φ sin φ  B=   , then θ – φ is cos φ sin φ sin 2 φ   π 2 (b) an odd multiple of π π (c) an even multiple of 2 (d) 0 (a) an odd multiple of

1 2 4    2  3 −5  2 4  (c)    3 −5 

1 2 (d) – 1

(b)

(c) 1

 4 2 2  1 −1 1  26. Let A =  2 1 −3  and (10) B =  −5 0 α  If B      1 −2 3  1 1 1 

(b) –1 (d) 5

27. If A is a singular matrix, then adj A is

(b) for no value of α (d) for α = – 4

(b)

1 2

is the inverse of matrix A, then α is

1 0 1 2  and B =  21. Let A =   and X be a matrix  0 2  3 −5  such that A = BX then X is equal to (a)

(a) –

(a) 2 (c) –2

α 0  1 0  20. If A =  1 1  , and B = 5 1  then A2 = B is true:     (a) for α = 1 (c) for α = 4

0 0  a 0 0 a 

1  −2 4    2  3 5

(d) None of these

 0 0 −1 22. Let A =  0 −1 0  . The only correct statement about    −1 0 0  the matrix A is (a) A–1 does not exist (b) A = (–1)I, where I is a unit matrix (c) A is a zero matrix (d) A2 = I

(a) non-singular (c) symmetric

(b) singular (d) not defined

28. If A, B are two n × n non-singular matrices, then (a) AB is non-singular (b) AB is singular (c) (AB)–1 = A–1B–1 (d) (AB)–1 does not exist 29. If A is 3 × 4 matrix and B is a matrix such that A′B and BA′ are both defined, then B is of the type (a) 3 × 4 (c) 4 × 4

(b) 3 × 3 (d) 4 × 3

 1 2 −1    30. If A =  −1 1 2  , then det. (adj (adj A)) is  2 −1 1  (a) (14)4 (c) (14)2  ab 31. If A =  − a 2 

(b) (14)3 (d) (14)1 b2   − ab  , then A

(a) idempotent (c) nilpotent

(b) involutary (d) scalar

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a 0 0 23. Let A =  0 a 0  , then An is equal to  0 0 a 

Matrices

a b α β  15. If A =  , A2 =    , then b a β α

748

3 8 1 2 32. What must be matrix X if 2X +  =  ?   7 2 3 4 

Objective Mathematics

1 3  (a)    2 −1

1 − 3 (b)    2 − 1

2 6  (c)    4 − 2

 2 − 6 (d)    4 − 2

33. If B is a non-singular matrix and A is a square matrix, then det (B–1AB) is equal to (a) det (A–1) (c) det (A)

(b) det (B–1) (d) det (B)

34. If A is a square matrix such that A2 = I, then A–1 is equal to (a) A + I (c) O

(b) A (d) 2A

 10 0  35. For any 2 × 2 matrix A, if A (adj A) =   , then  0 10  | A | is equal to (a) 20 (c) 10

(b) 100 (d) 0

36. If A and B are two matrices such that A + B and AB are both defined, then (a) A, B are square matrices of same order (b) number of columns of A = number of rows of B (c) A and B can be any matrices (d) None of these 2 1  −3 2  1 0 37. If   A  5 −3  =  0 1  , then the matrix 3 2       A is equal to 1 0  (a)   1 1 

 0 1 (b)    1 1

1 1  (c)   1 0 

 1 1 (d)    0 1

38. If A = (aij)2 × 2, where aij = i + j, then A is equal to 1 1 (a)   2 2

1 2  (b)   1 2 

1 2 (c)   3 4

2 3 (d)   3 4

3 2 39. If A =   , then A (adj A) is equal to 1 4  10 1  (a)    1 10 

 10 0  (b)    0 10 

 0 10  (c)    10 0 

(d) none of the above

(a) det A = 0 and det B = 0 (b) det A + det B = 0 (c) det A = 0 or det B = 0 (d) None of these 41. If a 3 × 3 matrix A has its inverse equal to A, then A2 is equal to 0 1 0 (a)  1 1 1     0 1 0 

1 0 1 (b)  0 0 0     1 0 1 

1 0 0 (c)  0 1 0     0 0 1 

1 1 1 (d)  1 1 1     1 1 1 

 cos α sin α  42. The matrix   is  − sin α cos α  (a) symmetric (c) orthogonal

(b) unique (d) scalar

43. If A, B are square matrices, such that A2 = A, B2 = B and A, B commute then (a) (AB)2 = I (c) (AB)2 = BA

(b) (AB)2 = AB (d) None of these

 3 1 −1  44. If A =   , then AA′ is 0 1 2  (a) symmetric matrix (c) orthogonal matrix

(b) skew-symmetric matrix (d) None of these

 1 −1  a 1  45. If A =   , B =  b −1  and 2 1     (A + B)2 = A2 + B2 + 2AB, then values of a and b are (a) a = 1, b = – 2 (c) a = – 1, b = 2

(b) a = 1, b = 2 (d) a = – 1, b = – 2

 cos α − sin α  46. If Aα =   , then  sin α cos α  (a) Aα ⋅ A(– α) = I (c) Aα ⋅ Aβ = Aα + β

(b) Aα ⋅ A(– α) = 0 (d) Aα ⋅ Aβ = Aα – β

1 3 2 1 47. If [1 x 1]  0 5 1   1  = 0, then x is equal to     0 3 2   x  (a) 1 (b) – 1 −9 ± 53 (d) None of these (c) 2 48. If A is an orthogonal matrix, then (a) det A = 1 (c) det A = – 1

(b) det A = 0 (d) None of these

49. If A is orthogonal matrix, then

40. If A and B are any 2 × 2 matrices, then det (A + B) = 0 implies

(a) At must be orthogonal (b) At may not be orthogonal (c) A–1 must be orthogonal (d) A–1 may not be orthogonal

(a) A2 = A (c) AB = 0

(b) A2 = I (d) BA = 0

 1 1 3  52. The matrix A =  5 2 6  is    −2 −1 −3  (a) idempotent (b) involutory (c) nilpotent (d) None of these

(b) involutory (d) None of these

54. If A is a square matrix of order n, then det (adj A) = (a) (det A)n – 1 (c) (det A)n

(b) (det A)n – 2 (d) None of these

 y+a 55. If the matrix A =  a  a then (a) y ≠ (a + b + c) (c) y ≠ 0

b y+b b

c  c  has rank 3, y + c 

(b) y ≠ 1 (d) y ≠ – (a + b + c)

56. The system of equations 2x + 6y = – 11, 6x + 20y – 6z = – 3, 6y – 18z = – 1 has (a) a unique solution (b) infinite number of solutions (c) no solution (d) None of these 57. The system of equations x + y + z = 8, x – y + 2z = 6, 3x + 5y – 7z = 14 has (a) a unique solution (b) infinite number of solutions (c) no solution (d) None of these

 3n −4n  (a)    n −n   3n (−4) n  (c)  1n (−1) n   

2+n 5−n (b)  − n   n (d) None of these

61. If A and B are two matrices such that AB = B and BA = A, then A2 + B2 = (a) 2AB (c) A + B

(b) 2BA (d) AB

62. The system of linear equations x + y + z = 2, 2x + y – z = 3, 3x + 2y + kz = 4 has a unique solution if (a) k ≠ 0 (c) – 2 < k < 2

(b) – 1 < k < 1 (d) k = 0

 1 0 1 0 63. If A =  and I =    , then the value of k  −1 7  0 1 so that A2 = 8A + kI is

 −5 −8 0  53. The matrix A =  3 5 0  is    1 2 −1  (a) idempotent (c) nilpotent

(a) k = 7 (c) k = 0

(b) k = – 7 (d) None of these

2 0 1 64. Let f (x) = x2 – 5x + 6 and A =  2 1 3  , then f    1 −1 0  (A) is equal to  1 −1 −3  (a)  −1 −1 −10   −5 4 4 

 −1 1 −3  (b)  1 1 −10   −5 4 4 

 1 1 −3  (c)  −1 1 −10   −5 4 −4 

(d) None of these

 3 −5  65. If A =   , then A2 – 5A – 14I is equal to  −4 2   2 −3  0 0 (a)  (b)    1 0   0 0  1 −3  (c)  (d) None of these   −1 0  66. If A is a square matrix, then (a) A + A′ is symmetric (b) A + A′ is skew symmetric (c) A – A′ is symmetric (d) A – A′ is skew symmetric

58. If the system of equations λ x + 2y – 2z = 1, 4x + 2λ y – z = 2, 6x + 6y + λ z = 3 has a unique solution, then 67. If A is a square matrix, then (a) λ ≠ 1 (b) λ ≠ 2 (a) AA′ is symmetric (b) AA′ is skew-symmetric (c) λ ≠ 3 (d) None of these (c) A′A is symmetric (d) A′A is skew symmetric 59. The value of λ so that the equations 2x + y + 2z = 0, 68. Which of the following is correct x + y + 3z = 0, 4x + 3y + λ z = 0 have non-zero (a) B′ AB is symmetric if A is symmetric solution is (b) B′ AB is skew-symmetric if A is symmetric (a) 2 (b) 4 (c) B′ AB is symmetric if A is skew-symmetric (c) 8 (d) None of these (d) B′ AB is skew-symmetric if A is skew-symmetric

749

 3 −4  n 60. If X =   , the value of X is  1 −1 

Matrices

 0 2β γ  50. If the matrix  α β − γ  is orthogonal, then  α −β γ  1 1 (b) β = ± (a) α = ± 6 2 1 (d) all of these (c) γ = ± 3 51. If B is an idempotent matrix, and A = I – B, then

750

69. Which of the following is wrong :

76. The system of linear equations x + y + z = 2, 2x + y – z = 3, 3x + 2y + kz = 4 has a unique solution, if

Objective Mathematics

(a) The elements on the main diagonal of a symmetric matrix are all zero. (b) The elements on the main diagonal of a skew-symmetric matrix are all zero. 77. 1 (A + A′ ) is symmetric (c) For any square matrix A, 2 1 (d) For any square matrix, A, (A – A′ ) is skew2 symmetric 78. 70. If A is an invertible matrix, then (a) adj A′ = (adj A)′ (c) adj A′ = A′

(b) adj A′ = adj A (d) None of these

71. Let A and B be two non-null square matrices. If the product AB is a null matrix, then (a) A is singular (b) B is singular (c) A is non-singular (d) B is non-singular 72. If A is a non-singular matrix, then (a) A–1 is symmetric if A is symmetric (b) A–1 is skew-symmetric if A is symmetric (c) | A–1 | = | A | (d) | A–1 | = | A |–1  0 2 −3 73. If A =  −2 0 −1 , then A is a    3 1 0 

(a)  symmetric matrix (c)  diagonal matrix

(b)  skew symmetric matrix (d)  none of the above

 6 8 5 74. If A =  4 2 3 is the sum of a symmetric matrix B 9 7 1  and skew symmetric matrix C, then B is 6 6 7  (a)    6 2 5  7 5 1 

 0 2 −2    (b)    −2 5 −2   2 2 0 

6 6 7 (c)    −6 2 −5  −7 5 1 

 0 6 −2    (d)    2 0 −2   −2 −2 0 

x  1 1 1   x  0  75. If 1 −2 −2   y  = 3  , then  y  is equal to       z  1 3 1   z   4  0  (a)   1  1 

 1   (b)    2   −3 

 5 (c)    −2     1 

 1 (d)    −2   3

(a)  k ≠ 0 (c)  –2 < k < 2

(b)  – 1 < k < 1 (d)  k = 0

If A is any square matrix, then which of the following is not symmetric? (a)  A + At (c)  AAt

(b)  –(A – At) (d)  AtA

 1 2 –1 Let A =   and A = xA + yl, then the values of  −5 1  x and y are (a)   x =

−1 2 ,y= 11 11

(b)   x =

−1 −2 ,y= 11 11

(c)   x =

1 2 ,y= 11 11

(d)   x =

1 −2 ,y= 11 11

 1   79.  −1  [ 2 1 – 1] is equal to  2  2  (a)    −1     −2 

 2 1 −1   (b)    −2 −1 1   4 2 −2 

(c)  [– 1]

(d)  not defined

1 2  –1 80. If A =   , then A is equal to 3 −5  −5 −2 (a)      −3 1 

5 / 11 2 / 11  (b)     3 / 11 −1 / 11

 −5 / 11 −2 / 11 (c)      −3 / 11 −1 / 11

5 2  (d)     3 −1

 1 2 2 81. If A =  2 1 2 , then A2 – 4A is equal to  2 2 1  (b)  3I3 (a)  2I3

(c)  4I3

(d)  5I3

 2 x 0  1 0 –1 82. If A =   and A =  −1 2 , then x equals  x x   (b)   − 1 2 (c)  1 (d)   1 2 83. Which one of the following is not true? (a)  2

(a)  Matrix addition is commutative (b)  Matrix addition is associative (c)  Matrix multiplication is not commutative (d)  Matrix multiplication is associative 84. Let A be a square matrix all of whose entries are integers. Then, which one of the following is true?

85. Let a, b, c are positive real numbers. The following system of equations x2 y 2 z 2 x2 y 2 z 2 x2 y 2 z 2 + 2 − 2 = 1, 2 − 2 + 2 = 1, − 2 + 2 + 2 = 1 2 a b c a b c a b c in x, y and z has (a)  infinite solution (b)  unique solution (c)  no solution (d)  finite number of solutions 86. A sphere S1 impings directly on an equal sphere S2 at rest. If the coefficient of restitution is e, then the velocities of v1 and v2 are in the ratio

(b)   1 − e 1+ e

(c)   e − 1 e +1

(d)   e + 1 e −1

A particle is projected from the top of tower 5 m high and at the same moment another particle is projected upward from the bottom of the tower with a speed of 10 m/s, meet at distance ‘h’ from the top of tower, then h is equal to (a)  1.20 m (b)  2.5 m (c)  1.25 m (d)  None of these

88. If a falling body covers 75 m in its last second, then the height from which it is falling is (take g = 10 m/ s2) (in metre) (a)  320 (c)  350

(b)  300 (d)  360

solutions 1. (a) Since aij = – aji, hence it is skew-symmetric.  a2  2. (c) Cofactor matrix =  0  0 

0 a2 0

0   0  2  a 

 a2  ∴ adj A = (cofactor Matrix)′ =  0  0  a2 ∴ | adj A | = 0 0

0 a2 0

0 0 a2

0 a2 0

0   0  a 2 

= a 6.

3. (c) Since the determinant of a skew symmetric matrix of odd order is zero, ∴ the matrix is singular. 1 2 3 4. (d) We have, | A | = 1 −1 4 2 1 7

= 1 (– 11) – 2 (– 1) + 3 (3) = 0.



 −11 1 3  ′  −11 −11 11  adj A =  −11 1 3  =  1 1 −1    11 −1 −3   3 3 −3 

 −11 −11 11   1   0  1 −1   0  =  0  = 0. ∴ (adj A) B =  1  3 3 −3   1   0 

∴ The given system of equations is consistent and has infinitely many solutions.  a11  a  21  5. (c) Let A =  and B = [b11 b12 b13 ... b1n] be two       am1  non-zero column and row matrices respectively.  a11b11 a11b12 a b  21 11 a21b12 ∴ AB =      am1b11 am1b12

 a11b1n   a21b1n       am1b1n 

Since A, B are non-zero matrices, ∴ matrix AB will be a non-zero matrix. The matrix AB will have atleast one non-zero element obtained by multiplying corresponding non-zero elements of A and B. All the two-rowed minors of AB clearly vanish. ∴ rank of AB = 1.  1   2  6. (a)  −2  × [4 5 2] ×  −3       3   5  5 2  2   4  3     =  −8 −10 −4   −3  =  −6  .    12 15 6   5   9   1 −1 1   1 2 1  7. (b) Let A =  2 1 0  , then A′ =  −1 1 −1     1 −1 2   1 0 2 

751

(a)   1 + e 1− e

Matrices

(a)  If det (A) = ± 1, then A–1 exists but all its entries are not necessarily integers (b)  If det (A) ≠ ± 1, then A–1 exists and all its entries are non-integers (c)  If det (A) = ± 1, then A–1 exists and all its entries are integers 87. (d)  If det (A) = ± 1, then A–1 need not exist

752

 1 −1 1   1 2 1  ∴ AA′ =  2 1 0   −1 1 −1   1 −1 2   1 0 2 



Objective Mathematics

a 8. (b) Let A =  h  g A11 = bc

h b f



g f c

  , then   2 – f  , A22 = ac – g2, A33 = ab – h2



A12 = – (hc – g f) = g f – hc = A21,



A13 = hf – bg = A31



A23 = – (af – hg) = gh – af = A32

A–1 =

1  4 − 4 1  −1 2   − 2 3  , B–1 =   4 2  0 −2 

 1  2 =  1 −  2



∴ A–1 is symmetric.



1 2a a 1 3b b = 0 1 4c c

=0

⇒ (3b – 2a) (c – a) – (4c – 2a) (b – a) = 0 ⇒ 3bc – 3ba – 2ac + 2a2 = 4bc – 2ab – 4ac + 2a2 ⇒ 4ac – 2ab = 2ab – 3bc + 3ab ⇒ 2ac = bc + ab Dividing by abc, we get 2 1 1 = + b a c i.e., a, b, c, are in H.P. 10. (c) We have,

2  i 0  i 0 i  A2 =  =   0 i  0 i   0



 −1 0   −1 0  1 0 A4 =    0 −1  =  0 1  0 − 1     

0   −1 0   i 2  =  0 −1  .

1 0 1 0 1 0  0 1   0 1  ...  0 1      = 1 0 . ∴ A4n =  0 1   n times  3 4   −2 −2   1 2  11. (d) We have, A + B =  + =   2 4   0 −2   2 2 

| A + B | =

1 2 2 2

   1 − 4 0

a b  α β  12. (b) A =   , A2 =  β α  b a   

a b  a b  A2 = A.A =     b a  b a 



 a 2 + b 2 ab + ba   a 2 + b 2 A2 = ba + ab b 2 + a 2   =   2ab   

R 2 → R 2 – R 1, R 3 → R 3 – R 1 1 2a a 0 3b − 2a b − a 0 4c − 2a c − a

   −1 1

∴ (A + B)–1 ≠ A–1 + B–1.

9. (c) The system of linear equations has a non-zero solution, then



−1  1 − 3 + 2   4  0

 1 ∴ A–1 + B–1 =  1 −  2

Since A12 = A21, A13 = A31, A23 = A32



 −1 1  1  2 −2   1.  −2 1  =  1 −   2   2

∴ (A + B)–1 = –

3 1 4 =  1 5 1   4 1 4 



 2 −2  Also, adj (A + B) =    −2 1 

= 2 – 4 = – 2.

a 2 + b2 α β  =  2ab ⇒     β α

2ab   a 2 + b2 

2ab   a 2 + b2 

⇒ α = a2 + b2, β = 2ab 13. (a) Since BA = B, ∴ (BA) B = BB = B2 ⇒ B (AB) = B2 ⇒ BA = B2 ⇒ B = B2.

(∵ AB = A) (∵ BA = B)

'

 1 −2   1 −2  14. (d) adj A =  =  .  −2 1   −2 1  a b a b 15. (c) A2 =     b a b a

 a 2 + b2 =  2ab 

2ab  α β   a 2 + b 2  =  β α 

∴ a2 + b2 = α, 2ab = β. 1 0 0 0 0 0 16. (a) If A =   , B =  0 1  , then AB =  0 0  . 0 0       Here A ≠ 0, B ≠ 0 but AB = 0. 17. (b) A–1 = | A |–1 is not true, as LHS is a matrix and RHS is a number. 18. (c) Trace of A = a11 + a22 + ... ann = k + k + ... + k = nk.

 cos 2 θ cos θ sin θ  cos 2 φ cos φ sin φ  =    2 sin θ  cos φ sin φ sin 2 φ  cos θ sin θ

cos 2 θ cos 2 φ + cos θ cos φ sin θ sin φ =  cos θ sin θ cos 2 φ + sin 2 θ cos φ sin φ  cos 2 θ cos φ sin φ + cos θ sin θ sin 2 φ   cos θ cos φ sin θ sin φ + sin 2 θ sin 2 φ 

 cos θ cos φ cos θ sin φ  = cos (θ – φ)    sin θ cos φ sin θ sin φ  Since AB = 0, ∴ cos (θ – φ) = 0 π ∴ θ – φ is an odd multiple of . 2 20. (b) A2 = B



 α 0   α 0  1 0  ⇒   =   1 1   1 1  5 1   α 2 0  1 0  ⇒ α + 1 1  = 5 1      ∴ α2 = 1, α + 1 = 5   i.e., α = 1, α = 4 ∴ Above is true for no value of α. a b 21. (a) Let X =   . Since A = BX, c d 1 2  1 0 a b   a b  =  =  ∴        3 −5  0 2 c d   2c 2d  3 ∴ a = 1, b = 2, 2c = 3, 2d = – 5 i.e., c = , 2 5 . d=– 2 1 2  1 2 4   5  = ∴ X =  3  . − 2  3 −5  2 2 22. (d) (i) Clearly A is not a zero matrix  −1 0 0  (ii) (–1)I =  0 −1 0  ≠ A    0 0 −1 (iii) | A | = 1 ≠ 0 ⇒ A–1 exist. 1 0 0  0 0 −1  0 0 −1 (iv)  0 −1 0   0 −1 0  =  0 1 0  .       0 0 1  −1 0 0   −1 0 0  a 0 0 a 0 0 23. (c) A2 =  0 a 0   0 a 0   0 0 a   0 0 a 



 a2  =  0  0 

0 a2 0

0  a 0 0  0   0 a 0  2  a   0 0 a 

 a3  =  0  0 

0 a3 0

0  0 a 3 

0 a2 0

0   0  2  a 



 an  ∴  An =  0  0 

753

 a2  A =  0  0  3

0 an 0

0   0 . a n 

24. (b) We have, AB = A and BA = B. Now, AB = A ⇒ (AB) A = A ⋅ A ⇒ A (BA) = A2 ⇒ AB = A2 ( BA = B) ⇒ A = A2 ( AB = A). Again, BA = B ⇒ (BA) B = B2 ⇒ B (AB) = B2 ⇒ BA = B2 ( AB = A) ⇒ B = B2 ( BA = B) Thus, A2 = A, B2 = B. ∴ A and B are idempotent matrices. 1 3 2  x  25. (b) We have, [1 x 1]  0 5 1   1  = 0  0 3 2   −2   x  ⇒ [1 5x + 6 x + 4]  1  = 0  −2  ⇒ x + 5x + 6 – 2x – 8 = 0 1 ⇒ 4x – 2 = 0 ⇒ x = . 2  4 2 2 26. (d) Given,  −5 0 α  = 10A–1    1 −2 3   4 2 2   1 −1 1  ⇒  −5 0 α   2 1 −3  = 10I     1 −2 3   1 1 1 



10 0 0  =  0 10 0     0 0 10 

⇒ – 5 + α = 0 ⇒ α = 5 (Equating the element of 2nd row and first column). 27. (b) Let A be of order n × n. Since A ⋅ adj A = | A | I ∴ | A ⋅ adj A | = | A |n ⇒ | A | | adj A | = | A |n ∴ | adj A | = | A |n – 1 Since A is singular, ∴ | A | = 0. ∴ | adj A | = 0. Hence, adj A is singular. 28. (a) Since A and B are non-singular ∴ | A | ≠ 0 and | B | ≠ 0. ⇒ | AB | = | A | | B | ≠ 0, ∴ AB is non-singular.

Matrices

19. (a) We have, AB

754

29. (a) Since A is 3 × 4 matrix, therefore A′ is 4 × 3 matrix

2 1 ⇒ A =   3 2

Objective Mathematics

Now, A′B defined ⇒ B is 3 × p Again B3 × p A′4 × 3 is defined ⇒ p = 4 ∴ B is of the type 3 × 4

=

30. (a) We know that adj (adj A) = | A |n – 2 A if | A | ≠ 0, provided order of A is n.

38. (d) A =



b2   − ab 

 1 −1   −1 1  40. (d) Let A =  and B =    1 0   −1 0  b 2   ab  − ab   − a 2

 a 2b 2 − b 2 a 2 =  3 3  −a b + a b

b2   − ab 

0 0 ⇒ A + B =  . 0 0

ab3 − ab3   0 0  =   − a 2b 2 + a 2b 2   0 0

∴ A is nilpotent.

But det A = 1, det B = 1, ∴ det A ≠ 0, det B ≠ 0. ∴ det (A + B) ≠ det A + det B. ⇒ I = A2 i.e., A2 = I.

 3 8  1 2  3 − 1 8 − 2 −   or  2X =  2X =      7 2  3 4   7 − 3 2 − 4

 cos α sin α  42. (c) Let A =  .  − sin α cos α   cos α sin α   cos α − sin α  Then, AA′ =      − sin α cos α   sin α cos α 

1 3  2 6  1 3  or 2X =   = 2  2 −1 ⇒ X =  2 −1 . 4 − 2       

33. (c) det (B AB) = det (B ) det A det B –1

= det (B–1) ⋅ det B ⋅ det A = det (B–1B) ⋅ det A = det (I) ⋅ det A = 1 ⋅ det A = det A. 34. (b) A2 = I ⇒ A2A–1 = IA–1 ⇒ A = A–1.

 cos 2 α + sin 2 α − sin α cos α + sin α cos α  =   − sin α cos α + cos α sin α sin 2 α + cos 2 α   1 0 =   .  ∴ A is orthogonal. 0 1



43. (b) (AB)2 = (AB)⋅(AB) = A (BA) B (Associative law)

35. (c) Since A (adj A) = | A | I,  10 0  1 0  | A| 0  ∴   = | A |  0 1  =  0 | A|  0 10       ∴ | A | = 10. 36. (a) Since A + B is defined, ∴ A, B are of the same type, say, m × n.

= A (AB) B (AB = BA) = (A ⋅ A) (B ⋅ B) (Associative law) = A2 ⋅ B2 = AB. 44. (a) We have,  3 0  3 1 −1    AA′ =    1 1 0 1 2     −1 2 

Since AB is defined, ∴ n = m. ∴ A, B must be square matrices of the same order. 2 1  −3 2  1 0 37. (c) We have,  A  =     3 2  5 −3  0 1 2 1  1 0   −3 2  ⇒  A =      3 2  0 1   5 −3 

∴ det (A + B) = 0.

41. (c) Since A–1 = A ⇒ AA–1 = A2

3 8 1 2 32. (b) Given that, 2X +   =  7 2 3 4     It can be written as,

–1

 a11 a12  1+1 1+ 2  2 3 =  =  a     2 +1 2 + 2 3 4  21 a22 

 10 0  39. (b) A (adj A) = | A | I2 = 10 I2 =  .  0 10 

∴ det (adj (adj A)) = (14)4.

 ab ∴ A2 =  − a 2 

−1

 2 −1   3 2  1  2 −1  1  −3 −2  =         1  −3 2  −1  −5 −3   −3 2   5 3 

∴ det (adj (adj A)) = | A |3 det A = | A |4.

 ab 31. (c) A =  2 −a

 1 0   −3 2   0 1   5 −3     

4−3   6−5 1 1  =   = 1 0  . − 9 + 10 − 6 + 6    

∴ adj (adj A) = | A | A (as n = 3)  1 2 −1  But | A | =  −1 1 2  = 14.  2 −1 1 

−1

−1



 9 +1+1 0 +1− 2   11 −1  =   =  −1 5  0 + 1 − 2 0 + 1 + 4     Clearly AA′ is symmetric as (AA′ )′ = AA′.

45. (d) Given (A + B)2 = A2 + B2 + 2AB ⇒ (A + B) (A + B) = A2 + B2 + 2AB

⇒ BA = AB



 1 −1   a 1   a 1   1 −1  ⇒    2 1  =  2 1   b −1  b − 1        

⇒ (AA′)′ = (A′A)′ = I⇒ (A′ )′ A′ = A′ ⋅ (A′)′ = I

 he corresponding elements of equal matrices are T equal.

a + 2 = a – b, – a + 1 = 2 ⇒ a = – 1



b – 2 = 2a + b, – b – 1 = 1 ⇒ b = – 2.

 0 2β γ   0 α α    50. (d) Let A =  α β − γ  , A′ =  2β β −β     α −β γ   γ − γ γ  Since A is orthogonal, ∴ AA′ = I  0 2β γ   0 α α  1 0 0 ⇒  α β − γ   2β β −β  =  0 1 0     α −β γ   γ − γ γ   0 0 1 

⇒ a = – 1, b = – 2.  cos α − sin α   cos α sin α  46. (a), (c) Aα ⋅ A(– α) =     sin α cos α   − sin α cos α  



 4β 2 + γ 2 ⇒  2β 2 − γ 2  −2β 2 + γ 2 

 cos 2 α + sin 2 α sin α cos α − sin α cos α  =   sin α cos α − sin α cos α sin 2 α + cos 2 α   1 0 =   =I 0 1



 cos α − sin α   cos β − sin β  Also Aα ⋅ Aβ =      sin α cos α   sin β cos β  



= A α + β.

1 3 2 1 47. (c) We have, [1 x 1]  0 5 1   1  = 0  0 3 2   x 



AA′ = I ⇒ | A ⋅ A′ | = 1 ⇒ | A | ⋅ | A′ | = 1 ⇒ | A | ⋅ | A | = 1 ( | A′ | = | A | ) ⇒ | A |2 = 1 ⇒ | A | = 1.

4β 2 + γ 2 = 1  1 ,γ=± ⇒β=± 6 2β 2 + γ 2 = 0 1 α2 + β2 + γ2 = 1 ⇒ α2 + + 6

1 3 1 =1 3

1 . 2 51. (a), (c), (d) Since B is an idempotent matrix, ∴ B2 = B. Now, A2 = (I – B)2 = (I – B) (I – B) = I – IB – BI + B2 = I – B – B + B2 = I – 2B + B2 = I – 2B + B = I – B = A. ∴ A is idempotent.

1 ⇒ [1 5x + 6 x + 4]  1  = 0  x 

−9 ± 81 − 28 −9 ± 53 ⇒ x = = . 2 2 48. (a) Since A is an orthogonal matrix, therefore,

1 0 0 = 0 1 0    0 0 1 

⇒ α = ±

1 ⇒ [1 3 + 5x + 3 2 + x + 2]  1  = 0  x 

⇒ 1 + 5x + 6 + x2 + 4x = 0. ⇒ x2 + 9x + 7 = 0

−2β 2 + γ 2   α2 − β2 − γ 2  2 2  2 α +β + γ 

2β 2 − γ 2 α + β2 + γ 2 α2 − β2 − γ 2 2

Equating the corresponding elements, we have

 cos α cos β − sin α sin β cos α sin β + sin α cos β  =    sin α cos β + cos α sin β − sin α sin β + cos α cos β   cos (α + β) sin (α + β)  =    sin (α + β) cos (α + β) 

...(1)

⇒ At is orthogonal. Also, from (1), we get (AAt)–1 = (AtA)–1 = I ⇒ (At)–1 ⋅ A–1 = A–1 ⋅ (At)–1 = I ⇒ (A–1)t ⋅ A–1 = A–1 ⋅ (A–1)t = I Hence A–1 is orthogonal.

 a + 2 −a + 1   a − b 1+1  ⇒   =  2a + b 2 − 1  b − 2 − b − 1      a + 2 −a + 1   a−b 2 ⇒  =     b − 2 −b − 1   2a + b 1 

AA′ = A′A = I 

Again AB = (I – B) B = IB – B2 = B – B2 = B2 – B2 = 0. Similarly, BA = B (I – B) = BI – B2 = B – B = 0.  1 1 3  52. (c) A2 = A ⋅ A =  5 2 6     −2 −1 −3 

 0 0 0  =  3 3 9     −1 −1 −3 

 1 1 3   5 2 6     −2 −1 −3 

755

49. (a), (c) Since A is orthogonal matrix, therefore

Matrices

⇒ A2 + AB + BA + B2 = A2 + B2 + 2AB

756

 1 1 3  0 0 0  Again, A = A · A =  5 2 6   3 3 9      −2 −1 −3   −1 −1 −3  3

Objective Mathematics



2

0 0 0 =  0 0 0  = 0.  0 0 0  Thus 3 is the least positive integer such that A3 = 0. So A is nilpotent with index 3.

 −5 −8 0  53. (b) A =  3 5 0   1 2 −1  2

 25 − 24 + 0 40 − 40 + 0 0 + 0 + 0  =  −15 + 15 + 0 −24 + 25 + 0 0 + 0 + 0     −5 + 6 − 1 −8 + 10 − 2 0 + 0 + 1 





 −5 −8 0   3 5 0     1 2 −1 

1 0 0 =  0 1 0  = I.    0 0 1  Hence the matrix A is involutory.







0 −11  2 6 =  6 20 −6 −3   0 6 −18 −1 



 2 6 0 −11  ~  0 2 −6 −30   0 6 −18 −1 

[R2 → R2 – 3R1]



 2 6 0 −11  ~  0 2 −6 30   0 0 0 −91 

[R3 → R3 – 3R2]

Rank of C = 3 but rank of A = 2. Hence equations are not consistent. 57. (a) The augmented matrix C = [A B]

 0  0  0 = | A |n    | A|

⇒ | A | | adj A | = | A |n (∵ | AB | = | A | | B | )







⇒ | adj A | = | A |n–1. 55. (c), (d) Here the rank of A is 3 Therefore, the minor of order 3 of A ≠ 0. ⇒

y+a a a

b y+b b

c c y+c

≠0

1 b ⇒ (y + a + b + c) 1 y + b 1 b

c c y+c

c 0 y

1 8  1 1 1 8  1 1  1 −1 2 6   0 −2 1 −2  =   ~   3 5 −7 14   0 2 −10 −10  [R2 → R2 – R1, R3 → R3 – 3R1] 8  1 1 1  0 −2 1 −2  =    0 0 −9 −12 

[R3 → R3 + R2]

Rank of A = 3, rank of C = 3. So, rank of A = rank of C = 3 = number of unknowns. Hence the equations are consistent with unique solution. 58. (b) The given system of equations can be written in a single matrix equation :

≠0

[Applying C1 → C1 + C2 + C3 and taking (y + a + b + c) common from C1] 1 b ⇒ (y + a + b + c) 0 y 0 0

x  −11   y  =  −3       z   −1 

Augmented matrix C = [A B]

54. (a) We know that | A| 0 0 0 | A| 0 | A ⋅ adj A | = 0 0 | A|    0 0 0

0  2 6  6 20 −6     0 6 −18 

≠0

[Applying R2 → R2 – R1, R3 → R3 – R1] ⇒ (y + a + b + c) (y2) ≠ 0 [Expanding along C1] ⇒ y ≠ 0 and y ≠ – (a + b + c). 56. (c) The given system of equations can be written in a single matrix equation as

1  λ 2 −2   x   4 2 λ −1   y  =  2  .        3   6 6 λ   z  The given system of equations will have a unique solution if and only if the coefficient matrix is nonsingular i.e., iff



 λ 2 −2   4 2 λ −1  ≠ 0.    6 6 λ 

⇒ λ (2λ2 + 6) – 2 (4λ + 6) – 2 (24 – 12λ) ≠ 0 or λ3 + 11λ– 30 ≠ 0. or (λ – 2) (λ2 + 2λ + 15) ≠ 0 or λ ≠ 2 (only real root).





2

Now, A = AA

[R2 → R2 – 2R1, R3 → R3 – 4R1] 3 0 1 1 ~  0 −1 −4 0   0 0 λ − 8 0 

[R3 → R3 – R2]

 or infinite solutions, ρ (C) = ρ (A) < number of F unknowns. or ρ (C) = ρ (A) = 2 ∴ λ – 8 = 0 ⇒ λ = 8. 60. (d) We have,



 5 −8  =    2 −3  For n = 2, matrices in (a), (b) and (c) do not match  5 −8  with  .  2 −3 

61. (c) We have, A 2 + B 2

= AA + BB = A (BA) + B (AB) ( AB = B and BA = A) = (AB) A + (BA) B = BA + AB = A + B ( AB = B and BA = A)

62. (a) The given system of equations has a unique solution if



1 1 1 2 1 −1 ≠ 0 ⇒ k ≠ 0. 3 2 k

 −10 0 −5  6 0 0  5 −1 2  =  9 −2 5  +  −10 −5 −15  +  0 6 0         −5 5  0 0 6   0 −1 −2  0   1 −1 −3  =  −1 −1 −10  .    −5 4 4 

65. (b) We have,



 8 0  k 0 =   + 0 k  − 8 56    



0  8+ k =  . − 8 56 + k   Thus, A2 = 8A + kI

 3 −5   3 −5   29 −25  =  A2 =       −4 2   −4 2   −20 24 

∴ A2 – 5A – 14 I 



 29 −25   −15 25   −14 0  =   +  20 −10  +  0 −14  − 20 24       0 0 =  . 0 0

66. (a), (d) Let P = A + A′. Then, P′ = (A + A′ )′

 1 0  1 0  1 0  =  63. (b) We have, A2 =       −1 7   −1 7   −8 49   1 0 1 0 and 8A + kI = 8  +k     −1 7  0 1

2 0 1 2 0 1 = 2 1 3 2 1 3      1 −1 0   1 −1 0 

 5 −1 2    =  9 −2 5   0 −1 −2  2 ∴ f (A) = A – 5A + 6I



 3 −4   3 −4  X2 = X ⋅ X =      1 −1   1 −1 

and 56 + k = 49 ⇒ k = – 7.

64. (a) We have, f (A) = A2 – 5A + 6I

3 0 1 1  ~ 0 −1 −4 0    0 −1 λ − 12 0 





⇒ 1 = 8 + k

757

2 1 2 0 1 1 3 0   = 1 1 3 0 ~  2 1 2 0  [R1 ↔ R2]      4 3 λ 0   4 3 λ 0 

0   1 0  8+ k =  ⇒     −8 49   −8 56 + k 

= A′ + (A′ )′ [ (A + B)′ = A′ + A [ = A + A′ = P. ∴ P is symmetric. Also, let Q = A – A′. Then, Q′ = (A – A′ )′ = A′ – (A′ )′ [ (A + B)′ = A′ – A  [ = – (A – A′ ) = – Q. ∴ Q is skew-symmetric.

= A′ + B′ ] (A′ )′ = A]

= A′ + B′ ] (A′ )′ = A]

67. (a), (c) We have,



(AA′ )′ = (A′ )′ ⋅ A′  = AA′  ⇒ AA′ is symmetric. Also, (A′ A)′ = A′ ⋅ (A′ )′  = A′ ⋅ A  ⇒ A′A is symmetric.

[ (AB)′ = B′ A′ ] [ (A′ )′ = A] [ (AB)′ = B′ ⋅ A′ ] [ (A′ )′ = A]

Matrices

59. (c) The augmented matrix C = [A B]

758

68. (a), (d) Let A be a symmetric matrix.

Objective Mathematics



Then A′ = A. Now, (B′ AB)′ = B′ A′ (B′ )′ [ (AB)′ = B′ A′ ] = B′ A′ B [ (B′ )′ = B] = B′ AB [ A′ = A] ⇒ B’ AB is a symmetric matrix. Now, let A be a skew-symmetric matrix. Then A′ = – A. ∴ (B′ AB)′ = B′ A′ (B′ )′  [ (AB)′ = B′ A′ ] = B′ A′ B [ (B′ )′ = B] = B′ (– A) B  [ A′ = – A] = – B′ AB ∴ B′ AB is a skew-symmetric matrix.

69. (a) Let A = [aij] be a skew-symmetric matrix.

⇒ A = O. But A is a non-null matrix. Hence B is a singular matrix. Similarly it can be shown that A is a singular matrix. 72. (a), (d) Since | A | ≠ 0, therefore A–1 exists. Now, AA–1 = I = A–1A ⇒ (AA–1)′ = I′ = (A–1A)′ ⇒ (A–1)′ A′ = I = A′ (A–1)′ ( A′ = A) ⇒ (A–1)′ A = I = A (A–1)′ –1 –1 –1 ⇒ A = (A )′ ⇒ A is symmetric. Also, since | A | ≠ 0, ∴ A–1 exists such that AA–1 = I = A–1A ⇒ | AA–1 | = | I | ( | AB | = | A | | B | ) ⇒ | A | | A–1 | = 1  ⇒ | A–1 | =

1 . | A|

Then, aij = – aji for all i, j 73. (b) Here, At = – A, therefore A is a skew symmetric ⇒ aii = – aii for all values of i matrix. ⇒ 2 aii = 0 ⇒ aii = 0 for all i Now, let A be any square matrix, then  6 8 5  4 2 3 1 1 74. (a) We have, A = (A + A′ )′ = [A′ + (A′ )′ ]   2 2 9 7 1  [ (A + B)′ = A′ + B′ ] 1 ∴ Symmetric matrix B = A + A ' = (A′ + A) [ (A′ )′ = A] 2 2 1 (A + A′ ) is symmetric. ⇒   6 8 5 6 4 9    6 6 7  2 1   B =   4 2 3 + 8 2 7   =  6 2 5  1 1 2  (A – A′ )′ = [A′ – (A′ )′ ] Also,  9 7 1  5 3 1   7 5 1  2 2 1 75. (b) We have, (A′ – A) = 2 1 1 1   x  0  1 1 −2 −2   y  = 3  (A – A′ ) =–      2 1 3 1   z   4  1 (A – A′ ) is skew-symmetric. ⇒ 2  x + y + z  0  70. (a) Since A is invertible matrix, therefore | A | ≠ 0. ⇒  x − 2 y − 2 z  = 3  Since | A | = | A′ | ∴ | A′ | ≠ 0 i.e., A′ is invertible.  x + 3 y + z   4  Now, A adj A = | A | In On comparing both sides, we get ⇒ (A adj A)′ = ( | A | In)′ x + y + z = 0 …(i) ...(1) ⇒ (adj A)′ A′ = | A | In x – 2y – 2z = 3 …(ii) Also, (adj A′ ) A′ = | A′ | In and x + 3y + z = 4 …(iii) ⇒ (adj A′ ) A′ = | A | In  ( | A | = | A′ | ) ...(2) On solving Eqs. (i), (ii) and (iii), we get From (1) and (2), we have (adj A′ ) A′ = (adj A )′ x = 1, y = 2 and z = – 3 A′ x   1  ⇒ adj A′ = (adj A)′  (by cancellation law) ∴  y  =  2  –1 71. (a), (b) Let B be non-singular, then B exists.      z   −3  Now, AB = O (given) ⇒ (AB) B–1 = OB–1 1 1 1  (post-multiplying both sides by B–1) 76. (a) For unique, solution  2 1 −1 ≠ 0   (by associativity) ⇒ A (BB–1) = O   3 2 k  ( BB–1 = In) ⇒ AIn = O  ⇒ 1(k + 2) – 1(2k + 3) + 1(4 – 3) ≠ 0 ⇒ k ≠ 0

77. (b) (A – At)t = At – (At)t

∴ A–1 =

 1  2x 0 x   1 ⇒A = =  2  2 x − x 2 x  1 −  2 x –1

1 1 −2    1 + 10 5 1 

 1  2x ⇒  − 1  2 x

1 1 −2    11 5 1  Also, A–1 = xA + yI =

⇒

1 1 −2   x 2 x   y  = + 11 5 1   −5 x x   0

0  y

⇒ x + y = 1 , 2 x = −2 11 11 −1 2 ⇒ x = 11 , y = 11

 1  2 1 −1     79. (b)  −1  [2 1 − 1] =  −2 −1 1   2   4 2 −2 

1 2  80. (b) Since, A =   3 −5

⇒

 0  1 x 

 0  1 0 = (given) 1   −1 2 x 

1 1 =1⇒ x = 2x 2

83. (c) 84. (c) As det (A) = ± 1, A–1 exists and A–1 =

1 (ajd A) = ± adj A det ( A)

All entries in adj (A) are integers. ∴ A–1 has integer entries. 2 2 2 85. (b) Let x = X , y = Y and z = Z , then given equaa2 b2 c2

tion will be

1 2 ∴ |A| = = – 5 – 6 = – 11 3 −5

X + Y – Z = 1, X – Y + Z = 1, – X + Y + Z = 1

 −5 −2 and adj (A) =    −3 1 

 1 1 −1   A =  1 −1 1   −1 1 1 

Hence, A–1 = =

−1  −5 −2 1 · adj(A) =   11 | A|  −3 1 

1 5 2  11 3 −1

5 / 11 2 / 11  =  3 / 11 −1 / 11 81. (d) We have,  1 2 2 A =  2 1 2    2 2 1   1 2 2   1 2 2  9 8 8  ⇒ A =  2 1 2   2 1 2  = 8 9 8        2 2 1   2 2 1  8 8 9 2

9 8 8   4 8 8  ∴ A2 – 4A = 8 9 8 − 8 4 8      8 8 9 8 8 4  5 0 0 = 0 5 0 = 5 I 3 0 0 5

Matrices

 1 2 A=    −5 1 

78. (a)

759

 2 x 0 82. (d) Given, A =    x x

= At – A = –(A – At)

The coefficient matrix of these equations is

Now, |A| = – 4 ≠ 0 Therefore, the given system of equation has unique solution. 86. (b) Let m be the mass of each sphere. Let u be the velocity of the first sphere before impact and v1 and v2 be their velocities after impact, then v1 – v2 = – eu and mv1 + mv2 = mu ⇒ v1 + v2 = u

…(i) …(ii)

On solving Eqs. (i) and (ii), we get v1 = u (1 − e) and v2 = u (1 + e) 2 2 v1 1 − e = v2 1 + e 87. (c) Since, h = 1 gt2 2 and h1 = ut – 1 gt2 2 On adding Eqs. (i) and (ii), we get

…(i)

760

h = 1 gt2 2

h

Objective Mathematics

⇒ h = 1 × 10 × 1 = 1.25 m 2 4

Sn

88. (a) Let the distance covered by the body in t th second is 75 m

h1

∴ 75 = 0 + 10 (2t – 1) 2 (∵ u = 0 and g = 10 m/s2) ⇒ 15 = 2t – 1 ⇒ t = 8 Also, h = ut + 1 gt2 2

h + h1 = ut ⇒ 5 = 10t ⇒ t = 1 s 2

Exercises for self-practice 1 0 0    1. If A =  0 1 0  , then A2 is equal to  a b −1 (a) null matrix (c) –A

(b) unit matrix (d) A

a b –1 2. If A =   such that ad – bc ≠ 0, then A is c d 1  d −b  (a)   ad − bc  −c a 

1  d b (b)   ad − bc  −c a 

 d −b  (c)    −c a 

(d) None of these

3. If A is an orthogonal matrix, then A–1 is equal to (a) A2 (c) A

(b) A′ (d) None of these

a b  4. If A =   , then adj A is c d   −d (a)   −c d (c)  c

−b   a b  a

c  a

5. The number of values of k for which the system of equations

1 1 7. The value of ∆ = 1 ω 1 ω2

(k + 1)x + 8y = 4k kx + (k + 3)y = 3k – 1 has infinitely many solutions, is (a) 0 (b) 1 (c) 2 (d) infinite

1 ω2 ω

is

(a) − 3i

(b) 3 3i

(c) −3 3i

(d)

3i

8. If A is a square matrix such that | A | = 2, then for any +ve integer n, | An | is equal to (a) 2n (b) n2 (c) 0 (d) 2n 9. The value of a for which the system of equations a3x + (a + 1)3 y + (a + 2)3 z = 0, ax + (a + 1) y + (a + 2) z = 0, x + y + z = 0 has a non zero solution is (a) 0 (c) 1

 d −b  (b)    −c a  d (d)  b

6. If A is of order 2 × 3, then each AB is of order (a) 3 × 2 (b) 3 × 3 (c) 2 × 3 (d) 2 × 2

(b) –1 (d) None of these

10. If (A) = (aij) m × n, B = (bjk)n × p and C = (ckl)p × q, then the product ABC will be of order. (a) m × q (c) p × q

(b) n × p (d) m × p

0 2   0 3a  11. If A =   , KA =   , then values of k, a, 3 − 4  2b 24  b are (a) – 6, – 12, – 18 (b) – 6, – 4, 9 (c) – 6, – 4, – 9 (d) – 6, 12, 18.

(a)

1 2 − 2   10  2 3 

2 − 2 (c)   2 3 

3 − 2 (b)   2 2  (d)

1  3 2   10  − 2 2 

13. The element in the first row and third column of the  1 2 − 3   inverse of the matrix 0 1 2  is : 0 0 1  (a) – 2 (c) 1

(b) 0 (d) 7

2 0 0   14. If A =  0 2 0  , then A5 =  0 0 2  (a) 5A (c) 16A

(b) 10A (d) 32A

15. If A and B are square matrices of order 3 such that | A | = – 1, | B | = 3, then | 3AB | (a) – 9 (c) – 27   1 16. If A =   − tan θ  2 (a) cos2 (c) cos2 

θ  · A 2 θ  · I 2

(b) – 18 (d) 81 θ tan  2  and AB = I, then B = 1   θ (b) cos2  · A′ 2 (d) None of these

9   17. If for AX = B, B = 52  and A–1 =  0   3 −1 / 2 −1 / 2    5 / 4  , then X is equal to − 4 3 / 4  2 −1 / 4 − 3 / 4 

18. If A =  0 i  , then A40 =   − i 0

761



0 −1 – 1 –1 – 1   , then [B  A ] = 1 0  

0 1  (a)   1 0 

1 0  (b)   0 1 

Matrices

 2 2 12. If A =  , B = − 3 2

−1 1  (c)    0 −1

(d) None of these

19. If A = 1 k  , then An =   0 1   n nk  (a)   0 n 

n k n  (b)   0 n 

1 nk  (c)   0 1 

1 k n  (d)   0 1 

20. The matrix (AB)–1 is equal to (b) BA (a) B–1A–1 (c) AB (d) A–1B–1 21. If A and B are matrices conformable for multiplication then (AB)t equals (a) At Bt (b) Bt At (c) At + Bt (d) Bt – At 22. In a system of linear equations, if A is the coefficient matrix and D is the constant matrix such that | A | = 0 and (adj A) D = 0, then the system of equations has (a) two solutions (b) no solution (c) infinite number of solutions (d) unique solution. 23. If A is any matrix, then the product AA i.e., A2 is defined only when A is a matrix of order (a) m > n (b) m = n (c) m < n (d) m ≥ n 1 0 0 24. If A =  0 1 0  , then    0 0 1 

(a) adj A = A–1 (b) adj A = A′ (c) adj A = A (d) all the statements are correct

25. A matrix A = (aij)m × n is said to be row matrix if

1    (a) 3 5

 −1 / 2    (b)  −1 / 2   2 

(a) m = 1 (c) m ≥ n

− 4 (c)  2     3 

 3    (d)  3 / 4   − 3 / 4 



(b) m ≤ n (d) m = n

26. If A, B are square matrices of order 3, then (a) AB = 0 ⇒ | A | = 0 or | B | = 0 (b) AB = 0 ⇒ | A | = 0 and | B | = 0 (c) (A + B)–1 = A–1 + B–1. (d) adj (AB) = (adj A) (adj B)

762

 −2 6  32. If A =   , then adj A:  −5 7 

1 2  27. If A =   , then adj A is equal to  3 −5 

Objective Mathematics

 −5 −2  (b)    3 −1 

5 2  (a)    3 −1  −5 −2  (c)    −3 1 

5 −2  (d)  . −  3 1 

28. If I3 is identity matrix of order 3, then (I3)–1 = (a) 0 (c) I3



b1 b2 b3

c1 c2 c3

= 0, then the system has

= 0 ⇒ | B | = 0 =0 ⇒ / | B | = 0 = | A |–1 A | = 2 | A |

(b) it is associative (d) None of these

 3 −2 4  1 34. If matrix A = 1 2 −1 and A– 1 = k adj (A), then   0 1 1  k is: (b) –7 (d) –11

0 3 –1 35. If A =   and A = λ (A), then λ: 2 0 −1 6 −1 (c) 3

(a)

1 3 1 (d) 6 (b)

1 0 −k  36. Matrix A =  2 1 3  is invertible for:    k 0 1 

31. If A = 1 1 , then A100:   1 1 (a) 2100 (c) 2101

(d) None of these

(a) 7 1 (c) 7

(a) more than two solutions (b) one trivial and one non-trivial solution (c) no solution (d) only trivial solution (0, 0, 0)

(a) | AB | (b) | AB | (c) | A–1 | (d) | A +

7 −5 (c)    5 −2 

(a) it is commutative (c) both

30. If A and B are 3 × 3 matrices and | A | ≠ 0, then which of the following are true ?

7 −6  (b)    5 −7 

33. Which is true about matrix multiplication:

(b) 3I3 (d) not necessarily exists

29. Consider the system of equations a1x + b1 y + c1z = 0, a2x + b2 y + c2z = 0, a3x + b3 y + c3z = 0. a1 If a2 a3

7 −6  (a)    5 −2 

(b) 299 (d) None of these

(a) k =1

(b) k = –1

(c) k = 0

(d) all real k

Answers

1. 11. 21. 31.

(b) (c) (b) (b)

2. 12. 22. 32.

(a) (c) (c) (c)

3. 13. 23. 33.

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

4. 14. 24. 34.

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

5. 15. 25. 35.

(b) (b) (a) (a)

6. 16. 26. 36.

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

7. (c) 17. (a) 27. (c)

8. (a) 18. (b) 28. (c)

9. (b) 19. (c) 29. (a)

10. (a) 20. (a) 30. (a), (b)

20

Determinants

CHAPTER

Summary of conceptS Second-order Determinant neous linear equations

Consider the two homoge-

a1x + b1 y = 0, a2x + b2 y = 0; multiplying the first equation by b2, the second by b1, subtracting and dividing by x, we obtain a1b2 – a2b1 = 0. This result can be written as a1 b1 a2 b2 = 0, and the expression on the left is called a determinant. The four numbers a1, a2, b1 and b2 are called the elements of the determinant. The elements in the horizontal line are said to form a row and the elements in the vertical line are said to form a column of the determinant. The above determinant is second order as it contains 2 rows and 2 columns. The value of the second-order determinant is given as: a1 a2

b1 = a1b2 – a2b1. b2

third-order Determinant mogeneous linear equations

b2 b3

c2 c2 + b1 c3 c3

a2 a2 + c1 a3 a3

b2 = 0. b3

This result can be written as a1 a2 a3

b1 b2 b3

c1 c2 = 0, c3

and the expression on the left is called a determinant of order three because it consists of three rows and three columns.

mInorS anD cofactorS minor of an element of a Determinant If we take an element of the determinant and delete the row and the column containing that element, the determinant left is called the minor of that element. It is denoted by Mij.

a13 a23 . a33 a22 a32

Then the minor of a11 is M11 = the minor of a12 is M12 =

a23 ; a33

a21 a23 and so on. a31 a33

cofactor of an element of a Determinant The cofactor Cij of an element aij (the element in the ith row and jth column) is defined as Cij = (– 1)i + j Mij. It is denoted by Cij. Thus, M ij , when i + j is even . Cij =  − M ij , when i + j is odd For example, the cofactor of a12 in the 3 × 3 determinant a11 a12 a21 a22 a31 a32 is C12 = (– 1)1 + 2

By eliminating x, y, z, we obtain a1 (b2c3 – b3c2) + b1 (c2a3 – c3a2) + c1 (a2b3 – a3b2) = 0, a1

a11 a12 a21 a22 a31 a32

Let us now consider the ho-

a1x + b1 y + c1z = 0 a2x + b2 y + c2z = 0 a3x + b3 y + c3z = 0.

or

For example, given the 3 × 3 determinant

a13 a23 a33

a21 a23 a a = – 21 23 . a31 a33 a31 a33

expanSIon of a DetermInant of orDer tHree Consider the following determinant: a1 a2 a3

b1 b2 b3

c1 c2 c3

.

We can find the value of this determinant by various methods. Method 1 Step 1 : Write the elements of the first row with alternatively positive and negative sign, the first element always has positive sign before it. Step 2 : Multiply each signed element by the determinant of second order obtained after deleting the row and the column in which that element occurs. i.e.,

764

Objective Mathematics

a1 a2 a3

b1 b2 b3

c1 b2 c2 = a1 b3 c3

c2 a2 – b1 c3 a3

c2 a2 + c1 c3 a3

c2 c3

The value of the given determinant = (2) (7) (6) + (– 1) (0) (4) + (3) (5) (1) – (4) (7) (3) – (1) (0) (2) – (6) (5) (– 1) = 84 + 15 – 84 + 30 = 45.

= a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2)

remark: Sarrus rule does not work for determinants of order greater than 3.

For example, 1 −3 4 2 5 + (– 3) (– 1) 0 2 5 = (1) 6 3 −2 6 3 0 5 0 2 + (4) −2 3 −2 6 = 1 (6 – 30) + 3 (0 + 10) + 4 (0 + 4) = – 24 + 30 + 16 = 22. We can also expand the determinant along any row or any column. For example, if we had expanded the above determinant along the first column, then 1 −3 4 −3 4 2 5 + 0 + (– 2) 0 2 5 = (1) 2 5 6 3 −2 6 3 = 1(6 – 30) – 2 (– 15 – 8) = – 24 + 46 = 22, as before. Method 2: Using Sarrus Rule The following diagram called sarrus diagram, enables us to write the value of the determinant of order 3 very convenientlty.

a1 Let ∆ = a2

a3

b1 b2 b3

c1 c2 be a determinant of order 3. c3

Write the elements as:

Multiply the elements joined by arrows. Assign the positive sign to an expression if it is formed by a downward arrow and negative sign to an expression if it is formed by an upward arrow. Note that the first two columns are repeated in the above table to complete the process. The value of the given determinant is: a1b2c3 + b1c2a3 + c1a2b3 – a3b2c1 – b3c2a1 – c3a2b1.

2 −1 3 7 0 , write the elements as 4 1 6

For example, to evaluate 5

Method 3: Using Cofactors A determinant can also be evaluated by multiplying the entries of any row (or column) by their cofactors and summing the resulting products. Let ∆=

a1 a2 a3

b1 b2 b3

c1 c2 , then c3

∆ = a1C11 + b1C12 + c1C13 = a1M11 – b1M12 + c1M13.

1 2 0 −1 −1 4 1 3 −1 4 1 For example, = 1 0 −3 3 − 2 0 −3 3 3 1 2 4 3 1 2 3 4 1 – 2 − 2 −3 3 4 1 2

3 −1 4 − + 0 + 1 2 0 −3 4 3 1

= 54 – 94 + 13 = 27. remark: It is a fact that determinant of a matrix is unique and does not depend on the row or column chosen for its evaluation.

propertIeS of DetermInantS Properties of determinants of order three only are stated below. However these properties hold for determinants of any order. These properties help a good deal in the evaluation of determinants. 1. The value of the determinant remains unchanged if rows are changed into columns and columns are changed into rows, i.e., a11 a12 a13 a11 a21 a31 a21 a22 a23 = a12 a22 a32 a31 a32 a33 a13 a23 a33 . 2. If two adjacent rows (columns) of a determinant are interchanged, the value of the determinant so obtained is the negative of the value of the original determinant, i.e., a11 a12 a13 a21 a22 a23 a21 a22 a23 = – a11 a12 a13 a31 a32 a33 a31 a32 a33 . 3. If two rows or columns of a determinant are identical then its value is zero, i.e., a11 a12 a11 a12 a31 a32

a13 a13 = 0. a33

ka11 ka12 a21 a22 a31 a32

ka13 a11 a12 a23 = k a21 a22 a33 a31 a32

a13 a23 . a33

1 1 1 ∆ = a b c = k (a – b) (b – c) (c – a). a 2 b2 c2 In order to find the value of k, give values to a, b and c such 5. If any two rows or columns of a determinant are proporthat calculations are easy and the two sides do not vanish. tional, then its value is zero, i.e., For example, assume a = 0, b = 1, c = 2, we get a11 a12 a13 a11 a12 a13 1 1 1 ka11 ka12 ka13 = k a11 a12 a13 = 0. 0 1 2 = k (0 – 1) (1 – 2) (2 – 0) a31 a32 a33 a31 a32 a33 0 1 4 6. If each element of a row (or column) of a determinant is or 2 = 2k (on solving the determinant along first column) the sum of two or more terms, then the determinant can be Thus k = 1. Hence, expressed as the sum of the two or more determinants, i.e., 1 1 1 a11 a12 a13 a b c = (a – b) (b – c) (c – a). a21 + c1 a22 + c2 a23 + c3 a 2 b2 c2 a31 a32 a33 Note:  In general, if r rows (or r columns) become identical when a is substituted for x, then (x – a)r – 1 is a factor of a11 a12 a13 a11 a12 a13 given determinant. = a21 a22 a23 + c1 c2 c3 . 10. The sum of the products of the elements of any row (or cola31 a32 a33 a31 a32 a33 umn) of a determinant with the corresponding co-factors is equal to the value of determinant, i.e., if 7. If each element of a row (column) of a determinant is multiplied by a constant k and then added to the corresponding elements of some other row (column), then the value of the determinant remains the same, i.e., a11 a12 a21 a22 a31 a32 =

a13 a23 a33

a11 a12 a21 a22 a31 + ka21 a32 + ka22

a11 a12 ∆ = a21 a22 a31 a32

a13 a23 , then a33

a11C11 + a12C12 + a13C13 = ∆ and so on.

a13 . a23 a33 + ka23

11. The sum of the products of elements of any row (or column) of a determinant with the co-factors of the corresponding elements of any other row (or column) is zero, i.e., if a11 a12 a13 ∆ = a21 a22 a23 , then a31 a32 a33

8. If each element of a row (column) of a determinant is zero, then its value is zero. a11C31 + a12C32 + a13C33 = 0 and so on. 9. If the elements of a determinant that involve x are polynomials in x, and if the determinant is equal to zero Note: In ∆ = | aij | is a determinant of order n, then the when a is substituted for x, the x – a is a factor of given value of the determinant | Aij |, where Aij is the cofactor determinant. of aij is∆n – 1. 2 1 1 1 − a 2 ab ac o c b For example, let ∆ = a b c . If we consider For example, ab −b 2 bc = c o a . a 2 b2 c2 b a o ac bc −c 2 a = b in ∆, then we have 1 a a2

1 b b2

1 c =0 c2

( first and second column are identical) This implies that (a – b) must be a factor of ∆. Similarly, (b – c) and (c – a) are also factors of ∆. Since the product of the diagonal elements of ∆ is 1.a.c,2 which is a third

EVALUATION OF DETERMINANTS USING ELEMENTARY OPERATIONS To evaluate determinants of higher order, we should always try to introduce zeros at the maximum number of places in a particular row (column) by using the properties of the determinant. We denote the rows of the determinant by R1, R2, R3, ... and columns by C1, C2, C3, ...

765

degree expression, ∆ is a polynomial of degree 3. But (a – b) (b – c) (c – a) is a factor of ∆ which is of degree 3 itself. Therefore, the only other factor of ∆ can be a constant, say k.

Determinants

4. If each element of a row or column of a determinant is multiplied by a constant k then the value of the new determinant is k times the value of the original determinant, i.e.,

766

We shall use the following notations to evaluate a determinant.

and

D2 =

a1 a2

c1 , provided D ≠ 0. c2

Objective Mathematics

(i) The operation of interchanging the ith row and jth row will 2. Cramer’s Rule: Solution of system of linear equabe denoted by Ri ↔ Rj. (ii) The operation of multiplying each element of the ith row tions in three unknowns  The solution of the system of equations by a number k will be denoted by Ri → kRi. (iii) The operation of adding to each element of the ith row, k a1x + b1 y + c1z = d1 times the corresponding elements of the jth row (j ≠ i) will a2x + b2 y + c2z = d2 be denoted by Ri → Ri + kRj. a3x + b3 y + c3z = d3 Similar notations are used for operations on columns re- is given by placing R by C. D D D x = 1 , y = 2 and z = 3 , where D D D

PRODUCT OF DETERMINANTS OF SAME ORDER a1 Let ∆1 = a2 a3

b1 b2 b3

c1 α1 β1 c2 and ∆2= α 2 β2 c3 α 3 β3

γ1 γ2 γ3

Then row by row multiplication of ∆1 and ∆2 is given by a1 α1 + b1 β1 + c1 γ1 a1 α 2 + b1 β2 + c1 γ 2 ∆1 ⋅ ∆2 = a2 α1 + b2 β1 + c2 γ1 a2 α 2 + b2 β2 + c2 γ 2 a3 α1 + b3 β1 + c3 γ1 a3 α 2 + b3 β2 + c3 γ 2

a1 α 3 + b1β3 + c1γ 3 a2 α 3 + b2β3 + c2 γ 3 a3 α 3 + b3β3 + c3 γ 3

Multiplication can also be performed row by column; column by row or column by column as required in the problem.

a1 D = a2 a3

b1 b2 b3

c1 d1 c2 ;  D1 = d 2 c3 d3

b1 b2 b3

c1 c2 c3

a1 D2 = a2 a3

d1 d2 d3

a1 c1 c2 ;  D3 = a2 a3 c3

b1 b2 b3

d1 d2 ; d3

provided D ≠ 0. Conditions for Consistency  The following cases may arise: (i) If D ≠ 0, then the system is consistent and has a unique solution, which is given by Cramer’s rule:

D1 D2 D ,y= ,z= 3 . D D D Note:  If ∆′ is the determinant obtained by replacing all the ele- (ii) If D = 0 and atleast one of the determinants D , D , D is 1 2 3 ments of determinant ∆ of order n by their corresponding cofacnon-zero, the given system is inconsistent, i.e., it has no n–1 tors, then ∆' = ∆ . solution. (iii) If D = 0 and D1 = D2 = D3 = 0, then the system is consistent SOLUTION OF LINEAR EQUATIONS BY and dependent, and has infinitely many solutions.

DETERMINANTS

x=

Homogeneous and Non-Homogeneous

1. Cramer’s Rule: Solution of system of linear equa- System tions in two unknowns  The solution of the system of If D1 = D2 = D3 = 0, then the system is said to be, homogeneous, equations otherwise it is called non-homogeneous. a1x + b1 y = c1 If the system of equations is homogeneous, then D1 = D2 = a2x + b2 y = c2 D3 = 0 (value of the determinant is zero, if one column has all elements = 0). Thus, is given by x = D1 and y = D 2 , where D D (i) if D ≠ 0, the system has only trivial solution (x = y = z = 0), a1 b1 c1 b1 and D= , D1 = (ii) if D = 0, the system has infinitely many solutions. a2 b2 c2 b2

Determinants

767

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: cos 2 x cos x ⋅ sin x − sin x 1. If f (x) = cos x sin x sin 2 x cos x , then for all x 0 sin x − cos x (a) f (x) = 0 (c) f (x) = 2

(b) f (x) = 1 (d) None of these

2

5

5

15 + 26

5

10  is

3 + 65

15

5

r =1

r

is

respectively of a 1 1 is equal to 1

5

(a) 0 (c) – 1

5

b 2 − ab b − c bc − ac ab − a 2 a − b b 2 − ab is bc − ac c − a ab − a 2

(a) 1 (c) ­– 1

(a) 1 (c) 2

log l log m log n

(b) 2 (d) 4

5. The value of the determinant a2 a 1 cos nx cos (n + 1) x cos (n + 2) x is sin nx sin (n + 1) x sin (n + 2) x (a) independent of n (c) independent of x

7. If ∆ r =

n 2n 2 3n3

6 4n − 2 3n 2 − 3n

equal to (b) 2 (d) 0

n

, then

is

p l q 1 equals r 1 (b) 2 (d) zero

A1 ∆ = A2 A3

B1 B2 B3

C1 C 2 is C3

(a) divisible by k (c) divisible by 2k

is (b) 2x (d) None of these

r −1 (r − 1) 2 (r − 1)3

42 52 62 72

12. If A1B1C1, A 2B2C2 and A3B3C3 are three three-digit numbers, each of which is divisible by k, then

(b) independent of a (d) None of these

(a) x (c) 0

32 42 52 62

(b) 0 (d) None of these

(a) 3 (c) 1

6. If a, b, c are in A. P., then the value of the determinant x +1 x + 2 x + a x+2 x+3 x+b x+3 x+4 x+c

22 32 42 52

11. If l, m, n are the pth, qth and rth term of a G.P. being all positive, then

= K abc, then K is equal to

(a) 1 (c) 3

(b) 1 (d) None of these

12 22 10. The value of the determinant 2 3 42

(b) 0 (d) None of these

b+c c b c c+a a 4. If b a a+b

(a) 1 (c) 3

n

∑∆

9. If a, b, c are the pth, qth and rth terms log a p geometric progression, then log b q log c r

3  (5 – 6 ) (b) – 5 3  (5 + 6 ) 3  ( 6 – 5) (d) None of these 3. The value of the determinant (a) – (c) –

x 2n − 1 y 3n − 1 , then z 5n − 1

(a) independent of x (b) independent of y (c) independent of z (d) independent of n

2. The value of the determinant 13 + 3

2r − 1 8. If ∆r = 2.3r − 1 4.5r − 1

∑∆ r =1

r

is

(b) divisible by k2 (d) None of these

x m n 1 a x n 1 13. The roots of the equation = 0 are a b x 1 a b c 1 (a) independent of m, n (b) independent of m, n, a, b (c) independent of a, b (d) None of these

768

Objective Mathematics

a 2 + 2a 2a + 1 1 14. The determinant 2a + 1 a + 2 1 is 3 3 1 (a) > 0 if a > 1 (c) < 0 if a < 1

(b) = 0 if a = 1 (d) all of these

15. If A, B, and C are the angles of a triangle, then the value of the determinant

x a a a a x a a 21. If a a x a a a a x

(a) 0 (c) ­– 1

16. If

22. For all values of A, B, C and P, Q, R, the value of the cos (A − P) cos (A − Q) cos (A − R) determinant cos (B − P) cos (B − Q) cos (B − R) is cos (C − P) cos (C − Q) cos (C − R) (a) 1 (c) 2

(b) 1 (d) 2

a b c a − b b − c c − a = a3 + b3 + c3 + K abc, then K b+c c+a a+b (b) – 3 (d) – 4

a + b + 2c a b = K (a + b + c)3, 17. If c b + c + 2a b c a c + a + 2b

(a) – 9 (c) 7 0 24. If x + a x+b (b) x = ±

(b) 2 (d) None of these

18. The value of the determinant



1+ x 2 3 4 1 2+ x 3 4 1 2 3+ x 4 1 2 3 4+ x (a) x2 (x + 10) (c) x4 (x + 10)

2bc − a 2 c2 19. If b2

is

b2 a2 2ab − c 2

= (a3 + b3 + c3 + kabc)2, then k is equal to (a) 2 (b) – 2 (c) 3 (d) – 3

(a) 0 (c) – 10

(b) 10 (d) None of these

x−b x−c 0

= 0, a ≠ b ≠ c, then

b (a + c) − ac if b (a + c) > ac

25. If the three digit numbers A28, 3B9 and 62C, where A, B and C are integers between 0 and 9, are divisible by A 3 6 a fixed integer k, then the determinant 8 9 C is 2 B 2 (a) divisible by k (c) divisible by 2k

(b) divisible by k2 (d) None of these

cos 2θ cos θ sin θ − sin θ sin 2 θ cos θ , then for all θ, 26. If f (θ) = cosθ sinθ sin θ − cos θ 0 (a) f (θ) = 1 (c) f (θ) = 3

(b) f (θ) = 2 (d) None of these

27. The value of the determinant

λ 2 + 3λ λ − 1 λ + 3 20. If λ + 1 1 − 2λ λ − 4 = pλ4 + qλ3 + rλ2 + sλ + t λ−2 λ+4 3λ be an identity in λ, where p, q, r, s and t are constants, then the value of t is

x−a 0 x+c

(c) x = 0, ± b (a + c) − ac if b (a + c) > ac (d) None of these

(b) x3 (x + 10) (d) None of these c2 2ca − b 2 a2

(b) 2 (d) all of these

(a) x = 0 if b (a + c) ≤ ac

then K is equal to (a) 1 (c) 4

(b) 0 (d) None of these

x 3 7 23. If 2 x 2 = 0 then x is equal to 7 6 x

is equal to (a) 3 (c) 4

(b) (x – a)3 (d) (x + a)3

(a) (x – a)2 (c) (x + a)2

−1 + cos B cos C + cos B cos B cos C + cos A −1 + cos A cos A is −1 + cos B −1 + cos A −1



= (x + 3a) f (x), then f (x) is equal to

(a1 − b1 ) 2 (a2 − b1 ) 2 (a3 − b1 ) 2 (a4 − b1 ) 2

(a1 − b2 ) 2 (a2 − b2 ) 2 (a3 − b2 ) 2 (a4 − b2 ) 2

(a1 − b3 ) 2 (a2 − b3 ) 2 (a3 − b3 ) 2 (a4 − b3 ) 2

(a1 − b4 ) 2 (a2 − b4 ) 2 is (a3 − b4 ) 2 (a4 − b4 ) 2

(a) dependent on ai, i = 1, 2, 3, 4 (b) dependent on bi, i = 1, 2, 3, 4 (c) dependent on ai, bi, i = 1, 2, 3, 4 (d) 0

1 f ′ ( x)  2 x , then lim   is x →0  x  1

(a) 2 (c) 1

(b) – 2 (d) – 1

x + a a2 29. If x + b b 2 x + c c2

a3 b3 c3

(a) x = a (c) x = c

= 0 and a ≠ b ≠ c then x is equal to

(a) abc/(ab + bc + ca) (b) – abc/(ab + bc + ca) (c) (ab + bc + ca)/(abc) (d) – (ab + bc + ca)/(abc) x 2 1 + x3 y 2 1 + y 3 = 0, z 2 1 + z3

(b) xyz = 1 (d) xyz = 2

sec x cos x sec 2 x + cot x cosec x 2 2 cosec 2 x 31. If f (x) = cos x cos x , then 1 cos 2 x cos 2 x π/ 2

∫

f ( x) dx is equal to

0

(a) (15π + 32)/60 (c) (15π + 32)/4 32. The value x 1 1 x by 1 1 ... ...

(b) – (15π + 32)/60 (d) None of these

of the determinant of nth order, being given 1 ... 1 ... , is x ... ... ...

(a) (x – 1)n – 1 (x + n – 1) (b) (x – 1)n (x + n ­– 1) (c) (1 – x)n – 1 (x + n – 1) (d) None of these 33. If f (x) , g (x) and h (x) f ( x) 2 and ∆ (x) = f ′ ( x) f ′′ ( x) polynomial of degree (a) 2 (c) at most 2

(b) 9, – 2, 7 (d) None of these

37. The value of λ for which the equations x + y – 3 = 0, (1 + λ) x + (2 + λ) y – 8 = 0, x – (1 + λ) y + (2 + λ) = 0 are consistent is (a) 1 (c) – 5/3

(b) 5/3 (d) None of these

a b a−b 38. The determinant b c b − c is equal to zero if a, b, c are in: 2 1 0 (a) G.P. (c) H.P.

(b) A.P. (d) None of these

39. If a, b, c are all different and the equations ax + a2y + (a3 + 1) = 0, bx + b 2y + (b3 + 1) = 0, cx + c 2y + (c 3 + 1) = 0 are consistent, then (a) abc = – 1 (c) abc = – 2

(b) abc = 1 (d) abc = 2

40. Let λ and α be real. The set of all values of λ for which the system of linear equations λ x + (sin α) y + (cos α) z = 0 x + (cos α) y + (sin α) z = 0 – x + (sin α) y – (cos a) z = 0 has a non-trivial solution, is (a) [0,

2 ]

(b) [– 2 , 0]

(d) None of these 2 ] 41. The value of k, for which the system of equations x + ky + 3z = 0, 3x + ky – 2z = 0, 2x + 3y – 4z = 0 possess are three polynomials of degree a non-trivial solution over the set of rationals, is g ( x) h ( x) 33 33 g ′ ( x) h′ ( x) , then ∆ (x) is a (b) (a) − 2 2 g ′′ ( x) h′′ ( x) (c) 11 (d) None of these (c) [–

(b) 3 (d) atmost 3

34. If a, b, c are positive and not all equal, then the value a b c b c a is of the determinant c a b (a) non-negative (c) negative

(b) x = b (d) x = 0

x 3 7 36. The three roots of the equation 2 x 2 = 0 are 7 6 x (a) – 9, 2, 7 (c) 9, 2, ­– 7

x 30. If x, y, and z are all different and y then z (a) xyz = – 1 (c) xyz = – 2

35. If a ≠ b ≠ c, one value of x which satisfies the equation 0 x−a x−b 0 x+a x − c = 0 is given by 0 x+b x+c

(b) non-positive (d) positive

2,

42. If x = cy + bz, y = az + cx, z = bx + ay where x, y, z are not all zero, then (a) a2 + b2 + c2 + 2abc = 0 (b) a2 + b2 + c2 – 2abc = 1 (c) a2 + b2 + c2 + 2abc = 1 (d) None of these 43. Let α1, α2 and β1, β2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of equations α1 y + α2z = 0 and β1 y + β2z = 0 has a non-trivial solution, then

769

x x2 x

Determinants

cos x sin x 2 28. If f (x) = tan x

770

ab c2 = 2 pq r

Objective Mathematics

(a)

ac b2 = 2 pr q

(b)

(c)

bc a2 = qr p2

(d) None of these

44. a, b, c are G.P. with common ratio r1 and α, β, r are in G.P. with common ratio r 2. If the equations ax + αy + z = 0, bx + βy + z = 0, cx + γy + z = 0 have only trivial solution, then (a) a, α = 0 (c) r1, r2 ≠ 1

(b) r1, r2 = 1 (d) r1 = r2

45. For what value of k the following system of linear equations will have infinite solutions x – y + z 2x + y – z – 3x – 2ky + 6z (a) k ≠ 2 (c) k = 3

=3 =2 =3 (b) k = 0 (d) k ε [2,3]

46. The value of the following determinant equals to

y+z y z

x z+x z

x y x+ y

(a) 6 xyz (c) 4 xyz

(b) xyz (d) xy + yz + zx

x +1 x + 2 x + 4 x + 3 x + 5 x + 8 is x + 7 x + 10 x + 14 (a) – 2 (c) 2

(b) x2 + 2 (d) None of these

48. In a third order determinant, each element of the first column consists of sum of two terms, each element of the second column consists of sum of three terms and each element of the third column consists of sum of four terms. Then it can be decomposed into n determinants, where n has the value. (a) 1 (c) 16

(b) 9 (d) 24

49. If the value of a third order determinant is 11, then the value of the determinant formed by its cofactors will be (a) 11 (c) 1331

(b) 121 (d) 14641

x b b x b a x b and ∆2 = are the given deter 50. If ∆1 = a x a a x minants, then (a) ∆1 = 3 (∆2)2 (c)

d ∆ = 3∆22 dx 1

(a) 1 (c) 4

a b c = k b c a then k = c a b (b) 2 (d) 8 p−q 0 r−q

0 52. The value of the determinant q − p r−p (a) 0 (c) pqr

p−r q−r = 0

(b) (p – q) (q – r) (r – p) (d) 3pqr

53. If ω (≠ 1) is a cube root of unity, then 1 1 + i + ω2 ω2 1 − i −1 ω2 − 1 equals −i −i + ω − 1 −1 (a) 0 (c) i

(b) 1 (d) ω

54. The system of linear equation x + y + z = 2, 2x + y – z = 3, 3x + 2y + kz = 4 has a unique solution, if

47. The value of the determinant

a+b b+c c+a 51. If b + c c + a a + b c+a a+b b+c

d (b) ∆1 = 3∆2 dx (d) ∆1 = 3 (∆2)3/2

(a) k ≠ 0 (b) – 2 < k < 2

(b) – 1 < k < 1 (c) k = 0

55. For positive numbers x, y, z, the numerical value of the 1 log x y log x z 1 log y z determinant log y x 1 log z x log z y (a) 0 (c) 2

is

(b) 1 (d) None of these

56. The determinant

a b aα + b b c bα + c aα + b bα + c 0

is equal

to 0, if (a) a, b, c are in A.P. (b) a, b, c are in G.P (c) a, b, c are in H.P. (d) (x – α) is a factor of ax2 + 2bx + c x 3 4 57. The sum of two non-integral roots of 5 x 5 = 0 is 4 2 x (a) 4 (c) – 25

58. If D1 =

(b) – 4 (d) None of these 1 x2 x

(a) D1 = D2 (c) D1 = – 2D2

1 y2 y

1 z2 z

and D2 =

1 yz x

(b) D1 = ­– D2 (d) D2 = 2D1

1 xz y

1 xy , then z

C3 C4 9 Cn

9

C5 C6 10 Cn + 2 9

C7 C8 becomes zero is 11 Cn + 4 10

(a) n = 2 (c) n = 4

60. The value of the determinant

(52 x + 5− 2 x ) 2 (62 x + 6− 2 x ) 2 is (7 2 x + 7 − 2 x ) 2

(a) 2102x c) 0

(b) 210 – 2x (d) None of these

(

61. If f (x)

(a) α (c) θ

(b) β (d) a

67. If b2 – ac < 0 and a > 0 then the value of the determinant a b ax + by b c bx + cy is ax + by bx + cy 0



(a) positive (c) zero

(b) negative (d) b2 + ac.

68. If

1 2 ( x − 1) 3 ( x − 1) ( x − 2) = ( x − 1) ( x − 1) ( x − 2) ( x − 1) ( x − 2) ( x − 3) , ( x − 1) x ( x − 1) ( x − 2) x x then f (41) = (a) 41 (c) 0 x3 + 4 x 62. If x − 2 x−3

sin (α − β) θ cos (α − β) θ is independent of sin αθ cos αθ sin (α − β) θ cos (α − β) θ



(b) n = 3 (d) None of these

1 (52 x − 5− 2 x ) 2 1 (6 2 x − 6 − 2 x ) 2 1 (7 2 x − 7 − 2 x ) 2

1 a a2

10

(b) – 41 (d) None of these x+3 x−2 5x x −1 x + 2 4x

= ax5 + bx4 + cx3 + dx2 + ex + f, be an identity in x, where a, b, c, d, e, f are independent of x, then the value of f is (a) 0 (b) 15 (c) 17 (d) None of these

−1 = i and ω is a nonreal cube root of unity then x +1 ω ω2 2 is the value of the determinant ω x+ω 1 2 ω x+ω 1

(a) x3 (c) x3 + 1

69. If Un =

1 k k 2n k2 + k + 1 k2 + k 2n − 1 k2 k2 + k + 1

K

and

∑U

n =1

(a) 8 (c) 6

(b) x3 – 1 (d) None of these

n

= 72 then k = (b) 9 (d) None of these 4

x3 2x cos 2 x π/ 2 63. Let α, β, γ be the cube roots of unity, then the value of 5 1 sec 2 x then ∫ f ( x) dx = 70. If f  (x) = tan x α β γ − π/ 2 x4 5 sin 3 x the determinant β γ α is γ α β (a) 2 (b) ­– 2 (c) 0 (d) None of these (a) 1 (b) – 1 71. The value of the determinant (c) i (d) 0 sin α cos β cos α cos β − sin α sin β x α 1 sin α sin β cos α sin β sin α cos β is 64. The roots of the equation β x 1 = 0 are independent 0 cos α − sin α of β γ 1 (a) is independent of α (a) α (b) β (b) independent of β (c) γ (d) all of these (c) independent of α and β (d) None of these 65. If the system of equations 1 x + ay + a2 = 0, x + by + b2 = 0, x + cy + c2 = 0 is consistent x n − 1 cos x then the equations. x+3 x + bcy + (b + c) z = 0 nπ (−1) n n! 0 cos x + cay + (c + a) z = 0 72. If f (x) = then 2 3n + 1 x + aby + (a + b) z = 0 have α α3 α5 (a) a unique solution (b) infinite number of solutions dn (c) no solution [ f ( x)]x = 0 = (d) None of these dx n

771

8

8

66. The value of the determinant

Determinants

59. The value of n for which the determinant

772

(a) 1 (c) 0

(b) – 1 (d) None of these

Objective Mathematics

73. The value of the determinant

79. If 1, ω, ω2 are the cube roots of unity, then

1 + α 1 + αz 1 + αz 1 + β 1 + βz 1 + βz 2 is 1 + γ 1 + γz 1 + γz 2 2



(b) a (d) None of these

(a) 0 (c) – b 1 2n ω D= ωn

ωn 1 ω2 n

ω2 n ωn has the value 1

(a) 0 (c) ω2

(a) (α – β) (β – γ) (γ – α) (b) 0. (c) αβγ (d) None of these

(b) ω (d) 1

yz + 2 x

z

2z ;

80. If α, β and γ are the roots of the equation x3 + α β = 0, then the value of the determinant β γ γ α (a) q (b) 0 (c) p (d) p2 – 2q

y + xz

yz

z

81. If the determinant

74. The value of the determinant x+

y

z

2

z

px + q γ α is β

b−c c−a a−b a b c b′ − c′ c′ − a′ a′ − b′ = m a′ b′ b′ b′′ − c′′ c′′ − a′′ a′′ − b′′ a′′ b′′ c′′

where x, y, z are positive real numbers, is (a) z ( 2 y − z y ) (b) y  ( 2z − y z )

then the value of m is (a) 0 (c) ­– 1

(c) x ( 2 y − z y ) (d) None of these

(b) 2 (d) 1

82. If the determinant α β γ k k 2.3 16.9 26.27 k 75. Let Dk = then the value 10 10 (3 − 1) 2 (9 − 1) ( 2710 − 1)



(a) a, b, c are in H.P (b) α is root of 4ax2 + 12bx + 9c = 0 or a, b, c are in G.P. (c) a, b, c are in G.P. only (d) a, b, c are in A.P.

10

of

∑D

k =1

k

is (b) αβ + αγ + βγ (d) 0

(a) 2 (α + β + γ) (c) αβγ

76. If

α −β 0 0 α β = 0 then β 0 α

(a) α/β is one of the cube roots of unity (b) α is one of the cube roots of unity (c) β is one of the cube roots of unity (d) None of these

77. If f (x) = (a) 3xλ2 (c) xλ2

x+λ x x

x x+λ x

x x , then f (3x) – f (x) = x+λ (b) 6xλ2 (d) None of these

78. If α, β and γ are the roots of the equation x3 + ax + b = 0, then the value of the determinant α −β− γ 2α 2α is β−γ−α 2β 2β γ −α −β 2γ 2γ

a b 2aα + 3b b c 2bα + 3c = 0, then 2aα + 3b 2bα + 3c 0

83. If [ ] denotes the greatest integer less than or equal to the real number under consideration, and – 1 ≤ x < 0; 0 ≤ y < 1; 1 ≤ z < 2, then the value of the determinant [ x] + 1 [ y ] [ z] [ x] [ y ] + 1 [ z ] is [ x] [ y] [ z] + 1 (b) [ y] (d) None of these

(a) [z] (c) [x]

84. If a, b, c are different, then the value of x satisfying

0 2 x +a x4 + b (a) c (c) b

x2 − a 0 x−c

x3 − b x2 + c 0

= 0 is (b) c (d) 0

π 85. The value of θ lying between θ = 0 and θ = and 2 satisfying the equation 1 + sin 2 θ cos 2 θ 4 sin 4θ 2 2 = 0 is sin θ 1 + cos θ 4 sin 4θ sin 2 θ cos 2 θ 1 + 4 sin 4θ

(b)

7π 11π or 24 24

(c)

3π 4

(d)

π 24

A1 ∆′ = A 2 A3

B1 B2 B3

C1 C 2 is C3

773

5π 2

(a) 0 (c) ∆2

(b) 2∆ (d) ∆

(a) x = 0 (c) x = 2

(b) x = 1 (d) x = a2 + b2 + c2

3x − 8 3 3 = 0 is 3 3x − 8 3 a−x c b 3 3 3x − 8 93. If a + b + c = 0, one root of =0 c b−x a is which of the following ? b a c−x

86. One root of the equation

(a)

2 3

(c)

16 3

8 3 1 (d) 3

(b)

94. The system of linear equations ax + by = 0, cx + dy = 0 has a non-trivial solution if

87. The value of ‘a’ for which the equations a3 x + (a + 1)3y + (a + 2)3 z = 0, ax + (a + 1) y + (a + 2) z = 0, x + y + z = 0 has a non-trivial solution is (a) – 1 (c) 0

(b) 1 (d) None of these

λ 2 + 3λ λ − 1 λ + 3 88. If pλ4 + qλ3 + rλ2 + sλ + t = λ +1 2 − λ λ − 3 λ − 3 λ + 4 3λ then t is equal to (a) 21 (c) 23

89. If | A | =

q −b y a b c x y z and | B | = − p a − x , then r −c z p q r

b+c a a b c+a b 90. The value of c c a+b

b1 b2 b3

(b) ab + bc + ca (d) 0

x z y

y x is z

(a) z – x (c) y – z

(b) x – y (d) None of these

(b) 0 (d) 4abc −1 x − 3 = 0 is x+2

(b) {2, – 3, 1} (d) {– 3, 1, 5}

c1 c2 , then the value of c3

(b) G.P (d) None of these

98. If x, y, z are in A.P., then the value of the determinant A, where

is

92. If the capital letters denote the cofactors of the corresponding small letters in the determinant a1 ∆ = a2 a3

y+z z+x x+ y

(a) H.P (c) A.P

x −6 91. Solution set of the equation 2 − 3 x − 3 2x



(a) a + b + c (c) abc

is

97. If the three linear equations x + 4ay + ax = 0, x + 3by + bz = 0 and x + 2cy + cz = 0 have a non-trivial solution, then a, b, c are in

(a) | A | = 2 | B | (b) | A | = | B | (c) | A | = – | B | (d) None of these

(a) {2, 0, 1} (c) {2, 1, 5}

0 a−b a−c 95. The value of the determinant b − a 0 b−c 0 c−a c−b

96. The repeated factor of the determinant

(b) 31 (d) 33

(a) abc (c) a + b + c

(a) ad – bc = 0 (b) ac + bd = 0 (c) ad – bc < 0 (d) ad – bc > 0

4 5 6 x  5 6 7 y   , is A=  6 7 8 z    x y z 0 (a) 2 (c) 0

(b) 1 (d) None of these

99. The only integral root of the equation 2− y 2 3 = 0 is 2 5− y 6 3 4 10 − y (a) y = 2 (c) y = 1

(b) y = 3 (d) None of these

Determinants

(a)

774

Objective Mathematics

a1 100. If D = a2 a3

b1 b2 b3

c1 c2 c3

a1 + pb1 , b1 + qc1 D’ = a + pb b + qc 2 2 2 2 a3 + pb3 b3 + qc3

c1 + ra1 c2 + ra2 , then c3 + ra3

(a) D’ = D (1 + pqr) (b) D’ = D (c) D’ = D (1 – pqr) (d) D’ = D (1 + p + q + r)

(b) α = 1 (d) None of these

102. The value of the determinant

a−b b−c c−a x− y y−z z−x p−q q−r r− p

is

(a) abc + pqr + xyz (c) 0 1 1 + ac 1 + bc 103. ∆ = 1 1 + ad 1 + bd 1 1 + ae 1 + be (a) a + b + c (c) 1

(b) (a – x) (y – z) (r – p) (d) None of these

= (b) 3 (d) 0

104. If a ≠ b ≠ c , one value of x which satisfies the equation 0 x+a x+b

x−a 0 x+c

x−b x − c = 0 is given by 0

(a) x = 0 (c) x = b

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

105. If Tp, Tq, Tr are pth, qth, rth terms of an A.P, then Tp p 1

Tq q 1

5 5α α    109. Let A = 0 α 5α  5  0 0

If |A2| = 25, then |α| equals (b) 1 (a) 52 1 (c) (d) 5 5 110. A is a square matrix of order 4 and I is a unit matrix, then it is true that (a) det (2A) = 2 det (A) (b) det (2A) = 16 det (A) (c) det (– A) = – det (A) (d) det (A + I) = det (A) + I x 3 7 111. If one root of the determinant 2 x 2 = 0 , is – 9, 7 6 x then the other two roots are

(a) 2, 7 (c) – 2, 7

(b) 2, – 7 (d) – 2, – 7

a b aα + b 112. The determinant b c bα + c is equal to 0 aα + b bα + c zero for all values of α, if

(b) a, b, c are in GP (d) None of the above

113. For how many value(s) of x in the closed interval [–4, –1], (b) 0 (d) 1

106. If A + B + C = π, then the value of sin ( A + B + C ) sin B cos C 0 tan A is equal to − sin B cos ( A + B ) 0 − tan A

(a) 0 (c) 2 sin B tan A cos C

(b) 3 ω(ω – 1) (d) 3 ω(1 – ω)

(a) divisible by neither x not y (b) divisible by both x and y (c) divisible by x but not y (d) divisible by y but not x

(a) a, b, c are in AP (c) a, b, c are in HP

Tr r is equal to 1

(a) p + q + r (c) – 1

(a) 3 ω (c) 3 ω2

1 1 1 1 for x ≠ 0, y ≠ 0, then D is 108. If D = 1 1 + x 1 1 1+ y

a−b b−c c−a 101. If α = b − c c − a a − b , then c−a a−b b−c (a) α = – 1 (c) α = 0

1 3 , then the value of the determinant +i 2 2 1 1 1 1 −1 − ω2 ω2 is 1 ω2 ω4

107. Let ω = −

and

(b) 1 (d) none of these

−1 + x  3 is the matrix  3 −1   x + 3 −1

(a) 2 (c) 3

2  x + 2 singular? 2 

(b) 0 (d) 1

114. If a ≠ b ≠ c, the value of x which satisfies the equation 0 x−a x−b 0 x+a x − c = 0 , is 0 x+b x+c

115. Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx and z = bx + ay. Then, a2 + b 2 + c 2 + 2abc is equal to (a) 2 (c) 0

(b) – 1 (d) 1

cos 2 θ cos θ.sin θ − sin θ sin 2 θ cos θ , then for all θ 116. f (θ) = cos θ.sin θ sin θ − cos θ 0 (a) f (θ) = 1 (c) f (θ) = 0

(b) f (θ) = 2 (d) none of these.

2( x − 1) ( x − 1)( x − 2) x( x − 1)

1 117. If f(x) = ( x − 1) x

3( x − 1)( x − 2) ( x − 1)( x − 2)( x − 3) ( x − 1)( x − 2)

then f (49) = (a) 49 (c) 0

(b) – 49 (d) none of these.

118. The coefficient of x in x f(x) = 1 x2

1 + sin x cos x log (1 + x) 2 , – 1 < x ≤ 1, is 1 + x2

0

(a) 1 (c) –1

(b) – 2 (d) 0

sOLUTIONs 1. (b) We have, cos x cos x ⋅ sin x − sin x sin 2 x cos x f (x) = cos x sin x 0 sin x − cos x

b 2 − ab b − c bc − ac 2 2 3. (b) We have, ab − a a − b b − ab bc − ac c − a ab − a 2

2

=

[Applying C1 → C1 – sin x ⋅ C3 and C2 → C2 + cos x . C3] 1 0 − sin x 1 cos x = 0 0 − cos x sin 2 x

= (b –­ a)



2

[Taking (b –­ a) common from C1 and C3]

= (b –­ a)2

 [Applying R3 → R3 – sin x⋅ R1]  sin2x + cos2x = 1 for all x. [Expanding along C1] =

b b−c c a a−b b c c−a a b−c b−c c a−b a−b b c−a c−a a



[Applying C1 → C1 – C3]

= 0 [ C1 and C2 are identical]. 13 + 3 2. (a) We have,

= ( 5 )2



5

15 + 26

5

10

3 + 65

15

5

13 + 3

2

15 + 26

5

2

3 + 65

3

5

− 3

2

0

5

2

0

3

5

 =–

5

[Taking

=5

2

3  (5 –

b+c c b c c+a a b a a+b

0 − 2a − 2a a = c c+a b a a+b

1 is

 5 common from C2 and C3]

[Applying R1 → R1 – (R2 + R3)]

0 1 1 = – 2a c c + a a b a a+b

1

[Applying C2 → C1 – 5

4. (d) We have,

 3 C2 –

6 ) [Expanding along C1].

13 C3]

[Taking – 2a common from R1]

0 0 1 c c a [Applying C2 → C2 – C3] = – 2a b −b a + b

775

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

Determinants

(a) x = 0 (c) x = b

776

= (– 2a) (– bc – bc) [Expanding along R1]

8. (a), (b), (c), (d) We have, n

= 4abc

Objective Mathematics

∑2

∴ K = 4.

r −1

x 2n − 1

r =1

a2 a 1 5. (a) We have, cos nx cos (n + 1) x cos (n + 2) x sin nx sin (n + 1) x sin (n + 2) x

n

∑ ∆r = r =1

n

∑ 2.3

r −1

y 3n − 1

r −1

z

r =1 n

∑ 4.5

5n − 1

r =1

a 2 − 2a cos x + 1 a 1 0 cos (n + 1) x cos (n + 2) x = 0 sin (n + 1) x sin (n + 2) x  [Applying C1 → C1 + C3 – 2 cos x C2] = (a2 – 2a cos x + 1) sin x [Expanding along C1], which is independent of n. 6. (c) We have,

x +1 x + 2 x + a x+2 x+3 x+b x+3 x+4 x+c

2n − 1 x 2n − 1 = 3n − 1 y 3n − 1 5n − 1 z 5n − 1  n 2n − 1 2 n −1 , = 2n – 1, ∵ ∑ ∆ r = 1 + 2 + 2 + ... + 2 = 2 1 − 1 r =  n 3n − 1 similarly ∑ 2 ⋅ 3r − 1 = 2 ·   = 3n – 1 3 −1 r =1 n

∑4 ⋅ 5

and

r −1

= 4 · 

r =1

0 0 a + c − 2b = x+2 x+3 x+b x+3 x+4 x+c

= 0 ∴

= 0 [Since a, b, c are A.P. ∴ a + c = 2b or a + c – 2b = 0]. 7. (d) We have, n

∑ (r − 1)

n

6

2

2n 2

4n − 2

3

3n3

3n 2 − 3n

r =1

n

∑∆ r =1

n

r

=

∑ (r − 1) n

r =1

[ t he terms in C1 are dependent on r whereas the terms in C2 and C3 are constant] 1 (n − 1) n 2 =

n

1 (n − 1) n (2n − 1) 2n 2 6 1 (n − 1) 2 n 2 3n3 4

6

r

3n 2 − 3n 6 2 (2n − 1) 3n (n − 1)

[Taking 1  n (n – 1) common from C1 and n com12 mon from C2] [ C1 and C3 are identical].

C1 and C3 are identical]

is independent of x, y, z and n. p 1 q 1 r 1

log A + ( p − 1)log R = log A + (q − 1)log R log A + (r − 1)log R

p 1 q 1 r 1

[Let A be the first term and R the common ratio of G.P., then a = Tp = AR p – 1, ∴ log a = log A + (p – 1) log R. b = Tq = AR q – 1,  ∴ log b = log A + (q – 1) log R. c = Tr = AR r – 1,∴ log c = log A + (r – 1) log R.] 1 p 1 = log A 1 q 1 + log R 1 r 1

4n − 2

6 1 1 2 = n (n − 1) 2 (2n − 1) 2n 12 3n (n − 1) 3n 2

= 0.

∑∆

log a 9. (a) We have, log b log c

r =1

∑ (r − 1)

[ n

r =1

[Applying R1 → R1 + R3 – 2R2]



 5n − 1 = 5n − 1. 5 −1 

= 0 + log R

0 0 0

p 1 q 1 r 1

p −1 p 1 q −1 q 1 r −1 r 1

[Applying C 1 → C 1 – C 2 + C 3]

 = 0. 10. (b) We have,



1 4 9 4 9 16 9 16 25 16 25 36

1 16 4 25 = 9 36 16 49

3 5 7 5 7 9 7 9 11 9 11 13

1 4 = 9 16

3 5 7 9

2 2 2 2

2 2 2 2

⇒ (x – a) (x – b) (x – c) = 0 [Expanding along C1] [ C3 and C4 are identical]

= 0.  11. (d) l = AR

m = AR

⇒ log l = log A + ( p – 1) log R

q–1

⇒ log m = log A + (q – 1) log R

n = Arr–1 ⇒ log n =log A + (r – 1) log R log l Now, log m log n

p l q l r 1

log A + ( p − 1)log R = log A + (q − 1)log R log A + (r − 1)log R

p l q l = 0. r l

=

n 1k

100A2 + 10B2 + C2

=

n 2k

100A3 + 10B3 + C3

=

n 3k

where n1, n2, n3 are integers.

A1 = A2 A3

C1 C2 C3

n1k n2 k n3k

A1 = k A2 A3

B1 B2 B3

n1 n2 n3

= k∆1

⇒ ∆ is divisible by k [since elements of ∆1 are integers, ∴ ∆1 is an integer]. x m n 1 a x n 1 13. (a) We have, =0 a b x 1 a b c 1

⇒

x−a m−x 0 x−b n− x 0 x−c 0 0 a b c

[Applying R1 → R1 – R3, and R2 → R2 – R3]

=

a 2 + 2a − 3 2a − 2 2a − 2 a −1

=

(a + 3) (a − 1) 2 (a − 1) a+3 2 = (a – 1)2 2 (a − 1) a −1 2 1

[Expanding along C3]

= (a – 1)2 ⋅ (a + 3 – 4) = (a – 1)3. Clearly, ∆ > 0 if a > 1; ∆ = 0 if a = 1 and ∆ < 0 if a < 1. −1 + cos B cos C + cos B cos B cos C + cos A −1 + cos A cos A −1 + cos B −1 + cos A −1

B1 100A1 + 10B1 + C1 B2 100A 2 + 10B2 + C 2 B3 100A 3 + 10B3 + C3

B1 B2 B3

a 2 + 2a 2a + 1 1 14. (d) Let ∆ = 2a + 1 a + 2 1 3 3 1

15. (a) We have,

[Applying C3 → C3 + 10C2 + 100C1] A1 = A2 A3

So, the roots are independent of m, n.



100A1 + 10B1 + C1

B1 B2 B3

⇒ x = a, b, c.

a 2 + 2a − 3 2a − 2 0 2a − 2 a −1 0 = 3 3 1

12. (a) Since A1B1C1 , A 2B2C2 and A 3B3C3 are divisible by k, therefore

A1 A Now ∆ = 2 A3

x−a m−x 0 x − b n − x = 0 [Expanding along C4] 0 x−c 0 0

⇒

[Applying C3 → C3 – C2, C4 → C4 – C3] p – 1

0 0 =0 0 1

−1 cos C cos B = cos C −1 cos A −1 cos B cos A  =

[Applying C1 → C1 – C3 and C2 → C2 – C3] 1 a

1 = a

−a cos C cos B −1 cos A a cos C −1 acos B cos A 0 cos C cos B 0 −1 cos A 0 cos A −1 [Applying C1 → C1 + b C2 + c C3]

 = 0. 16. (b) We have,

a b c a b c −c −a a − b b − c c − a = −b b+c c+a a+b b+c c+a a+b

777

[Applying R1 → R1 – R2, R2 → R2 – R3  and R3 → R3 – R4]

Determinants

[Applying C2 → C2 – C1, C3 → C3 – C2, C4 → C 4 – C 3]

778

[Applying R2 → R2 – R1]

Objective Mathematics

a b c =– b c a  c a b

[Applying R3 → R3 + R2]

= – (3abc – a – b – c ) = a + b + c – 3abc 3

3

3

3

3

3

∴  k = – 3. 17. (b) We have, a + b + 2c a b c b + c + 2a b c a c + a + 2b



a b c −a c b = b c a × −b a c c a b −c b a a b c = b c a c a b

20 (b) Putting λ = 0 in the given identity, we get





1 a b = 2 (a + b + c) 0 b + c + a 0 0 0 c+a+b

0 −1 3 1 1 −4 0 −2 4



18. (b) We have,





1 2 3 4 1 2+ x 3 4 = (x + 10) 1 2 3+ x 4 1 2 3 4+ x [Applying C1 → C1 + C2 + C3 + C4 and taking (x + 10) common from C1]

1 0 = (x + 10) 0 0

2 x 0 0

3 0 x 0

4 0 0 x

3 (4 + 2) = 10.

x a a a a x a a a a x a a a a x

1 1 = (x + 3a) 1 1



1 a 0 x−a = (x + 3a) 0 0 0 0

a 0 x−a 0

c2 2ca − b 2 a2

b2 a2 2ab − c 2

a 0 0 x−a

[Applying R2 → R2 – R1, R3 → R3 – R1 and R4 → R4 – R1] = (x + 3a) (x – a)3 [Expanding along C1] ∴ f (x) = (x – a)3. 22. (b) We have, cos (A − P) cos (A − Q) cos (A − R) = cos (B − P) cos (B − Q) cos (B − R) cos (C − P) cos (C − Q) cos (C − R) cos P sin P 0 cos A sin A 0 cos Q sin Q 0 = 0.0 = cos B sin B 0 × cos R sin R 0 cos C sin C 0

= 0. [Applying R 2 → R 2 – R 1 , R 3 → R 3 – R 1 x 3 7  and R4 → R4 – R1] 23. (d) We have, 3 x 2 =0 2 = x  (x + 10) [Expanding along C1] x 7 6 19 (d) We have, 2bc − a 2 c2 b2

a a a x a a   a x a a a x

[Applying C1 → C1 + C2 + C3 + C4 and taking (x + 3a) common from C1]

[Expanding along C1]

1+ x 2 3 4 1 2+ x 3 4 1 2 3+ x 4 1 2 3 4+ x

= t ⇒ t = 0 + 1 (0 – 8) +

21. (b) We have,

[Applying R2 → R2 – R1 and R3 → R3 – R1] = 2 (a + b + c)3 ∴ k = 2.

2

= [a (bc – a2) + b (ac – b2) + c (ab – c2)]2 = [a3 + b3 + c3 – 3abc]2 ∴ k = – 3.

1 a b = 2 (a + b + c) 1 b + c + 2a b 1 a c + a + 2b [Applying C1 → C1 + C2 + C3 and taking 2 (a + b + c) common from C1]

a b c a b c b c a = b c a c a b c a b

⇒

x+9 x+9 x+9 x 2 2 x 7 6

=0

⇒



=0

[Applying C1 → C1 – sin θ C3, C2 → C2 + cos θ C3]

[Applying C2 → C2 – C1 and C3 → C3 – C1]

⇒ (x + 9) (x – 2) (x – 7) = 0 [Expanding along R1]

=

⇒ x = – 9, 2, 7. 24. (a), (c) We have,

0 x+a x+b

x−a 0 x+c

x−b x−c 0

1 0 −sin θ 0 1 cos θ 0 0 1

 [Applying R3 → R3 – sin θ R1 + cosθ R2] = 1 for all θ. 27. (d) We have, the given determinant

=0

⇒ (x – a) (x + b) (x – c) + (x – b) (x + a)  (x + c) = 0  [Expanding along R1] ⇒ 2x (x2 + ac – ab – bc) = 0

a12 − 2a1b1 + b12 a22 − 2a2b1 + b12 = a32 − 2a3b1 + b12 a42 − 2a4b1 + b12

If b (a + c) > ac, we have three roots 0, ±



b (a + c) − ac. If b (a + c) ≤ ac, we have only one real root x = 0. 25. (a) Since A28, 3B9 and 62C are divisible by k ∴ A28 = n1k = 100A + 20 + 8

...(1)



3B9 = n2k = 300 + 10B + 9

...(2)



62C = n3k = 600 + 20 + C

...(3)

where n1, n2 and n3 are integers. A 3 6 Now, 8 9 C 2 B 2

[Applying R2 → R2 + 100R1 + 10R3] 3 n2 k B

A 3 = k  n1 n2 2 B

6 n3k 2

− 2a1 − 2 a2 − 2a3 − 2 a4

1 1 1 1

0 1 b1 0 1 b2 × 0 1 b3 0 1 b4

cos x − tan x 2 sin x = tan x 

0 x2 x

x x2 x

6 n3 , which is divisible by k. 2

26. (a) We have, cos 2θ cos θ sin θ − sin θ f (θ) = cosθ sinθ sin 2 θ cos θ sin θ − cos θ 0

b12 b22 b32 b42

0 0 0 0

1 2x 1

0 2x  1

[Applying R1 → R 1 – R 3]

= (cos x – tan x) (x2 – 2x2) = – x2 (cos x – tan x)

[Using (1), (2) and (3)

a12 − 2a1b4 + b42 a22 − 2a2b4 + b42 a32 − 2a3b4 + b42 a42 − 2a4b4 + b42

= 0 × 0 = 0. cos x 28. (b) We have, f (x) = 2 sin x tan x

A 3 6 = 100 A + 20 + 8 300 + 10B + 9 600 + 20 + C 2 B 2

A = n1k 2

a12 a2 = 22 a3 a42

a12 − 2a1b2 + b22 a22 − 2a2b2 + b22 a32 − 2a3b2 + b22 a42 − 2a4b2 + b22 a12 − 2a1b3 + b32 a22 − 2a2b3 + b32 a32 − 2a3b3 + b32 a42 − 2a4b3 + b32

⇒ x = 0 or x2 = b (a + c) – ac.



779

x+9 0 0 x−2 2 0 −1 x − 7 7

1 0 − sin θ 1 cos θ = 0 sin θ − cos θ 0

[Expanding along R1]

∴ f ′ (x) = – 2x (cos x – tan x) – x2 (– sin x – sec2 x) f ′ ( x) = lim – 2 (cos x – tan x) + x ∴ lim x→0 x→0 x   (sin x + sec2 x) = – 2 × 1 = – 2. x + a a2 29. (b) We have, x + b b 2 x + c c2

⇒

x a2 x b2 x c2

a3 b3 c3

a3 b3 c3

a a2 + b b2 c c2

=0

a3 b3 c3

=0

Determinants

[Applying R1 → R1 + R2 + R3]

780

Objective Mathematics

1 a a2 a3 2 3 b   + abc 1 b b 2 1 c c c3

1 a2 ⇒ x 1 b 2 1 c2

=0

⇒ x (a – b) (b – c) (c – a) (ab + bc + ca) + abc (a – b) (b – c) (c – a) = 0



⇒ x = – abc/(ab + bc + ca). x y z

30. (a) We have, x y z

⇒

x2 1 y2 1 + z2 1

1 x 1 y 1 z

⇒

x 2 1 + x3 y 2 1 + y3 z 2 1 + z3

x2 y2 y2

x y z

=0

x2 y2 z2

x3 y3 y3

x2 y2 y2

=0 x2 y2 y2

=0

[Applying R2 → R2 – R1, R3 → R3 – R1

 = ( x − 1)

sec x cos x sec 2 x + cot x cosec x 2 2 cosec 2 x f  (x) = cos x cos x 2 1 cos x cos 2 x cos x − cos x sin 2 x cosec 2 x − cos 4 x cos 2 x

sec 2 x +

[Applying R1 → R1 – R3 sec x and

[Expanding along R1]

= x ( x − 1)

n −1

n −1

0

1 π 4⋅2 = − ⋅ − 2 2 5 ⋅ 3 ⋅1 −π 8 = = – (15π + 32)/60. − 4 15

0

[1 + 1 + ...+ (n – 1) times]

(x + n – 1).

f (x) = a 0 x2 + a1 x + a2 g (x) = b0x2 + b1x + b2 h (x) = c0x2 + c1x + c2

=x

f ( x) g ( x) h ( x) 2a0 2b0 2c0 + 2a0 2b0 2c0

=0+2

f ( x) g ( x) h ( x) a1 b1 c1 2a0 2b0 2c0

f ( x) g ( x) h ( x) a1 b1 c1 a0 b0 c0

= 2 [(b1c0 – b0c1) f (x) – (a1c0 – a0c1) g (x)  + (a1b0 – a0b1) h (x)] Hence degree of ∆ (x) ≤ 2.

a  R2 → R2 – R3 cos2 x] 34. (c) We have, b Expanding along C1, we get cos x c  1  − cos x  f (x) = – cos2 x sin2 x  2 + 2  cos x sin x  [Applying R1 = – (sin2 x + cos3 x – cos3 x sin2 x)  = ­– (sin2 x + cos3 x – cos3 x + cos5 x) = – sin2 x – cos5 x. = (a + b π/ 2 π/ 2 π/ 2 2 5 ∴ ∫ f ( x) dx = ­– ∫ sin x dx − ∫ cos x dx 0

+ ( x − 1)

n −1

f ( x) g ( x) h ( x) Then, ∆ (x) = 2a0 x + a1 , 2b0 x + b1 2c0 x + c1 2a0 2b0 2c0

31. (b) We have,

= 0 cos 2 x − cos 4 x 1 cos 2 x

   R n → R n – R 1]

 = x (x – 1)n – 1 +

33. (c) Let

=0

∴ 1 + xyz. = 0.

0

... ... ... ...

( x − 1) n − 1 + ( x − 1) n − 1 + ... + ( x − 1) n − 1 (n − 1) times

⇒ (1 + xyz) (x – y) (y – z) (z – x) = 0 Since x, y, z are all different, ∴ x ≠ y ≠ z,

0

... ... ... ...

x 1 1 0 (1 − x) ( x − 1) = 0 (1 − x) ( x − 1) ... ... ... 

1 x + xyz 1 y 1 z

1 x ⇒ (1 + xyz) 1 y 1 z

( a ≠ b ≠ c).

x 1 1 1 x 1 32 (a) We have, 1 1 x ... ... ...

b c c a a b

1 1 1 = (a + b + c) b c a c a b

→ R1 + R2 + R3 and taking (a + b + c) common from R1] 1 0 0 + c) b c − b a − b c a−c b−c

[Applying C2 → C2 – C1 and C3 → C3 – C1] = (a + b + c) [– (b – c)2 – (a – c) (a – b)]  [Expanding along R1] = – (a + b + c) (a2 + b2 + c2 – ab – ac – bc)

which is clearly negative [ not all equal]

35. (d) Let ∆ =

0 x+a x+b

x−a 0 x+c

a, b, c are + ve and

0 0 a−b 2 0 a a − c = (a + c) (a + b) (a – c) ∆= 0 a+b a+c Clearly ∆ ≠ 0 on expansion along second column, so that x = a does not satisfy the equation ∆ = 0. Similarly x = b and x = c also do not satisfy. Now, put x = 0, we get 0 − a −b ∆ = a 0 − c = 0. 0 b c

1 1 1 x 3 7 36. (a) We have, 2 x 2 = 0  ⇒ (x + 9) 2 x 2 = 0 x 3 7 7 6 x



[Applying R1 → R1 + R2 + R3 and taking (x + 9) common from R1]

[Applying C2 → C2 – C1 and C3 → C3 – C1]



1 1 −3 1+ λ 2 + λ −8 = 0 1 −1− λ 2 + λ 1 0 0 1+ λ 1 − 5 + 3λ 1 −2−λ 5+λ

39. (a) For consistency of three equations in two unknowns, we must have D = 0 a a2 b b2 c c2

i.e.,

a3 + 1 b3 + 1 = 0 c3 + 1

⇒ 1 + abc = 0 ∴ abc = – 1.

therefore

2 cos   2 α − π   4

π Since – 1 ≤ cos   2 α −  ≤ 1 ∨ α ∈ R  4 ∴ –

2 ≤λ≤

1 k 3 3 k −2 2 3 −4



[Applying C2 → C2 – C1, C3 → C3 + 3C1]

⇒ (5 + λ) + (2 + λ) ( – 5 + 3λ) = 0 ⇒ 3λ2 + 2λ – 5 = 0 5 . 3

2 i.e., λ ∈ [– 2 ,

2 ].

41. (b) For the given system to have a non-trivial solution, we must have =0

1 k ⇒ 0 − 2k 2 3 − 2k

=0

⇒ (λ – 1) (3λ + 5) = 0 ⇒ λ = 1. −

λ sin α cos α 1 cos α sin α = 0 −1 sin α − cos α

⇒ λ  (– cos2 α – sin2 α) –  (– sin α cos α – sin α cos α) – (sin2 α – cos2 α) = 0 ⇒ – λ + sin 2α + cos 2α = 0 ⇒ λ = sin 2α + cos 2α

=0

⇒ (x + 9) (7 – x) (x – 2) = 0 ⇒ = – 9, 2, 7.

⇒

⇒ ac + 2b2 + ab – 2ca – ab – b2 = 0 ⇒ b2 – ca = 0 ⇒ b2 = ca ∴  a, b, c are in G.P.

⇒ λ =

37. (a), (c)  Here the equations are in two variables x and y. If they are consistent then the values of x and y obtained from first two equations should satisfy the third equation and hence D = 0. i.e., ⇒

[Apply C3 → C3 + C1]

40. (c) Since the system has a non-trivial solution,

Hence, x = 0 satisfies the equation ∆ = 0.

1 0 0 0 ⇒ (x + 9) 2 x − 2 x 3− x 7− x

781

a b a ⇒ b c b = 0 2 1 1

x−b x−c . 0

On putting x = a, we get

 

a b a−b b c b−c = 0 2 1 0

38. (a)

3 −11 = 0 −10

[Applying R2 → R2 – 3R1, R3 → R3 – 2R1]

⇒ 20k + 11 (3 – 2k) = 0 ⇒ k = 42. (c) We have, x – cy – bz = 0 cx – y + az = 0 bx + ay – z = 0

.

Determinants

= – 1 (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2], 2

782

 ince x, y, z are not all zero, so the system will S have a non-trivial solution. Therefore,

Objective Mathematics

1 −c −b c −1 a = 0 b a −1 ⇒ 1 ⋅ (1 – a2) + c (– c – ab) – b (ac + b) = 0 ⇒ a2 + b2 + c2 + 2abc = 1. 43. (a) Since α1 ,α2 and β1, β2 are the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively, therefore α1 + α2 =

c −b , α 1α 2 =  a a

...(1)

and β1 + β2 =

−q r , β 1β 2 =  p p

...(2)



 ince the given system of equations has a nonS trivial solution ∴

α1 α 2 β1 β2

⇒

pb = qa

⇒

b2 ac = . q2 pr

α1α 2 β1β2

pc ra

44. (c) Since a, b, c are in G. P. with common ratio r1 and α, β, γ are in G. P. with common ratio r2, therefore a ≠ 0, α ≠ 0, b = ar1, c = a r12 , β = αr 2, γ = α r22 Also, the system of equations have only trivial solution, so

⇒

a α 1 b β 1 ≠0 c γ 1 a ar1 ar12



0 0 r2 − r1 1 − r1 r22 − r12 1 − r12

1 1 r2 1 ≠ 0 r22 1

≠0

45. (c) We have x–y+z=3

∆

y+z y = z

x z+x z

x y x+ y

−2 z z+x z

−2 y y x+ y



0 = y z



= 0 + 2z (xy + y2 – yz) – 2y (yz – z2 – zx) = 2xyz – 2zy2 – 2yz2 – 2y2z + 2yz2 + 2xyz = 4 xyz.

47. (a) We have, x +1 x + 2 x + 4 x+3 x+5 x+8 x + 7 x + 10 x + 14 

=

0 0 1 1 r2 + r1 1 + r1

x +1 1 2 2 1 1 4 1 1



[Applying R2 → R2 – R1, R3 → R3 – R2] x +1 1 1 2 1 0  2 0 0

[Applying R3 → R3 – R2] [Expanding along R3].

48. (d) n = 2 × 3 × 4 = 24. 49. (b) We know that ∆c = ∆3

≠0

x +1 1 2 x+3 2 3 x+7 3 4

[Applying C2 → C2 – C1, C3 → C3 – C2]

= 2 (0 – 1) = – 2.

⇒ aα (r2 – r1) (1 – r1) (1 – r2) ≠ 0 ⇒ r1 ≠ r2, r1 ≠ 1, r2 ≠ 1.

46. (c) We have,

=

[Applying C2 → C2 – C1 C3 → C3 – C1]

1 ⇒ aα (r2 – r1) (1 – r1) r1 r12

1 −1 = 0 6

⇒ 1(6 – 2k) + 1(12 – 3) + 1 (– 4k + 3) = 0 ⇒ 6 – 2k + 12 – 3 – 4k + 3 = 0 ⇒ – 6k + 18 = 0 ⇒ k = 3 and for the value at k = 3, ∆1, ∆2 and ∆3 will be zero.

=

1 α 1 αr2 1 ≠ 0 ⇒ aα r1 r12 αr22 1

1 ⇒ aα r1 r12

1 −1 ∴ ∆ = 2 1 −3 −2k

Operating R1 → R1 – (R2 + R3)

= 0 i.e., α1β2 – α2β1 = 0

α1 α2 α + α2 = = 1 = or β1 β2 β1 + β2

2x + y – z = 2 – 3x – 2ky + 6z = 3 for, the system to have infinite number of solutions, we must have ∆ = 0, ∆1 = 0, ∆2 = 0 and ∆3 = 0

–1

= ∆2 = (11)2 = 121.

x b b 1 0 0 x b b d 0 1 0 ∆  =  a x b  +   +  a x b 50. (b) dx 1 a a x a a x 0 0 1  =

x b x b x b + + = 3∆2. a x a x a x





1 1 1 1 0 0 2 1 −1 ≠ 0 ⇒ 2 −1 − 3 ≠ 0 3 2 k 3 −1 k − 3



[Applying C2 → C2 – C1 and C3 → C3 – C1]

⇒ – (k – 3) – 3 ≠ 0 i.e., k ≠ 0.

[Applying C1 → C1 + C2 + C3]

55. (a) We have, 1

a + b + c − a −b = 2 a + b + c −b −c   a + b + c −c −a 

1 log x y log x z 1 log y x log y z 1 log z x log z y

[Applying C2 → C2 – C1, C3 → C3 – C1]

c − a −b = 2  a − b − c  b −c −a

[Applying C1 → C1 + C2 + C3]

c a b a b c = 2 ⋅ (­– 1) (– 1) a b c = 2 b c a b c a c a b

=

p−q = q− p r−p

p−q 0 r−q

p−q 0 r−q

p−q = q− p r−p 

1

=0

=0

[Applying R3 → R3 – (α R1 + R2)]

⇒ – (aα + 2bα + c) (ac – b2) = 0 ⇒ ac – b2 = 0 or aα2 + 2bα + c = 0 ⇒ a, b, c are in G. P or (x – α) is a factor of ax2 + 2bx + c. 2

0 p−r p−r

57 (b) We have,

[Applying C2 → C2 – C1, C3 → C3 – C1] = 0.

[ C2 and C3 are identical].

1 1 + i + ω2 ω2 53. (a) We have, 1 − i −1 ω2 − 1 −i −i + ω − 1 −1 0 0 0 −1 ω2 − 1 = 1− i −i −i + ω − 1 −1  = 0.

log x log y log z log x log y log z log x log y log z

a b aα + b b c bα + c aα + b bα + c 0



p−q 0 0 = ( p – q) (p – r) q − p 1 1 r−p 1 1 

log y log z

a b aα + b b c bα + c =0 0 0 − (aα 2 + 2bα + c)

[Applying R1 → R1 – R2] 0 p−q p−q

1

56. (b), (d) We have,

p−r q−r 0

p−q q−r 0



log z log x log z log y

[ all rows are identical].

⇒ k = 2. 0 52. (a) We have, q − p r−p

1 log x ⋅ log y ⋅ log z

log x log y log x log z

=

log y log x

[Applying R1 → R1 – R2 + R3]



x−4 3 4 x 3 4 x 5 0 5 x 5 =0 ⇒ 4− x 2 x 4 2 x

=0

[Applying C1 → C1 – C3]

1 3 4 ⇒ (x – 4) 0 x 5 = 0 −1 2 x 1 3 4 5 =0 ⇒ (x –­ 4) 0 x 0 5 4+ x 

[Applying R3 → R3 + R1]

⇒ (x – 4) [x (4 + x) – 25] = 0 [Expanding along C1] ⇒ (x – 4) (x2 + 4x – 25) = 0 ⇒ the two non-integral roots are the roots of the equation x2 + 4x – 25 = 0 ∴ Sum of roots = – 4.

Determinants

(a + b + c) b + c c + a = 2  (a + b + c) c + a a + b (a + b + c) a + b b + c

783

54. (a) The given system has a unique solution if

a+b b+c c+a 51. (b) We have, b + c c + a a + b c+a a+b b+c

784

58. (a) We have,

Objective Mathematics

1 yz x

D2 =





1 xz y

=

xyz xyz

=

1 x2 x

x 1 x2

8

=

1 z2 z

z 1 z2

9

C5 C6 10 Cn + 2

8

C5 C6 10 Cn + 2

C3 C4 9 Cn 9

= ­–

1 x x2

y xyz y2

z xyz z2

1 y y2

1 z z2

0 3 −2 62. (c) We have, − 2 0 −1 −3 2 0

63. (d) α, β, γ are the roots of the equation x3 – 1 = 0. ∴ α + β + γ = 0 Now, we have,

C7 C8 11 Cn + 4 10

9

 = 0. [using (1)].

C5 C8 11 Cn + 4 11

 Applying R 2 → R 2 + R 1 and using n C r + n C r + 1 = n + 1C r + 1 



 Applying R 2 → R 2 + R 1 and using C r + C r + 1 =  n

n

60. (c) The given determinant

65. (b) Since the equations x + ay + a2 = 0, x + by + b2 = 0, x + cy + c2 = 0 are consistent

− 2x 2

1 (5 − 5 ) 1 = 4  1 (62 x − 6− 2 x ) 2 1 = 0 1 (7 2 x − 7 − 2 x ) 2 1 

[Applying R1 → R1 – R2, R2 → R2 – R3]

Clearly, the roots are independent of α.

[Applying C3 → C3 – C2] 2x



x −β α − x 0 x−γ 0 = 0 0 β γ 1

⇒ (x – β) (x – γ) = 0 [Expanding along C3]

4 ⋅ 52 x ⋅ 5− 2 x 4 ⋅ 62 x ⋅ 6− 2 x 4 ⋅ 72 x ⋅ 7− 2 x



⇒

C r + 1 

n +1

For n = 4, R2 and R3 become identical and therefore the value of the determinant becomes zero.

1 (52 x − 5− 2 x ) 2 = 1 (6 2 x − 6 − 2 x ) 2 1 (7 2 x − 7 − 2 x ) 2

[Applying R1 → R1 + R2 + R3]

x α 1 64. (a) We have, β x 1 = 0 β γ 1

10

10

...(1)

α β γ α+β+γ α+β+γ α+β+γ β γ α = β γ α γ α β γ α β

10

9

= f  [Putting x = 0 on both sides]

 ∴ f = 17.

= D 1.

C3 C4 9 Cn 8

x xyz x2

1 xyz

=

y 1 y2

1 y2 y

59. (c) We have,



1 xy z

[

∴

C1 and C3 are identical].

61. (c) We have, f (x) = (x – 1) ⋅ (x – 1) (x – 2)

1 a a2 1 b b2 1 c c2

1 a a2 Also, 1 b b 2 1 c c2

= 0.

1 bc b + c = 1 ca c + a 1 ab a + b

=0

Therefore, the system of equations x + bcy + (b + c) z = 0, x + cay + (c + a) z = 0, x + aby + (a + b) z = 0

1 2 3 x −1 x − 2 x − 3 x x x

has infinite number of solutions.

[Taking (x – 1) and (x – 1) (x – 2) common from 66. (a) We have,  C2 and C3 respectively] 1 sin (α − β) θ cos (α − β) θ sin αθ cos αθ a 1 2 3 2 sin ( ) cos ( α − β) θ a α − β θ 2 = (x – 1) ⋅ (x – 2) x − 1 x − 2 x − 3 1  = 0. ∴  f (41) = 0.

2

3

[Applying R3 → R3 – R2]

=

[ R1 and R3 are identical] 

1 sin (α − β) θ cos (α − β) θ a sin αθ cos αθ a2 − 1 0 0 [Applying R3 → R3 – R1]

= – (a – 1) sin βθ, which is independent of α. 67. (b) We have, a b ax + by b c bx + cy ax + by bx + cy 0 a b 0 = b c 0 ax + by bx + cy − (ax 2 + 2bxy + cy 2 )

4



− x3 2x cos 2 x f (– x) = − tan 5 x 1 sec 2 x = ­– f (x) − sin 3 x 5 x4

∴

∫

sin α cos β cos α cos β − sin α sin β sin α sin β cos α sin β sin α cos β 0 cos α − sin α

= – (ax2 + 2bxy + cy2) (ac – b2) 1 (b2 – ac) [(ax + by)2 + y2 (ac – b2)] a

=

[ b2 – ac < 0 and a > 0]

< 0 68. (a) The given determinant 1 = (x + 1 + ω + ω ) ω ω2 2

 

1 x + ω2 1

1 1 x+ω

[Applying R1 → R1 + R2 + R3 and taking (x + 1 + ω + ω2) common from R1]

1 = x  ω ω2 

0 0 2 x+ω −ω 1− ω x + ω − ω2 1 − ω2

f ( x) dx = 0.

71. (b) We have,

[Applying C3 → C3 – xC1 – y C2]



π/ 2

− π/ 2

sin α sin β = sin α sin β cos α sin β sin α cos β − sin α 0 cos α 0

−

0

 cos β  R2   Applying R 1 → R 1 − sin β   sin α 2 = − (­– sin α sin β – cos2α sinβ) sin β = sin α, which is independent of β. 72. (c) We have, nπ   0 cos  x +  2  

dn [ f ( x)] = dx n

[Applying C2 → C2 – C1 and C3 → C3 – C1]

= x [{x2 – (ω – ω2)2} – (1 – ω) (1 – ω)2]

0

cos

α

= x3 [Using 1 + ω + ω2 = 0 and ω3 = 1]

(−1) n n! ( x + 3) n + 1 (−1) n n! 3n + 1 α5

nπ 2

α3

69. (a) k

∑1

k

k

k2 + k + 1

k2 + k

k2

k2 + k + 1

n =1 k

∑ Un = n =1

k

2 ∑n





k

k

n =1

n =1

k k k 2 k2 + k = k (k + 1) k + k + 1 k2 k2 k2 + k + 1 k k 0 2 k2 + k = k +k 1 k2 0 k2 + k + 1

 ⇒

∴

n =1

2 ∑ n − ∑1

= k (k + k + 1) – k = 72 (given) k = 8.

dn [ f ( x)]x = dx n

0

73. (b) The given determinant



1 α 1 1 1 = 1 β 0 × 1 z 1 γ 0 1 z2

0 0 = 0. 0

74. (a) We have, 3

= k (k + 1)

(−1) n n! 3n + 1 (−1) n n! = 0. 3n + 1 5 α

( R1 and R2 are identical).

[Applying C2 → C2 – C1] 2

nπ 2 nπ = 0 cos 2 α α3 0 cos

x+

y

z

z

yz + 2 x

z

2z

y + xz

yz

z

2

785

70. (c) We have,

2

Determinants

= (a2 – 1) [sin (α – β)θ cosαθ – cos (α – β)θ sin αθ]

786

x+ = z 

y

2

Objective Mathematics

yz + 2 x

z

2

y + xz

y

z

 − y

2

0

z

2

0

y

z



∴ f (3x) – f (x) = λ2 [(9x + λ) – (3x + λ)] = 6xλ2. 78. (a) We have,

1

[Applying C1 → C1 – y ⋅ z (z –

=–

y C2 –

x C 3]

2 y ) = z ( x  ⋅ y – z y ).

75. (d) We have, α

∑D

k =1

β

10

10

k

γ

10

k = 2 ∑3 k =1

10

16 ⋅ ∑ 9k

26 ⋅ ∑ 27 k

2 (910 − 1)

( 2710 − 1)

k =1

310 − 1 α

γ

 310 − 1   910 − 1   2710 − 1  = 2  16 ⋅   26 ⋅    3 −1   9 −1   27 − 1  2 (910 − 1) (2710 − 1) 310 − 1 α β γ 10 10 10 = 3 − 1 2 (9 − 1) (27 − 1) 310 − 1 2 (910 − 1) ( 2710 − 1)

=0

α −β 0 0 α β = 0 ⇒ α3 – β3 = 0 β 0 α

α ⇒   β of unity.

3

= 1 ⇒

77. (b) We have, f (x) =





[Applying R1 → R1 + R2 + R3]

0 0 0 2 β β − γ − α 2 β = 0. = 2γ 2γ γ −α −β [Since α, β, γ are roots of x3 + ax + b = 0 ∴ α + β + γ = 0] 79. (a) Expanding along R1, we get ∆ = 1 (1 – ω3n) – ωn (ω2n – ω2n) + ω2n (ω4n – ωn) = (1 – 1) – 0 + ω2n ( ωn – ωn) [Using ω3 = 1] = 0.

∴ α + β + γ = 0. α+β+γ β γ α β γ So, β γ α = α + β + γ γ α α+β+γ α β γ α β 



[Applying C1 → C1 + C2 + C3] 0 β γ = 0 γ α 0 α β

= 0.

α is one of the cube roots β 81. (a) Operating C1 → C1 + C2 + C3 on the LHS, we get

x+λ x x

x x+λ x

1 = (3x + λ) x x

x x x+λ

1 x+λ x

1 x x+λ

[Applying R1 → R1 + R2 + R3 and taking (3x + λ) common from R1]



80. (b) Since α, β, γ are the roots of the equation x3 + px+ q = 0

[ R2 and R3 are identical]



α+β+ γ α+β+ γ α+β+ γ β−γ−α 2β 2β = γ −α −β 2γ 2γ

k =1

β

76. (a) We have,

α −β− γ 2α 2α β−γ−α 2β 2β γ −α −β 2γ 2γ

z common from C2 and C3]

[Taking

=z

= (3x + λ) λ2 [Expanding along R1]

1

1 0 0 = (3x + λ) x λ 0 x 0 λ [Applying C2 → C2 – C1, C3 → C3 – C1]



0 c−a a b c a−b 0 c′ − a′ a′ − b′ = m a′ b′ c' 0 c′′ − a′′ a′′ − b′′ a′′ b′′ c′′

⇒ m = 0. 82. (b) Operate R 3 → R 3 – 2α R1 – 3R 2, we get

a b 2aα + 3b b c 2bα + 3c 0 0 − 4aα 2 − 6bα − 6bα − 9c

=0

⇒ (4aα2 + 12bα + 9c) (ac – b2) = 0 ⇒ α is a root of 4ax2 + 12bx + 9c = 0 or a, b, c are in G. P. 83. (a) Since, – 1 ≤ x < 0 ∴ [x] = – 1.

89. (c) | B | = –

0 0 1 ∴ Given determinant = −1 1 1 −1 0 2

= 1 = [z].

84. (d) Since for x = 0, the determinant reduces to the determinant of a skew-symmetric matrix of odd order which is always zero. Hence x = 0 is the solution of the given equation.



=–

p a q b r c

85. (b) Operate R1 → R1 – R 3 and R 2 → R 2 – R 3, we get



a = b c

p q r



=



= – | A |.

1 0 −1 =0 0 1 −1 sin 2 θ cos 2 θ 1 + 4 sin 4 θ



⇒ 1 + 4 sin 4 θ + cos2θ + sin2θ = 0 ⇒ 2 + 4 sin 4 θ = 0 ⇒ sin 4 θ = − ⇒ 4θ =

1 2

7π 7π 11π 11π or ⇒θ = or . 24 6 6 24

3x − 2 3x − 2 3x − 2 =0 3 3x − 8 3 3 3 3x − 8



⇒ 

( a + 2) 3 a+2 1

a3 a 1

(a + 1)3 − a 3 1 0

⇒ ⇒  ⇒ ⇒

[Applying R2 ↔ R3]

0 − 2c − 2b = b c+a b c c a+b

(Applying R1 → R1 – R2 – R3) 0 − 2c − 2b 1   0 c (c + a − b ) b (c − a − b ) c c c a+b



=0

2 [(a + 1) – a ] = (a + 2) – a 2 (a3 + 3a2 + 3a + 1 – a3) = (a3 + 8 + 6a2 + 12a) – a3 2 2 6a + 6a + 2 = 6a + 12a + 8 6a = – 6 ⇒ a = – 1. 3

3

⇒ ⇒ ⇒ =0 ⇒

0 −1 3 t = 1 2 − 3 = 1 (0 – 9) + 3 (4 + 6) −3 4 0

+ 6 (2x + 4 + 3x – 9) – 1 (4x – 9x) = 0 – 5x3 + 30x – 30 + 5x = 0 5x3 – 35x + 30 = 0 ⇒ x3 – 7x + 6 = 0 (x – 1) (x2 + x – 6) = 0 ⇒ (x – 1) (x + 3) (x – 2) x = 1, 2, – 3.

a1 92. (c) ∆∆′ = a2 a3



88. (a) Putting λ = 0, we shall get

= – 9 + 30 = 21.

a b c x y z p q r

91. (b) We have, x (– 3x2 – 6x – 2x2 + 6x) =0

( a + 2) 3 − a 3 2 0

3

a b c p q r =– x y z

(Applying R2 → cR2 – bR3) 1 = ⋅ c (2bc) [– (c – a –­ b) + (c + a – b)] c = (2bc) (2a) = 4abc. [Expanding along C3]

[Applying C2 → C2 – C1, C3 → C3 – C1] 3

[Applying C1 ↔ C2]



87. (a) For a non-trivial solution (a + 1)3 a +1 1

y x z

[Applying R1 ↔ R2]

x y  z

b+c a a b c+a b c c a+b

=

⇒ x = 2/3 is one of the roots.

a3 a 1

x y  z



1 1 1 ⇒ (3x – 2) 3 3 x − 8 =0 3 3 3 3x − 8

=

q b p a r c

90. (d) We have, .

86. (a) Operate R1 → R1 + R 2 + R 3, we get



y x z



b1 b2 b3

c1 c2 c3

A1 A2 A3

B1 B2 B3

C1 C2 C3

a1A1 + b1B1 + c1C1 a1A 2 + b1B2 + c1C 2 = a2 A1 + b2 B1 + c2C1 a2 A 2 + b2 B2 + c2C 2 a3A1 + b3B1 + c3C1 a3A 2 + b3B2 + c3C 2 a1A 3 + b1B3 + c1C3 a2 A 3 + b2 B3 + c2C3 a3A 3 + b3B3 + c3C3

787

q −b p −a r −c

Determinants

0 ≤ y < 1 ∴ [y] = 0 1 ≤ z < 2 ∴ [z] = 1.

788

Objective Mathematics

∆ 0 0 = 0 ∆ 0 = ∆ 3. 0 0 ∆



= (x + y + z)

Thus, ∆∆′ = ∆3 and hence ∆′ = ∆2 (provided ∆ ≠ 0). 93. (a) Operate C1 → C1 + C2 + C3, we get 1 c b (a + b + c – x) 1 b − x a 1 a c−x



=0

=0

98. (c) We have,

0 a−b a−c = – b−a 0 b−c 0 c−a c−b

=

x z y

=– ∆

y x z

2( x + y + z ) x + y + z z+x z x+ y y



= (x + y + z)

1 1 z x y z

[Applying R1 → R1 – 2R2 + R3]



0 0 0 0 5 6 7 y = 6 7 8 z x y z 0



[ = 0.

x, y, z are in A. P. ∴ x – 2y + z = 0]

(2 – y) (50 – 15y + y2) – 2 (20 – 2y – 18)  + 3 (8 – 15 + 3y) = 0 3 2 ⇒ – y + 17y – 67y + 75 = 0 or y3 – 17y2 + 67y – 75 = 0. Clearly, y = 3 satisfies it, ∴ we have ( y – 3) ( y2 – 14y + 25) = 0 14 ± 196 − 100 14 ± 96 = = 7 ± 24 2 2 which are not integers Hence only integral root is y = 3. ⇒ y =

x+ y+z x z

[Applying R1 → R1 + R2 + R3] 2 z+x x+ y

0 0 0 x − 2y + z 5 6 7 y = 6 7 8 z 0 x y z

99. (b) We have,

∴ 2∆ = 0 ⇒ ∆ = 0. y+z z+x x+ y

4 5 6 x 5 6 7 y | A | =   6 7 8 z x y z 0 

[Interchanging rows and columns]

96. (a) We have,

[Applying R2 → R2 – R1, R3 → R3 – R1]

∴ a, b, c are in H. P.

= 0.

0 b−a c−a = a−b 0 c−b 0 a−c b−c



1 4a a 1 4a a = 0 ⇒ 0 3 4 b − a b − a =0 1 3b b 0 2c − 4a c − a 1 2c c

⇒ (3b – 4a) (c – a) – (2c – 4a) (b – a) = 0 2 1 1 = + ⇒ bc + ab – 2ac = 0 ⇒ b a c

0 a−b a−c 95. (d) Let ∆ = b − a 0 b−c 0 c−a c−b



[Expanding along C1]

97. (a) For a non-trivial solution, we must have,



94. (a) System of given equations has a non-trivial solua b = 0 i.e., if ad – bc = 0. tion if c d



[Applying C1 → C1 – C2 – C3]

∴ repeated factor = (z – x).

∴ x = 0 is one of the roots of a−x c b c b−x a b a c−x

1 1 z x y z

= (x + y + z) (x – z)2

1 c b i.e., x  1 b − x a 1 a c−x

0 0 x−z

a1 100. (a) D′ = a2 a3

b1 b2 b3

c1 c2 c3

b1 + pqr b2 b3

c1 c2 c3

a1 a2 a3

 [All other determinants will vanish] = (1 + pqr) D.



0 0 0 α = b−c c−a a−b c−a a−b b−c

= 0.

102. (c) Operate C1 → C1 + C2 + C3, we get the value of the determinant = 0. 103. (d) Operate C2 → C2 – C1, C3 → C3 – C1, we get 1 ac bc ∆ = 1 ad bd 1 ae be



1 c = ab 1 d 1 e

c d = ab (0) = 0. e

104. (a) For x = 0, the determinant reduces to the determinant of a skew symmetric matrix of odd order which is always zero. Hence x = 0 is the solution of the given equation. 105. (b) We have, Tp = a + (p – 1) d, Tq = a + (q – 1) d, Tr = a + (r – 1) d, where a is the first term and d is the common difference. ∴ The given determinant a + ( p − 1) d p = 1

=

=a

a a a p q r  1 1 1

a + (q − 1) d q 1

a + (r − 1) d r 1

[Applying R1 → R1 – (R2 – R3) d]

0 = − sin B − cos C

789

1 0 0  = 1 x 0  = xy 1 0 y  Hence, D is divisible by both x and y.

109. (c) We have, 5 5α α  A = 0 α 5α  0 0 5  5 5α α  5 5α α  ∴ A2 = 0 α 5α  0 α 5α     0 0 5  0 0 5 

 25 25α + 5α 2 10α + 25α 2    5α 2 + 25α  α2 =0 0  0 25   25 25α + 5α 2 10α + 25α 2 ⇒ |A2| = 0 5α 2 + 25α α2 0 0 25 = 25

25 25α + 5α 2 = 625α2 0 α2 1 . 5

110. (b) Since, det (k A) = k n det (A),

Since, it is a determinant of a skew symmetric matrix of odd order, therefore, its value should be zero. 1 ω2 ω4

where n is a order of the matrix. ∴ det (2A) = 24 det (A) = 16 det (A) 111. (a) We have, x 3 7 2 x 2 =0 7 6 x

⇒ x(x2 – 12) – 3 (2x – 14) + 7(12 – 7x) = 0 ⇒ x3 – 67x + 126 = 0 ⇒ (x + 9) (x2 – 9x + 14) = 0 ⇒ (x + 9) (x – 2) (x – 7) = 0 ⇒ x = – 9, 2, 7 Hence, the other two roots are 2, 7. 112. (b) We have,

Applying C1 → C1 + C2 + C3

 3 1 = 0 −1 − ω2 0 ω2

On applying C2 → C2 – C1 and C3 → C3 – C1

∴ 625α2 = 25 ⇒ |α| =

sin B cos C 0 tan A 0 − tan A

1 1 107. (b) 1 −1 − ω2 1 ω2

1  1 1 1 1 + x 1  D=  1 1 1 + y 

But |A2| = 25

1 1 1 p q r = 0. 1 1 1

sin ( A + B + C ) sin B cos C 0 tan A − sin B cos ( A + B ) 0 − tan A

106. (a)

108. (b) We have

1 ω2 ω4

= 3 (ω2 – ω) = 3ω (ω – 1)

a b aá + b b c bá + c = 0 0 aá + b bá + c



Applying R3 → R3 – (α R1 + R2), we get

Determinants

101. (c) Operate R1 → R1 + R 2 + R 3, we get

790

a b aα + b b c bα + c =0 0 0 − aα 2 − 2 bα − c

Objective Mathematics

1 − c −b c −1 a = 0 b a −1

⇒ 1(1 – a2) + c (– c – ab) – b (ac + b) = 0 ⇒ 1 – a2 – c2 – abc – abc – b2 = 0 ⇒ a2 + b2 + c2 + 2abc = 1

⇒ – (aα2 + 2bα + c) (ac – b2) = 0 ⇒ b2 = ac ⇒ a, b, c are in G.P. 113. (d) The given matrix is singular, if 3 −1 + x 3 −1 x+3 −1

116. (a)

2 x+2 =0 2

117. (c) f (x) =

Applying R2 → R2 – R1, → R3 → R3 – R1, we get





3 −1 + x 2 0 x =0 −x 2 x −x

⇒

3( x − 1)( x − 2) ( x − 1)( x − 2)( x − 3) x( x − 1)( x − 2)

R3 → R3 – R1 – R2 2( x + 1) ( x − 1)( x − 2)

1 = ( x − 1)

Applying C1 → C1 + C2 + C3, we get



2( x − 1) ( x − 1)( x − 2) x( x − 1)

1 ( x − 1) x

0

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

0

0

=0 ∴ f (49) = 0

x + 4 −1 + x 2 0 x =0 −x 0 0 −x

⇒ (x + 4) (x2) = 0 ⇒ x = 0, – 4 Hence, x = – 4 is the only solution which lies in the given interval.

x 118. (b) Given, f(x) = 1 x2

1 + sin x cos x log (1 + x) 2 1 + x2

0

= x{– 2 (1 + x )} – (1 + sin x)(– 2x ) 2

2

+ cos x {1 + x2 – x2 log (1 + x)}

114. (a) O n putting x = 0, the given determinant becomes zero.



115. (d) For non-zero solution

∴ Coefficient of x in f(x) = – 2

= – 2x – 2x3 + 2x2 + 2x2 sin x + cos x{1 + x2 – x2 log (1 + x)}

Exercises for self-practice 1. Given ai2 + bi2 + ci2 = 1(i = 1, 2, 3) and aiaj + bibj + cicj = 0 (a) 2 (c) – 4 a a a 1

(i ≠ j, i, j = 1, 2, 3) then the value of b 1 c1

1 (a) 2 (c) 2

2

b2 c2

(b) 0 (d) 1

2 8 4 2. The value of determinant − 5 6 −10 is 1 7 2 (a) – 440 (c) 328 − a2 3. If ab ac

(b) 0 (d) 488 ab − b2 bc

ac bc − c2

= ka2b2c2, then k =

3

b3 is c3

(b) 4 (d) 8

a b aα − b 1 4. If b c bα − c = 0 and α ≠ , then 2 2 1 0

5.



(a) a, b, c are in A. P. (b) a, b, c are in G. P. (c) a, b, c are in H. P. (d) None of these If a ≠ b ≠ c, the value of x which satisfies the equation 0 x+a x+b

x−a 0 x+c

(a) x = a (c) x = c

x−b x − c = 0 is 0 (b) x = b (d) x = 0

13. The value of the determinant a −b−c 2a 2a is b−c−a 2b 2b c−a−b 2c 2c

cos α sin α 1 (a) cos β sin β 1 cos γ sin γ 1

(a) (a + b + c)3 (b) (a – b – c) (a2 + b2 + c2) (c) (a + b + c) (ab + bc + ca) (d) None of these If every element of a third order determinant of value ∆ is multiplied by 5, then the value of the new determinant is (a) ∆ (b) 5∆ (c) 25∆ (d) 125∆

(b) zero

cos α sin α 0 (c) sin β cos β (d) None of these 0 cos γ sin γ 0 7.

y+z x x

y x+ z y

z z x+ y

is equal to (b) xyz (d) x2y2z2

(a) 4xyz (c) 4x2y2z2 8. If a, b, x+3 x + 4 x+5

c are in A.P., then : x+4 x+a x + 5 x + b is equal to x+6 x+c

(a) 0 (b) (a + b + c) – 9x (c) 9x + a + b + c (d) None of these

a −b −c 9. If − a b − c −a −b c (a) – 2 (c) 4

10.

y+z y z

+ λabc = 0, then λ is : (b) 2 (d) – 4

x z+x z

x is equal to y x+ y

(a) xyz (c) 4xyz

(b) x2y2z2 (d) 4x2y2z2

11. The number of distinct reat roots of

14.

1 2 3 15. Τhe value of the determinant 3 5 7 is 8 14 20 (a) 20

(b) 10

(c) 0

(d) 5

16. The system of equations a1x + b1y + c1z = 0, a2x + b2y + a1 a c2z = 0, a3x + b3y + c3z = 0, if 2 a3

(a) α = β (b) α ≠ β (c) α = 2β (d) β = 2α 18. The system of equations λx + y + z = 0, – x + λy + z = 0, – x – y + λz = 0 will have a non-zero solution if real values of λ are given by (a) 0 (b) 1 (c) – 1 (d) None of these

(a) 0 (c) 1

(a) a + b + c (c) b3

1 ω ω2

ω ω2 1

(a) 1 (c) ω

ω 1 ω

6 (b) 0 (d) ω2 + 1

= 0 then abc =

(b) 0 (d) ab + bc

20. The value of the determinant

2

is

= 0, has

a b c a+b b+c c+a 17. If α = b + c c + a a + b and β = b c a , then c a b c+a a+b b+c

a 2b 2c 19. If a (≠ 6 ), b, c satisfy 3 b c 4 a b

12. If ω is a cube root of unity, then the value of

c1 c2 c3

(a) more then two solutions (b) one trivial and one non-trivial solution (c) no solution (d) only trivial solution (0, 0, 0)

sin x cos x cos x −π π cos x sin x cos x = 0 in the interval ≤x≤ is 4 4 cos x cos x sin x (b) 2 (d) 3

b1 b2 b3

2i

12

3 +i 8

18

2 + i 12

(a) Complex (c) Irrational

3+ 6 3 2+i 6

is equal to

27 + 2i (b) imaginary (d) None of these

Determinants

2

791

1 cos (α − β) cos (α − γ ) 6. cos (α − β) 1 cos (β − γ ) = 1 cos (α − γ ) cos (β − γ )

792

Objective Mathematics

a b ac + b 21. The determinant ∆ = b c bx + c ax + b bx + c 0 a, b, c are in: (a) A.P. (c) H.P.

= 0, if

22. If x > 0 and ≠ 1, y > 0 and ≠ 1, z > 0 and ≠ 1, then the

(a) 1 (c) 0

(a) x = 1 (c) x = a2 + b2 + c2 x

−x 2

(a + a ) 24. (b x + b − x ) 2 (c x + c − x ) 2

x

(a) G.P. (c) H.P. 26.

is:

(b) –1 (d) None of these

23. If a + b + c = 0, one root of

(b) 2abc (d) None of these

a b a−b 25. The determinant b c b − c is equal to zero if a, b, c 2 1 0 are in :

(b) G.P. (d) None of these

1 log x y log x z 1 log y z value of log y x 1 log z x log z y

(a) 0 (c) a2b2c2

a−x c b is: c b−x a b a c−x

(b) x = 2 (d) x = 0

(b) A.P. (d) None of these

If 2x + 3y + 4z = 9 4x + 9y + 3z = 10 5x + 10y + 5z = 11 then find the value of x:

9 3 4 2 3 4 (a) 10 9 3 ÷ 4 9 3 11 10 5 5 10 5 9 4 3 2 3 4 (b) 10 3 9 ÷ 4 9 3 11 5 10 5 10 5 9 4 9 3 2 4 (c) 10 3 3 ÷ 9 4 3 11 5 10 10 5 5

−x 2

(a − a ) 1 (b x − b − x ) 2 1 = (c x − c − x ) 2 1



(d) None of these

Answers

1. (d) 11. (c) 21. (b)

2. (b) 12. (b) 22. (c)

3. (b) 13. (a) 23. (d)

4. (b) 14. (d) 24. (a)

5. (d) 15. (c) 25. (a)

6. (b) 16. (a) 26. (a)

7. (a) 17. (c)

8. (a) 18. (a)

9. (c) 19. (c)

10. (c) 20. (a)

21

Trigonometric Ratios and Identities

CHAPTER

Summary of concepts Measurement of Angles

1º = π radians ; 1 radian = 180 degrees 180 π

There are three systems for measurement of an angle:

1g = π radians ; 1 radian = 200 grades. 1.  Sexagesimal System or English System  In this sys200 π tem an angle is measured in degrees, minutes and seconds. A complete rotation describes 360º. Thus if the measure of an angle in degrees, grades and 1 right angle = 90º, (read as 90 degrees) radians be D, G and θ respectively, then 1º = 60′ (read as 60 minutes) D G θ = = . 1' = 60′′ (read as 60 seconds) 180 200 π 2. Centesimal or French System  In this system an angle is measured in grades, minutes and seconds. 1 right angle = 100g (read as 100 grades) 1g = 100′ (read as 100 minutes) 1' = 100′′ (read as 100 seconds) Note: 1' of centesimal system ≠ 1' of sexagesimal system 1'' of centesimal system ≠ 1'' of sexagesimal system 3. Radian or Circular Measure  A radian is a constant angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle and is denoted by 1c. ∠AOB = 1 radian.

Relation between sides and interior angles of a regular polygon 1. Sum of interior angles of polygon of n sides = (2n – 4) × 90º 2. Each interior angle of a regular polygon of n sides =

2n − 4 × 90º. n

Fundamental Identities 1. 2. 3.

sin2θ + cos2θ = 1 or cos2θ = 1 – sin2θ or sin2θ = 1 – cos2θ 1 + tan2θ = sec2θ or sec2θ – tan2θ = 1 1 + cot2θ = cosec2θ or cosec2θ – cot2θ = 1.

Note: Since sin2θ + cos2θ = 1, | sinθ | ≤ | and | cosθ | ≤ | Note: ⇒

0 ≤ sin2θ ≤ 1, 0 ≤ cos2θ ≤ 1.

This angle does not depend upon the radius of the circle from which it is derived. Note:  Radian is a unit to measure angle and it should not be inter preted that π stands for 180°, π is a real number whereas πc stands for 180°. Remember:  π radians = 180° = 200g.

Relation between Different Systems of Measurement of Angles 10 10 1º = grades ; 1g = degrees 9 9

–1 ≤ sinθ ≤ 1 and –1 ≤ cosθ ≤ 1;

Since cosec θ = 1/sin θ, cosec θ ≥ 1 or cosec θ ≤ – 1 Also, since sec θ = 1/cos θ, sec θ ≥ 1 or sec θ ≤ – 1. Sign of Trigonometric Ratios Quadrants I

II

III

IV

Trigonometric Ratios sin, cosec

+

+

–

–

cos, sec

+

–

–

+

tan, cot

+

–

+

–

794

Domain and range of trigonometric ratios Functions

Objective Mathematics

Domain

Range

sin x, cos x (– ∞, ∞)

[–1, 1]

trigonometric ratios of Standard angles Angles 0º

30º

sin x

0

cot x

(– ∞, ∞) – {nπ | n ∈ I}

cos x

1

3 2

tan x

0

1 3

cosec x

Undefined 2

sec x

1

2 3

cot x

Undefined

3

(– ∞, ∞)

  π (− ∞, ∞) − (2n + 1) n ∈ I (– ∞, – 1] ∪ 2   [1, ∞)

{

}

(− ∞, ∞) − nπ n ∈ I

1 2

3 2

1 2

  π (− ∞, ∞) − (2n + 1) n ∈ I (– ∞, ∞) 2  

cosec x

60º

90º

T-Ratios

tan x

sec x

45º

(– ∞, – 1]∪ [1, ∞)

1 2

1 2 1

2 2

0

3

Undefined

2 3

1 Undefined

2

1

1

1 3

0

trigonometric ratios of allied angles (their sum or Difference is a multiple of 90º) sin θ

–θ

90º – θ

90º + θ

180º – θ

180º + θ

270º – θ

270º + θ

360º – θ

360º + θ

– sin θ

cos θ

cos θ

sin θ

– sin θ

– cos θ

– cos θ

– sin θ

sin θ

cos θ

cos θ

sin θ

– sin θ

– cos θ

– cos θ

tan θ

– tan θ

cot θ

– cot θ

– tan θ

tan θ

Working Rule to Find Allied Angles Case I: When the angle is nπ ± θ, where n ∈ I and θ is acute. (a) There is no change in trigonometric function i.e., sin remains sin, cos remains cos and tan remains tan. Angle associated becomes θ. (b) The sign is affixed according to the quadrant in which the angle lies.

 nπ



+±θθ, Case II: When the angle is  where n is an odd integer and  2  θ is acute. (a) The trigonometric function is replaced by its cofunction i.e., sin changes to cos, tan changes to cot and sec changes to cosec and vice-versa. Angle associated becomes θ. (b) T he sign is affixed according to the quadrant in which the angle lies. Note that the sign is always decided on the basis of the operating function.

Some uSeful reSultS on allieD angleS 1. sin nπ = 0, cos nπ = (– 1)n. 2. sin (nπ + θ) = (– 1)n sin θ, cos (nπ + θ) = (– 1)n cos θ n −1  2 ( − 1 ) cos θ, if n is odd,  nπ  3. sin  + θ =  n (−1) 2 sin θ, if n is even.  2  

– sin θ cot θ

sin θ

cos θ

cos θ

– cot θ

– tan θ

tan θ

n +1  2 ( − 1 ) sin θ, if n is odd,  nπ  cos  + θ =  n (−1) 2 cos θ, if n is even.  2  

aDDition anD Subtraction formulae 1. sin (A + B) = sin A cos B + cos A sin B 2. cos (A + B) = cos A cos B – sin A sin B 3. tan (A + B) = tan A + tan B 1 − tan A tan B 4. sin (A – B) = sin A cos B – cos A sin B 5. cos (A – B) = cos A cos B + sin A sin B 6. tan (A – B) = tan A − tan B 1 + tan A tan B 7. cot (A + B) = cot A cot B − 1 cot A + cot B 8. cot (A – B) = cot A cot B + 1 cot B − cot A 9. sin (A + B) sin (A – B) = sin2A – sin2B 10. cos (A + B) cos (A – B) = cos2A – sin2B 11. sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C or = cos A cos B cos C (tan A + tan B + tan C – tan A tan B tan C)

13. tan (A + B + C) =

tan A + tan B + tan C − tan A tan B tan C 1 − tan A tan B − tan B tan C − tan C tan A

795

sin (C − D) cos C cos D sin (C + D) sin C sin D

8. cot C – cot D = sin (D − C) sin C sin D

π 1 + tan A 14. tan   + A  = 4  1 − tan A

Trigonometric Ratios of Multiple Angles

π 1 − tan A 15. tan   − A  = . 4 1 + tan A  

(An Angle of the form nθ, n ∈ I)

16. sin (A1 + A2 + ... + An) = cos A1 cos A2 ... cos An  (S1 – S3 + S5 – ...)

2 tan θ 1 + tan 2 θ 2 2 2. cos 2θ = cos θ – sin θ = 2 cos2θ – 1 1. sin 2θ = 2 sin θ cos θ =

17. cos (A1 + A2 + ... + An) = cos A1 cos A2 ... cos An  (1 – S2 + S4 – S6 + ...) 

S1 − S3 + S5 − ... where 1 − S2 + S4 − S6 + ... S1= Σ tan A1, S2 = Σ tan A1 tan A2,  S3 = Σ tan A1 tan A2 tan A3 and so on. 19. sin α + sin (α + β) + sin (α + 2β) + ... + sin (α + (n – 1) β) β  cos  α + (n − 1)  2  nβ   = sin   β  2  sin 2 20. cos α + cos (α + β) + cos (α + 2β) + ... + cos (α + (n – 1) β) 18. tan (A1 + A2 + ... + An) =

β  cos  α + (n − 1)  2  nβ   sin   = β  2  sin 2

Transformation  Formulae Product into Sum or Difference

1. 2. 3. 4.

2 sin A cos B = sin (A + B) + sin (A – B), A > B 2 cos A sin B = sin (A + B) – sin (A – B), A > B 2 cos A cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos (A + B)

Sum and Difference into Product C+D C−D 1. sin C + sin D = 2 sin    cos    2   2  C+D C−D 2. sin C – sin D = 2 cos    sin    2   2  C+D C−D 3. cos C + cos D = 2 cos    cos    2   2  C+D C−D 4. cos C – cos D = – 2 sin    sin    2   2  5. tan C + tan D = sin (C + D) cos C cos D

Trigonometric Ratios and Identities

12. cos (A + B + C) = cos A cos B cos C – sin A sin B cos C  – sin A cos B sin C – cos A sin B sin C 6. tan C – tan D = or = cos A cos B cos C (1 – tan A tan B 7. cot C + cot D =  – tan B tan C – tan C tan A)

3. tan 2θ =

2 = 1 – 2 sin2θ = 1 − tan θ 1 + tan 2 θ

2 tan θ 1 − tan 2 θ

2 4. cot 2θ = cot θ − 1 2 cot θ

5. 1 + cos 2θ = 2 cos2θ, cos2θ = 1 (1 + cos 2θ) 2 6. 1 – cos 2θ = 2 sin2θ, sin2θ = 1 (1 – cos 2θ) 2

7. sin 3θ = 3 sin θ – 4 sin3θ, sin3θ = 1 (3 sin θ – sin 3θ) 4 3 3 8. cos 3θ = 4 cos θ – 3 cos θ, cos θ = 1 (cos 3θ + 3 cos θ) 4 3 9. tan 3θ = 3 tan θ − tan θ 1 − 3 tan 2 θ

3 10. cot 3θ = cot θ − 3 cot θ 3 cot 2 θ − 1 n 11. cos A cos 2A cos 22A ... cos 2n – 1A = sin 2 A n 2 sin A

Trigonometric Ratios of Submultiple Angles (An Angle of the form

θ , n ∈ I) n

1. sin θ = 2 sin  θ  cos  θ = 2 tan θ / 2 2 2 1 + tan 2 θ / 2 2. cos θ = cos2 θ – sin2 θ = 2 cos2 θ – 1 = 1 – 2 sin2 θ 2 2 2 2 1 − tan 2 θ 

3. tan θ =

= 2 tan θ

2 1 − tan 2 θ

2

1 + tan 2 θ

θ cot 2 − 1 2 4. cot θ = 2 cot θ 2

2 2

796

Objective Mathematics

5. cos2 θ = 1 + cos θ 2 2 1 − cos θ 2 θ 7. tan = 2 1 + cos θ 9. 1 − cos θ = tan θ sin θ 2

6. sin2 θ = 2 2 θ 8. cot = 2 10. 1 + cos θ sin θ

Greatest and Least Values of the Expression

1 − cos θ 2 1 + cos θ 1 − cos θ = cot θ 2

a sin θ + b cos θ Let

a2 + b2 = r2 or r = a 2 + b 2 Then a sin θ + b cos θ = r (sin θ cos α + cos θ sin α) = r sin (θ + α) But – 1 ≤ sin (θ + α) ≤ 1, so – r ≤ r sin (θ + α) ≤ r

Trigonometric Ratios of some special angles 3 −1 2 2 3. tan 15º = 2 – 3 1. sin 15º =

3 +1 2 2 4. cot 15º = 2 + 3 2. cos 15º =

or

10 + 2 5 4

12. cos 36º =

5 +1 4

14. cos 9º

11. sin 36º = 13. sin 9º=

25 − 10 5 5

16. tan 36º =

a 2 + b 2 ≤ a sin θ + b cos θ ≤

a 2 + b2 .

Some  Useful  Identities If A + B + C = π, then

10 − 2 5 4

3+ 5 − 5− 5 4

3+ 5 + 5−5 4

15. tan 18º =

–

Thus, the greatest and least values of a sin θ + b cos θ are a 2 + b 2 and – a 2 + b 2 respectively.

1 5. sin 22 1 = 2 ( 2− 2) 2 o o 1 1 1 = 7. tan  22 = 2 –1 6. cos 22 ( 2 + 2 ) 2 2 2 o 5 −1 8. cot 22 1 = 2 + 1 9. sin 18º = 4 2 o

10. cos 18º =

a = r cos α, b = r sin α, then

(i) (ii) (iii) (iv)

sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C cos 2A + cos 2B + cos 2C = – 1 – 4 cos A cos B cos C tan A + tan B + tan C = tan A tan B tan C cot A cot B + cot B cot C + cot C cot A = 1 A B C A B C (v) cot  + cot  + cot = cot cot cot 2 2 2 2 2 2 A B B C C A (vi) tan   tan  + tan tan + tan tan = 1. 2 2 2 2 2 2

5−2 5

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1 1 − 1. The expression tan 3A − tan A cot 3A − cot A equal to (a) cot 2A (c) cot 3A

is

(b) tan 2A (d) tan 3A

1 then the value of cosec θ + cot θ 2. If cosec θ = x + 4x is (a) 2x 1 (c) 2x

(b) – 2x 1 (d) – 2x

3. If 3 sin θ + 5 cos θ = 5, then the value of 5 sin θ – 3 cos θ is (a) 3 (c) 5

(b) – 3 (d) – 5

5. The value of the expression

tan A + sec A − 1 is tan A − sec A + 1

(a)

1 − sin A cos A

(b)

1 + sin A cos A

(c)

cos A 1 − sin A

(d)

cos A 1 + sin A

6. If tan2θ = 1 – e2, then sec θ + tan3θ cosec θ = (a) (1 – e2)3/2 ( c) (2 – e2)3/2

(b) (2 – e2)1/2 (d) None of these

7. The value of the expression 2 sin2β + 4 cos (α + β) sin α sin β + cos 2 (α + β) is (a) sin α (c) cos α

(b) sin 2α (d) cos 2α

4. If cos (α + β) sin (γ + δ) = cos (α – β) sin (γ – δ), then 8. The value of the expression cot α cot β cot γ = 2 (sin6A + cos6A) – 3 (sin4A + cos4A) + 1 is (a) 0 (b) 1 (a) cot δ (b) – cot δ (c) – 1 (d) None of these (c) tan δ (d) – tan δ

(a) cos2A (c) sin2A 10. If cos θ + sin θ =

(b) cos2B (d) sin2B 2 cos θ, then cos θ – sin θ =

(a)

2 sin θ

1 (b) cos θ 2

(c)

1 sin θ 2

(d) None of these

11. The expression 1 + cos 56º + cos 58º – cos 66º is equal to (a) 4 cos 28º cos 29º sin 33º (b) 4 sin 28º cos 29º cos 33º (c) 4 cos 28º sin 29º cos 33º (d) None of these 4 xy 12. The equation sec2θ = ( x + y ) 2 is possible for real values x and y only if (a) x + y = 2xy (c) x + y = 1

(b) x + y = – 2xy (d) x = y

3 cos 2β − 1 tan α , then = 13. If cos 2α = 3 − cos 2β tan β (a) 1 (c) 2

(b) – 1 (d) – 2

14. The value of tan 1º tan 2º ... tan 89º is (a) 0 (c) – 1 15. If

(b) 1 (d) None of these

cos α sin α = n and = m, then (m2 – n2) sin2β = cos β sin β

(a) n2 (c) n2 – 1

(b) 1 – n2 (d) m2 – 1

16. The value of the expression cos3θ + cos3 (120º + θ) + cos3 (240º + θ) is (a) cos 3θ (c)

3  cos 3θ 4

(b)

1  cos 3θ 4

(d) None of these

17. If x sin3θ + y cos3θ = sin θ cos θ and x sin θ = y cos θ, then x2 + y2 = (a) 1 (c) 0

(b) 2 (d) None of these

3π 5π 7π π + cos4 + cos4 + cos4 18. The value of cos4 8 8 8 8 is (a)

1 2

(b)

3 2

5 (d) None of these 2 19. If sin x + sin2x = 1, then cos8x + 2 cos6x + cos4x =

(a) 1 (c) 3

21. If sin θ + cos θ = m and sec θ + cosec θ = n, then

(b) – 1 (d) 1

(b) 2m = n (m2 – 1) (d) None of these

 22. The value of sin  nπ + (−1) n  (a) 0 (c) –

π , n ∈ I is 4 

(b) 1 2

1 2

(d) None of these

23. If A + B = 45º, then (1 + tan A) (1 + tan B) = (a) 1 (c) 2

(b) – 1 (d) None of these

24. The value of tan A + tan (60º + A) – tan (60º – A) is (a) tan 3A (c) 3 tan 3A 25. The value of

(b) 2 tan 3A (d) None of these cos 6θ + 6 cos 4θ + 15 cos 2θ + 10 is cos 5θ + 5 cos 3θ + 10 cos θ

(a) sin θ (c) cos θ

(b) 2 sin θ (d) 2 cos θ

26. If p sin θ + q cos θ = a and p cos θ – q sin θ = b, then p+a q−b + is equal to q+b p−a (a) 1 (c) 0

(b) 2 (d) None of these n

 cos A + cos B   sin A + sin B  27. The expression  +  sin A − sin B   cos A − cos B 

n

=

A − B if n is even (a) 2 cotn   2  (b) 0 if n is even A − B if n is even (c) 2 cotn   2  (d) 0 if n is odd 28. If m = cosec θ – sin θ and n = sec θ – cos θ, then m2/3 + n2/3 = (a) (mn)–2/3 (c) (mn)–1/3 29. If 3 tan θ tan φ = 1, then

(c)

(a) 0 (c) 2

(b) 2 (d) None of these

(a) 2n = m (n2 – 1) (c) 2n = m (m2 – 1)

(b) (mn)2/3 (d) (mn)1/3 cos (θ − φ) = cos (θ + φ)

(a)

1 2

(b) 2

(c)

1 3

(d) 3

797

π 3π 5π 7π + cos2 + cos2 + cos2 20. The value of cos2 16 16 16 16 is

Trigonometric Ratios and Identities

9. The value of the expression cos2(A – B) + cos2B – 2 cos (A – B) cos A cos B is

798

30. If three angles A, B and C are such that

Objective Mathematics

sin A − sin C cot B = , then A, B and C are in cos C − cos A (a) A.P. (c) H.P.

(b) G.P. (d) None of these π 9π 3π 5π cos + cos + cos is 13 13 13 13

31. The value of 2 cos (a) 1 (c) – 1 32. If tan θ = –

4 , then sin θ is 3

4 4 but not 5 5

(a) – (c)

(b) 0 (d) None of these

4 4 but not – 5 5

33. The value of cos

4 4 or 5 5

(d) None of these

2π 4π 6π + cos + cos is 7 7 7

(a) 0 (c)

(b) –

1 2

1 . 2

34. The value of the sin 20º sin 40º sin 80º is (a) (c)

3 8 1 16

(b)

(a)

sin φ sin θ

(b)

sin θ sin φ

(c)

sin φ 1 − cos θ

(d)

sin θ 1 − cos φ

39. If in a ∆ABC, cos A =

1 8

40. If sin θ = n sin (θ + 2α), then tan (θ + α) = 1− n  tan α 1+ n 1+ n  tan α (c) 1− n

(d) None of these

1 2 (a + b2 + 2) 2 1 2 (a + b2 + 4) (c) 2 (d) None of these

k +1 sin α k −1 k −1 cos α (c) k +1

k −1 sin α k +1

(d) None of these 2π 4π 8π 14π cos cos cos is 15 15 15 15 1 8

1 4

(b)

(c)

1 16

(d) None of these

44. The value of

(b) –

1 4

(d) None of these

37. If α + β = 90º, then the maximum value of sin α sin β is

3 2

(b)

(a)

1 3 1 (c) 2

36. The value of cos273º + cos247º + cos 73º cos 47º is

(c)

(d) None of these

(a)

(a)

(b)

(a) 1

n +1  tan α n −1

cos (α + β) 43. If cot α cot β = 2, then cos (α − β) =

1 2 (a + b2 – 2) 2

1 4 3 (c) 4

(b)

(a)

42. The value of cos

cos (x – y) is equal to

(a)

sin B , then it is 2 sin C

(a) an isosceles triangle (b) an equilateral triangle (c) a right angled triangle (d) None of these

35. If sin x + sin y = a and cos x + cos y = b, then

(a)

x sin φ y sin θ x and tan φ = , then = 1 − x cos φ 1 − y cos θ y

41. If θ + φ = α and tan θ = k tan φ, then sin (θ – φ) =

(b) 1 (d) –

38. If tan θ =

(b)

1 2

(d) None of these

(a) 4 (c) – 2

1 3 1 (d) – 2

(b) –

3 cosec 20º – sec 20º is (b) 12 (d) None of these

45. If 2 tan β + cot β = tan α, then 2 tan (α – β) = (a) tan β (c) tan α 46. sin

π 13π ⋅ sin is equal to 10 10

1 2 1 (c) – 4

(a)

(b) cot β (d) None of these

(b) – (d) 1

1 2

1 (a) 8 1 (c) 32

1 (b) 16 (d) None of these

 2π   2π  − θ  + cos2  + θ 48. The value of cos2θ + cos2   3   3  is (a)

1 2

(b) –

1 2

3 (d) 2

(c) 1

cot (α + β) 49. If b sin β = a sin (2α + β) then = cot α b−a (a) b+a (c)

b+a (b) b−a

a−b a+b

(d) none of these

50. The value of sin 36º sin 72º sin 108º sin 144º is (a)

1 16

(b)

3 16

(c)

5 16

(d) None of these

3π 5π 7π π 51. The value of cos2 8 + cos2 8 + cos2 8 + cos2 8 is (a) 1 (c) – 1 52. The expression

(b) 2 (d) None of these

4 xy is true if and only if ( x + y)2

55. sin2θ =

(a) x + y ≠ 0 (c) x = y 56. If cos (α + β) = 0 and

(b) x = y, x ≠ 0 (d) x ≠ 0, y ≠ 0 4 5 , sin (α – β) = and α1β lie between 5 13

π , then tan 2α = 4

(a)

56 33

(b)

(c)

33 48

(d) None of these

57. If cos θ =

33 56

1 1 1   a +  , then cos 3θ = k  a 3 + 3  , where 2 a  a 

k is equal to 1 4 (c) 1

1 2 (d) None of these

(b)

(a)

58. If sin x + sin y = a and cos x + cos y = b, then tan x− y =  2 

(a) ±

4 + a 2 + b2 a 2 + b2

(b) ±

2 − a 2 − b2 a 2 + b2

(c) ±

4 − a 2 − b2 a 2 + b2

(d) None of these

59. If 2 tan α = 3 tan β, then tan (α – β) =

sec 8A − 1 is equal to sec 4A − 1

(a)

sin 2β 5 − cos 2β

(b)

(c)

sin 2β 5 + cos 2β

(d) None of these

(a)

tan 8A tan 2A

(b)

tan 2A tan 8A

(c)

tan 4A tan 2A

(d)

tan 2A tan 4A

53. The value of 3π   7π   9π  π   1 + cos  1 + cos  1 + cos  1 + cos  is 8 8 8 8      1 8

(a)

1 4

(b)

(c)

1 16

(d) None of these

54. The value of 2 sin2θ + 4 cos (θ + α) sin α sin θ + cos 2 (α + θ) is (a) cos θ + cos α (c) independent of α

(b) independent of θ (d) None of these

cos 2β 5 − cos 2β

60. If sin x + cos x = a, then sin6x + cos6x = (a)

1 [4 + 3 (a2 – 1)2] 4

(b)

1 [4 – 3 (a2 – 1)2] 4

1 [4 – 3 (a2 + 1)2] 4 (d) None of these (c)

61. If tan2θ = 2 tan2φ + 1, then cos 2θ + sin2φ = (a) 0 (c) 2

(b) 1 (d) None of these

62. If tan A – tan B = x and cot B – cot A = y, then cot (A – B) =

799

π  3π   3π   9π   1 + cos  1 + cos  1 + cos  1 + cos  is  10   10   10   10 

Trigonometric Ratios and Identities

47. The value of

800

(a)

Objective Mathematics

(c)

1 1 − x y

(b)

70. The value of 1 + cos 56º + cos 58º – cos 66º is

1 1 + x y



1 1 − y x

(d) None of these

(a) 4 cos 28º cos 29º sin 33º (b) 4 sin 28º cos 29º cos 33º (c) 4 cos 28º sin 29º cos 33º (d) None of these

3 cos 2β − 1 , 71. If A + B + C = π, then 63. If α and β be acute angles and cos 2α = 3 − cos 2β then tan α = cos A cos B cos C = + + (a) tan β (b) 2 tan β sin B sin C sin A sin C sin A sin B (c) tan β (d) None of these (a) 1 (b) 2 64. If α and β are the solutions of a cos θ + b sin θ = c, (c) – 1 (d) – 2. then 72. If sin (y + z – x), sin (z + x – y) and sin (x + y – z) are 2bc (a) sin α + sin β = 2 in A.P. then tan x, tan y and tan z are in 2 a +b (a) A.P. (b) G.P. c2 − a2 (c) H.P. (d) None of these (b) sin α sin β = 2 a + b2 73. If A + B + C = 180º, then Σ cot A cot B = (c) sin α + sin β =

2ac b + c2

(d) sin α ⋅ sin β =

a 2 − b2 b2 + c2

(a) 1 (c) 2

2

74. If A + B + C = 180º and tan 3A + tan 3B + tan 3C = k tan 3A tan 3B tan 3C then k =

65. The value of sin 47º + sin 61º – sin 11º – sin 25º is (a) sin 7º (c) sin 14º 66. If cos θ = θ is 2

(b) cos 7º (d) cos 14º

(a) tan

α β cot 2 2

(b) tan

(c) cot

α β ⋅ tan 2 2

(d) None of these

α β tan 2 2

1 3 is − sin 10º cos10º

(a) 1 (c) 4

(a) 1 (c) 2

A

B

=

(b) – 1 (d) None of these

69. If an angle θ be divided into two parts such that the tangent of one part is m times the tangent of the other, then their difference φ is given by

(c) sin φ =

m −1  cos θ m +1

(d) cos φ =

m −1 sin θ m +1

(cot B + cot C) ⋅ (cot C + cot A) ⋅ (cot A + cot B) = (a) sin A sin B sin C (b) cos A cos B cos C (c) sec A sec B sec C (d) cosec A cosec B cosec C 76. The smallest positive value of x for which tan (x + 100º) = tan (x + 50º) tan x tan (x – 50º) is (a) 30º (c) 55º

(b) 45º (d) None of these n

∑c

m=0

∑ tan 2 tan 2

m −1 (a) cos φ =  cos θ m +1 m −1 sin θ (b) sin φ = m +1

(b) 2 (d) None of these

77. Suppose that sin3x sin 3x =

(b) 2 (d) None of these

68. If A + B + C = π, then

(a) 1 (c) 4 75. If A + B + C = π, then

cos α − cos β , then one of the values of tan 1 − cos α cos β

67. The value of

(b) – 1 (d) None of these

m

cos mx is an identity in

x, where c0, c1, c2, ..., cn are constants and cn ≠ 0, then the value of n is (a) 4 (c) 6

(b) 5 (d) None of these

78. The value of cos 20º – sin 20º is (a) positive (c) 0

(b) negative (d) 1

79. The value of cos 1º cos 2º cos 3º ... cos 179º is 1 2 (c) 1 (a)

(b) 0 (d) None of these

80. Which of the following is correct ? (a) sin 1º > sin 1 (b) sin 1º < sin 1 (c) sin 1º = sin 1 π (d) sin 1º = sin 1 180

82. If sin (α + β ) = 1, sin (α – β ) =

(a) 2 (c) 3

(b) 1/2 (d) 4 cosα cosβ cosγ

π < α < π, then the expression 84. Given 2

2 cos α

95. If k = sin of k is

85. If the angle A of a triangle ABC is given by the equation 5 cos A + 3 = 0, then sin A and tan A are the roots of the equation (b) 15x – 8 2 x + 16 = 0 (d) 15x2 + 8x – 16 = 0 2

(a) 15x – 8x – 16 = 0 (c) 15x2 – 8x + 16 = 0

86. If cosec θ – sin θ = a and sec θ – cos θ = b , then a2b2 (a2 + b2) = 3

(a) 1 (c) 2 87.

3

(b) – 1 (d) None of these

sec 2 θ − tan θ lies between sec 2 θ + tan θ 1 and 2 2 1 and 4 (c) 4

(a)

(d) None of these

1 and 2 4 (d) None of these

(b)

π 5π 7π , then the numerical value ⋅ sin ⋅ sin 18 18 18

(a)

1 8

(b)

(c)

1 2

(d) 1

1 4

96. Let α, β be such that π < α – β < 3π. If sin α + sin β = – 21/65 and cos α + cosβ = – 27/65, then the value α−β is of cos 2 6 3 (a) (b) 65 130 (c) –

1 (b) and 3 3

1 2

(d) None of these

1 and 1 4 (c) 0 and 1

(d) None of these

2

(b) 8

(a)

(b) –

2 (c) cos α

1 2 1 (c) 9 2 (a) 7

94. sin6x + cos6x lies between

1 − sin α 1 + sin α + = 1 + sin α 1 − sin α 1 cos α

(b) 1 (d) 4

93. The value of sin25º + sin210º + sin215º + ... + sin290º is

83. cosα sin (β – γ) + cosβ sin (γ – α) + cosγ sin (α – β ) =

(a)

(b) G.P. (d) None of these

92. If sin x + sin2x + sin3x = 1, then cos6x – 4 cos4x + 8 cos2x =

1 then 2

(b) –1 (d) None of these

(a) 0 (c) 1

cos (A + C) , then tan A, tan B, tan C are in cos (A − C)

(a) A.P. (c) H.P.

tan (α + 2β ) tan (2α + β) is equal to (a) 1 (c) zero

91. If cos 2B =

3 130

(d)

−6 65

97. The value of 2 sin 2 β + 4 cos (α + β) sin α sin β + cos 2 (α + β) depends

(a) only on α (b) only on β holds for 88. The inequality 2sin θ + 2cos θ ≥ 2 (c) independent of α and β (d) None of these (a) 0 ≤ θ < π (b) π ≤ θ < 2π (c) for all real θ (d) None of these π , then 98. If e–π/2 < θ < 89. If c cos3θ + 3c cos θ sin2θ = m and c sin3θ + 3c cos2θ sin 2 2/3 2/3 θ = n then (m + n) + (m – n) = (a) cos log θ < log cos θ 2/3 2/3 (b) cos log θ > log cos θ (a) c (b) 2c (c) cos log θ ≤ log cos θ (c) 4c2/3 (d) None of these (d) None of these tan x 90. The value of wherever defined never lies be- 99. If u = a 2 cos 2 θ + b 2 sin 2 θ + tan 3 x tween a 2 sin 2 θ + b 2 cos 2 θ then the difference between the 1 1 and 2 (b) and 3 (a) maximum and minimum values of u2 is given by 3 2 (a) (a + b)2 (b) 2 1 and 4 (d) None of these (d) 2 2 (c) 2(a + b ) (d) (a – b)2 4  1  1−   2

801

(a) tan 3A tan 2A tan A (b) – tan 3A tan 2A tan A (c) tan A tan 2A – tan 2A tan 3A – tan 3A tan A (d) None of these

Trigonometric Ratios and Identities

81. The value of tan 3A – tan 2A – tan A is equal to

802

100. Let 0 < x ≤

π , then (sec 2x – tan 2x) equals 4

Objective Mathematics

π  (a) tan2  x +   4

π  (b) tan   x +   4

π (c) tan   − x  4 

π  (d) tan   x −   4

101. In a ∆ABC, ∠A =

π , then cos2B + cos2C equals 2

(a) – 2 (c) 1

(b) – 1 (d) zero

102. If y = 4 sin2θ – cos 2θ, then y lies in the interval

(a) (– 1, 5) (b) [– 1, 5] (c) (– ∞, – 1) ∪ (5, ∞) (d) None of these

103. If x +

(b) cos nθ (d) 2 cos nθ, n ∈ Z+

   3π  – 2 sin 6  + α  + sin 6 (5π − α) is equal to   2   (b) sin 4α + cos 6α (d) None of these

105. sin 47º + sin 61º – sin 11º – sin 25º is equal to (a) cos 7º (c) cos 36º

(b) sin 7º (d) sin 36º

106. The minimum value of the expression sin α + sin β + sin γ, when α, β, γ are real numbers satisfying α + β + γ = π is (a) – 3 (c) positive

(b) negative (d) zero.

107. If sin α and cos α are the roots of the equation px2 + qx + r = 0, then (a) p2 + q2 – 2 pr = 0 (c) p2 – q2 + 2pr = 0

(b) (p – r)2 = q2 + r2 (d) (p + r)2 = q2 – r2.

108. If tan θ = n tan φ (n > 0), then (n − 1) 2 4n ( n − 1) 2 (b) tan2 (θ – φ) ≥ 4n (a) tan2 (θ – φ) >

(n − 1) 2 (c) tan  (θ – φ) ≤ 4n (d) None of these 2

109. If θ lies in the third quadrant, then the expression

π π (a)  ,  3 2

 π 2π  (b)  ,  2 3 

 2π 5π  , (c)    3 6 

 5π  (d)  , π   6 

111. The value of cot 5º cot 10º ... cot 85º is (a) 1

(b)

1 2

1 2

(d) 0

112. The greatest value of cos (xe| x | + 7x2 – 3x), x ∈ [– 1, ∞) is

   3π  104. The expression 3 sin 4  − α  + sin 4 (3π + α)  2   

(a) 1 (c) 3

(b) 2 (d) None of these

110. If θ and φ are acute angles satisfying sin θ = 1/2, cos φ = 1/3, then θ + φε

(c)

1 1 = 2 cos θ then xn + n is equal to x x

(a) 2 sin nθ (c) sin nθ

(a) 1 (c) 4

 π θ (4 sin 4 θ + sin 2 2θ) + 4 cos2  −  is equal to  4 2

(a) 0 (c) – 1

(b) 1 (d) None of these

113. The value of tan 56º – tan 11º – tan 56º tan 11º is (a) 0 (c) – 1

(b) 1 (d) None of these

114. In a ∆ABC, if angle C is obtuse, then (a) tan A tan B < 1 (c) tan A tan B > 1

(b) tan A tan B ≤ 1 (d) None of these

115. The greatest value of sin x cos x is (a) – 1

(b) 2

1 2

(d) 1

(c)

116. log tan 1º + log tan 2º + ... + log tan 89º = (a) 1 (c)

(b) 0

π 4

(d) None of these

π 5π π , x and tan are in A.P. and tan , y and 9 18 9 7π are also in A.P., then tan 18

117. If tan

(a) 2x = y (c) x = y

(b) x > y (d) None of these

118. If sin (x – y), sin x and sin (x + y) are in H.P. then sin x ⋅ sec (a) 2 (c) 1

y is equal to 2 (b) 2 (d) None of these

119. If cos (θ – φ), cos θ and cos (θ + φ) are in harmonic φ progression then cos θ sec = 2 (a) 2 (b) 2 (c) – 2 (d) None of these

(a) 0 (c) 2

(b) 1 (d) 4

121. If cosec θ + cot θ = (a)

21 22

(c)

44 117

129. If x = y cos

11 , then tan θ = 2 15 (b) 16 (d)

117 44

122. If A = cos2θ + sin4θ, then for all values of θ (a) 1 ≤ A ≤ 2 (c)

3 13 ≤A≤ 4 16

(b)

13 ≤ A ≤1 16

(d)

3 ≤ A ≤1 4

2a θ a (1 − θ2 ) , where a is a 2 , y = 1+ θ 1 + θ2 constant, is the parametric equation of the curve :

123. The equation x = (a) x2 – y2 = a2 (c) x2 + 4y2 = 4a2 124. If tan A =

(b) x2 + y2 = a2 (d) x = 2y.

a 1 and tan B = , then the value of a +1 2a + 1

A + B is (a) 0 (c)

π 3

(b)

π 2

(d)

π 4

125. If x1, x2, x3, ..., xn are in A.P. whose common difference is α, then the value of sin α [sec x1 sec x2 + sec x2 sec x3 + ... + sec xn – 1 sec xn] is equal to (a)

sin nα cos x1 cos xn

(b)

(c)

sin (n + 1) α cos x1 cos xn

(d) None of these

126. Minimum value of 1 (a) 12 (c)

1 is 3 sin θ − 4 cos θ + 7 5 (b) 12

7 12

127. If n =

sin (n − 1) α cos x1 cos xn

(d)

1 6

, then tan α ⋅ tan 2α ⋅ tan 3α ... tan (2n – 1)

α is equal to (a) 1 (c) ∞

(b) – 1 (d) None of these

128. If x = r sin θ cos φ, y = r sin θ sin φ and z = r cos θ, then the value of x2 + y2 + z2 is independent of

(a) – 1 (c) 1

2π 4π = z cos , then xy + yz + zx = 3 3 (b) 0 (d) 2

130. If cos α + cos β = 0 = sin α + sin β, then cos 2α + cos 2β = (a) – 2 sin (α + β) (c) 2 sin (α + β)

(b) – 2 cos (α + β) (d) 2 cos (α + β)

131. The ratio of the greatest value of 2 – cos x + sin2x to its least value is (a)

1 4

(b)

9 4

(c)

13 4

(d) None of these

132. The expression 2sin θ + 2– cos θ is minimum when θ is equal to π 7π (b) 2nπ + ,n∈I (a) 2nπ + , n ∈ I 4 4 π (d) None of these (c) nπ ± , n ∈ I 4 133. If (sec A + tan A) (sec B + tan B) (sec C + tan C) = (sec A – tan A) (sec B – tan B) (sec C – tan C), then each side is equal to (a) 1 (c) 0

(b) – 1 (d) None of these

134. In a triangle tan A + tan B + tan C = 6 and tan A tan B = 2, then the values of tan A, tan B and tan C are (a) 1, 2, 3 (c) 1, 2, 0

(b) 2, 1, 3 (d) None of these

135. Which of the following is a rational number ? (a) sin 15º (c) sin 15º cos 15º

(b) cos 15º (d) sin 15º cos 75º

1 , then the quadratic equation whose roots 3 θ θ and cot is are tan 2 2

136. If sin θ =

(a) x2 – 6x + 1 = 0 (c) x2 – 6x – 1 = 0

(b) x2 + 6x + 1 = 0 (d) x2 + 6x – 1 = 0

137. Let f (θ) = sin θ (sin θ + sin 3θ). Then f (θ) (a) ≥ 0 only when θ ≥ 0 (b) ≤ 0 for all real θ (c) ≥ 0 for all real θ (d) ≤ 0 only when θ ≤ 0 π and β + γ = α then tan α equals 2 (a) 2 (tan β + tan γ) (b) tan β + tan γ (c) tan β + 2 tan γ (d) 2 tan β + tan γ

138. If α + β =

139. The maximum value of cos α1 ⋅ cos α2 ... cos αn , under π and cot α1 ⋅ cot the restrictions 0 ≤ α1, α2, ... αn ≤ 2 α2 ... cot αn = 1 is

803

(b) r, θ (d) r

(a) θ, φ (c) r, φ

Trigonometric Ratios and Identities

120. Find the number of real solutions of 2sin (ex) = 5x + 5–x in [0,1]

804

(a)

1 2n/ 2

(b)

Objective Mathematics

1 2n

(c)

1 2n

(d) 1

π and cos A + cos B = 1 then cos (A – B) 140. If A + B = 3 is (a) –

1 3

1 (c) 4

(b)

(c)

1 6

1 (d) – 4

(b) 1

(b) un + 1 (d) None of these

143. The value of tan 9º – tan 27º – tan 63º + tan 81º is (a) 4 (c) 2

(b) 3 (d) 1

the interval 1  (a)  , 1 2 

 1  (b)  ,1  2 

1  (c)  − , 1  2 

(d) (0, 1)

145. If sin6θ + cos6θ + k cos2 2θ = 1, then k is equal to 1 tan22θ 2

(c) 4 cot22θ

(b) (d)

1 tan22θ 4 3 tan22θ 4

146. If sin θ + cosec θ = 2, then sin2θ + cosec2θ is equal to (a) 1 (c) 2 147. If cosec θ – cot θ = equal to 5 3 3 (c) – 5 (a)

(b) 4 (d) None of these 1 π , 0 < θ < , then cos θ is 2 2 (b)

3 5

(d) –

5 3

148. For m ≠ n, if tan mθ = tan nθ, then the different values of θ are in (a) A.P. (c) G.P.

1 3

(b)

3

(c)

2 3

(d)

3 . 2

3 tan 20º tan 40º is equal to

(a)

3 4

(b)

3 2

(c)

3

(d) 1

151. If sin (A + B + C) = 1, tan (A – B) =

1 and 3

(a) A = 90º, B = 60º, C = 30º (b) A = 120º, B = 60º, C = 0º (c) A = 60º, B = 30º, C = 0º (d) None of these 152. sin x +

3 cos x is maximum when

(a) x = 60º (c) x = 30º

(b) x = 45º (d) x = 0º

153. Which of the following is correct ?

π  π 144. If θ ∈  0,  then the value of cos  θ −  lies in  4  2

(a)

(a)

sec (A + C) = 2, then (d) None of these

142. If un = 2 cos nθ then u1 un – un – 1 is equal to (a) un – 2 (c) 0

1 + tan 2 θ is equal to 2 tan θ

150. tan 20º + tan 40º +

1 3

141. If in a ∆ABC, tan A + tan B + tan C has the value 6 then the value of cot A ⋅ cot B ⋅ cot C is equal to (a) 6

149. If θ = 60º, then

(b) H.P. (d) None of these

(a) tan 1 = tan 2 2 tan 2 (b) tan 1 = 3 (c) tan 1 > tan 2 (d) tan 1 < sin 2 154. sin 200º + cos 200º is (a) positive (c) zero

(b) negative (d) zero or positive

155. If xy + yz + zx = 1, then (a)

4 xyz

(c) xyz

x+ y

∑ 1 − xy

(b)

=

1 xyz

(d) None of these

1 3 , cos B = , where A, B are positive 5 10 acute angle then A + B is equal to

156. If sin A = π 2 π (c) 4 (a)

π 3 π (d) 6 (b)

157. If 3 sin 2θ = 2 sin 3θ and 0 < θ < π, then value of sin θ is (a)

2 5

(b)

15 4

(c)

3 5

(d)

2 3

(a) 8 cos x (c) 8 sin x

(b) cos x (d) sin x

159. If sin x + sin2x = 1, then the value of cos12x + 3 cos10x + 3cos8x + cos6x – 1 is equal to (a) 2 (c) 0

(b) 1 (d) – 1

160. If sin x + cos x = (a) – (c)

4 3 or – 3 4

4 5

1 , 0 ≤ x ≤ π, then tan x is equal to 5 4 (b) 3 (d) None of these

161. A and B are positive acute angles satisfying the equa3 sin A 2 cos B , then tions 3 cos2A + 2 cos2B = 4 and = sin B cos A A + 2B is equal to π 3 π (c) 6 (a)

(b)

π 2

(d)

π 4

1 2 1 (c) 2 2

163. If tan θ + sec θ = 5π 6 π (c) 6

(a)

(b) (d)

2

3 , 0 < θ < π, then θ is equal to 2π 3 π (d) 3

(b)

(b) sin 36º, cos 18º (d) sin 36º, sin 18º

165. If sin (π cos θ) = cos (π sin θ), then the value of π  cos  θ +   4 2 2 1 (c) – 2 2 (a)

1 2 1 (d) 2 2

(b)

166. If α and β be between 0 and

56 65 16 (d) 15

(b)

(c) 0 167. The expression to

cos 6 x + 6 cos 4 x + 15 cos 2 x + 10 is equal cos 5 x + 5 cos 3 x + 10 cos x

(a) cos2x (c) cos 2x 168. If sin2 θ =

(b) 1 + cos x (d) 2 cos x x2 + y 2 + 1 , then x must be 2x

(a) – 3 (c) 1

(b) –2 (d) None of these

169. If an angle θ is divided into two parts A and B such that A – B = K and tan A : tan B = K : 1, then the value of sin K is (a)

K +1 sin θ K −1

(b)

K sin θ K +1

(c)

K −1 sin θ K +1

(d) none of these

cos θ3 = (a) 0 (c) 2

(b) 1 (d) 3

171. The value of cos2θ + sec2θ is always (a) equal to 1 (b) less than 1 (c) greater than or equal to 2 (d) greater than 1 but less than 2

1 2

164. The roots of the equation 4x2 – 2 5 x + 1 = 0 are (a) cos 18º, cos 36º (c) sin 18º, cos 36º

64 65

170. If sin θ1 + sin θ2 + sin θ3 = 3, then cos θ1 + cos θ2 +

3π 162. If tan (π cos θ) = cot (π sin θ), 0 < θ < , then 4 π sin  θ +  equals  4 (a)

(a)

805

x x x x cos cos cos is equal to 8 8 2 4

π and if cos (α + β) = 2

12 3 and sin (α – β) = , then sin 2α is equal to 13 5

172. If A + B + C = 180º, then sec A (cos B cos C – sin B sin C) is equal to (a) 0 (c) 1 173. If

(b) – 1 (d) None of these

3π < α < π, then 4

(a) 1 – cot α (c) – 1 + cot α 174. Let 0 < x
t2 > t3 > t4 (c) t3 > t1 > t2 > t4

(d)

(1 − 7) 4

(b) t4 > t3 > t1 > t2 (d) t2 > t3 > t1 > t4

θ + α θ− α 190. If cos θ = cos α cos β, then tan  tan   2   2  is equal to (a) tan 2

α 2

(b) tan 2

β 2

θ β (d) cot 2 2 2 1 191. If tan θ = – 1 and cos θ = , then θ is (n ∈ I) 2 π π (b) 2nπ ± (a) nπ − 4 4 7π (c) 2nπ + (d) none of these 4 a 192. If tan x = , then the value of a cos 2x + b sin 2x is b (a) a (b) a – b (c) a + b (d) b 193. If x +

1 8

184. sin 163º sin 347º +  sin 73º sin 167º = (a) 0

(a)

(c) tan 2

π 2π 4π cos cos = 7 7 7

(a) 0

1 , then tan x is 2 (4 + 7) (b) − 3

188. If 0 < x < π and cos x + sin x =

(cot θ)tan θ and t4 = (cot θ)cot θ, then

181. The value of cos 52º + cos 68º + cos 172º is

(a) –

(b) 210 (d) 10

(a) 2 (c) 2a

 π 189. Let θ ∈  0,  at t1 = (tan θ)tan θ, t2 = (tan θ)cot θ, t3 =  4

3 and tan A tan B = 2, then 5

1 5 2 (c) sin A sin B = – 5 (a) cos A cos B =

(b) y ≤ 1 (d) 1 < y < 2

187. If sin θ + cosec θ = 2, the value of sin10 θ + cosec10 θ is

b , then the value of a cos 2x + b sin 2x is a

182. If sin α = –

(d) 150 cm

186. If y = cos2 x + sec2 x, then

(a)

178. If tan x =

(b) 770 cm

1 2

(a) 2 sin nθ (c) sin (2nθ) 194. If cos θ =

(b) 2 cos nθ (d) cos (2nθ)

8 and θ lies in the Ist quadrant, then the value 17

of cos (30º + θ) + cos (45º – θ) + cos (120º – θ) is (a)

23  3 − 1 1  +  17  2 2

(b)

23  3 + 1 1  +  17  2 2

(c)

23  3 − 1 1  −  17  2 2

(d)

23  3 + 1 1  −  17  2 2

(d) None of these

185. An arc of a circle of radius 77 cm subtends an angle of 10º at the centre. The length of the arc is

1 1 = cos θ, then x n + n is equal to x x

2 tan β

(a) (c)

(b)

tan 2 β 2

a2

(c)

tan β 2

(a + b )

2

b2 a + (1 − b) 2

(d)

2

(d) tan β

196. Given tan A and tan B are the roots of x2 – ax + b = 0. The value of sin2(A + B) is

π  197. If sin 4A – cos 2A = cos 4A – sin 2A,  0 < A <  , 4  then the value of tan 4A is

(a)

a2 a + (1 − b) 2

(a) 1

(b)

a2 a 2 + b2

(c)

2

807

3cos 2β − 1 , then tan α = … 3 − cos 2β

(b) 3

(d)

1 3 3 −1 3 +1

Solutions 1. (a) We have, 1 1 − tan 3A − tan A cot 3A − cot A 

⇒

cos A cos 3A sin 3A sin A − = sin 3A cos A − cos 3A sin A cos 3A sin A − sin 3A cos A =

⇒

cos (3A − A) = = cot 2A. sin 2A On adding, we get 1 2 cosec θ = k + k

1 k

Let 5 sin θ – 3 cos θ = x

...(1)

(9 sin2θ + 25 cos2θ + 30 sin θ cos θ) + (25 sin2θ + 9 cos2θ – 30 sin θ cos θ) = 25 + x2 ⇒ 9 (sin2θ + cos2θ) + 25 (sin2θ + cos2θ) = 25 + x2 ⇒ 34 = 25 + x2 ⇒ x2 = 9 ∴ x = ± 3, i.e., 5 sin θ – 3 cos θ = ± 3. 4. (a) We have, cos (α + β) sin (γ + δ) = cos (α – β) sin (γ – δ) ⇒

sin ( γ + δ) cos (α − β) = sin ( γ − δ) cos (α + β)

tan A + sec A − (sec 2 A − tan 2 A) tan A − sec A + 1 [ sec2A = 1 + tan2A]

=

(tan A + sec A) − (sec A − tan A) (sec A + tan A) tan A − sec A + 1

=

(tan A + sec A)(1 − sec A + tan A) (1 − sec A + tan A)

...(2)

Squaring and adding (1) and (2), we get

tan A + sec A − 1 tan A − sec A + 1

5. (b) We have,



⇒ 2x +

2 sin γ cos δ 2 cos α cos β = 2 cos γ sin δ 2 sin α sin β

⇒ cot δ = cot α cot β cot γ.

=

1 1   or 2   x +  = k + k   4x 1 1 =k+ ⇒ k = 2x. k 2x 3. (a), (b)  We have, 3 sin θ + 5 cos θ = 5

cos (α − β) + cos (α + β) cos (α − β) − cos (α + β) [Applying componendo and dividendo]

cos A cos 3A sin 3A sin A + sin (3A − A) sin (3A − A)

2. (a) Let cosec θ + cot θ = k ⇒ cosec θ – cot θ =

=

sin ( γ + δ) + sin ( γ − δ) sin ( γ + δ) − sin ( γ + δ)

= tan A + sec A =

1 + sin A sin A 1 + = . cos A cos A cos A

6. (c) We have, sec θ + tan3θ cosec θ cosec θ   3 = sec θ 1 + tan θ  sec θ   = sec θ (1 + tan3θ ⋅ cot θ) = sec θ (1 + tan2θ) = sec θ sec2θ = sec3θ 3

3

= (sec 2 θ) 2 = (1 + tan 2 θ) 2 3

3

= (1 + 1 − e 2 ) 2 = (2 − e 2 ) 2 

[ tan2θ = 1 – e2]

Trigonometric Ratios and Identities

195. cos 2α =

808

7. (d) We have,

= 2 cos 28º (2 cos 57º cos 29º)

2 sin β + 4 cos (α + β) sin α sin β + cos 2 (α + β) 2

Objective Mathematics

= (1 – cos 2β) + 4 cos (α + β) sin α sin β + 2 cos2 (α + β) – 1 = – cos 2β + 2 cos (α + β) [2 sin α sin β + cos (α + β)] = – cos 2β + 2 cos (α + β) [2 sin α sin β + cos α cos β – sin α sin β] = – cos 2β + 2 cos (α + β) cos (α – β) = – cos 2β + (cos 2α + cos 2β) = cos 2α. 8. (a) sin6A + cos6A = (sin2A)3 + (cos2A)3 a3 + b3 = (a + b) (a2 + b2 – ab) ∴ sin6A + cos6A = (sin2A + cos2A) (sin4A + cos4A – sin2A cos2A) = sin4A + cos4A – sin2A cos2A ( sin2A + cos2A = 1) ∴ 2 (sin6A + cos6A) – 3 (sin4A + cos4A) + 1 = 2 (sin4A + cos4A) – 2 sin2A cos2A – 3 (sin4A + cos4A) + 1 4 4 2 = – sin A – cos A – 2 sin A cos2A + 1 = – (sin4A + cos4A + 2 sin2A cos2A) + 1 = – (sin2A + cos2A)2 + 1 = – (1)2 + 1 = – 1 + 1 = 0.

= 4 cos 28º cos 29º sin 33º [ cos 57º = cos (90º – 33º) = sin 33º] 12. (d) We have, sec2θ =

sec2θ ≥ 1

⇒

4 xy ≥1 ( x + y)2

⇒ (x + y)2 ≤ 4xy ⇒ (x + y)2 – 4xy ≤ 0 ⇒ (x – y)2 ≤ 0 But for real values of x and y, (x – y)2 sin θ for 0 ≤ θ < 45º ∴ cos 20º – sin 20º > 0. 79. (b) One of the factors of the given expression is cos 90º which has the value 0. 80. (b) Since 1 radian = 57º 30′ approx. and

sin 57º 30′ > sin 1º, ∴ sin 1º < sin 1.

3A = 2A + A

⇒ tan 3A = tan (2A + A)

(tan 2A + tan A) = (1 − tan 2A tan A)

π 2

1 π ⇒α– β= 2 6 π π ∴ α = ,β= 3 6 2π 5π So, tan (α + 2β) tan (2α + β) = tan tan 3 6 π  π  =  − cot   − cot  = 1.  6  3

83.

84.

85.

86.

sin (α – β) =

a (a 2 + b 2 )

.

Substituting in (1), we get

⇒ tan 3A – tan 3A tan 2A tan A = tan 2A + tan A ⇒ tan 3A – tan 2A – tan A = tan 3A tan 2A tan A. 82. (a) sin (α + β) = 1 ⇒ α + β =

817



b3 b 3 . 3 = tan θ or tan θ = a a b ∴ sin θ = and cos θ = (a 2 + b 2 ) ∴



a3 =

⇒ a3b

(a 2 + b 2 ) b − 2 b (a + b 2 ) ( a 2 + b 2 ) = a2 + b2 – b2 = a2

⇒ a2b2 (a2 + b2) = 1. 87. (b) Let y =

sec 2 θ − tan θ sec 2 θ + tan θ

1 + tan 2 θ − tan θ 1 + tan 2 θ + tan θ ⇒ ( y – 1) tan2θ + ( y + 1) tan θ + ( y – 1) = 0. For tan θ to be real, discriminant ≥ 0 i.e., ( y + 1)2 – 4 (y – 1) (y – 1) ≥ 0 (a) cos α sin (β – γ) + cos β sin (γ – α) + cos γ sin ⇒ ( y + 1)2 – (2y – 2)2 ≥ 0 (α – β) ⇒ ( y + 1 + 2y – 2) ( y + 1 – 2y + 2) ≥ 0. 1 1 1 ⇒ (3y – 1) (3 – y) ≥ 0 = 2 sin 0 + 2 sin 0 + 2 sin 0 1  [Put α = β = γ = 60°] ⇒ (3y – 1) ( y – 3) ≤ 0 ⇒ ≤ y ≤ 3. 3 =0+0+0= 0 1 [2sin θ + 2cos θ] ≥ 2sin θ ⋅ 2cos θ 88. (c) We have, (1 − sin α) + (1 + sin α) 2 [ ∵ A.M. ≥ G.M.] (b) The given expression = [(1 + sin α)(1 − sin α)] ⇒ 2sin θ + 2cos θ ≥ 2. 2(sin θ + cos θ)/2 ...(1) Now (sin θ + cos θ) 2 2 2 π = (cos 2 α) 2 = | cos α | = –  cos α  = 2 sin  θ +  ≥ – 2 for all real θ. 4  π . sin θ cos θ (sin ∴ 2 +2 ≥ 2 ⋅ 2 θ + cos θ)/2 ∵ 2 < α < π ⇒ cos α < 0 ⇒ | cos α |= − cos α  1−(1/ 2 ) > 2 ⋅ 2− 2 / 2 = 2 (d) We are given 5 cos A + 3 = 0 1−(1/ 2 ) Thus, 2sin θ + 2cos θ ≥ 2 , for all real θ. 3 ⇒ cos A = – . 89. (b) From the given relations, we have 5 Now, A being the angle of ∆ABC, sin A cannot m + n = c (cos3θ + sin3θ + 3 cos θ sin2θ + 3 cos2θ sin θ) be negative. ⇒ m + n = c (cos θ + sin θ)3 3 So, cos A = – ⇒ sin A ∴ (m + n)2/3 = c2/3 (cos θ + sin θ)2 ...(1) 5 4 4 Also, m – n and then tan A = – .  = 5 3  = (cos3θ – sin3θ + 3 cos θ sin2θ – 3 cos2θ sin θ) 8 ∴ sin A + tan A = – ⇒ m – n = c (cos θ – sin θ)3 15 ∴ (m – n)2/3 = c2/3 (cos θ – sin θ)2 ...(2) 16 and sin A tan A = – Adding (1) and (2), we get 15 (m + n)2/3 + (m – n)2/3 = 2 c2/3. Hence, sin A and tan A are the roots of the equation tan x 8  16  x2 +   x –   = 0 or 15x2 + 8x – 16 = 0. 90. (b) Let y = tan 3 x 15 15 tan x 1 − 3 tan 2 x = (a) We have, ⇒ y = 3 3 − tan 2 x  3 tan x − tan x  2   2 1 cos θ  1 − 3 tan x  a3 = – sin θ =  ...(1) sin θ sin θ ⇒ 3y – y tan2x = 1 – 3 tan2x 2 1 sin θ ⇒ (y – 3) tan2x + (1 – 3y) = 0. b3 = – cos θ =  ...(2) cos θ cos θ ⇒ y =

Trigonometric Ratios and Identities

81. (a) We have,

818

Objective Mathematics

3y −1 y −3

⇒ tan2x =

For tan x to be real

3 = sin 10º  − sin 2 10º  4  3y −1 ≥0 y −3

= 1 (3 sin 10º – 4 sin3 10º) 4 1 sin (3 ⋅ 10º) = 1 sin 30º = 1 . = 8 4 4

⇒ (3y – 1) ( y – 3) ≥ 0 and y ≠ 3 ⇒ y ≤ 1 or y > 3. 3 cos 2B cos (A + C) 91. (b) We have, . = 1 cos (A − C)

−21   27  96. (c)  = (sin α + sin β)2 + −  65   65  + (cos α + cosβ)2 1170 α−β ⇒ = 2 + 2cos (α – β) = 4 cos2 4225 2 1170 1170 9 α − β ⇒ = –  = = (130) 2 130 4 × (65) 2 2 3 ⇒ cos α − β = –   [ π < α – β < 3π]. 130 2 2

Applying componendo and dividendo, we get

1 − cos 2B cos (A − C) − cos (A + C) = 1 + cos 2B cos (A − C) + cos (A + C)

⇒

2 sin 2 B 2 sin A sin C = 2 cos 2 B 2 cos A cos C

⇒ tan2B = tan A tan C

97. (a) The given expression

⇒ tan A, tan B, tan C are in G.P.

= 2 sin2β + 4 cos (α + β) sin α sin β + 2 cos2 (α + β) – 1 2 = 2 sin β + 2 cos (α + β) [2 sin α sin β + cos (α + β)] – 1 = 2 sin2β + 2 cos (α + β) [cos α cos β – sin α sin β + 2 sin α sin β] – 1 = 2 sin2β + 2 cos (α + β) cos (α – β) – 1 = 2 sin2β + 2 (cos2α – sin2β) – 1 = 2 cos2α – 1 = cos 2α.

92. (d) Given : sin x + sin2x + sin3x = 1



2

⇒ sin x + sin3x = 1 – sin2x ⇒ sin x [1 + sin2x] = cos2x ⇒ sin x [2 – cos2x] = + cos2x Squaring we have, (1 – cos2x) (2 – cos2x)2 = cos4x or 4 + cos4x – 4 cos2x – 4 cos2x – cos6x + 4 cos4x = cos4x 6 4 2 ⇒ cos x – 4 cos x + 8 cos x = 4.

98. (b) We have, e–π/2 < θ
log cos θ.

sin6x + cos6x = (sin2x)3 + (cos2x)3 = (sin2x + cos2x)3 – 3 sin2x cos2x (sin2x + cos2x) = 1 – 3 sin2x cos2x 3 (sin 2x)2 = 1 – 3  ⋅ 4 sin2x cos2x = 1 – 4 4 ⇒ maximum value of sin6x + 3 ×0=1  cos6x is 1 – 4 and minimum value is 1 – 3 × 1 = 1 . 4 4 95. (a) We have, K = sin 10º sin 50º sin 70º = sin 10º sin (60º – 10º) sin (60º + 10º) = sin 10º (sin260º – sin210º)

π 2

99. (d) 1 − sin 2 x cos 2 x 2 2 cos x + sin x − 2 sin x cos x = cos 2 x − sin 2 x

100. (c) We have, sec 2x – tan 2x =

=

(cos x − sin x) 2 (cos x + sin x)(cos x − sin x)

=

cos x − sin x 1 − tan x π  = = tan  − x  . 4  cos x + sin x 1 + tan x

2

2

2

102. (b) We have, y = 4 sin2θ – cos 2θ = 4 sin2θ – (1 – 2 sin2θ) = 6 sin2θ – 1 y +1 ∴ = sin2θ 6 But 0 ≤ sin2θ ≤ 1. y +1 ≤ 1 ⇒ 0 ≤ y + 1 ≤ 6 6 ⇒ – 1 ≤ y ≤ 5 ⇒ y ∈ [– 1, 5].

β α γ , cos , cos are all +ve 2 2 2 α β γ cos cos is +ve. ∴ min. value of 4 cos  2 2 2 Hence, minimum value of sin α + sin β + sin γ is + ve. ∴ cos

107. (c) Since sin α and cos α are the roots of given equation. ∴ sin α + cos α = –

∴ 0 ≤

and



Squaring (i), we get

103. (d) We have, 1 x + = 2 cos θ ⇒ x2 – 2x cos θ + 1 = 0 x ⇒ x = cos θ ± i sin θ ⇒ xn = cos nθ ± sin nθ Now,

1 1 = = cos θ ± i sin θ cos θ ± i sin θ x

1 = cos nθ ± i sin nθ xn 1 ∴ xn + n = 2 cos nθ. x 104. (a) We have, ∴

 3π  sin4  − α  = cos4α,  2  sin4 (3π + α) = sin4α,  3π  sin6  + α  = cos6α, and sin6 (5π – α) = sin6α  2  ∴ The given expression = 3 [cos4α + sin4α] – 2 [cos6α + sin6α] = 3 [1 – 2 sin2α cos2α] – 2 [1 – 3 sin2α cos2α] = 1. 105. (a) Given value

...(i)

r p

...(ii)

q2 sin2α + cos2α + 2 sin α cos α = 2 p 2r q2 ⇒ 1 + = 2 ⇒ p2 + 2 pr = q2 p p

⇒ p2 – q2 + 2 pr = 0.

 108. (c) We have, tan (θ – φ) =

tan θ − tan φ 1 + tan θ tan φ

n tan φ − tan φ = 1 + n tan φ tan φ =

=

[ ∵ tan θ = n tan φ]

(n − 1) tan φ n −1 = 1 + n tan 2 φ cot φ + n tan φ

∴ tan2 (θ – φ)

=

(n − 1) 2 (cot φ + n tan φ) 2

(n − 1) 2 (n − 1) 2 ≤ . 2 (cot φ − n tan φ) + 4n 4n [∵ square of a real number is never negative].

109. (b) We have, (4 sin 4 θ + sin 2 2θ) = =

= (sin 47º + sin 61º) – (sin 11º + sin 25º)

sin α cos α =

q p

(4 sin 2 θ (sin 2 θ + cos 2 θ)

(4 sin 2 θ) = 2 | sin θ |.

= 2 sin 54º cos 7º – 2 sin 18º cos 7º

But, since θ lies in the third quadrant, ∴ sin θ < 0 and | sin θ | = – sin θ.

= 2 cos 7º (sin 54º – sin 18º)

Hence

= 2 cos 7º 2 cos 36º sin 18º = 2 cos 7º

2 sin 18º cos18º cos 36º cos18º

= cos 7º

2 sin 36º cos 36º cos18º

= cos 7º

sin 72º = cos 7º. [since sin 72º = cos 18º] cos18º

106. (c) We have, α β γ sin α + sin β + sin γ = 4 cos  cos  cos 2 2 2 α β γ π , is less than ∵ each of 2 , 2 2 2

and

(4 sin 4 θ + sin 2 2θ) = – 2 sin θ   π θ 4 cos2  −  = 2 1 + cos  4 2 

 Hence the given expression = – 2 sin θ + 2 + 2 sin θ = 2. 110. (b) sin θ =

1 1 , cos φ = 2 3

⇒  cos θ =

3 2 2 , sin φ = 2 3

∴  sin (θ + φ) =

1+ 2 6 = 0.9 6

π  − θ  2  = 2 + 2 sin θ.

819

π  cos B + cos C = cos B + cos  − B    2 = cos2B + sin2B = 1. 2

Trigonometric Ratios and Identities

101. (c) We have,

820

Objective Mathematics

3−2 2 = – 0.2 6

and cos (θ + φ) =

 π 2π  Hence, it satisfies  2 , 3  .   111. (a) The given expression = (cot 5º cot 85º) (cot 10º cot 80º) ... and so on = (cot 5º tan 5º) (cot 10º tan 10º) ... and so on. = 1 ⋅ 1 ⋅ 1 ... 1 = 1. 112. (b) Since cos θ ≤ 1 for all real θ, ∴ the given expression ≤ 1 for all x. Hence, the greatest value = 1. 113. (b) We have, 1 = tan 45º = tan (56º – 11º) =

tan 56º − tan 11º 1 + tan 56º tan 11º

∴ 1 + tan 56º tan 11º = tan 56º – tan 11º ⇒ tan 56º – tan 11º – tan 56º tan 11º = 1. 114. (a) Since A + B + C = π, ∴ A + B = π – C ⇒ tan (A + B) = tan (π – C) ⇒

tan A + tan B = – tan C 1 − tan A tan B

tan A + tan B > 0 ...(1) 1 − tan A tan B [ ∵ angle C is obtuse ∴ tan C < 0] But since C is obtuse angle, so A and B will both π be less than . 2 ∴ Both tan A and tan B are positive. Hence from (1), 1 – tan A tan B > 0  ⇒ tan A tan B < 1.

⇒

1 1 (2 sin x cos x) = sin 2x. 115. (c) sin x cos x = 2 2 Since the greatest value of sin 2x is 1, \ the greatest 1 1 value of sin x cos x is ×1= . 2 2 116. (b) We have, log tan 1º + log tan 2º + ... + log tan 89º = log (tan 1º ⋅ tan 2º ... tan 89º) = log (tan 1º ⋅ tan 2º ... tan 44º tan 45º ⋅ cot 44º ⋅ cot 43º ... cot 1º) = log 1 = 0. 5π π 117. (a) Since tan , x and tan are in A.P. 18 9 ∴ 2x = tan 20º + tan 50º =

sin 70º = sec 50º cos 20º cos 50º

= cosec 40º ...(1) 7π π Since tan , y and tan  are in A.P. 18 9 sin 90º ∴ 2y = tan 20º + tan 70º = cos 20º ⋅ cos 70º

1 2 = cos 20º sin 20º sin 40º From (1) and (2), we have 2x = y.

=

...(2)

118. (b) Since sin (x – y), sin x and sin (x + y) are in H.P., 2sin ( x − y )sin( x + y ) ∴ sin x = sin( x − y ) + sin( x + y ) 

=

2 (sin 2 x − sin 2 y ) 2 sin x cos y

⇒ sin2x ⋅ cos y = sin2x – sin2y ⇒ sin2x (1 – cos y) = sin2y y y y = 4 sin2 cos2 ⇒ sin2x ⋅ 2 sin2 2 2 2 ⇒

sin x ⋅ sec

y 2

=

2.

119. (b), (c) Since cos (θ – φ), cos θ, cos (θ + φ) are in H.P., ∴ cos θ =

2 cos (θ − φ)cos (θ + φ) 2 [cos 2 θ − sin 2 φ = cos (θ − φ) + cos (θ + φ) 2 cos θ ⋅ cos φ

⇒ cos2θ ⋅ cos φ = cos2θ – sin2φ ⇒ sin2φ = cos2θ (1– cos φ) φ φ 4 sin 2 cos 2 2 2 2 φ 2 sin 2

sin 2 φ ⇒ cos θ = = 1 − cos φ 2

⇒ cos2θ = 2 cos2 ⇒ cos θ ⋅ sec

φ φ or, cos2θ ⋅ sec2 =2 2 2

φ =± 2

2.

120. (b) We have 2 sin (ex) =5x + 5x or

sin(e x ) =

5x + 5x 2

...(1)

Now l.h.s of (1) always less than equal to 1. Because sin φ always lies between – 1 and + 1. But in r.h.s geometrical mean =

5 x.5− x = 5 x / 5 x =1

∴ a.m ≥ g.m ∴ There is only one real solution in the interval [0, 1] 121. (c) We have, cosec θ + cot θ = 1 + cos θ 11 = sin θ 2 θ 2 cos 2 2 = 11 ⇒ θ θ 2 2 sin cos 2 2 ⇒

11 2

θ 2

2 44 11 = = ∴ tan θ = . 4 117 2 θ 1 − tan 1− 2 121 2 tan

2×

⇒ A ≤ cos θ + sin θ ⋅ 1 ⇒ A ≤ 1. Again, A = (1 – sin2θ) + sin4θ 2

( sin θ ≤ 1)

32 + 42 + 7 = 25 + 7 = 5 + 7 = 12 ∴  Minimum value of

3 ≤ A ≤ 1. 4



4a θ = 4a2 2 (1 + θ2 ) 2  a− y + 1  a + y 

128. (a) We have, x2 + y2 + z2 = r2 sin2θ cos2φ + r2 sin2θ sin2φ + r2 cos2θ = r2 sin2θ (cos2φ + sin2φ) + r2 cos2θ = r2 (sin2θ + cos2θ) = r2, which is independent of θ and φ.

124. (d) We have, tan (A + B)

2π 4π = z cos = k (say) 3 3 4π 2π cos 1 1 1 1 cos 3 = , = = ⇒ 3 , x k y z k k

a 1 + tan A + tan B a + 1 2 a +1 = = 1 − tan A tan B 1 − a ⋅ 1 a + 1 2a + 1

∴

2a 2 + 2a + 1 = 1. 2a 2 + 2a + 1

125. (b) We have,

=

129. (b) We have, x = y cos

2a 2 + a + a + 1 2a 2 + a + 2a + 1 − a

∴ A + B =

=

π . 4

sin α cos x1 cos x2

1 1 1 + + x y z

=

1 k

2π 4π   + cos  1 + cos 3 3 

1  1 1 1 − −  = 0 k  2 2

⇒ xy + xz + yz = 0. 130. (b) We have, cos α + cos β = sin α + sin β sin ( x2 − x1 ) = cos x1 cos x2

sin x2 cos x1 − cos x2 sin x1 cos x1 cos x2

= tan x2 – tan x1. ∴ sin α [sec x1 sec x2 + sec x2 sec x3 + ... to n terms]

 = (tan x2 – tan x1) + (tan x3 – tan x2) +

... + (tan xn – tan xn – 1] = tan xn – tan x1

= tan α ⋅ tan  π − α  = tan α cot α = 1. 2 

∴ The given expression = 1.

⇒ x2 = (a – y) (a + y) = a2 – y2 or x2 + y2 = a2.

=

 π  − 1  α = tan α ⋅ tan   2α 



a − y  a + y 

2 2

1 is 1/12. 3 sin θ − 4 cos θ + 7

127. (a) We have, tan α ⋅ tan (2n – 1) α

a (1 − θ2 ) a− y 123. (b) We have, y = ⇒ θ2 = 1 + θ2 a+ y

=

sin ( xn − x1 ) sin (n − 1) α = . cos x1 cos xn cos x1 cos xn

=

1  1  2 3 ⇒ A =  sin θ −  + 1 −  ⇒ A ≥ . 2  4 4 

∴ x2 =

=

3 sin θ – 4 cos θ + 7 = maximum value of 3 sin θ – 4 cos θ + 7

2

2

Hence

sin xn cos x1 − cos xn sin x1 cos xn cos x1

126. (a) Minimum value of

122. (d) We have, A = cos2θ + sin2θ sin2θ 2

=

Squaring both sides, we get cos2α + cos2β + 2 cos α cos β  = sin2α + sin2β + 2 sin α sin β 2 2 ⇒ (cos α – sin α) + (cos2β – sin2β) = – 2 (cos α cos β – sin α sin β) ⇒ cos 2α + cos 2β = – 2 cos (α + β). 131. (c) We have, 2 – cos x + sin2x = 2 – cos x + 1 – cos2x = – (cos2x + cos x) + 3 2 2  1  1 13  1 −  cos x +  . = –   cos x +  −  + 3 = 2  4  4  2  

821

θ 11 θ 2 = or tan = 2 2 2 11

Trigonometric Ratios and Identities

⇒ cot

822

1 ∴ Maximum value occurs at cos x = – and it 2 13 is and minimum value occurs at cos x = 1 and 4 it is 1 ∴ ratio is 13 . 4

Objective Mathematics

2sin θ + 2− cos θ ≥ 2

1+

1 (sin θ − cos θ ) 2

 π 1+ sin  θ −   4 2 1

= 2

Thus, 2sin θ + 2– cos θ is minimum π π  when sin  θ −  = – 1 i.e., θ – = 2nπ +  4 4

1 sin 2α1 ⋅ sin 2α2 ... sin 2αn. 2n But each of sin 2αi ≤ 1. 1 . 2n

∴ (cos α1 ⋅ cos α2 .. cos αn)2 ≤

(sec A + tan A) (sec B + tan B) (sec C + tan C) = x

∴ cos α1 ⋅ cos α2 ... cos αn = sin α1 ⋅ sin α2 ... sin αn Now, (cos α1 ⋅ cos α2 ... cos αn)2 = (cos α1 ⋅ cos α2 ... cos αn) (cosα1 ⋅ cosα2 ....cosαn) = (cos α1 ⋅ cos α2 .. . cos αn) ⋅ (sin α1 ⋅ sin α2 ... sin αn) =

7π or θ = 2nπ + , n ∈ I. 4 133. (a), (b) Let

 

π  = tan  2α −  = – cot 2α. 2 

139. (a) Here (cot α1) ⋅ (cot α2) ... (cot αn) = 1

2sin θ ⋅ 2− cos θ

⇒ 2sin θ + 2 – cos θ ≥ 2

3π 2

∴ tan β = cot α and tan γ

But tan α = cot α – 2 cot 2α. ∴ tan α = tan β + 2 tan γ.

132. (b) Since A.M. ≥ G.M. ∴

π π Solving, we get β = 2 – α and γ = 2α – 2 .

...(i)

But each of cos αi is positive.

...(ii)

∴ cos α1 ⋅ cos α2 .. cos αn ≤

 

then, (sec A – tan A) (sec B – tan B) (sec C – tan C) = x Multiplying (i) and (ii) we get (sec2A – tan2A) (sec2B – tan2B) (sec2C – tan2C) = x2 or x2 = 1. ∴ x = ± 1. Hence each side = ± 1. 134. (a), (b)  In a triangle tan A + tan B + tan C = tan A tan B tan C ⇒ 6 = 2 tan C or tan C = 3. ∴ tan A + tan B = 3 and tan A tan B = 2 ∴ tan A = 1 or 2 and tan B = 2 or 1. 1 1 sin 30º = . 2 4 θ θ 136. (a) We have, tan + cot 2 2 2 θ 2 θ sin + cos 2 2= 2 = 2 =6 = 1 θ θ sin θ sin cos 3 2 2 θ θ  ⋅ cot = 1. and tan 2 2 135. (c) sin 15º cos 15º =

∴ The required equation is x2 – 6x + 1 = 0. 137. (c) We have, f (θ) = sin θ (sin θ + sin 3θ) = sin θ (2 sin 2θ cos θ) = sin2 2θ ≥ 0, for all real θ. π 138. (c) We have, β + γ = – β= α 2 π So, 2β + γ = and β + γ = α 2

1 1 = n/2 . n 2 2

140. (a) Given : cos A + cos B = 1 A + B A − B ⇒ 2 cos   cos  =1  2   2  ⇒ 2 cos 

∴ cos 

A − B π ⋅ cos  =1  2  6

 1  A−B =   3 2

∴ cos (A – B) = 2 cos2 =2 ⋅

1 1 –1=– . 3 3

A−B –1 2

141. (c) In a ∆ABC,

tan A + tan B + tan C = tan A tan B tan C = 6 1 ∴ cot A ⋅ cot B ⋅ cot C = . 6 142. (b) u1 un – un – 1 

= 2 cos θ ⋅ 2 cos nθ – 2 cos (n – 1) θ = 2 [cos (n – 1) θ + cos (n + 1) θ] – 2 cos (n – 1) θ = 2 cos (n + 1) θ = un + 1.

143. (a) We have, tan 9º – tan 27º – tan 63º + tan 81º = tan 9º + tan 81º – (tan 27º + tan 63º) =

sin 90º sin 90º − cos 9º ⋅ cos 81º cos 27 º ⋅ cos 63º

=

2 2 − sin 18º sin 54º

= 2 ⋅ 

149. (c)

sin 54º − sin 18º 2 cos 36º ⋅ sin 18º = 2 ⋅  = 4. sin 18º ⋅ cos 36º sin 18º ⋅ cos 36º

144. (b) When θ =

π π , cos  θ −  = cos 0 = 1 = maxi4  4

mum value. π π π π ⇒– cos  ±  =   4 4 ∴

1 2

⇒ (sin2θ + cos2θ) (sin4θ – sin2θ ⋅ cos2θ + cos4θ) + k cos2 2θ = 1 ⇒ sin θ + cos θ – sin θ cos θ + k cos2 2θ = 1 2

2

⇒ (sin θ + cos θ) – 3 sin θ ⋅ cos θ + k cos 2θ = 1 3 sin2 2θ ⇒ 3 sin2θ ⋅ cos2θ = k cos2 2θ or 4 = k cos2 2θ 3 ∴ k = tan2 2θ. 4 2

2

2

2

2

2

146. (c) Given : sin θ + cosec θ = 2 Squaring both sides, we get sin2θ + cosec2θ + 2 sin θ cosec θ = 4 ⇒ sin2θ + cosec2θ + 2 sin θ ⋅ 

=4

⇒ sin2θ + cosec2θ = 4 – 2 = 2. 1 2 Also, cosec2θ – cot2θ = 1 ⇒ cosec θ + cot θ = 2 1 5 = ∴ 2 cosec θ = 2 + 2 2 4 ⇒ sin θ = 5 1 3 Also, 2 cot θ = 2 – = 2 2

147. (b) We have, cosec θ – cot θ =

⇒ cot θ =

3 –

3 tan 40º tan 20º = tan 40º + tan 20º

Hence tan 40º + tan 20º +

3 tan 40º tan 20º =

151. (c) We have, sin (A + B + C) = 1, 1 and sec (A + C) = 2 tan (A – B) = 3 ⇒ A + B + C = 90º, A – B

152. (c) sin x +

145. (d) We have, sin6θ + cos6θ + k cos2 2θ = 1

4

∴

= 30º and A + C = 60º ⇒ B = 30º, A = 60º and C = 0º.

1 π < cos  θ −  ≤ 1.  2 4

4

2 1 + tan 2 θ = cosec 2θ = cosec 120º = . 3 2 tan θ 150. (c) We have, 3 = tan 60º = tan (40º + 20º) tan 40º + tan 20º = 1 − tan 40º tan 20º

823

1 1 − cos 9º ⋅ sin 9º cos 27 º ⋅ cos 27 º

3 3 3 4 3 ⇒ cos θ = sin θ = × = . 4 4 4 5 5

148. (a) We have, tan mθ = tan nθ ⇒ mθ = nθ + kπ, k ∈ Z kπ , k ∈ Z. ∴ θ = m−n π These values of θ are in A.P. with first term m −n π . and common difference m−n

3 cos x = r sin (x + α) where 1 = r cos α and 3 = r sin α Its maximum value = r = 2 3 ⇒ α = 60º ⇒ sin α = 2 ∴ x + α = 90º gives x = 30º.

153. (c) Since 1 radian = 57º (nearly) ∴ 2 radians = 114º (nearly) This lies in 2nd quadrant ∴ tan 1 > 0 and tan 2 < 0 ∴ tan 1 > tan 2. 154. (b) Since 200º lies in IIIrd quadrant ∴ sin 200º, cos 200º are both – ve, ∴ their sum is – ve. 155. (b) Put x = tan A, y = tan B, z = tan C ∴ tan A tan B + tan B tan C + tan C tan A = 1 ⇒ tan C [tan A + tan B] = 1 – tan A tan B tan A + tan B x+ y ∑ 1 − xy = ∑ 1 − tan A tan B =

tan A + tan B tan B + tan C tan C + tan A + + 1 − tan A tan B 1 − tan B tan C 1 − tan C tan A

=

1 1 1 + + tan C tan A tan B

=

tan A tan B + tan B tan C + tan C tan A tan A tan B tan C

=

1 1 = . tan A tan B tan C xyz

156. (c) We have, sin A = ⇒ cos A = and

sin B =

1 and cos B = 5

1 − sin 2 A =

2 5

1 − cos 2 B =

1 10

3 10

3.

Trigonometric Ratios and Identities

=

824

∴ sin (A + B) = sin A cos B + cos A sin B 1 3 2 1 5 ⋅ + ⋅ = = 5 10 5 10 5 10

Objective Mathematics

1 π = = sin 2 4 π ∴ A + B = . 4 157. (b) We have, 3 sin 2θ = 2 sin 3θ

⇒ tan A tan 2B = 1 ∴ sin A sin 2B = cos A cos 2B 1 4 (∵ θ ≠ 0)

⇒ 4 cos3θ – 3 cos θ – 1 = 0 ⇒ cos θ = 1,

∴ cos θ = 158. (d) We have, 8 sin

1 ⇒ sin θ = 4

2 = tan 2B ⋅ (2 cos B − 1) 3 cos 2 A 2 = tan 2B ⋅ (4 − 3 cos A − 1) = tan 2B ⋅ tan2A 2 3 cos A

⇒ 6 sin θ cos θ = 2 (3 sin θ – 4 sin3θ)

Since cos θ ≠ 1.

⇒ tan A = sin 2B = sin 2B ⋅ cos 2B 3 cos 2 A cos 2B 3 cos 2 A

15 . 4

x x x cos cos x cos 2 8 8 4

x x x x =  2 sin cos  4 cos cos  8 8 2 4 x x x cos cos 4 4 2 x x = 2 sin cos  = sin x. 2 2

⇒ cos (A + 2B) = 0 ∴ A + 2B = π . 2 162. (c) We have, π  tan (π cos θ) = cot (π sin θ) = tan  − π sin θ  2  π ⇒ π cos θ = – π sin θ 2 π ⇒ π [cos θ + sin θ] = 2 1 ...(1) ⇒ cos θ + sin θ = 2 π ∴ sin   θ +   4

= 4 sin

159. (c) Since, sin x + sin2x = 1, (Given) ∴ sin x = cos x ...(1) Now, the given expression = cos6x (cos6x + 3 cos4x + 3 cos2x + 1) – 1 = cos6x (cos2x + 1)3 – 1 = sin3x (sin x + 1) – 1 [Using (1)] = (sin2x + sin x)3 – 1 = (1)3 – 1 = 1 – 1 = 0.

= sin θ ⋅  =

2



1 5 1 ∴ sin2x + cos2x + 2 sin x cos x = 25

160. (a) Since sin x + cos x =

or 1 + sin 2x = ⇒

1 1 or sin 2x = – 1 = – 24 25 25 25

2 tan x = – 24 1 + tan 2 x 25

⇒ – 25 tan x = 12 + 12 tan2x ⇒ 12 tan2x + 25 tan x + 12 = 0 ⇒ 12 tan2x + 16 tan x + 9 tan x + 12 = 0 ⇒ 4 tan x (3 tan x + 4) + 3 (3 tan x + 4) = 0 ⇒ (4 tan x + 3) (3 tan x + 4) = 0 4 3 or − . ∴ tan x = – 3 4 161. (b) We have,

3 sin A 2 cos B = sin B cos A

⇒ 3 sin A = 2 cos B ⋅ sin B cos A cos A cos A

1 2

= sin θ cos 

1 + cos θ 2

π π + cos θ sin  4 4

1 = (sin θ + cos θ) 2

1 . 2

1 2

[Using (1)]

163. (c) We have, sec θ + tan θ =

3.

Since sec θ – tan θ = 1 1 ∴ sec θ – tan θ = 3 2

2

∴ 2 tan θ = ⇒ tan θ =

3 −1 1 = = 3 3

3 –

2 3

1 3

∴ θ = 30º = π . 6 164. (c) The given equation is 4x2 – 2 5 x + 1 = 0 2 5 5 = 4 2

Sum of the roots =

and product of the roots =

1 4

Since, sin 18º cos 36º 

=

1 5 −1 5 +1 5 −1 ⋅ = = 4 4 4 16

and sin 18º + cos 36º 2 5 5 5 −1 5 +1 = + = 4 2 4 4 Hence, roots are sin 18º, cos 36º. 

=

=

2 cot α + 1 + cot 2 α 3π < α < π, 4 ∴ | 1 + cot α | = – 1 – cot α]

[since cot α < – 1 when

1 2

174. (b) We have, sec 2x – tan 2x

1 1 π π – sin θ ⋅ sin = ⋅ ⇒ cos θ ⋅ cos  2 2 4 4 π π 1   ∵cos 4 = sin 4 = 2    π 1  ⇒ cos   θ + 4  = 2 2 .

=

1 sin 2 x 1 − sin 2 x (cos x − sin x) 2 − = = cos 2 x cos 2 x cos 2 x cos 2 x − sin 2 x

=

cos x − sin x 1 − tan x π = = tan  − x  . cos x + sin x 1 + tan x 4 

175. (c) We have, cosec A (sin B cos C + cos B sin C) = cosec A sin (B + C) = cosec A sin (π – A) = cosec A sin A = 1.

166. (b) sin 2α = sin [(α + β) + (α – β)] = sin(α + β) cos (α – β) + cos (α + β) sin (α – β)

176. (b) sin6A + cos6A + 3 sin2A cos2A

5 4 12 3 56 = 13 ⋅ 5 + 13 ⋅ 5 = 65 . 167. (d) Given value (cos 6 x + cos 4 x) + 5 (cos 4 x + cos 2 x) + 10 (cos 2 x + 1)  = cos 5 x + 5 cos 3 x + 10 cos x =

1 = 2 cot α + cosec 2α sin 2 α

= | 1 + cos α | = – 1 – cot α

π ⇒ π (cos θ – sin θ) = 2 ⇒ cos θ – sin θ =

2 cot α +

2 cos 5 x cos x + 10 cos 3x cos x + 20 cos 2 x = 2 cos x. cos 5 x + 5 cos 3x + 10 cos x

168. (d) Since we have one equation and two variables, therefore, its impossible to find the value of x. 169. (c) A + B = θ and A – B = K tan A K = tan B 1

= (sin2A)3 + (cos2A)3 + 3 sin2A cos2A  (sin2A + cos2A) 2 2 3 3 = (sin A + cos A) = (1) = 1. 177. (b)

sec 8A − 1 1 − cos 8A cos 4A × = sec 4A − 1 cos 8A 1 − cos 4A

2 sin 2 4A × cos 4A = cos 8A × 2 sin 2 2A



=



sin 8A cos 2A tan 8A = cos 8A ⋅ sin 2A = tan 2A .

(2 sin 4A cos 4A) ⋅ 2 sin 2A cos 2A cos 8A ⋅ (2 sin 2 2A)

178. (a) a cos 2x + b sin 2x

By componendo and dividendo Rule ⇒

tan A + tan B K + 1 sin ( A + B ) K + 1 = = ⇒ tan A − tan B K − 1 sin ( A − B ) K − 1

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

⇒

sin θ K + 1 = sin K K − 1

 1 − b 2 /a 2   2 b/a   Put tan x = b    +b  = a 2 2  a  1 + b 2 /a 2    1 + b /a 

∴ sin K =

K −1 sin θ . K +1

170. (a) Since, sin θ1 + sin θ2 + sin θ3 = 3 ∴ sin θ1 = sin θ2 = sin θ3 = 1 ⇒ θ1 = θ2 = θ3 = ∴ cos θ1 + cos θ2 + cos θ3 = 0. 171. (c) cos2θ + sec2θ = (cos θ – sec θ)2 + 2 ≥ 2. 172. (b) We have, sec A (cos B cos C – sin B sin C) = sec A cos (B + C) = sec A cos (180º – A) = sec A (– cos A) = – 1.

= π 2

a (a 2 − b 2 ) 2ab 2 a (a 2 − b 2 + 2b 2 ) + 2 = a. 2 2 2 = a +b a +b a 2 + b2

179. (d) Given equation is cos θ = x + or

x2 – x cos θ + 1 = 0

1 x

For real values of x, b2 > 4ac a = 1, b = – cos θ, c = 1 ∴ cos2θ > 4 (1) (1) or | cos θ | > 2 which is not possible as | cos θ | ≤ 1. ∴ No real value of θ is possible.

825

173. (d)

π sin (π cos θ) = cos (π sin θ) = sin   + π sin θ  2  π ⇒ π cos θ = + π sin θ 2

Trigonometric Ratios and Identities

165. (d) We have,

826

Objective Mathematics

180. (a) Given tan A ⋅ tan B = 2 3 and cos (A – B) = 5 3 ⇒ cos A cos B + sin A sin B = 5

...(1)

3 ⇒ cos A cos B (1 + tan A tan B) = 5 3 ⇒ cos A cos B (1 + 2) = [Using (1)] 5 1 ⇒ cos A cos B = . 5 181. (a) cos 52º + cos 68º + cos 172º = = = = =

(cos 172º + cos 52º) + cos 68º 2 cos (112º) cos 60º + cos 68º cos 112º + cos 68º cos (180º – 68º) + cos 68º – cos 68º + cos 68º = 0.

182. (a) Given sin α = – ∴ cos α = –

3 5

4 3π  π < α <  5 2 

α 2 ∵ 2 cos 2 = 1 + cos α 1  4 α α = 1 +  −  or cos2 = ∴ 2 cos2  5 10 2 2 1 α ∴ cos =± 10 2 α α But, 90 < < 135º means lies in II quadrant. 2 2 1 ∴ cos α = – . 10 183. (d) cos







π 2π 4π ⋅ ⋅ cos 7 7 7 =

=

π π 2π 4π  2 sin ⋅ cos  ⋅ cos ⋅ cos π   7 7 7 7 2 sin 7 1

2π  4π  2π sin ⋅ cos  ⋅ cos π   7 7 7 2 sin 7 1

4π 4π sin ⋅ cos = π 7 7 22 sin 7 π  sin  π +   1 8π 1 1 7 sin = = =– . π π 7 8 8 3 2 sin sin 7 7 1

184. (b) sin 163º cos 347 + sin 73º sin 167

= sin (180º – 17º) cos (360º – 13º) + sin (90º – 17º) sin (180º – 13º) = sin 17º cos 13º + cos 17º sin 13º = sin (17º + 13º) = sin 30º = 1 . 2

185. (a) We have  π  121 cm l = rθ = 77 =   =  18  9 186. (c) We have, y = cos2 x + sec2 x ∴ y = (cos x – sec x)2 + 2 …(i) As (cos x – sec x)2 is zero or positive. Therefore Eq. (i) becomes, ⇒ y≥2 187. (a) We have, sin θ + cosec θ = 2 …(i) On squaring both sides, we get sin2 θ + cosec2 θ + 2 = 4 ⇒ sin2 θ + cosec2 θ = 2 …(ii) Again squaring Eq. (ii), we get sin4 θ + cosec4 θ = 2 …(iii) Again cubing Eq. (ii), we get  (sin2 θ + cosec2 θ)3 = 23 ⇒ sin6 θ + cosec6 θ + 3 sin2 θ cosec2 θ (sin2 θ  + cosec2 θ) = 8 6 6 ⇒ sin θ + cosec θ + 3·2 = 8 …(iv) ⇒ sin6 θ + cosec6 θ = 2 On multiplying Eqs. (iv) and (iii), we get  (sin4 θ + cosec4 θ) (sin6 θ + cosec6 θ) = 4 10 ⇒ sin θ + sin4 θ cosec6 θ + cosec4 θ sin6 θ  + cosec10 θ = 4 ⇒ sin10 θ + sin4 θ cosec4 θ (sin2 θ + cosec2 θ)  + cosec10 θ = 4 ⇒ sin10 θ + cosec10 θ = 4 – 2 = 2 188. (b) We have, 1 , 0 < x < π 2 1 1 1 cos x + sin x = 2 2 2 2

cos x + sin x = ⇒

…(i)

π 1 ⇒ cos  x −  =  4 2 2 ⇒ π < x − π < π 4 4 2 π 3π 1 ∴ (tan θ)cot θ < 1 and (cot θ)tan θ > 1 ∴ t4 > t1, which only holds in option (b). 190. (b) We have, θ + α θ− α ⋅ tan  tan   2   2  θ α tan 2 − tan 2 2 2 = 2 θ 2 α 1 − tan tan 2 2 =

2(cos α − cos θ) 2(cos α + cos θ)

194. (a) Since, cos θ = ⇒ sin θ =

8 and 0 < θ < π 17 2

827

π  , tan θ > cot θ 4

82 15 = 17 2 17

Trigonometric Ratios and Identities

189. (b) When θ ∈  0, 

1−

The value of the given expression = cos 30º cos θ – sin 30º sin θ + cos 45º cos θ + sin 45º sin θ + cos 120º cos θ + sin 120º sin θ  3 1 1 1 1 3 = cos θ  + −  − sin θ  − − 2  2 2 2  2 2 =

8  3 1 1  15  3 1 1 + − +  + −  17  2 2 2  17  2 2 2

=

23  3 − 1 1  +  17  2 2

(∵ given cos θ = cos α cos β) cos α(1 − cos β) β cos 2α 3cos 2β − 1 = = tan 2 195. (a) We have, = cos α(1 + cos β) 2 1 2 cos 2β 191. (c) Since, tan θ < 0 and cos θ > 0, therefore θ lies in Applying componendo and dividendo rule the fourth quadrant. cos 2α + 1 3cos 2β − 1 + 3 − cos 2β ∴ tan θ = – 1 = cos 2α − 1 3cos 2β − 1 − (3 − cos 2β) 1 7π ⇒θ= and cos θ = 2cos 2 α 2 + 2cos 2β 2 4 ⇒ = 7π −2sin 2 α 4cos 2β − 4 ∴ The most general value of θ is 2nπ + . 4 − cos 2 α 1 + cos 2β 2cos 2 β ⇒ = = b 2 192. (a) We have, tan x = sin α 2(cos 2β − 1) −4sin 2 β a 196. (d) Given that tan A and tan B are the roots of x2 – ax ∴ a cos 2x + b sin 2x + b = 0. 1 − tan 2 x 2 tan x = a⋅ + b ⋅ ∴ tan A + tan B = a and tan A tan B = b 1 + tan 2 x 1 + tan 2 x tan A + tan B a b2 b = Now, tan (A + B) = 1− 2 2⋅ 1 − tan A tan B 1 − b a +b⋅ a = a⋅ 1 b2 b2 2 Now, sin (A + B) = [1 – cos 2(A + B)] 1+ 2 1+ 2 2 a a 2 2 2 2 1  1 − tan ( A + B )  a(a − b ) 2b a = 1 − = + 2  2  1 + tan 2 ( A + B)  a 2 + b2 a + b2 a 3 + ab 2 = 2 =a 1  1 + tan 2 ( A + B ) − 1 + tan 2 ( A + B )  =  a + b2  2 1 + tan 2 ( A + B )  193. (b) We have,  tan 2 ( A + B )  a 2 /(1 − b) 2 1 = = …(i)  x + = 2cos θ 2 2 2 x  1 + tan ( A + B )   a + (1 − b)  2   2 1 1   Now, x 2 + 2 =  x +  − 2 = 4cos 2 θ − 2  (1 − b)  a2  x x = 2 a + (1 − b) 2 ⇒ x 2 + 1 = 2cos 2θ x2 197. (c) Given that, sin 4A + sin 2A = cos 4A + cos 2A On cubing Eq (i), we get ⇒ 2 sin 3A cos A = 2 cos 3A cos A 

1 1  + 3  x +  = 8cos3 θ  x3 x 1 3 ⇒ x + = 8 cos3 θ – 6 cos θ = 2 cos 3θ x3 1 Similarly, x n + n = 2 cos n θ x x3 +

∴ tan 3A = 1 π ⇒ A = 12

∴ tan4A = tan π 3 = 3

828

Exercises for self-practice

Objective Mathematics

7. The minimum value of sin θ + cos θ will be

1. For A = 133º, 2 cos Α is equal to 2 (a) − 1 + sinA − 1 − sinA (b) − 1 + sinA + 1 − sinA (c)

1 + sinA − 1 − sinA

(d)

1 + sinA + 1 − sinA

2. If sin(x – y) = cos(x + y) = 1 , then the values of x and 2 y lying between 0° and 90º are given by (a) x = 15°, y = 25° (b) x = 65º, y = 15º (c) x = 45°, y = 45° (d) x = 45º, y = 15º

2

8. If α + β + γ = π , then the value of 2 tan α tan β + tan β tan γ + tan γ tan α will be 1 (a) 1 (b) 2 3 (c) (d) 2 2

(a) 4 (c) 3

1 2

(b) a – b (d) a 2 + b 2 (b) 5 (d) None of these

11. Value of 2 (sin6θ + cos6θ) – 3 (sin4θ + cos4θ) + 1 is :

4. cos 52º + cos 68º + cos 172° =

(a) 2 (c) 4

(b) 1 (d) None of these

(b) 0 (d) 6

12. If α is a root of 25 cos 2 θ + 5cosθ – 12 = 0, π/2 < α < π, then sin 2α is equal to:

5. The value of log10tan 1º + log10 tan 3º + ... + log10 (tan 89º) is given by

(a) 24/25 (c) 13/18

(b) –24/25 (d) –13/18

13. If sinθ + cosceθ = 2, then sinnθ + coscenθ is equal to:

(a) 45 log10 (tan 89º) (b) 45 (log10 tan 1º) (c) 0 (d) 1

(b) 2n (d) None of these

(a) 2 (c) 2n–1

14. If A = cos2θ + sin4θ, then for all values of θ:

6. The minimum value of 3 cos x + 4 sin x + 5 is (a) 5 (c) 0

2

10. The maximum value of 12 sin θ – 9 sin2θ is

(d) 2

(a) 0 (c) 2

(d) –

(a) a + b (c) | a | + | b |

(b)

(c) 1

(b) – 1

1 (c) – 2

9. The maximum value of a cos x + b sin x is

3. The maximum value of sin x + cos x is (a)

(a) 0

(a) 1 ≤ A ≤ 2 (c) 3/4 ≤ A ≤ 13/16

(b) 10 (d) 7

(b) 13/16 ≤ A ≤ 1 (d) 3/4 ≤ A ≤ 1

Answers

1. (c) 11. (b)

2. (d) 12. (b)

3. (a) 13. (a)

4. (a) 14. (d)

5. (c)

6. (c)

7. (d)

8. (a)

9. (d)

10. (a)

22

Trigonometric Equations

CHAPTER

Summary of conceptS trIGonometrIc eQuatIon

a + b is not Note: If | c | ≤ statisfied, then no real solution exists. 2

An equation involving one or more trigonometric ratios of unknown angles is called a trigonometric equation. 1 For example, 2 cos θ + 3 cos 2θ = 0, cos2θ + sin θ = , etc. 3 are trigonometric equations in unknown angle θ.

2

Since the trigonometric functions are periodic, a solution generalised by means of periodicity is known as the general solution. For example, general solution of the equation: sin θ = 0 is θ = n π, where n is any integer. General Solutions of Trigonometric Equations Trigonometric Equation

General Solution

•  sin θ = 0

θ = nπ, n ∈ I

•  cos θ = 0

θ = (2n + 1) (π/2), n ∈ I

•  tan θ = 0

θ = nπ, n ∈ I

•  sin θ = sin α

θ = nπ + (– 1) α, n ∈ I

•  cos θ = cos α

θ = 2nπ ± α, n ∈ I

n

,

a 2 + b2 c

and cos β =

a 2 + b2

Some Useful Hints for Solving Trigonometric Equations (i) Squaring should be avoided as far as possible. If squaring is done, then check for extra solutions. For example, consider the equating sin θ + cos θ =1. On squaring, we get 1 + sin 2θ = 1 or sin 2θ = 0 ⇒ θ =

nπ , n = 0, ± 1, ± 2, ... 2

The values of the angle, θ = π and θ =

GeneraL SoLutIon

a + b2 2

b

sin α =

SoLutIon of a trIGonometrIc eQuatIon A value of the unknown angle which satisfies the given equation is called a solution of the equation. For example, consider the π equation 2 sin θ = 1. The of value the angle, θ = and θ = 4 3π π 3π satisfy this equation. Therefore, and are solutions of 4 4 4 the given equation 2 sin θ = 1.

a

where cos α =

3π do not satisfy the 2

given equation. So, we get extra solutions. Thus, if squaring is must, verify each of the solutions. (ii) Never cancel a common factor containing ‘θ’ from the two sides of an equation. For example, consider the equation tan θ =

2 sin θ. If we divide 1 both sides by sin θ, we get cosθ = , which is clearly not equiv2

alent to the given equation as the solutions obtained by sin x = 0 are lost. Thus, instead of dividing an equation by a common factor, take this factor out as a common factor from all terms of the equation. (iii) Make sure that the answer should not contain any value of unknown ‘θ’ which makes any of the terms undefined. (iv) If tan θ or sec θ is involved in the equation, θ should not be an odd multiple of π/2.

•  tan θ = tan α

θ = nπ + α, n ∈ I

•  sin2θ = sin2α

θ = nπ ± α, n ∈ I

(v) If cot θ or cosec θ is involved in the equation, θ should not be a multiple of π or 0.

•  cos2θ = cos2α

θ = nπ ± α, n ∈ I

(vi) The value of

•  tan θ = tan α

θ = nπ ± α, n ∈ I

•  a cos θ + b cos θ = c

θ = 2nπ + α ± β

2

2

f (θ) is always positive. For example,

cos θ = | cos θ | and not ± cos θ. 2

(vii) All the solutions should satisfy the given equation and lie in the domain of the variable of the given equation.

830

MULTIPLE-CHOICE QUESTIONS

Objective Mathematics

Choose the correct alternative in each of the following problems: 1. The general solution of the equation 7cos2θ + 3 sin2θ = 4 is π 3

(a) θ = 2nπ ± (c) θ = nπ ±

π 3

(b) θ = 2nπ ±

2π 3

(d) None of these

3 cos x + 1 = 0 is

(a) x = 2nπ ±

7π 6

(b) x = 2nπ ±

3 = 2 ( 3 + 1) cosθ, then θ =

π 3 π (c) 2nπ ± 6

(b) 2nπ ±

(a) 2nπ ±

π 4

(d) None of these

9. The solution of the equation cos2θ + sin θ + 1 = 0, lies in the interval

2. The general solution of the equation 2 sin 2 x +

8. If 4cos2θ +

5π 3

 −π π  (a)  ,   4 4

 π 3π  (b)  ,  4 4 

 3π 5π  (c)  ,   4 4 

 5π 7 π  (d)  ,  4 4 

5π (d) None of these 10. The general solution of the equation 6 3. The solution of the equation sin 2x + sin 4x = 2 sin 3x sin x + cos x = 2 cos A is is π π ± A (b) x = 2nπ + ±A (a) x = 2nπ + nπ 3 4 (b) x = nπ (a) x = 3 π (c) x = 2nπ + ± A (d) None of these (c) x = 2nπ (d) None of these 6 (c) x = 2nπ ±

4. The general value of θ satisfying the equation 3 tan (θ – 15º) = tan (θ + 15º), is π π nπ (b) + (– 1)n (a) nπ + (– 1)n 2 4 4 π nπ (c) + (– 1)n 6 2

(d) None of these

1 5. The values of θ satisfying sin2θ – cosθ = in the interval 4 0 ≤ θ ≤ 2π, are π 4π , 3 3 π 5π (c) , 3 3

(a)

π 7π , 3 3 π 5π (d) , 6 6 (b)

6. The solution set of 2 sin2θ = 3 cos θ, in the interval 0 ≤ θ ≤ 2π, is π (a)   3  π 5π  (c)  ,  3 3 

  π 5π (b)  , , cos −1 (−2)  3 3 (d) None of these

7. The general solution of 8 tan2  −1  (a) x = 2nπ ± cos– 1   3 π (b) x = 2nπ ± 6 1 (c) x = 2nπ ± cos–1   3 (d) None of these

x = 1 + sec x is 2

11. The general solution of cos θ cos 2θ cos 3θ = (a) θ = (2n + 1)

π 4

(b) θ = (2n + 1)

π 8

1 is 4

π (d) None of these 3 12. The general solution of the equation (c) θ = nπ ±

tan 2 x + (1 − 3 ) tan x – 3 = 0 is π π (b) nπ – 4 4 π π (c) nπ + (d) nπ – 3 3 13. The solution of the equation tan 2θ tan θ = 1 is (a) nπ +

nπ π nπ π (b) θ = + + 3 6 3 4 nπ π − (c) θ = (d) None of these 3 6 14. Number of solutions of the equation tan x + sec x = 2cos x, lying in the interval [0, 2π], is (a) θ =

(a) 0 (c) 2

(b) 1 (d) 3

15. The solution set of the equation 5cos 2θ + 2 cos2 = 0, in the interval – π < θ < π , is π π  3   3  (a)  , π − cos −1    (b)  , π − cos −1       5  6 5 3    π  3  (c)  , π − cos −1    (d) None of these  5  4 

θ +1 2

tan θ + tan 2θ +

3 tan θ tan 2θ = 3 is π nπ (b) θ = (a) θ = (3n + 1) 9 9 (c) θ = (3n – 1)

π 9

(d) None of these

17. The set of values of x for which

tan 3 x − tan 2 x =1 1 + tan 3 x tan 2 x

is (a) φ

4

(b) nπ (d) None of these

19. The general value of θ satisfying the equation 2sin 2θ – 3sin θ – 2 = 0 is π π (a) nπ + (– 1)n (b) nπ + (­– 1)n 6 2 5 π 7 π (c) nπ + (– 1)n (d) nπ + (­– 1)n 6 6 20. The most general value of θ which satisfies both the 1 1 and tan θ = , is equations sin θ = ­– 3 2 π 6 11π (c) 2nπ + 6 (a) 2nπ +

(b) 2nπ +

7π 6

(d) None of these

(a) 2nπ ±

π π + 4 12

(b) nπ + (­– 1)n

π π + 4 12

(c) 2nπ ±

π π − 4 12

(d) nπ + (– 1)n

π π − 4 12

22. The most general value of θ which satisfies both the 1 and tan θ = 1 is equations cos θ = – 2

5π (c) 2nπ + 4

π 2

(d) None of these

(b) 2nπ +

π , tan x + tan y = 1 is 4 π π – nπ, y = nπ (b) x = – nπ, y = nπ (a) x = 2 4 π – nπ, y = 2nπ (d) None of these (c) x = 4 27. If tan (cot x) = cot (tan x), then (a) sin 2x =

2 4 (b) sin x = (2n + 1) π (2n + 1) π

(c) sin 2x =

4 (c) None of these (2n + 1) π

28. The 2sec 2α = tan β + cot β, then one of the values of  α + β is (a)

π 4

7π 4

(d) None of these

(b)

π 2

(d) nπ –

π ,n∈I 4

29. The value of θ satisfying cos θ + 5π 3 2π (c) 3 (a)

( 3 − 1)  sin θ + ( 3 + 1)  cos θ = 2 is

3π 4

(b) (2n + 1)

x+y=

(c) π

21. General solution of the equation

(a) 2nπ +

(d) None of these

26. Solution of the system of equations

18. The general solution of 4 sin x + cos x = 1 is

2 5

(c) no real solution

π 2 π (c) (2n + 1) 4

4

(c) nπ ± sin– 1

(b) x = sin­– 1 log (2 − 5 )

(a) (2n – 1)

π   (d) 2nπ + : n = 1, 2, 3... 4   π 2

(a) x = 0

25. The common roots of the equations 2 sin2x + sin22x = 2 and sin 2x + cos 2x = tan x are

π (b)   4 π   (c) nπ + : n = 1, 2, 3... 4  

(a) (2n + 1)

nπ nπ (b) θ = 4 12 nπ (c) θ = (d) None of these 6 24. The solution of the equation esin x – e– sin x – 4 = 0 is (a) θ =

3 sin θ = 2 is

4π 3 π (d) 3 (b)

30. The quadratic equation 8 sec2θ – 6secθ + 1 = 0 has (a) exactly two roots (b) exactly four roots (c) infinitely many roots (d) no roots 31. The equation (cos p – 1) x2 + (cos p) x + sin p = 0, in the variable x has a real root. then p can take any value in the interval (a) (0, π)

 −π π  (b)  ,   2 2

(c) (– π, 0)

(d) (0, 2π)

831

23. The general of the equation tan θ + tan 4θ + tan 7θ = tan θ tan 4θ tan 7θ is

Trigonometric Equations

16. The general solution of the equation

832

32. cot θ = sin 2θ (θ ≠ nπ, n integer), if θ equals

Objective Mathematics

(a) 45º or 90º (c) 90º only 33. If 5cos 2θ + 2 cos2

(b) 45º or 60º (d) 45º only θ + 1 = 0, – π < θ < π, then θ is 2

equal to 3 (a) cos– 1    5

π −1  3  (b) , π − cos   5 3

π 3 (c) , cos −1   5 3

π (d) 3

1 34. The most general solution of tan θ = – 1 and cos θ = is 2 7π 4 7π (c) 2nπ + 4

n (b) nπ + (−1)

(a) nπ +

7π 4

(d) None of these

35. The number of points of intersection of 2y = 1 and y = sin x, – 2π ≤ x ≤ 2π, is (a) 2 (c) 4

(b) 3 (d) 1

36. The general value of θ,obtained from the equation cos 2θ = sin α, is π α (a) θ = nπ ±  −  4 2

nπ + (−1) n π (b) θ = 2

π π  (c) θ = 2nπ ±  − α  (d) 2θ = − α 2 2  37. The general value of x satisfying the equation 3 sin x + cos x =

3 , is given by

π π + 4 3 π π (b) x = nπ + (– 1)n − 3 6 π (c) x = nπ ± 6 π (d) x = nπ ± 3 (a) x = nπ + (– 1)n

38. The number of solutions of the equation tan x + sec x = 2cos x lying in the interval [0, 2π], is (a) 3 (c) 1

(b) 2 (d) 0

39. For m ≠ n, if tan mθ = tan nθ, then the different values of θ are in (a) A.P (c) G.P

(b) H.P (d) no particular sequence.

40. The number of integral values of k for which the equation 7cos x + 5sin x = 2k + 1 has a solution is (a) 4 (c) 10

(b) 8 (d) 12

41. If

tan 3 x − tan 2 x = 1, then x is equal to : 1 + tan 3 x + tan 2 x

(a) φ (b) π/4 π   (c)  nπ + , n = 1, 2, 3 ... 4   π (d)  2nπ + , n = 1, 2, 3 ...   4 1 x π 42. The equation 2 cos2   . sin2x = x2 + 2 , 0 ≤ x ≤ 2 x 2 has (a) one real solution (b) no solution (c) more than one real solution (d) None of these 43. The number of the solutions of the equation

(

)

cos π x − 4 cos ( π x ) = 1 is (a) >2 (c) 1

(b) 2 (d) 0

44. If cot θ – tan θ = sec θ, then θ is equal to 3π 2 π (c) nπ + 2 (a) 2nπ +

(b) nπ + (­–1)n

π 6

(d) None of these

45. The solution of (2cos x – 1) (3 + 2cos x) = 0 in the interval 0 ≤ x ≤ 2π is π π 5π (b) , 3 3 3 π 5π −1  − 3  (c) , , cos   (d) None of these 3 3 2

(a)

46. The smallest positive angle which satisfies the equation 2 sin2θ + 3 cos θ + 1 = 0 is 5π 6 π (c) 3

2π 3 π (d) 6

(b)

(a)

47. The general solution of the trigonometric equation tan θ = cot α is π – α 2 π + α (c) θ = nπ + 2 (a) θ = nπ +

π +α 2 π (d) θ = nπ – –α 2 (b) θ = nπ –

48. If cos pθ = cos qθ, p ≠ q, then (a) θ = 2nπ (c) θ =

nπ p+q

(b) θ =

2nπ p±q

(d) None of these

57. If 32 tan8θ = 2cos2α – 3cos α and 3cos 2θ = 1, then the general value of α is

then sin 2α is equal to (a) (c)

24 25

(b) – 

13 18

(d) –

24 25 13 18

50. The real roots of the equation cos7x + sin4 x = 1, in the interval (– π, π), are (a) 0,

π π ,− 3 3

(b) 0,

π π (c) 0, , − 2 2

π 3

(b)

π (c) 6

π 4

7π (d) 18

(b) a = 1 1 (d) < a < 1 2

53. The number of real solutions of the equation sin (ex) = 5x + 5 – x is (a) 0 (c) 2

(a) 45º, 60º, 75º (c) 20º, 60º, 100º

(b) 30º, 60º, 90º (d) None of these

(a) x = nπ (c) x = (2n + 1)

π 2

(d) None of these

 π 5π  (a)  ,  6 6 

5π   (b)  −1,   6 

(c) (– 1, 2)

π  (d)  , 2  6 

60. The number of points of intersection of the two curves y = 2sin x and y = 5x2 + 2x + 3 is (a) 0 (c) 2

(b) 1 (d) ∞

61. The number of distinct values of θ satisfying 0 ≤ θ ≤ π and satisfying the equation sin θ + sin 5θ = sin 3θ,is (a) 6 (c) 8

(b) 7 (d) 9

62. The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos 2x + a3 sin2x = 0 for all x is (a) 0 (c) 3

(b) 1 (d) infinite

63. If sin4x + cos4y + 2 = 4 sin x cos y and 0 ≤ x, y ≤

π 2

then sin x + cos y is equal to (a) – 2 (c) 2

(b) 0 (d) None of these

64. The equation sin4x – (k + 2) sin2x – (k + 3) = 0 possesses a solution if

2π , 55. The solution set of the system of equations x + y = 3 3 , where x and y are real, is cos x + cos y = 2 π (a) x = − nπ, y = nπ (b) φ 3 65. π (c) x = nπ, y = − nπ, (d) None of these 3 56. Solution of the equation 4sin4 x + cos 4 x = 1 is

2π 3

n π π + cos = , then 2 2n 2n (a) 6 ≤ n ≤ 8 (b) 4 < n ≤ 8 (c) 4 ≤ n < 8 (d) 4 < n < 8.

(b) 1 (d) infinitely many

54. ABC is a triangle such that sin (2A + B) = sin (C – A) 1 . If A, B and C are in A.P., = – sin (B + 2 C) = 2 then the values of A, B, and C are

(c) 2nπ ±

sin

52. The equation sin4x + cos4x = a has a solution for (a) all of values of a 1 (c) a = 2

(b) 2nπ ± cos­– 12

59. Let n be a positive integer such that

π which satisfy the equation 51. The values of x, 0 ≤ x ≤ 2 2 2 81sin x + 81cos x = 30 are (a)

π 3

58. Let 2sin 2 x + 3sinx – 2 > 0 and x2 – x – 2 < 0 (x is measured in radians). Then x lies in the interval

π π ,− 4 4

(d) None of these

(a) 2nπ ±

(a) k > – 3 (b) k < – 2 (c) – 3 ≤ k ≤ – 2 (d) k is any positive integer The least difference between the roots, in the first π  quadrant  0 ≤ x ≤  , of the equation 2  4cos x (2 – 3sin 2 x) + (cos 2x + 1) = 0, is

 3 (b) x = 2nπ ± cos– 1    5

(a)

π 6

(b)

π 4

(d) None of these

(c)

π 3

(d)

π 2

833

π 6 (d) a ∈ [2, 6]

78. The equation a sin x + b cos x = c where

70. Solution of the equation 3 tan (θ – 15º) = tan (θ + 15º) is (a) θ = nπ –

π 4

(c) nπ +

π 2 π 2

(b) 2nπ + (d) nπ +

π 3 π 3

81. The set of all x in (– π, π) satisfying | 4sin x – 1 | < is given by  π 3π  (a)  − ,   10 10 

 π  (b)  − , π   10 

(c) (– π, π)

3π   (d)  − π,  10  

(b) θ =

5

(a) 3 (c) 5

(b) 4 (d) None of these

(a)   θ = nπ ±

83. The number of solutions of the equation tan x + sec x = 2 cos x, which lie in the interval [0, 3π], is (a)  2 (c)  3

(b)  1 (d)  4

84. The general solution of the equation cot θ – tan θ = sec θ is (n ∈ I) (a)   nπ + (−1) n (c)   2nπ +

π 6

(b)   nπ +

π 6

(c)   θ = 2nπ ± π + π 4 6

(a)  kπ, k ∈ I (c)  

(b)  1 (d)  4

(b)  two (d)  zero

π π − 4 6

θ is 2

kπ , k ∈I 2

(d)  None of these

(b)  2n π – α (d)  n π – α

92. 2 sin 2θ + √3 cos θ + 1 = 0. Then θ = ….. π 2 3π (c)   2

π 6 5π (d)   6

(b)  

(a)  

93. The solution of the equation 4 sin4 x + cos 4 x = 1 is

87. Let (sin a)x2 + (sin a)x + (1 – cos a) = 0. The value of a for which roots of this equation are real and distinct, is (b)  (0, 2π/3) (d)  (0, 2π)

88. The number of solutions of the pair of equations 2 sin 2 θ – cos 2θ = 0 and 2 cos2 θ – 3 sin θ = 0 in the interval [0, 2π] is (a)  zero (c)  two

(d)   θ = 2nπ ±

(b)  2kπ, k ∈ I

(a)  2n π + α (c)  n π + α

86. The number of solutions of the equation |cos x| = 2[x], where [·] is the greatest integer, is

(a)  (0, 2 tan –1 1/4) (c)  (0, π)

π π − 4 6

91. The most general value of θ satisfying the equations sin θ = sin α and cos θ = cos α is

85. The number of values of x in the interval [0, 3π] satisfying the equation 2 sin 2 x + 5 sin x – 3 = 0 is

(a)  one (c)  infinite

(b)   θ = nπ ±

90. The roots of the equation 1 – cos θ = sin θ · sin

π 6

(d)  z

(a)  6 (c)  2

π π + 4 6

(n − 1)π 4 2nπ (c)   3

(b)  nπ

(a)  

(d)  (2n + 1)π/2

94. The number of solutions of the pair of equations 2 sin 2θ – cos2θ = 0 2 cos2θ – 3 sinθ = 0 in the interval [0, 2π] is (a)  zero (c)  two

(b)  one (d)  four

(b)  one (d)  four

solutions 1. (a), (b)  We have, 7 cos2θ + 3sin 2θ = 4 ⇒ 7cos2θ + 3 (1 – cos2θ) = 4 1 ⇒ 4 cos2θ = 1 ∴ cos θ = ± 2 Taking positive sign, π π 1 = cos ⇒ 0 = 2nπ ± cos θ = 3 3 2 Taking negative sign, 2π 2π 1 cos θ = – = cos ⇒ θ = 2nπ ± 3 3 2 ∴ θ = 2nπ ±

π 2π , 2nπ ± , 3 3

2. (c) We have, 2sin2x + ⇒ 2 (1 – cos2x) + ⇒ 2cos x – 2

where n ∈ I.

3 cos x + 1 = 0 3 cos x + 1 = 0

3 cos x – 3 = 0

∴ cos x =

3 ± 3 + 24 = 4

3±3 3 4

3 2 Since 3 is greater than 1, ∴ cos x = 3 is not possible as cos x cannot be greater than 1. =

3 or −

3 π = ­– cos = cos 2 6 5π . = cos 6 5π . ∴ x = 2nπ ± 6

∴ cos x = −

π  π −  6 

3. (a), (c)  We have, sin 2x + 4x = 2sin 3x ⇒ (sin 2x + sin 4x) – 2sin 3x = 0 ⇒ 2sin 3x cos x – 2sin 3x = 0

835

3 cos θ + sin θ = 2 is

Trigonometric Equations

82. The number of the solutions of the equation sin 5x cos 3x 89. The most general solution of = sin 6x cos 2x in the interval [0, π], is

836

⇒ 2sin 3x (cos x – 1) = 0

Objective Mathematics

nπ and 3 when cos x = 1 = cos 0 ⇒ x = 2nπ ± 0 = 2nπ.

When sin 3x = 0 ⇒ 3x = nπ or x = nπ ∴ x = 3

or

2nπ, n ∈ I.

⇒

tan (θ + 15º ) 3 = tan (θ − 15º ) 1

⇒

⇒ cos θ = 1/2 = cos π/3 or cos θ = 3 /2 = cos π/6.

sin (θ + 15º + θ − 15º ) =2 sin (θ + 15º − θ + 15º )

⇒ 2 sin 2θ = 2 ⇒ sin 2θ = 1 = sin

or

θ = 2nπ ± π/6.

9. (d) We have, cos2θ + sin θ + 1 = 0

[By componendo and dividendo]

⇒ 2θ = nπ + (– 1)n

⇒ (2cos θ – 1) (2cos θ – 3 ) = 0.

∴ θ = 2nπ ± π/3

4 tan (θ + 15º ) + tan (θ − 15º ) = 2 tan (θ + 15º ) − tan (θ − 15º )



8. (a), (c) We have, 4cos2θ + 3 = 2 ( 3 + 1) cos θ ⇒ 4cos2θ – 2 ( 3 + 1) cos θ – 3 = 0

4. (b) We have, 3 tan (θ – 15º) = tan (θ + 15º) ⇒

1 = cos α (say) ⇒ x = 2nπ ± α 3 1 ∴ x = 2nπ ± cos– 1   , where n ∈ I. 3 ⇒ cos x =

∴ sin 3x = 0 or cos x = 1

π 2

⇒ ⇒ ⇒ ⇒ ⇒

1 – sin2θ + sin θ + 1 = 0 sin2θ – sin θ – 2 = 0 (sin θ + 1) (sin θ – 2) = 0 sin θ = – 1 ( sin θ ≠ 2) θ = 3π/2 ∈ (5π/4, 7π/4). ­

10. (a) We have, sin x + cos x = 2 cos A

π 2

Dividing throughout by 2 , we get

nπ π + (−1) n , n ∈ I. ∴ θ = 2 4 1 4

5. (c) We have, sin2θ – cos θ = ⇒ 4cos θ + 4 cos θ – 3 = 0 2



1 1 sin x + cos x = cos A 2 2

⇒

1 1 cos x + sin x = cos A 2 2 π π = cos A + sin x sin 4 4

⇒ (2cos θ + 3) (2cos – 1) = 0

⇒ cos x cos

But cos θ = – 3/2 (rejected)

π  ⇒ cos  x −  = cos A; 4  π ∴ x – = 2nπ ± A, n ∈ I 4 π ⇒ x = 2nπ + ± A, n ∈ I. 4

1 π = cos ∴ cos θ = 2 3

∴ θ = 2nπ ± π/3.

We have to choose values of θ s.t. 0 ≤ θ ≤ 2π ∴ θ = π/3, 2π–­ π/3, i.e., θ =

π 5π . , 3 3

6. (c) We have, 2sin2θ = 3 cos θ ⇒ 2cos2θ + 3 cos θ – 2 = 0 ⇒ (cos θ + 2) (2cos θ – 1) = 0 ∴ ⇒ We ∴

1 π cos θ = – 2 (rejected) or cos θ = = cos 2 3 θ = 2n π ± π/3, n ∈ I. have to choose values of θ s.t. 0 ≤ θ ≤ 2π. θ = π/3, 2π – π/3, ∴ θ = π/3, 5π/3.

7. (c) We have, 8tan2

x = 1 + sec x 2

 1 − cos x  1 1 + cos x ⇒ 8  =  =1+ 1 + cos x cos x cos x   ⇒ ⇒ ⇒ ⇒

8cos x – 8cos2x = (1 + cos x)2 8cos x – 8cos2x = 1 + cos2x + 2cos x 9cos2x – 6cos x + 1 = 0 (3cos x – 1)2 = 0 ⇒ 3 cos x – 1 = 0

11. (b), (c) We have, 4 cos θ cos 2θ cos 3θ = 1 ⇒ (2cos 3θ cos θ) ⋅ 2 cos 2θ = 1 ⇒ (cos 4θ + cos 2θ) 2cos 2θ – 1 = 0 ⇒ 2 cos 4θ cos 2θ + 2 cos22θ – 1 = 0 ⇒ 2cos 4θ cos 2θ + cos 4θ = 0 ⇒ cos 4θ [2cos 2θ + 1] = 0 If cos 4θ = 0, 4θ = (2n + 1) ⇒ θ = (2n + 1) ⇒ cos 2θ = – ∴ θ = nπ ±

π 8

If

π 2

2cos 2θ + 1 = 0,

1 2π 2π = cos ⇒ 2θ = 2nπ ± 2 3 3

π . 3

Hence, θ = (2n + 1)

π π , nπ ± , n ∈ Ι. 8 3

⇒ (tan x + 1) (tan x – 3 ) = 0. π ⇒ tan x = – 1 = − tan = tan (– π/4). 4 ∴ x = nπ – π/4, n ∈ I.

18. (b), (c) We have, 4 sin4x + cos4x = 1 ⇒ 4 sin4x = 1 –cos4x = (1 – cos2x) (1 + cos2x) ⇒ sin2x [4 sin2x – 1 – (1 – sin2x)] = 0 ⇒ sin2x (5 sin2x – 2) = 0 ∴ sin x = 0 or sin x = ± 2 / 5 ⇒ x = nπ or x = nπ ± α, where sin α = 2 / 5 , n ∈ I.

or tan x = 3 = tan (π/3), ∴ x = nπ + π/3, n ∈ I. 13. (a) We have, tan 2θ tan θ = 1 ⇒

sin 2θ sin θ ⋅ =1 cos 2θ cos θ

19. (d) We have, 2sin2θ – 3sin θ – 2 = 0

⇒ sin 2θ ⋅ sin θ = cos 2θ ⋅ cos θ

⇒ (2 sin θ + 1) (sin θ – 2) = 0

⇒ cos 2θ cos θ – sin 2θ sin θ = 0 ⇒ cos (2θ + θ) = 0 ⇒ cos 3θ = 0 = cos ∴ 3θ = nπ +

π nπ π + . ⇒θ= 2 3 6

⇒ sin θ = – 1/2 = sin (– π/6),

π 2

14. (c) Clearly, the given equation is meaningful when cos x ≠ 0 i.e., sin x ≠ 1 or – 1. We have, tan x + sec x = 2cos x

20. (b) We have sin θ = –

⇒ sin x + 1 = 2 cos2x = 2 (1 – sin2x)

Also, tan θ = 1/ 3 = tan (π/6)or tan (π + π/6) ∴ θ = π/6, 7π/6. Hence, the value of θ between 0 and 2π which satisfies both the equations is 7π/6. Hence, the general value of θ is 2nπ + 7π/6, where n ∈ I.

⇒ (1 + sin x) (2sin x – 1) = 0 ( sin x ≠ – 1)

⇒ x = π/6, 5π/6 in [0, 2π]. ­ Hence the number of solutions is 2. 15. (b) We have, 5cos 2θ + 2cos2

θ +1=0 2

21. (a) Let

⇒ 5 (2 cos θ – 1) + (1 + cos θ) + 1 = 0 2

3 + 1 = r cos α, and

⇒ 10 cos θ + cos θ – 3 = 0 ⇒ (5cos θ + 3) (2cos θ – 1) = 0 16. (a) We have, tan θ + tan 2θ + 3 tan θ tan 2θ 3

3 – 1 = r sin α

∴ r2 = ( 3 + 1)2 + ( 3 – 1)2 = 8,

2

∴ θ = π/3, π – cos– 1 (3/5).

π 1 π  = – sin = sin  π +  2 6 6 

or sin (2π – π/6) ∴ θ = 7π/6 ; 11π/6

⇒ 2 sin2x + sin x – 1 = 0 ⇒ sin x = 1/2, 

sin θ ≠ 2

⇒ θ = nπ – (– 1)n (π/6). Also, we can take sin θ = – 1/2 = sin (7π/6) ⇒ θ = nπ + (– 1) n (7π/6).

and tan α =

3 −1 1−1/ 3 π π = = tan  −  4 6 3 +1 1+1/ 3

= tan (π/12), i.e., α = π/12. From the given equation, we get r cos (θ –­ α) = 2

⇒ tan θ + tan 2θ = 3 – 3 tan θ tan 2θ

⇒ cos (θ – π/12) = 1 / 2 = cos (π/4)



∴ θ – π/12 = 2nπ ± π/4 or θ = 2nπ ± π/4 + π/12.

⇒

= 3 (1 – tan θ tan 2θ) tan θ + tan 2θ = 3 1 − tan θ tan 2θ

π π ⇒ tan 3θ = tan 3 3 π (3n + 1) π ∴ 3θ = nπ + , ⇒ θ = , n ∈ I. 3 9 ⇒ tan (θ + 2θ) = tan

tan 3 x − tan 2 x =1 17. (a) We have, 1 + tan 3 x tan 2 x ⇒ tan (3x – 2x) = 1 ⇒ tan x = 1 ∴ x = nπ + π/4  ut this value does not satisfy the given equation B as

22. (c) We have, cos θ = – 1 / 2 = – cos (π/4) = cos (π – π/4) or cos (π + π/4) ∴ θ = 3π/4, 5π/4 Also, tan θ = 1 = tan (π/4) or tan (π + π/4) ∴ θ = π/4, 5π/4 Hence, the value of θ between 0 and 2π which satisfies both the equations is 5π/4. ∴ General value is 2nπ + 5π/4. 23. (b) We write the given equation as tan θ + tan 4θ = – tan 7θ (1 – tan θ tan 4θ) ⇒ (tan θ + tan 4θ)/(1 – tan θ tan 4θ) = – tan 7θ ⇒ tan (θ + 4θ) = – tan 7θ

837

tan 2x = tan π/2 = ∞ and it reduces to indeterminate form. So the solution is φ.

3 =0

Trigonometric Equations

12. (b), (c) We have, tan2x + (1 – 3 ) tan x –

838

28. (a) Given, 2 sec 2α = tan β + cot β

Objective Mathematics

⇒ tan 5θ = tan (– 7θ) ∴ 5θ = nπ + (– 7θ) or 12θ = nπ ∴ θ = nπ/12, n ∈ I.

⇒ 2sec 2α =

24. (c) Put esin x = t ∴ t2 – 4t – 1 = 0. ⇒ t = e

sin x



=2± 5

∴ sin x = loge (2 + 5 ), loge (2 – 5 )

=

1 + tan 2 β sec 2 β = tan β tan β

2 = 2 cosec 2 β 2cos β ⋅ sin β

π  ∴ sec 2α = sec  − 2β  2 

2+ 5 >e

⇒ sin x > 1, and 2 – 5 = – ive and logex i s defined only when x is + ive. Hence there does not exist any real solution.

π  ⇒ 2α = 2nπ ±  − 2β  2  Taking +ve sign, we have 2 (α + β) = 2nπ +

25. (c) The first equation can be written as or α + β = nπ +

sin22x = 2 – 2 sin2x ⇒ 4sin2x cos2x = 2cos2x

⇒ cos2x (2 sin2x – 1) = 0



⇒ cos2x cos 2x = 0 

⇒

⇒ tan x (2 cos2x – 1) + cos 2x = 0 ...(2)

 rom (1) and (2), we find that the common roots of F the two given equations are given by cos 2x = 0 ⇒ 2x = nπ + π/2 or x = (2n + 1) π/4, n ∈ I. 26. (b) We have, tan (x + y) = ⇒ 1 =

tan x + tan y 1 − tan x tan y

1  1 − tan x tan y

⇒ sin (30º + θ) = 1 ⇒ 30º + θ = 90º ⇒ θ = 60º =

π . 3

6 ± 36 − 32 6±2 1 1 = = or 16 6 2 4 ⇒ cos θ = 2 or 4.

30. (d) We have, sec θ =

Both are not possible ( | cos θ | ≤ 1) [Using given equations]

⇒ tan x tan y = 0 ⇒ Either tan x = 0 or tan y = 0 π π −x = ∴ Either x = nπ and y = – nπ 4 4 π or y = nπ and x = – nπ. 4 π  27. (c) We have, tan (cot x) = tan  − tan x  2 

1 (2n + 1) π = sin 2 x 4

 ∴ sin 2x =

p sin p 2

⇒ cos θ = 2 sin2θ cos θ ⇒ cos θ [2sin2θ – 1] = 0 1 ⇒ cos θ = 0 or sin2θ = ⇒ θ = 90º or 45º . 2 33. (b) The given equation can be written as

cos 2 x + sin 2 x π ⇒ = (2n + 1) sin x cos x 2

or

⇒ cos2p ≥ – 8 sin2

32. (a) Given cot θ = sin 2θ ⇒ cos θ = sin θ sin 2θ

π cos x sin x = (2n + 1) + 2 sin x cos x

π 1 = (2n + 1) 2 × 2 2sin x cos x

For equation to have real roots, cos2p – 4 (cos p – 1) sin p ≥ 0 ⇒ cos2p ≥ 4 (cos p – 1) sin p

3π π or – , R.H.S. > 1 but L.H.S. < 1, 2 2 ∴ the choices (b), (c) and (d) are ruled out. The correct alternative is (a)

π ⇒ cot x + tan x = nπ + 2

⇒

31. (a) Discriminant = cos2 p – 4 (cos p – 1)sin  p

For p =

π  ⇒ cot x = nπ +  − tan x  2 

⇒

29. (d) We have, cos θ + 3 sin θ = 2 1 3 cos θ + sin θ = 1 2 2 ⇒ sin 30º cos θ + cos 30º sin θ = 1

sin 2x + cos 2x = tan x

⇒ 2sin x cos x – tan x + cos 2x = 0 ⇒ cos 2x (tan x + 1) = 0 

π , n ∈ I. 4

For n = 0, α + β = π/4. ...(1)

The second equation is

π 2

4 . (2n + 1)π

5 [2cos2θ – 1] + (1 + cos θ) + 1 = 0

⇒ 10 cos2θ + cos θ – 3 = 0 ⇒ (5cos θ + 3) (2cos θ – 1) = 0 3 1 ⇒ cos θ = or cos θ = − 5 2 π 3 ⇒ θ = or θ = π – cos– 1   in – π < θ < π. 3 5

The value of θ lying between 0 and 2π and satisfy7π ing these two is 4 7π ∴ θ = 2nπ + . 4 35. (c) Since 2y = 1 and y = sin x 1 π π = sin ⇒ x = nπ + (– 1)n ⋅ 2 6 6 ∴ values of x in the interval – 2π ≤ x ≤ 2π are ∴ sin x =



x=

π 5π 7 π 11π , ,− ,− . 6 6 6 6

and common difference

2 2 2 2 40. (b) − 7 + 5 ≤ 7 cos x + 5 sin x ≤ 7 + 5 .

So, for solution − 74 ≤ 2k + 1 ≤ 74 or –8.6 ≤ 2k + 1 ≤ 8.6  or  –9.6 ≤ 2k ≤ 7.6 or –4.8 ≤ k ≤ 3.8. So, integral values of k are –4, –3, –2, –1, 0, 1, 2, 3 (eight values). 41. (a) We have,

π  36. (a) We have, cos 2θ = sin α = cos  − α  2 

3  sin x + cos x = 3

3 1 sin x + cos x = ⇒ 2 2 ⇒ sin x cos

3  2

π π + sin cos x = 6 6

[dividing by 2] 3 2

π π  ⇒ sin  x +  = sin 3 6  π π = nπ + (– 1)– n , n ∈ I 6 3 π π ⇒ x = nπ + (– 1)n − , n ∈ I. 3 6 38. (b) We have, tan x + sec x = 2cos x sin x + 1 ⇒ = 2cos x cos x [cos x ≠ 0 i.e., sin x ≠ 1 or – 1] ⇒ 2cos2x = sin x + 1 ⇒ 2 [1 – sin2x] – (1 + sin x) = 0 ⇒ (1 + sin x) [2 (1 – sin x) – 1] = 0

⇒ x = or

[ sin x ≠ – 1]

π 6

5π in [0, 2π] 6

Hence, the number of solutions is 2. 39. (a) We have, tan mθ = tan nθ ⇒ mθ = nθ + Kπ ∴ θ =

kπ , where k ∈ I. m−n

π , but for this value 4

π  of x tan 2x = tan  2n π +  = ∞ which does 2  not satisfy the given equation as it reduces to indeterminate form. 42. (b) Since x2 + x– 2 = (x – x– 1)2 + 2 ≥ 2 x sin2x ≤ 2, and 2cos2 2 ∴ the given equation is valid only if x sin2x = 2 2cos2 2 x = cosec x = 1, which cannot be true. 2 43. (c) Clearly, x ≥ 4 (Since x − 4 is real) so that x is also real. ⇔ cos

∴ x +

⇒ (1 + sin x) (1 – 2 sin x) = 0 1 ⇒ sin x =  2

tan 3 x − tan 2 x = 1, 1 + tan 3 x + tan 2 x

⇒ tan x = 1  ⇒  x = n x +

π π α  ⇒ 2θ = 2nπ ±  − α  ∴ θ = nπ ±  −  a 2 4 2  37. (b) We have,

π . m−n

π m−n

Again, if cos (π x ) < 1 then cos (π x − 4 ) >1 and if cos (π x ) >1 then cos ( x − 4 ) < 1 (since their product = 1). But both of these are not possible (since cos θ cannot be greater than 1). ∴ cos (π x − 4 ) = 1 and cos (π x ) = 1 ∴ x ­– 4 = 0 and x = 0.  ut x = 0 is not possible, ∴ x = 4 is the only B solution. 44. (b) We have, cot θ – tan θ = sec θ ⇒

cos θ sin θ 1 − = [cos θ ≠ 0, sin θ ≠ 0] sin θ cos θ cos θ

⇒ cos2θ – sin2θ = sin θ ⇒ 1 – 2sin2θ = sin θ ⇒ 2sin2θ + sin θ – 1 = 0 ⇒ (2sin θ – 1) (sin θ + 1) = 0 1 π = sin  2 6 π ⇒ θ = nπ + (– 1)n . 6 ⇒ sin θ =

[ sin θ ≠ – 1]

839

These values of θ are in A.P with first term

Trigonometric Equations

34. (c) tan θ = – 1, cos θ = 1 / 2

840

Objective Mathematics

45. (b) We have (2cos x – 1) (3 + 2 cos x) = 0. 1 π 5π If 2cos x – 1 = 0, then cos x = ,∴x= , 2 3 3 3 If 3 + 2 cos x = 0, then cos x = − , 2 which is not possible ( – 1 ≤ cos x ≤ 1) 46. (a) We have, 2sin 2θ +

3 cos θ + 1 = 0

⇒ 2 (1 – cos2θ) + 3 cos θ + 1 = 0 ⇒ 2cos2θ – 3 cos θ – 3 = 0 ⇒ (cos θ – 3 ) (2 cos θ + 3 ) = 0 ⇒ cos θ =

3 2



81sin

π  47. (a) Given equation is tan θ = cot α = tan  − α  2   π −α. ∴ θ = nπ + 2 48. (b) Given equation is cos pθ = cos qθ ∴ pθ = 2nπ ± qθ 2nπ p±q

⇒ 81sin

2x

∴ 25 cos2 α + 5 cos α – 12 = 0 −5 ± 25 + 1200 50 4 5 24 . 25

50. (c) The given equation can be written as cos7x + sin4x – 1 = 0

⇒ cos7x – (1 – sin2x) (1 + sin2x) = 0 ⇒ cos2x (cos5x – 1 – sin2x) = 0 ⇒ cos2x (cos5x + cos2x – 2) = 0 ⇒ cos x = 0 or cos x + cos x = 2 But cos2x = 0 ⇒ cos x = 0 π ⇒ x = ± . 2 Since the maximum value of 2

[ 0 ≤ x ≤ π/2]

52. (b), (c), (d) The given equation can be written as 1 – 2 sin2x cos2x = a ⇒ sin22x = 2 (1 – a) ⇒ 2 (1 – a) ≤ 1 and 2 (1 – a) ≥ 0 ⇒ 1/2 ≤ a ≤ 1. ∴ sin (ex) ≥ 2, which is not possible for any real value of x. Thus, the given equation has no solution. 54. (a) We have, A + B + C = 180º Also, 2B = A + C  ( A, B, ∴ 3B = 180º or B = 60º. Let θ be the common difference, 2θ. 1 ∴ sin (C – A) = ⇒ sin 2θ = 2

C, are in A.P.) then C – A = 1 2

π 5π π 5π or ⇒ θ= or 6 6 12 12 π π π 5π or − − 3 12 3 12

π as A 2.  5 5 25  5 5  

⇒ (cos x – 1) (5cos x – 3) = 0 2

2

If cos2x = 1, x = nπ, n ∈ I.

 ince y = 2 sin x ≤ 2, so there cannot be any point S of intersection.

If 5cos2x – 3 = 0 ⇒ cos2x = (3/5) ⇒ cos x = ±

(3 / 5)

Therefore, x = 2nπ ± cos– 1

61. (a) We have, sin θ + sin 5θ = sin 3θ

(3 / 5) .

⇒ 2 sin 3θ cos 2θ = sin 3θ

1 − tan θ 1 1 ⇒ = 1 + tan 2 θ 3 3

⇒ sin 3θ (2cos 2θ – 1) = 0

2

57. (c) We have, cos 2θ =

⇒ sin 3θ = 0

⇒ tan2θ = 1/2 ⇒ tan8θ = 1/16. ∴ or, or, or, or,

⇒ θ = 0,

2cos2α – 3 cos α = 32 tan8θ = 2 2cos2α – 3cos α – 2 = 0 (2cos α + 1) (cos α – 2) = 0 cos α = – 1/2, [∴ cos α ≠ 2] α = 2nπ ± 2π/3, n ∈ I.

π 2π , , π 3 3

or

1 2

2θ =

π 5π , 3 3

Hence there are six values. 62. (d) We have, a1 + a2 cos 2x + a3 sin 2 x = 0 a3 (1 – cos 2x) = 0 2 which is zero for all values of x. a if a1 = – 3 = – ­a2 2 k k or a1 = – ,  a =  , a3 = k 2 2 2 ⇒ a1 + a2 cos 2x +

58. (d) We have, 2sin x + 3sin x – 2 > 0 2

⇒ (2sin x – 1) (sin x + 2) > 0 ⇒ 2sin x – 1 > 0

or cos 2θ =

[∴ sin x + 2 > 0 ∨ x ∈ R]

⇒ sin x > 1/2 ⇒ x ∈ (π/6, 5π/6) Also, x2 – x – 2 < 0

for any k ∈ R.

⇒ (x – 1) (x + 1) < 0 ⇒ – 1 < x < 2. 5π As 2 < , we obtain that x must lie in 6

Hence, the required number of triplets is infinite. .

63. (c) We have sin4 x + cos 4y + 2 = 4sin x cos y ⇒ sin4x + cos4y + 2 – 4 sin x cos y = 0 ⇒ (sin2x – 1)2 + (cos2y – 1)2 + 2sin2x + 2cos2y – 4sin x cos y = 0 ⇒ (sin2x – 1)2 + (cos2y – 1)2 + 2 (sin x – cos y)2 =0 which is true if sin2x = 1, cos2y = 1 and sin x = cos y. As, 0 ≤ x, y ≤ π/2, we get sin x = cos y = 1 ⇒ sin x + cos y = 2.

 π  π 59. (d) We have sin   + cos   =  2n   2n 

n 2 ⇒ sin2  π  + cos 2  π  + 2 sin  π  cos  π   2n   2n   2n   2n  n = 4 π n n−4 ⇒ sin = –1= n 4 4

64. (c) We have, sin4 x – (k + 2) sin 2 x – (k + 3) = 0

As sin (π/n) > 0 ∨ n > 1, we get

n−4 >0 ⇒n>4 4

Also, sin  π  + cos  π  =  2n   2n  ⇒

π π  2 sin  +  4 2n 

n π π = sin  +  4 2n  2 2

π π  Since sin  + < 1 ∨ n > 2, we get  4 2n 

n < 1 or n < 8 2 2 Thus, 4 < n < 8.

(k + 2) ± (k + 2) 2 + 4 (k + 3) 2 (k + 2) ± (k + 4) = 2 ⇒ sin2x = k + 3  ( sin2x = – 1 is not possible) Since 0 ≤ sin2x ≤ 1, ∴ 0 ≤ k + 3 ≤ 1 or – 3 ≤ k ≤ – 2. ⇒ sin2x =



65. (a) We have, 4cos x (2 – 3 sin 2 x) + (cos 2x + 1) = 0 ⇒ 4cos x (3cos2x – 1) + 2 cos2x = 0 ⇒ 2cos x (6cos2x + cos x – 2) = 0 ⇒ 2cos x (3cos x + 2) (2cos x – 1) = 0 ⇒ either cos x = 0 which gives x = π/2 or cos x = – 2/3,

841

2 3  60. (a) We have, y = 5x2 + 2x + 3 = 5  x 2 + x +  5 5 

⇒ 5cos4x – 5cos2x – 3 cos2x + 3 = 0

Trigonometric Equations

⇒ 5cos4x – 8cos2x + 3 = 0

842

which gives no value of x for which 0 ≤ x ≤ π/2

Objective Mathematics

So, the required difference = π/2 – π/3 = π/6.

x Then, y = 2cos2   sin2x  2 and y = x2 + x– 2  x From (1), y = 2cos2 . sin2x 2 = (1 + cos x). sin2x < 2

...(1)

1 =  x −  x 

2

[ x > 0]

+2 ≥2

∴ y ≥ 2 

...(4)

 ince no value of y can be obtained satisfying (3) S and (4) simultaneously, ∴ no real solution of the equation exists. 67. (a), (b) We have,

3 2 2 sin x = cos x + cos x.

Squaring both sides, we get



3 (1 – cos2x) = 4 (cos2x + 2 cos3x + cos4x)

⇒ 4cos4x + 8cos3x + 7cos2x – 3 = 0 ⇒ (cos x + 1) (2cos x – 1) (2cos x + 3cos x + 3) =0 When cos x = – 1 = cos π, x = 2nπ + π = (2n + 1)π. 1 π π When cos x = = cos , x = 2nπ ± . 2 3 3

tan θ + 3 tan θ − 3 + 1 − (3) tan θ 1 + (3) tan θ

tan θ (1 − 3 tan 2 θ) + (tan θ + 3 ) (1 + 3 tan θ) + (tan θ − 3 ) (1 − 3 tan θ) 1 − 3 tan 2 θ



=



=

tan θ − 3 tan 3 θ + 2 tan θ + 3 (2 3 tan θ) 1 − 3 tan 2 θ



=

9 tan θ − 3 tan 3 θ (3 tan θ − tan 3 θ) = 3. 2 1 − tan θ 1 − 3 tan 2 θ

= 3tan 3θ. Hence, the given equation reduces to 3tan 3θ = 3 π π . ∴ 3 θ = nπ + ⇒ tan 3θ = 1 = tan 4 4 nπ π or θ = . + 3 12 70. (d) We have, 3 tan (θ – 15º) = tan (θ + 15º) ⇒

tan A 3 = , where A = θ + 15º, B = θ – 15º tan B 1

⇒

tan A + tan B 3 +1 = tan A − tan B 3 −1



2



= tan θ +

...(2)

π   ∵for 0 < x ≤ 2 , 0 ≤ cos x < 1 and 0 < sin x ≤ 1   ∴ y < 2  ...(3) 1 2 – 2 2 Also, from (2), y = x + x­ = x + x 2

sin

π 2π  69. (c) We have, tan θ + tan  θ +  + tan  θ +  3 3   

 x 66. (c) Let y = 2cos2   sin 2 x = x2 + x– 2 2



A−B A−B =0 ∴ = nπ 2 2 ⇒ A – B = 2nπ ∴ A = 2nπ + B.

or cos x = 1/2, which gives x = π/3.

⇒

(Apply componendo and dividendo) sin (A + B) =2 sin (A − B)

1 π = 1 = sin 2 2 π π ⇒ 2θ = 2nπ + or θ = nπ + . 2 4 ⇒ sin 2θ = 2sin 30º = 2 ⋅

When 2cos2x + 3cos x + 3 = 0, the discriminant 71. (b), (d) Given : sin 6x + cos 4x + 2 = 0 = 9 – 4 ⋅ 2 ⋅ 3 < 0. or sin 6x + cos 4x = – 2  ...(1) ∴ This factor does not give any real values of ⇒ sin 6x = – 1 cos x. π and cos 4x = – 1 both should be satisfied Hence, x = (2n + 1) π, 2nπ ± , n ∈ I. simultaneouly. 3 68. (b) We have, sin A – sin B = 0 and cos A – cos B =  [ minimum value of sin 6x and cos 4x is – 1] 0 3π 3π ⇒ 6x = 2nπ + Now, sin 6x = – 1 = sin A−B A+B =0 cos ⇒ 2sin 2 2 2 2 nπ π ∴ x = + , where n ∈ I. A+B B−A 3 4 and 2sin =0 sin 2 2 ⇒ Values of x between 0 and 2π are A+B A−B π 7 π 11π 15π 19π 23π or – 2sin = 0. sin , , , , , 2 2 4 12 12 4 12 12 From the two we observe that the common factor Also, cos 4x = – 1 = cos π ⇒ 4x = 2nπ + π gives

⇒ Values of x between 0 and 2π are

and 3 (1 – sin x) = 0 

π 3π 5π 7 π , , , 4 4 4 4

Hence, values of x lying between 0 and 2π and π 5π . satisfying both the equations are , 4 4

∴

⇒ sin x (2cos x – 1)2 = 1 ≠ 0 Thus, (2) and (3) can never be satisfied simultaneously. Hence no solution is possible. 75. (b) Given equation is sin x + cos (k + x) + cos (k – x) = 2 ⇒ sin x + 2cos k. cos x = 2 ⇒ 2cos k ⋅ cos x + sin x = 2

72. (c) We have, 4cot 2θ = cot 2θ – tan 2θ 1 4 = – tan2θ ⇒ tan 2 θ tan 2θ Put tan 2θ =

2 tan θ 1 − tan 2 θ

For real solutions, | c | ≤ ∴ | 2 | ≤

1 − tan 4 θ 4 (1 − tan 2 θ) = tan 2 θ 2 tan θ

⇒ cos2k ≥

⇒ (1 – tan2θ) [2tan θ – (1 + tan2θ)] = 0 ⇒ (1 – tan2θ) (tan θ – 1)2 = 0 ⇒ sin2k –

π ⇒ tanθ = 1, – 1. ∴ θ = nπ ± . 4 73. (c) We have, tan 3θ + tan θ = 2tan 2θ

or

2 ≤

⇒ – cos2k ≤ – 3 4

1 + 4 cos 2 k

3 4

⇒ sin2k ≤

1 4

1 1  1 ≤ 0 ⇒  sin k +   sin k −  ≤ 0 2 2 4 

1 1 ≤ sin k ≤ 2 2 π π ≤ k ≤ nπ + , n ∈ I. ⇒ nπ – 6 6 76. (d) Since 2cos x, | cos x | and 1 – 3cos2 x are in G.P.

sin (3θ − 2θ) sin (2θ − θ) = cos 3θ cos 2θ cos 2θ cos θ

⇒ sin θ (cos θ – cos 3θ) = 0 ⇒ sin θ. 2sin θ sin 2θ = 0 When sin θ = 0, θ = nπ. nπ 2 But for odd values of n, the values of θ given by

When sin 2θ = 0, 2θ = nπ or θ =

nπ do not satisfy the given equation, 2 nπ is rejected. ∴ θ = nπ, n ∈ I. ∴ θ = 2 θ=

74. (a) We have sin x + 2sin 2x = 3 + sin 3x ⇒ sin 3x – sin x – 2sin 2x + 3 = 0 ⇒ 2sin x cos 2x – 4sin x cos x + 3 = 0

∴ cos2x = 2cos x (1 – 3cos2x) ⇒ 6cos3x + cos2x – 2cos x = 0 ⇒ cos x (2cos x – 1) (3cos x + 2) = 0 π π 2 1 2 −1  ∴ x = , , cos  −  ⇒ cos x = 0, , − 2 3 2 3  3 π π ∴ If α = ,β= , α, β are are positive) 2 3 π . then | α – β | = 6 77. (d) Given equation is cos 2x + a sin x = 2a – 7 ⇒ 1 – 2 sin2x + a sin x = 2a – 7 ⇒ 2sin2x – a sin x + 2a – 8 = 0 ⇒ sin x =

⇒ sin x (2cos 2x – 4cos x) + 3 = 0

a ± a 2 − 16 (a − 4) a ± (a − 8) = 4 4

a−4 or sin x = 2 (not possible) 2 a−4 a−4 ∴ sin x = . – 1 ≤ sin x ≤ 1, ∴ – 1 ≤ 2 2 ≤ 1 ⇒ 2 ≤ a ≤ 6. ⇒ sin x =

⇒ sin x {2 (2cos2x – 1) – 4cos x} + 3 = 0 ⇒ sin x (4cos2x – 4cos x – 2) + 3 = 0 ⇒ sin x [(2cos x – 1)2 – 3 ] + 3 = 0. ⇒ sin x (2cos x – 1)2 + 3 (1 – sin x) = 0

3 4

a 2 + b2

⇒ –

⇒ tan 3θ – tan 2θ = tan 2θ – tan θ



1 + 4 cos 2 k

⇒ 1 – cos2k | ≤ | –

⇒ (1 – tan2θ) (tan2θ – 2tan θ + 1) = 0

⇒

...(3) ⇒ cos x = 0

From (3), sin x = 1

...(1)

 e have to find the solutions in the interval 0 ≤ W 78. (c) Given equation is a sin x + b cos x = c x ≤ π. c a sin x b In this interval, 0 ≤ sin x ≤ 1 + ⇒ cos x = 2 2 2 2 2 a + b2 a +b a +b ∴ 1 – sin x ≥ 0,

843

 lso, square term ≥ 0, hence the equation (1) holds A true only if ...(2) sin x (2cos x – 1)2 = 0

nπ π + , where n ∈ Z 2 4

Trigonometric Equations

∴ x =

844

dividing by a 2 + b 2    

Objective Mathematics

Let

a a 2 + b2

...(1) b

= cos α , then

a 2 + b2

c a + b2 2

> 1,

(∵|c | >

a 2 + b2

)

Therefore, in the interval [0, π], the given equation 5π π π or . is satisfied if x = 0, π, , 6 6 2 83. (d) We have, tan x + sec x = 2 cos x ⇒ sin x + 1 = 2 cos2x

which is impossible. ∴ The given equation has no solution.

⇒ 1 + sin x = 2(1 – sin x) ⇒ sin x = –1, 1 2 ⇒ x = π , 5π , 2π + π , 2π + 5π 6 6 6 6 (when sin x = – 1, then cos x = 0 and tan x is not defined)

Since a2 – 6a + 11 = (a – 3)2 + 2 > 2 for all a ∴ sin θ – cos θ = 1 1 π = sin 2 4

π π ⇒ θ – = nπ + (– 1)n 4 4 π π ∴ θ = nπ + (– 1)n + , where n ∈ I. 4 4

84. (a) We have, cot θ – tan θ = sec θ ⇒ cos2 θ – sin2 θ = sin θ, sin θ ≠ 0, cos θ ≠ 0 ⇒ 2 sin2 θ + sin θ – 1 = 0 ⇒ sin θ = –1, 1 But sin θ ≠ – 1 2 (∵ when sin θ = – 1, cos θ = 0) ∴ sin θ = 1 ⇒ θ = nπ + (–1)n π 6 2

80. (c) We have, sin 50 x – cos50 x = 1 ⇒ sin 50 x = 1 + cos50 x Since sin50x ≤ 1 and 1 + cos50x ≥ 1, therefore, the two sides are equal only if sin50x = 1 = 1 + cos50x i.e., sin50x = 1 and cos50x = 0 π ∴ x = nπ + , n ∈ I. 2 81. (a) We have, | 4sin x – 1 |
0 ⇒ sin a > 0 or (sin a – 4 + 4 cos a) > 0 ⇒ a ∈ (0, π) or

845

91. (c) Since, sin θ = sin α

87. (a) Since the roots of equation

846

3. The solution of the equation cos2x – 2cosx = 4sinx – sin 2x 9. If tan θ = (0 ≤ x ≤ π) is

Objective Mathematics

1 (a) π − cot   2

(b) π – tan 2

−1  1  (c) π + tan  −  2

(d) None of these

−1

π 6 π (c) 2nπ ± 6

10. If sin θ = 1, then the standard value of θ will be

4. The number of integral values of k for which the equation 7 cosx + 5 sinx = 2k + 1 has a solution is (a) 4 (b) 8 (c) 10 (d) 12 5. The solution of the inequality log½ sin x > log½ cos x in [0, 2π] is  π π (b)  − ,  4 4

 π (c)  0,  4

(d) None of these

6.

7. 8.

π 3 π (d) 2nπ + 3

(b) nπ +

(a) nπ +

–1

 π (a)  0,  2

3 , then the value of θ is

(a) nπ + (– 1)n (c) 2nπ ±

π 2

π 2

π 2 π (d) 2nπ + 2

(b) nπ ±

11. If tan θ + tan 2θ + 3 tan θ tan 2θ =

3 , then:

(a) θ = (6n + 1) ⋅ π/18 ∀ n ∈ I (b) θ = (5n + 1) ⋅ π/9 ∀ n ∈ I (c) θ = (3n + 1) ⋅ π/9 ∀ n ∈ I (d) None of these

12. The equation 3 sin 2x + 10 cosx – 6 = 0 is satisfied if: (a) x = nπ ± cos–1(1/3) (b) x = 2nπ ± cos–1(1/3) nπ sin 2 x − 3 sin x + 2 If x ≠ and (cos x ) = 1, then all solutions (c) x = nπ ± cos–1(1/6) (d) x = 2nπ ± cos–1(1/6) 2 1 of x are given by 13. The most general solution of tanθ = –1 and cosθ = 2 π π is: (a) 2nπ + (b) (2n + 1) π – 2 2 7π 7π π (a) nπ + (b) nπ + (−1) n (c) nπ + (– 1)n (d) None of these 4 4 2 7π (c) 2nπ + (d) None of these The number of solutions of the equation 4 x 1 + sin x sin2  = 0 in [– π, π] is tan 3 x − tan 2 x 2 14. If = 1, then x: 1 + tan 3 x tan 2 x (a) zero (b) 1 (c) 3 (d) None of these (a) φ The number of solutions of the equation 2cos x = [sin x] (b) π 4 in [– 2π, 2π] is (c) {nπ + π/4; n = 1, 2, 3, ...} (a) One (b) Four

(c) Six

(d) {2nπ + π/4; n = 1, 2, 3, ...}

(d) Eight

Answers

1. (d) 11. (c)

2. (a) 12. (b)

3. (c) 13. (a)

4. (b) 14. (c)

5. (c)

6. (d)

7. (a)

8. (d)

9. (b)

10. (a)

23

Inverse Trigonometric Functions

CHAPTER

Summary of concepts Inverse Functions

Some Important Formulae

If f : X → Y is a function which is both one-one and onto, then we define its inverse function f –1 : Y → X is defined as:

1. (a) sin–1 (sin θ) = θ, –

y = f (x) ⇔ f –1 (y) = x, ∨ x ∈ X, ∨ y ∈ Y.

Inverse Trigonometric Functions Consider the sine function with domain R and range [– 1, 1]. This function is many-one and onto. So, its inverse does not ex π π ist. If we restrict its domain to the interval  − ,  , then the  2 2 function  π π sin :  − ,  → [– 1, 1] given by sin θ = x  2 2 is one-one and onto and therefore it is invertible. The inverse of sine function is defined as  π π sin–1 : [– 1, 1] →  − ,   2 2 such that sin–1x = θ ⇔ sin θ = x. Thus, if x is a real number between – 1 and 1, then sin–1x is π π an angle between – and whose sine is x, i.e., 2 2 π π ≤θ≤ sin–1x = θ ⇔ x = sin θ, where – 2 2 and – 1 ≤ x ≤ 1. The least numerical value among all the values of the angle whose sine is x, is called the principal value of sin–1x. Similar definition for cos–1x, tan–1x etc. can be given.

Domain and Range of Inverse Trigonometric Functions Domain

Range (Principal Value)

1. y = sin x

–1≤x≤1

2. y = cos–1x

–1≤x≤1

π π – ≤y≤ 2 2 0≤y≤π

3. y = tan–1x

(– ∞, ∞)

–

4. y = cosec–1x

x ≥ 1 or x ≤ – 1

5. y = sec–1x

x ≥ 1 or x ≤ – 1

Function –1

6. y = cot x –1

(– ∞, ∞)

π 1



 1 = cosec–1  2  1− x

1 − x2 )

(c) If x, y ≥ 0, x2 + y2 ≤ 1



sin–1x ± sin–1y = sin–1 (x 1 − y 2 ± y

(b) If x, y ≥ 0, x2 + y2 > 1



= cosec–1



7. (a) If x, y ≥ 0, x2 + y2 ≤ 1



1 − x2

 1 − x2 (b) cos–1x = sin–1 1 − x 2 = tan–1   x 



π  , x > 0, y > 0 (c) If xy = 1, tan x + tan y =  2 − π , x < 0, y < 0  2 –1

= sec–1



  8. The value of sin  tan −1 (− 3 ) + cos −1  − 3   is  2      (a) 1 (c) 0

(b) – 1 (d) None of these

π 3 π (c) 4

π 3 π (d) – 4

(b) all real values of a (d) |a | ≥

1 2

11. If 2 tan– 1 (cos x) = tan– 1 (cosec2 x), then x is equal to (a)

π 2

(b) π

(c)

π 6

(d)

(c)

π 3

(d) None of these

1 − x2 x

(c) x

(b)

x 1 − x2



(d) None of these

 π −1  1  14. The value of tan  2 tan   −  is 5 4    (a)

7 17

(b) –

7 17

(c)

2 3

(d) –

2 3

15. If sin–1x = cos x  is –1

849

π 3 π 3π (d) or – 4 4 (b)

18. cot–1 9 + cosec–1 π 3 π (c) 6 (a)

41 is equal to 4 π (b) 4

19. If sin–1x + sin–1y =

(d) None of these 2π , then cos–1x + cos–1y = 3 π (b) 3 (d) π

20. The value of cos (2 cos–1x + sin–1x) at x =

13. The value of sin [cot–1 {tan (cos–1x)}] is (a)

π 4

(d) None of these

2π 3 π (c) 6

4 (b) 5

1 5

(b)

(a)

 −1 3  12. The value of sin  cos  is 5  3 (a) 5

π 3 π (c) 6

(a)

(a)

10. The trigonometric equation sin–1x = 2 sin–1a, has a solution for 1 1 cot–1x holds for

Objective Mathematics

(a) x > 1 (c) x = 1

(b) x < 1 (d) all values of x

60. If θ and φ are the roots of the equation 8x2 + 22x + 5 = 0, then (a) both sin–1θ and sin–1φ are real (b) both sec–1θ and sec–1φ are real (c) both tan–1θ and tan–1φ are real (d) None of these (b) x ∈ (– ∞, – 1] ∪ [1, ∞) (d) None of these

62. sec–1 (sin x) is real if (a) x ∈ (– ∞, ∞)

(b) x ∈ [– 1, 1] π (c) x = (2n + 1) , n ∈ I (d) x = nπ, n ∈ Z 2

 −1  17 π   63. The value of sin cot  cot  is 3    

is (a) 2 (c) 7

3 (b) 2 (d) None of these

3π then αβ + αγ + βγ is 64. If sin α + sin β + sin γ = 2 equal to –1

–1

–1

(a) 1 (c) 3 65. If

2n

(b) 0 (d) – 3

∑ cos

−1

i =1

xi = 0, then

71. If we consider only the principal values of the inverse trigonometric functions, then the value of  −1 1 4  − sin −1 tan  cos  is 5 2 17   (a)

i =1

is

i

(a) n

(b) 2n

n (n + 1) (c) 2

(d) None of these

66. If

20

∑ sin i =1

(a) 20 (c) 0

−1

xi = 10π then

20

∑x i =1

i

π 4 π (c) 2

is equal to

17 36 36 (c) 59

(b) 10 (d) None of these 5 is 2 π (b) 3 (d) None of these

(b)

(b)

29 3

(d)

3 29

  x 2 x3 x4 x6 −1  −1  2 π 72. If sin  x − 2 + 4 − ...  + cos  x − 2 + 4 − ...  =     2 for 0 < | x |
tan–1 1 (b)  tan 1 < tan–1 1 (d)  None of these (c)  tan 1 = tan–1 1 91. Let cos(2 tan–1 x) = 1 / 2, then the value of x is 1 3

(a)   3

(b)  

(c)   1 − 3

(d)   1 −

1 3

92. If

−1 −1 83. sin cot {cos (tan x)} =

(a)

−1  y  −1 87. If cos   − cos b

(b)   ± 2

is equal to

π 6

3π  −1  80. The principal value of sin  sin  is 4   5π (a) 4 π (c) 4

1 1 85. 4 tan −1 − tan −1 is equal to 5 239 π (a) π (b) 2 π π (c) (d) 3 4 86. The values of x which satisfy the trigonometric equation  x −1   x +1  π + tan −1  = , are tan −1   x + 2   x + 2  4

–1 3 + i = (a + ib) (c + id), then the value of x is tan  b  + tan–1  d  has the value     a c

π + 2nπ, n ∈ I 3 π (c)  nπ – , n ∈ I 3

(a)  

93. If sin −1 of x is

π ,n∈I 6 π (d)  2nπ – ,n∈I 3

(b)  nπ +

2a 1 − b2 2x , then the value − cos −1 = tan −1 2 1+ a 1 + b2 1 − x2

(a)  a a+b (c)   1 − ab

(b)  b a−b (d)   1 + ab

853

π for some x ∈ [– 1, 1], then the value of 5

Inverse Trigonometric Functions

77. If sin–1x =

854

94. 3tan–1 a is equal to

Objective Mathematics

3a + a 3 (a)   tan −1 1 + 3a 2 (c)   tan −1

3a − a 3 (b)   tan −1 1 + 3a 2

3a + a 3 1 − 3a 2

(d)   tan −1

3a − a 3 1 − 3a 2

π x −1  5  95. If sin–1  5  + cos ec  4  = 2 , then a value of x is (a)  1 (c)  4

96. If tan–1 x + tan–1 y + tan–1 z = π, then the value of x + y + z is

3 17

(c)  

4 17

(d)  

5 17

1 1 + y2 + cot–1 z = π, then x + y

y + z is equal to (a)  xyz

(b)  2xyz (d)  x2yz

100. If 0 < x < 1, then

97. If tan–1(x – 1) + tan–1 x + tan–1(x + 1) = tan–1 3x, then x is 1 2 1 (c)   0, − 2

(b)  

(c)  xyz

(d)  0

1 2 1 (d)   0, ± 2

(a)   ±

6 17

2

(b)  xyz

1 xyz

(c)  

(a)  

99. If sec −1 1 + x 2 + cos ec −1

(b)  3 (d)  5

(a)  – xyz

2  −1 5 + tan −1  is 98. The value of cot  cos ec  3 3

1 + x 2 {x cos(cot −1 x) + sin(cot −1 x)}2 − 1 x

(a)  

1 + x2

(b)   0,



1/ 2

is equal to

(b)  x

(c)   x 1 + x 2

(d)   1 + x 2

solutions 1. (a) We have,

π −1  1   sin  − sin  −   3  2   [



1 π = sin  + sin −1  3 2 

sin–1 (– x) = – sin–1x, x > 0]

π π π = sin  +  = sin = 1. 3 6 2  

2. (a) cot–1 ( cos α ) – tan–1 ( cos α ) = x  1  ⇒ tan–1   – tan–1( cos α ) = x  cos α  ⇒ tan

–1

⇒ tan–1

1 − cos α cos α =x 1 1+ . cos α cos α 1 − cos α 1 − cos α = x ⇒ tan x = 2 cos α 2 cos α

2 cos α 1 + cos α ⇒ cot x = or cosec x = . 1 − cos α 1 − cos α or sin x =

1 + cos α 1 − (1 − 2 sin 2 α / 2) = 1 − cos α 1 + 2 cos 2 a / 2 − 1

or sin x = tan2

α 2

 1 2  +   2 1 –1 –1 3. (d) tan + tan = tan  4 9  1 2 9 4  1 − ×  4 9  1 = tan–1   . 2 –1

4. (a) sin[cot–1(x + 1)] = cos(tan–1x) ⇒

1 (1 + x) + 1 2

=

1 1+ x

2

x= −

1 . 2

 3 3  = θ ⇒ sin θ = – 5. (a) Let sin–1  − 2 2   π π π and – ≤ sin θ ≤ ∴θ=– 2 2 3  3 π  = – ∴ sin–1  − 2 3   π  3  −1   ∴ sin  − sin  −  2  2  

π π 5π = sin  +  = sin 6 2 3



π  π 1 = sin  π −  = sin = . 6 6 2 

π  π = sin  π −  = sin = 3 3 

3 . 2



  3  π   3  π 7. (b) cos cos −1 − = cos  π − cos −1   +   2  + 6    2  6      cos–1 (– x) = π – cos–1x]

[

cos–1x = sin–1

(

1 − x2 )

4 4  . = sin  sin −1  = 5 5 

13. (c) Let y = sin [cot–1 {tan (cos–1x)}]

   1 − x2 = sin cot −1  tan  tan −1  x    

π π  = cos  π − +  = cos π = – 1. 6 6 

       

 1 − x2 ∵cos −1 x = tan −1  x 

  3  8. (a) sin  tan −1 (− 3 ) + cos −1  −     2  



 3 = sin  − tan −1 3 + π − cos −1  2  

 1 − x2  = sin cot −1  x  

   

...(1)

 [ tan–1 (– x) = – tan–1x and cos–1 (– x) = π – cos–1x] π  π π = 1. = sin  − + π −  = sin 3 6 2    π   3π  9. (d) tan–1  tan    = tan–1  tan  π −   4      4 

π  = tan–1  − tan  = tan–1 (– 1) = – tan–11 4  (



=–

tan–1 (– x) = – tan–1x)

π . 4

⇒ –

π π ≤ 2sin–1 x ≤    2 2



π π π  π  ≤ sin–1 a ≤ ⇒ sin  −  ≤ a ≤ sin 4 4 4 4   1 1 1 ≤a≤ ⇒|a|≤ 2 2 2



  −1 −1  2 x  ∵ 2 tan x = tan  2  1 − x   



− 1  2 cos x  ⇒ tan   = tan–1 (cosec2x) 2  1 − cos x 

2 cos x 1 = ⇒ 2 cos x = 1 sin 2 x sin 2 x

   1      2⋅ 5  − tan −1 1 = tan  tan −1  2    1     1 −       5  π  −1 −1 2 x and tan −1 1 =  ∵ 2 tan x = tan 2 1 4 − x 

11. (d) 2 tan–1 (cos x) = tan–1 cosec2x

⇒

1 − x2 x

  1  π 14. (b) tan  2 tan −1   −   5 4 

π π ∴ – ≤ 2sin–1 a ≤ 2 2 ⇒ –

1 − x2 = θ, then cot θ = x

⇒ sin θ = x ∴ From (1), y = sin θ = x.

10. (c) We have, sin–1x = 2sin–1a Since –

Let cot–1

5 = tan  tan −1 − tan −1 1 12   = tan tan

–1

 5   12 − 1     1 + 5 ⋅ 1   12    −1 −1 −1  x − y  ∵ tan x − tan y = tan    1 + xy   

∴ x = π/3.



7  7 = tan tan–1  −  = – . 17 17  

Inverse Trigonometric Functions

cos–1 (– x) = π – cos–1x)

(

855

 9  3  12. (b) sin   cos −1  = sin  sin −1 1 −  25  5  

 1  1   6. (c) sin arc cos  −   = sin  π − cos −1  2  2   

20. (b) cos (2 cos–1x + sin–1x)

856

Objective Mathematics

π π π 3π – sin–1x = – = . 2 2 5 10

15. (a) cos–1x =

16. (b) We have, A = tan–1 2 ⇒ tan A = 2. and B = tan–1 3 ⇒ tan B = 3 since A, B, C are angles of a triangle, A+B +C= π ⇒ C = π – (A + B) Now, A + B = tan–1 2 + tan–1

...(1)

 2+3  3 = π + tan–1   1 − 2 ⋅ 3 



  −1 −1 −1  x + y  ∵ tan x + tan y = π + tan   1 xy −    for x > 0, y > 0 and xy > 1   





= cos (cos–1x + cos–1x + sin–1x)



π  = cos  + cos −1 x  = – sin (cos–1x) 2 



= – sin (sin–1



=–

21. (c) sin −1

1−

17. (c) tan–1

x x− y   – tan–1   y   x+ y

= tan–1

 x x− y   y − x+ y  π  = tan–1 1 =  . x x y − 4  1+ ⋅   y x+ y   41 4

18. (b) cot–1 9 + cosec–1

−1 = tan

= cot–1 9 + cot–1

= tan–1

5 1 4 = tan–1 + tan–1 4 9 5

 1 4   9+5     1 − 1 ⋅ 4  9 5 



−1 = tan

4 3 + tan −1 3 4



−1 = tan

4 4 π + cot −1 = . 3 3 2

22. (b) Let cos–1x = α, cos–1y = β and cos–1z = γ ⇒ cos α = x, cos β = y and cos γ = z. Given α + β + γ = π ∴ cos (α + β) = cos (π – γ) ⇒ cos α cos β – sin α sin β = – cos γ 1 − x2 1 − y 2 = – z 1 − x2 1 − y 2

⇒ x2y2 + z2 + 2xyz = 1 – x2 – y2 + x2y2 (Squaring both sides) ⇒ x + y + z + 2xyz = 1. 2

2

2

1 2 23. (c) We have, sin −1 + sin −1 = sin–1x 3 3 1 4 2 1 ⇒ sin −1  1 − + 1− 3 9 3 9 

π  41  = tan–1   = tan–1 1 = . 4  41  2π 3

19. (b) We have, sin–1x + sin–1y =

π π 2π −1  −1  ⇒  − cos x  +  − cos y  = 2 2 3     ⇒ π – (cos–1x + cos–1y) = ⇒ cos–1x + cos–1y = π –

4  2/3  + tan −1   3 1−1 / 9 

⇒ xy + z =

∵cosec −1 x = cot −1 x 2 − 1   

1 2 6 =– . 25 5

x   −1 −1 ∵sin x = tan  2 − 1 x    −1 −1 2 x  and 2 tan x = tan  1 − x2  

⇒ xy –

41 −1 16

= cot–1 9 + cot–1

2π 3

2π π = . 3 3

1 − x2

4 1 + 2 tan −1 5 3

= π + tan–1 (– 1) = π – tan–1 1 π 3π =π– = 4 4 3π π ∴ From (1), C = π – = . 4 4

1 − x2 ) = –

  = sin–1x 

1 5 2 8  + ⋅ ⇒ sin −1  ⋅  = sin–1x 3 3 3 3    5 +4 2  ( 5 + 4 2) . ⇒ sin −1   = sin–1x ∴ x =  9 9   24. (d) Given expression

cos − 1



=

1 1 π π + 2 sin − 1 = + 2× 2 2 3 6

π π 2π + = . 3 3 3

= tan–1

  x+ y  1 − xy + z    1− x + y ⋅ z   1 − xy   

= sin [tan–1 (cot 2θ) + cos–1 (cos 2θ)]



 −1   π   = sin  tan  tan  − 2θ   + 2θ    2  



π π  = sin  − 2θ + 2θ  = sin = 1. 2 2 

30. (a) Let sin–1 (sin 10) = θ ⇒ sin θ = sin 10

 x + y + z − xyz  = tan–1    1 − xy − xz − yz  π = π or 2 ∴

(Given)



 35π  = sin    11 



2π   = sin  3π +  11  



= – sin

π x + y + z − xyz = tan π or tan 2 1 − ( xy + xz + yz )

Hence x + y + z – xyz = 0 or 1 – (xy + zx + yz) = 0 i.e., x + y + z = xyz or xy + xz + yz = 1. π  32 − 1   π  26. (b) cot  − 2 cot −1 3  = cot  − cot −1   4   2 × 3  4 4 π = cot  − cot −1  = 3 4

π 4 ⋅ +1 4 3 4 π − cot 3 4

cot

4 +1 = 3 = 7. 4 −1 3 27. (a) We have, sin −1

2a 2b + sin −1 = 2 tan–1x 1 + a2 1 + b2

⇒ 2 tan–1a + 2 tan–1b = 2 tan–1x a+b  a+b  –1 ⇒ tan–1  .  = tan x ⇒ x = 1 − ab  1 − ab  cos– 1 p + cos– 1 q + cos– 1 r = π

We know that, if y = cos– 1 x, then – 1 ≤ x ≤ 1 and 0 ≤ y ≤ π, H ence the given equation will hold only when each is π ∴ p = q = r = cos π = – 1 ∴ p2 + q2 + r2 + 2pqr = (–­ 1)2 + (– 1)2 + (– 1)2 + 2 (– 1)(– 1)(– 1) = 1 + 1 + 1 – 2 = 3 – 2 = 1. 29. (a) The given expression

 −1 1 − tan 2 θ 1 − tan 2 θ  + cos −1 = sin  tan  2 tan θ 1 + tan 2 θ   [Putting x = tan θ]

 2π  2π = sin  −  11  11 

2π ⇒ θ = – = 3π – 10. 11

[

sin (– θ) = – sin θ]

35π 2π   ∵3π − 10 = 3π − 11 = − 11  .   31. (c) The given expression

1 1  = tan  ⋅ 2 tan −1 x + 2 tan −1 x  = tan (2 tan–1x) 2 2 



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

p q  p2   q2   32. (b) cos −1  ⋅ − 1 − 2  1 − 2   = α a   b   a b   ⇒

28. (b) We have,

 22  ∵ 7 radian = π    7 π ∴1 radian =  ,   22   7 35 π π ∴10 radian =  × 10 =  22 11 

 pq p2   q2  − 1 − 2  1 − 2  = cos α ab a   b  

 pq  − cos α  ⇒   ab  ⇒ ⇒

2

= 1−

p 2 q 2 p 2q 2 − + a 2 b 2 a 2b 2

p 2q 2 2 pq + cos2α – cos α a 2b 2 ab = 1−

p 2 q 2 p 2q 2 − + a 2 b 2 a 2b 2

p 2 2 pq q2 − cos α + 2 = 1 – cos2α = sin2α 2 a ab b

2 pq . ab 33. (c) The given expression ∴ k = –

= 2 tan–1 [cosec ⋅ tan–1x – tan cot–1x]

857



Inverse Trigonometric Functions

25. (a), (c)  tan–1x + tan–1y + tan–1z x+ y + tan–1z = tan–1 1 − xy

858

Objective Mathematics

 1 + x2 1 = 2 tan–1 cosec cosec −1 − tan⋅ tan −1  x x    1 + x2 1  2   −  = 2 tan–1  1 + x − 1  = 2 tan–1  x   x x    sec θ − 1  = 2 tan–1    tan θ 

θ   2 sin 2  2  = 2 tan–1   θ  2 sin ⋅ cos θ  2 2  θ 2 ⋅ = θ = tan–1x. 2

1 − cos θ  = 2 tan–1    sin θ  = 2 tan–1 ⋅ tan

[Putting x = tan θ]

θ = 2

a a 34. (d) Let cos–1   = θ so that cos θ = b b Then, the given expression π θ π θ = tan  +  + tan  −   4 2  4 2

35. (a) The given equation can be written as

⇒

1+ x + 1− x 2 1 + x2 2 1− x

2

=

= tan α =

sin α cos α

cos α + sin α cos α − sin α

[Applying componendo and dividendo] 1+ x 1 + sin 2α ⇒ =  1 − x2 1 − sin 2α 2



= tan–1n + tan–1

37. (c), (d)  We have, tan −1

1 n +1

x −1 x +1 = tan–11 + tan −1 x−2 x+2

x −1 x +1 = tan–11 – tan–1 x +1 x+2 x +1 1− x −1 x +2 –1 –1 ⇒ tan = tan x +1 x−2 1+ x+2 x −1 1 –1 –1 ⇒ tan = tan x−2 2x + 3 x −1 1 ⇒ = ⇒ (x – 1) (2x + 3) = x – 2. x−2 2x + 3 ⇒ tan–1

–1

[Squaring both sides]

1 . 2

1 1 2 + tan −1 = tan −1 2 1 + 2x 4x + 1 x

1 1    1 + 2x + 4x + 1  −1 2   = tan 2 1 1 x 1 −  ×  1 + 2x 4x + 1 

⇒

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

⇒

6x + 2 2 = 2 x 8x2 + 6x

2 (3x + 1) x2 = 2 (8x2 + 6x) 3x3 + x2 = 8x2 + 6x ⇒ 3x3 – 7x2 – 6x = 0 x (3x2 – 7x – 6) = 0 x (x – 3) (3x + 2) = 0 2 . Hence, x = 0, 3, – 3

⇒ ⇒ ⇒ ∴

39. (a), (b)  We have, ⇒

tan–1 (x – 1) + tan–1x + tan–1 (x + 1) = tan–1 3x tan–1 (x – 1) + tan–1 (x + 1) = tan–1 3x – tan–1x

⇒ tan–1 ⇒

x2 = sin 2α.

36. (c) tan–1n + cot–1 (n + 1)

= tan–1 (n2 + n + 1).

2

θ θ θ 1 + tan 2   cos 2   + sin 2   2 2 2 =2 ⋅ =2 ⋅ 2θ 2θ 2θ 1 − tan   cos   − sin   2 2 2 1 1 2b =2 ⋅ =2 ⋅ = . cos θ a/b a

2



⇒ tan

2

2

= tan–1

38. (d) We have, tan −1

  θ    θ  1 + tan  2   + 1 − tan  2          =  θ   1 − tan 2   2

1 + x2 − 1 − x2



⇒ 2x2 + x – 3 = x – 2 ⇒ 2x2 = 1∴ x = ±

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



1    n + n +1   n2 + n + 1    = tan–1    1 − n ⋅ 1   n +1− n  n +1  

( x − 1) + ( x + 1) 3x − x = tan–1 1 − ( x 2 − 1) 1 + 3x ⋅ x

2x 2x = 2 − x2 1 + 3x 2

⇒ x + 3x3 = 2x – x3 ⇒ 4x3 – x = 0 1 . ⇒ x (4x2 – 1) = 0. ∴ x = 0, ± 2

∴ 2 tan–1 a – 2 tan–1 b = 2 tan–1x ⇒ tan–1 a – tan–1 b = tan–1x  a−b  –1 ⇒ tan–1   = tan x  1 + ab  41. (b), (c)  We have, tan–1

⇒ tan

–1

∴

x =

a−b . 1 + ab

1 1 1 = tan −1 + tan −1 2 a −1 x a − x +1

1 = tan–1 a −1

1 1 + x a2 − x + 1 1 1− 2 x (a − x + 1)

⇒ x2 – x (a2 + 1) + 1 + (a – 1) (a2 + 1) = 0 ⇒ (x2 – a2) – (a2 + 1) (x – a) = 0 ⇒ (x – a) (x + a – a2 – 1) = 0

1 1/ 5 π + tan −1 = x 4 1 1− 5

1 − x2 = x

⇒

2 5

⇒ 5 (1 – x2) = 4x2

 5 ⇒ 9x2 = 5, ∴ x = ±   .  3  xy + 1 yz + 1 xz + 1 + cot −1 + cot −1 x− y y−z x−z

45. (c) cot −1

x− y y−z x−z + tan −1 + tan −1 1 + xy 1 + yz 1 + xz

= (tan–1x – tan–1y) + (tan–1y – tan–1z) + (tan–1x – tan–1z) = 0. 3π 4

⇒ 6x2 – 5x – 1 = 0 ⇒ (x – 1) (6x + 1) = 0. ∴ x = 1, –

1 1 + tan −1 = tan–1 1 x 2 1 1 = tan–1 1 – tan–1 ⇒ tan −1 x 2

1 . 6

 α  π β  47. (d) We have, 2 tan −1  tan tan  −   2  4 2  

1    1− 2     1 + 1 ⋅ 1  2 

α π β tan  −   4 2 2 = tan π β 2 α 1 − tan tan 2  −   4 2 2 2 tan

⇒ tan −1

1 = tan–1 x

⇒ tan −1

1 1 = tan–1   ⇒ x = 3. x 3



2 cos x = tan–1 (2 cosec x) 1 − cos 2 x

⇒ 2 cos x ⋅ cosec2x = 2 cosec x ⇒ cosec x (cot x – 1) = 0 ⇒ cot x = 1 [

−1

π β α α π β 2 sin   cos   sin  −  cos  −   4 2 2  2   4 2 = tan β π β 2 α 2 π 2 α cos   cos  −  − sin   sin 2  −   4 2 2  4 2 2 −1

43. (b) We have, 2 tan–1 (cos x) = tan–1 (2 cosec x)

cosec x ≠ 0]

π ∴ x = nπ + , n ∈ I. 4 1  44. (b) We have, tan (cos–1x) = sin  cot −1  2    1 − x2 ⇒ tan  tan −1   x  

 2     = sin  sin −1   5  

3π  2 x + 3x  ⇒ tan–1   = nπ + 1 − 2 x ⋅ 3 x 4   5x 3π ⇒ = tan =–1 1 − 6x2 4

1 π = 4 5

⇒ tan −1

⇒ tan–1

  1 − x2 ⇒ tan  tan −1   x  

46. (a), (c)  We have, tan–1 2x + tan–1 3x = nπ +

∴ x = a or a2 – a + 1.

−1 ⇒ tan

  −1 2    = sin  sin   1+ 4   

−1 = tan

a2 + 1 1 ⇒ = 2 (a + 1) ⋅ x − x 2 − 1 a −1

42. (a)  We have, cot–1x + sin–1

  1 − x2 ⇒ tan  tan −1   x  

   = sin (tan–1 2)  

π  sin α ⋅ sin  − β  2  = tan  α β π  α β π  2 cos  − +  cos  + −    2 2 4    2 2 4 −1

= tan −1

sin α cos β sin α cos β = tan −1 . cos α + sin β π  cos α + cos  − β  2 

3 4 48. (b)  We have, cos −1 − sin −1 = cos–1x 5 5  9  −1 4 ⇒ sin −1  1 − = cos–1x  − sin 25  5 

859

 (1 − x 2 )  2 tan–1x = cos–1  2   (1 + x ) 

Inverse Trigonometric Functions

40. (a)

860

Objective Mathematics

x 1    = cot–1  − cot  = cot–1 cot  π − x  2 2    1 =π– x. 2

4 4 − sin −1 = cos–1x 5 5 ⇒ cos–1x = 0 ⇒ x = cos 0 = 1 ∴ x = 1. ⇒ sin −1

49. (b) The given expression y a1 − x –1 + (tan–1a2 – tan–1a1) + = tan y 1 + a1 ⋅ x (tan–1a3 – tan–1a2) + ... + (tan–1an – tan–1an – 1) + tan–1

1 an

y 1 + tan–1an – tan–1a1 + tan–1 x an y = (tan–1an + cot–1an) – tan–1 x π x −1 y −1 y = cot = tan1  . = − tan x 2 x y = tan–1a1 – tan–1

50. (c)  Let cot–1x = θ ⇒ cot θ = x 1 = ∴ sin (cot x) = sin θ = cosecθ 1 = 1 + x2

1 1 ⇒ tan x = 7 7 1 1 and y = tan–1 ⇒ tan y = 3 3

53. (a) x = tan–1



1 + cot 2 θ

 1  = cos  tan −1  1 x2  + 

=

1 = secφ

1

1+

1 1 + x2

=

1 + x2 . 2 + x2



   5 5 / 11  −1 −1  + sin = sin sin 4  5  1+   11  



 5 5 = sin sin −1 + sin −1  4 4  



 5  5 5 = sin  2 sin −1 1−   = sin sin–1  2 4 4 16    

1 − x2 1 1 − x2 < ≤1 2 < π ⇒ 1+ x 2 1 + x2 1 ⇒– 3

1 1.

1 5 ,– . 4 2 1 5 –1



2n

∑x

66. (a) Since –

59. (a) We have, tan–1x > cot–1x



π =– 3

3 . 2 π π 64. (c) Since – ≤ sin–1x ≤ , 2 2 π π π , sin–1β = , sin–1γ = ∴ sin–1α = 2 2 2

∴ α = β = γ = 1. Thus, αβ + αγ + βγ = 3.

1 . 2

π – tan–1x 2 π π ⇒ x > tan ⇒ tan–1x > 4 4

 −1   π   = sin cot  cot  −     3    

65. (b) Since 0 ≤ cos–1 xi ≤ π, ∴ cos–1 xi = 0 for all i.

 π 3π ⇒ x = sin or sin  −    4  4 1 or – 2



5π 2 8

π 5π 2 −1  ⇒ (sin–1x)2 +  − sin x  = 8 2  ⇒ 2 (sin–1x)2 = π sin–1x –

62. (c) sec–1 (sin x) is real if sin x ≤ – 1 or sin x ≥ 1. But – 1 ≤ sin x ≤ 1. ∴ sec–1 (sin x) is real only if π , n ∈ I. sin x = – 1 or 1 ⇒ x = (2n + 1) 2  −1  π      17 π   63. (a) sin cot −1  cot   = sin cot  cot  6π − 3    3        

 5  1 sec  −  exists but sec–1  −   2  4 does not exist, –1

 1  5 tan–1  −  and tan–1  −  both exist. 4   2 

61. (b) Since cot–1x exists for all x ∈ R and cosec–1x is real when x ≤ – 1 or x ≥ 1, ∴ f (x) is real for x ∈ (– ∞, – 1] ∪ [1, ∞).

= tan–1

20

∑x i=1

i

= 20.

5 1 1 = tan −1 + tan −1 2 3 2

1 1 + 3 2 = tan–1 1 = π . 1 1 4 1− ⋅ 3 2

1 1   68. (b) sin 2  cos −1  + cos 2  sin −1  2 3   1 1   = 1 − cos 2  cos −1  + 1 − sin 2  sin −1  2 3   2

2

1 1 59 1 1 = 1−   +1−   = 2 – – = . 4 9 36 2 3 π x 69. (b) We have, tan −1 < 3 π π x  ⇒ tan  tan −1  < tan 3 π   x ⇒ < 3 ⇒ x < 3 π = 5.5 (approx.) π ∴ The maximum value of x is 5.

861

3 (3 tan θ − tan 3 θ) = a tan 3θ (1 − 3 tan 2 θ)

Inverse Trigonometric Functions

⇒

862

70. (c) tan–1x and cot–1x exist for all x ∈ R.

2  −1 3 4 2 + tan −1  ∴ tan  cos −1 + tan −1  = tan   tan 4 3  5 3 

Objective Mathematics

cos–1 (2 – x) exists if – 1 ≤ 2 – x ≤ 1 i.e., 1 ≤ x ≤ 3 So, the given equation holds for 1 ≤ x ≤ 3.  1 4  − sin −1 71. (d) tan  cos −1  5 2 17  



 −1  7  −1  4   = tan  tan   − tan    1  1  



7−4   = tan (tan–1 7 – tan–1 4) = tan   tan −1 1 + 7 × 4   3  3  = tan   tan −1  = . 29  29  π for | x | ≤ 1. 72. (b) Since sin–1x + cos–1x = 2



∴ x – x

x 2 x3 x4 x6 + −... = x2 – + −... 2 4 2 4 x 2

=

x2 x2 1+ 2



⇒ 1 +



x x2 = ⇒ ⇒ 2x + x3 = 2x2 + x3 2 + x 2 + x2



⇒ x2 = x ⇒ x = 0, 1.



But x ≠ 0,

( 0 < | x |
45º ∴ tan 1 > tan 45º ⇒ tan 1 > 1 Also, tan–1 1 = π < 1 Hence, tan 1 > tan4–1 1 91. (b) We have, cos(2 tan–1 x) = 1 2

⇒ cos(2 tan–1 x) = cos π 3 π ⇒ tan–1 x = 6 π 1 ⇒ x = tan = 3 3 92. (b) We have, ( 3 + i) = (a + ib) (c + id) = (ac – bd) + i (ad + bc) On comparing the real and imaginary part on both sides, we get

Inverse Trigonometric Functions

1

Let cot–1

863

87. (c) We have,

∴ sin [cot–1 {cos (tan–1 x)}]  −1  1  = sin   cot (1 + x 2 )  

864



ac – bd =

3 and ad + bc = 1

97. (d) Given tan–1(x – 1) + tan–1 x + tan–1(x – 1) = tan–13x

Objective Mathematics

b d bc + ad  Now, tan −1   + tan −1   = tan −1  a c  ac − bd 

⇒ tan–1(x – 1) + tan–1 x = tan–1 3x – tan–1(x + 1)  ( x − 1) + x  −1  3 x − ( x + 1)  ⇒ tan −1   = tan   1 − ( x − 1) x  1 + 3 x ( x + 1) 

π  1  tan  = nπ + , n ∈ I  3  6 −1



⇒ (1 + 3x2 + 3x) (2x – 1) = (1 – x2 + x) (2x – 1)

93. (d) We have, sin −1

⇒ (2x – 1) (4x2 + 2x) = 0

2a 1− b 2x − cos −1 = tan −1 1 + a2 1 + b2 1 − x2 2

⇒ 2tan–1 a – 2tan–1 b = 2tan–1 x

⇒ x = 0, ±

5 3 98. (a) Since, cosec–1   = tan −1   3 4

a−b  a−b ⇒ tan −1  = tan −1 x ⇒ x =  1 + ab  1 + ab

2  −1 3 + tan −1  ∴ cot  tan 4 3

94. (d) We have,

  17   3 2   4 + 3 −1  12   = cot tan  = cot tan  1  1   1−    2   2 

3 tan 3A = 3tan A − tan A 2 1 − 3tan A

−1

 3tan A − tan 3 A  ⇒ 3A = tan–1   2  1 − 3tan A  Put tanA = a

  17   6 = cot  tan −1    =  6   17 

or A = tan–1 a

 3a − a 3  ∴ 3tan–1 a = tan–1  2   1 − 3a 

99. (a) sec −1 1 + x 2 + cos ec −1

1 1 + y2 + cot–1 =π z y

⇒ tan–1 x + tan–1 y + tan–1 z = π

95. (b) We have, π x −1  5  sin–1   + cos ec   = 5 4 2 ⇒ sin–1  x  + sin −1  4  = π 5 5 2 x π   −1 −1  4  ⇒ sin   = − sin   5 2 5 −1  x  −1  3  ⇒ sin   = sin   5 5

1 2

⇒ x = 3.

−1  x + y + z − xyz  ⇒ tan  =π  1 − xy − yz − zx  ⇒ x + y + z = xyz

100. (c) 

1 + x 2 ( x cos cot −1 x + sin cot −1 x) 2 − 1

1/ 2

2   x 1  −1 −1  = 1 + x  x cos cos + sin sin − 1    1 + x2 1 + x2  2

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

96. (b) We have, tan–1 x + tan–1 y + tan–1 z = π

1/ 2

2

 x + y + z − xyz  ⇒ tan–1  =π 1 − xy − yz − zx  ⇒ x + y + z = xyz

= 1 + x 2 ( x 2 + 1 − 1)1/ 2 = x 1 + x 2

EXERCISES FOR SELF-PRACTICE −1  x + y  1. If α = tan – 1 x + tan – 1 y and β = tan  1 − xy  ;   x < 0, y < 0, xy > 1 then



(a) α (b) α (c) α (d) β

– β +β +β – α

= = = =

π π – π π

 x 2  x −1  1 2 2 2. cos  x + 1 − x ⋅ 1 −  = cos −1 − cos −1 x 2 4 2   holds for

(a) | x | ≤ 1 (b) x ∈ R (c) 0 ≤ x ≤ 1 (d) – 1 ≤ x ≤ 0

1/ 2

(c)

π (b) – 3 −1

tan 3

−1 −1 4. If 2 tan x + sin

(d) None of these 2x is independent of x, then 1 + x2

(a) x ∈ [1, + ∞) (c) x ∈ (– ∞, – 1]

(b) x ∈ [– 1, 1] (d) None of these

π is equal to 2

 1 (a) f  −   2

(b) f (k2 – 2k + 3), k ∈ R

 1  (c) f  2  , k ∈ R  1+ k 

(d) f (– 2)

x     x3 x5 x2 x4 + + ...  = . 11. If sin −1  x − + + ...  + cos −1 1 − 2 3 5 3 5    

5. If α ≤ sin – 1 x + cos – 1 x + tan – 1 x ≤ β then π 3π (a) α = ,β= 4 4 (c) α = 0, β = π

10. Let f (x) = sin–1x + cos–1x. Then

(b) α = – π, β = 2π

(a) 1 (c) both

(b) –1 (d) None of these

(d) None of these

12. 2 tan–2 (cosx) = tan–1 (cosec2x) then x:

(a) 0

(b) 1

π 2 π (c) 6

(c) 2

(d) infinite

13. If 4 sin–1x + cos–1x = π, then x is equal to:

6. The number of real solutions of the equation

1 + cos 2 x = 2 sin (sin x) , – π ≤ x ≤ π is

7. The solution of the inequality (cot – 1 x)2 – 5 cot – 1 x + 6 > 0 is

(b) π

(a)

−1

(d)

(a) 0 (c) − 3 / 2

(a) (cot 3, cot 2) (b) (– ∞, cot 3) ∪ (cot 2, ∞) (c) (cot 2, ∞) (d) None of these

π 3

(b) 1/2 (d) 1 / 2

14. Which of the following is not the value of 2 tan –1x ?  2x  (a) tan–1  1 + x 2   

 2x  (b) tan–1  2   1− x 

8. The number of real solutions of (x, y), where | y | = sin  1 − x2   2x  (c) cos–1  (d) sin–1  x, y = cos–1 (cos x), – 2π ≤ x ≤ 2π, is 2 . 2  x 1 +  1+ x    (a) 2 (b) 1 (c) 3 (d) 4  3 15. The principal value of sin −1  − 1 5  is –1 –1  9. tan   + tan   equals  2  4 3 3 4π 5π 1 3  (b) (a) (b) 2  tan −1 + tan −1  (a) tan–14 + tan–1 5 3 3 4 4   −1 1 −1 3  2π π (c) 3  tan + tan  (d) None of these (c) – (d) – 4 5  3 3

Answers

1. (d) 11. (a)

2. (c) 12. (d)

3. (d) 13. (b)

4. (a) 14. (a)

5. (a) 15. (d)

6. (c)

7. (b)

8. (c)

9. (c)

10. (a), (c)

865

π (a) 3

Inverse Trigonometric Functions

3. The value of x that satisfies tan – 1 (tan 3) = tan2 x is

24

Properties and Solutions of Triangles

CHAPTER

Summary of concepts Properties of Triangles In a triangle ABC, we shall denote the angles BAC, CBA and ACB by A, B and C respectively and the corresponding sides, i.e., the sides opposite to them by a, b and c respectively.

SOME BASIC FORMULAE 1. Law of Sines or Sine Rule  The sides of a triangle are proportional to the sines of the opposite angles. That is, in a ∆ABC, we have

a b c = = . sin A sin B sin C 2. Law of Cosines or Cosine Formula  In any ∆ABC,

b2 + c2 − a 2 (i) cos A = 2bc c2 + a 2 − b2 (ii) cos B = 2ac a 2 + b2 − c2 (iii) cos C = 2ab 3. Projection Formula  In any ∆ABC, (i) a = b cos C + c cos B (ii) b = a cos C + c cos A (iii) c = a cos B + b cos A 4. Law of Tangents or Tangent Rule (Napiers Analogy)  In any ∆ABC,

A  B−C   b−c  (i) tan   2  =  b + c  cot 2    

(ii) cos

A = 2

B s ( s − a) = , cos 2 bc

cos

C = 2

s ( s − c) ab

s ( s − b) ac

A B ( s − b)( s − c) ( s − c)( s − a ) = (iii) tan 2 = s ( s − a ) , tan 2 s ( s − b) C ( s − a )( s − b) tan 2 = s ( s − c) 6. Area of a Triangle  The area of any triangle ABC is given by ∆=

1 1 1 bc sin A = ca sin B = ab sin C 2 2 2

Heron’s Formul  In a ∆ABC, if a + b + c = 2s, then its area is given by, ∆=

s ( s − a )( s − b)( s − c) .

Circles connected with a given triangle Circumcircle of a Triangle The circle which passes through three vertices of a triangle is called the circumcircle of the triangle. The centre of the circle is called circum-centre, usually denoted as O and its radius is called circumradius, usually denoted by R.

C  A −B   a −b  (ii) tan   2  =  a + b  cot 2     B  C−A   c−a  (iii) tan   2  =  c + a  cot 2 .     5. Half Angle Formulae or Semi-sum Formulae  In any triangle ABC, if a + b + c = 2s, then (i) sin 

A (s − b) ( s − c) B (s − c) ( s − a ) = , sin  = 2 bc 2 ac

sin

C (s − a ) ( s − b) = 2 ab

The circumcentre is the point of intersection of right bisectors of the sides of a triangle.

(ii) R =

abc 4∆

In any ∆ABC, ∆ ∆ ∆ (i) r1 = , r = , r = s−a 2 s −b 3 s−c

(iii) r1 = a

r3 = c

(iv)

incircle of a triangle The circle which is drawn within a triangle such that the three sides touch this circle is called the incircle. The centre of this circle is called incentre, usually denoted by I, and its radius is called inradius, usually denoted by r. The incentre is the point of concurrence of the bisectors of the three (internal) angles of the triangle. In any triangle ABC, ∆ (i) r = s A B C (ii) r = (s – a) tan = (s – b) tan = (s – c) tan 2 2 2 (iii) r =

A B C , r2 = s tan , r = s tan 2 2 3 2

(ii) r1 = s tan

C A B C cos cos cos 2 2 2 2 ,r =b , 2 B A cos cos 2 2

cos

A B cos 2 2 C cos 2

cos

r1 = 4R sin

A B C cos cos , 2 2 2

r2 = 4R cos

A B C sin cos , 2 2 2

r3 = 4R cos

A B C cos sin . 2 2 2

orthocentre of a triangle Let the perpendiculars AL, BM and CN from the vertices A, B and C on the opposite sides BC, CA and AB of ∆ABC, respectively, meet at O′. Then O′ is the orthocentre of the ∆ABC. The triangle LMN is called the pedal triangle of the ∆ABC.

B A B C A C csin sin sin bsin sin 2 2 2 2 ,r= 2 2 ,r= C A B cos cos cos 2 2 2

a sin

(iv) r = 4R sin

A B B sin sin . 2 2 2

escribed circles of a triangle

tHe diStanceS of tHe ortHocentre

The circle which lies outside the triangle and touches the side BC from tHe verticeS and also the sides AB and AC produced is called escribed circle The distances of the orthocentre of the triangle from the vertices or excircle opposite to angle A. Its centre is called excentre and are 2R cos A, 2R cos B, 2R cos C and its distances from the sides its radius is called exradius. Similarly, there are two more exare 2R cos B cos C, 2R cos C cos A, 2R cos A cos B. circles, one opposite to angle B and one opposite to angle C. The three excentres are usually denoted as I1, I2, I3 and the three ex-radii are usually denoted as r1, r2, r3. Important Results R (i) Circumradius of the pedal triangle = . 2 (ii) Area of the pedal triangle = 2 ∆cos A cos B cos C.

(iii) Lengths of the medians AL, BM and CN of ∆ABC are given by

1 2 2 1 b + c + 2bc cos A = 2b 2 + 2c 2 − a 2 2 2 1 2 1 c + a 2 + 2ac cos B = 2c 2 + 2 a 2 − b 2 BM = 2 2 1 2 1 a + b 2 + 2ab cos C = 2a 2 + 2b 2 − c 2 CN = 2 2

AL =

Excentre is the point of concurrence of internal bisector of angle A and external bisectors of angles B and C.

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a b c = = (i) R = 2 sin A 2 sin B 2 sin C

Properties and Solutions of Triangles

In any triangle ABC,

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(iv) Circumcentre, Centroid and orthocentre are collinear and G divides OO′ in the ratio 1 : 2.

Objective Mathematics

(v) Distance between the circumcentre O and the incentre I is OI = R

1 − 8 sin

A B C sin sin 2 2 2

(vi) OI1 = R

1 + 8 sin

A B C cos cos 2 2 2

OI2 = R

1 + 8 cos

A B C sin cos 2 2 2

A B C cos sin OI3 = R 1 + 8 cos 2 2 2 where I1, I2, I3 are the centres of the escribed circles opposite to the angles A, B and C respectively and O is the circumcentre.

CYCLIC QuadrILateraL

2. In a regular polygon the centroid, the circumcentre and the incentre are same. 3. Area of a regular polygon =

π nR 2 2π na 2 cot = sin 4 2 n n

π , where a is length of side, n is number of sides n of polygon, R is radius of circumscribing circle and r is radius of incircle of the polygon.

2 = nr tan

radius of circumscribing circle a π = cosec , π 2 n 2 sin n where a is the length of each side of regular polygon of n sides. R=

a

radius of inscribed circle

A quadrilateral is a cyclic quadrilateral if its vertices lie on a circle.

r=

a π cot 2 n

area of the polygon ∆=

1 π 1 2π π 2 na2 cot = nR sin = nr2 tan n 4 n 2 n

where a is the length of each side, R is the circumradius and r is the inradius of the regular polygon of n sides.

Solution of triangleS 1. Area of cyclic quadrilateral ABCD is =

( s − a )( s − b)( s − c)( s − d )

where 2s = a + b + c + d. 2. Circumradius of cyclic quadrilateral

A triangle has six elements: the three sides a, b, c and the three angles A, B, C. If three of the elements are given, atleast one of which must be a side, then the other three elements can be uniquely determined. The process of finding the remaining three elements of the triangle is called the solution of the triangle.

1 (ab + cd )(ad + bc)(ac + bd ) R = 4 ( s − a )( s − b)( s − c)( s − d ) 3. cos B =

a 2 + b2 − c2 − d 2 and similarly other angles. 2 (ab + cd )

4. ptolemy’s theorem If ABCD is a cyclic quadrilateral, then AC ⋅ BD = AB ⋅ CD + BC ⋅ AD

Solution of a right angled triangle

It is possible to find the remaining parts of a right triangle if in i.e., in a cyclic quadrilateral the product of the diagonals is equal addition to the right angle, one side and any other part (side or to the sum of the products of the lengths of opposite sides. angle) are known.

regular polygon A polygon is called a regular polygon if all its sides are equal and its angles are equal. Note: 1. If a polygon has ‘n’ sides, sum of its internal angles is π (n – 2)π and each angle is (n − 2) . n

or log a = log b + log sin A – log sin B Case 1: When the three sides are given. In this case only one triangle is possible and so angles can be de- which gives the value of a. termined uniquely. The following formulae are generally used : Here it is assumed that b > c. But if c > b, the Napier’s C−B c−b A = cot analogy, tan can be applied. A (s − b)( s − c) 2 c + b 2 = (i) sin 2 bc Case 3:  When two angles and one side are given. In this case only one triangle is possible and so remaining one B (s − c)( s − a ) = (ii) sin angle and two sides can be uniquely determined. Let one side a 2 ca and two angles B and C of ∆ABC are given. Then third angle A C (s − a )( s − b) is given by = (iii) sin 2 ab A = 180º – (B + C).

A (s − b)( s − c) (iv) tan 2 = s (s − a) (v) tan

B = 2

(s − c)( s − a ) s ( s − b)

(vi) tan

C = 2

(s − a )( s − b) s ( s − c)

A = (vi) cos 2

s ( s − a) bc

(vii) cos

B = 2

s ( s − b) ac

(ix) cos

C = 2

s ( s − c) ab

(x) cos A =

b2 + c2 − a 2 2bc

(xi) cos B =

a 2 + c2 − b2 2ac

(xii) cos C =

a 2 + b2 − c2 2ab

Case 2: When two sides and the included angle between them are given. In this case only one triangle is possible and so remaining two angles and one side can be uniquely determined. Let b, c be the two given sides and A the given included angle.

The other sides can be found out by using sine formula. b=a

sin B sin C and c = a . sin A sin A

Case 4: When three angles are given. In this case sides cannot be determined uniquely. Only ratio of sides can be determined by sine rule and hence there will be infinite number of such triangles. Case 5: When two sides and the angle opposite to one of them are given. In this case either no triangle or one triangle or two triangles are possible. For this reason, it is called the ambiguous case. Let b, c and B be given. The following possibilities are there : (i) when B is acute and b < c sin B, no triangle is possible. (ii) When B is acute and b = c sin B, then only one triangle is possible which is right angled. (iii) When B is acute and b > c sin B, two triangles are possible if b < c and only one triangle if b ≥ c. (iv) When B is obtuse, there is no triangle, if b < c and only one triangle if b > c. Remarks 1. The mid point of the hypotenuse of a right angled triangle is equidistant from the 3 vertices of the triangle. 2. In a right angled triangle, the orthocentre coincides with the vertex containing the right angle. 3. In a right angled triangle, the mid point of the hypotenuse is the circumcentre of the triangle.

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(i) To find angles B and C, proceed as follows: B−C b−c A b−c b a b = cot = tan = ⇒a= 2 2 (b + c) tan A b+c sin B sin 90º sin B 2 B−C A c or log tan = log (b – c) – log (b + c) – log . Also, = cot B, so that c = b cot B. 2 2 b B−C (ii) If the sides b and c are given, then a = b 2 + c 2 , Since b, c and A are given, the value of log tan  2 b B+C A tan B = and C = 90 – B. c can be found. Again 2  = 90º − 2  is known as A is   given. Solution of an Oblique Triangle By adding and subtracting, we obtain values of B and C. A triangle which is not a right angled triangle is called  oblique (ii) To find side a, we use the sine formula: triangle. Following cases are possible in solving oblique a b bsin A = b or a = triangles. sin A sin B sin B

Properties and Solutions of Triangles

(i) If A = 90º, side b and angle B are given, then

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mULTIPLE-CHOICE QUESTIONS

Objective Mathematics

Choose the correct alternative in each of the following problems:

1. In ∆ABC

(a) 90° (c) 120°

A B C cos 2 cos 2 2 + 2 + 2 =0 a b c

cos 2

(a)

s 2abc

(b)

(c)

s2 3abc

(d) None of these

2

10. In a triangle ABC, medians AD are BE are drawn. If AD = 4, ∠DAB = π/6 and ∠ ABE = π/3, then the area of the ∆ABC is

s abc 2

2. If b + c = 3a, then cot (a) 3 (c) 4

(a) 16/ 3 3 (c) 32/3

C   A  3b a cos 2   + c cos 2   = 2 2 2 then the sides a, b and c

(b) 1 (d) 2

1 2 2 3 (c) 2 2

1 2

(b)

(a)

A = 2

(d) None of these

4. Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn + 1 – Tn = 21 then n = (a) 5 (c) 6

(b) 7 (d) 4 A− B+C is equal to 2 (b) c2 + a2 – b2 (d) c2 – a2 – b2

5. In a triangle ABC, 2ca sin (a) a2 + b2 – c2 (c) b2 – c2 – a2 6. In any ∆ABC, 1 – tan a a+b+c 3a (c) a+b+c

(a)

a+b−c a+b+c a+b−c (c) 2 (a + b + c)

8. In ∆ABC, b cos 2 (a) 3s (c) s

2a a+b+c

(d) None of these

7. In any ∆ABC, tan (a)

A B tan = 2 2 (b)

(b) 32/ 3 3 (d) 64/3

11. If in a triangle ABC

B C cot = 2 2

3. In ∆ABC, if a = 16, b = 24, c = 20, then sin

(b) 60° (d) 150°

A B tan = 2 2 a+b+c (b) a+b−c

(a) are in A.P. (c) are in H.P.

(b) are in G.P. (d) satisfy a + b = c

12. In any ∆ABC, if (a + b + c) (b + c – a) = 3bc, then A = (a) 60º (c) 30º

(b) 45º (d) None of these

A B C  13. In any ∆ABC, 4 bc cos 2 + ca cos 2 + ab cos 2  = 2 2 2  (a) (a + b + c)2 (c) 2 (a + b + c)2

(b) a + b + c (d) None of these

14. If the cotagents of half the angles of a triangle are in A.P., then its sides are in (a) A.P. (c) H.P.

(b) G.P. (d) None of these

15. In a ∆ABC, if 3 tan

A C tan = 1, then sides a, b, c 2 2

are in (a) A.P. (c) H.P. 16. If a cos 2

(b) G.P. (d) None of these C A 3b , then the sides of the triangle + c cos 2 = 2 2 2

are in (a) G.P. (c) H.P. 17. In a ∆ABC, 2 sin

(b) A.P. (d) None of these A C B sin = sin , if a, b, c are in 2 2 2

(d) None of these

C B = + c cos 2 2 2 (b) 2s (d) None of these

9. The sides of a triangle are sin α, cos α and 1 + sin α cos α for some 0 < α < π/2. Then the greatest angle of the triangle is

(a) H.P. (c) A.P.

(b) G.P. (d) None of these

18. In a trangle ABC, a = 4, b = 3, ∠A = 60°, then c is the root of the equation (a) c2 – 3c – 7 = 0 (c) c2 + 3c + 7 = 0 (c) c2 – 3c + 7 = 0 (d) c2 + 3c – 7 = 0

20. In any ∆ABC 4∆ (cot A + cot B + cot C) = (a) 3 (a + b + c ) (c) a2 + b2 + c2 2

2

2

(b) 2 (a + b + c ) (d) None of these 2

2

2

A B C sin sin = 21. In any ∆ABC, abc s sin 2 2 2 (a) ∆3 (c) ∆2

(b) 3∆2 (d) None of these

22. In any ∆ABC, 4∆ cot A = (a) 2 (b2 + c2 – a2) (c) b2 + c2 – a2

(b) a2 + b2 + c2 (d) None of these

2abc A B C = cos cos cos 23. In any ∆ABC, a+b+c 2 2 2 (a) ∆ (c) 3∆

(b) 2∆ (d) None of these

24. If the area of a triangle is 75 square metres and two of its sides are 20 and 15 metres, then the included angle is (a) 30º (c) 150º

(b) 60º (d) None of these

25. Which of the following pieces of data does not uniquely determine an acute angled ∆ABC (R = circumradius)? (a) a, sin A, sin B (c) a, sin B, R

(b) a, b, c (d) a, sin A, R

(a) 2r (c) 3r

A B C sin sin = 2 2 2 (b) r (d) None of these

33. The sides of a triangle are 4, 5 and 6 cm. The area of the triangle is equal to : 15  cm2 4 15 (b) 7  cm2 4 4 (c) 7  cm2 15 (d) None of these (a)

34. If the angles of a triangle are in ratio 4 : 1 : 1, then the ratio of the longest side and perimeter of triangle is (a)

3 1+ 3

(b)

3 2+ 3

(c)

2 3

(d)

1 6

s s s 35. In any ∆ABC, 4  − 1  − 1  − 1 = a  b  c  r R 3r (c) R (a)

(b)

2r R

(d) None of these

26. If ∆ is the area of the ∆ABC, s its semi-perimeter, then A B C cot cot = 36. In any ∆ABC, r2 cot A B C 2 2 2 cos cos = abc cos 2 2 2 (a) ∆ (b) 2∆ (a) 2s∆ (b) s∆ (c) 3∆ (d) None of these (c) 3s∆ (d) None of these A B C cos  cos = 37. In any ∆ABC, 4R r cos  27. In any ∆ABC, if a = 18, b = 24, c = 30, then r = 2 2 2 (a) 6 (b) 9 (a) 3∆ (b) ∆ (c) 12 (d) None of these (c) 2∆ (d) None of these 28. If a = 13 cms, b = 14 cms, c = 15 cms, then R = 1 1 1 + + 1 1 = 38. In any ∆ABC, r1 r2 r3 (b) 8 cms (a) 7 cms 8 8 1 3 1 (b) (a) (c) 6 cms (d) None of these 8 r r 2 2 29. In any ∆ABC, 2R sin A sin B sin C = (c) (d) None of these r (a) 2∆ (b) 3∆ A B C (c) ∆ (d) None of these cos cos = 39. In any ∆ABC, 4R sin 2 2 2 30. The ratio of the sides of a triangle ABC is 1 : 3 : 2 . The ratio A : B : C is (a) 3 : 5 : 2 (c) 3 : 2 : 1

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

31. In a triangle ABC if b = 2, B = 30º then the area of the circumcircle of triangle ABC in square units is (a) π (c) 4π

(b) 2π (d) 6π

(a) r1 (c) 3r1

(b) 2r1 (d) None of these

40. If ABC is a triangle right angled at C, then R + r 1 (a + b) 2 1 (a + b) (c) 4

(a)

(b)

1 (a + b) 3

(d) None of these

871

32. In any ∆ABC, 4R sin

Properties and Solutions of Triangles

5 2 A C = , tan = , then 6 5 2 2 (a) a, c, b are in A.P. (b) a, b, c are in A.P. (c) b, a, c are in A.P. (d) a, b, c are in A.P.

19. In a ∆ABC, tan

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41. In any ∆ABC,

1 1 1 1 + + + = r12 r22 r32 r 2

Objective Mathematics

a 2 + b2 + c2 2∆ 2 a 2 + b2 + c2 (c) 3∆ 2

(b)

(a)

a 2 + b2 + c2 ∆2

(d) None of these

42. In any ∆ABC, rr1 + r2 r3 = (a) ab (c) bc

(b) ac (d) None of these

43. In any ∆ABC, r2 r3 + r3 r1 + r1 r2 = (a) s (c) 3s2 2

2

(b) 2s (d) None of these

44. In any ∆ABC, 16R rr1 r2 r3 = 2

(a) 2a2b2c2 (c) 3a2b2c2

(b) a2b2c2 (d) None of these

45. If p1, p2, p3 are the perpendiculars from the angular points of a triangle on the opposite sides, then 1 1 1 + + = p1 p2 p3 1 r 3 (c) r

(a)

(b)

2 r

(d) None of these

(b)

2 A

(d) None of these

(b) G.P. (d) None of these sin A , then the triangle is 2 sin C

(a) equilateral (b) isosceles (c) right angled (d) None of these

cos A cos B cos C and the side a = 2, = = a b c then the area of the triangle is (a) 1

3 2 53. If in a ∆ABC, (c)

(b) 2 (d)

3

2 cos A cos B 2 cos C a b , then the triangle is + + = + a b c bc ca (a) right angled (c) equilateral

(b) isosceles (d) None of these

(c) b + c – a

a+b+c 3 (d) None of these

(b)

A = 2

(b) a2 (d) None of these

(a) a + c – b (c) a + c – 2b

(b) a + b – c (d) None of these

57. In any ∆ABC, a (b cos C – c cos B) = (a) 2 (b2 – c2) (c) 3 (b2 – c2)

(b) b2 – c2 (d) None of these

58. In any ∆ABC, 2 (bc cos A + ca cos B + ab cos C) = (a) a2 + b2 + c2 (c) 3 (a2 + b2 + c2)

a 2 + b2 + c2 2 (d) None of these

(b)

59. The length of the sides of a triangle are x, y and x 2 + y 2 + xy . The measure of the greatest angle is (a) 120º (c) 135º

(b) 150º (d) None of these

60. If in a ∆ABC, ∠B = 90º, then,

50. In ∆ABC, if cot A, cot B, cot C are in A.P. then a2, b2, c2 are in (a) G.P. (c) H.P.

(d) None of these

C A  56. In any triangle ABC, 2  a sin 2 + c sin 2  =  2 2

48. In a ∆ABC, r1, r2, r3 are in H.P. if a, b, c are in

49. In a ∆ABC, if cos B =

b2 + c2 − a 2 2 (a 2 + b 2 − c 2 )

52. In a ∆ABC, if

(a) 2a2 (c) 3a2

47. If A, A1, A2, A3 be the areas of the incircle and encircles, 1 1 1 + + = then A1 A2 A3

(a) A.P. (c) H.P.

(b)

55. In any ∆ABC, (b + c)2 – 4bc cos2

(d) None of these

(c)

b2 + c2 − a 2 a 2 + b2 − c2 b2 + c2 − a 2 (c) 3 (a 2 + b 2 − c 2 )

(a)

(a) a + b + c

(a) r1 = r2 = r3 = 2r (b) r1 = r2 = r3 = r (c) r1 = r2 = r3 = 3r

1 A 3 A

tan C = tan A

B  2 C + c cos 2  = 54. In any ∆ABC, 2  b cos  2 2

46. In an equilateral triangle

(a)

51. In any ∆ABC,

(b) A.P. (d) None of these

A 2 A (c) 3 tan 2

(a) tan

(b) 2 tan

b−c = b+c A 2

(d) None of these

(a) right angled (c) isosceles

(b) equilateral (d) None of these

b +c a2 + c2 2 2 a +c (c) 2 a + b2 2

(a) right angled (c) equilateral

(b)

(a)

2

(d) None of these

63. In any ∆ABC, a2 sin 2B – b2 sin 2A = (a) 2ab sin (A – B) (b) 2b sin (A – B) (c) 2a sin (B – A) (d) None of these

a  π cot    2n  2

(b)

 π (c) a cot    2n 

a  π (d) cot    2n  4

65. In a triangle ABC, a = 5, b = 7 and sin A =

3 , how 4

(d) π

(a) 1

(b) 2

3 (c) 2

(d)

(a) 0º (c) 60º

A−B 2 (b) A+B tan 2

(a) 3 (c) 5

3

(b) c2 + 3c + 7 = 0 (d) c2 + 3c – 7 = 0.

(b) 30º (d) 90º

(b) 90º (d) None of these 2 ab C sin then C = a−b 2 (b) (a – b) sec θ (d) None of these

(b) 4 (d) 6

76. If the radius of the circumcircle of an isosceles ∆PQR is equal to PQ (= PR), then the angle P is

A−B 2 (c) (d) None of these A+B cos 2 67. The greatest angle of a triangle whose sides are x2 + x + 77. 1, 2x + 1 and x2 – 1, is

(a) (a – b) cos θ (c) (a – b) cosec θ

π 2

74. If in ∆ABC, 3 sin A = 6 sin B = 2 sin C, then angle 3 A is

tan

cos

68. In any ∆ABC, if tan θ =

π 3

75. If the radius of the incircle of a triangle with its sides 5k, 6k and 5k is 6, then k is equal to

a−b = a+b

(a) 60º (c) 120º

(c)

(a) c2 – 3c – 7 = 0 (c) c2 – 3c + 7 = 0

(b) 0 (d) infinite.

A−B 2 (a) A+B sin 2

(b)

73. In a ∆ABC, a = 4, b = 3, ∠A = 60º. Then c is the root of the equation

many such triangles are possible

sin

π 6

cos A cos B cos C = = , and the side a = a b c 2, then the area of the triangle is

π (a) a cot    n

66. In any ∆ABC,

(a)

72. In a ∆ABC, if

64. The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a is

(a) 1 (c) 2

(b) obtuse angled (d) isosceles

71. The perimeter of a ∆ABC is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then the angle A is

a +b a2 + c2 2

(a)

π 6

(b)

π 3

(c)

π 2

(d)

2π 3

If a circle is inscribed in an equilateral triangle of side a, then area of the square inscribed in the circle is (a)

a2 6

(b)

a2 3

(c)

2a 2 5

(d)

2a 2 3

78. The area of the circle and the area of the regular polygon of n sides and of perimeter equal to that of circle are in the ratio of 69. If ∆ denotes the area of any triangle and s its  semiperimeter, then π π π π (a) tan  n  : n (b) cos  n  : n s2 s2     (b) ∆ > (a) ∆ < 2 4 π π π π s2 (c) sin  n  : n (d) cot  n  : n (c) ∆ < (d) None of these     4

873

A B C = cos = cos , then the triangle a b c

is

1 + cos (A − B) cos C = 62. In any ∆ABC, 1 + cos (A − C) cos B 2

70. If in ∆ABC, cos

Properties and Solutions of Triangles

61. In any ∆ABC, if a cos A = b cos B, then triangle ABC is

874

79. In a triangle ABC, AB = 2, BC = 4, CA = 3. If D is the mid point of BC, then the correct statement (s) is/are

Objective Mathematics

(a) cos B =

11 16

(c) AD = 2 ⋅ 4

(b) cos C =

7 8

(d) AD2 = 2 ⋅ 5

(b) 45º (d) None of these

81. In a ∆ABC, if a cos A = b cos B then triangle is (a) right angled (c) equilateral

R 2 (c) R

(b) 2R (d) None of these

83. If D is the mid point of the side BC of a triangle ABC and AD is perpendicular to AC, then (a) b = a – c (c) 3b2 = a2 – c2 2

2

2

(b) a + b = 5c (d) 3a2 = b2 – 3c2 2

2

2

84. The angle A of ∆ABC, in which

85. cos2 

(b) 45º (d) 120º

A B C + cos2   + cos2  = 2 2 2

r R r (c) 2 + 2R (a) 2 –

(b) 2 –

93. In a right angled triangle, the hypotenuse is four times as long as the perpendicular drawn to it from the opposite vertex. One of the acute angle is

(b) 60º (d) None of these (b) 0 (d) None of these

(a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 3 : 1 : 2 (d) None of these 95. If R is the radius of the circumcircle of the ∆ABC, and ∆ is its area then a+b+c ∆ abc (c) R = 4∆ (a) R =

a+b+c 4∆ abc (d) R = ∆

(b) R =

(a)

b2 ac

(b) a + b

(c)

a2 bc

(d)

(a) – 2 (c) 1

1 and sec 3

90. If in a ∆ABC, 2 cos A = sin B cosec C, then

π , then cos2 B + cos2C equals 2 (b) – 1 (d) zero

99. If the sides of a right angled triangle are in A.P. then tangents of the acute angled triangle are

(c)

(a) A = 90º, B = 60º, C = 30º (b) A = 120º, B = 60º, C = 0º (c) A = 60º, B = 30º, C = 0º (d) None of these (b) c = a (d) b = c

c2 ab

(b) λ > 4 (d) 0 < λ < 4

98. In a ∆ABC, ∠A =

(a)

(b) c2a2 (d) s4

89. If sin (A + B + C) = 1, tan (A – B) = (A + C) = 2, then

(a) 2a = bc (c) a = b

(b) 30º (d) None of these

94. In an equilateral ∆, r : R : r1 =

88. If c2 = a2 + b2, then 4s (s – a) (s – b) (s – c) = (a) a2b2 (c) b2c2

(b) b cot A (d) a cot A

(a) λ < 0 (c) λ > 0

(d) None of these

87. In a triangle ABC, a (b cos C – c cos B) is equal to (a) b2 – c2 (c) a2

(a) c cot A (c) a cot B

97. If, in a ∆ABC, (a + b + c) (b + c – a) = λ bc, then

r 2R

86. In a ∆ABC, a = 2b and ∠A = 3∠B, then angle A is (a) 90º (c) 30º

92. If H is the orthocentre of ∆ABC, then AH is equal to

96. ∆ABC is right angled at C, then tan A + tan B =

(a + b + c) (b + c – a) = 3bc is (a) 30º (c) 60º

(b) tan 2º (d) cot 44º

(a) 45º (c) 15º

(b) isosceles (d) None of these

82. If in a triangle ABC, AD, BE and CF and the altitudes and R is the circum radius, then the radius of the circle DEF is (a)

3 + cot 76º cot 16º is cot 76º + cot 16º

(a) cot 60º (c) tan 44º

80. Let the angles A, B, C of ∆ABC be in A.P. and let b : c = 3 : 2 . Then angle A is (a) 75º (c) 60º

91. The value of

1 3+ , 2 1 3, 3

3−

1 5− , 2 4 3 (d) , 4 3

1 (b) 2

5+

1 2

100. With usual notations, in a ∆ABC, b2 − c2 c2 − a 2 a 2 − b2 + + = a sec C b sec C c sec C (a) 1 (c) abc

(b) 0 (d) None of these

101. If r1, r2, r3 in a triangle be in H.P., then the sides are in (a) H.P. (c) G.P.

(b) A.P. (d) None of these

(a) cos θ (c) tan θ

111. If in a ∆ABC, then ∠A =

(b) cot θ (d) sin θ

π π and ∠C = and D divides BC 3 4 sin ∠BAD equals internally in the ratio 1 : 3. Then, sin ∠CAD

103. In a ∆ABC, ∠B =

2 3 1 (c) 6 (a)

1 3 1 (d) 3 (b)

104. In a ∆ABC, a = 13 cm, b = 12 cm and c = 5 cm. The distance of A from BC is 144 13 60 (c) 13

(a)

65 4 (c) 24

67 8 (d) 4

(c)

π 2

(b) 3.0 (d) 2.5

π (b) 3 2π (d) 3

(a) 4, 6, 8 (b) 3, 9, 11 (c) 6, 8, 10 (d) None of these

(a) h  ( 3rh − h 2 )

(b) h  ( 2rh − h 2 )

(c) 2h  ( 2rh − h 2 )

(d) None of these

113. In a triangle, the lengths of two larger sides are 10 and 9 respectively. If the angles are in A.P., then the length of the third side can be

6

(b) 3 3 (d) 5 + 6

π . If r is the inradius and R is the 2 circumradius of the triangle, then 2 (r + R) is equal to (a) a + b (c) c + a

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

(a)

π 2

(b)

5π 6

(c)

2π 3

(d)

3π 4

116. Let the angles A, B, C of ∆ABC be in AP and let b : c ::

3 : 2 then angle A is

(a)  45º (c)  60º

(b)  75º (d)  30º

tan

B C , tan are in 2 2

(a)  AP (c)  HP

A , 2

(b)  GP (d)  AGP

118. In a triangle ABC, if b + c = 2a and ∠A = 60º, then ∆ABC is

109. The radius of the incircle of triangle whose sides are 18, 24 and 30 cms, is (b) 4 (d) 9

110. If in a ∆ABC, a = 6, b = 3 and cos (A – B) = its area is (a) 8 sq. units (c) 6 sq. units

the triangle has perimeter P = 2  ( 2hr − h 2 + 2hr ) and area A =

117. In a triangle ABC, if a, b, c are in AP, then tan

108. The radii r1, r2, r3 of escribed circles of the triangle ABC are in H.P. If its area is 24 sq. cm and its perimeter is 24 cm, then the lengths of its sides are

(a) 2 (c) 6

112. ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is the altitude from A to BC then

115. If the lengths of the sides of triangle are 3, 5, 7 then the largest angle of the triangle is

(b)

107. In a ∆ABC, ∠A > ∠B. If the measure of angles A and B satisfy the equation 3 sin x – 4 sin3x – k = 0, 0 < k < 1, then the measure of angle C is 5π (a) 6

(b) 45º (d) None of these

114. In a ∆ABC, let ∠C =

106. If the lengths of the sides of a triangle are 3, 4 and 5 units then R is (a) 3.5 (c) 2.0

(a) 90º (c) 60º

(a) 5 – (c) 5

65 12 25 (d) 13 (b)

105. If the sides of a triangle are 13, 14, 15 then radius of its incircle is (a)

2 cos A cos B 2 cos C a b + + + = a b c bc ca

875

2 bc A  sin  = b−c 2

(b) 9 sq. units (d) None of these

(a)  equilateral (c)  isosceles

(b)  right angled (d)  scalene

119. Inradius of a circle which is inscribed in a isosceles triangle one of whose angle is 2π/3 is of triangle is

4 then 5

(a)   4 3

(b)  12 – 7 3

(c)  12 + 7 3

(d)  none of these

3 , then area

120. In a ∆ABC, if (sin A + sin B + sin C) (sin A + sin B – sin C) = 3 sin A sin B, then angle C is equal to

Properties and Solutions of Triangles

102. If a = (b – c) sec θ, then

876

Objective Mathematics

π 2 π (c)   4

π 3 π (d)   6

(a)  

(b)  

π R 121. In ∆ABC, ∠A = , b = 4, c = 3, then the value of 2 r is equal to 5 2 9 (c)   2

7 2 35 (d)   24

(b)  

(a)  

122. In a triangle ABC, right angled at C, the value of cot A + cotB is (a)  

c2 ab

(b)  a + b

(c)  

a2 bc

(d)  

b2 bc

123. The sides of triangle are respectively 7 cm, 4 3 cm and 13 cm, then the smallest angle of the triangle is π 6 π (c)   4

π 3 π (d)   5

(a)  

125. In ∆ABC, if sin2 a, b, c will be in

A B C , sin2 , sin2 be in HP. Then 2 2 2

(a)  AP (b)  GP (c)  HP (d)  None of these 126. Angle of a triangle are in the ratio 4 : 1 : 1. The ratio between its greatest side and perimeter is (a)  

3 2+ 3

(b)  

1 2+ 3

(c)  

3 3+2

(d)  

2 2+ 3

1 1 1 1 + + + is r12 r2 2 r32 r 2

127. The value of (a)  0 (c)  

∆2 a + b2 + c2 2

(b)  

a 2 + b2 + c2 ∆2

(d)  

a 2 + b2 + c ∆

(b)  

A B − tan 2 2 is equal to 124. In any triangle ABC, A B tan + tan 2 2 a−b a−b (a)   (b)   ] a+b c a−b c (c)   (d)   a+b+c a+b tan

128. If in a ∆ABC, 2b = a2 + c2, then

(a)  

c2 − a2 2ca

(b)   2

sin 3B is equal to sin B

c2 − a2 ca

 c2 − a2   (d)     2ca 

 c2 − a2   (c)     ca 

2

129. In a ∆ABC, if ( 3 – 1)a = 2b, A = 3B, then C is (a)  60º (b)  120º (c)  30º (d)  45º

sOLUTIONS

1. (b) We have,

A B C cos 2 cos 2 2 + 2 + 2 a b c

cos 2

 s ( s − a )   s ( s − b)   s ( s − a )   bc   ca   ab  =   + + c   a   b         =

s ( s − a + s − b + s − c) s [3s − (a + b + c)] = abc abc

=

s (3s − 2 s ) s2 . = abc abc

2. (d) We have, cot

B C cot = 2 2

=

 s ( s − b)   s ( s − c)   ×   ( s − c )( s − a )    ( s − a)( s − b) 

=

 s ( s − b) s ( s − c)  ×   ( s − c )( s − a ) ( s − a )( s − b)  

=

s 2a = =2 s − a 2a − a

[ b + c = 3a ⇒ a + b + c = 4a ∴ 2s = 4a; s = 2a] 3. (a) Here s = ∴ sin

a + b + c 16 + 24 + 20 = 30 = 2 2 A = 2

 (s − b)( s − c)    bc

2



 6 × 10   24 × 20 

then cos C =

1 1 .   = 8 2 2

=

C3 – C3 = 21  ⇒ 

n+1

n

1 + sin α cos α

sin α cos α a 2 + b2 − c2 1 =– =– 2 sin α cos α 2ab 2

∴ C = 120°.

4. (b) Tn = nC3  ∴  Tn + 1 – Tn = 21 ⇒

9. (c) Let a = sin α, b = cos α, c =

10. (b) Given AD = 4 and BD = DC

C2 + C3 – C3 = 21

n

n

n

In ∆ABG, tan

n(n − 1) = 21 ⇒ nC2 = 21 ⇒ 2

π AG = 3 BG

⇒ BG = AG cot

⇒ n2 – n – 42 = 0 ⇒ (n – 7)(n + 6) = 0

8 π 8 1 = × = 3 3 3 3 3

⇒ n = 7 as n ≥ 1. 5. (b) We have A + C = π – B and

A− B+C π = −B 2 2

∴ Given expression 2ac cosB = 2ca 

c2 + a 2 − b2 2 ca

2

2

6. (b) We have, 1 – tan 

=1–

A B tan  2 2

 (s − c)( s − a )     s ( s − b) 

7. (a) We have, tan = =

B = 2

C and cos = 2 ∴ b cos2

= b ⋅ 



=



(s − c)( s − a ) s ( s − b)

s − c 2 s − 2c a + b + c − 2c a + b − c = . = = s 2s a+b+c a+b+c

8. (c) Since, cos



A B tan = 2 2

(s − b)( s − c) s ( s − a)



 (s − a)( s − b)     s ( s − c) 

( s − a) s − ( s − a ) a 2a 2a =1– = . = = = s s s 2s a + b + c

 s ( s − b)   ac   s ( s − c)   ab  ,

C B + cos2 2 2

s ( s − c) s ( s − b) + c ⋅  ab ac

s ( s − c ) s ( s − b) + a a s s = [(s – c) + (s – b)] = (2s – b – c) a a s s (a + b + c – b – c) = (a) = s. = a a

1 × BG × AD 2 1 8 16 ×4= = × 2 3 3 3 3 Since median divides a triangle into two triangles of equal area. Therefore Area of ∆ABC  = 2 × area of ∆ADB 16 32 = = 2× . 3 3 3 3 Area of ∆ADB  =

=a +c –b 2



C A 3b + c cos 2 = 2 2 2 s ( s − c ) s  ( s − a )  3b   = + c ⇒ a    bc  2  ab 

11. (a) a cos 2

⇒ ⇒ ⇒ ⇒ ⇒

s [ s − c + s − a ] 3b = b 2 2s[2s – c – a] = 3b2 2s[a + b + c – c – a] = 3b2 (a + b + c)b = 3b2 ⇒ a + b + c = 3b 2b = a + c  i.e.,  a, b, c are in A.P.

12. (a) We have, (a + b + c) (b + c – a) = 3bc ⇒ (a + b + c) (a + b + c – 2a) = 3bc ⇒ 2s (2s – 2a) = 3bc ⇒ 4s (s – a) = 3bc s (s − a) s (s − a) 3 3 = ⇒ = bc bc 2 4 A A ⇒ cos  = cos 30º or = 30º ∴ A = 60º. 2 2

⇒

A B C  13. (a) We have, 4 bc cos 2 + ca cos 2 + ab cos 2  2 2 2  s (s − a) s ( s − b) s ( s − c)  = 4 bc ⋅ + ca ⋅ + ab ⋅ bc ca ab   = 4s [(s – a + s – b + s – c)]

877

 (30 − 24)(30 − 20)    = 24 × 20

=

Properties and Solutions of Triangles



878



= 4s [3s – (a + b + c)] = 4s (3s – 2s) = 4s (s) = 4s2 = (2s)2 = (a + b + c)2.

Objective Mathematics

14. (a) We have, cot

A C B + cot = 2 cot 2 2 2

⇒ cot

Multiplying both sides by

⇒ ( s − a )( s − b)( s − c) s

⇒ (s – a) + (s – c) = 2 (s – b) ⇒ 2b = a + c. ⇒ a, b, c are in A.P. 15. (a) We have, 3 tan

A C tan =1 2 2

⇒ 3

( s − b)( s − c) ( s − a )( s − b) ⋅ =1 s (s − a) s ( s − c)

⇒ 3 ⋅

s−b = 1 ⇒ 3s – 3b = s s

⇒

s−b 1 = ⇒ 3s – 3b = s ⇒ 2s = 3b s 3

⇒ 2b = c + a ⇒ a, b, c are in A.P. 20. (c) We have, 4∆ (cot A + cot B + cot C) 1 cos A 1 cos B = 4 ⋅ bc sin A ⋅ + 4 ⋅ ca sin B ⋅ 2 sin A 2 sin B 1 cos C + 4 ⋅ absin C ⋅ 2 sin C = 2bc cos A + 2ca cos B + 2ab cos C

a 2 + b2 − c2 2ab = b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2 = a 2 + b 2 + c 2.

C A 3b + c cos 2 = 2 2 2

21. (c) We have, abc s sin

⇒ a + a cos C + c + c cos A = 3b

= abc ⋅

⇒ a + c + (a cos C + c cos A) = 3b

s ⋅

⇒ a + c + b = 3b [By projection formula a cos C + c cos A = b] ⇒ a + c = 2b ⇒ a, b, c are in A.P.

if



=

( s − c)( s − a ) ca

or if

A B C sin sin 2 2 2

( s − b)( s − c) ( s − c)( s − a ) ( s − a )( s − b) ⋅ ⋅ bc ca ab ( s − a )( s − b)( s − c) abc = s (s – a) (s – b) (s – c) = ∆2. = abc s

22. (c) We have, 4∆ cot A = 4 ⋅

( s − b)( s − c) ( s − a )( s − b) ⋅ bc ab

s − b ( s − c)( s − a) = b ca



A C B sin = sin 2 2 2

2 ⋅ 

or if 2 ⋅ 

b2 + c2 − a 2 c2 + a 2 − b2 + 2ca ⋅ 2bc 2ca + 2ab ⋅

 1 + cos A  3b  1 + cos C  ⇒ a   =  + c   2 2 2

17. (c) We have, 2 sin

1 A C = tan 3 2 2

( s − b) ( s − c) ( s − b)( s − a ) ⋅ s (s − a) s ( s − c)

= 2bc ⋅

⇒ 3s – s = 3b ; 2s = 3b ⇒ a + b + c = 3b or a + c = 2b ∴ a, b, c are in A.P. 16. (b) We have, a cos 2

5 2 A C = , tan  = 6 5 2 2

⇒ tan

s ( s − b) ( s − c)( s − b)

=2⋅

19. (b) tan

s (s − a) s ( s − c) + ( s − b)( s − c) ( s − a )( s − b)

⇒



A B C , cot , cot are in A.P. 2 2 2

b2 + c2 − a 2 2bc 9 + c 2 − 16 ⇒ cos 60° = ⇒ c2 – 3c – 7 = 0. 6c

18. (a) cos A =



= 2 bc ⋅ cos A = 2 bc ⋅ 



= b 2 + c 2 – a 2.

23. (a) We have, ( s − c)( s − a ) ca

2( s − b) = 1 or if 2s – 2b = b b

or if a + b + c – 2b = b or if a + c = 2b i.e., if a, b, c are in A.P.

1 cos A bcsin A ⋅ 2 sin A b2 + c2 − a 2 2bc

2abc A B C cos cos cos a+b+c 2 2 2

=

2abc s ( s − a ) s ( s − b) s ( s − c) ⋅ ⋅ ⋅ 2 bc ca ab

=

abc s s ( s − a)( s − b)( s − c) ⋅ s abc

=

s ( s − a )( s − b)( s − c) = ∆.

∆ =

1 ab sin C 2

1 × 20 × 15 sin C 2 75 × 2 1 = ⇒ sin C = 20 × 15 2 1 ⇒ sin C = = sin 30º or sin (180º – 30º) 2

∴ 75 =

∴ C = 30º or 150º.

a+b+c a b c = = = 3 3 /2 1/2 1/2 +1 2

a = a+b+c

A B C cos cos 2 2 2

 s ( s − c)   s ( s − b)   s (s − a)  ×  ×  = abc × d   ab   ca   bc  abc = × s [ s ( s − a )( s − b)( s − c)] abc = s [ s ( s − a )( s − b)( s − c)] = s∆. 27. (a) We have, a = 18, b = 24, c = 30 a+b+c 18 + 24 + 30 = = 36 ∴ s = 2 2 ∴ ∆ =

s ( s − a )( s − b)( s − c)



=

36 (36 − 18)(36 − 24)(36 − 30)



=

36 × 18 × 12 × 6 =

36 × 6 × 3 × 3 × 4 × 6

= 6 × 6 × 3 × 2 = 216 ∆ 216 = 6. = ∴ r = s 36 a = 13 cms, b = 14 cms, c = 15 cms, ∴ s = ∴ ∆ =

21 × 8 × 7 × 6

7 ×3× 4× 2×7 ×3× 2



=



= 7 × 3 × 2 × 2 = 84

∴ R =



cos A =

3x 2 + 4 x 2 − x 6 3 = = 2 4 3 x2 4 3



cos B =

x 2 + 4 x 2 − 3x 2 2 1 = = , cos C = 0 4x2 4 2

31. (c) Here, b = 2, B = 30º b 2 2 = = ∴ R = 2 sin B 2 sin 30º 1 or R = 2 Area of circumcircle = πR2 = π × (2)2 = 4π sq. units. 33. (b) Given that a = 4 cm, b = 5 cm and c = 6 cm. a+b+c We know, s = 2 4 + 5 + 6 15 s = = 2 2 ∴ Area of the triangle = s ( s − a )( s − b)( s − c)

=

 15   15   15   15     − 4   − 5   − 6  2 2 2 2



=

15 7 5 3 15 × × × = 7  cm2. 2 2 2 2 4

34. (b) Angles are 120o, 30o.30° = A, B, C

⇒

a + b + c 42 = 21 = 2 2 s ( s − a )( s − b)( s − c) =

abc = ∆. 4R

We have, a b c = = sin A sin B sin C



28. (b)

=

⇒ A = 30º,   B = 60º,  C = 90º.

3 /2 3 . = 3+2 3+2 2

26. (b) We have, abc cos



By cosine rule,

We have, a b c = = sin A sin B sin C

∴

a b c = sin A; = sin B; = sin C] 2R 2R 2R

30. (d) Let the sides be x, 3 x, 2 x

25. (b) Angles are 120o, 30o.30° = A, B, C

⇒

⇒

13 × 14 × 15 65 1 abc = = =8 cms. 4 × 84 8 8 4∆

∴

a b c a+b+c = = = 3 /2 1/2 1/2 3 +1 2

a = a+b+c

3 /2 3 . = 3+2 3+2 2

35. (a) We have, 4  s − 1  s − 1  s − 1 a  b  c 

= 4⋅

( s − a )( s − b)( s − c) abc

879

We have to find ∠C.

29. (c) We have, 2R2 sin A sin B sin C a b c = 2R2 ⋅ ⋅ 2R 2R 2R a b c  [ = = =R 2sin A 2sin B 2sin C

Properties and Solutions of Triangles

24. (a), (c) Let a = 20, b = 15, ∆ = 75

880

Objective Mathematics



= 4⋅

s ( s − a )( s − b)( s − c) s ⋅ abc



= 4⋅

∆2 4∆ ∆ r = ⋅ = . s ⋅ abc abc s R

36. (a) We have, r2 cot 

A B C cot cot 2 2 2

s ( s − a) s ( s − b) s ( s − c) ⋅ ⋅ ( s − b)( s − c) ( s − c)( s − a) ( s − a)( s − b)

2 = r ⋅

=

41. (b) We have,

∆2 s2 ⋅ = ∆. s2 ∆

∆  ∵ r = s 

37. (b) We have, 4R r cos =4 ⋅

A B C cos cos 2 2 2

abc ∆ s ( s − a ) s ( s − b) s ( s − c) ⋅ ⋅ ⋅ ⋅ 4∆ s bc ca ab

=

abc s s ( s − a)( s − b)( s − c) ⋅ s abc

=

s ( s − a )( s − b)( s − c) = ∆.

∆2 ∆ = = = r 1. ∆ ⋅ ( s − a) s − a c c = 2 sin C 2 [ ∵ sin C = sin 90º = 1] C 2 [ ∵ r = (s – a) tan =s –c Adding, R + r = 

=

=

1 [4s2 – 2s ⋅ 2s + a2 + b2 + c2] ∆2

=

a 2 + b2 + c2 . ∆2

42. (c) We have, rr1 + r2 r3 ∆ ∆ ∆ ∆ = ⋅ + + s s−a s−b s−c  1  1 + = ∆2    s ( s − a ) ( s − b)( s − c) 



=

s ( s − a )( s − b)( s − c) abc s ( s − b)( s − c) = = ⋅ ∆ ⋅ ( s − a) ∆ abc

and r = (s – c) tan

1 [4s2 – 2s (a + b + c) + a2 + b2 + c2] ∆2

43. (a) We have, r2 r3 + r3 r1 + r1 r2 =

A B C cos cos 2 2 2



=

= 2s2 – s ⋅ 2s + bc = bc.

abc s ( s − b)( s − c) s ( s − b) s ( s − c) =4 ⋅ ⋅ 4∆ bc ca ab

40. (a) We have, R =

( s − a ) 2 ( s − b) 2 ( s − c ) 2 s 2 + + + 2 ∆2 ∆2 ∆2 ∆

 2 s 2 − s (a + b + c) + bc  = ∆2   ∆2  

3s − (a + b + c) 3s − 2 s s 1 = = = . ∆ ∆ ∆ r

39. (a) We have, 4R sin 

=

 ( s − b)( s − c) + s ( s − a )  = ∆2    s ( s − a )( s − b)( s − c) 

1 1 1 s−a s−b s−c + + = + + 38. (b) We have, r1 r2 r3 ∆ ∆ ∆ =

1 1 1 1 + + + r12 r22 r32 r 2

∆ ∆ ∆ ∆ ∆ ∆ ⋅ + ⋅ + ⋅ s−b s−c s−c s−a s−a s−b

 s−a+s−b+s−c  = ∆2    ( s − a)( s − b)( s − c)  = ∆2 ⋅

3s − (a + b + c) ( s − a )( s − b)( s − c)

= ∆2 ⋅

3s − 2 s ( s − a)( s − b)( s − c)

=

∆2s ∆2s2 = ( s − a )( s − b)( s − c) s ( s − a )( s − b)( s − c)

=

∆2s2 = s 2. ∆2

44. (b) We have, 16R2 rr1 r2 r3 A = ...] 2

C 90º   ∵ tan 2 = tan 2 = tan 45º = 1 c +s– c 2

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

= 16 ⋅ 

=

∆ ∆ a 2b 2 c 2 ∆ ∆ ⋅ ⋅ ⋅ ⋅ 16∆ 2 s s − a s − b s − c

a 2b 2 c 2 ∆ 2 a 2b 2 c 2 ∆ 2 = = a2 b2 c2. s ( s − a )( s − b)( s − c) ∆2

45. (a) We have, ∆ =

1 1 BC · AM = ap1 2 2

∴ p1 = 2∆ a



Similarly p2 =



∴



2∆ 2∆ , p3 = b c

∴ cos B =

a+b+c 2s s 1 = = = . 2∆ 2∆ ∆ r



r =



r1 =



r2 =



r3 =

a + a + a 3a = 2 2 ∆ ∆ 2∆ = = s 3a 3a 2 ∆ ∆ 2∆ 2∆ =3 ⋅ = 3r = = s−a a a 3a 2 ∆ ∆ = 3r = s−b s−a ∆ ∆ = 3r = s−c s−a

⇒ 2 cot B = cot A + cot C 2cos B cos A cos C = + sin B sin A sin C b2 + c2 − a 2 a 2 + b2 − c2 2 (c 2 + a 2 − b 2 ) + ⇒ = 2bc (ka ) 2ab (kc) 2ac (kb) ⇒

⇒ 2c2 + 2a2 – 2b2 = b2 + c2 – a2 + a2 + b2 – c2 ⇒ 2c2 + 2a2 = 4b2 ⇒ c2 + a2 = 2b2 ⇒ a2, b2, c2 are in A.P. 51. (a) We have,

47. (a) Area of a circle = π × [radius]2 ∴ A = π r , A1 = π r , A2 = π r , A3 = π r 2 1

2 2

=

1 1 1 1   + +  = π  r1 r2 r3 

=

1  3s − (a + b + c)  1 3s − 2 s = ⋅   ∆ ∆ π π 

=

2 3

1 1 1 1 1 1 + + = + + A1 A2 A3 r1 π r2 π r3 π

∴

1 s 1 ⋅ = = π ∆ r π

48. (a) r1, r2, r3 are in H.P. if

1 s − a s − b s − c + + ∆ ∆  π  ∆

1 πr

2

=

1 . A

1 1 1 , , are in A.P. r1 r2 r3

s−a s−b s−c are in A.P. , , ∆ ∆ ∆ if s – a, s – b, s – c are in A.P. [Multiplying each term by ∆] or if (s – b) – (s – a) = (s – c) – (s – b) or if a – b = b – c or if 2b = a + c i.e., if a, b, c are in A.P. or if

sin A 2 sin C

50. (b) Since, cot A, cot B, cot C, are in A.P.

∴ r1 = r2 = r3 = 3r. 2

[By sine formula]

i.e., c2 + a2 – b2 = a2 ⇒ b2 = c2 ⇒ b = c, Hence the triangle is isosceles.

46. (c) In an equilateral triangle, a = b = c ∴ s =

[By sine formula]

2 2 2 a 2 2 2 k ⇒ c +a −b = a ⇒ c + a − b = 2ca 2c 2ca 2⋅ c k a ⇒ c2 + a2 – b2 = × 2ca 2c

1 1 1 a b c + + = + + p1 p2 p3 2∆ 2∆ 2∆ =

[cosine formula]

tan C sin C cos A = ⋅ tan A cos C sin A

By Sine Formula, a c sin C c ⇒ = = sin A sin C sin A a ∴



 b2 + c2 − a 2   c  tan C 2bc = ⋅ 2 2 2  a a +b −c  tan A   2ab [using Cosine Formula] =

2ab c b2 + c2 − a 2 × × 2 2bc a a + b2 − c2

b2 + c2 − a 2 . a 2 + b2 − c2 cos A cos B cos C 52. (d) We have, = = a b c cos A cos B cos C ⇒ = = k sin A k sin B k sin C

=

⇒ ⇒ ∴ ∴

cot A = cot B = cot C A =B=C ABC is an equilateral triangle Area of equilateral ∆ 3 3 = (side)2 = (2)2 = 3 . 4 4



Properties and Solutions of Triangles

c2 + a 2 − b2 , 2ca a sin A = k c and sin C = k

cos B =

881

49. (b) We know that

882

58. (a) We have, 2bc cos A + 2ca cos B + 2ab cos C

53. (a) We have, 2 cos A + cos B + 2 cos C a b c

Objective Mathematics

2 (b + c − a ) a + c − b 2 (a + b − c ) + + 2abc 2abc 2abc 2

=

2

2

2

2

a b a +b + = bc ca abc 2

R.H.S. =

2

2

2

2

⇒ 2 (b2 + c2 – a2) + (a2 + c2 – b2) + 2 (a2 + b2 – c2) = 2a2 + 2b2 2 2 2 2 2 ⇒ a + 3b + c = 2a + 2b ⇒ a = b + c ∴ ∠A = 90º. 2

2

C B 54. (a) The given expression = 2   b cos 2 + c cos 2   2 2 C  = b  2 cos 2  + c  2

 2 B  2 cos  2

= b (1 + cos C) + c (1 + cos B) = (b cos C + c cos B) + b + c = a + b + c [ ∵ b cos C + c cos B = a] 55. (b) The given expression 1 + cos A = (b + c)2 – 4bc ⋅ 2 [ ∵ 2 cos2θ = 1 + cos 2θ] = b2 + c2 + 2bc – 2bc (1 + cos A) = b2 + c2 + 2bc – 2bc – 2 bc cos A = b2 + c2 – 2bc cos A = a2.   b2 + c2 − a 2 . ∵cos A = bc 2   2 2 2 ∴ a = b + c − 2bc cos A  C A 56. (a) We have, 2  a sin 2 + c sin 2   2 2  2 C = a  2 sin  + c  2

 a 2 + b2 − c2  + 2ab ⋅   2ab 

2

Cancelling abc from both sides, we get (a 2 + c 2 − b 2 ) (b2 + c2 – a2) + 2  + (a2 + b2 – c2) = a2 + b2

2

 c2 + a 2 − b2   b2 + c2 − a 2  = 2bc ⋅  + 2ca ⋅    2bc 2ca   

 2 A  2 sin  2



= b 2 + c 2 – a 2 + c 2 + a 2 – b 2 + a 2 + b 2 – c 2.



= a 2 + b 2 + c 2.

59. (a) Let a = x, b = y and

Then a2 = x2, b2 = y2 and c2 = x2 + y2 + xy. ∴ c is the length of the greatest side and hence ∠C is the greatest angle. ∴ By cosine rule

cos C =



=–

= ab cos C – ac cos B = = = =

a 2 + b2 − c2 c2 + a 2 − b2 ab ⋅ − ac ⋅ 2ab 2ca a 2 + b2 − c2 c2 + a 2 − b2 − 2 2 2 2 2 2 a + b − c − c − a 2 + b2 2 2 (b 2 − c 2 ) = b 2 – c 2. 2

x 2 + y 2 − x 2 − y 2 − xy a 2 + b2 − c2 = 2 xy 2ab

1   ∴  m ∠C = 120º. 2

60. (a) By Napier’s Analogy, b−c A B−C = cot b+c 2 2 But B = 90º, ∴ C + A = 90º

tan

⇒ C = 90º – A. ∴ From (1),

tan 90º − (90º − A) = b − c cot A 2 2 b+c

  b−c 1  A ⇒ tan = b + c  tan A  2   2 ⇒ tan2

b−c A = b+c 2

⇒ tan

A = 2

b−c . b+c

61. (a) By sine formula,

= a (1 – cos C) + c (1 – cos A) = (a + c) – (a cos C + c cos A) = a + c – b. 57. (b) We have, a (b cos C – c cos B)

x 2 + y 2 + xy .

c=

a b c = k (say) = = sin A sin B sin C ∴ a = k sin A

b = k sin B



c = k in C

Now a cos A = b cos B ∴ k sin A cos A = k sin B cos B ⇒ 2 sin A cos A = 2 sin B cos B ⇒ sin 2A = sin 2B = sin (π – 2B) Now sin 2A = sin 2B ∴ 2A = 2B, or A = B

∴ 2A = π – 2B ⇒ 2A + 2B = π π ∴ A + B = 2 ∴   C = 90º ∴ ∆ is right-angled. 62. (b) We have,

1 + cos (A − B) cos C 1 + cos (A − C) cos B

1 + cos (A − B) cos [180º − (A + B)] 1 + cos (A − C) cos [180º − (A + C)]

=

1 − cos (A − B) cos (A + B) 1 − cos (A − C) cos (A + C)

=

1 − (cos 2 A − sin 2 B) 1 − cos 2 A + sin 2 B = 2 2 1 − (cos A − sin C) 1 − cos 2 A + sin 2 C

=

k (sin A + sin B) sin A + sin B = 2 k (sin 2 A + sin 2 C) sin 2 A + sin 2 C

=

(k sin A) 2 + (k sin B) 2 a 2 + b 2 = . (k sin A) 2 + (k sin C) 2 a 2 + c 2

2

2

2

=

a π cot 2 2n

65. (b) We have, a = 5, b = 7 sin A =

=

2

π π    − 1 1 + cos  1 + 2 cos 2 a a n 2n =    =  2  sin π  2  2 sin π cos π    n  2n 2n 

2

63. (a) We have, a2 sin 2B – b2 sin 2A = (k sin A)2 (2 sin B cos B) – (k sin B)2 (2 sin A cos A) = k2 sin2A ⋅ 2 sin B cos B – k2 sin2B ⋅ 2 sin A cos A = 2k2 sin A sin B (sin A cos B – cos A sin B) = 2k2 sin A sin B (sin A cos B – cos A sin B) = 2k sin A ⋅ k sin B sin (A – B) = 2ab sin (A – B). 64. (b) AB = a ON ⊥ AB and AN = BN

883

Again, sin 2A = sin (π – 2B)

Properties and Solutions of Triangles

∴ ∆ is isosceles.

∴

3 4

sin A sin B = a b

3 7× b sin A 4 = 21 ⇒ sin B = = 5 20 a which is not possible. 66. (b) By sine formula, a b c = k (say) = = sin A sin B sin C ∴ a = k sin A, b = k sin B, c = k sin C a − b k sin A − k sin B = Now, a + b k sin A + k sin B A+B A−B sin A − sin B 2 cos 2 sin 2 = = sin A + sin B 2 sin A + B cos A − B 2 2 A−B A+B cos 2 2 = A−B A+B cos sin 2 2 sin



A−B A + B tan 2 A−B = = tan cot . A+B 2 2 tan 2 67. (c) Since the sides are +ve ∴ x > 1 or else c will be negative. a = x2 + x + 1 > x + x + 1 = 2x + 1 = b ∴ a > b Also, a > c. Hence a is the greatest side so A is the greatest angle. ∴ cos A =

π AN = n ON π a π ⇒ ON = AN cot = cot n 2 n π AN and sin = n OA π a π ⇒ OA = AN cosec  = cosec n 2 n In ∆AON, tan

Sum of the radii = ON + OA =

a  cos π /n 1  a π a π + cot + cosec =   2  sin π /n sin π /n  2 n 2 n

b2 + c2 − a 2 1 =– 2bc 2

∴ A = 120º. 68. (b) ∵ tan θ =

2 ab C sin a−b 2

∴ (a – b)2 tan2θ = 4ab sin2

C 2

or, (a – b)2 (sec2θ – 1) = 4ab sin2

C 2

or, (a – b)2 sec2θ = (a – b)2 + 4ab sin2

C 2

C  or, (a – b)2 sec2θ = a2 + b2 – 2ab ⋅  1 − 2 sin 2   2

884

or,

(a – b)2 sec2θ = a2 + b2 – 2ab cos C

Objective Mathematics

or, (a – b)2 sec2θ = c2

 a 2 + b2 − c2  ∵cos C =  2ab  

⇒ cot A = cot B = cot C ⇒ A = B = C = 60º ∴ triangle is equilateral and side = a = 2.

∴ c = (a – b) sec θ. 69. (c) Let a, b, c be the sides of the triangle. a+b+c and so 2 s, s – a, s – b , s – c will be positive.



∵ For positive quantities, A.M. > G.M.  [∵ s, s – a, s – b , s – c are not all equal] ∴ 

3.

b2 + c2 − a 2 2bc

73. (a) We have, cos A =

Clearly, s =

3 2 a = 4

∴ area of the triangle =

1 9 + c 2 − 16 = ⇒ c2 – 3c – 7 = 0. 2 2 ⋅ 3c

⇒ cos 60º =

74. (d) We have, 3 sin A = 6 sin B = 2 3 sin C sin A sin B sin C = = ⇒ 2 1 3 ⇒

a b c = = = k (say) 2 1 3

s + ( s − a ) + ( s − b) + ( s − c ) 4 > [s (s – a) (s – b) (s – c)]

⇒ a = 2k, b = k, c =

cos A cos B cos C = = k sin A k sin B k sin C

⇒ sin R = sin Q =

3 k.

b +c −a =0 ∴ cos A = 1 4s − (a + b + c) 2bc or, > (∆ 2 ) 4 4 ⇒ ∠A = 90º. 1 2s ∵ 2 or, [ a + b + c = 2s] > ∆ 1 4 75. (b) We have, s = (a + b + c) = 8k 2 2 s >∆ or, 4 ∆ = s ( s − a )( s − b)( s − c) = 12k2 [∵ both s and ∆ are positive, squaring will not ∆ 12k 2 affect the inequality] r = ⇒6= ⇒ k = 4. s 8k s2 . i.e., ∆ < 76. (d) In ∆PQR, radius of circumcircle is PQ = PR. 4 PQ QR PR cos A cos B cos C = = ∴ PQ = PR = 70. (c) We have, = = 2 sin R 2 sin P 2 sin Q a b c ⇒

⇒ cot A = cot B = cot C ⇒ A = B = C = 60º ∴ the triangle is equilateral.

1 71. (a) Given that a + b + c = 6 ⋅ (sin A + sin B + sin C) 3 ⇒ k (sin A + sin B + sin C) = 2 (sin A + sin B + sin C) a b c = = where k = sin A sin B sin C ⇒ (k – 2) (sin A + sin B + sin C) = 0 ⇒ k – 2 = 0 ⇒ k = 2 a =k=2 ⇒ sin A 1 ⇒ sin A = 2 π ⇒ A = . 6 cos A cos B cos C 72. (d) Given = = a b c cos A cos B cos C = = ⇒ k sin A k sin B k sin C

2

2

⇒ ∠R = ∠Q =

2

1 2

π 6

⇒ ∠P = π – ∠R – ∠Q = 77. (a) We have, s =

∆=

2π . 3

1 3π (a + b + c) = 2 3

3a 2 . 4

∴ r =

∆ a = s 2 3

I f x is the length of a side of the square inscribed in in-circle of the triangle then ( a = 1)

x2 + x2 = (diameter)2 = (2r)2

⇒ x2 = 2 r2 =

a2 a2 ⇒ Area of square = . 6 6

78. (a) Let r be the radius of the circle. ∴ its area A1 = π r2. Length of one side of a regular polygon of n sides = (perimeter of the circle)/n 2πr = a (say). = n

= A2 =

π π 1 π2 r 2 π a2 cot   n  = c cot   n      4 n

∴ A1 : A2 = π r2 : 79. (a), (b), (d)  cot B =



π π π π2 r 2 cot  n  = tan  n  : .     n n

AB2 + BC 2 − AC 2 11 = 2AB ⋅ BC 16

∴ radius of circle DEF FE a cos A = = 2 sin ∠FDE 2 sin (180º −2A)

= 22 + 22 – 2 ⋅ 2 ⋅ 2 ⋅ 

∴ A + C = 2B ⇒ A + B + C = 2B ⇒ A + C = 120º

∴ sin C =

2 sin 60º = 3

R since a = R  .   2sin A 2 

85. (d) cos2

2 3 :

2]

2 3 1 ⋅ = . 3 2 2

∴ C = 45º ∴ A = 180º – 60º – 45º = 75º. 81. (a) Since a cos A = b cos B ∴ K sin A cos A = k sin B cos B ⇒ sin 2A = sin 2B ⇒ 2A = 2B or 2A = π – 2B. π π – B i.e., A + B = 2 2 ⇒ a = b or triangle is right angled. ⇒ A = B or A =

But a ≠ b ∴ triangle is right angled. 82. (a) Circle on BC as diameter passes through E, F (since ∠BEC = ∠CFB = 90º) and radius of this a . circle = 2 Also, ∠FBE = 90º – A FE = 2R sin A = a cos A  and ∠FDE = 180º – 2A,

=

b 2b = a a 2

2b a 2 + b 2 − c 2 ⇒ a2 – c2 = 3b2. = a 2ab

⇒ cos A =

⇒ 180º = 3B ⇒ B = 60º

[since b : c =



⇒ ab + ac – a2 + b2 + bc – ab + bc + c2 – ac = 3bc b2 + c2 − a 2 1 = ⇒ b2 + c2 – a2 = bc ⇒ 2bc 2

80. (a) Since A, B, C are in A.P.



a cos A 2R sin A ⋅ cos A = 2 sin2A 2 ⋅ 2 sin A cos A

84. (c) (a + b + c) (b + c – a) = 3bc

11 = 2 ⋅ 5 16

3 :

=

⇒

Obviously AD ≠ 2 ⋅ 4.

Now sin B : sin C =



83. (c) From right angled ∆CAD, cos C =

AC 2 + BC 2 − AB2 7 = cos C = 2AC ⋅ BC 8

In ∆ABD, AD2 = AB2 + BD2 – 2 AB ⋅ BD cos B

Properties and Solutions of Triangles



885

∴ Area of the polygon

1 = cos 60º ⇒ A = 60º. 2

A B C + cos2 + cos2 2 2 2

1 [1 + cos A + 1 + cos B + 1 + cos C] 2 3 1 r r [since cos A + cos B + cos C = ] = + 2 2R R 3 r . = + 2 2R a b = 86. (a) By sine formula sin A sin B =

∴

2b b = sin 3B sin B

∴ 2 sin B = 3 sin B – 4 sin3B ⇒ 4 sin3B – sin B = 0 ⇒ sin B [4 sin2B – 1] = 0 1 1 ⇒ either B = 0 or sin2B =   i.e.,  sin B = 4 2 ∴ B = 30º ∴ A = 90º. 87. (a) a (b cos C – c cos B)  a 2 + b2 − c2 c2 + a 2 − b2  −c⋅ = a b ⋅  2a 2ca   a 2 + b2 − c2 c2 + a 2 − b2  − =a   2ab 2a 

886



Objective Mathematics



1 [a2 + b2 – c2 – c2 – a2 + b2] 2 1 [2b2 – 2c2] = b2 – c2. = 2

=

88. (a) Since c2 = a2 + b2 ∴ ∆ABC is right angled with ∠C = 90º 1 ab 2 s ( s − a )( s − b)( s − c) = 1 ab ∴ 2 ⇒ 4s (s – a) (s – b) (s – c) = a2 b2. ∴ area =

89. (c) We have,

90. (b) 2 cos A = sin B = b sin C c b2 + c2 − a 2 b = 2bc c

3 + cot 76º cot 16º cot 76º + cot 16º

=

2 sin 76º sin 16º + cos (76º −16º ) sin (76º +16º )

sin (92º )

1 1 cos 60º − cos 92º + 2= 2 sin 92º

1 − cos 92º 2 sin 2 46º = sin 92º 2 sin 46º cos 46º

⇒ AH =



[Since A = B = C = 60º] = r1 = 4R sin

A B C cos cos 2 2 2

∴ r : R = 1 : 2 Also R : r1 = 2 : 3 ∴ r : R : r1 = 1 : 2 : 3. 95. (c) We have, a = 2R sin A = 2R sin  ∴ a =

4R∆  bc a b , tan B = b a

c cos A c cos A a = = cos A sin (180º − C) sin C sin A a c   since sin A = sin C 

AM 93. (c) = tan α ⇒ MB = AM cot α MB AM = cot α ⇒ MC = AM tan α MC

∴ tan A + tan B = Since a2 + b2 = c2 ∴ tan A + tan B =

A A  ⋅ cos  2 2 ∴  R =

AH c = sin (90º − A) sin (A + B)

= a cot A.

R 2

1  3   3  3R = 4R    =  2   2   2  2

96. (d) tan A =

= tan 46º = cot 44º. 92. (d) From ∆AHB,

A B C sin sin 2 2 2

1 1 1 = 4R       2 2 2



3 sin 76º sin 16º + cos 76º cos16º cos 76º sin 16º + sin 76º cos16º

2 sin 76º sin 16º +

1 1 ⇒ sin 2α = sin 2α 2 ⇒ 2α = 30º ⇒ α = 15º. i.e., 2 =



=

=

 1 + tan 2 α  4AM = 2 AM   2 tan α 



cos 76º cos16º sin 76º sin 16º = 3 + cos 76º cos16º + sin 76º sin 16º

=

i.e., BC = AM  1 + tan α   tan α 

94. (a) r = 4R sin

⇒ b2 + c2 – a2 = b2 ⇒ c2 – a2 = 0 ⇒ c = a. 91. (d)

(MB + MC) = AM [cot α + tan α]

i.e.,

A + B + C = 90º, A – B = 30º and A + C = 60º ⇒ B = 30º, A = 60º, C = 0º

⇒ 2



a 2 + b2 ab c2 . ab

abc . 4∆



b2 + c2 − a 2 λ − 2 λ−2 = ⇒ = cos A 2bc 2 2 But – 1 < cos A < 1 λ−2 ∴ – 1 < < 1 ⇒ 0 < λ < 4. 2

⇒ 2bc (1 – cos A) = (b – c)2 tan2θ

2

2

or

98. (c) cos2B + cos2C = cos2B + cos2 = cos2B + sin2B = 1. 99. (d) Let sides of the given triangle be a – d, a, a + d Now (a + d)2 = (a – d)2 + a2

= (b2 + c2 – 2bc) (1 + tan2θ)

⇒ 2bc 2sin2

A = (b – c)2 tan2θ 2

A = (b – c)2 tan2θ 2 A ⇒ (b – c) tan θ = 2 bc sin  2 2 bc A ⇒ × sin = tan θ. b−c 2

⇒ 2bc 2 sin2

103. (c) By sine formula, In ∆ABC, we have BD AD = sin ∠BAD sin π 3 By sine formula, In ∆ADC, we have



...(i)

DC AD = sin ∠DAC sin π 4 Divide (i) by (ii), we get

...(ii)

4ad = a2 or a = 4d a − d 4d − d 3 Now tan A = = = a 4d 4 or



tan C =

a 4d 4 = = . a − d 4d − d 3

100. (b) We have, b2 − c2 c2 − a2 a 2 − b2 ⋅ cos A + ⋅ cos B + ⋅ cos C a b c b2 − c2 b2 + c2 − a 2 c2 − a 2 c2 + a 2 − b2 = ⋅ + ⋅ 2bc 2ac a b +  =

a 2 − b2 a 2 + b2 − c2 ⋅ 2ab c

1 [b4 – c4 – a2 b2 + a2 c2 + c4 – a4 – b2 c2 2abc + a2 b2 + a4 – b4 – a2 c2 + b2 c2] = 0.

101. (b) Since r1, r2, r3 are in H.P. ∴ ∴

1 1 1 , , are in A.P. r1 r2 r3 1 1 1 1 ∆ − = − where r1 = etc. r2 r1 r3 r2 s−a

s−b s−a s−c s−b = − − ∆ ∆ ∆ ∆ ⇒ a – b = b – c ⇒ 2b = a + c ⇒ a, b, c are in A.P. ∴

102. (c) We have, a = (b – c) sec θ

⇒ a2 = (b – c)2 sec2θ = (b2 + c2 – 2bc) sec2θ



π BD sin ∠DAC sin 4 = ⋅ = DC sin ∠BAD sin π 3

∴

sin ∠BAD 6 BD = = ⋅ sin ∠DAC 2 DC

1 2 = 2 3 6 2 6 1 1 ⋅ = . 2 3 6

104. (c) Since a = 13, b = 12, c = 5 ∴ ∆ABC is right angled traingle 1 (5) (12) = 30 cm2 2 Hence required distance Area =



=

2Area 2 × 30 60 cm. = = Base 13 13

105. (d) 2s = 13 + 14 + 15 = 42 ∴ s = 21

∆ =

s ( s − a )( s − b)( s − c)

=

21(21 − 13)(21 − 14)(21 − 15)

=

21 × 8 × 7 × 6 = 21 × 4 × 84

∴ r =

∆ 84 = 4. = s 21

887

⇒ b2 + c2 – 2bc cos A

or b + c – a = (λ – 2) bc 2

Properties and Solutions of Triangles

97. (d) Given equation becomes (b + c)2 – a3 = λ bc

888

106. (d) Since R =

Objective Mathematics

and ∆ =

abc 4∆

111. (a) We have, 2 cos A + cos B + 2 cos C = a + b a b c bc ca

1 (3) (4) = 6, (since ∆ is right angled) 2

3× 4×5 4×6 ∴ R = 2.5. ∴ R =

or 

2 (b 2 + c 2 − a 2 ) c 2 + a 2 − b 2 2 (a 2 + b 2 − c 2 ) + + 2bca 2bac 2abc

a 2 + b2 abc ⇒ 2b2 + 2c2 – 2a2 + c2 + a2 – b2 + 2a2 + 2b2 – 2c2 = 2a2 + 2b2 or 3b2 + c2 + a2 = 2a2 + 2b2  or  b2 + c2 = a2 Hence ∠A = 90º. =

107. (d) We have, 3 sin x – 4 sin2x – k = 0 ⇒ sin 3x = k [0 < k < 1] Since A, B both satisfy this equation ∴ sin 3A = sin 3B π 3 π 2π ⇒ C = 180º – (A + B) = π – = . 3 3

⇒ 3A = π – 3B ⇒ A + B =

112. (b) In ∆ABC, AB = BC and altitude AD = h. Let r be the radius of circumscribed circle. ∴ PA = PB = PC = r ∴ PD = h – r.

108. (c) r1, r2, r3 are in H.P.

s −a s −b s −c are in A.P. , , S S S or s – a, s – b, s – c are in A.P. or a, b, c are in A.P.

or 

Now a + b + c = 24 ⇒ b = 8, s = 12, c = 4 – a ∴ 12 (12 – a) (12 – 8) (12 – c) = 242 = 576 or a2 – 16a + 60 = 0 or a = 10, 6

Now BD =

If a = 10, b = 6 and if a = 6, b = 10. ∴ Sides of the triangle are 6, 8 and 10 cm.



109. (c) s = a + b + c = 18 + 24 + 30 = 72 = 36 2 2 2 S r= s S=

s ( s − a )( s − b)( s − c)

S=

36 (36 − 18)(36 − 24)(36 − 30)

= =

36 × 18 × 12 × 6

So radius of the incircle r=

S 6×6×3× 2 = 6 cms. = s 36

C 1 6−3 C  A−B a −b ∴ tan   = a + b cot 2 or 3 = 6 − 3 cot 2 2   or cot

C =1 2

or C =

Hence area of ∆ABC =

π 2

1 × 6 × 3 = 9 sq. units. 2

=

2rh − h 2

∴ BC = 2BD = 2 2rh − h 2 Thus area of ∆ABC = 

=

1 × DC × DA 2

2 1 × 2 2rh − h 2 × h = h 2rh − h . 2

113. (a), (c)  Since the angles A, B, C of ∆ABC are in A.P., thus ∠B = 60º

∴

a 2 + c2 − b2 2ac

1 102 + c 2 − 92 = 2 20c

⇒c =

 A−B 4 1 ⇒ tan  110. (b) cos (A – B) =  =  2  5 3

r 2 − (h − r )2

=

cos B =

6 × 6 × 6 × 3 × 4 × 3 × 6 = 6 × 6 × 3 × 2.

BP 2 − PD 2

⇒ c2 – 10c + 19 = 0

10 ± 100 − 76 =5± 2

6

so (A) and (C) are the correct solutions. 114. (a) The hypotenuse of the right triangle ABC is AB. We take C at the origin and CB along x-axis and a b CA along y-axis. The mid-point M  2 , 2  of AB   is the circum-centre of the triangle. Therefore, R2 = MC2 =

1 1 2 C (a2 + b2) = c ⇒R= 4 4 2

C π = (s – c) tan =s–c 2 4

 sin A sin B sin C  ∵ a = b = c = k   

889

Next, r = (s – c) tan

Properties and Solutions of Triangles

B+C B−C 2 3 cos = 2 2 2 B − C ⇒ 2 sin 60º cos = 3 2 ⇒ cos B − C = 1 ⇒ ∠B = ∠C 2 ∴ ∠A = ∠B = ∠C ⇒ ∆ABC is an equilateral. ⇒ 2sin

119. (c) Let AB = AC and ∠A = 120º

Thus, 2 (r + R) = 2r + 2R = 2s – 2c + c

A

= a + b + c – 2c + c = a + b.

60 º 60º



115. (c) Since the largest angle is opposite the largest side, we have

cos C =

a

2π 3 +5 −7 15 1 . =− =− ⇒ C = 3 2 (3)(5) 30 2 2

2

2

B

116. (b) Since the angles A, B, C are in A · P ∴ 2B = A + C ∴ A + B + C = B + 2B = 180º ⇒ B = 60º ⇒ A + C = 120º Also, sin B : sin C :: 3 : 2

Since ∠A + ∠B + ∠C = 180º ∴ ∠B + ∠C = 120º Also, b + c = 2a (given) ⇒ sin B + sin C = 2 sin 60º

 1 + tan 45 tan 30  3  o o   tan 45 − tan 30 

=1+

 3 +1 3    3 −1

∴ a=4+ 2 3

 3 1 (4 + 2 3) 2   = 12 + 7 2 3  2  120. (b) Given, (sin A + sin B + sin C) × (sin A + sin B – sin C) = 3 sin A sin B

2 2 2 ⇒ a + b − c = 1 2ab 2

⇒ cos C = 1 2

121. (a) ∵ ∠A = π 2 a and = 2R  sin A ∆ bc Also, r = = s a+b+c ∴

[from Eq. (i)]

Hence, tan A , tan B , tan C are in HP. 2 2 2 118. (a) Given: ∠A = 60º

=1+

3 cot 15º

o

⇒ (a + b + c) (a + b – c) = 3ab ⇒ a2 + b2 – c2 = ab

s ( s − a ) ( s − b) ( s − c )

= –k{a + c – 2b} = 0

3 tan 30º +

∴ Area of ∆ =

…(i)

= k{s – a + s – c – 2(s – b)}



C

o

s( s − a) s ( s − c) s ( s − b) + −2 ( s − b) ( s − c ) ( s − a ) ( s − b) ( s − a) ( s − c)

where k =

O

15º 15º

Here a = AD + BD =

117. (c) Since, a, b and c are in AP.

=

√3

∴ Area of ∆ = 1 a 2 sin 120º 2

1 ⇒ sin C = sin 60° × 2 = 3 2 ⇒ C = 45º ∴ A = 180º – 60º – 45º = 75º ∴ 2b = a + c 1 1 2 Now, + − A C B tan tan tan 2 2 2 A C B = cot + cot − 2cot 2 2 2

a

D

⇒ C = π 3

⇒ a = 2R and a = 5

R a(a + b + c) 5 × 12 5 = = = r 2bc 2× 4×3 2

122. (a) cot A + cot B =

cos A sin B + cos B sin A sin A sin B

=

sin ( A + B) sin A sin B

=

=

1 sin A sin B

[∵ C = 90º]

sin C sin A sin B …(i)

890

126. (c) Since, angles of a triangle are in the ratio 4 : 1 : 1, ∴assume that angles are 4x, x and x.

Applying sine rule,

Objective Mathematics

sin A sin B sin C = = a b c sin A sin B 1 ⇒ = = a b c a ⇒ sin A = , sin B = b c c

[∵ C = 90º]

1

∴ From Eq. (i), cot A + cotB =

=

c2 ab

a b ⋅ c c 123. (a) Let a = 7 cm, b = 4 3 cm and c = 13 Since, the smallest side is c, therefore the smallest angle will be C. ∴ cos C = =

a 2 + b 2 − c 2 (7) 2 + (4 3) 2 − ( 13) 2 = 2ab 2×7× 4 3

49 + 48 − 13 3 = 2 56 3

A B − tan 2 2 124. (b) A B tan + tan 2 2 tan



=



=

⇒ C = π. 6

( s − b) ( s − c ) ( s − a ) ( s − c) − s( s − a) s ( s − b) ( s − b) ( s − c ) ( s − a ) ( s − c) + s( s − a) s ( s − b)

( s − b) s ( s − c ) − ( s − a ) s ( s − c ) ( s − b) s ( s − c ) + ( s − a ) s ( s − c ) s ( s − c)( s − b − s + a ) s ( s − c)( s − b + s − a )

=

a−b c

125. (c) Since, sin2 A , sin2 B , sin2 C are in HP. 2 2 2 1

1 1 , , A C are in AP. 2 B sin sin sin 2 2 2 2 1 1 1 1 − = − ⇒ A 2 C 2 B 2 B sin sin sin sin 2 2 2 2 2 ab ac − ⇒⇒ ( s − a ) ( s − b) ( s − a ) ( s − c ) ∴



2

=

ac ac − ( s − a ) ( s − c) ( s − a ) ( s − c)

 a   b( s − c ) − c ( s − b)  ⇒     s − a   ( s − b) ( s − c ) 

 c   a ( s − b) − b( s − a )      s − c   ( s − a ) ( s − b) 

⇒ abs – abc – acs + abc = acs – abc – bcs + abc ⇒ ab – ac = ac – bc ⇒ ab + bc = 2ac ⇒ 1 + 1 = 2 ⇒ a, b, c are in HP. c a b

We know that, in ∆ABC ∠A + ∠B + ∠C = 180º o ⇒ x = 180 = 30º 6 ∴ Angles are 120º, 30º and 30º ∴ Ratio of sides = sin A : sin B : sin C = sin 120º : sin 30º : sin 30º

⇒ 4x + x + x = 180º

=



3 1 1 : : = 3 :1:1 2 2 2

Thus, the required ratio = 127. (b) Since,

3 3 . = 1+1+ 3 2 + 3

1 ( s − a ) 2 1 ( s − b) 2 , = , 2 = r12 ∆2 r2 ∆2

1 ( s − c) 2 1 s2 and = = r32 ∆2 r 2 ∆2 ∴

1 1 1 1 + + + r12 r2 2 r32 r 2

s 2 + ( s − a ) 2 + ( s − b) 2 + ( s − c ) 2 ∆2 2 2 2 4s + a + b + c 2 − 2s (a + b + c) = ∆2 3 128. (d) sin 3B = 3sin B − 4sin B sin B sin B =

=

a 2 + b2 + c2 . ∆2

= 3 – 4 sin2 B = 3 – 4(1 – cos2 B) 2

 a2 + c2  2   2  ( a 2 + c 2 ) 2 − 4a 2 c 2  c 2 − a 2  = = = −1 +    (ac) 2 4(ac) 2  2ac  129. (b) Since, ( 3 – 1)a = 2b ⇒

a 2 = b 3 −1

and A = 3B We know that sin A = sin B a b ⇒

sin 3B = sin B

2 3sin B − 4sin 3 B ⇒ = sin B 3 −1

2 3 −1

⇒ 3 – 4 sin2B = 2( 3 + 1) 2 ⇒ 3 – 4 sin2B =

3 + 1

3 – 1 = 4 sin2B

⇒3– 2

⇒ 2 − 3 = sin2B 4

 3 −1   = sin2B ⇒   2 2 

3 −1 ⇒ B = 15º and A = 45º 2 2 ∴ C = 180º – A – B = 120º ⇒ sin B =

(b) 5 − 6

(a) 5 ± 6 (c) 3 3

(d) 5

2. Which of the following pieces of data does not uniquely determine an acute angled ∆ABC(R = circumradius)? (a) a, sin A, sin B (c) a, sin B, R

(b) a, b, c (d) a, sin A, R

3. Suppose the angles of a triangle ABC are in A.P. and sides b and c satisfy b : c = 3 : 2 , then the angle A is (a) 90º (c) 60°

(b) 75º (d) 45°

(a) b2 (b) b2 (c) b2 (d) b2

= = = =

c2 c2 c2 c2

+ + + +

a2 a2 a2 a2

– – + +

2 ca sin B 2 ca cos B 2ca cos B 2 ca tan B.

π 2

(b) b sin A > a, A >

π 2

π 2 π , b > a (d) b sin A < a, A < 2 π , b = a. (e) b sin A < a, A > 2 (c) b sin A > a, A
2

π π P ( B )

(c) P(B) < P ( A)

(d) None of these

4. The letters of word ‘SOCIETY’ are placed at random in a row. The Probability that three vowels come together, is

6 7 3 (c) 7

(a)

(b)

1 7

(d) None of these

5. The letters of the word ‘ARTICLE’ are arranged at random. The Probability that the vowels may occupy the even places, is 1 35 7 (c) 35

(a)

(b)

4 35

(d) None of these

6. Three distinct numbers are selected from first 100 natural numbers. The probability that all the three numbers are divisible by 2 and 3 is

4 25 4 (c) 55 (a)

4 35 4 (d) 1155 (b)

1 3 7. Given that P(B) = , P(A ∩ B ∩ C ) = , P A ∩ B ∩ C 3 4 1 = , then the value of P(B ∩ C) is: 3

(

(a) 1/15 (c) 1/9

(b) 1/6 (d) 1/12

)

908

Objective Mathematics

8. Two numbers are chosen, one by one (without replacement) from the set of numbers A = {1, 2, 3, 4, 5, 6}. Then the probability that minimum value of two numbers chosen is less than 4, is: (a) 14/15 (c) 1/5

(b) 1/15 (d) 4/5

9. The Probability that at least one of the events A and B occurs is 0.7 and they occur simultaneously with Probability 0.2. Then P (A) + P (B) = (a) 1.8 (c) 1.1

(b) 0.6 (d) 1.4

10. A coin and a dice is thrown. If A denotes the event head and even face and B denotes the event tail and multiple of 3, then 1 4 1 (c) P (A) = 3 (a) P (A) =

1 5 1 (d) P (B) = 6 (b) P (B) =

11. A number is chosen form each of two sets A = {1, 2, 3, 4, 5, 6, 7, 8} and B = {8, 7, 6, 5, 4, 3, 2, 1}. If p1 is the probability that the sum of the numbers is 9 and p2 is the probability that sum of the numbers is 7, then p1 + p2 is equal to 5 52 9 (c) 32

(a)

3 16 7 (d) 32 (b)

12. The probability of India winning a test match against 1 West-Indies is . Assuming independence from match 2 to match the probability that in a match series India’s second win occurs at the third test is (a)

1 8

(c)

1 2

1 4 2 (d) 3

(b)

13. The Probability of getting the product a perfect square (square of a natural number), when two dice are thrown together, is 2 9 1 (c) 3

(a)

(b)

4 9

(d) none of these

15. A room has 3 lamps. From a collection of 10 light bulbs of which 6 are not good, a person selects 3 at random and puts them in a socket, the Probability that he will have light, is 5 1 (b) 6 6 1 (c) (d) None of these 2 16. A die is tossed thrice. If event of getting an even number is a success, then the probability of getting atleast two sucesses is (a)

7 1 (b) 8 4 2 1 (c) (d) 3 2 17. If x = 33n , n is a positive integral value, then what is the probability that x will have 3 at its unit’s place? (a)

(a) 1/3 (c) 1/5

(b) 1/4 (d) 1/2

18. One of the two events must happen. Given that the chance of one is two-third of the other, the odds in favour of the other are (a) 3 : 5 (c) 3 : 2

(b) 2 : 5 (d) None of these

19. A, B, C are three mutually exclusive and exhaustive events associated with a random experiment. If P (B) 3 1 P (A) and P (C) = P (B), then P (A) = = 2 2 4 13 8 (c) 13 (a)

(b)

6 13

(d) none of these.

20. 100 Students appeared for two examinations, 60 passed the first, 50 passed the second and 30 passed both. The Probability that a student selected at random has failed in both examinations, is (a) 0.4 (c) 0.3

(b) 0.2 (d) None of these

21. A biased coin with probability p, 0 < p < 1 of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even 2 , then p equals is 5 1 2 (b) (a) 3 3 2 3 (c) (d) 5 5

14. There are 100 students in a collage class of which 36 are boys studying statistics and 13 girls not studying statistics. If there are 55 girls in all, the Probability 1 5 1 22. If P (A ∩ B) = , P(A ∪ B) = and P(A) = then that a boy picked up at random is not studying, is 3 6 2 which one of the following is correct? 3 2 (b) (a) (a) A and B are independent events 5 5 (b) A and B are mutually exclusive events 1 (c) P (A) = P (B) (c) (d) None of these 5 (d) P (A) < P (B)

9 13 5 (c) 13 (a)

(b)

7 13

(d) None of these

24. A police-man fires six bullets on a dacoit. The Probability that the dacoit will be killed by one bullet is 0.6. The Probability that dacoit is still alive is (a) 0.04096 (c) 0.4096

(b) 0.004096 (d) None of these

25. A class consists of 100 students, 25 of them are girls and 75 boys, 20 of them are rich and remaining poor, 40 of them are fair complexioned. The Probability of selecting a fair complexioned rich girl is (a) 0.02 (c) 0.05

(b) 0.04 (d) 0.08

(a)

4 9

(b)

2 9

(c)

2 9

(d) None of these

32. The Probability that a leap year selected at random contains 53 Sundays is (a)

7 366

(b)

26 183

(c)

1 7

(d)

2 7

33. If P (A) = 0.65, P (B) = 0.15, P (A) + P (B) =

(a) 1.5 (b) 1.2 (c) .8 (d) None of these 26. A speaks truth in 60% of the cases and B in 90% of the cases. The percentage of cases they are likely to 34. The Probability that on event A happens in a trial is 0.4. contradict each other in stating the same fact is These independent trials are made. The Probability that A happens at least once is (a) 36% (b) 48% (c) 42% (d) None of these (a) 0.936 (b) 0.216 (c) 0.784 (d) 0.064 27. Two dice are thrown. The Probability that the number appeared have a sum 8 if it is known that the second 35. In a class of 125 students 70 passed in Mathematics, dice always exhibits 4, is 55 in Statistics and 30 in both. The Probability that a student selected at random from the class, has passed 5 1 (b) (a) in only one subject is 6 6 13 3 2 (b) (a) (c) (d) None of these 25 25 3 28. A bag contains 6 red and 3 white balls. Four balls are drawn one by one and not replaced. The Probablity that they are alternatively of different colours is 4 (a) 42 7 (c) 42

5 (b) 42

(c)

34 90 53 (c) 90

4

2 (a)   7

(d) None of these

(b)

61 90

(d) None of these

30. The odds that A speaks the truth are 3 : 2 and the odds that B speaks the truth are 5 : 3. The percentage of cases they are likely to contradict each other on an identical point is (a) 47.5% (c) 42.5%

(b) 32.5% (d) None of these

(d)

8 . 25

36. Seven chits are numbered 1 to 7. Four are drawn one by one with replacements. The Probability that the least number on any selected chit is 5, is

29. A student takes his examination in four subjects α, β, 4 , γ, δ. He estimates his chance of passing in α as 5 5 2 3 , in γ as and in δ as . To qualify he 37. in β as 6 3 4 must pass in α and at least two other subjects. The Probability that he qualifies is (a)

17 25

4

2 (b) 4 ⋅   7

3

3 (c)   7

(d) None of these

A box contains 5 brown and 4 white socks. A man takes out two socks. The probability that they are of the same colour is 5 108 5 (c) 18 (a)

1 6 4 (d) 9

(b)

38. Three identical dice are rolled. The probability that the same number will appear on each of them is

1 6 1 (c) 36 (a)

(b)

1 18

(d) None of these

909

31. A bag contains 3 red and 7 black balls. Two balls are selected at random without replacement. If the second selected ball is given to be red, the Probability that the first selected ball is also red, is

Probability

23. A card is drawn from a pack of 52 cards. The Probability of getting a king or a heart or a red card is

910

39. For any two independent events E1 and E2, P {(E1 ∪ E2) ∩ ( E1 ) ∩ ( E 2 )} is

Objective Mathematics

1 4 1 (c) ≥ 2

(a)


1 4

(d) None of these

40. If the Probabilities that A and B will die within a year are p and q respectively then the Probability that only one of them will be alive at the end of the year is (a) p + q (c) p + q – pq

(b) p + q – 2pq (d) p + q + pq.

41. The Probability of solving a problem by three students 1 1 1 respectively. The probability that A, B, C is , , 2 3 4 the problem will be solved is 1 4 3 (c) 4 (a)

1 2 1 (d) 3 (b)

42. In order to get atleast once a head with probability ≥ 0.9, the number of times a coin needs to be tossed is

(a) 0.2 ≤ P (B ∩ C) ≤ 0.35 (b) 0.5 ≤ P (B ∩ C) ≤ 0. 85 (c) 0.1 ≤ P (B ∩ C) ≤ 0.35 (d) None of these 47. A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that value of the determinant chosen is positive is 5 16 3 (c) 16 (a)

(b)

7 16

(d) None of these

48. Three faces of a fair dice are yellow, two faces red and one blue. The dice is tossed three times. The probability that the colours, yellow, red and blue appear in the first, second and third tosses respectively is

5 36 1 (c) 36 (a)

(b)

7 36

(d) None of these

49. An unbiased dice with faces 1, 2, 3, 4, 5 and 6 is round 4 times. Out of four face values obtained the probability that the minimum face value obtained the probability 43. A man alternately tosses a coin and throws a dice beginthat the minimum face value is not less than 2 and the ning with the coin. The probability that he gets a head maximum face value is not greater than 5 is before he gets 5 or 6 in the dice is 16 1 (b) (a) 3 1 81 81 (b) (a) 4 2 80 65 (c) (d) 1 81 81 (c) (d) None of these 3 50. A pack of cards contains 4 aces, 4 kings, 4 queens and 44. You are given a box with 20 cards in it. 10 of these 4 jacks. Two cards are drawn at random. The probability cards have the letter I printed on them. The other that atleast one of them is an ace is ten have the letter T printed on them. If you pick up 1 3 3 cards at random and keep them in the same order, (b) (a) 5 16 the probability of making the word IIT is 9 1 9 1 (c) (d) (b) (a) 20 9 80 8 (a) 3 (c) 5

(c)

4 27

(b) 4 (d) None of these

(d)

5 38

45. The probability that Krishna will be alive 10 year hence 7 7 and Hari will be alive is . The probability is 15 10 that both Krishna and Hari will be dead 10 years hence is 21 150 49 (c) 150

(a)

24 150 56 (d) 150 (b)

46. The probabilites of three events A, B and C are 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 P (A ∪ B ∪ C) ≥ 0.85, then

51. A man is known to speak truth 3 out of 4 times. He throws a dice and reports that it is six. The probability that it is actually six is 3 8 3 (c) 5

(a)

(b)

1 5

(d) None of these

52. A bag contains 10 mangoes out of which 4 are rotten, two mangoes are taken out together. If one of them is found to be good, the probability that other is also good is 1 3 5 (c) 18 (a)

8 15 2 (d) 3

(b)

(a) 0.8750 (c) 0.0624

(b) 0.0875 (d) 0.0250

54. Ram and Shyam throw with one dice for a prize of Rs. 88 which is to be won by the player who throws 1 first. If Ram starts, then mathematical expectation for Shyam is (a) Rs. 32 (c) Rs. 48

(b) Rs. 40 (d) None of these

55. Out of 13 applicants for a job, there are 5 women and 8 men. It is desired to select 2 persons for the job. The probability that atleast one of the seleced persons will be a woman is 25 39 5 (c) 13 (a)

14 39 10 (d) 13

(b)

56. Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with three vertices is equilateral equals 1 2 1 (c) 10

(a)

1 5 1 (d) 20

(b)

57. Two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Then the probability that neither A nor B occurs is (a) 0.28 (c) 0.61

(b) 0.39 (d) 0.72

58. In a certain town, 40% of the people have brown hair, 25% have brown eyes and 15% have both brown hair and brown eyes. If a person selected at random from the town, has brown hair, the probability that he also has brown eyes is 3 (b) 8 2 (d) 3

1 (a) 5 1 (c) 3

59. A fair die is tossed eight times. The probability that a third six is observed on the eight throw is C2 × 55 67 7 5 (c) C2 × 5 (a)

7

66

(b)

7

C2 × 55 68

(d) None of these

60. An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses is needed is

2 5 2 (d) 3 (b)

61. The mean and the variance of binomial distribution are 4 and 2, respectively. Then the probability of 2 sucesses is (a) 128/256 (c) 7/64

(b) 219/256 (d) 28/256

62. Three rifle-men take one shot each at the same target. The probability of the first rifle-man hitting the target is 0.4, the probability of the second rifle-man hitting the target is 0.5 and the probability of the third rifle-man hitting the target is 0.8. The Probability that exactly two of them hit the target, is (a) 0.32 (c) 0.54

(b) 0.44 (d) None of these

63. If E and F are the complementary events of events E and F respectively and if 0 < P (F) < 1, then (a) P (E | F) + P ( E  | F) = 1 (b) P (E | F) + P (E | F  ) = 1 (c) P ( E  | F) + P (E | F  ) = 1 (d) P (E |  F  ) + P ( E  | F ) = 1 64. 6 boys and 6 girls sit in a row randomly. The probability that all 6 girls sit together, is 1 64 1 (c) 132 (a)

(b)

1 8

(d) None of these

65. A random varible X has the probability distribution X 1 2 3 4 5 6 7 8 P(X) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05 For the events E = ( X is a prime number ) and F = (X < 4), the probability P (E ∪ F) is (a) 0.35 (c) 0.87

(b) 0.77 (d) 0.50

66. Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is 4 5 1 (c) 5 (a)

3 5 2 (d) 5 (b)

67. There are 4 envelopes corresponding to 4 letters. If the letters are placed in the envelopes at random. The probability that all the letters are not placed in the right envelopes, is 18 24 17 (c) 24 (a)

(b)

23 24

(d) None of these

911

1 2 1 (c) 5 (a)

Probability

53. India plays two matches each with West Indies and Australia. In any match the probabilites of India getting points 0, 1, 2 are 0.45, 0.05 and 0.50 and respectively. Assuming that outcomes are independent, the probability of India getting at least 7 points is

912

Objective Mathematics

68. The odds in favour of standing first of three student 75. appearing in an enamination are 1 : 2, 2 : 5 and 1 : 7 respectively. The probability that either of them will stand first, is 125 168 32 (c) 168 (a)

75 168 4 (d) 168 (b)

69. A, B, C are three events such that P (A) = 0.3, P (B) = 0.4, P (C) = 0.8, P (AB) = 0.08, P (AC) = 0.28, P (ABC) = 0.09. If P (A ∪ B ∪ C) ≥ 0.75, then (a) 0.23 ≤ P (BC) ≤ 0.32 (b) 0.41 ≤ P (BC) ≤ 0.52 (c) 0.23 ≤ P (BC) ≤ 0.48 (d) None of these 70. The probability that certain electronic conponent fails when first used in 0.10. If it does not fail immediately, the probability that it lasts for one year is 0.99. The probability that a new component will last for one year is (a) 0.891 (c) 0.92

(b) 0.692 (d) None of these

71. Three groups A, B, C are contesting for position on the Board of Directors of a company. The probabilities of their winning are 0.5, 0.3, 0.2 respectively. If the group A wins, the probability of introducing a new product is 0.7 and the corresponding probabilities for group B and C are 0.6 and 0.5 respectively. The probability that the new product will be introduced, is (a) 0.52 (c) 0.74

(b) 0.63 (d) None of these

95 94 , P (E2/E1) = , 100 99 93 92 , P (E4/E1E2E3) = and P (E5/E1E2E3E4) P (E3/E1E2) = 98 97 91 , then P (E) = = 96

72. If E = E1 E2 E3 E 4 E5 and P (E1) =

(a)

91 ⋅ 92 ⋅ 93 ⋅ 94 97 ⋅ 98 ⋅ 99 ⋅ 100

(b)

91 ⋅ 92 ⋅ 93 ⋅ 94 ⋅ 95 96 ⋅ 97 ⋅ 98 ⋅ 99 ⋅ 100

(b)

4 7

Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose that each of them independently and with equal probability, can leave the cabin at any floor beginning with the first. The probability of all five persons leaving at different floors, is (a)

8 p5



(b)

(c)

7 p5



(d) None of these

74

75

9 p5 76

76. The probability that A speaks truth is 4/5, while this probability for B is 3/4. The probability that they contradict each other when asked to speak on a fact is (a) 7/20 (c) 3/20

(b) 1/5 (d) 4/5

77. A student is given a true-false exam with 10 questions. If he gets 8 or more correct answers he passes the exam. Given that he guesses at the answer to each question, the probability that he passes the exam, is 6 128 7 (c) 128 (a)

(b)

9 128

(d) None of these

78. 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 all the correct answers. The candidate decides to tick answers at random. If he is allowed upto three chanes to answer the question, the probability that he will get marks in the question, is 1 2 1 (c) 3

(b)

(a)

1 5

(d) None of these

79. An article manufactured by a company consists of two parts X and Y. In the process of manufacture of part X, 9 out of 104 parts may be defective. Similarly, 5 out 94 ⋅ 95 ⋅ 96 (c) (d) None of these of 100 are likely to be defective in the manufacture of 98 ⋅ 99 ⋅100 the part Y. The probability that the assembled product 73. If a coin is tossed n times, the probability that head will will not be defective, is appear on odd number of times, is 253 361 (b) (a) 1 1 416 416 (b) (a) 3 2 322 (c) (d) None of these 2 416 (c) (d) None of these 3 80. The chance of an event happening is the square of the 74. A die is loaded so that the probability of face i is prochance of a second event but the odds against the portional to i, i = 1, 2,...,6. The probability of an even first are the cubes of the odds against the second. The number occuring when the die is rolled, is chances of happening of each event are 2 7 3 (c) 7

(a)

(d) None of these

1 , 9 1 (c) , 3 (a)

1 3 1 6

(b)

1 1 , 6 9

(d) None of these

m −1 2m − 1 (c) m/2m

(a)

m 2m − 1 (d) None of these

(b)

82. Four tickets marked 00, 01, 10 and 11 respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on the ticket is 15, is 3 1024 7 (c) 1024

(a)

(b)

5 1024

(d) None of these

83. If X and Y are the independent random variables  1  1 B  5,  and B  7,  , then P (X + Y ≥ 1) =  2  2 4095 (a) 4096 4032 (c) 4096

309 (b) 4096 (d) None of these

84. A fair coin is tossed 99 times. If X is the number of times heads occurs P (X = r) is maximum when r, is

(b)

9 16

(d) None of these

89. If A and B are such events that P (A) > 0 and P (B) ≠ 1, then P (A/B) is equal to (a) 1 – P (A/B) (c)

1 − P (A ∪ B) P (B)

(b) 1 – ( P (A/B) (d)

P (A) . P (B)

90. An ordinary cube has four blank faces, one face marked 2 another marked 3. Then the probability of obtaining a total of exactly 12 in 5 throws is 5 1296 5 (c) 2592 (a)

(b)

5 1944

(d) None of these

91. Probability that in the toss of two dice we obtain an even sum or a sum less than 5, is 1 2 2 (c) 3 (a)

1 6 5 (d) 9 (b)

92. A person draws a card from a pack of playing cards, replaces it and shuffles the pack. He continues doing this until he shows a spade. The chance that he will fail the first two times is 85. Suppose X follows a binomial distribution with parameters n and p, where 0 < p < 1. If P (X = r)/P (X = n – r) is 9 1 (b) (a) independent of n and r then p is equal to 64 64 1 1 1 9 (b) (a) (c) (d) 3 2 16 16 1 93. Three letters are written to different persons, and ad(c) (d) None of these 4 dresses on three envelopes are also written. Without looking at the addresses, the probability that the letters 86. Numbers are selected at random one at a time, from the go into right envelopes is numbers 00, 01, 02,...,99 with replacement. An event E occurs if and only if the product of the two digits of 1 1 (b) (a) a selected number is 18. If four numbers are selected, 27 6 then the probability that E occurs at least 3 times, is 1 (c) (d) None of these 97 68 9 (b) (a) 390625 390625 94. Probbility is 0.45 that a dealer will sell at least 20 televi72 sion sets during a day, and the probability is 0.74 that (d) None of these (c) 390625 he will sell less than 24 televisions. The probability (a) 49 (c) 51

(b) 50 (d) None of these

87. A man takes a step forward with probability 0.4 and backward with probability 0.6. The probability that at the end of eleven steps he is one step away from the starting point, is (a) 0.37 (c) 0.54

(b) 0.32 (d) None of these

88. The mean and variance of a binomial variable X are 2 and 1 respectively. The probability that X takes values greater than 1, is

that he will sell 20, 21, 22, or 23 televisions during the day, is (a) 0.19 (b) 0.32 (c) 0.21 (d) None of these

95. A, B and C are three mutually exclusive and exhaustive 1 1 P (C) = P (A) = P (B), then P (B) = events. If 3 2

913

5 16 11 (c) 16 (a)

Probability

81. In the game odd man out, each of m ≥ 2 person tosses a coin to determine who will buy refreshments for the entire group. The odd man out is the one with a different outcome from the rest. The probability that there is a loser in any game is

914

(a)

1 3

(b)

Objective Mathematics

1 5

(c)

1 6

(d) None of these

96. The probability that a leap year selected at random will contain either 53 Thursdays or 53 Fridays, is (a)

3 7

5 (c) 7

(b)

2 7

(d) None of these

97. An investment consultant predicts that the odds ageinst the price of a certain stock will go up during the next week are 2 : 1 and the odds in favour of the price remaining the same are 1 : 3. The probability that the price of the stock will go down during the next week, is (a)

4 12

(b)

5 12

(c)

7 12

(d) None of these

102. A is known to tell the truth in 5 cases out of 6 and he states that a white ball was drawn from a bag containing 8 black and 1 white ball. The probability that the white ball was drawn, is 7 13 9 (c) 13 (a)

(b)

5 13

(d) None of these

103. The letters of the word ‘longest’ are arranged at random. The probability that the vowels may occupy only odd positions, is 2 7 1 (c) 7 (a)

(b)

5 7

(d) None of these

104. n persons are seated on n chairs at a round table. The probability that two specified persons are sitting next to each other, is (a)

1 n −1

(b)

2 n −1

98. From a sales force of 150 persons, one will be selected 5 to attend a special sales meeting. If 52 of them are (c) (d) None of these n −1 3 of the unmarried, 72 are college graduates, and 105. An article manufactured by a company consits of two 4 52 that are un married are college graduates, the probparts A and B. In the processs of manufacutre of part ability that a sales person selected at random will be A, 9 out of 100 are likely to be defective. Similarly neither single nor a college graduate, is 5 out of 100 are likely to be defective in the manufacture of part B. The probability that the assembled part 13 14 (b) (a) will not be defective, is 30 30 (a) 0.8645 17 (c) (d) None of these (b) 0.6243 30 (c) 0.9645 99. An MBA applies for a job in two firms X and Y. The (d) None of these probability of his being selected in firim X is 0.7 and 106. The sum of two postive quantities is equal to 2n. The being rejected at Y is 0.5. The probability of at least 3 times probability that their product is not less than one of his applications being rejected is 0.6. The prob4 ability that he will be selected in one of the firms, is their greatest product is (a) 0.6 (c) 0.8

(b) 0.4 (d) None of these

3 4 1 (c) 4 (a)

(b)

1 2

100. It is 8 : 5 against a person who is now 40 years old (d) None of these living till he is 70 and 4 : 3 against a person now 50 living till he is 80. The probability that at least one of these persons will be alive 30 years hence, is 107. At the college entrance examination each candidate is admitted or rejected according to whether he has 63 48 (b) (a) passed or failed the tests. Of the candidates who are 91 91 really capable, 80% pass the tests and of the incapable, 59 25% pass the test. Given that 40% of the candidates are (c) (d) None of these 91 really capable, then the proportion of capable college students is about 101. A certain player, say X, is known to win with probability 0.3 if the track is fast and 0.4 if the track is slow. For Monday, there is a 0.7 probability of a fast track and 0.3 probability of a slow track. The probability that player X will win a Monday, is (a) 0.22 (c) 0.33

(b) 0.11 (d) None of these

(a) 68% (c) 73%

(b) 70% (d) 75%.

108. Four digit numbers are formed using each of the digits 1, 2,...,8 only once. One number from these is picked up at random. The probability that the selected number contains unity is

(b)

(c)

1 4

(d) None of these

(a) P (A ∩ B) ≤ P (A) + P (B) (b) P (A ∪ B) ≤ P (A) + P (B) (c) P (A ∪ B) = P (A) + P (B) (d) P (A ∪ B) ≥ P (A) + P (B).

109. A book contains 1,000 pages. A page is chosen at random. The probabilities that the sum of the digits 116. The probabilities of three events A, B and C are P (A) = 0.6, P (B) = 0.4, P (C) = 0.5, also P (A ∪ B) = 0.8, of the marked number on the page is equal to 9, P (A ∩ C) = 0.3, P (A ∪ B ∪ C) ≥ 0.85, P (A ∩ B is ∩ C) = 0.2 and P (B ∩ C) = p1. Then 23 11 (b) (a) (a) p1 ≥ 0.35, (b) p1 ≤ 0.2 500 200 (d) None of these (c) 0.2 ≤ p1 ≤ 0.35 7 (c) (d) None of these 117. If A and B are independent events such that 0 < P (A) 100 < 1 and 0 < P (B) < 1, then

110. A binary number is made up of 8 digits. Suppose that the probability of an incorrect digit appearing in p and that of errors in different digits are independent of each other. The the probability of forming an incorrect number, is (a)

p 8

118. If A and B are two independent events such that

(b) p8

(c) (1 – p)

P (A ∩ B) = 3/25 and P(A′ ∩ B) = 8/25, then P (A) is

(d) 1 – (1 – p)

8

8

111. The following table represents a probability distribution for a random variable X : X :

1

2

p (X = x) 0.1 2k Then, the value of k is (a) 0.1 (c) 0.3

3

4

5

6

k

0.2

3k

0.3

(b) 0.2 (d) 0.4

112. The probability that a man aged x years will die in a year is p. The probability that out of n men A1, A 2, A 3,...A n , each aged x, A1 will die and be first to die, is (a)

1 n2

(b) 1 – (1 – p)n

(c)

1  (1 – (1 – p)n) n2

(d)

1 (1 – (1 – p)n). n

113. You are given a box with 20 cards in it. 10 of these cards have letter I printed on them. The other ten have the letter T printed on them. If you pick up 3 cards at random and keep them in same order, the probability of making the word IIT, is (a)

9 80

(b)

1 8

(c)

4 27

(d)

5 38

(a) A and B and mutually exclusive (b) A and  B are independent (c) A and B are independent (d) P (A/B) + P (A/B) = 1.

114. For two events A and B (a) P (A ∩ B) / P (A) + P (B) (c) P (A ∩ B) = P (A) + P (B) – P (A ∪ B) (d) P (A ∩ Β) ≤ P (A) + P (B) . 2

(a) 11/25 (c) 3/11

(b) 7/25 (d) 9/11

119. If two events A and B are such that P (A′ ) = 0.3, P (B) = 0.4 and P (A ∩ B′ ) = 0.5, then P (B/A ∪ B′ ) equals (a) 3/4 (c) 1/4

(b) 5/6 (d) 3/7.

120. If (1 + 4p)/4, (1 – p)/2 and (1 – 2p)/2 are the probabilities of three mutually exclusive events, then value of p is (a) 1/2 (c) 1/4

(b) 1/3 (d) None of these

121. Let 0 < P(A) < 1, 0 < P (B) < 1 and P (A ∪ B) = P (A) + P (B) – P (A) P (B), then (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). 122. An elevator starts with m passengers and stops at n floors (m ≤ n). The probability that no two passengers alight at the same floor is (a) (c)

n

Pm mn

n

Cm mn

(b) (d)

n

Pm nm

n

Cm nm

123. If ten objects are distributed at random among then persons, the probability that at least one of them will not get any things is (a)

1010 − 10 1010

(b)

1010 − 10! 1010

(c)

1010 − 1 1010

(d) None of these

915

1 2

Probability

115. If A and B are arbitarary events then

1 8

(a)

916

124. Five coins whose faces are marked 2, 3 are thrown. The probability of obtaining a total of 12 is

Objective Mathematics

5 32 5 (c) 16

(a)

11 16 10 (d) 16

(b)

125. Four tickets marked 00, 01, 10, 11 respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on tickets thus drawn is 23, is 25 256 231 (c) 256

(a)

(b)

100 256

(d) None of these

126. For a B.D., the parameters n and p are 16 and respectively. The its S.D. σ is equal to

1 2

(a) 2 (b) 2 (c) 2 2 (d) 4. 127. n (≥ 3) persons are sitting in a row. Two of them are selected at random. The probability that they are not together is 2 n 1 (c) 1 − n

(a) 1 −

(b)

2 n +1

(d) None of these

128. If A and B are any two events, the probability that exactly one of them occurs is (a) P (A) + P (B) – 2P (A ∩ B) (b) P (A) + P  (B) – 2P (A ∩ B) (c) P (A ∩ B ) + P ( A ∩ B) (d) P (A) + P (B) – P (A ∪ B). 129. A bag contains four tickets with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let E i (i = 1, 2, 3) denote the event that i-th digit on the drawn ticket is 2. Then (a) E1, E2, E3 are pair wise independent (b) E1, E 2 are independent (c) E 2 and E 3 are not independent (d) E1, E2, E3 are mutually independent. 130. For two events A and B if P (A) = P (A | B) = 1/4 and P (B | A) = 1/2, then (a) A and B are mutually exclusive (b) A and B are independent (c) A is subevent to B (d) P (A′ | B) = 3/4. 131. 5 grils and 10 boys sit at random in a row having 15 chairs numbered as 1 to 15. The probability that end seats are occupied by the girls and between any two girls odd number of boys sit, is

(a)

20 × 10! × 5! 15!

(b)

20 × 10! 15!

(c)

20 × 5! 15!

(d) None of these

132. An artillery target may be either at point I with prob8 1 or at point II with probability . We have ability 9 9 21 shells each of which can be fired either at point I or II. Each shell may hit the target independently of 1 . The number of the other shell with probability 2 shells that must be fired at point I to hit the target with maximum probability, is (a) 6 (c) 12

(b) 17 (d) 5.

133. A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is (a) 15/28 (c) 15/213

(b) 2/15 (d) None of these

134. A and B are two events. Odd against A are 2 : 1. Odds in favour of A ∪ B are 3 : 1. If x ≤ P (B) ≤ y, then the ordered pair (x, y) is  5 3 (a)  ,   12 4 

2 3 (b)  ,  3 4

1 3 (c)  ,  3 4

(d) None of these

135. Let A, B and C be three events such that P (A) = 0.3, P (B) = 0.4, P (C) = 0.8, P (A ∩ B) = 0.08, P (A ∩ C) = 0.28, P (A ∩ B ∩ C) = 0.09, If P (A ∪ B ∪ C) ≥ 0.75, then (a) 0.23 ≤ P (B ∩ C) ≤ 0.48 (b) 0.23 ≤ P (B ∩ C) ≤ 0.75 (c) 0.48 ≤ P (B ∩ C) ≤ 0.75 (d) None of these 136. The probability that the 13th day of a randomly chosen month is a Friday, is 1 12 1 (c) 84 (a)

(b)

1 7

(d) None of these

137. A person draws a card from a pack, replaces it, shuffles the pack, again draws a card, replaces it and draws again. This he does until he draws a heart. The probability that he will have to make at least four draws is 27 256 27 (c) 64

(a)

(b)

175 256

(d) None of these

1 2 1 (c) p = 4

(b) p =

(a) p =

1 3

(d) None of these

139. The mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is 2 3 7 (c) 8

4 5 11 (d) 16 (b)

(a)

140. A contest consists of predicting the results win, draw or defeat of 7 football matches. A sent his entry by predicting at random. The probability that his entry will contain exactly 4 correct predictions is

(a) 1/30 (c) 1/3

147. If the mean of a binomial distribution is 25, then its standard deviation lies in the interval given below: (a) [0, 5) (c) [0, 25)

(a) 3/4 (c) 1/3

(b) 1/2 (d) None of these

149. If X follows a binomial distribution with parameters n = 8 and p = 1/2, then P ( | X – 4| ≤ 2) =

(a) 8/3 (c) 280/37

(b) 16/3 (d) 560/37

(a) 10 (c) 12

(b) 11 (d) 13

(a) 1/2 (c) 50/101

7

(b) (0, 5] (d) (0, 25]

148. A man alternately tosses a coin and throws a die beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is

119 128 29 (c) 128

7

(b) 1/10 (d) 1/15

(a)

116 128

(b)

(b) 49/101 (d) 51/101

(d) None of these 141. A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the 150. One hundred identical coins, each with probability reamaining (n + 1) coins are fair. A coin is picked up p of showing up heads, are tossed. If 0 < p < 1 and at random from the bag and tossed. If the probability the probability of heads showing on 50 coins is equal that the toss results in a head is 31/42, then n is equal to that of heads showing on 51 coins, the value of p to is

142. A boy is throwing stones at a target. The probability of hitting the target at any trial is 1/2. The probability of hitting the target 5th time at the 10th throw is (a) 5/210 (c) 10C5/210

(b) 63/29 (d) None of these

143. Two players A and B toss a fair coin cyclically in the following order A, A, B, A, A, B, ... till a head shows. Let α (β) denote the probability that A (B) gets the head first. Then (a) α = 6/7 (c) β = 1/7

(b) α = 5/7 (d) β = 2/7

144. Three squares of a chess board are chosen at random, the probability that two are of one colour and one of another is 16 8 (b) 21 21 32 (c) (d) None of these 12 145. A box contains tickets numbered 1 to 20. 3 tickets are drawn from the box with replacement. The probability that the largest number on the tickets is 7 is (a)

(a)

2 19

(b)

7 20

3

 7  (c) 1 −    20 

(d) None of these

3 151. If A and B are two eveents such that P (A ∪ B) ≥ 4 1 3 ≤ P (A ∩ B) ≤ , then and 8 8 11 8 7 (c) P (A) + P (B) ≥ 8

(a) P (A) + P (B) ≤

(b) P (A) ⋅ P (B) ≤

3 8

(d) None of these

152. A point is selected at random from the interior of a circle. The probability that the point is closer to the centre than the boundary of the circle is 3 4 1 (c) 4 (a)

(b)

1 2

(d) None of these

153. Let A = [2, 3, 4,...,20]. A number is chosen at random from the set A and it is found to be a prime number. The probability that it is more than 10 is 9 10 1 (c) 5 (a)

(b)

1 10

(d) None of these

154. 6 ordinary dice are rolled. The probability that at least half of them will show at least 3 is

917

146. The value of C for which P (X = k) = Ck2 can serve as the probability function of a random variable X that takes value 0, 1, 2, 3, 4 is

Probability

138. Suppose X follows a binomial distribution with parameters n and p, where 0 < p < 1. If P (x = r) / P (X = n – r) is independent of n and r, then

918

(a) 41 ×

24 36

(b)

Objective Mathematics

24 36

(c) 20 ×

24 36

(d) None of these

155. From a box containing 20 tickets of value 1 to 20, four tickets are drawn one by one. After each draw, the ticket is replaced. The proability that the largest value of tickets drawn is 15 is 4

3 (a)   4 (c)

27 1280

(b)

162. For a poisson distribution whose mean is λ, the standard deviation will be (a) λ2

1 λ (d) λ. (b)

(c) λ 163. A person draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues doing this until be draws a spade, the chance that he will fail in the first two draws is

27 320

1 16 9 (c) 64 (a)

(d) None of these

9 16 1 (d) 64 (b)

156. If the integers m and n are chosen at random between 1 164. In tossing 10 coins the proability of getting exactly 5 heads is and 100 then the probability that a number of the form 7m + 7n is divisible by 5 is 193 9 (b) (a) 1 1 256 128 (b) (a) 5 7 63 1 (c) (d) 1 256 1 2 (c) (d) 49 4 165. Four tickets marked 00, 01, 10, 11 respectively are placed in a bag. A ticket is drawn at random five times, being 157. A three digit number is selected at random from the replaced each time, the probability that the sum of the set of all three-digit numbers. The probability that numbers on tickets thus drawn is 23, is the number selected has all the three digits same, is 100 231 (b) (a) 256 256 (a) 1/9 (b) 1/10 (c) 1/50 (d) 1/100 25 (c) (d) None of these 256 158. Let A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8}. An element (a, b) of their cartesian product A × B is chosen 166. A and B are two independent events. The probability at random. The probability that a + b = 9, is 1 and the probability that that both A and B occur is (a) 1/5 (b) 2/5 6 (c) 3/5 (d) 4/5 1 . The probability of the neither of them occurs is 159. If A and B are such that P (A) > 0 and P (B) ≠ 1, then 3 occurrence of the events A is P ( A/B ) is equal to (a)

1 − P (A ∪ B) P (B)

(c) 1 – P (A/B)

(b)

P (A) P (B)

(d) 1 – P ( A /B).

160. A and B are any two mutally exclusive events, then (a) P (A) < P (B) (b) P (A) > P ( B ) (c) P (A) ≤ P ( B ) (d) None of these

2 5 (b) 3 6 1 (c) (d) None of these 2 167. A random Variable X has the following proability distribution : (a)



X : 0 1 2 3 4 5 6 7 8 P (X = x)  : a 3a 5a 7a 9a 11a 13a 15a 17a then the value of a is :

7 5 161. In an entrance test there are multiple choice questions. (b) (a) There are four possible answers to each question of 81 81 which one is correct. The probability that a student 2 1 (c) (d) knows the answer to a questions is 90%. If he gets 81 81 the correct answer to a question, then the probability 168. Three identical dice are rolled. The probability that the that he was guessing is same number will appear on each of them is 1 36 (b) (a) 1 3 9 37 (b) (a) 18 28 1 47 (c) (d) 1 1 37 40 (c) (d) 36 6

(a) A is subevent of B (b) A and B are mutually exclusive A' 3 (c) A and B are independent and P   = B 4 (d) None of the above 170. Out of 13 applicants for a job there are 5 women and 8 men. It is desired to select 2 persons for the job. The probability that at least one of the selected persons will be a women is 14 (a) 39 25 (c) 39

5 (b) 15 10 (d) 23

171. If A and B are two independent events such that 7 8 , P (B′ ) = α and P (A ∪ B) = , then α is P (A′) = 10 10 2 5 (b) (a) 7 7 (c) 1 (d) None of these 172. Four positive integers are taken at random and are multiplied together. Then the probability that the product ends in an odd digit other than 5 is 3 5 16 (c) 625

609 625 2 (d) 5

(b)

(a)

(a) 0.4 (c) 0.8

(b) 0.5 (d) None of these

177. Sixteen players S1, S2, ... 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. Assuming that all the palyers are to equal strength. The probability that the players S1 is among the eight winners is 1 2 2 (c) 3 (a)

(b)

1 3

(d) None of these

178. Three numbers are chosen at random without replcement from {1, 2,...,10}. The probability that the minimum of the chosen numbers is 3, or their maximum is 7, is

7 40 11 (c) 40

(a)

(b)

5 40

(d) None of these

179. Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently, equals 1 2 2 (c) 15 (a)

7 15 1 (d) 3

(b)

180. 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 1 1 173. If P (A ∩ B) = , P  A ∩ B = and 2P (A) = P (B) 2 white and 1 black ball will be drawn is 2 2 13 1 = p, then the value of p is equal to (b) (a) 32 4 2 1 (b) (a) 1 3 3 2 (c) (d) 32 16 1 1 (c) (d) 181. There are four machines and it is known that exactly 3 4 two of them are faulty. They are tested one by one, 174. A five digit number is formed by the digits 1, 2, 3, 4, in a random order till both the faulty machines are 5 without repetition. The probability that the number identifed. Then the probability that only two tests are formed is divisible by 4, is needed is 1 4 (b) (a) 1 1 (b) (a) 5 5 3 6 3 (c) (d) None of these 1 1 (c) (d) 5 2 4 175. A committee consists of 9 experts taken from three institutions A, B and C, of which 2 are from A, 3 from 182. A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of head appearing B and 4 from C. If three experts resign, then the probon the fifth toss equals ability that they belong to different institutions is :

(

1 729 1 (c) 21

(a)

)

1 24 2 (d) 7 (b)

1 2 31 (c) 32

(a)

1 32 1 (d) 5 (b)

919

176. Four cards are drawn from a pack of 52 cards. The probability of drawing exactly one pair is

Probability

1 A 169. For two events A and B if P (A) = P   = and 4 B 1 B   , then P  = 2 A

920

183. An unbaised die is tossed until a number greater than 4 appears. The probability that an even number of tosses is needed is

Objective Mathematics

(a) 1/2 (c) 1/5

(b) 2/5 (d) 2/3

184. The probability of India winning a test match against 1 West Indies is . Assuming independence from match 2 to match the probability that in a 5 match series India’s second win occurs in the third test is

1 (a) 8 (c)

1 2

1 (b) 4 2 (d) 3

1 ≤ P( B) ≤ 2 1 (c)   ≤ P ( B ) ≤ 4

(a)  

3 4 3 5

(b)  

5 3 ≤ P( B) ≤ 12 4

(d)  none of these

190. A person draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues doing this until he draws a spade. The chance that he will fail in the first two draws, is (a)  

9 64

(b)  

1 64

(c)  

1 16

(d)  

9 16

191. Three coins are tossed simultaneously. Then the values of a, b, c and d in the following probability distribution 185. If two events A and B are such that P (A ) = 0.3, P (B) are respectively = 0.4 and P (AB c) = 0.5, then P [B\ (A ∪ B c)] = Number of heads 0 1 2 3 1 1 (b) (a) Probability a b c d 3 2 1 1 1 1 3 3 1 1 (b)   , , , (a)   0, , , (c) (d) None of these 4 4 2 8 8 8 8 4 c

1 1 1 1 3 1 1 3 186. An unbiased coin is tossed. If the result is a head, a (c)   , , , (d)   , , , 4 4 4 4 8 8 8 8 pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If 192. If M and N are any two events. The probability that the result is a tail, a card from a well shufled pack of exactly one of them occurs, is eleven cards numbered 2, 3, 4,...,12 is picked and the (a)  P(M) + P(N) – P(M ∩ N) number on the card is noted. The probability that the (b)  P(M) + P(N) + P(M ∩ N) noted number is either 7 or 8, is (c)  P(M) + P(N) 193 164 (b) (a) (d)  P(M) + P(N) – 2P(M ∩ N) 792 792

(c)

231 792

(d) None of these

187. 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,

16 81 80 (c) 81

(a)

1 81 65 (d) 81 (b)

193. If four dice are thrown together. Probability that the sum of the number appearing on them is 13, is (a)  

35 324

(b)  

5 216

(c)  

11 216

(d)  

11 432

194. A coin is tossed 4 times. The probability that atleast one head turn up, is (a)  

1 16

(b)  

2 16

14 15 (c)   (d)   188. These are four balls of different colours and four boxes 16 16 of colours, same as those of the balls. The number of ways in which the balls, one each in a box, could be 195. A pair of dice is thrown, if 5 appears on atleast one placed such that a ball does not go to a box of its own of the dice, then the probability that the sum is 10 or colour is greater, is

5 8 1 (c) 8

(a)

(b)

3 8

(d) None of these

(a)  

11 36

(b)  

2 9

(c)  

3 11

(d)  

1 12

189. Given two events A and B. If odds agains A are 196. For two events A and B, if P(A) = P(A/B) = 1/4 and 2 : 1 and those in favour of A ∪ B are as 3 : 1, 1 P ( B / A ) = , then then 2

1 197. If P(A) = P(B) = x and P ( A ∩ B ) = P ( A′ ∩ B′) = , then 3 x is equal to 1 2 1 (c)   4

(a)  

1 3 1 (d)   6

(b)  

198. The records of a hospital show that 10% of the cases of a certain disease are fatal. If 6 patients are suffering from the disease, then the probability that only three will die, is (a)  8748 × 10–5 (c)  1458 × 10–6

(b)  1458 × 10–5 (d)  41 × 10–6

199. Out of 40 consecutive natural numbers, two are chosen at random. Probability that the sum of the number is odd, is 14 29 1 (c)   2

(a)  

(b)  

20 39

(d)  none of these

205. 5 persons A, B, C, D and E are in 9 queue of a shop. The probability that A and E always together is 1 4 2 (c)   5

(a)  

921

(a)  P(E1 – E2) = P(E2) – P(E1 ∩ E2) (b)  P(E1 – E2) = P(E2) + P(E1 ∩ E2) (c)  P(E1 – E2) = P(E1) – P(E1 ∩ E2) (d)  P(E1 – E2) = P(E1) + P(E1 ∩ E2)

Probability

204. If E1 and E2 be two events, then

(a)  A and B are independent 3 (b)   P ( A′ / B ) = 4 1 (c)   P ( B′ / A′ ) = 2 (d)  all of the above

2 3 3 (d)   5

(b)  

206. A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is 3 8 3 (c)   4

(a)  

(b)  

1 5

(d)  None of these

207. Twelve tickets are numbered from 1 to 12. One ticket is drawn at random, then the probability of the number to be divisible by 2 or 3, is 2 3 5 (c)   6

(a)  

7 12 3 (d)   4

(b)  

200. A pair of fair dice is thrown independently three times. 208. Two cards are drawn successively with replacement from a well shuffled deck of 52 cards, then the mean The probability of getting a score of exactly 9 twice of the number of aces is is (a)  1/729 (c)  8/729

(b)  8/9 (d)  8/243

1 13 2 (c)   13

(a)  

(b)  

3 13

201. Two aeroplanes I and II bomb a target in succession. (d)  None of these The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that 209. If A and B are events such that P ( A ∪ B ) = 3 , P ( A ∩ B ) = 1 , P (A 4 4 the target is hit by the second plane is 3 1 2 A P ( ∩ B ) then is equal to P ( A ∪ B ) = , P ( A ∩ B ) = , P (A ) = , (a)  0.06 (b)  0.14 4 4 3 (c)  0.32 (d)  0.7 5 3 (b)   (a)   202. Let Ec denote the complement of an event E. Let E, F, 12 8 G be pairwise independent events with P(G) > 0 and 5 1 c c P(E ∩ F ∩ G) = 0. Then P(E ∩ F | G) equals (c)   (d)   8 4 (a)  P(Ec) + P(Fc) (b)  P(Ec) – P(Fc) 210. A die is thrown. Let A be the event that the number (c)  P(Ec) – P(F) (d)  P(E) – P(Fc) obtained is greater than 3. Let B be the event that the 203. One Indian and four American men and their wives are number obtained is less than 5. Then, P(A ∪ B) is to be seated randomly around a circular table. Then the 3 conditional probability that the Indian man is seated (a)   (b)  0 5 adjacent to his wife given that each American man is seated adjacent to his wife, is 2 (c)  1 (d)   5 1 1 (a)   (b)   2 3 211. It is given that the events A and B are such that 2 1 1  A 1 B 2 (c)   (d)   P ( A) = , P   = and P   = . Then, P(B) is 5 5 4 B 2  A 3

922

Objective Mathematics

1 6 2 (c)   3

1 3 1 (d)   2

(a)  

(b)  

212. Let A = { 1, 3, 5, 7, 9}, B = { 2, 4, 6, 8}, if a Cartesian product A × B is choosen at random, the probability of a + b = 9 is (a)  3/2 (c)  1

(b)  3/4 (d)  1/5

213. One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

(a)  1/2 (c)  2/5

(b)  1/3 (d)  1/5

214. Let E c denote the complement of an event E. Let E, F, G be pairwise independent events with P(G) > 0 and P(E∩F∩G) = 0. Then P(EC∩FC | G) equals (a)  P(EC) + P(FC) (c)  P(EC) – P(F)

(b)  P(EC) – P(FC) (d)  P(E) – P(FC)

215. 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, 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

solutions 1. (a) P(E1) =

1 1 1 , P(E2) = and P (E2) = 3 2 4

5. (a) The word ‘article’ has seven alphabets.

 hey can be arranged among themselves in 7! T ways. P(E1 ∪ E1 ∪ E2) = 1 – P ( E1 ) P ( E2 ) P ( E3 ) In this word ‘article’ there are 3 vowels viz. a, i and e and four consonants. These three vowels have to 1 1 1 = 1– 1 −  1 −  1 −  be placed in three even places 2nd, 4th and 6th. The  2  3  4 three vowels can occupy these 3 places in 3! ways 2 1 3 3 and the 4 consonants can occupy the remaining 4 =1– × × = . 3 2 4 4 places in 4! ways. Thus the number of ways favourable to the event are 2. (b) T he number of ways in which either player can 3! × 4!. choose a number from 1 to 25 is 25, so the total 3! × 4! 3 × 2 × 1 1 number of ways of choosing numbers is 25 × 25 . ∴ Required probability = = = 7! 7 × 6 × 5 35 = 625. So the probability that they will not win a 6. (d) Total numbers which are divisible by 2 and 3 = 16 prize in a single trial is 1 24 . = 25 25 3. (d) Since, A and B mutually exclusive events, therefore, =1–



A ∩ B = φ  ⇒  A ⊆ B or B ⊆ A



⇒ P (A) ≤ P( B ) or P(B) ≤ P( A ).

4. (b) T here are 7 letters in the word ‘SOCIETY’ which can be arranged in 7! ways.

∴ Required probability

(

)

(

16

C3 4 = . C3 1155

100

)

(

7. (d)  P B ∩ C = P A ∩ B ∩ C + P A ∩ B ∩ C

)

1 1 2 . + = 3 3 3 P B ∩ C = P(B) ­– P(B ∩ C)

=

(

)

3 2 = – P(B ∩ C) 4 3 9−8 3 2 1 = . ∴ P ( B ∩ C) = − = 12 4 3 12

⇒

 his is the exhaustive number of cases. Considering T the three vowels viz. i.e., in the word ‘SOCIETY’ as one letter, we can arrange 5 letters in a row in 5! ways. Also three vowels 0, i, e can themselves 8. (d) Total number of ways of choosing two numbers out be arranged in 3! ways. 6×5 = 15 of six = 6C2 = ∴ The total numbers of arrangements in which 2 three vowels come together are 5! × 3!. For favourable ways: Hence, the required probability If smaller number is chosen 3 then greater has choices 4, 5, 6 = 3 choices 5! × 3! 3 × 2 × 1 1 If smaller is 2, greater is 3, 4, 5, 6 = 4 choices = = = . 7! 7×6 7 If smaller is 1, greater is 2, 3, 4, 5, 6 = 5 choices

Required probability =

9. (c) P (A ∪ B) = 0.7 and P (A ∩ B) = 0.2 P (A ∪ B) = P (A) + P (B) – P (A ∩ B) ∴ 0.7 = P (A) + P (B) – 0.2 ⇒ P (A) + P (B) = 0.7 + 0.2 = 0.9 ⇒ 1 – P ( A ) + 1 – P ( B ) = 0.9 ⇒ P ( A ) + ( B ) = 2 – 0.9 = 1.1 10.  (a), (d) S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6),   (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

A = {(H, 2), (H, 4), (H, 6)} B = {(T, 3), (T, 6)}.



n (A) = 3, n (B) = 2, n (S) = 12



2 1 3 1 = . ∴ P (A) = and P (B) = = 12 6 12 4

11.  (d) A = {1, 2, 3, 4, 5, 6, 7, 8} B = {8, 7, 6, 5, 4, 3, 2, 1} Since the number is chosen from each of two sets, total number of outcomes = 8 × 8 = 64. Favourable ways in which the sum of 9 can be obtained as (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1) = 8 ways 8 1 ∴ p1 = = . 64 8  avourable ways in which the sum of 7 can be F obtained as (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) = 6 ways. 6 3 . ∴ p2 = = 64 32

Hence p1 + p2 =

12. (b) Required probability

1 3 4+3 7 + = = . 8 32 32 32



P ( A1 A2 A3 ) + P ( A1 A2 A3 )



= P(A1) P( A2 ) P(A3) + P( A1 ) P(A2) P(A3)



1 1 2 1 1 1 =   +  = + = = . 2 2 8 8 8 4

3

923

12 4 = . 15 5

14. (c) Number of students = 100 Number of girls = 55 ∴ Number of boys = 100 – 55 = 45 Out of 45 boys 36 boys are studying Statistics. ∴ N  umber of boys not studying Statistics = 45 – 36 =9 ∴ Probability that a boy picked up at random is 9 1 not studyng Statistics = = . 45 5 15. (a) There are 10 bulbs in all and of these 3 can be selected in C3 =

10 × 9 × 8 = 120 ways 3 × 2 ×1



10



∴ n (S) = 120

Let A be the event of getting light. Then A denotes the event of having no light. Now 6 bulbs are not good. Hence the number of ways in which all not good bulbs are chosen are

6

C3 =

6×5×4 = 20 1× 2 × 3



Thus n ( A ) = 20



∴ P ( A ) =

n (A) 20 1 = = n (S) 120 6

1 5 = . 6 6 16. (a) A die is tossed thrice, then in a single throw of die, E = {2, 4, 6} [ A and B are independent events]

∴ P (A) = 1 – P ( A ) = 1 –

n(E) = 3 n(S) = 6 1 3 1  = i.e., probability of success =  2 6 2  1 1 Now, probability of failure = 1 – = 2 2 ∴ Probability of atleast two success is

∴ P(E) =

P(X = 1) + P(X = 2) + P(X = 3) [where X = no. of successes]

3

2

2

1

3



11 1 1 1 1 = 3C1     + 3C2     + 3C3     22 2 2 2 2

∴ n (S) = 36.



=3×

 et ‘A’ represent ‘‘the product of two numbers be L a perfect square’’.



=

A = {(1, 1), (1, 4) (2, 2), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)}

17. (b) Given that

13. (a) Two dice can be thrown in 6 × 6 = 36 ways



n (A) = 8.



∴ P (A) =

n (A) 8 2 = = . n (S) 36 9

0

1 1 1 1 1 × +3× × + 2 4 4 2 8

3 3 1 7 + + = . 8 8 8 8

x = 33n, where n is a positive integral value. Here only four digits may come at the units place i.e., 9, 7, 1, 3

Probability

Total favourable choices = 3 + 4 + 5 = 12

21. (a) Let q = 1 – p. Since, head appears first time in an even throw 2 or 4 or 6.

924

∴ n(E) = 1 ∴ P(E) =

Objective Mathematics

n( E ) 1 = n( S ) 4

18. (c) Let the given events be E1 and E2 2 P (E2) 3 The events E1 and E2 are exhaustive ∴ P (E1 ∪ E2) = 1 Then P (E1) =

Since E1 and E2 are mutually exclusive P (E1) + P (E2) = 1 2 5 ⇒ P (E2) + P (E1) = 1 ⇒ P (E2) = 1 3 3 3 ⇒ P (E2) = 5 3 2 3 2 ∴ P (E2) = and P (E1) = × = 5 3 5 5 Since P (E2) =

3 , odds in favour of E2 are 3 : 5 – 3 5

i.e., 3 : 2. 19.  (a) Let P (A) = p, then P (B) =

3 3 P (A) = p 2 2

3 1 1 3 P (B) = × p= and P (C) = 4 2 2 2 ∵ A  , B, C are mutually exclusive and exhaustive events ∴ (A ∪ B ∪ C) = S ∴ P (A ∪ B ∪ C) = P (S) ⇒ P (A) + P (B) + P (C) = 1 3 3 ⇒ p + p+ p=1 2 4 4 13 p = 1 ∴ p = . ⇒ 13 4

20. (b) Let A be the event of passing the 1st examination and B the event of passing the second. Therefore n (A) = 60, n (B) = 50, n (A ∩ B) = 30 and n (S) = 100 P (A) =

n (A) 60 n (B) 50 , P (B) = = = n (S) 100 n (S) 100

n (A ∩ B) 30 and P (A ∩ B) = = n (S) 100  hus, the probability of passing atleast one examinaT tion i.e., P (A ∪ B) = P (A) + P (B) – P (A ∩ B) 60 50 30 80 + − = = = 0.8. 100 100 100 100 ∴ Probability that a student selected at random has failed in both examinations = 1 – P (A ∪ B) = 1 – 0.8 = 0.2.

∴

2 = qp + q3p + q5p +... 5

∴

qp 2 (1 − p ) p 2 1− p =   ⇒ = = 5 1 − q2 5 1 − (1 − p ) 2 2 − p



2 1− p 1 =   ⇒ 4 – 2p = 5 – 5p, 3p = 1 ⇒ p = . 5 2− p 3

22. (a) Given that P (A ∩ B) = 1/3, P (A ∪ B ) = 5/6 and P (A) = 1/2 ∴ by using the formula P(A ∩ B) = P(A) + P(B) – P(A ∩ B) ⇒

5 1 1 = + p (B) − 6 2 3

⇒ P(B) =

5 1 1 4 2 + − = = 6 3 2 6 3

1 2 1 = P (A ∩ B), × = 3 3 3 ∴  A and B are independent events.

P(A) × P(B) =

23.  (b) A : ‘The card drawn is a king’ B : ‘The card drawn is a heart’ C : ‘The card drawn is a red card’ Then A, B and C are not mutually exclusive. H ere A ∩ B = The card drawn is the king of hearts. B ∩ C = The card drawn is a heart A ∩ C = The card drawn is heart or a red king. and A ∩ B ∩ C = The card drawn is the king of hearts. P (A) =

4 13 26 , P (B) = , P (C) = 52 52 52

P (A ∩ B ∩ C) = P (A ∩ C) =

1 13 , P (B ∩ C) = , 52 52

2 1 , P (A ∩ B ∩ C) = 52 52

∴ P (A ∪ B ∪ C) = P (A) + P (B) + P (C) –  P (A ∩ B) – P (B ∩ C) – P (A ∩ C) + P (A ∩ B ∩ C) =

4 13 26 1 13 2 1 28 7 = . + + − − − + = 52 52 52 52 52 52 52 52 13

24.  (b) Let ‘A’ be the event that the dacoit is killed by the ith bullet (1 ≤ i ≤ 6) ∴ P (Ai) = 0.6 and P ( A i ) = 1 – 0.6 = 0.4 P (Dacoit is still alive) 

= P ( A

1

A

2

A

3

A

4

A

5

A 6)

= P ( A 1) P ( A 2) P ( A 3) P ( A 4) P ( A 5) P ( A 6) = 0.4 × 0.4 × 0.4 × 0.4 × 0.4 × 0.4 = 0.004096.

40 2 = 100 5

25 1 = 100 4 Since the events are independent,

20 1 = 100 5

Probability of selecting a girl =

∴ probability of selecting a fair complexioned rich girl =

3 6 2 5 × × × 9 8 7 6 Since these two are mutually exclusive ways. Required probability =

Probability of selecting a rich student =



925

=

I f white ball is drawn first, the probability of drawing the balls alternatively

Probability

25. (a) Probability of selecting a fair complexioned student

2 1 1 = 0.02. × × 5 5 4

= =

5 5 10 5 . + = = 84 84 84 42 4 4 1 , P ( α ) = 1 – = 5 5 5 3 3 1 P (β) = , P ( β ) = 1 – = 4 4 4

29.  (b) P (α) =

26. (c) Let E be the event that A speaks the truth and F be the event that B speaks the truth.



∴ E and F are the complementary events in which A and B tell a lie.



60 3 2 =  ; P ( E ) = 1 – P (E) = 100 5 5

6 3 5 2 3 6 2 5 × × × + × × × 9 8 7 6 9 8 7 6

P (γ) =

5 5 1 , P ( γ ) = 1 – = 6 6 6

(i) A speaks truth and B tells lie i.e., E ∩ F

1 2 2 , P ( δ ) = 1 – = . 3 3 3 Different possibilities to qualify are (i) passes in α, β, γ and fails in δ (ii) passes in α, β, δ and fails in γ (iii) passes in α, γ, δ and fails in β (iv) passes in all the four sujbects α, β, γ and δ. These are mutually exclusive possibilities.

(ii) A tells lie and B speaks the truth i.e., E ∩ F

∴ Required probability

Reqd. probability = P (E ∩ F ) + P ( E ∩ F)

 4 3 5 1  4 3 2 1 =  × × × + × × ×   5 4 6 3  5 4 3 6

Here P (E) =

P (F) =

90 9 1 , P ( F ) = 1 – P (F) = = 100 10 10

 and B will contradict each other when one speaks A the truth and the other tells a lie and this can happen in he following two mutually exclusive ways :

= P (E) P ( F ) + P ( E ) P (F) =

27.  (b) Let A : Second dice always exhibits 4 = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)} ∴ n (A) = 6 B : Sum of the numbers on both dice is 8 = {(4, 4)}

Similarly, A ∩ B = A ∩ {(4, 4)} = {(4, 4)} ∴ n (A ∩ B) = 1 n (A ∩ B) 1 = . ∴ P (B | A) = n (A) 6 28.  (b) Four balls can be drawn alternatively in the following two ways RWRW or WRWR If red ball is drawn first, the probability of drawing the balls alternatively =

6 3 5 2 × × × 9 8 7 6

P (δ) =

4 5 2 1 4 3 5 2 +  × × × + × × ×  5 6 3 4 5 4 6 3

3 1 2 9 21 × × × = 5 10 5 10 50

Hence, they are expected to contradict each other in 21 × 100 = 42%. 50





=

1 1 1 1 + + + 6 15 9 3

=

15 + 6 + 10 + 30 61 . = 90 90

30.  (a) T here are two mutually exclusive ways in which A and B can contradict each other on an identical point. (i) A speaks the truth and B tells a lie. (ii) A tells a lie and B speaks the truth. Let us define the events. E1 : A speaks the truth. E2 : B speaks the truth. The required probability = P (E1 ∩ E 2 )+ P ( E1 ∩ E2) = P (E1) × P ( E 2 ) + P ( E1 ) × P (E2) =

3 3 2 5 9 10 = 0.475. × + × = + 5 8 5 8 40 40

 ence A and B contrdict each other on an identical H point in 47.5% of the cases.

926

31. (b) Let ‘A’ = {a red ball is selected at first draw}

Objective Mathematics

‘B’ = {a red ball is selected at second draw} then P (A ∩ B) = P (Red ball and Red ball) =

3 2 1 × = 10 9 5

P (B) = P (Red ball and Red ball or Black ball and Red ball) =P   (Red ball and Red ball) + P (Black ball and Red ball)

Number of ways of drawing 2 brown socks = 5C2 ways = 10 ways Number of ways of drawing 2 white socks = 4C2 ways = 6 ways ∴ Probability of drawing 2 socks of the same colour = P (2 brown socks) + P (2 white socks) =

10 6 16 4 + = = . 36 36 36 9

38. (c) Three dice can be rolled in 6 × 6 × 6 = 216 ways. Favourable ways of getting the same number on each dice = (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), P (A ∩ B) 1 / 5 10 2 = = . ∴  P (A | B) = = (6, 6, 6) P (B) 3 / 10 3 × 15 9 i.e., 6 ways. 32. (d) Leap year consists of 366 days i.e., weeks having 6 1 = . ∴ Required probability = 52 Sundays and 2 days. These two days can be as 216 36 follows. S & M, M & T, T & W, W & Th, Th & Fri, Fri & Sat, 39. (a) Since E1 ∩ E 2 = E1 ∪ E 2 Sat & S. Out of these seven events, Sundays occurs in and (E1 ∪ E2) ∩ (E1 ∪ E 2 ) = φ two 2 events. =

3 2 7 3 3 × + × = 10 9 10 9 10

2 . 7 33. (b) P ( A ) = 1 – P (A) = 1 – 0.65 = 0.35 Hence required probability =

P ( B ) = 1 – P (B) = 1 – 0.15 = 0.85 ∴  P ( A ) + P ( B ) = 0.35 + 0.85 = 1.2 34. (c) Here P (A) = 0.4 ; ∴  P ( A ) = 1 – 0.4 = 0.6 P (A happens at least once) = 1 – P (non occurrence in all the three trials) = 1 – (0.6)3 = 1 – 0.216 = 0.784. 35. (a) Let ‘A’ be the event that ‘a student passed in mathematics’ and ‘B’ be the event that ‘a student passed in Statistics’. 70 55 30 , P (B) = , P (A ∩ B) = Then, P (A)= 125 125 125 Required probability = P (A ∩ B′) + P (A′ ∩ B) = P (A) – P (A ∩ B) + P (B) – P (A ∩ B) = P (A) + P (B) – 2P (A ∩ B) =

65 13 70 55 60 70 + 55 − 60 = = . + − = 125 25 125 125 125 125

3 36. (c) P (5 or 6 or 7) in one draw = 7 ∴ Probability that in each of 3 draws, the chit bears 5 or 3

3 6 or 7 =   . 7 37. (d) Total number of socks in the box = 5 + 4 = 9 2 Socks can be drawn from the box in 9C2 ways = 36 ways.

∴ P [(E1 ∪ E2) ∩ ( E1 ∩ E 2 )] = P (φ) = 0 < 1 . 4 40. (b) P (only one of A and B will die in a year) = P ( A B) + P (A B ) = P ( A ) P (B) + P (A) P ( B ) [Since A and B are independent] = (1 – p) q + p (1 – q) = p + q – 2pq. 41. (c) The probability that the problem is not solved

 = 1 − 

1  1   1 −  1 − 2  3 

1 2 3 1 1  = 2 × 3 × 4 = 4 4

1 ∴ The probability that problem is solved = 1 – 4 3 . = 4 42. (b) Probability of getting atleast one head in n tosses. n

1 = 1 –   ≥ 0.9 2 n

1 ⇒   ≤ 0.1 2 ⇒ 2n ≥ 10 ⇒ n ≥ 4 Hence least value of n = 4. 1 and probability of 2 1 2 throwing 5 or 6 with a dice = = . 3 6 He starts with a coin and alternatively tosses the coin and throws the dice and he will win if he gets a head before he gets 5 or 6. Hence Probability 1 1 2 1 1 2 1 2 1 = +  ⋅  ⋅ +  ⋅  ⋅  ⋅  × + ... 2 2 3 2 2 3 2 3 2

43. (a) Pobability of getting head =

1 1 1 3 3 = ⋅ = × = . 2 1− 1 3 2 4 3 44. (d) Required probability = P (I in Ist draw) × P (I in 2nd draw) × P (T in 3rd draw) 10 9 10 C C C 10 9 10 5 × × = 20 1 × 19 1 × 18 1 = = . C1 C1 C1 20 19 18 38 45. (b) Required probability = ( Probability that Krishna will be dead 10 years hence) × (Probability that Hari will be dead 10 years hence) 7 7 8 3 24  × = = 1 −  1 −  = .  15   10  15 10 150 46.  (a) P (A ∩ B) = P (A) + P (B) – P (A ∪ C) = 0.6 + 0.4 – 0.8 = 0.2 P (A ∪ B ∪ C) = P (A) + P (B) + P (C) – P (A ∩ C) + P (A ∩ B ∩ C) – P (A ∩ B) – P (B ∩ C)

50.  (c) P (both are aces) = P (one is ace) =

4 3 1 × = 16 15 20

1 12 12 4 2 ⋅ + ⋅ = 4 15 16 15 5

∴ P (at least one is ace) =

1 2 9 . + = 20 5 20

51.  (a) Let A denotes the event that a six occurs and B the event that the man reports that it is a six.  hen the probability that it is actually a six is T given by P (A ∩ B) P (A | B) = P (B) Now P (A ∩ B) =

1 3 3 × = 6 4 24

P (B) = P (A ∩ B) + P ( A ∩ B)

=

1 3 5 1 8 × + × = 6 4 6 4 24

Hence P (A/B) =

3 / 24 3 = . 8 / 24 8

...(1) 52. (c) Number of ways of selecting 2 good mangoes = 6C2 = 15 The number of ways that at least one of the two ∴  (1) ⇒ P (B ∩ C) ≤ 1.2 – 0.85 selected mangoes is to be good and P (B ∩ C) ≥ 1.2 – 1 = 6C1 × 9C1 = 6 × 9 = 54 ⇒ 0.2 ≤ P (B ∩ C) ≤ 0.35. 15 5 . Required probability = = 47. (c) Since each element of the determinant can be placed 24 18 4 in two ways 0 or 1, total number of ways = 2 = 16. 53.  (b) Matches played by India = 4 Since value of the determinant is + ve, so we have P robability of getting 7 points (3 win and one only 3 cases : drawn)  1 0   1 1 1 0  = 4C3 (.5)3 (0.05) = 0.0250 , ,  0 1   0 1 1 1        Probability for getting 8 points (all 4 win) 3 = (.5)4 = 0.0625 Hence the required probability = . 16 Required probability (getting at least 7 points i.e, 48. (c) Required probability 7 points or 8 points) = P (yellow in Ist toss) × P(Red in 2nd toss) × = 0.0250 + 0.0625 = 0.0875. P (Blue in 3rd toss) 54.  (b) Probability of winning of Shyam 3 C1 2 C1 1 C1 3 2 1 1 1 1 1 = 6 × 6 × 6 5 1 5 5 5 1 × × × × = × + × × × + ... C1 C1 C1 = 6 6 6 = 2 3 6 = 36 . 6 6 6 6 6 6 49. (a) In a single throw the favourable points are 2, 3, 4 2  5  25  25  and 5 whose number is 4. + + + ... 1    = 36   36  36  ∴ All possible outcomes are 6 ∴ P = Probability that in a single throw the 5 1 5 36 5 ⋅ = × = = minimum face value is not less than 2 and the 36 1 − 25 / 36 36 11 11 maximum face value is not greater than 5 = 5 4 2 Mathematical expectation for Shyam = Rs. × 88 = 11 6 3 = Rs. 40. Since the dice is rolled four times and all the four throws are independent events therefore the required 55.  (a) Total ways of selecting 2 persons = 13C ways. 2 probability 4 F  avourable ways for selection of no woman i.e, both 2 16 8 selected are men in C ways. =   = . 2 3 81 ⇒ P (B ∩ C) = 1.2 – P (A ∪ B ∪ C) ∵   0.85 ≤ P (A ∪ B ∪ C) ≤ 1

927

 1  1 2  1 + +   + ...  3  3  

Probability

1 = 2

928

∴P  robability of selecting two persons without a single woman is C 14 = 13 2 = C2 39 8

⇒ q = 1/2 and p 1/2. ∴ n ×

Objective Mathematics

2

Hence required probability = P (at least one woman is selected) 14 25 =1– . = 39 39 56. (c) 3 vertices out of 6 can be chosen in 6C3 ways =

61. (c) np = 4 and npq = 2

6! 6×5×4 = = 20 3! 3! 3 × 2 ×1

∴  Favourable ways = 2

62. (b) Let A, B, C be the three rifle-man. We have P (A) = 0.4, P (B) = 0.5 and P (C) = 0.8 P (exactly two succesess) = P (ABC) + P (ABC) + P ( A BC) + = + =

P ( A ) ⋅ P (B) ⋅ P (C) 0.4 × 0.5 × (1 – 0.8) + 0.4 × (1 – 0.5) × 0.8 (1 – 0.4) × 0.5 × 0.8 0.040 + 0.160 + 0.240 = 0.44.

63. (a) P (E/F) + P (

2 1 = . 20 10

57. (b) 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 ∴ P (neither A nor B occurs)

=

15 100

15 3 = . 40 8 59. (b) Required probability =

2

5

7 C2 × 55 1 5 1 . = 7C 2     × = 6 6 6 68

60. (b) p = probability of successs (S) =

2 1 = 6 3

1 2 = 3 3 Probability that the success occurs in even number of tosses = P (FS) + P (FFFS) + P (FFFFFS) + ...∴ q = probability of failure (F) = 1 –

= qp + q3p + q5p + ... =

P (E ∩ F) P (E ∩ F) + P (F) P (F)

=

P (E ∩ F) + P (E ∩ F) P (F)

[ E ∩ F and

∩ F are disjoint]

=

P (S ∩ F} P{(E ∪ E) ∩ F} = P (F) P (F)

=

P (F) = 1. P (F)

64. (c) Here random experiment is : Seating 6 boys and 6 girls in a row.

P (A ∩ B) 15 / 100 = P (A) 40 / 100

P (B | A) =

=



40 25 , P (B) = 100 100

P (A ∩ B) =

 | F)

P {(E ∩ F) ∪ (E ∩ F)} P (F)

= P ( A ∪ B ) = 1 – P (A ∪ B) = 1 – 0.61 = 0.39. 58. (b) P (A) =

6

∴ P(2) = 8C2  1   1  = 7 . 2 2 64

= P (A) ⋅ P (B) ⋅ P ( C ) + P (A) ⋅ P ( B ) ⋅ P (C)

 nly 2 equilateral triangles can be formed, ∆AEC O and ∆BDF.

Required probability =

1 =4⇒n=8 2

qp 1 − q2

2 / 3 ×1/ 3 2/9 2 9 2 × = = = = . 1 − (2 / 3) 2 1− 4 / 9 9 5 5

Let S = the sample space and E = the event that all the 6 girls sit together. Then n (S) = Total number of ways of seating 6 boys and 6 girls in row = 12! Regarding all the six girls as one person, we have only 7 persons. These 7 person can be seated in a row in 7! ways. But the 6 girls can be arranged among themselves in 6! ways ∴ n (E) = number of ways of seating 6 boys and 6 girls in a row so that all the 6 girls sit together = 7! 6! ∴  Required probability, P (E) = =

n (E) 7! 6! = n (S) 12!

720 1 . = 12 × 11 × 10 × 9 × 8 132

65. (b) Prime numbers are 2, 3, 5, 7 ∴ P(E) = 0.23 + 0.12 + 0.20 + 0.70 = 0.62

∴  P (F′) = 1 – P (F) = 0.90 and let P (Y) = Probability of new component to last for one year Obviously the two events are mutually exclusive and exhaustive. ∴ P (Y/F) = 0 and P (Y/F′) = 0.99

66. (d) T he probability that Mr A selected the loosing horse 4 3 3 × = = 5 4 5

Y Y ∴ P (Y) = P (F) ⋅  P   + P (F′) ⋅ P   F  F' 

The probability that Mr A selected the winning horse 3 2 =1– = . 5 5 67. (b) Here random experiment is: Placing of 4 letters in 4 envelopes.

= 0.10 × 0 + 0.90 × 0.99 = 0 + (0.9) (0.99) = 0.891. 71.  (b) Given P (A) = 0.5, P (B) = 0.3 and P (C) = 0.2 ∴ P (A) + P (B) + P (C) = 1 So the events A, B, C are exhaustive. If P (E) = Probability of introducing a new product, then as given

Let S = the sample space and E = the event that all the 4 letters are placed in the corresponding envelopes. ∴ E' = the event that at least one letter is placed in the wrong envelop = the event that all the 4 letters are not placed in the right envelope. Then n (s) = 4! and n (E) = 1 n (E) 1 1 = = ∴ P (E) = n (S) 4! 24 ∴ Required probability, P (E′) = 1 – P (E) =

23 . 24

68. (a) Let the three student be P, Q and R  et A, B, C denote the events of stading first of the L three students P, Q, R respectively Given, odds in favour of A = 1 : 2 odds in favour of B = 2 : 5 and odds in favour of C = 1 : 7 1 1 2 2 = , ∴  P (A) = = , P (B) = 1+ 2 3 2+5 7

P (C) =

1 1 = 1+ 7 8

Since events A, B, C are mutually exclusive ∴ P (A ∪ B ∪ C) = P (A) + P (B) + P (C) 1 2 1 = + + 3 7 8 =

56 + 48 + 21 125 = 168 168

69. (c) Let P (BC) = x Since P (A ∪ B ∪ C) = P (A) + P (B) + P (C) – P (AB) – P (BC) – P (CA) + P (ABC) ∴  = 0.3 + 0.4 + 0.8 – 0.08 – x – 0.28 + 0.09 = 1.23 – x But given P (A ∪ B ∪ C) ≥ 0.75 and P (A ∪ B ∪ C) ≤1 ∴  0.75 ≤ 1.23 – x ≤ 1 ⇒ – 0.75 ≥ – 1.23 + x ≥ – 1 or 1.23 – 0.75 ≥ x ≥ 1.23 – 1 or 0.23 ≤ x ≤ 0.48

P (E/A) = 0.7, P (E/B) = 0.6 and P (E/C) = 0.5 E E E ∴ P (E) = P (A) ⋅ P   + P (B) ⋅ P   + P (C) · P     C B A = 0.5 × 0.7 + 0.3 × 0.6 + 0.2 × 0.5 = 0.35 + 0.18 + 0.10 = 0.63 72. (b) ∵ P (E1 ∩ E2 ∩ E3 ∩ E 4 ∩ E5) 

or P (E1 E2 E3 E 4 E5) = P (E1) · P (E2/E1) · P (E3/E1E2) · P (E4/E1E2E3) · P (E5/E1E2E3E4) ∴ P (E) = =

95 94 93 92 91 × × × × 100 99 98 97 96

91 ⋅ 92 ⋅ 93 ⋅ 94 ⋅ 95  96 ⋅ 97 ⋅ 98 ⋅ 99 ⋅ 100

(∵ E = E1 E2 E3 E4 E5)

73.  (b) Here number of trials = n p = probability of occurrence of head when coin 1 1 1 is tossed once = and q = 1 – p = 1 – = 2 2 2 Now probability of occurrence of head exactly r times in n tirals is given by, P (r) = nCr prqn – r ∴ Probability of occurrence of head odd number of times in n tirals = P (1) + P (3) + P (5) + ... 3 n−3 1 n −1 1 1 = nC1  1   1  + nC 3     + ... 2 2 2 2 1 =   2

n

[nC1 + nC3 + nC5 + ...] n

1 =   × 2 n – 1 2 =

1 . 2

[∵ nC1 + nC3 + ... = 2n –1]

929

70. (a) Given : probability that electronic component fails when first used = 0.10 i.e., P (F) = 0.10

P(F ) = P(X = 1) + P(X = 2) + P(X = 3) = 0.15 + 0.23 + 0.12 = 0.5 P(E ∩ F ) = P(X = 2) + P(X = 3) = 0.23 + 0.12 = 0.35 P(E ∪ F) = P(E) + P (F ) – P(E ∩ F) 0.62 + 0.5 – 0.35 = 0.77.

Probability



930

74. (b) Since the probability of the faces are proportional to the numbers on them, we can take the probabilities of faces, 1, 2,... 6 as k, 2k, ..., 6k, respectively.

Objective Mathematics

Since one of the faces must occur, we have k + 2k + 3k + 4k + 5k + 6k = 1 1 or k = . 21 ∴ The probability of an even number

= 2k + 4k + 6k = 12k 1 4 = . = 12 × 21 7

Now the assembled product will not be defective if both the parts are not defective and hence the probability of this event. 95 19 19 × 19 361 × = = . = 104 20 104 × 4 416 80. (a) Let the chance of the second event be p. Then the chance of the first event is p 2. ∴ Odds against the first event are as 1 – p2 : p2 and odds against the second event are 1 – p : p

75. (c) Besides the ground floor, there are seven floors. The total number of ways in which each of the five persons can leave cabin at any of the 7 floors = 75.

Hence according to the condition given in the question, we have 1 − p2 1 − p  = p2  p 

3

(1 − p )(1 + p ) (1 − p )3 = p2 p3

And the favourable number of ways, that is, the number of ways in which 5 persons leave at different floors is 7P5.

or p (p + 1) = (1 – p)2

∴ The required probability = 7P5/75.

or p2 + p = p2 – 2p + 1 or 3p = 1

76. (a) Required probability =

4 5

 3 3  4 1 −  + 1 −  4 4 5

4 1 3 1 7 . ⋅ + ⋅ = 5 4 4 5 20 77. (c) n = total no. of ways = 210 = 1024. =

Since each answer can be true or false. and m = favourable number of ways = 10C8 + 10C9 + 10C10 = 45 + 10 + 1 = 56. since to pass the exam, he must give 8 or 9 or 10 true answers, Hence, m 56 7 . = = n 1024 128 78. (b) T he total number of ways of ticking the answers in any one attempt = 24 – 1 = 15. p=

It is reasonable to assume that in order to derive maximum benefit, the three solutions which he will submit must be all different. ∴  n = total no. of ways =

15

C3

m = the no. of ways in which the correct solution is excluded = 14C3. Hence the required probability 14 C3 4 1 = 1 – 15 = 1 − = . C3 5 5 79. (b) Since in the process of manufacture of part X, 9 out of 104 parts may be defective, therefore the probability of X to be defective = 9/104 and so its not being defective = 1 – (9/104) = 95/104. Similarly the probability of the part Y to be defective = 5/100 = 1/20 and not to be defective = 1 – (1/20) = 19/20.

or

∴ p =

1 1 and p2 = 3 9

1 9 1 and the probability of the second event = . 3 81. (b) Let S = the sample space, then n (S) = 2 m Hence the probability of the first event =

Let E1 = the event that one persons gets a head whereas each of the remaining (m – 1) persons get a tail and E2 = the event that one persons gets a tail whereas each of the remaining (m – 1) persons get a head. The required event E = E1 ∪ E2 ∴  n (E) = n (E1) + n (E2) = mC1m – 1 Cm – 11· 1m – 1 + mC1m – 1 Cm – 11.1m – 1 = m + m = 2m n ( E ) 2m m = = . ∴ Required probability, P (E) = n (S) 2m 2m − 1 82. (b) Let S be the sample space and E be the required event. Now, n (S) = total total number of cases = 45 = 1024. and n (E) = coefficient of x15 in (x0 + x + x10 + x11)5



= coefficient of x15 in [(1 + x)5 (1 + x10)5] = coefficient of x15 in (1 + 5x + 10x2 + 10x3 + 5x4 + x5) × (1 + 5x10 + 10x20 + ...) = 5 n (E) 5 ∴ Required probability, P (E) = n (S) = 1024 .

= 1 – P (X = 0, Y = 0) = 1 – P (X = 0) P (Y = 0) [ ∵ X and Y are independent] 5

1 1 = 1 – 5C0   · 7C0   2 2

7

1 = 1 –   2

12

=

4095 . 4096

84. (a), ( b) Putting n = 99 and p = 1/2, we have (n + 1) 1 p = (100)   = 50, and (n + 1) p – 1 = 49 2 For maximum value of P (X = r), (n + 1) p –­ 1 ≤ r ≤ (n + 1) p ⇒ 49 ≤ r ≤ 50 ∴ r = 49 and 50 Hence the maximum value of P (X = r) occurs at r = 50 and 49. 85.  (b)

P (X = r ) = P (X = n − r ) =

1  For  − 1 p 

C r p r (1 − p ) n − r n C n − r p n − r (1 − p ) r

⇒

to be independent of n and r,

1 =2 ⇒p= p

931



= 11C6 p6 q5 + 11C5 p5q6 = 11C5 (p6 q5 + p5 q6)



=

11⋅10 ⋅8 ⋅ 7 {(0 · 4)6 (0 · 6)5 + (0 · 4)5 (0 · 6)6} 1 ⋅ 2 ⋅ 3⋅ 4 ⋅ 5

11.10.9.8.7 (0.24)5 = 0 · 37. 1.2.3.4.5 Hence the required probability = 0.37.

=

88. (c) Given: mean np = 2  and

...(1)

Variance npq = 1

...(2)

1 Dividing (2) by (1), then q = 2 1 ∴  p = 1 – q = 2 1 From (1), n × = 2, 2 ∴

n=4

2

1 1 = 4C 2     2 2 1 −1 = 1 p

.

86. (a) Out of the numbers 00, 01, 02, ..., 99, those numbers the product of whose digits is 18 are 29, 36, 63, 92 i.e, only 4. p = P (E) =

∴ Required probability = P (X = 6 or X = 5)

4

Now, P (X > 1) = P (X = 2) + P (X = 3) + P (X = 4)

n − 2r

n − 2r

and q = Probability of failure = 0.6

1 1 The binomial distribution is  +  2 2

n

 (1 − p ) n − 2 r  1 =  − 1 pn − 2r p 

∴ p = Probability of success = 0.4

4 1 1 24 = = , q = P ( E ) = 1 – 100 25 25 25

=

2

1

3

1 1 1 + 4C 3     + 4C 4   2 2 2

4

6 + 4 + 1 11 = . 16 16

89. (c) P (A/B) = P (A ∩ B) P (B) =

P (A ∪ B) 1 − P (A ∪ B) = . P (B) P (B)

90. (c) n = Total number of ways = 65

To find the favourable no. of ways, a total of 12 in 5 throws can be obtained in the following two Let X be the random variable, showing the number ways only : of times E occurs in 4 selectios. (i) One blank and four 3′s. Then P (E occurs at least 3 times) = P (X = 3 or X = 4) or (ii) Three 2′s and two 3′s. = P (X = 3) + P (X = 4) The no. of ways in case (i) = 5C1 = 5 4 3 1 4 4 0 3 4 = C3 p q + C4 p  q = 4p q + p and the no. of ways in case (ii) = 5C2 = 10. 3 4 ∴ m = the favourable no. of ways. 1 24  1  97 = 4 ×   × . +  =  25  = 5 + 10 = 15. 25  25  390625 15 5 Hence the required probability = 5 = . 87. (a) Since the man is one step away from the starting 6 2592 point, ∴ either 91. (d) Let A be the event of obtaining an even sum and (i)  man has taken 6 steps forward and 5 steps B the event of obtaining a sum less than five. backward. Then we have to find P (A ∪ B). Since, A, B are or (ii)  man has taken 5 steps forward and 6 steps not mutually exclusive, we have P (A ∪ B) = P (A) backward. + P (B) – P (A ∩ B) Taking, movement 1 step forward as success and 1 18 6 4 5 = + − = step backward as failure. 36 36 36 9

Probability

1 83. (a) For random variable X, n = 5, p = and for 2 1 random variable Y, n = 7, p = 2 Now P (X + Y ≥ 1) = 1 – P (X + Y < 1) = 1 –­P (X + Y = 0)

932

Objective Mathematics

since there are 18 ways to get an even sum and 6 ways to get a sum < 5, viz. (1, 3), (3, 1), (2, 2), (1, 2), (2, 1), (1, 1) and 4 ways to get an even sum less than 5, namely, (1, 3), (3, 1), (2, 2), (1, 1). 92. (a) Required probability =

39 52

C1 × C1

39 52

C1 × C1

13 52

C1 3 3 1 9 . = × × = C1 4 4 4 64

93. (b) ∵ T  otal no. of ways = 3! = 6 and favourable no. of ways = 1. 1 . ∴ Required probability = 6 94. (a) Let A be the event that the sale is at least 20 televisions, i.e., 20, 21, 22, ... and B be the event that sale is less than 24, i.e., 0, 1, 2, 3, ..., 23. Then A ∩ B will denote the sale of 20 21, 22 and 23 televisions. We are given P (A) = 0.45 and P (B) = 0.74. It is required to find P (A ∩ B). Also P (A ∪ B) = P (sale of 0, 1, 2, 3, ..., 20, 21, 22, 23, televisions) = P (S) = 1. From addition rule, required probability is P (A ∩ B) = P (A) + P (B) – P (A ∪ B) = 0.45 + 0.74 – 1 = 0.19. 95. (c) Since A, B and C are mutualiy exclusive and exhaustive events, P (A ∪ B ∪ C) = P (A) + P (B) – P (C) = 1 ⇒ 2 P (B) + P (B) + 3P (B) = 1

[ ∵ P (A) = 2P (B), P (C) = 3 P (B)]

⇒ 6P (B) = 1 ∴ P (B) = 1/6.

Hence probability that stock price will go down is given by P  (A ∩ B) = 1 – P (A ∪ B) = 1 – (7/12) = 5/12. 98. (a) Let A denotes the event that the salesperson selected is unmarried and B the event that he is a college graduate. Then in usual notations, we are given : 52 72 , P (B) = , P (A ∩ B) = P (A) = 100 50 3 52 39 × = 4 100 150 The probability that a salesperson selected at random will be neither single nor a college graduate is ;



P  (A ∩ B) = 1 – P (A ∪ B) = 1 – {P (A) + P (B) – P (A ∩ B) 72 39  13  52 + − = = 1−  .  150 150 150  30 99. (c) Let A and B denote the events that the person is selected in firms X and Y respectively. Then in the usual notations, we are given : P (A) = 0.7 ⇒ P ( A ) = 1 – 0.7 = 0.3 P ( B ) = 0.5 ⇒ P (B) = 1 – 0.5 = 0.5 and P (A ∩ B) = 0.6 The probability that the persons will be selected in one of the two firms X or Y is given by : P (A ∪ B) = 1 – P (A ∩ B) = 1 – {P ( A ) + P ( B ) – P (A ∪ B) } = 1 – (0.3 + 0.5 – 0.6) = 0.8.

100.  (c) Let A and B respectively denote the events of the 96. (a) A leap year consists of 366 days, i.e., 52 full weeks, first person and second person living 30 years hence. and two over-days. These two over-days can be Then any one of the following possible outcomes ; (i) 5 5 3 3 = , P (B) = = Monday and Tuesday, (ii) Tuesday and Wednesday, P (A) = 8 + 5 13 4+3 7 (iii) Wednesday and Thursday, (iv) Thursday and 8 4 Friday, (v) Friday and Saturday (iv) Saturday and P ( A ) = and P ( B ) = Sunday, (vii) Sunday and Monday. 13 7 The event that at least one of the persons is alive Let A and B be the events that a leap year contains 30 years hence is given by A ∪ B. 53 Thursdays and Fridays respectively. Then 2 2 1 , P (B) = and P (A ∩ B) = 7 7 7 Required probability is given by: P (A) =

P (A ∪ B) = P (A) + P (B) – P (A ∩ B) 2 2 1 3 + − = . 7 7 7 7 97. (b) Let A denote the event ‘stock price will go up’, and B be the event ‘stock price will remain same’. =

Then P (A) = 1/3 and P (B) = 1/4. ∴P   (Stock price will either go up or remain same) is : P (A ∪ B) = P (A) + P (B) =

1 1 7 + = 3 4 12

∴ required probability = P (A ∪ B) = 1 ­– P [none of A and B is alive 30 years hence]

= 1 – P (A ∩ B) = 1– P ( A ) · P ( B )



[By multiplication rule, since A and B and consequently A and B are independent] 8 4 59 = 1− × = . 13 9 91 101. (c) Let us define the events : W : Player wins on Monday F : Track is fast S : Track is slow We have given : P (F) = 0.7, P (S) = 0.3, P (W/F) = 0.3, P (W/F) = 0.3, P (W/S) = 0.4.

102.  (b) Let W denote the event that A draws a white ball and T the event that A speaks truth. In the usual 1 , notations, we are given that P (W) = 9 P (T \ W) =

5 so that P ( 6

)=1–

1 8 = and 9 9

5 1 = . 6 6 Using Baye’s theorem required probability is given by P (W ∩ T) P (W\T) = P (T)

P (T \

) = 1−

P ( A 1 ∩ A 2) = P ( A 1) · ( A 2) (By multiplication rule, since A1 and A2 are independent and consequently A 1 and A 2 are independent) ∴ P ( A 1 ∩ A 2) = 0.91 × 0.95 = 0.8645. 106. (b) Let one of the quantities be x. Thus the other is 2n – x. Then product will be greatest when they are equal i.e., each is n in which case the product is n2. By the given condition­

3 2 n 4 2 ⇒ 8nx – 4x ≥ 3n2 ⇒ 4x2 – 8nx + 3n2 ≤ 0 P (W) × P (T\W) n 3 = ⇒ (2x – 3n) (2x ­– n) ≤ 0 ⇒ n. ≤x≤ P (W) × P (T\W) + P (W) × P (T\W) 2 2 (1\9) × (5\6) 5 3n n = n. ∴ favourable number of cases = = . − = (1\9) × (5\6) + (8\9) × (1 \ 6) 13 2 2 n 1 103. (a) There are 7 distinct letters in the word ‘longest’ and Hence required probability = = . 2 n 2 they can be arranged among themselves in 7! ways, 107. (a) Let A be the event that a really able candidate passes which gives the exhaustive number of cases. the test and let B be the event that any candidate In the word ‘longest’ there are 2 vowels, viz. o, e and passes this test. Then we have 5 consonants l, n, g, s, t. These 2 vowels are to be P (B/A) = 0.8, P (B/Ac ) = 0.25 placed in 4 odd places, viz, Ist, 3rd, 5th, and 7th and P (A) = 0.4, P (Ac ) = 1 – 0.4 = 0.6 this can be done in 4C2 ways. Further, these 2 vowels By Baye’s formula can be arranged among themselves in 2! ways and the remaining 5 consonants can be arranged among P (A) P (B/A) themselves in 5! ways. Amongst all these operations, P (A/B) = P (A) P (B/A) + P(A c ) P (B/A c ) total number of favourable cases for the consonants to occupy only odd positions is : 4C2 × 2! × 5!. Hence the 0.32 32 = = = 68%. required probability 0.32 + 0.15 47 4 C2 × 2 ! × 5! 6 × 2 × 5 ! 2 = = = . 108. (a) Required probability 7! 7 × 6 × 5! 7 7×6×5 7 104.  (b) Since n persons can be seated in n chairs at a round C3 4 1 1× 2 × 3 = = . table in (n – 1) ! ways, the exhaustive number of = 8 = C4 8 × 7 × 6 × 5 8 2 cases = (n – 1) !. 1× 2 × 3× 4 Assuming the two specified persons A and B who 109. (b) The number of numbers whose sum is 9 is: sit togehter as one, we get (n – 1) persons in all, who can be seated at a round table in (n – 2) ! ways. One digit number = 1 Further, since A and B can interchange their posiTwo digit numbers = 9 tions in 2! ways, total number of favourable cases 9 × 10 Three digit numbers = 9 + 8 + 7 + ... + 1 = of getting A and B togehter is (n – 2) ! × 2!. 2 ( n − 2) ! × 2 ! 2 . ∴ Required probability = = = 45 (n − 1) ! n −1 55 11 = . ∴ Required probability = 1000 200 105. (a) Let A1 be the event that manufacturing defect occurs in the part A of the article and A 2 denote the event 110. (d) Probability for an incorrect digit = p that the part B of the article is defective. Then we ∴ Probability for a correct digit = 1 – p are given : ∴ Probability for 8 correct digits = (1 – p)8 9 = 0.09 ⇒ P ( A 1) =1 – P (A1) = 0.91. P (A1) = 100 Hence required probability = 1 – (1 – p)8. x (2n – x) ≥

933

5 = 0.05 ⇒ P ( A 2) =1 – P (A2) = 0.95. 100 The assembled part will not be defective if both the parts A and B are non-defective, i.e., the event A 1 ∩ A 2 occurs. Thus the probability that assembled part is not defective is given by : P (A2) =

Probability

∴ P (player X will win on Monday) = P (W) = P (W ∩ F) + P (W ∩ S) = P (F) × P (W/F) + P (S) × P (W/S) = 0.7 × 0.3 + 0.3 × 0.4 = 0.33.

934

111. (a) We must have 0.1 + 2k + k + 0.2 + 3k + 0.1 = 1

Objective Mathematics

or 6k = 1 – 0.4 = 0.6 0.6 1 = 0.1. = ∴ k = 6 10

P (A/B) + P ( A /B) = P (A) + P ( A  ) = 1. ∴  (d) is also true. 118. (c) As A and B are independent events. P (A ∩ B) = P (A) P (B) and

112. (d) The probability that a man does not die in a year

P (A′ ∩ B) = P (A′) P (B)

=1– p

Therefore, P (A) P (B) = 3/25

∴ Probability that none of the n men dies in a year = (1 – p)n

and [1 – P (A)] P (B) = 8/25 11 3 . Thus P (A) = . ⇒ P (B) = 25 11

∴ The probability that at least one man dies in a year = 1 – (1 – p)n Since every man can die first, the chance that A1, 1 . Hence the probability that A1, will die first is n will die within a year and he will be first to die

119. (c) P (B | A ∪ B′) =

P[(B ∩ A) ∪ (B ∩ B′ )] = P (A) + P(B′ ) − P (A ∩ B′ )

1 [1 – (1 – p)n]. n 113. (d) Required probability =

=

But P [(B ∩ A) ∪ (B ∩ B′)

114. (a), (b), (c), (d) P (A ∩ B) = P (A) + P (B) – P (A ∪ B)

P (A) + P (B) . 2 115. (a), (b) P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= P (B ∩ A) = P (A) – P (A ∩ B′) = 0.2

and P (A) + P (B′  ) – P (A ∩ B′) = 0.7 + 0.6 – 0.5 0.2 1 = . = 0.8 ∴ P (B | A ∪ B′) = 0.8 4

10 9 10 1 × 5 5 × × = = . 20 19 18 19 × 2 38

∵ P (A ∪ B) = 1. So P  (A ∩ B) = P (A) + P (B) – 1 So P (A ∩ B) / P (A) + P (B) P (A) + P (B) = P (A ∪ B) + P (A ∩ B) P (A ∪ B) ≥ P(A ∩ B). 2P (A ∩ B) ≤ P (A) + P (B)

120. (a) As

1+ 4p 1− p 1− 2p , and are probabilities of , 4 2 2

three mutually exclusive events, we must have 0≤

1− p 1− 2p 1+ 4p ≤ 1, 0 ≤ ≤ 1, 0 ≤ ≤1 2 2 4

and 0 ≤

⇒ P (A ∩ B) ≤

⇒ −

So P (A ∪ B) ≤ P (A) + P (B) Also P (A ∩ B) = P (A) + P (B) – P (A ∪ B) So P (A ∩ B) ≤ P (A) + P (B).



116. (c) Given P (A) = 0.6, P (B) = 0.4, P (C) = 0.5,

P[B ∩ (A ∪ B′ )] P (A ∪ B′ )

−

1+ 4p 1− p 1− 2p + + ≤1 4 2 2

1 3 ≤ p ≤ , – 1 ≤ p ≤ 1, 4 4 1 1 1 5 ≤ p≤ and ≤ p ≤ 2 2 2 2

−1 1   1 3 1 5 ⇒ max  − , −1, ,  ≤ p ≤ min   , 1, ,  2 2  4 4 2 2

1 ⇒ 1/2 ≤ p ≤ 1/2 ⇒ p = . P (A ∪ B) = 0.8, P (A ∩ C) = 0.3, 2 P (A ∪ B ∪ C) = 0.2 121. (c), (d)  P (A ∪ B) = P (A) + P (B) – P (A) P (B) and 0.85 ≤ P (A ∪ B ∪ C) ≤ 1. Now P (A ∩ B) = P (A) + P (B) – P (A ∪ B) = 0.2 ⇒ P (A) + P (B) – P (A ∩ B) P (A ∪ B ∪ C) = 0.6 + 0.4 + 0.5 – 0.2 – 0.3 – p1 = P (A) + P (B) – P (A) P (B) + 0.2 ⇒ P (A ∩ B) = P (A) P (B) ⇒ p1 = 1.2 – P (A ∪ B ∪ C). ∴ A and B are independent events. Since 0.85 ≤ P (A ∪ B ∪ C) ≤ 1, ⇒ P (B ∩ A′) = P (B) P (A′ ) ≠ P (B) – P (A) we have 0.2 ≤ p1 ≤ 0.35. 117. (b), (c), (d)  Since mutually exclusive events can not be independent and vice versa so (a) is not true. Obviously (b) and (c) are true by definition and since P (A/B) = P (A), P ( A /B) = P ( A ),

and P (A ′ ∪ B ′ ) = P (A ∩ B)′  = 1 – P (A ∩ B) = 1 – P (A) P (B) ≠ 1 – P (A) + 1 – P (B) = P (A′ ) + P (B′ ). Also P [(A ∪ B)′ ] = P (A′ ∩ B′ ) = P (A′ ) P (B′ ) Since A and B are independent P (A | B) = P (A).

m − times

= nm. m – times The number of ways in which all passengers can alight at different floors is nCm × m ! = n Pm. n

Hence, required probability =

Pm . nm

123. (b) Since each object can be given to any one of ten persons. So, ten objects can be distributed among 10 persons in 1010 ways. Thus, the total number of ways = 1010. The number of ways of distribution in which each gets only one thing is 10 !. So, the number of ways of distribution in which at least one of them does not get any thing is 1010 – 10 !. 1010 − 10! . Hence, required probability = 1010 124. (c) The total number of ways in which 5 coins can fall = 25 = 32 The favourable number of ways = co-eff. of x12 in the expansion of (x2 + x3)5 = co-eff. of x12 in x10 (1 + x)5 = co-eff. of x2 in (1 + x)5 = 10. 10 5 . = Hence the required probability = 32 16 125. (a) The total number of ways in which 4 tickets can be drawn 5 times = 45 = 1024. Favourable number of ways of getting a sum of 23 = co-eff. of x23 in (x00 + x01 + x10 + x11)5 = co-eff. of x23 in [(1 + x) (1 + x10)]5 = co-eff. of x23 in (1 + x)5 (1 + x10)5 = co-eff. of x23 in (1 + 5x + 10x2 + 10x3 + 5x4 + x5) × (1 + 5x10 + 10x20 + 10x30 + ...) = 100 ∴ required probability = 126. (a) S. D σ =

npq = 16 ×

100 25 = . 1024 256

1 1 × = 4 = 2. 2 2

127. (a) The total number of ways of choosing 2 persons n (n − 1) . out of n in nC2 = 2 The no. of ways in which two chosen persons are together is (n – 1) ∴ the no. of favourable ways = nC2 – (n – 1) =

n (n − 1) n (n − 1) (n − 2) – (n – 1) = 2 2

=

(n − 1) (n − 2) / 2 n (n − 1) / 2

n−2 2 =1– . n n

128. (a), (b), (c)  P (exactly one of A and B occur) = P{(A ∩ B) or (A ∩ B)} = P (A ∩ B) + P (A ∩ B) = P (A) – P (A ∩ B) + P (B) – P (A ∩ B) = P (A ∪ B) – P (A ∩ B) = 1 − P (A ∩ B) − 1 + P (A ∪ B) = P (A) + P (B) − 2P (A ∩ B) . 129. (a), (b) P (E1) = 2/4 = 1/2, P (E2) = 2/4 = 1/2. Similarly P (E3) = 1/2 P (E1 E2) = 1/4, P (E1 E3) = 1/4 P (E2 E3) = 1/4, P(E1 E2 E3) = 1/4 ∴ P (E1 E2) = P(E1) · P(E2) P(E2 E3) = P(E2) ⋅ P(E3) P (E1 E3) = P(E1) · P(E3). ∴ E1, E2, E3 are pair-wise independent, so (a) is correct. Also P (E1E 2 ) = 1/4 = P (E1 ) ⋅ P (E 2 ) So E 1 and E 2 are independent, so (b) is also true. P (E 2 E 3 ) = 1/4 = P (E 2 ) ⋅ P (E 3 ) ∴  E 2 and E 3 are independent, i.e., (c) is not correct. Finally P (E1 E2 E3) = 1/4 ≠ P (E1) · P (E2) · P (E3). So, E1, E2, E3, are not mutually independent. 130. (d), (b)  P (A) = P (A | B) = P (A ∩ B)/P (B) ⇒ P (A ∩ B) = P (A) P (B) Thus, A and B are independent. Also, P (A ′  | B) =

P (A′ ∩ B) P (B) − P (A ∩ B) 3 = = . P (B) P (B) 4

131. (a) There are four gaps in between the girls where the boys can sit. Let the number of boys in these gaps be 2a + 1, 2b + 1, 2c + 1, 2d + 1 = 0, then 2a + 1 + 2b + 1 + 2c + 1 + 2d + 1 = 10 or a + b + c + d = 3 The number of solutions of above equation = coefficient of x3 in (1 – x)– 4 = 6C3 = 20 Thus boys and girls can sit in 20 × 10 ! × 5 ! ways Total ways = 15 ! 20 × 10! × 5! Hence, the required probability = . 15!

935

∴ required probability =

Probability

122. (b) Since a person can alight at any one of n floors. Therefore, the number of ways in which m pasn × n × n × ... × n   sengers can alight at n floors is 

936

Objective Mathematics

132. (c) L  et A denote the event that the target is hit when 135. (a) Since P (A ∪ B ∪ C) ≥ 0.75, therefore x shells are fired at point I. Let E1 (E2) denote the 0.75 ≤ P (A ∪ B ∪ C) ≤ 1 event. ⇒ 0.75 ≤ P (A) + P (B) + P (C) – P (A ∩ B) 8 1 – P (B ∩ C) – P (A ∩ C) , P (E2) = . We have P (E1) = 9 9 + P (A ∩ B ∩ C) ≤ 1 x 1 ⇒ 0.75 ≤ 0.3 + 0.4 + 0.8 – 0.08 ⇒ P (A/E1) = 1 –   and P (A/E2) = 1 2 – P (B ∩ C) – 0.28 + 0.09 ≤ 1 21 − x ⇒ 0.75 ≤ 1.23 – P (B ∩ C)) ≤ 1 –  1  2 x 21 − x ⇒ – 0.48 ≤ – P (B ∩ C) ≤ – 0.23  8  1  1  1 Now P (A) = 1 −    + 1 −    ⇒ 0.23 ≤ P (B ∩ C) ≤ 0.48. 9   2   9   2   136. (c) Any month out of 12 months, can be chosen with x  1   1  21 − x  1 dP (A) 8   1  probability = . ⇒ log 2 =    log 2 +  −   12   dx 9   2  9 2    There are 7 possible ways in which the month can start and it will be a Friday on 13th day if the first dP (A) Now, we must have =0 1 dx day of the month is Sunday. Its probability = . 7 2 P (A) d ⇒ x = 12, also 0 . p

and r if

1 1 . − 1 = 1 or p = p 2

139. (d) We have E (X) = np = 2

5 Again, P (B) = + P (A ∩ B) 12

var (X) = npq = 1 1 ∴ q= and 2

⇒ P (B) ≤ 5/12 + P (A) [∵ P (A ∩ B) ≤ P (A)]

5 1 3 ⇒ P (B) ≤ + = 2 3 4

cr p r (1 − p ) n − r (1 − p ) n − 2 r = n−r r cn − r p (1 − p ) pn − 2r

∴ the ratio will be independent of n

⇒



n

n

1 − p  =   p 

1 3 = + P (B) − P (A ∩ B) 3 4

5

P (X = r ) P (X = n − r )

138. (a) We have

⇒ x + y = n] ∴ P (X = 2) =

4

...(ii)

From (i) and (ii), we obtain 5/12 ≤ P (B) ≤ 3/4 Hence, x = 5/12 and y = 3/4.

∴ p = 1 –

1 1 = . 2 2

1 ∴ n    = 2 ⇒ n = 4. 2 Now P (X > 1) = 1 – P (X ≤ 0)

1 = 1 –   2

3

 6 and P (X ≤ 6) =   .  20 

4

5 11 = (1 + 4) = 1 – . 16 16 140. (c) P (correct prediction) = 1/3

Then the required probability 3

P (wrong prediction) = 2/3 P(exactly 4 right predictions) = 7C4 (1/3)4 · (2/3)3 = 280/37. 141. (a) Both heads appear on n coins and head and a tail appear on (n + 1) coins so P (head) =

n

C1 ⋅1 + 2n +1 C1

n +1

C1 1 31 ⋅ = 2n +1 C1 2 42

n n +1 31 ⇒ + = 2n + 1 2 (2n + 1) 42 ⇒ 2n + n + 1 = (2n + 1) (31/21) ⇒ 63n + 21 = 62n + 31 ⇒ n = 10. 142. (b) T he probability of hitting the target 5th time at the 10th throw = P (the probability of hitting the target 4 times in the first 9 throw) × (the probability of hitting the target at the 10th throw)  9  1 4 =  C4   2 

1  2 

5

10  1 9!  1  63  =   = 9 .    4!5! 2 2  2

143. (a), (c)  P (A wins the game) = P (H or TH or TTTH or TTTTH or TTTTTTH or ...) = P (H) + P (TH) + P (TTTH) + P (TTTTH) + ... =

1 1 1 1 1 + 2 + 4 + 5 + 7 + ... 2 2 2 2 2

1 1 1 1 1  1  =  + 4 + 7 + ... +  2 + 5 + 8 + ... 2 2  2  2 2 2 =

3

7 6 = P (X = 7) =   −   .  10   10  146. (a)

4

∑ P (X = k )

=1

k =0



⇒

4

∑ Ck

2

=1

k =0

⇒ C (12 + 22 + 32 + 42) = 1 1 . ⇒ C = 30 147. (a) We have, np = 25. Now, 0 ≤ p < 1 and 0 ≤ q ≤ 1 ⇒ 0 ≤ n pq ≤ np ⇒ 0 ≤

npq ≤ np

⇒ 0 ≤ S.D. ≤ 5. But p ≠ 0, therefore 0 ≤ S.D.< 5 ⇒ S.D. ∈ (0, 5). 148. (a) P robability of getting a head in a single toss of a coin = 1/2 (= p) say Probability of getting 5 or 6 in a single throw of a die = 2/6 = 1/3 (= q) say Required probability = p + (1 – p) (1 – q) p + (1 – p) (1 – q) (1 – p) (1 – q) p + ... = p + (1 – p) (1 – q) p + (1 – p)2 (1 – q)2 p + ... =

p 1/ 2 3 = = . 1 − (1 − p ) (1 − q ) 1 − 1 / 2 × 2 / 3 4

149. (a) We have

1/ 2 1/ 4 4 2 6 + = + = . 1−1/ 8 1−1/ 8 7 7 7

P (|X – 4| ≤ 2) = P ( – 2 ≤ X – 4 ≤ 2) = P (2 ≤ X ≤ 6) = 1 – [P (X = 0) + P (X = 1) + P (X = 7)

∴  β = 1 – α = 1 – 6/7 = 1/7. 144. (a) Three squares can be chosen out of 64 squares in 64C3 ways. Two squares of one colour and one another colour can be chosen in two mutually exclusive ways : (i) two white and one black, and (ii) two black and one white. Thus, the favourable number of cases = 32C2 × 32C1 + 32C1 × 32C2 Hence, the required probability =

3

937

 7  P (X ≤ 7) =   20

2 32C1 ⋅ 32C 2 16 . = 64 C3 21

145. (d) Let X denotes the largest number on the 3 tickets drawn. Then

+ P (X = 8)] 8 8 8 8  1 1 1 1  = 1 –    8 C0   +8 C1   +8 C7   + 8C8    2 2 2  2    8

1 18 119 = 1 –   (1 + 8 + 8 + 1) = 1 – 8 = . 2 2 128 150. (d) Let X ~ B (100, p) be the number of coins showing heads and let q = 1 – p. Then, since P (X = 51) = P (X = 50), we have

100

C51 (p51) (q49) =

100

C50 (p50) (q50)

Probability

 4  1  4 4  1 3  1  C + C1      = 1 –  0  2   2   2   

938

Objective Mathematics

⇒

p  100!   51! 49!  = q  50! 50!   100! 

⇒

p 51 = 1 − p 50

⇒ p =

⇒ 50p = 51 – 51p

51 . 101

151. (a), (c)  P (A ∪ B) = P (A) + P (B) – P (A ∩ B) ∴  1 ≥ P (A) + P (B) – P (A ∩ B) ≥

3 4

As the minimum value of P (A ∩ B) = we get P (A) + P (B) – ⇒ P (A) + P (B) ≥

1 , 8

1 3 ≥ 8 4

1 3 7 + = 8 4 8

3 As the maximum value of P (A ∩ B) = , we 8 get 1 ≥ P (A) + P (B) –

3 8

⇒ P (A) + P (B) ≤ 1 +

155. (b) T he probability of drawing a number less than or 15 3 equal to 15 in a draw = = . 20 4 The probability of drawing the ticket of value 15 1 . in a draw = 20 3 ∴  the required probability = 4 C ⋅ 1 ⋅  3  = 27 .   1 20 4 320 156. (a) We know 7k , k ∈ N, has 1, 3, 9, 7 at the units place for k = 4p, 4p – 1, 4p – 2, 4p – 3 respectively, where p = 1, 2, 3, ... Clearly, 7m + 7n will be divisible by 5 if 7m has 3 or 7 in the unit place and 7n has 7 or 3 in the units place or 7m has 1 or 9 in the units place and 7n has 9 or 1 in the units place. ∴ For any choice of m, n the digit in the units place of 7m + 7n is 2, 4, 6, 0 or 8. It is divisible by 5 only when this digit is 0. 1 ∴  the required probability = . 5 157. (d) Three digit numbers are 100, 1012, ... , 999 Total numbers = 900

3 11 = . 8 8

Three digit numbers (which have all same digits)

152. (c) n (S) = the area of the circle of radius r.

111, 222, 333, ... 999 ∴ Favourable numbers = 9 ∴ Required probability =

n (E) = the area of the circle of radius ∴ the required probability

r . 2

2

r π⋅  2 1 n (E) = = . = πr 2 4 n (S) 153. (c) The required probability = (probability of drawing a prime number) × (probability of drawing more than 10 from the prime numbers) 8 4 1 = × = . 20 8 5 154. (a) The probability of getting at least 3 in a throw 4 2 = = . 6 3 ∴ the required probability 2 = 6 C3 ⋅   3 4

= 41

3

3

4

1  2 1 ⋅   + 6C 4 ⋅     3  3 3 5

2

9 1 . = 900 100

158. (a) A × B = {1, 3, 5, 7, 9} × {2, 4, 6, 8} = (1, 2), (1, 4), (1, 6), (1, 8), (3, 2), (3, 4), (3, 6), (3, 8), (5, 2), (5, 4), (5, 6), (5, 8), (7, 2), (7, 4), (7, 6), (7, 8), (9, 2), (9, 4), (9, 6), (9, 8) Total ways = 5 × 4 = 20 Favourable cases = (1, 8), (3, 6), (5, 4), (7, 2) ( ∵ a + b = 9) ∴ no. of favourable cases = 4 4 1 = . ∴ Required probability = 20 5 159. (a) P (A/B) = =

P (A ∩ B) P (B)

P (A ∩ B) 1 − P (A ∪ B) = P (B) P (B)

160. (c) We have, A = (A ∩ B) ∪ (A ∩ B) = φ ∪ (A ∩ B) = A ∩ B ⇒ A ⊆ B and ∴ P (A) ≤ P ( B ). 161. (c) We define the following events

6

2 2 1 2 + 6 C5 ⋅   ⋅ + 6 C 6 ⋅   .   3 3 3 36

A1 : He knows the answer A1 : He does not know the answer

1 4

P (E/A1) = 1, P (E/A2) =

⇒

∴ required probability = P (A2/E)

∴  P (A) +

39 39 3 3 9 . × = × = 52 52 4 4 16

⇒ x =

r

1 1 Cr     2 2

a + 3a + 5a + 7a + 9a + 11a+ 13a

10 − r

5

∴ required probability =

10

1 1 C5     2 2

5

10 × 9 × 8 × 7 × 6  1  63 63 ⋅  = 8 = .   1× 2 × 3× 4 × 5 2 2 256 10

=

165. (c) The total number of ways in which 4 tickets can be drawn 5 times = 45 = 1024. Favourable number of ways of getting a sum of 23 = co-eff. of x23 in (x00 + x01 + x10 + x11)5 = co-eff. of x23 in [(1 + x) (1 + x10)]5 = co-eff. of x23 in (1 + x)5 (1 + x10)5 = co-eff. of x23 in (1 + 5x + 10x2 + 10x3 + 5x4 + x5) × (1 + 5x10 + 10x20 + 10x30 + ...) = 100 100 25 ∴ required porbability = . = 1024 256 166. (c) We have, P (A ∩ B) = i.e., P (A ∪ B) =

1 1 and P (A ∩ B) = 6 3

1 3

1 ⇒ 1 – P (A ∪ B) = 3

5 ± 25 − 24 5 ± 1 6 = = . 12 12 12

167. (d) We must have

1 1 i.e., p = 2 2

1 ∴  q = in a toss. 2 The probability that out of 10 coins, r coins show head 10

[If x = P (A)]

⇒ 6x2 – 5x + 1 = 0

163. (b) The required probability

=

1 5 = 6 P (A) 6

⇒ 6 [P (A)]2 + 1 = 5 P (A)

mean for P. D.

164. (d) Probability for a head =

2 1 4 +1 5 + = = 3 6 6 6

Again P (A ∩ B) = P (A) P (B) 1 ⇒ = P (A) P (B) 6

1 1 ⋅ 1 1 10 4 = = = . 9 1 1 36 + 1 37 ⋅1 + ⋅ 10 10 4

=

2 1 = P (A) + P (B) – 3 6

⇒ P (A) + P (B) =

P (A 2 ) ⋅ P (E/A 2 ) = P (A1 ) P (E/A1 ) + P (A 2 ) P (E/A 2 )

162. (c) S. D =

2 3 Now P (A ∪ B) = P (A) + P (B) – P (A ∩ B) ⇒ P (A ∪ B) =



+ 15a + 17a = 1

9 (a + 17 a ) = 1 ∴ 2  [ ∵ sum of the probability = 1] 9 × 18a or =1 2 2 1 ∴ a = 9 × 18 = 81 . 168. (c) The required probability =

6 1 = . 6 × 6 × 6 36

169. (c) P (A) = P (A/B) = P (A ∩ B)/P (B) ⇒ P (A ∩ B) = P (A) P (B) Thus, A and B are independent Also, P (A′/B) =

P (A′ ∩ B) P (B) − P (A ∩ B) = P (B) P (B)

= 1 – P (A/B) = 1 –

1 3 = . 4 4

170. (c) Total no. of equally likely cases = 13C2 =

13 × 12 = 78 1× 2

Favourable cases are Women Men 1 1 2 0 ∴  number of favourable cases = 5C1 × 8C1 + 5C2 × 8C0 = 40 + 10 = 50 ∴ required probability =

10 25 . = 78 39

939

9 9 1 Thus P (A1) = , P (A2) = 1 – = 10 10 10

Probability

E : He gets the correct answer

940

171. (a) Since A, B are independent

Objective Mathematics

∴ P (A′ ∩ B′  ) = P (A′ ) P (B′ ) ⇒ 1 – P (A ∪ B ) = P (A′  ) P (B′  ) ⇒ 1 –

2 1 1 1 2 1 1 1 2

8 7 2 . =  α ⇒ α =   10 10 7

172. (c) In any number the last digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, We want that the last digit in the product is an odd digit other than 5 i.e., it is any one of the digits 1, 3, 7, 9. This means that the product is not divisible 2 or 5. The probability that a number 6 is divisible by 2 or 5 is 10 [∵ in that case the last digit can be one of 0, 2, 4, 5, 6, 8] ∴  The probability that the number is not divisible 6 4 2 = by 2 or 5 = 1 – = 10 10 5 In order that the product is not divisible by 2 or 5 none of the constituent numbers should be divisible 4 16 2 by 2 or 5 and its probability =   = . 5 625 173. (b) P (A ∩ B) = P (A ∪ B) = 1 – P (A ∪ B) 1 = 1 – P (A ∪ B) 2 1 ⇒ P (A ∪ B) = 2 1 Also P (A ∩ B) = 2 Now P (A ∪ B) = P (A) + P (B) – P (A ∩ B) 1 p 1 ∴  = + p− 2 2 2

⇒

3p = 1, 2 2 ∴ p = . 3 174. (a) Given digits = 1, 2, 3, 4, 5 Total numbers formed by these digits = 5 ! = 120 A number will be divisible by 4 if the last two digits are divisible by 4 Therefore last two digits can be 12, 24, 32, 52 They can be filled in 4 ways These four ways are possible by 3 ! ways Hence number of ways in which the number formed is divisible by 4 = 4 × 3 ! = 4 × 3 × 2 × 1 = 24 24 1 = . ∴ Required Probability = 120 5 ∴ 

175. (d) Probability = =

denominations, 2 cards from 1, and one each from other two, e.g. a b c

2

C1 ⋅ 3C1 ⋅ 4C1 9 C3

2 ⋅ 3⋅ 4 ⋅ 6 2 = . 9 ⋅8 ⋅ 7 7

176. (d) We have 13 denominations 1, 2, 3, .. , 10, J, Q, K For selecting exactly one pair, we select first any 3

Thus favourable ways = Total ways = 52C4. ∴ probability = =

13

C3  ⋅ 3 ⋅ 4C2  ⋅ 4C1  ⋅ 4C1

13⋅12 ⋅11⋅ 3⋅ 6 ⋅ 4 ⋅ 4 ⋅ 24 6 ⋅ 52 ⋅ 51⋅ 50 ⋅ 49

48 × 132 6336 = 17 × 25 × 49 20825

= 0.3042 ≈ 0.3. 177. (a) Since all player are of equal strength, with whom S1 is paired, will have no bearing on the probability of his winning. The number of ways of choosing 8 winners out of 16 : 16C8. The number of ways of choosing S1 and 7 other winners out of 15 is 15C7. ∴ Probability that S1 will win 15! 8!8! 1 C7 = = . 7!8! 16 ! 2 C8 178. (a) Let A (B) denote the event that minimum (maximum) number on the chosen tickets is 3 (7). We have P (A) = P (choosing 3 and two other numbers from 4 to 10) 7 7×6×3 C 7 = = 10 2 = 10 × 9 × 8 C3 40 

=

15

16

P (B) = P (choosing 7 and choosing two other numbers from 1 to 6) 6

=

6×5×3 C2 1 = = 10 × 9 × 7 C3 8

10

P (A ∩ B) = P (choosing 3 and 7 and one other number from 4 to 6) =

10

3×3× 2 3 1 = = 10 × 9 × 8 C3 40

∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = 11/40. 179. (b) T he number of ways of placing 3 black ball is 10 C3. The number of ways in which two black balls are not together is equal to the number of ways of choosing 3 places marked with X out of eight places X W X W X W X W X W X W X W X This can be done in 8C3 ways. Thus, probability of 8

the required event =

8×7×6 C3 7 = = . 10 × 9 × 8 C3 15

10

941

186. (a) The possible outcomes on the pair of dice when the sum of the numbers on them is 7 are (1, 6), (2, 5) = P (W1 W2 B3 (3, 4), (4, 3), (5, 2), (6, 1), the probability for which 6 or W1 B2 W3 or B1 W2 W3) . is 36 = P (W1) P (W2) P (B3) + P (W1) P (B2) P (W3) The possible outcomes on the dice when the sum is  + P (B1) P (W2) P (W3) 8 are (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) the prob5 ability for which is 3 2 3 3 2 1 1 2 1 36 =    +   +    4 4 4 4 4 4 4 4 4 ∴ The probability that on the coin head comes and on the dice the sum is 7 or 8 1 13 (9 + 3 + 1) = = . 1 6 5  11 32 32 =  +  =  2 36 36  72 181. (b) T he probability that only two tests are needed = (probability that the first machine tested is faulty) The probability that from the pack of 11 cards when one card is chosen then number on it is 7 × (probability that the second machine tested is 2 faulty given the first machine tested is faulty) or 8 is 11 2 1 1 × = = . ∴ The probability that tail comes on the coin and 4 3 6 then on the chosen card number appearing is 182. (a) The event that the fifth toss results in a head is 7 or 8 independent of the event that the first four tosses 1 2 1 result in tails. = . × = 2 11 11 ∴  probability of the required event = 1/2. Thus, the required probability 183. (b) We have P (s) = P [5, 6] = 2/6 = 1/3. 121 + 72 193 ⇒ P (f) = 2/3. 11 1 + = 11 × 72 = . = P (an even number of tosses is needed) 72 11 792 = P (F S or F F F S 187. (a) Probability that the outcome of a single throw of a or  F F F F F S or ...) 4 2 = P (F) P (S) + P (F)3 P (S) + P (F)5 P (S) + ... die is any one of 2, 3, 4 and 5 is equal to = . 6 3 2/9 P (F) P (S) 2 If the die is rolled four times the required probability = = . = 4 1− 4 / 9 1 − P (F) 2 5 2 16 is equal to   = . 3 81 184. (b) Let the win in materials I to V be denoted by A1, A 2, A 3, A4, A5 188. (b) The exhaustive cases are 4! = 24 Required probability = P (A1 A2 A3) + P (A1 A2 A3) 1 1 1 The favourable cases are 4  − +  = P (A1) P (A2) P (A3) + P (A1) P (A2) P (A3)  2! 3! 4! 3 3 = 12 – 4 + 1 = 9 1 1 1 . =   +   = 3 9 2 2 4 = . ∴  the required probability = 8 24 185. (c) P (A) = 1 – P (A c ) = 1 – 0.3 = 0.7, P (B) = 0.4 We know that P (A) = P (A ∩ B) + P (A ∩ Bc ) 189. (b) We have, ⇒ 0.7 = P (A ∩ B) + 0.5 1 3 P ( A) = and P ( A ∪ B ) = or P (A ∩ B) = 0.7 – 0.5 = 0.2 3 4 P (A ∪ Bc ) = P (A) + P (Bc ) – P (A ∩ Bc ) Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.7 + (1 – 0.4) – 0.5 = 0.8 ≤ P(A) + P(B) P[B ∩ (A ∪ Bc ) c  Next P [B | P (A ∩ B )] = 3 1 5 ⇒ ≤ + P( B) ⇒ P( B) ≥ P[A ∩ Bc ] 4 3 12 From set theory Also, B ⊆ A ∪ B B ∩ (A ∩ Bc ) = (B ∩ A) ∪ (B ∩ Bc ) 3 = (B ∩ A) ∪ φ = B ∩ A ⇒ P(B) ≤ P(A ∪ B) = 4 P (A ∩ B) 0.2 1 ∴ Required Probability = = = . 5 3 c P (A ∪ B ) 0.8 4 ∴ ≤ P( B) ≤ 12 4

Probability

180. (a) P (2 white and 1 black)

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Objective Mathematics

39 39 3 3 9 × = × = 52 52 4 4 16

190. (d) Required probability =

191. (b)  1 + 1  = 3 C0  1  + 3 C1  1   1  2 2 2 2 2        

3

3

∴

11 1 + 3 C2     + 3 C3   22 2 3

C0 ⋅

1 = a, 8

3

C1 ⋅

3

1 = b, 8

 B′  1 and P   = .  A′  2 197. (a) ∵ P(A) = P(B) = x and P ( A ∩ B ) = P ( A′ ∩ B′) =

192. (d) The probability that exactly one of them occurs =P(M ∩ N ) + P( M ∩ N)

∴

193. (a) Since four dice are thrown together, ∴ n(S) = 64 = 1296 Total permutations of: (1, 1, 5, 6) =

4! = 12 2!

(1, 2, 4, 6) = 4! = 24 4! = 12 2! (1, 2, 5, 5) = 12,

(1, 3, 3, 6) =

⇒

and (4, 4, 3, 2) = 12 n(A) n(S)

12 + 24 + 12 + 12 + 24 + 12 + 12    + 12 + 4 + 4 + 12  =  1296 140 35 = = 1296 324

194. (d) T he probability of getting head and tail in a toss 1 is . 2 ∴ P(atleast one H) = 1 – P(four tails) 4

 1  15 =1−   =  2  16

195. (d) Favourable cases of getting 10 or greater than 10, if 5 is one of dice = {(5, 6), (6, 5), (5, 5)} 3 1 ∴ Required probability = = 36 12

1 B 1 196. (d) Given that, P   = ⇒ P ( B ∩ A) = 8  A 2 1  A 1 and P   = ⇒ P ( B ) = 2 B 4

2 1 = x+x− 3 3

1 2 198. (b) T he percentage of suffering from a diseases is 10%.

ie,  p =

(3, 3, 2, 5) = 12, (3, 3, 3, 4) = 4, (4, 4, 4, 1) = 4



2 3

⇒ x =

10 1 and probability of not suffering = 100 10

from a diseases is

(1, 3, 5, 4) = 24, (2, 2, 6, 3) = 12, (2, 2, 5, 4) = 12,



P( A ∪ B) =

1 3

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

=P(M) + P(N) – 2 P(M ∩ N)



 A′  P ( A′ ∩ B ) 3 P  = = P( B) 4 B

Now,

Hence, option (d) is correct.

1 1 C2 ⋅ = c and 3 C3 ⋅ = d 8 8 1 3 3 1 ⇒ a = , b = , c = , d = 8 8 8 8 3

∴ Required probability, P ( A) =

1 = P ( A) ⋅ P ( B ) 8

∴ Events are independent.

2

2



∴ P( A ∩ B) =



q =1−

1 9 = 10 10

Total number of patients, n = 6 3

 1   9  6 ∴ Required probability = C3      10   10 

=

3

6⋅5⋅4 1 9×9×9 × × 3 ⋅ 2 ⋅ 1 1000 1000

20 × 729 = 1458 × 10−5 106 199. (b) T he sum of two numbers is odd only when one is odd and other is even.



=

∴ Required probability = =

20

C1 ⋅20 C1 40 C2

20 × 20 40 × 39 = 20 2 ×1 39

200. (d) Probability of getting score 9 in a single throw

=

4 1 = . 36 9

Probability of getting score 9 exactly twice 2



8 1 8 = 3 C2 ×   × = .  9  9 243

201. (c) The required probability = (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + … = 0.14[1 + (0.56) + (0.56)2 + …] 1   0 ⋅ 14 = 0 ⋅ 14   = 0 ⋅ 44 = 0 ⋅ 32 − ⋅ 1 0 56  

209. (a) Since, P ( A ∪ B ) =

 E c ∩ F c  P( E c ∩ F c ∩ G ) P = 202. (c) We have,  G P (G )  

P (G ) − P ( E ∩ G ) − P (G ∩ F ) = P (G )



= 1 – P(E) – P(F) = P(E ) – P(F)

and P (A ) = ∴

204. (c)

2 . 5

⇒

∴

2 ⋅ 4! 2 Hence, required probability = = 5! 5

P( E ) =

3 1 and P ( A / E ′ ) = 4 4

From Baye’s theorem, P( E / A ) =

P( E ) ⋅ P( A / E ) P ( E ) ⋅ P ( A/E ) + P ( E ′) ⋅ P ( A / E ′ )

1 3 × 3 6 4 = = 1 3 5 1 8 × + × 6 4 6 4 207. (a) Let E be the event of numbers to be divisible by 2 or 3. Then, E = {2, 3, 4, 6, 8, 9, 10, 12} ⇒ n(E) = 8 and n(S) = 12 Hence, required probability =

n( E ) 8 2 = = n( S ) 12 3

208. (c) Let X denote a random variable which is the number of aces. Clearly, X takes values 1, 2 ∴



P=

4 1 1 12 = , q =1− = 52 13 13 13

 1   12  24 P ( X = 1) = 2 ×   ×   =  13   13  169

943

2 1 = 3 3

3 1 1 = + P( B) − 4 3 4 3 1 1 1 2 − + =1− = 4 3 4 3 3

P (A ∩ B ) = P ( B ) − P ( A ∩ B ) =

2 1 5 − = 3 4 12

210. (c) Since, A = {4, 5, 6} and B = {1, 2, 3, 4} ∴ A ∩ B = {4}

1 5 , P ( E ′) = 6 6

P( A / E ) =

P ( A) = 1 − P (A ) = 1 −

⇒ P ( B ) =

205. (c) Total number of ways = 5! Favourable number of ways = 2 · 4!

206. (a) Let E denote the event that a six occurs and A the event that the man reports that it is a ‘b’ we have

3 1 , P( A ∩ B) = 4 4

We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

c

203. (c) ∴ Required probability =

2 3

Probability



2

2  1   12  P ( X = 2) = 2   ⋅   =  13   13  169 24 2 2 Mean = Σ PiXi = + = 169 169 13

∵ P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ P ( A ∪ B ) =

3 4 1 + − =1 6 6 6

211. (b) Given that,

P ( A) =

1  A 1 ,P = 4  B  2



B 2 and P   =  A 3

 A  P( A ∩ B) We know, P   =  P( B) B

…(i)

 B  P ( B ∩ A) and P   =  P ( A)  A

…(ii)

∴

B  2  1  P   ⋅ P ( A)    1 A 3 4  =    = P( B) = 3  A 1 P  2 B  

212. (d) We have, A × B = {(1, 2), (1, 4), (1, 6), (1, 8), (3, 2), (3, 4), (3, 6), (3, 8), (5, 2), (5, 4), (5, 6), (5, 8), (7, 2), (7, 4), (7, 6), (7, 8), (9, 2), (9, 4), (9, 6), (9, 8)} ∴  n(s) = 20 Cartesian product which satisfies a + b = 9 is {(1, 8), (3, 6), (5, 4), (7, 2)}

944

∴ n(E) = 4

Objective Mathematics

∴ P(E) =

=

n( E ) 4 1 = = n( S ) 20 5

P (G )(1 − P( E ) − P( F )) P (G )



213. (c) Alternative

[∵ P(G) ≠ 0]

= 1 – P(E) – P(F) Fixing four American couples and one Indian man in = P(Ec) – P(F) between any two couples; we have 5 different ways in which his wife can be seated, of which 2 cases are 215. (d) By using option (D) favorable. 4 5 2 2 P(A ∩ B) = × = ∴ required probability = . 10 10 10 5

4 10 4 c c c c 214. (c) P  E ∩ F  = P ( E ∩ F ∩ G ) = P (G ) − P ( E ∩ G ) − P (G ∩ F ) P(A ∩ B) = 10 × 10 = 10 P (G ) P (G )  G  Hence option (D) is correct E c ∩ F c  P ( E c ∩ F c ∩ G ) P (G ) − P ( E ∩ G ) − P (G ∩ F ) = = G P (G ) P (G ) 

EXERCISEs FOR SELF-PRACTICE 1. In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective. 5

9 (a)   10

(b)

(c) 10–5

(d)  1  2

9 10

(a)

142 145

(b)

(c)

143 145

(d) None of these

144 145

16 3. If in a Binomial Distribution n = 4, P(X = 0) = , then 81 P(X = 4) =

(c)

1 16

(b)

1 27

(d)

1 81 1 8

(a)

11 20

(b)

(c)

1 10

(d) None of these

2

2. There number are chosen form 1 to 30. The probability that they are consecutive is

(a)

6. A natural number x is chosen at random form the first 100 > 50 100 natural numbers. The probability that x + x is

11 50

7. The following table represents a probability distribution for a random variable X: X p(X = x)

: :

1 2 0.1 2k

(a) 0.2 (c) 0.4

3 k

4 5 0.2 3k

6 0.1

(b) 0.1 (d) 0.3

8. A bag contains 50 tickets numbered 1, 2, 3, ... 50 of which five are drawn at random and arranged in ascending order of magnitude (x1 < x2 < x3 < x 4 < x5). The probability that x3 = 30 is 20

(a) 29

(c)

C 2 × 29 C 2 50 C5 C2 C5

20

(b)

50

C2 C5

(d) None of these 4. A rifleman is firing at a distance target and has only 10% chance of hitting it. The number of rounds, he must fire in order to have 50% chance of hitting it at 9 Two dices are thrown simultaneously. The probability of getting a pair of aces is least once is 1 1 (a) 11 (b) 9 (b) (a) 36 3 (c) 7 (d) 5 1 5. The probability of sure event is (c) (d) None of these 6 1 (a) 2 (b) 10. If E1 and E2 are independent events such that 2 0 < P (E1) < 1 and 0 < P (E2) < 1, then (c) 1 (d) unlimited. 50

11. For any two independent events E1 and E2 in a space S, P [(E1 ∪ E2) ∩ ( E1 ∩ E1 )] is 1 1 (b) ≥ 2 2 1 1 (c) > (d) ≤ 4 4 12. A number is chosen at random among the first 120 natural numbers. The probability that the number chosen being a multiple of 5 or 15 is (a) >

1 6 1 (c) 8

(b)

(a)

1 5

13. Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacment. The probability that the largest number appearing on a selected coupon is 9, is 7

1 (b)    15 

7

8 (c)    15 

7

(d) None of these

14. From a deck of 52 cards, the probability of drawing a court card is

3 13 4 (c) 13

1 4 1 (d) 13

(a)

(b)

15. A Binomial Probability Distribution is symmetrical if p, the probability of success in a single trial is (a) greater than

18. A and B are two independent events. The probability 1 and the probability that both A and B occurs is 6 1 that neither of them occurs is . Then the probability 3 of the two events, respectively, are (a)

2 1 and 3 4

(b)

1 1 and 6 2

1 1 1 1 and (d) and 3 5 6 2 19. For any two events A and B defined on the same sample space, P (A ∪ B) – P (A) – P (B) equals

1 2

1 2 (d) less than q where q = 1 – p 1 (d) equal to 2 16. Two players of equal skill are playing a set of games. They leave off when A requies 3 point to win and B require 2 points to win. If the stake was Rs. 32, what share would each take (b) less than

(a) 16, 16 (b) 9, 23 (c) 10, 22 (d) None of these 17. Among 52 cards, there are 12 face cards. Probability that a card drawn at random is not a face card, is

(a) – P (AB) (c) P (AB)

(b) 1 – P (AB) (d) 0

20. Three persons work independently on a problem. If the 1 1 respective probabilities that they will solve it are , 3 4 1 and , then the probability that none can solve it is 5

1 5 1 (c) 3 (a)

(b)

2 5

(d) None of these

21. It M and N are any two events, the probability that exactly one of them occurs is (a) P (M ∩ N°) + P (M° ∩ N) (b) P (M°) + P (N°) – 2 P (M° ∩ N°) (c) P (M) + P (N) – P (M ∩ N) (d) P (M) + P (N) – 2P (M ∩ N) 22. If mean of a Binomial Distribution is 20 and standard deviation is 4, then number of events is (a) 80 (c) 25

(b) 100 (d) 50.

23. Four persons are chosen at random from a group containing 3 men, 2 women, and 4 children. The chance that exactly 2 of then well be children is 11 10 (b) 21 21 2 9 (c) (d) 9 21 24. If n integers taken at random are multiplied together, then the probability that the last digit of the product is 2, 4, 6 or 8 is (a)

4n − 2n 5n 8n − 2 n (c) 5n (a)

(b)

8n 5n

(d) None of these

945

3 4 3 (d) 13

(b)

(c)

(d) None of these

3 (a)   5

1 4 10 (c) 13 (a)

Probability

(a) P (E1) + (E01) = P (E2) (b) P (E2) + (E02) = P (E1) (c) P (E1/E2) + P (E01/E2) = 1 (d) E1 and E2 are mutually exclusive

946

25. A single letter is selected at random from the word ‘‘PROBABILITY’’. The probability that it is a vowel is

Objective Mathematics

(a)

4 11

(b)

33. In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective ? (a) (10)– 5 (b) (1/2)5 (c) (9/10)5 (d) 9/10

3 11

2 (d) 0 11 26. Two dice are thrown, the probabilities that the sum of the points on two dice will be 7 is, (c)

5 (a) 36 (c)

7 36

6 (b) 36 (d)

8 36

27. The probability that Krishna will be alive 10 years hence 7 7 is and Hari will be alive is . The probability that 5 10 both Krishna and Hari will be dead 10 year hence, is

24 (a) 150 (c)

6 150

12 (b) 150 (d) None of these

28. Two friends Ashok and Baldev have equal number of sons. There are 3 tickets for a cricket match which are to be distributed among the sons. The probability that 2 tickets go to the sons of the one and one ticket go to the sons of the other is 6/7. The number of sons each of the two friends have (a) 3 (c) 5

(b) 4 (d) None of these

34. Three dice are thrown together. The probability of getting the same digits will be (a) 1/6 (c) 1/18

35. The odds against any event is 5 : 2 and odds in favour of another event in 6 : 5. If both event are independent then the probability that atleast one happens is

50 77 63 (c) 88 (a)

(a) 1/18 (c) 1/9

(b) 1/12 (d) 1/6

52 77 25 (d) 88 (b)

36. For any two events A and B, P (A ∩ B ) is equal to (a) P (A) – P (A ∪ B) (b) P (A) – P ( A ∩ B) (c) P (A) + P ( A ∩ B ) (d) P (A) – P (A ∩ B). 37. For two events A and B which of the following statement is true (a) P (A ∩ B) ≥ P (A) + P (B) – 1 (b) P (A ∩ B) ≤ P (A) + P (B) – 1 (c) P (A ∩ B) ≥ P (A) + P (B) (d) P (A ∩ B) ≤ P (A) + P (B).

 A  38. P  is equal to  A ∪ B 

29. The rifle-men take one shot each at the same target. The probability of the first rifle-men hitting the target is 0.4, the probability of the second rifle-men hitting the target is 0.5 and the probability of the third rifle-men hitting the target is 0.8. The probability of exactly two of them hitting the target, is 39. (a) 0. 44 (b) 0. 41 (c) 0. 49 (d) None of these 30. Two dice are thrown simultaneusly, the probability of obtaining a total score of 5 is

(b) 1/36 (d) 3/28

P (B) (a) P (A ∪ B)

P (A) (b) P (A ∪ B)

P (A ∩ B) (c) P (A ∪ B)

P (A) (d) P (A ∪ B)

One card is drawn from a pack of 52 cards. The probability that it will be an ace or black king or queen of heart is (a)

1 52

(b)

6 52

(c)

7 52

(d)

3 52

31. If A and B are two independent events in a sample space 40. If 4 integers are selected from the postive integers and then P (A A ) equals they are multiplied to each other then the probability getting 1, 3, 7 or 9 at the unit place of multiplied (a) 1 – P (A/B) (b) 1 – P B /B) number is (c) 1 – P (B) (d) 1 – P (A). 32. A purse contains 4 copper coins, 3 silver coins, the second purse contains 6 copper coins, and 2 silver coins. A coin in taken out from any purse the probability that it is a copper coins in (a) 4/7 (c) 3/7

(b) 3/4 (d) 37/56

16 625 2 (c) 625

(a)

8 625 7 (d) . 625 (b)

41. 15 coins are tossed together then the probability to come tail on 10 coins and to come head on 5 coins is

42. Three number are selected one by one from whole numbers 1 to 20. The probability that they are consecutive inegers is : (a)



(c)

43. When all letter of ‘CEASE’ are written in a line then the probability of comming two E′s together is : (b) 2/5 (d) 7/5

44. The sum of two positive numbers is 100. Probability that their product is greater than 1000 is, (a) 7/9 (c) 2/5

(b) 7/10 (d) None of these

45. Let p be the probability of happening of an event, and q its failure, then the total chance of r successes in n trials is (a) nCr pr qn – r (c) nCr pr + 1 qr – 1

(b) nCr pr – 1 qr + 1 (d) nCr pr qr

46. The items produced by a firm are supposed to contain 5% defective items. The probability that a sample of 8 items will contain less than 2 defective items is

27 (a) 20 (c)

7

 19    20

153  1    20  20  7

6

533  19  (b)   400  20  (d)

35  1    16  20 

6

47. The probability that a leap year selected at random will contain 53 Sundays is

1 7 4 (c) 7

(a)

2 7 4 (d) 7 (b)

48. The probability of happening of an impossible event, i.e., P (φ) is (a) 1 (c) 2

(b) 0 (d) – 1

49. A coin is tossed three times in succession. If E is the event that there are at least two heads and F is the event in which first throw is a head, then P (E/F) = 3 4 1 (c) 2 (a)

3 8 1 (d) 8 (b)

(b) 1/5 (d) 9/10

52. In two events P(A ∪ B) = P(A) = 5/6, P(B) = 2/3, then A and B are:

(d)

(a) 1/5 (c) 3/5

(b) 49/64 (d) 24/64

51. The probability that in a random arrangement of the letters of the word ‘UNIVERSITY’, the two I’s do not come together is: (a) 4/5 (c) 1/10

(b)

(a) 64/64 (c) 40/64

(a) independent (b) mutually exclusive (c) mutually exhaustive (d) dependent 53. Four boys and three girls stand in queue for an interview, probability that they will stand in alternate position is: (a) 1/34 (c) 1/17

(b) 1/35 (d) 1/68

54. 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, then p(B) is equal to: (a) 5/7 (c) 3/4

(b) 4/3 (d) 7/5  

55. Among 15 players, 8 are batsmen and 7 are bowlers, find the probability that a team chosen is of 6 batsmen and 5 bowlers: 8

(a) (c)

C6 × 7C5 15 C11

15 28

C6 + 7C5 (b) 15 C11 8

(d) None of these

56. Two cards are drawn from a well-shuffled pack. Find the probability that one of them is an ace of heart:

1 25 1 (c) 52

(a)

(b)

1 26

(d) None of these

57. Exactly 2 machines out of 4 are faulty. They are rested. Then, the probability that only two tests are required is: (a)

1 3

(b)

2 3

(c)

1 6

(d)

5 6

58. Let A and B be two events such that P(A) = .3, P(A ∪ B) = .8, if A and B are independent events, then P(B):

5 7 1 (c) 3

(a)

(b)

5 13

(d)

1 2

947

50. A box contains 10 good articles and 6 with defects. One item is drawn at random. The probability that it is either good or has a defect, is:

3003 32768 511 (d) 32768 (b)

Probability

3005 32768 1001 (c) 32768 (a)

948

59. Find the probability that the two digit number formed by digits 1, 2, 3, 4, 5 is divisible by 4:

1 30 1 (c) 40

(b)

Objective Mathematics

(a)

1 20

60. Three integers are choosen at random from the first 20 integers. The probability that their product is even, is: (a) 2/19 (b) 3/29 (c) 17/19 (d) 4/29

(d) None of these

Answers

1. 11. 21. 31. 41. 51.

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

2. 12. 22. 32. 42. 52.

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

3. 13. 23. 33. 43. 53.

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

4. 14. 24. 34. 44. 54.

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

5. 15. 25. 35. 45. 55.

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

6. 16. 26. 36. 46. 56.

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

7. 17. 27. 37. 47. 57.

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

8. 18. 28. 38. 48. 58.

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

9. 19. 29. 39. 49. 59.

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

10. 20. 30. 40. 50. 60.

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

27

Statistics

CHAPTER

Summary of conceptS meaSureS of central tendency For a given data, a single value of the variable which describes its characteristics is identified. This single value is known as the average. An average value generally lies in the central part of the distribution and therefore, such values are called the measures of central tendency. The commonly used measures of central tendency are: (a) (b) (c) (d) (e)

Arithmetic Mean. Geometric Mean. Harmonic Mean. Median. Mode.

In such cases, the deviation di = xi – A, of variates xi from the assumed mean A are divided by the common factor. The A.M is then obtained by the following formula: fd x = A + ∑ i i × h; N = ∑ f i . N where A = assumed mean, xi − A di = = the deviation of any variate from A, h h = the width of the class - interval.

Weighted arithmetic mean If w1, w2, w3, ...., wn are the weights assigned to the values x1, x2, x3, ...., xn respectively, then the weighted average is defined as :

arithmetic mean

w1 x1 + w2 x2 + ...... + wn xn w1 + w2 + ....... + wn

Weighted A. M. =

mean of unclassified data Let x1, x2, ...., xn be n observations, then their arithmetic mean is given by x1 + x2 + ..... + xn 1 n = ∑ xi x = n n i =1

combined mean If we are given the A.M. of two data sets and their sizes, then the combined A.M. of two data sets can be obtained by the formula: x12 =

mean of Grouped data

n1 x1 + n2 x2 n1 + n2

Let x1, x2, x3, ...., xn be n observations and let f1, f2, where, x12 = Combined mean of the two data sets 1 and 2 ....., fn be their corresponding frequencies, then their x1 = Mean of the first data arithmetic mean is given by n

x =

f1 x1 + f 2 x2 + ... + f n xn = f1 + f 2 + .... + f n

∑fx i =1 n

i i

∑f i =1

i

Short cut method For the given data, we suitably choose a term, usually the middle term and call it the assumed mean, to be denoted by A. We find the deviation, di = (xi – A) for each term. Then the arithmatic mean is given by x =A+

∑ fi di ∑f

.

i

Step deviation method When the class intervals in a grouped data are equal, then the calculations can be simplified further by taking out the common factor from the deviations. This common factor is equal to the width of the class interval.

x2 = Mean of the second data = Size of the first data = Size of the second data.

Properties of A. M. (i) In a statistical data, the sum of the deviations of individual values from A.M. is always zero, i.e., n

∑f i =1

i

( xi − x ) = 0,

where fi is the frequency of xi (1 ≤ i ≤ n) (ii) In a statistical data, the sum of squares of the deviations of individual values from A.M. is least, i.e., n

∑f i =1

i

( xi − x ) 2 is least.

950

(iii) If each of the n given observations is doubled, then their mean is doubled.

Objective Mathematics

(iv) If x is the mean of x1, x2, ..... , xn, then the mean of ax1, ax2, ....., axn where a is any number different from zero, is a x .

calculation of median (a)

median of an individual Series Let n be the number of observations. (i) Arrange the data in ascending or descending order. (ii) (a) If n is odd, then 1 (n + 1) th observation Median = value of the 2 (b) If n is even, then n n  Median = mean of the   th and  + 1 th 2 2 observation.

(b)

median of a discrete Series (i) Arrange the values of the variate in ascending or descending order. (ii) Prepare a cumulative frequency table. (iii) (a) If n is odd, then n +1 th term. Median = size of the 2 (b) If n is even, then n n    2  +  2 + 1  Median = size of the   th term. 2    

Geometric mean If x1, x2, x3, ..., xn are n observations, none of them being zero, then their geometric mean is defined as 1

n G.M. = ( x1.x2 .x3 ...xn )

 log x1 + log x2 + ....... + log xn  G.M. = antilog   n In the case of a grouped data, geometric mean of n observations x1, x2, ......., xn, is given by

(

fn f1 f2 G.M. = x1 . x2 ....... xn

)

n

1/ N

, where N =

∑f i =1

i

 n  f i log xi  ∑ i =1  G.M. = antilog  N    

or

Note: In the case of continuous or grouped frequency distribution, the values of the variate x are taken to be the values corresponding to the mid points of the class intervals.

harmonic mean The harmonic mean of n observation x1, x2, ...., xn is defined as: H.M. =

n 1 1 1 + + ...... + x1 x2 xn

If x1, x2, x3, ...., xn are n observations which occur with frequencies f1, f2, ..., fn respectively, then, their H.M. is given by n

H.M. =

∑f i =1

n

i

 fi 

∑  x  i =1

i

relation BetWeen a.m, G.m. and h.m.

median of a continuous Series (i) Prepare the cumulative frequency table. n (ii) Find the median class, i.e., the class in which the   th 2 observation lies. (iii) The median value is given by the formula n    − c f Median = l +  2 f  

   × h, where  

l = lower limit of the median class n = total frequency f = frequency of the median class h = width of the median class cf = cumulative frequency of the class preceding the median class.

QuartileS, decileS and percentileS

The arithmetic mean (A.M.), geometric mean (G.M.) and harQuartile Just as the median divides a set of observations monic mean (H.M.) for a given set of observations are related (when arranged in ascending or descending order of magnitudes), as under: into two equal parts, similarly Quartile divides the observations A.M, ≥ G.M. ≥ H.M. into four equal parts. The value of the item midway, between the first item and the median is known as first or lower quartile Equality sign holds only when all the observations are equal. and is denoted be Q1. The value of the item midway between the last item and the median is known Third or Upper Quartile and median is denoted Q3. The median is known as the Second Quartile and Median is the middle most or the central value of the variate in a is denoted by Q . The methods for finding the values of Q and 2 1 set of observations, when the observations are arranged either in Q are as similar to that of the median. In the case of ungrouped 3 ascending or in descending order of their magnitudes. It divides data, when arranged in ascending or descending order of magnithe arranged series in two equal parts. tudes Q1, Q3 can be obtained as follows:

For a frequency distribution, Q1 and Q3 are given by [(n / 4) − C] Q1 = l +  × h, f Q3 = l +

[(3n / 4) − C]  × h, f

  (b) Mode of Discrete Series  In the case of discrete frequency distribution, mode is the value of the variate corresponding to the maximum frequency.   (c) Mode of Continuous Series (i) Find the modal class, i.e., the class which has maximum frequency. The modal class can be determined either by inspection or with the help of grouping table. (ii) The mode is given by the formula

fm − fm −1 l = lower limit of the class in which a particular Mode = l +  × h, 2 f − fm −1 − fm +1 m quartile lies, f = frequency of the class-interval in which a par-    where ticular quartile lies, l = the lower limit of the modal class i = class-interval of the class in which a particular h = the width of the modal class quartile lies, fm – 1 = the frequency of the class preceding modal class cf = cumulatively frequency of the class preceding = the frequency of the modal class f the class in which the particular quartile lies. m fm +1 = the frequency of the class succeeding modal [(nh / 4) − c f ] class. In general, Qi = l +  × h, i = 1, 2, 3, 4. f In case, the modal value lies in a class other than the one Decile  The value of the variable which divides the series, containing maximum frequency, we take the help of the followwhen arranged in ascending or descending order, into 10 equal ing formula; parts is called decile. There are 9 deciles denoted by D1, D2 ... fm +1 Mode = l +  × h, D9. When the series is ungrouped the deciles are calculated as fm −1 + fm +1 follows: where symbols have usual meaning. n×h A distribution in which mean, median and mode coincide Di = , i = 1, 2, ..., 9. 10 is called a symmetrical distribution. If the distribution is moderWhen the data is classified or grouped, ately skewed, then mode can be calculated as follows: [(nh / 10) − c f ] Mode = 3 median – 2 Mean. Di = l +  × h f where

where symbols have their usual meaning.

Measures of Dispersion

Percentile  The value of the variable which divide the series, when arranged in ascending or descending order, into 100 equal parts is called percentile. There are 99 percentiles denoted by P1, P2, P3, P4, ..., P99 respectively. When the series is ungrouped the percentiles are calculated by the following formula:

The degree to which numerical values in the set of values tend to spread about an average value is called the dispersion or variation. The commonly used measures of dispersion are:

n×h Pi = , h = 1, 2, ..., 99. 100 When the data is classified or grouped, the percentiles are calculated by the formula Pi = l +

[(nh / 100) − c f ] f

× h, i = 1, 2, ...., 99,

where symbols have their usual meanings.

Mode Mode is that value in a series which occurs most frequently. In a frequency distribution, mode is that variate which has the maximum frequency.

Computation of Mode   (a) Mode of Individual Series  In the case of individual series, the value which is repeated maximum number of times is the mode of the series.

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

Range Quartile Deviation or Semi-interquartile range Mean Deviation Standard Deviation.

Range  It is the difference between the greatest and the smallest observations of the distribution. If L is the largest and S is the smallest observation in a distribution, then its Range = L – S. Also, L−S Coefficient of range = L + S .

Quartile Deviation Quartile deviation or semi - interquartile range is given by Q. D. =

1  (Q3 – Q1) 2

Q3 − Q1 Coefficient of Q.D. = Q + Q . 3 1

951

n +1 3(n + 1) th item, Q3 = th item. 4 4

Statistics

Q1 =

952

Mean Deviation  For a frequency distribution, the mean deviation from an average (median, or arithmetic mean) is given by

Objective Mathematics

n

∑ fi xi − x M.D. =

Probable error of standard deviation =  quartile range. Quartile deviation =

i =1

n

∑f i =1

Coefficient of M.D. =

2 σ = Semi-inter3

5 M.D. 6

From these relationships, we have 4 S. D. = 5 M.D. = 6 Q.D.

i

Mean deviation . Corresponding average

Standard Deviation

Coefficient of S.D. (C.V.)  For comparing two or more series for variability, the relative measure, called coefficient of variation (C.V.) is used. This measure is defined as:

σ The standard deviation of a statistical data is defined as the posiC.V. =  × 100. x tive square root of the squared deviations of observations from The coefficient of variation is also represented as percentthe A.M. of the series under consideration. (a) Standard deviation (also denoted by σ) for ungrouped set age. The square of S.D. is called the variance of the distribution of observations is given by and is denoted by σ2. n S.D. σ =

∑ (x

i

i =1

− x )2

n (b) Standard deviation for frequency distribution is given by, n



S.D. =

∑ f (x i =1

i

i

Computation of Standard Deviation (a) Direct Method σ=

− x )2

∑x n

2

∑x −   n 

2

(b)  Short-cut Method N 2 where fi is the frequency of xi (1 ≤ i ≤ n). ∑ d 2 −  ∑ d  , for ungrouped data When the values of the variable are given in the form of σ=  n  n   classes, then their respective mid-points are taken as the values of the variable. where A is assumed mean and d = x – A. Statand Deviation of n Natural Numbers: 1 2  σ =  (n − 1)   12 

1/ 2

σ=

.

∑ fd N

2

 ∑ fd  −  , for grouped data  N  2

Standard deviation shows the limits of variability by which where N = ∑ f . the individual observation in a distribution will vary from the mean. For a symmetrical distribution with mean x , the follow- (c)  Step-Deviation Method ing area relationship holds good : 2 x ± σ covers 68.27 % observations. ∑ fd' 2 −  ∑ fd'  ; d′ = x − A σ = h  N  x ± 2σ  covers 95.45 % observations. h N   x ± 3σ  covers 99.73 % observations. These limits are illustrated by the following curve known Combined Standard Deviation as Normal Curve. Let A1 and A2 be two series having n1 and n2 observations respectively. Let their A.M. be x 1 and x 2, and standard deviations be σ1 and σ2. Τhen the combined standard deviation σ or σ12 of A1 and A2 is given by n1σ12 + n2σ 22 + n1d12 + n2 d 22 σ12 or σ = n1 + n2 Empirical Relationships  If the data is moderately nonsymmetrical, then the following empirical relationships hold: 4 Mean deviation = σ 5 Semi-Inter-quartile range =

2 σ. 3

=

n1 (σ12 + d12 ) + n2 (σ 22 + d 22 ) n1 + n2

where d1 = ( x1 − x12 ), d2 = ( x2 − x12 ), and

x12 =

n1 x1 + n2 x2 is the combined mean. n1 + n2

953

There are two types of distributions.

Methods of calculating Correlation Coefficient

Statistics

Correlation Analysis

1. Univariate distribution:  These are the distributions (a)  Karl Pearson’s Correlation Coefficient in which there is only one variable such as the heights of cov ( x, y ) ∑ ( xi − x ) ( yi − y ) the students of a class or marks obtained by the students r= = var( x ) . var y ( ) of a class in a school. ∑ ( xi − x )2 . ∑ ( yi − y )2 2. Bivariate distribution:  Distribution involving two discrete variables is called a Bivariate Distribution. For example, the heights and weights of the students of a class in a school.

( dx ) ( dy ) ∑ (dx . dy ) − ∑ n ∑  (∑ dx )   (∑ dy )  ∑ (dx ) −   ∑ (dy ) − i

=

2

i

i

i

i

2

i

2

2

  

,

i i n n   Concept of Correlation  Correlation is a statistical tool which studies the relationship between two variables and correwhere dxi = xi – A and dyi = yi – B and A, B are the assumed lation analysis involves various methods and techniques used for means of x-series and y-series respectively. studying and measuring the extent of the relationship between (b) Coefficient of Rank Correlation  This formula is two variables. applied to the problems in which data cannot be measured quantitatively but qualitative assessment is possible such as Covariance beauty, honesty etc. In this case the best individual is given The covariance between two variables x and y with n pairs of rank number 1, next rank 2 and so on. The coefficient of observations (x1, y1), (x2, y2), ....., (xn , yn) is defined as rank correlation is given by the formula ( x − x ) ( y − y ) 6∑ Di2 ∑ i l cov  (x, y) = R=1– , n n (n 2 − 1)

where x = and y =

∑x

 ∑ xi yi  =  − x y ,  n 

i

n

∑y

i

n

.

where Di is the difference of corresponding rank and n is the number of pairs of observations.

Regression Analysis It is the method used for estimating the unknown values of one variable, corresponding to the known values of another variable.

Correlation Coefficient

Line of Regression of y on x

The number showing the degree or extent to which x and y are related to each other, is called the correlation coefficient. It is denoted by ρ (x, y) or rxy or simply r.

The line of regression of y on x gives the best estimate of the value of y for given value of x and is given by y – y = byx ( x − x ) ; byx = r ⋅ 

Characteristics of Correlation Coefficient (i) – 1 ≤ r ≤ 1 (ii) If r = – 1, there is perfect negative correlation between x and y, i.e., corresponding to an increase (or decrease) in one variable, there is a proportional decrease (or increase) in the other variable. (iii) If r = 1, there is perfect positive correlation between x and y, i.e., corresponding to an increase (or decrease) in one variable, there is a proportional increase (or decrease) in the other variable. (iv) If r = D, then x and y are not correlated, i.e., the changes in one variable are not followed by changes in the other. (v) If 0 < r ≤ 1, there is a positive correlation between x and y, i.e., an increase (or decrease) in one variable corresponds to an increase (or decrease) in the other. (vi) If – 1 ≤ r < 0, there is negative correlation between x and y, i.e., an increase (or decrease) in one variable corresponds to a decrease (or increase) in the other.

σy σx

Line of Regression of x on y The line of regression of x on y gives the best estimate of the value of x for given value of y and is given by

x – x = bxy ( y − y ) ;



bxy = r ⋅ 

σx σy

Coefficient of Regression of y on x The coefficient of regression of y on x denoted by byx and is given by

σy byx = r ⋅  σ x

=

n∑ xi yi − ∑ xi ∑ yi cov ( x,y ) = n x 2 − ( x )2 σ 2x ∑i ∑i

954

coefficient of regression of x on y

Objective Mathematics

The coefficient of regression of x on y denoted by bxy and is given by r σx n∑ xi yi − ∑ xi ∑ yi cov ( x, y ) bxy = = = σy σ 2y n∑ yi2 − (∑ yi ) 2 Properties of Regression Coefficients (i) Both regression coefficients have the same sign, i.e., either both are positive or both are negative. (ii) The sign of correlation coefficient is same as that of regression coefficient, i.e., r > 0 if bxy > 0 and byx > 0; and r < 0 if bxy < 0 and byx < 0. (iii) The coefficient of correlation is the geometric mean between the two regression, coefficients r = ± byx × bxy The sign to be taken outside the square root is that of the regression coefficients. (iv) Both the regression coefficients cannot be numerically greater than one.

(v) Regression coefficients are independent of change of origin but not scale. (vi) A.M. of the regression coefficients is greater than the correlation coefficient.

Properties of Lines of Regression The two lines of regression have the following properties: (i) The two lines of regression pass through the point ( x , y ) . (ii) Slope of the line of regression of y on x = byx (iii) Slope of the line of regression of x on y =

1 bxy

(iv) The angle θ between two lines of regression is given by

 1 − r 2  σ xσ y tan θ = ±   r  σ 2x + σ 2y π i.e., if two variables are uncor2 rected, the lines of regression are perpendicular to each other. If r = ± 1, then tan θ = 0 ⇒ θ = 0 or π, i.e., in the case of perfect correlation (r = ± 1) the two lines of regression coincide. If r = 0, then tan θ = ∞ ⇒ θ =

multiple-choice QueStionS choose the correct alternative in each of the following problems: 1. Mean of 25 observations was found to be 78.4. But later on it was found that 96 was misread as 69. The correct mean is (a) 79.48 (c) 81.32

(b) 76.54 (d) 78.4

2. The mean height of 15 students is 154 cm. It is discovered later on that while calculating the mean the reading 175 cm. was wrongly read as 145 cm. The correct mean height is (a) 145 cm (c) 156 cm

(b) 170 cm (d) None of these

3. A firm of readymade garments make both men’s and women’s shirts. Its profit average is 6 % of sales. Its profits in men’s shirts average 8% of sales and women’s shirts comprise 60% of output. The average profit per sales rupee in women’s shirts is (a) 0.0466 (c) 0.0666

(b) 0.0166 (d) None of these

4. In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? (a) 73 (c) 68

(b) 65 (d) 74

5. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations in the set is increased by 2, then the median of the new set

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

is increased by 2 is decreased by 2 is two times the original median remains the same as that of the original set

6. If the value of mode and mean is 60 and 66 respectively, then the value of median is (a) 6 0 (c) 6 8

(b) 64 (d) None of these

7. In an experiment with 15 observations on x, the following results were available ∑x2 = 2830, ∑x = 170 One observation that was 20 was found to be wrong and was replaced by the correct value 30. Then the corrected variance is (a) 78.00 (b) 188.66 (c) 177.33 (d) 8.33 8. Consider the following statements (1) Mode can be computed from histogram (2) Median is not independent of change of scale (3) Variance is independent of change of origin and scale. Which of these is/are correct? (a) only (1) and (2) (c) only (1)

(b) only (2) (d) (1), (2) and (3)

9. The weighted mean of the first n natural numbers if their weights are the same as the numbers, is

2n + 1 3

(d) None of these

10. If the frequencies of first four numbers out of 1, 2, 4, 6, 8 are 2, 3, 3, 2 respectively, then the frequency of 8 if their A.M. is 5, is (a) 4 (c) 6

(b) 5 (d) None of these

11. The mean weight of 120 students in the second year class of a college is 56 kg. If the mean weights of the boys and that of the girls in the class are 60 kg and 50 kg respectively, then the number of boys and girls separately in the class are (a) 72, 64 (c) 72, 48

(b) 38, 64 (d) None of these

12. The mean of 10 numbers is 12.5; the mean of the first six is 15 and the last five is 10. The sixth number is (a) 15 (c) 18

(b) 12 (d) none of these

13. In a family, there are 8 men, 7 women and 5 children whose mean ages separately are respectively 24, 20 and 6 years. The mean age of the family is (a) 17.1 years (c) 19.1 years

(b) 18.1 years (d) None of these

(b) x + n + 1 (a) x + 2n (c) x + 2 (d) x + n. 20. The arithmetic mean of first n odd natural numbers is (a) n (c) (n – 1)

(b) (n + 1)/2 (d) None of these

21. The arithmetic mean of the series 1, 2, 22, ..... 2 n – 1 is (a) 2n /n (c) (2n + 1)/n

(b) (2n – 1)/n (d) None of these

22. The arithmetic mean of the series nC0, nC1, nC2 ..., nC n is : (a) 2n /(n +1) (c) (2n–1/(n+ 1)

(b) 2n /n (d) None of these

23. The variance of first n natural numbers is n2 + 1 12 (n + 1) (2n + 1) (c) 6 (a)

(b)

n2 − 1 12

(d) None of these

24. The AM of n observations is M. If the sum of n – 4 observations is a, then the mean of remaining 4 observations is nM − a 4 nM − a (c) 4 (a)

(b)

nM + a 2

(d) nM + a.

14. The mean of 100 items is 50 and their S.D. is 4. the sum 25. The weighted mean of first n natural numbers whose of all the items and also the sum of the squares of the weights are equal to the squares of corresponding items is numbers is (a) 5000, 251600 (b) 4000, 251600 n +1 3n (n + 1) (c) 5000, 261600 (d) None of these (b) (a) 2 2 (2n + 1) 15. In a series of 2n observations, half of them equal a and remaining half equal –a. If the standard deviation of (n + 1) (2n + 1) n (n + 1) (c) (d) . the ovservations is 2, then | a | equals 6 2 (a) 2 (b) 2 26. Mean of 100 items is 49. It was discovered that three 1 2 items which should have been 60, 70, 80 were wrongly (c) (d) read as 40, 20, 50 respectively. The correct mean is n n 16. The coefficient of variation of two series are 58% and 1 (a) 48 (b) 82 69%. If their standard deviations are 21.2 and 15.6, 2 then their A.Ms are (c) 50 (d) 80 (a) 36.6, 22.6 (b) 34.8, 22.6 5 7 (c) 36.6, 24.4 (d) None of these 27. If a variable takes the discrete values α + 4, α – , α– ­, 2 2 17. If n = 10, x = 12, ∑ x 2 = 1530, then the coefficient of 1 1 , α –  , α +5 (α > 0), then α – 3, α – 2, α + variation is 2 2 (a) 36 % (b) 41 % the median is (c) 25 % (d) None of these 1 5 (a) α –  (b) α –  18. The mean of the squares of first n natural numbers is 2 4 1 1 2 5  n (b)  n (n + 1) (a) (c) α – 2 (d) α +  8 2 4 1 1 (c)  n (2n + 1) (d) (n + 1) (2n + 1). 28. The geometric mean of numbers 7, 72, 73, ...., 7n is : 6 6 (a) 7 7 / n (b) 7 n / 7 19. If the mean of the set of number x1, x2, ........, xn is , ( n − 1) / 2 (c) 7 (d) 7 ( n + 1) / 2 then the mean of the numbers xi + 2i, 1 ≤ i ≤ n is

955

(b)

Statistics

n +1 3 2n − 1 (c) 3

(a)

956

29. Mean deviation of the series a, a + d, a + 2d, a + 2nd from its mean is

Objective Mathematics

(n + 1) d (a) (2n + 1)

nd (b) 2n + 1

n (n + 1) d (c) (2n + 1)

(d)

(2n + 1) d n (n + 1)

30. In any discrete series (when all values are not same), the relationship between M.D. about mean and S.D. is

32.

33.

34.

(a) 6 (c) 8

(b) 7 (d) None of these

39. The quartile deviation for the data x : 2 f : 3

3 4

4 8

5 4

6 1

is

1 (b) 4

(a) 0

1 (d) 1 2 40. If 25 % of the items are less than 20 and 25 % are more The mean of n items is X . If the first item is increased then 40, the quartile deviation is by 1, second by 2 and so on, then the new mean is (a) 20 (b) 30 n (c) 40 (d) 10 (b) X + (a) X + n 2 41. If the regression coefficient of X on Y and Y on X are n +1 (c) X + (d) None of these – 0.4 and – 0.9 respectively, then the correlation coef2 ficient between X and Y is The mean deviation from the mean for the set of obser(a) – 0.6 (b) 0.6 vations – 1, 0, 4 is (c) 0.3 (d) – 0.3 (a) less than 3 (b) less than 4 (c) greater than 2.5 (d) greater than 4.9. 42. If ∑ xi = 60, ∑ yi = 95, ∑ xi yi = 574, n = 10 then If the s.d of a set of observations is 4 and if each obCov. (x, y) is equal to servation is divided by 4, the s.d of the new set of (a) – 0.2 (b) 0.2 observations will be (c) – 0.4 (d) 0.4 (a) 4 (b) 3 43. If Cov. (X, Y) = – 16.5, Var. (X) = 2.89 and Var. (Y) = (c) 2 (d) 1 100 then coefficient of correlation ρ (X, Y) is equal to The variable x takes two values x1 and x2 with frequen(a) 0.97 (b) – 0.97 cies f1 and f2 respectively. If σ denotes the standard (c) 0.74 (d) – 0.74 deviation of x, then 44. If X and Y are independent variables, then ρ (X, Y) is 2  f x + f 2 x2  f x 2 + f 2 x22 equal to (a) σ2 = 1 1 –  1 1  f1 + f 2  f1 + f 2  (a) 1 (b) – 1 (c) 0 (d) None of these f f 1 2 (b) σ2 =  (x – x )2 45. If the two regression coefficients are 0.8 and 0.2, then ( f1 + f 2 ) 2 1 2 the coefficient of correlation is ( x − x2 ) 2 (c) σ2 = 1 (a) 0.4 (b) – 0.4 f1 + f 2 (c) 0.8 (d) – 0.8 (d) None of these 46. If the equations of two regression lines are 4x + 3y + 7 (a) M.D. = S.D. (c) M.D. < S.D.

31.

38. The variance of the data 2, 4, 6, 8, 10 is :

(b) M.D. ≥ S.D. (d) M.D. ≤ S.D

35. A sample of 35 observations has the mean 80 and s.d. as 4. A second sample of 65 observations from the same population has mean 70 and s.d. 3. The s.d. of the combined sample is (a) 5.85 (c) 3.42

(b) 5.58 (d) None of these

36. If µ is the mean of a distribution, then

∑ f (y i

i

− µ) is equal to

(a) M.D. (c) 0

(b) S.D. (d) None of these

(c)

= 0 and 3x + 4y + 8 = 0, then 4 7

11 7 4 11 (d) mean of y = (c) mean of x = 7 7 47. The line of regression of y on x and x on y are respectively y = x + 5 and 16x – 9y = 94. If the variance of y is 16, then the variance of x is (a) mean of x = −

(a) 9 (c) 6

(b) mean of y = −

(b) 3 (d) None of these

37. The means of five observations is 4 and their variance 48. If ax + by + c = 0 is a line of regression of y on x and a1 x + b1y + c1 = 0 is a line of regression of x on y, is 5.2. If three of these observations are 1, 2, and 6, then then the other two are: (a) ab1 ≥ a1b (b) ab1 ≤ a1b (a) 2 and 9 (b) 3 and 8 (c) 4 and 7 (d) 5 and 6. (c) a1b1 ≥ ab (d) a1b1 ≤ ab

The height of a student A who weights 200 units is

59. Let r > 0 and m be the A. M. of the regression coefficient of y on x and x on y, then

(a) 68.75 (c) 71.75

(b) 70.75 (d) 111.6

(a) 1 (c) 0

(b) – 1 (d) None of these

(a) m = r (b) m ≤ r 50. The coefficient of correlation between two variables x (c) m ≥ r (d) None of these and y is 0.8 and their covariance is 20. If variance of 60. If M.D. is 12, the value of S.D. will be x series is 16, then the S.D. of y series is (a) 15 (b) 12 (a) 6.25 (b) 4.25 (c) 24 (d) None of these (c) 5.25 (d) None of these 61. If regression coefficient of y on x is 2, then the regres 51. For the data : 3x = y; 8y = 6x and σx = 4 sion coefficient of x on y is (a) σy = 6 (b) r = 0.5 1 (a) 2 (b) (c) r = – 0.5 (d) σy = 4. 2 1 52. If x = 15, y = 80, σx = 12, σy = 12 and r = 0.75 then the (c) ≤ (d) None of these estimated values of y corresponding to x = 55 is 2 2 (a) 110 (b) 120 62. If Cov (u, v) = 3, σu = 4.5, σv2 = 5.5 then ρ (u. v) is (c) 100 (d) None of these (a) 0.121 (b) 0.603 53. The co-efficient of rank correlation of marks obtained by 10 students in Statistics and Accountancy was found 63. to be 0.2. It was later discovered that the difference in ranks in the two subjects obtained by one of the students was wrongly taken as 9 instead of 7. The correct value of co-efficient of rank correlation is 64. (a) 0.61 (b) 0.39 (c) 0.42

(d) None of these

54. If the two lines of regression are 3x + 12y – 19 = 0 and 9x + 3y – 46 = 0, then 1 (a) bxy = − 3 1 (c) byx = − 4

1 3 1 (d) byx = 4 (b) bxy =

55. If the lines of regression of Y on X and X on Y are respectively y = Kx + 4 and x = 4y + 5, then (a) 0 ≤ K ≤ 4 (c) K >

1 4

1 (b) 0 ≤ K ≤ 4 (d) None of these

56. If the correlation coefficient between x and y is 0.6 and Y −3 X−5 and v = if u = 3 2 then the coefficient of correlation between u and v is (a) 0.8 (b) 0.6 (c) 0.3 (d) 0.2 57. If x and y are two random variables with the same standard deviation and coefficient of correlation r , then the coefficient of correlation between x and x + y is 1+ r (a) 2

(b)

1− r 2

(d)

(c)

1+ r 2 1− r 2

(c) 0.07

(d) 0.347

The two regression lines for a bivariate data are x + y + 50 = 0 and 2x + 3y + k = 0. If x = 0, then y is (a) 50 (b) k – 100 (c) – 50 (d) 50 + k If the mean of numbers 27, 31, 89, 107, 156 is 82 , then the mean of 130, 126, 68, 50, 1 is (a) 80 (c) 157

(b) 82 (d) 75

65. Suppose values taken by a variable X are such that a ≤ xi ≤ b where xi denotes the value of X in the ith case for i = 1, 2, .... n then (a) (b – a)2 ≥ var (X) (c) a2 ≤ var (X) ≤ b2

a2  ≤ var (X) 4 (d) a ≤ var (X) ≤ b

(b)

66. The A.M. of a set of 50 numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is (a) 36 (c) 37.5

(b) 36.5 (d) 38.5

67. If x and y are two variables such that u = – ax and v = by where a and b are positive constants, then (a) r (x, y) = – r (u, v) (c) r (x, y) = abr (u, v)

(b) r (u, v) = abr (x, y) (d) r (x, y) = r (u, v).

68. The average of n numbers x1, x2, x3, ..... xn is M. If xn is replaced by x′, then the new average is (a) M − xn + x' n (c)

nM − xn + x' n

(b)

(n − 1) M + x' n

(d) M – xn + x′.

2 and 5 the geometric mean 24. The greater number between them is

69. The harmonic mean between two numbers is 14

957

58. The two lines of regression are parallel to the axes if r =

Statistics

49. Given the following results for the height (x) and weight (y) in appropriate units of 1000 students: x = 68, y = 150, σx = 2.5, σy = 20, r = 0.60.

958

(a) 72 (c) 36

(b) 54 (d) None of these

Objective Mathematics

70. For a bivariate distribution (x, y) if

∑ x = 50 , ∑ y = 60 , ∑ xy = 350 ,

x = 5, y = 6.

Variance of x is 4, variance of y is 9, then r xy equals 11 11 (b) 3 18 5 5 (c) (d) 6 36 71. For a symmetrical distribution Q1 = 20 and Q3 = 40. The value of 50 th percentile is (a)

(a) 20 (c) 40

(b) 30 (d) None of these

72. The mean deviation of the numbers 3, 4, 5, 6, 7 is (a) 25 (c) 1.2

(b) 5 (d) 0

73. If byx and bxy are the regression coefficient of y on x and x on y respectively, then (a) byx + bxy ≥ 2r (x, y)

77. A person purchases one kg of tomatoes from each of the 4 places at the rate of 1 kg, 2 kg, 3 kg, 4 kg per rupee respectively. On the average he has purchased x kg of tomatoes per rupee, then the value of x is (a) 2 (c) 1.92

(b) 2.5 (d) None

78. The co-efficient of correlation between two variables x and y is 0.8 while the regression coefficient of y on x is 0.2, Then the regression co-efficient of x on y is (a) – 3.2 (c) 4

(b) 3.2 (d) 0.16

79. Karl pearson’s coefficient of correlation between the heights (in inches) of teachers and students corresponding to the given data :

Heights of teachers  x : 66 67 Heights of students  y : 68 66 1 (a) (b) 2 2 1 (c) − (d) 0 2

68 69

69 72

70 70

(b) byx + bxy = 2r (x, y)

80. Two numbers within the bracket denote the ranks of 10 students of a class in two subjects -

(d) none of the above.

(1, 10), (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 2), (10, 1), then rank correlation coefficient is

(c) byx + bxy  0

(b) c > 2 (d) c = ± 3 67. Distance between two parallel planes 2x + y + 2z = 8 58. The radius of the circular section of the sphere | r | = 5 and 4x + 2y + 4z + 5 = 0 is by the plane r ⋅  (iˆ + ˆj + kˆ) = 3 3 is (a) 7/2 (b) 5/2 (a) 16 (b) 8 (c) 3/2 (d) 9/2 (c) 4 (d) None of these 68. A line with direction cosines proportional to 2, 1, 2 meets 59. The equation of the sphere whose centre has the position each of the lines x = y + a = z and x + a = 2y = 2z. vector (3iˆ + 6 ˆj − 4kˆ) and which touches the plane The co-ordinates of each of the points of intersection are given by r ·(2iˆ − 2 ˆj − kˆ) = 10 is (a) (3a, 2a, 3a), (a, a, 2a) (a) |r − (3iˆ + 6 ˆj - 4kˆ)| = 4 (b) (3a, 2a, 3a), (a, a, a) (b) | r − (3iˆ + 6 ˆj + 4kˆ)| = 4 (c) (3a, 3a, 3a), (a, a, a) ˆ ˆ ˆ (d) (2a, 3a, 3a), (2a, a, a) (c) | r − (3i + 6 j − 4k )| = 2 (d) None of these 60. The equation of the line passing through the point with position vector 2iˆ − 3 ˆj − 5kˆ and perpendicular to the plane r .(6iˆ − 3 ˆj − 5kˆ) + 2= 0 is (a) r = 2iˆ − 3 ˆj − 5kˆ + λ (− 6iˆ + 3 ˆj + 5kˆ) (b) r = 2iˆ + 3 ˆj − 5kˆ + λ (− 6iˆ + 3 ˆj + 5kˆ) (c) r = 2iˆ − 3 ˆj − 5kˆ + λ (6iˆ + 3 ˆj + 5kˆ) (d) None of these 61. The equation | r |2 – r . (2iˆ + 4 ˆj − 2kˆ) − 10 = 0 represents a (a) circle (c) sphere of radius 4 (e) None of these

(b) plane (d) sphere of radius 3

69. The ratio in which the plane r . (iˆ − 2 ˆj + 2kˆ) = 17 divides the line joining the points − 2iˆ + 4 ˆj + 7 kˆ and 3iˆ − 5 ˆj + 8kˆ  is (a) 3 : 5 (c) 3 : 10

(b) 1 : 10 (d) 1 : 5

70. The equation of the plane containing the line r = iˆ + ˆj + λ (2iˆ + ˆj + 4kˆ) is (a) r . (− iˆ − 2 ˆj + kˆ) = 3 (b) r . (iˆ + 2 ˆj − kˆ) = 0 (c) r . (iˆ + 2 ˆj − kˆ) = 3 (d) None of these x −1 y +1 z − 3 = = and the plane 3 2 −1 π : x – 2y = 0. Of the following assertions, the only

71. Gives the line L :

one that is always true is

72 . The equation | r |2 – r · (2iˆ + 4 ˆj − 2kˆ) − 10 = 0 represents a (a) sphere of radius 3 (b) sphere of radius 4 (c) plane (d) circle

r •(iˆ − 2 ˆj + 4kˆ) + 8 = 0 (b) 4 (d) 2

74. The line of intersection of the planes r ⋅ (3iˆ − ˆj + kˆ) = 1 and r ⋅ (iˆ + 4 ˆj − 2kˆ) = 2 is parallel to the vector (a) 2iˆ + 7 ˆj + 13kˆ (b) −2iˆ − 7 ˆj + 13kˆ (c) 2iˆ + 7 ˆj + 13kˆ

(a) – 1 (c) 2

(a) 30º (c) 75º

(b) 60º (d) 45º

76. The position vector of the centre of the circle | r | = 5, r ⋅ (iˆ + ˆj + kˆ) = 3 3 is (a) 3(iˆ + ˆj + kˆ) (b) iˆ + ˆj + kˆ (c) 3(iˆ + ˆj + kˆ) (d) None of the above

(a) – 1/2 (c) –2

x −1 y −1 z −1 (a) = = 1 2 1 x −1 y −1 z −1 (b) = = −1 1 −1 x −1 y −1 z −1 = = (c) 3 2 1 x −1 y −1 z −1 = = (d) 2 3 1 79. The distance between the line r = 2iˆ − 2 ˆj + 3kˆ + λ (iˆiˆ −– ˆjˆj +4k4ˆ k)ˆand

(

)

the plane r ⋅ (iˆ + 5 ˆj + kˆ) = 5 is 10 9 10 (c) 3

(a)

(b)

10 3 3

(d) None of these

(b) – 1 (d) 0

83. The intercepts of the plane 2x – 3y + 4z = 12 on the coordinate axes are given by (a) 3, – 2, 1.5 (c) 6, – 4, – 3

(b) 6, – 4, 3 (d) 2, – 3, 4

84. The locus of x2 + y2 + z2 = 0 is (a) a circle (c) (0, 0, 0)

(b) a sphere (d) None of these

85. The angle between the lines x = 1, y = 2 and y = – 1, z = 0 is (a) 30º (c) 90º

(b) 60º (d) 0º

86. Lines r = a1 + tb1 and r = a2 + sb2 lie in a plane if

77. Equation of the sphere with centre (1, – 1, 1) and radius equal to that of sphere 2x2 + 2y2 + 2z2 – 2x + 4y – 6z = 1 is 87. (a) x2 + y2 + z2 – 2x + 2y – 2z + 1 = 0 (b) x2 + y2 + z2 + 2x – 2y + 2z + 1 = 0 (c) x2 + y2 + z2 – 2x + 2y – 2z – 1 = 0 (d) none of the above 78. Equation of the line passing through (1, 1, 1) and parallel to the plane 2x + 3y + z + 5 = 0 is

(b) x + y + 2z = 2 (d) x + 2y + 4z = 10

82. If the straight lines x = 1 + s, y = –3 – λ s, z = 1 t + λ s and x = , y = 1 + t, z = 2 – t, with parameters 2 s and t, respectively, are co-planar, then λ equals

(d) −2iˆ + 7 ˆj + 13kˆ

75. A straight line which makes an angle of 60º with each of y and z-axis, inclines with x-axis at an angle

(b) 3 (d) – 2

81. The equation of the plane through (2, 3, 4) and parallel to the plane x + 2y + 4z = 5 is (a) x + 2y + 4z = 24 (c) x + 2y + 4z = 3

2 73. Radius of the circle r + r •(2iˆ − 2 ˆj − 4kˆ) − 19 = 0

(a) 5 (c) 3

80. The points (5, 2, 4), (6, – 1, 2) and (8, – 7, k) are collinear if k is equal to

(a) b1 × b2 = 0 (c) a1 × a2 = 0

(b) (a2 – a1) . (b1 × b2) = 0 (d) None of these

The coordinates of a point which is equidistant from the points (o, o, o), (a, o, o), (o, b, o) and (o, o, c) are given by a b c (a)  , ,  2 2 2

 − a −b c  , (b)  ,  2 2 2 

 a −b − c  , (c)  ,  2 2 2 

 − a b −c  (d)  , ,   2 2 2 

88. Perpendicular distance of the point (3, 4, 5) from the y-axis, is (a) 34 (c) 4

(b) 41 (d) 5

89. If r be position vector of any point on a sphere and a and b are respectively position vectors of the extremities of a diameter, then (a) r · (a – b) = 0 (b) r ·  (r – a) = 0 (c) (r + a) · (r + b) = 0 (d) (r – a) . (r – b) = 0 90. The angle between the lines 2x = 3y = – z and 6x = – y = – 4z, is (a) 0º (c) 45º

(b) 30º (d) 90º

1031

(b) L lies in π (d) None of these

Three-Dimensional Geometry

(a) L is ⊥ to π (c) L is parallel to π

1032

91. The number of straight lines that are equally inclined to three dimensional coordinate axes, is

Objective Mathematics

(a) 2 (c) 6

(b) 4 (d) 8

98. The equation of the plane containing the two lines x −1 y +1 z x y − 2 z +1 is = = and = = 2 −1 3 2 3 −1

(a) 8x + y – 5z – 7 = 0 (b) 8x + y + 5z – 7 = 0 92. The ratio in which the line joining the points (a, b, c) (c) 8x – y – 5z – 7 = 0 (d) None of these and (– a, – c, – b) is divided by the xy-plane is 99. If a plane passes through the point (1, 1, 1) and is per(a) a : b (b) b : c x −1 y −1 z −1 = = , then its pendicular to the line (c) c : a (d) c : b 3 0 4 93. The intersection of the spheres x 2 + y 2 + z 2 + 7x – 2y – z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8 is the same as the intersection of one of the spheres and the plane (a) x – y – 2z = 1 (b) x – 2y – z = 1 (c) x – y – z = 1 (d) 2x – y – z = 1 x − x1 y − y1 z − z1 = = is parallel to the plane l m n ax + by + cz + d = 0, then a b c = = l m n a b c (c) + + = 0 l m n

(b) al + bm + cn = 0

2 2 2 , , 3 3 3 (c) 2, – 2, 1

(d) None of these

100. The direction ratios of the line (a) 3, 1, – 2 3 1 −2 , , (c) 14 14 14

(d) 1, 2, 3

x−4 y−2 z−k lies in the plane = = 1 1 2 2x – 4y + z = 7, then the value of k is (b) –7 (d) No real value

(b) 2, – 4, 1 2 −4 1 , , (d) 41 41 41

x −1 y−2 z+3 = = 2 1 −2 and the plane x + y + 4 = 0, is:

101. The angle between the line

(b) 30º (c) 90º

102. If projection of a line on x, y and z axis are 6, 2 and 3 respectively, then direction cosines of the line are (a)  

6 2 3 , , 7 7 7

(b)  

(c)  

1 2 3 , , 7 7 7

(d)  none of these

(b) 1, 1, 1

96. If the line

(a) 4 (c) 7

(b) 4/3 (d) 1

(a) 0º (c) 45º

95. The direction ratios of the diagonals of a cube which joins the origin to the opposite corner are (when the three concurrent edges of the cube are coordinate axes) (a)

(a) 3/4 (c) 7/5

x – y + z – 5 = 0 = x – 3y – 6 are:

94. If line

(a)

perpendicular distance from the origin is

3 5 6 , , 5 7 7

103. The shortest distance form the point (1, 2, – 1) to the surface of the sphere x2 + y 2 + z2 = 54 is (a)   3 6 (c)   6

(b)   2 6 (d)  2 →

∧

∧

∧

104. The angle between r = ì(1)+ 2 (2i +ì ) + (2ìj + 1) − k x −1 y +1 z −1 x−3 y−k = = and = = 97. If the lines and the plane 3x – 2y + 6z = 0 (where µ is a scalar) 2 3 4 1 2 is z intersect, then k = 1  15   16  (b)  cos –1   (a)  sin–1      21  21 2 9 (b) (a)  16  9 2 π (c)  sin –1   (d)    21  (c) 0 (d) – 1 2

SOLUTIONS 1. (a) Any plane pasing through (3, 2, 0) a (x – 3) + b (y – 2) + c (z – 0) = 0

...(1)

It passes through (4, 7, 4)

a + 5b + 4c = 0

Normal to plane (i) is perpendicular to given line

a + 5b + 4c = 0 a b c = = 1 −1 1 = k so a = k, b = –k, c = k Putting the value of a, b, c in Equation (1) we get, x – y + z = 1

...(2)

∴  The co-ordinates of P are  −3k + 1 4k + 2 −5k + 3   k +1 , k + 1 , k + 1  Since the point P lies upon xy-plane ∴ z-co-ordinate of P must be ZERO. 3 −5k + 3 = 0 ⇒ – 5k + 3 = 0 ⇒ k = 5 k +1 3 Hence required ratio is  : 1 i.e., 3 : 5 internally. 5 3. (d) AD is the bisector of ∠BAC. BD AB ∴ = DC AC ∴

Now AB = =

4 +1+ 4 =

AC = =

(5 − 3) 2 + (3 − 2) 2 + (2 − 0) 2 9 =3

(−9 − 3) 2 + (6 − 2) 2 + (−3 − 0) 2

144 + 16 + 9 = 13

BD 3 = ∴  DC 13 ⇒ D divides BC in the ratio 3 : 13. ∴  The co-ordinates of D are  3 (−9) + 13 (5) 3(6) + 13(3) 3(−3) + 13(2)  , ,   3 + 13 3 + 13 3 + 13    19 57 17  =  , , .  8 16 16  4. (c) Let the required radio be k : 1 The co-ordinates of the point which divides the join of (– 2, 4, 7) and (3, – 5, 8) in the ratio k : 1 are  3k − 2 −5k + 4 8k + 7   k + 1 , k + 1 , k + 1  Since this point lies on the plane x – 2y + 3z – 17 = 0  3k − 2   − 5k + 4   8k + 7  – 2  + 3 – 17 = 0 ∴   k + 1   k + 1   k + 1  ⇒ (3k – 2) – 2 (–5k + 4) +3 (8k + 7) = 17k – 17 ⇒ 3k + 10k + 24k – 17k = 17 + 2 + 8 – 21 ⇒ 37k – 17k = 6 ⇒ 20k = 6 ;

k=

3 6 = 10 20

5. (a), (b)  We have, α = 120º and β = 60º 1 ∴  cos α = cos 120º = –  2 1 and cos β = cos 60º = 2

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But cos2α + cos2β + cos2λ = 1  −1   1  ∴   +   + cos2λ = 1  2  2 2

2

1 1 1 − = 4 4 2 1 ∴ cos2λ = ± 2 ∴ λ = 45º or 135º. cos2λ = 1 –



6. (c), (d)  Let r be inclined at an angle α to each axis, then

l = m = n = cosα Since l2 + m2 + n2 = 1 ⇒ 3cos2α =1 1 ⇒ cosα = ± 3 1 If α is acute : l = m = n =  : | r | = 6 3 r = | r | (liˆ + mjˆ + nkˆ)  1 ˆ 1 ˆ 1 ˆ i+ j+ k = 6   3 3 3 



= 2 3 (iˆ + ˆj + kˆ) .



If α is obtuse l = m = n = – 



r = | r | (liˆ + mjˆ + nkˆ)



 1 ˆ 1 ˆ 1 ˆ i− j− k = 6   − 3 3 3 

1  : | r | = 6 3

= – 2 3 (iˆ + ˆj + kˆ) . 7. (b) If the given plane contains the given line then the normal to the plane, must be perpendicular to the line and the condition for the same is

al + bm + cn = 0. 8. (b) Let l, m, n be the d.c’s of the given line segment PQ. ∴ l = cos α, m = cos β, n = cos γ where α, β, γ are the angles which the line segment PQ makes with the axes. Suppose length of line segment PQ = r. Thus, projection of line segment PQ on x-axis = PQ cos α = rl. Also the projection of line segment PQ on x-axis = 3 (given). ∴ lr = 3 Similarly, mr = 4, nr = 5. Now squaring and adding these equations, we get (lr)2 + (mr)2 + (nr)2 = 32 + 42 + 53 ⇒ r2 (l2 + m2 + n2) = 9 + 16 + 25 ⇒ r2 = 50 [ l2 + m2 + n2 = 1] ⇒ r =

50 = 5 2 .

Three-Dimensional Geometry

2. (b) Let A ≡ (1, 2, 3) and B ≡ (– 3, 4, – 5). Let the point P where the line AB meets the xy-plane divides it in the ratio k:1.

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9. (d) Suppose the line is drawn through the origin and its length be r.

Objective Mathematics

∴ If l, m, n are its direction cosines, then

i.e., 1, k – 1, – 2 and 1, 2, – 2.

∴ l2r2 = 36, m2r2 = 4 and n2r2 = 9.

Since the lines are parallel, we have

On adding these, we get

1 k − 1 −2 = = 1 2 −2 i.e., k – 1 = 2 ⇒ k = 3.

l2r2 + m2r2 + n2r2 = 36 + 4 + 9 ( l2 + m2 + n2 = 1)

∴ cos α = cos β = cosγ ⇒ l = m = n.

10. (b) If l, m, n be the direction cosines of the line perpendicular to the given lines, we have

These give

l m n = = = −1 5 3

14. (c), (d) If a line makes angles α, β, γ with the axes, we have α = β = γ.

6 2 3 ∴ l = ,m = ,n= . 7 7 7

l – m + 2n = 0 and

5 – 4, k – 1, 0 – 2 and 3 – 2, 3 – 1, –1 – 1

lr = 6, mr = 2 and nr = 3

⇒ r2 (l2 + m2 + n2) = 49 ⇒ r2 = 49 ⇒ r = 7

13. (a) The direction ratios of the lines are

2l + m – n = 0. l +m +n 2

2

1 + 25 + 9

2

=



l2 + m2 + n2 = 1,

∴ l2 + l2 + l2 = 1 ⇒ 3l2 = 1, ∴ l2 =

1 . 35

1 1 or l = ± 3 3

1 1   1 ,± ,± ∴ The d.c’s of the line are  ± .  3 3 3

5 3   −1 , , Hence the direction cosines are  .  35 35 35  15. (b) We know that l2 + m2 + n2 = 1 11. (a) The direction cosines l, m, n of the line are given by

l m n = = = 6 2 3

l 2 + m2 + n2 62 + 22 + 32

=

1 1 = 7 49

6 3 2 ,m = ,n = 7 7 7 The required projection is given by = l (x2 – x1) + m ( y2 – y1) + n (z2 – z1) ∴ l =

6 [2 – (– 7 6 ×3+ = 7 18 10 + = 7 7 =

16. (d) The equation of xy plane is z = 0. ∴ direction cosines of its normal are 0, 0, 1.

2 3 (5 – 0) + (1 – 3) 7 7 2 3 ×5+ ×–2 7 7 22 18 + 10 − 6 6 – = = . 7 7 7

1)] +

12. (c) Let the line makes an angle γ with the z-axis, then, since cos2α + cos2β + cos2γ = 1 ⇒ cos2

π π + cos2 + cos2γ = 1 4 4

 2

 1  1  or  +  + cos2γ = 1   2   2 2

1 1 + + cos2γ = 1 2 2 ⇒ cos2γ = 0 π π ⇒ cosγ = 0 = cos  ⇒γ= . 2 2

π  ∵α = β = 4 

z-axis.

17. (a) Let A (5, 2, 4), B (6, –1, 2) C (8, –7, k) be the given points Direction ratios of AB are 6 – 5, – 1 – 2, 2 – 4 i.e., 1, – 3, – 2 Direction ratios of BC are 8 – 6, – 7 + 1, k – 2 i.e., 2, – 6, k – 2 Since A, B, C are collinear 2 −6 k − 2 = = 1 −3 −2 ∴ k – 2 = – 4 ⇒ k = 2 – 4 = – 2. ∴

18. (d) Let l, m, n be the DC's of r. Then l = m = n (given). ∴ l2 + m2 + n2 = 1 ⇒ 3l2 = 1 ⇒ 1 =

1 =m =n 3

∴ r = | r | (liˆ + mjˆ + nkˆ) = 6  1 ˆ 1 ˆ 1 ˆ i+ j+ k   3 3 3 

or

Hence the line makes an angle of

∴ cos2α + cos2β + cos2γ = 1. Changing cosines into sines, we get (1 – sin2α) + (1 – sin2β) + (1 – sin2γ) = 1 ⇒ sin2α + sin2β + sin2γ = 3 – 1 = 2.

π with the 2

= 2 3 (iˆ + ˆj + kˆ) .

19. (a) DRs of r are 2, –3, 6. Therefore its DCs are 2 −3 6 , m= , n= l= 7 7 7

= 6iˆ − 9 ˆj + 18kˆ . 20. (c) Let the equation of the plane be

x y z + + = 1. α β γ

Then points A, B, C are (α, 0, 0), (0, β, 0) and (0, 0, γ) respectively, Now (a, b, c) are the coordinates of centroid of triangle ABC. 0+β+0 0+0+γ α+0+0 ,b= ,c = 3 3 3 i.e., α = 3a, β = 3b, γ = 3c ∴ The equation of plane is x y z x y z + + + + =3 = 1 i.e., 3a 3b 3c a b c ∴ a =

∴ k = 3. 21. (b) P is (2, 3, –1) and O is (0, 0, 0) ; direction cosines of OP are proportional to [2, 3, –1]. Therefore these will be the coefficients of x, y, z in a plane which is perpendicular to OP. So let the plane be 2x + 3y – z = λ. The plane also passes through P (2, 3, – 1). ∴ 4 + 9 + 1 = λ ∴ λ = 14. Hence the required plane is 2x + 3y – z = 14. 22. (a) Points are (3, 4, –1) and (2, –1, 5). Hence the direction ratios of the line joining them are [3 – 2, 4 + 1, – 1 – 5], i.e., [1, 5, – 6].

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we get – 2b + 4c + d = 0 and – 4b + 5c + d = 0 giving b = c = d . −1 −2 6 Thus the plane is y + 2z = 6. 25. (d) Direction ratios of OC are [1, – 1, 2]. Since the plane passes through (1, 0, 1), let its equation be a (x – 1) + b ( y – 0) + c (z – 1) = 0 ...(1) Its normal is ⊥ to OC. ∴ 1.a + (– 1) b + 2 c = 0, i.e., a – b + 2c = 0  ...(2) Also since (1) passes through (3, 1, 2), we have 2a + b + c = 0.  ....(3) From (2) and (3), a b c a b c = = , i.e., = = −1 − 2 4 − 1 1 + 2 −1 1 1 Thus the plane is (x – 1) – 1 (y – 0) – (z – 1) = 0 or x – y – z = 0. 26. (c) Centre of the given sphere is (–1, 1, 2) and radius = (−1) 2 + (1) 2 + (2) 2 + 19 = 25 = 5 Length of the perpendicular from centre O on the given plane is ON =

−1 × 1 + 1 × 2 + 2 × 2 + 7 1 +2 +2 2

2

2

=

12 =4 3

Therefore any plane perpendicular to this line will have 1, 5,– 6 as the coefficients of x, y and z respectively. Let the plane be x + 5y – 6z = λ. It passes through (2, – 3, 1). ∴ λ = – 19. Hence the required plane is x + 5y – 6z + 19 = 0. 23. (c) Any plane parallel to x-axes is by + cz + d = 0. If it passes through (2, 3, 1) and (4, –5, 3), then 3b + c + d = 0 and – 5b + 3c + d = 0, i.e.,

c d b c d b = = , i.e., = = . 4 −7 1− 3 −5 − 3 9+5 1

Hence the plane parallel to x-axis is y + 4z – 7 = 0. 24. (c) Let the plane be ax + by + cz + d = 0, ...(1) The yz-plane is x = 0 or l.x + 0.y + 0.z = 0. ...(2) Since (1) and (2) are perpendicular to each other, we have a · 1 + b · 0 + c · 0 = 0, i.e., a = 0. ∴ The plane (1) reduces to by + cz + d = 0. Now since it passes through (1, – 2, 4) and (3, – 4, 5),

In ∆OBN, OB2′ = ON2 + NB2 52 = 42 + NB NB2 = 25 – 16 NB2 = 9 NB = 3. 27. (c) The lines x−2 y−3 z−4  = = 1 1 −k x −1 y − 4 z − 5  = = k 2 1 are coplanar ∴

1 −1 −1 1 1 −k = 0 k 2 1

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

Three-Dimensional Geometry

3 6  2 ∴ r = | r | (liˆ + mjˆ + nkˆ) = 21  iˆ − ˆj + kˆ  7 7 7 

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Objective Mathematics

⇒ 1(1 + 2k) + 1(1 + k2) – (2 – k) = 0 ⇒ 2k + 1 + 1 + k2 – 2 + k = 0 ⇒ k = 0 or –3. 28. (a) The equation of the plane through (– 1, 3, 2) is a (x + 1) + b (y – 3) + c (z – 2) = 0. Since the plane is ⊥ to the two given planes, we have a + 2b + 3c = 0 and 3a + 3b + 2c = 0 a c −b a b c = = = = . or giving 4−6 2−6 3−6 2 −4 3

31. (a) Let a, b, c be the intercepts when Ox, Oy, Oz are taken as axes ; then the equation of the plane is x y z + + = 1 ...(1) a b c Also let a', b', c' be the intercepts when OX, OY, OZ are taken as axes ; then in this case equation of the same plane is

29. (d) Equation of the line

i.e.,

x = ay + b, z = cy + d can be written as or

...(1)

Also, x = a'y + b', z = c'y + d' can be written as x − b' z−d' = y, =y a' c' x − b' y−0 z−d' ⇒ = = a 1 c'



...(2)

Lines (1) and (2) will be perpendicular to each other if l 1l 2 + m 1m 2 + n 1n 2 = 0 ⇒ aa' + 1 × 1 + c × c' = 0 aa' + cc' + 1 = 0 30. (a), (c)  Let the plane be

x y z + + = 1. α β γ

It passes through (a, b, c) ; ∴

a b c + + = 1. α β γ

....(2)

Now (1) and (2) are equations of the same plane and in both the cases the origin is same. Hence length of the perpendicular drawn from the origin to the plane in both the case must be the same,

Thus, the required plane is 2 (x + 1) – 4 (y – 3) + 3 (z – 2) = 0 or 2x – 4y + 3z + 8 = 0.

z−d x−b = y, =y c a x−b y−0 z−d = =  ⇒ a 1 c

X Y Z + + = 1 a' b' c'

...(1)

Now co-ordinates of the points A, B, C are (α, 0, 0), (0, β, 0) and (0, 0, γ) respectively.

1 = 1 1 1 + + a 2 b2 c2

1 1 1 1 + + a '2 b '2 c ' 2

1 1 1 1 1 1 + 2 + 2 = a '2 + b '2 + c ' 2 2 a b c

∴k=1

32. (b) Let x + 1 = y + 3 = z − 2 = r. 1 3 −2 Then any point on the line is (r – 1, 3r – 3, –2r + 2). If this point lies on the plane 3x + 4y + 5z = 25, then 3 (r – 1) + 4 (3r – 3) + 5 (– 2r + 2) = 25 or 5r = 30, i.e., r = 6. Putting this value of r in the co-ordinates of the point, the required point is (5, 15, – 10). 33. (a), (c)  Let

x + 1 y − 12 z − 7 = r. = = −1 5 2

Any point on the line is (– r – 1, 5r + 12, 2r + 7) for every value of r. If this point lies on the surface 11x2 – 5y2 + z2 = 0, then 11 (– r – 1)2 – 5 (5r + 12)2 + (2r + 7)2 = 0. i.e., 110r2 + 550r + 660 = 0, i.e., r2 + 5r + 6 = 0 i.e., (r + 3) (r + 2) = 0, i.e., r = – 3, – 2 For these two values of r, the two points in which the given line cuts the surface are (2, – 3, 1) and (1, 2, 3).

Equations of the planes through A, B, C parallel to co-ordinate planes are x = α ...(2) 34. (c) Any point on the line is (r + 3. 2r + 4, 2r + 5). y = β ...(3) It lies on the plane x + y + z = 17, and z = γ. ...(4) ∴ (r + 3) + (2r + 4) + (2r + 5) = 17, The locus of their point of intersection will be obtained i.e., r = 1. by eliminating α, β, γ from these with the help of the Thus the point of intersection of the plane and the line relation (1). is (4, 6, 7). We thus get Required distance = distance between (3, 4, 5) and a b c (4, 6, 7) + + = 1,  i.e., ayz + bxz + cxy = xyz. x y z = {(4 − 3) 2 + (6 − 4) 2 + (7 − 5) 2 } = (1 + 4 + 4) = 3. which is the required locus.

x −1 y + 2 z −1 = = = r (say). 2 3 −6

...(1)

Then any point on (1) is  (2r + 1, 3r – 2, – 6r + 3). If this point lies on the plane x – y + z = 5, then (2r + 1) – (3r – 2) + (– 6r + 3) = 5 ⇒ – 7r + 6 = 5, 1 i.e., r = . 7  9 11 15  Hence the point is  , − ,  . 7 7 7 9 11 15 Distance between (1, – 2, 3) and  , − ,  7 7 7 =

9 36   4 +  =  + 49 49 49 

 49    = 1. 49

3l – m + n = 0 and 5l + m + 3n = 0, l m n = = i.e., l = m = n . ∴ −3 − 1 5 − 9 3 + 5 1 1 −2

b c a = = , 1 − 15 −12 − 2 −10 − 4

a b c = = . 1 1 1 Hence the plane (1) becomes i.e.,

(x + 1) + (y – 3) + (z + 2) = 0, i.e., x + y + z = 0. 39. (b) Any plane through the given line is a (x – 1) + b (y + 1) + c (z – 3) = 0,

...(1)

where 2a – b + 4c = 0

...(2)

If this plane is perpendicular to x + 2y + z = 12, then their normals are also perpendicular to each other. ∴ a + 2b + c = 0 ...(3) a b c , = = −1 − 8 4 − 2 4 + 1

i.e., a = b = c . −9 2 5 ∴ plane (1) becomes i.e., 9x – 2y – 5z + 4 = 0. 40. (b) Any plane through (2, – 1, 0) is

37. (a) Any plane through the given line 2x – y + 3z + 1 + λ (x + y + z + 3) = 0 (From S + λS' = 0) x y z = = , If this plane is parallel to the line 1 2 3 then the normal to the plane is also perpendicular to the above line or (2 + λ) 1 + (λ – 1) 2 + (3 + λ) 3 = 0.

a (x – 2) + b (y + 1) + cz = 0 ; it will pass through (3, – 4, 5) if a – 3b + 5c = 0

3 This gives λ = –  and the required plane is 2 x – 5y + 3z – 7 = 0.

...(2)

Also (1) will be parallel to 2x = 3y = 4z, i.e.,

if a · 

z y x = = 1, 1 1 4 3 2 1 1 + b ·  + c ·  1 = 0 3 2 4

or 6a + 4b + 3c = 0.  From (2) and (3), i.e.,

(From l1l2 + m1m2 + n1n2 = 0)

...(1)

...(3)

b c a = = , 30 − 3 4 + 18 −9 − 20

a b c . = = 29 −27 −22

Hence the plane is  29 (x – 2) – 27 (y + 1) – 22z = 0. 41. (a) The vector equation of the plane passing through points a, b, c is

x +1 y−3 z+2 = = is −3 2 1

a (x + 1) +b (y – 3) + c (z + 2) = 0,

...(1)

where – 3a + 2b + c = 0.

...(2)

r ⋅ (a × b + b × c + c × a) = [a b c]. Therefore, the length of the ⊥ from the origin to this plane is given by [a b c] . |a × b + b × c + c × a |

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...(3)

– 9 (x – 1) + 2 (y + 1) + 5 (z – 3) = 0.

Now a plane ⊥ to PQ will have l, m, n as the coefficients of x, y and z. Hence the plane ⊥ to PQ is x + y – 2z = λ. It passes through (2, 1, 4) ; ∴ 2 + 1 – 2.4 = λ i.e., λ = – 5. Hence the required plane is x + y – 2z = – 5.

If the plane through (0, 7, – 7), then

From (2) and (3),

From (2) and (3),

36. (b) Let [l, m, n] be the direction -cosines of PQ, then

38. (c) Any plane containing

a + 4b – 5c = 0

Three-Dimensional Geometry

35. (a) Equation of the line through (1, – 2, 3) parallel to x y z −1 the line = = is 2 3 −6

1038

Objective Mathematics

42. (b) The lines r = a + λ (b × c) and r = b + µ (c × a) pass through points a and b respectively and are parallel to vector b × c and c × a respectively. Therefore, they intersect if a – b, b × c and c × a are coplanar and so (a – b) ⋅ {(b × c) × (c × a)} = 0 ⇒ (a – b) ⋅ ([b c a] c – [b c c] a) = 0 ⇒ ((a – b) ⋅ c) [ b c a] = 0 ⇒ a ⋅ c – b ⋅ c = 0 ⇒ a ⋅ c = b ⋅ c. 43. (b) The required plane passes through the points having position vectors a1 and a2 and is parallel to the vector b. Therefore, if r is the position vector of any variable point on the plane, then the vectors r ­– a1, a2 – a1 and b are coplanar. ∴ (r – a1) ⋅ ((a2 – a1) × b) = 0 ⇒ r ⋅ (a2 – a1) × b – a1 ⋅ (a2 × b) = 0 ⇒ r ⋅ (a2 – a1) × b = [a1 a2 b]. 44. (c) A straight line r = a + λb meets the plane r ⋅ n = 0 in P for which λ is given by (a + λb) ⋅ n = 0 ⇒λ= −

a⋅n b⋅n

−8 − 3 − 40 AP ⋅ (6iˆ + 3 ˆj − 4kˆ) = = 61 |6iˆ + 3 ˆj − 4kˆ |

=

∴ PN =

AP 2 − AN 2 =

61

110 − 61 = 7.

48. (a) The position vector of any point on the given line is iˆ + ˆj + t  ( 2iˆ + ˆj + 4kˆ ) = (1 + 2t)  iˆ + (1 + t) ˆj + 4t ˆ k This lies on r ⋅  (iˆ + 2 ˆj − kˆ) = 3 if (1 + 2t) ⋅ 1 + (1 + t) ⋅ 2 + 4t (– 1) = 3 i.e., if 1 + 2t + 2 + 2t – 4t = 3. i.e., if 3 = 3 which is true. Hence the plane r ⋅  (iˆ + 2 ˆj − kˆ) = 3 contains the given line. 49. (a) The line of intersection of the planes r ⋅  (3iˆ − ˆj + kˆ) = 1 and r ⋅  (iˆ + 4 ˆj − 2kˆ) = 2 is ⊥ to each of the normal vectors n1 = 3iˆ − ˆj + kˆ and n2 = iˆ + 4 ˆj − 2kˆ ∴ It is parallel to the vector n × n = (3iˆ − ˆj + kˆ) × (iˆ + 4 ˆj − 2kˆ) 1

2

= –  2iˆ + 7 ˆj + 13kˆ .

Thus, the position vector of P is a⋅n r=a– b b⋅n  [Putting the value of λ in r = a + λb] 45. (c) The position vectors of two given points are a = iˆ − ˆj + 3 kˆ and b = 3 iˆ + 3 ˆj + 3 kˆ and the equation of the given plane is r ⋅  (5iˆ + 2 ˆj − 7 kˆ) + 9 = 0 or r ⋅ n + d = 0. We have, a ⋅ n + d = (iˆ − ˆj − 3kˆ) ⋅ (5iˆ + 2 ˆj − 7 kˆ) + 9 = 5 – 2 – 21 + 9 < 0 and b · n + d = (3iˆ + 3 ˆj − 3kˆ) ⋅ (5iˆ + 2 ˆj − 7 kˆ) + 9 = 15 + 6 – 21 + 9 > 0 So, the points a and b are on the opposite sides of the plane. 46. (c) Lines lie in a plane if (a2 – a1) ⋅ (b1 × b2) = 0 b1 × b2 is a vector ⊥ to b1, b2. 47. (a) We have AP = −3iˆ − ˆj + 10kˆ ∴ | AP | =



9 + 1 + 100 =

50. (c) The given plane passes through a and parallel to the vectors b – a and c. ∴ it is normal to (b – a) × c. Hence the equation is (r – a) .[(b – a) × c] = 0 or r . [b × c + c × a] = [a b c] ∴  The length of the ⊥ from the origin to this plane is

[abc] . |b × c + c × a | 51. (c) Centre of the given sphere is (–2, 1, 3) Length of the perpendicular from centre on the plane is =

−2 × 12 + 1 × 4 + 3 × 3 − 327 −11 − 327 = = 26 13 144 + 16 + 9

Shortest distance between the plane and sphere = 26 – radius of sphere = 26 – 4 + 1 + 9 + 155 = 26 – 13 = 13. 52. (d) Consider OX, OY, OZ and Ox, Oy, Oz are two system of rectangular axes.

110

Equation of the plane corresponding to OX, OY, OZ as axes is X Y Z + + = 1 a b c

...(1)

Similarly, equation of the plane corresponding to Ox, Oy, Oz as axes is AN = Projection of AP on 6iˆ + 3 ˆj − 4kˆ

x y z + + a ' b ' c ' = 1

...(2)

1 1 1 1 1 1 ⇒ 2 + 2 + 2 = 2 + 2 + 2 a b c a' b' c' ⇒

57. (a) Let A (a, b, c) be the fixed point on the variable plane π.

1 1 1 1 1 1 = 0. + + − − − a 2 b 2 c 2 a '2 b '2 c '2

53. (a) Here l = cosθ, m = cosβ, n = cosθ Now l2 + m2 + n2 = 1⇒ 2 cos2θ + cos2β = 1 ⇒ 2cos2θ = sin2β ...(1) Given : sin2β = 3sin2θ ⇒   2cos2θ = 3sin2θ ⇒ 5cos2θ = 3. 54. (c) The distance between the parallel planes r ⋅  (2iˆ − ˆj + 3kˆ) = 4 13 and r ⋅ (2iˆ − ˆj + 3kˆ) = –  3 ∵r ⋅ (6iˆ − 3 ˆj + 9kˆ) + 13 = 0    so r ⋅ (2iˆ − ˆj + 3kˆ) = − 13   3 

is

 −13  4−  3  22 + (−1) 2 + 32 13 3 4 +1+ 9 = 4+

=



 |d − k | ∵ Reqd. distance =  |n |  

Now D.R.’s of OM are x – 0, y – 0, z – 0 i.e., x, y, z D.R.’s of MA are x – a, y – b, z – c. Since OM ⊥ MA, x (x – a) + y (y – b) + z (z – c) = 0 ⇒ x2 + y2 + z2 – ax – by – cz = 0 which is the required locus and represents a sphere. 58. (c) The sphere | r | = 5 has centre at the origin and radius 5. Distance of the plane r ⋅  (iˆ + ˆj + kˆ) = 3 3 from the origin.

25 3 = 25 . 3 14 14

55. (d) Any plane through the intersection of r ⋅ a = p and r ⋅ b = q is r  · (a – λb) = p – λq Since it passes through the origin ∴ 0 ⋅ (a – λb) = p – λq

= ....(1)

p . q Putting this value of λ in (1), we get

⇒ p – λq = 0 ⇒ λ =

 p  p r ⋅   a − b  = p − q = 0 q  q  i.e., r ⋅ (aq – pb) = 0 This is the required equation. 56. (b) The equation of the sphere is | r − 2iˆ − ˆj + 6kˆ | = 18 ⇒ Its centre is at the point (2iˆ + ˆj − 6kˆ) , i.e., at (2, 1, – 6). Co-ordinates of A are (3, 2, – 2). Lets the co-ordinates of B be (α, β, γ)

3 3 = | iˆ + ˆj + kˆ |

3 3 12 + 12 + 12

= 3.

Thus, in figure OP = 5, ON = 3. ∴ NP2 = OP2 – ON2 = (5)2 – (3)2 = 16, ∴ NP = 4 Hence, the radius of the circular section = NP = 4. 59. (a) The centre of the sphere has the position vector 3iˆ + 6 ˆj − 4kˆ . Radius = The distance of the centre whose position vector a = 3iˆ + 6 ˆj − 4kˆ from the plane r ⋅  (2iˆ − 2 ˆj − kˆ) = 10 =

|(3iˆ + 6 ˆj − 4kˆ) ⋅ (2iˆ − 2 ˆj − kˆ) − 10| |a ⋅n − d | = |n | 22 + (−2) 2 + (−1) 2

=

|6 − 12 + 4 − 10| 12 | −12 | = = = 4. 3 4 + 4 +1 3

1039

3+α 2+β −2 + γ = 2, = 1  and = – 6. 2 2 2 ⇒ α = 1, β = 0, γ = – 10. Hence B ≡ (1, 0, – 10). Then,

Three-Dimensional Geometry

Length of perpendicular from origin to (1) and (2) must be same 1 1 i.e, = 1 1 1 1 1 1 + + + + a 2 b2 c2 a '2 b '2 c ' 2

1040

64. (b) The sphere x2 + y2 + z2 = 49 has centre at the origin (0, 0, 0) and radius 7.

Objective Mathematics

Centre = (3, 6, – 4); Radius = 4. Required equation of the sphere is | r − (3iˆ + 6 ˆj − 4kˆ)| = 4.

Distance of the plane 2x + 3y – z – 5 14 = 0 from the origin.

60. (a) Equation of plane is r ⋅  (6iˆ − 3 ˆj − 5kˆ) = – 2 ⇒ r ⋅  (−6iˆ + 3 ˆj + 5kˆ) = 2

=

2(0) + 3(0) − (0) − 5 14 22 + 32 + (−1) 2

A vector normal to the plane is −6iˆ + 3 ˆj + 5kˆ . Direction ratios of normal to the plane are – 6, 3, 5.

−5 14

Since the required line is perpendicular to the plane, therefore d-ratios of line are – 6, 3, 5.

=

=

Hence vector equation of the line passing through the point whose position vector is 2iˆ − 3 ˆj − 5kˆ and having direction ratios – 6, 3, 5 is

Thus in Fig.

14

5 14 = 5. 14

r  =  2iˆ − 3 ˆj − 5kˆ + λ (−6iˆ + 3 ˆj + 5kˆ) . 61. (c) Since the equation | r |2 – 2 (r ⋅ a) + λ = 0 represents a sphere of radius | a |2 − λ , therefore | r |2 – r ⋅  (2iˆ + 4 ˆj − 2kˆ) – 10 = 0 represents a sphere of radius =

| iˆ + 2 ˆj − kˆ |2 + 10 =

6 + 10 = 4.

OP = 7, ON = 5 0NP2 = OP2 – ON2 = (7)2 – (5)2 = 49 – 25 = 24 ∴ NP = 2 6 .

62. (a) Let the co-ordinates of the points A, B, C be (a, 0, 0), (0, b, 0), (0, 0, c) respectively. Then equation of the sphere OABC is

Hence the radius of the circle = NP = 2 6 . 65. (a) The point A (6, 7, 7) is on the line. Let the perpendicular from P meet the line in L. Then

x2 + y2 + z2 – ax – by – cz = 0 Radius of this sphere is equal to k ∴ a2 + b2 + c2 = 4k2

...(1)

Co-ordinates (x, y, z) of the centroid of ∆ABC are given by 1 1 1 x = a, y = b, z = c, so that 3 3 3 a = 3x, b = 3y, c =3z. ...(2) Eliminating a, b, c from (1) and (2), we have

AP2 = (6 – 1)2 + (7 – 2)2 + (7 – 3)2 = 66. Also

(3x)2 + (3y)2 + (3z)2 = 4k2.

AL = projection of AP on line 3 2 −2   , ,  actual d.c.'s  17 17 17 

or 9 (x2 + y2 + z2) = 4k2. 63. (c) Let co-ordinates of the points A, B, C be (a, 0, 0), (0, b, 0) and (0, 0, c) respectively. Then equation of the sphere OABC is x2 + y2 + z2 – ax – by – cz = 0.

=

Radius of this sphere is equal to 2k. ∴ a2 + b2 + c2 = 4 (2k)2 = 16k2 ...(1) Also let (x, y, z) be the co-ordinates of the centroid of the tetrahedron OABC, then 1 1 1 x = a, y = b, z = c, 4 4 4 so that a = 4x,

b = 4y, c = 4z

= (6 – 1) ·

...(2)

Eliminating a, b, c from (1) and (2), the required locus is

17 .

∴ ⊥ distance d of P from the line is given by d2 = AP2 – AL2 = 66 – 17 = 49, so that d = 7. 66. (d) Since d.c. of line are < ∴

1 1 1 , , > c c c

1 1 1 + 2 + 2 = 1 ⇒ c2 = 3 ⇒ c = ± 3 . 2 c c c

67. (a) Given planes are 4x + 2y + 4z – 16 = 0 and4x + 2y + 4z + 5 = 0 Distance between planes (1) and (2) is

16x2 + 16y2 + 16z2 = 16k2 or x2 + y2 + z2 = k2.

−2 3 2 + (7 – 2) ⋅  + (7 – 3) ⋅  17 17 17

=

−16 − 5 21 7 = . = 16 + 4 + 16 6 2

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

It’s radius =

1 + 4 + 1 + 10 = 4.

73. (b) Given circle is intersection of sphere

CD :

x+a y z = = 2 1 1



...(1)

– 2i + 4j + 7k and 3i – 5j + 8k in the ratio t : 1 at the point P. −5t + 4 8t + 7 3t − 2 j + k. i + t +1 t +1 t +1

This lies on the given plane, 3t − 2 −5t + 4 8t + 7  ⋅ 1 + (– 2) + (3) = 17 t +1 t +1 t +1

⇒ 3t – 2 + 10t – 8 + 24t + 21 = 17t + 17

∴ Reqd. ratio is 3 : 10.

=

−1 − 2 + 4 + 8 9 = =3 3 1+ 4 + 4 1 + 1 + 4 + 19 = 5

R = Radius of the sphere =

=

69. (c) Let the plane r ⋅ (i – 2j + 3k) = 17 divide the line joining the points

∴ 20t = 17 – 21 + 10 = 6 ⇒ t =

...(ii)

Radius of the circle

(ii) and (iii) ⇒ r – a = 2a (i) and (iii) ⇒ λ = a  ∴  r = 3a, λ = a ∴ p ≡ (3a, 2a, 3a) and Q ≡ (a, a, a).

∴

and plane x – 2y + 2z + 8 = 0

p = Length of the ⊥ from, (– 1, 1, 2) upon (ii)

r−λ−a r−λ r − 2λ + a = = 1 2 2 (i) (ii) (iii)

∴ P is

...(i)

Centre of sphere is (– 1, 1, 2).

Let p ≅ (r, r – a, r) and Q = (2λ – a, λ, λ) Direction ratios of PQ are r – 2λ + a, r – λ – a, r – λ According to question direction ratios of PQ are (2, 1, 2) ∴

x2 + y2 + z2 + 2x – 2y – 4z – 19 = 0

3 6 = 10 20

R2 − p2 =

25 − 9 =

16 = 4.

74. (d) The line of intersection of the planes r ⋅ (3i – j + k) = 1 and r ⋅ (i + 4j – 2k) = 2 is ⊥ to each of the normal vectors n1 =  3i – j + k and n2 =  i + 4j – 2k ∴ it is parallel to the vector n1 × n2 = (3i – j + k) × (i + 4j – 2k) = – 2i + 7j + 13k. 75. (d) Clearly cos2 60º + cos2 60º + cos2 α = 1 where α is the angle which the straight line makes with Xaxis. 1 1 1 ∴ cos2 α = 1 – − = 4 4 2 ⇒ cos α =

1 ⇒ α = 45º. 2

76. (a) The equation of ON is r = λ (i + j + k)

.....(i)

70. (c) Any point on the given line is i + j + λ (2i + j + 4k) i.e., (1 + 2λ) i + (1 + λ) j + 4λk which lies on the plane r ⋅ (i + 2j – k) = 3. Hence the plane r ⋅ (i + 2j – k) = 3 contains the given line. 71. (b) Since 3 (1) + 2 (– 2) + (– 1) (– 1)

Since it passes through the origin and is parallel to the vector (i + j + k), any pt. on it is λ (i + j + k). If this pt. lies on the plane r · (i + j + k) = 3 3

= 3 – 4 + 1 = 0,

then λ (i + j + k) ⋅ (i + j + k) = 3 3

∴ given line is ⊥ to the normal to the plane i.e., given line is parallel to the given plane.

or λ (1 + 1 + 1) = 3 3 ∴ λ = 3

Also (1, – 1, 3) lies on the plane x – 2y – z = 0 if 1 – 2 (– 1) – 3 = 0 i.e., 1 + 2 – 3 = 0 which is true

∴ L lies in plane π.

Putting the value of λ in (i), we get the position vector N i.e., centre of the circle as 3 (i + j + k). 77. (c) Radius of the sphere 2x2 + 2y2 + 2z2 – 2x + 4y – 6z = 1

1041

72. (b) Taking, r = xiˆ + yjˆ + zkˆ , the given equation can be written as x2 + y2 + z2 – 2x – 4y + 2z – 10 = 0, which represents a sphere.

Three-Dimensional Geometry

68. (b) Given AB : x = y + a = z 1 1 1

1042

i.e., x2 + y2 + z2 – x + 2y – 3z –

Objective Mathematics

2

1 = 0 is 2

and x =

i.e., x − 0 = y − 1 = z − 2  1 2 −2

2

1 3  1 2   + (−1) +   −  −  2 2 2 

t , y = 1 + t, z = 2 – t 2

=

Lines (1) and (2) are coplanar,

1 9 1 +1+ + = 2 4 4 2

⇒

∴ Reqd. sphere is (x – 1)2 + (y – 1)2 + (z – 1)2 = (2)2 or x2 + y2 + z2 – 2x + 2y – 2z – 1 = 0.

79. (b) The given line is r = a + tb

x y z + + =1 6 −4 3 ∴ Intercepts are 6, – 4, 3.

83. (b) Plane can be written as

84. (c) x2 + y2 + z2 = 0

where a = 2i – 2j + 3k, b = i – j + 4k and given plane is r ⋅ n = p, where n = iˆ + 5 ˆj + kˆ , p = 5.

1 − 0 −3 − 1 1 − 2 1 −λ λ =0 1 2 −2

⇒ –5λ – 10 = 0 ⇒ λ = –2.

78. (b) If the direction ratios of the line are l, m, n then it is perpendicular to the normal to the plane. ∴ 2l + 3m + n = 0. And the only values of l, m, n that satisfy this equation are – 1, 1, – 1. ∴ (b) is the correct answer.

...(2)

⇒  x = 0, y = 0, z = 0.

85. (c) Lines are

z y +1 x −1 y − 2 z x and = = = = 0 0 0 0 1 1

∴ cos θ = 0.1 + 0.0 + 1.0 = 0 Since b · n = 1 – 5 + 4 = 0, ∴ θ = 90º. ∴ given line is parallel to the given plane ∴ the distance between the line and the plane is equal 86. (b) Lines lie in a plane if (a2 – a1) · (b1 ×  b2) = 0, to length of the perpendicular from the point a = since b1 × b2 is a vector ⊥ to b1, b2. 2i – 2j + 3k on the line to the given plane. 87. (a) Sphere passing through (a, 0, 0) (0, b, 0) (0, 0, c) (2i − 2 j + 3k ) ⋅ (i + 5 j + k ) − 5 and (0, 0, 0) is x2 + y2 + z2 –ax – by – cz = 0. Its ∴ Reqd. distance = a b c 1 + 25 + 1 centre  , ,  is equidistant from given points. 2 2 2 2 − 10 + 3 − 5 10 = = . 88. (a) Distance of (α, β, γ) from y-axis is given by 27 3 3 d = α2 + γ 2 80. (d) Let A (5, 2, 4), B (6, – 1, 2), and C (8, – 7, k) be the ∴ Distance (d) of (3, 4, 5) from y-axis is given points. Direction ratios of AB are i.e., Direction ratios of BC are i.e., Since A, B, C are collinear ∴

d=

9 + 25 =

34 .

89. (d) Let ends of diameter be A and B and P be any point on sphere whose position vector is r. Then AP & BP will be mutually perpendicular ∴ AP × BP = 0

2 −6 k − 2 = = 1 −3 −2

...(1)

But AP = r – a, BP = r – b ∴ From (1), (r – a) · (r – b) = 0.

∴ k – 2 = – 4 ⇒ k = – 4 + 2 = – 2. 81. (a) Any plane parallel to the plane x + 2y + 4z = 5 is x + 2y + 4z = k This passes through (2, 3, 4), ∴ 2 + 6 + 16 = k ⇒ k = 24 ∴ Required plane is x + 2y + 4z = 24.

90. (d) The given lines may be written as z x y x y z and = = . = = −1 /4 1/ 6 −1 1 / 2 1 / 3 −1 1 1 1 Direction ratios of two lines are , , – 1 and , 3 2 6 −1 – 1, . 4 Angle between two lines is given by

82. (c) Given lines are x = 1 + s, y = – 3 – λ s, z = 1 + λ s i.e., x − 1 = y + 3 = z − 1  −λ 1 λ

32 + 52 =

...(1)

cosθ =

a1a2 + b1b2 + c1c2 a + b12 + c12 2 1

a22 + b22 + c22

Then, coordinates of corner (P) opposite to origin are (a, a, a).

=0

∴ Direction ratios of diagonal OP are a – 0, a – 0, a – 0 i.e., a, a, a. i.e., 1, 1, 1.

⇒ θ = 90º.

91. (b) If α, β, γ are the angles made by the line with x, 96. (c) The point (4, 2, k) on the line also lies on the plane 2x – 4y + z = 7. y, and z-axis respectively, then So, 8 – 8 + k = 7 ⇒ k = 7. cos2α + cos2β + cos2γ = 1 Given α = β = γ, ∴ 3cos2α = 1 or cos α = ±

97. (b) 98. (a) Equation of any plane through the first line is

1 3

1 1   1 ,± ,± Possible direction cosines are  ± .  3 3 3  1 1 1  , Different sets of Dc’s are  , ,  3 3 3   1 1 −1  ,   1 , −1 , 1  , ,     3 3 3  3 3 3

k ( −b) + c  m1 z2 + m2 z1  z =  Using z = m + m  k +1 1 2

If XY plane divides the join of (a, b, c) and (– a, – c, – b) at this point then z = 0 −bk + c c = 0 or k = . k +1 b

S1 – S2 = 0 ⇒ x2 + y2 + z2 + 7x – 2y – z – 13 – (x2 + y2 + z2 – 3x + 3y + 4z – 8) = 0 ⇒ 10x – 5y – 5z – 5 = 0 ⇒ 2x – y – z = 1. 94. (b) Angle between line

2

2

or

al + bm + cn a + b + c2 2

2

al + bm + cn = 0.

0+0−7 9 + 16

=

7 . 5

100. (a) If l, m, n are the d.c's of the line, then 1 ⋅ l – 1 ⋅ m + 1 ⋅ n = 0 and 1 ⋅ l – 3 ⋅ m + 0 ⋅ n = 0 l m n = = ∴ 0 + 3 1 − 0 −3 + 1



If line is parallel to plane then θ = 0º ∴ sin θ =

3 · (x – 1) + 0 · ( y – 1) + 4 · (z – 1) = 0 or 3x + 4y – 7 = 0 Distance of plane from the origin

∴ Their d.c’s are 2/3, 1/3, – 2/3

l +m +n 2

99. (c) Equation of plane through (1, 1, 1) and perpendicular to line

101. (c) Direction ratios of the line and the normal to the plane are 2, 1, – 2 and 1, 1, 0 respectively.

al + bm + cn 2

⇒ 8x + y – 5z – 7 = 0.

Hence the d.r’s of the line are 3, 1, – 2.

x − x1 y − y1 z − z1 = = l m n and plane ax + by + cz + d = 0 is a +b +c

a b c = = 8 1 −5

8 (x – 1) + (y + 1) – 5z = 0

=

93. (d) Equation of plane of intersection of the two spheres is

2

...(3)

x −1 y −1 z −1 is = = 3 0 4

−bk + c k +1

2

i.e., if – a + 3b – c = 0 

Hence equation of required plane is

92. (d) z-coordinate of a point which divides the join of (a, b, c) & (– a, – c, – b) in the ratio k : 1 is

sin θ =

It will pass through the second line if (0, 2, – 1), a point on the second line lies on it

a b c = =   or  −8 −1 5

Thus four lines are equally inclined to axes.

or

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

Solving (2) and (3), we get

and  −1 , 1 , 1    3 3 3

or z =

a (x – 1) + b ( y + 1) + cz = 0 where 2a – b + 3c = 0

l 2 + m2 + n2

and 1 / 2 , 1 / 2 , 0

If θ is the angle between the line and the plane, then =0

cos (90 – θ) =

2 1 ⋅ + 3 2

1 1 −2 ⋅ +   ⋅ 0  3  3 2

1043

95. (b) Let the length of side of cube be a.

Three-Dimensional Geometry

1 1 1  −1  × + × (−1) +   × (−1)  4 2 6 3 ∴ cos θ = 1 1 1 1 + +1 +1+ 4 9 36 16

1044

⇒ sin θ =

1 2

and surface of the sphere = 3 6 – = 2 6

Objective Mathematics

⇒ θ = 45º.

104. (c) The given line is

102. (a) We have, →

∧

→ ∧

6

r = 6 i + 2 j + 3k Let the direction cosines be l, m and n 6 6 6 = l= → = 2 2 2 ∴ 7 6 +2 +3 |r| 2 2 3 3 = = m= → and n = → |r| 7 |r| 7 So direction cosine of the line are

∧

∧

∧

r = (1 + 2µ) i + (2 + µ) j + (2µ − 1) k

∧

∧

∧

∧

∧

∧

∧

= ( i + 2 j − k ) + µ (2 i + j + 2 k ) ∴ Equation of line in cartesian form is x −1 y − 2 z +1 = = 2 1 2 ∴ Angle between line and a plane is

6 2 3 , , 7 7 7

sin θ =

103. (b) The equation of sphere is x2 + y2 + z2 = 54 The centre and radius of this sphere are (0, 0, 0) and 54 , ie, 3 6 .

=

a1 a2 + b1 b2 + c1 c2 a + b12 + c12 a22 + b22 + c22 2 1

2 × 3 + 1 × (−2) + 2 × 6 4 + 1 + 4 9 + 4 + 36

6 − 2 + 12 16 = 21 21  16  ⇒ θ = sin–1    21  =

Distance between (1, 2, – 1) and (0, 0, 0) is 6 . ∴ Shortest distance between point (1, 2, – 1)

Exercises for self-practice 1. Two lines with direction cosines and are at right angles if (a) l1l2 + m1m2 + n1n2 = 1 (b) l1l2 + m1m2 + n1n2 = 0 (c)

l1 m1 n1 = = l2 m2 n2

(d) l1 = l2, m1 = m2, n1 = n2 2. The equation of the line joining the points (– 2, 4, 2) and (7, – 2, 5) are x+2 y−4 z−2 (a) = = 3 −2 1 (b)

x y z = = −2 4 2

x + 2y – 3z + 4 = 0 are

(c)

3 2 1 , − (b) − , 14 14 14 1 1 2 3 , , (d) , − 14 14 14 14 ,

z−4 x +1 y+3 = = 5. The angle between two lines −1 2 2 x−4 y+4 z +1 and = + is 1 2 2 2 (a) cos −1   9

−1  4  (b) cos   9

1 (c) cos −1   9

−1  3  (d) cos   9

(a)

3. The direction cosine of the normal to the plane 1 14

(a) ax + cz + d = 0 (b) by + cz + d = 0 (c) ax + by + d = 0 (d) ax + by + cz + d = 0

6. The straight line through (a, b, c) and parallel to x-axis are

x z y = = 7 5 −2 (d) None of the above (c)

(a)

4. Equation of the plane parallel to x-axis is

x−a y−b z−c = = 1 1 1

x−a y−b z−c = = 0 0 1 y−b z−c x−a = = (c) 0 0 1

(b) 2 , 14 2 , 14

3 14 3 14

(d)

x−a z−c y−b = = 0 0 1

are

(a) coincident (c) intersecting

(b) skew (d) parallel

8. The angle between the two planes 3x – 4y + 5z = 0 and 2x – y – 2z = 5 is π (b) 6 π (d) 3

π 4 π (c) 2 9. The direction cosines of the line joining the points (4, 3, – 5) and (– 2, 1, – 8) are (a)

(a)

6 2 3 , , 7 7 7

(b)

(c)

(d) None of these

10. The equation | r | – 2 (r . a) + λ = 0 represents a 2

(a) sphere (c) plane

(b) straight line (d) None of these

11. In three dimensional space, the equation 3y + 4z = 0 represents (a) a plane containing z-axis (b) a line with direction numbers 0, 3, 4 (c) a plane containing x-axis (d) a plane containing y-axis 12. The radius of the sphere x2 + y 2 + z2 – 6x + 8y – 10z + 1 = 0 is (a) 5 (c) 7

(b) 15 (d) 2

13. The angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l2 + m2 + n2 = 0, is given by 2π π (b) (a) 3 6 (c)

5π 6

(d)

π 3

14. The equation of a plane which passes th rough (2, – 3, 1) and is normal to the line joining the points (3, 4, – 1) and (2, – 1, 5) is given by (a) x + 5y – 6z + 19 = 0 (b) x – 5y + 6z – 19 = 0 (c) x + 5y + 6z + 19 = 0 (d) x – 5y – 6z – 19 = 0 15. The equation of straight line passing through the points (a, b, c) and ( a – b, b – c, c – a), is x−a y−b z−c (a) a − b = b − c = c − a x−a y−b z−c (b) = = b c a x−a y−b z−c (c) = = a b c x−a y−b z−c (d) 2a − b = 2b − c = 2c − a

16. If the planes 3x – 2y + 2z + 17 = 0 and 4x + 3y – kz = 25 are mutually perpendicular then k = (a) 3 (c) 9

(b) – 3 (d) – 6

17. The equation of a plane which cuts equal intercepts of unit length on the axes, is : (a) x + y + z = 0 (c) x + y – z = 1

(b) x + y + z = 1 x y z (d) + + = 1 a a a

18. The coordinates of the point which divides the join of the points (2, – 1, 3) and (4, 3, 1) in the ratio 3 : 4 internally are given by (a)

2 20 10 , , 7 7 7

(b)

15 20 3 , , 7 7 7

(c)

10 15 2 , , 7 7 7

(d)

20 5 15 , , 7 7 7

x−6 19. The coordinates of the points where the line = −1 y +1 z + 3 = meets the plane x + y – z = 3, are 0 4 (a) (2, 1, 0) (c) (1, 2, – 6)

(b) (7, – 1, – 7) (d) (5, – 1, 1)

20. The acute angle between the line joining the points x −1 y = (2, 1, – 3), (– 3, 4, 7) and a line parallel to 3 4 z+3 through the point (– 1, 0, 4) is = 4  7  (a) cos − 1   5 10 

−1  7  (b) cos   10 

 3  (c) cos − 1   5 10 

 1  (d) cos − 1   5 10 

21. If the plane x – 3y + 5z = d passes through the point (1, 2, 4) then the lengths of intercepts cut by it on the axes of x, y, z are respectively (a) 15, – 5, 3 (c) – 15, 5, 3

(b) 1, – 5, 3 (d) 1, – 6, 20

22. If the length of perpendicular drawn from origin on a plane is 7 units and its direction ratios are – 3, 2 and 6, then that plane is (a) – 3x + 2y + 6z – 7 = 0 (b) – 3x + 2y + 6z – 49 = 0 (c) 3x – 2y + 6z + 7 = 0 (d) – 3x + 2y – 6z – 49 = 0. 23. If a line makes the angle a, b and g with the axes respectively, then cos 2α + cos 2β + cos2γ = (a) – 2 (c) 1

(b) – 1 (d) 2

24. The angle between the lines 2x = 3y = ­– z and (a) 0º (c) 45º

(b) 30º (d) 90º

25. The equation of the sphere thro’ (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1) is

1045

z z−3 x y x −1 y−2 = = and = = 3 −6 1 2 −2 −4

Three-Dimensional Geometry

7. The lines

1046

(a) x2 + y2 + z2 – x + y – z = 0 (b) x2 + y2 + z2 – x – y + z = 0 (c) x2 + y2 + z2 + x – y + z = 0 (d) x2 + y2 + z2 – x – y – z = 0

28. The sine of the angle between the straight line

Objective Mathematics

=

26. The equation of the plane which bisects the line joining (2, 3, 4) and (6, 7, 8) is

(a) x + y + z + 15 = 0 (b) x – y – z – 15 = 0



(c) x – y + z – 15 = 0 (d) x + y + z – 15 = 0

27. The equation of the sphere concentric with the sphere x2 + y 2 + z2 – 4x – 6y – 5 = 0 and which passes thro’ (0, 1, 0) is

y−3 z−4 = and the plane 2x – 2y + z = 5 is 4 4

(a)

10 6 5

(b)

4 5 2

(c)

2 10

(d)

2 3 5

29. The plane x – 2y + z – 6 = 0 and the line

z x y = = 3 1 2

are related as (a) meet the plane obliquely

(a) x2 + y2 + z2 – 4x – 6y – 5z + 3 = 0 (b) x2 + y2 + z2 – 4x – 6y – 5z + 2 = 0 (c) x2 + y2 + z2 – 4x – 6y – 8z + 1 = 0 (d) x2 + y2 + z2 – 4x – 6y – 8z + 5 = 0

(b) lies in the plane (c) at right angles to the plane (d) parallel to the plane

Answers

1. (b) 11. (c) 21. (a)

2. (a) 12. (c) 22. (b)

3. (a) 13. (a) 23. (b)

4. (b) 14. (a) 24. (d)

5. (b) 15. (b) 25. (d)

6. (c) 16. (a) 26. (d)

x−2 3

7. (d) 17. (b) 27. (d)

8. (c) 18. (d) 28. (c)

9. (b) 19. (d) 29. (d)

10. (a) 20. (a)

30

Statics

CHAPTER

Summary of concepts DEFINITIONS

Statics

Matter  Anything that occupies space and is perceived by our senses is matter. Table, cup, air, water, etc. are examples of matter.

It is that branch of mechanics which deals with the action of forces on bodies, the forces being so arranged that the bodies are at rest.

Body  A body is a portion of matter occupying finite space. It has, therefore, a definite volume and a definite mass.

EQUILIBRIUM OF FORCES

Particle  A particle is a body indefinitely small in size, so that the distance between its different parts is negligible. It may be regarded as a mathematical point associated with mass. Rigid Body  A body is said to be rigid when it does not change its shape and size when subjected to external forces i.e., a rigid body is a body in which the distance between any two points always remains the same.

A system of forces acting on a body is said to be in equilibrium if it produces no change in the motion of the body. Equilibrium of Two Forces  Two forces acting at a point are in equilibrium if and only if they, (i) are equal in magnitude (ii) act along the same line (iii) have opposite directions.

Force  Force is an agent which changes or tends to change the state of rest or uniform motion of a body. Force is a vector quantity.

Forces in Statics

Representation of a force  A force is completely known if we know:

1. Action and Reaction  Whenever one body is in contact with another body, there act equal and opposite forces at the point of contact. Such forces are called action and reaction.

(i) its magnitude (ii) its direction (iii) its point of application.

Some Important Cases 

Thus, we can completely represent a force by a straight line AB drawn through the point of application A along the line of action of the force, the length of the line AB representing the magnitude of the force and the order of the letters A, B specifying the direction.

Mechanics It is the science which deals with moving bodies or bodies at rest under the action of some forces.

Dynamics It is that branch of mechanics which deals with the action of forces on bodies in motion.

(i) When a rod AB rests with one end B upon a smooth plane, the reaction is along the normal to the plane at the point of contact. (ii) When a rod rests over a smooth peg, the reaction at the point of contact is ⊥ to the rod. (iii) If one point of a body is in contact with the surface of another body, the reaction at the point of contact is ⊥ to the surface. For example, the equilibrium of a ladder in contact with the ground and a wall [both being smooth]. (iv) When a rod rests completely within a hollow sphere, the reactions at the extremities of the rod are along the normals at those points and will pass through centre of the hollow sphere.

1048

Objective Mathematics

2. Weight  Everybody is attracted towards the centre of the earth with a force proportional to its mass (the quantity of matter in the body). This force is called the weight of the body. If m is the mass of the body and g, the acceleration due to gravity, then its weight W = mg Note: (i) Weight always acts vertically downwards. (ii) Weight always acts at the centre of gravity of a body. (iii) Centre of gravity of a smooth rod is its middle point. (iv) By ‘a light body’ we mean, a body without weight. 3. Tension  or  Thrust  Whenever a string is used to support a weight or drag a body, there is a force of pull along the string. This force is called tension. Similarly, if some rod be compressed, a force will be exerted. This type of force is called thrust. Notes:

(iv) If D is the point of intersection of the diagonals of the ||gm OABC, then

→

→

→

→

OA + OB = OC = 2 OD

since diagonals of a ||gm bisect each other. Hence, if two forces acting at a point be represented in magnitude and direction by the two sides of a triangle (∆OAB), drawn from a point, their resultant is completely represented by twice the median drawn from that point to the third side.

Resultant of Two Forces  If two concurrent forces P and Q are inclined at an angle α to each other, then the magnitude R of their (i) The tension in a string is the same throughout. When two resultant is given by strings are knotted together, the tensions in the two portions are different. (ii) When a weight W hangs by a string, the tension in the string must be equal to the weight suspended, i.e., T = W. (iii) The tension of a string always acts in a direction diverging away from the body under consideration and acts along the 2 2 string. R = P + Q + 2PQ cos α .

FORCES ACTING AT A POINT

If R makes an angle θ with the direction of P, then Q sin α tan θ = P + Q cos α .

Resultant and Components  If a number of forces P1, P2, P3, ..., Pn act on a rigid body and if a single force R can be found such that the effect of R on the body is the same as the combined effect of the forces P1, P2,..., Pn then the force R is called the re- Particular  Cases: sultant of the forces P1, P2, P3, ..., Pn and the forces P1, P2, P3, ..., (a) If P and Q are at right angles to each other, i.e., α = 90º, Pn are called components of the force R. then Parallelogram Law of Forces  If two forces acting at a Q 2 2 . point, be represented in magnitude and direction by the sides of R = P + Q and tan θ = P a parallelogram drawn from the point, their resultant is represented both in magnitude and direction, by the diagonal of the (b) If P = Q, then R = 2P cos α and θ = α . 2 2 parallelogram drawn through that point. (c) If P and Q are in the same direction, i.e., α = 0, then R is in the same direction as P and Q and R = P + Q. This is called the greatest resultant of the two forces. (d) If P and Q are in the opposite directions (i.e., α = 180º) and P > Q, R is in the direction of P and R = P – Q. This is called the least resultant of the two forces. Let P and Q be the forces represented in magnitude and direction by the sides AB and AD of a parallelogram ABCD, then Components of a Force in Two Given Directions  The their resultant R is represented in magnitude and direction by the components of a force F along directions making angles α and diagonal AC. In vector notation β with it are → → → F sin β F sin α AB+ AD = AC . and sin ( α + β ) sin (α + β) respectively Notes: (i) When the forces P and Q are collinear and act in the same direction then their resultant is P + Q. (ii) When the forces P and Q are collinear and act in the op. posite directions (P > Q) then their resultant is P – Q in the direction of P. (iii) For two given forces P and Q, the greatest resultant force is Resolved Parts of a Force  When the components of a force are at right angles to each other they are called the resolved P + Q and the least resultant force is P – Q. parts of the force.

(a) Choose two mutually perpendicular directions OX and OY along which the forces are to be resolved. (b) Find the algebraic sum of the resolved parts of the forces along these lines. Denote these sums by X and Y respectively. (c) The magnitude of the resultant R is given by R = X2 + Y2 . (d) If the resultant R makes an angle α with OX, then Y tan θ = X This gives the direction of the resultant force R.

Condition of Equilibrium of a Number of Coplanar Concurrent Forces

P Q R = = . sin α sin β sin γ Converse of Lami’s Theorem  If three forces acting at a point be such that each is proportional to the sine of the angle between the other two, then the three forces are in equilibrium. λ – µ Theorem The resultant of two forces, acting at a point O along OA and OB and represented in magnitude by λ ⋅ OA and µ ⋅ OB, is represented by a force (λ + µ) ⋅ OC, where C is a point on AB such

A given number of forces acting at a point are in equilibrium if and only if the algebraic sum of their resolved parts in each of the two perpendicular directions OX and OY vanish separately. Triangle Law of Forces  If three forces, acting at a point, represented in magnitude and direction by the three sides of triangle, taken in order, then they are in equilibrium. Remark:  Since, by triangle law,

→

→

→

AB+ BC+ CA = 0 that  λ CA = µ CB, i.e., C divides AB in the ratio µ : λ. In vector notation the above statement can be written as: →

→

→

λ ⋅ OA + µ ⋅ OB = (λ + µ) ⋅ OC →

→

→

→

→

→

therefore  AB+ BC  = – CA or AB+ BC = AC . Thus, the triangle law of forces can also be stated as:

where C is a point on AB dividing it in the ratio µ : λ. Remarks →

→

→

If two forces acting at a point be represented in magnitude (i) In the λ – µ Theorem, if λ = µ = 1, then OA + OB = 2OC , and direction by the two sides of a triangle, taken in order, where C is the mid point of AB, i.e., the resultant of two → → → their resultant is given in magnitude and direction by the forces OA and OB is 2 OC , where C is the mid point of third side of the triangle taken in the opposite order. AB. Converse of the Triangle Law of Forces  If three forces (ii) It is important to note that the force along OA must contain → acting at a point are in equilibrium, they can be represented in OA with it, e.g. if the force along OA is given to be P, it magnitude and direction by the sides of any triangle drawn in → → P P  ⋅  OA = λ ⋅ OA , where λ = , must be written as such a way as to have its sides parallel to the forces taken in OA OA order. before using λ – µ Theorem. Note: If three coplanar forces acting upon a rigid body keep it in equilibrium, they are either parallel or concurrent (all meeting PARALLEL FORCES in a point). Polygon Law of Forces  If a number of forces, acting at a Like and Unlike Parallel Forces  Two parallel forces are point, be represented in magnitude and direction by the sides of a said to be like when they act in the same direction and unlike polygon, taken in order, the forces shall be in equilibrium. when the act in the opposite directions.

1049

Statics

The resolved part of a force in a given direction is obtained Lami’s Theorem by multiplying the given force by the cosine of the angle be- If three forces acting at a point be in equilibrium, then each force tween the line of action of the given force and the given direc- is proportional to the sine of the angle between the other two. tion, i.e., the resolved part of a force F in the direction making an Let P, Q and R be three forces and α, β, γ be the angles angle θ with the direction of the force is F cos θ. between Q, R; R, P and P, Q respectively. If the forces are in Resultant of a Number of Coplanar and Concurrent equilibrium, then we have Forces  The resultant of a number of concurrent forces is obtained using following steps.

1050

Objective Mathematics

Resultant of Two Like Parallel Forces The resultant of It is equal to the product of the force and the ⊥ distance of two like parallel forces P and Q acting at the points A and B the point from the line of action of the force. respectively is given by If F be a force and p be the ⊥ distance of its line of action from the fixed point O, then moment of F about O = Fp. Sign of moment The moment of a force F about a point O is +ve or –ve according as the force rotates the body about O in the anti-clockwise or clockwise direction. R = P + Q. This resultant is parallel to P and Q and acts in the same direction as P and Q, at the point C between A and B such that P ⋅ AC = Q ⋅ BC That is, the point of application of the resultant force divides AB internally in the inverse ratio of the forces. Note.

Working rule to decide the sign of moment If we walk along the line of action in the direction of the force and find that the point about which we are taking moments falls to out left, then

(i) The line of action of the resultant force is nearer to the the moment of the force about that point will be +ve and if it falls to the greater force. right, then the moment is negative. P Q P+Q R = = = (ii) i.e., each force P, Q and R BC AC AC+BC AB Notes: respectively is proportional to the distance between the (i) If the moment of a force about a point is zero, then either other two. the force is zero, or it passes through the point, i.e., F p = 0 resultant of two unlike parallel forces The resultant ⇒ Either F = 0 or p = 0. of two unlike parallel forces P and Q (P > Q) acting at the points (ii) The moment of a force about a point on its line of action is A and B is given by zero, because in such a case p = 0. Geometrical Interpretation of the moment of a force about a point Let a force F be represented in magnitude, direction and line of action by the line AB. Let O be any point about which the moment is to be taken. R = P – Q. This resultant is parallel to P and Q and acts in the same direction as the larger force P, at the point C which is on the line BA produced such that P ⋅ AC = Q ⋅ BC

1 AB ⋅ OL 2 = 2 area of ∆OAB. Thus the moment of a force about a point can be repre(i) If three parallel forces acting on a rigid body are in equilib- sented geometrically by twice the area of triangle whose base is rium, each is proportional to the distance between the other the line completely representing the force and whose vertex is the point about which moment is to be taken. two. (ii) The algebraic sum of the resolved parts of any two parallel Varignon’s theorem of moments The algebraic sum of forces (not forming a couple) along any direction is equal moments of any number of coplanar forces about any point in to the resolved part of the their resultant along the same their plane is equal to the moment of their resultant about the direction. same point.

That is, the point of application of the resultant force divides BA produced externally in the inverse ratio of the forces. Notes:

Moment of F about O = F ⋅ OL = AB ⋅ OL = 2 ⋅

resultant of a System of coplanar forces acting on a rigid Body Let OX and OY be moments of a force about a point The moment of a two ⊥ directions and P , P , P , ..., P be 1 2 3 n force about a point is the measure of its turning effect about that n coplanar forces acting on a rigid body point. making angles α1, α2, α3, ..., αn respectively with OX. Let R be their resultant making an angle θ with OX. Resolving the forces along OX, we have

momentS anD coupLeS

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

If X > 0 and Y > 0, then θ is an acute angle. If X < 0 and Y > 0, then 90º < θ < 180º If X < 0 and Y < 0, then 180º < θ < 270º If X > 0 and Y < 0, then 270º < θ < 360º.

1051

Statics

R cos θ = P1 cos α1 + P2 cos α2 + ... + Pn cos αn = X (say) ...(1) Resolving the forces along OY, we have R sin θ = P1 sin α1 + P2 sin α2 + ... + Pn sin αn = Y (say) ...(2) Squaring and adding these, we get 2 2 R2 = X2 + Y2 ⇒ R = X + Y , which gives the magnitude of the resultant. Again on dividing, we get R sin θ Y Y = or tan θ = , R cos θ X X which gives the direction of the resultant. For finding θ, the following points are useful.

Important Results (i) The algebraic sum of the moments of the two forces forming a couple about any point in their plane is constant and equal to the moment of the couple. (ii) The moment decides the tendency to produce rotation. The greater the moment, the greater is the tendency for rotatory motion. (iii) The sum of the resolved parts of the forces forming a couple in

any direction is zero. To determine the line of action of the resultant, let us suppose that it meets OX in A at a distance x from O. (iv) The resultant of a number of coplanar couples is equivalent to a single couple whose moment is equal to the algebraic sum of the Since A lies on the line of action of the resultant, the algemoments of the given couples. braic sum of moments of forces about A is equal to zero, which gives the value of x and hence the position of A. (v) If three forces acting on a rigid body be represented in magnitude, direction and line of action by the sides of a triangle, taken Hence, the resultant is a force of magnitude R and its line in order, then they are equivalent to a couple whose moment is of action passes through A, at a distance x from O and makes an represented by twice the area of the triangle. angle θ with OX. Note: If the forces are in equilibrium, then R = 0 and the algebraic sum of moments of the forces about O is equal to zero. frIctIon i.e., X = 0, Y = 0 and algebraic sum of moments of forces about O is zero. friction and force of friction The property by virtue of which a resisting force is created between two rough bodies which prevents the sliding of one body over the other is called Conditions of Equilibrium the friction and this force which always acts in the direction op(i) X = 0 i.e., the sum of the resolved parts of the forces in any direc- posite to that in which the body has a tendency to slide or move tion OX in their plane is zero. is called force of friction. (ii) Y = 0 i.e., the sum of the resolved parts of the forces in a ⊥ direction OY in their plane is zero.

(iii) The algebraic sum of the moments of the forces about O is zero.

Limiting friction When one body is just on the point of sliding on another body, the force of friction called into play attains its maximum value and is called limiting friction and the equilibrium then is said to be limiting equilibrium.

couple Two equal and unlike parallel forces with different lines of action are said to form a couple. Static friction When a body in contact with another body is in any position of equilibrium but not limiting equilibrium, then arm of the couple The ⊥ distance between the lines of the friction exerted is called static friction. Thus, static friction action of the forces forming the couple is known as the arm of is less than the limiting friction. the couple. Note: A couple, each of whose forces is P and the length of Dynamic friction When motion ensues by one body slidwhose arm is p, is usually denoted by (P, p). ing on the other, the friction exerted between the bodies is called dynamic friction. moment of a couple The moment of a couple is obtained in magnitude by multiplying the magnitude of one of the forces forming the couple and the arm of the couple. Thus, the moment of the couple (P, p) is P p, i.e., Moment of a couple = Force × Arm of the couple. Note: Moment of a couple can never be zero. Sliding friction If the body is sliding, the force of friction Sign of the moment of a couple The moment of a that comes into play is called sliding friction. couple is regarded as positive or negative according as it has a tendency to turn the body in the anti-clockwise or clockwise rolling friction If the body is rolling, the force of friction direction. that comes into play is called rolling friction.

1052

Laws of Friction  The following laws govern the different kinds of friction i.e., static, limiting and dynamic friction.

Objective Mathematics

Laws of Static Friction

Cone of Friction  The right cone described with its vertex at the point of contact of two rough bodies and having the common normal at the point of contact as axis and the angle of friction as the semi-vertical angle, is called the cone of fraction.

(i) The direction of friction is opposite to the direction in which the body tends to move. (ii) The magnitude of the force of friction is just sufficient to prevent the body from moving.

Laws of Limiting Friction (i) Limiting friction is equal in magnitude and opposite in direction to the force which tends to produce motion. (ii) The magnitude of limiting friction at the point of contact between two bodies bears a constant ratio to the normal reaction at the point. (iii) The constant ratio depends entirely on the nature of the material of which the surfaces in contact are composed of and is independent of their extent and shape.

Least Force on Horizontal Plane  The least force required to pull a body of weight W on the rough horizontal plane is W sin λ. Least Force on Inclined Plane  Let α be the inclination of rough inclined plane to the horizontal and λ, the angle of friction.

(i) If α = λ, then the body is in limiting equilibrium and is just on the point of moving downwards. Laws of Dynamic Friction (ii) If α < λ, then the least force required to pull a body of weight W down the plane is W sin (λ – α). (i) The direction of dynamic friction is opposite to that in (iii)  If α > λ, then the body cannot rest on the plane under its which the body is moving. own weight and reaction of the plane. So, the question of (ii) The magnitude of dynamic friction bears a constant ratio finding the least force does not arise. to the normal reaction on the body but this ratio is slightly less than the coefficient of friction in the case of limiting Note:  The least force required to pull a body of weight W up an inclined rough plane is W sin (α + λ). friction. (iii) The dynamic friction is independent of the velocity of motion. Coefficient of Friction  When a rough body is on the verge of sliding on another, the friction exerted bears a constant ratio to the normal reaction. This ratio of the limiting friction to the normal reaction is called the coefficient of friction. It is usually denoted by µ. If F be the limiting friction and N, the normal reaction between the two bodies, then for the equilibrium to be limiting, we F have = µ or F = µN. N

CENTRE OF GRAVITY The centre of gravity of a body or system of particles rigidly connected together, is that point through which the line of action of the weight of the body always passes. Centre of Gravity of a Number of Particles Arranged in a Straight Line  If n particles of weights w1, w2, w3, ..., wn be placed at points A1, A2, A3, ..., An on the straight line OAn such that the distances of these points from O are x1, x2, ..., xn respectively. Then, the distance of their centre of gravity G (say) from O is given by

As friction is maximum when the equilibrium is limiting, µN is the maximum value of friction. Angle of Friction  When one body, placed on another body is in limiting equilibrium, the friction exerted is the limiting friction. In this case, the angle which the resultant of the force of friction and the normal reaction makes with the normal reaction at the point of contact is called the angle of friction and is usually denoted by λ. If N and µN (= F) are the resolved parts of R, we have R cos λ = N and R sin λ = µN This means tan λ = µ Hence, the coefficient of friction is equal to the tangent of the angle of friction.

OG = x =

Σ wi xi . Σ wi

Centre of Gravity of a Number of Weights Placed at Points in a Plane  If w1, w2, ..., wn be the weights of the particles placed at the points A1 (x1, y1), A2 (x2, y2), ..., An (xn , yn) respectively, then the centre of gravity G ( x , y ) of these particles is given by

Σ wi xi Σ wi yi ,y= x= Σ wi Σ wi .

1053

Statics

Position of Centre of Gravity in Some Special Cases Centre of Gravity of a Compound Body  Let G1, G2 be (a) Uniform Rod:  At its mid point. the centres of gravity of the two parts of a body and let w1, w2 be (b) Parallelogram, Rectangle or Square:  At the intersection of the diagonals. their weights. Let G be the centre of gravity of the whole body. Then, at (c) Triangular lamina:  At the centre. (a sin α) G, acts the whole weight (w1 + w2) of the body. from the centre on (d) Circular Arc:  At a distance Join G1G2; then G must lie on G1G2. Let O be any fixed α point on G1G2. Let OG1 = x1, OG2 = x2 and OG = . Taking mothe symmetrical radius. where a = radius and 2α = angle subtended by the arc at ments about O, we have the centre. 2a sin α ⋅ from the (e) Sector of a Circle:  At a distance 3 α centre on the symmetrical radius. where a = radius and 2α = angle subtended at th centre. 4a (w1 + w2) x = w1x1 + w2x2 from the centre on (f) Semi-Circular Arc:  At a distance 3π w x + w2 x2 the symmetrical radius, where a is the radius. x= 1 1 or w1 + w2 . 3a (g) Hemisphere:  At a distance from the centre on the 8 Centre of Gravity of the Ramainder  Let w be the symmetrical radius, where a is the radius. weight of the whole body. Let a part B of the body of weight w1 be removed so that a part A of weight w – w1 is left behind. a Let G be the centre of gravity of whole body and G1, the C.G. of (h) Hemispherical Shell:  At a distance 2 from the centre portion B which is removed. Let G2 be the C.G. of the remaining on the symmetrical radius, where a is the radius. portion A. Let O be a point on G1G2 and let it be regarded as orih (i) Solid Cone:  At a distance from the base on the axis, gin. Let OG1 = x1, OG = x, OG2 = x2 Taking moments about O, 4 (w – w1) x2 + w1x1 = wx where h is the height of the cone. h wx − w1 x1 from the base on the (j) Conical Shell:  At a distance or x2 = x2 = w − w . 3 1 axis, where h is the height of the cone.

mULTIPLE-CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. Forces equal to 3Q, 5Q and 7Q acting at a point are in equilibrium. The angle between the forces 3Q and 5Q is (a) 45º (c) 30º

(b) 60º (d) None of these

2. If two forces acting at right angles have their resultant 10 kg. wt. and when they act at an angle of 60º, the resultant is 13 kg. wt., then the forces are (a) 3 kg. wt., 1 kg. wt. (b) 2 kg. wt., 2 kg. wt. (c) 4 kg. wt., 1 kg. wt. (d) None of these 3. If a body is in equilibrium under the action of three co-planar forces then

(a) they must act in a straight line (b) they must meet in a point or must be parallel (c) their horizontal and vertical components are equal (d) None of these 4. A uniform beam of length 2a rests in equilibrium against a smooth vertical plane and over a smooth peg at a distance h from the plane. If θ be the inclination of the beam to the vertical then sin 3θ is (a)

h a

(b)

h2 a2

(c)

a h

(d)

a2 h2

1054

Objective Mathematics

5. A uniform rod AB, 17 m long whose mass is 125 kg rests with one end against a smooth vertical wall and the other end on a smooth horizontal floor, this end being tied by a chord 8 metre long, to a peg at a bottom of the wall, the tension of the chord is (a) 32 kg wt (c) 64 kg wt

(a) tanθ = (b – a) (b + a) tanα (b) tanθ =

(b + a ) tan α (b − a ) 1 (d) tanθ = . (b − a )(b + a ) (c) tanθ =

(b) 16 kg wt (d) 8 kg wt

6. The greatest and least resultant of two forces of given magnitudes acting at a point are 16 gm wt. and 4 gm wt. respectively. The resultant of the forces when they act at an angle of 60º with one another is

13. Forces forming a couple are of magnitude 12N each and the arm of the couple is 8 m. The force of an equivalent couple whose arm is 6m is of magnitude (a) 8N (c) 12N

(a) 14 gm. wt. (b) 12 gm. wt. (c) 10 gm. wt. (d) None of these 7. A heavy uniform rod 15 cm long is suspended from a fixed point by a string fastened to its ends, their lengths being 9 and 12 cm. If θ be the angle at which the rod is inclined to the vertical then, sinθ = (a)

4 5

(b)

8 9

(c)

19 20

(d)

24 25

8. Two forces P and Q have resultant R. If P be increased, the new resultant bisects the angle between R and P. The increase in P is given by R 2

(a) 3R

(b)

(c) 2R

(d) R

9. Two forces equal to P and 2P respectively act on a particle. If the first be doubled and second be increased by 10 lbs. wt., the direction of the resultant is unaltered. The value of P is (a) 6 lbs. wt. (b) 5 lbs. wt. (c) 7 lbs. wt. (d) None of these

(a) G.M. of Q and Q′ (b) A.M. of Q and Q′ (c) H.M. of Q and Q′ (d) None of these

(b) 16N (d) 4N

14. Three forces P, Q, R act along the sides BC, CA, AB of a triangle ABC, taken in order. If their resultant passes through the incentre of ∠ABC, then P Q R =0 + + a b c

(a) P + Q + R = 0

(b)

(c) aP + bQ + cR = 0

(d) None of these

P and 15. If R and R′ are the resultant of two forces Q Q (P > Q) according as they are like or unlike such P that R : R′ = 25 : 7, then P : Q = (a) 2 : 1 (c) 4 : 3

(b) 3 : 4 (d) 1 : 2

16. Two unlike parallel forces P and Q act at points 5 metre apart. If the resultant force is 9N and acts at a distance of 10 metre from the greater force P, then (a) P = 16N, Q = 7N (c) P = 27N, Q = 18N

(b) P = 15N, Q = 7N (d) P = 18N, Q = 9N

17. A given force P is resolved into two components inclined 2P at 45º and αº. If the latter component is , then 3 the other component is

10. If the resultant of two forces P and Q is at right angles to P and the resultant of P and Q′ acting at the same angle, is at right angles to Q′, then P is

(a)

3 +1 P 3

(b)

3 +1 P 2

(c)

3 +1 P 6

(d) None of these

18. If AD is the altitude of ∆ABC, then the force AD has components along AB and AC equal to

11. Two forces act an angle of 120º. If the greater force is represented by 80 kg and the resultant is at right angles to the smaller, then the smaller force is (a) 10 kg (c) 20 kg

(b − a) tan α (b + a )

(b) 15 kg (d) 40 kg

12. A beam whose centre of gravity divides it into two portions a and b, is placed inside a smooth horizontal 19. sphere. If θ be its inclination to the horizontal in the position of equilibrium and 2α be the angle subtended the by beam at the centre of the sphere, then

(a)

a 2 + b2 − c2 c2 + a 2 − b2 AB , AC 2 2a 2a 2

(b)

a 2 + b2 − c2 c2 + a 2 − b2 AB , AC 2 2b 2b 2

a 2 + b2 − c2 c2 + a 2 − b2 AB , AC 2 2c 2c 2 (d) None of these (c)

Two like parallel forces P and 3P act on a rigid body at point A and B respectively. If the forces are interchanged in position, the resultant will be displaced through the distance of

27. If the force P is applied as shown in the figure and the block is in limiting equilibrium, then θ is

20. If the sides AB and AC of a ∆ABC are bisected in points D and E respectively, then the resultant of forces represented by BE and DC is represented in magnitude and direction by (a)

3 → BC 2 →

(c) BC

(b)

1 → BC 2

(d) None of these

21. If P be any point in the plane of the ∆ABC and G is the centroid of the triangle. Then the resultant of PA, PB and PC is equal to (a) PG (c) 3PG

(b) 2PG (d) None of these

22. Three like parallel forces P, Q, R act at the corner points of a triangle ABC. Their resultant passes through the circumcenter if P Q R = = a b c (c) P + Q + R = 0 (a)

(b) P = Q = R (d) None of these

23. If the position of the resultant of two like parallel forces P and Q is unaltered, when the positions of P and Q are interchanged, then

−1  3  (a) cos   5

−1  2  (b) cos   5

−1  3  (c) sin   5

(d) None of these

28. The resultant of three forces represented in magnitude and direction by the sides of a triangle ABC taken in order with BC = 5 cm, CA = 5 cm and AB = 8cm, is a couple of moment (a) 12 units (c) 36 units

(b) 24 units (d) 16 units

29. A rod can turn freely about one of its ends, which is fixed. At the other end a horizontal force equal to half the weight of the body is acting. In the position of equilibrium, the rod is inclined to the vertical at an angle (a) 30° (c) 60°

(b) 45° (d) None of these

30. A couple is of moment G and force forming the couple is P. If P is turned through a right angle, the moment of the couple thus formed is H. If instead, the forces 24. A force P is applied as shown in the figure. Then the P is turned through an angle α, then the moment of magnitude of P if the mass is in limiting equilibrium couple becomes is (a) G sin α – H cos α (b) H cos α + G sin α (c) G cos α + H sin α (d) H sin α – G cos α (a) P = Q (c) 2P = Q

(a) 10 kg (c) 6 kg

(b) P = 2Q (d) None of these

(b) 8 kg (d) 4 kg

25. A force P is applied on a block placed on a rough horizontal plane. If P = 20 kg and weight of block is 40 kg, then coefficient of friction is (a) 0.3 (c) 0.5

(b) 0.4 (d) 0.6

26. If force P = 10 kg is applied at an angle of 60° from the vertical to a block of 20 kg and the block is in limiting equilibrium, then µ is 1 (a) 2 (c)

3 4

3 (b) 5 3 (d) 2

1055

1 AB 3 3 (d) AB 4 (b)

Statics

1 AB 2 1 (c) AB 4 (a)

31. A weight W hangs by a string and is drawn aside by a horizontal force until the string makes an angle of 60º with the vertical. Then the horizontal force and the tension in the string are (a) 3 W, W (c) 2W, W

(b) 3 W, 2W (d) None of these

32. The greatest weight which can be supported by two light strings making angles 60º, 45º with the vertical, it being given that either string will break under a tension of W lbs. wt., is (a)

3 +1 W 6

(b)

3 −1 W 6

(c)

3 +1 W 2 6

(d) None of these

33. O is a point within a triangle ABC. Forces P, Q, R acting along OA, OB, OC are in equilibrium. If O is the incentre of ∆ABC, then P : Q : R is equal to

1056

of motion up the plane when under the influence of a force of 36 kg parallel to the plane, then the coefficient of friction is

Objective Mathematics

A B C : cos : cos 2 2 2 A B C : sin : sin (b) sin 2 2 2 (c) cos A : cos B : cos C (d) None of these (a) cos

(a)

3 14

(b)

3 15

3 3 (c) (d) 34. O is a point within a triangle ABC. Forces P, Q, R act17 7 ing along OA, OB, OC are in equilibrium. If O is the 42. Forces equal to P each act at a point, parallel to the orthocentre of ∆ABC, then P : Q : R is equal to sides of length 3, 4, 5 m of a triangle taken in order. (a) sin A : sin B : sin C The magnitude of their resultant is (b) a : b : c 2P P (b) (a) A B C : cos : cos (c) cos 5 5 2 2 2 3P (d) None of these (c) (d) None of these 5 35. A force P is applied on a block of 50 kg placed on 30º degree inclined plane at 30°, if µ = 0.25, then P is 43. Forces P, 2P, 3P, 4P and 2 2 P act at a point in direction AB, BC, CD, DA and AC, where ABCD is a square. (a) 100 kg (b) 200 kg The magnitude of their resultant is (c) 300 kg (d) None of these (a) 0 (b) 3 P 36. A force P = 20 kg is applied on a block of 20 kg on (c) P (d) None of these 2 30 degree inclined plane. If block is in limiting equilibrium, then the coefficient of friction 44. A mass of M is placed on a hemispherical bowl. Then the angle θ (where mass will be in limiting equilibrium 1 1 (b) (a) 1 5 2 ) is and µ = 3 1 (c) (d) None of these π π 3 (b) (a) 6 4 37. A body of weight W is free to slide on a smooth vertical π circular wire and is connected by a string, equal in (c) (d) None of these 3 length to the radius of the circle, to the highest point of the circle. Then the tension of the string and the 45. The height of the particle is, where it can rest inside a reaction  of  the  circle  are hollow sphere of radius a, if the coefficient of friction (a) W, W (c) 2W, 2W

(b) 2W, W (d) None of these

is

1 3

38. A uniform plane lamina in the form of a rhombus, one of (a) 0.5 a (b) 0.134 a whose angles is 120º, is supported by two forces P, Q (c) 0.15 a (d) 0.25 a applied at the centre in the direction of the diagonals, so that one side of the rhombus is horizontal. If P > 46. Forces 2, P, 5, Q, 3 lbs wt. act along the lines AB, CA, AD, AE and FA respectively of a regular hexagon. The Q, then magnitudes of P and Q in order that the system may 2 2 2 2 (b) Q = 3P (a) P = 2Q be in equilibrium, are (c) P2 = 3Q2 (d) Q2 = 2P2 (a) P = 4 3 lbs. wt, Q = 3 lbs. wt. 39. The minimum angle of an inclined plane when a mass (b) P = 3 lbs. wt., Q = 4 3 lbs. wt. placed on it is in limiting equilibrium, is (λ is a friction angle) (c) P = 2 3 lbs. wt., Q = 2 3 lbs. wt. (a) λ (c) 3λ

(b) 2λ (d) None of these

40. Forces P – Q, P, P + Q act at a point in directions parallel to the sides of an equilateral triangle, taken in order. Then their resultant is

(d) None of these 47. Three forces of magnitude 30, 60 and P acting at a point are in equilibrium. If the angle between the first two is 60°, then the value of P is

(a) 30 7 (b) 30 3 (c) 20 6 (d) 25 2 (c) 3 Q (d) 2 3 P 48. Forces M and N acting at a point O make an angle 150°. Their resultant acting at O, has magnitude 2 units and 41. A weight of 60 kg is on the point of motion down a is perpendicular to M. Then, in the same unit, the rough inclined plane when supported by a force of magnetises of M and N are 24 kg acting parallel to the plane and is on the point (a) 2 3 Q

(b)

3P

(c) 3, 4

49. Three weightless strings AC, BC and AB are knotted together to form an isosceles triangle whose vertex is C. If a weight W be suspended from C and the whole be supported, with AB horizontal, by the equal forces P bisecting the angles A and B, then the tension T′ of the string AB is W B sec (a) 2 2 C (c) W sec 2

W C (b) sec 2 2 (d) None of these

50. If the resultant of two forces of magnitude P and 2P is perpendicular to P, then the angle between the forces is 2π 3 4π (c) 5

3π 4 5π (d) 6 (b)

(a)

51. The resultant R of two forces P and Q act at right angles to P. Then the angle between the forces is −1 (a) cos

−1 (c) sin

P Q P Q

P −1  (b) cos  −   Q P −1  (d) sin  −  Q  

52. If the line of action of the resultant of two forces P and Q divides the angle between them in the ratio 1: 2 then the magnitude of resultant is (a)

P +Q P

(b)

P +Q Q

(c)

P2 − Q2 P

(d)

P2 − Q2 Q

2

2

2

2

53. The side of a square lamina ABCD is 2a metres. Along AB, CB, CD, AD and BD act forces of magnitudes 1, 2, 3, 4 and 5 kg wt. respectively. Then the algebraic sum of their moments about the centre of the square is (a) – 2a kg.m (c) 3a kg.m

(b) – 4a kg.m (d) None of these

54. Forces 3, 4, 5 kg wt. act along the sides BC, CA, AB of an equilateral ∆ABC of side 2m in directions indicated by the order of letters. Then the moment of the system about the C.G. of the triangle is (a) 2 3 kg m (c) 4 3 kg m

(b) 3 3 kg m (d) None of these

55. The side of a regular hexagon ABCDEF is 2 3 m; along the sides AB, CB, DC, DE, EF and FA act forces of magnitude 1, 2, 3, 4, 5, 6 kg. wt. respectively. Then the algebraic sum of the moments of the forces about the centre of the hexagon is

(b) 15 kg m (d) None of these

56. With two forces acting at a point, the maximum effect is obtained when their resultant is 4 N. If they act at right angles, then their resultant is 3 N. The forces are 1   2  N and (a)  2 + 2  

1   2   N 2 − 2  

(b) (2 +

3 ) N and (2 –

3 ) N

(c) (2 +

2 ) N and (2 –

2 ) N

1   3  N and (d)  2 + 2  

1   3   N 2 − 2  

57. Three forces P, Q, R act along the sides BC, CA, AB of ∆ABC. Their resultant passes through the incentre if (a) P + Q + R = 1 (b) P + Q + R = 0 (c) P = Q = R (d) None of these 58. The maximum resultant of two forces is P and the minimum resultant is Q, the two forces are at right angles, the resultant is (a) P + Q (c)

(b) P – Q

1 2 P + Q2 2

(d)

P2 + Q2 2

59. Two equal forces act at a point. If the square of the magnitude of their resultant is three times the product of their magnitudes, the angle between the forces is (a) 30° (c) 90°

(b) 45° (d) 60°

60. There forces P, Q, R are acting at a point in a plane. The angle between P and Q, Q and R are 150° and 120° respectively, then for equilibrium, force P, Q, R are in the ratio (a) 1 : 2 : 3 (c) 3 : 2 : 1

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

61. If the resultant of two forces acting on a particle be at right angles to one of them and its magnitude be one third of the other. The ratio of the larger forces to the smaller is (a) 5 : 2 2 (c) 3 : 2 2

(b) 2 : 2 2 (d) None of these

P Q R = = , where R is the resultant of two forces P 2 3 4 and Q acting at a point, then angle between R and P is

62. If

(a) tan −1 (c) tan −1

12 11 15 11

(b) tan −1

3 15 11

(d) None of these

1057

(a) 30 kg m (c) 33 kg m

3 ,2 2 (d) 4, 5

(b)

Statics

(a) 2 3 , 4

1058

63. The resultant of forces P and Q is R. If Q is doubled then R is doubled. If the direction of Q is reversed, then R is again doubled. then P2 : Q2 : R 2 is

Objective Mathematics

(a) 3 : 1 : 1 (c) 1 : 2 : 3

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

64. Le R1 and R 2 respectively be the maximum ranges up and down an inclined plane and R be the maximum range on the horizontal plane. Then R1, R, R 2 are in (a) Arithmetic-Geometric Progression (A. G. P) (b) A.P (c) G.P (d) H.P 65. Forces P, 3P, 2P and 5P act along the sides taken in order of a square. The magnitude of the resultant is (a) (c)

3 P 5 P

(b)

3P (d) None of these

66. Three parallel forces 3 gm, 4 gm, 5 gm act at the angular points A, B, C of square ABCD each of whose sides is 4 cm. The position of the centre of these forces is 5 (a)  , 3

5  3

5  (c)  , 3  3 

 (b)  3, 

5  3

(d) None of these

67. Parallel forces P, P, P act at angular points of a right angled triangle ABC and parallel forces 2P, 2P, 2P act at the middle points of the sides. If the sides of the triangle including the right angle are 6 cm and 8 cm respectively. Then, the distance of the centre of these forces from C, the right angle, is 10  cm 3 5 (c)  cm 3 (a)

(b)

8  cm 3

(d) None of these

68. A uniform bar of weight 3 kg and length 4m, rests on a peg and is supported in a horizontal position by a force equal to 1 kg. wt. acting vertically upwards at the other end. Then the distance of the peg from the centre of the beam is (a) 2m (c) 3m

(b) 1m (d) None of these

(a) 5 (c) 3

(b) 4 (d) 9

71. A heavy rod ACDB, where AC = a and DB = b rests horizontally upon two smooth pegs C and D. If a load P were applied at A, it would just disturb the equilibrium. Similar would do the load Q applied to B. If CD = c, then the weight of the rod is Pa + Qb c Pa + Qb (c) 2c (a)

(b)

Pa − Qb c

(d) None of these

72. A pole 20 ft. long is placed with its lower end on a horizontal plane and is pulled by a string attached to its upper end and inclined at 30º to the horizontal. The tension in the string is equal to 30 lbs. wt. The horizontal force applied at a point 4 ft above the ground that will keep the pole upright, is (b) 50 3 lbs. wt. (a) 25 3 lbs. wt. (c) 75 3 lbs. wt. (d) None of these 73. Three forces P, Q and R acting along IA, IB and IC, where I is the incentre of a ∆ABC, are in equilibrium. Then P : Q : R is (a) sec

A B C : sec : sec 2 2 2

(b) sin

A B C : sin : sin 2 2 2

(c) cos

A B C : cos : cos 2 2 2

(d) cosec

A B C : cosec : cosec 2 2 2

74. A body of weight 80 kg rests on a rough horizontal plane while a force of 20 kg is acting on it in a direction making an angle of 60º with the horizontal. Then the force of friction is (a) 20 kg (c) 10 kg

(b) 10 kg (d) None of these

75. A force of 30 kg acting at an angle of 30º with the horizontal is about to drag a body of weight 60 kg lying on the floor. Then the coefficient of friction is

2 (b) (a) 3 69. A uniform rod, 4m in length and weight 2 kg turns 3 1 freely about a point distant one metre from one end (c) (d) None of these and from that end a weight of 10 kg is suspended. The 3 weight that must be placed at the other end to produce 76. A particle of weight 100 lbs is at rest on an inclined equilibrium is plane of inclination 30º. Then the force of friction and 2 4 the reaction of the plane are (b)  kg (a) 2 kg 3 3 (a) 50 lbs.; 50 3 lbs. (b) 30 lbs.; 50 3 lbs. 1 (d) None of these (c)  kg (c) 50 lbs.; 30 3 lbs. (d) None of these 3 77. The base of an inclined plane is 4m in length and height 70. In a right angle ∆ABC, ∠A = 90° and sides a, b, c are is 3m. A force of 8 kg acting parallel to the plane, will respectively, 5 cm, 4 cm and 3 cm. If a force F has just prevent a weight of 20 kg from sliding down. Then moments 0, 9 and 16 N cm. units respectively about the coefficient of friction is vertices A, B and C, the magnitude of F is

84. The angle between two forces each equal to P when their resultant is also equal to P is 

1 2

(a)

(d) None of these

2π 3

π 3 π (d) 2

(b)

(c) π 78. The least pull P inclined at an angle α to the horizontal that will cause a block of 1000 lbs lying on a rough table to slide, the coefficient of friction being 0.2, 85. A sphere of weight W is in equilibrium on a smooth plane of inclination α to the horizontal, being supis ported by a string which is in length equal to the radius 1000 lbs (a) P = and is fastened to two points, one on the sphere and 5sin α + cos α the other on the plane. Then the tension of the string 1000 is lbs (b) P = sin α − 5 cos α 2 3 (b) W sin α (a) 3 W sin α 1000 3 (c) P = sin α + 5 cos α 2 3 W cos α (d) None of these (c) (d) None of these 3 79. A uniform ladder rests in limiting equilibrium with its lower end on a rough horizontal plane and its upper 86. A sphere of radius r and weight W rests against a smooth vertical wall, to which is attached a string of length l end against a smooth vertical wall if θ is the angle of whose one end is fastened to a point on the wall and inclination of the ladder to the vertical wall and µ is the other end to the surface of sphere. Then the tension the coefficient of friction, then tan θ is equal to in the string is (a) µ (b) 2µ W (l − r ) W (l + r ) 3µ (b) (a) (d) µ + 1. (c) 2 l + 2rl l 2 + 2rl 2 2W (l + r ) 80. Three forces P, Q, R are acting at a point in a plane. If (c) (d) None of these l 2 + 2rl the angle between P, Q and Q, R are 150º and 120º respectively then for equilibrium; forces P, Q, R are 87. A body of mass 10 kg is suspended by two strings 7 cm in the ratio, and 24 cm long, their other ends being fastened to the (a) 1 : 2 : 3 (c) 3 : 2 : 1

(b) 1 : 2 : (3)1/2 (d) 31/2 : 2 : 1 →

→

81. In triangle ABC three forces of magnitude 3 AB , 2 AC →

and 6 CB are acting along the sides AB, AC and CB respectively. If the resultant meets AC in D, then the ratio DC : AD will be equal to (a) 1 : 1 (c) 1 : 3

(b) 1 : 2 (d) 1 : 4

82. If a system of coplanar forces acting on a rigid body is represented in magnitude, direction and sense by the sides of a polygon taken in order, then the system is equivalent to (a) a single non-zero force (b) a zero force (c) a couple where moment is equal to the area of the polygon (d) a couple where moment is twice the area of polygon. 83. A body is in equilibrium on a rough inclined plane of 1 . The angle of which the coefficient of friction is 3 inclination of the plane is gradually increased. The body will be on the point of sliding downwards when the inclination of the plane reaches (a) 15º (c) 45º

(b) 30º (d) 60º

extremeties of a rod of length 25 cm. If the rod be so held that the body hangs immediately below its middle point, the tensions in the strings are 3 4 kg and 2 kg 5 5 3 4 (c) 6 kg and 4 kg 5 5

(a) 9

3 4 (b) 8 kg and 3 kg 5 5 (d) None of these

88. The sum of two forces is 18 and their resultant perpendicular to the lesser of the forces is 12, then the lesser one is (a) 5

(b) 3 3 (c) 7 (d) 2 89. The resultant of two forces 3P and 2P is R, if first force is doubled the resultant is also doubled. Then the angle between two forces is (a) 30º (c) 120º

(b) 60º (d) 150º

90. One end of a heavy uniform rod of weight W rests on a smooth horizontal plane and a string tied to the other end is fastened to a fixed point above the plane. Then the tension of the string is W 2 (c) 2W (a)

(b) W (d) None of these

1059

(b)

Statics

1 4 1 (c) 3 (a)

1060

Objective Mathematics

91. A horizontal force F is applied to a small object P of mass m on a smooth plane inclined to the horizontal at an angle θ. If F is just enough to keep P in equilibrium, the magnitude of F is (a) mg cos2 θ (c) mg cos θ sin θ

97. The vertics of a pentagon are A (0, 0), B (4, 3), C (10, 11), D (10, 12) and E (0, 12). Forces of magnitudes 10, 15, 8, 12 and 16N act along the sides AB, BC, CD, DE and EA respectively. The magnitude of resultant force is

(b) mg sin2 θ (d) mg tan θ

92. The foot of a uniform ladder is on a rough horizontal ground and the top rests against a smooth vertical wall. The weight of the ladder is 400 units. A man weighing 800 units stands on the ladder one quarter of its length from the bottom. If the inclination of the ladder to the horizontal is 30º, the reaction at the wall is

(a) 3 5 N (c) 2 5 N

(b) 5 5 N (d) None of these

98. If the resultant of two forces of magnitudes P and 2P is perpendicular to P, then the angle between the forces is 2π 3 4π (c) 5

(a)

3π 4 5π (d) 6 (b)

(a) 0 (b) 1200 3 (c) 800 3 (d) 400 3 93. A horizontal rod AB is suspended at its end by two 99. If the angle α between two forces of equal magnitude is π vertical strings. The rod is of length 0.6 metre and reduced to α – , then the magnitude of their resultant weights 3 units. Its centre of gravity G is at a distance 3 0.4 metre from A. Then the tension of the string at A becomes 3 times the earlier one. The angle α is is π 2π (b) (a) (a) 0.2 (b) 1.4 2 3 (c) 0.8 (d) 1.0 π 4π (c) (d) 94. A uniform one metre long rod AB of 17 kg. weight is 4 5 suspended horizontally from fixed supports by two vertical strings attached to points C and D on the rod 100. If the greatest and the least resultants of two forces are P and Q respectively, then the resultant of these at distances of 12 cm and 16 cm from A and B respecforces, when acting at right angles, will be tively. The strings at C and D can support weights of 10 kg and 9 kg respectively without breaking. Then the (P 2 − Q 2 ) P−Q (a) (b) position where a weight of 2 kg can be attached to the 2 2 rod without breaking either of the string is P+Q (P 2 + Q 2 ) (a) 10 cm from A (b) 12 cm from A (c) (d) 2 2 (c) 13 cm from A (d) None of these 95. A light triangular frame ABC is placed with its plane 101. vertical and the base BC horizontal on two pegs situated just below the vertices B and C. A weight of 289 kg is suspended from the vertex A. If AB = 15m, BC = 17m and CA =  8m, then (a) Reaction at B is 64 kg (b) Reaction at C is 225 kg (c) Reaction at B is 43 kg (d) Reaction at C is 123 kg.

(b) Tension in portion AC is

(c) Tension in AB is (d) All of these

(a) 3, 15 (c) 5, 13

(b) 4, 14 (d) 6, 12

102. P and Q (P > Q) are like parallel forces. If P is moved parallel to itself through a distance x, then the resultant of P and Q moves through a distance

96. A weight W is hung from a point C of an endless string ABC. The points A and B are tied to strings AD and BE whose ends D and E are fixed. The lengths BC, CA and AB are 4, 3 and 5 respectively and AB is not horizontal. If ∠DAB = 135º and ∠EBA = 150º, then (a) Tension in position BC is

The sum of magnitudes of two forces acting at a point is 18 and the magnitude of their resultant is 12. If the resultant is at 90º with the force of smaller magnitude, then their magnitudes are

W 2 11 − 6 3 (3 3 − 4) W 2 11 − 6 3

(3 3 − 4) W 10 ( 11 − 6 3 )

(a)

Px (P + Q)

(b)

Px (P − Q)

(c)

Px (P + 2Q)

(d) None of these

103. If the forces 6W, 5W acting at a point (2, 3) in cartesian rectangular coordinates are parallel to the positive x and y axis respectively, then the moments of the resultant force about the origin is (a) 8W (c) 3W

(b) – 3W (d) – 8W

104. Two weights of 10 gms and 2 gms hang from the ends of a uniform lever one meter long and weighing 4 gms. The point in the lever about which it will balance is from the weight of 10 gms at a distance of

105. A light rod AB of length 30 cm rests on two pegs 15 cm. apart. At what distance from the end A the pegs should be placed so that the reaction of pegs may be equal when weight 5W and 3W are suspended from A and B respectively ?

x−3 y−4 = , 2 −1 the moment of the force about the point (4, 1) along z-axis is

112. A force

5 units act along the line

(a) zero (b) 5 5 (c) – 5 (d) 5 113. A uniform plank AB of length 2a and weight W is (a) 1.75 cm, 15.75 cm hinged smoothly at A and rests horizontally on a rough (b) 2.75 cm, 17.75 cm wheel. The wheel runs clockwise. AC = c, where C is (c) 3.75 cm, 18.75 cm the point of contact of the plank with the wheel.G is (d) None of these C.G. of the plank. If the coefficient of friction between 106. A blacksmith carries a hammer on his shoulder and the wheel and plank is µ, then the horizontal pull and holds it at the other end of its light handle in his hand. the vertical weight on the hinge A are If he changes the point of support of the handle at the a c−a c c shoulder and if x is the distance between his hand and µW, W (b) µW, W (a) c c a c − a the point of support, then the pressure on his shoulder is proportional to a c µW, W (d) None of these (c) (a) x (b) x2 c c−a 1 1 114. A ladder, 10 metre long, rests with one end against (c) (d) 2 x x smooth vertical wall and the other end on the ground 1 107. At what height from the base of a vertical pillar, a . which is rough; the coefficient of friction being string of length 6 meters be tied, so that a man sitting 2 on the ground and pulling the other end of the string Then, the height of a man whose weight is 4 times has to apply minimum force to overturn the pillar ? that of the ladder can ascend before it begins to slip, the foot of the ladder being 2 metre from the wall, is (a) 1.5 meters (b) 3 2 meters 5 (c) 3 3 meters (d) 4 2 meters (10 6 – 1) metres (a) 4 108. A string ABC has its extremities tied to two fixed points 5 A and B in the same horizontal line. If a weight W (10 6 + 1) metres (b) 4 is knotted at a given point C, then the tension in the portion CA is (where a, b, c are the sides and ∆ is the 5 (10 6 – 1) metres (c) area of the ∆ABC) 4 (d) None of these Wb Wb (a2 + b2 + c2) (b) (b2 + c2 – a2) (a) 4c∆ 4c∆ 115. A horizontal beam AB of weight W is standing with the end B on a horizontal floor and end A leaning Wb Wb (a2 + c2 – b2) (d) (a2 + b2 – c2). (c) against a vertical wall. The beam stands in a vertical 4c∆ 4c∆ plane perpendicular to the wall inclined at 45º to the 109. A 2m long uniform rod ABC is resting against a smooth vertical, and is in the position of limiting equilibrium. vertical wall at the end A and on a smooth peg at a If the two points of contact are equally rough, then the point B. If distance of B from the wall is 0.3m, then coefficient of friction is (a) AB < 0.3m (a) 2 + 1 (b) 2 (b) AB < 1.0m (c) 2 – 1 (d) None of these (c) AB > 0.3m 116. A uniform ladder rests in limiting equilibrium with its (d) AB > 1.0m. lower end on a rough horizontal plane and its upper 110. Three force P, Q, R are acting at a point in a plane. end against a smooth vertical wall. If θ is the angle The angle between P and Q, and Q and R are 150º and of inclination of the ladder to the vertical wall and µ 120º respectively, then for equilibrium, forces P, Q, R is the coefficient of friction, then tan θ is equal to are in the  ratio, (a) µ (b) 2µ (a) 1 : 2 : 3 (b) 1 : 2 : 31/2 3µ 1/2 (d) µ + 1 (c) (c) 3 : 2 : 1 (d) 3 : 2 : 1 2 111. Forces 2, 3 , 5, 3 and 2 kg respectively act at one 117. Two uniform solid spheres composed of the same mateangular point of a regular hexagon towards the five other rial and having their radii 6 cm and 3 cm respectively angular points, then the magnitude of the resultant is are firmly united. The distance of the centre of gravity (a) 3 kg (c) 10 kg

(b) 5 kg (d) None of these

of the whole body from the centre of the larger sphere is

1061

(b) 25 cm (d) 65 cm

Statics

(a) 5 cm (c) 45 cm

1062

(a) 4 cms (c) 2 cms

(b) 3 cms (d) 1 cms

Objective Mathematics

118. Two uniform solid spheres of diameters 4 cm and 8 cm and densities respectively 3 and 5 are placed in contact with each other on a horizontal plane. Then the height of the combined centre of gravity of the spheres from the plane is 37 cm 43 37 cm (c) 5 43 (a) 4

(b) 3

37 cm 43

(d) None of these

119. The centre of mass of a rod of length ‘a’, whose density varies as the square of the distance from one end, will be at a distance k from that end, where k = 1 a 4 3 a (c) 4

(a)

(b)

3 a 2

(d) None of these

120. Two like parallel forces P and Q [P > Q] act on a rigid body at A and B respectively. If P and Q be interchanged in position show that the point of application of the resultant will be displaced through a distance x where x= (a)

P+Q AB P−Q

P − 2Q AB (c) P + 2Q

(b)

P−Q AB P+Q

(d) None of these

121. Two unlike parallel force P and Q [P > Q] act at A and B respectively. If P and Q are both increased by ‘x’, then the resultant will move through a distance (a)

xAB P+Q

x ⋅ AB (c) P−Q

(b)

2x ⋅ AB P−Q

(d) None of these

122. A uniform rod AB of length 6m and weight 5 kg rests in a horizontal position over a peg Q and under a peg P. If AP = 1m and BQ = 4m, then the reactions of the pegs are (a) 5 kg. wt., 10 kg. wt. (b) 7 kg. wt., 10 kg. wt. (c) 5 kg. wt., 8 kg. wt. (d) None of these

124. A heavy uniform bar 100 cm long balances at a point 30 cm from one end when a weight of 200 gm is suspended from that end. The weight of the bar is (a) 200 gm (b) 300 gm (c) 100 gm (d) None of these 125. Two men carry a heavy basket of 224 kg hanging from the light pole of length 8m each end of which rests on a shoulder of one of the men. The point from which the basket is hung lies 2m nearer to one man than to the other. The pressure on each shoulder is (a) 140 kg. wt., 84 kg. wt. (b) 130 kg. wt., 84 kg. wt. (c) 140 kg. wt., 80 kg. wt. (d) None of these 126. A uniform rod of weight 50 lbs and length 18 ft is carried on the shoulders of two men at distances of 2 ft and 3 ft respectively from the two ends. A weight of 80 lbs is also hung from the middle point of the rod. Then the total weight carried by each is (a) 60 lbs, 70 lbs (b) 50 lbs, 70 lbs (c) 70 lbs, 80 lbs (d) None of these 127. P and Q are two unlike parallel forces. When P is doubled, it is found that the line of action of Q is mid way between the line of action of the old and new resultants. The ratio of P and Q is (a) 2 : 4 (c) 4 : 3

(b) 1 : 4 (d) 3 : 4 →

→

→

128. Three forces P, Q, R are acting at a point in a plane. →

→

→

→

The angle between P and Q, Q and R are 150º and 120º respectively, then for equilibrium, forces P, Q, R are in the ratio (a)  1 : 2 : 3

(b)  1 : 2 :

3 (d)   3 : 2 : 1

(c)  3 : 2 : 1 →

→

→

129. If forces P, Q, R acting at a point can be represented by the sides of a triangle taken in order, then →

→

→

→

→

→

→

→

(a)   P + Q + R = 0 (c)   P + Q − R = 0

→

→

→

→

→

→

→

→

(b)   P − Q + R = 0 (d)   P − Q − R = 0

123. A heavy uniform rod AB of mass 12 kg is suspended 130. The magnitude of the two forces forming a couple from a fixed point by means of a fine light string atis 36 N each and the arm of the couple is 4 m. The tached to the rod at P such that AP is one third of AB. magnitude of each force of an equivalent couple whose The mass that must be suspended from A to keep the arm 9 m, is (in Newtons) rod horizontal is (a)  18 (b)  26 (a) 3 kg. wt. (b) 4 kg. wt. (c)  16 (d)  15 (c) 6 kg. wt. (d) None of these

1. (b) Since the forces are in equilibrium, the resultant of any two of them and the third force must act along the same line, in opposite directions, and be equal in magnitude.

In particular, the resultant of 3Q and 5Q is a force 7Q in magnitude. Let α be the angle between 3Q and 5Q. Then (7Q)2 = (3Q)2 + (5Q)2 + 2 (3Q) (5Q) cos α ⇒ 49 Q2 = 34 Q2 + 30 Q2 cos α ⇒ cos α =

(ii) The reaction S at C perpendicular to AB Since the rod is in equilibrium. So the three forces are concurrent at O h In ∆ACK, we have sinθ = AC In ∆ACO, we have sinθ =

AC AO

In ∆AGO, we have sinθ =

AO a

h . a 5. (a) Let T be the tension in the chord AC. Let R and S be the normal reaction at A and B respectively. Resolving the forces horizontally, we have, T = S, R = 120 Hence sin3θ =

1 2

∴ α = 60º. 2. (a) Let the forces be P and Q (P > Q) The resultant of P and Q acting at right angles is 10  kg. wt. ⇒ P2 + Q2 = 10 ...(i)  [P2 + Q2 = R2 when α = 90º] The resultant of P and Q acting at an angle of 60º is 3   kg. wt. ⇒ P2 + Q2 + 2PQ cos 60º = 13 ⇒ 10 + PQ = 13 [using (i)] ⇒ PQ = 3 ...(ii) Now (P + Q)2 = P2 + Q2 + 2PQ = 10 + 6 = 16 (P – Q)2 = P2 + Q2 – 2PQ = 10 – 6 = 4 ∴ P + Q = 4 and P – Q = 2 Hence P = 3 kg. wt. and Q = 1 kg. wt.

Taking moment about A, we have S. BC = 120. AG cos θ 17 ⇒ S × 17 sinθ = 120 × cosθ 2 ⇒ S = 60cotθ AC ⇒ S = 60 × BC ⇒ S = 60 ×

8 17 − 82 2

= 32

Hence, T = 32 kg wt.

3. (b) Force must meet in a point or they must be paral- 6. (a) Let the two forces be P and Q [P > Q] lel. Greatest resultant = P + Q = 16 gm. wt. 4. (a) Let AB be a rod of length 2a and weight W. It rests Least resultant = P – Q = 4 gm. wt. against a smooth vertical wall at A and over peg C Adding, P = 10 gm. wt. at a distance h from the wall. Subtracting, Q = 6 gm. wt. The rod is in equilibrium under the following forces: When P and Q act at an angle of 60º, let R be their resultant. (i) the weight W acting at G R= =

(ii) the reaction R at A

100 + 36 + 2 (10)(6)cos 60º 136 + 60 =

196 = 14 gm. wt.

7. (d) Applying m-n Theorem in ∆OAB, we have 12 9 7 − = 2 cotθ = cotα – tanα = 9 12 12 24 ⇒ sinθ = 25

1063

Statics

solutions

1064

10. (a) Let the angle between the forces be α.

Objective Mathematics

The resultant of P and Q is at right angles to P. Q sin α ∴ tan 90º = P+Q cos α or ∞ =

Q sin α ⇒ P + Q cos α = 0 P+Q cos α

P  ...(i) Q The resultant of P and Q′ is at right angles to

⇒ cos α = – 8. (d) Let the forces P and Q be represented by OA and OB. Complete the parallelogram OACB. Then the resultant R is represented by the diagonal OC. Let the force P be increased by the force represented by AA′. Complete the parallelogram OA′ C′ B. The new resultant, represented by OC′, bisects the angle between R and P ∴ ∠1 = ∠2

Q′

∴ tan 90º = ⇒ ∞ =

P sin α Q' +P cos α

P sin α ⇒ Q′ + P cos α = 0 Q' +P cos α

⇒ cos α = –

Q'  P

...(ii)

From (i) and (ii) –

P Q' =– ⇒ P2 = QQ′ Q P

∴ P is geometric mean of Q and Q′. But ∠1 = ∠3 (alternate angles) ∴ ∠2 = ∠3 ∴ OC = CC′ = AA′ ∴ Required increase in P (represented by AA′ ) = R.

11. (d) Let the smaller force be P. Here α = angle between the forces = 120º The resultant makes an angle of 90º with P

9. (b) Let the forces P and 2P be represented by OA and OB. Complete the parallelogram OACB. The resultant is represented by OC. Let OA′ and OB′ represent the forces 2P and (2P + 10) respectively. Complete the parallelogram OA′ C′ B′. Since the direction of the resultant remains unaltered OCC′ is a straight line. Now in ∆OA′ C′, AC || A′ C′

⇒ tan 90º = ⇒ ∞ =

80 sin 120º P + 80 cos120º

80 sin 120º ⇒ P + 80 cos 120º = 0 P + 80 cos120º

 1 ⇒ P + 80  −  = 0 ⇒ P = 40 kg.  2 12. (b) Applying m-n theorem in ∆ABO, we get (AG + GB) cot ∠OGB = GB cot ∠OAB–AG cot ∠OBG

∴

OA AC P 2P = ⇒ = OA' A'C' 2P 2P+10

1 2P = ⇒ 2P + 10 = 4P 2 2P+10 ⇒ P = 5 lbs. wt. ⇒

π – θ) = b cot 2

17. (c) F2 =

2P (given) 3

1065

⇒ (a + b) cot (

Statics

π  π   − α  − a cot  − α  2  2 



⇒ (a + b) tanθ = b tanα – a tan α (b − a ) tan α . ⇒ tanθ = (b + a ) 13. (b) Since the two couples are equivalent. Therefore, their moments are equal. Hence, 12 × 8 = F × 6 ⇒ F = 16N.

⇒

14. (a) Let I be the incentre of ∆ABC and r be the inradius.

⇒

Psin 45º = sin (45º +α)

2P 3

1 2 =  ⇒ sin (45º + α) = 2 sin (45º +α) 3

⇒ sin (45º + α) = sin 60º or ⇒ 45º + α = 60º or 120º ⇒ α = 15º ∴ F1 =

= Then, ID = IE = IF = r Since the resultant of the given forces passes through I, therefore, moment of the resultant of about I = sum of the moments of the given forces about I ⇒ P . ID + Q . IE + R . IF = 0 ⇒ 0 = P . r + Q . r + R . r ⇒ P + Q + R = 0. 15. (c) We have,

P +Q R = 2 P − Q2 R' 2

⇒ ⇒ ⇒

2

P2 + Q2 P2 25 32 = 2 = 2 ⇒ P −Q Q2 7 18 P 4 = . Q 3

16. (c) The resultant of P and Q acts at C. Therefore,

sin 120º

75º

Psin α Psin 15º Psin 75º = or sin (45º + α) sin 60º sin 120º

3 −1 P or 6

3 +1 P. 6 →

18. (a) Component of AD along AB

P Q P Q + − = R and = R′ Q P Q P

P 2 +Q 2 P2 − Q2 ⇒ = R and = R′ PQ PQ

or

3 2

=



AD cos C AD sin (90º − C) = sin A sin A

b a 2 + b2 − c2 AB sin B cos C k a 2 + b2 − c2 2ab AB ⋅ AB = = = a sin A 2a 2 k →

Component of AD along AC =

AD sin (90º − B) AD cos B AC sin C cos B = = sin A sin A sin A

c c2 + a 2 − b2 ⋅ c2 + a 2 − b2 2ca k ⋅ AC = AC . = a 2a 2 k 19. (a), (d)  Case I: If the resultant acts at C, then

 AB  5 AC =   Q ⇒ 10 =   Q ⇒ Q = 18 P − Q   9 Putting Q = 18 in P – Q = 9, we get P = 27.

 AB  3 AC =   3P = AB 4  P+3P  Cases II: If the resultant act, at D, C then  AB  1 1 AD =   P = AB ⇒ CD = AC – AD = AB . 4 2  3P+P 

1066

→

→

→

→

→

→

26. (c) P cos30° = µR

20. (a) BE = BC+ CE

⇒ 10 ×

Objective Mathematics

DC = DB+ BC

3 = µ × 20 ⇒ µ = 2

3 4

27. (c) P cosθ = µR = 0.2 × 20 = 4 4 5 cosθ = 4 ⇒ cosθ = 5 −1  4  −1  3  ⇒ θ = cos   = sin   5   5 →

→

→

→

Hence (c) is the correct answer.

→

∴ BE + DC = 2 BC+ CE + DB → 1 → 1 → = 2 BC+ CA + AB 2 2 →

→ 1 → (CA + AB) 2 → → 1 → 1 → 3 → = 2 BC+ CB = 2 BC− BC = BC . 2 2 2

= 2 BC+

21. (c) Let D be the mid point of BC

28. (b) We know that the resultant of three forces represented in magnitude and direction by the sides of a triangle taken in order is a couple of moment equal to twice the area of the triangle. Therefore, Resultant is a couple of moment = 2(Area of ∆ABC) = 2 s ( s − a )( s − b)( s − c) , where  2s = a + b + c = 5 + 5 + 8

= 2 9 × 4 × 4 × 1 = 24 units.

29. (b) Let AB be the rod which can turn freely about its fixed end B. Let R be the reaction at B. →

→

Since the rod in equilibrium, so the algebraic sum of the moments about any point is zero. Taking moments about B, we get

→

Now PA = PG + GA →

→

→

→

→

→

PB = PG + GB PC = PG + GC →

→

→

→

→

→

→

∴ PA + PB + PC = 3PG + GA + GB + GC →

→

→

→

→

= 3 PG + GA + 2 GD ( D is mid point of BC)

 →

= 3PG + GA + AG  →

→

( AG = 2GD)

→

= 3PG + O = 3PG . 22. (c) Since the resultant passes through the circumecentre of ∆ABC, therefore, the algebraic sum of the moments about it is zero. Hence, P + Q + R = 0  AB   AB  23. (a) We have AC =   Q and AD =  P + Q  P   P+Q Since the position of the resultant does not alter ⇒ AC = AD ⇒ P = Q. 24. (b) P = µR = 0.8 × 10 = 8 kg Hence (b) is the correct answer. 25. (c) P = µR ⇒ 20 = µ × 40 2 1 = = 0.5 ⇒µ = 4 2 Hence (c) is the correct answer.

W BC = 0 2



W . BG sinθ –

⇒

AB 1 sin θ − ABcos θ = 0 2 2 [since BG =

1 AB and BC = AB cosθ] 2

⇒ sinθ = cosθ ⇒ tanθ = 1 ⇒ θ = 45º. 30. (c) Let AB = d. AM is the prependicular distance from A to P.

1

T1 sin 45º = = T2 sin 60º

3

∴ T2 > T1.

AM = d sin θ Moment of couple G = P ⋅ AM = P ⋅ d cos θ

...(1)

When P is turned a right angle, then ....(2)

When P is turned through an angle α, then Moment of couple

2

When greatest weight is supported to O, tension T2 attains the greatest value which is given to be W. ⇒ X is greatest when T2 = W

W sin 105º = X = sin 60º

W⋅

= P ⋅ d sin (α + θ) = P ⋅ d [sin α cos θ + cos α sin θ]

33. (a) ∠OBC =

= (P ⋅ d cos θ) sin α + (P ⋅ d sin θ) cos α = G cos α + H sin α

2 F2) be two forces, then

= Algebraic sum of moments of forces about O (0, 0). = 5W ⋅ OL – 6W ⋅ OM = 5W ⋅ 2 – 6W ⋅ 3 = – 8W. 104. (b) Let the rod balances at the point P at distance x cm from A.

P = F1 + F2 and Q = F1 – F2 ⇒ F1 =

(P + Q) (P − Q) and F2 = 2 2

Required resultant =

(P 2 + Q 2 ) . 2

(F12 + F22 ) =

101. (c) Let P and Q be the forces and α the angle between them ∴ P + Q = 18 and (12)2 = P2 + Q2 + 2PQ cos α

∴ Taking moment about P, we get 10 ⋅ PA = 4 ⋅ PG + 2 ⋅ PB ⇒ 10x = 4 (50 – x) + 2 (110 – x) ⇒ x = 25 cm.

...(1) ...(2) 105. (c) Let R, R be the reactions at the pegs P and Q such that AP = x. Since resultant is at 90º with the force P (say smaller force) ∴ tan 90º =

Q sin α P + Q cos α

⇒ P + Q cos α = 0 ⇒ cos α = –

P Q

∴ (2) ⇒ 144 = P2 + (18 – P)2 – 2P2

Resolving all forces vertically, we get R + R = 8W ⇒ R = 4W.

⇒ 36P = 182 – 144 ⇒ P = 5 ⇒ Q = 13.

Now taking moment of forces about A, we get

∴ the forces are 5 and 13. 102. (a) Let the resultant of P at A and Q at B act at C. ∴ P ⋅ AC = Q ⋅ BC

...(1)

When P is shifted to the point A′ s.t. AA′ = x, let resultant of P at A′ and Q at B act at C′, s.t. CC′ = y.



R ⋅ AP + R ⋅ AQ = 3W ⋅ AB

⇒ 4W ⋅ x + 4W ⋅ (x + 15) = 3W ⋅ 30 ⇒ x = 3.75 cm. ∴ AP = x = 3.75 cm and AQ = 18.75 cm which are given in (c). 106. (c) Let W be the weight of the hammer at A, F the force applied by the hand at B and P the pressure on the shoulder at C.

∴ P ⋅ A′ C′ = Q ⋅ BC′ ⇒ P ⋅ (AC – x + y) = Q ⋅ (BC – y) ∴ (2) – (1) ⇒ P (y – x) = Q (– y) ⇒y=

...(2)

Px . (P + Q)

103. (d) M oment of the resultant force about origin O (0, 0).

Taking moment about B, P ⋅ BC = W ⋅ BA AB ⇒ P = W ⋅  x ⇒P α

1 , which is given in (c). x

107. (b) Let the string be tied at the point C of the vertical pillar, so that AC = x. Now moment of F about A   = F ⋅ AL = F ⋅ AP sin θ

1079

α 3 ⇒ tan   = 2

Statics

α 3 1 tan   = + 2 2 2

⇒

1080

  = F ⋅ 6 cos θ sin θ = 3F sin 2θ.

Objective Mathematics

To overturn the pillar with minimum (fixed) force F, moment is maximum if sin 2θ = 1 (max) ⇒ 2θ = 90º

i.e., θ = 45º

or

or

∴ AC = PC sin 45º = 3 2 , which is given in (b).

R P Q = = 12 ( 3 2) 1 P Q R = = 3 2 1

∴ P : Q : R = W T1 108. (c) By Lami’s theorem at C, = sin (90º + B) sin C

3 : 2 : 1.

111. (c) Let the forces 2, 3 , 5, 3 and 2 act at the angular points A towards the angular points B, C, D, E and F respectively. Taking AB and AE as axis of X and Y respectively, we get on resolving the forces

⇒ T1 =

W cos B W ⋅ (c 2 + a 2 − b 2 ) = 2ca sin C sin C



Wb (c 2 + a 2 − b 2 ) 4c∆

=

∴ ∆ =

1 ab sin C. 2

109. (b), (c) For equilibrium, the reaction R at A, and reaction S at B meet at O, so the line of action of weight W also passes through O.

X = 2 +

3 cos 30º + 5 cos 60º – 2 cos 60º

= 2 +

3  ⋅ 

3 1 1 + 5 ⋅  – 2 ⋅  = 5 2 2 2

Y =

3 sin 30º + 5 sin 60º +

=

3  ⋅ 

R =

(X 2 + Y 2 ) =

1 3 + 5 ⋅  + 2 2

3 + 2 sin 60º

3 + 2 ⋅ 

3 =5 3. 2

[52 + (5 3 ) 2 ] = 10.

112. (d) First note that direction of the moment of force is perpendicular to the plane of the force. Since the force lies in the xy plane, the moment of the force Clearly AB > BM i.e., AB > 0.3 m which is given in (c) 1 AC Also AB < AG ⇒ AB < 2 ⇒ AB < 1.0 m which is given in (b). 110. (d) Forces are as marked in the figure. By Lami’s theorem, we have Q R P = = sin 90º sin 150º sin 120º

5 about (4, 1) will be along Z-axis. So we have to find the moment in the usual sense. Now the equation of the line may be written as x + 2y – 11 = 0 whose perpendicular distance from (4, 1) is – 5 i.e.,

5 . Hence, the required moment is F ⋅ p =

5 ⋅ 

  5 = 5. 113. (a) Let AC = c where C is the point of contact of the plank with the wheel. If G is the C.G. of the plank, then AG = BG = a where 2a is the length of the plank. Forces are as marked in the figure. Resolving

Hence from (3) and (4), 1  2S (1 + µ) = S  + µ  µ  or 2µ + 2µ2 = 1 + µ2  or  µ2 + 2µ – 1 = 0 ∴ µ =

X = µR, Y + R = W and R ⋅ c = W ⋅ a

−2 ± (4 + 4) –1± 2

2

Since µ cannot be negative, we have

Solving these, we get a c−a X = µW   and  Y = W. c c 1 114. (a) AB  = 10m, OA = 2m so that cos θ = . 5

µ=

2 – 1.

116. (b) Forces are as marked in the figure. Resolving horizontally and vertically and taking moments about A, we get, S = µR R = W and S ⋅ 2a cos θ = W ⋅ a sin θ, where 2a is the length of the ladder

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

Let the man be at P and let AP = x metres. Resolving the forces horizontally, vertically and taking moments about A, we get 1 R, R = W + 4W = 5W 2

S=

and W ⋅ 5 cos θ + 4W ⋅ x cos θ = S ⋅ 10 sin θ. 5 From first two of these relations, S = W and 2 substituting this value of S in third, we get

(5W + 4Wx) cos θ =

or

2S 2µW = by (1) and (2) W W

Hence, tan θ = 2µ.

5 W × 10 sin θ 2

117. (d) Weights are proportional to the volume of the spheres. Let x be the distance of C.G. of the whole body

4x = 25 tan θ – 5 = 25 × 2 6 – 5

5 (10 6 – 1) metres. 2 115. (c) Resolving horizontally and vertically and taking moments about B. ∴ x =

S = µR R + µS = W

or tan θ =

from the centre O of the larger sphere. Now OO′ 4 π ⋅ 63d and weight at O′ = 9 cm. Weight at O is 3 1 π ⋅ 33d. the centre of the smaller sphere is 3

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

4 4 π ⋅ 63 d ⋅ 0 + π ⋅ 33 d ⋅ 9 3 3 Hence x = = 1 cm. 4 4 3 π ⋅ 6 d + π ⋅ 33 d 3 3

1081

S ⋅ 2a cos 45º + µS ⋅ 2a sin 45º = W ⋅ a sin 45º or 2S (1 + µ) = W ...(3) S From (1) and (2), + µS = W ...(4) µ

Statics

horizontally and vertically and taking moments about A, we get

1082

4 π 23 ⋅ 3 = 32π 3

118. (b) Weight of the smaller sphere =

Objective Mathematics

and the height of its C.G. above the plane = 2 cm. Weight of the larger sphere =

4 1280 π ⋅ 43 ⋅ 5 = π 3 3

and the height of its C.G. from the plane = 4 cm. Hence, if y denotes the height of the C.G. of the combined body, we have y =

1280 π⋅4 166 37 3 = cm = 3 cm. 1280 43 43 32π + π 3

32π ⋅ 2 +

119. (c) Element of mass of the rod at a distance x from one end is Aδ x ⋅ kx2 where A is the area of the crosssection of the rod, kx2 is the density of the element δ x. Hence if x denotes the distance of C.G. from that end, then a

x =



∫ x ⋅ kx 0

a

∫ kx

2

2

⋅ A dx

⋅ A dx

0

1 = 4 1 3

a

 x 4  0 a

 x  0 3

⇒ BC =

In the second case, P+x Q+ x (P + x) − (Q + x) = = BC' AC' AB



⇒ BC′ =

3 a. 4

(P + x) ⋅ AB P−Q

Resultant moves through CC′ = BC′ – BC

=

P ⋅ AB P−Q

=

(P + x) ⋅ AB P ⋅ AB x ⋅ AB − = . P−Q P−Q P−Q

122. (a)  the rod AB is uniform, its weight acts at its midpoint G.

120. (b) Let the resultant in the two cases act at C and C′ respectively. In the first case,

AB = 6m, BQ = 4m AQ = 2m But AP = 1m ∴ PQ = 1m and GQ = 1m

P Q P+Q = = BC AC AB ∴ BC =

P  ⋅ AB P+Q

Let R and S be the reactions of the pegs P and Q. ...(i)

In the second case, Q P Q P+Q = ∴ BC′ =  ⋅ AB (iii) A= P +Q AC' BC' AB Resultant is displaced through x = CC′ = BC – BC′ [(i) – (ii)] =

P−Q  ⋅ AB. P+Q

121. (c) Proceeding as in previous problem. P Q P−Q = = BC AC AB

Then, since the rod is in equilibrium, R 5 S = = GQ PQ PG or

R S 5 = = 1 2 1

∴ R = 5 kg. wt., S = 10 kg. wt. 123. (c)  the rod is uniform, its weight acts at its mid-point G. Let T be the tension in the string and W the weight suspended from A.

Now

W T 12 = = PG AG AP

or

W T 12 = = a 3a 2a

∴ W = 6 kg. wt. 124. (b) As usual

W 200 = AC CG

⇒ W = 200 ⋅ 

AC 30 = 200 ⋅  = 300 gm. CG 20

P Q Q−P = = BC AC AB

...(i)

When P is doubled, let the new resultant 2P – Q act at D. Then BC = BD (given) Also,

2P Q 2P − Q = = BD AD AB

2P Q 2P − Q = =  BC AD AB

or 125. (a) Let AB be the light pole and C, the point from where a weight of 224 kg is suspended.

...(ii)

AB Q−P 2P − Q = = BC P 2P

From (i) and (ii),

or 2Q – 2P = 2P – Q ∴ P : Q = 3 : 4.

or

4P = 3Q

128. (d) By Lami’s theorem P Q R = = Sin 120° sin 90° sin 150° Q

AB = 8m

150º

Let AC = x m then BC = (x + 2) metres ∴ x + x + 2 = 8; x = 3 AC = 2m, BC = 5m

126. (a)

Now

R S 224 = = BC AC AB

or

R S 224 = = = 28 5 3 8

120º

P 90º

R

(∵ angle between P and R = 360º – 150º – 120º = 90º) P Q R = = 3 / 2 1 1/ 2

∴ R = 140 kg. wt.; S = 84 kg. wt.

⇒

R S 130  =   =  GD CG CD

⇒ P : Q : R =

3 :2:1 →

→

→

129. (a) By triangle law of forces, forces P, Q and R will be in equilibrium. →

→

→

→

∴ P + Q + R = 0

130. (c) Since, F1 = 36 N, d1 = 4 m, F2 = F (say) R S 130 ⇒  = = = 10 6 7 13 ⇒  R = 60 lbs, S = 70 lbs.

1083

Statics

127. (d) Let P and Q, the two unlike parallel forces act at A and B. Let their resultant Q – P act at C. Then

Let AB = 6a, then AG = 3a 1 AP = AB = 2a ∴ PG = a 3

and d2 = 9 m Also, F1 × d1 = F2 × d2 ⇒ 36 × 4 = F × 9 ⇒ F = 16 N

1084

EXERCISEs FOR SELF-PRACTICE

Objective Mathematics

1. The greatest and least magnitude of the resultant of two forces of constant magnitude are F and G. When the forces act at an angle 2α, the resultant in magnetises is equal to: (a)

F2cos 2α + G 2sin 2α

(b)

F sin α + G cos α

(c)

F +G

2

(d)

F −G

2

2

2

2

2

2

−g 4 −g (d) 2 (b)

6. Two parallel like forces 4 N and 8N are acting 18m apart, find the position of the resultant from smaller force:

2

(a) 12 (c) 8

(b) 10 (d) None of these

7. If the force represented by 3 ˆj + 2kˆ is acting through the point 5iˆ + 4 ˆj − 3kˆ , then its moment about the point

2. If a couple is acting on 2 particles of mass 1kg attached with a rigid rod of length 4 m, fixed at centre acting at the end and the angular acceleration of system about centre is 1 rad/s2, then magnitude of force is: (a) 2N (c) 1N

g 4 g (c) 2 (a)

(b) 4N (d) None of these

3. If an angle θ is divided into 2 parts A and B such that A – B = x and tan A : tan B = k : 1, then the value of sin k is:

(1, 3, 1) is: (a) 14iˆ − 8 ˆj + 12kˆ

(b) – 14iˆ − 8 ˆj + 12kˆ

(c) – 6iˆ + ˆj − 9kˆ

(d) 6iˆ + ˆj − 9kˆ

8. If N is resolved in two components such that first is twice of other, the components are: (a) 5N, 5 2N

(b) 10N, 10 2N

N 2N , (d) None of these 5 5 9. A man carries a hammer on his shoulder and holds it at the other end of its light handle in his hand. If he k −1 changes the point of support of the handle at the shoulsin θ (c) k + 1 (d) None of these der and if x is the distance between his hand and the point of support, then the pressure on his shoulder is 4. A weight of 10 N is hanged by two ropes as shown in proportional to: fig., find tensions T1 and T2: (a) x (b) x2 (c) 1/x (d) 1/x2

k +1 (a) k − 1 sin θ

k (b) k + 1 sin θ

(c)

10. A man falls vertically under gravity with a box of mass m on his head. Then the reaction force between his head and the box is,

(a) 5N, 5 3 N

(b) 5 3 N, 5N

(c) 5N, 5N

(d) 5 3 N, 5 3 N

5. Two masses are attached to the pulley as shown in fig. Find acceleration of centre of mass:

(a) mg (b) 2mg (c) 0 (d) 1.5 mg 11. The resultant of a force of 5 newtons acting along x-axis and a couple 20 newton-metre with axis perpendicular to x-y plane is (a) a single force of 5 newtons parallel to x-axis (b) a single force of 5 newtons parallel to y-axis (c) a single force of 5 newtons along the line y = x (d) a single couple of 10 newton metre 12. A circular cylinder of radius r and height h rests on a rough horizontal plane with one of its flat ends on the plane. A gradually increasing horizontal force is applied through the centre of the upper end. If the coefficient of friction is µ. The cylinder will topple before sliding if (a) r < µ h (c) r ≥ 2µ h

(b) r ≥ µ h (d) r = 2µ h

Answers 1. (a) 11. (d)

2. (a) 12. (b)

3. (c) 13. (d)

4. (b) 14. (b)

5. (b) 15. (a)

6. (a) 16. (d)

7. (b) 17. (a)

8. (c)

9. (c)

10. (c)

1085

5 units acts along the line

Statics

(a) 0 (b) r y−4 x−3 = , n r (d) nr (c) −1 2 2 the moments of the force about the point (4, 1) along z-axis is 16. A resultant of two forces P and Q is R, if Q is doubled, R is doubled and if Q is reversed R is again doubled. If (a) zero (b) 5 5 the ratio of P2 : Q2 : R 2 = 2 : 3 : x, then x is equal to (c) – 5 (d) – 5 (a) 5 (b) 4 (c) 3 (d) 2 14. Forces of 1, 2, units act along the lines x = 0 and y = 0. The equation of the line of action of the resultant is 17. The length of an inclined plane is 5 feet and the height is 3 feet; a force of 3 lb. weight acting parallel to the (a) y – 2x = 0 (b) 2y – x = 0 plane will just prevent a weight 10 lb from sliding (c) y + x = 0 (d) y – x = 0 down. Then the coefficient of friction is 15. An odd number of n forces act along the radius vector 3 2 of a circle with radius r, from its centre to the circum (b) (a) ference. If the angle between any two adjacent forces 8 8 is same, then the magnitude of the resultant force 1 (c) (d) None of these is 8 13. A force

31

Dynamics

CHAPTER

Summary of concepts Dynamics

Displacement

Dynamics is that branch of mechanics which deals with the mo- The displacement of a moving point is the distance covered by it tion of bodies under the action of given forces in a definite direction.

Kinematics  Kinematics is the study of geometry of motion without reference to the cause of motion. It deals with displacement, velocity, acceleration etc. and we establish relations between these and time without reference to the cause of motion.

For example if a point, moving along a curve AB is at P at any time t and moves to Q in subsequent interval of time δ t, then Kinetics  Kinetics is the study of relationship between the its displacement in time δ t is chord (not arc) PQ. Again let O be a fixed point on a straight line OX. Let P be forces and the resulting motion of bodies on which they act. the position of the particle at a time t1 and Q be the position of the particle at a time t2. The n PQ is called the change in posiRest and Motion tion or the displacement of the particle in time t2 – t1. A particle is said to be at rest if it does not change its position with respect to its surroundings and is said to be in motion if it changes its position with respect to its surroundings. A displacement has two fundamental characteristics-magnitude and direction. So, it is a vector quantity.

Path

The straight line or the curve along which an object moves is Velocity and Acceleration called its path. If the path is a straight line, the object is said to have Velocity  The rate of change of displacement of a moving parrectilinear motion and if the path is a curve (in plane or in ticle is called its velocity. Velocity has magnitude as well as direction. So, it is a vecspace), the object is said to have curvilinear motion. tor quantity. In M.K.S. system, its magnitude is measured in m/sec. and Speed in C.G.S. system, its magnitude is measured in cm./sec. The speed of a moving point is the rate at which it describes 1× 100 5 its path. Speed is a scalar quantity. Velocity of 1 Km/hr = m/sec. = m/sec. The 60 × 60 18 Average Speed  Average speed of a moving object in a time velocity measured in Km/hr. is expressed as velocity in m/sec. interval is defined as the distance travelled by the object during 5 by multiplying it by . that time interval divided by the time interval. 18 Thus, average speed in a time interval 22 ft/sec. Velocity of 1 mile/hr = 15 distance travelled in the given time interval For the converse, i.e., to convert ft/sec. into miles/hr. mul= time interval 15 . tiply by 22

Let Q be the position of the particle at time (t + δ t) and let OQ = x + δ x. The displacement of the particle in time (t + δ t) – t = δ t is PQ and PQ = OQ – OP = (x + δ x) – x = δ x. This means that the displacement of the particle in time δ t in δ x. δx Average displacement during the interval δ t = . δt δ x dx = . Velocity at time t is given by, v = δlim t →0 δ t dt This is known as instantaneous velocity at t. Notes:

Thus, a =

dv d 2 x dv = =v . dt dt 2 dx

Notes: (i) If a = 0, the particle is said to be moving with uniform velocity. (ii) If a > 0, then v increases with time. (iii) If a < 0, then v decreases with time. (iv) If v and a have the same sign, then the speed of the particle is increasing. (v) If v and a have the opposite signs, then the speed of the particle is decreasing.

(i) If we do not take into consideration the direction of the Uniform Acceleration  A particle is said to be moving with displacement, the velocity becomes speed. In other words, uniform acceleration, if equal changes in velocity take place in equal intervals of time, however small these intervals may be. the magnitude of velocity is called speed. (ii) If a particle moves in the direction of its displacement, then v > 0 i.e., v > 0 in the direction of x increasing. Also, Equations of Motion v < 0 in the direction of x decreasing. If a particle moves along a straight line with initial velocity u and Uniform Velocity  A particle is said to move with uniform constant acceleration f, then the following relations are known as velocity if it moves in a constant direction and covers equal dis- equations of motion : tances in equal intervals of time, however small these intervals 1 2 (i) v = u + f t   (ii) s = ut + f t   (iii)  v2 – u2 = 2 f s may be. 2 Constant Velocity  A particle is said to move with constant where ‘v’ is the velocity after time t and s, the distance travelled in this time. velocity if it covers equal distances in equal intervals of time, even if it is not moving in a constant direction. Notes: Average Velocity  The average velocity of a particle in a given interval of time is given by the ratio of the total displace- (i) If a particle moves along a straight line with retardation f, then changing f to – f, the corresponding formulae take the ment undergone to the total time taken. form: 1 2 Total displacement (a) v = u – f t   (b) s = ut – ft   (c) v2 = u2 – 2fs Average velocity = 2 Total time (ii) If the particle starts from rest (u = 0), the above relations Acceleration and Retardation take the simple form 1 2 The acceleration of a particle is the rate of increase in its velocity (a) v = f t     (b) s = f t     (c) v2 = 2 f s 2 while retardation is the rate of decrease in velocity. Thus, retar (iii) The average velocity during an interval of time is the mean dation is negative acceleration. of the initial and final velocities and is equal to the velocity Units of acceleration are cm/sec2 and m/sec2. at the middle of the interval. Acceleration has magnitude as well as direction. So, it is a vector quantity. u+v 1 Average velocity = = Velocity at t. 2 2

Expressions for Acceleration

(i) If v is the velocity of a moving particle at time t, then the acceleration at time t is given by a=

dv d  dx  d x (ii) a = dt = dt  dt  = dt 2   2

dv . dt

 dx  ∵ dt = v   

(iv) Displacement in time t = product of time and the mean of initial and final velocities.

u+v i.e.,  s = t   2  = (average velocity) × t.   (v) The distance travelled by a particle in the nth second of its motion in a straight line is given by

Sn = u +

1 f (2n – 1). 2

1087

(Chain Rule)

Dynamics

Velocity at a Point  Consider the motion of a particle along dv dv dx the straight line OX and let O be a fixed point on it. Let P be the (iii) a = dt = dx ⋅ dt  position of the particle at time t and let OP = x. dv ∵ dx = v  = v    dx  dt

1088

Objective Mathematics

(ii) Time to Reach the Greatest Height  At the highest point, velocity is zero, so u 0 = u – gt  or  t = . g

Motion Under Gravity

When a body is allowed to fall towards the earth, it will move vertically downwards with an acceleration which is always the same at the same place on the earth but varies slightly from place to place. This acceleration is called acceleration due (iii) Time of Flight  It is the total time taken by the particle to reach the greatest height and then return to the starting to gravity. Its value in F.P.S. system is 32 ft/sec2, in C.G.S. syspoint. tem is 981 cms/sec2 and in M.K.S. system is 9.8 m/sec2. It is denoted by g. When the particle returns to the starting point, h = 0. ThereThe acceleration due to gravity always acts vertically fore, from the result downwards. If the body moves downwards, then the effect of 1 2 acceleration due to gravity is to increase its velocity. If the body h = ut – gt , 2 moves upwards, then the effect of acceleration due to gravity is that of retardation, i.e., the velocity of the body decreases. 2u 1 2 Hence, g is taken positive for the downwards motion and nega- we have, 0 = ut – 2 gt ⇒ t = 0 or t = g . tive for the upward motion of a body. Since t = 0 corresponds to the instant when the body starts Downward Motion  If a body is projected vertically downcorresponds to the time when the body after attaining wards from a point at a height h above the earth’s surface with and t = velocity u, the equations of its motion are the greatest height reaches the starting point, therefore 1 2 2u (i) v = u + gt (ii) h = ut + gt 2 t= g 1 gives the required time of flight. (iii) v2 = u2 + 2gh (iv) hn = u + g (2n – 1) 2 (iv) Time of descent = time of flight – time of ascent where hn denotes the distance covered in the nth second. 2u u u = – = . g g g Notes:

Time of ascent = time of descent and each is equal to half (i) In particular, if the body starts from rest i.e., u = 0, the the time of flight. above equations become 1 2 (v) Time to Reach a Given Height  Let t be the time taken by (a) v = gt (b) h =  gt the particle to reach a given height h. Then 2 (c) v2 = 2gh

(d) hn =

1 g (2n – 1). 2

(ii) The velocity acquired by a body in falling from rest through a distance h is given by v=

2gh .

This is usually known as the velocity due to a free fall from a height h.

h = ut – or gt 2 – 2ut + 2h = 0

1 2 gt 2

This is a quadratic equation in t and on solving gives two positive values of t, say t1 and t2. If t1 < t2, then t1 corresponds to the time when the particle attains the height h in the upward motion and t2 is the time from the point of projection to the highest point and back to the given height h, i.e., t1 is the time from O to A while t2 is the time from O to B and B to A.

Upward Motion  When a body is projected vertically up2h wards with initial velocity u, then it moves in a straight line with Also, t1 t2 = product of roots = . g constant retardation g. So, the equations of motion in this case are: (vi) Velocity at a Given Height  Let v be the velocity of the 1 2 particle at a given height h. Then (i) v = u – gt (ii) h = ut – gt 2 v 2 – u2 = – 2gh  or  v = ± u 2 − 2 gh . 1 Thus, at a given height h, the particle has two velocities (iii) v2 = u2 – 2gh (iv) hn = u – g (2n – 1). 2 which are equal in magnitude but opposite in sign. One of these Notes:

represents the velocity in the upward direction and the other in the downward direction. (i) Greatest  Height  Attained  At the highest point, velocity (vii) Velocity on Reaching the Ground  Let v be the velocis zero and let H be the maximum height reached, thereity of the body on reaching the ground. When the body fore, from v2 = u2 – 2gh, we get is again at O, displacement is zero, therefore, from v2 = u2 – 2gh, we get u2 . 0 = u2 – 2gH ⇒ H = v2 = u2 – 2g ⋅ 0 ⇒ v2 = u2,  i.e.,  v = ± u. 2g

Newton’s Laws of Motion

Gravitational Units

First Law of Motion

The units of force which depend on the value of ‘g’ are called gravitational units of force.

Everybody continues in its state of rest or of uniform motion in (a) In CGS system, 1 gm.wt. = g dynes = 981 dynes a straight line unless it is compelled by an external force (called (b) In MKS system, 1 kg.wt. = g Newtons = 9.8 N. impressed force) to change that state. (c) In FPS system, 1 lb.wt. = g Poundal = 32 poundals. This law asserts that a force is necessary to change Note:  While using the formula P = m f, P is always measured in (a) the velocity of a body absolute units i.e., in poundals or dynes or Newtons. (b) the velocity of a body (c) the direction of motion. Inertia  A body has no tendency of itself to change its state of rest or motion if it is kept free from the action of external forces. The inability of a body to change its position by itself is called Inertia. Newton’s first law of motion is also, therefore, called the Law of Inertia. Momentum of a Body  It is the quantity of motion possessed by a body and is equal to the product of its mass and velocity with which it moves. Thus, if m be the mass of a body moving with velocity v, then its momentum is mv. Momentum is a vector quantity. The direction of the momentum is the same as that of velocity. The units of momentum are kg m/sec or gm cm/sec.

Third Law of Motion To every action, there is an equal and opposite reaction. This law asserts that forces occur in pairs. When a book is placed on the table, the book presses the table with a certain force (which is action) and the table in turn presses the book with an equal but opposite force (which is reaction). Motion of a Lift Moving Upwards  Suppose a body of mass m is carried by a lift upwards with an acceleration f.

Second Law of Motion The rate of change of momentum is directly proportional to the impressed force and it takes place in the direction in which the force acts.

The forces acting upon the body are: Relation between Force, Mass and Acceleration If a force P acting on a body of mass m sets it in motion under (i) The normal reaction R of the plane of the lift acting vertically upwards. acceleration f, the force P is given by P = m f. That is, (ii)  the weight mg of the body acting vertically downwards. Force causing motion in absolute units = mass × acceleration. This is called the fundamental equation of dynamics. Since the lift is moving upwards, we have R > mg, therefore Total upward force acting on the body = R – mg. Note:  The equation P = m f is valid only when the mass of the body Equation of motion is, R – mg = m f  or  R = m (g + f ). in motion remains constant whereas the force may be constant or variable. If the force is constant, the acceleration is uniform. Motion of a Lift Moving Downwards  Let the lift be moving vertically downwards. Then R < mg. Units of Force So that total downward force = mg – R. Absolute Units (a) Dyne : In CGS system, the unit of force is dyne. 1 dyne is the force which produces an acceleration of 1 cm/sec2 in a mass of 1g. (b) Newton : In MKS system, the unit of force is newton. 1N is the force which produces an acceleration of 1 m/sec2 in a mass of 1 kg. (c) Poundal : In FPS system, the unit of force is Poundal. 1 Poundal is the force which produces an acceleration of 1 ft/ sec2 in a mass of one pound.

1089

Note:  These units of force are called absolute units because their values are the same everywhere and do not depend upon the value of ‘g’ which varies from place to place on the Earth’s surface.

Dynamics

But v = u corresponds to the starting position, so we get v = – u. Thus, the magnitude of the velocity on reaching the ground is equal to the magnitude of velocity of projection and its direction is vertically downwards.

1090

Objective Mathematics

3. One Particle on an Inclined Plane  Two particles of masses m1 and m2 (m1 > m2) are connected by a light inextensible mg – R = m f string. The mass m2 is placed on a smooth plane inclined at an or R = m (g – f ). angle α to the horizontal, the string passes over a light pulley Note:  When f = 0, R = mg i.e., when the lift is at rest or moves at the edge of the plane and m is hanging freely. Then, for the 1 with a constant velocity, the reaction is equal to the weight of system the body. Equation of motion is

Motion of Two Particles Connected by a String 1. Two Particles Hanging Vertically  If two particles of masses m1 and m2 (m1 > m2) are attached to the ends of a light inextensible string, which passes over a smooth fixed pulley, then for the system

Acceleration f =

m1 − m2 sin α g m1 + m2

Tension in the string T = Pressure on the pulley =

m1m2 (1 + sin α) g m1 + m2

2 (1 + sin α) T 3



Projectiles 

(m1 − m2 ) g Acceleration f = m1 + m2

A particle projected in any direction is called a projectile. Trajectory  The path described by the particle is called its trajectory. In figure, the curve OAB is the trajectory of the projectile.

Tension in the string, T=

2m1m2 g m1 + m2

Pressure on the pulley = 2T =

=

2 m1m2 (1 + sin α) 2 g . m1 + m2

4m1m2 g . m1 + m2

2. One Particle on Smooth Horizontal Plane  Two particles of masses m1 and m2 (m1 > m2) are connected by a light inextensible string. The mass m2 is placed on a smooth horizontal plane, the string passes over a light pulley at the edge of the plane and m1 is hanging freely. Then, for the system Velocity of Projection  The velocity with which the particle is projected is called the velocity of projection. In figure, u is the velocity of projection. Angle of Projection  It is the angle which the direction of projection makes with the horizontal. In figure, α is the angle of projection. Range  The distance between the point of projection and the point where the projectile hits a given plane through the point of projection is called its range. When the given plane is horizontal, it is called the horizontal range. In figure, OB is the horizontal range of the projectile.

m1 g Acceleration f = m + m 1 2 m1m2 g Tension of the string T = m + m 1 2 Pressure on the pulley = 2 T =

2 m1m2 g m1 + m2

Time of Flight  The time taken between the instant of projection and the instant when the projectile meets a fixed plane through the point of projection is called the time of flight. In figure, the time taken by the projectile in moving from O to B is the time of flight.

• Latus rectum =

Some Important Results Let a particle be projeted from a point O with velocity u. Let the horizontal and vertical lines OX and OY in the plane of motion be taken as axes of reference. Let α be the angle of projection. Let P (x, y) be the position of the particle at any time t and v be the velocity of the particle at P. Let the direction of v makes an angle θ with the horizontal. Then

=

u2 . 2g

1091

• Its directrix is y =

2u 2 cos 2 α g

2 [horizontal component of velocity]2 g

It is the reciprocal of the numerical coefficient of x2 in the equation of the trajectory.

 u 2 sin 2α u 2 sin 2 α   Vertex is  2 g , 2g  .  3. Time of flight= velocity]

2u sin α g

=

2 g

[vertical component of

4. Horizontal Range, R= 1. Horizontal distance covered in time t = x = u cos α ⋅ t Vertical distance covered in time t = y = u sin α ⋅ t – Horizontal component of velocity at time t = Vertical component of velocity at time t = Resultant velocity at time t 2

=

=

1 g t2 2

dx = u cos α dt

2 [horizontal component of velocity] g × [vertical component of velocity]

2 2 5. Greatest height attained = u sin α 2g

dy = u sin α dt

=

1 [vertical component of velocity]2 2g

6. Maximum horizontal range =

2

 dx   dy  2 2 2  dt  +  dt  = u − 2ugt ⋅ sin α + g t .    

u2 . g

π . 4 8. For a given range, there are two directions of projection which are complements of each other i.e., α and 90º – α. 7. Angle of projection for maximum horizontal range =

 u sin α − gt    u cos α 

Its direction is, θ = tan–1 

9. Locus of the focus of trajectory is x2 + y2 =

2. Equation of trajectory is y = x tan α –

u 2 sin 2α 2u cos α ⋅ u sin α = g g

g x2 . It is a parabola. 2u 2 cos 2 α

u2 . 4g 2

10. Locus of the vertex of trajectory is x2 + 4 y2 =

2 2  2  • Its vertex is A  u sin α cos α , u sin α  . 2g  g 

11. Velocity of the projectile at a height h =

 u 2 sin 2 α − 2 gh Its direction is, θ = tan–1   4 cos α 

2 2 • Its focus is S  u sin 2α , −u cos 2α  2 g 2g  

2u 2 y . g

u 2 − 2 gh .  .  

multiple-choice QueStionS choose the correct alternative in each of the following problems: 1. The average speed of a bicycle over a journey of 20 km if it travels the first 10 km at 15 km/hr and the second 10 km at 10 km/hr, is (a) 12 km/hr (c) 15 km/hr

(b) 10 km/hr (d) None of these

2. A train covers a distance of 250 km in 3 hours, first 195 km is covered in 2 hours and 16 minutes. The average speed with which it should cover the remaining distance so that it reaches at correct time is (a) 67 km/hr (c) 71 km/hr

(b) 75 km/hr (d) None of these

Dynamics

Greatest Height The maximum height reached by the particle above the point of projection during its motion is called the greatest height. In figure, the distance AN is the greatest height.

1092

Objective Mathematics

3. A man walks a distance of 3 km towards the north east at the rate of 6 km/h and then walks a distance of 4 km towards the south east at the rate of 2 km/h. The average speed of the man for the whole journey is (a) 2.7 km/h (c) 2.8 km/h

(b) 3.1 km/h (d) None of these

4. A car travels first half distance between two places with a speed of 40 km/h and the rest half distance with a speed of 60 km/h. The average speed of the car is (a) 48 km/h (c) 44 km/h

(b) 14m (d) None of these

(b) 3.9 km/h (d) None of these

(a) 4 km/h (c) 3 km/h

(b) 8 km/h (d) None of these

(b)

f cos α u

(d)

u cos α f

12. If the relation between v cm/sec and s is given by s , then the acceleration when s = 900 cm is v = 10 + 15 2 cm/sec2 3 1 (c) 8 cm/sec2 2

(a) 4

(b) 6

1 cm/sec2 3

(d) None of these

13. Two stones are projected from the top of a cliff h meters high, with the same speed u so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle θ to the horizontal then tan θ equals

6. A particle moves towards east from A to B at the rate of 4 km/hr and then towards north from B to C at the rate of 5 km/h. Finally, it reaches C. If AB = 12 km and BC = 10 km. The average velocity of the particle for the whole journey is (a) 4.4 km/h (c) 4.7 km/h

u sin α f

(c) u sin α

(b) 37 km/h (d) None of these

5. A particle moves in a straight line with velocity 3 m/s for 3S and then moves in a direction at right-angles to the original direction with velocity 4 m/sec for 2 sec. The distance travelled by the particle from the starting point is (a) 17m (c) 21m

(a)

(a)

2u gh

(c) 2h u h

(b) 2g (d) u

u h 2 gh

14. A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, 7. The velocity of a man who swims across a river 400 m it moves with constant acceleration f and in the secwide and flowing with a velocity of 1 km/h, so that ond part with constant retardation r. the value of t is he may be carried away down the stream by 50 m is given by  1 1 (a) 2 s  +   f r

8. A steamer takes a time t1 to travel a distance ‘a’ up a river and a time t2 to return. The speed of the streamer in still water is (a)

a (t1 + t2 ) t1t2

a (t1 + t2 ) (c) 3 t1t2

(b)

a (t1 + t2 ) 2 t1t2

(d) None of these

9. For a moving point, s = 3t3 – 2t cm is the relation between distance S and time t. Then the velocity after 4 seconds is (a) 142 cm/sec (c) 94 cm/sec

(b) 131 cm/sec (d) None of these

10. The distance S metres moved by a particle travelling in a straight line in t seconds is given by S = 45 + 12 t – t3 . Then the time t, when the acceleration is zero is (a) 2 seconds (c) 6 seconds

(c)

2s ( f + r )

2s (b) 1 + 1 f r (d)

 1 1 2s  +   f r

15. A particle moves along a straight line, the law of motion being S = A cos (nt + k). Then the acceleration (a) is positive (b) is negative (c) varies as the distance (d) varies as the square of the distance. 16. The velocity of a train increase from 15 km/hr to 1 60km/hr while it moves through a distance of km . If 8 the acceleration is uniform, find its magnitude.

(b) 4 seconds (d) None of these

11. Two particles start simultaneously from the same point and move along two straight lines, one with uniform 17.  velocity u and the other from rest with uniform accel eration f . Let α be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time

(a)

26 m/sec2 25

(c)

24 m/sec2 25

27 m/sec2 25 28 (d) m/sec2 25 (b)

Initial velocity of a body moving with a constant acceleration of 2 cm/sec2 is 6 cm/sec. Then the time required in covering a distance of 16 m is (a) 4 sec. (c) 6 sec.

(b) 2 sec. (d) None of these

(a) 20 sec., 0.7 m/s2 (c) 20 sec., 0.5 m/s2

(b) 10 sec., 0.5 m/s2 (d) None of these

19. Initial velocity of a body with constant acceleration 6 m/sec2 is 10 cm/sec. The distance moved by it in 5th second of its motion is (a) 41 metres (c) 63 metres

(b) 3 m/sec2, 5 sec. (d) None of these

29 If T be the time of flight, R be the horizontal range and α the angle of projection of a particle, then tan α = gT 2R gT 2 (c) 2R

(a)

gT 2 R T2 (d) 2 Rg

(b)

30. The least velocity with which a circket ball can be thrown 20 m horizontally is

(b) 18 m/s2 (d) None of these

21. A body moving with uniform acceleration of 4 cm/sec2 describes 60 cm in 8 seconds. Its initial velocity is (a) 24 cm/s (c) 30 cm/s

(a) 4 m/sec2, 5 sec. (c) 4 m/sec2, 10 sec.

(b) 54 metres (d) None of these

20. A particle starts from rest and moves with a constant acceleration. It describe in 10th second a distance of 171 metres. Its acceleration is (a) 15 m/s2 (c) 12 m/s2

(a) 28 m/sec (c) 21 m/sec

(b) 14 m/sec (d) 42 m/sec

31. The horizontal range of a projectile is 4/ 3 times its maximum height. The angle of projection is

(b) 19 cm/s (d) None of these

(a) 30° (c) 60°

(b) 45° (d) None of these

22. If r and r′ be the maximum ranges up and down the 32. A train is running at 5 m/s and a man jumps out of it inclined plane respectively and R be maximum range with a velocity 10 m/s in a direction making an angle on horizontal plane, then r, R, r′ are in of 60º with the direction of the train. The velocity of the man relative to the ground is equal to (a) A.P (b) G.P (c) H.P (d) None of these (a) 12.24 m/s (b) 11.25 m/s 23. A train which is moving at the rate of 36 km/hr is brought to rest in 2 minutes with a uniform retardation. The 33. retardation and also the distance that the train travels before coming to rest are 1 m/sec2, 600 m 6 1 m/sec2, 400 m (c) 12 (a)

(b)

1 m/sec2, 600 m 12

(d) None of these

24. The horizontal range of a projectile is 4 3 times its maximum height. The angle of projection is (a) 30° (c) 60°

(b) 45° (d) None of these

25. A body moving with a speed of 36 km/hr is brought to rest in 10 seconds. The negative acceleration and the distance travelled by the body before coming to rest are (a) 4 m/sec2, 50m (c) 1 m/sec2, 50m

(b) 1 m/sec2, 32m (d) None of these

26. The maximum horizontal range of a projectile when the velocity of projection is 28 m/sec is (a) 40 m (c) 100 m

(b) 80 m (d) None of these

27. The least velocity with which a cricket ball can be thrown 80 m horizontally is (a)

40g m/sec

(b)

(c)

10g m/sec

(d) None of these

80g m/sec

28. If a body moving with uniform acceleration passes over 200 m while the velocity increases from 30 to

(c) 14.23 m/s

(d) 13.23 m/s

A body moving with a uniform acceleration describes 55 m in the sixth second from rest. How much distance will it move in the 8th second (a) 55 meters (c) 75 meters

(b) 65 meters (d) 85 meters

34. Find the distance described during the tenth second by a particle having initial velocity of 60 m/sec and moving with a retardation of 2 m/sec2. Find the time in which it comes to rest. (a) 31 m, 20 sec (c) 21 m, 20 sec

(b) 41 m, 30 sec (d) 51 m, 20 sec

35. A partical describes 650 meters in 10 seconds and its velocity at the end of that time is 80 m/sec. Find the initial velocity and acceleration (a) 50 m/sec, 3 m/sec2 (c) 10 m/sec, 5 m/sec2

(b) 10 m/sec, 30 m/sec2 (d) 5 m/sec, 10 m/sec2

36. A body has an initial velocity of 200 cm/sec and is subject to a retardation of 2 cm/sec2. At what time will its velocity be zero and how far will it then travel. (a) 10 sec, 100 meter (c) 20 sec, 200 meter

(b) 100 sec, 100 meter (d) 200 sec, 2000 meter

37. If x1 and x2 be distances described by a particle during the pth and ( p + q)th second of its motion respectively, then its acceleration is x2 − x1 3q x −x (c) 2 1 q (a)

(b)

x2 − x1 2q

(d) None of these

1093

50 m/sec. Then the acceleration and time of motion are

Dynamics

18. After moving a distance of 100 m from rest, a body acquires a velocity of 36 km/hr. Then the time taken and the acceleration is

1094

Objective Mathematics

38. A train which is moving at the rate of 36 km/hr is brought to rest in 2 minutes with a uniform retardation. Find the retardation and also the distance that the train travels before coming to rest. (a) 500 meters (c) 400 meters

(a) 60 m (c) 50 m

(b) 600 meters (d) 700 meters

39. A man walking at v1 km/hr towards east finds a car which is actually moving at v 2 km/hr to be moving at w1 km/hr. But, when he walks at v 2 km/h, the car appears to be moving at w 2 km/h in a direction at right angles to what it appeared earlier to him. Then (a) 3v1 = v2 (c) 4v1 = v2

45. A body moving with a speed of 36 km/hr is brought to rest in 10 seconds. What is the negative acceleration and distance travelled by the body before coming to rest

(b) 2v1 = v2 (d) v1 = v2

40. An aeroplane A is flying at a level height towards North with a speed of 800 km/hr as observed from the ground. Its pilot observes that another plane B at the same level height is flying at 600 km/hr towards west. The velocity of the aeroplane B as observed from the ground is 4 (a) 1000 km/hr, tan −1   west of north 3 3 (b) 500 km/hr, tan −1   west of north 4

46. A body whose velocity is being constantly accelerated has at certain instant a velocity of 22 m/sec and in the following minute, it travels a distance of 10320 m. Find the acceleration (a) 4 m/sec2 (c) 6 m/sec2

(a) v 3 , perpendicular to the third v , perpendicualr to the third 3 (c) v, perpendicular to the third (d) None of these

(b)

(b) 5 m/sec2 (d) 3 m/sec2

47. A shot fired from a gun on top of a tower, 272 feet high hits the ground at a distance of 4352 feet in 17 seconds. The velocity and direction of projection are (a) 356, 30°

(b) 256 2 , 45°

(c) 256 3 , 60°

(d) None of these

48. A particle is projected with a velocity V0 so that its range on a horizontal plane is twice the greatest height attained. The range is

(c) 1000 km/hr, tan −1  3  west of north   4 (d) None of these 41. There points P, Q, R move with same velocity v along    the sides BC , CA , AB of an equilateral triangle ABC. The velocity of any point relative to any other is

(b) 70 m (d) 40 m

(a)

5 2 V0 4g

(b)

4 2 V0 5g

(c)

4 2 V0 3

(d)

3 2 V0 5

49. A body starts from rest and moves in a straight line with uniform acceleration F, the distances covered by it in second, fourth and eighth seconds are (a) in arithmetic progression (b) in geometric progression (c) in the ratio 1 : 3 : 7 (d) in the ratio 3 : 7 : 15.   50. If v1 be the relative velocity of A with respect to B, v2  be the relative velocity of C with respect to B and v3 be the relative velocity of A with respect to C then       (a) v2 = v1 + v3 (b) v1 = v2 + v3       (c) v3 = v1 + v2 (d) v1 = v2 − v3

42. If the distance between any two moving points at any time be s, their velocity be v and the components of relative velocity along and perpendicular to the direction s be v1 and v 2 respectively then the distance between 51. A body is projected vertically upwards with a velocity them when they are nearest is of 98 m/sec. The maximum height achieved by it is sv1v2 v1v2 (b) (a) (a) 360 metres (b) 475 metres v sv (c) 490 metres (d) None of these sv sv2 52. A particle is projected vertically upwards from a point (c) (d) 1 v v with a velocity of 9.8 m/sec. The time in which the particle return to the same point, given that g = 9.8 m/sec2, is

43. A bullet fired into a target loses half its veliocity after penetrating 3 cm. How much further will it penetrate. (a) 1 cm. (c) 3 cm

(a) 2 seconds (c) 4 seconds

(b) 2 cm (d) 4 cm

44. A particale, starting with initial velocity of 50 m/sec, 53. moves in a straight line with a uniform acceleration of 40 m/sec2. Then velocity of the particale after 4 seconds is (a) 210 m/sec (c) 150 m/sec

(b) 110 m/sec (d) 160 m/sec

(b) 3 seconds (d) None of these

A hammer falls freely from a height of 1.25 m. Then its velocity just before striking is (Take g = 10 m/s2) (a) 7 cm/sec. (c) 5 cm/sec

(b) 10 cm/sec (d) None of these

54. If the ratio of the space described by a body in the fifth second of its fall from rest to the space described

(a) 7 seconds (c) 5 seconds

(b) h =

(b) 9 seconds (d) None of these

55. A jet plane is rising vertically with a velocity of 10 m/s. It has reached a certain height when the pilot drops a coin, which takes 4 seconds to hit the ground. Assuming that there is no resistance to the motion of the coin, 62. the height of the plane and the velocity of the coin on impact with the ground are

1 g t t 2 12

(c) velocity of projection is

1 g (t1 + t2) 2

(d) None of these In order to keep a body in air above the earth for 12seconds, the body should be thrown vertically up with a velocity of (a)

(a) 38.4 m, 29.2 m/sec (b) 38.4 m, 28.7 m/sec. (c) 26.5 m, 13.5 m/sec. (d) None of these

1095

(a) h = g t1t2

(6g )  m/sec

(c) 6g m/sec

(b)

(12g ) m/sec

(d) 12g m/sec.

63. Two particle of m1 and m2 gms are projected upwards such that the velocity of projection of m1 is double that 56. Two bodies are thrown vertically upwards with their of m2. If the maximum heights to which m1 and m2 rise velocities in the ratio 2 : 5. Then the ratio of the maxibe h1 and h2 respectively, then mum heights attained by them is (b) 2 h1 = h2 (a) h1 = 2 h2 (a) 3 : 4 (b) 4 : 5 (c) h1 = 4 h2 (d) 4 h1 = h2. (c) 3 : 5 (d) None of these 64. AB is the vertical diameter of a circle in the vertical plane. 57. A juggler keeps three balls going with one hand, so that Another diameter CD makes an angle of 60º with AB, at any instant, two are in the air and one in hand. If then the ratio of the time taken by a particle to slide each ball rises to a height of x metres, then each ball along AB to the time taken by it to slide along CD is stays in the hand for (a) 1 : 1 (b) 2 : 1 (a) 0.45 x seconds (b) 0.33 x seconds (d) 31/ 4 : 21/ 2 (c) 1 : 2 (c) 0.24 x seconds (d) None of these 58. Two bodies are dropped on earth from different heights. 65. A bullet of mass 0.01 kg is fired from a rifle of mass 20 kg with a speed of 100 m/sec. Velocity of recoil of If their heights are in the ratio h1 : h2, then the ratio the rifle (in m/sec) is of the time taken by them to reach the ground is (a) 1 (b) 0.05 (a) h1 : 2 h2 (b) h1 : h2 (c) 2 0 (d) 0.01 (c) 2 h1 : h2 (d) None of these 66. Two particles A and B are dropped from the heights of 5m and 20m respectively. Then the ratio of time taken 59. A particle is dropped from a height h and another is simulby A to that taken by B, to reach the ground is taneously thrown upwards. If the two meet at a height (a) 1 : 4 (b) 2 : 1 3 h, then the height to which the latter will rise is (c) 1 : 2 (d) 1 : 1 4 67. Two balls are projected simultaneously with the same 3 3 velocity from the top of a tower, one vertically upwards h (b) h (a) 4 7 and the other vertically downwards. If they reach the ground in times t1 and t2 respectively, then the height 3 (d) None of these (c) h of the tower is 8 60. A particle falls from rest and in the last second of its fall passes through 224 ft. The height from which it fell and the time of its fall are 1 (a) 900 ft, 7 seconds 2 (b) 900 ft, 6 seconds 1 (c) 862 ft, 7 seconds 2

(a)

1 g t  t 2 1 2

(c)

1 g ( t22 − t12 ) 2

1 g ( t12 + t22 ) 2 1 (d) g (t1 + t2)2. 2 (b)

68. A body is projected vertically upwards from the bottom of a mine, whose depth is 30g metres, with a velocity 89 metres/sec. The time in which the body, after rising to its greatest height, will return to the surface of the earth is

(d) None of these 61. A heavy particle is projected vertically upwards. If t1, t2 are the times at which it passes a point at a height h 69. above the point of projection in ascending and descending, then

(a) 10 sec (c) 20 sec

(b) 15 sec (d) None of these

A constant force act on a body of mass 20 gm and produces a velocity of 27 cm/sec. in 6 seconds. If the body was initially at rest, the force is

Dynamics

in the last second but three is 9 : 11, then the body fell for

1096

(a) 65 dynes (c) 90 dynes

(b) 85 dynes (d) None of these

Objective Mathematics

70. A 1000 kg car goes from 10 m/s to 20 m/s in 5 sec. The force acting on it is (a) 2000 N (c) 1500 N

(b) 1000 N (d) None of these

71. While launching a rocket of mass 2 × 10 4 kg, a force of 5 × 105 N is applied for 20 sec. Then, the velocity attained by the rocket at the end of 20 seconds is (a) 330 m/s (c) 500 m/s

(b) 440 m/s (d) None of these

72. A force of 100 dynes acts on a body of mass 25 gms. The time in which the body will move 32 centimetres from rest is (a) 2 seconds (c) 4 seconds

(b) 3 seconds (d) None of these

73. An empty truck whose mass is 2000 kg has a maximum acceleration of 1 m/s2. When it carries a load of 1000 kg, its maximum acceleration is 1 m/s2 3 4 (c)  m/s2 3

(a)

(b)

2  m/s2 3

(d) None of these

74. A force of 1 kg acts on a body continuously for 10 seconds and causes it describe 10 m in that time. The mass of the body is (a) 49 kg (c) 54 kg

(b) 45 kg (d) None of these

75. A mass of 80 gm is rolled on grass with a velocity of 98 1 th of the weight, then cm/sec. If the resistance be 10 the body will move S cms where S = (a) 49 cm (c) 45 cm

(b) 34 cm (d) None of these

76. A mass of 10 kg has a velocity of 20 m/sec. A force of 100 N is applied to the mass in the direction of motion and increases the velocity to 35 m/s. The time for which the force is applied is (a) 3.2 seconds (c) 2.1 seconds

(b) 1.5 seconds (d) None of these

77. The amount of force that is needed to accelerate a train whose mass is 106kg from rest to a velocity of 6 m/s in 2 minutes is (a) 50N (c) 41N

(b) 39N (d) None of these

78. A force of 6N acts for 2 seconds on a mass of 3kg which is originally moving with a velocity of 10 m/sec in the direction of the force. The velocity and the change in momentum of the body are (a) 14 m/s, 12 kg m/s (c) 14 m/s, 21 kg m/s

(b) 12 m/s, 14 kg m/s (d) None of these

79. A 2.5 kg body at rested is acted upon by a constant external force of 10N until a displacement of 30m is

obtained. The speed attained by the body after this displacement is (a) 13.5 m/sec (c) 15.5 m/sec

(b) 14.5 m/sec (d) None of these

80. A car of mass 1000 kg is moving up at a velocity of 10 m/s and is acted upon by a forward force of 1000 N due to engine and a retarding force of 500 N due to friction. Its velocity after 10 sec is (a) 15 m/s (c) 21 m/s

(b) 13 m/s (d) None of these

81. A mass of 10 kg is travelling with a velocity of 3 m/s when a constant force is applied for 4 sec in the opposite direction. At this time, the velocity of the body is – 2 m/sec. The magnitude of the applied force is (a) 10 Newton (c) 12.5 Newton

(b) 15 Newton (d) None of these

82. A mass of 10 kg has a velocity of 20 m/s. A force of 100 N is applied to the mass in the direction of motion and increases the velocity to 35 m/s. The time for which the force is applied, is (a) 2.3 seconds (c) 3.1 seconds

(b) 1.5 seconds (d) None of these

83. A cannon ball of mass 1000 gm is discharged with a velocity of 500 m/sec from cannon the length of whose barrel is 2.5 m. The mean force exerted on the ball during the explosion is (a) 5 × 109 dynes (c) 8 × 109 dynes

(b) 7 × 109 dynes (d) None of these

84. A mass of 4 kg falls freely from rest at a height of 4.9 m. The momentum of the body when it hits the ground, is (a) 42.3 kg m/s (c) 39.2 kg m/s

(b) 39.8 kg m/s (d) None of these

85. A force of 6N acts for 2 seconds on a mass of 3 kg which is originally moving with a velocity of 10 m/s in the direction of the force. The final velocity and the change in momentum of the body are (a) 14 m/sec, 12 kg (c) 14 m/sec, 10 kg

(b) 12 m/sec, 12 kg (d) None of these

86. A mass of 2 kg hits the ground with a velocity of 10 m/s and rebounds in the same vertical line with a velocity of 5 m/s. The change in the momentum of the body is (a) 15 kg m/s (c) 21 kg m/s

(b) 30 kg m/s (d) None of these

87. A 100 kg man slides down a rope. The rope can at the most withstand a tension of 50 kg. The acceleration of the man so that he reaches the ground in longest possible time, is (a) 3.8 m/sec2 (c) 4.9 m/sec2

(b) 4.2 m/sec2 (d) None of these

88. A balloon of mass m is rising with acceleration f. The fraction of weight of the balloon that must be detached in order to double its acceleration is

mf 2f + g

(b)

mf f + 2g

 gT 2   (a) tan–1   2R 

(c)

2mf 2f + g

 gT 2  (b) tan –1    R 

(d) None of these

 gT 2  (c) tan –1    3R 

(d) None of these

89. Two masses of 48 and 50 gms are attached to the string 98. If R is the range of a projectile on a horizontal plane and of an Atwood’s machine and start from rest. The greater h its maximum height for a given angle of projection, mass falls 10 cms in 1 second. The acceleration due to then the maximum horizontal range with the velocity gravity is of projection u is (a) 243 cms/sec2 (b) 460 cms/sec2 u2 u2 (c) 980 cms/sec2 (d) None of these (b) (a) 2g g 90. Masses 260 gm and 230 gm are connected to a string 2 passing over a smooth peg. The distance and the veloc(c) 2u (d) None of these ity at the end of 3 seconds are g (a) 163 cm, 180 cm/sec 99. If the focus of a trajectory lies on much below the (b) 270 cm, 184 cm/sec horizontal plane through the point of projection as the (c) 270 cm, 180 cm/sec vertex is above, then the angle of projection is given (d) None of these by 91. A body moves up a smooth inclined plane starting with a 1 1 velocity of 48 ft/sec. If the plane is inclined at an angle (b) sin α = (a) sin α = 3 of 30º, the velocity of the body when it has moved a 2 distance of 64 ft from the starting point is 1 (c) sin α = (d) None of these (a) 16 ft/sec (b) 32 ft/sec 3 (c) 41 ft/sec (d) None of these 92. The least velocity with which a cricket ball can be thrown 100. A velocity 1 m/s is resolved into two components along 4 200 metres is OA and OB making angles 30° and 45° respectively with (a) 43.2 m/sec. (b) 44.3 m/sec. the given velocity. Then the component along OB is (c) 45.4 m/sec. (d) None of these 93. If the maximum horizontal range of a particle is R, then the greatest height attained is (a)

1 R 4

(b)

1 R 2

(c)

1 R 3

(d) None of these

94. If a boy can throw a cricket ball 100 metres, then the ball is in the air for (a) 3.2 sec. (c) 4.9 sec

(b) 4.5 sec (d) None of these

95. A particle is projected with a velocity ‘v’ so that its range on a horizontal plane is twice the greatest height attained. Then the range is (a)

2v 2 3g

2 (c) 4v 5g

2 (b) 5v 3g

(d) None of these

96. If the velocity of projection is 56 ft/sec, then the two directions of projection to give a range of 49 ft. are

(a)

1 m/s 4

(b)

1 ( 3 – 1) m/s 4

(c)

1 m/s 8

(d)

1 ( 6 – 8

2 ) m/s

101. A particle possess two velocities simultaneously at an 12 to each other. Their resultant is 15 m/s. angle of tan −1 5 If one velocity is 13 m/s, then the other will be (a) 5 m/s (c) 12 m/s

(b) 4 m/s (d) 13 m/s

102. Two balls are projected from the same point in directions inclined at 60º and 30º to horizontal. If they attain same height, then the ratio of their velocities of projection is (a) 2: 3

(b) 4: 3

(c) 1: 3

(d) None of these

103. The maximum horizontal range of a ball projected with a velocity of 39.2 m/s is (take g = 9.8 m/s2) (a) 100 m (c) 157 m

(b) 127 m (d) 177 m

104. A ball is thrown vertically upwards from the ground with velocity 15 m/s and rebounds from the ground with one-third of its striking velocity. The ratio of its 97. If the time of flight of a bullet over a horizontal range R greatest heights before and after striking the ground is T secs, the inclination of the direction of projection is equal to to the horizontal is (a) 15º (c) 60º

(b) 75º (d) None of these

Dynamics

1097

(a)

1098

(a) 4:1 (c) 5:1

(b) 9:1 (d) 3:1

Objective Mathematics

105. The velocity at the maximum height of a particle is half of its initial velocity u. Then, the range on the horizontal plane is u2 3 2g

(a)

u2 3 (c) 4g

(b)

109. If t1 and t2 are the times of fight of two particles having the same inital velocity u and range R on the horizontal, 2 then t1 + t22 is equal to

u2 3 g

(a)

u2 2g

(b)

(c)

u2 g

(d) 1

4u 2 g2

(d) None of these

110. The velocity of a particle when at its greatest height 2 of its velocity when at half its greatest height. is 106. A particle is projected so as to have range R on the 5 horizontal plane through the point of projection. If α, Then, the angle of projection is β are the possible angles of projection and t1, t2 the corresponding times of flight, then (a) 60º (b) 45º (c) 30º (d) None of these 2 2 sin (α − β) t +t (a) 12 22 = 111. If a projectile having horizontal range of 24 acquires sin (α + β) t1 − t2 a maximum height of 8, then its initial velocity and 2 2 sin ( α − β ) t −t the angle of projection are (b) 12 22 = sin (α + β) t1 + t2 (a) 24 g , sin–1 (0.6) (b) 8 g , sin–1 (0.8) t1 − t2 sin (α − β) = (c) (c) 5 g , sin–1 (0.8) (d) 5 g , sin–1 (0.6) t1 + t2 sin (α + β) (d) None of these 112. A projectile is projected with the speed of 10 5 m/sec 107. A number of particles are projected simultaneously at an angle of 60º from the horizontal. The velocity from the same point with equal velocities in the same of the projectile when it reaches the height of 10m is vertical plane, in different direction. The locus of the (g = 9.8 m/sec2) foci of their path is a (a) 4 19 m/sec (b) 179 m/sec (a) circle (b) ellipse (c) 15 m/sec (d) 5 15 m/sec (c) parabola (d) None of these 108. If a number of particles start simultaneously from the 113. same point in all directions in a vertical plane with the same speed u. Then, after a time t, they will all lie on (a) circle (c) ellipse

(b) parabola (d) None of these

From the top of a tower of height 100m, a ball is projected with a velocity of 10 m/sec. It takes 5 seconds to reach the ground. If g = 10 m/sec2, then the angle of projection is : (a) 30º (c) 60º

(b) 45º (d) 80º

Solutions 1. (a) Time taken by the bicycle to travel the first 10 km 10 2 hours = hours. = 15 3 Again time taken by the bicycle to travel the second 10 km 10 hours = 1 hour = 10 Total distance travelled = 10 + 10 = 20 km. 2 5 +1= hours. 3 3 ∴ Average speed of bicycle

Total time taken =

 =

= 20 × 3 = 12 km/hr. 5

Distance 20 = km/hr. Time 5/3

2. (b) In 3 hours train covers = 250 km. In 2 hour 16 minutes, it travelled = 195 km. Remaining distance = 250 – 195 = 55 km. 34   11 Remaining time =  3 −  hours = hours 15  15  ∴ To cover the remaining distance of 55 km, its average speed =

55 55 km/hr. = × 15 = 75 km/hr. (11 / 15) 11 3+4 Total distance 7 = 3 4 = 5 Total time + 2 6 2 14 = km/hr. = 2.8 km/hr. 5

3. (c) Average speed =





=

∴ Average speed

Total distance = Total time

x 80 × 120 x x + = = = 48 km/hr. 120 + 80 80 120 5. (a) Let AB and BC be the distances travelled by the particle with velocities 3 m/s and 4 m/s respectively.

Dynamics

x2 x = hour 60 120

1099

4. (a) Let the total distance travelled be x km. Then the time taken to travel the first half distance x2 x = hour = 40 80 Time taken to travel the rest half distance

∴ Time taken by the man to flow from A to B Distance 50 1 = = hr. = velocity 1000 20 In this time the man swims 400 m across the river. 400 Distance ∴ velocity = = 1 Time 20 8000 m/hr. = 8 km/hr. 8. (b) Let u be the velocity of the steamer in still water and let v be the velocity of the current, then velocity up-stream = u – v and velocity down-stream = u + v a ∴ a = t1 (u – v) ⇒ u – v = t1

∴ Distance = velocity × time ∴ AB = 3 × 3 = 9m and BC = 4 × 2 = 8 m Hence the distance travelled by the particle from the starting point = AB + BC = 9m + 8m = 17m. 6. (a) Time taken by the particle to move from A to B

and

a = t2 (u + v) ⇒ u + v =

a t2

Adding these equations, we get a a a (t1 + t2 ) + ⇒ 2u = 2u = t1 t2 t1t2 Hence, u =

a (t1 + t2 ) . 2 t1t2

9. (a) Here s = 3 t3 – 2 t ds = 9 t2 – 2 Differentiating w.r.t t, we get dt After t seconds, velocity ds = (9 t2 – 2) cm/sec. dt After 4 seconds, velocity = 9 (4)2 – 2 = (144 – 2) cm/sec = 142 cm/sec.

v =

10. (b) Here s = 45 t + 12 t2 – t3  Distance 12 = = = 3 hrs. Speed 4 Time taken by the particle to move from B to C Distance 10 Speed = 5 = 2 hrs. 12+10 22 = = 4.4 km/hr. ∴ Average velocity = 3+2 5 =

7. (b) Both the velocities work independently i.e., each velocity has the same effect as if it were acting alone.

Differentiating (1) w.r.t. t, we get ds = 45 + 24 t – 3 t2  v= dt Again differentiating (2) w.r.t. t, we get dv d 2s = = 24 – 6t f = dt dt 2 As acceleration, f = 0 ⇒ 24 – 6t = 0 ∴ t = 24 ÷ 6 = 4 seconds. 11. (d) We have R2 = u2 + f 2t2 + 2u f t cos(180 – α) R2 = u2 + f 2t2 – 2u f t cosα

...(1)

...(2)

1100

R = ut and h =

1 2 gt 2

Objective Mathematics

2h 2h u ⇒ R = g  g

R = ut and t =

when stone is projected and an angle of θ to the horizontal then R = u cosθ × t ...(2) 1 2 ...(3) and h = − u sin θ × t + gt  2

Let V = u2 + f 2t2 – 2uft cosα dV = 0 + 2f 2t – 2uf cosα dt

using (1) in (2), we get

d 2v = 2f 2 = +ve dt 2

u

i.e., velocity will be least after a time t. dV = 0 = 2f 2t – 2uf cos α dt 2uf cos α ⇒ 2f 2t = 2uf cosα ⇒ t = 2f 2 u cos α t= . f ⇒



=

2 s + 3 225

13. (d) When stone is projected horizontally then 1 2 h = u sin 0 × t + gt 2

h=

− u sin θ cos θ

2h 1 1  2h  + g   g 2 cos 2 θ  g 

⇒ h = − u tan θ

2h h + g cos 2 θ

2h 2 u ⇒ tan θ = g gh .

14. (d) Here x1 + x2 = s t1 + t2 = t

10 s + = 15 225

∴ Acceleration when s = 900 cm 2 900 2 2 + = +4=4 cm/s2. = 3 225 3 3

R = u cos 0 × t

...(4)

Using (4) in (3), we get

⇒ h(– tan2θ) = −u tan θ

s   dv 1 = 10 +  × 15  ds 15 



2h 1 2h g = u cos θ × t ⇒ t = cos θ g 

2h 1   −u tan θ ⇒ h 1 − 2  = g  cos θ 

s 15 Differentiating w.r.t. s, we get dv 1 1 =0+ ×1= ds 15 15

12. (a) Here v = 10 +

Acceleration = v

...(1)

We have from A to C v2 = u2 + 2fx1

v2 v2 = 0 + 2fx1 ⇒ x1 = 2 f  and v = u + ft v v = 0 + f t1 ⇒ t1 = f  from C to B v2 = u2 + 2fs v′ 2 = v2 – 2rx2 ⇒ 0 = v2 – 2rx2 ⇒ x2 = and v′ = v – rt2

v 0 = v – rt2 ⇒ t2 =  r Adding (1) and (3) v2  1 1   +  2  f r  1 1 2s = v 2  +    f r Adding (2) and (4)

v2  2r

...(1)

...(2)

...(3)

...(4)

x1 + x2 =

...(5)



t = v  1 + 1   f r



 1 1 t 2 = v 2  +    f r

⇒ (t + 8) (t – 2) = 0 ⇒ t – 2 = 0  [ t + 8 ≠ 0] ⇒ t = 2 seconds Hence particle takes 2 seconds to cover distance of 16 cm. 18. (c) Here u = 0, v = 36 km/hr. = 10 m/s,

2

...(6)

Dividing (6) by (5)

102 – 0 = 2 × f × 100 ⇒ f =

2

t2 = 2s



t=

 1 1 v2  +   f r  1 1 =  +    1 1  f r v2  +   f r

∴ 10 = 0 + 0.5 × t ⇒ t =

...(i)

⇒

d 2s = – n2s  [ s = A cos (nt + k) from (i)] dt 2

But

d 2s = f, the acceleration. dt 2

∴ f = – n2s i.e., negative Since the acceleration is –ve, ∴ it is directed towards the origin and varies as s i.e., the distance. 25 m/sec, v 6 50 m/sec = 60 km/hr = 3 1 km = 125 m s= 8

16. (c) u = 15 km/hr =

So v2 = u2 + 2fs 2

2

 50   25  ⇒   =   + 2 × f × 125 3    6 



24  m/sec2 ⇒ f = 25 Hence (c) is correct answer. 17. (b) Let the required time be t seconds. 1 f t 2 Using s = ut + 2 Here u = 6 cm/s, f = 2 cm/s2, s = 16 cm. Substituting these values, we get

16 = 6 t +

10 = 20 seconds. 0.5

19. (c) Distance moved in tth seconds is

 1 1 2s  +  .  f r

Differentiating both sides of (i) w.r.t. ‘t’ ds = – nA sin (nt + k) dt Differentiating again w.r.t. ‘t’, we get d 2s = – n2A cos (nt + k) dt 2



10 × 10 = 0.5 m/s2 2 × 100

Also v = u + f t

15. (b), (c) We have, s = A cos (nt + k) 



s = 100 m, f = ?, t = ? Using v2 – u2 = 2 f s

1 × 2 t2 ⇒ t2 + 6 t – 16 = 0 2

1 f (2t – 1). 2 Here u = 36 m/s, f = 6 m/s2, t = 5 seconds Substituting these values, we get

s = u +



s = 36 +

1 × 6 (2 × 5 – 1) 2

= 36 + 3 × 9 = 36 + 27 = 63 metres. Hence distance travelled by the particle in 5th second of its motion is 63 metres. 20. (b) Let f be the acceleration of the particle 1 f (2t – 1) Using s = u + 2 Here u = 0, s = 171 m, t = 10 seconds ∴ 171 = 0 + 

1 f (2 × 10 – 1) 2 171 × 2 = 18 m/s2. ⇒f= 19

21. (c) Let the initial velocity be u cm/s 1 f (2t – 1) 2 Here s = 60 cm, f = 4 cm/s2, t = 8 seconds Substituting these values, we get Hence, the initial velocity of the particle is 30 cm/s. Using s = u +

22. (c) Consider ⇒

1 1 g = 2 (1 + sin β + 1 − sin β) + r r′ u

2 1 1 2 1 1 1 + = u2 / g = ⇒ , , are in A.P r r′ R r R r′

⇒ r, R, r′ are in H.P Hence (c) is the correct answer. 23. (b) We have, 36 × 1000 m/sec. = 10 m/sec. 3600



u = 36 km/hr. =



v = 0, t = 2 minutes = 120 seconds, f = ?, s = ?

1101

 1 1 t1 + t2 = v  +   f r

Dynamics



1102

Objective Mathematics

Using v = u + f t, we get

⇒ 400 f = 1600 ⇒ f = 4 m/sec2.

10 1 = m/sec2 0 = 10 + f (120) ⇒ f = – 120 12 1 ∴ Retardation = – m/sec2 12

Now, v = u + f t ⇒ 50 = 30 + 4 t

Using v2 – u2 = 2 fs, we get

 1 (0)2 – (10)2 = 2  −  s   12 

⇒ 100 =

1 s ⇒ s = 600 ms 6

1 m/sec2 12 and distance travels before coming to rest = 600 m. 2u 2 sin α cos α u 2 sin 2 α 4 3 = 2g g

1 ⇒ α = 30°. 3 Hence (a) is the correct answer. ⇒ tan α =

25. (c) We have, u = 36 km/hr = 

36 × 1000 m/sec 3600 = 10 m/sec

v = 0; t = 10 seconds ∴ v = u + f t ⇒ 0 = 10 + f × 10 ⇒ f = –

29. (c) We have, T = ⇒

Hence, retardation =

24. (a) Given that

⇒ 4 t = 20 ⇒ t = 20 ÷ 4 = 5 seconds.

10 = – 1 m/sec2 10

∴ Retardation = 1 m/sec2 (ii) Let s be the distance travelled by the body before coming to rest Now v2 – u2 = 2 f s ⇒ (0)2 – (10)2 = 2 (– 1) s 10 × 10 = 50 m. ⇒ 2s = 10 × 10; s = 2 u2 (28) 2 26. (b) M aximum horizontal range = g = = 9.8 80 m Hence (b) is the correct answer. 27. (b) Let the required velocity be u m/sec. This means that with this velocity the ball covers the maximum horizontal range i.e., Maximum horizontal range = 80 meters u2 ⇒ g = 80 ⇒ u = 80g m/sec. Hence (b) is the correct answer. 28. (a) Here u = 30 m/s, v = 50 m/sec, s = 200 m Let f be the uniform acceleration and t, the time taken Using v2 – u2 = 2 f s ⇒ (50)2 – (30)2 = 2 × f × 200 ⇒ 2500 – 900 = 400 f

2u sin α 2u 2sin α cos α and R = g g

4u 2sin 2α g T2 2 tan α × 2 = = 2 g 2 u sin α cos α R g

⇒ tan α =

gT 2 . 2R

30. (b) Let the required veloccity be u m/sec. Then, Maximum horizontal range = 20 m u2 2 g = 20 ⇒ u = 20 × 9.8 ⇒ u = 14 m/sec u 2 sin 2α (given) 31. (c) Horizontal range (R) = g ⇒



Maximum height



(H) =

R=

u 2 sin 2 α (formula) 2g 4 u 2 sin 2α H⇒ = 3 g

∴ 2 sin α cos α = ⇒ sin α cos α ×

sin α (cos α ×



cos α ×

4 u 2 sin 2 α 3 2g

4 sin 2 α 3 2g 3 = sin2 α 3 – sin α) = 0 [ α ≠ 0]

3 – sin α = 0

3 = tan 60° ⇒ α = 60° 32. (d) Let the velocity of the man, relative to the ground be v.

tan α =

∴ v =

52 + 102 + 2 × 5 × 10 × cos 60º



=

25 + 100 + 100 ×



=

175 = 13.23 m/s.

1  2

1  ∵ cos 60º = 2   

1 1 f (2t − 1) ⇒ 55 = 0 + × f (12 − 1) 2 2 ⇒ f = 10 m/sec2 1 1 Distance = u + f (2t − 1) = 0 + × 10 × (2 × 8 − 1) 2 2 = 75 meters. Hence (c) is correct answer.

33. (c) S =  u +

1 1 f (2t − 1) = 60 − × 2 × (2 × 10 − 1) 2 2 = 60 – 19 = 41 meters And v = u – f t ⇒ 0 = 60 – 2 × t  ⇒ t = 30 seconds Hence (b) is correct answer.

34. (b) Sn = u −

1103

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

Dynamics

1 × f × 102 ⇒ u + 5f = 65 2 and 80 = u + 10f

35. (a) 650 = 10u +

So from equation (1) and (2), f = 3 m/sec2 and u = 50 m/sec Hence (a) is correct answer. 36. (b) v = u – f t = 0 ⇒ 0 = 200 – 2t ⇒ t = 100 sec And v2 = u2 – 2fs ⇒ (200 × 10–2 )2  = 2 × 2 × 10–2 × s ⇒ s = 100 meters. Hence (b) is correct answer.



37. (c) Let u be the initial velocity and f, the uniform acceleration of the particle.



1 f (2n – 1) 2 = x1,S(p + q)th = x2

Using Snth = u + Here Spth

41. (a) VQP =

1 f (2p – 1) = x1 ...(i) 2 1 f [2 (p + q) – 1] = x2 ...(ii) u+ 2 Subtracting (i) from (ii) [to eliminate u], we get

Hence, f =

−1  3  θ = tan   with the north 4

Hence (c) is the correct answer.

∴ u +

1 f [2q] = x2 – x1 2

vBA 3 tan θ = v = 4 A

v 2 + v 2 − 2v 2cos120° =

3V

v sin 120° 1 = − i.e., θ = 150° v − v cos120° 3 ∴  Perpendicular to the third side Hence (a) is the correct answer

tan θ =

or fq = x2 – x1

( x2 − x1 ) . q

38. (b) u = 36 km/hr = 10 m/sec v = u – f t ⇒ 0 = u – ft ⇒f=



10 1 =  m/sec2 120 12

42. (c) v cos θ = v1

1 2 Now, v2 = u2 – 2fs ⇒ 02 = (10) − 2 × × s 12 ⇒ s = 600 meters. Hence (b) is correct answer.

v sin θ = v2

∴ v =

v2 v12 + v22 and tan θ = v 1

39. (d)

 w1 =  w2 =   w1 ⋅ w2

  v2 − v1    v2 + v1 



...(2)

    = 0 ⇒ (v2 − v1 ) ⋅ (v2 + v1 ) = 0 ⇒ v2 = v1

Hence (d) is the correct answer 40. (c) vB =

...(1)

(vBA ) 2 + (vA ) 2 + 2(vBA )(vA )cos α (800) 2 + (600) 2 (α = π/2)



=



= 1000 km/hr

shortest distance = BC = s sin θ 

= s

v2 v + v22 2 1

=

Hence (c) is the correct answer 2 u2 u 43. (a) v2 = u2 – 2 fs ⇒   = u2 – 2f × 3 ⇒ f = 8 2 After penetrating 3 cm, initial velocity will u become = 2

1104

u u So, v = u – 2fs ⇒ 0 =   − 2 × × s 8 2

Objective Mathematics

⇒ s = 1 cm. Hence (a) is correct answer.

2

2

   v3 = vA − vC      From (1) and (2), v1 − v2 = vA − vC       ⇒ v1 − v2 = v3 or v1 = v2 − v3

2

2

Hence (b) is the correct answer.

44. (a) v = u + f t ⇒ v = (50 + 40 × 4) = 210 m/sec

51. (c) Given u = 98 m/s, g = 9.8 m/sec2

Hence (a) is correct answer. 45. (c) u = 36 km/ hr = 10 m/sec, (v = 0 final velocity) v = u – f t ⇒ u = ft 10 ⇒ f = = 1 m/sec2 10 so v2 = u2 – 2f s ⇒ 0 = (10)2 – 2 × 1 × s  ⇒ s = 50 meter. Hence (c) is correct answer.

Maximum height achieved u2 (98) 2 98 × 98 980 = 2g = = = = 490 metres. 2 × 9.8 2 × 9.8 2 52. (a) The velocity of projection (u) = 9.8 m/sec Also g = – 9.8 m/sec2 and at the highest point v = 0 Using v = u + gt, we have 9.8 = 1 second. 9.8 ∴ The particle will return to the same point in (2 × 1) seconds = 2 seconds.

1 2 ft 2 2(10320 − 1320) ⇒ f = = 5 m/sec2 3600

46. (b) s = ut +



Hence (b) is correct answer. 47. (b) Given that time t = 17 second, the shot goes down 272 ft and describes horizontal distance 4352 ft. 1 2 ⇒ 272 = − u sin α ⋅ t + gt 2  ⇒ 17 u sin α = 16 × 17(17 – 1) ⇒ u sin α = 256 and 4352 = u cos α t = 17 u cos α  ⇒ u cos α = 256 ⇒ u = 256 2   and tan α = 1 ⇒ α = 45°.

1 gt 2  ...(1) 2 Here, h = 1.25, u = 0, g = 10 m/s2 Substituting these values in (1) 1 2 × 1.25 × 10 t2 ⇒ t 2 = = 0.25 1.25 = 0 (t) + 2 10

53. (c) h = ut +

∴ t 2 = 0.5 × 0.5 ⇒ t = 0.5 seconds Again v = u + gt ∴ v = 0 + 10 + 0.5 = 5 cm/s. Here u = 0. Now snth = u +

Range = 2(Greatest height attained) 2  V 2 sin 2α  ⇒ V0 sin 2α = 2  0  ⇒ tan α = 2 g  2g 

0+ We have, 0+

2 hence range = V0 sin 2α g

V02  2 tan α  4 2 V0 .   = = g  1 + tan 2 α  5g

49. (d) If s2, s4 and s8 are the distances covered in second, fourth and eighth second, then

u=0 s2 = 0 +



1 2 1 s8 = 2 s2 : s4 :   = vA − vB    = vC − vB 

∴  50. (b) v1  v2

s4 =



⇒

1 g (2n – 1). 2 1 g (2 × 5 − 1) 2

1 g [2 (t − 3) − 1] 2

=

⇒

9 11 9 g 2 1 g (2t − 7) 2

=

9 11

9 9 = ⇒ 2t – 7 = 11 ⇒ t = 18 ÷ 2 = 9 2t − 7 11

Hence the body fell for 9 seconds.

1 3 F (2.2 – 1) = F, 2 2 7 F (2.4 – 1) = F, 2 15 F (2.8 – 1) = F. 2 s6 = 3 : 7 : 15.



0 = 9.8 – 9.8 × t ⇒ t =

54. (b) Let the body fell for t seconds.

48. (b) Let α be the angle of projection. Then,



...(3)

55. (a) Since the plane was moving vertically upward when the coin was dropped, the initial velocity of coin is – 10 m/sec. ∴ u = – 10 m/sec, t = 4 seconds

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

Let velocity of coin when it reaches the ground be v m/sec. 1 Using h = ut + gt 2, we have 2

∴ Height of plane at the time of dropping of coin = 38.4 m. Let v be the velocity of the coin on impact with the ground. Using v = u + gt, we get v = – 10 + 9.8 × 4  ⇒ v = – 10 + 39.2 = 29.2 m/sec. 56. (b) Let the bodies be thrown vertically upwards with velocities 2u and 5u respectively. Let h1 and h2 be the maximum heights attained by the bodies then,

(2u ) 2 h1 = 2g

and

h2 =

 u2  ∵ Maximum height =  2g  

(5u ) 2 . 2g

∴ The ratio of the maximum height attained by them (5u ) 2 (2u ) 2 : ⇒ h1 : h2 = 4 : 5. = 2g 2g 57. (a) Taking upward direction +ve. If velocity of projection of any ball be u m/sec, then velocity after ascending a height x metre is 0 ∴ 0 – u2 = – 2gx ⇒ 0 = u2 – 2 × 9.8 x  ⇒ u2 = 19.6x ∴ u = 4.43 x m/sec. 2u 2 × 4.43 x = 0.9   Hence, Time of flight = g = 9.8 x seconds. This time is to be divided into two equal intervals for the two balls in the air. Hence the time a ball stays in the hand 0.9 x = = 0.45 x seconds. 2 58. (b) Let the time taken by two bodies to reach the earth be t1 and t2 seconds For the motion of Ist body 1 1  g t12 ⇒ h1 =  g  t12  h1 = 0 × t1 + 2 2

...(1)

For the motion of second body 1 1  g  t22 ⇒ h2 =  g t22  ...(2) h2 = 0 × t2 + 2 2 Dividing (1) by (2), we get 1 2 g t1 h1 h1 t2 t = 2 = 12 ⇒ 1 = 1 h2 t2 h2 t2 g t22 2  ⇒ t1 : t2 = h1 : h2 . 59. (c) Let u be the velocity of projection and t be the time after which the particles meet.

first particle, we have 1 h= g t 2 ...(1) 2 second particle, we have 1 h = ut – g t 2 ...(2) 2

1105

For the 2 3 For the 1 2

Dynamics

1 h = – 10 × 4 + × 9.8 × 16 2  ⇒ h = – 40 + 78.4 = 38.4 m.

Adding (1) and (2), we get 1 1 1 1 h+ h= g t 2 + ut – g t 2 2 2 2 2  ⇒ h = ut ⇒ t = 2 1 gh 2 3 From (1), h= ⇒ u2 = gh. 3 2 u2 4



.

Hence the greatest height attained by the second particle 3 gh u2 3 = 2g = 4 = h. 2g 8 60. (a) Let the required height be h ft. and the time of its fall be n seconds. 1 Using Snth = u + g (2n – 1) 2 we get, 0 +

1 g (2n – 1) = 224 2

224 × 2 224 × 2 = g 32  [ g = 32 ft./sec2] = 14 15 1 ⇒ 2n = 15 i.e., n = =7 . 2 2

⇒ 2n – 1 =

∴ Time of fall = 7 Also h = ut +

1 seconds. 2

 15  1  15  1 gt 2 = 0   + × 32 ×   2  2 2  2

2

225 = 900 ft. 4 61. (b), (c) Let u be the velocity of projection and particle be at height h after time t. 1 g t2 ∴ h = ut – 2

= 16 ×

⇒ g t 2 – 2 ut + 2h = 0 ...(1) Let t1 and t2 be its roots 2h . (i) ∴ t1 t2 = g 1 g t1 t2. 2 2u 1 ⇒u= g (t1 + t2). (iii) t1 + t2 = g 2

(ii) From (2), h =

62. (c) Let the body be projected upwards with velocity u m/sec. Since the body remains above the earth for

1106

Objective Mathematics

12 sec, so displacement of the body in 12 sec above the earth is zero. 1 2 f t , we get ∴ Using s = ut + 2 1 0 = u ⋅ 12 – g ⋅ 122 2 ⇒ u = 6g m/sec. 63. (d) If u and 2u are the velocities of the two particles in the vertically upward direction, then using ⇒ ⇒

v2 = u2 + 2gs 0 = u2 – 2gh1 and 0 = (2u)2 – 2gh2 2gh2 = 4 u2 = 4 ⋅ 2gh1 h2 = 4 h1.



Using h = ut –

1 g t 2, we have 2

1 30g = 8g t – g t 2 or t 2 – 16 t + 60 = 0 2 ∴ t = 6 or 10 Thus the body will be at the surface of the earth after 6 sec. while on its way up and after 10 sec. while on its way down. Hence, the required time = 10 sec. 69. (c) We have, u = 0, t = 6 secs., v = 27 cm/sec Now v = u + f t ⇒ 27 = 0 + f × 6 ⇒ 6 f = 27 ⇒ f =

27 9 = cm/sec2 6 2

64. (c) The acceleration on the particle to slide on diameter Mass of the body (m) = 20 gm AB and CD are g and g cos 60º respectively. If t1 9 and t2 are the two times then = 90 dynes. ∴ Force = m × f = 20 × 2 1 gt 22 and CD AB = 0 + 70. (a) Here u = 10 m/s, v = 20 m/s, t = 5s, f = ? 2 1 Using v = u + f t ⇒ 20 = 20 + f × 5  =0+ g (g cos 60º) t 22 10 2 ⇒ f = = 2 m/s2 5 But AB = CD. Now m = 1000 kg, f = 2 m/s2 1 1 2 2 Using P = m f = 1000 × 2 = 2000 N ∴ g t1 = g t1 ⇒ t1 : t2 = 1 : 2 . 2 4 Hence force of 2000 N is acting on the car. 65. (b) Mass of bullet × velocity 71. (c) Here P = 5 × 105 N, m = 2 × 104 kg = Mass of rifle × velocity of recoil of the rifle Using P = m f, we get (0.01 × 100) ⇒ velocity of recoil of the rifle = = P 5 × 105 5 × 10 20 = = 25 m/s2. f = 4 = 0.05 m/sec. m 2 × 10 2 66. (c) Let t1 and t2 be the two times, then Now u = 0, v = ?, f = 25 m/s2, t = 20 s 1 1 v = u + f t ⇒ v = 0 + 25 × 20 = 500 m/s.  g t 2 and 20 = 0 +  g t 2 5=0+ 2 2 2 1 Hence velocity attained by the rocket at the end of 20 seconds is 500 m/s. 1 t2 Dividing, 12 = ⇒ t1 : t2 = 1 : 2. 4 72. (c) Here P = 100 dynes, m = 25 gramms. t2 P 100 Since P = m f ⇒ f = = = 4 cm/sec2. 67. (a) Let the two balls be projected vertically upwards m 25 and downwards with the same velocity u from the The body starts from rest, u = 0. top of a tower of height h. Taking positive direction downwards from the top of the tower and using s Now u = 0, f = 4 cm/sec2, s = 32 cm. 1 2 1 2 1 = ut +  f t f t ⇒ 32 = 0 + × 4 × t2 Using s = ut + 2 2 2 1 For first ball, h = – ut1 + g 2  ...(1) ⇒ t2 = 32 ÷ 2 = 16 ⇒ t = 4 seconds. 2 t1 For second ball, h = ut2 +

1 g t22  2

...(2)

Multiplying (1) by t2 and (2) by t1 and adding, we get g t12t2 + t22t1 1 h = 2 ⋅ t +t = g t1 t2. 2 1 2 68. (a) Let t be the time at which the body is at a height of 30 g from the bottom of the mine i.e., when it is at the surface of the earth.

73. (b) m = 2000 kg, f = 1 m/s2 P = m f ⇒ P = 2000 × 1 = 2000 N Again m = (2000 + 1000) kg = 3000 kg P = 2000 N P 2000 2 ∴ P = m f ⇒ f = = m/s2 = m/s2. m 3000 3 Hence maximum acceleration when truck carries a 2 m/s2. load of 1000 kg = 3



∴ 10 = 0 + ⇒ f =

1 f × 100 2

20 1 m/sec2 = m/sec2 100 5

Force, P = 1 kg-wt = 9.8 N. P Now, P = m f ⇒ m = f ⇒ m =

9.8 = 9.8 × 5 = 49 kg. 1/ 5

75. (a) Here u = 98 cm/sec, v = 0 Mass, m = 80 gm 1 × 80 gm.wt = 8 gm-wt. = 8 Resistance, P = 10 × 980 dynes Using P = m f Retardation =

8 × 980 = 98 cm/sec2 80

∴ Acceleration f = – 98 cm/sec2 Using v2 – u2 = 2 fs, we have 0 – 98 × 98 = 2 (– 98) s 98 × 98 = 49 cm. ⇒ s = 2 × 98 76. (b) Here m = 10 kg, P = 100 N Using P = m f, we get 100 = 10 m/s2 10 Again u = 20 m/s, v = 35 m/s f = 10 m/s2, t = ? Using, v = u + f t ⇒ 35 = 20 + 10 × t

100 = 10 × f ⇒ f =

15 = 1.5 seconds. ⇒ t = 10 Hence the force is applied for 1.5 seconds. 77. (a) Here u = 0, v = 6 m/s t = 2 minutes = 120 seconds, f = ? Using v = u + f t ⇒ 6 = 0 + f × 120 ∴ f =

6 1 = m/s2 120 20

1 Now m = 10 kg, f = m/s2, P = ? 20 6

Using P = m f ⇒ P = 106 ×

1 100 × 10000 = N = 50 N. 20 20 × 1000

6 = 2 m/s2 . Using P = m f ⇒ 6 = 3 × f ; f = 3 Now, u = 10 m/s, v = ? t = 2 seconds, f = 2 m/s2 Using v = u + f t, we get v  = 10 + 2 × 2 = 14 m/s Change in momentum = m (v – u) = 3 (14 – 10) = 12 kg m/s Hence final velocity = 14 m/s and change in momentum = 12 kg m/s. 79. (c) m = 2.5 kg, P = 10N Using, P = m f, we get 10 = 2.5 × f 10 = 4 m/s2 ⇒ f = 2.5 Now u = 0 m/sec, f = 4 m/s2, s = 30 m 1 2 f t , Using S = ut + 2 30 = 0 (t) +

1 × 4 t 2 2

30 × 2 = 15 ⇒ t = 4 Using v = u + f t, we get ⇒ t2 =

=0+4×

15 seconds

15 = 4 15 m/sec = 15.5 m/sec.

80. (a) Forward force = 1000 N Retarding force = 500 N ∴ P = (1000 – 500) N = 500 N Mass (m) = 1000 kg Using P = m f, we get 500 1 m/sec2 = m/s2 500 = 1000 × f ⇒ f = 1000 2 Using v = u + f t, we get v = 10 +

1 × 10 = (10 + 5) m/sec = 15 m/sec. 2

81. (c) Here u = 3 m/sec, v = – 2 m/sec, t = 4 sec Using v = u + f t, we get – 2 = 3 – f × 4

[Here f is retardation]

5 ⇒ – 4 f = – 5, f = m/s2 4 5 25 Using P = m f, we get P = 10 × = = 12.5 4 2 Newton. 82. (b) Using P = m f, we get 100 = 10 × f ⇒ f = 100 ÷ 10 m/s2 = 10 m/s2. Given u = 20 m/s, v = 35 m/s. Using v = u + f t, we get 35 = 20 + 10 t 15 = 1.5 seconds. ⇒ t = 10

1107

1 2 f t 2

s = ut +

78. (a) P = 6N, m = 3 kg

Dynamics

74. (a) Here u = 0, t = 10 seconds, s = 10 m

1108

83. (a) Final velocity in the barrel

Objective Mathematics

In the second case, let a mass m1 be removed so that the remaining mass is (m – m1). The equation of motion is ...(2) R – (m – m1) g = (m – m1) ⋅ 2 f

(v) = 50,000 cm/s. Distance travelled (s) = 250 cm Using v2 – u2 = 2 f s ⇒ (50,0000)2 – 02 = 2 × f × 250 50, 000 × 50, 000 ⇒ f = i.e., f = 5 × 106 cm/s2. 500 ∴ Mean force exerted on the ball of mass 1000 gm is P = m × f = 1000 × 5 × 106 = 5 × 109 dynes. 84. (c) Let the body which falls freely from rest at a height of 4.9 m, hits the ground with a velocity of v m/s ∴ u = 0, v = ?, g = 9.8 m/s2, h = 4.9 Using v2 = u2 + 2gh ⇒ v2 = 0 + 2 × 9.8 × 4.9 ⇒ v2 = 9.8 × 9.8 v = 9.8 m/s ∴ Momentum of body when it hits the ground = mass × velocity = 4 × 9.8 = 39.2 kg m/s. 85. (a) Using P = m f, we get

89. (c) m1 = 50 gms, m2 = 48 gms

u = 0, s = 10 cms, t = 1 sec. 1 Using s = ut + f t 2, we have 2 1 × f × 1 ∴ f = 20 cms/sec2. 2 m1 − m2 Also, f = m + m g 1 2



10 = 0 +

∴ 20 =

6 = 3 × f ⇒ f = 3 m/s2 Now u = 10 m/sec, t = 2 sec, v = ? Using v = ut + f t, we get v = 10 + 2 × 2 = 14 m/sec

50 − 48 2 g or 20 = g 50 + 48 98

g = 980 cms/sec2.

90. (c) Here m1 = 260 gm, m2 = 230 gm (m1 > m2) m1 − m2 ∴ Acceleration f = m + m g 1 2 30 × 980 = 60 cm/sec2. 490 Now u = 0, f = 60 cm/sec2, t = 3 sec. Distance (s) at the end of 3 seconds 1 2 1 f t = (60) (3)2 = 270 cm. = ut + 2 2 Velocity (v) at the end of 3 seconds = u + f t = 60 × 3 = 180 cm/sec.

Change in momentum = 3 × 14 – 3 × 10 = 3 (14 – 10) = 3 × 4 = 12 kg.



86. (b) Taking downward motion positive initial momentum = mass × velocity = 2 × 10 m/s = 20 km/s Final momentum = 2 × – 5 = – 10 kg m/s ∴ Change in momentum of the body = 20 – (– 10) = 30 kg m/s.

=

91. (a) Here u = 48 ft/sec, f = – g sin α = – g sin 30º

87. (c) Let T be the tension in the rope and let  the man descends with acceleration f m/ sec2.

g = – 16 ft/sec2 [ α = 30º] 2 s = 64 ft, v = ? Using v2 – u2 = 2 f s, we have v2 – (48)2 = 2 (– 16) ⋅ 64 v2 = 48 × 48 – 32 × 64 = 16 × 16 (3 × 3 – 2 × 4) = 256 ∴ v = 16 ft/sec.

The equation of motion is 100g – T = 100 f ⇒ T = 100 (g – f ) Newtons But T ≤ 50 kg = 50g N (given) ∴ 100 (g – f ) ≤ 50g or 2 (g – f ) ≤ g or g ≤ 2 f g ∴ f ≥ 2 The man will reach the ground in longest possible time if his downward acceleration is the least. g 9.8 = = 4.9 m/sec2. ∴ Required acceleration = 2 2 88. (a) Let R be the upthrust of the air in each case. In the first case, the equation of motion is R – mg = m f 

Subtracting (1) from (2), m1 g = m f – 2m1 f ⇒ m1 (g + 2 f) = m f mf ∴ m1 = 2 f + g . 

=–

92. (b) Let u be the least velocity of projection. The cricket ball can be thrown 200 m means max. horizontal range = 200 m. u2 ∴ g = 200 or u2 = 200 × 9.81 = 1962 ∴ u = 44.3 m/sec. 93. (a) Let u be the velocity of projection.

...(1)

u2 R = max. horizontal range = g 

...(i)

u2 1 1 u2 1 ⋅ = 2g 2 = 4 ⋅ g = R. 4

94. (b)

]

∴ 2h + [ of (i)]

the boy can throw the ball 100 m. ∴ max. horizontal range = 100 m. 

The ball is in the air for time = time of flight π 2 × 31.3 × sin 2u sin α 4 = 4.5 sec. (nearly) = = g 9.81 95. (c) Let α be the angle of projection. Range of horizontal plane = 2 × greatest height attained [given] 2 2 2v sin α cos α v sin 2α ∴ =2× ⇒ tan α = 2 g 2g v 2 sin α 2v 2 sin α cos α Now reqd. range = = g g v 2 2 tan α v2 2 × 2 4v 2 ⋅ ⋅ = = = . 2 g 1 + tan α g 1+ 4 5g 96. (a), (b) Here u = 56 ft/sec. Horizontal range = 49 ft. u 2 sin 2α ⇒ = 49 g

u 2 sin 2 α u 2 sin 2α ,h= 2g g u 2 sin 2 α 1 u 4 sin 2 2α 2g R2 + ⋅ ⋅ 2 2 = 2 g 8 g u sin α 8h



=

u 2 sin 2 α 1 u 4 ⋅ 4 sin 2 α cos 2 α 2g + ⋅ ⋅ 2 2 2 g 8 g u sin α



=

u2 u 2 sin 2 α u 2 cos 2 α + = g g g



= Max. horizontal range.

...(i)

π Let u be the velocity of projection, α = 4 u2 ∴ from (i), = 100 2 ∴ u = 100 × 9.81 = 31.3 m/sec.

⇒

R=

99. (c) When two points are equidistant from x-axis; one above and the other below it, their ordinates are equal and opposite, so that their sum = 0. Let u be the velocity of projection and α the angle of projection. u 2 sin 2 α ; Ordinate of vertex = 2g u 2 cos 2α Ordinate of focus = – 2g By virtue of given condition, we have u 2 sin 2 α u 2 cos 2α =0 − 2g 2g



or sin2α – cos 2α  = 0 or sin2α – (1 – 2 sin2α) = 0 1 . ∴ 3 sin2α = 1 or sin α = 3 100. (d) v = 1/4 m/sec

56 × 56 sin 2α = 49 32 1 98

⇒ sin 2α = 49 ×

1 = sin 30º or sin (180º – 30º)  = 2 ∴ 2α = 30º or 150º ∴ α = 15º or 75º. 97. (a) Let u be the velocity of projection and α, the angle of projection. Then R=

u 2 sin 2α 2u 2 sin α cos α 2u sin α = ;T= g g g

∴

gT g 4u sin α g = ⋅ ⋅ 2 = tan α 2R 2 g2 2u sin α cos α 2

2

w v = sin 30° sin 105° ⇒ w =



=

1 2 = 3 +1 2 2



=

1 v ( 3 − 1) = 4 4 2



=

v.

2

 gT  . Hence α = tan–1   2R  2

vsin 30° sin 105° 2v = 3 +1

1 m/s. 8 ( 6 − 2)

2v( 3 − 1) 3 −1

1109

Dynamics

98. (b) Let u be the velocity of projection and α, the angle of projection. Then

u 2 sin 2 α Also, greatest height = 2g π u 2 sin 2 4 =  [ for max. H.R. α = 2g

1110

101. (b) Let the particle possess other velocity

Objective Mathematics

⇒ 0 = (15)2 – 2gh1



225 ∴ h1 = 2 g metres

= x m/sec Angle between the two velocities is −1 θ = tan

∴ cos θ =

After striking the ground (i.e., rebound) 2

12 (given) 5



5 13

2

3 = 121.32 m/s. 5 102. (c) Let the velocities of projection of the two balls be u1 and u2.

= 202.2 ×

Angles of projection are α1 = 60º and α2 = 30º. Since both the particles attain the same height and u 2 sin 2 α the greatest height is given by . 2g

or

u12 sin 2 α1 u22 sin 2 α 2 = 2g 2g u1 sin α 2 sin 30º u2 = sin α1 = sin 60º =

Hence u1 : u2 = 1 :

3.

103. (c) Maximum horizontal range

5×5 25 2 g = 2g metres

...(2)

∴ From equation (1) and (2), we get

5 ⇒ 15 = 13 + x + 2 × 13 × x × 13 ⇒ x2 + 10x – 56 = 0 ⇒ (x – 4)(x + 14) = 0 ∴ x = 4 m/s. ∴ Velocity at highest point = horizontal component of velocity  [which is constant] = u cos α 2

 15  0 =   − 2 gh2  3

∴ h2 =

∴ R2 = P2 + Q2 + 2PQ cos θ 2

...(1)



225 25 h1 : h2 = 2 g : 2 g = 9 : 1.

105. (a) Let α be the angle of projection. Velocity at max. height = u cos α 1 u By the given condition u cos α = 2 1 , ∴ α = 60º ⇒ cos α = 2 Range on horizontal plane =

u 2 sin 2α u 2 sin 120º u2 3 = = . g g 2g

106. (b) Let u be the velocity of projection. For a range R, the two possible angles of projection are α and β. π  ...(1) ∴ α + β = 2 ∴ t1 =

1 2 = 3 2

1 3

2 2u sin β t1 2u sin α sin 2 α , t2 = 2 = t g g 2 sin 2 β

By componendo and dividendo, we have sin (α + β) sin (α − β) t12 − t22 sin 2 α − sin 2 β 2 = 2 2 2 π  t1 + t2 sin α + sin β sin 2 α + sin 2  − α  2   π sin sin (α − β) sin (α − β) 2 = = sin 2 α + cos 2 α 1

u2 (39.2) 2 = g =  [ u = 39.2 m/s (given)] 9.8 sin (α − β) π =  [ sin (α + β) = sin = 1.] 1536.64 sin (α + β) = 156.8 = 157 m (approx.). = 2 9.8 104. (b) Let the greatest height attained by the ball before 107. (a) Let (x , y ) be the co-ordinates of the focus of any 1 1 and after striking the ground be h1 metre and h2 trajectory, where u and α are the velocity and angle metre respectively. The the final velocity (v) at the of projection respectively. Then highest point will be zero. u 2 sin 2α u 2 cos 2α Before striking the ground, , y = – x1 = 1 2g 2g v2 = u2 – 2gh 1

108. (a) Let a particle be projected in a direction making an angle α with the horizontal, then its co-ordinates after t secs are given by

x = u cos α ⋅ t



y = u sin α ⋅ t –

1 g t 2  2

2

109. (b) For the same velocity of projection and same range there are two direction of projections α and 90° – α 2u sin α 2u cos α and t2 = ∴ t1 = g g 4u 2 ∴ t12 + t22 = g 2

...(1)

where u is the velocity of projection and α the angle of projection. 1 y Let v1, v2 be the velocities at heights y and 2 respectively.

1111

2 2 ⋅ v2, ∴ 5 v1 = 2v22 5 ∴ 5 (u2 – 2g y) = 2 (u2 – g y)

or or

3u2 = 8 g y = 8g 3 = 4 sin2α

111. (c) Range =

3 4

u 2 sin 2 α  2g

⇒`sin α =

3 . Hence α = 60º. 2

2u 2 sin α cos α = 24  g

u 2 sin 2 α =8  2g 4 4 (2) ⇒ tan α = ⇒ sin α = = 0.8 3 5 (1) ⇒ α = sin–1 (0.8). ∴ (2) ⇒ u2 = 25 g ⇒ u = 5 g , Max. height =

i.e.,

u=5 g

of (1)

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

and α = sin–1 (0.8).

112. (a) Velocity at height h = ∴ Required velocity =

(u 2 − 2 gh) {(10 5 ) 2 − 2 × 9.8 × 10}

= 4 (19) m/sec.

110. (a) Let y be the greatest height, then u 2 sin 2 α  2g

1  y = u2 – g y 2

Since v1 =

...(1)

1   x 2 +  y + g t 2  = u 2 t2 2   1 2  which is a circle with centre at  0, − g t  and 2   radius ut.

y=

v22 = u2 – 2g ⋅

∴ sin2α =

1 g t 2 = u sin α ⋅ t  ...(2) or y + 2 Eliminating the arbitrary constant α, by squaring and adding, we have





Dynamics

Squaring and adding, [to eliminate α], we have u4 x12 + y12 = . 4g 2 u4 , which ∴ Required locus of foci is x2 + y2 = 4g 2 u2 and the point of projection is a circle of radius 2g as its centre.

v12 = u2 – 2g y

113. (a) Let the angle of projection be α, then the vertical upward component of the velocity at the top of the tower = 10 sin α. 1 2 f t , we get Then, using s = ut + 2 1 × (10) × 52 – 100 = (10 sin α) × 5 – 2 ⇒ sin α =

1 ⇒ α = 30º. 2

Exercises for self-practice 1. A body is projected through an angle α from vertical so that its range is half of maximum range, α is:

3. A train whose mass is 16 metric tons, moves at the rate of 72 km/hr. After applying breaks it stops in 500 metre. What is the force exerted by breaks obtaining it to be uniform?

(a) 60° (b) 75° (c) 30° (d) 22.5° 2. A particle is thrown with velocity u at an angle of 30° (a) 800 N (b) 1600 N from horizontal line when it becomes ⊥ar to its original (c) 3200 N (d) 6400 N position: 2u 4. A person travelling on a straight line moves with uniform (b) 2ug (a) g velocity v1 for some time and with uniform velocity v 2 for the next equal time. The average velocity ‘v’ is u 3 given by: (c) (d) None of these g

1112

Objective Mathematics



(a) v =



(c)

v1 + v2 2

2 1 1 = + v v1 v2

(b) v = (d)

7. The range of a projectile fixed at an angle of 15° is 50 m, if it is fixed with the same speed at an angle of 45°, then the range will be:

v1v2

1 1 1 = + . v v1 v2

5. A particle is projected from the top of tower 5 m high and at the same moment another particle is projected upward from the bottom of the tower with a speed of 10 m/s, meet at distance ‘h’ form the top of tower, then h:

(a) 1.25 m

(b) 2.5 m



(c) 3 m

(d) None of these

6. A ball falls from a height h upon fixed horizontal plane, e is the coefficient of restitution, the whole distance described by the ball before it comes to rest is:

1 + e2 (a) 1 − e 2 h

(b)



1 + e2 (c) (1 − e 2 )h

1 − e2 (d) (1 + e 2 )h

1 − e2 h 1 + e2



(a) 50 m

(b) 100 m



(c) 150 m

(d) None of these

8. A man weighing 60 kg jumps off a railway train running on horizontal rails at 20 km/h with a packet weighing 10 kg in his hand. The thrust of the packet on his hand is

(a) 0 (c) 50 kg wt.

(b) 10 kg wt. (d) 70 kg wt

9. The maximum range for a given particle is possible only when the angle of projection is (a) 0º (b) 30º (c) 45º (d) 60º 10. From the top of a hill of height 150m, a ball is projected with a velocity of 10 m/sec. It takes 6 seconds to reach the ground. The angle of projection of the ball is

(a) 15º (c) 45º

(b) 30º (d) 60º

Answers

1. (b)

2. (d)

3. (d)

4. (a)

5. (a)

6. (a)

7. (b)

8. (c)

9. (c)

10. (b)

32

Set Theory

CHAPTER

Summary of concepts SET A set is any collection of objects such that given an object, it is possible to determine whether that object belongs to the given collection or not. For example, the collection of all students of Delhi University, is a set, whereas, collection of all good books on mathematics, is not a set, since a mathematics book considered good by one person might be considered bad or average by another. Notations  The sets are usually denoted by capital letters A, B, C, etc. and the members or elements of the set are denoted by lower-case letters a, b, c etc. If x is a member of the set A, we write x ∈ A (read as ‘x belongs to A’) and if x is not a member of the set A, we write x ∉ A (read as ‘x does not belong to A’). If x and y both belong to A, we write x, y ∈ A.

REPRESENTATION OF A SET Usually, sets are represented in the following two ways : (i) Roster form or Tabular form (ii) Set Builder form or Rule Method

Roster Form In this from, we list all the member of the set within braces (curly brackets) and separate these by commas. For example, the set A of all odd natural numbers less that 10 in the roster from is written as: A = {1, 3, 5, 7, 9} Note:

The symbol ‘|’ stands for the words ‘such that’. Sometimes, we use the symbol ‘:’ in place of the symbol ‘|’.

TYPES OF SETS Empty Set or Null Set  A set which has no element is called the null set or empty set. It is denoted by the symbol Φ. For example, each of the following is a null set : (a) The set of all real numbers whose square is – 1. (b) The set of all rational numbers whose square is 2. (c) The set of all those integers that are both even and odd. A set consisting of atleast one element is called a nonempty set. Singleton Set  A set having only one element is called singleton set. For example, {0} is a singleton set, whose only member is 0. Finite and Infinite Set  A set which has finite number of elements is called a finite set. Otherwise, it is called an finite set. For example, the set of all days in a week is a finite set whereas, the set of all integers, denoted by { ..., – 2, – 1, 0, 1, 2, ...} or {x | x is an integer}, is an infinite set. An empty set φ which has no element, is a finite set. The number of distinct elements in a finite set A is called the cardinal number of the set A and it is denoted by n (A).

Equal Sets Two sets A and B are said to be equal, written as A = B, if every element of A is in B and every element of B is in A.

(i) In roster form, every element of the set is listed only Equivalent Sets once. Two finite sets A and B are said to be equivalent, if n (A) = n (ii) The order in which the elements are listed is immaterial (B). For example, each of the following sets denotes the same Clearly, equal sets are equivalent but equivalent sets need set {1, 2, 3}, {3, 2, 1}, {1, 3, 2}. not be equal. For example, the sets A = {4, 5, 3, 2} and B = {1, 6, 8, 9} Set-Builder Form are equivalent but are not equal. In this form, we write a variable (say x) representing any member of the set followed by a property satisfied by each member of the set. For example, the set A of all prime numbers less than 10 in the set-builder form is written as A = {x | x is a prime number less that 10}

Subset Let A and B be two sets. If every elements of A is an element of B, then A is called a subset of B and we write A ⊆ B or B ⊇ A (read as ‘A is contained in B’ or B contains A’). B is called superset of A.

1114

Notes:

Objective Mathematics

For example , if A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7, 9}, then A – B = {2, 4} and B – A = {7, 9}. (i) If A ⊆ B and A ≠ B, we write A ⊂ B or B ⊃ A (read as : A is a proper subset of B or B is a proper superset of A). (ii) Every set is a subset and a superset of itself. Important Results (iii) If A is not a subset of B, we write A ⊄ B. (a) A – B ≠ B – A (iv) The empty set is the subset of every set. (b) The sets A – B, B – A and A ∩ B are disjoint sets (v) If A is a set with n (A) = m, then the number of subsets of (c) A – B ⊆ A and B – A ⊆ B m m A are 2 and the number of proper subsets of A are 2 –1. (d) A – φ = A and A – A = φ For example, let A = {3, 4}, then the subsets of A are φ, {3}, {4}, {3, 4}. Here, n (A) = 2 and number of subsets of A = Symmetric Difference of Two Sets 22 = 4. The symmetric difference of two sets A and B, denoted by Also, {3} ⊂ {3, 4} and {2, 3} ⊄ {3, 4} A ∆ B, is defined as A ∆ B = (A – B) ∪ (B – A).

Power Set

For example, if A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7, 9} then The set of all subsets of a given set A is called the power set of A A ∆ B = (A – B) ∪ (B – A) = {2, 4} ∪ {7, 9} = {2, 4, 7, 9}. and is denoted by P(A). For example, if A = {1, 2, 3}, then Complement of a Set P (A) = {φ, {1}, {2}, {3}, {1, 2} {1, 3}, {2, 3}, {1, 2, 3}} Clearly, if A has n elements, then its power set P (A) contains If U is a universal set and A is a subset of U, then the compleexactly 2n elements. ment of A is the set which contains those elements of U, which are not contained in A and is denoted by A′ or Ac. Thus,

OPERATIONS ON SETS Union of Two Sets

The union of two sets A and B, written as A ∪ B (read as ‘A union B’), is the set consisting of all the elements which are either in A or in B or in both. Thus, A ∪ B = {x : x ∈ A or x ∈ B} Clearly, x ∈ A ∪ B ⇒ x ∈ A or x ∈ B, and x ∉ A ∪ B ⇒ x ∉ A and x ∉ B. For example, if A = {a, b, c d} and B = {c, d, e, f }, then A ∪ B = {a, b, c, d, e, f}

Intersection of Two Sets



A′ = {x : x ∈ U and x ∉ A}

For example, if U = {1, 2, 3, 4, ...} and A = {2, 4, 6, 8, ...}, then, A′ = {1, 3, 5, 7, ...}

Important Results (a) U ′ = φ (c) A ∪ A ′ = U

(b) φ ′ = U (d) A ∩ A ′ = φ.

ALGEBRA OF SETS 1. Idempotent Laws  For any set A, we have (a) A ∪ A = A

(b) A ∩ A = A

The intersection of two sets A and B, written as A ∩ B (read as 2. Identity Laws  For any set A, we have ‘A intersection B’) is the set consisting of all the common ele(a) A ∪ φ = A (b) A ∩ φ = φ ments of A and B. Thus, (c) A ∪ U = U (d) A ∩ U = A A ∩ B = {x : x ∈ A and x ∈ B} 3. Commutative Laws  For any two sets A and B, we Clearly, x ∈ A ∩ B ⇒ x ∈ A and x ∈ B, and have x ∉ A ∩ B ⇒ x ∉ A or x ∉ B. (a) A ∪ B = B ∪ A (b) A ∩ B = B ∩ A For example, if A = {a, b, c, d} and B = {c, d, e, f}, then A ∩ B = {c, d}. 4. Associative Laws  For any three sets A, B and C, we have

Disjoint Sets

(a) A ∪ (B ∪ C) = (A ∪ B) ∪ C (b) A ∩ (B ∩ C) = (A ∩ B) ∩ C Two sets A and B are said to be disjoint, if A ∩ B = φ, i.e., A and B have no element in common. 5. Distributive Laws  For any three sets A, B and C, For example, if A = {1, 2, 5} and B = {2, 4, 6}, then we have A ∩ B = φ, so A and B are disjoint sets. (a) A ∪ (B ∩ C) = (A ∪ B) ∩ ( A ∪ C) (b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Difference of Two Sets

If A and B are two sets, then their difference A – B is defined as:

A – B = {x : x ∈ A and x ∉ B}.



B – A = {x : x ∈ B and x ∉ A}.

Similarly,

6. For any two sets A and B, we have (a) P (A) ∩ P (B) = P (A ∩ B) (b) P (A) ∪ P (B) ⊆ P (A ∪ B), where P (A)  is  the  power set of A. 7. If A is any set, we have (A′ ) ′ = A.

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

(A ∪ B)′ = A′ ∩ B′ (A ∩ B)′ = A′ ∪ B′ A – (B ∪ C) = (A – B) ∩ (A – C) A – (B ∩ C) = (A – B) ∪ (A – C).

Some Useful Results on Cartesian Product If A, B, C are three sets, then (i) A × (B ∪ C) = (A × B) ∪ (A × C) (ii) A × (B ∩ C) = (A × B) ∩ (A × C) (iii) A × (B – C) = (A × B) – (A × C) (iv) (A × B) ∩ (S × T) = (A ∩ S) × (B ∩ T),

Important Results on Operations on Sets A ⊆ A ∪ B, B ⊆ A ∪ B, A ∩ B ⊆ A, A ∩ B ⊆ B A – B = A ∩ B′ (A – B) ∪ B = A ∪ B (A – B) ∩ B = φ A ⊆ B ⇔ B′ ⊆ A′ A – B = B′ – A′ (A ∪ B) ∩ (A ∪ B′ ) = A A ∪ B = (A – B) ∪ (B – A) ∪ (A ∩ B) A – (A – B) = A ∩ B A–B=B–A⇔A=B A∪B=A∩B⇔A=B A ∩ (B ∆ C) = (A ∩ B) ∆ (A ∩ C)

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii)

Some Basis Results about Cardinal Number If A, B and C are finite sets and U be the finite universal set, then (i) n (A′ ) = n (U) – n (A) (ii) n (A ∪ B) = n (A) + n (B) – n ( A ∩ B) (iii) n (A ∪ B) = n (A) + n (B), where A and B are disjoint non-empty sets (iv) n (A ∩ B′ ) = n (A) – n (A ∩ B) (v) n (A ′ ∩ B ′) = n (A ∪ B) ′ = n (U) – n (A ∪ B) (vi) n (A ′ ∪ B ′) = n (A ∩ B) ′ = n (U) – n (A ∩ B) (vii) n (A – B) = n (A) – n (A ∩ B) (viii) n (A ∩ B) = n (A ∪ B) – n (A ∩ B ′ ) – n (A ′ ∩ B) (ix) n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C) (x) If A1, A2, A3, ... An are disjoint sets, then n (A1 ∪ A2 ∪ A3 ∪ ...∪ An) = n (A1) + n (A2) + n (A3) + ... + n (An)

(xi) n (A ∆ B) = number of elements which belong to exactly one of A or B

carteSIan proDuct of tWo SetS If A and B are any two non-empty sets, then cartesian product of A and B is defined as A × B = {(a, b) : a ∈ A and b ∈ B} Notes:

where S and T are two sets. (v) If A ⊆ B, then (A × C) ⊆ (B × C) (vi) If A ⊆ B, then (A × B) ∩ (B × A) = A2 (vii) If A ⊆ B and C ⊆ D then A × C ⊆ B × D (viii) If A ⊆ B, then A × A ⊆ (A × B) ∩ (B × A) (ix) If A and B are two non-empty sets having n elements in common, then A × B and B × A have n2 elements in common. (x) A × B = B × A if and only if A = B (xi) A × (B ′ ∪ C ′ ) ′ = (A × B) ∩ (A × C) (xii) A × (B ′ ∩ C ′ ) ′ = (A × B) ∪ (A × C)

reLatIonS Let A, B be any two non-empty sets, then every subset of A × B defines a relation from A to B and every relation from A to B is a subset of A × B. If R is a relation from A to B and if (a, b) ∈ R, then we write a R b and say that ‘a is related to b’ and if (a, b) ∉ R, then we write a R b and say that a is not related to b.

Important Results (a) Every subset of A × A is said to be a relation on A. (b) If A has m elements and B has n elements, then A × B has mn elements and total number of different relations from A to B is 2mn. (c) Let R be a relation from A to B, i.e., R ⊆ A × B, then Domain of R = {a : a ∈ A, (a, b) ∈ R for a some b ∈ B} Range of R = {b : b ∈ B, (a, b) ∈ R for some a ∈ A} For example, let A = {1, 3, 4, 5, 7}, B = {2, 4, 6, 8} and R be the relation ‘is one less than’ from A to B, then R = {(1, 2), (3, 4), (5, 6), (7, 8)}. Here, domain of R = {1, 3, 5, 7} and range of R = {2, 4, 6, 8}.

Identity relation R is an identity relation if (a, b) ∈ R iff a = b, a ∈ A, b ∈ A. In other words, every element of A is related to only itself.

universal relation Let A be any set and R be the set A × A, then R is called the Universal Relation in A.

(i) If A = φ or B = φ, then we define A × B = φ. Void universal relation (ii) A × B ≠ B × A (iii) If A has n elements and B has m elements then A × B has φ is called Void Relation in a set. mn elements. (iv) If A1, A2, ..., Ap are p non-empty sets, then their cartesian product, is defined as ai ∈ Ai for all i}

p

∏ Ai = {(a1, a2, a3, ...ap); i =1

1115

For any three sets A, B and C,

Inverse relation

Let R ⊆ A × B be a relation from A to B. Then R– 1 ⊆ B × A is defined by

Set Theory

8. Demorgan’s Laws we have

1116



R– 1 = {(b, a) : (a, b) ∈ R}

Thus, (a, b) ∈ R ⇔ (b, a) ∈ R , ∨ a ∈ A, b ∈ B. – 1

Objective Mathematics

Notes: (i) dom (R– 1) = range (R) and range (R– 1) = dom (R) (ii) (R– 1)– 1 = R. For example, if R = {(1, 2), (3, 4), (5, 6)} then R–1 = {(2, 1), (4, 3), (6, 5)} and (R–1)–1 = {(1, 2), (3, 4), (5, 6) = R. dom (R) = {1, 3, 5}, range (R) = {2, 4, 6} and dom (R– 1) = {2, 4, 6}, range (R– 1) = {1, 3, 5} So, dom (R– 1) = range (R) and range (R– 1) = dom (R).

(ii) a R b ⇒ b R a, ∨ a, b ∈ A and (iii) a R b, b R c ⇒ a R c, ∨ a, b, c ∈ A. For example, let I be the set of all integers, m be a positive integer. Then the relation, R on I is defined by R = {(x, y) : x, y ∈ I, x – y is divisible by m}. Consider any x, y, z ∈ I.

(i) Since x – x = 0 = 0 . m ⇒ x – x is divisible by m ⇒ (x, x) ∈ R ⇒ R is reflexive. (ii) Let (x, y) ∈ R ⇒ x – y is divisible by m ⇒ x – y = mq, for some q ∈ I ⇒ y – x = m (– q) ⇒ y – x is divisible by m ⇒ ( y, x) ∈ R TYPES OF RELATIONS ON A SET Thus, (x, y) ∈ R ⇒ ( y, x) ∈ R ⇒ R is symmetric. (iii) Let (x, y) ∈ R and ( y, z) ∈ R Let A be a non-empty set, then a relation R on A is said to be : ⇒ x – y is divisible by m and y – z is divisible by m (a) Reflexive  If a R a, ∨ a ∈ A, i.e., if ⇒ x ­– y = mq and y – z = mq ′ for some q, q ′ ∈ I ⇒ (x – y) + ( y – z) = m (q + q ′ ) (a, a) ∈ R, ∨ a ∈ A ⇒ x – z = m (q + q ′ ), q + q ′ ∈ I (b) Symmetric  If a R b ⇒ b R a, ∨ a, b ∈ A, i.e., if ⇒ (x, z) ∈ R Thus, (x, y) ∈ R and ( y, z) ∈ R ⇒ (x, z) ∈ R, so R is transi (a, b) ∈ R ⇒ (b, a) ∈ R, ∨ a, b ∈ A tive. (c) Anti-Symmetric  If a R b and b R a ⇒ a = b, ∨ a, Hence the relation R is reflexive, symmetric and transitive b∈A and it is also an equivalence relation. Note: It is important to note that every identity relation is reflex (d) Transitive  If a R b and b R c ive but every reflexive relation need not be an identity relation. ⇒ a R C, ∨ a, b, c ∈ A Also, identity relation is reflexive symmetric and transitive. i.e., (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∨ a, b, c ∈ A

EQUIVALENCE RELATION

CONGRUENCE  MODULO  m

Let m be a positive integer and x, y ∈ I, then x is said to be A relation R on a non-empty set A is called an equivalence relacongruent to y modulo m, written as x ≡ y (mod m), iff x – y is tion if and only if it is divisible by m. (i)  reflexive  (ii)  symmetric and  (iii)  transitive. That For example, 155 ≡ 7 (mod 4) as is, R satisfies following properties: 155 − 7 148 = 37 (integer). = (i) a R a, ∨ a ∈ A 4 4

mULTIPLE CHOICE QUESTIONS Choose the correct alternative in each of the following problems: 1. X and Y are two sets such that n (X) = 17, n (Y) = 23, n (X ∪ Y) = 38 then n (X ∩ Y) is (a) 4 (c) 6

(b) 2 (d) None of these

2. If A and B are two sets such that A has 12 elements, B has 17 elements and A ∪ B has 21 elements, then number of elements in A ∩ B are (a) 6 (c) 8

(b) 4 (d) None of these

3. If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements; then number of elements in X ∩ Y are (a) 5 (c) 6

(b) 8 (d) None of these

4. If n (U) = 700, n (A) = 200, n (B) = 300, n (A ∩ B) = 100, then n (A ′ ∩ B ′ ) is equal to (a) 400 (c) 300

(b) 240 (d) None of these

5. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, then number of elements S ∪ T has (a) 42 (c) 48

(b) 50 (d) None of these

6. X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, then number of elements Y has (a) 58 (c) 30

(b) 40 (d) None of these

(a) 60 (c) 38

(b) 40 (d) None of these

8. In a group of 70 people, 37 like coffee, 52 like tea and each person like atleast one of the two drinks. The number of persons liking both coffee and tea is (a) 16 (c) 19

(b) 13 (d) None of these

9. Which of the following is the empty set? (a) {x / x is a real number and x2 – 1 = 0} (b) {x / x is a real number and x2 + 1 = 0} (c) {x / x is a real number and x2 – 9 = 0} (d) {x / x is a real number and x2 = x + 2} 10. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. The number of persons liking tennis only and not cricket is (a) 21 (c) 15

(b) 25 (d) None of these

11. In a group of 1000 people, there are 750 people who can speak Hindi and 400 who can speak English. Then number of persons who can speak Hindi only is (a) 300 (c) 600

(b) 400 (d) None of these

12. If the sets A and B are given by A = {1, 2, 3, 4, }, B = {2, 4, 6, 8, 10} and the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, then (a) (A ∪ B) ′ = {5, 7, 9} (b) (A ∩ B) ′ = {1, 3, 5, 6, 7} (c) (A ∩ B) ′ = {1, 3, 5, 6, 7, 8, 9, 10} (d) None of these 13. If A = {1, 2, 3, 4}, B = {2, 3, 5, 6} and C = {3, 4, 6, 7}, then (a) A – (B ∩ C) = {1, 3, 4} (b) A – (B ∩ C) = {1, 2, 4} (c) A – (B ∪ C) = {2, 3} (d) A – (B ∪ C) ={1}. 14. If X and Y are two sets and X ′ denotes the complement of X, then X ∩ (X ∪ Y) ′ equals (a) X (c) φ

(b) Y (d) None of these

15. Let A = {2, 3, 4} and X = {0, 1, 2, 3, 4}, then which of the following statements is correct (a) {0} ∈ A ′ in X (c) {0} ⊂ A ′ w.r.t X

(b) φ ∈ A ′ w.r.t. X (d) 0 ⊂ A ′ w.r.t X.

16. If n (U) = 60, n (A) = 35, n (B) = 24 and n (A ∪ B) ′ = 10 then n (A ∩ B) is (a) 9 (c) 6

(b) 8 (d) None of these

17. If f : R → R, defined by f(x) = x2 + 1, then the values of f–1(17) and f–1(–3) respectively are

(b) {3, –3}, φ (d) {4, –4}, φ

18. Let ρ be the relation on the set R of all real numbers 1 defined by setting a ρ b iff | a − b | ≤ . Then, ρ is 2 (a) reflexive and symmetric but not transitive (b) symmetric and transitive but not reflexive (c) transitive but neither reflexive nor symmetric (d) none of these 19. In a statistical investigation of 1,003 families of Calcutta, it was found that 63 families had neither a radio nor a T.V, 794 families had a radio and 187 had a T.V. The number of families in that group having both a radio and a T.V is (a) 36 (c) 32

(b) 41 (d) None of these

20. In a city, three daily newspapers A, B, C are published. 42% of the people in that city read A, 51% read B and 68% read C. 30% read A and B; 28% read B and C; 36% read A and C; 8% do not read any of the three newspapers. The percentage of persons who read all the three papers is (a) 25% (c) 20%

(b) 18% (d) None of these

21. If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, B = {2, 4, ...., 18} and N is the universal set, then A′ ∪ ((A ∪ B) ∩ B′) is (a) A (c) B

(b) N (d) none of these

22. The composite mapping fog of the maps f : R → R, f(x) = sinx; g : R → R, g(x) = x2 is (a) sinx + x2

(b) (sinx)2 sin x (d) x2

(c) sinx2

23. If X and Y are two sets, then x ∩ (Y ∪ X)′ equals (a) X (c) φ

(b) Y (d) None of these

24. If X = {8n – 7 n – 1/n ∈ N} and Y = {49 (n – 1) /n ∈ N}, then (a) X ⊂ Y (c) X = Y

(b) Y ⊂ X (d) None of these

25. Suppose A1, A2,... A30 are thirty sets, each with five elements and B1, B2, ... , Bn are n sets each with three n

30

elements Let

A =  B i =1

i

j =1

j

=S

If each elements of S belongs to exactly ten of the A i ’s and exactly nine of the Bj’s then n = (a) 45 (c) 40

(b) 35 (d) None of these

26. Let A = {x : x is a multiple of 3} and B = {x : x is a multiple of 5}. Then A ∩ B is given by (a) {3, 6, 9...} (c) {15, 30, 45,...}

(b) {5, 10, 15, 20, ...} (d) None of these

1117

(a) φ, {4, –4} (c) φ, {3, –3}

Set Theory

7. In a committee 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. The number of persons speaking at least one of these two languages is

1118

27. If R be a relation < from A = {1, 2, 3, 4} to B = {1, 3, 5} i.e., (a, b) ∈ R iff a < b, then RoR–1 is

Objective Mathematics

(a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (b) {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} (c) {(3, 3), (3, 5), (5, 3), (5, 5)} (d) {(3, 3), (3, 4), (4, 5) 28. If Y ∪ {1, 2} = {1, 2, 3, 5, 9}, then (a) The smallest set of Y is {3, 5, 9} (b) The smallest set of Y is {2, 3, 5, 9} (c) The largest set of Y is {1, 2, 3, 4, 9} (d) The largest set of Y is {2, 3, 4, 9}. 29. Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of second set. The value of m and n are (a) 7, 6 (c) 5, 1

(b) 6, 3 (d) 8, 7

(a) surjective but not injective (b) injective but not surjective (c) bijective (d) neither surjective nor injective 38. Let n be a fixed positive integer. Let a relations R be defined on I (the set of all integers) as follows : a R b iff n/(a – b), that is iff a – b is divisible by n. Then, the relation R is (a) reflexive only (c) transitive only

39. Let R be a relation defined as a R b iff | a – b | > 0. Then , the relation R is (a) reflexive (c) transitive

32. If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6} then

(b) symmetric (d) None of these

40. Let R be a relation defined as a R b iff 1 + ab > 0. Then, the relation R is

30. If A = {1, 2, 3}, B = {a, b}, then A × B mapped on A to B is 41. (a) {(1, a), (2, b), (3, b)} (b) {(1, b), (2, a)} (c) {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)} (d) {(1, a), (2, a), (2, b), (3, b)} 42. 31. The solution of 3x2 – 12x = 0 when (a) x ∈ N is {4} (b) x ∈ I is {0, 4} (c) x ∈ S = {a + ib : b ≠ 0, a, b ∈ R} is φ (d) all of these

(b) symmetric only (d) an equivalence relation.

(a) Reflexive (c) transitive

(b) symmetric (d) None of these

Let R be a relation defined as a R b if | a | ≤ b. Then, relation R is (a) reflexive (c) transitive

(b) symmetric (d) None of these

N is the set of natural numbers. The relation R is defined on N × N as follows (a, b) R (c, d) ⇔ a + d = b + c. Then, R is (a) reflexive only (c) transitive only

(b) symmetric only (d) an equivalence relation

43. Let R be a relation defined on the set of natural numbers N as

(A × B) ∩ (B × C) is equal to

R = {(x, y) : x ∈ N, y ∈ N, 2x + y = 41}. Then

(a) {(3, 4)} (c) {(1, 4), (2, 3)}

(a) Domain of R = {1, 2, 3, ..., 19, 20} (b) Range of R = {39, 37, 35, 9, 7, 5, 3, 1} (c) R is reflexive (d) R is symmetric.

(b) {(1, 4), (3, 4)} (d) None of these

33. If A has 3 elements and B has 6 elements, then the minimum number of elements in the set A ∪ B is (a) 6 (c) φ

(b) 3 (d) None of these

34. If a N = {ax : x ∈ N}, then 3N ∩ 7N = (a) 3N (c) N

(b) 7N (d) 21 N

35. Let R be a relation defined as: aRb iff b is divisible by a where a and b are natural numbers. α R β iff α is perpendicular to β where α, β are straight lines in a plane, then the relation R is (a) reflexive (c) transitive

(b) symmetric (d) None of these

44. n/m means that n is a factor of m, then the relation ‘/’ is (a) reflexive and symmetric (b) transitive and reflexive (c) reflexive, transitive and symmetric (d) reflexive, transitive and not symmetric. 45. Let A = {x : x ∈ R, | x | < 1}, B = {x : x ∈ R, | x – 1 | ≥  1} and A ∪ B = R – D, then the set D is (a) {x : 1 < x ≤ 2} (c) {x : 1 ≤ x ≤ 2}

(b) {x : 1 ≤ x < 2} (d) None of these

46. Consider the set of all determinants of order 3 with entries 0 or 1 only. Let B be subset of A consisting of 36. The function f : N → N (N is set of natural numbers) all determinants with value 1. Let C be the subset of defined by f(n) = 2n + 3 is the set of all determinants with value – 1. Then (a) surjective (b) not surjective (a) C is empty (c) injective (d) none of these (b) B has as many elements as C 37. Let R be the set of real numbers, If f : R → R is defined by f(x) = ex, then f is

(c) A = B ∪ C (d) B has twice as many elements as C.

(a) reflexive (c) transitive

(b) symmetric (d) None of these

48. (i) Let R be the relation on the set R of all real numbers 1 . Then R is defined by setting a R b iff | a – b | ≤ 2 (a) reflexive and symmetric but not transitive, (b) symmetric and transitive but not reflexive, (c) transitive but neither reflexive nor symmetric, (d) None of these 49. If sets A and B are defined as A = {(x, y) : y = ex, x ∈ R} B = {(x, y) : y = x, x ∈ R} then (a) B ⊂ A (c) A ∩ B = φ

(b) A ⊂ B (d) A ∪ B = A

50. If n (U) = 700, n (A) = 200, n (B) = 300, n (A ∩ B) = 100, then n (A ′ ∩ B ′ ) = (a) 300 (c) 400

(b) 350 (d) None of these

51. Assume R and S (non-empty) relations in a set A. Which of the relations given below is false ? (a) If R and S are transitive, then R ∪ S is transitive, (b) If R and S are transitive, then R ∩ S is transitive (c) If R and S are symmetric, then R ∪ S is symmetric (d) If R and S are reflexive, then R ∩ S is reflexive 52. If R is a relation ‘ . 8 2

but

3 1 − 4 8

Thus

3 1 3 1 1 1 ρ and ρ but (~ ρ) 4 8 4 3 3 8

=

5 1 < 24 2

Hence (a) is the correct answer. 19. (b) Let R be the set of families having a radio and T, the set of families having a T. V., then

Then we have n (A) = 42, n (B) = 51, n (C) = 68; n (A ∩ B) = 30, n (B ∩ C) = 28, n (A ∩ C) = 36 n (A ∪ B ∪ C) = 100 – 8 = 92 Using n (A ∪ B ∪ C) = n (A) n (B) – n (A ∩ B) – n (B ∩ C) – n (A ∩ C) + n (A ∩ B ∩ C) Substituting the above values, we have 92 = 42 + 51 + 68 – 30 – 28 – 36 + n (A ∩ B ∩ C) ⇒ n (A ∩ B ∩ C) = 92 – 161 + 94 ⇒ n (A ∩ B ∩ C) = 92 – 67 = 25 Hence, 25% of the people read all the three papers. 21. (b) We have, (A ∪ B) ∩ B′ = A ((A ∪ B) ∩ B′) ∪ A′ = A ∪ A′ = N. Hence (b) is the correct answer. 22. (c) We have (fog) (x) = f(g(x)) = f(x2) = sinx2 Hence (c) is the correct answer. 23. (a) We have, X ∩ (Y ∪ X)′ = X ∩ (Y′ ∩ X′) ∩ Y′ = (X ∩ X′) ∩ Y′ φ ∩ Y′ = φ. 24. (a) We have, 8n – 7n – 1 = (7 + 1)n – 7n – 1 = (nC272 + nC373 + ... + nCn7n) = 49(nC2 + nC37 + ... + nCn7n – 2) for n ≥ 2 For n = 1, 8n – 7n – 1 = 0 Thus, 8n – 7 n – 1 is a multiple of 49 for n ≥ 2 and 0 for n = 1. Hence X consists of all positive integral multiple of 49 of the form 49 Kn. where Kn = nC2 + nC27 + ... + nCn7n – 2 together with zero. Also Y consists of all positive integral multiples of 49 including zero. Therefore, X ⊂ Y. Hence (a) is the correct answer. 25. (a) Given A’s are thirty sets with five elements each, so 30

∑ n (A ) i =1

i

= 5 × 30 = 150

...(1)

If the m distinct elements in S and each element of S belongs to exactly 10 of the Ai’s, we have n (R ∪ T) = The no of families having at least one of radio and T.V. = 1003 – 63 = 940 n (R) = 794 and n (T) = 187. Let x families had both a radio and a T. V. i.e., n (R ∩ T) = x. The no. of families who have only Radio = 794 – x and the no. of families who have only T.V.= 187 – x From Venn diagram, 794 – x + x – 187 – x = 940 ⇒ 981 – x = 940 or x = 981 – 940 = 41

30

∑ n (A ) i =1

i

= 10m

...(2)

∴ From (1) and (2), we get 10m = 150 ∴ m = 15 30

Similarly

∑ n (B ) j =1

j

...(3)

30

= 3n and

∑ n (B ) = 9m j =1

j

∴ 3n = 9m ⇒ n =

9m = 3m = 3 × 15 = 45 3

Hence, n = 45.

[from (3)]

∴  R–1 = { (3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} Hence R O R–1 = {(3, 3), (3, 5), (5, 3), (5, 5)} Hence (c) is the correct answer.

28. (a), (c) Since the set on the right hand side has 5 elements, ∴ smallest set of Y has three elements and largest set of Y has five elements, ∴ smallest set of Y is {3, 5, 9} and largest of Y is {1, 2, 3, 4, 9}. 29. (b) We are given 2m – 2n = 56 By trial m = 6 and n = 3 Hence (b) is the correct answer. 30. (c) Given that, A = {1, 2, 3} and B = {a, b} then, A × B = {1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}. 31. (d) We have, 3x2 – 12x = 0 ⇒ 3x (x – 4) = 0 ⇒ x = 0, 4. Now, if x ∈ N, then the solution set is {4}. Also, if x ∈ I, then the solution set is {0, 4}. Further, since there is no root of the form a + ib, where a, b are real and b ≠ 0, ∴ if x ∈ S = {a + ib : b ≠ 0, a, b ∈ R} then the solution set is φ. 32. (a) We have, A × B = {(1, 3), (1, 4), (2, 3) (2, 4), (3, 3), (3, 4)} and B × C = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)} ∴ (A × B) ∩ (B × C) = {(3, 4)}. 33. (a) Clearly the number of elements in A ∪ B will be minimum when A ⊂ B. Hence the minimum number of elements in A ∪ B is the same as the number of elements in B, that is, 6. 34. (d) We have 3N = {3x : x ∈ N} = {3, 6, 9, 12,...} and 7N = {7x : x ∈ N} = {7, 14, 21, 28, 35, 42...} Hence 3N ∩ 7N = {21, 42, 63,...} = {21x : x ∈ N} = 21N. 35. (b) R is not reflexive since no line can be perpendicular to itself. R is symmetric since if a line α is perpendicular to another line β, then β is also perpendicular to α. R is not transitive since if a line α is perpendicular to another line β and β is perpendicular to a line γ, then α is parallel to γ so that α is not perpendicular to γ.

x1 ≠ x2

1

2

⇒ ex1 ≠ ex2 ⇒ f(x1) ≠ f(x2) f is not surjective since ex > 0 for all x and so no negative real number can be the image of any real number. For example, there is no real x such that f(x) = –2 Hence (b) is the correct answer. 38. (d) R is reflexive since for any integer a we have a – a = 0 and 0 is divisible by n. Hence a R a ∨ a ∈ I. R is symmetric, let a R b. Then by definition of R, a – b = nk where k ∈ I. Hence b – a = (– k) n where – k ∈ I and so b R a. Thus we have shown that a R b ⇒ b R a. R is transitive, let a R b and b R c. Then by definition of R, we have a – b = k1n and b – c = k2n, where k1, k2 ∈ I. It then follows that a – c = (a – b) + (b – c) = k1n + k2n = (k1 + k2)n where k1 + k2 ∈ I. 39. (b) R is not reflexive since | a – a | = 0 and so | a – a | >/ 0. Thus a a for any real number a. R is symmetric since if | a – b | > 0, then | b – a | = | a – b | > 0. Thus a R b ⇒ b R a R is not transitive. For example, consider the numbers 3, 7, 3. Then we have 3 R 7 since | 3 – 7 | = 4 > 0 and 7 R 3 since | 7 – 3 | = 4 > 0. But 3 3 since | 3 – 3 | = 0 so that | 3 – 3 | >/ 0. 40. (a), (b)  Here relation R is reflexive since 1 + a ⋅ a > 0 ∨ real numbers a. It is symmetric since 1 + ab > 0 ⇒ 1 + ba > 0. However R is not transitive : consider three 1 and – 2. We have real numbers 2, – 6  1 2  1 4 1 + 2 ×  −  = > 0 and 1 +  −  (– 2) = > 0. 3  6 3  6  1  1 Hence 2 R  −  and  −  R (– 2).  6  6 But 2 R – 3 since 1 + 2 (– 2) = – 3 >/ 0. 41. (c) R is not reflexive, if – a is any negative real number, then | – a | > – a so that – a R – a. R is not symmetric consider the real numbers a = – 2 and b = 3. Then a R b since |– 2 | < 3. But b R a since | 3 | > – 2.

1123

Set Theory

26. (c) Since x ∈ A ∩ B ⇔ x ∈ A and x ∈ B ⇔ x is 36. (b), (c)  f is injective, that is, one-one since a multiple of 3 and x is a multiple of 5 ⇔ x is a n, m ∈ N, n ≠ m ⇒ 2n + 3 ≠ 2m + 3 multiple of 15. ⇒ f(n) ≠ f(m) Hence A ∩ B = {x | x is a multiple of 15} = {15, f is not surjective, that is, f is not onto since for 30, 45, ...} example 1 cannot be the image of any natural Hence (c) is the correct answer. number. Hence (b), (c) are the correct answers. 27. (c) We have R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} 37. (b) f is injective (i.e., one -one) since x  ⋅ x ∈ R and

1124

Objective Mathematics

R is transitive: let a, b, c be three real numbers 47. (b) Since R and R′ are not disjoint, there is at least one such that ordered pair, say, (a, b) in R ∩ R′. | a | ≤ b and | b | ≤ c. But (a, b) ∈ R′ ⇒ (a, b) ∈ R and (a, b) ∈ R′ But | a | ≤ b ⇒ b ≥ 0, and so | b | ≤ c ⇒ b ≤ c. Since R and R′ are symmetric relations, we get It follows | a | ≤ c. (b, a) ∈ R and (b, a) ∈ R′ Thus a R b and b R c ⇒ a R c. and consequently (b, a) ∈ R ∩ R′. 42. (d) We have, (a, b) R (a, b) for all (a, b) ∈ N × N Similarly, if any other ordered pair (c, d) ∈ R ∩ R′, since a + b = b + a. then we must also have (d, c) ∈ R ∩ R′. Hence R is reflexive. Hence R ∩ R′ is symmetric. R is symmetric : we have (a, b) R (c, d) ⇒ a + d = b + c ⇒ d + a = c + b 1 for all a ∈ 48. (a) R is reflexive since | a – a | = 0 < ⇒ c + b = d + a ⇒ (c, d) R (a, b) 2 R. R is transitive : let 1 (a, b) R (c, d) and (c, d) R (e, f ). R is symmetric since | a – b | < 2 Then by definition of R, we have a + d = b + c and c + f = d + e, ⇒ | b – a | < 1 ⇒ a + d + c + f = b + c + d + e or a + f = b + 2 e. R is not transitive : we take three numbers Hence (a, b) R (e, f ). 3 1 1 , , , then Thus (a, b) R (c, d) and (c, d) R (e, f ) ⇒ (a, b) 4 3 8 R (e, f ). 43. (a), (b)  The relation R can be written as R = {(1, 39), (2, 37), (3, 35), ...(10, 21), (11, 19), ... (19, 3), (20, 1)} ∴ Domain of R = {1, 2, 3, ..., 19, 20} Range of R = {39, 37, 35, 9, 7, 5, 3, 1} R is not reflexive since (x, x) ∉ R ∨ x ∈ N. For example, (1, 1) ∉ R R is not symmetric since (1, 39) ∈ R but (39, 1) ∉ R.



3 1 5 1 1 1 5 1 − = < − < and and 4 3 12 2 3 8 24 2

3 1 5 1 − = > 4 8 8 2 3 1 R Thus and 1 R 1 but 3 4 3 3 8 4 R But

1. 8

x 2 x3 + + .... ∴ ex > x ∨ x 2! 3! ∈ R so that the two curves given by y = ex and y = x do not intersect in any point Hence, there is no common point, so that A ∩ B = φ.

49. (c) Since y = ex = 1 + x +

44. (b) ‘/’ is reflexive since every natural number is a factor of itself, that is n/n for n ∈ N. ‘/’ is transitive : if n is a factor of m and m is a factor of p, then n is surely a factor of p. Thus ‘n/m’ and ‘m/p’ ⇒ ‘n/p’. However ‘/’ is not symmetric : for example, 50. (a) n (A′ ∩ B′ ) = n (A ∪ B)′ = n (U) – n (A ∪ B) 2 is factor of 4 but 4 is not a factor of 2. = n (U) – n (A) – n (B) + n (A ∩ B) 45. (b) We have = 700 – 200 – 300 + 100 = 300. A = {x : x ∈ R, – 1 < x < 1} 51. (a) Let A = {a, b, c}, R = {(a, b), (b, c), (a, c)} and B = {x : x ∈ R, x – 1 ≤ – 1 or x – 1 ≥ 1} and S = {(b, c), (c, a), (b, a)}. = {x : x ∈ R, x ≤ 0 or x ≥ 2} ∴ A ∪ B = R – D, where D = {x : x ∈ R, 1 ≤ x < 2}.

Then R and S are both transitive but

R ∪ S = {(a, b), (b, c), (a, c), (c, a), (b, a)}

is not transitive since (a, b) ∈ R ∪ S and (b, a) ∈ 46. (b) We know that the interchange of two adjacent rows R ∪ S but (a, a) ∉ R ∪ S. (or columns) changes the value of a determinant only in sign and not in magnitude. Hence corresponding 52. (c) Here R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} to every element ∆ of B there is an element ∆ ′ in ∴ R– 1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} C obtained by interchanging two adjacent rows (or Hence R 0 R– 1 = {(3, 3), (3, 5), (5, 3), (5, 5)}. columns) in ∆. It follows that n (B) ≤ n (C). that is, the number of elements in B is less than or equal to the number of elements in C. Similarly n (C) ≤ n (B) Hence n (B) = n (C), that is. B has as many elements as C.

53. (c) A ∩ B = φ [ y=

1 1 , y = – x meet when – x = x x ⇒  x2 = – 1 which does not give any real value of x]

n

= 7 + C1 7

n – 1

n

n

n

+ ...... + Cn – 1 7 + Cn –7 n–1 = nC2 72 + nC3 73 + .... + nCn 7n n

n

+ C2 7

n – 2

n

n

( C0 = Cn, C1 = Cn – 1 etc.) ∴ For For ∴

n

n

n

n – 2

= 49 [ C2 + C3 (7) + .... + Cn 7 ] 8n – 7n – 1 is a multiple of 49 for n ≥ 2. n = 1, 8n – 7n – 1 = 8 – 7 – 1 = 0 n = 2, 8n – 7n – 1 = 64 – 14 – 1 = 49 8n – 7n – 1 is multiple of 49 for all n ∈ N.

55. (d) Not a well defined collection [ intelligency is not defined for students in a class] 56. (a) Number of non-empty subsets of {1, 2, 3, 4} 57. (c) n (P) = 25%, n (C) = 15%, n (P ∩ Cc ) = 65%, n (P ∩ C) = 2000 c

Since n (P ∩ C ) = 65% c

c 

∴ n (P ∪ C)c = 65%

∴ n (P ∪ C) = 35%

Now n (P ∪ C) = n (P) + n (C) – n (P ∩ C) ∴ 35 = 25 + 15 – n (P ∩ C) ∴ n (P ∩ C) = 40 – 35 = 5 Thus n (P ∩ C) = 5% But n (P ∩ C) = 2000 ∴ 5% of the total = 2000 ∴ total no. of families 2000 × 100 = 40000 ∴ n (P ∪ C) = 35%, = 5 total no. of families = 40,000 and n (P ∩ C) = 5%. 58. (a), (b)  | a – a | = 0 < 1 so a R a ∨ a ∈ R′. ∴ R is reflexive a R b ⇒ | a – b | ≤ 1 ⇒ | b – a | ≤ 1 ⇒ b R a. ∴ R is symmetric 2 R 3/2, 3/2 R 2 but 2 ≠ 3/2 so R is not antisymmetric. Finally 1 R2, 2 R 3 but 1 R 3 as

63. (c) Let the universal set be U = {x1, x2, x3 ....xn} We know every set is a subset of itself. Therefore, inclusion of a subset is reflexive Now the elements of the set {x1} are included in the set {x1, x2} but converse is not true i.e., {x1} ⊂ {x1, x2} but {x1, x2} ⊄ {x1} Hence, the inclusion of a subset is not symmetric. Thus, the inclusion of a subset is not an equivalence relation. 64. (a) R × (Pc ∪ Qc )c = R × [(P c)c ∩ (Q c)c]

= R × (P ∩ Q) = (R × P) ∩ (R × Q)

65. (c) Clearly A = φ = {}

= 24 – 1 = 15.

There is no value of x which satisfies both the above equations. Thus the set A contains no element. ∴ A = ϕ.

| 1 – 3 | = 2 > 1.

59. (b) As A has p elements and B has q elements so, A × B has pq elements. 60. (d) A ∪ (A ∩ B) = A [ A ∩ B ⊆ A] 61. (c) n (A ∪ B) = n (A) + n (B) – n (A ∩ B) = 3 + 6 – n (A ∩ B) Since maximum number of elts. in A ∩ B = 3, ∴ minimum no. of elts in A ∪ B = 9 – 3 = 6. 62. (a) We have x2 = 16 ⇒ x = ± 4. Also, 2x = 6 ⇒ x = 3.

66. (b) B ∪ C = {a, b, c, d, e} A ∩ (B ∪ C) = {a, b, c} ∩ {a, b, c, d, e} = {a, b, c} = A. 67. (c) B ∩ C = {4} ∴ A ∪ (B ∩ C) = {1, 2, 3, 4}. 68. (b) Since R ∩ R′ are not disjoint, there is at least one ordered pair, say, (a, b) in R ∩ R′. But (a, b) ∈ R ∩ R′ ⇒ (a, b) ∈ R and (a, b) ∈ R′ since R and R′ are symmetric relations, we get

(b, a) ∈ R and (b, a) ∈ R′

and consequently (b, a) ∈ R ∩ R′ similarly if any other ordered pair (c, d) ∈ R ∩ R′, then we must also have, (d, c) ∈ R ∩ R′ Hence R ∩ R′ is symmetric Hence (b) is the correct answer. 69. (b) Since the two finits sets have m and n elements, so number of subsets of these sets well be 2 m and 2n respectively. According to the question 3m – 2n = 56 putting m = 6, n = 3 we get 26 – 23 = 56 or 64 – 8 = 56. 70. (b) The relation ‘less than’ is only transitive because ⇒ ∴ ⇒

x x x x

< y, y < z < z, x, y, z ∈ N. R y, y R z R z.

71. (d) Since A   ⊄ B, ∃ x ∈ A such that x ∉ B Then x ∈ B′. ∴ A ∩ B′ ≠ φ. 72. (d) Given that S = {1, 2, 3, 4, 5} and A = S × S A relation R on A is defined as follows: “(a, b) R (c, d)” if and only if ad = cb (i) R is reflexive, since ab = ba ⇒ ba = ab, therefore,

1125

n

Set Theory

54. (a) Since 8n – 7n – 1= (7 + 1)n – 7 n – 1

Objective Mathematics

73. (c) We have, bN = the set +ve integral multiples of b cN = the set of +ve integral multiples of c

77. (a) Clearly, from the Venn-Euler’s diagram U

(A U B)'

(A' B ) U

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(a, b) R (b, a) ∨ a, b ∈ S. (ii) R is symmetric, since (a, b) R (c, d) ⇒ ad = cb ⇒ cd = da ⇒ (c, d) R (a, b) ∨ a, b ∈ S. (iii)  R is transitive (a, b) R (c, d) and (c, d) R (e, f ) ⇒ ad = cb and cf = ed ⇒ adcf = cb ed ⇒ cd (af ) = cd (be) ⇒ af = eb ⇒ (a, b) R (e, f ) ∨ a, b, c, d, e, f ∈ S.

B

A

(A ∪ B)´ ∪ (A´ ∩ B) = A´ 78. (c) Given, n(C) = 63, n(A) = 76, and n(C ∩ A) = x

∴ bN ∩ cN = the set of +ve integral multiples of bc, = bcN ∴ d = bc.

We know that, n(C ∪ A) = n(C) + n(A) – n(C ∩ A) ⇒ x = 139 – 100 = 39 ⇒ 100 = 63 + 76 – x and dn(C ∩ A) ≤ n(C) ⇒ x ≤ 63 ∴ 39 ≤ x ≤ 63

74. (b) (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B).

79. (d) Since, (1, 2) ∈ S but (2, 1) ∈ S ∴ S is not symmetric. Hence, S is not equivalence relation. Given, T = {(x, y) : x – y ∈ I} Now, x – x = 0 ∈ I, it is reflexive relation x – y ∈ I ⇒ y – x ∈ I, it is symmetric relation. Let x – y = I1 and y – z = I2 Now, x – z = (x – y) + (y – z) = I1 + I2 ∈ I ∴T is also transitive.

75. (a) n (C) = 224, n (H) = 240, n (B) = 336

n (H ∩ B) = 64, n (B ∩ C) = 80 n (H ∩ C) = 0, n (C ∩ H ∩ B) = 24



n (Cc ∩ Hc ∩ Bc) = n [(C ∪ H ∪ B)c]



= n (U) – n (C ∪ H ∪ B)



= 800 – [n (C) + n (H) + n (B) – n (H ∩ C)



– n (H ∩ B) – n (C ∩ B) + n (C ∩ H ∩ B)]



= 800 – [224 + 240 + 336 – 64 – 80 – 40 + 24]



= 800 – [824 – 184] = 984 – 824 = 160.

80. (c) A = {x/x is a real no. x2 = 16 and 2x = 6} =

{x/x is a real no. x = ± 4 and x = 3} = φ

81. (d) According to given pattern 26 51 76  S25 =  , , , ... upto 25 terms   25 25 25  26 Here, a = , n = 25, d = 1 25

76. (c) n (M) = 55, n (P) = 67, n (M ∪ P) = 100 Now, n (M ∪ P) = n (M) + n (P) – (M ∩ P) ⇒ 100 = 55 + 67 – n (M ∩ P) n (M ∩ P) = 122 – 100 = 22

∴ S25 = 25  52 + 24  = 326  2  25 

EXERCISEs FOR SELF-PRACTICE 1. Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A to B is: (a) 144 (c) 24

(b) 12 (d) 64

3. If sets A and B are defined as: A = {(x, y), y = ex, x ∈ R} B = {(x, y), y = x, x ∈ R} then: (a) B ⊂ A (c) A ∩ B = φ

(b) A ⊂ B (d) A ∪ B = A

2. If A, B, C are 3 non-zero sets, then (A ∩ B) ∩ (B ∩ C) 4. If A, B and C are any three sets, then A – (B ∪ C) is ∩ (C ∩ A) is equal to: equal to (a) A ∩ B ∩ C (b) A ∪ B ∪ C (a) (A – B) ∩ C (b) (A – B) ∪ C (c) φ (d) None of these (c) (A – B) ∩ (A – C) (d) (A – B) ∪ (A – C)

(a) {3} (c) {1, 2, 5, 6}

(b) {1, 2, 3, 4} (d) {1, 2, 3, 4, 5, 6}

6. If X and Y are two sets, then X ∩ (X ∪ Y) equals (a) X (c) φ

(b) Y (d) None of these

(a) {3, 5} ∈ {1, 3, 5} (c) 3 ∈ {1, 3, 5}

f = {(a, a), (b, b) (c, c)}, then f is

(b) 2n (d) 2n + 1

(b) A ∪ B (d) A′ ∪ B′

15. If A, B and C are any three sets, then A – (B ∩ C) is equal to (b) (A – B) ∪ C (d) (A – B) ∪ (A – C)

16. If A, B C be three sets such that

(a) a subset of A × B (b) a set equivalent of A × B (c) a set equal to A × B (d) a universal set of A × B

A ∪ B = A ∪ C and A ∩ B = A ∩ C, then (a) A = B = C (c) B = C

10. If A is any set, then

17. If Q = {x : x =

(b) A ∪ A ′ = φ (d) None of these

2 ∈ Q 3 (c) 0 ∈ Q

11. If A ⊆ B, then B ∪ A is equal to (a) B ∩ A (c) B

(a) A ∩ B (c) A′ ∩ B′

(a) (A – B) ∩ C (c) (A – B) ∩ (A – C)

9. A relation from set A to B is

(a) A ∩ A ′ = X (c) A ∪ A ′ = X

(b) {3} ∈ {1, 3, 5} (d) 3 ⊆ {1, 3, 5}

14. If A, B be any two sets then (A ∪ B)′ is equal to

(b) Symmetric (d) None of these

8. If a set A has n elements then the number of elements in the power set of A is (a) n2 (c) 22n

(a) (A ∩ B) × (A ∩ C) (b) (A ∪ B) (A ∪ C) (c) (A × B) ∩ (A × C) (d) (A × B) ∪ (A × C) 13. Which of the following statements is true ?

7. Let f : R → R be a relation given by (a) Reflexive (c) Transitive

(a)

(b) A (d) None of these

(b) A = C (d) A = B

1 , where y ∈ N}, then y (b) 2 ∈ Q (d) 1 ∈ Q

Answers

1. (c) 11. (c)

2. (a) 12. (c)

3. (c) 13. (c)

4. (c) 14. (c)

5. (b) 15. (d)

6. (a) 16. (d)

7. (b), (c) 8. (b) 17. (a)

9. (a)

10. (c)

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12. If A, B and C an any three sets, then A × (B ∩ C) is equal to

Set Theory

5. Given the set A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, then A ∪ (B ∩ C) is

33

Numerical Methods

CHAPTER

sUMMARY OF cONCEPTS NUMERICAL METHODS OF SOLVING AN EQUATION OF THE TYPE f(x) = 0

This procedure is repeated till the root is obtained to the desired degree of accuracy.

Note:  This method may give a false root or may not converge if either a and b are not sufficiently close to Step 1: Find an interval [a, b] such that f (a) and f (b) are each other or f (x) is discontinuous on [a, b]. of opposite signs i.e., f (a) ⋅ f (b) < 0.

Successive Bisection Method

First approximation is given by x1 = 0, it is the required root.

Newton-Raphson Method a+b . If f (x1) = 2 Step 1: Find an interval [a, b] such that f (a) and f (b) are

Step 2: If f (x1) ≠ 0, then consider the interval (a, x1) or

(x1, b) according as f ( a) f (x1)