Complete Companion for JEE Main 2020 Mathematics 9789353435097

Table of contents :
Cover
Half Title
Title
Copyright
Contents
Preface
Mathematics Trend Analysis (2011 To 2019)
Chapter 1 Set Theory
SET
REPRESENTATION OF A SET
TYPES OF SETS
OPERATIONS ON SETS
ALGEBRA OF SETS
CARTESIAN PRODUCT OF TWO SETS
RELATIONS
TYPES OF RELATIONS ON A SET
EQUIVALENCE RELATION
CONGRUENCE MODULO M
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 2 Complex Numbers
IMAGINARY NUMBERS
INTEGRAL POWERS OF i
COMPLEX NUMBERS
CONJUGATE OF A COMPLEX NUMBER
MODULUS OF A COMPLEX NUMBER
SQUARE ROOTS OF A COMPLEX NUMBER
ARGAND PLANE AND GEOMETRICAL REPRESENTATION OF COMPLEX NUMBERS
POLAR FORM OF A COMPLEX NUMBER
PARTICULAR CASES OF POLAR FORM
EULERIAN REPRESENTATION OF A COMPLEX NUMBER
LOGARITHM OF A COMPLEX NUMBER
VECTORIAL REPRESENTATION OF A COMPLEX NUMBER
ROOTS OF A COMPLEX NUMBER
GEOMETRY OF COMPLEX NUMBERS
PRACTICE EXERCISES
Chapter 3 Quadratic Equations and Expressions
QUADRATIC EQUATION
COMMON ROOTS
SYMMETRIC FUNCTION OF THE ROOTS
GRAPH OF A QUADRATIC EXPRESSION
GREATEST AND LEAST VALUES OF AQUADRATIC EXPRESSION
NATURE OF ROOTS OF A QUADRATICE QUATION WITH RESPECT TO ONE OR TWO REAL NUMBERS
RELATION BETWEEN ROOTS AND COEFFICIENTS OF A POLYNOMIAL EQUATION
FORMATION OF A POLYNOMIAL EQUATION FROM GIVEN ROOTS
SIGN OF A POLYNOMIAL EXPRESSION
RATIONAL ALGEBRAIC EXPRESSION
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 4 Permutations and Combinations
FACTORIAL NOTATION
FUNDAMENTAL PRINCIPLES OF COUNTING
PERMUTATION
COMBINATION
KEY RESULTS ON COMBINATION
DERANGEMENT
EXPONENT OF PRIME p IN n!
NUMBER OF DIVISORS
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 5 Mathematical Induction
MATHEMATICAL INDUCTION
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 6 Binomial Theorem
BINOMIAL EXPRESSION
BINOMIAL THEOREM
SPECIAL CASES
PASCAL’S TRIANGLE
MIDDLE TERM IN THE BINOMIAL EXPANSION
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 7 Sequence and Series
SEQUENCE
SERIES
PROGRESSIONS
ARITHMETIC PROGRESSION (A.P.)
SUM OF n TERMS OF AN A.P.
PROPERTIES OF A.P.
ARITHMETIC MEAN (A.M.)
GEOMETRIC PROGRESSION (G.P.)
GEOMETRIC MEAN (G.M.)
SOME SPECIAL SEQUENCES
ARITHMETICO-GEOMETRIC PROGRESSION (A.G.P.)
METHOD FOR FINDING SUM OF A.G. SERIES
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 8 Limits
LIMIT OF A FUNCTION
INDETERMINATE FORMS
ALGEBRA OF LIMITS
EVALUATION OF LIMITS
ALGEBRAIC LIMITS
LIMIT OF AN ALGEBRAIC FUNCTIONWHEN X → ∞
TRIGONOMETRIC LIMITS
EXPONENTIAL AND LOGARITHMIC LIMITS
EVALUATION OF LIMITS USING L’HOSPITAL’S RULE
PRACTICE EXERCISES
Chapter 9 Differential Equations
DIFFERENTIAL EQUATION
LINEAR AND NON-LINEAR DIFFERENTIAL EQUATIONS
INITIAL VALUE PROBLEMS
HOMOGENEOUS DIFFERENTIAL EQUATIONS
SOLUTION BY INSPECTION
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 10 Coordinates and Straight Lines
DISTANCE FORMULA
SECTION FORMULAE
AREA OF A TRIANGLE
CONDITION FOR COLLINEARITY OF THREE POINTS
STAIR METHOD FOR FINDING THE AREA
AREA OF A QUADRILATERAL
AREA OF A POLYGON
STAIR METHOD
LOCUS
TRANSLATION OF AXES
ROTATION OF AXES
REFLECTION (IMAGE) OF A POINT
GENERAL EQUATION OF A STRAIGHT LINE
SLOPE OF A LINE
INTERCEPT OF A LINE ON THE AXES
EQUATION OF A STRAIGHT LINE IN VARIOUS FORMS
REDUCTION OF THE GENERAL EQUATION TO DIFFERENT STANDARD FORMS
ANGLE BETWEEN TWO INTERSECTING LINES
CONDITION FOR TWO LINES TO BE COINCIDENT, PARALLEL, PERPENDICULAR OR INTERSECTING
EQUATION OF A LINE PARALLEL TO A GIVEN LINE
EQUATION OF A LINE PERPENDICULAR TO A GIVEN LINE
POINT OF INTERSECTION OF TWO GIVEN LINES
CONCURRENT LINES
POSITION OF TWO POINTS RELATIVE TO ALINE
LENGTH OF PERPENDICULAR FROM A POINTON A LINE
DISTANCE BETWEEN TWO PARALLEL LINES
EQUATIONS OF STRAIGHT LINES PASSING THROUGH A GIVEN POINT AND MAKING A GIVEN ANGLE WITH A GIVEN LINE
REFLECTION ON THE SURFACE
IMAGE OF A POINT WITH RESPECT TO A LINE
EQUATIONS OF THE BISECTORS OF THE ANGLES BETWEEN TWO LINES
EQUATIONS OF LINES PASSING THROUGH THE POINT OF INTERSECTION OF TWO GIVEN LINES
STANDARD POINTS OF A TRIANGLE
ORTHOCENTRE
COORDINATES OF NINE POINT CIRCLE
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 11 Circles
CIRCLE
STANDARD EQUATION OF A CIRCLE
GENERAL EQUATION OF A CIRCLE
CONDITIONS FOR AN EQUATION TO REPRESENT A CIRCLE
EQUATION OF A CIRCLE IN SOME SPECIALCASES
EQUATION OF A CIRCLE IN DIAMETER FORM
INTERCEPTS MADE BY A CIRCLEON THE AXES
PARAMETRIC EQUATIONS OF A CIRCLE
POSITION OF A POINT WITH RESPECTTO A CIRCLE
CIRCLE THROUGH THREE POINTS
INTERSECTION OF A LINE AND A CIRCLE
LENGTH OF INTERCEPT MADE BY A CIRCLEON A LINE
THE LEAST AND GREATEST DISTANCE OF APOINT FROM A CIRCLE
CONTACT OF TWO CIRCLES
TANGENT TO A CIRCLE AT A GIVEN POINT
EQUATION OF THE TANGENT IN SLOPE FORM
CONDITION OF TANGENCY
TANGENTS FROM A POINT OUT SIDE THE CIRCLE
LENGTH OF THE TANGENT FROM A POINT TOA CIRCLE
NORMAL TO THE CIRCLE AT A GIVEN PO
PAIR OF TANGENTS
COMMON TANGENTS TO TWO CIRCLES
POWER OF A POINT WITH RESPECT TO A CIRCLE
DIRECTOR CIRCLE
EQUATION OF CHORD OF CONTACT
EQUATION OF CHORD IF ITS MID POINTIS KNOWN
COMMON CHORD OF TWO CIRCLES
DIAMETER OF A CIRCLE
ANGLE OF INTERSECTION OF TWO CIRCLES
ORTHOGONAL INTERSECTION OF TWO CIRCLES
FAMILY OF CIRCLES
IMAGE OF THE CIRCLE BY THE LINE MIRROR
PRACTICE EXERCISES
Chapter 12 Conic Sections (Parabola, Ellipse and Hyperbola)
CONIC SECTION
IMPORTANT TERMS
SECTION OF A RIGHT CIRCULAR CONE BYDIFFERENT PLANES
EQUATION OF CONIC
GENERAL EQUATION
CENTRE OF CONIC
PARABOLA
SOME TERMS RELATED TO PARABOLA
INTERSECTION OF A LINE AND A PARABOLA
EQUATION OF A CHORD
POINT OF INTERSECTION OF TANGENTS
POSITION OF A POINT WITH RESPECT TO APARABOLA
NUMBER OF TANGENTS DRAWN FROM APOINT TO A PARABOLA
EQUATION OF THE PAIR OF TANGENTS
EQUATIONS OF NORMAL IN DIFFERENT FORMS
POINT OF INTERSECTION OF NORMALS
CO-NORMAL POINTS
CHORD OF CONTACT
CHORD WITH A GIVEN MID POINT
ELLIPSE
POSITION OF A POINT WITH RESPECT TO ANELLIPSE
EQUATION OF NORMAL IN DIFFERENT FORMS
EQUATION OF THE PAIR OF TANGENTS
CHORD WITH A GIVEN MID POINT
OPTICAL PROPERTY OF PARABOLA
EQUATION OF A HYPERBOLA IN STANDARD FORM
SOME TERMS AND PROPERTIES RELATED TO A HYPERBOLA
CONJUGATE HYPERBOLA
POSITION OF A POINT WITH RESPECT TO A HYPERBOLA
EQUATION OF THE PAIR OF TANGENTS
CHORD WITH A GIVEN MID POINT
CHORD OF CONTACT
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 13 Vector Algebra
SCALARS AND VECTORS
REPRESENTATION OF VECTORS
TYPES OF VECTORS
EQUAL VECTORS
FIXED VECTORS
FREE VECTORS
ANGLE BETWEEN TWO VECTORS
ADDITION (SUM OR RESULTANT) OF TWO VECTORS
POSITION VECTOR
COMPONENT OF A VECTOR
LINEAR COMBINATION
LINEARLY DEPENDENT AND INDEPENDENT SYSTEM OF VECTORS
COLLINEARITY OF THREE POINTS
COPLANARITY OF FOUR POINTS
SOME RESULTS ON LINEARLY DEPENDENTAND INDEPENDENT VECTORS
PRODUCT OF TWO VECTORS
SCALAR PRODUCT OF TWO VECTORS
SOME USEFUL IDENTITIES
WORK DONE BY A FORCE
VECTOR PRODUCT OF TWO VECTORS
MOMENT OF A FORCE ABOUT A POINT
SCALAR TRIPLE PRODUCT
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 14 Measures of Central Tendency and Dispersion
MEASURES OF CENTRAL TENDENCY
ARITHMETIC MEAN
GEOMETRIC MEAN
HARMONIC MEAN
MEDIAN
QUARTILES, DECILES AND PERCENTILES
MODE
SYMMETRIC DISTRIBUTION
PRACTICE EXERCISES
Chapter 15 Trigonometric Ratios and ldentities
ANGLE
MEASUREMENT OF ANGLES
RELATION BETWEEN DIFFERENT SYSTEMS OFMEASUREMENT OF ANGLES
RELATION BETWEEN SIDES AND INTERIORANGLES OF A REGULAR POLYGON
FUNDAMENTAL TRIGONOMETRICIDENTITIES
SIGNS OF TRIGONOMETRIC RATIOS INDIFFERENT QUADRANTS
INCREASE AND DECREASE OF TRIGONOMETRIC FUNCTIONS
DOMAIN AND RANGE OF TRIGONOMETRICRATIOS
TRIGONOMETRIC RATIOS OF STANDARDANGLES
TRIGONOMETRIC RATIOS FOR SOME SPECIAL ANGLES
TRIGONOMETRIC RATIOS OF ALLIED ANGLES
TABLE FOR TRIGONOMETRIC RATIOS OF ALLIED ANGLES
TRIGONOMETRIC RATIOS IN TERMS OF EACH OTHER
ADDITION AND SUBTRACTION FORMULAE
TRANSFORMATION FORMULAE
PRODUCT INTO SUM OR DIFFEREN
SUM AND DIFFERENCE INTO PRODUCT
TRIGONOMETRIC RATIOS OF MULTIPLE ANGLES
TRIGONOMETRIC RATIOS OFSUBMULTIPLE ANGLES
GREATEST AND LEAST VALUES OF THE EXPRESSION
CONDITIONAL IDENTITIES
GRAPHS OF TRIGONOMETRIC FUNCTIONS
PRACTICE EXERCISES
Chapter 16 Trigonometric Equations
TRIGONOMETRIC EQUATION
SOLUTION OR ROOT OF A TRIGONOMETRIC EQUATION
METHOD FOR FINDING PRINCIPAL VALUE (SOLUTION)
SOLUTION OF AN EQUATION OF THE FORM
SOLUTIONS OF BASIC TRIGONOMETRIC INEQUALITIES
NCERT EXEMPLARS
PRACTICE EXERCISES
Chapter 17 Heights and Distances
SOME TERMINOLOGY RELATED TO HEIGHT AND DISTANCE
NORTH EAST
BEARINGS OF A POINT
SOME PROPERTIES RELATED TO TRIANGLE
SOME PROPERTIES RELATED TO CIRCLE
SOME IMPORTANT RESULTS
PRACTICE EXERCISES
Chapter 18 Mathematical Reasoning
MATHEMATICAL REASONING
STATEMENT
THE CONNECTING WORD ‘OR’
QUANTIFIERS
IMPLICATIONS
NCERT EXEMPLARS
PRACTICE EXERCISES

Citation preview

F IF T H ED IT IO N

Complete Companion for

FIFTH EDITION

JEE Main 2020

Complete Companion for

MATHEMATICS Complete Companion for



Dinesh Khattar I Rohan Sinha

JEE Main 2020

The Complete Resource Book for JEE Main series has been designed to be an independent resource to enable faster and effective learning. This series includes three separate books on Physics, Chemistry and Mathematics where the core objective of each book is to provide ‘effective preparation via modular and graded content’. Developed by highly experienced and qualified faculties, these books would act as trusted content for aspirants who are aiming to clear the JEE (Joint Entrance Examinations) and other key engineering entrance examinations.

 Complete coverage of JEE Main curriculum with emphasis on important concepts  Provides varieties of solved example after each concept for better understanding  Key points highlighted within text along with features like – ‘Concepts at a glance’,  Includes ample number of solved examples along with NCERT Exemplar  Updated Practice Exercises as per recent patterns of JEE Main for result-oriented preparation

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‘info boxes’, and ‘concept notes’ to aid in quick last-minute revision

MATHEMATICS 

Y L L D U F VISE RE

HIGHLIGHTS  Coverage of key topics along with solved examples

MATHEMATICS

H I G H LI G H TS

JEE Main 2020

 Includes practice problems with complete solutions  Chapter-wise Previous 18 years’ AIEEE/JEE Main questions  Fully-solved JEE Main 2019 questions (April & Jan)  Includes online tests based on JEE Main pattern

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ISBN: 9789353435097

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Dinesh Khattar Rohan Sinha

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About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide. Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world. We believe learning opens up opportunities, creates fulfilling careers and hence better lives. We hence collaborate with the best of minds to deliver you classleading products, spread across the Higher Education and K12 spectrum. Superior learning experience and improved outcomes are at the heart of everything we do. This product is the result of one such effort. Your feedback plays a critical role in the evolution of our products and you can contact us - [email protected]. We look forward to it.

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FIFTH EDITION

Complete Companion for

JEE Main 2020

MATHEMATICS 

Dinesh Khattar Rohan Sinha

This page is intentionally left blank.

FIFTH EDITION

Complete Companion for

JEE Main 2020

MATHEMATICS 

Dinesh Khattar Rohan Sinha

Copyright © 2019 Pearson India Education Services Pvt. Ltd Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128. No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time.

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Contents Prefacexi Mathematics Trend Analysis (2011 to 2019)

Chapter 1  Set Theory

xiii

1.1

Set 1.1 Representation of a Set 1.1 Types of Sets 1.1 Operations on Sets 1.2 Algebra of Sets 1.4 Cartesian Product of Two Sets 1.4 Relations 1.9 Types of Relations on a Set 1.10 Equivalence Relation 1.10 Congruence Modulo m1.10 NCERT Exemplars 1.14 Practice Exercises 1.17

Chapter 2  Complex Numbers

2.1



Imaginary Numbers

2.1



Integral Powers of i

2.1



Complex Numbers

2.2



Conjugate of a Complex Number

2.3



Modulus of a Complex Number

2.4



Square Roots of a Complex Number

2.6

Chapter 3 Quadratic Equations and Expressions3.1 Quadratic Equation Common Roots Symmetric Function of the Roots Graph of a Quadratic Expression Greatest and Least Values of a Quadratic  Expression Nature of Roots of a Quadratic Equation with   Respect to One or Two Real Numbers Relation Between Roots and Coefficients   of a Polynomial Equation Formation of a Polynomial Equation From   Given Roots Sign of a Polynomial Expression Rational Algebraic Expression NCERT Exemplars Practice Exercises

3.1 3.7 3.9 3.9 3.9 3.10 3.11 3.11 3.11 3.16 3.19 3.23

Chapter 4 Permutations and Combinations 4.1 Factorial Notation 4.1 Fundamental Principles of Counting 4.1 Permutation 4.4 Combination 4.7 Key Results on Combination 4.8 Derangement 4.13 Exponent of prime p in n!4.14 Number of Divisors 4.15 NCERT Exemplars 4.16 Practice Exercises 4.19

Argand Plane and Geometrical Representation   of Complex Numbers

2.7



Polar Form of a Complex Number

2.7



Particular Cases of Polar Form

2.9



Eulerian Representation of a Complex Number 2.10



Logarithm of a Complex Number



Vectorial Representation of a Complex Number 2.10



Roots of a Complex Number

2.12

Chapter 5  Mathematical Induction



Geometry of Complex Numbers

2.15



Mathematical Induction

5.1

2.21



NCERT Exemplars

5.4



Practice Exercises

5.5



Practice Exercises

2.10

5.1

viii  Contents

Chapter 6  Binomial Theorem

Binomial Expression

6.1

Chapter 10 Coordinates and Straight Lines10.1



Binomial Theorem

6.1



Distance Formula

10.1



Special Cases

6.1



Section Formulae

10.3



Pascal’s Triangle

6.2



Area of a Triangle

10.5



Middle Term in the Binomial Expansion 

6.9



Condition for Collinearity of Three Points

10.5



NCERT Exemplars

6.14



Stair Method for Finding the Area

10.6



Practice Exercises

6.16



Area of a Quadrilateral

10.8



Area of a Polygon

10.8



Stair Method

10.8

Chapter 7  Sequence and Series

7.1

Sequence

7.1

Locus

Series

7.1



Translation of Axes

Progressions

7.1



Rotation of Axes

10.10



7.1



Reflection (Image) of a Point

10.10

General Equation of a Straight Line

10.12

Arithmetic Progression (A.P.)

10.9 10.9



Sum of n Terms of an A.P.

7.3





Properties of A.P.

7.4



Slope of a Line

10.12

Intercept of a Line on the Axes

10.12

Equation of a Straight Line in Various forms

10.13



Arithmetic Mean (A.M.)

7.7





Geometric Progression (G.P.)

7.9





Geometric Mean (G.M.)

7.14



Some Special Sequences

7.16



Arithmetico-Geometric Progression (A.G.P.)

7.18



Method for Finding Sum of A.G. Series

7.18



NCERT Exemplars

7.21



Practice Exercises

7.24

Chapter 8  Limits

8.1

Reduction of the General Equation to Different   Standard Forms 10.17

Angle Between Two Intersecting Lines

10.17

Condition for Two Lines to be Coincident,   Parallel, Perpendicular or Intersecting

10.18



Equation of a Line Parallel to a Given Line

10.18



Equation of a Line Perpendicular to a Given Line 10.19



Point of Intersection of Two Given Lines

10.19



Limit of A Function

8.1



Concurrent Lines

10.19



Indeterminate Forms

8.2



Position of Two Points Relative to a Line

10.19



Algebra of Limits

8.3



Length of Perpendicular from a Point on a Line 10.20



Evaluation of Limits

8.3



Distance between Two Parallel Lines



Algebraic Limits

8.3



Limit of an Algebraic Function when x → ∞

8.4



Trigonometric Limits

8.6

Equations of Straight Lines Passing through   a Given Point and Making a Given   Angle with a Given Line

10.21



Exponential and Logarithmic Limits

8.8



Reflection on the Surface

10.21

Image of a Point with Respect to a Line

10.22 10.23

10.21



Evaluation of Limits using L’Hospital’s Rule

8.12





Practice Exercises

8.16

Equations of the Bisectors of the Angles   Between Two Lines

9.1

Equations of Lines Passing through the Point   of Intersection of Two Given Lines

10.24



10.24

Chapter 9  Differential Equations

Differential Equation

9.1



Linear and Non-linear Differential Equations

9.3



Initial Value Problems

9.7



Homogeneous Differential Equations

9.10



Solution by Inspection

9.13



NCERT Exemplars

9.17



Practice Exercises

9.26

Standard Points of a Triangle

Orthocentre

10.26



Coordinates of Nine Point Circle

10.26



NCERT Exemplars

10.28



Practice Exercises

10.33

Contents  ix

Chapter 11  Circles

11.1

Circle

11.1



Standard Equation of a Circle

11.1



General Equation of a Circle

11.1



Conditions for an Equation to Represent a Circle11.1



Equation of a Circle in some Special Cases

11.2



Equation of A Circle In Diameter Form

11.4



Intercepts Made by a Circle on the Axes

11.5



Parametric Equations of a Circle

11.5



Position of a Point with Respect to a Circle

11.6



Circle Through Three Points

11.7



Intersection of a Line and a Circle

11.8



Length of Intercept Made by a Circle on a Line 11.8

The Least and Greatest Distance of a Point   from a Circle

11.9



Centre of Conic

12.3

Parabola

12.3



Some Terms Related to Parabola

12.3



Intersection of a Line and a Parabola

12.5



Equation of a Chord

12.5



Point of Intersection of tangents

12.5



Position of a Point with Respect to a Parabola

12.6

Number of Tangents Drawn from a Point   to a Parabola

12.6



Equation of the Pair of Tangents

12.6



Equations of Normal in Different Forms

12.6



Point of Intersection of Normals

12.6



Co-normal Points

12.7



Chord of Contact

12.7



Chord with a Given Mid Point

12.7

Ellipse

12.14



Contact of Two Circles

11.10



Tangent to a Circle at a Given Point

11.11



Equation of the Tangent in Slope form

11.11



Condition of Tangency

11.11



Tangents From a Point Outside the Circle

11.11



Length of the Tangent From a Point to a Circle 11.12



Normal to the Circle at a Given Point

11.12



Pair of Tangents

11.13



Common Tangents to Two Circles

11.13



Power of a Point with Respect to a Circle

11.15



Director Circle

11.15



Equation of Chord of Contact

11.16



Equation of Chord if its Mid Point is Known

11.16



Common Chord of Two Circles

11.17



Diameter of a Circle

11.17



Angle of Intersection of Two Circles

11.18



Orthogonal Intersection of Two Circles

11.18



Family of Circles

11.19

Chapter 13



Image of the Circle by the Line Mirror

11.20



Scalars and Vectors

13.1



Practice Exercises

11.21



Representation of Vectors

13.1



Types of Vectors

13.2



Equal Vectors 

13.2



Fixed Vectors 

13.2

Free Vectors 

13.2

Chapter 12 Conic Sections (Parabola, Ellipse and Hyperbola) 12.1



Position of a Point with Respect to an Ellipse 12.16



Equation of Normal in Different Forms

12.18



Equation of the Pair of Tangents

12.18



Chord with a Given Mid Point

12.19



Optical Property of Parabola

12.25



Equation of a Hyperbola in Standard Form

12.26

Some Terms and Properties Related to a  Hyperbola

12.26



12.27

Conjugate Hyperbola

Position of a Point with Respect to a  Hyperbola

12.28



Equation of the Pair of Tangents

12.29



Chord with a Given Mid Point

12.29



Chord of Contact

12.29



NCERT Exemplars

12.35



Practice Exercises

12.38

Vector Algebra

13.1



Conic Section

12.1





Important Terms

12.1



Angle between Two Vectors

13.2



Addition (Sum or Resultant) of Two Vectors

13.3

Section of a Right Circular Cone by Different  Planes

12.1



Position Vector 

13.4



Equation of Conic

12.2



Component of a Vector

13.8



General Equation

12.2



Linear Combination

13.8

x  Contents Linearly Dependent and Independent   System of Vectors



Table for Trigonometric Ratios of Allied Angles

15.5

13.8



Trigonometric Ratios in Terms of Each Other

15.6



Collinearity of Three Points

13.8



Addition and Subtraction Formulae

15.6



Coplanarity of Four Points

13.9



Transformation  Formulae

15.7



Product into Sum or Difference

15.7



Sum and Difference into Product

15.7



Trigonometric Ratios of Multiple Angles

15.7



Trigonometric Ratios of Submultiple Angles

15.8



Greatest and Least Values of the Expression

15.9

Some Results on Linearly Dependent and   Independent Vectors

13.9



Product of Two Vectors

13.10



Scalar Product of Two Vectors

13.10



Some Useful Identities

13.10



Work Done by a Force

13.10



Vector Product of Two Vectors

13.14



Moment of a Force about a Point

13.15



Scalar Triple Product

13.18



NCERT Exemplars

13.21



Practice Exercises

13.24

Chapter 14 Measures of Central Tendency and Dispersion 14.1

Conditional Identities

15.9



Graphs of Trigonometric Functions

15.9



Practice Exercises

Chapter 16  Trigonometric Equations

15.11

16.1



Trigonometric Equation

16.1



Solution or Root of a Trigonometric Equation

16.1



Method for finding Principal Value (Solution)

16.1



Solution of an Equation of the Form

16.5



Measures of Central Tendency

14.1



Solutions of Basic Trigonometric Inequalities

16.6



Arithmetic Mean

14.1



NCERT Exemplars

16.9



Geometric Mean

14.2



Practice Exercises

16.15



Harmonic Mean

14.2

Median

14.7



14.7

Quartiles, Deciles and Percentiles

Chapter 17  Heights and Distances

17.1

Mode

14.8

Some Terminology Related to   Height and Distance

17.1



Symmetric Distribution

14.9



North East

17.1



Practice Exercises

14.13



Bearings of a Point

17.1



Some Properties Related to Triangle

17.2



Some Properties Related to Circle

17.4



Some Important Results

17.7



Practice Exercises

Chapter 15 Trigonometric Ratios and Identities

15.1

Angle

15.1



15.1

Measurement of Angles

Relation between Different Systems of   Measurement of Angles

15.2

Relation between Sides and Interior Angles of   a Regular Polygon

15.2



15.2

Fundamental Trigonometric Identities

Chapter 18  Mathematical Reasoning

Mathematical Reasoning

17.12

18.1 18.1

Statement

18.1



18.2

The Connecting Word ‘OR’

Quantifiers

18.2

15.2

Implications

18.3

Increase and Decrease of Trigonometric   Functions 



NCERT Exemplars

18.5

15.3



Practice Exercises

18.8



Domain and Range of Trigonometric Ratios

15.3



Trigonometric Ratios of Standard Angles

15.4



Trigonometric Ratios for Some Special Angles

15.4



Trigonometric Ratios of Allied Angles

15.5

Signs of Trigonometric Ratios in   Different Quadrants

Preface About the Series Complete Companion for JEE Main series is a must have resource for students preparing for Joint Entrance Examination. There are three subjective books—Physics, Chemistry, and Mathematics; the core objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations. It provides class-tested course material and numerical applications that will supplement any ready material available as student resource. To ensure high level of accuracy and practicality, this series has been authored by highly qualified and experienced faculties for all three titles. About the Book Complete Companion for JEE Main 2020 Mathematics, Volume 1 particularly developed for class 11th students, so that they can start their preparation from the early days. This title is designed as per the latest JEE Main syllabus, where the important topics are covered in 18 chapters. It has been structured in user friendly approach such that each chapter begins with topic-wise theory, followed by sufficient solved examples and then practice questions along with previous years' questions. The chapter-end exercises are structured in line with JEE questions with ample of questions on single choice correct question (SCQ) for extensive practice. Previous 18 years’ questions from JEE Main and AIEEE are also added in every chapter. Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and numerical applications. Because of its comprehensive and in-depth approach, it will be especially helpful for those students who prefers self-study than going for any classroom teaching. Series Features

• • • • • •

Complete coverage of topics along with ample number of solved examples Large variety of practice problems with complete solutions Chapter-wise Previous 18 years’ AIEEE/JEE Main questions Fully solved JEE Main 2019 (Jan/Apr) questions added in opening section of the book Includes 5 Mock Tests papers based on JEE Main pattern in the book Free Online Mock Tests as per the recent JEE Main pattern

It would have been difficult to prepare this book without aid and support from a number of different quarters. I shall be grateful to the readers for their regular feedback. I am deeply indebted to my parents without whose encouragement this dream could not have been translated into reality. The cherubic smiles of my daughters, Nikita and Nishita, have inspired me to treat my work as worship. Anuj Agarwal from IIT-Delhi, Ankit Katial from National Institute of Technology (Kurukshetra) and Raudrashish Chakraborty from Kirori Mal College, University of Delhi, with whom I have had fruitful discussions, deserve special mention. I earnestly hope that the book will help the students grasp the subject well and respond with a commendable score in the JEE Main examination. There are a plethora of options available to students for Mathematics, however, ever grateful to them and to the readers for their candid feedback. Despite of our best efforts, some errors may have crept into the book. Constructive comments and suggestions to further improve the book are welcome and shall be acknowledged gratefully. Best of luck! Dinesh Khattar

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Mathematics Trend Analysis (2011 to 2019) S. No. Chapters 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

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CHAPTER

1

Set Theory

LEARNING OBJECTIVES After reading this chapter, you will be able to:   Learn the definition of set and how is it denoted   Know how the sets are represented and what are its types

 Understand the operations on sets and Identify some key results  Establish a relation between the two sets and study about its types

SET A set is a well-defined 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: 1. Roster form or tabular form 2. Set builder form or rule method

Roster Form In this form, we list all the members of the set within braces (curly brackets) and separate these by commas. For example, the set A of all odd natural numbers less than 10 in the roster form is written as: A = {1, 3, 5, 7, 9}

I M P O R TA N T P O I N T S In roster form, every element of the set is listed only once. ■ The order in which the elements are listed is immaterial For example, each of the following sets denotes the same set {1, 2, 3}, {3, 2, 1}, {1, 3, 2}. ■

Set-builder Form 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 than 10} 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 f. For example, each of the following is a null set: 1. The set of all real numbers whose square is –1. 2. The set of all rational numbers whose square is 2. 3. The set of all those integers that are both even and odd. A set consisting of atleast one element is called a ­non-empty set.

1.2  Chapter 1

A set having only one element is called singleton set. For example, {0} is a singleton set, whose only ­member is 0.

For example, let A = {3, 4}, then the subsets of A are f, {3}, {4}, {3, 4}. Here, n(A) = 2 and number of subsets of A = 22 = 4. Also, {3} ⊂ {3, 4} and {2, 3} ⊄ {3, 4}

Finite and Infinite Set

Power Set

A set which has finite number of elements is called a finite set. Otherwise, it is called an infinite 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 f 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).

The set of all subsets of a given set A is called the power set of A and is denoted by P(A). For example, if A = {1, 2, 3}, then

Singleton Set

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.

Equivalent Sets Two finite sets A and B are said to be equivalent, if n(A) = n(B).

P(A) = [f, {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 exactly 2n elements.

QUICK TIPS Number of elements in P {P [P(f)]} is 4 or Cardinal Number of P{P[P(f)]} = 4 Since, P (f) = {f} Also, P [P (f)] = {f, {f}} and P {P[P (f)]} = {f, {f}, [{f}], [f, {f}]} Hence, n [P {P[P (f)]}] = 4

Euler–Venn Diagrams

ERROR CHECK Equal sets are equivalent but equivalent sets need not be equal. For example, the sets A = {4, 5, 3, 2} and B = {1, 6, 8, 9} are equivalent but are not equal.

Subset Let A and B be two sets. If every element 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.

I M P O R TA N T P O I N T S 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). ■ Every set is a subset and a superset of itself. ■ If A is not a subset of B, we write A ⊄ B. ■ The empty set is the subset of every set. ■ If A is a set with n (A) = m, then the number of subsets of A are 2m and the number of proper subsets of A are 2m–1. ■

To express the relationship among sets, we represent them pictorially by means of diagrams, known as Euler–Venn Diagrams or simply Venn diagrams. In Venn diagrams, the universal set U is represented by the rectangular region and its subsets are represented by closed bounded circles inside this rectangular region.

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, ⇒ or and ⇒ and

x ∈ A ∪ B x ∈ A x ∈ B, x ∉ A ∪ B x ∉ A 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 }.

Set Theory  1.3 U

Similarly,

B – A = {x: x ∈ B and x ∉ A}

For example,

A∪B

A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7, 9},

if A

B

then

A – B = {2, 4} and B – A = {7, 9}

FIGURE 1.1 U

Intersection of Two Sets The intersection of two sets A and B, written as A ∩ B (read as ‘A intersection B’) is the set consisting of all the common elements of A and B. Thus,

A–B

B–A

A

A ∩ B = {x: x ∈ A and x ∈ B}

Clearly,

x ∈ A ∩ B

⇒

x ∈ A and x ∈ B,

and

x ∉ A ∩ B

⇒

x ∉ A or x ∉ B.

B

■ ■

A–B≠B–A The sets A – B, B – A and A ∩ B are disjoint sets A – B ⊆ A and B – A ⊆ B A – ϕ = A and A – A = ϕ

Symmetric Difference of Two Sets The symmetric difference of two sets A and B, denoted by A D B, is defined as A D B = (A – B) ∪ (B – A).

A∩B

A

A

Info Box! ■

U

B FIGURE 1.4  (a–b)

■

For example, if A = {a, b, c, d} and B = {c, d, e, f }, then A ∩ B = {c, d}.

U

For example, if A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7, 9} then A D B = (A – B) ∪ (B – A) = {2, 4} ∪ {7, 9} = {2, 4, 7, 9}.

B

FIGURE 1.2

Disjoint Sets Two sets A and B are said to be disjoint, if A ∩ B = f, i.e., A and B have no element in common. For example, if A = {1, 2, 5} and B = {2, 4, 6}, then A ∩ B = f, so A and B are disjoint sets.

B A

U FIGURE 1.5

Complement of a Set A

B FIGURE 1.3

Difference of Two Sets

If U is a universal set and A is a subset of U, then the complement 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, A′ = {x: x ∈ U and x ∉ A}

If A and B are two sets, then their difference A – B is defined as

A – B = {x: x ∈ A and x ∉ B}

For example,

1.4  Chapter 1 U = {1, 2, 3, 4, …} and A = {2, 4, 6, 8, …},

if

A′ = {1, 3, 5, 7, …}

then,

U

A A′ FIGURE 1.6

Info Box!

■ ■ ■ ■

U′ = f f′ = U A ∪ A′ = U A ∩ A′ = f

(b) (A ∩ B)′ = A′ ∪ B′ (c) A – (B ∪ C) = (A – B) ∩ (A – C) (d) A – (B ∩ C) = (A – B) ∪ (A – C)

Key Results on Operations on Sets 1. A ⊆ A ∪ B, B ⊆ A ∪ B, A ∩ B ⊆ A, A ∩ B ⊆ B 2. A – B = A ∩ B′ 3. (A – B) ∪ B = A ∪ B 4. (A – B) ∩ B = f 5. A ⊆ B ⇔ B′ ⊆ A′ 6. A – B = B′ – A′ 7. (A ∪ B) ∩ (A ∪ B′) = A 8. A ∪ B = (A – B) ∪ (B – A) ∪ (A ∩ B) 9. A – (A – B) = A ∩ B 10. A – B = B – A ⇔ A = B 11. A ∪ B = A ∩ B ⇔ A = B 12. A ∩ (B D C) = (A ∩ B) D (A ∩ C)

Some Results about Cardinal Number If A, B and C are finite sets and U be the finite universal set, then

ALGEBRA OF SETS 1. Idempotent Laws: For any set A, we have (a) A ∪ A = A (b) A ∩ A = A 2. Identity Laws: For any set A, we have: (a) A ∪ f = A (b) A ∩ f = f (c) A ∪ U = U (d) A ∩ U = A 3. Commutative Laws: For any two sets A and B, we have (a) A ∪ B = B ∪ A (b) A ∩ B = B ∩ A 4. Associative Laws: For any three sets A, B and C, we have (a) A ∪ (B ∪ C) = (A ∪ B) ∪ C (b) A ∩ (B ∩ C) = (A ∩ B) ∩ C 5. Distributive Laws: For any three sets A, B and C, we have (a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 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. 8. Demorgan’s Laws: For any three sets A, B and C, we have (a) (A ∪ B)′ = A′ ∩ B′

1. n (A′) = n (U) – n (A) 2. n (A ∪ B) = n (A) + n (B) – n (A ∩ B) 3. n (A ∪ B) = n (A) + n(B), where A and B are disjoint non-empty sets 4. n (A ∩ B′) = n (A) – n (A ∩ B) 5. n (A′ ∩ B′) = n (A ∪ B) ‘ = n (U) – n (A ∪ B) 6. n (A′ ∪ B′) = n (A ∩ B) ‘ = n (U) – n (A ∩ B) 7. n (A – B) = n (A) – n (A ∩ B) 8. n (A ∩ B) = n (A ∪ B) – n (A ∩ B′) – n (A′ ∩ B) 9. n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n (A ∩ B) – n (B ∩ C) – n (C ∩ A) + n (A ∩ B ∩ C) 10. If A1, A2, A3, … An are disjoint sets, then n (A1 ∪ A2 ∪ A3 ∪ … ∪ An) = n (A1) + n (A2) + n (A3) + … + n (An) 11. n (A D 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]

ERROR CHECK A×B≠B×A

Set Theory  1.5 ⇒ 50 = 35 + 24 – n (A ∩ B)

QUICK TIPS If A = f or B = f, then we define A × B = f. ■ If A has n elements and B has m elements then A × B has mn elements. ■ If A1, A2, …, Ap are p non-empty sets, then their cartesian ■

p

product, is defined as for all i]

∏ Ai

= [(a1, a2, a3, …, ap); ai ∈ Ai

i =1

Key Results on Cartesian Product If A, B, C are three sets, then 1. A × (B ∪ C) = (A × B) ∪ (A × C) 2. A × (B ∩ C) = (A × B) ∩ (A × C) 3. A × (B – C) = (A × B) – (A × C) 4. (A × B) ∩ (S × T) = (A ∩ S) × (B ∩ T), where S and T are two sets. 5. If A ⊆ B, then (A × C) ⊆ (B × C) 6. If A ⊆ B, then (A × B) ∩ (B × A) = A2 7. If A ⊆ B and C ⊆ D then A × C ⊆ B × D 8. If A ⊆ B, then A × A ⊆ (A × B) ∩ (B × A) 9. 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. 10. A × B = B × A if and only if A = B 11. A × (B′ ∪ C ′)′ = (A × B) ∩ (A × C) 12. A × (B′ ∩ C ′)′ = (A × B) ∪ (A × C)

SOLVED EXAMPLES 1. If n (U) = 60, n (A) = 35, n (B) = 24 and n (A ∪ B)′ = 10 then n (A ∩ B) is (A) 9 (B) 8 (C)  6 (D)  None of these Solution: (A) We have,

n(A ∪ B) = n (U) – n(A ∪ B)′ = 60 – 10 = 50

Now, n (A ∪ B) = n (A) + n(B) – n(A ∩ B)

⇒

2. 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 (B) f ∈ A′ with respect to. X (C) {0} ⊂ A′ with respect to X (D) 0 ⊂ A′ with respect to X. Solution: (C) We have, A′ in X = The set of elements in X which are not in A = {0, 1} {0} ∈ A′ in X is false, because {0} is not an e­ lement of A′ in X. f ⊂ A′ in X is false, because f is not an element of A′ in X {0} ⊂ A′ in X is correct, because the only element of {0} namely 0 also belongs to A′ in X. 0 ⊂ A′ in X is false, because 0 is not a set. 3. If X = {8n – 7 n – 1/n ∈ N} and Y = {49 (n – 1)/n ∈ N}, then (A) X ⊂ Y (B)  Y⊂X (C) X = Y (D)  None of these Solution: (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 multiples of 49 of the form 49 Kn. where Kn = nC2 + n C37 + … + nCn7n – 2 together with zero. Also, Y consists of all positive integral multiples of 49 including zero. Therefore, X ⊂ Y. 4. The set (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∩ C′ is equal to (A) A ∩ B (B)  A ∩ C ′ (C) B ∩ C ′ (D)  B′ ∩ C ′ Solution: (C) (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C ′)′ ∩ C ′

U (A ∪ B)′ A

B

A∩B

n(A ∩ B) = 59 – 50 = 9.



= (A ∪ B ∪ C) ∩ (A′ ∪ B ∪ C) ∩ C ′



= [(A ∩ A′) ∪ (B ∪ C)] ∩ C ′



= (f ∪ B ∪ C) ∩ C ′ = (B ∪ C) ∩ C ′



= (B ∩ C ′) ∪ (C ∩ C ′)



= (B ∩ C ′) ∪ f = B ∩ C ′

1.6  Chapter 1 5. If A, B and C are non-empty subsets of a set, then (A – B) ∪ (B – A) equals (A) (A ∩ B) ∪ (A ∪ B) (B) (A ∪ B) – (A ∩ B) (C) A – (A ∩ B) (D) (A ∪ B) – B Solution: (B) (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B)

A

A−B

B

A∩B

B−A

6. Let A and B two non-empty subsets of a set X such that A is not a subset of B then (A) A is subset of the complement of B (C) B is a subset of A (C) A and B are disjoint (D) A and the complement of B are non-disjoint Solution: (D) Since A ⊄ B, $ x ∈ A such that x ∉ B Then \

x ∈ B ′. A ∩ B ′ ≠ f

7. Two finite sets have m and n elements, then total number of subsets of the first set is 56 more that the total number of subsets of the second. The values of m and n are, (A)  7, 6 (B)  6, 3 (D)  5, 1 (D)  8, 7 Solution: (B) Since the two finite sets have m and n elements, so number of subsets of these sets will be 2m and 2n respectively. According to the question



= {x : (x ∈ R and x ≥ 2) or (x ∈ R and x ≤ 0)}



= {x : x ∈ R and x ≥ 2} ∪ {x : x ∈ R and x ≤ 0}

Similarly, B′ = {x : x ∈ R and x ≤ 1} ∪ {x : x ∈ R and x > 3} \

A ∪ B = {x : x ∈ R and 0 < x ≤ 3},



A ∩ B = {x : x ∈ R and 1 < x < 2}



A – B = {x : x ∈ R and 0 < x ≤ 1}

9. 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} Solution:  (A and 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 set of Y is {1, 2, 3, 4, 9}. 10. If A has 3 elements and B has 6 elements, then the minimum number of elements in the set A ∪ B is (A) 6 (B) 3 (C) f (D)  None of these Solution: (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. 11. Suppose A1, A2, … A30 are thirty sets, each with five elements and B1, B2, …, Bn are n sets each with three 30

elements. Let

∪ Ai

=

i =1

n

∪ Bj

=S

j =1

2m – 2n = 56 putting m = 6, n = 3, we get

If each element of S belongs to exactly ten of the Ai′s and exactly nine of the Bj′s then n =

26 – 23 = 56  or  64 – 8 = 56

(A) 45 (C)  40

8. Let U = R (the set of all real numbers) If A = {x : x ∈ R, 0 < x < 2}, B = {x : x ∈ R, 1 < x ≤ 3}, then (A) A ∪ B = {x : x ∈ R and 0 < x ≤ 3} (B) A ∩ B = {x : x ∈ R and 1 < x < 2} (C) A – B = {x : x ∈ R and 0 < x ≤ 1} (D)  All of these Solution: (D) We have

A′ = R – A = {x : ∈ R and x ∉ A}

(B) 35 (D)  None of these

Solution: (A) Given Ai′s are thirty sets with five elements each, so 30



∑ n ( Ai )

= 5 × 30 = 150

(1)

i =1

If there are m distinct elements in S and each element of S belongs to exactly 10 of the Ai′s, we have 30



∑ n ( Ai )

i =1

= 10 m

(2)

Set Theory  1.7 \ From Eq. (1) and (2), we get

∵

10 m = 150

65 = 40 + n (B) – 10

\ Similarly

m = 15

(3)

30

30

j =1

j =1

∑ n ( B j ) = 3n and ∑ n ( B j ) = 9 m

\ 3 n = 9 m ⇒ Hence,

9m n= = 3 m = 3 × 15 = 45 3 n = 45.

[from (3)]

12. 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 (B)  N (C) B (D)  None of these Solution: (B) We have,

n (B) = 65 – 40 + 10 = 35 Number of people who like only tennis = n(B) – n (A ∩ B) = 35 – 10 = 25 \ Number of people who like tennis only and not cricket = 25. 15. 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 (B) 400 (C)  600 (D)  None of these Solution: (C) Here

(A ∪ B) ∩ B′ = A [(A ∪ B) ∩ B′] ∪ A′ = A ∪ A′ = N. 13. If X and Y are two sets and X′ denotes the complement of X, then X ∩ (X ∪ Y)′ equals (A) X (B)  Y (C) f (D)  None of these

Using

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

1000 = 750 + 400 – n (H ∩ E) ⇒

n (H ∩ E) = 1150 – 1000 = 150. U H

X ∩ (X ∪ Y)′ = X ∩ (X ′ ∩ Y ′) [∵ By De-Morgan’s Law (A ∪ B)′ = (A′ ∩ B′)]

600

E 150

250

= (X ∩ X ′) ∩ Y ′ = f ∩ Y ′ = f

14. 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 (B) 25 (C)  15 (D)  None of these Solution: (B) Let A be the set of people who like cricket and B the set of people who like tennis. Then

n(H ∪ E) = 1000, n (H) = 750,

n (E) = 400

Solution: (C)



n (A ∪ B) = n(A) + n(B) – n(A ∩ B)

n (A) = 40, n(A ∩ B) = 10 U

30

B 10

25



= n(H ∩ E′) = n(H) – n(H ∩ E)



= 750 – 150 = 600.

16. If f : R → R, defined by f (x) = x2 + 1, then the values of f –1(17) and f –1(–3) respectively are (A) f, {4, –4} (C) f, {3, –3}

n(A ∪ B) = 65

A

Number of people who can speak Hindi only

(B)  {3, –3}, f (D)  {4, –4}, f

Solution: (D) Let y = x2 + 1. Then for y = 17, we have x = ± 4 and for y = –3, x becomes ­imaginary that is, there is no value of x. Hence, and

f (17) = {–4, 4} –1

f (–3) = f

1.8  Chapter 1 17. In a statistical investigation of 1,003 families of Kolkata, it was found that 63 families had neither a radio nor a TV, 794 families had a radio and 187 had a TV. The number of families in that group having both a radio and a TV is (A) 36 (B) 41 (C)  32 (D)  None of these Solution: (B) Let R be the set of families having a radio and T, the set of families having a TV, then n (R ∪ T) = The no. of families having at least one of radio and TV = 1003 – 63 = 940



n (R) = 794 and n(T) = 187 Let x families had both a radio and a TV i.e., ∪

x

187 – x

T 794 – x

R

Using n(A ∪ B ∪ C) = n (A) + n(B) + n(C) – 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. 19. Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. The number of boys who did not play any game is (A) 160 (B) 240 (C) 216 (D) 128 Solution: (A)

n (C) = 224, n (H) = 240, n (B) = 336

n (H ∩ B) = 64, n (B ∩ C) = 80 n (H ∩ C) = 40, n (C ∩ H ∩ B) = 24

n (R ∩ T ) = x The number of families who have only radio = 794 – x and the number of families who have only TV = 187 – x From Venn diagram, ⇒

794 – x + x + 187 – x = 940 981 – x = 940 or x = 981 – 940 = 41

Hence, the required no. of families having both a radio and a TV = 41. 18. 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% (B)  18% (C) 20% (D)  None of these Solution: (A) Let the no. of persons in the city be 100. 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

n (C c ∩ H  c ∩ B c) = 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.

20. In a certain town 25% families own a phone and 15% own a car, 65% families own neither a phone nor a car. 2000 families own both a car and a phone. Consider the following statements in this regard: 1. 10% families own both a car and a phone. 2. 35% families own either a car or a phone. 3.  40,000 families live in the town. Which of the above statements are correct? (A)  1 and 2 (B)  1 and 3 (C)  2 and 3 (D)  1, 2 and 3 Solution: (C) n(P) = 25%, n(C) = 15%, n (P ′ ∩ C ′) = 65%, n (P ∩ C) = 2000 Since,

n (P ′ ∩ C ′) = 65%

Set Theory  1.9 \ n(P ∪ 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 5

n(P ∪ C) = 35%,

Total no. of families = 40,000 and n(P ∩ C) = 5%. 21. If P, Q and R are subsets of a set A, then R × (P ′ ∪ Q′)′ equals (A) (R × P) ∩ (R × Q) (B) (R × Q) ∩ (R × P) (C) (R × P) ∪ (R × Q) (D)  None of these Solution: (A) R × (P′ ∪ Q′)′ = R × [(P ′)′ ∩ (Q ′)′] = R × (P ∩ Q) = (R × P) ∩ (R × Q) 22. 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 (B)  A⊂B (C) A ∩ B = f (D)  A∪B=A Solution: (C) Since

x 2 x3 + y=e =1+x+ + … 2 ! 3! x

\ ex > x ∀ x ∈ 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 = f. 23. 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 f (D)  All of these Solution: (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 f.

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.

Key Results on Relations 1. Every subset of A × A is said to be a relation on A. 2. 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. 3. 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 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}.

Info Box!

Info Box! Domain of a relation from A to B is a subset of A and its range is a subset of B

Identity Relation R is an identity relation if (a, b) ∈ R if 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.

Void Relation f is called void relation in a set.

QUICK TIPS The void and the universal relations on a set A are respectively the smallest and the largest relations on A

1.10  Chapter 1

Inverse Relation Let R ⊆ A × B be a relation from A to B. Then R–1 ⊆ B × A is defined by R–1 = [(b, a): (a, b) ∈ R] Thus, (a, b) ∈ R ⇔ (b, a) ∈ R– 1, ∀ a ∈ A, b ∈ B

I M P O R TA N T P O I N T S dom (R–1) = range (R) and range (R–1) = dom (R) –1 –1 ■ (R ) = R. For example, if R = [(1, 2), (3, 4), (5, 6)] then R–1 = [(2, 1), (4, 3), (6, 5)] –1 –1 and (R ) = [(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) ■

1. Since x – x = 0 = 0 . m ⇒ x – x is divisible by m ⇒ (x, x) ∈ R ⇒ R is reflexive. 2. 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 Thus, (x, y) ∈ R ⇒ (y, x) ∈ R ⇒ R is symmetric. 3. Let (x, y) ∈ R and (y, z) ∈ R ⇒  x – y is divisible by m and y – z is divisible by m ⇒  x ­– y = mq and y – z = mq′ for some q, q′ ∈ I ⇒ (x – y) + (y – z) = m (q + q′) ⇒  x – z = m (q + q′), q + q′ ∈ I ⇒ (x, z) ∈ R Thus, (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R, so R is transitive. Hence the relation R is reflexive, symmetric and transitive and therefore it is an equivalence relation.

ERROR CHECK Every identity relation is reflexive but every reflexive relation need not be an identity relation. Also, identity relation is reflexive, symmetric and transitive.

TYPES OF RELATIONS ON A SET Let A be a non-empty set, then a relation R on A is said to be: 1. Reflexive: If a R a, ∀ a ∈ A, i.e., if (a, a) ∈ R, ∀ a ∈ A 2. Symmetric: If a R b ⇒ b R a, ∀ a, b ∈ A, i.e., if (a, b) ∈ R ⇒ (b, a) ∈ R, ∀ a, b ∈ A 3. Anti-symmetric: If a R b and b R a ⇒ a = b, ∀ a, b ∈ A 4. Transitive: If a R b and b R c ⇒ a R c, ∀ a, b, c ∈ A i.e., (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A

EQUIVALENCE RELATION A relation R on a non-empty set A is called an equivalence relation if and only if it is 1. reflexive 2. symmetric and 3. transitive That is, R satisfies following properties: 1. a R a, ∀ a ∈ A 2. a R b ⇒ b R a, ∀ a, b ∈ A 3. 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.

CONGRUENCE MODULO M Let m be a positive integer and x, y ∈ I, then x is said to be congruent to y modulo m, written as x ≡ y (mod m), if x – y is divisible by m. For example, 155 ≡ 7 (mod 4) as 155 − 7 148 = = 37 (integer) 4 4 27 − 5 22 But 27 ≡/ 5 (mod 4) as = (Not an integer) 4 4

I M P O R TA N T P O I N T S The universal relation on a non-void set is reflexive The identity and the universal relations on a non-void set are symmetric and transitive ■ The identity relation on a set is an anti-symmetric relation –1 ■ The relation R on a set A is symmetric if and only if R = R ■ If R and S are two equalvalence relations on a set A, then R ∩ S is also an equivalence relation on A. ■ The union of two equivalence equivalence relations on a set is not necessarily an equivalence realtion on the set. ■ The inverse of an equivalence relation is an equivalence relation. ■ ■

Set Theory  1.11 1 Hence, 2 R ⎛⎜ − ⎞⎟ and ⎛⎜ − ⎝ 6⎠ ⎝

ERROR CHECK A reflexive relation on a set is not necessarily symmetric.

Equivalence Classes of an Equivalence Relation Let R be an equivalence relation in A (≠ f), Let a ∈ A. Then the equivalence class of a, denoted by [a] or { a }is defined as the set of all those points of A which are related to a under the relation R. Thus, [a] = {x ∈ A : x Ra} It is easy to see that 1. b ∈ [a] ⇒ a ∈[b] 2. Two equivalence classes are either disjoint or identical

SOLVED EXAMPLES 24. Let R be a relation defined as a R b if | a – b | > 0. Then, the relation R is (A) Reflexive (C)  Transitive

(B) Symmetric (D)  None of these

Solution: (B) R is not reflexive since | a – a | = 0 and so | a – a | w 0. Thus a R 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 R 3 since | 3 – 3 | = 0 so that | 3 – 3 | >/ 0. 25. Let R be a relation defined as a R b if 1 + ab > 0. Then, the relation R is (A) Reflexive (B) Symmetric (C)  Transitive (D)  None of these Solution: (A) and (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 real 1 numbers 2, – and –2. We have 6

and

2 ⎛ 1⎞ 1 + 2 × ⎜ − ⎟ = > 0 ⎝ 6⎠ 3 ⎛ 1 + ⎜− ⎝

4 1⎞ > 0 ⎟⎠ (– 2) = 3 6

1⎞ ⎟ R (– 2) 6⎠

But 2 R – 2 since 1 + 2 (– 2) = – 3 >/ 0 26. Let R be a relation defined as a R b if | a | ≤ b. Then, relation R is (A) Reflexive (B) Symmetric (C)  Transitive (D)  None of these Solution: (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. R is transitive: let a, b, c be three real numbers such that | a | ≤ b and | b | ≤ c. But | a | ≤ b ⇒ b ≥ 0, and so | b | ≤ c ⇒ b ≤ c. It follows | a | ≤ c. Thus a R b and b R c ⇒ a R c. 27. 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 (B)  Symmetric only (C)  Transitive only (D)  an equivalence relation Solution: (D) We have, (a, b) R (a, b) for all (a, b) ∈ N × N since a + b = b + a. Hence, R is reflexive. R is symmetric: we have (a, b) R (c, d) ⇒ a + d = b + c ⇒ d + a = c + b ⇒

c + b = d + a ⇒ (c, d) R (a, b)

R is transitive: let (a, b) R (c, d) and (c, d) R (e, f ). Then by definition of R, we have a + d = b + c and c + f = d + e, ⇒

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

or

a + f = b + e.

Hence, (a, b) R (e, f ) Thus, (a, b) R (c, d) and (c, d) R (e, f ) ⇒ (a, b) R (e, f ) 28. Let S = {1, 2, 3, 4, 5} and let A = S × S. Define the relation R on A as follows (a, b) R (c, d) if and only if ad = cb. Then, R is

1.12  Chapter 1

(A)  reflexive only (B)  symmetric only (C)  transitive only (D)  equivalence relation Solution: (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

(A) R is reflexive, since ab = ba ⇒  ba = ab, therefore, (a, b) R (b, a) ∀ a, b ∈ S (B) R is symmetric, since (a, b) R (c, d) ⇒  ad = cb ⇒ cd = da ⇒ (c, d) R (a, b) ∀ a, b ∈ S (C) 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 29.

The relation ‘less than’ in the set of natural numbers is (A)  only symmetric (B)  only transitive (C)  only reflexive (D)  equivalence relation Solution: (B) The relation ‘less than’ is only transitive because x < y, y < z ⇒ x < z, x, y, z ∈ N

\ x R y, y R z ⇒ x R z 30. If R and R′ are symmetric relations (not disjoint) on a set A, then the relation R ∩ R′ is (A) reflexive (B) symmetric (C)  transitive (D)  None of these Solution: (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 31. With reference to a universal set, the inclusion of a subset in another, is relation which is (A) symmetric only (B) equivalence (C)  reflexive only (D)  None of these

Solution: (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. 32. Let R be the relation on the set R of all real numbers defined by a R b if | a – b | ≤ 1. Then R is (A) reflexive (B) symmetric (C) transitive (D) anti-symmetric Solution:  (A and 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 anti-symmetric. Finally, 1 R2, 2 R 3 but 1 R 3 as | 1 – 3 | = 2 > 1. 33. If R is a relation ‘ 1) with two sides along the positive direction of X-axis and Y-axis. Then, (A)  R = {( x, y ) : 0 ≤ x ≤ a, 0 ≤ y ≤ b} (B)  R = {( x, y ) : 0 ≤ x < a, 0 ≤ y ≤ b} (C)  R = {( x, y ) : 0 ≤ x ≤ a, 0 < y < b} (D)  R = {( x, y ) : 0 < x < a, 0 < y < b}

(A)  A (B)  B (C) ϕ (D)  A∩ B 12. If A = {1, 3, 5, 7, 11, 13, 15, 17}, B = {2, 4, ….., 18} and N the set of natural numbers is the universal set, then ( A ’∪ ( A ∪ B ) ∩ B ’) is (A) ϕ

(B) N

(C) A

(D) B

13. If {x|x is a positive multiple of 3 less than 100} and P = {x|x is a prime number less than 20}. Then, n(S) + n(P) is equal to (A) 34 (B) 31 (C) 33 (D) 41 14. If X and Y are two sets and X’ denotes the complement of X, then X ∩ ( X ∪ Y ) ’ is equal to (A) X

(B) Y

(C) ϕ (D)  X ∩Y

Set Theory  1.15

ANSWER K EYS 1. (C) 11. (A)

2. (A) 12. (B)

3. (B) 13. (D)

4. (D) 14. (C)

5. (C)

6. (D)

7. (B)

8. (A)

9. (C) 10. (C)

HINTS AND EXPLANATIONS A1 ∪ A2 ∪ A3 ∪,.... ∪ A30 is 30 × 5. But each element is used 10 times, so S =

30 × 5 = 15 10

So, F1 is either of F1, F2, F3 and F4. ∴ F1 = F2 ∪ F3 ∪ F4 ∪ F1 5. The given sets can be represented in Venn diagram as shown below

T

If elements in B1, B2,…….., Bn are not repeated, then total numberof elements is 3n but each element is repeated 9 times, so

C

3n 3n S= ⇒ 15 = 9 9 2. Since, number of subsets of a set containing m elements is 112 more than the subsets of the set containing n elements. ∵ 2m − 2n = 112

(

)

⇒ 2n. 2m− n − 1 = 24 .7

S

It is clear from the diagram that, S ∪ T ∪ C = S . 6. Since, R be the set of points inside the rectangle. ∴ R = {( x, y ) : 0 < x < a and 0 < y < b}

⇒ 2n = 24 and 2m− n − 1 = 7 ⇒ n = 4 and 2m−1 = 8

⇒ 2 m − n = 23 ⇒ m − n = 3

b

3. We know that,

0, 0 a 7. Let H be the set of persons who read Hindi and E be the set of persons who read English. Then, n (∪) = 840, n ( H ) = 450, n ( E ) = 300, n ( H ∩ E ) = 200

( A ∩ B ) ′ = ( A′ ∪ B′ ) and ( A′)′ = A

Number of persons who read neither = n ( H ′ ∩ F ′ )

⇒ m − 4 = 3⇒ m = 4 + 3 ∴ m=7

= ( A ∩ B′ ) ′ ∪ ( B ∩ C )

∴

=  A′ ∪ ( B′ ) ′  ∪ ( B ∩ C )





= ( A′ ∪ B ) ∪ ( B ∩ C ) = A′ ∪ B 4. Every rectangle, rhombus, square in a plane is a parallelogram but every trapezium is not a parallelogram.



= n(H ∪ E )′

= n (U ) − n ( H ∪ E )

= 840 − n ( H ) + n ( E ) − n ( H ∩ E )

= 840 – (450 + 300 – 200) = 840 – 550 = 290 8. X = 8n − 7n − 1 n ∈ N = 0, 49, 490,..... } { } {

HINTS AND EXPLANATIONS

1. If elements are not repeated, then number of elements in

1.16  Chapter 1 Y = {49n − 49 n ∈ N } = {0, 49, 98,147,.....}

12. A′ ∪ ( A ∪ B ) ∩ B   

Clearly, every elements of X is in Y but every element of Y is not in X. ∴ X ⊂Y 9. Let A be the set of percentage of those people who watch a news channel and B be the set of percentage of those people who watch another channel.

∵ A ∩ B ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) = A′ ∪ ( A ∩ B ′ ) ∪ ( B ∩ B ′ )  = A′ ∪ ( A ∩ B ′ ) ∪ φ  = A′ ∪ ( A ∩ B′ )

n ( A) = 63, n ( B ) = 76, and n ( a ∩ B ) = x

=

∵ n ( A ∪ B ) ≤ 100

= A′ ∪ B ′ = N ∩ ( A′ ∪ B ′ )



⇒ n ( A) + n ( B ) − n ( A ∩ B ) ≤ 100



⇒ 63 + 76 − x ≤ 100 ⇒ 139 − x ≤ 100



⇒ 139 − 100 ≤ x ⇒ x ≤ 63



∵ x ( A ∩ b) ≤ n ( A) ⇒ x ≤ 63



∴ 39 ≤ x ≤ 63

= A′ ∪ B′ = ( A ∩ B ) ′  ∵ A ∩ B = φ  =φ=N 13.  S = {x|x is a positive multiple of 3 less than 100} ∴ n(S) = 33 and P = {x|x is a prime number less than 20} ∴ n(P) = 8 n(S) + n(P) = 38 + 8 = 41 14. X ∩ ( X ∪ Y ) ′ = X ∩ ( X ′ ∩ Y ′ )

10. Let x ∈ R We know that, − x ≠ 1

x

∴

( A′ ∪ A) ∩ ( A′ ∪ B′ )

∵ ( A ∪ B ) ′ = A′ ∩ B′ 

A∩ B =φ

= ( X ∩ X ′) ∩ ( X ∩Y ′) = φ ∩ ( X ∩Y ′) = φ

HINTS AND EXPLANATIONS

A ∩ ( A ∪ B) = A 11.  A

B

A ∩ ( A ∪ B)

U

 ∵ φ ∩ A = φ 

Set Theory  1.17

PRACTICE EXERCISES Single Option Correct Type

2. (i) Let R be the relation on the set R of all real numbers 1 defined by setting a R b if | a – b | ≤ . Then R 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 3. 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. 4. 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} (B)  {x: 1 ≤ x < 2} (C) {x: 1 ≤ x ≤ 2} (D)  None of these 5. Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be subset of A consisting of all determinants with value 1. Let C be the subset A of consisting of all determinants with value –1. Then (A) C is empty. (B) B has as many elements as C. (C) A = B ∪ C. (D) B has twice as many elements as C. 6. Let A and B be two sets then (A ∪ B)′ ∪ (A′ ∩ B) is equal to (A) B′ (B)  B (C)  A (D)  A′ 7. If A is the set of even natural numbers less than 8 and B is the set of prime numbers less then 7, then the number of relations from A to B is (A) 29 (b) 92 (C) 32 (D) 29 – 1 8. For real numbers x and y, define a relation R, x R y if and only if x – y + 2 is an irrational number. Then the relation R is

(A) reflexive (B) symmetric (C) transitive (D)  an equivalence relagtion 9. If A = { x : x2 = 1} and B { x : x4 = 1}, then A D B is equal to: (A) {i, – i} (B)  {–1, 1} (C)  {–1, 1, i,–i} (D)  None of these 10. Which of the following is a singleton set? (A) {x : | x | < 1, x ∈ Z} (B) {x : | x | = 5, x ∈Z} (C) {x : x2 = 1, x ∈Z} (D) {x : x2 + x + 1 = 0, x ∈R} 11. Consider the following relations: (1) A– B = A – (A ∩ B) (2) A = (A ∩ B) ∪ (A– B) (3) A – (B ∪ C) = (A – B) ∪ (A – C) Which of these is/are correct? (A) 1 and 3 (B)  2 only (C)  2 and 3 (D)  1 and 2 12. Let R be a reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then (A) m ≥ n (B)  m≤n (C) m = n (D)  None of these 13. If two sets A and B are having 99 elements in common then the number of elements common to each of the sets A × B and B × A are (A) 299 (B) 992 (C) 100 (D) 18 14. The relation R defined on the set A = [1, 2, 3, 4, 5] by R = [(x, y): |x2 – y2| < 16] is given by, (A)  [(1, 1), (2, 1) (3, 1), (4, 1), (2, 3)] (B)  [(2, 2), (3, 2) (4, 2), (2, 4)] (C)  [(3, 3), (3, 4) (5, 4), (4, 3), (3, 1)] (D)  None of these 15. Let L denotes the set of all straight lines in a plane. Let a relation R be defined by a R b ⇔ a ^ b, a, b ∈ L. Then R is (A) reflexive (B) symmetric (C) transitive (D)  None of these

PRACTICE EXERCISES

1. Let F1 be the set of all parallelograms, F2 the set of rectangles, F3 the set of rhombuses, F4 the set of squares and F5 the set of trapeziums in a plane then F1is equal to (A) F2 ∩ F3 (B)  F2 ∪ F3 ∪ F4 ∪ F1 (C) F3 ∩ F4 (D)  None of these

1.18  Chapter 1 16. Let R = [(2,3), (3,4)] be a relation defined on the set of natural numbers. The minimum number of ordered pairs required to be added in R so that enlarged relation be comes an equivalence relation is (A) 3 (B) 5 (C) 7 (D) 9 17. The solution of 8x ≡ 6 (mod 14) is

(A)  [8], [6] (C)  [6], [13]

(B)  [6], [14] (D)  [8], [14], [16]

18. Let X = {1, 2, 3, 4, 5}. The number of different ordered pairs (Y, Z) that can formed such that Y ⊆ X, Z ⊆ X and Y ∩ Z is empty, is (A) 52 (B) 35 (C) 25 (D) 53

Previous Years’ Questions 19. Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is [2004] (A) a function (B) reflexive (C) not symmetric (D) transitive

24. If A, B and C are three sets such that A ∩ B = A ∩ C and A ∪ B = A ∪ C, then [2009] (A) A = B (B)  A=C (C) B = C (D)  A∩B=f

20. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12} be a relation the set A = {3, 6, 9, 12}. The relation is  [2005] (A)  reflexive and transitive only (B)  reflexive only (C)  an equivalence relation (D)  reflexive and symmetric only

25. Let S be a non-empty subset of R. Consider the following statement: [2010] P: There is a rational number x ∈ S such that x > 0. Which of the following statements is the negation of the statement P ? (A) There is no rational number x ∈ S such that x ≤ 0 (B)  Every rational number x ∈ S satisfies x ≤ 0 (C) x ∈ S and x ≤ 0 ⇒ x is not rational (D)  There is a rational number x ∈ S such that x ≤ 0

PRACTICE EXERCISES

21. Let W denote the words in the English dictionary. Define the relation R by: [2006] R = {(x, y) ∈ W × W | the words x and y have at least one letter in common}. Then R is (A)  not reflexive, symmetric and transitive (B)  reflexive, symmetric and not transitive (C)  reflexive, symmetric and transitive (D)  reflexive, not symmetric and transitive 22. The set S = {1, 2, 3, …, 12) is to be partitioned into three sets A, B, C of equal size. Thus, A ∪ B ∪ C = S, A ∩ B = B ∩ C = A ∩ C = f. The number of ways to partition S is [2007] 12 ! 12 ! (A)  (B)  3!( 4 !)3 3!(3!) 4 (C) 

12 !

12 ! (D)  ( 4 !) (3!) 4 3

23. Let R be the real line. Consider the following subsets of the plane R × R. S = {(x, y) : y = x + 1 and 0 < x < 2}, T = {(x, y) : x - y is an integer}. Which one of the following is true?  [2008] (A) neither S nor T is an equivalence relation on R (B) both S and T are equivalence relations on R (C) S is an equivalence relation on R but T is not (D) T is an equivalence relation on R but S is not

26. Let R be the set of all real numbers. [2011] Statement 1: A = {(x, y) ∈R × R : y − x is an integer} is an equivalence relation on R. Statement 2: B = {(x, y) ∈ R × R : x = α y for some rational number α} is an equivalence relation on R. (A) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (B)  Statement 1 is true, Statement 2 is false. (C)  Statement 1 is false, Statement 2 is true. (D) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1 27. Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B, each having at least three elements is (A) 220 (B) 219 (C) 211 (D) 256 28. Two sets A and B are as under:

[2018]

A = {(a, b) ∈ R × R: |a – 5| < 1 and |b – 5| < 1}; B = {(a, b) ∈ R × R: 4(a – 6)2 + 9(b – 5)2 ≤ 36}. Then: (A) B ⊂ A (B) A ⊂ B (C) A ∩ B = ϕ (an empty set) (D) neither A ⊂ B nor B ⊂ A

Set Theory  1.19 29. Let S = {x ∈ R: x ≥ 0 and 2| x – 3| + + 6 = 0}. Then S

x ( x – 6) [2018]



(A)  is an empty set (B)  contains exactly one element (C)  contains exactly two elements (D)  contains exactly four elements

ANSWER K EYS Single Option Correct Type 1. (B) 11. (D)

2. (A) 12. (A)

3. (D) 13. (B)

4. (B) 14. (D)

5. (B) 15. (B)

6. (D) 16. (C)

7. (A) 17. (C)

8. (A) 18. (B)

9. (A)

10. (A)

22. (C)

23. (D)

24. (C)

25. (B)

26. (B)

27. (B)

28. (B)

19. (C) 29. (C)

20. (A)

21. (B)

PRACTICE EXERCISES

Previous Years’ Questions

1.20  Chapter 1

HINTS AND EXPLANATIONS Single Option Correct Type 1. Since every rectangle, rhombus and square is a parallelogram so, F1 = F2 ∪ F3 ∪ F4∪ F1 The correct option is (B) 1 2. R is reflexive since | a – a | = 0 < for all a ∈ R. 2 1 1 R is symmetric since | a – b | < ⇒|b–a|< 2 2 R is not transitive: we take three numbers 3 1 1 , , , then 4 3 8

HINTS AND EXPLANATIONS

3 1 1 1 5 1 5 1 − = < and = < − 12 2 24 2 4 3 3 8 But

3 1 5 1 = > − 4 8 8 2

Thus

3 1 1 1 1 3 R R and R but 4 4 3 3 8 8

The correct option is (A) 3. ‘/’ 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, 2 is factor of 4 but 4 is not a factor of 2. The correct option is (D) 4. We have

A = {x : x ∈ R, – 1 < x < 1}

6. (A ∪ B)′ ∪ (A′ ∩ B) = (A′ ∩ B′) ∪ (A′ ∩ B) A′ ∩ (B′ ∪ B) A′ ∩ U = A′ The correct option is (D) 7. A = {2,4,6}, B = {2,3,5} Number of relations from A to B = 23×3 = 29 The correct option is (A) 8. Clearly x R x as x – x + 2 = 2 is an irrational number. Thus, R is reflexive. Also ( 2 , 1) ∈ R as 2 –1 + 2 = 2 2 – 1 is an irrational number but (1, 2 ) ∉ R as 1– 2 + 2 = 1 is a rational number. So, R is not symmetric, 1R

2 and 2 2 R

2 . So R is

9. A = {–1, 1}, B = {–1, 1, –i, i} A – B = f, B–A= {–i, i} (A – B) ∪ (B – A) ={–i, i} The correct option is (A) 10. | x | < 1 ⇒ –1 < x < 1  (x = 0 integer satisfies it) The correct option is (A) 11. A – B = A – (A ∩ B) is correct A = (A ∩ B) ∪ (A - B) is correct A

and  B = {x : x ∈ R, x – 1 ≤ – 1 or x – 1 ≥ 1}

2

Since, 1 R 2 and 2 R 2 but 1 is not related to not transitive. The correct option is (A)

B

= {x : x ∈ R, x ≤ 0 or x ≥ 2}

\ A ∪ B = R – D, where D = {x : x ∈ R, 1 ≤ x < 2} The correct option is (B) 5. We know that the interchange of two adjacent rows (or ­columns) changes the value of a determinant only in sign and not in magnitude. Hence corresponding to every element D of B there is an element D′ in C obtained by interchanging two adjacent rows (or columns) in D. 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. The correct option is (B)

A–B

A–(A ∩ B)

(3) is false. \ (1) and (2) are true. The correct option is (D) 12. The set consists of n elements and for relation to be reflexive it must have at least n ordered pairs. It has m ordered pairs therefore m ≥ n. The correct option is (A) 13. n [(A × B) ∩ (B × A)] = n [(A ∩ B) × (B ∩ A)] = n (A ∩ B) ⋅ n (B ∩ A) = n (A ∩ B) . n (A ∩ B) = (99)(99) = 992 The correct option is (B)

Set Theory  1.21 14. Here, R = {(x, y) : |x2 – y2| < 16} and A = { 1, 2, 3, 4, 5} \ R = [(1,2), (1,3), (1,4); (2,1), (2,2), (2,3), (2,4); (3,1), (3,2) (3,3), (3,4); (4,1), (4,2), (4,3); (4,4), (4,5); (5,4), (5,5)] The correct option is (D) 15. Here a ^ b ⇒ b ^ a. Hence, R is symmetric. The correct option is (B) 16. To make it reflexive, we need to add (2, 2), (3, 3), (4,4). To make symmetric, it requires (3, 2), (4, 3) to be added. To make transitive, (2, 4) and (4, 2) must be added, so, the relation.

R = [(2,2), (3,3), (4,4) (2,3),(3, 2), (3, 4), (4,3), (2,4),(4,2)] The correct option is (C) 17. Since 8x ≡ 6 (mod 14) i.e, 8x –6 = 14 k for k ∈ I. The values 6 and 13 satisfy this equation, while 8, 14, and 16 do not. The correct option is (C) 18. Every element has 3 options. Either set Y or set Z or none, so number of ordered pairs = 35. The correct option is (B)

Previous Years’ Questions

20. R is Reflexive and transitive only so not an equivalence relation. e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive] (3, 6), (6, 12), (3, 12) [Transitive]. The correct option is (A) 21. Clearly (x, x) ∈ R ∀ x ∈ W. So, R is reflexive. Let (x, y) ∈ R, then (y, x) ∈ R as x and y have at least one letter in common. So, R is symmetric. But R is not transitive for example Particularly if x = DELHI, y = DWARKA and z = PARK, then (x, y) ∈ R and (y, z) ∈ R but (x, z) ∉ R. The correct option is (B) 12! 22. The required number of ways is 12C4 × 8C4 × 4C4 = ( 4!)3 The correct option is (C) 23. Given T = {(x, y) : x - y ∈ I} As 0 ∈ I, T is a reflexive relation. If x - y ∈ I ⇒ y - x ∈ I \ T is symmetrical also If x - y = I1 and y - z = I2 Then x - z = (x - y) + (y - z) = I1 + I2 ∈ I \ T is also transitive. Hence T is an equivalence relation. Clearly x ≠ x + 1 ⇒ (x, x) ∉ S \ S is not reflexive. The correct option is (D)

26. x − y is an integer x – x = 0 is an integer ⇒ A is Reflexive x − y is an integer ⇒ y − x is an integer ⇒ A is symmetric x − y, y − z are integers As sum of two integers is an integer. ⇒ (x − y) + (y − z) = x − z is an integer ⇒ A is transitive. Hence statement 1 is true. x Also = 1 is a rational number ⇒ B is reflexive x x y = α is rational ⇒ need not be rational x x i.e.,

0 1 is rational ⇒ is not rational 1 0

Hence B is not symmetric ⇒ B is not an equivalence relation. The correct option is (B) 27. n(A) = 4, n(B) = 2 n(A × B) = 8 therefore symbol Number of subsets having at least 3 elements = 28 − (1 + 8C1 + 8C2) = 219 The correct option is (B) 28. The figure is represented as A

B

24. The correct option is (C) 25. P: there is a rational number x ∈ S such that x > 0 ~P: Every rational number x ∈ S satisfies x ≤ 0 The correct option is (B)

The region A fully falls in the region B, hence A is the subset of B Put a = x and b = y

HINTS AND EXPLANATIONS

19. (2, 3) belongs to R but (3, 2) is not. Hence R is not symmetric. The correct option is (C)

1.22  Chapter 1 2 2 B represents an ellipse ( x − 6) + ( y − 5) ≤ 1 9 4 A represent a rectangle lines −1 < x − 5 < 1 4< x 0 z− z =±

2

where b > 0

{

⎫ a2 + b2 + a ⎬ ⎭ z +a

} ⎫ a2 + b2 − a ⎬ t ⎭

⎧ 2⎨ ⎩

}

z − a t

SOLVED EXAMPLES

Equating real and imaginary part, we get

Now,

⎧ 2⎨ ⎩

a + tb + a − tb = ± where b > 0

13. If 3 a − ib = x – iy, then 3 a + ib = (A) x + iy (B)  x – iy (C) y + ix (D)  y – ix Solution: (A) We have,

3

a − ib = x – iy

⇒ a – ib = (x – iy)3 = x3 – 3x2 ⋅ iy + 3x (iy)2 – (iy)3 



\

= (x3 – 3xy2) – i (3x2y – y3) a + ib = (x3 – 3xy2) + i (3x2y – y3)



= x3 + 3x2 ⋅ (iy) + 3x (iy)2 + (iy)3



= (x + iy)3

\ 3 a + ib = x + iy.

Complex Numbers  2.7 14. The complex number z satisfying the equations |z – i| = |z + 1| = 1 is (A) 0 (B) 1 + i (C) –1 + i (D)  1–i

1 = – i i Now, z = i ⇒ |z| = |i| = 1 and z2 = – i ⇒ |z2| = | –i| ⇒ |z|2 = 1 ⇒ |z| = 1

Solution:  (A, C) Let z = x + iy. Then,

Thus, in both cases |z| = 1.

|(x + iy) – i| = |(x + iy) + 1| = 1 or

ARGAND PLANE AND GEOMETRICAL REPRESENTATION OF COMPLEX NUMBERS

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

(1) (2)

From Eq. (1) and (2), x2 + x2 + 2x = 0; or x (x + 1) = 0 \

x = 0, –1;

\

y = 0, 1

\

z = x + iy = 0, –1 + i.

15. The complex number z satisfying the equations |z| – 4 = |z – i| – |z + 5i| = 0, is (A)  3 – i (C)  – 2 3 – 2i

(B) 2 (D) 0

3 – 2i

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,

|z| – 4 = 0 and |z – i| – |z + 5i| = 0

OM = x and PM = y. Join OP

Putting z = x + iy, these equations become |x + iy| = 4 i.e., x2 + y2 = 16

(1)

(2)

Putting y = –2 in (1), x2 + 4 = 16  or  x = ±2. Hence, the complex numbers z satisfying the given equations are z1 = 2 – 2i, and z2 = – 2 – 2i. 16. If i z3 + z2 – z + i = 0, then (A) |z| < 1 (B)  |z| > 1 (C) |z| = 1 (D)  |z| = 0 Solution: (C)

⇒

iz3 + z2 – z + i = 0 i z2 (z – i) – (z – i) = 0 2

⇒ (z – i) (i z – 1) = 0 ⇒ z = i

Let

OP = r and ∠XOP = q.

Then

and |x + iy – i| = |x + iy + 5i| or x2 + (y – 1)2 = x2 + (y + 5)2 i.e. y = –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

Solution:  (B, C) We have two equations

Given,

z2 =

or

z = x + iy = r (cos q + i sin q)

This form of z is called polar or trigonometric form. Comparing real and imaginary parts, we get x = r cos q(1) and y = r sin q(2) Squaring Eq. (1) and (2) and adding, we get r2 = x2 + y2 or r =

x 2 + y 2 = |z|

Thus, r is known and is equal to the modulus of the ­complex number z. Substituting the value of r in Eq. (1) and (2), we get cos q =

x 2

x +y

2

y

and sin q =

Dividing Eq. (2) by (1), we get tan q =

2

x +y2 y . x

(3)

2.8  Chapter 2 The form z = r (cos q + i sinq ) = reiq of the complex number z is called exponential form. Any value of q satisfying (3) is known as amplitude or argument of z and written as q = arg (z) or q = amp z.

I M P O R TA N T P O I N T S The unique value of q such that – p < q ≤ p for which x = r cos q and y = r sin q, is known as the principal value of the argument. The general value of the argument is (2np + q), where n is an integer and q is the principal value of arg (z). While reducing a complex number to polar form, we always take the principal value. The complex number z = r (cos q + i sin q) can also be written as rcisq. r

( c os θ +

i

s in θ )

rcis θ

z 3. arg ⎛⎜ ⎞⎟ = 2 arg z ⎝z ⎠ 4. arg (zn) = n arg z ⎛z ⎞ ⎛z ⎞ 5. If arg ⎜ 2 ⎟ = q, then arg ⎜ 1 ⎟ = 2kp – q where k ∈ I ⎝ z1 ⎠ ⎝ z2 ⎠ 6. arg z = – arg z

( ) = arg (positive real number) = 0

7. arg ( z z ) = arg z

2

SOLVED EXAMPLES 17. The inequality |z – 4 | < |z – 2| represents the region given by, (A)  Re (z) > 0 (B)  Re (z) < 0 (C)  Re (z) > 3 (D)  None of these Solution: (C)

QUICK TIPS If x > 0, y > 0 (i.e., z is in first quadrant), then

Given |z – 4 |2 < |z – 2|2 ⇒ |(x – 4) + iy |2 < | (x – 2) + iy |2

■

⎛ y⎞ arg z = q = tan– 1 ⎜ ⎟ . ⎝ x⎠ ■ If x < 0, y > 0 (i.e., z is in second quadrant), then ⎛ y⎞ arg z = q = p – tan–1 ⎜ ⎟ . x⎠ ⎝|| ■ If x < 0, y < 0 (i.e., z is in third quadrant), then ⎛ y⎞ arg z = q = – p + tan–1 ⎜ ⎟ . ⎝ x⎠ If x > 0, y < 0 (i.e., z is in fourth quadrant), then y⎞ ⎛|| arg z = q = – tan–1 ⎜ ⎟ . ⎝ x⎠ ■ Argument of the complex number 0 is not defined. ⎧0, if x > 0 ■ arg (x + i0) = ⎨ ⎩π , if x < 0

⇒ (x – 4)2 + y2 < (x – 2)2 + y2 ⇒

– 4x < – 12 ⇒ 4x > 12; x > 3

⇒ Re (z) > 3. ⎛π⎞ 18. If zr = cos ⎜ r ⎟ + i sin ⎝3 ⎠

⎛π⎞ ⎜⎝ r ⎟⎠ , r = 1, 2, 3, …, then 3

z1 z2 z3 … ∞ = (A) i (B)  –i (C) 1 (D) –1

■

⎧π /2, if y > 0 arg (0 + iy) = ⎨ . ⎩3π /2, if y < 0

Solution: (A) Since

π π⎞ ⎛ π π⎞ ⎛ = ⎜ cos + i sin ⎟ ⎜ cos 2 + i sin 2 ⎟ ⎝ 3 3⎠ ⎝ 3 3 ⎠

1. arg (z1z2) = arg (z1) + arg (z2) ⎛z ⎞ 2. arg ⎜ 1 ⎟ = arg z1 – arg z2 ⎝z ⎠ 2

⎛π⎞ ⎜⎝ r ⎟⎠ , 3

r = 1, 2, 3, … we have, z1 · z2 · z3 … ∞

■

Properties of Argument

⎛π⎞ zr = cos ⎜ r ⎟ + i sin ⎝3 ⎠



π π⎞ ⎛ ⎜⎝ cos 3 + i sin 3 ⎟⎠ ...∞ 3 3



⎛π π π ⎞ ⎛π π π ⎞ = cos ⎜ + 2 + 3 + ...⎟ + i sin ⎜ + 2 + 3 + ...⎟ ⎝3 3 ⎠ ⎝3 3 ⎠ 3 3

Complex Numbers  2.9





22. Let zk (k = 0, 1, 2, …, 6) be the roots of the equation

⎛ π ⎞ ⎛ π ⎞ ⎜ 3 ⎟ ⎜ ⎟ = cos ⎜ + i sin ⎜ 3 ⎟ 1⎟ 1 ⎜1− ⎟ ⎜1− ⎟ ⎝ 3⎠ ⎝ 3⎠  π π = cos + i sin = 0 + i ⋅ 1 = i 2 2 10

19. The value of

⎛

∑ ⎜⎝ sin

k =1

(A) 1

3 (B)  2 Solution: (C)

(C) i

10

⇒ (zk + 1)7 = – zk7 ⇒ |zk + 1|7 = |zk|7 ⇒ |zk + 1| = |zk| ⇒ |xk + iyk + 1|2 = |xk + iyk|2 2π k 2π k ⎞ − i cos 11 11 ⎟⎠

⎛

∑ ⎜⎝ sin

k =1

 2π k 2π k ⎞ ⎛ 2 = ∑ ⎜ −i sin − i cos ⎝ 11 11 ⎟⎠ k =1

⇒ (xk + 1)2 + y k2 = xk2 + yk2  1 ⇒ 2xk + 1 = 0  or  xk = – 2

10



= –i

10

⎛

∑ ⎜⎝ cos k =1



2π k 2π k ⎞ + i sin =–i 11 11 ⎟⎠



⎤ ⎡ 10 i 2π k = – i ⎢ ∑ e 11 − 1⎥ ⎢ k =0 ⎥ ⎣ ⎦ = – i (sum of 11th roots of unity – 1)



= – i (0 – 1) = i.



20. The argument of

1− i 3

Thus,

10 i 2π k e 11

∑

k =1





1+ i 3

=

k =0

k =0

7 . 2

(A) p (B)  –p π π (C) – (D)  2 2

we take z = r [cos (– q ) + i sin (– q )] (D) –

2π 3

(1 − i 3 ) 4

2

=

3 1 = − − i . 2 2

π π (A) p (B)  (C) – (D) 0 2 2 Solution: (B) Since b > 0, bi represents a point on the positive side of the imaginary axis on which the argument of every π point is . 2

= r (cos q – i sinq ) (–z) r

−2 − 2 3 i 4 

⎛ 1 3 ⎞ 2π i⎟ = – (p – tan–1 3) = – \ arg ⎜ − − 3  ⎝ 2 2 ⎠ 21. arg bi (b > 0) is

6

∑ Re ( zk ) = ∑ x k = –

23. If arg (z) < 0, then arg (–z) – arg (z) =

Solution: (D) 1− i 3

6

Solution: (A) As –q = arg (z) < 0,

is

1+ i 3 π 2π 4π (A)  (B)  (C)  3 3 3



7 7 (D)  2 2

(zk + 1)7 + z k7 = 0

we have

(D) –i

(C) –

is equal to

zk = xk + iyk,

Let

Solution: (C) We have,

∑ Re (z k )

k =0

(A) 0

2π k 2π k ⎞ − i cos is 11 11 ⎟⎠

(B) – 1

6

(z + 1)7 + z7 = 0, then

π –θ

O

–θ r (Z)

⇒ – z = r (– cos q + i sinq )

= r [cos (p – q ) + i sin (p – q )]

\ arg (–z) = p – q Thus, arg (–z) – arg (z) = p – q + (q ) = p

PARTICULAR CASES OF POLAR FORM 1. 1 = 1 + i0 = cos 0 + i sin 0 2. –1 = – 1 + i0 = cos p + i sin p

2.10  Chapter 2 3. i = 0 + i1 = cos

π π + i sin 2 2

⎛ π⎞ ⎛ π⎞ 4. –i = 0 + i (– 1) = cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎝ 2⎠ ⎝ 2⎠ 5. 1 – i = 6. –1 – i =

= 2⋅e



π 6

= 2⋅e

= 2 ⋅ ep   ( e

\ arg z =

2 nπ i

2npi

i/6



⎡ ⎛ π⎞ ⎛ π⎞⎤ 2 ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥ ⎝ ⎠ ⎝ 4⎠⎦ 4 ⎣

i (12 n +1)

πi ⋅e 6



= 1)

π . 6

LOGARITHM OF A COMPLEX NUMBER

⎡ ⎛ 3π ⎞ ⎛ 3π ⎞ ⎤ 2 ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥ ⎝ ⎠ ⎝ 4 ⎠⎦ 4 ⎣

log (x + iy) = loge (reiq) = loge eiq = loge r + iq



= loge



EULERIAN REPRESENTATION OF A COMPLEX NUMBER

⎛ y⎞ ( x 2 + y 2 ) + i tan −1 ⎜ ⎟ ⎝ x⎠

loge (z) = loge |z| + i arg (z)

Since eiq = cos q + i sin q, thus any non zero complex number z = x + iy = r (cos q + i sin q ) can be represented in Eulerian form as



ERROR CHECK

z = reiq = r (cos q + i sin q ),

Since the argument of a complex number is not unique, the log of a complex number cannot be unique. In general, loge (z) = loge |z| + i [2kp + arg (z)], k ∈ I

where |z| = r and q = arg (z).

Info Box! cos q =

e iθ − e − iθ e iθ + e − iθ and sin q = 2i 2

Info Box! iπ

⎛ iπ ⎞ iπ , log (log i) = log ⎜ ⎟ ⎝ 2⎠ 2 iπ ⎛π⎞ = log i + log ⎜ ⎟ = + log (p/2). ⎝ 2⎠ 2

log i = log e 2 =

SOLVED EXAMPLE 24. For any integer n, the argument of z =

( 3 + i ) 4 n +1 (1 − i 3 ) 4 n

π π (B)  6 3 π 2π (C)  (D)  2 3 (E)  All of the above (A) 



z=

=

VECTORIAL REPRESENTATION OF A COMPLEX NUMBER If P is the point (a, b) on the argand plane corresponding to the complex number z = a + ib. Then

Solution: (A) We have,

is

\

( 3 + i ) 4 n +1

and

(1 − i 3 ) 4 n  ⎛ iπ ⎞ ⎜ 2e 6 ⎟ ⎜⎝ ⎟⎠

4 n +1

⎛ −i π ⎞ ⎜ 2e 3 ⎟ ⎜⎝ ⎟⎠

4n

=

2 4 n +1e

i ( 4 n +1)

π − i 4n 3 24n e

π 6

 ˆ aiˆ + bjˆ , OP i=  OP = a 2 + b 2 = |z|

 ⎛ b⎞ arg(z) = direction of the vector OP = tan–1 ⎜ ⎟ . ⎝ a⎠

De’Moivre’s Theorem 

If n is any integer, then (cos q + i sin q)n = cos nq + i sin nq

Complex Numbers  2.11

QUICK TIPS If n is any rational number, then cos nq + i sin nq is one of the values of (cos q + i sinq)n. –n ■ (cos q + i sinq ) = cos (–n)q + i sin (–n)q = cos nq – i sin n q n n ■ (cos q – isinq ) = [cos (–q ) + i sin (–q )] = cos (–nq ) + isin (–nq ) = cos nq – isin nq 1 = (cos q + isin q )–1 = cos q – i sin q ■ cos θ + i sinθ n ■ The theorem cannot be applied to (cos q + isinf) i.e., q must be same with cos and sin both. n ■ The theorem is not directly applicable to (sin q + i cosq ) , rather n ⎡ ⎛π ⎞⎤ ⎛π ⎞ (sin q + icosq )n = ⎢cos ⎜ − θ ⎟ + i sin ⎜ − θ ⎟ ⎥ ⎝2 ⎠⎦ ⎝2 ⎠ ⎣ ■

⎛π ⎞ ⎛π ⎞ = cos n ⎜ − θ ⎟ + i sin ⎜ − θ ⎟ ⎝2 ⎠ ⎝2 ⎠



(cosq1 + i sinq1) (cos q2 + isin q2) … (cosqn + isin qn) = cos (q1 + q2 + … + qn) + isin (q1 + q2 + … + qn) ■

SOLVED EXAMPLES 25. If z = cos q + i sin q, then 1 (A) z n + n = 2 cos nq z 1 (B) z n + n = 2n cos nq z (C)  z n −

(D) z n −

1

= 2i sin nq

zn 1 zn

= (2i)n sin nq

Solution:  (A, C) We have, 1 1 = = cos q – i sin q. z cos θ + i sin θ zn = (cos q + i sinq)n = cos nq + i sin nq, 1 = (cos q – i sinq)n = cos nq – i sin nq zn

\ and

n

Hence,  z + n

and  z −

1 zn

1 zn

= 2 cos nq

= 2i sin nq.

26. If z = cos q + i sin q, then

z 2n − 1

= z 2n + 1 (A) i cot nq (B)  i tan nq (C) tan nq (D)  cot nq (n is an integer) Solution: (B) We have,

z 2n − 1

=

(cos θ + i sin θ )2 n − 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) z 2n + 1



=



=



=

(1 − 2 sin 2 nθ ) + 2i sin nθ cos nθ − 1 (2 cos2 nθ − 1) + 2i sin nθ cos nθ + 1 

i sin nθ cos nθ + i 2 sin 2 nθ 2

cos nθ + i sin nθ cos nθ

 ( i2 = –1)

i sin nθ (cos nθ + i sin nθ ) = i tan nq. cos nθ (cos nθ + i sin nθ )

27. If a = cos a + i sin a, b = cos b + i sin b, a b c c = cos g + i sin g and + + = – 1, then b c a cos (b – g ) + cos (g – a) + cos (a – b ) = (A) 0 (B) 1 (C)  –1 (D)  None of these Solution: (C) We have, 1 a a Now b a or b b Similarly, c c and a

= cos a – i sin a,

1 = cos b – i sin b b

= (cos a + i sina) (cos b – i sinb ) = cos (a – b ) + i sin (a – b ) = cos (b – g ) + i sin (b – g ) = cos (g – a) + i sin (g – a)

a b c Putting these values in + + = –1, b c a we get

[cos (a – b ) + cos (b – g ) + cos (g – a)]



+ i [sin (a – b ) + sin (b – g ) + sin (g – a)] = –1 = – 1 + 0 i

Comparing real part on both sides, we get cos (a – b ) + cos (b – g ) + cos (g – a) = –1

2.12  Chapter 2 28. If n is a positive integer, then ( 3 + i)n + ( 3 – i)n is equal to nπ nπ (A) 2n cos (B) 2n + 1 cos 6 6 nπ n–1 (C) 2 cos (D)  None of these 6 Solution: (B) Let 3 = r cos q and 1 = r sin q so that 1 π r2 = 4 and tan q = ⇒ r = 2, q = 6  3 n π π⎞ ⎛ \ ( 3 + i)n = 2n ⎜ cos + i sin ⎟ ⎝ 6 6⎠  ⎧ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎫ or ( 3 + i)n = 2n ⎨cos ⎜ ⎟ + i sin ⎜ ⎟ ⎬ (1) ⎝ ⎠ ⎝ 6 ⎠⎭ 6 ⎩ Similarly,

⎧ ⎛ nπ ⎞ ⎛ nπ ⎞ ⎫ ( 3 – i)n = 2n ⎨cos ⎜ ⎟ − i sin ⎜ ⎟ ⎬ (2) ⎝ ⎠ ⎝ 6 ⎠⎭ 6 ⎩ Adding Eq. (1) and (2), we obtain ⎛ nπ ⎞ ( 3 + i)n + ( 3 – i)n = 2 ⋅ 2n cos ⎜ ⎟ ⎝ 6 ⎠  ⎛ nπ ⎞ n+1 =2 cos ⎜ ⎟ ⎝ 6⎠  29. If (sin q1 + i cos q1) (sin q2 + i cos q2) … (sin qn + i cos qn) = a + ib, then a2 + b2 = (A) 4 (B) 2 (C)  1 (D)  None of these Solution: (C) Given expression =



n

⎛

⎛π

⎛π

⎞

 n n ⎛π ⎞ ⎛π ⎞ = cos ∑ ⎜ − θ r ⎟ + i sin ∑ ⎜ − θ r ⎟ ⎝ ⎠ ⎝ ⎠ r =1 2 r =1 2 = cos a + i sin a,

where

= a + ib







a2 + b2 = cos2 a + sin2 a = 1.

\ 2

2

–2

30. If z – 2zcosθ + 1 = 0, then z + z is equal to (A) 2cos2θ (B) 2sin2θ (C)  2 cosθ (D)  2 sinθ Solution: (A) We have, z2 – 2zcosθ + 1 = 0

2 cos θ ± 4 cos 2 θ − 4 = cos θ ± cos 2 θ − 1 2 

= cos θ ± − sin 2 θ = cos θ ± i 2 sin 2 θ  = cosθ ± isinθ.



z = cosθ + isinθ

When

z2 + z–2= cos2θ + isin2θ + (cos2θ – isin2θ) = 2cos2θ z = cosθ – isinθ,

and when 2

–2

z + z = cos2θ – isin2θ + cos2θ + isin2θ = 2cos2θ



ROOTS OF A COMPLEX NUMBER If z = r (cos q + i sinq ) and n is a positive integer, then 1 zn

=

1 rn

⎡ ⎛ 2k π + θ ⎞ ⎛ 2k π + θ ⎞ ⎤ + i sin ⎜ , ⎟ ⎢ cos ⎜⎝ ⎠ ⎝ n n ⎟⎠ ⎥⎦ ⎣

where k = 0, 1, 2, 3, … (n – 1).

Cube Roots of Unity z = 11/3 or z3 – 1 = 0

Let

⇒ (z – 1) (z2 + z + 1) = 0 z = 1,

i.e.,

−1 + i 3 −1 − i 3 ,  2 2

Put

w=

−1 + i 3 , 2

then

w2 =

−1 − i 3 . 2

Thus cube roots of unity are 1, w, w2.

n

⎛π ⎞ a = ∑ ⎜ − θr ⎟ ⎝ ⎠ 2 r =1

z=

⎞⎞

∏ ⎜⎝ cos ⎜⎝ 2 − θ r ⎟⎠ + i sin ⎜⎝ 2 − θ r ⎟⎠ ⎟⎠ r =1



⇒

Properties of Cube Roots of Unity 1. 1 + w + w2 = 0 2. w3 = 1 3. w3n = 1, w3n + 1 = w, w3n + 2 = w2 2π i 4. ω = w2 and ( ω )2 = w, w ω = w3, w = e 3 , 2

−

2π i 3

w = e 5. If a + bw + cw2 = 0, then a = b = c provided a, b, c are real. 6. If these roots are marked on the argand plane, then these are vertices of an equilateral triangle with ­circumcentre at origin, as shown in the Fig. 3.1.

Complex Numbers  2.13 = [(1 + w) (1 + w) … to n factors]

Imaginary axis

[(1 + w2) (1 + w2) … to n factors]

–1 , 3 2 2

O

2π 3 2π 3

(1, 0)

= (1 + w)n (1 + w2)n = [(1 + w) (1 + w2)]n



= (1 + w + w2 + w3)n = (0 + 1)n = 1

( 1 + w + w2 = 0, w3 = 1).

Real axis

32. If 1, w, w2 are the three cube roots of unity, then (1 – w + w2) (1 – w2 + w4) (1 – w4 + w8) … to 2n factors = (A) 2n (B)  22n 4n (C) 2 (D)  None of these

–1 , 3 2 2 FIGURE 2.1

Solution: (B) We have,

Fourth Roots of Unity

(1 – w + w2) (1 – w2 + w4) (1 – w4 + w8)

The four, fourth roots of unity are given by the solution set of the equation x4 – 1 = 0

(1 – w8 + w16) … to 2n factors



⇒ (x2 – 1) (x2 + 1) = 0 ⇒ x = ± 1, ± i

= (1 – w + w2) (1 – w2 + w) (1 – w + w2)

Fourth roots of unity are vertices of a square which lies on coordinate axes.

(1 – w2 + w) … to 2n factors.



[ w4 = w, w8 = w2, w16 = w and so on]

Some Useful Relations 1. x2 + y2 = (x + iy) (x – iy) 2. x3 + y3 = (x + y) (x + yw) (x + yw 2) 3. x3 – y3 = (x – y) (x – yw) (x – yw 2) 4. x2 + xy + y2 = (x – yw) (x – yw 2), in particular, x2 + x + 1 = (x – w) (x – w 2) 5. x2 – xy + y2 = (x + yw) (x + yw2), in particular, x2 – x + 1 = (x + w) (x + w2) 6. x2 + y2 + z2 – xy – xz – yz = (x + yw + zw2) (x + yw2 + zw) 7. x3 + y3 + z3 – 3xyz = (x + y + z) (x + w y + w2z) (x + w2y + w z)



= (– 2w) (– 2w2) (– 2w) (– 2w2) … to 2n factors



= (22 w3) (22 w3) … to n factors

[ (– 2w) (– 2w2) = 22 w3 = 22] = (22)n = 22n.



− 1 − − 1 − − 1 − ... to ∞ =

33.

(A) 1

(B) –1

(C) w (D)  w2

Solution:  (C, D)

SOLVED EXAMPLES

x=

Let

2

31. If 1, w, w be the three cube roots of unity, then (1 + w) (1 + w2) (1 + w4) (1 + w8) … to 2n factors = (A) 1 (B) –1 (C)  0 (D)  None of these Solution: (A) We have, 2



4

8

(1 + w) (1 + w ) (1 + w ) (1 + w ) … to 2n factors

 − 1 − x  or x = – 1 – x 2

x=

Then or

− 1 − − 1 − − 1 − ... to ∞

2

x + x + 1 = 0

\

x= =



−1 ± − 3 − 1 ± 1 − 4 ⋅1⋅1 = 2 2 ⋅1  − 1 ± 3i = w or w2. 2

= (1 + w) (1 + w2) (1 + w3 ⋅ w) (1 + w6 ⋅ w2) ... to 2n factors = (1 + w) (1 + w2) (1 + w) (1 + w2) … to 2n factors

⎛ i − 3⎞ ⎛ 3 + i⎞ +⎜ 34. ⎜ ⎟ = ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠

( w3 = w6 = … = 1)

(A) –2

6

6

(B) 2

(C) –1

(D) 1

2.14  Chapter 2

nth Roots of Unity

Solution: (A) We have,

and

Since 1 = cos 0 + i sin 0, therefore,

3+i i 3+i = 2 2i

2

⎛ −1 + 3 i ⎞ =–i ⎜ ⎟ = –iw 2 ⎠ ⎝

⎛ −1 − 3 i ⎞ i− 3 i2 − i 3 2 = =–i ⎜ ⎟ = –iw 2 2 2i ⎠ ⎝ 6

Hence,

= i6(w6 + w12)



= –1 (1 + 1) = –2.

⇒ (z + 1) (z2 + z + 1) = 0 Its roots are – 1, w and w2. The root z = – 1 does not satisfy the equation z1985 + z100 + 1 = 0 but z = w and z  =  w2 satisfy it. Hence, w and w2 are the common roots. 36. (i + 3 )100 + (i – 3 )100 + 2100 = (A) 1 (B) – 1 (C)  0 (D)  None of these Solution: (C) −1 + 3 i 2 2ω ⋅ = 3 = 2 i i 

−1 − 3 i 2 2ω 2 ⋅ = and i– 3 = 2 i i  \ (i + 3 )100 + (i – 3 )100 + 2100

⎛ 2ω ⎞ = ⎜ ⎝ i ⎟⎠ 100

2

100

⎛ 2ω 2 ⎞ +⎜ ⎟ ⎝ i ⎠



= 2100 (w + w2) + 2100



= –2100 + 2100 = 0.

2 rπ n

; r = 0, 1, 2, …, (n – 1) , e(i4p/n), …, e[i2(n – 1)p/n]



= 1, e



= 1, a, a2, a3, …, a n – 1,

Properties of nth Roots of Unity 1. 1 + a + a2 + … + a n – 1 = 0 2. 1 ⋅ a ⋅ a2 ⋅ … a n–1 = (–1)n–1

4. nth roots of unity form a G.P. with common ratio e(i2p/n).

SOLVED EXAMPLES 37. If r is non-real and r = 5 1, then the value of 1 + r + r 2 + r −2 − r −1 is equal to

(

)

(A) 2 (C)  8

(B) 4 (D)  None of these

Solution: (B) |1 + r + r2 + r–2 – r–1| = |1 + r + r2 + r3 – r4| [ r5 = 1 ⇒ r3 ⋅ r2 = 1 or r–2 = r3 and

r4 ⋅ r = 1  or  r–1 = r4] = |1 + r + r2 + r3 + r4 – 2r4|



1 − r5 − 2r 4 = 0 − 2r 4 ( r5 = 1) 1− r

100

+2

(w100 + w200) + 2100

=

i

(i2p/n)

100



i100

=e



3. The n, nth roots of unity lie on the unit circle |z| = 1 and form the vertices of a regular polygon of n sides.

z3 + 2 z2 + 2 z + 1 = 0

i+



where a = e(i2p/n)

35. The common roots of the equations z3 + 2 z2 + 2 z + 1 = 0 and z1985 + z100 + 1 = 0 are (A) –1, w (B)  –1, w2 2 (C) w, w (D)  None of these

We have,

2π r + 0 2π r + 0 + i sin ; r = 0, 1, 2, …, (n – 1) n n 2π r 2π r = cos + i sin ; r = 0, 1, 2, …, (n – 1) n n

= cos



6

⎛ i − 3⎞ ⎛ 3 + i⎞ 6 2 6 ⎜ 2 ⎟ + ⎜ 2 ⎟ = (–iw) + (–iw ) ⎠ ⎠ ⎝ ⎝

Solution: (C) We have,

(1)1/n = (cos 0 + i sin 0)1/n



\

= 2|r|4 = 2(1) = 2 2|1+ r + r

2

+ r −2 − r −1 |

( |r| = 1 as r5 = 1)

= 22 = 4

38. The values of (16)1/4 are (A) ±2, ±2 i (B)  ±4, ±4 i (C) ±1, ±i (D)  None of these

Complex Numbers  2.15 Solution: (A) We have (16)1/4 = (24)1/4 = 2 (1)1/4 = 2 (cos 0 + i sin 0)1/4 

1 1 ⎧ ⎫ = 2 ⎨cos (2 k π + 0) + i sin (2 k π + 0) ⎬ , 4 4 ⎩ ⎭ k = 0, 1, 2, 3 = 2 × 1, 2 × i, 2 × –1, 2 × –i = ±2, ±2i

GEOMETRY OF COMPLEX NUMBERS 1. Distance Formula: The distance between two points P(z1) and Q(z2) is given by PQ = |z2 – z1| = |affix of Q – affix of P| (see Fig. 3.2)

4. Equation of a Straight Line  (i) Parametric form: Equation of a straight line joining the points having affixes z1 and z2 is z = t z1 + (1 – t)z2, where t ∈ R (ii) Non-parametric form: Equation of a straight line joining the points having affixes z1 and z2 is z z1 z2

z 1 z1 1 = 0 z2 1

⇒  z ( z 1 − z 2 ) − z ( z 1 − z 2 ) + z 1z 2 − z 2 z 1 = 0

QUICK TIPS Three points z1, z2 and z3 are collinear if, z1 z1 1 z2 z2 1 = 0 z3 z3 1

■

Q(z2)

If three points A(z1), B(z2), C(z3) are collinear then slope of AB = slope of BC = slope of AC z1 − z2 z2 − z3 z1 − z3 ⇒ = = z1 − z2 z2 − z3 z1 − z3

■

P(z1) FIGURE 2.2

2. Section Formula: If R(z) divides the line segment joining P(z1) and Q(z2) in the ratio m1 : m2(m1, m2 > 0) then m z + m2z1  (i)  For internal division, z = 1 2 m1 + m 2 m1z 2 − m 2 z 1 (ii)  For external division, z = m1 − m 2 3. Equation of the Perpendicular Bisector: If P(z1) and Q(z2) are two fixed points and R(z) is moving point (see Fig. 3.3) such that it is always at equal distance from P(z1) and Q(z2) then locus of R(z) is perpendicular bisector of PQ i.e., PR = QR or |z – z1| = |z – z2| ⇒ |z – z1|2 = |z – z2|2

(iii) General equation of a straight line: The general equation of a straight line is of the form az + az + b, where a is complex number and b is real number.  (iv) Slope of a line: The complex slope of the line a coeff. of z az + az + b = 0 is – =– and real a coeff. of z Re(a) slope of the line az + az + b is – = –i Im( a) (a + a ) . (a − a )  (v) Length of perpendicular: The length of perpendicular from a point z1 to the line az + az + b = 0 is given by

P(z1)

az 1 + az 1 + b a+a

or

2a

.

5. Equation of a circle   (i) The equation of a circle whose centre is at point having affix z0 and radius r is |z – z0 | = r. (ii) If the centre of the circle is at origin and radius r, then its equation is |z| = r (see Fig. 3.4).

R(z)

P(z) Q(z2) FIGURE 2.3

After solving,

az 1 + az 1 + b

z ( z1 − z2 ) + z ( z1 − z2 ) = |z1|2 – |z2|2

r C(z0)

FIGURE 2.4

2.16  Chapter 2 (iii) |z – z0 | < r represents interior of a circle |z – z0| = r whereas |z – z0| > r represents exterior of the circle |z – z0| = r.

||z1| – |z2|| ≤ |z1 + z2| ≤ |z1| + |z2| Thus |z1| + |z2| is the greatest possible value of |z1 + z2| and ||z1| – |z2|| is the least possible value of |z1 + z2|.

■

  A C = AB eiθ  or (z3 – z1) = (z2 – z1)eiq



QUICK TIPS

z3 − z1

or

z2 − z1

= eiq

If z +

■

a + a2 + 4 − a + a2 + 4 and . 2 2 ■ The area of the triangle whose vertices are z, iz and z + iz 1 is |z|2. 2 ■ The area of the triangle with vertices z, wz and z + wz is 3 |z|2. 4 ■ If z1, z2, z3 be the vertices of an equilateral triangle and z0 be the circumcentre, then z12 + z22 + z32 = 3z02 . ■ If z1, z2, z3 be the vertices of a triangle, then the triangle is equilateral iff (z1 – z2)2 + (z2 – z3)2 + (z3 – z1)2 = 0 respectively

 (iv) If A, B and C are three points in argand plane such that AC = AB and ∠CAB = q then use the rotation about A to find eiq, but if AC ≠ AB use coni method.   (v)  If four points z1, z2, z3 and z4 are con-cyclic then (z 4 − z 1 ) (z 2 − z 3 ) = real (z 4 − z 2 ) (z 1 − z 3 )



⎛ (z − z 3 ) (z 4 − z 1 ) ⎞   or arg ⎜ 2 = ±p, 0 ⎝ ( z 1 − z 3 ) ( z 4 − z 2 ) ⎟⎠

QUICK TIPS If z is a variable point and z1, z2 are two fixed points in the argand plane, then 1. |z – z1| = |z – z2| ⇒ Locus of z is the perpendicular bisector of the line segment joining z1 and z2. 2. |z – z1| + |z – z2| = constant (≠ |z1 – z2|) ⇒ Locus of z is an ellipse 3. |z – z1| + |z – z2| = |z1 – z2| ⇒ Locus of z is the line segment joining z1 and z2 4. |z – z1| – |z – z2| = |z – z2| ⇒ Locus of z is a straight line joining z1 and z2 but z does not lie between z1 and z2. 5. |z – z1| – |z – z2| = constant (≠ |z1 – z2|) ⇒ Locus of z is a hyperbola. 6. |z – z1|2 + |z – z2|2 = |z1 – z2|2 ⇒ Locus of z is a circle with z1 and z2 as the extremities of diameter. 7. |z – z1| = k |z – z2|, (k ≠ 1) ⇒ Locus of z is a circle. ⎛ z − z1 ⎞ 8. arg ⎜ = a (fixed) ⇒ Locus of z is a segment of ⎝ z − z2 ⎟⎠ circle. ⎛ z − z1 ⎞ π 9. arg ⎜ = ± ⇒ Locus of z is a circle with z1 and ⎝ z − z2 ⎟⎠ 2 z2 as the vertices of diameter. ⎛ z − z1 ⎞ 10. arg ⎜ = 0 or p ⇒ Locus of z is a straight line ⎝ z − z ⎟⎠

1 = a, the greatest and least values of |z| are z

or

z12 + z22 + z32 = z1z2 + z2z3 + z3z1

or

1 1 1 =0 + + z1 − z2 z2 − z3 z3 − z1

The equation |z – z|2 + |z – z2|2 = k (where k is a real 1 (z1 + z2) number) will represent a circle with centre at 2

■

and radius

1 2

2k − | z1 − z2 |2 provided k ≥

1 |z – z |2. 2 1 2

The one and only one case in which |z1| + |z2| + … + |zn| = |z1 + z2 + … + zn| is that the numbers z1, z2, … zn have the same amplitude. ■ If three points z1, z2, z3 are connected by relation az1 + bz2 + cz3 = 0 where a + b + c = 0, then the three points are collinear. z ■ If z is a complex number, then e is periodic. ■ If three complex numbers are in A.P., then they lie on a straight line in the complex plane. ■

SOLVED EXAMPLES 39. The centre of a square ABCD is at z = 0. If A is z1, then the centroid of triangle ABC is (A) 

2

­passing through z1 and z2.



(B) 

z1 3

π π⎞ ⎛ ⎜⎝ cos 2 + i sin 2 ⎟⎠

z1 (cos p + i sin p) 3

Complex Numbers  2.17 41. If z = x + iy and ‘a’ is a real number such that |z – ai| = |z + ai|, then locus of z is (A) x-axis (B)  y-axis (C) x = y (B)  x2 + y2 = 1

π π (C) z1 ⎛⎜ cos + i sin ⎞⎟ ⎝ 2 2⎠ (D)  None of these Solution: (A) π Since A is z1 and ∠AOB = 2 π π ⎛ ⎞ \ B is z1 cos + i sin ⎜⎝ 2 2 ⎟⎠

Solution: (A) |z – ai| = |z + ai|

We have,

⇒ |x + i (y – a)|2 = |x + i (y + a)|2 ⇒

B

A(z1)

O (0, 0)

C

D

x2 + (y – a)2 = x2 + (y + a)2

⇒ 4ay = 0; y = 0, which is x-axis. 42.

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

Also, c is z1(cos p + i sin p) \ Centroid of DABC is

Solution: (C) We have,

z1 ⎛ π π ⎞ 1 + cos + i sin + cos π + i sin π ⎟ ⎠ 3 ⎜⎝ 2 2  z1 z1 = (1 + 0 + i – 1 + 0) = i 3 3

⇒ |(x – 1) + iy| = |x + i (y + 1)|



=

z1 ⎛ π π⎞ cos + i sin ⎟ ⎜ 3⎝ 2 2⎠ 

|z – 1| = |z + i|

⇒ (x – 1)2 + y2 = x2 + (y + 1)2 ⇒

x + y = 0, which is a line through the origin.

43. The centre of a regular polygon of n sides is located at the point z = 0, and one of its vertex z1 is known. If z2 be the vertex adjacent to z1, then z2 is equal to

40. If z1 and z2 (≠ 0) are two complex numbers such that z1 − z2 = 1, then z1 + z2

2π 2π ⎞ ⎛ ± i sin ⎟ (A) z1 ⎜ cos ⎝ n n⎠

(A) z2 = ikz1, k ∈ R (C) z2 = z1



(B) z2 = kz1, k ∈ R (D)  None of these

Solution: (A) We have, z1 − z 2 z /z − 1 =1⇒ 1 2 = 1 z1 + z 2 z 1/z 2 + 1 ⇒

z1 z −1 = 1 +1 z2 z2

 z1 lies on the perpendicular bisector of the z2 ­segment joining A (–1 + 0i) and B (1 + 0i). ⇒

\ ⇒ \

π π⎞ ⎛ (B)  z1 ⎜ cos ± i sin ⎟ ⎝ n n⎠

π π⎞ ⎛ ± i sin ⎟ (C) z1 ⎜ cos ⎝ 2n 2n ⎠ (D)  None of these Solution: (A) Let A be the vertex with affix z1. There are two possi2π bilities and can be obtained by rotating z1 through either in clockwise or in anti-clockwise direction. n z2 = z1e

±

12π n

  ( |z2| = |z1|) O

z1 = ai for some a ∈ R z2 z2 1 −i = = z1 ai a  z2 = i kz1 for some k ∈ R

C(z2)

A(z1)

B(z2)

2.18  Chapter 2 44. The locus of the complex number z in the Argand 1− iz plane if = 1, is z −i (A)  a circle (B)  x-axis (C) y-axis (D)  None of these Solution: (B) Let z = x + iy 1− iz z −i

Given,

= 1

⇒

1− i ( x + iy ) = 1 x + iy − i

⇒

1 + y − ix = 1 x + i ( y − 1)

⇒

(1 + y )2 + x 2 x 2 + ( y − 1)2

= 1

45. The equation |z – 1|2 + |z + 1|2 = 4 represents on the Argand plane (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

2

(1)

2

2 (x2 + y2 + 1) = 4

\ ⇒ |z| = 1

x2 + y2 = 1 or |z|2 = 1 (since |z| cannot be –ve)

Thus, the Eq. (1) represents all points z on the circle with centre origin and radius unity. 46. The locus of the point z satisfying the condition z −1 π arg = is z +1 3 (A)  a straight line (B)  circle (C)  a parabola (D)  None of these

2 3

y – 1 = 0, which is a circle.

Solution: (D) n

We have,

⎛ z −i ⎞ ⎛ z −i ⎞ w= ⎜ = ⎜ ⎝ 1 + iz ⎟⎠ ⎝ i ( z − i ) ⎟⎠ ⎛ 1⎞ = ⎜ ⎟ ⎝ i⎠



n

n



= (–i)n

\ |w | = |(–i)n| = |–i|n = 1 for all n.

⇒ (x – 1) + y + (x + 1) + y = 4 (Putting z = x + iy) ⇒

x2 + y2 – n

⇒ y = 0, which is the equation of x-axis.

Solution: (D) We have, |z – 1|2 + |z + 1|2 = 4

arg

⎛ z −i ⎞ 47. If w = ⎜ , n integral, then w lies on the unit ⎝ 1 + iz ⎟⎠ circle for (A)  only even n (B)  only odd n (C)  only positive n (D) all n

⇒ 1 + y2 + 2 y + x2 = x2 + y2 – 2 y + 1 ⇒ 4 y = 0

2

z −1 π = 3 z +1 x + iy − 1 π ⇒ arg = (Putting z = x + iy) x + iy + 1 3 y y π ⇒ tan −1 − tan −1 = x −1 x +1 3 ⎛ ⎞ z1 ⎜⎝∵ Arg z = Arg z1 − Arg z2 ⎟⎠ 2  y y − π ⇒ tan −1 x − 1 x2 + 1 = 3 y 1+ 2 x −1 2y π ⇒ = tan = 3 2 2 3 x + y −1 We have,

⇒

⇒ (1 + y)2 + x2 = x2 + (y – 1)2

2

Solution: (B)

\ w lies on unit circle for all n. 48. The equation z z + a z + a z + b = 0, b ∈ R ­represents a circle (not point circle) if (A) |a|2 > b (B)  |a|2 < b (C) |a| > b (D)  |a| < b Solution: (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 This represents a circle (not point circle) if |a|2 > b. 49. If z4 = (z – 1)4, then the roots are represented in the argand plane by the points that are

Complex Numbers  2.19 53. 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 (B) 20 (C)  30 (D)  None of these

(A) collinear (B) concyclic (C)  vertices of a parallelogram (D)  None of these Solution: (A) We have, z4 = (z – 1)4 ⇒

⎛ z − 1⎞ 1/4 ⎜⎝ z ⎟⎠ = 1 =

Solution: (B)

2 nπ i e 4

Area of the triangle on the argand plane formed by the 3 complex numbers – z, iz, z – iz is |z|2. 2 3 2 \ |z| = 600 ⇒ |z| = 20 2

,  n = 0, 1, 2, 3

Since for all these values of z, z −1 = 1 so they lie on the line bisecting perpendicz ularly the join of z = 1 and z = 0. 50. The equation z2 + z 2 – 2|z|2 + z + z = 0 represents a (A) straight line (B) circle (C) hyperbola (D) parabola

54.

Solution: (C) We have, |z + z | + |z – z | = 8

Solution: (D) 2

z +z

We have, 2

2

2

– 2|z| + z + z = 0

⇒ (x + iy) + (x – iy) – 2(x2 + y2) + x + iy + x – iy = 0 (Putting z = x + iy)

⇒ 2|x| + 2|y| = 8 or |x| + |y| = 4

2

⇒ 2x2 + 2 (iy)2 – 2x2 – 2y2 + 2x = 0 1 ⇒ – 4 y2 + 2x = 0  or  y2 = x, 2 which is a parabola. 51. 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 (B) 3 (C) 2 (D)  2

⎛ z + 2i ⎞ 55. If Im ⎜ = 0, then z lies on the curve ⎝ z + 2 ⎟⎠ (A) x2 + y2 + 2x + 2y = 0 (B) x2 + y2 – 2x = 0 (C) x + y + 2 = 0 (D)  None of these Solution: (C) Let z = x + iy Then,

Solution: (B) We have, |z – ω |2 + |z – ω2|2 = λ ⇒

If |z + z | + |z – z | = 8, then z lies on (A)  a circle (B)  a straight line (C)  a square (D)  None of these



λ = |ω – ω2|2 = |ω2 + ω4 – 2ω3|

2

= |ω + ω – 2| = |– 1 – 2| = 3

52. 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 (B)  imaginary axis (C) y = x (D)  y=–x Solution: (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.

z + 2i x + iy + 2i x + ( y + 2) i = = x + iy + 2 ( x + 2) + iy  z+2 [ x + ( y + 2) i ] [( x + 2) − iy ] = ( x + 2) 2 + y 2  =

( 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 + 2 ⎟⎠ which represents a straight line. 56.

The cube roots of unity (A)  lie on the circle |z| = 1 (B)  are collinear (C)  form an equilateral triangle (D)  None of these



2.20  Chapter 2 Solution:  (A, C) Clearly, cube roots of unity 1, w, w2 satisfy |z| = 1. 2

2

Also, ⇒

⎛ 3⎞ ⎛ 3⎞ |1 – w | = ⎜ ⎟ + ⎜ ⎟ = 3 ⎝ 2⎠ ⎝ 2 ⎠ 2

|1 – w | =

3

|w – w2| = | 3 i| = and



(A)  (0, 8) (C)  [1, 9]

(B)  [0, 8] (D)  [5, 9]

Solution: (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).

3

Y

⎛ 1 i 3⎞ |1 – w2| = 1 − ⎜ − −  2 ⎟⎠ ⎝ 2 3 i 3 = + = 3 . 2 2

Therefore, 1, w, w2 form an equilateral triangle. 57. If |z – 1| + |z + 3| ≤ 8, then the range of values of |z – 4| is

(4, 0) (–5, 0)

(–3, 0)

O (1, 0)

(3, 0)

X

Now, |z – 4| is distance of z from (4, 0). Minimum ­distance is 1 and maximum is 9.

Complex Numbers  2.21

PRACTICE EXERCISES Single Option Correct Type 1. If a, b, c, p, q, r are three complex numbers such that p q r a b c + + = 1 + i and + + = 0, then the value a b c p q r p2 q2 r 2 of 2 + 2 + 2 is a b c

8. If ( 3 + i)100 = 299 (a + ib), then b = (A)  3 (B)  2 (C)  1 (D)  None of these

2. The complex numbers sin x + i cos 2x and cos x – i sin 2x are conjugate to each other, for

9. The real value of a for which the expression 1 − i sin α is purely real is 1 + 2i sin α π π (A) (2n + 1) (B) (n + 1) 2 2 (C) np (D)  None of these

(A) x = np (B)  x=0

10. The locus of z which satisfies the inequality

1⎞ ⎛ (C) x = ⎜ n + ⎟ p ⎝ 2⎠



(A) 2i (B)  i (D)  None of these

(D)  no value of x

3. The number of solutions of the equation z2 + |z|2 = 0, where z ∈ C is (A) one (C) three

(B) two (D) infinitely many

4. If w is the nth root of unity, then

(1 + w + w2 + … + wn –1) is

(A) 2

(B) 0

(C) 1

z+

(D) –1

2 |z + 1| + i = 0 is

(A)  2 – i (B)  –2 – i (C) 2 + i (D)  –2 + i 6. z1, z2 are two non-real complex numbers such that z1 z 2 + = 1. Then z1, z2 and the origin z 2 z1 (A)  are collinear (B)  form right angled triangle (C)  form right angle isosceles triangle (D)  form an equilateral triangle a − ib ⎤ ⎡ 7. tan ⎢i log is equal to a + ib ⎥⎦ ⎣ a2 − b 2 (A)  2 (B)  2ab a + b2 2ab

(C) 

2ab 2

a −b

2

(A) x + y > 0 (C) x + y < 0

(B)  x – y < 0 (D)  x – y > 0

11. If centre of a regular hexagon is at origin and one of the vertices on argand diagram is 1 + 2i then its ­perimeter is (A) 2

5 (B) 6

2 (C) 4

5 (D) 6

5

12. If z1, z2, z3 are three complex numbers, then z1 Im ( z2 z3 ) + z2 Im ( z 3 z 1 ) + z3 Im ( z 1z 2 ) is equal to

5. The complex number which satisfies the equation

log0.3 |z – 1| > log0.3 |z – i| is given by,

(D)  ab

(A) 1 (B) –1 (C)  0 (D)  None of these 2z 1 z − z2 13. If is purely imaginary number, then 1 3z 2 z1 + z 2 is equal to 3 2 4 (A)  (B) 1 (C)  (D)  2 3 9

4

14. If x6 = (4 – 3i)5, then the product of all of its roots is (where q = – tan–1 (3/4)) (A) 55 (cos 5q + i sin 5q) (B) –55 (cos 5q + i sin 5q) (C) 55 (cos 5q – i sin 5q) (D) –55 (cos 5q – i sin 5q) 15. |z1 + z2| = |z1| + |z2| is possible if

1 (B)  z2 = z1 (C) arg z1 = arg z2 (D) |z1| = |z2| (A) z2 = z1

16. If z = x + iy, x, y real, then |x| + |y| ≤ k |z|, where k is equal to

PRACTICE EXERCISES

(C) –2i

2.22  Chapter 2 (A)  1 (B)  2 (C)  3 (D)  None of these 17. If (1 + i) (1 + 2i) (1 + 3i) … (1 + ni) = a + ib then 2 × 5 × 10 … (1 + n2) = (A) a – ib (B)  a2 – b2 2 2 (C) a + b (D)  None of these 18. Let z1 = a + ib, z2 = p + iq be two unimodular complex numbers such that Im ( z 1z 2 ) = 1. If w1 = a + ip, w2 = b + iq, then (A)  Re (w1 w2) = 1 (B)  Im (w1 w2) = 1 (C)  Re (w1 w2) = 0

(D)  Im (ω1 ω 2 ) = 1

a b 19. If a + ib = x + iy, then + = x y 2 2 (A)  4 (x + y ) (B)  4 (x2 – y2) 2 2 (C)  2 (x – y ) (D)  None of these 3

20. If z = a + ib where a > 0, b > 0, then 1 1 (A) |z| ≥ (a – b) (B) |z| ≥ (a + b) 2 2 1 (C) |z| < (a + b) (D)  None of these 2 21. The complex numbers z1, z2 and z3 satisfying

z1 − z 3 = z2 − z3

1 − 3i are the vertices of a triangle which is 2

PRACTICE EXERCISES

(A)  of area zero (C)  equilateral

(B)  right angled isosceles (D)  obtuse angled isosceles

22. If (1 + x + x2)n = a0 + a1 x + a2 x2 + … + a2n x2n, then a0 + a3 + a6 + … = (A) 3n (B)  3n – 1 n–2 (C) 3 (D)  None of these 23. 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 (B)  n (C) n – 1 (D)  None of these 24. The closest distance of the origin from a curve given as = 0 (a is a complex number) is (A) 1

(B) 

|a| Re a Im a (C)  (D)  2 |a| |a|

25. The roots of the equation z4 + 1 = 0 are (A) (± 1 ± i) (B)  (± 2 ± 2i) 1 (C)  (± 1 ± i) (D)  None of these 2

26. The integral solution of the equation (1 – i)n = 2n is (A) n = 0 (B)  n=1 (C) n = – 1 (D)  None of these 27. The greatest value of the moduli of complex numbres z satisfying the equation (A)  (C) 

z−

4 = 2 is z

5 (B)  5 –1 5 + 1 (D)  None of these

28. The locus of the complex number z in an argand plane satisfying the equation π Arg (z + i) – Arg (z – i) = is

2

(A)  boundary of a circle (C)  exterior of a circle

(B)  interior of a circle (D)  None of these

z2 is always real, then z −1 (A) z lies only on a circle (B) z lies only on the real axis (C) z lies either on the real axis or on a circle (D)  None of these z − 2z 2 30. z1 and z2 are two complex numbers such that 1 2 − z 1z 2 is unimodular whereas z2 is not unimodular. Then |z1| = (A) 1 (B) 2 (C) 3 (D) 4 29.

31. If for the complex numbers z1 and z2, |z1 + z2| = |z1 – z2|, then amp z1 ~ amp z2 = π (A) p (B)  2 π (C)  (D)  None of these 4 32. The locus of the complex number z in an argand plane satisfying the inequality ⎛ | z − 1| + 4 ⎞ log1/ 2 ⎜ >1 ⎝ 3 | z − 1| − 2 ⎟⎠

2⎞ ⎛ ⎜⎝ where | z − 1| ≠ 3 ⎟⎠ is

(A)  a circle (B)  interior of a circle (C)  exterior of a circle (D)  None of these

33. The equation z3 + iz – 1 = 0 has

(A)  three real roots (B)  one real root (C)  no real roots (D)  no real or complex roots

Complex Numbers  2.23

35. Let z1 and z2 be two complex numbers such that z1 z2 + = 1, then z2 z1 (A) z1, z2 are collinear (B) z1, z2 and the origin from a right angled triangle (C) z1, z2 and the origin form an equilateral triangle (D)  None of these 36. If S (n) = i n + i –n, where i = −1 and n is a positive integer, then the total number of distinct values of S (n) is (A) 1 (B) 2 (C) 3 (D) 4 37. If z1 ≠ –z2 and |z1 + z2| =

1 1 + , then z1 z 2

(A)  at least one of z1, z2 is unimodular (B) z1 × z2 is unimodular (C) both z1, z2 are unimodular (D)  None of these 38. If z = x + iy satisfies amp (z – 1) = amp (z + 3i) then the value of (x – 1) : y is equal to (A)  2 : 1 (B)  – 1 : 3 (C)  1 : 3 (D)  None of these 39. Let z be a complex number with modulus 2 and argu2π ment , then z is equal to 3 (A) –1 + i 3

(B)  1 – i 3

1 i 3 (C) − + 2 2

(D)  None of these

⎛ | z |2 − | z | +1⎞ 40. If log 3 ⎜ ⎟ < 2, then the locus of z is ⎝ 2 + |z | ⎠ (A) |z| < 5 (B)  |z| = 5 (C) |z| > 5 (D)  None of these 41. If z1 and z2 are complex numbers, such that z1 + z2 is a real number, then (A) z1 = – z2 (B) z2 = z 1 (C) z1 and z2 are any two complex numbers (D) z1 = z1 , z2 = z 2

42. The locus of the points representing the complex numbers which satisfy |z| – 2 = 0, |z – i| – |z + 5i| = 0 is:

(A)  a circle with centre at origin (B)  a straight line passing through origin (C)  the single point (0, –2) (D)  None of these

43. Let the affix of 2 – 4i be P. Then OP is rotated about O through an angle of 180° and is stretched 5/2 times. The complex number corresponding to the new position of P is (A)  5 – 10i (C) –5 + 10i

(B) 5 + 10i (D)  None of these

44. If P, P′ represent the complex number z1 and its additive inverse respectively then the complex equation of the circle with PP′ as a diameter is (A) 

z ⎛z ⎞ = 1 z 1 ⎜⎝ z ⎟⎠

(B)  zz + z 1z 1 = 0

(C)  zz1 + zz1 (D)  None of these 45. If a, b, c, p, q, r are three non-zero complex numbers p q r a b c such that + + = 1 + i and + + = 0, then a b c p q r 2 2 2 p q r value of 2 + 2 + 2 is a b c (A) 0 (B) –1 (C) 2i (D) –2i z1 − z2 =1 z1 + z2 and tz1 = kz2 where k ∈  , then the angle between (z1 – z2) and (z1 + z2) is

46. If z1, z2 are two complex numbers such that

⎛ 2k ⎞ (A) tan–1 ⎜ 2 ⎟ ⎝ k + 1⎠

⎛ 2k ⎞ (B) tan–1 ⎜ ⎝ 1 − k 2 ⎟⎠

(C)  –2 tan–1(k) (D)  2 tan–1(k) 24

2

1⎞ ⎛ 47. 1 + x = 3x , then ∑ ⎜ x n − n ⎟ is equal to ⎝ x ⎠ n =1 (A) 48 (B)  – 48 (C) ±48(w – w2) (D) 1 ± 48w 2

48. For all complex numbers z1, z2 satisfying |z1| = 12 and |z2 – 3 – 4i| = 5, the minimum value of |z1 – z2| is (A) 0 (B) 2 (C) 7 (D) 17

PRACTICE EXERCISES

34. If all the roots of z3 + az2 + bz + c = 0 are of unit ­modulus, then (A) |a| ≤ 3 (B)  |b| > 3 (C) |c| ≤ 3 (D)  None of these

2.24  Chapter 2 49. For any two complex numbers z1 and z2 with |z1| ≠ |z2| 2

2 z 1 + i 3z 2 +

3z 1 + i 2 z 2

2

is

50. If the roots of (z – 1)25 = 2w2(z + 1)25 where w is a complex cube root of unity are plotted in the argand plane, they lie on (A)  a straight line (B)  a circle (C)  an ellipse (D)  None of these 51. Let A0A1A2A3A4A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0A1, A0A2 and A0A4 is 3 4



(B)  3 3

(C) 3

(D) 

3 3 2

52. If z1 and z2 are the two complex roots of equal magniπ tude and their arguments differ by , of the quadratic 2 equation ax2 + bx + c = 0 (a ≠ 0) then a (in terms of b and c) is b 2 2c b (C)  2c (A) 

PRACTICE EXERCISES

53. 54.

(B) 

(B)  d2 – abd + b2c = 0 (D)  None of these

57. If |z – i| ≤ 2 and z0 = 5 + 3i, the maximum value of |iz + z0| is (A) 7 (B) 9 (C)  13 (D)  None of these

(A)  less than 5 |z1|2 + |z2|2 (B)  greater than 10 |z1z2| (C)  equal to 2|z1|2 + 3 |z2|2 (D) zero

(A) 

(A) d2 – abd – c2 = 0 (C) d2 + abd + c2 = 0

b2 c

(D)  None of these

–1 ⎡1

⎤ sin ⎢ ( z − 1) ⎥ , where z is non-real, can be the angle ⎣i ⎦ of a triangle if (A)  Re (z) = 1, Im (z) = 2 (B)  Re (z) = 1, –1 ≤ Im (z) ≤ 1 (C)  Re (z) + Im (z) = 0 (D)  Re (z) = Im (z) 2 5 1⎞ ⎛ If x2 – x + 1 = 0 then the value of ∑ ⎜ x n + n ⎟ is ⎝ x ⎠ n =1

(A) 8 (B) 10 (C)  12 (D)  None of these 1+ i 55. The triangle formed by the points 1, and i as 2 ­vertices in the Argand diagram is (A) scalene (B) equilateral (C) isosceles (D) right-angled 56. If the quadratic equation z2 + (a + ib)z + c + id = 0, where a, b, c, d are non-zero real numbers, has a real root, then

(

)

58. The solutions of the equation z z − 2i = 2 (2 + i) are

(A) 3 + i, 3 – i (C) 1 + 3i, 1 – i

(B) 1 + 3i, 1 – 3i (D)  1 – 3i, 1 + i

z2 is real, then the point represented by z −1 the complex number z lies (A)  either on the real axis or on a circle passing through the origin. (B)  on a circle with centre at the origin. (C) either on the real axis or on a circle not passing through the origin. (D)  on the imaginary axis. 59. If z ≠ 1 and

60. Two circles in complex plane are C1: |z – i| = 2 C2: |z – 1 – 2i| = 4. Then (A) C1 and C2 touch each other. (B) C1 and C2 intersect at two distinct points. (C) C1 lies within C2. (D) C2 lies within C1. 61. If z1z2 ∈ C, z21 + z22 ∈ R, z1(z21 – 3z22) = 2 and z2(3z21 – z22) = 11, then the value of z21 + z22 is (A) 2 (B) 3 (C) 4 (D) 5 62. If 1 − C 2 = nc – 1 and z = eiq, then

c ⎛ (1 + nz ) ⎜1 + ⎝ 2n

n⎞ = z ⎟⎠

(A) 1 + c cosq (C) 1 + 2c cosq

(B)  1 – c cosq (D)  1 – 2c cosq

63. Let ‘a’ be a complex number such that | a | < 1 and z1, z2,..., zn be the vetices of a polygon such that zk = 1 + a + a2 + … + ak, then the vertices of the polygon lie within the circle 1 1 (A) | z | = (B)  |z – a | = |1 − a | |1 − a | (C)  z −

1 1 = |1 − a | 1− a

(D)  None of these

Complex Numbers  2.25

4

4

65. If z = (z – 1) , 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 66. If a, b, c are real, a2 + b2 + c2 = 1 and b + ic = (1 + a)z, 1 + iz then = 1 − iz a − ib (A)  1+ c

(B) 

a + ib 1+ c

a + ib (C)  1− c

(D) 

a − ib 1− c



67. If z1, z2 are two complex numbers and wk, k = 0, 1, …, n −1

n – 1 are the nth roots of unity, then

∑

k =0

(A)  < n (|z1|2 + |z2|2) (C)  > n (|z1|2 + |z2|2)

| z1 + z2 w k |2

(B)  = n (|z1|2 + |z2|2) (D)  can’t say

68. The equation |z – z1|2 + |z – z2|2 = k, k ∈ R represents a circle if 1 1 (A) k ≥ | z 1 − z 2 |2 (B)  k ≤ | z1 − z2 |2 2 2 1 1 (C) k ≥ | z 1 + z 2 |2 (D)  k ≤ | z 1 + z 2 |2 2 2 69. f (z) when divided by z – i gives remainder i; when divided by z + i gives remainder i + 1. When f (z) is divided by z2 + 1, the remainder is i 1⎞ ⎛ (A)  z + ⎜ i − ⎟ ⎝ 2 2⎠

i 1⎞ ⎛ (B)  z − ⎜ i + ⎟ ⎝ 2 2⎠

i 1⎞ ⎛ (C)  z + ⎜ i + ⎟ ⎝ 2 2⎠

−i 1⎞ ⎛ (D)  z + ⎜ i + ⎟ ⎝ 2 2⎠ 2

3

70. The value of the expression (w – 1) (w – w ) (w – w ) … (w – wn–1), where w is the nth root of unity, is (A)  nwn–1 (B)  nwn n (C) (n – 1) w (D) (n – 1) wn–1

π 71. If |z – i| = 1 and arg (z) = q, θ ∈ ⎛⎜ 0, ⎞⎟ , then the value ⎝ 2⎠ 2 of cot θ − is equal to z (A) 0 (B) i (C) – i (D) 1 4 + 3i 72. The reflection of the complex number in the + 2i 1 straight line iz = is (A) 2 + i (B)  2–i (C) 1 + 2i (D)  1 – 2i 73. If i = −1 , then ⎛ 1 i 3⎞ 4 + 5 ⎜ − + 2 ⎟⎠ ⎝ 2

334

⎛ 1 i 3⎞ +3 ⎜− + 2 ⎟⎠ ⎝ 2

365

is equal to

(A)  1 – i 3 (B)  – 1 + i 3 (C)  i 3 (D)  – 3i 74. Let b z + b z = c, b ≠ 0, be a line in the complex plane, where b is the complex conjugate of b. If a point z1 is the reflection of a point z2 through the line, then z 1 b + z2 b = (A) 4c (B)  2c (C) c (D)  None of these 75. Let z1 and z2 be roots of the equation z2 + pz + q = 0, where the coefficients p and q may be complex ­numbers. Let A and B represent z1 and z2 in the complex plane. If ∠AOB = a ≠ 0 and OA = OB, where O is α the origin, then p2 = k cos2 , where k = 2 (A) q (B)  2q (C) 4q (D)  None of these 76. If z1, z2, z3 are complex numbers such that |z1| = |z2| = |z3| =

1 1 1 + + = 1, then |z1 + z2 + z3| is z1 z2 z3

(A)  equal to 1 (C)  greater than 3

(B)  less than 1 (D)  equal to 3

77. If |z| ≤ 1, |w | ≤ 1, then |z – w |2 (A) ≤ (|z| – |w |)2 – (Arg z – Arg w)2 (B) ≤ (|z| – |w |)2 + (Arg z – Arg w)2 (C) ≤ (|z| – |w |)2 + 2 (Arg z – Arg w)2 (D)  None of these 78. 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

PRACTICE EXERCISES

64. All the roots of the equation a1z3 + a2z2 + a3z + a4 = 3, where |ai| ≤ 1, i = 1, 2, 3, 4 lie outside the circle with centre origin and radius 1 2 (A)  (B)  3 3 (C)  1 (D)  None of these

2.26  Chapter 2 79. If k = k

3n , where n is an even positive integer, then 2

∑ (−3)r −1 ⋅ 3n C2r −1 = r =1

(A) 0 (C)  – 1

(B) 1 (D)  None of these

80. If a and b are real numbers between 0 and 1 such that the points z1 = a + i, z2 = 1 + bi and z3 = 0 form an equilateral triangle, then a and b are (A) 2 + 3 , 2 – 3 (C)  2 –

3 , 2 + 3

(B)  2 –

3,2– 3

(D)  None of these

81. Let z1 and z2 be complex numbers such that z1 ≠ z2 and |z1| = |z2|. If z1 has positive real part and z2 has negative z + z2 imaginary part, then 1 may be z − z2 (A)  0 1 (B)  real and positive (C)  real and negative (D)  purely imaginary 82. 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 (B) 2 (C)  4 (D)  None of these

PRACTICE EXERCISES

83. If the origin and the two points represented by complex numbers A and B form vertices of an equilateral A B triangle, then + = B A (A) 1 (B) – 1 (C)  2 (D)  None of these 84. If 2 2 x 4 = ( 3 – 1) + i ( 3 + 1), then 1 1 x = cos (2np + k) + i sin (2np + k); 4 4 n = 0, 1, 2, 3, where k = π 5π (A)  (B)  12 12 (C) 

7π 12

(D)  None of these p

⎡ 10 ⎛ 2q π 2q π ⎞ ⎤ ⎥ = 85. (3 p + 2) ⎢ ⎜ sin − i cos 11 11 ⎟⎠ ⎥ ⎢⎣ q =1 ⎝ p =1 ⎦ 32

∑

∑

(A)  8 (1 – i) (B)  16 (1 – i) (C)  48 (1 – i) (D)  None of these 86. The three vertices of a triangle are represented by the complex numbers 0, z1 and z2. If the triangle is ­equilateral, then

(A) z 12 + z 22 + z 1 z 2 = 0

(B)  z 12 + z 22 = z1 z2

(C) z 22 − z 12 = z1 z2

(D) z 12 − z 22 = z1 z2

87. If |z – 25i| ≤ 15, then |maximum amp (z) – minimum amp (z)| is equal to ⎛ 3⎞ ⎛ 3⎞ (A) sin–1 ⎜ ⎟ – cos–1 ⎜ ⎟ ⎝ 5⎠ ⎝ 5⎠

π ⎛ 3⎞ (B)  + cos −1 ⎜ ⎟ ⎝ 5⎠ 2

3 (C)  p – 2 cos–1 ⎛⎜ ⎞⎟ ⎝ 5⎠ ⎛ 3⎞ (D) cos–1 ⎜ ⎟ ⎝ 5⎠ 88. If z2 + (p + iq) z + r + is = 0 where p, q, r, s are n­ on-zero, has real roots, then (A) pqs = s2 + q2r (B)  pqr = r2 + p2s 2 2 (C) prs = q + r p (D)  qrs = p2 + s2q 89. If z1 and z2 are any two complex numbers, then z 1 + z 12 − z 22 + z 1 − z 12 − z 22 is equal to (A) |z1 + z2| (B)  |z1| (C) |z2| (D)  None of these z 90. If z = x + iy lies in IIIrd quadrant, then also lies in z the IIIrd quadrant if (A) y > x > 0 (B)  y < x < 0 (C) x < y < 0 (D)  x > y > 0 91. If in an argand plane points z1, z2, z3 are the vertices of an isosceles triangle right angled at z2, then (A)  z12 + 2 z22 + z32 = 2 z2 (z1 + z3) (B) z 12 + z 22 + z 32 = 2 z2 (z1 + z3) (C) z 12 + z 22 + 2 z 32 = 2 z2 (z1 + z3) (D) 2 z12 + z22 + z32 = 2 z2 (z1 + z3) 92. In the Argand diagram, if O, P and Q represent respectively the origin and the complex numbers z and z + iz, then the ∠OPQ is π π π 2π (A)  (B)  (C)  (D)  4 3 2 3 93. If z satisfies |z + 1| < |z – 2|, and w = 3z + 2 + i, then (A) |w + 1| < |w – 8| (B)  |w + 1| < |w – 7| (C) w + ω > 7 (D)  |w + 5| < |w – 4| 94. 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

Complex Numbers  2.27

⎛ 8π ⎞ ⎛ 8π ⎞ 95. If a = cos ⎜ ⎟ + i sin ⎜ ⎟ , then ⎝ 11 ⎠ ⎝ 11 ⎠ Re (a + a2 + a3 + a4 + a5) is equal to 1 1 (A)  (B)  – 2 2 (C)  0 (D)  None of these 96. If A, B, C are the angles of a triangle and eiA, eiB, eiC are in A.P., then the triangle must be (A)  right angle (B)  isosceles triangle (C)  equilateral (D)  None of these 97.

e

2 mi cot −1 p

⎛ pi + 1⎞ ⋅⎜ ⎝ pi − 1⎟⎠

m

(A) 0 (C)  – 1

= (B) 1 (D)  None of these

98. If z1 and z1 represent adjacent vertices of a regular Im ( z 1 ) polygon of n sides and if = 2 – 1, then n Re ( z 1 ) is equal to (A) 4 (C)  16

(B) 8 (D)  None of these

102. If z1, z2 are two complex numbers and c > 0 such that |z1 + z2|2 ≤ (1 + c) |z1|2 + k |z2|2, then k = (A)  1 – c (B)  c – 1

(C)  1 + c–1 (D)  1 – c–1

103. If |z – 4 + 3i| ≤ 1 and m and n are the least and greatest x4 + x2 + 4 values of |z| and k is the least value of on x the interval (0, ∞), then k is equal to (A)  m (B)  n (C)  m + n (D)  None of these 104.

If n > 1, then the roots of zn = (z + 1)n lie on a (A) circle (B)  straight line (C) parabola (D)  None of these

105. 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

106. If 1, a1, a2, a3 and a4 be the roots of x5 – 1 = 0, then

ω −α ω −α ω −α ω −α 2 1 ⋅ 2 2 ⋅ 2 3 ⋅ 2 4 = ω − α1 ω − α 2 ω − α 3 ω − α 4 (A)  1 (B)  w (C)  w2 (D)  None of these

99. If z1, z2, z3 are non-zero, non-collinear complex num2 1 1 bers such that = + , then the points z1, z2, z3 z1 z 2 z 3 lie (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

107. If z1 and z2 both satisfy the relation z + z = 2|z – 1| π and arg (z1 – z2) = , then the imaginary part of 4 (z1 + z2) is

100. If |z – 25 i| ≤ 15, then the least positive value of arg z is 4 4 (A)  p – tan– 1 (B) tan– 1 3 3 –1 4 (C) –p + tan (D)  None of these 3

108. If

101. If |z – 4 + 3i| ≤ 2, then the least and the greatest ­values of |z| are (A)  3, 7 (B)  4, 7 (C)  3, 9 (D)  None of these

1 1 1 1 2 (C)  + + + =– 2 2 2 2 2 a+ω b +ω c +ω d +ω ω (D)  abc + bcd + abd + acd = 4

(A) 0 (C)  2

(B) 1 (D)  None of these

1 1 1 1 + + + =, where a, b, c are a+ω b +ω c +ω d +ω real and w is a non-real cube root of unity, then (A)  a + b + c + d = – 2abcd 1 1 1 1 (B)  + + + =2 1+ a 1+ b 1+ c 1+ d

PRACTICE EXERCISES

(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

2.28  Chapter 2 109. If z1 + z2 + z3 = A, z1 + z2 w + z3 w2 = B and z1 + z2 w2 + z3 w = C, where 1, w, w2 are the three cube roots of unity, then |A|2 + |B|2 + |C|2 = (A)  3 (|z1|2 + |z2|2 + |z3|2) (B)  2 (|z1|2 + |z2|2 + |z3|2) (C) (|z1|2 + |z2|2 + |z3|2) (D)  None of these 1 110. If a, b are the roots of z + = 2(cos q + sin q ) Then, z (A) |a – i| > |b – i| (B) |a – i| < |b – i| (C) |a – i| = |i – b | (D) |a – i| = |b – i| 111. If at least one value of the complex number z = x + iy satisfies the condition z + inequality z + i

2 = a2 – 3a + 2 and the

2 < a2 , then

(A)  a > 2 (B)  a=2 (C)  a < 2 (D)  None of these 112. If a is the nth root of unity, then 1 + 2a + 3a2 + … to n terms is equal to n n − (A) − (B)  1−α (1 − α ) 2 2n 2n (C) − (D)  − 1−α (1 − α ) 2 113. Let O, A, B be three collinear points such that

OA  ·  OB  = 1. If O and B represent the complex ­numbers o and z, then A represents 1 1 (A)  (B)  z z (C) z (D)  z2 114. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. If the points D and M represent the complex numbers 1 + i and 2  –  i, respectively, then A represents the complex number i i i 3 (A) 3 − or 3 + (B)  3 + or 1 + i 2 2 2 2 (C)  3 – i or 1 – 3i

(D)  None of these

115. The locus represented by the complex equation ⎛π ⎞ |z – 2 – i| = | z | sin ⎜ − arg z ⎟ is the part of ⎝4 ⎠ (A)  a pair of straight lines (B)  a circle (C)  a parabola (D)  a rectangular hyperbola 116. If z1, z2, z3 are three points lying on the circle |z| = 2, then the minimum value of |z1 + z2|2 + |z2 + z3|2 + |z3 + z1|2 is equal to (A) 6 (B) 12 (C) 15 (D) 24

PRACTICE EXERCISES

Previous Year’s Questions 117. If ω is an imaginary cube root of unity, then (1 + ω – ω 2)7 equals: [2002] (A) 128 ω (B)  –128 ω (C) 128 ω2 (D) –128 ω2 118. Let z1 and z2 be 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  [2003] (A) a2 = b (B)  a2 = 2b (C) a2 = 3b (D)  a2 = 4b 119. If z and ω are two non-zero complex numbers such π that | zω | = 1, and Arg (z) − Arg (ω) = , then Z ω 2 is equal to [2003] (A) 1 (B) –1 (C) i (D)  −i x

⎛1+ i⎞ 1 20. If ⎜ = 1 , then ⎝ 1 − i ⎟⎠

[2003]

(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 121. Let z, w be complex numbers such that z + iw = 0 and arg zw = π. Then arg z equals [2004] (A)

π 5π (B)  4 4

(C)

3π π (D)  4 2



122. If z = x − iy and

1 z3

= p + iq , then

 (A) 1 (C) 2

⎛ x y⎞ ⎜⎝ p + q ⎟⎠ ( p2 + q2 )

is equal to

[2004] (B) –2 (D) –1

Complex Numbers  2.29 If | z2 –1| = | z |2+ 1, then z lies on (A)  the real axis (B)  an ellipse (C)  a circle (D)  the imaginary axis.

[2004]

124. If the cubes roots of unity are 1, ω, ω2 then the roots of the equation [2005] 3 (x − 1) + 8 = 0, are (A)  −1, −1 + 2ω, −1 −ω2 (B)  −1, −1, −1 (C)  −1, 1 − 2ω, 1 − 2ω2 (D)  −1, 1 + 2ω, 1 + 2ω2 125. If z1 and z2 are two non-zero complex numbers such that | z1 + z2 | = | z1 | + | z2 | then argz1− argz2 is equal to  [2005] π (A)  (B)  −π 2 π (C) 0 (D)  − 2 z 126. If w = and | w | = 1, then z lies on [2005] 1 z− i 3 (A)  an ellipse (B)  a circle (C)  a straight line (D)  a parabola

*131. If Z − 4 = 2 , then the maximum value of | Z | is Z equal to [2009] (A)  3 + 1

(B) 

(C) 2

(D) 2 +

[2006]

133. Let α, β be real numbers and z a complex number. If z2 + α z+ β = 0 has two distinct roots on the line Re(z) = 1, then it is necessary that [2011] (A)  β ∈(−1, 0) (B)  | β | = 1 (C)  β ∈(1,∞) (D)  β ∈(0, 1) 134. If ω (≠ 1) is a cube root of unity, and (1 + ω)7 = A + Bω. Then (A, B) equals [2011] (A)  (1, 1) (B)  (1, 0) (C)  (–1, 1) (D)  (0, 1)

128. If z + z + 1 = 0, where z is a complex number, then the value of 2

2

2

2

1⎞ 1⎞ 1⎞ 1⎞ ⎛ ⎛ ⎛ ⎛ ⎜ z + ⎟ + ⎜ z 2 + 2 ⎟ + ⎜ z 3 + 3 ⎟ + ... + ⎜ z 6 + 6 ⎟ ⎝ ⎝ ⎝ ⎝ z⎠ z ⎠ z ⎠ z ⎠ is [2006] (A) 18 (B) 54 (C) 6 (D) 12 129. If | z + 4 | ≤ 3, then the maximum value of |z + 1| is  [2007] (A) 4 (B) 10 (C) 6 (D) 0 1 130. The conjugate of a complex number is Then the i − 1 [2008] complex number is −1 (A)  i −1 −1 (C)  +1 i

1 (B)  i +1 1 (D)  i +1

z2 is real, then the point which is repz −1 resented by the complex number z lies [2012] (A) either on the real axis or on a circle passing through the origin (B)  on a circle with centre at the origin (C) either on the real axis or on a circle not passing through the origin (D)  on the imaginary axis

135. If z ≠ 1 and

2

2

132. The number of complex numbers z such that | z− 1| = |z + 1| = |z−i| equals [2010] (A) 1 (B) 2 (C)  ∞ (D) 0

10

2k π 2k π ⎞ ⎛ 127. The value of ∑ ⎜ sin + i cos ⎟ is ⎝ 11 11 ⎠ k =1  (A)  i (B) 1 (C)  –1 (D)  −i

5+1

136. If z is a complex number of unit modulus and argu⎛ 1+ z ⎞ ment θ , then ⎜ equals [2013] ⎝ 1 + z ⎟⎠ π (A)  − θ (B)  θ 2 (C)  π − θ (D)  −θ 137. If z is a complex number such that z ≥ 2 , then the 1 minimum value of z +  [2014] 2 5 2 (B)  lies in the interval (1, 2) (A)  is equal to

(C)  is strictly greater than

5 2

(D)  is strictly greater than

3 5 but less than 2 2

PRACTICE EXERCISES

123.

2.30  Chapter 2 138. A complex number z is said to be unimodular if | z | = 1. Suppose z1 and z2 are complex numbers such z − 2 z2 that 1 is unimodular and z2 is not unimodular. 2 − z1 z2 Then the point z1 lies on a [2015] (A)  straight line parallel to y-axis. (B)  circle of radius 2. (C)  circle of radius 2. (D)  straight line parallel to x-axis. 139. A value of θ for which nary, is

2 + 3i sin θ is purely imagi1 − 2i sin θ [2016]

⎛ 1 ⎞ (A) sin −1 ⎜ ⎟ ⎝ 3⎠

π (B)  3

⎛ 3⎞ π (C) (D)  sin −1 ⎜ ⎟ 6 ⎝ 4 ⎠ 5

5

    140. Let z =  3 + i  +  3 − i  . If R(z) and I(z)  2 2  2 2     respectively denote the real and imaginary parts of z, then: [2019] (A) R(z) = – 3 (B) R(I) < 0 and I(z) > 0 (C) I(z) = 0 (D) R(z) > 0 and I(z) > 0

ANSWER K EYS

PRACTICE EXERCISES

Single Option Correct Type   1. (A) 2. (D) 3. (D) 4.  (B) 5. (B) 6.  (D) 7. (C) 8. (A) 9. (C)  10.  (D)   11. (D) 12.  (C) 13. (B) 14.  (B) 15. (C) 16.  (B) 17. (C) 18.  (D) 19. (B)  20.  (B)   21. (C) 22.  (B) 23. (B) 24. (B) 25.  (C) 26. (A) 27.  (C) 28. (A) 29.  (C)   30. (B)   31. (B) 32. (C) 33.  (C) 34. (A) 35.  (C) 36. (C) 37.  (B) 38. (C)  39. (A) 40.  (A)   41.  (B) 42. (C) 43.  (C) 44. (A) 45. (C) 46. (C) 47.  (B) 48. (B)  49.  (B) 50.  (B)   51. (C) 52. (A) 53.  (B)   54. (A) 55.  (C) 56. (B) 57.  (A)   58. (C) 59. (A) 60. (C)   61.  (D) 62. (A) 63.  (C) 64.  (B) 65. (A) 66.  (B) 67. (B)  68.  (A) 69. (C) 70.  (A)   71 (B) 72. (D) 73. (C) 74.  (C) 75. (C) 76.  (A) 77. (B)  78.  (A) 79. (A) 80.  (B)   81. (D) 82.  (B) 83. (A) 84.  (B) 85. (C) 86.  (B) 87. (C)  88. (A) 89.  (D) 90. (C)   91.  (B) 92. (C) 93.  (A) 94. (D) 95. (B) 96.  (C) 97. (B) 98.  (B) 99. (B) 100.  (B) 101. (A) 102.  (C) 103. (B) 104.  (B) 105. (D) 106.  (B) 107. (C) 108. (B) 109. (A) 110. (D) 111.  (A) 112. (B) 113.  (A) 114. (A) 115.  (C) 116. (B)

Previous Years’ Questions 1 17. (D) 118. (C) 119.  (D) or (C) 120. (A) 121. (C) 122. (B) 123. (D) 124. (C) 125. (C) 126. (C) 127. (D) 128. (D) 129. (C) 130. (C) 131. (B) 132. (A) 133. (C) 134. (A) 135. (A) 136. (B) 137. (B) 138. (B) 139. (A) 140. (C)

Complex Numbers  2.31

HINTS AND EXPLANATIONS

p q r + + =1+i a b c

which is not possible for any value of x.

\ x2 + 4x + 4 = 0 ⇒ (x + 2)2 = 0 or x = –2 \ z = –2 – i. The correct option is (B) z 6. If 1 = z, the given equation becomes z2 z2 – z + 1 = 0 i.e., z = – w and – w2 z or, 1 = – w ⇒ z1 = – z2w z2 OB = |z2 – 0| = |z2| OA = |z1 – 0| = |– z2w | = |z2| |–w | = |z2| and AB = |z2 – z1| = |z2 + z2w | = |z2| |1 + w | = |z2| |–w2| = |z2| Thus z1, z2 and origin form an equilateral triangle. The correct option is (D) 7. Let a = r cos q and b = r sin q b \ tan q = a a − ib r (cos θ − i sin θ ) Now = = (cos q – i sinq)2 a + ib r (cos θ + i sin θ )

The correct option is (D)



1. We have,

2

⎛ p q r⎞ ⇒ ⎜ + + ⎟ = 1 – 1 + 2i = 2i ⎝ a b c⎠ ⇒ ⇒ ⇒

p2 a

2

p a

2

2

p

2

+ + +

q2 b

2

q2 b q

2

2

+ + +

r2

⎛ qr rp pq ⎞ + 2⎜ + + = 2i ⎝ bc ca ab ⎠⎟ c 2

r2 c

2

+

2 pqr ⎛ a b c ⎞ + + = 2i abc ⎜⎝ p q r ⎟⎠

r2

= 2i a2 b2 c2 The correct option is (A) 2. sin x + i cos 2x and cos x – i sin 2x are conjugate of each other if 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,

3. Let z = x + iy, then z2 + |z|2 = 0 ⇒ (x + iy)2 + |x + iy|2 = 0 ⇒ x2 – y2 + 2ixy + x2 + y2 = 0 ⇒ 2x2 + 2ixy = 0 ⇒ 2x2 = 0 and 2xy = 0 ⇒ x = 0 and xy = 0

= cos 2q – i sin 2q = e– 2iq a − ib \ i log = i log e– 2iq = i (– 2 iq) = 2q a + ib b 2 a − ib ⎤ 2 tan θ ⎡ a \ tan ⎢i log = tan 2q = = a + ib ⎥⎦ 1 − tan 2 θ b2 ⎣ 1− 2 a 2ab = 2 2 a −b The correct option is (C)

Clearly y can be any real number. Hence, we will get infinitely many solutions.

8. Since ( 3 + i)100 = 299 (a + ib)

The correct option is (D)

\ ( 3 + i)100 – ( 3 – i)100 = 299 (2ib) = 2100 (ib) n

4. As w is the nth root of unity so, w – 1 = 0 ⇒ (w – 1) (1 + w + w2 +, …, + wn – 1) = 0 Hence, 1 + w + w2 +, …, + wn – 1 = 0 or w – 1 = 0 i.e., w = 1 The correct option is (B) 5. Since z + 2 |z + 1| + i = 0 \ x + i (y + 1) + 2  |x + iy + 1| = 0 \ y + 1 = 0 ( |x + iy + 1| is real) \ y = – 1 \ x + 2 |x – i + 1| = 0 ⇒ x2 = 2 [(x + 1)2 + 1] = 2 (x2 + 2x + 2)

\ ( 3 – i)100 = 299 (a – ib) ⇒ i100 [1 – 3 i]100 – i100 [–1 +

3 i]100 = 2100 (ib)

or (– 2w)100 – (2w2)100 = 2100 (ib) or w – w2 = (ib) or

3 i = ib

\ b = 3 The correct option is (A) 9. We have, 1 − i sin α (1 − i sin α ) (1 − 2i sin α ) = (1 + 2i sin α ) (1 − 2i sin α ) 1 + 2i sin α

=

1 − 3i sin α − 2 sin 2 α 1 − 4i 2 sin 2 α

HINTS AND EXPLANATIONS

Single Option Correct Type

2.32  Chapter 2 =

(1 − 2 sin 2 α ) − 3i sin α

x = 55/6 (cos 5q + i sin 5q)1/6

,

1 + 4 2 sin α which is purely real if sin a = 0, i.e., if a = np, where n is an integer. The correct option is (C) 10. By the given condition, |z – 1| < |z – i| ⇒ |x – 1 + iy| < |x + i (y – 1)| ⇒ (x – 1)2 + y2 < x2 + (y – 1)2 ⇒ –2x < – 2y ⇒ x > y ⇒ x – y > 0 The correct option is (D)

Squaring, |z1 + z2|2 = (|z1| + |z2|)2

HINTS AND EXPLANATIONS

12. Let z1 = x1 + iy1 z2 = x2 + iy2 z3 = x3 + iy3 z2 z3 = (x2 – iy2) (x3 + iy3) = (x2x3 + y2y3) + i (x2y3 – x3y2) ⇒ Im ( z2 z3 ) = x2y3 – x3y2 \ z1 Im ( z2 z3 ) = (x1 + iy1) (x2y3 – x3y2) Similarly z2 Im ( z3 z1 ) = (x2 + iy2) (x3y1 – x1y3) z3 Im ( z1z2 ) = (x3 + iy3) (x1y2 – x2y1) Adding we get the result = 0 The correct option is (C)

\

z1 − z2 z1 + z2



=

=

3 ib − 1 2

4

3 ib + 1 2

4

The correct option is (B)

=

⇒ 2Re |z1| | z2 | = 2 |z1| |z2| ⇒ |z1| | z2 | cos (q1 – q2) = |z1| |z2| ⇒ cos (q1 – q2) = 1 ⇒ q1 – q2 = 0  or  q1 = q2 or arg z1 = arg z2 The correct option is (C) a2

\ |a|2 = a2 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 z1 is purely an imaginary number, 3 z2 2z z 3 \ let 1 = ib  or  1 = ib 2 3 z2 z2 4

⇒ |z1|2 + |z2|2 + 2Re |z1| |z2| = |z1|2 + |z2|2 + 2 |z1| |z2|

16. For every a ∈ R, |a| =

\ |x| + |y| ≤ 2 |z|. The correct option is (B)

13. ∵

z1 3 −1 ib − 1 z2 = 2 3 z1 ib + 1 +1 2 z2

x1x2 … x6 = 55 [cos (5p + 5q) + i sin (5p + 5q)] = 55 (– cos 5q – i sin 5q) = – 55(cos 5q + i sin 5q) The correct option is (B) 15. Given |z1 + z2| = |z1| + |z2|

11. Perimeter = 6 × distance of vertex 1 + 2i  from the centre (0, 0). = 6 [(1 – 0) + i (2 – 0)] = 6 5 The correct option is (D)

4

⎡ ⎛ 2kπ + 5θ ⎞ ⎛ 2kπ + 5θ ⎞ ⎤ ⎟⎠ + i sin ⎜⎝ ⎟⎠ ⎥ = 55/6 ⎢cos ⎜⎝ 6 6 ⎣ ⎦

(1)

4

⎛ 9b 2 ⎞ + 1⎟ ⎜ ⎜⎝ 4 ⎟⎠

4

⎛ 9b 2 ⎞ + 1⎟ ⎜ ⎜⎝ 4 ⎟⎠

4

=1

14. x6 = (4 – 3i)5 ⎛ 4 3i ⎞ x6 = 56 ⎜ − ⎟ = 55 (cos q + i sin q)5 ⎝5 5⎠ ⎛ 3⎞ where  q = –tan–1 ⎜ ⎟ = 55 (cos 5q + i sin 5q) ⎝ 4⎠

17. We have, |1 + i|2 × |1 + 2i|2 × |1 + 3i|2 … |1 + ni|2 = |a + ib |2 ⇒ (1 + 1) × (1 + 4) × (1 + 9) … (1 + n2) = a2 + b2 ⇒ 2 × 5 × 10 … (1 + n2) = a 2 + b 2. The correct option is (C) 18. |z1| = 1, |z2| = 1 ⇒ a2 + b2 = 1 p2 + q2 = 1

(1) (2)

Im ( z1 z2 ) = Im (a + ib) (p – iq) = bp – aq \ bp – aq = 1

(3)

ω1ω 2 = (a + ip) (b – iq) = (ab + pq) + i (bp – aq) = (ab + pq) + i (1)

(

)

\ Im ω1ω 2 = 1 The correct option is (D) 19. We have,

3

a + ib = x + iy

a + ib = (x + iy)3 = (x3 – 3xy2) + i (3x2y – y3)

Complex Numbers  2.33 \ a = x3 – 3xy2 and b = 3x2y – y3 a b \ = x2 – 3y2 and = 3x2 – y2 x y a b \ + = x2 – 3y2 + 3x2 – y2 = 4 (x2 – y2) x y The correct option is (B) 20. As (a – b)2 ≥ 0, a2 + b2 ≥ 2ab(1) a 2 + b 2 ; so from Eq. (1), |z|2 ≥ 2ab

\ |z|2 + a2 + b2 ≥ a2 + b2 + 2ab or |z|2 + |z|2 ≥ (a + b)2; \ 2 |z|2 ≥ (a + b)2 \

2 |z| ≥ a + b as |z| is positive. 1 \ |z| ≥ (a + b) 2 The correct option is (B) z3

z1 − z3 1 − i 3 = z2 − z3 2 ⎛ −π ⎞ ⎛ −π ⎞ = cos ⎜ = e– i p/3 + sin ⎜ ⎝ 3 ⎟⎠ ⎝ 3 ⎟⎠ 21.

\

z1 − z3 = e − i π /3 = 1 z2 − z3

z2

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. The correct option is (B) 24. The closest distance = length of the perpendicular from the origin on the line a z + a z + a a = 0 a( 0 ) + a | 0 | + a a | a |2 | a| = = = 2| a | 2 2| a | The correct option is (B)

5π 5π 7π 7π + i sin , cos + i sin 4 4 4 4 1 1 1 1 = (1 + i), (– 1 + i), (– 1 – i), (1 – i). 2 2 2 2 1 Hence, the four roots of z4 + 1 = 0 are (±1 ± i). 2 The correct option is (C) 26. Put 1 = r cos q and – 1 = r sin q π ⇒ r = 2 and q = – 4 Then given equation takes the form cos

n

π and angle between z1 – z3 and z2 – z3 is . 3 \ triangle is equilateral. The correct option is (C) 22. Putting x = 1, w, w2 respectively, 3n = a0 + a1 + a2 + … + a2n (1 + w + w2)n = a0 + a1 w + a2 w2 + … + a2n w2n (1 + w2 + w4)n = a0 + a1 w2 + a2 w4 + … + a2n w4n Adding these, 3n + (1 + w + w2)n + (1 + w2 + w4)n = 3a0 + a1 (1 + w + w2) + a2 (1 + w2 + w4) + a3 (1 + w3 + w6) + … \ 3n + 0n + 0n = 3a0 + 3a3 + 3a6 + … \ 3n – 1 = a0 + a3 + a6 + … The correct option is (B) 23. Let n 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) =

π π 3π 3π + i sin , cos + i sin , 4 4 4 4

⎡ ⎛ π⎞ ⎛ π⎞⎤ ( 2 )n ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥ = 2n ⎝ ⎠ ⎝ 4⎠⎦ 4 ⎣

π 3

z1

⇒ z = cos

nπ nπ ⎤ ⎡ or 2n/2 ⎢cos − i sin = 2n 4 4 ⎥⎦ ⎣ Equating real and imaginary parts, we get nπ nπ cos = 2n/2 and – sin =0 4 4 These are satisfied only for n = 0. Hence, n = 0 is the only solution. The correct option is (A) 27. We have, 2 = z − ⇒ |z| –

4 4 ≥ |z| – z z

4 ≤2 z

⇒ |z|2 – 2 |z| – 4 ≤ 0 or (|z| – 1)2 – 5 ≤ 0 ⇒ (|z| – 1)2 ≤ 5 or |z| – 1 ≤ ⇒ |z| ≤ + 1 Hence the greatest value of |z| is The correct option is (C) 28. We have, Arg (z + i) – Arg (z – i) =

5 + 1.

π 2

π ⎛ z + i⎞ ⇒ Arg ⎜ = ⎝ z − i ⎟⎠ 2 ⎛ z + i⎞ \ Re ⎜ =0 ⎝ z − i ⎟⎠ ⎛ z + i⎞ ⎛ z + i⎞ ⎜⎝ ⎟ +⎜ ⎟ z − i⎠ ⎝ z − i⎠ = 0 ⇒ ⎛ z + i⎞ + ⎛ z + i ⎞ = 0 ⇒ ⎜⎝ ⎟ z − i ⎠ ⎜⎝ z − i ⎟⎠ 2 ⇒ (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|2 = 1

HINTS AND EXPLANATIONS

But |z| =

25. We have, z4 + 1 = 0 ⇒ z4 = –1 ⇒ z = (cos p + i sin p)1/4 1 1 ⇒ z = cos (2kp + p) + i sin (2kp + p), k = 0, 1, 2, 3. 4 4

2.34  Chapter 2 ⇒ |z| = 1 The equation represents a circle centred at origin and radius 1 unit. The correct option is (A) 29. Let z = x + iy

The correct option is (B) 32. We have,

⇒

( x + iy ) 2 =k x + iy − 1

⇒ x2 + 2i xy – y2 = k (x – 1) + i ky ⇒ x2 – y2 = k (x – 1) 2xy = ky Eq. (2) gives either y = 0 or k = 2x If k = 2x, then x2 – y2 = 2x2 – 2x ⇒ x2 + y2 – 2x = 0, which is a circle. Thus, lies either on the real axis y = 0 or on a circle. The correct option is (C)

| z − 1| + 4 1 < 6 ⇒ |z – 1| > 3 which is an exterior of a circle. The correct option is (C) 33. Suppose x is a real root Then x3 + ix – 1 = 0 ⇒ x3 – 1 = 0 and x = 0 There is no real number satisfying these two equations.

z − 2 z2 30. Clearly 1 =1 2 − z1z2

The correct option is (C) 34. Let a, b, g  be the roots \ a + b + g  = –a \ |–a| = |a + b + g | ≤ |a| + |b | + |g | = 1 + 1 + 1 \ |a| ≤ 3 The correct option is (A)

⎛ z − 2 z2 ⎞ ⎛ z1 − 2 z2 ⎞ ⇒ ⎜ 1 =1 ⎝ 2 − z1z2 ⎟⎠ ⎜⎝ 2 − z1z2 ⎟⎠ ⇒ z1z1 − 2 z1z2 − 2 z1z2 + 4 z2 z2 = 4 − 2 z1z2 − 2 z1z2 + z1z2 z1z2

HINTS AND EXPLANATIONS

π . 2

⎛ | z − 1| + 4 ⎞ ⎛ 1⎞ log1/ 2 ⎜ > 1 = log1/ 2 ⎜ ⎟ ⎝ 2⎠ ⎝ 3 | z − 1| − 2 ⎟⎠

z2 Since is real = k (say), where k ∈ R z −1 \

\ |amp z1 – amp z2| =

⇒ |z1|2 + 4 |z2|2 = 4 + |z1|2 |z2|2

35. We have,

⇒ |z1|2 [1 – |z2|2] = 4 [1 – |z2|2] ⇒ |z1|2 = 4

( |z2| ≠ 1)

z1 z2 + =1 z2 z1

⇒ z12 + z22 = z1 z2

⇒ |z1| = 2

⇒ z12 + z22 + z32 = z1 z2 + z2 z3 + z3 z1,

The correct option is (B)

where z3 = 0. ⇒ z1, z2 and the origin form an equilateral triangle. The correct option is (C)

31. Let z1 = x1 + iy1 and z2 = x2 + iy2 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 y1 y Now amp z1 – amp z2 = tan– 1 x – tan– 1 2 1 x 2



= tan– 1

y1 y2 − x1 x2 y y 1+ 1 ⋅ 2 x1 x2



= tan– 1

x2 y1 − y2 x1 x1 x2 + y1 y2



= tan–1 ∞, by (1)

36. We have, S (n) = in + i–n = in + (1)

=

( −1) n + 1

1

=

i 2n + 1

in i , n = 1, 2, 3, 4, … n

in \ values of S (n) are 0, – 2, 2, 0, – 2, 2, … \ Total number of distinct values of S (n) is 3. The correct option is (C) 37. We have, |z1 + z2| =

1 1 + z1 z2

=

z1 + z2 z1 z2

⎛ 1 ⎞ ⇒ |z1 + z2| ⎜1 − =0 ⎝ | z1 z2 | ⎟⎠ ⇒ |z1 z2| = 1.( z1 ≠ –z2) The correct option is (B)

Complex Numbers  2.35

⇒ tan–1

y y+3 = tan– 1 x −1 x

⇒ xy = (x – 1) (y + 3) ⇒ 3 (x – 1) = y \ (x – 1) : y = 1 : 3. The correct option is (C) 39. Given: |z| = 2 and arg (z) =

2π . 3

2π \ If z = r (cos q + i sinq), then r = 2 and q = 3 ⎛ 1 2π 2π ⎞ 3⎞ ⎛ \ z = 2 ⎜ cos + i sin ⎟⎠ = 2 ⎜ − + i ⎟ ⎝ 3 3 2 ⎠ ⎝ 2 = (–1 + i 3) The correct option is (A) 40. log

3

⎛ | z |2 − | z | +1⎞ ⎜ 2 + | z| ⎟ < 2 ⎝ ⎠

| z |2 − | z | +1 ⇒ < ( 3 )2 2 + | z| ⇒ |z|2 – |z| + 1 < 6 + 3 |z| ⇒ |z|2 – 4 |z| – 5 < 0 ⇒ (|z| – 5) (|z| + 1) < 0 ⇒ |z| – 5 < 0, since |z| + 1 > 0 ⇒ |z| < 5 The correct option is (A) 41. Let z1 = x1 + iy1 \

z1 = x1 – iy1

Now z2 = z1 ⇒ z1 + z2 = z1 + z1 = x1 + iy1 + x1 – iy1 = 2x1, which is real. Hence, result holds if z2 = z1 . The correct option is (B) 42. |z – i| = |z + 5i| represents equation of perpendicular bisector of points (0, 1) and (0, –5) i.e., y = –2, now |z| = 2 is x2 + y2 = 4 ⇒ x2 + 4 = 4 ⇒ x = 0 \ z represents a single point (0, –2). The correct option is (C) 43. If a complex number z is rotated through an angle 180º, then it’s new position is –z. So, 2 – 4i is –2 + 4i and stretching it 5/2 times means modulus 5/2 times of previous complex number 5 (–2 + 4i) = –5 + 10i 2 The correct option is (C) i.e.,

44. Midpoint of P and P′ is centre of circle C such that z + ( − z1 ) 1 =0 2 \ Centre of circle lies at origin. Now, the equation of circle with centre at origin and radius |z1| or |–z1| is |z – 0| = |z1| P(–z1) ⇒ |z|2 = |z1|2 ⇒ z · z = z1 · z1 \

z z ⎛z ⎞ = 1 = ⎜ 1⎟ ⎝ z⎠ z z1

P(z1)

The correct option is (A) 45. We have ⎛ p q r⎞ (1 + i)2 = ⎜ + + ⎟ ⎝ a b c⎠ p2

⇒ 1 – 1 + 2i = ⇒ 2i = =

⇒

p

2

p2 a p a

2

q

+ 2

b

2

q

2

b

2

2

+

+

r

+

+ r2 c

2

r2 c

2

q2 b

2

+

r2 c2

⎛ qr rp pq ⎞ + 2⎜ + + ⎝ bc ca ab ⎟⎠

+

2abc ⎛ a b c ⎞ + + pqr ⎜⎝ p q r ⎟⎠

+

2abc (0) pqr

2

= 2i a b c2 The correct option is (C) 2

+

2

q2

+

2

a

2

2

46. Interpretating according to Coni’s Theorem. Let the angle between the lines joining z1, z2 and z1, –z2 be a z −z \ 1 2 = cos a + t sin a z1 + z2 Using Componendo and Dividendo, we have 2z 1 + cos α + i sin α 1 = −2 z2 cos α − 1 + i sin α z ⇒ 1 = z2 ⇒

α α ⎛α⎞ 2 cos 2 ⎜ ⎟ + i 2 sin cos ⎝ 2⎠ 2 2 α α α ⎛ ⎞ −2 sin 2 ⎜ ⎟ + i sin cos ⎝ 2⎠ 2 2

z1 α = i cot 2 z2

⇒ iz1 = –cot

α z2 2

α ⎡ ⎤ But iz1 = kz2  ⎢ where, k = − cot 2 (say)⎦⎥ ⎣ α Now, k = –cot 2 2k α ⇒ cot = –k ⇒ tan a = 2 k −1 2

HINTS AND EXPLANATIONS

38. We have, amp (z – 1) = amp (z + 3i)

2.36  Chapter 2 Since A · M > G · M for |z1| ≠ |z2| The correct option is (B)

2k

⇒ tan a = –

1− k2 2k ⎞ ⎛ ⇒ a = tan–1 ⎜⎝ − 1 − k 2 ⎟⎠ = –2 tan–1 k Now,

50. If z is a root of (z – 1)25 = 2w2(z + 1)25, then ⎛ z − 1⎞ ⎜⎝ ⎟ z + 1⎠

z1 − z2 = cos a + i sin a z1 + z2

47. x2 – 3x + 1 = 0 3 ± 3−4 3 i π π = ± = cos + i sin 2 2 2 6 6

⇒ x =

⎛ n 1⎞ x ⎜⎝ − n ⎟⎠ x ⇒

24

2

⎛

1 nπ ⎛ ⎞ = ⎜ x 2 n + 2 n − 2⎟ = –2 + 2 cos ⎝ ⎠ 3 x 1⎞

∑ ⎜⎝ x n − x n ⎟⎠

2

n =1

=

24

⎛

∑ ⎜⎝ −2 + 2 cos

n =1

nπ ⎞ ⎟ 3⎠

2π 24π ⎤ π ⎡ + ... + cos = –48 + 2 ⎢cos + cos 3 3 3 ⎥⎦ ⎣

HINTS AND EXPLANATIONS

⎡ ⎛ 2π 23π ⎞ ⎛ 24π ⎞ ⎤ +2 ⎢cos ⎜ + ⎟⎠ .sin ⎜⎝ ⎟ ⎝ 3 6 6 ⎠ ⎥⎦ ⎣ = – 48 π sin 6 = 0 – 48 = – 48 The correct option is (B) 48. C1(0, 0) is centre of bigger circle and C2 (3, 4) is centre of smaller circle C1B = r1 = 12 (radius of bigger circle) C2A = r2 = 5 (radius of smaller circle) C1C2 = 5, minimum value of |z1 – z2| = AB

C1 C2

A

B

Now, C1C2 + C2A + AB = C1B ⇒ AB = 12 – 5 – 5 = 2 The correct option is (B) 49.

2

2 z1 + i 3 z2 +

( 2z + i + ( 3z +i =

1

1

)( 2 z )( 3 z2

2

3 z1 + i 2 z2

) 2z )

2 z1 − i 3 z2 3 z1 − i

= 5 (|z1|2 + |z2|2) > 5 · 2

z1

2

2

z2

2

= 10 |z1 z2|

z −1 z +1

25

= 2 |w2| = 2

As 21/25 ≠ 1, we get z lies on a circle. The correct option is (B) 1 3 51. We have (A0  A1)2 = + = 1 4 4 ⇒ A0A1 = 1 2

2

⎛ 3⎞ 9+3 ⎛ 3⎞ A0A1 = ⎜ ⎟ + ⎜ ⎟ = =3 ⎝ 2⎠ 4 ⎝ 2 ⎠ Imaginary axis –1 , 2

3 2

A2

A1

–1 2

1, 2

3 2

A0

A3 (–1, 0)

(1, 0)

3 2

A4

A5

–1 2

Real axis

3 2

⇒ A0A2 = 3 Similarly, A0A4 = 3 Thus, (A0A1) (A0A2) (A0A4) = 3 The correct option is (C) b 52. z1 + z2 = – (1) a c z1z2 = (2) a z2 = iz1(3) From Eq. (1) and (2) −b a −b b2 −b 2 ⇒ z1 = (1 – i) ⇒ z12 = (–2i) = i 2 2a 4a 2a 2 From Eq. (2) and (3) z1(1 + i) =

c −c −b 2 c b2 = i⇒ 2 i= i⇒a= 2c ai a a 2a The correct option is (A) z −1 53. should be real i x + iy − 1 i.e., = y – i (x – 1) is real i z12 =

2

= 2w2 ⇒

z −1 = 21/25 z +1

⇒

where a is the angle between (z1 – z2) and (z1 + z2). ⇒ a = –2tan–1k is the angle between (z1 – z2) and (z1 + z2). The correct option is (C)

25

Complex Numbers  2.37

5

1 ⎞ ⎛ \ S = ∑ ⎜ ω n + n ⎟ ⎝ ⎠ ω n =1 5

(

⇒ S = ∑ ω n + ω 2 n n =1

)

(

2

(given)

2

2

⇒ S = (–1)2 + (–1)2 + 22 + (–1)2 + (–1)2(\ S = 8) Case II:  x = w2

5

(

⇒ S = ∑ ω 2 n + ω n n =1

)

2

2

=8

The correct option is (A) 55. Points A, B and C are at equal distance from origin O and as ∠ BOA = 45º clearly BC = AB. So, triangle is isoceles. Imaginary axis C (0, 1)

B

1, 1 2 2

45° O

2

d2

ad − +c=0 b b2 or d2 – abd + b2c = 0 The correct option is (B) 57. |iz + z0| = |iz – i2 + z0 – 1| = |i (z – i) + 5 + 3i – 1| = |i (z – i) + (4 + 3i)| \ |iz + z0| ≤ |i (z – i)| + |4 + 3i| ≤ 1.2 + 5 = 7 \ Maximum value of |iz + z0| is 7 The correct option is (A) ⇒

z2 z2 = z −1 z −1

⇒ zzz − z 2 = zz z − z 2 ⇒ | z |2 ( z − z ) − ( z − z )( z + z ) = 0 ⇒ ( z − z )[| z |2 − ( z + z )] = 0 Either z = z ⇒ real axis or |z|2 = z + z ⇒ zz − z − z = 0 represents a circle passing through origin. The correct option is (A) 60. C1: |z – i| = 2 ⇒ x2 + (y – 1)2 = 22 C2: |z – 1 – 2i| = 4 ⇒ (x – 1)2 +(y – 2)2 = 42 C1C2 = 2 r1 = 2, r2 = 4 ⇒ C1C2 < |r1 – r2| ⇒ one circle lies inside the other. Now point (2, 2) lies outside circle C1 and inside circle C2. ⇒ C1 lies inside C2 The correct option is (C) 61. We have, z1(z21 – 3z22) = 2 2

2

and, z2(3z 1 – z 2) = 11 Multiplying (2) by i and adding it to (1), we get Real axis

A (1, 0) The correct option is (C) 56. Let a be a real root of the given equation. Then, a2 + (a + ib) a + c + id = 0 ⇒ a2 + aa + c = 0 and ba + d = 0

⎛ d⎞ ⎛ b⎞ ⇒ ⎜ − ⎟ + a ⎜ − ⎟ + c = 0 ⎝ b⎠ ⎝ d⎠

2

x + y – 2y = 4 and 2x = 2, on equating the real and imaginary parts. \ y2 – 2y – 3 = 0 giving y = 3, –1 The solutions are 1 + 3i and 1 – i. The correct option is (C) 59.

n 5 ⎡ ⎛ 1 ⎞ ⎤ \ S = ∑ ⎢ω 2 n + ⎜ 2 ⎟ ⎥ ⎝ω ⎠ ⎥ n =1 ⎢ ⎣ ⎦

)

58. z z − 2i = zz + 2iz = 2 (2 + i) gives

(1) (2)

z31 – 3z22z1 + i(3z21z2 – z32) = 2 + 11i ⇒ (z1 + iz2)3 = 2 + 11i(3) Again, multiplying (2) by i and subtracting it from (1), we get z31 – 3z22z1 – i(3z21z2 – z32) = 2 – 11i ⇒ (z1 – iz2)3 = 2 – 11i(4) Multiplying (3) and (4), we get (z21 + z22)3 = 4 + 121 ⇒ z21 + z22 = 5. The correct option is (D) 62. We have,

1 − c 2 = nc – 1

⇒ 1 – c2 = n2c2 + 1 – 2nc ⇒ Now, =

c 1 = (1) 2n 1 + n2 c ⎛ (1 + nz ) ⎜1 + ⎝ 2n

n⎞ ⎟ z⎠

1⎞ ⎫ 1 ⎧ ⎛ 2 ⎨1 + n + n⎜ z + ⎟ ⎬ ⎝ z⎠⎭ 1 + n2 ⎩

HINTS AND EXPLANATIONS

⇒ x – 1 = 0 i.e., x = 1 z −1 \ sin–1 = sin–1 y i \ –1 ≤ y ≤ 1 The correct option is (B) 54. Since, x2 – x + 1 = 0 \ Solving for x, we have x = –w and x = –w2 Case I:  x = w

2.38  Chapter 2

=

1 1 + n2

(

1 + n2 + n( 2 cos θ )



)

⎛ 2n ⎞ = 1+ ⎜ ⎟ cos θ = 1 + ccosq ⎝ 1 + n2 ⎠ The correct option is (A) 63. We have, zk = 1 + a + a2 + … ak = ⇒ zk – ⇒

1 −ak + 1 = 1− a 1− a

zk −

(Using (1))

1 − ak + 1 1− a

1 1 | a |k + 1 < = ( |a| < 1) 1− a |1 − a | |1 − a |

Therefore, vertices z1, z2, …, zn of the polygon lie within the circle. 1 1 z − = |1 − a | 1− a The correct option is (C)

HINTS AND EXPLANATIONS

( |ai| ≤ 1)

3

2

4

65. We have, z = (z – 1)

2 nπ i 4

, n = 0, 1, 2, 3.

z −1 = 1 so they lie on the line bisecting perpendicularly z the join of z = 1 and z = 0. The correct option is (A) 66. We have, b + ic 1+ i 1 + iz 1 + a = 1 + a − c + ib = b + ic 1 − iz 1 + a + c − ib 1− i 1+ a

=

[1 + a − c + ib] [1 + a + c + ib] [1 + a + c − ib] [1 + a + c + ib] z0 (1 + a + c) 2 + b 2



+ z1z2

n −1

∑ wk

k =0

n −1

∑ w −k

k =0

n −1 ⎡ n −1 ⎤ = n (| z1 |2 + | z2 |2 ) ⎢ ∑ w k = 0 = ∑ w − k ⎥ . k =0 ⎣k =0 ⎦

The correct option is (B) 68. We have, |z – z1|2 + |z – z2|2 = k ⇒ 2 | z |2 + | z1 |2 + | z2 |2 − 2 Re( z z1 ) − 2 Re( z z2 ) = k

Since for all these values of z,



⎡⎣ w k = ei ( 2π k / n) ⇒ w k = e − i ( 2π k / n) = w − k ⎤ ⎦⎥

k =0

⇒ 3 < 1 + |z| + |z| + |z| < 1 + |z| + |z| + … ∞ 1 1 ⇒ 3 < ⇒ 1 – |z| < 1−| z| 3 2 \ |z| > 3 The correct option is (B)

=

= ( z1 + z2 w k ) ( z1 + z2 w − k )

n −1

⇒ 3 ≤ |z|3 + |z|2 + |z| + 1



)

∑ | z1 + z2 w k |2 = n (| z1 |2 + | z2 |2 ) + z1z2

⇒ 3 ≤ |a1| |z3| + |a2| |z2| + |a3| |z| + |a4|

⎛ z − 1⎞ ⇒ ⎜ = 11/4 = e ⎝ z ⎠⎟

(

|z1 + z2 wk|2 = ( z1 + z2 w k ) z1 + z2 w k

Therefore, we have,

⇒ 3 ≤ |a1z3| + |a2z2| + |a3z| + |a4|

4

= 2a2 + 2a = 2a (1 + a) and, Im(z0) = 2b (1 + a). Thus, z0 = 2 (1 + a) (a + ib) Also, denominator of (1) = 1 + a2 + c2 + 2a + 2c + 2ac + b2 = 2 + 2a + 2c + 2ac = 2 (1 + a) (1 + c) 1 + iz a + ib Therefore, = 1+ c 1 − iz The correct option is (B) 67. We have,

= | z1 |2 + | z2 |2 + z1 z2 w k + z1z2 w − k

64. We have, |3| = |a1z3 + a2z2 + a3z + a4|

2

= (1 + a)2 – c2 – b2 = 1 + a2 + 2a – (1 – a2)

(1)

Now,  R (z0) = (1 + a – c) (1 + a + c) – b2

⇒ 2 | z |2 − 2 Re { z ( z1 + z2 )} = k − (| z1 |2 + | z2 |2 ) ⇒ | z |2 − Re { z ( z1 + z2 )} = z−

z1 + z2 2

2

⇒

z−

z1 + z2 2

2

⇒

1 ( k − | z1 |2 − | z2 |2 ) 2

−

1 1 | z1 + z2 |2 = ( k − | z1 |2 − | z2 |2 ) 4 2

=

1 1 k− | z1 |2 + | z2 |2 − 2 Re( z1 z2 )} 2 4

=

{

1 1 k − | z1 − z2 |2 2 4

2 z1 + z2 = 1 ( 2k − | z − z |2 ) 1 2 4 2 z + z2 which will represent a real circle having centre at 1 2

⇒

z−

1 1 2k − | z1 − z2 |2 , provided k ≥ | z1 − z2 |2 2 2 The correct option is (A) and radius =

Complex Numbers  2.39

The correct option is (C) 70. We have, xn – 1 = (x – 1) (x – w) (x – w2) … (x – wn–1) xn − 1 = (x – 1) (x – w2) … (x – wn–1) x−w Putting x = w on both sides, we have xn − 1 ⎛ 0 ⎞ (w – 1) (w – w2) … (w – wn–1) = lim ⎜ ⎟ x→w x − w ⎝ 0 ⎠ ⇒

= lim



x→w

nx n −1 = nwn–1 1

The correct option is (A) 71. We have, z – i = eia ⇒ z = i + eia = cos a + i (1 + sin a) \ q = arg (z) = tan

−1

⎛ 1 + sin α ⎞ ⎜⎝ cos α ⎟⎠

1 + sin α cos α 2 cos α 2 − \ cot θ − = z 1 + sin α cos α + i (1 + sin α ) ⇒ tan q =

=

cos α 2 [cos α − i (1 + sin α )] − 1 + sin α cos 2 α + (1 + sin α ) 2

cos α 2 [cos α − i (1 + sin α )] = − 1 + sin α 2 (1 + sin α ) = i. The correct option is (B) 72. We have, 4 + 3i ( 4 + 3i ) (1 − 2i ) z1 = = (1 + 2i ) (1 − 2i ) 1 + 2i 10 − 5i = =2–i 5 which represents the point whose coordinates are (2, –1) Also, we have,

iz = z ⇒ i(x + iy) – (x – iy) = 0 [Putting z = x + iy] ⇒ i(x + y) – (x + y) = 0 ⇒ (i – 1) (x + y) = 0 which represents the line y = –x Hence, reflection of the point (2, –1) in the line y = –x gives the point (1, –2) which is equivalent to 1 – 2i in the argand plane. The correct option is (D) 1 3 73. We know that w = − + i 2 2 ⎛ 1 i 3⎞ Thus, 4 + 5 ⎜ − + 2 ⎟⎠ ⎝ 2

334

⎛ 1 i 3⎞ + 3 ⎜− + 2 ⎟⎠ ⎝ 2

= 4 + 5w334 + 3w365 = 4 + 5 (w3)111 w + 3 (w3)123 w2 = 4 + 5 (1)111 w + 3 (1)123 w2 = 4 + 5w + 3w2 = 1 + 2w + 3 (1 + w + w2) = 1 + 2w + 3 (0) = 1 + 2w = 1 – 1 + 3 i = The correct option is (C) 74. The given line is b z + b z = c(1) Let A (z1) be a reflection of B (z2) in the line (1). Let P (z) be any point on the line (1). We have, AP = BP ⇒ |AP|2 = |BP|2

365

3 i.

⇒ |z – z1|2 = |z – z2|2 ⇒ (z – z1) ( z – z1 ) = (z – z2) ( z – z2 ) ⇒ ( z2 – z1 ) z + (z2 – z1) z + z1 z – z2 z2 = 0 Since (1) and (2) represent the same line, we get b c b = = = k (say) z2 − z1 z1 z1 − z2 z2 z2 − z1 ⇒ k ( z2 – z1 ) = b , k (z2 – z1) = b, k (z1 z1 – z2 z2 ) = c Now, z1 b + z2 b

= z1 {k (z2 – z1)} + z2 {k ( z2 – z1 )}



= k { z1 z2 – z1 z1 + z2 z2 – z2 z1 }

= k (z2 z2 – z1 z1 ) = c. The correct option is (C) 75. We have, z1 + z2 = – p and z1 z2 = q We know that

z1 | z1 | = (cos a + i sin a) z2 | z2 |

Since |z1| = |z2|  ( OA = OB) we get

z1 cos α + i sin α = z2 1

(2)

HINTS AND EXPLANATIONS

69. Let a z + b be the remainder when f (z) is divided by z2 + 1. Then, we have f (z) = (z2 + 1) g(z) + Az + B Given: f (z) when divided by z – i gives remainder i ⇒ f (i) = i ⇒ (i2 + 1) g(i) + Ai + B = i ⇒ Ai + B = i (1) Also, f (z) when divided by z + i gives remainder i + 1 ⇒ f (– i) = i + 1 ⇒ (i2 + 1) g(– i) – B = i + 1 ⇒ –Ai + B = i + 1 (2) Solving equations (1) and (2), we get A = i/2 and B = i + 1/2 1⎞ ⎛ i⎞ ⎛ \ remainder is ⎜ ⎟ z + ⎜ i + ⎟ ⎝ 2⎠ ⎝ 2⎠

2.40  Chapter 2 Applying Componendo and Dividendo, we get z1 + z2 cos α + i sin α + 1 = z1 − z2 cos α + i sin α − 1 α α α 2 cos 2 + 2i sin cos 2 2 2 = α α α −2 sin 2 + 2i sin cos 2 2 2

α ⎡ cos 2 ⎢⎣ = α ⎡ 2i sin ⎢cos 2 ⎣ p α ⇒ = i cot z1 − z2 2 2 cos

α + i sin 2 α + i sin 2

α⎤ α 2 ⎥⎦ = –i cot α⎤ 2 2 ⎥⎦

2

≤ (r1 – r2)2 + 4 × 1 × 1 ⎛⎜ α − β ⎞⎟ ⎝ 2 ⎠ or, |z – w|2 ≤ (|z| – |w|)2 + (Arg z – Arg w)2. The correct option is (B) 78. A(z1), B(z2), C(z3) lie on |z| = 2 whose centre is at O(0, 0) and radius 2. π z1 = 1 + 3 i hence |z| = 2 and Arg (z1) = 3

Squaring we obtain p2 α = –cot2 2 ( z1 + z2 ) 2 − 4 z1z2

HINTS AND EXPLANATIONS

p2

α ⇒ = –cot2 2 2 p − 4q α 2 2 2 α ⇒ p = – p cot + 4q cot2 2 2 2 ⎛ 2 α 2 α⎞ ⇒ p ⎜1 + cot ⎟ = 4q cot ⎝ 2⎠ 2 α ⇒ p2 cosec2 = 4q cot2 2 α ⇒ p2 = 4q cos2 2 \ k = 4q The correct option is (C) 76. Since |z1| = |z2| = |z3| = 1, we get, z1 z1 = z2 z2 = z3 z3 = 1. Now, 1 =

1 1 1 + + z1 z2 z3

= | z1 + z2 + z3 |

= z1 + z2 + z3 = |z1 + z2 + z3|. The correct option is (A) 77. Let z = r1 (cos a + i sin a) and, w = r2 (cos b + i sin b) ⇒ Arg z = a, Arg w = b |z| = r1, |w| = r2. Given: |z| ≤ 1, |w | ≤ 1 ⇒ r1 ≤ 1, r2 ≤ 1 Now, consider |z – w |2 = |(r1 cos a – r2 cos b) + i (r1 sin a – r2 sin b )2| = (r1 cos a – r2 cos b)2 + (r1 sin a – r2 sin b)2 = r12 (cos2a + sin2a) + r22 (cos2b + sin2b) –2 r1 r2 (cos a cos b + sin a sin b) = r12 + r22 – 2 r1 r2 cos (a – b) = (r1 – r2)2 + 2 r1 r2 [1 – cos (a – b)] ⎛ α − β⎞ = (r1 – r2)2 + 4 r1 r2 sin2 ⎝⎜ 2 ⎠⎟

In turn |z2| = |z3| = 2 and Arg (z2) = Arg (z1) + 120º = 180º \ z2 = –2 Further, Arg (z3) = Arg (z2) + 120º = 300º ⎡ π⎞ π⎞⎤ ⎛ ⎛ Hence,  z3 = 2 ⎢cos ⎜ 2π − ⎟ + i sin ⎜ 2π − ⎟ ⎥ ⎝ ⎠ ⎝ 3 3⎠⎦ ⎣ ⎛ 1 i 3⎞ π π⎤ ⎡ = 2 ⎢cos − i sin ⎥ = 2 ⎜ − 3 3⎦ 2 ⎟⎠ ⎝2 ⎣ = 1 – 3 i Thus, z2 = –2 and z3 = 1 – i 3 The correct option is (A) 79. We know that, (1 + x)3n = 1 + 3nC1 x + 3nC2 x2 + … + 3nC3n x3n(1) (1 – x)3n = 1 – 3nC1 x + 3nC2 x2 + … + (– 1)3n 3nC3n x3n(2) Subtracting (2) from (1) gives (1 + x)3n – (1 – x)3n = 2 [3nC1 x + 3nC3 x3 + 3nC5 x5 + …] = 2x [3nC1 + 3nC3 x2 + 3nC5 x4 + …] Putting x = i 3 , we get (1 + i 3 )3n – (1 – i 3 )3n = 2i 3 [3nC1 – 3 × 3nC3 + 32 × 3nC5 …] Therefore, 3nC1 – 3 3nC3 + 32 × 3nC5 … 1 = [(1 + i 3 )3n – (1 – i 3 )3n] 2i 3 3n ⎡⎛ 1 1 i 3 ⎞ ⎛ 1 i 3 ⎞ ⎥⎤ = × 23n ⎢⎜ + − − ⎢⎝ 2 2 ⎟⎠ ⎜⎝ 2 2 ⎟⎠ ⎥ 2i 3 ⎣ ⎦ 23n −1 = [(cos np + i sin np) – (cos np – i sin np)] i 3 23n −1

2i sin np = 0 as n is an integer. i 3 The correct option is (A) =

Complex Numbers  2.41 80. We know that the triangle with vertices z1, z2, z3 is an ­equilateral if z12 + z22 + z32 = z1 z2 + z2 z3 + z3 z1 \ The triangle with vertices z1 = a + i, z2 = 1 + bi and z3 = 0 will be equilateral if (a + i)2 + (1 + bi)2 + 0 = (a + i) (1 + bi) + 0 + 0 ⇒ a2 – 1 + 2ai + 1 – b2 + 2bi = (a – b) + i (1 + ab) ⇒ a2 – b2 = a – b (1) and 2 (a + b) = 1 + ab (2) [Equating real and imaginary parts] (1) ⇒ (a – b) [a + b – 1] = 0 ⇒ a = b or a = 1 – b. Subsitituting the value of a – b in (2), we get 2 (a + a) = 1 + a2 ⇒ a2 – 4a + 1 = 0

⇒ (z1 – z2)2 = 2 [z1 – z3] [z3 – z2] The correct option is (B) 83. Let A = z1, B = z2 and C = z3, where A, B, C are vertices of equilateral triangle. Given that third point C is origin, so z3 = 0. Let z2 – z3 = a, z3 – z1 = b, z1 – z2 = g or, z2 = a, – z1 = b, z1 – z2 = g \ a + b + g = z2 – z1 + z1 – z2 = 0 or, α + β + γ = 0 Since the triangle is equilateral triangle, \ BC = CA = AB or, |(z2 – 0)| = |0 – z1| = |z1 – z2| or, |a| = |b | = |g |

4 ± 16 − 4 =2± 3 2 Since 0 < a < 1 and 0 < b < 1,

or, |a|2 = |b |2 = |g |2

\ a = b = 2 – 3 . Substituting a + b = 1 in (2), we get 2 = 1 + a (1 – a) ⇒ a2 – a + 1 = 0 which gives imaginary values of a and b. Hence, a = b = 2 – 3 The correct option is (B) 81. Let z1 = x1 + iy1 and z2 = x2 + iy2 where, x1 ≠ x2, y1 ≠ y2 and x12 + y12 = x22 + y22

\ α =

⇒ a =

[( x1 + x2 ) + i ( y1 + y2 )] [( x1 − x2 ) − i ( y1 − y2 )] 2

( x1 − x2 ) + ( y1 − y2 ) [( x12

= =

−

x22 ) + ( y12

2

− y22 )] + i [ x1 y1 − y1x2 + y2 x1

− y2 x2 − x1 y1 + x1 y2 − x2 y1 + x2 y2 ] 2

( x1 − x2 ) + ( y1 − y2 ) ( x1 − x2 ) 2 + ( y1 − y2 ) 2

x y = a purely imaginary or 0 if 1 = 1 . x2 y2 y x If 1 = 1 then x1 + iy1 = k (x2 + iy2) y x2 2

If k = 1, z1 = z2, which is not true and if k ≠ 1, |z1| ≠ |z2|. \

Substituting values of α , β and γ in (1), we get k k k + + =0 α β γ or,

1 1 1 + + =0 α β γ

or,

1 1 1 + + =0 z2 − z1 z1 − z2

or,

z1 ( z1 − z2 ) − z2 ( z1 − z2 ) + z1 z2 =0 z1 z2 ( z1 − z2 )

or, z12 – z1 z2 – z1 z2 + z22 + z1 z2 = 0 or, z12 + z22 = z1 z2

2

2 i ( x1 y2 − y1x2 )

z1 + z2 is purely imaginary. z1 − z2

The correct option is (D) 82. Since |CA| = |CB| and ∠ACB = 90º \ (z2 – z3) = ± i (z1 – z3) ⇒ (z2 – z3)2 = – (z1 – z3)2 ⇒ z22 + z32 − 2 z2 z3 = − z12 − z32 − 2 z1 z3 z12 + z22 − 2 z1 z2 = 2 ( z1 z3 − z1 z2 − z2 z3 + z22 )

k k k , β = , γ = α β γ

or,

z12 z2 + 2 =1 z1 z2 z1 z2

or,

z1 z2 + =1 z2 z1

A B = 1. + B A The correct option is (A) or,

84. Given 2 2 x4 = ( 3 – 1) + i ( 3 + 1) ⎛ 3 − 1⎞ ⎛ 3 + 1⎞ ⇒ x4 = ⎜ ⎟ + i⎜ ⎟ ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ 2

⎛ 3 − 1⎞ ⎛ 3 + 1⎞ ⇒ |x4|2 = ⎜ ⎟ +⎜ ⎟ ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ \ |x4| = 1

⎡⎛ 3 + 1⎞ and,  arg (x4) = tan– 1 ⎢⎜ ⎟ ⎢⎣⎝ 2 2 ⎠

2

=1 ⎛ 3 − 1⎞ ⎤ ⎜ ⎟⎥ ⎝ 2 2 ⎠ ⎥⎦

HINTS AND EXPLANATIONS

=

or, α α =β β =γ γ = k (say)

z1 + z2 ( x + x2 ) + i ( y1 + y2 ) = 1 ( x1 − x2 ) + i ( y1 − y2 ) z1 − z2

Now,

(1)

2.42  Chapter 2 ⎛ 3 + 1⎞ 5π = tan– 1 ⎜ ⎟ = 75º = ⎝ 3 − 1⎠ 12



⇒ z12 + z22 = z1 z2. The correct option is (B) 87. We have, max. amp (z) = amp (z2), min. amp (z) = amp (z1).

⎡ 5π ⎞ 5π ⎞ ⎤ ⎛ ⎛ \ x4 = 1 ⎢cos ⎜ 2nπ + ⎟ + i sin ⎜ 2nπ + ⎟ ⎥ ⎝ ⎠ ⎝ 12 12 ⎠ ⎦ ⎣ n Using (cos q + i sin q) = cos nq + i sin nq, we have 1⎛ 5π ⎞ 1⎛ 5π ⎞ x = cos ⎜ 2nπ + ⎟ + i sin ⎜ 2nπ + ⎟ ; 4⎝ 12 ⎠ 4⎝ 12 ⎠ n = 0, 1, 2, 3 The correct option is (B) 10

85.

⎛

∑ ⎜⎝ sin q =1

2qπ 2qπ ⎞ − i cos ⎟ 11 11 ⎠

2π 4π ⎧ ⎫ = ⎨sin + sin + ... + 10 terms⎬ 11 11 ⎩ ⎭ 2π 4π ⎧ ⎫ i ⎨cos + cos + ... + 10 terms⎬ 11 11 ⎩ ⎭ 10π 10π ⎛ 2π 9π ⎞ ⎛ 2π 9π ⎞ sin ⎜ + sin cos ⎜ + ⎟ sin ⎝ 11 11 ⎟⎠ ⎝ ⎠ 11 11 11 11 −i = π π sin sin 11 11 = 0 – i (– 1) = i (1) \ S =

HINTS AND EXPLANATIONS



=

⎡ 10 ⎛ 2qπ 2qπ ⎞ ⎤ ∑ (3 p + 2) ⎢∑ ⎜⎝ sin 11 − i cos 11 ⎟⎠ ⎥ ⎢⎣ q =1 ⎥⎦ p =1 32

32

∑ (3 p + 2) i p

=3

p =1

32

32

p =1

p =1

p

∑ pi p + 2 ∑ i p

= 3A + 2B(2) Now,  A = i + 2i2 + 3i3 + … + 32i32 ⇒ Ai = i2 + 2i3 + … + 31i32 + 32i33 ⇒ A (1 – i) = i + i2 + … + i32 – 32i33 i 32 − 1 – 32i = – 32i [∵ i32 = 1] i −1 −32i 32i (1 + i ) \ A = = = 16 (1 – i)(3) 1− i 2

=

and,  B = i + i2 + … + i32 = 0 Hence,  S = 3 × 16 (1 – i) = 48 (1 – i). The correct option is (C) 86. Since the triangle is equilateral \ |z1 – 0| = |z2 – z1| = |0 – z2|

(4)

⇒ |z1|2 = |z2 – z1|2 = |z2|2 ⇒ z1 z1 = (z2 – z1) × ( z2 – z1 ) = z2 z2 z z z −z Now,  z1 z = z2 z2 ⇒ 1 = 2 = 1 2 . 1 z2 z1 z2 − z1

(z − z ) z Also,  z2 z2 = (z2 – z1) ( z2 – z1 ) = (z2 – z1) 1 2 2 z1 \ z1 z2 = (z2 – z1) (z1 – z2) = z2 z1 – z12 − z22 + z1 z2

⎛ 15 ⎞ ⎛ 3⎞ Now, amp (z1) = q1 = cos–1 ⎜ ⎟ = cos–1 ⎜ ⎟ ⎝ 25 ⎠ ⎝ 5⎠ π and,  amp (z2) = + q2 2 π ⎛ 15 ⎞ = + sin–1 ⎜ ⎟ ⎝ 25 ⎠ 2

=

π ⎛ 3⎞ + sin–1 ⎜ ⎟ ⎝ 5⎠ 2

\ |max. amp (z) – min. amp (z)| = 3 3 π π + − − cos −1 − cos −1 2 2 5 5 ⎛ 3⎞ = p – 2 cos–1 ⎜ ⎟ ⎝ 5⎠

π 3 3 + sin −1 − cos −1 2 5 5

=

The correct option is (C) 88. Putting z = x + iy, we get (x + iy)2 + (p + iq) (x + iy) + r + is = 0 ⇒ (x2 – y2 + px – q y + r) + i (2xy + py + qx + s) = 0 ⇒ x2 – y2 + px – q y + r = 0 and, 2xy + py + qx + s = 0 If the roots are real, then y = 0 \ (1) gives x2 + px + r = 0 and (2) gives pqx + s = 0 s From (4), x = – q s 2 ps Putting in (3), we get 2 − +r =0 q q

(1) (2) (3) (4)

or, r2 – pqs + rq2 = 0 ⇒ pqs = s2 + rq2. The correct option is (A) 89. We know that |z1 + z2|2 + |z1 – z2|2 = 2 [|z1|2 + |z2|2](1) Now, ⎡ z1 + z12 − z22 + z1 − z12 − z22 ⎤ ⎢⎣ ⎥⎦ = z1 + z12 − z22

2

+ z1 − z12 − z22

= 2 |z1|2 + 2 | z12 − z22 | + 2 | z22 |

2

2

+ 2 | z12 − ( z12 − z22 )| [By (1)]

Complex Numbers  2.43 z −z z −z z1 − z2 z −z = – 1 2 ⇒ 1 2 = 2 3 (2) z1 − z2 z3 − z2 z3 − z2 z3 − z2

= 2 |z1|2 + 2 |z2|2 + 2 | z12 − z22 |

⇒

= |z1 + z2|2 + |z1 – z2|2 + 2 |z1 + z2| |z1 – z2|

(1) × (2) ⇒ (z1 – z2)2 = – (z2 – z3)2

= (|z1 + z2| + |z1 – z2|)2 Taking square root of both sides, we get z1 + z12 − z22 + z1 − z12 − z22

= |z1 + z2| + |z1 – z2|.

The correct option is (D) 90. Since z = x + iy lies in the third quadrant \ x < 0, y < 0. Again z = x – iy z x − iy ( x − iy )( x − iy ) x 2 − y 2 − 2ixy = = = 2 2 x + iy z x +y x2 + y2 =



x2 − y2 x2 + y2

where,  A =

−

x2 − y2 x2 + y2

2 xy x2 + y2

\

x2 + y2

and B = –

0) 1 ⇒ c |z1|2 + |z2|2 ≥ 2 |z1| |z2| c 1 \ |z1|2 + |z1|2 + 2 |z1| |z2| ≤ |z1|2 + |z2|2 + c |z1|2 + |z2|2 c ⇒ |z1|2 + |z2|2 + 2 |z1| |z2| ≤ (1 + c) |z1|2

+ (1 + c–1) (|z2|2)(2)

From (1) and (2), we get |z1 + z2|2 ≤ (1 + c) |z1|2 + (1 + c–1) |z2|2 \ k = 1 + c–1 The correct option is (C)

⇒ z =



=

1 rπ ⎞ ⎡ rπ rπ ⎤ ⎛ i ⎜ 2 sin + i sin ⎟ cos ⎝ n ⎠ ⎢⎣ n n ⎥⎦ 1 rπ ⎞ ⎛ i ⎜ 2 sin ⎟ ⎝ n⎠

rπ rπ ⎤ ⎡ ⎢cos n − i sin n ⎥ ⎣ ⎦

i rπ 1 1 − ⇒x=– cot 2 n 2 2 1 Hence, all the points lie on the line x = – 2 The correct option is (B) ⇒ x + iy = −

105. We have, z2 + z + 1 = 0 ⇒ (z – 1) (z2 + z + 1) = 0, \ z3 = 1. If n is not a multiple of 3, then we can write n = 3m + r, where m ∈ I and r = 1 or 2,

(1)

HINTS AND EXPLANATIONS

⎛ z + 1⎞ ⇒ ⎜ = 1 = cos 0 + i sin 0 ⎝ z ⎟⎠ z +1 ⇒ = (cos 2p r + i sin 2p r)1/n z 2π r 2π r = cos + i sin n n where, r = 0, 1, 2, …, n – 1.

2.46  Chapter 2 then 2n = 6m + 2r If r = 1, then 2r = 2 \ zn + z2n = (z3)m × zr + (z3)2n × 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 Hence, zn + z2n = – 1 The correct option is (D) 106. Since 1, a1, a2, a3, a4 are the roots of the equation x5 – 1 = 0. \ (x5 – 1) = (x – 1) (x – a1) (x – a2) (x – a3) (x – a4) x5 − 1 ⇒ = (x – a1) (x – a2) (x – a3) (x – a4)(1) x −1 Putting x = w in (1), we get ω5 − 1 = (w – a1) (w – a2) (w – a3) (w – a4) ω −1 ω2 −1 ⇒ = (w – a1) (w – a2) (w – a3) (w – a4)(2) ω −1 and putting x = w2 in (1), we get

ω10 − 1 2 = (w2 – a1) (w2 – a2) (w2 – a3) (w2 – a4) ω −1 ω −1 = (w2 – a1) (w2 – a2) (w2 – a3) (w2 – a4)(3) ω2 −1 Dividing (2) by (3), we get ⇒

HINTS AND EXPLANATIONS

ω −α ω −α ω −α ω −α (ω 2 − 1) 2 2 1 ⋅ 2 2 ⋅ 2 3 ⋅ 2 4 = ω − α1 ω − α 2 ω − α 3 ω − α 4 (ω − 1) 2 =

ω 4 + 1 − 2ω 2 ω + 1 − 2ω 2 = ω 2 + 1 − 2ω ω 2 + 1 − 2ω

=

− ω 2 − 2ω 2 − 3ω 2 = =w − ω − 2ω − 3ω

The correct option is (B)

\ y1 – y2 = x1 – x2,(4) \ From (3) and (4) we get y1 + y2 = 2 \ Im (z1 + z2) = 2. The correct option is (C) 1 1 1 1 2 + + + = a+ω b+ω c+ω d +ω ω \ w is the root of the equation 108. Since,

2 1 1 1 1 = + + + x a+ x b+ x c+ x d + x ⇒ 2x4 + (S a) x3 + 0 × x2 – (S abc) x – 2 abcd = 0 Let a, b, g  be the other roots, then Σa w + a + b + g = – (1) 2 Sab = 0 ab + aw + bw + g w + bg + g a = 0 (2) Σ abc (3) 2 abg w = –abcd(4) Since complex roots occurs in conjugate pairs \ g = ω = w2. \ From (2), ab + w (a + b) + w × w2 + w2 (a + b) = 0 ⇒ ab + (w + w2) (a + b) + w3 = 0 ⇒ ab + (– 1) (a + b) + 1 = 0 ⇒ ab – a – b + 1 = 0 ⇒ a (b – 1) – (b – 1) = 0 ⇒ (a – 1) (b – 1) = 0 \ either a = 1 or b = 1 Hence one root is unity 1 1 1 1 \ + + + =2 a +1 b +1 c +1 d +1 The correct option is (B) Sabg =

107. Let z = x + iy We have, z + z = 2 |z – 1| z+z ⇒ = |z – 1| 2 ⇒ x = |x + iy – 1| ⇒ x = |(x – 1) + iy| ⇒ x2 = (x – 1)2 + y2 ⇒ 2x = 1 + y2. If z1 = x1 + iy1 and z2 = x2 + iy2

109. We have, |A|2 + |B|2 + |C|2 = A A + B B + C C (1)

then, 2x1 = 1 + y12 (1)

B B = (z1 + z2 w + z3 w2) ( z1 + z2ω + z3ω 2 )

and, 2x2 = 1 + y22 (2) Subtracting (2) from (1), we get 2 (x1 – x2) = y12 – y22 ⇒ 2 (x1 – x2) = (y1 + y2) (y1 – y2)(3) π But given arg (z1 – z2) = 4 ⎛ ⎞ y − y y1 − y2 π 2 i.e., tan– 1 ⎜ 1 = ⇒ =1 ⎟ 4 x −x ⎝x −x ⎠



1

2

1

But  A A = (z1 + z2 + z3) ( z1 + z2 + z3 ) = z1z1 + z2 z2 + z3 z3 + z1 ( z2 + z3 ) + z2 ( z3 + z1 ) + z3 ( z1 + z2 ) = | z1 |2 + | z2 |2 + | z3 |2 + z1 ( z2 + z3 ) + z2 ( z3 + z1 ) + z3 ( z1 + z2 )

2

= (z1 + z2 w + z3 w2) ( z1 + z2ω 2 + z3ω )

[∵ ω = w2 and (ω 2 ) = w] = z1z1 + z2 z2ω 3 + z3 z3ω 3 + z1 (z2 w + z3 w2)

+ z2 (z3 w4 + z1 w2) + z3 (z1 w + z2 w2) = | z1 |2 + | z2 |2 + | z3 |2 + z1 ( z2ω + z3ω 2 ) + z2 ( z3ω + z1ω 2 ) + z3 ( z1ω + z2ω 2 ) (2)

Similarly, C C = | z1 |2 + | z2 |2 + | z3 |2 + z1 ( z2ω 2 + z3ω )

Complex Numbers  2.47 + z2 ( z3ω 2 + z1ω ) + z3 ( z1ω 2 + z2ω ) (3) 2

2

2

A A + B B + C C = 3 [|z1| + |z2| + |z3| ]

2 = a2 – 3a + 2 represents a circle with centre

)

(

at A −

Adding (1), (2) and (3), we get 2

Since z +

a 2 − 3a + 2 and z +

2 , 0 and radius

2i

< a represents the interior of the circle with centre at

)

(

2

(−

2 −0



+ z1 [z2 (1 + w + w2) + z3 (1 + w2 + w)]

B 0,



+ z2 [z3 (1 + w + w2) + z1 (1 + w + w2)]



+ z3 [z1 (1 + w + w2) + z2 (1 + w2 + w)]

plex 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, i.e., if

= 3 [|z1|2 + |z2|2 + |z3|2][∵ 1 + w + w2 = 0] \ From (1) and (2), we conclude 2

2

2

2

2

2

|A| + |B| + |C| = 3 [|z1| + |z2| + |z3| ].

⇒ 2 ± a
0, \ a > 2 The correct option is (A)

1 z + = 2 (cos q + i sin q ) = 2eiq z

⇒ z = eiθ ±



and radius a. Therefore, there will be a com-

1 iθ z 1 1 = e = | z | eiθ = 2 zz z |z| |z|

1 z The correct option is (A) 114. Let ABCD be the rhombus and M be the point of intersection of the diagonals AC and BD \ A is

HINTS AND EXPLANATIONS



2.48  Chapter 2 Similarly, taking clockwise rotation we get another possible position of A as

Let point D be z1 = 1 + i and point M be z2 = 2 – i Also, let point A be z3 Then, z2 – z1 = 1 – 2i and |z2 – z1| = 5 = MD As given, 5 1 1 AC = BD ⇒ AM = DM ⇒ AM = 2 2 2 ⇒ AD = |z3 – z1| =

=

( 5)

2

DM 2 + AM 2

⎛ 5⎞ +⎜ ⎟ ⎝ 2 ⎠

2

=

5 2

Therefore, in D AMD,

HINTS AND EXPLANATIONS

5 5/2 2 1 cos q = = and sin q = = 5/2 5 / 2 5 5

z3 − z1 |z −z | = 3 1 e − iθ z2 − z1 | z2 − z1 |

3 ⎛1 − i⎞ ⇒ z3 = ⎜ (1 − 2i ) + (1 + i ) ⇒ z3 = 1 − i ⎝ 2 ⎟⎠ 2 i 3 So, A represents the complex numbers 3 − or 1 − i 2 2 The correct option is (A) 115. Let z = x + iy = r (cosq + i sin q), then the equation is ⎛ 1 ⎞ 1 cos θ − |(x – 2) + i (y – 1)| = r ⎜ sin θ ⎟ 2 ⎝ 2 ⎠ =

or,

1

+ z2 z3 + z2 z3 + z3 z1 + z1z3 )

= 24 + ( z1z2 + z1z2 + z2 z3 + z2 z3 + z3 z1 + z3 z1 ) (1) Also, ⇒ z1z2 + z1z2 + z2 z3 + z2 z3 + z3 z1 + z3 z1 ≥ – 12 \ |z1 + z2|2 + |z2 + z3|2 + |z3 + z1|2 ≥ 12 The correct option is (B)

Previous Year’s Questions

118. Given z12 + z22 − z1z2 = 0 ⇒ (z1 + z2)2 – 3z1z2 = 0 ⇒ a2 = 3b. The correct option is (C)

( x − y)

= 2 (| z1 |2 + | z2 |2 + | z3 |2 ) + ( z1z2 + z1z2

| z1 + z2 + z3 |2 ≥ 0

117. Key Idea : If w is a cube root of unity, then 1 + ω + ω2 = 0 and ω3 = 1 (l + ω + w2) = 0 and ω3 = 1 (1 + ω – ω2)7 = (−ω2 − ω2)7 (∵1 + ω + ω2 = 0) 2 7 = (− 2ω ) = –27⋅ω14 = –128(ω3)4ω2 = − 128 ω2(∵ ω3= 1) The correct option is (D)

2

116. |z1 + z2|2 + |z2 + z3|2 + |z3 + z1|2

z − (1 + i ) 5/2 ⇒ 3 = (cos θ + i sin θ ) 1 − 2i 5

i 2+i (1 − 2i ) + (1 + i ) ⇒ z3 = 3 − 2 2

( r cos θ − r sin θ )

which is the part of a parabola with focus (2, 1) and ­directrix x – y = 0. The correct option is (C)



⇒ z3 =

2

( x − 2) 2 + ( y − 1) 2 =

z −z Now, by rotation of complex numbers we know that 3 1 z2 − z1 |z −z | = 3 1 eiθ (anticlockwise rotation) | z2 − z1 |

z − (1 + i ) i ⇒ 3 = 1 + (using values of cosq and sinq) 1 − 2i 2

1

119.

⎛ z⎞ π arg( z ) − arg(ω ) = arg ⎜ ⎟ = ⎝ω⎠ 2

⇒ | zω | = 1 ⇒ z ω = −i or + i . The correct option is (D) or (C) 120.

1 + i (1 + i ) 2 = =i 1− i 2 x

⎛1+ i⎞ ⇒ ⎜ = ix ⎝ 1 − i ⎟⎠ ⇒ x = 4 n . The correct option is (A)

Complex Numbers  2.49 Z ⎛ z⎞ ⇒ arg ⎜ z ⎟ = π ⎝ i⎠ i

121. Here ω =

⇒ 2arg( z ) − arg(i ) = π

π =π 2 3π . ⇒ arg( z ) = 4 The correct option is (C) ⇒ 2 arg( z ) −

122. z = (p + iq)3 = p(p2− 3q2)− iq(q2− 3p2) x y + x y p q 2 2 2 2 ⇒ = p − 3q and = q − 3 p ⇒ 2 = −2 . p q ( p + q2 )

128. The given equation z2 + z + 1 = 0 ⇒ z = ω or ω2. 1 1 So, z+ = ω + ω2 = −1, z2 + 2 z z 1 = ω2 + ω = −1, z3 + 3 = ω3 + ω3 = 2, z 1 1 1 z4 + 4 = –1, z5 + 5 = –1 and z6 + 6 = 2. z z z ∴ The given sum = 1 + 1 + 4 + 1 + 1 + 4 = 12 The correct option is (D) 129. From the Argand diagram, maximum value of | z + 1| is 6. Alternative: | z + 1| = | z + 4 − 3| ≤ | z + 4| + |−3| = 6.

The correct option is (B)

(

)

2

123. Since | z 2 − 1 |2 = | z |2 +1 , we have

( z − 1)( z 2



2

2

)

− 1 = | z |4 + 2 | z |2 +1

(–7, 0)

⇒ z + z + 2 zz = 0 ⇒ z + z = 0

⇒ R (z) = 0 ⇒ z lies on the imaginary axis. The correct option is (D) 124. Given equation (x − 1)3 + 8 = 0 implies that (x − 1) = (−2) (1)1/3 ⇒ x − 1 = −2 or −2ω or −2ω2 Or n = −1 or 1 − 2ω or 1 − 2ω2. The correct option is (C) 125. | z1 + z2| = | z1 | + | z2| ⇒ z1 and z2 are collinear and are to the same side of origin; hence argz1− argz2 = 0. The correct option is (C) 126. As given w =

z 1 z− i 3

⇒| w |=

z 1 z− i 3

= 1 ⇒ distance of z

Hence z lies on the bisector of the line joining points (0, 0) and (0, 1/3). Hence z lies on a straight line. The correct option is (C) 127. Given sum 10



⎛

2kπ

2kπ ⎞

sin + i cos ∑ ⎟ 10 ⎜ ⎝ 211 kπ 211 kπ ⎠⎞ k =1 ⎛ sin + cos i ∑1⎜⎝0 11 10 11 ⎟ 2kπ 2⎠kπ k =1 = sin + i ∑ cos ∑ 10 10 11 2kπ k =1 211 kπ k =1 = ∑ sin + i ∑ cos 11 11 k =1

k =1

= 0 + i (−1) = −i The correct option is (D)

The correct option is (C) 130. Put −i in place of i Hence, the complex conjugate is The correct option is (C)

−1 i +1

4⎞ 4 4 4 ⎛ 131. One can write Z = ⎜ Z − ⎟ + ⇒ Z = Z − + ⎝ Z Z Z⎠ Z ⇒ Z ≤ Z−

⎛ 1⎞ from origin and point ⎜ 0, ⎟ is same. ⎝ 3⎠



(–4, 0) (–1, 0)

2



4 4 4 + ⇒ Z ≤ 2+ Z Z Z

2

⇒ Z −2 Z −4≤0 ∴ Z − 5 + 1 Z − 1 − 5 ≤ 0

( (

))(

(

))

⇒1 − 5 ≤ Z ≤ 5 + 1 The correct option is (B) 132. Let z = x + iy |z− 1| = |z + 1| ⇒ Re z = 0 ⇒ x = 0 |z−1| = |z−i| ⇒ x = y |z + 1| = |z−i| ⇒ y = −x Only (0, 0) will satisfy all conditions. ⇒ Number of complex number z = 1 The correct option is (A) 133. Suppose roots are 1 + pi, 1 + qi Sum of roots 1 + pi + 1 + qi = −α which is real ⇒ roots of 1 + pi,1 − pi Product of roots = β = 1 + p2 ∈(1, ∞)

HINTS AND EXPLANATIONS



2.50  Chapter 2 p ≠ 0 since roots are distinct. The correct option is (C) 134. 1 + ω = −ω2

(1 + ω )7 = ( −ω 2 ) = −ω14 = −ω 2 = 1 + ω = A + Bω ⇒ ( A, B ) = (1,1)

The correct option is (A) 135. Let z = x + iy



z 2 = ( x 2 − y 2 ) + i( 2 xy )



z2 is real ⇒ its imaginary part = 0 z −1



⇒ 2 xy( x − 1) − y( x 2 − y 2 ) = 0



⇒ y( x 2 + y 2 − 2 x ) = 0



⇒ y = 0; x 2 + y 2 − 2 x = 0

1+ z 1+ z = =z. 1+ z 1+ 1 z The correct option is (B)

HINTS AND EXPLANATIONS

2

⇒ z1 + 4 z2 − z1 ⇒ z1

z =

2

2

2

2

z2 2

z2 − 4 = 0

(1 − z ) − 4 (1 − z ) = 0 2

2

2

2

2 + 3i sin θ 1 − 2i sin θ

⇒ z = =

( 2 + 3i sin θ )(1 + 2i sin θ ) 1 + 4 sin 2 θ

( 2 − 6 sin 2 θ ) + 7i sin θ

1 + 4 sin 2 θ For z to be purely imaginary, we have Re (z) = 0 ⇒ 2 − 6 sin 2 θ = 0 ⇒ sin θ = ±

z ≥2

1 1 1 3 z+ ≥ z − ≥2− ≥ . 2 2 2 2 Hence, minimum distance between z and The correct option is (B)



= 4 − 2 z1z2 − 2 z1z2 + z1

z1 − 2 z2 =1 2 − z1z2

⇒ ( z1 − 2 z2 ) ( z1 − 2 z2 )

2

⇒ z1 = 2( as z2 ≠ 1) The correct option is (B) 139. We have,

∴

138. Given that

⇒ z1 − 2 z2 − z1 − 2 z2 z1 + 4 z2

2



∴z lies either on real axis or on a circle through origin. The correct option is (A) 136. Given z = 1 ⇒ zz = 1

137.

2



(∴ x ≠ 1 as z ≠ 1 )

= ( 2 − z1z2 ) ( 2 − z1z2 )

3 ⎛ 1 ⎞ ⎜⎝ − , 0⎟⎠ is 2 2

⎛ 1 ⎞ ⇒ θ = ± sin −1 ⎜ ⎟ ⎝ 3⎠ The correct option is (A) 140. Let a =

3 i + 2 2

()

then z = a5 + a So, lm (z) 0

5

()

= 2 Re a

5

1 3

Quadratic Equations and Expressions

CHAPTER

3

LEARNING OBJECTIVES After reading this chapter, you will be able to:   Know about quadratic equations and their roots

 Learn how to represent quadratic equation in a graph and what are rational algebraic expressions

QUADRATIC EQUATION An algebraic expression of the form: ax2 + bx + c, where a (≠0), b, c ∈ R is called a real quadratic expression. An equation of the form: ax2 + bx + c = 0, where a (≠0), b, c ∈ R is called a real quadratic equation. The numbers a, b, c are called the coefficients of the quadratic equation and the expression b 2 − 4 ac is called its discriminant. Discriminant of a quadratic equation is usually denoted by D or D.

Roots of the Quadratic Equation A root of the quadratic equation

ax2 + bx + c = 0

(1) 2

is a number a (real or complex) such that aa + ba + c = 0. The roots of the quadratic Eq. (1) are given by, x=

−b ± b 2 − 4 ac 2a

Nature of Roots of the Quadratic Equation 1. If D < 0, then roots a, b are imaginary 2. If D > 0, then roots α, β are real and distinct 3. If D = 0, then roots α, β are real and equal

QUICK TIPS For the quadratic equation ax2 + bx + c = 0 ■ One root will be reciprocal of the other if a = c. ■ One root is zero if c = 0. ■ Roots are equal in magnitude but opposite in sign if b = 0. ■ Both roots are zero if b = c = 0. ■ Roots are positive if D > 0, a and c are of same sign and b is of opposite sign. ■ Roots are of opposite sign if a and c are of opposite sign. ■ Roots are negative if D > 0 and a, b, c are of the same sign. ■ Roots are rational ⇔ D is a perfect square ■ Roots are irrational ⇔ D is positive but not a perfect square. c ■ If a + b + c = 0, then 1, are the roots of the equation a ax2 + bx + c = 0 c and if a – b + c = 0, then the roots are –1 and – . a 2 ■ If ax + bx + c = 0 is satisfied by more than two values, it is an identity and a = b = c = 0 and vice-versa. 2 ■ If ax + bx + c = 0, where a, b, c ∈ R, has one root p + iq, then the other root will be p – iq. Hence, the imaginary roots occur in conjugate pair.

3.2  Chapter 3 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. ■ 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). ■ If a = 1, b, c ∈ Z and the roots are rational numbers, then these roots must be integers. 2 ■ The condition that the roots of the equation ax + bx + c = 2 0 may be in the ration m : n is mnb = ac(m + n). 2 3 ■ If sum of roots of ax + bx + c = 0 is equal to the sum of c b a their reciprocals, then ab2, bc2, ca2 are in A.P. or , , b a c are in H.P. ■

Solution: (C) Let g(x) = x2 + lx + m, then g(x) is an integer for every integer x. g(0) = m ⇒ m is an integer

g(1) = 1 + l + m ⇒ l is an integer

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 (B) irrational (C) imaginary (D) equal Solution: (A) We have,

( m is integer)

4. The equation 125x + 45x = 227x has (A)  no solution (B)  one solution (C)  two solutions (C)  more than two solutions Solution: (B) The given equation can be written as

D = (c + a – b)2 – 4 (b + c – a) (a + b – c)

(5/3)3x + (5/3)x = 2 Putting (5/3)x = t, the equation becomes

= (a + b + c – 2b)2 – 4 (a + b + c – 2a) (a + b + c – 2c) 2

t3 + t – 2 = 0

2



= (– 2b) – 4 (­­– 2a) (– 2c) = 4 (b – 4ac)



= 4 [(– a – c)2 – 4ac] = 4 (a – c)2



= [2 (a – c)]2 = perfect square

t3 – 1 + (t – 1) = 0

⇒

⇒ (t – 1) (t2 + t + 1) + (t – 1) = 0 ⇒ (t – 1) (t2 + t + 2) = 0

∴ Roots are rational

⇒

2. The number of values of the pair (α, β ) for which the equation α (x + 1)2 + β (x2 – 3x – 2) + x + 1 = 0, ∀ x ∈ R is (A) 1 (B) 0 (C)  infinite (D)  None of these

t = 1 2

t + t + 2 = 0

or 2

But t + t + 2 = 0 does not have real solutions. Therefore,

t = 1 ⇒ (5/3)x = 1 ⇒ x = 0.

5. For a > 0, the roots of the equation logax a + logx a2 + log a x a3 = 0, are given by:

Solution: (B) The equation

2

(A) a–4/3 (B)  a–3/4 (C)  a1/2 (D)  a–1

(α + β)x2 + (2α – 3β + 1)x + (α + 2β + 1) = 0 is an identify in x if

α + β = 0

(1)

2α – 3β + 1 = 0

(2)

α – 2β + 1 = 0

(3)

and

3. If x2 + lx + m is an integer for every integer x, then (A) l is always an integer but m need not be an integer (B) m is always in integer but l need not be an integer (C) l and m, both are always integer (D)  None of these

Therefore, Also,

SOLVED EXAMPLES



1 1 and β = . 5 5 But these values do not satisfy Eq. (3). Hence the given equation cannot be an identify for any values of α and β.

Solving Eq. (1) and (2), we get α = −

Solution: (A) We have, log a a 2 log a a 3 log a a + + =0 log a a + log a x log a x 2 log a a + log a x

Quadratic Equations and Expressions  3.3 ⇒ ⇒

1 2 3 = 0  (let loga x = t) + + 1+ t t 2 + t 2

2

2

2t + t + 2t + 6t + 4 + 3t + 3t =0 t (1 + t ) ( 2 + t ) 2

⇒ 6t + 11t + 4 = 0 ⇒ 6t + 8t + 3t + 4 = 0

⇒

Clearly for these values of a, D > 0. Hence, 1 < a < 2.

⇒ (2t + 1) (3t + 4) = 0 1 4 ⇒ t=– ,– 2 3 1 4 ⇒ loga x = – , – 2 3 –1/2 –4/3 ∴ x=a ,a

9. The number of real solutions of the equation 271/x + 121/x = 2 × 81/x is (A) one (B) two (C) infinite (D) zero Solution: (D) The given equation can be written as

6. The number of solutions of the equation sin (ex) = 5x + 5– x is (A) 0 (B) 1 (C) 2 (D) infinite Solution: (A) Put 5x = y. Then the given equation becomes

⎛ 3⎞ ⎜⎝ ⎟⎠ 2 ⎛ 3⎞ Put ⎜ ⎟ ⎝ 2⎠

⎛ 1 1 ⎞ x = ⎜ y− ⎟ + 2  (∵ 5 > 0) y y⎠ ⎝ sin (ex) ≥ 2.

Which is not possible for any real value of x. Hence, the given equation has no real solution. 2/3

1/3

3

Solution: (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) 3

x – 6x + 12x – 8 = –6 + 6x.

∴

x3 – 6x2 + 6x = 2.

1/ x

= 2.

1/ x

= t, then the equation becomes

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. 10. For all real values of x,

8. 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 (B)  a ∈ (2, ∞) (C) 1 < a < 3 (D)  None of these Solution: (A) Roots are of opposite sign if (a) roots are real and ­distinct, (b) product is negative.

12 x 4x2 + 9

(A) ≤1 (B)  ≤2 (C)  >1  (D)  >2 Solution: (A) 12 x

Let

4x2 + 9

= y,

Now, 4yx2 – 12x + 9y = 0 As x is real,  D = 144 – 4 ⋅ 4y ⋅ 9y ≥ 0  ⇒  1 – y2 ≥ 0

2

or

⎛ 3⎞ +⎜ ⎟ ⎝ 2⎠

2

7. If x = 2 + 2 + 2 then the value of x – 6x + 6x is (A) 3 (B) 2 (C)  1 (D)  None of these



3/ x

t3 + t – 2 = 0  ⇒ (t – 1) (t2 + t + 2) = 0.

2

⇒

a 2 − 3a + 2 0

So,

y2 ≤ 1;

⇒

∴ |y| ≤ 1. Hence,

12 x 4x2 + 9

≤ 1.

11. 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

3.4  Chapter 3 Solution: (B) Since x2 – 3x + 2 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 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. 12. If the roots of x2 – bx + c = 0 are two consecutive ­integers, then b2 – 4c is (A) 1 (B) 0 (C)  2 (D)  None of these Solution: (A) The roots of the equation are given by, 2

x =

α=

If

b ± b − 4c 2  2

b + b − 4c 2  2

b − b − 4c 2  α – β = 1

β=

and Then, ⇒

b 2 − 4c = 1

⇒

b2 – 4c = 1. 2

13. If p (q – r) x + q (r – p) x + r (p – q) = 0 has equal 2 roots, then = q 1 1 (A) p + (B)  +r r p 1 1 (C) p + r (D)  + p r Solution: (D) Since p (q – r) + q (r – p) + r (p – q) = 0 ∴ one root is 1 ∴

r ( p − q) other root = . p (q − r) 

Since roots are equal ∴

rp − rq = 1 pq − pr

⇒

rp – rq = pq – pr

⇒ 2rp = q (p + r) 2 p+r 1 1 = = + . q p r  pr p a b 14. If c ≠ 0 and the equation = + has two 2x x+c x−c equal roots, then p can be ∴

(A) ( a − b ) 2 (B)  ( a + b )2 (C) a + b (D)  a–b Solution:  (A, B) We can write the given equation as or

p ( a + b) x + c ( b − a) = 2x x 2 − c2  p (x2 – c2) = 2 (a + b) x2 – 2c (a – b) x

or (2a + 2b – p) x2 – 2c (a – b) x + pc2 = 0 For this equation to have equal roots c2(a – b)2 – pc2 (2a + 2b – p) = 0 ⇒ (a – b)2 – 2p (a + b) + p2 = 0

( c2 ≠ 0)

⇒ [  p – (a + b)]2 = (a + b)2 – (a – b)2 = 4ab ⇒ ⇒

p – (a + b) = ± 2 ab  p = a + b ± 2 ab = ( a ± b ) 2 

15. If (7 − 4 3 ) x − 4 x + 3 + (7 + 4 3 ) the value of x is given by 2

(A)  2, 2 ± (C) 3 ±

2 2 , 2

x2 − 4 x +3

(B) 2 ±

= 14, then

3,3

(D)  None of these

Solution: (A) (7 + 4 3 ) (7 − 4 3 ) = 1,

Since

∴ the given equation becomes 1 y+ = 14 y

y = (7 − 4 3 ) x

where ⇒ Now ⇒

⇒

− 4x + 3

2

y – 14y + 1 = 0 ⇒ y = 7 ± 4 3  y= 7±4 3 x2 – 4x + 3 = –1

⇒ Also,

2

x = 2, 2 y= 7−4 3 x – 4x + 3 = 1 ⇒ x = 2 ± 2  2



Quadratic Equations and Expressions  3.5

Sum and Product of the Roots

17. If the ratio of the roots of lx2 + nx + n = 0 is p : q, then

If α and β are roots of ax2 + bx + c = 0, then

(A) 

q + p

p + q

l =0 n

(B) 

p + q

q + p

n =0 l

(C) 

q + p

p + q

l =1 n

(D) 

p + q

q + p

n =1 l

−b Coefficient of x =– a Coefficient of x 2  c Constant term Product of roots = αβ = = a Coefficient of x 2 

Sum of roots = α + β =

Formation of Equation with Given Roots If α and β are roots of 2



f (x) = ax + bx + c = 0,

then

f (x) = (x – α) (x – β ) = 0 = x2 – (α + β )x + αβ = 0



i.e., x2 – (sum of the roots) x + (product of the roots) = 0.

A quadratic equation with all odd integer coefficients cannot have rational roots.

SOLVED EXAMPLES 16. If r be the ratio of the roots of the equation ( r + 1) 2 ax2 + bx + c = 0, then = r a2 b2 (A)  (B)  bc ca 2 c (C)  (D)  None of these ab Solution: (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 and rα2 = (3) a From Eq. (2), b α=– (4) a ( r + 1) Putting the value of α in Eq. (3), we get or,

a ( r + 1)

2

=

c a

b2 ( r + 1) 2 = ac r 

p + q

Now,



rb 2

−n n ; αβ = ; l l α p = β q

α+β=

Then and

ERROR CHECK

2

Solution: (B) Let the roots be α and β.

=

q + p

α + β + αβ αβ

=

n = l −

α β + + αβ β α

n n + l l = 0 n l

18. In a quadratic equation with leading coefficient 1, a student reads the coefficient 16 of x wrongly as 19 and obtain the roots as – 15 and – 4. The correct roots are (A)  6, 10 (B)  –6, –10 (C)  –7, –9 (D)  None of these Solution: (B) Since coefficient of x = 16, ∴ sum of roots = –16 Since constant term = (–15) (–4) = 60, ∴ correct answer is –6, –10. 1 1 1 + = are x+a x+b c equal in magnitude but opposite in sign, then their product is

19. If the roots of the equation

−1 2 (B)  (a + b2) 2 −1 (C)  ab (D)  ab 2 1 (A)  (a2 + b2) 2 −1 × ⎡⎣(ω 33 )3ω 2 + ω 214 ⎤⎦

Solution: (B) We have, ⇒

((x + b) + (x + a)c = (x + a) (x + b) x2 + bx + ax – 2cx + ab – bc – ca = 0

3.6  Chapter 3 Now, let roots be α and β, then

α + β = 0, αβ = ab – bc – ac



α + β = 0 ⇒ b + a = 2c

and ⇒ ⇒ ∴

αβ = ab – (b + a) c ( a + b) 2 αβ = ab – 2  1 2 2 αβ = (– a – b ) 2 1 αβ = – (a2 + b2) 2

20. If sin θ and cos θ are the roots of the equation ax2 + bx + c = 0, then (A) (a – c)2 = b2 – c2 (B) (a – c)2 = b2 + c2 (C) (a + c)2 = b2 – c2 (D)  (a + c)2 = b2 + c2 Solution: (D) Since sin θ and cos θ are the roots of the equation ax2 + bx + c = 0 c b ∴ sin θ + cos θ = − and sin θ cos θ = a a Now (sin θ + cos θ )2 = 1 + 2 sin θ cos θ ∴ ⇒ ⇒ Hence,

2

2c a + 2c = a a  a 2 b = a (a + 2c) = a2 + 2ac

b

2

=1 +

b2 + c2 = a2 + 2ac + c2 = (a + c)2 (a + c)2 = b2 + c2

21. 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 (B)  4, 6 (C)  4, 5 (D)  None of these Solution: (A) In case of the first student, product of the roots = 3 × 10 = q. So the correct value of q is 30. In case of the second student, sum of the roots = 4 + 7 = –p. So the correct value of p is –11. 2

∴ The correct equation is x – 11x + 30 = 0 or (x – 5) (x – 6) = 0; ∴

x = 5, 6.

∴ Roots of the correct equation are 5, 6.

22. 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

Solution: (B) We have, α + β = 2p;



αβ = q, γ + δ = 2r and γ δ = s   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; s – q = r2 – p2

or

23. 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 (B)  a = – 13, b = 8 (C) a = 13, b = 8 (D)  a = – 13, b = ­– 8 Solution: (A)

One root =

1 4+ 3

×

4− 3 4− 3

=

4− 3 13 

4+ 3 13  ∴ The quadratic equation is ∴

other root =

⎛ 4 + 3 4 − 3⎞ 4+ 3 4− 3 ⋅ + x+ x 2 − ⎜ = 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.

24. 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 + (A) x2 + 4ax – 2b = 0 (C) x2 – 4ax + 2b = 0

α 2 + β 2 is (B)  x2 + 4ax + 2b = 0 (D)  x2 – 4ax – 2b = 0

Solution: (B) Since α, β are roots of x2 + 2ax + b = 0



α + β = –2a and αβ = b

y = α + β + α 2 + β2  ⇒ (y + 2a)2 =α2 + β 2 = (α + β)2 – 2αβ = 4a2 – 2b Let

⇒ y2 + 4ay + 2b = 0 So, the required equation is x2 + 4ax + 2b = 0.

Quadratic Equations and Expressions  3.7 25. 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 (B)  a, c (C)  b, d (D)  a, b Solution: (D) (x – a) (x – b) – k = 0

We have, ⇒

x2 – (a + b) x + ab – k = 0

(1)

Since the roots of Eq. (1) are c and d ∴

c + d = a + b,(2) cd = ab – k(3)

and

Now (x – c) (x – d) + k = 0 ⇒

x2 – (c + d) x + cd + k = 0 2

⇒ x – (a + b) x + ab = 0  [Putting the values of a + b and  ab from Eqs (2) and (3)] ⇒ (x – a) (x – b) = 0 ⇒ x = a, b. 26. 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 (B) 1 (C) 0 (D) –4 Solution: (D) We know that if α, β are roots of the equation Ax2 + Bx + C = 0, B 2 − 4 AC A  Equating the value of α – β from both the given ­equations, we get

α–β=

then

⇒

b 2 − 4c = c 2 − 4b  b2 – 4c = c2 – 4b

⇒

b2 – c2 = –4 (b – c)



⇒ (b – c) (b + c + 4) = 0 ⇒ b + c = –4  (∵ b ≠ c) 27. 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 Solution: (D) b2 – 4ac < 0

Here ∴

α=

− b + i 4 ac − b 2 2a 

− b − i 4 ac − b 2  2a

and

β=

∴

α = β .

COMMON ROOTS One Root Common If α is a common root of the equations a1x2 + b1x + c1 = 0(1) and a2x2 + b2x + c2 = 0(2) then we have and These give

a1α 2 + b1α + c1 = 0 a2α 2 + b2α + c2 = 0

α2 α = b1c2 − b2 c1 c1a2 − c2 a1

 1 = ( a1b2 − a2 b1 ≠ 0). a1b2 − a2 b1



Thus, the required condition for one common root is ( a1b2 − a2 b1 ) (b1c2 − b2 c1 ) = (c1a2 − c2 a1 ) 2 and the value c a − c2 a1 b c − b2 c1 of the common root is α = 1 2 or 1 2 . a1b2 − a2 b1 c1a2 − c2 a1

Both Roots Common If the Eq. (1) and (2) have both roots common, then these equations will be identical. Thus the required condition for both roots common is a1 b c = 1 = 1 a2 b2 c2

QUICK TIPS 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. ■ If two quadratic equations with real coefficients have an imaginary root common, then both roots will be common and the two equations will be identical. The required condition is a1 b c = 1 = 1 a2 b2 c2 ■

If two quadratic equations have an irrational root common, then both roots will be common and the two equations will be identical. The required condition is a1 b c = 1 = 1 a2 b2 c2 ■

3.8  Chapter 3 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. ■ 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. ■

30. 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 (B)  a=b=–c (C) a = b = c (D)  None of these Solution: (C) We have,

x3 + 3x2 + 3x + 2 = 0

⇒ (x + 1)3 + 1 = 0 ⇒ (x + 1 + 1) [(x + 1)2 – (x + 1) + 1] = 0

SOLVED EXAMPLES 2

28. The value of k so that the equations x – x – 12 = 0 and kx2 + 10x + 3 = 0 may have one root in common, is 43 43 (A)  (B) 3 (C) – 3 (D)  16 18 Solution:  (B, D) Let α be the common root Then, α2 – α – 12 = 0  and  kα2 + 10α + 3 = 0 Solving the two equations, we get ⇒ ⇒

α2 α 1 = = 117 − 12k − 3 10 + k  (– 12k – 3)2 = 117 (10 + k) 9 (4k + 1)2 = 117 (10 + k) 2

⇒ (4k + 1) = 13 (10 + k) ⇒ 16k2 + 8k + 1 = 130 + 13k ⇒ 16k2 – 5k – 129 = 0 ⇒ 16k2 – 48k + 43k – 129 = 0 ∴

k = 3  or  k = 2

− 43 16 

2

29. If the equations ax + bx + c = 0 and x + 2x + 3 = 0 have a common root, then a : b : c = (A)  2 : 4 : 5 (B)  1 : 3 : 4 (C)  1 : 2 : 3 (D)  None of these Solution: (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 both roots will be common. Hence both the equations are identical. a b c = = ∴ 1 2 3 i.e.

a : b : c = 1 : 2 : 3.

⇒ (x + 2) (x2 + x + 1) = 0 − 1 ± 3i 2  x = –2, ω, ω 2

⇒ x = –2, ⇒

Since a, b, c ∈ R, ax2 + bx + c = 0 cannot have one real and one imaginary root. Therefore, two common roots of ax2 + bx + c = 0 and x3 + 3x2 + 3x + 2 = 0 are ω, ω2. b Thus, − = ω + ω2 = –1 a ⇒ a = b c and = ω ⋅ ω2 = 1 ⇒ c = a a ⇒ a = b = c 31. 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 (B) 1 (C)  –1 (D)  None of these Solution: (A) The two equations can be written as x2 (6k + 2) + rx + (3k – 1) = 0

(1)

and x2 (12k + 4) + px + (6k – 2) = 0 Divide by 2, we get p x + (3k – 1) = 0 x2 (6k + 2) + 2 Comparing Eq. (1) and (3), we get p r = 2 ∴ 2r – p = 0

(2) (3)

32. If the equations x2 – ax + b = 0 and x2 + bx – a = 0 have a common root, then (A) a + b = 1 (B) a = b (C) a – b = 2 (D) a + b = 0 or a – b = 1

Quadratic Equations and Expressions  3.9 Solution: (D) Let α be a common root of the given equations. Then

= ⎛ p2 − ⎜⎝

α2 – aα + b = 0  and  α2 + bα – a = 0

2

⎞ +2+ 2 ≥ 2+ 2  2⎟ 2p ⎠ 1

Therefore, minimum value of α4 + β4 is 2 + 2

⇒ (a + b) α – (a + b) = 0

GRAPH OF A QUADRATIC EXPRESSION

⇒ (a + b) (α – 1) = 0

We have, y or f (x) = ax2 + bx + c where a, b, c ∈ R, a ≠ 0.

a + b = 0 or α = 1

⇒

α = 1,

If

1 – a + b = 0 ⇒ a – b = 1.

then

SYMMETRIC FUNCTION OF THE ROOTS A function of α and β is said to be a symmetric function if it remains unchanged when α and β are interchanged. For example, α 2 + β 2 + 2αβ is a symmetric ­function of α and β whereas α 2 – β 2 + 3αβ is not a symmetric ­function of α and β.

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 a > 0, D > 0 x-axis x-axis

a > 0, D > 0

QUICK TIPS



In order to find the value of a symmetric function of α and β, express the given function in terms of α + β and αβ. The following results may be useful. 2 2 2 ■ α + β = (α + β ) – 2αβ 3 3 3 ■ α + β = (α + β ) – 3αβ (α + β ) 4 4 3 3 2 2 ■ α + β = (α + β ) (α + β ) – αβ (α + β ) 5 5 3 3 2 2 2 2 ■ α + β = (α + β ) (α + β ) – α β (α + β ) |α – β | =

■

2

2

(α + β ) − 4αβ

FIGURE 3.1 (A)

6. For D = 0, parabola touches x-axis in one point. a > 0, D = 0

a > 0, D = 0

x-axis

x-axis

FIGURE 3.2 (A)

FIGURE 3.2 (B)

7. For D < 0, parabola does not cut x-axis.

2

α – β = (α + β ) (α – β ) 3 3 2 ■ α – β = (α – β ) [(α + β ) – αβ] 4 4 2 2 ■ α – β = (α + β ) (α – β ) (α + β ) ■

FIGURE 3.1 (B)

a < 0, D < 0

a > 0, D < 0

SOLVED EXAMPLE



33. If α and β be the roots of the equation x2 + px –

(A)  2 (B)  2+ 2 (C) 2 − 2 (D)  2 Solution: (B)

FIGURE 3.3 (A)

FIGURE 3.3 (B)

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

α 4 + β 4 = (α 2 + β 2) – 2α2β 2 2

x-axis

1

= 0, 2 p2 where p ∈ R, then the minimum value of α4 + β 4 is

2

x-axis

2



= [(α + β) – 2αβ] – 2(αβ )



1 ⎞ 1 1 ⎛ = ⎜ p2 + 2 ⎟ − = p4 + +2 4 2 p4 ⎝ p ⎠ 2p 

4 ac − b 2 b  at x = – 4a 2a

2. If a < 0, then the quadratic expression y = ax2 + bx + c has no least value but it has greatest value

2



4 ac − b 2 b  at x = − 4a 2a

3.10  Chapter 3

Sign of Quadratic Expression We have, y or f (x) = ax2 + bx + c where a, b, c ∈ R, a ≠ 0. 1. If a > 0 and D < 0, then f (x) > 0 for all x ∈ R i.e., f (x) is positive for all real values of x. 2. If a < 0 and D < 0, then f (x) < 0 for all x ∈ R i.e., f (x) is negative for all real values of x. 3. If a > 0 and D = 0, then f (x) ≥ 0 for all x ∈ R i.e., f (x) is positive for all real values of x except at vertex, where f (x) = 0. 4. If a < 0 and D = 0, then f (x) ≤ 0 for all x ∈ R i.e. f (x) is negative for all real values of x except at vertex, where f (x) = 0. 5. If a > 0 and D > 0, let f (x) = 0 have two real roots α and β(α < β ), then f (x) > 0 for all x ∈ (–∞, α) ∪ (β, ∞) and f (x) < 0 for all x ∈ (α, β ). 6. If a < 0 and D > 0, let f (x) = 0 have two real roots α and β(α < β ). Then f (x) < 0 for all x ∈ (–∞, α) ∪ (β, ∞) and f (x) > 0 for all x ∈ (α, β ).

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.

Nature of Roots with Respect to One Real Number 1. If both the roots of f (x) = 0 are greater than k, then b D ≥ 0, a f (k) > 0 and k < – 2a α

k

β

2. If both the roots of f (x) = 0 are less than k, then D ≥ 0, b a f (k) > 0 and k > – 2a α

β

k

3. If one root is less than k and other is greater than k, then D > 0 and a f (k) < 0 α

β

k

Roots with Respect to Two Real Numbers 1. If exactly one root of f (x) = 0 lies in the interval (k1, k2), then D > 0 and f (k1) · f (k2) < 0 k1

α

k2

β

2. If both roots of f (x) = 0 lie between k1 and k2, then α+β D ≥ 0, a f (k1) > 0, a f (k2) > 0 and k1 < < k2 2 k1

α

β

k2

3. If k1 and k2 lie between the roots of f (x) = 0, then D ≥ 0, a f (k1) < 0 and a f (k2) < 0. α

k1

k2

β

QUICK TIPS 1. Let f(x) = 0 be a polynomial equation. Let p and q be two real numbers, p < q. (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. (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) 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

Quadratic Equations and Expressions  3.11

RELATION BETWEEN ROOTS AND COEFFICIENTS OF A POLYNOMIAL EQUATION

FORMATION OF A POLYNOMIAL EQUATION FROM GIVEN 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,

If α1, α2, α3, … αn are the roots of a polynomial equation of degree n, then the equation is

1. Sum of all roots σ1 = α1 + α2 + … + αn a a = Σα1 = – 1 = (–1)1 1 a0 a0

where σr = Σα1α2 … αr.

2. Sum of the product of two roots σ2 = α1α2 + α1α3 + … a a = Σα1α2 = 2 = (–1)2 2 a0 a0 3. Sum of the product of three roots σ3 = α1α2α3 + α2α3α4 + … a a = Σα1α2α3 = – 3 = (–1)3 3 and so on. a0 a0 a In general, σr = Σα1α2 … αr = (–1)r r . a0

Particular Cases Quadratic Equation If α, β are roots of the quadratic equation ax2 + bx + c = 0, then b c α + β = – and αβ = a a

Cubic Equation If α, β, γ are roots of the cubic equation ax3 + bx2 + cx + d = 0, then b σ1 = α + β + γ  = – a c σ2 = αβ + αγ + βγ =  a d σ3 = αβγ = – a

Biquadratic Equation If α, β, γ, δ are roots of the biquadratic equation ax4 + bx3 + cx2 + dx + e = 0, then b σ1 = α + β + γ  + δ = – a c σ2 = αβ + αγ + αδ + βγ + βδ + γδ = a d σ3 = αβγ + αβδ + αγδ + βγδ = – a e σ4 = αβγδ = a

xn – σ1xn – 1 + σ2xn – 2 – σ3xn – 3 + … + (–1)nσn = 0

Particular Cases Quadratic Equation If α, β are the roots of a quadratic equation, then the ­equation is x2 – (α + β)x + αβ = 0.

Cubic Equation If α, β, γ are the roots of a cubic equation, then the equation is 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 x4 – σ1x3 + σ2x2 – σ3x + σ4 = 0 or x4 – (α + β + γ + δ )x3 + (αβ + αγ + αδ + βγ + βδ + γ δ )x2 – (αβγ + αβδ + αγδ + βγδ)x + αβγδ = 0

SIGN OF A POLYNOMIAL EXPRESSION Step 1: Factorize the given polynomial expression as f (x) = ( x − α1 ) k ( x − α 2 ) k ( x − α 3 ) k … ( x − α n −1 ) k ( x − α n ) k 1

3

2

n −1

n

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. –∞

α1

α2

α 3 ....

αn

∞

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. Step 6: The solution of f (x) > 0 is the union of all the intervals in which there is a ‘+’ sign and the solution of f (x) < 0 is the union of all the intervals in which there is a ‘–’ sign.

3.12  Chapter 3 For example, consider the polynomial expression 40

31

f (x) = (x + 2) (x + 1) +

–

+

–2

Thus,

⎛ ⎜⎝ x −

–1

3⎞ ⎟ 2⎠

58

(x – 4)37

– 3/2

+ 4

⎛ f (x) > 0 if x ∈ (–∞, –2) ∪ ⎜ − 1, ⎝

3⎞ ⎟ ∪ (4, ∞) 2⎠

⎛3 ⎞ f (x) < 0 if x ∈ (–2, –1) ∪ ⎜ , 4⎟ . ⎝2 ⎠

and

1⎞ ⎛ 34. If the expression ⎜ ax − 1 + ⎟ is non-negative for all ⎝ x⎠ positive real x, then the minimum value of a must be 1 2

(A) 0

(B) 

1 (C)  4

(D)  None of these

ax – 1 + ⇒

1 ax 2 − x + 1 ≥0⇒ ≥ 0 x x

2

ax – x + 1 ≥ 0 as x > 0 a > 0 and 1 – 4a ≤ 0 ⇒ a ≥

1 4

Therefore, the minimum value of a is

1 . 4

2

35. If α, β are the roots of the equation x – 3x + a = 0, a ∈ R and α < 1 < β, then a belong to 9⎞ ⎛ (A)  ⎜ − ∞, ⎟ ⎝ 4⎠

(B) (–∞, 2)

⎛9 ⎞ (C) (2, ∞) (D)  ⎜⎝ , ∞⎟⎠ 4 Solution:  (B, C) Since 1 lies between the roots of the given equation, therefore, D > 0 and f (1) < 0 ⇒ 9 – 4a > 0 and 1 – 3 + a 4 Solution: (A) We can write the given equation as (x – a)2 = 3 – a

It will hold if a > 0 and D ≤ 0 ⇒

x + 3 − 4 x − 1 +

where t ≥ 0. ⇒ |t – 2| + |t – 3| = 1,

SOLVED EXAMPLES

Solution: (C) We have,

36. The equation

This shows that a ≤ 3 and x = a ± 3 − a Both the roots of the given equation will be less than 3 if the larger of the two roots is less than 3, that is, if a +

3 − a < 3

⇒

3 − a – (3 – a) < 0

⇒

3 − a (1 − 3 − a ) < 0

⇒

a < 3 and 1 − 3 − a < 0

But Thus,

3 − a > 1 ⇒ 3 – a > 1 or a < 2 a < 3 and a < 2 ⇒ a < 2

38. If f (x) = x2 + 2bx + 2c2 and g(x) = – x2 – 2cx + b2 such that min f (x) > max g(x), then the relation between b and c is: (A) |c| < |b| 2 (B) 0 < c < b 2 (C) |c| < |b| 2

(D)  |c| > |b| 2

Quadratic Equations and Expressions  3.13 Solution: (D) 2

2

D 4b − 8c =– 4 4a  2 2 = –(b – 2c ) (upward parabola)

min f (x) = –

D 4c 2 + 4b 2 max g(x) = – = 4a 4  2 2 =b +c  (downward parabola) Now 2c2 – b2 > b2 + c2 c2 > 2b2 ⇒ |c| >

⇒ 39.

2 |b|

Solution: (B) ∴ equation is

∴

 x < 0

We have,

 x ≥ 0

x = 2 is the solution.

Hence, x = 2, –2 are the solutions and their sum is zero. 40. If a ≤ 0, then the root of the equation x2 – 2a | x – a | – 3a2 = 0 is ( − 1 + 6 )a (A) (1 − 2 )a (B)  − (1 + 6 )a (C) (1 + 2 )a (D)  Solution:  (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

(C) infinite   (D) 2

⎛ 1⎞ f (x) + f ⎜ ⎟ = 1 ⎝ x⎠ 1 ⎡1⎤ − = 1 x ⎢⎣ x ⎥⎦ 1 ⎡1⎤ x + − 1 = [x] + ⎢ ⎥ x ⎣x⎦  2 x +1− x = (integer) k (say) x

⇒ x – [x] +

⇒

⇒

2

∴

Solution: (C)

⇒

∴ equation is

∴

(B) 1

x2 – (k + 1) x + 1 = 0

Since x is real, so (k + 1)2 – 4 ≥ 0

x ≥ 0, |x| = x,

x + x – 6 = 0 ⇒ x = 2, –3 

x = (1 − 2 ) a.

41. If f (x) = x – [x], x (≠0) ∈ R, where [x] is the greatest integer less than or equal to x, then the number of solu⎛ 1⎞ tions of f (x) + f ⎜ ⎟ = 1 are ⎝ x⎠

⇒

x = –2 is the solution

For,

∴

⇒

x < 0, |x| = – x

x2 – x – 6 = 0 ⇒ x = – 2, 3 

x2 – 2ax – a2 = 0 x = (1 + 2 ) a, (1 − 2 ) a   x ≥ a and a ≤ 0

(A) 0

For the equation |x2| + |x| – 6 = 0, the roots are (A)  real and equal (B)  real with sum 0 (C)  real with sum 1 (D)  real with product 0 For,

⇒ ⇒ 

 x < a ≤ 0

x = ( − 1 + 6 ) a

If x – a ≥ 0, |x – a| = x – a ∴ the equation becomes x2 – 2a (x – a) – 3a2 = 0

k2 + 2k – 3 ≥ 0  ⇒ (k + 3) (k – 1) ≥ 0 k ≤ – 3  or  k ≥ 1

Therefore, number of solutions is infinite. 42. If (log5x)2 + log5x < 2, then x belongs to the interval ⎛1 1 ⎞ ⎛ 1 ⎞ (A)  ⎜ , 5⎟ (B)  ⎜⎝ 5 , ⎟ ⎝ 25 ⎠ 5⎠ (C) (1, ∞)

(D)  None of these

Solution: (A) We have, (log5 x)2 + log5 x < 2 Put log5 x = a then a2 + a < 2 ⇒

a2 + a – 2 < 0

⇒ (a + 2) (a – 1) < 0 ⇒ –2 < a < 1 or –2 < log5 x < 1 ∴ 5–2 < x < 5 1 i.e., < x < 5 25 43. The greatest negative integer satisfying x2 – 4x – 77  4 is (A) –4 (B) –7 (C)  –6 (D)  None of these

3.14  Chapter 3 Solution: (C)

46. The solution set of the inequality log ⎛ π ⎞ (x2 – 3x sin ⎜ ⎟ + 2) ≥ 2 is ⎝ 3⎠

We have, x2 – 4x – 77 < 0 and x2 – 4 > 0 ⇒ (x + 7) (x – 11) < 0 and (x – 2) (x + 2) > 0 ⇒ –7 < x < 11 and x < – 2 or x > 2 ∴

– 7 < x < –2

44. The solution of the inequation 4–x + 0.5 – 7.2–x < 4, x ∈ R, is (A) (–2, ∞) (B)  (2, ∞) ⎛ 7⎞ (C)  ⎜ 2, ⎟ ⎝ 2⎠

(D)  None of these

⎛1 ⎞ ⎛ 5⎞ (A)  ⎜ , 2⎟ (B)  ⎜⎝1, ⎟⎠ ⎝2 ⎠ 2 ⎡1 ⎞ ⎛ 5⎤ (C)  ⎢ , 1⎟ ∪ ⎜ 2, ⎥ ⎣2 ⎠ ⎝ 2⎦

(D)  None of these

Solution: (C) We have, log ⇒

⎛π⎞ sin ⎜ ⎟ ⎝ 3⎠

(x2 – 3x + 2) ≥ 2 x2 – 3x + 2 ≤

3 4

5 ≤ 0 4 ⇒ 4x2 – 12x + 5 ≤ 0

Solution: (A) The given inequation is

⇒

4–x + 0.5 – 7.2–x < 4, x ∈ R Let 2–x = t ∴ 2t2 – 7t < 4

⇒ 4x2 – 10x – 2x + 5 ≤ 0 ⇒ (2x – 5) (2x – 1) ≤ 0 1 5 ⇒ ≤ x ≤ (1) 2 2 2 Also, x – 3x + 2 > 0

⇒ 2t2 – 7t – 4 < 0 ⇒ (2t + 1) (t – 4) < 0 1 ⇒ – < t < 4 2 but 2–x > 0 so 0 < t < 4

⇒ (x – 1) (x – 2) > 0 ⇒

⎡1 ⎞ ⎛ x ∈ ⎢ , 1⎟ ∪ ⎜ 2, ⎝ ⎠ 2 ⎣

⇒ –2 < x < ∞ or x ∈ (–2, ∞) x

x

⎛ 1⎞ ⎛ 1⎞ 45. The real values of x for which 372 ⎜ ⎟ ⎜ ⎟ > 1, ⎝ 3⎠ ⎝ 3⎠ are (A) x ∈ [0, 64] (B)  x ∈ (0, 64) (C) x ∈ [0, 64) (D)  None of these Solution: (C) The given inequation is valid only when x ≥ 0 The given inequation can be written in the form 372 − x −

⇒

72 – x –

⇒

x+

x

(1)

> 1

x > 0

x – 72 < 0

⇒ ( x + 9) ( x – 8) < 0 But

x + 9 > 0 for all x ≥ 0

∴

x –8 2

(2)

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

⇒ 0 < 2–x < 4



x2 – 3x +

(∵ 3 > 1)

5⎤ 2 ⎥⎦

47. The values of a which make the expression x2 – ax + 1 – 2a2 always positive for real values of x, are 2 2 ≤a≤ 3 3 2 (C) a < 1 (D)  0 4

⇒

Solution: (A) Since the roots of the given equation are real 2

∴

B – 4AC ≥ 0 2

2

⇒ 4a – 4 (a + a – 3) ≥ 0 ⇒ –a + 3 ≥ 0  or  a ≤ 3

(1)

Since the root is less than 3, so f (3) > 0 2

a < 2  or  a > 3

(2)

From Eq. (1) and (2), we have a < 2. 49. 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 (B)  k < 2 (C) 2 < k < 3 (D)  k > 5

t2 – 2 2 t + 1 ≥ 0

⇒

[t − ( 2 − 1)] [t − ( 2 + 1)] ≥ 0

⇒

t≤

2 − 1  or t ≥

2 +1 

t > 0 2 − 1  or 2x ≥

but ⇒ 0 < 2x ≤

⎡1 ⎞ ∴ x ∈ (–∞, log2 ( 2 − 1) ] ∪ ⎢ , ∞⎟ ⎣2 ⎠  2 51. If a < b, then the solution of x + (a + b) x + ab < 0, is given by (A) x < –b or x < –a (B)  a 0

or

x + a > 0, x + b < 0

⇒

x < –a, x > –b

or

x > ­–a, x < –b

Now, f (3) = 9 + (1 – 2k) 3 + k – k – 2 2

2

= 10 – 7k + k = k – 7k + 10



f (3) < 0

Hence

k2 – 7k + 10 < 0

⇒

⇒ (k – 2) (k –­5) < 0 ⇒ 2 < k < 5. 50. Solution of 2x + 2|x| ≥ 2

2 is

(A) (–∞, log2 ( 2 + 1) (B)  ⎣⎡log 2

(

)

2 + 1 , ∞)

⎛1 ⎞ (C)  ⎜ , log 2 ( 2 − 1)⎟ ⎝2 ⎠ ⎡1 ⎞ (D) (–∞, log2 ( 2 − 1) ] ∪ ⎢ , ∞⎟ ⎣2 ⎠ Solution: (D) We have, 2x + 2x ≥ 2 2 (x ≥ 0) ⇒ 2x ≥

2 ⇒x≥

2 +1 

x ≥ log2 ( 2 + 1)  (but not acceptable as x < 0)

Solution: (A) Let f (x) = x2 + (1 – 2k) x + k2 – k – 2

2



⇒

or 

a2 – 5a + 6 > 0 or (a – 2) (a – 3) > 0

⇒

1 ≥ 2 2 (where t = 2x) t

⇒ –∞ < x ≤ log2 ( 2 − 1) 

2

⇒ 3 – 2a (3) + a + a – 3 > 0 ⇒

t+

1 2

and 2x + 2–x ≥ 2 2 (x < 0)

⇒ (x + a) (x + b) < 0

⇒ –b < x < –a or –a < x < –b Since ∴

a < b –a > –b

Hence –b < x < –a. 52. The conditions that the equation ax2 + bx + c = 0 has both the roots positive is that (A) a and b are of the same sign (B) a, b and c are of the same sign (C) a and c are of the same sign opposite to that of b (D) b and c are of the same sign opposite to that of a Solution: (C) Since both the roots are positive −b c ∴ >0, >0 a a  b c ⇒ 0 a a  ∴ a and c have same sign opposite to that of b.

3.16  Chapter 3 53. The smallest value of x2 – 3x + 3 in the interval 3⎞ ⎛ ⎜⎝ − 3, ⎟⎠ is 2 (B) –15 3 (C)  5 (D)  4 Solution: (D) 2 3⎞ 9 ⎛ We have, x2 – 3x + 3 = ⎜ x − ⎟ + 3 − ⎝ ⎠ 2 4

To Find the Values of a Rational Expression in x, Where x is Real QUICK TIPS

(A) –20

⎛ = ⎜x − ⎝



2

3⎞ 3 ⎟⎠ + 2 4

3 ∴ smallest value = , which lies in the interval 4 3 ⎛ ⎞ ⎜⎝ − 3, ⎟⎠ . 2

RATIONAL ALGEBRAIC EXPRESSION P (x ) where P(x) and Q(x) are Q (x ) polynomials and Q(x) ≠ 0, is known as a rational algebraic expression. An expression of the form

Sign Scheme for a Rational Algebraic Expression in x Step 1: Factorise the numerator and denominator of the given rational expression into linear factors. Make the coefficient of x positive in all factors. Step 2: Find the real values of x by equating all the factors to zero. Step 3: If n distinct real values of x are obtained then the entire line will be divided into (n + 1) parts. 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. –∞

–

+

–

+

–

+

∞

ERROR CHECK 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: ■ |x| = k ⇔ x = ±k ■ |x| < k ⇔ = – k < x < k ■ |x| > k ⇔ = x < – k or x > k.

Put the given rational expression equal to y and form the quadratic equation in x. ■ Find the discriminant D of the quadratic equation obtained in step 1. ■ Since x is real, therefore, put D ≥ 0. We get an inequation in y. ■ Solve the above inequation for y. The values of y so obtained determine the set of values attained by the given rational expression. ■

QUICK TIPS The general quadratic expression ax2 + 2hxy + by2 + 2gx + 2fy + c in x and y may be resolved into two linear rational factors if abc + 2fgh – af 2 – bg2 – ch2 = 0 a h g h b f =0 g f c

or

If sum of coefficients of a polynomial equation a0 + a1x + a2x2 + … + an xn = 0 is zero, then x = 1 is always atleast one root of equation e.g., if a(b – c) x2 + b(c – a) x + c (a – b) = 0, then as Σa (b – c) = 0, x = 1 is atleast one root of this equation. 2 2 2 ■ Least value of the expression (x – y) + (y – z) + (z – x) is 0. n n–1 ■ Sum of real roots of the equation an |x| + an – 1 |x| + … + a0 = 0 is 0, e.g. if |x| = 2 satisfies the equation, then x = 2 and x = –2 are real roots, their sum is 0. 1 2 ■ Length of latus rectum of parabola y = ax + bx + c is . a

■

SOLVED EXAMPLES 54. The sum of the real roots of the equation

|x – 2|2 + |x – 2| – 2 = 0 is

(A) 2

(B) 6

(C) 4

(D) 8

Solution: (C) Put |x – 2| = t. The given equation becomes t2 + t – 2 = 0 or (t + 2) (t – 1) = 0

Quadratic Equations and Expressions  3.17 Since

t + 2 = |x – 2| + 2 > 0

∴ we get

t – 1 = 0

57. If x is real, then the maximum value of 3 – 6x – 8x2 is 17 33 (B)  8 8 21 (C)  (D)  None of these 8 (A) 

⇒ |x – 2| = 1 ⇒ x – 2 = ±1 ⇒

x = 3, 1.

Thus, the sum of roots is 4. x 2 + 2 x − 11 55. If x is real, the expression takes all real x −3 values except those which lie between a and b, then a and b are (A)  –12, –4 (B)  –12, 2 (C)  4, 12 (D)  –4, 4 Solution: (C) Let y =

⇒

x2 + (2 – y) x + (3y – 11) = 0

∴ 62 – 4 ⋅ 8 (y – 3) ≥ 0, 36 – 32y + 96 ≥ 0

(D ≥ 0)

(2 – y) – 4 (3y – 11) ≥ 0

⇒ 4 + y2 – 4y – 12y + 44 ≥ 0 ⇒

y2 – 16y + 48 ≥ 0

⇒

y2 – 12y – 4y + 48 ≥ 0

132 32  33 or y≤ 8  33 Hence, maximum value of y = . 8 ∴

2

⇒ (y – 4) (y – 12) ≥ 0 y ≤ 4  or  y ≥ 12

x 2 − 3x + 4 56. For real values of x, the expression 2 lies x + 3x + 4 between 1 1 (A) − and 7 (B)  and 7 7 7 1 (C)  and 3 (D)  None of these 3 Solution: (B) x 2 − 3x + 4 Let y= 2 x + 3x + 4  ⇒ yx2 + 3xy + 4y = x2 – 3x + 4 2

⇒ (y – 1) x + 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 2

Since x is real,

or 32y ≤ 132

xy – 3y = x2 + 2x – 11

⇒

then 8x2 + 6x + y – 3 = 0.

or

x 2 + 2 x − 11 x −3 

⇒ ⇒

Solution: (B) Let y = 3 – 6x – 8x2

2

⇒ –7y + 50y – 7 ≥ 0 ⇒ 7y – 50y + 7 ≤ 0 1 1⎞ ⎛ ⇒ (y – 7) ⎜ y − ⎟ ≤ 0 ⇒ ≤ y ≤ 7. ⎝ 7 7⎠

y≤

58. For all real x, the maximum value of 1 (B)  3 Solution: (B)

1 + x + x2 (C) 1 (D) 3

(A) 0

Let ⇒

z=

1 − x + x2

1 − x + x2

1 + x + x2  z + zx + zx2 = 1 – x + x2

⇒

zx2 – x2 + zx + x + z – 1 = 0

⇒

x2 (z – 1) + x (z + 1) + (z – 1) = 0

For real x, B2 – 4AC ≥ 0



⇒ (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

⇒

– 3z (z – 3) + 1 (z – 3) ≥ 0

⇒ (z – 3) (– 3z + 1) ≥ 0 ⇒

1 ≤ z ≤ 3 3

∴ minimum value of z =

1 . 3

is

3.18  Chapter 3 x2 − 2x + 4

59. Given that, for all real x, the expression 2 x + 2x + 4 1 lies between 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 (B)  –1 and 1 1 (C)  –2 and 0 (D)  and 3. 3 Solution: (D) Given

1 x2 − 2x + 4 < < 3 for all x ∈ R. 3 x2 + 2x + 4

⇒

1 x2 + 2x + 4 < < 3 for all x ∈ R.(1) 3 x2 − 2x + 4

Let 3x + 1 = y Then y ∈ R for all x ∈ R. ∴

9 ⋅ 32 x + 6 ⋅ 3x + 4 9 ⋅ 32 x − 6 ⋅ 3x + 4



= =

32 x + 2 + 2 ⋅ 3x + 1 + 4 32 x + 2 − 2 ⋅ 3x + 1 + 4  y2 + 2 y + 4 y2 − 2 y + 4 

From Eq. (1),

1 y2 + 2 y + 4 < < 3 3 y2 − 2 y + 4

∴

1 9 ⋅ 32 x + 6 ⋅ 3x + 4 < < 3. 3 9 ⋅ 32 x − 6 ⋅ 3x + 4

Quadratic Equations and Expressions  3.19

NCERT EXEMPLARS 1. sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for (A) x = nπ (B)  x =  n + 1  π 2 2  (C) x = 0 (D)  No value of x 1 − i sin α 2. The real value of a for which expression is 1 + 2i sin α purely real is π π (A)  ( n + 1) (B)  ( 2n + 1) 2 2 (C) nπ (D)  None of these 3. If z = x + iy lies in the third quadrant, then in the third quadrant, if (A) x > y > 0 (B)  x < y < 0 (C) y < x < 0 (D)  y > x > 0

z also lies z

4. The value of (z + 3) ( z + 3) is equivalent to (A) |z + 3|2 (B)  |z – 3| (C) z2 + 3 (D)  None of these x

11. The complex number z which satisfies the condition i+z = 1 lies on i−z (A) circle x2 + y2 = 1 (C) the Y-axis

(B)  the X-axis (D)  the line x + y = 1

12. If z is a complex number, then (A)  z 2 > z (B)  z2 = z (C)  z 2 < z

2



(D)  z 2 ≥ z

2

2

6. A

real value of x satisfies the equation  3 − 4ix  2 2  3 + 4ix  = α − i β (α , β ∈ R ) , if α + β is equal to   (B) – 1 (D) – 2

7. Which of the following is correct for any two complex number z1 and z2? (A)  z1 z2 = z1 z2 (B)  arg (z1z2) = arg (z1) · arg (z2) (C)  z1 + z2 = z1 + z2 (D)  z1 + z2 ≥ z1 − z2 8. The point represented by the complex number (2 – i) is π rotated about origin through an angle in the clock2 wise direction, the new position of point is (B)  – 1 – 2i (D)  – 1 + 2i

1 z2 = (A)  z2 + z1 (B)  z1 (C)  arg (z1) = arg (z2)

(D)  z1 = z2

14. The real value of θ for which the expression is a real number is

1 + i cos θ 1 − 2i cos θ

π n π (B)  nπ + ( −1) 4 4 π (C)  2nπ ± (D)  None of these 2 (A) nπ +

15. The value of arg(x), when x < 0 is

π (A)  0 (B)  2 (C)  π (D)  None of these 16. If ƒ ( z ) = (A) 

z 2

7− z , where z = 1 + 2i, then ƒ ( z ) is equal to 1 − z2

z (B) 

(C)  2 z

(D)  None of these

NCERT EXEMPLARS

(A) x = 2n + 1 (B)  x = 4n (C) x = 2n (D)  x = 4n + 1

(A)  1 + 2i (C)  2 + i

10. If a + ib = c + id, then (A) a2 + c2 = 0 (B) b2 + c2 = 0 (C) b2 + d2 = 0 (D) a2 + b2 = c2 + d2

13. z1 + z2 = z1 + z2 is possible, if

1+ i 5. If  1 − i  = 1, then

(A) 1 (C) 2

9. If x, y ∈R, then x + iy is a non-real complex number, if (A) x = 0 (B)  y = 0 (C) x ≠ 0 (D)  y≠0

3.20  Chapter 3

ANSWER K EYS   1. (D) 2. (C)   11. (B) 12.  (B)

3. (B) 13. (C)

4.  (A) 14.  (C)

5. (B) 15. (C)

6.  (A) 16.  (A)

7. (A)

8. (B)

9. (D)  10.  (D)

HINTS AND EXPLANATIONS 1. Let z = sin x + I cos2x and z = sin x − i cos 2 x  Given that, z = cos x − i sin 2 x  ∴ sin x – I cos2x = cos x – i sin2x ⇒ sin x = cos x and cos2x = sin2x ⇒ tan x = 1 and tan 2x = 1

π ⇒ tan x = tan π and tan 2 x = tan 4 4 π π ⇒ x = nπ + and 2 x = nπ + 4 4

HINTS AND EXPLANAT I O N S

⇒ 2x – x = 0 ⇒ x = 0 1 − i sin α 1 + 2i sin α

2. Given expression,

Z=

=

(1 − i sin α ) (1 − 2i sin α ) (1 + 2i sin α ) (1 − 2i sin α )

=



=

1 − i sin α − 2i sin α + 2i 2 sin 2 α 1 − 4i 2 sin 2 α 1 − 3i sin α − 2i sin 2 α 1 + 4 sin 2 α

2 = 1 − 2 sin α − 3i sin α 1 + 4 sin 2 α 1 + 4 sin 2 α It is given that z is a purely real.

∴

−3 sin α 1 + 4 sin 2 α

=0

⇒ – 3 sin α = 0 ⇒ sin α = 0 α = nπ 3. Given that, z = x + iy lies in third quadrant. x 0 So, x < y < 0 4. Given that, (z + 3) ( z + 3) Let z = x + iy

x2 + y2

< 0 and

(

x2 + y2

– c), for a > c, b > c is (c + x ) (A) ( a − b + c − b ) 2 (B) ( a − c + b − c ) 2 (C) ( a − c − b − c ) 2 (D)  None of these 72. If the ratio of the roots of a1x2 + b1x + c1 = 0 be equal a to the ratio of the roots of a2x2 + b2x + c2 = 0, then 1 , a2 b1 c1 , are in b2 c2 (A)  A.P. (B)  G.P. (C)  H.P. (D)  None of these 73. 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

(

p + 6 q + 4 q1/ 4 (A) 1

p+2 q

) , where k is equal to k

1 1 1 (B)  (C)  (D)  2 3 4

74. If a, b are the roots of the equation x2 + px + 1 = 0 and c, d are the roots of the equation x2 + qx + 1 = 0, then (a – c) (b – c) (a + d) (b + d) = (A) p2 – q2 (B)  q2 – p2 (C) p2 + q2 (D)  2 (p2 – q2) 75. 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 (B) p2 – 2q = 0 (C) 4p3 + 27q + 1 = 0 (D) 4p3 + 27q = 0 76. If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the squares of their a b c reciprocals, then , and are in c a b (A)  arithmetic progression (B)  geometric progression (C)  harmonic progression (D)  arithmetico-geometric progression 77. If both the roots of the quadratic equation x2 – 2kx + k2 + k – 5 = 0 are less than 5, then k lies in the interval (A) (–∞, 4) (B)  [4, 5] (C) (5, 6] (D) (6, ∞) 78. If for real number a, the equation (a– 2) (x– [x])2 + 2  (x  – [x]) + a2 = 0 (where [x] denotes the greatest integer ≤ x) has no integral solution and has exactly one solution in (2, 3), then a lies in the interval (A)  (–1, 2) (B)  (0, 1) (C)  (–1, 0) (D)  (2, 3) 79. Let a, b, c be distinct positive numbers such that each of the quadratics ax2 + bx + c, bx2 + cx + a and cx2 + ax  + b is non-negative for all x ∈ R. If a2 + b2 + c2 R= , then ab + bc + ca (A) 1 ≤ R < 4 (B)  1 < R ≤ 4 (C) 1 ≤ R ≤ 4 (D)  1 < R < 4 80. The set of values of a for which the equation (x2 + x + 2)2 – (a – 3) (x2 +x + 2) (x2 + x + 1) + (a – 4) (x2 + x + 1)2 = 0 has at least one real root is ⎛ 19 ⎞ (A) ⎜ 5, ⎟ ⎝ 3⎠

⎡ 19 ⎤ (B)  ⎢5, 3 ⎥ ⎣ ⎦

19 ⎡ ⎛ 19 ⎤ ⎞ (C) 5 ⎜⎝ 5, ⎥ ⎢ , 3 ⎟⎠ (D)  3⎦ ⎣

81. If all real values of x obtained from the equation 4x – (a – 3)2x + a – 4 = 0 are non-positive, then a belongs to

PRACTICE EXERCISES

(A)  real and distinct (B) rational (C)  rational but not integer (D) imaginary

3.28  Chapter 3 (A)  [4, 5] (C)  [ 4, 5)

(B)  (4, 5] (D)  (4, 5)

82. Let f (x) = x2 + ax + b be a quadratic polynomial, where a and b are integers. If for a given integer n, f (n) f (n + 1) = f (m) for some integer m, then the value of m is (A)  n (n + 1) + an + b (B)  n (n + 1) + a + bn (C)  n (n + 1) + a + b (D)  None of these x 2 + nx − 2 ≤ 2 , then n 83. If for any real x, we have –1 ≤ 2 x − 3x + 4 belongs to (A) [− 40 + 6, −1] (B) [− 40 + 6, 40 − 6] (C) [−1, 40 − 6] 84.

(D)  None of these If b > a, then the equation (x – a) (x – b) –1 = 0 has (A)  both roots in (–∞, a) (B)  one root in (–∞, a) and other in (b, ∞) (C)  both roots in (b, ∞) (D)  both roots in [a, b]

85. The quadratic equation

PRACTICE EXERCISES

( x + b)( x + c) ( x + c)( x + a) ( x + a)( x + b) + + =1 (b − a)(c − a) (c − b)( a − b) ( a − c)(b − c) has (A)  two real and distinct roots (B)  imaginary roots (C)  equal roots (D)  infinite roots 86. The equation ax4 – 2x2 – (a – 1) = 0 will have real and unequal roots if (A)  a < 0, a ≠ 1 (B)  a > 0, a ≠1 (C) 0 < a < 1 (D)  None of these 2

2

87. If the equation x + [a – 5a + b + 4] x + b = 0 has roots –5 and 1, where [a] denotes the greatest integer less than or equal to a, then the set of values of a is ⎛ 5−3 5 5 + 3 5⎞ , (A) ⎜ 2 ⎟⎠ ⎝ 2

⎛ 5 + 3 5⎞ (B)  ⎜ 0, 2 ⎟⎠ ⎝

⎛ 5 − 3 5 ⎤ ⎡5 + 3 5 ⎞ (C) ⎜ −1, , 6⎟ ⎥∪⎢ 2 ⎦ ⎣ 2 ⎠ ⎝ (D)  None of these 88. Let α1, β be the roots of the equation x2 – ax + p = 0 and γ, δ be the roots of the equation x2 – bx + q = 0. If α, β, √, δ are in increasing G.P., then the value of q+ p is equal to q− p b2 − a2 b2 + a2 (A)  2 (B)  b + a2 b2 − a2 b+a b−a (C)  (D)  b−a b+a 1 89. If tn denotes the nth term of an A.P. and tp = and tq q 1 = , then which of the following is necessarily a root p of the equation (p + 2q – 3r) x2 + (q + 2r – 3p) x + (r + 2p – 3q) = 0 (A)  tp (B)  tq (C)  tpq (D)  tp + q 90. If the roots of the equation 4x2 + 4ax + b = 0 are real and differ at most by a, then b lies in ⎛ a2 2 ⎞ ⎛ a2 ⎞ (A) ⎜ 0, ⎟ (B)  ⎜ 2 ,a ⎟ ⎝ ⎠ ⎝ 2⎠ (C) [0, a2] (D)  (0, a2) 91. The roots of the equation ax2 + bx + c = 0, where a  ∈  R+, are two consecutive odd positive integers, then (A) |b| ≤ 4a (B)  |b| ≥ 4a (C) |b| = 2 a (D)  None of these 92. If a, b, c, d are real numbers, then the number of real roots of the equation (x2 + ax – 3b) (x2 – cx + b) (x2 – dx + 2b) = 0 are (A) 3 (B) 4 (C)  6 (D)  at least 2

Previous Year’s Questions 93. If α ≠ β with a2 = 5α − 3 and β2 = 5β − 3, then the equation having α/β and β/α as its roots, is [2002] (A) 3x2 + 19x + 3 = 0 (B)  3x2− 19x + 3 = 0 2 (C) 3x − 19x− 3 = 0 (D)  x2− 16x + 1 = 0

94. The number of real roots of 32 x − 7 x + 7 = 9 is  [2002] (A) Zero (B) 2 (C) 1 (D) 4 2

Quadratic Equations and Expressions  3.29

2

96. The number of real solutions of the equation x − 3 | x | + 2 = 0 is [2003] (A) 2 (B) 4 (C) 1 (D) 3 97. 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 [2003] 2 2 (A)  (B)  − 3 3 1 1 (C)  (D)  − 3 3 98. If (1 − p) is a root of quadratic equation x2 + px + (1 − p) = 0, then its roots are [2004] (A)  0, 1 (B)  –1, 2 (C) 0, –1 (D)  –1, 1 99. 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 [2004] 49 (A)  (B)  4 4 (C) 3 (D) 12 100. If 2a + 3b + 6c = 0, then at least one root of the equation ax2 + bx + c = 0 lies in the interval [2004] (A)  (0, 1) (B)  (1, 2) (C)  (2, 3) (D)  (1, 3) 101. The values of α for which the sum of the squares of the roots of the equation x2− (a − 2)x − a − 1 = 0 assume the least value is [2005] (A) 1 (B) 0 (C) 3 (D) 2 102. If roots of the equation x2− bx + c = 0 be two ­consectutive integers, then b2− 4c equals [2005] (A)  −2 (B)  3 (C) 2 (D) 1

104. All the values of m for which both roots of the equations x2− 2mx + m2− 1 = 0 are greater than –2 but less than 4, lie in the interval [2006] (A)  –2 < m < 0 (B)  m > 3 (C)  –1 < m < 3 (D)  1 < m < 4 105. If x is real, the maximum value of

is 3x 2 + 9 x + 7 [2006] (A) 1/4 (B) 41 (C) 1 (D) 17/7 106. If the difference between the roots of the equation x2 + ax + 1 = 0 is less than 5 , then the set of possible values of a is [2007] (A) (−3, 3) (B)  (−3, ∞) (C) (3, ∞) (D)  (−∞, −3) 107. 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 [2008] (A) 1 (B) 4 (C) 3 (D) 2 108. If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is [2009] (A)  greater than 4ab (B)  less than 4ab (C)  greater than − 4ab (D)  less than − 4ab 109. If a and β are the roots of the equation x2−x + 1 = 0, then the value of α2009 + β2009 = [2010] (A)  −1 (B)  1 (C)  2 (D)  −2 110. The equation esin x –e–sin x – 4 = 0, for x real, has  [2012] (A)  infinite number of roots (B)  no roots (C)  exactly one root (D)  exactly four roots 111. The real number k for which the equation,

2

103. If both the roots of the quadratic equation x − 2kx + k2 + k− 5 = 0 are less than 5, then k lies in the interval  [2005] (A) (5, 6] (B) (6, ∞) (C) (−∞, 4) (D)  [4, 5]

3 x 2 + 9 x + 17



2 x 3 + 3 x + k = 0 has two distinct real roots in [0, 1]  [2013] (A)  lies between 2 and 3 (B)  lies between −1 and 0 (C)  does not exist (D)  lies between 1 and 2

PRACTICE EXERCISES

95. If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the squares of a b c their reciprocals, then , and , are in [2003] c a b (A)  arithmetic progression. (B)  geometric progression. (C)  harmonic progression. (D) arithmetic-geometric-progression.

3.30  Chapter 3 112. If the equations x2 + 2x + 3 = 0 and 2 ax + bx + c = 0, a, b, c ∈ R have a common root, then a : b : c is [2013] (A)  3 : 2 : 1 (B)  1 : 3 : 2 (C)  3 : 1 : 2 (D)  1 : 2 : 3

115. If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1) (x + 2) + … + (x + n – 1)(x + n) = 10n has two consecutive integral solutions, then n is equal to [2017] (A) 9 (B) 10 (C) 11 (D) 12

116. 113. If a ∈ R and the equation −3 ( x − [ x ]) + 2 ( x − [ x ]) + a = 0 2 3 ( x − [ x ]) + 2 ( x − [ x ]) + a 2 = 0 (where [ x ] denotes the greatest integer ≤ x ) 117. has no integral solution, then all possible values of a 2

lie in the interval (A) − ( 1, 0) ∪ (0,1)

(B) 1 ( , 2)

2

[2014]

(C) − ( 2, −1) (D)  ( −∞, −2) ∪ ( 2, ∞ ) 114. Let α and β be the roots of equation x 2 − 6 x − 2 = 0. a − 2a8 If an = α n − β n, for n ≥ 1,then the value of 10 2a9 is equal to [2015] (A) − 6 (B)  3 (C) − 3 (D)  6

If α, β ∈ C are the distinct roots, of the equation x2 – x + 1 = 0, then α101 + β107 is equal to [2018] (A) –1 (B) 0 (C) 1 (D) 2

The value of λ such that sum of the squares of the roots of the quadratic equation, x2 + (3 – λ) x + 2 =λ has the least value is: [2019] (A) 2 (B) 1 15 4 (C)  (D)  8 9

ANSWER K EYS

PRACTICE EXERCISES

Single Option Correct Type 1. (C) 11. (A) 19. (A) 29. (A) 39. (C) 49. (B) 59. (D) 69. (A) 79. (D) 89. (C)

2. (A) 12. (B) 20. (C) 30. (D) 40. (B) 50. (C) 60. (B) 70. (B) 80. (D) 90. (C)

3. (C) 13. (A) 21. (B) 31. (B) 41. (B) 51. (A) 61. (A) 71. (B) 81. (B) 91. (B)

4. (D) 14. (A) 22. (D) 32. (C) 42. (C) 52. (A) 62. (A) 72. (B) 82. (A) 92. (D)

5. (B) 15. (A) 23. (D) 33. (B) 43. (B) 53. (A) 63. (C) 73. (D) 83. (C)

6. (A) 16. (C) 24. (C) 34. (C) 44. (C) 54. (A) 64. (B) 74. (B) 84. (B)

7. (B) 17. (C) 25. (B) 35. (C) 45. (B) 55. (B) 65. (A) 75. (D) 85. (D)

8. (D) 18. (B) 26. (B) 36. (B) 46. (A) 56. (C) 66. (A) 76. (C) 86. (C)

9. (B)

10. (C)

27. (A) 37. (D) 47. (D) 57. (D) 67. (B) 77. (A) 87. (C)

28. (D) 38. (A) 48. (C) 58. (D) 68. (C) 78. (C) 88. (B)

Previous Years’ Questions   93. (A) 94. (B) 103. (C) 104. (C) 113. (A) 114. (B)

95. (C) 96. (B) 105. (C) 106. (A) 115. (C) 116. (C)

97. (A) 98. (C) 99. (A) 100. (A) 101. (A) 102. (C) 107. (D) 108. (C) 109. (B) 110. (B) 111. (C) 112. (D) 117. (A)

Quadratic Equations and Expressions  3.31

HINTS AND EXPLANATIONS Single Option Correct Type

⎛ ⎞ 1 ⇒ (x + 2) ⎜ log 2 + log 3⎟ = 0 x −1 ⎝ ⎠ ⇒ x = –2 or x = 1 –

log 3 . log 2

The correct option is (C) 2. Since a, b, c are all +ve ∴ ax2 + b | x | + c > 0 for all real x ∴ ax2 + b | x | + c ≠ 0 for any real x ∴ no real solution is possible. The correct option is (A) 3. Since x2 – x + 1 = 0 ∴ (x – 1) (x2 – x + 1) = 0 ⇒ x3 – 1 = 0 ⇒ x3 = 1, ∴ x3n = 1 The correct option is (C) 4. The given equation can be written as 1⎤ ⎡ 2 x +1 ⎢ x 2 − ⎥ = 2| x − 3 | ⋅ 4 [4 x 2 − 1] 4⎦ ⎣

⎡ 2 1⎤ = 16 · 2|x – 3| ⎢⎣ x − 4 ⎥⎦

⇒ 2x – 3 = 2|x – 3| 1 [ x2 = does not give negative integral value] 4 ⇒ x – 3 = ± (x – 3) ⇒ either x – 3 = x – 3 or x – 3 = – x + 3 ⇒ 2x = 6 or x = 3 ∴ Given equation does not give any negative integral solution. The correct option is (D) 5. We have α + β = – b, aβ = c As c < 0, b > 0, we get α 0 ⇒ a ∈ (–∞, –1) ∪ (0, ∞) The correct option is (D) 9. (a + c)2 + 4b2 – 4b (a + c) ≤ 0 ⇒ (a – 2b + c)2 ≤ 0 ⇒ a – 2b + c = 0 ⇒ 2b = a + c ⇒ a, b, c are in A.P. The correct option is (B) 10. 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; aβ = c, α ′ + β ′ = – q, α ′ β ′ = r. α α′ α+β α ′ + β′ It is given that = ⇒ = β β′ α −β α ′ − β′ 2 2 (α + β ) (α ′ + β ′ ) ⇒ = (α − β ) 2 (α ′ − β ′ ) 2 ⇒

⇒

b2

=

2

q2 2

b − 4c q − 4r The correct option is (C)

⇒ b2r = q2c.

16 11 Let 16sin x = y, then 16cos x = 161−sin x = y The given equation becomes 16 y+ = 10 ⇒ y2 – 10y + 16 = 0 or y = 2, 8 y Now, 16sin x = 2 ⇒ 24 sin x = 2(1) 2 ⇒ 4 sin x = 1 π π⎞ ⎛ 1 ∴ sin x =± ⇒x=  ⎜⎝∵0 ≤ x < ⎟⎠ 6 2 2 sin x and 16 =8 2

2

2

2

2

2

⇒ 24 sin

2

x

= 23

⇒ sin x = ±

π 3 ⇒x=  3 2

The correct option is (A)

π⎞ ⎛ ⎜⎝∵ 0 ≤ x < ⎟⎠ 2

HINTS AND EXPLANAT I O N S

1. We have, 2x + 2 ⋅ 33x /(x – 1) = 9 = 32 3x ⇒ (x + 2) log 2 + log 3 = 2 log 3 x −1 ⎞ ⎛ 3x ⇒ (x + 2) log 2 + ⎜ − 2⎟ log 3 = 0 ⎝ x −1 ⎠

3.32  Chapter 3 12. The given condition is fulfilled if and only if f (2) = 4 – 2 (p + 1) + p2 + p – 8 < 0 ⇒ (p – 3) (p + 2) < 0 ⇒ –2 < p < 3 The correct option is (B) 2

13. We have, |x – 3x + 2| + |x – 1| = x – 3 ⇒ x ≥ 3 ⇒ x2 – 3x + 2 + x – 1 = x – 3 ⇒ x2 – 3x + 4 = 0 2

3⎞ 9 ⎛ ⇒ ⎜ x − ⎟ + 4 − = 0 ⎝ 2⎠ 4 ⇒ No solution. The correct option is (A) 14. We have, 3 x 2 + 6 x + 7 = and

3 ( x + 1) 2 + 4 ≥ 2

5 x 2 + 10 x + 14 =

5( x + 1) 2 + 9 ≥ 3

∴ L.H.S. ≥ 5 R.H.S. = 4 – 2x – x2 = 5 – (x + 1)2 ≤ 5 ∴ the equation holds only when L.H.S. = R.H.S. = 5 ∴ x = – 1. The correct option is (A) 15. Equating the coefficients of similar powers of x, we get ⎫ ⎪ ⎬ ⎪ 2 a − 4 a + 3 = 0 ⇒ a = 1, 3⎭

HINTS AND EXPLANAT I O N S

a 2 − 1 = 0 ⇒ a = ±1 a − 1 = 0 ⇒ a = 1

∴ common value of a = 1. The correct option is (A) 16. 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. The correct option is (C) 17. Sinceα, β are roots of the equation x2 + px + q = 0 ∴ α + β = – p and aβ = q 2

⎛α⎞ α Now q ⎜ ⎟ + ( 2q − p 2 ) + q β ⎝ β⎠ =

1

β

2

Thus,

[q (α + β)2 – p2ab] =

1

β2

(qp2 – p2q) = 0

α is a root of the equation β

qx2 + (2q – p2) x + q = 0 The correct option is (C)

18. Since f (x) has no real roots, f (x) has same sign for every x ∴ f (0) · f (1) > 0 The correct option is (B) 19. Let x1 and x2 be two roots of ax2 + bx + c = 0 1 < x1 < 2 and 1 < x2 < 2 b c⎞ ⎛ Now a (5a + 2b + c) = a2 ⎜⎝ 5 + 2 a + a ⎟⎠

= a2(5 + 2(–1) (x1 + x2) + x1x2) = a2[(x1 – 2) (x2 – 2) + 1] > 0 Hence a and 5a + 2b + c are of same sign The correct option is (A) 20. Since a < 0, in case of positive root of the equation x > a ∴ The equation is x2 – 2a (x – a) – 3a2 = 0 ⇒ x2 – 2ax – a2 = 0 2a ± 4 a 2 + 4 a 2 2a ± 2 2a Thus, the roots are = 2 2 = a(1 + 2 ) or a(1 − 2 ) ∴ the only positive root possible is a(1 − 2 ) . The correct option is (C) 21. Let α + iβ, α – iβ be the roots r Then, α 2 + β 2 = >0 p ∴ p, r must be of the same sign. Since p + r > 0 ∴ p, r are both positive. If q < r, p – q + r > 0 If q > 0, (p + r)2 – (p – r)2 = 4pr ≥ q2  ( roots are non-real) ∴ (p + r)2 ≤ q2 + (p – r)2 ≥ q2 ∴ p + r ≥ q The correct option is (B) 22. Let f (x) = lx2 – mx + 5 Since lx2 – mx + 5 = 0 does not have two distinct real roots, therefore either f (x) ≥ 0 ∀ x ∈ R, or f (x) ≤ 0 ∀ x ∈ R But f (0) = 5 > 0 ∴ f (x) ≥ 0 ∀ x ∈ R ∴ f (–5) ≥ 0 ⇒ 25l + 5m + 5 ≥ 0 ⇒ 5l + m ≥ –1 Hence, minimum value of 5l + m is –1. The correct option is (D) 23. Let α, β be the roots of the given equation, then α + β = λ2 – 5λ + 5 < 1 ⇒ λ2 – 5λ + 4 < 0 aβ = 2λ2 – 3λ – 4 < 1 ⇒ 2λ2 – 3λ – 5 < 0 ∴ (λ – 4) (λ – 1) < 0 or 1 < λ < 4 and (2λ – 5) (λ + 1) < 0 or –1 < λ
0 and 1 lies between the roots of 3x2 – 3sin θ – 2cos2 θ = 0 ∴ f (1) < 0 ⇒ 3 – 3sin θ – 2cos2θ < 0 ⇒ 1 + 2(1 – cos2 θ) – 3sin θ < 0 ⇒ 2sin2 θ – 3sin θ + 1 < 0 ⇒ (2sin θ – 1) (sin θ – 1) < 0 ⇒

1 < sin θ < 1 2

QUICK TIPS If one root is less than k and other is greatter than k, then D > 0 and af (k) < 0, where f(x) = ax2 + bx + c, a, b, c ∈ R, a ≠ 0 The correct option is (C) n

25.

∑ Sr

= (α + β ) + (α 2 + β 2) + … + (α n + β n)

r =1

= (α + α 2 + … + α n) + (β + β 2 + … + β n)



n

Lt

n→∞

∑ Sr

= (α + α 2 + … + ∞) + (β + β 2 + … + ∞)

r =1



=

α β + 1− α 1− β



=

α − αβ + β − αβ 1 − (α + β ) + αβ



=

α + β − 2αβ 1 − (α + β ) + αβ

25 4 + 29 1 375 375 = = = 25 2 12 348 1− − 375 375 The correct option is (B) 26. If x = n ∈ Z, (x)2 + (x + 1)2 = 25 ⇒ n2 + (n + 1)2 = 25 ⇒ 2n2 + 2n – 24 = 0 ⇒ n2 + n – 12 = 0 ⇒ n = 3, – 4 ∴ x = 3, – 4 If x = n + k, n ∈ Z, 0 < k < 1, then (x)2 + (x + 1)2 = 25 ⇒ (n + 1)2 + (n + 2)2 = 25 ⇒ 2n2 + 6n – 20 = 0 ⇒ n2 + 3n – 10 = 0 ⇒ n = 2, –5

∴ x = 2 + k, –5 + k, where 0 < k < 1 ∴ x > 2, x > –5 ∴ Solution set is (–5, –4] ∪ (2, 3] The correct option is (B) 27. log2 (9 – 2x) = 3 – x ⇒ 9 – 2x = 23–x 8 ⇒ 9 – 2x = x 2 ⇒ 9 · 2x – 22x = 8 ⇒ 22x – 9 · 2x + 8 = 0 ⇒ (2x – 8) (2x – 1) = 0 ⇒ x = 3  or  x = 0 But x = 3 is not the solution of original equation, ∴ x = 0. The correct option is (A) 28. Since ax2 + bx + 6 = 0 does not have two distinct real roots. ∴ b2 – 24a ≤ 0 Let 3a + b = y ∴ 3a = y – b ∴ b2 – 8(y – b) ≤ 0 i.e., b2 + 8b – 8y ≤ 0 Since b is real ∴ 64 + 32y ≥ 0 ⇒ y ≥ –2 ∴ Min. value of y i.e., 3a + b = –2. The correct option is (D) 29. Here α + β = –p ⇒ α + β = γ + δ γ + δ = –p Now (α – γ) (α – δ) = α 2 – α(γ + δ) + γ δ = α 2 – α (α + β ) + r = –αβ + r = –(–q) + r = q + r By symmetry (β – γ ) (β – δ ) = q + r ∴ Ratio is 1. The correct option is (A) 30.

x+2 2

x +1

⇒ ⇒

−

1 >0 2

− x2 − 1 + 2x + 4 2( x 2 + 1)

>0

3 + 2x − x2

>0 2( x 2 + 1) Since denominator is positive ∴ 3 + 2x – x2 > 0 ⇒ –1 < x < 3 ⇒ x = 0, 1, 2 The correct option is (D) 31. Let α, β be the roots of λx2 + μx + v = 0 μ v ∴ α + β = – , aβ = λ λ

HINTS AND EXPLANAT I O N S

Quadratic Equations and Expressions  3.33

3.34  Chapter 3

∴

(α + β ) αβ

2

μ2 2 μ2 = λ = v λv λ

α β μ2 ⇒ + +2=  β α λv

(1)

Let γ, δ be the roots of x2 + x + 1 = 0 ∴ γ + δ = –1, γ δ = 1 ∴

(γ + δ ) 2 γ δ = 1 ⇒ + + 2 = 1 γδ δ γ

(2)

α γ = , β δ ∴ from Eq. (1) and (2) Since

HINTS AND EXPLANAT I O N S

μ2 = 1 ⇒ μ2 = λv λv ∴ λ, µ, v are in G.P. The correct option is (B) 32. Given equation can be written as bx2 + x – bcx – bdx + bcd – a = 0 ⇒ bx (x – c) – bd (x – c) + x – a = 0 ⇒ b (x – c) (x – d) + (x – a) = 0 Let f (x) = b (x – c) (x – d) + (x – a) f (c) = c – a < 0; f (d) = d – a > 0 Hence, one root of the given equation lies between c and d. The correct option is (C) b α α +1 α α +1 c + ⋅ 33. We have, = − and = a α −1 α α −1 α a ⇒

2α 2 − 1

α2 − α 2

= −

loge ( x − 4) loge ( x 2 − 5 x + 6) > loge 3 loge 9 2 2 ⇒ x – 5x + 6 > x – 8x + 16 10 ⇒ 3x – 10 > 0 ⇒ x >  3 2 Also, x – 5x + 6 > 0 ⇒ x > 3 or x < 2  and  x – 4 > 0 ⇒ x > 4  Common solution from Eqs (1), (2) and (3) is x > 4 The correct option is (B) 37. Let f (x) = a2x2 + 2bx + 2c = 0 Given: a2α 2 + bα + c = 0 and a2β 2 – bβ – c = 0 Now,  f (α) = a2α 2 + 2bα + 2c = (a2α 2 + bα + c) + (bα + c) = bα + c = – a2α 2 36.

b c+a and α = a c−a

⇒ (c + a) + 4ac = –2b(c + a) ⇒ (c + a)2 + 2b(c + a) + b2 = b2 – 4ac ⇒ (a + b + c)2 = b2 – 4ac The correct option is (B) 34. Here α + β = p; aβ = 1 ⇒ aβ = γ δ γ + δ = q; γ δ = 1 Now, (α – γ ) (β – γ) (α + δ ) (β + δ ) = [aβ – γ (α + β ) + γ 2] [αβ + δ(α + β ) + δ 2] = [1 + γ p + γ 2] [1 – pδ + δ 2] = [(γ 2 + 1) + γ p] [(δ 2 + 1) – pδ] = (–qγ + γ p) (–qδ – pδ ) = γ δ (q2 – p2) = 1 (q2 – p2) The correct option is (C) 35. Let y = [x] ∴ The given equation y2 + ay + b = 0 must have integral roots which is not possible as a and b are odd integers. ∴ Discriminant can’t be perfect square. The correct option is (C)

(1) (2) (3)

⎫ f (β ) = a2β 2 + 2bβ + 2c ⎬ ⎭ = 3bβ + 3c = 3 (bβ + c) = 3a2β 2 Since 0 < α < β ∴ α, β are real ∴ f (α) < 0, f (β) > 0 ∴ f (γ ) = 0 where α < γ < β

QUICK TIPS Let f(x) = 0 be a polynomial equation. Let p and q be two real numbers, p < q. If f(p) ⋅ f(q) < 0, then the equation f(x) = 0 has atleast one real roots between p and q. The correct option is (D) 38. Let us see the graph of y = x2 – 2 and y = [x] Y

O

If [x] = –1 We have x2 – 2 + 2 = 0 ⇒ x = 0 not possible [x] = 0 ⇒ x = ± 2 not possible [x] = 1 ⇒ x = ± 4 = ± 2 not possible [x] = 2 ⇒ x = ± 6 ⇒ x = 6 is the only solution. The correct option is (A)

X

Quadratic Equations and Expressions  3.35 2 Z Therefore, the given equation becomes 2 Z– – 1 = 0 ⇒ Z = 2 or Z = –1 Z ⇒ 2sin x = 2 or 2sin x = –1 (not possible) π ⇒ sin2 x = 1 ⇒ x = (2n + 1) , n ∈ 1 2 The correct option is (C)

⇒ k = ±2 ⇒ k = 2 The correct option is (B) 44. x – x2 – 9 = –(x2 – x + 9)

40. Given: |α – β | >



2

= Z ⇒ 2cos

x

2

2

x

=

2

3p

If α, β are the roots of x2 + px + 1 = 0, then α + β = –p, aβ = 1 ∴ (α – β)2 > 3p ⇒ (α + β)2 – 4aβ > 3p ⇒ (–p)2 – 4 · 1 > 3p ⇒ p2 – 3p – 4 > 0 ⇒ (p – 4) (p + 1) > 0 ⇒ p > 4, p > –1 or p < 4, p < –1 ⇒ p > 4 or p < –1 But p is not –ve ( If p is –ve, then 3 p is not real) ⇒ p > 4 The correct option is (B) 41. If f (x) = x2 + ax + b f  (x + c) = x2 + (2c + a) x + c2 + ac + b ∴ Roots of the given equation are 0 and d – c.  (since roots of x2 + ax + b = 0 are c and d.) The correct option is (B) 42. Let f (x) = x2 + 2 (k + 1) x + 9k – 5. Let α, β be the roots of f (x) = 0. The equation f (x) = 0 will have both negative roots if and only if   (i) Disc. ≥ 0  (ii)  α + β < 0 and (iii)  f (0) > 0 Now, discriminant ≥ 0 ⇒ 4 (k + 1)2 – 36k + 20 ≥ 0  ⇒  k2 – 7k + 6 ≥ 0 ⇒ (k – 1) (k – 6) ≥ 0 ⇒ k ≤ 1 or k ≥ 6  (1) (α + β) < 0 ⇒ –2 (k + 1) < 0 ⇒ k + 1 > 0 ⇒ k > –1 (2) and aβ > 0 ⇒ 9k – 5 > 0 5 ⇒ k >  (3) 9 From Eqs (1), (2) and (3) we get k ≥ 6. The correct option is (C) 43. Product of roots = 2 e2 ln k – 1 = 7 (given) ⇒ 2e ln k − 1 = 7 ⇒ 2k2 – 1 = 7 2

(Since ln k is defined for k > 0)

2 ⎡⎛ 1⎞ 35 ⎤ = – ⎢⎜ x − ⎟ + ⎥ < 0 for all x ∈ R ⎝ ⎠ 2 4 ⎥⎦ ⎢⎣ ∴ no. solution i.e., solution set = ϕ



⎡ ⎛ 3⎞ x ⎤ ⎢∵ ⎜ ⎟ > 0 for all x ∈ R ⎥ ⎠ ⎝ 5 ⎢⎣ ⎥⎦

The correct option is (C) 45. Let esin x = t ⇒ t2 – 4t – 1 = 0 4 ± 16 + 4 2 ⇒ t = esin x = 2 ± 5 ⇒ t =

⇒ esin x = 2 –

sin x

e

=2–

5 , esin x = 2 +

5

5 < 0,

⇒ sin x = ln(2 + 5 ) > 1 So both rejected. Hence no solution The correct option is (B) 46. Let f (x) = x3 – px + q Now for maxima/minima f ′(x) = 0 ⇒ 3x2 – p = 0 –(p/3) –(p/3)

⇒ x2 =

p 3

p 3 The correct option is (A) 47. Let α and 4β be roots of x2 – 6x + a = 0 and α, 3β be the roots of x2 – cx + 6 = 0, then α + 4β = 6 and 4αβ = a α + 3β = c and 3αβ = 6. We get αβ = 2 ⇒ a = 8 So the first equation is x2 – 6x + 8 = 0 ⇒ x = 2, 4 If α = 2 and 4β = 4 then 3β = 3 If α = 4 and 4β = 2, then 3β = 3/2 (non-integer) ∴ common root is x = 2. The correct option is (D) ∴ x = ±

HINTS AND EXPLANAT I O N S

39. Let 2sin

3.36  Chapter 3 48. bx2 + cx + a = 0 Roots are imaginary ⇒ c2 – 4ab < 0 ⇒ c2 < 4ab ⇒ –c2 > –4ab Given expression has minimum value =

4(3b 2 )( 2c 2 ) − 36b 2c 2 4 + 3b 2 12b 2c 2

= –c2 > –4ab. 12b 2 The correct option is (C) =–

49. We have, 3sec ⇒ 3sec

2

x

⇒ 3sec

2

x

2

x −1

y2 −

9 y2 − 6 y + 2 ≤ 1 2 2 y+ ≤1 3 9 2

1⎞ 1 ⎛ ⎜⎝ y − ⎟⎠ + ≤ 1 3 9

Now, sec2x ≥ 1 ⇒ 3sec2x ≥ 3 and

2

1⎞ 1 1 ⎛ ⎜⎝ y − ⎟⎠ + ≥ , so we 3 9 3

1 = 0. 3 1 ⇒ x = 0, π, 2π, 3π and y = . 3 ∴ There are 4 solutions. The correct option is (B) 50. Let f (x) = (x – n)m + (x – n2)m + (x – n3)m + … + (x – nm)m. ⇒ f ′(x) = m(x – n)m–1 + m(x – n2)m–1 + … + m(x – nm)m–1 Since m is odd, (m – 1) is even. Therefore, f ′(x) = 0 has no real root. ⇒ f (x) = 0 has one real and (m – 1) imaginary roots. The correct option is (C) 51. Since the given equation has imaginary roots ⇒ D < 0 or (a + b + c)2 – 4(ab + bc + ca) < 0 ⇒ (a2 + b2 + c2 – 2ab – 2bc + 2ac) < 4ac ⇒ (a – b + c)2 < 4ac

HINTS AND EXPLANAT I O N S

must have sec2x = 1 and y –

⇒ − 2 ac < a – b + c ⇒ ⇒

( a + c + 2 ac ) > b ( a + c ) > b or

Similarly,

2

a + c > b.

b + c > a and

a + b > c.

Therefore, a , b , c can be the sides of a triangle. The correct option is (A) 52. We have, xn + ax + b = (x – x1) (x – x2) … (x – xn) x n + ax + b ⇒ (x – x2) (x – x3) … (x – xn) = x − x1 x n + ax + b ⇒ (x1 – x2) (x1 – x3) … (x1 – xn) = lim x→ x x − x1 = nx1n –1 + a. The correct option is (A) 1

53. Since the roots of the given equation are real ∴ B2 – 4AC ≥ 0 ⇒ 4a2 – 4 (a2 + a – 3) ≥ 0 ⇒ –a + 3 ≥ 0 or a ≤ 3.  (1) Since the root is less than 3, so f (3) > 0 ⇒ 32 – 2a (3) + a2 + a – 3 > 0 ⇒ a2 – 5a + 6 > 0 or (a – 2) (a – 3) > 0 ⇒ a < 2 or a > 3 (2) From (1) and (2), we have a < 2. The correct option is (A) 54. Let f (x) = x2 + (1 – 2k) x + k2 – k – 2 The number 3 lies between the roots of the given equation, if f (3) < 0 Now, f (3) = 9 + (1 – 2k) 3 + k2 – k – 2 = 10 – 7k + k2 = k2 – 7k + 10 Hence, f (3) < 0 ⇒ k2 – 7k + 10 < 0 ⇒ (k – 2) (k – 5) < 0 ⇒ 2 < k < 5. The correct option is (A) 55. If x = n ∈ I, |n – 2n| = 4 ∴ n = ± 4, If x = n + k, n ∈ I, 0 < k < 1 then |n – 2 (n + k)| = 4 ∴ |–n – 2k| = 4. 1 It is possible if k = . 2 The correct option is (B) 56. Since a < 0, in case of positive root of the equation x > a ∴ The equation is x2 – 2a (x – a) – 3a2 = 0 ⇒ x2 – 2ax – a2 = 0 2a ± 2 2a 2a ± 4 a 2 + 4 a 2 = 2 2 = a(1 + 2 ) or a(1 − 2 ) ∴ the only positive root possible is a(1 − 2 ) . The correct option is (C) 57. The required a satisfies the inequality 2a2 – 2(2a + 1)a + a(a + 1) < 0 ⇒ a(a + 1) > 0 ⇒ a ∈ (–∞, –1) ∪ (0, ∞) The correct option is (D) Thus, the roots are

58. We have, 2x + 2x ≥ 2 2 (x ≥ 0) 1 2 and, 2x + 2–x ≥ 2 2 (x < 0) 1 ⇒ t + ≥ 2 2 (where t = 2x) t ⇒ t2 – 2 2 t + 1 ≥ 0 ⇒ 2x ≥

2 ⇒x≥

⇒ (t − ( 2 − 1)) (t − ( 2 + 1)) ≥ 0 ⇒ t ≤

2 − 1 or t ≥ x

⇒ 0 < 2 ≤

2 + 1 but t > 0

2 − 1 or 2x ≥

2 +1

Quadratic Equations and Expressions  3.37

or, x ≥ log2 ( 2 + 1) 

(but not acceptable as x < 0)

⎡1 ⎞ ∴ x ∈ (–∞, log2 ( 2 − 1) ] ∪ ⎢ , ∞⎟ ⎣2 ⎠ The correct option is (D) D 4b 2 − 8c 2 =– 4a 4 = –(b2 – 2c2)

59. min. f (x) = –

D 4c 2 + 4b 2 = 4a 4 = b2 + c2 Now, 2c2 – b2 > b2 + c2

(upward parabola)

max. g(x) = –

(downward parabola)



⇒ c2 > 2b2 ⇒ |c| > 2 |b| The correct option is (D) 60.

| x + 1| | x + 1 |2 + | x + 1| = |x| |x|

⎧ 1 | x + 1 |⎫ +1− ⇒ | x + 1 | ⎨ ⎬ =0 |x| ⎭ ⎩| x | ∴ |x + 1| = 0 or 1 + |x| – |x + 1| = 0. |x + 1| = 0 ⇒ x = –1. If x < – 1, 1 + |x| – |x + 1| = 0 ⇒ 1 – x + x + 1 = 0 ⇒ 2 = 0 (absurd) If –1 ≤ x < 0, 1 + |x| – |x + 1| = 0 ⇒ 1 – x – (x + 1) = 0 ⇒ x = 0 (not possible) If x ≥ 0, 1 + x – (x + 1) = 0 ⇒ 0 = 0 ⇒ x can have any value in the interval ∴ x = –1, x > 0.( x ≠ 0) The correct option is (B) 61. As α, β are roots of ax2 + bx + c = 0, we have α + β = –b/a, aβ = c/a Now, (α – β )2 = (α + β )2 – 4aβ 2



4c b 2 − 4 ac ⎛ b⎞ = ⎜− ⎟ − = (1) ⎝ a⎠ 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 )

=

B2

−

A2 From (1) and (2), we get

4C B 2 − 4 AC = (2) A A2

b 2 − 4 ac B 2 − 4 AC 2 = . a A2 The correct option is (A) 62. Given equation is ax2 + bx + c = 0 Since α, β are the roots of the given equation c −b ∴ α + β = , aβ = . a a

Also, since α < – 1, β > 1 ∴ aβ < – 1 c c ⇒ < – 1 or +1 0, ∴ (a + b + c) (a – b + c) > 0 or, (a + c)2 – b2 > 0 or (a + c)2 > b2 2

c⎞ ⎛ ⎛ b⎞ or, ⎜1 + ⎟ > ⎜ ⎟ ⎝ a⎠ ⎝ a⎠



2

c⎞ b ⎛ ⇒ ⎜1 + ⎟ < −  ⎝ a⎠ a

⎡ ⎛c ⎞ ⎤ ⎢∵ ⎜⎝ a + 1⎟⎠ < 0 ⎥ ⎣ ⎦

c b + 1 ⎝ n−r r⎠ ⎝ n − r ⎟⎠ r

∴ a > b ⇒ D > 0 ⇒ roots of given equation are real and distinct. The correct option is (A) 1 67. Let α and β be the roots of x – px + q = 0 and α and be β the roots of x2 – ax + b = 0. Then,  α + β = p and aβ = q. 1 α Also,  α + = a and = b. β β 2

2

⎛ ⎛ α⎞ 1⎞ Now, (q – b)2 = ⎜ αβ − ⎟ = α 2 ⎜ β − ⎟ β⎠ β⎠ ⎝ ⎝

2

⎡ ⎛ α 1⎞⎤ ⋅ βα ⎢(α + β ) − ⎜ α + ⎟ ⎥ β β ⎝ ⎠⎦ ⎣ 2 = bq (p – a) . The correct option is (B)

=

69. We have, 3x2 + x – 5 = 0. Its 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 c b ∴ = = = k (say) 1 3 −5 ⇒ a = 3k; b = k, c = –5k. ∴ 3a + b + 2c = 9k + k – 10k = 10k – 10k = 0. The correct option is (A) 70. 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 )

n

( r + 1)! ⋅ ( n − r − 1)! n! =

68. 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 pair. Hence both the roots will be common, i.e., two equations will be identical. So their coefficients will be proportional b a c i.e., = = , 2 4 −3 ∴ 6a = –4b = 3c. The correct option is (C)

2

⇒ (ca1 – ac1)2 = 4 (bc1 – b1c) (ab1 – ba1)  a b c ∵ , , are in A.P., a1 b1 c1 ∴

b a c b − − = = k (say) b1 a1 c1 b1

⇒ ab1 – a1b = – ka1b1 and bc1 – b1c = – kb1c1. Also, 2k = k + k =

ca − ac1 b a c b − + − = 1 a1c1 b1 a1 c1 b1

or, ca1 – ac1 = 2ka1c1. ∴ From (1), 4 k 2 a12c12 = 4 (–ka1b1) (–kb1c1) or, a1c1 = b12. Hence a1, b1, c1 are in G.P. The correct option is (B) ( a + x ) (b + x ) (c + x ) ⇒ x2 + (a + b) x + ab = cy + xy ⇒ x2 + (a + b – y) x + ab – cy = 0. For real x, B2 – 4AC ≥ 0 ⇒ (a + b – y)2 – 4ab + 4cy ≥ 0 ⇒ (a + b)2 + y2 – 2 (a + b) y – 4ab + 4cy ≥ 0 ⇒ (a – b)2 + y2 – 2 (a + b – 2c) y ≥ 0 71. Let y =

(1)

Quadratic Equations and Expressions  3.39 ⇒ y2 – 2 (a + b – 2c) y + (a – b)2 ≥ 0 ⇒ [ y − ( ( a − c) − (b − c)) ] × [ y − ( ( a − c) + (b − c)) 2 ] ≥ 0 ∴ y ≤ ( ( a − c) − (b − c)) 2 or, y ≥ ( ( a − c) + (b − c)) 2 . Hence, the minimum value of y is ( ( a − c) + (b − c)) 2 The correct option is (B) 72. Let the ratio of the roots be k. Then, the roots of a1x2 + b1x + c1 = 0 are α, kα and the roots of a2x2 + b2x + c2 = 0 are β, kβ. −b1 ∴ α + kα = (1) a1 c1 α ⋅ kα = (2) a1 −b2 (3) a2 c β ⋅ kβ = 2 .(4) a2 β + kβ =

Dividing (1) by (3), we get ba ba α (1 + k ) α = 1 2 , or = 1 2 .(5) β (1 + k ) β a1b2 a1b2 Dividing (2) by (4), we get ca kα 2 2 = 1 2 ; a1c2 kβ ⎛α⎞ or ⎜ ⎟ ⎝ β⎠

2

=

c1a2 a1c2

2

⎛ba ⎞ ca or, ⎜ 1 2 ⎟ = 1 2  a1c2 ⎝ a1b2 ⎠ ⎛b ⎞ ⇒ ⎜ 1 ⎟ ⎝b ⎠

(Using (5))

2

c1a2 a12 ca a c × 2= 11 = 1⋅ 1 a c c a a a 2 2 2 1 2 2 c2 2 a1 b1 c1 ∴ , , are in G.P. a2 b2 c2 The correct option is (B) 73. Since α, β are the roots of the equation x2 – px + q = 0 ∴ α + β = p and aβ = q. =

b c and aβ = a a 1 1 a2 + β 2 Also,  α + β = 2 + 2 = α β α 2β 2 then, α + β = –



Now, (α1/4 + β1/4)4 = [(α1/4 + β1/4)2]2 = [α1/2 + β1/2 + 2 (aβ)1/4]2



= ⎡ α + β + 2 αβ + 2(αβ )1/ 4 ⎤ ⎢⎣ ⎥⎦



= ⎡ p + 2 q + 2 ( q)1/ 4 ⎤ ⎣⎢ ⎦⎥ =p+6

1/ 4

q + 4q

=

(α + β ) 2 − 2αβ (αβ ) 2 2





∴ α1/4 + β1/4 = ⎡ p + 6 q + 4 q1/ 4 p + 2 q ⎤ ⎢⎣ ⎥⎦ The correct option is (D) 74. Since a and b are the roots of the equation x2 + px + 1 = 0 ∴ a + b = –p(1) and, ab = 1 (2) Also, since c and d are the roots of the equation x2 + qx + 1 = 0 ∴ c + d = –q(3) and, cd = 1 (4) Now, (a – c) (b – c) (a + d) (b + d) = (ab – bc – ac + c2) (ab + db + ad + d2) = [ab – c (b + a) + c2] ⋅ [ab + d (a + b) + d2] = (1 + cp + c2) (1 – pd + d2) [Putting the values of a + b and ab] = 1 + cp + c2 – pd – cdp2 – c2pd + d2 + cpd2 + c2d2 = 1 + (c2 + d2) + c2d2 – cdp2 + p (c – d) + cpd (d – c) = 1 + [(c + d)2 – 2cd] + c2d2 – cdp2 + p (c – d) + cpd (d – c) = 1 + (q2 – 2) + 1 – p2 + p (c – d) + p (d – c) [Putting the values of c + d and cd] = 2 – 2 + q2 – p2 = q2 – p2. The correct option is (B) 75. Let f (x) = x3 + px2 + q = 0 (1) Since f (x) = 0 has a root of multiplicity 2 ∴ f (x) = 0 and f ′ (x) = 3x2 + 2px = 0 have a common root. The roots of 3x2 + 2px = 0 are x = 0 and x = –2p/3. But x = 0 is not a root of f (x) = 0 ( q ≠ 0) ∴ common root of f (x) = 0 and f ′ (x) = 0 is x = –2p/3. ∴ (–2p/3)3 + p (–2p/3)2 + q = 0 ⇒ 4p3 + 27q = 0. The correct option is (D) 76. Given equation is ax2 + bx + c = 0 Let α, β be the roots of this equation.

2

p+2 q

2

⎛ α + β⎞ 2 ⇒ α + β = ⎜ – αβ ⎝ αβ ⎟⎠ 2

2

2 b 2a ⎛ b⎞ ⎛ − b /a ⎞ ⎛ b⎞ ⇒ ⎜ − ⎟ = ⎜ – ⇒– = ⎜ ⎟ – ⎝ a⎠ ⎝ c /a ⎟⎠ ⎝ c⎠ c /a a c 2

⇒

2a b 2a b ⎡b ⎛ b⎞ = ⎜ ⎟ + ⇒ = + ⎝ c⎠ c a c c ⎢⎣ c

c⎤ a ⎥⎦

HINTS AND EXPLANAT I O N S

1/ 4

2

3.40  Chapter 3 2a b c a b c = + ⇒ , , are in A.P. b c a b c a a b c ⇒ , , are in H.P. c a b The correct option is (C) 77. Discreminent equals –4(k – 5) ≥ 0 ⇒ k ≤ 5. The quadratic equation at x = 5 must be positive and sum of the roots must be less than 10. These conditions imply k2 – 9k + 20 > 0. So, k < 4. The correct option is (A) 78. Let y = x – [x] ∴ the given equation becomes f (y) = (a – 2) y2 + 2y + a2 = 0 (1) Since x is not an integer, ∴ y = x – [x] ≠ 0 Then,  a ≠ 0 [ of (1)] When 2 < x < 3, [x] = 2 ⇒ 0 < x – [x] 0 = ⇒ R > 1 2 The correct option is (D) 80. The given equation can be written as (z + 1)2 – (a – 3) z (z + 1) + (a – 4) z2 = 0 [Putting x2 + x + 1 = z] ⇒ (1 + 3 – a + a – 4) z2 + (2 + 3 – a) z + 1 = 0 1 ⇒ (5 – a) z + 1 = 0  or  z = a−5 1 ⇒ x2 + x + 1 – =0 a−5 a−6 ⇒ x2 + x + =0 a−5 whose roots will be real if discriminant ≥ 0 4( a − 6 ) 3a − 19 ⇒ 1 – ≥0⇒ ≤0 a−5 a−5 19 ∴ 5 < a ≤ 3 The correct option is (D)

HINTS AND EXPLANAT I O N S

⇒

81. Given equation can be written as (2x)2 – ( a– 4)2x – 2x + a– 4 = 0 ⇒ (2x – 1) (2x – a + 4) = 0 ⇒ 2x = 1, 2x = a – 4 Since x ≤ 0 and 2x = a– 4 ( x is non-positive) ∴ 0 < a – 4 ≤ 1 i.e., 4 < a ≤ 5 i.e., a ∈ (4, 5] The correct option is (B) 82. Let f (x) = (x – α) (x – β)(1) Now, f (n) f (n + 1) = (n – α) ( n– β) ( n + 1– α) ( n + 1 – β) = (n – α) ( n + 1– β) (n– β) (n + 1 – α) = {n (n + 1) – n (α + β) – α + αβ} {n (n + 1) – n (α + β) –β + αβ} = {n (n + 1) + na + b–α } {n (n + 1) + na + b – β} = (m – α) (m – β), where m = n (n + 1) + an + b The correct option is (A) 83. We have,

x 2 + nx − 2 x 2 − 3x + 4

⇒

−2 ≤0

x 2 − ( n + 6) x + 10 x 2 − 3x + 4

≥0

⇒ x2 – (n + 6) x + 10 ≥ 0 [ x2 – 3x + 4 >0 ∀ x ∈ R, as its D < 0 and a > 0] The above inequality will be true for all real x if its discriminant ≤ 0 i.e., (n + 6)2 – 40 ≤ 0 ⇒ –( 40 – 6) ≤ n ≤ ( 40 – 6)

(1)

Also, we have,

x 2 + nx − 2 2

x − 3x + 4

+ 1≥ 0 ⇒

2 x 2 + ( n − 3) x + 2 x 2 − 3x + 4

≥0

⇒ 2x2 + (n – 3) x + 2 ≥ 0 The above inequality will be true for all real x if its discriminant ≤ 0 ⇒ (n – 3) 2 – 16 ≤ 0 ⇒ 1 ≤ n ≤ 7  (2) Drawing the number line for inequalities (1), (2) and taking their intersection we get n ∈ [ −1, 40 − 6] The correct option is (C) 84. Let f (x) = (x – a) ( x – b) –1 ⇒ f (a) = –1 and f (b) = –1. Also, the coefficient of x2 = 1 > 0. Hence, a and b both lie between the roots of the equation f (x) = 0 ∴ The equation (x – a) (x – b) – 1 = 0 has one root in (–∞, a) and other in (b, ∞)[ b > a] The correct option is (B)

Quadratic Equations and Expressions  3.41

⇒ a ≤

5 − 45 5 + 45  or a ≥ and – 1 < a < 6 2 2

⇒ – 1 < a ≤

5−3 5 5+3 5  or  ≤a 1) be the common ratio of the increasing G.P. α, β, γ, δ then β = ar, γ = ar2 and δ = ar3 The above equations then reduce to α (1 + r) = a, a2r = p and,  ar2 (1 + r) = b, a2 r5 = q b q ⇒ r2 = and r4 = a p 2 q ⎛ b⎞ ⇒ = ⎜ ⎟ ⎝ a⎠ p q+ p b2 + a2 Hence, we have = 2 . q− p b − a2

The correct option is (B) 89. The sum of the coefficients of the equation = 0 ∴ x = 1 is a root of the equation. Let a be the first term and d be the common difference of given A.P. 1 tp = a + (p – 1)d = (1) q 1 and, tq = a + (q –1) d = (2) p Solving (1) and (2), a = d =

1 pq

∴ tpq = a + (pq – 1)d = 1 ∴ tpq is the root of the given equation. The correct option is (C) 90. Let α, β be the roots of the equation 4x2 + 4ax + b = 0, then we have, b α + β = –a and aβ = 4 According to the given condition, |α – β | ≤ a ⇒ (α + β)2 – 4 α β ≤ a2 ⇒ a2 – b2 ≤ a2 ⇒ b ≥ 0 Also, we have for real roots (4a)2 – 16 b ≥ 0 i.e., b ≤ a2 Therefore, 0 ≤ b ≤ a2 The correct option is (C) 91. Let the roots be α and α + 2, where α is an odd positive integer. Then, aa2 + bα + c = 0  (1) and a (α + 2)2 + b (α + 2) + c = 0 ⇒ aa2 + bα + c + (4 aα + 4a + 2b) = 0 ⇒ 2a (1 + α) + b = 0

[using (1)]

⇒ b = – 2a (1 + α) ⇒ b2 = 4a2 (1 + α)2 ⇒ b2 ≥ 4a2 (1 + 1)2 

[ α ≥ 1as α is odd positive integer]

⇒ b2 ≥ 16 a2 or |b| ≥ 4a The correct option is (B) 92. The discriminants of the given quadratic equations are, D1 = a2 + 12b, D2 = c2– 4b and D3 = d2 – 8b ∴ D1+ D2 + D3 = a2 + c2 + d2 ≥ 0 ⇒ At least one of D1, D2, D3 is non-negative. Hence, the equation has at least two real roots. The correct option is (D)

Previous Year’s Questions 93. Key Idea : The equation having α and β as its roots, is x2− (α + β )x + aβ = 0. Since a2 = 5α − 3 ⇒ α2− 5α + 3 = 0 and β 2 = 5β − 3 ⇒ β 2− 5β + 3 = 0

The above two equations imply that α and β are the roots of the equation x2−5x + 3 = 0. α + β = 5 and αβ = 3

HINTS AND EXPLANAT I O N S

85. The given quadratic equation is satisfied by x = –a, x = –b and x = –c, Hence, the quadratic equation has three roots, which is only possible if it is an identity hence it has infinite roots. The correct option is (D) 86. Putting x2 = y, the given equation in x reduces to ay2 – 2y – (a – 1) = 0  (1) The given biquadratic equation will have four real and distinct roots, if the quadratic equation (1) has two distinct and positive roots. For that, we must have D > 0 ⇒ a2 – a + 1 > 0, which is true ∀ a ∈ R Product of roots > 0 ⇒ 0 < a < 1 sum of roots > 0 ⇒ a > 0 Hence, the acceptable values of a are 0 < a < 1. The correct option is (C) 87. Since –5 and 1 are the roots. Product of roots = –5 × 1 = b ⇒ b = –5 and, sum of roots = –5 + 1= –[ a2 – 5a + b + 4] ⇒ [a2– 5a – 1] = 4 ⇒ 4 ≤ a2 – 5a – 1< 5 ⇒ a2 – 5a – 5 ≥ 0 and a2 – 5a – 6 < 0

3.42  Chapter 3 α β α 2 + β 2 (α + β ) 2 − 2αβ + = = β α αβ αβ 25 − 6 19 = = 3 3 α β and . = 1 β α α β Thus, the equation having and as its roots is given by α β Now

(3a − 1) 2 2

2

=

1 2



⇒ 2x2 − 7x + 7 = 2

a − 5a + 3 9( a − 5a + 3) 2 ⇒a= . 3 The correct option is (A) 98. (1 − p)2 + p (1 − p) + (1 − p) = 0(since (1 − p) is a root of the equation x2 + px + (1 − p) = 0) ⇒ (1− p)(1− p + p + 1) = 0 ⇒ 2(1− p) = 0 ⇒ (1 − p) = 0 ⇒ p = 1 Now, the sum of roots is α + β = -p and the product αβ = 1 − p = 0 (where β = 1 − p = 0) ⇒ α + 0 = –1 ⇒ α = –1 ⇒ Roots are 0, –1 The correct option is (C) 99. Since 4 is one of the root of x2 + px + 12 = 0, we have 16 + 4p + 12 = 0 ⇒ p = −7 And the equation x2 + px + q = 0 has equal roots 49 ⇒ D = 49 − 4q = 0 ⇒ q = . 4 The correct option is (A)



⇒ 2x2 − 7x + 5 = 0

100. Let f  ′(x) = ax2 + bx + c, then f ( x ) =



⎛α β⎞ α β x2 − ⎜ + ⎟ x + ⋅ = 0 β α ⎝ β α⎠ ⇒ x2 −

19 x +1= 0 3

⇒ 3 x 2 − 19 x + 1 = 0 The correct option is (A) 94. Key Idea : If the discriminant of ax2 + bx + c = 0 is positive, then this equation has two real roots. We have 32 x − 7 x + 7 = 32

2

D = b 2 − 4 ac = ( −7) 2 − 4 × 2 × 5 = 49 − 40 =9 Since the discriminant is positive, the equation has two real roots. The correct option is (B) 95. Let α, β be the roots 1 1 α+β = 2 + 2 α β ∴

HINTS AND EXPLANAT I O N S

So,

α+β =

(α + β )2 − 2αβ (αβ ) 2



2 ⎛ b ⎞ b − 2ac ⎜⎝ − ⎟⎠ = a c2

⇒ 2a 2c = b(c 2 + ba)

a b c ⇒ , , are in HP c a b The correct option is (C) 96. x2 – 3 | x | + 2 = 0 ⇒ (| x | –1) (| x | – 2) = 0, as | x |2 = x2 ⇒ x = ±1, ± 2 . The correct option is (B) 97. β = 2α 1 − 3a 3α = 2 a − 5a + 3

2α 2 =

2 2

a − 5a + 3

ax 3 bx 2 + + cx + d 3 2

1 ⇒ f ( x ) = ( 2ax 3 + 3bx 2 + 6cx + 6 d ), Now f (1) = f (0) = d, 6 then according to Rolle’s theorem 2 ⇒ f  ′(x) = ax + bx + c = 0 has at least one root in (0, 1) The correct option is (A) 101. x2− (a − 2)x − a − 1 = 0 Sum of roots, α + β = a − 2 Product, αβ = − (a + 1) α2 + β2 = (α + β)2− 2αβ = a2− 2a + 6 = (a − 1)2 + 5 ⇒ a = 1 The correct option is (A) 102. Let α, α + 1 be the roots of the equation, then α + (α + 1) = b α(α + 1) = c ∴  b2− 4c = (2α + 1)2 − 4α (α + 1) = 1 The correct option is (D) −b = 2k < 10 ⇒ k < 5 a c or, = k 2 + k − 5 < 25 ⇒ ( k + 6)( k − 5) < 0 a ⇒k − 1 and m < 3 ⇒ − 1 < m < 3. The correct option is (C) 3x 2 + 9 x + 7



, then

106. Given equation x2 + ax + 1 = 0 The sum and product of the roots α + β = −a αβ = 1 Now, | α − β | = (α + β ) 2 − 4αβ a2 − 4

⇒ a2 − 4 < 5 2 ⇒ a − 4 < 5 ⇒ a2− 9 < 0 ⇒ a ∈ (−3, 3) The correct option is (A) 107. Let α and 4β be roots of x2− 6x + a = 0 and α, 3β be the roots of x2− cx + 6 = 0, then α + 4β = 6 and 4αβ = a α + 3β = c and 3αβ = 6 We get αβ = 2 ⇒ a = 8 So the first equation is x2− 6x + 8 = 0 ⇒ x = 2, 4 Now, if α = 2 and 4β = 4 then 3β = 3 If α = 4 and 4β = 2, then 3β = 3/2 (non-integer) ∴ common root is x = 2. The correct option is (D) 108. Given equation bx2 + cx + a = 0 has imaginary roots ⇒ c2− 4ab < 0 ⇒ c2< 4ab ⇒ –c2 > −4ab Since 3b2 > 0, the expression 3b2x2 + 6bcx + 2c2 has ­minimum value. Minimum value 4(3b 2 )( 2c 2 ) − 36b 2c 2 12b 2c 2 = =− = − c 2 > −4 ab. 2 4(3b ) 12b 2 The correct option is (C) 109.



x2 − x + 1 = 0 x=

α=

⇒x=

1± 1− 4 2

110.

esin x − e − sin x = 4 ⇒ esin x = t



1 t− −4 t



β = cos

π π − i sin 3 3

t 2 − 4t − 1 − 0 ⇒ t = ⇒t =

4 ± 16 + 4 2

4±2 5 ⇒t = 2± 5 2

esin x = 2 ± 5 −1 ≤ sin x ≤ 1

1 ≤ esin x ≤ e e

sin x = 2 + 5 not possible e sin x e = 2 − 5 not possible ∴ Hence no solution The correct option is (B) 111. If 2 x 3 + 3 x + k = 0 has 2 distinct real roots in [0, 1], then f  ′ (x) will change sign. But f  ′(x) = 6x2 + 3 > 0 So, no value of k exists. The correct option is (C) 112. Both the roots of the equation x 2 + 2 x + 3 = 0 are imaginary. Since, a, b, c ∈ R. So, if one root is common then both roots are common

a b c = = 1 2 3 a : b : c = 1 : 2 : 3. The correct option is (D) 113. a 2 = 3t 2 − 2t For non-integral solution 2 0 < a < 1 a ∈ ( −1, 0) ∪ (0,1). [Note: It is assumed that a real solution of given equation exists.] The correct option is (A) 114. x 2 = 6 x + 2 ⇒ α 2 = 6α + 2 Hence,

α 10 = 6α 9 + 2α 8  ⇒ and β10 = 6β 9 + 2β 8  Subtract (2) from (1)

1 ± 3i 2 1 3 +i , 2 2

2π ⎛ 1⎞ = −2 ⎜ − ⎟ = 1 ⎝ 2⎠ 3 The correct option is (B) = −2 cos

3x2 (y − 1) + 9x (y − 1) + 7y − 17 = 0 Now, D ≥ 0 (∵ x is real) implies that 81(y − 1)2 − 4x3 (y − 1) (7y − 17) ≥ 0 ⇒ (y − 1) (y − 41) ≤ 0 ⇒ 1 ≤ y ≤ 41 The correct option is (C)

⇒ |α − β | =

2π ⎞ 2π ⎤ ⎡ ⎛ = 2 cos ⎢668π + π + = 2 cos ⎜ π + ⎟ ⎝ 3 ⎥⎦ 3⎠ ⎣

⇒ a10 = 6 a9 + 2a8 ⇒

a10 − 2a8 = 3. 2a9

The correct option is (B)

(1) (2)

HINTS AND EXPLANAT I O N S

105. Let y =

3 x 2 + 9 x + 17

⎛π⎞ α 2009 + β 2009 = 2 cos 2009 ⎜ ⎟ ⎝ 3⎠

3.44  Chapter 3 115. Rearranging equation, we get nx2 +(1 + 3 + 5 + … + (2n + 1)}x + {1.2 + 2.3 + … (n – 1)n} = 10n ( n − 1)n( n + 1) ⇒ nx2 + n2x + = 10n 3

HINTS AND EXPLANAT I O N S

⎛ n2 − 31⎞ ⇒ x2 + nx + ⎜ ⎟ =0 ⎝ 3 ⎠ Given difference of roots = 1 ⇒ | α – β | = 1 ⇒ D = 1 4 ⇒ n2 – (n2 – 31) = 1 3 So, n = 11 Hence, the correct option is (C) 116. Given x2 – x + 1 = 0 Roots are α and β

 α + β = 1 αβ=1 1 + ω + ω2 = 0  ω3 = 1  –ω – ω2 = 1 α = –ω  β = –ω2 To calculate α101 + β107 ⇒ (–ω)101 + (–ω2)107

[cube roots fully]

⇒ −1 × ⎡⎣(ω 33 )3ω 2 + ω 214 ⎤⎦



⇒ –[ω2 + (ω3)71 ⋅ ω] ⇒ –[ω + ω2] = –1 × –1 = 1 117. Sum of square of roots is (α + β)2 – 2αβ = λ2 – 4λ + 5 which is minimum for λ = 2.

CHAPTER

4

Permutations and Combinations

LEARNING OBJECTIVES After reading this chapter, you will be able to:  Know about factorial notation and fundamental principles of counting  Learn to define permutation and combination, and how are they denoted

 Grasp the knowledge of some key results on combination and some useful results for geometrical problems   Be acquainted with exponent of prime p in n!

FACTORIAL NOTATION We often come across products of the form 1 ⋅ 2, 1 ⋅ 2 ⋅ 3, 1 ⋅ 2 ⋅ 3 ⋅ 4, … Instead of writing all the factors of such a product in full, it is convenient to use a special notation. We write 1! = 1, 2! = 1 ⋅ 2, 3! = 1 ⋅ 2 ⋅ 3, ………………..... n! = 1 ⋅ 2 ⋅ 3 … n. “n!” denotes the product of the first n natural numbers. We read ‘n!’ as ‘n factorial’. n! is also written as ‘ |−− n ’ and read as ‘factorial n’. It is easy to see that 1! = 1, 2! = 1 ⋅ 2 = 2, 3! = 1 ⋅ 2 ⋅ 3 = 6, 4! = 1 ⋅ 2 ⋅ 3 ⋅ 4. = 24, and so on.

I M P O R TA N T P O I N T S  We know that n! = n (n – 1) (n – 2) (n – 3) … 3 · 2 · 1 = n (n – 1)! = n (n – 1) (n – 2)! = n (n – 1) (n – 2) (n – 3)! and so on. Thus, if m, n ∈ N and m > n, then m! can be expressed in terms of n! For example 8! = 8 · 7 · 6! 10! = 10 · 9 · 8 · 7 · 6 · 5! Also, m! = n! if and only if m = n  Putting n = 1 in n! = n (n – 1)!, we have 1! = 1 · 0! \ 0! = 1 n  (2n)! = 2 · n! [1 · 3 · 5 … (2n – 1)]

ERROR CHECK The factorial is defined only for whole numbers. We do not define the factorial of proper fractions or ­negative integers.  (2n)! ≠ 2(n)!  (m + n)! ≠ m! + n!  (mn)! ≠ m! n! 

FUNDAMENTAL PRINCIPLES 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.

Illustration Anu wishes to buy a birthday card for the brother Manu and send it by post. Five different types of cards are available at the card-shop, and four different types of postage stamps are available at the post-office. In how many ways can she choose the card and the stamp?

Solution She can choose the card in five ways. For each choice of the card she has four choices for the stamp. Therefore, there are 5 × 4 ways, i.e., 20 ways of choosing the card and the stamp.

4.2  Chapter 4

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.

Info Box! The above two principles can be extended for any finite number of operations.

(C) a1 + a2 + … + ak + 1 (D)  None of these Solution: (A) Total number of wrong answers

= 1 · (a1 – a2) + 2 ⋅ (a2 – a3) +, …,



+ (k – 1) (ak – 1 – ak) + kak = a1 + a2 + a3 +, …, + ak

17 1 (B)  7 4 Solution: (C) The word BAC cannot be spelt if the m selected ­coupons do not contain atleast one of A, B and C. Number of ways of selecting m coupons which are A or B = 2m. This also includes the case when all the m coupons are A or all are B. Number of ways of selecting m coupons which are B or C = 2m. This also includes the case when all the m coupons are B or all are C. Number of ways of selecting m coupons which are C or A = 2m. This also includes the case when all the m coupons are C or all are A. Number of ways of selecting m coupons when all are A = 1m. Number of ways of selecting m coupons when all are B = 1m. Number of ways of selecting m coupons when all are C = 1m.

(A) 

Illustration Suppose there are 5 gates in a stadium, 2 on one side and 3 on the other. Sohan has to go out of the stadium. He can go out from any one of the 5 gates. Thus, the number of ways in which he can go out is 5. Hence, the work of going out through the gates on one side will be done in 2 ways and the work of going out through the gates on other side will be done in 3 ways. The work of going out will be done when the Sohan goes out from side I or side II. Thus, the work of going out can be done in (2 + 3) = 5 ways.

E D

A Side B I

C

I eI

Sid

FIGURE 4.1

3x 2 + 9x + 17 is 3x 2 + 9x + 7 (C) 41 (D) 1

2. If x is real, the maximum value of

\ Required number = 2m + 2m + 2m – (1m + 1m + 1m)

Info Box! Addition theorem of counting is also true for more than two works.

SOLVED EXAMPLES 1. 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 (B) a1 + a2 + … + ak – 1



= 3 ⋅ 2m – 3 ⋅ 1m = 3 (2m – 1)

3. 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 (B) 1024 (C)  1072 (D)  None of these Solution: (B) Each man can vote for any one of the 4 candidates and this can be done in 4 ways. Similar is the case with every other man. ( Repetition is allowed) Hence 5 men can vote in 45 i.e., 1024 ways.

Permutations and Combinations  4.3 4. 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 (B)  18 10 (C) 2 (D)  1023 Solution: (D) Each 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.) 5. 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

Solution: (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 6. 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

Solution: (A) Each stall can be filled in 3 ways as there are three types of animals (animals of each category being not less than 10). Shipload, i.e., filling up of 10 stalls, can be made in 3 × 3 × … up to 10 times = 310 = 59049 7. 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

Solution: (B) Total number of coins = 2 + 2 + 3 + 1 = 8. 2 coins are 10 paise, 2 are 20 paise, 3 are 25 paise and 1 is of 50 paise. \ Required no. of ways 8! = 2 ! × 2 ! × 3! × 1! 



8 × 7 × 6 × 5 × 4 × 3 × 2 ×1 2 ×1× 2 ×1× 3 × 2 ×1×1

=



= 8 × 7 × 6 × 5 = 1680



8. 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 (B) 800 (C)  900 (D)  None of these Solution: (C) There are in all seven letters in the word BALLOON in which L occurs 2 times and O occurs 2 times. \ The number of arrangements of the seven letters 7! of the word = = 1260. 2! x 2! If two L’s always come together, taking them as one letter, we have to arrange 6 letters in which O occurs 2 times. \ The no. of arrangements in which the two L’s come together 6! = = 6 × 5 × 4 × 3 = 360. 2! Hence, the required no. of ways in which the two L’s do not come together = 1260 – 360 = 900. 9. A Letter lock contains 5 rings each marked with four different letters. The number of all possible unsuccessful attempts to open the lock is (A) 625 (B) 1024 (C) 624 (C) 1023 Solution: (B) 1

1

1

1

1

2

2

2

2

2

3

3

3

3

3

4

4

4

4

4

Number of options on 1st Ring = 4 Number of options on 2nd Ring = 4 Number of options on 3rd Ring = 4 Number of options on 4th Ring = 4

4.4  Chapter 4 Number of options on 5th Ring = 4 \ Total number of options/arrangements

The grand children wish to occupy the 4 seats on either side of the table = 4! ways

= 4 × 4 × 4 × 4 × 4 = 1024 10. We have (n + 1) white balls and (n + 1) black balls. In each set the balls are numbered from 1 to (n + 1). If these balls are to be arranged in a row so that two consecutive balls are of different colours, then the number of these arrangements is (A) (2n + 2)! (B)  2 × (2n + 2) (C) 2 × (n + 1)! (D)  2 [(n + 1)!]2 Solution: (D) Between (n + 1) white balls there are (n + 2) gaps in which (n + 1) black balls are to arranged. Number of required arrangements = (n + 1)! (n + 1)! = [(n + 1)!]2 Now between (n + 1) black balls (n + 1) white balls are to be filled in number of ways = (n + 1)! (n + 1)! \ Required number of ways = 2[(n + 1)!]2 11. Number of points having position vectoriˆ aiˆ + bjˆ + ckˆ, where a, b, c ∈ {1, 2, 3, 4, 5} such that 2a + 3b + 5c is divisible by 4 is (A) 140 (B) 70 (C)  100 (D)  None of these Solution: (B)

= 24 ways and grand father can occupy a seat in (5 – 1) ways = 4 ways (Since 4 gaps between 5 sons and daughters) and the remaining seats can be occupied in 5! ways = 120 ways (5 seats for sons and daughters) Hence, required number of ways

= 24 × 4 × 120 = 11520

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.

I M P O R TA N T P O I N T S 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.

4m = 2a + 3b + 5c = 2a + (4 – 1)b + (4 + 1)c 4m = 4k + 2a + (–1)b + (1)c \

a = 1, b = even, c = any number a ≠ 1, b = odd, c = any number

\ Required number = 1 × 2 × 5 × 4 × 3 × 5 = 70 12. 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 child on either side of him. The number of ways in which the family can be made to sit is (A) 11360 (B) 11520 (C)  21530 (D)  None of these Solution: (B) The total number of seats

= 1 grand father + 5 sons and daughters + 8 grand children = 14

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).

QUICK TIPS n! = n(n – 1) (n – 2) … [n – (r – 1)], (n − r )! 0 ≤ r ≤ n. 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! 1. nPr =

Permutations and Combinations  4.5 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

(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



r ⋅ n – 1Pr – 1

SOLVED EXAMPLES 13. The number of positive terms in the sequence

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



All those numbers whose last three digit number is ­divisible by 8 are divisible by 8.  All those numbers the sum of whose digits is divisible by 9 are themselves divisible by 9.  All those numbers whose last two digits are divisible by 25 are themselves divisible by 25. 

n–1

Pr

(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 anti-clockwise arrangements are not different, i.e., when observa1 tion can be made from both sides is: (n − 1)! . 2 (c) Number of circular permutations of n different things, taken r at a time, when clockwise, and anti-clockwise n P orders are taken as different, is = r . r (d) Number of circular permutations of n different things, taken r at a time, when clockwise and anti-clockwise n P orders are not different, is = r . 2r

QUICK TIPS Numbers Divisible by 2, 3, 4, 5, 8, 9, 25  All those numbers having their last digit as an even number (i.e., 0, 2, 4, 6 or 8) are divisible by 2.  All those numbers the sum of whose digits is divisible by 3 are themselves divisible by 3.  All those numbers whose last two digits are divisible by 4 are themselves divisible by 4.  All those numbers whose last digit is either 0 or 5 are divisible by 5.

xn =

195 n

4 Pn

−

n+3 n +1

Pn + 1

(A) 2 (C)  4 Solution: (C) We have, xn =

195 n

4 ⋅ pn

−

P3

, n ∈ N is

(B) 3 (D)  None of these

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!  2 195 − 4 n − 20 n − 24 = 4 ⋅ n!  171 − 4 n2 − 20 n = 4 ⋅ n!  xn is positive. \

171 − 4 n2 − 20 n > 0 4 ⋅ n!

⇒ 4n2 + 20n – 171 < 0 which is true for n = 1, 2, 3, 4. Hence, the given sequence has 4 positive terms. 14. 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 (B) 1470 (C) 1370 (D)  None of these Solution: (A) There are 5 consonants and 2 vowels in the word STRANGE. Out of 7 places for the 7 letters, 4 places are odd and 3 places are even.

4.6  Chapter 4 Two vowels can be arranged in 4 odd places in P (4, 2) ways = 12 ways and then 5 consonants can be arranged in the remaining 5 places in P(5, 5) ways = 5 × 4 × 5 × 3 × 2 × 1 = 120 ways.

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

Hence, the required number of ways = P (4, 2) × P (5, 5) = 12 × 120 = 1440 15. How many numbers greater than 0 and less than a million can be formed with the digits from 0 to 9? (A) 25927 (C) 22759

(B) 27925 (C) 72925

Solution: (A) Any number between 1 and 1000000 must be of less than seven digits. Therefore, it must be of the form a1 a2 a3 a4 a5 a6 where a1, a2, a3, a4, a5, a6 ∈ {0, 1, 2, …, 9} According to question, sum of the digits = 18 Thus,

17. 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 (B) 6 (C)  8 (D)  None of these Solution: (A) As shown in figure, 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 X

a1 + a2 + a3 + a4 + a5 + a6 = 18

where 0 ≤ ai ≤ 9, i = 1, 2, 3, …, 9. Required number

= coefficient of x18 in (1 + x + x2 + … + x9)6



⎛ 1 − x 10 ⎞ = coefficient of x in ⎜ ⎟ ⎝ 1− x ⎠



6

 = coeff. of x18 in [(1 – x10)6 (1 – x)– 6] 18

6

10

–6

= coeff. of x in [(1 – C1 x ) (1 – x) ] 18

–6

= coeff. of x in (1 – x)

6

– C1.

coeff. of x8 in (1 – x)– 6

= 6 + 18 – 1C18 – 6 ⋅ 6 + 8 – 1C18 = 23C5 – 6 ⋅ 13C8



=



2

3

4

18

(leaving terms containing powers of x greater than 18)

1

23 ⋅ 22 ⋅ 21⋅ 20 ⋅19 13 ⋅12 ⋅11⋅10 ⋅ 9 −6 120 120  = 33649 – 7722 = 25927

16. 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 (B) 108 (C) 36 (D) 18 Solution: (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.

Y

= 2! = 2 × 1 = 2 ways.



Also, 3 and 4 can be arranged in P (2, 2) ways = 2! = 2 × 1 = 2 ways.



Hence, required number of permutations = 2 × 2 = 4



18. The number of divisors of 9600 including 1 and 9600 are (A) 60 (B) 58 (C) 48 (D) 46 Solution: (C)  9600 = 27 × 3 × 52 \

No. of divisors = (7 + 1) × (1 + 1) × (2 + 1) = 8 × 2 × 3 = 48



QUICK TIPS If N = p1a ⋅ p2a … pka , where p1, p2, …, pk are different primes and a1, a2,…, ak are natural numbers, then the total number of divisors of N including, and N is (a1 + 1) (a2 + 1) … (ak + 1). 1

2

k

Permutations and Combinations  4.7 19. 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 teachers, middle teachers and secondary teachers are always in a set. The number of ways in which they can do so, is (A) 34 (B) 48 (C)  52 (D)  None of these Solution: (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!

Solution: (B) Four lines intersect each other in 4C2 = 6 points and 4 circles intersect in 4P2 = 12 points. Each line cuts 4 circles into 8. \ 4 lines cut four circles into 32 points. \ Required number = 6 + 12 + 32 = 50. 22. 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 n (A) nPm (B)  Cm × (m – 1)! 1 n n –1 (C)  . Pm (D)  Pm 2 Solution: (A) Since the seats are numbered, \ the arrangement is not circular. \ the required number of ways = the number of arrangements of n things

= 2 × 1 ways = 2 ways These three sets can be arranged in themselves in

taken m at a time = nPm

= 3! = 3 × 2 × 1 = 6 ways Hence, the required number of ways = 2 × 2 × 2 × 6 = 48 20. 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 ! (A)  (B)  4! 6! 4! 6! (C) 

8! 8! 6 ! 6! 4!

(D)  None of these

Solution: (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. Hence the total number of ways in which the ­persons can be arranged = 8P4 × 8P2 × 10! =

8! 8! 10 ! 4! 6!

21. The maximum number of points into which 4 circles and 4 straight lines intersect is (A) 26 (B) 50 (C) 56 (D) 72

COMBINATION Each of the different groups or selections which can be made by taking some or all of a number of things (irrespective of order) is called a combination.

I M P O R TA N T P O I N T S Combination of things means selection of things. Obviously, in selection of things order of things has no importance. Thus, with the change of order of things selection of things does not change.

Notations The number of combinations of n different things taken r at a time is denoted by nCr or C(n, r). Thus, n

Cr =

n Pr n! = (0 ≤ r ≤ n) r ! ( n − r )! r!

n ( n − 1) ( n − 2) … ( n − r + 1) r ( r − 1) ( r − 2) … 3 ⋅ 2 ⋅1  r > n, =

If then

n

Cr = 0.

Info Box! Selecting things without any order in called combination and arrangement of things in some order is called permutation.

4.8  Chapter 4

KEY RESULTS ON COMBINATION 1. nCr = nCn – r , 0 ≤ r ≤ n 2. nC0 = nCn = 1, nC1 = n 3. If nCx = nCy then either x = y or x + y = n. 4. nCr + nCr–1 = n + 1Cr , 1 ≤ r ≤ n 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 = ,1≤r≤n r C r −1

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  n C n −1 2

2

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). 12. nC0 + nC1 + nC2 + … + nCn = 2n. 13. nC0 + nC2 + nC4 + … = nC1 + nC3 + nC5 + … = 2n – 1. 14. Number of combinations of n different things taken r at a time (a) when p particular things are always included = n–p Cr–p. (b) when p particular things are never included = n–p Cr. (c) when p particular things are not together in any selection = nCr – n–pCr–p. 15. (a) Number of selections of r consecutive things out of n things in a row = n – r + 1. (b) Number of selections of r consecutive things out of n things along a circle 10. nCr =

⎧n, when r < n ⎨ ⎩1, when r = n



16. (a) Number of selections of zero or more things out of n different things

n

C0 + nC1 + nC2 + … + nCn = 2n

(b)  Number of combinations of n different things selecting at least one of them is

n

C1 + nC2 + … + nCn = 2n – 1

(c) Number of selections of r things (r ≤ n) out of n identical things is 1. (d) Number of selections of zero or more things out of n identical things = n + 1.

(e) Number of selections of one or more things out of n identical things = n. (f ) If out of (p + q + r + t) things, p are alike of one kind, q are alike of second kind, r are alike of third kind and t are different, then the total number of selections is (p + 1) (q + 1) (r + 1) 2t – 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 alike of second kind and rest are alike of third kind is [(p + 1) (q + 1) (r + 1)] – 1.

Division into Groups 1. (a) Number of ways of dividing m + n different things in two groups containing m and n things respectively (m ≠ n) is

m+n

Cm =

( m + n)! m ! n!

(b) Number of ways of dividing m + n + p different things in three groups containing m, n and p ( m + n + p)! things respectively (m ≠ n ≠ p) is , m ! n! p ! if the order of the groups is not important and ( m + n + p )! × 3! , if the order of the groups is m ! n! p ! important. (c) Number of ways of dividing 2m different things in two groups, each containing m things and the order ( 2n)! of the groups is not important, is , of the 2 ! ( n !) 2 ( m + n + p)! groups is not important and × 3!, if m ! n! p ! the order of the groups is important. (d) Number of ways of dividing 2m different things in two groups, each containing m things and the ( 2n)! order of the groups is important, is . ( n !) 2 (e) Number of ways of dividing 3m different things in three groups, each containing m things and the (3m)! order of the groups is not important, is . 3! ( m !)3 (f ) Number of ways of dividing 3m diferent things in three groups, each containing m things and the (3m)! order of the groups is important, is . ( m !)3 2. (a) Number of ways of dividing n identical things into r groups, if blank groups are allowed is

n+r–1

Cr – 1

Permutations and Combinations  4.9 (b) Number of ways of dividing n identical things into r groups, if blank groups are not allowed is n–1 Cr–1. (c) Number of ways of dividing n identical things into r groups such that no group contains less than m things and more than k (m < k) things is coefficient of xn in the expansion of (xm + xm + 1 + … + xk)r 3. The number of ways of selecting r things out of n things of which p are alike and are of one kind, q are alike and are of second, s are alike and are of third kind and so on, is = coefficient of xr in [(1 + x + x2 + … + xp) (1 + x + x2 + … + xq) × (1 + x + x2 + … + xs) …] 4. The number of ways of selecting r things out of n things of which p are a like and are of one kind, q are alike and are of second kind and rest (n – p – q) things are all different is = coefficient of xr in [(1 + x + x2 + … + xp) (1 + x + x2 + … + xq) × (1 + x)n–p–q] 5. The number of ways of selecting r things out of n things of which p are alike and are of one kind, q are alike and are of second kind, s are alike and are of third kind when each thing is taken at least once = coefficient of xr – 3 in [(1 + x + x2 + … + xp–1) (1 + x + x2 + … + xq–1) × (1 + x + x2 + … + xs–1) …] 6. 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 xr in (1 + x + x2 + … + xr)n = coefficient of xr in (1 – x)–n = n + r – 1Cr 7. The number of non-negative integral solutions of the equation x1 + x2 + … + xr = n is n + r – 1Cr. 8. The number of terms in the expansion of r

(a1 + a2 + a3 + … + an) is



SOLVED EXAMPLES 23.

47

C4 +

5

∑ 52− j C3

=

j =1

(A) 52C4 (B) 51C4 (C) 52C3 (D)  None of these

n+r–1

Cr

Solution: (A) We have, 5

C4 + ∑

47

52 − j

j =1

C3



47 = C4 + 51C3 + 50C3 + 49C3 + 48C3+ 47C3



= (47C4 + 47C3) + 48C3 + 49C3 + 50C3 + 51C3



= 48C4 + 48C3 + 49C3 + 50C3 + 51C3



= 49C4 + 49C3 + 50C3 + 51C3 = 50C4 + 50C3 + 51C3



= 51C4 + 51C3 = 52C4

24. 15C8 + 15C9 – 15C6 – 15C7 = (A) 8 (C)  6

(B) 0 (D)  None of these

Solution: (B) We have, 15

C8 + 15C9 – 15C6 – 15C7



= (15C8 + 15C9) – (15C6 + 15C7)



= 16C9 – 16C7( nCr + nCr + 1 = n + 1Cr + 1)



= 16C9 – 16C9( nCr = nCn – r)



= 0

⎛ n⎞ ⎛ n ⎞ ⎛ n ⎞ 25. For 2 ≤ r ≤ n, ⎜ ⎟ + 2 ⎜ = + ⎝ r⎠ ⎝ r − 1⎟⎠ ⎜⎝ r − 2⎟⎠ ⎛ n + 1⎞ ⎛ n + 1⎞ (A)  ⎜ (B)  2⎜ ⎝ r − 1⎟⎠ ⎝ r + 1⎟⎠ ⎛ n + 2⎞ ⎛ n + 2⎞ (C) 2 ⎜ (D)  ⎜⎝ r ⎟⎠ ⎟ ⎝ r ⎠ Solution: (D) n

Cr + 2nCr–1 + nCr–2 = (nCr + nCr–1) + (nCr–1 + nCr–2)



= n+1Cr + n+1Cr–1



⎛ n + 2⎞ = n+2Cr = ⎜ ⎝ r ⎟⎠

 26. 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! (B)  9! 40 12 ! 13! (C)  (D)  40 40 (A) 

4.10  Chapter 4 Solution: (D) The number of ways in which 13 guests may be divided 13! into groups of 8 and 5 = 13C5 = . 5! 8! Now, corresponding to one such group, the 8 guests may be seated at one round table in (8 – 1)! i.e., 7! ways and the five guests at the other table in (5 – 1)! i.e., 4! ways. But each way of arranging the first group of 8 persons can be associated with each way of arranging the second group of 5, therefore, the two processes can be performed together in 7! × 4! ways. Hence required number of arrangements 13! 13! 13! = × 7! × 4! = × 7! × 4! = 5! 8! 5 ⋅ 4 ! 8 ⋅ 7! 40 27. 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 (B) 48 (C)  72 (D)  None of these Solution: (A) Total number of handshakes =T  he number of ways of selecting 2 persons from among 12 persons

= 12C2 =

12 x 11 = 66 2x 1

28. 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 (B) 178 (C)  202 (D)  None of these Solution: (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! 8x 7x 6x 5 8 C4 = = = 70 ways 4! 4! 4x 3x 2 (ii) If oldest is excluded, then we are to select 5 candidates from 9 which can be done in 9! 9x 8x 7x 6 9 C5 = = = 126 ways 5! 4 ! 4x 3x 2x 1  Hence, the total number of ways in which ­committee can be formed = 126 + 70 = 196

29. How many non-negative integers solutions does the equation 3x + y + z + w = 30 have? (A) 2151 (C) 1215

(B) 1521 (C) 1512

Solution: (C) Let w be a non-negative integer such that 3x + y + z + w = 30 a = x – 1, b = y – 1, c = z – 1, d = w,

Let then

3a + b + c + d = 25, where a, b, c, d ≥ 0

(1)

Clearly, 0 ≤ a ≤ 8. If a = k, then b + c + d = 25 – 3. (2) Number of non-negative integral solutions of Eq. (2)

= n + r – 1Cr



= 3 + 25 – 3k – 1C25 – 3k



= 27 – 3kC25 – 3k = 27 – 3kC2



=

(27 − 3k ) (26 − 3k ) 2 

= 3 (3k2 – 53k – 234) 2 \ Required number





=

=

8

3 (3k 2 − 53k + 234) 2k =0

∑



3 ⎛ 8 × 9 × 17 8×9 ⎞ 3 − 53 + 234 × 9⎟ = 1215 ⎜ ⎠ 2⎝ 6 2

30. The number of different ways in which 8 persons can stand in a row so that between two particular persons A and B there are always two persons, is (A)  60 (5!) (B) 4! × 5! (C)  15 (4!) × 5! (D)  None of these Solution: (A) The number of selections of 4 persons including A, B considering these four as a group, the number = 6C2 of arrangement with the other four = 5!. But in each group of four, number of arrangements  = 2! × 2! therefore, the required number of ways = 6C2 × 5! × 2! × 2!.

Permutations and Combinations  4.11 31. There are 10 points in a plane of which no three points are collinear but 4 points are concyclic. The number of different circles that can be drawn through atleast 3 of these points is (A) 110 (B) 112 (C) 116 (D) 117

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 (B) 640 (C)  820 (D)  None of these

Solution: (D) Since a unique circle can be drawn through three points, therefore a selection of three points results in a circle. So, the maximum number of circles using 10 points is 10C3. Now, out of these 10 points 4 are concyclic, hence 4C3 circles are actually single circle. \ Required number of circles = 10C3 – 4C3 + 1 = 117.

Solution: (A) A candidate can attempt 5 questions from group I and 2 from group II or 4 from group I and 3 from group II or 3 from group I and 4 from group III or 2 from group I and 5 from group II. This can be done in

32. The number of ways in which a committee of 3 ladies and 4 gentlemen can be appointed from a meeting consisting of 8 ladies and 7 gentlemen, if Mr X refuses to serve in a committee if Mr Y is a member is (A) 1960 (B) 1540 (C)  3240 (D)  None of these Solution: (B) 3 ladies out of 8 can be selected in 8C3 ways and 4 ­gentlemen out of 7 in 7C4 ways. Now each way of selecting 3 ladies is associated with each way of selecting 4 gentlemen. Hence, the required number of ways = 8C3 × 7C4 = 56 × 35 = 1960 We now find the number of committees of 3 ladies and 4 gentlemen in which both Mrs X and Mr Y are members. In this case, we can select 2 other ladies from the remaining 7 in 7C2 ways and 3 other gentlemen from the remaining 6 in 6C3 ways. \ The number of ways in which both Mrs X and Mr Y are always included = 7C2 × 8C3 = 21 × 20 = 420. Hence, the required number of committes in which Mrs X and Mr Y do not serve together = 1960 – 420 = 1540. 33. 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 16 16 20 (A) 16C11 (B)  C5 (C)  C9 (D)  C9 Solution: (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. So the remaining 9 players are to be selected from the remaining 16 players. This can be done in 16C9 ways. 34. A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each

6

C5 × 6C2 + 6C4 × 6C3 + 6C3 × 6C4 + 6C2 × 6C5



= 6 × 15 + 15 × 20 + 20 × 15 + 15 × 6 = 90 + 300 + 300 + 90 = 780

35. 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 (A) 41 (B) 32 (C)  51 (D)  None of these Solution: (A) The following are the different possibilities in which three books can be borrowed:   (i)  When Chemistry part II is selected, then Chemistry part I is also borrowed and the third book is selected from the remaining 6 books. (ii) When Chemistry part II is not selected, in this case he has to select the three books from the remaining 7 books. First choice can be made in 6C1 = 6 ways. Second choice can be made in 7×6×5 7 C3 = = 35 ways 1× 2 × 3 Total number of ways in which he can choose the three books to be borrowed = 6 + 35 = 41. 36. The number of words that can be formed from the ­letters a, b, c, d, e, f, taken 3 at a time, each word containing at least one vowel is (A) 96 (B) 84 (C) 106 (D)  None of these Solution: (A) The total number of words

= (2C1 × 4C2 + 2C2 × 4C1) 3!



= (12 + 4) × 6 = 96.

4.12  Chapter 4 37. In an examination a candidate has to pass in each of the papers to be successful. If the total number of ways to fail is 63, how many papers are there in the examination? (A) 6 (B) 8 (C)  14 (D)  None of these Solution: (A) Let the number of papers be n. \ Total number of ways to fail or pass n

C0 + nC1 + nC2 + … + nCn = 2n

But there is only one way to pass, i.e., when he fails in none. \ Total number of ways to fail = 2n – 1 \ From question, 2n – 1 = 63; \ 2n = 64 = 26 \ n = 6. 38. The total number of selections from 4 boys and 3 girls if each selection has to contain at least one boy is (A) 106 (B) 120 (C)  240 (D)  None of these Solution: (B) Number of selections of at least one boy from 4 boys = 4C1 + 4C2 + 4C3 + 4C4 = 24 – 1 Number of selections of any number of girls from 3 girls 3

3

3

3

3

= C0 + C1 + C2 + C3 = 2

\ Required number of selections of at least one boy from 4 boys and 3 girls = (24 – 1)23 = 15 × 8 = 120. 39. A boat is to be manned by eight men of whom 2 can only row on bow side and 1 can only row on stroke side; the number of ways in which the crew can be arranged is (A) 4360 (B) 5760 (C)  5930 (D)  None of these Solution: (B) First we have to select 2 men for bow side and 3 for stroke side. \ The number of selections of the crew for two sides = 5C2 × 3C3 For each selection, there are 4 persons each on both sides who can be arranged in 4! × 4! ways. \ Required number of arrangements 5x 4 = 5C2 × 3C3 × 4! × 4! = x 1 x 24 x 24 = 5760 2

40. The number of ways in which a mixed doubles game in tennis can be arranged from 5 married couples, if no husband and wife play in the same game, is (A) 46 (B) 54 (C)  60 (D)  None of these Solution: (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. After choosing the two husbands their wives are to be excluded (since no husband and wife play in the same game). So we are to choose 2 wives out of remaining 5 – 2 = 3 wives, i.e., 3C2 = 3 ways. Again two wives can interchange their sides A and B in 2! = 2 ways. Therefore, the required number of ways = 10 × 3 × 2 = 60. 41. 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 (B) 36 (C)  54 (D)  None of these Solution: (A) Considering CC as single object, U, CC, E can be arranged in 3! ways × U × CC × E × Now the three S are to be placed in the four available places. Hence required number of ways = 3! · 4C3 = 24. 42. Four boys picked up 30 mangoes. The number of ways in which they can divide them if all mangoes be identical, is (A) 5456 (B) 3456 (C)  5462 (D)  None of these Solution: (A) Clearly, 30 mangoes can be distributed among 4 boys such that each boy can receive any number of mangoes. Hence, total number of ways = 30 + 4 – 1C4 – 1 = 33C3 =

33.32.31 = 5456 1.2.3

QUICK TIPS Number of ways of dividing n idential things into r groups, if blank groups are allowed is n + r – 1

Cr – 1

Permutations and Combinations  4.13 43. The number of ways of choosing 10 balls from infinite white, red, blue and green balls is (A) 70 (B) 84 (C) 286 (D) 86

3

Solution: (C)

C1 + 3C2 + 3C3 = 23 – 1 = 7

⎧⎪ Coefficient of x ⎫⎪ Required ways = ⎨ ⎬ 2 4 ⎩⎪in (1 + x + x + ...) ⎭⎪ 10

Number of ways in which at least one question can be selected out of 4 are 4

C1 + 4C2 + 4C3 + 4C4 = 24 – 1 = 15

4



⎛ 1 ⎞ = Coefficient of x in ⎜ ⎝ 1 − x ⎟⎠



= Coefficient of x10 in (1 – x)–4



= Coefficient of x10 in 5.4 2 4.5.6 3 ⎛ ⎞ ⎜⎝1 + 4 x + 2 ! x + 3! x + …⎟⎠

10





= Coefficient of x10 in (1 + 4C1x + 5C2x2 + 6C3x3 + 7C4x4 + 8C5x5 + 9C6x6 + … 13C10x10) \ Required number of ways = 13C10 =

13.12.11 = 286 3.2.1

44. 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 (B) 15 (C)  17 (D)  None of these Solution: (B) Suppose the two players did not play at all so that the remaining (n – 2) players played n – 2C2 matches. Since these two players played 3 matches each, hence the total number of matches is n–1

C2 + 3 + 3 = 84  (given)

or

( n − 2)( n − 3) = 78 or n2 – 5n + 6 = 156 1.2 n2 – 5n – 150 = 0 or (n – 15) (n + 10) = 0

\

n = 15 (n ≠ –10)

or

45. An examination paper, which is divided into two groups consisting of 3 and 4 questions respectively carries the note: It is not necessary to answer all the questions. One question atleast should be answered from each group. The number of ways can an examinee select the questions is (A) 22 (B) 105 (C) 3P3 × 4P4 (D) 3C3 × 4C4

Solution: (B) Number of ways in which at least one question can be selected out of 3 are

\ Total number of ways = 7 × 15 = 105 46. Given 5 different green dyes, 4 different blue dyes and 3 different red dyes, the number of combinations of dyes that can be chosen by taking atleast one green and one blue dye is (A) 248 (B) 120 (C) 3720 (D) 465 Solution: (C) Number of ways of selecting at least one green dye out of 5 different green dyes 5

C1 + 5C2 + 5C3 + 5C4 + 5C5 = 25 – 1

Also, at least one blue dye can be selected out of 4 blue dyes in 4

C1 + 4C2 + 4C3 + 4C4 = 24 – 1

Again, 3 different red dyes can be selected in 3

C0 + 3C1 + 3C2 + 3C3 = 23 ways

\ Required ways = (25 – 1) (24 – 1) (23) = 3720 47. 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 (B) 144 (C)  153 (D)  None of these Solution: (B) The number of st. lines = 18C2 – (5C2 – 1) = 144

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 1 n ! ⎜1 − + − + … + ( −1) n ⎟ = n ! ∑ ( −1) r ⎝ 1! 2 ! 3! n !⎠ r ! r =0

4.14  Chapter 4 For example, if 4 letters are taken out of 4 different ­envelopes, then the number of ways in which they can be reinserted in the envelopes so that no letter goes in to its original envelope 1 1 1⎞ ⎛ = 4 ! ⎜1 − 1 + − + ⎟ ⎝ 2 ! 3! 4 !⎠  1 1 1⎞ ⎛ = 24 ⎜1 − 1 + − + ⎟ = 9 ⎝ 2 6 24 ⎠

Some Useful Results for Geometrical Problems 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. Number of Rectangles and Squares: (a) Number of rectangles of any size in a square of size n × n is n

size is

n

∑ r 3 and number of squares of any

r =1

∑ r 2. r =1

(b) Number of rectangles of any size in a rectangle of np size n × p (n < p) is (n + 1) (p + 1) and number 4 n of squares of any size is ∑ ( n + 1 − r ) ( p + 1 − r ) . r =1

6. If m parallel lines in a plane are intersected by a family of other n parallel lines. Then, total number of parallelograms so formed is mC × nC 2

i.e.,

2

mn ( m − 1) ( n − 1) 4 

SOLVED EXAMPLES 48. The number of diagonals in a polygon of n sides is n ( n − 3) n ( n − 1) (B)  2 2 ( n − 1) ( n − 2) (C)  (D)  None of these 2 (A) 

Solution: (A) The number of diagonals + number of sides = number of selections of two vertices from n vertices \ the number of diagonals. = nC2 – n =

n ( n − 1) n 2 − n − 2n n ( n − 3) −n = = . 2 2 2

49. If each of 10 points on a straight line be joined to each of 10 points on a parallel line then the total number of triangles that can be formed with the given points as vertices, is (A) 860 (B) 900 (C)  920 (D)  None of these Solution: (B) A triangle is formed for each selection of 2 points from one line and 1 point from the other line. \ The number of triangles

= 10C2 × 10C1 + 10C1 × 10C2



=

10 × 9 10 × 9 × 10 + 10 × = 900 2 2

EXPONENT OF PRIME p IN n! Exponent of a prime p in n! is denoted by Ep (n!) and is given by, ⎛ n⎞ ⎛ n⎞ ⎛ n ⎞ ⎛ n ⎞ Ep (n!) = ⎜ ⎟ + ⎜ 2 ⎟ + ⎜ 3 ⎟ + .... + ⎜ k ⎟ , ⎝ p⎠ ⎝ p ⎠ ⎝ p ⎠ ⎝p ⎠ where pk < n < pk + 1 ⎛ n⎞ and  ⎜⎝ p ⎟⎠ denotes the greatest integer less than or n equal to . p For example, exponent of 3 in (100)! is ⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ E3 (100!) = ⎜ + + + ⎝ 3 ⎟⎠ ⎜⎝ 32 ⎟⎠ ⎜⎝ 33 ⎟⎠ ⎜⎝ 34 ⎟⎠  = 33 + 11 + 3 + 1 = 48

Permutations and Combinations  4.15

NUMBER OF DIVISORS a

a

a

a

Let N = p11 . p 22 . p33 ... p k k . where p1, p2, p3, … pk are different primes and a1, a2, a3, …, ak are natural numbers then: 1. The total number of divisors of N including 1 and N is = (a1 + 1) (a2 + 1) (a3 + 1) … (ak + 1). 2. The total number of divisors of N excluding 1 and N is = (a1 + 1) (a2 + 1) (a3 + 1) … (ak + 1) – 2. 3. The total number of divisors of N excluding 1 or N is = (a1 + 1) (a2 + 1) (a3 + 1) … (ak + 1) – 1. 4. The susm of these divisors is = ( p10 + p11 + p12 + ... + p1a ) a2

( p20 + p12 + p22 + ... + p2 ) a

( p k0 + p1k + p k2 + ... + p k k )

5. The number of ways in which N can be resolved as a product of two factors is ⎧1 ⎪⎪ 2 (α1 + 1)(α 2 + 1)… (α k + 1), If N is not a perfect square ⎨ ⎪ 1 [(α + 1)(α + 1)… (α + 1) + 1], If N is a perfect square 2 k ⎪⎩ 2 1 6. The number of ways in which a composite number N can be resolved into two factors which are relatively prime (or co-prime) to each other is equal to 2n–1, where n is the number of different factors in N.

4.16  Chapter 4

NCERT EXEMPLARS 1. If n C12 = n C8 , then n is equal to (A) 20

(B) 12

(C) 6

(D) 30

2. The number of possible outcomes when a coin is tossed 6 times is (A) 36 (B) 64 (C) 12 (D) 32 3. The number of different four-digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is (A) 120 (B) 96 (C) 24 (D) 100 4. The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is (A) 432 (B) 108 (C) 36 (D) 18 5. The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is (A) 60 (B) 120 (C) 7200 (D) 720 6. If a five-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions, then the total number of ways this can be done is (A) 216 (B) 600 (C) 240 (D) 3125

NCERT EXEMPLARS

7. Everybody in a room shakes hands with everybody else. If the total number of hand shakes is 66, then the total number of persons in the room is (A) 11 (B) 12 (C) 13 (D) 14 8. The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is (A) 105 (B) 15 (C) 175 (D) 185

10. The number of ways in which a team of eleven players can be selected from 22 player always including 2 of them and excluding 4 of them is 16 (A) 16C11 (B)  C5 16 20 (C) C9 (D)  C9 11. The number of 5-digit telephone numbers having atleast one of their digits repeated is (A) 90000 (B) 10000 (C) 30240 (D) 69760 12. The number of ways in which we can choose a committee from four men and six women, so that the committee includes atleast two men and exactly twice as many women as men is (A) 94 (B) 126 (C)  128 (D)  None of these 13. The total number of 9-digit numbers which have all different digits is (A) 10! (B) 9! (C)  9 × 9! (D)  10 × 10! 14. The number of words which can be formed out of the letters of the word ‘ARTICLE’, so that vowels occupy the even place is (A) 1440 (B) 144 4 C 4 × 3 C3 (C)  7! (D)  15. Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking atleast one green and one blue dye is (A) 3600 (B) 3720 (C) 3800 (D) 3600

9. 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 (B) 18 (C) 12 (D) 9

ANSWER K EYS   1. (A) 11. (D)

2.  (B) 12.  (A)

3. (C) 4.  (B) 13. (C) 14.  (B)

5. (C) 15. (B)

6.  (A)

7. (B)

8.  (D)

9. (B)

10. (C)

Permutations and Combinations  4.17

HINTS AND EXPLANATIONS n

⇒ nCn−12 = n C8

∵ nC = n C  r n− r  

⇒ n − 12 = 8

C12 = n C8 ,

7. Let the total number of person in the room is n. We know that, two person form 1 hand shaken. ∴ Required number of hand shakes n ( n − 1) n! = 2!( n − 2 )! 2

⇒ n = 12 + 8 = 20 aα

= nC2 =

2. Number of outcomes when tossing a coin 1 times = 2 (head or tail) ∴ Total possible outcomes when a coin tossed 6 times = 26

According to the question,

∵ 2n for n time tossed coin   

= 64

3 Given, digits 2, 3, 4 and 7, we have to form four-digit numbers using these digits. ∴ Required number of ways = 4 P4 = 4! = 4 × 3 × 2! = 24 4. If we fixed 3 at units place. Total possible number is 3!, i.e., 6. Sum of the digits in unit place of all these numbers = 3! × 3 Similarly, if we fixed 4, 5 and 6 at units place, in each case total possible numbers are 3!. Required sum of unit digits of all such numbers = (3 + 4 + 5 + 6) × 3! = 18 × 3! = 18 × 6 =108 5. Given that, number of vowels = 4 and total number of consonants = 5 Total number of words formed by 2 vowels and 3 consonants

3

2

= 66

⇒ n ( n − 1) = 132 ⇒ n2 − n − 132 = 0 ⇒ ( n − 12) ( n + 11) = 0 ⇒ n = 12, − 11 [inadmissible] ∴ n = 12 8. Total number of triangles formed from 12 points taking 3 at time = 12C3 But out of 12 points 7 are collinear. So, these 7 points constitute a straight line mean no triangle is formed by joining these 7 points. ∴

Required

number

of

triangles

= 12C3 −7 C3

= 220 − 35 = 185a 9. To form parallelogram we required a pair of line from a set of 4 lines another pair of line from another set of 3 lines. = 4C2 × 3C2

= 6 × 3 = 18

0, 1, 2, 4, 5 = 4 × 4 × 3 × 2 × 1 = 96 4

2

∴ Required number of parallelograms

4! 5! × 2! 2! 3! 2! 4 × 3! 5 × 4 × 3 × 2! 4 × 5 × 4 × 3 = = × 4 2! 2! 3! × 2!

= 4 C2 ×5 C3 =

= 5 × 4 × 3 = 60 Choose what order they appear in 5!, i.e., 120. So, total number of words = 60 × 120 = 7200 6. We know that, a number is divisible by 3, when sum of digits in the number must be divisible by 3. So, if we consider the digits 0, 1, 2, 4, 5, then (0 + 1 + 2 + 4 + 5) = 12 We see that, sum is divisible by 3. Therefore, five-digit numbers using the digit 4

n ( n − 1)

1

and if we consider the digit 1, 2, 3, 4, 5, then (1 + 2 + 4 + 5 = 15) This sum is also divisible by 3. So, five-digit number can be formed using the digit 1, 2, 3, 4, 5 in 5! ways. Total number of ways = 96 + 5! = 96 + 120 = 216

10. Total number of players = 22 We have to select a team of 11 players. Selection of 11 players when 2 of them ia always included and 4 are never included. Total number of player = 22 – 2 – 4 = 16 ∴ Required number of selections = 16C9 11. If all the digits repeated, then number of 5 digit telephone numbers can be formed in 105 ways and if no digit repeated, then 5-digit telephone numbers can be formed in 10P5 ways. ∴ Required number of ways 105 −10 P5 = 100000 −

10! 5!

= 100000 – 10 × 9 × 8 × 7 × 6 = 100000 – 30240 = 69760 12. ∴ Number of men = 4 and number of women = 6 It is given committee includes two men and exactly twice as many women as men Thus, possible selection is given in following table

HINTS AND EXPLANATIONS

1. Given that,

4.18  Chapter 4 Men

Women

2

4

3

6

Required number = 4 C2 × 6C4 + 4C3 × 6C6

of

committee

Since, it is given that vowels occupy even place, therefore the arrangement of vowel, consonant can be understand with the help of following diagram.

formed

HINTS AND EXPLANATIONS

= 6 × 15 + 4 × 1 = 94 13. (C) We have to form 9-digit numbers with the digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 cannot be placed at the first place from left. So, first place from left can be filled in 9 way. Since, repetition is not allowed, so remaining 8 places can be filled in 9! ways. ∴ Required number of ways = 9 × 9! 14. (B) Total number of letters in the word article is 7, out of which A, E, I are vowels and R, T, C, L are consonants.

1 2 3 4 5 6 7 Now, vowels can be placed at 2, 4 and 6th position. Therefore, number of arrangement = 3P3 = 3! = 6 ways And consonants can be placed at 1, 3, 5 and 7th position. ∴ Total number of words = 6 × 24 = 144 15. (B) Possible number of choosing green dyes = 25 Possible number of choosing blue dyes = 24 Possible number of choosing red dyes = 23 If at least one blue and one green dyes are selected.

(

)(

)

Then, total number of selection = 25 − 1 24 − 1 × 23 = 3720

Permutations and Combinations  4.19

PRACTICE EXERCISES Single Option Correct Type

6! 3! × × 4! 4! × 2! 2! × 1!

2. If m = number of distinct rational numbers

∈

(0, 1) such that p, q ∈ {1, 2, 3, 4, 5} and n = number of mappings from {1, 2, 3} onto {1, 2}, then m – n is (A) 1 (B) –1 (C)  0 (D)  None of these

9. If a represents the number of permutations of (x + 2) things taken together, b represents the number of permutations of 11 things taken together out of x things, and c represents the number of permutations of (x – 11) things taken together so that a = 182bc, then x = (A) 15 (B) 12 (C) 10 (D) 18 10. 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? (A) 16 (B) 36 (C) 60 (D) 180

3. The letters of the word RANDOM are written in all possible orders and these words are written out as in a dictionary then the rank of the word RANDOM is (A) 614 (B) 615 (C) 613 (D) 616

11. For a game in which two partners play against two other partners, six persons are available. If every possible pair must play with every other possible pair, then the total number of games played is (A) 90 (B) 45 (C) 30 (D) 60

4. 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 (A) 40320 (B) 2520 (C)  20160 (D)  None of these

12. A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is (A) 216 (B) 600 (C) 240 (D) 3125

5. The number of permutations of letters a, b, c, d, e, f, g so that neither the pattern beg nor cad appears is (A) 

6! × 3! ( 2!)

2

= 1080

(C)  4806

7! (B)  2 ! 3! 3! (D)  None of these

6. The number of ways of selecting 10 balls from the unlimited number of red, green, white and yellow balls, if selection must include 2 red and 3 yellow balls, is (A) 36 (B) 56 (C)  112 (D)  None of these 7. Let A = {1, 2, 3, 4} and B = {1, 2}. Then, the number of onto functions from A to B is: (A) 8 (B) 14 (C)  12 (D)  None of these 8. Given five line segments of lengths 2, 3, 4, 5, 6 units. Then the number of triangles that can be formed by joining these lines is (A) 5C3 – 3 (B)  5C3 – 1 5 5 (C)  C3 (D)  C3 – 2

13. 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 (B) 64 (C)  128 (D)  None of these 14. 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 (B) 84844 (C)  93324 (D)  None of these 15. If the number of ways in which n different things can be distributed among n persons so that at least one person does not get any thing is 232. Then n is equal to (A) 3 (B) 4 (C)  5 (D)  None of these 16.

m

Cr + 1 + =

n

∑ k Cr =

k =m

n +1 (A) nCr + 1 (B)  Cr + 1 n (C)  Cr (D)  None of these

17. Two straight lines intersect at a point O. Points A1, A2, …, An are taken on one line and points B1, B2, …,

PRACTICE EXERCISES

1. Let y be an element of the set A = {1, 2, 3, 5, 6, 10, 15, 30} and x1, x2, x3 be integers such that x1x2x3 = y, then the number of positive integral solutions of x1x2x3 = y is (A) 64 (B) 27 (C)  81 (D)  None of these

4.20  Chapter 4 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) (B)  n (n – 1)2 2 (C) n (n – 1) (D)  n2 (n – 1)2 18. If the letters of the word MOTHER are written in all possible orders and these words are written out as in a dictionary, then the rank of the word MOTHER is (A) 240 (B) 261 (C) 308 (D) 309

PRACTICE EXERCISES

19. The number of divisors a number 38808 can have, excluding 1 and the number itself is (A) 70 (B) 72 (C)  71 (D)  None of these

for 6 of the animals, then the number of ways of caging the animals is (A) 304800 (B) 504800 (C)  604800 (D)  None of these 28. If n is even and n C0 < nC1 < nC2 < … < nCr > nCr+1 > … > nCn then r = n n −1 (A)  (B)  2 2 n−2 n+2 (C)  (C)  2 2

20. The number of positive integral solutions of 15 < x1 + x2 + x3 ≤ 20, is equal to (A) 785 (B) 685 (C)  1150 (D)  None of these

29. 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 (A) 230 (B) 210 (C)  340 (D)  None of these

21. The number of different 7 digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is 7 (A) 7P225 (B)  C2 25

30. The number of ordered pairs (m, n), m, n ∈ {1, 2,…, 50} such that 6n + 9m is a multiple of 5 is (A) 6250 (B) 1250 (C)  1875 (D)  None of these

(C) 7C252

22. The tensdigit of 1! + 2! + 3! + … + 49! is (A) 1 (B) 2 (C) 3 (D) 4

31. A set contains (2n + 1) elements. The number of subsets of the set which contain at most n elements is (A) 2n (B)  2n+1 2n–1 (C) 2 (D)  22n

23. Let S be the set of all functions from the set A to the set A. If n (A) = k then n (S ) is (A) k! (B)  kk (C) 2k – 1 (d) 2k

32. 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

24. There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum umber of triangles with vertices at these points is

(A) 

( n −1) n ( n − 1) ( n − 2) (B)  2 2

(A) 3pC3 (B)  p2 (p – 1) (C) p2 (4p – 1) (D)  p2 (4p – 3)

(C) 

n( n +1) ( n + 1) ( n + 2) (D)  2 2

25. The number of ways in which thirty five apples can be distributed among 3 boys so that each can have any number of apples, is (A) 1332 (B) 666 (C)  333 (D)  None of these

33. 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 (B) 7 (C) 8 (D) 9

(D)  None of these

26. The number of non-negative solutions of x1 + x2 + x3 +, …, + xn ≤ n (where n is positive integer) is (A) 2nCn – 1 (B)  2n–1Cn – 1 2n–1 (C) 2n+1Cn – 1 (D)  Cn–1 – 1 27. Eleven animals of a circus have to be placed in eleven cages one in each cage. If 4 of the cages are too small

34. If all permutations of the letters of the word AGAIN are arranged as in dictionary, the forty ninth word is (A) NAAGI (B)  NAGAI (C)  NAAIG (D) NAIAG 35. The number of ways of choosing n objects out of (3n + 1) objects of which n are identical and (2n + 1) are distinct, is

Permutations and Combinations  4.21

36. In a group of boys, two boys are brothers and in this group 6 more boys are there. In how many ways they can sit if the brothers are not to sit along with each other (A) 4820 (B) 1410 (C)  2830 (D)  None of these 37. If 20% of three subsets (i.e., subsets containing exactly three elements) of the set A = {a1, a2,…, an} contain a1, then the value of n is (A) 15 (B) 16 (C) 17 (C) 18 38. The number of two digit numbers which are of the form xy with y < x are given by (A) 45 (B) 55 (C)  17 (D)  None of these 39. A crocodile is known to have not more than 68 teeth. The total number of crocodiles with different set of teeth is (A) 68 (B) 68! (C) 1617 (D) 6868 40. For x ∈ R, let [x] denotes the greatest integer ≤ x, then the value of 1   1 2   1  1 −  3  +  − 3 − 100  +  − 3 − 100         1 99  +, ..., +  − − is  3 100  (A) –100

(B) – 123

(C) –135

(D) –153

41. The total number of ways in which a beggar can be given at least one rupee from four 25 p. coins, three 50 p. coins and 2 one rupee coins is (A) 54 (B) 53 (C) 51 (D) 48 42. A student is allowed to select atmost n books from a collection of (2n + 1) books. If the total number of ways in which he can select books is 63, then n = (A) 4 (B) 3 (C) 7 (D) 8 43. In a certain test there are n questions. In this test 2k ­students gave wrong answers to at least (n – k) questions, where k = 0, 1, 2,…, n. If the total number of wrong answers is 4095, then value of n is (A) 11 (B) 12 (C) 13 (D) 15 44. The number of permutations of the letters a, b, c, d such that b does not follow a, c does not follow b, and d does not follow c, is (A) 12 (B) 14 (C) 13 (D) 11

45. If S =

m

∑ n + r C k , then

r=0

(A) S + nCk+1 = n+mCk+1 (B) S + nCk+1 = n+m+1Ck+1 (C) S + nCk = n+mCk (D)  None of these 46. The number of ways of dividing 15 men and 15 women into 15 couples, each consisting of a man and a woman, is (A) 1240 (B) 1840 (C) 1820 (D) 2005 47. Suman writes letters to his five friends. The number of ways can be letters be placed in the envelopes so that atleast two of them are in the wrong envelopes are (A) 119 (B) 120 (C)  125 (D)  None of these 48. Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is (A) 880 (B) 629 (C) 630 (D) 879 49. The number of 4-digit numbers with distinct digits is (A) 504 (B) 4536 (C) 4634 (D) 5040 50. In a shop there are five types of ice-creams available. A child buys six ice-creams. Statement 1: The number of different ways the child can buy the six ice-creams is 10C5. Statement 2: The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row. (A) Statement 1 is false, Statement 2 is true (B) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 (C) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (D)  Statement 1 is true, Statement 2 is false 51. In an examination a candidate has to pass in each of the papers to be successful. If the total number of ways to fail is 63, how many papers are there in the examination? (A) 6 (B) 8 (C)  14 (D)  None of these 52.

∑ ∑

0 ≤ i ≤ j ≤ 10

(A) 310

10

C j j C i is equal to (B) 310 – 1 (C)  210

(D) 210 – 1

PRACTICE EXERCISES

(A) 22n (B)  22n+1 2n (C) 2 – 1 (D)  None of these

4.22  Chapter 4 53. The number of ways of choosing n objects out of (3n + 1) objects of which n are identical and (2n + 1) are distinct, is (A) 22n (B)  22n+1 (C) 22n – 1 (D)  None of these 54.

2n

Cr (0 ≤ r ≤ 2n) is greatest when r is equal to

n n +1 (A)  (B)  2 2 (C) r = n (D)  None of these

PRACTICE EXERCISES

55. The number of even numbers greater than 100 that can be formed by the digits 0, 1, 2, 3 (no digit being repeated) is (A) 20 (B) 30 (C)  40 (D)  None of these

(A) 38664 (C)  58664

(B) 48664 (D)  None of these

63. The number of four digit numbers that can be formed from the digits 0, 1, 2, 3, 4, 5 with at least one digit repeated is (A) 420 (B) 560 (C)  780 (D)  None of these 64. The number of odd 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 (A) 864 (B) 932 (C)  766 (D)  None of these

56. The number of positive numbers less than 1000 and divisible by 5 (no digit being repeated) is (A) 150 (B) 154 (C)  166 (D)  None of these

65. A table has provision for 7 seats, 4 being on one side facing the window and 3 being on the opposite side. The number of ways in which 7 people can be seated at the table if 2 people, X and Y, must sit on the same side, is (A) 3260 (B) 2160 (C)  3350 (D)  None of these

57. 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 (B) 7200 (C)  9200 (D)  None of these

66. There are four oranges, five apples and six mangoes in a fruit basket. The number of ways in which a person can make a selection of fruits among the fruits in the basket, is (A) 210 (B) 330 (C)  209 (D)  None of these

58. The total number of ways of selecting five letters from the letters of the word INDEPENDENT, is (A) 4200 (B) 3320 (C)  3840 (D)  None of these

67. The number of zeros at the end of 100! is (A) 36 (B) 18 (C)  24 (D)  None of these

59. The sum of five digit numbers which can be formed with the digits 3, 4, 5, 6, 7 using each digit only once in each arrangement, is (A) 5666600 (B) 6666600 (C)  7666600 (D)  None of these 60. The sum of all the numbers that can be formed by ­writing all the digits 3, 2, 3, 4 only once is (A) 39996 (B) 49996 (C)  57776 (D)  None of these 61. The sum of all numbers greater than 10000 formed by using the digits 1, 3, 5, 7, 9, no digit being repeated in any number, is (A) 4666600 (B) 5666600 (C)  6666600 (D)  None of these 62. The sum of all numbers greater than 1000 formed by using the digits 0, 1, 2, 3, no digit being repeated in any number, is

68. The largest integer n such that 33! is divisible by 2n is (A) 30 (B) 31 (C)  32 (D)  None of these 69. The number of non-negative integral solutions of x1 + x2 + x3 + 4x4 = 20 is (A) 436 (B) 536 (C)  602 (D)  None of these 70. The product of r consecutive positive integers is divisible by (A)  r! (B)  (r – 1)! (C) (r + 1)! (D)  None of these 71. The number of ordered triplets of positive integers which are solutions of the equation x + y + z = 100 is (A) 5081 (B) 6005 (C)  4851 (D)  None of these 72. The number of words that can be formed, with the letters of the work ‘Pataliputra’ without changing the relative order of the vowels and consonants, is

Permutations and Combinations  4.23 (B) 4200 (D)  None of these

73. On a new year day every student of a class sends a card to every other student. The postman delivers 600 cards. The number of students in the class are (A) 42 (B) 34 (C)  25 (D)  None of these 74. For any positive integers m, n (with n ≥ m), let

( ) = C , then ( ) + ( ) + ( ) + ... + ( ) = (A) ( ) (B)  ( ) (C) ( ) (D)  None of these and

n

n m

n+1 m

m

n m

n −1 m

n− 2 m

m m

n +1 m +1

n m +1

75. The number of 7 digit numbers the sum of whose ­digits is even, is (A) 35 × 105 (B) 45 × 105 5 (C) 50 × 10 (D)  None of these 76. The number of ways of choosing m coupons out of an unlimited number of coupons bearing the letters A, B and C so that they cannot be used to spell the word BAC, is (A)  3 (2m – 1) (B)  3 (2m – 1 – 1) m (C)  3 (2 + 1) (D)  None of these 77. Six X ’s have to be placed in squares of the figure given below, such that each row contains at least one X. The number of different ways in which this can be done is (A) 26 (B) 28 (C)  18 (D)  None of these 78. The number of ways in which 16 identical things can be distributed among 4 persons if each person gets at least 3 things, is (A) 33 (B) 35 (C)  38 (D)  None of these 79. 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 (B) 117280 (C)  116280 (D)  None of these 80. In an examination the maximum marks for each of the three papers are 50 each. Maximum marks for the fourth paper are 100. The number of ways in which the candidate can score 60% marks in aggregate is (A) 110256 (B) 110456 (C)  110556 (D)  None of these

81. The number of integers between 1 and 1000000 that have the sum of the digits 18, is (A) 25927 (B) 25827 (C)  24927 (D)  None of these 82. The number of non-negative integral solutions to the system of equations x + y + z + u + t =20 and x + y + z = 5 is (A) 336 (B) 346 (C)  246 (D)  None of these 83. The number of positive integral solutions of the inequality 3x + y + z ≤ 30, is (A) 1115 (B) 1215 (C)  1315 (D)  None of these 84. In a city no person has identical set of teeth and there is no person without a tooth. Also, no person has more than 32 teeth. If we disregard the shape and size of tooth and consider only the positioning of the teeth, then the maximum population of the city is (A) 232 (B)  232 – 1 32 (C) 2 + 1 (D)  None of these 85. Eleven scientists are working on a secret project. They wish to lock up the documents in a cabinet such that cabinet can be opened if six or more scientists are present. Then, the smallest number of locks needed is (A) 460 (B) 461 (C)  462 (D)  None of these 86. The number of numbers greater than 106 that can be formed using the digits of the number 2334203, if all the digits of the given number must be used, is (A) 360 (B) 420 (C) 260 (D)  None of these 87. If ‘n’is an integer between 0 and 21, then the minimum value of n! (21 – n)! is (A)  9! 2! (B)  10! 11! (C) 20! (D) 21! 88. In how many ways can 20 oranges be given to four children if each child should get at least one orange? (A) 869 (B) 969 (C)  973 (D)  None of these 89. The total number of 5-digit numbers of different digits in which the digit in the middle is the largest is

PRACTICE EXERCISES

(A) 3600 (C)  3680

4.24  Chapter 4 9

(A) ∑

n=4 9

(C) ∑

n=4

(

n

P4 −

n −1

n −1

)

9

∑

P3 (B) n P4

P3

n=4

(D)  None of these

90. A train is going from Delhi to Indore, stops at nine intermediate stations. Six persons enter the train during the journey with six different tickets. The number of different sets of tickets possessed by them is 50 54 (A)  C6 (B)  C6 45 (C)  C6 (D)  None of these

91. 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) 5 (B) 6 (C) 7 (D) 8 92. If a, b, c are three natural numbers in A.P. such that a + b + c = 21, then the possible number of values of a, b, c is (A) 13 (B) 14 (C) 15 (D) 16 93. The number of ways in which a mixed doubles game can be arranged from amongst n couples such that no husband and wife play in the same game, is 1n 1 n n (A)  P4 (B)  C4 (C)  P4 (D)  nC4 2 2

Previous Year’s Questions 94. 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 [2003] (A) 140 (B) 196 (C) 280 (D) 346 95. If nCr denotes the number of combinations of n things taken r at a time, then the expression nCr + 1 + nCr-1+ 2 × nCr equals [2003]

PRACTICE EXERCISES

n+2 (A)  Cr n+1 (C)  Cr

(B)  n + 2Cr + 1 (D)  n + 1Cr + 1

96. How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order? [2004] (A) 120 (B) 480 (C) 360 (D) 240 97. The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is [2004] 8 (A)  5 (B)  C3 (C) 38 (D) 21 98. If the letters of word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number [2005] (A) 601 (B) 600 (C) 603 (D) 602

99. At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is [2006] (A) 5040 (B) 6210 (C) 385 (D) 1110 100. How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent? [2008] (A) 8 ⋅ 6C4 ⋅ 7C4

(B) 6 ⋅ 7 ⋅ 8C4

(C) 6 ⋅ 8 ⋅ 7C4 (D) 7 ⋅ 6C4 ⋅ 8C4 101. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then the number of such arrangements is [2009] (A)  less than 500 (B)  at least 500 but less than 750 (C)  at least 750 but less than 1000 (D)  at least 1000 102. There are two urns. Urn I has 3 distinct red balls and Urn II has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is [2010] (A) 36 (B) 66 (C) 108 (D) 3

Permutations and Combinations  4.25

Statement-2: The number of ways of choosing any 3 places from 9 different places is 9C3. (A)  Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 (B)  Statement-1 is true, Statement-2 is false. (C)  Statement-1 is false, Statement-2 is true. (D)  Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 104. Assuming the balls to be identical except for difference in colors, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is [2012] (A) 880 (B) 629 (C) 630 (D) 879 105. Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1 − Tn = 10, then the value of n is [2013] (A) 5 (B) 10 (C) 8 (D) 7 106. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is: [2015]

(A) 192 (C) 72

(B) 120 (D) 216

107. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is: [2016] th th (A) 58 (B)  46 (C) 59th (D)  52nd 108. A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is [2017] (A) 468 (B) 469 (C) 484 (D) 485 109. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is [2018] (A)  at least 1000 (B)  less than 500 (C)  at least 500 but less than 750 (D)  at least 750 but less than 1000

ANSWER K EYS Single Option Correct Type   1. (A) 2.  (D) 11. (B) 12.  (A) 21.  (B) 22. (A) 31.  (D) 32. (B) 41. (A) 42.  (B) 51.  (A) 52. (A) 61.  (C) 62. (A) 71. (C) 72. (A) 81. (A) 82. (A) 91.  (C) 92. (A)

3. (A) 4.  (C) 13. (B) 14.  (C) 23.  (B) 24. (D) 33. (B) 34. (A) 43.  (B) 44. (D) 53.  (A) 54.  (C) 63. (C) 64. (A) 73. (C) 74. (B) 83.  (B) 84. (B) 93.  (C)

5. (C) 15. (B) 25.  (B) 35.  (A) 45.  (B) 55. (A) 65. (B) 75. (B) 85.  (C)

6.  (B) 7. (B) 16. (B) 17. (C) 26. (A) 27.  (C) 36. (D) 37.  (A) 46. (A) 47.  (B) 56.  (B) 57. (A) 66. (C) 67.  (C) 76.  (A) 77. (A) 86. (A) 87.  (B)

8.  (A) 18. (D) 28. (A) 38. (A) 48. (D) 58.  (B) 68. (B) 78. (B) 88. (B)

9. (B) 19.  (A) 29.  (B) 39.  (C) 49.  (B) 59. (B) 69.  (B) 79. (C) 89.  (A)

10. (C) 20. (B) 30. (B) 40. (C) 50. (A) 60. (A) 70. (A) 80. (C) 90. (C)

Previous Years’ Questions   94. (B) 95. (B) 96. (C) 97. (D) 98. (A) 99. (C) 100. (D) 101. (D) 102. (C) 103. (D) 104. (D) 105. (A) 106. (A) 107. (A) 108. (D) 109. (A)

PRACTICE EXERCISES

103. Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3 [2011]

4.26  Chapter 4

HINTS AND EXPLANATIONS Single Option Correct Type 1. Number of solutions of the given equation is the same as the number of solutions of the equation x1x2x3x4 = 30 = 2 × 3 × 5 Here, x4 is there because if x1x2x3 = 15, then x4 = 2 and if x1x2x3 = 5, then x4 = 6 etc. x4 is in fact a dummy variable. Each of 2, 3 and 5 will be a factor of exactly one of x1, x2, x3, x4 in 4 ways. ∴ Required number = 43 = 64 The correct option is (A)

HINTS AND EXPLANATIONS

2. Now, n = 23 – 2 = 6 ⎛ 2 1⎞ Also, m = 4 + 3 + 2 + 1 – 1 = 9 ⎜ as = ⎟ ⎝ 4 2⎠ ∴ m – n = 3 The correct option is (D) 3. A D M N O R (in order) Number of words beginning with A _ _ _ _ _ = 5! D _ _ _ _ _ = 5! M _ _ _ _ _ = 5! N _ _ _ _ _ = 5! O _ _ _ _ _ = 5! R A D _ _ _ = 3! R A M _ _ _ = 3! R A N D M O = 1 R A N D O M = 1 ∴ Rank of word RANDOM = 614 The correct option is (A) 4. Let A, B be the corresponding speakers. Without any restriction the eight persons can be arranged among themselves in 8! ways, but the number of ways in which A speaks before B and the number of ways in which B speaks before A make up 8!. Also number of ways in which A speaks before B is exactly same as the number of ways in which B speaks before A. 1 ∴ the required number of ways = (8!) = 20160. 2 The correct option is (C) 5. Total number of permutations = 7! Let A be the property that ‘beg’ occurs. B be the property that cad occurs. Number of permutations with A = 5! = that of with B and n (A ∩ B) = 3! ∴ n(A ∪ B) = 5! + 5! – 3! = 234 ∴ Required number = 7! – 234 = 4806

The correct option is (C) 6. Number of 10 balls selections = coefficient of x10 in (x2 + x3 + …) (1 + x + x2 +…) (1 + x + x2 +…) (x3 + x4 +…) = coefficient of x5 in (1 + x + x2 +…)4 = coefficient of x5 in (1 – x)–4 (4 + 5 – 1) = C5 = 8C5 = 56 ways. The correct option is (B) 7. Since each element of A can be associated with elements of 8 in two ways, therefore the total number of functions from A to B is 2 × 2 × 2 × 2 = 16. Out of these functions, the functions which are not onto are f (x) = 1, ∀ x ∈ A and f (x) = 2 ∀ x ∈ A. Thus, the number of onto functions = 16 – 2 = 14. The correct option is (B) 8. We know that in any trinagle the sum of two sides is always greater than the third side. ∴ The triangle will not be formed if we select segments of lengths (2, 3, 5), (2, 3, 6) or (2, 4, 6). Hence number of triangles formed = 5C3 – 3. The correct option is (A) 9. a = x+2Px+2, b = xP11, C = x–11Px–11 x! ⇒ a = (x + 2)!, b = , c = (x – 11)! ( x − 11)! Now, a = 182 bc x! ⇒ (x + 2)! = 182 (x – 11)! ( x − 11)! ⇒ (x + 2) (x + 1)x! = 182x! ⇒ (x + 2) (x + 1) = 182 = 14 × 13 ⇒ x + 1 = 13 ∴ x = 12 The correct option is (B) 10. The four digits 3, 3, 5, 5, can be arranged at four even places 4! in = 6 ways and the remaining digits viz., 2, 2, 8, 8, 8 2! 2! 5! can be arranged at the five odd places in = 10 ways. 2! 3! Thus, the number of possible arrangements is (6) (10) = 60. The correct option is (C) 11. For one game four persons are required. This can be done in 6C4 = 15 ways. Once a set of four persons are selected, 4 C2 number of games possible will be = 3 games. 2 ∴ Total number of possible games = 3 × 15 = 45. The correct option is (B) 12. We know that a number is divisible by 3 if the sum of its digits is divisible by 3.

Permutations and Combinations  4.27

13. The number of ways of selecting 3 balls out of total 9 (2 white, 3 black, 4 red balls) is 9C3 9×8×7 i.e., = 84 6 The number of ways of selecting 3 balls out of non-black six balls is 6C3 6×8×4 i.e., = 20 3 × 2 ×1 Therefore, the number of ways of selecting 3 balls out of 9 balls so as to include atleast one black ball = 84 – 20 = 64. The correct option is (B) 14. 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 required total = 84 (1 + 10 + 102 + 103) = 84 (1111) = 93324. The correct option is (C) 15. Total number of ways = nn Number of ways so that each person gets at least one thing (here exactly one thing) = n! Given, nn – n! = 232 11 – 1! = 0, 22 – 2! = 2, 33 – 3! = 21, 44– 4! = 232 ∴ n = 4 The correct option is (B) 16. mCr + 1 + m

n

∑

k=m m

k

Cr

= Cr + 1 + Cr + =

m+1

=

m+2



Cr + 1 +

2n ( 2n − 1) ( 2n − 2) 2n ( n − 1) ( n − 2) − 6 6 1 = n (n – 1) (3n) = n2 (n – 1) 3 The correct option is (C) 18. E H M O R T (alphabetical order) Number of words beginning with E _ _ _ _ _ = 5! H _ _ _ _ _ = 5! M E _ _ _ _ = 4! M H _ _ _ _ = 4! M O E _ _ _ = 3! M O H _ _ _ = 3! M O R _ _ _ = 3! M O T E _ _ = 2! M O T H E R = 1! ∴ Rank of word MOTHER = 309 The correct option is (D) 19. Factorizing the given number, we have 38808 = 23 ⋅ 32 ⋅ 72 ⋅ 11 Therefore the total number of divisors = (3 +1) (2 + 1) (1 + 1) – 1 = 71. But this includes the division by the number itself. Hence, the required number of divisors = 71 – 1 = 70 =

QUICK TIPS Let N = p1a ⋅ p2a ⋅ p3a … pka where p1, p2, p3 …pk are different primes and a1, a2, …, ak are natural numbers then total ­number of divisors of N excluding 1 and N is (a1 + 1) (a2 + 1) … (ak + 1) – 2 1

Cr + … +

Cr + … +

n–1

n–1

Cr + Cr

Cr + nCr

Cr + 1 + m + 2Cr + … + n – 1Cr + nCr    

= nCr + 1 + nCr = n + 1Cr + 1 The correct option is (B)

n

3

2

k

The correct option is (A) 20. We have, 15 < x1 + x2 + x3 ≤ 20 ⇒ x1 + x2 + x3 = 16 + r, r = 0, 1, 2, 3, 4. Now, number of positive integral solutions of x1 + x2 + x3 = 16 + r is 16 + r – 1C3 – 1 = 15 + r C2 Thus, total number of solutions 4

m+1

m+1

17. No. of triangles = 2nC3 – nC3 – nC3

= ∑ 15+ r C2 r=0

20 15 C3 – C3

15

C2 + 16C2 + 17C2 + 18C2 + 19C2

= 685

QUICK TIPS The total number of positive integral solutions of the equation x1 + x2 + … + xr = n is n – 1Cr – 1 The correct option is (B)

HINTS AND EXPLANATIONS

Now the sum of the digits 1, 2, 3, 4 and 5 is 15, which is divisible by 3. ∴ All the five digit numbers formed by the digits 1, 2, 3, 4, 5 are divisible by 3 and their number = 5! = 120. When we include 0, the four other digits whose sum is divisible by 3 are 1, 2, 4 and 5. ∴ The number of numbers in this case = 4 × 4! = 4 × 24 = 96. Hence the required number of numbers = 120 + 96 = 216 The correct option is (A)

4.28  Chapter 4 21. Other than 2, remaining five places are to be filled by by 1 and 3 ∴ 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. The correct option is (B) 22. We know that 1! + 2! + 3! + 4! = 33. Also, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880. Thus, tens digit of 1! + 2! + … + 9! is 1. Also, note that n! is divisible by 100 for all n ≥ 10. Therefore, the tens digit of 1! + 2! + … + 49! is 1. The correct option is (A) 23. Each element of the set A can be given the image in the set A in k ways. ∴ the required number of functions, i.e., n (S) = k × k × … (k times) = kk. The correct option is (B) 24. pC × pC × pC + pC × 2 pC × 3C 1

1

1

2

1

1

HINTS AND EXPLANATIONS

2

= p (4p – 3) The correct option is (D) 37! 25. Number of ways = 37C2 = = 666 ways 35! × 2! For the distribution equation. x1 + x2 + x3 + … + xr = n The number of ways in which n things can be distributed among r in such a way each can receive none, one or more or all of n items are n+r–1Cr–1 The correct option is (B) 26. In general, we know that For the distribution equation x1 + x2 + x3 + … + xn ≤ n Let required ways = W ⎧ No. of ways of ⎫ ⎧ No. of ways of ⎫ ⎪ ⎪ ⇒ W = ⎪⎨ distributing ⎪⎬ + ⎨ distributing ⎬ ⎪ ⎪ ⎪ ⎪ 2 items 1 item ⎭ ⎩ ⎭ ⎩ ⎧ No. of Ways of ⎫ ⎪ ⎪ + … + ⎨ distributing ⎬ ⎪ ⎪ n items ⎩ ⎭ = 1+n–1Cn–1 + 2+n–1Cn–1 + … + n+n–1Cn–1 =

Cn–1 + n+1Cn–1 + … + 2n–1Cn – 1

n

= (nCn–1

+ nCn) + n+1Cn–1 + … + 2n–1Cn–1 – nCn

+ n+1Cn–1) + … + 2n–1Cn–1} – nCn --------------------------------- ---------------------------------= {(n+1Cn

= (2n–1Cn + 2n–1Cn–1) – nCn =

2n

∴

Cn – nCn W = 2nCn – 1

The correct option is (A) 27. 6 large animals can be caged in 7 large cages in 7P6 = 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 The correct option is (C) 28. By the given condition, it is clear that nCr is the greatest among nC0, nC1, …, nCn. Since n is even, n ∴ nCr is the greatest for r = 2 n ∴ r = 2 The correct option is (A) 29. For each pair of stations, two different types of tickets are required, Now, the number of selections of 2 stations from 15 stations = 15C2. ∴ Required number of types of tickets 15! = 2 15C2 = 2 = 15 × 14 = 210 2!13! The correct option is (B) 30. All the numbers of the form 6n will end with 6 and 9m will end with 9, if m is odd and will end with 1, if m is even. Therefore, 6n + 9m will end with 5 if n is any number and m is odd. Thus, number of ordered pairs = 50 × 25 = 1250. The correct option is (B) 31. Number of subsets = 2n+1C0 + 2n+1C1 + … + 2n+1Cn = N (say) ∴ N = 22n+1 – N ⇒ 2N = 22n+1 ⇒ N = 22n The correct option is (D) 32. The number of triangles = number of selections of 2 lines from the (n – 1) lines which are cut by the last line ( n − 1)! ( n − 1) ( n − 2) = n – 1C2 = = 2!( n − 3)! 2 The correct option is (B) 33. Distinct n digit 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  or  n ≥ 7 The correct option is (B) So, the least value of n is 7. 34. Starting with the letter A and arranging the other four letters, there are 4! = 24 words. These are the first 24 words. Then, starting with G and arranging A, A, I and N in different ways,

Permutations and Combinations  4.29

35. If we choose k (0 ≤ k ≤ n) identical objects, then we must choose (n – k) distinct objects. This can be done in 2n+1Cn–k ways. Thus, the required number of ways n

=

∑

k=0 2n

2 n +1

2n+1

Cn − k =

Cn +

2n+1

Cn–1 + … +

C0

The correct option is (A) 36. 6 boys can sit in 6! ways × B1 × B2 × B3 × B4 × B5 × B6 × Now, two brothers can sit in any of the 7 cross marked (×) places Therefore, required number of ways = 6! × 7C2 × 2! = 30240 The correct option is (D) 37. The number of subsets of A containing exactly three elements is nC3 whereas the number of three subsets of A that contain a1 is n–1C2. We are given, 20 n C2 = ( C3) 100 ( n − 1) ( n −2) 1 n ( n − 1) ( n −2) ⇒ = 6 2 5 ⇒ n = 15 The correct option is (A) n–1

38. When zero is in the place of y then there are nine numbers and when 1 is in the place of y then there are 8 numbers and

2

⎛ 1

r ⎞

∑ ⎜⎝ − 3 − 100 ⎟⎠

The correct option is (C) 41. The following four cases are there: (A) At least one one rupee coin and any number of other coins = 2 × 4 × 5 = 40 ways. (B) At least two 50 p. coins and any number of 25 p. coins = 2 × 5 = 10 ways. (C) One 50 p. coin and at least two 25 p. coins = 1 × 3 = 3 ways. (D) Four 25 p. coins in one way only ∴ total = 54. The correct option is (A) 42. The number of ways in which at most n books can be selected out of a collection of (2n + 1) books is 2n + 1 C1 + 2n + 1C2 + … 2n + 1Cn

1 2n + 1 [( C1 + 2n + 1C1) + (2n + 1C2 + 2n + 1C2) + … 2 … + (2n + 1Cn + 2n + 1Cn)] = 63 ⇒

⇒ (2n + 1C1 + 2n + 1C2n) + (2n + 1C2 + 2n + 1C2n – 1) + … … + (2n + 1Cn + 2n + 1Cn + 1) = 126 ⇒

2n + 1

C1 + 2n + 1C2 + … + 2n + 1C2n = 126

⇒ 22n + 1 = 128 = 27

68

r 2 < 100 3

2 r n, etc. All these statements are concerned with n, which takes values 1, 2, 3, … Such statements are usually denoted by p(n). By giving particular values to n, we get particular statement. For example, if the statement 32n – 1 is divisible by 8 is denoted by p(n), then p(4) is the statement 32.4 – 1 is divisible by 8.

The Principle of Mathematical Induction This principle states: If p(n) is a statement involving natural number n, then 1. if p(1) is true, and 2. if p(k + 1) is true whenever p(k) is true, then p(n) is true for all natural numbers n.

  Understand the principle of mathematical induction

Thus, in order to prove a statement p(n) to be true for all natural numbers, we have the following working rule:

WORKING RULE Prove that p(1) is true; i.e., p(n) is true for n = 1. Assume p(k) to be true; i.e., p(n) is true for n – k. ■ Prove that p(k + 1) is also true; i.e., p(n) is also true for n = k + 1. ■ ■

Info Box! It is important to note that for the proof by mathematical induction both the conditions (i) and (ii) as stated above must be fulfilled. The result obtained may be fallacious if only one of these conditions is satisfied. Even if we prove a certain statement for a large number of values of n, say n = 1, 2, …, 100, we cannot say that the statement is true for all values of n unless we also establish the condition (ii).

SOLVED EXAMPLES 1. If n ∈ N, then 72n + 33n –3 ⋅ 3n – 1 is always divisible by (A) 25 (B) 35 (C)  45 (D)  None of these Solution: (A) Putting n = 1 is 72n + 23n–3.3n–1, we get 72.1 + 23.1–3 ⋅ 31–1 = 7220.30 = 49 + 1 = 50 Also, for n = 2

(1)

72.2 + 23.2–3 ⋅ 32–1 = 2401 + 24 = 2425 (2) From Eq. (1) and (2), it is always divisible by 25.

5.2  Chapter 5 2. For every natural number n, (n2 – 1) is divisible by (A) 4 (B) 6 (C)  10 (D)  None of these Solution: (B) We have, n, (n2 – 1) = (n – 1) n(n + 1). It is product of three consecutive natural numbers, so by Lagrange’s theorem it is divisible by 3! i.e., 6. 3. For a positive integer n, 1 1 1 1 Let a(n) = 1 + + + + … + n .Then 2 3 4 2 −1 (A) a (100) ≤ 100 (B)  a (100) > 100 (C) a (200) ≤ 100 (D)  a (200) > 100 Solution: (A) n It can be proved by induction that > a( n) ≤ n . 2 200 < a( 200) ⇒ a(200) > 100 and a(100) ≤ 100. ∴ 2 4. Let p(n) denote the statement that n2 + n is odd. It is seen that p(n) ⇒ p(n + 1), Pn is true for all (A) n > 1 (B) n (C) n > 2 (D)  None of these Solution: (D) Since the square of any odd number is always odd an sum of two odd numbers is always even, so for no ‘n’ this statement is true. 5. If n ∈ N, then 10n + 3 (4n + 2) + 5 is divisible by (A) 7 (B) 5 (C) 9 (D) 17 Solution: (C) For n = 102 + 3(44) + 5 = 100 + 768 + 5 = 873, which is divisible by 9. 6. The value of the natural numbers n such that the inequality 2n > 2n + 1 is valid, for (A) n ≥ 3 (B) n < 3 (C) no n (D)  any n Solution: (A) Check through options, the condition 2n > 2n + 1 is valid for n ≥ 3. 7. Let p(n) be statement 2n < n!, where n is a natural number, then p(n) is true for: (A) all n (B)  all n > 2 (C) all n > 3 (D)  None of these Solution: (C) Let  p(n) : 2n < n! Then, p (1) : 2! < 1!, which is not true  p (2) : 22 < 2!, which is not true

p (3) : 23 < 3!, which is not true p (4) : 24 < 4!, which is true. Let p(k) be true if k ≥ 4, i.e., 2k < k !, k ≥ 4 ⇒ 2.2k < 2 (k!) ⇒ 2k + 1 < k(k!) ( k ≥ 4 >2) ⇒ 2k + 1 (k + 1)! ⇒ p(k + 1) is true. Hence, we conclude that p(n) is not true for n = 2,3 but hold true for n ≥ 4. 8. x(xn–1 – nan – 1) + an(n – 1) is divisible by (x – a)2 for (A) n > 1 (B) n > 2 (C) all n ∈ N (D)  None of these Solution: (C) Check the options. The condition is satisfied for all n ∈ N. 9. For natural number n, 2n(n – 1)! < nn, if (A) n < 2 (B) n > 2 (C) n ≥ 2 (D)  for no n Solution: (B) Check the options. The condition is satisfied for n > 3 n

⎛ n + 1⎞ ≥ n ! is true 10. If n is a natural number then ⎜ ⎝ 2 ⎟⎠ when (A) n > 1 (B) n ≥ 1 (C) n > 2 (D) n ≥ 2 Solution: (B) Check the options. The condition is true for n ≥ 1. 11. Statement 1: For every natural number n ≥ 2, 1 1 1 + +… + > n n 1 2 Statement  2: For every natural number n ≥ 2, n( n +1) < n + 1. (A)  Statement 1 is false, Statement 2 is true (B) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 (C) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (D)  Statement 1 is true, Statement 2 is false Solution: (C) P(n) =

P(2) =

1 1 1 1

+ +

Let us assume that P(k) = is true.

1 2 1 2

+…+

1 n

> 2 1 1

+

 1 2

+… +

1 k

> k

Mathematical Induction  5.3 ∴ P(k + 1) =

1 1

+

1 2

+…+

1 k

+

has to be true. L.H.S. > Since ∴ Let

k +

1 k +1

=

1 k +1

> k +1

k ( k + 1) + 1 k +1

k ( k + 1) > k (∀ k ≥ 0) k ( k + 1) + 1 k +1 P(n) =

>

k +1 k +1

n( n + 1) < n + 1

=

k +1 

Statement 1 is correct. P(2) = 2 × 3 < 3 If

P(k) =

k ( k + 1) < (k + 1) is true

Now  P(k + 1) = ( k + 1)( k + 2) < k + 2 has to be true. Since (k + 1) < k + 2 ∴

( k + 1)( k + 2) < (k + 2)

Hence, Statement  2 is not a correct explanation of Statement 1.

5.4  Chapter 5

NCERT EXEMPLARS 1. If 10n + 3.4n + 2 + k is divisible by 9, for all n Î N, then the least positive integral value of k is (A) 5 (B) 3 (C) 7 (D) 1

3. If xn – 1 is divisible by x – k, then the least positive integral value of k is (A) 1 (B) 2 (C) 3 (D) 4

2. For all n Î N, 3.52n+1 + 23n+1 is divisible by (A) 19 (B) 17 (C) 23 (D) 25

ANSWER K EYS 1. (A)

2.  (B), (C) 3. (A)

HINTS AND EXPLANATIONS 1. Let P(n) : 10n + 3.4n+2 + k is divisible by 9, for all n Î N. For n = 1, the given statement is also true 101+3.41+2+k is divisible by 9. ∴ = 10 + 3.64 + k = 10 + 192 + k = 202 + k If (202 + k) id divisible by 9, then the least value of k must be 5. 202 + 5 = 207 is divisible by 9

⇒

207 = 23 9

NCERT EXEMPLARS

Hence, the least value of k is 5.

2. Given that, 3.52n+1 + 23n+1 For n = 1, 3.52(1)+1 + 23(1) + 1 = 3.53 + 24 = 3 × 125 + 16 = 375 + 16 = 391 Now, 391 = 17 × 23 Which id divisible by both 17 and 23. 3. Let P(n) : xn –1 is divisible by (x – k). For n = 1, x1 – 1 is divisible by (x – k). Since, if x – 1 is divis ible by x – k. Then, the least possible integral value of k is 1.

Mathematical Induction  5.5

PRACTICE EXERCISES Single Option Correct Type 1. For each natural number n, 3n > n3 for (A) n > 2 (B) n ≥ 3 (C) n ≥ 4 (D) n < 4

(C)  an odd positive integer (D)  None of these

n5 n3 7 + + n is 5 3 15 (A)  an integer (B)  a natural number (C)  a positive fraction (D)  None of these

6. For every positive integer n,

n n7 n5 2n3 + + − is 7 5 3 105

2. For n ∈ N ,



⎡ cos θ sin θ ⎤ n 3. If A = ⎢ ⎥ , then for n ∈ N, A is equal to ⎣ − sin θ cos θ ⎦

7. 10n + 3.4n+2 + k is divisible by 9 for n ∈ N . Then, the least positive integral value of k is (A) 1 (B) 3 (C) 5 (D) 7

⎡ cos n θ sin n θ ⎤ ⎡ cos nθ sin nθ ⎤ (A)  ⎢ ⎥ (B)  ⎢ − sin nθ cos nθ ⎥ n n ⎣ ⎦ ⎢⎣ − sin θ cos θ ⎥⎦ − sin nθ ⎤ ⎡ n cos θ (D)  ⎢ − n sin θ cos nθ ⎥⎦ ⎣

n sin θ ⎤ n cos θ ⎥⎦

8. The sum of the cubes of three consecutive natural numbers is divisible by (A) 2 (B) 4 (C) 6 (D) 9 9. For all n ∈ N, 1 +

1

+

1

+… +

1

2 3 4 (A) > n (B)  < n

1 1 1 + + +… to n terms = 1.4 4.7 7.10 1 1 (A)  (B)  5n − 1 3n − 1 n n (C)  (D)  3n + 1 5n − 1 4.

≥ n (C) ≤ n (D)  10. The statement p(n): 1 × 1! + 2 × 2! + 3 × 3! + … + n × n! = (n + 1)! – 1 is (A)  true for all n > 1 (B)  not true for any n (C)  true for all n ∈ N (D)  None of these

5. x2n–1 + y2n–1 is divisible by x + y if n is (A)  a positive integer (B)  an even positive integer

ANSWER K EYS Single Option Correct Type 1. (C)

2.  (B)

3. (B)

4.  (C)

5. (A)

6.  (A)

7. (C)

8.  (D)

9. (D) 10.  (C)

PRACTICE EXERCISES



⎡cos nθ (C)  ⎢ ⎣ sin nθ

(A)  an integer (B)  a rational number (C)  an odd integer (D)  a negative real number

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CHAPTER

6

Binomial Theorem

LEARNING OBJECTIVES After reading this chapter, you will be able to:  Learn the definition of binomial theorem and binomial expression  Be familiar with pascal’s triangle and methods to find general term, independent term and greatest term in the expansion of (1 + x)n

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.

BINOMIAL THEOREM This theorem gives a formula by which any power of a binomial expression can be expanded. It was first given by Sir Isaac Newton.

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) n

i.e., (x + y)n = ∑ n Cr x n−r y r r =0

Here nC0, nC1, nC2, ..., nCn are called binomial coefficients. For the sake of convenience, we may denote nCr by Cr ⋅ nCr  n may also be denoted as  .  r 

SPECIAL CASES 1. 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) 2. Replacing x by 1 and y by x, we get (1 + x)n = nC0 + nC1 x + nC2 x2 + ... + nCn xn

  Know how to calculate middle term in the binomial expansion

3. Replacing x by 1 and y by – x, we get (1 – x)n = nC0 – nC1 x + nC2 x2 – ... + (– 1)n nCn xn. 4. Adding (1) and (2), we get (x + y)n + (x – y)n = 2 (xn + nC2 xn – 2 y2 + n C4 xn – 4 y4 + ...) = 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. 5. Subtracting Eq. (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.

QUICK TIPS The coefficient of (r + 1)th term in the expansion of (1 + x)n is nCr. r n n  The coefficient of x in the expansion of (1 + x) is C . r 

I M P O R TA N T P O I N T S The positive integer n is called the index of the binomial.  Number of terms in the expansion of (x + y)n is n + 1, i.e., one more than the index n.  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. 

6.2  Chapter 6

The binomial coefficients of the terms equidistant from the beginning and the end are equal.  If n is odd, then number of terms in (x + a)n + 

(x – a)n and (x + a)n – (x – a)n are equal to

1

n

2

.

If n is even, then the number of terms in the expansion of



 n + 2  and that in the expan(x + a)n + (x – a)n are   2    sion of (x + a)n – (x – a)n are  n   2 

xn + yn is divisible by x + y if n is odd as xn + yn = (x + y) (xn – 1 – xn – 2y + xn – 3y2 – ... + yn–1)  xn – yn is divisible by x – y if n is even as xn – yn = (x – y) (xn – 1 + xn–2y + xn–3y2 + ... + yn – 1) –1  The coefficient of xn in the expansion of (x + 1) (x + 2) ... 

(x + n) =

n ( n 1) . 2

The coefficient of xn – 1 in the expansion of (x – 1) (x – 2)



... (x – n) = –

n ( n 1) . 2

PASCAL’S TRIANGLE The coefficients nC0, nC1, nC2, ... , nCn or simply C0, C1, C2 ..., Cn are called binomial coefficients and they can be evaluated with the help of Pascal’s triangle as below: Exponent of Binomial     Coefficients of successive    terms in Binomial Theorem n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 I. Each row starts and ends in 1. II. The coefficients of an expansion are obtained from the coefficients of the previous expansion. III. Each coeffcient is the sum of the two coefficients which lie on either side of it in the previous row.

General Term in the Expansion of (x + y)n Info Box! Students are advised to remember the following values: n ( n −1) n C0 = 1, nC1 = n, nC2 = , 

2! n ( n 1 ) ( n 2 ) n C3 = , 3! n( n − 1) ( n − 2) ... ( n − r + 1) n Cr = r!

and nCn = 1. Also, nCr = nCn–r for 1 ≤ r ≤ n. i.e., nC0 = nCn, nC1 = nCn – 1, nC2 = nCn – 2, ... i.e., coefficients of terms equidistant from the beginning and end are equal. n

n 2 ∑ ( Ci ) = 2nCn



i=0 n

n ∑ i ( Ci ) = n ⋅ 2n – 1



i=0 n

n n–2  ∑ i (i − 1) ⋅ Ci = n(n – 1)2 i=2 n

n ∑ i(i − 1) ... (i − k + 1) ⋅ Ci = n(n – 1) ...



i=k

(n – k + 1)⋅ 2n – k n

∑ i ( Ci ) = n ⋅ 2n – 1Cn – 1



i =1

n

2

In the binomial expansion of (x + y)n, the (r + 1)th term from the beginning is usually called the general term and it is denoted by Tr + 1, i.e., Tr + 1 = nCr xn – r yr

SOLVED EXAMPLES 1. The sum of the coefficients in the expansion of (1 + 5x – 7x3)3165 is (A) 1 (B) 23165 3164 (C) 2 (D) –1 Solution (D)

Putting x = 1 in (1 + 5x – 7x3)3165, the required sum of coefficients = (1 + 5 – 7)3165 = (– 1)3165 = –1. 12

 2  2. The 8th term of 3 x + 2  , when expanded in  3x  ascending power of x, is 228096 (A) 3 x

(B) 

328179 (C) 9 x

(D)  none of these

228096 x9

Solution (A) 12

 2  When 3 x + 2  is expanded, the power of x goes  3x  on decreasing as the terms proceed. Hence, it is

Binomial Theorem  6.3 12

  2 expanded in descending powers of x. So  2 + 3 x  ,   3x when expanded, will be in ascending powers of x. 12

  2 Now, t8 in  2 + 3 x  = 12C7   3x

12−7

 2    3 x 2 

⋅ (3x)7

n

 1 5. The term independent of x in (1 + x)m 1 +  is  x m+n (A) m + nCm (B)  Cn (C) m + nCm – n (D)  none of these Solution (B)

We have,

5

12 !  2  = 7 ! 5! ⋅  2  ⋅ (3x)7  3x  =

12´11´10 ´ 9´8 25 × 32 ⋅ 5´ 4 ´3´ 2 x3

228096 x3 3. If A is the sum of the odd terms and B the sum of even terms in the expansion of (x + a)n, then A2 – B2 = (A) (x2 + a2)n (B)  (x2 – a2)n 2 2 n (C) 2 (x – a ) (D)  none of these =

Solution (B)



Solution (C) 10

1  2 When  + y  is expanded, the powers of y go on y  increasing as the terms proceed. Hence it is expanded 10  2 1 in ascending powers of y. So  y +  , when y  expanded, will be in descending powers of y. 6

 1  10 ´ 9´8´ 7  Hence, t7 = 10C6 (y2)4  y  = y2   4 ´3´ 2´1 = 210 y2

n

 x + 1   x 

(1+ x ) m+ n = x– n (1 + x)m + n xn ∴ Required term independent of x = coefficient of x0 in x– n (1 + x)m + n = coefficient of xn in (1 + x)m + n m+n = Cn

=

6. The coefficient of x53 in the expansion 100

∑

m=0

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 = (nC0 xn + nC2 xn – 2 a2 + ...) – (nC1 xn – 1 a1 + nC3 xn – 3 a3 + ...) = A – B ∴ A2 – B2 = (A + B) (A – B) = (x + a)n (x – a)n = (x2 – a2)n 10 1  2 4. The 7th term in  + y  , when expanded in descendy  ing power of y, is y2 210 (a) 2 (b)  210 y 2 (c) 210 y  (d)  none of these

n

 1  (1 + x) 1 +  = (1 + x)m  x m

100

Cm ( x − 3)100−m ⋅ 2m is

(A) 100C47 (C) – 100C53

100 (B)  C53 (D) –100C100

Solution (C) 100

We have,

∑

100

Cm ( x − 3)100−m ⋅ 2m

m=0 100

= (x – 3) + 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 7. The coefficient of xm in (1 + x)m + (1 + x)m + 1 +, ..., + (1 + x)n, m ≤ n is n (A) nCm (B)  Cm + 1 (C) n + 1Cm + 1 (D)  none of these Solution (C)

The coefficient of xm in (1 + x)m + (1 + x)m + 1 + (1 + x)m + 2 + ... + (1 + x)n = mCm + m + 1Cm + m + 2Cm + ... + nCm = m + 1Cm + 1 + m + 1Cm + m + 2Cm + ... + nCm (  mCm = m + 1Cm + 1 = 1) = m + 2Cm + 1 + m + 2Cm + ... + nCm (  nCr + nCr + 1 = n + 1Cr + 1) = m + 3Cm + 1 + ... + nCm = n + 1Cm + 1. 8. The coefficient of x3 in the expansion of (1 – x + x2)6 is (A) 50 (B)  – 50 (C) 68 (D)  none of these Solution (B)

(1 – x + x2)6 = [1 – x (1 – x)]6 = 6C0 – 6C1 x (1 – x) + 6C2 x2 (1 – x)2

6.4  Chapter 6 – 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 ∴ Coefficient of x3 = – 2 ⋅ 6C2 – 6C3, (collecting coefficients of x3 from each term) 6! 6! =–2 = – 50 2 ! 4 ! 3! 3! log x 5 9. The value of x in the expression ( x + x 10 ) , if the third term in the expansion is 10,00,000, is (A) 10– 1 (B) 101 – 5/2 (C) 10 (D) 105/2

Solution (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 n n n C C C 10. The value of 1 + 3 + 5 + ... is 2 4 6 2n 1 2n + 1 (A) (B)  n n 2n + 1 (D)  n +1

2n − 1 (C) n +1

11. The coefficient of x5 in the expansion of (1 + x2)5 (1 + x)4 is (A) 40 (B)  50 (C) – 50 (D)  60 Solution (D)

We have, (1 + x2)5 (1 + x)4 = (1 + 5C1 x2 + 5C2 x4 + ...) (1 + 4C1 x + 4C2 x2 + 4C3 x3 + 4C4 x4) = (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. 12. If (1 + x – 2x2)6 = 1 + a1 x + a2 x2 + ... + a12 x12, then a2 + a4 + a6 + ... + a12 = (A) 21 (B)  11 (C) 31 (D)  none of these Solution (C)

Given (1 + x – 2x2)6 = 1 + a1x + a2x2 + ... + a12x12 Putting x = 1, we get 0 = 1 + a1 + a2 + ... + a12...(1) Putting x = – 1, we get 64 = 1 – a1 + a2 – ... + a12...(2) Adding Eq. (1) and (2), we get 64 = 2 (1 + a2 + a4 + ...) ∴ a2 + a4 + a6 + ... + a12 = 31 13. If 7103 is divided by 25, then the remainder is (A) 20 (B)  16 (C) 18 (D)  15

Solution (C)

Solution (C)

The rth term of the given expression is n C2 r - 1 Tr = 2r 1 n 1 n +1 ⋅ Cr = ⋅ C r +1 Since r +1 n +1

We have, 7103 = 7 (49)51 = 7 (50 – 1)51 = 7 (5051 – 51C1 5050 + 51C2 5049 – ... – 1) = 7 (5051 – 51C1 5050 + 51C2 5049 – ...) – 7 + 18 – 18 = 7 (5051 – 51C1 5050 + 51C2 5049 – ...) – 25 + 18 = k + 18 (say)  Q  k is divisible by 25, ∴  remainder is 18.

∴ Tr =

n

C2 r - 1

=

1 n +1 ⋅ C2 r n +1

2r n n n C C C1 ∴  + 3 + 5 + ... 2 4 6 1 n +1 ( C2 + n + 1C4 + ...)   = n +1 1 ( 2n + 1 − 1 − n + 1C0 ) = n +1 =

2n − 1 n +1

14. The sum of rational terms in the expansion of ( 2 + 31/5 )10 is (A) 31 (B)  41 (C) 51 (D)  none of these Solution (B)

(r + 1)th term in the given expansion is given by tr + 1 = 10Cr 102-r 5r , where r = 0, 1, 2, ..., 10 3 2 For rational terms r = a multiple of 5 = 0, 5, 10 ...(1)

Binomial Theorem  6.5 10 – r = a multiple of 2 = 0, 2, 4, 6, 8, 10 ...(2) From Eq. (1) and (2) possible values of r are : 0 and 10 ∴  sum of rational terms = t1 + t11 = 10C0 ( 2 )10 (31/5)0 + 10C10 ( 2 )0 (31/5)10 = 25 + 32 = 32 + 9 = 41 15. 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 (B) (x + a)n + (x – a)n 2n 2n (C) (x + a) – (x – a) (D)  none of these

For denominator, 31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243 6 C1 = 6, 6C2 = 15, 6C3 = 20 6 C4 = 6C2 = 15, 6C5 = 6C1 = 6, 6C6 = 1 ∴  denominator = 36 + 6C1 35 ⋅ 21 + 6C2 34 ⋅ 22 + 6C3 33 ⋅ 23 + 6C4 32 ⋅ 24 + 6C5 3 ⋅ 25 + 6C6 26 This is clearly the expansion of (3 + 2)6 = 56 = (25)3 ∴ 

Solution (c)

We have, (x + a)n = xn + nC1 xn – 1 a + nC2 xn – 2 a2 + nC3 xn – 3 a3 + ... = (xn + nC2 xn – 2 a2 + ...) + (nC1 xn – 1 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 16. If C0, C1, C2, ..., Cn are the coefficients of the expansion n C of (1 + x)n, then the value of ∑ k is 0 k +1 2n 1 (A) 0 (B)  n n +1 2 −1 (C) (D)  none of these n +1 Solution (C)

n

∑ 0

Ck k +1

1 = (n + 1C1 + n + 1C2 + n + 1C3 + ... + n + 1Cn + 1) n +1 2n+1 −1 1 = (2n + 1 – n + 1C0) = n +1 n +1 17. The value of (183 + 73 + 3 ⋅18 ⋅ 7 ⋅ 25) is 3 + 6 ⋅ 243 ⋅ 2 + 15 ⋅ 81⋅ 4 + 20 ⋅ 27 ⋅ 8 + 15 ⋅ 9 ⋅16 + 6 ⋅ 3 ⋅ 32 + 64 6

(A) 0 (C) 2

(B)  1 (D)  none of these

Solution (B)

The numerator is of the form a3 + b3 + 3ab (a + b) = (a + b)3 where a = 18 and b = 7 ∴  Numerator = (18 + 7)3 = 253.

18. Larger of 9950 + 10050 and 10150 is (A) 10150 (B) 9950 + 10050 (C) both are equal (D)  none of these Solution (A)

We have, 10150 = (100 + 1)50 50 × 49 = 10050 + 50 ⋅ 10049 + ⋅ 10048 + ... 1× 2 and 9950 = (100 – 1)50 = 10050 – 50 ⋅ 10049 + 50 × 49 ⋅ 10048 – ... 1× 2 Subtracting, we get 10150 – 9950 = 2 (50 ⋅ 10049 +

× 10047 + ...)

50 × 49 × 48 ⋅ 10047 + ... > 10050 1× 2 × 3 Hence, 10150 > 9950 + 10050.

= 10050 + 2 ⋅

19. For all n ∈ N, 24n – 15n – 1 is divisible by (A) 225 (B)  125 (C) 325 (D)  none of these

n

Cr 1 Here, tr + 1 = = ⋅ nCr r +1 r +1 1 = ⋅ n + 1Cr + 1 n +1 Putting r = 0, 1, 2, ... n and adding, we get

(25)3 Numerator = ( 25)3 = 1 Denominator

Solution (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. 20. When 599 is divided by 13, the remainder is (A) 8 (B)  9 (C) 10 (D)  none of these Solution (A)

We have, 599 = 53 ⋅ 596 = (125) (625)24 = [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.

6.6  Chapter 6 (32)

21. The last digit of the number (32)32 is (A) 4 (B)  6 (C) 8 (D)  none of these

24. When 32(32) is divided by 7, the remainder is (A) 4 (B)  6 (C) 8 (D)  none of these

Solution (B)

Solution (A)

(32) = (2 + 3 × 10) = 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. 32

32

22. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then C C1 C + 23 ⋅ 2 + ... + 2n + 1 n = n +1 2 3 3n 1 3n+1 −1 (A) (B)  n n +1 2C0 + 22 ⋅

25. The number of non zero terms in the expansion of (1 + 3 2 x )9 + (1 − 3 2 x )9 is

n+ 2

3 −1 (C) n+2

(D)  none of these

(A) 9 (C) 5

Solution (A)

We have,

(B)  0 (D)  10

Solution (C) n

Cr 1 = 2r + 1 ⋅ ⋅ n + 1Cr + 1 r +1 n +1 Putting r = 0, 1, 2, ..., n and adding, we get the required sum 1 = (2 ⋅ n + 1C1 + 22 ⋅ n + 1C2 + ... + 2n + 1 ⋅ n + 1Cn + 1) n +1 3n+1 −1 1 = [(1 + 2)n + 1 – n + 1C0] = . n +1 n +1 tr + 1 = 2r + 1

23. For integer n > 1, the digit at units place in the number 100

(32)32 = (25)32 = 2160 = (3 – 1)160 = 160C0 3160 – 160C1 ⋅ 3159 + ... + 160C159 ⋅ 3 + 160C160 ⋅ 30 = 3k + 1, where k ∈ N ( 32 ) Now, 32( 32 ) = (32)3k + 1 = (25)3k + 1 = 215k + 5 = 23(5k + 1) ⋅ 22 = (23)5k + 1 ⋅ 4 = 4(7 + 1)5k + 1 = 4[5k + 1C0 75k + 1 + 5k + 1C1 75k + ... + 5k + 1C5k7 + 5k + 1C5k + 1 ⋅ 70] = 4(7n + 1), where n ∈ N = 28n + 4. ( 32 ) Therefore, when 32( 32 ) is divided by 7, the remainder is 4.

n

2 ∑ r ! + 2 is

(B)  1 (D)  3

Solution (A)

Since the digit at units place in each of 5!, 6!, ..., 100! is 0 and 0! + 1! + 2! + 3! + 4! = 34. 100

therefore the digit at units place in

∑ r ! is 4. r=0

n

Now, 22 = 24k, k ∈ N (2n is a multiple of 4 form n > 1) n ∴  The digit at units place in 22 = 24k = (16)k is 6. 100

Thus, the digit at units place in

26. The expression [x + (x3 – 1)1/2]5 + [x – (x3 – 1)1/2]5 is a polynomial of degree (A) 5 (B)  6 (C) 7 (D)  8 Solution (c)

[x + (x3 – 1)1/2]5 + [x – (x3 – 1)1/2]5 = 2 [5C0 x5 + 5C2 x3 (x3 – 1) + 5C4 x (x3 – 1)2] = 2 [x5 + 10x3 (x3 – 1) + 5x (x3 – 1)2] = 5x7 + 10x6 + x5 – 10x4 – 10x3 + 5x which is a polynomial of degree 7.

r=0

(A) 0 (C) 2

In the expansion of (1 + 3 2 x)9 + (1 – 3 2 x)9 2nd, 4th, 6th, 8th and 10th terms get cancelled. ∴  Number of non-zero terms in 2 [9C0 + 9C2 (3 2x )2 + ... + 9C8 (3 2x)8] is 5.

∑ r! + 2 r=0

2n

is 0.

27. The value of x, for which the 6th term in the expansion 7   1  log2 ( 9x−1 +7 )  + 1 of  2  is 84, is equal to log 2 ( 3x−1 +1)   5 2   (A) 4 (C) 2 Solution (C,D)

The given expression

(B)  3 (D)  1

Binomial Theorem  6.7  x−1  1  =  9 + 7 + x−1 1/ 5  ( 3 1 ) +   Given, T6 = 84

∴  Number of rational terms = 5. ∴  Number of irrational terms = 46 – 5 = 41. 31. In the expansion of (1 + x + x3 + x4)10, the coefficient of x4 is 10 (A) 40C4 (B)  C4 (C) 210 (D)  310

7

5

⇒ C5 ( 9 7

x −1

+ 7)

7−5

  1    (3x−1 + 1)1/ 5  = 84

Solution (D)

1 = 84 (3 + 1) ⇒ 9x – 1 + 7 = 4 (3x – 1 + 1) ⇒ 32x – 12 ⋅ 3x + 27 = 0 ⇒ (3x – 3) (3x – 9) = 0 ⇒ 3x = 3, 9 ⇒ x = 1, 2 ⇒ 7C5 (9x – 1 + 7) ⋅

x−1

28. When 337 is divided by 80, the remainder is (A) 3 (B)  4 (C) 6 (D)  none of these Solution (A)

We have, 337 = 34.9 ⋅ 3 = 3(81)9 = 3(80 + 1)9 = 3(9C0 ⋅ 809 + 9C1808 + ... + 9C9) Thus, when 337 is divided by 80, the m remainder is 3. n

 a  29. If the second term in the expansion 13 a +  is  a−1  n 14 a5/2, then the value of (A) 8 (C) 4

n

C3 C2

is

(B)  12 (D)  none of these

Solution (C)

Given : T2 = 14 a5/2 ⇒ C1 (a n

)

1/13 n – 1

1

 a  ⋅  −1/ 2  = 14 a5/2 a 

⇒ 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 ∴

14

n

C3 n C2

=

C3 14 C2

= 4

30. The number of irrational terms in the expansion of (41/5 + 71/10)45 is (A) 40 (B)  5 (C) 41 (D)  none of these Solution (C)

Total number of terms in the expansion of (41/5 + 71/10)45 is 45 + 1,  i.e., 46. The general term in the expansion is r Tr + 1 = 45Cr ⋅ 455−r ⋅ 10 7 4 Tr + 1 is rational if r = 0, 10, 20, 30, 40.

∴ 

 (1 + x + x3 + x4)10 = [(1 + x) (1 + x3)]10 = (1 + x)10 (1 + x3)10 = (1 + 10C1 x + 10C2 x2 + 10C3 x3 + 10C4 x4 ...) × (1 + 10C1 x3 + 10C2 x6 ...) Coefficient of x4 = (10C1) (10C1) + 10C4



= 100 +

10 9 8 7 = 100 + 210 = 310 12 3 4

32. If A = 2nC0 · 2nC1 + 2nC1 2n – 1C1 + 2nC2 2n – 2C1 + ..., then A is (A) 0 (B)  2n 2n (C) n 2 (D) 1 Solution (C)

A = coeff. of x in [2nC0(1 + x)2n + 2nC1 (1 + x)2n – 1 + ...)] = coeff. of x in (1 + (1 + x))2n = coeff. of x in (2 + x)2n 2n



 x = coeff. of x in 22n 1 +  = n · 22n 

2 

33. The greatest integer which divides the number 101100 – 1 is (A) 100 (B)  1000 (C) 10000 (D)  100000 Solution (C)

By Binomial theorem   n ( n −1) 2 (1 + x)n = 1 + nx + ⋅ x ... + x n    2 or (1 + x)n – 1 = nx + n ( n −1) x2 ... + xn 2 n 2 If x = n, (1 + n) – 1 = n + n ( n −1) n2 ... nn 2   n ( n − 1) (1 + n)n – 1 = n2 1 + ... + nn−2    2 Put n = 100,  100 (100 −1)  (1 + 100)100 – 1 = (100)2 1 + ... + 100 98    2  100 × 99  (101)100 – 1 = (100)2 1 + ... + 100 98  2   Clearly (101)100 – 1 is divisible by (100)2 = 10000

6.8  Chapter 6 2n

 2 1 34. If x  occurs in the expansion of  x +  , its coeffi x cient is 2n 2n (A) C 4 n- p (B)  C 2 n- p p

3 2n (C) C 4 n - p

3

(D)  none of these

3

Solution (A)

Let tr + 1 contains x p.

r

1 Then, tr + 1 = 2nCr (x2)2n – r   = 2nCr x4n – 3r  x  4n − p 3 2n p Hence, the coefficient of x = C 4n − p ∴ 4n – 3r = p; or r =

35. Given positive integers r > 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 (A) , n even (B)  2 2 (C) n (D) 1 Solution (A)

We have, t3r = 2nC3r – 1 x3r – 1 and tr + 2 = 2nCr + 1 xr + 1 2n Given, C3r – 1 = 2nCr + 1 ⇒ 3r – 1 = r + 1; or (3r – 1) + (r + 1) = 2n ⇒ 2r = 2 ; or 4r = 2n n ⇒ r = 1 (impossible); or r = . 2 But r is a positive integer greater than 1. So the value n of r is provided n is an even integer (> 2), other2 wise r has no value. 36. If the last term in the binomial expansion of log 8 n  1  3   3 2 − 1  is   , then the 5th term is  3.3 9   2  (A) 2 ⋅ 10C6 (B) 4 ⋅ 10C4 1 2

10

C6

(D)

10

C6

Solution (D) log3 8

n  1   1  The last term of  3 2 −  =  3   2   3⋅ 9  log 8

1 = 2–5 =    2

⇒  n = 10

n  1   1  3 ⇒  nCn ⋅ −  =    2   3 ⋅ 3 9  n/2 log 8 5     3 ⇒  (−1) n ⋅  1   =  1  = 3− 3 ⋅ 3 log3 2 5 / 3   2   3 

10

 1  Therefore, 5th term in  3 2 −  is  2  4 10 − 4   − 1  T5 = T4 + 1 = 10C4 ( 3 2 )  2  = 10C4 ( 4) 1 = 10C4 = 10C6 4 2 n 37. If (1 – x + x ) = a0 + a1 x + a2 x2 + ... + a2n x2n, then a0 + a2 + a4 + ... + a2n is equal to

(A) 3n2 +

3

(C)

5



1 2

3n -1 (C) 2 Solution (D)

(B) 

1- 3n 2

3n + 1 (D)  2

Putting x = – 1, 1 successively in the given equation and adding, we shall get the result.

Method for Finding the Independent Term or Constant Term Step I

 rite down the general term in the expansion of W (x + a)n i.e., (r + 1) th term ⇒ tr+1 = nCrxn–rar Step II Separate the constants and variables. Also group them separately. Setp II Since, we need to find the term independent of x in the given binomial expansion, equate to zero the index of x and accordingly we will get the value of r for which there exists a term independent of x in the expansion.

Greatest Term (Numerically) in the Expansion of (1 + x)n Method 1 1. Let Tr (the rth term) be the greatest term. 2. Find Tr – 1, Tr , Tr + 1 from the given expansion. Tr T 3. Put ≥ 1 and r ≥ 1. This will give an inequalTr+1 Tr-1 ity from where value or values of r can be obtained. 4. Then, find the rth term Tr which is the greatest term. Method 2 1. Find the value of k =

( n + 1)| x | 1+| x |

2. If k is an integer, then Tk and Tk + 1 are equal and both are greatest terms.

Binomial Theorem  6.9 3. If k is not an integer, then T(k) + 1 is the greatest term, where (k) is the greatest integral part of k.

Solution (A)

We have,

QUICK TIPS n

 y  (x + y) = x 1+  and then find the greatest term in  x n   1+ y  .  x  n

| x | ( n + 1) (| x | + 1)

∴ m =

To find the greatest term in the expansion of (x + y)n, write

11

n

 1  − < 0   3

The greatest terms in the expansion are T3 and T4 ∴  Greatest term (when r = 2) = 311 | T2 + 1 | 2

38. The greatest term (numerically) in the expansion of 3 (2 + 3x)9, when x = , is 2 5 313 5 311 (A) (B)  2 2 7 313 2

=3

∴ m =



9

 ∵ x = 

3   2 

x ( n + 1) ( x + 1)

 9    (9 + 1)  4  9 90 12 =   + 1 = =6 ≠ Integer 4 13 13

The greatest term in the expansion is T[m] + 1 = T6 + 1 = T7 Hence, the greatest term = 29 ⋅ T7 6 9 = 29 ⋅ T6 + 1 = 29 ⋅ 9C6    4  6



= 311

11⋅10 1 × = 55 × 39 1⋅ 2 9

=3

11

 1 C3 −   3 

11

= 311

11⋅10 ⋅ 9 1 = 55 × 39 ×− 1⋅ 2 ⋅ 3 27

From above, we see that the values of both greatest terms are equal.

We have, 9

 1 C2 −   3 

11

3

(D)  none of these

 3x   9 (2 + 3x)9 = 29 1 +  = 29 1 +    4   2

11

and greatest term (when r = 3) = 311 | T3 + 1 |

Solution (C)



  ∵ x = 1   5 

 1   −  (11 + 1)  3  = =3  1   − + 1   3



SOLVED EXAMPLES

(C)

11

 5x   1 (3 – 5x)11 = 311 1−  = 311 1−    3  3



13 9 9 × 8 × 7 312 = 29 ⋅ 9C3   = 29 ⋅ ⋅ = 7´3  4  12 1× 2 × 3 2 2

39. The greatest term (numerically) in the expansion of 1 (3 – 5x)11 when x = is 5 (A) 55 × 39 (B)  46 × 39 6 (C) 55 × 3 (D)  none of these

MIDDLE TERM IN THE BINOMIAL EXPANSION  The middle term in the binomial expansion of (x + y)n depends upon the value of n. 1. If n is even, then there is only one middle term i.e.  n   + 1 th term.  2  2. If n is odd, then there are two middle terms, i.e.,  n + 1  n + 3     th and   2   2  th terms.

I M P O R TA N T P O I N T S When there are two middle terms in the expansion, their binomial coefficients are equal.  Binomial coefficient of the middle term is the greatest binomial coefficient. 

SOLVED EXAMPLE 2n

 1 40. The greatest coefficient in the expansion of  x +   x is

6.10  Chapter 6

Properties of nCr

2n ! 1⋅ 3 ⋅ 5...( 2n −1) ⋅ 2n (B) 2 ( n!) n!

(A)

If 0 < r < n, n, r ∈ N, then 1. r ⋅ nCr = n ⋅ n – 1Cr – 1

n! 2

 n   (C)   !  2  

(D)  none of these

Solution (A, B)

Since the middle term has greatest coefficient, ∴ greatest coefficient = coefficient of the middle term ( 2n)! = 2nCn = n! n! =

2n ( 2n −1) ( 2n − 2) ( 2n − 3)...4 ⋅ 3 ⋅ 2 ⋅1 n! n!

= [( 2n −1) ( 2n − 3)...3 ⋅1] [2n ( 2n − 2) ( 2n − 4)...4 ⋅ 2] n! n! =

[1⋅ 3 ⋅ 5...( 2n −1)] 2n [n ( n −1) ( n − 2)...2 ⋅1] n! n!

=

1⋅ 3 ⋅ 5...( 2n −1) 2n n ! 1⋅ 3 ⋅ 5...( 2n −1) 2n = . n! n! n!

pth Term from the End in the Binomial Expansion  of  (x + y)n pth term from the end in the expansion of (x + y)n is (n – p + 2)th term from the beginning.

n

Cr 2. = r +1 n 3. Cr =

n r

n +1

Cr +1 n +1 Cr – 1

n–1

n Cr n − r +1 4. = n Cr 1 r n Cr r +1 5. = n Cr +1 n − r

6.  nCr – 1 + nCr = n + 1Cr

n 7. Cx = nCy ⇒ x = y or x + y = n n 8. Cr = nCn – r. n / 2 if n is even  n 9. Cr is greatest if r =  n − 1 n + 1 or if n is odd  2  2

Thus, if n is even, the greatest coefficient is nCn/2 and if n is odd, the greatest coefficient is nC n 1 or nC n + 1, 2 2 both being equal. 10. The greatest term in (1 + x)2n has the greatest coeffi-

Properties of Binomial Coefficients

cient if

In the binomial expansion of (1 + x)n, the coefficients nC0, C1, nC2, ... nCn are denoted by C0, C1, C2, ... Cn respectively. n 1. If n is even, then greatest coefficient = Cn / 2

n n +1 1, n > 2 and the coefficients of (3r)th and (r + 2)th term are equal in the expansion (1 + x)2n. 2n 3r −1 Then, T3r = T3r −1+1 = C3r −1x

2n r +1 and Tr + 2 = Tr +1+1 = Cr +1x



2n

C3r −1 = 2 n Cr +1

∵ nC = nC ⇒ x + y = n x y  

⇒ 3r – 1 + r + 1 = 2n ⇒ 4 r = 2n ⇒ n =

4r 2

3. Let two successive terms in the expansion of (1 + x)24 are (r + 2)th terms. ∴ Tr +1 = 24Cr x r and Tr + 2 = 24Cr +1x r +1

⇒

(24)! r !( 24 − r )! 1 = 4 (24)! 1 24 1 r + ! − r − ! ( )( )

Binomial Theorem  6.15

(r + 1) r !(23 − r )! = 1 ⇒ r !( 24 − r ) ( 23 − r )! 4

⇒ 6 n − 6 = 6 + n2 − 3n + 2

r +1 1 ⇒ = ⇒ 4 r + 4 = 24 − r 24 −r 4

⇒ n2 − 7n − 2n + 14 = 0

2 ⇒ n − 9n + 14 = 0

⇒ n ( n − 7) − 2 ( n − 7) = 0

⇒ 5r = 20 ⇒ r = 4

⇒ ( n − 7) ( n − 2) = 0

∴ T = T and T4 + 2 = T6 4 +1 5 Hence, 5th and 6th terms. 4. ∵ Coefficient of xn in the expansion of (1 + x )2 n = 2 nCn And coefficient of xn in the expansion of (1 + x )2 n−1 = 2 n−1Cn ∵

Cn

( 2 n) ! n! n!

=

(2n − 1)! ! n n!( n − 1)! (2n)!ni (n − 1)! = n!n!(2n − 1)! 2 n−1

Cn



=

=

Since, n = 2 is not possible. ∴ n=7 6. Since, the coefficient of xn in the expansion (1 + x)2n is 2nCn. ∴ A = 2nCn Now, the coefficient of xn in the expansion of (1 + x)2n – 1 is 2n – 1 Cn. ∴ B = 2n – 1Cn Now,

2n ( 2n − 1)! n!( n − 1)!

A = B

2n

Cn

2 n −1

Cn

2n 2 = = 2 :1 n 1

∴ Coefficient of 2nd term = n C1, Coefficient of 3rd term nC2, and coefficient of 4th term = nC3. Given that, n C1, nC2 and nC3 are in AP .

 1 10 −5 5 10 ∴ T6 = T5+1 = C5   ( x sin x ) x 63 10 = C5 x –5 x 5 sin 5 x 8 63 10.9.8.7.6.5! 5 ⇒ = sin x 8 5.4.3.2.1.5! ⇒ 63 = 3 ⋅ 2 ⋅ 7 ⋅ 6 sin 5 x 8 ⇒

n r n ∴ 2 C2 = C1 + C3

⇒ sin 5 x = 1 32

 ( n)!  ( n) ! + ( n)! = ⇒ 2  ( n − 2) 2! (n − 1)! 3!( n − 3)!

5 ⇒ sin 5 x =  1  2

2.n ( n − 1) ( n − 2)!

( n − 2) ! 2 !

⇒ n ( n − 1) = n +

=

n ( n − 1)!

(n − 1)!

n ( n − 1) ( n − 2) 6

+

2 =2 1

1 10 7. Given expansion is  + x sin x  . x 

n! n ( n − 1)!( 2n − 1)

5. The expansion of n (1 + x ) is nC0 + nC1x + nC2 x 2 + nC3 x 3 + ...... + nCn x n

⇒

=

n ( n − 1) ( n − 2) ( n − 3)! 3.2.1( n − 3)!

⇒ sin x =

1 2

∴ x = nπ + ( −1) π / 6 n

HINTS AND EXPLANATIONS

2n

∴ n = 2 or n = 7

6.16  Chapter 6

PRACTICE EXERCISES Single Option Correct Type 1. The coefficient of x17 in the expansion of (x – 1) (x – 2) (x – 3) ... (x – 18) is 171 (A) (B)  342 2 (C) – 171 (D)  684 24n 2. The fractional part of 15 is 2 1 (A) (B)  15 15

PRACTICE EXERCISES

4 (C) 15

8. The interval in which x must lie so that the numerically greatest term in the expansion of (1 – x)21 has the greatest coefficient is, (x > 0). 5 6  5 6   ,  (A)  ,  (B)   6 5  6 5  4 5  4 5  ,  (C)  ,  (D)  5 4  5 4  9. If Cr stands for nCr, then the sum of the series  n  n 2  !  !  2   2  [C02 − 2C12 + 3C22 − n!

... + (− 1) n ( n + 1) Cn2 ],

(D)  none of these ... + (− 1) n ( n + 1) Cn2 ], where n is an even positive integer, is (A) 0 (B)  (–1)n/2 (n + 1) n/2 2003 (C) (–1) (n + 2) (D)  (– 1)n n  2  3. If {x} denotes the fractional part of x, then   is 2n  17  (A) 2/17 (B)  4/17  10. If (1 + 2x + x2)n = ∑ ar x r , then ar = r =0 (C) 8/17 (D)  16/17 n (A) (nCr)2 (B)  Cr ⋅ nCr + 1 2n 2n (C) Cr (D)  Cr + 1 4. The sum of the coefficients of all the integral powers of n n 80  1 + 4 x + 1  1 − 4 x + 1   x in the expansion of (1 + 2 x ) is 1   11. If     −      2 2 4 x + 1  1 80 1 80    (A) (3 + 1) (B)  (3 - 1) 5 2 2 = a0 + a1x + ... + a5x , then n equals (A) 11 (B)  9 (C) (380 + 1) (D)  (380 – 1) (C) 10 (D)  none of these 5. If [x] denotes the greatest integer less than or equal to  p m 10 20  x, then [(6 6 + 14) 2 n+1 ]  , (where   = 0 if p < q) is 12. The sum ∑    q    i m − i (A) is an even integer (B)  is an odd integer i = 0 (C) depends on n (D)  none of these maximum when m is 6. The two consecutive terms in the expansion of (A) 5 (B)  10 (3x + 2)74, whose coefficients are equal, are (C) 15 (D)  20 (A) 20th and 21st (B)  30th and 31st 13. The number of distinct terms in the expansion of (C) 40th and 41st (D)  none of these n   3  x + 1 + 1  ; x ∈ R+ and n ∈ N is n T  x 1   3 x 3  7. If in the expansion of 2 + x  , T = 7 and the sum  4  2 (A) 2n (B)  3n of the coefficients of 2nd and 3rd terms is 36, then the (C) 2n + 1 (C)  3n + 1 value of x is 14. The number of terms with integral coefficients in the 1 1 (A) - (B)  expansion of (171/3 + 351/2x)600 is 3 2 (A) 100 (B) 50 1 1 (C) 150 (D) 101 (C) (D)  3 2

Binomial Theorem  6.17  3 i   3 i    15. If z =  2 + 2  +  2 – 2  , then    

24. The interval in which x (> 0) must be so that the greatest term in the expansion of (1 + x)2n has the greatest coefficient is

(A) Re (z) = 0 (C) Re (z) > 0, Im (z) > 0

 n − 1 n   , (A)   n n − 1

 n n + 1   (B)   n + 1 , n 

 n n + 2   , (C)  n  n + 2

(D)  none of these

5

(B)  Im (z) = 0 (D)  Re (z) > 0, Im (z) < 0

16. The greatest value of the term independent of x in the expansion of (x sinα + x–1 cosα)10, α ∈ R, is 10 ! (A) 10 ! (B)  5 ( 5!) 2 2 (C)

1 10 ! 25 (5!) 2

(D)  none of these

17. If coefficient of xn in (1 + x)101 (1 – x + x2)100 is nonzero, then n can not be of the form (A) 3t + 1 (B)  3t (C) 3t + 2 (D)  4t + 1

25. If n is positive integer and k is a positive integer not exceeding n, then 2 n   3  Ck  n  k   C  , where Ck = Ck, is k =1 k −1

∑

(A) n( n + 1) ( n + 2) 12 (C)

n( n + 1) 2 ( n + 2) 6

2 (B)  n( n + 1) ( n + 2) 12

(D)  none of these

18. The sum of the last ten coefficients in the expansion of (1 + x)19 when expanded in ascending powers of x is (A) 218 (B)  219 1  (219 – 1) (C) 218 – 19C10 (D)  2

  1 1/12    + x 26. If the fourth term in the expansion of  x log x +1   is equal to 200 and x > 1, then x is equal to

19. The number of integral terms in the expansion of ( 2 5 + 6 7 )642 is (A) 105 (B)  107 (C) 321 (D)  108

27. The coefficient of λnµn in the expansion of [(1 + λ) (1 + µ) (λ + µ)]n is

20. The number of positive terms in the sequence xn = 

195 − n Pn

( n + 3) ( n + 1)

P3 is Pn+1

(A) 14 (B)  11 (C) 12 (D)  13 21. The digit at unit’s place in the number 171995 + 111995 –71995 is (A) 0 (B)  1 (C) 2 (D)  3 22. The positive integer which is just greater than (1 + 0.0001)1000 is (A) 3 (B)  4 (C) 5 (D)  2 23. The coefficient of xn in the polynomial (x + nC0 ) (x + 3 n C1) (x + 5 nC2)... (x + (2n + 1) nCn) is (A) n.2n (B)  n.2n+1 n (C) (n +1).2 (D) n.2n–1

6

(A) 10 2 (B)  10 (C) 104 (D)  none of these

n

∑

(A)

n

∑

Cr2 (B)  Cr2+ 2

r=0 n

∑C

(C)

r=0

r=0

2 r +3

n

∑

Cr3 (D)  r=0

n−3

 1 28. If there is a term containing x2r in  x + 2  , then  x  (A) n –2r is a positive integral multiple of 3 (B) n – 2r is even (C) n – 2r is odd (D) none of these 29. If Pn denotes the product of the binomial coefficients Pn+1 in the expansion of (1 + x)n, then equals Pn (A) 

( n + 1) n nn (B)  n! n!

(C) 

( n + 1) n ( n + 1) n+1 (C)  ( n +1)! ( n +1)!

PRACTICE EXERCISES

5

6.18  Chapter 6 30. The coefficient of the term independent of x in the 10  x +1 x − 1   − expansion of  2 / 3  is  x − x1/ 3 + 1 x − x1/ 2  (A) 210 (B)  105 (C) 70 (D)  112 1 10 2 n 10 2 2 n 31. The value of n − n C2 + n C2 81 81 81 3 2n 10 10 − n 2 nC3 + ... + n is 81 81 (A) 2 (B)  0 (C) 1/2 (D)  1

4 39. The value of 2(nC0) + 3  (nC1) +  (nC2) + 5  (nC3)...is 3 2 4 2n (1 − n) − 1 (A) n +1 (C)

32. If n is an even integer and a, b, c are distinct, the number of distinct terms in the expansion of (a + b + c)n + (a + b – c)n is 2

2  n + 1  n  (A)   (B)   2   2  2

33. Coefficient of t 24 in (1 + t 2)12 (1 + t 12) (1 + t 24) is (A) 12C6 + 3 (B)  12C6 + 1 12 12 (C) C6 (D)  C6 + 2

PRACTICE EXERCISES

34. (mC0 + mC1 – mC2 – mC3) + (mC4 + mC5 – mC6 – mC7) + ... = 0 if and only if for some positive integer k, m = (A) 4k (B)  4k + 1 (C) 4k – 1 (D)  4k + 2 35. If the sum of the coefficients in the expansions of (1 + 2x)m and (2 + x)n are respectively 6561 and 243, then the position of the point (m, n) with respect to the circle x2 + y2 – 4x – 6y – 32 = 0 (A) is inside the circle (B) is outside the circle (C) is on the circle (D) can not be fixed 36. Let n(> 1) be a positive integer. Then largest integer m such that (nm + 1) divides 1 + n + n2 + ... + n255 is (A) 128 (B)  63 (C) 64 (D)  32

∑

2n ( n + 3) − 1 (B)  n +1

2n − 1 2n + 2 (D)  n +1 n −1

40. Which of the following expansions will have term containing x3 ? 1 24 3 25  3  −1  x 5 + 2 x− 5   5 5  x + 2 x (A)  (B)        23

2

 n + 2   n + 3   (C)  (D)     2   2 

n

38. The value of the sum of the series 3nC0 – 8nC1 + 13n C2 – 18nC3 + ... upto (n + 1) terms is (A) 0 (B)  3n n (B) 5 (D)  none of these

22

1  3 −   5 5 (C)  x − 2 x   

1  3  x 5 + 2 x− 5  (D)    

41. The coefficient of x7 in the expansion of (1 – x – x2 + x3)6 is (A) 132 (B)  144 (C) –132 (D)  –144 42.

If n is a positive integer, then ( 3 + 1) − ( 3 − 1) (A) an irrational number (B) an odd positive integer (C) an even positive integer (D) a rational number other than positive integers 2n

2n

is

43. If ai(i = 0, 1, 2, ..., 16) be real constants such that for every real value of x, (1 + x + x2)8 = a0 + a1x2 + a2x2 ... + a16x16, then a5 is equal to (A) 502 (B)  504 (C) 506 (D)  508 n

44. Statement-1: 

∑ (r + 1)

n

Cr = ( n + 2)2n − 1

n

Cr x r = (1 + x ) n + nx(1 + x ) n − 1

r=0 n



Statement-2:

∑ (r + 1) r=0

+ 1)+nC3)r xn r –= (1 + x ) n + nx(1 + x ) n − 1 37. The coefficient of xn in the expansion( r(2x =0 2 (2x + 3)n–1 (5 – 2x) + (2x + 3)n–2 (5 –r 2x) + ... + (– 1)n (A) Statement-1 is false, Statement-2 is true n (5 – 2x) is (B) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1 1 n (A) 2 (B)  (n + 1)2n (C) Statement-1 is true, Statement-2 is true; Statement-2 8 is not a correct explanation for Statement-1 (C) (n + 1)2n–3 (D)  – (n + 1)2n–2 (D) Statement-1 is true, Statement-2 is false

Binomial Theorem  6.19

(C)

log10 4

9 ( - log10 3

D)  log10 4

4 - log10 3

46. The remainder left out when 82n – (62)2n+1 is divided by 9 is (A) 0 (B)  2 (C) 7 (D)  8 47. If C0, C1, C2, ..., Cn denote the binomial coefficients in the expansion of (1 + x)n, then 1 + r log e 10 = (1 + log e 10 n ) r (A) 2 (B)  1 (C) 0 (D)  none of these n

r n ∑ (−1) ⋅ Cr ⋅ r=0

48. If C0, C1, C2, ..., Cn are the coefficients of the expansion Ck is 0 k +1 2 n -1 (A) 0 (B)  n 2n+1 −1 (C) (D)  none of these n +1 n

of (1 + x)n, then the value of ∑

49. The greatest coefficient in the expansion of (x + y + z + w)15 is 15! 15! (A) (B)  3!( 4 !)3 (3!)3 4 ! (C)



(D)  none of these 10

50. The sum of the series ∑ 20 Cr is r= 0 (A) 219 – 1 ⋅ 20C10 (B) 219 + 1 ⋅ 20C10 2 2 (C) 219 (D)  220 51.

C2 + 2 [2C2 + 3C2 + 4C2 + ... + nC2] =

n+1

n ( n + 1) ( 2n + 1) (A) 6 n ( n -1) ( 2n -1) (C) 6

(B)  n ( n +1) 2 (D)  none of these

52. Given positive integers r > 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 (A) n , n even (B)  2 2 (C) n (D)  1 53. Let n be a positive integer such that (1 + x + x2)n = a0 + a1x + a2x2 + ... + a2nx2n, then ar = (A) an–r, 0 ≤ r ≤ 2n (B)  a2n, 0 ≤ r ≤ 2n (C) a2n – r, 0 ≤ r ≤ 2n (D)  none of these 54. If Cr stands for nCr, then the sum of the series  n  n 2  !  !  2   2  [C02 − 2C12 + 3C22 − n! ... + (− 1) n ( n + 1) Cn2 ], where n is an even positive integer, is (A) 0 (B)  (–1)n/2 (n + 1) n/2 (C) (–1) (n + 2) (D)  (– 1)n n 55. The sum of the series 1 1.4.1 1.4.7 1 1+ 2 + + + ... is 3 1.2.34 1.2.3 36 1

 3 3 3   (A) (B)   2  2 1

 1 3 1   (C) (D)   3  3 56. The digit at unit’s place in the number 171995 + 111995 –71995 is (A) 0 (B)  1 (C) 2 (D)  3 57. The coefficient of xn in the polynomial (x + nC0 ) (x + 3 n C1) (x + 5 nC2)... (x + (2n + 1) nCn) is (A) n.2n (B)  n.2n+1 (C) (n +1).2n (D) n.2n–1 58. Let n(> 1) be a positive integer. Then, largest integer m such that (nm + 1) divides 1 + n + n2 + ... + n255 is (A) 128 (B)  63 (C) 64 (D)  32 59. The coefficient of λnµn in the expansion of [(1 + λ) (1 + µ) (λ + µ)]n is n (A) n Cr2 (B)  Cr2+2 n 3 (C) n Cr2+3 (D)  Cr

PRACTICE EXERCISES

 1 45. In a binomial distribution B n, p =  , if the prob 4 ability of at least one success is greater than equal to 9 , then n is greater than 10 1 1 (A)   (B)  4 3 4 log10 - log10 log10 + log10 3

6.20  Chapter 6 60. The sum to (n + 1) terms of the series

1001 (A) 1000C50 (B)  C50 1002 (C) C50 (D)  none of these

C C C C 0 − 1 + 2 − 3 +... is 2 3 4 5

1 1 (A) (B)  n ( n + 1) n+2 1 (D)  none of these n +1 61. Let R = (5 5 + 11)2n + 1 and f = R – [R] where [ ] denotes the greatest integer function. Then R f = (A) 22n + 1 (B)  W24n + 1 2n + 1 (C) 4 (D)  none of these (C)

62. Let n and k be positive integers such that n ≥ k ( k +1) 2 . The number of solutions (x1, x2, ..., xk), x1 ≥ 1, x2 ≥ 2, ..., xk ≥ k, all integers, satisfying x1 + x2 + ... + xk = n, is m (A) mCk – 1 (B)  Ck m (C) Ck + 1 (D)  none of these 1 where m = (2n – k2 + k – 2) 2 n

r =0

(A) 2 sin (n – 1) x (C) 2n – 1 sin nx

(B) 2 sin nx (D)  none of these

n–1

n

64. nCn + n + 1Cn + n + 2Cn + ... + n + kCn = n+k (A) n + k – 1Cn + 1 (B)  Cn + 1 n+k+1 (C) Cn + 1 (D)  none of these 65. If Sn = 1 + q + q + q + ... + q and 3

n

PRACTICE EXERCISES

2

1 1 (C) 2m 2n 1

n

      S’n = 1 +  q + 1 +  q + 1 + ... +  q + 1 , q ≠ 1, then  2   2   2  n+1 C1 + n + 1C2 ⋅ S1 + n + 1C3 ⋅ S2 + ... + n + 1Cn + 1 ⋅ Sn = (A) 2n – 1 ⋅ S’n (B)  2n ⋅ S’n n+1 (C) 2 ⋅ S’n (D)  none of these 66. If (1 + x)15 = C0 + C1 x + C2 x2 + ... + C15 x15, then the value of C2 + 2 C3 + 3 C4 + ... + 14 C15 is (A) 219923 (B) 16789 (C) 219982 (D)  none of these 67. If a0, a1, a2, ..., a2n be the coefficients in the expansion of (1 + x + x2)n in ascending powers of x, then a02 − a12 + a22 − a32 + ... − a22n−1 + a22n = (A) a2n (B)  an (C) a0 (D)  none of these 68. The coefficient of x50 in the expression (1 + x)1000 + 2x (1 + x)999 + 3x2 (1 + x)998 + ... + 1001 x1000 is

(D)  none of these

70. If (1 + x)n = C0 + C1 x + C2 x2 + ... + Cn xn, then for n even, C02 − C12 + C22 − ... + (−1) n Cn2 is equal to (A) 0 (B)  ( 1) n / 2 nCn 2 (C) n Cn 2 n

71. ∑

k =0

63. ∑ n C sin rx cos ( n − r ) x = r

2

69. The sum of the series r r r   n ∑ (−1) r ⋅ nCr  1 + 3 + 7 + 15 + ... to m terms 2 3 4 r r r r 2  r =0 2 2 2   is 1 1 1 1 mn 2mn (A) 2 (B)  n m 2 1 2 1

(D)  none of these

n

Ck = ( k + 1) ( k + 2)

2n+1 − n − 3 (A) ( n + 1) ( n + 2)

2n+ 2 − n − 3 (B)  ( n + 1) ( n + 2)

2n+ 2 − n + 3 (C) ( n + 1) ( n + 2)

(D)  none of these

72. For all n ∈ N, the integer just above ( 3 + 1)2n is divisible by (A) 2n + 1 (B)  2n + 1 n + 1 (C) 2 + 1 (D)  none of these 73. If C0, C1, C2, ..., Cn be the coefficients in the expansion of (1 + x)n, then 22 ⋅ C0 23 ⋅ C1 2 n+ 2 ⋅ Cn + + ... + is equal to 1⋅ 2 2⋅3 ( n + 1) ( n + 2) 3n+1 − 2n − 5 (A) ( n + 1) ( n + 2) n+ 2 (C) 3 + 2n − 5 ( n + 1) ( n + 2)

3n+ 2 − 2n − 5 (B)  ( n + 1) ( n + 2) (D)  none of these

74. mCr + mCr – 1 ⋅ nC1 + mCr – 2 ⋅ nC2 + ... + mC1 ⋅ nCr – 1 + nCr = m+n (A) m + nCr – 1 (B)  Cr m+n (C) Cr + 1 (D)  none of these 75. If a, b, c and d are any four consecutive coefficients of a+b b+c c+d any binomial expansion, then , , are a b c in

Binomial Theorem  6.21

76. The last two digits of the number 3400 are (A) 38 (B)  27 (C) 01 (D)  none of these C0 C1 C C − + 2 − 3 + ... to (n + 1) terms is 1.2 2.3 3.4 4.5 1 2n (A) (B)  ( n + 2) ( n + 2) 77. The sum

2 n −1 (C) ( n + 2)

(D)  none of these

78. The sum of the series (1.2) C2 + (2.3) C3 +......+ (n – 1.n) Cn is (A) n (n –1)2n –1 (B)  n (n –1)2n –2 n (C) n (n –1)2 (D)  none of these 79. If n is an even positive integer and k = 3n , then 2 k ∑ (−3) r −1 3nC2 r −1 = r =1

(A) 1 (B)  –1 (C) 0 (D)  none of these 80. The coefficient of x301 in the expansion of (1 + x)500 + x (1 + x)499 + x2 (1 + x)498 +.... .+ x500 is (A) 501C301 (B) 500C301 (C) 501C300 (D) none of these

( 6)

2n

81. The fractional part of

5

, n ∈ N is equal to

1 1 (A) (B)  3 5 1 (C) (D)  none of these 6 82. The coefficient of xn in the expansion of (x + C0) (x –3C1) (x + 5C2)..... up to (n + 1) terms, where Cr = n Cr is equal to (A) 0 (B)  1 (C) –1 (D)  none of these 83. The number of irrational terms in the expansion of

(

8 5 + 6 2

)

100

is

(A) 96 (B)  97 (C) 98 (D)  none of these 84. Let n be an odd natural number greater than 1. Then, the number of zeros at the end of the sum 99n + 1 is (A) 2 (B)  3 (C) 4 (D)  none of these 85. ∑

1 = ( 2r )!( 2n − 2r )!

(A)

22n 22 n-1 (B)  ( 2n)! ( 2n)!

(C)

22 n + 1 ( 2n)!

n

r=0

(D)  none of these

86. The coefficient of xn in polynomial (x + 2n + 1C0) (x + 2n +1C1)(x + 2n +1C2)....(x + 2n + 1Cn) is (A) 22n + 1 (B)  22n 2n – 1 (C) 2 (D)  none of these

Previous Year’s Questions 87. The coefficient of x5 in (1 + 2x + 3x2 + ...)−3/2 is: (A) 21 (B)  25[2002] (C) 26 (D)  none of these

90. The coefficient of the middle term in the binomial expansion in powers of x of (1 + αx)4 and of (1–αx)6 is the same if α equals [2004]

88. If | x | < 1, then the coefficient of xn in expansion of (1 + x + x2 + x3 + ...)2 is : [2002] (A) n (B)  n−1 (C) n + 2 (D)  n + 1

(A) −

89. The number of integral terms in the expansion of [2003] ( 3 8 5) 256 is (A) 32 (B)  33 (C) 34 (D)  35

(C)

5 3

3 (B)  5

3 10

10 (D)  3

91. The coefficient of xn in expansion of (1 + x) (1 − x)n is (A) (n – 1) (B)  (– 1)n (1 − n) [2004] n–1 2 n–1 (C) (–1) (n– 1) (D) (–1) n

PRACTICE EXERCISES

(A) A.P. (B)  G.P. (C) H.P. (D)  none of these

6.22  Chapter 6 92. If the coefficients of rth, (r + 1)th and (r + 2)th terms in the binomial expansion of (1 + y)m are in A. P., then m and r satisfy the equation [2005]

(A) m2 − m(4r − 1) + 4r2 − 2 = 0 (B) m2 − m(4r + 1) + 4r2 + 2 = 0 (C) m2 − m(4r + 1) + 4r2 − 2 = 0 (D) m2 − m(4r − 1) + 4r2 + 2 = 0 93. The value of 50C4

6

56 r

r 1

[2005]

C3 is

56 (C) 56C3 (D)  C4

1 bx

coefficient of x−7 in ax 1

1 bx2

isfy the relation (A) a − b = 1 (B) a = 1 b

a equals b 5 6 (A) (B) n 4 n 5

of 5th and 6th terms is zero, then

(C) n 5 (D) n 4 6 5

11

equals the

11

, then a and b sat[2005]

(B) a + b = 1 (D) ab = 1

1 20 (A) − 20C10 (B)  C10 2 2 0 (C) 0 (D)  C10 1 , if the proba4 bility of at least one success is greater than or equal to 9 , then n is greater than  [2008] 10

100. In a binomial distribution B n, p

(A)

1 4 3 log10 log10

(B)

1 4 3 log10 + log10

(C)

9 4 3 log10 log10

(D)

4 4 3 log10 log10

95. If x is so small that x3 and higher powers of x may be neglected, then mated as

PRACTICE EXERCISES

(A) 1 − (C)

(1

x)

3/ 2

1

1 x 2

3

may be approxi[2005]

(1 x )1/ 2

3 2 x (B) 3x + 3 x 2 8 8

3 2 x x (D) 8 2

(C)

a n+1 b

an an (B) a b b n+1 a

(D)

b n+1 b

bn  a

101. The remainder left out when 82n − (62)2n +1 is divided by 9 is [2008] (A) 0 (B) 2 (C) 7 (D) 8 102. The coefficient of x7 in the expansion of the expression (1 – x − x2 + x3)6 is [2011]

3 2 x 8

96. If the expansion in powers of x of the function 1 is a0 + a1x + a2x2 + a3x3 + … , then an is (1 ax )(1 bx ) bn (A) b

[2007]

99. The sum of the series [2007] 20 20 20 20 20 C0 − C1 + C2 − C3 + … − … + C10 is

55 (A) 55C4 (B)  C3

94. If the coefficient of x7 in ax 2

98. In the binomial expansion of (a - b)n, n ≥ 5, the sum

[2006]

a n+1 a

97. For natural numbers m, n if (1 - y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10, then (m, n) is (A) (20, 45) (B)  (35, 20) [2006] (C) (45, 35) (D)  (35, 45)

(A) –132 (B) –144 (c)132 (D) 144 103. If n is a natural number, then ( 3 1) 2 n (1) an irrational number

( 3 1) 2 n is [2012]

(2) an odd positive integer (3) an even positive integer (4) a rational number other than positive integers 104. If x = -1 and x = 2 are extreme points of [2013] f ( x ) α log x β x 2 x , then (A) α (C) α

6, 2,

1 2

(B) α

1 2

(D) α

1 2

6,

2,

1 2

Binomial Theorem  6.23 105. If the coefficients of x3 and x4 in the expansion of (1 ax bx 2 )(1 2 x )18, in powers of x, are both zero, then (a, b) is equal to [2014] (A) 16,

251 3



(B) 14, 251 3

3n 1: n

9( n 1) : n

N and Y

(x + N ,

where N is the set of natural numbers, then the set Y is equal to [2014] (A) N (B) Y -X (C) X (D) Y X

107. The sum of the coefficients of integral powers of x in

(

the binomial expansion of 1 2 x

)

50

[2015]

is:

(A) 1 (350 ) (B) 1 (350 2 2

1)

(C) 1 (250 + 1) (D) 1 (350 + 1) 2 2

(A) 221 - 210 (B)  220 - 29 20 10 (C) 2 - 2 (D)  221 - 211 109. The sum of the coefficients of all odd degree terms in the expansion of [2018]

(C) 14, 272 (D) 16, 272 3 3 n 106. If X = 4

108. The value of (21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4) + … + (21C10 – 10C10) is [2017]

x3 − 1

(A) –1 25

110. If

∑{ r =0

50

) +(x − 5

)

5

x 3 − 1 , ( x > 1)

(B) 0 Cr .50 − r C25− r

(C) 1

} =K(

(D) 2

C25), then K is equal to :

50

[2019]

 (A) (25)2 (B)  224 25 (C) 2 –1 (D)  225

111. The positive value of l for which the co-efficient of x2 10

in the expression x2  x + λ   x2   (A)  4 (B)  5

[2019]

(C)  2 2 (D)  3

ANSWER K EYS 1. (C) 2.  (B) 3.  (C) 4.  (A) 5.  (A) 6.  (A) 7.  (A) 8.  (B) 9.  (C) 10.  (C) 11.  (A) 12.  (C) 13.  (C) 14.  (D) 15.  (B) 16.  (C) 17.  (C) 18.  (A) 19.  (D) 20.  (B) 21.  (B) 22.  (D) 23.  (C) 24.  (B) 25.  (B) 26.  (B) 27.  (D) 28.  (A) 29.  (A) 30.  (A) 31.  (D) 32.  (C) 33.  (D) 34.  (C) 35.  (A) 36.  (A) 37.  (A) 38.  (A) 39.  (B) 40.  (A) 41.  (D) 42.  (A) 43.  (B) 44.  (B) 45.  (A) 46.  (B) 47.  (C) 48.  (C) 49.  (A) 50.  (B) 51.  (A) 52.  (A) 53.  (C) 54.  (C) 55.  (B) 56.  (B) 57.  (C) 58.  (A) 59.  (D) 60.  (D) 61.  (C) 62.  (A) 63.  (C) 64.  (C) 65.  (B) 66.  (A) 67.  (B) 68.  (C) 69.  (B) 70.  (B) 71.  (B) 72.  (A) 73.  (B) 74.  (B) 75.  (C) 76.  (C) 77.  (A) 78.  (B) 79.  (C) 80.  (A) 81.  (B) 82.  (A) 83.  (B) 84.  (A) 85.  (B) 86.  (B)

Previous Years’ Questions   87. (D) 88. (D) 89. (B) 90.  (C) 91. (B) 92. (C) 93. (D) 94. (D) 95.  (C) 96. (D)   97. (D) 98. (D) 99. (B) 100.  (A) 101. (B) 102. (B) 103. (A) 104. (C) 105.  (D) 106. (D) 107. (D) 108. (C) 109. (D) 110. (D) 111. (A)

PRACTICE EXERCISES

Single Option Correct Type

6.24  Chapter 6

HINTS AND EXPLANATIONS Single Option Correct Type 5. Let l + f = (6 6

= – 9 × 19 = – 171 The correct option is (C)

Now, I + f – f′ = (6 6

4n

2. We have,

n

16 2 (1 + 15) = = 15 15 15



n

8 , 17

 22003   8  =  17  17  

∴

2

(14)3

...]

 1  C3  log x +1  ( x1/12 )3 = 200  x 10 



⇒



⇒ 20. x 2(log10 x+1)



⇒



⇒



⇒ y = – 4 or y = 1 ⇒ log10 x = – 4 or log10 x = 1 ⇒ x = 10–4 or 10

6

3

+

3 2 (log10 x + 1)

1 4

= 200 ⇒ x

+

 3 1    2 (log x +1) + 4   

= 10

1 1 = logx 10 = 4 log10 x

1 3 1 + = where y = log10x y 2 ( y + 1) 4

T3 = 7 (given) T2

7. Since, n

C2 ( 2 x ) n 2 ( 4 x ) 2 =7 C1 ( 2 x ) n 1 . ( 4 x )



⇒



⇒   2

n

 n − 1 1  . x 3 = 7  (2 )

Also, nC2 + nC1 = 36

C0 + 80C2 ⋅ 22 + 80C4 ⋅ 24 + ... + 80C80 ⋅ 280 = 1 (380

80

The correct option is (A)

C3 ( 6 6 ) 2 n

⇒ x = 10  (∴ x > 1) The correct option is (B)

∴  

The correct option is (C) 4. The coefficients of the integral powers of x are 80 C0, 80C2 ⋅ 22, 80C4 ⋅ 24, ..., 80C80 ⋅ 280 Now, (1 + 2)80 = 80C0 + 80C1 ⋅ 2 + 80C2 ⋅ 22 + ... + 80C80 ⋅ 280  ...(1) and  (1 – 2)80 = 80C0 – 80C12 + 80C2 ⋅ 22 – ... + 80C80 ⋅ 280 ...(2) Adding Eq. (1) and (2), we get 380 + 1 = 2(80C0 + 80C2 ⋅ 22 + 80C4 ⋅ 24 + ... + 80C80 ⋅ 280)

2n 1

3

where k = (17)499 – 500C1(17)498 + ... + 500C499 such that k is an integer

2 n +1 – (6 6 − 14)

6. Given, T4 = 200

3. 22003 = (24)500. 23 ⇒ 22003 = 8 (16)500 ⇒ 22003 = 8 (17 – 1)500 ⇒ 22003 = 8[(17)500 – 500C1(17)499 + ... – 500C499 (17) + 1]

22003 = 8k 17

1

⇒ I + f – f′ = 2 (Integer) = even ...(2) Now, 0 ≤ f < 1 Also, 0 ≤ f – f′ < 1 ∴ 0 ≤ f – f′ < 0 ⇒ f – f′ = 0 Substituting respective values in (2), we get I = even integer The correct option is (A)

The correct option is (B)

⇒

14) 2 n

⇒ I + f – f′ = 2 [ 2 n 1 C1 (6 6 ) 2 n 141



1 + nC115 + nC2 152 + ... + nCn 15n 15 1 + 15 k = , where k ∈ N 15 1 +k = 15 24n 1 ∴  Fractional part of is . 15 15



1

2 n +1 Assuming, f′ = (6 6 − 14) ...(1)

=

HINTS AND EXPLANATIONS

14) 2 n

1. Coefficient of x17 = – (1 + 2 + 3 + ... + 18) 18 =– (1 + 18) 2

2

1)



⇒ n( n − 1) + n = 36



⇒ n2 + n – 72 = 0 ⇒ n = 8, – 9

2

...(1)

Binomial Theorem  6.25 n = – 9 is not possible as in Eq. (1), n – 1 should be positive. Substituting n = 8 in Eq. (1), we get



1 =2–1 2



⇒ 3x = – 1 ⇒  x = −

= a0 + a1x + a2x2 + ... + a5x5

1 3

n

The correct option is (A) 8. If n is odd, then numerically greatest coefficient in the n

expansion of (1 – x)n is

Cn 2

n 1

or

Cn 2

1

⇒







5 6 6 5 and x >   ⇒  x ∈  ,   6 5  5 6 2 1

2 2



=n

C1 1

n

2n

    

4x + 1 2

3   4 x + 1    + ...  2    

C3

1 (4 x 2n

n

1)

C5

1 (4x + 1)2 2n

r −1 1 2 ( 4 x + 1 ) + ... 2n The expansion contains a term x5 if r −1 = 5  or  r = 11. 2

+ ... + nCr

The correct option is (A) 2 3

n

9. C − 2C + 3C − 4C + ... + (− 1) ( n + 1)C

n−3

1 + C3    3  n

21! 21! x> (Q x > 0) 11! 10 ! 9 ! 12 !

2 0

n−1   n C  1   1  2  4 x + 1 

1

=



The correct option is (B)



Now, 

21! 21! > x and 10 ! 11! 9 ! 12 !

⇒ x

1 4 x + 1  −  −  2   2

.

Therefore in (1 – x)21, the numerically greatest coefficient is 21 C10 or 21C11. So, the numerically greatest term 21 =  C11x11 or 21C10x10 and 21 |  C10x10 | > | 21C9.x9 |



n n  1 + 4 x + 1  1 − 4 x + 1    −        2 2 4 x + 1   

1

2 2 2 2 n 2 = [C0 − C1 + C2 − C3 + ... + (− 1) Cn ]



− [C12 − 2C22 + 3C32 − ... + (− 1) n nCn2 ]





= (− 1) n / 2





n −1 n n! n Cn − (− 1) 2  n   n  2 2  !  !  2   2  n !  n n/ 2 1 +  = (− 1) n n  2  !  !   2 2

2 n

m

12.

10  20 

∑  i  m − i i=0

= Coefficient of xm in (1 + x)10 (1 + x)20 = 30Cm m = 15 (for maximum value) The correct option is (C) n  3 1  1   3 13.  x + 1 + 3  = 1 +  x + 3     x  x  

 

= n C0 + nC1  x 3 + 

n

n

 3 1  1 n  + ... + Cn  x + 3  3  x  x 

All the terms are distinct with powers (x3)0, (x3), (x3)2, ... (x3)

n  n   n  , (x3)–n,... (x3)–1. Hence, (2n+1) terms.   ∴  2  2 !  2 ! [C02 − 2C12 + 3C22 − ... The correct option is (C) n 2 n/ 2 ! n + ( − ) + n 1 ( 1 ) C ] = ( − 1 ) ( n + 2 ) n 600−r r [C02 − 2C12 + 3C22 − ... 14. (d).  t = 600 r r+1

+ (− 1) n ( n + 1) Cn2 ] = (− 1) n / 2 ( n + 2)

The correct option is (C)

As 0 ≤ r ≤ 600 and 2n

10. We have, (1 + 2x + x2)n = ⇒ (1 + x)2n =

∑a x

∑a x

r

r

r =0

2n



Cr 17

r

r

3

35 2 x

r r and 200 − are integers  ⇒  r should 3 2

be a multiple of 6 ∴  r = 0, 6, 12,... 600 The correct option is (D)

r =0

2n



⇒ 

∑ r =0

2n

Cr x r =

2n r ∑ a x ⇒ ar = 2nCr. r r=0

The correct option is (C)

5 3   3  i 4   3  i 2 5     3  5   + C  + 2 C 4 15. z =   2  16    2   2  4     2  

= Purely real number

HINTS AND EXPLANATIONS

23x =

11. Given:

6.26  Chapter 6 Hence, Im (z) = 0 The correct option is (B)

Thus, the units place digit is 1. The correct option is (B)

16. Term independent of x = 10C5 (sinα)5 (cosα)5

22. (1 + 0.0001)1000

=

= 1 + 1000 × 10–4 +

1 25

C5 sin5 2α

10

Hence, the greatest value =

1 10 ! 25 (5!) 2

The correct option is (C) 17. (1 + x)101 (1 – x + x2)100 = (1 + x) (1 + x3)100 =  (1 + x) (C0 + C1x3 + C2x6 + ... + C100x300) Clearly in this expression xλ will be present if λ = 3t, or λ = 3t + 1 So, λ can not be of the form 3t + 2. The correct option is (C) 18. Sum of last 10 coefficients, 19 C10 + 19C11 + ... + 19C19 = S (say) Also, 19C0 + 19C1 + ... + 19C9 = 19C19 + 19C18 + ... + 19C10 = S  ∴  (nCn = nCn–r) 19



∴ 2 S =

∑

19

So, the integer just greater than the given expression must be 2. The correct option is (D) 23. Given polynomial is (x + nC0) (x + 3. nC1) (x + 5. nC2) ...(x + (2n + 1). nCn) =  xn+1 + xn [nC0 + 3. nC1 + 5. nC2 + ... + (2n + 1). nCn] + xn–1 (...) + ... ∴  Coefficient of xn in the expression is n

n

∑

n ( 2r + 1) Cr = r=0

∑

n

2r n Cr +

r=0

n

=  2

∑r r=0

Cn = 219 ⇒ S = 218

n=0

HINTS AND EXPLANATIONS

1000 × 999 −8 1000 10 + C3 10−12 + ... 2 1 1 1 1 10 (n + 2) (n + 3). Hence, n ≤ 11 ∴  n can take the values 1, 2, 3,..., 11. ∴  Number of positive terms = 11 The correct option is (B) 21. We have   171995 + 111995 – 71995 =  (7 + 10)1995 + (1 + 10)1995 – 71995 = [71995 + 1995C1. 71994. 101 + 1995C2. 71993. 102 + ... + 1995C1995. 101995] + [1995C0 + 1995C1. 101 + 1995C2. 102 + ... + 1995C1995. 101995] – 71995 = [1995C1. 71994. 101 + ... + 101995] + [1995C1. 101 + ... + 1995C1995. 101995] + 1 =  a multiple of 10 + 1.

Cr

r=0

Cr −1 + 2n

The correct option is (A)

195 ( n + 3) ( n + 2) ( n + 1) − >0 n! ( n + 1)!

n

n−1

=  2n

20. xn =

∑

r=0

24. Greatest coefficient in the expansion of (1 + x)2n is 2nCn. We are given 2nCn xn is the greatest term. ∴ 2nCn–1 xn–1 < 2nCn xn and 2nCn+1 xn+1 < 2nCn xn 2n



⇒ 



⇒ 



⇒ 

Cn−1 0, we then have 

1⎞ ⎛ ⎛ 1⎞ 5 1 a = 5t + + t 2 + = ⎜t2 + 2 ⎟ + 5 ⎜t + ⎟ 2 ⎝ ⎠ ⎝ t⎠ t t t

or

1 a = ⎛⎜ t − ⎞⎟ + 2 + 5 ⎝ t⎠

2 ⎤ ⎡⎛ 1⎞ ⎢ ⎜ t − ⎟ + 2⎥ t⎠ ⎢⎣⎝ ⎥⎦

2



Solution: (A) Let d be the common difference of the A.P., then 4 = abc = (b – d)b (b + d) = b(b2 – d2) b3 = 4 + bd2 ≥ 4

⇒

b ≥ 22/3

7. There are four numbers of which the first three are in 1 G.P. whose common ratio is and the last three are 2 in A.P. If the last number is two less than the first, then the four members are 1 1 1 (A)  3, 1, , – (B)  2, 1, , 0 3 3 2 1 1 (C)  4, 1, , – (D)  None of these 4 2

= ⎛⎜ t − 1⎞⎟ + 5 ⎛ t − 1 ⎞ + 12 ≥ 12. ⎜⎝ ⎟ ⎝ t⎠ t⎠

Solution: (B) a a The numbers can be taken as a, , , a – 2. 2 4 a a By question, , , a – 2 are in A.P. 2 4

Thus, values of a are given by the inequality a ≥ 12. 5. If 1, logy x, logz y, –15logx z are in A.P., then (A) x = y–3 (B)  y = z–2 3 (C) x = z (D)  None of these

\

Solution: (C) Let d be the common difference of the A.P.

2·

a a = + a – 2; or a = 2. 4 2

Hence, the numbers are 2, 1,

1 + d

Then, logy x = 1 + d ⇒ x = y

logz y = 1 + 2d ⇒ y = z1 + 2d –(1 + 3d)/15

and –15logx z = 1 + 3d ⇒ z = x \ ⇒

x = y1 + d = z(1 + 2d) (1 + d)

⇒ 6d3 + 11d2 + 6d + 16 = 0 ⇒ (d + 2)(6d2 – d + 8) = 0 ⇒ d = –2 \ x = y1 + d = y–1, y = z1 + 2d = z–3 and x = (z–3)–1 = z3. 6. If three positive real numbers a, b, c are in A.P. such that abc = 4, then the minimum possible value of b is (A) 22/3 (B)  21/3 5/3 (C) 2 (D)  None of these

1 , 0. 2

SUM OF n TERMS OF AN A.P. The sum of n terms of an A.P. with first term ‘a’ and common difference ‘d’ is given by n Sn = [2a + (n – 1) d] 2

= x–(1 + d)(1 + 2d)(1 + 3d)/15 (1 + d)(1 + 2d)(1 + 3d) = –15

( b > 0, d2 ≥ 0)

Thus, the minimum possible value of b is 22/3.

2

2

⇒

QUICK TIPS If Sn is the sum of n terms of an A.P. whose first term is ‘a’ and last term is l, then

■

Sn =



n (a + l) 2

If common difference d, number of terms n and the last term l, are given then

■

Sn =



n [2l – (n – 1) d] 2

tn = Sn – Sn – 1. The sum of an A.P. consisting of odd number of terms = n (middle term), where n is number of terms.

■ ■

7.4  Chapter 7

PROPERTIES OF A.P. 1. 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 2. 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. a a a (d)  1 , 2 , 3 , … may not be in A.P. b1 b2 b3 3. If a1, a2, a3, …, an are in A.P., then (a)  a1 + an = a2 + an – 1 = a3 + an – 2 = … ar − k + ar + k , 0 ≤ k ≤ n – r. 2 In other words, in an A.P., the sum of two terms equidistant from the beginning and end is a constant and is equal to the sum of first term and last term 4. If nth term of a sequence is a linear expression is n then the sequence is an A.P. 5. If the sum of first n terms of a sequence is a quadratic expression in n, then the sequence is an A.P. 6. If a1, a2, a3, a4, . . . , an are in A.P., then terms taken at regular intervals from this A.P. are also in A.P. e.g., a1, a4, a7, a10, . . . also form an A.P. (b)  ar =

SOLVED EXAMPLES 8. If

14 24 34 n4 + + + ... + = 1.3 3.5 5.7 (2n − 1) (2n + 1)

1 n f ( n) + , then f (n) is equal to 48 16( 2n + 1) (A) n(4n2 + 3n + 2) (C) n(4n2 + 5n + 6)

(B)  n(4n2 + 6n + 5) (D)  None of these

Solution: (B) We have, tn =

\ Sn =



=

n2 1 1 + + 4 16 16( 4 n2 − 1)



=

4n2 + 1 1 ⎛ 1 1 ⎞ + − 16 32 ⎜⎝ 2 n − 1 2 n + 1⎟⎠

=

n =1

4 n2 + 1 1 + 16 32 n =1 n

∑

n

⎛

1

1

⎞

∑ ⎜⎝ 2n − 1 − 2n + 1⎟⎠

n =1

1 n( n + 1) (2 n + 1) 1 1 + n+ 4 6 16 32 ⎛ 1 1 1 1 1 ⎞ ⎜⎝1 − 3 + 3 − 5 + ... + 2n − 1 − 2n + 1⎟⎠



=

n 1 ⎛ 1 ⎞ 1− ( 4 n2 + 6 n + 5) + ⎜ 2n + 1⎟⎠ 48 32 ⎝



=

n( 4 n2 + 6 n + 5) n + 48 16( 2n + 1)

\ f (n) = n(4n2 + 6n + 5) 9. The maximum sum of the series 1 2 20 + 19 + 18 + 18 + … is 3 3 (A) 310 (B) 290 (C)  320 (D)  None of these Solution: (A) The given series is arithmetic whose first term = 20, common difference = – . 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. Now ⎛ 2⎞ tn = 20 + (n – 1) ⎜ − ⎟ ≥ 0 if 60 – 2 (n – 1) ≥ 0; ⎝ 3⎠ or 62 ≥ 2n or 31 ≥ n \ The first 31 terms are non-negative. \ Maximum sum = S31 =

n4 (2 n − 1) (2 n + 1)

=

n

∑ tn

=

31 2

⎡ ⎛ 2⎞ ⎤ ⎢ 2 × 20 + (31 − 1) ⎜⎝ − 3 ⎟⎠ ⎥ ⎣ ⎦

31 (40 – 20) = 310 2

10. The sum to n terms of the sequence log a, log ar, log ar2, … is n (A)  log a2 rn – 1 (B)  n log a2 rn – 1 2 (C) 

3n log a2 rn – 1 2

(D)  None of these

Sequence and Series  7.5 Solution: (A) The given sequence can be expressed as log a, (log a + log r), (log a + 2 log r) … which is clearly an A.P. whose first term is log a and common difference is log r. The nth term = log a + (n – 1) log r n Since sum to n terms = (a1 + an) 2 n \ Sn = [log a + log a + (n – 1) log r] 2 n n  = [2 log a + (n – 1) log r] = log a2 rn – 1 2 2 11. 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 (B) 2 (C)  3 (D)  None of these Solution: (B) We have, and \ 

\ First 24 terms are positive. \ Sum of the positive terms 24 ⎛ −3 ⎞ 2 × 10 + 23 × ⎟ = S24 = ⎜ ⎝ 2 7⎠ 852 69 ⎞ ⎛ = 12 ⎜ 20 − ⎟ = ⎝ 7 7⎠ 13. The minimum number of terms from the beginning of 2 1 the series 20 + 22 + 25 + …, so that the sum may 3 3 exceed 1568, is (A) 25 (B) 27 (C) 28 (D) 29 Solution: (D) 2 8 It is in A.P. for which a = 20, d = 2 = 3 3 Now, Sn > 1568 ⇒ ⇒

an = Sn – Sn – 1(1) an – 1 = Sn – 1 – Sn – 2(2) d = an – an – 1

⇒ ⇒

n ⎡ 8⎤ 40 + ( n − 1) ⎥ > 1568 2 ⎢⎣ 3⎦ n 112 + 8n > 1568 × 2 3 6 n2 + 14n > × 1568 = 1176 8 n2 + 14n – 1176 > 0,

or (n + 42) (n – 28) > 0

= (Sn – Sn – 1) – (Sn – 1 – Sn – 2) [From (1) and (2)] = Sn – 2Sn – 1 + Sn – 2.

12. The sum of positive terms of the series 4 1 10 + 9 + 9 + … is 7 7

14. If the first, second and last terms of an A.P. are a, b and 2a respectively, then its sum is ab ab (A)  (B)  2(b − a) b−a (C) 

352 437 (A)  (B)  7 7 852 (C)  (D)  None of these 7 Solution: (C) 3 Here, a = 10, d = – . 7 3 Then, tn = 10 +(n – 1) ⎛⎜ − ⎞⎟ ⎝ 7⎠ 3 tn is positive if 10 + (n – 1) ⎛⎜ − ⎞⎟ ≥ 0; ⎝ 7⎠ or 70 – 3 (n – 1) ≥ 0 or 73 ≥ 3n; or 24

As n is positive, n – 28 > 0 i.e., n > 28 \ Minimum value of n = 29.

1 ≥ n 3

3ab 2 ( b − a)

(D)  None of these

Solution: (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 a \ n–1= b−a a a+b−a b or n= +1= = b−a b−a b−a n b 3ab . \ Sum = (a1 + an) = (a + 2a) = 2 ( b − a) 2 2(b − a)

7.6  Chapter 7 15. 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 odd places, then

S1 S2

=

2n n (B)  n +1 n +1 n +1 n +1 (C)  (D)  2n n (A) 

a, a + d, a + 2d, a + 3d, a + 4d, ….., a + (n – 1) d

S2 = a + (a + 2d) + (a + 4d) + … to

n +1 terms 2

\

S1 2n = S2 n +1

16. 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 (B) 25 (C) 20 (D) 30

Here a = 1st term = 7 years, d = 3 months =

Sn = 250

\ 250 =

n 2

1⎤ ⎡ ⎢ 2 × 7 + ( n − 1) × 4 ⎥ ⎣ ⎦

n ⎛ n + 55 ⎞ ⎜ ⎟ 2⎝ 4 ⎠ ⇒ 2000 = n2 + 55n ⇒ 250 =

⇒

=



n2 + 55n – 2000 = 0

⇒ (n – 25) (n + 80) = 0 ⇒ n = 25. \ Number of members in the club = 25.

k [( 2a − d ) + kxd ] ( 2a − d ) + xd

Skx to be independent of x, 2a – d = 0 or 2a = d. Sx

18. 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 = 1 mn (m + 1) (n + 1) 4 1 (B)  mn (m + 1) (n + 1) 2 (C) mn (m + 1) (n + 1) (D)  None of these (A) 

n +1 (2a + (n – 1) d) 4

Solution: (B) n Sn =  [2a + (n – 1) d] 2

kx [ 2a + (kx − 1) d ] Skx = 2 x Sx [ 2a + ( x − 1) d ] 2

We have,

For

⎤ n +1 ⎡ ⎛ n +1 ⎞ 2a + ⎜ − 1⎟ × 2d ⎥ = ⎢ ⎝ ⎠ 2×2 ⎣ 2 ⎦ =

S

the sum to k terms of an A.P., then for k x to be indeSx pendent of x (A) a = 2d (B)  a=d (C) 2a = d (D)  None of these Solution: (C)

Solution: (A) Let the odd number of terms of an arithmetic series be

Then, n S1 = {2a + (n – 1) d} 2

17. If a is the first term, d the common difference and Sk

Solution: (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 1 year, 4

= 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

19. If the first, second and the last terms of an A.P. are a, b, c respectively, then the sum is ( a + b) ( a + c − 2b) 2 ( b − a) ( b + c ) ( a + b − 2c ) (B)  2 ( b − a) (A) 

Sequence and Series  7.7 Solution: (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 sum

( a + c ) ( b + c − 2a) 2 ( b − a) (D)  None of these (C) 

Solution: (C)

13 + 17 + 21 + … + 97

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+d ⇒ c = a + (n – 1) d ⇒ n = . d (b + c − 2a) ⇒ n= (∵ d = b – a) (b − a) 12

n

∑ a4 k +1 = 416

k =0

(a + l) = 2 [2a + ( n − 1)4d ] = 416

\ Sum of n terms Sn =

20. Four different integers form an increasing A.P. If one of these numbers is equal to the sum of the squares of the other three numbers, then the numbers are (A)  – 2, – 1, 0, 1 (B)  0, 1, 2, 3 (C)  – 1, 0, 1, 2 (D)  None of these Solution: (C) Let the numbers 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 2

2

⇒ 2d – 2d + 3a – a = 0 \

Since d is positive integer, 1 + 2a – 6a2 > 0 2 a12 + a22 + … + a17 = 140 n

⇒

17

0,

\ d = 1. Hence, the numbers are – 1, 0, 1, 2. 21. The sum of all two digit numbers which when divided by 4, yield unity as remainder, is (A) 1100 (B) 1200 (C)  1210 (D)  None of these

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.

7.8  Chapter 7

Inserting Single A.M. between Two given Numbers

SOLVED EXAMPLES

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, = 4 n1( n1 + 1)

A – a = b – A or 2A = a + b, or A =

( 2n1 + 1) 6

.

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. Now, b = (n + 2)th term of A.P. = a + (n + 2 – 1) d = a + (n + 1) d



S = n( n + 1)

d=

or

( 2n + 1) ( 2n + 1) + 4 n1( n1 + 1) 1 6 6

, where d is common difference of A.P. A = 20 × 21 ×

41 21 + 4 × 10 × 11 × = 4410 6 6

and A1 = a + d = a +

, B = 40 × 41 ×







, 

⎛ b − a⎞ An = a + nd = a + n ⎜ ⎝ n + 1 ⎟⎠



A1 + A2 + A3 + … + A2n = 2n + 1; 13n \ 2n + 1 = ; or 12n + 6 = 13n; 6 \ n = 6. \ The number of means = 2n = 2 × 6 = 12.



REMEMBER 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 ⎠⎟

QUICK TIPS Sum to n terms of the series of the form 1 1 1 is + + ... + t1t2 ... t k t2t3 ... t k +1 t n t n +1 ... t n + k −1

■

Sn =

Solution: (A) Let 2n arithmetic means be A1, A2, A3, …, A2n between a and b. a+b Then, A1 + A2 + A3 + … + A2n = × 2n 2 13 13n = 6 × 2n = 2 6 Given:

81 41 + 4 × 20 × 21 × = 22140 + 11480 6 6

A2 = a + 2d = a + 2



1 , an even 6 number of arithmetic means are inserted. If the sum of these means exceeds their number by unity, then the number of means are (A) 12 (B) 10 (C)  8 (D)  None of these 23. Between two numbers whose sum is 2

1 ⎛ ⎞ 1 1 − ⎜ (k − 1)(t2 − t1) ⎝ t1t2 ... t k −1 t n +1 t n +2 ... t n + k −1 ⎟⎠

■ Sum to n terms of the series of the form (t1t2 … tk) + (t2t3 … tk+1) + … + (tn tn+1 … tn+k–1) is 1 Sn = (tn tn+1 … tn+k – t0 t1 t2 … tk) (k + 1)(t2 − t1)

24. If a, b, c are in A.P. and p is the A.M. between a and b and q is the A.M. between b and c, then (A) a is the A.M. between p and q (B) b is the A.M. between p and q (C) c is the A.M. between p and q (D)  None of these Solution: (B)  a, b, c are in A.P., \ 2b = a + c

(1)

 p is the A.M. between a and b a+b \ p= (2) 2  q is the A.M. between b and c b+c \ q= (3) 2 Adding Eq. (2) and (3) a+b b+c a + c + 2b + p + q = = 2 2 2 2b + 2b = = 2b [Using (1)] 2 p+q \ 2b = p + q or b = 2 Hence, b is the A.M. between p and q.

Sequence and Series  7.9

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. 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, …, 2 4 8 = = = … = 2, which is a constant. 1 2 4 Thus, the sequence is a G.P. whose first term is 1 and the common ratio is 2.

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 = arn – 1

I M P O R TA N T P O I N T S 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, ar2, …, arn – 1, … (a ≠ 0) ■ 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 arm – n. ■ The nth term from the end of a G.P. with last term l and l common ratio r is ( n −1) . r a ■ Three numbers in G.P. can be taken as , a, ar; r Four numbers in G.P. can be taken as ■

a a , , ar, ar3; r3 r



Five numbers in G.P. can be taken as a a , , a, ar, ar2 r2 r



Three numbers a, b, c are in G.P. if and only if

■



b c = i.e. if and only if b2 = ac. a b

SOLVED EXAMPLES 25. 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 (B) – 1 (C)  0 (D)  None of these

Solution: (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 = log (ay – z . bz – x . cx –y) = log [(ARx – 1) y – z × (ARy – 1)z – x . (ARz – 1)x – y] = log [Ay – z + z – x + x – y . R(x – 1) (y – z) + (y – 1) (z – x) + (z – 1) (x – y) ] = log (A0 × R0) = log 1 = 0. 26. 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 (B) 2 (C)  – 1 (D)  None of these Solution: (A) Let d be the common difference of A.P. and R (≠ 0), the common ratio of G.P., then 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 = x– d . (xR)2d . (xR2)– d

= (x– d . x2d . x– d) (R2d × R– 2d )



= (x– d + 2d – d ) . (R2d – 2d )



= x0 . R0 = 1 × 1 = 1.

27. 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 (B)  (ab)1/2 1/4 (C) (ab) (D)  None of these Solution: (B) Let the G.P. be x, xy, xy2, … Then and

tp + q = xy p + q – 1 = a p–q–1

tp – q = xy 

Dividing Eqs (1) by (2), a y2q = ; b

= b

(1) (2)

7.10  Chapter 7 1

\

⎛ a ⎞ 2q y= ⎜ ⎟ ⎝ b⎠



\

From Eq. (1), ⎛ b⎞ x = a ⎜ ⎟ ⎝ a⎠ \



p + q −1 2q

or

⎛ b⎞ tp = xy p – 1 = a  . ⎜ ⎟ ⎝ a⎠ = =

p + q −1 p −1 1− + 2q 2q

a

1 1 a2b2

=

p + q −1 2q

×b

⎛ a⎞ ⎜⎝ b ⎟⎠

p + q −1 p −1 − 2q 2q

30. 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 consecutive terms of a G.P. The three numbers are (A)  3, 8, 14 (B)  2, 9, 14 (C)  5, 8, 12 (D)  None of these

p −1 2q



Solution: (C)

ab

Solution: (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 or

a = b + 12

(1)

a, b, b + 6 are in G.P. i.e., b2 = a (b + 6) b2 = (b + 12) (b + 6)

( a = b + 12)

or 18b = – 72 \ b = –4, From Eq. (1),

a = – 4 + 12 = 8.

Hence, the four numbers are 8, –4, 2 and 8. 29. If the first term of an infinite G.P. is 1 and each term is twice the sum of the succeeding terms, then the common ratio is 1 2 (A)  (B)  3 3 3 (C)  (D)  None of these 4 Solution: (A) Let the G.P. be 1, r, r2, r3, … Given:

a + b + c = 25

We have,

28. 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 (B)  1, 7, 13, 19 (C)  8, – 4, 2, 8 (D)  None of these

Given,

t n+1 ,( common ratio = r) 1− r t 1− r = n +1 ; or = r 2 tn 1 1 – r = 2r ; \ r = . 3 tn = 2 . 

or

tn = 2 (tn + 1 + tn + 2 + tn + 3 + … to ∞)

(1)

 2, a, b are in A.P., \ 2a = 2 + b

(2)

c2 = 18b(3)

 b, c 18 are in G.P., \ From Eq. (1) and (2),

3b + 2c = 48; or 3b = 48 – 2c \ From Eq. (3), c2 = 6 (48 – 2c) = 288 – 12c c2 + 12c – 288 = 0;

or

or c2 + 24c – 12c – 288 = 0 or (c + 24) (c – 12) = 0; \

c = 12, as c ≠ – 24.

\ From Eq. (3), b = 8 and from (2), a = 5. 31. Let an be the nth term of the G.P. of positive numbers. 100

Let

∑ a2n

= a and

n =1

100

∑ a2n−1

= b, such that a ≠ b, then

n =1

the common ratio is α β (A)  (B)  β α (C) 

α β (D)  β α

Solution: (A) Let a be the first term and r, the common ratio of the given G.P. Then

a=

100

∑ a2n n =1

⇒ a = a2 + a4 + … + a200

Sequence and Series  7.11 ⇒ ⇒ and

a = ar + ar3 + … + ar199 2

4

Sum of an Infinite G.P. when | r | < 1 198

a = ar (1 + r + r + … + r b=

100

∑ a2n−1 n =1

)(1)

⇒ b = a1 + a3 + … + a199

⇒

b = a + ar2 + … + ar198

⇒

b = a (1 + r2 + … + r198)(2)

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

1 1⎞ ⎛ 1 32. If a, b, c are in G.P., then a2 b2 c2 ⎜ 3 + 3 + 3 ⎟ = ⎝a b c ⎠ (A) a + b + c (B)  ab + ac + bc (C) a3 + b3 + c3 (D)  None of these

(1)

1 1⎞ ⎛ 1 Now, a2 b2 c2 ⎜ + + ⎟ 3 3 ⎝a b c3 ⎠



ac ⋅ c 2 (b 2 )2 a2 ⋅ ac + +  a b c

[Using (1)]

= a3 + b3 + c3.

Sum of n terms of a G.P. The sum of first n terms of a G.P. with first term a and common ratio r(≠ 1) is given by, Sn =

a ( r n − 1) a(1 − r n ) = r −1 1− r

SOLVED EXAMPLES 33. The sum Sn to n terms of the series 1 3 7 15 + + + + ... is equal to 2 4 8 16 (A) 2n – n – 1 (B)  1 – 2– n –n (C) 2 + n – 1 (D)  2n – 1 Solution: (C) We have, 1⎞ ⎛ 1⎞ ⎛ 1⎞ 1⎞ ⎛ ⎛ Sn = ⎜1 − ⎟ + ⎜ 1 − ⎟ + ⎜ 1 − 3 ⎟ + ... + ⎜1 − n ⎟ ⎝ ⎝ 2⎠ ⎝ 4⎠ ⎝ 2 ⎠ 2 ⎠ 1⎞ ⎛1 1 = n−⎜ + + ... + n ⎟ ⎝ 2 22 2 ⎠

2 2 2 2 2 2 = b c +a c +a b a b c

=

ERROR CHECK If r ≥ 1, then S∞ does not exist.

α = r. β

Solution: (C) Since a, b, c are in G.P. b c \ = ⇒ b2 = ac a b

The sum of an infinite G.P. with first term a and common ratio r is a S∞ = ; when | r | < 1, i.e., – 1 < r < 1. 1− r

=

n−

When r = 1,

n⎤

⎥ ⎥⎦ = n – 1 + 2–n

1 2 35. The minimum number of terms of the series 1 + 3 + 9 + 27 + … so that the sum may exceed 1000, is (A) 7 (B) 5 (C)  3 (D)  None of these 1−

Solution: (A) Let, the sum of n terms exceeds 1000. Then

QUICK TIPS

1 ⎡ ⎛ 1⎞ ⎢1 − 2 ⎢⎣ ⎜⎝ 2 ⎟⎠

1(1 − 3n ) 3n − 1 > 1000; or > 1000 1− 3 2

or 3n > 2001; but 36 = 729 and 37 = 2187;

■

Sn = a + a + a + … upto n terms = na. If l is the last term of the G.P., then

■

lr − a Sn = r≠1 r −1

\ n > 6; but n is a positive integer; \ n = 7, 8, 9, … \ The minimum number of terms = 7.

7.12  Chapter 7 x

36. The sum of series

1− x2 infinite terms, if | x | < 1 is

+

1− x4

x4 1 − x8

+ ... to

Solution: (A) n

n + 2 ⎛ 1⎞ 2( n + 1) − n ⎛ 1 ⎞ ⋅⎜ ⎟ = We have, tn = ⎜ ⎟ n( n + 1) ⎝ 2 ⎠ n ( n + 1) ⎝ 2 ⎠

1 x (B)  1− x 1− x



1 x (D)  1+ x 1+ x

\

(A)  (C) 

x2

+

=

1 ⎛ 1⎞ ⎜ ⎟ n ⎝ 2⎠

Sn =

∑ tn

Solution: (B) We have, tn = =



Therefore, Sn = 

x

2n − 1

1− x2

n −1

1 1− x

2n − 1

−

(1 + x

2n − 1

−1 n −1

) (1 − x 2 )

∑ tn

n =1



1



⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ = + − ⎢⎜ 1 − x − 2⎟ 2 ⎜ 1− x ⎠ ⎝ 1 − x 1 − x 4 ⎟⎠ ⎣⎝ 1 ⎞⎤ 1 ⎛ + ... + − n −1 n ⎟⎥ ⎜ ⎝1 − x2 1 − x 2 ⎠ ⎥⎦ 1 1 − = 1 − x 1 − x2 n

lim S n = = n→∞



1 −1 1− x

⎛ ⎞ x 2 = 0, as | x | < 1⎟ ⎜⎝∵ nlim ⎠ →∞

3

3 ⋅ 1 + 4 ⎛⎜ 1 ⎞⎟ + 5 ⎛⎜ 1 ⎞⎟ + ... is 1.2 2 2.3 ⎝ 2 ⎠ 3.4 ⎝ 2 ⎠ (A) 1 − (C) 1 −

( n + 1)2 n



1 ( n − 1)2 n − 1

= 1−



1 ( n + 1)2 n

38. If (1 – y) (1 + 2x + 4x2 + 8x3 + 16x4 + 32x5) = 1 – y6, (y ≠ 1), then a value of y/x is 1 (A)  (B)  2 2 1 (C)  (D)  4 4 Solution: (B) We can rewrite the given expression as

one of the possible values of y is clearly 2x. Therefore, one of the possible values of y/x is 2.

(B)  1−

⎛1 1 1 ⎞ log 5 ⎜ + + + ...∞⎟ ⎝ 4 8 16 ⎠

is

(A) 1 (B) 2 1 (C)  (D)  4 2

37. The sum to n terms of the series

1

n ⎡1 ⎛ 1⎞ n −1 1 ⎛ 1⎞ ⎤ + ... + ⎢ ⎜ ⎟ − ⎜ ⎟ ⎥ n + 1 ⎝ 2 ⎠ ⎥⎦ ⎢⎣ n ⎝ 2 ⎠

39. The value of (0.2)

x = 1− x

2



6 (1 – y) [1 − ( 2 x ) ] = 1 – y6, 1 − 2x

n



1 ⎛ 1⎞ ⎜ ⎟ n + 1 ⎝ 2⎠

⎡1 ⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 ⎤ ⎡ 1 ⎛ 1 ⎞ 1 1 ⎛ 1 ⎞ 2 ⎤ = ⎢ ⎜ ⎟ − ⎜ ⎟ ⎥+⎢ ⎜ ⎟ − ⎜ ⎟ ⎥ 3 ⎝ 2 ⎠ ⎥⎦ ⎢⎣1 ⎝ 2 ⎠ 2 ⎝ 2 ⎠ ⎥⎦ ⎢⎣ 2 ⎝ 2 ⎠



\ The sum to infinite terms

n =1

n

1− x2

n

−

n



n



1+ x

=

2n − 1

n −1

n

Solution: (D) 1

We have,

( n + 1)2n −1

(D)  None of these

\

1 1 1 + + + ...∞ = 4 8 16

(0.2)

⎛1 1 1 ⎞ log 5 ⎜ + + + ...∞⎟ ⎝ 4 8 16 ⎠

1 4

1 1− 2

⎛ 1⎞ = ⎜ ⎟ ⎝ 5⎠

=

1 = 2–1. 2

log 5 2−1

Sequence and Series  7.13 −1

= (5−1 )1/ 2

= 52 log



5

2

⎛ A − 1⎞ ⎛ B − 1⎞ \ a log r = log ⎜ and b log r = log ⎜ ⎝ A ⎟⎠ ⎝ B ⎟⎠

log5 2

2

= 5log5 ( 2 ) = 22 = 4 \

40. i – 2 – 3i + 4 … to 100 terms = (A)  50 (1 – i) (B) 25i (C)  25 (1 + i) (D)  100 (1 – i)

S = i + 2i2 + 3i3 +, …, + 100i100 S . i = i2 + 2i3 +, …, + 99i100 + 100i101



\ S(1 – i) = (i + i2 + i3 +, …, + 100 terms) – 100i101 =

B

S=

− 100i − 100 i (1 + i ) = = 50 (1 – i) 1− i 2

41. The largest value of the positive integer k for which nk + 1 divides 1 + n + n2 + … + n127 is divisible by (A) 8 (B) 16 (C) 32 (D) 64 Solution: (D) We have, 1 + n + n2 + … + n127 =

n128 − 1 n −1

( n64 − 1) ( n64 + 1) n −1 2 = (1 + n + n + … + n63) (n64 + 1)

=



\ k = 64 which is divisible by 8, 16, 32 and 64. 42. If A = 1+ ra + r2a + r3a + … ∞ and a If B = 1 + rb + r2b + r3b + … ∞, then is equal to b (A) logB A (B)  log1 – B (1 – A) ⎛ A − 1⎞ (C) log B − 1 ⎜ ⎝ A ⎟⎠

(D)  None of these

B

Solution:(C) A= B =

1 1− r

a

1 1− r b

⇒ 1 – ra =

1 1 A −1 ⇒ ra = 1 – = A A A

⇒ 1 – rb =

1 1 B −1 ⇒ rb = 1 – = B B B



43. Sum of the series : (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + … upto n terms is 1 1− x

⎡ x 2 (1 − x n ) ⎤ ⎢n − ⎥ 1− x ⎦ ⎣

1 1− x

⎡ x 3 (1 − x n ) ⎤ n − ⎢ ⎥ 1− x ⎦ ⎣

(A) 

i (1 − i100 ) i (1 − 1) − 100 i = − 100 i = – 100 i 1− i 1− i

\

⎛ A − 1⎞ log ⎜ ⎝ A ⎟⎠ ⎛ B − 1⎞ log ⎜ ⎝ B ⎟⎠

⎛ A − 1⎞ = log B − 1 ⎜ ⎝ A ⎟⎠



Solution: (A)

a = b

(B) 

1 ⎡ x(1 − x n ) ⎤ ⎢n − ⎥ 1− x ⎣ 1− x ⎦ (D)  None of these

(C) 



Solution: (A) We have, (1 + x) + (1 + x + x2) + (1 + x + x2 + x3)  + … upto n terms 1 − x 2 1 − x3 1 − x 4 = + + + … to n terms 1− x 1− x 1− x 1 = [(1 + 1 + 1 + … n terms) 1− x  – (x2 + x3 + x4 + … to n terms)] 1 ⎡ x 2 (1 − x n ) ⎤ = ⎢n − ⎥ 1 − x ⎢⎣ 1 − x ⎥⎦

Properties of G.P. 1. If a1, a2, a3, … are in G.P., then (a) a1k, a2k, a3k, … are also in G.P. a a a (b) 1 , 2 , 3 , … are also in G.P. k k k 1 1 1 (c) , , , … are also in G.P. a1 a2 a3 (d) ak1, ak2, a3k … are also in G.P. 2. If a1, a2, a3, … and b1, b2, b3, … are two G.P.s, then (a) a1b1, a2b2, a3b3, … are also in G.P. a a a (b) 1 , 2 , 3 … are also in G.P. b1 b2 b3 (c) a1 ± b1, a2 ± b2, a3 ± b3, … may not be in G.P.

7.14  Chapter 7 3. If a1, a2, a3, …, an are in G.P., then (a) a1 an = a2 an – 1 = a3 an – 2 = . . .

or (r – 4) (r2 + 5r + 21) = 0;

(b) ar =

\

4. 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. 5. If a1, a2, a3, …, an–2, an–1, an are in G.P., then a1an = (a2an–1) = (a3an–2) = … = aran–(r–1) In a G.P., the product of two terms equidistant from the beginning and end is a constant and is equal to the product of first term and last term. 6. If first term of a G.P. of n terms is a and last term is l, then the product of all terms of the G.P. is (al)n/2. 7. If there be n quantities in G.P. whose common ratio is r and Sm denotes the sum of the first m terms, then the r sum of their product taken two by two is Sn Sn–1. r +1 8. If a x , a x , a x , ..., a x are in G.P., then x1, x2, x3, …, xn will be in A.P. 9. Product of a G.P.  Case I: If number of terms is odd, then Product of terms = (middle term)No. of terms Case II: If number of terms is even, then Product of terms = (Geometric mean of middle terms)No. of 1

−5 ± 25 − 84 2 \ Real value of r is 4. Putting this value in (2),

ar − k ar + k , 0 ≤ k ≤ n – r.

2

3

n

terms

r = 4,

3 × 4 = 3. 4 \ The three numbers are, 3, 3 × 4, 3 × 42, i.e., 3, 12, 48. a=

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.

n-Geometric Means The numbers G1, G2, …, Gn are said to be the n geometric means between two given positive numbers a and b if a, G1, G2, …, Gn , b are in G.P. For example, since 1, 2, 4, 8, 16 are in G.P., therefore 2, 4, 8 are three geometric means between 1 and 16.

Inserting Single G.M. between Two given Numbers

REMEMBER

Let a and b be two given positive numbers and G be the G.M. between them. Then a, G, b are in G.P. Thus

Equal non zero numbers are in G.P.

G b = or G2 = ab, or G = a G

ab (∵ G > 0)

SOLVED EXAMPLE



44. If the sum of three numbers in G.P. is 63 and the prod3 uct of the first and the second term is of the third 4 term, then the numbers are

Inserting n-Geometric Means between Two given Numbers

(A)  3, 12, 48 (C)  2, 10, 50

(B)  4, 12, 36 (D)  None of these

Solution: (A) Let the three numbers be a, ar, ar2 Given and

a + ar + ar2 = 63 3 3 a . ar =  . ar2 or a = r 4 4

Putting in (1), or

3 3 . 3 . 2 r+ r   r + r   r = 63 4 4 4

r3 + r2 + r – 84 = 0

Let G1, G2, G3, …, Gn be the n geometric means between two given numbers a and b. Then, a, G1, G2, G3, …, Gn, b are in G.P. Now,

b = (n + 2)th term of G.P. = arn + 1, where r is the common ratio

(1) (2)

1

or r

n+1

b ⎛ b ⎞ n +1 = or r = ⎜ ⎟ ⎝ a⎠ a 1

and

⎛ b ⎞ n +1 G1 = ar = a ⎜ ⎟ ⎝ a⎠



Sequence and Series  7.15 2

⎛ b ⎞ n +1 G2 = ar = a ⎜ ⎟ ⎝ a⎠

i i i

For example, 0. 234 =

2











i

If R = 0.X Y and x denotes the number of digits in X and y denotes the number of digits in Y, then

■

2

⎛ b ⎞ n +1 Gn = ar = a ⎜ ⎟ ⎝ a⎠ n



R=





234 234 26 = = 111 103 − 1 999

XY − X 10 x + y − 10 x i i

For example, if R = 0.4362, then

QUICK TIPS

R=

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. If A and G are respectively arithmetic and geometric means between two positive numbers a and b then

■

A > G

(A)

(B)  the quadratic equation having a, b as its roots is x2 – 2Ax + G2 = 0 A2 − G 2 .

(C)  the two positive numbers are A ±

If number of terms of any A.P./G.P. is odd, then A.M./G.M. of first and last terms is middle term of series. ■ If number of terms of any A.P./G.P. is even, then A.M./G.M. of middle two terms is A.M./G.M./H.M. of first and last terms respectively. th th ■ If p , q and rth terms of a G.P. are in G.P., then p, q, r are in A.P. ■ If a, b, c are in A.P. as well as in G.P. then a = b = c. a b c ■ If a, b, c are in A.P., then x , x , x will be in G.P. (x ≠ ±1) ■ Value of recurring decimal ■

i

If R = 0.bbb … = 0. b, then R =

■

i

For example, 0.5 =

b 10′ − 1

5 5 = 10′ − 1 9 i i

If R = 0.ab ab ab … = 0. ab, then R =

■

i i

For example, 0. 37 =

37 37 = . 102 − 1 99

abc If R = 0.abc abc abc … , 0. abc then R = and so 3 −1 10 on.

■

i i i

SOLVED EXAMPLES 45. If two geometric means g1 and g2 and one arithmetic mean A be inserted between two numbers, then g2 g2 1 + 2 = g2 g1 (A) 4A (B)  3A (C) 2A (D)  A Solution: (C) Let the two numbers be a and b. a+b or 2A = a + b 2 Again, a, g1, g2, b are in G.P. g1 g b \ = 2 = g1 g2 a 2 g1 g g Now = 2 ⇒ 1 = a g1 a g2 \

and \

ab 102 − 1

4319 4362 − 43 = 9900 104 − 102

A=

(1)

g2 b g2 = ⇒ 2 = b g1 g2 g1 a+b=

g12 g2 + 2 (2) g2 g1

\ From Eqs (1) and (2), we get 2A =

g12 g2 + 2 g2 g1



46. Let a = 1 . 2 . 3 . 4 . 5. Then (A) 55 ≥ a (B)  35 ≥ 5! 5 (C) 5 ≥ 6a (D)  None of these

7.16  Chapter 7 Solution: (A, B) Since A.M. > G.M. 1+ 2 + 3 + 4 + 5 ≥ 5

⇒

⇒ 3 ≥

5

5

Solution: (B) The required sum = (1 + 2 + 3 + … + 199) – (3 + 6 + 9 + … + 198)

1⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5

 

5

a ⇒ 3 ≥ a = 5!

Also, 5 ≥ 1 . 2 . 3 . 4 . 5 = a. 5

47. 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 Solution: (A) Since A.M. ≥ G.M.

≥

Let

3

alog b − log c ⋅ blog c − log a ⋅ c log a − log b (1)

x = alog b – log c . blog c – log a . clog a – log b

⇒ log x = (log b – log c) log a + (log c – log a) log b 

+ (log a – log b) log c = 0 ⇒ x = 1.



SOME SPECIAL SEQUENCES 1. The sum of first n natural numbers n ( n +1) . 2 2. The sum of squares of first n natural numbers Sn = 1 + 2 + 3 + … + n =

n ( n + 1) ( 2n + 1) 6 3. The sum of cubes of the first n natural numbers Sn2 = 12 + 22 + 32 + … + n2 = .

2

⎡ n ( n + 1) ⎤ Sn = 1 + 2 + 3 + … + n = ⎢ ⎥ . ⎣ 2 ⎦ 3

Solution :(C) We have, tn = [nth term of 1, 2, 3, …] ×  [nth term of 3, 5, 7, …]2

3

= n (2n + 1)2 = 4n3 + 4n2 + n.

\ Sn = S tn = 4 S n3 + 4 S n2 + S n 2

alog b – log c + blog c – log a + clog a – log b ≥ 3.

3

= 10732



\ From (1),

3

= 199 × 100 – 33 × 201 – 39 × 100 + 13 × 105 49. Sum to 20 terms of the series 1.32 + 2.52 + 3.72 + … is (A) 178090 (B) 168090 (C)  188090 (D)  None of these

alog b − log c + blog c − log a + c log a − log b 3

\

– (5 + 10 + 15 + … + 195) + (15 + 30 + 45 + … + 195) 199 66 39 = (1 + 199) – (3 + 198) – (5 + 195) 2 2 2 13 + (15 + 195) 2

3

SOLVED EXAMPLES 48. The sum of all natural numbers less than 200, that are divisible neither by 3 nor by 5, is (A) 10730 (B) 10732 (C)  15375 (D)  None of these

n ( n + 1) ( 2n + 1) n ( n + 1) ⎡ n ( n + 1) ⎤ + 4⋅ + = 4 .  ⎢ ⎥ 6 2 ⎣ 2 ⎦

2 = n2 (n + 1)2 + n (n + 1) (2n + 1) 1 3 + n (n + 1); 2 2 \ S20 = 202 . 212 + × 20 . 21 . 41  3 1 +  . 20 . 21 = 188090 2 50. Number of terms in the sequence 1, 3, 6, 10, 15, …, 5050 is (A) 50 (B) 75 (C) 100 (D) 125 Solution: (C) Let S = 1 + 3 + 6 + 10 + 15 +, …, + tn(1) then  S = 1 + 3 + 6 + 10 +, …, + tn – 1 + tn (1) – (2) ⇒ 0 = (1 + 2 + 3 + 4 + … to n terms) – tn n ( n +1) 2 n ( n +1) Given, 5050 = ⇒ n2 + n – 10100 = 0 2 ⇒

tn =

(2)

Sequence and Series  7.17

⇒

n=



−1 ± 40401 = 2



=

\

QUICK TIPS

−1 ± 201 = – 101,100 2

n = 100.

n

51.

−1 ± 1 + 40400 2

( n is a positive integer)

j

i

∑∑∑ 1 =

SOLVED EXAMPLES

i =1 j =1 k =1

(A) 

n ( n + 1) ( 2n + 1) 6

(B) 

2

⎛ n ( n + 1) ⎞ (C)  ⎜ ⎝ 2 ⎟⎠

n ( n +1) 2

n ( n + 1) ( n + 2) (D)  6

Solution: (D) We have, n



i

j

n

i =1 j =1 k =1

j

i =1 j =1

n ⎤ n i (i + 1) 1 ⎡ n = ∑ 2 = 2 ⎢∑ i 2 + ∑ i ⎥ i =1 ⎦ i =1 ⎣ i =1



1 ⎡ n ( n + 1) ( 2n + 1) n ( n + 1) ⎤ + = ⎢ 2 ⎣ 6 2 ⎥⎦



=



n ( n + 1) ( n + 2) 6

n ( n +1) n ( n + 1) ( 2n + 1) (B)  2 6 n ( n + 1) ( 2n + 1) (C) – (D)  None of these 6 (A) –

Solution: (C) We have, (1 – 1 + 2 – 2 + 3 – 3 + … + n – n)2 = 12 + 12 + 22 + 22 + … + n2 + n2 + 2S,

where S is the required sum ⇒ 0 = 2 (12 + 22 + … + n2) + 2S S = – (12 + 22 + … + n2) = –

n ( n + 1) ( 2n + 1) 6

=

\

∞

∑ tr

5r − ( r − 1) r

5 ⋅ r ( r − 1)

=

1 5

r −1

( r − 1)

−

1 r

5 ⋅r

=

r=2

 

52. The sum of the products of the 2n numbers ± 1, ± 2, ± 3, …, ± 2n taking two at a time is

⇒

9 13 17 + 3 + 4 + ... to 53. The sum of the series 2 5 .2.1 5 .3.2 5 .4.3 infinite terms, is 2 1 (A)  (B)  5 5 (C)  1 (D)  None of these Solution: (B) The general term of the series is 4r + 1 tr = r , where r ≥ z 5 ⋅ r ( r − 1)

i

∑∑∑ 1 = ∑∑



If nth term of a sequence is Tn = an3 + bn2 + cn + d, then the sum of n terms is given by, Sn = STn = aSn3 + bSn2 + cSn + Sd, which can be evaluated using the above results.



1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 = ⎜ 1 − 2 ⎟ +⎜ 2 − 3 ⎟ +⎜ 3 − 4 ⎟ ⎝ 5 .1 5 .2 ⎠ ⎝ 5 .2 5 .3 ⎠ ⎝ 5 .3 5 .4 ⎠ + … to infinity 1 = ( terms tend to zero as n → ∞) 5

54. For any odd integer n ≥ 1, n3 – (n – 1)3 + … + (–1)n–113 = 1 (A)  (n – 1)2 (2n – 1) 2 1 (B)  (n – 1)2 (2n – 1) 4 1 (C)  (n + 1)2 (2n – 1) 2 1 (D)  (n + 1)2 (2n – 1) 4 Solution: (D) Since n is an odd integer (–1)n–1 = 1 and n – 1, n – 3, n – 5 etc., are even integers. We have

7.18  Chapter 7 n3 – (n – 1)3 + (n – 2)3 – (n – 3)3 + … + (–1)n–113

= n3 + (n – 1)3 + (n – 2)3 + … + 13 – 2[(n – 1)3 + (n – 3)3 + … + 23]



= n3 + (n – 1)3 + (n – 2)3 + … + 13

⎡⎛ n − 1⎞ 3 ⎛ n − 3 ⎞ 3 ⎤ 3  – 2 × 2 ⎝ 2 ⎟⎠ + ⎜⎝ 2 ⎟⎠ + ... + 1 ⎥ ⎢⎣ ⎥⎦ [ n –1, n – 3 are even integers] 3 ⎢⎜

2



⎡ 1 ⎛ n − 1⎞ ⎛ n − 1 ⎞ ⎤ ⎡ n( n + 1) ⎤ − 16 ⎢ ⎜ + 1⎟ ⎥ = ⎢ ⎟⎜ ⎥ ⎠⎦ ⎣ 2 ⎦ ⎣2 ⎝ 2 ⎠ ⎝ 2 2

1 2 ( n − 1) ( n + 1) n (n + 1)2 – 16 4 16 × 4

2

2



=



1 1 = (n + 1)2 [n2 – (n – 1)2] = (n + 1)2(2n – 1). 4 4

ARITHMETICO-GEOMETRIC PROGRESSION (A.G.P.) If a1, a2, a3, …, an, … is an A.P. and b1, b2, …, bn, … is G.P. then the sequence a1b1, a2b2, a3b3, …, anbn, … is said to be an arithmetico-geometric sequence. Thus, the general form of an arithmetico geometric sequence is a, (a + d) r, (a + 2d) r2, (a + 3d) r3, …

nth term of A.G.P. th

From the symmetry we obtain that the n term of this sequence is [a + (n – 1)d] rn–1. Also, let a, (a + d) r, (a + 2d) r2, (a + 3d) r3, … be an arithmetico-geometric sequence. Then, a + (a + d) r + (a + 2d) r2 + (a + 3d) r3 + … is an arithmetico-geometric series.

Sum of A.G.P. 1. Sum to n terms: The sum of n terms of an arithmetico-geometric sequence a, (a + d) r, (a + 2d) r2, (a + 3d) r3, … is given by, ⎧ a (1 − r n −1 ) [a + ( n − 1)d ]r n + dr − , when r ≠ 1 ⎪ 1− r (1 − r )2 Sn = ⎪⎨1 − r ⎪n ⎪⎩ 2 [2a + ( n − 1)d ], when r = 1 2. Sum to infinite terms: Let | r | < 1. Then rn, rn–1 → 0 as n → ∞ and it can also be shown that n . rn → 0 as a dr + n → ∞. So, we obtain that Sn → , as 1 − r (1 − r )2 n → ∞.

In other words, when | r | < 1 the sum to infinity of an a dr + arithmetico-geometric series is S∞ = . 1 − r (1 − r ) 2

METHOD FOR FINDING SUM OF A.G. SERIES 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–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 find the sum of the n terms of the sequence as S=

n

∑ ak

k =1

SOLVED EXAMPLES 55. Find the sum to n terms of the series: (A) 2 + 5 + 10 + 17 + … (B) 3 + 5 + 9 + 17 + … Solution (A) Here, the difference in consecutive terms are 3, 5, 7, … which are in A.P. Let Sn = 2 + 5 + 10 + 17 + … + tn–1 + tn Shifting every term one place to the right Sn = 2 + 5 + 10 + … + tn–1 + tn Subtracting, we get 0 = (2 + 3 + 5 + 7 + … to n terms) – tn ⇒ tn = 2 + [3 + 5 + 7 + … to (n – 1) terms] n −1 =2+ [2 × 3 + (n – 2) × 2] 2 n −1 =2+ (2n + 2) = 2 + (n – 1) (n + 1) 2 = 2 + (n2 – 1) = n2 + 1. Putting n = 1, 2, 3, … , n and adding, we get n n( )( ) Sn = ∑ k 2 + n = n + 1 2n + 1 + n 6 k =1 n n = (2n2 + 3n + 1 + 6) = (2n2 + 3n + 7) 6 6 (B) Here the differences of consecutive terms are 2, 4, 8, … which are in G.P. Let Sn = 3 + 5 + 9 + 17 + … + tn–1 + tn

Sequence and Series  7.19 Shifting every term one place to the right Sn = 3 + 5 + 9 + … + tn–1 + tn Subtracting, we get 0 = (3 + 2 + 4 + 8 + … to n terms) – tn ⇒ tn = 3 + [2 + 4 + 8 + … to (n – 1) terms] n−1 = 3 + 2 ( 2 − 1) = 3 + 2n – 2 2 −1 = 1 + 2n Putting n = 1, 2, 3, … , n, we get t1 = 1 + 21 t2 = 1 + 22 t3 = 1 + 23  tn = 1 + 2n Adding column-wise, we get Sn = n + (2 + 22 + 23 + … + 2n)



n = n + 2 ( 2 − 1) = 2n+1 + n – 2 2 −1

56. Sum to infinity of the series 2 5 2 11 − + − +… is 3 6 3 24 4 1 (A)  (B)  9 3 2 (C)  (D)  None of these 9 Solution: (C) 2 5 2 11 Let S= − + − + ... to ∞(1) 3 6 3 24 1 Multiplying both sides by – , the common ratio of 2 G.P. –

1 2 5 8 S=– + − + ... to ∞(2) 2 6 12 24

Subtracting Eqs (2) from (1), we have

3 2 3 3 3 S= − + − + ... to ∞ 2 3 6 12 24 2 ⎛1 1 1 ⎞ − − + + ... to ∞⎟ ⎠ 3 ⎜⎝ 2 4 8



=



1 2 2 1 1 2 = − = − = 3 ⎛ 1⎞ 3 3 3 1− ⎜ − ⎟ ⎝ 2⎠

\

S=

1 2 2 × = 3 3 9

57. 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 Solution: (A) Let Sn = 1 . 1! + 2 . 2! + 3 . 3! + 4 . 4!  ⇒

+ … + n × n!

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!] = (n + 1)! – 1! = (n + 1)! – 1.



58. If a, b, c are digits, then the rational number represented by 0 . cababab … is 99c + ab 99c + 10 a + b (B)  990 99 99c + 10 a + b (C)  (D)  None of these 990 (A) 

Solution: (C) R = 0 . cababab…

Let

⇒ 102R = ca . bababa … and 104R = caba . baba … ⇒ (104 – 102) R = caba – ca ⇒ R =

caba − ca 1000c + 100a + 10b + a − 10c − a = 9900 9900

99c + 10 a + b 990 59. The sum of first n terms of the series

=

2 12 + 2.22 + 32 + 2.42 + 52 + 5.62 + … is n ( n +1) 2 when n is even. When n is odd the sum is

n2 ( n + 1) n ( n +1) 2 (B)  2 2 2 ⎡ n ( n + 1) ⎤ n ( n +1) (C)  ⎢ ⎥ (D)  ⎣ 2 ⎦ 2 (A) 

7.20  Chapter 7 Solution: (A) When n is odd, last term will be n2, \ then the sum is 2

2

2

2

2

2

2

2

1 + 2.2 + 3 + 2.4 + 5 + 2.6 + … + 2 (n – 1) + n =

⎡ ( n − 1) n2 n ( n + 1) 2 ⎤ + n2 ⎢ Replacing n by n − 1 in ⎥ 2 2 ⎣ ⎦

3 2 2 n3 + n2 n2 ( n + 1) = n − n + 2n = = 2 2 2

60. The sum of the series 1 + 2 . 2 + 3 . 22 + 4 . 23 + 5 . 24 + … + 100 . 299 is (A) 99 . 2100 + 1 (B)  100 . 2100 100 . 2100 + 1 (C) 99 . 2 (D)  99 

Solution: (D) Let S = 1 + 2 . 2 + 3 . 22 + 4 . 23 + 5 . 24   + … + 100 . 299 \ 2S = 1 . 2 + 2 . 22 + 3 . 23 + … + 99 . 299   + 100 . 2100 Substracting, we get – S = 1 + 1 . 2 + 1 . 22 + … + 1 . 299 –100 . 2100

= (1 + 2 + 22 + … + 299) – 100 . 2100

1( 2100 − 1) – 100 · 2100 = 2100 – 1 – 100 . 2100 2 −1 \ S = 100 . 2100 – 2100 + 1 = 99 . 2100 + 1.

=

Sequence and Series  7.21

NCERT EXEMPLARS 1. If the sum of n terms of an AP is given by Sn = 3n + 2n2, then the common difference of the AP is (A) 3 (B) 2 (C) 6 (D) 4 2. If the third term of GP is 4, then the product of its first 5 terms is (A) 43 (B)  2 (C) 6 (D) 4 3. If 9 times the 9th term of an AP is equal to 13 times the 13th term, then the 22nd term of the AP is (A) 0 (B) 22 (C) 198 (D) 220 4. If x, 2y and 3z are in AP where the distinct numbers x, y and z are in GP, then the common ratio of the GP is 1 (A)  3 (B)  3 1 (C)  2 (D)  2 5. If in an AP, Sn = q n2 and Sm = qm2, where Sr denotes the sum of r terms of the AP, then Sq equals to a3 (A)  (B)  mnq 2

(C)  q3 (D)  (m + n)q2

6. Let Sn denote the sum of the first n terms of an AP, if S2n = 3Sn, then S3n : Sn is equal to

(A) 4 (C) 8

(B) 6 (D) 10

7. The minimum value of x x + 41− x , x ∈ R is (A) 2 (B) 4 (C) 1 (D) 0 8. Let Sn denote the sum of the cubes of the first n natural numbers and Sn denote the sum of the first n natural n S numbers, then ∑ r equals to r =1 S4 n ( n + 1) ( n + 2 ) n ( n + 1) (A)  (B)  6 2 2 n + 3n + 2 (C)  (D)  None of these 2 9. If tn denotes the nth term of the series 2 + 3 + 6 + 11 + 18 + …., then t50 is (A) 492 – 1 (B)  492 2 (C) 50 + 1 (D)  492 + 2 10. The lengths of three unequal edges of a rectangular solid block are in GP. If the volume of the block is 216 cm3 and the total surface area is 252 cm2, then the length of the longest edge is (A)  12 cm (B)  6 cm (C)  18 cm (D)  3 cm

ANSWER K EYS 2.  (C)

3. (A)

4.  (B)

5. (C)

6.  (B)

7.  (B)

8.  (A)

9.  (D)

10.  (A)

HINTS AND EXPLANATIONS 1. Given, Sn = 3n + 2n2, First term of the AP, 2 ∴ T1 = 3 ×1 + 2 (1) = 3 + 2 = 5 and T2 = S2 − S1 2 2 = 3 × 2 + 2 × ( 2 )  − 3 × 1 + 2 × (1)      = 14 – 5 = 9 ∴ Common difference (d) = T2 – T1 = 9 – 5 = 4 2. It is given that, T3 = 4 Let a and r the first term and common ratio, respectively. Then, ar2 = 4  (i) 2 3 4 Product of first 5 terms = a · ar · ar · ar · ar

= a5 r10 = (ar2)5 = (4)5 [using Eq. (i)] 3. Let the first term be a and common difference be d. According to the question, 9.T9 = 13.T13 ⇒ 9 ( a + 8d ) = 13 ( a + 12d ) ⇒ 9a + 72d = 13a + 156 d ⇒ ( 9a − 13a ) = 156 d − 72d ⇒ − 4 a = 84 d ⇒ a = −21d ⇒ a + 21d = 0 ∴ 22nd term i.e., T22 = [ a + 21d ]  (i) T22 = 0  [using Eq. (i)]

NCERT EXEMPLARS

1. (D)

7.22  Chapter 7 4. Given, x, 2y and 3z are in AP. x + 3z Then, 2y = 2 x + 3z ⇒ y= 4 ⇒ 4 y = x + 3z and x, y, z are in GP. 4y = x + 3z y z Then, = =λ x y

⇒ d=

2a 3n  2a + ( 3n − 1) d  6 a + ( 9n − 3) S3 n n +1 = = 2 Now, n 2a Sn  2a + ( n − 1) d  2a + ( n −11) 2 n +1 6 an + 6 a + 18an − 6 a = 2an + 2a + 2an − 2a 24 an S = = 3n = 6 4 an an

⇒ y = xλ and z = λ y = λ 2 x On substituting these values in Eq. (i), we get 2 4 ( xλ ) = x + 3 λ x

(

7. We know that, AM ≥ GM

)

⇒ 4λ x = x + 3λ 2 x



⇒ 3λ 2 = x + 3λ 2 x



4 x + 41− x ≥ 4 x ·41− x 2 ⇒ 4 x + 41− x ≥ 2 4

⇒ 4λ = 1 + 3λ 2



⇒ 4 x + 41− x ≥ 2.2

⇒ 3λ − 4λ + 1 = 0



⇒ 4 x + 41− x ≥ 4

2

⇒ ( 3λ − 1) ( λ − 1) = 0

n

S Sr Sr S2 S3 = + + + ...... + n S1 S2 S3 Sn r =1 Sr

Let Tn be the nth term of the above series.

5. Given, Sn = qn2 and Sm = qm2 Now,

 n ( n + 1)    Sn  2  = ∴ Tn = n ( n + 1) Sn 2 n ( n + 1) 1 2 = =  n + n  2 2 2

= S1 q= , S= 4 q, S3 9= q and S4 16 q 2

∴

T1 = q

∴

T2 = S2 − S1 = 4 q − q = 3q



T3 = S3 − S2 = 9q − 4 q = 5q



T4 = S4 − S3 = 16 q − 9q = 7q

1  ∑ n2 + ∑ n  2 1  n ( n + 1) ( 2n + 1) n ( n + 1)  1 n ( n + 1)  ( 2n + 1)  =  + + 1 = .  2 6 2  2 2  3  ∴ Sum of the above series = ∑ Tn =

So, the series is q, 3q, 5q, 7q, …….. Here,



a = q and d = 3q – q = 2q

q  2 × q + ( q − 1) 2q  2 q q = ×  2q + 2q 2 − 2q  = × 2q 2 = q3 2 2 6. Let first term be a and common difference be d. n Then, Sn =  2a + ( n − 1) d   2 ∴

Sq =

2n ∴  2a + ( 2n − 1) d  S2 n = 2  S2 n = n  2a + ( 2n − 1) d   3n S3n =  2a + ( 3n − 1) d   2 According to the question, S2n = 3Sn n ⇒ n  2a + ( 2n − 1) d  = 3  2a + ( n − 1) d  2 ⇒ 4 a + ( 4 n − 2 ) d = 6 a + ( 3n − 3) d ⇒ − 2a + ( 4 n − 2 − 3n + 3) d = 0 ⇒ − 2a + ( n + 1) d = 0

⇒

8. ∑

1 ∴ λ = ,λ = 1 3

HINTS AND EXPLANATIONS

2a  (iv) n +1

(i)



=

1 1  2n + 1 + 3  n ( n + 1)   = 4 × 3 n ( n + 1) ( 2n + 4 ) 4 3  



=

1 1 n ( n + 1) ( 2n + 4 ) = n ( n + 1) ( n + 2 ) 12 6

9. Let Sn be sum of the series 2 + 3 + 6 + 11 + 18 …….+ t50. ∴ Sn = 2 + 3 + 6 + 11 + 18 + ....... + t50 .  (i) (ii) (iii)

and Sn = 0 + 2 + 3 + 6 + 11 + 18 + .... + t 49 + t50 (ii) On subtracting Eq. (ii) from Eq. (i), we get 0 = 2 + 1 + 3 + 6 + 7 + …… + t50 ⇒ t50 = 2 + 1 + 3 + 5 + 7 + …… upto 49 terms ∴ t50 = 2 + [1 + 3 + 5 + 7 + ……. Utpo 49 terms] 49 = 2 + [ 2 × 1 + 48 × 2] 2 49 = 2 + [ 2 + 96 ] 2 = 2 + [ 49 + 49 × 48] = 2 + 49 × 49 = 2 + ( 49 )

2

Sequence and Series  7.23

∴ Volume = ⇒ a3 = 216 ∴ a=6

a × a × ar = 216 cm3 r ⇒ a3 = 63

 a2  Surface area = 2  . + a 2 r + a 2  = 252  r  1  ⇒ 2a 2  + r + 1 = 252 r   1+ r2 + r  ⇒ 2 × 36   = 252 r  

⇒

1+ r2 + r 252 = 2 × 36 r

126 21 r ⇒ 1+ r2 + r = r 36 6 ⇒ 6 + 6 r 2 + 6 r = 21r ⇒ 6 r 2 − 15r + 6 = 0 ⇒ 2r 2 − 5r + 2 = 0 ⇒ ( 2r − 1) ( r − 2 ) = 0 1 ∴ r = ,2 2 1 a 6×2 For r = : Length = = = 12 2 r 1 Breadth = a = 6 1 Height = ar = 6 × = 3 2 a 6 For r = 2 : Length= = =3 r 2 Breadth = a = 6 Height = ar = 6 × 2 = 12 ⇒ r + r2 + r =

HINTS AND EXPLANATIONS

10. Let the length, breadth and height of rectangular solid block a is , a and ar, respectively. r

7.24  Chapter 7

PRACTICE EXERCISES Single Option Correct Type 1. If a, b, c are positive numbers in A.P. such that their product is 64, then the minimum value of b (A) = 2 (B)  =4 (C) = 1 (D)  Does not exist 2. If three successive terms of a G.P. with common ratio r(r > 1) form the sides of a DABC and [r] denotes greatest integer function, then [r] + [–r] = (A) 0 (B) 1 (C)  –1 (D)  None of these 21

3. If

∑ aj j =1

= 693, where a1, a2, …, a21, are in A.P., then

10

∑ a2i +1

is

i=0

PRACTICE EXERCISES

(A) 361 (C) 363

(B) 396 (D) data insufficient

(A) 5 (C)  125 ∞

8. If

∑ x n−1

= a and

n =1

∞

∑ y n−1

= b where | x |, | y | < 1,

n =1

∞

then

(B) 25 (D)  None of these

∑ ( xy)n−1

=

n =1

a + b −1 (A) ab (B)  ab 1 ab (C)  (D)  1− ab a + b −1 9. Let p, q, r ∈ R+ and 27pqr ≥ (p + q + r)3 and 3p + 4q + 5r = 12 then p3 + q4 + r5 is equal to (A) 3 (B) 6 (C)  2 (D)  None of these 10. The sum of the series

4. Number of increasing geometrical progression(s) with first term unity, such that any three consecutive terms, on doubling the middle become an A.P. is (A) 0 (B) 1 (C) 2 (D) infinity

1 2 3 2 4 + + + … to n terms 2 4 2 1 + 1 + 1 1 + 2 + 2 1 + 3 + 34 is n( n2 + 1) n( n + 1) (A)  2 (B)  n + n +1 2( n2 + n + 1)

5. If a1, a2, a3 (with a1 > 0) are in G.P. with common ratio r, then the value of r for which the inequality 9a1 + 5a3 > 14a2 holds, cannot be in the interval

(C) 

⎡ (A)  ⎢1, ⎣

11. a, b, c are three distinct real numbers, which are in G.P. and a + b + c = xb. Then (A) x < –1 or x > 3 (B)  –1 < x < 3 (C) –1 < x < 2 (D)  0 < x < 1

9⎤ (B)  (– ∞, 0) 2 ⎥⎦

⎡ 9⎤ 5 (C)  ⎢⎡ , 1⎤⎥ (D)  ⎢1, 5 ⎥ ⎣ ⎦ ⎣9 ⎦ 6. Let Sn (1 ≤ n ≤ 9) denotes the sum of n terms of series 1 + 22 + 333 + … + 999999999, then for 2 ≤ n ≤ 9 1 (A) Sn – Sn–1 = (10n – n2 + n) 9 1 n (B) Sn = (10 – n2 + 2n – 2) 9 (C) 9(Sn – Sn–1) = n(10n – 1) (D)  None of these 7. If log 5 x + log5 x + log5 x + … upto 7 terms = 35, then x is equal to 1/3

1/4

n( n2 − 1)



2( n2 + n + 1)

(D)  None of these

12. The sum of the first hundred terms of an A.P. is x and the sum of the hundred terms starting from the third term is y. Then the common difference is y−x y−x (B)  50 2 y−x y−x (C)  (D)  100 200 (A) 

13. If λ =

∞

1

∑ i4 i =1

∞

, then

1

∑ (2i − 1)4 i =1

is

Sequence and Series  7.25 14 λ λ (B)  2 15 16 15 (C)  λ (D)  λ 15 16 (A) 

14. The sum of all possible products of the first n natural numbers taken two at a time is 1 (A)  [Sn2 – Sn] 2

1 (B) [(Sn)2 – Sn] 2

1 (C)  [Sn2 – Σ(n + 1)] 2

(D) 

1 [(Sn)2 – Sn2] 2

15. The minimum value of 8sin x/8 + 8cos x/8 is 1

3+ 2

(B)  2 2

1

(C) 2

3+ 2 / 2

3− 2

(D)  2 2

16. If log 2 a + log 2 a + log2 a + log2 a + … upto 20 terms is 840, then a is equal to (A) 2 (B) 1 (C)  4 (D)  2 1/2

1/4

1/6

nx n (A)  (B)  (1 + x ) (1 + nx ) (1 + x ) [1 + ( n + 1) x ] (D)  None of these

18. If a, b, c are distinct positive real numbers and a2 + b2 + c2 = 1, then ab + bc + ca is (A)  less than 1 (B)  equal to 1 (C)  greater than 1 (D)  any real number 19. The value of (n – 2)2 + (n – 4)2 + (n – 6)2 + … to n terms is n n (A)  (n2 + 2) (B)  (n2 + 3) 3 2 n (C)  (n2 – 2) 3

(D) 

n 2 (n – 3) 2

20. The sum to infinity of the series 1 + 2 ⎛⎜1 − 1 ⎞⎟ + 3 ⎛⎜1 − 1 ⎞⎟ ⎝ n⎠ ⎝ n⎠ by

2

21. a1, a2, a3, … are in A.P. with common difference not a multiple of 3. Then, maximum number of consecutive terms so that all the terms are prime numbers is (A) 2 (B) 3 (C) 5 (D) infinite 22. The coefficient of x49 in the product (x – 1) (x – 3) … (x – 99) is (A)  – 992 (B)  1 (C)  – 2500 (D)  None of these 23. If x, y, z are three real numbers of the same sign then x y z the value of + + lies in the interval y z x (A) [2, ∞) (B)  [3, ∞) (C) (3, ∞) (D)  (–∞, 3) 24. In a G.P. of alternating positive and negative terms, any term is the A.M. of the next two terms. Then the common ratio is (A) –1 (B) –3 −1 (C)  –2 (D)  2

1 1 is + (1 + x ) (1 + 2 x ) (1 + 2 x ) (1 + 3 x )

x (1 + x ) (1 + ( n − 1) x )

2

⎛ n − 1⎞ (C) n2 (D)  ⎜⎝ n ⎟⎠

18 /

17. Sum to n terms of the series

(C) 

2

+ … where n ∈ N, is given

25. If A = 1 + ra + r2a + r3a + .... as and B = 1+ rb + r2b + a r3b + …. as, then is equal to b (A) log BA

(1− A) (B)  log1− B

⎛ A −1 ⎞ (C) log B −1 ⎜ A ⎟⎠ B ⎝

(D)  None of these

26. If the sum of n terms of an A.P. is cn (n – 1), where c ≠ 0, then sum of the squares of these terms is (A) c2n2(n + 1)2 2 (B)  c2 n (n – 1) (2n – 1) 3 2

(C)  2c n (n + 1) (2n + 1) 3 (D)  None of these 27. If in an A.P., Sn = p.n2 and Sm = p.m2 where Sr denotes the sum of r terms of the A.P., then Sp is equal to

PRACTICE EXERCISES

(A) 2

3− 2 / 2

(A) n (n – 1)

⎛ 1⎞ (B)  n ⎜1 − ⎟ ⎝ n⎠

7.26  Chapter 7 1 (A)  p3 2

34. Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. If 3 a < b < c and a + b + c = , then the value of a is 2

(B)  mnp

(C) p3 (D)  (m + n) p2 28. If p, q, r are positive and are in A.P., the roots of quadratic equation px2 + qx + r = 0 are all real for r (A)  − 7 ≥ 4 3 p

p −7 ≥ 4 3 (B)  r

29. The sum to n terms of the series 1 5 19 65 + + + + … is 3 9 27 81 (A) n –

n

(3 − 2 )



2n

n

(B)  n –

n

2 (3 − 2 ) 3n

(C) 2n – 1 (D)  3n – 1 30. Sum to n terms of the series 1 + 1! + 2 ! + 3! + … is 5! 6 ! 7 ! 8! (A) 

2 1 − 5! ( n + 1)!

PRACTICE EXERCISES

1⎛ 1 3! ⎞ (C)  ⎜ − 4 ⎝ 3! ( n + 2)!⎟⎠

(B) 

1⎛ 1 n! ⎞ − 4 ⎜⎝ 4 ! ( n + 4)!⎟⎠

(D)  None of these

31. If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2) p2 – 2p (ab + bc + cd) + (b2 + c2 + d2) ≤ 0 then a, b, c, d are in (A) A.P. (B) G.P. (C) H.P. (D)  ab = cd 32. If a + b + c = 3 and a > 0, b > 0, c > 0, then the greatest value of a2 b3 c2 is 10

(A) 

3 ⋅2 7

7

4

39 ⋅ 24 (B)  77

8 4 (C)  3 ⋅ 2 77

(D)  None of these

a b aα − b 1 33. If b c bα − c = 0 and α ≠ , then 2 2 1 0 (A) a, b, c are in A.P. (C) a, b, c are in H.P.

1 2 2



1 1 (C)  − 2 3

(B)  (D) 

1 2 3 1 1 − 2 2

35. If a1, a2, …, an are in A.P. with common difference d ≠ 0, then sum of the series sin d [sec a1 sec a2 + sec a2 sec a3 + … + sec an–1 sec an] is (A) tan an – tan a1 (B) cot an – cot a1 (C) sec an – sec a1 (D) cosec an – cosec a1

(C) all p and r (D) no p and r

n

(A) 

(B)  a, b, c are in G.P. (D)  None of these

36. 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 common difference is , then k is equal to k − ( l + a) (A) S (B)  2S (C) 3S (D)  None of these 37. If a, b, c, d are in G.P., then (a2 + b2 + c2) (b2 + c2 + d2) = (A) (ab + ac + bc)2 (B) (ac + cd + ad)2 2 (C) (ab + bc + cd) (D)  None of these 38. 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) (B) (2p – q) (2q – p) (C) (4p – q) (4q – p) (D)  None of these 39. The product of n positive integers is 1, then their sum is a positive integer, that is (A)  equal to 1 (B)  equal to n + n2 (C)  divisible by n (D)  never less than n 40. A man saves ` 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ` 40 more than the saving of immediately previous months. His total saving from the start of service will be ` 11040 after (A)  21 months (B)  18 months (C)  19 months (D)  20 months 41. Statement-1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + …+ (361 + 380 + 400) is 8000. n

Statement-2: number n.

∑ (k 3 − (k − 1)3 ) = n3 , for any natural

k =1

Sequence and Series  7.27

42. If 100 times the 100th term of an AP with non-zero common difference equals the 50 times its 50th term, then the 150th term of this AP is (A) –150 (B)  150 times its 50th term (C) 150 (D) zero 43. If the sum of first n terms of two A.P.’s are in the ratio 3n + 8 : 7n + 15, then the ratio of their 12th terms is (A)  8 : 7 (B)  7 : 16 (C)  74 : 169 (D)  13 : 47 1 3 7 15 44. The sum of n terms of the series + + + + ... 2 4 8 16 is 1 2 (C) n + 2–n – 1 (A)  2n − n −

(B)  1 – 2–n 1 (D)  ( 2n − 1) 2

45. The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is (A) –4 (B) –12 (C) 12 (D) 4 46. The

sum to the infinity of 2 6 10 14 1 + + 2 + 2 + 4 + ... is 3 3 3 3 (A) 2 (B) 3 (C) 4 (D) 6

the

series

47. The sum of positive terms of the series 4 1 10 + 9 + 9 + … is 7 7

852 7

(A) 

n ( n + 1) n ( n + 1) (B)  − 2 2

(C) 

n ( n + 1) ( 2n + 1) 6

n ( n + 1) ( 2n + 1) 6 S

the sum to k terms of an A.P., then for kx to be indeSx pendent of x (A) a = 2d (B)  a=d (C) 2a = d (D)  None of these 50. Given that α, γ are roots of the equation Ax2 – 4x + 1 = 0 and β, δ are roots of the equation Bx2 – 6x + 1 = 0. If α, β, γ and δ are in H.P., then (A) A = 5 (B)  A=–3 (C) B = 8 (D)  B=–8 51. If three positive numbers a, b, c are in H.P., then an + cn (A) > 2bn (B)  = 2bn n (C) < 2b (D)  > bn 52. a, b, c are three distinct real numbers, which are in G.P. and a + b + c = xb. Then, (A) x < –1 or x > 3 (B)  –1 < x < 3 (C) –1 < x < 2 (D)  0 < x < 1 1 53. If a1, a2, a3, a4 are in H.P., then a 1a4 root of (A) x2 + 2x + 15 = 0 (C) x2 – 6x – 8 = 0

3

∑ ar ar +1

is a

r =1

(B)  x2 + 2x – 15 = 0 (D)  x2 – 9x + 20 = 0

54. 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 (B)  2 : 1 (C)  3 : 1 (D)  4 : 1 55. If

and

an

an = ( x )

(D)  None of these

(D) –

49. If a is the first term, d the common difference and Sk

1/2n

352 437 (A)  (B)  7 7 (C) 

48. The sum of the products of the 2n numbers ±1, ±2, ±3. …, ±n taking two at a time is

bn 1/2n

+ ( y)

be two sequences given by n

n ∈ N. Then, a1a2a3 … an is equal to (A) x – y (C) 

n

and bn = ( x )1/2 − ( y )1/2 for all

x− y bn

(B) 

x+ y bn

(D) 

xy bn

PRACTICE EXERCISES

(A)  Statement-1 is false, Statement-2 is true. (B)  Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1 (C)  Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (D)  Statement-1 is true, statement-2 is false.

7.28  Chapter 7 56. For any odd integer n ≥ 1, n3 – (n – 1)3 + … + (–1)n–113 =

62. 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

1 (A)  (n – 1)2 (2n – 1) 2

1 1 (B)  x≥– 2 2 1 (C) x < – (D)  None of these 2 (A) x ≤ –

1 (B)  (n – 1)2 (2n – 1) 4 1 (C)  (n + 1)2 (2n – 1) 2

63. If a1, a2, a3, …, an are in H.P., then a1 a2 , …, a2 + a3 + ... + an a1 + a3 + ... + an

1 (D)  (n + 1)2 (2n – 1) 4 57. For a positive integer n, let a (n) = 1+

1 1 1 1 + + + ... + n . Then 2 3 4 (2 ) − 1

(A) a (100) ≤ 100 (C) a (200) ≤ 100

(B)  a (100) > 100 (D)  a (200) > 100

PRACTICE EXERCISES

58. Let α, β, γ be the roots of the equation 3x3 – x2 – 3x + 1 = 0. If α, β, γ are in H.P. then | α – γ | = 1 2 (A)  (B)  3 3 4 (C)  (D)  None of these 3 59. Suppose a, b > 0 and x1, x2, x3 (x1 > x2 > x3) are roots b a x−a x−b of = + and x1 – x2 – x3 = c, + x − a x − b b a then a, b, c are in (A)  A.P. (B)  G.P. (C)  H.P. (D)  None of these 60. The coefficient of xn in the product (1 – x) (1 – 2x) (1 – 22 ⋅ x) (1 – 23 ⋅ x) … (1 – 2n ⋅ x) is equal to )

n ( n −1) 2 2

(B) (2n + 1 – 1)

n ( n −1) ⋅2 2

n+1

(A)  (1 – 2

n

⋅

n ( n −1) 2 2

(C)  (1 – 2 ) ⋅ (D)  None of these

61. If 0.272727…, x and 0.727272… are in H.P., then x must be (A) rational (B) integer (C)  irrational (D)  None of these

an are in a1 + a2 + ... + an −1 (A)  A.P. (B)  G.P. (C)  H.P. (D)  None of these 64. 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 (B) 193 (C)  931 (D)  None of these 65. The largest term of the sequence 4 9 16 1 , , , , … is 503 524 581 692 (A)  (C) 

4 16 (B)  524 692

49 1529



(D)  None of these

66. The coefficient of x99 and x98 in the polynomial (x – 1) (x – 2) (x – 3) … (x – 100) are (A)  – 5050 and 12482075 (B)  – 4050 and 12582075 (C)  – 5050 and 12582075 (D)  None of these 67. The three successive terms of a G.P. will form the sides of a triangle if the common ratio r satisfies the inequality 3 −1 and < bn > be two sequences given by an = x 2 + y 2 and bn = x 2 − y 2 ∀ n ∈ N, then the value of a1 a2 a3 ….an is −n

−n

−n

−n

x+ y x−y (A)  (B)  bn bn x2 + y2 x2 − y2 (C)  (D)  bn bn 89. The sixth term of an A.P. is equal to 2. The value of the common difference of the A.P. which makes the product a1 a4 a5 greatest, is 8 2 (A)  (B)  5 3 3 3 (C)  (D)  5 4 90. If the natural numbers are written as 1 2 3 4 5 6 7 8 9 10 ...................... ...................... Then, the sum of the terms of the nth row is n ( n2 − 1) (A)  2

n ( n2 + 1) (B)  4

n ( n2 + 1) (C)  2

(D)  None of these

Previous Year’s Questions 91. If 1, log3 (31− x + 2) , log3 (4 ⋅ 3x− l) are in AP, then x equals: [2002] (A) log3 4 (B)  l − log3 4 (C) 1 − log, 3 (D)  log4 3

92. The value of 21/4⋅ 41/8⋅ 81/16 . . . ∞ is: (A) 1 (C) 3/2

(B) 2 (D) 4

[2002]

Sequence and Series  7.31 93. Fifth term of a GP is 2, then the product of its 9 terms is: [2002] (A) 256 (C)  1024

(B) 512 (D)  None of these

94. 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 equals : [2002] (A) 5 (C) 6

3n( n + 1) (A)  2 n( n +1) 2 (C)  4 100. If x =

y=

n= 0

(B) 7 (D) 4

1 1 1 − + − ……… upto 95. The sum of the series 1.2 2.3 3.4 ∞ is equal to [2003] (B)  log2 2 − 1

(A)  2 loge 2

∞

∑ an ,



n2 ( n + 1) 2



(B) 



⎡ n( n + 1) ⎤ (D)  ⎢ ⎥ ⎣ 2 ⎦

∞

∞

n= 0

n= 0

∑ bn , z = ∑ cn

2

where a, b, c are

in A.P. and |a| < 1, |b|< 1, |c|< 1, then x, y, z are in  [2005] (A) G.P. (B) A.P. (C) Arithmetic − Geometric Progression (D) H.P.

(C) loge 2 (D)  loge ⎛ 4 ⎞ ⎜⎝ ⎟⎠ e

a1 + a2 + ...a p p 2 = 1 01. Let a1, a2, a3, … be terms of an A.P. If , p ≠ q, a1 + a2 ... + aq q 2 2 a1 + a2 + ...a p p a = 2 , p ≠ q, then 6 equals [2006] 96. If f : R → R satisfies f (x + y) = f (x) + f (y), for all x, y a1 + a2 ... + aq q a21 n

∑ f (r ) is

[2003]

r =1

7n (A)  2



(C) 7n( n + 1) 97. If Sn = 

n

1

∑ nC

r =0

1 (A)  n 2





7( n + 1) (B)  2



7n( n + 1) (D)  2

and t n = r



n

r

∑ nC

r =0

, then r

tn is equal to Sn [2004]

1 (B)  n − 1 2

2n − 1 (C)  n − 1 (D)  2 98. Let Tr be the r th term of an A.P. whose first term is a and common difference is d. If for some positive 1 1 integers m, n, m ≠ n, Tm = and Tn = , then a − d, n m equals [2004] (A) 0

(B) 1

1 (C)  mn

(D) 





1 1 + m n

99. The sum of the first n terms of the series 12 + 2 ⋅ 22 n( n +1) 2 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + ... is when n is 2 even. When n is odd the sum is [2004]

41 (A)  11 2 (C)  7



7 (B)  2

(D) 

11 41

102. If a1, a2, … , an are in H.P., then the expression a1 a2 + a2 a3 + … + an-1 an is equal to [2006] (A)  n(a1− an) (B) (n − 1) (a1−an) (C)  na1an (D) (n − 1)a1an 103. In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals [2007] 1 (A)  (1 − 5 ) 2



(C)  5





(B) 

1 5 2

(D) 

1 ( 5 − 1) 2

104. If p and q are positive real numbers such that p2 + q2 = 1, then the maximum value of (p + q) is [2007] (A) 2 (B) 1/2 1 (C)  (D)  2 2 105. The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is  [2008]

PRACTICE EXERCISES

∈ R and f (1) = 7, then

7.32  Chapter 7 (A)  −4 (B)  −12 (C) 12 (D) 4 106. The sum to the infinity of the series 2 6 10 14 1 + + 2 + 3 + 4 + ....... is 3 3 3 3 (A) 2 (B) 3 (C) 4 (D) 6

112. Let α and β be the roots of equation px 2 + qx + r − 0, p ≠ 0 . If p, q, r are in A.P. and [2009]

108. A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediate preceding month. His total saving from the start of service will be Rs. 11040 after [2011] (A)  19 months (B)  20 months (C)  21 months (D)  18 months 109. Statement 1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + ...... + (361 + 380 + 400) is 8000. n Statement 2: ∑ ( k 3 = ( k − 1)3 ) = n3 for any natural

PRACTICE EXERCISES

k =1

[2012]

(A)  Statement 1 is false, statement 2 is true (B) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (C) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (D)  Statement 1 is true, statement 2 is false 110. If 100 times the 100th term of an Arithmetic Progression with non zero common difference equals the 50 times its 50th term, then the 150th term of this A.P. is [2012] (A) –150 (B)  150 times its 50th term (C) 150 (D) zero 111. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, . . . is [2013]

(

)

(B) 

(

)

7 (D)  179 − 10 −20 81

7 (A)  99 − 10 −20 9 7 (C)  99 + 10 −20 9

61 9 34 (C)  9 (A) 

107. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ...... = a10 = 150 and a10, a11 . . . are in A.P. with common difference −2, then the time taken by him to count all notes is [2010] (A)  34 minutes (B)  125 minutes (C)  135 minutes (D)  24 minutes

number n.

1 1 + = 4 , then the value of α − β is α β

(

)

(

)

7 179 + 10 −20 81



(B) 

2 17 9

(D) 

2 13 9

[2014]

113. Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is  [2014] (A)  2 + 3 (B)  3 + 2 (C) 2 − 3 (D)  2+ 3 114. If (10)9 + 2(11)1 (10)8 + 3(11) 2 (10)7 + ..... + 10(11)9, = k(10)9 then k is equal to 121 (A)  10 (C) 100 115. The 3

sum 3

of 3

3

[2014] 441 100 (D) 110 (B) 

first

9

3

3

terms

of

1 1 +2 1 +2 +3 + + + ...... is: 1 1+ 3 1+ 3 + 5

the

series [2015]

(A) 96

(B) 142

(C) 192

(D) 71

116. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is  [2016] 7 8 (A)  (B)  4 5 4 (C)  (D)  1 3 117. If the sum of the first terms of the series 2

2

2

2

16 ⎛ 3⎞ ⎛ 2⎞ ⎛ 1⎞ 2 ⎛ 4⎞ ⎜⎝1 ⎟⎠ + ⎜⎝ 2 ⎟⎠ + ⎜⎝ 3 ⎟⎠ + 4 + ⎜⎝ 4 ⎟⎠ +…, is 5 5 5 5 5 m, then m is equal to (A) 99 (C) 101

[2016] (B) 102 (D) 100

118. For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then [2017]

Sequence and Series  7.33 (A)  b, c and a are in A.P. (B)  a, b and c are in A.P. (C)  a, b and c are in G.P. (D)  b, c and a are in G.P.

(A) 66 (C) 34 12

119. Let a1, a2, a3, …, a49 be in A.P. such that

∑ a4 k +1 = 416

k =0

2 and a9 + a43 = 66. If a12 + a22 +…+ a17 = 140 m , then m is equal to [2018]

(B) 68 (D) 33

120. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2 · 22 + 32 + 2 · 42 + 52 + 2 · 62 + …. If B – 2A = 100λ, then λ is equal to [2018] (A) 232 (B) 248 (C) 464 (D) 496

ANSWER K EYS Single Option Correct Type   1. (B)  2.  (C) 11.  (A) 12.  (D) 21.  (B) 22.  (C) 31.  (B) 32.  (A) 41.  (B) 42.  (D) 51.  (A) 52.  (A) 61.  (A) 62.  (B) 71.  (B) 72.  (A) 81.  (D) 82.  (C)

3. (C) 13.  (D) 23.  (B) 33.  (B) 43.  (B) 53.  (B) 63.  (C) 73.  (A) 83.  (B)

4.  (B) 14.  (D) 24.  (C) 34.  (D) 44.  (C) 54.  (D) 64.  (C) 74.  (B) 84.  (C)

7.  (A) 17.  (B) 27.  (C) 37.  (C) 47.  (C) 57.  (D) 67.  (B) 77.  (A) 87.  (A)

8.  (D) 9.  (A) 10. (B) 18.  (A) 19.  (A)   20.  (C) 28.  (B)   29.  (B) 30.  (B) 38.  (B) 39.  (D) 40.  (A) 48.  (D) 49.  (C) 50.  (C) 58.  (C) 59.  (C) 60.  (A) 68.  (A) 69.  (B) 70.  (C) 78.  (B) 79.  (C) 80.  (B) 88.  (B) 89.  (A) 90.  (C)

95.  (D) 96.  (D) 97.  (A) 105.  (B) 106.  (B) 107.  (A) 115.  (A) 116.  (C) 117.  (C)

98.  (A) 99.  (D) 100.  (D) 108.  (C) 109.  (B) 110.  (D) 118.  (A) 119.  (C) 120.  (B)

5. (D) 15.  (B) 25.  (C) 35.  (A) 45.  (B) 55.  (C) 65.  (C) 75.  (D) 85.  (B)

6.  (C) 16.  (C) 26.  (B) 36.  (B) 46.  (B) 56.  (D) 66.  (C) 76.  (C) 86.  (C)

  91.  (B) 92.  (B) 101.  (D) 102.  (D) 111.  (B) 112.  (D)

93.  (B) 103.  (D) 113.  (D)

94.  (B) 104.  (D) 114.  (C)

PRACTICE EXERCISES

Previous Years’ Questions

7.34  Chapter 7

HINTS AND EXPLANATIONS Single Option Correct Type 1. Given a + c = 2b a+b+c Also, ≥ 3 abc = 3 64 = 4 3 3b ⇒ ≥4 3 ⇒ b≥4 ∴ Minimum b=4 The correct option is (B) 2. Let sides of triangle be a, ar, ar2. Since r > 1, ∴ ar2 is greatest side ∴ a + ar > ar2 ⇒ r2 – r – 1 < 0

1+ 5 1− 5 0 and a1, a2, a3 are in G.P. such that a2 = a1r and a3 = a1r2 ∴ 9a1 + 5a3 > 14a2 ⇒ 9a1 + 5a1r2 > 14a1r and since a1 > 0, we get 9 + 5r2 > 14r 2 ⇒ 5r – 14r + 9 > 0 ⇒ (5r – 9) (r – 1) > 0

10 n [10(9n – 1) – (9n – 10)]– n 81



∴

log 5

+

log 51/3

+

log 51/4

+ … upto 7 terms

log x =1 log 5

⇒ log x = log 5 ⇒ x=5 The correct option is (A) ∞

8. Given

a = ∑ x n −1 = 1 + x + x2 + … = n =1

⇒

x=

Similarly y =

1 1− x

a −1 a b −1 ∴ b

∞

∑ ( xy)n−1 = 1 + (xy) + (xy)2 + …

n =1

Sequence and Series  7.35



=

ab ab = ab − ( ab − ( a + b) + 1) a + b −1

The correct option is (D) 9.

27pqr ≥ (p + q + r)3



⇒ (pqr)1/3 ≥

p +q +r ⇒p=q=r 3

Also, 3p + 4q + 5r = 12 ⇒ p=q=r=1 The correct option is (A) 10. Let Tn be the nth term of the series 1 2 3 2 4 + +… + 1 + 1 + 1 1 + 22 + 24 1 + 32 + 34 Then, Tn = =



=



n

n

=

1 + n2 + n4

(1 + n2 ) 2 − n2

n

n

∑ Tr

( n2 + n + 1) ( n2 − n + 1) 1⎛ 1 1 ⎞ − 2 ⎜⎝ n2 − n + 1 n2 + n + 1⎟⎠

1 + 1 + r = x, r is the common ratio of the G.P. r 2 ⇒ r + r (1 – x) + 1 = 0. Since r is real, therefore discriminent > 0 ⇒ (1 – x)2 – 4 > 0 ⇒ x2 – 2x + 1 – 4 > 0 ⇒ x2 – 2x – 3 > 0 ⇒ (x + 1) (x – 3) > 0 ⇒ x < –1 or x > 3 The correct option is (A) ⇒

100 [2a + 99d] = x 2 100 and [2(a + 2d) + 99d] = y 2 On subtraction, 200d = y – x y−x ⇒d= 200 The correct option is (D) 12. Given:

13. Given: … ∴

∞

1 14

+

1

+

24

1

∑ ( 2i − 1)4

1

1

=

+

34

14

+

1

=λ

44

1 34

1

+

54

+ ...

= 1 + 1 + 1 + 1 + ... ∞ − ⎡ 1 + 1 + ... ∞ ⎤ ⎢ 24 4 4 ⎥ 14 24 34 4 4 ⎣ ⎦ = λ−

1 ⎡1 1 1 ⎤ + 4 + 4 + ... ∞ ⎥ 4 ⎢ 4 2 ⎣1 2 3 ⎦

⎤ 1⎡ 1 1 − ⎢ ⎥ 2 ⎣1 + ( n − 1)n 1 + n( n + 1) ⎦

=λ–

=

1 ⎡1 1 ⎤ 1⎡ 1 1 ⎤ + − − 2 ⎢⎣1 1 + 1.2 ⎥⎦ 2 ⎢⎣1 + 1.2 1 + 2.3 ⎥⎦

14. Let A = 1.2 + 1.3 + … + 2.3 + 2.4 + … + (n – 1) · n Now, (1 + 2 + 3 + … + n)2 – (12 + 22 + 32 + … + n2) = 2A

+ +



a c +1+ =x b b

=

r =1





i =1

Now

Divide by b,

=

=

1⎡ 1 1 ⎤ +… − 2 ⎢⎣1 + 2.3 1 + 3.4 ⎥⎦

⎤ 1⎡ 1 1 − ⎢ ⎥ ( ) 2 ⎣1 + n − 1 n 1 + n( n + 1) ⎦

⎤ 1⎡ 1 ⎢1 − ⎥ 2 ⎣ 1 + n( n + 1) ⎦ n( n + 1) 2( n2 + n + 1)

.

3 1 Trick: Checking for n = 1, 2. S1 = and S2 = which are 3 7 given by (b). The correct option is (B) 11. a + b + c = xb

1 15 λ= λ 16 16 The correct option is (D)

⇒ A =

1 ⎡ n2 ( n + 1) 2 n( n + 1) ( 2n + 1) ⎤ − ⎢ ⎥ 2 ⎢⎣ 4 6 ⎥⎦

1 [(Sn)2 – Sn2] 2 The correct option is (D)

=

15. A.M. ≥ G.M. ⇒

⎛ x π⎞ 3 2 cos ⎜ − ⎟ ⎝ 8 4⎠

23 sin x/8 + 23 cos x/8 ≥ 2

Now maximum of So, A.M. ≥ 23/

2

2

⎛ x π⎞ 3 2 cos ⎜ − ⎟ ⎝ 8 4⎠

2

sin

⇒ 8

The correct option is (B)

x 8

cos

+8

x 8

=

23

2 .1

⎛ 3 ⎞ +1⎟ ⎜ 2 ⎠

≥ 2⎝

= 23/

2

HINTS AND EXPLANATIONS



1 1 ⎛ a − 1⎞ ⎛ b − 1⎞ 1− ⎜ ⎝ a ⎟⎠ ⎜⎝ b ⎟⎠ = 1 − xy =

7.36  Chapter 7 16. log 2 a + log 2 a + log2 a + log 2 a + …

= 2 log2 a + 4 log2 a + 6 log2 a + … + 40 log2 a

⎛ 1⎞ ∴ ⎜1 − ⎟ S = ⎝ n⎠



= log2 a [2 + 4 + 6 + … + 40]

(1) and (2)



20 = (2 + 40) log2 a 2

⇒

S =1+ n



=

1/2

1/4

1/6

18 /

= 420 log2 a = 840 (Given) ⇒ log2 a = 2 ⇒ a = 4 The correct option is (C) 17. If tr denotes the rth term of the series, then x x tr = (1 + rx ) (1 + ( r + 1) x ) 1 1 − = 1 + rx 1 + ( r + 1) x



n

n

⎡ 1

1

r =1 ⎣

r =1

1 1 − 1 + x 1 + ( n + 1) x

=



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

∑ tr r =1

HINTS AND EXPLANATIONS

a + b > a 2b 2 [A.M. > G.M. for unequal numbers] 2 ⇒ a2 + b2 > 2ab Similarly, b2 + c2 > 2bc and c2 + a2 > 2ca Hence, 2 (a2 + b2 + c2) > 2 (ab + bc + ca) ⇒ ab + bc + ca < 1 The correct option is (A)



=

∑ (n

⇒ S = n2 The correct option is (C) 21. In this type of A.P. it can be easily shown that exactly one out of any 3 consecutive terms will be multiple of 3. So at most 3 consecutive terms can be prime numbers. The correct option is (B)

1/3

24. Let the G.P. be a – ar + ar2 – ar3 + … with common ratio =–r By the given condition

2 a = ar − ar ⇒ 2a = ar2 – ar

2

2

⇒ 2 = r – r ⇒

r2 – r – 2 = 0

⇒ (r – 2) (r + 1) = 0

n

∑ ( n − 2r ) 2

⇒

r =1

2

1 ⎛ 1⎞ = n 1 − ⎜1 − ⎟ ⎝ n⎠

The correct option is (B)

2

n

2

⎛ x y z⎞ ⇒ ⎜ + + ⎟ ≥3 ⎝ y z x⎠

18. Since a and b are unequal,

19. The given series =

r = 2, –1

∴ Common ratio = –2 or 1 2

− 4 nr + 4 r )

Hence, common ratio = –2

1

( common ratio is –ve)

The correct option is (C)

n n ( n + 1) = n · n – 4n · + 4 · (n + 1) (2n + 1) 6 2 n 2 = (n + 2) 3 The correct option is (A) 2

2

20. Let

(2)

⎛ 1⎞ ⎛ 1⎞ + … ∞ ⎜⎝1 − n ⎟⎠ + ⎜⎝1 − n ⎟⎠

⎛ x y z⎞ ⎛ ⎞ ⇒ ⎜ + + ⎟ 3 ≥ x⋅ y⋅z ⎜⎝ y z x ⎟⎠ ⎝ y z x⎠

The correct option is (B)

2

+ …

23. A.M. ≥ G.M.

n (1 + x ) [1 + ( n + 1) x ]

=

2

22. The coefficient of x49 = – [1 + 3 + 5 + … + 99] = – 2500 The correct option is (C)

⎦



⇒

⎤

∑ ⎢1 + rx − 1 + ( r + 1) x ⎥

⇒ x ∑ t r =

⎛ 1⎞ ⎛ 1⎞ ⎜⎝1 − n ⎟⎠ + 2 ⎜⎝1 − n ⎟⎠

S = 1 + 2 ⎛⎜1 − 1 ⎞⎟ + 3 ⎛⎜1 − 1 ⎞⎟ +…(1) ⎝ n⎠ ⎝ n⎠

25. A =

1 1 1 A −1 ⇒ 1 − ra = ⇒ ra = 1 − = A 4 A 1 − ra

B=

1 1 1 B −1 ⇒ 1 − rb = ⇒ rb = 1 − = B B B 1 − rb

⎛ B −1 ⎞ ⎛ A −1 ⎞ ∴ a log r = log ⎜ ⎟ and b log r = log ⎜ B ⎟ A ⎝ ⎠ ⎝ ⎠

Sequence and Series  7.37

a = b

⎛ B −1 ⎞ log ⎜ ⎟ ⎝ B ⎠

⎛ A −1 ⎞ . = log B −1 ⎜ A ⎟⎠ B ⎝

The correct option is (C)

28. Since p, q, r are in A.P. so 2q = p + r . The roots of the equation px2 + qx + r = 0 are real if and only if 2

p + r⎞ q2 – 4pr ≥ 0 ⇒ ⎛⎜ - 4pr ≥ 0 ⎝ 2 ⎟⎠



p2

p2 + r2 - 14pr ≥ 0 ⇒

⇒

r

26. If tr be the rth term of the A.P., then tr = Sr – Sr–1 = cr (r – 1) – c (r – 1) (r – 2)



= c (r – 1) (r – r + 2) = 2c (r – 1)

We have,

+ t 22

⎛p ⎞ p − 7⎟ - 48 ≥ 0 ⇒ −7 ≥ 4 3 ⎠ r r

⇒ ⎜ ⎝

The correct option is (B)

+ … + tn

= 4c2[02 + 12 + 22 + … + (n – 1)2]



= 4c2



=

( n − 1) n (2 n − 1) 6

2 2 c n (n – 1) (2n – 1) 3

29.

1 5 19 65 + … to n terms + + + 3 9 27 81 ⎛

2⎞ 3

n [2a + (n – 1) d] = pn2(1) 2

m [2a + (m – 1) d] = pm2 2

2a + ( n − 1) d n ⇒ = 2a + ( m − 1) d m

⎛

=n–

⎛

⎛ 2⎞ 1− ⎜ ⎟ ⎝ 3⎠ 2 =n– · 3 1− 2 3

n

=n–

⇒ 2a – d = 0 ∴ d = 2a ∴ (1) gives, 2a + (n – 1) 2a = 2pn ⇒ 2an = 2pn ⇒ a = p ∴ d = 2p Now,

p Sp = · [2a + (p – 1) d] 2

2 3n

(3n – 2n)

The correct option is (B) 30. We have, tr =

r! ( r − 1)! and tr+1 = ( r + 5)! ( r + 4)!

⇒ 2a (m – n) + d (mn – m – nm + n) = 0 ⇒ 2a (m – n) + d (n – m) = 0

8 ⎞ ⎛ 16 ⎞ +… + 1− 27 ⎟⎠ ⎜⎝ 81⎟⎠

2 ⎤ 2 ⎡ 2 ⎛ 2⎞ ⎢1 + + ⎜ ⎟ + ... to n terms ⎥ 3 ⎢⎣ 3 ⎝ 3 ⎠ ⎥⎦



⇒ 2am + (n – 1) md = 2an + n (m – 1) d

4⎞ 9

= ⎜1 − ⎟ + ⎜1 − ⎟ + ⎜1 − ⎝ ⎠ ⎝ ⎠ ⎝

The correct option is (B)

Sm =

p +1≥0 r

2



27. Sn =

- 14

2



t12

2

r! r! – =0 ( r + 4)! ( r + 4)!

Now,

rtr – (r + 5)tr+1 =

⇒

rtr – (r + 1)tr+1 = 4tr+1

⇒

4 ∑ t r +1 =

n −1

r =1

n −1

∑ [rtr − ( r + 1) tr +1] r =1

⇒ 4(t2 + t3 + … + tn) = 1t1 - ntn

⎛ 0 !⎞ n( n − 1)! − ⎝ 5!⎟⎠ ( n + 4)!



=

p · [2p + (p – 1) 2p] 2

⇒



=

p · [2p + 2p2 – 2p] 2



=



=

p . 2 3 n 2p = p 2

⇒ t1 + t2 + … + tn =

1⎡1 n! ⎤ − ⎢ 4 ⎣ 4 ! ( n + 4)! ⎥⎦

The correct option is (C)

4 (t1 + t2 + … + tn) = 5t1 - ntn = 5 ⎜

The correct option is (B)

1 n! − 4 ! ( n + 4)!

HINTS AND EXPLANATIONS

∴

⎛ A −1 ⎞ log ⎜ ⎟ ⎝ A ⎠

7.38  Chapter 7 31. We have, (a2 + b2 + c2)p2 – 2p(ab + bc + cd) + (b2 + c2 + d2) ≤ 0 ⇒ (ap – b)2 + (bp – c)2 + (cp – d)2 ≤ 0 Therefore, ⇒

ap − b = 0 cp − d = 0 bp − c = 0 b = ap ⇒ d = cp ⇒ c = bp

b c d ⇒ = = =p a b c ∴ a, b, c, d are in G.P. The correct option is (B)

a a b b b c c , , , , , , we get 2 2 3 3 3 2 2

,

1 a b c 2 ⋅ + 3⋅ + 2 ⋅ ⎡⎛ a ⎞ 2 ⎛ b ⎞ 3 ⎛ c ⎞ 2 ⎤ 7 2 3 2 ≥ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎥ 7 ⎢⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 2 ⎠ ⎥

⎣

⎦

1

3 ⎛ a2b 3c 2 ⎞ 7 a2b 3c 2 37 ⇒ ≥ ⎜ ⇒ 7 ≥ ⎟ 7 7 ⎝ 22 3322 ⎠ 22 ⋅ 33 ⋅ 22 ⇒ a2 b3 c2 ≤

1

∴

d=

∴

a=b–d=

=

2

310 ⋅ 2 4

The correct option is (A)

77

36. We have, S =



l −a

= 2S

a+l

⇒ 2α (b2 - ac) - (b2 - ac) = 0

−1

l−a n −1

[Using (1)]

l 2 − a2 2S − ( l + a) ∴ k = 2S. The correct option is (B)

⇒ 2 (b2α - bc - acα + bc) - (abα - ac - abα + b2) = 0

=

37. As a, b, c, d are in G.P., therefore

or (b2 - ac) (2α - 1) = 0 b2 - ac = 0

n 2S (a + l) ⇒ = n 2 a+l

l = a + (n – 1) d ⇒ d =

.

b aα − b a aα − b +0=0 − c bα − c b bα − c

or

1 1 − 2 2

= (tan a2 – tan a1) + (tan a3 – tan a2)  + … + (tan an – tan an–1) = tan an – tan a1 The correct option is (A)

33. Expanding along R3, we get 2

( d > 0)

sin (a2 − a1 ) sin (a3 − a2 ) sin (an − an −1 ) + + .. + cos a1 cos a2 cos a2 cos a3 cos an −1 cos an

Also,

∴ greatest value of a2 b3 c2 =

HINTS AND EXPLANATIONS

2 1 1 1 1 = ⎛⎜ 1 − d 2 ⎞⎟ ⇒ − d 2 = ± ⇒ d = ± . 4 ⎠ ⎝ 16 4 2 4

⇒

310 ⋅ 2 4 77

2

The correct option is (D) 35. As a1, a2, a3, … an–1, an are in A.P., d = a2 - a1 = a3 - a2 = … = an - an–1 sin d [sec a1 sec a2 + sec a2 sec a3 + .. + sec an–1 sec an]

32. Taking A.M. and G.M. of 7 numbers



2

4 ⎛1 ⎞ ⎛1 ⎞ ⇒ ⎛⎜ 1 ⎞⎟ = ⎜ − d ⎟ ⎜ + d ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎝ 2⎠



[∵ (2α – 1) ≠ 0]



b c d = = = r (say) a b c

∴ b2 = ac ∴ a, b, c are in G.P. The correct option is (B)

⇒ b = ar, c = br = ar ⋅ r = ar2, d = cr = ar2 ⋅ r = ar3. (a2 + b2 + c2) (b2 + c2 + d 2)

= (a2 + a2 r2 + a2 r4) (a2 r2 + a2 r4 + a2 r6)

34. Let a = b – d and c = b + d,



= a4 r2 (1 + r2 + r4) (1 + r2 + r4)



= (a2 r + a2 r3 + a2 r5)2



= (a ⋅ ar + ar ⋅ ar2 + ar2 ⋅ ar3)2

then a + b + c =

1 3 ⇒β= . 2 2

1 1 1 −d, , +d 2 2 2 (d > 0 as a < b < c) Therefore, the number are

Now a2, b2, c2 are in G.P. ⇒ (b2)2 = a2c2



= (ab + bc + cd)2. The correct option is (C) 38. Let the two numbers be a and b,then G=

ab or G2 = ab

(1)

Sequence and Series  7.39

⎛ x + x 2 + ... + x n ⎞ 1/2 1/n ∴ ⎜ 1 ⎟⎠ ≥ (x1 · x2 …xn) = (1) = 1 ⎝ n

⇒ x1 + x2 + … + xn ≥ n. Hence x1 + x2 + … + xn can never be less than n. The correct option is (D) 40. Total savings = 200 + 200 + 200 + 240 + 280 + … to n months = 11040 n−2 ( 400 + ( n − 3) ⋅ 40) = 11040 2 ⇒ (n – 2)(140 + 20n) = 10640 ⇒ 20n2 + 100n – 280 = 10640 ⇒ n2 + 5n – 546 = 0 ⇒ (n – 21)(n + 26) = 0 ⇒ n = 21 as n ≠ –26 The correct option is (A) (( n − 1)3 − n3 ) 41. Tn = (n – 1)2 + (n – 1)n + n2 = ( n − 1) − n = n3 – (n – 1)3 T1 = 13 – 03 T2 = 23 – 13  T20 = 203 – 193 S20 = 203 – 03 = 8000 The correct option is (B) 42. 100(a + 99d) = 50(a + 49d) 2a + 198d = a + 49d a + 149d = 0 T150 = a + 149d = 0 The correct option is (D) ⇒ 400 +

43.

sn 3n + 8 = ⇒ sn′ 7n + 15

n ( 2a + ( n − 1)d ) 3n + 8 2 = n 7n + 15 ( 2a′ + ( n − 1)d ′ ) 2

⎛ n − 1⎞ a+⎜ d 3n + 8 a + 11d 3( 23) + 8 ⎝ 2 ⎟⎠ ⇒ = = = 7( 23) + 15 7n + 15 a′ + 11d ′ ⎛ n − 1⎞ a′ + ⎜ d′ ⎝ 2 ⎟⎠

=

77 7 = 176 16

The correct option is (B) 44. Sn =

1 3 7 15 + + + + … upto n terms 2 4 8 16

1 1 1 ⇒ Sn = ⎛⎜1 − ⎞⎟ + ⎛⎜1 − ⎞⎟ + ⎛⎜1 − ⎞⎟ + ... ⎝ 2⎠ ⎝ 4⎠ ⎝ 8⎠ 1⎛ 1⎞ ⎜⎝1 − n ⎟⎠ ⎛1 1 1 ⎞ 2 2 = n − ⎜ + + + …⎟ = n − ⎝2 4 8 ⎠ 1 1− 2 1 –n = n −1+ n = n + 2 – 1 2 The correct option is (C) 45. Let a, ar, ar2, … a + ar = 12 ar2 + ar3 = 48 dividing Eq. (2) by (1), we have

(1) (2)

ar 2 (1 + r ) =4 a( r + 1)



⇒ r2 = 4 if r ≠ –1 ∴ r = –2 Also, a = –12 The correct option is (B) 46. Let

S = 1+

[using (1)].

2 6 10 14 + + + + ... (1) 3 32 33 34

1 1 2 6 10 S = + + 3 + 4 + ... (2) 2 3 3 3 3 3



From (1) and (2),

⎛

1⎞

1

4

4

4

S ⎜1 − ⎟ = 1 + + + + + ... ⎝ 3 32 33 34 3⎠

2 4 4 ⎛ 1 1 ⎞ S = + 1 + + 2 + ...⎟ ⎜ 2 3 ⎝ ⎠ 3 3 3 3

⇒

2 + S = 3 32 ⎜ 1⎟ 3 ⎜⎝ 1 − ⎟⎠

4

4 ⎛ 1 ⎞ 3

⇒

4 4 3 4 2 6 + = + = 3 32 2 3 3 2 2 6 S = 3 3 =

⇒ S=3 The correct option is (B) 47. Here,

a = 10, d = – .

3 7

HINTS AND EXPLANATIONS

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). The correct option is (B) 39. Given x1 · x2 …xn = 1 Since A.M. ≥ G.M.

7.40  Chapter 7

Then,

⎛ 3⎞ 7

⎛ 3⎞ 7

tn is positive if 10 + (n – 1) ⎜ − ⎟ ≥ 0; ⎝ ⎠ or, 70 – 3 (n – 1) ≥ 0 or 73 ≥ 3n; or 24

HINTS AND EXPLANATIONS

∴ First 24 terms are positive. ∴ Sum of the positive terms 24 2

1 ≥n 3

−3 ⎤ ⎡ ⎢ 2 × 10 + 23 × 7 ⎥ ⎣ ⎦



= S24 =



69 ⎤ = 852 . = 12 ⎡ 20 − ⎢ 7 7 ⎥⎦ ⎣

1 + 2d = 3 (2) α 1 Solving (1) and (2), we get = 1 and d = 1. α 1 1 1 1 ∴ = 1, = 2, = 3 and = 4. α β γ δ ⇒

tn = 10 +(n – 1) . ⎜ − ⎟ ⎝ ⎠

Since,

1 1 = B ⇒ B = 8. = A ⇒ A = 3. Also, βδ αγ

The correct option is (C) 51. For the positive numbers a, c we have harmonic mean H = b { a, b, c are in H.P.} and geometric mean G = ac . But G > H; ∴ ac > b. (1) For the positive numbers an, cn, we have

The correct option is (C) 48. We have, (1 – 1 + 2 – 2 + 3 – 3 + … + n – n)2 2 = 1 + 12 + 22 + 22 + … + n2 + n2 + 25, where S is the required sum. ⇒ o = 2(12 + 22 + … + n2) + 2S n ( n + 1)( 2n + 1) ⇒ S = – (12 + 22 + … + n2) = – 6 The correct option is (D)

geometric mean =

kx [ 2a + ( kx − 1) d ] Skx 49. We have, = 2 x Sx [ 2a + ( x − 1) d ] 2

The correct option is (A) 52. a + b + c = xb Divide by b, a c + 1 + =x b b

For

=

k [( 2a − d ) + kxd ] ( 2a − d ) + xd

S kx to be independent of x, 2a – d = 0 or 2a = d. Sx

The correct option is (C) 50. α, β, γ and are in H.P. 1 1 1 1 , , , are in A.P. α β γ δ Let d be the common difference of the A.P. Since, a, γ are roots of Ax2 – 4x + 1 = 0 ⇒

∴

α +γ 4/A 1 1 = = 4 or + =4 α γ αγ 1/ A

1 1 1 ⇒ + + 2d = 4 or + d = 2 (1) α α α Also, β, δ are roots of Bx2 – 6x + 1 = 0 ∴

β +δ 1 1 6/B 1 1 = + = = 6 or + d + + 3d = 6 βδ β δ 1/ B α α

an c n

n n arithmetic mean = a + c ; 2

n n  A.M. > G.M., ∴ a + c

2

>

an c n (2)

From (1) and (2), we get

an + c n > ( ac)n > bn, ∴ an + cn > 2bn. 2

⇒ 1 + 1 + r = x, r is the common ratio of the G.P.

r

⇒ r2 + r (1 – x) + 1 = 0. Since r is real, therefore, discriminent > 0 ⇒ (1 – x)2 – 4 > 0 ⇒ x2 – 2x + 1 – 4 > 0 ⇒ x2 – 2x – 3 > 0 ⇒ (x + 1) (x – 3) > 0 ⇒ x < –1 or x > 3 The correct option is (A) 3

53.

∑ ar ar +1

= a1a2 + a2a3 + a3a4 = 3a1a4

1

Since

a1, a2, a3, a4 are in H.P.

1 1 1 1 , , , are in A.P. a1 a2 a3 a4

1 1 − = d ⇒ a1 – a2 = da1a2(1) a2 a1

Similarly, a2 – a3 = d a2a3(2) a3 – a4 = d a3a4 (3) On adding (1), (2) and (3), we get a1 – a4 = d[a1a2 + a2a3 + a3a4]

Sequence and Series  7.41 a − a4 ∴ a1a2 + a2a3 + a3a4 = 1

b0 x −y a1b1 = = bn bn bn The correct option is (C)

⇒ a1a2 + a2a3 + a3a4 = 3a1a4 ∴ Given expression = 3. It is a root of x2 + 2x – 15 = 0 The correct option is (B)

Geometric mean G = Given: or,

ab

2ab a+b

2 ab

2

⎡ 1 ⎛ n − 1⎞ ⎛ n − 1 ⎞ ⎤ n( n + 1) ⎤ = ⎡ + 1⎟ ⎥ ⎟⎜ ⎢ ⎥ − 16 ⎢ ⎜ ⎣

By componendo and dividendo ( a + b )2



9 = or 1

( a − b )2

a+ b

3 = a− b 1

3+1 Again, by componendo and dividendo = 3 −1 2 b a a 4 = 2 or = b b 1 2 a

The correct option is (D) n

n

n

55. an = x1/2 + y1/2 and bn = x1/2 − y1/2

(

n

Now, anbn = x1/2 + y1/2

n

) (x

1/2n

( ) − (y ) 1/2n

⇒ anbn = x

3



5 4

=

⎤ ⎡⎛ n − 1⎞ 3 ⎛ n − 3 ⎞ 3 3 – 2 × 2 ⎢⎜⎝ 2 ⎟⎠ + ⎜⎝ 2 ⎟⎠ + ... + 1 ⎥ ⎢⎣ ⎥⎦ [ n –1, n – 3 are even integers]



H 4 4 = , so 2 ab = G 5 5 a+b

a+b

56. Since n is an odd integer (–1)n–1 = 1 and n – 1, n – 3, n – 5 etc., are even integers. We have, n3 – (n – 1)3 + (n – 2)3 – (n – 3)3 + … + (–1)n–113 = n3 + (n – 1)3 + (n – 2)3 + … + 13  – 2[(n – 1)3 + (n – 3)3 + … + 23] = n3 + (n – 1)3 + (n – 2)3 + … + 13

2

1/2n

n −1

⇒ anbn = x 1/2 − y 1/2 Now, a1a2a3 … an

− y1/2

n

)



⎛ a a a ...a ⎞ = ⎜ 1 2 3 n ⎟ bn bn ⎝ ⎠



=

=

1 1 (n + 1)2 [n2 – (n – 1)2] = (n + 1)2(2n – 1). 4 4

The correct option is (D) 57. We have, a (n) = 1 +

1 1 1 1 + + + ... + n 2 3 4 (2 ) − 1

1⎞ ⎛ 1 1⎞ ⎛ 1 1⎞ ⎛ 1 = 1+ ⎜ + ⎟ + ⎜ +… + ⎟ + ⎜ +… + ⎟ ⎝ 2 3⎠ ⎝ 4 7⎠ ⎝ 8 15 ⎠ 1 ⎞ ⎛ 1 +… + ⎜ n −1 + … + n −1 ⎟ ⎝2 2 ⎠

n −1 < 1 + 2 + 4 + 8 + ... + 2 2 4 8 2n −1 1+… +1 =1+ = n. ( n + 1) times Thus, a (100) < 100 1 ⎛ 1 1⎞ ⎛ 1 1⎞ Next, a (n) = 1 + + ⎜ + ⎟ + ⎜ + … + ⎟ + … ⎝ ⎠ ⎝ 2 3 4 5 8⎠

= bn–1(1)

(a1a2 a3 ...an −1 ) (anbn ) bn



a a a ...a b = 1 2 3 n −1 n −1 



=

bn

( a1a2 a3 ...an − 2 ) ( an −1bn −1 ) bn



a a a ...a b = 1 2 3 n −1 n − 2



……………… ………………

bn

⎠⎦

1 2 ( n − 1)2 ( n + 1)2 n (n + 1)2 – 16 4 16 × 4



2

n −1

⎣2 ⎝ 2 ⎠ ⎝ 2

⎦

=



n

2

2

+ {using (1)}

1 2

n −1

n −1

> 1+ 1 + 2 + 4 +… + 2 − 1 2 4 8 2n 2n 1 1 1 1 1 + + +… + =1+ 2 2 2 2 – n    2 n times



1⎞ n ⎛ = ⎜1 − n ⎟ + ⎝ 2 ⎠ 2

1 ⎞ 200 > 100. Thus, a (200) > ⎛ 1 − + ⎜⎝ 100 ⎟ ⎠ 2

2

+1

+ ... +

1 n

2 −1

HINTS AND EXPLANATIONS

d

1 a −a 1 ⇒ = + 3d ⇒ 1 4 = 3a1a4 a4 d a1

54. Harmonic mean of a, b is H =

=



7.42  Chapter 7 The correct option is (D) 58. Let α =

and, ⋅

1 1 1 ,β= and γ  = . a−d a+d a

1 1 1 1 1 1 + ⋅ + ⋅ =–1 a−d a a+d a a−d a+d



= (– 1)n 2

⇒ (a – d) a (a + d) = – 3 (a + d ) + (a − d ) + a =–1 (a − d ) a (a + d ) ⇒ 3a = 3 or a = 1 and (1 – d) (1 + d) = – 3 or, 1 – d2 = – 3 or d2 = 4 ⇒ d = ± 2. and,

When

d = 2, α = – 1, β = 1, γ  =

When

d = – 2, α =

4 3 The correct option is (C)

1 . 3

1 , β = 1, γ  = – 1. 3

Therefore, | α – γ | =

HINTS AND EXPLANATIONS

n ( n +1) 2

1⎞ ⎛ 1⎞ 1⎞ ⎛ ⎛ ⋅ (x – 1) ⎜ x − ⎟ ⎜ x − 2 ⎟ … ⎜ x − n ⎟ ⎝ ⎝ 2⎠ ⎝ 2 ⎠ 2 ⎠ ∴ coefficient of xn = (– 1)n ⋅ 2

1 1 1 1 ⋅ ⋅ =– a−d a a+d 3

Then,

1⎞ 1⎞ ⎛ ⎛ ⎜⎝ x − 2 ⎟⎠ … ⎜⎝ x − n ⎟⎠ 2 2



= (–

n ( n +1) 2

1 1 1⎞ ⎛ ⋅ ⎜ −1 − − 2 ... n ⎟ ⎝ 2 2 2 ⎠

n ( n +1) 1)2n + 1 ⋅ 2 2

1 ⎞ ⎛ n ( n − 1) ⎜⎝1 − n+1 ⎟⎠ 2 ⋅ ⋅ = (1 – 2n + 1) ⋅ 2 2 . ⎛ 1⎞ ⎜⎝1 − ⎟⎠ 2

The correct option is (A) 61. Let R = 0.272727… 2 ⇒ 10 R = 27.2727… and, 104R = 2727.2727… ⇒ (104 – 102) R = 2727 – 27 ∴

R=

2700 3 =. 9900 11

59. We have,

x−a x−b b a + = + a b x−a x−b

Similarly, 0.727272… =

⇒

x−a b a x−b − = − x−a b a x−b

Since 0.272727…, x and 0.727272… are in H.P.

⇒

( x − a) 2 − b 2 a 2 − ( x − b) 2 = b ( x − a) ( x − b) a

(x − a − b ) (x − a + b ) ( x − b − a) ( x − b + a) =– b ( x − a) (x − b ) a ⇒ (x – a – b) [a (x – b) (x – a + b) + b (x – b + a) (x – a)] = 0 ⇒ x (x – a – b) [(a + b) x – (a2 + b2)] = 0

62. We have, | ai | = | ai – 1 + 1 | (∵ a, b > 0)

2 2 we take x1 = a + b, x2 = a + b and x3 = 0. a+b 2 2 Since x1 – x2 – x3 = c, we get a + b – a + b = c a+b 2ab ⇒ = c ⇒ a, b, c are in H.P. a+b The correct option is (C)

60. The given product = (– 1)n (x – 1) (2x – 1) (22 x – 1) (23 x – 1) … (2n x – 1) n

= (– 1) ⋅ 2

1+2+…+n

1⎞ ⎛ ⋅ (x – 1) ⎜ x − ⎟ ⎝ 2⎠

8 3 , x, are in H.P. 11 11 3 8 2⋅ ⋅ 11 11 = 48 ⇒ x = 3 8 121 + 11 11 ∴ x is rational. The correct option is (A) ⇒

⇒

2 2 ⇒ x = 0, x = a + b or x = a + b . a+b 2ab a2 + b 2 Since =a+b– < a + b a+b a+b

8 . 11

⇒ ai2 = ai2−1 + 2 ai −1 + 1 Putting i = 1, 2, 3, …, n + 1, we get a12 = 0

a22 = a12 + 2a1 + 1



a32 = a22 + 2a2 + 1 







an2



2 an+1 = an2 + 2an + 1



2 = an−1 + 2an – 1 + 1

n +1

On adding, we get

∑ ai2 i =1

=

n

n

i =1

i =1

∑ ai2 + 2∑ ai + n

Sequence and Series  7.43 65. The general term of the given sequence is

n

2 ≥–n ∑ ai = – n + an+1

a + a + ... + an 1 1 ⇒ 1 2 ≥– ⇒x≥– . n 2 2 The correct option is (B)



⇒

a + a + ... + an a1 + a2 + a3 + ... + an – 1, 1 2 – 1, …, a2 a1

a2 + a3 + ... + an a1 + a3 + ... + an ,, …, a1 a2 a1 + a2 + ... + an −1 are in A.P. an a2 a1 , , …, a2 + a3 + ... + an a1 + a3 + ... + an an are in H.P. a1 + a2 + ... + an −1

 The correct option is (C)

Then, according to the hypothesis, 100a + 10ar + ar2 + 792 = 100ar2 + 10ar + a ⇒ a (r2 – 1) = 8

(1)

2

and, a, ar + 2, ar are in A.P. 2 (ar + 2) = a + ar2



⇒ a (r2 – 2r + 1) = 4 Dividing (1) by (2),

⇒

a ( r 2 − 1) a ( r 2 − 2 r + 1) ( r + 1)( r − 1) ( r − 1)2

3

1, then greatest side will be ar2 and in this case triangle will be formed if a + ar > ar2 ⇒ r2 – r – 1 < 0 ⇒

1− 5 < r < 1+ 5 2 2

r < 1 + 5 [∵r > 1]  (2) 2 If r < 1, then greatest side will be a and triangle will be formed if ar + ar2 > a ⇒ r2 + r – 1 > 0 ⇒

⇒

r
−1 + 5 2 2

5 −1 < r < 1. [∵ 0 < r < 1]  (3) 2 From (1), (2) and (3), possible values of r are given by ⇒

5 −1 1. A

n −1 3 = 2 + 3 (1 − 3 ) = 2 – (1 – 3n – 1) 2 1− 3 1 1 = + ⋅ 3n; 2 2 1 1 ∴ Sn = Σ tn = Σ1+ Σ 3n 2 2





=

1 1 ⋅n+ (3 + 32 + 33 + … + 3n) 2 2

n 1 3 (1 − 3n ) n 3 n + ⋅ = + (3 − 1) . 2 2 1− 3 2 4 The correct option is (B) =



70. Let the A.P. be x, x + y, x + 2y … Then, a = x – (p – 1) y,(1) b = x + (q – 1) y (2) c = x + (r – 1) y (3) ∴ b – c = (q – r) y (4) c – a = (r – p) y (5) a – b = ( p – q) y (6) 2 Let the G.P. be u, uv, uv , … Then, a = uvp – 1(7)

b = uvq – 1,(8)



c = uvr – 1,(9)

HINTS AND EXPLANATIONS

Now, log (ab – c ⋅ bc – a ⋅ ca – b)

ar 2

ar



= (b – c) log a + (c – a) log b + (a – b) log c



= (q – r) y log (uvp – 1) + (r – p) y log (uvq – 1)

 

B

a 2 4

2

C 2 2

Now, a r = a + a r

or, r4 – r2 – 1 = 0 ∴ r2 = 1 ± 5 2



2 2 1+ 5 . ∵ r > 1 ∴ r > 1 ∴ r = 2 Angle C is the greater acute angle

Now, cos C =

a ar 2

=

1 r2

=

1 . 1+ 5

The correct option is (A) 69. Here, the series is 2 + (2 + 3) + (2 + 3 + 9) + (2 + 3 + 9 + 27) + … the 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]

+ ( p – q) y log (uvr – 1), using (4), (5), (6), (7), (8), (9). = y [(q – r) {log u + (p – 1) log v} + (r – p) {log u



+ (q – 1) log v} + (p – q) {log u + (r – 1) log v}



= y [log u (q – r + r – p + p – q) + log v {(q – r) (p – 1)



+ (r – p) (q – 1) + (p – q) (r – 1)}]



= y [log u × 0 + log v × 0] = 0.

Then, log [ab – c ⋅ bc – a ⋅ ca – b] = 0 ∴ ab – c ⋅ bc – a ⋅ ca – b = 1; ∴ ab bc ca = ac ba cb. The correct option is (C) 71. 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, …, ar2n – 1

Then,



n S1 = a + ar + ar2 + … + arn – 1 = a (1 − r ) 1− r

S2 = arn + arn + 1 + arn + 2 + … + ar3n – 1 n n = ar (1 − r ) 1− r

Sequence and Series  7.45 S3 = ar2n + ar2n + 1 + ar2n + 2 + … + ar3n – 1 n

= ar (1 − r ) 1− r



Now, (S2)2 = a2 r2n

74. Let S = n 2

(1 − r )

(1 − r ) 2

n a (1 − r n ) ⋅ ar2n (1 − r ) = S1S3 1− r 1− r Hence, S , S , S are in G.P. 1 2 3

The correct option is (B) n 72. Let r > 1. Then, Sn = a ( r − 1) r −1 un = S1 + S2 + S3 + … + Sn



=



⎤ a ⎡ r ( r n − 1) − n⎥ = ⎢ r − 1 ⎢⎣ r − 1 ⎥⎦

a [(r + r2 + r3 + … + rn) – n] r −1

1

+ ….up to ∞

54

π4 1 1 1 1 1 = + + + + + ….. 90 14 2 4 2 4 34 4 4



=S+



= S+



= S+

1

+

24

1 44

+

1 64

+ ...

1 ⎡1 1 1 ⎤ + + + ....⎥ 24 ⎢⎣14 24 34 ⎦

1 24

×

π4 90

π 1⎞ π4 ⎛ 1− ⎟ = ⎜ ⎝ ⎠ 96 90 16

The correct option is (B) 75. Let a and d respectively be the first term and common difference of the A.P.

n

a ( r − 1) ar ( r − 1) an (1 − r ) + (1 − r ) − r −1 r −1 (1 − r ) 2 n

⎛a ⎞ a a + n⎟ = ⎛⎜ + m ⎞⎟ ⎛⎜ + r ⎞⎟ ⎠ d ⎝d ⎠⎝d ⎠

⇒ ⎜ ⎝

⇒ (x + n)2 = (x + m) (x + r) ⇒ x2 + 2nx + n2 = x2 + (m + r) x + mr ⇒ (m + r – 2n) x = n2 – mr = n 2 −

2 2 73. We have, cos A = b + c − a



and, sin A = ka (k is a content)

∴x= −

2bc

2 2 2 ∴ cot A = b + c − a and similarly, we have

2abck

2 a2 + b2 − c2 cot B = a + c − b , cot C = 2abck 2abck

in A.P.

3

+

2mr (1) m+r and, (a + nd)2 = (a + md) (a + rd)

= – ar ( r − 1) + ar ( r − 1) + an = na 1− r 1− r The correct option is (A)

Given:

4

2

n

n

2

1

Given: n =

∴ r Sn + (1 – r) un



1

+

4

2 2 n = a ( r − 1) + a ( r − 1) + a ( r − 1) + ... + a ( r − 1) r −1 r −1 r −1 r −1

= r⋅

4

∴ S =





1

Given:

=



The correct option is (A)

n

2

b 2 + c 2 − a2 a2 + c 2 − b 2 a2 + b 2 − c 2 , , are 2abck 2abck 2abck

⇒ b2 + c2 –a2, a2 + c2 – b2, a2 + b2 –c2 are in A.P. (Multiplying each term by 2abck) ⇒ –2a2, – 2b2, – 2c2 are in A.P. (Subtracting a2 + b2 + c2 from each term) ⇒ a2, b2, c2 are in A.P. (dividing each term by –2]

=

a ⎡ ⎤ ⎢ Putting d = x ⎥ ⎣ ⎦

n( m + r )  2 [Using (1)]

n ( 2n − m − r ) 2

n , which is the required ratio. 2 The correct option is (D) 76. Let the n numbers in G.P. be a, ar, ar2,…, arn–1 Thus, we have,

⎛1− rn ⎞ ⎟ ⎝ 1− r ⎠

Sn = a ⎜

Also,

S2n = [a + ar + ar2 +….+ arn–1]2

⎛1− rn ⎞ = a2 + (ar)2 + (ar2)2 +…..+(arn–1)2 + 2S ⎟ ⎝ 1− r ⎠

⇒ a2 ⎜



where S denotes the sum of the product of the terms of the G.P. taken two at a time.

HINTS AND EXPLANATIONS



7.46  Chapter 7 where d is the common difference of the A.P. Since, a2n = a1 +(2n–1) d

2

⎛1− rn ⎞ ⎛ 1 − r 2n ⎞ ⇒ a ⎜ = a2 ⎜ ⎟ + 2S ⎟ ⎝ 1 − r2 ⎠ ⎝ 1− r ⎠ 2

a2 S= 2

HINTS AND EXPLANATIONS

⇒

a2 2

⇒

⎡ (1 − r n )2 1 − r 2 n ⎤ − ⎢ ⎥ 2 1 − r 2 ⎥⎦ ⎢⎣ (1 − r )

d=

a2 n − a1 2n − 1

Thus, we have from (1), required expression

⎛ 1 − r n ⎞ ⎡1 − r n 1 + r n ⎤ ⎜ 1− r ⎟ ⎢ 1− r − 1 + r ⎥ ⎝ ⎠ ⎢⎣ ⎥⎦



=



a (1 − r n )(1 + r ) − (1 + r n )(1 − r ) = Sn × × 2 (1 + r )(1 − r )



= Sn ×



r a(1 − r n −1 ) ⎛ r ⎞ = Sn × =⎜ S S × ⎝ 1+ r ⎟⎠ n n–1 1+ r 1− r



a − a2 n = 1 × n 2a1 + a2 n − a1



=

[

2n − 1

n ( a1 − a2 n )( a1 + a2 n ) n ( a12 − a22n ) = 2n − 1 2n − 1

The correct option is (C)

2( r − r n ) a × 2 (1 + r )(1 − r )

an + 1 =

80. We have,

r ∴ k= r +1 The correct option is (C)

∴ a2 =

77. Let the common difference of the given A.P. be t. Then, d = a2 + b2 + c2 ⇒ a + 3t = a2 + (a + t)2 + (a + 2t)2 ⇒ 5t2 + 3 (2a –1) t + 3a2– a = 0 (1)  t is real ⇒ D ≥ 0 ⇒ 9 (2a – 1)2–4 (5) (3a2– a) ≥ 0 ⇒ 24 a2 + 16a – 9 ≤ 0



⇒ a = –1, 0

[  a is integer] 3 When a = 0, from (1), t = 0, . Rejecting both these values 5 since t must be non zero 4 When, a = – 1, from (1), t = 1, ⇒t=1

78. 1 +

5

3 5 ⎛ 2n − 1⎞ + +… + ⎜ ⎝ n ⎟⎠ 2 3 ⎛

1⎞

⎛

1⎞

⎛

1⎞

1 1 1⎞ ⎛ = 2n – ⎜ 1 + + + ..... + ⎟ = 2n – Hn ⎝ 2 3 n⎠ The correct option is (B) 79. We have, a21 – a22 + a23 – a24 + ……+ a22n–1 – a22n = (a1– a2) [a1 + a2 + a3 +….+ a2n]

[ a1 – a2 = a3 – a4 = ……..a2n–1 – a2n]

=–d×

1 1− an

1 1 1 − a1

1 − a1 1 − a1 = 1 − a1 − 1 −a1

a3 = a1, ∴

Since

1

= 1 and a3 = 1 − a2 1− 1 − a1

1 − a1 = a1 −a1

a21 – a1 + 1 = 0 ⇒ a1 = – w or – w2

⇒

a5 =

1 1 = 1 1− 1 − a4

=

1 − a3

1 − a3 − a3

1 − a1 = = a1 = a3 and so on −a1

∴

a1 = a3 = a5 ………a2001

= (–w)2001 or (–w2)2001 = –1 The correct option is (B) Thus, (a 2001)

2001

81. Given: b2 = ac(1)

= (2–1) + ⎜ 2 − ⎟ + ⎜ 2 − ⎟ + ..... + ⎜ 2 − ⎟ ⎝ ⎝ 2⎠ ⎝ 3⎠ n⎠



=

Now,

1 ⇒ − – 70 < a < −1 + 70 3 12 3 12

∴ a + b + c + d = – 1 + 0 + 1 + 2 = 2 The correct option is (A)

]

2n [ 2a1 + ( 2n − 1)d ] (1) 2

⎛ 3b ⎞ ⎛ 5c ⎞ ⎛ a⎞ and, 2 log ⎜ ⎟ = log ⎜ ⎟ + log ⎜ ⎟ ⎝ 5c ⎠ ⎝ a⎠ ⎝ 3b ⎠

⎛ 5c a ⎞ ⎛ 5c ⎞ . = log ⎜ ⎟ = – log ⎝ 3b ⎠ a 3b ⎟⎠

= log ⎜ ⎝

⎛ 3b ⎞ ⎜⎝ 5c ⎟⎠

⎛ 3b ⎞ 5 = 0 or b = c (2) ⎟ ⎠ 5c 3

i.e., 3 log ⎜ ⎝

From (1) and (2), we have, 2 a = b = 25c

c

9

Now, we have,

Sequence and Series  7.47 5c 8c 25c +c= < =a 3 3 9

and hence, a, b, c cannot form the sides of a triangle. The correct option is (D) 82. Given: b2 = ac  ( a, b, c are in G.P.) and, 2(log 2b – log 3c) = log a – log 2b + log 3c – log a  ( given terms are in A.P)

Since d1, d2 and d3 are given to be in H.P, therefore, 1 1 1 1 − = − d2 d1 d3 d2 ⇒ 

[Using results (1), (2), (3)] S2 − S3 S1 − S2 ⇒ = ( S2 − n)( S1 − n) ( S2 − n)( S3 − n)

2

⎛ 2b ⎞ ⎛ 3c ⎞ = log ⎜ ⎟ ⇒ b = 3c ⎝ 2b ⎠ 3c ⎟⎠ 2

1 1 1 1 − = − S3 − n S2 − n S2 − n S1 − n

S1 − S 2 = S 2 − S 3 ⇒ n = 2S3S1 − S1S2 − S2 S3 S1 − 2S2 + S3 S3 − n S1 − n

⇒ log ⎜ ⎝

⇒

2 Now, a = b = 3b = 9c

The correct option is (C)

2

c

4

∴ a is the largest side

9c 2 81 + c2 − c2 + − b c a 4 16 Now, cos A = = negative = 3 2bc 2× c ×c 2 2

2

2

∴ A > 90º, ∴ triangle is obtuse. The correct option is (C)

2

83. The middle term of the 4n + 1 terms is the (2n + 1) th term. Let it be m. The middle term of (2n + 1) terms is the (n + 1) th term. Thus, the middle term of the A.P. is = m – (n + 1– 1)2 = m –2n and the middle term of the G.P is

1 =m ⎛ ⎞



⎜⎝ 2 ⎟⎠

n + 1−1

=

m 2n

According to the given condition, we have ⇒

m – 2n = m=

n.2n + 1 2n − 1

2( S1 − n) (1) n( n − 1)

Similarly, d2 = 2(S 2 − n )(2) and,

2 2 = m2 (2 m + 1)2 + 16 m ( m + 1)

4

n2 ( n2 + 3n + 1)  2 The correct option is (B) =

n( n − 1) 2 d3 = (S 3 − n )(3) n( n − 1)

[Put m =

86. Since p, q, r are in A.P. we have p – q = q – r = k (say) Given:

2n

84. Let the common difference of the three A.P.s be d1, d2 and d3 Then, we have n S1 = [ 2.1 + ( n − 1)d1 ] 2 d1 =

⎡ 2m( 2m + 1) ⎤ 3 3 3 3 = ⎢ ⎥ + 8 × 2{1 + 2 + 3 + .... + m } 2 ⎣ ⎦

m

The correct option is (B)

⇒

85. We have, Sn = 13 + 3.23 + 33 + 3.43 + 53 +… Let n = 2m. Then, S2m = (13 + 33 + 53 …..to m terms)  + 3 (23 + 43 + 63 + …..to m terms) 3 3 3 3 = {1 + 2 + 3 + 4 + ……+ (2m–1)3 + (2m)3}  – { 23 + 43 +…..+ (2m)3} +3{ 23 + 43 + 63 +….+ (2m)3}

⇒

n ] 2

(1)

a−x a− y a−z = = qy px rz ⎛a ⎞ ⎛a ⎞ ⎛a ⎞ ⎛a ⎞ ⎜⎝ x − 1⎟⎠ − ⎜⎝ y − 1⎟⎠ ⎜⎝ y − 1⎟⎠ − ⎜⎝ z − 1⎟⎠ = q−r p −q



(by componendo–dividendo)

⇒

a a a a − = −  x y y z

⇒

1 1 1 1 2 1 1 − = − ⇒ = + x z x y y y z

[Using equation (1)]

∴ x, y, z are in H.P. The correct option is (C) 87. We have, S = 1 + (1 + a)b + (1 + a + a2) b2 + (1 + a + a2 + a3) b3 +…∞ = (1 + b + b2 +….∞) + a (b + b2 + b3 +…..∞) + a2 (b2 + b3 + ……∞) +….∞

HINTS AND EXPLANATIONS

b+c=

7.48  Chapter 7

=

1 ab a 2b 2 1 + ....∞ = + + (1 − b)(1 − ab) 1− b 1− b 1− b

The correct option is (A) 88. We have, i.e.,

2

2 − n −1

2

− n −1

2 − n −1

2 − n −1

+y

2 − n −1

)

= 34 – 30 d > 0 for d =

x−y b0 b1 b2 bn−1 . . .... = bn b1 b2 b3 bn

89. Given: a1 + 5d = 2 Let y = a1 a4 a5 = a1 (a1 + 3d) (a1 + 4d) = (2 – 5d) (2 – 2d) (2 – d) (Putting a1 = 2 – 5d) = 2 (4 – 16d + 17d 2 – 5d3) The value of d at which y attains maxima is given by

dy = 0 dx

⇒ – 16 + 34 d – 15 d2 = 0

2 3

< 0 for d = 8 5

8 5

The correct option is (A)

The correct option is (B)

HINTS AND EXPLANATIONS

dx

2

Hence, y is maximum for d =

b an = n −1(Putting n in place of n – 1) bn



d 2y

34 ± 14 2 8 = , 3 5 30



2

= bn + 1 an + 1

∴ a1, a2… an =

d=

Now,

( ) − (y ) = (x − y ) (x

bn = x

2

− n −1

⇒

(by calculus)

90. Let tn denotes the nth term of the sequence 1, 2, 4, 7 Let S = 1 + 2 + 4 + 7 + ….+ tn Again, S = 1 + 2 + 4 + …..+ tn–1 + tn On subtracting, we get 0 = 1 + [1 + 2 + 3 +……(n – 1) terms] – tn ⇒

tn = 1 +

n ( n − 1) n2 − n + 2 = 2 2

which denotes the first term of the nth row which contains n terms in A.P., having common difference 1. Hence, we have, n ⎡ 2t n + ( n − 1)1⎤⎦ 2⎣ n n ( n2 + 1) = ⎡⎣ n2 − n + 2 + n − 1⎤⎦ = 2 2 The correct option is (C) S=



Previous Year’s Questions 91. ∵1,log3 31− x + 2 ,log3 ( 4 ⋅ 3x − 1) are in AP



∴ 2 log3 (31− x + 2)1/ 2 = log3 3 + log3 ( 4 ⋅ 3x − 1)

⇒ log3 (31− x + 2)1/ 2 = log3 3( 4 ⋅ 3x − 1) Let 3x = t, then 3 + 2 = 12t − 3 t ⇒ 12t2− 5t − 3 = 0 ⇒ (3t + l)(4t− 3) = 0

1 3 ⇒ t=− , 3 4 3 x ⇒ 3 = (neglecting the negative value) 4



⎛ 3⎞ ⇒ log3 ⎜ ⎟ = x ⎝ 4⎠



⇒ x = log3 3 − log3 4



⇒ x = 1 − log3 4

The correct option is (B) 92. The product 21/4 ⋅ 41/4 ⋅ 81/16 …



= 21/ 4 ⋅ 22 / 8 ⋅ 23/16...





= 24 ⎣



1⎡ 2 3 ⎤ ⎢1+ 2 ⋅ 22 + ...⎥ ⎦

⎡ 1 ⎤ ⎥ ⎢ 1⎢ 1 + 2 ⎥ 4 ⎢ 1 ⎛ 1⎞ 2 ⎥ ⎥ ⎢1− 1− 2 ⎜⎝ 2 ⎟⎠ ⎥⎦ 2 ⎢⎣



=



= 24

1



[ 2 + 2]

=2 The correct option is (B)

93. Since fifth term of a GP = 2 ∴ ar4 = 2 where a and r are the first term and the common ratio respectively of a GP. Now required product

Sequence and Series  7.49 = a × ar × ar2 × ar3 × ar4 × ar5 × ar6 × ar7 × ar8 = a9r36 = (ar4)9 = 29 = 512 The correct option is (B) 94. Key Idea : The number of triangles those can be formed using n points = nC3 ∵Tn = nC3 ∴ Tn +1 − Tn = 21 ⇒ n +1C3 − nC3 = 21





(∵ nC2 + nC3 =





⇒ nC2 = 21









n −1

C3 )

n( n − 1) = 21 ⇒ n2 − n − 42 = 0 2 ⇒ ( n − 7)( n + 6) = 0 ⇒

















96. f (1) = 7 f (1 + 1) = f (1) + f (1) ⇒ f (2) = 2 ×7 Also, f (3) = 3 ×7



n( n + 1) 2 The correct option is (D)

r=0



r

r

n−r

n

∑

r=0

t n n n 1 n ∑ = sn ⇒ Sn = 2 2 r = 0 nCr 2 n 1 = a + ( m − 1)d (1) n

∞

1

∑ an = 1 − a

n= 0 ∞

1 x

⇒a = 1−

1

1

∑ bn = 1 − b ⇒ b = 1 − y

n= 0 ∞

1

1

∑ cn = 1 − c ⇒ c = 1 − z



⎛ 1⎞ 1 1 2 ⎜1 − ⎟ = 1 − + 1 − y⎠ x y ⎝





p [2a + ( p − 1)d ] p 2 2a + ( p − 1)d p 2 1 = 2 ⇒ 1 = q 2a1 + ( q − 1)d q q [2a1 + ( q − 1)d ] 2 ⎛ p − 1⎞ a1 + ⎜ d ⎝ 2 ⎟⎠ p = q ⎛ q − 1⎞ a1 + ⎜ d ⎝ 2 ⎟⎠ For

=

r

n= 0 a, b, c are in A.P. 2b = a + c

=7

∑ nC

x=

z=



r =1

Also, t n =

r=0

1 = a + ( n − 1)d (2) m 1 1 From (1) and (2) we get a = ,d= mn mn Hence, a − d = 0 The correct option is (A)

n

∴ ∑ f ( r ) = 7(1 + 2 + ......... + n)

n−r

n

2 1 1 = + y x z ⇒ x, y, z are in H. P The correct option is (D) 101. Given condition implies that

The correct option is (D)

n

n

∑ nC

And Tn =



r=0

Cr

98. Given that Tm =



= 2 log 2 − log e

∑ nC

n

=

The correct option is (A)

y=

⎛1 1 1 ⎞ = 1 − 2 ⎜ − + − ........⎟ (Rearranging the terms) ⎝2 3 4 ⎠   ⎛ 1 1 1 ⎞ = 2 ⎜1 − + − + ........⎟ − 1 ⎝ 2 3 4 ⎠

97. Given t n =

r=0

⇒ tn =

100.

1 1 1 1 1 − + + − − ......... 2 2 3 3 4

n

r+n−r

( n − 1)n2 n2 ( n + 1) . + n2 = 2 2 The correct option is (D)

⎛ 4⎞ = log ⎜ ⎟ . ⎝ e⎠



n

∑

=

1 1 1 − + − .......... 1.2 2.3 3.4 = 1−

2t n =

99. If n is odd then (n − 1) is even and so the sum of odd terms

(∵ n ≠ −6) ⇒n=7 The correct option is (B) 95.



n−r n

Cr

(∵ C n

r

Adding above two equalities we write

= nCn − r

)

a6 , p = 11, q = 41, a21

a6 11 = a21 41

The correct option is (D)

HINTS AND EXPLANATIONS



⇒ nC2 + nC3 − nC3 = 21



7.50  Chapter 7

102. Given that

1 1 1 1 1 1 − = − = .... = − = d ( say ) a2 a1 a3 a2 an an −1

Then a1a2 =

a −a a −a a1 − a2 , a2 a3 = 2 3 ,...., an−1an = n−1 n d d d

∴ a1a2 + a2 a3 + .... + an−1an = Also,



a1 − an d

1 1 = + ( n − 1)d an a1

a −a ⇒ 1 n = ( n − 1)a1an d

The correct option is (D) 103. Given that arn-1 = arn + arn + 1 ⇒ 1 = r + r2 5 −1 ∴r = 2 The correct option is (D) 104. Using the condition A.M. ≥ G.M., we write p2 + q2 ≥ pq 2 1 ⇒ pq ≤ 2 (p + q)2 = p2 + q2 + 2pq

HINTS AND EXPLANATIONS





2 4 4⎛ 1 1 ⎞ S = + 2 ⎜1 + + 2 + .... ⎟ 3 3 3 ⎝ 3 3 ⎠ ⎛ ⎞ 2 4 4⎜ 1 ⎟ 4 4 3 4 2 6 2 6 ⇒ S= + 2⎜ ⎟= + 2 = + = ⇒ S = 1 3 3 3 ⎜1− ⎟ 3 3 2 3 3 2 3 3 ⎜ ⎟ ⎝ 3⎠ ⇒S =3 The correct option is (B) 107. Till 10th minute number of counted notes = 1500 n 3000 = [2 × 148 + ( n − 1)( −2)] = n[148 − n + 1] 2

n2−149n + 3000 = 0 n = 125, 24

n = 125 is not possible. Total time = 24 + 10 = 34 minutes. The correct option is (A) 108. 1 2 3 4 5 6 …… 200 200 200 240 280 ……. …….. Sum = 11040 120 + 80 + 160 + 40 + 200 + 240 + … = 11040



n [240 + ( n − 1)40] = 10920 2

⇒ p+q≤ 2

The correct option is (D) 105. Let a, ar, ar2, ar3 be the first four terms of a G.P., then a + ar = 12    (1) ar2 + ar3 = 48  (2) dividing (2) by (1), we have ar 2 (1 + r ) =4 a( r + 1)

⇒ r2 = 4  if r ≠ −1 ∴ r = −2 also, a = −12  (using (1)) The correct option is (B) 2 6 10 10 1 06. Let S = 1 + + 2 + 3 + 4 + ......  (1) 3 3 3 3 1 1 2 6 10 S = + 2 + 3 + 4 + ....  (2) 3 3 3 3 3

n [2a + ( n − 1)d ] + 80 + 40 = 11040 2

n [6 + n − 1] = 546 n (n + 5) = 546 n = 21

The correct option is (C) 109. Statement 1 has 20 terms whose sum is 8000 And statement 2 is true and supporting statement 1. th 2 2 2 ∵ k bracket is (k – 1) + k(k – 1) + k = 3k – 3k + 1. The correct option is (B) 110. 100(T100 ) = 50(T50 )



⇒ 2[a + 99d ] = a + 49d





⇒ a + 149d = 0

⇒ T150 = 0 The correct option is (D)

Subtracting (2) from (1), we get

111. tr = .7 + .77 + . . .r times

1 4 4 4 ⎛ 1⎞ S ⎜1 − ⎟ = 1 + + 2 + 3 + 4 + .... ⎝ 3⎠ 3 3 3 3





( 7 = (1 − 10 ) 9

= 7 10 −1 + 10 −2 + 10 −3 + ..... + 10 − r −r





)

Sequence and Series  7.51 Now,

The correct option is (C)

⎞ 7⎛ S20 = ∑ t r = ⎜⎜ 20 − ∑10 − r ⎟⎟ 9 r =1 r =1 ⎝ ⎠ 7 7⎛ 1 ⎞ 179 + 10 −20 = ⎜ 20 − 1 − 10 −20 ⎟ = 81 9⎝ 9 ⎠ 20

(

The correct option is (B) 112.











)

(

)

2q = p + r ⇒ −2(α + β ) = 1 + αβ ⎛ 1 1⎞ 1 ⇒ −2 ⎜ + ⎟ = +1 ⎝ α β ⎠ αβ 1 = −9 αβ

Equation having roots α , β is 9 x 2 + 4 x − 1 = 0

tr =

∑ r 3 = r 2 ( r + 1)2 = 1 (r + 1)2 4 4r 2 ∑ (2r − 1)

Now, S9 =

1 1 + =4 α β

⇒



115.

α,β =

−4 ± 16 + 36 2×9

2 13 α −β = . 9 The correct option is (D) 113. Let numbers be a, ar , ar 2 Now, 2( 2ar ) = a + ar 2 [ a ≠ 0 ]

=

1 9 ∑ ( r + 1)2 , let t = r + 1 4 r =1 1 ⎛ 10 2 ⎞ ∑ t − 1⎟⎠ = 96 . 4 ⎜⎝ t =1

  The correct option is (A) 116. Let ‘a’ be the first term and d be the common difference 2nd term = a + d, 5th term = a + 4d, 9th term = 4 + 8d a + 4 d a + 8d 4 d 4 = ∴ Common ratio = = = a+d a + 4 d 3d 3 The correct option is (C) 117. Given series is S = =

42 52

82 5

2

+

122 5

2

+

16 2 52

+ …10 terms

( 22 + 32 + 4 2 + …10 terms)

16 ⎛ 11.12.23 ⎞ 16 × 505 = ⎜ − 1⎟ = ⎠ 25 25 ⎝ 6 ∴ m = 101





⇒ 4r = 1 + r 2

The correct option is (C)





⇒ r 2 − 4r + 1 = 0





⇒r = 2± 3

118. 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c) ⇒ (15a)2 + (3b)2 + (5c)2 – 45ab – 15bc – 75ac = 0 ⇒ (15a – 3b)2 + (3b – 5c)2 + (15a – 5c)2 = 0 It is possible when 15a – 3b = 0 and 3b – 5c = 0 and 15a – 5c = 0 15a = 3b = 5c a b c = = 1 5 3 \ b, c, a are in A.P. Hence, the correct option is (A) 119. Let the first term be ‘a’ and common difference be ‘d’ a9 + a43 = 66 (a + 8d) + (a + 42d) = 66 2a + 50d = 66 a + 25d = 33 (i)

r = 2 + 3 (Positive value) The correct option is (D) 9 1 8 9 114. S = 10 + 2 ⋅ 11 ⋅ 10 + ... + 10 ⋅ 11









11 ⋅ S = 111 ⋅ 108 + ... + 9 ⋅ 119 + 1110 10 ⇒

1 S = 109 + 111 ⋅ 108 + 112 ⋅ 107 + ... + 119 − 1110 10

⎛ ⎛ 11 ⎞10 ⎞ ⎜ ⎜ ⎟ −1 ⎟ 1 10 ⎟ − 1110 ⇒ S = 109 ⎜ ⎝ ⎠ ⎜ 11 ⎟ 10 −1 ⎟ ⎜ ⎜ 10 ⎟ ⎝ ⎠ 1 10 10 10 ⇒ S = 11 − 10 − 11 10



11   S = 10



9   S = 100 ⋅ 10



  ⇒ k = 100 .

12



∑ a4 k +1 = 416

k =0

a1 + a5 + a9 + … + a49 = 416 Number of terms of the above series 1 + (n – 1)4 = 49 (n – 1)4 = 48

(A)

HINTS AND EXPLANATIONS

20

7.52  Chapter 7 S = 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + … S = S1 + S2 S1 = 12 + 22 + 32 + 42 + … S2 = 22 + 42 + 62+ … Sum of n terms of series S1 ( 2n + 1) S1n = n( n + 1) 6 Sum of n1 terms of series S2 where

n = 13 Common difference of series A is 4d a1 = a n [2a + ( n − 1)4 d ] = 416 2 13 [2a + 48d ] = 416 2 13[a + 24d] = 416 a + 24d = 32 Solving equation (i) and (ii), we get d = 1 and a = 8





(ii)

2 In the series a12 + a22 + … + a17 = 140 n

The nth term is (8 + (n – 1) × 1)2 Tn = (n + 7)2 17

17

n =1

n =1

∑ Tn = ∑ ( n + 7)

⇒

HINTS AND EXPLANATIONS

17

17

17

n =1

n =1

n =1

= ∑ n2 + 14∑ n + 49∑1 n( n + 1)( 2n + 1) 14 n( n + 1) ⎫ + + 49n ⎬ n = 17 6 2 ⎭

17 × 18 × 35 18 + 14 × 17 × + 49 × 17 = 140 m 6 2 ⇒ 1785 + 2142 + 833 = 140m ⇒ 140m = 4760 m = 34 120. We are required to find the sum of 20 terms and 40 terms, therefore the number of terms are even ⇒

n1 =



S2 n = ∑ ( 2n1 ) 2 = 4∑ n12 1

= 4 n1 ( n1 + 1)



( 2n1 + 1) 6

( 2n + 1) ( 2n + 1) + 4 n1( n1 + 1) 1 6 6 A = S where n = 20 and n1 = 10 Solving, we get,

2

n 2







S = n( n + 1)

A = 20 × 21 ×

B=S Solving, we get

B = 40 × 41 ×

41 21 + 4 × 10 × 11 × = 4410 6 6

81 41 + 4 × 20 × 21 × = 22140 + 11480 6 6

= 33620 B – 2A ⇒ 33620 – 2 × 4410 = 100λ ⇒ 24800 = 100λ λ = 248

CHAPTER

Limits

8

LEARNING OBJECTIVES After reading this chapter, you will be able to:  Understand the limit of a function and the method to calculate right and left hand limit of a function

 Learn about algebra of limits, evaluation of limits, methods of factorization and rationalization, trigonometric limits, exponential and logarithmic limits  Know how to evaluate limits using L’hospital’s rule

LIMIT OF A FUNCTION

Method for Finding Right Hand Limit

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

To evaluate lim f ( x )

lim f ( x ) = l

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.

ERROR CHECK A number is said to be a limiting value only if it is finite and real, otherwise we say that the limit does not exist.

x → a+

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

Left Hand 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

x→ a−

or

f (a – 0) = l2

Method for Finding Left Hand Limit To evaluate lim f ( x ). 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

Right Hand 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 lim f ( x ) = l1 or f (a + 0) = l1

x → a+

ERROR CHECK For finding lim f( x ), we study the behaviour of the function f x →a

in the neighbourhood of ‘a’ and not at ‘a’. Thus, the function f may or may not be defined at x = a.

8.2  Chapter 8

SOLVED EXAMPLES

and

x 2 − 9 x + 20 = x→5 x − [ x] (A) 1 (B) 0 (C)  Does not exist (D)  Cannot be determined

\

1. lim

Solution: (C) x 2 − 9 x + 20 ( x − 4)( x − 5) = lim x→5 x→5 x − [ x] x − [ x] ( x − 4)( x − 5) LHL = lim x → 5− x − [ x]  (5 − h − 4)(5 − h − 5) = lim (h > 0) h→0 (5 − h) − [5 − h] lim

= lim



h→0

\ LHL = 0

(1 − h)( − h) (1 − h)( − h) = lim h→0 5−h−4 (1 − h) 

( x − 4)( x − 5) x→5 x − [ x] 

Also, RHL = lim

= lim



= lim

h→0

h→0

\ RHL = 1

(5 + h − 4)(5 + h − 5) (5 + h) − [5 + h] 

(1 + h)( h) (1 + h)h = lim h→0 5+h−5 h 

As LHL ≠ RHL \ Limit does not exist. x − ai 2. If Ai = , i = 1, 2, …, n and if a1 < a2 < a3 < … < an. | x − ai | Then lim ( A1 A2 ... An ) , 1 ≤ m ≤ n x → am



(A)  is equal to (–1)m (B)  is equal to (–1)m + 1 (C)  is equal to (–1)m – 1 (D)  Does not exist Solution: (D) We have, Ai = and

x − ai , i = 1, 2, …, n | x − ai |

a1 < a2 < … an – 1 < an.

Let x be in the left neighbourhood of am. Then, x – ai < 0 for i = m, m + 1, … n

and

x – ai > 0  for  i = 1, 2, …, m – 1 x − ai = –1, for i = m, m + 1, …, n − ( x − ai ) x − ai Ai = = 1, for i = 1, 2, …, m – 1 x − ai Ai =

Similarly, if x is in the right neighbourhood of ai Then x – ai < 0 for i = m + 1, …., n and x – ai > 0 for i = 1, 2, …, m x − ai \ Ai = = – 1 for i = m + 1, … n − ( x − ai ) x − ai and Ai = = 1 for i = 1, 2, …, m x − ai Now, and

lim ( A1 A2 ... An ) = (–1)n – m + 1

x → am−

lim ( A1 A2 ... An ) = (–1)n – m

x → am+

Hence, lim ( A1 A2 ... An ) Does not exist. x → am

INDETERMINATE FORMS If a unique value cannot be assigned to f (a), then f (x) is said to be inderminate at x = a. 0 Most general of all indeterminate forms is , others 0 being 1 1 0−0 0 1. ∞ – ∞ = − = = which is indeterminate 0 0 0 0 and hence is (∞ – ∞) ∞ 1/ 0 0 = = which is indeterminate and hence is 0 ∞ 1/ 0 ⎛ ∞⎞ ⎜⎝ ⎟⎠ ∞ 1 0 3. 0 × ∞ = 0. = which is indeterminate and hence is 0 0 (0 × ∞) 4. 1∞ Let y = 1∞ ⇒ log y = ∞ log 1 = ∞ × 0 which is indeterminate and hence is 1∞ 2.

5. 00 Let y = 00 ⇒ log y = 0 ⋅ log 0 = 0 × ∞ which is indeterminate and hence is 0º 6. ∞º Let y = ∞º ⇒ log y = 0 ⋅ log ∞ = 0 × ∞ which is indeterminate and hence is ∞º.

Limits  8.3

ALGEBRA OF LIMITS

Solution: (B)

If lim f ( x ) = l and lim g ( x ) = m, then following results

Since, 0 ≤ (rx) < 1 for r = 1, 2, 3, …, n

are true:

⇒ 0 ≤

x→a

x→a

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

x→a

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

⇒ 0 ≤

x→a

x→a

x→a

In particular,

r =1

lim g ( x) x→a

x→a

= em.

n

0



n

⇒

2

≤

∑ (rx)

r =1

n

x→a


0. x →∞ x In order to find the limit of a function of the type

QUICK TIPS ⎞ ⎟ is x + 6 − 3⎠ x −3

n( n + 1) . 2

∞, ⎧ ⎪ 1, ⎪ ■ lim a n = ⎨ 0, n→∞ ⎪ ⎪⎩does not exist,

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

Limits  8.5

lim

■

x→∞

p

p −1

+ … … + ap − 1 x + ap

q

q −1

+ … … + bq − 1 x + bq

a0 x + a1x b0 x + b1x

⎧ a0 ⎪b , 0 = ⎪⎪0, ⎨ ⎪∞, ⎪ ⎪⎩



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

Some Useful Summations n ( n + 1) 2 n ( n + 1) ( 2n + 1) 2. S n2 = 12 + 22 + 32 + … + n2 = 6 1. S n = 1 + 2 + 3 + … + n =

⎡ n ( n + 1) ⎤ 3. S n3 = 13 + 23 + 33 + … + n3 = ⎢ ⎥ 2 ⎣ ⎦ n–1

2

4. S ar = a + ar + ar + … + ar ­provided r < 1.

n–1

2

a (1 − r n ) = ; 1− r

SOLVED EXAMPLES 7. The value of ⎤ ⎛ n −1 ⎞ ⎛n−2 ⎞ 1 ⎡ ⎛ n ⎞ ⎢ ⎥ 1 + 2 + 3 + ... + ⋅ k k k n 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∑ ∑ ∑ n → ∞ n4 ⎢ ⎝ ⎥⎦ ⎝ k =1 ⎠ ⎝ k =1 ⎠ ⎣ k =1 ⎠ will be 1 1 1 1 (A)  (B)  (C)  (D)  24 12 6 3 lim

Solution: (A) The (r + 1)th term of the series is n−r

\

S=

1 n 3 ∑ [r − 2nr 2 + (n2 − 2n − 1)r + n2 ]  2 r =1

⇒

S=

1 2

 Solving and rearranging, we have 1 4 S = ( n − 11n3 − 19n2 + 6 n)  24 1 ⎛ n4 − 11n3 − 19n2 + 6 n ⎞ S ∴ lim 4 = lim ⎜ ⎟ n→∞ n n → ∞ 24 ⎝ n4 ⎠

k =1  tr + 1 = (r + 1)[1 + 2 + 3 + … (n – r) terms]

⇒

tr + 1 =

⇒

tr + 1 =

⇒

tr + 1 =

⇒

tr + 1 =

Now,

S=

1 ( r + 1) ( n − r )( n − r + 1) 2  1 2 ( r + 1)( n − rn + n − rn + r 2 − r ) 2  1 ( r + 1)( r 2 − (1 + 2n)r + n2 ) 2  1 3 ( r − 2nr 2 + ( n2 − 2n − 1)r + n2 ) 2  n −1

∑ tr + 1

r=0



1 ⎛ 11 19 6 ⎞ = lim ⎜1 − − 2 + 3 ⎟  24 n → ∞ ⎝ n n n ⎠

∴

1

n −1 ⎧

⎪

n−r

⎪⎫

r =0 ⎩

k =1

⎭

∑ ⎨⎪(r + 1) ∑ k ⎬⎪ n→∞ n 4 lim

n

=

1  24

⎛ r 3 − 1⎞

∏ ⎜⎝ r 3 + 1⎟⎠ n→∞

8. lim

r=3

1 6 (A)  (B)  3 7 2 (C) − (D)  None of these 3 Solution: (B) (n – 2)th factor of the series is tn =

n − 1 n2 + n + 1 ⋅ n + 1 n2 − n + 1

Therefore, required limit = lim t3t 4 t5 ... t n − 2 t n − 1t n n→∞



⎡⎛ 2 3 4 n − 3 n − 2 n − 1⎞ = lim ⎢⎜ ⋅ ⋅ ... ⋅ ⋅ n→∞ ⎝ 4 5 6 n −1 n n + 1⎟⎠ ⎣  2 ⎛ 13 ⎞ 21 31 n + n +1 ⎤ ⋅ ⎜ ⎟ ⋅ ⋅ ... 2 ⎥ ⎝ 7 ⎠ 13 21 n − n + 1⎦



= lim



tr + 1 = ( r + 1) ∑ k ⇒

2

⎡ ⎧ n( n + 1) ⎫ ⎧1 ⎫ ⎢ ⎨ 2 ⎬ − 2n ⎨ 6 n ( n + 1)( 2n + 1)⎬ ⎩ ⎩ ⎭ ⎭ ⎣ ⎤ ⎧1 ⎫ + ( n2 − 2n − 1) ⎨ n( n + 1)⎬ + n2 ( n) ⎥ ⎩2 ⎭ ⎦

2 ⋅ 3 n2 + n + 1 6 ⋅ = . n → ∞ n( n + 1) 7 7

9. If [x] denotes the integral part of x, then 1 ⎛ n 2 ⎞ ⎜ ∑ [k x ]⎟ = n→∞ n3 ⎝ ⎠ k =1 x (A)  0 (B)  2 x x (C)  (D)  3 6

lim

8.6  Chapter 8 Solution: (C)

Solution: (C) ⎞ 1 ⎛ 2 ⎜ ∑ [k x ]⎟ 3 n→∞ n ⎝ ⎠ k =1 n

L = lim

Since k2x – 1 ≤ (k2x) < k2x n

∑ (k 2 x − 1) ≤

⇒

k =1

⇒

⎛ ⎞ x ⎜ ∑ k 2 ⎟ − ∑ (1) ≤ ⎝ k =1 ⎠ k =1 n

n

n

∑ (k 2 x)
0, a ≠ 1) x→0 x 7. lim

x →∞

⎛ 9. lim ⎜1 + x →∞ ⎝

x

a⎞ a ⎟⎠ = e x

⎛ 1 ⎞ 10. lim ⎜1 + x →∞ ⎝ f ( x ) ⎟⎠

= e, where f (x) → ∞ as x → ∞.

x→a

Some Useful Expansions

x 2 x3 + − ... to ∞, – 1 < x ≤ 1 2 3

x 2 x3 − − ... to ∞, – 1 ≤ x < 1 2 3 ( x log a) 2 5. ax = ex log a = 1 + x log a + + ... to ∞ 2! n ( n − 1) 2 x + ... to ∞, – 1 < x < 1, 6. (1 + x)n = 1 + nx + 2! n being any negative integer or fraction. The expansion formulae mentioned above can be used with advantage in simplification and evaluation of limits.

lim

x→0

Is the above statement true? No. If fact,

lim [ − x ] = lim [0 − x ] = lim − 1 = –1

x →0

x →0

x→ 0

Thus, limit must be applied only after removing [⋅] sign.

cos x − 1 + x4

x sin{x} , where {x} denotes the fractional part of x, x →1 x − 1 is equal to (A) –1 (B) 0 (C)  1 (D)  Does not exist 19. lim

x sin{x} x −1  Let x = 1 – h, as x → 1, h → 0 (1 − h) sin{1 − h} ⇒ LHL = lim h→0 h  (1 − h) sin (1 − h) ⇒ LHL = lim h→0 h  (1 − h) \ LHL = lim sin (1) = ∞ h→0 h x sin ( x ) Now, RHL = lim x → 1+ x − 1  Let x = 1 + h, as x → 1, h → 0 x → 1−

4. loge (1 ­– x) = – x –

For example,

x →0

LHL = lim

x x 2 x3 + − + ... to ∞ 1! 2 ! 3!

3. loge (1 + x) = x –

If [⋅] denotes the greatest integer function, then lim [ − x ] = [0] = 0

Solution: (D)

x x 2 x3 + + + ... to ∞ 1! 2 ! 3!

2. e–x = 1 –

ERROR CHECK

SOLVED EXAMPLES

f ( x)

11. lim (1 + f ( x )1/ f ( x ) = e.

1. ex = 1 +



x2 2

(1 + h) sin (1 + h) h  (1 + h) sin h ⇒ RHL = lim = lim (1 + h) h→0 h→0 h  ⇒ RHL = lim

h→0

\ RHL = (1 + 0) = 1 

Since LHL ≠ RHL, \ the limit of the function Does not exist at x = 1.

8.10  Chapter 8 ⎛ b⎞ ⎡ x ⎤ lim ⎜ ⎟ ⎥ where a > 0, b > 0 and [x] denotes x →0 ⎝ x⎠ ⎢ ⎣a⎦ greatest integer less than or equal to x is 1 b (A)  (B)  b (C)  (D) 0 a a 20.

+

Solution: (D)

where [x] denotes the greatest integer less than or equal to x, then lim f ( x ) is equal to x →0

1 (A) – (B)  1 2 π (C)  (D)  Does not exist 4 Solution: (D)

⎛ b ⎞ ⎡0 + h⎤ ⎛ b⎞ ⎡ x ⎤ lim ⎜ ⎟ ⎥ = lim ⎜ x →0 ⎝ x⎠ ⎢ h → 0 a ⎝ 0 + h ⎟⎠ ⎢⎣ a ⎥⎦ ⎣ ⎦ 



+

⎛ b⎞ ⎡ h⎤ = lim ⎜ ⎟ ⎢ ⎥ = 0 h → 0 ⎝ h⎠ ⎣ a ⎦





⎧ sin[ x ] , [ x] ≠ 0 ⎪ 21. If f (x) = ⎨ [ x ] , where [x] denotes the ⎪0 0 , [ x ] = ⎩ greatest integer ≤ x, then lim f ( x ) equals

π /4 π = . 1 4 tan −1 ([h] + h) RHL = lim f (0 + h) = lim h→ 0 h→ 0 [h] − 2h 

(B) –1 (D)  None of these

Solution: (A) sin[ − h] sin( −1) = lim = sin 1. h → 0 [ − h] h→ 0 ( −1)

lim f (0 − h) = lim h→ 0

lim f (0 + h) = lim h→ 0

h→0

= 1



sin[h] [ h] 

Solution: (C) As 0 ≤ x – [x] < 1 ∀ x ∈ R, 0 ≤ f (x) < 1. n→∞

= 0

[ f ( x)]2n − 1 n→∞ f ( x ) 2 n + 1 [ ] 

Thus, for x ∈ R, g (x) = lim

=

1 tan −1 ( h) =– h→ 0 −2h 2

QUICK TIPS If lim f( x ) = A > 0 and lim g( x ) = B, then

(B) 1 (D)  None of these

2n

= lim

■

2n f ( x)] − 1 [ lim , then g (x) = n→∞ f ( x ) 2 n + 1 [ ]

lim [ f ( x ) ]



[∵ h → 0 ⇒ (h) → 0]

22. Let f (x) = x – [x], where [x] denotes the greatest ­integer

\

=

Since LHL ≠ RHL \ lim f ( x ) Does not exist.

x →0

(A) 0 (C)  –1



x →0

\ lim f ( x ) does not exist.

≤ x and g (x) =

tan −1 (1 + h) h→ 0 (1 − 2h) 

= lim



x →0

(A) 0 (C)  1

tan −1 ([ − h] − h) h→ 0 h→ 0 [ − h] + 2 h  −1 tan ( −1 − h) = lim h→ 0 ( 2h − 1) 

LHL = lim f (0 − h) = lim



0 −1 = –1 0 +1

⎧ tan −1 ([ x ] + x ) , [ x] ≠ 0 ⎪ 23. If f (x) = ⎨ [ x ] − 2 x ⎪0 , [ x] = 0 ⎩

x →a

x →a

lim [f( x )]g ( x ) = AB



x →a

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

■

x →a

x →a

lim [f( x )]g( x ) = e



lim g( x )[ f ( x )−1] x→a

x →a

SOLVED EXAMPLES 24. If lim

729 x − 243x − 81x + 9 x + 3x − 1 x3

x→0

then k = (A) 4 (C)  6

= k(log 3)3,

(B) 5 (D)  None of these

Solution: (C) Required limit

= lim

x→0

= lim

x→0

243x (3x − 1) − 9 x (32 x − 1) + (3x − 1) x3 (3 − 1) {( 243) − ( 27) x − 9 x + 1} x



x

x3



Limits  8.11 = lim



(3x − 1) {( 243) x − ( 27) x − 9 x + 1} 3

x→0



x x x (3 − 1) (9 − 1) ( 27 x − 1) = lim x→0 x x x  = log 3 ⋅ log 9 ⋅ log 27



= log 3 ⋅ 2log 3 ⋅ 3log 3



⎛ 1 + tan x ⎞ 27. lim ⎜ ⎟ x → 0 ⎝ 1 + sin x ⎠



and

Clearly f (x) → 1 and g (x) → ∞ as x → 0.

lim (1 + ax 2 + bx + c)1 ( x −α ) is \

a (a – b)

(A)  log |a (a – b )|

(B)  e

(C) ea (b – a)

(D)  None of these

lim (1 + ax 2 + bx + c)1 ( x −α ) = e

1 lim ⎡(1+ ax 2 + bx + c ) −1⎤⎦ x →α ( x −α ) ⎣

x →α

g (x)

[Using lim [f (x)]

= e

lim g ( x )[ f ( x ) −1] x→a

provided f (x) → 1 and g (x) → ∞ as x → a] x→a

 = e



( ax 2 + bx + c ) lim x →α ( x −α )

= e

a ( x −α )( x − β ) lim x →α ( x −α )

 [∵ a, b are roots of ax2 + bx + c = 0]



⎛ sin x − sin a ⎞ = lim ⎜1 + ⎟ x→a ⎝ sin a ⎠







sin a ⎡ ⎧ ⎛ sin x − sin a ⎞ ⎫ sin x − sin a ⎥ ⎢ = lim ⎢ ⎨1 + ⎜ ⎟⎬ ⎥ ⎝ x→a ⎩ sin a ⎠ ⎭ ⎢⎣ ⎥⎦ sin x − sin a = lim e ( x − a) sin a 



2 ⎛ x + a⎞ ⎛ x − a⎞ 1 ⋅ cos ⎜ sin ⎜ ⎝ 2 ⎟⎠ ⎝ 2 ⎟⎠ sin a x−a  ⎛ x + a⎞ ⎡ ⎛ x − a⎞ ⎛ x − a⎞ ⎤ 1 = lim cos ⎜ sin ⎝ 2 ⎟⎠ ⎢⎣ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎥⎦ sin a e =

=e

tan

πx 2a

= e

lim x →0

lim g ( x ) [ f ( x ) −1] x→a

1− cos x cos x (1+ sin x )

}

= e0 = 1

is equal to

tan

πx 2a

= e

⎛ π x⎞ ⎛ x ⎞ ⋅ 2 − −1 lim tan ⎜ x→a ⎝ 2 a ⎟⎠ ⎜⎝ a ⎟⎠

{Using lim [ f ( x)]

g( x)

x→a

=e

lim g ( x ) [ f ( x ) −1] x→a

⎫ as f ( x ) → 1 and g ( x ) → ∞ as x → a⎬ ⎭ = e

⎛ x⎞ ⎛ π x⎞ lim ⎜1− ⎟⎠ tan ⎜⎝ ⎟ x→a ⎝ 2a ⎠ a

= e

−1 / a ⎛ π x⎞ π − cosec 2 ⎜ ⎟ ⋅ ⎝ 2a ⎠ 2a

= e

lim x→a

= e

(1− x a ) cot (π x 2 a )



2 2 ⎛ π x⎞ lim sin ⎜ x→a ⎝ 2 a ⎟⎠ π

= e2/p

x →0



(A) e − a b (B)  e ab (C)  e a b (D)  e−b a 2

2

Solution: (C)

2

a

lim (cos x + a sin bx ) x = e

2

a lim (cos x + a sin bx −1) x→ 0 x

x →0

e x→a

cot a

= e

1 ⎛ tan x − sin x ⎞ ⎜ ⎟ sin x ⎝ 1+ sin x ⎠

=e

a

= lim

cos a e sin a

g( x)

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

x→a

x→a

lim x →0

lim x→a

sin x − sin a ⎤ ( x − a )sin a

1 ⎛ 1+ tan x ⎞ − 1⎟ ⎜ sin x ⎝ 1+ sin x ⎠

Solution: (B)

Solution: (A) ⎛ sin x ⎞ lim ⎜ ⎟ x → a ⎝ sin a ⎠

lim x →0

(A) ep/2 (B)  e2/p (C)  e–2/p (D)  e–p/2



1 x −a

= e

x→a

x⎞ ⎛ 28. lim ⎜ 2 − ⎟ x→a ⎝ a⎠

1

(A) ecot a (B)  etan a (C)  esin a (D)  ecos a

1 /sin x

{Using lim [ f ( x)]

x⎞ ⎛ lim ⎜ 2 − ⎟ x→a ⎝ a⎠

⎛ sin x ⎞ x − a 26. lim ⎜ , a ≠ np, n is an integer, equals ⎟ x → a ⎝ sin a ⎠

1 x −a

⎛ 1 + tan x ⎞ lim ⎜ ⎟ x → 0 ⎝ 1 + sin x ⎠



= ea (a – b)



1 + tan x 1 + sin x  1 g (x) = . sin x f (x) =

Let

25. If a and b be the roots of ax + bx + c = 0, then

Solution: (B)

(B) 1 (D)  None of these

Solution: (B)

2

x→α

is equal to

(A) 0 (C)  –1

= 6(log 3)3 = k(log 3)3(given) k = 6

\

1/sin x

⎡ Using lim f ( x ) ϕ ( x ) = e lim ϕ ( x )[ f ( x ) −1] [ ] ⎢ x→a ⎣  as f (x) → 1 and f (x) → ∞ as x → a] x→a





= e

lim x →0

a ( − sin x + ab cos bx ) 1

2

= ea b 

8.12  Chapter 8 sin x

2

⎛ sin x ⎞ x − sin x 30. The value of lim ⎜ is ⎟ x →0 ⎝ x ⎠ (A) 1 (B) –1 (C)  0 (D)  None of these

=e

sin x

sin x ⎛ sin x ⎞ −1⎟ ⎠ x

lim ⎜ ⎛ sin x ⎞ x − sin x lim ⎜ = e x − sin x ⎝ ⎟ x →0 ⎝ x ⎠ x →0

x→a



sin x → 1 and g ( x ) x sin x sin x x = = → ∞ as x → 0⎤⎦ sin x x − sin x 1− x



= e

as

f ( x) =

− lim x→ 0

sin x x

= e– 1

⎛ x 4 + x 2 + x + 1⎞ 31. lim ⎜ ⎟ x →−1 ⎝ x2 − x + 1 ⎠

1− cos( x +1) ( x +1) 2

⎛ 2⎞ (B)  ⎜ ⎟ ⎝ 3⎠

(A) 1

1/ 2

1/ 2

(D) e1/2





1− cos( x +1) ( x +1) 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 indeterminate forms: ⎛ 0⎞ 1. ⎜ ⎟ form: If lim f ( x ) = 0 and lim g ( x ) = 0, then ⎝ 0⎠ x→a x→a f ( x) f ′ ( x) lim = lim , provided the limit on the x → a g ( x) x → a g ′ ( x) R.H.S. exists. Here, f ′ is derivative of f.



QUICK TIPS

⎛ x +1⎞ 2 sin 2 ⎜ ⎝ 2 ⎟⎠ 2 x + 1⎞ ( x +1)

⎛ x4 + x2 + = lim ⎜ ⎟ x →−1 ⎝ x2 − x + 1 ⎠

= e0 = 1

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

Solution: (B) ⎛ x 4 + x 2 + x + 1⎞ lim ⎜ ⎟ x →−1 ⎝ x2 − x + 1 ⎠

x2 4n

⎛ ∞⎞ 2. ⎜ ⎟ form: If lim f ( x ) = ∞ and lim g ( x ) = ∞, ⎝ ∞⎠ x→a x→a f ( x) f ′ ( x) then lim = lim , provided the limit on x → a g ( x) x → a g ′ ( x) the R.H.S. exists.

is equal to

⎛ 3⎞ (C) ⎜ ⎟ ⎝ 2⎠

−2 × lim n→∞

EVALUATION OF LIMITS USING L’HOSPITAL’S RULE

⎡ Using lim f ( x ) g ( x ) = e lim g ( x )[ f ( x ) −1] [ ] x→a ⎣⎢



= e



Solution: (B)

⎛ ⎛ x ⎞⎞ ⎜ sin ⎜⎝ 2 n ⎟⎠ ⎟ x 2 ⎜ ⎟ ⋅ 2 ⋅n −2 lim n→∞ x ⎜ ⎟ 4n ⎜⎝ 2 n ⎟⎠

f( x ) L’Hospital’s Rule is applicable only when becomes of g( x ) 0 ∞ . the form or 0 ∞ 0 ∞ ■ If the form is not or , simplify the given expression till 0 ∞ 0 ∞ it reduces to the form and then use L’Hospital’s or 0 ∞ rule. ■



2

⎛ ⎛ x +1⎞ ⎞ sin 1 ⎜ ⎜⎝ 2 ⎟⎠ ⎟ ⎜ ⎟ x + 1⎞ 2 ⎜⎜ ⎛⎜ x +1⎞⎟ ⎟⎟ ⎟ ⎝ ⎝ 2 ⎠ ⎠

⎛ x4 + x2 + = lim ⎜ x →−1 ⎝ x2 − x + 1 ⎠

⎛ 2⎞ = ⎜ ⎟ ⎝ 3⎠

1/ 2



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

■

x⎞ ⎛ 32. lim ⎜ cos ⎟ n→∞ ⎝ n⎠

n

is equal to

(A) e1 (B)  e–1 (C)  1 (D)  None of these Solution: (C) ⎛

n



x

ERROR CHECK L’ Hospital’s rule cannot be applied in every problem. Consider the example, lim

⎞

n→∞

= e

⎛ x⎞ − n⋅2 sin 2 ⎜ ⎟ lim n→∞ ⎝ 2n⎠

3 x + sin 2 x ⎛ ∞⎞ ⎜ form ⎟⎠ . − sin 2 x ⎝ ∞

x →0 3x

lim n ⎜ cos −1⎟ x⎞ ⎛ lim ⎜ cos ⎟ = e ⎝ n ⎠  n→∞ ⎝ n⎠

Here, if we apply L’ Hospital’s rule, we get



lim

3 + 2 cos 2 x 3 x + sin 2 x = lim x →∞ 3 − 2 cos 2 x − sin 2 x

x →∞ 3 x

Limits  8.13

Now, both the numerator and denominator are undefined because lim cos 2 x Does not exist. x →∞ We can find the above limit as: ⎛ sin 2 x ⎞ 3 + 2⎜ ⎝ 2 x ⎟⎠ = 3 + 2(0) 3 x + sin 2 x = lim lim 3 − 2(0) x →∞ 3 x − sin 2 x x →∞ ⎛ sin 2 x ⎞ 3 − 2⎜ ⎝ 2 x ⎟⎠ sin 2 x =0 2x

= 1, since lim



x →∞

e x − e − x + 2 sin x  x →0 24 x



= lim



= lim

e x + e − x + 2 cos x 4 1 = = . x →0 24 24 6

35. If f (2), g (x) be differentiable functions and f (1) = f (1) g ( x ) − f ( x ) g (1) − f (1) + g (1) g (1) = 2 then lim x →1 g( x) − f ( x) is equal to (A) 0 (B) 1 (C)  2 (D)  None of these

SOLVED EXAMPLES

Solution: (C)

33. If a is a repeated root of ax2 + bx + c = 0, then

lim

lim

tan ( ax 2 + bx + c) ( x − α )2

x →α

is

(A) a (B)  b (C)  c (D) 0 Solution: (A) lim

tan ( ax 2 + bx + c) (x − α)

x →α

( 2ax + b) sec 2 ( ax 2 + bx + c) x →α 2(x − α)

⎛0 ⎞ form as α being a repeated root of ax 2 + bx + c = 0, ⎜0 ⎟ ⎜ ⎟ 2aα + b = 0 ⎠ ⎝ 2a sec 2 ( ax 2 + bx + c) + ( 2ax + b) 2 ×



2 sec 2 ( ax 2 + bx + c) tan( ax 2 + bx + c) = lim x →α 2  2a = = a. 2 x

34. lim

e +e

−x

+ 2 cos x − 4

is equal to

x4 (B) 1

x →0

(A) 0

1 (C)  6

Solution: (C) lim

e x + e − x + 2 cos x − 4 x4

x →0



= lim



= lim

x →0

x →0

4x

3

e x + e − x − 2 cos x 12 x 2

1 (D) – 6 ⎛0 ⎞ ⎜⎝ form⎟⎠ 0



e x − e − x − 2 sin x

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

f (1) g’( x ) − f’( x ) g (1) = lim x →1 g’( x ) − f’( x )  g’( x ) − f’( x ) = 2 lim x →1 g’( x ) − f’( x ) 



= 2.



= lim



x →1

f (1) g ( x ) − f ( x ) g (1) − f (1) + g (1)  g( x) − f ( x)

2

⎛0 ⎞ 2 ⎜⎝ form as aα + bα + c = 0⎟⎠ 0



⎛0 ⎞ ⎜⎝ form⎟⎠ 0



⎛0 ⎞ ⎜⎝ form⎟⎠ 0



⎛0 ⎞ ⎜⎝ form⎟⎠ 0

36. Let f (x) be a twice differentiable function and f  ″ (0) = 5, 3 f ( x ) − 4 f (3 x ) + f ( 9 x ) then lim is equal to x →0 x2 (A) 30 (B) 120 (C)  40 (D)  None of these Solution: (B) lim

3 f ( x ) − 4 f (3 x ) + f ( 9 x )

x →0

x

⎛0 ⎞ ⎜⎝ form⎟⎠ 0



3 f’( x ) − 12 f’(3 x ) + 9 f’(9 x )  x →0 2x



= lim



= lim



=



2

x →0

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

3 f" ( x ) − 36 f" (3 x ) + 81 f" (9 x ) 2 

3 f" (0) − 36 f" (0) + 81 f" (0) = 24 f  ″(0) 2 = 24 ⋅ 5 = 120.

37. If f (9) = 9 and f ′ (9) = 1, then lim

x →9

to (A) 0 (C)  –1

3−

f ( x)

3− x

is equal

(B) 1 (D)  None of these

Solution: (B) lim

x →9

3−

f ( x)

3− x



⎛0 ⎞ ⎜⎝ form⎟⎠ 0

8.14  Chapter 8 0− = lim



x →9

Solution: (B)

1 ⋅ f ′( x ) 2 f ( x)  1 0− 2 x



[Using L’Hospital’s Rule] 3 ⋅ f’( x ) = × f ′ (9) = 1. 3 f ( x) x

= lim



x →9

38. If lim



⎛ ln cos x ⎞ ln [1 + (cos x − 1) ] lim ⎜ = lim ⎟ x →0 ⎝ 4 1 + x 2 − 1⎠ x →0 4 1 + x 2 − 1 



= 4 lim

⎡ ⎢∵ ln [1 + (cos x − 1) ] ~ (cos x − 1) and ⎣

sin 2 x + a sin x

be finite, then the value of a and x3 the limit are given by (A)  – 2, 1 (B)  – 2, –1 (C)  2, 1 (D)  2, –1 x →0



k = lim

x →0

= lim

sin 2 x + a sin x x3

 2 cos 2 x + a cos x 3x 2

x →0

  [Using L’Hospital’s Rule] We require 2 cos2x + a cos x = 0 for x = 0 as denominator is zero. \ a = –2. 2 cos 2 x − 2 cos x ⎛ 0⎞ Hence, k = lim  ⎜⎝ ⎟⎠ 2 x →0 0 3x

−4 sin 2 x + 2 sin x = lim  x →0 6x



= lim

⎛ 0⎞ ⎜⎝ ⎟⎠ 0

−8 cos 2 x + 2 cos x −8 + 2 = = –1. x →0 6 6

x →0

(A) 0 (C)  – 1 Solution: (B) Let y = lim x x x →0

(B) 1 (D)  None of these

x2/ 2

x →0

= –2



x2

)

1 + x2 − 1 ~

x2 ⎤ ⎥ 4⎦

⎡ x2 ⎤ ⎢∵(1 − cos x ) ~ ⎥ 2⎦ ⎣



2 − x2 x

41. lim

is equal to x x − 22 log 2 − 1 log 2 + 1 (A)  (B)  log 2 + 1 log 2 − 1 (C) 1 (D) –1 Solution: (A) 2x − x2 ⎛0 ⎞ lim x ⎜⎝ form⎟⎠ x → 2 x − 22 0  x 2 log 2 − 2 x 4 log 2 − 4 log 2 − 1 = lim x = = . x → 2 x (1 + log x ) 4 (1 + log 2) log 2 + 1 42. lim x→

π 2

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

(A) 1

(B) 2

(C) 3

(D) 4

Solution: (B) Let sin x = h, then as x → p/2, h → 1



⎛∞ ⎞ ⎜⎝ form⎟⎠ ∞

h − hh 1 − hh − hh ln h = lim h → 1 1 − h + ln h h →1 −1 + 1 / h 

lim =

(Using L’ Hospital Rule) − h − 2h ln h − hh − 1 − hh (ln h) 2 h



= lim



−1 − 1 = = 2 −1

y = e = 1.

− 1 / h2

h →1

x→a

⎛ ln cos x ⎞ 40. lim ⎜ ⎟ is equal to x →0 ⎝ 4 1 + x 2 − 1⎠ (C) 1

h

43. The value of lim a 2 − x 2 cot

0

(B) –2



log x ⇒ log y = lim x log x = lim  x →0 x →0 1 x 1/ x = lim = – lim x = 0 x → 0 −1/ x 2 x →0

(A) 2

= –4 lim



(

4



\ given limit

39. lim x x is equal to

⇒

x2

x→2

Solution: (B) Let

cos x − 1

x →0

2a 2a (B)  – π π 4a 4a (C)  (D)  – π π (A) 

(D) –1

π a−x is 2 a+ x



Limits  8.15 −2 x

Solution: (C)

π a−x lim a − x cot  x→a 2 a+ x 2



= lim

x→a

2

a2 − x 2

π a−x tan 2 a+ x



(0 × ∞ form)



⎛0 ⎞ ⎜⎝ form⎟⎠ 0



= lim

x→a

− sec 2

π 2 =

2 a2 − x 2  a−x π 2a × × a + x 2 2( a + x ) a 2 − x 2 4a π 

8.16  Chapter 8

PRACTICE EXERCISES Single Option Correct Type ⎛ ⎡100 x ⎤ ⎡ 99 sin x ⎤⎞ 1. The value of lim ⎜ ⎢ ⎥ + ⎢ x ⎥⎟ , where [⋅] x → 0 ⎝ ⎣ sin x ⎦ ⎣ ⎦⎠ represents greatest integer function, is (A) 199 (B) 198 (C)  0 (D)  None of these 2. If f (x) = sin x, = 2, where n ∈Z and g(x) = x2 + 1, = 3, then lim g [ f ( x ) ] is

x ≠ np, x = np

is 5x (A) 1 (C)  0 x →∞

(B) –1 (D)  None of these

1

9. lim (cos x + sin x ) x is equal to x →0

10. The value of lim x→

x →0

(B) 0 (D)  Does not exist

⎛ ⎞ 3. The value of lim ⎜ x + x + x − x ⎟ is ⎠ x →∞ ⎝ 1 (A)  1 2 (C)  0 (D)  None of these x +1 π ⎤ ⎡ − 4. The value of lim x ⎢ tan −1 is x →∞ ⎣ x + 2 4 ⎥⎦

PRACTICE EXERCISES

x5

(A) e (B) e2 (C)  e–1

x ≠ 2, x = 2.

(A) 1 (C)  3

8. The value of lim

(A) 

π 4

2 2 − (cos x + sin x ) 1 − sin 2 x

(D) 1

3

is

3

2 1 (B)  (C)  (D) 2 3 2 2

11. The value of lim

ln(1 + 2h) − 2 ln(1 + h)

h→ 0

(A) 1 (C)  0

h2

is

(B) – 1 (D)  None of these

⎛ 1 e1/ n e 2 / n e( n −1)/ n ⎞ 12. The value of lim ⎜ + + + ... + is n→∞ ⎝ n n n n ⎟⎠ (A) 1

(B) 0

(C) e– 1

(D)  e + 1

5. lim cos π n2 + n , n ∈ Z is equal to

x sin ( x − [ x ]) , where [⋅] denotes the greatest x −1 ­integer function, is equal to (A) 1 (B) –1 (C) ∞ (D)  Does not exist

(A) 0 (C)  – 1

14. If f (x) =

1 (A)  2

n→∞

(

1 (B) – 2

(C) 1

(D) – 1

)

(B) 1 (D)  None of these

nk sin 2 ( n !) 0 < k < 1, is equal to n →∞ n+2 (A) ∞ (B)  1 (C)  0 (D)  None of these 6. lim

1 − cos 2( x − 1) x →1 x −1 (A)  exists and it equals 2 (B)  exists and it equals – 2 (C)  Does not exist because (x – 1) → 0 (D) Does not exist because left hand limit is not equal to right hand limit

7. lim

13. lim

x →1

∫

(A) 0

2 sin x − sin 2 x x3 (B) ∞

dx , x ≠ 0, then lim f’( x ) is x →0

(C) –1

(D) 1

⎡x⎤ ⎢⎣ 2 ⎥⎦ 15. 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 16. lim lim (1 + cos 2 m n !π x ) is equal to m→∞ n→∞

(A) 2 (C)  0

(B) 1 (D)  None of these

Limits  8.17

18. 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

(B)  a = –1, b = 1 (D)  a = 2, b = –1

⎛ 1 1 1 1 ⎞ + + + ... + 19. lim ⎜ is equal to n→∞ ⎝ 1 ⋅ 2 2 ⋅ 3 3⋅ 4 n( n + 1) ⎟⎠ (A) 1 (C)  0 20. lim

(log x ) 2 xn

x →∞

(A) 1

(B) –1 (D)  None of these

, n > 0 is equal to (B) 0

(C) – 1

(D) ∞

21. If the rth term, tr, of a series is given by n r tr = 4 , then lim ∑ tr is n→∞ r + r2 + 1 r =1 1 (A)  1 (B)  2 1 (C)  (D)  None of these 3

(A) 1 26. lim

(B) –1

2 x + 23 − x − 6

x→2

2 − x / 2 − 21− x

(C) 0

(D) ∞

is equal to

(A) 8 (C)  2

(B) 4 (D)  None of these

8 ⎛ x2 x2 ⎞ x2 x4 1 − cos − cos + cos cos ⎟ is equal ⎜ 8 x →0 x ⎝ 2 4 2 4⎠ to 1 1 1 1 (A)  (B) – (C)  (D) – 16 16 32 32 27. lim

28. lim [log n −1 ( n) ⋅ log n ( n + 1) ⋅ log n +1 ( n + 2)… log n n→∞

k

−1 ( n

k

)]

is equal to (A) ∞ (B)  n (C) k (D)  None of these ⎤ ⎡ 1 1 1 1 + + + ... + 29. lim ⎢ is equal n→∞ ⎣1 ⋅ 3 3 ⋅ 5 5 ⋅ 7 ( 2n + 1) ( 2n + 3) ⎥⎦ to 1 (A)  1 (B)  2 1 (C) – (D)  None of these 2 nx ⎡11/ x + 21/ x + 31/ x + ... + n1/ x ⎤ 30. The value of lim ⎢ ⎥ is x →∞ n ⎣ ⎦ (A) n! (B)  n (C) (n – 1)!   (D)  0

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

31. lim (1 + x) (1 + x2) (1 + x4) … (1 + x2n), | x | < 1, is

than or equal to x, is equal to (A) (–1)n (B)  (–1)n – 1 (C)  0 (D)  Does not exist

equal to 1 1 (A)  (B)  x −1 x −1

x→n

· 23. lim

x →1 y →0

y3 3

2

x − y −1

as (x, y) → (1, 0) along the line y =

x – 1 is given by (A)  1 (B)  ∞ (C)  0 (D)  None of these 24. lim

n →∞

1 − 2 + 3 − 4 + 5 − 6 + ... − 2n n2 + 1 + 4 n2 − 1

is equal to

1 1 (A)  (B)  – 3 3 1 (C) – (D)  None of these 5 ⎡ x 4 sin (1/ x ) + x 2 ⎤ 25. The value of lim ⎢ ⎥ is x→−∞ 1 + | x |3 ⎣ ⎦

n→∞

(C)  1 – x (D)  x–1

xn

= 0, (n integer), for ex (A)  no value of n (B)  all values of n (C)  only negative values of n (D)  only positive values of n

32. lim

x →∞



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

33. The value of lim

n ( n +1) (B) 0 2 (C)  1 (D)  n (A) 

34. If tr =

12 + 22 + 32 + ... + r 2 3

3

3

1 + 2 + 3 + ... + r lim Sn is given by

n→∞

3

and Sn=

n

∑ ( −1)r ⋅ tr, then r =1

PRACTICE EXERCISES

⎡ sin([ x − 3]) ⎤ 17. lim ⎢ ⎥ , where [⋅] represents greatest x → 0 ⎣ [ x − 3] ⎦ ­ integer function, is (A) 0 (B) 1 (C)  Does not exist (D)  sin 1

8.18  Chapter 8 2 (A)  3 35. If lim

1 2 (B) – (C)  3 3

(1 + a3 ) + 8e1 x

x → 0 1 + (1 − b3 ) e1 x

(B)  a = 1, b = 31/3 (D)  None of these

36. If a = min {x2 + 4x + 5, x∈R} and b = lim

∑a

then the value of (A)  (C)  37. lim

2n +1 − 1 4⋅2

n

2n +1 − 1 3 ⋅ 2n

r

⋅b

n− r

n

∑ ai

44. If Sn=

= 2, then

(A) a = 1, b = (–3)1/3 (C) a = –1, b = –(3)1/3 n

(A) 0 (C)  – 1

1 (D) – 3

1 − cos 2θ

θ →0

is

θ2

(B) 2n + 1 – 1



(D)  None of these

and lim an = a, then lim n→∞

n3 (A) 1 1 (C)  3

,

is equal to

n→∞

(B) –1 (D)  None of these

2

n→∞

n

is

∑i

equal to (A)  0 (B)  a (C)  2a (D)  None of these 3 45. The value of lim ⎡ n2 − n3 + n⎤ is ⎢ ⎥⎦ n→∞ ⎣ 1 1 2 (A)  (B) – (C)  3 3 3

2

log (1 + x + x ) + log (1 − x + x ) is equal to x →0 sec x − cos x (A) 1 (B) –1 (C) 0 (D) ∞

(A) 1

(D) –

2 3

3

4

n5 + 2 − n2 + 1

n→∞ 5

n 4 + 2 − 2 n3 + 1

46. The value of lim

1⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + ... + n ( n + 1)

Sn + 1 − Sn i =1

r=0



i =1

(B) 1 (D)  None of these

(B) 0

is

(C) –1

47. The integer n for which lim

(D) ∞

(cos x − 1)(cos x − e x )

x→0

a finite non-zero number, is (A) 1 (B) 2 (C) 3

xn

is

(D) 4

38. lim

ln x − 1 is equal to | x−e| 1 1 (A)  (B)  – e e (C) e (D)  Does not exist 39. lim

PRACTICE EXERCISES

x →e

40. If x1 = 3 and xn + 1= 2 + x n , n≥ 1, then lim xn is n→∞ equal to (A) –1 (B) 2 (C)  5 (D) 3 41. The value of lim

3x +1 − 5 x +1

x →∞

3x − 5 x

is

1 (A)  5 (B)  5 (C)  –5 (D)  None of these 1⎛ 42. lim ⎜1 + e1/ n + e 2 / n + ... + e n→∞ n ⎝

n −1 ⎞ n

⎟ is equal to ⎠

(A) e (B)  –e (C) e– 1 (D)  1 – e 43. lim

x →∞

x + sin x = x − cos x

48. The value of lim

x →∞

2

(A) 

3 1

(C) 

49. lim

3

(

x→0 3

(A) 

2 x +3 3 x +5 5 x 3x − 2 + 3 2 x − 3

is

(B)  3

x

3

z 2 − ( z − x)2

8 xz − 4 x 2 + 3 8 xz z

211/ 3

(D)  None of these

)

4

is equal to

1 (B)  223/ 3 ⋅ z

(C) 221/3 z

(D)  None of these

50. In a circle of radius r, an isosceles triangle ABC is inscribed with AB = AC. If the D ABC has perimeter P = 2 ⎡ 2 hr − h2 + 2 hr ⎤ and area A = h 2 hr − h2 , ⎣⎢ ⎦⎥ A where h is the altitude from A to BC, then lim 3 is h→ 0 P equal to 1 (A) 128 r (B)  128 r 1 (C)  (D)  None of these 64 r +

Limits  8.19 ⎛ 1 − cos{2( x − 2)} ⎞ 51. lim ⎜ ⎟ x→2⎝ x−2 ⎠

(C) equals

1 2 1



2



(B)  Does not exist (D) equals − 2

x x x x⎞ ⎛ 52. lim ⎜ cos cos cos ... cos n ⎟ = n→∞ ⎝ 2 4 8 2 ⎠ x sin x (A)  (B)  sin x x (C)  0 (D)  None of these 53. The value of lim

n→∞

⎤ ⎛ n −1 ⎞ ⎛n−2 ⎞ 1 ⎡ ⎛ n ⎞ ⎢ 1 + 2 + 3 k k ⎜∑ ⎟ ⎜∑ ⎟ ⎜ ∑ k ⎟ + ... + n ⋅ 1⎥ 4 n ⎢⎣ ⎝ k = 1 ⎠ ⎥⎦ ⎝ k =1 ⎠ ⎝ k =1 ⎠

will be 1 1 1 1 (A)  (B)  (C)  (D)  24 12 6 3 54. If [x] denotes the integral part of x, then ⎞ 1 ⎛ n lim 3 ⎜ ∑ [k 2 x ]⎟ = n→∞ n ⎝ ⎠ k =1 x x x (A) 0 (B)  (C)  (D)  2 3 6

1 1 θ 1 θ θ⎞ ⎛ 55. lim ⎜ tan θ + tan + 2 tan 2 + … + n tan n ⎟ = n→∞ ⎝ 2 2 2 2 2 2 ⎠ 1 1 (A)  (B)  − 2 cot 2θ θ θ (C)  2 cot 2q

(D)  None of these

43n − 2 − 9 n + 1

= 82 n − 1 − 9 n − 1 1 (A)  (B)  81 2 (C)  Does not exist (D)  None of these 56. lim

n→ 0

x − ai , i = 1, 2, …, n and if a1 < a2 < a3 … < an. | x − ai | Then, lim ( A1 A2 ... An ), 1 ≤ m ≤ n

57. If Ai =

x → am



(A)  is equal to (–1)m (B)  is equal to (–1)m +1 (C)  is equal to (–1)m – 1 (D)  Does not exist

represents greatest integer function, is (A) 199 (B) 198 (C)  0 (D)  None of these 5 x 59. The value of lim x is x →∞ 5 (A) 1 (B) –1 (C)  0 (D)  None of these 60. If the rth term, tr, of a series is given by n r tr= 4 , then lim t r is ∑ n→∞ r + r2 + 1 r =1 1 (A)  1 (B)  2 1 (C)  (D)  None of these 3 nx

⎡11/x + 21/x + 31/x + ... + n1/x ⎤ 61. The value of lim ⎢ ⎥ is x →∞ n ⎣ ⎦ (A) n! (B)  n (C) (n – 1)!   (D)  0 62. lim (1 + x )(1 + x 2 )(1 + x 4 )… (1 + x 2 n ), | x | < 1, is equal to n→∞

1 1 (B)  x −1 1− x (C)  1 – x (D)  x–1 (A) 

xn

= 0, (n integer), for ex (A)  no value of n (B)  all values of n (C)  only negative values of n (D)  only positive values of n 63. lim

x →∞

64. If a = min {x2 + 4x + 5, x ∈R} and b = lim n

then the value of (A)  (C) 

2 2

n +1

−1

4⋅2

n

n +1

−1

3 ⋅ 2n

∑ a ⋅b r

r=0

n−r

1 − cos 2θ

θ →0

is

θ2



(B) 2n + 1 – 1



(D)  None of these

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) ∞ 65. lim

4

n5 + 2 − 3 n2 + 1

n→∞ 5

n4 + 2 − 2 n3 + 1

66. The value of lim (A) 1

(B) 0

(C) –1

is (D) ∞

PRACTICE EXERCISES

(A) equals

⎛ ⎡100 x ⎤ ⎡ 99 sin x ⎤⎞ ⎢ ⎥+ 58. The value of xlim , where [⋅] →0 ⎜ ⎝ ⎣ sin x ⎦ ⎢⎣ x ⎥⎦⎟⎠

8.20  Chapter 8

67.

lim

x →0

x

3

z 2 − ( z − x)2

( 8xz − 4 x 3

(A) 

z 11/3

2

2

+ 3 8 xz

)

4

x→ 0

is equal to

(A) 

1 (B)  223/3 ⋅ z

21/3

(C) 2

74. lim

(C) 

(D)  None of these

z

68. In a circle of radius r, an isosceles triangle ABC is inscribed with AB= AC. If the DABC has perimeter P = 2 ⎡ 2hr − h2 + 2hr ⎤ and area A = h 2hr − h2 , ⎣⎢ ⎦⎥ A whereh is the altitude from A to BC, then lim 3 is h→ 0 P equal to 1 (A) 128r (B)  128r 1 (C)  (D)  None of these 64r +

69.

π⎞ ⎛ cos ⎜ x + ⎟ ⎝ 6⎠

lim

2/3

x →π /3 (1 −

(B) –1 (D)  None of these

ln ( 2 − cos 2 x )

is equal to ln 2 (sin 3 x + 1) 2 2 (A)  (B)  – 9 9 (C)  0 (D)  None of these x →0

PRACTICE EXERCISES

71.

lim

1 − cos (cx 2 + bx + a)

(1 − xα ) 2 ax  + bx+ c = 0, is equal to x → 1/α 2

(A)  (C) 

b 2 − 4 ac 2α 2 4 ac − b

72. lim

2α 2

, where a is a root of

b 2 − 4 ac (B) 2 α

2



1 − cos x

(D)  None of these

x 1 1 (A)  (B)  – 2 2 (C)  Does not exist (D)  None of these 73. lim

x→ 2 3

x + 7 − 3 2x − 3 x + 6 − 2 3x − 5 3

=

17 34 (B)  23 23 (C)  1 (D)  None of these (A) 

m

m2 1 m2

−

m −1

1 n2 −



n

1



n2n −1

1 1 (B)  m + n m2 n2 (D)  None of these

(cos θ ) x − (sin θ ) x − cos 2θ = x→ 4 x−4 (A) cos4 q lncos q – sin4 q ln sin q (B) cos4 q lncos q + sin4 q ln sin q (C) cos4 q ln sin q – sin4 q lncos q (D)  None of these 75. lim

1/ x

⎛ x − 1 + cos x ⎞ 76. lim ⎜ ⎟⎠ x→ 0 ⎝ x

=

(A) e1/2 (B)  e–1/2 1/4 (C) e (D)  None of these ⎤ ⎡ e ⎥ = 77. lim ⎢ x →∞ ⎢ 1 + 1/ x x ⎥ ) ⎦ ⎣( (A) e (B)  e–1 (C)  e1/2 (D)  e–1/2 ⎡ a sin x ⎤ ⎡ b tan x ⎤ 78. lim ⎢ + , where a, b are integers and x→ 0 ⎣ x ⎥⎦ ⎢⎣ x ⎥⎦ [ ] denotes integral part, is equal to (A) a+ b (B)  a+b–1 (C) a– b (D)  a – b – 1 [ x ] + [2 x ] + [3 x ] + ... + [nx ] = 1 + 2 + 3 + ... + n (A) x (B)  2x (C)  0 (D)  None of these 79. lim

n→∞

(

)

80. lim n2 x1/n − x1/n + 1 , x > 0 is equal to n→∞

(A)  0 (B)  ex (C) lnx (D)  None of these 1/ x

1/ x

=

x→ 0

1

x

=

2 cos x ) (A) 1 (C)  0 70. lim

( 2m + x )1/m − ( 2n + x )1/n is equal to x

f ( x) ⎤ ⎡ f ( x) ⎤ 81. If lim ⎢⎡1 + x + = e3, then lim ⎢1 + ⎥ x → 0 x ⎥⎦ x→ 0 ⎣ x ⎦ ⎣ (A) e (B)  e2 3 (C) e (D)  None of these x

82. If y = x +

x

x+

x ... ∞ (A) 1 (C)  0

, then lim

x →∞

x is equal to y

x+

(B) –1 (D)  None of these

=

Limits  8.21 (A) e (B)  e2 4 (C) e (D)  None of these y x x −y 1− k = 93. If lim x , then k = x→ y x − y y 1+ k

(B) 1 (D)  None of these

(A) log y (B)  ey (C) y (D)  None of these

(tan x ) tan x − tan x 84. The value of lim is x → π /4 ln (tan x ) − tan x + 1 (A) –2 (B) 1 (C)  0 (D)  None of these 85.

(

lim 11/cos

x → π /2

2

x

+ 21/cos

2

x

+ ... + n1/cos

2

x

)

cos 2 x

1⎞ ⎛ 3 r −r + ⎜ r ⎟ is equal to 94. lim ∑ cot −1 ⎜ ⎟ n→∞ 2 r =1 ⎜ ⎟ ⎝ ⎠ (A)  0 (B)  p π (C)  (D)  None of these 2 ⎛ ⎡ n sin x ⎤ ⎡ n tan x ⎤⎞ 95. The value of lim ⎜ ⎢ , where [·] + n→∞ ⎝ ⎣ x ⎥⎦ ⎢⎣ x ⎥⎦⎟⎠ denotes the greatest integer function, is (A) n (B)  2n + 1 (C) 2n – 1 (D)  None of these n

=

n ( n + 1) (A) n (B)  2 (C) n! (D)  None of these n

⎛

3⎞

∑ cot −1 ⎜⎝ r 2 + 4 ⎟⎠ n→∞

86. lim

=

r =1

(A) 0 π (C)  4 87. The value of

lim

x→ 5π / 4

(B) tan–12 (D)  None of these

[sin x + cos x], where [·] denotes

the greatest integer function, is (A) 2 (B) –2 (C) 1

(D) –1

88. lim lim

⎤ ⎡ x2 96. lim ⎢ ⎥ , where [·] denotes the greatest x → 0 sin x tan x ⎣ ⎦ ­integer function, is (A) 0 (B) 1 (C)  2 (D)  Does not exist

m→∞ n→∞

⎛ 1 + n 1n + 2n + n 2n + 3n + n 3n + 4 n + … + n ( m − 1) n + m n ⎞ ⎜ ⎟ ⎜⎝ ⎟⎠ m2 1 (A) 0 (B) 1 (C) –1 (D)  2 ⎡ n 1⎤ 89. The value of lim ⎢ ∑ r ⎥ , where [·] denotes the n→∞ ⎢ ⎣ r = 1 2 ⎥⎦ greatest integer, is 1 (A) 0 (B) 1 (C) –1 (D)  2 [cos x ] 90. The value of lim | x | , where [·] denotes the x →∞

greatest integer, is (A) 0 (C)  –1

(B) 1 (D)  Does not exist _ n ≥ 2, then 91. If a1 = 1 and an= n (1 + an – 1), ∨ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ lim ⎜1 + ⎟ ⎜1 + ⎟ ... ⎜1 + ⎟ = n→∞ ⎝ a1 ⎠ ⎝ a2 ⎠ ⎝ an ⎠

2

1⎞ ⎛ 1⎞ ⎛ 1 ⎞⎤ ⎟⎠ ⎜⎝ n + 2 ⎟⎠ ... ⎜⎝ n + n −1 ⎟⎠ ⎥ 2 2 2 ⎦

θ→ 0

cos 2 (1 − cos 2 (1 − cos 2 (1 ... cos 2 θ )) = ⎛ π ( θ + 4 − 2) ⎞ sin ⎜ ⎟ θ ⎝ ⎠

(A) 1 98. Let f (x) =

(B) 0

(C)  2

tan x − sin{tan −1 (tan x )} tan x + cos 2 (tan x )

(D) – 2

, then lim f ( x ) = x→

(A) 1 (B) –1 (C) 0 (D)  Does not exist

n

π 2

−1 ⎧ ⎡ 1/2 1/2 −1 ⎤ 2( ax )1/4 ⎪ ⎢⎛ a + x ⎞   99. lim ⎨ ⎜ 1/2 1/4 ⎟ − 3/4 1/4 1/2 1/2 1/4 3/4 ⎥ x → a ⎢⎝ a − x ⎠ x −a x +a x −a ⎥ ⎪⎣ ⎦ ⎩ ⎫⎪ −2log ⎬ = ⎪⎭ (A)  a3/4 (B)  a (C)  a2 (D)  None of these 4 a

(A)  0 (B)  e (C) e2 (D)  Does not exist ⎡ ⎛ 92. lim n − n ⎢( n + 1) ⎜ n + ⎝ n→∞ ⎣

97. lim

100. lim

x →1

(log (1 + x ) − log 2) (3.4 x −1 − 3 x ) {(7 + x )1/3 − (1 + 3 x )1/2 } sin π x

=

PRACTICE EXERCISES

cos x − (cos x )cos x 83. lim = x → 0 1 − cos x + ln (cos x ) (A) 0 (C)  2

8.22  Chapter 8 9 4 3 4 (A)  log (B)  log π e π e 9 2 (C)  log (D)  None of these π e 101. lim

(1 − x ) (1 − x 2 ) ... (1 − x 2 n )

x →1 [(1 −

x ) (1 − x 2 ) ... (1 − x n )]2

=

k

∑ cos −1 α r =

r =1

and q  = (A) 1

(1 + x 2 )1/3 − (1 − 2 x )1/4

r =1

x + x2

1 (C)  2

(B) –1

x→

(D) –

is

1 2

1 − cos( cx 2 + bx + a)

1 α

c ⎛ 1 1⎞ − αβ ⎜⎝ α β ⎟⎠

(D)  None of these

2(1 − α x ) 2

⎛ ⎞ f ( x )⎟ is The value of cot ⎝ xlim ⎠ →0 –1 ⎜

(A) 0 (C)  –1 105. If lim

(B) 1 (D)  None of these

x a sin b x

, a, b, c∈ R – {0} exists and has sin ( x c ) ­non-zero value, then (A)  a, b, c are in A.P. (B)  a, c, b are in A.P. (C)  a, c, b are in G.P. (D)  None of these x →0

103. If a and b are the roots of the quadratic equation ax2 + bx+ c = 0, then lim

(C) 

2

k

 

c ⎛ 1 1⎞ (B)  2β ⎜⎝ α − β ⎟⎠

⎧ tan 2 {x} , x>0 ⎪ 2 2 ⎪ x − [ x] ⎨ , x=0 f (x) = ⎪ 1 ⎪ {x} cot {x} , x < 0 ⎩

kπ for any k ≥ 1 2

∑ (α r )r , then xlim →θ

c ⎛ 1 1⎞ − 2α ⎜⎝ α β ⎟⎠

104. Given a real valued function f  such that

( 2n)! (A)  n! (B)  n! ( 2n)! (C)  2 (D)  None of these ( n !) 102. If

(A) 

=

Previous Year’s Questions 106. lim

1 − cos 2 x

[2002]

is 2x (A) l (C)  Zero

PRACTICE EXERCISES

x →∞

(B) − 1 (D)  Does not exist

x

⎛ x 2 + 5x + 3⎞ 107. lim ⎜ 2 ⎟ is equal to x →∞ ⎝ x + x + 2 ⎠ (A) e4 (C) e3

[2002]

(B)  e2 (D)  e 2

⎛ x − 3⎞ 108. For x ∈ R, lim ⎜ ⎟ is equal to x →∞ ⎝ x + 2 ⎠ (A) e (B)  e−1 5 (B) e− (D)  e5

[2002]

xf ( 2) − 2 f ( x ) 109. Let f (2) = 4 and f  ′ (2) = 4. Then lim is x→2 x−2 given by [2002] (A) 2 (B) − 2 (C) − 4 (D) 3

⎡ ⎛ x⎞ ⎤ ⎢1 − tan ⎜⎝ 2 ⎟⎠ ⎥ [1 − sin x ] ⎦ 1 10. lim ⎣ is x →π / 2 ⎡ ⎛ x⎞ ⎤ 3 ⎢1 + tan ⎜⎝ 2 ⎟⎠ ⎥ [π − 2xx ] ⎣ ⎦

[2003]

1 (A)  (B)  0 8 1 (C)  (D)  ∞ 32 log(3 + x ) − log(3 − x ) 111. If lim = k , the value of k is x  x →0 [2003] 1 (A)  0 (B)  − 3 2 2 (C)  (D)  − 3 3 112. 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) = 4 , then the lim x→a g( x) − f ( x) value of k is [2003]

Limits  8.23 118. Let f (x) be a forth degree polynomial having extreme

(B) 2 (D) 0

2x

⎛ a b⎞ 113. If lim ⎜1 + + 2 ⎟ = e 2 , then the values of a and b, x →∞ ⎝ x x ⎠ are  [2004] (A) a ∈ R, b ∈ R (B)  a = 1, b ∈ R (C) a ∈ R, b = R

(D)  a = 1 and b = 2

114. Let α and β be the distinct roots of ax2 + bx + c = 0, then lim

1 − cos( ax 2 + bx + c) ( x − α )2

x →α

is equal to

[2005]

a (α − β ) 2 2

(C) −

a2 (α − β ) 2 2

(B) 0 1 (D) (α − β ) 2 2

115. Let f : R → R be a positive increasing function such f (3 x ) f (2 x) = 1 . Then, lim = that lim [2010] x →∞ f ( x ) x →∞ f ( x ) 2 3 (A)  (B)  3 2 (C) 3 (D) 1 ⎛ 1 − cos{2( x − 2)} ⎞ 1 16. Limit of ⎜ ⎟ as x tends to 2 [2011] x−2 ⎝ ⎠ (A) equals (C) equals

2 1 2



119. The value of lim

(1 − cos 2 x ) (3 + cos x )

x →0

x tan 4 x

is equal to

 (A) 3 (B) 2 1 (C)  (D)  4 2

[2015]

1

2

(A) 

f ( x) ⎤ ⎡ values at x = 1 and x = 2. If lim ⎢1 + 2 ⎥ = 3 , then x →0 ⎣ x ⎦ f (2) is equal to [2015] (A) -4 (B)  0 (C)  4 (D)  -8

(B)  equals – 2 (D)  does not exist

(1 − cos 2 x ) (3 + cos x ) is equal to 117. The value of lim x →0 x tan 4 x  [2013] 1 (A)  (B)  1 2 1 (C)  2 (D)  − 4

120. Let p = lim (1 + tan 2 x ) 2 x then log p is equal to x →0 +

 1 (A)  (B)  2 4 1 (C)  1 (D)  2 cot x − cos x 121. lim equals π (π − 2π )3 x→

[2016]

[2017]

2

1 1 1 1 (A)  (B)  (C)  (D)  16 6 4 24 122. For each t ∈ R, let [t] be the greatest integer less than or equal to t. Then





⎛⎡1⎤ ⎡2⎤ ⎡15 ⎤ ⎞ lim x ⎜ ⎢ ⎥ + ⎢ ⎥ +…+ ⎢ ⎥ ⎟ x x ⎣ x ⎦⎠ ⎝⎣ ⎦ ⎣ ⎦ (A)  is equal to 0. (B)  is equal to 15. (C)  is equal to 120. (D)  does not exist (in R).

[2018]

x →0 +

PRACTICE EXERCISES

(A) 4 (C) 1

8.24  Chapter 8

ANSWER K EYS Single Option Correct Type   1. (B) 2. (A) 3.  (A) 4. (B)   11.  (B) 12. (C) 13.  (D) 14. (D)   21.  (B) 22. (D) 23. (C) 24. (B)   31.  (B) 32. (B) 33.  (A) 34. (B)   41.  (A) 42. (C) 43.  (B) 44. (A)   51. (B) 52.  (B) 53. (A) 54. (C)   61. (A) 62.  (B) 63. (B) 64. (B)   71.  (A) 72. (C) 73. (B) 74.  (C)   81. (B) 82.  (A) 83. (C) 84. (A)   91.  (B) 92. (B) 93.  (A) 94. (C) 101.  (C) 102. (C) 103. (A) 104. (D)

5.  (A) 15. (C) 25.  (B) 35.  (A) 45.  (A) 55.  (B) 65.  (A) 75. (A) 85.  (A) 95. (C) 105. (D)

6. (C) 16.  (a, b) 26. (A) 36. (B) 46. (B) 56. (C) 66. (B) 76.  (B) 86. (B) 96. (A)

7.  (D) 17.  (C) 27.  (C) 37.  (C) 47. (C) 57.  (D) 67.  (B) 77. (C) 87. (B) 97. (C)

8. (C) 18. (A) 28. (C) 38. (A) 48. (A) 58.  (B) 68. (B) 78.  (B) 88. (D) 98. (A)

9.  (A) 10. (A) 19.  (A) 20. (B) 29.  (B) 30. (A) 39. (D) 40. (B) 49.  (B) 50. (B) 59. (C) 60.  (B) 69. (C) 70. (A) 79. (A) 80.  (C) 89. (A) 90. (B) 99.  (C) 100. (A)

Previous Years’ Questions

PRACTICE EXERCISES

106. (D) 107. (A) 116. (D) 117. (C)

108. (C) 109. (C) 118. (A) 119.  (B)

110. (C) 111. (C) 112. (A) 113. (B) 114. (A) 115. (D) 120.  (D) 121.  (A) 122. (C)

Limits  8.25

HINTS AND EXPLANATIONS Single Option Correct Type 1. We know that lim

x→0

sin x x → 1– and lim → 1+ x → 0 sin x x

⎡ x ⎤ ⎡ sin x ⎤ lim ⎢100 = 100 + 98 = 198. ⎥ + lim 99 x→0⎣ sin x ⎦ x → 0 ⎢⎣ x ⎥⎦ The correct option is (B) 2. g [ f (x)] = [ f (x)]2 + 1, f (x) ≠ 2 3, f (x) = 2 2 ∴ g [ f (x)] = sin x + 1, x ≠ nπ 3, x = nπ RHL = lim g [ f (0 + h)] = lim(sin 2 h + 1) = 1.

∴

h→ 0

h→ 0

h→ 0

h→ 0

LHL = lim g [ f (0 − h)] = lim (sin 2 h + 1) = 1. ∴

lim g [ f ( x )] = 1

The correct option is (B) 1 ⎤ ⎡ ⎛ 1⎞ 2 5. lim cos ⎡π n2 + n ⎤ = lim cos ⎢ nπ ⎜1 + ⎟ ⎥ ⎣⎢ ⎦⎥ n→∞ n→∞ ⎢ ⎝ n⎠ ⎥ ⎣ ⎦

⎡ ⎛ 1 1 ⎞⎤ = lim cos ⎢ nπ ⎜1 + − 2 + ...⎟ ⎥ ⎠⎦ n→∞ ⎣ ⎝ 2 n 8n



π π ⎛ ⎞ = lim cos ⎜ nπ + − + ...⎟ ⎝ ⎠ n→∞ 2 8n π ⎛ ⎞ = – lim sin ⎜ nπ − + ...⎟ ⎝ ⎠ n→∞ 8n π = – lim ( −1) n −1 sin ⎛⎜ − ...⎞⎟ ⎝ ⎠ n→∞ 8n

x→0

= 0

The correct option is (A)

⎛ π ⎞ ⎜⎝∵ − ... → 0 as n → ∞⎟⎠ 8n

The correct option is (A) ⎡ ⎤ 3. lim ⎢ x + x + x − x ⎥ x→∞ ⎣ ⎦ x→∞

= lim

x→∞

= lim

x→∞

=

x+ x+ x −x x+ x+ x + x x+ x x+ x+ x + x 1 ⎞ ⎛ x ⎜1 + ⎟ ⎝ x⎠

a finite quantity ∞ [∵ sin2 (n !) always lies between 0 and 1. Also,  since 1 – k > 0, \  n1 – k→ ∞ as n → ∞] = 0. The correct option is (C)

1/ 2 ⎤ ⎡⎛ 1 1 ⎞ ⎥ 1+ 1 x ⎢⎜1 + + ⎟ ⎢⎝ ⎥ x x⎠ ⎣ ⎦

7. lim

x →1

The correct option is (A)

⎛ x +1 ⎞ −1 ⎟ ⎡ −1 x + 1 −1 ⎜ x + 2 −1 ⎤ = lim x ⎢ tan − tan 1⎥ = lim x tan ⎜ ⎟ x + 1 x →∞ ⎣ x →∞ x+2 ⎦ ⎜1+ ⎟ ⎝ x + 2⎠ −1 ⎞ = lim x tan ⎜ = – lim ⎝ 2 x + 3 ⎟⎠ x →∞ x →∞ 1 1 =– 2 2

1 − cos 2( x − 1) 2 sin 2 ( x − 1) = lim x →1 x −1 x −1 = lim



x +1 π ⎤ ⎡ 4. lim x ⎢ tan −1 − x →∞ ⎣ x + 2 4 ⎥⎦

−1 ⎛

=



1/ 2

1 1 = 1+1 2

= –1 ×

sin 2 ( n!) n → ∞ 1− k ⎛ 2⎞ n ⎜1 + ⎟ ⎝ n⎠

= lim



⎛ 1 ⎞ tan −1 ⎜ ⎝ 2 x + 3 ⎟⎠ ⎛ 1 ⎞ ⎜⎝ ⎟ 2 x + 3⎠

⋅

1 3⎞ ⎛ ⎜⎝ 2 + ⎟⎠ x

x →1

LHL = lim x →1

−

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



= lim



=– 2⋅1=– 2

h→ 0

RHL = lim x →1

+

2 sin( x − 1) ( x − 1)

2 sin( x − 1) 2 sin(1 + h − 1) = lim h→ 0 ( x − 1) (1 + h − 1)

HINTS AND EXPLANATIONS

= lim

nk sin 2 ( n!) nk sin 2 ( n!) = lim n→∞ n→∞ 2⎞ n+2 ⎛ n ⎜1 + ⎟ ⎝ n⎠

6. lim

8.26  Chapter 8 2 sin h = h→0 h Since LHL ≠ RHL,

sin h = 2 hlim →0 h

= lim



2⋅1=

2

⎛ h2 ⎞ ln ⎜1 + ⎟ ⎝ 1 + 2h ⎠ = – lim h→ 0 ⎛ h2 ⎞ ⎜ 1 + 2h ⎟ (1 + 2h) ⎝ ⎠

The correct option is (D) x5

x →∞ 5

= lim

x

x →∞ e

x5 x log 5

= lim

x5

x →∞ e kx

,

where k = log 5 = lim

x5

x →∞ ⎛

= lim

x →∞

⎞ k x k x k 4 x 4 x 5k 5 k 6 x 6 ⎜1 + kx + 2! + 3! + 4! + 5! + 6! + ...⎟ ⎝ ⎠ 2 2

3 3

1 2 ⎡⎛ 1 1 k 1 k3 1 k4 1⎞ ⋅ + ⋅ + ⋅ ⎢⎜ 5 + k ⋅ 4 + 2! x 3 3! x 2 4! x ⎟⎠ x ⎢⎣⎝ x ⎞⎤ k5 ⎛ k6 + + ⎜ x + ...⎟ ⎥ 5! ⎝ 6! ⎠ ⎥⎦

⎡1 + e1 n + (e1 n ) 2 + ... + (e1 n ) n −1 ⎤ = lim ⎢ ⎥ n→∞ ⎢ n ⎥⎦ ⎣

1 =0 ∞ The correct option is (C) 1

HINTS AND EXPLANATIONS

x→0

1 lim (cos x + sin x −1) x→0 x lim x→0

( − sin x + cos x ) 1

(Using L’Hospital’s Rule)

2 2 − (cos x + sin x )  1 − sin 2 x

= lim x→



π 4

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

−3 (cos x + sin x ) ( − sin x + cos x ) −2 cos 2 x (Using L’Hospital’s Rule)

−3 (cos x + sin x )(cos x − sin 2 x ) = lim π −2 cos 2 x x→ 4

x→

π 4

= lim

π x→ 4

−3 (cos x + sin x )cos 2 x −2 cos 2 x 3 ⎛ 1 1 ⎞ 3 3 (cos x + sin x ) + = ⋅⎜ ⎟ = 2 ⎝ 2 2⎠ 2 2

The correct option is (A) 11. lim



2

2

= lim

13. RHL = lim

h→ 0

3

π x→ 4

1 ⋅ ⎡⎣(e1 n ) n − 1⎤⎦ 1 = (e – 1) lim 1 n n→∞ ⎛ e n→∞ n ( e1 n − 1) − 1⎞ ⎜ 1n ⎟ ⎝ ⎠ = (e – 1) × 1 = (e – 1). The correct option is (C) = lim

= e  = e1 = e The correct option is (A) 10. lim

⎛ h2 ⎞ ln ⎜1 + ⎟ ⎝ 1 + 2h ⎠ 1 = – lim = –1 ⋅ 2 h→ 0 1 + 2h ⎛ h ⎞ ⎜ 1 + 2h ⎟ ⎝ ⎠ log(1 + x ) ⎤ ⎡  = 1⎥ ⎢ Using xlim →0 x ⎣ ⎦ The correct option is (B) ⎡ 1 e1 n e 2 n e( n −1) n ⎤ 12. lim ⎢ + + + ... + ⎥ n→∞ ⎢ n n n n ⎥⎦ ⎣

=

9. lim (cos x + sin x ) x = e

ln(1 + 2h) − 2 ln(1 + h)

h→ 0

= – lim

h→ 0

h2 ln(1 + h) 2 − ln(1 + 2h) h2

h2

h→ 0

1 − cos 2( x − 1) ∴ lim does not exist. x →1 x −1

8. lim

= – lim

⎛ (1 + h) 2 ⎞ ln ⎜ ⎟ ⎝ 1 + 2h ⎠

(1 + h)sin (1 + h − [1 + h]) 1+ h −1

(1 + h)sin (1 + h − 1) h sin h = lim(1+ h) =1 h→ 0 h = lim

h→ 0

LHL = lim

h→ 0

(1 − h)sin (1 − h − [1 − h]) 1− h −1

(1 − h)sin (1 − h) = –∞ −h Since LHL ≠ RHL, x sin ( x − [ x ]) ∴ lim does not exist. x →1 x −1 The correct option is (D)

= lim

h→ 0

14. f (x) =

∫

2 sin x − sin 2 x

dx x3 d 2 sin x − sin 2 x 2 sin x − sin 2 x ⇒ f ′ (x) = dx = dx ∫ x3 x3 ∴

lim f ′( x ) = lim

x →0

x→0

= lim

x→0

2 sin x − sin 2 x

 x3 2 cos x − 2 cos 2 x 3x

2

⎛0 ⎞ ⎜⎝ form⎟⎠ 0 

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

Limits  8.27 −2 sin x + 4 sin 2 x  6x −2 cos x + 8 cos 2 x = lim x→0 6 6 = =1 6 The correct option is (D)

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

= lim

x→0

⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ 1 ⎞⎤ ⎛1 ⎢⎜ − 2 ⎟⎠ + ⎜⎝ 2 − 3 ⎟⎠ + ⎜⎝ 3 − 4 ⎟⎠ + ... + ⎜⎝ n − n + 1⎟⎠ ⎥ = nlim →∞ ⎣ ⎝ 1 ⎦ 1 ⎤ ⎡ = lim ⎢1 − = 1 – 0 = 1. n→∞ ⎣ n + 1⎥⎦ The correct option is (A) 20. lim

π 15.  < 1, 4 ⎛π⎞ ∴ ⎜ ⎟ = 0 ⎝ 4⎠

(log x ) 2 x

x →∞

n

⎛∞ ⎞ ⎜⎝ ∞ form⎟⎠



1 x = lim 2 log x  lim n − 1 x →∞ n x x →∞ n x n 2 = lim 2 n = 0 x →∞ n x The correct option is (B) 2 log x ⋅

⎛ x⎞ ⎜⎝ ⎟⎠ 2 ∴ lim =0 x → π / 2 ln(sin x ) The correct option is (C)

r

⎛∞ ⎞ form⎟ ⎝⎜ ∞ ⎠

r

16. We know that |cosθ | ≤ 1 for all θ. So, if |cos n! px| < 1,

21. tr =

lim lim (1 + cos 2 m n!π x ) = (1 + 0) = 1

=

1⎛ 1 1 ⎞ − ⎜ ⎟ 2 ⎝ r 2 − r + 1 r 2 + r + 1⎠

=

⎤ 1⎡ 1 1 − ⎢ ⎥ 2 ⎣ r ( r − 1) + 1 ( r + 1) r + 1⎦

∴

∑ tr

m →∞ n→∞

|cosn! px| = 1,

and if

lim lim (1 + cos 2 m n!π x ) = lim lim (1 + 12 m ) m →∞ n→∞

m →∞ n→∞

= lim lim (1 + 1) = 2



m →∞ n→∞

The correct option is (A) and (B)

n

=

r =1

n

( r 2 + 1) 2 − r 2

1

∑ 2 [ f ( r ) − f ( r + 1)] , r =1

where f (r) =

⎡ sin([ x − 3]) ⎤ ⎡ sin( − 4) ⎤ 17. LHL = lim ⎢ ⎥ = ⎢ ⎥ x → 0 ⎣ [ x − 3] ⎦ ⎣ −4 ⎦

=

r4 + r2 + 1

1 1 = [ f (1) − f ( n + 1)] r ( r − 1) + 1 2

−



⎡ sin 4 ⎤ = ⎢ ⎥ = –1 ⎣ 4 ⎦

3π   π < 4 < 2

⎡ sin[ x − 3] ⎤ ⎡ sin( − 3) ⎤ RHL = lim ⎢ ⎥ = ⎢ ⎥ x → 0 ⎣ [ x − 3] ⎦ ⎣ −3 ⎦

⎛ x2 + 1 ⎞ 18. We have, lim ⎜ − ax − b⎟ = 0 x →∞ ⎝ x + 1 ⎠ ⇒ lim

x →∞

2

( x + 1) − ( ax + b)( x + 1) =0 x +1

x 2 (1 − a) − ( a + b) x − b + 1 =0 x →∞ x +1 ⇒ 1 – a = 0 and a + b = 0 ⇒ a = 1 and b = –1. The correct option is (A) ⇒ lim

⎡ 1 1 1 1 ⎤ + + + ... + 19. lim ⎢ ⎥ n→∞ ⎣1 ⋅ 2 2 ⋅ 3 3⋅ 4 n( n + 1) ⎦

⎤ 1⎡ 1 1 ⎢1 − ⎥ → as n → ∞ 2 2 ⎣ ( n + 1)n + 1⎦

The correct option is (B) 22. LHL = lim( −1)[ n − h] = lim( −1) n −1 = (–1)n – 1

+

⎡ sin 3 ⎤ = ⎢ ⎥ = 0 ⎣ 3 ⎦ The correct option is (C)

=



h→ 0

π   < 3 < π. 2

h→ 0

RHL = lim( −1)[ n + h] = lim( −1) n = (– 1)n



h→ 0

h→ 0



Since LHL ≠ RHL

∴

lim( −1)[ x ] does not exist.

x→n

The correct option is (D) 23. Since y = x – 1, \ x = y + 1. As (x, y) → (1, 0) along the line y = x – 1, x = y + 1 holds throughout. ∴

lim

x →1 y→0

lim

y→0

y3 3

2

x − y −1

= lim

y3 3

2

y + 2 y + 3y

y→0 ( y

= lim

The correct option is (C)

y3 + 1) − y 2 − 1

y→0

3

y2 2

y + 2y + 3

=

0 =0 3

HINTS AND EXPLANATIONS



8.28  Chapter 8 24. lim

1 − 2 + 3 − 4 + 5 − 6 + ... − 2n

n→∞

n2 + 1 + 4 n2 − 1

= lim

[1 + 3 + 5 + 7 + ... + ( 2n − 1)] − ( 2 + 4 + 6 + ... + 2n)

n→∞

n 1+

1 1 +n 4− 2 n2 n

n n [2 ⋅1 + ( n − 1) ⋅ 2] − [2 ⋅ 2 + ( n − 1) ⋅ 2] 2 = lim 2 n→∞ ⎛ 1 1⎞ n ⎜ 1+ 2 + 4 − 2 ⎟ n n ⎠ ⎝ n n ⋅ 2n − 2 ( n + 1) 2 2 = lim n→∞ ⎛ 1 1⎞ n ⎜ 1+ 2 + 4 − 2 ⎟ n n ⎠ ⎝ n2 − n2 − n

= lim

⎛ 1 1⎞ n ⎜ 1+ 2 + 4 − 2 ⎟ n n ⎠ ⎝ −n = lim n→∞ ⎛ 1 1⎞ n ⎜ 1+ 2 + 4 − 2 ⎟ n n ⎝ ⎠ n→∞

−1

= lim

1 1 + 4− 2 n2 n The correct option is (B) n→∞

HINTS AND EXPLANATIONS

25.

1+

2 x + 23 − x − 6

( 22 x + 23 − 6 ⋅ 2 x ) / 2 x 1 2 x→2 − x x/2 2 2

= lim

2x / 2 − 2

x→2

= lim

x→2

= lim

(2

x/2

+ 2) ( 2 (2

x/2

8 x

8

⋅ 2 sin 2

x2 x2 ⋅ 2 sin 2 4 8 2

2

2⎞ ⎛ ⎛ x2 ⎞ 2 sin x 2 sin 2 ⎟ ⎛ x2 ⎞ ⎟ ⎛x ⎞ ⎜ 32 ⎜ = lim 8 ⎜ 24 ⎟ ⎜ ⎟ ⎜ 28 ⎟ ⎜ ⎟ x→0 x ⎜ x ⎟ ⎝ 4⎠ ⎜ x ⎟ ⎝ 8⎠ ⎜⎝ ⎟ ⎝ 4 ⎠ 8 ⎠ 1 = 32

28. lim ⎡⎣log n −1( n) ⋅ log n ( n + 1) ⋅ log n +1( n + 2)...log n n→ ∞

k

−1 ( n

k

)⎤⎦

⎛ log m ⎞ ⎜⎝ Using log n m = log n ⎟⎠ log nk log n = k lim  n→∞ log( n − 1) n→∞ log( n − 1)

= lim

= k lim

1/ n  −1

⎛∞ ⎞ form⎟ ⎝⎜ ∞ ⎠

(Using L’Hospital’s Rule)

n→∞ 1 / n

⎛ 1⎞ = k lim ⎜1 − ⎟ = k n→∞ ⎝ n⎠ The correct option is (C) ⎡ 1 ⎤ 1 1 1 + + + ... + 29. lim ⎢ ⎥ n→∞ ⎣1 ⋅ 3 ( 2n + 1)( 2n + 3) ⎦ 3⋅5 5⋅ 7

n→∞

( 2 x − 4 ) ( 2 x − 2) ( 2 x / 2 − 2) x

− 2) ( 2 − 2)

− 2)

x→0

= lim

x→2

x/2

= lim

⎡ 1 ⎛ 1 1⎞ 1 ⎛ 1 1⎞ 1⎛ 1 1 ⎞⎤ = lim ⎢ ⎜ − ⎟ + ⎜ − ⎟ ... + ⎜ − ⎟ n→∞ ⎣ 2 ⎝ 1 3 ⎠ 2 ⎝ 3 5⎠ 2 ⎝ 2n + 1 2n + 3 ⎠ ⎥⎦

2 − x / 2 − 21 − x

22 x − 6 ⋅ 2 x + 8

x2 ⎞ x4 ⎛ x2 ⎞ ⎤ 8 ⎡⎛ − − − cos cos cos 1 1 ⎢ ⎥ ⎜ x → 0 x8 ⎢⎝ 2 ⎟⎠ 4 ⎜⎝ 2 ⎟⎠ ⎥⎦ ⎣ 8 ⎛ x2 ⎞ ⎛ x2 ⎞ = lim 8 ⎜1 − cos ⎟ ⎜1 − cos ⎟ x→0 x ⎝ 2 ⎠⎝ 4⎠ = lim



The correct option is (B)

= lim

x2 x2 x2 x2 ⎤ 8 ⎡ ⎢1 − cos − cos + cos cos ⎥ 8 x→0 x ⎢ 2 4 ⎥⎦ 2 4 ⎣

27. lim

⎡ log n log( n + 1) log( n + 2) log( nk ) ⎤ ⋅ ⋅ = lim ⎢ ... ⎥ n→∞ ⎢ log( n − 1) log n log( n + 1) log( nk − 1) ⎥⎦ ⎣

1 − y 4 sin + y 2 y = lim y→∞ 1+ y 3 (Putting x = – y; as x→ – ∞, y→ ∞) 1⎞ ⎛ sin ⎜ y⎟ 1 −⎜ ⎟+ 1 ⎜ ⎟ y ⎝ y ⎠ −1 + 0 = lim = = –1 1 y→∞ 1+ 0 1+ 3 y

x→2

The correct option is (A)

The correct option is (C)

−1 −1 = = 1+ 2 3

⎛ x 4 sin (1 / x ) + x 2 ⎞ lim ⎜ ⎟ x→−∞ ⎝ 1 + | x |3 ⎠

26. lim

( 2 x / 2 + 2) ( 2 x − 2) = (2 + 2) ⋅ (4 – 2) = 8 = xlim →2

1⎛ 1 ⎞ 1 1 (1 – 0) = ⎜1 − ⎟ = 2 2 2 ⎝ 2n + 3 ⎠

The correct option is (B) 1 1 1 ⎞ ⎛ 1x 1 + 2 x + 3 x + ... + n x 30. lim ⎜ ⎟ x →∞ ⎜ n ⎟⎠ ⎝

nx

Limits  8.29 

(Using L’Hospital’s Rule) n ( n + 1) = n + (n – 1) + … + 2 + 1 = 2 The correct option is (A)

n

y y⎞ y y ⎛ y = lim 1 + 2 + 3 + ... + n ⎜ ⎟ y→0 ⎝ n ⎠

= e =

34. tr =

⎛ 1y + 2 y + 3 y + ... + n y − n ⎞ lim y→0 ⎜ y ⎝ ⎠⎟

⎡ (1y −1) ( 2 y −1) ( 3 y −1) ( n y −1) ⎤ lim + + + ...+ ⎥ y→0 ⎢ y y y y ⎦ ⎣ e

12 + 22 + 32 + ... + r 2 13 + 23 + 33 + ... + r 3

=

r ( r + 1)( 2r + 1) ⎛ 2 ⎞ ⋅⎜ 6 ⎝ r ( r + 1) ⎟⎠

=

2⎛1 1 ⎞ ⎜ + ⎟ 3 ⎝ r r + 1⎠

= e (log 1 + log 2 + log 3 + … + log n) log (1 ⋅ 2 ⋅ 3 … n) =e = n! The correct option is (A)

∴ Sn =

31. lim (1 + x) (1 + x2) (1 + x4) … (1 + x2n)



(1 − x )(1 + x )(1 + x 2 )(1 + x 4 )...(1 + x 2 n ) n→∞ 1− x

= lim

(1 − x 2 )(1 + x 2 )(1 + x 4 )...(1 + x 2 n ) n→∞ 1− x . . . . . . . . . = lim

e

x

x → 0 0 + (1 − b3 ) e1 x ( −1

= lim



e

∴ From Eq. (1), 2 = lim

x

x

= … = lim

n! x

−m

= lim e x x →∞ e x (Putting n = – m, where m is a positive integer) 1 1 = lim m x = = 0. x →∞ x e ∞ Case III:  n = 0 xn 1 1 lim x = lim x = = 0. x →∞ e x →∞ e ∞ n x Hence, lim x = 0 for all values of n. x →∞ e The correct option is (B) lim

1 + 4e1 x

3

x →∞ e x →∞ e  (Using L’Hospital’s Rule repeatedly) =0 Case II:  n is a negative integer. n

(1 + a3 ) + 8e1 x

x→0

x

x →∞

x n + x n −1 + x n − 2 + ... + x 2 + x − n  x →1 x −1

33. lim

nx n −1 + ( n − 1) x n − 2 + ... + 2 x + 1 = lim x →1 1

x2 ) (Using L’Hospital’s Rule)



x

n ( n − 1) x n − 2

 

 ⇒ 1 – b3 = 4 ⇒ b3 = – 3 ⇒ b = (–3)1/3

nx n −1

x →∞

⎛∞ ⎞ ⎜⎝ ∞ form⎟⎠ (1)

0 + 8e1 x ( −1 x 2 )

⇒ 2 = lim

32. Case I:  n is a positive integer x →∞

(1 + a3 ) + 8e1 x

x → 0 1 + (1 − b3 ) e1 x

The correct option is (B)

= lim

=

35. We have 2 = lim

1 − x 4n + 2 1 = lim = for |x| < 1 n→∞ 1− x 1− x

xn

2 ⎡ ⎛ 1 ⎞ ⎛ 1 1⎞ ⎛ 1 1 ⎞ 1 ⎞⎤ ⎛1 − ⎜1+ ⎟ + ⎜ + ⎟ − ⎜ + ⎟ + … ± ⎜ + ⎝ n n + 1⎟⎠ ⎥⎦ 3 ⎢⎣ ⎝ 2 ⎠ ⎝ 2 3 ⎠ ⎝ 3 4 ⎠

2⎛ 1 ⎞ ⎜ −1 ± ⎟ 3⎝ n + 1⎠ 2 ∴ lim Sn = – n→∞ 3 The correct option is (B)

n→∞

lim

2

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

⇒ 1 + a = 2 i.e., a = 1 Hence a = 1 and b = (–3)1/3. The correct option is (A) 36. x2 + 4x + 5 = (x + 2)2 + 1 ≥ 1. So, a = 1. Also, b = lim ∴

n

1 − cos 2θ

θ →0

∑ ar ⋅ bn − r

θ

2

= lim

2 sin 2 θ

θ →0

θ2

= 2.

= bn + abn – 1 + a2 bn – 2 + … + an

r=0



⎡ ⎛ a ⎞ n +1 ⎤ ⎡ ⎛ 1 ⎞ n +1 ⎤ b n ⎢1 − ⎜ ⎟ ⎥ 2n ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ b ⎠ ⎥⎦ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ = = a 1 1− 1− b 2



=

2n +1 ( 2n +1 − 1) 2n +1

The correct option is (B) 37. lim

1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + ... + n ( n + 1) n3

n→∞

= lim

n→∞

Σ n ( n + 1) n3

= lim

n→∞

Σ n 2 + Σn n3

= (2n + 1 – 1)

HINTS AND EXPLANATIONS

= e

n ⎛ 1y + 2 y + 3 y + ... + n y ⎞ lim −1⎟ y→0 ⎠ y ⎝⎜ n

8.30  Chapter 8 = lim

n→∞

1 ⎡ n ( n + 1)( 2n + 1) n ( n + 1) ⎤ + 6 2 ⎥⎦ n3 ⎢⎣

⎡1 ⎛ 1⎞ ⎛ 1⎞ 1 ⎛ 1 1 ⎞ ⎤ = lim ⎢ ⎜1 + ⎟ ⎜ 2 + ⎟ + ⋅ ⎜ + 2 ⎟ ⎥ n→∞ ⎣ 6 ⎝ n⎠ ⎝ n⎠ 2 ⎝ n n ⎠ ⎦ 1 1 = ×1×2= . 6 3 The correct option is (C) log(1 + x + x 2 ) + log(1 − x + x 2 ) 38. lim x→0 sec x − cos x 2⎤

log ⎡⎣(1 + x ) − x ⎦ x → 0 (1 − cos 2 x ) cos x 2 2

= lim

log(1 + x 2 + x 4 ) ⎛0 ⎞ ⎜⎝ form⎟⎠ x→0 sin x tan x 0  log[1 + x 2 (1 + x 2 )] 2 1 = lim ⋅ x (1 + x 2 ) ⋅ sin x tan x 2 x→0 x 2 (1 + x 2 ) ⋅ ⋅x x x log(1 + x ) ⎤ ⎡ = 1. ⎢as lim = 1⎥ x ⎣ x→0 ⎦ The correct option is (A) ⎛ h⎞ loge e ⎜1 − ⎟ − 1 ⎝ ln(e − h) − 1 e⎠ 39. LHL = lim = lim h→ 0 | e − h − e | h→ 0 | −h | = lim



⎛ h⎞ log e + log ⎜1 − ⎟ − 1 ⎝ e⎠ = lim h→ 0 h

HINTS AND EXPLANATIONS

2



h h − − 2 − ... 1 = lim e 2e =– h→ 0 | e + h − e | e ln (e + h) − 1 h→ 0 | e + h − e |

RHL = lim





⎛ h⎞ log e ⎜1 + ⎟ − 1 ⎝ e⎠ = lim h→ 0 |h| ⎛ h⎞ log e + log ⎜1 + ⎟ − 1 ⎝ e⎠ = lim h→ 0 h

h h2 − 2 + ... 1 = lim e 2e = h→ 0 h e Since LHL ≠ RHL ln x − 1 ∴ lim does not exist. x→e | x − e | The correct option is (D) 40. We have x1 = 3, xn + 1 = 2 + xn x2 =

2 + x1 =

2+3 =

5

x3 = 2 + x2 = 2 + 5 ∴ x1 > x2 > x3 It can be easily shown by mathematical induction that the sequence x1, x2, …xn, … is a monotonically decreasing sequence bounded below by 2. So it is convergent. Let lim xn = x. Then n→∞

xn + 1 =

⇒

2 + xn

lim xn + 1 =

2 + lim xn

n→∞

n→∞

⇒ x = 2 + x ⇒ x2 – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0 ⇒ x = 2 The correct option is (B) 41. lim

3x +1 − 5 x +1 x

3 −5

x →∞

= lim

x

x →∞



= lim



= 5.

( xn > 0 ∀n, ∴ x > 0)

3 ⋅ 3x − 5 ⋅ 5 x 3x − 5 x x ⎛ 3⎞ 3⋅ ⎜ ⎟ − 5 ⎝ 5⎠

x →∞

⎛ 3⎞ ⎜⎝ ⎟⎠ − 1 5

The correct option is (A) 1⎛ 1/ n 2/ n ⎜1 + e + e + ... + e n→∞ n ⎝

42. lim

x

=

−5 −1

⎛ ⎞ a n = 0, if − 1 < a < 1⎟ ⎜⎝∵ nlim ⎠ →∞ n −1 ⎞ n

⎟ ⎠

( )

n

1/ n 1 1− e 1− e = lim ⋅ = lim n→∞ n n→∞ ⎛ 1 1 1 ⎞ 1 − e1/ n n ⎜1 − 1 − − ⋅ 2 ...⎟ ⎝ n 2! n ⎠ 1− e 1− e = lim = =e–1 1 1 n→∞ −1 −1 − ⋅ ... 2! n The correct option is (C)

sin x 1+ x + sin x x = lim = cos x x →∞ x − cos x 1− x

43. lim

x →∞

1+ 0 = 1. 1− 0

⎡ ⎤ ⎛1⎞ sin x = lim y sin ⎜ ⎟ = O × (a finite quantity) ⎥ ⎢∵ lim y →0 ⎝ y⎠ ⎢ x →∞ x ⎥  ⎢ ⎥ cos x ⎢ = 0. Similarly lim =0 ⎥ x →∞ x ⎣ ⎦ The correct option is (B) 44. Sn=

n

∑ ai ,

i =1

lim an = a

n→∞

Sn + 1 – Sn = an + 1

Limits  8.31

So, n → ∞

an + 1

(Dividing the numerator and denominator by the highest power x1/2) 2 = 3 The correct option is (A)

2a

=0 n( n + 1) = nlim → ∞ n( n + 1) 2

The correct option is (A) 13 ⎡⎛ ⎤ 1⎞ 3 45. lim ⎡ n2 − n3 + n⎤ = lim n ⎢⎜ −1 + ⎟ + 11 3 ⎥ ⎢ ⎥⎦ ⎠ n→∞ ⎣ n→∞ ⎢ ⎝ n ⎥⎦ ⎣

= lim n ⋅



n→∞

23

⎛1 ⎞ + 1 − ⎜ − 1⎟ ⎝n ⎠

1

= lim

n→∞ ⎛ 1

⎞ ⎜⎝ − 1⎟⎠ n

23

⎛1 ⎞ + 1 − ⎜ − 1⎟ ⎝n ⎠

13

=

3

4 n→∞ 5

1 1 = 1+1+1 3

3

2 n2 3 1 n5 4 4 1+ 5 − 3 2 3 1+ 2 32 n n n n = lim 4 5 n→∞ n 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 = lim = = 0. n→∞ 0 −1 1 5 2 2 1 1+ 4 − 1+ 3 n7 10 n n The correct option is (B) 47. Minimum power in numerator on x is 3. So n = 3. The correct option is (C)

=

=

5

2 x +3 x +5 x

x→∞

lim

3

2−

3 5 + x1/ 6 x 3/10 2 1 3 3 − + 1/ 6 3 2 − x x x 2+

x→∞

2 3 + x x

(

x4 3

43 3

3

2z

3

8z

1

)

3

2z − x

8z − 4 x + 3 8z

3

)

x )4

4

4

23 3

+

A P

3

h→0



= lim



= lim

h→0

h→0

=



h 2 hr − h2

= lim

(

+

8

( 2 hr − h

8⋅ h ⋅ h

+ 2 hr

8

(

(

2r − h + 2r

2r − h 2r − h + 2r

2r

8 2 2r

)

3

=

)

3

)

3

)

3

1 128r

The correct option is (B) ⎛ 1 − cos 2( x − 2) ⎞ 51. lim ⎜ ⎟ x → 2⎝ x−2 ⎠ 2 | sin( x − 2) | x−2 which doesn’t exist as LHL = − 2 whereas = lim

x→2

RHL =

2

The correct option is (B) x x x x cos 2 cos 3 ... cos n 2 2 2 2

⎧ x x ⎞⎫ x x ⎛ cos ... cos n − 1 ⎜ 2 sin n cos n ⎟ ⎬ x ⎨⎩ ⎝ ⎠⎭ 2 2 2 2 2 sin n 2 ⎧ x ⎛ x x ⎞⎫ 1 = lim cos ... ⎜ 2 cos n − 1 sin n − 1 ⎟ ⎬ x ⎨⎩ ⎝ n→∞ 2 2 2 2 ⎠⎭ 2 sin n 2 ………………………………………………………….. ………………………………………………………….. = lim

3 x

2

h ⋅ h 2r − h

n→∞

2 x +33 x +55 x x 3−

=

(2

x

8z − 4 x + 3 8z

52. Required limit= lim cos

3x − 2 + 3 2 x − 3 lim

=

h→0

2

n4 + 2 − n3 + 1

2 1 − n2 3 3 1 + 2 5 n n = lim n→∞ 4 5 2 1 n 5 1 + 4 − n3 2 2 1 + 3 n n

x→∞

x→0

50. lim

n5 4 4 1 +

48. lim

= lim

3

x

2 ⋅z The correct option is (B)

n5 + 2 − n2 + 1

3

x 2 xz − x 2

x→0 ( 3

13

The correct option is (A) 46. lim

( 8 xz − 4 x 2 + 3 8 xz ) 4

= lim

⎛ a3 + b3 ⎞ ⎜ Using a + b = 2 ⎟ a − ab + b 2 ⎠ ⎝



z 2 − ( z − x )2

3

x→0 3

⎛1 ⎞ ⎜⎝ − 1⎟⎠ + 1 n ⎛1 ⎞ ⎜⎝ − 1⎟⎠ n

x

49. lim

n→∞

1

HINTS AND EXPLANATIONS

lim

8.32  Chapter 8

= nlim →∞

x x⎞ ⎛ 2 cos sin ⎟ x ⎜⎝ ⎠ 2 2 n 2 sin n 2 1

⇒

⇒

n−r

⇒ tr + 1 = (r + 1){1 + 2 + 3 + … (n – r) terms} 1 ⇒ tr + 1 = ( r + 1) ( n − r )( n − r + 1) 2 1 ⇒ tr + 1 = ( r + 1)( n2 − rn + n − rn + r 2 − r ) 2

HINTS AND EXPLANATIONS

1 ∑ {r 3 − 2nr 2 + ( n2 − 2n − 1)r + n2} 2 r =1 ⎡ ⎧ n( n + 1) ⎫ ⎧1 ⎫ ⎢ ⎨ 2 ⎬ − 2n ⎨ 6 n ( n + 1)( 2n + 1)⎬ ⎩ ⎭ ⎩ ⎭ ⎣

⎤ ⎧1 ⎫ + ( n2 − 2n − 1) ⎨ n( n + 1)⎬ + n2 ( n)⎥ ⎩2 ⎭ ⎦ Solving and rearranging, we have, 1 4 S= {n − 11n3 − 19n2 + 6 n} 24 S 1 ⎛ n4 − 11n3 − 19n2 + 6 n ⎞ ∴ lim 4 = lim ⎜ ⎟ n→∞ n n → ∞ 24 ⎝ n4 ⎠

n→∞

n

4

1 ⎛ 11 19 6 ⎞ lim ⎜1 − − 2 + 3 ⎟ n n 24 n → ∞ ⎝ n ⎠

n −1⎧

⎪

n−r

⎫⎪

∑ ⎨( r + 1) ∑ k ⎬ =

r = 0⎪ k =1 ⎭ ⎪ ⎩ The correct option is (A)

1 24

1 ⎛ n 2 ⎞ ⎜ ∑ [k x ]⎟ n → ∞ n3 ⎝ ⎠ k =1

54. Let L = lim Since

k =1

[k 2 x ] n3


0 for i = 1, 2, …, m – 1 x − ai ∴ Ai = = –1, for i = m, m + 1, …, n − ( x − ai ) x − ai and,  Ai = = 1, for i = 1, 2, …, m – 1 x − ai Similarly, if x is in the right neighbourhood of ai Then,  x – ai< 0 for i = m + 1, …, n and  x – ai> 0 for i = 1, 2, …, m x − ai ∴ Ai = = –1 for i = m + 1, …n − ( x − ai ) x − ai and,  Ai = = 1 for i = 1, 2, …, m x − ai

and, 

n–m+1 lim ( A1 A2 ... An ) = (–1)

x → am−

lim ( A1 A2 ... An ) = (–1)n – m

x → am+

x→0

x



lim

x → 0 sin x



∴

59. lim

8−1 ⋅ 64 n − 9−1 ⋅ 9n

n→ + ∞

Now, 

58. We know that lim

sin x → 1– x

→ 1+

⎡ lim ⎢100

x ⎤ ⎡ sin x ⎤ = 100 + 98 = 198. ⎥ + lim 99 sin x ⎦ x → 0 ⎢⎣ x ⎥⎦

x→0⎣

x5

x →∞ 5 x

4 − 2 ⋅ 64 n − 9 ⋅ 9n

= lim

=

m

The correct option is (D)

The correct option is (B)

4 3n − 2 − 9 n + 1 2n − 1

( A1 A2 ... An ) does not exist. Hence, xlim →a







x5

= lim

x →∞ e

= lim

x log 5

x5

x →∞ e kx

, where k = log 5 x5

= lim

x →∞ ⎛

⎞ k 2 x 2 k 3 x 3 k 4 x 4 x 5k 5 k 6 x 6 ⎜1 + kx + 2! + 3! + 4! + 5! + 6! + ...⎟ ⎝ ⎠ 1 = lim x →∞ ⎡ ⎛ 1 1 k2 1 k3 1 k4 1⎞ ⋅ + ⋅ + ⋅ ⎢⎜ 5 + k ⋅ 4 + 2! x 3 3! x 2 4! x ⎟⎠ x ⎢⎣⎝ x ⎞⎤ k5 ⎛ k6 + + ⎜ x + ...⎟ ⎥ 5! ⎝ 6! ⎠ ⎥⎦ 1 = =0 ∞ The correct option is (C) 60. tr =

r 4

2

r + r +1

=

r 2

( r + 1) 2 − r 2

=

1⎡ 1 1 ⎤ − 2 ⎢ 2 2 ⎣ r − r + 1 r + r + 1⎥⎦

=

⎤ 1⎡ 1 1 − ⎢ ⎥ 2 ⎣ r ( r − 1) + 1 ( r + 1) r + 1⎦

∴

∑ tr

n

r =1

=

n

1

∑ 2 [ f ( r ) − f ( r + 1)] , r =1

1 r ( r − 1) + 1 1 = [ f (1) − f ( n + 1)] 2

where,  f (r) =

=

⎤ 1⎡ 1 1 ⎢1 − ⎥ → as n→ ∞ 2 2 ⎣ ( n + 1)n + 1⎦

The correct option is (B) ⎡11/x + 21/x + 31/x + ... + n1/x ⎤ 61. lim ⎢ ⎥ x →∞ ⎢ n ⎥⎦ ⎣

nx

HINTS AND EXPLANATIONS

∴

8.34  Chapter 8



⎡ ⎛ a ⎞ n +1 ⎤ ⎡ ⎛ 1 ⎞ n +1 ⎤ b n ⎢1 − ⎜ ⎟ ⎥ 2n ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ b ⎠ ⎥⎦ ⎢⎣ ⎝ 2 ⎠ ⎥⎦ = = a 1 1− 1− b 2



=

n

⎡1y + 2 y + 3 y + ... + n y ⎤ y = lim ⎢ ⎥ y→0 ⎢ n ⎥⎦ ⎣ n ⎡1 + 2 + 3 + ... + n ⎤ lim −1⎥ y→0 y ⎢⎣ n ⎦ y

= e

y

y

y

=

⎡1y + 2 y + 3 y + ... + n y − n ⎤ lim ⎥ y→0 ⎢ y ⎦ e ⎣

=

⎡ (1y −1) ( 2 y −1) ( 3 y −1) ( n y −1) ⎤ lim + + + ...+ ⎥ y→0 ⎢ y y y y ⎦ ⎣ e

2n +1 The correct option is (B)

= e (log 1 + log 2 + log 3 + … + log n) log (1 ⋅ 2 ⋅ 3 …. n) =e = n! The correct option is (A) 62. lim (1 + x) (1 + x2) (1 + x4) … (1 + x2n)

log(1 + x + x 2 ) + log(1 − x + x 2 ) x→0 sec x − cos x

= lim

(1 − x )(1 + x )(1 + x 2 )(1 + x 4 )...(1 + x 2 n ) = lim n→∞ 1− x 2

4

2n

(1 − x )(1 + x )(1 + x )...(1 + x ) 1− x . . . . . . . . . = lim

n→∞

HINTS AND EXPLANATIONS

x →∞

x

ex

= lim

x →∞

= lim



nx

ex

= … = lim

n!

x →∞ e x

−m

x x = lim e x x →∞ e x [Putting n = – m, where m is a positive integer] 1 1 = lim m x = =0 x →∞ x e ∞ Case III:  n = 0 xn 1 1 lim x = lim x = =0 x →∞ e x →∞ e ∞ xn Hence, lim x = 0 for all values of n. x →∞ e The correct option is (B) 64. x2 + 4x + 5 = (x + 2)2 + 1 ≥ 1. So, a = 1 2 sin 2 θ 1 − cos 2θ Also, b = lim = lim =2 2 θ →0 θ 2 θ →0 θ lim

x →∞

n

n n–1 2 n–2 n ∑ ar ⋅ bn − r = b + ab + a b + … + a

r=0

log(1 + x 2 (1 + x 2 )) x 2 (1 + x 2 )

⋅ x 2 (1 + x 2 ) ⋅

1 sin x tan x 2 ⋅ ⋅x x x

3

4

n5 + 2 − n2 + 1

n→∞ 5

n4 + 2 − n3 + 1

66. lim

 [Using L’Hospital’s rule repeatedly] =0 Case II:  n is a negative integer.

∴

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

The correct option is (A)

ex

x →∞

n

log(1 + x 2 + x 4 ) x→0 sin x tan x 

= lim

log(1 + x ) ⎞ ⎛ = 1. ⎜ as lim = 1⎟ ⎝ x→0 ⎠ x

n −1

n ( n − 1) x n − 2

(1 − cos 2 x ) cos x

x→0

1 − x 4n + 2 1 = for |x| < 1 n→∞ 1− x 1− x The correct option is (B) 63. Case I:  n is a positive integer lim

log ⎡⎣(1 + x 2 ) 2 − x 2 ⎤⎦

x→0

= lim

= lim

n

= (2n + 1 – 1)

65. lim

n→∞

2

2n +1 ( 2n +1 − 1)

2

2 1 − n2 3 3 1 + 2 5 n n = lim n→∞ 4 5 2 1 3 2 n 5 1+ 4 − n 2 1+ 3 n n n5 4 4 1 +

2 n2 3 1 n5 4 4 1+ 5 − 3 2 3 1+ 2 32 n n n n = lim 4 5 n→∞ n 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 = lim = = 0. n→∞ 0 −1 1 5 2 2 1 1 + − 1 + n7 10 n4 n3 The correct option is (B)

67. lim

x

3

z 2 − ( z − x )2

x→0 3

( 8 xz − 4 x 2 + 3 8 xz ) 4 3

= lim

x→0 ( 3

x 2 xz − x 2 x

3

8z − 4 x + 3 8z

3

x )4

Limits  8.35

x→0

3

2z − x

x 4 3 ⎡⎣ 3 8 z − 4 x + 3 8 z ⎤⎦

4

=

3

2z

⎡2 3 8z ⎤ ⎣ ⎦

z→ 0

223 3 ⋅ z The correct option is (B) A

68. lim h→0

P

+

3

h→0

= lim



h→0

= lim



h→0

=



=

+

8 ⎡ 2hr − h2 + 2hr ⎤ ⎣⎢ ⎦⎥

3

=

h ⋅ h 2r − h 8⋅ h ⋅ h ⎡⎣ 2r − h + 2r ⎤⎦

1 a2 ( y − α )2 ( y − β )2 lim 2 y→ α y 2 ( y − α )2

=

3

a 2 (α − β ) 2 2α

2r

(

8 2 2r

)

3

3

1 − cos x

72. lim

x

x→ 0

= lim

x→ 0

(

)

(

cos π

+z

)

6 = lim 2 69. lim 2/3 x →π /3 (1 − 2 cos x ) 2/3 z→ 0 ⎡ 1 − 2 cos π + z ⎤ 3 ⎣ ⎦ [putting x –π/3 = z] − sin z

= lim

z→ 0

(1 − cos z +

3 sin z

)

(

)

2/3

z→ 0

⎡ ⎛ z⎞⎤ ⎢ 2 sin ⎜⎝ 2 ⎟⎠ ⎥ ⎣ ⎦

2/3

⎡ ⎛ z⎞ ⎛ z⎞⎤ ⎢sin ⎜⎝ 2 ⎟⎠ + 3 cos ⎜⎝ 2 ⎟⎠ ⎥ ⎣ ⎦

2/3

x→ 0

ln (sin 3 x + 1)

= lim



lim

1 − cos 2 x (sin 3 x ) 2

= lim

x→ 0

2x

(1 − xα ) 2

1 − cos (cx 2 + bx + a) (cx 2 + bx + a) 2 ⋅ x → 1/α (cx 2 + bx + a) 2 (1 − xα ) 2

= lim

2

(3 x ) 2

1 − cos (cx 2 + bx + a)

x → 1/α

x→ 2

ln (1 + sin 3 x )

The correct option is (A) 71.

= lim

2

x→ 0

( )

−

( )

x + 7 − 3 2x − 3

x→ 2 3

ln {1 + (1 − cos 2 x )}

x→ 0

1 + cos x

x 1 −1 sin 2 ⋅ ⋅ x→ 0 x 2 2 1 + cos x 1 1 −1 =− ⋅ = 2 2 2

LHL = lim

73. lim

The correct option is (C) = lim

Now, we have,

1

x 1 sin 2 1 ⋅ x→ 0 2 x 1 + cos x 2 1 1 1 = ⋅ = 2 2 2 Hence, limit does not exist. The correct option is (C)

( )

2

( 2) ⋅ 2(x ) 2

2 sin x

+

1/3

ln ( 2 − cos 2 x )

x→ 0

1 − cos x 1 ⋅ x 1 + cos x

RHL = lim

⎡ ⎛ z⎞⎤ ⎛ z⎞ −21/3 ⎢sin ⎜ ⎟ ⎥ cos ⎜ ⎟ ⎝ 2⎠ ⎦ ⎝ 2⎠ −21/3 ⋅ 0 ⋅1 ⎣ = lim = =0 2/3 2 / 3 z→ 0 ⎡ ⎛ z⎞⎤ ⎛ z⎞ 3 ⎢sin ⎜⎝ 2 ⎟⎠ + 3 cos ⎜⎝ 2 ⎟⎠ ⎥ ⎣ ⎦

70. lim

= lim



⎛ z⎞ ⎛ z⎞ −2 sin ⎜ ⎟ cos ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

= lim

a2 ⎡ (α + β ) 2 − 4αβ ⎤⎦ 2α 2 ⎣

⎛ b 2 4c ⎞ b 2 − 4 ac − = ⎜ a ⎟⎠ 2α 2 ⎝ a 2 2α 2 a

The correct option is (B) cos x + π

=

The correct option is (A)

1 128r

=

2

2

2r − h 8 ⎡⎣ 2r − h + 2r ⎤⎦

( ay 2 + by + c) 2

[If α, β are roots of ax2 + bx+ c = 0 then ax2 + bx+ c = a (x – α) (x – β)]

h 2 hr − h2

= lim

⋅ lim

y →α z2 y 2 ( y − α )2 [putting cx2 + bx+ a = z and x = 1/y]

4

1

=

1 − cos z

= lim

=

2 9

x + 6 − 2 3 3x − 5 ⎡ a2 − b2 ⎤ × ⎢ using a − b = ⎥ a + b ⎥⎦ x + 7 + 3 2x − 3 ⎢⎣

( x + 7) − 9( 2 x − 3)

( x + 6) 2/3 + 2( x + 6)1/3 (3 x − 5)1/3 + 4(3 x − 5) 2/3 ( x + 6) − 8 (3 x − 5) ⎡ a3 − b3 ⎤ ⎢ using a − b = 2 ⎥ a + ab + b 2 ⎥⎦ ⎢⎣

 = lim

x→ z

×

−17 ( x − 2) x + 7 + 3 2x − 3

( x + 6) 2/3 + 2( x + 6)1/3 (3 x − 5)1/3 + 4(3 x − 5) 2/3 −23 ( x − 2)

HINTS AND EXPLANATIONS

x4 3

= lim

8.36  Chapter 8 82/3 + 2.81/3 + 4 17 12 34 = = ⋅ − 23 6 23 9+3 1 23 The correct option is (B) −17

=

74. lim

x→ 0

⋅

( 2m + x )1/m − ( 2n + x )1/n x

( 2m + x )1/m − 2 ( 2n + x )1/n − 2 − lim x→ 0 x→ 0 x x a−2 b−2 = lim m − lim n a → 2 a − 2m b → 2 b − 2n  [Putting 2m+ x = am and 2n+ x = bn]

Now, we have,

1 ln e − ln (1 + y ) ⎤ e 1 ⎡ y ⎥ = lim lim ln ⎢ y→ 0 y→ 0 y y ⎢⎣ (1 + y )1/ y ⎥⎦ y − ln (1 + y ) = lim y→ 0 y2

= lim

1

=

−

1

x→ 4

y→ 0

(cos θ ) y + 4 − (sin θ ) y + 4 − (cos 4 θ − sin 4 θ ) y

[Putting x – 4 = y and cos 2θ = cos4 θ – sin4 q] ⎡ (sin θ ) − 1⎤ ⎡ (cos θ ) − 1⎤ 4 = lim cos 4 θ ⎢ ⎥ − sin θ ⎢ ⎥ y→ 0 y y ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ y

y

= cos4 q lncos θ – sin4 q lnsin θ The correct option is (A)

HINTS AND EXPLANATIONS

⎛ x − 1 + cos x ⎞ 76. lim ⎜ ⎟⎠ x→ 0 ⎝ x Now, 

lim

x→ 0

1/ x

= lim e

1 ⎛ x − 1 + cos x ⎞ ln ⎜ ⎟⎠ x x ⎝

x→ 0

1 ⎛ 1 − cos x ⎞ ln ⎜1 − ⎟⎠ x ⎝ x

x→ 0

−2 sin 2 ( x/2) 4( x/2) 2

z→ 0

lim

x→ 0

ln (1 + z ) −1 = 2 z

⎤ cos x − 1 −1 ⎛ cos x − 1⎞ = lim x ⎜ ⎟ = 0 ⋅ = 0⎥ n→ 0 ⎝ 2 x x2 ⎠ ⎦

Hence, the required limit is e–1/2. The correct option is (B)

= lim

y→ 0

while

tan x tan x ⎤ → 1 but > 1⎥ x x ⎦

= (a – 1) + b =a+b–1 The correct option is (B) [ x ] + [2 x ] + [3 x ] + ... + [nx ] 1 + 2 + 3 + ... + n Now, we have, x + 2 x + 3 x + ... + nx =x f (x) ≤ 1 + 2 + 3 + ... + n 79. Let f (x) =

( x − 1) + ( 2 x − 1) + (3 x − 1) + ... + ( nx − 1) 1 + 2 + 3 + ... + n

2 x Σn − n =x− Σn n +1

[∵ x − 1 ≤ [ x ] < x ∀x ∈ R]



Thus, we have, 2 x– < f (x) ≤ x n +1 Now, we have, 2 lim x − = x and lim x = x n→∞ n→∞ n +1 Hence, by Sandwich Theorem, we have n→∞

⎡ ⎡ ⎤ ⎤ e e ⎥ = lim ⎢ ⎥ 77. lim ⎢ x 1 / y x →∞ ⎢ 1 + 1/ x ⎥ ) ⎦ y → 0 ⎢⎣ (1 + y ) ⎥⎦ ⎣(



lim f ( x ) = x 1/ y

x

y→ 0

1 y y2 1 − + − ... = 2 3 4 2

sin x sin x ⎡ as ⎢ x → 0, x → 1 but x < 1 ⎣

=

cos x − 1 ⎡ , we can see that ⎢ Putting z = x ⎣

 

⋅ lim

= lim

and,  f (x) >

⎛ cos x − 1⎞ ln ⎜1 + ⎟⎠ ⎝ cos x − 1 x = lim ⋅ cos x − 1 x→ 0 x2 x = lim



y2

y→ 0

⎡ a sin x ⎤ ⎡ b tan x ⎤ + 78. lim ⎢ x→ 0 ⎣ x ⎥⎦ ⎢⎣ x ⎥⎦

(cos θ ) x − (sin θ ) x − cos 2θ x−4

= lim

= lim

Hence, the required limit is e1/2. The correct option is (C)

m 2m − 1 n 2n − 1 The correct option is (C) 75. lim



⎛ ⎞ y 2 y3 y 4 y−⎜y− + − + ...⎟ 2 3 4 ⎝ ⎠

⎤ 1 ⎡ e ⎥ ln ⎢ y ⎢⎣ (1 + y )1/ y ⎥⎦ e

The correct option is (A)

(

)

1 1 ⎛ 1 1 ⎞ − 80. lim n2 x1/n − x n + 1 = lim n2 ⋅ x n + 1 ⎜⎝ x n n + 1 − 1⎟⎠ n→∞

n→∞

1



(

1

)

= lim x n + 1 x n ( n + 1) − 1 n2 n→∞

Limits  8.37

x = nlim →∞

⋅

x

1 n ( n + 1)

−1

1 n ( n + 1)

⋅

2

n n ( n + 1)

83. We have, cos x − (cos x )cos x lim x → 0 1 − cos x + ln (cos x )

= 1 · lnx · 1 = lnx The correct option is (C)

⎡ 1 − (cos x )cos x −1 ⎤ = lim cos x ⎢ ⎥ x→ 0 ⎢⎣1 − cos x + ln (cos x ) ⎥⎦

81. We have,

= lim

t→ 0

1/ x

f ( x) ⎤ ⎡ lim ⎢1 + x + x→ 0 ⎣ x ⎥⎦

⇒ lim e

3

=e

f ( x) ⎡ ⎤ ⎢ Putting x = g ( x )⎥ ⎣ ⎦

= e3

x→ 0

ln [1 + x + g ( x )] =3 x Since, the denominator approaches zero, the numerator should also approach zero for a finite limit to exist. Thus, we have, lim g ( x ) = 0 ⇒ lim

x→ 0

x→ 0

Now, using L’Hospital’s rule, the above equation reduces to 1 + g ′( x ) =3 x→ 0 1 + x + g( x) i.e., 1 + g′(0) = 3  ⇒  g′(0) = 2 Hence, we have, 1/ x

= lim e x→ 0

t 3 (t − 1) t 4 (t − 1) (t − 2) + + ... 2! 3! = lim =2 2 3 4 t→ 0 t t t − + − ... 2 3 4 The correct option is (C)

= lim

x→ 0

84. We have, lim

x → π /4

(tan x ) tan x − tan x ln (tan x ) − tan x + 1

tt − t ⎛ 0⎞ ⎜ ⎟ t → 1 ln t − t + 1 ⎝ 0 ⎠

= lim = lim

ln [1 + g ( x )] x e

= e g ′( 0 )

⎛ 1⎞ t t (1 + ln t ) 2 + t t ⎜ ⎟ ⎝t⎠ 1+1 = lim = = –2 −1 t→ 1 −1 t2 The correct option is (A)

= e2 The correct option is (B)

85. We have,

(

lim 11/cos x

82. Let y = x +

x+ i.e.,  y2 – xy–

=x+

x

x+

x y

x → π /2

x ... ∞

2

x

+ 21/cos

2

+ ... + n1/cos

x

(

= lim 1y + 2 y + ... + n y y→∞

x =0

= n (0 + 0 + … + 1) = n The correct option is (A)

x + x2 + 4 x y= 2 Hence, the required limit is

86. lim

2x x = lim = lim x →∞ y x →∞ x + x2 + 4 x 2 −3/2

=

1 + 1 + 4x The correct option is (A)

1/n

x

)

cos 2 x

1 ⎡ ⎤ ⎢ Putting cos 2 x = y ⎥ ⎣ ⎦

 1/ y

x ± x2 + 4 x 2 We can see y is a positive quantity for positive x, therefore

x →∞

)

2

y ⎡⎛ 1 ⎞ y ⎛ 2 ⎞ y ⎛ n⎞ ⎤ = lim n ⎢⎜ ⎟ + ⎜ ⎟ + ... + ⎜ ⎟ ⎥ ⎝ n⎠ ⎝ n⎠ ⎥ y →∞ ⎢ ⎝ n ⎠ ⎣ ⎦

i.e.,  y =

= lim

[Putting tan x = t]

t t (1 + ln t ) − 1 ⎛ 0 ⎞ ⎜⎝ ⎟⎠ 1 t→ 1 0 −1 t

lim

g ′( x ) 1 + g( x)

[Putting cos x – 1 = t]

t2 +

ln [1 + x + g ( x )] x

f ( x) ⎤ ⎡ lim ⎢1 + x→ 0 ⎣ x ⎥⎦

1 − (1 + t )t  ln (1 + t ) − t

2 =1 1+1

n

⎛

3⎞

∑ cot −1 ⎜⎝ r 2 + 4 ⎟⎠ n→∞ r =1

n

⎡ ( r − 1/2) ( r + 1/2) + 1⎤

∑ cot −1 ⎢⎢ ( r + 1/2) − ( r − 1/2) ⎥⎥ n→∞

= lim

r =1

⎣

⎦

n

1⎞ 1⎞ ⎛ ⎛ = lim ∑ cot −1 ⎜ r − ⎟ − cot −1 ⎜ r + ⎟ ⎝ ⎠ ⎝ n→∞ 2 2⎠ r =1 

⎡ −1 ⎛ ab + 1⎞ −1 −1 ⎤ ⎢∵cot ⎜ ⎟⎠ = cot a − cot b ⎥ b − a ⎝ ⎣ ⎦

HINTS AND EXPLANATIONS



1 n +1

8.38  Chapter 8 89. We know,

⎡ 1 3 3 5 = lim ⎢cot −1 ⎛⎜ ⎞⎟ − cot −1 ⎛⎜ ⎞⎟ + cot −1 ⎛⎜ ⎞⎟ − cot −1 ⎛⎜ ⎞⎟ + ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ n→∞ ⎣ 1⎞ ⎛ ⎛ ... + cot −1 ⎜ n − ⎟ − cot −1 ⎜ n + ⎝ ⎝ 2⎠



1⎞ ⎤ ⎟ 2 ⎠ ⎥⎦

⎡ 1 1 ⎤ = lim ⎢cot −1 ⎛⎜ ⎞⎟ − cot −1 ⎛⎜ n + ⎞⎟ ⎥ ⎝ 2⎠ ⎝ n→∞ ⎣ 2⎠ ⎦ ⎛ 1⎞ = cot–1 ⎜⎝ 2 ⎟⎠ – 0 = tan–1 2

For x →

⎡ n 1⎤ lim ⎢ ∑ r ⎥ = 0 n→∞ ⎢ ⎣ r = 1 2 ⎥⎦ The correct option is (A) ∴

π⎞ ⎛ 2 sin ⎜ x + ⎟ ⎝ 4⎠

5π π⎞ ⎛ + h, 2 sin ⎜ x + ⎟ → − 2 , ⎝ 4 4⎠

90. Here, 0 < cos x < 1; if 0 – h < x < 0 + h ∴ [cos x] = 0

But greater than – 2 ⎡ π⎞⎤ ⎛ ∴ ⎢ 2 sin ⎜ x + ⎟ ⎥ = –2 (1) ⎝ 4⎠⎦ ⎣ 5π π⎞ ⎛ Also, for x → − h, 2 sin ⎜ x + ⎟ → − 2 , but greater ⎝ 4 4⎠ than – 2 ⎡ π⎞⎤ ⎛ ∴ ⎢ 2 sin ⎜ x + ⎟ ⎥ = –2 ⎝ 4⎠⎦ ⎣ From (1) and (2), we get

Hence, lim | x | [cos x ] x →0

= lim | x |0 = lim 1 = 1 x →0

x→0

The correct option is (B) 91. We have, an – 1 + 1 =

(2) ∴

lim [sin x + cos x ]

an (1) n

⎛ a + 1⎞ ⎛ a + 1⎞ ⎛ a + 1⎞ lim ⎜ 1 ⎟ ⎜ 2 ⎟ ... ⎜ n ⎟ ⎝ a1 ⎠ ⎝ a2 ⎠ ⎝ an ⎠

n→∞

= lim − 2 = –2

1 ⎛a ⎞ ⎛a ⎞ ⎛a ⎞ ⎛ a ⎞ = lim ⎜ 2 ⎟ ⎜ 3 ⎟ ⎜ 4 ⎟ ... ⎜ n +1 ⎟ ⋅ n→∞ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ⎝ n + 1⎠ a ⋅ a ... a n 1 2

The correct option is (B)

= lim

x → 5π /4

x→ 5π /4

HINTS AND EXPLANATIONS

1

⎡ n 1⎤ Thus, ⎢ ∑ r ⎥ → 0 as n → ∞ ⎢⎣ r = 1 2 ⎥⎦

The correct option is (B) 87. Here, sin x + cos x =

1⎛ 1⎞ ⎜⎝1 − n ⎟⎠ 2 2 ∑ r = i.e., sum of n terms of G.P. 1 r =12 1− 2 which tends to one as n → ∞ but always remains less than one. n

n→∞

88. lim lim

m →∞ n→∞

⎧⎪1 + n 1n + 2n + n 2n + 3n + ... + n ( m − 1) n + m n ⎫⎪ ⎬ ⎨ m2 ⎪⎭ ⎪⎩ = lim m→∞

n n n ⎧ ⎫ ⎛ 1⎞ ⎛ 2⎞ ⎛ ( m − 1) ⎞ ⎪ 1 + 2 n ⎜ ⎟ + 1 + 3n ⎜ ⎟ + 1 + ... + m n ⎜ + 1⎪ ⎟ ⎝ 2⎠ ⎝ 3⎠ ⎝ m ⎠ ⎪ ⎪ ⎨ lim ⎬ 2 n→∞ m ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 1 + 2 + 3 + ... + m = lim m→∞ m2 n n ⎛ ⎛ 1⎞ ⎛ 2⎞ ∵ → 0 as n → ∞ ; ⎜ ⎜ ⎟ ⎜⎝ ⎟⎠ → 0 as n → ∞; ... ; 3 ⎝ ⎝ 2⎠

⎞ ⎛ m − 1⎞ ⎜⎝ ⎟⎠ → 0 as n → ∞⎟ m ⎠ n

 m ( m + 1)

1⎛ 1⎞ 1 = lim ⎜1 + ⎟ = m →∞ 2 ⎝ m⎠ 2 2m 2 The correct option is (D) = lim

m →∞

an +1 1 + an = lim  n→∞ n! ( n + 1)!

[using (1)]

⎛1 a ⎞ = lim ⎜ + n ⎟ n→∞ ⎝ n! n! ⎠ ⎡1 ⎤ a 1 = lim ⎢ + + n −1 ⎥  n→∞ ⎣ n! ( n − 1)! ( n − 1)! ⎦

[using (1)]

⎡1 1 1 a⎤ 1 1 = lim ⎢ + + + ... + + + 1 ⎥ n→∞ ⎣ n! ( n − 1)! ( n − 2)! 2! 1! 1! ⎦ ⎡1 1 1 1 1 1⎤ = lim ⎢ + + + ... + + + ⎥ n→∞ ⎣ n! ( n − 1)! ( n − 2)! ( 2)! 1! 1⎦ [  a1 = 1; given] 1 1 1 ⎡ ⎤ = e ⎢as, e = 1 + + + + ... ∞ ⎥ 1! 2! 3! ⎣ ⎦ The correct option is (B) ⎡ 1⎞ ⎛ 1 ⎞⎤ ⎛ 92. lim n − n ⎢( n + 1) ⎜ n + ⎟ ... ⎜ n + n −1 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ n→∞ 2 2 ⎣ 2

⎡ ⎛ ⎢ ( n + 1) ⎜⎝ n + = lim ⎢ n→∞ ⎢ ⎢ ⎣

1⎞ ⎛ 1 ⎞⎤ ⎟ ... ⎜ n + n −1 ⎟⎠ ⎥ 2⎠ ⎝ 2 ⎥ ⎥ nn ⎥ ⎦

n

n

Limits  8.39 n

1 ⎞ ⎛ 1⎞ ⎛ n + n −1 ⎜ ⎟ n + 1⎞ ⎜ n + 2 ⎟ ⎛ 2 = lim ⎜ ... ⎜ ⎟ ⋅⎜ ⎟ ⎟ n→∞ ⎝ n ⎠ n n ⎜ ⎟ ⎟ ⎜ ⎠ ⎝ ⎝ ⎠

95. We know n ≤ [x] < n + 1  ⇒ [x] = n

n

n

n

n

⎛ = lim ⎜1 + n→∞ ⎝

2n

n

1⎞ ⎛ 1⎞ 2 ⎛ 1 ⎞ ⎟⎠ ⋅ ⎜⎝1 + ⎟⎠ ... ⎜⎝1 + n −1 ⎟⎠ n 2n 2 n

= e1 · e1/2 · e1/4 … e1/2n–1

an ⎫⎪ ⎧⎪ ⎛ 1⎞ … ⎨ using; lim ⎜1 + ⎟ = e a ⎬ n→∞ ⎝ n⎠ ⎩⎪ ⎭⎪

 1 1−

2n −1. n 2n − 1

1

= e(1 + 1/2 + 1/4 + …) = e 2 = e2 The correct option is (B) 93. lim

⎛0 ⎞ ⎜⎝ form⎟⎠ 0

x→ y

= lim

x→ y

(applying L-Hospital’s rule)

yx y −1 − y x log y

n tan x → n as x → 0 but more than n x

⎡ n sin x ⎤ Thus,  n – 1 ≤ ⎢ ⎥ < n as x → 0 ⎣ x ⎦ ⎡ n sin x ⎤ ⇒ ⎢ ⎥ =n–1 ⎣ x ⎦ ⎡ n tan x ⎤ Again,  n ≤ ⎢ ⎥ < n + 1 as x → 0 ⎣ x ⎦

Thus, 

y y (log y + 1)

1 − log y 1 + log y The correct option is (A) 1⎞ n ⎛ 3 94. lim ∑ cot −1 ⎜ r − r + r ⎟ n→∞ ⎜⎝ ⎟⎠ r =1 2 =

= (n – 1) + (n) = (2n – 1) The correct option is (C) 96. We know,

Thus,

The correct option is (A)

2r ⎛ ⎞ = lim ∑ tan −1 ⎜ 2 4⎟ ⎝ ⎠ n→∞ − r + r 1 r =1

97. lim

θ→ 0

⎛ ⎞ 2r = lim ∑ tan −1 ⎜ ⎟ 2 2 n→∞ ( ) − − ( r + r ) 1 r r ⎝ ⎠ r =1 n

⎛ (r 2 + r) − (r 2 − r) ⎞ = lim ∑ tan −1 ⎜ ⎟ n→∞ ⎝ 1 − (r 2 + r) (r 2 − r) ⎠ r =1 n

cos 2 (1 − cos 2 (1 − cos 2 (1 ... cos 2 θ )) ⎛ π ( θ + 4 − 2⎞ sin ⎜ ⎟ θ ⎝ ⎠

= lim

θ→ 0

n

∑ ⎡⎣tan −1( r 2 + r ) − tan −1( r 2 − r )⎤⎦ n→∞ r =1

= lim

θ→ 0

= lim [tan–1 2 – tan–1 0) + (tan–1 6 – tan–1 2) + (tan–1 12 – = lim {tan–1 (n2 + n) – tan–1 (0)} n→∞

∴

n

⎛

r =1

⎝

∑ cot −1 ⎜⎜ r n→∞ lim

3

π 2

−r+

2 The correct option is (C)

cos 2 (sin 2 (sin 2 ... (sin 2 θ )) cos 2 0 = = π ⎛ ⎞ sin θ ⎟ sin ⎜ π lim 4 ⎜⎝ θ → 0 θ θ + 4 + 2 ⎟⎠

)

The correct option is (C) 98.

lim f ( x ) = lim x→

1⎞ π r⎟ = ⎟⎠ 2

cos 2 (sin 2 (sin 2 ... (sin 2 θ )) ⎛ π ( θ + 4 − 2⎞ sin ⎜ ⎟ θ ⎝ ⎠

(

n→∞

tan–1 6) + … + {tan–1 (n2 + n) – tan–1 (n2 – n)}]

= tan–1 (∞) – tan–1 (0) =

x2 → 1 as x → 0, but less than 1. sin x tan x

⎡ ⎤ x2 x2 Hence, lim ⎢ < 1 as x → 0 ⎥ = 0 as 0 ≤ x → 0 ⎢ sin x tan x ⎥ sin x tan x ⎣ ⎦

n

= lim

⎛ ⎡ n sin x ⎤ ⎡ n tan x ⎤⎞ + lim ⎜ ⎢ ⎝ ⎣ x ⎥⎦ ⎢⎣ x ⎥⎦⎟⎠

x →∞

x → 1 , as x → 0, but less than 1 sin x x Also, → 1 , as x → 0 but less than 1 tan x

x x (1 + x log x )

y ⋅ y y −1 − y y log y

=

Also,

⎡ n tan x ⎤ ⇒ ⎢ ⎥ =n ⎣ x ⎦

x y − yx

xx − y y  yx y −1 − y x log y = lim x  x → y x (1 + x log x ) − 0

n sin x → n as x → 0 but less than n x



π − 2

x→

π − 2

= lim

π x→ − 2

tan x − sin{tan −1(tan x )} tan x + cos 2 (tan x ) tan x − sin x tan x + cos 2 (tan x )

2

HINTS AND EXPLANATIONS

n

1 1 1 ⎞ = lim ⎛⎜1 + ⎞⎟ ⋅ ⎛⎜1 + ⎞⎟ ... ⎛⎜1 + (1∞ form) n −1 ⎟ ⎝ ⎠ n→∞ ⎝ n⎠ ⎝ 2n ⎠ 2 n

Here,

8.40  Chapter 8 [ tan–1(tan x) = x, where x < π/2]

= lim



x→

=



π 2

sin x tan x cos 2 (tan x ) 1+ tan x 1−

1− 0 =1 1+ 0

101.

⎡ ⎤ cos 2 (tan x ) finite quantity = = 0⎥ ⎢∵ lim π tan x ∞ ⎢ x→ − ⎥ 2 ⎣ ⎦



Similarly, 

sin x tan x cos 2 (tan x ) 1+ tan x 1+

lim f ( x ) = lim

π x→ + 2

π x→ + 2

lim f ( x ) = 1

1+ 0 =1 1+ 0

π x→ 2

99. Put c = a1/4 and z = x1/4, we get the function whose limit is required, as

HINTS AND EXPLANATIONS

−1 ⎧⎡ 2 1 ⎤ 2 −1 log 2cz ⎪ ⎢⎛ c + z ⎞ 2 ⎥ − 2 − ⎨ ⎜ ⎟ 3 2 2 3 z − cz + c z − c ⎥ ⎪ ⎢⎣⎝ c − z ⎠ ⎦ ⎩ −1 ⎧⎡ c2 + z 2 ⎫ ⎤ 2cz ⎪ ⎪ − = ⎨⎢ − c ⎥ ⎬ ( z − c) ( z 2 + c 2 ) ⎥⎦ ⎪⎩ ⎢⎣ c − z ⎪⎭ 8 8 2 = (c – z – c) = z = x Hence, required limit as x → a = a2 The correct option is (C)

lim

⎫

4

8

c4 ⎪

⎬ ⎪ ⎭

= lim

t →0

lim

[log ( 2 + t ) − log 2] [3.4t − 3(t + 1) 1/2

{(8 + t ) − ( 4 + 3t ) } sin π (t + 1) [By putting x = 1 + t]



t⎞ ⎛ log ⎜1 + ⎟ (3 ( 4t − 1) − 3t ) ⎝ 2⎠ = − lim 1/2 1/3 t →0 ⎡ 3t ⎞ ⎤ t⎞ ⎛ ⎛ ⎢ 2 ⎜1 + ⎟ − 2 ⎜1 + ⎟ ⎥ sin π t ⎝ 8⎠ 4 ⎠ ⎥⎦ ⎢⎣ ⎝ 1 = − lim ⋅ t →0 π

⎛ log ⎜1 + ⎝ t 2

[(1 − x )(1 − x 2 )…(1 − x n )]2

x →1

(1 − x n +1 )(1 − x n + 2 )...(1 − x 2 n ) x )(1 − x 2 ) (1 − x 3 ) ... (1 − x n )]

(1 − x n +1 )(1 − x n + 2 )...(1 − x 2 n ) x →1 (1 − x )(1 − x ) ... (1 − x )

= lim

×

t⎞ ⎟ t ⎤ 2 ⎠ ⎡ 3 ( 4 − 1) ⋅⎢ − 3⎥ t ⎢⎣ ⎦⎥

(1 − x ) (1 − x ) ... (1 − x )

(1 − x ) (1 − x 2 ) ... (1 − x n ) ( 2n)! 1 1 1 = (n + 1) (n + 2) … 2n · · ... = ( n!) 2 1 2 n The correct option is (C) 102. Since maximum value of cos–1x = k

∑ cos −1 α r = r =1

cos–1 a r = ∴ θ =

π 2

kπ is possible if and only if each 2

π ⇒  a r = 0 2 

k

∑ (α r )

r

=0

r =1

∴

{(7 + x )1/3 − (1 + 3 x )1/2} sin π x 1/3

x ) (1 − x 2 ) ... (1 − x n )]2

(1 − x )(1 − x 2 )…(1 − x n )(1 − x n +1 )(1 − x n + 2 )…(1 − x 2 n )

8

(log (1 + x ) − log 2) (3.4 x −1 − 3 x )

x →1

x →1 [(1 −



The correct option is (A)

100.

(1 − x ) (1 − x 2 ) ... (1 − x 2 n )

x →1 [(1 −

[ tan (tan x) = x – π, if x > π/2] =

lim

= lim

–1

∴

t π ⋅ t 3t sin π t ⎡ ⎤ 2 3 ⎢1 + 24 − 1 − 8 + terms containing t , t , etc.⎥ ⎣ ⎦ 3 9 4 = ·(3 log 4 − 3) = log π π e The correct option is (A)



lim

(1 + x 2 )1/3 − (1 − 2 x )1/4

x→θ

= lim

x→ 0

x + x2 (1 + x 2 )1/3 − (1 − 2 x )1/4 x + x2

1 2 x ⎛ ⎛ 4 ⎞ 2 ⎞ ⎜⎝1 + x + O ( x )⎟⎠ − ⎜⎝1 − + O ( x )⎟⎠ 3 2 = lim x→ 0 x (1 + x )

1 1 + x + O ( x2 ) 1 = lim 2 3 = x→0 1+ x 2 [O (x2) means terms containing x2, x3, x4, …] The correct option is (C) 103.  ax2 + bx+ c = 0 has roots α and β, therefore 1 a b 1 2 + + c = 0 i.e., cx2 + bx+ a = 0 has roots and α β x x b a⎞ 1⎞ ⎛ 1⎞ ⎛ ⎛ ⇒ c ⎜ x 2 + + ⎟ = c⎜x − ⎟ ⎜x − ⎟ ⎝ ⎝ α⎠ ⎝ β⎠ cx c ⎠

Limits  8.41

Now, lim1 x→

α

⎡1 − cos (cx 2 + bx + a) ⎤ ⎢ ⎥ 2 (1 − α x )2 ⎢⎣ ⎥⎦

c ⎛ 1 1⎞ 1· ⎜ − ⎟ 2 ⎝ α β⎠ = −α

1/ 2

⎧ 2 ⎛ cx 2 + bx + a ⎞ ⎫ ⎪ sin ⎜ ⎟⎪ = lim ⎨ 2 ⎝ ⎠⎬ 1 x→ ⎪ ⎪ 2 α ⎩ (1 − α x ) ⎭

The correct option is (A) 104.

⎛ cx + bx + a ⎞ sin ⎜ ⎟ = lim 2 ⎝ ⎠ 1 x→ α 1 − αx 2

lim x→

1⎞ c⎛ ⎜ x − ⎟⎠ 2⎝ α

h→0

= lim (1 − h) cot (1 − h) = h→0

lim f ( x ) = lim x → 0+

h→0

= lim



⎛c ⎛ 1⎞ sin ⎜ ⎜ x − ⎟ α⎠ ⎝2 ⎝

1 α

lim f ( x ) = lim {− h} cot {− h}

x → 0−



⎛c ⎛ 1⎞ ⎛ 1⎞⎞ sin ⎜ ⎜ x − ⎟ ⎜ x − ⎟ ⎟ α⎠ ⎝ β⎠⎠ ⎝2 ⎝ = lim 1 1⎞ ⎛ x→ −α ⎜ x − ⎟ α ⎝ α⎠

=

c ⎛ 1 1⎞ − 2α ⎜⎝ α β ⎟⎠

=

⎛ 1⎞⎞ 1⎞ c⎛ x− ⎟ ⎜⎝ x − β ⎟⎠ ⎟ 2 ⎜⎝ β⎠ ⎠ · lim 1 −α ⎛ 1⎞ x→ x − α ⎟ ⎜⎝ β⎠

∴

h→0

cot 1

tan 2 {h} h2 − [h]2 tan 2 h2 h2

=1

lim f ( x ) does not exist,

x→0

The correct option is (D) 105.

lim

x→0

b c ⎞ ⎛ sin x ⎞ ⎛ x = lim x a + b − c ⎜ ⎟ ⎜ ⎝ x ⎠ ⎝ sin ( x c ) ⎟⎠ x→0 sin ( x )

x a sin b x c

The above limit is non-zero if a + b – c = 0 The correct option is (D)

Previous Year’s Questions 106. Key Idea: Limit of a function exists only, if LHL = RHL. Now, lim

1 − cos 2 x 2x

x→0



= lim

1 − 1 + 2 sin 2 x

 = lim

x→0

sin x Let f ( x ) = x sin( 0 − h) LHL = lim h→ 0 0−h Now, sin h = lim = −1 h→ 0 − h sin( 0 + h) 0+h and sin h = lim =1 h→ 0 h ∵ LHL ≠ RHL RHL = lim

h→ 0

sin x does not exist. x The correct option is (D)

2x

x→0

2 sin x 2x

= lim

x→0

sin x x

4x + 1 ⎞ ⎛ = lim ⎜1 + 2 ⎟ x →∞ ⎝ x + x + 2⎠



⎡ 1 4x + 1 ⎞ ⎛ = lim ⎢⎜1 + 2 ⎟ x →∞ ⎢⎝ x + x + 2⎠ ⎢⎣

⎥ ⎥⎦

⎛ 1⎞ ⎜⎝ 4 + x ⎟⎠ lim x →∞ 1 2 1+ − 2 x x e

⎛ x − 3⎞ 108. The limit lim ⎜ ⎟ x →∞ ⎝ x + 2 ⎠

x

x



5 ⎤ ⎡ = lim ⎢1 − x →∞ ⎣ x + 2 ⎥⎦



⎛ 5 ⎞ ⎤ ⎜⎝ x + 2 ⎟⎠ ⎡ 1/ ⎜ ⎟ ⎢⎛ ⎛ −5 ⎞ ⎞ ⎝ x + 2 ⎠ ⎥ = lim ⎢⎜1 + ⎜ ⎟ ⎟ ⎥ x →∞ ⎝ ⎝ x + 2 ⎠⎠ ⎢ ⎥ ⎣ ⎦

⎛ −5 x ⎞

x→0

x →∞

=

( 4 x +1) x ( 4 x +1) ⎤ x 2 + x +2 x2 + x + 2 ⎥

= e4 The correct option is (A)

∴ lim

107. Key Idea : lim (1 + λ x )1/ x = e λ

x





x

HINTS AND EXPLANATIONS

⎛ x 2 + 5 x + 3⎞ Now, the limit lim ⎜ 2 ⎟ x →∞ ⎝ x + x + 2 ⎠

8.42  Chapter 8 111. By applying L’Hopital Rule, the given limit equals

5 ⎞ ⎛ lim ⎜ − 1+ 2 / x ⎠⎟ x →∞ ⎝

=e = e −5 Alternative Method:



⎛ x − 3⎞ lim ⎜ ⎟ x →∞ ⎝ x + 2 ⎠

= lim

x →∞ ⎛

The correct option is (C)

x

⎛ 3⎞ ⎜⎝1 − x ⎟⎠ 2⎞ ⎜⎝1 + x ⎟⎠

112. Applying L. Hospital’s Rule

x



=

e −3

HINTS AND EXPLANATIONS

xf ( 2) − 2 f ( x ) + 2 f ( 2) − 2 f ( x ) x→2 x−2 f ( 2)( x − 2) − 2{ f ( x ) − f ( 2)} x−2



= lim



= f ( 2) − 2 lim



= f ( 2) − 2 f ′( 2) = 4 − 2 × 4 = −4

x→2

f ( x ) − f ( 2)} x−2

Alternative Solution:

x→2

=f (2) − 2f  ′(2) =4 − 2 × 4 = − 4 The correct option is (C) 110. The limit, by applying L’Hopital rule,





⎛ π x⎞ tan ⎜ − ⎟ (1 − sin x ) ⎝ 4 2⎠ lim x →π / 2 ⎛ π x⎞ 4 ⎜ − ⎟ (π − 2 x ) 2 ⎝ 4 2⎠ ( − cos x ) = lim x →π / 2 8( −2)(π − 2 x ) − sin x = lim x →π / 2 16( −2) =

1 . 32

The correct option is (C)



2x

=

⎛ 1 ⎞ a b ⎜ a b ⎟ × 2 x × ⎛⎜ + 2 ⎞⎟ ⎝x x ⎠ 2 ⎟ ⎠



a b ⎞⎜ + ⎛ lim ⎜1 + + 2 ⎟ ⎝ x x x →∞ ⎝ x x ⎠ ⇒ a = 1, b ∈ R

= e2a

The correct option is (B) 114. Let L =

lim

1 − cos a( x − α )( x − β ) ( x − α )2

x →α

⎧ xf ( 2) − 2 f ( x ) ⎫ lim ⎨ ⎬ x→2 ⎩ x−2 ⎭ = lim{ f ( 2) − 2 f ′( x )} ( by L′ Hopital’s Rule)

f ( a) g ′( a) − g ( a) f ′( a) =4 g ′( a) − f ′ ( a)

a b 113. The limit lim ⎛⎜1 + + ⎞⎟ x →∞ ⎝ x x2 ⎠

= lim

x→2

x→ 2a

k ( g ′( a) − ff ′( a)) =4 ( g ′( a) − f ′( a)) k = 4. The correct option is (A)

xf ( 2) − 2 f ( x ) 1 09. The limit lim x→2 x−2

lim

x

= e −5 e2 The correct option is (C)



1 1 + + x − x =2. 3 3 lim x→0 1 3

= lim

x →α

2 (x − α)

2

×

= lim

⎛ ( x − α )( x − β ) ⎞ 2 sin 2 ⎜ a ⎟⎠ ⎝ 2 ( x − α )2

x →α

⎛ ( x − α )( x − β ) ⎞ sin 2 ⎜ a ⎟⎠ ⎝ 2 2

2

a (x − α) (x − β) 4

2

a 2 (α − β ) 2 . 2 The correct option is (A) Then, the limit L =

115. f (x) is a positive increasing function ⇒ 0 < f (x) < f (2x) < f (3x) f ( 2 x ) f (3 x ) < f ( x) f ( x)



⇒ 0 0 is a constant and T is dt the total life in years of the equipment. Then the scrap value V(T) of the equipment is

I (A) e–kT (B)  T2 – k (C) I – 35. If

kT 2 k (T − t ) 2 (D)  I− 2 2

dy = y + 3 > 0 and y(0) = 2, then y (ln 2) is equal to dx

(A) –2

(B) 7

(C) 5

(D) 13

36. The population p(t) at time t of a certain mouse dp(t ) species satisfies the differential equation dt = 0.5 p(t) – 450. If p(0) = 850, then the time at which the population become zero is (A)  2 ln 18 (B)  ln 9 (C)  1/2 ln 18 (D)  ln 18

Differential Equations  9.29 1 − y2 dy 37. The differential equation = determines a y dx family of circles with (A)  variable radius and fixed centre (B)  variable radius and variable centre (C)  fixed radius and variable centre on x-axis (D)  fixed radius and variable centre on y-axis dy x+ y 38. The solution of the differential equation = x dx satisfying the condition y(1) = 1 is 2 (A) y = ln x + x (B)  y = x ln x + x (C) y = xe(x – 1) (D)  y = x ln x + x 39. The differential equation of the family of circles with fixed radius 5 units 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 40. The differential equation which represents the family of curves y = c1e c x, where c1 and c2 are arbitrary constants is (A) y′ = y2 (B)  y″ = y′y (C) y″ = y′ (D)  yy″ = (y′)2 2

(A) ey= e x − 1 + ce − e (B)  ey = e x − 1 + ce e x

(C) ex = e y − 1 + ce − e y

x

(D)  None of these 2

dy ⎛ dy ⎞ 45. Solution of the equation x ⎜ ⎟ + ( y − x ) −y ⎝ dx ⎠ dx = 0 is (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 46. The differential equation of the family of general ­circles is (A) y′′′ (1 + y′ 2) – 3y′y″ 2 = 0 (B) y′′′ (1 + y′ 2) + 3y′y″ 2 = 0 (C) y′′′ (1 + y′ 2) – 3y″y′ 2 = 0 (D)  None of these 47. The equation of the family of curves which intersect the hyperbola xy = 2 orthogonally is

xdy − ydx x2 + y2

= 0 is

(A) cos



2⎞

⎛c + x + y (B) x = y tan ⎜ ⎟ 2 ⎝ ⎠

c y =1+ x x

(B)  x2 = (c + x2) tan

y x

1 y =c– 2x 2x2 1 y (D) tan = c + x x

(C) tan

⎛ c − x2 − y2 ⎞ (C) y = x tan ⎜ ⎟ 2 ⎝ ⎠



49. Solution of the differential equation

(D)  None of these ⎛ f ( y / x) ⎞ 43. Solution of the equation x dy = ⎜ y + x dx is ⎝ f’ ( y / x ) ⎟⎠ ⎛ y⎞ (B)  f ⎜ ⎟ = cx ⎝ x⎠

⎛ x + y − 1 ⎞ dy ⎛ x + y + 1 ⎞ ⎜⎝ x + y − 2 ⎟⎠ dx = ⎜⎝ x + y + 2 ⎟⎠ , given that y = 1 when ( x + y)k − k = 0, where k = k (B) 2 (C) 3 (D) 4

x = 1, is k(y – x) + log (A) 1

PRACTICE EXERCISES

48. The solution of the differential equation dy y x 2 − xy = 1 + cos is dx x

⎛ c + x2 + y2 ⎞ (A) y = x tan ⎜ ⎟ 2 ⎝ ⎠

⎛ x⎞ (A)  f ⎜ ⎟ = cy ⎝ y⎠

dy = ex – y (ex – ey) is dx

(A) y =

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

2

44. Solution of the equation

(D)  None of these

x3 x2 + C (B)  y= +C 6 4 x3 x2 (C) y = − +C (D)  y= − +C 6 4

41. The general solution of the differential equation

42. Solution of the equation xdx + ydy +

⎛ y⎞ (C)  f ⎜ ⎟ = cxy ⎝ x⎠

9.30  Chapter 9 50. Solution of the equation xdy – [ y + xy3 (1 + log x)] dx = 0 is −x 2 2 x3 ⎛ 2 ⎞ (A)  2 = ⎜⎝ + log x ⎟⎠ + C 3 3 y (B)  (C) 

x2 y2

=

−x 2 y2

2 x3 ⎛ 2 ⎞ ⎜⎝ + log x ⎟⎠ + C 3 3 x3 ⎛ 2 ⎞ = ⎜⎝ + log x ⎟⎠ + C 3 3

(D)  None of these dy 51. Solution of equation = dx

y

d (ϕ ( x )) − y2 dx is ϕ( x)

ϕ( x) + c ϕ( x) (B)  y = +c x x ϕ( x) (C) y = (D)  y = f (x) + x + c x+c (A) y =

52. Solution of the differential equation ydx + (x + x2y) dy = 0 is 1 (A) log y = Cx (B) – + log y = C xy

PRACTICE EXERCISES

1 (C)  + log y = C xy

1 (D) – =C xy

53. The equation of the curve for which the square of the ordinate is twice the rectangle contained by the abscissa and the x-intercept of the normal and passing through (2, 1) is (A) x2 + y2 – x = 0 (B) 4x2 + 2y2 – 9y = 0 2 2 (C) 2x + 4y – 9x = 0 (D)  4x2 + 2y2 – 9x = 0 x dx + y dy 54. Solution of = x dy − y dx

1 − ( x2 + y2 ) x2 + y2

2

(C) sin–1 x + y = tan–1 (D)  None of these

is

sin x + xy = c x sin x (C)  + y = c x

58. If the curve y = f (x) passing through the point (1, 2) and satisfies the differential equation xdy+ (y + x3y2) dx = 0, then 1 (A) xy = (B)  x3y = 2 2 1 (C)  = 2 (D)  None of these xy 59. The solution of the differential equation





x dx + y dy = x dy − y dx

a2 − x 2 − y 2 x2 + y2

⎛ x2 ⎞ (2x log y) dx + ⎜ + 3 y 2 ⎟ dy = 0 is ⎝ y ⎠ (B) y3 log x + x3 = c (D)  None of these

61. The solution of the equation

y x +c

(B) 

is

y⎫ ⎧ (A)  x 2 + y 2 = a cos ⎨c + tan −1 ⎬ x⎭ ⎩ y⎫ ⎧ (B)  x 2 + y 2 = a sin ⎨c + tan −1 ⎬ x ⎩ ⎭ ⎫ ⎧ x (C)  x 2 + y 2 = a sin ⎨c + tan −1 ⎬ y⎭ ⎩ (D)  None of these

(A) x2 log y + y3 = c (C) x2 log y – y3 = c y + sin x cos 2 ( xy )

55. The solution of the differential equation (x cos x – sin x + yx2) dx + x3dy = 0 is equal to (A) 

57. The equation of the curve, passing through (2, 5) and having the area of triangle formed by the x-axis, the ordinate of a point on the curve and the tangent at the point 5 square units, is (A) xy = 10 (B)  x2 = 10 y (C) y2 = 10x (D)  xy1/2 = 10

60. The solution of the equation

2 2 (A) sin–1 x + y = c y (B) tan–1 x = c 2

56. Solution of the differential equation [y (1 + x–1) + sin y] dx + (x + log x + x cos y) dy = 0 is (A) xy+ y log x = c (B)  xy+ x sin y = c (C) xy+ y log x + x sin y = c (D)  None of these

sin x +x=c x

(D)  None of these



2

cos ( xy )

⎛ ⎞ x dx + ⎜ + sin y ⎟ dy = 0 is 2 ⎝ cos ( xy ) ⎠

(A)  tan (xy) + cos x – cos y = c (B)  tan (xy) – cos x – cos y = c (C)  tan (xy) + cos x + cos y = c (D)  None of these

62. The solution of the equation

dy y = is dx 2 y log y + y − x

Differential Equations  9.31 c y

(C) x = –y log y +

c y

63. Solution of the equation cos2x when | x | < (A) y = (C) y =

(B) y = x log x +

(D)  None of these dy – y tan 2x = cos4x, dx

π ⎛π⎞ 3 3 and y ⎜ ⎟ = , is ⎝ 6⎠ 4 8 sin 2 x

)

2

sin 2 x 2 (1 + tan 2 x )

(



)

(D)  None of these

dy 64. The solution of the equation + x(x + y) = 3 3 dx x (x + y) – 1 is (A) (x + y)–3 = cex2 + x2 + 1 (B) (x + y)–2 = cex2 – x2 + 1 (C) (x + y)–2 = cex2 + x2 + 1 (D)  None of these dy 65. The solution of the equation sin y = cos y (1 – x dx cos y) is (A) sec y = (1 + x) + cex (B) tan y = (1 + x) + cex (C) sec y = (1 + x) + ce–x (D)  None of these 66. The solution of the differential equation x(y2exy + ex/y)dy = y(ex/y– y2exy) dx is (A) xy = ln(ey/x+ c) (B)  xy = ln(ex/y+ c) y x (C)  = ln(exy+ c) (D)  = ln(exy+ c) x y





x

0

0

x ∫ y(t ) dt = ( x + 1) ∫ ty(t ) dt , x > 0 is c

e

−

1 x

3

x − y2 1 2

x − y2 1 2

x − y2 (D)  None of these

=0 =0 =0

dy 70. Solution of the differential equation 2y sin x = dx ⎛π⎞ 2 sin x cos x – y2cos x satisfying y ⎜ ⎟ = 1 is given by ⎝ 2⎠ 2 (A) y = sin x (B)  y = sin2x 2 (C) y = cos x + 1 (D)  y2 sin x = 4 cos2x 71. The solution of the equation y (2x2y + ex) dx – (ex+ y3) dy = 0, if y(0) = 1, is (A) 6ex– 4x3y – 3y3 – 3y = 0 (B) 6ex+ 4x3y – 3y3 – 3y = 0 (C) 6ex+ 4x3y + 3y3 – 3y = 0 (D)  None of these 72. The solution of the equation ye–x/y dx – (xe–x/y + y3) dy = 0 is (A) 3e–x/y+ y2 = c (B) 2e–x/y+ y2 = c –x/y 2 (C) 2e + y = c (D)  None of these 73. The differential equation corresponding to y =

3

∑ ci e m x , i

i =1

where ci’s are arbitrary constants and

(A) y3 – 7y1 + 6y = 0 (C) y3 – 7y1 – 6y = 0

(B)  y3 + 7y1 + 6y = 0 (D)  None of these

1

c (B)  y = 3 ex x3 x 1 1 c − c (C) y = e x (D)  y = ex x x 68. The solution of the differential equation (1 + tan y) (dx– dy) + 2xdy = 0 is (A) x (sin y + cos y) = sin y + ce–y (B) x (sin y – cos y) = sin y + ce–y (C) x (sin y + cos y) = cos y + ce–y (D)  None of these (A) y =

(C) ( 2 x − 3) +

1 2

m1, m2, m3 are roots of the equation m3 – 7m + 6 = 0, is

67. Solution of the equation x

(A) ( 2 x − 1) + (B) (3 x − 2) +

sin 2 x (B)  y= 2 tan x − 1 2 1 − tan 2 x

(

69. The equation of the curve satisfying the differential equation dy x − y = | x2 − y2 | dx and passing through the point (1, 0) is

c x



3

74. If y = c1e2x+ c2ex+ c3e–xsatisfies the differential equation

then

d3 y dx

3

+a

d2 y dx

2

+b

dy + cy = 0, dx

a3 + b3 + c 3 is equal to abc

1 1 1 (A)  (B)  − (C)  4 4 2

(D) –

1 2

PRACTICE EXERCISES

(A) x = y log y +

9.32  Chapter 9

Previous Year’s Questions 75. The order and degree of the differential equation 2/3 d3 y dy ⎞ ⎛ 1 + 3 = 4 are [2002] ⎜⎝ ⎟ dx ⎠ dx 3 ⎛ 2⎞ (A)  ⎜1, ⎟ (B)  (3, 1) ⎝ 3⎠ (C)  (3, 3) 76. The solution of the equation 

d y dx 2

= e −2 x is [2002]

e −2 x e −2 x (B)  + cx + d 4 4 1 1 (C)  e −2 x + cx 2 + d (D)  e −2 x + c + d 4 4 77. The differential equation of all non-vertical lines in a plane is [2002] d2 y d2x (A)  2 = 0 (B)  =0 dy 2 dx dy dx =0 (C)  = 0 (D)  dx dy 78. The degree and order of the differential equation of the family of all parabolas whose axis is x-axis, are respectively [2003] (A)  2, 1 (B)  1, 2 (C)  3, 2 (D)  2, 3

PRACTICE EXERCISES

solution

of

(1 + y ) + ( x − e

tan −1 y

2

the differential dy ) = 0 , is dx

− tan (A) ( x − 2) = k e −1

−1

1 (C)  + log y = C xy

(

(A) 

79. The

1 1 = C (B)  − + log y = C xy xy (D)  log y = Cx

82. The differential equation representing the family of curves y 2 = 2c x + c where c > 0, is a parameter,

(D)  (l, 2) 2

(A) −

equation [2003]

y

−1

tan 2 tan +k (B) 2 x e y = e (C) x e tan y = tan −1 y + k

)

is of order and degree as follows: [2005] (A)  order 1, degree 2 (B)  order 1, degree 1 (C)  order 1, degree 3 (D)  order 2, degree 2 dy = y(log y − log x + 1) , then the solution of the dx equation is [2005]

83. If x

⎛ x⎞ (A) y log ⎜ ⎟ = cx ⎝ y⎠

⎛ y⎞ (B)  x log ⎜ ⎟ = cy ⎝ x⎠

⎛ y⎞ (C) log ⎜ ⎟ = cx ⎝ x⎠

⎛ x⎞ (D) log ⎜ ⎟ = cy ⎝ y⎠

84. The differential equation whose solution is Ax2 + By2 = 1, where A and B are arbitrary constants is of  [2006] (A)  second order and second degree (B)  first order and second degree (C)  first order and first degree (D)  second order and first degree 85. The differential equation of all circles passing through the origin and having their centres on the x-axis is  [2007] dy dy (A) x 2 = y 2 + xy (B)  x 2 = y 2 + 3 xy dx dx dy dy 2 2 2 2 (C) y = x + 2 xy (D)  y = x − 2 xy dx dx

−1

(D) x e

2 tan −1 y

=e

tan −1 y

+k

80. The differential equation for the family of curves x2 + y2− 2ay = 0, where a is an arbitrary constant is  [2004] 2 2 (A) 2(x − y )y′ = xy (B) 2(x2 + y2)y′ = xy (C) (x2 − y2 )y′ = 2xy (D) (x2 + y2 )y′ = 2xy 81. The solution of the differential equation y dx + (x + x2y) dy = 0 is [2004]

86. The solution of the differential equation satisfying the condition y (1) = 1 is

dy x + y = dx x [2008]

(A)  y = ln x + x (B)  y = x ln x + x2 (x−1) (C)  y = xe (D)  y = x ln x + x 87. The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is  [2008] (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

Differential Equations  9.33

2

89. Solution to the differential equation cos x dy = y(sin π x−y) dx, 0 < x < is [2010] 2 (A)  y secx = tan x + c (B)  y tan x = sec x + c (C) tan x = (sec x + c)y (D) sec x = (tan x + c)y 90. Let l be the purchase value of an equipment and V(t) be the value of equipment after it has been used for t years. The value V(t) depreciates at a rate given by the dV (t ) differential equation = k (T − t ) , where k > 0 is dt a constant and T is the total life in years of the equipment. Then, the scrap value V(T) of the equipment is  [2011] kT 2 k (T − t ) 2 (B)  l− 2 2 l (C)  e–kT (D)  T 2 − k 91. The population p(t) at time t of a certain mouse s­ pecies dp(t ) satisfies the differential equation = 0.5 p(t ) dt – 450 with initial condition p(0) = 850, then the value of t for which p(t) = 0 is [2012] (A)  2 ln 18 (B)  ln 9 1 (C)  ln 18 (D)  ln 18 2 92. At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P with respect to additional number of workers x is given dP by = 100 − 12 x . If the firm employs 25 more dx workers, then the new level of production of items is  [2013] (A) 3000 (B) 3500 (C) 4500 (D) 2500 (A) l −

93. Let the population of rabbits surviving at a time t be governed by the differential equation dp(t ) 1 = p(t ) − 200 . If initially p(0) = 100, then p(t) dt 2 equals [2014] t/2 (A)  400 – 300 e (B)  300 – 200 e–t/2 t/2 (C)  600 – 500 e (D)  400 – 300 e–t/2 94. Let y(x) be the solution of the differential equation dy ( x log x ) + y = 2 x log x,( x ≥ 1) . Then y(e) is equal dx

to: (A) 0 (B) 2 (C) 2e (D)  e

[2015]

95. If a curve y = f (x) passes through the point (1, –1) and satisfies the differential equation, y(1+ xy )dx = x dy. ⎛ 1⎞ then f ⎜ − ⎟ is equal to: [2016] ⎝ 2⎠ 4 2 (A)  (B)  − 5 5 2 4 (C) − (D)  5 5 96. If (2 + sin x)

dy + (y + 1) cos x = 0 and y(0) = 1, then dx

⎛π⎞ y ⎜ ⎟ is equal to ⎝ 2⎠ 2 1 (A) – (B)  – 3 3 4 1 (C)  (D)  3 3

[2017]

97. Let y = y(x) be the solution of the differential equation dy π  + y cos = 4x, x ∈ (0, π). If y   = 0, then dx 2 π ⎛ ⎞ y ⎜ ⎟ is equal to [2018] ⎝6⎠ 4 2 −8 2 π π (A)  (B)  9 3 9 3 sin x

8 (C)  − π 2 9

4 (D)  − π 2 9

98. Let f be a differentiable function such that f'(x) = 7 1 f x xf   : – 3 ( ) , ( x > 0 ) and f(1) ≠ 4. Then lim x x → 01 4 x  [2019] (A) does not exist. (B) exists and equals 4. 4 (C) exists and equals 7 (D) exists and equals 0. 99. The curve amongst the family of curves represented by the differential equation, (x2 – y2) dx + 2xy dy = 0 which passes through (1, 1) is: [2019] (A) a circle with centre on the x-axis. (B) a circle with centre on the y-axis. (C) an ellipse with major axis along the y-axis. (D) a hyperbola with transverse axis along the x-axis.

PRACTICE EXERCISES

88. The differential equation which represents the family of curves y = c1e c x, where c1 and c2 are arbitrary constants is [2009] (A)  y′ = y2 (B)  y′′ = y′y (C)  yy′ = y′ (D)  yy′ = (y′)2

9.34  Chapter 9 x

100. If

∫ 0

1

f ( t ) dt = x 2 + ∫ t 2 f ( t ) dt , then f' (1/2) is: x

[2019]

6 24 (A)  (B)  25 25 4 18 (C)  (D)  5 25

ANSWER K EYS Single Option Correct Type 1. (C) 11.  (C) 21.  (A) 31. (A) 41. (B) 51. (C) 61. (B) 71. (B)

2. (D) 12. (D) 22. (B) 32.  (C) 42. (C) 52. (B) 62. (A) 72. (B)

3. (A) 13.  (B) 23.  (C) 33. (B) 43. (B) 53. (D) 63. (B) 73. (A)

4. (B) 14. (C) 24. (A) 34. (C) 44. (A) 54. (C) 64. (C) 74. (B)

5.  (B) 15. (C) 25.  (D) 35. (B) 45. (A) 55. (A) 65. (A)

6. (A) 16. (D) 26. (A) 36.  (A) 46. (A) 56. (C) 66. (B)

7.  (C) 17.  (D) 27.  (C) 37. (C) 47. (A) 57. (A) 67. (A)

8. (D) 18. (B) 28. (C) 38. (D) 48. (C) 58. (B) 68. (A)

9.  (A) 19.  (C) 29.  (A) 39.  (C) 49. (B) 59. (B) 69. (C)

10. (D) 20. (C) 30. (A) 40. (D) 50. (A) 60. (A) 70. (A)

78. (B) 88. (D) 98. (B)

79. (B) 80. (C) 89. (D) 90. (A) 99. (A) 100. (B)

81. (B) 91. (A)

82. (C) 92. (B)

83. (C) 93. (A)

84. (A) 94. (B)

Previous Years’ Questions

PRACTICE EXERCISES

75. (C) 85. (C) 95. (A)

76. (B) 86. (D) 96. (D)

77. (A) 87. (C) 97. (C)

Differential Equations  9.35

HINTS AND EXPLANATIONS Single Option Correct Type 1. Given equation can be re-written as



dy ϕ ′( x ) y2 =– −y ϕ( x) ϕ( x) dx

⇒ –

1 dy 1 ϕ ′( x ) 1 · + · = (1) ϕ( x) y 2 dx y ϕ ( x )

Putting

1 1 dy du = u, we get – 2 = y y dx dx

Now (1) becomes

du ϕ ′( x ) 1 = +u ϕ( x) ϕ( x) dx

ϕ ′( x )

I.F. = e

∫ ϕ ( x ) dx

d ϕ( x) [uf (x)] = 1 ⇒ y = x+c dx

The correct option is (C) 2.

y2 = 1 ⇒ d (log y1) = 1 y1

dy + 2x tan y = x3 dx Let tan y = v dy dv ⇒ sec2 y = dx dx 3. sec2 y

3

x2

2

2

dx + c

1 t 1 te dt + c = et (t – 1) + c 2∫ 2 1 x = e (x2 – 1) + c 2

⇒

v · ex =



⇒

v · ex

2

2

1 2 (x – 1) + ce − x 2 The correct option is (A)

\ tan y =

= e–log y = y–1



\ Solution is t(y–1) = ∫ ( y −1 ) dy + c



⇒

1 1 1 · = log y + c ⇒ log y – =C x y xy

5. For the family of curves represented by the first differential equation, the slope of the tangent at any point (x, y) is given by x2 + x + 1 ⎛ dy ⎞ = ⎝⎜ dx ⎠⎟ c y2 + y + 1

For the family of curves represented by the second differential equation, the slope of the tangent at any point is given by 1

y2 + y + 1 ⎛ dy ⎞ = – ⎝⎜ dx ⎠⎟ c x2 + x + 1

1

x = t ⇒ dt = 2x dx



Integrating factor = e

2

Hence, the two curves are orthogonal. The correct option is (B)

2

Let

1

∫ − y dy

⎛ dy ⎞ ⎛ dy ⎞ Since, ⎜ ⎟ × ⎜ ⎟ = –1 ⎝ dx ⎠ c ⎝ dx ⎠ c

Hence, the solution of the differential equation is

∫x ·e

dx dt x–1 = t ⇒ –x–2 dy = , dy ⎛ 1⎞ dt dt ⎛ 1 ⎞ get, – + t ⎜ ⎟ = –1 ⇒ − t =1 dy dy ⎜⎝ y ⎟⎠ ⎝ y⎠ Put

2

dv + 2xv = x3 dx 2 xdx Now, I.F. = e ∫ = ex 2

dx −1 ⎛ 1 ⎞ ⇒ x–2 dy + x ⎜⎝ y ⎟⎠ = –1



The given equation becomes

v · ex =



The correct option is (B)

⇒ log y1 = x + c ⇒ y1 = kex ⇒ y = kex+ B Since it passes through (0, 0), k + B = 0 Thus, family is y = k (ex– 1). The correct option is (D)



⇒

It is linear in t.

= f (x).

Multiplying both sides by I.F., we have

dx x dx x =– – x2⇒ + = –x2, dy y dy y



2

6. The given equation can be written in the linear form as follows: dy + yf ′(x) = f (x) f ′(x) dx ϕ ′ ( x ) dx The integrating factor of this equation is e ∫ = ef (x). d (yef (x)) = f (x) f ′(x) ef (x) dx Integrating, we have yef (x) = ∫ tet dt + c , (where t = f (x)) Hence

= tet– et+ c Hence, y = (f (x) – 1) + c e–f (x) The correct option is (A)

HINTS AND EXPLANATIONS



4. ydx + (x + x2y)dy = 0

9.36  Chapter 9 7. x =

⇒



⇒



⇒

e

yx

dy dx

dy dx 1 ∫ ydy = ∫ x ln xdx

⇒ ln x = xy

y2 (ln x ) 2 = +k 2 2 y=±

(ln x ) 2 + c

The correct option is (C) 8. Family of semi-cubical parabolas is given by ay2 = x3(1) a is variable parameter \ Differentiating the above semi-cubical equation, we have dy 2ay = 3x2(2) dx Eliminating a from (1) and (2) 2 x3 dy dy 3 · y = 3x2⇒ 2x = 3y(3) dx dx y dy dx Now for orthogonal trajectories changing → –  in (3) dx dy dx = 3y ⇒ 3y dy + 2x dx = 0 dy Integrating we have

3y2 x2 c2 + = ⇒ 3y2 + 2x2 = c2 2 1 2 The correct option is (D) 9. The given equation can be written as

HINTS AND EXPLANATIONS

\ –2x

⇒

2y sin x

dy + y2 cos x = sin 2x dx

d 2 (y sin x) = sin 2x dx 1 On integrating, we get y2 sin x = – cos 2x + c 2 π 1 Putting x = and y = 1, we get c = . 2 2 1 2 Hence, y sin x = (1 – cos 2x) 2 2 2 ⇒ y sin x = sin x or y2 = sin x. The correct option is (A)

⇒

dy 10. Here y – x + log x = 0 dx i.e., x i.e.,

dy – y = log x dx

dy y log x − = dx x x



\ e ∫



\ solution is yx–1 =

i.e.,

= e∫

Pdx

y = x

−1/ x dx

−1

= e(– log x) = e log x = x–1

∫

log x 1 ⋅ dx + c x x

1

∫ x 2 log x dx + c = ∫ e

Put log x = t \

−t

⋅ t dt + c

1 dx = dt x

Also, x = et y e −t e −t \ = t −1· +c −1 −1 x = e–t[–t + 1] + c = e–log x(– log x + 1) + c 1 = (1 – log x) + c x \ y = (1 – log x) + cx The correct option is (D) 11. The given diff. eqn. can be written as (sin y dx + x cos y dy) x2 sin2 y = 2y sin x dy + y2 cos x dx ⇒ d (x sin y) (x sin y)2 = d (y2 sin x) Integrating, we get

( x sin y )3 = y2 sin x + c 3 The correct option is (C) 12. Equation of the normal at (x, y) is Y – y = –

\ x-intercept = y

dy + x dx

dx (X – x) dy (Putting Y = 0)

dy Given: y2 = 2x ⎛⎜ y + x⎞⎟ ⎝ dx ⎠

dy y2 − 2x2 = . Put y = vx. Then dx 2 xy

⇒



v+x 2v

∫ v 2 + 2 dv + ∫

dy v2 − 2 dv −( 2 + v 2 ) = ⇒x = dx 2v dx 2v dx = log k x



⇒



⇒ log (v2 + 2) + log x = log k



⇒

x (v2 + 2) = k ⇒ y2 + 2x2 – kx = 0

9 2 Then, the equation becomes 4x2 + 2y2 – 9x = 0. The correct option is (D) If this passes through (2, 1), i =

dy 4 dx = x2 + 2y2 + y 13. Given: 2 dy x y−x dx x+ y

Differential Equations  9.37

⇒

d ( x2 + y2 ) ( x 2 + y 2 )2

=2

⎛ x⎞ ⎜⎝ y ⎟⎠

2

Integrating, we get 1 1 −1 y = +c⇒c= − 2 x/ y x x + y2 x2 + y2 The correct option is (B) 14. The given equation can be written as (2x – 2y + 5)dy = (x – y + 3)dx –



⎛ dx ⎞ ⎜⎝ dy ⎟⎠

dy x− y+3 = 2( x − y ) + 5 dx

⇒



dV V +3 dV V +2 = or = 2V + 5 2V + 5 dx dx

1 ⎞ 2V + 5 ⎛ dV or dx = ⎜ 2 + dV V +2 V + 2 ⎟⎠ ⎝ On integrating, we get x = 2V + log(V + 2) + c ⇒ x = 2(x – y) + log(x – y + 2) + c Therefore, 2y – x = log(x – y + 2) + c, is the required solution. The correct option is (C) 15. Put x = r cos q, y = r sin q \ dx = – r sin q dq + dr cos q dy = r cos q dq + dr sin q \ We have

⇒

dx =

r cos θ ( − r sin θ dθ + dr cos θ ) + r sin θ ( r cos θ dθ + dr sin θ ) = ( r cos θ ) ( r cos θ dθ + dr sin θ ) − r sin θ ( − r sin θ dθ + dr ⋅ cos θ )

1 − r2 r

2

− r 2 sin θ cos θ dθ + r cos 2 θ dr

⇒

+ r 2 sin θ cos θ dθ + r sin 2 θ dr 2

2

2

2

r cos θ dθ + r cos θ sin θ dr

=

1 − r2 r

+ r sin θ dθ − r sin θ cos θ dr

⇒

⇒

r dr r 2 dθ dr 1 − r2

=

1 − r2 r

= dq

⇒ sin–1 r = q + c y ⇒ sin–1 x 2 + y 2 = tan–1 x + c. The correct option is (C)

a ⎛ dy ⎞ ⇒ ⎜ ⎟ ⎝ dx ⎠ x

2

=

x dy ⇒ =± a dx 2 3 a

x a

x3/2(1)

4 3 x ⇒ 9a(y + c)2 = 4x3 (2) 9a From (1) and (2), all of the first three given options represent required equations. The correct option is (D)

dy dV = dx dx Therefore, the given equation becomes 1–

=

Integrating we get, y + c = ±

x–y=V⇒1–

Put

2

⇒ (y + c)2 =

⎛ x + y − 1 ⎞ dy ⎛ x + y + 1⎞ 17. ⎜ = ⎜ ⎝ x + y − 2 ⎟⎠ dx ⎝ x + y + 2 ⎟⎠ Put x + y = t 1+ ⇒

dy dt dt ⎛ t + 1⎞ ⎛ t − 2⎞ = ⇒ –1= ⎜ ⎝ t + 2 ⎟⎠ ⎜⎝ t − 1 ⎟⎠ dx dx dx ⎛ t 2 − t − 2⎞ dt = ⎜ 2 ⎟ +1 dx ⎝ t + t − 2⎠

⎛ ( x + y )2 − 2 ⎞ On solving, we get 2(y – x) + log ⎜ ⎟ =0 2 ⎝ ⎠ The correct option is (D) 18. The given differential equation is [y + x xy (x + y)] dx + (y xy (x + y) – x) dy = 0

⇒ y dx – x dy + xy (x + y) (x dx + y dy) = 0



⇒ x dx + y dy =



⇒



⇒ x2 + y2 = 4 tan–1

x dy − y dx ( x + y ) xy

⎛ 1 y⎞ d (x2 + y2) = 2 d ⎜ tan −1 2 x ⎟⎠ ⎝ y x + c.

The correct option is (B) 19. Given:

dx = cos2 p x. Differentiate with respect to t, dt

d2x dt 2

= –2p sin 2px = –ve

HINTS AND EXPLANATIONS



16. The family of curves which are orthogonal (i.e., intersect at right angles) to a given system of curves is obtained by dx dy substituting – for in the differential equation of the dy dx given system. 2 a ⎛ dy ⎞ The given differential equation is ⎜ ⎟ = ⎝ dx ⎠ x dy dx Replacing by – , we get dx dy

⎛ x⎞ d⎜ ⎟ ⎝ y⎠

9.38  Chapter 9



d2x

=0

2

dt ⇒ 2p sin 2px = 0 ⇒ sin 2px = sin p



1 2 The correct option is (C) ⇒ 2px = p ⇒ x =

20. Given: Cartesian sub-tangent ∝ i.e.,

1 square of abscissa

y k dy x2 = 2 or = dx dy/dx y k x

Integrating, log y =

x3 + log c 3k

3



⇒



⇒

HINTS AND EXPLANATIONS

dy = y (x log y – y) dx

⎛ x ⎞ dy y ⎜⎝ log x − y ⎟⎠ dx = log y – x y dy x dy + log x ⋅ = log y + x dx y dx

d d (y log x) = (x log y) dx dx ⇒ y log y = x log y + log c ⇒ log xy= log yx+ log c ⇒ xy= c yx The correct option is (A)

⇒

ax + 3 dy = 2y + f dx ⇒ (2y + f  )dy = (ax + 3)dx

2 y2 ax 2 + fy = + 3x + c 2 2 For the curve to be circle, a = –2 where f 2 + 9 + 4c > 0. The correct option is (B) 23. Given diff. equation can be written as

⇒



⇒

x dy − y dx x2 + y2 d ( y/ x ) 2

= – dx ⇒

+ dx = 0

⎛ y⎞ 1+ ⎜ ⎟ ⎝ x⎠ Integrating, we get

y +x=c x The correct option is (C) tan–1

dy − cos x dx = (1 + y ) 2 + sin x

x dy − y dx = – dx ⎡ ⎛ y⎞ 2⎤ x 2 ⎢1 + ⎜ ⎟ ⎥ ⎢⎣ ⎝ x ⎠ ⎥⎦

⇒

dy 2ay = dx x ( y − a)

y−a 2a dy = dx y x

On integrating both sides, we get a log | y | – y = –2a log | x | + loc c ⇒ ya × x2a = cey Since, the curve passes through (1, 1), therefore 1 e So, the equation of the curve is ya . x2a = ey – 1 The correct option is (D) 1 = ce ⇒ c =

26. We have (x cos x – sin x + yx2) dx + x3dy = 0 x cos x − sin x



⇒



⎛ sin x ⎞ ⇒ d ⎜ + d (xy) = d (c) ⎝ x ⎟⎠

x2

dx + ydx + xdy = 0

sin x + xy = c x The correct option is (A) 27. Dividing the given equation by y2, we get

22.

⇒

Integrating both sides ⇒ ln (1 + y) = – ln (2 + sin x) + c Put x = 0 and y = 1 ⇒ ln (2) = – ln 2 + c ⇒ c = ln 4. π 4 1 Put x = , ln (1 + y) = ln 3 + ln 4 = ln ⇒ y = 2 3 3 The correct option is (A)



dy y ( x log y − y ) = x ( y log x − x ) dx

⇒ x (y log x – x)

dy ⎛ 2 + sin x ⎞ = – cos x, y(0) = 1 dx ⎜⎝ 1 + y ⎟⎠

25. We have,

or y = ce x /3k The correct option is (C) 21.

24.

⇒

( )

x y dx − x dy d ⎛ x⎞ d x = – 3x2 e dx ⇒ =– e ⎜ ⎟ 2 dx ⎝ y ⎠ dx y 3

On integrating, we get,

3

x x = – ex + c ⇒ + ex = c y y 3

3

The correct option is (A) 28. The given diff. equation can be written as

⎛y ⎞ ⇒ (y dx + x dy) + ⎜ dx + log x⎟ dy ⎝x ⎠

 + sin y dx + x cos y dy = 0 ⇒ d (xy) + d (y log x) + d (x sin y) = 0 Integrating, we get xy + y log x + x sin y = c. The correct option is (C)

Differential Equations  9.39 29. The equation of the normal at any point (x, y) is given by, dx Y–y=– (X – x). dy This passes through (h, k) dx dx ⇒k–y=– (h – x) ⇒ (y – k) = (h – x) . dy dy

Using (1, 2) in (1), we get

The correct option is (A)

is the required curve. The correct option is (B)

30. 3xy2dy + y3dx + sin (xy) (xdy + ydx) = 0 or d (xy3) + sin (xy) d (xy) = 0 On integrating, we get xy3 – cos xy = c The correct option is (A) 31. We have Cartesian subtangent + abscissa = constant ⇒

∫

T

dV (t ) =

∫

− k (T − t )dt

t=0

I

T

y dy dy dx +x=a⇒y +x=a⇒ = a−x dy/dx dx y



⎡ (T − t ) 2 ⎤ ⇒ V(T) – I = k ⎢ ⎥ ⎣ 2 ⎦0



⎛T2⎞ ⇒ V(T) – I = – k ⎜ ⎟ ⎝ 2⎠

kT 2 2 The correct option is (C)

⇒

V(T) = I −

35. We have dy =y+3 dx



1 dt 1 g ′( x ) 1 + = (1) g( x) t 2 dx t g ( x ) 1 Let = z t 1 dt dz \  2 = dx t dx Now by (1), we have

1 dy = dx y+3

⇒





⇒ ln|(y + 3)| = x + k, where k is a constant of integration ⇒ (y + 3) = c ex Initially when x = 0, y = 2 ⇒c=5 Finally the required solution is y + 3 = 5ex ⇒ y(ln 2) = 5eln 2 – 3 = 10 – 3 = 7 The correct option is (B)

dz g ′( x ) 1 +z = g( x) dx g( x)

I.F. = e

∫

g ′( x ) dx g( x)

= g(x)

Therefore, the solution is z g(x) = x + c ⇒ g(x) g( x) x+c The correct option is (C)

⇒

1 =x+c t

t=

33. xdy + (y + x3y2) dx = 0 ⇒ xdy + ydx = –x3y2dx x dy + ydx ⇒ = –x dx x2 y2 ⇒

d ( xy )

= –x dx ( xy ) 2 Integrating, we get

1 x2 2 = − ⇒ y = 3 or x3y = 2 2 xy x

V (t )

dt g ′( x ) t2 −t 32. =– g( x) dx g( x)



\ −

34.

Integrating, we get log y + log (x – a) = log c \ y(x – a) = c As the curve passes through the point (2a, a), we have c = a2 Hence the required curve is y (x – a) = a2. The correct option is (A)





1 1 = − + c ⇒ c = 0. 2 2

⇒ –

2

1 x =– + c(1) 2 xy

⇒

36. 2

dp(t ) = –dt 900 − p(t )

–2 ln [900 p(t)] = –t + c when t = 0, p(0) = 850 –2 ln (50) = c

50 ⎛ ⎞ \ 2 ln ⎜ = –t ⎝ 900 − p(t ) ⎟⎠

900 – p(t) = 50 et/2 p(t) = 900 – 50 et/2 let p(t1) = 0 0 = 900 – 500 et / 2 \ t1 = 2 ln 18 The correct option is (A) 1

HINTS AND EXPLANATIONS



−

9.40  Chapter 9

37.

∫

ydy 1− y

2

=

∫ dx

⇒ −

1 −2 y dy =x+c 2 ∫ 1 − y2

⇒ − 1 − y 2 = x + c ⇒ 1 – y2 = x2 + c2 + 2cx

⇒ x2 + y2 + 2cx + c2 – 1 = 0 Which is a circle with centre (–c, 0) and radius 1. The correct option is (C) 38. y = vx dv dy = v + x dx dx dv v+x =1+v dx



HINTS AND EXPLANATIONS









⇒



(1)

⇒ (x – h) = –(y – 2) 2

⎛ dy ⎞ ( y − 2) 2 ⎜ ⎟ + ( y − 2) 2 = 25 ⎝ dx ⎠ (y – 2)2y′2 = 25 – (y – 2)2 The correct option is (C)

y =c x ⎛ c − x2 − y2 ⎞ \ y = x tan ⎜ ⎟ is the required solution. 2 ⎝ ⎠ ⇒ x2 + y2 + 2 tan– 1

⎛ xf ( y /x ) ⎞ dx 43. We have, x dy = ⎜ y + f ′ ( y /x ) ⎟⎠ ⎝

⇒

f ( y /x ) dy y = + which is homogeneous. dx x f ′ ( y /x )

Put y = Vx so that

dy dV =V+x , dx dx

we obtain, dV f (V ) f (V ) dx V+x =V+ ⇒ dV = dx f’ (V ) f’ (V ) x Integrating, we get log f (V) = log x + log c

y′ = c2 c e c x



2

2

y′ = c2 y(2) y′′ = c2 y′ From Equation (2)

⎛ y⎞ ⇒ log f (V) = log cx ⇒ f ⎜ ⎟ = cx ⎝ x⎠ The correct option is (B) dy = ex – y (ex – ey) dx

44. We have,

dy ⇒ ey dx + ex . ey = e2x.

( y ′)2 So, y′′ = ⇒ yy ″ = ( y ′ ) 2 y



The correct option is (D)

Putting ey = V so that e y

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 41. We have,

=0

1 2 c y (x + y2) + tan– 1 = 2 2 x

40. y = c1e c x (1)

y′ c2 = y

x2 + y2

The correct option is (C)

dy ⇒ 2(x – h) + 2(y – 2) =0 dx

1

x dy − y dx

1 ⎛ y⎞ d (x2 + y2) + d tan– 1 ⎜ ⎟ = 0 ⎝ x⎠ 2

Integrating,

dy dx Substituting in Equation (1), we have



42. We have, x dx + y dy +

⇒ dv =

39. (x – h)2 + (y – 2)2 = 25

⇒

The correct option is (B)

dx x \ v = log x + c y ⇒ = log x + c x Since, y(1) = 1, we have y = x log x + x The correct option is (D)

dV dV = (1 – V) g′ (x) ⇒ = g′ (x) dx 1−V dx dV ⇒ ∫ = ∫ g’( x ) dx 1−V ⇒ – log (1 – V) = g (x) – C ⇒ g (x) + log (1 – V) = C \ g (x) + log [1 + y – g (x)] = C



dy dV = , we get dx dx

dV + ex × V = e2x, which is linear in V. dx e dx I.F. = e ∫ = ee . x



x

So, the solution is

Differential Equations  9.41 x



∫e

ex

x

2x

⋅ e dx + c

⇒ e y ⋅ e e =

∫e

z

Integrating, we get y =

⋅ z dz + c

The correct option is (A)

[Putting ex = z ⇒ ex dx = dz]

⇒ e y ⋅ e e = (z – 1) ez + c = (e x − 1) e e + c x

x

⇒ ey = ex – 1 + ce–ex The correct option is (A) 2

dy ⎛ dy ⎞ 45. We have, x ⎜ ⎟ + (y – x) –y=0 ⎝ dx ⎠ dx

⎛ dy ⎞ ⎛ dy ⎞ ⇒ ⎜ −1 x + y⎟ = 0 ⎝ dx ⎟⎠ ⎜⎝ dx ⎠



⇒



⎛ y ⎞ ⎡ 2 dy ⎤ ⇒ sec2 ⎜⎝ 2 x ⎟⎠ ⋅ ⎢⎣ x dx − xy ⎥⎦ = 2



1 ⇒ sec 2 2



⇒

d dx

⎛ y⎞ ⎜⎝ ⎟⎠ ⋅ 2x

x

dy dy dx = – y i.e., = 0 is + dx y x

y⎞ 1 ⎛ ⎜⎝ tan ⎟⎠ = 3 ⋅ 2x x

dy dt = . dx dx Therefore, the given equation becomes 49. Let x + y = t⇒ 1 +

log (xy) = log c i.e., xy = c. Hence, general solution is (x – y + c) (xy – c) = 0. The correct option is (A) 46. The equation of the general circle is given by x2 + y2 + 2gy + 2fy + c = 0 (1) Differentiating with respect to x, we get 2x + 2yy′ + 2g + 2fy′ = 0 (2) Differentiating again, we get 1 + y′2 + yy″ + fy″ =0(3) Differentiating again, we have 2y′y′′ + yy′′′ + y′ y′′ + fy′′′ = 0 (4) Eliminating f from (3) and (4), we get y′′′ (1 + yy′′ + y′2) – y′′ (yy′′′ +3y′ y′′) = 0 ⇒ y′′′ (1 + y′2) – 3y′ y′′2 = 0, which is the required differential equation. The correct option is (A)

⎛ t − 1 ⎞ ⎛ dt t +1 ⎞ ⎜ − 1⎟ = ⎜ ⎟ ⎝ ⎠ t+2 ⎝ t − 2 ⎠ dx

⇒

⎛ t + 1⎞ ⎛ t − 2⎞ dt −1 = ⎜ dx ⎝ t + 2 ⎟⎠ ⎜⎝ t − 1 ⎟⎠



⇒

dt t2 − t − 2 = 2 +1 dx t +t −2



⇒

dt 2t 2 − 4 = 2 dx t +t −2



⇒



⎛ t ⎞ ⇒ ⎜1 + 2 ⎟ dt = 2dx t − 2⎠ ⎝

t2 + t − 2 t2 − 2

dt = 2dx

On integrating, we get 1 log t 2 − 2 = 2x + c 2

47. Let m1 =

dy for required family of curves at (x, y). dx



Let m2 =

dy for the hyperbola xy = 2. dx



⇒ ( x + y ) +



⇒ 2 ( y − x ) + log

Then, m2 =

dy −2 = 2 ⋅ dx x

Since the required family of curves is orthogonal to the hyperbola, \ m1 × m2 = – 1

⇒

dy −y 1 dx = 3 2 x x

Integrating, we get tan

dy = 1 is y = x + c dx

and solution of x

y dy y – xy = 1 + cos = 2cos2 ⋅ x dx 2x

48. We have, x2

1 y =c − 2 , 2x 2x which is the required solution. The correct option is (C)

dy dy = 1 or x =–y dx dx

The solution of

x3 + C, which is the required family. 6

2

2

dy ⎛ −2 ⎞ dy x x ⋅ ⇒ dy = ×⎜ ⎟ =–1⇒ = dx 2 2 dx ⎝ x 2 ⎠ dx

t+

1 log ( x + y ) 2 − 2 = 2c(1) 2 Given, y = 1, when x = 1, therefore log 2 = 2c. Substituting the value of c in (1), we get 2(y – x) + log | (x + y)2 –2 | = log 2 ( x + y )2 − 2 = 0. \ k = 2. 2

The correct option is (B)

HINTS AND EXPLANATIONS

V . e e =

9.42  Chapter 9 50. We have, x dy – y dx = xy3 (1 + log x) dx

⎛ ydx − xdy ⎞ ⇒ – ⎜ ⎟ = xy (1 + log x) dx y2 ⎝ ⎠



⎛ x⎞ ⇒ – d ⎜ ⎟ = xy (1 + log x) dx ⎝ y⎠



⇒ −

x ⎛ x⎞ d = x2 (1 + log x) dx y ⎜⎝ y ⎟⎠

Integrating, we get



⇒



⇒

⎛ x⎞ −⎜ ⎟ ⎝ y⎠

−

= (1 + log x)

x2 2y −

2

x2 y

2

x3 x3 1 −∫ ⋅ dx 3 3 x

⇒ –

HINTS AND EXPLANATIONS



\ x-intercept = y



⇒

2 x3 ⎛ 2 ⎞ ⎜ + log x⎟⎠ + C 3 ⎝3



du ϕ ′( x ) 1 = +u ϕ( x) ϕ( x) dx

ϕ ′( x )

= f (x). Multiplying both sides by I.F., we have

d ϕ( x) [u f (x)] = 1 ⇒ y = x+c dx The correct option is (C) 52. ydx + (x + x2y)dy = 0

dx x dx x ⇒ =– – x2⇒ + = –x2, dy y dy y dx −1 ⎛ 1 ⎞ x–2 dy + x ⎝⎜ y ⎠⎟ = –1

dx dt Put x–1 = t, –x–2 dy = dy We get, –

dy + x dx

(Putting Y = 0)



⎛ 1⎞ dt dt ⎛ 1 ⎞ + t ⎜ ⎟ = –1 ⇒ − t =1 dy dy ⎜⎝ y ⎟⎠ ⎝ y⎠

It is linear in t. Integrating factor = e

1 ∫ − y dy

= e–log y = y–1

dy y2 − 2x2 = . Putting y = vx. Then, dx 2 xy dy v2 − 2 dv −( 2 + v 2 ) = ⇒x = dx 2v dx 2v 2v

dx = log K x ⇒ log (v2 + 2) + log x = log K ⇒ x (v2 + 2) = K ⇒ y2 + 2x2 – Kx = 0 ⇒

∫ v 2 + 2 dv + ∫

9 2 2 Then, the equation becomes 4x + 2y2 – 9x = 0. The correct option is (D) 54. Put x = r cos q, y = r sin q \ dx = – r sin q dq + dr cos q dy = r cos q dq + dr sin q \ We have If this passes through (2, 1), K =

r cos θ ( − r sin θ dθ + dr cos θ ) + r sin θ ( r cos θ dθ + dr sin θ ) = ( r cos θ ) ( r cos θ dθ + dr sin θ ) − r sin θ ( − r sin θ dθ + dr cos θ )

1 − r2 r2

− r 2 sin θ cos θ dθ + r cos 2 θ dr

⇒

+ r 2 sin θ cos θ dθ + r sin 2 θ dr r 2 cos 2 θ dθ + r cos θ sin θ dr

=

1 − r2 r

+ r 2 sin 2 θ dθ − r sin θ cos θ dr

It is Bernoulli’s form. Divide by x2

dx (X – x) dy

53. Equation of the normal at (x, y) is Y – y = –

=

1 1 dy du = u, we get – 2 = . y dx y dx

∫ ϕ( x)

1 1 1 · = log y + c ⇒ log y – =C x y xy

The correct option is (B)

v+x

Now, (1) becomes I.F. = e

⇒

x3 x3 C (1 + log x ) − + 3 9 2

1 dy 1 ϕ ′( x ) 1 · + · = (1) ϕ( x) y 2 dx y ϕ ( x )

Putting



=

The correct option is (A) 51. Given equation can be rewritten as dy ϕ ′( x ) y2 =– −y ϕ( x) ϕ( x) dx

\ Solution is t(y–1) = ∫ ( y −1 ) dy + c

⎛ dy ⎞ Given, y2 = 2x ⎜ y + x⎟ ⎠ ⎝ dx

2

2



r dr

dr

⇒



y 2 2 ⇒ sin–1 r = q + c ⇒ sin–1 x + y = tan–1 x + c.

2

r dθ

=

1 − r2 ⇒ r



1 − r2

= dq

The correct option is (C) 55. We have (x cos x – sin x + yx2) dx + x3dy = 0

⇒

x cos x − sin x x2

dx + ydx + xdy = 0

Differential Equations  9.43 59. Let x = r cos q, y = r sin q so that x2 + y2 = r2(1)

⎛ sin x ⎞ ⇒ d ⎜ + d (xy) = d (c) ⎝ x ⎟⎠

sin x + xy = c x The correct option is (A)

and, tan q =

⇒

From (1), we have d(x2 + y2) = d(r2)

56. The given differential equation can be written as

⎛y ⎞ ⇒ (y dx + x dy) + ⎜ dx + log x⎟ dy ⎝x ⎠

 + sin y dx + x cos y dy = 0 ⇒ d (xy) + d (y log x) + d (x sin y) = 0 Integrating, we get xy + y log x + x sin y = c The correct option is (C) 57. Y – y =

⇒ 10

dy 1 dx (X – x) ⇒ 5 = y 2 dx 2 dy dy y2

P (x, y) θ

x

The correct option is (A) 58. xdy + (y + x3y2) dx = 0 ⇒ xdy + ydx = –x3y2dx x dy + ydx x2 y2

= –x dx ⇒

d ( xy ) ( xy ) 2

= –x dx

Integrating, we get –

(3) ⎛ y⎞ From (2), we have d ⎜ ⎟ = d(tan q) ⎝ x⎠ x dy − y dx i.e., = sec2q dq x2 i.e., x dy – y dx = x2 sec2q dq = r2 cos2q sec2q dq(4) Using (3) and (4) in the given equation, we get; rdr 2 = r dθ

1 x2 =– + c(1) 2 xy

or,

a2 − r 2 r2

i.e.,

dr 2

a − r2

= dθ

1 1 − = − +c ⇒c=0 2 2 1 x2 2 \ − = − ⇒ y = 3 or x3y = 2 2 xy x

is the required curve. The correct option is (B)

x 2 + y 2 = a sin [c + tan–1(y/x)]

The correct option is (B) 60. The given differential equation can be written as (log y) 2x dx +

x2 dy + 3y2dy = 0 y



⇒ (log y) d(x2)+ x2d(log y) + d(y3) = 0



⇒ d(x2 log y) + d(y3) = 0

⇒ x2 log y + y3 = c (Integrating both sides) The correct option is (A) 61. The given differential equation can be written as; y dx + x dy 2 + sin x dx + sin y dy = 0 cos ( xy ) ⇒ sec2(xy) d(xy) + sin x dx + sin y dy = 0 ⇒ tan(xy) – cos x – cos y = c (Integrating both sides) The correct option is (B) 62. The equation can be written as dx 2 y ln y + y − x x = = ( 2 ln y + 1) − dy y y i.e.,

Using (1, 2) in (1), we get



⇒ x dx + y dy = rdr

or, r = a sin (q + c)

= – dx

y

⇒



⎛ r⎞ i.e., sin–1 ⎜⎝ a ⎟⎠ = q + c

10 ⇒ − + x = c y Since it passes through (2, 5), therefore, c = 0. Thus, the equation of curve is xy = 10



y (2) x

dx 1 + · x = (2 ln y + 1) dy y 1



∫ \ I.F. = e ∫ P dy = e y



\ The solution is

dy

= e ln y = y

xy = ∫ ( 2 ln y + 1) ⋅ y dy = y 2ln y + c c y The correct option is (A)

⇒ x = y lny +

HINTS AND EXPLANATIONS



9.44  Chapter 9 Dividing by cos2y, we get dy tan y · sec y · – sec y = –x dx dy dv Let sec y = v ⇒ sec y tan y = dx dx So, we get

63. The given equation can be written as; dy − sec 2 x ⋅ tan 2 x ⋅ y = cos 2 x dx − tan 2 x sec I.F. = e ∫ dt ∫ e t

2

x dx

= e

2 tan x sec 2 x dx 2 x −1

∫ tan

where t = tan2x – 1



=



= e ln | t | = | t | = |tan 2 x − 1|

It is given that | x | < p/4 and for this region tan2x < 1 \ I.F. = 1 – tan2x \ The solution is; 2

y(1 – tan x) =

∫ cos

2

(

x 1 − tan x dx



=

∫ (cos



=

sin 2 x +c 2

2



\

)

∫ cos 2x dx

x − sin 2 x dx =

Now, when x = p/6, y = 3 3 \ 8

)

2

1 3 ⎛ 1⎞ +c⇒c=0 ⎜⎝1 − ⎟⎠ = · 3 2 2 sin 2 x y= 2 1 − tan 2 x

)

HINTS AND EXPLANATIONS



⎛ dy ⎞ + 1⎟ + x( x + y ) = x3(x + y)3 ⎜⎝ dx ⎠



d ( x + y) ⇒ ( x + y )3 + x ( x + y ) −2 = x 3 dx

Let (x + y)–2 = z so that −2 ( x + y ) −3

d ( x + y ) dz = dx dx

The given equation reduces to

I.F. = e ∫ −2 x dx = e–x2 \ The solution is,

∫ −2 x · e 3

− x2

dx = ( x 2 + 1) e − x + c 2

2 1 = cex + x2 + 1 ( x + y)2 The correct option is (C) dy 65. We have, sin y = cos y (1 – x cos y) dx dy ⇒ sin y – cos y = –x cos2y dx

or,

−x

dx = xe − x + e − x + c

= e–x(x + 1) + c or, v = (1 + x) + cex or, sec y = (1 + x) + cex The correct option is (A)



⇒ y2 · exy(xdy + ydx) = ex/y(ydx – xdy)



⇒



⇒

⎛ ⎞ exy(xdy + ydx) = ex/y y dx − x dy ⎜ ⎟ y2 ⎝ ⎠ exy· d(xy) = ex/y(d(x/y))

x







x

xy ( x ) + 1 ∫ y(t ) dt = ( x + 1) xy ( x ) + 1 ∫ t y(t ) dt 0

x

⇒

∫ y(t ) dt 0

0

x

= x2y (x) + ∫ t y(t ) dt 0

Differentiating again with respect to x we get, y(x) = x2y′(x) = 2xy(x) + xy(x) x 2 dy( x ) dx dy ( x ) (1 − 3 x ) dx ⇒ = y ( x) x2 On Integrating, we get c y = 3 e −1/x x The correct option is (A)

−1 dz dz + xz = x3 i.e., − 2 xz = –2x3 2 dx dx

2

∫ −x e

67. Differentiating the given equation with respect to x, we get

d ( x + y) ⇒ + x ( x + y ) = x 3 ( x + y )3 dx

z · e–x =

ve–x =



⇒ d(exy) = d(ex/y) ⇒ exy= ex/y+ c or, xy = ln (ex/y+ c) The correct option is (B)

The correct option is (B) 64. The given equation can be written as

The solution is given by

66. The given equation is xy2exy dy + xex/ydy = yex/y dx – y3exydx

3 3 8

(

dv − v = –x, which is a linear differential equation with dx P = –1, Q = –x Pdx −1dx \ I.F. = e ∫ = e∫ = e–x

⇒ (1 – 3x) y(x) =

68. We have, (1 + tan y) (dx – dy) + 2xdy = 0 ⇒ (1 + tan y) dx = (1 + tan y – 2x) dy

⇒

dx 2 + x = 1, which is linear in x dy 1 + tan y

Differential Equations  9.45 2

∫ 1 + tan y dy ⎛

2 cos y

∫ sin y + cos y dy

=

cos y − sin y ⎞

∫ ⎜⎝1 + cos y + sin y ⎟⎠ dy



=



= y + log (cos y + sin y)

⇒

2



\ I.F. = e

∫ 1 + tan y dy

= ey · elog (cos y + sin y) y

= (cos y + sin y) e The solution is given by, x · ey(cos y + sin y) = eysin y + c or, x(sin y + cos y) = sin y + ce–y The correct option is (A)



dy = (x2 – y2)2 dx d ( x2 − y2 )

⇒

2 ( x 2 − y 2 )2

⇒

(x

−1 2

− y2

)

= dx

= 2x + c

1

2

)

−1 cos 2 x + c 2

π −1 , y = 1, we get c = 2 2

Hence, the solution is y2 sin x =

1 (1 − cos 2 x ) = sin 2 x 2

⇒ y2 = sin x The correct option is (A) x

x

⇒

ye x dx − e x dy y2

⇒

y dx − x dy

74. We have, 2 2

3

71. We have, ye dx – e dy + 2x y dx – y dy = 0

ex 2 3 y2 + x − =0 2 y 3

⇒

73. Given, y = c1em1x+ c2em2x+ c3em3x(1) ⇒ y1 = c1m1em1x+ c2m2em2x+ c3m3em3x = m1(y – c2em2x– c3em3x) + c2m2em2x + c3m3em3x  [from (1)] m2x m3x = m1y + c2(m1 – m2)e + c3(m3 – m1)e (2) Now, y2 = m1y1 + c2m2(m2 – m1)em2x + c3m3(m3 – m1)em3x = m1y1 + m2[y1 – m1y – c3(m3 – m1)em3x] + c3m3 (m3 – m1) em3x  [from (2)] = (m1 + m2) y1 – m1m2y + c3(m3 – m1) (m3 – m2)em3x(3) Further, y3 = (m1 + m2)y2 – m1m2y1 + c3m3(m3 – m1) (m3 – m2) em3x = (m1 + m2)y2 – m1m2y1 + m3[y2 – (m3 – m1) y1 + m1m2y]  [from (3)] = (m1 + m2 + m3)y2 – (m1m2+ m1m3 + m2m3) y1 + m1m2m3y = 0 .  y2 – (–7)y1 – 6y ⇒ y3 – 7y1 + 6y = 0 The correct option is (A)

d y 2 sin x = sin 2x dx On integrating, we get

Put x =



y2 +c 2 or, 2e–x/y + y2 = c The correct option is (B)

dy + y 2 cos x = sin 2x dx

y2 sin x =

t+

–e–x/y =

70. The given equation can be written as

(

2 x3 y 2 − =c 3 2



e − x/ y = y dy y2 ⇒ d(x/y)e–x/y = y dy On integrating, we get

=0 x − y2 The correct option is (C)

2y sin x

= dt y2 \ (1) becomes, dt + 2x2dx – y dy = 0 Integrating, we get,



Since it passes through (1, 0), \ –1 = 2 + c ⇒ c = –3 Thus, the curve is (2x – 3) +

ye x dx − e x dy

Put x = 0, y = 1, we get c = 1/2 Hence, the solution is 6ex+ 4x3y – 3y3 – 3y = 0 The correct option is (B) 72. The given equation can be written as (y dx – x dy)e–x/y – y3dy = 0

69. We have, x–y

dy ye x − e x ex dx = dt Let =t⇒ y dx y2

+ 2 x 2 dx − y dy = 0 (1)

dy = 2c1e2x+ c2ex– c3e–x dx d2y 2 = 4c1e2x+ c2ex– c3e–x dx

HINTS AND EXPLANATIONS

Now,

9.46  Chapter 9 d3y 3 = 8c1e2x+ c2ex– c3e–x dx Substituting in the given differential equation, we get 8 + 4a + 2b + c = 0 1+a+b+c=0

–1 + a – b + c = 0 ⇒ a = –2, b = –1, c = 2 Thus,

a3 + b3 + c 3 1 =− abc 4

The correct option is (B)

Previous Year’s Questions 75. The given differential equation is 2/3

⎛ d y⎞ = 4⎜ 3 ⎟ ⎝ dx ⎠ Making the above equation free from radical 2/3, the ­equation can be rewritten as dy ⎞ ⎛ ⎜⎝1 + 3 ⎟⎠ dx

2

⎛ d3y ⎞ dy ⎞ ⎛ ⎜⎝1 + 3 ⎟⎠ = 4 ⎜ 3 ⎟ dx ⎝ dx ⎠

d2y dx 2

HINTS AND EXPLANATIONS

dy e ⇒ = +c −2 dx e2 x ⇒y= + cx + d 4 The correct option is (B) 77. The general equation of all non-vertical lines in a plane is ax + by = 1, where b ≠ 0



⇒b ⇒

dy =0 dx

d2y dx 2

=0

d2y

=0 dx 2 which is the desired differential equation. The correct option is (A) 78. Equation y2 = 4a (x – h) ⇒ 2 yy ′ = 4 a ⇒ yy ′ = 2a ⇒ yy ′′ + ( y ′ ) 2 = 0 The correct option is (B) 79.

−1

(1 + y 2 ) + ( x − e tan y ) ⇒ (1 + y 2 )

xe tan

−1

xe tan

−1

xe tan

−1

−1

pdy

=e

= e tan

y

e 2 tan

−1

⋅

e tan

−1

y

1 + y2

y

1 + y2

1 = e 2 tan 2

−1

y

−1

y

⋅ dy + K1

dy + K1 + K1

−1

x 2 + y 2 − 2ay = 0

⇒ a=



⇒ y 2 = 2 yy ′ ( x + yy ′ ) [on putting value of c from (2) in (1)] On simplifying, we get (y − 2xy′)2 = 4yy′3 (3) Hence equation (3) is order 1 and degree 3. The correct option is (C) xdy = y(log y − log x + 1) dx

⇒

dy y ⎛ ⎛ y⎞ ⎞ log ⎜ ⎟ + 1⎟ = ⎝ x⎠ ⎠ dx x ⎜⎝

Substituting y = vx

e tan y 1 dx + x= 2 dy 1 + y 1 + y2

I.F = e ∫

=∫

−1

x + yy ′ (eliminating a) y ⇒ (x2 − y2)y′ = 2xy. The correct option is (C) 81. The equation ydx + xdy + x2y dy = 0 implies d ( xy ) 1 1 + dy = 0 ⇒ − + log = C . 2 2 y xy x y The correct option is (B) 82. y2 = 2c (x + √c)(1) Differentiating (1) we obtain 2yy′ = 2c ⋅1 or yy′ = c(2)



y

y

= ∫ e tan

Differentiating above equation: 2x + 2yy′ − 2ay′ = 0

83.

dx =0 dy

dx + x = e tan dy

1 ∫ 1+ y 2 dy

y

−1

−1

⇒

y

tan y = e 2 tan y + K 2 xe The correct option is (B)

80.

= e −2 x

−2 x

∴a + b

Therefore the required solutions is

3

This shows that the order and degree of given equation are 3 and 3 respectively. The correct option is (C) 76. The equation



3









dy xdv =v+ dx dx xdv ⇒v+ = v(log v + 1) dx

Differential Equations  9.47



xdv = v log v dx

⇒

dv dx ⇒ = k v log v x Put log v = z



dz dx ⇒ = z x ln z = ln x + ln c z = cx log v = cx ⎛ y⎞ log ⎜ ⎟ = cx ⎝ x⎠ The correct option is (C)



2

(2) 2

d2y

⎛ dy ⎞ + B ⎜ ⎟ = 0 (3) ⎝ dx ⎠ dx 2

From (2) and (3)





2 ⎧⎪ d2y dy ⎛ dy ⎞ ⎫⎪ x ⎨ − By 2 − B ⎜ ⎟ ⎬ + By =0 ⎝ dx ⎠ ⎪ dx dx ⎪⎩ ⎭

⇒ xy

d2y

2

dy ⎛ dy ⎞ + x⎜ ⎟ − y =0 2 ⎝ ⎠ dx dx dx

The correct option is (A)

85. General equation of all such circles is x2 + y2 + 2gx = 0. On differentiating, we get dy + 2g = 0 dx Equating the expression for g from above equation in the general equation, we get 2x + 2 y

dy ⎞ ⎛ x2 + y2 + ⎜ −2 x − 2 y ⎟ x = 0 ⎝ dx ⎠ dy dx The correct option is (C)

⇒ y2 = x2 + 2xy

86. Substituting, y = vx dy dv ⇒ =v+x dx dx dv ⇒ v+x = 1+ v dx

⇒ dv =

dx x

⇒ (x − h) = −(y − 2)

⎛ dy ⎞ (y − 2) ⎜⎝ dx ⎟⎠ + (y − 2)2 = 25

dy = 0, and dx

A + By

dy =0 dx

2

(1)



⇒ 2(x − h) + 2(y − 2)

(1)

dy dx Substitution in (1) implies that

84. Given equation Ax2 + By2 = 1 implies

y = log x + c x

Since, y (1) = 1, we have y = x log x + x The correct option is (D) 87. (x − h)2 + (y − 2)2 = 25

1 dv = dz v

Ax + By

∴ v = logx + c

(y − 2)2y′2 = 25 − (y − 2)2. The correct option is (C) 88. Given y = c1e c x (1) 2



y ′ = c2c1e c x .

⇒

2

So, y′ = c2y(2) ⇒ y′′ = c2y′ From (2)

c2 =



So, y ′′ =

y′ y ( y ′)2 ⇒ yy ′′ = ( y ′ ) 2 y

The correct option is (D) 89. cos x dy = y (sin x−y) dx

⇒

dy = y tan x − y 2 sec x dx



⇒

1 dy 1 − tan x = − sec x y 2 dx y Let

1 =t y







⇒ −

1 dy dt = y 2 dx dx



⇒ −

dy dt − t tan x = − sec x ⇒ + (tan x )t = sec x. dx dx







I.F = e ∫

tan xdx

= sec x

Solution is t (I.F) = ∫ (I.F)sec x dx 1 = sec x = tan x + c y The correct option is (D)

⇒

HINTS AND EXPLANATIONS



9.48  Chapter 9 dV = − k (T − t ) ⇒ dV = − k (T − t )dt dt Integrate 90.

V=

− k (T − t ) 2 k (T − t ) 2 + c ⇒V = +c 2 ( −2)

at t = 0 ⇒ V = I

kT 2 kT 2 kT 2 +c⇒c=l− ⇒ c = V (T ) = l − 2 2 2 The correct option is (A) l=







⇒ p − 400 = 300et / 2 ⇒ 400 − p = 300et / 2 (as p < 400)

⇒ p = 400 − 300et / 2 The correct option is (A) 2 x ln x dy y 94. + = dx x ln x x ln x

1

dx ∫ ∴ I.F. = e x ln x = e ln(ln x ) = ln x The solution is





d ( p(t )) 1 = p(t ) − 450 dt 2





y ln x = ∫ 2 ln xdx





⇒ y ln x = 2( x ln x − x ) + c

For x = 1, c = 2



d ( p(t )) p(t ) − 900 = dt 2





d ( p(t )) 2∫ = dt p(t ) − 900 ∫

Put x = e ⇒ y(e) = 2 . The correct option is (B)







91.







y ln x = 2( x ln x − x + 1)

2 ln p(t ) − 900 = t + c

95. We have ydx + xy 2 dx = xdy



t=0

⇒





⇒ 2 ln 50 = 0 + c





⇒ c = 2 ln 50











HINTS AND EXPLANATIONS





p( t ) = 0 ⇒ 2 ln 900 = t + 2 ln 50 ⎛ 900 ⎞ t = 2(ln 900 − ln 50) = 2 ln ⎜ = 2 ln18. ⎝ 50 ⎟⎠

The correct option is (A) 92. Given

∫

25

dP =

2000

∫ (100 − 12 0

( P − 2000) = 25 × 100 −

93.





dp p − 400 = dt 2 dp 1 = dt p − 400 2

1n p − 400 = 1 t + c 2 at t = 0, p = 100 ln 300 = c p − 400 t 1n = 300 2

)

x dx

12 × 2 ( 25)3/ 2 3

P = 3500. The correct option is (B)

y2

= xdx

⎛ x2 ⎞ ⎛ x⎞ ⇒ − d ⎜ ⎟ = d ⎜ ⎟ ⎝ 2⎠ ⎝ y⎠

∴ 2 ln p(t ) − 900 = t + 2 ln 50

p

xdy − ydx

Integrating, we get x x2 − = +C y 2 Since, it passes through (1, –1) 1 1 ∴ 1 = + C ⇒ C = 2 2 ⇒ x 2 + 1 + ∴

−2 x 2x =0⇒ y= 2 y x +1

⎛ 1⎞ 4 f ⎜− ⎟ = ⎝ 2⎠ 5

The correct option is (A) dy 96. (2 + sin x) + (y + 1)cos x = 0 dx ⎛π⎞ y(0) = 1, y ⎜ ⎟ = ? ⎝ 2⎠ 1 cos x dy + dx = 0 y +1 2 + sin x ln |y + 1 | + ln (2 + sin x ) = ln C (y + 1) (2 + sin x) = C Put x = 0, y = 1 (1 + 1) . 2 = C ⇒ C = 4 Now, (y + 1) (2 + sin x ) = 4 dy For, x = 1 − x 2 − xy = 1 dx

(

)

Differential Equations  9.49 (y + 1) (2 + 1) = 4 4 3

4 1 –1= 3 3 Hence, the correct option is (D)

98. Let y = f(x)

dy 97. sin x + y cos x = 4 x   x ∈ (0, π) dx

I.F. = e ∫ 4

y=

dy + y cot x = 4 x cosec x dx

⇒

dy + Py = Q where P and Q are function f(x) dx

⇒ e ∫

Pdx

Pdx

= ∫ Q ⋅ e∫

=

Pdx

cos x dx ∫ e sin x

dx + C

3 dx x

y.

= e ln sin x = sin x

x2 +C 2

7

3 4

x4 x = 7, +c 7/4

as

lim x→01

x.f (1/4) = 4

 y2  x

 y2 = +c ⇒ − x  x 

As it passes through (1, 1) so c = 2 \ x2 + y2 – 2x = 0 2

π π +C ⇒C = − 2 2 2

100. Differentiating both side f (x) = 2x – x2 f(x)

2 π ⎛π ⎞ π = 2⎜ ⎟ − 6 2 ⎝6⎠

y π2 π2 y −8π 2 = − ⇒ = 2 18 2 2 18

( (

) )

\ f(x) =

2 1− x2 2x f x ’ = ( ) 1+ x2 x2 +1

f(1/2) =

4  1  24 f’ = 5  2  25

2

⇒

3

99. x2dx = y2dzx – 2x ydy

π π ⇒ y sin x = 2 x − x= 6 2 ⇒ y sin

∫

⇒ – dx = d 

2

2

3

= x4

−3

π π2 = 2⋅ +C 2 4

⇒ 0 × 1 =

4x

\ f(x) = 4x + c. x 4

⇒ y sin x = 2x2 + C ⎛π ⎞ Given y ⎜ ⎟ = 0 ⎝2⎠ ⇒ yx sin

dx

So, y. x 4 = 7 x 4 dx

⇒ yx sin x = ∫ 4 x cosec x sin x dx + C ⇒ y sin x = 4 ⋅

\ dy + 3 y = 7

3

⇒

⇒ ye ∫

2

HINTS AND EXPLANATIONS

y+1=

⇒ y = − 8π 9

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Coordinates and Straight Lines

CHAPTER

10 LEARNING OBJECTIVES

After reading this chapter, you will be able to:  Be acquainted with distance formula and section formula  Know how to calculate area of triangle, area of quadrilateral, area of a polygon  Learn the working rule to find the locus of a point, translation of axes, rotation of axes and reflection of a point

DISTANCE FORMULA The distance between two points P(x1, y1) and Q(x2, y2) is given by

FIGURE 10.1

PQ = ( x2 − x1 ) 2 + ( y2 − y1 ) 2

 Be familiar with general equation of a straight line and the method to calculate the slope of a line  Understand various forms of the equation of a line and distance of a point from a given line

SOLVED EXAMPLES 1. Consider the point A ≡ (0, 1) and B ≡ (2, 0). Let P be a point on the line 4x + 3y + 9 = 0. Coordinates of the point P such that |PA – PB| is maximum, are  12 17  84 13 (A)  − ,  (B)  − ,   5 5  5 5 6 17   (C)  − ,  (D)  none of these  5 5 Solution: (A)

We have, |PA – PB | ≤ AB.

Info Box! Distance is always positive. Therefore, we often write PQ instead of |PQ|.  Distance MN between two points M(x , 0) and N(x , 1 2 0) on the x-axis is |x2 – x1|. Similarly, the distance between two points M(x1, y1) and x(x2, y1), (which lie on a line parallel to x-axis) is |x2 – x1|.  Distance AB between two points A(0, y ) and B(0, 1 y2) on the y-axis is |y2 – y1|. Similarly, the distance between two points A(x1, y1) and B(x1, y2), (which lie on a line parallel to y-axis) is |y2 – y1|.  Distance between the origin O(0, 0) and the point 

P(x, y) is OP =

x2 + y 2 .

Thus, for |PA – PB| to be maximum, point A, B and P must be collinear. The equation of line AB is x + 2y = 2

10.2  Chapter 10 84 13 Solving it with the given line, we get P ≡  − ,   5 5 2. 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 (B) 2(x2 + y2) = a2 2 2 2 (C) x + y = 2a (D) 4(x2 + y2) = a2 Solution: (D)

Let AB be the ladder. Let the cross section of the floor and wall be taken as the coordinates axes. Let P(x, y) be the mid-point of AB whose locus is required. Then the coordinates of A and B are (2x, 0) and (0, 2y) respectively. Given, AB = a ⇒ ( 2 x − 0) 2 + ( 0 − 2 y ) 2 = a ⇒ 4 x 2 + 4 y 2 = a 2 or 4(x2 + y2) = a2, which is the required locus.

3. The point (2t2 + 2t + 4, t2 + t + 1) lies on the line x + 2y = 1 for (A) all real values of t (B) some real values of t (C) t = −4 ± 7 8 (D) none of these Solution: (D) The point (2t2 + 2t + 4, t2 + t + 1) lies on the line 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. 4. 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 Solution: (A) If a and b are the lengths of perpendiculars, then |a| + |b | = 1 (given), whose graph is a square. 5. The condition to be imposed on b so that (0, b) lies on or inside the triangle having sides y + 3x + 2 = 0, 3y – 2x – 5 = 0 and 4y + x – 14 = 0 is

7 5 (B)  0 0 the straight x y line + = 1 is drawn so as to form with coordinate a b axes a triangle of area S. If ab > 0, then the least value of S is (A) ab (B)  2ab (C) 4ab (D)  none of these Solution: (B)

FIGURE 10.10

1 Thus, ∆ = [( x1 y2 − x2 y1 ) + ( x2 y3 − x3 y2 ) + ( x3 y1 − x1 y3 )] 2

The equation of the given line is x y + = 1 (1) a b

QUICK TIPS 

In an equilateral triangle

FIGURE 10.11

3 2 a 4 p2 (ii)  having length of perpendicular as ‘p’, area is . 3

(i)  having sides a, area is

This line cuts x-axis and y-axis at A(a, 0) and B(0, b) respectively. Since area of DOAB = S (Given) 1 ∴ ab = S or ab = 2S (∵ ab > 0) (2) 2

Coordinates and Straight Lines  10.7 Since the line (1) passes through the point P(a, b) α β α aβ [Using (2)] ∴ + = 1 or + =1 a b a 2S or a2b – 2aS + 2aS = 0. Since a is real, \ 4S2 – 8abS ≥ 0 1 or 4 S 2 ≥ 8αβ S or S ≥ 2αβ ∵ S = ab > 0 as ab > 0  2   Hence the least value of S = 2ab. 18. 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 DRPQ = 7, then the number of such points R is (A) 0 (B)  1 (C) 2 (D)  4 Solution: (C)

Since the angle RPQ is a right angle, \ slope of RP × slope of PQ = –1 1− y 5 −1 ⇒ × = −1 ⇒ 3 x + 4 y = 13 (1) 3− x 6 −3 Also, area of DRPQ = 7 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 Eq. (1) and (2) and (1) and (3), we get two different coordinates of the point R. So, there are two such points R. ⇒

19. If two vertices of an equilateral triangle have integral coordinates then the third vertex will have (A) coordinates which are irrational (B) atleast one coordinate which is irrational (C) coordinates which are rational (D) coordinates which are integers Solution: (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 x1 y1 1 1 area of ∆ = x2 y2 1 = retional. 2 x3 y3 1 But the area of an equilateral triangle 3 = = (side) 2 irrational 4 Thus, the two statements are contradictory. Therefore, both the coordinates of the third vertex cannot be rational.

20. Let P(2, – 4) and Q(3, 1) be two given points. Let R (x, y) be a point such that (x – 2) (x – 3) + (y – 1) (y + 4) = 13 0. If area of DPQR is , then the number of possible 2 positions of R are (A) 2 (B)  3 (C) 4 (D)  none of these Solution: (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.

Now, area of ∆PQR =

13 2

1 13 × 26 × (altitude) = 2 2 26 ⇒ altitude = = radius 2 ⇒  there are two possible positions of R. ⇒

21. The base of a triangle lies along the line x = a and is of length a. The area of the triangle is a2, if the vertex lies on the line (A) x = 0 (B)  x = –a (C) x = 3a (D)  x = –3a Solution: (B, C)

Let h be the height of the triangle. Since, the area of the triangle is a2 1 ∴ × a × h = a 2 ⇒ h = 2a 2 Since the base lies along the line x = a, the vertex lies on the line parallel to the base at a distance 2a from it. So, the required lines are x = a ± 2a i.e., x = –a or x = 3a 22. If a, c, b are three terms of a G.P., then the line ax + by +c=0 (A) has a fixed direction (B) always passes through a fixed point

10.8  Chapter 10 (C)  forms a triangle with the axes whose area is constant (D) always cuts intercepts on the axes such that their sum is zero. Solution: (C)

24. 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) (B)  (5, 0) or (1, 4) (C) (5, 0) or (4, 1) (D)  (0, 5) or (1, 4) Solution: (B)

Since a, c, b are in G.P., \ c2 = ab(1) The area of the triangle =

Let A ≡ (2, –1) and B ≡ (3, 2). Let the third vertex be C (a, b). Then, a + b = 5 (given) 2 −1 1 1 Area of ∆ABC = 3 2 1 = ± 4 2 α β 1

1  c  c × − × − 2  a   b 

(1) (given )

⇒ b – 3a = 1 or b – 3a = –15 Solving (1) and (2), we get, a = 1, b = 4 Solving (1) and (3), we get, a = 5, b = 0 Thus, the third vertex is either (5, 0) or (1, 4).

(2) (3)

AREA OF A QUADRILATERAL 1 c2 1 × =  2 ab 2



=



= constant

[Using (1)]

The area of a quadrilateral, whose vertices are A(x1, y1), B(x2, y2), C(x3, y3) and D(x4, y4), is =

23. 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

1  x1  2  x2

y1 x2 + y2 x3

y2 x3 + y3 x4

y3 x4 + y4 x1

y4   y1 

Info Box! The rule for writing down the area of a quadrilateral is the same as that of a triangle.

Solution: (A)

Thus, area of quadrilateral with vertices (xr, yr), r = 1, 2, 3, 4 is

x2 x3 y2 y3 Let = = r and = = r x1 x2 y1 y2

1 [( x1 y2 − x2 y1 ) + ( x2 y3 − x3 y2 ) + ( x3 y4 − x4 y3 ) + ( x4 y1 − x1 y4 )] 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) = 0 ( R2 and R3 are identical) Thus, (x1, y1), (x2, y2), (x3, y3) lie on a straight line.

AREA OF A POLYGON • The area of a polygon of n sides with vertices A1(x1, y1), A2(x2, y2), ..., An(xn, yn) is =

1  x1  2  x2

y1 x2 + y2 x3

y2 x +  + n −1 y3 xn

yn −1 xn + yn x1

yn   y1 

STAIR METHOD Repeat first coordinates one time in last. For down arrow use positive sign and for up arrow use negative sign.

∴

Coordinates and Straight Lines  10.9 100 200 > 27  ⇒  n > 7

Now, area of Sn < 1 ⇒ an2 < 1 ⇒ \ n = 8, 9, 10

∴ Area of polygon =

LOCUS 1 = {[( x1 y2 + x2 y3 +  + xn y1 ) − ( y1 x2 + y2 x3 +  + yn x1 )]} 2

SOLVED EXAMPLES 25. The area of the region enclosed by 4 |x| + 5 |y| ≤ 20 is (A) 10 (B)  20 (C) 40 (D)  none of these

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 Let the coordinates of the moving point P be (h, k). Using the given geometrical conditions, find the relation between h and k. This relation must contain only h, k and known quantities.  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).

Solution: (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.



TRANSLATION OF AXES 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. 1 × 10 × 8 = 40. 2 26. Let S1, S2, ... be squares such that for each n ≥ 1, the length of a side of Sn equals the length of a diagonal of Sn+1. If the length of a side of S1 is 10 cm, then for which of the following values of n is the area of Sn less than 1 cm2? (A) 7 (B)  8 (C) 9 (D)  10 \ Required area =

FIGURE 10.12

Solution: (B, C, D)

Let a be the side of the square, then diagonal d = a 2. Given : an = 2an +1 ⇒ an +1 = ∴ an +1 =

an 2

=

a1 ( 2)

n

an −1 ( 2)

2

=

an − 2 ( 2)

⇒ an =

3

= =

a1 ( 2)

n −1

=

a1 ( 2 )n 10

2

n −1 2

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 direction as OX and OY respectively. Let P be any point in the plane having coordinates (x, y) referred to old axes and (X, Y) referred to new axes. Then,

10.10  Chapter 10 x = OM = ON + NM = ON + O′ M′ = h + X = X + h  or  X = x – h and y = MP = MM′ + M′ P = NO′ + M′ P = k + Y = Y + k. or  Y = y – k Thus, the point whose coordinates were (x, y) has now the coordinates (x – h, y – k). QUICK TIPS If origin is shifted to point (h, k) without rotation of axes, then new equation of curve can be obtained by putting x + h in place of x and y + k in place of y

FIGURE 10.14

x = h + x′ cosθ – y′ sinθ and y = β + x′ sinθ + y′ cosθ

ROTATION OF AXES Rotation of Axes without Changing the Origin

REFLECTION (IMAGE) OF A POINT

Let OX, OY be the original axes and OX′, OY′ be the new axes obtained by rotating OX and OY through an angle θ in the anticlockwise sense. Let P be any point in the plane having coordinates (x, y) with respect to axes OX and OY and (x′, y′) with respect to axes OX′ and OY′. Then,

(i) x axis is (x, –y) (iii) origin is (– x, –y)

Let (x, y) be any point, then its image with respect to (ii)  y-axis is (–x, y) (iv) line y = x is (y, x)

QUICK TIPS If area is a rational number. Then the triangle cannot be equilateral.  If two opposite vertices of a rectangle are (x , y ) and (x , 1 1 2 y2), then its area is |(y2 – y1)(x2 – x1)|.  If two opposite vertices of a square are A(x , y ) and 1 1 C(x2, y2), then its area is 

=

1 2 1 AC = [( x2 − x1)2 + ( y2 − y1)2 ] 2 2

SOLVED EXAMPLES FIGURE 10.13

x = x′ cosθ – y′ sinθ; y = x′ sinθ + y′ cosθ and x′ = x cosθ + y sinθ;  y′ = – x sinθ + y cosθ The above relation between (x, y) and (x′, y′) can be easily obtained with the help of following table x↓

y↓

x′→

cosθ

sinθ

y′→

– sinθ

cosθ

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 co-ordinates of P(x, y) become (x′, y′), then the equations of transformation will be

27. The image of the point (3, –8) under the transformation (x, y) → (2x + y, 3x – y) is (A) (–2, 17) (B)  (2, 17) (C) (–2, –17) (D)  (2, –17) Solution: (A)

Let (x1, y1) be the image of the point (x, y) under the given transformation. Then, x1 = 2x + y = 2 (3) – 8 = –2 and y1 = 3x – y = 3 (3) – (–8) = 17 Hence, the image is (–2, 17). 28. Without changing the direction of coordinates axes, origin is transferred to (a, b) so that the linear terms in the equation x2 + y2 + 2x – 4y + 6 = 0 are eliminated. The point (a, b) is

Coordinates and Straight Lines  10.11 (A) (–1, 2) (C) (1, 2)

(B)  (1, –2) (D)  (–1, –2)

Solution: (A)

The given equation is x2 + y2 + 2x – 4y + 6 = 0 (1) Putting x = x′ + a, y = y′ + b in (1), we get x′2 + y′2 + x′(2a + 2) + y′(2b – 4) + (a 2 + b 2 + 2a – 4b + 6) = 0 To eliminate linear terms, we should have 2a + 2 = 0 and 2b – 4 = 0 ⇒  a = –1 and b = 2 \ (a, b) ≡ (–1, 2) 29. 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 60° 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

(C) 3 x + y − 5 3 = 0 (D) 3 x − y − 5 3 = 0 Solution: (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) ⇒

31. 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, 1 1 1 1 (A) 2 + 2 = 2 + 2 a b p q (B)

Solution: (A)

Since P is the image of the point (–3, 2) with respect to x-axis, therefore, the coordinates of P are (–3, –2). Let (x′, y′ ) be the coordinates of P with respect to new axes. Then, x′ = x cos a + y sin a = –3 cos (–60º) –2 sin (–60º) 3 2 3 −3 =− + 3= . 2 2 y′ = – x sin a + y cos a = 3 sin (–60º) – 2 cos (–60º) 3 3+2 = −  2     2 3 − 3  3 3 + 2  \ Coordinates of P are  . ,−  2    2    30. A ray of light travelling along the line x + 3 y = 5 is incident on the x-axis and after refraction it enters π the other side of the x-axis by turning away from 6 the x-axis. The equation of the line along which the refracted ray travels is (A) x + 3 y − 5 3 = 0 (B) x − 3 y − 5 3 = 0

1 1 1 1 − 2 = 2− 2 2 a b p q

 1 1 1 1  + 2 = 2 2 + 2  2 a b q  p (D) none of these (C)

Solution: (A)

Since the line L has intercepts a and b on the coordinate axes, therefore its equation is x y + = 1 (1) a b When the axes are rotated, its equation with respect to the new axes and same origin will become x y + = 1 (2) p q



In both the cases, the length of the perpendicular from the origin to the line will be same. ∴

⋅



y = − 3 ( x − 5) or 3 x + y − 5 3 = 0

or

1 1 1 + a2 b2

=

1 1 1 + p2 q2

1 1 1 1 + 2 = 2+ 2 2 a b p q

which is the required relation.

10.12  Chapter 10

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.

SLOPE OF A LINE π  If a line makes an angle θ  θ ≠  with the positive direc2  tion of x-axis, the slope or gradient of that line is usually denoted by m, i.e., tan θ = m. QUICK TIPS 







The slope of a line parallel to x-axis = 0 and perpendicular to x-axis is undefined. If three points A, B, C are collinear, then slope of AB = slope of BC = slope of AC. 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. Slope of the line joining two points (x1, y1) and (x2, y2) is given as m=

y1 − y2 y2 − y1 Difference of ordinates = = . bscissae x1 − x2 x2 − x1 Difference of ab

a  Slope of the line ax + by + c = 0, b ≠ 0 is − , i.e., b (Coefficient of x ) . (Coefficient of y )

SOLVED EXAMPLES 32. 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) (B)  (5, –3) (C) (–5, 3) (D)  (–5, –3) Solution: (B)

Let (x1, y1) be the image of the point P(3, 5) with respect to the line y = x. Then, x1 = 5, y1 = 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)

INTERCEPT OF A LINE ON THE AXES 1. 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. 2. 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.

SOLVED EXAMPLES 33. A square is constructed on the portion of the line x + y = 5 which is intercepted between the axes, on the side of the line away from origin. The equations to the diagonals of the square are (A) x = 5, y = –5 (B) x = 5, y = 5

Coordinates and Straight Lines  10.13 (C) x = –5, y = 5 (D) x – y = 5, x – y = –5 Solution: (B)

Clearly, the equations of the two diagonals are x = 5 and y = 5.

FIGURE 10.15

Point-Slope Form 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 34. If a straight line cuts intercepts from the axes of coordinates the sum of the reciprocals of which is a constant k, then the line passes through the fixed point 1 1 (A) (k, k) (B)  k,k    (C) (k, –k) (D)  (–k, k) Solution: (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

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

SOLVED EXAMPLES 35. In the above problem, coordinates of the point P such that |PA – PB| is minimum are  3 12  9 12 (A)  − , −  (B)   ,   20 5  5  20 9 12 (C)  , −  5  20

1 1 + = 1 or ak bk

1/k 1/k + = 1 (2) a b

From (2) it follows that the line (1) passes through the 1 1 fixed point  ,  . k k

EQUATION OF A STRAIGHT LINE IN VARIOUS FORMS 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 the equation of the line will become y = mx.

 9 12  (D)  − ,   20 5 

Solution: (A)

The minimum value of |PA – PB| is zero which is attained if PA = PB, i.e., P must lie on the perpendicular bisector of AB. The equation of perpendicular bisector of AB is y−

1 = 2( x − 1) or 2

y = 2x −

3 2

 9 12  Solving it with the given line, we get P ≡  − , −  5  20 36. Given the system of straight line a(2x + y – 3) + b(3x + 2y – 5) = 0, the line of the system farthest form the point (4, –3) has the equation (A) 3x – 4y + 1 = 0 (B)  4x + 3y – 5 = 0 (C) 7x – y + 4 = 0 (D)  none of these

10.14  Chapter 10 Solution: (A)

The given system of lines pass through (1, 1) So, the required line is the through (1, 1) and perpendicular to the join of (1, 1) and (4, –3). y −1 3 \ The equation of line is = , i.e., 3 x − 4 y + 1 = 0 x −1 4 37. The image of the point (–8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is (A) (16, –2) (B)  (–16, 2) (C) (16, 2) (D)  (–16, –2) Solution: (D)

Equation of the given line is 4x + 7y + 13 = 0 (1) Let Q (a, b) 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) ( y − 12) =

Solving Eq. (1) and (2), we get x = –12 and y = 5. \ C ≡ (–12, 5)

FIGURE 10.16

SOLVED EXAMPLES 38. Through the point (1, 1), a straight line is drawn so as to form with coordinate axes a triangle of area S. The intercepts made by the line on the coordinate axes are the roots of the equation (A) x2 – |S| x + 2 |S| = 0 (B) x2 + |S| x + 2 |S| = 0 (C) x2 – 2 |S| x + 2 |S| = 0 (D) none of these Solution: (C)

If a, b are the intercepts made by the line, then the x y equation of the line is + = 1. a b 1 1 + =1 Since it passes through (1, 1), ∴ a b a+b = 1 (1) ab Also, area of the triangle made by the straight line on the coordinate axes is S 1 ∴ ab = |S | i.e., ab = 2 |S | (2) 2 So, by (1), a + b = 2 |S|(3) From (2) and (3), the intercepts a and b are the roots of the equation x2 – 2 |S| x + 2 |S| = 0.

Since C is mid-point of PQ, α −8 β + 12 ∴ − 12 = and 5 = 2 2 ⇒ a = –16 and b = –2 \ Q ≡ (–16, –2).

⇒

39. A line passing through the point P(4, 2), meets the x-axis and y-axis at A and B respectively. If O is the origin, then locus of the centre of the circum circle of DOAB is (A) x–1 + y–1 = 2 (B)  2x–1 + y–1 = 1 –1 –1 (C) x + 2y = 1 (D)  2x–1 + 2y–1 = 1 Solution: (B)

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

Let the coordinates of A and B be (a, 0) and (0, b) respectively. Then, equation of line AB is x y + =1 a b

Coordinates and Straight Lines  10.15 Since, it passes through the point P(4, 2)

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



∴

4 2 + = 1. (1) a b

a b Now, centre of the circumcircle of ∆OAB =  ,  . 2 2 2 1 So, Eq. (1) can be written in the form + =1 a/2 b/2 2 1 \ locus of circumcentre is + = 1 or 2 x −1 + y −1 = 1 x y 40. 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 through the fixed point (A) (a, 0) (B)  (0, a) (C) (a, a) (D)  none of these Solution: (C) We take A as the origin and AB and AC as x-axis and y-axis respectively. Let AP = h, AQ = k. Equation of the line PQ is

FIGURE 10.17

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 anti-clockwise direction between 0 and 2π.

Parametric Form (or Symmetric Form) 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 given by x − x1 y − y1 = =r cos θ sin θ where r is the distance of the point (x, y) from the point (x1, y1).

FIGURE 10.18

QUICK TIPS 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



x y + = 1 (1) h k

Given, BP × CQ = AB2 ⇒ (h – a) (k – a) = a2 ⇒ hk – ak – ah + a2 = a2 or ak + ha = hk a a or + = 1 (2) h k From (2), it follows that line (1) i.e., PQ passes through the fixed point (a, a).

(x1 + r cosθ, y1 + r sinθ) or (x1 – r cosθ, y1 – r sinθ)

where the line is inclined at an angle θ with x-axis

SOLVED EXAMPLES 41. A line joining two points A(2, 0) and B(3, 1) is rotated about A in anticlockwise direction through an angle 15°. If B goes to C in the new position, then the coordinates of C are

10.16  Chapter 10  3  3 (A)  2, (B)   2, −    2  2     1 3 (C)  2 + ,   2 2   Solution: (C)

(D)  none of these

Also, y-axis is the bisector of the angle between the two lines. P1, P2 are two points on these lines, at a distance 5 units from A. Q is the foot of the ^ from P1 and P2 on the bisector (y-axis). Then, the coordinates of Q are (0, 2 + 5 cos 30º)

Slope of line AB =

 5 3  1  =  0, 2 +  =  0, ( 4 + 5 3 )   2   2  

0 −1 = 1 = tan 45° 2−3

\ ∠BAX = 45°. Given ∠CAB = 15°. \ ∠CAX = 60°. \ Slope of line AC = tan 60º = 3 . Now, line AC makes an angle of 60° with positive direction of x-axis and AC = AB = (3 − 2) 2 + (1 − 0) 2 = 2 \ Coordinates of C are ( 2 + 2 cos 60°, 0 + 2 sin 60°)

43. If the straight line drawn through the point P( 3 , 2) π and making an angle with the x-axis meets the line 6 3 x − 4 y + 8 = 0 at Q, then the length of PQ is (a) 4 (b)  5 (c) 6 (d)  none of these ⋅

Solution: (C)

The given line is

 1 3 ,  2 +  2 2  

i.e.,

42. 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. The coordinates of the foot of the perpendicular from P on the bisector of the angle between them are 1 1 (A) 0, ( 4 + 5 3 )  or 0, ( 4 − 5 3 )  depend 2   2  ing on which line the point P is taken  1  (B) 0, ( 4 + 5 3 )   2   1  (C) 0, ( 4 − 5 3 )  2   5 5 3  (D)  ,   2 2  Equation, of two lines are y = 3 x + 2, if y = − 3 x + 2, if

3 x − 4 y + 8 = 0 (1)



Let PQ = r Then, the coordinates of Q are  π π 3 r   3 + r cos 6 , 2 + r sin 6  or  3 + 2 r , 2 + 2     

Since the point Q lies on the given line, ∴

 r 3   r  − 42 +  + 8 = 0 3 3 +   2 2    

⇒ 6 + 3r – 16 – 4r + 16 = 0 or r = 6 Hence, PQ = 6

Solution: (B)

and

Clearly y ≥ 2.

x≥0 x≥0

44. A line is drawn from the point P(a, b), making an angle q with the positive direction of x-axis, to meet the line ax + by + c = 0 at Q. The length of PQ is

Coordinates and Straight Lines  10.17 aα + bβ + c aα + bβ + c (B)  a cos θ + b sin θ a2 + b2 aα + bβ + c (C) (D)  none of these a cos θ + b sin θ (A) −

QUICK TIPS 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.

Solution: (A)

Equation of a straight line passing through the point P(a, b) and making an angle q with positive direction of x-axis is x −α y − β = = r (say) cos θ sin θ Coordinates of any point on this line are (a + r cosq, b + r sinq) If it lies on the line ax + by + c = 0, then a(a + r cosq) + b(b + r sinq) + c = 0 aα + bβ + c ⇒ r=− a cos θ + b sin θ aα + bβ + c Thus, PQ = r = − a cos θ + b sin θ

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 where m = − and c = − B B Thus, slope of the line Ax + By + C = 0 is

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) CASE 1. If C < 0 or –C > 0, then divide both sides of Eq. (1) by A2 + B 2 , we get A A +B 2

A +B 2

C

y=−

2

A + B2 2

which is of the form x cos α + y sin α = p, A

where, cos α =

A +B 2

 p = −

and

2

, sin α =

B A + B2 2

C A + B2 2

If C > 0 or –C < 0, then divide Eq. (1) by − A + B 2 , we get CASE 2. 2

−A

y=−

cofficient of x A m= =− cofficient of y B

B

x+

2

A +B 2

2

x−

B A +B 2

2

y=

C A + B2 2

which is of the form x cos α + y sin α = p, where and

cos α = −  p =

A A +B 2

, sin α = −

2

B A + B2 2

C A + B2 2

Intercept Form The equation Ax + By + C = 0 can be reduced to the form x y + = 1 by expressing it as a b Ax + By = –C A B or − x − y = 1, where C ≠ 0 C C x y x y + = −1, which is of the form + = 1, or C C a b − − A B C C where a = − and b = − are intercepts on x-axis and A B y-axis, respectively.

ANGLE BETWEEN TWO INTERSECTING LINES The angle θ between two lines y = m1x + c1 and y = m2x + c2 is given by tan θ = ±

m1 − m2 , 1 + m1m2

provided no line is perpendicular to x-axis and the acute angle θ is given by tan θ =

m1 − m2 1 + m1m2

10.18  Chapter 10 QUICK TIPS If both the lines are perpendicular to x-axis, then the angle between them is 0º.  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.  The two lines are parallel if and only if m = m . 1 2  The two lines are perpendicular if and only if m × m = –1. 1 2 

SOLVED EXAMPLE 45. Number of equilateral triangles with y = 3 ( x − 1) + 2 and y = 3 x as two of its sides, is (A) 0 (B)  1 (C) 2 (D)  none of these Solution: (D)

The sides are, y = 3 ( x − 1) + 2 and y = − 3 x

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. WORKING RULE Keep the terms containing x and y unaltered.  Change the constant.  The constant k is determined from an additional condition given in the problem. Thus, the equation of any line parallel to 2x – 3y + 5 = 0 is 2x – 3y + k = 0. 

SOLVED EXAMPLES 46. The vertices of a DOBC 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 origin is , is 2 1 1 =0 (A) x + y + (B) x + y − = 0 2 2 1 x+ y+ =0 (C) x + y − 1 = 0 (D)  2 2 Solution: (A)

The two lines are at an angle of 60° to each other. Now any line parallel to obtuse angle bisector will make equilateral triangle with these lines as its two sides.

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 a2

b1 c1 = ; b2 c2

1 1. Coincident, if =

2. Parallel, if

a1 b1 c1 = ≠ ; a2 b2 c2

3. Perpendicular, if a1a2 + b1b2 = 0; 4. Intersecting, if dent nor parallel.

a1 b1 ≠ i.e., if they are neither coincia2 b2 ⋅

The equation of line BC is x+y+4=0 \ Equation of a line parallel to BC is x+y+k=0 1 This is at a distance from the origin. 2 k 1 ∴ = 2 2 ∴

k=±

1 2

Since BC and the required line are on the same side of the origin 1 ∴ k= 2 1 = 0. Hence, the required line is x + y + 2 47. The distance of the point (2, 3) from the line 3x + 2y = 17, measured parallel to the line x – y = 4 is (A) 4 2 (B)  5 2 (C) 2 (D)  none of these

Coordinates and Straight Lines  10.19 Solution: (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 cosq, 3 + r sinq), where tan q = 1.

WORKING RULE TO PROVE CONCURRENCY Following three methods can be used to prove that the three lines are concurrent:  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.  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 b1 c1 a2 b2 c2 = 0 a3 c3 c3 

If this point lies on the line 3x + 2y = 17, then  3(2 + r cosq) + 2(3 + r sinq) = 17 ⇒ 6 + 3r ⋅

1 2

+ 6 + 2r ⋅

1 2

= 17 ⇒

5 2

r =5

⇒ r= 2

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.

SOLVED EXAMPLE

EQUATION OF A LINE PERPENDICULAR TO A GIVEN LINE The equation of a line perpendicular to a given line ax + by + c = 0 is bx – ay + k = 0, where k is a constant. WORKING RULE Interchange the coefficients of x and y and change the sign of one of them.  Change the constant.  The value of k can be determined from an additional condition given in the problem. 

POINT OF INTERSECTION OF TWO GIVEN LINES Let the two given lines be a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0. Solving these two equations, the point of intersection of the given two lines is given by  b1c2 − b2 c1 c1a2 − c2 a1  ,    a1b2 − a2 b1 a1b2 − a2 b1 

CONCURRENT LINES The three given lines are concurrent if they meet in a point.

1 1 1 , , are in A.P., then the straight line a b c x y 1 + + = 0 always passes through a fixed point, a b c that point is (A) (–1, –2) (B)  (–1, 2) 1  (C) (1, –2) (D)  1, − 2    Solution: (C) 1 1 1 Since , , are in A.P. a b c 1 1 2 (1) ⇒ + = a c b x y 1 The given line is + + = 0 a b c x y 2 1 ⇒ + + − = 0 [Using (1)] a b  b a  48. If

1 1 ( x − 1) + ( y + 2) = 0 a a ⇒  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. ⇒

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

10.20  Chapter 10 expressions: ax1 + by1 + c and ax2 + by2 + c have same sign or opposite signs.

WORKING RULE Make the R.H.S. of the equation of the line zero by transposing every term to L.H.S.  On the L.H.S., replace x by x and y by y . 1 1 

SOLVED EXAMPLES 49. The point (1, b) lies on or inside the triangle formed by the lines y = x, x-axis and x + y = 8, if (A) 0 < b < 1 (B)  0 ≤ b ≤ 1 (C) 0 < b < 8 (D)  none of these Solution: (B)

The point (1, b) lies on the line x = 1, for all real b. Clearly, from the figure, it will lie on or inside the triangle formed by the given lines if 0 ≤ b ≤ 1.

Divide by (coeff. of x )2 + (coeff. of y )2 .  Take the modulus of the expression thus obtained. This will give the length of the perpendicular. 

SOLVED EXAMPLES 51. If P and Q are two points on the line 4x + 3y + 30 = 0 such that OP = OQ = 10, where O is the origin, then the area of the DOPQ is (A) 48 (B)  16 (C) 32 (D)  none of these Solution: (A)

Let OR ^ PQ. Then, OR =

| 4(0) + 3(0) + 30 | 16 + 9

=

30 = 6. 5

50. Let P(2, 0) and Q(0, 2) be two points and O be the origin. If A(x, y) is a point such that xy > 0 and x + y < 2, then (A) A cannot be inside the DOPQ (B) A lies outside the DOPQ (C) A lies either inside DOPQ or in the third quadrant (D) none of these Solution: (C)

∴ PR = OP 2 − OR 2 = 100 − 36 and

Since xy > 0, therefore the point A lies either in the first quadrant or in the third quadrant. Since x + y < 2, therefore the point A lies either inside the DOPQ or in the third quadrant.



LENGTH OF PERPENDICULAR FROM A POINT ON A LINE The length of the perpendicular from the point (x1, y1) to the line ax + by + c = 0 is given by p=

| ax1 + by1 + c |

. a2 + b2 Corollary The length of the perpendicular from the origin (0, 0) on the line ax + by + c = 0 is | a×0 + b×0 + c | a2 + b2

=

|c| a2 + b2

PQ = 2PR = 16

∴ Area of ∆OPQ = =

1 × PQ × OR 2 1 × 16 × 6 = 48 2

52. On the portion of the straight line x + y = 2 which is intercepted between the axes, a square is constructed away from the origin, with this portion as one of its side. If p denotes the perpendicular distance of a side of this square from the origin, then the maximum value of p is (A) 2 (B)  2 2 (C) 3 2 (D)  4 2 Solution: (C) p = ON = OM + MN = ^ distance from

Coordinates and Straight Lines  10.21 In the case of a rhombus, p1 = p2. Thus, 

Area of rhombus =

p12 . sinθ

1 d1d2 2 where d1 and d2 are the lengths of two perpendicular diagonals of a rhombus. Also, area of rhombus =

SOLVED EXAMPLE

O to the line AB + AD =

2 2

+2 2 = 2+2 2 =3 2

DISTANCE BETWEEN TWO PARALLEL LINES The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is given by d=

| c1 − c2 | a2 + b2

Info Box! 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.

QUICK TIPS 

Area of a parallelogram or a rhombus, equations of whose sides are given, can be obtained by using the following formula Area =

p1p2 , sinθ

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.

53. The diagonals of the parallelogram whose sides are lx + my + n = 0, lx + my + n′ = 0, mx + ly + n = 0, mx + ly + n′ = 0 include an angle π π (A) (B)  2 3 ⋅

⋅

 l 2 − m2 (C) tan −1  2 2 l +m

  

 2lm  (D) tan −1  2 2  l +m 

Solution: (B)

Since the distance between the parallel lines lx + my + n = 0 and lx + my + n′ = 0 is same as the distance between the parallel lines mx + ly + n = 0 and mx + ly + n′ = 0. Therefore, the parallelogram is a rhombus. Since the diagonals of a rhombus are at right angles, π therefore the required angle is . 2 ⋅

EQUATIONS OF STRAIGHT LINES 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

REFLECTION ON THE SURFACE Here, IP = Incident Ray PN = Normal to the surface PR = Reflected Ray

FIGURE 10.19

Then, ∠IPN = ∠NPR Angle of incidence = Angle of reflection

10.22  Chapter 10

IMAGE OF A POINT WITH RESPECT TO A LINE 1. The image of a point with respect to the line mirror. The image of A(x1, y1) with respect to the line mirror ax + by + c = 0 be B(h, k) given by,

h − x1 k − y1 −2( ax1 + by1 + c) = = a b a2 + b2

FIGURE 10.20

SOLVED EXAMPLE

 26  (C)  − , 0   7 

(D)  none of these

FIGURE 10.21

2. The image of a point with respect to x-axis: Let P(x, y) be any point and P′(x′, y′) its image after reflection in the x-axis, then x′ = x and y′ = – y, (  O′ is the mid point of PP′) \

54. A ray of light is sent along the line which passes through the point (2, 3). The ray is reflected from the point P on x-axis. If the reflected ray passes through the point (6, 4), then the coordinates of P are  26   26  (A)  , 0  (B)   0, 7     7 

Solution: (A)

Let P ≡ (a, 0). Let the reflected ray makes an angle q with +ve direction of x-axis, then the incident ray makes angle (p – q) with positive direction of x-axis. The slope of the incident ray is

FIGURE 10.22

\

0−3 3 = tan(π − θ ) i.e., tan θ = (1) α −2 α −2 The slope of the reflected ray is 4−0 4 (2) = = tan θ i.e., tan θ = 6 −α 6 −α =

3. The image of a point with respect to y-axis: P(x, y) be any point and P′(x′, y′) its image after reflection in the y-axis, then x′ = – x and y′ = y (  O′ is the mid point of PP′)

From Eq. (1) and (2), we get 3 4 = ⇒ 18 − 3α = 4α − 8 α − 2 6 −α ⇒ 7α = 26 or α =

26 7

 26  \ The coordiantes of A are  , 0  .  7 

FIGURE 10.23

Coordinates and Straight Lines  10.23 4. The image of a point with respect to the origin: Let P(x, y) be any point and P′(x′, y′) be its image after reflection through the origin, then x′ = –x and y′ = –y (  O is the mid-point of PP′)

a1 x + b1 y + c1 a +b 2 1

2 1

=±

a2 x + b2 y + c2 a22 + b22

\

Info Box! Any point on a bisector is equidistant from the given lines.

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 a2x + b2 y + c2 = 0

and FIGURE 10.24

5. The image of a point with respect to the line y = x: Let P(x, y) be any point and P′(x′, y′) be its image after reflection in the line y = x, then, x′ = y and y′ = x (  O′ is the mid-point of PP′)

(1) (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. FIGURE 10.25

6. The image of a point with respect to the line y = x tanθ : Let P(x, y) be any point and P′(x′, y′) be its image after reflection in the line y = x tanθ, then,

SOLVED EXAMPLE 55. 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) (B)  (–3, 3) (C) (3, 3) (D)  (–3, –3) Solution: (D)

Reflection in the y-axis gives the new position as (–3, 2).

FIGURE 10.26



x′ = x cos 2θ + y sin 2θ y′ = x sin 2θ – y cos 2θ, (  O′ is the mid-point of PP′) \

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

When it moves towards the negative side of y-axis through 5 units, then the new position is (–3, 2 – 5) i.e., (–3, –3).

10.24  Chapter 10

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 (a1x + b1 y + c1) + k (a2x + b2 y + c2) = 0, where k is a parameter. The value of k can be obtained by using one more condition which the required line satisfies.

SOLVED EXAMPLES 56. If a, b, c are three terms of an A.P., then the line ax + by + c = 0 (A) has a fixed direction (B) always passes through a fixed point (C) always cuts intercepts on the axes such that their sum is zero (D)  forms a triangle with the axes whose area is constant. Solution: (B) Let a, b, c be pth, qth and rth terms of an A.P. whose first term is A and common difference is d. The given line is ax + by + c = 0 ⇒ [A + (p – 1) d] x + [A + (q – 1) d] y + [A + (r – 1) d] = 0 ⇒  A(x + y + 1) + d((p – 1) x + (q – 1) y + r – 1) = 0 ⇒  The given line passes through the point of intersection of lines x + y + 1 = 0 and ( p – 1) x + (q – 1)y + r – 1 = 0, which is a fixed point. 57. The line (p + 2q) x + (p – 3q)y = p – q for different values of p and q passes through the fixed point 2 2 3 5 (A)  ,  (B)  5,5 2 2      2 3 3 3 (C)  ,  (D)  5,5   5 5 Solution: (D) The equation of the given line can be re-written as p(x + y – 1) + q(2x – 3y + 1) = 0 which, clearly, passes through the point of intersection of the lines x + y – 1 = 0 (1) and  2x – 3y + 1 = 0 (2)

for different values of p and q. Solving (1) and (2), we get the coordinates of the point  2 3 of intersection as  ,  . 5 5 58. If a + b + g = 0, the line 3a x + b y + 2g = 0 passes through the fixed point 2   2 (A)  2,  (B)   3 ,2 3     2  (C)  −2,  (D)  none of these 3  Solution: (B) The given line is 3a x + b y + 2g = 0 ⇒ 3a x + b y + 2 (–a –b) = 0 ( a + b + g = 0) ⇒  a (3x – 2) + b (y – 2) = 0 ⇒  the given line passes through the point of inter2  section of the lines 3x – 2 = 0 and y – 2 = 0 i.e.,  , 2  , 3  for all values of a and b. ∴

QUICK TIPS

59. The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer, is (A) 2 (B)  0 (C) 4 (D)  1 Solution: (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. Solving these, only integer values of m are –1 and –2. \ m = –1, –2

STANDARD POINTS OF A TRIANGLE Centroid of a Triangle The point of intersection of the medians of the triangle is called the centroid of the triangle. The centroid divides the medians in the ratio 2 : 1 (2 from the vertex and 1 from the opposite side).

FIGURE 10.27

Coordinates and Straight Lines  10.25 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  

SOLVED EXAMPLE 60. If the vertices P, Q, R of a DPQR are rational points, which of the following points of the DPQR is (are) always rational point(s)? (A) centroid (B)  incentre (C) circumcentre (D)  orthocentre (A rational point is a point both of whose coordinates are rational numbers) Solution: (A) Let P ≡ (x1, y1), Q ≡ (x2, y2); R ≡ (x3, y3), where xi, yi (i = 1, 2, 3) are rational numbers. Now, the centroid of DPQR is  x1 + x2 + x3 y1 + y2 + y3  ,   3 3   which is rational point. Incentre, circumcentre and orthocentre depend on sides of the triangle which may not be rational even if vertices are so. For example, for P(0, 1) and Q(1, 0); PQ = 2.

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. 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  ax − bx2 + cx3 ay1 − by2 + cy3  I2 =  1 , a − b + c   a−b+c  ax + bx2 − cx3 ay1 + by2 − cy3  I3 =  1 , 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.

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

FIGURE 10.29

QUICK TIPS The circumcentre of a right angled triangle is the mid point of its hypotenuse.  The circumcentre of the triangle formed by (0, 0), (x , y ) 1 1 and (x2, y2) is 

 y2( x12 + y12 ) − y1( x22 + y22 ) x2( x12 + y12 ) − x1( x22 + y22 )  ,   2( x1y2 − x2 y1) 2( x2 y1 − x1y2 )   FIGURE 10.28

 ax1 + bx2 + cx3 ay1 + by2 + cy3   a+b+c , a + b + c   QUICK TIPS The incentre of the triangle formed by (0, 0), (a, 0) and (0, b) is   ab ab ,   2 2 2 2 a b a b a b + + + + + a + b  

WORKING RULE TO PROVE CONCURRENCY 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  Let D, E and F be the mid-points of the sides BC, CA and AB of the ∆ABC respectively. 

10.26  Chapter 10 Then, OD ⊥ BC, OE ⊥ AC, OF ⊥ AB.

Solving any two of these, we can get coordinates of O.  (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.  If angles A, B and C and vertices A(x , y ), B(x , y ) and 1 1 2 2 C(x3, y3) of a ∆ABC are given, then orthocentre of ∆ABC is given by

FIGURE 10.30

slope of OD × slope of BC = –1 slope of OE × slope of AC = –1 slope of OF × slope of AB = –1 Solving any two of the above equations, we get the circumcentre (x, y).  (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.  If angles A, B, C and vertices A(x , y ), B(x , y ) and C(x , 1 1 2 2 3 y3) of a ∆ABC are given, then its circumcentre is given by  x1 sin 2 A + x2 sin 2B + x3 sin 2C ,  sin 2 A + sin 2B + sin 2C  y1 sin 2 A + y2 sin 2B + y3 sin 2C   sin 2 A + sin 2B + sin 2C 

 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 

QUICK TIPS 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.  The orthocentre of the triangle with vertices (0, 0), (x , 1 y1) and (x2, y2) is 

  x1 x2 − y1 y2   x1 x2 + y1 y2   ( y1 − y2 )   , ( x1 − x2 )  .  x2 y1 − x1 y2   x1 y2 − x2 y1    The orthocentre (O), centroid (G) and circum centre (C) of any triangle lie in a straight line and G divides the join of O and C in the ratio 2 : 1.  In an equilateral triangle, orthocentre, centroid, circumcentre and incentre coincide. 

ORTHOCENTRE The orthocentre of a triangle is the point of intersection of altitudes. QUICK TIPS Let O be the orthocentre. Since AD ⊥ BC, BE ⊥ CA and CF ⊥ AB, then OA ⊥ BC OB ⊥ CA and OC ⊥ AB 

FIGURE 10.32

COORDINATES OF NINE POINT CIRCLE

FIGURE 10.31

If a circle passes through the feet of perpendiculars (i.e., D, E, F), midpoints of sides BC, CA, AB respectively (i.e., H, I, J) and the midpoints of the line joining the orthocentre O to the angular points A, B, C (i.e., K, L, M), thus the nine points D, E, F, H, I, J, K, L, M, all lie on a circle.

Coordinates and Straight Lines  10.27 This circle is known as nine point circle and its centre is called the nine-point centre.

(A) orthocentre (C) centroid

(B)  incentre (D)  none of these

Solution: (A)

I M P O R TA N T P O I N T S The orthocentre (O). Nine point Centre (N), Centroid (G) and Circumcentre (C) all lie in the same striaght line.  The Nine point centre bisects the join of Orthocentre (O) and Circumcentre (C)  The radius of Nine Point Circle is half the radius of Circumcirle. 

SOLVED EXAMPLE 61. If the equations of the sides of a triangle are x + y = 2, y = x and 3 y + x = 0, then which of the following is an exterior point of the triangle?

The lines y = x and 3 y + x = 0, are inclined at 45° and 150°, respectively, with the positive direction of x-axis. So, the angle between the two lines is an obtuse angle. Therefore, orthocentre lies outside the given triangle, whereas incentre and centroid lie within the triangle (In any triangle, the centroid and the incentre lie within the triangle).

10.28  Chapter 10

NCERT EXEMPLARS 1. A line cutting off intercept – 3 from the Y-axis and the 3 tangent at angle to the X-axis is , its equation is 5 (A) 5y – 3x + 15 = 0 (B)  3y – 5x + 15 = 0 (C) 5y – 3x – 15 = 0 (D)  None of these 2. Slope of a line which cuts off intercepts of equal lengths on the axes is (A) – 1 (B) 0 (C)  2 (D)  3 3. The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is (A) x – y = 5 (B)  x + y = 5 (C) x + y = 1 (D)  x – y = 1 4. The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 is (A) y – x + 1 = 0 (B)  y – x – 1 = 0 (C) y – x + 2 = 0 (D)  y – x – 2 = 0 5. The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a respectively, is b −a (B) 

2 2 (A) a − b

2

ab 2 2 b (C)  − a 2ab

2

2

(D)  None of these

6. If the line x + y = 1 passes through the points (2, – 3) and

NCERT EXEMPLARS

a

b

(4, – 5), then (a, b) is (A)  (1, 1) (C)  (1, – 1)

(B)  (– 1, 1) (D)  (– 1, – 1)

7. The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is 130 13 (A)  (B)  17 29 7 29 130 (C)  (D)  None of these 7 8. The equation of the lines which pass through the point (3, – 2) and are inclined at 60° to the line 3x + y = 1 is (A) y + 2 = 0, 3x − y − 2 − 3 3 = 0 (B) x − 2 = 0, 3x − y + 2 + 3 3 = 0 (C)  3x − y − 2 − 3 3 = 0 (D)  None of the above 9. The equations of the lines passing through the point (1, 0) and at a distance

3 from the origin, are 2

(A)  3x + y − 3 = 0, 3x − y − 3 = 0 (B)  3x + y + 3 = 0, 3x − y + 3 = 0 (C) x + 3 y − 3 = 0, x − 3 y − 3 = 0 (D)  None of these 10. The distance between the lines y = mx + c1 and y = mx + c2 is (A)  c1 − c2

−

c1 c2 (B) 

m + 1 1 + m2 (C)  c2 − c1 (D)  0 1 + m2 2

11. The coordinates of the foot of perpendiculars from the point (2, 3) on the line y = 3x + 4 is given by (A)  37 , −1 

 1 37  (B)   − 10 , 10  10 10     (C)  10 , −10  (D)   2 , − 1   37  3 3 

.12. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be (A) 2x + 3y = 12 (B)  3x + 2y = 12 (C) 4x – 3y = 6 (D)  5x – 2y = 10 13. Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is (A) y + 2 = x + 1 (B)  y + 2 = 3(x + 1) (C) y – 2 = 3 (x – 1) (D)  y – 2 = x – 1 14. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are (A) y = x, y + x = 1 (B)  y = x, x + y = 2 1 (C) 2y = x, y + x = (D)  y = 2x, y + 2x = 1 3 15. For specifying a straight line, how many geometrical parameters should be known ? (A) 1 (B) 2 (C) 4 (D) 3 16. The point (4, 1) undergoes the following two successive transformations (i) Reflection about the line y = x (ii) Translation through a distance 2 units along the positive X-axis. Then, the final coordinates of the point are (A)  (4, 3) (B)  (3, 4) 7 7  

(C)  (1, 4) (D)   2,2

Coordinates and Straight Lines  10.29 17. A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is (A)  (1, – 1) (B)  (1, 1) (C)  (0, 0) (D)  (0, 1) 18. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is 2 (A) 1 (B)  3 3 4 (C)  1 (D)  3

19. The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is (A)  1 : 2 (B)  3 : 7 (C)  2 : 3 (D)  2 : 5 20. One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is (A)  (– 1, – 1) (B)  (2, 2) (C)  (– 2, – 2) (D)  (2, – 2)

AN SW ER K EYS 1. (A) 11. (B)

2. (A) 12. (A)

3. (B) 13. (C)

4. (B) 14. (A)

5. (C) 15. (B)

6. (D) 16. (B)

7. (A) 17. (C)

8. (A) 18. (D)

9. (A) 19. (B)

10. (B) 20. (C)

C = – 3 and m = 3 5 ∴ Equation of the line is y = mx + c    y = 3 x − 3 5 ⇒ 5 y = 3 x − 15 ⇒ 5 y − 3 x + 15 = 0 1. Given that,

2. Let equation of line be x + y = 1 a a ⇒ x+y=a ⇒    y = – x + a ∴ Required slope = – 1 3. Since, line passes through the point (3, 2) and perpendicular to the line y = x. ∵ Slope  (m) = – 1 [since, line is perpendicular to the line y = x] ∴ Equation of line which passes through (3, 2) is y – 2 = – 1 (x – 3) y–2=–x+3 x+y=5 4. Given point is (1, 2) and slope of the required line is 1. ∵ x + y + 1 = 0 ⇒ − x − 1 ⇒ m1 = −11 −1 ∴ slope of the line = =1 −1 ∴ Equation of required line is y – 2 = 1 (x – 1) ⇒ y − 2 = x −1 ⇒ y − x −1 = 0

5. Since, intercepts on the axes are a, – b then equation of the line is x − y = 1. a b y x ⇒ = −1 b a bx ⇒ y= −b a So, the lope of this line i.e., m1 = b . a Also, for intercepts on the axes as b and – a, then equation of the line is x y − =1 b a y x a ⇒ = −1 ⇒ y = x − a a b b and slope of this line ie., m2 = a b b a b2 − a2 − b2 − a2 ∴ tan θ = a b = ab = ab 2 2ab 1+ • ba 6. Given, line is x + y = 1 a b Since, the points (2, – 3) and (4, – 5) lies on this line.

(i)

2 3 ∴ − = 1  a b and 4 − 5 = 1  a b

(ii) (iii)

HINTS AND EXPLANATIONS

HINTS AND EXPLANATIONS

10.30  Chapter 10 On multiplying by 2 in Eq. (ii) and then subtracting Eq. (ii) from Eq. (ii), we get 6 5 − + =1 b b −1 ⇒ =1 b ∴ b = −1 On putting b = – 1 in Eq. (ii), we get 2 + 3 =1 a 2 ⇒ = −2 ⇒ a = −1 a ∴ ( a, b ) = ( −1, −1)

HINTS AND EXPLANATIONS

7. Given equation of lines 2x – 3y + 5 = 0  (i) and 3x + 4y = 0  (ii) −4 y From Eq. (ii), put the value of x = in Eq. (i), we get 3  −4 y  2 3 5 0 − y + =   3  ⇒ − 8 y − 9 y + 15 = 0 15 ⇒ y= 17 From Eq. (ii), 3 x + 4·15 = 0 17 − 60 − 20 ⇒ x= = 17.3 17 So, the point of intersection is  −20 , 15  .  17 17  ∴ Required distance from the line 5x – 2y = 0 is, d=

20  15  −5 × − 2   17  17  25 + 4

−100 30 − 130 17 17 = = 9 17 29

8. So, the given point A is (3, – 2). So, the equation of line 3 x + y = 1. ⇒ y = − 3x + 1

⇒

⇒ m2 = 3 \ Equation of line passing through (3, – 2) is y + 2 = 3 ( x − 3) y − 3 x + 2 + 3 3 = 0 ⇒ 3 x − y − 2 − 3 3 = 0  [taking negative sign from Eq. (i)]

(i)

⇒ 3 − 3m2 = 3 + m2 ⇒ m2 = 0

∴ The equation of line is y + 2 = 0 ( x − 3) (ii) ⇒ y + 2 = 0 So, the required equation of lines are 3 x − y − 2 − 3 3 = 0 and y + 2 = 0. 9. Let slope of the line be m. ∵ Equation of line passing through (1, 0) is y – 0 = m(x – 1) ⇒ y − mx + m = 0 Since, the distance from origin is 3 . 2 3 0−0+m Then, = 2 1 + m2 ⇒

3 m = 2 1 + m2

3 m2 ⇒ = 4 1 + m2 ⇒ m2 = 3 ⇒ m=± 3 and the second equation of line is y = 3 ( x − 1) ⇒

3x + y − 3 = 0

y = mx + c1  and y = mx + c2  ∴ Distance between them is given by c −c d= 1 2 1 + m2

 m1 − m2  ∵ tan θ =  1 + m1m2  

 − 3 − m2  ⇒ tan 60° = ±     1 − 3m2 

⇒ 2 3 = 2m2

10. Given, equation of the lines are

∴ Slope, m1 = − 3 Let slope of the required line be m2. − 3 − m2 1 − 3m2

3 − 3m2 = − 3 − m2

⇒ 3 + 3m 2 = 4 m 2

 ∵ distance of a point p ( x1 , y1 ) from the line  ax + by1 + c  ax + by + c = 0 is d = 1  a2 + b2  

∴ tan θ =

⇒

(i)

 − 3 − m2  3 =    [taking positive sign]  1 − 3m2 

(i) (ii)

11. Given, equation of the line is y = 3x + 4  ∴ Slope of this line, m1 = 3 1 So, the slope of line OP is − .  3

(i)

[∵

OP ⊥ AB ]

Coordinates and Straight Lines  10.31 ⇒ y=x And equation of AC is ⇒ ⇒



⇒



Put x =



10 x + 1 = 0 ⇒ x = −

1 10

−1 in Eq. (i), we get 10 −3 −3 + 40 37 y= +4= = 10 10 10

So, the foot of perpendicular is  − 1 , 37   10 10 

12. Since, the coordinates of the middle point are P(3, 2). 1.0 + 1.a ∴ 3= 1+1 a ⇒ 3= ⇒a=6 2 Similarly, b=4 ∴ ⇒

x y Equation of the line is + = 1 6 4 2x + 3y = 12

y=–x+1 x+y–1=0

1− 0 ( x − 1) 0 −1

15. Equation of straight line are y = mx + c, parameter = 2  (i) x y (ii) + = 1, parameter = 2  a b y – y1 = m (x – x1), parameter = 2  (iii) and x cos w + y sin w = p, parameter = 2  (iv) It is clear that from Eqs. (i), (ii), (iii) and (iv), for specifying a straight line clearly two parameters should be known. 16. Let the reflection of A(4, 1) in y = x is B (h, k). Now, mid-point of AB is  4 + h , 1 + k  which lies on y = x. 2   2 i.e., 4 + h = 1 + k ⇒ h − k = −3 (i) 2 2 So, the slope of line y = x is 1. h−4 ∴ Slope of AB = k −1 h − 4   ⇒ 1·  = −1  k −1  ⇒ h–4=1–k ⇒ h + k = 5 (ii) and h – k = – 3 2h = 2  ⇒ h = 1 On putting h = 1 in Eq. (ii), we get k=4 So, the point is (1, 4). Hence, after translation the point is (1 + 2, 4) or (3, 4).

13. Since, the line passes through (1, 2) and parallel to the line y = 3x – 1. So, slope of the required line m = 3. [∵ slope of y = 3x − 1 is 3] Hence, the equation of line is y – 2 = 3 (x – 1) 14. Equation of OB is 1− 0 y−0 = ( x − 0) 1− 0

17. The given equation of lines are 4x + 3y + 10 = 0  (i) ⇒ 5x – 12y + 26 = 0  (ii) ⇒ 7x + 24y – 50 = 0  (iii) Let the point (h, k) which is equidistant from these lines. 4 h + 3k + 10 Distance from line (i) = 16 + 9 5h + 12k + 26 Distance from line (ii) = 25 + 44 7h + 24 k − 50 Distance from the line (iii) = 72 + 24 2

HINTS AND EXPLANATIONS

\ Equation of line OP is 1 y − 3 = − ( x − 2) 3 ⇒ 3y − 9 = −x + 2 (ii) ⇒ x + 3 y − 11 = 0  Using the value of y from Eq. (i) in Eq. (ii), we get x + 3(3x + 4) – 11 = 0 ⇒ x + 9 x + 12 − 11 = 0

y−0 =

10.32  Chapter 10 So, the point (h, k) is equidistant from lines (i), (ii) and (iii). 4 h + 3k + 10 5h − 12k + 26 7h + 24 k − 50 ∴ = = 16 + 9 25 + 144 49 + 576 ⇒

4 h + 3k + 10 5

=

5h − 12k + 26 13

=

7h + 24 k − 50

25 10 26 50 Clearly, if h = 0, k = 0, then = = =2 5 13 25 Hence, the required point is (0, 0). 18. Given line is y = 3 – 3x. Then, slope of the required line = 1 . 3 ∵  Equation of the required line is 1 y − 2 = ( x − 2) 3 ⇒ 3y − 6 = x − 2 ⇒ x − 3y + 4 = 0 For y-intercept, put x = 0, 0 – 3y + 4 = 0 4 ⇒ y= 3

HINTS AND EXPLANATIONS

19. Let point A(x1, y1) lies on the line 3x + 4y + 5 = 0, then 3x1 + 4y1 + 5 = 0

Now, perpendicular distance from A to the line 3x + 4y + 2 = 0 −5 − 2 −7 3 x1 + 4 y1 + 2 ⇒ = = 9 + 16 9 + 16 5 Let point B(x2, y2) lies on the line 3x + 4y – 5 = 0 i.e., 3x2 + 4y2 – 5 = 0. Now, perpendicular distance from B to the line 3x + 4y + 2 = 0,

3 x2 + 4 y2 + 2

+5 − 2

3 = 9 + 16 5 3 7 Hence, the required ratio is : i.e., 3 : 7. 5 5

9 + 16

=

20. Let ABC be the equilateral triangle with vertex A(h, k) Let the coordinates of D are (α, β).

We know that, 2 : 1 from the vertex A. ∴ 0 = 2α + h and 0 = 2 β + k 3 3 ⇒ 2α = − h and 2 β = −k Also, D(α, β) lies on the line x + y – 2 = 0. ∴ α + β − 2 = 0   AD ⊥ BC Since, the slope of line BC i.e., mBC = – 1 And slope of the line AG i.e., mAG = k − 0 = k h−0 h k  ⇒ ( −1)· = −1 h  ⇒ h = k  From Eqs. (i) and (iii) 2 α = – h and 2β = – h ∴ α =β From Eq. (ii), 2α – 2 = 0 ⇒ α = 1 if α = 1, then β = 1 So, the vertex A is (– 2, – 2).

(i)

(ii)

(iii)

Coordinates and Straight Lines  10.33

PRACTICE EXERCISES Single Option Correct Type

2. Through the point P(α, β), where aβ > 0 the straight x y line + = 1 is drawn so as to form with coordinate a b axes a triangle of area S. If ab > 0, then the least value of S is (A) aβ (B)  2aβ (C) 4aβ (D)  none of these 3. A line joining two points A (2, 0) and B (3, 1) is rotated about A in anti-clockwise direction through an angle 15°. If B goes to C in the new position, then the coordinates of C are   3 3 (A)  2,  (B)   2, −   2 2     1 3 (C)  2 + ,   2 2  

(D)  none of these

4. 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. The coordinates of the foot of the perpendicular from P on the bisector of the angle between them are  1   1  (A) 0, ( 4 + 5 3 )  or 0, ( 4 − 5 3 )  depend2 2     ing on which line the point P is taken  1  (B) 0, ( 4 + 5 3 )   2  1 (C) 0, ( 4 − 5 3 )   2  5 5 3  (D)  ,   2 2  5. A string of length 12 units is bent first into a square PQRS and then into a right-angled DPQT by keeping the side PQ of the square fixed and other is one more than its side. Then, the area of PQRS equals

3 (A) ar (DPQT) (B)  ⋅ ar ( ∆PQT ) 2 (C) 2 ⋅ ar (DPQT) (D)  none of these 6. 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 5 7 (A) 0 < β < (B)  0 0 and x + y < 2, then (A) P lies either inside the triangle OAB or in the third quadrant (B) P cannot be inside the triangle OAB (C) P lies inside the triangle OAB (D) none of these 22. Consider the equation y – y1 = m(x – x1). In this equation, if m and x1 are fixed and different lines are drawn for different values of y1, then, (A) the lines will pass through a single point (B) there will be one possible line only (C) there will be a set of parallel lines (D) none of these 23. D is a point on AC of the triangle with vertices A(2, 3), B(1, –3), C(–4, –7) and BD divides ABC into two triangles of equal area. The equation of the line drawn through B at right angles to BD is (A) y – 2x + 5 = 0 (B)  2y – x + 5 = 0 (C) y + 2x – 5 = 0 (D)  2y + x – 5 = 0 24. If two points A(a, 0) and B(–a, 0) are stationary and if ∠A – ∠B = θ in ∆ABC, the locus of C is (A) x2 + y2 + 2xy tan θ = a2 (B) x2 – y2 + 2xy tan θ = a2 (C) x2 + y2 + 2xy cot θ = a2 (D) x2 – y2 + 2xy cot θ = a2 25. The straight line y = x – 2 rotates about a point where it cuts the x-axis and becomes perpendicular to the straight line ax + by + c = 0. Then, its equation is (A) ax + by + 2a = 0 (B) ax – by – 2a = 0 (C) by + ay – 2b = 0 (D) ay – bx + 2b = 0 26. If the point P(a2, a) lies in the region corresponding to the acute angle between the lines 2y = x and 4y = x, then (A) a ∈ (2, 6) (B)  a ∈ (4, 6) (C) a ∈ (2, 4) (D)  none of these

Coordinates and Straight Lines  10.35

⋅

28. A light ray emerging from the point source placed at P(2, 3) is reflected at point ‘θ’ on the y-axis and then passes through the point R(5, 10). Coordinates of ‘Q’ are (A) (0, 3) (B)  (0, 2) (C) (0, 5) (D)  none of these 29. 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 (A) (C)

2 3 1 3

⋅ a 2 + a + 1 a 2 + a + 1

(B)  (D) 

2 3 1 3

a2 − a + 1 a2 − a + 1

30. Two points A and B are given. P is a moving point on one side of the line AB such that ∠PAB – ∠PBA is a positive constant 2θ. The locus of the point P is (A) x2 + y2 + 2xy cot 2θ = a2 (B) x2 + y2 – 2xy cot 2θ = a2 (C) x2 + y2 + 2xy tan 2θ = a2 (D) x2 – y2 + 2xy cot 2θ = a2. 31. The four points A( p, 0), B(q, 0), C(r, 0) and D(s, 0) are such that p, q are the roots of the equation ax2 + 2hx + b = 0 and r, s are those of equation a′x2 + 2h′x + b′ = 0. If the sum of the ratios in which C and D divide AB is zero, then (A) ab′ + a′ b = 2hh′ (B) ab′ + a′b = hh′ (C) ab′ – a′b = 2hh′ (D)  none of these 32. The coordinates of a point P on the line 3x + 2y + 10 = 0 such that | PA – PB | is maximum where A is (4, 2) and B is (2, 4), are (A) (22, 28) (B)  (22, –28) (C) (–22, 28) (D)  (–22, –28)

33. A line through A(–5, –4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0 and x – y – 5 = 0 at the point B, C and D, 2

2

2

15   10   6  respectively. If   +  =  , the equa AB   AC   AD  tion of the line is (A) 2x + 3y + 22 = 0 (B)  2x – 3y + 22 = 0 (C) 3x + 2y + 22 = 0 (D)  3x – 2y + 22 = 0 34. A(0, 0), B(2, 1) and C(3, 0) are the vertices of a ∆ABC and BD is its altitude. If the line through D parallel to the side AB intersects the side BC at a point K, then the product of the areas of the triangles ABC and BDK is 1 (A) 1 (B)  2 1 (C) (D)  none of these 4 35. A line cuts the x-axis at A (7, 0) and y-axis at B (0, – 5). A variable line PQ is drawn ⊥ to AB cutting the x-axis in P and the y-axis in Q. If AQ and BP intersect at R, then the locus of R is (A) x(x – 7) + y(y + 5) = 0 (B) x(x – 7) – y(y + 5) = 0 (C) x(x + 7) + y(y – 5) = 0 (D) none of these 36. The point (2, 3) undergoes the following three transformations successively (i)  reflection about the line y = x (ii) translation through a distance 2 units along the positive direction of y-axis (iii) rotation through an angle of 45º about the origin in the anti-clockwise direction. The final coordinates of the point are 7   1 (A)  ,  2 2 

7   1 , (B)  −  2 2 

7   1 (C)  ,−  2  2

(D)  none of these

37. Lines L1 = ax + by + c = 0 and L2 = lx + my + n = 0 intersect at the point P and make an angle θ with each other. The equation of line L different from L2 which passes through P and makes the same angle θ with L1 is (A) 2 (al + bm) (ax + by + c) – (a2 + b2) (lx + my + n) =0 (B) 2 (al + bm) (ax + by + c) + (a2 + b2) (lx + my + n) =0 (C) 2 (a2 + b2) (ax + by + c) – (al + bm) (lx + my + n) =0 (D) none of these

PRACTICE EXERCISES

27. The point (4, 1) undergoes the following three successive transformations (A) Reflection about the line y = x – 1 (B) Translation through a distance 1 unit along the positive x-axis π (C) Rotation through an angle about the origin in 4 the anti-clockwise direction. Then, the coordinates of the final point are 7 7 (A) (4, 3) (B)  2, 2   (C) (0, 3 2 ) (D)  (3, 4)

10.36  Chapter 10 38. The equations of the perpendicular bisector of the sides AB and AC of a ∆ ABC are x – y + 5 = 0 and x + 2y = 0, respectively. If the point A is (1, –2) then the equation of the line BC is (A) 14x + 23y = 40 (B)  14x – 23y = 40 (C) 23x + 14y = 40 (D)  23x – 14y = 40 39. The equation of a family of lines is given by (2 + 3t) x + (1 – 2t) y + 4 = 0, where t is the parameter. The equation of a straight line, belonging to this family, at the maximum distance from the point (2, 3) is (A) 21x + 14y = 0 (B)  21x – 14y = 0 (C) 14x – 21y = 0 (D)  none of these 40. ABCD is a square whose vertices A, B, C and D are (0, 0), (2, 0), (2, 2) and (0, 2), respectively. This square is rotated in the X-Y plane with an angle of 30º in anti-clockwise direction about an axis passing through the vertex A. The equation of the diagonal BD of this rotated square is (A) 3 x + (1 − 3 ) y = 3 (B) (1 + 3 ) x − (1 − 2 ) = 2

PRACTICE EXERCISES

(C) ( 2 − 3 ) x + y = 2( 3 − 1) (D) none of these 41. The equations of the straight lines passing through (– 2, –7) and cutting an intercept of length three units between the straight lines 4x + 3y = 12 and 4x + 3y = 3 are 7 (A) x + 2 = 0, y + 7 = ( x + 2) 24 7 (B) x − 2 = 0, y + 7 = − ( x + 2) 24 (C) x + 2 = 0, y + 7 = − 7 ( x + 2) 24 7 (D) x + 2 = 0, y + 7 = − ( x + 2) 12 42. The coordinates of the point which is at unit distance from the lines L1 ≡ 3x – 4y + 1 = 0 and L2 ≡ 8x + 6y + 1 = 0 and lies below L1 and above L2 are 1 6 6 1 (A)  ,  (B)   ,−  5 10   5 10   6 1 6 1 (C)  ,  (D)   ,−  5 5 5 5 43. The vertices of a triangle are A(x1, x1 tan α), B(x2, x2 tan β) and C(x3, x3 tan γ). If the circumcentre of triangle ABC coincides with the origin and H(a, b) be its a orthocentre then = b

cos α + cos β + cos γ (A) cos α ⋅ cos β ⋅ cos γ sin α + sin β + sin γ (B) sin α ⋅ sin β ⋅ sin γ (C)

tan α + tan β + tan γ tan α ⋅ tan β ⋅ tan γ

cos α + cos β + cos γ (D) sin α + sin β + sin γ 44. OX and OY are two coordinate axes. On OY is taken a fixed point P and on OX any point Q. On PQ an equilateral triangle is described, its vertex R being on the side of PQ away from O, then the locus of R will be (A) straight line (B)  circle (C) ellipse (D)  parabola 45. If the vertices of a variable triangle are (3, 4), (5 cos θ, 5 sinθ) and (5 sinθ, –5 cosθ), then the locus of its orthocentre is (A) (x + y – 1)2 + (x – y – 7)2 = 100 (B) (x + y – 7)2 + (x – y + 1)2 = 100 (C) (x + y – 7)2 + (x – y – 1)2 = 100 (D) (x + y + 7)2 + (x + y – 1)2 = 100 46. If a right-angled isosceles triangle right-angled at origin has 3x + 4y = 6 as its base, then the area of the triangle is 11 (A) 7 (B)  25 12 36 (C) (D)  25 25 47. The line x + y = 1 meets x-axis at A and y-axis at B . P is the mid-point of AB . P1 is the foot of the perpendicular from P to OA; M1 is that from P1 to OP; P2 is that from M1 to OA and so on. If Pn denotes the foot of the nth perpendicular on OA from Mn–1, then OPn is equal to 1 1 (A) n (B)  2n−1 2 1 (D)  none of these 2n− 2 48. The line x + y = a meets x-axis at A. A triangle AMN is inscribed in the triangle OAB, O being the origin with right angle at N; M and N lie respectively on OB and 3 AB. If area of ∆AMN is of the area of triangle OAB, 8 AN then is equal to BN 1 (A) 3 (B)  3 (C)

2 (C) 2 (D)  3

Coordinates and Straight Lines  10.37

Previous Year's Questions



(A) isosceles and right angled (B) isosceles but not right angled (C) right angled but not isosceles (D) neither right angled nor isosceles

50. The equation of the directrix of the parabola y2 + 4y + 4x + 2 = 0 is:[2002] (A) x = -1 (B)  x=1 3 3 (C) x =− (D)  x= 2 2 51. The incentre of the triangle with vertices (1, 3 ), (0, 0) and (2, 0) is:[2002] 2 1  (A) 1, 3  (B)  3,  3   2   1  (C)  2 , 3  (D)  1,  3  3 2  52. Three straight lines 2x + 11y – 5 = 0, 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0:[2002] (A) form a triangle (B) are only concurrent (C) are concurrent with one line bisecting the angle between the other two (D) none of the above 53. A straight line through the point (2, 2) intersects the lines 3 x + y = 0 and 3 x − y = 0 at the points A and B. The equation to the line AB so that the triangle OAB is equilateral, is:[2002] (A) x – 2 = 0 (B)  y–2=0 (C) x + y – 4 = 0 (D)  none of these 54. If the equation of the locus of a point equidistant from the points (a1, b1) and (a2, b2) is (a1 – a2)x + (b1 – b2)y + c = 0, then the value of ‘c’ is[2003] (A) 1 ( a22 + b22 − a12 − b12 ) 2

55. Locus of centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, -b cos t) and (1, 0), where t is a parameter, is[2003] (A) (3x – 1)2 + (3y)2 = a2 – b2 (B) (3x – 1)2 + (3y)2 = a2 + b2 (C) (3x + 1)2 + (3y)2 = a2 + b2 (D) (3x + 1)2 + (3y)2 = a2 – b2 56. 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[2004] (A) 2x + 3y = 9 (B)  2x – 3y = 7 (C) 3x + 2y = 5 (D)  3x – 2y = 3 57. 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[2004] x y x y (A) + = −1 and + = −1 2 3 −2 1 x y x y (B) − = −1 and + = −1 2 3 −2 1 x y x y (C) + = 1 and + = 1 2 3 2 1 x y x y (D) − = 1 and + =1 2 3 −2 1 58. 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[2004] (A) 1 (B)  –1 (C) 2 (D)  –2 59. If one of the lines given by 6x2 – xy + 4cy2 = 0 is 3x + 4y = 0, then c equals[2004] (A) 1 (B)  –1 (C) 3 (D)  –3 60. Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid-point of PQ is[2005] (A) y2 – 4x + 2 = 0 (B)  y2 + 4x + 2 = 0 2 (C) x + 4y + 2 = 0 (D)  x2 – 4y + 2 = 0

(C) 1 ( a12 + a22 − b12 − b22 ) 2

61. The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a = 0, where (a, b) ≠ (0, 0) is[2005] 3 (A) below the x-axis at a distance of from it 2

(D) a12 + b12 − a22 − b22

(B) below the x-axis at a distance of

(B) a12 + a22 − b12 − b22

2 from it 3

PRACTICE EXERCISES

49. A triangle with vertices (4, 0), (–1, –1), (3, 5) is:[2002]

10.38  Chapter 10 3 (C) above the x-axis at a distance of from it 2 2 (D) above the x-axis at a distance of from it 3 62. If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex are (–1, 2) and (3, 2) then the centroid of the triangle is [2005] 7  −1 7  (A)  −1,  (B)   ,  3   3 3 7 1 7 (C) 1,  (D)   ,   3 3 3

69. 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 [2008] (A) 1 (B)  2 (C) –2 (D)  –4 x y + = 1 passes through the point 5 b (13, 32) and the line K which is parallel to L has the x y equation + = 1. Then, the distance between L and 3 c K is [2010]

63. A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is  [2006] (A) x + y = 7 (B)  3x - 4y + 7 = 0 (C) 4x + 3y = 24 (D)  3x + 4y = 25

70. The line L given by

64. The locus of the vertices of the family of parabolas a3 x 2 a 2 x y= + − 2a is [2006] 3 2 3 105 (A) xy = (B)  xy = 4 64

17 (A) 17 (B)  15

(C) xy =

35 64 (D)  xy = 16 105 x , 2 [2006]

65. If (a, a2) falls inside the angle made by the lines y =

PRACTICE EXERCISES

68. If one of the lines of my2 + (1 – m2)xy – mx2 = 0 is a bisector of the angle between the lines x = 0 and y = 0, then m is [2007] 1 (A) − (B)  –2 2 (C) 1 (D)  2

x > 0 and y = 3x, x > 0, then a belongs to 1 (A)  0,  (B)  (3,∞)  2 1 1  (C)  , 3  (D)   −3, −   2  2

66. Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which ‘k’ can take is given by [2007] (A) {1, 3} (B)  {0, 2} (C) {–1, 3} (D)  {–3, –2} 67. Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3 ) be three points. The equation of the bisector of the angle PQR  [2007] (A) 3 x + y = 0 (C)

3 x + y = 0 2

x+ (B) 

3 y=0 2

x + 3y = 0 (D) 

(C)

23 (D)  15 17

23

71. The lines L1: y – x = 0 and L2: 2x + y = 0 intersect the line L3: y + 2 = 0 at two respective points P and Q. The bisector of the acute angle between L1 and L2 intersect L3 at R. [2011] Statement - 1 : The ratio PR : RQ equals 2 2 : 5 . Statement - 2 : In any triangle, bisector of an angle divides the triangle into two similar triangles. (A) Statement - 1 is true, Statement-2 is true; Statement - 2 is not a correct explanation for Statement -1 (B) Statement - 1 is true, Statement- 2 is false. (C) Statement - 1 is false, Statement- 2 is true. (D) Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement -1 72. Equation of the ellipse which passes through the point (– 3, 1), whose axes are the coordinate axes and has 2 eccentricity is [2011] 5 (A) 5x2 + 3y2 – 48 = 0 (C) 5x2 + 3y2 – 32 = 0

(B)  3x2 + 5y2 – 15 = 0 (D)  3x2 + 5y2 – 32 = 0

73. If the line 2x + y = k passes through the point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2, then k equals [2012]

Coordinates and Straight Lines  10.39

74. A line is drawn through the point (1, 2) to meet the coordinate axes at points P and Q respectively such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is [2012] 1 (A) − (B)  –4 4 1 (C) –2 (D)  − 2 75. A ray of light along x + 3 y = 3 gets reflected upon reaching x-axis, the equation of the reflected ray is  [2013] (A) 3 y = x − 3

(B)  y = 3x − 3

(C) 3 y = x − 1

(D)  y = x+ 3

76. The abscissa of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is [2013] (A) 2 − 2 (B)  1+ 2 (C) 1 − 2 (D)  2+ 2 77. Let a, b, c and d be non-zero numbers. If the point of intersection of the line 4ax + 2ay + c = 0 with the line 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then [2014] (A) 2bc – 3ad = 0 (B)  2bc + 3ad = 0 (C) 3bc – 2ad = 0 (D)  3bc + 2ad = 0 78. 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 [2014] (A) 4x – 7y – 11 = 0 (B)  2x + 9y + 7 = 0 (C) 4x + 7y + 3 = 0 (D)  2x – 9y – 11 = 0 79. The number of points, having both co-ordinates as integers, which lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0), is: [2015] (A) 861 (B)  820 (C) 780 (D)  901

80. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k (x – 2y + 3) = 0, k ∈ R, is a: [2015] (A) straight line parallel to y-axis. (B) circle of radius 2. (C) circle of radius 3. (D) straight line parallel to x-axis. 81. Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then which one of the following is a vertex of this rhombus? [2016] 10 7 (A)  − , −  3  3

(B)  (–3, –9)

1 8 (C) (–3, –8) (D)   ,−  3 3 82. Let k be an integer such that the triangle with vertices (k, – 3k), (5, k) and (– k, 2) has area 28 sq. units. Then the orthocentere of this triangle is at the point [2017] 3⎞ ⎛ 3⎞ ⎛ 1, − ⎟ (A)  ⎜1, ⎟ (B)  ⎜ ⎝ 4⎠ ⎝ 4⎠ 1⎞ ⎛ 1⎞ ⎛ (C)  ⎜ 2, ⎟ (D)  ⎜⎝ 2, − ⎟⎠ ⎝ 2⎠ 2 83. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is [2018] (A) 3x + 2y = 6 (B)  2x + 3y = xy (C) 3x + 2y = xy (D) 3x + 2y = 6xy 84. Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant? [2019] (A) first (B)  second (C) fourth

(D) third

85. If the area of an equilateral triangle inscribed in the circle, x2 + y2 + 10x + 12y + c = 0 is 27 3 sq. units then c is equal to: [2019] (A) 13 (C) 25

(B) 20 (D) –25

PRACTICE EXERCISES

29 (A) (B)  5 5 11 (C) 6 (D)  5

10.40  Chapter 10

ANSWER K EYS Single Option Correct Type 1. (B) 11. (B) 21. (A) 31. (A) 41. (C)

2. (B) 12. (A) 22. (B) 32. (C) 42. (B)

3. (C) 13. (A) 23. (A) 33. (A) 43. (D)

4. (B) 14. (C) 24. (D) 34. (B) 44. (A)

5. (B) 15. (A) 25. (D) 35. (A) 45. (B)

6. (C) 16. (D) 26. (C) 36. (B) 46. (C)

7. (B) 17. (A) 27.  (C) 37. (A) 47. (A)

8. (C) 18. (A) 28. (C) 38. (A) 48. (A)

9. (A) 19. (A) 29. (B) 39. (D)

10. (B) 20. (B) 30. (D) 40. (C)

52. (C) 62. (C) 72. (D) 82. (C)

53. (B) 63. (C) 73. (C) 83. (C)

54. (A) 64. (A) 74. (C) 84. (B)

55. (B) 65. (C) 75. (A) 85. (C)

56. (A) 66. (C) 76. (A)

57. (D) 58. (C) 67. (A) 68. (C) 77. (C) 78. (B)

Previous Years’ Questions

PRACTICE EXERCISES

49. (A) 59. (D) 69. (D) 79.  (C)

50. (D) 60. (A) 70. (C) 80. (B)

51. (D) 61. (A) 71. (B) 81. (D)

Coordinates and Straight Lines  10.41

HINTS AND EXPLANATIONS Single Option Correct Type 1. The point (3, 0) does not lie on the diagonal x = 2y. Let the equation of a side through the vertex (3, 0) be y – 0 = m(x – 3) Since the angle between a side and a diagonal of a square is π , we have 4 m − 1/ 2 π 2m − 1 = ± tan = 4 1 + m(1/ 2) 2 + m ⋅

⇒ m = 3, – 1/3 Thus, the equation of a side through (3, 0) is y = 3(x – 3) or  1 y =  −  ( x − 3) and the one nearer to the origin is  3 3y + x – 3 = 0

∴ Slope of line AC = tan 60° = 3. Now, line AC makes an angle of 60º with positive direction of x-axis and AC = AB = (3 − 2) 2 + (1 − 0) 2 = 2 ∴ Coordinates of C are ( 2 + 2 cos 60°, 0 + 2 sin 60°)

2. The equation of the given line is x y + = 1 (1) a b

 1 3 , i.e.,  2 +   2 2   4. Equation of two lines are

y = 3 x + 2, if x ≥ 0

This line cuts x-axis and y-axis at A (a, 0) and B  (0, b) respectively. Since area of DOAB = S

(Given)

1 ab = S or ab = 2S (∵ ab > 0) (2) 2

∴

Since the line (1) passes through the point P (α, β) ∴

α β α aβ + = 1 or + = 1   [Using (2)] a b a 2S

or, a2β – 2aS + 2aS = 0. Since a is real, ∴ 4S2 – 8abS ≥ 0 1   or, 4 S 2 ≥ 8aβ S or S ≥ 2aβ ∵ S = ab > 0 as ab > 0  2   Hence, the least value of S = 2aβ. 3. Slope of line AB = ∴

0 −1 = 1 = tan 45° 2−3  ∠BAX = 45º.

Given, ∠CAB = 15º. ∴ ∠CAX = 60º.

Also, y-axis is the bisector of the angle between the two lines. P1, P2 are two points on these lines, at a distance 5 units from A. Q is the foot of the ⊥ from P1 and P2 on the bisector (y-axis). Then, the coordinates of Q are (0, 2 + 5 cos 30º)  5 3  1  =  0, 2 +  =  0, ( 4 + 5 3 )   2   2   5. Side of square = 3 unit ⇒  ar (PQRS) = (3)2 = 9 square unit One side of DPQT is the side PQ of the square i.e., 3 units The other is one more than its side, i.e., (3 + 1) = 4 units 1 ⇒ ar ( ∆PQT ) = (3)( 4) = 6 square unit 2

HINTS AND EXPLANATIONS

and, y = − 3 x + 2, if x ≤ 0 Clearly, y ≥ 2.

10.42  Chapter 10 3 ∴ ar ( PQRS ) = {ar( ∆PQT )} 2 6. Clearly, point (0, β) lies on y-axis. Drawing the graph of the three straight lines, we see that  7 Q ≡  0,  and  2

9. Let P ≡ (α, 0). Let the reflected ray makes an angle θ with positive direction of x-axis, then the incident ray makes angle (π – θ) with positive direction of x-axis. The slope of the incident ray is

 5 P ≡  0,  .  3

Therefore, the point (0, β) lies on or inside ∆ABC, when 5 7 ≤ β ≤ 3 2

=

0−3 3 (1) = tan(π − θ ), i.e., tan θ = α −2 α −2

The slope of the reflected ray is 4−0 4 (2) = tan θ , i.e., tan θ = 6 −α 6 −α From (1) and (2), we get =

3 4 = α − 2 6 −α

⇒ 18 − 3α = 4α − 8

⇒ 7α = 26 or α =

HINTS AND EXPLANATIONS

7. 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.

26 7

26 ∴ The coordiantes of P are  , 0  .  7  10. Let the coordinates of A and B be (a, 0) and (0, b), respectively. Then, equation of line AB is

x a

+

y = 1. b

8. 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) ⇒ y = − 3 ( x − 5) or 3x + y − 5 3 = 0

Since, it passes through the point P (4, 2) 4 2 ∴ + = 1. (1) a b a b Now, centre of the circumcircle of ∆OAB =  ,  . 2 2 So, equation (1) can be written in the form 2 1 =1 + a/2 b/2 ∴ locus of circumcentre is

Coordinates and Straight Lines  10.43

11. Clearly, the point (2 cosθ, 2 sinθ) lie on the circle

x2 + y2 = 4. The two lines represented by the equation y = |x – 2| are y = x – 2 and y = 2 – x. π π From the figure, θ can be vary from − to . 2 2 12. The lines y = x and 3 y + x = 0 are inclined at 45º and 150º, respectively, with the positive direction of x-axis. So, the angle between the two lines is an obtuse angle. Therefore, orthocentre lies outside the given triangle, whereas incentre and centroid lie within the triangle (In any triangle, the centroid and the incentre lie within the triangle). 13. Equation of a straight line passing through the point P (α, β) and making an angle θ with positive direction of x-axis is x −α y − β = = r (say) cosθ sin θ Coordinates of any point on this line are (α + r cosθ, β + r sinθ) If it lies on the line ax + by + c = 0, then a (α + r cosθ) + b (β + r sinθ) + c = 0 ⇒ r=−

Equation of the line PQ is x y   + = 1 (1) h k Given, BP ⋅ CQ = AB2 ⇒ (h – a) (k – a) = a2 ⇒ hk – ak – ah + a2 = a2 or, ak + ha = hk a a or, + = 1 (2) h k From (2), it follows that line (1), i.e., PQ passes through the fixed point (a, a). x2 x3 y2 y3 = r and = = r 15. Let = x1 x2 y1 y2 ⇒ x2 = x1r, x3 = x1r2, y2 = y1r and y3 = y1r2. We have, x1 ∆ = x2 x3

y1 1 x1 y2 1 = x1r y3 1 x1r 2

y1 1 x1 y1r 1 = 0 y1r 2 1 0

y1 1 0 1− r 0 1− r

[Applying R3 → R3 – rR2 and R2 → R2 – rR1] = 0 ( R2 and R3 are identical) Thus, (x1, y1), (x2, y2), (x3, y3) lie on a straight line. 16. The sides are, y = 3 ( x − 1) + 2 and

y = − 3x

aα + bβ + c a cosθ + b sin θ

Thus, PQ = r = −

aα + bβ + c a cosθ + b sin θ

14. We take A as the origin and AB and AC as x-axis and y-axis, respectively. Let AP = h, AQ = k.

The two lines are at an angle of 60° to each other. Now, any line parallel to obtuse angle bisector will make equilateral triangle with these lines as its two sides. 17. We have, d(x,y) = Max. {|x|, |y|} = k If |x| > |y|, then k = |x| ⇒ x = ±k If |y| > |x|, then k = |y| ⇒ y = ±k Hence, the locus represents a straight line. 18. Area of the triangle a 1 = b 2 c

x 1 a 1 y 1 = b−a 2 z 1 c−b

x 1 y−x 0 z−y 0

HINTS AND EXPLANATIONS

2 1 + = 1 or 2 x −1 + y −1 = 1. x y

10.44  Chapter 10 [Applying R2 → R2 – R1 and R3 → R3 – R2] 1 = [(b − a)( z − y ) − (c − b)( y − x )] 2 1 ∴ = d ( z − 2 y + x ) [ b – a = c – b = d] 2 1 = d ( xr 2 − 2 xr + x ) [∴ y = xr and z = xr2] 2 1 = dx( r − 1) 2 , which is independent of b. 2 19. Let ∠BAO = α.

HINTS AND EXPLANATIONS

Then, BC = 2 sin  α = CO and CA = 2 cosα. Therefore, the coordinates of A and B are

A ≡ (2 cos α – 2 sin α, 0) and, B ≡ (–2 sin α, 2 sinα). If P(x, y) is the mid point of AB, then 2x = 2 cos α – 4 sin α and 2y = 2 sin α ⇒ x = cos α – 2 sin α and y = sin α ⇒ cos α = x + 2y and sin α = y Squaring and adding, we get (x + 2y)2 + y2 = 1, which is the required locus. 20. It is clear that the lines lie on opposite sides of the origin O. Let the equation of any line through O be x y . If OP = r1 and OQ = r2 then the coordi= cosθ sin θ nates of P are (r1 cosθ, r1 sinθ) and that of Q are (–r2 cosθ, – r2sinθ). Since P lies on 4x + 2y = 9, 2r1 (2 cosθ + sinθ) = 9 and Q lies on 2x + y + 6 = 0, – r2(2 cosθ + sinθ) = –6 so that r1 9 3 = = and the required ratio is thus 3 : 4. r2 12 4 21. Since, xy > 0, therefore P lies either in the first quadrant or in the third quadrant. The inequality x + y < 2 represents all the points below the line x + y = 2. Therefore, xy > 0 and x + y < 2 imply that P lies either inside DOAB or in the third quadrant.

22. Since, y – y1 = m(x – x1), where m and x1 are fixed. Therefore, the system represents a set of parallel lines of slope m. Since all these parallel lines have fixed x coordinate (abscissa) ∴ there will only be one possible line. 23. Since the line BD divides the triangle into two parts of equal area, BD is a median and D is (–1, –2). Slope of 1 BD = − 2

So, the required line is y + 3 = 2(x – 1) ⇒  y – 2x + 5 = 0 k −0 24. Slope of line AC = mAC = = tan α h−a Slope of line BC = mBC = Since, ∠A = π – α

k −0 = tan β h+a

{from figure} k ∴ tan A = tan (π – α) = – tan α = (1) h−a

Coordinates and Straight Lines  10.45 Similarly, ∠B = β

k (2) h+a Also, A – B = θ {given} ∴ tan (A – B) = tan θ ∴ tan B = tan β =

⇒

tan A − tan B = tan θ 1 + tan A tan B

 −k   k   −  ⇒  h − a   h + a  = tan θ {using (1) and (2)}  −k   k  1+     h − a  h + a  − k ( h + a) − k ( h − a) −2hk = 2 h − k 2 − a2 ( h2 − a 2 ) + ( −k 2 ) ⇒ h2 – k2 + 2hk cot θ = a2 ∴ locus is x2 – y2 + 2xy cot θ = a2 ⇒ tan θ =

25. Slope of the line in the new position is b , since it is ⊥ to the a line ax + by + c = 0 and it cuts the x-axis at (2, 0). Hence, b the required line passes through (2, 0) and its slope is . a Therefore, its equation is

After translation through a distance 1 unit along the positive x-axis of the point. The coordinates of the final point are (0, 3 2 ) . 28. If P1 be the reflection of P in y-axis then P1 ≡ (– 2, 3) Equation of line P1R is 10 − 3 ( y − 3) = ( x + 2) 5+2

29. Let PQ = QR = RP = r and, ∠ PQX = θ, then b y − 0 = ( x − 2) a ⇒ ay = bx – 2b ⇒ ay – bx + 2b = 0 a2 a2 26. We have, a − > 0 and a − < 0 4 2

⇒ 0 < a < 4, a ∈ (– ∞, 0) ∪ (2, ∞) ⇒ a ∈ (2, 4) 27. If (α, β) be the image of (4, 1) w.r.t. y = x – 1 then (α, β) = (2, 3), say point Q.



∠RQX =

π +θ . 3

Given, PL = a, RN = 1. Now, a = PL = r sin θ

π  and, 1 = RN = r sin  + θ  3 

π π ⇒ 1 = r  sin cosθ + cos sin θ  3 3  

HINTS AND EXPLANATIONS

⇒  y = x + 5. It meets y-axis at (0, 5) ⇒ Q ≡ (0, 5)

10.46  Chapter 10   3 1 ⇒ r  cosθ + sin θ  = 1  2  2    3 a2 1 a  a  ⇒ r  1 − 2 + ⋅  = 1 ∵sin θ =   2  r 2 r r    ⇒ or

a 3 2 r − a2 = 1 − 2 2 r 2 − a2 =

⇒ r 2 = ∴ r =

( 2 − a) 2 4 + a 2 − 4 a = 3 3

4 + a2 − 4a 4 + 4a2 − 4a + a2 = 3 3 2 3

a − a + 1. 2

30. Let P (x, y) be the moving point whose locus is required. Let A ≡ (a, 0) and B (– a, 0).

b 2h and pq = (1) a a Also, r, s are the roots of the equation a′ x2 + 2h′ x + b′ = 0 2h′ b′ and rs = (2) ∴ r + s = − a′ a′ Let C divides AB in the ratio α : 1. αq + p p−r ⇒ α= Then, r = α +1 r−q ∴

p+q=−

Let D divides AB in the ratio β : 1. βq + p p−s Then, s = ⇒ β= β +1 s−q Given, α + β = 0 p−r p−s + ⇒ = 0 ⇒ (p + q) (r + s) – 2pq – 2rs = 0 r−q s−q  2h   2h′  2b 2b′ ⇒  −   − = 0 (Using (1) and (2)) − − a′  a   a′  a ⇒ ab′ + a′b = 2hh′.

HINTS AND EXPLANATIONS

32. Let P ≡ (x1, y1) and ∠APB = θ

Let ∠PAB = α and ∠PBA = β. ∴ In DPCA, tan α = and in DPBC, tan β =

PC y = CA a − x PC y = BC a + x

Given, ∠PAB – ∠PBA = 2θ (constant) ⇒ α – β = 2θ ⇒ tan (α – β) = tan 2θ ⇒

tan α − tan β = tan 2θ 1 + tan α tan β

y y − a − x a + x = tan 2θ ⇒ y y 1+ ⋅ a−x a+x ⇒

2 xy = tan 2θ a2 − x 2 + y 2

⇒ x2 – y2 + 2xy cot 2θ = a2, which is the required locus. 31. Since p, q are the roots of the equation ax2 + 2hx + b = 0

(PA) 2 + ( PB ) 2 − ( AB ) 2 2 PA ⋅ PB Since cos θ ≤ 1 (PA) 2 + ( PB ) 2 − ( AB ) 2 ⇒ ≤1 2 PA ⋅ PB Then, cosθ =

⇒ (PA)2 + (PB)2 – (AB)2 ≤ 2 PA ⋅ PB ⇒ (PA – PB)2 ≤ (AB)2 2 2 ⇒ | PA – PB | ≤ | AB | = ( 4 − 2) + ( 2 − 4) ⇒ | PA – PB | ≤ 2 2 . ⇒ Maximum value of | PA – PB | is 2 2 when θ = 0 i.e., P lies on the line AB as well as on the given line.  equation of AB is 4−2 y − 2 = (x – 4) or x + y = 6  (1) 2−4 and given line is 3x + 2y + 10 = 0 (2) Solving (1) and (2), we get P ≡ (–22, 28). 33. Suppose, the required line has slope tanθ. Its equation is x+5 y+4 = = r (1) cosθ sin θ Any point on this line has coordinates (–5 + r cos θ, –4 + r sin θ) Its distance from (–5, –4) is r.

Coordinates and Straight Lines  10.47

2

2

=

Hence, area of ∆ABC × area of ∆BDK = (2) (3) (4)

35. Equation of the line AB is x y + =1 7 −5 ⇒ 5x – 7y – 35 = 0. The equation of the variable line PQ perpendicular to AB is 7x + 5y + k = 0.  k  k ⇒ Coordinates of P are  − , 0  and that of Q are  0, −  .  7  5  x y + = 1 (1) 7 −k 5 x y and, equation of the line BP is + = 1 (2) k −5 − 7 Now, locus of R, the point of intersection of AQ and BP can be obtained from (1) and (2) by eliminating k from their equations. Equation (1) can be rewritten as y x 7− x =1− = k 7 7 − 5

2

−k 7y = 5 7− x

 7 y  (3) ⇒ k = −5   7− x x y 5+ y and, equation (2) can be rewritten as =1+ = k 5 5 − 7 k 5x 35 x ⇒ − = ⇒ k =− (4) 7 5+ y 5+ y ⇒

From (3) and (4), we get −

∴

8 1 Solving (1) and (2), we get coordinates of k as  ,  . 3 3  2 1 1 1 ∴ Area of ∆BDK = | 2 0 1 | 2 8 1 1 3 3

3 1 1 × = . 2 3 2

Now, equation of the line AQ is

15   10   6  Since   +  =  , we get  AB   AC   AD  (cosθ + 3 sinθ)2 + (2 cosθ + sinθ)2 = (cosθ – sinθ)2 ⇒ 4 cos2θ + 9 sin2θ + 12 sinθ cosθ = 0 ⇒ (2 cosθ + 3 sinθ)2 = 0 ⇒ 2 cosθ = – 3 sinθ 2 ⇒ tan θ = − . 3 ∴ From (1), the equation of line is 2 y + 4 = − ( x + 5) 3 or 2x + 3y + 22 = 0. 34. Area of ∆ABC = Area of ∆ABD + Area of ∆BDC 3 1 1 = ( 2) (1) + (1) (1) = . 2 2 2 1  slope of AB = slope of DK = , 2 1 ∴ equation of DK is y – 0 = (x – 2) or 2y = x – 2 2 Equation of BC is y – 0 = – 1 (x – 3) or y = – x + 3

1  1  8  2  1 2  0 −  − 1 2 −  + 1 − 0  = . 2  3  3  3  3

(1) (2)

35 y 35 x =− 7− x 5+ y

⇒ y(5 + y) = x (7 – x) ⇒ x(x – 7) + y (y + 5) = 0 is the locus of R. 36. Let Q be the position of the point P (2, 3) after first transformation. Then, Q ≡ (3, 2). Let R be the position of Q after second transformation. Then, R ≡ (3, 4). Let S be the new position of R after third transformation. Let S ≡ (x1, y1). If OR makes an angle α with x-axis, then 4 tan α = . 3

HINTS AND EXPLANATIONS

Coordinates of points B, C and D are B ≡ (–5 + AB cosθ, –4 + AB sinθ) C ≡ (–5 + AC cosθ, –4 + AC sinθ) D ≡ (–5 + AD cosθ, –4 + AD sinθ) Points B, C, D lie on the lines x + 3y + 2 = 0 2x + y + 4 = 0 x – y – 5 = 0 respectively. ∴ – 5 + AB cosθ + 3 (– 4 + AB sinθ) + 2 = 0 15 ⇒ = cosθ + 3 sin θ AB and, 2(– 5 + AC cosθ) + (– 4 + AC sinθ) + 4 = 0 10 ⇒ = 2 cosθ + sin θ AC and, (– 5 + AD cosθ) – (– 4 + AD sinθ) – 5 = 0 ⇒ 6 = cosθ − sin θ AD

10.48  Chapter 10

β +2 (1) α −1 The equation of the perpendicular bisector of AB is x – y + 5 = 0 (2) β +2  From (1) and (2), we have   (1) = −1  α −1  ⇒ α + β + 1 = 0  (3) Also, the mid point of AB lies on (2), ∴ the slope of AB =

(ii) (iii) ⇒ sin α =

4 3 and cos α = . Also, OR = 5 = OS. 5 5 x1 = OS1 = 5  cos  (α + 45º) =

Now,

5

– sin α)

2

(cos  α

 α +1  β − 2  ∴  − +5=0  2   2  ⇒ α – β + 13 = 0 Solving (3) and (4), we get α = –7 and β = 6. So, the coordinates of B are (–7, 6). 11 2 Similarly, the coordinates of C are  ,  .  5 5 ∴ The equation of the line BC is

5 3 4 1 =  − =− 25 5 2 and, y1 = SS1 = 5 sin (α + 45º) = =

5 2

(sin α + cos α )

5  4 3 7 .  + = 2  5 5 2

HINTS AND EXPLANATIONS

 1 7  Hence, the coordinates of S are  − , . 2 2  37. The equation of the line L passing through the intersection of L1 and L2 is (ax + by + c) + λ (lx + my + n) = 0 (1) By the given condition, L1 is the bisector of the angle between L and L2. Let Q (α, β) be any point on L1. ∴ Length of perpendicular from Q (α, β) on L2 and L must be equal, thus lα + mβ + n ( aα + bβ + c) + λ (lα + mβ + n) =± 2 (2) l + m2 ( a + λ l ) 2 + ( b + λ m) 2 Since Q (α, β) lies on L1 ∴ aα + bβ + c = 0 From (2) and (3), we get 1 λ =± 2 2 2 l +m ( a + λ l ) + ( b + λ m) 2

(3)

⇒ (a + ll)2 + (b + λm)2 = λ2 (l2 + m2) ⇒ a2 + b2 + 2λ al + 2λ bm = 0 a2 + b2 ⇒ λ = − (4) 2( al + bm) Substituting the value of λ from (4) in (1), we get a2 + b2 ( ax + by + c) − (lx + my + n) = 0 2 ( al + bm) ⇒ 2 (al + bm) (ax + by + c) – (a2 + b2) (lx + my + n) = 0, which is the required equation of the line L. 38. Let the coordinates of B be (α, β). Since coordinates of A are (1, –2),

(4)

2 −6 y − 6 = 5 ( x + 7) 11 +7 5 28 (x + 7) ⇒ 23(y – 6) + 14(x + 7) = 0 46 ⇒ 14x + 23y = 40. Hence, the equation of the line BC is 14x + 23y = 40. ⇒ y – 6 = –

39. We have, (2 + 3t) x + (1 – 2t) y + 4 = 0 ⇒ (2x + y + 4) + t (3x – 2y) = 0 ⇒  Every member of the given family of lines passes through the point of intersection of the lines

2x + y + 4 = 0 and, 3x – 2y = 0 Solving (1) and (2), we get 8 12 x = − and y = − . 7 7  8 12  So, the point of intersection is  − , −  . 7  7

(1) (2)

Coordinates and Straight Lines  10.49 8 12 The required line passes through the point  − , −  and 7  7  8 12  is ⊥ to the line joining (2, 3) and  − , −  . 7  7 8 12 Slope of the line joining (2, 3) and  − , −  is 7 7  12 3+ 7 = 33 = 3 . = 8 22 2 2+ 7 2 ∴ Slope of required line = − . 3 ∴ Equation of the required line is

If θ be the angle that the required line makes with the normal, then 3 4 cosθ − ∴ tan θ = . 5 3 Let θ be the angle that the required line makes with the nor3 mal whose slope is . Then, we have 4 3 m− −7 4 4 ± tan θ = ± = ⇒ m = ∞, 3 1 + 3m 24 4 Therefore, the equations of the required lines are −7 x + 2 = 0 and y + 7 = (x + 2). 24

12  2 8   y + 7  = −3 x + 7     or, 14x + 21y + 52 = 0.

42. If (x, y) be the coordinates of the required point, then we have | 3x − 4 y + 1 | = 1 (1) 5

40. Coordinates of vertices B, C and D are (2 cos 30°, 2 sin 30°), ( 8 cos75°, 8 sin 75°)

and,

⋅

and, (– 2 cos 60°, 2 sin 60°), respectively.

| 8x + 6 y + 1 | = 1 (2) 10

Since (x, y) lies below L1, therefore 3x − 4 y + 1 < 0, i.e., 3x – 4y + 1 > 0 −4 8x + 6 y + 1 > 0, i.e., 8x + 6y + 1 > 0 6 Removing the mod sign from equations (1) and (2), we have + (3x – 4y + 1) = 5 and, + (8x + 6y + 1) = 10 Solving, we get the coordinates of the required point as  6 −1   , .  5 10 

i.e., B ≡ ( 3 ,1), C ≡ ( 3 − 1, 3 + 1) and D ≡ ( −1, 3 ). Slope of BD =

3 −1

=

( 3 − 1) 2

−1 − 3 − ( 3 + 1) ( 3 − 1) ∴ Equation of diagonal BD is y − 1 = ( 3 − 2) ( x − 3 )

= 3 − 2.

⇒ ( 2 − 3 ) x + y = 2( 3 − 1). 41. The normal to the given parallel lines cuts an intercept =

9 5

43. Let R be the radius of the circumcircle and O be the origin, then AO = x12 + x2 2 tan 2 α ⇒ R = x1 sec α ⇒ x1 = R cos α. Similarly, x2 = R cos β and x3 = R cos γ So, the coordinates of vertices are A(R cos α, R sin α), B(R cos β, R sin β), C(R cos γ, R sin γ). Hence, the ∑ R cos α ∑ R sin α  coordinates of centroid G are  , . 3 3   Since the orthocentre H(a, b), circumcentre C(0, 0) and the centroid G are collinear, therefore Slope of OH = Slope of OG b R(sin α + sin β + sin γ ) . ⇒ = a R(cos α + cos β + cos γ ) 44. Let P be (0, c), c is a constant From the figure, h = OL = OQ + QL = c cot θ + QR cos (180 – 60 – θ) = c cot θ + PQ {cos 120 cosθ + sin 120 sinq}

HINTS AND EXPLANATIONS

and since it lies above L2, therefore

10.50  Chapter 10

OM = PQ 3 = c cot θ − cosθ + PQ sin θ 2 2 1 3 c   = c cot θ − c cosec θ cosθ + c cosecθ sinθ ∵ sin θ = 2 2 PQ   c = (cotθ + 3 ) (1) 2 Again, k = RL = RQ sin (180 – 60 – θ)

6 3 +4 2

2

=

6 . 5

6 π , therefore   AM = OM = . 4 5 36 Hence, area of ∆OAB is   ∆ = OM × AM = . 25 Since, ∠OAM =

1 2 1 1 1 P2 is mid-point of OP1, therefore OP2 = × = 2 2 2 2 47.  P1 is mid-point of OA, therefore OP1 =

1 3   = PQ sin θ . + cosθ .  2 2     1 3 cosθ  = c cosec θ  sin θ + 2 2   =

c (1 + 3 cot θ ) (2) 2

HINTS AND EXPLANATIONS

Eliminating cot θ from (1) and (2), we get k = 3 h − c ∴ The required locus is y = 3 x − c a straight line. 45. Circurmcentre of the triangle is (0, 0) and  3 + 5 cosθ + 5 sin θ 4 + 5 sin θ − 5 cosθ  centroid ≡  ,  3 3   ∴ Centroid divides the join of orthocentre and circumcentre in the ratio, 2 : 1 ∴ h = 3 + 5 cos θ + 5 sin θ and k = 4 + 5 sin θ – 5 cos θ, where (h, k) represents the orthocentre ⇒ sin θ =

h+k −7 h − k +1 and cosθ = 10 10

⇒ (h + k – 7)2 + (h – k + 1)2 = 100 ∴ Locus of orthocentre is (x + y – 7)2 + (x – y + 1)2 = 100. 46. Let OAB be the right-angled isosceles triangle. Then, ∠OAB = ∠OBA =

π 4

The perpendicular distance of the line 3x + 4y = 6 from the origin is

Proceeding like this, we have 1 OPn = n . 2 48. Let

AN =λ BN

Coordinates and Straight Lines  10.51 aλ   a Then, N ≡  ,  1+ λ 1+ λ  Slope of AB = – 1 ∴ slope of MN = + 1 Equation of MN is given by aλ a y − =x− (1) 1+ λ 1+ λ  a (λ − 1)  Hence, M ≡  0,  λ +1   [putting x = 0 in equation (1)] 1 Area of ∆AMN = × AN × MN 2

=



1 2aλ 2a a 2λ × = (1 + λ ) 2 2 1+ λ 1+ λ

Area of ∆ABC = Given,

1 1 × OA × OB = a 2 2 2

3 1 a 2λ = × a2 (1 + λ ) 2 8 2

⇒ 16λ = 3(1 + λ)2 1 ⇒ 3l2 – 10λ + 3 = 0  ⇒  λ = , 3 3 1 For λ = , M lies outside segment OB, hence the only 3 acceptable value for λ = 3.

Previous Year's Questions Then AB = ( −1 − 4) 2 + ( −1 − 0) 2 = 25 + 1 = 26 BC = (3 + 1) 2 + (5 + 1) 2 = 4 2 + 6 2 = 16 + 36 = 52 and CA = ( 4 − 3) 2 + (0 − 5) 2

= 1 + 25 = 26

CA + AB 2 = 2

(

26

) +( 2

26

)

2

∴ = 26 + 26 = 52 = BC2 ⇒ CA2 + AB2 = BC2 Thus, the triangle is isosceles and right angled triangle. 50. The equation of parabola is y2 + 4y + 4x + 2 = 0 ⇒ y2 + 4y + 4 = –4x – 2 + 4 1 ⇒ ( y + 2) 2 = − 4  x −  2  1 Transformation y + 2 = Y and x − = X gives 2 Y 2 = – 4X Here a = 1 ∴ Equation of directrix is X = 1 1 ⇒ x − = 1 2 3 ⇒ x = 2 51. Key Idea: If the triangle is equilateral, then the incentre coincides with the centroid of the triangle. Let A(1, 3 ), B(0, 0), C(2, 0) be the vertices of a triangle ABC. ∴ a = BC ( 2 − 0) 2 + (0 − 0) 2 = 2

b = AC ( 2 − 1) 2 + (0 − 3 ) 2 = 2 and c = AB (0 − 1) 2 + (0 − 3 ) 2 = 2 b = AC = -J (2 - 1f + (0 - V3)2 = 2 and c = AB = V(0 - 1)2 + (0 - 41) = 2 ∴ The triangle is an equilateral triangle. ∴ Incentre is same as centroid of the triangle. ⇒ Co-ordinates of incentre are 1+ 0 + 2 3 + 0 + 0   1  ,   i.e., 1,  3 3   3  52. Key Idea: Equations of angle bisectors of lines a1x + b1y + c1 = 0, and a2x + b2y + c2 = 0 are

a1 x + b1 y + c1 a12 + b12

=±

a2 x + b2 y + c2 a22 + b22

For the two lines 24x + 7y - 20 = 0 and 4x - 3y - 2 = 0, the angle bisectors are given by 24 x + 7 y − 20 4x − 3y − 2 =± 25 5 Taking positive sign, we get 2x + 11y - 5 = 0 Therefore, the given three lines are concurrent with one line bisecting the angle between the other two. 3 x + y = 0 makes an angle 120° with OX and 3 x + y = 0 makes an angle of 60° with OX. So, the required equation of line is y - 2 = 0. 54. Let p (x, y) be the point equidistant to the given points, then (x - a1)2 + (y - b1)2 = (x - a2)2 + (y - b2)2 1 ⇒ (a1 - a2) x + (b1 - b2) y + (b22 − b12 + a22 − a12 ) = 0 2 Hence, (A) is the correct answer. 53.

HINTS AND EXPLANATIONS

49. Let A(4, 0), B(–1, –1) and C(3, 5) be the vertices of a ∆ABC.

10.52  Chapter 10 55. Given parametric equations a cos t + b sin t + 1 a sin t − b cos t + 1 , y= x= 3 3  a2 + b2 1 . This implies that  x −  + y 2 =   3 9 Hence, (B) is the correct answer.  h ( k − 2)  56. If C be (h, k) then centroid is  ,  it lies on 2x + 3y = 1. 3  3 Therefore the locus is 2x + 3y = 9. x y 4 3 57. Point (4, 3) lies on + = 1 with a + b = -1 then + = 1 a b a b ⇒ a = 2, b = -3 or a = -2, b = 1. x y x y Hence − = 1 and + = 1. 2 3 −2 1

1− 3 + 5 1+ 3 + 3   7  ,  = 1,  3   3  3

∴ centroid is 

2

58. m1 + m2 = −

2c 7

and, m1m2 = −

1 7

⇒ m1 + m2 = 4m1m2  (given) ⇒ c = 2. 1 6 3 , m1m2 = and m1 = − . 4c 4c 4 Hence c = -3. 60. P = (1, 0) Let Q = (h, k) then k2 = 8h Let (α, β) be the midpoint of PQ, then 59. m1 + m2 =

HINTS AND EXPLANATIONS

α=

h +1 k +0 ,β = 2 2

⇒ 2α – 1 = h, 2β = k. ⇒ (2β)2 = 8(2a – 1)  ⇒  b 2 = 4α – 2 ⇒ y2 – 4x + 2 = 0. 61. Required equation is of the from ax + 2by + 3b + λ(bx – 2ay – 3a) = 0 ⇒ (a + bλ)x + (2b – 2aλ)y + 3b – 3la = 0

a b a ⇒ ax + 2by + 3b − (bx – 2ay – 3a) = 0 b 2a 2 3a 2 ⇒ ax + 2by + 3b – ax + =0 y+ b b  2a 2  3a 2 =0 ⇒ y  2b +  + 3b + b  b  a + bλ = 0 ⇒ λ = −

63. The required line intersects the axes at points (0, 8) and (6, 0) Hence, the equation of the line is

y=

−4 ( x − 6) 3

⇒ 4x + 3y = 24 64. Given parabola: y =

a2 x 2 a2 x + − 2a 3 2

Vertex: (α, β) implies

 a4  a3 1 8 −a 2 −  + 4 ⋅ ⋅ 2a  −  +  a4 3 3  = −  4 3 α = 23 = − , β =  4 4 3 4a 2a a3 a 4 3 3 3 35 a 35 =− ×3 = − a 12 4 16 3  35  105 ∴ αβ = −  −  a = 4 a  16  64 65. We must have a2 - 3a < 0 and a 2 −

⇒

a > 0 2

1 < a< 3 2

66. We have

1 ( k − 1) = ±1 2

⇒ k - 1 = ±2 ⇒ k = 3 or k = -1

 2b 2 + 2 a 2   3b 2 + 3a 2   = −  b b     2 2 −3( a + b ) −3 ⇒ y = = 2 2(b 2 + a 2 ) ⇒ y 

⇒ y = −

3 3 so it is units below x-axis. 2 2

62. Vertex of triangle is (1, 1) and midpoint of sides through this vertex is (–1, 2) and (3, 2) ⇒ vertices B and C come out to be (–3, 3) and (5, 3)

2π =− 3 3 Hence, the equation of line QM is y = − 3 x. 67. Slope of the line QM is tan

Coordinates and Straight Lines  10.53 32 b2 = 5 ∴ Required equation of ellipse 3x2 + 5y2 - 32 = 0.

 6 + 2 12 + 2  ,  5   5  8 14  p =  ,  5 5   8 14  lies on 2x + y = k p =  ,  5 5  16 14 ⇒ + =k 5 5 30 ⇒ = k = 6 5 73. Point p = 

 k +1 7  ,   2 2

Mid-point of PQ = 

7 ( k + 1)   = ( k − 1)  x −  2  2 

Put x = 0 and y = -4. ⇒ k = ±4. 70. Slope of line L = −

b 5

Slope of line K = −

3 c

b 3 = ⇒  bc = 15 5 c

∆=

m 2 + −2 2 m

∆

is least if

m 2 = 2 m

1 3

.

So, the slope of the reflected ray must be

13 32 32 8 ⇒ + =1 ⇒ =− 5 b b 5 3 ⇒ b = -20  ⇒  c = − 4 Distance between L and K =

m2 + 4 − 4m 2m

75. Slope of the incident ray = −

(13, 32) is a point on L

Equation of K: y - 4x = 3

∆=

⇒ m2 = 4 ⇒ m = ±2 ⇒ m = – 2

Link L is parallel to link k. ⇒

( m − 2) 2 2|m|

Area of ∆ OPQ =

Equation of bisector is y −

74. Equation of line passing through (1, 2) with slope m is y - 2 = m(x - 1)

1 3

.

Now, the point of incidence is ( 3 , 0) And so, the equation. of reflected ray is y =

| 52 − 32 + 3 | 17

71. P(-2, -2); Q = (1, -2) Equation of angular bisector OR is ( 5 + 2 2)x = ( 5 − 2) y

23 17

76. Abscissa = = =

∴ PR : RQ = 2 2 : 5 3 3a  2 72. b 2 = a 2 (1 − e 2 ) = a 2 1 −  = a 2 = 5 5  5 2 2 9 5 x y 2 + 2 = 1 ⇒ + =1 b a 2 3a 2 a 32 a2 = 3

=

2

1 3

(x −

3 ).

ax1 + bx2 + cx3 a+b+c

2× 2 + 2 2 × 0 + 2× 0 2+2+2 2 4 4+2 2

=

2 2+ 2

= 2 − 2.

Alternate Solution: Abscissa = r = (s - a) tan

A 2

4+2 2  π = − 2 2  tan = 2 − 2 

2



4

HINTS AND EXPLANATIONS

68. Equation of bisectors of line x = 0 and y = 0 are y = ±x. Put y = ±x in my2 + (1 - m2)xy - mx2 = 0, we get (1 - m2)x2 =0 ⇒ m = ±1. 69. Slope of the bisector = k - 1

10.54  Chapter 10 77. Let point of intersection is (h, - h)

 4 ah − 2ah + c = 0 ⇒  5bh − 2bh + d = 0 So,

P=

P

 13  ,1 , P ( 2, 2) 2  2 Slope = − 9 y +1 2 Equation will be =− x −1 9

⎛ 2m − 3 ⎞ R=⎜ , 3 − 2m ⎟ m ⎝ ⎠

9y + 9 + 2x - 2 = 0 2x + 9y + 7 = 0

x=

78. S 

x=

80. Let M be mid-point of BB′ and AM is ⊥ bisector of BB′ (where A is the point of intersection of the given lines) (x - 2) (x - 1) + (y - 2) (y - 3) = 0 h+2 h+2  k +3 k +3  ⇒  − 2  − 1 +  − 2  − 3 = 0  2  2   2  2 

7x − y − 5 5 2

1 8 Cleary  ,  lies on (1)  3 3 y −3 82. Any line passing through (2, 3) with slope in m = x−2 mx – 2m = y – 3 mx – y = 2m – 3

⇒ 3x – xy = –2y or 3x + 2y = xy k 1 5 2 −k

83. Area =

−3k 1 k 1 2 1

= 28

5k2 + 13k + 10 = ±56 5k 2 + 13k − 46 = 0 5 K 2 + 13K + 66 = 0 5k 2 + 13k − 46 = 0 −13 ± 169 + 920 10 = 2, –4.6 (reject) k=

For k = 2 A (2, –6)

8

(1)

=

⇒ 5(x – y + 1) = 7x – y – 5 and 5(x – y + 1) = –7x + y + 5 ∴ 2  x + 4y – 10 = 0  ⇒  x + 2y – 5 = 0 and 12 x – 6y = 0  ⇒ 2  x–y=0 Now equation of diagonals are (x + 1) + 2(y + 2) = 0  ⇒  x + 2y + 5 = 0 and 2(x + 1) – ( y + 2) = 0  ⇒ 2x – y = 0

−2 y 3− y

E

m

=±

⇒ x =

(k2 – 7k + 10) + 4k2 + 20k = ±56

81. Equation of angle bisector of the lines x – y + 1 = 0 and 7x – y – 5 = 0 is given by 2

3− y −3 (3 − y ) 2

k − 5 −4 k 0 5 + k k − 2 0 = ± 56 −k 2 1

⇒ (h - 2)(h) + (k - 1)(k - 3) = 0 ⇒ x2 - 2x + y2 - 4y + 3 = 0 ⇒ (x - 1)2 + (y - 2)2 = 2.



2m − 3 y = 3 – 2m m

2m = 3 – y Substituting the value of m in x

79. x + y < 41, x > 0, y > 0 is bounded region. Now, number of positive integral solutions of the equation x + y + k = 41 will be number of integral co-ordinates in the bounded region. =41-1C3-1 = 40C2 = 780.

x − y +1

R

Q

2m − 3 ,0 m Q = (0, 3 –2m)

c d =− 2a 3b

3bc - 2ad = 0

HINTS AND EXPLANATIONS

x y 2m − 3 + 3 − 2m = 1 m

(2) (5, 2) B

D

m = –2 C (–2, 2)

m=0

Equation of AD, x = 2

(i)

Coordinates and Straight Lines  10.55 Also equation BE,

AC = 4x + 3y = 6,

BC º y = 3

 3  , 3  4 

1 (x – 5) 2 2y – 4 = x – 5 y–2=

\ C = −

x – 2y – 1 = 0

(ii)

Solving (i) and (ii), 2y = 1 1 y= 2 ⎛ 1⎞ Orthocentre is ⎜ 2, ⎟ ⎝ 2⎠

85.

3 2 .a = 27 3 4

r=a´

3 2 × 2 3

\a= 6 3 \r=6

g2 + f2 – c = 36 \ c = 25

Hence, the correct option is (C) 84. As C is the orthocenter of triangle A H B A (0, 2)



• H (0, 0) C (h, k)

HINTS AND EXPLANATIONS

B (4, 3)

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CHAPTER

Circles

11 LEARNING OBJECTIVES

After reading this chapter, you will be able to:  Learn the definition of a circle and standard equation of a circle  Be familiar with the conditions and some special cases for an equation to represent a circle  Understand parametric equations of a circle along with position of a point with respect to a circle

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.

  Know how to find the equation of a circle passing through three non-collinear points, intersection of a line and a circle, contact of two circles, equation of the tangent in slope form

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

STANDARD EQUATION OF A CIRCLE The equation of a circle with the centre at (h, k) and radius a, is (x - h) + (y - k) = a 2

2

2

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

1. coefficient of x2 = coefficient of y2, i.e., a = b, 2. coefficient of xy is zero, i.e., h = 0.

To Find the Centre and Radius of a Circle whose Equation is GIven SHORT-CUT METHOD Make the coefficients of x2 and y2 equal to 1 and right hand side equal to zero.  The coordinates of centre will be (h, k), where 

1 h = − (coefficient of x ) 2

The general equation of a circle is of the form x2 + y2 + 2gx + 2 fy + c = 0,

(1)

where g, f and c are constants. The coordinates of its centre are (-g, -f  ) and radius is g2 + f 2 - c.

and

1 k = − (coefficient of y ) 2

Radius = h2 + k 2 − constant term



11.2  Chapter 11

Nature of the Circle 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. FIGURE 11.3

EQUATION OF A CIRCLE IN SOME SPECIAL CASES

5. If the circle touches x-axis at origin then its equation is x2 + (y ± k)2 = k2 ⇒ x2 + y2 ± 2ky = 0. (Two cases)

1. If the centre of the circle is (h, k) and it passes through origin then its equation is (x - h)2 + (y - k)2 = h2 + k2 ⇒  x2 + y2 - 2hx - 2ky = 0

FIGURE 11.4

6. If the circle touches y-axis at origin, the equation of circle is (x ± h)2 + y2 = h2 ⇒ x2 + y2 ⇒ 2xh = 0. (Two cases) FIGURE 11.1

2. If the circle touches x-axis then its equation is (x ± h)2 + (y ± k)2 = k2. (Four cases) 3. If the circle touches y-axis then its equation is (x ± h)2 + (y ± k)2 = h2. (Four cases) FIGURE 11.5

7. If the circle passes through origin and cuts intercepts a and b on the axes, the equation of circle is x2 + y2 - ax - by = 0 and centre is C(a/2, b/2). (Four cases)

FIGURE 11.2

4. If the circle touches both the axes then its equation is (x ± r)2 + (y ± r)2 = r2. (Four cases)

FIGURE 11.6

Circles  11.3

SOLVED EXAMPLES 1. The tangent to the circle x2 + y2 = 9, which is parallel to y-axis and does not lie in third quadrant, touches the circle at the point (A) (−3, 0) (B)  (3, 0) (C) (0, 3) (D)  (0, −3) Solution: (B)

Any line parallel to y-axis is x = k. If 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 | = 3 Þ 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). i.e.,

2. 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 (B)  x2 + y2 = 16a2 2 2 2 (C) x + y = 4a (D)  x2 + y2 = a2 Solution: (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 = 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 3 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 Solution: (B)

Let the radius of the circle be a. Then the centre is C ≡ (a, a). Also, the distance of C (a, a) from the line 12x + 5y = 60 is a.

∴

| 12a + 5a − 60 |

=a

or

| 17a − 60 | = a. 13

12 + 5 or 17a − 60 = ±13a or a = 15, 2 It is clear from the figure that a ≠ 15. \ a=2 \ The equation of the incircle is (x − 2)2 + (y − 2)2 = 22 or x2 + y2 − 4x − 4y + 4 = 0 4. The area of an equilateral triangle inscribed in the ­circle x2 + y2 + 2gx + 2fy + c = 0 is 3 3 2 3 3 2 (A) ( g + f 2 - c) ( g + f 2 - c) (B)  4 2 3 3 2 (C) ( g + f 2 + c) (D)  none of these 4

2

2

Solution: (B)

Given Circle is x2 + y2 + 2gx + 2fy + c = 0 (1) Let C be its centre and PQR be an equilateral triangle inscribed in the circle, then C ≡ (−g, −f ) and radius of the circle CQ = g 2 + f 2 - c .

From ∆QLC , QL = CQ sin 60° = \

QR = 2QL = 3 ⋅ g 2 + f 2 − c

Now, area of DPQR =

3 g2 + f 2 − c 2

3 3 × QR 2 = × 3( g 2 + f 2 - c) 4 4 3 3 2 = ( g + f 2 - c) 4

11.4  Chapter 11 5. 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 (B)  an ellipse (C) a hyperbola (D)  none of these Solution: (A) Take the centre of the square as origin and axes p­ arallel to its sides. Let side of square be 2a.

\ 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 7. The DPQR is inscribed in the circle x2 + y2 = 25. If Q and R have coordinates (3, 4) and (− 4, 3) respectively, then ∠QPR is equal to (A) p /2 (B)  p /3 (C) p /4 (D)  p /6 Solution: (C) 4 −3 and m2 = slope of OR = . 3 4 p As m1m2 = −1, ∠QOR = . 2

Let m1 = slope of OQ =

Thus, ÐQPR =

p 4

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 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. 6. If the equation of the incircle of an equilateral triangle is x2 + y2 + 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 Solution: (B) Given equation of the incircle is x2 + y2 + 4x − 6y + 4 = 0 Its incentre is (−2, 3) and inradius = 4 + 9 - 4 = 3. Since in an equilateral triangle, the incentre and the circumcentre coincide, \ Circumcentre ≡ (−2, 3) Also, in an equilateral triangle, circumradius = 2 (inradius)

(∵ angle subtended at the centre of a circle is double the angle subtended in the alternate segment).

EQUATION OF A CIRCLE IN DIAMETER FORM The equation of the circle drawn on the line segment joining two given points A(x1, y1) and B(x2, y2) as diameter is

FIGURE 11.7

(x - x1) (x - x2) + (y - y1) (y - y2) = 0.

æ x + x y + y2 ö Its Centre = ç 1 2 , 1 and 2 ÷ø è 2 2

æ x - x ö æ y - y2 ö Radius = ç 1 2 ÷ + ç 1 ÷ è 2 ø è 2 ø

2

Circles  11.5

SOLVED EXAMPLE

SOLVED EXAMPLES

8. Extremities of a diagonal of a rectangle are (0, 0) and (4, 3). The equations of the tangents to the circumcircle of the rectangle which are parallel to this diagonal are (A) 16x + 8y ± 25 = 0 (B)  6x − 8y ± 25 = 0 (C) 8x + 6y ± 25 = 0 (D)  none of these Solution: (B) Extremities of the diagonal OA of the rectangle are O (0, 0) and A(4, 3). Then OA is the diameter of the circumcircle, so equation of the circumcircle is x (x − 4) + y ( y − 3) = 0 i.e., x2 + y2 − 4x − 3y = 0

9. 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 (A) (x − 1)2 = 8y (B) (x + 1)2 = 8y (C) ( y − 1)2 = 8x (D)  none of these Solution: (A) The equation of any circle touching x-axis is of the form

2

2

3ö æ5ö æ ( x - 2) 2 + ç y - ÷ = ç ÷ (1) 2ø è2ø è m = slope of OA = 3/4 (2) \ Tangents parallel to the diagonal OA are

i.e.,

5 9 3 3 = ( x − 2) ± 1+ 2 4 2 16 That is 6x − 8y ± 25 = 0 y−

(x − h)2 + ( y − k)2 = k2 Let the coordinates of the other end of the diameter through P be (a, b )

a +1 b +2 = h and =k 2 2

Then, i.e.,

a = 2h − 1 and b = 2k − 2 2

2

INTERCEPTS MADE BY A CIRCLE ON THE AXES 1. The length of the intercept made by the circle x2 + y2 + 2gx + 2fc + c = 0 on

x-axis = AB = 2 g - c 2



y-axis = CD = 2 f - c 2. Intercepts are always positive. 3. If the circle touches x-axis then | AB | = 0 Thus, c = g2 4. If the circle touches y-axis, then | CD | = 0 Thus, c = f 2 5. If the circle touches both the axes, then | AB | = 0 = | CD | Thus, c = g2 = f 2 2

(1)

Also, (h − 1) + (k − 2) = (radius) = k 2

2

2

2

ö æ b +2ö æ a +1 ö æ b + 2 - 2÷ = ç Þ ç - 1÷ + ç ÷ è 2 ø è 2 ø è 2 ø

2

⇒ (a − 1)2 + (b − 2)2 = (b + 2)2 ⇒ (a − 1)2 = 8b \  Locus of (a, b ) is (x − 1)2 = 8y 10. A square is inscribed in the circle x2 + y2 − 2x + 4y + 3 = 0. Its sides are parallel to the coordinate axes. Then one vertex of the square is (A) (1 + 2 , -2) (B)  (1 - 2 , -2) (C) (1, -2 + 2 )

(D)  none of these

Solution: (D)

The 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. FIGURE 11.8

QUICK TIPS The circle x + y + 2gx + 2fy + c = 0 cuts the x-axis in real and distinct points, touches or does not meet in real points according as g2 > 0 = or < c. 2 2  Similarly, the circle x + y + 2gx + 2fy + c = 0 cuts the y-axis in real and distinct points, touches or does not meet in real points according as f2 >, = or , = or < 0, respectively.

(A) k∈ (-3, -2) ∪ (3, 4) (B) k∈ (-3, 4) (C) k∈ (-∞, -3) ∪ (4, ∞) (D) k∈ (-∞, -2) ∪ (3, ∞) Solution: (C) Since the point (2, k) lies outside the circle x2 + y2 + x - 2y - 14 = 0 ∴ 4 + k2 + 2 - 2k - 14 > 0  or  k2 - 2k - 8 > 0 or (k + 2) (k - 4) > 0 or  k∈ (-∞, -2) ∪ (4, ∞)(1) Also, the point (2, k) lies outside the circle x2 + y2 = 13. ∴ 4 + k2 - 13 > 0 or k2 - 9 > 0 or (k - 3) (k + 3) > 0 or k∈ (- ∞, - 3) ∪ (3, ∞)(2) The common solution of (1) and (2) is given by, k∈ (-∞, -3) ∪ (4, ∞) 13. If the point (k + 1, k) lies inside the region bounded by the curve and y-axis, then k belongs to the interval 2 x = 25 - y .

FIGURE 11.11

(A) (-1, 3) (B)  (-4, 3) (C) (-∞, -4) ∪ (3, ∞) (D)  none of these

Circles  11.7 Solution: (A)

From given three points taking any two as extremities of diameter of a circle S = 0 and equation of straight line passing through these two points is L = 0. then required equation of circle is S + λL = 0, where λ is a parameter, which can be found out by putting third point in the equation.  If the two lines a x + b y + c = 0 and a x + b y + c = 1 1 1 2 2 2 0 meet the coordinate axes in four distinct points, then those points are concyclic if a1a2 = b1b2. Also, the equation of the circle passing through those concyclic points is (a1x + b1y + c1) (a2x + b2y + c2) - (a1b2 + a2b1) xy = 0.  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. 

Since the point (k + 1, k) lies inside the region bounded by x = 25 - y 2 and y-axis, ∴ (k + 1)2 + k2 - 25 < 0 and k+1>0

⇒ 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

CIRCLE THROUGH THREE POINTS We can find a unique circle through three non-collinear points. To find the unique equation of circle, we can follow the following method. Step I: Assume the general equation of the circle as S : x2 + y2 + 2gx + 2fy + c = 0 Step II: The coordinates of three points P(x1, y1), Q(x2, y2) and R(x3, y3) (if they lie on the circle) will satisfy the equation of the circle and thus we shall get three simultaneous equations in g, f and c such that

S2 : x12 + y12 + 2 gx1 + 2 fy1 + c = 0



S2 : x + y + 2 gx2 + 2 fy2 + c = 0 2 2

SOLVED EXAMPLES 14. If a circle passes through the points of intersection of the coordinate axes with the lines lx - y + 1 = 0 and x - 2y + 3 = 0, then the value of λ is (A) 2 (B)  1 (C) -1 (D)  -2 Solution: (A) Let the lines cuts the x-axis at A and B, then OA = -

Also, if the lines cut the y-axis at C and D, then OC = 1 and OB =

3 æ 1ö OA ´ OB = OC ´ OD Þ ç - ÷ ( -3) = 1´ l 2 è ø

S3 : x32 + y32 + 2 gx3 + 2 fy3 + c = 0

QUICK TIPS The equation of the circle through three non-collinear points A(x1, y1), B(x2, y2) and C(x3, y3) is



x2 x12 x22 x32

+ + + +

y2 y12 y22 y32

x x1 x2 x3

y y1 y2 y3

1 1 =0 1 1

3 2

Now if the circle passes through A, B, C and D then

2 2

Step III: Solve the above three simultaneous equations in three variables to obtain the values of g, f and c. Step IV: Substitute the values of g, f and c obtained from step III in the equation assumed in step I to get the desired equation of the circle.

1 and OB = -3 × l



⇒  λ = 2

15. If the lines a1x + b1 y + c1 = 0 and a2x + b2y + c2 = 0 cut the coordinate axes in concyclic points, then (A) a1a2 = b1b2 (B) a1b1 = a2b2 (C) a1b2 = a2b1 (D) none of these Solution: (A) The line a1x + b1y + c1 = 0 cuts the coordinate axes at A(-c1/a1, 0) and B (0, -c1/b1) and the line a2x + b2y + c2 = 0 cuts the axes at C (-c2/a2, 0) and D (0, -c2/b2). So, AC and BD are chords along x-axis and y-axis respectively, intersecting at origin O.

11.8  Chapter 11 Since A, B, C, D are concyclic, therefore OA.OC = OB.OD ⇒

 −c1   − c2   − c1   − c2   a  ⋅  a  =  b  ⋅  b  1 2 1 2

or

a1a2 = b1b2

16. Two distinct chords drawn from the point (p, q) on the circle x2 + y2 = px + q y, where pq ≠ 0, are bisected by the x-axis. Then (A) | p | = | q | (B)  p2 = 8q2 2 2 (C) p < 8q (D)  p2 > 8q2 Solution: (D)

1. l intersects S in two distinct points iff d < a. 2. l intersects S in one and only point iff d = a, i.e., the line l touches the circle if perpendicular distance from the centre to the line l must be equal to radius of the circle. 3. 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 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 .

Given circle is x2 + y2 = px + qy.  p q Since the centre of the circle is,  ,  , so (p, q) and  2 2 (0, 0) are the end points of a diameter. As the two chords are bisected by x-axis, the chords will cut the circle at the points (x1, -q) and (x2, -q), where x1, x2 are real.

FIGURE 11.13

QUICK TIPS Note: If the points of intersection of a line l and a circle S are known, then the distance between these points is the required length of intercept and there is no need of using the above formula.



The length of the intercept cut off from the line y = mx + c a2(1 + m2 ) - c 2 by the circle x2 + y2 = a2 is 2 . 1 + m2



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

INTERSECTION OF A LINE AND A CIRCLE

FIGURE 11.12

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

If a2(1 + m2) - c2 > 0, line will meet the circle at two real and different points. 2 2 2  If c = a (1 + m ), line will touch the circle. 2 2 2  If a (1 + m ) - c < 0, line will meet the circle at two imaginary points. 

SOLVED EXAMPLES 17. If a chord of the circle x2 + y2 = 32 makes equal intercepts of length l on the coordinate axes, then (A) | l | < 8 (B)  | l | < 16 (C) | l | > 8 (D)  none of these Solution: (A) Since the chord makes equal intercepts of length l on the coordinate axes, so its equation can be written in the form x ± y = ±l. Since the chord intersects the given circle at two distinct points, therefore, the length of the ⊥ from the centre (0, 0) of the given circle to the chord must be less than the radius ±l i.e., < 32 Þ l 2 < 64 Þ | l |< 8 2

Circles  11.9 18. 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 Solution: (D)

Let C be the centre of the circle, then C ≡ (3, -1). Equation of line AB is 2x - 5y + 18 = 0 and AB = 6 ∴  AL = 3

∴  Equation of such circles is

2

9ö æ 2 ç x - 2 ÷ + ( y - k) è ø 2 25 ö æ9 = ç - 2 ÷ + ( k - 0) 2 = + k2 4 è2 ø or  x2 + y2 - 9x - 2ky + 14 = 0

20. The line y = mx + c intersects the circle x2 + y2 = r2 at the two real distinct points if (A) -r 1 + m 2 < c < r 1 + m 2 2 2 (B) -c 1 - m < r < c 1 + m (C) -r 1 - m 2 < c < r 1 + m 2 (D) none of these

Solution: (A)



CL = length of the ⊥ from C on AB =



| 2 ´ 3 - 5( -1) + 18 | ( 2) 2 + ( -5) 2

∴  radius of the circle AC =

= 29

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

19. 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

Given line is y = mx + c(1) and the given circle is x2 + y2 = r2 (2) Solving (1) and (2), we get (1 + m2)x2 + 2mcx + c2 - r2 = 0

For two real distinct points of intersection, both the roots of (3) must be real and distinct. ∴ 4m2c2 - 4(1 + m2) (c2 - r2) > 0 ⇒  c2 < r2 (1 + m2) ⇒  − r 1 + m 2 < c < r 1 + m 2

THE LEAST AND GREATEST DISTANCE OF A POINT FROM A CIRCLE Let S = 0 be a circle and A(x1, y1) be a point. If the diameter of the circle through A is passing through the circle at P and Q, then AP = |AC - r| = least distance; AQ = AC + r = greatest distance where ‘r’ is the radius and C is the centre of the circle.

Solution: (C)

It is clear from the figure that the coordinates of centre 9  of such circles are  , k  . 2 

(3)

FIGURE 11.14

11.10  Chapter 11

CONTACT OF TWO CIRCLES

5. One circle is contained in the other if AB < |r1 - r2|.

The two circles having centres at A(x1, y1) and B(x2, y2) and radii r1 and r2 respectively will

1. Intersect in two real distinct points if and only if | r1 - r2 | < AB < r1 + r2

FIGURE 11.19

SOLVED EXAMPLES FIGURE 11.15

2. Touch each other externally AB = r1 + r2 and their point of contact C is given by,

21. The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 - 8x + 12 = 0 is (A) 1 (B)  2 (C) 3 (D)  4 Solution: (C)

FIGURE 11.16

ærx +r x ry +r y ö C ºç 1 2 2 1, 1 2 2 1÷ r1 + r2 ø è r1 + r2 3. Touch each other internally if AB = | r1 - r2  |, and their point of contact C is given by,

The equations of the circles are x2 + y 2 = 4 (1) 2 2 and x + y - 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. 22. 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 (B) x 2 + y 2 + 4 3 y + 8 = 0 (C) x 2 + y 2 - 4 3 y + 8 = 0 (D) none of these Solution: (B, C)

The given circles are FIGURE 11.17

ærx -r x r y -r y ö C ºç 1 2 2 1, 1 2 2 1÷ r1 - r2 ø è r1 - r2 4. One circle lies outside the other if AB > r1 + r2.

x2 + y2 - 4x = 0, x > 0 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 Clearly, from the figure, the centres of the required circles are at (0, 12 ) and (0, 12 ). ∴  Equations of the required circles are ( x - 0) 2 + ( y ∓ 12 ) 2 = 22

FIGURE 11.18

i.e.,

x 2 + y 2 + 2 12 y + 8 = 0

and

x 2 + y 2 - 2 12 y + 8 = 0

Circles  11.11

EQUATION OF THE TANGENT IN SLOPE FORM 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 æ ö am a ,∓ çç ± ÷÷ 2 1 + m2 ø è 1+ m

CONDITION OF TANGENCY The straight line y = mx + c will be a tangent to the circle 23. The coordinates of the point at which the circles x + y2 - 4x - 2y - 4 = 0 and x2 + y2 - 12x - 8y -36 = 0 touch each other, are (A) (3, -2) (B)  (-2, 3) (C) (3, 2) (D)  none of these

x2 + y2 = a2 if c = ± a 1 + m 2

2

Solution: (D)

Given circles are x2 + y2 - 4x - 2y - 4 = 0 and x2 + y2 - 12x - 8y - 36 = 0 Centre of circle (1) is C1 ≡ (2, 1) and radius = r1 = 4 + 1 + 4 = 3 Centre of circle (2) is C2 ≡ (6, 4) and radius = r2 = 36 + 16 + 36 = 88 Also, d = distance between C1 and C2 = C1C2 = 16 + 9 = 5

(1) (2)

Since d ≠ r1 ± r2 ∴  the two circles do not touch each other.

TANGENT TO A CIRCLE AT A GIVEN POINT 1. Equation of the tangent to the circle x2 + y2 = a2 at the point (x1, y1) on it is xx1 + yy1 = a2. 2. Equation of the tangent to the circle

x + y + 2gx + 2fy + c = 0 2

2

at the point (x1, y1) on it is

xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0 3. Equation of the tangent to the circle x2 + y2 = a2 at the point (a cosθ, a sinθ) on it is

x cosθ + y sinθ = a (Parametric form of equation of tangent)

Notation: The equation of the tangent at the point (x1, y1) on the circle S = 0 is T = 0.

QUICK TIPS 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 condition that the line lx + my + n = 0 touches the circle x2 + y2 + 2gx + 2fy + c = 0 is 

(lg + mf - n)2 = (l2 + m2)(g2 + f  2 - c) Equation of tangent to the circle x2 + y 2 + 2gx + 2fy + c = 0 in terms of slope is y = mx + mg - f ± ( g2 + f 2 − c ) (1 + m2 ).



If the line lx + my + n = 0 is a tangent to the circle (x - h)2 + (y - k)2 = a2, then (hl + km + n)2 = a2(l2 + m2).



TANGENTS FROM A POINT OUTSIDE THE CIRCLE WORKING RULE  Let the point be (x , y ). 1 1 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) 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).  Obtain the value of m from the Eq. (2).  Substitute this value of m in Eq. (1) to obtain the required equation of tangent. 

SOLVED EXAMPLES 24. 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

11.12  Chapter 11 (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

| T | = PT =

x12 + y12 + 2 gx, + 2 fy1 + c = S1 QUICK TIPS

Solution: (A)

While calculating the length of tangent using the above formula, it must be noted that the coefficients of x2 and y2 must be unity.  If the point lies inside the circle, then S < 0 and |T | is 1 imaginary therefore we cannot have any tangent from a point inside the circle.  If the point lies on the circle, then S = 0 and hence |T | 1 = 0. So, if the point lies on the circle, then we have only one tangent having zero length.  If the point lies outside the circle, then S > 0 and |T | is 1 finite. 

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 ( y - 5) = ( x - 3) i.e., x + 2 y = 13 (1) 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). Also, radius of the circle = 36 + 9 = 45. ∴  Equation of the circle is ( x + 3) + ( y - 8) = ( 45 ) 2 That is, x + y2 + 6x - 16y + 28 = 0 2

2

2

No tangent can be drawn to a circle from a point lying within the circle.

25. The tangent to the circle x2 + y2 = 5 at the point (1, -2) also touches the circle x2 + y2 - 8x + 6y + 20 = 0. Then its point of contact is (A) (3, -1) (b)  (-3, 0) (C) (-1, -1) (d)  (-2, 1) Solution: (A)

Equation of tangent to the circle x2 + y2 = 5 at (1, - 2) is x - 2y - 5 = 0 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 Now (1) and (2) represent the same line x - 4 y1 + 3 -4 x1 + 3 y1 + 20 \ 1 = = 1 -2 -5

ERROR CHECK

(1)

SOLVED EXAMPLE 26. If the distances from the origin of the centres of the three circles x2 + y2 + 2aix = 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 + y2 = a2 are in (A) A.P. (B)  G.P. (C) H.P. (D)  none of these Solution: (B)

(2)

⇒  -2x1 + 8 = y1 + 3 or 2x1 + y1 - 5 = 0 Only the point (3, -1) satisfies it. Hence, the point of contact is (3, -1).

LENGTH OF THE TANGENT FROM A POINT TO A CIRCLE The length of the tangent that can be drawn from the point P(x1, y1) to the circle S = 0 is S1 , where the coefficients of x2 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 + c = 0 is given by

The centres of the three given circles are (-a1, 0), (-a2, 0) and (-a3, 0). The distances of the three points from the origin are a1, a2 and a3. Given: a1, a2 and a3 are in G.P. i.e., a 22 = a1a 3 (1) Now, coordinates of any point on the circle x2 + y2 = a2 are (a cosθ, a sinθ). ∴  The lengths of the tangents drawn from the point (a cosθ, a sinθ) to the three given circles are 2a1a cos q , 2a 2 a cos q and 2a 3 a cos q which, in view of (1), are in G.P.

NORMAL TO THE CIRCLE AT A GIVEN POINT The normal of a circle at any point is a striaght line perpendicular to the tangent at the point and always passes through the centre of the circle.

Circles  11.13 1. Equation of normal: •  The equation of normal to the circle x2 + y2 + 2gx + 2fy + c = 0 at any point (x1, y1) is y - y1

y +f x - x1 y - y1 = 1 ( x - x1 ) or = . x1 + g x1 + g y1 + f •  The equation of normal to the circle x2 + y2 = a2 at any point (x1, y1) is xy1 - x1y = 0 or

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)

x y = . x1 y1

FIGURE 11.22 FIGURE 11.20

SHORT-CUT METHOD Find the coordinates of centres C1, C2 and radii r1, r2 of two given circles.  Find the coordinates of the point P dividing C C in the 1 2 ratio r1 : r2 externally. Let P ≡ (h, k).  Write the equation of any line through the point P(h, k), i.e., ( y - k) = m (x - h)(1)  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.  Substituting these values of m in equation (1), the equations of two direct common tangents are obtained. 

2. Parametric form:  Since parametric coordinates of any point on the circle x2 + y2 = a2 is (a cosθ, a sinθ). ∴  equation of normal at (a cosθ, a sinθ) is

x y = = or y = x tanθ or y = mx, cos q sin q where m = tanθ, which is slope form of normal.

PAIR OF TANGENTS The equation of the pair of tangents drawn from the point P(x1, y1) to the circle

Transverse Common Tangents The transverse common tangents to the two circles intersect at a point (say P) which lies on the line joining the centres C1 and C2 of the two circles and divide C1C2 internally in the ratio of their radii r1 and r2. SHORT-CUT METHOD

FIGURE 11.21

S = 0 is SS1 = T2, where S : x2 + y2 + 2gx + 2fy + c, and

S1 = x12 + y12 + 2 gx1 + 2 f y1 + c

T : xx1 + yy1 + g(x + x1) + f (y + y1) + c. QUICK TIPS

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.

Find the coordinates of centres C1, C2 and radii r1, r2 of two given circles.  Find the coordinates of the point P dividing C C , in the 1 2 ratio r1 : r2 internally. Let P ≡ (h, k).  Write the equation of any line through the point P (h, k) 



( y - k) = m (x - h)(1)

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.  Substituting these values of m in eqn. (1), the equations of two transverse common tangents are obtained. 

11.14  Chapter 11 QUICK TIPS Two or More Circles in a plane

Direct common Tangents

0

1

2

3

2

Transverse ­common Tangents

0

0

0

0

2

Solution: (B)

SOLVED EXAMPLES 26. If the two circles x + y = 4 and x + y - 24x - 10y + a2 = 0, a ∈Ι, have exactly two common tangents, then the number of possible values of a is (A) 11 (B)  13 (C) 0 (D)  2 2

2

2

2

Clearly, from the figure, the radius of the smallest circle touching the given circles is

Solution: (B)

The equations of the circles are x2 + y2 = 4 and x2 + y2 - 24x - 10y + a2 = 0 Centre of (1) is C1 ≡ (0, 0) and radius r1 = 2

(1) (2)

Centre of (2) is C2 ≡ (12, 5) and radius r2 = 169 - a 2 d = distance between centres = C1C2

= 4 2 + 4 2 - 4 i.e., 4 2 - 4

= 144 + 25 = 13 If the two circles have exactly two common tangents, then 169 - a2 > 0 and r1 + r2 > d ⇒

( a − 13) ( a + 13) < 0 and 2 + 169 − a 2 > 13

⇒ ⇒ ⇒

-13 < a < 13 and 169 - a2 > 121 -13 < a < 13 and a2 - 48 < 0 −13 < a < 13 and − 48 < a < 48

⇒ − 48 < a < 48 Since a is an integer, ∴  a = -6, -5, -4, ... , 4, 5, 6 ∴  The number of possible values of a is 13. 27. 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 (A) 4( 2 + 1) (B)  4( 2 - 1) (C) 2( 2 - 1)

(D)  none of these

28. Locus of the centre of a circle of radius 4 which touches the circle x2 + y2 - 4x + 2y - 4 = 0 externally is (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 Solution: (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 Also, d = distance between the centres. = ( x1 - 2) 2 + ( y1 + 1) 2 Since the two circles touch each other externally, ∴ d = r1 + r2 ⇒

( x1 − 2) 2 + ( y1 + 1) 2 = 4 + 3

Circles  11.15 ⇒

That is for a number of secants PA . PB = PA1 . PB1 = PA2 . PB2 = ... = PT 2 = S1 where S1 = x12 + y12 + 2 gx1 + 2 f y1 + c

x12 + y12 − 4 x1 + 2 y1 + 5 = 49

∴ locus of (x1, y1) is x2 + y2 - 4x + 2y - 44 = 0 29. The number of common tangents to the circles x2 + y2 = 4 and x2 + y2 - 6x - 8y - 24 = 0 is (A) 0 (B)  1 (C) 3 (D)  4 Solution: (B)

Given circles are x2 + y2 - 4 = 0 2 2 and x + y - 6x - 8y - 24 = 0

(1) (2)

Centre of circle (1) is C1 ≡ (0, 0) and radius r1 = 2 Centre of circle (2) is C2 ≡ (3, 4) and radius r2 = 7 Also d = distance between the centres = C1C2 = 5 Since d = r2 - r1, therefore the given circles touch internally, as such they can have just one common tangent at the point of contact. 30. The number of tangents to the circle x2 + y2 - 8x - 6y + 9 = 0 which pass through the point (3, -2) is (A) 2 (B)  1 (C) 0 (D)  none of these

QUICK TIPS If the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 meet the coordinate axes in four distinct points, then these points are concyclic if a1a2 = b1b2. Also, the equation of the circle passing through these concyclic points is (a1x + b1y + c1)(a2x + b2y + c2) - (a1b2 + a2b1)xy = 0.  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 

DIRECTOR CIRCLE The locus of the point of intersection of two perpendicular tangents to a circle is called the Director circle. Let the circle be x2 + y2 = a2, then equation of director circle is x 2 + y 2 = 2a 2 = ( 2a ) 2 . Clearly, director circle is a concentric circle whose radius is 2 times the radius of the given circle.

Solution: (A)

Let S ≡ x2 + y2 - 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).

POWER OF A POINT WITH RESPECT TO A CIRCLE If from a point P(x1, y1), inside or outside the circle a secant be drawn intersecting the circle in two points A and B then PA . PB = constant. The product PA . PB is called power of the point P(x1, y1) w.r.t. the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0

FIGURE 11.24

REMEMBER Director circle of circle x2 + y2 + 2gx + 2fy + c = 0 is x2 + y2 + 2gx + 2fy + 2c - g2 - f  2 = 0.

SOLVED EXAMPLE 31. The coordinates of a point on the line y = 2 from which the tangents drawn to the circle x2 + y2 = 25 are perpendicular, are (A) ( 46 , 2) (B)  ( - 46 , 2) (C) ( 37 , 2) (D)  ( - 37 , 2) Solution: (A, B)

FIGURE 11.23

Let the point on the given line be (x1, 2). Since the tangents drawn from (x1, 2) to the given cir-

11.16  Chapter 11 cle are at right angles, so the point (x1, 2) must also lie on the director circle whose equation is x2 + y2 = 2.25  i.e., x2 + y2 = 50 2 ∴ x 1 + 4 = 50  ⇒  x1 = ± 46 So, the points are ( 46 , 2) and ( − 46 , 2).

EQUATION OF CHORD OF CONTACT The chord joining the points of contact of the two tangents to a conic drawn from a given point outside it, is called the chord of contact of tangents 1. Equation of chord of contact The equation of the chord of contact of tangents drawn from the point (x1 y1) to the circle x2 + y2 = a2 is xx1 + yy1 = a2. 2. The equation of chord of contact of tangent drawn from the point (x1, y1) to the circle x2 + y2 + 2gx + 2fy - c = 0 is T = 0.

⇒

xx1 + yy1 + g(x + x1) + f (y + y1) + c = 0

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 æ -3 1 ö ∴  Coordinates of M are ç , ÷ . è 2 2ø

EQUATION OF CHORD IF ITS MID POINT IS KNOWN The equation of the chord of the circle S ≡ x2 + y2 + 2gx + 2fy + c = 0 bisected at the point (x1, y1) is given by T = S1. That is xx1 + yy1 + g(x + x1) + f (y + y1) + c = x12 + y12 + 2 gx1 + 2 fy1 + c

SOLVED EXAMPLES

QUICK TIPS It is clear from above that the equation to the chord of contact coincides with the equation of the tangent, if the point (x1, y1) lies on the circle. 2 2  The length of chord of contact = 2 r − p , p being length of perpendicular from centre to the chord. a( x12 + y12 − a2 )3/2  Area of ∆APQ is given by . x12 + y12 

33. The locus of the mid point of the chord of the circle x2 + y2 - 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 Solution: (A)

SOLVED EXAMPLE 32. The coordinates of the middle point of the chord which the circle x2 + y2 + 4x - 2y - 3 = 0 cuts off on the line y = x + 2, are

Given equation of circle is x2 + y2 - 2x - 2y - 2 = 0 Let mid point of chord AB be (h, k) Its centre is (1, 1) and radius = 1 + 1 + 2 = 2. = OB.

æ3 1ö æ -3 1 ö (A) ç , ÷ (B)  ç2,2÷ è ø è 2 2ø æ -3 -1 ö æ 3 -1 ö (C) ç , ÷ (D)  ç2, 2 ÷ 2 2 è ø è ø Solution: (A)

Equation of chord PQ is y = x + 2

or

x - y + 2 = 0

(1)

In DOPB, ∠OBP = 30º. ∴ sin 30º = OP/2 or OP = 1 Since, OP = 1  ⇒ (h - 1)2 + (k - 1)2 = 1 or h2 + k2 - 2h - 2k + 1 = 0 ∴  Locus of mid point of chord is x2 + y2 - 2x - 2y + 1 = 0 34. The locus of the centres of circles passing through the origin and cutting the circle x2 + y2 + 6x - 4y + 2 = 0 orthogonally is

Circles  11.17 (A) 2x - 3y + 1 = 0 (C) 3x - 2y + 1 = 0

(B)  2x + 3y +1 = 0 (D)  none of these

SOLVED EXAMPLE

Solution: (C)

Let the equation of one of the circles be x2 + y2 + 2gx + 2f y + c = 0 Since it passes through origin, ∴  c = 0. So, the equation becomes x2 + y2 + 2gx + 2fy = 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

35. If the circle x2 + y2 + 6x + 8y + a = 0 bisects the circumference of the circle x2 + y2 + 2x - 6y - b = 0, then a + b is equal to (A) 38 (B)  -38 (C) 42 (D)  none of these Solution: (B)

Given circles are S1: x2 + y2 + 6x + 8y + a = 0 (1) 2 2 and S2 : x + y + 2x - 6y - b = 0 (2) The equation of common chord of the two circles is S1 - S2 = 0 i.e., 4x + 14y + (a + b) = 0 (3) Since the circle S1 bisects the circumference of circle S2, therefore, (1) passes through the centre of circle S2, i.e., (-1, 3) ∴  4 (-1) + 14 (3) + a + b = 0  ⇒  a + b = - 38

-6x + 4y = 2 or 3x - 2y + 1 = 0

COMMON CHORD OF TWO CIRCLES The chord joining the points of intersection of two given circles is called their common chord. 1. Equation of common chord:  The equation of the common chord of two circles

The locus of the middle points of a system of parallel chords of a circle is called a diameter of the circle.

FIGURE 11.26

FIGURE 11.25

S1 ≡ x2 + y2 + 2g1x + 2f1y + c1 = 0 and S2 ≡ x2 + y2 + 2g2x + 2f2y + c2 = 0 is 2x(g1 - g2) + 2y(f1 - f2) + c1 - c2 = 0 i.e., S1 - S2 = 0 2. Length of the common chord

DIAMETER OF A CIRCLE

(1) (2)

PQ = 2( PM ) = 2 C1 P 2 - C1 M 2 where C1P = radius of the circle S1 = 0 and C1M = length of the perpendicular from the centre C1 to the common chord PQ.

The equation of the diameter bisecting parallel chords y = mx + c (c is a parameter) of the circle x2 + y2 = a2 is x + my = 0. REMEMBER The diameter corresponding to a system of parallel chords of a circle always passes through the centre of the circle and is perpendicular to the parallel chords.

QUICK TIPS The length of the tangent drawn from any point on the circle x2 + y2 + 2gx + 2fy + c1 = 0 to the circle x2 + y2 + 2gx + 2fy + c = 0 is c − c1  If two tangents drawn from the origin to the circle x2 + y2 + 2gx + 2fy + c = 0 are perpendicular to each other, then g2 + f   2 = 2c.  The angle between the tangents from (α, β) to the circle ö æ a ÷. x2 + y2 = a2 is 2 tan-1 ç ç a 2 + b 2 - a2 ÷ ø è 

QUICK TIPS The coefficients of x2 and y2 in both the equations S1 = 0 and S2 = 0 must be unity.  If the circles S = 0 and S = 0 touch, then the common 1 2 chord S1 - S2 = 0 becomes the tangent to both of the circles and hence perpendicular from the centre of the either circle to it should be equal to the corresponding radius. 

11.18  Chapter 11 If OA and OB are the tangents from the origin to the circle x2 + y2 + 2gx + 2fy + c = 0 and C is the centre of the circle, then the area of the quadrilateral OACB is c( g2 + f 2 − c ). 2 2  The length of the common chord of the circles x + y + ax + by + c = 0 and x2 + y2 + bx + ay + c = 0 is 1 (a + b)2 − 4c . 2 2  If O is the origin and OP, OQ are tangents to the circle x + y2 + 2gx + 2fy + c = 0, then the circumcentre of the − g −f  triangle OPQ is  , .  2 2  

(a) α/2 (b)  α (c) 2α (d)  none of these Solution: (C)

Let the angle between the tangents be 2θ. From the figure,

The length of the chord intercepted by the circle x2 + y2 =



r2 on the line

æ r 2(a2 + b2 ) - a2b2 ö x y + = 1 is 2 ç ÷ a b a2 + b2 è ø

ANGLE OF INTERSECTION OF TWO CIRCLES The angle between the two circles is the angle between their tangents at their point of intersection.

sin q =

a sin a = sin a a

⇒ θ=α Thus, the required angle = 2θ = 2α

ORTHOGONAL INTERSECTION OF TWO CIRCLES FIGURE 11.27

The angle of intersection θ of two circles S ≡ x2 + y2 + 2g1x + 2f1 y + c1 = 0 and S′ ≡ x2 + y2 + 2g2x + 2f2 y + c2 = 0 is given by cos q = ±

2 g1 g 2 + 2 f1 f 2 - c1 - c2

Two circles are said to intersect orthogonally when they intersect at right angles. The condition for the circles x2 + y2 + 2g1x + 2f1 y + c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 to intersect orthogonally is given by, 2g1g2 + 2 f1 f2 = c1 + c2

2 g12 + f12 - c1 . g 22 + f 22 - c2

r12 + r22 - d 2 , where r1 and r2 are radii of the two 2r1r2 circles and d is the distance between their centres. or cos q =

QUICK TIPS cos  θ ∈ (-∞, -1) ∪ (1, ∞) ⇒ Circles do not intersect  cos  θ ∈ (-1, 1) ⇒ Circles intersect each other  cos  θ = 0 ⇒ Circles intersect each other orthogonally  cos  θ ∈ {-1, 1} ⇒ Circles touch each other internally or externally. 

SOLVED EXAMPLE 36. From any point on the circle x2 + y2 = a2 tangents are drawn to the circle x2 + y2 = a2 sin2 α. The angle between them is

FIGURE 11.28

SOLVED EXAMPLE 37. The locus of the centres of the circles which cut the circles x2 + y2 + 4x - 6y + 9 = 0 and x2 + y2 - 4x + 6y + 4 = 0 orthogonally is (A) 8x - 12y + 5 = 0 (B)  8x + 12y - 5 = 0 (C) 12x - 8y + 5 = 0 (D)  none of these

Circles  11.19 Solution: (A)

Let the equation of one of the circles be x2 + y2 + 2gx + 2fy + c = 0 Since it cuts the given circles orthogonally, ∴ 2g (2) + 2f (-3) = c + 9 and 2g (-2) + 2f (3) = c + 4 i.e., 4g - 6f = c + 9  and  -4g + 6f = c + 4 On subtracting, we get, 8g - 12f = 5 i.e., -8 (-g) + 12 (-f ) = 5 So the locus of (-g, -f ) is - 8x + 12y = 5.

FIGURE 11.32

x ( x - x1 ) ( x - x2 ) + ( y - y1 ) ( y - y2 ) + l x1 x2

(where λ is a parameter)

FAMILY OF CIRCLES 1. The equation of the family of circles passing through the point of intersection of two given circles S = 0 and S ′ = 0 is given as S + lS′ = 0, (where λ is a parameter, λ ≠ -1)

y 1 y1 1 = 0, y2 1

SOLVED EXAMPLES 38. The intercept on the line y = x by the circle x2 + y2 - 2x = 0 is AB. Equation of the circle with AB as a diameter is (A) x2 + y2 + x + y = 0 (B)  x2 + y2 - x - y = 0 2 2 (C) x + y + x - y = 0 (D)  none of these Solution: (B)

FIGURE 11.29

2. The equation of the family of circles passing through the point of intersection of circle S = 0 and a line L = 0 is given as

FIGURE 11.30

S + λL = 0, (where λ is a parameter) 3. The equation of the family of circles touching the circle S = 0 and the line L = 0 at their point of contact P is

Equation of any circle passing through the point of intersection of x2 + y2 - 2x = 0 and y = x is x2 + y2 - 2x + λ(y - x) = 0 or x2 + y2 - (2 + λ) x + ly = 0

æ 2 + l -l ö Its centre is ç , . 2 ÷ø è 2 For AB to be the diameter of the required circle, the centre must lie on AB, i.e., 2 + l -l = Þ l = -1 2 2 Thus, equation of required circle is x2 + y2 - x - y = 0 39. The distance from the centre of the circle x2 + y2 = 2x to straight line passing though the points of intersection of the two circles x2 + y2 + 5x - 8y + 1 = 0 and x2 + y2 - 3x - 7y - 25 = 0 is 1 (A) (B)  2 3 (C) 3 (D)  1 Solution: (B)

FIGURE 11.31

S + λL = 0, (where λ is a parameter) 4. The equation of the family of circles passing through two given points P(x1, y1) and Q(x2, y2) can be written in the form

The 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., 8x - 15y + 26 = 0 (1) Also, centre of the circle x2 + y2 - 2x = 0 is (1, 0).

11.20  Chapter 11 ∴  Distance of the point (1, 0) from the straight line (1) is =

8(1) - 15(0) + 26 64 + 225

=

34 =2 17

IMAGE OF THE CIRCLE BY THE LINE MIRROR Let the circle be S ≡ x2 + y2 + 2gx + 2fy + c = 0 and the line be L ≡ lx + my + n = 0

The radius of the image circle remains unchanged but centre changes. Let the centre of image circle be (x1, y1). Then, Slope of C1C2 × Slope of the line L = -1(1) and mid point of C1 (- g, - f ) and C2 (x1, y1) lies on the line lx + my + n = 0 æ y1 - f ö æx -gö i.e., lç 1 ÷ + m ç 2 ÷ + n = 0 (2) 2 è ø è ø Solving eqns. (1) and (2), to get value of (x1, y1). Then the required image circle is (x - x1)2 + (y - y1)2 = r2 where

FIGURE 11.33

r = g2 + f 2 - c

Circles  11.21

PRACTICE EXERCISES Single Option Correct Type

a (A) (B)  a 2 (C)

3a (D)  2 3

2a

2. Let an, n = 1, 2, 3, 4 represent four distinct positive real numbers other than unity such that each pair of the logarithm of an and the reciprocal of logarithm denotes a point on a circle, whose centre lies on y-axis. The product of these four numbers is (A) 0 (B)  1 (C) 2 (D)  13 3. If the tangents PA and PB are drawn from the point P(-1, 2) to the circle x2 + y2 + x - 2y - 3 = 0 and C is the centre of the circle, then the area of the quadrilateral PACB is (A) 4 (B)  16 (C) does not exist (D)  none of these 4. 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 (B)  1 (C) any real value of m (D)  none of these 5. The number of points on the circle x2 + y2 - 4x - 10y + 13 = 0 which are at a distance 1 from the point (-3, 2) is (A) 1 (B)  2 (C) 3 (D)  none of these 6. If the equations of four circles are (x ± 4) + ( y ± 4) = 42, then the radius of the smallest circle touching all the four circles is 2

2

4( 2 - 1) (A) 4( 2 + 1) (B)  (C) 2( 2 - 1)

(D)  none of these

7. The intercept on the line y = x by the circle x2 + y2 - 2x = 0 is AB. Equation of the circle with AB as a diameter is (A) x2 + y2 + x + y = 0 (B)  x2 + y2 - x - y = 0 2 2 (C) x + y + x - y = 0 (D)  none of these

8. The locus of the mid-point of the chord of the circle x2 + y2 - 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 9. A square is inscribed in the circle x2 + y2 - 2x + 4y + 3 = 0. Its sides are parallel to the coordinate axes. Then, one vertex of the square is (A) (1 + 2 , - 2)

(B)  (1 - 2 , - 2)

(C) (1, - 2 + 2 )

(D)  none of these

10. If the lines a1x + b1 y + c1 = 0 and a2x + b2y + c2 = 0 cut the coordinate axes in concyclic points, then (A) a1a2 = b1b2 (B)  a1b1 = a2b2 (C) a1b2 = a2b1 (D)  none of these 11. The circle x2 + y2 = 4 cuts the line joining the points BP A(1, 0) and B(3, 4) in two points P and Q. Let =a PA BQ = b . Then, α and β are roots of the quadratic and QA equation (A) 3x2 + 2x - 21 = 0 (B)  3x2 + 2x + 21 = 0 2 (C) 2x + 3x - 21 = 0 (D)  none of these 12. If the equation of the incircle of an equilateral triangle is x2 + y2 + 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 13. Two distinct chords drawn from the point (p, q) on the circle x2 + y2 = px + qy, where pq ≠ 0, are bisected by the x-axis. Then, (A) | p | = | q | (B)  p2 = 8q2 2 2 (C) p < 8q (D)  p2 > 8q2 14. For the two circles x2 + y2 = 16 and x2 + y2 - 2y = 0, there is/are (A) one pair of common tangents (B) two pairs of common tangents (C) three common tangents (D) no common tangent

PRACTICE EXERCISES

1. An isosceles ∆ABC is inscribed in a circle x2 + y2 = a2 with the vertex A at (a, 0) and the base angles B and C each equal to 75º then length of the base BC is

11.22  Chapter 11 15. Let AB be a chord of the circle x2 + y2 = r2 subtending a right angle at the centre. Then, the locus of the centroid of the DPAB as P moves on the circle is (A) a parabola (B) a circle (C) an ellipse (D) a pair of straight lines 16. The equation of the smallest circle passing through the intersection of the line x + y = 1 and the circle x2 + y2 = 9 is (A) x2 + y2 + x + y - 8 = 0 (B) x2 + y2 - x - y - 8 = 0 (C) x2 + y2 - x - y + 8 = 0 (D) none of these 17. If the circle x2 + y2 + 2gx + 2fy + c = 0 bisects the circumference of the circles x2 + y2 + 2g′x + 2f′y + c′ = 0, then (A) 2g′(g - g′) + 2f ′(f - f ′) = c - c′ (B) g′(g - g′) + f′(f - f ′) + c - c′ = 0 (C) 2g(g - g′) + 2f(f - f ′) = c - c′ (D) none of these 18. If (a, b) is a point on the circle whose centre is on the x-axis and which touches the line x + y = 0 at (2, -2), then the greatest value of a is (A) 4 + 2 2 (B)  2+2 2

PRACTICE EXERCISES

(C) 4 + 2

(D)  none of these

19. The equation (x + y - 6) (xy - 3x - y + 3) = 0 represents the sides of a triangle then the equation of the circumcircle of the triangle is (A) x2 + y2 - 5x - 9y + 20 = 0 (B) x2 + y2 - 4x - 8y + 18 = 0 (C) x2 + y2 - 3x - 5y + 8 = 0 (D) x2 + y2 + 2x - 3y - 1 = 0 20. If a > 2b > 0 then the positive value of m for which y = mx - b 1 + m 2 is a common tangent to x2 + y2 = b2 and (x - a)2 + y2 = b2 is (A) (C)

2b a 2 - 4b 2



a 2 - 4b 2 (B)  2b

b 2b (D)  a - 2b a - 2b

21. If the locus of a point which moves so that the line joining the points of contact of the tangents drawn from it to the circle x2 + y2 = b2 touches the circle x2 + y2 = a2, is the circle x2 + y2 = c2, then a, b, c are in (A) A. P. (B)  G. P. (C) H. P. (D)  none of these

22. A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is (A) (y - q)2 = 4px (B) (x - q)2 = 4py 2 (C) (y - p) = 4qx (D) (x - p)2 = 4qy 23. The point (1, 4) lies inside the circle x2 + y2 - 6x - 10y + p = 0 which does not touch or intersect the coordinate axes, then (A) 0 < p < 29 (B)  25 < p < 29 (C) 9 < p < 25 (D)  9 < p < 29 24. A circle C1 of radius 2 touches both x-axis and y-axis. Another circle C2 whose radius is greater than 2 touches circle C1 and both the axes. Then, the radius of circle C2 is (A) 6 - 4 2 (B)  6+4 2 (C) 6 - 4 3 (D)  6+4 3 25. The equation to the sides AB, BC, CA of a ∆ABC are x + y = 1, 4x - y + 4 = 0 and 2x + 3y = 6. Circles are drawn on AB, BC, CA as diameters. The point of concurrence of the common chords is (A) centroid of the triangle (B) orthocentre (C) circumcentre (D) incentre 26. The coordinates of the point on the circle x2 + y2 - 2x - 4y - 11 = 0 farthest from the origin are æ 8 4 ö (A) ç 2 + , 1+ ÷ 5 5ø è æ 4 (B) ç1 + , 2+ 5 è æ 8 (C) ç1 + , 2+ 5 è (D) none of these

8 ö ÷ 5ø 4 ö ÷ 5ø

27. If the line 3x + ay - 20 = 0 cuts the circle x2 + y2 = 25 at real, distinct or coincident points, then a belongs to the interval (A) [- 7 , 7 ] (B) ( - 7 , 7 ) (C) ( - ¥ - 7 ] È [ 7 , ¥) (D) none of these 28. The locus of centre of the circle which touches the circle x2 + (y - 1)2 = 1 externally and also touches x-axis is (A) {(x, y): x2 + (y - 1)2 = 4} ∪ {(x, y): y < 0} (B) {(x, y): x2 = 4y} ∪ {(0, y): y < 0}

Circles  11.23

29. Let PQ and RS be tangents at the extremeties of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals PQ + RS (A) PQ × RS (B)  2 (C)

2PQ × RS PQ 2 + RS 2 (D)  PQ + RS 2

30. Circles are drawn through the point (-5, 0) to cut the x-axis on the positive side and making an intercept of 10 units on the x-axis. The equation of the locus of the centre of these circles is (A) x + y = 0 (B)  x-y=0 (C) x = 0 (D)  y=0 31. The circle x2 + y2 - 4x - 8y + 16 = 0 rolls up the tangent to it at ( 2 + 3, 3) by 2 units, assuming the x-axis as horizontal, the equation of the circle in the new position is (A) x 2 + y 2 - 6 x - 2( 4 + 3 ) y + 24 + 8 3 = 0 (B) x 2 + y 2 + 6 x - 2( 4 + 3 ) y + 24 + 8 3 = 0 (C) x 2 + y 2 - 6 x + 2( 4 + 3 ) y + 24 + 8 3 = 0 (D) none of these 32. The equation of the circle, passing through the point (2, 8), touching the lines 4x - 3y - 24 = 0 and 4x + 3y - 42 = 0 and having x coordinate of the centre of the circle numerically less then or equal to 8, is (A) x2 + y2 + 4x - 6y - 12 = 0 (B) x2 + y2 - 4x + 6y - 12 = 0 (C) x2 + y2 - 4x - 6y - 12 = 0 (D) none of these x y 33. If the line + = 1 moves in such a way that a b 1 1 1 + 2 = 2 , where c is a constant, then the locus 2 a b c of the foot of the perpendicular from the origin on the straight line describes the circle (A) x2 + y2 = 4c2 (B)  x2 + y2 = 2c2 2 2 2 (C) x + y = c (D)  none of these 34. A circle touches both the x-axis and the line 4x - 3y + 4 = 0. If its centre is in the third quadrant and lies on the line x - y - 1 = 0, then the equation of the circle is (A) 9(x2 + y2) + 6x + 24y - 1 = 0 (B) 9(x2 + y2) + 6x - 24y + 1 = 0 (C) 9(x2 + y2) + 6x + 24y + 1 = 0 (D) none of these

35. The line Ax + By + C = 0 cuts the circle x2 + y2 + ax + by + c = 0 in P and Q. The line A′x + B′y + C′ = 0 cuts the circle x2 + y2 + a′x + b′y + c′ = 0 in R and S. If P, Q, R, S are concyclic points, then a + a′ b + b′ c + c ′ (A) A B C =0 A′ B′ C′ (B)

a − a′ b − b′ c − c ′ A B C =0 A′ B′ C′

A(a + a′ ) B(b + b′ ) C (c + c′ ) (C) =0 A B C A′ B′ C′ (D) none of these 36. If q1, q2 be the inclinations of tangents drawn from the point P to the circle x2 + y2 = a2 and cotq1 + cotq2 = k, then the locus of P is (A) k(y2 + a2) = 2xy (B)  k(y2 - a2) = 2xy 2 2 (C) k(y - a ) = xy (D)  none of these 37. A line meets the coordinate axes in A and B. A circle is circumscribed about the ∆AOB. If m, n are the distances 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) (B)  m+n (C) n(m + n) (D)  none of these 38. If the chord of contact of tangents from a point on the circle x2 + y2 = a2 to the circle x2 + y2 = b2 touches the circle x2 + y2 = c2, then a, b, c are in (A) A. P. (B)  G. P. (C) H. P. (D)  none of these 39. To which of the following circles, the line y - x + 3 = 3 3 ö æ 0 is normal at the point ç 3 + , ÷? 2 2ø è 2

2

3 ö æ 3 ö æ (A) ç x - 3 ÷ +ç y÷ =9 2 2ø è ø è 2

2

3 ö æ 3 ö æ (B) ç x ÷ =9 ÷ +ç y2 2ø è ø è (C) x2 + (y - 3)2 = 9 (D) (x - 3)2 + y2 = 9 40. 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 =

PRACTICE EXERCISES

(C) {(x, y): x2 = y} ∪ {(0, y): y < 0} (D) {(x, y): x2 = 4y} ∪ {(x, y): y < 0}

11.24  Chapter 11 (A) 1 (B)  -1 -1 1 (C) (D)  2 2 41. Let C be any circle with centre (0, 2 ). Then, on the circle C, there can be (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 42. If the tangents PQ and PR are drawn to the circle x2 + y2 = a2 from the point P(x1, y1), then the equation of the circumcircle of DPQR is (A) x2 + y2 - xx1 - yy1 = 0 (B) x2 + y2 + xx1 + yy1 = 0 (C) x2 + y2 - 2xx1 - 2yy1 = 0 (D) none of these

PRACTICE EXERCISES

43. A ray of light, incident at the point (-2, -1), gets reflected from the tangent at (0, -1) to the circle x2 + y2 = 1. The reflected ray touches the circle. The equation of the line along which the incident ray moved is (A) 4x - 3y + 11 = 0 (B)  4x + 3y + 11 = 0 (C) 3x + 4y + 11 = 0 (D)  none of these 44. The coordinates of a point P on the circle x2 + y2 - 4x - 6y + 9 = 0 such that ∠POX is minimum, where O is the origin and OX is the x-axis, are æ -36 15 ö æ 36 15 ö (A) ç , ÷ (B)  ç 13 , 13 ÷ è ø è 13 13 ø 14 12 ö æ (C) ç , ÷ (D)  none of these è 27 27 ø 45. 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 (B) x2 + cx = b2 + c2 2 2 2 (C) y + 2cy = b + c (D)  none of these 46. 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 p (B) cuts the circle S at an angle 4 (C) cuts the circle S orthogonally (D) none of these ×

47. 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 æ 5 -5 ö 5 5 (A) æç , ÷ö (B)  ç 12 , 12 ÷ è ø è 12 12 ø æ -5 5 ö (C) ç , ÷ è 12 12 ø

(D)  none of these

48. If the circles x2 + y2 = 1 and x2 + y2 - 4x - 6y + 12 = 0 cut off equal intercepts on a line which passes through the point (1, 1), then the slope of the line is (A) 1 (B)  -1 3 3 (C) (D)  2 2 49. Consider a curve ax2 + 2hxy + by2 = 1 and a point P not on the curve. A line drawn from the point P intersects the curve at points Q and R. If the product PQ.PR is independent of the slope of the line, then the curve is (A) an ellipse (B)  a hyperbola (C) a circle (D)  none of these 50. 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 + y2 - x + 3y = 0 on L1 and L2 are equal, then which of the following equations can represent L1? (A) x + y = 0 (B)  x-y=0 (C) 7y + 2x = 0 (D)  x - 7y = 0 51. A triangle has two of its sides along the axes. If the third side touches the circle x2 + y2 - 2ax - 2ay + a2 = 0, then the equation of the locus of the circumcentre of the triangle is (A) 2a(x + y) = 2xy + a2 (B) 2a(x - y) = 2xy + a2 (C) 2a(x + y) = 2xy - a2 (D) none of these 52. 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) centre of the circle coincides with that of square æ1 1ö (B) centre of the circle is ç , ÷ è2 2ø (C) radius of the circle is 2 (D) all the above are true 53. The range of values of a for which the line y + x = 0 bi­ æ 1 + 2a 1 - 2a ö , sects two chords drawn from a point çç ÷ 2 ÷ø è 2 to the circle 2x2 + 2y2 - (1 + 2a) x -(1 - 2a) y = 0 is (A) (-∞, -2) ∪ (2, ∞) (B)  (-2, 2) (C) (2, ∞) (D)  none of these 54. The locus of the centres of the circles which touch the two circles x2 + y2 = a2 and x2 + y2 = 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

Circles  11.25

56. A circle touches the line y = x at a point P such that OP = 4 2, where O is the origin. The circle contains the point (-10, 2) in its interior and the length of its chord on the line x + y = 0 is 6 2. The equation of the circle is (A) x2 + y2 + 18x - 2y + 32 = 0 (B) x2 + y2 - 18x - 2y + 32 = 0 (C) x2 + y2 + 18x + 2y + 32 = 0 (D) none of these 57. If S ≡ x2 + y2 + 2gx + 2f y + c = 0 is a given circle, then the locus of the foot of the perpendicular drawn from origin upon any chord of S which subtends a right angle at the origin, is (A) 2(x2 + y2) + 2gx + 2fy + c = 0 (B) 2(x2 + y2) + 2gx + 2fy - c = 0 (C) x2 + y2 + gx + fy + c = 0 (D) none of these 58. The equation of the circle, having the lines x2 + 2xy + 3x + 6y = 0 as its normals and having size just sufficient to contain the circle x(x - 4) + y( y - 3) = 0, is (A) x2 + y2 + 6x + 3y - 45 = 0 (B) x2 + y2 + 6x - 3y - 45 = 0 (C) x2 + y2 + 6x - 3y + 45 = 0 (D) none of these 59. 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 + y2 = 9, is (A) ( x 2 + y 2 - 18) + l ( 2 x + 4 y - 18) = 0 (B) ( x 2 + y 2 - 18) + l ( 4 x + 2 y - 18) = 0 (C) ( x 2 + y 2 - 16) + l ( 2 x + 4 y - 16) = 0 (D) none of these 60. The point on the straight line y = 2x + 11 which is nearest to the circle 16(x2 + y2) + 32x - 8y - 50 = 0 is æ 9 ö 9 (A) æç , 2 ö÷ (B)  ç - 2 , 2÷ è ø è2 ø 9 æ ö (C) ç , - 2 ÷ (D)  none of these è2 ø 61. Extremities of a diagonal of a rectangle are (0, 0) and (4, 3). The equations of the tangents to the circumcircle of the rectangle which are parallel to this diagonal are (A) 16x + 8y ± 25 = 0 (B)  6x - 8y ± 25 = 0 (C) 8x + 6y ± 25 = 0 (D)  none of these

62. The base AB of a triangle is fixed and its vertex C moves such that sin A = k sin B(k ≠ 1). If a is the length of the base AB, then the locus of C is a circle whose radius is equal to ak ak (A) (B)  2 k2) ( 1 (2 - k ) (C) 2ak (D)  none of these 1- k 2 63. The equation of the image of the circle x2 + y2 + 16x - 24y + 183 = 0 by the line mirror 4x + 7y + 13 = 0 is (A) x2 + y2 + 32x + 4y + 235 = 0 (B) x2 + y2 - 32x + 4y + 235= 0 (C) x2 + y2 + 32x + 4y - 235 = 0 (D) none of these 64. The locus of the centre of a circle touching the circle x 2 + y 2 - 4 y - 2 x = 2 3 - 1 internally and tangents on which from (1, 2) is making a 60º angle with each other, is (A) (x - 1)2 + ( y - 2)2 = 3 (B) ( x - 2) 2 + ( y - 1) 2 = 1 + 2 3 (C) x2 + y2 = 1 (D) none of these 65. The equation of locus of the point of intersection of tangents to the circle x2 + y2 = 1 at the points whose parametric angles differ by 60º is (A) 3x2 + 3y2 = 1 (B)  x2 + y2 = 3 2 2 (C) 3x + 3y = 4 (D)  none of these 66. If a square is inscribed in the circle x2 + y2 - 2x + 4y + 3 = 0 and its sides are parallel to the coordinate axes, then one vertex of the square is (1 - 2 , - 2) (A) (1 + 2 , - 2) (B)  (C) (1, -2 + 2 )

(D)  none of these

67. The equation of the chord of the circle x2 + y2 = a2 passing through the point (2, 3) and farthest from the centre is (A) 2x + 3y = 13 (B)  3x + 2y = 13 (C) 2x - 3y = 13 (D)  none of these 68. The range of values of p such that the angle θ between the pair of tangents drawn from the point (p, 0) to the ö æp circle x2 + y2 = 1 lies in ç , p ÷ is 3 è ø (A) (-2, -1) ∪ (1, 2) (B)  (-3, -2) ∪ (2, 3) (C) (0, 2) (D)  none of these 69. A circle whose centre coincides with the origin having radius ‘a’ cuts x-axis at A and B. If P and Q are two points on the circle whose parametric angles differ by 2θ, then the locus of the intersection point of AP and BQ is

PRACTICE EXERCISES

55. If a circle passes through the points of intersection of the coordinate axes with the lines lx - y + 1 = 0 and x - 2y + 3 = 0, then the value of λ is (A) 2 (B)  1 (C) -1 (D)  -2

11.26  Chapter 11 (A) x2 + y2 + 2ay tanθ = a2 (B) x2 + y2 - 2ay tanθ = a2 (C) x2 + y2 + 2ay cotθ = a2 (D) none of these 70. If a chord AB subtends a right angle at the centre of a given circle, then the locus of the centroid of the triangle PAB as P moves on the circle is a/an (A) parabola (B)  ellipse (C) hyperbola (D)  circle 71. If -3l2 - 6l -1 + 6m2 = 0 then the equation of the circle for which lx + my + 1 = 0 is a tangent is (A) (x + 3)2 + y2 = 6 (B)  (x - 3)2 + y2 = 6 2 2 (C) x + (y - 3) = 6 (D)  x2 + (y + 3)2 = 6

72. Let S1 and S2 be two circles with S2 lying inside S1. A circle S lying inside S1 touches S1 internally and S2 externally. The locus of the centre of S is a/an (A) parabola (B)  ellipse (C) hyperbola (D)  circle 73. S(x, y) = 0 represents a circle. The equation S(x, 2) = 0 gives two identical solutions x = 1 and the equation S(1, y) = 0 gives two distinct solutions y = 0, 2. The equation of the circle is (A) x2 + y2 + 2x + 2y + 1 = 0 (B) x2 + y2 + 2x + 2y - 1 = 0 (C) x2 + y2 - 2x - 2y + 1 = 0 (D) none of these

Previous Year's Questions

PRACTICE EXERCISES

74. The greatest distance of the point P(10, 7) from the circle x2 + y2 − 4x − 2y − 20 = 0 is [2002] (A) 10 unit (B)  15 unit (C) 5 unit (D)  none of these

(A) (x − p)2 = 4qy (C) (y − p)2 = 4qx

(B) (x − q)2 = 4py (D) (y − q)2 = 4px

75. The equation of the tangent to the circle x2 + y2 + 4x − 4y + 4 = 0 which make equal intercepts on the positive co-ordinate axes, is [2002] (A) x + y = 2 (B)  x+y= 2 2 (C) x + y = 4 (D)  x+y=8

80. 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 [2004] (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

76. If the two circles (x − 1)2 + (y − 3)2 = r2 and x2 + y2 − 8x + 2y + 8 = 0 intersect in two distinct points, then  [2003] (A) 2 < r < 8 (B)  r2

81. The intercept on the line y = x by the circle x2 + y2 − 2x = 0 is AB. Equation of the circle on AB as a diameter is  [2004] (A) x2 + y2 − x − y = 0 (B)  x2 + y2 − x + y = 0 (C) x2 + y2 + x + y = 0 (D)  x2 + y2 + x − y = 0

77. The lines 2x − 3y = 5 and 3x − 4y = 7 are diameters of a circle having area as 154 sq units. Then the equation of the circle is [2003] 2 2 (A) x + y + 2x − 2y = 62 (B) x2 + y2 + 2x − 2y = 47 (C) x2 + y2 − 2x + 2y = 47 (D) x2 + y2 − 2x + 2y = 62

82. If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 − 3ax + dy − 1 = 0 intersect in two distinct points P and Q then the line 5x + by − a = 0 passes through P and Q for [2005] (A) exactly one value of a (B) no value of a (C) infinitely many values of a (D) exactly two values of a

78. If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is [2004] (A) 2ax + 2by + (a2 + b2 + 4) = 0 (B) 2ax + 2by − (a2 + b2 + 4) = 0 (C) 2ax − 2by + (a2 + b2 + 4) = 0 (D) 2ax − 2by − (a2 + b2 + 4) = 0 79. A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is [2004]

83. A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is [2005] (A) an ellipse (B)  a circle (C) a hyperbola (D)  a parabola 84. If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is [2005]

Circles  11.27

85. If the lines 3x − 4y − 7 = 0 and 2x − 3y − 5 = 0 are two diameters of a circle of area 49π square units, the equation of the circle is [2006] (A) x2 + y2 + 2x − 2y − 47 = 0 (B) x2 + y2 + 2x − 2y − 62 = 0 (C) x2 + y2 − 2x + 2y − 62 = 0 (D) x2 + y2 − 2x + 2y − 47 = 0 86. Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of 2p the chords of the circle C that subtend an angle of 3 at its centre is [2006] ×

3 2 27 (C) x 2 + y 2 = 4 (A) x 2 + y 2 =

(B)  x2 + y2 = 1 (D)  x2 + y2 =

9 4

87. Consider a family of circles which are passing through the point (−1, 1) and are tangent to x-axis. If (h, k) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interval [2007] 1 1 (A) 0 < k < (B)  k³ 2 2 1 1 1 (C) − ≤ k ≤ (D)  k≤ 2 2 2 ⋅

88. The point diametrically opposite to the point P(1, 0) on the circle x2 + y2 + 2x + 4y − 3 = 0 is [2008] (A) (3, −4) (B)  (−3, 4) (C) (−3, −4) (D)  (3, 4) 89. If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y + 2p − 5 = 0 and x2 + y2 + 2x + 2y − p2 = 0, then there is a circle passing through P, Q and (1, 1) for [2009] (A) all values of p (B) all except one value of p (C) all except two values of p (D) exactly one value of p 90. Three distinct points A, B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of anyone of them from the point (1, 0) to 1 the distance from the point (−1, 0) is equal to Then 3 the circumcentre of the triangle ABC is at the point  [2009]

æ5 ö (A) (0, 0) (B)  ç 4 , 0÷ è ø æ5 ö 5 ö æ (C) ç , 0 ÷ (D)  ç 3, 0÷ è2 ø è ø 91. The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x − 4y = m at two distinct points then m satisfies [2010] (A) −35 < m < 15 (B)  15 < m < 65 (C) 35 < m < 85 (D)  −85 < m < −35 92. The two circles x2 + y2 = ax and x2 + y2 = c2 (c > 0) touch each other if [2011] (A) |a| = c (B)  a = 2c (C) |a| = 2c (D)  2|a| = c 93. The length of the diameter of the circle which touches the x-axis at the point (l, 0) and passes through the point (2, 3) is [2012] 10 3 (A) (B)  3 5 6 5 (C) (D)  5 3 94. The circle passing through (1, −2) and touching the x-axis at (3, 0) also passes through the point [2013] (A) (2, −5) (B)  (5, −2) (C) (−2, 5) (D)  (−5, 2) 95. Let C be the circle with centre at (l, 1) and with radius 1. If T is the circle centered at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to [2014] 3 (A) 3 (B)  2 2 1 1 (C) (D)  2 4 96. The number of common tangents to the circles x2 + y2 − 4x − 6y − 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0, is  [2015] (A) 2 (B)  3 (C) 4 (D)  1 97. If one of the diameters of the circle, given by the ­equation. x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is [2016] 5 2 (A) 10 (B)  (C) 5 3 (D)  5

PRACTICE EXERCISES

(A) x2 + y2 − 3ax − 4by + (a2 + b2 − p2) = 0 (B) 2ax + 2by − (a2 − β 2 + p2) = 0 (C) x2 + y2 − 2ax − 3by − (a2 − β 2 − p2) = 0 (D) 2ax + 2by − (a2 + b2 + p2) = 0

11.28  Chapter 11 98. Let the orthocentre and centroid of a triangle be A(–3,  5) and B(3, 3), respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is [2018] (A) 

3 5 2

(C)  3

(B)  2 10

5 2

(D) 

99. If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the value of c is  [2018] (A) 195 (B) 185 (C) 85 (D) 95

10

ANSWER K EYS Single Option Correct Type 1. (B) 11. (A) 21. (B) 31. (A) 41. (B) 51. (A) 61. (B) 71. (B)

2. (B) 12. (B) 22. (D) 32. (C) 42. (A) 52. (D) 62. (B) 72. (B)

3. (C) 13. (D) 23. (B) 33. (C) 43. (B) 53. (A) 63. (A) 73. (C)

4. (C) 14. (D) 24. (B) 34. (C) 44. (A) 54. (A) 64. (A)

5. (D) 15. (B) 25. (B) 35. (B) 45. (A) 55. (A) 65. (C)

6. (B) 16. (B) 26. (B) 36. (B) 46. (C) 56. (A) 66. (D)

7. (B) 17. (A) 27.  (C) 37. (B) 47. (A) 57. (A) 67. (A)

8. (A) 18. (A) 28.  (B) 38. (B) 48. (C) 58. (B) 68. (A)

9. (D) 19. (B) 29. (A) 39. (D) 49. (C) 59. (A) 69. (B)

10. (A) 20. (A) 30. (C) 40. (C) 50. (B) 60. (B) 70. (D)

77. (C) 87. (B) 97. (C)

78. (B) 88. (C) 98. (C)

79. (A) 89. (A) 99. (D)

80. (A) 90. (B)

81. (A) 91. (A)

82. (B) 92. (A)

83. (D) 93. (A)

Previous Years’ Questions

PRACTICE EXERCISES

7 4. (B) 84. (D) 94. (B)

75. (B) 85. (D) 95. (D)

76. (A) 86. (D) 96. (B)

Circles  11.29

HINTS AND EXPLANATIONS Single Option Correct Type 1. ∠B = ∠C = 75º ⇒ ∠BAC = 30º ⇒ ∠BOC = 60º

∴ (-1, 2) is an interior point of the circle. Thus, m can have any real value. 5. Given circle is x2 + y2 - 4x - 10y + 13 = 0 Its centre is C ≡ (2, 5) and radius = 4 Also, AC = ( 2 - 5) 2 + ( -3 - 2) 2

⇒ BOC is an equilateral triangle ⇒ BC = OB = the radius of the cirlce ⇒ BC = a. 2. Let (0, a) be the centre and r be the radius of the given circle, then its equation is (x - 0)2 + (y - a)2 = r2 ⇒ x2 + y2 - 2ay + a2 - r2 = 0 (1) æ 1 ö Since the point ç log an , ÷ ; n = 1, 2, 3, 4 lie on the above log an ø è 1 2a + a 2 - r 2 = 0, n circle, therefore (log an ) 2 + 2 (log a ) log a n n = 1, 2, 3, 4 ⇒ loga1, loga2, loga3, loga4 are the roots of the equation

λ4 + (a2 - r2)l2 - 2aλ + 1 = 0. ∴  Sum of the roots = 0 ⇒ loga1 + loga2 + loga3 + loga4 = 0 ⇒ log(a1a2a3a4) = 0  or  a1a2a3a4 = 1. 3. The given circle is S : x2 + y2 + x - 2y - 3 = 0. Since S]P  (-1, 2) = 1 + 4 - 1 - 4 - 3 = -3 < 0, the point P(-1, 2) lies inside the circle. Consequently, the tangents from the point P(-1, 2) to the circle do not exist. Thus, the quadrilateral PACB cannot be formed. 4. 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,

= 4 2 + 4 2 - 4   i.e.,  4 2 - 4. 7. Equation of any circle passing through the point of intersection of x2 + y2 - 2x = 0 and y = x is

x2 + y2 - 2x + λ(y - x) = 0

or,

x2 + y2 - (2 + λ) x + ly = 0.

æ 2 + l -l ö , . Its centre is ç 2 ÷ø è 2 For AB to be the diameter of the required circle, the centre must lie on AB, i.e., 2 + l -l = 2 2

Þ l = -1.

Thus, equation of required circle is x2 + y2 - x - y = 0. 8. Given equation of circle is x2 + y2 - 2x - 2y - 2 = 0. Let mid-point of chord AB be (h, k) Its centre is (1, 1) and radius = In DOPB, ∠OBP = 30º. ∴ sin 30º = OP/2 or OP = 1.

1 + 1 + 2 = 2 = OB

HINTS AND EXPLANATIONS

= 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). 6. Clearly, from the figure, the radius of the smallest circle touching the given circles is

11.30  Chapter 11 12. Given equation of the incircle is x2 + y2 + 4x - 6y + 4 = 0.

Since, OP = 1  ⇒ (h - 1)2 + (k - 1)2 = 1 or, h2 + k2 - 2h - 2k + 1 = 0 ∴  Locus of mid point of chord is x2 + y2 - 2x - 2y + 1 = 0. 9. The 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. 10. The line a1x + b1y + c1 = 0 cuts the coordinate axes at A(-c1/ a1, 0) and B(0, - c1/b1) and the line a2x + b2y + c2 = 0 cuts the axes at C(-c2/a2, 0) and D(0, -c2/b2). So, AC and BD are chords along x-axis and y-axis, respectively, intersecting at origin O. Since A, B, C, D are concyclic, therefore OA.OC = OB.OD

Its incentre is (-2, 3) and inradius = 4 + 9 - 4 = 3. Since in an equilateral triangle, the incentre and the circumcentre coincide, ∴ Circumcentre ≡ (-2, 3). Also, in an equilateral triangle, circumradius = 2 (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.

QUICK TIPS In an equilateral triangle  the incentre and the circumcentre coincide  circum radius = 2 (in radius) 13. Given circle is x2 + y2 = px + qy. æ p qö Since the centre of the circle is ç , ÷ , so (p, q) and (0, è 2 2ø 0) are the end points of a diameter. As the two chords are bisected by x-axis, the chords will cut the circle at the points (x1, -q) and (x2, -q), where x1 , x2 are real.

æ -c1 ö æ -c2 ö æ -c1 ö æ -c2 ö ç ÷. ç ÷=ç ÷. ç ÷ è a1 ø è a2 ø è b1 ø è b2 ø or, a1a2 = b1b2. 11. The equation of the line joining A(1, 0) and B(3, 4) is y = 2x  8 6 -2. This cuts the circle x2 + y2 = 4 at Q(0, -2) and P  ,  .  5 5

HINTS AND EXPLANATIONS

⇒

We have, BQ = 3 5, QA = 5, BP = ∴ a =

7 5

and PA =

3

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. 14. Given circles are

5



S1: x2 + y2 - 16 = 0

(1)

and,

S2: x + y - 2y = 0

(2)

BP 7 / 5 7 BQ 3 5 = = and b = = = -3 PA 3/ 5 3 QA - 5

∴ α, β are roots of the equation x2 - x(α + β) + aβ = 0 æ7 ö 7 i.e., x 2 - x ç - 3 ÷ + ( -3) = 0 è3 ø 3 or, 3x2 + 2x - 21 = 0.

2

2

Centre of S1 is C1: (0, 0) and radius r1 = 4 Centre of S2 is C2: (0, 1) and radius r2 = 1 ∴

C1C2 = 0 + 1 = 1

Since | C1C2 | < | r1 - r2 |,  ∴  S2 is completely within S1 and hence there are no common tangents to the two circles.

Circles  11.31 18. Since the slope of the given line is -1, ∴ ∠COP = 45º ∴ OP = 2 2 = CP ∴

a=

r + r cosq r + r sin q ,b = 3 3 2

or,

2

rö æ rö r2 æ ça - 3 ÷ + ç b - 3 ÷ = 9 è ø è ø 2

2

2

r  r   r ∴  The locus is  x −  +  y −  =   , which is a      3 3 3 circle. 16. The equation of a circle passing through the intersection of the given line and the circle is (x2 + y2 - 9) + k(x + y - 1) = 0 kö æ k Its centre is ç - , - ÷ . 2 2ø è kö æ k The circle is the smallest if the centre ç - , - ÷ lies on the 2 2ø è chord x + y = 1

OC = ( 2 2 ) 2 + ( 2 2 ) 2 = 4

The point on the circle with the greatest x-coordinate is A. ∴  a = OA = OC + CA = 4 + 2 2. 19. We have, x + y = 6 and, xy - y - 3x + 3 = 0 ⇒  y(x - 1) - 3(x - 1) = 0 ⇒ (x - 1)(y - 3) = 0 Equations of the sides of the triangle are x + y = 6 (1) y = 3 (2) x = 1 (3)

k k ∴  - - = 1 Þ k = -1 2 2 Thus, the equation of the smallest circle is (x2 + y2 - 9) - 1(x + y - 1) = 0 i.e., x2 + y2 - x - y - 8 = 0 17. The given circles are S1: x2 + y2 + 2gx + 2fy + c = 0 and, S2: x2 + y2 + 2g′x + 2f′y + c′ = 0 The equation of common chord of (1) and (2) is S1 - S2 = 0 i.e., 2(g - g′)x + 2(f - f ′ )y + (c - c′) = 0

(1) (2)

(3)

Since (1) bisects the circumference of (2), therefore common chord will be the diameter of circle (2) ∴ Centre (-g′, - f ′) of circle (2) lies on (3). ∴ -2(g - g′)g′ - 2(f - f ′)f ′ + c - c′ = 0 or, 2g′(g - g′ ) + 2f ′(f - f  ′ ) = c - c′.

QUICK TIPS If a circle bisects the circumference of another circle, then their common chord is the diameter of second circle

Shaded triangle is right angled at (1, 3). ∴  the circumcircle is the circle on (3, 3) and (1, 5) as ends of a diameter and its equation is (x - 3)(x - 1) + (y - 3) (y - 5) = 0, i.e., x2 + y2 - 4x - 8y + 18 = 0. 20. Clearly, the line y = mx - b 1 + m 2 will pass from the point (a/2, 0) (mid-point of the centres of the circles) Þ m=

2b 2

a – 4b 2

HINTS AND EXPLANATIONS

15. Let the centroid ≡ (α, β). Then,

11.32  Chapter 11 21. Let P(h, k) be any point on the locus. Equation of the chord of contact of P with respect to the circle x2 + y2 = b2 is hx + ky = b2. If it touches the circle x2 + y2 = a2, then −b 2 h2 + k 2

= a ⇒ a 2 ( h2 + k 2 ) = b 4

25. Since ADB = ADC = 90°, circles on AB and AC as diameters pass through D and therefore the altitude AD is the common chord. Similarly, the other two common chords are the other two altitudes and hence they concur at the ortho-centre.

So that the locus of P(h, k) is x2 + y2 = (b2/a)2



2

æ b2 ö ∴ c 2 = ç ÷ Þ ac = b 2 è aø ⇒ a, b, c are in G. P. 22. Let the variable circle be x2 + y2 + 2gx + 2fy + c = 0 ∴ p2 + q2 + 2gp + 2fq + c = 0 Circle (1) touches x-axis, ∴ g2 - c = 0 ⇒ c = g2. From (2) p2 + q2 + 2gp + 2fq + g2 = 0 Let the other end of diameter through (p, q) be (h, k), h+ p k+q = - g and =-f 2 2 Put in (3) 2 æ h+ pö æ k +qö æh+ pö p2 + q2 + 2 p ç + 2q ç +ç =0 ÷ ÷ ÷ 2 ø è 2 ø 2 ø è è

HINTS AND EXPLANATIONS



(1) (2)

(3) then

⇒ h2 + p2 - 2hp - 4kq = 0 ∴  Locus of (h, k) is x2 + p2 - 2xp - 4yq = 0  ⇒  (x - p)2 = 4qy 23. Since the circle does not touch or intersect the coordinates axes, the absolute values of x and y coordinates of the centre are greater than the radius of the circle. Coordinates of the centre of the circle are (3, 5) and the radius is 9 + 25 - p so that 3 > 9 + 25 - p

Þ

26. The equation of the given circle can be written as (x - 1)2 + (y - 2)2 = 16 = 42 So, the coordinates of any point P on the circle are (1 + 4 cosθ, 2 + 4 sinθ) whose distance from the origin is d = (1 + 4 cosq ) 2 + ( 2 + 4 sin q ) 2

= 21 + 8 cosq + 16 sin q

⇒ d = 21 + 8(cosq + 2 sin q ) = 21 + 8r cos(q - a ) where r cos α = 1, r sin α = 2 = 21 + 8 5 cos(q - a )



which is maximum when cos (θ - α) = 1, i.e., θ = α = tan-1 2. 1 2 ⇒ tanθ = 2 so sinq = = and cosq = 5 5 and, the coordinates of the required point are 8 ö 4 æ , 2+ ç1 + ÷. 5ø 5 è

p > 25 (1)

5 > 9 + 25 - p Þ p > 9 (2) and the point (1, 4) lies inside the circle ⇒ 1 + 16 - 6 - 10 × 4 + p < 0  ⇒  p < 29 (3) From (1), (2), (3) we get 25 < p < 29. 24. First circle touches both axes and radius is 2 unit. Hence, centre of circle is (2, 2). Let radius of other circle be a and this circle also touches both the axes. Hence, centre of circle is (a, a). This circle touches first circle Hence, ( a - 2) 2 + ( a - 2) 2 = a + 2 On squaring both the sides, we get a2 - 12a + 4 = 0 12 ± (12) 2 − 4 × 4 × 1 12 ± 128 ⇒ a= = =6±4 2 2 2 But a > 2, therefore a = 6 - 4 2 is neglected. Hence, a = 6 + 4 2

27. The length of the ⊥ from the centre (0, 0) of the given circle to the line 3x + ay - 20 = 0 is | 3(0) + a(0) - 20 | 9 + a2

=

20 9 + a2

.

Radius of the given circle = 5 Since the line cuts the circle at real, distinct or coincident points ∴ 

20 9 + a2

£ 5   ⇒  a2 + 9 ≥ 16  ⇒  a2 - 7 ≥ 0

⇒ ( a + 7 )( a - 7 ) ³ 0 ⇒ a Î ( - ¥, - 7 ] È [ 7 , ¥) 28. According to given condition ( h - 0) 2 + ( k - 1) 2 = 1+ | k | ⇒ h2 + (k - 1)2 = (1 + |k |)2 ⇒ h2 = 2k + 2|k |

Circles  11.33 Equation of tangent to the circle (1) at P( 2 + 3 , 3) is ( 2 + 3 ) x + 3 y - 2( x + 2 + 3 ) - 4( y + 3) + 16 = 0 or, 

3 x - y - 2 3 = 0 (2)

If line (2) makes an angle θ with the positive direction of x-axis, then tan q = 3,   ∴  θ = 60º. Hence locus is x2 = 2y + 2| y | Clearly, for y > 0, x2 = 4y and for y < 0, x2 = 0  ⇒  x = 0. PQ PQ = PR 2r æp ö RS Also, tan ç - q ÷ = è2 ø 2r 29. tan q =

i.e.,  cot q =

RS 2r

Let A and B be the centres of the circles in old and new positions, respectively, then A ≡ (2, 4) and B ≡ (2 + 2 cos 60º, 4 + 2 sin 60º)( AB = 2) Thus, B º (3, 4 + 3 ).

PQ × RS 4r 2 ⇒ 4r2 = PQ ⋅ RS

\

∴ tanq cot q =

Radius of the circle = 22 + 4 2 - 16 = 2. ∴  Equation of the circle in the new position is ( x - 3) 2 + ( y - 4 - 3 ) 2 = 22 or, x 2 + y 2 - 6 x - 2( 4 + 3 ) y + 24 + 8 3 = 0



∴  radius of the circle = (a - 2) 2 + ( b - 8) 2 . Since the circle touches the lines 4x - 3y - 24 = 0 and, 4x - 3y - 42 = 0, | 4a - 3b - 24 | | 4a + 3b - 42 | 2 2 = = (a - 2) + ( b - 8) 5 5 (i) (ii) (iii) From (i) and (ii), we get ∴  ⇒ 2r = ( PQ )( RS ). 30. Since the circles pass through (-5, 0) and makes an intercept of 10 units on the positive side of the x-axis, it also passes through (5, 0). ∴ if C(x, y) is the centre of such a circle, (x + 5)2 + y2 = (x - 5)2 + y2 (   CA = CB = radius) ⇒  x = 0.

4α - 3β - 24 = ± (4α + 3β - 42) ì6 b = 18 Þ b = 3 ( taking positive sign ) ï 33 ∴  í ïî8a = 66 Þ a = 4 ( taking negative sign). 33 . 4 Putting β = 3 in equations (i) and (iii) and equating, we get (4α - 33)2 = 25 [(α - 2)2 + 25] or, 16a2 - 264α + 1089 = 25a2 + 725 - 100α or, 9a2 + 164α - 364 = 0 Given, | α | ≤ 8 or -8 ≤ α ≤ 8,  ∴  a ¹

-164 ± (164) 2 + 36 ´ 364 18 -164 ± 200 -182 = = 2, . 18 9 But -8 ≤ α ≤ 8,  ∴  α = 2.

\

∴  a = 31. Given circle is x2 + y2 - 4x - 8y + 16 = 0 Let P = ( 2 + 3 , 3).

(1)

×

HINTS AND EXPLANATIONS

32. Let C (α, β) be the centre of the circle. Since the circle passes through the point (2, 8),

11.34  Chapter 11 Now, (radius)2 = (α - 2)2 + (3 - 8)2 = (2 - 2)2 + (3 - 8)2 = 25. Hence, the equation of the required circle is (x - 2)2 + (y - 3)2 = 25 2 2 or, x + y - 4x - 6y - 12 = 0. x y 33. The equation of the line is + = 1 (1) a b 1 1 1 where, 2 + 2 = 2 (2) a b c Here, a, b are parameters while c is a constant. x y - + k = 0. b a If it passes through the origin then k = 0. ∴  Equation of the line through the origin and ^r to (1) is x y - = 0 (3) b a The locus of the foot of the ^r from origin on (1), i.e., locus of the point of intersection of (1) and (3) is obtained by eliminating the parameters a and b between them. Squaring (1) and (3) and adding, we get Any line ⊥ r to (1) is



HINTS AND EXPLANATIONS

or,

1ö 2 æ 1 1 ö 2 æ 1 ç a2 + b2 ÷ x + ç b2 + a2 ÷ y = 1 è ø è ø 1 2 1 2 x + 2 y =1 c2 c

(Using (2)

or x2 + y2 = c2, which is clearly a circle with centre at origin and radius c. 34. Let the equation of the required circle be x2 + y2 + 2gx + 2f y + c = 0 (1) Its centre is C(-g, -f ) and radius is g 2 + f 2 - c . Since circle (1) touches the x-axis ∴  g2 - c = 0 or c = g2(2) Again, since circle (1) touches the line 4x - 3y + 4 = 0 (3) | - 4 g + 3 f + 4| 2 2 2 ∴  = g + f - c = f = |f | 5  [from (2)] or, - 4g + 3f + 4 = ± 5f ∴ 4g + 2f = 4 or 2g + f = 2 (4) - 4g + 8f = - 4  or  g - 2f = 1 (5) Again, since centre C(-g,- f ) lies on the line x-y-1=0 ∴  -g + f = 1 (6) 1 4 Solving (4) and (6), we get g = , f = . 3 3 æ -1 -4 ö Thus, C = ç , ÷ which lies in the third quadrant. è 3 3ø 1 Also, from (2), c = g 2 = . 9 Solving (5) and (6), we get f = -2, g = -3 ∴  C ≡ (3, 2) which lies in the first quadrant.

1 4 1 Thus, for the required circle g = , f = , c = . 3 3 9 ∴  Equation of the required circle is x2 + y2 +

2 8 1 x+ y+ =0 3 3 9

or, 9(x2 + y2) + 6x + 24y + 1 = 0. 35. Equation of any circle through the points of intersection P and Q of line Ax + By + C = 0 (1) and circle x2 + y2 + ax + by + c = 0 (2) is x2 + y2 + ax + by + c + λ (Ax + By + C) = 0 or, x2 + y2 + (a + λA) x + (b + λB) y + c + λC = 0 (3) Again, equation of any circle through the points of intersection R and S of line A′x + B′y + C′ = 0 (4) and circle x2 + y2 + a′ x + b′ y + c′ = 0 (5) 2 2 is x + y + a′ x + b′ y + c′ + µ(A′ x + B′ y + C′) = 0 or, x2 + y2 + (a′ + µA′) x + (b′ + µB′) y + c′ + µC′ = 0 (6) If circles (3) and (6) are same, then points P, Q, R, S will lie on the same circle, i.e., points P, Q, R, S will be concyclic. Comparing the coefficients in (3) and (6), we get b + lB 1 1 a + lA c + lC = = = = 1 1 a¢ + m A¢ b¢ + m B¢ b¢ + mC ¢ (i) (ii)  (iii)  (iv)  (v) From (i) and (iii), we get a - a′ + λA - µA′ = 0 From (i) and (iv), we get b - b′ + λB - µB′ = 0 From (i) and (v), we get c - c′ + λC - µC′ = 0  Eliminating λ and -µ from equations (7), (8) and (9) writing the result in determinant form, we get

(7) (8) (9) and

a − a′ A A′ b − b′ B B = 0 c − c′ C C ′ a − a′ b − b′ c − c ′ or,  A B C = 0. A′ B′ C′ 36. Equation of the circle is x2 + y2 = a2(1) Let P be the point (x1, y1). Equation of any tangent to (1) is y = mx + a 1 + m 2 If it passes through P(x1, y1), then y1 = mx1 + a 1 + m 2 or, y1 - mx1 = a 1 + m 2 . Squaring + 2mx1 y1 + m2 = a2 (1 + m2) or, (x21 - a2) m2 - 2x1 y1m + (y21 - a2) = 0 (2) This is a quadratic in m. If m1 and m2 are its roots, then these are the slopes of the tangents from P. Since inclinations of tangents are given to be q1 and q2,

Circles  11.35 ∴ let m1 = tan q1 and m2 = tan q2. 1 1 Here, cotq1 + cotq2 = k i.e.,  + =k tan q1 tan q 2 1 1 or, + = k  or m1 + m2 = km1m2. m1 m2

∴  the length of ⊥ from centre (0, 0) on (2) is = radius c. ⇒

2 x1 y1 y -a =k×  or 2x1 y1 = k ( y12 - a2) x12 - a 2 x - a2 ∴  Locus of P is k (y2 - a2) = 2xy. 37. Let A ≡ (a, 0) and B ≡ (0, b). Since ∠AOB = 90º,  ∴  AB is the diameter. æa bö 1 2 ∴  Centre of the circle is ç , ÷ and radius = a + b2 . 2 2 2 è ø 2 1 2 1

∴ 

2

∴  Equation of the circle is 2

b2 x + y12 2 1

=c

b2

=c a2 or, b2 = ac. Hence, a, b, c are in G.P. or,

[Using (1)]

3 3   , 39. The point  3 +  does not satisfy circles given in  2 2 (a) and (c). ∴ (a) and (c) cannot be the correct choices. The centre of  3 3  , circle given in (b) is  which does not lie on the  2 2  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).  1  40. The line 3kx - 2y -1 = 0 meets x-axis and y-axis at A  , 0  3k  1  and B  0, −  respectively and the line 4x - 3y + 2 = 0 cuts  2  1   2 x-axis and y-axis at C  − , 0 and D  0,  respectively.  2   3

2

or,  x2 + y2 - ax - by = 0. Equation of tangent to the circle at O(0, 0) is ax + by = 0 a2 ∴ m = length of ⊥ from A(a, 0) on (1) = a 2 + b2 and, n = length of ⊥ from (0, b) on (1) =

b2 a 2 + b2

(1)

.

∴ Diameter = a 2 + b 2 = m + n. 38. Let (x1, y1) be any point on the circle x2 + y2 = a2. then, x12 + y12 = a 2 (1) Equation of chord of contact of tangents from (x1, y1) to the circle x2 + y2 = b2 is xx1 + yy1 = b2 (2)

Since the four points are concyclic, therefore OA  ⋅ OC = OB ⋅ OD 1 1 1 2 1 ⋅ = ⋅ ⇒ | k |= . ⇒  3| k | 2 2 3 2 According to the given geometrical position (see figure), k must be positive, 1 ∴  k = . 2 41. Let the equation of the circle be x 2 + ( y − 2 )2 = a2 ⇒  x 2 + y 2 − 2 2 y = c, where, c = a2 - 2 → Rational number. Let (x1, y1), (x2, y2), (x3, y3) be three distinct rational points on the circle, then

  (2) touches the circle x2 + y2 = c2,



x12 + y12 − 2 2 y1 = c (1)

HINTS AND EXPLANATIONS

1 2 a  b  2  x −  +  y −  = ( a + b ) 2 2 4

\

11.36  Chapter 11 x22 + y22 − 2 2 y2 = c (2)



x32 + y32 − 2 2 y3 = c (3)

Comparing the irrational parts of the equations, we get y1 = y2 = y3(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, ± x3. ∴  There can be at the most two rational points on the circle C. 42. Given circle is x2 + y2 - a2 = 0 (1) Since PQ and PR are tangents to the circle (1), 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 x2 + y2 - a2 + k(xx1 + yy1 - a2) = 0 (3) Circle (3) will be circumcircle of DPQR if it passes through the point P(x1, y1).

4 ∴  Slope of the incident ray = − . 3 Hence, equation of the incident ray is −4 ( x + 2) i.e., 3(y + 1) = -4(x + 2) 3 or, 4x + 3y + 11 = 0. 44. Given circle is x2 + y2 - 4x - 6y + 9 = 0 (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.

( y + 1) =

Now, CP = 2, OC = 22 + 32 = 13. ∴ OP = OC 2 − CP 2 = 13 − 4 = 3. C ≡ (2, 3),   ∴  OL = 2. From the figure, OM = OL + LM = OL + HP

\



HINTS AND EXPLANATIONS

i.e., if x12 + y12 − a 2 + k ( x12 + y12 − a 2 ) = 0   ⇒  k = -1. Hence, from (3), equation of required circle is x2 + y2 - a2 - (xx1 + yy1 - a2) = 0 or, x2 + y2 - xx1 - yy1 = 0. 43. The equation of the reflected ray is ( y + 1) = m(x + 2) or, mx - y + 2m - 1 = 0

(1)

∴ OP cosθ = 2 + 2 sinθ or 3 cosθ = 2 + 2 sinθ or, 3 = 2 secθ + 2 tanθ or 3 - 2 tanθ = 2 secθ or, 9 + 4 tan2θ - 12 tanθ = 4 (1 + tan2θ) 5 or, 5 = 12 tanθ  ∴  tan q = 12 ∴ cosq =

Since it touches the circles x + y = 1. ∴ length of ⊥ from (0, 0) on (1) = radius 1 | m( 0 ) − 0 + 2 m − 1 | 2m − 1 ⇒ =1 ⇒ = ±1 2 1+ m 1 + m2 2

2

4 ⇒ (2m - 1)2 = (1 + m2)  ⇒ 3m2 - 4m = 0  ⇒  m = 0, . 3 ∴  Equation of the reflected ray is 4 ( x + 2)   or 4x - 3y + 5 = 0. 3 Let α be the angle between the reflected ray and the line y = -1. ( y + 1) =

4 -0 4 3 Then, tan a = =± . 4 3 1+ ×0 3

12 5 and sin q = . 13 13

 36 15  ∴ P ≡ (OP cosθ, OP sinθ) i.e., P ≡  ,  .  13 13  45. Let the centre of the circle be C(x1, y1). 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.

Circles  11.37 CB = radius = x12 + y12 . CM = length of ⊥ from C on AB = x1 - c. Now, AB = 2b (given). ∴ 2BM = 2b or  CB 2 − CM 2 = b 2 2 2 2 or, CB2 - CM2 = b2 or  x1 + y1 − ( x1 − c) = b

or, y12 + 2cx1 − c 2 = b 2 .



∴  Locus of (x1, y1) is y2 + 2cx = b2 + c2. 46. Let the coordinates of P and Q be (x1, y1) and (x2, y2), respectively. Let the given circle be S ≡ x2 + y2 - a2 = 0



(1)

Polar of P w.r.t. (1) is xx1 + yy1 = a2.   P, Q are conjugate points, \

∴ polar of P passes through Q

=

| 0 − 0 − m + 1| m +1 2

=

|1 − m | m2 + 1

.

So, the length of the intercept 2m (1 − m) 2 = 2 12 − 2 =2 . 2 + 1 +1 m m Also, the length of ⊥ from the centre (2, 3) of second circle | 2m − 3 − m + 1 | | m − 2 | = to the line (1) = m2 + 1 m2 + 1

⇒ x1x2 + y1 y2 = a2(2) Equation of circle on PQ as diameter is (x - x1) (x - x2) + (y - y1) ( y - y2) = 0

Let the equation of the second circle be

x2 + y2 + 2g x + 2f y + c = 0.

Since it passes through origin,  ∴  c = 0.

∴ the length of intercept in this case ( m − 2) 2 4m − 3 = 2 1− 2 =2 m +1 m2 + 1 Given, 2

So, the equation becomes

x2 + y2 + 2g x + 2f y + c = 0

⇒

(3)

49. Let the coordinates of point P be (x1, y1). Equation of any line through P can be written as

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

3 2m 4m − 3   ⇒ 2m = 3  or  m = . = 2 m2 + 1 m2 + 1

(2)

The equation of common chord of (1) and (2) is 2(g - 4) x + 2( f - 2) y + 5 = 0

2m 4m − 3 =2 2 m +1 m2 + 1

2 (g - 4) x + 2 (- g - 2) y + 5 = 0

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  48. Let the equation of line be y = mx + c. Since it passes through (1, 1), ∴ 1 = m + c, i.e., c = 1 - m. So, the line becomes y = mx + 1 - m i.e., mx - y - m + 1 = 0 (1) Length of ⊥ from the centre (0, 0) of first circle to the line (1)



x - x1 y - y1 = = r (1) cosq sin q

⇒ x = x1 + r cosθ, y = y1 + r sinθ. Coordinates of any point an (1) is of the form (x1 + r cosθ, y1 + r sinθ). This point will lie on ax2 + 2hxy + by2 = 1 if a (x1 + r cosθ)2 + 2h (x1 + r cosθ) ( y1 + r sinθ) + b (y1 + r sinθ)2 - 1 = 0 2 2 ⇒ r (a cos θ + 2h cosθ sinθ + b sin2θ) + 2r [x1 (a cosθ + h sinθ) + y1 (h cosθ + b sinθ)]

+ ax12 + 2hx1 y1 + by12 - 1 = 0 (2)

Let PQ = r1 and PR = r2. Then r1, r2 are the roots of (2). ∴

PQ × PR = r1r2 =

ax12 + 2hx1 y1 + by12 - 1 . a cos q + 2h cosq sin q + b sin 2 q 2

HINTS AND EXPLANATIONS

or, x2 + y2 - (x1 + x2) x - ( y1 + y2) y + (x1x2 + y1 y2) = 0 x2 + y2 - (x1 + x2) x - ( y1 + y2) y + a2 = 0 (3)  [Using (2)] Clearly, circles (1) and (3) cut each other orthogonally. 47. Given circle is x2 + y2 + 8x + 4y - 5 = 0 (1)

11.38  Chapter 11 We know rewrite the denominator. We have D = a cos2θ + 2h cosθ sinθ + b sin2θ. 1 = [( a + b) + ( a - b)cos 2q ] + h sin 2q 2 a+b 1 = + ( a - b)cos 2q + h sin 2q 2 2 1 Put ( a − b) = k  sinα, h = k cosα. 2 2

 a − b a−b + h2 and tan  a = . ⇒ k =   2  2h 2

∴

1 æa-bö 2 D = ( a + b) + ç ÷ + h sin( 2q + a ) 2 è 2 ø ax12 + 2hx1 y1 + by 2 - 1

Thus, PQ × PR =

1

2

1 æa-bö 2 ( a + b) + ç ÷ + h sin( 2q + a ) 2 è 2 ø For this to be independent of θ, we must have 2



 a − b 2   + h = 0   ⇒  a = b and h = 0. 2 

HINTS AND EXPLANATIONS

But this is the condition for the given curve to represent a circle. 50. 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 the circle. Centre of the given æ 1 -3 ö circle is ç , ÷ . Therefore, è2 2 ø 1 3 m 3 - -1 + 2 2 2 2 = 1+1 1 + m2

Þ

ö a 1 ö æ a æ 1 ∴ From (1), a 2 ç 2 + 2 ÷ = ç + - 1÷ 4d ø è 2g 2d è 4r ø

2

⇒ 2a(γ + δ ) – a2 = 2gδ So, the locus of (γ, δ) is 2a(x + y) = 2xy + a2. 52. Let the sides of the square be y = 0, 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 is  2 2 also the centre of the square) and radius 2. 53. Equation of the given circle can be written as  1 + 2a   1 − 2a  x2 + y2 −  x−   y = 0.  2   2  Since y + x = 0 bisects two chords of this circle, mid-points of the chords must be of the form (α, -α). Equation of the chord having (α, -α) as mid-point is æ 1 + 2a ö T = S1 i.e., ax + ( -a ) y - çç ÷÷ ( x + a ) è 4 ø æ 1 - 2a ö -ç (y +a) ç 4 ÷÷ ø è æ 1 - 2a ö æ 1 + 2a ö = a 2 + a2 - ç a - çç ÷÷ ( -a ) ç 2 ÷÷ ø è 2 ø è

4 2

=

m+3 1 + m2

⇒ 7m2 – 6m – 1 = 0 ⇒ (7m + 1) (m – 1) = 0 -1 . 7 Thus, two chords are y = x and 7y + x = 0. ⇒ m = 1,

x y + = 1. a b For the circle, centre ≡ (a, a) and radius = a. Since the third side touches the circle, 51. Let the third side be

⇒ 4 ax − 4 ay − (1 + 2a ) x − (1 − 2a ) y

a a + -1 a b ∴ a= (1) 1 1 + a2 b2

= 8a 2 − (1 + 2a )a + (1 − 2a )a .

Vertices of the triangle are (0, 0), (α, 0) and (0, β), ∴  if the circumcentre is (γ, δ) then

æ 1 + 2a ö - 4a 4a ç ç 2 ÷÷ ø è

g =

b a and d = . 2 2

 1 + 2a 1 − 2a  , This chord will pass through the point  if 2   2 æ 1 - 2a ö (1 + 2a) (1 + 2a) ÷÷ çç 2 è 2 ø -

(1 - 2a) (1 - 2a) 2

Circles  11.39

= 8a 2 - 2 2aa

1 2 2 ⇒ 2a éë1 + 2a - 1 + 2a ùû - 2 éë(1 + 2a) + (1 - 2a) ùû = 8a 2 - 2 2aa 1 ⇒ 4 2aa - é 2(1) 2 + 2( 2a) 2 ù = 8a 2 - 2 2aa û 2ë [ ∵ (a + b)2 + (a - b)2 = 2a2 + 2b2] ⇒ 8a 2 − 6 2a a + 1 + 2a 2 = 0. 2

æa + b ö 2  r 2 = AC 2 = CL2 + LA2 = ç ÷ + (3 2 ) è 2 ø

This quadratic equation will have two distinct roots if (6 2a) 2 − 4(8)(1 + 2a 2 ) > 0.

Since

⇒ 72a2 - 32 (1 + 2a2) > 0 ⇒ 9a2 - 4 - 8a2 > 0 ⇒ a2 - 4 > 0 ⇒ (a - 2) (a + 2) > 0 ⇒ a∈ (-∞, -2) ∪ (2, ∞). 54. Let x2 + y2 + 2g x + 2f y + c = 0 be the variable circle. Since it touches the given circles externally

⇒ a 2 + b 2 = r 2 + ( 4 2 ) 2 (2) 2

æa + b ö From (1), and (2), we get a 2 + b 2 = ç ÷ + 18 + 32 è 2 ø

( − g − 0) 2 + ( − f − 0) 2 = g 2 + f 2 − c + a 

(1)

and,

( − g − 2a) 2 + ( − f − 0) 2 = g 2 + f 2 − c + 2a 

(2)

Subtracting (1) from (2), we get

2

2 2 ⇒ (a + b ) -

(a + b ) = 50 ⇒ (α - β)2 = 100 2

⇒ α - β = ±10

(3) (a - b ) a -b 2 ⇒ r = (4) 2 2 2

2

2

Also, CP = r ⇒ r =

Squaring both sides, we get

From (3) and (4), we get r 2 = (5 2 ) 2



Substituting r = 5 2 in (1), we get

( g + 2a) 2 + f 2 = a 2 + g 2 + f 2 + 2a g 2 + f 2

⇒ r = (5 2 ) 2 .

2

⇒ 4 ag + 4 a 2 = a 2 + 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. 55. Let the lines cuts the x-axis at A and B, then

OA = -

1 and OB = -3. l

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 æ 1ö OA × OB = OC × OD Þ ç - ÷ ( -3) = 1 ´ ⇒ λ = 2. 2 è lø 56. Let C(α, β) be the centre of the circle touching OP at P and making intercept AB = 6 2 on the line x + y = 0 as shown in the figure. If r is the radius of the circle then



æa + b ö (5 2 ) 2 = ç ÷ + 18 è 2 ø 2

⇒ 32 = (a + b ) ⇒ (α + β)2 = 64 2 ⇒ α + β = ± 8 Form (3) and (5), we get α = 9, β = -1 or α = -9, β = 1 or α = 1, β = -9 or α = -1, β = 9. Since Q(-10, 2) lies in the interior of the circle,

(5)

∴ CQ must be less than 5 2. Thus, centre of the circle must be (-9, 1). ∴ The equation of the required circle is

( x + 9) 2 + ( y − 1) 2 = (5 2 ) 2

⇒ x2 + y2 + 18x - 2y + 32 = 0. 57. Let P be the foot of ⊥ from origin O on any chord of the circle S whose coordinates are (α, β). Then, the slope of OP b a is and thus the slope of chord is and its equation a b passing through (α, β) is

HINTS AND EXPLANATIONS

( g + 2a) + f = g + f + a. 2

2 2 2 OP = 4 2,   ∴  OC = CP + PO



∴

2

(1)

11.40  Chapter 11 a ( x - a)   ⇒  βy - β2 = - ax + a2 b ⇒ ax + βy = a2 + β 2  Now, homogenizing the equation of the given circle x2 + y2 + 2g x + 2f y + c = 0 with the help of (1), we get

9 225 − =0 4 4 ⇒ x2 + y2 + 6x - 3y - 45 = 0. 59. Centre of the circle x2 + y2 = 9 is (0, 0) and any tangent to the circle is x cosα + y sinα = 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 sinα - 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. ∴ Locus is S ≡ x2 + y2 - 18 = 0 (3) ⇒ x 2 + y 2 + 6 x − 3 y + 9 +

y - b = -

(1)

2

æax + b y ö æax + b y ö x 2 + y 2 ( 2 gx + 2 fy ) ç 2 + cç 2 = 0. (2) 2 ÷ 2 ÷ èa +b ø èa +b ø Now, equation (2) represents a pair of straight lines passing through origin. These lines will be at right angle if sum of the coefficients of x2 and y2 is zero. i.e., (a2 + b2)2 + (a2 + b2)2 + 2gα (a2 + b2) + 2β f (a2 + b2) + c (a2 + b2) ⇒ 2 (a2 + b2) + 2gα + 2f β + c = 0 (3) From equation (3), the locus of P(α, β) is 2(x2 + y2) + 2gx + 2f y + c = 0 which is the required locus. 58. Given equation of line is x2 + 2xy + 3x + 6y = 0 (1) ⇒ x(x + 2y) + 3(x + 2y) = 0 ⇒ (x + 2y) (x + 3) = 0 So, equations of normals are x + 3 = 0 (2) and, x + 2y = 0 (3)

Equation of tangent to (3) at ( 2 , 4) is p ≡ 2 x + 4 y − 18 = 0. ∴ System of coaxal circles is S + lP = 0. 60. Let P be the point on the line 2x - y + 11 = 0 which is nearest to the circle

(1)

3 Solving (2) and (3), we get x = -3, y = . 2

HINTS AND EXPLANATIONS

3 ∴ Coordinates of centre of the circle are  −3,  .  2 Given equation of circle is x(x - 4) + y(y - 3) = 0 ⇒ x2 + y2 - 4x - 3y = 0.

(4)

2

5  −3  Radius of the circle = ( −2) +   = and coordinates  2 2

x2 + y2 + 2x −

2

 3 of centre are  2,  .  2 Since the required circle is just sufficient to contain the circle (4), therefore the distance between the centres 3   3 −  3,  and  2,  = the difference of their radius.  2 2 Let the radius of the required circle be a. 2

∴

5 5  3 3 ( −3 − 2) 2 +  −  = a − ⇒ 5=a−  2 2 2 2 ⇒ a = 5+

Therefore, equation of required circle is 2



3   15  ( x + 3) 2  y −  =     2 2

2

5 15 = . 2 2

1 25 y− = 0 (2) 2 8

1  with centre C  −1,  .  4 Then, CP is ⊥ to the line (1) and CP > radius. [Note that if CP ≤ r, the line intersects or touches the circle and then the point of intersection or point of contact are required points] 1 + 11 35 67 4 Here, CP = (radius). = > 4 5 4 5 Now, equation of CP [⊥ to line (1)] is −2−



x + 2y = λ, where -1 +

1 1 = l = λ or l = - . 2 2

∴ Equation of CP is 2x + 4y + 1 = 0 Solving (1) and (3), we get y = 2, x = -9/2.  –9  Hence, the required point is  , 2 .  2 

(3)

Circles  11.41 61. Extremities of the diagonal OA of the rectangle are O(0, 0) and A (4, 3). Then, OA is the diameter of the circumcircle, so equation of the circumcircle is x(x - 4) + y(  y - 3) = 0 i.e., x2 + y2 - 4x - 3y = 0 2

2

3 5 i.e., ( x − 2) 2 +  y −  =   (1)   2 2 3 (2) 4 ∴ Tangents parallel to the diagonal OA are m = slope of OA =

3 3 5 9 = ( x − 2) ± 1+ 2 4 2 16

i.e., 6x - 8y ± 25 = 0. 62. Let the coordinates of C be (x1, y1) and the coordinates of A and B be (0, 0) and (a, 0), respectively. Given, k =

sin A BC = sin B AC

y1 − 12 4 × − = −1 ⇒ 7x - 4y + 104 = 0 1 1 x1 + 8 7

(3)

 x − 8 y1 + 12  , and mid point of C1C2 i.e.,  1  lie on (2)  2 2  i.e., 4  x1 − 8  + 7  y1 + 12  + 13 = 0  2   2  or, 4x1+ 7y1 + 78 = 0 (4) Solving (3) and (4), we get (x1, y1) ≡ (-16, -2). ∴ Equation of the imaged circle is (x + 16)2 + (y + 2)2 = 52 or, x2 + y2 + 32x + 4y + 235 = 0. 64. Let r and R be radius of required and given circle, respectively and let centre is (h, k), By given condition

⇒ BC2 = k2 AC2 ⇒ (x1 - a)2 + y12 = k2 (x12 + y12) ⇒ (1 − k 2 ) x12 + (1 − k 2 ) y12 − 2ax1 + a 2 = 0 2ax1 a2 + = 0, [ x 1− k2 1− k2 Hence, locus of C is ⇒ x12 + y12 −

\

x2 + y2 −

k ≠ 1]

a2 2a + = 0, 2 1− k 1− k2

 a  which is a circle whose centre is  ,0  1 − k 2  and radius =

a2 a2 ak − = . 2 2 (1 − k ) (1 − k 2 ) 1 − k 2

63. The given circle and line are x2 + y2 + 16x - 24y + 183 = 0 (1) and, 4x + 7y + 13 = 0 (2) Centre and radius of circle (1) are C1(-8, 12) and 5 respectively. Let the centre of the imaged circle be C2 (x1, y1). Then, slope of C1C2 × slope of given line

( h − 1) 2 + ( k − 2) 2 = R − r Now,

r = tan 30° AB

r = AB tan 30° = ( R − r ) ⇒

1



(AB = R - r)

( h − 1) 2 + ( k − 2) 2 = R −

 3  = R  1+ 3  1+ 3 

3

R

Now, R = 1 + 3 ⇒

( h − 1) 2 + ( k − 2) 2 = 3

∴ Locus is (x - 1)2 + (y - 2)2 = 3 65. The point of intersection of the tangents at the points P(q1) and Q(q2) on the circle x2 + y2 = 1 is given by

HINTS AND EXPLANATIONS

y−

⇒

11.42  Chapter 11

æq +q ö æq +q ö a sin ç 1 2 ÷ sin ç 1 2 ÷ 2 ø è è 2 ø = y = æ q1 - q 2 ö æ 60° ö cos ç cos ç ÷ ÷ è 2 ø è 2 ø 2 2 ⇒ (x cos 30º) + (y cos 30º) = 1 3 4 = 1 ⇒ x2 + y2 = 4 3 ⇒ 3x2 + 3y2 = 4 ⇒ ( x 2 + y 2 )

66. Centre is (1, -2). Radius = 1 + 4 − 3 = 2 . Since the sides are parallel to coordinate axes, vertices do not lie on horizontal and vertical lines through. (1, 2). ∴ the given points are not vertices.

∴1 1. r +1 2 (B) a hyperbola whose eccentricity is , when 0 < r < 1. r +1 1 (C) an ellipse whose eccentricity is , when r > 1. r +1 2 (D) a hyperbola whose eccentricity is , when 0 < r < 1. 1− r 99. The length of the chord of the parabola x2 = 4y having equation x – 2 y + 4 2 0 is: [2019] (A)  2 11 (B) 8 2 (C)  3 2 (D)  6 3

ANSWER K EYS Single Option Correct Type 1. (B) 11. (C) 21. (A) 31. (B) 41. (B) 51. (A) 61. (B)

2. (A) 12. (B) 22.  (D) 32. (B) 42. (B) 52. (B) 62. (A)

3. (B) 13. (D) 23.  (A) 33. (A) 43. (A) 53. (B) 63. (A)

4. (A) 14. (D) 24. (B) 34. (C) 44. (B) 54. (B) 64. (A)

5. (B) 15. (C) 25. (A) 35.  (B) 45. (D) 55. (B) 65. (A)

6. (B) 16. (C) 26. (A) 36. (C) 46. (A) 56. (B)

7. (A) 17. (C) 27. (C) 37. (A) 47. (C) 57. (B)

8. (A) 18. (D) 28. (C) 38. (A) 48. (C) 58. (A)

9. (C) 19. (D) 29. (C) 39. (A) 49. (A) 59. (B)

10. (C) 20. (B) 30. (B) 40. (C) 50. (B) 60. (A)

69. (A) 79. (B) 89. (C) 99. (D)

70. (C) 80. (B) 90. (B)

71. (D) 81. (B) 91. (D)

72. (A) 82. (B) 92. (B)

73. (B) 83. (D) 93. (A)

74. (A) 84. (A) 94. (A)

75. (D) 85. (D) 95. (A)

Previous Years’ Questions 6 6. (A) 76. (A) 86. (C) 96. (B)

67. (B) 77. (A) 87. (A) 97. (A)

68. (A) 78. (A) 88. (C) 98. (A)

PRACTICE EXERCISES

4 (A) 3 (B)  3 4 2 (C) (D)  3 3 92. Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the cente C of the circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is [2016] 2 2 (A) x + y – 4x + 9y + 18 = 0 (B) x2 + y2 – 4x + 8y + 12 = 0 (C) x2 + y2 – x + 4y + 12 = 0 x (D) x 2 + y 2 - + 2 y - 24 = 0 4 93. The centres of those circles which touch the circle, x2 + y2 – 8x – 8y – 4 = 0, externally and also touch the x-axis, lie on [2016] (A) A parabola (B) A circle (C) An ellipse which is not a circle (D) A hyperbola 94. The eccentricity of an ellipse whose centre is at the 1 origin is . If one of its directrices is x = –4, then the 2 ⎛ 3⎞ equation of the normal to it at ⎜1, ⎟ is [2017] ⎝ 2⎠

12.46  Chapter 12

HINTS AND EXPLANATIONS

HINTS AND EXPLANATIONS

Single Option Correct Type 1. The general equation of a parabola having its axis parallel to y-axis is y = ax2 + bx + c(1) Since the line y = x touches the parabola at x = 1, therefore slope of the tangent at (1, 1) = 1 æ dy ö i.e., ç ÷ = 1   or  2a + b = 1. è dx ø(1, 1) Also, x = ax2 + bx + c must have equal roots ⇒ (b - 1)2 = 4ac. Since, (1, 1) lies on (1), \ a + b + c = 1. Now, 2a + b = 1 and a + b + c = 1 ⇒ a - c = 0 or a = c. \ a + b + c = 1 ⇒ 2c + b = 1 \ ⇒ 2f(0) + f ′(0) = 1 ( f(0) = c and f ′(0) = b) 2. The equation of axis of the parabola is x - 4 = 0 which is parallel to y-axis. So, the ray of light is parallel to the axis of the parabola. We know that any ray parallel to the axis of a parabola passes through the focus after reflection. \ The ray must pass through the point (4, -1). 3. Clearly, the two tangents, having slopes m1 and m2, meet on the line x = -2, which is the directrix of the parabola y2 = 8x, therefore the two tangents must be at right angles, i.e., m1m2 = -1. 4. 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 æ1 ö parabola y2 = 4ax at the point Qç at 2 , at ÷ . è4 ø Coordinates of N are (at2, 0). 0 - at ( x - at 2 ), Equation of NQ is y = 1 2 2 at - at 4

⇒ ( y2 + 6y + 9) - 2(x + 2) = 0 or, ( y + 3)2 = 2(x + 2). Clearly, x = 2t2 - 2 and y = 2t - 3 satisfy it for all t. 6. Directirx of y2 = 4(x + 1) is x = -2 Any point on x = -2 is (-2, k) Now, mirror image (x, y) of (-2, k) in the line x + 2y = 3 is given by x+2 y-k æ -2 + 2k - 3 ö = = -2 ç ÷ 1 2 5 è ø

Þ x=

Also,

10 - 4 k 4k -2 Þ x =(1) 5 5 y=

20 - 8k (2) 5

From (1) and (2) 3 æ 5x ö y = 4 + ç ÷ 5è 4 ø or 4y = 16 + 3x is the equation of the mirror image of the directrix. 7. Coordinates of any point on the parabola y2 = -4ax are (-at2, 2at). Equation of the normal at (-at2, 2at) is y - xt = 2at + at3 If the normal passes through the point (h, k), then k - th = 2at + at3 or, at3 + (2a + h)t - k = 0, which is a cubic equation whose three roots t1, t2, t3 are the parameters of the feet of the three normals. 2 \ Sum of the roots = t1 + t 2 + t3 = - Coefficient of t = 0 Coefficient of t 3 \ Centroid of the triangle formed by the feet of the normals æ a ö 2a = ç - (t12 + t 22 + t32 ), (t1 + t 2 + t3 )÷ è 3 ø 3



æ a ö = ç - (t12 + t 22 + t32 ), 0 ÷ è 3 ø which, clearly, lies on the x-axis. which meets the tangent at the vertex, x = 0, at the point 4 y = at . 3 5. The equation of the parabola is 2 2 æ æ 2x + 5 ö 3ö 2 x ç + ÷ + ( y + 3) = ç ÷ è è 2 ø 2ø

8. The coordinates of any point on the parabola x2 = 4by are (2bt, bt2). dy x For the parabola x2 = 4by, . = dx 2b

9 é ù Þ 4 ê x 2 + + 3 x ú + 4[ y 2 + 9 + 6 y ] = ( 4 x 2 + 25 + 20 x ) 4 ë û



Slope of the normal at (2bt, bt2) = -

1 2b =2bt t

1 \ Equation of normal is y - bt2 = - ( x - 2bt ) t

Conic Sections (Parabola, Ellipse and Hyperbola)  12.47 x + 2b + bt 2 t It will touch the parabola y2 = 4ax if æ a aö 2b + bt 2 =   ç ∵ c = ÷ -1 / t è mø ⇒ bt2 + at + 2b = 0 For distinct real roots, discriminant > 0 ⇒ a2 - 8b2 = 0  or  a2 > 8b2 9. The coordinates of the focus of the given parabola are (a + 1, b). or, y = -

Since OX and OY are mutually perpendicular tangents to sliding ellipse for all its positions, therefore, O(0, 0) will lie on circle (1) \ a2 + b 2 = a2 + b2. Hence, locus of C(a, b) is x2 + y2 = a2 + b2. 12. P(a cosq, bsinq), then q is angle of a corresponding pont on auxilliary circle x2 + y2 = a2 i.e., (a cosq, a sinq). 13. Let S(5, 12), S′(24, 7) be the two foci. P(0, 0) is a point on the conic. SP =

25 + 144 = 13

S′P =

576 + 46 =

SS′ =

( 24 - 5) 2+ (7 - 12) 2

625 = 25

= 192 + 52 = 386 If the conic is an ellipse, then, SP + S′P = 2a and SS′ = 2ae SS¢ 386 386 = = SP + S¢P 13 + 25 38 14. Given ellipses are x2 + 4y2 = 4 x2 y2 i.e., 2 + 2 = 1 (1) 2 1 \ e=

11. Let the two given lines be taken as the coordinate axes. Let C(a, b) 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 semi-minor axes of the ellipse. Equation of the director circle of the ellipse is (x - a)2 + (y - b)2 = a2 + b2 (1)

and, x2 + 2y2 = 6  i.e.,  

x2 ( 6)

2

+

y2 ( 3 )2

= 1 (2)

Let R(a, b) be the point of intersection of the tangents to ellipse (2) at P and Q. then PQ will be chord of contact of R. \ its equation is ax b y + =1 6 3 i.e., ax + 2yb = 6 3 a x + (3) 2b b Since (3) touches (1) or, y = -



æ 3ö a2 \ ç ÷ = 22 × + 12 [c2 = a2m2 + b2] 4 b2 è bø



Þ



⇒ a 2 + b 2 = 9 \ locus of (a , b ) is

2

a2 a2 + b2 9 = 2 + 1= 2 b b b2

x2 + y2 = 9 = ( 6 ) + ( 3 ) i.e., director circle. \ tangent at P, Q meet at right angles. 15. Let S and S′ be the centres and a and b be the radii of the given circles. Let P be the centre and r be the radius of the circle which touches the given circles externally. Then, S′P = r + a and SP = r + b \ S′P - SP = (r + a) - (r + b) = a - b = constant. 2

2

HINTS AND EXPLANATIONS

Clearly, focus must lie to the opposite side of the origin w.r.t. the line x + y - 1 = 0 and same side as origin with respect to the line x + y - 3 = 0. Hence, a + b > 0 and a + b < 2. x2 y2 10. Let the equation of the ellipse be 2 + 2 = 1 . a b Let e be the eccentricity of the ellipse. Since distance between foci = 2h \ 2ae = 2h ⇒ ae = h(1) Focal distance of one end of minor axis say (0, b) is k. \ a + e(0) = k ⇒ a = k(2) From (1) and (2) b2 = a2(1 - e2) = k2 - h2. \ The equation of the ellipse is x2 y2 2 + 2 = 1. k k - h2

12.48  Chapter 12 Let P(a, b) be the mid point of any of the parallel chords whose slope is -7. \ equation of the chord in terms of mid-point P(a, b) is T = S1

Hence, the locus of P is a hyperbola whose foci are S and S′. 16. Let the two given ^ lines be the coordinate axes and let the equation of variable circle be x2 + y2 + 2gx + 2f y + c = 0 (1)

HINTS AND EXPLANATIONS

Then, 5 = 2 g 2 - c and 3 = 2 f 2 - c . Squaring and subtracting these, we get 4(g2 - c) - 4( f  2 - c) = 25 - 9 ⇒ g2 - f 2 = 4 or (-g)2 - (- f )2 = 4. Hence, locus of the centre (-g, -f ) of circle is x2 - y2 = 4, which is a rectangular hyperbola. 17. We have, 2x2 + 3y2 - 8x - 18y + 35 = k ⇒ 2(x2 - 4x) + 3( y2 - 6y) + 35 = k ⇒ 2[(x - 2)2 - 4] + 3[( y - 3)2 - 9] + 35 = k ⇒ 2(x - 2)2 + 3( y - 3)2 = k For k = 0, we get 2(x - 2)2 + 3(y - 3)2 = 0 which represents the point (2, 3). 18. Let P(asecq, btanq); Q(a secq, -btanq) be end points of double ordinate and C(0, 0), is the centre of the hyperbola. Now, PQ = 2b tanq CQ = CP = a 2 sec 2q + b 2 tan 2q since CQ = CP = PQ \ 4b2 tan2q = a2 sec2q + b2 tan2q ⇒ 3b2 tan2q = a2 sec2q

i.e.,

xa y b a2 b2 -1 = -1 3 7 3 7

i.e.,

ax b y a2 b2 = 3 7 3 7

7a = -7   ⇒  a + 3b = 0 3b \  locus of (a, b) is x + 3y = 0. 20. Clearly, the common tangent to the circle x2 + y2 = 1 and hyperbola x2 - y2 = 1 is x =1 [which is nearer to P(1/2, 1)]. æ1 ö Given, one focus at P ç , 1÷ . è2 ø \  equation of the directrix is x = 1. \  ellipse is Its slope =

æ 1ö 1 ç x - ÷ + ( y - 1) 2 = ( x - 1) è 2ø 2 2

On simplification, it becomes 2

1ö æ 9 ç x – ÷ + 12( y – 1) 2 = 1. 3ø è 21. Since the line CP passes through the origin, i.e., centre, let its equation be y = mx. The line CP meets the hyperbola x2 y2 2 - 2 = 1 in P whose abscissa is given by a b

⇒ 3b2 sin2q = a2 ⇒ 3a2(e2 - 1) sin2q = a2 ⇒ 3(e2 - 1)sin2q = 1 1 = sin 2 q < 1 (∴   sin2q < 1) Þ 2 3(e - 1)

1 1 < 3    Þ e 2 - 1 > e -1 3 x2 y2 19. Given hyperbola: = 1 (1) 3 7 Given chord: 7x + y - 2 = 0 (2) Its slope = -7

Þ

2

x 2 m2 x 2 a2b2 2 = 1 or x 2 = 2 2 b - a2 m2 a b a 2b 2 m 2 b2 - a2m2



\ y2 = m2x2 =



\ CP2 = x2 + y2

a 2b 2 + a 2b 2 m 2 b2 - a2m2 2 2 = a b (1 + m2) Since CQ ^ CP =

Replace m by -

1 , we get m

Conic Sections (Parabola, Ellipse and Hyperbola)  12.49 Dividing (1) by (2), we get t 2 = Putting in (1),

1 1 b2 - a2 = 2– 2 a b a 2b 2 22. Given, P(a secq, b tanq) and Q(a secf, b tanf). The equation of tangent at point P is x sec q y tan q =1 a b sec q b 1 b Slope of tangent = ´ = × tan q a a sin q Hence, the equation of perpendicular at P is a sin q ( x - a secq ) y - b tan q = b or, by -b2 tanq = -a sinq x + a2 tanq or, a sinq x + by = (a2 + b2) tanq(1) Similarly, the equation of perpendicular at Q is a sinf x + by = (a2 + b2) tanf(2) On multiplying (1) by sinf and (2) by sinq, we get a sinq sinf x + b sinf y = (a2 + b2) tanq sinf a sinf sinq x + b sinq y = (a2 + b2) tanf sinq On subtraction we get by (sinf - sinq) = (a2 + b2)(tanq sinf - tanf sinq)

=

a 2 + b 2 tan q sin f - tan f sin q × b sin f - sin q p p ∵ q +f = Þ f = -q 2 2 ⇒ sinf = cosq and tanf = cotq a 2 + b 2 tan q cosq - cosq sin q \ y=k = × b cosq - sin q a 2 + b 2 æ sin q - cosq ö (a2 + b2 ) = =ç ÷ b b è cosq - sin q ø 23. If coordinates of one end of focal chord are P(at2, 2at) then æ a - 2a ö the coordinates of other end will be Qç 2 , ÷. èt t ø

\ y=k =

\ SP = ( at 2 - a) 2 + ( 2at - 0) 2 = a(t2 + 1) = b   (given) æ a ö æ - 2a ö - 0÷ ç 2 - a÷ + ç èt ø è t ø 2

and, SQ =

(1)

2

æ1 ö = a ç 2 + 1÷ = k (given) t è ø

(2)

b . k

b b ab + 1= Þ k= . b- a k a

24. Let S ≡ x2 - 4y Since the point (2a, a) lies inside the parabola, \ S](2a, a) = 4a2 - 4a < 0 i.e., 4a(a - 1) < 0 or, a(a - 1) < 0 (1) Also, the vertex A(0, 0) and the point (2a, a) are on the same side of the line y = 1 (the equation of latus-rectum)

So, a - 1 < 0 i.e., a < 1 (2) From (1) and (2), we have a (a - 1) < 0 or, 0 < a < 1. 25. We know any side of the triangle is more than the difference of remaining two sides, such that |PR - PQ| ≤ RQ. ⇒ The required piont P will be the point of intersection of the line RQ with parabola which is (a, 2a) as RQ is a tangent to the parabola. 26. Normal at a point (m2, - 2m) on the parabola y2 = 4x is given by y = mx - 2m - m3. If this is normal to the circle also, then it will pass through centre of the circle so 6 = -3m - 2m - m3 ⇒ m = -1 Since shortest distance between parabola and circle will occur along common normal, shortest distance is 4 2 – 5 . 27. Tangent at P is ty = x + at2, which meets axis at T(-at2, 0). Normal at P is tx + y = 2at + at3, which meets axis at G(2a + at2, 0) p \ Ð TPG = , so TG is diameter of the circle. 2 Equation of the circle is (x + at2)(x - 2a - atI2) + (y - 0)(y - 0) = 0 ⇒ x2 + y2 - 2ax - at2(2a + at2) = 0 Tangent to above circle at P(at2, 2at) is xat2 + y . 2at - a(x + at2) - a2t2(1 + t2) = 0 1- t 2 or, (t2 - 1)x + 2ty - a(2 + t2) = 0. It has slope = 2t \ Angle q between two tangents is given by 1 1- t 2 2 2t = t (1+ t )   ⇒  q = tan- 1 (t) tan q = t 2 2 1 1- t t +1 1+ × t 2t

HINTS AND EXPLANATIONS

1 ö æ a 2b 2 ç1 + 2 ÷ 2 2 2 m ø a b ( m + 1) è CQ = = 2 2 2 c b m - a2 b2 - 2 m 1 1 b2 - a2m2 + b2m2 - a2 + = \ CP 2 CQ 2 a 2b 2 (1 + m 2 ) 2

12.50  Chapter 12 28. Equation of normal is y = mx - 2am - am3 Put y = 0, we get x1 = 2a + am12 x2 = 2a + am22 and, x3 = 2a + am33

HINTS AND EXPLANATIONS

1 3 t =0 2 Both are same if -2m - m3 = -4m - 1/2m3 ⇒ m = 0, ±2 So, points will be (4, 4) and (5, 2) Hence, shortest distance will be = 1+ 4 = 5 . 33. These are two common tangents to the circle x2 + y2 = 1 and the hyperbola x2 - y2 = 1. These are x = 1 and x = -1 y + t(x - 3) - t -

where x1, x2, x3 are intercepts on the axis of the parabola. The normal passes through (h, k). \ am3 + (2a - h)m + k = 0 Now, m1 + m2 + m3 = 0 2a - h m1m2 + m2m3 + m3m1 = a ⇒ m12 + m22 + m32 = (m1 + m2 + m3)2 - 2(m1m2 + m2m3 + m3m1) ( 2a - h) =-2 a ⇒ x1 + x2 + x3 = 6a - 2(2a - h) = 2(h + a) 29. The focus of the parabola y2 = 8ax is (2a, 0). So, the coordinates of the point on the axis of the 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 (  m ≠ 0) æ pö Þ m = ± 3 = tanç ± ÷ è 3ø 30. Let the three points A, B and C on the parabola are A(x1, y1), B(x2, y2) and C(x3, y3), respectively. 2 Also, y2 = y1 y3 {given} If T be the point of intersection of tangents at y1 and y3 then T is æ y 2 y + y3 ö æ y y y + y3 ö =Tç 2 , 1 Tç 1 3, 1 ÷ ÷ 2 ø 2 ø è 4a è 4a æ y + y3 ö or, T ç x2 , 1  (∵ y22 = 4 ax2 ) 2 ÷ø è This point lies on x = x2 which is a line through B(x2, y2) parallel to y-axis.

Out of these, x = 1 is nearer to the point P(1/2, 1). Thus, a directrix of the required ellipse is x = 1. If Q(x, y) is any pont on the ellipse, then its distance from the æ 1ö 2 ç x - ÷ + ( y - 1) and its distance from è 2ø the directrix x = 1 is |x - 1| By definition of ellipse, QP = e|x - 1| 2

focus is QP =

æ 1ö 1 2 ç x - ÷ + ( y - 1) = | x - 1 | è 2ø 2 ⇒ 3x2 - 2x + 4y2 - 8y + 4 = 0 2



Þ

æ 1ö or, çè x – 3 ÷ø ( y – 1) 2 + = 1. 1/ 9 1 / 12 34. Let the point of contact be 2

R º ( 2 cosq , sin q ) Equation of tangent AB is

31. y2 = 4ax, Normal: y = mx - 2am - am3 (1) y2 = 4c(x - b), Normal: y = m(x - b) - 2cm - cm3 (2) Since the two parabolas have a common normal, therefore (1) and (2) must be identical After comparing the coefficients we get m=±

2( a - c) - b  ( c - a)

\ -2-

b >0 Þ c-a

b >2 a–c

QUICK TIPS Shortest distance between two curves occurred along the common normal 32. Normal to y2 = 4x at (m2, 2m) is y + mx - 2m - m3 = 0 Normal to y2 = 2(x - 3) at (1/2t2 + 3, t) is

x cosq + y sin q = 1 2

Þ

A º ( 2 secq , 0 ) ; B º (0, cosecq )

Let the middle point Q of AB be (h, k) cosecq secq Þ h= ,k = 2 2

Conic Sections (Parabola, Ellipse and Hyperbola)  12.51 1 1 , sin q = 2k h 2 Þ 1 + 1 =1 2h 2 4 k 2 1 1 + = 1. \ Required locus is 2 2x 4 y2 Trick: The locus of mid-points of the portion of tangents to x2 y2 the ellipse 2 + 2 = 1 intercepted between axes is a2y2 + a b b2x2 = 4x2y2 a2 b2 1 1 = 1. i.e., + = 1  or  2 + 2 4x 4 y2 2x 4 y2 35. Point P goes to Q. Its direction with respect to x-axis is q in original position. In new position y-axis will play the role of major axis so its inclination with negative direction of y-axis will be same. So, new coordinates will be (b sinq, -a cosq ).

Þ cosq =

p is 4

37. Equation of the tangent at

×

æ 1 ö yç ÷ è 2 ø = 1 i.e. x + y - 2 = 0 (1) a b b p Equation of the normal at is 4 x y a b - = (2) b a b 2 a 2 p1 = length of the perpendicular from the centre to the æ 1 ö xç ÷ 2ø + è a

×

- 2 tangent =

1 1 = + a2 b2

2ab a + b2 2

p2 = length of the perpendicular from the centre to the a b a2 - b2 2 2= b a normal = . 1 1 2 a2 + b2 + a2 b2 ab( a 2 - b 2 ) . a2 + b2 38. If S1 = 0 and S2 = 0 are the equations, then, lS1 + S2 = 0 is a second degree curve passing through the points of intersection of S1 = 0 and S2 = 0. ⇒ (l + 4)x2 + 2(l + 1)y2 - 2(3l + 10)x - 12(l + 1)y + (23l + 35) = 0 (1) For it to be a circle, choose l such that the coefficients of x2 and y2 are equal: ⇒ l + 4 = 2l + 2 \ l = 2 This gives the equation of the circle as 6(x2 + y2) - 32x - 36y + 81 = 0 {(using (1))} 16 27 Þ x2 + y2 - x - 6 y + = 0. 3 2 æ8 ö Its centre is C ç , 3 ÷ and radius is è3 ø

36. Given,

x2 y2 + =1 a2 b2

b2 = a2 - b2 a2 So, two points on the minor axis are,

\ OS = ae = a 1-

S1 ( 0, a 2 - b 2 ) , S 2 ( 0, -

a2 - b2 )

64 27 1 47 +9– = . 9 2 3 2 39. Equation of the chord joining the points (acosq1, bsinq1) and (acosq2, b sinq2) is given by q +q q -q x y q +q cos 1 2 + sin 1 2 = cos 1 2 a 2 b 2 2 which passes through (ae, 0), we have q -q æq +q ö e cos ç 1 2 ÷ = cos 1 2 2 2 è ø q2 q1 q 2 ù é q1 Þ e êcos cos - sin sin ú 2 2 2 2û ë q1 q2 q1 q 2 = cos cos + sin sin 2 2 2 2 q1 q2 ù q q é Þ e ê1 - tan tan ú = 1 + tan 2 tan 2 2 2 2 2 ë û r=

Let tangent to the ellipse by y = mx + c = mx + a 2 m 2 + b 2 where m is parameter. Now, sum of the squares of ^’s on this tangent from the points S1 and S′ 1 is æ a2 - b2 - a2 m2 + b2 ö æ ç ÷ +ç ç ÷ ç 1+ m 2 è ø è 2

a 2 m 2 + b 2 ö÷ ÷ 1+ m 2 ø

a2 - b2 -

æ a 2 - b 2 + a 2 m 2 + b 2 ö 2a 2 (1 + m 2 ) = 2a 2 = 2ç ÷= 1 + m2 1 + m2 è ø

2

HINTS AND EXPLANATIONS

Area of the rectangle = p1 p2 =

12.52  Chapter 12 q1 q tan 2 2 2 q1 q2 e - 1 Þ tan × tan = which is (a) 2 2 e+ 1 40. The coordinates of the given point P are (a cosa, b sina). b sin a - 0 b \ tan b = = tan a a cos a - 0 a

Þ (e - 1) = (e + 1) tan



\ tan(a - b ) =

tan a - tan b 1 + tan a × tan b

tan a ( a - b) = l (say) a + b tan 2 a dl ( a + b tan 2 a )( a - b)sec 2 a - ( a - b) tan a × 2b tan a sec 2 a = da ( a + b tan 2 a ) 2 =

=

( a - b)sec 2 a ( a - b tan 2 a ) ( a + b tan 2 a ) 2

dl For extremum, = 0  ⇒  a - b tan2a = 0 da a Þ tan a = . b

HINTS AND EXPLANATIONS

41. Since x cosa + y sina = p subtends a right angle at the centre (0, 0), therefore x2 y2 making equation of hyperbola 2 - 2 = 1 homogeneous a b with the help of xcosa + ysina = p 2 x 2 y 2 æ x cos a + y sin a ö ÷ we get 2 - 2 = ç p a b è ø æ 1 cos 2 a ö 1 sin 2 a ö -2 sin a cos a xy 2æ =0 i.e., x 2 ç 2 ÷ + y ç- 2 ÷+ 2 p ø p2 ø p2 èa è b Coefficient of x2 + coefficient of y2 = 0 1 cos 2 a 1 sin 2 a Þ 2- 2= 0 a p2 b p2 ab 1 1 1 - 2 = 2 Þ p= . 2 2 a b p b - a2 Since p is also the length of the perpendicular from (0, 0) to the line x cos a + y sin a = p ab \ Radius of the circle = p = . 2 b - a2

Þ

42. Let the directrix be x = a/e and focus be S(ae, 0). Let P(a secq, btanq) be any point on the curve. Equation of tangent x secq y tan q = 1 . Let F be the intersection point at P is a b æ a b(secq - e) ö of tangent and the directrix, then F = ç , e tan q ÷ø èe b(secq - e) b tan q , mPS = Þ mSF = - e tan q ( a 2 - 1) a(secq - e) ⇒ mSF × mPS = -1. 43. Director circle is the locus of point of intersection of perpendicular tangents drawn to a curve:

i.e., x2 + y2 = a2 - b2 director circle of given hyperbola is x2 + y2 = -11, which is not possible. 44. Given two hyperbolas are x2 y2 = 1 (1) 9 16 y2 x2 = 1 (2) 9 16 Equation of tangent to (1), having slope m is and,

y = mx ± 9m 2 - 16 (3) Eliminating y using equation (2) and (3), we get 16 ( mx ±

9m 2 - 16 ) - 9 x 2 = 144 2

Þ (16 m 2 - 9) x 2 ± 32m 9m 2 - 16 x + (144 m 2 - 400) = 0 (4)

For it to be a tangent, we must have D = 0

\ ( 32m 9m 2 - 16 ) = 4(16m2 - 9)(144m2 - 400) ⇒ m2 = 1 ⇒ m = ±1 45. Since the asymptotes of rectangular hyperbola are mutually perpendicular, the other asymptote should be 4x + 3y + l = 0. Also, intersection point of asymptotes is also the centre of the hyperbola. Thus, intersection point of 4x + 3y + l = 0 and 3x - 4y - 6 = 0 æ 18 - 4l -12l - 96 ö , i.e., ç ÷ should lie on the line x - y - 1 = 0. 100 è 25 ø 18 - 4l 12l - 96 \ -1 = 0 25 100 ⇒ l = 17. Hence, the equation of other asymptote is 4x + 3y + 17 = 0 2

46. Equation of normal at any point (ct, c/t) is ct4 - xt3 + ty - c = 0 ⇒ Slope of normal = t2 Let P(h, k) be the point through which the normal is passing. Then, ct4 - ht3 + tk - c = 0 ⇒ Sti = h/c and Stitj = 0 Hence, sum of the slopes of the normal

=

åt

2 i

=

(å t ) i

2

= h2 = c 2 l

Therefore, required locus is x2 = lc 2 47. We have, for the given hyperbola 9 = 16(e2 - 1) Þ e =

5 4

Since (5, 0) satisfies the equation of the line 3 x + ( 5 - 4 2 ) y = 15, so the reflected ray must pass through (-5, 0) and P = ( 4 2 , 3)

Conic Sections (Parabola, Ellipse and Hyperbola)  12.53

QUICK TIPS The transverse axis is the bisector of the angle between the aymptotes containing the origin and the conjugate axis is the other bisector. 48. The equations of the bisectors of the angles between the asymptotes are 3x - 4 y - 1 4x - 3y - 6 =± 5 5 So, the equations of transverse and conjugate axis are x + y - 5 = 0 and x - y - 1 = 0. 49. Let ax + by = 1 be the chord (1) Making the equation of hyperbola homogeneous using (1), we get 3x2 - y2 + (-2x + 4y)(ax + by) = 0 or, (3 - 2a)x2 + (-1 + 4b)y2 + (-2b + 4a)xy = 0 Since the angle subtended at the origin is a right angle, so, coefficient of x2 + coefficient of y2 = 0 ⇒ (3 - 2a) + (-1 + 4b) = 0 ⇒ a = 2b + 1 \ The chords are (2b + 1)x + by - 1 = 0 or, b(2 + y) + (x - 1) = 0, which, clearly, pass through the fixed point (1, -2). 50. The coordinates of any point on the hyperbola are ( 24 secq , 18 tanq ) . Equation of tangent at this point is x sec q y tan g q = 1 (1) 24 18 The point is nearest to the line 3x + 2y + 1 = 0 (2) If (1) and (2) are parallel secq

18 3 1 =Þ sin q = 2 24 tan q 3 Thus, the point is (6, -3). 51. Let the point be (x1, y1), Equation of chord of contact of tangents drawn from the point (x1, y1) to the hyperbola x2 - y2 = a2 is xx1 - yy1 = a2(1) Equation of normal chord is x y + = 2a (2) secq tan q Since (1) and (2) are identical, comparing coefficients in (1) and (2), we get x - y1 a2 1 = = 1 / secq 1 / tan q 2a a -a Þ sec q = and tan q = 2 x1 2 y1 sec2q - tan2q = 1 a2 a2 =1 Þ 4 x12 4 y12 \ The required locus is a2( y2 - x2) = 4x2y2.

Þ

52. Centre and radius of the given circle is P(6, 0) and, respectively. Equation of normal for y2 = 4x at (t2, 2t) is y = -tx + 2t + t3. The normal must pass though (6, 0) in order that it gives minimum distance between the two curves.

\ 0 = t3 - 4t ⇒ t = 0 or t = ±2 \ A (4, 4) and C (4, -4) PA = PC = 20 = 2 5 \ required minimum distance = 2 5 - 5 = 5 . 53. Any point on the given parabola is (t2, 2t). The equation of the tangent at (1, 2) is x - y + 1 = 0 The image (h, k) of the point (t2, 2t) in x - y + 1 = 0 is given 2(t 2 - 2t + 1) h - t 2 k - 2t = =by = 1 1+ 1 -1 \ h = t2 - t2 + 2t -1 = 2t -1 and, k = 2t + t2 -2t + 1 = t2 + 1 Eliminating t from h = 2t -1 and k = t2 + 1. we get (h + 1)2 = 4 (k - 1) The required equation of reflection is (x + 1)2 = 4(y -1). 54. Given parabola is y2 = 4ax(1) Let QP and PR be the incident and reflected rays, respectively. Let PT be the tangent to the parabola at P and PN be the normal to the parabola at P.

×

Let ∠QPN = q, then ∠RPN = q. Let S be the focus of the parabola. Then, S ≡ (a, 0). Let P ≡ (at2, 2at). Equation of tangent PT is 1 yt = x + at2. Its slope = . t \ Slope of normal PN = -t. Slope of PQ = 0. Let slope of PR = m. Equating the two values of tanq, we get

HINTS AND EXPLANATIONS

\ equation of S′P is 2 y = x + 5.

12.54  Chapter 12 0+t -t - m = 1 + 0( -t ) 1 - m

-t - m 1- m 2t ⇒ t - t2m = - t - m ⇒ m = 2 . t -1 2t ( x – at 2 ). \ Equation of PR is y - 2at = 2 t –1 a2 55. Given circle is x 2 + y 2 = (1) 2 and given parabola is y2 = 4ax(2) Þ t=

HINTS AND EXPLANATIONS

Let PQ be a common tangent to the circle and the parabola. Let P ≡ (at2, 2at). Now, equation of PQ which is a tangent to the parabola at P is ty = x + at2  or  x - ty + at2 = 0 (3) Since PQ is also a tangent to circle (1) 0 - 0 × t + at 2

\



⇒ 2t4 - t2 - 1 = 0 ⇒ t2 = 1,

1+ t

2

=

a



2 -1 . 2

\ t2 = 1 ⇒ t = ±1. Hence, there will be two points P′(a, 2a) and P (a, -2a) on the parabola, the tangents at which will also be tangents to the circle. Now, equations of tangents to the parabola at P′(a, 2a) and P(a, -2a) will be y = x + a(4) and, -y = x + a(5) Solving (4) and (5), we get x = -a, y = 0 \ A ≡ (-a, 0). a2 \ Equation of QQ′ will be - ax + y × 0 = 2 or, 2x = -a(6) Equation of PP′ is y × 0 = 2a (x - a) or x = a(7) æ -a a ö Solving (4) and (6), we get Q′ ≡ ç , ÷ . è 2 2ø æ a - aö ÷ Solving (5) and (6), we get Q ≡ ç - , è 2 2 ø Solving (4) and (7), we get P′ ≡ (a, 2a). Solving (5) and (6), we get P ≡ (a, -2a). Clearly, QPP’ Q’ is a trapezium, therefore its area 1 1 æa ö = ( QQ¢ + PP ¢ ) × LM = ( a + 4 a) ç + a ÷ 2 2 è2 ø 1 a 15 2 = × 5a × 3 = a. 2 2 4

56. Given parabola is y2 - 16x - 8y = 0 (1) Let the coordinates of the feet of the normal from (14, 7) be P(a, b). Now, equation of the tangent at P(a, b) to parabola (1) is yb - 8(x + a) - 4(y + b) = 0 or, (b - 4)y = 8x + 8a + 4b(2) 8 . Its slope = b -4 Equation of normal to parabola (1) at (a, b) is 4-b y - b = ( x - a ). 8 It passes through (14, 7), 4-b 6b (14 - a ) or a = \ 7-b = (3) 8 b -4 Also, (a, b) lies on parabola (1), \  b 2 - 16a - 8b = 0 (4) Putting the value of a from (3) in (4), we get 96 b b2 - 8b = 0   ⇒  b(b 2 - 12b - 64) = 0 b -4 ⇒ b (b - 16) (b + 4) = 0. \ b = 0, 16, - 4. From (3), when b = 0, a = 0, when b = 16, a = 8 and when b = -4, a = 3. Hence, the feet of the normals are (0, 0) , (8, 16) and (3, - 4). 57. Given curve is ax2 + 2hxy + by2 = 1 (1) Let P ≡ (a, b). Let line PS make an angle q with the positive direction of x-axis. Coordinates of any point on line PS may be taken as (a + rcosq, b + rsinq).

If point (a + rcosq, b + rsinq) lies on curve (1), then a(a + rcosq)2 + 2h(a + rcosq) (b + rsinq) + b(b + rsinq)2 = 1 ⇒ (acos2q + 2hcosq sinq + bsin2q)r2 + 2(aacosq + hbcosq + hasinq + bbsinq)r + aa2 + 2hab + bb2 - 1 = 0 (2) Equation (2) will give two real values of r say r1 and r2 and corresponding to these two values of r we will get two points Q and R on curve (1). Also, PQ = | r1 | and PR = | r2 | Now, PQ × PR = | r1 | | r2 | = | r1r2 | aa 2 + 2hab + bb 2 - 1 = a cos 2 q + b sin 2 q + h sin 2q

Conic Sections (Parabola, Ellipse and Hyperbola)  12.55 From (3), it is clear that PQ × PR will be independent of slope of line PS, i.e., independent of q, if a = b and h = 0. Thus, equation of curve will become a(x2 + y2) = 1, which is a circle.

59. The given ellipse is

x2 y2 + = 1. a2 b2

58. We can write the ellipse x2 + 4y2 = 4 as x2 + y 2 = 1 (1) 4 Equation of any tangent to the ellipse (1) can be written as x cosq + y sin q = 1 (2) 2 Let A ≡ (acosq, -bsinq) Then, C ≡ (acosq, -bsinq) 1 D = Area of ΔABC = ´ AC ´ BD = AD × BD 2 = bsinq(a - acosq) 1 = ab( 2 sin q - sin 2q ) 2 x2 y2 + = 1 (3) 6 3 Suppose, the tangents at P and Q meet in A(h, k). Equation of the chord of contact of the tangents through A(h, k) is hx ky + = 1 (4) 6 3 Since (4) and (2) represent the same line h/6 k /3 1 = = \ cosq sin q 1 2 ⇒ h = 3 cosq and k = 3sinq. Thus, coordinates of A are (3cosq, 3sinq). The joint equation of the tangents at A is given by T2 = SS1

dD 1 = ab( 2 cosq - 2 cos 2q ) = 0 (1) dq 2 ⇒ cos2q = cosq ⇒ 2cos2q - cosq - 1 = 0 ⇒ (2cosq + 1) (cosq - 1) = 0 -1 ⇒ cosq = or cosq = 1 2 If q = 0, D = 0, which is not possible. \ q = 2p /3. Now,



\ D max =

3 3 ab [substitiuting the value of q in (1)] 4

60. Given ellipse is

x2 y2 + = 1 (1) a2 b2

2

æ x2 y2 ö æ h2 k 2 ö hx ky i.e., æç + - 1ö÷ = ç + - 1÷ ç + - 1÷  3 3 3 è 6 ø è 6 øè 6 ø 2 Let a = coefficient of x in (5) =

(5)

ö -k 2 1 h2 1 æ h2 k 2 + - ç + - 1÷ = 36 6 è 6 3 ø 18 6

and, b = coefficient of y2 in (5) ö -h2 1 - k 2 1 æ h2 k 2 + . - ç + - 1÷ = 9 3è 6 3 ø 18 3 1 1 -1 We have,  a + b = ( h2 + k 2 ) + + 18 6 3 =

=

1 -1 (9 cos 2 q + 9 sin 2 q) + 18 2

-1 1 (9) + = 0. 18 2 Thus, (5) represents two lines which are at right angles to each other. =

Its auxiliary circle is x2 + y2 = a2(2) Let P ≡ (acosa, bsina) Equation of tangent to the ellipse at P(acosa, bsina) is x cos a y sin a (3) + = 1 a b Making equation (2) homogeneous with the help of (3), we get

HINTS AND EXPLANATIONS

Equation of the second ellipse can be written as

12.56  Chapter 12 æ x cos a y sin a ö x 2 + y 2 - a 2ç + ÷ = 0 è a b ø 2



æ ö a2 Þ (1- cos 2 a) x 2 + ç 1- 2 sin 2 a÷ y 2 b è ø

(4)

a - 2 cos a sin a xy = 0 b (4) is the joint equation of OL and OM. Since ∠ LOM = 90º, \  coefficient of x2 + coefficient of y2 = 0 a2 2 sin a = 0 b2



Þ 1 - cos 2 a + 1 -



æ a2 ö Þ sin 2 a ç 2 - 1÷ = 1 èb ø

æ 1 ö 2 - 1÷ = 1 [ or, sin a ç 2 è1- e ø

∴



b2 = a2 (1 - e2)]

⇒ e2 sin2 a = 1 - e2   or  e2 (1 + sin2 a) = 1 \ e=

62. Let the orbit of the earth be the ellipse x2 y2 2 + 2 = 1 (1) a b Length of major axis = 2a = 186 × 106 miles (given) ⇒ a = 93 × 106 miles. 1 Also, eccentricity e = (given). 60 Let the sun be at the focus S(ae, 0). Then, the earth 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. \ Shortest distance of the earth from the sun = SA, where S is (ae, 0) and A is (a, 0) æ 1ö ÷ = 9145 × 104 miles and lon = a - ae = (93 × 106) ç 1è 60 ø gest distance of the earth from the sun = SA′, where S is (ae, 0) and A′ is (-a, 0) æ 1ö ÷ = 9455 × 104 miles. = a + ae = (93 × 106) ç 1+ è 60 ø 63. Let P ≡ (asecq, btanq)

1 1+ sin 2 a

HINTS AND EXPLANATIONS

61. Let P ≡ (acosa, bsina) and Q ≡ (acosb, bsinb)

The equation of the chord PQ is x æa + b ö y æa + b ö æa - b ö + sin ç = cos ç cos ç ÷ ÷ ÷ (2) a è 2 ø b è 2 ø è 2 ø Since it cuts the major axis of the ellipse at a distance d from the centre, \  it must pass through the point (d, 0), i.e., d æa + b ö æa - b ö = cos ç cos ç ÷ ÷ a è 2 ø è 2 ø



d æa + b ö æa - b ö cos ç = cos ç ÷ ÷ a è 2 ø è 2 ø

Þ

æa - b cos ç d -a è 2 = d+a æa - b cos ç è 2



ö æa ÷ - cos ç ø è ö æa ÷ + cos ç ø è

+b 2 +b 2

ö ÷ ø ö ÷ ø

[By componendo and dividendo]

2 sin a / 2 sin b / 2 = = tan a/2 × tan b/2. 2 cos a / 2 cos b / 2

\ tan a / 2× tan b / 2 =

d- a d+ a

Then, N ≡ (asecq, 0). Since Q divides AP in the ratio a2 : b2, \ coordinates of Q are æ ab 2 + a 2 secq a 2b tan q ö =ç , 2 ÷. 2 2 a + b2 ø è a +b Slope of

A′ P =

Slope of

QN =

b tan q a(secq + 1)

a 2b tan q ab 2 + a3 secq - a3 secq - ab 2 secq

a 2b tan q . ab 2 (1- sec q) a 2b 2 tan 2 q \ Slope of A′ P × slope of QN = = -1. - a 2b 2 tan 2 q \ QN is ^ to A′ P. 64. There are two common tangents to the circle x2 + y2 = 1 and the hyperbola x2 - y2 = -1. These are x = 1 and x = - 1. Out of these two, x = 1 is nearer to the point æ1 ö P ç , 1÷ . Thus, a directrix of the required ellipse is x = 1. è2 ø

=

Conic Sections (Parabola, Ellipse and Hyperbola)  12.57

If Q(x, y) is any point on the ellipse, then its distance from

from the directrix x = 1 is | x - 1 |. By definition of ellipse, QP = e | x - 1 | æ 1ö 1 2 ç x - ÷ + ( y - 1) = | x - 1 | è 2ø 2 2



Þ



⇒ x2 - x +



⇒ 3x2 - 2x + 4y2 - 8y + 4 = 0



2 ö æ Þ 3 ç x 2 - x ÷ + 4( y - 1) 2 = 0 3 ø è



2 éæ 1ö 1ù 2 Þ 3 êç x - ÷ - ú + 4( y - 1) = 0 3 ø 9 úû êëè



1ö 1 æ Þ 3 ç x - ÷ + 4( y - 1) 2 = 3ø 3 è



1ö æ ç x - 3 ÷ ( y - 1) 2 è ø + Þ = 1. 1/ 9 1 / 12

1 1 + y2 - 2y + 1 = ( x 2 - 2 x + 1) 4 4

2

2

65. Let P(h, k) be any point on the locus. Equation of the line through P and having slope 4 is

y - k = 4(x - h)(1) Suppose, this line meets xy = 1 (2) in A(x1, y1) and B(x2, y2). Eliminating y from (1) and (2), we get 1 - k = 4(x - h) x ⇒ 4x2 - (4h - k)x - 1 = 0 (3) Since x1 and x2 are the roots of (3) 4h - k \ x1 + x2 = (4) 4 -1 and, x1 x2 = (5) 4

8h + k ö æ 2 h + k ö 1 Also, çæ ÷ç ÷ = - = h or 2x1 + x2 = 3h(6) è

2

øè

2

ø

4

From (4) and (6), we get ( 4 h - k ) 8h + k x1 = 3h = 4 4 (8h + k ) - ( 2h + k ) and, x2 = 3h = 2 2 Putting these values in (5), we get 1 æ 8h + k ö æ 2 h + k ö ç ÷ç ÷=2 ø 4 è 2 øè ⇒ (8h + k) (2h + k) = 2 or 16h2 + 10hK + k2 = 2. Thus, equation of required locus is 16x2 + 10xy + y2 = 2.

Previous Year's Questions 66. The equation of an ellipse is x2 y2 + =1 16 9 Here a = 4, b = 3 ∴  Foci of the above ellipse are ( ± 7 , 0) ∴  Radius of required circle = ( 7 - 0) + (0 - 3) 2

2

= ( 7 + 9 = 16 = 4 unit

x2 y2 67. Key Idea :The foci of an ellipse 2 + 2 = 1 is given by a b (±ae, 0) 1 Since, e = , ae = 2 2 ⇒  a = 4 b2 = a2(1 – e2) ∴ 

æ 1ö = 16 ç1 - ÷ = 12 è 4ø

Thus, the equation of an ellipse is

x2 y2 + = 1. 16 12

HINTS AND EXPLANATIONS

æ 1ö 2 ç x - ÷ + ( y - 1) and its distance è 2ø 2

the focus is QP =

12.58  Chapter 12 æ x + x2 y1 + y2 ö 68. The mid-point of the chord is ç 1 , . 2 ÷ø è 2 The equation of the chord in terms of its mid-point is T = S1 æ y + y2 ö i.e., x ç 1 ÷+ è 2 ø

æ x + x2 ö æ x1 + x2 ö æ y1 + y2 ö yç 1 ÷=2ç 2 ÷ç 2 ÷ è 2 ø è øè ø

⇒  x (y1 + y2) + y (x1 + x2) = (x1 + x2)(y1 + y2)

x y + =1 x1 + x2 y1 + y2 69. is correct answer.

⇒ 

70.

x2 æ 12 ö ç 5÷ è ø

2

-

y2 æ9ö ç5÷ è ø

2

=1 5 4

b2 ´4 =3 16

⇒  b2 = 7 Hence, (C) is the correct answer. æ9 ö 71. Any point on the given parabola is of the form ç t 2 , 9t ÷ 2 è ø 2 Differentiating y = 18 x w.r.t. x

HINTS AND EXPLANATIONS

76. ∵  ∠FBF' = 90° \

1 dy 9 1 = = = 2 (given ) Þ t = 2 dx y t æ9 9ö ⇒  Point is ç , ÷ è8 2ø 72. Points of intersection of given parabolas are (0, 0) and (4a, 4a). And the equation of line passing through these points is y = x On comparing this line with the given line 2bx + 3cy + 4d = 0, we get d = 0 and 2b + 3c = 0  ⇒  (2b + 3c)2 + d2 = 0. a 73. The equation of directrix is x = = 4. So, a = 2b2 = a2 (1 e - e2) ⇒ b2 = 3 Hence the equation of ellipse is 3x2 + 4y2 = 12. 74. Area of rectangle ABCD = (2a cosθ) (2b sinθ) = 2ab sin2θ ⇒  Area of greatest rectangle is equal to 2ab When sin2θ = 1

y = mx ± a 2 m 2 - b 2

a 2e 2 + b 2

x2 y2 = 1 is a2 b2

) +( 2

a2 e 2 + b 2



⇒ 2(a2e2 + b2) = 4a2e2





Þ e2 =

)

2

= 2( ae) 2

b2 a2 b2 = 1 - e2 a2

1 . ⇒  2e2 = 1  ⇒ e = 2   77. We have a2 = cos2α and b2 = sin2α So, the coordinates of foci are (±ae, 0) ∴  b2 = a2 (e2 − 1)  ⇒  e = secα Hence, abscissae of foci remain constant when α varies.

78. Major axis of length 2a is along x-axis.0 a Now, - ae = 4 e 1ö æ Þ aç 2 - ÷ = 4 2ø è 8 Þ a= 3 79. We have 2ae = 6  ⇒  ae = 3 2b = 8  ⇒  b = 4 ∴  b2 = a2(1 − e2) ⇒ 16 = a2 − a2e2 ⇒  a2 = 16 + 9 = 25 ⇒  a = 5 3 3 \ e= = a 5 80. Vertex is (1, 0)

81.

75. Tangent to the hyperbola

(

Also, e 2 = 1 -

Eccentricity of hyperbola , e1 = Now, ae2 = 1 -

Given that y = αx + β is the tangent of hyperbola ⇒  m = α and a2m2 − b2 = β2 ∴  a2α 2 − b2 = β2 Locus is a2x2 − y2 = b2which is hyperbola.

Conic Sections (Parabola, Ellipse and Hyperbola)  12.59 x2 y2 + = 1   ⇒  a = 2, b 4 1

x2 y2 x2 y2 2 + 2 = 1 Þ + =1 42 b2 a b Now, (2, 1) lies on it 4 1 1 1 3 4 Þ + =1 Þ =1- = Þ b2 = b2 16 b 2 4 4 3

x2 y2 x2 3y2 + + =1 =1 Þ 16 æ 4 ö 16 4 ç3÷ è ø Þ x 2 + 12 y 2 = 16 82. The locus of perpendicular tangents is directrix i.e, x = −a; x = −1 83. y 2 = 16 3 x \

4 3 is tangent to parabola y = mx + m Which is tangent to ellipse ⇒ c2 = a2m2 + b2 48 Þ = 2m 2 + 4 m2 ⇒  m4 + 2m2 = 24 ⇒  m4 = 4 84. Semi minor axis b = 2 Semi major axis a = 4 2 x y2 Equation of ellipse = 2 + 2 = 1 a b 2

x y + =1 16 4 ⇒  x2 + 4y2 = 16. 5 85. Let the tangent to the parabola be y = mx + ( m ¹ 0) m Now, its distance from the center of the circle must be equal to the radius of the circle. Therefore,



Þ



5 5 1 + m2 = m 2

Þ (1 + m 2 )m 2 = 2

Þ m4 + m2 - 2 = 0 ⇒ (m − 1) (m + 2) = 0 ⇒ m = ± 1 And so, the common tangents are 2

y = mx ± 6 m 2 + 2 where m =

2

y = x + 5 and y = - x - 5 . 86. Foci ≡ (±ae, 0) We have, a2e2 = a2 − b2 = 7 Now, the equation of circle is

(



( x - 0) 2 + ( y - 3) 2 =



⇒  x2 +y2 − 6y − 7 = 0.

)





x Þ x 2 + 32 æç + t ö÷ = 0 èt ø 32 2 Þ x + x + 32t = 0 t 2

æ 32 ö Þ ç ÷ - 4(32t ) = 0 è t ø æ 32 ö Þ 32 ç 2 - 4t ÷ = 0 t è ø

⇒  t3=8  ⇒  t = 2.

⇒  Slope of tangent is

1 1 = . t 2

4t =t 4 2t 2 t 2 And, k = = 4 2 ⇒ x2 = 2y 90. a = 3, b = 5 89. h =

5 2 e = 1- = 9 3 Foci = (±2, 0) Tangent at P Þ

2x 5 y + =1 9 3.5

2x y + =1 9 3 ⇒ 2x + 3y = 9 ∴  Area of quadrilateral = 4 × (area of triangle QCR) æ1 9 ö = ç ´ ´ 3 ÷ ´ 4 = 27 è 2 2 ø

⇒ 

91. Given 2b 2 = 8 (1) a ⇒ 2b = ae(2) We know b2 = a2(e2 − 1) (3)

2

7 - 0 + (0 - 3) 2

h k

6 h2 h2 + k 2 Þ +2 = 2 k k 2 2 2 2 2 6h + 2k = (h + k ) So required locus is 6x2 + 2y2 = (x2 + y2)2. 88. Equation of tangent at A(t2, 2t) yt = x + t2 is tangent to x2 + 32y = 0 at B



x2 y2 + =1 4 2

2

87. Let the foot of perpendicular be P(h, k) Equation of tangent with slope m passing P(h, k) is

Putting

b e = from (2) in (3), we get a 2

HINTS AND EXPLANATIONS

Given ellipse x2 + 4y2 = 4  ⇒  = 1  ⇒  p = (2, 1) Required Ellipse has equation

12.60  Chapter 12 x2 y2 + =1 4 3

e2 = e 2 - 1 4 ⇒  4 = 3e2 2 Þ e= 3

Equation of normal is 3 x −1 y − 2 ⇒ 4x – 2y – 1 = 0 = 1 3 4 2×3

92. Circle and parabola are as shown: y2 = 8x a=2

Hence, the correct option is (A)

O P(2t 4t) (0, − 6)C

HINTS AND EXPLANATIONS

2 a2

−

3 b2 3

=1

∴ a2 = 1 y2 =1 3 ∴ Tangent at P( 2 , 3 ) is ∴ x2 =

C

Q

y2

⇒ b2 = 3

⇒  x2 + y2 – 4x + 8y + 12 = 0 93. Consider line L at a distance of 6 units below x axis ⇒  PC = PQ ⇒  P lies on a parabola, for which C is focus and L is directrix

94. x = –4

−

2 − =1 4 − b2 b2

\ CP = 4 + 4 = 2 2 ∴  equation of circle is ( x - 2) 2 + ( y + 4) 2 = ( 2 2 ) 2

r

2

and

Minimum distance occurs along common normal. Let the equation of normal to parabola y + tx = 2.2.t + 2t3 Since it passes through (0, –6) ∴  −6 = 4t + 2t3 ⇒ t3 + 2t + 3 = 0 ⇒  t = −1 (only real value) ∴  coordinates of point P are (2, −4).

6

x2

=1 a b2 a2 + b2 = 4 95.

2,

2x −

y 3

=1

Clearly it passes through ( 2 2 , 3 3 ) Hence, the correct option is (A) 96. Let the parabola be y2 = 16x; y2 = 4ax a = 4 Equation of tangent a y = mx + m 4 m It passes through (16, 16) 4 ⇒ 16 ± 16 m + m ⇒ y = mx +

P r L

P (16, 16)

⇒ 16 m = 16 m 2 + 4 A(–16, 0)

B (24, 0)

⇒ m 2 − m +

1 =0 4

1 2 Equation of tangent is x y = +8 2 x – 2y + 16 = 0 Equation of normal 2x + y = k x = 16 y = 16 k = 48 Solving, we get m =

e=

1 2

−a = –4 e –a = –4 × e a=2 Now, b2 = a2(1 – e2) = 3 Equation to ellipse

(i)

Conic Sections (Parabola, Ellipse and Hyperbola)  12.61 Normal equation 2x + y = 48 Axis of Parabola is y = 0 x – 2y + 16 = 0 y=0 x = –16 A = (–16, 0) 2x + y = 48 y=0 x = 24 B = (24, 0)

(ii)

T2 : y = 5 x + 3



Using T1 − 5x + y = 3



⇒

θ

⇒

B (24, 0)



AB is diameter as slope of AP × slope of PB = –1 C = (x, y)

2

−

y2

y = − 5x + 3



C (4, 0)

x2

= 1 at point a b2 yy ⎡ xx ⎤ (x1, y1) on Hyperbola  ⎢ 21 − 21 = 1⎥ b ⎣a ⎦ (x1, y1) are point of contact ∴ First point of contact is x1 = −3 5 and y1 = –12 Taking equation T2 Equation of tangent to hyperbola

P (16, 16)

A (–16, 0)

x y + =1 3 3 − 5

5x + y = 3

5x y ⎡x x y y ⎤ + = 1    ⎢ 22 − 22 = 1⎥ 3 3 b ⎣a ⎦

∴ We get (x2, y2) on comparison as x2 = 3 5 and y2 = –12

−16 + 24 x= =4 2 y=0 16 − 0 16 4 Slope of PC = = = 16 − 4 12 3

T (0, 3)



4 − ( −2) 4+6 10 tan θ = 3 = = =2 4 3−8 −5 1 + × −2 3

P (–3√5, –12)

∴ PQ = 2 × 3 5 = 6 5 Altitude = 3 – (–12) = 15 Area of triangle 1 PQT = ´ 6 5 ´ 15 2 = 45 5 unit

97. 4x2 – y2 = 36 is a hyperbola ⎡ x2 y2 ⎤ x2 y2 − = 1    ⎢ 2 − 2 = 1 form ⎥ 9 36 b ⎢⎣ a ⎥⎦ Equation of tangent ⇒

y = mx + a 2 m 2 − h2

98. If r Î (0, 1) it’s a hyperbola

1 0, 3

y = mx + 9m 2 − 36

Whose eccentricity is

It passes through (0, 3) Hence, 3 = 0 ± 9m 2 − 36 ⇒ 9m2 – 36 = 9 m2 = 5

m=± 5

Equation of Tangents are

T1 : y = 5 x + 3

P

Q (3√5, –12)

Q

1+

1− r = 1+ r

2 r +1

While if r > 1, then curve is an ellipse having eccentricity

1-

r -1 = r =1

2 r +1

99. Taking minor image of whole Scenario w.r.t. y = x then solving chord and parabola we have x2 – 10x + 16 = 0 As length of chord is |x1 – x2| 1 + m 2 = 6 1 + 2 = 6 3

HINTS AND EXPLANATIONS

Referring to figure

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CHAPTER

13

Vector Algebra

LEARNING OBJECTRIVES After reading this chapter, you will be able to:   Understand the concepts of scalar and vector quantities   Learn how to represent vectors and what are its types   Know how to calculate angle between two vectors, addition of two vector, and subtraction of two vectors, volume of tetrahedron and vector triple product  Be familiar with position vector, component of a vector, linear combination, collinearity of three points

SCALARS AND VECTORS Scalar Quantity A quantity which has only magnitude and no direction is called a scalar quantity or simply a scalar. Examples of scalars are mass, temperature, volume, work and so on. To specify a scalar, two things are needed. 1. a unit in terms of which it is measured 2. a real number (+ve, -ve or zero) Vector Quantity  A quantity which has magnitude as well as direction is called a vector quantity or simply a vector. Examples of vectors are displacement, velocity, acceleration, force and so on. To specify a vector, three things are needed. 1. a unit in terms of which it is measured 2. a real number (+ve, -ve or zero) 3. a particular direction

REPRESENTATION OF VECTORS The best way to represent a vector is with the help of a directed line segment. Suppose A and B are two points, then by the vector AB. , we mean a quantity whose magnitude is the length AB and whose direction is from A to B. Aand B are called the end points of the vector AB. In particular A is called the ini- FIGURE 13.1 tial point and B is called the terminal point.  Sometimes a vector AB. is expressed by a single letter a (which is always written in bold type, to distinguish it

 Grasp the knowledge on coplanarity of four points, some results on linearly dependent and independent vectors, product of the two vectors and scalar product of two product, vector product of two vectors, moment of a force about a point, scalar triple product

from a scalar). Sometimes, however, we write the vector a  as a or a .

I M P O R TA N T P O I N T S 

Every vector ( AB) has the following  three characteristics:

 (a)  Length:  The length of AB is denoted by | AB | or simply AB. (b)  Support:  The line of unlimited length of which AB  is segment is called thesupport of vector AB .  (c)  Sense:  The sense of AB is from A to B and that of  BA will be from B to A. The sense of directed line segment is from its initial point to the terminal point.

Modulus (or Magnitude) of a Vector  The positive real number which is the measure of the length of the vector, is called the modulus, length, magnitude, absolute value or norm of the vector. The modulus of a vector a or OA is usually denoted  by | a | or | OA | or by the corresponding letter ‘a’ (not in bold-faced type), i.e.,  | OA | = OA and | a | = a

Multiplication of a Vector by a Scalar The product of a scalar m and a vector a, is defined as a vector ma or am whose magnitude is the product of the magnitudes of m and a and whose direction is that of a or opposite to a accordingly as m is positive or negative.

13.2  Chapter 13

TYPES OF VECTORS Equal Vectors  Two vectors a and b are equal when they have (1) the same magnitude and (2) the same direction. Symbolically such vectors are written as: a = b. Unit Vectors  A vector whose magnitude is unity is called a unit vector. The unit vector having the same direction as that of given vector a is usually denoted by the symbol aˆ (read as ‘a cap’), i.e., A vector = Modulus of vector × Unit vector in its direction a = | a | aˆ

or

Also, Unit vector in a direction = aˆ =

vector in that direction modulus of vector a |a|

ERROR CHECK No units are to be attached with a unit vector, i.e., unit vector is dimensionless physical quantity Zero or Null Vector  A vector whose magnitude is zero, is called a zero vector. For such a vector, initial and terminal points are coincident so that its direction is indeterminate. A zero vector is denoted by the bold-faced symbol O or O. Collinear (or Parallel Vectors)  The vectors which are parallel to the same straight line are called collinear vectors. Vectors which are not parallel to the same line are called non-collinear vectors. Like and Unlike Vectors    Collinear vectors having the same direction are called like vectors and those having the opposite directions are called unlike vectors. Remark:  If two vectors a and b are collinear, then there exists a scalar m such that b = ma, m being positive or negative according as a and b are like or unlike vectors. Conversely, if b = ma be given, then a and b must be collinear (or parallel) vectors such that | b | = | m | | a |. Reciprocal Vector  Let | a | be the modulus of the given vector a. Then a vector whose direction is that of a but modulus is 1/| a | (reciprocal of the modulus of a) is called the reciprocal of a and is written as a-1. Thus, a −1 =

1 |a| a aˆ = 2 aˆ = 2 |a| |a| |a|

Coplanar and Non-coplanar Vectors  Three or more vectors are said to be coplanar when they are parallel to the same plane. Otherwise they are said to be non-coplanar vectors.

Co-initial Vectors  The vectors which have the same initial point are called co-initial vectors. Negative of a Vector  A vector having the same modulus as that of a given vector a and the direction opposite to that of a, is called the negative of a and is denoted by -a. Clearly, if OA = a, then AO = -a, and therefore, OA = -AO

EQUAL VECTORS Two vectors a and b are said to be equal when they have equal magnitudes and same direction. Geometrically, if head of one vector coincides with, the head of other and so do the tails coincide then the vectors are said to be equal. ERROR CHECK If a = b, then a = b, always. But if, a = b doesn’t always imply a = b

FIXED VECTORS Fixed vector is that vector whose initial point or tail is fixed. It is also called localised vector. For example, position vector and displacement vector are fixed vectors.

FREE VECTORS Free vector is that whose initial point or tail is not fixed. It is also known as non localised vector, For example, velocity vector of a particular moving particle along a straight line is a free vector.

d c

O

b a FIGURE 13.2

ANGLE BETWEEN TWO VECTORS The angle between two vectors a and b represented by OA and OB, is defined as the angle AOB which does not exceed π. This is also known as the inclination of given vectors a and b. If the angle AOB be θ, then 0 ≤ θ ≤ π.

Vector Algebra  13.3

OB = OA + AB = a + b

(1)

This method of addition of vectors is known as the triangle law of addition. Completing the parallelogram OABC. Since FIGURE 13.3

p If q = , then vectors are said to be perpendicular 2 or orthogonal and if θ = 0 or π, then vectors are said to be parallel or coincident.

I M P O R TA N T P O I N T S Whenever finding angle between two vectors, make sure that either their heads coincide or their tails coincide.

FIGURE 13.4

i.e., if heads coincide or tails coincide then internal angle is the angle between two vectors (whether acute or obtuse) as in (1), (2), (3) and (4).   If heads coincide with tall then external angle is the angle between the two vectors as in (5) and (6).

FIGURE 13.5

ADDITION (SUM OR RESULTANT) OF TWO VECTORS

FIGURE 13.6

Let a, b be two vectors. Take any point O and draw the vectors OA = a and AB = b such that the terminal point of the vector a is the initial point of vector b. Join OB. Then the vector OB is defined as the sum of a and b and is written as

AB = OC = b OB = OA + AB = OA + OC(2)

That is, the sum of two co-initial vectors is the vector represented by the diagonal of the parallelogram formed with the component vectors as adjacent sides. This method of addition of vectors is known as the parallelogram law of addition. Remark  From Equation (1), -BO = OA + AB or OA + AB + BO = O, showing that the sum of vectors determined by the sides of a triangle, taken in order, is zero.

Properties of Vector Addition 1. Vector addition is commutative  For any two vectors a and b, we have a+b=b+a 2. Vector addition is associative  For any three vectors a, b and c, we have (a + b) + c = a + (b + c) 3. Existence of additive identity  For every vector a, we have a+O=a=O+a where O is the null vector.

4. Existence of additive inverse  For a given vector a, there exists a vector -a such that a + (-a) = (-a) + a = 0 The vector -a is called the additive inverse of a.

Properties of Multiplication of Vector by a Scalar 1. If m = o, then ma = o 2. If m and n be two scalars, then m(na) = mna = n(ma) 3. If m and n be two scalars, then (m + n)a = ma + na 4. If a, b are any two vectors and m be any scalar, then m(a + b) = ma + mb

Subtraction (or Difference) of Two Vectors Subtraction of vectors    If a and b are two vectors, then their subtration a - b is defined as a - b = a + (-b) where -b is the negative of b having magnitude equal to that of b and direction opposite to b.

13.4  Chapter 13 Then a - b = (a1- b1) i + (a2 - b2) j + (a3 - b3)k

Since OP = r =



B b

a+b O

a

where l = A

mb + na m n = b+ a m+n m+n m+n = la + mb

n m and m = m+n m+n

Thus, p.v. of any point P on AB can always be taken as r = la + mb, where λ + µ = 1.

−b

a + (−b) = a − b

mb + na m+n ⇒ (m + n) r = mb + na ⇒  n ⋅ OA + m ⋅ OB = (n + m) OP, where P is a point on AB dividing it in the ratio m : n. In particular, if P is the mid-point of AB, then Since OP = r =



B FIGURE 13.7

Properties of Vector Subraction

OA + OB = 2OP

P

POSITION VECTOR If a point O is fixed as the origin in space (or plane) and P is any point, the OP is called the position vector of P with respect to O. If we say that P is the point r, then we mean that the position vector of P is r with respect to some origin O.

That is, twice the position vector of the middle point is equal to the sum of the vectors of the ends.

SOLVED EXAMPLES 1. In a regular hexagon ABCDEF, AB + AC + AD + AE + AF = kAD, where k is equal to (A) 1 (B)  2 (C) 3 (D)  none of these Solution (C)

AB + AC + AD + AE + AF = ED + AC + AD + AE + CD (

\

(i) a - b ≠ b - a (ii)  (a - b) - c ≠ a - (b - c) (iii) Since any one side of a trianr gle is less than the sum and greater than the difference of the other two sides, so for any two vectors a and b, we have (A) | a + b | ≤ | a | + | b | O(Origin) (B) | a + b | ≥ | a | - | b | FIGURE 13.8 (C) | a - b | ≤ | a | + | b | (D) | a - b | ≥ | a | - | b |

F

∴  AB = (Position vector of B) - (Position vector of A) = OB - OA = b - a 2. Position vector of a dividing point  The position vectors of the points dividing the line AB in the ratio m : n inter-

mb + na nally or externally are m+n mb − na or m−n

QUICK TIPS If P is the mid-point of AB, then it divides AB in the ratio 1 : 1. Thus position vector of P is given by



a+b OP = 2

D

E

1. AB in terms of the position vectors of points A and B: If a and b are position vectors of points A and B respectively. Then OA = a, OB = b

AB = ED, AF = CD)

C

A

B

= (AC + CD) + (AE + ED) + AD = AD + AD + AD = 3AD 2. If M and N are the mid points of the diagonals AC and BD respectively of a quadrilateral ABCD, then AB + AD + CB + CD = (A) 4NM (B)  4MN (C) 2MN (D)  none of these Solution (B)

In ∆ABD, N is the mid-point of BD, ∴ AB + AD = 2AN(1)

Vector Algebra  13.5

AB + AD + CB + CD = 2(AN + CN)(3)



In ∆ANC, M is the mid-point of AC ∴

AN + CN = 2MN

From (3), we get AB + AD + CB + CD = 2(2MN) = 4MN 3. Five forces AB, AC, AD, AE, AF act at the vertex A of a regular hexagon ABCDEF. If O is the centroid of the hexagon, then their resultant is a force given by (A) 4AO (B)  5AO (C) 6AO (D)  none of these Solution (C)

If R is the resultant of given forces, then

R = AB + AC + AD + AE + AF = ED + AC + AD + AE + CD ( AB = ED and AF = CD) = (AC + CD) + (AE + ED) + AD = AD + AD + AD = 3AD = 6AO. \

4. ABCD is parallelogram. If L and M are the middle points of BC and CD, then AL + AM = 1 3 (A) AC (B)  AC 2 2 (C) AC

(D)  none of these

Solution (B)

1 1 BC = AB + AD 2 2 1 1 AM = AD + DM = AD + DC = AD + AB 2 2

AL = AB + BL = AB +

Adding, AL + AM =

3 ( AB + AD ) 2



3 3 ( AB + BC ) = AC 2 2

=

5. ABCD is a quadrilateral and E the point of intersection of the lines joining the middle points of opposite sides. If O is any point, then the resultant of OA, OB, OC and OD is equal to (A) 2OE (B)  OE (C) 4OE (D)  none of these Solution (C)

Let P, Q, R, S be the mid-points of sides BC, CD, DA, AB respectively of a quadrilateral ABCD. By geometry the figure formed by joining the mid-points P, Q, R, S will be a parallelogram. Hence, its diagonals will bisect each other, say at E.

Now, P is the mid-point of BC ∴ OB + OC = 2OP(1) And R is the mid-point of AD ∴ OA + OD = 2OR(2) Adding (1) and (2), we have OA + OB + OC + OD = 2(OP + OR) = 2 ⋅ 2OE = 4OE ( E is mid-point of PR, ∴ OP + OR = 2OE) \

In ∆CBD, N is the mid point of BD, ∴ CB + CD = 2CN(2) Adding (1) and (2), we have

6. Two forces act at the vertex A of a quadrilateral ABCD represented by AB, AD and two at C represented by CB and CD. If E and F are the middle points of AC and BD respectively, then their resultant is represented by (A) EF (B)  2EF 3 (C) EF (D)  4EF 2 Solution (D)

We have, AB + AD = 2AF, where F is mid-point of BD

13.6  Chapter 13

Also, CB + CD = 2CF; ∴  AB + AD + CB + CD = 2(AF + CF) = -2(FA + FC) = -2 [2FE], where E is the mid-point of AC = - 4FE = 4EF. 7. Let OA = i + 3j - 2k and OB = 3i + j - 2k. The vector OC bisecting the angle AOB where C is a point on the line AB is (A) 2(i + j - k) (B) 4(i + j - k) (C) i + j - k (D)  none of these Solution (A)

Taking O as origin, the position vectors of A and B are a = i + 3j - 2k and b = 3i + j - 2k respectively. We have | a | = | b | = 14

So, the bisector OC of ∠AOB meets AB at its midpoint C. 1 \ OC = (OA + OB) = 2(i + j - k ) 2 8. The vector c, directed along the internal bisector of the angle between the vectors a = 7i - 4j - 4k and b = - 2i - j + 2k with | c | = 5 6 is 5 (A) (5i + 5 j + 2k ) 3 5 (C) ( -5i + 5 j + 2k ) 3

5 (B)  (i + 7 j + 2k ) 3 5 (D)  (i - 7 j + 2k ) 3

Solution (D)

The required vector c is given by or, ⇒

æ a b ö c = l ( a + b) = l ç + ÷ è |a| |b| ø 1 ì1 ü = l í (7i - 4 j - 4 k ) + ( -2i - j + 2k ) ý 9 3 î þ l ˆ (i − 7 j + 2 k ) 9 l l |c| = ± 1 + 49 + 4 = ± 54 . 9 9 c=

But

| c | = 5 6 (given)

\ ±

l 54 = 5 6 Þ l = ± 15. 9

Hence, c = ±

15 5 (i - 7 j + 2 k ) = ± (i - 7 j + 2 k ) 9 3

9. If the points P, Q, R, S have position vectors p, q, r, s such that p - q = 2(s - r), then (A) PQ and RS bisect each other (B) PQ and PR bisect each other (C) PQ and RS trisect each other (D) QS and PR trisect each other Solution (D)

We have, p - q = 2(s - r)

p + 2r q + 2 s = 1+ 2 1+ 2 ∴  Point dividing PR in the ratio 2 : 1 is same as the point dividing QS in the ratio 2 : 1 Hence QS and PR trisect each other. ⇒ p + 2r = q + 2 s Þ

10. If a and b are position vectors of A and B respectively, then the position vector of a point C in AB produced such that AC = 3AB is (A) 3a - b (B)  3b - a (C) 3a - 2b (D)  3b - 2a Solution (D)

AC = 3AB  ⇒  c - a = 3(b - a)  ⇒  c = 3b - 2a. 11. A vector a has components 2p and 1 w.r.t a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If w.r.t the new system, a has components p + 1 and 1, then 1 (A) p = 0 (B)  p = 1 or p = 3 1 (C) p = -1 or p = (D)  p = 1 or p = -1 3 Solution (B)

Let i, j be unit vectors along the co-ordinate axes ∴ a = 2pi + 1 ⋅ j(1) On rotation, let b be the vector having components p + 1 and 1. ∴ b = (p + 1) i + 1⋅j(2) where i, j are unit vectors along the new co-ordinate axes. But on rotation | b | = | a |  ⇒ | b |2 = | a |2 ⇒ ( p + 1)2 + 1 = (2p)2 + 1  ⇒ 3p2 - 2p - 1 = 0 ⇒ (3 p + 1)( p - 1) = 0 Þ

p = 1 or -

1 3

Vector Algebra  13.7 12. Let ABCDEF be a regular hexagon. If AD = xBC and CF = yAB, then xy = (A) 4 (B)  - 4 (C) 2 (D)  -2

1 1 (A) (B)  4 2 1 1 (C) (D)  6 8

Solution (B)

Solution (A)

Since ABCDEF is a regular hexagon, from plane geometry, we have

OM1 = OA + OB + OC + OD (given) = OM + MA + OM + MB + OM + MC + OM + MD = 4OM + (MA + MC) + (MB + MD) = 4OM ( MA = - MC, MB = - MD) 1 1 \ OM = OM1 \ l = 4 4 15. In a trapezoid the vector BC = λAD and p = µAD and p = AC + BD is collinear with AD. Then (A) µ = λ + 1 (B)  λ = µ + 1 (C) λ + µ = 1 (D)  µ = 2 + λ \

AD = 2 BC and FC = 2AB ∴ AD = 2BC and FC = 2AB(1) Given that AD = xBC. ∴ 2BC = xBC, by (1) ⇒ x = 2 (2) Again, given that CF = yAB or -FC = yAB. ∴ -2AB = yAB, using (2) ⇒ y = -2(3) From (2) and (3), xy = 2 (-2) = - 4. 15 13. A vector a is collinear with vector 6i - 8 j - k of 2 magnitude 50 making an obtuse angle with z-axis is (A) 24i - 32j - 30k (B)  -24i + 32j + 30k (C) 24i + 32j - 30k (D)  none of these

Solution (A)

We have, p = AC + BD = AC + BC + CD = AC + λAD + CD = (AC + CD) + λAD = AD + λAD = (1 + λ)AD Since p = µAD  ∴  µ = 1 + λ 16. AB = 3i + j - k and AC = i - j + 3k . If the point P on the line segment BC is equidistant from AB and AC, then AP is (A) 2i - k (B)  i - 2k (C) 2i + k (D)  none of these Solution (C)

A point equidistant from AB and AC is on the bisector of the angle BAC. A vector along the internal bisector of the angle BAC

Solution (A)

Let b = 6i − 8 j −

15 k. 2

15 2 (6i − 8 j − k ). A unit vector along b is ± 25 2 ∴  a = a vector of length 50 along b = ±4 (6i − 8 j −

15 k ). 2

Since a makes obtuse angle with z-axis, so we must have a ⋅ k < 0 Thus, a = 24i - 32j - 30k 14. Given a cube ABCDA1 B1 C1 D1 with lower base ABCD, upper base A1 B1 C1 D1 and the lateral edges AA1, BB1, CC1, and DD1; M and M1 are the centres of the faces ABCD and A1B1C1D1 respectively. O is a point on line MM1, such that OA + OB + OC + OD = OM1, then OM = λ. OM1 if λ =



=



=

AB AC + | AB | | AC | 3i + j - k

+

i - j + 3k

=

1

( 4i + 2 k ) 9 +1+1 1+1+ 9 11 ∴ AP = t (2i + k) ∴ BP = AP - AB = t (2i + k) - (3i + j - k) = (2t - 3) i - j + (t + 1) k Also BC = AC - AB = (i - j + 3k) - (3i + j - k) = - 2i - 2j + 4k. But BP = s BC. ∴ (2t - 3)i - j + (t + 1) k = s(- 2i - 2j + 4k) ∴ 2t - 3 = -2s, -1 = -2s, t + 1 = 4s

1 and t = 1 ∴ AP = 2i + k . 2 17. The position vectors a, b, c and d of four distinct points A, B, C and D lie on a plane are such that | a - d | = | b - d | = | c - d | then the point D is the ∴ s=

13.8  Chapter 13 (A) centroid of ∆ABC (B) orthocentre of ∆ABC (C) circumcentre of ∆ABC (D) none of these

= (6 j - 2 k ) \ BM = -

Solution (C)

We have, a - d = -AD, b - d = -BD and c - d = -CD. According to the given condition | AD | = | BD | = | CD |. Thus, D is the circumcentre of ∆ABC.

COMPONENT OF A VECTOR The component of a vector PQ on a line l is RS, where R and S are the feet of perpendiculars from P and Q on the line l. The vector component of PQ on l will be denoted by RS. 1. If θ is the angle between PQ and RS, then the component of PQ on l = PQ cos θ = | PQ | cos θ and the vector component of PQ on l = PQ cos θ. 2. If r is the position vector of a point P, having coordinates (x, y, z) then r = xi + yj + zk, where i, j, k are unit vectors along x, y and z axes respectively. 3. xi, yj, zk are the vector components of r on x, y and z axes respectively. 4. If a point P in space has coordinates (x, y, z), then its p.v. r is xi + yj + zk and | r | = | r |=

x +y +z . 2

2

2

SOLVED EXAMPLE 18. The triangle ABC is defined by the vertices A(1, -2, 2), B(1, 4, 0) and C(-4, 1, 1). Let M be the foot of the altitude drawn from the vertex B to side AC. Then BM = (A) (-20/7, -30/7, 10/7) (B)  (-20, -30, 10) (C) (2, 3, -1) (D)  none of these Solution (A)

Since MB is the component of AB ⊥ to AC

20 10 ( -5i + 3 j - k ) = ( 2i + 3 j - k ). 35 7

10 ( 2i + 3 j - k ). 7

LINEAR COMBINATION A vector r is said to be a linear combination of the given vectors a, b, c ..., and so on if there exist a system of scalars x, y, z, ..., and so on such that r = xa + yb + zc + ...

LINEARLY DEPENDENT AND INDEPENDENT SYSTEM OF VECTORS The system of n vectors a1, a2, ..., an is said to be linearly dependent, if there exist scalars x1, x2, ..., xn not all zero such that x1a1 + x2a2 + ... + xnan = 0 The same system of vectors is said to be linearly independent, if all scalers are zero, i.e. x1 = x2 = ... = xn = 0. REMARK When the system of n vectors a1, a2, ... , an is linearly dependent (or independent), then n vectors a1, a2, ..., an are said to be linearly dependent (or independent).

COLLINEARITY OF THREE POINTS The necessary and sufficient condition for three points with position vectors a, b and c to be collinear is that there exist three scalars x, y, z, not all zero, such that xa + yb + zc = 0, where x + y + z = 0 Test of Collinearity of Two Vectors  To prove that two vectors a and b are collinear, find a scalar m such that one of the vectors is m times the other. In case no such scalar m exists, then the two vectors will be non-collinear vectors.

Test of Collinearity of Three Points

MB = AB - AM = AB -

( AB × AC ) AC ( AC ) 2

= (6 j - 2 k ) -

{(6 j - 2k ) × ( -5i + 3 j - k )}( -5i + 3 j - k ) ( 25 + 9 + 1)

Method 1:  To prove that three points A, B, C are collinear, find the vectors AB and AC and show that there exists a scalar m such that AB = mAC. If no such scalar m exists, then the points are not collinear. Method 2:  To prove that three points A, B, C with position vectors a, b, c respectively are collinear, find three scalars x, y, z (not all zero) such that

Vector Algebra  13.9 xa + yb + zc = 0, where x + y + z = 0 If no such scalars x, y, z exist, then the points are not collinear.

SOLVED EXAMPLES 19. If the vectors a and b are non-collinear, then the value of x, for which, the vectors c = (x - 2) a + b and d = (2x + 1) a - b are collinear is 2 4 (A) (B)  3 3 1 (C) (D)  none of these 3 Solution (C)

The vector c is non-zero since the coefficient in b is different from zero, and so the vectors c and d are collinear if for some number y we have d = yc that is (2x + 1)a - b = y (x - 2)a + yb or ( yx - 2y - 2x - 1)a + (y + 1)b = 0 Since a, b are non-collinear, we must have yx - 2y - 2x - 1 = 0 and y + 1 = 0. Solving these equations, we get y = -1 and x = 1/3 20. With reference to a right handed system of mutually perpendicular unit vectors i, j, k α = 3i - j and β = 2i + j - 3k If β = b1 + b2, where b1 is parallel to α and b2 is perpendicular to α, then 3 1 i+ j 2 2 3 1 (B) b1 = i - j 2 2 1 3 (C) b 2 = i + j - 3k 2 2 1 3 (D) b 2 = i - j - 3k 2 2 (A) b1 =

Solution (B, C)

Since b1 is parallel to α, let b1 = lα, where λ is a scalar. β = b1 + b2  (Given) ∴ b2 = β - b1 = 2i + j - 3k - lα = 2i + j - 3k - λ(3i - j) = (2 - 3λ)i + (1 + λ)i - 3k Since b2 is perpendicular to α ∴ b2 ⋅ a = 0 ⇒ [(2 - 3λ)i + (1 + λ)i - 3i] ⋅ (3i - i) = 0 ⇒ 3(2 - 3λ) - (1 + λ) = 0

Þ 6 - 9l - 1 - l = 0 Þ 5 - 10l = 0, \ l =

1 2

1 3 1 (3i - i ) = i - j 2 2 2 3ö æ 1ö 1 3 æ b 2 = ç 2 - ÷ i + ç1 + ÷ j - 3k = i + j - 3k 2 2 2 2 è ø è ø ∴ 2i + j - 3k = lα + b2 Hence, β = b1 + b2 \ b1 = la =

COPLANARITY OF FOUR POINTS The necessary and sufficient condition for four points with position vectors a, b, c and d to be coplanar is that there exist scalars x, y, z and w, not all zero, such that xa + yb + zc + wd = 0 where x + y + z + w = 0. Test of Coplanarity of Three Vectors  To prove that three vectors a, b and c are coplanar, express one of these vectors as the linear combination of the other two such as c = xa + yb. Now, compare the coefficients from the two sides and find the values of x and y. If real values of scalars x and y exist, then the vectors are coplanar otherwise non-coplanar.

Test of Coplanarity of Four  Points Method 1:  To prove that four points A, B, C and D are coplanar, find the vectors AB, AC and AD and show that these three vectors are coplanar. Method 2: To prove that four points A, B, C and D with position vectors a, b, c and d respectively are coplanar, find four scalars x, y, z, w (not all zero) such that xa + yb + zc + wd = 0 where x + y + z + w = 0 If no such scalars x, y, z, w exist, then the points are non-coplanar.

SOME RESULTS ON LINEARLY DEPENDENT AND INDEPENDENT VECTORS 1. If a, b, c are non-coplanar vectors, then these are linearly independent and conversely if a, b, c are linearly independent, then they are non-coplanar. 2. If a = a1i + a2   j + a3k, b = b1i + b2   j + b3k and c = c1i + c2   j + c3k are three linearly dependent vectors, then a1 b1 c1 a2 b2 c2 = 0 a3 b3 c3 3. Let a, b, c be three non-coplanar vectors. Then, vectors x1a + y1b + z1c, x2a + y2b + z2c and x3a + y3b + z3c will be coplanar if

13.10  Chapter 13 x1 y1 z1

x2 y2 z2

x3 y3 = 0 z3

4. Two non-zero, non-collinear vectors are linearly independent. 5. Any two collinear vectors are linearly dependent. 6. Any three non-coplanar vectors are linearly independent. 7. Any three coplanar vectors are linearly dependent. 8. Any four vectors in 3-dimensional space are linearly dependent.

PRODUCT OF TWO VECTORS Between two vectors, two distinct kinds of products are defined. One being a pure number is called the scalar product while the other being a vector quantity is called the vector product.

SCALAR PRODUCT OF TWO VECTORS The scalar product or dot product of two vectors a and b is defined as: | a | | b | cos θ, where θ is the angle between them such that 0 ≤ θ ≤ π. It is denoted by placing a dot between the vectors a and b. Thus, a ⋅ b = | a | | b | cos θ If either a or b is O, we define a ⋅ b = O. B b

O

M

a

A

FIGURE 13.9

a ⋅ a = | a |2 = a2 5. Scalar product of two perpendicular vectors is zero, i.e., if a and b are two perpendicular vectors, then a ⋅ b = 0. However, if a ⋅ b = 0 ⇒ Either a = 0 or b = 0 or a ⊥ b. 6. Scalar product of mutually orthogonal unit vectors i, j, k: i ⋅ i = 1 = j ⋅ j = k ⋅ k and i ⋅ j = j ⋅ k = k ⋅ i = 0 7. Scalar porduct of two vectors in terms of components: If a = a1i + a2   j + a3k and b = b1i + b2   j + b3k, then a ⋅ b = a1b1 + a2b2 + a3b3 Thus, the scalar product of two vectors is equal to the sum of the products of their corresponding components. 8. Angle between two vectors in terms of the components of the given vectors. If θ is the angle between two vectors a = a1i + a2   j + a3k and b = b1i + b2   j + b3k, then a1b1 + a2 b2 + a3 b3 a×b cos q = = 2 | a|| b| a + a2 + a2 b2 + b2 + b2 1

2

3

1

2

3

I M P O R TA N T P O I N T S If q is acute, a.b is positive and if q is obtuse, a.b is negative

9. Components of a vector b along and perpendicular to vector a  a⋅b  Component of b along a =  2  a | a|   a⋅b  Component of b perpendicular to a = b −  a  | a | 2 10. Any vector r can be expressed as: r = (r ⋅ i)i + (r ⋅ j) j + (r ⋅ k)k

Key Results on Scalar Product

SOME USEFUL IDENTITIES

1. Scalar product is commutative. For any two vectors a and b we have a ⋅ b = b ⋅ a. 2. If m is any scalar and a, b be any two vectors, then (ma) ⋅ b = m(a ⋅ b) = a ⋅ (mb) 3. Scalar product is distributive with respect to vector addition, i.e., for any three vectors a, b and c, we have a ⋅ (b + c) = a ⋅ b + a ⋅ c. 4. Magnitude of a vector as a scalar product: For any vector a 

Since scalar product satisfies commutative and distributive laws, we have 1. (a + b)2 = a2 + b2 + 2a ⋅ b 2. (a - b)2 = a2 + b2 - 2a ⋅ b 3. (a + b) ⋅ (a - b) = a2 - b2

WORK DONE BY A FORCE Work done by a force F in displacing a particle from A to B is defined by W = F ⋅ AB

Vector Algebra  13.11

I M P O R TA N T P O I N T S (a + b + c)2 = |a|2 + |b|2 + |c|2 + 2|(a . b + b . c + c . a)  | iˆ + ˆ j + kˆ |= 3 

Cauchy - Schwarz Inequality (a . b) ≤ |a|2|b|2  If a number of forces are acting on a particle, then 

the sum of the works done by the separate forces is equal to the work done by the resultant force

  21. If a and b are unit vectors, then the greatest value of     | a + b | + | a - b | is (A) 2 2 (B)  2 (C) 2 (D)  4 2 Solution (A)

  Let θ be the angle between a and b .   Then, a . b = cos q       Now, | a + b | = | a |2 + | b |2 + 2a . b = 2 + 2 cos q q = 4 cos 2 2  q  Þ | a + b | = 2 cos . 2  q  Similarly, | a - b | = 2 sin 2   q qö   æ \ | a + b | + | a - b | = 2 ç cos + sin ÷ £ 2 2 2 2ø è 22. If a + b + c = 0 and | a | = 3, | b | = 5, | c | = 7, then the angle between a and b is p p (A) (B)  3 2 p p (C) (D)  6 4 ×

∴ From (1), we get

×

×

Solution (B)

Let θ be the angle between a and b ∴ a ⋅ b = |a| |b| cosθ a×b a×b a×b ∴ cos q = = = (1) | a || b | (3)(5) 15 Now a + b + c = 0 ∴ a + b = -c ⇒ |a + b| = |-c| = |c| ⇒ |a + b|2 = |c|2 ⇒ (a + b) ⋅ (a + b) = (7)2 ⇒ |a|2 + 2a ⋅ b + |b|2 = 49 ⇒ (3)2 + 2a ⋅ b + (5)2 = 49

15 2

15 1 = = cos 60° Þ q = 60° 2 ´ 15 2   23. a and c are unit vectors and | b | = 4. The angle      æ1ö between a and c is cos -1 ç ÷ . If b - 2c = l a , then è4ø λ is equal to cos q =

(A) 3, 4 (C) 3, -4

SOLVED EXAMPLES

×

⇒ 2a ´ b = 49 - 9 - 25 = 15 Þ a ´ b =

(B)  -3, 4 1 3 (D) , 4 4

Solution (C)

   Given: | a | = 1, | c | = 1 and | b | = 4. 1 1   Þ | a × c | = 1 ×1 × = 4 4        Now, b - 2c = l a Þ a × b - 2a × c = l a 2 1   Þ a ×b - 2× = l 4  1  Þ a ×b = l + 2          Again, b - 2c = l a Þ b × b - 2b × c = l b × a   1ö æ Þ 16 - 2b × c = l ç l + ÷ è 2ø 2   l l Þ b ×c = 82 4          Also, b - 2c = l a Þ b × c - 2c × c = l b × a

l2 l æ1ö - - 2(1) = l ç ÷ 2 4 è4ø 2 ⇒  l + λ - 12 = 0 Þ 8-

∴  λ = - 4, 3 24. The vectors a = xi - 3j - k and b = 2xi + xj - k include an acute angle and b and positive y-axis include an obtuse angle. Then values of x may be (A) -2 (B)  -3 (C) all x < 0 (D)  all x > 0 Solution (A, B, C)

According to the question a ⋅ b > 0 and b ⋅ j < 0 ⇒ 2x2 - 3x + 1 > 0 and x < 0 ⇒ (2x - 1) (x - 1) > 0 and x < 0 1   i.e.,  x < or x > 1 and x < 0 ⇒ x < 0   2 25. If the unit vectors a and b are inclined at angle 2θ (0 ≤ θ ≤ π) and |a - b| < 1, then θ lies in the interval

13.12  Chapter 13 é pö ép p ù (A) ê0, ÷ (B)  ê6 , 2ú ë 6ø ë û é p 5p ù (C) ê , ú (D)  none of these ë2 6 û

⇒

Similarly, | a - b |= 4 3 Hence the lengths of the diagonals are 4 3 and 4 3. 28. Let a = 2i - j + k, b = i + 2j - k and c = i + j - 2k be three vectors. A vector in the plane of b and c whose 2 projection on a is of magnitude is 3

Solution (A)

a ⋅ b = | a | | b | cos2θ ⇒ a ⋅ b = (1) (1) cos 2θ = cos 2θ. | a - b | < 1 ⇒ a2 + b2 - 2a ⋅ b < 1 ⇒ 1 + 1 - 2cosθ < 1 ⇒ 2(1 - cos 2θ) < 1 ⇒ 2(2sin2θ) < 1 Þ sin 2 q
0 and x < 0 ⇒ (x < 1/2, or x > 1) and x < 0 Hence, x < 0 is the required solution. 39. A unit vector in xy plane that makes an angle of 45º with the vector i + j and an angle of 60º with the vector 3i - 4j is (A) i (B)  (i + j ) / 2 (C) (i - j ) / 2 (D)  none of these Solution (D)

Let the required unit vector in the x-y plane be r = xi + yj , \ | r | = ( x 2 + y 2 ) = 1 (1) or x2 + y2 = 1 Since angle between r and vector i + j is 45º and the angle between r and vector 3i - 4j is 60º. ( xi + yj ) × (i + j ) cos 45° = \ xi + yj i + j

4iˆ - ˆj + 4 kˆ (B)  (D)  none of these

   Let vector r be coplanar to a and b    \ r = a + tb ⇒ r = (iˆ + 2 ˆj + kˆ ) + t (iˆ − ˆj + kˆ ) = iˆ(1 + t ) + ˆj ( 2 − t ) + kˆ(1 + t )

VECTOR PRODUCT OF TWO VECTORS

1   Then projection of r on c = 3   r ⋅c 1 ⇒  = |c | 3 |1⋅ (1 + t ) + 1.( 2 − t ) − 1.(1 + t )| 3

The vector product or cross product of two vectors a and b is defined as a×b

=

1 3

⇒ | 2 - t | = ±1  ⇒  t = 1 or 3  When t = 1, we have r = 2iˆ + ˆj + 2kˆ  When t = 3, we have r = 4iˆ − ˆj + 4 kˆ 38. The values of x for which the angle between the vectors a = xi - 3j - k and b = 2xi + xj - k is acute and the angle between the vector b and the y-axis lies between p and π are 2 1 1 (A) 1, (B)  0< x< 2 2 (C) all x < 0 (D)  x < 0 or x > 1 ×

Solution (C)

a ⋅ b > 0 and b ⋅ j < 0. ⇒ 2x2 - 3x + 1 > 0 and x < 0

( xi + yj ) × (3i - 4 j ) xi + yj 3i - 4 j

⇒ x + y = 1 (2) 3x - 4y = 5/2 (3) No real values of x and y exist satisfying equations (1), (2), and (3).

Solution (A, B)

⇒

cos 60° =

and

B

n

b

O

a

A

FIGURE 13.10

a × b = | a | | b | sin θ n 1. | a | | b | sin  θ is the modulus of a × b, θ being the angle between the directions of a and b and 0 ≤ θ ≤ π; 2. direction of a  ×  b is that of the unit vector n which is perpendicular to both a and b such that a, b and n form a right handed system.

REMEMBER By right handed system we mean that as the first vector a is turned towards the second vector b through an angle θ, n will point in the direction in which a right handed screw would advance if turned in a similar manner.



Vector Algebra  13.15 If either a or b is O, we have a ⋅ b = O.  |  a × b | = | a | | b | sin θ  A unit vector perpendicular to the plane of two given veca´b . tors a and b is given as n = |a ´ b |  a × b is perpendicular to the plane of a and b. 

Key Results on Vector Product 1. Vector product is not commutative. For any two vectors a and b a×b≠b×a In fact, a × b = - b × a 2. Vector product is associative with respect to a scalar. If m and n be any scalars and a, b any vectors, then m(a × b) = (ma) × b = a × (mb) = (a × b)m; (ma) × (nb) = (na) × (mb) = (mna) × b = a × (mnb) = mn(a × b) 3. Vector product is distributive with respect to addition. For any three vectors a, b and c a × (b + c) = a × b + a × c 4. If two vectors a and b are parallel, then a × b = 0. In particular, a × a = 0. 5. Vector product of mutually orthogonal unit vectors i, j, k:

2. The area of a triangle with adjacent sides a and b is 1 given by | a × b | . 2 3. The area of a triangle ABC is 1 1 1 | AB × AC | or | BC × BA | or | CB × CA | 2 2 2 4. The area of a parallelogram with diagonals a and b is 1 given by | a × b | . 2 5. The area of a quadrilateral ABCD is given by 1 | AC × BD | , where AC and BD are its diagonals. 2 6. Vector area of a ∆ABC, when a, b, c are the position vectors of A, B, C respectively is given by

1 ( a × b + b × c + c × a) 2

∆ ABC =

MOMENT OF A FORCE ABOUT A POINT The vector moment or torque M of a force F acting at a point A about the point O is given by M = r × F = OA × F where r = OA is the position vector of the point A with respect to the point O. F

i×i=j×j=k×k=O O

and  i × j = k = - j × i,

j × k = i = - k × j, k × i = j = - i × k.

r

6. Vector product in terms of components. Let

A

a = a1i + a2   j + a3k and  b = b1i + b2   j + b3k, then

a × b = (a2b3 - a3b2)i + (a3b1 - a1b3) j + (a1b2 - a2b1) k i = a1 b1

j a2 b2

k a3 b3

7. Angle between two vectors: If θ is the angle between | a´b | two vectors a and b, then sin q = . | a || b |

Geometrical Interpretation of Cross Product 1. The area of a parallelogram with adjacent sides a and b is given by | a × b | .

FIGURE 13.11

Info Box! The algebraic sum of the moments of a system of forces about any point is equal to the moment of their resultant about the same point.

QUICK TIPS Lagrange’s Identity If a and b are any two vectors, then

|a × b|2 + (a . b)2 = |a|2 |b|2

SOLVED EXAMPLES

    40. If a , b , c be three non-coplanar vectors and r be any arbitrary vector, then

13.16  Chapter 13             ( a × b ) × ( r × c ) + (b × c ) × ( r × a ) + (c × a ) × ( r × b ) is equal to   (A) 0 (B)  [a b c ]r     (C) 2[a b c ]r (D)  3[a b c ]r

\ cos q =

Solution (C)

    We have, ( a × b ) × ( r × c )         = (( a × b ) ⋅ c ) r − (( a × b ) ⋅ r ) c     = [a b c ]r − [a b r ]c         Similarly, (b × c ) × ( r × a ) = [b c a ]r − [b c r ]c          and, (c × a ) × ( r × b ) = [c a b ]r − [c a r ]b             \  ( a ´ b ) ´ ( r ´ c ) + (b ´ c ) ´ ( r ´ a ) + (c ´ a ) ´ ( r ´ b )         = 3[a b c ]r − ([b c r ]a + [c a r ]b + [a b r ]c )     = 3[a b c ]r − [a b c ]r   = 2[a b c ]r 41. If A = 2i + k, B = i + j + k and C = 4i - 3j + 7k, then a vector R which satisfies R × B = C × B and R ⋅ A = O, is (A) - i - 8j + 2k (B)  i - 8j + 2k (C) i + 8j + 2k (D)  none of these Solution (A)

Let R = xi + yj + zk ∴ R ⋅ A = 0  ⇒ 2x + z = 0 i j k i j k R´ B = C ´ B Þ x y z = 4 -3 7 1 1 1 1 1 1

(1)

⇒ (y - z)i + (z - x)j + (x - y)k = -10i + 3j + 7k ⇒ y - z = -10  (2) z - x = 3 (3) and x-y=7 Solving (1) and (2), we get x = -1, z = 2 ∴  From (2), y = -8. Hence R = - i - 8j + 2k 42. Let A(0, 0, 0), B(1, 1, 1), C(3, 2, 1) and D(3, 1, 2) be four points. The angle between the planes through the points A, B, C and through the points A, B, D is p p (A) (B)  6 2 p p (C) (D)  3 4 ×

×

n1 = AB × AC = - i + 2j - k n2 = AB × AD = i + j - 2k Let θ be the acute angle between the planes, then θ is the acute angle between their normals n1 and n2

×

×

Solution (D)

Let n1 and n2 be the vectors normal to the planes ABC and ABD respectively.

Þ =

-1+ 2 + 2 6× 6

=

3 1 p = = cos 2 2 3

p 3

43. Given A = ai + bj + ck, B = di + 3j + 4k and C = 3i + j - 2k. If the vectors A , B and C form a triangle such that A = B + C, then (A) a = -8, b = -4, c = 2, d = -11 (B) a = -8, b = 4, c = -2, d = -11 (C) a = -8, b = 4, c = 2, d = -11 (D) none of these Solution (C)

Here A, B, C are the vectors which represent the sides of the triangle ABC where A = ai + bj + ck B = di + 3j + 4k C = 3i + j - 2k Given that, A = B + C ∴ ai + bj + ck = (d + 3)i + 4j + 2k ⇒ a = d + 3, b = 4, c = 2. i j k B ´C = d 3 4 3 1 -2 = -10i + ( 2d + 12) j + ( d - 9)k ∴ Area of the DABC = ⇒

=

1 B ´C 2

1 [100 + ( 2d + 12) 2 + ( d − 9) 2 ] 2

= 5 6 (Given ) (5d 2 + 30 d + 325) = 10 6

⇒ 5d2 + 30d + 325 = 600 ⇒ 5d2 + 30d - 275 = 0 ⇒ d2 + 6d - 55 = 0 ⇒ (d + 11) (d - 5) = 0 ⇒ d = 5 or -11 When d = 5, a = 8, b = 4, c = 2 and when d = -11, a = -8, b = 4, c = 2. 44. If x + y = a, x × y = b and x ⋅ a = 1, then a + a´b ( a 2 - 1)a - a ´ b y = (A) x = (B)  a2 a2 2 b + a´b (b - 1)b - a ´ b y= (C) x = (D)  2 a a2

Vector Algebra  13.17 Solution (A, B)

Given x+y=a ⇒ y = a - x(1) x × y = b(2) x ⋅ a = 1 (3) From (1) and (2), we get x × (a - x) = b ⇒ x × a - x × x = b ⇒ x × a = b ⇒ a × (x × a) = a × b ⇒ (a ⋅ a) x - (a ⋅ x) a = a × b ⇒ | a |2 x - 1 ⋅ a = a × b [From (3)] Þ

x=

( a + a ´ b) a2

= 2 [a 2 b 2 sin 2 q ] = 2 ( a 2 b 2 - a 2 b 2 cos 2 q )



= 2 [16 - ( a × b) 2 ] (∵ | a | = | b | = 2)

47. If a ⋅ i = 4, then (a ⋅ j) × (2j - 3k) = (A) 12 (B)  2 (C) 0 (D)  -12 Solution (D)

We have, (a · j) × (2j - 3k) = a × ( j ´ ( 2 j - 3k )) = a × ( -3( j ´ k )

( a 2 - 1)a - a ´ b and y = a - x = a2 45. If a × (b × c) + (a ⋅ b) b = (4 - 2β - sin α) b + (b 2 - 1) c and (c ⋅ c) a = c, while b and c are non-collinear, then

p (A) a = , b = -1 2 p (C) a = , b = -1 3



p (B) a = , b = 1 2 p (D) a = , b = -1 3

Solution (B)

We have, a × (b × c) + (a · b) b = (4 - 2β - sinα) b + (b2 - 1)c(1) and (c ⋅ c) a = c(2) where b and c are non-collinear vectors and α, β are scalars From (2), (c ⋅ c) a ⋅ c = c ⋅ c ∴ a ⋅ c = 1 (3) From (1), we get (a ⋅ c)b - (a ⋅ b)c + (a ⋅ b)b = (4 - 2β - sinα) b + (b 2 - 1)c or [1 + (a ⋅ b)] b - (a ⋅ b) c = (4 - 2β - sinα) b + (b 2 - 1) c ⇒ 1 + (a · b) = 4 - 2β - sin α(4) and a ⋅ b = -(b2 - 1) (5) ∴ sin α = 1 + (1 - β)2 ⇒ β = 1, sin α = 1

p + 2np , n Î I . 2 46. If u = a - b, v = a + b and | a | = | b | = 2, then | u × v | = i.e., a =

(A) 2 [16 - ( a × b) 2 ]

(B)  [16 - ( a × b) 2 ]

(C) 2 [4 - ( a × b) 2 ]

(D)  [4 - ( a × b) 2 ]

Solution (A)

u × v = (a - b) × (a + b) = 2a × b ∴ | u × v| = 2| a × b |

= -3( a × i ) (∵

j ´ k = i)

= -3(4) = -12.

48. If a ⋅ b = β and a × b = c, then b is equal to (A) (ba - a × c)/a2 (B) (ba + a × c)/a2 2 (C) (bc - a × c)/a (D) (bc + a × c)/a2 Solution (A)

Here a and c = a × b are non-collinear vectors. ∴ Let b = xa + y(a × c)(1) ∴ β = a ⋅ b = a ⋅ [xa + y(a × c)] = x | a |2 + ya ⋅ (a × c) = xa2 ⇒ x = β/a2 And c = a × b = a × [xa + y (a × c)] = xa × a + ya × (a × c) = 0 + y (a ⋅ c) a - y(a ⋅ a)c = y [a ⋅ (a × b)]a - ya2c = -ya2c ⇒ y = -1/a2 ∴ from (1), b = (ba - a × c)/a2 49. Let OA = a, OB = 10a + 2b and OC = b where O, A, C are non-collinear. Let p denote the area of the quadrilateral OABC and q denote the area of the parallelop gram with OA and OC as adjacent sides. Then is q equal to (A) 4 (B)  6 1 a-b (C) (D)  none of these 2 | a| Solution (B)

Given, | a × b | = q 1 1 | b ´ (10 a + 2b)| + | a ´ (10 a + 2b)| = p 2 2 ∴ 5 | b × a | + | a × b | = p and

Þ 6 | a ´ b | = p Þ 6q = p Þ

p = 6. q

50. Two planes are perpendicular to one another. One of them contains a and b and the other contains vector c and d, then (a × b) ⋅ (c × d) is equal to (A) 1 (B)  0 (C) [a b c] d (D)  [b c d] a

13.18  Chapter 13 Solution (B)

SCALAR TRIPLE PRODUCT

a × b is a vector perpendicular to the plane of a and b and c × d is a vector perpendicular to the plane of c and d. Since, these planes are ⊥ to one another ∴ (a × b) ⋅ (c × d) = 0

Scalar triple product of three vectros  If a, b, c are three vectors, then their scalar triple product is defined as the dot product of two vectors a and b × c. It is generally denoted by a ⋅ (b × c) or [a b c].

51. If the vectors c, a = xi + yj + zk and b = j are such that a, c and b form a right handed system then c is (A) zi - xk (B)  0 (C) yj (D)  - zi - xk Solution (A)

Since a, c, b form a right-handed system, so c = λ(b × a), λ > 0 i.e., c = l i ´ ( xi + yj + zk ) = l ( zi - xk ) Taking λ = 1, c = zi - xk 52. ABCD is a quadrilateral with AB = a, AD = b and AC = 2a + 3b. If its area is α times the area of the parallelogram with AB, AD as adjacent sides, then α is equal to 5 (A) 5 (B)  2 1 (C) 1 (D)  2 Solution (B)

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD 1 1 = [a ´ ( 2a + 3b)] + [( 2a + 3b) ´ b] 2 2 1 5 = [3a ´ b + 2a ´ b] = a ´ b 2 2 5 5 = Area of ||gm ABCD. \ a = 2 2 53. Consider a tetrahedron with faces F1, F2, F3, F4. Let V1, V2, V3, V4 be the vectors whose magnitudes are respectively equal to areas of F1, F2, F3, F4 and whose directions are perpendicular to their faces in outward direction. Then | V1 + V2 + V3 + V4 | equals (A) 1 (B)  4 (C) 0 (D)  none of these Solution (C)

We have, 1 1 ( a ´ b), v2 = (b ´ c) 2 2 1 1 v3 = (c ´ a) and v4 = [(c - a) ´ (b - a)] 3 2 ∴ v1 + v2 + v3 + v4 = 0 ∴ | v1 + v2 + v3 + v4 | = 0 v1 =

Properties of Scalar Triple Product (i) If a, b, c are cyclicallly permuted, the value of scalar triple product remains the same, i.e., (a × b). c = (b × c) ⋅ a = (c × a). b or [a b c] = [b c a] =[c a b] (ii) The change of cyclic order of vectors in scalar triple product changes the sign of the scalar triple product but not the magnitude i.e., [a b c] = -[b a c] = -[c b a] = -[a c b] (iii) In scalar triple product the positions of dot and cross can be interchanged provided that the cyclic order of the vectors remains same i.e., (a × b) . c = a . (b × c) (iv) The scalar triple product of three vectors is zero if any two of them are equal. (v) For any three vectors a, b, c and scalar λ, [l a b c] = λ [a b c] (vi) The scalar triple product of three vectors is zero if any two of them are parallel or collinear. (vii) If a, b, c, d are four vectors, then [(a + b) c d ] = [a c d] + [b c d] (viii) The necessary and sufficient condition for three nonzero non-collinear vectors a, b, c to be coplanar is that [a b c] = 0. (ix) Four points with position vectors a, b, c and d will be coplanar, if [a b c] + [d c a] + [d a b] = [a b c]. (x) Volume of parallelopiped whose coterminous edges are a, b, c is [a b c] or a ⋅ (b × c).

Scalar Triple Product in Terms of Components (i) If a = a1 i + a2 j + a3k, b = b1i + b2   j + b3k and c = c1i + c2   j + c2   j + c3k be three vectros a1 then, [a b c] = a2 a3

b1 b2 b3

c1 c2 c3

(ii) If a = a1l + a2m + a3n, b = b1l + b2m + b3n and c = c1l + c2m + c3n, then a1 [a b c] = b1 c1

a2 b2 c2

a3 b3 [l m n] c3

Vector Algebra  13.19 (iii) For any three vectors a, b and c (A) [a + b b + c c + a] = 2[ a b c] (B) [a - b b - c c - a] = 0 (C) [a × b b × c c × a] = [ a b c]2 Tetrahedron  A tetrahendron is a three-dimensional figure formed by four triangles. OABC is a tetrahedron with DABC as the base. OA, OB, OC, AB, BC and CA are known as edges of the tetrahedron. OA, BC; OB, CA and OC, AB are known as the pairs of oppposite edges. A tetrahedron in which all edges are equal, is called a regular tetrahedron. Any two edges of regular tetrahedron are perpendicular to each other. A (a)

c

B (b)

C (c) FIGURE 13.12

Volume of Tetrahedron (i) The volume of a tetrahendron 1 = (area of the base)(corresponding altitude) 3 1 = [ AB BC AD ] 6 (ii) If a, b, c are position vectors of vertices A, B and C with respect to O, then volume of tetrahedron 1 OABC = [a b c] 6 (iii) If a, b, c, d are position vectors of vertices A, B, C, D of a tetrahedron ABCD, then its volume 1 = [b − a c − a d − a ] 6 Reciprocal system of vectors  Let a, b, c be three non-coa×b b×c c×a planar vectors, and let a′ = , b′ = , c′ = . [a b c] [a b c] [a b c] a′, b′, c′ are said to form a reciprocal system of vectors for the vectors a, b, c. If a, b, c and a′, b′, c′ form a reciprocal system of vectors then (i)  a, a′ = b b′ = c.c′ = 1 (ii)  a. b′ = a. c′ = 0; b.c′ = b.a′ = 0; c. a′ = c.b′ = 0 1 [a b c] (iv)  a, b, c are non-coplanar iff so are a′, b′, c′. (iii)  [a′ b′ c′ ] =

54. If i, j, k are the unit vectors and mutually perpendicular, then [i - j j - k k - i] = (A) 0 (B)  1 (C) -1 (D)  none of these Solution (A)

(i - j) · [(j - k) × (k - i)] = (i - j) ⋅ (j × k - j × i - k × k + k × i) = (i - j) ⋅ (i + k - 0 + j) =i·i+i⋅k+i⋅j-j⋅i-j⋅k-j⋅j = i2 + 0 + 0 - 0 - 0 - j2 = i2 - j2 = 1 - 1 = 0. 55. If e1 ′, e2 ′, e3 ′ are vectors reciprocal to the non-coplanar vectors e1 , e2 , e3  then [e1 ′, e2 ′, e3 ′ ] [e1  e2 e3 ] = −1 (A) (B)  1 2 (C) 0 (D)  4

a

b

SOLVED EXAMPLES

Solution (B)

1 [e1e2 e3 ] ∴ [e′1 e′2 e′3] [e1 e2 e3] = 1 Since [e1¢e2¢ e3¢ ] =

1 56. If r = λ (a × b) + µ (b × c) + v (c × a) and [a b c] = , then 8 λ + µ + v is equal to (A) r ⋅ (a + b + c) (B) 8r ⋅ (a + b + c) (C) 4r ⋅ (a + b + c) (D)  none of these Solution (B)

1 Clearly r × c = l[a b c] = l 8 1 r × a = m [a b c] = m 8 1 r ´ b = v [a b c] = v 8 1 \ r × ( a + b + c) = (l + m + n ) 8 ∴ λ + µ + v = 8r ⋅ (a + b + c)

Vector triple product Let a, b, c be any three vectors, then the vectors a × (b × c) and (a × b) × c are called vector triple product of a, b, c. Thus, a × (b × c) = (a.c) b - (a.b) c

Properties of Vector Triple Product (i) The vector triple product a × (b × c) is a linear combination of those two vectors which are within brackets. (ii) The vector r = a × (b × c) is perpendicular to a and lies in the plane of b and c.

13.20  Chapter 13 (iii) The formula a × (b × c) = (a .c) b - (a.b) c is true only when the vector outside the bracket is on the left most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula. Thus, (b × c) × a = - {a × (b × c)} = - {(a ⋅ c) b - (a ⋅ b) c} = (a . b) c - (a ⋅ c) b (iv)  Vector triple product is a vector quantity. (v)  a × (b × c) ≠ (a × b) × c

Rotation of a Vector About an Axis Let a = (a1, a2, a3). If system is rotated about (i)  x-axis through an angle α, then the new components of a are (a1, a2 cos α + a3 sin α, - a2 sinα + a3 cosα). (ii)  y-axis through an angle α, then the new components of a are (-a3 sinα + a1 cosα, a2, a3 cosα + a1 sinα). (iii)  z-axis through an angle α, then the new components of a are (a1 cosα + a2 sin α , -a1 sin α + a2 cosα, a3).

SOLVED EXAMPLES 57. Given three unit vectors a, b, c no two of which are col1 linear satisfying a ´ (b ´ c) = b. The angle between a 2 and b is

p (A) 3 p (C) 2 ×

p (B)  4 ×

(D)  none of these

Solution (C)

1 b 2 ⇒ (a ⋅ c) b - (a ⋅ b) c = 1/2b ⇒ (a ⋅ c - 1/2) b = (a ⋅ b) c But since b and c are non-parallel, so the only possibility is a ⋅ c = 1/2 and a ⋅ b = 0 Hence the angle between a and b is π/2. We have, a × (b × c) =

58. If a and b are two unit vectors, then the vector (a + b ) × (a × b) is parallel to the vector (A) a - b (B)  a+b (C) 2a - b (D)  2a + b Solution (A)

We have, (a + b) × (a × b) = a × (a × b) + b × (a × b) = (a ⋅ b) a - (a ⋅ a) b + (b ⋅ b) a - (b ⋅ a) b = (a ⋅ b) (a - b) + a - b (b ⋅ b = b2 = 1, a ⋅ a = a2 = 1 as a, b are unit vectors) = (a ⋅ b + 1) (a - b) = x (a - b) where x = a ⋅ b + 1 is a scalar. ∴ The given vector is parallel to a - b.

Vector Algebra  13.21

NCERT EXEMPLARS

(

)

(

)

2. The position  vector  of the point which divides the join of points 2a - 3b and a + b in the ratio 3 : 1, is     (A) 3a - 2b (B)  7a - 8b  2 4 (C) 3a (D)  5a 4 4 3. The vector having initial and terminal points as (2, 5, 0) and (– 3, 7, 4), respectively is (A) -i + 12 j + 4 k (B)  5i + 2 j - 4 k (C) -5i + 2 j + 4 k (D)  i + j + k   4. The angle between two vectors a and b with magni tudes 3 and 4, respectively and a.b = 2 3 is (A) p (B)  p 6 3 p (C)  (D)  5p 2 2

 5. Find the value of l such that the vectors a = 2i + l j + k  and b = i + 2 j + 3k are orthogonal.

(A) 0 (C) 3 2

(B) 1 (D)  -5 2

6. The value of λ for which the vectors 3i - 6 j + k and 2i - 4 j + l k are parallel, is (A) 2 (B)  3 3 2 5 (C)  (D)  2 5 2 7. The vectors from origin to the points A and B are   a = 2i - 3 j + 2k and b = 2i + 3 j + k respectively, then the area of ΔOAB is equal to (A)  340 (B)  25 1 (C)  229 (D)  229 2

 8. For any vector a, the value of    2 2 2 a ´ i + a ´ j + a ´ k is

( ) (

) (

)

2 2 (A) a (B)  3a 2 2 (C) 4 a (D)  2a      9. If a = 10, b = 2 and a·b = 12, then the value of a ´ b is (A) 5 (C) 14

(B) 10 (D) 16

 10. The vectors l i + j + 2k , i + l j - k and c = 2i - j + l k        11. If a, b and c and unit vectors such that a + b + c = 0,    then the value of a·b + b·c + c·a is (A) 1 (B) 3 3 (C) - (D)  None of these 2 



12. The projection vector of a on b is  æ   ö a b a · (A) ç  ÷ b (B)  ·b ç b ÷ b è ø  æö (C) a·b (D)  ç a·b2 ÷ b ç ÷ a ça ÷ è ø        13. If a, b and c are three vectors such that a + b + c = 0    and a = 2, b = 3 and c = 5, then the value of    a·b + b·c + c·a is (A) 0 (B) 1 (C) – 19 (D) 38   14. If a = 4 and -3 £ l £ 2, then the range of l a is (A)  [0, 8] (C)  [0, 12]

(B)  [–12, 8] (D)  [8, 12]

15. The number of vectors of unit length perpendicular to the





vectors a = 2i + j + 2k and b = j + k is

(A) one (C) three

(B) two (D) infinite

NCERT EXEMPLARS

1. The vector in the direction of the vector i - 2 j + 2k that has magnitude 9 is    (A) i - 2 j + 2k (B)  i - 2 j + 2k 3     (C) 3 i - 2 j + 2k (D)  9 i - 2 j + 2k

13.22  Chapter 13

ANSWER K EYS 1. (C) 11.  (C)

2.  (D) 12.  (A)

3. (C) 13.  (C)

4.  (B) 14.  (C)

6.  (A)

5. (D) 15.  (B)

7.  (D)

8.  (D)

9.  (D)

10.  (A)

HINTS AND EXPLANATIONS     1. Let a = i - 2 j + 2k  Any vector in the direction of a vector a is given by

=

i + 2 j + 2k 12 + 22 + 22

=

 a  a .

\

i - 2 j + 2k 3

 \ Vector in the direction of a with magnitude i - 2 j + 2k 3   = 3 i - 2 j + 2k

)

 2. Let the position  vector   of the point R divides the join of points 2a - 3b and a + b.     3 a + b + 1 2a - 3b \ Position vector R = 3 +1 Since, the position vector of a point R dividing the line seg-

HINTS AND EXPLANATIONS

(

) (

)

ment joining the points P and Q, whose position vectors are     p and q in the ratio m : n internally, is given by mq + n p . m+n  5a \ R= 4 3. Required vector = - ( -3 - 2 ) i + ( 7 - 5 ) i + ( 4 - 0 ) k = -5i + 2 j + 4 k Similarly, we can say that for having initial and terminal points as

(i) (4, 1, 1) and (3, 13, 5), respectively. (ii) (1, 1, 9) and (6, 3, 5), respectively. (iii) (1, 2, 3) and (2, 3, 4), respectively, we shall get (a), (b) and (d) as its correct options.    a = 3 , b = 4 and a.b = 2 3[given]    We know that, a·a = a b cosq

4. Here,

2 3 = 3.4.cosq 2 3 1 cosq = = 4 3 2 Þ p \ q= 3   5. (D) Since, two non-zero vectors a and b are orthogonal  i.e., a·b = 0. Þ

⇒ 2 + 2l + 3 = 0 -5 \ l = 6. Since, two2 vectors are parallel i.e., angle between them is zero. ∴ 3i - 6 j + k · 2i - 4 j + l k = 3i - 6 j + k · 2i - 4 j + l k     é∵ a·b = a b cos 0° Þ a·b = a b ù  ë û

(

 9 = 9.

(

( 2i + l j + k )·(i + 2 j + 3k ) = 0

)(

)

2 ⇒ 6 + 24 + l = 9 + 36 + 1 4 + 16 + l 2 ⇒ 30 + l = 46 20 + l 900 + l 2 + 60l = 46 20 + l 2 ⇒ [on squaring both sides] 2 2 ⇒ l + 60l - 46l = 920 - 900 2 ⇒ -45l + 60l - 20 = 0 2 ⇒ -45l + 30l + 30l - 20 = 0

(

)

⇒ -15l ( 3l - 2 ) + 10 ( 3l - 2 ) = 0 ⇒

(10 - 15l ) ( 3l - 2 ) = 0

2 2 \ l = , 3 3 Alternate Method   Let a = 3i - 6 j + k and b = 2i - 4 j + l k   Since, a b 3 -6 1 2 = = Þl = 3 2 -4 l    7. ∴ Area of DAOB = 1 O A ´ OB 2 1   = 2i - 3 j + 2k ´ 2i + 3 j + k 2 i j k 1 = 2 -3 2 2 2 3 1

⇒

(



1 2 1 = 2 =

)(

)

éi ( -3 - 6 ) - j ( 2 - 4 ) + k ( 6 + 6 ) ù ë û -9i + 2 j + 12k

∴ Area of DOAB = 1 2

(81 + 4 + 144 ) =

1 229 2

Vector Algebra  13.23  8. Let a = xi + y j + zk 2 a = x2 + y2 + z2 ∴

(



= i [0 ] - j [ - z ] + k [ - y ]



= z j - yk  2 a ´ i = z j - yk z j - yk

( ) (



(

Þ

)

Þ

= y +z   2 a ´ j = x2 + z2   2 a ´ k = x2 + z2 2

( (

Similarly, and 

)(

2 ± 12 Þ l = -2 or l = 2 ± 2 2 3 Þ l = -2 or l = = 1± 3 2    2 2 2 11. We have, a + b + c = 0 and a = 1.b = 1, c = 1       a+b+c a+b+c =0 ∵

2

) )





( a ´ i ) + ( a ´ j ) + ( a ´ k ) 2

2

Þ Þ

2

= y2 + z2 + x2 + z2 + x2 + y2 2  = 2 x 2 + y 2 + z 2 = 2a ∴

(

9. Here,

   a = 10, b = 2 and a•b = 12 

∴

   a·b = a b cosq 12 = 10 × 2 cos θ 12 3 cosq = = 20 5



⇒

⇒

sin q = 1 - cos 2 q = 1 -



sin q = ±

9 25

4 5     a ´ b = a b sin q

∴

= 10 ´ 2 ´

4 5

= 16     10. a = l i + j + 2k , b = i + l j - k and c = 2i - j + l k    For a, b and c to be coplanar,

l 1 2 1 l -1 = 0 2 -1 l

(

)

Þ l l - 1 - 1( l + 2 ) + 2 ( -1 - 2l ) = 0 Þ l 3 - l - l - 2 - 2 - 4l = 0 Þ l 3 - 6l - 4 = 0 2

)

[given]

⇒

)(

)

 2        2       2 a + a·b + a·c + b·a + b + b·c + c·a + c·b + c = 0

 2  2 2    a + b + c + 2 a·b + b·c + c·a = 0       é∵ a·b = b·c, b·c = c·b and c·a = a·c ù ë û    1 + 1 + 1 + 2 a·b + b·c + c·a = 0

(

)

(

)

   3 a·b + b·c + c·a = 2

   b  æ b   12. Projection vector of a on b is given by = a·  b = ç a·  ç b b è  14. We have, a = 4 and - 3 £ l £ 2   \ la = l a = l 4 Þ

ö ÷·b ÷ ø

 λ a = −3 4 = 12, at λ = 3  l a = 0 4 = 0, at l = 0

 l a = 2 4 = 8, at l = 2  So, the range of l a is = [0, 12]. Alternate Method Since, –3 £ l £ 2 0£ l £3 and

⇒

0 £ 4 l £ 12  l a Î [0, 12]

15. The number of vectors of unit length perpendicicular to the       vectors a and b is c (say) i.e., c = ± a ´ b .

(

)

So, therewill be two vectors of unit length perpendicular to the vectors a and b.

HINTS AND EXPLANATIONS

∴

i j k  a ´ i = x y z 1 0 0

∴

)

Þ ( l + 2 ) l 2 - 2l - 2 = 0

13.24  Chapter 13

PRACTICE EXERCISES Single Option Correct Type 1. a and b are mutually perpendicular unit vectors. If r is a vector satisfying r ⋅ a = 0, r ⋅ b = 1 and [r a b] = 1, then r is (A) a × b + b (B)  a + (a × b) (C) b + (a × b) (D)  a×b+a

7. If AB = 3i + j - k and AC = i - j + 3k . If the point P on the line segment BC is equidistant from AB and AC, then AP is

2. a, b, c are three vectors of magnitude, 3 , 1, 2 such that a × (a × c) + 3b = O. If θ is the angle between a and c, then cos2θ is equal to 1 3 (A) (B)  4 4

8. A, B, C, D are four points on a plane with position vectors a, b, c, d, respectively, such that (a - d)⋅ (b - c) = (b - d) ⋅ (c - a) = 0. For ∆ ABC, D is the

1 (C) 2

9. If a and b are two unit vectors, then the vector (a + b ) × (a × b) is parallel to the vector

(D)  none of these

PRACTICE EXERCISES

3. A vector a has components 2p and 1 w.r.t. a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise sense. If w.r.t. the new system, a has components p + 1 and 1, then 1 p = 1 or p = (A) p = 0 (B)  3 1 (C) p = -1 or p = (D) p = 1 or p = -1 3 4. If a ⋅ b = β and a × b = c, then b is equal to (A) (ba - a × c)/a2 (B) (ba + a × c)/a2 2 (C) (bc - a × c)/a (D) (bc + a × c)/a2 5. Let ABCDEF be a regular hexagon. If AD = x BC and CF = y AB, then xy = (A) 4 (B)  -4 (C) 2 (D)  -2 6. Given a cube ABCDA1 B1 C1 D1 with lower base ABCD, upper base A1B1C1D1 and the lateral edges AA1, BB1, CC1 and DD1; M and M1 are the centres of the faces ABCD and A1B1C1D1 respectively. O is a point on line MM1, such that OA + OB + OC + OD = OM1, then OM = λ OM1 if λ = 1 1 (A) (B)  4 2 1 1 (C) (D)  6 8

(A) 2i - k (C) 2i + k

(B)  i - 2k (D)  none of these

(A) incentre (B)  orthocentre (C) centroid (D)  none of these

(A) a - b (B)  a+b (C) 2a - b (D)  2a + b 1 10. If r = λ (a × b) + µ (b × c) + ν (c × a) and [a b c] = , 8 then λ + µ + ν is equal to (A) r ⋅ (a + b + c) (B) 8r ⋅ (a + b + c) (C) 4r ⋅ (a + b + c) (D)  none of these 11. In a parallelogram ABCD, | AB | = a, | AD | = b and | AC | = c. Then, DB ⋅ AB has the value 3a 2 + b 2 - c 2 a 2 + 3b 2 - c 2 (A) (B)  2 2 a 2 - b 2 + 3c 2 a 2 + 3b 2 + c 2 (C) (D)  2 2 12. Let a = a1i + a2   j + a3k, b = b1i + b2   j + b3k, and c = c1i + c2   j + c3k be three non-zero vectors such that c is a unit vector perpendicular to both vectors a and b. If the angle between vectors a and b is π/6, then a1 b1 c1

a2 b2 c2

a3 b3 c3

2

is equal to

(A) 0 (B) 1 1 (C) ( a12 + a22 + a32 ) (b12 + b22 + b32 ) 4 3 (D) ( a12 + a22 + a32 ) (b12 + b22 + b32 ) (c12 + c22 + c32 ) 4 (E) none of these

Vector Algebra  13.25 13. If a, b, c are non-coplanar unit vectors such that b+c a × (b × c) = , then the angle between a and b is 2 3p p (A) (B)  4 4 p (C) (D)  π 2 14. If the vectors a and b are perpendicular to each other, then a vector v in terms of a and b satisfying the equations v ⋅ a = 0, v ⋅ b = 1 and [v a b] = 1 is ×

×

×

2

a×b

21. A vector A has components A1, A2, A3 in a right-handed rectangular cartesian coordinate system Ox, Oy, Oz.

b a×b (B) | b | + | a × b |2 b a×b (C) | b |2 + | a × b | 15. Let the unit vectors a and b be perpendicular to each other and the unit vector c be inclined at an angle θ to both a and b. If c = xa + yb + z(a × b), then (A) x = cosθ, y = sinθ, z = cos2θ (B) x = sinθ, y = cosθ, z = -cos2θ (C) x = y = cosθ, z2 = cos2θ (D) x = y = cosθ, z2 = -cos2θ 16. If S is the circumcentre, O is the orthocentre of ∆ABC, then SA + SB + SC = (A) SO (B) 2SO (C) OS (D) 2OS 17. If a, c, d are non-coplanar vectors and d ⋅ {a × [b × (c × d) ]} is equal to (A) (b ⋅ d) [a c d] (B) (a ⋅ d) [a c d] (C) (c ⋅ d) [a c d] (D)  none of these 18. If 4a+ 5b + 9c = 0, then (a × b) × [(b × c) × (c × a)] is equal to (A) A vector perpendicular to the plane of a, b and c (B) A scalar quantity (C) 0 (D) none of these 19. If

∑ a = 0 where ∑ ∑a ⋅a i =1

1≤ i < j ≤ n

i

i

The coordinate system is rotated about the z-axis p through an angle . The components of A in the new 2 coordinate system are (A) A1, -A2, A3 (B)  A2, A1, A3 (C) A1, A2, -A3 (D)  A2, -A1, A3. ×

(D) none of these

n

P Q R = 1. (D) + OA OB OC

| ai | = 1 V i, then the value of is

j

n (A) - (B)  -n 2 n (C) (D)  n 2

22. In a ∆ OAB, E is the mid-point of OB and D is a point on AB such that AD : DB = 2 : 1. If OD and AE intersect at P, then the ratio OP : PD is (A) 1 : 2 (B)  2 : 1 (C) 3 : 2 (D)  2 : 3. 23. If a, b, c are three non-parallel unit vectors such that 1 a × (b × c) = b, then the angles which a makes with 2 b and c are (A) 90º, 60º (B)  45º, 60º (C) 30º, 60º (D)  none of these 24. If a = i + j - k, b = i - j + k and c is a unit vector perpendicular to the vector a and coplanar with a and b, then a unit vector d perpendicular to both a and c is (A) (C)

1 6 1 2

( 2i − j + k ) ( j + k )

1

(i + j ) 2 1 (D) (i + k ) 2 (B) 

25. If the pth, qth and rth terms of a G. P. are positive numbers a, b and c, respectively, then the angle between the vectors ilna + jlnb + klnc and i(q - r) + j(r - p) + k (p - q) is p p (A) (B)  3 6 ×

p (C) 2 ×

×

(D)  none of these

PRACTICE EXERCISES

1 1 b (A) b 2 + a × b

20. Forces P, Q act at O and have a resultant R. If any transversal cuts their lines of action at A, B, C, respectively, then P Q R + =0 (A) + OA OB OC P Q R + + =1 (B) OA OB OC P Q R + =0 (C) OA OB OC

13.26  Chapter 13 1  25. A vector a is collinear with vector b =  6, − 8, − 7   2 and make an acute angle with the positive direction of z-axis. If | a | = 50, then a = (A) (24, 32, 30) (B)  (24, -32, 30) (C) (-24, 32, 30) (D)  none of these 27. The perpendicular distance of a corner of a unit cube form a diagonal not passing through it is 6 (A) 6 (B)  3 3 (C) (D)  none of these 6 28. The vectors a, b and c are equal in length and taken pairwise, they make equal angles. If a = i + j, b = j + k, and c makes an obtuse angle with the base vector i, then c is equal to (A) i + k (B)  - i + 4j - k (C)

−1 4 1 i + j − k 3 3 3

1 −4 1 j + k. (D)  i + 3 3 3

29. If the four points a, b, c, d are coplanar, then [b c d ] + [c a d ] + [a b d ] = (A) 0 (B)  1 (C) -1 (D)  [a b c]

PRACTICE EXERCISES

30. A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(-1, 1, 2). Then, the angle between the faces OAB and ABC will be æ 71 ö -1 æ 19 ö cos -1 ç ÷ (A) cos ç ÷ (B)  è 35 ø è 31 ø (C) 30° (D)  90° 31. If a quadrilateral ABCD is such that AB = b, AD = d and AC = mb + pd (m + p ≥ 1), then the area of the quadrilateral is k(p + m) | b × d |, where k is equal to 1 1 (A) (B)  8 4 1 (C) (D)  none of these 2 32. Let a be a unit vector and b be a non-zero vector not parallel to a. If two sides of the triangle are represented by the vectors 3 ( a ´ b) and b - (a · b) a, then the angles of the triangle are (A) 30°, 90°, 60° (B) 45°, 45°, 90° (C) 60°, 60°, 60° (D)  none of these 33. Let u and v be unit vectors. If w is a vector such that w + (w × u) = v, then | (u × v) · w |

1 1 (B)  £ 3 2 1 1 (C) > (D)  ³ 3 2 (A) £

×

×

34. If b and c are any two non-collinear unit vectors and a is a ⋅ (b + c) any vector, then ( a ⋅ b)b + ( a ⋅ c)c + (b × c) = | b + c |2 (A) a (B)  b (C) c (D)  none of these 35. If the vector - i + j - k bisects the angle between 3i + 4j and vector c, then the unit vector along c is −11i − 10 j − 2k (A) 15 (C)

−11i + 10 j − 2k 15

(B) 

−11i + 10 j + 2k 15

(D)  none of these

36. If a, b and c are three unit vectors such that a + b + c is also a unit vector and q1, q2 and q3 are angles between the vectors a, b; b, c and c, a, respectively, then among q1, q2 and q3. (A) all are acute angles (B) all are right angles (C) at least one is obtuse angle (D) none of these 37. If x and y are two non-collinear vectors and ABC is a triangle with side lengths a, b, c satisfying  ( 20 a - 15b) x + (15b - 12c) y + (12c - 20 a) ( x ´ y ) = 0, then ∆ABC is (A) an acute-angled triangle (B) an obtuse-angled triangle (C) a right-angled triangle (D) an isosceles triangle 38.

If a ( a ´ b) + b (b ´ c) + g (c ´ a) = 0, then (A) a, b, c are coplanar if all of α, β, γ ≠ 0 (B) a, b, c are coplanar if any one of α, β, γ ≠ 0 (C) a, b, c are non-coplanar for any α, β, γ (D) none of these

39. If p × q = r and p . q = c, then q = cp − p × r cp + p × r (A) | p |2 (B)  | p |2 (C)

cr − p × r cr + p × r (D)  2 | p| | p |2

Vector Algebra  13.27 40. Let a = i + j and b = 2i - k. The point of intersection of the lines r × a = b × a and r × b = a × b is (A) - i + j + k (B) 3i - j + k (C) 3i + j - k (D)  i-j-k

41. The sides of a parallelogram are 2i + 4j - 5k and i + 2j + 3k. The unit vector parallel to one of the diagonals size is 1 1 (A) (3i + 6 j − 2k ) (B)  (3i − 6 j − 2k ) 7 7 1 1 (C) ( −3i + 6 j − 2k ) (D)  (3i + 6 j + 2k ) 7 7

Previous Year’s Questions

(A)

1 ˆ ˆ 1 ˆ ˆ (2i + j ) (i + j ) (B)  5 2

(C) ±

1 ˆ ˆ (i + k ) 2

(D)  none of these

43. The vector iˆ + xjˆ + 3kˆ is rotated through an angle θ and doubled in magnitude, then it becomes 4iˆ + ( 4 x - 2) ˆj + 2kˆ. The value of x are: [2002] ì1 ü ì 2 ü í , 2ý (A) í- , 2 ý (B)  î3 þ î 3 þ ì2 ü (C) í , 0 ý (D) {2, 7} î3 þ    44. If the vectors a , b and c from the sides BC, CA and AB respectively of a triangle ABC, then: [2002]       (A) a × b = b × c = c × b = 0       (B) a ´ b = b ´ c = c ´ a       (C) a × b = b × c = c × a = 0       (D) a ´ a + a ´ c + c ´ a = 0    45. If the vectors a - xiˆ + yjˆ + zkˆ and such that a , c and  [2002] b form a right handed system, then c is:  (A) ziˆ - xkˆ (B)  0 (C) yjˆ (D)  - ziˆ + xkˆ 46. The centre of the circle given by r ⋅ (iˆ + 2 ˆj + 2kˆ ) = 15 and | r ⋅ ( ˆj + 2kˆ ) | = 4 is: [2002] (A) (0, 1, 2) (C) (–1, 3, 4)

(B)  (1, 3, 4) (D)  none of these

1 + a3 1 + b3 = 0 and vectors (1, a, a2) (1, b, b2) 1 + c3 c2) are non-coplanar, then the product abc [2003]

(A) 2 (B)  –1 (C) 1 (D)  0    48. Let a , b and c be three non-zero vectors such that no  two of these are collinear.   If the vector a + 2b is col linear with c and b + 3c is collinear a (λ being  with   some non-zero scalar) then a + 2b + 6c equals [2004]   (A) l a (B)  lb  (C) l c (D)  0 49. A particle is acted upon by constant forces 4iˆ + ˆj - 3kˆ and 3iˆ + ˆj - kˆ which displace it from a point iˆ + 2 ˆj + 3kˆ to the point 5iˆ + 4 ˆj + kˆ. The work done in standard units by the forces is given by [2004] (A) 40 (B)  30 (C) 25 (D)  15 50. If a , b , c are non-coplanar vectors and λ is a real number, then the vectors a + 2b + 3c , l b + 4c and ( 2l - 1) c are non-coplanar for [2004] (A) all values of λ (B) all except one value of λ (C) all except two values of λ (D) no value or λ 51. Let u , v , w be such that | u | = 1, | v | = 2, |w | = 3. If the projection v along u is equal to that of w along u and v , w are perpendicular to each other then | u - v + w | equals [2004] 7 (A) 2 (B)  (C)

14 (D)  14

52. Let a , b and c be non-zero vectors such that ( a ´ b ) ´ c = - | b | c | a . If θ is the acute angle be­tween the vectors b and c then sinθ equals[2004]

PRACTICE EXERCISES

42. Given two vectors are iˆ - ˆj and iˆ + 2 ˆj the unit vector coplanar with the two vectors and perpendicular to first is: [2002]

a a2 47. If b b 2 c c2 and (1, c, equals

13.28  Chapter 13 2 1 (A) (B)  3 3 2 2 2 (C) (D)  3 3 53. If C is the mid point of AB and P is any point outside AB, then [2005]    (A) PA + PB = 2 PC    (B) PA + PB = PC    (C) PA + PB + 2 PC = 0    (D) PA + PB + PC = 0

PRACTICE EXERCISES

 54. The distance between the line r = 2iˆ - 2 ˆj + 3kˆ +l (iˆ + ˆj + 4 kˆ ) and the plane r ⋅ (iˆ + 5 ˆj + kˆ ) = 5 is  [2005] 10 10 (A) (B)  3 3 9 3 10 (C) (D)  10 3  55. For any vector a, the value of ( a × iˆ) 2 + ( a × ˆj ) 2 + ( a × kˆ ) 2 is equal to [2005] 2 2 (A) 3a (B)  a   (C) 2a 2 (D)  4a 2 56. If non-zero numbers a, b, c are in H.P., then the straight x y 1 line + + = 0 always passes through a fixed point. a b c That point is [2005] (A) (–1, 2) (B)  (–1, –2) 1ö æ (C) (1, –2) (D)  ç1, - 2 ÷ è ø 57. Let a, b and c be distinct non-negative numbers. If the vectors aiˆ + ajˆ + ckˆ, iˆ + kˆ and ciˆ + cjˆ + bkˆ lie in a plane, then c is [2005] (A) the Geometric Mean of a and b (B) the Arithmetic Mean of a and b (C) equal to zero (D) the Harmonic Mean of a and b    58. If a , b , c are non-coplanar vectors and λ, is a real        number then [l ( a + b ) l 2 b l c ] = [ab + cb ] for [2005] (A) exactly one value of λ (B) no value of λ (C) exactly three values of λ (D) exactly two values of λ    59. Let a = iˆ - kˆ, b = xiˆ + ˆj + (1 - x )kˆ and c = yiˆ + xjˆ    [2005] +(1 + x − y )kˆ. Then [a , b , c ] depends on

(A) only y (B) only x (C) both x and y (D) neither x nor y          60. If ( a ´ b ) ´ c = a ´ (b ´ c ), where a , b and c are any      three vectors such that a × b ¹ 0, b × c ¹ 0, then a and  [2006] c are p (A) inclined at an angle of between them 3 p (B) inclined at an angle of between them 6 (C) perpendicular (D) parallel ×

×

61. The values of a, for which the points A, B, C with position vectors 2iˆ - ˆj + kˆ, iˆ - 3 ˆj - 5kˆ and aiˆ - 3 ˆj + kˆ respectively are the vertices of a right-angled triangle p with C = are [2006] 2 (A) 2 and 1 (B)  -2 and -1 (C) -2 and 1 (D)  2 and -1 62. If uˆ and vˆ are unit vectors and θ is the acute angle between them, then 2uˆ ´ 3vˆ is a unit vector for[2007] (A) exactly two values of θ (B) more than two values of θ (C) no value of θ (D) exactly one value of θ 63. Let a = iˆ + ˆj + kˆ, b = iˆ - ˆj + 2kˆ and c = xiˆ + ( x - 2) ˆj - kˆ. If the vector c lies in the plane of a and b , then x equals [2007] (A) 0 (B)  1 (C) −4 (D)  −2  64. The vector a = a iˆ + 2 ˆj + b kˆ lies in the plane of the   vectors b = iˆ + ˆj and c = ˆj + kˆ and bisects the angle   between b and c . Then which one of the following gives possible values of α and β ? [2008] (A) α = 2, β = 2 (B)  α = 1, β = 2 (C) α = 2, β = 1 (D)  α = l, β = 1      65. The non-zero vectors a , b and c are related by a = 8b     and c = 7b . Then the angle between a and c is[2008] (A) 0 (B)  π /4 (C) π /2 (D)  π    66. If u , v , w are non-coplanar vectors and p, q are real       numbers, then the equality [3u pv pw ] - [ pv w qu ]    - [2w qv qu ] = 0 holds for [2009]

Vector Algebra  13.29 (A) exactly one value of (p, q) (B) exactly two values of (p, q) (C) more than two but not all values of (p , q) (D) all values of (p, q)

67. The projections of a vector on the three coordinate axis are 6, −3, 2 respectively. The direction cosines of the vector are [2009] 6 3 2 (A) 6, −3, 2 (B)  , - , 5 5 5 6 3 2 6 3 2 - ,- , (C) , - , (D)  7 7 7 7 7 7         68. Let a = j - k and c = i - j - k . Then, the vector b      satisfying a ´ b + c = 0 and a × b = 3 is [2010] (A) 2iˆ - ˆj + 2kˆ

(B)  iˆ - ˆj - 2kˆ

-iˆ + ˆj - 2kˆ (C) iˆ + ˆj - 2kˆ (D)  



69. If the vectors a = iˆ - ˆj + 2kˆ, bˆ = 2iˆ + 4 ˆj + kˆ and c = liˆ + ˆj + m kˆ are mutually orthogonal, then the tuple (λ, µ) =  [2010] (A) (2, -3) (B)  (-2, 3) (C) (3, -2) (D)  (-3, 2) 1

1 (3iˆ + kˆ ) and b = (2iˆ + 3 ˆj − 6 kˆ ), then the 7 10 value of the expression( 2 z - b).[( a ´ b) ´ ( a + 2b)] is  [2011] (A) −3 (B)  5 (C) 3 (D)  −5   71. The vectors a and b are not perpendicular and       c and d are two vectors satisfying: b ´ c = b ´ d and    a × d = 0. Then, the vector d is equal to [2011] 70. If a =

    73. Let ABCD be a parallelogram such that AB = q, AD = p  and BAD be an acute angle. If r is the vector which coincides with the altitude directed from the vertex B  to the side AD, then r is given by [2012]     3( p × q )  (A) r = 3q -   p ( p × p)     æ p×q ö  (B) r = - q + ç   ÷ p è p× p ø     æ p×q ö  (C) r = q - ç   ÷ p è p× p ø     3( p × q )  (D) r = - 3q +   p ( p × p)

  74. If the vectors AB = 3iˆ + 4 kˆ and AC = 5iˆ + 2 ˆj + 4 kˆ represent the sides of a triangle ABC, then the length of the median through A is[2013] (A) 72 (B)  33 (C) 45 (D)  18       75. If [a ´ bb ´ cc ´ a ] = l[abc ]2 , then the value of λ is equal to [2014] (A) 2 (B)  3 (C) 0 (D)  1    76. Let a , b and c be three non-zero vectors such that no    1    two of them are collinear and ( a ´ b ) ´ c = | b || c | a. 3   If θ is the angle between vectors b and c , then a value of sinθ [2015] - 2 2 (A) (B)  3 3 -2 3 2 2 (D)  3 3

æ a.c ö (A) c + ç ÷ b è a.b ø

æ b.c ö (B)  b + ç ÷c è a.b ø

(C)

æ a.c ö (C) c - ç ÷ b è a.b ø

æ b.c ö b-ç (D)  ÷c è a.b ø

   77. Let a , b and c be three unit vectors such that

72. Let aˆ and bˆ be two unit vectors. If the vectors  c = aˆ + 2bˆ and d = 5aˆ - 4bˆ are perpendicular to each other, then the angle between aˆ and bˆ is  [2012]

p p (B)  6 2 p p (C) (D)  3 4 (A)

×

×

×

×

 3       a ´ (b ´ c ) = (b + c ). if b is not parallel to c, then 2   the angle between a and b is [2016] 3p 5p (A) (B)  4 6 ×

×

p 2p (C) (D)  2 3 ×

×

PRACTICE EXERCISES



13.30  Chapter 13    78. Let a = 2iˆ + ˆj − 2kˆ and b = iˆ + ˆj . Let c be a vector      such that | c − a |= 3, | ( a × b ) × c |= 3 and the angle      between c and a × b be 30º. Then a . c is equal to  (A) 2

[2017] (B) 5

1 25 (C)  (D)  8 8  79. Let u be a vector co-planar with the vectors    a = 2iˆ + 3 ˆj − kˆ and b = ˆj + kˆ. If u is perpendicular    2 to a and u ⋅ b = 24, then u is equal to [2018] (A) 336

(B) 315

(C) 256

r

r

r

1

r

1

80. Let a = (l – 2) a + b and b = (4l – 2) a + 3 b be two 1

1

given vectors where vectors a and b are non-collinear. 1

1

The value of l for which vector a and b are collinear, is: [2019] (A) 4 (B) 3 (C)  – 3 (D)  – 4 81. Two sides of a parallelogram are along the lines, x + = 3 and x – y + 3 = 0. If its diagonals intersect at (2, 4) then one of its vertex is: [2019] (A)  (2, 1) (B)  (3, 6) (C)  (2, 6) (D)  (3, 5)

(D) 84

ANSWER K EYS Single Option Correct Type 1. (A) 2. (B) 3. (B) 4. (A) 5. (B) 6. (A) 7. (C) 8. (B) 9. (A) 10. (B) 11. (A) 12. (C) 13. (A) 14. (A) 15. (D) 16. (A) 17. (A) 18. (C) 19. (A) 20. (C) 21. (D) 22. (C) 23. (A) 24. (C) 25. (C) 26. (C) 27. (B) 28. (C) 29. (D) 30. (A) 31. (C) 32. (A) 33. (B) 34. (A) 35. (A) 36. (C) 37. (C) 38. (B) 39. (A) 40. (C) 41. (A)

PRACTICE EXERCISES

Previous Years’ Questions 42. (A) 52. (A) 62. (D)   72. (C)

43. (A) 53. (A) 63. (D) 73. (B)

44. (B) 54. (B) 64. (D) 74. (B)

45. (A) 55. (C) 65. (D) 75. (D)

46. (B) 56. (C) 66. (A) 76. (D)

47. (B) 57. (C) 67. (C) 77. (A)

48. (D) 58. (B) 68. (B) 78. (A)

49. (A) 59. (D) 69. (D) 79. (A)

50. (C) 60. (D) 70. (D) 80. (D)

51. (C) 61. (A) 71. (C) 81. (B)

Vector Algebra  13.31

HINTS AND EXPLANATIONS Single Option Correct Type 1. Let r = x1a + x2b + x3 (a × b) Then, r ⋅ a = x1 |a|2, r ⋅ b = x2 |b|2 and, r ⋅ (a × b) = x3 |a × b|2 ⇒ x1 = 0, x2 = 1, x3 = 1 ∴  r = a × b + b 2. We have, a × (a × c) + 3b = 0 ⇒ (a ⋅ c)a - (a ⋅ a)c + 3b = 0



Þ |( 2 cosq ) a - 3 c |2 = | - 3 b |2 Þ

4 cos 2 q | a |2 +3 | c |2 - 4 3 cosq ( a × c) = 3 | b |2

Þ 12 cos 2 q + 12 - 4 3 cosq ´ 3 ´ 2 cosq = 3 ⇒ 12 cos2θ + 9 - 24cos2θ = 0 Þ 12 cos 2 q = 9 Þ cos 2 q =

9 3 = . 12 4

3. Let i, j be unit vectors along the coordinate axes ∴ a = 2pi + 1 · j(1) On rotation, let b be the vector having components p + 1 and 1. ∴ b = (p + 1) i + 1 · j(2) where i, j are unit vectors along the new coordinate axes. But on rotation | b | = | a | ⇒ | b |2 = | a |2 ⇒ ( p + 1)2 + 1 = (2p)2 + 1 ⇒ 3p2 - 2p - 1 = 0 Þ (3 p + 1)( p - 1) = 0 Þ

1 p = 1 or - . 3

4. Here, a and c = a × b are non-collinear vectors. ∴ Let b = xa + y (a × c)(1) ∴ β = a ⋅ b = a ⋅ [xa + y(a × c)] = x | a |2 + ya ⋅ (a × c) = xa2 ⇒ x = β /a2. And, c = a × b = a × [xa + y(a × c)] = xa × a + ya × (a × c) = 0 + y(a ⋅ c) a - y (a · a) c = y {a ⋅ (a × b)} a - ya2c = -ya2c ⇒ y = -1/a2 ∴ from (1), b = (ba - a × c)/a2. 5. Since ABCDEF is a regular hexagon, from plane geometry, we have

From (ii) and (iii), xy = 2 (-2) = -4. 6. OM1 = OA + OB + OC + OD (given) = OM + MA + OM + MB + OM + MC + OM + MD = 4OM + (MA + MC) + (MB + MD) = 4OM( MA = -MC, MB = -MD) 1 1 \ OM = OM1 , \ l = . 4 4 7. A point equidistant from AB and AC is on the bisector of the angle BAC. A vector along the internal bisector of the angle BAC AB AC = + | AB | | AC | i − j + 3k 1 = + ( 4i + 2 k ) 9 +1+1 1+1+ 9 11 ∴ AP = t(2i + k) ∴ BP = AP - AB = t(2i + k) - (3i + j - k) = (2t - 3) i - j + (t + 1) k Also, BC = AC − AB = (i − j + 3k ) − (3i + j − k ) = - 2i - 2 j + 4k. But BP = s BC. ∴ (2t - 3) i - j + (t + 1) k = s (- 2i - 2 j + 4k) ∴ 2t - 3 = -2s, -1 = -2s, t + 1 = 4s 1 ∴ s = and t = 1 ∴ AP = 2i + k . 2 =

3i + j − k

8. Since (a - d) ⋅ (b - c) = 0 ∴ DA · CB = 0 ∴ AD ⊥ BC.

HINTS AND EXPLANATIONS

Þ ( 2 cosq ) a - 3 c + 3 b = 0

AD = 2BC and FC = 2AB ∴ AD = 2BC and FC = 2AB.(1) Given that AD = xBC. ∴ 2BC = xBC, by (i) ⇒ x = 2. (2) Again, given that CF = yAB or -FC = yAB. ∴ -2AB = yAB, using (ii) ⇒ y = -2.(3)

\

Þ ( 2 3 cosq ) a - 3c + 3b = 0

13.32  Chapter 13 Since (b - d) ⋅ (c - a) = 0, ∴ DB ⋅ AC = 0 ∴ BD ⊥ CA. Then, D is the intersection of the altitudes through A and B. Therefore, D is the orthocentre of the triangle ABC. 9. We have, (a + b) × (a × b) = a × (a × b) + b × (a × b) = (a ⋅ b) a - (a ⋅ a) b + (b ⋅ b) a - (b ⋅ a) b = (a ⋅ b) (a - b) + a - b (b ⋅ b = b2 = 1, a ⋅ a = a2 = 1 = (a ⋅ b + 1) (a - b) ∴ The given vector is parallel to a - b. 1 10. Clearly, r × c = l [a b c] = l 8 1 r × a = m [a b c] = m 8 1 r × b = n [abc] = n 8 1 \ r × ( a + b + c) = (l + m + n ) 8 ∴ λ + µ + ν = 8r ⋅ (a + b + c). 11.   DB = DA + AB or, DA = DB - AB

\

HINTS AND EXPLANATIONS

c

B

a

∴ (DA)2 = (DB)2 + (AB)2 - 2DB ⋅ AB(1) In parallelogram 2(a2 + b2) = c2 + DB2 ∴ (DB)2 = 2a2 + 2b2 - c2 ∴ From (1), b2 = 2a2 + 2b2 - c2 + a2 - 2AB ⋅ DB 3a 2 + b 2 − c 2 . 2

12. We have, 2

= [a b c]2 = [( a × b) ⋅ c]2



= [| a | | b | sin π /6 c ⋅ c]2



= | a |2 | b |2 sin2 π /6 · 1

[ |c| = 1]

= ( a + a + a )(b + b + b ) ´ 1 / 4 2 1

2 2

2 3

2 1

2 2

1  1    ⇒ a⋅c −  b −  a ⋅ b +  c=0  2 2 Since a, b, c are non-coplanar so a, b, c are linearly independent. Hence, 1 . a⋅b = − 2

2 3

= 1 / 4 ( a12 + a22 + a32 )(b12 + b22 + b32 ).

a×b 1 = a×b = | a| |b| 2

1 (4) | a ´ b |2 From (1), (2), (3) and (4), we get 1 1 v = 2 b+ a × b. |b| | a × b |2

C

b

a3 b3 c3

( b + c ) = a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b) c

or, z =

D

a1 a2 b1 b2 c1 c2

2

(θ is the angle between a and b) ⇒ θ = 3π /4. 14. We have a ⊥ b ⇒ a, b, a × b are linearly independent ⇒ v can be expressed uniquely in terms of a, b and a × b. Let v = xa + yb + za × b(1) Given that, : a · b = 0, v · a = 0, v · b = 1, [v, a, b] = 1 ∴ v · a = xa2 or xa2 = 0 or x = 0 (2) 2 v · b = xv · b + yb + zb · (a × b) 1 or, yb 2 = 1 or y = 2 (3) b v · a × b = x ⋅ 0 + y ⋅ 0 + z |a × b|2 or, z |a × b|2 = 1

= x(a - b), where x = a ⋅ b + 1 is a scalar.

∴ AB ⋅ DB =

1

\ cosq =

as a, b are unit vectors)

A

13.

15. We have, c = xa + yb + z (a × b) ⇒ c ⋅ a = x and c ⋅ b = y ⇒ x = y = cosθ Now, c ⋅ c = | c |2 ⇒ [xa + yb + z(a × b)] ⋅ [xa + yb + z(a × b)] = | c |2 ⇒ 2x2 + z2 | a × b |2 = 1 ⇒ 2x2 + z2 [ | a |2 | b |2 - (a ⋅ b)2] = 1 ⇒ 2 x 2 + z 2 [1 − 0] = 1

[∵ a ⊥ b ∴ a ⋅ b = 0]

⇒ 2x + z = 1 ⇒ z = 1 - 2x2 = 1 - 2cos2 θ = -cos2θ. 16. From geometry 2 SD = AO, where D is the mid-point of BC. ∴ SA + SB + SC = SA + SD + DB + SD + DC ∴ = SA + 2 SD[ DB + DC = 0] = SA + AO = SO ∴ SA + SB + SC = SO. 2

2

2

Vector Algebra  13.33

n  n   n  19.  ∑ ai  ⋅  ∑ ai  = ∑ | ai |2 + 2 ∑ ∑ ai ⋅ a j  i= 1   i = 1  i = 1 1≤ i < j≤ n

⇒ 0 = n+2

∑ ∑a ⋅a

1≤ i < j≤ n

i

j

⇒

∑ ∑a ⋅a

1≤ i < j≤ n

i

j

=−

n 2

20. Let O be taken as the origin of reference. Let a, b, c be the position vectors of A, B, C respectively so that OA = a, OB = b, OC = c.

Then, the vector representing the force P along OA OA P = P (unit vector along OA) = P = a. OA OA Similarly, the vector representing the force Q along OB Q = b and the vector representing the force R along OC OB R c. OC Since R is the resultant of P and Q, we have (P/OA) a + (Q/OB) b = (R/OC) c or, (P/OA) a + (Q/OB) b - (R/OC) a = 0.(1) Since, a, b, c are the position vectors of three collinear points, therefore, we have the algebraic sum of the coefficients of a, b, c in (i) = 0 i.e., (P/OA) + (Q/OB) - (R/OC) = 0 P Q R Hence, + = . OA OB OC p , the new x-axis is along old 21. When rotated through 2 y-axis and new y-axis is along the old negative x-axis; z-axis remains same as before. Hence, the components of A in the new system are A2, -A1, A3. 22. Let a and b be the position vectors of A and B respectively w.r.t. O as the origin. Then p.v. of E = OE 1 1 = OB = b 2 2 1 and, p.v. of D = OD = ( 2b + a), ∵ AD : DB = 2 :1 3

=

×

Now let OD and AE intersect at P such that OP/OD = λ and AP/AE = µ. ∴ OP = lOD and AP = µAE. 1 or, OP = lOD = ( 2b + a) l . 3 æ1 ö and, AP = m AE = m (OE - OA) = m ç b - a ÷ . è2 ø In ∆ OAP, OA + AP = OP æ1 ö 1 Þ a + m ç b - a ÷ = ( 2b + a ) l è2 ø 3 1 ö 2 ö æ æ1 Þ ç1 - m - l ÷ a + ç m - l ÷ b = 0 3 ø 3 ø è è2 1 1 2 Þ 1 - m - l = 0 and m - l = 0, 3 2 3 ∴ a and b are non-collinear. 4 3 Solving these equations, we get m = , l = . 5 5 Now, OP = lOD, PD = OD - OP = (1 - λ) OD 3/5 3 OP l \ = = = PD 1 - l 1 - 3 / 5 2 ⇒ OP =

3 3 PD ⇒ OP = PD ⇒ OP : PD = 3 : 2. 2 2

1 b 2 1 ∴ ( a ⋅ c ) b − ( a ⋅ b) c = b 2 23. We have, a × (b × c) =

1ö æ Þ ç a × c - ÷ b - ( a × b) c = 0 (1) 2ø è Since b and c are non-parallel, (Given) ∴ (i) exists if coefficients of b and c vanish separately i.e., a ⋅ c −

1 = 0 and a ⋅ b = 0. 2

1 ⇒ a ⋅ c = and a ⋅ b = 0. 2 Let q1 and q2 be the angles which a makes with b and c, respectively. Also a, b, c are unit vectors, 1 \ cosq 2 = and cosq1 = 0 2 i.e., q2 = 60º and q1 = 90º Hence, q1 = 90º and q2 = 60º. 24. Let c = xa + yb = x(i + j - k) + y(i - j + k) = (x + y)i + (x - y)  j - (x - y)k Since c is ⊥ to a ∴ x + y + x - y + x - y = 0 ⇒ 3x = y Since | c | = 1, ∴ (x + y)2 + (x - y)2 + (x - y)2 = 1 ∴ (4x)2 + (-2x)2 + (-2x)2 = 1.

HINTS AND EXPLANATIONS

17. We have, d ⋅ {a × [b × (c × d)]} = d ⋅ (a × ((b ⋅ d)c - (b ⋅ c)d )) = d ⋅ ((b ⋅ d) (a × c) - (b ⋅ c) (a × d)) = (b ⋅ d) d ⋅ (a × c) - (b ⋅ c) d ⋅ (a × d) = (b ⋅ d) [a c d][∴ d ⋅ (a × d) = 0]. 18. 4a + 5b + 9c = 0 ⇒ Vectors a, b and c are collinear ⇒ (b × c) × (c × a) = 0.

13.34  Chapter 13 Þ 24 x 2 = 1. \ x =

1 2 6

2

∴ c = 4 xi + 2 xj + 2 xk = i.e., c =

6

i−

1 6

j+

1 6

k

1

( 2i − j − k ) 6 Let d = d1i + d2   j + d3k Since d is ⊥ to a and c ∴ d · a = 0 ⇒ d1 + d2 - d3 = 0 and, d ⋅ c = 0 ⇒ 2d1 - d2 + d3 = 0 ⇒ 3d1 = 0, ∴ d1 = 0, ∴ d2 = d3 Since | d | = 1, \ d12 + d22 + d32 = 1 1 1 1 , \ d2 = d3 = . Þ d = , or d2 = 2 2 2 2 2

HINTS AND EXPLANATIONS

1

a×k - (15 / 2) t = . | a| | k | | a|

∵ q < 90° \ cosq =

1 3

(i + j + k ).

= k ⋅e = k ⋅

1

(i + j + k ) =

3 ∴ CN2 = OC2 - ON2

1 3

. [In right triangle ∆ OCN]

2

( j + k ). 2 25. Let x be the first term and y the common ratio of G.P. Then, a = pth term = xy p - 1 b = qth term = xy q - 1 c = rth term = xy r - 1 Now, (ilna + jlnb + klnc) ⋅ [i(q - r) + j(r - p) + k (p - q)] = (q - r) lna + (r - p) lnb + (p - q) lnc = (q - r) ln (xyp - 1) + (r - p) ln (xyq - 1) + (p - q)ln (xyr - 1) = (q - r) {lnx - (p - 1) lny} + ... = {(q - r) + (r - p) + (p - q)} lnx + {(q - r) (p - 1) + (r - p) (q - 1) + (p - q) (r - 1)} lny = 0. Hence, the vectors are perpendicular. 26. Since a is collinear with b = 6i - 8j - (15/2) k. ∴ there exists a scalar t, s.t. a = tb = 6ti - 8tj - (15/2) tk. Now, if a makes an angle θ with positve z-axis, then cosq =

=

i+ j+k i+ j+k = |i + j + k | (1 + 1 + 1)

∴ ON = projection of OC on OE

Þ 0 + d22 + d22 = 1 Þ 2d22 = 1

Hence, d =

e=

-15t > 0 Þ t < 0. 2 |a|

Now, | a |= [{36 + 64 + ( 225 / 4)}t 2 ] = 50 ⇒ t2 = 16 ⇒ t = -4, ∴ [Rejecting t = 4, t < 0] Hence, a = - 24i + 32  j + 30k = (-24, 32, 30). 27. Let the unit vectors i, j and k be denoted by the coterminous edges OA, OB and OC, respectively of the unit cube. Let CN be the perpendicular drawn from C on the diagonal OE of the cube which does not pass through C. Here, OE = i + j + k. Let e be the unit vector along OE. Then,

1 2 æ 1 ö = 12 - ç ÷ =1- = . 3 3 è 3ø \ CN =

2 2 3 6 = ´ = 3 3 3 3

28. The length of a is | a | = 12 + 12 = 2 . Similarly, | b | = 2 . Since the three vectors have equal length, ∴ | c | = 2 . Let c = c1i + c2   j + c3k. Then, since c makes an obtuse angle with i, we must have c · i = c1 < 0. We are also given that the angles between the vectors are equal, i.e., cos −1

a⋅b a⋅c b⋅c = cos −1 = cos −1 | a| |b| | a| |c| |b| |c|

Now, a ⋅ b = 1, a ⋅ c = c1 + c2 and b ⋅ c = c2 + c3, so cos −1

c +c a⋅b 1 = cos −1 = cos −1 1 2 2 2 | a| |b| = cos -1

c2 + c3 . 2

This gives the equations c1 + c2 = 1 and c2 + c3 = 1, from which we get c3 = c1 and c2 = 1 - c1. Putting these values in c12 + c22 + c32 = 2, we get c12 + (1 - c1 ) 2 + c12 = 2 Þ 3c12 - 2c1 - 1 = 0 1 Þ (c1 - 1)(3c1 + 1) = 0 Þ c1 = 1, - . 3 Since c1 must be less than zero, we rule out the solution c1 4 = 1, giving c1 = -1/3 = c3 and c2 = 1 - c1 = . Hence, the 3 required vector is 1 4 1 c = − i + j − k. 3 3 3

Vector Algebra  13.35 29. Let the given points be A, B, C, D, respectively. Now, if the points A, B, C, D are coplanar, then AB, AC, AD are coplanar. ∴ AB ⋅ (AC × AD) = 0 ⇒ (b - a) ⋅ {(c - a) × (d - a)} = 0. ∴a×a=0 ⇒ (b - a) ⋅ {c × d - c × a - a × d} = 0 ⇒ b ⋅ (c × d) - b ⋅ (c × a) - b ⋅ (a × d) - a ⋅ (c × d) + a ⋅ (c × a) + a ⋅ (a × d) = 0 ⇒ [b c d] - [b c a] - [b a d] - [a c d] + 0 + 0 = 0 ⇒ [b c d] - [a b c] + [a b d] + [c a d] = 0 ⇒ [b c d] + [c a d] + [a b d] = [a b c].

Hence, the area of the quadrilateral ABCD 1 1 = ( p + m) | b × d | . ∴ k = . 2 2



= 3[( a × b) ⋅ b − ( a ⋅ b)( a × b) ⋅ a]

30. Vector ⊥ to face OAB



= 3 [ 0 - 0 ] = 0.

i.e., AB = 3 ( a × b) and, AC = b - (a ⋅ b)a ∴ AB ⋅ AC = 3( a × b) ⋅ [b − ( a ⋅ b)a]

Therefore, ∠ BAC = 90°

j k

= OA × OB = 1 2 1 = 5i − j − 3k (1) 2 1 3

AB 2 = [ 3 ( a × b)]2 = 3( a × b) 2 (1) AC 2 = [b − ( a ⋅ b)a]2

= (b) 2 + ( a ⋅ b) 2 a 2 − 2(b ⋅ a)( a ⋅ b)

i j k = AB × AC = 1 −1 2 = i − 5 j − 3k (2) − 2 −1 1



= (b) 2 + ( a ⋅ b) 2 − 2( a ⋅ b) 2



= ( b) 2 − ( a ⋅ b) 2



= (b) 2 - | a |2 | b |2 cos 2 q

Since the angle between the faces = angle between their normals



= (b) 2 [1- | a |2 cos 2 q ] = (b) 2 (1 - cos 2 q )



= (b) 2 sin 2 q = | a |2 | b |2 sin 2 q = ( a ´ b) 2 . (2)

Vector ⊥ to the face ABC

\ cosq =

5+5+9 35 35

=

19 35

æ 19 ö \ q = cos -1 ç ÷ . è 35 ø 31. We have, BC = BA + AC = - b + mb + pd = (m - 1) b + pd and, CD = CA + AD = - mb - pd + d = - mb + (1 - p) d 1 Area of the triangle ABD = | b × d | . 2

d b

Area of the triangle BCD 1 = |( m − 1)b + pd ) × ( − mb + (1 − p)d )| 2 1 = | − pmd × b + (1 − p)( m − 1)b × d | 2 1 1 = | m − 1 + p | | b × d | = ( p + m − 1) | b × d | 2 2

Dividing (i) by (ii), we get

AB 2 3 ( a × b) 2 = AC 2 ( a × b) 2

⇒

AB 2 = 3 ⋅ AC 2

⇒

AB = 3 AC .

AB 3 AC = = 3 , \ ÐC = 60° AC AC So ∴ ∠ A = 180 - 90° - 60° = 30°. Hence, angles of the triangle are 30°, 90° and 60°. tan C =

33. Given equation is w + (w × u) = v(1) Taking cross-product with u, we get u × [w + (w × u)] = u × v ⇒ u × w + u × (w × u) = u × v ⇒ u × w + (u ⋅ u)w - (u · w)u = u × v ⇒ u × w + w - (u ⋅ w)u = u × v(2) Taking scalar product of (1) with u we get u ⋅ w + u ⋅ (w × u) = u ⋅ v ∴ ⇒ u ⋅ w = u ⋅ v [   u ⋅ (w × u) = 0] (3) Taking scalar product of (1) with v we get v ⋅ w + v ⋅ (w × u) = v ⋅ v ⇒ v ⋅ w + [vwu] = 1 ⇒ (u × v) ⋅ w = 1 - v ⋅ w(4) Taking scalar product of (2) with w we get (u × w) ⋅ w + w ⋅ w - (u ⋅ w) (u ⋅ w) = (u × v) ⋅ w ⇒ 0 + | w |2 - (u ⋅ w)2 = (u × v) ⋅ w(5) ⇒ (u × v) · w = |w|2 - (u · w)2. Taking scalar product of (1) with w we get

HINTS AND EXPLANATIONS

i

32. Let ABC be a triangle in which the given vectors are represented by the sides AB and AC.

13.36  Chapter 13 w ⋅ w + (w × u) ⋅ w = v ⋅ w ⇒ | w |2 = v ⋅ w ⇒ | w |2 = 1 - (u × v) ⋅ w From (5) we get (u × v) ⋅ w = | w |2 - (u ⋅ w)2 = 1 - (u × v) ⋅ w - (u ⋅ w)2 ⇒ 2(u × v) ⋅ w = 1 - (u ⋅ v)2

But x2 + y2 + z2 = 1 2

[using (4)]

(6)

[Using (3)]

1 Thus, |(u × v ) ⋅ w | = |1 − (u ⋅ v ) 2 | 2

1 ≤ [∴ (u ⋅ v ) 2 ≥ 0]. 2

34. Let I be a unit vector in the direction of b, J in the direction of c. Note that b = I and c = J. We have b × c = | b | | c | sin α k = sin α k, where k is a unit vector perpendicular to b and c. Þ | b ´ c | = sin a

Þ k=

36. Since | a + b + c | = 1 ⇒ ( a + b + c) ⋅ ( a + b + c) = 1 b´c |b ´ c|

Any vector a can be written as a linear combination of I, J and k. Let a = a1i + a2 J + a3k

HINTS AND EXPLANATIONS

Now, a ⋅ b = a ⋅ I = a1 , a ⋅ c = a ⋅ J = a2 and, a ⋅ b × c = a ⋅ k = a3 |b × c| Thus, ( a ⋅ b) b + ( a ⋅ c) c + = a1b + a2c + a3

æ 3t + 5 ö æ 5 - 4t ö 1 \ ç÷ +ç ÷ + 2 =1 5t ø è 5t ø t è ⇒ 9t2 + 30t + 25 + 25 + 16t2 - 40t + 25 = 25t2 75 15 Þ 75 - 10t = 0, \ t = = 10 2 45 +5 - 55 11 Hence, x = - 2 = =75 75 15 2 5 - 30 - 50 -10 = = y= 75 75 15 2 1 2 and, z = =15 15 2 ∴ unit vector along c −11i 10 j 2k 11i + 10 j + 2k = − − or − . 15 15 15 15

⇒ 1 + 1 + 1 + 2 ( a ⋅ b + b ⋅ c + c ⋅ a) = 1 ⇒ a ⋅ b + b ⋅ c + c ⋅ a = −1 ⇒ cosq1 + cosq2 + cosq3 = -1 So, at least one of cosq1, cosq2 and cosq3 must be negative. 37. Since x and y are linearly independent, a b c 20 a - 15b = 15b - 12c = 12c - 20 a = 0 Þ = = 3 4 5 ⇒ c2 = a2 + b2  ⇒  ∆ABC is right-angled.

a ⋅ (b × c) (b × c) | b × c |2

b×c |b × c|

= a1I + a2 J + a3k = a. 35. Let c = xi + yj + zk , where x2 + y2 + z2 = 1 1 Unit vector along 3i + 4 j = (3i + 4 j ) 5 ∴ equation of the bisector of these two is   3i + 4 j   r = t ( xi + yj + zk ) +   5   

38. We have, α (a × b) + β (b × c) + γ (c × a) = 0 Taking dot product with c, we have a [a b c] + b [b c c] + g [c a c] = 0 i.e., a [a b c] + 0 + 0 = 0 i.e., a [a b c] = 0 Similarly, taking dot product with b and c, we have γ [a b c] = 0, b [a b c] = 0 Now, even if one of α, β, γ ≠ 0, then we have [a b c] = 0  ⇒  a, b, c are coplanar. 39. Since p × q = r ∴ p × ( p × q) = p × r ⇒ ( p ⋅ q) p − ( p ⋅ p) q = p × r

But the bisector is - i + j - k.

⇒ cp − ( p) 2 q = p × r

3 4   ∴ t  x +  i + t  y +  j + tz = − i + j − k   5 5

⇒ ( p) 2 q = cp − p × r

1 3 3t + 5 , Þ x=- - =5t t 5

1 4 5 - 4t 1 y= - = and z = t 5 5t t

⇒ q=

cp − p × r . | p |2

 40. r × a = b × a ⇒ ( r − b) × a = 0 ⇒ r − b is paralllel to a.

Vector Algebra  13.37 \ r - b = l a i.e. r = b + l a (1) Similarly, r × b = a × b can be written as r = a + m b (2) For point of intersection of the two lines (1) and (2), we get b + l a = a + m b Þ l = m = 1 Hence, the required point of intersection is given by r = a + b = i + j + 2i − k = 3i + j − k . 41. Let a = 2i + 4 j - 5k , b = i + 2 j + 3k ∴ diagonals of the parallelogram are

p = a + b and q = b - a i.e., p = 3i + 6 j - 2k and, q = - i - 2 j + 8k ∴ unit vectors along the diagonals are - i - 2 j + 8k and 9 + 36 + 4 1 + 4 + 64 3i + 6 j - 2k - i - 2 j + 8k i.e., and 7 69 ∴ unit vector parallel to one of the diagonals is 1 (3i + 6 j - 2k ) 7

3i + 6 j - 2k

Previous Year’s Questions

ˆ ˆ ˆ ˆ a ⊥ (i − j ) ⇒ a ⋅ (i − j ) = 0 ⇒ ( xiˆ + yjˆ ) ⋅ (iˆ − ˆj ) = 0 ⇒ x − y = 0 ⇒ x = y   ∴ a = xiˆ + xjˆ and | a | − x 2 + x 2 = x 2 a x(iˆ + ˆj ) Required unit vector = = |a| x 2 1 ˆ ˆ ∴ = (i + j ) 2

43. The vector, iˆ + xjˆ + zkˆ if it is doubled in magnitude becomes 4iˆ + ( 4 x − 2) ˆj + 2kˆ ∴ 2 | iˆ + xjˆ + 3kˆ | = | 4iˆ + (4 x − 2) ˆj + 2kˆ | ⇒ 2 1 + x 2 + 9 = 16 + ( 4 x − 2) 2 + 4 ⇒ 40 + 4x2 = 20 + (4x - 2)2 ⇒ 3x2 - 4x - 4 = 0 ⇒ 3x2 - 6x + 2x - 4 = 0 ⇒ 3x (x - 2) + 2(x - 2) = 0 ⇒ (x - 2) (3x + 2) = 0 2 ⇒ x = 2, − 3    44. Key Idea: If the vectors a , b and c represent the sides of a    triangle, then a + b + c = 0    ∵ a+b+c =0    ⇒ a + b = −c      ⇒ (a + b ) × c = − c × c

    ⇒ a×c +b ×c =0     ⇒ b ×c =c ×a     Similarly, a × b = b × c       Hence a × b = b × c = c × a.      45. Given that a = xi + yjˆ + zkˆ and b = ˆj are such that a, c and b form a right handed system. iˆ ˆj kˆ    ∴ c =b×a= 0 1 0 x y z ˆ − xkˆ = iz 46. The equation of a line through the center represented by vector ˆj + 2kˆ and normal to the given plane is  r = ˆj + 2kˆ + l (iˆ + 2 ˆj + 2kˆ ) (1) This meets the plane at a point for which we must have [( ˆj + 2kˆ ) + l (iˆ + 2 ˆj + 2kˆ )] ⋅ (iˆ + 2 ˆj + 2kˆ ) = 15 ⇒ 6 + 9λ = 15 ⇒ λ = 1 On putting λ = 1 in Eq. (i), we get  r = iˆ + 3 ˆj + 4 kˆ ∴ Centre of the circle is (1, 3, 4). a a2 1 1 a a2 47. b b 2 1 + 1 b b 2 = 0 c c2 1 1 c c2 a a2 1 ⇒ (1 + abc) b b 2 1 = 0 c c2 1 ⇒ abc = −1. Hence, (8) is the correct answer

HINTS AND EXPLANATIONS

42. Given two vectors lie in xy-plane. Hence a vector coplanar with them is    a = xi + yj

13.38  Chapter 13    48. ( a + 2b ) = t1c (1)    And b + 3c = t 2 a (2)   (1) - 2 ´ ( 2) Þ a (1 + 2t 2 ) + c ( -t1 - 6) = 0 Þ 1 + 2t 2 = 0 Þ t 2 = -1 / 2 & t1 = -6.   (since a and c are non-collinear) Putting of t1 and t2 in (1) and (2), we get  the value   a + 2b + 6c = 0.       49. Work done by the forces F1 and F2 is ( F1 + F2 ) ⋅ d where d is the displacement. Now, according to question F1 + F2 = (4iˆ + ˆj − 3kˆ ) + (3iˆ + ˆj − kˆ ) = 7iˆ + 2 ˆj − 4 kˆ and d = (5iˆ + 4 ˆj + kˆ ) − (iˆ + 2 ˆj + 3kˆ ) = 4iˆ + 2 ˆj − 2kˆ. Hence    ( F1 + F2 ) ⋅ d is 40. 50. Condition for given three vectors to be coplanar is 1 2 3 0 l 4 = 0 Þ l = 0, 1 / 2. 0 0 2l - 1

HINTS AND EXPLANATIONS

Hence given vectors will be non coplanar for all real values of λ. except for 0 and 1/2.     v ⋅u w⋅u     51. Projection of v along u and w along u is  and  |u| |u| respectively Now, according to question     v ⋅u w⋅u       ⇒ v ⋅ u = w ⋅ u ⋅ and v ⋅ w = 0  =  |u| |u|         | u - v + w |2 = | u |2 + | v |2 + | w |2 -2u × w = 14    Þ | u - v + w |= 14    1          1    52. ( a × b ) × c = | b || c | a ⇒ ( a ⋅ c ) b − (b ⋅ c )a = | b || c | a 3 3

53. By triangle law,    PA + AC + CP = 0    PB + BC + CP = 0 Adding, we get      PA + PB + AC + BC + 2CP = 0   Since AC = − BC   & CP = − PC    ⇒ PA + PB − 2 PC = 0.

 54. Distance between the line with equation r = 2iˆ - 2 ˆj + 3kˆ + l (iˆ − ˆj + 4 kˆ ) and the plane r ⋅ (iˆ + 5 ˆj + kˆ ) = 5 equivalently x + 5 y + z = 5 is equal to the perpendicular distance of point (2, –2, 3) from the plane i.e.,

2 − 10 + 3 − 5 1 + 52 + 1

=

10 3 3

.

 55. Let a = xiˆ + yjˆ + zkˆ, then  a × iˆ = zjˆ − ykˆ 2 2 2 ⇒ ( a × iˆ) = y + z Similarly ( a × ˆj ) 2 = x 2 + z 2 2 2 2 2 2 2 and ( a × kˆ ) = x + y ⇒ ( a × iˆ) = y + z Similarly ( a × kˆ ) 2 = x 2 + y 2

⇒ ( a × iˆ) 2 + ( a + ˆj ) 2 + ( a × kˆ ) 2 = 2( x 2 + y 2 + z 2 ) = 2a 2 . 56.

Given that a, b, c are in H.P. 2 1 1 ⇒ − − =0 b a c x y 1 + + = 0 a b c x y 1 ⇒ = = −1 2 −1 ∴ x = 1, y = -2 57. Vectors aiˆ + ajˆ + ckˆ, iˆ + kˆ and ciˆ + cjˆ + bkˆ are coplanar a a c Therefore, 1 0 1 = 0 ⇒ c 2 = ab c c b ∴ a, b, c are in G.P.       58. [l ( a + b ) l 2b l c ] = [a b + c b ] ⇒

l l 0 1 0 1 2 0 l 0 = 0 1 1 0 0 l 0 1 0

⇒ λ4 = –1 Hence no real value of λ.   59. Given a = iˆ − kˆ, b = xiˆ + ˆj + (1 − x ) kˆ and  c = yiˆ + xjˆ + (1 + x − y ) kˆ.      [abc ] = a ⋅ (b × c )

Vector Algebra  13.39

- ˆj ( x + x 2 - xy - y + xy ) + kˆ ( x 2 - y )   aˆ ⋅ (b × c ) = 1 Which does not depend on x and y.           60. Given condition a × b × c = a × b × c , a ⋅ b ≠ 0, b ⋅ c ≠ 0             ⇒ ( a ⋅ c ) b − b ⋅ c a = ( a ⋅ c ) b − a ⋅ b c       ⇒ a ⋅ b c = b ⋅ c a   ⇒ a || c

(

)

( ) ( ) ( )

( ) ( )

61. The vectors  BA = iˆ − 2 ˆj + 6 kˆ CA = (2 − a)iˆ + 2 ˆj CB = (1 − a)iˆ − 6 ˆj imply   CA ⋅ CB = 0 ⇒ ( 2 − a)(1 − a) = 0 ⇒ a = 2,1 62. Given that | 2uˆ × 3vˆ | = 1 ⇒ 6 | uˆ || vˆ ||sin q | = 1 1 Þ sin q = 6 Hence there is exactly one value of θ for which 2uˆ × 3vˆ is a unit vector.    63. a = iˆ + ˆj + kˆ, b = iˆ − ˆj + 2kˆ and c = xiˆ + ( x − 2) ˆj − kˆ

x Therefore, 1 1

x − 2 −1 1 1 =0 −1 2

⇒ 3x + 2 − x + 2 = 0 ⇒ 2 x = −4 ⇒ x = −2. 64. Let for some l , a = l (bˆ + cˆ), then iˆ + 2 ˆj + kˆ a iˆ + 2 ˆj + b kˆ = l ( ) 2 Þ l = 2a and l = 2 and l = 2 b Þ a = 1 and b = 1.     65. Since a = 8b , c = −7b .

    ∴ a and b are like vectors and b and c are unlike.   ⇒ a and c will be unlike   Hence, the angle between a and c = p .    66. (3 p 2 − pq + 2q 2 )[u v w ] = 0       But since u , v , w are non-coplanar, [u v w ] ≠ 0 which implies that 3p2 − pq + 2q2 = 0 7q 2 q Þ 2 p 2 + p 2 - pq + ( ) 2 + =0 2 4 q 7 Þ 2 p2 + ( p - )2 + q2 = 0 = 0 2 4 q 2 This is possible only when p = 0, q = 0 and therefore exactly one value of (p, q). ⇒

p = 0, q = 0, p =

67. Projection of a vector on coordinate axes are x2 − x1, y2 − y1, z2 − z1 So, ( x2 − x1 ) 2 + ( y2 − y1 ) 2 + ( z2 − z1 ) 2 = 36 + 9 + 4 = 7 6 3 2 The D.C’s of the vector are , − , respectively. 7 7 7    68. c = b × a   ⇒ b ⋅c = 0 ⇒ (b1iˆ + b2 ˆj + b3kˆ ) ⋅ (iˆ − ˆj − kˆ ) = 0 ⇒ b1 − b2 − b3 = 0   and a ⋅ b = 3 ⇒ b2 − b3 = 3 ⇒ bl = b2 + b3 = 3 + 2b3 ⇒ b = (3 + 2b3 )iˆ + (3 + b3 ) ˆj + b3kˆ   69. a ⋅ b = 0,   b ⋅ c = 0,   c ⋅a = 0 Þ 2l + 4 + m = 0l - 1 + 2 m = 0 Solving we get: l = -3, m = 2.

70. ( 2a - b ).{( a ´ b ) ´ ( a + 2b )} = ( 2a - b ).{2a - b ){[a .( a + 2b )]b - [b .( a + 2b )a ]} = -5( a )2(b )2 + 5( ab )2 = -5 71. b × c = b × d ⇒ a × (b × c ) = a × (b × d ) ⇒ ( a .c )b − ( a .b )c = ( a .d )b − ( a .b )d ⇒ ( a .c )b − ( a .b )c = −( a .b )d a .c ∴ d = c −( )b a .b

HINTS AND EXPLANATIONS

iˆ ˆj kˆ   ⇒ b ´ c = x 1 1 - x = iˆ (1 + x - x - y - x 2 ) y x 1+ x - y

13.40  Chapter 13        72. c ⋅ d = 0 ⇒ 5 | a |2 + 6 a ⋅ b − 8 | b |2 = 0     1   p Þ 6a × b = 3 Þ a × b = Þ (a × b ) = 2 3   on q p 73. AE = vector component of   ( p ⋅ q)  AE =   p ( p ⋅ q)    ∴ From ∆ABE ; AB + BE = AE      ( p ⋅ q) p ⇒ q+r =   ( p ⋅ q)     ( p × q)  Þ r =-q+   p ( p × q)

HINTS AND EXPLANATIONS

74. Let M be the median, then    AB + AC AM = 2  ˆ AM = 4i = − ˆj + 4 kˆ  ∴ | AM | = 16 + 16 + 1 = 33       75. [a ´ bb ´ cc ´ a ] = l [abc ]2 λ=1      1    76. Since ( a × c )b - (b × c )a = | b || c | a 3  1   ⇒ − b ⋅ c = | b || c | 3   1   Þ - | b || c | cosq = | b || c | 3 1 Þ cosq = 3 Þ sin q =

2 2 . 3

ö æ ö æ  77. Given expression is ç a × c - 3 ÷ b - ç a × b + 3 ÷ c = 0 2 ø 2 ø è è   Þ a × b = cosq = - 3 / 2 Þ q = 5p / 6    78. | ( a × b ) × c | = 3   ˆ ˆ ˆ a × b = 2i − 2 j + k    ⇒ | a × b || c | sin 30° = 3    |a|=3=|a×b|  ⇒ | c | = 2   | c − a |= 3     ⇒ | c |2 + | a |2 −2( a . c ) = 9

  9−3− 2 a .c = =2 2

Hence, the correct option is (A)

   79. If u is coplanar with a = 2iˆ + 3 ˆj − kˆ and b × ˆj + kˆ, then    u ⋅ (a × b ) = 0 iˆ ˆj kˆ   a × b = 2 3 −1 0 1 1 = iˆ[3 + 1] − ˆj[2 − 0] + kˆ[2 − 0] = 4iˆ − 2 ˆj + 2kˆ



   u ⋅ ( a × b ) = 4u1 − 2u2 + 2u3 = 0 (i)  Where u = u1iˆ + u2 ˆj + u3kˆ   u is perpendicular to a   Hence, u ⋅ a = 0 2u1 + 3u2 – u3 = 0 (ii)   u ⋅b = 0 u2 + u3 = 24 (iii) Equation (ii) + equation (iii) 2u1 + 4u2 = 24 (iv) Equation (i) – 2(equation (iii)) 4u1 – 4u2 = –48 u1 – u2 = –12 (v) Solving equation (iv) and equation (v), we get u2 = 8 u1 = –4 u3 = 16  u = −4iˆ + 8 ˆj + 16 kˆ

2 u = 4 2 + 82 + 16 2 = 336

r

r

80. As a + b are collinear

l -2 1 = \l=–4 4l - 2 3 81.

C (4, 5)

D m (2, 4)

•

y=x-3 A (0, 3) Slope of BC = 1 Þ

B (h, 3 - h)

3- h-5 = 1 h-4

B º (1, 2) and D = (3, 6)

\h=1

CHAPTER

14

Measures of Central Tendency and Dispersion

LEARNING OBJECTRIVES After reading this chapter, you will be able to:  Learn about the measures of central tendency and arithmetic mean  Know about geometric mean, harmonic mean, median and quartiles

 Understand deciles and percentiles and mode and symmetric distribution

MEASURES OF CENTRAL TENDENCY

Short Cut Method

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:

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 arithmetic mean is given by

1. Arithmetic Mean 2. Geometric Mean 3. Harmonic Mean 4. Median 5. Mode

Step Deviation Method

Mean of Unclassified Data Let x1, x2, ...., xn be n observations, then their arithmetic mean is given by, x1 + x2 + ..... + xn 1 n = ∑ xi n n i =1

Mean of Grouped Data Let x1, x2, x3, ...., xn be n observations and let f1,  f2, . . . , fn be their corresponding frequencies, then their arithmetic mean is given by n

f x + f 2 x2 + ... + f n xn x = 1 1 = f1 + f 2 + .... + f n

∑ fi di ∑ fi

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. 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:

ARITHMETIC MEAN

x =

x =A+

∑ fi xi

i =1 n

∑

i =1

fi

∑ fi di

× h; N = ∑ fi N where A = assumed mean, x −A di = i = the deviation of any variate from A, h h = the width of the class - interval. x =A+

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: Weighted A.M. =

w1 x1 + w2 x2 + ...... + wn xn w1 + w2 + ....... + wn

14.2  Chapter 14

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

where, x12 x1 x2 n1 n2

n x + n2 x2 = 1 1 n1 + n2 = Combined mean of the two data sets 1 and 2 = Mean of the first data = Mean of the second data = Size of the first data = Size of the second data

(

G.M. = x1f ⋅ x2f ....... xnf where

N=

1

2

)

1/ N

n

,

n

∑ fi

i =1



⎛ n ⎞ ⎜ ∑ fi log xi ⎟ i =1 ⎟ or G.M. = antilog ⎜ ⎜ ⎟ N ⎜ ⎟ ⎜⎝ ⎟⎠



I M P O R TA N T P O I N T S

Properties of A. M. 1. In a statistical data, the sum of the deviations of individual values from A.M. is always zero, i.e.,

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.

n

∑ fi ( xi − x )

= 0 i =1 where fi is the frequency of xi (1 ≤ i ≤ n) 2. In a statistical data, the sum of squares of the deviations of individual values from A.M. is least, i.e., n

∑ fi ( xi − x )2

is least.

Some Points About Geometric Mean • It is based on all items of the series. • It is most suitable for constructing index number, average ratios, percentages etc. • G.M. cannot be calculated if the size of any of the items is zero or negative.

i =1

3. If each of the n given observations is doubled, then their mean is doubled. 4. 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 .

Some Points About Arithmetic Mean • Of all types of averages, the arithmetic mean is most commonly used average. • It is based upon all observations. • If the number of observations is very large, it is more accurate and more reliable basis for comparison.

The harmonic mean of n observations 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. =

GEOMETRIC MEAN If x1, x2, x3, ..., xn are n observations, none of them being zero, then their geometric mean is defined as G.M. = ( x1 ⋅ x2 ⋅ x3 ... xn

HARMONIC MEAN

1 )n

⎛ 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

∑ fi

i =1 n

⎛ f ⎞

∑ ⎜⎝ xi ⎟⎠

i =1

i

Some Points About H.M. • It is based on all item of the series. • This is useful in problems related with rates, ratios, time etc. • A.M. ≥ G.M. ≥ H.M. and also (G.M.)2 = (A.M.) (H.M.). • A.M. gives more weightage to larger values whereas G.M. and H.M. give more weightage to smaller values.

Measures of Central Tendency and Dispersion  14.3

Relation Between AM, GM and HM The arithmetic mean (AM), geometric mean (GM) and harmonic mean (HM) for a given set of observations are related as under: AM ≥ GM ≥ HM Equality sign holds only when all the observations are equal.

SOLVED EXAMPLES 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 (B) 76.54 (C) 81.32 (D) 78.4 Solution: (A) We know that the mean is given by

∑x

∑ x = nx



x =

Here

x = 78.4, n = 25

\

Sx = 25 × 78.4 = 1960

n

or

But this Sx is incorrect as 96 was misread as 69. Correct Sx = 1960 – 69 + 96 = 1987 \

Correct mean =

1987 = 79.48. 25

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 (B)  170 cm (C)  156 cm (D)  None of these Solution: (C) Total height of 15 students = Sx = 154 × 15 = 2310 cm. It was found that 175 cm was wrongly read as 145 Correct sum = 2310 – 145 + 175 = 2340 cm. 2340 Correct mean = = 156 cm. 15 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 (B) 0.0166 (C)  0.0666 (D)  None of these

Solution: (A) Here x = 6, x1 = 8, n1 = 40, n2 = 60. Assuming that the total output is 100, we are required to find out x2, we know that n x + n2 x2 40 × 8 + 60 × x2 x = 1 1 = n1 + n2 40 + 60 320 + 60 x2 100 600 − 320 280 14 ⇒ x2 = = = = 4.66. 60 60 3 Thus, the average profit in womens shirt is 4.66 % of sales or Re 0.0466 per sale rupee. ⇒ 6 =

4. The weighted mean of the first n natural numbers if their weights are the same as the numbers, is n +1 2n + 1 (B)  3 3 2n − 1 (C)  (D)  None of these 3 Solution: (B) Here the numbers are 1, 2, 3, ....... , n and their weights also are respectively 1, 2, 3, ....... , n. So weighted (A) 



Aw =



=



=

∑ wx ∑w

=

1.1 + 2.2 + 3.3 + ...... + n.n 1 + 2 + 3 + ...... + n

12 + 22 + 32 + ..... + n2 1 + 2 + 3 + ..... + n n( n + 1) ( 2n + 1) 2n + 1 = n( n + 1) 3 6. 2

5. 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 (B) 5 (C)  6 (D)  None of these Solution: (C) Here mean A = 5. Let the frequency of 8 be x. Then by the formula A = 5 =

∑ xf ∑f



1.2 + 2.3 + 4.3 + 6.2 + 8. x 32 + 8 x = 2+3+3+ 2+ x 10 + x

or 18 = 3x; \ x = 6.

14.4  Chapter 14 6. 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 (B)  38, 64 (C)  72, 48 (D)  None of these Solution: (C) We know that the combined mean n A + n2 A2 A = 1 1 (1) n1 + n2 Here A1 = mean weight of boys = 60 kg.

8. 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 (B)  18.1 years (C)  19.1 years (D)  None of these Solution: (B) Here we have three collections for which A1 = 24, n1 = 8; A2 = 20, n2 = 7 and A3 = 6, n3 = 5. Their combined mean is the required mean. By the formula A=



8 × 24 + 7 × 20 + 5 × 6 A= 8+7+3

A2 = mean weight of girls = 50 kg. A = combined mean = 56 kg. and n1 + n2 = 120

\ (2)

So, from (1) and (2), n1 ⋅ 60 + n2 ⋅ 50 ; 120 \ 56 × 120 = n1 ⋅ 60 + (120 – n1) 50; 56 = \

120 (56 – 50) = 10n1;

\ n1 = 72, n2 = 48. Thus, the number of boys = 72 and the number of girls = 48. 7. 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 (B) 12 (C)  18 (D)  None of these

192 + 140 + 30 362 = = = 18.1 20 20 \ The mean age of the family = 18.1 years. 9. The mean of 100 items is 50 and their S.D. is 4. The sum of all the items and also the sum of the squares of the items is (A)  5000, 251600 (B)  4000, 251600 (C)  5000, 261600 (D)  None of these Solution: (A) Here n = 100, A = 50, σ = 4. ∑ x ; \ x = nA = 100 × 50 = 5,000. Now, A = ∑ n Again, from the formula,

Solution: (A) Let the mean of the last four be A2. Then by the formula for combined mean, 12.5 =

6 × 15 + 4 × A2 ; 6+4

or 125 = 90 + 4 A2; 35 \ A2 = 4

Let the sixth number = x ; then taking the sixth number as a collection, the combined mean of this collection and the collection of the last four is 10, by question. \ By definition of combined mean 35 1× x + 4 × 4 ; 10 = 1+ 4 \ 50 = x + 35; \ x = 15. \

Sixth number = 15.

n1 A1 + n2 A2 + n3 A3 n1 + n2 + n3

σ 2 + A2 = \

∑ x2

∑ x 2 , we get n

∑ x2

= n(σ 2 + A2)

= n(σ 2 + A2) = 100(16 + 2500) = 2,51,600

10. If the mean of the set of number x1, x2, ........, xn is x , then the mean of the numbers xi + 2i, 1 ≤ i ≤ n is (A)  x + 2n

(B)  x +n+1

(C)  x + 2

(D)  x + n.

Solution: (B) n

x =

⇒

n

∑ xi

i =1

∑ xi

i =1

n

= nx



Measures of Central Tendency and Dispersion  14.5 n

\

∑ ( xi + 2i)

i =1

n



=

∑ xi + 2 (1 + 2 + .... + n)

n n ( n + 1) nx + 2 2 = = x + (n + 1). n

a + 4 X1 nM − a ⇒ X1 = . ( n − 4) + 4 4

12. The weighted mean of first n natural numbers whose weights are equal to the squares of corresponding numbers is (A) 

n +1 3n ( n + 1) (B)  2 2 ( 2n + 1)

(C) 

( n + 1) ( 2n + 1) 6

(D) 

n ( n + 1) 2

Solution: (B) Weighted Mean =





\ New mean =

i =1

11. The A.M. of n observations is M. If the sum of n – 4 observations is a, then the mean of remaining 4 observations is nM – a nM – a (A)  (B)  4 2 nM – a (C)  (D)  nM + a 4 Solution: (A) Let the mean of the remaining 4 observations be X1 . Then, M =

New Sum = 4900 – 110 + 210 = 5000

n

1.12 + 2.22 + ...... + n.n2 12 + 22 + ....... + n2

n ( n + 1) n ( n + 1) Σn3 2 2 = = 2 1 2 ( + ) ( + 1) n n n Σn 6 3n ( n + 1) . = 2 ( 2n + 1)

13. Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is 1 (A) 48 (B) 82 2 (C) 50 (D) 80 Solution: (C) Sum of 100 items = 49 × 100 = 4900 Sum of items added = 60 + 70 + 80 = 210 Sum of items replaced = 40 + 20 + 50 = 110

5000 = 50 100

14. The mean of n items is X . If the first item is increased by 1, second by 2 and so on, then the new mean is n (A)  X + n (B)  X + 2 n +1 (C)  X + (D)  None of these 2 Solution: (C) Let x1, x2, ...xn be n items. 1 Then X = Σxi . n Let

y1 = x1 + 1, y2 = x2 + 2

y3 = x3 + 3, ..... , yn = xn + n Then the mean of the new series is 1 1 1 1 Σy = Σ( xi + i ) = Σxi + (1 + 2 + 3 + ..... + n) n i n n n = X+

1 n( n + 1) n +1 ⋅ = X+ . 2 n 2

15. The number of observations in a group is 40. If the average of first 10 is 4.5 and that of the remaining 30 is 3.5, then the average of the whole group is 15 1 (A)  (B)  4 5 (C) 8 (D) 4 Solution: (A) ⇒

x1 + x2 + ..... + x10 = 4.5 10 x1 + x2 + ..... + x10 = 45

Also,

x11 + x12 + ..... + x40 = 3.5 30

⇒

x11 + x12 + ..... + x40 = 105

\

x1 + x2 + ..... + x40 = 150

x1 + x2 + ..... + x40 150 15 = = 40 40 4 16. 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 (B) 2.5 (C)  1.92 (D)  None of these \

14.6  Chapter 14 Solution: (C) Since we are given rate per rupee, harmonic mean will give the correct answer 4 48 4 × 12 H.M. = = = 1 1 1 1 25 12 + 6 + 4 + 3 + + + 1 2 3 4 = 1.92 kg per rupee. 17. The A.M. of 2nC0, 2nC2, 2nC4,..., 2nC2n is (A) 

2 2 (B)  ( n + 1) ( n + 1)

(1 + x)2n = 2nC0 + 2nC1x + 2nC2x2 + 2nC3x3 +, ..., + 2nC2nx2n Put x = – 1 2n

C0 – 2nC1 + 2nC2 – 2nC3 +, ..., + 2nC2n = 0

(1)

Now put x = 1 2n

C0 + 2nC1 + 2nC2 + 2nC3 +, ..., + 2nC2n = 22n(2)

Adding (1) and (2), we get 2n

C0 + 2nC2 +, ..., + 2nC2n = 22n–1 22 n −1 ( n + 1)

18. The A.M. of 2n+1C0, 2n+1C1, 2n+1C2, ... , 2n+1Cn is 2n 2n (B)  n n +1

(C) 

22n 22n (D)  n ( n + 1)

Solution: (D) 2n+1

C0 + 2n+1C1 + 2n+1C2 +, ...,

+ 2n+1C2n + 2n+1C2n+1 = 22n+1 2n+1

C0 = 2n+1C2n+1, 2n+1C1 = 2n+1C2n,...

Now

Cr = 2n+1C2n–r+1

or

C0 +

2 n +1

or 

C0 +

2 n +1

C1 +

2n+1

2 n +1

C2 +, ..., +

C1 + C2 +, ..., + n +1

20. In a factory, workers work in three shifts say shift 1, shift 2 and shift 3 and they get wages in the ratio 4 : 5 : 6 depending on the shift 1, 2 and 3 respectively. Number of workers in the shifts are in the ratio 3 : 2 : 1. If total number of workers working is 1500 and wages per worker in Ist shift is Rs 400. Then mean wage of a worker is (A)  Rs. 467 (B)  Rs. 500 (C)  Rs. 600 (D)  Rs. 400 Solution: (A) Workers in Ist shift = 750 Wages in Ist shift = Rs. 400 Workers is IInd shift = 500 Wages in IInd shift = Rs. 500 Workers in IIIrd shift = 250 Wages in IIIrd shift = Rs. 600 750 × 400 + 500 × 500 + 250 × 600 \ Mean = 1500 Rs. 467 per worker 21. If a variable takes values 1, 2, 3, ... , n with frequencies 12, 22 , ... , n2, then the mean is Σn3 (A) Sn (B)  Σn 2 Σn3 (C)  (D)  None of these Σn Solution: (B) 12 + 22 + ... + n2

So sum of first (n + 1) terms = Sun of last (n + 1) terms 2n+1

n n ⋅ 2n−1 = n 2 2

1.12 + 2 . 22 + ... + n . n2

2n+1

2n+1

C0 + nC1 + nC2 + ... + nCn =

Solution: (C)

(A) 

C1 + 2 ⋅ nC2 + 3 nC3 + ... + n nCn n

22 n −1 2n −1 (C)  (D)  ( n + 1) ( n + 1)

A.M. of 2nC0, 2nC2 +, ..., + 2nC2n =

2n 2n +1 (B)  ( n + 1) n( n + 1) n +1 n (C)  (D)  2 2 Solution: (D) (A) 

n

2n

n

19. The mean of the values 0, 1, 2, 3, ... , n with the ­corresponding weights nC0, nC1,..., nCn respectively is

2n+1

2n

Cn = 2

2 n +1

Cn =

22n ( n + 1)

=

Σn3 Σn 2

22. Mean of n items is x. If these n items are increased by 12, 22, 32, ..., n2 successively, then mean gets increased by ( n + 1) ( 2n + 1) n( n + 1) ( 2n + 1) (A)  (B)  6 6 2 n (C)  (D)  Remains same 2

Measures of Central Tendency and Dispersion  14.7 Solution: (A)  ⇒ ⇒

x1 + x2 + x3 + ... + ... + xn n 2 x1 + 1 + x2 + 22 + x3 + 32 + ... + xn + n2 x′ = n x1 + x2 + ... + xn 12 + 22 + ... + n2 + x′ = n n ( n + 1) ( 2n + 1) x′ = x + 6 x =

MEDIAN Median is the middle most or the central value of the variate in a set of observations, when the observations are arranged either in ascending or in descending order of their magnitudes. It divides the arranged series in two equal parts.

Calculation of Median Median of an Individual Series Let n be the number of observations. 1. Arrange the data in ascending or descending order. 2. (A) If n is odd, then 1 Median = value of the (n + 1)th observation 2 (B) If n is even, then ⎛n ⎞ ⎛ n⎞ Median = mean of the ⎜ ⎟ th and ⎜ + 1⎟ th ⎝2 ⎠ ⎝ 2⎠ observation.

Median of a Discrete Series 1. Arrange the values of the variate in ascending or descending order. 2. Prepare a cumulative frequency table. 3. (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 ⎜ ⎟ ⎜⎝ ⎟⎠

Median of a Continuous Series 1. Prepare the cumulative frequency table.

⎛ n⎞ 2. Find the median class, i.e., the class in which the ⎜ ⎟ ⎝ 2⎠ th observation lies.

3. The median value is given by the formula ⎛ ⎛ n⎞ ⎜ ⎜⎝ 2 ⎟⎠ − c f Median = l + ⎜ 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

Some Points About Median • It is an appropriate average in dealing with qualitative data, like intelligence, wealth etc. • The sum of the deviations of the items from median, ignoring algebraic signs, is less than the sum from any other point.

QUARTILES, DECILES AND PERCENTILES Quartile Just as the median divides a set of observations (when arranged in ascending or descending order of magnitudes), into two equal parts, similarly Quartile divides the observations into four equal parts. The value of the item midway, between the first item and the median is known as first or lower quartile 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 is denoted Q3. The median is known as the Second Quartile and is denoted by Q2. The methods for finding the values of Q1 and Q3 are similar to that of the median. In the case of ungrouped data, when arranged in ascending or descending order of magnitudes Q1, Q3 can be obtained as follows: Q1 =

n +1 3( n + 1) th item, Q3 = th item. 4 4

For a frequency distribution, Q1 and Q3 are given by, Q1 = l + Q3 = l +

[( n / 4) − C f ] f [(3n / 4) − C f ] f

× h, × h,

where l = lower limit of the class in which a particular quartile lies, f =  frequency of the class-interval in which a ­particular quartile lies,

14.8  Chapter 14 i = class-interval of the class in which a particular quartile lies, cf = cumulatively frequency of the class preceding the class in which the particular quartile lies. [( nh / 4) − c f ] In general, Qi = l + × h, i = 1, 2, 3, 4 f

Decile The value of the variable which divides the series, when arranged in ascending or descending order, into 10 equal parts is called decile. There are 9 deciles denoted by D1, D2 ... D9. When the series is ungrouped the deciles are calculated as follows: n×h , i = 1, 2, ..., 9 10 When the data is classified or grouped, Di =

Di = l +

[( nh / 10) − c f ] f

×h

where symbols have their usual meaning.

Percentile The value of the variable which divides 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: n×h , h = 1, 2, ..., 99. 100 When the data is classified or grouped, the percentiles are calculated by the formula Pi =

Pi = l +

[( nh / 100) − c f ] f

× h, i = 1, 2, ...., 99.

where symbols have their usual meanings.

SOLVED EXAMPLES 7 23. If a variable takes the discrete values a + 4, a – , 2 5 1 1 a ­ – ←, a – 3, a – 2, a + , a – , a +5 (a > 0), 2 2 2 then the median is 5 1 (A) a – (B)  a– 4 2 5 (C) a – 2 (D)  a+ 4

Solution: (A) 7 5 1 Arrange the data as a – , a – 3, a – , a – 2, a – , 2 2 2 1 a + , a + 4, a + 5. 2 1 5 α −2+α − 2α − 2 2 = a – 5. Median = = 2 2 4 24. Median of 2nC0, 2nC1, 2nC2, 2nC3,..., 2nCn (when n is even) is 2n (A) 2nCn/2 (B)  C(n+1)/2 2n (C)  C(n–1)/2 (D)  None of these Solution: (A) 2n C0, 2nC1, 2nC2, ... , 2nCn is odd number of binomial coefficients (when n is even) and middle binomial coefficient is 2nCn/2. 25. Median of 2nC0, 2nC1, 2nC2, 2nCn (when n is odd) is

(

1 2n C( n −1)/ 2 + 2 (B) 2nCn/2 (C) 2nCn (D)  None of these (A) 

2n

C( n +1)/ 2

)

Solution: (A) 2n C0, 2nC1, 2nC2, ... , 2nCn is even number of binomial coefficients (when n is odd), and then middle terms are 2nCn–1/2 and 2nCn+1/2. So median is 2n

Cn −1/ 2 + 2

2n

Cn +1/ 2

.

5 26. If a variable takes the discrete values a + 4, α − 2 7 1 1 α − , a – 3, a – 2, α + , α − , a + 5 (a > 0), 2 2 2 then the median is 1 5 (B)  α− 2 4 5 (C) a – 2 (D)  α + 4 Solution: (B) 1⎛ 1⎞ Median = ⎜ α − 2 + α − ⎟ 2⎝ 2⎠ 5 =α− 4 (A)  α −

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.

Measures of Central Tendency and Dispersion  14.9

Computation of Mode

The commonly used measures of dispersion are:

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.

Mode of Discrete Series In the case of discrete frequency distribution, mode is the value of the variate corresponding to the maximum frequency.

Mode of Continuous Series 1. 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. 2. The mode is given by the formula Mode = l +

f m − f m −1 2 f m − f m −1 − f m + 1

× h,

where l = the lower limit of the modal class h = the width of the modal class fm – 1 =  the frequency of the class preceding modal class fm = the frequency of the modal class fm +1 = the frequency of the class succeeding modal class In case, the modal value lies in a class other than the one containing maximum frequency, we take the help of the following formula; f m +1 ×h , Mode = l = f m −1 + f m + 1 where symbols have usual meaning.

1. Range 2. Quartile Deviation or Semi-interquartile range 3. Mean Deviation 4. 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, Coefficient of range =

Quartile deviation Quartile deviation or semi-interquartile range is given by 1 Q.D. = (Q3 – Q1) 2 Q − Q1 Coefficient of Q.D. = 3 Q3 + Q1

Mean deviation For a frequency distribution, the mean deviation from an average (median, or arithmetic mean) is given by, n

M.D. =

Measures of Dispersion The degree to which numerical values in the set of ­values tend to spread about an average value is called the d­ ispersion or variation.

i =1

xi − x

n

∑ fi



Mean deviation Coefficient of M.D. = Corresponding average

Standard deviation The standard deviation of a statistical data is defined as the positive square root of the squared deviations of observations from the A.M. of the series under consideration. 1. Standard deviation (also denoted by σ ) for ungrouped set of observations is given by

SYMMETRIC DISTRIBUTION A distribution is a symmetric distribution if the values of mean, mode and median coincide. In a symmetric distribution frequencies are symmetrically distributed on both sides of the centre point of the frequency curve.

∑ fi

i =1

Some Points about Mode • It is not based on all items of the series • It is not necessary that a distribution has unique mode. • As compared to other averages mode is affected to a large extent by fluctuations of sampling. • It is not suitable in a case where the relative importance of items have to be considered.

L−S . L+S

n

∑ (xi − x )2



σ=

i =1

n

2. Standard deviation for frequency distribution is given by, n

∑ fi (xi − x )2

S.D. =

i =1

N

14.10  Chapter 14 where fi is the frequency of xi (1 ≤ i ≤ n). When the values of the variable are given in the form of classes, then their respective mid-points are taken as the values of the variable.

Standard Deviation of n Natural Numbers ⎛1 ⎞ σ = ⎜ ( n2 − 1)⎟ ⎝ 12 ⎠

C.V. =

The coefficient of variation is also represented as percentage. The square of S.D. is called the variance of the distribution and is denoted by σ 2.

Computation of Standard Deviation

1/ 2

Direct Method

Standard deviation shows the limits of variability by which the individual observation in a distribution will vary from the mean. For a symmetrical distribution with mean x , the following area relationship holds good: x ± σ covers 68.27 % observations.

σ=

These limits are illustrated by the following curve known as Normal Curve.

2

68.27%

where N = x + 3σ

∑ d2 n

⎛ ∑ fd ⎞ −⎜ ⎟ , for grouped data ⎝ N ⎠

N

∑f.

Step-deviation Method σ =h

Empirical relationships

4 S.D. = 5 M.D. = 6 Q.D

2

∑ fd 2

FIG. 14.1

If the data is moderately non-symmetrical, then the following empirical relationships hold: 4 Mean deviation = σ 5 2 Semi-Inter-quartile range = σ 3 2 Probable error of standard deviation = σ 3 = Semi-inter-quartile range. 5 Quartile deviation = M.D. 6 From these relationships, we have

2

⎛ ∑d⎞ −⎜ ⎟ , for ungrouped data ⎝ n ⎠

where A is assumed mean and d = x – A.

σ=

x + 2σ

n

⎛ ∑ x⎞ −⎜ ⎟ ⎝ n ⎠

Short-cut Method

x ± 3σ covers 99.73 % observations.

x – 3σ x – 2σ

∑ x2

σ=

x ± 2σ covers 95.45 % observations.

95.44% 99.73% x – 1σ x x + 1σ

σ × 100 x

∑ fd ′ 2 N

2

⎛ ∑ fd ′ ⎞ x−A −⎜ ⎟ ; d′ = h ⎝ N ⎠

Combined Standard Deviation Let A1 and A2 be two series having n1 and n2 observations respectively. Let their A.M. be x1 and x2 , and standard deviations be σ1 and σ2. Then the combined standard deviation σ or σ12 of A1 and A2 is given by



Coefficient of S.D. (C.V.)

where

For comparing two or more series for variability, the ­relative measure, called coefficient of variation (C.V.) is used. This measure is defined as

and

σ12 or σ =

n1σ12 + n2σ 22 + n1d12 + n2 d22 n1 + n2



=

n1 (σ12 + d12 ) + n2 (σ 22 + d22 ) n1 + n2



d1 = ( x1 − x12 ), d2 = ( x2 − x12 ), x12 =

n1 x1 + n2 x2 is the combined mean. n1 + n2

Measures of Central Tendency and Dispersion  14.11

I M P O R TA N T P O I N T S Quartile deviation is less affected by extreme values of the series. ■ Mean deviation is based on all the items of series. It is therefore more representative than the range or quartile deviation. ■ Mean deviation from the median is less than that measured from any other mean. 2 ■ Standard deviation ≤ Range i.e., variance ≤ (Range) . ■

S.D. of first n natural numbers is

■

n2 − 1 . 12

SOLVED EXAMPLES 27. The coefficient of variation of two series are 58% and 69%. If their standard deviations are 21.2 and 15.6, then their A.Ms are (A)  36.6, 22.6 (B)  34.8, 22.6 (C)  36.6, 24.4 (D)  None of these Solution: (A) We know that

σ × 100 x σ x = × 100 C.V.

C.V. = or

\ Mean of first series =

21.2 × 100 = 36.6 58

Mean of second series =

15.6 × 100 = 22.6 69

29. The mean deviation from the mean for the set of ­observations – 1, 0, 4 is (A)  less than 3 (B)  less than 4 (C)  greater than 2.5 (D)  greater than 4.9 Solution: (A and B) −1 + 0 + 4 = 1. 3 1 \ Mean Deviation = (| – 1 – 1| + | 0 – 1 | + | 4 – 1 | ) 3 =2 x =

30. If the S.D of a set of observations is 4 and if each observation is divided by 4, the S.D of the new set of observations will be (A) 4 (B) 3 (C) 2 (D) 1 Solution: (D) We know that if y = x/h when σ y = σx /| h |. \ The S.D. of new set of observations will be 4/4 = 1. 31. 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 (B) 5.58 (C)  3.42 (D)  None of these Solution: (A) Here n1 = 35, x1 = 80, σ1 = 4, n2 = 65, x2 = 70, σ 2 = 3. \

x12 =

28. Mean deviation of the series a, a + d, a + 2d, a + 2nd from its mean is



σ12 =

( n + 1) d nd (B)  ( 2n + 1) 2n + 1



=

(A) 

n ( n + 1) d (C)  ( 2n + 1)

( 2n + 1) d (D)  n ( n + 1)

Solution: (C)

2n + 1 .( a + a + 2nd ) 2 x = = a + nd ( 2n + 1)

Σ x − x = 2d (1 + 2 + ......+ n) = n (n + 1) d

\ M.D. =

n ( n + 1)d 2n + 1

35 × 80 + 65 × 70 = 73.5. 35 + 65 ⎡ 35(16 + 42 × 25) + 65(9 + 12 × 25) ⎤ ⎢ ⎥ 100 ⎣ ⎦ 34.21 = 5.85

32. If μ is the mean of a distribution, then ∑ fi ( yi − μ ) is equal to (A) M.D. (C)  0

(B) S.D. (D)  None of these

Solution: (C) We have,

μ=

Σfi yi Σf i

⇒

Σfi yi – S fi μ = 0

⇒

S fi (yi – μ) = 0



14.12  Chapter 14 33. The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6, then the other two are (A)  2 and 9 (B)  3 and 8 (C)  4 and 7 (D)  5 and 6

34. If 25 % of the items are less than 20 and 25 % are more then 40, the quartile deviation is (A) 20 (B) 30 (C) 40 (D) 10 Solution: (D)

Solution: (C)

Q.D. =

x1 = 4, N = 5

and

Σ(x − x ) 2 = 5.2 N

⇒

Σ(x - x ) 2 = (5.2) 5

\

Σ(x - x ) 2 = 26 2

2

35. The sum of squares of deviations for 10 observations taken from mean 50 is 250. The coefficient of variation is (A) 10 % (B) 40 % (C) 50 % (D)  None of these 2

\ (1 – 4) + (2 – 4) + (6 – 4) +  (a – 4)2 + (b – 4)2 = 26 where a, b are the other two observatios. \ 9 + 4 + 4 + (a – 4)2 + (b – 4)2 = 26 \ (a – 4)2 + (b – 4)4 = 9 Also, \

40 − 20 = 10. 2

1+ 2 + 6 + α + β = 4 5

a + b = 20 – 9 = 11

Clearly 4, 7 only satisfy the above equation in a, b. Hence reqd. numbers are 4, 7.

Solution: (A) Co-efficient of variation σ σ = × 100 = × 100 ( x = 50) = 2σ x 50 Σ ( xi − 50) 2 250 = = 25 = 5 10 n \ Co.efficient of variation = 2 × 5 = 10%. Also, σ =

Measures of Central Tendency and Dispersion  14.13

PRACTICE EXERCISES Single Option Correct Type

5. The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are (A)  14.98, 39.95 (B)  39.95, 14.98 (C)  39. 95, 224.5 (D)  None of these 6. If Q.D. is 16, the most likely value of S.D. will be (A) 24 (B) 42 (C)  10 (D)  None of these 7. If a variable x takes values 0, 1, 2, ..., n with fre­ quencies proportional to the binomial coefficients nC0, n C1, nC2,..., nCn, then the Var (x) is

9. If the standard deviation of n observations x1, x2,..., xn is 4 and another set of n observations y1, y2,..., yn is 3. The standard deviation of n observations x1 – y1, x2 – y2,..., xn – yn is 2 (A)  1 (B)  3 (C) 5 (D) Data insufficient 1 n ∑ ( x − x)2 be the n − 1i =1 i S.D. of a set of observations x1, x2,..., xn, then

10. Let r be the range and S 2 =

(A) S ≤ r

n n −1

n (B)  S= r n −1

n (C) S ≥ r n −1

(D)  None of these

11. The A.M. of n numbers of a series is x . If the sum of the first (n – 1) term is k, them the nth number is (A)  x − k (B)  nx − k (B)  x − nk (D)  nx − nk 12. If a variable takes values 0, 1, 2,..., n with frequencies n n( n − 1) n − 2 2 qn, q n −1 p , q p ,..., pn, where p + q = 1, 1 1. 2 then the mean is (A) np (B)  nq (C) n(p + q) (D)  None of these 13. The S.D. of a variate x is σ. The S.D. of the variate ax + b where a, b, c are constants, is c a ⎛ a⎞ (A)  ⎜ ⎟ σ (B)  σ ⎝ c⎠ c

n2 − 1 n (A)  (B)  12 2

⎛ a2 ⎞ (C)  ⎜ 2 ⎟ σ ⎝c ⎠

n (C)  (D)  None of these 4 8. The sum of squares of deviations for 10 observations taken from mean 50 is 250. The coefficient of variation is (A) 50% (B)  10% (C) 40% (D)  None of these

14. Consider any set of observations x1, x2, x3,..., x101; it being given that x1 < x2 < x3 < ... < x100 < x101; then the mean deviation of this set of observations about a point k is minimum when k equals (A) x1 (B)  x51 x + x2 + ... + x101 (C)  1 (D)  x50 101

(D)  None of these

PRACTICE EXERCISES

1. The average of n numbers x1, x2, x3,..., xn is M. If xn is replaced by x′, then new average is nM − x + x ′ (A) M – xn + x′ (B)  n n M − xn + x ′ ( n − 1) M + x ′ (C)  (D)  m n 2. The standard deviation of 25 numbers is 40. If each of the numbers is increased by 5, then the new standard deviation will be (A) 40 (B) 45 21 (C) 40 + (D)  None of these 25 3. If M.D. is 12, the value of S.D. will be (A) 15 (B) 12 (C)  24 (D)  None of these 4. The mean weight of 9 items is 51. If one more item is added to the series the mean becomes 16. The value of the 10th item is (A) 35 (B) 30 (C) 25 (D) 20

14.14  Chapter 14 15. The mean of the numbers 50

50

50

50

C0 C50 C2 C4 equals , , ..., 3 5 51 1 250 249 (A)  (B)  51 51 49 2 (C)  (D)  None of these 39 × 17 16. The standard deviation of a distribution is 30 and each item is raised by 3, then new S.D. is (A) 32 (B) 28 (C)  27 (D)  None of these 17. For three numbers a, b, c product of the average of the 1 1 1 numbers a2, b2, c2 and 2 , 2 , 2 cannot be less than a b c (A) 1 (B) 3 (C)  9 (D)  None of these 18. The variance of a, b and g is 9, then variance of 5a, 5b and 5g is (A) 45 (B) 9/5 (C) 5/9 (D) 225 19. Mean of the numbers 1, 2, 3,..., n with respective weights 12 + 1, 22 + 2, 32 + 3,..., n2 + n is 3n( n + 1) 2n + 1 (A)  (B)  2( 2n + 1) 3 3n + 1 3n + 1 (C)  (D)  4 2

PRACTICE EXERCISES

20. The G.M. of the number 3, 32, 33,..., 33n is 3n

n

(A) 3 2 (B)  32 (C) 3

3n+1 2

n+1



(D)  32

23. The mean weight per student in a group of seven students is 55 kg If the individual weights of 6 students are 52, 58, 55, 53, 56 and 54; then weight of the seventh student is (A)  55 kg (B)  60 kg (C)  57 kg (D)  50 kg 24. If the mean of a set of observations x1, x2, x3,..., xn is x , then mean of observations xi + 3i ∀ i = 1, 2, 3,... n equals (A)  x + 3(n + 1) (C)  x +

n +1 2n

(B)  x +

3( n + 1) 2

(D)  None of these

25. The weighted mean of the square of 1st n natural numbers whose weights are corresponding numbers, equals ( n + 1) ( 2n + 1) n( n + 1) (A)  (B)  2 2 (C) 

n +1 2

(D)  None of these

26. If the variate of a distribution takes the values 1, 2, 3, ...n with frequencies n, n – 1, n – 2, ... 3, 2, 1, then mean value of the distribution is (A) 

n( n + 2) n( n + 1) ( n + 2) (B)  3 6

n+2 ( n + 1) ) ( n + 2) (D)  3 6 27. The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two are (A)  2 and 9 (B)  3 and 8 (C)  4 and 7 (D)  5 and 6 (C) 

(A) 

2( 2n + 1) 3n( n + 1) (B)  3n( n + 1) n( 2n + 1)

28. If the variate takes the values 0, 2, 4, 8, ... 2n with fre91 × 8 quencies nC0, nC1, nCn and if the mean is , then 2n n equals (A) 4 (B) 6 (C)  5 (D)  None of these

(C) 

3n( n + 1) 2n + 1

29. The mean of n items is x . If each item is successively increased by 3, 32, 33,... 3n, then new mean equals

21. The reciprocal of the weighted mean of first n natural numbers whose weights are equal to the squares of the corresponding numbers is

(D)  None of these

22. 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) 38.5 (B) 37.5 (C) 36.5 (D) 36

(A)  x +

3n +1 (3n − 1) (B)  x +3 n 2n

(C)  x +

(3n − 1) 3n (D)  x +3 2n n

Measures of Central Tendency and Dispersion  14.15

The mean of the nth row is (A) 

n3 ( 2n2 + 1) 3

n3 ( 4 n2 + 2) (B)  6

(C) 

n( n − 1) ( 2n − 1) 6

(D) 

n( 2n2 + 1) 3

31. The arithmetic mean of a set of observation is . If each observation is divided by a and then is increased by 10, the means of the new series is x x + 10 (B)  α α x + 10α (C)  (D)  α x + 10 α (A) 

32. The average salary of male employees in a firm was Rs. 520 and that of females was Rs. 420. The mean salary of all the employees was Rs. 500. The percentage of male employees is (A) 80

(B) 60

(B) 40

(D) 20

33. The average weight of students in a class of 35 students is 40 kg. If the weight of the teacher be included, 1 the average rises by kg; the weight of the teacher is 2 (A)  40.5 kg (B)  50 kg (C)  41 kg (D)  58 kg 34. An automobile driver travels from plane to a hill station 120 km distant at an average speed of 30 km per hour. He then makes the return trip at an average speed of 25 km per hour. He covers another 120 km distance on plane at an average speed of 50 km per hour. His average speed over the entire distance of 360 km will be 1

30 + 25 + 50 km/h (B)  (30 ⋅ 25 ⋅ 50) 3 3 3 (C)  km/h (D)  None of these 1 1 1 + + 30 25 50 (A) 

35. If the mean deviation about the median of the numbers a, 2a, ..., 50a is 50, then | a | equals (A) 5 (B) 2 (C) 3 (D) 4 36. Let x1, x2, ..., xn be n observations, and let x be their arithmetic mean and σ2 be the variance Statement-1: Variance of 2x1, 2x2, ..., 2xn is 4σ2. Statement-2: Arithmetic mean 2x1, 2x2, ..., 2xn is 4x. (A)  Statement-1 is false, Statement-2 is true. (B)  Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (C)  Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (D)  Statement-1 is true, statement-2 is false. 37. If the mean of a set of observations x1, x2, ..., x10 is 20 then the mean of x1 + 4, x2 + 8, x3 + 12, ..., x10 + 40 is (A) 34 (B) 42 (C) 38 (D) 40 38. The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b? (A) a = 0, b = 7 (B)  a = 5, b = 2 (C) a = 1, b = 6 (D)  a = 3, b = 4 39. If the mean deviation of number 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then the d is equal to (A) 10.0 (B) 20.0 (C) 10.1 (D) 20.2 40. Statement-1: The variance of first n even natural n2 − 1 ­numbers is 4 Statement-2: The sum of first n natural numbers is n( n + 1) and the sum of squares of first n natural 2 n( n + 1) ( 2n + 1) ­numbers is 6 (A)  Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (B)  Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (C)  Statement-1 is true, Statement-2 is false. (D)  Statement-1 is false, Statement-2 is true.

Previous Year’s Questions 41. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set [2003]



(A)  is increased by 2 (B)  is decreased by 2 (C)  is two times the original median (D)  remains the same as that of the original set

PRACTICE EXERCISES

30. A sequence of odd positive integers is written as 1 3  5  7  9 11, 13, 15, 17, 19, 21, 23, 25, 27

14.16  Chapter 14 42. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation [2004] (A) x2 + 18x + 16 = 0 (B) x2 - 18x - 16 = 0 (C) x2 + 18x - 16 = 0 (D) x2 - 18x + 16 = 0 43. Consider the following statements [2004] (A)  Mode can be computed from histogram (B)  Median is not independent of change of scale (C) Variance is independent of change of origin and scale. Which of these is/are correct? (A)  only (A) (B)  only (B) (C)  only (A) and (B) (D)  (A), (B) and (C) 44. In a series of 2n observations, half of them equal a and remaining half equal -a. If the standard deviation of the observations is 2, then |a| equals [2004] 1 (A)  (B)  2 n 2 (C)  2 (D)  n

PRACTICE EXERCISES

45. If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately  [2005] (A) 22.0 (B) 20.5 (C) 25.5 (D) 24.0 46. Let x1, x2, …, xn be n ∑ xi2 = 400 and ∑ xi = 80. n among the following is (A) 15 (C) 9

observations such that Then a possible value of [2005] (B) 18 (D) 12

47. Suppose a population A has 100 observations 101, 102, … , 200, and another population B has 100 observations 151, 152, … , 250. If VA and VB represent the variances of the two populations, respectively, then VA is [2006] VB (A) 1 (B) 9/4 (C) 4/9 (D) 2/3 48. The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is  [2007] (A) 40 (B) 20 (C) 80 (D) 60

49. The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b? [2008] (A) a = 0, b = 7 (B) a = 5, b = 2 (C) a = 1, b = 6 (D) a = 3, b = 4 50. If the mean deviation of number 1, 1 + d, 1 + 2d, ….. , 1 + 100d from their mean is 255, then the d is equal to  [2009] (A) 10.0 (B) 20.0 (C) 10.1 (D) 20.2 51. For two data sets, each with size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is [2010] 11 (A)  (B)  6 2 13 5 (C)  (D)  2 2 52. If the mean deviation about the median of the numbers a, 2a . . . 50a is 50, then |a| equals [2011] (A) 3 (B) 4 (C) 5 (D) 2 53. Let x1, x2 . . . xn be n observations, and let x be their arithmetic mean and σ 2 be their variance. Statement-1: Variance of 2x1, 2x2 . . . 2xn is 4 σ 2 . Statement-2: Arithmetic mean of 2x1, 2x2 . . . 2xn is 4 x . [2012] (A)  Statement-1 is false, statement-2 is true (B) Statement-1 is true, statement-2 is true; statement 2 is a correct explanation for statement 1 (C) Statement-1 is true, statement-2 is true; statement 2 is not a correct explanation for statement 1 (D)  Statement-1 is true, statement-2 is false 54. All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to entire class. Which of the following statistical measures will not change even after the grace marks were given? [2013] (A) median (B) mode (C) variance (D) mean 55. The variance of the first 50 even natural numbers is  [2014] 833 (A)  (B)  833 4 437 (C)  437 (D)  4

Measures of Central Tendency and Dispersion  14.17 56. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is  [2015] (A) 16.0 (B) 15.8 (C) 14.0 (D) 16.8

(A) 3a 2 − 23a + 44 = 0     (B) 3a 2 − 26 a + 55 = 0

57. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?  [2016]

(A) 9

(C) 3a 2 − 32a + 84 = 0    (D) 3a 2 − 34 a + 91 = 0 9

58. If

∑ ( xi − 5) = 9 and i =1

9

∑ ( xi − 5)2 = 45,

then the stan-

i =1

dard deviation of the 9 items x1, x2, …, x9 is (B) 4

(C) 2

[2018]

(D) 3

ANSWER K EYS Single Option Correct Type 1. (B) 2. (A) 11.  (B) 12.  (A) 21. (A) 22.  (B) 31. (C) 32. (A)

3.  (A) 13. (B) 23. (C) 33. (D)

4. (C) 14.  (B) 24. (B) 34.  (C)

5.  (B) 15. (C) 25. (B) 35.  (D)

6. (A) 16.  (D) 26.  (C) 36.  (D)

7.  (C) 17. (A) 27. (C) 37.  (B)

8. (B) 18.  (D) 28.  (B) 38.  (D)

9. (D) 19. (C) 29. (B) 39.  (C)

44. (C) 54. (C)

45. (D) 55. (B)

46. (B) 56. (C)

47. (A) 57. (C)

48. (C) 58. (C)

49. (D) 40. (C)

10. (A) 20.  (C) 30.  (D) 40.  (D)

41. (C) 51. (A)

42. (D) 52. (B)

43. (C) 53. (D)

PRACTICE EXERCISES

Previous Years’ Questions

14.18  Chapter 14

HINTS AND EXPLANATIONS Single Option Correct Type 1. M =

x1 + x2 + x3 ... xn n

7. We have,

HINTS AND EXPLANATIONS

nM = x1 + x2 + x3 + ... xn–1 + xn0 i.e., nM – xn = x1 + x2 + x3 + ... xn – 1 nM − xn + x ′ x + x2 + x3 + ... xn −1 + x ′ = 1 n n nM − xn + x ′ \ New average = n The correct option is (B) 2. If each item of a data is increased or decreased by the same constant, the standard deviation of the data remains unchanged. The correct option is (A) 3. We know that 5 5 Q.D. = × M.D. = × 12 = 10 6 6 3 3 \ S.D. = × Q.D. = × 10 ⇒ S.D. = 15 2 2 The correct option is (A) 4. Let the values of 9 items be x1, x2, ..., x9 Therefore, mean of x1, x2, ..., x9 is x + x2 + ... + x9 15 = 1 9 ⇒ x1 + x2 + ... + x9 = 15 × 9 = 135 Let x10 be the 10th item. The mean of x1, x2, ... x9, x10 is 16. x + x2 + ... + x9 + x10 ⇒ 1 = 16 10 ⇒ x1 + x2 + ... + x9 + x10 = 160 ⇒ 135 + x10 = 160 \ x10 = 160 – 135 = 25 The correct option is (C) 5. Corrected Sx = 40 × 200 – 50 + 40 = 7990 \ Corrected x = 7990 / 200 = 39.95





\ Corrected σ =

364100 − (39.95) 2 200





(1820.5 − 1596)

=

= 224.5 = 14.98 The correct option is (B) 3 6. We know that, S.D. = Q.D. 2 3 \ S.D. = × 16 = 24 2 The correct option is (A)

n

C0 + nC1 + nC2 + ... + nCn

n



⇒ x =

r=0 n

∑ r⋅





⇒ x =

2

n

Cr

n

n

1 n n r⋅ n ∑ r 2 r=0

⇒ x =

n

r=0

∑ n −1Cr −1

n

=

n−1

Cr −1

n 2

2n −1 =

2n

r =1

⎡ n ⎤ ⎢∵ ∑ n −1Cr −1 = 2n −1 ⎥ ⎢⎣ r = 1 ⎥⎦

 n

and

1 ∑ fi xi2 = n

∑ ∑ fi

fi xi2

=

∑ r 2nCr

r=0 n

∑ n Cr

r=0 n

∑ r2

n r

n −1

Cr −1



1 ⇒ ∑ fi xi2 = n



⇒

n n 1 fi xi2 = n ∑ ( r − 1 + 1) ∑ n 2 r=0



⇒

n ⎤ 1 n ⎡ n fi xi2 = n ⎢ ∑ ( r − 1) n −1Cr −1 + ∑ n −1Cr −1 ⎥ ∑ n 2 ⎢⎣ r = 0 ⎥⎦ r=0







⇒





2



0 ⋅ nC0 + 1⋅ nC1 + 2 ⋅ nC2 + ... + n ⋅ nCn

∑ r ⋅ nCr

Incorrect Sx2 = n[σ 2 + x ] = 200[152 + 402] = 365000 Correct Sx2 = 365000 – 2500 + 1600 = 364100

x =

=

r=0

2n n −1

Cr −1

( n − 1) n ⎡ n ( r − 1) ⎢ n ∑ ( r − 1) 2 ⎢⎣ r = 0

⎤ Cr − 2 + 2n −1 ⎥ ⎥⎦

n−2

1 n ∑ fi xi2 = 2n [(n – 1)2n–2 + 2n–1] n =

Now, Var (x) =

n 2n

[(n – 1 + 2)2n–2] =

n 2n

(n + 1)

1 ∑ fi xi2 − x 2 n

n( n + 1) n2 n = − 4 4 4 The correct option is (C)

\ Var (x) =

8. S.D. (σ) =

250 = 10

25 = 5

Hence, coefficient of variation =

σ × 100 mean

Measures of Central Tendency and Dispersion  14.19



5 × 100 = 10%. 50

=

The correct option is (B) 9. S.D = σ x2− y



= =

1 n ∑ ( xi − yi − x + y)2 n i =1 n



=

+



⇒ x =

σ 2y



⇒ x =

n

2 ∑ ( xi − x) ( yi − y) n i =1

− 2 cov ( x, y )



10. We have r = max | xi – xj |



=

x + x2 + ... + xn ⎞ ⎛ Now, ( xi − x ) = ⎜ xi − 1 ⎟⎠ ⎝ n =

 n2

1 n2





⇒ ( xi − x ) ≤ r2 ⇒

2

[(xi ­– x1) + (xi – x2) + ... + (xi – xi – 1) + (xi – xi + 1) + ... + (xi – xn)]

n

∑ ( xi − x)2 ≤ nr 2

i =1

nr 2 nr 2 1 n ⇒ ( xi − x ) 2 ≤ ⇒ S2 ≤ ∑ n − 1i =1 ( n − 1) ( n − 1)

n n −1 The correct option is (A) 11. Let the numbers be x1, x2,..., xn. Then, ⇒ S≤r

x =

r=0 n

∑ n Cr q n− r p r n

∑ r⋅ r

n−1

Cr −1 q n−r p ⋅ p r −1

r=0

n

∑ n Cr q n− r p r

⎛ n ⎞ np ⎜ ∑ n −1Cr −1 p r −1q( n −1) − ( r −1) ⎟ ⎝r =1 ⎠ n

r=0

[( n − 1)r ]2 ( | xi – xj | ≤ r)





∑ r ⋅ nCr qn−r pr

∑ n Cr q n − r p r

2



⇒ x =

1 S = ∑ ( xi − x)2 n − 1i =1

1

C0 q n p0 + nC1 q n−1 p1 + ... + nCn q n−n p n

n

2

≤

n

r=0



i≠ j



n n −1 n( n − 1) n − 2 2 q p+ q p + ... + p n 1 2!

0 ⋅ nC0 q n p0 + 1⋅ nC1q n−1 p + ... + n ⋅ nCn q0 p n

n

The correct option is (D)



qn +

r=0

As cov (x, y) is not known, therefore we cannot find σ x2− y or σ x − y . Hence data is insufficient.

and

+ ... + n . p n

n

1 1 ∑ ( xi − x)2 + n ∑ ( yi − y) n i =1 i =1

σ x2

⇒ x =

n

−



1 n ∑ xi n i =1







⇒ x =

x1 + x2 + ... + xn −1 + xn n



⇒ x =

k + xn [ x1 + x2 + ... + xn–1 = k] n

\ xn = nx − k The correct option is (B) 12. The required mean is n n( n − 1) n−2 2 0 ⋅ q n + 1⋅ q n−1 p + 2 ⋅ q p 1 2!



⇒ x =

np( q + p) n −1

( q + p) n \ x = np ( q + p = 1) The correct option is (A) ax + b a b i.e., y = x + c c c b a i.e., y = Ax + B, where A = , B = c c 13. Let y =



\ y = Ax + B



\ y − y = A( x − x ) ⇒ ( y − y ) 2 = A2 ( x − x ) 2



⇒ Σ( y − y ) 2 = A2 Σ( x − x ) 2



⇒ n ⋅σ y2 = A2 ⋅ nσ x2



⇒ σ 2y = A2σ x2



⇒ σ y = | A |σx ⇒ σ y =

a σx c

a σ. c The correct option is (B) 14. Mean deviation is minimum when it is considered about the item, equidistant from the beginning and the end i.e., the 101 + 1 median. In this case median is th i.e., 51st item i.e., 2 x51. The correct option is (B) Thus, new S.D. =

15. Consider (1 + x)50 = 50C0 + 50C1x1 + ... + 50C50x50(1)

HINTS AND EXPLANATIONS



14.20  Chapter 14 and (1 – x)50 = 50C0 – 50C1x1 + ... + 50C50x50(2) Adding (1) and (2), we get on integrating with limits 0 to 1, 50



C0 x +

50

C2 3 x + 3



=

50

C4 5 x + ... + 5

50

1

C50 x 51 ⎤ ⎥ 51 ⎥⎦

50

50

\

50

50

C0 + 1



C2 + 3

C0 +

50

C2 + 3



C4 + ... + 5

50

C4 + ... + 5 26

C50 51

16. S.D. of a series is unaltered if each item is raised (reduced) by the same scalar quantity, S.D. is independent of change of origin. Hence S.D. will be same as it was already. \ S.D. = 30 The correct option is (D)

HINTS AND EXPLANATIONS

2

2

2

17. a + b + c ≥ (a2 b2 c2)1/3 ( A.M ≥ G.M.) 3 1 1 1 + 2 + 2 1/ 3 2 a b c and ≥ ⎛⎜ 1 1 1 ⎞⎟ 3 ⎝ a2 b2 c2 ⎠ On multiplying, we get







=

3n2 + 7n + 2 (3n + 1) ( n + 2) 3n + 1 = = 4 ( n + 2) 2( 2n + 4) 4

1

50

=



The correct option is (C) 20. G.M. of 31, 32, 33,..., 33n

50

2 1 249 × = 51 26 39 × 17 The correct option is (C)

i =1 n

n( n + 1) ⎡ n( n + 1) 2n + 1⎤ + ⎢ 2 3 ⎥⎦ 2 ⎣ = n( n + 1) ⎡ 2n + 1 ⎤ ⎢ 3 + 1⎥ 2 ⎣ ⎦

50

51 C50 2 = 12 = 51 51 2 51 50 50 Now number of terms from C0 to C50 are 26 (items) \ Required mean



i =1 n



0

50

n

n2 ( n + 1) 2 n( n + 1) ( 2n + 1) + 4 6 = = n( n + 1) ( 2n + 1) n( n + 1) 2 + ∑i + ∑i 6 2 i =1 i =1

51 ⎤1

1 ⎡ (1 + x )51 (1 − x ) − ⎢ ⎥ 2 ⎢⎣ 51 51 ⎥⎦ 0

n

∑ i3 + ∑ i 2

1 1⎞ ⎛ 1 ⎛ a2 + b2 + c2 ⎞ ⎜ a2 + b2 + c2 ⎟ ⎟ ≥1 ⎜ ⎟ .⎜ 3 3 ⎝ ⎠ ⎜ ⎟ ⎠ ⎝



= (3.32 ⋅ 33...33n ) 3n =



= 3

3n( 3n +1) 3n ⋅ 2

Weighted mean =

13 + 23 + 33 + ... + n3



=



=



\ the reciprocal of the weighted mean is

12 + 22 + 32 + ... + n2 3n( n + 1) n2 ( n + 1) 2 6 = × 2( 2n + 1) 4 n( n + 1) ( 2n + 1)

22. Given,

18. When each item of a data is multiplied by l, variance is multiplied by l2. Hence, new variance = 52 × 9 = 225. The correct option is (D)



Σwi xi = Σwi

w1x1 + w2 x2 + ... + wi xi Σwi xi = w1 + w2 + ... + wi Σwi

w ( x ) = weighted mean 1.1 + 2.22 + 3.32 + ... + n.n2 = 12 + 22 + 32 + ... + n2

1 1 1 \ Product of the averages of a2, b2, c2 and 2 , 2 , 2 a b c cannot be less than 1. The correct option is (A)

Hence, the required mean =

3n+1

3 2 The correct option is (C) 21. Given x 1 2 3 4 ...n fw 1 22 32 42 ...n2

2( 2n + 1) 3n( n + 1)

The correct option is (C)



19. Here, for each xi = i, weight wi = i2 + i

=

1

(31+ 2 + 3 + ...3n ) 3n

n

∑ i (i 2 + i )

i =1 n

∑ (i 2 + i )

i =1

Σxi = 38, \ Sxi = 1900 50

New value of Sxi = 1900 – 55 – 45 = 1800, n = 48 1800 = 37.5 48 The correct option is (B) 23. Total weight of 7 students is = 55 × 7 = 385 kg Sum of weights of 6 students = 52 + 58 + 55 + 53 + 56 + 54 = 328 kg \ Weight of seventh student = 385 – 328 = 57 kg. The correct option is (C) \ New mean =

24. Given: n x = x1 + x2 + x3 + ... + xn Now new observation are x1 + 3, x2 + 3.2,... xn + 3.n

Measures of Central Tendency and Dispersion  14.21 \ new mean



=

( x + 3) + ( x + 3. 2) + ( x + 3. 3) + ... + ( x + 3. n) 1 2 3 n n



=

x1 + x2 + x3 + ... + xn 3(1 + 2 + 3 + ... + n) + n n

x = 4, y = 7 or x = 7, y = 4 The correct option is (C) 28. Given mean = 91 × 23 – n xi fi fixi 0 nC0 0 21 nC1 2 ⋅ nC1 22 nC2 22 ⋅ nC2 23 nC3 23 ⋅ nC3

3( n + 1) 2 The correct option is (B) 25. x 1 2 3 4 ...n x2 1 4 9 16 ...n2 fi 1 2 3 4 ...n



= x+

x =

Σfi xi2 Σf i

=

2

2n

2

2

1.1 + 2.2 + 3.3 + ... + n.n 1 + 2 + 3 + ... + n

\

n

∑ xi fi





= 1⋅ n + 2 (n – 1) + 3(n – 2) +, ..., + (n – 2) 2 + n.1 =



n

= ( n + 1) ∑ r − r =1

=

( n + 1) ( n) ( n + 1) n( n + 1) ( 2n + 1) − 2 6





=

n( n + 1) ( n + 2) 6

Sfi = 1 + 2 +, ..., + n =

n( n + 1) 2

Σfi xi 2 n( n + 1) ( n + 2) n+2 = × = Σf i 6 n( n + 1) 3 The correct option is (C) 27. Let the two unknown items be x and y, then Mean = 4 Mean

=

1+ 2 + 6 + x + y =4 5 ⇒ x + y = 11 and variance = 5.2

( x1 + 3) + ( x2 + 32 ) + ( x3 + 33 ) + ... + ( xn + 3n ) n



=

( x1 + x2 ... + xn ) 31 + 32 + ... + 3n + n n

⇒

2 2 2 2 2 ⇒ 1 + 2 + 6 + x + y – (mean)2 = 5.2 5 ⇒ 41 + x2 + y2 = 5[5.2 + (4)2] ⇒ 41 + x2 + y2 = 106 or x2 + y2 = 65 Solving (1) and (2) for x and y, we get

n = x + 3(3 − 1) 2n The correct option is (B) 30. The number of numbers in the nth row = n2 Sequence of first terms in different row is 1, 3, 11, 29, 61, ... 1 \ Tn of 1, 3, 11, 29, 61,... = (2n3 – 3n2 + n + 3) = first th 3 element of n row. Similarly, sequence of last terms of each row = 1, 9, 27, 59,... 1 \ tn = [2n3 + 3n2 + n – 3] 3 = last element of the nth row. Hence, in the nth row elements can be written as



r =1



Also,

=

n

∑ r2



Σfi xi Σf i



n

∑ (n + 1)r − r 2

r =1



Σfi xi = 3n − 1

3 n ⇒ 91 × 2 = 3 − 1 2n 2n n 6 \ 3 = 3 or n = 6 The correct option is (B) 29. Let n items be denoted by x1, x2, x3,... xn \ new items are x1 + 3, x2 + 32, x3 + 33,... xn + 3n \ new mean

i =1



Σf i = 2

n



2

The correct option is (B) 26. xi : 1 2 3 4 ... n – 1 n fi : n n – 1 n – 2 n – 3 ... 2 1

2n nCn

Cn

Now mean =

2

n ( n + 1) 2 n( n + 1) =⋅ × 4 n( n + 1) 2

=

n

(1)



(2)

1 1 (2n3 – 3n2 + n + 3), ... (2n3 + 3n2 + n – 3) 3 3 (Note: adding 2 in the preceding number to get the succeeding number) \ sum of the elements of nth row (using sum of n terms of A.P.) 3 N n2 ⎛ 4 n3 + 2n ⎞ ( A + L) = = n ( 2n2 + 1) ⎜ ⎟ 2 ⎝ 3 2 ⎠ 3



=



\ mean of the numbers in the nth row

HINTS AND EXPLANATIONS



14.22  Chapter 14



=

n3 ( 2n2 + 1) 3 × n2

2 = n( 2n + 1) 3

⎧ N = n2 , A = Tn , L = t n ⎫ ⎪ ⎪ Here ⎨ 3 4 n + 2n ⎬ ⎪∴ A + L = Tn + t n = ⎪ 3 ⎩ ⎭ The correct option is (D) 31. Let x1, x2, ... xn be n observations. 1 Then, x = Σxi n xi + 10 Let yi = α Then, y =

1 n 1 ⎛1 n ⎞ 1 yi = ⎜ ∑ xi ⎟ + (10 n) ∑ n i =1 α ⎝ n i =1 ⎠ n

1 x + 10α x + 10 = α α The correct option is (B) 32. x1 = 520, x2 = 420 and x = 500

\ y=

Also, we know x =

n1x1 + n2 x2 n1 + n2

⇒ 500 (n1 + n2) = 520n1 + 420n2 ⇒ 20n1 = 80n2 ⇒ n1 : n2 = 4 : 1 Hence, the percentage of male employees in the firm

HINTS AND EXPLANATIONS



⎛ 4 ⎞ = ⎜ × 100 = 80% ⎝ 4 + 1⎟⎠

The correct option is (A) 33. Let the weight of the teacher be w kg, then



40 +

1 30 × 40 + w = 2 35 + 1

⇒ | a | = 4 The correct option is (D) 2 x + 2 x2 + ... + 2 xn 36. A.M. of 2x1, 2x2, ..., 2xn is 1 n ⎛ x1 + x2 + ... + xn ⎞ = 2⎜ ⎟⎠ = 2x ⎝ n So statement-2 is false. Variance (2xi) = 22 variance (xi) = 4σ2 so statement-1 is true. The correct option is (D) ( x1 + 4) + ( x2 + 8) + ( x3 + 12) + ... + ( x10 + 40) 10

37. Mean =

x1 + x2 + ... + x10 4(1 + 2 + 3 + ... + 10) + 10 10 = 20 + 22 = 42 The correct option is (B) 38. Mean of a, b, 8, 5, 10 is 6



=

a + b + 8 + 5 + 10 =6 5 ⇒ a + b = 7 Given that Variance is 6.8 Σ( X i − A) 2 \ Variance = n

⇒

(1)

2 2 = ( a − 6) + (b − 6) + 4 + 1 + 16 = 6.8 5 ⇒ a2 + b2 = 25 a2 + (7 – a)2 = 25 [Using (1)] ⇒ a2 – 7a + 12 = 0 \ a = 4, 3 and b = 3, 4 The correct option is (D)





39. Mean ( x ) =

sum of quantities n

n ( a + 1) 1 = 2 = [1 + 1 + 100 d ] = 1 + 50d n 2

1 = 35 × 40 + w ⇒ w = 58 2 \ Weight of the teacher = 58 kg. The correct option is (D)



120 + 120 + 120 34. Average speed = 120 120 120 + + 30 25 50



⇒ 255 =

1 (50 d + 49d + 48d + ... + d + 0 + d + ... + 50 d ) 101





2d ⎛ 50 × 51⎞ ⎜ ⎟ 101 ⎝ 2 ⎠





⇒ 36 × 40 + 36 ×



3 = 1 1 1 km/h. + + 30 25 50

The correct option is (C) 25 + 26 a = 25.5a 2 Required mean deviation about median 35. From the given data, median =



=

2 | 0.5 + 1.5 + 2.5 + ... + 24.5 | | a | = 50 50

M.D. =

1 ∑ xi − x n

=

255 × 101 = 10.1 50 × 51 The correct option is (C)

⇒ d =

40. Statement-2 is true Statement-1: Sum of n even natural numbers = n (n + 1) Mean ( x ) =

n( n + 1) =n+1 n

Measures of Central Tendency and Dispersion  14.23 ⎡1 ⎤ Variance = ⎢ ∑ ( xi ) 2 ⎥ − ( x ) 2 ⎣n ⎦





=

( n + 1)[2( 2n + 1) − 3( n + 1)] 3



=

( n + 1)[4 n + 2 − 3n − 3] 3





=

1 2 [2 + 4 2 + ... + ( 2n) 2 ] − ( n + 1) 2 n







=

1 2 2 2 (1 + 22 + ... + n2 ) − ( n + 1) 2 n







=

4 n( n + 1)( 2n + 1) − ( n + 1) 2 n 6

2 = ( n + 1)( n − 1) = n − 1 3 3 \ Statement 1 is false. The correct option is (D)



Previous Year’s Questions 3x + 1 1 − x 1 − 2 x + + ≤1 3 4 2



⇒ 12x + 4 + 3 - 3x + 6 - 12x ≤ 1 ⇒ 0 ≤ 13 - 3x ≤ 12 ⇒ 3x ≤ 13 1 ⇒x≥ . 13 The correct option is (C) 42. Let the numbers be a and b then a + b = 18 and ab = 4 ⇒ ab = 16 So, a and b are roots of the equation x2 - 18x + 16 = 0. The correct option is (D) 43. Mode can be computed from histogram and median is dependent on the scale. Hence statements (a) and (b) are correct. The correct option is (C) 44. xi = a for i = 1, 2,...., n and xi = -a for i = n+1,...., 2n





S.D =

1 ∑ ( xi − x )2 2n i =1

⇒2=

2n ⎞ 1 2n 2 ⎛ xi ⎜ Since ∑ xi = 0⎟ ∑ 2n i =1 ⎝ ⎠ i =1





2n

46. We have that

∑

n

⎛ ∑ xi ⎞ ≥⎜ ⎟ ⎝ n ⎠

52x + 42y = 50 (x + y) ⇒ 2x = 8y ⇒

x 4 x 4 = and = y 1 x+ y 5

\ % of boys = 80. The correct option is (C)

49. Mean of a, b, 8, 5, 10 is 6



a + b + 8 + 5 + 10 =6 5 ⇒ a + b = 7 ⇒

47. The variance is given by σ x2 =

∑ ( X i − A)2



\ Variance =







 ⇒ a2 + b2 = 25

=

n ( a − 6) 2 + (b − 6) 2 + 4 + 4 + 16 = 6.8 5

  a2 + (7 - a)2 = 25 (Using (1))

⇒ a2 - 7a + 12 = 0



\ a = 4, 3 and b = 3, 4.

The correct option is (D)

2

⇒ n ≥ 16 . The correct option is (B)

taken from the mean)

48. According to question

Given that Variance is 6.8

1 ⇒2= ⋅ 2na 2 ⇒ | a |= 2 2n The correct option is (C) 45. We have Mode + 2 Mean = 3 Median ⇒ Mode = 3 × 22 - 2 × 21 = 66 - 42 = 24 The correct option is (D) xi2

Since A and B both has 100 consecutive integers, therefore both have same standard deviation and hence the variance. V ∴ A = 1 (As ∑ di2 is same in both the cases) . VB The correct option is (A)

Σdi2 (Here deviations are n

n/ sum of quantities 2 ( a + l ) = 50. Mean ( x ) = n/ n 1 = [1 + 1 + 100 d ] = 1 + 50 d 2 Now, since M.D. about mean is 255, we have

(1)

HINTS AND EXPLANATIONS

41. 0 ≤

14.24  Chapter 14 1 ∑ xi − x n 1 ⇒ 255 = [50 d + 49d + 48d + ... + d + 0 + d + ..... + 50 d ] 101 2 d ⎡ 50 × 51 ⎤ = 101 ⎢⎣ 2 ⎥⎦ M.D. =



255 × 101 = 10.1 50 × 51 The correct option is (C) 51. σx2 = 4 σy2 = 5 x=2 y=4



⇒ d=



∑ xi = 2 5

(

HINTS AND EXPLANATIONS

)

2

=

∑ y−y n

2

=σ .

⎛ ∑ xi2 ⎞ 2 55. σ 2 = ⎜ ⎟−x ⎝ n ⎠ 50

)





x=

∑ 2r r =1

50

= 51

50

∑ 4r 2

− (51) = 833 50 The correct option is (B)

σ2 =

r −1

56. New sum

2

∑ yi = (16 × 16 − 16) + (3 + 4 + 5) = 252

Number of observation = 18 ⇒ New mean 252 y= = 14 . 18 The correct option is (C)

25a + 26 a = 25 − 5a 2 1 Mean deviation = { a − 25.5a + 2a − 25.5a } 50 2 = {( 24.5a + 23.5a ) + ... (0.5a )} 50 A = Median =

57.

2 {312.5a} = 50 Given 50

S.D. =

∴

⇒ 625a = 2500 ⇒ a = 4 The correct option is (B) xi 2 ⎛ xi ⎞ ∑ n ⎟⎠ n ⎜⎝

∑ y + 10 − y − 10

n The correct option is (C)

1 52. ∑ xi − A n

53. σ 2 = ∑

Statement-2 is false. The correct option is (D) 54. Variance is not changed by the change of origin. 2 ∑ x−x σ= n

σ1 =

1 ⎛1 ⎞ σ x2 = ⎜ ∑ xi2 ⎟ − ( x ) 2 = ∑ yi2 − 16 ⎝ ⎠ 2 5 2 x = 40 ∑ i y 2 = 105 ∑ i 2 1 ⎛ x + y⎞ 2 2 σ z2 = x y − + ∑ i ∑ i ⎜⎝ 2 ⎟⎠ 10 1 145 − 90 = ( 40 + 105) − 9 = 10 10 55 11 = = 10 2 The correct option is (A)

=

2 x1 + 2 x2 + ... + 2 xn n ⎛ x1 + x2 + .... + xn ⎞ = 2⎜ ⎟⎠ = 2 x ⎝ n

A.M. of 2 x1, 2 x2 ,....., 2 xn =

Therefore, y = x + 10 ⇒ y + x + 10

∑ xi = 10; ∑ yi = 20

(

Statement-1 is true.

Σxi2 ⎛ Σxi ⎞ −⎜ ⎝ n ⎟⎠ n

2

49 4 + 9 + a 2 + 121 ⎛ 16 + a ⎞ = −⎜ ⎝ 4 ⎟⎠ 4 4

2

⇒ 3a 2 − 32a + 84 = 0 The correct option is (C)

2

( xi − x ) 2 n i =1 n

Variance of 2 x1, 2 x2 ,...., 2 xn = ∑

(2 xi )2 − ⎛ n

⎜⎝ ∑

2 xi ⎞ n ⎟⎠

2

⎡ x 2 ⎛ x ⎞ 2⎤ = 4 ⎢ ∑ i − ⎜ ∑ i ⎟ ⎥ = 4σ 2 n ⎝ n ⎠ ⎥⎦ ⎢⎣

58. Formula of variance, σ 2 = ∑

n

Formula of Standard deviation, σ =

n=9

∑ ( xi − x )2 i =1

n

Measures of Central Tendency and Dispersion  14.25



9

9

i =1

i =1

∑ xi − 5∑1 = 9 9



∑ xi − 45 = 9

⇒ ⇒



∑ xi = 54

⇒

x=

9

∑ ( xi − 5)

Given

2

9

9

9

i =1

i =1

i =1

i =1

i =1

9

∑ ( xi − 6)2 = 45 − 2 × 54 + 99 9

∑ ( xi − 6)2 = 45 − 108 + 99 = 36 i =1

i=1



9

i =1

i =1 9

9

∑ xi2 − 12∑ xi + 36∑1 = 45 − 2∑ xi + 11∑1

54 = 6 [ x = Mean ] 9

⇒

= 45

Hence, standard deviation, σ = 4 = 2

9

∑ i =1

( xi − 6) = 4 = σ 2   [VARIANCE] 9

i =1

9

9

9

i =1

i =1

i =1

∑ xi2 − 10∑ xi + 25∑1 = 45

HINTS AND EXPLANATIONS

⇒

This page is intentionally left blank.

Trigonometric Ratios and Identities

CHAPTER

15 LEARNING OBJECTIVES

After reading this chapter, you will be able to:   Learn about the angles and their measurements   Grasp various relations between different systems of measurement of angles and sides and interior angles of a regular polygon

  Understand fundamental trigonometric identities and signs of trigonometric ratios in different quadrants  Know domain and range of trigonometric ratios and learn various transformation formulae

ANGLE Let a revolving line starting from OX revolves about its end point O on a plane in the direction of arrow and occupy the position OP. It is said to trace out an angle XOP. Here OX is called the initial position and OP, the terminal position. The fixed point O is called the vertex.

FIGURE 15.2

(a)

(b)

There are three systems for measurement of an angle:

FIGURE 15.1

An angle is considered as the figure traced by rotating a given ray about its end point.

MEASUREMENT OF ANGLES

I M P O R TA N T P O I N T S If the rotation is in anticlockwise sense, the angle measured is positive and if the rotation is in clockwise sense, the angle measured is negative.

1. Sexagesimal System or English System In this system an angle is measured in degrees, minutes and seconds. A complete rotation describes 360°. 1 right angle = 90°, (read as 90 degrees) 1° = 60′ (read as 60 minutes) 1′ = 60′′ (read as 60 seconds) 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)

15.2  Chapter 15 1g = 100′ (read as 100 minutes) 1′ = 100′′ (read as 100 seconds)

D G θ = = 180 200 π

ERROR CHECK 1′ of centesimal system ≠ 1′ of sexagesimal system 1″ of centesimal system ≠ 1″ of sexagesimal system

RELATION BETWEEN SIDES AND INTERIOR ANGLES OF A REGULAR POLYGON 1.  Sum of interior angles of polygon of n sides

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.

= (2n – 4) × 90° 2. Each interior angle of a regular polygon of n sides =

2n − 4 × 90° n

QUICK TIPS The angle between two consecutive digits in a clock is 30°.  The hour hand rotates through an angle of 30° in one  1 ° hour or   in one minute 2  The minute hand rotates through an angle of 6° in one minute. 

FIGURE 15.3

∠AOB = 1 radian. This angle does not depend upon the radius of the circle from which it is derived.

FUNDAMENTAL TRIGONOMETRIC IDENTITIES

ERROR CHECK Radian is a unit to measure angle and it should not be interpreted that π stands for 180°, π is a real number whereas p c stands for 180°.

REMEMBER

1. sin2θ + cos2θ  = 1 or cos2θ  = 1 – sin2θ or sin2θ = 1 – cos2θ 2. 1 + tan2θ = sec2θ or sec2θ – tan2θ = 1 3. 1 + cot2θ = cosec2θ or cosec2θ – cot2θ = 1 QUICK TIPS

π radians = 180° = 200g.

Since sin θ + cos θ = 1, | sinθ | ≤ | and | cosθ | ≤ | ⇒ –1 ≤ sinθ ≤ 1 and –1 ≤ cosθ ≤ 1; 0 ≤ sin2θ ≤ 1, 0 ≤ cos2θ ≤ 1. Since cosecθ = 1/sinθ, cosecθ ≥ 1 or cosecθ ≤ –1 Also, since secθ = 1/cosθ, sec θ ≥ 1 or secθ ≤ –1 2

RELATION BETWEEN DIFFERENT SYSTEMS OF MEASUREMENT OF ANGLES 1°=

10 9 grades; 1g = degrees 9 10

1°=

π radians = 0.0172 radians; 180



1 radian = 1g =

2

⋅

180 degrees = 57° 17′ 45′′ π

π 200 radians; 1 radian = grades 200 π

Thus if the measure of an angle in degrees, grades and radians be D, G and θ respectively, then

SIGNS OF TRIGONOMETRIC RATIOS IN DIFFERENT QUADRANTS The following table describes the signs of various t-ratios in different quadrants. Also, refer to, the figure given below the table. Quadrant

I

II

III

IV

MP = y

+ve

+ve

-ve

-ve

OM = x

+ve

-ve

-ve

+ve

Trigonometric Ratios and Identities  15.3 Quadrant

I

III

IV

Quadrant: →

I

II

III

IV

y sinθ = r

+ ve + ve = + ve = + ve + ve + ve

− ve = − ve + ve

− ve = − ve + ve

t-ratios which are + ve

All

sinθ cosecθ

tanθ cotθ

cosθ secθ

x r

− ve + ve = − ve = + ve + ve + ve

− ve = − ve + ve

+ ve = + ve + ve

+ ve + ve = − ve = + ve −ve + ve

− ve = + ve − ve

− ve = − ve + ve

cosθ =

y , x x≠0

tanθ =

II

REMEMBER    

In the first quadrant, all are +ve.  In the second quadrant, sin and cosec are +ve In the third quadrant, tangent and cotangent are +ve In the fourth quadrant, cosine and secant are +ve.

FIGURE 15.5

Simple rule to remember:

FIGURE 15.4

The signs of other t-ratios can be found by using reciprocal relations, i.e. 1 1 1 cosecθ = and cot θ = . So, we have , sec θ = sin θ cos θ tan θ

add – sugar – to – coffee or after – school – to – college In the above, ‘a’ stands for ‘all’, ‘s’ stands for ‘sine’, ‘t’ stands for ‘tan’ and ‘c’ stands for ‘cos’. The reciprocals of these ratios are also positive in the respective quadrants.

INCREASE AND DECREASE OF TRIGONOMETRIC FUNCTIONS We can discuss the way of increase and decrease of trigonometric functions as described in the following tables: I quadrant

II quadrant

III quadrant

IV quadrant

sin θ

increases from 0 to 1

decreases from 1 to 0

decreases from 0 to -1

increases from -1 to 0

cos θ

decreases from 1 to 0

decreases from 0 to -1

increases from -1 to 0

increases from 0 to 1

tan θ

increases from 0 to ∞

decreases from ∞ to 0

increases from 0 to ∞

increases from -∞ to 0

cot θ

decreases from ∞ to 0

increases from 0 to ∞

decreases from ∞ to 0

decreases from 0 to -∞

sec θ

increases from 1 to ∞

increases from -∞ to -1

decreases from -1 to -∞

decreases from ∞ to 1

cosec θ

decreases from ∞ to 1

increases from 1 to ∞

increases from -∞ to -1

decreases from -1 to -∞

DOMAIN AND RANGE OF TRIGONOMETRIC RATIOS Functions

Domain

Range

sin x, cos x

(-∞, ∞)

[-1, 1]

tan x cot x

  π ( −∞, ∞) − ( 2n + 1) n ∈ I  2  

(-∞, ∞) - {nπ | n ∈ I}

(-∞, ∞) (-∞, ∞)

Functions

Domain

Range

sec x

  π ( − ∞, ∞) − ( 2n + 1) n ∈ I  2  

(-∞, -1] ∪ [1, ∞)

cosec x

( − ∞, ∞ ) − nπ n ∈ I

{

}

(-∞, -1] ∪ [1, ∞)

15.4  Chapter 15

TRIGONOMETRIC RATIOS OF STANDARD ANGLES Angles 0°

30°

45°

60°

Angles 0°

0

cos x

1

tan x

0

1

1 2

3 2

2

3 2 1 3

1

1 2

2

1

3

45°

60°

90°

T-Ratios

90°

T-Ratios sin x

30°

1 0

cosec x

Undefined 2

sec x

1

cot x

Undefined

2

2

2

2

3

3

2

Undefined

1

1

3

1

3

0

Undefined

TRIGONOMETRIC RATIOS FOR SOME SPECIAL ANGLES Angle

sin θ

cos θ

tan θ

cot θ

sec θ

coses θ

0°/0

0

1

0

∞

1

∞

15° π /2

2 2

2 2

18° π /10

5 −1 4

10 + 2 5 4

22.5° π /8

2− 2 2

2+ 2 2

30° π /6

3−1

1 2

3+1

3 2

36° π /5

10 − 2 5 4

5+1 4

45° π /4

1

1

54° 3π /10 60° π /3

2

2

5+1 4

10 − 2 5 4

3 2

1 2

3−1

3+1

3+1

3−1

5 −1

4

5 −1

10 + 2 5

2 +1

1 3

10 − 2 5

10 − 2 5

5+1

1

10 − 2 5

10 − 2 5

5+1 1

2+ 2 2

2− 2 2

72° 2π/ 5

10 + 2 5 4

5 −1 4

10 + 2 5

5 −1

5 −1

10 + 2 5

3 +1

3 −1

3+1

3−1

3−1

3+1

72° 90° π /2

2 2

1

2 2

0

∞

4

5 −1

10 − 2 5

2 3

2 −1

0

10 − 2 5 2

2

3

67.5° 3π /8

5π/12

4

2

5+1

2 +1

4+ 2 2

5 −1

1

3

5+1

2

3

5+1

3−1

4− 2 2 2

3

2 2

3+1

10 + 2 5

10 + 2 5 2 −1

2 2

4− 2 2

4+ 2 2

4

5+1 2 2

10 + 2 5 2 2

3−1

∞

3+1

1

Trigonometric Ratios and Identities  15.5

TRIGONOMETRIC RATIOS OF ALLIED ANGLES Two angles are said to be allied when their sum or difference is either zero or a multiple of 90°.

i.e. the angles –θ, 90° ± θ, 180° ± θ, 270° ± θ and 360° ± θ are called allied angles.

FIGURE 15.6

TABLE FOR TRIGONOMETRIC RATIOS OF ALLIED ANGLES -θ

90° - θ

90° + θ

180° - θ

180° + θ

270° - θ

270° + θ

360° - θ

360° + θ

sinθ

-sinθ

cosθ

cosθ

sinθ

-sinθ

-cosθ

-cosθ

-sinθ

sinθ

cosθ

cosθ

sinθ

-sinθ

-cosθ

-cosθ

-sinθ

sinθ

cosθ

cosθ

tanθ

-tanθ

cotθ

-cotθ

-tanθ

tanθ

cotθ

-cotθ

-tanθ

tanθ

WORKING RULE TO FIND ALLIED ANGLES Case I: When the angle is nπ ± θ, where n ∈ I and θ is acute.  There is no change in trigonometric function i.e., sin remains sin, cos remains cos and tan remains tan. Angle associated becomes θ.  The sign is affixed according to the quadrant in which the angle lies. nπ Case II: When the angle is ± θ , where n is an odd inte2 ger and θ is acute.

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.  The 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. 

15.6  Chapter 15

I M P O R TA N T P O I N T S

n +1  2  nπ  (−1) sin θ , if n is odd, cos  +θ  =  n  2   2 cos θ , if n is eveen. (−1)

sin nπ = 0, cos nπ = (–1)n. n n  sin (nπ + θ) = (–1) sinθ, cos (nπ + θ) = (–1) cosθ  For odd integer n, 

sin

nπ

To find ratios for nπ – θ and , replace θ by –θ in all of 2 the above

n −1 nπ nπ = (−1) 2 , cos =0 2 2

⋅

n −1  (−1) 2 cos θ , if n is odd, nπ     sin +θ  =   n  2   2 sin θ , if n is eveen. (−1)

TRIGONOMETRIC RATIOS IN TERMS OF EACH OTHER sin θ sinθ cosθ

sinθ 1 − sin2θ sinθ

cos θ

cotθ

1− sin2θ sinθ

1 − cos2θ

secθ

1 1 − sin2θ

cosecθ

1 sinθ

1 + tan2θ 1 1 + tan2θ

cosθ 1 − cos2θ cosθ cosθ

cot A cot B − 1 cot A + cot B

8. cot( A − B ) =

cot A cot B + 1 cot B − cot A

cotθ 1 + cot2θ

sec2θ − 1 secθ 1 secθ

cosec θ 1 cosecθ cosec2θ − 1 cosecθ

1 cotθ

sec2θ − 1

1 cosec2θ − 1

1 tanθ

cotθ

1 sec2θ − 1

cosec2θ − 1

1 1 − cos2θ

ADDITION AND SUBTRACTION FORMULAE

7. cot( A + B ) =

1 1+ cot2θ

sec θ

tanθ

1 cosθ

1. sin(A + B) = sinA cosB + cosA sinB 2. cos(A + B) = cosA cosB – sinA sinB tan A + tan B 3. tan( A + B) = 1 − tan A tan B 4. sin(A – B) = sinA cosB – cosA sinB 5. cos(A – B) = cosA cosB + sinA sinB tan A − tan B 6. tan( A − B) = 1 + tan A tan B

cost θ

tanθ

1 − cos2θ

1 − sin2θ

tanθ

tan θ

1 + tan2θ

1 + cot2θ cot

1+ sin2θ tanθ

1 + cot2θ

secθ secθ sec2θ − 1

cosecθ cosec2θ − 1

cosecθ

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 (Σ (one sine and two cos) – three sines) or = cosA cosBcosC(tanA + tanB + tanC – tanA tanB tanC) 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 (three cos – Σ(one cos and two sine)) or = cos A cosB cosC(1 – tanA tanB – tanB tanC – tanC tanA)

Trigonometric Ratios and Identities  15.7 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 14. tanA + tanB + tanC – tanA tanB tanC sin ( A + B + C ) = cos A cos B cos C

13. tan( A + B + C ) =

15. 1 – tanA tanB – tanB tanC – tanC tanA =

cos ( A + B + C ) cos A cos B cos C

TRANSFORMATION  FORMULAE PRODUCT INTO SUM OR DIFFERENCE 1. 2 sinA cosB = sin(A + B) + sin(A – B), A > B 2. 2 cosA sinB = sin(A + B) – sin(A – B), A > B 3. 2 cosA cosB = cos(A + B) + cos(A – B) 4. 2 sinA sinB = cos(A – B) – cos(A + B)

SUM AND DIFFERENCE INTO PRODUCT

 1 + tan A π 16. tan  + A  = 4  1 − tan A

C + D  C −D  1. sin C + sin D = 2 sin  cos     2   2 

 1 − tan A π 17. tan  − A  = 4   1 + tan A 18. sin(A1 + A2 + ... + An) = cosA1 cosA2 ... cosAn (S1 – S3 + S5 – ...) 19. cos(A1 + A2 + ... + An) = cosA1 cosA2 ... cosAn (1 – S2 + S4 – S6 + ...)

C + D  C −D  sin C − sin D = 2 cos  2.  sin    2   2 

S1 − S3 + S5 −  20. tan( A1 + A2 +  + An ) = 1 − S2 + S4 − S6 +  where S1 = Σ tanA1, S2 = Σ tanA1 tanA2, S3 = Σ tanA1 tanA2 tanA3 and so on. 21. sin α + sin (α + β) + sin (α + 2β) + ... + sin (α + (n – 1) β)

β  sin  α + ( n − 1)  2   nβ   = sin   β  2  sin 2 If β = α, then

C+D  C −D  cos  3. cos C + cos D = 2 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

6. tan C − tan D =

sin (C − D ) cos C cos D

7. cot C + cot D =

sin (C + D ) sin C sin D

8. cot C − cot D =

sin ( D − C ) sin C sin D

nα  n +1 × sin sin   2  2  sin α + sin 2+, … , + sin nα = α sin 2

TRIGONOMETRIC RATIOS OF MULTIPLE ANGLES

22. cos α + cos(α + β) + cos(α + 2β) + ,..., + cos(α + (n – 1)β )

1. sin 2θ = 2 sin θ cos θ =

β  cos  α + ( n − 1)  2  nβ   = sin   β  2  sin 2 If β = α, then



nα α  cos  ( n + 1)  × sin 2 2  cos α + cos 2α +, … , + cos nα = α sin 2

(An Angle of the form nθ, n ∈ I) 2 tan θ 1 + tan 2 θ

2. cos 2θ = cos2θ – sin2θ = 2 cos2θ – 1 1 − tan 2 θ = 1 − 2 sin 2 θ = 1 + tan 2 θ 3. tan 2θ =

2 tan θ 1 − tan 2 θ

4. cot 2θ =

cot 2 θ − 1 2 cot θ

15.8  Chapter 15 1 (1 + cos 2θ ) 2 1 6. 1 − cos 2θ = 2 sin 2 θ , sin 2 θ = (1 − cos 2θ ) 2 5. 1 + cos 2θ = 2 cos 2 θ , cos 2 θ =

1 (3 sinθ – sin3θ) 4 = 4 sin(60° – θ)sinθ sin(60° + θ) 1 8. cos 3θ = 4 cos3θ – 3 cosθ, cos3 θ = (cos 3θ + 3 cosθ) 4 = 4 cos(60° – θ) cosθ cos(60° + θ) = 4 cos(120° – θ) cosθ cos (120° + θ) 3 7. sin 3θ = 3sin θ – 4 sin3θ, sin θ =

3 tan θ − tan 3 θ 1 − 3 tan 2 θ = tan (60° – θ) tanθ tan (60° + θ)

9. tan 3θ =

10. cot 3θ =

cot 3 θ − 3 cot θ 3 cot 2 θ − 1

11. cos A cos 2A cos 22A ... cos 2n – 1 A =

1 − cos θ θ = tan 9. sin θ 2 1 + cos θ θ 10. = cot 2 sin θ θ θ π   π 11. sin ± cos = 2 sin  ± θ  = 2 cos  θ ∓  2 2 4 4    QUICK TIPS sin A + cos A = 1 + sin A 2 2 A A or sin + cos = ± 1 + sin A 2 2



+, If 2nπ − π /4 ≤ A/2 ≤ 2nπ + 3π /4 i.e.,  −, other wise 

sin 2n A 2n sin A

sin

A A − cos = 1 − sin A 2 2

TRIGONOMETRIC RATIOS OF SUBMULTIPLE ANGLES

A A  or  sin − cos  = ± 1 − sin A 2 2   +, If 2nπ − π /4 ≤ A/2 ≤ 2nπ + 5π /4 i.e.,  −, other wise

(An Angle of the form



θ , n ∈ I) n

(i) tan

2 A ± tan A + 1 − 1 = 2 tan A

2 tan θ /2 θ θ sin θ = 2 sin cos = 1. 2 2 1 + tan 2 θ /2

θ θ θ θ − sin 2 = 2 cos 2 − 1 = 1 − 2 sin 2 2 2 2 2 2θ 1 − tan 2 = 1 + tan 2 θ 2

2. cos θ = cos 2

3. tan θ =

=±

where  A ≠ (2n + 1)π (ii) cot

1 + cos A A =± 2 1 − cos A

2 tan θ

2 1 − tan 2 θ

1 − cos A sin A

=



=

2

1 − cos A 1 + cos A



1 + cos A sin A

θ −1 2 4. cot θ = 2 cot θ 2 1 + cos θ 2 θ = 5. cos 2 2 cot 2

6. sin 2

θ 1 − cos θ = 2 2

7. tan 2

θ 1 − cos θ = 2 1 + cos θ

θ 1 + cos θ 8. cot = 2 1 − cos θ 2

FIGURE 15.7



where A ≠ 2nπ

The ambiguities of signs are removed by locating the quadrant in which

A lies or you can follow the adjoining figure. 2

Trigonometric Ratios and Identities  15.9 3. tan(C + A) = tan B, cot A = –cot(B + C) A+ B A+ B C C 4. cos = sin , cos = sin 2 2 2 2 B+C C+A B A = cos , sin = cos 5. sin 2 2 2 2 C+A B +C A B = cot , tan = cot 6. tan 2 2 2 2 REMEMBER If A + B + C = π, then  sin2A + sin2B + sin2C = 4sinA sinB sinC  cos2A + cos2B + cos2C = – 1 – 4 cos A cosB cosC  tan A + tanB + tanC = tanA tanB tanC  cot A cotB + cotB cotC + cotC cotA = 1

FIGURE 15.8

GREATEST AND LEAST VALUES OF THE EXPRESSION a sinθ + b cosθ Let a = r cosα, b = r sin α, then a2 + b2 = r2 or r = a 2 + b 2 Then a sinθ + b cosθ = r(sinθ cosα + cosθ sinα) = r sin(θ +α) But –1 ≤ sin(θ + α) ≤ 1, so –r ≤ r sin(θ + α) ≤ r



cot

A B C A B C + cot + cot = cot cot cot 2 2 2 2 2 2



tan

A B B C C A tan + tan tan + tan tan = 1 2 2 2 2 2 2



sin A + sin B + sin C = 4 cos



cos A + cos B + cos C = 1 + 4 sin

A B C cos cos 2 2 2 A B C sin sin 2 2 2

GRAPHS OF TRIGONOMETRIC FUNCTIONS 1. Graph of y = sin x

or − a 2 + b 2 ≤ a sin θ + b cos θ ≤ a 2 + b 2 Thus, the greatest and least values of a sinθ + b cosθ are a 2 + b 2 and –  a 2 + b 2 respectively. QUICK TIPS sin x + cosec x ≥ 2, for every real x cos2x + sec2x ≥ 2, for every real x tan2x + cot2x ≥ 2, for every real x 2

2

FIGURE 15.9

2. Graph of y = cos x

CONDITIONAL IDENTITIES When the angles A, B and C satisfy a given relation, many interesting identities can be established connecting the trigonometric functions of these angles. In providing these identities, we require the properties of complementary and supplementary angles. For example, if A + B + C = π, then 1. sin(B + C) = sinA, cosB = –cos(C + A) 2. cos(A + B) = –cosC, sin C = sin(A + B)

FIGURE 15.10

3. Graph of y = tanx

15.10  Chapter 15

FIGURE 15.11

4. Graph of y = cotx

FIGURE 15.14

6. Graph of y = 3 sin 2x

FIGURE 15.12

5. Graph of y = secx and y = cosecx

FIGURE 15.13

FIGURE 15.15

Since the period of sinx is 2π. Therefore, the period of 2π sin 2x is = π . Also, –3 ≤ 3 sin2x ≤ 3. The graph 2 of y = 3 sin2x is drawn for a period 0 ≤ x ≤ π. The complete graph is simply the repetition of the portion. In a similar way the graphs of other trigonometric functions can be drawn.

Trigonometric Ratios and Identities  15.11

PRACTICE EXERCISES Single Option Correct Type

1 1 (A) 99 (B)  2 2999 (C)

1 29999

9. sin θ =

1 (D)  1999 2

π  2. Let a1 =  tan  8  tan

tan

π 8

π  , a2 =  tan  8 

π

cot

cot

π 8

,

π

π 8 π 8   a3 =  cot  , a4 =  cot  Then, 8 8   (A) a4 > a3 > a2 > a1 (B) a3 > a4 > a2 > a1 (C) a4 > a3 > a1 > a2 (D) a3 > a1 > a2 > a4 3. If x cos23θ + y cos4θ = 16 cos6θ + 9 cos2θ be an identity, then (A) x = –1, y = 24 (B)  x = 1,  y = 24 (C) x = 24, y = 1 (D)  none of these 4. |tanθ + secθ | = |tan θ | + |secθ |, 0 ≤ θ ≤ 2π is possible only if π  (A) θ ∈ [0, π ] −   (B)  θ ∈ [0, p] 2  π (C) θ ∈ 0,   2

(D)  none of these

5. If sinθ, sinφ and cosθ are in G.P., then the roots of the equation x2 + 2x cot φ + 1 = 0 are always (A) real (B)  imaginary (C) equal (D)  greater than 1 6. If cos 25° + sin 25° = k, then cos 50° is equal to − 2−k (A) k 2 − k (B)  2

2

(C) 2 − k 2 (D)  −k 2 − k2 7. If

2 sin α 1 − cos α + sin α =… = x then 1 + cos α + sin α 1 + sin α

1 (A) (B)  x x (C) 1 – x (D)  1+x

8. If e–π/2 < θ < π/2, then (A) cos logθ < log cosθ (B)  cos logθ > log cosθ (C) cos logθ ≤ log cosθ (D) none of these 1 x +  2  y

y  necessarily implies x 

(A) x > y (B) x < y (C) x = y (D) both x and y are purely imaginary x+ y = 10. If xy + yz + zx = 1, then ∑ 1 − xy (A)

4 1 (B)  xyz xyz

(C) xyz

(D)  none of these

11. cos 12° cos 24° cos 36° cos 48° cos 72° cos 96° equals 1 1 (A) − 6 (B)  2 28 1 1 (C) 7 (D)  − 7 2 2 sin (α + β + γ )  π 12. If α , β , γ ∈  0,  , then is sin α + sin β + sin γ  2 (A) < 1 (B)  > 1 (C) = 1 (D)  none of these

{

13. If cos θ sin θ + sin 2θ + sin 2α

} ≤ k , then the value

of k is (A) 1 + cos 2 α

(B)  1 + sin 2α

(C) 2 + sin 2α

(D)  2 + cos 2α

14. The maximum value of (cos a1) (cos a2) . . .(cos an) π under the restrictions 0 ≤ a1, a2, . . . , α n ≤ and (cot 2 a1) (cot a2) . . . (cot an) = 1 is 1 1 (A) n (B)  2n 2 2 (C)

1 (D)  1 2n

PRACTICE EXERCISES

1. The value of cos a cos 2a cos 3a ....cos 999 a, where 2π a= , is 1999

15.12  Chapter 15  1  1− 

23. If sin θ + cos θ =

15. The inequality 2sin θ + 2cos θ ≥ 2 2  holds for (A) 0 ≤ θ < π (B)  π ≤ θ < 2π (C) for all real θ (D)  none of these

equals

16. The expression 2sinθ + 2–cosθ is minimum when θ is equal to 7π π (A) 2nπ + , n ∈ I (B) 2nπ + , n∈ I 4 4 π (C) nπ ± , n ∈ I (D)  none of these 4   1 − x02 1 + xn  ( −1 < x0 < 1) , then cos   x1 x2 x3 ....to ∞  2   is equal to (A) x0 (B)  1/x0 (C) 1 (D)  –1 17. If xn +1 =

18. If 0 < θ < π, then (A) 1 + cot θ ≤ cot

θ 2

(B)  1 + cot θ ≥ cot

θ 2

θ θ (D) 1 + cot ≤ cot θ ≥ cot θ 2 2 19. If cos (θ – α) = a and sin (θ – β) = b (0 < θ – α, θ –β < π/2), then cos2 (α – β) + 2ab sin (α –β) is equal to (A) a2 – b2 (B)  a2 + b2 2 2 (C) 2a b (D)  a2 b2 (C) 1 + cot

A B C , tan and tan are in 2 2 2 B harmonic progression, then the least value of cot is 2 (A) 2 (B)  3 (C) 2 (D)  none of these

PRACTICE EXERCISES

20. If in the triangle ABC , tan

7 θ  and 0 < θ < π/6, then tan   2 2

1 (A) 7 − 2 (B)  ( 7 − 2) 3 1 (C) 2 − 7 (D)  (2 − 7 ) 3

π  24. If sin (θ + α) = a and sin (θ + β = b  0 < α , β , θ <  2  2 then cos (α – β) – 4ab cos (α – β) is equal to (A) 1 – 2a2 – 2b2 (C) 1– a2 – b2

(B)  1+ 2 a2 + 2b2 (D)  none of these

25. If sin x + cosec x + tan y + cot y = 4, where x and y  π y ∈ 0,  , then tan is a root of the equation 2 2   (A) a2 2+ 2α + 1 = 0 (C) 2a – 2α – 1 = 0

(B)  a2 + 2α – 1 = 0 (D)  none of these

26. The value of 2 sin2θ + 4 cos (θ + α) sin α sin θ + cos 2 (α + θ) is (A) cosθ + cosα (B)  independent of θ (C) independent of α (D)  none of these 27. The value of cos θ ⋅ cos 2θ ⋅ cos 22θ ... cos 2n – 1θ for π is θ= n 2 +1 1 (A) 1 (B)  2n (C) 2n (D)  none of these

21. If x sin a + y sin 2a + z sin 3a = sin 4a x sin b + y sin 2b + z sin 3b = sin 4b x sin c + y sin 2c + z sin 3c = sin 4c, then the roots of the equation z y+2 z−x t3 − t2 − + = 0; a, b, c ≠ nπ; are 2 4 8 (A) sin a, sin b, sin c (B) cos a, cos b, cos c (C) sin 2a, sin 2b, sin 2c (D) cos 2a, cos 2b, cos 2c

28. The sum of the series sinθ ⋅ sec3θ + sin 3θ ⋅ sec32θ + sin 32θ sec33θ + ... up to n terms is 1 (A) (tan 3n θ − tan θ ) (B) (tan 3nθ – tanθ) 2 (C) tan3nθ – tan3n – 1θ (D)  none of these

22. If a sin x + b cos (x + θ) + b cos (x – θ) = d, then the minimum value of | cosθ | is 1 1 (A) d 2 − a 2 (B)  d 2 − a2 2 |a| 2 |b|

sin nα (A) cos x1 cos xn

1 (C) a 2 − d 2 2 |b|

(D)  none of these

29. 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

(C)

sin ( n +1) α cos x1 cos xn

sin ( n −1) α (B)  cos x1 cos xn (D)  none of these

Trigonometric Ratios and Identities  15.13

Previous Year's Questions A− B+C

30. In a triangle ABC, 2ca sin is equal to: 2  [2002] (A) a2 + b2 − c2 (B)  c2 + a2 − b2 (C) b2 − c2 − a2 (D)  c2 − a2 − b2 4 xy

2 31. sin θ = ( x + y ) 2 is true if and only if: [2002] (A) x + y ≠ 0 (B)  x = y, x ≠ 0, y ≠ 0 (C) x = y (D)  x ≠ 0, y ≠ 0

1 − tan 2 15° is: 1 + tan 2 15°

[2002]

3 (D)  2 2 4 3

33. If tanθ = − , then sinθ is: 4 5

4 5 4 4 (C) but not − 5 5

4 5

− or (B) 

1 2

(2α + β) is equal to: [2002] (A) 1 (B)  −1 (C) zero (D)  none of these 2 2 35. If y = sin θ + cosec θ, θ ≠ 0 then: [2002] (A) y = 0 (B)  y≤2 (C) y ≥ −2 (D)  y≥2 36. In a triangle ABC, a = 4, b = 3, ∠A = 60°, then c is the root of the equation: [2002] (A) c2 − 3c − 7 = 0 (B)  c2 + 3c + 7 = 0 (C) c2 − 3c + 7 = 0 (D)  c2 + 3c − 7 = 0 [2002]

(B)  a, b, c are in AP (D)  a, b, c are in GP

38. The equation a sin x + b cos x = c where | c | > a 2 + b 2 has: [2002] (A) a unique solution (B) infinite number of solutions (C) no solution (D) none of the above 39. If α is a root of 25 cos 2 θ + 5 cosθ − 12 = 0 sin 2α is equal to :

27 α −β , then the value of cos is 2 65

[2004]

(A) −

3 130

3

(B)  130

6 6 − (D)  65 65

34. If sin (α + β ) = 1, sin (α − β ) = , then tan(α + 2β) tan

(A) a, c, b are in AP (C) b, a, c are in AP

21 65

(C)

4 5

(D)  none of these

A 5 C 2 37. In a ∆ABC , tan = , tan = , then: 2 6 2 5

41. Let α, β be such that π < α − β < 3π. If sin α + sin β = −

 [2002]

(A) − but not

[2003] (B)  are in G.P. (D)  satisfy a + b = c

the sides a, b and c (A) are in A.P. (C) are in H.P.

and cosα + cos β = −

3 (A) 1 (B) 

(C)

13 13 − (D)  18 18 C  A  3b 40. If in a triangle ABC a cos 2   + c cos 2   = then  2  2  2,

(C)

π < α < π then 2

[2002]

42. If u = a 2 cos 2 θ + b 2 sin 2 θ + a 2 sin 2 θ + b 2 cos 2 θ , then the difference between the maximum and minimum values of u2 is given by [2004] (A) 2 ( a 2 + b 2 ) (B)  2 a2 + b2 (C) (a + b)2 (D)  (a − b)2 43. The sides of a triangle are sinα, cosα and 1 + sin α cos α π

for some 0 < α < . Then the greatest angle of the tri2 angle is [2004] (A) 60° (B)  90° (C) 120° (D)  150° 44. In a triangle PQR, ∠ R =

 P  Q π . If tan  and tan  are  2  2 2

the roots of ax2 + bx + c = 0, a ≠ 0 then (A) a = b + c (B)  c=a+b (C) b = c (D)  b=a+c

[2005]

π

45. In a triangle ABC, ∠ C = If r is the inradius and R is 2 the circumradius of the the triangle ABC, then 2 (r + R) equals [2005] (A) b + c (B)  a+b (C) a + b + c (D)  c+a 46. If the roots of the quadratic equation x2 + px + q = 0 are tan 30° and tan l5°, respectively then the value of 2 + q − p is [2006] (A) 2 (B)  3 (C) 0 (D)  1

PRACTICE EXERCISES

32. The value of

24 24 − (B)  25 25

(A)

15.14  Chapter 15 47. The number of values of x in the interval [0, 3π] satisfying the equation 2 sin2x + 5sinx − 3 = 0 is [2006] (A) 4 (B)  6 (C) 1 (D)  2

53. If A = sin2x + cos4x then, for all real values of x,  [2011]

48. A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length x. The maximum area enclosed by the park is [2006]

(C) ″ A ″

3 3 2 8 1 2 (C) 2 x (D)  π x2 2 x (A) x (B) 

1 2

49. If 0 < x < and cos x + sin x = , then tan x is

PRACTICE EXERCISES

⋅

π 4

⋅

⋅

[2006]

3π 4

55. ABCD is a trapezium such that AB and CD are parallel and BC ⊥ CD . If angle ADB = B , BC = P and CD = q, then AB is equal to [2013]

[2009]

(B) p 2 cosθ + q 2 sin θ

( p 2 + q 2 )sin θ ( p cosθ + q sin θ ) 2

(D) 

(C)

tan A

(A) A is true and B is false (B) A is false and B is true (C) both A and B are true (D) both A and B are false Let

p2 + q2

p 2 + q 2 cosθ p cosθ + q sin θ

(A)

3 2

( p 2 + q 2 )sin θ p cosθ + q sin θ

cot A

+ 56. The expression can be written as 1 − cot A 1 − tan A  [2013] (A) secA cosecA + 1 (B)  tanA + cotA (C) secA + cosecA (D) sinA cosA + 1 1 k

57. Let f k ( x ) = (sin k x + cos k x ) where x ∈ R and k ≥ 1 then,

π 4 5 and sin (α − β ) = , where 0 ≤ α , β ≤ , 5 13 4

then tan 2α =

[2010]

56 19 (B)  33 12

(A) (C)

π 5π (B)  6 6

(C) (D) 

If cos( β − λ ) + cos( β − α ) + cos(α − β ) = − , then

cos(α + β ) =

13 3 ″ A″ 1 (D)  16 4

54. In a ΔPQR, if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to [2012] ⋅

(4 + 7 ) (1 + 7 ) (C) − (D)  4 3

51.

3 4

(A)

(4 − 7 ) (1 − 7 ) (A) (B)  3 4

50. Let A and B denote the statements A: cosα + cosβ + cosλ = 0 B: sinα + sinβ + sinλ = 0

13 ″ A ″ 1 (B)  1 ″ A″ 2 16

(A)

20 25 (D)  7 16

52. For a regular polygon, let r and R be the respective radii of the inscribed and the circumscribed circles. A false statement among the following is [2010] (A) There is a regular polygon with

r 1 = R 2

(B) There is a regular polygon with

r 2 = R 3

(C) There is a regular polygon with

3 r = R 2

(D) There is a regular polygon with

r 1 = R 2

the value of f 4 ( x ) − f 6 ( x ) equals

[2014]

1 6

1 3

1 4

1 12

(A) (B)  (C) (D)  58. If 5(tan2x – cos2x) = 2 cos 2x + 9, then the value of cos 4x is [2017] 1 2 7 3 (A)  (B)  (C) – (D) – 3 9 9 5

Trigonometric Ratios and Identities  15.15

ANSWER K EYS Single Option Correct Type 1. (B) 11. (A) 21. (B)

2. (C) 12. (A) 22. (B)

3. (B) 13. (B) 23. (B)

4. (A) 14. (A) 24. (A)

5. (A) 15. (C) 25. (B)

6. (A) 16. (B) 26. (B)

7. (B) 17. (A) 27. (B)

8. (B) 18. (A) 28. (A)

9. (C) 10. (B) 19. (B) 20. (B) 29. (B)

33. (B) 43. (C) 53. (D)

34. (A) 44. (B) 54. (B)

35. (D) 45. (B) 55. (D)

36. (A) 46. (B) 56. (A)

37. (A) 47. (A) 57. (D)

38. (C) 39. (B) 48. (C) 49. (C) 58. (C)

30. (B) 40. (A) 50. (C)

31. (B) 41. (A) 51. (A)

32. (C) 42. (D) 52. (B)

PRACTICE EXERCISES

Previous Years’ Questions

15.16  Chapter 15

HINTS AND EXPLANATIONS Single Option Correct Type ⇒ sinθ ≥ 0 and cosθ ≠ 0

1. Let P = cos a cos 2a cos 3a ... cos (999a) Let Q = sin a sin 2a sin 3a ... sin (999 a) Thus, 2999 PQ = 2 sin a cos a) (2 sin 2a cos 2a) (2 sin 3a cos 3a)... (2 sin (999 a) cos (999 a) = sin 2a sin 4a sin 6a ... sin (1998 a) = (sin 2a sin 4a sin 6a ... sin 998 a) [–sin (2π – 1000 a)] [–sin (2π – 1002a)].... [–sin (2π –1998a)] = sin 2a sin 4a ... sin 998 a sin 999 a sin 997 a ... sin a = Q [ ∴ 2π = 1999a] Clearly, Q ≠ 0

π  ⇒ θ ∈ [ 0, π ] −   2 5. We have, sinθ, sinφ and cosθ are in G.P. ⇒ sin 2 φ = sin θ cosθ

 cos 2 φ − sin 2 φ  4 cot 2 φ − 4 = 4   sin 2 φ   = 4 cos 2φ × cosec2φ ≥ 0

HINTS AND EXPLANATIONS

π π < 8 4

∴

(   of (i))

⇒ cos 25° − sin 25° = 2 − k 2

π π π ∴ cot > cot > tan 8 4 8 ⇒ a4 > a3(1) For α < 1, the function y = ax is a decreasing function. ∴  a2 < a1(2) π π > 1 > tan > 0 8 8

∴  a1 < 1 and a3 > 1  ∴  a1 < a3 From (1), (2) and (3), we get a2 < a1 < a3 < a4

(3)

3. We have, x cos2 3θ + y cos4θ  = 16 cos6θ + 9 cos2θ ⇒  x(4 cos3θ – 3 cosθ)2 + y cos4θ = 9 cos2θ  + 16 cos6θ ⇒  x(16 cos6θ + 9 cos2θ – 24 cos4θ) + y cos4θ = 9 cos2θ  + 16 cos6θ ⇒ 9x cos2θ + (y – 24x) cos4θ + 16x cos6θ = 9 cos2θ + 16 cos6θ On comparing, we get 9x = 9 and y – 24x = 0 ∴  x = 1 and y = 24. 4. | tan θ + sec θ | = | tan θ | + | sec θ | only if sec θ and tanθ both have same sign i.e., secθ ⋅ tanθ ≥ 0 or

(i)

∴  Roots of the given equation are always real. 6. We have, cos 50° = cos225° – sin225° = (cos 25° + sin 25°) (cos25° – sin 25°) = k(cos25° – sin 25°). But (cos 25° + sin 25°)2 + (cos25° – sin 25°)2 = 2

2. For α >1, the function y = ax is an increasing function

Again, cot

1 − cos 2φ sin 2θ = 2 2

⇒  cos 2φ = 1 – sin 2θ ≥ 0. Now, the discriminant of the given equation is

1 ∴ 2999 PQ = Q ⇒ P = 999 2

Since 0
log cosθ.

(1) (2)

Trigonometric Ratios and Identities  15.17

{

x >0 y

13. Let u = cosθ sin θ + sin 2 θ + sin 2α 2

1  1 t+ = t −  +2≥2 t  t equality holding iff t = 1 1 Also, t + = 2 sin θ ≤ 2, so that t should necessarily be 1. t i.e.,  x = y. 10. 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 Now, ∑

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 11. cos

2π 3π 4π 6π 8π  π π  cos cos cos cos cos = 12°  15 15 15 15 15 15  15 

2π 4π 8π  3π π   3π  =  cos cos cos cos  cos cos 2   15 15 15 15  15   15  π   3π  sin 24   sin 22   1 15    15  = × 2 π 3π  2  4 2 sin sin   15  15   12π  16π sin  sin  1  15  15 = × 2 π 2 3π 16 sin sin 15 15 1 = 64

3π  π    sin  π +  sin  π − 1 15  15    × =− 6 π 3π 2 sin sin 15 15

12. We have, sin α + sin β + sin γ – sin (α + β + γ) = sin α + sin β + sin γ – sin α cos β cos γ – cos α sin β cos γ – cos α cos β sin γ + sin α sin β sin γ = sin α[1 – cos β cos γ] + sin β [1 – cos α cos γ] + sin γ [1 – cos α cos β] + sin α sin β sin γ > 0 ∴ sin α + sin β + sin γ > sin (α + β + γ) sin (α + β + γ ) 2 ⋅ 2−



2 /2

= 21− (1 /

2)

Thus, 2sin θ + 2cos θ ≥ 21− (1 /

2)

, for all real θ.

16. Since A.M. ≥ G.M. ∴

2sin θ + 2− cos θ ≥ 2sin θ ⋅ 2− cos θ 2 1

1+ (sin θ − cos θ ) ⇒ 2sin θ + 2− cos θ ≥ 2 2

π

1+ 2 sin θ − 4  =2 Thus, 2sinθ + 2– cosθ is minimum 1

π π 3π  when sin  θ −  = −1 i.e., θ − = 2nπ + 4 4 2  7π , n∈ I. or, θ = 2nπ + 4 17. Let x0 = cosθ, then x1 =

1 (1 + cosθ ) 2

= cos θ/2, x2 = cos (θ/22), x3 = cos (θ/23),...and so on.  1 − x02   ∴   x1 x2 x3 …to ∞ 

HINTS AND EXPLANATIONS

9. Put t =

15.18  Chapter 15 sin θ θ θ θ cos cos 2 … cos n … ∞ 2 2 2 θ θ 2 sin cos 2 2 = θ θ θ cos cos 2 … cos n … ∞ 2 2 2

= a 2 (1 − b 2 ) + b 2 (1 − a 2 ) + 2ab (1 − a 2 ) (1 − b 2 )

=

+ 2a 2b 2 − 2ab (1 − a 2 ) (1 − b 2 ) = a 2 + b 2 . 20. Given, A + B + C = π A B C A B C ⇒ cot cot cot = cot + cot + cot (i) 2 2 2 2 2 2 A B C A B C But tan , tan , tan are in H. P. ⇒ cot , cot , cot 2 2 2 2 2 2

θ θ cos 2 2 2 2 = θ θ θ cos 2 cos 3 … cos n … ∞ 2 2 2 22 sin

A C B + cot = 2 cot (ii) 2 2 2 A B C B From (i) and (ii), we get cot ⋅ cot ⋅ cot = 3 cot 2 2 2 2 are in A. P. ⇒ cot

θ 2n = lim n→∞ θ cos n+1 2 2n sin



θ   sin 2n = lim θ  n→∞  θ  2n

A C ⋅ cot = 3 (iii) 2 2 A C cot + cot Now, 2 2 ≥ cot A cot C 2 2 2 ∴ cot

  1 =θ   cos θ n +1 2 

⇒

 1− x   = cosθ = x0 . ∴ cos   x1 x2 … ∞  2 0

θ cot 2 − 1 θ θ 2 18. 1 + cot θ − cot = 1 + − cot θ 2 2 2 cot 2

HINTS AND EXPLANATIONS

2

=

2 cot

θ θ θ  cot θ − 1 + cot 2 − 1 − 2 cot 2  2  2 2 2 = ≤0 θ θ 2 cot 2 cot 2 2

θ π  ∵ 0 < 2 < 2    θ ⇒ 1 + cot θ ≤ cot 2 19. We have, sin(α – β) = sin(α – θ + θ – β) = sin[ (θ – β) – (θ – α)] = sin(θ – β) cos(θ – α) – cos(θ – β) sin (θ – α) = ba − 1 − b 2 1 − a 2

2 cot

B 2 ≥ 3

[From (ii) an (iii)] 2 B ∴ cot ≥ 3 2 21. The first equation can be written as x sin a + y × 2 sin a cos a + z sin a(3– 4 sin2 a) = 2 × 2 sin a cos a cos 2a ⇒  x + 2y cos a + z (3 + 4 cos2 a – 4) = 4 cos a (2 cos2 a –1) as sin a ≠ 0 ⇒  8 cos3 a – 4z cos2 a – (2y + 4) cos a + (z – x) = 0 z y+2 z−x ⇒ cos3 a − cos 2 a − cos a + =0 2 4 8 which shows that cos a is root of the equation z y+2 z−x t3 − t2 − t+ =0 2 4 8 Similarly, from second and third equations we can verify that cos b and cos c are the roots of the above equation 22. a sin x + b cos(x + θ) + b cos(x – θ) = d ⇒  a sin x + 2b. cos x. cos θ = d

⇒ | d | ≤ a 2 + 4b 2 ⋅ cos 2 θ

⇒

d 2 − a2 ≤ cos 2 θ 4b 2

d 2 − a2 2 |b| 23. From the given condition, we have

and cos(α – β) = cos[θ – β) – (θ – α)] = cos(θ – β) cos(θ – α) + sin(θ – β) sin(θ – α)

⇒ |cosθ | ≥

= a 1 − b2 + b 1 − a2



Substituting these values in the given expression, we get cos2(α – β) + 2ab sin(α – β)

α α α   ⇒ 2  2 tan + 1 − tan 2  = 7 1 + tan 2  2 2 2    α α ⇒ ( 7 + 2) tan 2 − 4 tan + ( 7 − 2) = 0. 2 2

= ( a 1 − b 2 + b 1 − a 2 ) 2 + 2ab [ab − (1 − a 2 ) (1 − b 2 )]



2 tan (α / 2) 1 − tan 2 (α / 2) 7 + = 2 1 + tan (α / 2) 1 + tan 2 (α / 2) 2

Trigonometric Ratios and Identities  15.19 = 2sin θ (sin 2θ ⋅ cos 2θ ⋅ cos22θ ... cos2n – 1θ)

This quadratic has the roots 2( 7 + 2)

=

4±2 2 ( 7 + 2)

That is, tan (α / 2) = 3 /( 7 + 2) or 1/( 7 + 2). The given condition on α implies 0 < α/2 < π/12. Therefore, we get

α π α < tan ⇒ 0 < tan < 2 − 3. 2 12 2 Only the second root satisfies this condition, because 0 < tan





3 7+2

and,

= 7 −2>2− 3 1

7+2

1 ( 7 − 2). 3

= 1 − a 2 1 − b 2 + ab, the given expression can be written as 2 cos2 (α – β) – 1 – 4ab cos(α – β) = 2( 1 − a 2 1 − b 2 + ab) 2 −1 − 4 ab ( 1 − a 2 1 − b 2 + ab) = 2 [ (1 – a2) (1 – b2) – a2b2] – 1 = 1 − 2a2 − 2b2. 25. Given that sin x + cosec x + tan y + cot y = 4

π π and y = 2 4 ⇒ tan y = 1 ⇒ x=

2 tan y / 2 y y = 1 ⇒ tan 2 + + 2 tan − 1 = 0. 1 − tan 2 y / 2 2 2

26. We have, 2sin2θ + 4 cos(θ + α) sinα sinθ + cos 2(α + θ) = 2sin2θ + 2cos(θ + α) 2sinα sinθ + cos 2(α + θ) = 2sin2θ + 2cos(θ + α) [cos(θ – α) – cos(θ + α)] + cos 2(α + θ) = 2sin2θ + 2cos(θ + α) cos(θ – α) – 2cos2(θ + α) + cos2(α + θ) = 2sin2θ + 2(cos2θ – sin2α) – 2cos2(θ + α) + [2 cos2(α + θ) – 1] = 2sin2θ + 2cos2θ – 2sin2α – 1 = 2(sin2θ + cos2θ) – 2sin2α – 1 = 2 – 2sin2α – 1 = 1 – 2sin2α = cos2α, which is independent of θ. 27. We have, cos θ ⋅ cos 2θ ⋅ cos 22θ ... cos 2n – 1θ =

=

1 sin 2n − 1θ ⋅ cos 2n − 1θ 2n−1 sin θ 1 1 ( 2 sin 2n− 1θ cos n−1 θ ) = n sin 2nθ 2n sin θ 2 sin θ

π 1   [sin (π − θ )] ∵ θ = n , ∴ 2nθ + θ = π  2n sin θ 2 +1   1 1 = n ⋅ sin θ = n . 2 sin θ 2 =

24. Using, cos(α – β) = cos[(α + θ) – (θ + β)] = cos(θ + α) cos(θ + β) + sin(θ + α) sin(θ + β)

⇒

=

1 = ( 7 − 2) < 2 − 3. 3

Hence, the required value of tan (α/2) is

1 ( 2 sin 2θ ⋅ cos 2θ ⋅ cos 22θ … cos 2n− 1θ ) 22 sin θ 1 (sin 22θ cos 22θ … cos 2n− 1θ ) = 2 2 sin θ .......................................... .......................................... =

1 ( 2 sin θ ⋅ cosθ ⋅ cos 2θ ⋅ cos 22θ … cos 2n − 1θ ) 2 sin θ

θ 2 28. We have, tan θ = 2θ 1 − tan 2 2 tan

2 ⇒

θ θ 2 cot cot 1 2 2 = = cot θ 1 − 1 2θ cot − 1 θ 2 cot 2 2

θ −1 2 ⇒ cot θ = θ 2 cot 2 cot 2

θ cot 2 − 1 θ θ 2 Now, cot − 1 − cot θ = cot − 1 − θ 2 2 2 cot 2 =

2 cot 2

θ θ θ − 2 cot − cot 2 + 1 2 2 2 θ 2 cot 2 2

=

cot 2

 θ  θ θ − 2 cot + 1  cot − 1 2   . 2 2 = θ θ 2 cot 2 cot 2 2

θ π Now, since 0 < θ < π, ∴ 0 < < 2 2 positive. Also, square of a real number ≥ 0 ∴ cot

⇒ cot

θ θ − 1 − cot θ ≥ 0 ⇒ cot ≥ 1 + cot θ . 2 2

29. We have,

sin ( x2 − x1 ) sin α = cos x1 cos x2 cos x1 cos x2

θ cot is 2

HINTS AND EXPLANATIONS

4 ± 16 − 4 ( 7 + 2)( 7 − 2)

α tan = 2

15.20  Chapter 15 =

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 x – tan x ] n n –1 = tan xn – tan x1 =

sin xn cos x1 − cos xn sin x1 cos xn cos x1

=

sin ( xn − x1 ) sin ( n − 1) α = cos x1 cos xn cos x1 cos xn

Previous Year's Questions 30. We know that: The sum of angles A + B + C = π ⇒A + C = π − B A− B +C π ⇒ = −B 2 2  A− B +C  π  ∴ 2ca sin   = 2ca sin  2 − B  2      a2 + c2 − b2  = 2ac cos B = 2ac   2ac  2 2 2 =a +c −b 31. ∵ sin θ < 1, sin 2θ < 1 4 xy ⇒ ≤1 ( x + y)2 ⇒ 0 ≤ ( x + y ) − 4 xy

∴ h=

p 2 + b 2 = 16 + 9 = 25

⇒h=5

p 4 = h 5 But tanθ is negative which is possible only if θ lies in second or fourth quadrant. 4 4 ∴ sin θ may be or = − . 5 5 ∴ sin θ =

34. ∵ sin(α + β ) = 1 ⇒ sin(α + β ) = sin ⇒ α +β =

π (i) 2

and sin (α − β ) =

2

HINTS AND EXPLANATIONS

π 2

1 2

π (ii) 6 Solving eqs. (i) and (ii), we find

⇒ x 2 + y 2 + 2 xy − 4 xy ≥ 0

⇒ α −β =

⇒ ( x − y)2 ≥ 0 which is true for all real x and y provided x + y ≠ 0, otherwise 4 xy will be meaningless. ( x + y)2

∴ tan(α + 2 β ) tan( 2α + β )

Info Box! sec2θ =

4 xy is possible only, if x = y. ( x + y )2

32. Key Idea: Cosine in terms of tangent 1 − tan 2 θ cos 2θ = 1 + tan 2 θ 1 − tan 2 15° = cos 30° 1 + tan 2 15° 3 2 33. Key Idea: tanθ is negative in second and fourth quadrants. 4 ∴ tan θ = − 3 Here p = 4 and b = 3 ∴ =

 2π = tan   3

  5π   tan  6    

π  π  =  − cot   − cot  6  3  = 3×

2 3

=1

2 2 35. ∵ y = sin θ + cos ec θ

= (sin θ − cos ecθ ) 2 + 2 ⇒

y≥2

But at θ = 0, y becomes meaningless ∴y≥2 36. Using cos A =

b2 + c2 − a2 2bc

We write for a = 4, b = 3 and ∠A = 60° c 2 + 9 − 16 ∴ cos60° = 2×3×c

Trigonometric Ratios and Identities  15.21 1 c2 − 7 = 2 2 × 3c

⇒ c2 − 7 = 3c ⇒ c2 − 3c − 7 = 0 Thus, c is the root of above equation. 37. Key Idea: tan Now tan

⇒

A ( s − b)( s − c) = 2 s( s − a )

A C 5 2 tan = × 2 2 6 5

( s − b)( s − c) ( s − a)( s − b) 1 ⋅ = s( s − a ) s( s − c ) 3

s−b 1 ⇒ = s 3 ⇒ 3s − 3b = s ⇒ 2s = 3b ⇒ a + b + c = 3b ⇒ a + c = 2b a, b, c are in AP.

2 + 2 cos(α − β ) =

1170 (65) 2

9 α − β  α − β  ⇒ cos 2   = 130 ⇒ cos  2  2     −3  π α − β 3π  = ∵ 2 < 2 < 2  . 130   42. We have u = a 2 cos 2 θ + b 2 sin 2 θ + a 2 sin 2 θ + b 2 cos 2 θ =

a2 + b2 a2 − b2 a2 + b2 b2 − a2 + cos 2θ + + cos 2θ 2 2 2 2 2

2

 a2 + b2   a2 − b2  2 ⇒ u 2 = a2 + b2 + 2   −  cos 2θ  2   2  Now, min value of u2 = a2 + b2 + 2ab And the max. value of u2 = 2(a2 + b2)

38. Using − a 2 + b 2 ≤ a sin x + b cos x ≤ a 2 + b 2 . Given | c | > a 2 + b 2 which inplies c = a sin x + b cos x > a + b , or 2

2

c = a sin x + b cos x < − a 2 + b 2 which are reverse inequalities. Then no solution exists for a sin x + b cos x = c. 39. ∵ θ = α satisfies the equation 25 cos2θ + 5 cosθ − 12 = 0 ∴ 25cos2α + 5cosα − 12 = 0 ⇒ 25cos2α + 20cosα 15cosα − 12 = 0 ⇒ 5cosα(5cosα + 4) − 3(5cosα + 4) = 0 4 3 ⇒ cos α = − , 3 5 But

−21 −27 and cos α + cos β = . 65 65 Squaring and adding, we get 41. Given that sin α + sin β =

π 1 and if cos (p x ) > 1 then cos(p x - 4 ) < 1 (since their product = 1). But both of these are not possible (since cosθ cannot be greater than 1). \ cos(p x - 4 ) = 1 and cos(p x ) = 1 ∴  x – 4 = 0 and x = 0 But x = 0 is not possible, ∴ x = 4 is the only solution. 7. The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos2x + a3 sin2x = 0 for all x is (A) 0 (B)  1 (C) 3 (D)  infinite

×

Solution (A)

We have, 4cosx (2 – 3 sin2x) + (cos 2x + 1) = 0 ⇒ 4cosx (3cos2x – 1) + 2 cos2x = 0 ⇒ 2cosx (6cos2x + cosx – 2) = 0 ⇒ 2cosx (3cosx + 2) (2cosx – 1) = 0 ⇒ either cosx = 0 which gives x = π/2 or cosx = –2/3, which gives no value of x for which 0 ≤ x ≤ π/2 or cosx = 1/2, which gives x = π/3 So, the required difference = π/2 – π/3 = π/6. 9. Solution of the equation 4cot 2θ = cot2θ – tan2θ is p p (A) q = np ± (B)  q = np ± 3 2 (C) q = np ± p (D)  none of these 4 Solution (C)

We have, 4cot 2θ = cot2θ – tan2θ 4 4 Þ = - tan 2q 2 tan 2q tan q Put \

tan 2q =

2 tan q 1 - tan 2q

4(1 - tan 2q ) 1 - tan 4q = 2 tanq tan 2q

⇒ (1 – tan2θ) [2tanθ – (1 + tan2θ)] = 0 ⇒ (1 – tan2θ) (tan2θ – 2tanθ  + 1) = 0

16.4  Chapter 16 ⇒ (1 – tan2θ) (tanθ – 1)2 = 0

(D) irrational of the form integer

p Þ tan q = 1, -1. \ q = np ± 4

Solution (C)

10. Let α, β be any two positive values of x for which 2cosx, | cosx | and 1 – 3 cos2x are in G.P. .The minimum value of | α – β | is p p (A) (B)  3 4 p (C) (D)  none of these 2 ×

We have, sinπ (x2 + x) = sinpx2 ⇒ π (x2 + x) = nπ + (–1)npx2 ∴ Either x2 + x = 2m + x2  ⇒  x = 2m ∈ Z. or x2 + x = k – x2, where k is an odd integer -1 ± 1 + 8k Þ 2x2 + x - k = 0 Þ x = 4 For least positive non-integral solution,

×

×

Solution (D)

Since 2cosx, | cosx | and 1 – 3cos2x are in G.P. 2 ∴ cos x = 2cosx(1 – 3cos2x) ⇒ 6cos3x + cos2x – 2cosx = 0 ⇒ cosx(2cosx – 1) (3cosx + 2) = 0 p p 1 2 æ 2ö Þ cos x = 0, , \ x = , , cos -1 ç - ÷ 2 3 2 3 è 3ø

p p ,b = , ( 2 3

then

∴

∴ If a =

α, β are are positive)

11. The most general values of θ for which sinθ – cosθ = a ∈ R (1, a2 – 6a + 10) are given by p p p p (A) np + ( -1) n - (B)  np + ( -1) n + 4 4 4 4 p (C) 2np + (D)  none of these 4 Solution (B)

We have, min sinθ – cosθ =  a Î R {1, a2 – 6a + 11}. Since a2 − 6a + 11 = (a – 3)2 + 2 > 2 for all a

pö 1 p æ \ sin q - cos q = 1 Þ sin ç q - ÷ = = sin 4 4 2 è ø p p Þ q - = np + ( -1) n 4 4 p p + , where n ∈ Z. 4 4

12. The least positive non-integral solution of the equation sinπ(x2 + x) = sin px2 is (A) rational (B) irrational of the form p (C) irrational of the form integer

p - 1 where p is an odd integer. -1 + 1 + 8k = , 4 4 p p 13. In the interval êé - , úù , the equation logsinθ(cos2θ) ë 2 2û = 2 has x=



(A) no solution (B) a unique solution (C) two solutions (D) infinitely many solutions Solution (B)

p p £q £ 2 2 ∴ –1 ≤ sinθ ≤ 1, here 0 < sinθ < 1 Now, logsinθcos2θ = 2 ⇒ cos2θ = sin2θ ⇒ 1 – 2sin2θ = sin2θ 1 Þ 3 sin 2q = 1 Þ sin 2q = 3 1 Þ sinq = (∵ 0 < sin q < 1) 3 The given equation has a unique solution. We have, -

p |a - b | = 6

\ q = np + ( -1)n

p +1 where p is an even , 4

p -1 where p is an odd , 4

14. The number of solutions of the equation sinx = cos 3x in [0, p] is (A) 1 (B)  2 (C) 3 (D)  4 Solution (C)

The given equation can be written as sinx = 4cos3x – 3cosx i.e., sec2x tanx + 3sec2x – 4 = 0 In terms of tanx, this leads to the equation tan3x + 3tan2x + tanx – 1 = 0 ⇒ (tan x + 1)(tan2x + 2tanx – 1) = 0 ⇒ tanx = –1 or tan2x = 1 3p p 5p i.e., x= , , 4 8 8 15. The number of values of x in [0, 2p] satisfying the equation | cos x - sin x | ³ 2 is,

Trigonometric Equations  16.5 (A) 0 (B)  1 (C) 2 (D)  3

æ p+qö æ p-qö Þ cos ç ÷ x cos ç 2 ÷ x = 0 2 è ø è ø ( 2n + 1)p æ p+qö cos ç ÷ x = 0 Þ x = p + q , n = 0, ± 1, ± 2, … è 2 ø 2p which forms an A.P. with common difference = p+q ( 2n + 1)p æ p-qö cos ç ÷ x = 0 Þ x = p - q , n = 0, ± 1, ± 2, … 2 è ø 2p which forms an A.P. with common difference = p-q

Solution (C)

Given equation is | cos x - sin x | ³ 2 Since | cos x - sin x | £ 1 + 1 = 2 ∴ We must have | cos x - sin x | = 2

pö pö æ æ cos ç x + ÷ = 1 Þ cos ç x + ÷ = 1, - 1 4 4ø è è ø p \ x + = 0, 2p , 4p , 6p , . . .  or  π, 3π, 5π. . . 4 3p 7p \ x= , 4 4

Þ

3 1 sin 2 x - sin x + 2 2

16. If |cos x |

In each of the above cases, n ∈ Z, where Z is the set of integers

= 1, then possible values of x are:

(A) np or np + ( -1) n

p ,nÎ I 6

SOLUTION OF AN EQUATION OF THE FORM

p p (B) np or 2np + or np + ( -1) n , n Î I 2 6 p (C) np + ( -1) n , n Î I 6 (D) nπ, n ∈ I Solution (C, D)

The equation holds if | cos x | = 1 i.e., if x = nπ , n ∈ I 3 1 If | cosx | ≠ 1, then sin 2 x - sin x + = 0 2 2 1 Þ sin x = 1 or 2 sin x ≠ 1, as in that case cos x = 0 1 p \ sin x = Þ x = np + ( -1) n 2 6 17. A set of values of x satisfying the equation æ1 ö æ1 ö cos 2 ç px ÷ + cos 2 ç qx ÷ = 1 2 è ø è2 ø form an arithmetic progression with common difference 2 2 (A) (B)  p -q p+q (C)

ERROR CHECK

p p+q

(D)  none of these

Solution (D)

The given equation can be written as 1 + cospx + 1 + cosqx = 2

acosθ + bsinθ = c The given equation is acosθ + bsinθ = c.(1)

Divide throughout by



i.e., by



(coeff. of cosq ) 2 + (coeff. of sinq ) 2 , we get

a a +b 2

a2 + b2

2

cosq +

b a +b 2

2

sinq =

c a + b2 2

(2)

Let α be the least +ve angle such that a b = cos a and = sin a 2 2 2 a +b a + b2 ∴  (2) becomes c cos a cos q + sin a sin q = 2 a + b2 ⇒ cos(θ – α) = cosβ (say), c where cos b = 2 a + b2 ∴ θ – α = 2nπ ± β ⇒ θ = 2nπ ± β + α, where n ∈ Z.

WORKING RULE TO SOLVE AN EQUATION OF THE FORM acosq + bsinθ = c 1. Divide throughout by

a2 + b2

i.e., by (coeff. of cosq ) 2 + (coeff. of sinq ) 2 2. Write L.H.S. as a single cosine. 3. Use the formula

θ  = 2nπ ± α, n ∈ Z  for cosθ  = cosα

16.6  Chapter 16

ERROR CHECK Check that |c| £ a2 + b2 . If it is not satisfied, no real solution exits.

QUICK TIPS Squaring should be avoided as far as possible. If squaring is done, then check for extra solutions. For example, consider the equation sinθ + cosθ = 1. On squaring, we get 

1 + sin2θ  = 1 or sin 2θ  = 0 np , n = 0, ±1, ±2, ... ⇒  q = 2 3p do not satisfy The values of the angle, θ = π and q = 2 the given equation. So, we get extra solutions. Thus, if squaring is must, verify each of the solutions. Never cancel a common factor containing ‘θ ’ from the two sides of an equation. For example, consider the equation tan q = 2 sin q . If 1 , which we divide both sides by sinθ, we get cos q = 2 is clearly not equivalent to the given equation as the solutions obtained by sinx = 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.  Make sure that the answer should not contain any value of unknown ‘θ ’ which makes any of the terms undefined.  If tanθ or secθ is involved in the equation, θ should not be an odd multiple of π/2.  If cotθ or cosecθ is involved in the equation, θ should not be a multiple of π or 0.  The value of f ( q) is always positive. For example,

cos2 q = |cos q | and not ± cos q .

All the solutions should satisfy the given equation and lie in the domain of the variable of the given equation.  Check the validity of the given equation, e.g., 2cosθ + sinθ = 3 can never be true for any θ as the value (2cosθ + sinθ) can never exceed 22 + 12 = 5. So there is no solution for this equation. 



ERROR CHECK x a then always write general solution first in terms of the argument and then find x. If the argument in inequalities is not x but x + a or ax or

SOLUTIONS OF BASIC TRIGONOMETRIC INEQUALITIES Inequality

Solution (n ∈ Z)

sinx > α, where |α| < 1

sin–1α + 2nπ < x < π – sin–1α + 2nπ

sinx < α, where |α| < 1

–π – sin–1α + 2nπ < x < sin–1α + 2nπ

cosx > α, where |α| < 1

–cos–1α + 2nπ < x < cos–1α + 2nπ

cosx < α, where |α| < 1

cos–1α + 2nπ < x < 2π – cos–1α + 2nπ π tan−1α + nπ < x < + nπ

tanx > α, where –∞ < α < ∞ tanx < α, where –∞ < α < ∞

2 π −1 − + nπ < x < tan α + nπ 2

cotx > α, where –∞ < α < ∞

nπ < x < cot–1α + nπ

cotx < α, where –∞ < α < ∞

cot–1α + nπ < x < π + nπ Solution (C)

SOLVED EXAMPLES

We have, | cosx | ≤ sinx  ⇒ sinx ≥ 0

é pù ép p ù (A) x Î ê0, ú (B)  xÎê , ú ë 4û ë4 2û é p p ù é p 3p ù (C) x Î ê , ú È ê , ú ë4 2û ë2 4 û

(D)  none of these

(

⇒ x ∉ (π, 2π)

∴

18. If 0 ≤ x ≤ 2π and | cosx | ≤ sinx, then

| cosx | ≥ 0)

If x = 2π, | cos2π | ≤ sin2π which is not possible, ∴ x ∈ [0, p] é pù If x Î ê0, ú , then | cos x |£ sin x Þ cos x £ sin x ë 2û

Trigonometric Equations  16.7

æp ö x Î ç , p ÷ , then è2 ø | cosx | ≤ sinx  ⇒  – cosx ≤ sinx ⇒ tanx ≤ –1 ( cosx < 0) If

∴

Þ

æ p 3p ù é p p ù æ p 3p ù x Îç , ú . \ x Î ê , ú È ç , ú 2 4 è û ë4 2û è 2 4 û

19. The equation (cosp – 1) x2 + (cosp) x + sinp = 0, in the variable x has a real root. then p can take any value in the interval æ -p p ö (A) (0, π) (B)  ç 2 .2÷ è ø (B) (–π, 0) (C)  (0, 2π) Solution (A)

Discriminant = cos2p – 4 (cos p – 1)sin  p For equation to have real roots, cos2p – 4(cosp – 1) sinp ≥ 0 ⇒ cos2p ≥ 4(cosp – 1) sinp p ⇒ cos2p ≥ –8 sin2 sinp 2 3p p For p = 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) 20. The equation sin4x + cos4x = a has a solution for (A) all of values of a (B) a = 1 1 1 (C) a = (D)  < a –3 (B) k < –2 (C) –3 ≤ k ≤ –2 (D) k is any positive integer Solution (C)

We have, sin4x – (k + 2) sin2x – (k + 3) = 0 ( k + 2) ± ( k + 2) 2 + 4( k + 3) 2 ( k + 2) ± ( k + 4) = 2

Þ sin 2 x =

⇒ sin2x = k + 3 ( sin2x = –1 is not possible) Since 0 ≤ sin2x ≤ 1, ∴ 0 ≤ k + 3 ≤ 1 or –3 ≤ k ≤ –2. ∴

ép p ù xÎê , ú ë4 2û

Þ



22. The equation cos2x + asinx = 2a – 7 has a solution for (A) all a (B)  a>6 (C) a < 2 (D)  a ∈ [2, 6] Solution (D)

Given equation is cos2x + asinx = 2a – 7 ⇒ 1 – 2sin2x + asinx = 2a – 7 ⇒ 2sin2x – asinx + 2a – 8 = 0 a ± a 2 - 16( a - 4) a ± ( a - 8) = 4 4 a-4 Þ sin x = or sinx = 2 (not possible) 2 a-4 \ sinx = 2 a-4 ∵ -1 £ sin x £ 1, \ - 1 £ £1 Þ 2 £ a £ 6 2 Þ sin x =

23. The set of all x in (–π, π) satisfying | 4 sin x - 1 | < 5 is given by æ p 3p ö (A) ç - , ÷ è 10 10 ø

æ p ö (B)  ç - 10 , p ÷ è ø

3p ö æ (C) (–π, π) (D)  ç -p , 10 ÷ è ø Solution (A)

We have, |4 sin x - 1| < 5 Þ - 5 < 4 sin x - 1 < 5 æ 5 -1 ö 5 +1 Þ -ç < sin x < ç 4 ÷÷ 4 è ø p p Þ - sin < sin x < cos 10 5 æ -p ö æp p ö Þ sin ç ÷ < sin x < sin ç 2 - 5 ÷ è 10 ø è ø æ -p ö æ 3p ö Þ sin ç ÷ < sin x < sin ç 10 ÷ 10 è ø è ø æ p 3p ö Þ x Î ç - , ÷ [∵ x Î ( -p , p )] è 10 10 ø 24. The number of values of x in [0, 2p] satisfying the equation |cos x - sin x| ³ 2 , is (A) 0 (B)  1 (C) 2 (D)  3 Solution (C)

Given equation is | cos x - sin x | ³ 2 .

16.8  Chapter 16 Since | cos x - sin x | £ 1 + 1 = 2 , ∴  we must have |cos x - sin x| = 2 Þ

pö pö æ æ cos ç x + ÷ = 1 Þ cos ç x + ÷ = 1, -1. 4ø 4ø è è

p = 0, 2p , 4p , 6p , ..., π, 3π, 5π, .... 4 3p 7p \ x= , 4 4 \ x+

25. The set of all x in (–π, π) satisfying |4 sin x - 1| < 5 is given by æ p 3p ö æ p 3p ö (A) ç - , ÷ (B)  ç 10 , 10 ÷ è ø è 10 10 ø

p -3p ö (C) çæ , ÷ è 10 10 ø

(D)  none of these

Solution (A)

Since | 4 sin x - 1 | < 5 Þ - 5 < 4 sin x - 1 < 5 æ 5 -1 ö 5 +1 Þ -ç < sin x < ç 4 ÷÷ 4 è ø Þ - sin

p p < sin x < cos 10 5

æ p ö æp p ö Þ sin ç - ÷ < sin x < sin ç - ÷ è2 5ø è 10 ø 3p æ p ö Þ sin ç - ÷ < sin x < sin 10 è 10 ø æ p 3p ö \ x Îç- , ÷ è 10 10 ø

Trigonometric Equations  16.9

NCERT EXEMPLARS

2 2 2. If ¦ ( x ) = cos x + sec x, then

(A) f(x) < 1 (C)  2 < f (x) < 1

(B)  f(x) = 1 (D)  f (x) ≥ 2

1 1 3. If tan q = and tan f = , then the value of θ + ϕ is 2 3

p (B)  π 6 p (C)  0 (D)  4

(A) 

4. Which of the following is not correct ? 1 5 1 (C)  secq = - 2 (A) sin θ = −

(B)  cos q = 1 (D)  tan q = 20

5. The value of tan°tan2°tan3°……tan89° is (A) 0 (B) 1 1 (C)  (D)  Not defined 2 1 - tan 2 15° is t + tan 2 15° (A)  1 (B)  3 6. The value of

(C) 

3 (D)  2 2

7. The value of cos1° cos2° cos3° ….. cos179° is 1 (A) (B)  0 2 (C) 1

(D) – 1

10. Which of the following is correct? (A)  sin 1° > sin 1 (B)  sin 1° < sin 1 p sin 1 (C) sin 1° = sin 1 (D)  sin 1° = 18° 11. If tan a =

m 1 and tan b = , then α + β is m +1 2m + 1

equal to

p p (B)  2 3 p p (C)  (D)  6 4 (A) 

12. The minimum value of 3 cos x + 4 sin x + 8 is (A) 5 (B) 9 (C) 7 (D) 3 13.

The value of tan 3A – tan 2A – tan A is (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 the above

14. The value of sin (45° + θ) – cos (45° – θ) is (A)  2 cos θ (B)  2 sin θ (C) 1

(D) 0

æp ö æp ö 15. The value of cot ç + q ÷ cot ç - q ÷ is è4 ø è4 ø (A) – 1 (B) 0 (C)  1 (D)  None of these 2 2 16. cos 2q cos 2f + sin (q - f ) - sin ( +f ) to

cos 2 (q + f ) (A)  sin 2 (q + f ) (B)  cos 2 (q - f ) (C) sin 2 (q - f ) (D) 

8. If tan θ = 3 and θ lies in third quadrant, then the value of sin θ is 1 1 (A)  (B)  10 10 3 -3 (C)  (D)  10 10

17. The value of cos 12° + cos 84° + cos 156° + cos 132° is

9. The value of tan 75° – cot 75° is

18. If tan A =

(A) 2 3 (B)  2+ 3 (C) 2 - 3 (D)  1

1 (A)  (B)  1 2 1 1 (C)  - 2 (D)  8 1 1 and tan B = , then tan(2A + B) is equal to 2 3 (A) 1 (B) 2 (C) 3 (D) 4

NCERT EXEMPLARS

1. If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to (A) 1 (B) 4 (C)  2 (D)  None of these

16.10  Chapter 16 19. The value of sin

p 13p is sin 10 10

7p 4p + sin (B) 1 18 9 p 3p p p (C) cos + cos (D)  cos + sin 6 7 9 9 (A) sin

1 1 (A)  (B)  2 2 1 (C) - (D)  1 4

26. If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of 2 cot A – 5 cos A + sin A is -53 23 (A)  (B)  10 10

20. The value of sin 50° – sin 70° + sin 10° is (A) 1 (B) 0 1 (C)  (D)  2 2

(C) 

21. If sinθ + cos θ = 1, then the value of sin2θ is 1 (A)  1 (B)  2 (C) 0 22. If a + b =

27. The value of cos2 48º – sin2 12º is

(B) 2 (D)  Not defined

-4 and θ lies in third quadrant, then the 5 p value of cos is 2 1 1 (A)  (B)  5 10 1 1 (C)  (D)  5 10 23. If sin q =

NCERT EXEMPLARS

5 +1 5 -1 (B)  8 8 5 +1 5 +1 (C)  (D)  2 2 5 (A) 

(D) –1

p , then the value of (1 + tanθ) (1 + tanβ) is 4

(A) 1 (C)  – 2

37 7 (D)  10 10

24. The number of solutions of equation tan x + sec x = 2 cos x lying in the interval [0, 2 π] is (A) 0 (B) 1 (C) 2 (D) 3 p p 2p 5p 25. The value of sin + sin + sin + sin is 18 9 9 18

1 1 and tan b = , then cos 2 α is equal to 7 3 (A)  sin 2β (B)  sin 4β (C)  sin 3β (D)  cos 2β 28. If tan a =

a 29. If tan q = , then b cos 2θ + a sin 2θ is equal to b (A) a (B)  b a (C) (D)  None of these b 1 30. If for real values of x, cos q = x + , then x (A) θ is an acute angle (B) θ is right angle (C)  θ is an obtuse angle (D)  No value of θ is possible

ANSWER K EYS   1. (C) 2. (D)   11. (D) 12.  (D)   21. (C) 22.  (B)

3. (D) 13. (A) 23. (C)

4.  (C) 14.  (D) 24. (C)

5. (B) 15. (C) 25.  (A)

6.  (C) 16.  (B) 26. (B)

7. (B) 17. (C) 27.  (A)

8. (C) 18.  (C) 28. (B)

9. (A)  10.  (B) 19. (C)  20.  (B) 29.  (B)   30. (D)

Trigonometric Equations  16.11

HINTS AND EXPLANATIONS

1 1 3+ 2 + ⇒ 5 2 3 tan (q + f ) = Þ tan (q + f ) = 6 = = 1 1 1 6 -1 5 1- • 2 3 6 p ⇒ tan (q + f ) = tan 4 p ∴ q +f = 4 4. We know that, the range of secθ is R – (– 1, 1). Hence, secθ cannot be equal to 1 . 2 5. Given expression, tan 1° tan2° tan3° …..tan 89° = tan 1° tan2° ….. tan 45°·tan(90° – 44°) tan (90° – 43°)  ….. tan(90° – 1°) = tan 1°·cot 1°· tan 2°·cot 2°·cot 2° …. Tan 89°·cot 89° = 1·1 ….. 1·1 = 1 6. Given expression, 1 - tan 2 15° 1 + tan 15° Let q = 15° 1 - tan 2 q We know that, cos 2q = 1 + tan 2 q 1 - tan 2 15° ∴ cos 30° = 1 + tan 2 15° 2

⇒

1 - tan 2 15° 3 =  1 + tan 2 15° 2

⇒

sec2 θ = 1 + tan2 θ secq = 1 + 9 = ± 10

⇒

secq = - 10 1 secq = 10

⇒

sin q = ± 1 -

 ∴

sin q = -

⇒

1 9 3 =± =± 10 10 10 [since, θ lies in third quadrant]

3 10 9. Given expression, tan 75° – cot 75° sin 75° cos 75° = cos 75° sin 75°



sin 2 75° - cos 2 75° sin 75° • cos 75° 2 cos15° = sin150° -2 cos ( 90° + 60° ) = sin ( 90° + 60° ) +2 sin 60° = cos 60° =

=

2.

3 2 =2 3 1 2

10. We know that, if θ is increasing, then sin θ is also increasing. ∴ sin 1° < sin 1  [∵ 1 rad = 57°30 ’] m 1 and tan b = 2m + 1 m +1 tan a + tan b tan (a + b ) = 1 - tan a •tan b

11. Given that, tan q = Now,

1 m + m + 1 2 m +1 ⇒ tan (a + b ) = æ m öæ 1 ö 1- ç ÷ç ÷ è m + 1 ø è 2m + 1 ø ⇒ tan (a + b ) = é 3ù ê∵ cos30° = ú 2 êë úû

7. Given expression, cos1° cos2° cos3° …. cos 179° = cos 1° cos2° ….. cos90° ….. cos 179° =0 8. Given that, tan θ = 3

⇒

[∵ cos 90° = 0]

m ( 2m + 1) + m + 1 ( m + 1) ( 2m + 1) - m

2m 2 + m + m + 1 2m 2 + 2m + m + 1 - m 2m 2 + 2m + 1 ⇒ tan (a + b ) = Þ tan (a + b ) = 1 2m 2 + 2m + 1 p ⇒ a + b = 4 12. Given expression, 3 cos x + 4 sin x + 8 Let y = 3 cos x + 4 sin x + 8 ⇒ y – 8 = 3 cos x + 4 sin x ⇒ tan (a + b ) =

HINTS AND EXPLANATIONS

1. Given that, sin θ + cosec θ = 4 ⇒ sin2θ + cosec2θ + 2sinθ . cosecθ = 4 ⇒ sin2θ + cosec2θ = 4 – 2 ⇒ sin2θ + cosec2θ = 2 2. Given that, f (x) = cos2 x + sec2x We know that, AM ≥ GM cos 2 x + sec 2 x ³ cos 2 x • sec 2 x 2 2 2 [∵ cos x • sec x = 1] ⇒ cos x + sec x ³ 2  f ( x) ³ 2 ⇒ 1 1 3. Given that, tan q = and tan f = 2 3 tan q + tan f Now, tan (q + f ) = 1 - tan q • tan f

16.12  Chapter 16 ∴ Minimum value of y - 8 = - 9 + 16 ⇒ y – 8 = – 5   ⇒  y = – 5 + 8 ⇒ y=3 Hence, the minimum value of 3 cos x + 4 sin x + 8 is 3. 13. Let 3A = A + 2A tan 3A = tan (A + 2A) tan A + tan 2 A ⇒ tan 3 A = 1 - tan A•tan 2 A ⇒ tan A + tan 2A = tan 3A – tan 3A·tan 2A · tan A ⇒ tan 3A – tan 2A – tan A = tan 3A · tan 2A · tan A 14. Given expression, sin(45° + θ) – cos(45° – θ)

= sin 45° · cos θ + cos 45° · sin θ – cos 45° · cos θ · – sin45° . sin θ =

1 1 1 1 .cosq + • sin q • cosq sin q 2 2 2 2

=0 15. Given expression,

HINTS AND EXPLANATIONS

æp æp ö ö cot ç + q ÷ - cot ç - q ÷ è4 è4 ø ø p p æ ö ö æ ç cot 4 cot q - 1 ÷ ç cot 4 cot q + 1 ÷ =ç ÷•ç ÷ ç cot p + cot q ÷ ç cot q - cot p ÷ 4 4 ø è ø è æ cot q - 1 ö æ cot q + 1 ö =ç ÷•ç ÷ è cot + 1 ø è cot q - 1 ø =1

16. Given expression, cos 2q cos 2f + sin 2 (q - f ) - sin 2 ( +f ) = cos 2q • cos 2f + sin q - f + q + f • sin q - f - q - f ( ) ( ) = cos 2q • cos 2f - sin 2q • sin 2f

= cos ( 2q + 2f ) = cos 2 (q + f )

17. Given expression, cos 12° + cos 84° + cos 150° + cos 132° = cos 12° + cos 150° + cos 84° + cos 132° æ 12° + 150° ö æ 12° - 150° ö •cos ç = 2 cos ç ÷ ÷ 2 2 è ø è ø 84 132 ° + ° ° ° 84 – 132 æ ö æ ö + 2 cos ç ÷·cos ç ÷ 2 2 è ø è ø = 2 cos 84° cos 72° + 2 cos 108° · cos 24º = 2 cos 84º cos (90° – 18°) + 2 cos (90° + 18°) · cos 24° = 2 cos 84° sin 18° – 2 sin 18° · cos 24° = 2 sin 18° (cos 84° – cos 24°) æ 84° + 24° ö æ 84° - 24° ö = 2 sin18° • 2 sin ç ÷ • sin ç ÷ 2 2 è ø è ø = 4 sin 18° · sin 54° sin 30° æ 5 -1 ö 1 = -4 çç ÷÷·cos 36°· 4 2 è ø = – 4 sin 18° · sin 54° sin 30°

æ 5 -1 ö 1 = -4 çç ÷÷·cos 36°· 4 2 è ø æ 5 +1ö 1 æ 5 - 1 ö -4 -1 = - 5 - 1 çç = ÷÷· = - ç ÷= 2 4 2 è 8 ø 8 è ø

(

)

1 1 and tan B = 3 2 tan 2 A + tan B Now, tan ( 2 A + B ) = 1 - tan 2 A•tan B 1 2. 2 tan A 4 Also, tan 2 A = = 2 = 1 - tan 2 A 1 - 1 3 4 4 1 4 1 5 + + 3 3 From Eq. (i), tan ( 2 A + B ) = = 3 3 = 3 =3 4 1 9-4 5 1- • 3 3 9 9 18. Given that, tan A =

19. Given expression, sin p sin 13p = sin p sin æç p + 3p ö÷ 10 10 10 è 10 ø 3p p sin = - sin18° • sin 54° 10 10 = - sin18° • cos 36° = - sin

æ 5 -1 ö æ 5 +1 ö [since, put this value here] = - çç ÷÷  ÷÷ çç è 4 øè 4 ø 1 æ 5 -1 ö = -ç ÷=-4 è 16 ø 20. Given expression, sin 50° – sin 70° + sin 10°

æ 50° + 70° ö æ 50° - 70° ö = 2 cos ç ÷ • sin ç ÷ + sin10° 2 2 è ø è ø = – 2cos 60° sin10° + sin10° 1 = -2. sin10° + sin10° = 0 2 21. Given that, sinθ + cos θ = 1 On squaring both sides, we get sin2θ + cos2θ + 2sinθ · cosθ = 1 ⇒ 1 + sin2θ = 1 ∴ sin 2θ = 0

22. Given that, a + b = p 4 Now, (1+ tan α) (1 + tanβ) = 1 + tanα + tanβ + tanα tanβ We know that, tan (a + b ) = tan a + tan b 1 - tan a •tan b tan a + tan b ⇒ 1 = 1 - tan a • tan b ⇒ tan a + tan b = 1 - tan a tan b From Eq. (i), (1 + tanα) (1 + tanβ) = 1 + 1 tan α · tanβ + tan α · tanβ =2

Trigonometric Equations  16.13 sin q =

q 3 ⇒ 2 cos = 1 2 5 2

⇒ 2 cos 2 q = 2 q ∴ cos = ± 2 q ⇒ cos = 2

2 5 1 5 1 5 24. Given equation,

[since, θ lies in third quadrant] tan + sec x = 2 cos x

sin x 1 + = 2 cos x cos x cos x ⇒ 1 + sin x = 2 cos2 x ⇒ 1 + sin x = 2 (1 – sin2 x) ⇒ 1 + sin x = 2 – 2sin2 x ⇒ 2 sin2 x + sin x – 1 = 0 ⇒ 2 sin2 x + 2 sin x – sin x – 1= 0 ⇒ 2 sin2 x (sin x + 1) – 1 (sin x + 1) = 0 ⇒ (sin x + 1) (2 sin x – 1) = 0 ⇒ sin x + 1 = 0 or (2 sin x – 1) = 0 ⇒ sin x = - x,sin x = 1 2 3p p ∴ x = ,x = 2 6 Hence, only two solutions possible. ⇒

2p 5p p p + sin + sin + sin 18 9 9 18 = sin10° + sin 20° + sin 40° + sin 50°

25. Given expression, sin

= sin 50° + sin10° + sin 40° + sin 20° = sin130° + sin10° + sin140° + sin 20° = 2 sin 70° cos 60° + 2 sin 80° • cos 60° x+ y x- yù é  ê∵ sin x + sin y = 2 sin 2 • cos 2 ú ë û 1 é ù 1 1 = 2· sin 70° + 2· sin 80°  ê∵ cos60° = 2 ú 2 2 ë û 1 1 = 2. sin 70° + 2. sin 80° 2 2 7p 4p = sin 70° + sin 80° = sin + sin 18 9 26. Given equation, 3tan A + 4 = 0 ⇒ 3tan A = – 4

-4 3 -3 ⇒ cot A = 4 ⇒ tan A =

16 25 5 = =± 9 9 3

⇒ sec A = 1 + -5 3 -3 ⇒ cos A = 5 ⇒ sec A =



sin A = 1 -



4 sin A = 5

[ since, A lies is second quadrant]

9 25 - 9 4 = =± 25 25 5 [since, A lies in second quadrant]

æ -3 ö æ -3 ö 4 ∴ 2 cot A - 5 cos A + sin A = 2 ç ÷ - 5 ç ÷ + è 2 ø è 5 ø 5 6 4 = +3+ 4 5 30 + 60 + 16 46 = = 20 20 23 = 10 27. Given expression, cos2 48° – sin2 12° = cos ( 48° + 12° ) - cos ( 48° - 12° )



= cos 60° • cos 36° 1 5 +1 = · 2 4 5 +1 = 8

28. Given that,

tan =

1 1 and tan b = 7 3

1 48 49 49 cos 2a = = 1 50 1+ 49 49 48 24 = = 50 25 24 ⇒ cos 2a =  25 We know that, sin 4 b = 2 tan 22 b  1 + tan 2 b 1 2´ 2 tan b 3 and tan 2 b = = 1 - tan 2 b 1 - 1 9 2 2´9 3 =3= = 8 3´ 8 4 9 From Eq. (ii), 1-

(i) (ii)

HINTS AND EXPLANATIONS

-4 5 3 16 25 - 16 cosq = 1 = =± 25 25 5 -3 [since, θ lies in third quadrant] cosq = 5 -3 2q ⇒ 2 cos - 1 = 2 5 23. Given that,

16.14  Chapter 16 3 6 6 ´ 16 4 sin 4 b = = 4 = 25 4 ´ 25 9 1+ 16 16 24 ⇒ sin 4 b = 45 2´



⇒ sin 4 b = cos 2a  ⇒ cos 2a = sin 4 b

[from Eq. (i)]

29. Given that, tan q = a b

HINTS AND EXPLANATIONS



æ a2 ç 1- 2 = b ç b2 ç1+ a ç è b2

ö æ 2 tan q ö ÷ + aç ÷ 2 ø è 1 + tan q ø

ö æ 2a ÷ ç ÷ + aç b 2 ÷ ç1+ a ÷ ç ø è b2

ö ÷ ÷ ÷ ÷ ø

( (



=b

30. Here,

cosq = x +

⇒

æ 1 - tan 2 q ∴ b cos 2 + a sin 2q = b ç 2 è 1 + tan q

æ b 2 - a 2 ö 2a 2b = bç 2 + 2 2 2 ÷ èb +a ø a +b a2 + b2 b b 2 2 2 é ù = 2 b a + 2 a = û a + b2 ë a2 + b2

) )

1 x x2 +1 cosq = x

x2 – x cos θ + 1 = 0 For real value of x, (– cos θ)2 – 4 × 1 × 1 = 0 cos2 θ = 4 cos θ = ± 2 which is not possible.  [∵ - 1 £ cosq £ 1] which is not possible.

Trigonometric Equations  16.15

PRACTICE EXERCISES Single Option Correct Type ×

2. The general solution of the equation sin50 x – cos50x = 1 is p p (A) 2nπ + (B)  2nπ + 2 3 ×

(C) nπ +

×

p p (D)  nπ + 2 3

×

×

3. General solution of the equation ( 3 - 1) sin q + ( 3 + 1) cos q = 2 is (A) 2nπ ±

p p + 4 12

p p (B) nπ + (-1) + 4 12 n

(C) 2nπ ±

p p 4 12

p p (D) nπ + (- 1)n 4 12 4. The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos 2x + a3 sin2 x = 0 for all x is (A) 0 (B)  1 (C) 3 (D)  infinite 5. The equation sin4 x - (k + 2) sin2 x - (k + 3) = 0 possesses a solution if (A) k > -3 (B) k < -2 (C) - 3 ≤ k ≤ -2 (D) k is any positive integer 6. The least positive non-integral solution of the equation sin π (x2 + x) = sin p x2 is (A) rational (B) irrational of the form p

(C) irrational of the form integer (D) irrational of the form integer

p −1 , where p is an odd 4 p +1 , where p is an even 4

7. If sin2x - 2sinx - 1 = 0 has exactly four different solutions in x ∈ [0, np], then minimum value of n can be (n ∈ N) (A) 4 (B)  3 (C) 2 (D)  1 8. A set of values of x satisfying the equation 1  1  cos 2  px  + cos 2  qx  = 1 2  2  form an arithmetic progression with common difference 2 2 (A) (B)  p-q p+q (C)

p p+q

(D)  none of these

9. If 0 ≤ x ≤ 2π and | cos x | ≤ sin x, then é pù (A) x Î ê0, ú ë 4û

ép p ù (B)  xÎê , ú ë4 2û

é p 3p ù (C) x Î ê , ú ë4 4 û

(D)  none of these

10. The general solution of the equation 1 - sin x +  + ( -1) n sin n x +  1 - cos 2 x = 1 + cos 2 x 1 + sin x +  + sin n x +  x ≠ (2n + 1)

p , n ∈ Z is 2

×

æp ö (A) ( -1) n ç ÷ + np è3ø

æp ö (B) ( -1) n ç ÷ + np è6ø

æp ö æp ö (C) ( -1) n +1 ç ÷ + np (D)  ( -1) n -1 ç ÷ + np è6ø è3ø 11. The general solution of the equation n

å cos r 2q sin rq = r =1

1 is 2

PRACTICE EXERCISES

1 p  x 1. The equation 2cos2   . 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

16.16  Chapter 16 2k + 1 p 4k - 1 p (A) , k Î Z (B)  , k Î Z n( n + 1) 2 n( n + 1) 2 (C)

4k + 1 p , k ÎZ n( n + 1) 2

12. The solution of sin8x + cos8x =

(D)  none of these 17 is 32

p np p (A) ± (B)  np ± 4 2 8 p (D)  no solution 8 13. The general solution of the equation 2cos2θ + 1 = 3.2– sin2θ is p (A) 2np ± , np , n Î Z 2 p (B) np ± , 2np , n Î Z 2 (C) np ±

p , np , n Î Z 2 (D) none of these (C) np ±

14. The solution of the inequality log1/2 sinθ > log1/2 cosθ in [0, 2p] is

æp p ö æ pö (A) ç 0, ÷ (B)  ç4, 2÷ è ø è 2ø

PRACTICE EXERCISES

æ pö (C) ç 0, ÷ è 4ø

(D)  none of these

7p ö æ = -2 then x = 15. If cos 3 x + sin ç 2 x 6 ÷ø è

p p (A) (6 k + 1) (B)  (6 k - 1) 3 3 p (C) ( 2k + 1) 3 where k ∈ Z.

(D)  none of these

2x , where 1+ x2 x, y ∈ R, then least positive value of y is

16. If tan2[π (x + y)] + cot2 [π (x + y)] = 1 +

5 1 (A) (B)  4 4 3 (C) (D)  2 4 17. The general value of y satisfying the equation 1 - 2x - x2 = tan2(x + y) + cot2(x + y) is

p p (B)  np ± 4 4 n p p (C) ± (D)  none of these 2 4 (A) 2np ±

18. If [sin x] + [ 2 cos x ] = -3, x ∈ [0, 2p] ([ . ] denotes the greatest integer function) then x belongs to é 5p ù æ 5p ö (A) ê , 2p ú (B)  ç 4 , 2p ÷ ë 4 û è ø é 5p ù æ 5p ö (C) ç p , (D)  ÷ êp , 4 ú 4 ø ë û è 19. The number of solutions of the equation æ px ö 2 sin çç ÷÷ = x - 2 3 x + 4 è2 3ø (A) forms an empty set (B) is only one (C) is only two (D) is more than 2 20. The number of solutions of the equation | cos x | = 2[x], where [ . ] is the greatest integer, is (A) one (B)  two (C) infinite (D)  nil 21. The general solution of sin x - 3 sin 2x + sin 3x = cos x - 3cos 2x + cos 3x is np p p + (A) np + (B)  2 8 8 3 np p (C) ( -1) n (D)  2np + cos -1 + 2 2 8 22. sin x + 2 sin 2x = 3 + sin 3x, 0 ≤ x ≤ 2π has (A) 2 solutions in I quadrant (B) one solution in II quadrant (C) no solution in any quadrant (D) one solution in each quadrant 23. The solution of the equation 1 + sin2ax = cos x, where a is irrational, is np (A) x = 0 (B)  x= a (C) x = 2nπ (D)  none of these 24. The values of α for which the equation sin4x + cos4x + sin 2x + α = 0 may be valid, are 1 3 (A) - ≤ α ≤ 1 (B)  0 ≤ α ≤ 2 2 1 3 (C) - ≤ α ≤ (D)  none of these 2 2 25. If α and β be two distinct values of θ lying between 0 and 2π, satisfying the equation 3 cos θ + 4 sin θ = 2, then the value of sin (α + β) is

Trigonometric Equations  16.17 12 24 (A) (B)  25 25 13 (C) (D)  none of these 25 26. | tanx + secx | = | tanx | + | secx |, x ∈ [0, 2p], if and only if x belongs to the interval (A) (π, 2p] (B) [0, p]

é p ö æp ù (C) ê0, ÷ È ç , p ú ë 2ø è2 û é 3p ö æ 3p ù (D)  êp , ÷ È ç , 2p ú ë 2 ø è 2 û 27. | cosx | = cosx - 2sinx if (A) x = nπ p (B) x = 2nπ or (2n + 1) π + 4 p (C) x = n π + 4 p (D) x = n π or nπ + 4

28. A solution of the equation (1 – tanθ) (1 + tanθ) sec2 θ æ p pö + 2 tan2θ = 0, where θ lies in the interval ç - , ÷ is è 2 2ø given by p p (A) θ = 0 (B)  q = or 3 3 p p (C) q = (D)  q =6 6 29. If p cosx – 2sinx = p∈

2 + 2 − p has a solution, then

(A)  5 + 1, 4 

 5 − 1, 2 (B) 

(C)  3 + 1, 3

(D)  none of these

30. The value of ‘b’ such that the equation b cos x b + sin x = 2 2 cos 2 x - 1 (cos x - 3 sin 2 x ) tan x possess solutions, belongs to the set 1ö 1  æ (A) ç -¥, ÷ (B)   , ∞ 2 è 2ø

×

×

1  (C) ( −∞, ∞) (D)   −∞,  ∪ (1, ∞) 2

×

31. If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0, is: [2016] (A) 9 (B)  3 (C) 5 (D)  7

2 (A)  3

(B) 

13 9

8 9

(D) 

20 9



(C) 

32. If sum of all the solutions of the equation ⎛ ⎛π ⎞ ⎛π ⎞ 1⎞ 8 cos x ⋅ ⎜ cos ⎜ + x ⎟ ⋅ cos ⎜ − x ⎟ − ⎟ = 1 in [0, π] ⎝6 ⎠ ⎝6 ⎠ 2⎠ ⎝ is kπ, then k is equal to

[2018]

ANSWER K EYS Single Option Correct Type 1. (B) 11. (C) 21. (B)

2. (C) 12. (A) 22. (C)

3. (A) 13. (C) 23. (A)

Previous Years’ Questions 31. (D)

32. (D)

4. (D) 14. (C) 24. (C)

5. (C) 15. (A) 25. (B)

6. (C) 16. (B) 26. (C)

7. (B) 17. (D) 27. (B)

8. (D) 18. (C) 28. (B)

9. (C) 10. (B) 19. (B) 20. (D) 29. (B) 30. (A)

PRACTICE EXERCISES

Previous Year's Questions

16.18  Chapter 16

HINTS AND EXPLANATIONS Single Option Correct Type

×

3. Let

3 + 1 = r cosα, and

3 - 1 = r sinα

∴ r 2 = ( 3 + 1) 2 + ( 3 − 1) 2 = 8, 3 - 1 1 - 1/ 3 æp p ö = = tan ç - ÷ 3 + 1 1 + 1/ 3 è4 6ø = tan(π/12), i.e., α = π/12. From the given equation, we get r cos(θ − α) = 2 ⇒ cos(θ − π/12) = 1/ 2 = cos(π/4) ∴ θ − π/12 = 2nπ ± π/4 or θ = 2nπ ± π/4 + π/12. 4. We have, a1 + a2cos2x + a3 sin2x = 0 a ⇒ a1 + a2cos2x + 3 (1 − cos2x) = 0 2 which is zero for all values of x, a if a1 = - 3 = −a2 2 k k or a1 = - , a2 = , , a3 = k 2 2 for any k ∈ R. Hence, the required number of triplets is infinite. 5. We have, sin4x − (k + 2) sin2x − (k + 3) = 0 Þ sin 2 x =

( 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. 6. We have, sinπ (x2 + x) = sinpx2 ⇒ π (x2 + x) = nπ + (–1)npx2

∴

HINTS AND EXPLANATIONS

and, tan a =

- 1 ± 1 + 8k . 4 For least positive non-integral solution, ⇒ 2x2 + x − k = 0 ⇒ x =

x=

- 1 + 1 + 8k = 4

p -1 , where p is an odd integer. 4

7. We have, sin2x − 2sin x − 1 = 0 ⇒ (sin x − 1)2 = 2 ⇒ sin x − 1 = ± 2 ⇒ sinx = 1 - 2 as sinx >/ 1. There are 2 solutions in [0, 2p] and two more in [2π, 4p]. Thus, minimum value of n is 4. 8. The given equation can be written as 1 + cospx + 1 + cosqx = 2 æ p+qö æ p-qö Þ cos ç ÷ x cos ç 2 ÷ x = 0 2 è ø è ø ( 2n + 1)p æ p+qö cos ç ÷ x = 0 Þ x = p + q , n = 0, ± 1, ± 2 … which 2 è ø forms an A. P. with common difference = æ cos ç è

2p p+q

p-qö ( 2n + 1)p x=0 Þ x= , n = 0, ± 1, ± 2,… 2 ÷ø p-q

which forms an A. P. with common difference = 9. We have, | cos x | ≤ sin x ⇒ sin x ≥ 0 [ | cos x | ≥ 0 | ⇒ x ∉ (π, 2π) If x = 2π, | cos 2π | ≤ sin 2π, which is not possible ∴ x ∈ [0, p] é pù If x ∈ ê0, ú , then | cos x | ≤ sin x ë 2û ép p ù ⇒ cos x ≤ sin x ⇒ x ∈ ê , ú ë4 2û p é ù If x ∈ then | cos x | ≤ sin x ê 2 , p ú, ë û ⇒ –cos x ≤ sin x ⇒ tan x ≤ –1 ( cos x < 0)

∴

2. We have, sin50x − cos50x = 1 ⇒ sin50x = 1 + cos50x 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 p ∴ x = nπ + , n ∈ Z. 2

∴  Either x2 + x = 2m + x2 ⇒ x = 2m ∈ Z. or, x2 + x = k − x2, where k is an odd integer

∴

1. Since x2 + x−2 = (x − x− 1)2 + 2 ≥ 2 x and, 2cos2 sin2x ≤ 2, 2 ∴ the given equation is valid only if x 2cos2 sin2x = 2 2 x ⇔ cos = cosecx = 1, which cannot be true. 2

æ p 3p ù ⇒ x ∈ ç , ú è2 4 û Combining two results, we get é p p ù æ p 3p ù xÎê , ú Èç , ú ë4 2û è 2 4 û é p 3p ù or, x Î ê , ë 4 4 úû

2p p-q

Trigonometric Equations  16.19 1 - sinx +  + ( -1) n sin n x +  1 - cos 2 x = 1 + sinx +  + sin n x +  1 + cos 2 x

Þ

1 1- sin x 2 sin 2 x ´ = as –1 < sin x < 1 1+ sin x 1 2 cos 2 x

sin 2 x (1+ sin x ) 1- sin 2 x 2 ⇒ (1 − sin x) = sin2x ⇒ 1 − 2 sin x = 0

If z =1, then 2−sin2θ = 20 ⇒ sin2θ = 0 ⇒ sinθ = 0 ⇒ θ = nπ p Thus, q = np , np ± 2 14. sin θ > 0 ⇒ θ ∈ (0, π)(1)

⇒ 1 − sin x =

æ p ö æ 3p ö cosθ > 0 ⇒ q Î ç 0, ÷ È ç , 2p ÷ (2) è 2ø è 2 ø

p 1 p ⇒ sin x = = sin ⇒ x = np + ( -1) n 2 6 6

p From (1) and (2), q Î çæ 0, ÷ö . è 2ø Also, log1/2sinθ > log1/2 cosθ

11. We have,

n

1

å cos r q sin r q = 2 2

æ pö æ pö ⇒ sinθ < cosθ in ç 0, ÷ Þ q Î ç 0, ÷ . 2 è ø è 4ø

r =1

Þ

n

7p ö æ 15. We have cos 3x + sin ç 2 x = –2 6 ÷ø è

å 2 cos r q sin r q = 1 2

r =1

Þ

n

å [sin r( r + 1)q - sin r( r - 1)q ] = 1 r =1

⇒  sin n(n + 1)θ =1 ⇒ q = 12. Given, sin8x + cos8x =

4k + 1 p × ,k Î Z n( n + 1) 2

17 32

Þ 2 cos 2

⇒ (sin4x + cos4x)2 − 2sin4x cos4x =

17 32

⇒ (1 − 2sin2x cos2x)2 − 2sin4x cos4x = 17 32 ⇒ 1 − 4 sin2x cos2x + 2sin4x cos4x = ⇒ 1 − (2sinx cosx)2 +

17 32

1 17 ( 2 sin x cos x ) 4 = 8 32

1 4 17 sin 2 x = 0 8 32 ⇒ 4sin42x − 32sin22x + 15 = 0 ⇒ 1 − sin22x +

15 ù é ⇒ 2sin22x − 1 = 0 ê∵ sin 2 2 x ¹ ú 2û ë ⇒ sin22x =

p 1 ⇒ sin22x = sin 2 4 2

⇒ 2x = np ±

7p ö æ ⇒ 1 + cos3x + 1 + sin ç 2 x 6 ÷ø è 2p æ ⇒ (1 + cos3x) + 1 − cos ç 2 x 3 è

×

p , n∈I 4

np p ± , n∈I 2 8 13. We have, 2cos2θ + 1 = 3.2− sin2θ ⇒ 21−2sin2θ + 1 = 3.2−sin2θ ⇒ x =

1 ⇒ 2z2–3z + 1 = 0, where z = 2−sin2θ Þ z = ,1. 2

p 1 If z = , then 2−sin2θ = 2−1 ⇒ sin2θ =1 ⇒ θ = np ± 2 2

⇒ cos

=0 ö ÷ =0 ø

pö 3x æ + 2 sin 2 ç x - ÷ = 0 2 3ø è

3x pö æ = 0 and sin ç x - ÷ = 0 2 3ø è

3 x p 3p p = , ,… and x - = 0, π, 2π… 2 2 2 3 p Þ x= 3 Therefore, the general solution of Þ

cos

3x p pö æ = 0 and sin ç x - ÷ = 0 is x = 2kp + 2 3ø 3 è =

p (6k + 1), where k ∈ Z 3

16. Minimum value of L.H.S. is 1 æ ö 2 ç x + form where x > 0 ÷ and 1 + x è ø

2 x+

1 x

£2

Equality is possible only when both sides are 2 at x = 1 p

i.e., tan2 [π (x + y)] = 1 ⇒ π (x + y) = nπ ± 4 5p ⇒ π (x + y) = (for least value of y) 4 1 ∴ y = 4 ×

×

17. The given equation can be written as 3 − 2x − x2 = 1 + tan2 (x + y) + 1 + cot2 (x + y) ⇒ 4 − (x + 1)2 = sec2(x + y) + cosec2 (x + y)

HINTS AND EXPLANATIONS

10.

16.20  Chapter 16 ⇒ cos2(x + y) sin2(x + y) [4 − (x + 1)2] = 1 ⇒ sin2(2x + 2y) [4 − (x + 1)2] = 4 Since sin2(2x + 2y) ≤ 1 and 4 − (x + 1)2 ≤ 4 ∴ (1) holds only if sin2(2x + 2y) = 1 and, 4 − (x + 1)2 = 4 From (3), we get x = –1 Putting in (2), we get sin (2y − 2) = ±1 ⇒ 2y − 2 = np ±

(1) (2) (3)

p , n∈Z 2

p ⇒ y = 1 + ( 2n ± 1) , n ∈ Z 4 p ∴ x = –1, y =1 + ( 2n ± 1) , n ∈ Z 4 18. We have, [sinx] + [ 2 cos x ] = –3 ⇒ [sin x] = –1 and [ 2 cos x ] = –2 ⇒ –1 ≤ sin x < 0  or  x ≤ (π, 2π) and, -2 £ 2 cos x < -1 æ 3p 5 p ö or x Î ç , ÷ è 4 4 ø 2 æ 5p ö So x Î ç p , 4 ÷ø è -1

æ px ö 2 19. sin çç ÷÷ = x - 2 3 x + 4 è2 3ø = ( x - 3 )2 + 1 R.H.S > 1. So, the solution exists if and only if x - 3 = 0 Þ x = 3 and then equation is obviously satisfied. 20. From the graph, | cos x | and 2 [x] do not cut each other for any real value of x. Hence, number of solutions is nil.

∴

2 ± ( 4 + 8 + 8a ) 2 ± (12 + 8a ) = 2 2 2{1± (3 + 2 a)} = = 1± (3 + 2 a) 2 =

If sin 2 x = 1 + (3 + 2a ) > 1 which is not possible 21. sin x − 3sin 2x + sin 3x = cos x − 3cos 2x + cos 3x ⇒ 2sin2xcosx − 3sin2x − 2cos2xcosx + 3cos2x = 0 ⇒ sin2x(2cosx − 3) − cos2x (2cosx − 3) = 0 ⇒ (sin2x − cos2x) (2cosx − 3) = 0 ⇒ sin2x = cos2x ( cosx ≠ 3/2)

∴

HINTS AND EXPLANATIONS

or, cos x
−π è ø max è ø min

p+4

⇒ p ≥ 0, 2 − p ≥ 0

and,

b -a ≠ 0 because β ≠ α and 2

p cos x - 2 sin x £

and,

)

2

p + 4 ⇒ p ≥ 0, 2 − p ≥ 0 ≤ p + 4 ⇒ p ≥ 0, p ≤ 2

4 - 2p ≤ p+4

⇒ 0 ≤ p ≤ 2 and 2 4 - 2 p ≤ 2p ⇒ p2 + 2p − 4 ≥ 0 ⇒ 0 ≤ p ≤ 2 and (p + 1)2 ≤ 5

a +b a +b - 4 cos =0 2 2 a +b 4 = or, tan 2 3 \ 3 sin

⇒ 0 ≤ p ≤ 2 and p ∈ (– ∞, − 5 − 1] ∪ [ 5 −1, ∞) ⇒ p ∈ [ 5 − 1, 2]

4 a +b 2´ 2 tan 3 = 24 . 2 \ sin(a + b ) = = 16 25 2a + b 1 + tan 1+ 2 9

30. For the domain of definition of the given equation, we have, p (i) 2 cos2x − 1 ≠ 0 ⇒ x ≠ nπ ± 6 np p (ii) tan ≠ 0 ⇒ x ≠ ± [For odd multiples of , tan x is 2 2 not defined] p (iii) cos2x − 3sin2x ≠ 0 ⇒ x ≠ n π ± , 6 Also, 2cos2x − 1 = 2(cos2x − sin2x) − (cos2x + sin2x) = cos2 x − 3sin2x Now, the given equation reduces to bsin x = b + sinx b Þ sin x = , b -1 ×

×

–1 ≤ sin x ≤ 1 \ - 1 £ ∴

26. | tan x + sec x | = | tan x | + | sec x | iff sec x and tan x both have sin x same sign. ⇒ sec x. tan x ≥ 0 ⇒ ³ 0, but cos x ≠ 0 ⇒ cos 2 x é p ö æp ù x ∈ ê0, ÷ È ç , p ú ë 2ø è2 û 27. | cos x | = cos x − 2 sin x ⇒ cos x = cos x − 2sin x if cos x ≥ 0 ⇒ sin x = 0 ⇒ x = 2nπ (as cos x ≥ 0).n ∈ I Also, |cos x| = cos x –2sin x ⇒ −cos x = cos x –2 sin x if cos x < 0 ⇒ cos x − sin x = 0 ⇒ tan x = 1 5p Now, cos x < 0 and tan x = 1 ⇒ tan x = tan 4 p æ 5p ö ⇒ x = 2nπ + ç ÷ = (2n + 1)π + 4 è 4 ø

Þ

b b + 1 £ 0 and -1 £ 0 b -1 b -1

Þ

2b - 1 1 ³ 0 and £0 b -1 b -1

×

28. We have, (1– tanθ) (1 + tanθ) sec2θ + 2 tan2θ = 0 ⇒ (1 − tan4θ) + 2 tan2θ = 0 Put tan2θ = x, ∴ (1 − x2) + 2x = 0 ⇒ 2x = x2 − 1 = y (say) ∴ y = 2x and y = x2 –1 By inspection, x = 3, ∴ tan2θ = 3 p ⇒ tan θ = ± 3 ⇒ θ = ± 3

b £1 b -1

Þ b£

1 1 or b > 1 and b < 1 Þ b £ 2 2

1 When b = , sin x = 1, which is not possible 2 \ b