CBSE Question Bank Class 12 Mathematics Book Based On Latest Board Sample Paper Released On 16th Sep 2022 (For 2023 Exam) [2022 ed.] 9789355955647

Table of contents :
Cover
Title Cover
Copyright Page
Contents
Syllabus
Solved Paper-2022
UNIT – I : RELATIONS AND FUNCTIONS
1. Relations and Functions
2. Inverse Trigonometric Functions
 Self Assessment Paper- 1
UNIT – II : ALGEBRA
3. Matrices
4. Determinants
 Self Assessment Paper- 2
UNIT – III : CALCULUS
5. Continuity & Differentiability
6. Applications of Derivatives
7. Integrals
8. Applications of the Integrals
9. Differential Equations
 Self Assessment Paper- 3
UNIT – IV : VECTORS & THREE-DIMENSIONAL GEOMETRY
10. Vectors
11. Three Dimensional Geometry
 Self Assessment Paper- 4
UNIT – V : LINEAR PROGRAMMING
12. Linear Programming
 Self Assessment Paper- 5
UNIT – VI : PROBABILITY
13. Probability
 Self Assessment Paper- 6
 Practice Paper- 1
 Practice Paper- 2

Citation preview

MATHEMATICS (INCLUDES COMPETENCY BASED QUESTIONS)

(1)

18th EDITION

ISBN SYLLABUS COVERED

YEAR 2022-23 “978-93-5595-564-7”

CENTRAL BOARD OF SECONDARY EDUCATION DELHI

PUBLISHED BY

COPYRIG HT

RESERVED

BY THE PUBLISHERS

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without written permission from the publishers. The author and publisher will gladly receive information enabling them to rectify any error or omission in subsequent editions.

OSWAAL BOOKS & LEARNING PVT. LTD. 1/11, Sahitya Kunj, M.G. Road, Agra - 282002, (UP) India

1/1, Cambourne Business Centre Cambridge, Cambridgeshire CB 236DP, United kingdom

0562-2857671

[email protected]

www.OswaalBooks.com

D I SC L A IM ER

Oswaal Books has exercised due care and caution in collecting all the data before publishing this book. In spite of this, if any omission, inaccuracy or printing error occurs with regard to the data contained in this book, Oswaal Books will not be held responsible or liable. Oswaal Books will be grateful if you could point out any such error or offer your suggestions which will be of great help for other readers. Kindle ( 2Edition )

TABLE OF CONTENTS g

Latest CBSE Circular released on 20th May 2022 for Academic Year 2022-2023



(CBSE Cir. No. Acad 57/2022)

g

Latest CBSE Syllabus released on 21st April 2022 for Academic Year 2022-2023



(CBSE Cir. No. Acad 48/2022)

g

CBSE Solved Board Papers 2022 Term-II Examination





7 - 7



8 - 11

14 - 32

(To download Solved paper for Term-I 2021-22 & Latest Topper’s Answers 2020, scan the QR Code given on Page 32)



UNIT-I : RELATIONS & FUNCTIONS 1. Relations and Functions

Topic - 1 : Relations



Topic - 2 : Functions

1 - 22

2. Inverse Trigonometric Functions

23 - 41

 Self Assessment Paper- 1

42 - 43

UNIT-II : ALGEBRA 3. Matrices

Topic - 1 : Matrices and Operations



Topic - 2 : Symmetric, Skew Symmetric and Invertible Matrices

4. Determinants

Topic - 1 : Determinants, Minors & Co-factors



Topic - 2 : Solutions of System of Linear Equations

 Self Assessment Paper- 2

44 - 66

67 - 87

88 - 90

UNIT-III : CALCULUS 5. Continuity & Differentiability

Topic - 1 : Continuity



Topic - 2 : Differentiability

6. Applications of Derivatives

Topic - 1 : Rate of Change of Bodies



Topic - 2 : Increasing/Decreasing Functions



Topic - 3 : Maxima and Minima

7. Integrals

Topic - 1 : Indefinite Integral



Topic - 2 : Definite Integral

8. Applications of the Integrals

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91 - 115

116 - 145

146 - 185

186 - 207

TABLE OF CONTENTS 9. Differential Equations

Topic - 1 : Basic Differential Equations



Topic - 2 : Variable Separable Methods



Topic - 3 : Linear Differential Equations



Topic - 4 : Homogeneous Differential Equations

 Self Assessment Paper- 3

208 - 237

238 - 240

UNIT-IV : VECTORS & THREE DIMENSIONAL GEOMETRY 10. Vectors Topic - 1 : Basic Algebra of Vectors Topic - 2 : Dot Product of Vectors Topic - 3 : Cross Product 11. Three Dimensional Geometry Topic - 1 : Direction Ratios and Direction Cosines Topic - 2 : Lines & Its Equations in Different forms  Self Assessment Paper- 4

241 - 268

269 - 290

291 - 293

UNIT-V : LINEAR PROGRAMMING 12. Linear Programming  Self Assessment Paper- 5

294 - 314 315 - 317

UNIT-VI : PROBABILITY 13. Probability Topic - 1 : Conditional Probability and Multiplication Theorem on Probability Topic - 2 : Bayes’ Theorem Topic - 3 : Random Variable and its Probability Distributions  Self Assessment Paper- 6  Practice Paper- 1  Practice Paper- 2

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318 - 348

349 - 350 351 - 354 355 - 359 qq

PREFACE Who is this book for? We are delighted to present the updated edition of the CBSE Questions Banks for the 2022-23 academic session for Class 10. This edition is based on the latest CBSE Syllabus and Scheme of Assessment and Evaluation Practices of the Board for the Session 2022-23. National Education Policy 2020 has affirmed the need to move from rote to competency based learning. The Board has taken multiple steps towards the implementation of Competency Based Education (CBE) to equip the learning to meet 21st Century challanges proactively.

Why is this book different? The structural changes in Education that are being brought in by the latest Syllabus and the new National Educational Policy (NEP) resonate with our approach of focusing on deeper understanding instead of rote learning. Available for all subjects, these books have been designed to help students achieve high scores in their Board Examinations. Prepared by the Editorial Board of Oswaal Books, comprising leading subject matter experts, it consists of CBE pedagogical features such as Multiple Choice Questions, Case based Questions, Source based Integrated Questions alongwith Objective Types, Short & Long Answer Types Questions - all of which reinforce learning and improve conceptual understanding. To make these books 100% exam-centric, we've also includedl



l l l l l l l

Latest Board Exam Papers (all sets of Delhi & Outside Delhi) of CBSE Term-II 2022 Topic-wise/Concept wise segregation of chapters Important Keywords for quick recall of concepts Fundamental Facts to enhance knowledge Practice Questions within the chapters for better practice Reflection to ask about your learnings Self-Assessment Tests & Practice Papers for Self-Evaluation Art Integration for Experiential learning

The new way of learning; Blended Learning

The pandemic introduced us all to a phenomenon Blended Learning. In no time e-learning has become mainstream. Oswaal Books identified this like an opportunity and thus, we decided to introduce Oswaal360 A platform that simplifies learning with Chapter-wise e-assessments articulated for students to crack Board Exams.

Other Important features of this book such as-Art Integrated activities, Mind Maps, Concept Videos and Academically Important (AI) Questions make it an indispensable resource for students desiring to secure top marks in the examination.

How to use this book? You should use this book to supplement your learning and test your conceptual understanding. The rich instructive pedagogy encourages critical thinking and reflection and acts as a handy resource for students to learn, course correct, and test their knowledge. Making this book a part of your day-to-day exam preparation routine will help you sharpen your conceptual understanding, enhance your application to exam questions, and help you improve your test-taking abilities.

We wish you great success for your Board Examinations in 2023 and hope this book helps you achieve exemplary success. Team Oswaal Books (5)

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CBSE CIRCULAR 2022-23 dsUæh; ek/;fed f'k{kk cksMZ

CENTRAL BOARD OF SECONDARY EDUCATION CBSE/DIR(ACAD)/2022/

Dated: 20.05.2022



Circular No. ACAD-57/2022

All Heads of schools affiliated to CBSE Subject : Assessment and Evaluation Practices of the Board for the Session 2022-23

National Education Policy 2020 has affirmed the need to move from rote to competency-based learning. This will equip the learners with key competencies to meet the challenges of the 21st century proactively. Accordingly, the Board has taken multiple steps towards the implementation of Competency Based Education (CBE) in schools. These range from aligning assessment to CBE, development of exemplar resources for teachers and students on CBE pedagogy and assessment and continued teacher capacity building.



In this context the Board has released Circular No. Acad-05/2019 dated 18.01.2019; Circular No. Acad-11/2019 dated 06.03.2019; Circular No. Acad-18/2020 dated 16.03.2020; Circular No. Acad-32/2020 dated 14.05.2020 and Circular No. Acad-31/2020 dated 22.04.2021. In continuation to these circulars, the Board is initiating further corresponding changes in the Examination and Assessment practices for the year 2022-23 to align assessment to Competency Based Education. Therefore, in the forthcoming sessions a greater number of Competency Based Questions or questions that assess application of concepts in real-life/ unfamiliar situations will be part of the question paper.



The changes for classes IX-XII (2022-23) internal year-end/Board Examination are as detailed: (Classes IX-X) Year End Examination/ Board Examination (Theory)

(2021-22) Existing (As per Special Scheme of Assessment for Board Examination – Circular No. Acad51/2021 dated 05.07.2021)

(2022-23) Modified (Annual Scheme)

Composition

• Term I – Multiple Choice Question including case based and assertion reasoning type MCQs – 100% (30% questions competency based) • Term II – Case based/ Situation based, Open Ended- short answer/long answer questions (30% questions competency based)

• Competency Based Questions would be minimum 40% These can be in the form of Multiple Choice Questions, Case based Questions, Source based Integrated Questions or any other types. • Objective Type Questions will be 20% • Remaining 40% short

Composition

• Term I – Multiple Choice Question including case based and assertion reasoning type MCQs – 100% (30% questions competency based) • Term II – Case based/ Situation based, Open Ended- short answer/long answer questions (30% questions competency based)

• Competency Based Questions would be minimum 40% These can be in the form of Multiple Choice Questions, Case based Questions, Source based Integrated Questions or any other types. • Objective Type Questions will be 20% • Remaining 40% short answer/long answer questions (as per existing pattern)

Internal Assessment : No change Internal Assessment: End of year examination = 20:80 Year End Examination/ Board Examination (Theory)



Curriculum document released by the Board vide circular No.Acad-50/2022 dated 28th April, 2022 and the forthcoming Sample Question Papers may be referred for the details of changes in the QP design of individual subjects.



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(Dr. Joseph Emmanuel) Director (Academics)

SYLLABUS Latest Syllabus issued by CBSE for Academic Year 2022-23

MATHEMATICS (Code No. 041) CLASS–XII

One Paper No. Units I. Relations and Functions II. Algebra III. Calculus IV. Vectors and Three - Dimensional Geometry V. Linear Programming VI. Probability

Total Internal Assessment

No. of Periods 30 50 80 30 20 30 240

Max Marks : 80 Marks 08 10 35 14 05 08 80 20

Unit I : Relations and Functions 1.

Relations and Functions : 15 Periods Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions. 2. Inverse Trigonometric Functions 15 Periods Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit II : Algebra 1. Matrices

25 Periods



Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operation on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Oncommutativity of multiplication of matrices, and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). 2. Determinants 25 Periods Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit III : Calculus 1. Continuity and Differentiability : 20 Periods Continuity and differentiability, chain rule, derivative of inverse trigonometric functions, like sin–1 x, cos–1 x and tan–1 x, derivative of implicit functions. Concept of exponential and logarithmic functions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives. 2. Applications of Derivatives 10 Periods Applications of derivatives: rate of change of bodies, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as reallife situations).

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SYLLABUS 3.



Integrals 20 Periods Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

∫x

2

dx dx dx dx dx , , , ± a 2 ∫ x 2 ± a 2 ∫ a 2 − x 2 ∫ ax 2 + bx + c ∫ ax 2 + bx + c

∫ ax ∫

px + q dx , + bx + c

2

∫

px + q ax + bx + c 2

dx , ∫ a 2 ± x 2 dx , ∫ x 2 − a 2 dx

ax 2 + bx + c dx ,

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals. 4. Applications of the Integrals 15 Periods Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only). 5. Differential Equations 15 Periods Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

dy + py = q , where p and q are functions of x or constants. dx dx + px = q , where p and q are functions of y or constants. dy

Unit-IV: Vectors and Three-Dimensional Geometry 1. Vectors 15 Periods Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors. 2. Three - dimensional Geometry 15 Periods Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines. Unit-V: Linear Programming 1.

Linear Programming

20 Periods



Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints). Unit-VI: Probability

1. Probability

30 Periods

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

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SYLLABUS QUESTION PAPER DESIGN Mathematics (Code No. 041) Class XII Time 3 Hours

Max. Marks : 80

S. No. Typology of Questions 1.

3.

%

Marks

Weightage

44

55

20

25

16

20

80

100

Remembering : Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers. Understanding : Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas

2.

Total

Applying : Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way. Analysing : Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations Evaluating : Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria. Creating : Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions TOTAL

1. No chapter wise weightage. Care to be taken to cover all the chapters. 2. Suitable internal variations may be made for generating various templates keeping the overall weightage to different form of questions and typology of questions same. Choice(s) : There will be no overall choice in the question paper. However, 33% internal choices will be given in all the sections. INTERNAL ASSESSMENT

20 MARKS

Periodic Tests ( Best 2 out of 3 tests conducted)

10 Marks

Mathematics Activities

10 Marks

Note : For activities NCERT Lab Manual may be referred. Conduct of Periodic Tests: Periodic Test is a Pen and Paper assessment which is to be conducted by the respective subject teacher. The format of periodic test must have questions items with a balance mix, such as, very short answer (VSA), short answer (SA) and long answer (LA) to effectively assess the knowledge, understanding, ( 10 )

SYLLABUS



application, skills, analysis, evaluation and synthesis. Depending on the nature of subject, the subject teacher will have the liberty of incorporating any other types of questions too. The modalities of the PT are as follows: a) Mode: The periodic test is to be taken in the form of pen-paper test. b) Schedule: In the entire Academic Year, three Periodic Tests in each subject may be conducted as follows: Test

Pre Mid-term (PT-I)

Mid-Term (PT-II)

Post Mid-Term (PT-III)

Tentative Month July-August November December-January This is only a suggestive schedule and schools may conduct periodic tests as per their convenience. The winter bound schools would develop their own schedule with similar time gaps between two consecutive tests. c) Average of Marks: Once schools complete the conduct of all the three periodic tests, they will convert the weightage of each of the three tests into ten marks each for identifying best two tests. The best two will be taken into consideration and the average of the two shall be taken as the final marks for PT. d) The school will ensure simple documentation to keep a record of performance as suggested in detail circular no.Acad-05/2017. e) Sharing of Feedback/Performance: The students’ achievement in each test must be shared with the students and their parents to give them an overview of the level of learning that has taken place during different periods. Feedback will help parents formulate interventions (conducive ambience, support materials, motivation and morale-boosting) to further enhance learning. A teacher, while sharing the feedback with student or parent, should be empathetic, non- judgmental and motivating. It is recommended that the teacher share best examples/performances of IA with the class to motivate all learners. Assessment of Activity Work: Throughout the year any 10 activities shall be performed by the student from the activities given in the NCERT Laboratory Manual for the respective class (XI or XII) which is available on the link: http:// www.ncert.nic.in/exemplar/labmanuals.htmla record of the same may be kept by the student. An year end test on the activity may be conducted The weightage are as under:  The activities performed by the student throughout the year and record keeping : 5 marks  Assessment of the activity performed during the year end test: 3 marks  Viva-voce: 2 marks Prescribed Books: 1) Mathematics Textbook for Class XI, NCERT Publications 2) Mathematics Part I - Textbook for Class XII, NCERT Publication 3) Mathematics Part II - Textbook for Class XII, NCERT Publication 4) Mathematics Exemplar Problem for Class XI, Published by NCERT 5) Mathematics Exemplar Problem for Class XII, Published by NCERT 6) Mathematics Lab Manual class XI, published by NCERT 7) Mathematics Lab Manual class XII, published by NCERT qq

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What is ART INTEGRATION? “When I see the universe through my songs, I recognize it, then only I know it.” - ”Gitabitan”by Ravindranath Tagore

A rt has always been a great part of our lives. It is usually defined as the creative

expression of an individuals but more often, Art defines us. From the care paintings to great artifacts found in Harappan Civilisation to the latest animation software, Art has always been there surrounding every aspect of our lives from the beginning of time. Art, not only help us understand things around us but at times we understand through it. As Rabindranath Tagore once said, "…. only man knows……. objects". Over the centuries, the ways of learning has gone through many refinements and as a result it has become a great deal easier for students to obtain knowledge dividing the ways of learning. Remember when the visuals of rips apples helped us learn about the color 'red' and the 'twinkle-twinkle' melody first introduced us to the Celestial bodies in our space. Even in the primary education, schools promote Art, Dance, Crafts, Theatre. But as students progress the higher classes and core academic subjects take the stage, they are somehow left with scribbled words and symbole and Art is either forgotten or relegated to insignificant background. It is with hopes of connecting "ever-growing" minds once again with their artistic expression and ability to not only understand and learn them but truly recognise them, a cross-circulation approach to teaching and learning is introduced by CBSE, known as Art Integration.

Art Integration is an art-based approach to make learning joyful alongwith allowing students to have greater appreciation and understanding of the subject and its application. Based on the collaboration between the teaching of subject and subject of Arts, it aims at better understanding of subject with arts and its multiple forms being the primary pathway of learning it. We at Oswaal, have always valued the improtance of simplifying the learning process and have welcomed this creative approach of learning in our books, with open arms. We have incorporated Art Integration in our Question Banks in forms of subject related fun-case based studies and various interactive visual videos to make learning not only fun but allowing deeper understanding of subjects in students mind. Now, that's learning truly made simple.

 

For experiential learning through Art Integrated Activities, scan the code below-

OR SCAN

Scan the QR code

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WRITING YOUR NOTES Just in case you have forgotten today, takedown your notes! But why is it so important? Tools for the hands are tools for the brain writes Hetty Roessingh. Handwritten notes are a powerful tool for encrypting embodied cognition and in turn supporting the brain’s capacity for recuperation of information. If that sounds so scientific then in simple words: Writing notes by hand help you in: u

Increasing your comprehension �u Strengthening your memory �u Igniting your creativity u Engaging your mind �� u Increasing your attention span Are these reasons enough to get you started?

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SOLVED PAPER

2022

Mathematics

Class–XII

Delhi/Outside Delhi Time : 2 Hours

Max. Marks : 40

General Instructions : (i) This question paper contains three Sections - A, B and C. (ii) Each section is compulsory. (iii) Section-A has 6 short answer type-I questions of 2 marks each. (iv) Section-B has 4 short answer type-II questions of 3 marks each. (v) Section-C has 4 long answer type questions of 4 marks each. (vi) There is an internal choice in some questions. (vii) Question 14 is a case study based question with two sub parts of 2 marks each. Delhi Set-I

65/5/1

SECTION - A Question numbers 1 to 6 carry 2 marks each. dx 1. Find  4 x  x 2 2. Find the general solution of the following differential equation:

2 2

dy  exy  x2ey dx



3. Let X be a random variable which assumes values x1, x2, x3, x4 such that 2P(X = x1) = 3P(X = x2) = P(X = x3) = 5P(X = x4). 2 Find the probability distribution of X.      4. If a  i  j  k , a.b  1 and a  b  j  k , then find | b |.



5. If a line makes an angle a, b, g with the coordinate axes, then find the value of cos2a + cos2b + cos2g. 1 7 1 6. (a) Events A and B are such that P( A ) = , P( B) = and P( A ∪ B) = 2 12 4

2 2

Find whether the events A and B are independent or not. OR (b) A box B1 contains 1 white ball and 3 red balls. Another box B2 contains 2 white balls and 3 red balls. If one ball is drawn at random from each of the boxes B1 and B2, then find the probability that the two balls drawn are of the same colour.

SECTION - B Question numbers 7 to 10 carry 3 marks each. /4

7. Evaluate:

 0

dx 1  tan x

3

        8. (a) If a and b are two vectors such that | a  b |  | b |, then prove that ( a + 2b ) is perpendicular to a .

3

[ 15

SOLVED PAPER - 2022

OR    1   (b) If a and b are unit vectors and q is the angle between them, then prove that sin  | a  b |. 3 2 2  9. Find the equation of the plane passing through the line of intersection of the planes r.(i  j  k )  10 and  r.( 2i  3 j  k )  4  0 and passing through (–2, 3, 1). 3 10. (a) Find:

∫e

x

.sin 2 xdx

3

OR

(b) Find:

2x

 ( x 2  1)( x 2  2) dx

3



SECTION - C Question numbers 11 to 14 carry 4 marks each. 11. Three persons A, B and C apply for a job a manager in a private company. Chances of their selection are in the ratio 1 : 2 : 4. The probability that A, B and C can introduce chances to increase the profits of a company are 0.8, 0.5 and 0.3 respectively. If increase in the profit does not take place, find the probability that it is due to the appointment of A. 4 12. Find the area bounded by the curve y = |x – 1| and y = 1, using integration.

4

2

13. (a) Solve the following differential equation: (y – sin x)dx + tanx dy = 0

4

OR (b) Find the general solution of the differential equation: (x3 + y3)dy = x2ydx

4

Case Study Based Question



14. Two motorcycles A and B are running at the speed more than the allowed speed on the roads represented by the   lines r  (i  2 j  k ) and r  ( 3i  3 j )  ( 2i  j  k ) respectively. 2×2=4

Based on the above information, answer the following questions:

(a) Find the shortest distance between the given lines.

2

(b) Find the point at which the motorcycles may collide.

2

 Delhi Set-II

65/5/2

Note: Except these all other Questions are from Set-I.

SECTION - A 1. Find the vector equation of a line passing trough a point with position vector 2i  j  k and parallel to the line joining the points i  4 j  k and i  2 j  2 k .



2

SECTION - B       9. (a) Let a  i  j , b  i  j and c  i  j  k . If n is a unit vector such that a.n = 0 and b.n = 0, then find | c.n |. 3

16 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII

OR   (b) If a and b are unit vectors inclined at an angle 30° to each other, then find the area of the parallelogram with     ( a + 3b ) and ( 3a + b ) as adjacent sides. 3 /2

10. Evaluate:

 0

1 dx 1  (tan x )2 / 3

SECTION - C 13. In a factory, machine A produces 30% of total output, machine B produces 25% and the machine C produces the remaining output. The defective items produced by machines A, B and C are 1%, 1.2%, 2% respectively. An item is picked at random from a day's output and found to be defective. Find the probability that it was produced by machine B? 4 Delhi Set-III

65/5/3

Note: Except these all other Questions are from Set-I.

SECTION - A 2x  1 y  2 z  3   12 2 3 Fidn the direction cosines of a line parallel to line AB. 1. The Cartesian equation of a line AB is:

2

SECTION - B 3

9. Evaluate:

∫ 1

x x + 4−x

dx .

3

10. Find the distance of the point (2, 3, 4) measured along the line

x 4 y 5 z1   from the plane 3x + 2y + 2z 3 6 2

+ 5 = 0.

SECTION - C 13. There are two boxes, namely box-I and box-II. Box-I contains 3 red and 6 black balls. Box-II contains 5 red and 5 black balls. One of the two boxes, is selected at random and a ball is drawn at random. The ball drawn is found to be red. Find the probability that this red ball comes out from box-II.

 Outside Delhi Set-I

65/3/1

SECTION - A Question Nos. 1 to 6 carry 2 marks each. dx 1. Find :  2 x  6 x  13

2

2. Find the general solution of the differential equation : edy/dx = x2.       3. Write the projection of the vector b  c on the vector a , where a  2i  2j   k , b  i  2j  2 k and c  2i  j  4 k.

 

4. If the distance of the point (1, 1, 1) from the plane x – y + z + l = 0 is

5 3

, find the value(s) of l.

2

2 2

5. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spade cards. 2 6. A pair of dice is thrown and the sum of the numbers appearing on the dice is observed to be 7. Find the probability that the number 5 has appeared on atleast one die. 2

[ 17

SOLVED PAPER - 2022

OR The probability that A hits the target is

1 2 and the probability that B hits it, is . If both try to hit the target 3 5 2

independently, find the probability that the target is hit.

SECTION - B Question Nos. 7 to 10 carry 3 marks each. 2x

dx

 1  e sin x

7. Evaluate :

0

3



8. Find the particular solution of the differential equation x

dy  y  x 2 .e x , given y(1) = 0. dx

3

OR Find the general solution of the differential equation x

dy  y(log y  log x  1). dx

k and i − 2j − 3 k . Find the unit 9. The two adjacent sides of a parallelogram are represented by vectors 2i  4j  5 vector parallel to one of its diagonals, Also, find the area of the parallelogram. 3 OR       a  2i  2j  3 k , b  i  2j   k and c  3i  j are such that the vector a   b is perpendicular to vector c , then If





find the value of l.

3

10. Show that the lines :

3

1x y 3 z x − 4 2y − 2   and = = z − 1 are coplanar. 2 4 1  3 −4

SECTION - C Question Nos. 11 to 14 carry 4 marks each. 11. Find the area of the region bounded by curve 4x2 = y and the line y = 8x + 12, using integration. 12. Find :

x2

 ( x 2  1)(3x 2  4 ) dx 1



Evaluate :

4 4

OR

5  4 x  x 2 dx

2

 k   3i  4j  2 k and the point 13. Find the distance of the point (1, –2, 0) from the point of the line r  4i  2j  7  r. i  j   k  10. 4 Case Study Based Problem: 14. A shopkeeper sells three types of flower seeds A1, A2, A3. They are sold is the form of a mixture, where the proportions of these seeds are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. 4









18 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII

Based on the above information : (a) Calculate the probability that a randomly chosen seed will germinate: (b) Calculate the probability that the seed is of type A2, given that a randomly chosen seed germinates.

2 2

 Outside Delhi Set-II

65/3/2

Note: Except these all other Questions are from Set-I.

SECTION - A  dy  2. Find the general solution of the differential equation : log    ax  by.  dx 

2

SECTION - B               7. If a , b , c are three vectors such that a . b = a . c and a  b  a  c , a  0 , then show that b = c.

3

OR           If= a 3= , b 5, c = 4 and a  b  c  0, then find the value of a.b  b.c  c.a .

3



2

8. Evaluate :

|x

3



 x | dx

3



1

SECTION - C 12. Using integration, find the area of the region bounded by the curves x2 + y2 = 4, x =

3y and x-axis lying in the 4

first quadrant.

 Outside Delhi Set-III

65/3/3

Note: Except these all other Questions are from Set-I.

SECTION - A 3. Find the general solution of the differential equation :

dy 3e 2 x  3e 4 x  x dx e  e x

2

SECTION - B 7. Find the shortest distance between the following lines :   r  3i  5j  7 k   i  2j   k and r = −i − j −  k + µ 7i − 6j +  k .





(

) (

)

3

 2

10. Evaluate :

  sin| x |  cos| x | dx

 2

3



SECTION - C 13. Find the area of the region enclosed by the curves y2 = x, x =

1 , y = 0 and x = 1, using integration. 4

4



[ 19

SOLVED PAPER - 2022

Solutions Delhi Set-I

65/5/1 Þ 61k = 30 30 Þ k = 61

I =



=



=



1. Let

= =

 



SECTION - A dx

The required probability distribution is:

4x  x2 dx

P(X = x1)

P(X = x2)

P(X = x3)

P(X = x4)

2

30 15  61  2 61

30 10  61  3 61

30 61

30 6  61  5 61

( x  4 x ) dx 2

( x  4 x  2 2  2 2 ) dx ( x  2 )2  2 2 dx 2 2  ( x  2 )2

−1  x − 2  = sin   +C 2 

  dx  x = sin −1   + C  ∵ ∫ 2 2   a a −x  

2. Given differential equation is dy = ex – y + x2e–y dx Þ

dy = e–y(ex + x2) dx

dy Þ = dx(ex + x2) e−y Þ eydy = exdx + x2dx On integrating both sides, we get x3 ey = e x + +c 3 3. Given, 2P(X = x1) = 3P(X = x2) = P(X = x3) = 5P(X = x4) Let 2P(X = x1) = 3P(X = x2) = P(X = x3) = 5P(X = x4) = k k \ P(X = x1) = ...(i) 2 k ...(ii) 3



P(X = x2) =



P(X = x3) = k ...(iii) k P(X = x4) = ...(iv) 5



On adding eqs. (i) – (iv), and equating sum of all probabilities is equal to 1, we get k k k + +k+ =1 2 3 5 15k + 10 k + 30 k + 6 k Þ =1 30

 a = i + j + k  a.b = 1   a × b = j − k

4. Given, and

    b = ai + b j + ck  a.b = 1

Let Now,

      Þ (i + j + k )( b1 i + b2 j + b3 k ) = 1

Þ

b1 + b2 + b3 = 1   a × b = j − k

and Þ

Þ

i 1 b1

j 1 b2

k 1 b3

...(i)

= j − k

i( b − b ) − j( b − b ) + k ( b − b ) 3 2 3 1 2 1

= j − k

On comparing both sides, we get –(b3 – b1) = 1 and b2 – b1 = –1 Þ b3 – b1 = –1 and b2 – b1 = –1 Þ b3 = –1 + a and b2 = –1 + b1 ...(ii) Now from eq. (i), we get b1 + (–1 + b1) + (– 1 + b1) = 1 Þ 3b1 = 3 Þ b1 = 1 From eq. (ii), we get b2 = 0 and b3 = 0  b = i \  Therefore, |b| = 1 5. We have, cos2a + cos2b + cos2g = 2cos2a – 1 + 2cos2b – 1 – 2cos2g – 1 = 2(cos2a + cos2b + cos2g) – 3 = 2 × 1 – 3 [ cos2a + cos2b + cos2g = 1] =2–3 = –1 1 7 6. (a) Given P(A) = , P(B) = 2 12 and P(A È B) =

1 4

20 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII



For A and B are independent P(A Ç B) = P(A).P(B) Now, P(A È B) = P(A Ç B) Þ P(A È B) = 1 – P(A Ç B) Þ P(A Ç B) = 1 – P(A È B) 1 3 Þ P(A Ç B) = 1 − = 4 4

...(i)

and

...(ii)

1 7 7 Now, P(A).P(B) = × = ...(iii) 2 12 24 Since from eqs. (ii) & (iii) P(A Ç B) ¹ P(A).P(B) Therefore, events A and B are not independent. OR (b) Box B1

1 White Balls 3 Red Balls

Box B2 2 White Balls 3 Red Balls

\ P(Required) = P(Both are white) + P(Both are red) 1 2 3 3 =    4 5 4 5 2 9 11 + = = 20 20 20



SECTION - B /4

dx 1  tan x

/4

dx sin x 1 cos x

/4

cos xdx cos x  sin x

I =

0

=

0

=

0

=

1  / 4 2 cos x dx 2 0 cos x  sin x

7. Let

=

1 π / 4 cos x + sin x + cos x − sin x dx 2 ∫0 cos x + sin x

=

π / 4 cos x − sin x 1  π / 4 cos x + sin x  dx dx + ∫ 0 2  ∫ 0 cos x + sin x cos x + sin x 

=

π / 4 cos x − sin x 1  π/4  dx 1dx + ∫ ∫  0 0 2 cos x + sin x 

=

1 ( I1 + I 2 ) 2

where, I1 =

π/4

∫0

and I2 =

Let cosx + sinx = t

\ I2 =

2

π ,t= 4 2/ 2

∫1

2

dt t 2

= [log t]1 2 = log = log

2 2 2 2

− log 1 −0

= log23/2 3 = log 2 2 \

I =

1 ( I1 + I 2 ) 2

1 3   log 2  2  4 2     | a + b| = |b| 8. (a) Given, On squaring both sides, we get    | a + b |2 = | b |2  2  2  2   Þ | a | + | b | +2 | a || b | = | b |    | a |2 + 2 | a || b | = 0 Þ    | a |.(| a | + 2 | b |) Þ = 0    a.( a + 2b ) = 0 Þ    Since, dot product of a and a + 2b is zero, thus or

I =

vectors are perpendicular. Hence Proved OR   | a | = 1 and | b | = 1 Given, Now, we take       | a − b |2 = ( a − b ).( a − b )  2    = | a |  | b | 2 | a |.| b |   = 1  1  2 | a |.| b |cos 

    = 1   1  2 sin 2   2    π 4

− sin x dx cos x + sin x

π / 4 cos x

∫0

x =

= 2 – 2 × 1 × 1cosq = 2(1 – cosq)

1dx

π/4 = [ x]0 =

Þ (–sinx + cosx)dx = dt when x = 0, t = 1

 2  = 2  2 sin  2  = 4 sin 2

θ 2

[ 21

SOLVED PAPER - 2022

or,

  θ | a − b |2 sin = 2 4   θ |a − b| sin = 2 2 2

Þ Hence proved 9. Given planes are:  r.(i + j + k ) = 10     r .( 2 i  3 j  k )  4 and = 0 Changing to cartesian form, the equations of these planes are x + y + z – 10 = 0 and 2x + 3y – z + 4 = 0 Equation of any plane through the intersection of these planes is (x + y + z – 10) + l(2x + 3y – z + 4) = 0 ...(i) Now, it passes through the point (–2, 3, 1) \ (–2 + 3 + 1 – 10) + l[2 × (–2) + 3(3) – 1 + 4] = 0 Þ –8 + l(–4 + 9 + 3) = 0 Þ –8 + 8l = 0 Þ 8l = 8 l = 1 On substituting the value of l in eq. (i), we get (x + y + z – 10) + 1(2x + 3y – z + 4) = 0 Þ 3x + 4y – 6 = 0, which is the required equation of plane in cartesian form. In vector form, required equation of plane is:  r.( 3i  4 j )  6 = 0 10. (a) Let

I =

∫e

x

sin 2 xdx

Applying integration by parts I = ∫ e x sin 2 xdx I II d x  x = e  sin 2 xdx    ( e ) sin 2 xdx  dx dx   1 x   cos 2 x    e x cos 2 xdx = e   2   2 =

1 x 1 d   ( e cos 2 x )   e x  cos 2 xdx    ( e x ) cos 2 xdx  dx  dx 2 2   

=

 1 1  e x sin 2 x 1 x (  e x cos 2 x )     e sin 2 xdx  2 2 2 2 

=

1 1 1 ( − e x cos 2 x ) + ( e x sin 2 x ) − ∫ e x sin 2 xdx + K 2 4 4

\ 4I = –2excos2x + exsin2x – I + K or 5I = –2excos2x + exsin2x + K 1 K I = ( e x sin 2 x − 2 e x cos 2 x ) + 5 5 or I =

1 x ( e sin 2 x  2 e x cos 2 x )  c 5 OR

Let

I =

2x

 ( x 2  1)( x 2  2) dx

1 A B + 2 let = 2 2 ( x + 1)( x 2 + 2 ) + 1 +2 x x 2 Þ 1 = A(x + 2) + B(x2 + 1) Þ 1 = (A + B)x2 + (2A + B) On comparing both sides, we get A + B = 0 and 2A + B = 0 On solving above equations, we get A = 1 and B = –1 1   1  2 \ I =   2  2 xdx x 1 x 2 2x

2x

 x 2  1 dx   x 2  2 dx



I =



I = log|x2 + 1| – log|x2 + 2| + C



I = log

x2 + 1 +C x2 + 2

SECTION - C 11. Let E1 = Person A gets the job E2 = Person B gets the job E3 = Person C gets the job A = No change takes place The changes of selection of A, B and C are in the ratio 1 : 2 : 4 1 2 4 Hence, P(E1) = , P(E2) = , P(E3) = 7 7 7 A A  2 5 Also, given P   = 0.2 = , P   = 0.5 = E1  E2  10 10   A  7 P   = 0.7 = and E3  10  Required probability is A P   .P( E1 )  E1   E1  P  = A  A A A P   .P( E1 )  P   .P( E 2 )  P   .P( E3 )  E1   E2   E3 

2 1  10 7 = 2 1 5 2 7 4      10 7 10 7 10 7 2 2 1 70 = = = 2 10 28 40 20 + + 70 70 70

\ If no change takes palace, the probability that it is 1 due to appointment of person A is . 20

22 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII

12.

y.If = ∫ Q.If dx + C1



y.sinx = ∫ (sin x cos x sin x )dx + C1



y.sinx =



y.sinx = I + C1 ...(i)

where

I =

∫ sin ∫ sin

2

2

x cos xdx + C1 x cos dx

let sinx = t Þ cosxdx = dt



We have, y = (x – 1) y = x – 1, if x – 1 ³ 0 y = –x + 1, if x – 1 < 0 Required Area = Area of shaded region

A =

=

2

∫0 ydx 1

2

0 (1  x )dx  1 ( x  1)dx 1

2

  x2   x2 =  x      x  2 2 1  0 

1  0 4   1  =  1     0      2     1  2 2 2 2         =

1 1 + 2 2

= 1 sq. unit 13. (a) Given differential equation is (y – sin2x)dx + tanxdy = 0 (y – sin2x)dx = –tanxdy

dy y − sin 2 x = dx − tan x 2



dy sin x − y = dx tan x 2

\

2 I = ∫ t dt =

or

I =



y.sinx =

sin 3 x + C2 + C1 3

or

y.sinx =

sin 3 x +C 3

OR Given differential equation is (x3 + y3)dy = x2ydx

v+y



dv = dy dv v+y = dy dv y = dy dv y = dy



v2dv =





dy = sinxcosx – ycotx dx



P = cot x Q = sin x cos x ∫ Pdx ∫ cot xdx Here, If = e = e = elog|sinx| = sinx \ Solution is given by

...(i)

dv ( vy )3 + y 3 = dy ( vy )2 y

v+y



where

(where C = C1 + C2)

dx x3 + y3 = dy x2y \ Put x = vy dv dx = v+y Þ dy dy from eq. (i), we have

dy sin x y = − dx tan x tan x

which is a linear differential equation of the form dy + Py = Q dx

sin 3 x + c2 3

from eq. (i), we have



dy + y cot x = sinxcosx dx

t3 + C2 3

v3 y 3 + y 3 v2 y3 v3 + 1 v2 v3  1 v v2 1 v2 dy y

(variable seponation method) Integrating both sides, we get dy 2 ∫ v dv = ∫ y

v3 = logy + C 3

[ 23

SOLVED PAPER - 2022

Putting v =

x , we get y

x3 = logy + c 3y3 14. (a) Given, lines are:   r  (i  2 j  k ) and r  ( 3i  3 j )  ( 2i  j  k ) We know that, shortest distance between the lines        r  a  b1 and r  a2   b1 is 1         |( a2  a1 ).( b1  b2 )|     d = | b1  b2 |     a1 = 0, a2 = ( 3i + 3 j ) Here,   b1 = i  2 j  k   b2 = 2i + j + k and     a2 − a1 = ( 3i  3 j )  0 = 3i + 3 j \ i j k     b1 × b2 = 1 2 −1 2 1 1 = i( 2  1)  j(1  2 )  k (1  4 )

= 3i − 3 j − 3k     | b1 × b2 | = 3 2  ( 3)2  ( 3)2 and = 9+9+9 = 3 3         Also, ( a2  a1 ).( b1  b2 ) = ( 3i  3 j ).( 3i  3 j  3k ) = 9 – 9 = 0 0 d = =0 3 3 Thus, distance between lines is 0.  (b) We have, r = (i  2 j  k ) and

...(i)  r = 3i  3 j  ( 2i  j  k )  r = ( 3  2 )i  ( 3   ) j   k ...(ii)

or Now, from eq. (i) & eq. (ii), we get (i  2 j  k ) = ( 3  2 )i  ( 3   ) j   k On comparing both sides, we get 3 + 2m = l, 3 + m = 2l and m = –l On solving for values of l and m, we get l = 1 and m = –1  from eq. (i), we get r = i  2 j  k xi + y j + zk = i  2 j  k So, required point is (1, 2, –1).

 Delhi Set-II

65/5/2

SECTION - A 1. Let A, B and C be the points with position vectors 2i  j  k ,  i  4 j  k and i + 2 j + 2 k , respectively. We have to find the equation of a line passing through the point A and parallel to vector BC. Now,   BC = position vector of C – position vector of B (i  2 j  2 k )  ( i  4 j  k ) = 2i  2 j  k = We know that, the equation of a line passing   through a position vector a and parallel to vector b is    r = a  b  r = ( 2i  j  k )  ( 2i  2 j  k ) \ is the required equation of line in vector from.   [ Here, BC = b]

SECTION - B 9. Given,

  a = i + j , b = i − j

and Also, given and

 c = i + j + k

 a.n = 0  b.n = 0

Here,

  n = a × b |a × b|

Here,

i j   a × b = 1 1 1 −1

=

k 0 0

i( 0  0 )  j( 0  0 )  k ( 1  1)

= −2k −2 k \ n = = −k ( −2 )2  Therefore, | c.n | = |(i  j  k ).(  k )| = |–1| = 1 OR We know, Area of parallelogram with adjacents

24 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII

  sides p and q is given by   A = | p × q |     Here, Area = |( a  3b )  ( 3a  b )|

SECTION - C

        = | 3( a  a )  ( a  b )  9( b  a )  3( b  b )|     = | 3  0  ( a  b )  9( a  b )  3  0 |         [∵ a × a = 0 = b × b and b × a = − a × b]   | 8( a  b )| =   = 8| a × b |   8| a |.| b |sin θ =   = 8.1. 1. sin 30° [Given, | a | = 1 = | b | and q = 30°] 1 = 8. 2 = 4 sq. units 10. Let

I =



I =

∫0

1 dx 1 + (tan x )2 / 3

∫0

I =



I =



I =



π / 2 1 + (tan x )

∫0

π/2

I=

∫0

2I =

∫0

1 + (tan x )3 / 2

1.dx − I

π/2

1.dx

π/2 2I = [ x]0

π

2I = 2

π

I= 4

3/2



a

dx

0 f ( x )dx  0 f ( a  x )dx]

π/2

(tan x )2 / 3 dx (tan x )2 / 3 + 1 2/3

+1−1 dx 2/3 (tan x ) + 1

π / 2 (tan x )

∫0

dx − ∫

2/3

a

1 dx 1 + (cot x )2 / 3

∫0



 π  1 + tan  − x    2 

π/2

∫0

...(i)

1

π/2

[ Using property



I=

π/2

π/2 0

1 dx 1 + (tan x )3 / 2

[From eq.(i)]

13. Let E1 = choosing machine A E2 = choosing machine B E3 = choosing machine C A = Producing a defective output 30 Given, P(E1) = 30% = = 0.3 100 P(E2) = 25% =

25 = 0.25 100

P(E3) = [100 – (30 + 25)]% = 45% 45 = = 0.45 100 A and P    E1  = P(Producing defective output from machine A) 1 = 1% = = 0.01 100 A P  E  2  = P(Producing defective output from machine B) 1.2 = 1.2% = = 0.12 100 A P  E  3  = P(Producing defective output from machine C) 2 = 2% = = 0.02 100  E2  Required probability = P   A = P(The found defective item is produced by machine B) Using Bayes' theorem, E  P 2  A   A P( E 2 ).P    E2  = A A A P( E1 ).P    P( E 2 ).P    P( E3 ).P   E E  E3   1  2 0.25  0.12 = ( 0.3  0.01)  ( 0.25  0.12 )  ( 0.45  0.02 ) 300 300 1 = = = 300 + 300 + 900 1500 5 Thus, required probability is



1 . 5



[ 25

SOLVED PAPER - 2022

Delhi Set-III

65/5/3

1. We have,

2x − 1 y+2 z−3 = = 12 2 3

The equation of line AB can be rewritten as 1 x− 2 = y − ( −2 ) = z − 3 6 2 3 Thus, direction ratios of the line parallel to AB are proportional to 6, 2, 3. Hence, the direction cosines of the line parallel to AB are proportional to 6 2 3 , , 2 2 2 2 2 2 2 6 +2 +3 6 + 2 2 + 32 6 + 2 + 3 6 2 3 , , or 49 49 49 6 2 3 or , , 7 7 7

SECTION - B 9. Let Using property

I = b

a

x

3

1

b

a

3

1

4x 4x  x

dx

On adding eqs. (i) and (ii), we get

2I =

3

1

x  4x x  4x

d= =



...(ii)

3

3 = [ x]1 = 3 – 1 = 2 \ I = 1 10. Here, Equation of line PQ is x2 y3 z4    ( let ) 6 2 3

9 + 36 + 4 =

49 = 7 units

13. Let E1 = Selecting Box-I E2 = Selecting Box-II A = getting a red ball from the selected box 1 1 Here, P(E1) = , P(E2) = 2 2



A 3 1 P  = = 3  E1  9



A 5 1 P  = = 2  E 2  10

 E2  Required probability = P   A = P(Red ball comes out from Box-II) Using Bayes' theorem,

dx

= ∫1 1dx

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

SECTION - C

...(i)

x  4x

f ( x )dx   f ( a  b  x )dx , we get

I =

\ Q(3l + 2, 6l + 3, 2l + 4) Point Q lies on the plane 3x + 2y + 2z + 5 =0 \ 3(3l + 2) + 2(6l + 3) + 2(2l + 4) + 5 = 0 Þ 9l + 6 + 12l + 6 + 4l + 8 + 5 = 0 Þ 25l = –25 Þ l = –1 So, Q(–1, –3, 2) Now, distance between points P(2, 3, 4) and Q(–1, –3, 2) is given by

SECTION - A



A P( E 2 )P   E    E2  P 2  = A A  A  P( E1 ).P    P( E 2 ).P   E  1  E2 

1 1  2 2 = 1 1 1 1    2 3 2 2 =

1 4

1 1 + 6 4

1 1 24 3 4 = = = × 10 4 10 5 24 Thus, probability that the red ball comes out form 3 Box-II is . 5



26 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII

Outside Delhi Set-I

65/5/1

SECTION - A

=

 3i  j  2k . 2i  2j  k  ( 2 )2  ( 2 )2  (1)2

1. Given integral is

I =

dx

=

dx

=

 ( x  3)2  13  9

=

 ( x  3 )2  4

=

622 3 = 2 4. We know that, distance of (x1, y1, z1) from a plane ax + by + cz + d = 0 is given by

 x 2  6x  13 dx



dx

 ( x  3 )2  2 2 Given, p =

1 1  x  3  = tan  C 2  2  1 1 x   dx  tan 1  C   Using  2 2 a a x a  

dy Þ = 2 logx dx Þ dy = 2 logx dx On integrating both sides, we get

[ loge = 1]

y = 2 ∫ 1.log x dx

Þ

d   y =  log x  1 dx   (log x )  1. dx dx  dx  





[Using integration by parts] 1   y = 2  log x( x )   ( x )dx  x  

Þ y = 2[xlogx – x] + C Þ y = 2x(logx – 1) + C 3. Given vectors,  k a = 2i  2j    k b = i  2j  2  k c = 2i  j  4   k  2i  j  4 k b + c = i  2j  2   b + c = 3i + j + 2 k or,      b  c .a  Projection of b  c on a =  a



 

 

 

3 5 3

=

=

a2 + b2 + c 2 5

and (x1, y1, z1) = (1, 1, 1)

3

111  12  12  12 1 3

Þ l + 1 = ±5 Þ l = 4 or –6 5. Let X denote the number of spades in a sample of 2 drawn cards from a well shuffle pack of 52 cards. Then, X can take the values 0, 1, 2. Now, P(X = 0) = P(no spade) 39

Þ

Þ

5

ax1 + by1 + cz1 + d

=

∫ dy = 2∫ log x dx



\

Þ

2. Given differential equation is edy/dx = x2 Taking log both sides, we get dy log e = 2 logx dx



p =

52

=

741 1326

=

19 34



P(X = 1) = P(one spade card) 13

=

C1 × 39C1 52

=

507 1326

=

13 34



C2

P(X = 2) = P(both cards are spade) 13



C2

C2

=

52

C2 C2

=

78 1326

=

1 17

[ 27

SOLVED PAPER - 2022

Thus, the probability distribution of X is given by X

P(X)

0

19 34

1 2



8. Given differential equation is dy − y = x2.ex x dx

13 34



Þ

1 17



OR

P(A) = P(A hits target) =

1 3



P(B) = P(B hits target) =

2 5

=

562 15

=

9 3 = 15 5

=

1 + e sin x

1

π

1



∫ 0  1 + e sin x + 1 + e sin( 2 π − x )  dx

 2 a f ( x )dx = a f ( x ) + f ( 2 a − x ) dx  }  ∫0 {  ∫ 0

= = =

π

1

 π

1

∫ 0  1 + e sin x ∫ 0  1 + e sin x 

π1+

e

sin x







e

∫

−1 dx x

log

1

y.

1 = x y = x

 xe

x

1  dx  C x

 e dx  C x

y = ex + C x y = ex x



Given y = 0 when x = 1 from eq (i), we get 0 = 1.e1 + C.1 Þ C = –e OR y = xex – ex Given differential equation is dy x = y(logy – logx + 1) dx dy y y  log  1  = dx x  x 

Þ

e sin x  +  dx 1 + e sin x 

Þ

  dx 1 + e − sin x 

y.I.F. = ∫ Q × I.F. dx + C



Put

∫ 0 1 + e sin x dx

=

= e–logx

1

+

1 and Q = xex x

pdx I.F. = e ∫





dx

2π

P = −

= The solution is given by

SECTION - B ∫0

Here,

dy y − = xex, which is of the form dx x dy + Py = Q dx

e x

Now, P(A È B) = P(target will be hit) = P(A) + P(B) – P(A Ç B) = P(A) + P(B) – P(A).P(B) [ A and B are independent] 1 2 1 2 =    3 5 3 5

I =

π

∫ 0 1.dx

=  x 0  

6. Let E = event that 5 has appeared on atleast one die \ E = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5), (4, 5), (3, 5), (2, 5), (1, 5)} Let F = event that sum of no. on die is 7. \ F = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} E Ç F = {(2, 5), (5, 2)} \ n(E Ç F) = 2 P E  F  n E  F  2 1 E = = = Now, P   = P( F ) n( F ) 6 3 F

7. Let

=

y = vx Þ v+x

dy dv = v+x dx dx

dv = v(logv + 1) dx

dv dx = v log v x On integrating both sides, we get Þ

...(i)

28 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII

∫



dx = x

∫

10. Given lines are : 1−x y3 z x  4 2y  2    z 1 = and 3 4 1 4 2

dx x

Þ log(logv) = logx + logC Þ log(logv) = logCx Þ log(y/x) = Cx 9. Given two adjacent sides of a parallelogram are  a = 2i  4j  5 k

or

 b = i − 2j − 3 k



 Let c be the diagonal of given parallelogram.    c = a + b



 

k  i  2j  3 k = 2i  4j  5



k = 3i  6j  2  c = ( 3)2  ( 6 )2  ( 2 )2  7 \   c 3i  6j  2 k Unit vector in direction of c =  = 7 c

x −1 y3 z0 x  4 y 1 z 1    = and −2 4 1 3 2 1

These lines will be coplanar if x 2 − x1 y 2 − y1 z2 − z1 a1 b1 c1 =0 a2 b2 c2 4 −1 1−3 1−0 3 −2 1 − 2 4 − 1 − = 2 4 −1 = 0 3 −2 1 3 −2 1 [Since, R1 = R3] Thus, given lines are coplanar,

SECTION - C 11. Given curve is 4x2 = y and line is y = 8x + 12 On solving both equation, we get

  Now, Area of parallelogram = a × b i j  k   a × b = 2 −4 5 1 −2 −3

\

k = 12  10 i   6  5 j   4  4   = 22i + 11j

( 22 )2 + (11)2

=

484  121  605 sq. units



=

OR   Given vectors are a  2i  2j  3 k , b  i  2j   k and  c  3i  j         Now, a   b = 2i  2 j  3k   i  2 j  k =  2   i   2  2 j   3   j



 



   If a   b is perpendicular c , then    a   b .c = 0      Þ  2    i   2  2  j   3    k . 3i  j = 0









Þ

(2 – l).3 + (2 + 2l).1 + (3 + l).0 = 0

Þ

6 – 3l + 2 + 2l = 0

Þ Þ

–l + 8 = 0 l = 8



  Therefore, Area of parallelogram = a × b

Þ Þ Þ

4x2 = 8x + 12 x2 = 2x + 3 2 x – 2x – 3 = 0 x = 3, –1 Required area =

1 8x  12   4 x 3

3



2

dx



2 = 4 1 2 x  3  x dx 3

 x3  = 4  x 2  3x   3  1   1   = 4  9  9  9    1  3    3    1  = 4  9  2   3  1  = 4  11   3  = 4 

32 128  sq. units 2 2

[ 29

SOLVED PAPER - 2022

12. Let

x2

Þ 4 + 3l – 2 – 4l + 7 + 2l = 10 Þ l + 9 = 10 Þ l = 1 Now, from (i), we get  k  1 3i  4j  2 k r = 4i  2j  7  k or, r = 7i + 6j + 9  = 7i + 6j + 9 k or, xi + yj + zk \ Required point is (7, 6, 9) Now, distance between (7, 6, 9) and (1, –2, 9) is

 ( x 2  1)(3x 2  4 ) dx

I =

Put t = x2 t A B + = (t + 1)( 3t + 4 ) + 1 3 +4 t t t = A(3t + 4) + B(t + 1) t = (3A + B)t + (4A + B) On comparing both sides, we get 3A + B = 1 and 4A + B = 0 1 4 dx   2 dx \ I =  2 x 1 3x  4





= 

1 x2  1

dx  4 

1

=

dx

3x 2  4

1 dx  4  =  2 x  12





2

 22



(7  1)2  ( 6  2 )2  ( 9  9 )2 62 + 82 + 02

= 36 + 64

dx

= 100 = 10 units 14. (a)

 3x  1 4 x C tan 1    tan 1   2  1 2   2 3  

1 =  tan x 

d =

=

1

3x



 

 3x  C tan 1   2  3  

2

OR 5  4 x  x 2 dx

1

( x 2  4 x  5)dx

1

( x 2  4 x  2 2  2 2  5)dx

1

  x  2   9 dx

1

32  ( x  2 )2 dx

Let I =

2

=

2

=

2

=

2

=

2





1

1

x  2 2 3  x  2  3  ( x  2 )2  sin 1  =   2  3   2  2 2

9  = 0  .  ( 0  0 ) 2 2 =

9π 4

13. Given line is  r = 4i  2j  7 k   3i  4j  2 k ...(i) and plane is  r i − j +  k = 10 ...(ii) from (i) and (ii), we get 4i + 2j + 7 k + λ 3i + 4j + 2 k . i − j +  k = 10       Þ ( 4 + 3λ ) i + ( 2 + 4 λ ) j + 7 + 2 k. i − j + k = 10 Þ (4 + 3l).1 + (2 + 4l)(–1) + (7 + 2l).1 = 10



(



)

(

(

)(

)(

4 4 2 , P(E2) = , P(E3) = 10 10 10

A  A A 45 60 35 P  = , P  = , P  = E E E 100 100 100  2  3  1  A  A A \ P(A) = P( E1 ).P    P( E2 ).P    P( E3 ).P   E E  E3   2  1



2

Here, P(E1) =

)

)

=

4 45 4 60 2 35 × + × + × 10 100 10 100 10 100

=

180 240 70 + + 1000 1000 100

=

490 = 4.9 1000

E  (b) Required probability = P  2   A  A P  E2  .P    E2  = P(A ) 4 60 × 10 100 = 490 1000 =

240 24 = 490 49



30 ]

Oswaal CBSE Chapterwises & Topicwise Question Bank, MATHEMATICS, Class–XII

Outside Delhi Set-II

65/5/1

SECTION - A 2. Given differential equation is  dy  log   = ax + by  dx  Þ

dy = eax + by dx

Þ

dy = eax.eby dx

dy Þ = eax dx e by Þ e–by dy = eax dx On integrating both sides, we get

e



 by

e − by e ax +C = −b a

Þ

dy = ∫ e ax dx

ax

 by

e e  C = 0 a b

SECTION - B     a.b = a.c     a . b − a . c Þ = 0    a. b  c = 0 Þ      Þ either b = c or a  b  c     Also, given a × b = a × c     a  b  a  c = 0 Þ    a b  c = 0 Þ      Þ either a || b  c or b = c  But vector a cannot be both parallel and perpendicular   to vector b − c . 7. Given,

      Þ ( a + b + c ).( a + b + c ) = 0          Þ a.a + a.b + a.c + b.a + b.b + b.c + c.a + c.b + c.c = 0 2    2    2 Þ a + a.b + c.a + a.b + b + b.c + c.a + b.c + c = 0       ∵ a.b  b.a , a.c  c.a , b.c  c.b    2 2 2    a  b  c  2 a.b  b.c  c.a = 0    2 2 2 Þ ( 3)  ( 5)  ( 4 )  2 a.b  b.c  c.a = 0    Þ 9  25  16  2 a.b  b.c  c.a = 0    Þ 50  2 a.b  b.c  c.a = 0    Þ a.b + b.c + c.a = –25 8. Let

  Hence, vector b = c. OR

Given, and \ Þ

   a = 3, b = 5, c = 4     a + b + c = 0     a+b+c = 0   2 a+b+c =0

2

1 x

3

2

 x dx

1 x( x

=

1 x( x  1)( x  1) dx

2

 1) dx

2

Here, x3 – x = 0, when x = 0, 1, –1

 

( )

I =

   

=

 

   

   

Þ

Value of x

Value of (x3 – x)

–1 < x < 0

+ve

0 0, b > 0 and b ¹ 1. Also note that logb 1 is equal to zero i.e., 0. 2.  Range: If a function is expressed in the form y = f(x), then range of f means set of all possible real values of y corresponding to every value of x in its domain.

Remember the following points: (a) At first find the domain of the given function. (b)  If the domain does not contain an interval, then find the values of y putting these values of x from the domain. The set of all these values of y obtained will be the range. (c)  If domain is the set of all real numbers R or set of all real numbers except a few points, then express x in terms of y and from this find the real values of y for which x is real and belongs to the domain. 3. Function as a special type of relation: A relation f from a set A to another set B is said be a function (or mapping) from A to B if with every element (say x) of A, the relation f relates a unique element (say y) of B. This y is called f - image of x. Also x is called pre-image of y under f.

These questions are for practice and their solutions are available at the end of the chapter

12

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

4. Difference between relation and function: A relation from a set A to another set B is any subset of A × B; while a function f from A to B is a subset of A × B satisfying following conditions: (a) For every x Î A, there exists y Î B such that (x, y) Î f. (b) If (x, y) Î f and (x, z) Î f then, y = z. S. No.

(i)

Function

Relation

Each element of A There may be must be related to some elements of some element of B. A which are not related to any element of B.

5.

S. No.

Function

Relation

(ii)

An element of A should not be related to more than one element of B.

An element of A may be related to more than one element of B.

Real valued function of a real variable: If the domain and range of a function f are subsets of R (the set of real numbers), then f is said to be a real valued function of a real variable or a real function.

6. Some important real functions and their domain & range S. No.

Function

(i) Identity function

Representation

Domain

Range

I(x) = x " x Î R

R

R

− x , if x < 0

(ii) Modulus function or Absolute value function

f(x) = |x| = x , if x ≥ 0 

R

[0, ¥)

(iii) Greatest integer function or Integral function or Step function

f(x) = [x] "x Î R

R

Z

(iv) Smallest integer function

f(x) = [x] " x Î R

R

Z

R

{– 1, 0, 1}

(v) Signum function

| x |  , if x ≠ 0 f(x) =  x i.e., f(x) = 0 , if x = 0

1, x > 0  0 , x = 0 −1, x < 0

(vi) Exponential function

f(x) = ax, " a > 0, a ¹ 1

R

(0, ¥)

(vii) Logarithmic function

f(x) = loga x, " a ¹ 1, a > 0 and x > 0

(0, ¥)

R

7. Types of Function (a) One-one function (Injective function or Injection): A function f : A ® B is one-one function or injective function if distinct elements of A have distinct images in B.

(b)  Onto function (Surjective function or Surjection): A function f : A ® B is onto function or a surjective function if every element of B is the f - image of some element of A. That implies f(A) = B or range of f is the co-domain of f.

Thus, f : A ® B is one-one Û f(a) = f(b) Þ a = b, "a, b Î A

Thus, f : A ® B is onto Û f(A) = B i.e., range of f = co-domain of f.

Û a ≠ b Þ f(a) ≠ f(b) " a, b Î A.   If A and B are two sets having m and n elements respectively such that m ≤ n, then total number of one-one functions from set A to set B is nCm × m! i.e., nPm.  If n(A) = n, then the number of injective functions defined from A onto itself is n!. ALGORITHM TO CHECK THE INJECTIVITY OF A FUNCTION STEP 1: Take any two arbitrary elements a, b in the domain of f. STEP 2: Put f(a) = f(b). STEP 3: Solve f(a) = f(b). If it gives a = b only, then f is a one-one function.

ALGORITHM TO CHECK SURJECTIVITY OF A FUNCTION

THE

STEP 1: Take an element b Î B, where B is the co-domain of the function. STEP 2: Put f(x) = b. STEP 3: Solve the equation f(x) = b for x and obtain x in terms of b. Let x = g(b). STEP 4: If for all values of b Î B, the values of x obtained from x = g(b) are in A, then f is onto. If there are some b Î B for which values of x, given by x = g(b), is not in A. Then f is not onto.



RELATIONS AND FUNCTIONS

Mnemonics



Types of functions Indian Syndicate Bank Interpretations Indian – injective Syndicate – surjective Bank - Bijective

Also note that a bijective function is also called a one-to-one function or one-to-one correspondence.

If f : A ® B is a function such that, (i) f is one-one Þ n(A) ≤ n(B). (ii) f is onto Þ n(B) ≤ n(A). For an ordinary finite set A, a one-one function f : A ® A is necessarily onto and an onto function f : A ® A is necessarily one-one for every finite set A. (d) Identity function: The function IA : A ® A; IA (x) = x, " x Î A is called an identity function on A. Note:  Domain (IA) = A and Range (IA) = A. (e) Equal function: Two functions f and g having the same domain D are said to be equal if f(x) = g(x) for all x Î D. Constant and Types of Variables



(a) Constant: A constant is a symbol which retains the same value throughout a set of operations. So, a symbol which denotes a particular number is a constant. Constants are usually denoted by the symbols a, b, c, k, l, m, ... etc.

8.

(b) Variable: It is a symbol which takes a number of values i.e., it can take any arbitrary values over the interval on which it has been defined. For example, if x is a variable over R (set of real numbers) then we mean that x can denote any arbitrary real number. Variables are usually denoted by the symbols x, y, z, u, v, … etc. (i)  Independent variable: The variable which can take an arbitrary value from a given set is termed as an independent variable.

(ii)  Dependent variable: The variable whose value depends on the independent variable is called a dependent variable. 9.

Defining a Function

Consider A and B be two non-empty sets, then a rule f which associates each element of A with a unique element of B is called a function or the mapping from A to B or f maps A to B. If f is a mapping from A to B, then we write f : A ® B which is read as ‘f is mapping from A to B’ or ‘f is a function from A to B’. If f associates a Î A to b Î B, then we say that ‘b is the image of the element a under the function f’ or ‘b is the f - image of a’ or ‘the value of f at a’ and denotes it by f(a) and we write b = f(a). The element a is called the pre-image or inverse-image of b.

Thus for a bijective function from A to B,

(a) A and B should be non-empty.

(b) Each element of A should have image in B.

(c) No element of A should have more than one image in B.

(d) If A and B have respectively m and n number of elements then the number of functions defined from A to B is nm.

10. Domain, Co-domain and Range of A function The set A is called the domain of the function f and the set B is called the co- domain. The set of the images of all the elements of A under the function f is called the range of the function f and is denoted as f(A).

Thus range of the function f is f(A)={f(x) : x Î A}.

Clearly f(A) = B for a bijective function. Note:  It is necessary that every f-image is in B; but there may be some elements in B which are not the f-images of any element of A i.e., whose pre-image under f is not in A.  Two or more elements of A may have same image in B.  f : x ® y means that under the function f from A to B, an element x of A has image y in B.  Usually we denote the function f by writing y = f(x) and read it as ‘y is a function of x’.

Example 1 Determine whether the function f : A ® B defined by f(x) = 4x + 7, x Î is one-one. Show that no two elements in domain have same image in codomain.

13

Solution : Given, f : A ® B defined by f(x) = 4x + 7, x Î A Let, x1, x2 Î A, such that f(x1) =f(x2) Þ 4x1 + 7 = 4x2 + 7 Þ 4x1 = 4x2 Þ x1 = x2 So, f is one-one function.

14

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

OBJECTIVE TYPE QUESTIONS (A) Surjective function (B) Injective function (C) Bijective function (D) None of these [CBSE TERM-I SQP 2021-22]

Multiple Choice Questions

Q. 1. Let X = {x2 : x Î N} and the relation f : N ® X is defined by f(x) = x2, x Î N. Then, this function is (A) injective only (B) not bijective (C) surjective only (D) bijective [CBSE TERM-I 2021-22] Ans. Option (A) is correct. Explanation: Let x1, x2 Î N f(x1) = f(x2) Þ x12 = x22 2 Þ x1 – x22 = 0 Þ (x1 + x2)(x1 – x2) = 0 Þ x1 = x2 {x1 + x1 ¹ 0 as x1, x2 Î N} Hence, f(x) is injective. Also, the elements like 2 and 3 have no pre-image in N. Thus, f(x) is not surjective. Q. 2. A function f : R ® R defined by f(x) = 2 + x2 is (A) not one-one (B) one-one (C) not onto (D) neither one-one nor onto [CBSE TERM-I 2021-22] Ans. Option (D) is correct. Explanation: f(x) = 2 + x2 For one-one, f(x1) = f(x2) Þ 2 + x12 = 2 + x22 Þ x12 = x22 Þ x1 = ±x2 Þ x1 = x2 or x1 = –x2 Thus, f(x) is not one-one. For onto Let f(x) = y such that y Î R \ x2 = y – 2 Þ

x = ± y − 2

Put y = –3, we get Þ

x = ± −3 − 2 = ± −5

Q. 3. A function f : R ® R defined by f(x) = 2 + x3 is : (A) One-one but not onto (B) Not one-one but onto (C) Neither one-one nor onto (D) One-one and onto [CBSE TERM-I SQP 2021-22] Ans. Option (D) is correct. Explanation: f(x) = x3 is a bijective function. Q. 4. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as :

Ans. Option (B) is correct. Explanation: F is injective since every element in set B has atmost one pre-image in set A. Q. 5. If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is (A) 720 (B) 120 (C) 0 (D) None of these Ans. Option (C) is correct. Explanation: We know that, if A and B are two non-empty finite sets containing m and n elements, respectively, then the number of one-one and onto mapping from A to B is n! if m = n 0, if m ¹ n Given that, m = 5 and n = 6 \m¹n Number of one-one and onto mapping = 0 Q. 6. Let A = {1, 2, 3, ...n} and B = {a, b}. Then the number of surjections from A into B is (A) nP2 (B) 2n – 2 n (C) 2 – 1 (D) None of these Ans. Option (B) is correct. Explanation: Total number of functions from A to B = 2n Number of into functions = 2 Number of surjections from A to B = 2n – 2 1 Q. 7. Let f : R ® R be defined by f(x) = , " x Î R. Then x f is (A) one-one (B) onto (C) bijective (D) f is not defined Q. 8. Which of the following functions from Z into Z are bijections? (A) f(x) = x3 (B) f(x) = x + 2 (C) f(x) = 2x + 1 (D) f(x) = x2 + 1 Ans. Option (B) is correct. Explanation: For bijection on Z, f(x) must be oneone and onto. Function f(x) = x2 + 1 is many-one as f(1) = f(–1) Range of f(x) = x3 is not Z for x Î Z. Also f(x) = 2x + 1 takes only values of type = 2k + 1 for x Î k Î Z But f(x) = x + 2 takes all integral values for x Î Z Hence f(x) = x + 2 is bijection of Z. Q. 9. Let f : R ® R be defined as f(x) = x4. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto

A



RELATIONS AND FUNCTIONS

Ans. Option (D) is correct. Explanation: We know that f : R ® R is defined as f(x) = x4. Let x, y Î R such that f(x) = f(y) Þ x4 = y4 Þ x = ± y \ f(x) = f(y) does not imply that x = y. For example, f(1) = f(–1) = 1 \ f is not one-one.

15

Consider an element 2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = 2. \ f is not onto. Hence, function f is neither one-one nor onto. Q. 10. Let f : R ® R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type (1 mark each) Questions Q. 1. Check whether the function f : R ® R defined as



f(x) = x3 is one-one or not. Q. 2. A relation R in the set of real numbers R defined as R − {( a , b) : a = b} is a function or not. Justify R&U [CBSE SQP - 2021]

Sol. Since

a is not defined for a Î(−¥, 0)

\

1

a = b is not a function [CBSE SQP Marking Scheme 2021]

Q. 3. If A = {1, 2, 3}, B = {4, 5, 6, 7} and f = {(1, 4), (2, 5), (3, 6)} is a function from A to B. State whether f is one-one or not. R [CBSE SQP 2020-21]

Short Answer Type Questions-I (2 marks each) 2 Q. 1. Show that the function f in A = R –   defined 3 as f(x) =

4 x +3 is one-one. 6x - 4 

A

2 Q. 2. Show that the function f in A = R –   defined 3 as f(x) =

4 x +3 is onto. 6x - 4

Þ



x = ± y

Now, for y = –2 Î R, x = Ï R So, f is not onto. Hence, given function is neither one-one nor onto. Q. 4. Show that the function f : N ® N, given by f(x) = 2x is one-one but not onto. Sol. Given, a function f : N ® N, defined as f(x) = 2x For one-one Let, x1, x2 Î N, such that f(x1) = f(x2) Þ 2x1 = 2x2 Þ x1 = x2 So, f is one-one For onto Let y Î N (codomain) be any arbitrary element. Then, y = f(x) Þ y = 2x y Þ x = 2

Now, for y = 1, x =

1 Ï N. 2

Thus, y = 1 Î N (codomain) does not have a preimage in domain (N). So, f is not onto.

Short Answer Type Questions-II (3 marks each)

A

Q. 3. Show that the function f : R ® R defined as f(x) = x2 is neither one-one nor onto. Sol. Given, a function f : R ® R defined as f(x) = x2 For one-one Here, at x = 1, f(1) = 1 and at x = –1, f(–1) = (–1)2 = 1 Thus, f(1) = f(–1) = 1, but 1 ¹ –1 So, f is not one-one.

For onto Let y Î R (codomain) be any arbitrary element. Then, y =f(x) Þ y =x2 Þ x =x2

Q. 1. Show that the function f : R ® R defined by f(x) = x , ∀ x Î R is neither one-one nor onto. x2 + 1 Sol. Checking for one-one: æ 1ö æ 1ö here f(x) = f ç ÷ . For example f(2) = f ç ÷ è 2ø è xø

These questions are for practice and their solutions are available at the end of the chapter

16

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

\ f is not one-one Checking for onto: Let y = 1 Î R (co-domain) . Then

x = 1 x +1 2 Þ x – x + 1 = 0, which has no real roots. \ Rf ¹ co-domain Þ f is not onto. 1½ [CBSE SQP Marking Scheme 2020 (Modified)] y = f(x) Þ

Long Answer Type Questions (5 marks each)

1½

2

Q. 2. Show that the function f : N ® N, given by f(1) = f(2) = 1 and f(x) = x – 1 for every x > 2, is onto but not one-one. Sol. We have a function f : N ® N, defined as f(1) = f(2) = 1 and f(x) = x – 1, for every x > 2. For one-one Since f(1) = f(2) = 1, therefore 1 and have same image, namely 1. So, f is not one-one. For onto Note that y = 1 has two pre-images, namely 1 and 2. Now, let y Î N, y ¹ 1 be any arbitrary element. Then, y = f(x) Þ y = x – 1 Þ x = y + 1 > 2 for every y Î N, y ¹ 1. Thus, for every y ÎN, y ¹ 1, there exists x = y + 1 such that f(x) = f(y + 1) = y + 1 – 1 = y Hence, f is onto. Q. 3. Prove that the function f : N ® N, defined by f(x) = x2 + x + 1 is one-one but not onto. R [CBSE Delhi Set III-2019]

Sol. For one-one. Let x1 , x2 Î N. f ( x1 ) = f ( x 2 ) Þ x12 + x1 + 1 = x 22 + x 2 + 1 Þ ( x1 - x 2 )( x1 + x 2 + 1) = 0 Þ x1 = x 2 as x1 + x 2 + 1 ¹ 0 (∵ x1 , x 2 Î N )  1½ Þ f is one-one. For not onto. for y = 1 Î N, there is no x Î N for which f(x) = 1  1½  [CBSE Marking Scheme, 2019] Detailed Solution: Given, f(x) = x2 + x + 1 for x1, x2 Î N f(x1) = f(x2) x12 + x1 + 1 = x22 + x2 +1 x12 – x22 + x1 – x2 = 0 (x1 – x2) (x1 + x2) + (x1 – x2) = 0 (x1 – x2) (x1 + x2 + 1) = 0 Therefore, the given function is one-one. Also, f is not onto as for 1 Î N, there does exist any 'x' in f(x) = 1.

Q. 1. Check which of the following function is onto or into. (i) f : A ® B, given by f(x) = 3x, where A = {0, 1, 2} and B = {0, 3, 6}. (ii) f : Z ® Z, given by f(x) = 3x + 2, where Z = set of integers. Sol. (i) We have a function f : A ® B, given by f(x) = 3x, where A = {0, 1, 2} and B = {0, 3, 6} Let y Î B be any arbitrary element. y Then, y = f(x) Þ y = 3x Þ x − 3 Now,

at y = 0, x =

0 = 0 ∈A 3



At y = 3, x =

3 = 1 ∈A 3



At y = 6, x =

6 = 2 ∈A 3

Thus, for each element y of B, there is a pre-image in A. (ii) We have a function f : Z ® Z, given by f(x) = 3x + 2. Let y Î Z, (codomain of f) be any arbitrary element. Q. 2. Let R be the set of all non-zero real number. Then, 1 show that f : R ® R, given by f(x) = is one-one x and onto. 1 Sol. Given, f ( x ) = x For one-one Let x1, x2 Î R, such that f(x1) =f(x2) 1 1 1  = Þ put x1 and x 2 in f ( x ) = x  x1 x2   Þ x1 = x2 So, f is one-one For onto Let y Î R be any arbitrary element. Then, y = f(x) 1 Þ y = x Þ

x =

1 [expressing x in terms of y] y

It is clear that for every y Î R(codomain), x Î R(domain) Thus, for each y Î R(codomain), there exist 1 x = ∈ R (domain), such that f ( x ) = y

 1 1 f  = =y  y  1  y 

[i.e., every element of codomain has pre-image in domain] So, f is onto.

These questions are for practice and their solutions are available at the end of the chapter



RELATIONS AND FUNCTIONS

17

COMPETENCY BASED QUESTIONS Case based MCQs

Attempt any four sub-parts from each question.

Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same: A general election of Lok Sabha is a gigantic exercise. About 911 million people were eligible to vote and voter turnout was about 67%, the highest ever Let I be the set of all citizens of India who were eligible to exercise their voting right in general election held in 2019. A relation ‘R’ is defined on I as follows:

Q. 4. The above defined relation R is __________ (A) Symmetric and transitive but not reflexive (B) Universal relation (C) Equivalence relation (D) Reflexive but not symmetric and transitive Ans. Option (C) is correct. Explanation: R is reflexive, since every person is friend or itself. i.e., (F1, F2) Î R Further, (F1, F2) Î R Þ F1 is friend of F2 Þ F2 is friend of F1 Þ (F2, F1) Î R Þ R is symmetric

ONE – NATION ONE – ELECTION FESTIVAL OF DEMOCRACY GENERAL ELECTION - 2019

Þ F1 is friend of F2 and F2 is friend of F3. Þ F1 is a friend of F3. Þ (F1, F3) Î R Therefore, R is an equivalence relation. Q. 5. Mr. Shyam exercised his voting right in General Election - 2019, then Mr. Shyam is related to which of the following? (A) All those eligible voters who cast their votes (B) Family members of Mr. Shyam (C) All citizens of India (D) Eligible voters of India Ans. Option (A) is correct. II. Read the following text and answer the following questions on the basis of the same:





R = {(V1, V2) : V1, V2 Î I and both use their voting right in general election – 2019} [CBSE QB 2021] Q. 1. Two neighbours X and Y Î I. X exercised his voting right while Y did not cast her vote in general election - 2019. Which of the following is true? (A) (X, Y) Î R (B) (Y, X) Î R (C) (X, X) Ï R (D) (X, Y) Ï R Ans. Option (D) is correct. Explanation: (X, Y) Ï R.  X exercised his voting right while, Y did not cast her vote in general election-2019 And R = {(V1, V2) : V1 V2 Î I and both use their voting right in general election-2019} Q. 2. Mr. ’X’ and his wife ‘W’ both exercised their voting right in general election -2019, Which of the following is true? (A) both (X, W) and (W, X) Î R (B) (X, W) Î R but (W, X) Ï R (C) both (X, W) and (W, X) Ï R (D) (W, X) Î R but (X, W) Ï R Ans. Option (A) is correct. Q. 3. Three friends F1, F2 and F3 exercised their voting right in general election-2019, then which of the following is true? (A) (F1, F2) Î R, (F2, F3) Î R and (F1, F3) Î R (B) (F1, F2) Î R, (F2, F3) Î R and (F1, F3) Ï R (C) (F1, F2) Î R, (F2, F2) Î R but (F3, F3) Ï R (D) (F1, F2) Ï R, (F2, F3) Ï R and (F1, F3) Ï R Ans. Option (A) is correct.

Moreover, (F1, F2), (F2, F3) Î R

Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible outcomes of the throw every time belongs to set {1, 2, 3, 4, 5, 6}. Let A be the set of players while B be the set of all possible outcomes.

A = {S, D}, B = {1, 2, 3, 4, 5, 6} [CBSE QB 2021]

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Q. 1. Let R : B ® B be defined by R = {(x, y) : y is divisible by x} is (A) Reflexive and transitive but not symmetric (B) Reflexive and symmetric but not transitive (C) Not reflexive but symmetric and transitive (D) Equivalence Ans. Option (A) is correct. Explanation: R is reflexive, since every element of B i.e., B = {1, 2, 3, 4, 5, 6} is divisible by itself. i.e., (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) Î R further, (1, 2) Î R but (2, 1) Î R Moreover, (1, 2), (2, 4) Î R Þ (1, 4) Î R Þ R is transitive. Therefore, R is reflexive and transitive but not symmetric. Q. 2. Raji wants to know the number of functions from A to B. How many number of functions are possible? (A) 62 (B) 26 (C) 6! (D) 212 Ans. Option (A) is correct. Q. 3. Let R be a relation on B defined by R = {(1, 2), (2, 2), (1, 3), (3, 4), (3, 1), (4, 3), (5, 5)}. Then R is (A) Symmetric (B) Reflexive (C) Transitive (D) None of these Ans. Option (D) is correct. Explanation: R = {(1, 2), (2, 2), (1, 3), (3, 4), (3, 1), (4, 3), (5, 5)} R is not reflexive. (3, 3) Ï 4 Since, (1, 1), (3, 3), (4, 4), (6, 6) Î R R is not symmetric. Because, for (1, 2) Î R there (2, 1) Ï R. R is not transitive. Because for all element of B there does not exist, (a, b) (b, c) Î R and (a, c) Î R. Q. 4. Raji wants to know the number of relations possible from A to B. How many numbers of relations are possible? (A) 62 (B) 26 (C) 6! (D) 212 Ans. Option (D) is correct. Q. 5. Let R : B ® B be defined by R = {(1, 1), (1, 2), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, then R is (A) Symmetric (B) Reflexive and Transitive (C) Transitive and symmetric (D) Equivalence Ans. Option (B) is correct. III. Read the following text and answer the following questions on the basis of the same: An organization conducted bike race under 2 different categories–boys and girls. Totally there



were 250 participants. Among all of them finally three from Category 1 and two from Category 2 were selected for the final race. Ravi forms two sets B and G with these participants for his college project. Let B = {b1, b2, b3} G = {g1, g2} where B represents the set of boys selected and G the set of girls who were selected for the final race. [CBSE QB 2021]

Ravi decides to explore these sets for various types of relations and functions Q. 1. Ravi wishes to form all the relations possible from B to G. How many such relations are possible? (A) 26 (B) 25 (C) 0 (D) 23 Ans. Option (A) is correct. Q. 2. Let R : B ® B be defined by R = {(x, y) : x and y are students of same sex}, Then this relation R is_______ (A) Equivalence (B) Reflexive only (C) Reflexive and symmetric but not transitive (D) Reflexive and transitive but not symmetric Ans. Option (A) is correct. Explanation: R : B ® B be defined by R = {(x, y) : x and y are students of same sex} R is reflexive, since, (x, x) Î R R is symmetric, since, (x, y) Î R and (y, x) Î R R is transitive. For a, b, c Î B $ (a, b) (b, c) Î R and (a, c) Î R. Therefore R is equivalence relation. Q. 3. Ravi wants to know among those relations, how many functions can be formed from B to G? (A) 22 (B) 212 2 (C) 3 (D) 23 

18

Ans. Option (D) is correct. Q. 4. Let R : B ® G be defined by R = {(b1, g1), (b2, g2),

(b3, g1)}, then R is__________

(A) Injective (B) Surjective (C) Neither Surjective nor Injective (D) Surjective and Injective Ans. Option (B) is correct.



RELATIONS AND FUNCTIONS

Explanation: R : B ® G be defined by R = {(b1, g1), (b2, g2), (b3, g1)} R is surjective, since, every element of G is the image of some element of B under R, i.e., For g1, g2 Î G, there exists an elements b1, b2, b3 Î B, (b1 g1) (b2, g2), (b3, g1) Î R. Q. 5. Ravi wants to find the number of injective functions from B to G. How many numbers of injective functions are possible? (A) 0 (B) 2! (C) 3! (D) 0! Ans. Option (A) is correct. IV. Read the following text and answer the following questions on the basis of the same: Students of Grade 9, planned to plant saplings along straight lines, parallel to each other to one side of the playground ensuring that they had enough play area. Let us assume that they planted one of the rows of the saplings along the line y = x − 4. Let L be the set of all lines which are parallel on the ground and R be a relation on L. [CBSE QB 2021]

19

Q. 2. Let R = {(L1, L2) : L1 ^ L2 where L1, L2 Î L} which of the following is true? (A) R is Symmetric but neither reflexive nor transitive (B)  R is Reflexive and transitive but not symmetric (C)  R is Reflexive but neither symmetric nor transitive (D) R is an Equivalence relation Ans. Option (A) is correct. Explanation: R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1, L1) Ï R. R is symmetric as (L1, L2) Î R As, L1 is perpendicular to L2 and L2 is perpendicular to L1 (L2, L1) Î R R is not transitive. Indeed, it L1 is perpendicular to L2 and L2 is perpendicular to L3, then L1 can never be perpendicular to L3. In fact L1 is parallel to L3, i.e., (L1, L2) Î R, (L2, L3) Î R but (L1, L3) Ï R i.e., symmetric but neither reflexive nor transitive. Q. 3. The function f : R ® R defined by f(x) = x − 4 is___________ (A) Bijective (B) Surjective but not injective (C) Injective but not Surjective (D) Neither Surjective nor Injective



Q. 1. Let relation R be defined by R = {(L1, L2) : L1 || L2 where L1, L2 Î L} then R is______ relation (A) Equivalence (B) Only reflexive (C) Not reflexive (D) Symmetric but not transitive Ans. Option (A) is correct. Explanation: Let relation R be defined by R = {(L1, L2) : L1 || L2 where L1, L2 Î L}. R is reflexive, since every line is parallel to itself. Further, (L2, L1) Î R Þ L1 is parallel to L2 Þ L2 is parallel to L1 Þ (L2, L1) Î R Hence, R is symmetric. Moreover, (L1, L2), (L2, L3) Î R Þ L1 is parallel to L2 and L2 is parallel to L3 Þ L1 is parallel to L3 Þ (L1, L3) Î R Therefore, R is an equivalence relation

Ans. Option (A) is correct. Explanation: The function f is one-one, for f(x1) = f(x2) Þ x1 – 4 = x2 – 4 Þ x1 = x2 Also, given any real number y in R, there exists y + 4 in R Such that f(y + 4) = y + 4 – 4 = y Hence, f is onto Hence, function is both one-one and onto, i.e., bijective. Q. 4. Let f : R ® R be defined by f(x) = x − 4. Then the range of f(x) is ________ (A) R

(B) Z

(C) W

(D) Q

Ans. Option (A) is correct. Explanation: Range of f(x) is R Q. 5. Let R = {(L1, L2) : L1 || L2 and L1 : y = x – 4} then which of the following can be taken as L2? (A) 2x – 2y + 5 = 0 (B) 2x + y = 5 (C) 2x + 2y + 7 = 0 (D) x + y = 7 Ans. Option (A) is correct. Explanation: Since, L1 || L2 then slope of both the lines should be same. Slope of L1 = 1 Þ Slope of L2 = 1 And 2x – 2y + 5 = 0

20

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

–2y = –2x – 5 5 y = x + 2 Slope of 2x – 2y + 5 = 0 is 1 So, 2x – 2y + 5 = 0 can be taken as L2.

Case based Subjective Questions (4 mark each) (Each Sub-part carries 2 marks) I. Read the following text and answer the following questions on the basis of the same: Rohan is confused in the Mathematics topic 'Relation and equivalence relation'. To clear his concepts on the topic, he took help his elder brother. He has following notes on this topic. Relation : A relation R from a set A to a set B is a subset of the cartesian product A × B obtained by describing a relationship between first element x and the second element 'y' of the ordered pairs in A × B. A relation R in a set A is called. : Reflexive : If (a, a) Î R " a Î A. Symmetric : If (a1, a1) Î R Þ (a2, a1) Î R " a1, a2 Î R. Transitive : If (a1, a1) Î R and (a2, a3) Î R Þ (a1, a3) Î R " a1, a2, a3 Î A Equivalence Relation : A relation R in a set A is an equivalence relation if R is reflexive, symmetric and transitive.

Q. 1. Show that relation defined by R1 = {(x, y) | x2 = y2} x, y Î R is an equivalence relation. Sol. Given relation R1 = {(x, y) | x2 = y2} Reflexive : For all x Î R, x2 = x2, so, (x, x) Î R1 Hence, R1 is reflexive relation. Symmetric : For all x, y Î R If x2 = y2 then y2 = x2 Hence, R1 is symmetric relation. 1 Transitive : For all x, y Î R, x2 = y2 and for all y, z Î R y2 = z2 \ x2 = y2 = z2, for all x, y, z Î R Hence, R1 is transitive. Thus, R1 is an equivalence relation. 1 Q. 2. Check whether the relation (R) 'x greater than y' for all x, y Î N is reflexive, symmetric or transitive. Sol. Given, x greater than y, " x, y Î N Þ x > y " x, y Î N Reflexive : Now, for (x, x) Î R Therefore, x > x is not true for any x Î N Thus, R is not reflexive. Symmetric : Now, let (x, y) Î R, then x > y If x > y, then y ≤ x for any x, y Î N

Thus, R is not symmetric. Transitive : Now, let (x, y) Î R and (y, z) Î R Þ x > y and y > z Therefore, x => (x, z) Î R for all x, y, z Î N Thus, R is transitive.

1

Solutions for Practice Questions (Topic-1) 6. Option (C) is correct. Explanation: Consider that aRb, if a is congruent to b, " a, b Î T. Then, aRa Þ a @ a, Which is true for all a Î T So, R is reflexive, ...(i) Let aRb Þ a @ b Þb@a Þ bRa So, R is symmetric. ...(ii) Let aRb and bRc Þ b @ b and b @ a Þ a @ c Þ aRc So, R is transitive ...(iii) Hence, R is equivalence relation. 7. Option (B) is correct. Explanation: aRb Þ a is brother of b. This does not mean b is also a brother of a as b can be a sister of a. Hence, R is not symmetric. aRb Þ a is brother of b and bRc Þ b is a brother of c. So, a is brother of c. Hence, R is transitive.

Very Short Answer Type Questions [(1, 3)] = {(x, y) Î A × A : x + 3 = y + 1} = {(x, y) Î A × A : y – x = 2} = {(1, 3), (2, 4)} 1 [CBSE Marking Scheme 2017-18]

3.

Short Answer Type Questions-I 2. (i) 1, 2 Î R such that 1 < 2 Þ (1, 2) Î R, but since 2 is not less than 1 Þ (2, 1) Ï R. Hence R is not symmetric. 1 (ii) Let (a, b) Î R and (b, c) Î R, \ a < b and b < c Þ a < c Þ (a, c) R \ R is transitive. 1 [CBSE SQP Marking Scheme 2020] Detailed Answer: (i) It is not symmetric because if a < b then b < a is not true. (ii) Here, if a < b and b < c then a < c is also true for all a, b, c Î Real numbers. Therefore R is transitive.

Commonly Made Error





Multiple Choice Questions

Students use examples to show that the relation is transitive which is wrong.



RELATIONS AND FUNCTIONS

Commonly Made Error



Use only arbitrary transitivity.

elements

to

prove





Answering Tip



5. Let (a, b), (b, c) Î R, f(a) = f(b), f(b) = f(c) Þ f(a) = f(c), (a, c) Î R. Thus, Relation is transitive. 5.

Reflexive: R is reflexive, as 1 + a.a = 1 + a2 > 0 Þ (a, a) Î R ∀ a Î R 1 Symmetric: If (a, b) Î R then, 1 + ab > 0 Þ 1 + ba > 0 Þ (b, a) Î R Hence, R is symmetric. 1 Transitive: 1 Let a = – 8, b = –1, c = 2



Since, 1 + ab = 1 + (– 8)(– 1) = 9 > 0 \ (a, b) Î R 1 1 also, 1 + bc = 1 + ( −1)   = > 0 2 2

\ (b, c) Î R

1 1 + ac = 1 + ( −8 )   = − 3 < 0 2

But,

Hence, R is not transitive. 1 [CBSE Marking Scheme 2018 (modified)]

Students use counter example to prove reflexive and symmetric.

Answering Tip





Short Answer Type Questions-II



21

Counter examples can be used only to show exceptions.

Long Answer Type Questions 2. Given R = {(a, b) : |a – b| is divisible by 2} and

A = {1, 2, 3, 4, 5}



R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)

(1, 3), (1, 5), (2, 4), (3, 5), (3, 1), (5, 1), (4, 2), (5, 3)} 2 (i) " a Î A, (a, a) Î R, \ R is reflexive. 1 [As {(1, 1), (2, 2), (3, 3), (4, 4) (5, 5)} Î R] (ii) " (a, b) Î A, (b, a) Î R, \ R is symmetric. 1 [As {(1, 3), (1, 5), (2, 4), (3, 5) (3, 1), (5, 1), (4, 2), (5, 3)} Î R] (iii) " (a, b), (b, c) Î R, (a, c) Î R \ R is transitive. 1 [As {(1, 3), (3, 1) Î R Þ (1, 1) Î R and similarly others] \ R is an equivalence relation. [CBSE Marking Scheme 2015 (Modified)]

Solutions for Practice Questions (Topic-2) Multiple Choice Questions 7. Option (D) is correct. 1 Explanation: We have, f(x) = , " x Î R x For x = 0, f(x) is not defined.

Very Short Answer Type Questions 1. Let f(x1) = f(x2) for some x1, x2 Î R 1 Þ (x1)3 = (x2)3 Þ x1 = x2 Hence f(x) is one-one.

Hence, f(x) is a not defined function.

[CBSE SQP Marking Scheme 2021]

10. Option (A) is correct. Explanation: f : R ® R is defined as f(x) = 3x.

Commonly Made Error

Þ

3x = 3y

Þ

x = y





Let x, y Î R such that f(x) = f(y)

\ f is one-one.

\ f is onto. Hence, function f is one-one and onto.

Answering Tip





Also, for any real number (y) in co-domain R, there y y y exists in R such that f = 3 = y. 3 3 3

Students get confused between one-one and many-one functions.

Injectivity should be determined considering the domain and co-domain. A function which is one-one in a domain may not be one-one in another domain.

22

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

3. Given, A = {1, 2, 3}, B = {4, 5, 6, 7} and f : A ® B is defined as f = {(1, 4), (2, 5), (3, 6)} i.e. f(1) = 4, f(2) = 5 and f(3) = 6.

It can be seen that the images of distinct elements of A under f are distinct. So, f is one-one. 1

Short Answer Type Questions-I 1. Given

Let,

f(x) =

4x + 3 6x − 4 f(x1) = f(x2),

4 x1 + 3 4 x2 + 3 then = ½ x − x2 − 4 6 4 6 1 or (4x1 + 3)(6x2 – 4) = (6x1 – 4)(4x2 + 3) ½ or 24x1x2 – 16x1 + 18x2 – 12 = 24x1x2 + 18x1 – 16x2 – 12 or – 16x1 + 18x2 = 18x1 – 16x2 ½ or –16x1 – 18x1 = – 18x2 – 16x2 or – 34x1 = – 34x2 or x1 = x2 or f is one-one. ½

2.

Let,



\



or

or or or or

y = f(x) 4x + 3 y = 6x − 4 y(6x – 4) = 4x + 3 6xy – 4y = 4x+3 6xy – 4x = 4y + 3 x(6y – 4) = 4y + 3

½

{}

2 4y + 3 ÎB=R– 1 3 6y − 4 2 or For every value of y except y = , there is a 3

or

2 y Î B = R - ìí üý î3 þ

x =

{}

pre-image x =

4y + 3 = g(y). 6y − 4

or x Î A \ f is onto.

½

REFLECTIONS •

In this chapter we have covered the different types of relations and functions. Look around you and pick some real life relations say ‘is the father of ’,

‘is the friend of ’ etc and check whether they are reflexive, symmetric and transitive.



INVERSE TRIGONOMETRIC FUNCTIONS

CHAPTER

2



Syllabus

Definition, range, domain, principal value branch, Graphs of inverse trigonometric functions.

In this chapter you will study

 Domain

of inverse trigonometric functions. of inverse trigonometric functions.  Principal branch value of inverse trigonometric functions.  Graph of different inverse trigonometric functions.  Range

Revision Notes As we have learnt in class XI, the domain and range of trigonometric functions are given below: S. No.

Function

Domain

Range

(i)

sine

R

[– 1, 1]

(ii)

cosine

R

[– 1, 1]

(iii)

tangent

(iv)

cosecant

(v)

secant

(vi)

cotangent

{

}

π R − x : x = ( 2n + 1) ; n ∈ Z 2

R – (– 1, 1)

R – {x : x = np, n Î Z}

{

}

π R − x : x = ( 2n + 1) ; n ∈ Z 2 R – {x : x = np, n Î Z}

R

R – (– 1, 1) R

24

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII



INVERSE TRIGONOMETRIC FUNCTIONS

25

1. Inverse function  We know that if function f : X ® Y such that y = f(x) is one-one and onto, then we define another function g : Y ® X such that x = g(y), where x Î X and y Î Y, which is also one-one and onto.

Key Words

Scan to know more about this topic

One-one function: One to one function or one to one mapping states that each element of one set, say set A is mapped with a unique element of another set, say set B, where A and B are two different sets. In terms of function, it states as if f(x) = f(y) Þ x = y, then f is one to one.

Introduction of Inverse Trigonometric Functions



In such a case, Domain of g = Range of f and   Range of g = Domain of f g is called the inverse of f g = f –1 or Inverse of g = g –1 = (f –1)–1 = f The graph of sine function is shown here:



Onto function: If A and B are two sets, if for every element of B, there is atleast one or more element matching with set A, it is called onto function.



Principal value branch of function cos–1: The graph of the function cos–1 is as shown in figure. Domain of the function cos–1 is [–1, 1]. Its range in one of the intervals (– p, 0), (0, p), (p, 2p), etc. is one-one and onto with the range [– 1, 1]. The branch with range (0, p) is called the principal value branch of the function cos–1. Thus, cos–1 : [– 1, 1] ® [0, p]



Scan to know more about this topic

Principal value of branch function sin–1: It is a function with domain [– 1, 1] and range  −3 π − π   − π π   2 , 2  ,  2 , 2 

 π 3π  or  ,  and so on 2 2 

corresponding to each interval, we get a branch of the function sin–

1

Scan to know more about this topic

x. The branch with range

 −π π   2 , 2  is called the principal value branch. Thus, sin–1 : [–1, 1]  −π π  ®  , .  2 2

Inverse Trigonometric Functions

Inverse Trigonometric Trick

26

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

cosec–1 is defined on a function whose domain is R – (– 1, 1) and the range is any one of the interval,



 −3 π − π   −π π   π 3π   2 , 2  − {π},  2 , 2  − {0},  2 , 2  − {π},...

Principal value branch function tan–1: The function tan–1 is defined whose domain is set of real numbers and range is one of the intervals,  −3 π π   − π π   π 3 π  , , ,  ,  ,  ,.....  2 2  2 2 2 2  Graph of the function is as shown in the figure:





The function corresponding to the range  −π π   2 , 2  − {0} is called the principal value branch of cosec–1.

 π π Thus, cosec −1 : R − ( −1, 1) →  − ,  − {0} .  2 2



 π π The branch with range  − ,  is called the  2 2

Principal value branch of function sec–1: The graph of function sec–1 is shown in figure. The sec–1 is defined as a function whose domain R – (– 1, 1) and π  3π   −π  range is [– p, 0] −   , [0 , π] −   , [ π, 2 π] −   , 2 2     2

principal value branch of function tan–1. Thus,  −π π  tan −1 : R →  ,  2 2  .  Principal value branch of function cosec–1 : The graph of function cosec–1 is shown in the figure. The

etc. Function corresponding to range [0 , π] −

{}

known as the principal value branch of sec–1. π Thus, sec −1 : R − ( −1, 1) → [0 , π] −   2

π 2

is



INVERSE TRIGONOMETRIC FUNCTIONS

27



The principal value branch of function cot–1: The graph of function cot–1 is shown below: Y



2 1 X'

–π

π 0 2 –1

π 2

π

3π 2 2π

X

–2

y=cot x Y'



The cot–1 function is defined on function whose domain is R and the range is any of the intervals, (– p, 0), (0, p), (p, 2p), ....

The function corresponding to (0, p) is called the principal value branch of the function cot–1.



Then, cot–1 : R ® (0, p)



The principal value branch of trigonometric inverse functions is as follows: Inverse Function

Domain

Principal Value Branch

sin–1

[ – 1, 1]

cos–1

[– 1, 1]

 −π π   2 , 2  [0, p]

cosec–1

R – (– 1, 1)

 −π π   2 , 2  − {0}

sec–1

R – (– 1, 1)

tan–1

R

 −π π  ,   2 2

cot–1

R

(0, p)

[0 , π] −

{} π 2

28

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Key Facts

a−x or a+x

• Inverse trigonometric functions were considered early in the 1700s by Daniel Bernoulli, who use 'A.sin' for the inverse sine of a number, and in 1736. Euler wrote "At" for the inverse tangent. • Inverse trigonometric functions one used to find the elevation of sun to the ground. The angle of tilt of the building can be found using inverse trigonometric functions. • Inverse trigonometric functions help in identifying the angles of bridges to build scale models. • Inverse trigonometric functions are often called 'arc functions', since given a value of a trigonometric function, they produce the length of arc needed to obtain that value. (3)







Principal Value: Numerically smallest angle is known as the principal value. Finding the principal value: For finding the principal value, following algorithm can followed : STEP 1: First draw a trigonometric circle and mark the quadrant in which the angle may lie. STEP 2: Select anti-clockwise direction for 1st and 2nd quadrants and clockwise direction for 3rd and 4th quadrants. STEP 3: Find the angles in the first rotation. STEP 4: Select the numerically least (magnitude wise) angle among these two values. The angle thus found will be the principal value. STEP 5: In case, two angles one with positive sign and the other with the negative sign qualify for the numerically least angle then, it is the convention to select the angle with positive sign as principal value. The principal value is never numerically greater than p.

(4)

To simplify inverse trigonometric expressions,



following substitutions can be considered: Expression

Substitution

a2 + x2 or

a2 + x 2

x = a tan q or x = a cot q

a2 – x2 or

a2 − x 2

x = a sin q or x = a cos q

x2 – a2 or

x 2 − a2

x = a sec q or x = a cosec q

a2 − x 2 or a2 + x 2

a+x a−x

x = a cos 2q

a2 + x 2 a2 − x 2

x2 = a2 cos 2q

x or a−x

a−x x

x = a sin2 q or x = a cos2 q

x or a+x

a+x x

x = a tan2 q or x = a cot2 q

Note the following and keep them in mind:  The symbol sin–1 x is used to denote the smallest angle whether positive or negative, such that the sine of this angle will give us x. Similarly cos–1 x, tan–1 x, cosec–1 x, sec–1 x and cot–1 x are defined.  You should note that sin–1 x can be written as arcsin x. Similarly, other Inverse Trigonometric Functions can also be written as arccos x, arctan x, arcsec x etc.  Also note that sin–1 x (and similarly other Inverse Trigonometric Functions) is entirely different from (sin x)–1. In fact, sin–1 x is the measure of an angle in Radians whose sine is x whereas (sin x)–1 is

1 sin x

(which is obvious as per the laws of exponents).  Keep

in mind that these inverse trigonometric relations are true only in their domains i.e., they are valid only for some values of ‘x’ for which inverse trigonometric functions are well defined.

Mnemonics Inverse trigonometric ratio can be used to find the angle of a right triangle when given two sides of the triangle. opposite   sin1 SOH hypotenuse CAH

  cos 1

adjacent hypotenuse

TOA

  tan1

opposite adjacent



INVERSE TRIGONOMETRIC FUNCTIONS

Key Formulae TRIGONOMETRIC FORMULAE (ONLY FOR REFERENCE):  Relation between trigonometric ratios: tan θ = (a)

sin θ cos θ

(d) cosec θ =



1 sin θ

1 cot θ

(e) sec θ =

1 cos θ

(c) cot θ =

cos θ sin θ

Trigonometric Identities:

(a) sin2 q + cos2 q = 1 

(b) tan θ =

(b) sec2 q = 1 + tan2 q

(c) cosec2 q = 1 + cot2 q

Addition/subtraction/ formulae & some related results: (b) cos (A ± B) = cos A cos B ∓ sin A sin B

(a) sin (A ± B) = sin A cos B ± cos A sin B

(c) cos (A + B) cos (A – B) = cos A – sin B = cos B – sin2 A 2



2

2

(d) sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A (f) cot (A ± B) =

cot B cot A ∓ 1 cot B ± cot A

(a) sin 2A = 2 sin A cos A

(b) sin A = 2 sin

A A cos 2 2

(c) cos 2A = cos2A – sin2A

(d) cos A = cos2

A A – sin2 2 2

(e) cos 2A = 2 cos2A – 1 (g) cos 2A = 1 – 2sin2A

(f) 2 cos2 A = 1 + cos 2A (h) 2 sin2 A = 1 – cos 2A

(e) tan (A ± B) = 

tan A ± tan B 1 ∓ tan A tan B

Multiple angle formulae involving A, 2 A & 3 A:

1 − tan 2 A 1 + tan 2 A

(i) sin 2A =

2 tan A 1 + tan 2 A

(j) cos 2A =

(k) tan 2A =

2 tan A 1 − tan 2 A

(l) sin 3A = 3 sin A – 4 sin3 A



(m) cos 3A = 4 cos3 A – 3 cos A

(n) tan 3A =

3 tan A − tan 3 A 1 − 3 tan 2 A



Transformation of sums/differences into products & vice-versa: C+D C+D C−D C−D sin C + sin D = 2 sin cos (b) sin C − sin D = 2 cos sin (a) 2 2 2 2 C+D C−D cos 2 2 (e) 2 sin A cos B = sin (A + B) + sin (A – B) (g) 2 cos A cos B = cos (A + B) + cos (A – B) (c) cos C + cos D = 2 cos



C+D C−D sin 2 2 (f) 2 cos A sin B = sin (A + B) – sin (A – B) (h) 2 sin A sin B = cos (A – B) – cos (A + B) (d) cos C − cos D = −2 sin

Relations in different measures of Angle:

(a) Angle in Radian Measure = (Angle in degree measure) ×

π rad 180°



(b) Angle in Degree Measure = (Angle in radian measure) ×

180° π



(c) q (in radian measure) =



Also following are of importance as well:



l arc = r radius

(a) 1 right angle = 90° (c) 1° =

π = 0·01745 radians (Approx.) 180°

(b) 1° = 60’, 1’ = 60” (d) 1 radian = 57°17’45” or 206265 seconds.

29

30

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII



General Solutions: (a) sin x = sin y Or, x = np + (–1)n y, where n Î Z. (b) cos x = cos y Or, x = 2np ± y, where n Î Z. (c) tan x = tan y Or, x = np + y, where n Î Z. 

Relation in Degree & Radian Measures:



Angles in Degree

0°

30°

45°

60°

90°

180°

270°

360°

Angles in Radian

0°

π   6

π   4

π   3

π   2

(p)

 3π    2

(2p)

Trigonometric Ratio of Standard Angles: Degree



0°

30°

45°

60°

90°

3 2

1

1 2

0

sin x

0

1 2

1 2

cos x

1

3 2

1 2

tan x

0

1 3

1

3

¥

cot x

¥

3

1

1 3

0

cosec x

¥

2

2

2 3

1

sec x

1

2 3

2

2

¥

Trigonometric Ratios of Allied Angles: Angles (®)

p −q 2

p +q 2

p–q

p+q

3p −q 2

3p +q 2

2p – q or – q

2p + q

sin

cos q

cos q

sin q

– sin q

– cos q

– cos q

– sin q

sin q

cos

sin q

– sin q

– cos q

– cos q

– sin q

sin q

cos q

cos q

tan

cot q

– cot q

– tan q

tan q

cot q

– cot q

– tan q

tan q

cot

tan q

– tan q

– cot q

cot q

tan q

– tan q

– cot q

cot q

sec

cosec q

– cosec q

– sec q

– sec q

– cosec q

cosec q

sec q

sec q

cosec

sec q

sec q

cosec q

– cosec q

– sec q

– sec q

– cosec q

cosec q

T – Ratios (¯)

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions

 1  1 Q. 1. The principal value of cos1    sin 1    is 2 2  (A)

π 12

(B) p

(C)

π 3

(D)



π [CBSE Board 2021] 6

Ans. Option (A) is correct.  1  1 Explanation: cos1    sin 1    2 2    1   –1 –1 = cos1  cos   sin 1   [sin (–q) = – sin q] 3   2 π     =  sin 1  sin  =  = 12 3 4 3 4  



INVERSE TRIGONOMETRIC FUNCTIONS

9  1  Q. 2. The principal value of tan  tan  is: 8   (A)

π 8

 (C)  8

(B)

3π 8

(D) 



3 8

   4 2   (C)  cos1 x 4 2 (A)

Ans. Option (A) is correct.     9 Explanation: tan 1  tan  = tan 1 tan      8  8       1  = tan  tan   8 8         ,  Q    8  2 2  Q. 3. What is the domain of the function cos–1 (2x – 3)? (A) [–1, 1] (B) (1, 2) (C) (–1, 1) (D) [1, 2] [CBSE Board 2021] Ans. Option (D) is correct. Explanation: Let, f(x) = cos–1 (2x – 3) –1 £ 2x – 3 £ 1 Þ 2 £ 2x £ 4 Þ 1 £x£2 \ x Î [1, 2] or domain of x is [1, 2]. −1 3 − cot −1 ( − 3 )] is: Q. 4. The principal value of [tan

(A) p

 (B)  2

(C) 0

(D) 2 3 [CBSE Board 2021]

Ans. Option (B) is correct. Explanation: We have, tan −1 3 − cot −1 ( − 3 )



π −1  −1 = tan  tan  − π + cot (cot 3 )  3

       =  2 3  6

[cot–1(–q) = p – cot–1q]

 1 x  1 x  Q. 5. The simplest form of tan 1   is:  1  x  1  x 

[CBSE Board 2021]



=

31

(B)

   4 2

(D)

   cos1 x 4 2 [CBSE Board 2021]

Ans. Option (C) is correct.  1  1  x  1  x  Explanation: We have, tan  1  x  1  x   Put x = cos2q, so that q =



1 cos–1x 2

 1  cos 2  1  cos 2  tan 1   1  cos 2  1  cos 2   

 2 cos2   2 sin 2   1  = tan   2 cos2   2 sin 2     1  cos   sin   = tan    cos   sin   1 − tan θ = tan −1 1 + tan θ

Divide by cos q

 π  = tan −1 tan  − θ    4 

=

 1  cos1 x 4 2

13  1  Q. 6. The principal value of cos  cos is: 6   (A)

13π 6

(B)

π 2

(C)

π 3

(D)

π 6



[CBSE Board 2020] Ans. Option (D) is correct. Explanation:

Topper Answer, 2020 Sol.

3   Q. 7. The value of sin–1  cos  is: 5   (A)

p 10

(B)

(C) 3p 5

p 10

Ans. Option (C) is correct.

-3p 5 [CBSE OD Set-I 2020]

(D)

32

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Explanation:

pö -1 æ = - sin ç sin ÷ è 10 ø

æ 3p ö ù -1 é = sin êcos ç ÷ ú è 5 øû ë

[ sin–1 (–x) = –sin–1 x]

æ p p öù -1 é = sin êcos ç + ÷ ú è 2 10 ø û ë pö -1 æ = sin ç - sin ÷ è 10 ø

= -

p 10 é æ -p p ö ù -1 êQsin (sin x ) = x , x Î çè 2 , 2 ÷ø ú ë û

ù é æp ö êQcos çè 2 + x÷ø = - sin x ú ë û

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type (1 mark each) Questions é æ 17 p ö ù . Q. 1. Find the value of sin -1 ê sin ç è 8 ÷ø úû ë

Detailed Answer: tan 1 3  cot 1 (  3 )



  1 3) = tan  tan   (   cot 3 

R&U [CBSE Delhi SET-I, 2020]

pö ù æ 17 p ö ù æ -1 é Sol. sin êsin ç ÷ ú = - sin êsin çè 2 p + 8 ÷ø ú ½ ë è 8 øû ë û -1 é

p = - 8

½

=

    cot 1 3

=

4  7   3 6 6

Short Answer Type (2 marks each) Questions-I

[CBSE Marking Scheme 2020]





Commonly Made Error Most of the students write the answer as

-17 p . 8

é -p p ù sin -1 (sin q) = q, iff q Î ê , ú ë 2 2û



p æ cos x ö -3p 0   2

−1

Answering Tip





π π x π  π x  As, − > − > −  1 = − 2 4 4 2 4 2 [CBSE Marking Scheme 2015 (Modified)]

5π 2 , find x. 8



R&U [Delhi I, II, III 2015]



[Foreign set III 2017]

2



2 π  or (tan–1x)2 +  − tan −1 x  = 5 π 2  8



or 2 (tan–1x)2 – π tan −1 x +

π2 5π2 − =0 4 8



or 2(tan–1x)2 – π tan −1 x −

3π2 =0 8



or



or



\

R&U [NCERT Exemplar]

 6x - 8x Q. 2. Prove that tan -1   1 - 12 x 2

3

 4x  -1   - tan  1 - 4 x 2 

1 -1 . = tan 2 x ;|2 x |< 3 R&U [Outside Delhi Set I, II, III, Comptt. 2016]

π ± π2 + 3π2 4 π π 3 − tan–1 x = , 4 4

tan–1 x =

Case based MCQs Attempt any four sub-parts from each question. Each sub-part carries 1 mark.



I. Read the following text and answer the following questions on the basis of the same: These questions are for practice and their solutions are available at the end of the chapter

1

2 1

x = –1 1 [CBSE Marking Scheme 2015 (Modified)]

COMPETENCY BASED QUESTIONS



5π2 8

Sol. (tan–1x)2 + (cot–1x)2 =

 1+ x 2 + 1 - x 2     1+ x 2 - 1 - x 2 

π 1 = + cos- 1 x 2 ; -1< x < 1 4 2

The right substitution gives the answer in a few steps.

Q. 3. If (tan–1 x)2 + (cot–1 x)2 =

Long Answer Type Questions (5 marks each) Q. 1. Prove that tan

Students apply the formula for tan–1 x + tan–1 y and makes it complicated, even though the answer is obtained.

½

 π x  −1  = tan  tan  −    4 2  

-1



 2 x 2  2 cos 2 + 2 sin −1 = tan   2 cos2 x − 2 sin 2  2

2x = tan q 1 3   tan tan tan 3 θ − θ 2 θ   LHS = tan −1  2 − tan −1   1 − tan 2 θ   1 − 3 tan 2 θ  = tan–1(tan 3q) – tan–1(tan 2q) 1 = 3q – 2q = q or tan–1 2x 1 \ LHS = RHS [CBSE Marking Scheme 2016 (Modified)]

Sol. Let,



Sol.



INVERSE TRIGONOMETRIC FUNCTIONS



Two men on either side of a temple of 30 metres high observe its top at the angles of elevation a and b respectively. (as shown in the figure above). The distance between the two men is 40 3 metres and the distance between the first person A and the temple is 30 3 meters. [CBSE QB-2021] Q. 1. ÐCAB = a = æ 2 ö (A) sin−1 ç è 3 ÷ø

æ 1ö (B) sin−1 ç ÷ è 2ø

(C) sin−1(2)

æ 3ö (D) sin−1 ç ÷ è 2 ø

Ans. Option (B) is correct. Explanation: In D BDA BD sin a = AB AB222 AB

Now, Now, Now,



iii...eee...

AD 222 AD

Explanation: DC = AC - AD = 40 3 - 30 3 = 10 3 m In DBDC tan b =

Q. 4. ÐABC =

= AB = = AD + BD = (( 30 30 3 )22 + (( 30 30 )22 = = ( 30 33 ))2 + + ( 30 ))2 22 = ( 60 ))2 = = (( 60 60 ) AB = 60 m AB 60 m AB = = 60 m 30 30 sin a a= = 30 sin sin a = 60 60 60 1 sin a a= = 11 sin sin a = 22 2 æ 1ö Ð CAB CAB = =a = sin sin ---111 ææçç 11 öö÷÷ a= Ð Ð CAB = a = sin çèèè 222 ÷øøø

p 4

(B)

p 6

(C)

p 2

(D)

p 3

1 2 sin a = sin 30° sin a =

i.e.,

\ we, have

æ 1ö (A) cos−1 ç ÷ è 5ø

æ 2ö (B) cos−1 ç ÷ è 5ø

æ 3ö (C) cos−1 ç ÷ è 2 ø

(D) cos−1 æç 4 ö÷ è 5ø

æ ö -1 æ 3 3ö a a= = cos cos -1 çç 2 ÷÷ èè 2 øø æ ö -1 æ 3 3ö Ð ÐCAB CAB = =a a= = cos cos -1 çç 2 ÷÷ èè 2 øø

Q. 3. ÐBCA = b = æ 1ö (A) tan−1 ç ÷ è 2ø

(B) tan−1(2)

æ 1 ö (C) tan−1 ç è 3 ÷ø

(D) tan−1

Ans. Option (D) is correct.

(A)

Ans. Option (C) is correct. Explanation: Since,

+ BD BD 222 +

Ans. Option (C) is correct. Explanation: In D BDA AD AD cos cos a a= = AB AB 30 33 30 cos a a= = cos 60 60



30 BD = = 3 DC 10 3

ÐBCA = b = tan -1 ( 3 )



Q. 2. ÐCAB = a =

\ \

37

( 3)



a = 30°

1ù é êQ sin 30° = 2 ú ë û

tan b =

3 tan b = tan 60° b = 60°

\ Now , In DABC ÐABC + ÐBCA + ÐCAB = 180° ÐABC + 60° + 30° = 180° ÐABC = 90° p \ ÐABC = 2

Q. 5. Domain and Range of cos−1 x is: (A) (−1, 1), (0, p) (B) [−1, 1], (0, p) é p pù (C) [−1, 1], [0, p] (D) (−1, 1), ê - , ú ë 2 2û Ans. Option (C) is correct. II. Read the following text and answer the following questions on the basis of the same: The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. "A" is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. "C" is the height of 10 metres from the ground level. For the viewer A, the angle of elevation of "D" is double the angle of elevation of "C". The angle of elevation of "E" is triple the angle of elevation of "C" for the same viewer. Look at the figure given and based on the above information answer the following:

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

[CBSE QB 2021]



38

Q. 1. Measure of ÐCAB is: 1  1  (B) tan   2



(C) tan–1(1) (D) tan–1(3) Ans. Option (B) is correct. Explanation: Let ÐCAB = q, therefore ÐDAB = 2q and ÐEAB = 3q

In right DABC, we have



Þ \

tanq =

ÐCAB = tan −1

Q. 2. Measure of ÐDAB is: 1  3  (A) tan   4 1  4  (C) tan   3



\

1 2 ...(i) 1 2

(B) tan–1(3) (D) tan–1(4)

Ans. Option (C) is correct. Explanation: From eq. (i), we have tan q = tan 2q =

Explanation: We have tanq =

BC 10 1 = = AB 20 2

q = tan −1

(B) tan–13 1  11  (D) tan    2

Ans. Option (D) is correct.



(A) tan–1(2)

Q. 3. Measure of ÐEAB is: (A) tan–1(11) 1  2  (C) tan    11 

\

tan 3q =

3 tan   tan 3  1  3 tan 2  3

=

1 2

1 1  2  2 

1 1  3  2

3

2

3 1 11 − 2 8 = = 8 3 1 1− 4 4 11 11 = ×4 = 8 2 1  11  \ 3q = tan    2

1 2 2 tan  1  tan 2 

1 2  2 = 1 = 4 = 2 1 3 1 1− 1  4 2

1  4  Þ 2q = tan   3



\

1  4  ÐDAB = tan   3



Þ

1  11  ÐEAB = tan    2

Q. 4. A' is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ÐCAB and ÐCA'B is: 1 1 (A) tan 1   (B) tan 1    12  8 2 11 (C) tan 1   (D) tan 1   5  21  Ans. Option (A) is correct. Explanation: Let ÐCA'B = a

39



INVERSE TRIGONOMETRIC FUNCTIONS



 Principal value of branch function sin–1: It is a function with domain [– 1, 1] and range  −3 π − π   − π π   2 , 2  ,  2 , 2 

or

 π 3π   2 , 2 

and

so

on

corresponding to each interval, we get a branch of



 −π π  the function sin– 1 x. The branch with range  ,   2 2 In DA'CB,

is called the principal value branch. Thus, sin–1 : [–1,

BC 10 tan a = = A'B 20 + 5



10 2 Þ tan a = = 25 5



Þ

1  2  a = tan   5



Þ

1  2  ÐCA'B = tan   5

 −π π  1] ®  ,  .  2 2

1  1  ÐCAB = tan   2 [Calculated above] 1  1  1  2  ÐCAB – ÐCA'B = tan    tan   2 5

Also,



\

 1 2   25   = tan   1  1  2  2 5  1  1  = tan    12  Therefore, the required difference between ÐCAB 1 and ÐCA'B in tan 1   . 12  

Q. 5. Domain and Range of tan–1x is:         (A) R ,   ,  (B) R ,   ,   2 2  2 2      (C) R,   ,  (D) R,  0 ,   2 2  2 Ans. Option (C) is correct. Explanation: Domain and Range of tan–1x are     respectively (–¥, ¥) and   ,  i.e., R,   ,  .  2 2  2 2

Case based Subjective (4 marks each) Questions

Read the following text and answer the following questions, on the basis of the same. (Each Sub Part Caries 2 Marks)



Today in the class of Mathematics, Mrs. Agrawal in explaining the inverse trigonometric function. She draws the graph of the sin–1x and and write down the following about the principal value of branch function sin–1:



1

  1  Q. 1. Find the value sin   sin 1     .  2  3

[2]

 1  1   Sol. sin   sin      2  3 π  1  = sin  + sin −1     2  3 

[ sin–1(–q) = –sinq]

π π   = sin  + sin −1  sin    3 6 

1

  = sin    3 6  2    = sin    6   3  = sin    6   = sin   = 1  2  −1 x − 1 and sin–1[x]. Q. 2. Find the domain of sin

Sol. Let

−1 x −1 f(x) = sin

Þ 0 £x–1£1

1

40

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Þ 1 £x£2 \

\ Domain of sin–1[x] is {x : –1 £ [x] £ 1}

x Î [1, 2] [Q x  1  0 and  1  x  1  1] 1

But

We know that,

 −1 ∀ − 1 ≤ x < 0  [x] =  0 ∀ 0 ≤ x < 1  1 ∀ 1≤ x < 2 

\ Domain of sin–1[x] is [–1, 2)

Domain of sin–x is [–1, 1]

1

Solutions for Practice Questions

LHS =

=

9p 9 1 - sin -1 8 4 3 9 ép 1ù 9 1 - sin -1 ú = cos -1 4 êë 2 3û 4 3

1

1

[CBSE Marking Scheme 2020]



sin2 q =

8 9





sin q =

2 2 3













Now, from equation (i) =





L.H.S. =





=

[Using, sin x + cos x = –1



=

= R.H.S.





Hence proved.

Apply the properties to minimise the working.

9p 9 1 - sin -1 8 4 3

Answering Tip

9 ép 1ù - sin -1 ú 4 êë 2 3û



p p Þ cos–1x = – sin–1x] 2 2 9 æ 1ö cos -1 ç ÷ è 3ø 4

æ2 2ö 9 sin -1 ç ÷ 4 è 3 ø

Commonly Made Error

æ2 2ö 9p 9 9 æ 1ö - sin -1 ç ÷ = sin -1 ç ÷ è 3ø 8 4 4 è 3 ø

–1





Detailed Answer: To prove:





2 2 3 æ 2 2ö æ 1ö cos -1 ç ÷ = sin -1 ç ÷ è 3ø è 3 ø q = sin -1









2ö æ æ ö 9 9 æ 1ö -1 -1 2 2 = sin ç 1 - ç ÷ ÷ = sin ç ÷ = RHS è 3ø ÷ 4 4 çè è 3 ø ø 2ö æ æ2 2ö 9 9 æ 1ö sin -1 ç = sin -1 ç 1 - ç ÷ ÷= ÷ = RHS è 3ø ÷ 4 4 çè è 3 ø ø

1 9



3.

sin2 q = 1 -



Short Answer Type Questions-I

Some students convert sin to tan and makes it complicated.

Put x = cos q in R.H.S. ½ 1 as £ x £ 1, RHS = cos–1(4cos3 q – 3cos q) 2 = cos–1 (cos 3 q) = 3q ½+½ = 3 cos–1x = LHS ½ [CBSE Marking Scheme 2018] 7.

...(i)

Short Answer Type Questions-II







1. Let

cos -1



1 cosq = 3



sin2q + cos2q = 1



æ 1ö sin q + ç ÷ è 3ø 2



1 =q 3

2



Given equation can be written as p -1 æ 5 ö -1 æ 12 ö sin ç ÷ = - sin ç ÷ è xø è xø 2

Þ

æ 12 ö -1 æ 5 ö sin -1 ç ÷ = cos ç ÷ è xø è xø



\

æ x 2 - 25 ö æ 12 ö -1 ÷ sin -1 ç ÷ = sin ç çè ÷ø x è xø



Þ

=1

12 = x

x 2 - 25 x

1



INVERSE TRIGONOMETRIC FUNCTIONS

Þ x2 – 25 = 144 Þ x = ± 13, 1½ Since, x = –13 does not satisfy the given equation, \ required solution is x = 13. ½ [CBSE Marking Scheme 2020 (Modified)] 5.

Long Answer Type Questions 1. Put x2 = cos 2 q or θ =



 cos θ + sin θ  −1  1 + tan θ  = tan −1   = tan    cos θ − sin θ   1 − tan θ 

3 π tan −1 x = 2 4



 π  π = tan −1  tan  + θ   = + θ 4  4 

–1



1 − x  1 tan −1  = tan −1 x  1 + x  2 1 or tan–11 – tan–1x = tan −1 x 2



or or

or \



1 cos −1 x 2 2

 2 cos2 θ + 2 sin 2 θ   LHS = tan −1   2 cos2 θ − 2 sin 2 θ   

Given

1½

π tan x = ½ 6 π x = tan   ½ 6 1 x = ½ 3 [CBSE Marking Scheme 2015]

41

=

π 1 + cos−1 x 2 , −1 < x < 1 = RHS 4 2

1

1

1½

1

½

[CBSE Marking Scheme 2017 (Modified)]

REFLECTIONS • •

Inverse trigonometric ratios are used by carpenters to create a desired cut angle. Inverse trigonometric ratio one used to measure the angle of depth and angle of inclination.

•

Architects used the inverse trigonometric ratios to calculate the angle of bridge and the supports.



42

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

SELF ASSESSMENT PAPER - 01 Time: 1 hour

MM: 30

UNIT-I (A) OBJECTIVE TYPE QUESTIONS: I. Multiple Choice Questions

[1×6 = 6]

Q. 1. The function f(x) = x2 + 4x + 4 is (A) odd function (C) neither odd or even function

(B) even function (D) periodic function

Q. 2. What is the domain of sin–1(x + 1) ? (A) [–1, 1]

(B) [–2, 0]

(C) [–2, 0)

(D) [–2, 2]

(C) x2 – 2

(D) 4x2 – 2

(C) 2p + 6

(D) None of these

1 1  2 Q. 3. If f  x +  = x + 2 , then f(x) is  x x (A) x2 + 1

(B) 2 – x2

Q. 4. The value of tan–1{tan(–6)} is (A) 2p

(B) 2p – 6

17   Q. 5. The value of sin  sec −1  = ...........  15 



8 (A) 17

(B)

15 17

(C)

17 8

(D)

8 15

(D)

π 6

 −1  Q. 6. Principal value of sin −1  is  2 



(A)

π 4

π (B) − 4

(C) −

π 6

II. Case-Based MCQs

[1×4 = 4]

Attempt any 4 sub-parts from each questions. Each question carries 1 mark. Read the following text and answer the questions on the basis of the same. Reena is preparing notes on the topic ‘Relations’ which she studied in the last Mathematics class. She write down the following in her note book. A relation R on a set A is said to be an equivalence relation on A iff it is • Reflexive i.e., (a, a) Î R " a Î A. • Symmetric i.e., (a, b) Î R Þ (b, a) Î R Î a, b " A • Transitive i.e., (a, b) Î R and (b, c) Î R Þ (a, c) Î R " a, b, c Î A Q. 7. If the relation R = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} defined on the set A = {1, 2, 3}, then R is (A) Reflexive (B) Symmetric (C) Transitive (D) Equivalence Q. 8. If the relation R = {(1, 2), (2, 1), (1, 3), (3, 1)} defined on the set A = {1, 2, 3}, then R is (A) Reflexive (B) Symmetric (C) Transitive (D) Equivalence Q. 9. If the relation R on a set of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}, then R is (A) Reflexive (B) Symmetric (C) Transitive (D) Equivalence Q. 10. If a relation R on set A = {1, 2, 3, ......., 13, 14} defined as R = {(x, y) : 3x – y = 0}, then R is (A) Reflexive (B) Symmetric (C) Transitive (D) None of these



SELF ASSESSMENT PAPER

43

Q. 11. If the relation on set A = {1, 2, 3} defined as R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, then R is (A) Reflexive (B) Symmetric (C) Transitive (D) Equivalence (B) SUBJECTIVE TYPE QUESTIONS: III. Very Short Answer Type Questions

[3×1 = 3]

Q. 12. If f(x) = 2x – x , then find the value of f(x + 2) – f(x – 2) at x = 0. 2

Q. 13. Check function f : R ® R defined as f(x) = 5x2 – 8 is one-one or not. −1  1  −1 Q. 14. Prove that cos   = sec x " x Î (–¥, –1] È [1, ¥). x

IV. Short Answer Type Questions–I

[2×3 = 6]

Q. 15. Consider a function f : [0, p/2] ® R, given by f(x) = sin x and g : [0, p/2] ® R, given by g(x) = cos x. Show that f and g and are one-one, but f = g is not one-one. Q. 16. Find the principal value of the following.  1  (i) sin −1   2  2 ( p − q ) , if Q. 17. If function l, is defined by l(p, q) =  if  p + q ,

−1  1  (ii) cot   3

p≥q λ( −( −3 + 2 ),( −2 + 3)) . then find the value of expression p n.  −5 −1 e.g :  8 −9  4 0 3×2  3. EQUALITY OF MATRICES: Two matrices A and B are said to be equal and written as A = B, if they are of the same order and their corresponding elements are identical i.e. aij = bij i.e., a11 = b11, a22 = b22, a32 = b32 etc. 4. ADDITION OF MATRICES: If A and B are two m × n matrices, then another m × n matrix obtained by adding the corresponding elements of the matrices A and B is called the sum of the matrices A and B and is denoted by ‘A + B’. Thus if A = [aij], B = [bij], or A + B = [aij + bij]. Properties of matrix addition: • Commutative property: A + B = B + A



MATRICES

• Associative property: A + (B + C) = (A + B) + C • Cancellation laws: (i) Left cancellation: A + B = A + C Þ B = C (ii) Right cancellation: B + A = C + A Þ B = C. 5. MULTIPLICATION OF A MATRIX BY A SCALAR: If a m × n matrix A is multiplied by a scalar k (say), then the new kA matrix is obtained by multiplying each element of matrix A by scalar k. Thus, if A = [aij] and it is multiplied by a scalar k, then kA = [kaij], i.e. A = [aij] or kA = [kaij]. e.g : A =  2 −4  or 3A =  6 −12   5 6  15 18  6. MULTIPLICATION OF TWO MATRICES: Scan to know Let A = [aij] be a m × n matrix more about and B = [bjk] be a n × p matrix such that this topic the number of columns in A is equal to the number of rows in B, then the m × p matrix C = [cik] such that [cik] = Σ nj =1 aij bjk is said to be the product of the matrices A Multiplication of Matrices and B in that order and it is denoted by Part-1 AB i.e. “C = AB”. Properties of matrix multiplication: • Note that the product AB is defined only when the number of columns in matrix A is equal to the number of rows in matrix B. • If A and B are m × n and n × p matrices, respectively, then the matrix AB will be an m × p matrix i.e., order of matrix AB will be m × p. • In the product AB, A is called the pre-factor and B is called the post-factor. • If two matrices A and B are such that AB is possible then it is not necessary that the product BA is also possible. • If A is a m × n matrix and both AB as well as BA are defined, then B will be a n × m matrix. • If A is a n × n matrix and In be the unit matrix of order n, then A In = In A = A . • Matrix multiplication is associative i.e., A(BC) = (AB)C. • Matrix multiplication is distributive over the addition i.e., A.(B + C) = AB + AC .

• Matrix multiplication is not commutative.

47

7. IDEMPOTENT MATRIX:

A square matrix A is said to be an idempotent

matrix if A2 = A.

For example,



1 0 0  2 −2 −4  4 A = 0 1 0  , A =  −1 3 0 0 1  1 −2 −3     



are idempotent matrices.

Scan to know more about this topic

8. TRANSPOSE OF A MATRIX: If A = [aij]m × n be a m × n matrix, then the matrix obtained by interchanging the Multiplication of Matrices rows and columns of matrix A is said to be Part-2 a transpose of matrix A. The transpose of A is denoted by A’ or AT i.e., if AT = [aji]n × m. For example, 0   5 T  5 − 4 1 =  − 4 5   0  5 3 1 3     PROPERTIES OF TRANSPOSE OF MATRICES: (i) (A + B)T = AT + BT

(ii) (AT )T = A



(iii) (kA)T = kAT, where k is any constant



(iv) (AB)T = BT AT



(v) (ABC)T = CT BT AT

Mnemonics Types of Matrices Ram Charan Says Drink Sprite ¯ ¯ ¯ ¯ ¯ Row Column Square Diagonal Scalar Matrix Matrix Matrix Matrix Matrix and Nescafe Ice Tea ¯ ¯ ¯ Null Identity Triangular Matrix Matrix Matrix Matrix Multiplication No. of columns of first matrix = No. of rows of second matrix u v  a b c  w x  d e f   y z  R×C

R×C

Class Representative

48

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

OBJECTIVE TYPE QUESTIONS A

2 0 0 =  0 2 0  0 0 2   

Multiple Choice Questions

Q. 1. The number of all possible matrices of order 2 × 3 with each entry 1 or 2 is (A) 16 (B) 6 (C) 64 (D) 24 [CBSE Board 2021] Ans. Option (C) is correct. Explanation: The order of the matrix = 2 × 3 The number of elements = 2 × 3 = 6 Each place can have either 1 or 2. So, each place can be filled in 2 ways. Thus, the number of possible matrices = 26 = 64

1 1 1 Þ X2 – X = 2 1 1 1 = 2I 1 1 1    3 4  −1 2 1 Q. 4. For two matrices P =  −1 2  and QT =  1 2 3 ,    0 1   P – Q is: 2  (A) −3 0  4 (C)  −0  −1 

3 c  6 a  d   12 2  Q. 2.  a  d 2  3b    8 4  are equal, then value     of ab – cd is: (A) 4

(B) 16

(C) –4

(D) –16 [CBSE Board 2021]

Ans. Option (A) is correct.

 0 1 1 Q. 3. For the matrix X =  1 0 1 , (X2 – X) is: 1 1 0   (A) 21

(B) 31

(C) 1

(D) 51 [CBSE Board 2021]

Ans. Option (A) is correct. Explanation: 0 1 1  0 1 1  Here X2 =  1 0 1   1 0 1  1 1 0 1 1 0    2 1 1 Þ X2 =  1 2 1  1 1 2    2 1 1  0 1 1 Þ X2 – X =  1 2 1  −  1 0 1  1 1 2 1 1 0    

3 4   3 0 − (B)  −1 −2    2 3  (D)  0 −3  0 −3  

3 −3  −2 

[CBSE Board 2021] Ans. Option (B) is correct. Explanation:

3c + 6 a − d   12 2  Explanation: Given,  a + d 2 − 3b  =  −8 −4      \ 3c + 6 = 12 ...(i) a – d = 2 ...(ii) a + d = –8 ...(iii) 2 – 3b = –4 ...(iv) From eq. (i), we get c = 2 On solving eqs. (ii) and (iii), we get a = –3 and d = –5 from eq. (iv), we get b = 2 Now, ab – cd = (–3)2 – 2(–5) Þ ab – cd = –6 + 10 = 4

3 0 −3

 −1 1  Q = (QT)T =  2 2   1 3  

Here,

Now,

 3 4   −1 1  P – Q =  −1 2  −  2 2   0 1  1 3    

3 4 =  −3 0   −1 −2    Q. 5. A matrix A = [aij]3 × 3 is defined by





2i + 3 j , i < j  , i= j aij =  5 3i − 2 j , i > j



The number of elements in A which are more than 5, is: (A) 3 (B) 4 (C) 5 (D) 6 [CBSE Board 2021] Ans. Option (B) is correct.  5 8 11 Explanation: Here, A =  4 5 13 7 5 5    Thus, number of elements more than 5, is 4. Q. 6. If A is a square matrix such that A2 = A, then (I – A)3 + A is equal to: (A) I (B) 0 (C) I – A (D) I + A [CBSE Board 2021] Ans. Option (A) is correct.



MATRICES

49

Explanation:

Topper Answer, 2020 Sol.

 1 0 Q. 7. If [x 1]   = 0, then x equals:  2 0 

(A) 0

(B) –2

(C) –1

(D) 2





X = [1 2 3] ,



é2 ù Y = ê 3 ú ê ú êë 4 úû

[CBSE Delhi Set - II 2020] Ans. Option (D) is correct. Explanation:

2 3     AB + XY = [ 2 −3 4]  2  + [1 2 3]  3   4   2 

 1 0 x 1   = [0 0]  −2 0 

Þ

ëéx - 2 0 ùû = [0 0]

Þ

x – 2 = 0 [By def. of equality]

Þ

x = 2

3  Q. 8. If A = [2 –3 4], B =  2  , X = [1 2 3] and Y =  2  then AB + XY equals: (A) [28]

(B) [24]

(C) 28

(D) 24

2  3  ,    4 

[CBSE OD Set - I 2020] Ans. Option (A) is correct. Explanation: Given,

é3 ù B = ê 2 ú , ê ú êë 2 úû

A = [ 2 -3 4 ] ,

= [6 – 6 + 8] + [2 + 6 + 12] = [8] + [20] = [28] Q. 9. Suppose P and Q are two different matrices of order 3 × n and n × p, then the order of the matrix P × Q is? (A) 3 × p (B) p × 3 (C) n × n (D) 3 × 3 [CBSE SQP 2019-20] Ans. Option (A) is correct. Q. 10. A = [aij]m×n is a square matrix, if (A) m < n (B) m > n (C) m = n (D) None of these Ans. Option (C) is correct. Explanation: It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. Therefore, A = [aij]m×n is a square matrix, if m = n.

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type (1 mark each) Questions Q. 1. If A and B are matrices of order 3 × n and m × 5 respectively, then find the order of matrix 5A – 3B, given that it is defined. R [CBSE SQP 2020-21]

Sol. 3 × 5 1 [CBSE Marking Scheme 2020] Detailed Solution: 5[A]3 × n – 3[B]m × 5 = [5A – 3B]3 × 5 é1 0 ù Q. 2. If A = ê ú , then find A3. ë1 1 û [CBSE OD SET-II 2020] R

50

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

é1 0 ù é1 0 ù , A3 = ê Sol. A = ê ½+½ ú ú ë2 1 û ë3 1 û [CBSE Marking Scheme 2020] 2

Detailed Solution: é1 A = ê ë1 2 A = A×A



matrix A.

é1 0 ù ê1 1 ú ë û 0 + 0ù 0 + 1 úû



A = 3 1  1 2 

1

[CBSE Marking Scheme 2019] Detailed Solution: é5 3A – B = ê ë1 é5  4 3 or 3A –  2 5 = ê   ë1 Given

0 ù é1 0 ù 1 úû êë1 1 úû 0ù 1 úû

é5 3A = ê ë1 9 3A =  3 

or or

Some students cube the elements in A and obtain the answer.

Answering Tip



é 4 3ù ê 2 5 ú , then find the ë û

R&U [CBSE Delhi Set III-2019]

Sol.

Commonly Made Error



R&U [CBSE SQP 2020-21]

é5 0 ù Q. 4. If 3A – B = ê ú and B = ë 1 1û

é1 0 ù = ê ú ë2 1 û A3 = A2×A é1 = ê ë2 é1 A3 = ê ë3

ì 1 if i ¹ j aij = í î 0 if i = j

0ù 1 úû

é1 0 ù = ê ú ë1 1 û é1 + 0 = ê ë1 + 1

Q. 3. Find the value of A2, where A is a 2 × 2 matrix whose elements are given by

or

1 Q. 5. If 2  0

3 y + x   1

(x + y).

Matrix multiplication is different from normal multiplication.

Sol.

0ù 1 úû 0ù  4 3 +  1 úû  2 5 3 6 

½

3 1  A = 1 2  ½   0  5 6 write the value of = , 2 1 8

x + y = 6 1 [CBSE Marking Scheme 2019]

Topper Answer, 2019

These questions are for practice and their solutions are available at the end of the chapter

é4 3ù ê2 5ú ë û

R&U [CBSE Delhi Set II-2019]

Detailed Solution:

Sol.

0ù ,B= 1 úû



MATRICES

 1 2 2 Q. 6. If A =  2 1 x  is a matrix satisfying AA’ = 9I,    −2 2 −1 find x.

æ 1ö æ 1ö = ( -2 - 1 1 + 3 - 2 + 3) ç 0 ÷ = ( -3 4 1) ç 0 ÷ ç ÷ ç ÷ çè -1÷ø çè -1÷ø

[CBSE Delhi Comptt. Set I, II, III, 2018]

R&U [Delhi Set I, II, III, 2016][NCERT Exemplar]

Sol. Given A2 = I

\ (A – I)3 + (A + I)3 – 7A



= A3 – I3 – 3A2I + 3AI2 + A3 + I3 + 3A2I + 3AI2 – 7A ½



= AI – I – 3I + 3A + AI + I + 3I + 3A – 7A



= A + 3A + A + 3A – 7A = A

= ( -3 - 1) = ( -4 )1 ´1

\ Order of matrix A is 1 × 1. 1 Q. 11. Write the element a23 of a 3 × 3 matrix A = (aij) whose elements aij are given by

Sol.

=

½

0 Q. 12. If A =  2

= 34





Answering Tip Give ample practice on problems based on multiplication of two matrices.

Q. 10. If (2

1

 −1 0 −1  1  3)  −1 1 0   0  = A , then write the  0 1 1   −1

order of matrix A. 

U [Foreign 2016]

[NCERT Exemplar]

½

 0 4 a 3  and kA =   find the -5   - 8 5b 

Q. 1. Find a matrix A such that 2A – 3B + 5C = O, where  −2 2 0  B=   and C =  3 1 4

 2 0 −2  7 1 6  .  

R [CBSE All Set-III, 2019]

Sol. 

Mostly candidates commit errors while calculation as they confuse in multiplication of two matrices.

1 2

Short Answer Type Questions-I (2 marks each)

x = 1, y = – 2, z = 1 x + y + z = 0 [CBSE Marking Scheme 2016] 1

Commonly Made Error

½

R&U [Outside Delhi Set II 2015]

R&U [Delhi Set I, II, III Comptt. 2016]

Sol. or

| 2 − 3| 2

values of k and a.

= 81 \ 81 matrices of order 2 × 2 are possible with each entry 1, 2 or 3. ½ é 1 0 0ù é x ù é 1ù ê úê ú ê ú Q. 9. If ê 0 y 0 ú ê -1ú = ê 2 ú , find x + y + z. êë 0 0 1 úû êë z úû êë 1 úû

A [CBSE March 2015]

[CBSE Marking Scheme 2015]

R&U [O.D. Set I, II, III 2016]

\ Number of ways to write 1, 2 or 3 at 4 places



|i − j| 2 a23 =

½

Sol. Number of elements of 2 × 2 matrix = 4

aij =



Q. 8. Write the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3.

æ -1 0 -1ö æ 1 ö A = ( 2 1 3) ç -1 1 0 ÷ ç 0 ÷ ç ÷ç ÷ çè 0 1 1 ÷ø çè -1÷ø

Sol. We have,

R&U [S.Q.P. 2018-19]

Q. 7. If A is a square matrix such that A2 = I, then find the simplified value of (A – I)3 + (A + I)3 – 7A.

51



é -6 6 0 ù é10 0 -10 ù é0 0 0 ù 2A - ê ú+ê ú=ê ú ë 9 3 12 û ë 35 5 30 û ë0 0 0 û é -8 3 5 ù Þ A= ê ú ë -13 -1 -9 û

1 1

[CBSE Marking Scheme, 2019]

 Detailed Solution:

−2 2 0   and C = Given, B =   3 1 4



Also, given 2A – 3B + 5C = O



Þ



2 0 −2  −2 2 0    –5  Þ 2A = 3  7 1 6   3 1 4



−6 6 0  −10 0 10  Þ 2A =   +   9 3 12 −35 −5 −30

These questions are for practice and their solutions are available at the end of the chapter

2 0 −2    7 1 6 

2A = 3B – 5C

½ ½

52

−16 6 10  2A =   −26 −2 −18



Þ



A =

1 −16 6 10    2 −26 −2 −18

 −8 3 5  A =   −13 −1 −9

Þ



½

½

R [CBSE Delhi Set-II, 2019]



R [CBSE OD Set-I, 2019]

1  2 2  1 2  Sol. ( A − 2 I )( A − 3I ) = − −  − −   1 1  1 2  0 0  1 = =  0 0 0    [CBSE Marking Scheme, 2019]

Detailed Solution:  4 2 A =   −1 1   4 2  1 0    − 2 A – 2I =  −1 1  0 1



\







 4 − 2 2 + 0 = −1 + 0 1 − 2   



2 2 =   −1 −1





= 0 = RHS

½ ½

é3 1 ù Q. 4. If A = ê ú , show that A2 – 5A + 7I = 0. ë -1 2 û Hence find A–1. [CBSE SQP 2020-21]

é 4 2ù Q. 3. If A = ê ú, ë -1 1 û show that (A – 2I) (A – 3I) = 0.

Given,



½

Hence Proved.





2−2 4−4  0 0   =  = −1 + 1 −2 + 2 0 0 



é 2 0 1ù ê ú Q. 2. If A = ê 2 1 3 ú , then find (A2 – 5A). êë 1 -1 0 úû



2 2 1 2 =      −1 −1  −1 −2 



5ù é3 1 ù é3 1 ù é8 A2 = ê ú ê -1 2 ú = ê -5 3 ú 1 2 û ë ûë û ë

Sol.

5ù é3 1 ù é3 1 ù é8 =ê ê -1 2 ú ê -1 2 ú= ú 5 3 ë ûë û ë û



é15 5 ù é7 0 ù , 7I = ê 5A = ê ú ú ë -5 10 û ë0 7 û

Þ

éO O ù A2 – 5A + 7I = ê ú = O 1 ëO O û

Þ A– 1 (A2 – 5A + 7I) = A–1O Þ A – 5I + 7A–1 = O Þ 7A–1 = 5I – A Þ

A– 1 =

1 æ é 5 0 ù é 3 1 ùö ê ú-ê ú 7 çè ë0 5 û ë -1 2 û÷ø

Þ

A– 1 =

1 é 2 -1ù ê ú 1 7 ë1 3 û

é 4 2 ù é -2 0 ù = ê ú+ê ú ë -1 1 û ë 0 -2 û

[CBSE SQP Marking Scheme 2020-21]

Commonly Made Error





Þ

½

After proving the equation, some students find the inverse using formula.



Answering Tip

 4 2  1 0  A – 3I =   − 3  −1 1  0 1



and



 4 2  −3 0  =  +  −1 1   0 −3



 4 − 3 2 + 0 =   −1 + 0 1 − 3 



1 2 =   −1 −2 



Now,

LHS = (A – 2I) (A – 3I)







Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Learn to find inverse from a matrix equation.

Short Answer Type (3 marks each) Questions-II æ 1 2 3ö æ -7 -8 -9ö Q. 1. Find matrix X so that X ç = è 4 5 6÷ø çè 2 4 6 ÷ø ½

R&U [Delhi Set I, II, III 2017] [NCERT] a b  1 Sol. Let X =   c d

These questions are for practice and their solutions are available at the end of the chapter



MATRICES

equating and solving to get a = 1, b = −2 , c = 2 , d = 0 1½  1 −2  X = 2 0  

½

Commonly Made Error



éa b c ù Some students take X = ê ú but this is ëd e f û wrong find the order of X using the order of A and B. XA = B (Xp×q) (A)2×3 = B2×3 p = 2, q = 2









Students get wrong with the order while forming matrices.

Answering Tip



Take the first matrix as a row matrix and the second one as a column matrix.

Q. 3. To promote the making of toilets for women, as organization tried to generate awareness through (i) house calls (ii) letters and (iii) announcements. The cost for each mode per attempt is given below : (i) ` 50 (ii) ` 20 (iii) ` 40 The number of attempts made in three villages X, Y and Z are given below: (i)

(ii)

(iii)

X

400

300

100

Y

300

250

75

Z

500

400

150

Find the total cost incurred by the organization for the three villages separately, using matrices.

Answering Tip



Commonly Made Error



æ a b ö æ 1 2 3ö æ -7 -8 -9ö then, æ a b ö æ 1 2 3ö =æ -7 -8 -9ö çèçècc dd÷ø÷øçèçè44 55 66÷ø÷ø = çèçè22 44 66 ÷ø÷ø æ a + 4 b 2 a + 5b 3a + 6b ö æ -7 -8 -9ö Þæça + 4 b 2 a + 5b 3a + 6b ö÷ ==æç-7 -8 -9ö÷ Þ Þ çèècc++44dd 22cc++55dd 33cc++66dd÷øø çèè22 44 66 ÷øø 1 a + 4b = – 7 3a + 6b = – 9 c + 4d = 2 2a + 5b = – 8 3c + 6d = 6 2c + 5d = 4

R&U [All India, 2015]

Learn the properties of matrix multiplication.

Q. 2. A trust fund has ` 35,000 is to be invested in two different types of bonds. The first bond pays 8% interest per annum which will be given to orphanage and second bond pays 10% interest per annum which will be given to an N.G.O. (Cancer Aid Society). Use matrix multiplication, determine how to divide ` 35,000 among two types of bonds if the trust fund obtains an annual total interest of ` 3,200. R&U [O.D. Set I, II, III Comptt. 2015] Sol. Let investment in first type of bond be ` x. \ The investment in second type of bond = ` (35,000 – x)  8    \ [ x 35, 000 − x ]  100  = [3,200] 10    100 

53

Long Answer Type Questions (5 marks each) é 2 0 1ù ê ú 2 Q. 1. If A = ê 2 1 3 ú , find A – 5A + 4I and hence êë 1 -1 0 úû find a matrix X such that A2 – 5A + 4I + X = 0. R&U [Delhi, 2015] Sol. Getting 2 0 = 2 1  1 −1 

1 1½

8 10 x + ( 35, 000 − x ) or = 3,200 100 100 or x = ` 15,000 1 \ Investment in first bond = ` 15,000 and Investment in second bond = `(35,000 – 15,000) = ` 20,000 ½ [CBSE Marking Scheme 2015]

1½

 5 −1 2  2 0 1 \ A2 – 5A + 4I = 9 −2 5  − 5  2 1 3  0 −1 −2   1 −1 0      1 0 0 + 4 0 1 0  1 0 0 1   



These questions are for practice and their solutions are available at the end of the chapter

A2 = A·A 1 2 0 1  5 −1 2  3   2 1 3  = 9 −2 5  0 −1 −2  0   1 −1 0   

 5 −1 2   −10 0 −5  = 9 −2 5  +  −10 −5 −15 0 −1 −2   −5 5 0    

4 0 0  + 0 4 0  0 0 4   

54

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

 −1 −1 −3  =  −1 −3 −10   −5 4 2   \ or

Sol. Clearly order of X is 3 × 2 a b  Let X =  c d  1 e f     −7 −8 −9  a b  1 2 3   1 2 4 6  = So  c d      4 5 6     11 10 9   e f  a + 4 b = −7 c + 4 d = 2 e + 4 f = 11   2 a + 5b = −8 2 c + 5d = 4 2 e + 5 f = 10  1 3a + 6b = – 9, 3c + 6d = 6, 3e + 6f = 9 Solving we get 2 a = 1, b = −2 , c = 2 d = 0 , e = −5 f = 4

1½

X = – (A2 – 5A + 4I) 3 1 1 X =  1 3 10  1  5 −4 −2    [CBSE Marking Scheme 2015]

Commonly Made Error Most of the students attempt this question incorrectly. They make errors while calculating A2. Sometimes they add two matrices to solve A2.



−2  1   Thus X =  2 0  1  −5 4    [CBSE Marking Scheme 2017]

Answering Tip Stress upon developing logical and reasoning skills to apply the correct property of matrix.

Commonly Made Error



-1ö æ2 æ -1 -8 -10ö ç ÷ 0 A = ç 1 -2 -5 ÷ Q. 2. Find matrix A, if 1 ç ÷ ç ÷ çè -3 4 ÷ø çè 9 22 15 ÷ø



Students fail to find the order of the matrix X correctly.

Answering Tip

R&U [NCERT Exemplar] [Delhi Comptt. 2017]



æ - 7 - 8 - 9ö æ 1 2 3ö ç Q. 3. Find matrix X if: X ç = 2 4 6 ÷ ÷ è 4 5 6÷ø çç è 11 10 9 ÷ø











Learn the confirm ability of matrix multiplication with the order of the product matrix.

R&U [Delhi Comptt. set II 2017]

Topic-2

Symmetric, Skew Symmetric and Invertible Matrices Concepts Covered 



Symmetric Matrix,  Skew Symmetric Matrix,  Invertible Matrix

Uniqueness Theorem

Revision Notes  Symmetric matrix: A square matrix A = [aij] is said to be a symmetric matrix if AT = A. i.e., if A = [aij], then AT = [aji] = [aij] or AT = A.

For example :



a h g 

h b f

g  2 + i 1 3  f , 1 3 + 2i  2 c   3 3 + 2i 4 

Scan to know more about this topic

AT = –[A] i.e., if A = [aij], then AT = [aji] = – [aij] or AT = – A.

Symmetric Matrix

Skew symmetric matrix: A square matrix A = [aij] is said to be a skew symmetric matrix if

 0 1 −5  0 2 5 , For example :  −1 0  5 −5 0   −2 0   

Orthogonal matrix: A said to be orthogonal where AT is transpose of A.

matrix if A.AT

A =

is I,

 Invertible Matrix: An invertible matrix is a matrix for which matrix inversion operation exists, given that it satisfies the requisite conditions. Any given square matrix A of order n × n is called invertible if there exists another n × n square matrix B such

These questions are for practice and their solutions are available at the end of the chapter



MATRICES



that, AB = BA = In, where In is on identity matrix of order n × n. 1 2 Example: Let matrix A =  2 5  and matrix B =  



 5 −2   −2 1 



 1 2   5 −2  1 0 Now, AB =  2 5   −2 1  = 0 1      



 5 −2   1 2  1 0 and BA =  −2 1   2 5  = 0 1      



Hence, A–1 = B and B is called the inverse of A. So, A can also be the inverse of B or B–1 = A.

55

Uniqueness of Inverse of Matrix If there exists an inverse of a Scan to know more about square matrix, it is always unique. this topic Proof: Let A be a square matrix of order n × n. Let us assume matrices B and C be inverses of matrix A. Uniqueness of Now, AB = BA = I, since B is the Inverse inverse of matrix D. Similarly, AC = CA = I But, B = BI = B(AC) = (BA)C = IC = C This proves B = C, or B and C are the same matrices.

Key Fact

 Note that [aji] = – [aij] or [aii] = – [aii] or 2[aii] = 0 (Replacing j by i). i.e., all the diagonal elements in a skew symmetric matrix are zero.



 For any matrices, AAT and ATA are symmetric matrices.



 For a square matrix A, the matrix A + AT is a symmetric matrix and A – AT is always a skew-symmetric matrix.



 Also note that any square matrix can be expressed as the sum of a symmetric and a skew

A + AT symmetric matrix i.e., A = P + Q where P =  is a symmetric matrix 2 A - AT is a skew symmetric matrix. and Q = 2

OBJECTIVE TYPE QUESTIONS

Multiple Choice Questions cos   sin   Q. 1. If A =   , then A + A’ = I, then the  sin  cos  

value of a is: p (A) 6

p (B) 3

(C) p

3p (D) 2



1 2

Þ

cos a =

Þ

cos a = cos

\

a =

p 3

p 3

Q. 2. Matrices A and B will be inverse of each other only if:

Ans. Option (B) is correct. Explanation: cos α − sin α Given that, A =    sin α cos α 

Also A + A’ = I cos α − sin α  cos α sin α  1 ⇒  = +  sin α cos α  − sin α cos α 0 2 cos α 0  1 ⇒  = 0 2 cos α  0 

Equating corresponding entries, we have Þ 2 cos a = 1

(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

Ans. Option (D) is correct. Explanation: We know that if A is a square matrix of order m, and if there exists another square matrix B

0  1 0  1

of the same order m, such that AB = BA = I, then B is said to be the inverse of A. In this case, it is clear that A is the inverse of B. Thus, matrices A and B will be the inverse of each other only if AB = BA = I.

56

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Q. 3. If A and B are symmetric matrices of same order, then AB – BA is a: (A) Skew-symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrix Ans. Option (A) is correct. Explanation: Given that,

Ans. Option (B) is correct.

1 0 0    Explanation: A = 0 2 0  0 0 4 



1 0 0    \ ∴ A ' = 0 2 0  = A 0 0 4 



So, the given matrix is a symmetric matrix.

A and B are symmetric matrices.

Þ

A = A’ and B = B’ Now, (AB – BA)’ = (AB)’ – (BA)’...(i)



Þ (AB – BA)’ = B’A’ – A’B’ [By reversal law] [From Eq. (i)] Þ (AB – BA)’ = BA – AB Þ (AB – BA)’ = –(AB – BA) Þ (AB – BA) is a skew-symmetric matrix. Q. 4. If the matrix A is both symmetric and skewsymmetric, then: (A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of these [CBSE Board 2021] Ans. Option (B) is correct. Explanation: If A is both symmetric and skew symmetric, then we have, A ' = A and A ' = − A ⇒ A = −A ⇒ A+A=0 ⇒ 2A = 0 ⇒ A=0



[Since, in a square matrix A, if A’ = A, then A is called symmetric matrix.]

5 8  0  Q. 6. The matrix 5 0 12  is a:   8 12 0 

(A) diagonal matrix (B) symmetric matrix (C) skew symmetric matrix (D) scalar matrix Ans. Option (C) is correct. Explanation: We know that, in a square matrix, if bij = 0 when i ≠ j then it is said to be a diagonal matrix. Here, b12, b13…. ≠ 0 so the given matrix is not a diagonal matrix.  -5 8 ù é0 ê ú Now, B=ê5 0 12 ú êë -8 -12 0 úû

Therefore, A is a zero matrix.

1 0 0  Q. 5. The matrix 0 2 0  is a:   0 0 4 

é 0 5 -8 ù ê ú B ' = ê -5 0 -12 ú êë 8 12 0 úû -5 8 ù é0 ê ú 0 12 ú = -ê 5 êë -8 -12 0 úû

= -B So, the given matrix is a skew-symmetric matrix, since we know that in a square matrix B, if B’ = −B, then it is called skew-symmetric matrix.

(A) identity matrix (B) symmetric matrix (C) skew-symmetric matrix (D) None of these

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. A square matrix A is said to be symmetric if ................. .

Topper Answer, 2020 Sol.

[CBSE Board 2020]



MATRICES

 0 a −3 Q. 2. If the matrix A =  2 0 −1 is skew symmetric. b 1 0    Find the values of ‘a‘ and ‘b‘. R&U [Delhi & O.D. 2018]

Sol. a = – 2, b = 3

57

Q. 5. Give an example of a skew symmetric matrix of order 3. R&U [CBSE S.Q.P, 2015-16] é 0 1 -3 ù Sol. ê -1 0 -2 ú . ê ú êë 3 2 0 úû

½+½

Short Answer Type Questions-I (2 marks each)

[CBSE Marking Scheme 2018] Detailed Solution:

Q. 1. A is skew-symmetric matrix of order 3, then prove

Topper Answer, 2018

Sol.

é 0 2 b -2 ù Q. 3. Matrix A = êê 3 1 3 úú is given to be symmetric, êë 3 a 3 -1 úû find values of a and b. R&U [Delhi Set I, II, III 2016] Sol. As A is a symmetric matrix, or A’ = A ½ 0 3 3 a é ù é 0 2b -2 ù ê ú ê ú 2 b 1 3 \ ê ú = ê3 1 3ú ê -2 3 -1úû ë êë3a 3 -1 úû −2 3 \ By equality of matrices, a = and b = . ½ 3 2 





Commonly Made Error Few students commit error in solving the problems based on symmetric matrix.





Answering Tip Learn the difference between symmetric and skew symmetric matrices.

Q. 4. Write a 2 × 2 matrix which is both symmetric and skew symmetric. R&U [Delhi Set I Comptt. 2014]

that det A = 0. [OD 2017] Q. 2. Show that A'A' and AA' are both symmetric matrices for any matrix A. Sol. Let, P = A'A \ P' = (A'A)' = A'(A')' [ (AB')' = B'A'] = A'A = P So, A'A is symmetric matrix for any matrix A. Similarly, Let Q = AA' Q' = (AA')' = (A')'A' = AA' = A So, AA' is symmetric matrix for any matrix A.  1 0 0 2 0 0 Q. 3. If A = 0 −1 0  and B =  0 3 0  then find 0 0 2   0 0 −1     3A + 4B. 2 0 0  1 0 0 Sol. We have, A = 0 −1 0  and B =  0 3 0  0 0 2   0 0 1    

2 0 0  1 0 0    3 0 − 1 0 4 0 3 0 + Now, 3A + 4B = 0 0 2   0 0 −1    



0  3 0 0  8 0 = 0 −3 0  + 0 12 0 0 0 6  0 0 −4     



0 0  3 + 8 −3 + 12 =  0 0   0 0 6 − 4  



11 0 0  =  0 0 9  0 0 2  

Q. 4. If A =  4 4  , B =  1 2  and C =  1 4  , then  2 7   −1 3 3 −5 find A – B – C. Sol. Here, A, B and C are the three matrices of same order 2 × 2. Now, A – B – C = A + (–1)B + (–1)C



These questions are for practice and their solutions are available at the end of the chapter

4 4   1 2 1 4 =  2 7  + ( −1)  −1 3 + ( −1) 3 −5      

58

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

 4 4   −1 −2   −1 −4  =  2 7  +  1 −2  +  −3 5       4 − 1 − 1 4 − 2 − 4 =  2 + 1 − 3 7 − 3 + 5    2 −2  = 0 9  



symmetric and skew symmetric matrix. A [O.D. Set I, II, III Comptt. 2015]



[NCERT Exemplar]

Sol.



é 2 4 -6 ù ê ú 5ú A = ê7 3 êë 1 -2 4 úû

2 7 1 Then A’ =  4 3 −2   −6 5 4   

1



11 5 −  2 2 3   3 2  3 4  2 

1



 0 3 Q' = –  2 7  2

3 − 2 7 − 2 −

3 2

0 7 − 2

7 −  2 7   =–Q 2  0  

 2 4 −6  5 = 7 3  1 −2 4   

−

3 2

0 −

7 2

7 −  2 7   2  0  

=A

1

[CBSE Marking Scheme 2015] Q. 2. Express the following matrix as a sum of a symmetric and skew-symmetric matrices and verify your result: é 3 -2 -4 ù ê ú A ê 3 -2 -5 ú êë -1 ú 1 2û Q. 3. Show that the matrix BT AB is symmetric or skewsymmetric accordingly when A is symmetric or skew-symmetric. AE [NCERT] Sol. Case I: Let A be a symmetric matrix. Then AT = A.

Now,

(BT AB)T = BTAT(BT)T

[By reversal law]

T



or (BTAB)T = BTAB [Q AT = A]



\ BT AB is a symmetric matrix.



Case II: Let A be a skew-symmetric matrix. Then, AT = – A.



Now,



or (B AB) = B A B [Q (BT)T = B] or (BTAB)T = BT(– A)B [Q AT = – A] or (BTAB)T = – BTAB 2



\ BT AB is a skew-symmetric matrix.

(BT AB)T = BTAT(BT)T T



T

T

2

[By reversal law]

T

é1 3 2 ù ê ú Q. 1. If A = ê 2 0 -1ú , then show that êë 1 2 3 úû

7 −  2 7   2  0  

0

11 5  −  0 2 2 3  3 + 3 2  2 3 7 4   2  2

Long Answer Type Questions (5 marks each)

0 −3 −7  1 1 7 Q = ( A − A ') =  3 0 2 7 −7 0  2  

 0  = 3 2 7  2

Q’ = – Q

= B A B [Q (BT)T = B]

Since P’ = P \ P is a symmetric matrix. Let

\ Q is a skew symmetric matrix.

T

1 Let P = ( A + A ') 2  4 11 −5 1 =  11 6 3   2  −5 3 8    2  11 =   2 − 5  2

Since



  2  11 Also P+Q=   2 − 5   2

Short Answer Type Questions-II (3 marks each) é 2 4 -6 ù ê ú 5 ú as the sum of a Q. 1. Express the matrix A = ê 7 3 êë 1 -2 4 úû



A3 – 4A2 – 3A +11I = 0, Hence find A–1.

R&U [CBSE OD Set I, II, III-2020]

[OD Set I, II, III COMPTT. 2016] 1

Sol.

These questions are for practice and their solutions are available at the end of the chapter

é9 7 5 ù ê ú A2 = ê 1 4 1 ú êë8 9 9 úû

1



MATRICES

é 28 37 26 ù ê ú A = ê10 5 1 ú 1½ êë 35 42 34 úû

é 28 37 26 ù é36 28 20 ù ê ú ê ú = ê10 5 1 ú - ê 4 16 4 ú êë 35 42 34 úû êë32 36 36 úû

LHS = A3 – 4A2 – 3A + 11I

é 3 9 6 ù é11 0 0 ù ê ú ê ú - ê6 0 -3 ú + ê 0 11 0 ú ê 3 6 9 úû êë 0 0 11úû ë 0ù ú 0ú 1 úû

0 0 0    = 0 0 0  = 0 1½ 0 0 0  Now,

A–1 = -

1 ( A 2 - 4 A - 3I ) 11

é 2 -5 -3 ù 1 ê ú 5 ú 1 = - ê -7 1 11 êë 4 1 -6 úû [CBSE Marking Scheme 2018] Detailed Solution: é1 3 2 ù ê ú A = ê 2 0 -1ú êë 1 2 3 úû 2

A2 = A×A é1 3 2 ù é1 3 2 ù ê úê ú = ê 2 0 -1ú ê 2 0 -1ú êë 1 2 3 úû êë 1 2 3 úû é9 7 5 ù ê ú = ê 1 4 1 ú êë8 9 9 úû A3 = A2×A é9 7 5 ù ê ú = ê 1 4 1 ú êë8 9 9 úû é 28 37 ê = ê10 5 êë 35 42 3 2 \ A – 4A – 3A + 11I é 28 37 ê = ê10 5 êë 35 42

é1 3 2 ù ê ú ê 2 0 -1ú êë 1 2 3 úû 26 ù ú 1ú 34 úû 26 ù é9 7 5 ù ú ê ú 1 ú - 4 ê1 4 1 ú êë8 9 9 úû 34 úû

é1 0 0 ù é1 3 2 ù ê ú ê ú -3 ê 2 0 -1ú + 11 ê0 1 0 ú ê 1 2 3 úû êë0 0 1 úû ë

é 28 - 36 - 3 + 11 37 - 28 - 9 + 0 26 - 20 - 6 + 0 ù ú = ê 10 - 4 - 6 + 0 5 - 16 - 0 + 11 1-4+3+0 ú ê êë 35 - 32 - 3 + 0 42 - 36 - 6 + 0 34 - 36 - 9 + 11úû

0 0 0  = 0 0 0  = 0   0 0 0 



A3 – 4A2 – 3A + 11I = 0 Hence Proved. 3 –1 2 –1 –1 –1 –1 A ×A – 4A ×A – 3A×A + 11IA = 0×A A2I – 4AI – 3I + 11A–1 = 0 A2 – 4A – 3I + 11A–1 = 0 11A–1 = 3I + 4A – A2 é 1 3 2 ù é9 7 5 ù é1 0 0 ù ê ú ê ú ê ú 3 ê0 1 0 ú + 4 ê 2 0 -1ú - ê 1 4 1 ú = êë0 0 1 úû êë 1 2 3 úû êë8 9 9 úû é 3 + 4 - 9 0 + 12 - 7 0 + 8 - 5 ù 11A–1 = êê 0 + 8 - 1 3 + 0 - 4 0 - 4 - 1 úú êë0 + 4 - 8 0 + 8 - 9 3 + 12 - 9 úû é -2 5 3 ù ê ú 11A–1 = ê 7 -1 -5 ú êë -4 -1 6 úû

é 2 ê - 11 ê 7 A–1 = ê ê 11 ê ê- 4 êë 11

5 11 1 11 1 11

3 ù 11 ú é 2 -5 -3 ù ú 1 ê 5ú ú 5ú = - ê -7 1 ú 11 11 êë 4 1 -6 úû ú 6 ú 11 úû

Commonly Made Error





é 28 37 26 ù é9 7 5 ù = ê10 5 1 ú - 4 ê 1 4 1 ú ê ú ê ú êë 35 42 34 úû êë8 9 9 úû é1 0 é1 3 2 ù ê ú ê - 3 ê 2 0 -1ú + 11 ê0 1 êë 1 2 3 úû êë0 0

After proving the matrix equation, students use the formula to find the inverse which is wrong.

Answering Tip



These questions are for practice and their solutions are available at the end of the chapter



3

59

Learn to find inverse from elementary operations, matrix multiplication, from equations and using formula.

60

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

COMPETENCY BASED QUESTIONS Case based MCQs

(4 marks each)

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.



I. Read the following text and answer the following questions on the basis of the same: A manufacture produces three stationery products Pencil, Eraser and Sharpener which he sells in two markets. Annual sales are indicated below Market

Products (in numbers) Pencil

Eraser

Sharpener

A

10,000

2,000

18,000

B

6,000

20,000

8,000

Q. 3. Cost incurred in market A: (A) ` 13,000 (B) ` 30,100 (C) ` 10,300 (D) ` 31,000 Ans. Option (D) is correct. Explanation: Cost incurred in market A  2.00    = [10 , 000 2 , 000 18 , 000] 1.00  0.50 

= 2.00 × 10 , 000 + 1.000 × 2 , 000 + 0.50 × 18 , 000 = ` 31, 000

Q. 4. Profits in market A and B respectively are: (A) (`15,000, `17,000) (B) (`17,000, `15,000) (C) (`51,000, `71,000) (D) (`10,000, `20,000) Ans. Option (A) is correct. Explanation: Cost incurred in market B



If the unit Sale price of Pencil, Eraser and Sharpener are `2.50, `1.50 and `1.00 respectively, and unit cost of the above three commodities are `2.00, `1.00 and ` 0.50 respectively, then, [CBSE QB 2021] 





Q. 1. Total revenue of market A:

é 2.00 ù éê 2.00 ùú = [6 , 000 20 , 000 8 , 000] ê1.00 ú = [6 , 000 20 , 000 8 , 000] ê1.00 ú 0.50 êëë0.50 úûû = [ 2.00 ´ 6 , 000 + 1.00 ´ 20 , 000 + 0.50 ´ 8 , 000] = [ 2.00 ´ 6 , 000 + 1.00 ´ 20 , 000 + 0.50 ´ 8 , 000] = [36 , 000] = [36 , 000] Profit of market A & B = total revenue of A and B – Cost increased in market A and B é A ù é 46 , 000 ù é31, 000 ù êB ú = ê 50 , 000 ú - ê36 , 000 ú ë û ë û ë û é15, 000 ù = ê ú ë17 , 000 û

(A) ` 64,000

(B) ` 60,400



(C) ` 46,000

(D) ` 40600

i.e., (`15,000, `17,000) Q. 5. Gross profit in both markets is: (A) `23,000 (B) `20,300 (C) `32,000 (D) `30,200 Ans. Option (C) is correct. Explanation: Gross profit in both markets = Profit in A + Profit in B = 15,000 + 17,000 = `32,000

Ans. Option (C) is correct. Explanation: Total revenue of  2.50    = [10 , 000 2 , 000 18 , 000] 1.50  1.00 

= 2.50 × 10 , 000 + 1.500 × 2 , 000 + 1.00 × 18 , 000 = ` 46 , 000

Q. 2. Total revenue of market B is: (A) ` 35,000 (B) ` 53,000 (C) ` 50,300 (D) ` 30,500 Ans. Option (B) is correct. Explanation: Total revenue of market B  2.50    = [6 , 000 20 , 000 8 , 000] 1.50  1.00 



= 2.50 × 6 , 000 + 1.50 × 20 , 000 + 1.00 × 8 , 000 = ` 53, 000



II. Read the following text and answer the following questions on the basis of the same:



Three schools DPS, CVC and KVS decided to organize a fair for collecting money for helping the flood victims. They sold handmade fans, mats and plates from recycled material at a cost of `25, `100 and `50 each respectively. The numbers of articles sold are given as [CBSE QB 2021]



MATRICES

DPS

CVC

KVS

Handmade fans

40

25

35

Mats

50

40

50

Plates

20

30

40

Q. 1. What is the total money (in Rupees) collected by the school DPS? (A) `700 (B) `7,000 (C) `6,125 (D) `7,875 Ans. Option (B) is correct. Explanation: The funds collected by the schools can be obtained by matrix multiplication :

é7000 ù  40 50 20   25      = ê6125 ú 25 40 30 100 ê ú    êë7875 úû 35 50 40   50 



Funds collected by school DPS = 7000



Funds collected by school, CVC = 6125



Funds collected by school KVS = 7875

Q. 2. What is the total amount of money (in `) collected by schools CVC and KVS? (A) 14,000

(B) 15,725

varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B [CBSE QB 2021]



School /Article

61

September sales (in Rupees).

é 10 , 000 20 , 000 30 , 000 ù Ramakrishan A=ê 50 , 000 30 , 000 10 , 000 úû Gurccharan ë October sales (in Rupees) é 5 , 000 10 , 000 6 , 000 ù Ramakrishan B=ê 20 , 000 10 , 000 10 , 000 úû Gurchaaran ë Q. 1. The total sales in September and October for each farmer in each variety can be represented as _______. (A) A + B (B) A – B (C) A > B (D) A < B Ans. Option (A) is correct.

(C) 21,000 (D) 13,125 Ans. Option (A) is correct. Explanation: Total amount of money collected by school = 6125 + 7875 = 14000 Q. 3. What is the total amount of money collected by all three schools DPS, CVC and KVS? (A) `15,775 (B) `14,000 (C) `21,000 (D) `17,125 Ans. Option (C) is correct. Explanation: Total amount of money collected by all school DPS, CVC and KVS = 7000 + 7875 + 6125 = 21000

Explanation: Combined sales in September and October for each farmer in each variety is given by A+B= Basmati Permal Naura

Q. 4. If the number of handmade fans and plates are interchanged for all the schools, then what is the total money collected by all schools? (A) `18,000 (B) `6,750 (C) `5,000 (D) `21,250 Ans. Option (D) is correct. Q. 5.  How many articles (in total) are sold by three schools? (A) 230 (B) 130 (C) 430 (D) 330 Ans. Option (D) is correct. Explanation: 110 + 95 + 125 = 330

(A) A + B

(B) A – B

(C) A > B

(D) A < B



(A) `100, `200 and `120

III. Read the following text and answer the following questions on the basis of the same:

Two farmers Ramakishan and Gurucharan Singh cultivate only three varieties of rice namely Basmati, Permal and Naura. The sale (in `) of these

é15, 000 30 , 000 36 , 000 ù Ramkrishan ê70 , 000 40 , 000 20 , 000 ú Gurcharan singh ë û

Q. 2. What is the value of A23? (A) 10,000

(B) 20,000

(C) 30,000

(D) 40,000

Ans. Option (A) is correct. Explanation: A23 = 10,000 Q. 3. The decrease in sales from September to October is given by _______ .

Ans. Option (B) is correct.

Explanation: Change in sales from September to October is given by A-B= Basmati Permal Naura é 5000 10 , 000 24 , 000 ù Ramkishan ê30 , 000 20 , 000 0 úû Gurcharan Singh ë

Q. 4. If Ramkishan receives 2% profit on gross sales, compute his profit for each variety sold in October. (B) `100, `200 and `130 (C) `100, `220 and `120 (D) `110, `200 and `120

62

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Ans. Option (A) is correct. Explanation: 2 2% of B = 2 ´ B 2% of B = 100 ´ B 100 = 0.02×B = 0.02×B = 0.02 = 0.02 Basmati Permal Naura Basmati Permal Naura 10 , 000 6000 ù Ramkishan é 5000 éê 5000 10 , 000 6000 ùú Ramkishan , 000 10 , 000 10 , 000 ûú Gurcharn Singh ëê 20 ë 20 , 000 10 , 000 10 , 000 û Gurcharn Singh Thus, in October Ramkishan receives ` 100, ` 200 and ` 120 as profit in the sale of each variety of rice, respectively. Q. 5. If Gurucharan receives 2% profit on gross sales, compute his profit for each variety sold in September: (A) `100, `200, `120

(B) `1000, `600, `200

(C) `400, `200, `120 (D) `1200, `200, `120 Ans. Option (B) is correct. Explanation: 2 22% % of of A A= = 2 ´ ´A A 100 100 = 0.02× 0.02× A A = = 0.02 = 0.02 Basmati Permal Permal Naura Naura Basmati



10 ,, 000 000 20 20 ,, 000 000 30 30 ,, 000000 ùù Ramkishan Ramkishan éé10 êê 50 , 000 30 , 000 10 , 000 úú Gurcharn Singh ëë 50 , 000 30 , 000 10 , 000 ûû Gurcharn Singh Basmati Basmati Per Perrrmal mal Naura Naura Ramkishan 200 400 600 ùù Ramkishan é 200 400 600 é = êê = ú Gurcharn Singh ú 1000 600 200 600 200 ûû Gurcharn Singh ëë 1000 Thus, in September Gurucharan receives ` 1000, ` 600 and ` 200 as Profit in the sale of each variety of rice, respectively. IV. Read the following text and answer the following questions on the basis of the same:

On her birthday, Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got `10 more. However, if there were 16 children more, everyone would have got `10 less. Let the number of children be x and the amount distributed by Seema for one child be y (in `).

Q. 1. The equations in terms x and y are: (A) 5x – 4y = 40 (B) 5x – 4y = 40 5x – 8y = –80 5x – 8y = 80 (C) 5x – 4y = 40 (D) 5x + 4y = 40 5x + 8y = –80 5x – 8y = –80 Ans. Option (A) is correct. Explanation: Let number of children = x Amount distributed by Seema for one child = `y Now, Total money = xy and Total money will remain the same. Given that, if there were 8 children less, everyone would have got `10 more. Total money now = Total money before (x – 8) × (y + 10) = xy Þ x(y + 10) – 8(y + 10) = xy Þ xy + 10x – 8y – 80 = xy Þ 10x – 8y – 80 = 0 Þ 10x – 8y = 80 Þ 5x – 4y = 40 Also, if there were 16 children more, everyone would have got `10 less. Total money now = Total money before (x + 16) × (y – 10) = xy Þ x(y – 10) + 16(y – 10) = xy Þ xy – 10x + 16y – 160 = xy Þ –10x + 16y – 160 = 0 Þ 10x – 16y + 160 = 0 Þ 5x – 8y = –80 Thus, required equations are: 5x – 4y = 40 ...(i) 5x – 8y = –80 ...(ii) Q. 2. Which of the following matrix equations represent the information given above?  5 4   x   40  (A)  5 8   y  =  −80      

5 (B)  5  5 (C)  5 

−4   x   40  = −8   y   80  −4   x   40  = −8   y   −80 

 5 4   x   40  (D)  5 −8   y  =  −80       Ans. Option (C) is correct. Explanation: Writing eq. (i) & eq. (ii) in matrix form, we get  5 −4   x   40   5 −8   y  =  −80 

[CBSE QB-2021]

Q. 3. The number of children who were given some money by Seema, is: (A) 30 (B) 40 (C) 23 (D) 32 Ans. Option (D) is correct. Explanation: On solving eqs. (i) & (ii) for x, we get x = 32.



MATRICES



Q. 4. How much amount is given to each child by Seema? (A) `32 (B) `30 (C) `62 (D) `26 Ans. Option (B) is correct. Explanation: On solving eqs. (i) & (ii) for y, we get y = 30 i.e., y = `30 Q. 5. How much amount Seema spends in distributing the money to all the students of the Orphanage? (A) `609 (B) `960 (C) `906 (D) `690 Ans. Option (B) is correct. Explanation: Total amount = xy = 32 × 30 = `960

Case based Subjective Questions (2 marks each) I. Read the following text and answer the following questions on the basis of the same: (Each Sub-part carries 2 marks) In a city there are two factories A and B. Each factory produces sports clothes for boys and girls. There are three types of clothes produced in both the factories type I, type II and type III. For boys the number of units of types I, II and III respectively are 80, 70 and 65 in factory A and 85, 65 and 72 are in factory B. For girls the number of units of types I, II and III respectively are 80, 75, 90 in factory A and 50, 55, 80 are in factory B.

Sol. In factory A, number of units of type I, II and III for boys are 80, 70, 65 respectively and for girls number of units of type I, II and III are 80, 75, 90 respectively. Boys Girls



I 80 80  P = II 70 75  1 III 90 65  In factory B, number of units of type I, II and III for boys are 85, 65, 72 respectively and for girls number of units of type I, II and III are 50, 55, 80 respectively. Boys Girls



I 85 50  \ Q = II 65 55  III 72 80 



\

\ X = [ 80 70 65]



I II III 1 85 65 72 ] Y =[ Now, total production of sports clothes of each type for boys = X + Y = [80 70 65] + [85 65 72] = [165 135 137] Similarly, for girls, let matrix S represents the number of units of each type produced by factory A and matrix T represents the number of units of each type produced by factory B. I II III



\ S = [ 80 75 90 ]



II III



T = [ 50 55 80 ]



Now, required matrix = S + T = [80 75 90] + [50 55 80] = [130 130 170] 1

Solutions for Practice Questions (Topic 1) Very Short Answer Type Questions



Þ





3.

Commonly Made Error

é0 1 ù é0 1 ù é1 0 ù A2 = ê úê ú=ê ú 1 1 ë1 0 û ë1 0 û ë0 1 û [CBSE SQP Marking Scheme 2020-21]

Students fail to find the matrix A correctly.

Answering Tip







I

é0 1 ù A = ê ú ë1 0 û

1

Q. 2. Find the total production of sports clothes of each type for boys and girls. Sol. Let matrix X represent the number of units of each type produced by factory A for boys and matrix Y represents the number of units of each type produced by factory B for boys. I II III



Q. 1. Write the matrices P and Q, if P represents the matrix of number of units of each type produced by factory A for both boys and girls; and Q represents the matrix of number of units of each type produced by factory B for both boys are girls.

63

Practice questions on matrix formation and get familiar with the elemental notations.

6.

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

 1 2 −2 A’ =  2 1 2  and getting x = – 2  2 x −1

2x = – 4 4 x = 2 x = – 2 0 3  A =    2 −5



64

½+½

So,



[CBSE Marking Scheme 2018]

12.

Detailed Solution:

æ 1 2 -2ö A' = ç 2 1 2 ÷ ç ÷ çè 2 x -1÷ø

Since, AA' = 9I æ 1 2 2 ö æ 1 2 -2ö æ 1 0 0ö ç 2 1 x ÷ ç 2 1 2 ÷ = 9 ç 0 1 0÷ ç ÷ç ÷ ç ÷ çè -2 2 -1÷ø çè 2 x -1÷ø çè 0 0 1÷ø æ 1 + 4 + 4 2 + 2 + 2 x -2 + 4 - 2ö æ 9 0 0ö ç ÷ 2 -4 + 2 - x÷ = ç 0 9 0÷ ç 2 + 2 + 2x 4 + 1 + x ç ÷ çè 0 0 9÷ø çè -2 + 4 - 2 -4 + 2 - x 4 + 4 + 1 ÷ø 2x + 4 0 ö æ 9 æ 9 0 0ö ç ÷ 2 ç 0 9 0÷ 2 4 5 2 x + x + x = ½ ç ÷ ç ÷ çè 0 0 9÷ø çè 0 9 ÷ø -x - 2 Now, x2 + 5 = 9 x2 = 9 – 5 x2 = 4 Also,

x =

4

x = ± 2 2x + 4 = 0





0 kA =  2k



But given

 0 4 a kA =    −8 5b 



\

0 2k 

5 1 2      A 2 =9 2 5     0 1 2   

2.

5  A  5 A=9  0   5    1   5  2

½

½

   

1

1 2  10 0 5     2 5   10 5 15     1 2   5 5 0  1 3   7 10   4 2   1 [CBSE Marking Scheme, 2019]

Topper Answer, 2020



3k   0 4 a =   −5k  −8 5b 

Short Answer Type Questions-I

Detailed Solution:

Sol.

3k  −5k 

On equating individual terms, 2k = – 8 Þ k = – 4 3k = 4a 3 × (– 4) = 4a Þ a = – 3



Given,

æ 1 2 2ö A = ç 2 1 x ÷ ç ÷ çè -2 2 -1÷ø

½

65





MATRICES

Short Answer Type Questions-II

300 250 400

100 ù ú 75 ú 150 úû

é 50 ù and B = ê 20 ú ê ú êë 40 úû

1

Now, the cost incurred by the organisation for three villages can be represented as. é 400 300 100 ù é 50 ù AB = ê 300 250 75 ú ê 20 ú 1 ê úê ú êë 500 400 150 úû êë 40 úû é 400 ´ 50 + 300 ´ 20 + 100 ´ 40 ù = êê300 ´ 50 + 250 ´ 20 + 75 ´ 40 úú êë 500 ´ 50 + 400 ´ 20 + 150 ´ 40 úû



Students get wrong when multiplying the matrices together.

Answering Tip



é 20000 + 6000 + 4000 ù é30000 ù = ê15000 + 5000 + 3000 ú = ê 23000 ú ê ú ê ú êë 25000 + 8000 + 6000 úû êë39000 úû



1

Thus, the total cost incurred by the organisation for



é 400 A = ê 300 ê êë 500

Commonly Made Error

3. Let A be the matrix that represent the number of attempts made in three villages X, Y and Z and let B be the matrix that represent the cost for each mode per attempt. Then, the matrices A and B can be represented as House Calls Letters Announcements

the three villages X, Y and Z separately are ` 30000, ` 23000 and ` 39000, respectively. 1

The answer can be obtained by separately multiplying the matrices also.

Long Answer Type Questions 2. Clearly order of A is 2 × 3 1 a b c   1 Let AA==a b c  dd ee f f 10 22 −−11  a b c  −−11 −−88 −−10  1 00 a b c ==11 −−22 −−55  So 1  d e f   −−33 44  d e f  99 2222 15 15  gives 2 a − d = −1, 2b − e = −8 , 2 c − f = −10   a = 1, b = −2 , c = −5  2 ⇒ d = 3, e = 4 , f = 0 1  1 −2 −5  Thus A =  1  3 4 0 

Solutions for Practice Questions (Topic 2) Very Short Answer Type Questions 4.

Short Answer Type Questions-I

0 0  0 0  is a 2 × 2 matrix which is symmetric as well as skew symmetric matrix. 1 [CBSE Marking Scheme 2014]

Topper Answer, 2018 1.

66

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

1 -5 ù é 6 1ê ú -4 ú which is symmetric 1 4 2ê êë -5 -4 4 úû . é 3 - 3 -2 - 3 -4 + 1ù 1 1ê ú ( A - A ') = ê 3 + 2 -2 + 2 -5 - 1ú 2 2 êë -1 + 4 1 + 5 2 - 2 úû 0 −5 −3  1 =  5 0 −6  which is skew-symmetric. 2 3 6 0  1 ( A − A ') 2 −5  6 1 0 −5 −3  1 1 ∴ A = 1 −4 −4  +  5 0 −6  2 2  −5 −4 4  3 6 0  =

Short Answer Type Questions-II

2. We know that 1 1 A = ( A + A ') + ( A − A ') 2 2

1 ( A + A ') is symmetric matrix 2 (A – A') is skew symmetric matrix. 3 é 3 é 3 -2 -4 ù ê ú ê Now, A = ê 3 -2 -5 ú Þ A ' = ê -2 -2 êë -1 êë -4 -5 1 2 úû −2 + 3 −4 − 1 3 + 3 1 1 ∴ ( A + A ') = 3 − 2 −2 − 2 −5 + 1  2 2  −1 − 4 1 − 5 2 + 2  Here,

and -1ù ú 1ú 2 úû

1

1

1

1 / 2 −5 / 2  0 −5 / 2 −3 / 2  3    ⇒ A = 1 / 2 −2 −2  +  5 / 2 0 −3   −5 / 2 −2 2  3 / 2 3  0  é 3 -2 -4 ù ê ú = ê 3 -2 -5 ú êë -1 1 2 úû



1

REFLECTIONS •

•

Matrices are used in cryptography. In cryptography, the process of encryption is carried out with the help of invertible key. In this method, matrices are used. Wireless signals are modelled and optimized using matrices.

• •

In the realm of graphics, matrices are used to project three-dimensional images into two-dimensional planes. Matrices are applied in the study of electrical circuits, quantum mechanics and optics, in the calculation of battery power outputs and resistor conversion of electrical energy into another useful energy.



CHAPTER

4



Syllabus

DETERMINANTS

Determinant of a square matrix (up to 3 × 3 matrices), minors, cofactors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples. Solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.





In this chapter you will study  Determinant

of matrix  Minors, Cofactors of matrix  Area of triangle using determinant  Inverse of matrix using adjoint method  Solution of Linear equations in two or three variables  Consistency or inconsistency of system of linear equations.

Topic-1



List of Topics Topic-1: Determinants, Minors & Co-factors Page No. 67

Topic-2: Solutions of System of Linear Equations  Page No. 76

Determinants, Minors & Co-factors Concepts Covered 



Determinant value of a matrix,  Co-factor and Minor of a matrix

Inverse of matrix using Adjoint method,  Area of triangle with the help of determinant

Revision Notes Determinants, Minors & Co-factors (a) Determinant: A unique number (real or complex) can be associated to every square matrix A = [aij] of order m. This number is called the determinant of the square matrix A, where aij = (i, j)th element of A.

a b  For instance, if A =  c d  then, determinant   a b of matrix A is written as |A| = c d = det (A) and its value is given by ad – bc.



(b) Minors: Minors of an element aij of a determinant (or a determinant corresponding to matrix A) is the determinant obtained by deleting its ith row and jth column in which aij lies. Minor of aij is denoted by Mij. Hence, we can get 9 minors corresponding to the 9 elements of a third order (i.e., 3 × 3) determinant.



(c) Co-factors : Cofactor of an element aij, denoted by Aij, is defined by Aij = (–1)(i + j) Mij, where Mij is minor of aij. Sometimes Cij is used in place of Aij to denote the co-factor of element aij.

68

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII



DETERMINANTS

1. ADJOINT OF A SQUARE MATRIX: Let A = [aij] be a square matrix. Also, assume B = [Aij], where Aij is the cofactor of the elements aij in matrix A. Then the transpose BT of matrix B is called the adjoint of matrix A and it is denoted by “adj (A)”. To find adjoint of a 2 × 2 matrix: Follow this, A = a b   d −b   c d  or adj A =  − c a  .

For example, consider a square matrix of order 3 as



1 2 3 A =  2 3 4  , then in order to find the adjoint 2 0 5  

matrix A, we find a matrix B (formed by the co-factors of elements of matrix A as mentioned above in the definition)  15 −2 −6  i.e., B =  −10 −1 4  . Hence, adj A = BT =  −1 2 −1  

2. SINGULAR MATRIX AND NON-SINGULAR MATRIX: Scan to know more about (a) Singular matrix: A square matrix this topic A is said to be singular if |A|= 0 i.e., its determinant is zero. 1 2 3  e.g. A =  4 5 12  1 1 3  Inverse of a   = 1(15 – 12) – 2(12 – 12) + 3(4 – 5) =3–0–3=0 \ A is singular matrix.  −3 4  B =  = 12 – 12 = 0  3 −4  \ B is singular matrix. (b) Non-singular matrix: A square matrix A is said to be non-singular if |A| ≠ 0 . 0 1 1 e.g. A =  1 0 1  1 1 0   = 0 (0 – 1) – 1(0 – 1) + 1(1 – 0) =0+1+1=2≠0 \ A is non-singular matrix.

3.



4.

 15 −10 −1  −2 −1 2   −6 4 −1  





3 × 3 Matrix using Adjoint Scan to know more about this topic



•

A square matrix A is invertible if and only if A is non-singular. ALGORITHM TO FIND A–1 BY DETERMINANT METHOD: STEP 1: Find |A|. STEP 2: If |A| = 0, then, write “A is a singular matrix and hence not invertible”. Else write “A is a non-singular matrix and hence invertible”. STEP 3: Calculate the co-factors of elements of matrix A. STEP 4: Write the matrix of co-factors of elements of A and then obtain its transpose to get adj.A (i.e., adjoint A). STEP 5: Find the inverse A–1 of = A by using the relation 1 ( adj A ) . | A|

Scan to know more about this topic

Matrices Determinant Part - 2 Scan to know more about this topic

Matrices Determinant Part - 3

AREA OF TRIANGLE: Area of a triangle whose vertices are (x1, y1), (x2 , y2) and (x3, y3) is given by, x y1 1 sq. units 1 1 D = x2 y2 1 2 x y 1 3 3 •

Since area is a positive quantity, we take absolute value of the determinant. • If the points (x1, y1), (x2 , y2) and (x3, y3) are collinear, then Δ = 0. • The equation of a line passing through the points (x1, y1) and (x2, y2) can be obtained by the expression given here: x y 1 x 1 y1 1 = 0 x y2 1 2

Key Facts Matrices Determinant Part - 1

69

• In mathematics, the determinant is a scalar value that is a function of the entries of a square matrix. • There are 10 main properties of determinants which include reflection property, all-zero property, proportionality or repetition property, switching property, scalar multiple property, sum properly, invariance properly, factor properly, triangle properly and w-factor properly.

70

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions

Q. 1. Three points P(2x, x + 3), Q(0, x) and R(x + 3, x + 6) are collinear, then x is equal to: (A) 0

(B) 2

(C) 3

(D) 1

Ans. Option (D) is correct. Explanation: As points are collinear Þ Area of triangle formed by 3 points is zero. 1 ( x1 − x 2 ) ( x 2 − x 3 ) Þ =0 2 ( y1 − y 2 ) ( y 2 − y 3 ) 1 ( 2 x − 0 ) {0 − ( x + 3)} Þ =0 2 ( x + 3 − x ) {x − ( x − 6 )} 2 x −( x + 3) =0 3 −6

Þ –12x + 3(x + 3) = 0 Þ –12x + 3x + 9 = 0 Þ –9x = –9 Þ x = 1 Q. 2. If Cij denotes the cofactor of element Pij of the  1 −1 2  matrix P =  0 2 −3 , then the value of C31.C23    3 2 4  is: (A) 5 (C) –24

(B) 24 (D) –5 [CBSE Term-I 2021]

Ans. Option (A) is correct. Explanation: Here, C31 = ( −1)3 + 1 and C23 = ( −1)2 + 3

the value of a is: (A) ±3

(B) –3

(C) ±1

(D) 1 [CBSE Term-I 2021]

[CBSE Term-I 2021]

Þ

 α −2  Q. 4. If for the matrix A =   , |A3| = 125, then  −2 α 

−1 2 = 3 – 4 = –1 2 −3 1 −1 = –(2 + 3) = –5 3 2

Thus, C31.C23 = (–1)(–5) = 5 Q. 3. If A is a square matrix of order 3 and |A| = –5, then |adj A| is: (A) 125 (B) –25 (C) 25 (D) ±25 [CBSE Term-I 2021] Ans. Option (C) is correct. Explanation: We know that, |adj A| = |A|n – 1 where n is the order of the matrix \ |adj A| = [5]3 – 1 = (–5)2 = 25

Ans. Option (A) is correct.  α −2  A =    −2 α 

Explanation:

Þ |A| = a2 – 4 Also, given |A3| = 125 Þ |A|3 = 125 Þ |A| = 5 Þ a2 – 4 = 5 Þ a2 = 9 Þ a = ±3

...(i)

[from eq. (i)]

2 0 0 Q. 5. The inverse of the matrix X =  0 3 0  is:    0 0 4  1 2  (A) 24  0  0 

(C)

0 1 3 0

 0  0  1  4 

1 0 0 1 0 1 0  (B)  24  0 0 1  1 2  (D)  0   0 

2 0 0  1   0 3 0 24   0 0 4 

0 1 3 0

 0  0  1  4 

[CBSE Term-1 2021] Ans. Option (D) is correct. Explanation: The inverse of a diagonal matrix is obtained by replacing each element in the diagonal with its reciprocal. Since,

2 0 0    X =  0 3 0   0 0 4 

1 2  –1 Therefore X =  0  0 

0 1 3 0

 0  0  1  4 



DETERMINANTS

On expanding along R1



Q. 6. If A is a 3 × 3 matrix such that |A| = 8, then |3A| equals (A) 8 (B) 24 (C) 72 (D) 216 [CBSE Delhi Set-I 2020] Ans. Option (D) is correct. Explanation: Here |A| = 8 Then |3A| = 33|A| = 27 × 8 = 216 Q. 7. If A is skew symmetric matrix of order 3, then the value of |A| is: (A) 3 (B) 0 (C) 9 (D) 27 [CBSE Delhi Set-III 2020] Ans. Option (B) is correct. Explanation: Determinant value of skew symmetric matrix is always '0'. 2 3 Q. 8. If x x 4 9

2 x  3  0 , then the value of x is: 1

(A) 3 (C) – 1

(B) 0 (D) 1 [CBSE OD Set-I 2020]

Ans. Option (C) is correct. Explanation:



71

2(x – 9x) – 3(x – 4x)+ 2(9x – 4x) + 3 = 0 2(–8x) – 3(–3x) + 2(5x) + 3 = 0

– 16x + 9x + 10x + 3 = 0

3x + 3 = 0 3x = –3 x =

3 3

x = –1

a 0 0    Q. 9. If A   0 a 0  , then det (adj A) equals:  0 0 a 

(A) a27 (C) a6

(B) a9 2 (D) a [CBSE OD Set-III 2020]

Ans. Option (C) is correct. Explanation: éa 0 0ù A = ê0 a 0 ú ê ú êë0 0 a úû

Det(A) = a(a × a – 0 × 0) – 0 + 0 = a3 Det(adj A) = (a3)2 = a6



2 3 2 x x x +3 =0 4 9 1

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. If A a non-singular square matrix of order 3 and A2 = 2A, then find the value of |A|. U [CBSE Delhi Set-II 2020]

Topper Answer, 2020

é 3 -4 ù Q. 2. For A = ê ú write A–1. ë 1 -1 û R [CBSE Delhi Set-I 2020]

é 1 -2 ù Q. 3. Find the co-factors of all the elements of ê ú. ë4 3 û R [CBSE Delhi Set-II 2020] Sol. A11 = 3, A12 = –4, A21 = 2, A22 = 1 1 [CBSE Marking Scheme 2020]

Sol.





Commonly Made Error Students may find minors instead of co-factors.





These questions are for practice and their solutions are available at the end of the chapter



Answering Tip Learn to differentiate between minors and co-factors.



72

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

é 2 -1ù Q. 4. Find adj A, if A = ê ú. ë4 3 û

Þ \



R [CBSE OD Set-I, II, III 2020]

Q. 9. If A and B are square matrices of the same order 3,

Q. 5. Given that A is a square matrix of order 3 × 3 and |A| = –4. Find |adj A|.

(x – 1) (x +1) = 0 Þ x = ± 1 Value of A = ± 1

such that A = 2 and AB = 2I, write the value of B . U [CBSE Delhi Set-I, 2019]

R&U [CBSE SQP 2020-21]

Sol. AB = 2I Þ |AB| = |2I| Þ |A| · |B| = 23|I| ½ Þ 2 × |B| = 8 Þ |B| = 4 ½ [CBSE Marking Scheme, 2019]

Detailed Solution: |adj A| = |A|n–1 where n = order of matrix A |adj A| = (–4)3–1 =(–4)2 = 16

Detailed Solution: Given,

Q. 6. Let A = [aij] be a square matrix of order 3 × 3 and |A| = –7. Find the value of a11A21 + a12A22 + a13A23 where Aij is the cofactor of element aij. [CBSE SQP 2020-21] Sol. 0 Q. 7. Check whether (l+ m + n) is a factor of the determinant

l+m m+n n+l n l m 2 2 2

Sol. Apply R1 → R1 + R2

l+m+n m+n+l n+l+m n l m 2 2 2







= 2 (l + m + n) n

1 1



or







Since,



\





3 5 0 0  .  

Q. 11. If for any 2 × 2 square matrix A, A(adj A) = 8 0  0 8  , then write the value of |A|.  

1 [CBSE SQP Marking Scheme 2020]

R [OD Set I 2017]

Sol. |A| = 8 [CBSE Marking Scheme 2017] Detailed Solution:

Topper Answer, 2017

U [CBSE OD Set-I, 2019]

\     |A| = 1 or |A| = –1

B = AB = 8 = 4 A 2

A [CBSE OD Set-II, 2019]

1

Sol. |A'| |A| = |I| ⇒ |A|2 = 1

AB = A B

0 −1 Q. 10. Find |AB|, if A = 0 2  and B =  

Q. 8. If A is a square matrix satisfying A’ A = I, write the value of |A|.

AB = 2 (4 – 0) – 0 + 0 = 8

\

[given, A = 2]

l m ; yes (l + m + n) is a 1 1 1

factor.

AB = 2I 1 0 0   2 0 0      AB = 2 0 1 0 = 0 2 0  0 0 1 0 0 2

or not. Give

A [CBSE SQP-2020]

reason.



Sol. |adj A| = (–4)3–1 = 16 1 [CBSE Marking Scheme 2020]

½

Sol.

½

[CBSE Marking Scheme, 2019]

 Detailed Solution:



Let the value of A = x Since, A = A' and I = 1 Given, AA’ = I AA' = I \ A A' = I [ AA' = A A' ] Þ Þ x.x = 1 Þ (x2 – 1) = 0

Q. 12. Find the maximum value of 1 1 1 1 1 + sin θ 1 1 1 1 + cos θ R&U [Delhi Set I, II, III 2016]



These questions are for practice and their solutions are available at the end of the chapter



DETERMINANTS

Sol. Let the maximum value be 1 1 1 1 D = 1 1 + sin θ 1 1 1 + cos θ Expanding, we get D = sin q . cos q 2 sin θ.cos θ 1 = = sin 2θ ½ 2 2 1 1 \ Maximum value = × 1 = 2 2 1 = [as − 1 ≤ sin 2θ ≤ 1] ½ 2

Q. 14. Evaluate x if : 2 4 = 2 x 6 5 1



R&U [Delhi Set I, II, III Comptt. 2016]

Sol.

4 . x

73

1

2 – 20 = 2x2 – 24 x = ± 3

or

[CBSE Marking Scheme 2016]  4 2 5 Q. 15. Given A =  2 0 3  , write the value of det.  - 1 1 0  (2AA–1). A [Outside Dec. Set I, II, III Comptt. 2016]

x + 3 −2 Q. 13. If x Î N and −3 x 2 x = 8, then find the value of x. R&U [O.D. Set I, II, III 2016]

Sol.

|2AA–1| = (2)3

[Q AA–1 = I]

= 8

1

[CBSE Marking Scheme 2016]

Q. 16. If A is a square matrix such that |A| = 5, write the value of |AAT|. Sol.

R&U [OD Set II 2016]

Topper Answer, 2016



 −3 2  Q. 1. Find the inverse of the matrix  .  5 −3 Hence, find the matrix P satisfying the matrix  −3 2   1 2  equation P  = .  5 −3  2 −1



R&U [S.Q.P. 2017-18]  Q. 2. If A is a skew-symmetric matrix of order 3, then prove that det A = 0. R&U [OD Set I 2017]

Sol. Any skew symmetric a é0 ê A = ê - a 0 êë - b - c

matrix of order 3 is bù ú cú 0 úû



or Þ | A | = - a( bc ) + a( bc ) = 0 1 Since A is a skew-symmetric matrix \ AT = –A   |AT| = |–A| = (–1)3. |A| ½



or



or



Short Answer Type Questions-I (2 mark each)



or

|A| = – |A| 2|A| = 0 or |A| = 0.

½

[CBSE Marking Scheme 2017] Detailed Solution:

Topper Answer, 2017 Sol.

Q. 3. If A and B are square matrices of order 3 such that

|A | = – |A| T

These questions are for practice and their solutions are available at the end of the chapter

|A| = –1, |B| = 3, then find the value of |2AB|. R&U [Foreign 2017]

74

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

p 2 3 Q. 4. If A =   and |A | = 125, then find the values 2 p R [CBSE OD Set-II, 2019]

of p.





Sol. |A| = p2 – 4 ½ |A3| = 125 ⇒ |A|3 = 125 ⇒ |A| = 5 1 \ p2 – 4 = 5 ⇒ p = ± 3  ½  [CBSE Marking Scheme, 2019] Detailed Solution: p 2 3 A =   and |A | = 125 2 p



Given,



Since, |An| = |A|n \ |A3| = |A|3



Now,



According to given condition, |A3| = 125 Þ |A|3 = 125 Þ (p2 – 4)3 = 125 Þ (p2 – 4)3 = 53 Þ p2 – 4 = 5 Þ p2 = 9 Þ p = ± 3 Hence, values of p = ± 3.

|A| =

p 2 = p2 − 4 2 p





Commonly Made Error Some students find A3 first and then take its determinant which is time consuming.





Answering Tip Learn all the properties of determinants thoroughly.

Q. 5. If A is a square matrix of order 3 such that A2 = 2A, then find the value of |A|. R&U [CBSE SQP 2020-21]

Sol. Þ Þ

A2 = 2A |AA| = |2A| |A| |A| = 8|A| ( |AB| = |A| |B| and |2A| = 23|A| ) ½ Þ |A| (|A| – 8) = 0 1 Þ |A| = 0 or 8 ½ [CBSE SQP Marking Scheme 2020-21]







Commonly Made Error Students make mistakes in applying the property |kA|= kn|A|. Instead they take it as |kA|= k|A|.



Answering Tip Practice more problems involving properties of determinants.

Long Answer Type Questions (5 & 6 mark each)  −1 −2 −2  1 −2  Q. 1. Find the adjoint of the matrix A =  2  2 −2 1  and hence show that A(adj A) = |A|I3.  −1 −2 −2  1 −2  Sol. We have, A =  2  2 −2 1  Let Cij be the cofactor of the element aij of |A|. Now, cofactros of |A| are 1 −2 C11 = (–1)1+1 −2 1 = 1 – 4 = –3 2 −2 C12 = (–1)1+2 2 1 = –2(2 + 4) = –6 2 1 C13 = (–1)1+3 2 −2 = – 4 – 2 = – 6 −2 −2 C21 = (–1)2+1 −2 1 = – (–2 – 4) = 6 C22 = (–1)2+2

−1 −2 2 1 = –1 + 4 = 3

C23 = (–1)2+3

−1 −2 2 −2 = –(2 + 4) = –6

C31 = (–1)3+1

−2 −2 1 −2 = 4 + 2 = 6

C32 = (–1)3+2

−1 −2 2 −2 = –(2 + 4) = –6

C33 = (–1)3+3

−1 −2 2 1 = –1 + 4 = 3





Now, the adjoint of the matrix A is given by 6  −3 6  C11 C21 C31  3 −6  adj(A) = C12 C22 C32  =  −6  −6 −6 C  3  13 C23 C33   Now, |A| =

−1 −2 −2 2 1 −2 2 −2 1

= –1(1 – 4) + 2(2 + 4) – 2(– 4 – 2) = –1(–3) + 2(6) – 2(–6) = 3 + 12 + 12 = 27 6  −1 −2 −2   −3 6 3 −6  1 −2   −6 and A·(adj A) =  2  2 −2 3 1  −6 −6 

These questions are for practice and their solutions are available at the end of the chapter



DETERMINANTS

75

[interchange rows and columns] 0 0 1  1 1 A–1 = adj(A) = 0 1 2  | A| 1  1 1 1 

 3 + 12 + 12 −6 − 6 + 12 −6 + 12 − 6  =  −6 − 6 + 12 12 + 3 + 12 12 − 6 − 6   −6 + 12 − 6 12 − 6 − 6 12 + 12 + 3  



Now,

 27 0 0   1 0 0 =  0 27 0  − 27 0 1 0   0 0 27  0 0 1    



Þ



which is the required inverse of given matrix A.



= 27I3 = |A|I3

Q. 2. Find the inverse of following matrices. [NCERT]  1 −1 1 A =  2 −1 0     1 0 0  Sol. Let

 1 1 1  A =  2 1 0     1 0 0  1 0 1 1 1 1 2 1 0 0 0 0 1 0



Then |A| =



= 1(0) – 2(0) + 1(0 + 1) [expanding along C1] = 1 ¹ 0 Thus, A is a non-singular matrix, so A–1 exists. Now, co-factors corresponding to each element of determinant A are 1  1 1 0  1( 0  0 )  0 C11 = ( 1) 0 0





1 2 C12 = ( 1)

2 0  1( 0  0 )  0 0 0





1 3 C13 = ( 1)

2 1  1( 0  1)  1 1 0





2 1 C21 = ( 1)

1 1  1( 0  0 )  0 0 0





2+2 C22 = ( −1)

1 1 = 1( 0 − 1) = ( −1) 1 0





23 C23 = ( 1)

1 1  1( 0  1)  1 1 0





3 1 C31 = ( 1)

1 1  1( 0  1)  1 1 0





32 C32 = ( 1)

1 1  1( 0  2 )  2 2 0





33 C33 = ( 1)

1 1  1( 1  2 )  1 2 1



Thus, matrix of co-factors C11 C12 C13  0 0 1  = C21 C22 C23   0 1 1 C31 C32 C33   1 2 1 





C11 C12  and adj(A) = C21 C22 C31 C32

C13  0 0 1  C23   0 1 1 C33   1 2 1 

0 0 1  A–1 = 0 1 2   1 1 1 

Q. 3. Find the area of the triangle, whose vertices are (3, 8), (–4, 2) and (5, 1). Sol. Given vertices of a triangle are (3, 8), (–4, 2) and (5, 1). Let (x1, y1) = (3, 8), (x2, y2) = (–4, 2) and (x3, y3) = (5, 1) x1 y1 1 3 8 1 1 1 Then, area of triangle = x 2 x 2 1 = − 4 2 1 2 2 x3 y3 1 5 1 1

=

4 1 4 2  1 2 1 8 1 3  5 1  5 1 2  1 1



=

1 [3( 2  1)  8( 4  5)  1( 4  10 )] 2



=

1 61 [3  72  14] = sq units 2 2

Q. 4. If the area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq units, then find the value of k. Sol. Given area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq units. −3 0 1 −3 0 1 1 We have, 3 0 1 = ±9 Þ 3 0 1 = ±18 2 0 k 1 0 k 1 Þ –3(0 – k) – 0 + 1(3k – 0) = ±18 Þ 3k + 3k = ±18 Þ 6k = ±18 Þ k = ±3 Q. 5. Find the equation of line joining P(11, 7) and Q(5, 5) using determinants. Also, find the value of k, if R(–1, k) is the point such that area of DPQR is 9 sq m. Sol. Let A(x, y) be any point on line PQ. Then, the points P, Q and A are collinear. 11 7 1 So, 5 5 1 =0 x y 1 Applying R2 ® R2 – R1 and R3 ® R3 – R1, we get 11 7 1 −6 −2 0 = 0 x − 11 y − 7 0 Expanding determinant along C3, we get −6 −2 1 =0 x − 11 y − 7 Þ  –6(y – 7) + 2(x – 11) = 0 Þ  –6y + 42 + 2x – 22 = 0 Þ  2x – 6y + 20 = 0

76



Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Þ  x – 3y + 10 = 0 [dividing by 2] which is the required equation of line joining points P and Q. Now, according to the question, Area of DPQR = 9 sq m 11 7 1 1 \  5 5 1 = ±9 2 −1 k 1  area of a triangle with vertices (x1 , y1 ),    x1 y1 1   1 (x 2 , y 2 ) and (x3 , y3 ) = x 2 y 2 1  2   x3 y3 1   Applying R2 ® R2 – R1 and R3 ® R3 – R1, we get

Topic-2



7 1 11 1 −6 −2 0 = ±9 2 −12 k − 7 0 Expanding the determinant along C3, we get −2 1 −6 = ±9 2 −12 k − 7 1 Þ [ 6 k  42  24] = ±9 2

(–6k + 18) = ±18 Þ –6k = 18 ± 18 18  18 For positive sign, k = =0 6 Þ

−18 − 18 =6 −6



and for negative sign, k =



Hence, the required values of k are 0 and 6.

Solutions of System of Linear Equations Concepts Covered



Unique Solution,  Consistent System,  Inconsistent System

Revision Notes SOLVING SYSTEM OF EQUATIONS BY MATRIX METHOD [INVERSE MATRIX METHOD]



(a) Homogeneous and Nonhomogeneous system : A system of equations AX = B is said to be a homogeneous system if B = O. Otherwise it is called a non-homogeneous system of equations. a1x + b1y + c1z = d1, a2x + b2y + c2z = d2, a3x + b3y + c3z = d3

Matrices – System of Linear Equations (Part 1)

STEP 1 : Assume  a1 A =  a2 a  3

b1 b2 b3

c1  c2  , B = c3 

Key Words

Scan to know more about this topic

 d1  d2  and X = d   3

x y . z  

Consistent System: A system is considered to be consistent if it has atleast one solution. Inconsistent System: If a system has no solution, it is said to be inconsistent. In order to check proceed as follow: Þ Find (adj A) B. Now, we Scan to know more may have either (adj A) B ¹ about this topic O or (adj A) B = O. • If (adj A)B = O, then the given system may be consistent or inconsistent. To check, put z = k in Matrices – System the given equations and of Linear Equations (Part 2) proceed in the same manner in the new two variables system of equations assuming di – cik, 1 £ i £ 3 as constant. • And if (adj A) B ¹ O, then the given system is inconsistent with no solutions.





STEP 2 : Find |A|. Now there may be following situations : (i) |A| ¹ 0 Þ A–1 exists. It implies that the given system of equations is consistent and therefore, the system has unique solution. In that case, write AX = B Mnemonics-1 Þ X = A–1B   1 −1 Inverse of a Square Matrix ( adj A )  where A = A  

Then by using the definition of equality of matrices, we can get the values of x, y and z. (ii) |A| = 0 Þ A–1 does not exist. It implies that the given system of equations may be consistent or inconsistent.



DETERMINANTS

77

≠0 (Zero)

Mnemonics-2 Singular Matrix A square matrix is said to be singular matrix if determinant of matrix denoted by |A| is zero otherwise it is non pingular matrix Inverse Of a Matrix

"If Determined Artist is Not Optimistic then ADJust Below International Adjoint

By

Musicians"

Determinant

Determinant

"a Determined Artist Can become a Singer, if he is Optimistic.

A–1 =

"a Determined Artist Can Never be Singer Non Singular if | A | = O, then A is Singular Otherwise, A is non-Singular

Matrix

"A is non-singular i.e. | A | ≠ 0 then

(Zero)

if he is Not Optimistic

Inverse

1 . (adj A) |A|

Interpretation : Singular & Non Singular Matrix if |A| = 0, then A is singular. Otherwise A is nonsingular Inverse of a Matrix– Inverse of a Matrix exists if A is non- singular i.e |A| ¹ 0,and is given by A

−1

=

1

|A|

adj A

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions

Q. 1. The system of linear equations 5x + ky = 5, 3x + 3y = 5; will be consistent if: (A) k ¹ –3 (B) k = –5 (C) k = 5 (D) k ¹ 5 [CBSE Term-1 2021] Ans. Option (D) is correct. Explanation: We have, 5x + ky – 5 = 0 and 3x + 3y – 5 = 0 For consistent system 5 k ¹ 3 3 Þ k ¹ 5 Q. 2. Consider the system, each consisting of m linear equations in n variables. (i) If m < n, then all such system have a solution (ii) If m > n, then none of these systems has a solution (iii) If m = n, then there exists a system which has a solution which one of the following is correct? (A) (i), (ii) and (iii) are true (B) only (ii) and (iii) are true (C) only (iii) is true (D) none of them is true

Ans. Option (C) is correct. Q. 3. The solution to the system of equation  2 5  x   2   −4 3  y  =  −30  is:       (A) 6, 2 (C) –6, –2 Ans. Option (D) is correct. Explanation:

(B) –6, 2 (D) 6, –2

 2 5  x   2   −4 3   y  =  −30       2x + 5y = 2 –4x + 3y = –30 On solving eqs. (i) and (ii), we get

...(i) ...(ii)

x = 6 and y = –2 Q. 4. The pair of equations 3x – 5y = 7 and 6x – 10y = 14 have: (A) a unique solution (B) infinitely many solution (C) no solution (D) two solutions Ans. Option (B) is correct. Explanation: Given equations are: 3x – 5y = 7 ...(i) and 6x – 10y = 14 or 3x – 5y = 7 ...(ii) Equations (i) and (ii) are same. Hence it will have infinitely many solutions.

78

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Q. 5. For what value of k system of equation 3x + 5y = 0 and kx + 10y = 0 has a non-zero solution? (A) 2 (B) 2 (C) 6 (D) 8 Ans. Option (C) is correct.

Explanation: for a non-zero solution 3 5   k 10  = 0  

Þ



30 – 5k = 0 Þ k = 6

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. For what values of k, the system of linear equations x + y + z = 2 2x + y – z = 3 3x + 2y + kz = 4 has a unique solution ? A [Outside Delhi 2016] Q. 2. If the system of equations : x – ky – z = 0, kx – y – z = 0, x + y – z = 0 has a non-zero solutions, then find the possible values of k Sol. For the given homogeneous system to have nonzero solution determinant of coefficient matrix should be zero. 1 − k −1 k −1 −1 = 0 1 1 −1



2 – k2 + k – k – 1 = 0

Þ

k2 = 1 Þ k = ±1

1

y  x + y Þ  2 x x − y  2x + y = 3 and 3x + y = 2   On solving above equations, we get x = –1 and y = 5.

2x + 3y = 8, 7x – 5y + 3 = 0, 4x – 6y + l = 0

5 4 =    −2 −6 

1



2A–1 = 9I – A. 2 3  Q. 2. If A =  , be such that A–1 = kA, then find  5 −2  the value of k. A [Comptt. Set I, II, III, 2018] Sol. Finding

Sol. We know that if XAY = I, then A = X–1Y–1 = (YX)–1 In this case,



 −3 2   2 1  YX =     5 −3 7 4 

½



Þ



Þ

−1



½

Short Answer Type Questions-I (2 marks each)



4 – 12 + l = 0

 2 1   −3 2   1 0  Q. 4. If   A =  , then find matrix A.  7 4   5 −3   0 1



½

5  4 5 4 A2 =    −2 −6  − 2 − 6   

 6 −10  =    4 −26 

Solving first two equations, we get x = 1, y = 2 substituting this value in third equation, we get l = 8

5   −1 + 5 A =  ( − ) − 1 − 5 2 1 

\

 2 −3 Q. 1. Given A =  , compute A–1 and show that  −4 7 

Sol. Given system of equations :

Þ

2 3  −1 =  2     

Sol. Here,

Q. 3. Find the value of l, if the system of equations 2x + 3y = 8, 7x – 5y + 3 = 0, 4x – 6y + l = 0 is solvable. U 



y  2  3 B =   and C =   .  x − y  −1  2

If AB = C, then find A2.

Thus,

Þ 1(1 + 1) + k(–k + 1) – 1(k + 1) = 0 Þ

x + y Q. 5. Let A =   2x

5 5  8  7 [YX] =   =  −11 −8  ½ − 11 − 7     –1

These questions are for practice and their solutions are available at the end of the chapter

A–1 =

−1  −2 −3 19  −5 2 

−1  −2 −3  2 k = 19  −5 2   5k

3k  −2 k 



1

½

1 ½ 19 [CBSE Marking Scheme 2018]

k =



DETERMINANTS

Short Answer Type Questions-II (3 marks each)  1 −2 Q. 1. If A =  then using A–1, solve the following  2 1  system of equations : x – 2y = – 1, 2x + y = 2. R&U [S.Q.P. 2016-17]

Sol.

|A| = 5  1 2 adj A = −2 1  



A–1 =

adj A 1  1 2  = | A | 5  − 2 1 

Given system of equations is AX = B, where  1 −2  x A =   , X = y 2 1    

½

 −1 B =   2

 1 −2   x   −1 2 1   y =  2       x 1 é 1 2 ù é -1ù  y  = ê -2 1 ú ê 2 ú   5ë ûë û 3 5 x 1  −1 + 4   y  =   = 4 2 + 2 5       5

½

 2 −3 1 A–1 =  −1 2    The given system of equations are equivalent to A’X = C, x  where X =   1 y 2 1 4  and C =   , A’ =   1   3 2 X = (A’)–1C = (A–1)’C x  2 −1  4  Þ  y  =  −3 2   1 

x  7  Þ  y  =  −10  \ x =7 and y = – 10 1 [CBSE Marking Scheme 2015] (Modified) Q. 3. Ishan wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50m and breadth is increased by 50m, then its area will remain the same. But if length is decreased by 10m and breadth is decreased by 20m, then its area will decrease by 5300m2. Using matrices, find the dimensions of the plot. [CBSE OD Set-I, 2016]

1

3 4 and y = 1 5 5 [CBSE Marking Scheme 2016] (Modified) x =

é2 3ù é 4 -6 ù Q. 2. Let A = ê ú , B = ê -2 4 ú . Then compute AB. 1 2 ë û ë û Hence, solve the following system of equations : 2x + y = 4, 3x + 2y = 1. A [S.Q.P. 2015-16]

Sol.



Let length be x m and breadth be y m \ (x – 50)(y + 50) = xy ⇒ 50x – 50y = 2500 or x – y = 50 and (x – 10)(y – 20) = xy – 5300 ⇒ 2x + y = 550 æ 1 -1ö æ x ö æ 50 ö çè 2 1 ÷ø çè y÷ø = çè 550÷ø

⇒





æ xö 1 æ 1 1ö æ 50 ö çè y÷ø = 3 çè -2 1÷ø çè 550÷ø

Sol.

A =  2 3   1 2 



⇒

æ xö 1 æ 50 + 550 ö çè y÷ø = 3 çè -100 + 550÷ø



 4 −6  B =   −2 4 





1 æ 600ö æ xö çè y÷ø = 3 çè 450÷ø 





æ 200ö æ xö çè y÷ø = çè 150 ÷ø



\ x = 200 m and y = 150 m

and

then



Þ

AB =  2 3   4 −6   1 2   −2 4  AB =  2 0  = 2 I  0 2  1  A  B = I 2 

On multiplying by A–1 1 A–1 = B 2

1

79

½ ½ ½ 1

[CBSE Marking Scheme 2016] Q. 4. On her birthday Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got `10 more. However if there were 16 children more,

These questions are for practice and their solutions are available at the end of the chapter

80

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

everyone would have got `10 less. Using matrix method, find the number of children and the amount distributed by Seema. [CBSE Delhi Set-I, 2016] Sol. Try yourself similar to Q. No. 3 of 3 marks.

Long Answer Type Questions (5 marks each) é1 2 0ù ê ú Q. 1. If A = ê -2 -1 -2 ú , find A−1. Hence êë 0 -1 1 úû Solve the system of equations: x – 2y = 10 2x – y – z = 8 –2y + z = 7

é 12 ù 1 ê ú -6 Þ = 6 ê ú ê ë 24 úû



Þ









Q. 2. Evaluate the product AB, where 2 -4 ù é2 ê ú ê -4 2 -4 ú êë 2 -1 5 úû

y + 2z = 7

Sol.

2 -4 ù é 1 -1 0 ù é 2 AB = êê 2 3 4 úú êê -4 2 -4 úú êë 0 1 2 úû êë 2 -1 5 úû



Þ



Þ



AB = 6I

æ1 ö A ç B÷ = I è6 ø 1 Þ A–1 = ( B) 6



The given equations can be written as



é 1 -1 0 ù é x ù é3ù ê úê ú ê ú y 2 3 4 = ê úê ú ê17 ú ê ú ê ú z 0 1 2 ë û ë û êë 7 úû AX = D



where

é3ù D = ê17 ú ê ú êë 7 úû



Þ

X = A–1D

Þ



1½

2 -4 ù é 3 ù é2 éx ù 1ê úê ú ê ú ê y ú = 6 ê -4 2 -4 ú ê17 ú êë 2 -1 5 úû êë 7 úû êë z úû





R&U [CBSE SQP 2020-21]

é6 0 0 ù = êê0 6 0 úú êë0 0 6 úû

After finding the product, some students find the inverse using formula which is wrong.

Answering Tip

Hence solve the system of linear equations x–y=3 2x + 3y + 4z = 17

é2ù éx ù ê ú ê ú -1 êyú = ê ú êë z úû êë 4 úû x = 2 y = –1 z = 4 1½ [CBSE SQP Marking Scheme 2020-21]

Commonly Made Error

R&U [CBSE SQP 2020-21]

é 1 -1 0 ù A = ê 2 3 4 ú and B = ê ú êë 0 1 2 úû

1

é2 Q. 3. If A = ê 1 ê ëê 0

3 4ù - 1 0 úú , find A– 1. Hence, solve the 1 2 úû

system of equations : x – y = 3; 2x + 3y + 4z = 17; y + 2z = 7. R&U [CBSE SQP, 2020-21]

Sol. 1

Learn to find inverse from the product of two matrices.



2  A= 1   0

3 4 −1 0   1 2 

|A| = 2 (– 2) – 3 (2 – 0) + 4(1 – 0) = – 6 ¹ 0

½

\ A exists –1

Co-factors A11 = – 2

A12 = – 2

A13 = 1

A21 = – 2

A22 = 4

A23 = – 2

A31 = 4

A32 = 4

A33 = – 5



1  −2 −2 Adj A =  − 2 4 − 2    4 4 − 5  

4  −2 −2  −2 4 4    1 − 2 − 5 

Adj A =

These questions are for practice and their solutions are available at the end of the chapter

T

2



DETERMINANTS



4  −2 −2 Adj A 1  − 2 4 4 = A =  | A| −6   1 − 2 − 5  –1

Now, ⇒

⇒

⇒

⇒

⇒

3 4 −1 0  , X =  1 2 

x y , B =    z 

é1 3 4ù

17  3   ½  7 

AX = B X = A–1 B





 − 12  1  6 X =  −6   − 24  x  2  y  =  −1  X =      z   4  

Sol. |A| = 4 ≠ 0 ⇒ A–1 exists.  −2 0 2    Adj A =  5 − 2 − 1  1 2 − 1



(AB)–1.

10  2 3  4 -6 5  find A–1. Using A–1 solve the Q. 7. In A =    6 9 -20

½

½

6 9 20 + = – 4 x y z Sol.

½ written as

½ 3  1     2 

1

2 3 10 4 6 5 + + = 2 ; - + = 5 ; x y z x y z U&A [Delhi Set-II 2017]



Here |A| = 1200 Co-factors are  C11 = 75, C21 = 150, C31 = 75 C12 = 110, C22 = – 100, C32 = 30  C13 = 72, C23 = 0, C33 = – 24   75   75 150 1   A–1 = 110 −100 30  1200   72 0 −24   Given equation in matrix from is : 1    2 3 10   x  é2ù  1 ê ú  4 −6 5   y  = ê 5 ú  6 9 −20    êë -4 úû  1  z 



or



or



2

 −2 0 2  1 1 ⋅ adj A =  5 − 2 − 1  4 A 1 2 − 1

1 3 3  1 4 3  , compute 1 3 4   

R&U [Comptt. 2018 Set I, II, III]

system of equations:

of equations: x + y + z = 6, x + 2z = 7, 3x + y + z = 12. R&U [CBSE Delhi Set III 2019]



x + 3y + 4z = 8

5 0 4 Q. 6. Given A =  2 3 2  , B–1 =  1 2 1  

 − 34 − 6 + 28  1  − 34 + 12 + 28  X =  −6   17 − 6 − 35 

Given system of equations can be AX = B x  6      where, X =  y  , B = 7   z  12  \ X = A–1 ⋅ B 0 2  6   −2 1  = 5 − 2 − 1  7=  4  1 2 − 1 12 

Hence solve the system of equations:

R&U [CBSE OD Set I 2019]

 1 1 1 1 0 2 –1 Q. 4. If A =   , find A . Hence, solve the system 3 1 1



êë 5 1 1 úû

and 5x + y + z = 7

17  3    7 

½ ⇒ x = 2, y = – 1, z = 4  [CBSE SQP Marking Scheme 2020] (Modified)

∴A −1 =

ê ú Q. 5. If A = ê 2 1 2 ú , find A–1.

2x + y + 2z = 5

4  −2 −2 1  −2 4 4 X=  −6   1 − 2 − 5 

\ x = 3, y = 1, z = 2 ½ [CBSE Marking Scheme, 2019] (Modified)

1

System of equations can be written as AX = B

2  where A=  1  0

81





These questions are for practice and their solutions are available at the end of the chapter

or

1 1

½

1

AX = B X = A–1B 1   x 1 y =   1  z 

1 2    −1  3   1  5 

1½



or x = 2, y = – 3, z = 5 [CBSE Marking Scheme 2017] (Modified)

82

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

COMPETENCY BASED QUESTIONS Case based MCQs

(4 marks each)

Attempt any four sub-parts from each question. Each sub-part carry 1 mark.

I. Read the following text and answer the following questions on the basis of the same: Manjit wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50 m, then its area will remain same, but if length is decreased by 10 m and breadth is decreased by 20 m, then its area will decrease by 5300 m2 [CBSE QB 2021]

Explanation: We have,









Q. 1. The equations in terms of x and y are: (A) x – y = 50, 2x – y = 550 (B) x – y = 50, 2x + y = 550 (C) x + y = 50, 2x + y = 550 (D) x + y = 50, 2x + y = 550 Ans. Option (B) is correct. Explanation: (x – 50)(y + 50) = xy x – y = 50 ...(i) (x – y)(y – 20) = xy – 5300 2x + y = 550 ...(ii) Q. 2.  Which of the following matrix equation is represented by the given information? é 1 -1ù é x ù é 50 ù (A) ê = 1úû êë y úû êë 550 úû ë2







é 1 1ù é x ù é 50 ù (B) ê úê ú=ê ú ë 2 1û ë y û ë 550 û



é 1 1ù (C) ê ú ë 2 -1û

é x ù é 50 ù ê y ú = ê 550 ú ë û ë û

é 1 1ù é x ù é -50 ù (D) ê úê ú=ê ú ë 2 1û ë y û ë -550 û Ans. Option (A) is correct. Q. 3. The value of x (length of rectangular field) is: (A) 150 m (B) 400 m (C) 200 m (D) 320 m Ans. Option (C) is correct.



é 11 -11ù é xx ù = éé 50 50 ù ééêê 1 -1ùùúú ééêê x ùùúú = éêê 50 ùùúú 550 úû êëë 222 111 úûû êëë yyy úûû = êëë 550 ë û ë û ë 550 ûû éé 11 -11ù Let A = éêê 1 -1ùùúú Let A = Let A = êë 22 11 úû ëë 2 1 ûû é 50 50 ù = B ééêê 50 ùùúú ,, = B B = êë 550 550 úû , ëë 550 ûû éé xx ùù X= êé x úù X = X = êëê yy úûú ëë y ûû Now AX = B Now AX B = Now AX = B -1 X= A -1B X = A X= A -1B B éé 11 11ùù Adj( A )) = Adj = éêê -12 11ùúú Adj(( A A) = êëë -2 11úûû ë -2[ 2 ´ û -1)] | A | 1 = || A = 11 - [[ 22 ´ ´ ((( -11))]] A || = 11 + 22 = = + = 1+ 2 3 = = = 33 Adj(( A ) -1 -1 = Adj A Adj( A A )) A | A -1 = = || A A || |A 11 éé 11 11ùù = 1 êé 1 1úúù = 2 1û = 33 êëê -22 11ûúû 3 ëë 50 ùù 11 éé 11 11ùù éé 50 X 50 ùú = X= 112 111úúù êêé 550 50 131 êêëéê X = ûú ëëê 550 úúûû 2 1 X = 33 ë û ë -2 1û ë 550 û éé3 11−2 111ùù  550  éêê 11 11 ùúú é 50 ù 50 ù êê 333 333 úú ééê 50 = = 50 ùúú êêë 550 2 1 3 3 = ê ú -2 1 550 550 úûû = ê êêë −322 113 úúúû ëë 550 û êëë 33 33 úûû é 350  550 ù 50 +3550 550 50 ééêê 50 ùùúú + 550  3 3 + 3 + 33 ú = = êêê -100 ú 33 3550 = -100 100 550 úú = ê + 550 + êëê −100 ú 3 + 550 3 ûú ëëê 333 + 333 úûû éé xx ùù éé 200 ù 200 ù x = 200 = éêê150 200 ùú éêêêë xy ùúúúû = ë ê 150 úúûû ëë yyy ûû = ëë150  xx = 150 û 200 = 200 200 x = ⇒ 200 yx = = 150 150 = yyy = = 150 150

Q. 4. The value of y (breadth of rectangular field) is: (A) 150 m (B) 200 m (C) 430 m. (D) 350 m Ans. Option (A) is correct. Q. 5. How much is the area of rectangular field? (A) 60000 sq.m. (B) 30000 sq.m. (C) 30000 m (D) 3000 m Ans. Option (B) is correct. Explanation: Area of rectangular field = xy = 200 × 150 = 30000 sq m.





83

DETERMINANTS

II. Read the following text and answer the following questions on the basis of the same:

The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to kept the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. The sum of the number of awardees for honesty and supervision is twice the number of awardees for helping.



Q. 1. x + y + z = _______. (A) 3 (B) 5 (C) 7 (D) 12 Ans. Option (D) is correct. Explanation: x + y + z = 12 2x + 3y + 3z = 33 x – 2y + z = 0 é12 ù éx ù é1 1 1ù ê ú ê ú ê ú A = ê 2 3 3 ú , B = ê33 ú , X = ê y ú êë 1 -2 1 úû êë 0 úû êë z úû |A| = 1(3 + 6) – 1(2 – 3) + 1(– 4 – 3) = 9 + 1 – 7 = 3 1 ( adj A ) A–1 = | A|

é 9 ù é3 ù 1ê ú ê ú = 12 = 4 3ê ú ê ú êë15 úû êë 5 úû

Þ x = 3, y = 4, z = 5 x + y + z = 12 Q. 2. x – 2y = _______. (A) z (B) – z (C) 2z (D) –2z Ans. Option (B) is correct. Explanation: x – 2y =– z [from (iii)] Q. 3. The value of z is _______. (A) 3 (B) 4 (C) 5 (D) 6 Ans. Option (C) is correct. Explanation: z = 5 Q. 4. The value of x + 2y = _______. (A) 9 (B) 10 (C) 11 (D) 12 Ans. Option (C) is correct. Explanation: x + 2y = 3 + 8 = 11 Q. 5. The value of 2x + 3y + 5z = _______. (A) 40 (B) 43 (C) 50 (D) 53 Ans. Option (B) is correct. Explanation: 2x + 3y + 5z = 6 + 12 + 25 = 43

[from (i)]

Case based Subjective Questions (4 marks each) ...(i) ...(ii) ...(iii)

I. Read the following text and answer the following questions on the basis of the same: (Each Sub-part carries 2 marks) The Palace of Peace and Reconciliation, also known as the Pyramid of peace and Accord is a 62-metre high Pyramid in Mursultan, the capital of Kazakhstan, that serves as a non-denominational national spiritual centre and an event venue. It is designed by faster and partners with a stained glass apex. It has 25 smaller equilateral triangles as shown in the figure.

é 9 -3 0 ù 1ê ú 0 -1ú = ê 1 3 êë -7 3 1 úû X = A–1B é 9 -3 0 ù é12 ù 1ê úê ú A -1B = ê 1 0 -1ú ê33 ú = 3 êë -7 3 1 úû êë 0 úû =

é 9 ù é3 ù 1ê ú ê ú 12 = 4 3ê ú ê ú êë15 úû êë 5 úû





(

)

Q. 1. If the vertices of one triangle are (0, 0), 3 , 3 and

( 3, − 3 ) then find the area of such triangle area of and a face of the Pyramid.

84

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Sol.

(

)



1 = 1 −3 3 − 3 3  2



6 3 = 3 3 sq units = 2



Since, a face of the Pyramid consists of 25 smaller equilateral triangles.



\ Area of a face of the Pyramid = 25 × 3 3 = 75 3

sq. units. 1 Q. 2. Find the length of a altitude of a smaller equilateral triangle. Sol. Area of equilateral triangle =

3 (side)2 4



3 \ (side)2 3 3 = 4 [As calculated above area of equilateral triangle is



Q. 1. It x, y and z respectively denotes the quantity (in tons) of first, second and third product produced, then construct the system of equations and write it in matrix form. Sol. According to given information, the system of linear equations is: x + y + z = 45 z = x + 8 x + z = 2y or, x + y + z = 45 –x + z = 8 x – 2y + z = 0 1½ In matrix form,

0 0 1 1 Required Area = 3 3 1 2 3 − 3 1

1 1 1  Q. 2. If  1 0 −2     1 −1 1 

Þ



Þ



Let h be the length of the altitude of a smaller equilateral triangle. 1 Then, × base × height = 3 3 2



or,



or,



(side)2 = 12 side = 2 3 units

1

3 3×2 2 3

= 3 units

1

II. Read the following text and answer the following questions on the basis of the same: A factory produces three products every day. Their production on a particular day is 45 tons. It is found that production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product.

½

2 2 2 1  =  3 0 −3 , then find the 6  1 −2 1 

1 1 1  A = 1 0 −2    1 −1 1 

Sol. Let

Then

1 × side × height = 3 3 2

−1

 45   8  0 

1 1 1  inverse of  1 0 −1 .    1 −2 1 

3 3 sq. units]



 1 1 1  x      −1 0 1  y  =  1 −2 1  z 











2 2 2  1  A =  3 0 −3 6  1 −2 1  –1

We know that, (A')–1 = (A–1)' 1  1 1 3 3 3    1 1 0 −  \ (A–1)' =  2 2  1 −1 1    6   6 3

1 1 1 3 2 6   1 1 − 0 =  3 3  1 −1 1     3 2 6 

1 1 1  \ 1 0 −1   1 −2 1 

−1

1 3  1 =  3 1   3

or, (A') = (A )' –1

–1

1 2 0 −1 2

1 6  −1  3 1  6 

1





DETERMINANTS

85

Solutions for Practice Questions (Topic-1)

2.

½

|A| = 1 é -1 4 ù A–1 = ê ú ë -1 3 û



10. det A = det B = 0



Very Short Answer Type Questions

⇒ AB = 0



½



Students use elementary operations for finding the inverse.

Answering Tip

Find the inverse using formula for very short answer questions.

é 3 1ù adj A = ê 4. 1 ú ë -4 2 û [CBSE Marking Scheme 2020]

Given,



\

0 −1  3 5  AB =    0 2  0 0 



or

0 + 0 0 + 0  AB =   0 + 0 0 + 0 



or

0 0  AB =  =0 0 0 



Hence,





é 2 -1ù A = ê ú ë4 3 û C11 = 3, C12 = –4





C21 = 1, C22 = 2

Given,

T

C12 ù éC é3 -4 ù adj A = ê 11 =ê ú ú ëC21 C22 û ë1 2 û é 3 1ù adj A = ê ú ë -4 2 û





2x2 + 6x – 6x = 8 8 2

or

x2 =

or

x2 = 4 x = ± 2

or

T



½

½

Since x Î N, so, x = 2



Short Answer Type Questions-I

Students take the minors instead of co-factors for finding adj A.

-1

1.

é -3 2 ù ê 5 -3 ú ë û



é 1 2 ù é 3 2 ù é13 8 ù \ P=ê úê ú=ê ú ë 2 -1û ë 5 3 û ë 1 1 û

=

1 é -3 -2 ù é 3 2 ù ê ú=ê ú 9 - 10 ë -5 -3 û ë 5 3 û

1+½

½

[CBSE Marking Scheme 2017-18] 3. |2AB| = 23 × |A| × |B| 1 = 8 × (–1) × 3 = –24 1 [CBSE Marking Scheme 2017]

Answering Tip



x + 3 −2 =8 −3x 2 x

or 2x(x + 3) – (– 2) (– 3x) = 8

or

Commonly Made Error



|AB| = 0

13. Given

Detailed Solution:

3 5 0 −1 A =   and B = 0 0  0 2    



Commonly Made Error



½ [CBSE Marking Scheme, 2019]

Detailed Solution:

[CBSE Marking Scheme 2020]



½

Learn the difference between minors and co-factors.

Solutions for Practice Questions (Topic-2) Very Short Answer Type Questions 1.

For a unique solution 1 1 1 2 1 − 1 ¹ 0 3 2 k

1

k + 2 – 2k – 3 + 4 – 3 ¹ 0 Þ k ¹ 0 [CBSE Marking Scheme 2016]

86

Learn the solution of equations with the different conditions.







|A| = 2, 1 7 3  A–1 =  4 2  2  7 3  LHS = 2A–1 =  4 2  ,  



1

 1 0   2 −3 RHS = 9 0 1  −  −4 7     



é -3 -2 -4 ù ú = êê 2 1 2ú 1 3 úû êë 2 The given equations can be written as :



Short Answer Type Questions-I 1.





Answering Tip



é10 ù é 1 -2 0 ù é x ù ê úê ú = ê 8 ú y 2 1 1 ê ú ê úê ú êë 0 -2 1 úû ëê z úû êë 7 úû Which is of the form A’X = B



Þ



Þ



7 3  =  4 2   

1

LHS = RHS

[CBSE Marking Scheme 2018] Detailed Solution:

Topper Answer, 2018

A is non-singular, therefore A–1 exists

é -3 -2 -4 ù ú Adj A = ê 2 1 2ú ê êë 2 1 3 úû 1 ( Adj A ) Þ A–1 = |A|



Students get confused with the conditions for unique and infinite solutions.







Commonly Made Error



Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

1½

½

X = (A’)–1B = (A–1)’B éx ù ê ú = êyú êë z úû

1

é -3 2 2 ù é10 ù ê úê ú ê -2 1 1 ú ê 8 ú êë -4 2 3 úû êë 7 úû

é0ù = êê -5 úú êë -3 úû Þ x = 0 y = –5 z = –3 1½ [CBSE SQP Marking Scheme 2020-21]



Sol.





Commonly Made Error Students get confused with how to use the given matrix to solve the equation.









|A| = 1(–1 – 2) – 2(–2 – 0) = –3 + 4 = 1 ¹ 0

The coefficient matrix is the transpose of the given matrix.

2  −1 1 5. = A 11; Adj = (A)  8 −19 6  1+2  −3 14 −5  −1

∴A =

Long Answer Type Questions 1.



Answering Tip

½

2  −1 1 1 1   . Adj= A 8 −19 6   A 11   −3 14 −5 

x  Taking; = X = y  ; B  z 

8  5   7 

½



DETERMINANTS



The system of equations in matrix form is

A⋅X = B ∴X= A ⋅B \ Solution is : 2  8  1 x   −1 1 1  y =    1 1 8 − 19 6   11   5 =    z   −3 14 −5  7  1 −1

\ x = 1, y = 1, z = 1 ½ [CBSE Marking Scheme, 2019] (Modified)



Detailed Solution: Try yourself similar to Q. No. 4 of LATQ. 6.

|A| = 5(– 1) + 4(1) = – 1 C11 = – 1 C21 = 8 C31 = – 12

1



C12 = 0 C13 = 1



C22 = 1 C32 = – 2 C23 = – 10 C33 = 15  1 −8 12  A–1 =  0 −1 2   −1 10 −15  

87 2 1



(AB)–1 = B–1A–1 1 3 3   1 −8 12  = 1 4 3   0 −1 2  1 3 4   −1 10 −15     −2 19 −27  =  −2 18 −25   −3 29 −42   

1

[CBSE Marking Scheme 2018] (Modified)

REFLECTIONS •

Here you practiced of finding area of triangle using determinants?

•

Will you be able to find the volume of parallelopiped using determinants? qq

88

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

SELF ASSESSMENT PAPER - 02 Time: 1 hour

MM: 30

UNIT-II (A) OBJECTIVE TYPE QUESTIONS: I. Multiple Choice Questions

[1×6 = 6]

Q. 1. A 2 × 2 matrix whose elements aij are given by aij =  0 (A)  1  2

1 2  0 

1  (B)  2 0  

 0  1 2 

 2 3 −5  Q. 2. If A = [aij] =  1 4 9  and B = [bij] =    0 7 −2  (A) 10

(i − j ) is : 2 2

 0 (C)   −1  2  2 −1    −3 4  ,  1 2 

(B) 15

(C) 20

 x − y z   −1 4  Q. 3. If  =  , then values of x and y are 2x − y w  0 5 (A) 2, 3 Q. 4. Value of

(B) 2, 1

−1  1  2 (D)  2  0 0    then value of a11b11 + a22b22 is :

:

−1  2  1 2 

(D) 25

(C) 1, 2

(D) 3, 2

(C) –1

(D) None of these

(C) 25

(D) –15

(C) –1

(D) 2

1 log b a is : log a b 1

(A) 1

(B) 0

 1 3 Q. 5. If A =   , then |A2 – 2A| = ............  2 1 (A) 15

(B) 20

x+y Q. 6. Find the value of z 1 (A) 0

y+z z+x x y . 1 1 (B) 1

II. Case-Based MCQs

[1×4 = 4]

Attempt any 4 sub-parts from each questions. Each question carries 1 mark. Read the following text and answer the questions on the basis of the same. A bolt manufacturing company produces three types of bolts x, y and z which he sells in two markets. Annual sales (in `) are indicated in the following table. Markets I II

x 10000 6000

Products y 2000 20000

x 18000 8000



SELF ASSESSMENT PAPER

89

It unit sales price of x, y and z are `2.50, `1.50 and `1.00 respectively, then answer the following questions using the concept of matrices. Q. 7. The total revenue collected from market I is : (A) `44,000

(B) `48,000

(C) `46,000

(D) `53,000

(C) `53,000

(D) `49,000

Q. 8. Total revenue collected from market II is : (A) `46,000

(B) `51,000

Q. 9. If the unit costs of the above three types of bolts are `2, `1 and `0.50 respectively, then the gross profit from the market is : (A) `46,000

(B) `25,000

(C) `32,000

(D) `20,000

Q. 10. If matrix A = [aij]2 × 2, where aij = 1 if i ¹ j and aij = 0 if i = j, then A2 is : (A) I

(B) A

 2 −1 Q. 11. If A =   and B = 3 4 

5 2 7 4  , then AB is:  

3 0 (A)    43 22 

0 3 (B)    22 43

(C) 0

(D) none of these

 43 22  (C)   0 3

 22 43 (D)   3 0

(B) SUBJECTIVE TYPE QUESTIONS: III. Very Short Answer Type Questions

[1×3 = 3]

1 2  Q. 12. If A =   , then find A + AT. 3 4   1 −5 7  Q. 13. Find the trace of matrix A =  0 7 9  .   11 8 9  Q. 14. If A is a skew symmetric matrix of odd order n, then |A| = 0. IV. Short Answer Type Questions–I

[2×3 = 6]

Q. 15. Using determinants, find the area of the triangle whose vertices are (–2, 4), (2, –6) and (5, 4). Are the given points collinear? 3 1 Q. 16. For the matrix A =   , find x and y so that A2 + xI = yA. 7 5  2 3  1 Q. 17. If A =  A.  , then show that A–1 = 5 − 2 19   V. Short Answer Type Questions-II

[3×2 = 6]

1 2 0 Q. 18. Find the minors and cofactors of the elements of first row of determinant 3 5 −1 . 4 7 8 tan x   1 , show that ATA–1 = Q. 19. If A =  1   − tan x

cos 2 x − sin 2 x   sin 2 x cos 2 x  .  

90

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

VI. Long Answer Type Questions

[5×1 = 5]

 −4 4 4   1 −1 1     Q. 20. Determine the product  −7 1 3   1 −2 −2  and use it to solve the system of equation :  5 −3 −1  2 1 3  x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1

qq

UNIT – III : CALCULUS

CHAPTER

5



Syllabus

CONTINUITY & DIFFERENTIABILITY

Continuity and differentiability, chain rule, derivative of inverse trigonometric functions like sin–1x, cos–1x and tan–1x derivative of implicit functions. Concept of exponential and logarithmic functions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.







 Continuity

of functions  Differentiability of functions like implicit, inverse trigonometric, exponential,logarithmic, etc.  Chain Rule  Higher order derivatives

Topic-1



Topic-1: Continuity Page No. 91

Topic-2: 

Continuity Concepts Covered

List of Topics



In this chapter you will study

Differentiability Page No. 97

Left hand Limit,  Right Hand Limit

Revision Notes FORMULAE FOR LIMITS: (a) lim cos x = 1

Scan to know more about this topic

x →0



sin x =1 (b) lim x →0 x



tan x lim =1 (c) x →0 x



Definition Continuity

sin −1 x =1 (d) lim x x →0

tan −1 x (e) lim =1 x →0 x



ex − 1 =1 x →0 x

(g) lim

log e (1 + x ) =1 x

x n − an = nan −1 x→a x − a

(i) lim

 For a function f(x), lim f(x) exists if lim f(x) = x →m

x → m−

x → m+

 A function f(x) is continuous at a point x = m

ax − 1 = log e a , a > 0 x →0 x

x →0

lim f(x).



(f) lim

(h) lim

if,

lim f ( x ) = lim f ( x ) = f ( m) ,

x → m−

x → m+

where

lim f ( x ) is Left Hand Limit of f(x) at x = m

x → m−

f t ,y

g t be two functions of parameter 't'. dy dy dt dy dy dx  dx  . Then, or dx dx dt dx dt  dy  0  dt dy g ' t (provided f '( t ) π 0) Thus, dx f ' t dx y a eg : if x a then – a sin and d a cos dy dy dy / d a cos , and so – cot . d dx dx / d a sin

(v)

1

1

1 1 x2

x x2

–

(viii)

First Level

Second Level

Third Level

d log x dx

d (vi) cos e c 1 x dx 1 – x x2 1 d x e ex (vii) dx

Trace the Mind Map

d sec 1 x dx

d cot 1 x dx



(iv)

1 1 x2

d (iii) tan 1 x dx



1 – x2

–

1

1 – x2

1

d cos 1 x (ii) dx

d sin -1 x (i) dx



Let y

1 x

f x then dy f ' x , if f ’ x is dx d dy d differentiable, then f ' x i.e., dx dx dx 2 d y f " x is the second order derivative of y w.r.t.x. dx 2 eg : if y = 3x2+2, then y'= 6x and y"= 6.

Let x

Derivatives of functions in parametric form

ty

ntinuity and o C erentiabi li iff

Some Standard derivatives

Suppose f is a real function and c is a

y

u x

v x

v x

u' x

v ' x log [u(x)] v ' x log u x

1 u' x u x

v x log u x

e.g. : Let y ax Then log y = x log a 1 dy . log a y dx dy y log a a x log a. dx

dy dx

1 dy . y dx

log y

Let y=f(x)=[u(x)]v(x)

dv dt . . dt dx

or,

dx

dy

=– sin x – cos y.

dx

dy

or,

dx

dy

(2n+1)

dy d d = cos x – sin y dx dx dx =– sin x / 1 cos y , where y

For example: Let y =cos x – sin y, then

If two variables are expressed by some relation then one will be the implicit function of other, is called Implicit function.

Derivatives of Implicit functions

Logarithmic differentiation

Chain Rule

df dt dv exists, then , dx dx dt

but the converse is not true.

point in its domain. The derivative of f c h –f c f at c is f ' c lim h 0 h Every differentiable function is continuous,

If f=vou, t=u(x) and if both

Differentiability

Algebra of continuous functions

Continuous Function

Suppose f and g are two real functions continuous at a real number c, then, f+g, f–g, f.g and f are continuous at g x=c[g(c)

Suppose f is a real function on a subset of the real numbers and let 'c' be a point in the domain of f. f x f c Then f is continuous at c if lim x c A real function f is said to be continuous if it is continuous at every point in the domain of f. 1 eg: The function f x , x 0 is continuous x 1 1 1 Let 'c' be any non-zero real number, then lim f ( x ) lim . For c 0, f c So lim f x f c x c x c x c x c c and hence f is continuous at every point in the domain of f.

Second order derivative

D

92 Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII



CONTINUITY & DIFFERENTIABILITY

and

lim f ( x ) is Right Hand Limit of f(x) at



x → m+

x →m

x = m. Also f(m) is the value of function f(x) at x = m.

 A function f(x) is continuous at x = m (say) if, f(m) = lim f(x) i.e., a function is continuous at x →m

a point in its domain if the limit value of the function at that point equals the value of the function at the same point.

 For a continuous function f(x) at x = m, lim f(x) can



be directly obtained evaluating f(m).  Indeterminate forms meaningless forms:

by

93

Scan to know more about this topic

or

0 ∞ , , 0 × ∞ , ∞ − ∞ , 1∞ , 0 0 , ∞0 . 0 ∞

Example : continuity at a point

OBJECTIVE TYPE QUESTIONS A

Explanation: Since, f(x) is continuous at x = 0, then LHL = RHL = f(0) or LHL = RHL = k e 3( 0  h )  e 5( 0  h ) Now, LHL = lim h 0 0h

Multiple Choice Questions

Q. 1. If a function f defined by   k cos x    2 x , if x  2 f (x)     3 if x   2

e −3 h − e 5 h h→0 −h

= lim

π is continuous at x = , then the value of k is: 2

 e 5h  1   e 3 h  1     lim  = lim  h 0  h  h 0  h 

(A) 2 (C) 6

 e 5h  1   e 3 h  1     5 lim  = 3 lim  h 0 h 0  5h   3h 

(B) 3 (D) –6 [CBSE Term-I 2021]

Ans. Option (C) is correct. Explanation: Since, f(x) is continuous at x =  lim f ( x ) = f    2 x 2 k cos x lim =3    2x x 2   sin   x  2  =3 k lim    x 2 2  x  2 

π 2

Therefore, Þ

Þ

Þ Þ

  sin   x  k 2   =3 lim 2 x     2   x 2  k ×1 = 3 Þ k = 6 2

 e3 x  e 5 x , if x  0 Q. 2. The function f(x) =  x  if x  0 k 

is continuous at x = 0 for the value of k, as: (A) 3 (B) 5 (C) 6 (D) 8 [CBSE Term-I 2021] Ans. Option (D) is correct.

Thus,

=3×1+5×1=8 k=8

x2  1 then which of the 2 following can be a discontinuous function? (B) f(x) – g(x) (A) f(x) + g(x) g( x ) (D) (C) f(x).g(x) f (x)

Q. 3. If f ( x )  2 x and g ( x ) 

Ans. Option (D) is correct. x2 Explanation: Since f ( x= +1 ) 2x and g ( x=) 2 are continuous functions, then by using the algebra of continuous functions, the functions f(x) + g(x), f(x) – g(x), f(x).g(x) are also continuous functions g (x) is discontinuous function at x = 0. but f (x) Q. 4. The function f ( x ) 

4  x2

is: 4 x  x3 (A) discontinuous at only one point (B) discontinuous at exactly two points (C) discontinuous at exactly three points (D) none of these Ans. Option (C) is correct. Explanation: Given that, 4 − x2                   f ( x ) = , 4x − x 3 then it is discontinuous if

94

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

⇒ ⇒

p then 2 (A) m = 1, n = 0

4x − x 3 = 0

(

x 4−x

2

0 )=

⇒ x ( 2 + x )( 2 − x ) = 0 ⇒ x= 0, − 2, 2 Thus, the given function is discontinuous at exactly three points. Q. 5. The function f(x) = cot x is discontinuous on the set: (A) {x = np ; n Î Z}



(B) {x = 2np ; n Î Z}

p ì ü (C) íx = ( 2n + 1) ; n Î Z ý 2 î þ

np +1 2 p (D) m = n = 2 (B) m =

mp 2 Ans. Option (C) is correct. Explanation: Given that, p ì ïïmx + 1 if x £ 2 f (x) = í ïsin x + n, if x > p ïî 2 (C) n =

is continuous function at x =

np ì ü (D) íx = ; n Î Zý 2 î þ Ans. Option (A) is correct.

Þ Þ

Explanation: Given that, f ( x ) = cot x =

cos x sin x

p , then 2

LHL = RHL LHL lim f ( x=) =RHL lim f ( x ) pp+ xlim ® x ® f ( x ) = lim f (x) 2 2 x®

p-

p+

ö æp ö x® æ p Þ lim f ç 2 - h÷ = lim 2f ç + h÷ èæ p øö h ® 0 æè p 2 2 öø h ® 0 Þ lim f ç - h÷ = lim f ç + h÷ h®0 è 2 æ p ø ö h®0 æ p è 2 ö ø Þ lim m ç - h÷ + 1 = lim sin ç + h÷ + n øö èæ p øö h®0 è h®0 2 2 æp Þ lim m ç - h÷ + 1 = lim sin ç + h÷ + n èp ø è ø h®0 æ h ® 0 2 2 ö Þ lim m ç - h÷ + 1 = lim cos h + n èp h®0 æ h®0 2 öø Þ lim m ç - h÷ + 1 = lim cos h + n ø h®0 è 2 æ p ö h®0 Þ mç ÷ + 1 = 1 + n èæ p 2 øö Þ mç ÷ + 1 = 1 + n è 2ø mp Þ n= m2p Þ n= 2

It is discontinuous at sin x = 0 Þ x = np , n Î Z Thus, the given function is discontinuous at {x = np : n Î Z}.   mx  1 if x  2 , is continuous at x = Q. 6. If f ( x )   sin x  n , if x    2

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. If the function f defined as ì x2 - 9 ï , f ( x) = í x - 3 ïk , î

Q. 3. Determine the value of the constant ‘k’ so that the  kx , if x < 0  function f(x) = | x | is  3 , if x ≥ 0  is continuous at x = 0.

x¹3 x=3

Sol.

is continuous at x = 3, find the value of k.

Q. 2. Determine the value of ‘k’ for which the following function is continuous at x = 3:  ( x + 3)2 − 36  , x≠3 f ( x) =  x−3  , x=3 k  R&U [OD, 2017]

lim f(x) = lim k ( − x ) = – k x →0− | x |

x→0 −



R&U [CBSE Delhi Set-I, II, III 2020]

R&U [Delhi, 2017]

½

k = – 3 [CBSE Marking Scheme, 2017] ½

Detailed Solution:

Since f is continuous at x = 0,



lim f(x) = lim+ f(x) = f(0)

x→0 −

x→0

Here f(0) = 3,

These questions are for practice and their solutions are available at the end of the chapter

LHL = lim− f(x) x→0

½

x →0

k( − x ) kx = xlim =–k − → 0 |x| |x|

\ – k = 3 or k = –3. ½ Q. 4. If the following function f(x) is continuous at x = 0, then write the value of k.  3x  sin 2 , x≠0 f(x) =   x  k , x = 0 R&U [OD Comptt., 2017]

3x 3x sin 3 2 2 Sol. lim = lim . ½ x →0 x x →0 2 3x 2 3 or k = ½ 2 [CBSE Marking Scheme, 2017] sin

Short Answer Type Questions-I (2 marks each) Q. 1. Find the value(s) of k so that the following function is continuous at x = 0 ì 1 - cos kx if x ¹ 0 ïï f ( x ) = í x sin x ï1 if x = 0 ïî 2 R&U [CBSE SQP 2020-21]



Sol.

1 - cos kx Lt = Lt x® 0 x sin x x® 0

æ kx ö 2 sin 2 ç ÷ è 2ø x sin x

æ kx ö 2 sin 2 ç ÷ è 2ø 2 x = Lt x sin x x® 0 x2 æ kx ö 2 sin 2 ç ÷ 2 è 2 ø æ kö ´ç ÷ Lt 2 è 2ø x ®0 æ kx ö çè 2 ÷ø = sin x Lt x ®0 x k2 2 ´1´ 4 = 1½ 1  f(x) is continuous at x = 0 Lt f ( x ) = f(0) \ x®0

Þ Þ Þ

k2 1 = 2 2 k2 = 1 k = ±1 ½ [CBSE Marking Scheme 2020-21]

95

Commonly Made Error





= lim−

Some students do not know how to evaluate 0 limits of the form . 0

Answering Tip











CONTINUITY & DIFFERENTIABILITY

Learn to evaluate the indeterminate forms of limits.

Q. 2. Find the value of k for which the function.  x 2 +3 x − 10  ,x ≠ 2 f(x) =  x−2  k ,x=2  is continuous at x = 2. R&U [Delhi Comptt. 2017] Q. 3. Find the value of p for which the function  1 − cos4 x  ,x ≠0 f(x) =  x2  , x=0 p 

is continuous at x = 0. R&U [Delhi Comptt. 2017] lim f(x) = f(0)

Sol.

x→0

4 × 2 sin 2 2 x

= p

1



æ sin 2 x ö 8 lim ç ÷ = p x ®0 è 2x ø

½



p = 8

½

lim

x →0



4x2 2

[CBSE Marking Scheme, 2017] Q. 4. Find the value of k for which the function ì sin x - cos x ï f(x) = ïí 4 x - p ï k ïî

p x® 4

lim x®



lim

These questions are for practice and their solutions are available at the end of the chapter

R&U [Delhi Comptt. 2017]

lim

Sol.



p 4 is continuous at x = x . 4 p ,x = 4 ,x ¹

x®

p 4

p 4

sin x - cos x æ pö = fç ÷ è 4ø 4x - p

1 æ 1 ö 2ç sin x cos x÷ è 2 ø 2 = k 4x - p

p pö æ 2 ç sin x cos - cos x.sin ÷ è 4 4ø = k 4x - p

96

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

lim x®

RHL

pö æ 2 sin ç x - ÷ è 4ø = k pö æ 4çx - ÷ è 4ø

p 4

p ( at x= ) 2

Lt

p+ ( x® ) 2

. f (x)

q(1 − sin x ) Lt 2 + π (π − 2x ) x→ =

2 = k 4



=

k =

2

1 2 2

[CBSE Marking Scheme, 2017]

Short Answer Type Questions-II (3 marks each)

x =

Put





π π + h. As x → , h → 0 2 2

= q Lt

π  1 − sin  + h 2 

h→0 



π   π − 2  + h    2 

2

Q. 1. Find the values of p and q, for which  1 - sin 3 x p , if x <  2 2 3cos x   p if x = , f ( x) =  p, 2  p  q(1 - sinx )  ( p - 2 x )2 , if x > 2  p is continuous at x = . 2



=

Since f(x) is continuous at x =

π− 2

= Lt −

3

(1 − sin x ) 3 cos2 x

π x→ 2

=

(13 − sin 3 x )

Lt x→

=

2

2

π− 2

½

2

3(1 − sin x )

π−

Lt x→

(1 − sin x )(1 + sin x + sin 2 x ) 3(1 − sin x )(1 + sin x )

= limp x® 2



\



\

½

π 2

 π LHL = RHL = f    2 q 1 = = p 8 2

or

q = 4 and p =

1 2

½

Commonly Made Error



1 + sin x + sin x 3 (1 + sin x )

½

Answering Tip



-

p p - h. As x ® , h®0 2 2 æp ö æp ö 1 + sin ç - h÷ + sin 2 ç - h÷ è2 ø è2 ø lim h®0 æ ö p æ ö 3 ç 1 + sin ç - h÷ ÷ è2 øø è

Sufficient time needs to be spent on this topic.

Q. 2. For what value of k is the following function π continuous at x = − ? 6  3 sin x + cos x  π x+ f(x) =   6   k, 

1 + cos h + cos2 h h®0 3 (1 + cos h ) lim

1+1+1 1 = = 3(1 + 1) 2

Many student commit errors in finding the Left hand limit and Right hand limit.

2

Put x =



4 h2

h→0

h 2



(1 – sin x) ¹ 0





Lt f ( x ) x→

= q Lt

Also, f  π  = p  2 



LHL

4 h2

2 sin 2

h2  sin q  2 × 1 = q Lt = 2 h→0  h  4 8   2 

A [Delhi Set I, II, III 2016] π   at x=  2

1 − cos h

h→0



Sol.

= q Lt

½

½



,x≠−

x=−

π 6

π 6

A [SQP 2016-17]



CONTINUITY & DIFFERENTIABILITY

3 sin x + cos x p p π x ®x →− x+ 6 6 6 æ 3 1ö 2ç sin x + cos x· ÷ 2ø è 2 lim f ( x ) = lim p p π x ®x →− x+ 6 6 6 π  2 sin  x +   6 lim f ( x ) = lim 1 π π π x →− x →− x+ 6 6 6   sin h p p  x ® - ® - + h = 2 lim  h →0  h  6 6 = 2 1  π f  −  = k ½  6 π For the continuity of f(x) at x = − , 6  π  = lim f ( x ) f −  π  6 x →− 6 or k = 2 ½ [CBSE Marking Scheme, 2016] (Modified) Sol.

lim f ( x ) = lim

Q. 3. Determine the values of 'a' such that the following function is continuous at x = 0 : ì x + sin x ï sin( a + 1) x , if - p < x < 0 ïï f(x) = í 2 , if x = 0 ï sin bx -1 ï 2e , if x > 0 bx îï

97

x + sin x Sol. lim− f(x) = lim− x→0 x →0 sin( a + 1)x sin x x = lim x →0 − sin( a + 1)x ( a + 1) ( a + 1)x 1+

=

2 a+1

lim f(x) = lim 2 x→0 + x →0 +

½ e sin bx − 1 bx

½

e sin bx − 1 sin bx = lim+ 2 × x →0 sin bx bx

½

f(0) = 2.

½



For the function to be continuous at 0, we must have lim+ f(x) = lim− f(x) = f(0) x→0



i.e., we must have

x→0

2 = 2 or a = 0; b may be any a+1 1

real number other than 0.

[CBSE SQP Marking Scheme, 2017-18] (Modified) p ì ïï k sin 2 ( x + 1), Q. 4. Find k, if f(x) = í tan x - sin x ï , ïî x3 is continuous at x = 0.

x£0 x>0 R&U [OD 2016]

R&U [SQP, 2017-18]

Topic-2

Differentiability Concepts Covered 



Left Hand Derivative,  Right Hand Derivative,

Relation between Continuity and Differentiability

Derivative of Some Standard Functions: d n (a) ( x ) = nx n −1 dx d x (c) ( a ) = a x log e a , a > 0 dx

(b)

d ( k ) = 0 , where k is any constant dx

(d)

d x (e ) = ex dx

1 1 d d 1 (log a x ) = = log a e (f) (e) (log e x ) = dx x log e a x x dx d d (cos x ) = − sin x (g) (sin x ) = cos x (h) dx dx

d (tan x ) = sec 2 x dx d (k) (cot x ) = −cosec 2 x dx

(i)

(j)

d (sec x ) = sec x tan x dx

(l)

d ( cosec x ) = − cosec x cot x dx

These questions are for practice and their solutions are available at the end of the chapter

Scan to know more about this topic

Introduction to Differentiation Part - 1

98

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

d 1 d 1 (sin −1 x ) = , x ∈( −1, 1) (n) , x ∈( −1, 1) (cos−1 x ) = − (m) dx dx 1 − x2 1 − x2



d 1 d 1 (cot −1 x ) = − , x ∈R (tan −1 x ) = , x ∈ R (p) (o) dx 1 + x2 dx 1 + x2



d 1 -1 , where x Î( -¥ , -1) È (1, ¥ ) (q) dx (sec x ) = x x2 - 1



d 1 -1 , where x Î( -¥ , -1) È (1, ¥ ) (r) dx ( cosec x ) = x x2 - 1



Following derivatives should also be memorized by you for quick use:



(i)

d dx

(ii)

1 d  1 =− x2 dx  x 



( x ) = 2 1x

Scan to know more about this topic

 Left Hand Derivative of f(x) at x = m, Lf ’(m) = lim

x → m−

Differentiability

f ( x ) − f ( m) and, x−m

Key Word Discontinuous Function: A discontinuous function is a function in algebra that has a point where either the function is not defined at that point or the LHL and RHL of the function are equal but not equal to the value of the function at that point or the limit of the function does not exist at the given point.

Right Hand Derivative of f(x) at x = m,

Rf ’(m) = lim

x → m+



 Derivative of y w.r.t. x:

Key Facts

f ( x ) − f (m) x−m

For a function to be differentiable at a point, LHD and RHD at that point should be equal. dy δy = lim . dx δx → 0 δx





• All differentiable functions happen to be continuous but not all continuous functions can said to be differentiable. • A function is said to be continuously differentiable if the derivative exists and

Also, for very-very small value h, f ’(x) = f ( x + h) − f (x ) , (as h ® 0) h

• f(x) = 0 is a continuous function because

Relation between Continuity and Differentiability:

it is an unbroken line, without holes or

(i) If a function is differentiable at a point, it is continuous at that point as well.

• If f(0) = ¥, then function is continuous

(ii) If a function is not differentiable at a point, it may or may not be continuous at that point.

is itself a continuous function.

jumps. at 0. • All polynomial functions are continuous

(iii) If a function is continuous at a point, it may or may not differentiable at that point.

functions.

(iv) If a function is discontinuous at a point, it is not be differentiable at that point.

Mnemonics



Rules of Derivatives:

Quotient Rule of Derivative



 Product or Leibniz's rule of derivatives:

Ho D Hi Minus Hi D Ho Over Ho Ho

d d d (uv ) = u ( v ) + v (u) dx dx dx



 Quotient Rule of derivatives:

d d d  u  v dx (u) − u dx ( v ) vu '− uv ' . = = dx  v  v2 v2

In mathematical notation,

Ho D Hi - Hi D Ho ho ho

where, Ho ® function in numerator Hi ® function in denominator D ® derivative of



CONTINUITY & DIFFERENTIABILITY

99

OBJECTIVE TYPE QUESTIONS A

Multiple Choice Questions

(C)

cos a sin 2 y

(D)



− cos a sin 2 y [CBSE Term-I 2021]

Q. 1. Differential of log[log(log x5)] w.r.t x is: (A)

5 5 (B) x log( x 5 )log(log x 5 ) x log(log x 5 )

(C)

5x 4 log( x )log(log x 5 ) 5

(D)



5x 4 log x log(log x 5 ) 5

Ans. Option (A) is correct. Explanation: Given, siny = xcos(a + y) sin y Þ x = cos( a + y ) Differentiating with respect to y, we get d d cos( a + y ) (sin y ) − sin y {cos( a + y )} dx dy dy = dy cos2 ( a + y )

[CBSE Term-I 2021] Ans. Option (A) is correct. Explanation: Let y = log[log(log x5)] dy dy 1 \ [log(log x 5 )] = dx log(log x 5 ) dx

(By Chain Rule) 1 1 d . log x 5 = 5 5 log(log x ) log x dx



1 1 d 5 = (x ) 5 5 5 log( x )log(log x ) x dx



5 = 5 5 x log( x )log(log x )

(A)

y + 4 x( x 2 + y 2 ) 4 y( x 2 + y 2 ) − x

dy is: dx



y − 4 x( x + y ) (C) 4 y( x 2 + y 2 ) − x 2

2

(B)

y − 4 x( x 2 + y 2 ) x + 4( x 2 + y 2 )



dy  4 x 2 y + 4 y 3 − x  +  4 x 3 + 4 xy 2 − y  = 0 dx  −  4 x 3 + 4 xy 2 − y  dy =  2 dx  4 x y + 4 y 3 − x  y − 4 x( x 2 + y 2 ) dy = 4 y( x 2 + y 2 ) − x dx

(A)

cos a cos2 ( a + y )



dx is: dy

(B)

cos[( a + y ) − y] cos2 ( a + y ) cos a cos2 ( a + y )

− cos a cos2 ( a + y )

Sometimes students commit errors in

u rule v

Answering Tip



Ans. Option (C) is correct. Explanation: Given, (x2 + y2)2 = xy 4 2 2 Þ x + 2x y + y4 – xy = 0 Differentiating w.r.t. x, we get dy  dy  dy   − y + x  = 0 4 x 3 + 2  2 xy 2 + 2 x 2 y  + 4 y 3 dx  dx  dx  

Q. 3. If sin y = x cos(a + y), then

cos( a + y )cos y + sin y sin( a + y ) cos2 ( a + y )

for differentiating the function.

4 y( x 2 + y 2 ) − x (D) y − 4 x( x 2 + y 2 ) [CBSE Term-I 2021]

or

cos( a + y )cos y − sin y[ − sin( a + y )] cos2 ( a + y )

Commonly Made Error



Q. 2. If (x2 + y2)2 = xy, then

dx = dy dx Þ = dy dx Þ = dy dx Þ = dy Þ

Differentiation rules for different functions and forms need continuous revision practice.

Q. 4. If y = sin(msin–1 x), then which one of the following equations is true ? 2 (A) (1 − x )

d2y dy +x + m2 y = 0 dx 2 dx

(B) (1 − x 2 )

d2y dy −x + m2 y = 0 2 dx dx

2 (C) (1 + x )

d2y dy −x − m2 y = 0 dx 2 dx

2 (D) (1 + x )

d2y dy +x − m2 y = 0 2 dx dx [CBSE Term-I 2021]

Ans. Option (B) is correct. Explanation: Given, y = sin(m(sin–1x)) ...(i) Differentiating both sides w.r.t x, we get

100

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

m dy −1 = cos( m sin x ) × dx 1 − x2



m cos( m sin −1 x ) dy = dx 1 − x2 −1 m cos( m sin x ) y' = 1 − x2

Þ Þ Þ

(

...(ii)

)

)

−1

= − m sin(m sin x ) 2

(C)

...(ii)

1 − x 2 y ' = mcos(msin–1 x) Differentiating again w.r.t. 'x', we get ( −2 x ) yn 1 − x 2 + y ' 2 1 − x2

(

 x2  d2 y equals: Q. 7. If y = loge  2  , then e  dx 2   1 1 (A) (B) - 2 x x

Þ Þ

1 − x2

Þ Þ

Q. 5. If x = 2cosq – cos2q and y = 2sinq – sin2 q, then dy is: dx (A)

cos θ + cos 2θ sin θ − sin 2θ

(B)

cos θ − cos 2θ sin 2θ − sin θ

(C)

cos θ − cos 2θ sin θ − sin 2θ

(D)

cos 2θ − cos θ sin 2θ + sin θ

dy = 2cosq – 2cos2q dθ

and

dy 2 cos θ − 2 cos 2θ = dx −2 sin θ + 2 sin 2θ dy cos θ − cos 2θ = dx sin 2θ − sin θ

\ or

Q. 6. If y = Ae5x + Be–5x, then

d2 y is equal to: dx 2

(A) 25y

(B) 5y

(C) – 25y

(D) 15y [CBSE Delhi Set-I 2020]

Ans. Option (A) is correct. Explanation: y = Ae5x + Be–5x



Þ Þ





dy = 5Ae5x – 5Be–5x dx d2y = 25Ae5x + 25Be–5x d 2x = 25y



y = 2loge x – logee2 y = 2loge x – 2 dy 2 = dx x -2 d2y = 2 2 dx x

Q. 8. The set of points where the function f given by f ( x )  | 2 x  1|sin x is differentiable is: ì1 ü (A) R (B) R – í ý î2 þ (C) ( 0 , ¥ ) (D) none of these Ans. Option (B) is correct. Explanation: Given that, f ( x ) = | 2 x - 1|sin x

[CBSE Term-I 2021] Ans. Option (B) is correct. Explanation: Given, x = 2cosq – cos2q and y = 2sinq – sin2q dx Therefore, = –2sinq + 2sin2q dθ

2 (D) - 2 x [CBSE Delhi Set-III 2020]



Ans. Option (D) is correct. Explanation: æ x2 ö Given, y = log e ç 2 ÷ èe ø

1

Þ yn(1 – x2) – xy' = –m2y Þ yn(1 – x2) – xy' + m2y = 0 d2y dy or (1 − x 2 ) 2 − x + m2 y = 0 dx dx

2 2 x

The function sin x is differentiable. The function | 2x − 1| is differentiable, except 2x - 1 = 0 1 Þ x= 2 1  Thus, the given function is differentiable R –   . 2 |x| Q. 9. The function f(x) = e is (A) continuous everywhere but not differentiable at x=0 (B) continuous and differentiable everywhere (C) not continuous at x = 0 (D) none of these Ans. Option (A) is correct.



Explanation: Given that, f ( x ) = e|x| The functions e x and| x | are continuous functions for all real value of x. Since e x is differentiable everywhere but | x | is nondifferentiable at x = 0. |x| Thus, the given function f ( x ) = e is continuous everywhere but not differentiable at x = 0.

Q. 10. Let f ( x ) = |sin x |, then: (A) f is everywhere differentiable (B) f is everywhere continuous differentiable at x = np, n Î Z.

but

not



CONTINUITY & DIFFERENTIABILITY

(C) f is everywhere continuous but not differentiable p at x = (2n + 1) , n Î Z. 2 (D) none of these Ans. Option (B) is correct. Explanation: Given that,

1

(C)

4-x

 1 −x 2  y = log    1 + x2 

The functions |x| and sin x are continuous function for all real value of x. Thus, the function f ( x ) = |sin x |, is continuous function everywhere. Now, |x| is non-differentiable function at x = 0. Since f ( x ) = |sin x |,is non-differentiable function at sin x = 0 Thus, f is everywhere continuous but not differentiable at x = np, n Î Z.

(

(B)



)

(

)

Differentiate with respect to x, we have dy d  d  log 1 − x 2  − log 1 + x 2  =  dx   dx dx  − 2x 2x = − 1 −x2 1 + x2

(

)

(

 2 = − 2x   1 − x 2 1 + x 2

(

1 x  dy , then dx is equal to: Q. 11. If y  log  2    x 1  

1 - x4

1 - x4

y = log 1 −x 2 −log 1 + x 2 .

⇒

2

(A)

-4 x 3

(D)



Ans. Option (B) is correct. Explanation: Given that,

f (x ) =|sin x |

4x3

4

101

= −

-4 x 1 - x4

)(

)

)

  

4x 1 −x 4

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each)



Q. 1. Let f(x) = x|x|, for all x Î R check its differentiability at x = 0.



Commonly Made Error

R&U [CBSE Delhi Set-III 2020]





Answering Tip

ìï - x 2 , x < 0 , differentiable at x = 0. ½+½ Sol. f ( x ) = í 2 îïx , x ³ 0 , [CBSE Marking Scheme 2020]

Mostly students find difficulty in differentiating modulus functions.

Learn the modulus function with its properties. dy for x < 0. dx

Q. 2. If y = x|x|, find

Detailed Solution: Here,





ìï - x 2 f(x) = í 2 ïî x

if x < 0 if x ³ 0

R [CBSE OD Set-I, 2019]

Q. 3. Differentiate e



x ®0+

Short Answer Type Questions-I (2 marks each)

x2 - 0 =0 x-0

f ( x ) - f (0) L.f ' (0) = lim x-0 x ®0-







-x 2 - 0 = lim =0 x ®0- x - 0







Therefore f(x) is differentiable at x = 0.

R.f ' (0) = L.f ' (0)

, with respect to x. R [CBSE OD Set-II, 2019]

f ( x ) - f (0) R.f ' (0) = lim + x-0 x ®0 = lim

3x

d2 y

dy

Q. 1. If y = ae2x + be–x, then show that - 2y dx 2 dx = 0.

[CBSE SQP-2020-21]

Q. 2. Find the value of y = sin q – sin 2q.

These questions are for practice and their solutions are available at the end of the chapter

dy p at q = , x = cos q – cos 2q, dx 3

R&U [CBSE OD Set I, II, III-2020]

102

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Sol.

\ \

Q. 4. Find the differential of sin2x w.r.t. ecosx.

dx = –sinq + 2 sin 2q dq dy = cosq – 2 cos 2q dq dy cos q - 2 cos 2q = dx - sin q + 2 sin 2q dy dx

q=

p 3

½



\ ½

= 3



½

dy ù dx úû q = p

æ pö æ pö cos ç ÷ - 2 cos 2 ç ÷ è 3ø è 3ø = æ pö æ pö 2 sin 2 ç ÷ - sin ç ÷ è 3ø è 3ø æ 1ö æ -1 ö - 2ç ÷ èç 2 ø÷ è 2ø = æ 3ö æ 3ö 2ç ÷ -ç ÷ è 2 ø è 2 ø







1 3 +1 2 = = 2 = 3 3 2 2



2 sin x cos x dy = sin x e cos x dz -2 cos x = or –2 cos x e–cos x e cos x ½+½ \

[CBSE Marking Scheme 2020] Detailed Solution Here, suppose u = sin2x, v = ecosx Then, we need to differentiate u w.r. to v du du dx i.e. = . dv dx dv







Þ

Now,

log u = (log x)2

Þ

1 du 1 = 2 log x. u dx x

Þ Again, Þ

Find the derivative first and then substitute the values. d2 y

d(sin 2 x ) du dx = dx = d( e cos x ) dv dx dx 2 sin x cos x 2 cos x du = cos x = - cos x e .( - sin x ) dv e

Q. 5. Find the derivative of xlog x w.r.t. log x.

3

Some students first substitute the value and then take the derivative which is wrong.

Q. 3. If x = a cos q; y = b sin q, then find

dz = –sin x.ecos x ½+½ dx

Sol. Let u = xlog x and v = log x

Answering Tip



y = sin2 x and z = ecosx dy = 2 sin x cos x dx

[CBSE OD Set-II 2020]

Commonly Made Error



and

[CBSE Marking Scheme 2020]

3



Sol. Let

½

Detailed Solution: Given, x = cosq – 2 cos2q y = sinq – sin2q Differentiate x = cosq – 2cos2q w.r.t q dx = –sinq + 2sin2q ...(i) dq Differentiate y = sinq – sin2q w.r.t q dy = cosq – cos2q ...(ii) dq On dividing eq(ii) by eq(i) dy cos q - 2 cos 2q dq = dx - sin q + 2 sin 2q dq cos q - 2 cos 2q dy = 2 sin 2q - sin q dx

[CBSE Delhi Set I, II, III-2020]

. dx 2 [CBSE Delhi Set-I,II,III 2020]

\

du 2 log x log x = 1 .x dx x v = log x dv 1 = ½ dx x du = 2.xlog x log x ½ dv [CBSE Marking Scheme 2020]v

Detailed Solution: Let u = xlogx, v = logx and u = xlogx taking log on both sides logu = logxlogx logu = logxlogx logu = (logx)2 Differentiate w.r.t. x

These questions are for practice and their solutions are available at the end of the chapter



CONTINUITY & DIFFERENTIABILITY









1 du 1 = 2 log x ´ u dx x du 2 log x = u· dx x







On dividing eq(i) by eqn. (ii) 2 log x du x log x . x dx = 1 dv x dx



The derivative of xlogx w.r.t. to logx is 2logx.xlogx.

6 1 1 − 9x 2 [CBSE Marking Scheme, 2017]



x.

Sol. From the given equation

 dy    x = 1 = dx

y=

p 4



dy dx

\

p 4

=

4

(

p

dy

find dx .

(6 x

)

1 − 9 x2 , −

1 3 2

½



Þ

½

π −x 4

dy = −1 dx

½

[CBSE Marking Scheme, 2018]

Short Answer Type Questions-II (3 marks each) 1 2 1 2

Q. 1. If y = e x sin

)

2 −1

0 2 (x − a) + 2 ( y − b) ⋅

dy dy x−a =⇒ 0 = −  dx dx y−b

1



( y − b ) − ( x − a) ⋅ dy d2y dx = − dx 2 ( y − b )2



=

−c 2

( y − b)

3

dy    By substitution ' dx ' 

=

( y − b) −

3

= − c 1

c2

( y − b)

3



Þ

  dy 2  1 +      dx   y−b

x−a dy = − y − b    dx

...(i)

...(ii)

...(iii)

2

\

( x − a)  dy 2 1+  = 1+ 2  dx  ( y − b)

Þ

 dy 2 ( x − a) + ( y − b ) 1+  = 2  dx  ( y − b)

2

2

2







Q. 12. If (x – a)2 + (y – b)2 = c2, for some c > 0, Prove that é æ dy ö 2 ù ê1 + ç ÷ ú êë è dx ø úû d2 y dx 2

c3

2

log (sin y ) + y tan x dy = dx log (cos x ) − x cot y

Þ





d2y = − dx 2 From eq (ii), we have



dy = log (siny) + y tanx dx

3/2

Detailed Solution: Given, (x – a)2 + (y – b)2 = c2 Differentiating w.r.t. x, we get dy 2(x – a) + 2(y – b) =0 dx dy Þ (x – a) + (y – b) = 0 dx Differentiating again w.r.t. x, we get 2  2 1 + (y – b) d y +  dy  = 0 dx 2  dx2   dy  2  d y − 1 +    Þ (y – b) = dx 2   dx  



dy dy + log(sin y ) = x cot y dx dx

 ( x − a )2  1 +  2  ( y − b )  = c2 − 3 ( y − b)

which is a constant independent of ‘a’ and ‘b’. ½  [CBSE Marking Scheme, 2019] (Modified)

dy y ⋅ tan x + log(sin y ) 1 ⇒ = dx log(cos x ) − x cot y 

d2y dx 2

 

dy ⇒ ⋅ log ( cos x ) + y ( − tan x ) 1 dx dy = log ( sin y ) + x ⋅ cot y ⋅ dx

3/2

107

c  dy 2 1 + 2   = Þ  dx  ( y − b) [from eq (i)] ...(iv)  3 / 2   dy 2 3 / 2 c2  c 3 ...(v) =  = Þ 1 +    3 2 ( y − b)   dx    ( y − b )  From eqns. (iii) and (iv), we get   dy 2  1 +    2 c 2 / ( y − b)   dx   d2y =– = – y−b ( y − b) dx 2 −c 2 = 3 ...(vi) ( y − b) From eqns. (v) and (vi), we obtain

½

These questions are for practice and their solutions are available at the end of the chapter

  dy 2 3 / 2 c3 1 +    3   dx   ( y − b) = = –c, which is independent 2 −c d2y 3 dx 2 ( y − b) of a and b. Hence Proved.

108

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

æ yö Q. 13. If log (x2 + y2) = 2 tan–1 ç ÷ , show that è xø dy x+y . = U [CBSE Delhi Set-III, 2019] dx x-y Sol. log(x2+y2) = 2 tan-1  y  x  

 dy   x ⋅ dx − y ⋅ 1  1  dy  1 2 2 2 x + y = ⋅ ⋅  2 2  x 2 + y 2  dx  x2  y    1+    x 2  dy  2x 2 1  dy  1 ⇒ 2 ⋅ ⋅ x − y  x+y = x + y 2  dx  x 2 + y 2 x 2  dx  ⇒ (x + y) = (x − y)

Differentiating both sides w.r.t. x,

dy dy x + y ⇒ =  dx dx x − y

1

[CBSE Marking Scheme, 2019] Detailed Solution:

Topper Answer, 2019 Sol.

t Q. 14. If x = cos t + log tan , y = sin t, then find the 2 π d2 y d2 y value of and at t = . 4 dt 2 dx 2

R&U [CBSE OD Set I, II, III-2019]

Q. 15. If x = a (2q – sin 2q) and y = a (1 – cos 2q), find π when q = . 3

dy dx

R&U [Delhi/OD, 2018]

Long Answer Type Questions (5 marks each) Q. 1. Find

dy .  sin 2 x  −1 1 − x  , if y = e  2 tan 1 + x  dx  A [S.Q.P. Dec. 2016-17] [HOTS]

 −1 1 − x  Sol. Putting x = cos 2q in 2 tan  , we get 1 1+ x 

These questions are for practice and their solutions are available at the end of the chapter

2 tan

−1

1 − cos 2θ 1 + cos 2θ



CONTINUITY & DIFFERENTIABILITY



i.e., 2 tan −1

2 sin 2 θ



2

2 cos θ

= 2 tan–1(tan q) = 2q = cos–1x sin 2 x

−1

1



Hence,



or log y = sin2x + log(cos–1x) 1 −1 1 dy × × or = 2 sin x cos x + −1 y dx cos x 1 − x2





y = e

cos

x

1 = sin 2 x − −1 cos x 1 − x 2

or

109

  dy 1 sin 2 x −1 = e cos x sin 2 x −  1 1 2 − dx  cos x 1 − x  [CBSE Marking Scheme, 2016] (Modified)

Q. 2. If x cos (a + y) = cos y, then prove that dy cos2 ( a + y ) d2 y = . Hence, show that sin a + dx sin a dx 2 sin2 (a + y)

dy = 0. dx

2

COMPETENCY BASED QUESTIONS Case based MCQs

(4 marks each)

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same: Ms. Remka of city school is teaching chain rule to her students with the help of a flow-chart The chain rule says that if h and g are functions and f(x) = g(h(x)), then

Q. 3. d (sin3 x ) = _______. dx (A) cos3 x (B) 3sin xcos x (C) 3sin2 x cos x (D) –cos3 x Ans. Option (C) is correct. Explanation: d d (sin 3 x ) = 3 sin 2 x (sin x ) dx dx = 3 sin 2 x cos x Q. 4.

d sin x3 _______. dx



(A) cos (x3) (B) –cos (x3)

Q. 1. Find

d (cos x 5 ). dx

(A) x4sinx5 (B) –5x4sinx5 4 5 (C) 5x sinx (D) 4x5sinx4 Ans. Option (B) is correct. d d Explanation: (cos x 5 ) = − sin x 5 ( x 5 ) dx dx = –sinx5(5x4) = –5x4sinx5 d Q. 2. Find sin(cos x ). dx (A) cos(cosx) (B) sinxcos(cosx) (C) –sinx.cos(cosx) (D) cosxsin(cosx) Ans. Option (C) is correct. d d Explanation: sin(cos x ) = cos(cos x ) (cos x ) dx dx = cos(cosx)(–sinx) = –sinx.cos(cosx)

(C) 3x2 sin (x3) (D) 3x2 cos (x3) Ans. Option (D) is correct. Explanation: d d (sin x 3 ) = cos x 3 (x3 ) dx dx = 3x 2 cos x 3 Q. 5. d (sin 2 x ) at x   is _______. 2 dx (A) 0 (B) 1

(C) 2 (D) –2

Ans. Option (D) is correct. Explanation: d d (sin 2 x ) = cos 2 x ( 2 x ) dx dx = 2 cos 2 x d  d p 2´ = = 2–2 cos p At x (sin,2 x ) (sin 2=x2) cos = 2cosp 2 dx 2 dxx = p 4

II. Read the following text and answer = 2 ( -1)the following questions on the basis of the same: = -2 A potter made a mud vessel, where the shape of the pot is based on f(x) = |x – 3| + |x – 2|, where f(x) represents the height of the pot. [CBSE QB-2021]

These questions are for practice and their solutions are available at the end of the chapter

110

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Case based Subjective (4 marks each) Questions



Q. 1. When x > 4 what will be the height in terms of x? (A) x – 2 (B) x – 3 (C) 2x – 5 (D) 5 – 2x Ans. Option (C) is correct. Explanation: The given function can be written as ì5 - 2 x , if x < 2 ï f ( x ) = í1, if 2 £ x < 3 ï2 x - 5, if x ³ 3 î When x > 4, f(x) = 2x – 5 Q. 2. Will the slope vary with x value? (A) Yes (B) No (C) Can’t say (D) Incomplete data Ans. Option (A) is correct. Explanation: ì -2 , if x < 2 ï f ¢( x ) = í 0 , if 2 £ x < 3 ï2 , if x ³ 3 î Q. 3. What is (A) 2

dy at x = 3: dx

(B) –2 (C) Function is not differentiable (D) 1 Ans. Option (C) is correct. Explanation: f(x) is not differentiable at x = 2 and x = 3. Q. 4. When the value of x lies between (2, 3) then the function is: (A) 2x – 5 (B) 5 – 2x (C) 1 (D) 5 Ans. Option (C) is correct. Explanation: In (2, 3), f(x) = 1 Q. 5. If the potter is trying to make a pot using the function f(x) = [x], will he get a pot or not? Why? (A) Yes, because it is a continuous function (B) Yes, because it is not continuous (C) No, because it is a continuous function (D) No, because it is not continuous [CBSE QB 2021] Ans. Option (D) is correct. Explanation: [x] is not continuous at integral values of x.

I. Read the following text and answer the following questions on the basis of the same: (Each Sub-part carries 2 marks) Reena started to read the notes on the topic 'differentiability' which she has prepared in the class of mathematics. She wanted to solve the questions based on this topic, which teacher gave as home work. She has written following matter in her notes : Let f(x) be a real valued function, then its Left Hand derivative (LHD) is : f ( a − h) − f ( a) Lf '(a) = lim h→0 −h



Right Hand Derivative (RHD) is : f ( a + h) − f ( a) Rf '(a) = lim h→0 h Also, a function f(x) is said to be differentiable at x = a if its LHD and RHD at x = a exist and one equal. x≥1  | x − 3 |,  For the function, f(x) =  x 2 3x 13 + , x 0 + – + –∞ ∞ 2 3 —

Q. 1. The function y = x2e–x is decreasing in the interval (A) (0, 2) (B) (2, ¥) (C) (–¥, 0) (D) (–¥, 0) È (2, ¥) [CBSE Board 2021] Ans. Option (D) is correct. Explanation: We have, f(x) = y = x2e–x dy \ = 2x e–x + x2(–1)e–x = xe–x(2 – x) dx – – + –∞ ∞ 0 2

(A) (–¥, 2) È (3, ¥) (C) (–¥, 2] È [3, ¥)

—

Multiple Choice Questions

or, x2 – 5x + 6 > 0 or, (x – 3)(x – 2) > 0 \ x Î (– ¥, 2] È [3, ¥) Q. 3. The interval on which the function f ( x )  2 x 3  9 x 2  12 x  1 is decreasing is: (A) [–1, ¥) (B) [–2, –1] (C) (–¥, –2] (D) [–1, 1] Ans. Option (B) is correct. Explanation: Given that, f ( x ) = 2 x 3 + 9 x 2 + 12 x - 1 f ¢( x ) = 6 x 2 + 18 x + 12 = 6( x 2 + 3x + 2 ) = 6( x + 2 )( x + 1)  decreasing. So, f '(x) < 0, for 

123



APPLICATIONS OF DERIVATIVES

On drawing number lines as below: +ve – –2

æ -p p ö Hence, f ¢( x ) > 0 , when cos x > 0 , i.e., x Î ç , . è 2 2 ÷ø

+ve

–1

æ -p p ö So, f ( x ) is increasing when x Î ç , è 2 2 ÷ø

We see that f ’ ( x ) is decreasing in [−2, −1]. 2

Q. 4. y  x( x  3) decreases for the values of x given by: (A) 1 < x < 3 (B) x < 0 3 (C) x > 0 (D) 0 < x < 2 Ans. Option (A) is correct.

æ p 3p ö and f ¢( x ) < 0 , when cos x < 0 , i.e., x Î ç , ÷ . è 2 2ø æ p 3p ö Hence, f ¢( x ) is decreasing when x Î ç , ÷ è 2 2ø

Explanation: Given that, \ \

æ p ö æ p 3p ö Since ç , p÷ Î ç , è 2 ø è 2 2 ÷ø

- 33))22 yy = = xx(( xx dy dy = - 33). ).11 + + (( xx - 33))22 ..11 = xx..22(( xx dx dx = 22 xx 22 - 66 xx + + xx 22 + + 99 - 66 xx = = = = = = =

æp ö Hence, f (x) is decreasing in çè , p÷ø 2

33xx 22 - 12 12 xx + + 99 22 33(( xx - 33xx - xx + + 33)) - 33)( )( xx - 11)) 33(( xx -

+ve

Q. 6. Which of the following functions is decreasing on    0, 2 .  

–

–2

+ve –1

y x ( x − 3) decreases for (1, 3). So,= [Since, y ’ < 0 for all x ∈ (1, 3) , hence y is decreasing on (1, 3)] 3 2 Q. 5. The function f ( x )  4 sin x  6 sin x  12 sin x  100 is strictly æ 3p ö (A) increasing in ç p , è 2 ÷ø 2

æp ö (B) decreasing in ç , p÷ è2 ø -p p ö (C) decreasing in æç , è 2 2 ÷ø æ pö (D) decreasing in ç 0 , ÷ è 2ø Ans. Option (B) is correct. Explanation: Given that, f ( x ) = 4 sin 3 x - 6 sin 2 x + 12 sin x + 100

(A) sin 2x (C) cos x Ans. Option (C) is correct.

(B) tan x (D) cos 3x

π 2



Explanation: In the given interval  0,   

f(x) = cos x On differentiating with respect to x, we get f´(x) = – sin x



π

which gives f ’ ( x ) < 0 in  0,   2



π

Hence, f ( x ) = cos x is decreasing in  0,  .  2 Q. 7. The function f ( x )  tan x  x (A) always increases (B) always decreases (C) never increases (D) sometimes increases and sometimes decreases Ans. Option (A) is correct. Explanation: We have, f ( x ) = tan x - x On differentiating with respect to x, we get

On differentiating with respect to x, we get f ′( x ) = 12 sin 2 x.cos x − 12 sin x.cos x + 12 cos x

f ′( x ) = sec 2 x − 1

= 12 cos x[sin 2 x − sin x + 1]

f ′( x ) = tan 2 x

⇒ f ′( x ) = 12 cos x[sin 2 x + 1(1 − sin x )]

Þ

2

⇒ 1 − sin x ≥ 0 and sin x ≥ 0

f ¢( x ) > 0 , " x Î R

So, f(x) always increases.

⇒ sin 2 x + 1 − sin x ≥ 0

SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) Q. 1. Show that the function given by f(x) = sin x is π  strictly decreasing in  , π . 2 

Sol. Consider,

f(x) = sin x f ’(x) = cos x

π  cos x < 0 for each x Î  , π 2 

\ f(x) < 0 [from (i)]

.......(i)

124

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

π  Hence, function is strictly decreasing in  , π . 1 2 

Q. 3. Find the interval for which the function f(x) = cot–1 x + x increases. f(x) = cot–1x +x -1 −1 + 1 + x 2 = + 1 1 + x2 1 + x2

Here x ¹ 0 The function decreases in the interval (–3, 0) È (0, 3)

x2 = ≥ 0 for all x Î R. 1 + x2 \ f increases for (–¥, ¥).

the intervals (– 3, 0) È (0, 3).

f ¢(x) =

1 3 3 x2

1 3 for decreasing, f ¢(x) < 0 Þ - 2 < 0 3 x



½ ½

Þ x2 < 9 Þ –3 < x < 3 since f(x) is not defined at x = 0. so f(x) decreasing in (– 3, 0) È (0, 3). 1 [CBSE Marking Scheme 2020] Detailed Solution:

Detailed Solution:

x 3 + 3 x 1 3 f ’(x) = - 2 3 x f(x) =



x 3 decreases in + 3 x

R&U [CBSE OD Set-III 2020]

Sol.

Commonly Made Error

1

Short Answer Type Questions-I (2 marks each) Q. 1. Show that the function f(x) =

f’(x) = 0 1 3 =0 3 x2 x2 - 9 =0 3x 2 x2 – 9 = 0, x not equal to 0 (x + 3)(x – 3) = 0 x = – 3, 3



f ’(x) =









Sol. Given,



Students add the function and apply quotient rule for finding the derivative which consumes time.

Answering Tips







Q. 2. Show that function y = 4x – 9 is increasing for all x Î R.



Practice more problems on finding increasing/ decreasing intervals.

Q. 2. Show that the function f defined by f(x) = (x – 1)ex + 1 is an increasing function for all x > 0. R&U [CBSE OD Set-I 2020]

Q. 3. Show that the function f(x) = x3 – 3x2 + 6x – 100 is increasing on R. R&U [NCERT] [O.D. Set I 2017] Sol. f(x) = x3 – 3x2 + 6x – 100 f ’(x) = 3x2 – 6x + 6 ½ = 3[x2 – 2x + 2] = 3[(x – 1)2 + 1] 1 since f’(x) > 0; xÎR \ f(x) is increasing on R. ½ [CBSE Marking Scheme, 2017]

Topper Answer, 2017 Sol.

These questions are for practice and their solutions are available at the end of the chapter

125







APPLICATIONS OF DERIVATIVES

Q. 4. Show that the function f(x) = 4x3 – 18x2 + 27x – 7 is always increasing or R. R&U [Delhi 2017]

Q. 5. Show that the function f given by f(x) =  π π tan–1 (sin x + cos x) is decreasing for all x ∈  ,  .  4 2 R&U [Foreign 2017]

Sol.

f '(x) =

cos x − sin x 1 + (sin x + cos x )2



1

1 + (sin x + cos x)2 > 0; x ÎR π π and < x < or cos x < sin x or cos x – sin x < 0 4 2

½

π π or f’(x) < 0 or f(x) is decreasing in  ,  .  4 2

½

[CBSE Marking Scheme 2017]

Short Answer Type Questions-II (3 marks each) Q. 1. Find the intervals in which the function f given by æ pö f(x) = tanx − 4x, x Î ç 0 , ÷ is è 2ø (a) strictly increasing (b) strictly decreasing R&U [CBSE SQP 2020-21]

Q. 2. Find the intervals in which the function f (x) = x4 - x 3 - 5 x 2 + 24 x + 12 is (a) strictly increasing, 4 (b) strictly decreasing. R&U [Delhi/OD 2018] f ‘(x) = x3 – 3x2 – 10x + 24 Sol. = (x – 2) (x – 4)(x + 3) f ‘(x) = 0 Þ x = –3, 2, 4. sign of f ‘(x): –



–∞

+ –3

– 2

1

+ 4

∞



\ f(x) is strictly increasing on (–3, 2) È (4, ¥) 1 and f(x) is strictly decreasing on (–¥, –3) È (2, 4) 1



[CBSE Marking Scheme, 2018] (Modified)

Sol. f(x) = –2x3 – 9x2 – 12x + 1 Now, f’(x) = – 6x2 – 18x – 12 = – 6[x2 + 3x + 2] = – 6[x2 + 2x + x + 2] f’(x) = –6(x + 1) (x + 2) 1 Put, f ’(x) = 0 Þ x = –2, x = –1 Þ Intervals are (–¥, –2), (–2, –1) and (–1, ¥) Getting f’(x) > 0 in (–2, –1) and f ’(x) < 0 in (–¥, –2) È (–1, ¥) 1 ⇒ f ( x ) is strictly increasing in (−2 , −1)

and strictly decreasing in (−∞, −2 ) ∪ (−1, ∞)

1 [CBSE Marking Scheme, 2018] (Modified)

Q. 4. Find the intervals in which the function given by f(x) = sin x + cos x, 0 £ x £ 2p is (i) increasing (ii) decreasing. R&U [NCERT] [Delhi Comptt., 2017] Sol. f(x) = sinx + cosx, 0 £ x £ 2p f’(x) = cosx – sinx f’(x) = 0 or cos x = sinx π 5π \ x = , 4 4

Sign of f’(x) +ve

–ve  4

0

1 1 1

+ve 5 4

2

 π   5π  So f(x) is strictly increasing in 0 ,  ∪  , 2 π    4  4  π 5π  and strictly decreasing in  ,  4 4 

1

[CBSE Marking Scheme, 2017] (Modified) Detailed Solution: f(x) = sin x + cos x or f’(x) = cos x – sin x Now, f’(x) = 0 gives sin x = cos x which gives π 5π x = , as 0 £ x £ 2p 4 4 tan x = 1 π 5π The points x = , divides the interval 4 4

Q. 3. Find the intervals in which the function f(x) = –2x3 – 9x2 – 12x + 1 is (i) Strictly increasing (ii) Strictly decreasing R&U [O.D Comptt Set I, II, III 2018] These questions are for practice and their solutions are available at the end of the chapter

π [0, 2p] into 3 disjoint intervals, 0 ,  ,  π , 5 π  and  4 4 4   5π   , 2 π  . 4 



126

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

 π   5π  Note that f’(x) > 0 if x Î 0 ,  ∪  , 2 π  or f(x) is  4  4 

Answering Tip

strictly increasing in intervals 0 , π  and  5 π , 2 π  .  4   4   π 5π  Also f’(x) < 0 if x Î  ,  4 4  or f(x) is strictly decreasing in this interval.



Sign of f ’(x)

Nature of function

 π 0 , 4  

+ ve

f is strictly increasing

 π 5π   ,  4 4 

– ve

f is strictly decreasing

 5π   , 2 π  4 

+ ve

Interval



f is strictly increasing

1

1

dy to 0 and getting critical point as cos q dθ



π = 0 i.e., q = 2



For all q, 0 £ q £



 π Hence, y is an increasing function of q on 0 ,  . 1  2

π dy , ³ 0 2 dθ

or f’(q) =

8 cos θ + 4 − 4 − 4 cos θ − cos2 θ ( 2 + cos θ)2

or f’(q) =

4 cos θ − cos2 θ ( 2 + cos θ)2







or here f’(q) is increasing when f’(q) ³ 0 cos θ( 4 − cos θ) i.e., ³0 ( 2 + cos θ)2 or cos q ³ 0

1

  4 − cos θ ≥ 0 ∀ θ ∈ R ∵ 2  ( 2 + cos θ)  or

 π q Î 0 ,   2



(4 – cos q) is always greater than 0.



Since – 1 £ cos q £ 1, (2 + cos q)2 > 0.

1

Long Answer Type Questions (5 marks each) Q. 1. Find the intervals on which the function f(x) = (x – 1)3(x – 2)2 is (a) strictly increasing (b) strictly decreasing. Sol.

Sometimes candidates do not have any basic knowledge of applications of derivatives while few candidates do not able to differentiate the functions with respect to q.

cos θ( 4 − cos θ) f’(q) = ( 2 + cos θ)2

\



1

[CBSE Marking Scheme, 2016] (Modified)

Commonly Made Error

( 2 + cos θ)4 cos θ − 4 sin θ( − sin θ) −1 ( 2 + cos θ)2

8 cos θ + 4(cos2 θ + sin 2 θ) − ( 2 + cos θ)2 or f’(q) = 1 ( 2 + cos θ)2



cos θ( 4 − cos θ) dy = ( 2 + cos θ)2 dθ

4 sin θ −θ 2 + cos θ

8 cos θ + 4 cos2 θ + 4 sin 2 θ −1 or f’(q) = ( 2 + cos θ)2

 π function of q on 0,  . R&U [Outside Delhi 2016]  2 [NCERT] [HOTS]



f(q) =

Now, f’(q) =

4sin θ − θ is an increasing Q. 6. Prove that y = 2 +cos θ

Equating

Give adequate practice on problems based on applications of derivatives.

Detailed Solution:

Q. 5. Determine for what values of x, the function 1 f(x) = x 3 + (x ¹ 0) is strictly increasing or strictly x3 decreasing. A [SQP Dec. 2016-17] [NCERT][HOTS]

Sol. Getting







Þ

R&U [CBSE OD Set I, II, III-2020]

f(x) = (x – 1)3 (x – 2)2 f ¢(x) = (x – 1)2 (x – 2) (5x – 8)



2



(a) for strictly increasing, f ¢(x) > 0



Þ (x – 1)2 (x – 2) (5x – 8) > 0



Þ

(x – 2)(5x – 8) > 0



Þ

8ö æ x Î ç -¥ , ÷ È (2, ¥) (as x ¹ 1) è 5ø

These questions are for practice and their solutions are available at the end of the chapter

(as x ¹ 1)



APPLICATIONS OF DERIVATIVES



Interval

æ 8ö x Î (– ¥, 1) È çè 1, ø÷ È (2, ¥) 1½ 5 (b) for strictly decreasing, f ¢(x) < 0

\

æ8 ö x Î çè , 2÷ø 1½ 5 [CBSE Marking Scheme 2020] (Modified)

Þ



Detailed Solution: 2

dv du   d ∵ dx (uv ) = u dx + v dx   



f’(x) = (x – 1)3. 2(x – 2) + (x – 2)2. 3(x –1)2 2 = (x – 1) (x – 2) [2(x – 1) + 3(x – 2)] = (x – 1)2 (x – 2) (2x – 2 + 3x – 6) or, f’(x) = (x – 1)2 (x – 2) (5x – 8) Now, put f’(x) = 0 or, (x – 1)­2 (x – 2) (5x – 8) = 0 Either (x – 1)­­2 = 0 or x – 2 = 0 or 5x – 8 = 0 +

–

1

8 5

–ve

x>2

(+)(+)(+)

+ve

æ (a) increasing on intervals ç - ¥ , è



æ8 ö (b) decreasing on interval ç , 2÷ è5 ø

8ö È [2, ¥) 5 ÷ø

Commonly Made Error





8 x = 1, , 2 5



\



Now, we find intervals and check in which interval f(x) is strictly increasing and strictly decreasing. Interval

f’(x) = (x – 1)2 (x – 2) (5x – 8)

Sign of f’(x)

x 0 such that f(c) > f(x), for all x in (c – h, c + h). The value f(c) is called the local maximum value of f. Maxima and (ii)  c is called a point of local Minima word minima if there exists a Problems number h > 0 such that f(c) < f(x), for all x in (c – h, c + h). The value f(c) is called the local minimum value of f. 2. Critical points It is a point c (say) in the domain of a function f(x) at which either f ’(x) vanishes i.e., f’(c) = 0 or f is not differentiable.

3. First Derivative Test: Consider y = f(x) be a well defined function on an open interval I. Now proceed as have been mentioned in the following algorithm: dy STEP 1: Find . dx dy = 0. dx Suppose c Î I (where I is the interval) be any critical point point and f be continuous at this point c. Then we may have following situations : dy   changes sign from positive to negative as x dx increases through c, then the function attains a local maximum at x = c. dy  changes sign from negative to positive as dx x increases through c, then the function attains a local minimum at x = c. dy   does not change sign as x increases dx through c, then x = c is neither a point of local maximum nor a point of local minimum.

STEP 2: Find the critical point(s) by putting

Rather in this case, the point x = c is called the point of inflection.

4. Second Derivative Test: Consider y = f(x) be a well defined function on an open interval I and twice differentiable at a point c in the interval. Then we observe that:  x = c is a point of local maxima if f ’(c) = 0 and f”(c) < 0. The value f(c) is called the local maximum value of f.  x = c is a point of local minima if f ’(c) = 0 and f”(c) > 0 The value f(c) is called the local minimum value of f. This test fails if f ’(c) = 0 and f”(c) = 0. In such a case, we use first derivative test as discussed above. 5. Absolute maxima and absolute minima: If f is a continuous function on a closed interval I, then f has the absolute maximum value and f attains it atleast once in I. Also f has the absolute minimum value and the function attains it atleast once in I. ALGORITHM STEP 1: Find all the critical points of f in the given interval, i.e., find all the points x where either f’(x) = 0 or f is not differentiable. STEP 2: Take the end points of the given interval. STEP 3: At all these points (i.e., the points found in STEP 1 and STEP 2) calculate the values of f. STEP 4: Identify the maximum and minimum value of f out of the values calculated in STEP 3. This maximum value will be the absolute maximum value of f and the minimum value will be the absolute minimum value of the function f. Absolute maximum value is also called as global maximum value or greatest value. Similarly absolute minimum value is called as global minimum value or the least value.

Key Facts • Rate of change itself is something that we use daily, like comparing one's salary, the weather or even how long it takes for a car to arrive some place. • When a cycle moves along a road, then the road becomes the tangent at each point when the wheel rolls on it. • Maxima and minina is used to solve optimization problems such as maximizing profit, minimizing the amount of material used in manufacturing or finding the maximum height a rocket can reach.



APPLICATIONS OF DERIVATIVES

129

OBJECTIVE TYPE QUESTIONS Multiple Choice Questions Q. 1. A function f : R ® R is defined as f(x) = x3 + 1. Then the function has (A) no minimum value (B) no maximum value (C) both maximum and minimum values (D) neither maximum value nor minimum value [CBSE Board 2021] Ans. Option (D) is correct. Explanation: Given, f(x) = x3 + 1 \

f '(x) = 3x2 and f "(x) = 6x

Put

f '(x) = 0

Þ 3x = 0 Þ x = 0 x = 0, f "(x) = 0

Thus, f(x) has neither maximum value nor minimum value. Q. 2. The absolute maximum value of the function f(x) 9  1 = 4 x − x 2 in the interval  −2 ,  is 2  2 (A) 8 (C) 6 Ans. Option (A) is correct.

(B) 9 (D) 10 [CBSE Board 2021]

1 2 Explanation: Given, f(x) = 4 x − x 2 \

On differentiating with respect to x, we get f ¢( x ) = 2 x - 8 So, f ¢( x ) = 0 Þ 2x - 8 = 0 Þ 2x = 8 \  x =4

So, x = 4 is the point of local minima. Minimum value of f(x) at x = 4 f(4) = 4×4−8×4+17 = 1 3 2 Q. 4. The function f ( x )  2 x  3 x  12 x  4 , has (A) two points of local maximum (B) two points of local minimum (C) one maxima and one minima (D) no maxima or minima Ans. Option (C) is correct. Explanation: We have,

1 f '(x) = 4 − ( 2 x ) = 4 − x 2

put f '(x) = 0 Þ 4 – x = 0 Þ x = 4 Then, we evaluate the f at critical point x = 4 and at 9  the end points of the interval  −2 ,  . 2 

1 f (4x) = 16 − (16 ) = 16 – 8 = 8 2



1 f(–2) = −8 − ( 4 ) 2

= –8 – 2 = –10

f ( x ) = x 2 - 8 x + 17

Now, Again on differentiating with respect to x, we get f ²( x ) = 2 > 0 , "x

2

At

Q. 3. If x is real, the minimum value of x 2  8 x  17 is (A) –1 (B) 0 (C) 1 (D) 2 Ans. Option (C) is correct. Explanation: Let,

9 9 1 9 f   = 4   −   2 2 2 2

= 18 −

2

81 = 7.875 8

9  Thus, the absolute maximum value of f on  −2 ,  2  is 8 occurring at x = 4.

f ( x ) = 2x 3 − 3x 2 − 12x + 4 f '( x ) = 6x 2 − 6x − 12 f '( x ) = 0

Now, 2

Þ 6( x - x - 2 ) = 0 Þ 6( x + 1)( x - 2 ) = 0 Þ x = -1 and x = +2 On number line for f’(x), we get + –

+

–1 2 Hence, x=−1 is point of local maxima and x = 2 is point of local minima. So, f(x) has one maxima and one minima. x

Q. 5. The maximum value of  1  is : x (A) e (B) ee 1/e

(C) e1/e

Ans. Option (C) is correct.

æ 1ö (D) ç ÷ è ø e [CBSE Board 2021]

y dx

1 è x ø x 1 = -1 + log x

Explanation: æ 1ö y = ç ÷ è xø

Let

x

1 x 1 dy 1 æ 1ö 1 = x. . ç - 2 ÷ + log .1 . 1 è x ø x y dx x 1 = -1 + log x

Þ

log y = x.log

\

x

dy 1 ö æ 1ö -1 . çè dx x ÷ø çè x ÷ø dy Now, = 0 dx 1 Þ log = 1 = log e x 1 Þ = e x 1 Þ x = e æ 1ö Hence, the maximum value of f ç ÷ = ( e )1 / e . è eø

æ Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII \ = log

130

x

x

dy 1 ö æ 1ö æ = ç log - 1÷ . ç ÷ è dx x ø è xø dy Now, = 0 dx 1 Þ log = 1 = log e x 1 Þ = e Very Short xAnswer Type 1 (1 mark each) Þ Questions x = e æ 1ö Q. 1. Find the smallest value of the polynomial x3 – 18x2 Hence, the maximum value of f ç ÷ = ( e )1 / e . è eø + 96x in [0, 9]. Sol. Given that, the smallest value of polynomial is f(x) = x 3 − 18x 2 + 96x On differentiating with respect to x, we get \

SUBJECTIVE TYPE QUESTIONS



Q. 2. Find the maximum value of sin x. cos x. Q. 3. Find the maximum slope of the curve y2 = –x3 + 3x2 + 9x – 27. Sol. Given that, y = - x 3 + 3x 2 + 9 x - 27 dy = -3x 2 + 6 x + 9 dx = Slope of the curve

\

f ¢( x ) = 3x 2 - 36 x + 96

and

So,

d2y = -6 x + 6 = -6( x - 1) dx 2 d2y 2 = 0 dx -6( x - 1) = 0 x = 1> 0

ff ¢¢(( xx )) = = 00 Þ Þ 3x - 36 36 xx + + 96 96 = = 00 2 Þ Þ 33(( xx 2 - 12 12 xx + + 32 32 )) = = 00 Þ ( x 8 )( x 4 ) = Þ ( x - 8 )( x - 4 ) = 00 Þ xx = Î[[00 ,, 99]] Þ = 88 ,, 44 Î

\

We shall now calculate the value of f(x) at these points and at the end points of the interval [0, 9], i.e., at x = 4 and x = 8 and at x = 0 and at x = 9.

So, the maximum slope of given curve is at x = 1.  dy  ∴ = −3 × 12 + 6 × 1 + 9  dx   ( x =1)

3x 22

Þ Þ Now,

f ( 4 ) = 4 3 − 18 × 4 2 + 96 × 4 =64 − 288 + 384 = 160  

f ( 8 ) = 8 3 − 18 × 8 2 + 96 × 8 =512 − 1152 + 768 = 128 f ( 9 ) = 9 3 − 18 × 9 2 + 96 × 9 =729 − 1458 + 864 = 135

and f ( 0 ) = 0 3 − 18 × 0 2 + 96 × 0 =0 Thus, we conclude that absolute minimum value of f(x) in [0, 9] is 0 occurring at x = 0. 1

3 d y = -6 < 0 dx 3

= 12

1

Short Answer Type Questions-I (2 marks each) Q. 1. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question? R&U [Delhi/OD 2018]

These questions are for practice and their solutions are available at the end of the chapter



APPLICATIONS OF DERIVATIVES

Sol. Let side of base = x and depth of tank = y

V V = x y Þ y = 2 , x 2





(V = Quantity of water = constant) Cost of material is least when area of sheet used is minimum.

4V A (Surface area of tank) = x + 4xy = x + x dA 4V = 2x – 2 dx x dA = 0 dx 2

2



Þ Þ







½



x3 = 2V, x3 = 2x2y, y =

½

½+½ \ Area is minimum, thus cost is minimum when

x

y = 2 1 Value: Any relevant value. [CBSE Marking Scheme, 2018] (Modified)

Topper's Answer, 2018



x3 x [as V = x2y] 2 = 2x 2

d2A 8V = 2 + 3 > 0, dx 2 x  

Detailed Solution:

Sol.

131

132

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII



Þ x = 25 when x = 25, y = 50 – 25 = 25 Hence, adjacent sides are x = 25 and y = 25 Q. 3. Find the minimum value of 4e2x + 9e–2x. Sol. Let f(x) = 4e2x + 9e–2x \ f '(x) = 8e2x – 18e–2x Put f '(x) = 0 Þ 8e2x – 18e–2x = 0





Commonly Made Error In most of the cases, candidates do not read the question attentively which results incorrect variable in area.

Answering Tip

½





1

Give adequate practice on mensuration related to concept and problems.



Q. 2. Calculate the adjacent sides of a rectangle with a given perimeter as 100 cm and enclosing the maximum area. U Sol. Let x and y be the adjacent sides of the rectangle, \ 2x + 2y = 100 Þ x + y = 50 ...(i) Let A be the area of rectangle. A \ A = xy Þ y = x

Using (i), we get



Þ



\ dA For maximum area = 0 Þ 50 – 2x = 0 dx

x+

A = 50 x

A = 50x – x2 dA = 50 – 2x dx

1

½



Þ Again

3  3 2 e = Þ x = log   2 2 2x

1

f"(x) = 16e2x + 36e–2x > 0 1 1 1/ 2    3 2 3 2. log   −2 log    3 2  2 2  +9×e Now, f log   = 4e 2     3 2 = 4× +9× 2 3

=6+6 = 12 1 Q. 4. It is given that x = 2, the function x3 – 12x2 + kx – 8 attains maximum value, on the interval [0, 3]. Find the value of k. U Sol. Let f(x) = x3 – 12x2 + kx – 8 \ f '(x) = 3x2 – 24x + k 1 It is given that function attains its maximum value of interval [0, 3] at x = 2 \ f '(2) = 0 Þ 3(2)2 – 24(2) + k = 0 Þ k = 36 1

Long Answer Type Questions (5 marks each) Q. 1. Show that the height of the right circular cylinder of greatest volume which can be inscribed in a right circular 4 cone of height h and radius r is one-third of the height of the cone, and the greatest volume of the cylinder is 9 Sol. times the volume of the cone. [CBSE Board 2020] Sol.

Topper Answer, 2020

133

Q. 2. Find the minimum value of (ax + by), where xy = c2. A [CBSE Delhi Set I, III-2020] Sol. Let S = ax + by, where y =

\

c2 x

S = ax +

bc 2 x



dS bc 2 = a – 2 dx x



dS = 0 dx

Þ

x2 =

or

x =













APPLICATIONS OF DERIVATIVES

d 2S

b .c a

1

2bc 2

1



1

é a 1ù ú >0 = 2bc2 ê êë b c úû

dx

2

x=

b .c a

=

bc 2 a

for a, b, c > 0 and x =

x3 3

b .c a

134

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

\ minimum value = a



b c2 .c + b. a c

2

a b

= 2 ab .c

æ q2 ö 2 S2 = ç - 2÷ + ( q - 1) è 4 ø



Then



To find nearest point let us suppose S2 = T 2

[CBSE Marking Scheme 2020] (Modified) Q. 3. Find the point on the curve y2 = 4x which is nearest to the point (2, 1). A [CBSE Delhi Set-II 2020] Sol. Let Q(x, y) be the point on curve y = 4x, which is nearest to the point P(2, 1)

æ q2 ö 2 T = ç - 2÷ + ( q - 1) è 4 ø



Then



Þ For critical points

2

æ q2 ö 2q + 2( q - 1) T’ = 2 ç - 2÷ ´ è 4 ø 4 T’ = 0

æ q2 ö 2q + 2( q - 1) = 0 2 ç - 2÷ ´ è 4 ø 4



Þ



æ q3 ö Þ ç 4 - 2 q÷ + ( 2 q - 2 ) = 0 è ø Þ q3 = 8 Þ q = 2 To find the maxima or minima

...(ii)

3q 2 4 3´2´2 Þ T” at (q = 2) = = 3 > 0 4 Therefore, T is least From (i) q2 = 4p Þ 22 = 4p Þ p = 1 Therefore, the required point is K(1, 2). Q. 4. Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its side. Also, find the maximum volume. T” =

2

æ y2 ö - 2÷ + (y – 1)2 Let (PQ) = S, then S = ç è 4 ø



1

2

æ y2 ö 2y y3 - 8 dS + 2( y - 1) = - 2÷ = 2 ç 1 4 dy è 4 ø 4 dS = 0 dy Þ \



y = 2 x = 1 2

1

d S 3y >0 2 = 1 dy 4 \ (1, 2) will be at minimum distance from (2, 1). ½ 2

[CBSE Marking Scheme 2020 (Modified)] Detailed Solution: Suppose the required point on the curve is K(p, q) and the given point is A(2, 1). \ q2 = 4p ...(i)

Then



Þ

AK =

R&U [CBSE OD Set I, II, III-2020] Q. 5. A given quantity of metal is to be cast into a solid half circular cylinder with a rectangular base and semi-circular ends. Show that in order that total surface area is minimum, the ratio of length of cylinder to the diameter of semi-circular ends is AE [CBSE SQP-2020-21] p : p + 2.

Sol. Let r be the radius and h be the height of half cylinder





Þ (PQ)2 = (x – 2)2 + (y – 1)2 ½



Volume =

1 2 πr h = V(constant) 2

( p - 2 )2 + ( q - 1)2 2





æ q2 ö 2 ç 4 - 2÷ + ( q - 1) è ø

½

[Let S = AK from (i) p =



S =

...(1)

q2 ] 4

Total surface area of half cylinder is

These questions are for practice and their solutions are available at the end of the chapter

1



2 S = 2  πr  + πrh + 2 rh 2 

...(2)



APPLICATIONS OF DERIVATIVES

From (1) put the value of h in (2)

 2V   2V  S = ( πr ) + πr  2  + 2 r  2  1 π πr  r 2



 1   4V



2 + 2V S = ( πr ) +    r   π 

 − 1  2  r

dS = ( 2 πr ) + dr dS For maxima/minima dr

 4V   π + 2V  ...(3)

=0

 − 1   4V

Þ



½

 1   4V + 2V π   π 2 + π

Þ

pr3 = V   ½  π  

Þ

V = ...(4) π+2

Þ



π2r3

π2r3 1 2 πr h = 2 π+2 π h = 2r π + 2

For maxima/minima

6h2R – 4h3 = 0 6R = 4h (h ¹ 0)

3R

h = 1 2 Differentiating (3) w.r.t. h Þ

d2Z = 12hR – 12h2 dh 2



d2Z  = 12  dh 2  h = 3 R

 3R  R − 12  3R      2 2

2

2 = 18R2 – 27R2 = –ve

1

 4V   π + 2V  = positive

so Z = A2 is maximum when h = when h =

(as all quantities are +ve) 1 So, S is minimum when height : diameter = p : (p + 2) Hence Proved ½ [CBSE SQP Marking Scheme, 2020] (Modified) Q. 6. Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral AE [CBSE SQP-2020-21] triangle.



Sol. Let 2r be the base and h be the height of triangle, which is inscribed in a circle of radius R

1 (base) (height) 2 1 A = (2r) (h) = rh  2

Area of triangle =

dZ =0 dh

Þ

Þ

Þ Þ height : diameter = p : (p + 2) Differentiating (3) with respect to r

d 2S  2 (2π) +  3  2 = r  dr

dZ = 6h2R – 4h3...(3)½ dh

Þ Þ

(2pr) =  2    r

From (1) and (4),

Area being positive quantity, A will be maximum or minimum if A2 is maximum or minimum. Z = A2 = r2 h2 ...(2) Now in triangle OLB, BL2 = OB2 – OL2 In DOBD, Z = A2 = r2 h2, r2 = R2 – (h – R)2 Þ r2 = 2hR – h2 put in (2), Z = h2 (2hR – h2) Þ Z = (2h3R – h4) Þ

+ 2V  = 0 Þ ( 2 πr ) +  2    π  r

135

...(1)

3R = 2r2x – x3 1 2

3R  3R  3R 2 −  , r = 2hR – h2 = 2 R .  2  2 2 r2 =

2

3R 2 4

3R 1 2 3R π h 1 = 2 = 3,θ = tan q = 3 r 3R 2 r =

Triangle ABC is equilateral triangle. Hence Proved. [CBSE SQP Marking Scheme, 2020] (Modified) Q. 7. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere 4r of radius r is . Also show that the maximum 3 8 volume of the cone is of the volume of the 27

sphere. ½

A [NCERT], [CBSE Delhi Set I-2019],



[Delhi Set I, II, III-2016], [Delhi Set I, II, III Comptt. 2016] These questions are for practice and their solutions are available at the end of the chapter

136

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Q. 8. Find the point on the curve y2 = 4x, which is nearest to the point (2, – 8).



Now,

AE [CBSE OD Set-II, 2019]





Sol. Let P(x, y) be any point on the curve y = 4x



y3 + 16 4 For maximum or minimum value of z, we have



dz = 0 dy

=



2

(x − 2) + ( y + 8) 2

 

z = AP =

2

1



2

 y2  2 let s = z 2 =  − 2  + ( y + 8 )   4 

d 2s 3y2 =  dy 2 4

 



y3 Þ + 16 = 0 4 Þ y3 + 64 = 0 2 Þ (y + 4)(y – 4y + 16) = 0 Þ y = – 4 [ y2 – 4y + 16 = 0 gives imaginary values of y ] d2z 1 3 = × 3y2 = y2 Now, dy 2 4 4



 y2  y  ds y3   = 2  − 2   + 2 ( y + 8) = + 16  dy 4  4  2 

1

½

ds Let = 0 ⇒ y3 = −64 ⇒ y = −4  dy

1

d s 3(16) = >0  dy 2  y = −4 4  y2 = 4 4



\ s or z is minimum at y = –4; = x



\ The nearest point is P(4, –4) 1 [CBSE Marking Scheme, 2019] (Modified)

Detailed Solution: The equation of the given curve is y2 = 4x



y2 ...(i) 4 Let P(x, y) be a point on the curve, which is nearest to point A(2, – 8). Now, distance between the points A and P is given by  : \





x =

AP =



Thus, z is minimum when y = – 4



Substituting y = – 4 in the equation of the curve y2 = 4x, we obtain x = 4.



Hence, the point (4, – 4) on the curve y2 = 4x is nearest to the point (2, – 8).

=











Let

A [NCERT] [CBSE Delhi Set II-2019]

Sol. Let the radius and height of cylinder be r and h respectively \ V = pr2h ...(i) h2 But r2 = R2 – ½ 4

=

 y4  2 2  16 − y + 4 + ( y + 16 y + 64 )

=

y4 + 16 y + 68 16

z = AP2 =

y4 + 16 y + 68 16

  h2  h3  πh  R 2 −  = π  R 2 h −  4 4   2   h 3 dV 2 = π R − or 4   dh For maximum or minimum dV 4R2 \ = 0 or h2 = 3 dh \



... using equ. (i)



y =−4

Q. 9. Prove that the height of the cylinder of maximum volume, that can be inscribed in a sphere of radius R 2R . Also find the maximum volume. is 3



2

3 = (−4 )2 = 12 > 0 4

for

( x − 2 )2 + ( y + 8 )2  y2  2  4 − 2 + ( y + 8 )

d2z dy 2



2

½

dz 1 = × 4 y 3 + 16 dy 16

1 ½



APPLICATIONS OF DERIVATIVES



and





h =

2R 3

 6h  d 2V = π −  < 0 1 2 dh  4 



=

3 3

cubic units



dV = 0 or h = x or h = 3x dx

i.e.,

 2R 1  2R  3  −  Maximum volume = π· R 2 ·   3 4  3    4 πR 3

dV = ptan2a (h – x) (h – 3x) dx

1

or

x = d 2V

1

Hence Proved.



R&U [SQP 2018-19] walls. Q. 11. Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle a, is one-third that of the cone. Hence find the greatest volume of the cylinder. A [NCERT][Delhi Comptt., 2017]

Sol.

dx 2 d 2V

h 3

1

= ptan2a(6x – 4h) h 3



\



\ V is maximum, at x =



and maximum volume is V =

[CBSE Marking Scheme, 2019] (Modified) Q. 10. A cuboidal shaped godown with square base is to be constructed. Three times as much cost per square meter is in curved for constructing the roof as compared to the walls. Find the dimensions of the godown if it is to enclose a given volume and minimize the cost of constructing the roof and the

137

dx 2

< 0 at x =

1

h 3 4 ph3 tan2 a 27 Hence Proved. 1

[CBSE Marking Scheme, 2017] (Modified) Q. 12. A wire of length 34 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a rectangle whose length is twice its breadth. What should be the lengths of the two pieces, so that the combined area of the square and the rectangle is minimum ? A [Foreign 2017] Sol. Let the length of one piece be x m, then length of the other piece = (34 – x) m x \ Side of square is m and perimeter of rectangle 4



 l 2  l +  = 34 – x  2 2



r = tan a h−x

1

r = (h – x) tan a ½ Volume of cylinder V = pr2x V = p(h – x)2 x tan2 a ½ dV = ptan2a {(h – x)2 + 2x(h – x)(–1)} dx

2

æ xö æ 34 - x ö Area (A) = ç ÷ + 2 ç è 4ø è 6 ÷ø

½

or

dA x 1 = - ( 34 - x ) dx 8 9

1



dA = 0 or x = 16 dx

½





Now,

also,

d2 A

1 1 = + >0 dx 8 9 2

so, A is minimum when x = 16

= ptan2a (h – x) (h – x – 2x)

These questions are for practice and their solutions are available at the end of the chapter

1

\ Lengths of the two pieces are 16 m and 18 m. 1 [CBSE Marking Scheme, 2017] (Modified)

138

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

COMPETENCY BASED QUESTIONS Case based MCQs

(4 marks each)

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:

In a residential society comprising of 100 houses, there were 60 children between the ages of 10-15 years. They were inspired by their teachers to start composting to ensure that biodegradable waste is recycled, For this purpose, instead of each child doing it for only his/her house, children convinced the Residents welfare association to do it as a society initiative. For this they identified a square area in the local park. Local authorities charged amount of `50 per square metre for space so that there is no misuse of the space and Resident welfare association takes it seriously. Association hired a labourer for digging out 250 m3 and he charged `400 x (depth)2. Association will like to have minimum cost.

Put

dC = 0 dh



−12500 + 800 h = 0 h2

\

Þ

800h3 = 12500 125 h3 = 8

Þ Þ Q. 3.

h =

5 = 2.5 m 2

d 2C is given by dh2

(A)

25000 + 800 h3

(C)

100 + 800 h3

500 + 800 3 h 500 (D) 3 + 2 h (B)

Ans. Option (A) is correct. Explanation: dC −12500 + 800 h \ = h2 dh \



d 2C −( −2 ) × 12500 + 800 = h3 dh 2 d 2C 25000 + 800 = h3 dh 2

Þ

Q. 4. Value of x (in m) for minimum cost is (A) 5

[CBSE Board 2021] Q. 1. Let side of square plot is x m and its depth is h metres, then cost C for the pit is 50 12500 + 400 h 2 + 400 h 2 (A) (B) h h 250 + h2 h Ans. Option (B) is correct. (C)

250 + 400 h 2 h

(D)

250 × 50 + 400 × h 2 h

Explanation:

C =

Þ

12500 + 400 h 2 C = h



dC Q. 2. Value of h (in m) for which = 0 is dh (A) 1.5 (B) 2 (C) 2.5 (D) 3 Ans. Option (C) is correct. 12500 + 400 h 2 Explanation: C = h \



dC −12500 + 800 h = dh h2

(C) 5 5 Ans. Option (D) is correct.

(B) 10

5 3

(D) 10

dC Explanation: For minimum cost, put = 0, we get dh At

h = 2.5 m d 2C h = 2.5, 2 > 0 dh (Hence, minimum)

Value of x at minimum cost x =



=

400 × ( 2.5)2 250 2500 = 10 m 250

Q. 5. Total minimum cost of digging the pit (in `) is (A) 4,100 (B) 7,500 (C) 7,850 (D) 3,220 Ans. Option (B) is correct. Explanation: Total minimum cost, 12500 + 400 h 2 (At 2.5) C = h Þ



C =

12500 + 400( 2.5)2 2.5



APPLICATIONS OF DERIVATIVES

Þ Þ

C = 5000 + 2500 C = `7500

II. Read the following text and answer the following questions on the basis of the same: The shape of a toy is given as f(x) = 6(2x4 – x2). To make the toy beautiful 2 sticks which are perpendicular to each other were placed at a point (2, 3), above the toy. [CBSE QB-2021]

139

Ans. Option (B) is correct. Explanation: We have dy ù = 360 dx úû( 2 , 3 ) \

dy (x - x¢) dx ( y - 3) = 360 ( x - 2 ) y - 3 = 360 x - 7220 y = 360 x - 717

( y - y¢) =

Q. 4. Find the second order derivative of the function at x = 5. (A) 598 (B) 1,176 (C) 3,588 (D) 3,312 Ans. Option (C) is correct. Explanation: f ( x ) = 6( 2 x 4 - x 2 ) f ¢( x ) = 6 [ 8 x 3 - 2 x ] f ²( x ) = 6 [ 24 x 2 - 2] f ²( 5) = 6 [ 24 ´ 25 - 2] = 6 [600 - 2] = 3588



Q. 1. Which value from the following may be abscissa of critical point? (A) ± 1/4 (B) ± 1/2 (C) ± 1 (D) None of these Ans. Option (B) is correct. Q. 2. Find the slope of the normal based on the position of the stick. (A) 360

(B) –360

1 (C) 360

-1 (D) 360

Ans. Option (D) is correct. Explanation: Slope of the normal based on the position of the stick -1 = f ¢( x ) f ¢( x ) = 6 [ 8 x 3 - 2 x ] f ¢( 2 ) = 6 [ 8 ´ 8 - 2 ´ 2 ] = 6 [64 - 4] = 360 \

Slope =

-1 360

Q. 3. What will be the equation of the tangent at the critical point if it passes through (2, 3)? (A) x + 360y = 1082 (B) y = 360x – 717 (C) x = 717y + 360 (D) None of these

Q. 5. At which of the following intervals will f(x) be increasing? -1 ö æ 1 ö æ -1 1 (A) ç - ¥ , ÷ È ç , ¥÷ (B) æç , 0ö÷ È æç , ¥ö÷ è è 2 ø è2 ø 2 ø è2 ø 1 1 (C) æç 0 , ö÷ È æç , ¥ö÷ è 2ø è 2 ø

-1 ö æ 1 ö æ (D) ç - ¥ , ÷ È ç 0 , ÷ è 2 ø è 2ø

Ans. Option (B) is correct. Explanation: For increasing f ¢( x ) > 0 6 (8x 3 - 2x ) > 0 i.e., Þ and

Þ Þ and

x( 4 x 2 - 1) > 0 4x2 - 1 > 0 x > 0 4x2 > 1 1 x2 > 4 1 x > 2 x > -

1 2

æ -1 ö æ 1 ö i.e., x Î ç , 0÷ È ç , ¥÷ è 2 ø è2 ø III. Read the following text and answer the following questions, on the basis of the same: The relation between the height of the plant (y in cm) with respect to exposure to sunlight is 1 governed by the following equation y = 4x – x2 2 where x is the number of days exposed to sunlight. [CBSE QB 2021]

140

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII

Q. 1. The rate of growth of the plant with respect to sunlight is ______ . 1 (A) 4 x –   x 2 2

(B) 4 – x

(C) x – 4

(D) x –

Ans. Option (B) is correct. Explanation:

1 2 x 2

1 y = 4x - x2 2 \ rate of growth of the plant with respect to sunlight dy = dx d é 1 ù = 4x - x2 ú dx êë 2 û = ( 4 - x ) cm / day

Q. 2. What is the number of days it will take for the plant to grow to the maximum height? (A) 4 (B) 6 (C) 7 (D) 10 Ans. Option (A) is correct. Explanation: dy = 4-x dx The number of days it will take for the plant to grow to the maximum height, dy  0 dx 4x  0 x  4 days. Q. 3. What is the maximum height of the plant? (A) 12 cm (B) 10 cm (C) 8 cm (D) 6 cm Ans. Option (C) is correct. Explanation: We have, number of days for maximum height of plant = 4 Days \ Maximum height of plant 1 y( x = 4 ) = 4 ´ 4 - ´ 4 ´ 4 Þ 2 = 16 - 8 = 8 cm

Q. 5. If the height of the plant is 7/2 cm, the number of days it has been exposed to the sunlight is ______ . (A) 2 (B) 3 (C) 4 (D) 1 Ans. Option (D) is correct. Explanation: 7 Given, y = 2 1 2 7 i.e., 4x − x = 2 2 8x − x 2 = 7 x 2 − 8x + 7 = 0 x2 − 7x − x + 7 = 0 x( x − 7 ) − ( x − 7 ) = 0 x = 1, 7 We will take x = 1, because it will take 4 days for the plant to grow to the maximum height i.e. 7 8 cm and cm is not maximum height so, it will 2 take less than 4 days. i.e., 1 day. IV. Read the following text and answer the following questions, on the basis of the same: P(x) = –5x2 + 125x + 37500 is the total profit function of a company, where x is the production of the company. [CBSE QB 2021]

Q. 1. What will be the production when the profit is maximum? (A) 37,500 (B) 12.5 (C) – 12.5 (D) – 37,500 Ans. Option (B) is correct.



Q. 4. What will be the height of the plant after 2 days? (A) 4 cm (B) 6 cm (C) 8 cm (D) 10 cm Ans. Option (B) is correct. Explanation: Height of plant after 2 days 1 = y( x = 2 ) = 4 ´ 2 - ´ 2 ´ 2 2 = 8-2 = 6 cm

5x 2 − 125x + 750 = 0



x 2 − 25x + 150 = 0 x ( x − 15) − 10 ( x − 15) = 0 ( x − 10 ) ( x − 15) = 0 APPLICATIONS OF DERIVATIVES x = 10 , 15

Explanation: We, have

P( x ) = −5x 2 + 125x + 37500

P( x ) = -5x 22 + 125x + 37500 P ¢( x ) = -10 x + 125 For maximum profit P ¢( x ) = 0 -10 x + 125 = 0 -10 x = -125 125 x = 10 = 12.5 Q. 2. What will be the maximum profit? (A) ` 38,28,125 (B) ` 38,281.25 (C) ` 39,000 (D) None of these Ans. Option (B) is correct. Explanation: Maximum profit = P (12.5)

P(10 ) = −5 × 10 2 + 125 × 10 + 37500 = −500 + 1250 + 37500 = ` 38 , 250 Hence, production of company is 10 units when the profit is `38250.

Case based Subjective Questions (4 marks each) I. Read the following text and answer the following questions on the basis of the same:

= -5 (12.5)2 + 125 ´ 12.5 + 37500 = -781.25 + 1562.5 + 37500 = 38 , 281.25



Q. 3.  Check in which interval the profit is strictly increasing. (A) (12.5, ∞) (B) for all real numbers (C) for all positive real numbers (D) (0, 12.5) Ans. Option (D) is correct. Q. 4. When the production is 2 units what will be the profit of the company? (A) 37,500 (B) 37,730 (C) 37,770 (D) None of these Ans. Option (B) is correct. Explanation: When production is 2 units, then profit of company = P(2) = -5 ´ 2 2 + 125 ´ 2 + 37500 = -20 + 250 + 37500 = 37 , 730 Q. 5. What will be production of the company when the profit is ` 38,250? (A) 15 (B) 30 (C) 10 (D) data is not sufficient to find Ans. Option (C) is correct. Explanation: Profit 38 ,, 250 250 Profit = = 38 ii..ee.,., − −55xx 22 + + 125 + 37 125xx + 37 ,, 500 500 55xx 22 − 125 x − 125x + + 750 750 x 22

+ 150 − 2255xx + x − 150 xx (( xx − − 15 ) − ( x − 10 15) − 10 ( x − 15 15)) (( xx − − 10 ) ( x − 15 10 ) ( x − 15)) xx P(( xx )) P

= = = =

38 38 ,, 250 250 00

= 00 = = 00 = = 00 = = 10 10 ,, 15 15 = =− −55xx 22 + + 125 125xx + + 37500 37500 =

P( 10 )) = =− −55 × × 10 10 22 + + 125 125 × × 10 10 + + P(10 37500 37500

141



(Each Sub-part carries 2 marks) Revathi is going to her best friend's birthday party. She has made a beautiful painting for her friend as a birthday gift. Now, she wants to prepare a hand made gift box for that painting. For making lower part of the box, she takes a square piece of cardboard of side 30 cm.

Q. 1. If x can be the length of each side of the square cardboard which is to be cut off from corners of the square piece of side 30 cm, then write the expression for volume of the open box formed by folding up the cutting corners. Also, find

dV . dx

Sol. Given, side of square is of length 30 cm. Also, height of the open box = x cm Length of open box = 30 – 2x and width of open box = 30 – 2x \ Volume (V) of open box = x × (30 – 2x) × (30 – 2x) = x(30 – 2x)2 dV Now, = x[2(30 – 2x)(–2)] + 1(30 – 2x)2 1 dx



or,

dV = –4x(30 – 2x) + (30 – 2x)2 dx





or,

dV = (30 – 2x) (–4x + 30 – 2x) dx





or,

dV = (30 – 2x) (30 – 6x) dx

1

142

Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII



Q. 2. Revathi is inserted in maximising the volume of the box. So, what should be the side of square to be cut off so that volume of the box is maximum? dV Sol. Since, = (30 – 2x)(30 – 6x) dx [from above]

\



or,



or,

d 2V = –2(30 – 2x) + (30 – 2x)(–6) dx 2 d 2V = –2(30 – 6x + 90 – 6x) dx 2 d 2V = –2(120 – 12x) dx 2

d 2V = –24(10 – x) dx 2 dV put =0 dx



or,



Now,



Þ (30 – 2x)(30 – 6x) = 0 Þ x = 15 or x = 5



d 2V At x = 15, = –24(10 – 15) = –24 × (–5) 1 dx = 120 > 0 d 2V At x = 5, = –24(10 – 15) = –24 × 5 dx = –120 < 0 So, volume will be maximum when x = 5.



1

Solutions for Practice Questions (Topic-1)



Here, R(x) = 3x + 36x, where x is the number of sets of books dR = 6x + 36 dx Þ

7.

So the marginal cost of food for 1200 such sets. = `7,236 f(x) = x3 + 6x2 + 5x + 3 df ( x ) = 3x2 + 12x + 5 dx



 df ( x )  = 3 × (3)2 + 12 × 3 + 5  dx    x =3 = 27 + 36 + 5 = 68 units

1







\



Þ





Now,







3. Let AB be the ladder and OB be the wall.

Marginal contentment

Let x denote the edge of cube, V denote the volume and s denotes the surface area of cube at instant t. dV = 9 cm2/sec dt



1

Short Answer Type Questions-II

1

Short Answer Type Questions-I 2.



 dR   dx  x =1200 = 6(1200) + 36 = 7,236





12 × 9 36 = 3 × 10 10 = 3.6 cm2/sec =



2



3.



Very Short Answer Type Questions

Þ



V = x3 dV dx = 3x2 dt dt 9 dx = 3x 2 dt s = 6x2 dx ds = 12x × dt dt  dx  9 = 12 × 10 ×   dt x =10 3 × 10 × 10

1



At any instant, let OA = x m and OB = y m



Also, AB = 5 m \ x2 + y2 = 52 = 25 Differentiating (i) w.r.t. 't', we get dx dy 2x + 2y =0 dt dt



Þ





x

dx dy +y =0 dt dt



When x = 4, then from (i), we get y2 = 25 – x2 = 25 – 16 = 9







dx Also, given = 2 cm/sec = 0.02 m/sec dt



Using these values in (ii), we get







Þ

(given) ...(i) 1

y = 3

4 × 0.02 + 3

dy =0 dt dy 0.08 8 = − m/sec = − cm/sec dt 3 3

1



APPLICATIONS OF DERIVATIVES



Let r be the radius and l be the slant height of water surface π r sin = 4 l



Long Answer Type Questions 1.

1 2 πl = 2 dl dS 1 π.2l How, = dt dt 2



Given that



dl − 2 = dt πl

Þ

 dl  − 2 − 2 = cm/s   = dt l = 4 π.4 4π



1

1

Commonly Made Error

1 1

dS = –2 cm2/s dt

1

At l = 4 cm





1 r = l. or 2 Curved surface, S = prl 1 .l \ S = π.l. 2

π dl .2 l \ –2 = 2 dt

Many candidates considered volume as function instead of surface area. Majority of candidates did not apply proper sign through it was specified in the question that find the rate of decrease of slant height of the water.

Answering Tip







Thus, the height of the ladder on the wall is 8 decreasing at the rate of cm/sec. 1 3

143

The difference between rate of increase and rate of decrease need to be understood carefully. Practice a number of problems based on application of derivatives.

Solutions for Practice Questions (Topic-2) Very Short Answer Type Questions 2. Given,

Short Answer Type Questions-II

y = 4x – 9 dy = 4 > 0 for all x Î R. dx

Hence, function is increasing for all x Î R.

1

Short Answer Type Questions-I 2. f ¢(x) = xex 1 Now x > 0 and ex > 0 for all x ½ \ f ¢(x) > 0 Þ f is an increasing function ½ [CBSE Marking Scheme 2020] Detailed Solution: f(x) = (x – 1)ex + 1, x > 0 f’(x) = (x – 1)ex + ex(1 – 0) + 0 = xex – ex + ex = xex For all x > 0 x.ex > 0.e0 xex > 0 f’(x) > 0 Hence, function f(x) = (x – 1)ex + 1 is an increasing function for all x > 0. 4. f(x) = 4x3 – 18x2 + 27x – 7 f’(x) = 12x2 – 36x + 27 ½ 2 = 3(2x – 3) ³ 0; x ÎR 1 \ f(x) is increasing on R. ½ [CBSE Marking Scheme, 2017]

1. f(x) = tanx – 4x f ’(x) = sec2x – 4 (a) For f(x) to be strictly increasing f ’(x) > 0 2 Þ sec x – 4 > 0 Þ sec2x > 4 1 Þ cos2x < 4



æ 1ö Þ cos2x < ç ÷ è 2ø Þ

-

½

2

1 1 < cosx < 2 2

p p 4 1 Þ cos2x > 4 Þ

æ 1ö Þ cos2x > ç ÷ è 2ø

2

1½



Oswaal CBSE Question Bank Chapterwise & Topicwise, MATHEMATICS, Class-XII





144

Þ cosx >

Þ

1 2

é æ pö ù ê∵ x Î çè 0 , 2 ÷ø ú ë û

0 < x
0 if x > 1 or x < – 1, and f’(x) < 0 if – 1 < x < 1







or



or x < – 1; and strictly decreasing in



(– 1, 0) È (0, 1)

1 1

[CBSE SQP Marking Scheme, 2016] (Modified)

Long Answer Type Questions f(x) = sin 3x – cos 3x, 0