Oswaal NTA CUET (UG) Question Bank Chapterwise & Topicwise Mathematics/Applied Math (For 2024 Exam) 9789359587226, 9359587222

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For 2024 Exam

Highly Recommended

CHAPTER-WISE

CHAPTER-WISE

QUESTION BANK

QUESTION BANK Includes SOLVED PAPERS (2021 - 2023)

MATHEMATICS / APPLIED MATTHEMATICS

MATHEMATICS/ APPLIED MATHEMATICS Section-II (Domain Specific Subject) Strictly as per the Latest Exam Pattern issued by NTA

The ONLY book you need to #AceCUET(UG)

SECTION II

1

2

3

4

5

100% Updated

Previous Years’ Questions

Revision Notes

Concept Videos

800+ Questions

With 2023 CUET Exam Paper

(2021-2023) for Better Exam Insights

for Crisp Revision with Smart Mind Maps

for Complex Concepts Clarity

for Extensive Practice

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1st EDITION

I S BN

YEAR 2024 "9789359587226"

CUET (UG)

SYLLABUS COVERED

PUBLISHED BY OSWAAL BOOKS & LEARNING PVT. LTD.

COP Y RIGHT

RESERVED

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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.

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DI SCL A IM ER

This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/ guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

Kindle Edition (2)

Contents l Know your CUET(UG) Exam

5 - 5

l Latest CUET (UG) Syllabus

6 - 8

l Examination Paper CUET 2023



Chapter-1 – Relations and Functions

11 - 16 1 - 11

Chapter-2 – Inverse Trigonometric Functions

12 - 26

Chapter-3 – Matrices

27 - 37

Chapter-4 – Determinants

38 - 50

Chapter-5 – Continuity and Differentiability

51 - 63

Chapter-6 – Applications of Derivatives

64 - 75

Chapter-7 – Integrals

76 - 92

Chapter-8 – Application of Integrals 

93 - 109

Chapter-9 – Differential Equations

110 - 122

Chapter-10 – Vectors 

123 - 134

Chapter-11 – Three-Dimensional Geometry

135 - 148

Chapter-12 – Linear Programming

149 - 161

Chapter-13 – Probability and Probability Distributions

162 - 176

Chapter-14 – Numbers, Quantification and Numerical Application

177 - 188

Chapter-15 – Index Numbers and Time-Based Data

189 - 201

Chapter-16 – Inferential Statistics

202 - 213

Chapter-17 – Financial Mathematics

214 - 224 qqq

LOOK OUT FOR 'HIGHLY LIKELY QUESTIONS'

These questions are selected by Oswaal Editorial Board. These are highly likely to be asked in the upcoming examinations.

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Preface Welcome to the ultimate resource for your Common University Entrance Test (CUET) preparation! The Common University Entrance Test (CUET) marks a significant shift in the admission process for UG programs in Central Universities across India. The introduction of CUET aims to create a level playing field for students nationwide, regardless of their geographical location, and revolutionize the way students connect with these prestigious institutions. CUET (UG), administered by the esteemed National Testing Agency (NTA), is a prestigious all-India test that serves as a single-window opportunity for admissions. The NTA consistently provides timely notifications regarding the exam schedule and any subsequent updates. The curriculum for CUET is based on the National Council of Educational Research and Training (NCERT) syllabus for class 12 only. CUET (UG) scores are mandatory required while admitting students to undergraduate courses in 44 central universities. A merit list will be prepared by participating Universities/organizations. Universities may conduct their individual counselling on the basis of the scorecard of CUET (UG) provided by NTA. Oswaal CUET (UG) Question Bank is your strategic companion designed to elevate your performance and simplify your CUET journey for success in this computer-based test.

Here’s how this book benefits you: • 100% Updated with 2023 CUET Exam Paper • Previous years Questions (2021-2023)for Better Exam insights • Revision Notes for Crisp Revision with Smart Mind Maps • Concept Videos for complex concepts clarity • 800+ questions for Extensive Practice Almost 1.92 million candidates registered for CUET (UG) in 2023. Candidates have been quite anxious about appearing for CUET (UG), however, with the right preparation strategy and resources, you can secure a good rank in CUET (UG). We believe that with dedication, hard work, and the right resources, you can conquer CUET and secure your place in the Central Universities of your choice. Good luck with your preparations, with this trusted companion on your journey to academic success! All the best! Team Oswaal Books

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2 Languages + 6 Domain Specific Subject + General Test

OR

3 Languages + 5 Domain Specific Subjects + General Test

Subject/Language Choice

Objective Type with MCQs

CBT

Mode of Test

Test Pattern

Know Your CUET (UG) Exam SECTIONS

SECTION I (A) 13 Languages

Tested through reading Comprehension (i) Factual (ii) Literary (iii) Narrative

SECTION III SECTION I (B)

SECTION II

20 Languages

Domain Specific Subjects ( 27 Subjects)

General Test (Compulsory)

INCLUDES : NCERT Model syllabus (only of 12th Standard) is available on all the Subjects

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• • • • • •

General Knowledge Current Affairs General Mental Ability Numerical Ability Quantitative Reasoning Logical & Analytical Reasoning

Latest CUET (UG) Syllabus Mathematics/Applied Mathematics (319) Note: There will be one Question Paper which will contain Two Sections i.e. Section A and Section B [B1 and B2]. Section A will have 15 questions covering both i.e. Mathematics/Applied Mathematics which will be compulsory for all candidates Section B1 will have 35 questions from Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Applied Mathematics out of which 25 question will be attempted.

SECTION A 1. Algebra (i) Matrices and types of Matrices (ii) Equality of Matrices, transpose of a Matrix, Symmetric and Skew Symmetric Matrix (iii) Algebra of Matrices (iv) Determinants (v) Inverse of a Matrix (vi) Solving of simultaneous equations using Matrix Method 2. Calculus (i) Higher order derivatives (ii) Tangents and Normals (iii) Increasing and Decreasing Functions (iv) Maxima and Minima 3. Integration and its Applications (i) Indefinite integrals of simple functions (ii) Evaluation of indefinite integrals

4. 5. 6.

(iii) Definite Integrals (iv) Application of Integration as area under the curve Differential Equations (i) Order and degree of differential equations (ii) Formulating and solving of differential equations with variable separable Probability Distributions (i) Random variables and its probability distribution (ii) Expected value of a random variable (iii) Variance and Standard Deviation of a random variable (iv) Binomial Distribution Linear Programming (i) Mathematical formulation of Linear Programming Problem (ii) Graphical method of solution for problems in two variables (iii) Feasible and infeasible regions (iv) Optimal feasible solution

Section B1: Mathematics UNIT I: RELATIONS AND FUNCTIONS 1. Relations and Functions Types of relations: Reflexive, symmetric, transitive and equivalence relations. One to one and onto functions, composite functions, inverse of a function. Binary operations. 2. Inverse Trigonometric Functions Definition, range, domain, principal value branches. Graphs of inverse trigonometric functions. Elementary properties of inverse trigonometric functions. UNIT II: ALGEBRA 1. Matrices Concept, notation, order, equality, types of matrices, zero matrix, transpose of a matrix, symmetric and skew symmetric matrices. Addition, multiplication and scalar multiplication of matrices, simple properties of addition, multiplication and scalar multiplication. Noncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Concept of elementary row and column operations. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries). 2. Determinants Determinant of a square matrix (up to 3 × 3 matrices), properties of determinants, 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. UNIT III: CALCULUS 1. Continuity and Differentiability



2.

Continuity and differentiability, derivative of composite functions, chain rule, derivatives of inverse trigonometric functions, derivative of implicit function. Concepts of exponential, logarithmic functions. Derivatives of log x and ex. Logarithmic differentiation. Derivative of functions expressed in parametric forms. Second-order derivatives. Rolle’s and Lagrange’s Mean Value Theorems (without proof) and their geometric interpretations. Applications of Derivatives Applications of derivatives: Rate of change, increasing/ decreasing functions, tangents and normals, approximation, 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). Tangent and Normal.

3. Integrals

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, only simple integrals of the type—

∫x ∫



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∫ ∫

dx dx dx dx , , , , ± a 2 ∫ x 2 ± a 2 ∫ a 2 − x 2 ∫ ax 2 + bx + c ( px + q ) ( px + q ) dx dx , dx , ∫ ,∫ 2 2 ax 2 + bx + c ax + bx + c ax + bx + c

2

a 2 ± x 2 dx and

∫

a 2 − x 2 dx , ∫ ax 2 + bx + c dx and

ax 2 + bx + c dx

to be evaluated. Definite integrals as a limit of a sum. Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

Contd... 4.

Applications of the Integrals



Applications in finding the area under simple curves, especially lines, arcs of circles/parabolas/el- lipses (in standard form only), area between the two above said curves (the region should be cleraly identifiable).

5.

Differential Equations Definition, order and degree, general and particular solutions of a differential equation. Formation of differential equation whose general solution is given. Solution of differential equations by method of separation of variables, homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type—



addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) product of vectors, projection of a vector on a line. Vector (cross) product of vectors, scalar triple product. 2.

Three-dimensional Geometry



Direction cosines/ratios of a line joining two points. Cartesian and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane.

Unit V: Linear Programming

dy + Py = Q , where P and Q are functions of x or constant dx dx + Px = Q , where P and Q are functions of y or dy



constant UNIT IV: VECTORS GEOMETRY

AND

THREE-DIMENSIONAL

Unit VI: Probability

1. Vectors

Introduction, related terminology such as constraints, objective function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P. problems, graphical method of solution for problems in two variables, feasible and infeasible regions, feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constrains).



Vectors and scalars, magnitude and direction of a vector. Direction cosines/ratios of vectors. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector,

Multiplications theorem on probability. Conditional probability, independent events, total probability, Baye’s theorem. Random variable and its probability distribution, mean and variance of haphazard variable. Repeated independent (Bernoulli) trials and Binomial distribution.

Section B2: Applied Mathematics Unit I: Numbers, Quantification and Numerical Applications A. Modulo Arithmetic l Define modulus of an integer l Apply arithmetic operations using modular arithmetic rules B. Congruence Modulo l Define congruence modulo l Apply the definition in various problems C. Allegation and Mixture l Understand the rule of allegation to produce a mixture at a given price l Determine the mean price of a mixture l Apply rule of allegation D. Numerical Problems l Solve real life problems mathematically E. Boats and Streams

l Distinguish between upstream and downstream



l Express the problem in the form of an equation

F.

Pipes and Cisterns



l Determine the time taken by two or more pipes to fill or

I. Numerical Inequalities

l Describe the basic concepts of numerical inequalities



l Understand and write numerical

inequalities

UNIT II: ALGEBRA A. Matrices and types of matrices

l Define matrix



l Identify different kinds of matrices

B.

Equality of matrices, Transpose of a matrix, Symmetric and Skew symmetric matrix



l Determine equality of two matrices



l Write transpose of given matrix



l Define symmetric and skew symmetric matrix

UNIT III: CALCULUS A. Higher Order Derivatives

l Determine second and higher order derivatives



l Understand differentiation of parametric functions and

implicit functions Identify dependent and independent variables

B.

Marginal Cost and Marginal Revenue using derivatives



l Compare the performance of two players w.r.t. time,



l Define marginal



l distance taken/ distance covered/ Work done from the



l Find marginal cost and marginal revenue

G. Races and Games

given data

H. Partnership

l Differentiate between active partner and sleeping

partner



l Determine the gain or loss to be divided among the



l consideration of the time volume/ surface area for solid

partners in the ratio of their investment with due

cost and marginal revenue

C.

Maxima and Minima



l Determine critical



l Find the point(s) of local



l Find the absolute maximum and absolute minimum

maxima and local minima and corresponding local maximum and local minimum values value of a function

formed using two or more shapes

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points of the function

Contd... UNIT IV: PROBABILITY DISTRIBUTIONS

C.

Time Series analysis for univariate data

A. Probability Distribution



l Solve practical problems based on statistical data and



l Understand the concept of Random Variables and its



l Find probability distribution of discrete random variable

Interpret UNIT VIII: FINANCIAL MATHEMATICS A. Perpetuity, Sinking Funds l Explain the concept of perpetuity and sinking fund l Calculate perpetuity l Differentiate between sinking fund and saving account B. Valuation of Bonds l Define the concept of valuation of bond and related terms l Calculate value of bond using present value approach C. Calculation of EMI l Explain the concept of EMI l Calculate EMI using various methods D. Linear method of Depreciation l Define the concept of linear method of Depreciation l Interpret cost, residual value and useful life of an asset from the given information l Calculate depreciation UNIT IX: LINEAR PROGRAMMING A. Introduction and related terminology l Familiarize with terms related to Linear Programming Problem B. Mathematical formulation of Linear Programming Problem l Formulate Linear Programming Problem C. Different types of Linear Programming Problems l Identify and formulate different types of LPP D. Graphical Method of Solution for problems in two Variables l Draw the Graph for a system of linear inequalities involving two variables and to find its solution graphically E. Feasible and Infeasible Regions l Identify feasible, infeasible and bounded regions F. Feasible and infeasible solutions, optimal feasible solution l Understand feasible and infeasible solutions l Find optimal feasible solution

Probability Distributions

B.

Mathematical Expectation



l Apply arithmetic mean of frequency distribution to find

C.

Variance



l Calculate the Variance and S.D. of a random variable

the expected value of a random variable

UNIT V: INDEX NUMBERS AND TIME BASED DATA A. Index Numbers

l Define Index numbers as a special type of average

B.

Construction of Index numbers



l Construct different type of index numbers

C.

Test of Adequacy of Index Numbers



l Apply time reversal test

UNIT VI: UNIT V: INDEX NUMBERS AND TIME BASED DATA A. Population and Sample l Define Population and Sample l Differentiate between population and sample l Define a representative sample from a population B. Parameter and Statistics and Statistical Interferences l Define Parameter with reference to Population l Define Statistics with reference to Sample l Explain the relation between Parameter and Statistic l Explain the limitation of Statistic to generalize the estimation for population l Interpret the concept of Statistical Significance and Statistical Inferences l State Central Limit Theorem l Explain the relation between Population-Sampling Distribution-Sample UNIT VII: INDEX NUMBERS AND TIME-BASED DATA A. Time Series l Identify time series as chronological data B. Components of Time Series l Distinguish between different components of time series

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Trimurti Book World, 9437034735

0808

BARIPADA

UTTAR PRADESH DEHRADUN

( 10 )

CUET (UG) Exam Paper 2023 National Testing Agency Held on 26th May 2023

MATHEMATICS/APPLIED MATHEMATICS

[This includes Questions pertaining to Domain Specific Subject only] Time Allowed: 45 Mins. 

Maximum Marks: 200

General Instructions : (i)

Section A will have 15 questions covering both i.e., Mathematics/Applied Mathematics which will be compulsory for all candidates. (ii) Section B1 will have 35 questions from Applied Mathematics out of which 25 questions need to be attempted. Section B2 will have 35 questions purely from Mathematics out of which 25 questions will be attempted. (iii) Correct answer or the most appropriate answer : Five marks (+ 5) (iv) Any incorrect option marked will be given minus one mark (– 1). (v) Unanswered/Marked for Review will be given no mark (0). (vi) If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options. (vii) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question. (viii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (ix) Calculator / any electronic gadgets are not permitted.

Section - A



Mathematics/Applied Mathematics

1. The area enclosed by the ellipse

x2 92

+

y2

62

= 1 is:

(1) 15p

(2) 54p 3 (3) 18p (4) p π 2 2. If A is a square matrix of order 3, B=kA and |B|=x|A| then, (1) x=2k (2) x=k2 (3) x=k3 (4) x=3k  0 1 3  3. The matrix A=  1 0 0  is a  3 0 0  (1) Diagonal matrix (2) Symmetric matrix (3) Skew-symmetric matrix (4) Scalar matrix

4. The degree of the differential equation  d2y    2  is:  dx  dx   

dy   1   

1. 3.

4

2

2. 3 4. 4 5. The differential equation dy + x = 0, represents dx y the family of curves: x (1) x2−y2=C (2) =C y (3) xy=C (4) x2+y2=C 1 2

2 3x  0 , then the values of x are: x 1 (1) 1 and 3 (2) 1 and 2 (3) 2 and 3 (4) 3 and 0

6. If

15

7.  [x]dx where

[x] denotes the greatest integer

0

function ≤x is equal to: 1 1 (1) (2) (3) 1 4 2

(4) 0

8. If matrix A is of order 2 × 3 and B of order 3 × 2, then

(1) AB, BA both are defined and are equal (2) AB is defined but BA is not defined (3) AB is not defined but BA is defined (4) AB, BA are defined but are not equal 9. In a box containing 100 bulbs. 10 are defective. Then the probability, that out of a sample of 5 bulbs none is defective, is: 5

1(1) 10−1

1 (2)   2 2

5  9  (3)  9  (4)   10  10   

d y 1 , then at x=2 is: x +1 dx 2 3 2 2 (1) (2) (3) 2 9 27 11. Match List I with List II

10. If y =

LIST I

(4)

3 8

LIST II

A Maximum value of f(x)=−|x +1|+3

I 6

B Minimum value of f(x)=(2x−1)2+5

II 5

12

OSWAAL CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH.

C Maximum value of f(x) = 6−x2

III no maximum value

D Maximum value of f(x) =x3+1

IV 3

Choose the correct answer from the options given below: (1) A-IV, B-II, C-I, D-III (2) A-III, B-IV, C-I, D-II (3) A-I, B-II, C-III, D-IV (4) A-II, B-III, C-IV, D-I

12. The mean number of heads in two tosses of a coin is:

(1) 2

1 3 (3) 1 (4) 2 2 , then for x > 1, f(x) is:

(2)

13. If f(x) =

1 1−x

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

decreasing constant increasing neither decreasing nor increasing

14. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed. and X= 1 if he is in favour. Then. E (X) is : 1 7 1 7 (1) (2) 2 (3) (4) 10 3 11 15. The solution of a LPP with basic feasible solutions (0, 0), (10, 0), (0, 20). (10, 15) and objective function Max Z=2x+3y is: (1) x =0, y=20, Max Z= 60 (2) x=10, y=15, Max Z=65 (3) x=10, y=20. Max Z=70 (4) x=15, y=10, Max Z=60

Section - B1



Mathematics

16. The linear constraints, for which the shaded area in the figure is the feasible region of an LPP. are: y 80 70 60 50 40 30 (30, 20)

20 10

x

x' 10

20

30 40 50

60

y'

(1) (3)

x + y ≥ 50 2x + y ≤ 80 x, y ≥ 0 x + y ≤ 50 2x + y ≤ 80 x, y ≥ 0

(2) (4)

x + y ≤ 50 2x + y ≥ 80 x, y ≥ 0 x + y ≥ 50 2x + y ≥ 80 x, y ≥ 0

17. Two dice are thrown simultaneously. If X denotes the number of sixes, then the variance of X is: 5 7 1 2 (1) (2) (3) (4) 18 18 3 3 18. If f : R→R is defined by f(x) = sin x + x. then f(f (x)) is: (1) (2) (3) (4)

2sin x + 2x sin2 x + x2 sin (sin x + x) + sin x + x sin2 x + 2sin x + x

19. Which

of the following statements is NOT CORRECT. (1) A row matrix has only one row. (2) A diagonal matrix has all diagonal elements equal to zero.

(3) A symmetric matrix is a square matrix satisfying certain conditions. (4) A skew-symmetric matrix has all diagonal elements equal to zero. 20. The angle between the line x  2  y  3  z  5 3 2 6 and the plane 2x + 10y − 11z = 5 is: 1  8  1  8  (1) cos   (2) sin  21  21     1  21  1  21  (3) cos   (4) sin    82   82  21. Match List I with List II LIST I LIST II sin x −1 dx A.  I. etan x + C 1  cos x 1 dx II. log (log x +1) + C B.  1  tan x C.

etan

1

x

dx III. −log |1+cos x| + C 1  x2 1 dx IV. x − 1 log|cos x −sin x + C D.  x  x log x 2 2 Choose the correct answer from the options given below: (1) A-II, B-III, C-IV, D-I (2) A-III, B-IV, C-I, D-II (3) A-I, B-II, C-III, D-IV (4) A-IV, B-I, C-III, D-II x −1 22. The function f(x)= 2 , x≠1, f(1)=1, is x( x − 1) discontinuous at:



(1) Exactly one point (2) Exactly two points (3) Exactly three points (4) No point

23. The

area of the region bounded by the lines x=2y +3, x=0, y=l and y=−1 is: (1) 4 sq. units (2) 6 sq. units 3 (3) 8 sq. units (4) sq. units 2

13

CUET (UG) Exam Paper 2023

24. Match List I with List II A.

LIST I The area of parallelogram determined by vectors 2i and 3 j

LIST II I. 2

B.

The value of i  j  k  j  k  i

II.

4

C.

The value of a for which the vectors 2i  3 j  4 k and ai  6 j  8 k are

III.

0

 





symmetric, then :

The value of l for which the vectors 2i + j + k and 2i  4 j   k are

IV.

6

perpendicular

Choose the correct answer from the options given

below: (1) A-I, B-II, C-III, D-IV (2) A-II, B-I, C-III, D-IV (3) A-III, B-IV, C-II, D-I (4) A-IV, B-I, C-II, D-III  cos  sin   2 25. If the matrix A=   , then A is equal  sin  cos    to:  cos 2 sin 2  (1)     sin 2 cos 2   cos2  sin 2    (2)    sin 2  cos2    cos 2 sin 2   (3)    sin 2 cos 2   cos   sin  cos   sin   (4)   sin   sin  cos   sin  

(2) 12

(4) 32

28. Match List I with List II LIST I

LIST II

tan−1

C.

13   cos−1  cos  6  



3 −cos−1  3



I.



 2

II.



 6

III.

π 2

π  1 IV. sin−1    6  2 Choose the correct answer from the options given below: (1) A-III, B-I, C-IV, D-II (2) A-IV, B-I, C-II, D-III (3) A-II, B-III, C-IV, D-I (4) A-I, B-II, C-III, D-IV D.

are

31. If a line makes angles 90°. 60° and q with X, Y and Z axis respectively, where q is acute, then value of q is: π π π π (1) (2) (3) (4) 6 4 2 3 32. The area of the region bounded by the parabola y2 = 4ax and its latus rectum is: 4a2 sq. units 3

(2)

8a2 sq. units 3

ex+e−y=2 (2) e−x+ey=2 x y e +e =2 (4) e−x+e−y=2 dy 34. Solution of dx = (1+x2)(1+y2) is: x3 (1) tan−1 y=x+ +c 3 x3 (2) tan−1 y=x+ +c 3 x3 (3) tan−1 y=x2+ +c 3 x3 (4) tan−1 y=x2+ +c 3 (1) (3)

on increasing the side by 4% is: (1) 1.04a3 m3 (2) 1.004a3 m3 (3) 1.12a3 m3 (4) 1.12a2 m3

B.

z=0 z=0 z=0 z = −1

9a2 2a2 sq. units (4) sq. units 5 3 33. Particular solution of the differential equation log  dy  =x+y given that when x= 0, y= 0 is:    dx 

(3) 16

sin−1 x+cos−1 x, x ∈[−1,1]

y = 1, y = 2, y = −1, y =−1,

(3)

27. The approximate volume of a cube of side a metres

A.

x = 2, x = 2, x = −2, x = −2,

x  y 1  z 2  is skew2 0 

a1 b1 1 collinear and D= a2 b2 1 then: a3 b3 1 (1) D=0 (2) D=±1 (3) D2=0 or 1 (4) D= (a1+a2+a3)−(b1+b2+b3)

(1)

26. The maximum slope of the curve y = −x3 + 3x2 + 9x − 27 is: (1) 0

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

 0   3  x  y

30. If three points A(a1, b1), B(a2, b2),C(a3, b3) and

collinear. D.

29. If the matrix A =

35. A manufacturer can sell x items at a price of `3x+5 each. The cost price of x items is `x2+5x. If x is the number of items she should sell to get no profit and no loss. then: (1) x=10 (2) x=30 (3) x=0 (4) x=−10 dy 36. If 1  x 2  1  y 2 = a(x−y), then = dx (1)

(3)

1 − x2 1 − y2 1  x2 1  y2



(2)



(4)

1 − y2 1 − x2 1  x2 1  y2

14

OSWAAL CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH.



37.   1  x 2x

2

 1 x

(1) x = etan

 tan 1 x e dx =  

−1

x

(3) −1

(2) etan x−x+c

+c

tan−1 x

−1

(3) e +c (4) xetan x+c 38. A.  Equation of the line passing through the point (1, 2, 3) and parallel to the vector x 1 y  2 y 3 3i  2 j  2 k is   . 3 2 2 B.  Equation of line passing through (1, 2, 3) and parallel to the line given by x3 4 y z8 x 1 y  2 z  3 . is     3 5 6 3 5 6 C. Equation of line passing through the origin and (5, −2, 3) is x  y  z 5 2 3 D. Equation of plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratio’s 2, 3, −1 is 2(x−1)+3(y−2)−1(z−3)=0

E. Equation of plane with intercepts 2, 3 and 4 on X, Y and Z-axes respectively is 2x+3y+4z=1.

Choose the correct answer from the options given

below: (1) A, E only (2) A, C, D only (3) C, D, E only (4) E only  cos  sin  0  39. If A=   sin  cos  0  and B is a square matrix    0 0 1  of order 3. then |AB| is equal to: (1) |B|2 (2) |B| (3) sin2q|B| (4) cos2q|B|

40. Value of

 sin  sin e e

 , x ∈[−1,1], is: x

sin tan 1 x  cot 1 x 1

x  cos1

π  (3) 1 (4)  2 2 41. Probabilities to solve a specific problem by A, B and 1 1 1 C are , respectively. Probability that at and 2 3 4 least one will solve the problem is: 1 3 23 1 (1) (2) (3) (4) 4 4 24 24 (1) 0

42.

(2)

The derivative of sin (tan−1 e2x) with respect to x is:

(1)

2e

2x

sin (tan 1e

4x

1 2 x

e )

(2)

2e 2 x cos (tan 1 e 2 x ) 1  e4 x

2e 2 x sin (tan 1 e 2 x ) 1  ex

2

(4)

2e 2 x cos (tan 1 e 2 x ) 1  e2 x

43. If A is a square matrix of order 3, then |adj A| is equal to: (1) |A|

(2) |A|2

(3) |A|3

(4) 3|A|

44. The

feasible region of an LPP Max Z = 3x + 2y subject to x≥0, y≥0, x−2y≤3 is: (1) Bounded in first quadrant but has no solution (2) Unbounded in first quadrant but has a solution (3) Unbounded in first quadrant and has no solution (4) Bounded and has a solution x=0, y=0, Z=0

45. Let

A={1.2.3}. Consider the relation R={(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is (1) reflexive only (2) reflexive and transitive (3) symmetric and transitive (4) neither symmetric nor transitive 1 46. The range of the function f(x) = 3 − sin 4 x is 1  1 1 (1)  ,  (2)  , 1 2  4 2 (3)  1 , 3  (4)  1 , 3  4 4 2 4     47. The equation of the tangent. to the curve = x2 − 2x − 3 which is perpendicular to the line x+2y+3=0, is: (1) 4x−2y=7 (2) 2x−y=7 (3) 2x−y=5 (4) 4x−2y=5

48. The solution of the differentiable equation

dy 2x +y=14x3, x>0, is: dx (1) y = 2x3 + cx½ (2) y = x3 + cx½ 3 −½ (3) y = 2x + cx (4) y = x3 + cx−½     49. Let the vectors a  i  3 j  2 k , b  2i  j  k and  c  3i  5 j  2 k be coplanar. Then l is equal to: (1) −1 (2) 1 (3) −2 (4) 2

50. A coin is tossed 7 times. The probability of getting at least 4 heads is: 5 (1) 8 1 (3) 4

3 4 1 (4) 2 (2)

Section - B2

51. Consider the following data: Commodity A B C

Applied Mathematics

52. The present value of a perpetuity of `1200 payable

Price Price Quantity Quantity year 2010 year 2016 year 2010 year 2016 1 2 10 13 5 10 12 16 6 10 15 18

The Laspeyre’s price index number for year 2016 with year 2010 as base year is: (1) 160 (2) 200 (3) 150

(4) 170

at the beginning of each year, if money is worth 5% per annum is: (1) `25,500 (2) `24,000 (3) `24,200 (4) `25,200

53. Pure

honey costs `300 per litre. A shopkeeper adds water to 10 litres of pure honey and sells the resulting syrup at `250 per litre. The quantity of water added by the shopkeeper is : (1) 2 litres (2) 5 litres (3) 3 litres (4) 1.5 litres

15

CUET (UG) Exam Paper 2023

54. Three persons A, B and C enter into a partnership to run a business. They invested their capitals in the ratio 4 : 5 : 6 . After 5 months B increases his 3 2 5

share by 40%. If the total profit at the end of a year is `50,550, then A’s share in the profit is: (1) `8,000 (2) `10,000 (3) `20,000 (4) `12,000

55. Two positive numbers x and y whose sum is 25 and 3 2 the product x y is maximum are: (1) x=10, y=15 (2) x=15, y=10 (3) x =12, y=13 (4) x=16, y=9 56. If the function f(x)=alog x+ b +x has extreme x values at x=1 and x=3, then (a, b) is: 3  1 (1)   ,   (2) (4, 3) 2 2  (3) (−2, −1) (4) (−4, −3)

(3) x=−3, y=2, z=6 (4) x=−2, y=3, z=8

65. Consider the following hypothesis test: H0 : m≥20 H1 : m 8 cm (3) Shortest-side ≤ 8 cm (4) Shortest-side ≥ 8 cm

16

OSWAAL CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH.

71. An

asset costing `2,00,000 is expected to have a useful life of 10 years and a final scrap valve of `40,000. The book value of the machine at the end of sixth year is: (1) `1,36,000 (2) `1,04,000 (3) `1,20,000 (4) `88,000

76. If

 a 2 5b  the matrix  2 0 15  is skew-symmetric.   15 3c 0 

then the value of a2 + b2 + c2 is: (1) 15 (2) 34 (3) 25

72. The point on the straight line 3x + 4y = 8, which is 77. The closest to the origin is:  13 (1)  ,  24  5 (3)  ,  24

 24 32  (2)  ,   25 25   5 (4)  1,   4

17   24  7   24 

78. Match List I with List II

73. Match List I with List II LIST I LIST II A. A special characteristic of a I. statistic population is known as a B. A special characteristic of a II. Confidence sample is known as a interval C. The uncertainty of a sampling III. Estimation process is expressed by D. The process by which one IV. Parameter makes the inferences about a population based on the information obtained from a sample is known as Choose the correct answer from the options given below: (1) A-II, B-III, C-IV, D-I (2) A-I, B-IV, C-II, D-III (3) A-IV, B-I, C-II, D-III (4) A-IV, B-I, C-III, D-II

74. Match List I with List II LIST I A. The solution set of the inequality 3x+7>12 B. The solution set of the inequality 3x + 5 ≥1, x ∈ R 2

I.

LIST II [−1, ∞]

II.

  17  8 ,   

C. The solution set of the inequality III. 2x+5 0 then the condition on p and q so that the minimum of Z occurs at (2, 3) and (7, 0) is: (1) 7p=4q (2) 5p=3q (3) 4p=q (4) 3p=3q



Course of Action

Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

1

RELATIONS AND FUNCTIONS

  Revision Notes

Let A be a non-empty set and R ∈ A × A. Then, R is called a relation on A. If (a, b) ∈ R, then say that ‘a’ is related to ‘b’ and write aRb. Types of Relation: Scan to know more about  Reflexive Relation: A relation R defined this topic on set A is said to be reflexive, if (x, x)∈R, x ∈A i.e., x R x x ∈A.  For example: N is the set of all natural numbers and the relation R = {(a, b) : a = b} Type of Relation is a reflexive relation.  Symmetric Relation: A relation R defined on set A is said to be symmetric, if (x, y)∈ R → (y, x) ∈R,  x, y ∈ A i.e., x R y → y R x  x, y ∈ A.  For example: N is the set of all natural Scan to know more about numbers and the relation R = {(a, b) : a = b} this topic is a symmetric relation because whenever a = b, then it obviously means that b = a.  Transitive Relation: A relation R defined on set A is said to be transitive, if (x, y) ∈ R What is a Relation? and (y, z) ∈R → (x, z) ∈R, x, y, z ∈A i.e., x R y and y R z → x R z x, y, z ∈ A.  For example: N is the set of all natural numbers and the relation R = {(a, b) : a = b} is a transitive relation because whenever a = b and b = c then it obviously means that a = c.  Equivalence Relation: A relation R on a set A is called an equivalence relation, if it is reflexive, symmetric and transitive.  For example: We have already seen that the relation R = {(a, b) | a = b} on the set of natural numbers is reflexive, symmetric, and transitive and hence it is an equivalence relation.  Equivalence Classes: Consider, an arbitrary equivalence relation on R on an arbitrary set X, R divides X into mutually disjoint subsets Ai called partitions or sub-divisions of X, satisfying (i) all elements of Ai are related to each other for all i.

(ii) no element of Ai is related to any element of Aj; i ≠ j. (iii) Ai∪ Aj = X and Ai ∩ Aj = , i ≠ j. Then, subset Ai are called equivalence classes.

 Domain and Range of a Relation: In domain and range of a relation, if R be a relation from set A to set B, then,  The

set of all first components of the ordered pairs belonging to R is called the domain of R.  

Thus, Dom(R) = {a ∈ A: (a, b) ∈ R for some b ∈ B}.

The set of all second components of the ordered pairs belonging to R is called the range of R.  Thus,

range of R = {b ∈ B: (a, b) ∈ R for some a ∈ A}. Therefore, Domain (R) = {a : (a, b) ∈ R} and Range (R) = {b : (a, b) ∈ R}  For example: State the domain and range of the following relation: {(4, 5), (-1, 8), (5, 3), (0, 7), (3, 2)}?  Solution: The domain is the first component of the ordered pairs whereas, range is the second component of the ordered pairs. We need to remove the duplicate numbers if any present.  Domain = {4, -1, 5, 0, 3}, Range = {5, 8, 3, 7, 2}  Function  Let A and B be two non-empty sets. Then, a relation ƒ from A to B which associates each element x ∈ A, to a unique element of f (x) ∈ B is called a function from A to B Scan to know and we write f : A → B. Here, A is called the more about this topic domain of f i.e., dom (f); A and B are called the codomains of f. Also, {f(x): x ∈ A} ⊆ B is called the range of f. Note: Every function is a relation but Types of every relation is not a function. Functions  Types of Function 1. One-one (Injective) Function: A function f : A → B is said to be one-one, if distinct element of A have distinct images in B. ie., if f (x1)= f (x2) ⇒ x1 = x2 or x1 ≠ x2 ⇒ f (x1) ≠ ƒ(x2) ∀, x1, x2 ∈ A. 2. Onto (Surjective) Function: A function f : A → B is said to be onto, if every element in B has its pre-image in A i.e., if for each y ∈ B, there exists an element x ∈ A, such that f(x) = y. 3. One-one and Onto (Bijective) Function: A function f : X → Y is said to be one-one and onto, if ƒ is both one-one and onto. 

Every

f

cd

a b

b

a

Types of Functions

A function f : X → Y is defined to be invertible if there exists a function g: Y → Y such that gof = IX and fog = IY. The function g is called the inverse of

1 2 3

f is onto. f

Then f is surjective

y ∈ Y, x∈X such that f(x) = y,

f : X → Y is onto if every

2 3

1

f is many- one, f is one-one.

∀ x1, x2 ∈ X. Otherwise,

f(x1) = f(x2) ⇒ x1 = x2

Relations d Functions n a

R

m Sy

met

if aRb ⇒ bRa ∀ a, b∈A

A relation R : A → A is symmetric

A relation R : A → A is transitive

ric relation

A relation R : A → A is reflexive if aRa ∀ a∈A

First Level

Second Level

Trace the Mind Map Third Level

The binary operations * on a non-empty set A are functions from A × A to A. The binary operation, *: A × A → A, is an operation of two elements of the set whose domains and co-domains are in the same set.

equivalence.

is congruent to T 2 }.Then, R is

R : T → T defined by R = {(T1, T2)} : T1

the set of all triangles in a plane and

and transitive relations) e.g., Let T =

Equivalence relation (If a relation has reflexive, symmetric

if aRb and bRc ⇒ aRc ∀ a,b,c ∈A. Tra n si t ive relation

Types of Relations

e

tion rela e xiv fle



f : X → Y is one-one if



A composite function is a function that comprises two functions. Composition of function is the process or operation that combines two or more functions together into a single function. Suppose we take two functions f(x) and g(x) which both take x as input values and give the specific output then the composition of function f(x) and g(x) when f(x) is first computed is, g(f(x)) or (gof)(x).



+

+

2 Oswaal CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH.

3

RELATIONS AND FUNCTIONS

4. Inverse of a Function: A function f : X → Y is defined to be invertible if there exists a function g: Y → X such that gof = IX and fog = IY. The function g is called the inverse of f and is denoted by f−1. An important note is that, if f is invertible, then f must be oneone and onto and conversely if f is one-one and onto, then ƒ must be invertible. For example: Let us find inverse of f(x) = x + 7 Put f -1 (x) in place of x in the given equation, ⇒ f(f -1 (x)) = f -1 (x) + 7 Since, f(f -1 (x)) = x, ⇒ x = f -1 (x) + 7 ⇒ f -1 (x) = x - 7 Here, f(x) = x + 7 is an invertible function as it is a bijective function. 5. Composite Function: A composite function is a function that comprises of two functions. Composition of function is the process or operation that combines two or more functions together into a single function. Suppose we take two functions f(x) and g(x) which both take x as input values and give the specific output then the composition of function f(x) and g(x) when f(x) is first computed is, g(f(x)) or (gof)(x). For example: ℤ+ ℤ ℚ ℝ

+ closed closed closed closed

× closed closed closed closed

If, f(x) = x2 and g(x) = x + 3. Then calculate the composition g(f(x)) and f(g(x)).  Here, g(f(x)) = g(x2) = x2 + 3 Similarly, we can easily calculate the f(g(x)) or (fog)(x) where g(x) is computed first. f(g(x)) = f(x + 3) = (x + 3)2 Scan to know more about  Binary Operations: this topic  Just as we get an answer when two numbers are either added or subtracted or multiplied or are divided, the binary operations associate Binary any two elements of a set. The resultant of Operations the two elements are in the same set. Binary operations on a set are the calculations that combine two elements of the set to produce another element of the same set.  The binary operations * on a non-empty set A are functions from A × A to A. The binary operation, *: A × A → A, is an operation of two elements of the set whose domains and codomains are in the same set.  Properties of Binary Operations:  Closure property: Let S be a non-empty set. A binary operation * on S  is said to be a closed binary operation on S, if a * b ∈ S, ∀a, b ∈ S

− not closed closed closed closed

Associative property: Let S be a subset of Z. A binary operation * on S is said to be associative, if (a * b) * c = a * (b * c), ∀ a, b, c ∈ S.  Commutative property: Let S be a non-empty set. A binary operation * on S is said to be commutative, if a * b = b * a, ∀ a, b ∈ S.  Identity: A non-empty set S with binary operation *, is said to have an identity e ∈ S, if e * a = a * e = a, ∀ a ∈ S.  Distributive property: Let S be a non-empty set. Let *1 and *2 be two different binary operations on S. Then *1 is said to be distributive over *2 on S if a*1(b*2 c) = (a*1 b)*2(a*1c), ∀ a, b, c, ∈ S 

÷ not closed not closed closed (only when 0 is not included) closed (only when 0 is not included)

Example: Determine whether or not * given below gives a binary operation. (i)  On Z+, * : ab = a - b Here, Z+ denotes the set of all non-negative integers and R denotes the set of real numbers. Solution: Given: On Z+, * : ab = a-b If a = 1 and b = 2 in Z+, then a* b = a - b = 1 - 2 = -1 -1 ∉ Z+ (as Z+ is the set of non-negative integers) For a = 1 and b = 2, a * b ∉ Z+ Thus, * is not a binary operation on Z+.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1.  Consider a binary operation * on N defined as a * b = a3 + b3, choose the correct answer: [CUET 2023] (1) * is both associative and commutative (2) * is associative but not commutative (3) * is commutative but not associative (4) * is neither commutative nor associative 2. The Greatest Integer Function f: R → R given by f(x) = [x], x ∈ R and [x] denotes the greatest integer less than or equal to x is:  [CUET 2023] (1) one-one (2) onto (3) both one-one and onto (4) neither one-one nor onto

3.  The relation R in the set A = {1, 2, 3} given by R = {(1, 2), (2, 1)} is: [CUET 2023] (1) Reflexive (2) Symmetric (3) Reflexive but not symmetric (4) Equivalence Relation 4.  Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is: (1) reflexive but not transitive (2) transitive but not symmetric (3) equivalence relation (4) none of these

4 Oswaal CUET (UG) Chapterwise Question Bank 5. C  onsider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is (1) symmetric but not transitive (2) transitive but not symmetric (3) neither symmetric nor transitive (4) both symmetric and transitive 6. If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is (1) reflexive (2) transitive (3) symmetric (4) none of these 7. The maximum number of equivalence relations on the set A = {1, 2, 3} are (1) 1 (2) 2 (3) 3 (4) 5 8. Let us define a relation R in R as aRb if a ≥ b. Then R is (1) an equivalence relation (2) reflexive, transitive but not symmetric (3) symmetric, transitive but not reflexive (4) neither transitive nor reflexive but symmetric. 9. Let A = {1, 2, 3} and consider the relation R = (1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is (1) reflexive but not symmetric (2) reflexive but not transitive (3) symmetric and transitive (4) neither symmetric, nor transitive 10. Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer: (1) R is reflexive and symmetric but not transitive (2) R is reflexive and transitive but not symmetric (3) R is symmetric and transitive but not reflexive (4) R is an equivalence relation 11. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is (1) 1 (2) 2 (3) 3 (4) 4 12. 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 (1) 720 (2) 120 (3) 0 (4) none of these 13. Which of the following functions from Z into Z are bijections? (1) f(x) = x3 (2) f(x) = x + 2 (3) f(x) = 2x + 1 (4) f(x) = x² + 1 14. Let A = {1, 2, 3,...n} and B = {a, b}. Then the number of surjections from A into B is (1) nP2 (2) 2n − 2 (3) 2n − 1 (4) None of these 1 15. Let f : R → R be defined by f  x   x  R . Then f is x (1) one-one (2) onto (3) bijective (4) f is not defined 16. Let f : R → R be defined as f(x) = x4. Choose the correct answer. (1) f is one-one onto (2)  f is many-one onto (3) f is one-one but not onto (4)  f is neither one-one nor onto 17. Let f : R → R be defined as f(x) = 3x. Choose the correct answer. (1) f is one-one onto (2)  f is many-one onto (3) f is one-one but not onto (4)  f is neither one-one nor onto

MATHEMATICS/APP. MATH.

18. If A = {1, 2, 3}, B = (x, y) then the number of functions from A to B is: (1) 3 (2) 6 (3) 8 (4) 12 3  3x + 2 19. Let f : R −   → R be defined by f ( x ) = , Then 5x − 3 5  (1) f −1 ( x ) = f ( x ) (2)  f −1 ( x ) = − f ( x ) 1 (3) ( fof ) x = − x (4)  f 1  x   f  x 19

20. Let f : A → B and g: B → C be the bijective functions. Then (g o f)−1 is [CUET 2022] (1) f−1 o g−1 (2)  f o g

(3)  g−1 o f−1  (4) g o f

21. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is  [NCERT Exemplar Pg 8 Ex 17] (1) Reflexive and symmetric (2) Transitive and symmetric (3) Equivalence (4) Reflexive, transitive but not symmetric 22. Let L denote the set of all straight lines in a plane. Let a relation R be defined by lRm if and only if l is perpendicular to m ∀ l, m ∈ L. Then R is  [NCERT Exemplar Pg 8 Ex 18] (1) reflexive (2) symmetric (3) transitive (4) none of these 23. Let N be the set of natural numbers and the function f : N → N be defined by f(n) = 2n + 3 ∀ n ∈ N. Then f is  [NCERT Exemplar Pg 8 Ex 19] (1) surjective (2) injective (3) bijective (4) none of these 24. Let ƒ : R → R be defined by ƒ (x) = sin x and g : R → R be defined by g(x) = x², then f o g is  [NCERT Exemplar Pg 8 Ex 21] (1) x² sin x

(2) (sin x)2    (3) sin x²   (4) 

sin x x2

25. Let ƒ: R → R be defined by f ( x ) = 3 x − 4. Then f −1 ( x ) is given by

[NCERT Exemplar Pg 9 Ex 22]

x+4 x -4 (1)  (2) 3 3 (3)  3x + 4 (4) None of these 26. Let f : R → R be defined by f (x) = x² + 1. Then, pre-images of 17 and − 3, respectively, are  [NCERT Exemplar Pg 9 Ex 23] (1) φ, {4, −4} (2) {3, −3}, φ (3) {4, −4}, φ (4) {4, −4}, {2, −2} 27. For real numbers x and y, define xRy if and only if x − y + √2 is an irrational number. Then the relation R is  [NCERT Exemplar Pg 9 Ex 24] (1) reflexive (2) transitive (3) symmetric (4) none of these 28. Let * be binary operation defined on R by a * b = 1 + ab, ∀� a, b ∈ R. Then the operation * is  [NCERT Exemplar Pg 13 Q27] (1) commutative but not associative (2) associative but not commutative (3) neither commutative nor associative (4) both commutative and associative

5

RELATIONS AND FUNCTIONS

29. The identity element for the binary operation * defined on

Reason (R): The range of the function defined f  x   x  1 is 0,   .

 [NCERT Exemplar Pg 14 Q34] (1) 1 (2) 0 (3) 2 (4) none of these

[C] COMPETENCY BASED QUESTIONS

ab Q ~ {0} as a ∗ b = ∀a, b ∈ Q ∼ {0} is: 2

30. Let f: R→ R be defined by f ( x ) = 3x 2 − 5 and g: R→ R by x Then g of is: g ( x) = 2 +1 x  [NCERT Exemplar Pg 15 Q38] (1) (3)

3x 2 − 5 9 x 4 − 30 x 2 + 26 3x

2

x + 2x − 4 4

2

2 (2) 3x − 5

9 x 4 − 6 x 2 + 26

3x (4)

2

9 x + 30 x 2 − 2 4

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is not the correct explanation of A (3) A is true but R is false (4) A is false and R is true 1.  Let R be the relation in the set of integers Z given by R = {(a, b): 2 divides a - b}. Assertion (A): R is a reflexive relation. Reason (R): A relation is said to be reflexive if xRx, ∀x ∈ Z. x 2. Consider the function f : R → R defined as f  x   2 . x 1 Assertion (A): f(x) is not one-one. Reason (R): f(x) is not onto. 3. Assertion (A): Given a relation R = {x, y); (x, y ∈ Z; x² + y² ≤ 9}, the domain of R is {−3, −2, −1, 0, 1, 2, 3}. Reason (R): For domain of R, put y = 0, then x² ≤ 9. 4. Assertion (A): Let R = {(a, a³) : a is a prime number less than 9}, then range of R = {8, 27, 125, 343}. Reason (R): Here, R = {(1, 1), (2, 8), (3, 27), (4, 64), (5, 125), (6, 216), (7, 343), (8, 512). 5. Assertion (A): The possible number of reflexive relations of a set A whose n(A) = 4 is 212. Reason (R): Number of reflexive relation on a set contain n elements is 2n²-n. 6. Assertion (A): Let A = {1, 5, 8, 9}, B = {4, 6} and f = {(1, 4), (5, 6), (8, 4), (9, 6)}, then f is a bijective function. Reason (R): Let A = {1, 5, 8, 9}, B = {4, 6} and ƒ = {(1, 4), (5, 6), (8, 4), (9, 6)}, then ƒ is a surjective function. 7. Assertion (A): A = {(1, 5), (1, 5), (7, −8), (7, −8), (7, −8)} is function. Reason (R): A function is a relation which describes that there should be only one output for each input (or), we can say that a special kind of relation (a set of ordered pairs), which follows a rule i.e., every x-value should be associated with only one y-value is called a function. 8. Assertion (A): The range of the function f (x) = 2 - 3x, x ∈ R, x > 0 is R. Reason (R): The range of the function f (x) = x2 + 2 is [2, ∞). 9. Assertion (A): Let A = {1, 2, 3, 5} , B = {4, 6, 9} and R = {( x, y ) :| x − y | is odd, x ∈ A, y ∈ B} . Then, domain of R is {1, 2, 3, 5}. Reason (R): |x| is always positive ∀x ∈ R . 10. Assertion (A): The domain of the real function f defined by f ( x ) = x − 1 is R − {1}.

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: ONE - NATION ONE - ELECTION FESTIVAL OF DEMOCRACY GENERAL ELECTION - 2019 R = {(V₁, V₂) : V₁, V₂ ∈ I and both use their voting right in general election - 2019} 1. Two neighbours X and Y ∈ I. X exercised his voting right while Y did not cast her vote in general election - 2019. Which of the following is true? (1) (X, Y) ∈ R (2) (Y, X) ∈ R (3) (X, X) ∉R (4) (X, Y) ∉ R 2. Mr. ‘X’ and his wife ‘W’ both exercised their voting right in general election - 2019. Which of the following is true? (1) both (X, W) and (W, X) ∈ R (2) (X, W) ∈ R but (W, X) ∉ R (3) both (X, W) and (W, X) ∉ R (4) (W, X) ∈ R but (X, W) ∉ R 3.  Three friends F1, F2 and F3 exercised their voting right in general election - 2019, then which of the following is true? (1) (F1, F2) ∈ R‚ (F2, F3) ∈ R and (F1, F3) ∈ R (2) (F1, F2) ∈ R, (F2, F3) ∈ R and (F1, F3) ∉ R (3) (F1, F2) ∈ R, (F2, F2) ∈ R but (F3, F3) ∉ R (4) (F1, F2) ∉ R, (F2, F3) ∉ R and (F1, F3) ∉ R 4. In the above question no. 3 defined relation R is __________ (1) Symmetric and transitive but not reflexive (2) Universal relation (3) Equivalence relation (4) Reflexive but not symmetric and transitive 5. Let the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : a - b is a multiple of 4}. Then [1], the equivalence class containing 1, is: (1) {1, 5, 9} (2) {0, 1, 2, 5} (3) φ (4) A II. Read the following text and answer the following questions on the basis of the same: An organisation 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. Ravi decides to explore these sets for various types of relations and functions.

6 Oswaal CUET (UG) Chapterwise Question Bank

6.  Ravi wishes to form all the relations possible from B to G. How many such relations are possible? (1) 26 (2) 25 (3) 0 (4) 23 7. Let R : B → B be defined by R = {(x, y): x and y are students of same sex}, Then this relation R is __________ (1) Equivalence (2) Reflexive only

MATHEMATICS/APP. MATH.

(3) Reflexive and symmetric but not transitive (4) Reflexive and transitive but not symmetric 8. Ravi wants to know among those relations, how many functions can be formed from B to G? (1) 22 (2) 212 (3) 32 (4) 23 9.  Let R : B → G be defined by R = {(b1, g1), (b2, g2), (b3, g1), then R is _________ (1) Injective (2) Surjective (3) Neither Surjective nor Injective (4) Surjective and Injective 10. If R = {(x, y); x, y ∈ Z, x² + y² ≤ 4} is a relation is set Z, then domain of R is (1) {0, 1, 2} (2) {−2, −1, 0, 1, 2} (3) {0, −1, −2} (4) {−1, 0, 1}

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (4)

3. (2)

4. (3)

5. (2)

6. (2)

7. (4)

8. (2)

9. (1)

10. (2)

11. (1)

12. (3)

13. (2)

14. (2)

15. (4)

16. (4)

17. (1)

18. (3)

19. (1)

20. (1)

21. (4)

22. (2)

23. (2)

24. (3)

25. (1)

26. (3)

27. (1)

28. (1)

29. (3)

30. (1)

8. (4)

9. (2)

10. (4)

8. (4)

9. (2)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (2)

3. (1)

4. (3)

5. (1)

6. (4)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (4)

2. (1)

3. (1)

4. (3)

5. (1)

6. (1)

7. (1)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

∴

1. Option (3) is correct.

∴ f is not one-one

Explanation: Check Commutative: * is commutative if a * b = b * a

a * b = a3 + b3



and b * a = b3 + a3 = a3 + b3

f (4.2) = f (4.9), but 4.2 ≠ 4.9

Now, consider 0.7 ∈ R I t is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈R such that f(x) = 0.7 ∴ f is not onto.

⇒ * is commutative

3. Option (2) is correct.

Check associative:

xplanation: The given relation R in the set {1, 2, 3} given by E R = {(1, 2), (2, 1)}.

* is associative if (a * b)* c = a * (b * c)  (a * b) * c = (a3 + b3) * c

(1, 1) ∉R. Hence, R is not reflexive.

3

(1, 2) ∈R and (2, 1) ∈R. Hence, R is symmetric.

 a * (b * c) = a * (b3+ c3)

⇒ a3 + (b3 + c3)3

( 1, 2) and (2, 1) ∈R, but (1, 1) ∉R. Hence, R is not transitive, thus not equivalent.

Since, (a3 + b3)3 + c3 ≠ a3 + (b3 + c3)3

4. Option (3) is correct.

* is not associative.

Explanation: Consider that aRb, if a is congruent to b, ∀ a, b ∈ T.

2. Option (4) is correct.

Then,

Explanation:  f(x) = [x]

which is true for all a ∈ T

It is seen that f(4.2) = [4.2] = 4,

So, R is reflexive.



Let aRb ⇒ a ≅ b

3

3 3

⇒ (a + b ) + c

f(4.9) = [4.9] = 4.

aRa ⇒ a ≅ a,

7

RELATIONS AND FUNCTIONS

⇒

  b≅a

⇒

bRa

10.  Option (2) is correct. Explanation: Let R be the relation in the set {1, 2, 3, 4} is given by:

So, R is symmetric.

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

Let aRb and bRc

(1) (1, 1), (2, 2), (3, 3), (4, 4) ∈ R

⇒

a ≅ b and b ≅ c

Therefore, R is reflexive.

⇒

 a ≅ c ⇒ aRc

(2) (1, 2) ∈ R but (2, 1) ∉ R.

So, R is transitive.

Therefore, R is not symmetric.

Hence, R is equivalence relation.

(3) If (1, 3) ∈ R and (3, 2) ∉ R then (1, 2) ∈ R.

5. Option (2) is correct.

Therefore, R is transitive.

Explanation: aRb ⇒ a is brother of b.

11.  Option (1) is correct.

This does not mean b is also a brother of a as b can be a sister of a.

Explanation: The given set is A = {1, 2, 3}.

Hence, R is not symmetric. aRb ⇒ a is brother of b

he smallest relation containing (1, 2) and (1, 3), which is T reflexive and symmetric, but not transitive is given by:

and bRc ⇒ b is a brother of c.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}

So, a is brother of c.

This is because relation R is reflexive as

Hence, R is transitive.

(1, 1), (2, 2), (3, 3) ∈ R.

6. Option (2) is correct.

Relation R is symmetric since (1, 2), (2, 1) ∈ R and

Explanation: R on the set {1, 2, 3} is defined by

(1, 3), (3, 1) ∈ R.

R = {(1, 2)} That shows neither reflexive nor symmetric but transitive. 7. Option (4) is correct. Explanation: Given that, A = {1, 2, 3} Now, number of equivalence relations are as follows: R1 = {(1, 1), (2, 2), (3, 3)} R2 = ((1, 1), (2, 2), (3, 3), (1, 2), (2, 1))

But relation R is not transitive as (3, 1), (1, 2) ∈ R. but (3, 2) ∈ R. ow, if we add any two pairs (3, 2) and (2, 3) (or both) to N relation R, then relation R will become transitive. Hence, the total number of desired relations is one. 12.  Option (3) is correct.

R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}

xplanation: We know that, if A and B are two non-empty finite E sets containing m and n elements, respectively, then the number of one-one and onto mapping from A to B is

R5 = {(1, 2, 3) ⇔ A × A = A²}

n! if m = n

R3 = ((1, 1), (2, 2), (3, 3), (1, 3), (3, 1))

Maximum number of equivalence relations on the set \  A = {1, 2, 3} = 5 8. Option (2) is correct. Explanation: Given that, aRb if a ≥ b Now, aRa ⇒ a ≥ a, which is true Let aRb, a ≥ b, then b ≥ a which is not true as R is not symmetric. Now, aRb and bRc ⇒  a ≥ b and b ≥ c ⇒  a ≥ c Hence, R is transitive. 9. Option (1) is correct. Explanation: Given that A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.  (1, 1), (2, 2), (3, 3) ∈ R Hence, R is reflexive. (1, 2) ∈ R but (2, 1) ∈ R Hence, R is not symmetric. (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R Hence, R is transitive.

0, if m ≠ n Given that, m = 5 and n = 6 \ m ≠ n Number of one-one and onto mapping = 0 13.  Option (2) is correct. Explanation: For bijection on Z, f(x) must be one-one 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 ∈ Z. Also f(x) = 2x + 1 takes only values of type = 2k + 1 for x ∈ k ∈ Z But f(x) = x + 2 takes all integral values for x ∈ Z Hence, f(x) = x + 2 is bijection of Z. 14.  Option (2) 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

8 Oswaal CUET (UG) Chapterwise Question Bank 15.  Option (4) is correct.

1 Explanation: We have, f  x   x  R x For x = 0, f(x) is not defined.

MATHEMATICS/APP. MATH.

20.  Option (1) is correct. xplanation: Given that f : A → B and g : B → C be the E bijective functions.

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

(f −1 o g−1) o (g o f ) = f −1 o (g−1 o g o f )

16.  Option (4) is correct.

= f −1 o (g−1 o g o f ) [As composition of functions is associative.]

Explanation: We know that f : R → R is defined as f(x) = x4. Let x, y ∈ R such that f(x) = f(y)

(f −1 o g−1) o (g o f ) = ( f −1 o IB o f ) [where IB is identity function on B]

⇒

x4 = y4

⇒

  x = ±y



= (f −1 o IB) o f

\

 f(x) = f(y)



= f −1 o f

 



does not imply that x = y.

For example, f(1) = f(−1) = 1 \  f is not one-one. 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. 17.  Option (1) is correct. Explanation: f : R → R is defined as f(x) = 3x. Let x, y ∈ R such that f(x) = f(y) ⇒

3x = 3y

⇒

x=y

= IA −1

\  f is one-one.

y Also, for any real number y in co-domain R, there exists in 3  y  y R such that f    3    y. 3 3 \  f is onto.

 Thus, (g o f ) = f −1 o g–1 21.  Option (4) is correct. Explanation: Let there be a natural number N. We know that N divides N, ⇒ NRN So, every natural number is related to itself in relation R. ⇒ R is reflexive. ow, let there are three natural numbers; a, b and c such that N aRb, bRc a Rb means a divides b bRc means b divides c ⇒ a divides c ⇒ aRc; thus relation is transitive. Now, let there be two natural numbers a and b such that aRb. a Rb implies that a divides b but it does not mean that b divides a. ⇒  R is not symmetric.

Hence, function f is one-one and onto.

22.  Option (2) is correct.

18.  Option (3) is correct.

xplanation: If line l is perpendicular to line m, line m must E also be perpendicular to line l, So, lRm ⇔ mRl and thus Relation R is symmetric.

Explanation: Given A  = {1, 2, 3} \

n(A) = 3

and

 

\

  n(B) = 2

23. Option (2) is correct.

B = (x, y)

Explanation: f(n) = 2n + 3 is a linear function.

\  The number of function from A to B is = 23 = 8 19.  Option (1) is correct.

3x  2 xplanation: Given that, f  x   E 5x  3 3x  2 Let y 5x  3 3 x  2  5 xy  3 y  x  3  5 y   3 y  2

  

x

 3 y  2   5 y  3

3y  2 x 5y  3 3x  2 f 1  x   5x  3

f 1  x   f  x 

⇒

Hence f(n1​) = f(n2​)  n1 = n2

ere domain is N but range is set of all odd numbers −{1, 3} H Hence, f(n) is injective or one-to-one function. 24.  Option (3) is correct. Explanation: We are given f(x) = sin x and g(x) = x2 So, fog = f (g(x)) = f(x2) = sin x2 25.  Option (1) is correct. Explanation: f(x) = 3x − 4

So, y = 3x − 4

o find the inverse of a function, we need to find x in terms of y. T  So, y + 4 = 3x y4 x ⇒ 3 x  4 Hence, f 1  x   3

9

RELATIONS AND FUNCTIONS

26.  Option (3) is correct.

[B] ASSERTION REASON QUESTIONS

Explanation: For pre-image of 17, we have:

1.  Option (1) is correct.

2

Explanation: By definition, a relation in Z is said

f(x) = x + 1 = 17 2

to be reflexive if xRx, ∀ x ∈ Z. So, R is true.

⇒

x=±4

a – a = 0 ⇒ 2 divides a − a ⇒ aRa.

⇒

x ∈ {−4, 4}

Hence, R is reflexive and A is true.

⇒

x = 16

R is the correct explanation for A.

For pre-image of −3, we have 2



2.  Option (2) is correct.

 f(x) = x + 1 = −3 2

⇒ x = −4 which is not possible ⇒

x∈

27.  Option (1) is correct. Explanation: For every value of x belongs to R, x – x + 2 i.e., 2 is an irrational number. Therefore, it is Reflexive. ow Let’s say x = 2 and y = 2 then x – y + N which is irrational but when y = 2 and x = 2,

2 =2

2 ​−2

x – y + 2 is not irrational; ∴ It is not symmetric. 28.  Option (1) is correct. xplanation: Given that * is a binary operation defined on R E by a * b = 1 + ab, a, b ∈ R So, we have a * b = ab + 1 = ba + 1 = b * a

Explanation: Given, f : R → R x f  x  x2 1     1 Taking x1 = 4, x2 = ∈ R 4 4 f ( x1 ) = f ( 4 ) =   17 1 4 ( x1 ≠ x2 )   f ( x2 ) = f   =  4  17 ∴ f is not one-one. A is true. Let y ∈ R (co-domain) f(x) = y

⇒

Now, a * (b * c ) = a * (1 + bc) = 1 + a (1 + bc) = 1 + a + abc Also, (a * b) * c = (1 + ab) * c = 1 + (1 + ab) c = 1 + c + abc ence, * is not associative. H Therefore, * is commutative but not associative.

⇒

x

1  1  4 y2 2y

1 − 4 y2 ≥ 0 1 1     y  2 2

So, Range ( f ) ∈  − 1 , 1 

29.  Option (3) is correct.

 2 2

ab , ∀ a, b ∈Q ∼{0} 2 Let  e be the identity elements for * ae ∴ a * e = (a ∗ e = e ∗ a = a) 2 ae ⇒   a = 2 Explanation: Given that a * b =

Range

( f ) ≠ R (Co-domain)

∴ f is not onto. is true. R R is not the correct explanation for A. 3.  Option (1) is correct.

e=2

Explanation: We have, R = {(x, y); x, y ∈ Z; x² + y² ≤ 9}

30.  Option (1) is correct. Explanation: Given that f  x   3 x 2  5 and g  x  







gof  g f  x   g 3 x  5 =





Since x ∈R ∴



y

y 1  x2  x

⇒

Thus, a * b * c    a *b  * c





1  x2

⇒ yx 2  y  x  0

So, * is a commutative binary operation.

⇒

x

⇒

(

3x 2 − 5 3x 2 − 5

)

2

= +1 =

2



3x 2 − 5 9 x 4 − 30 x 2 + 25 + 1 3x 2 − 5 9 x 4 − 30 x 2 + 26

x x2  1

.

Put y = 0, then x2 ≤ 9 ⇒ x = 0, ± 1, ± 2, ± 3 Thus, domain of R = {−3, −2, −1, 0, 1, 2, 3} 4.  Option (3) is correct. xplanation: Given, R = {(a, a³) : a is a prime number less E than 9} Assertion is correct as: ⇒ R = {(2, 8), (3, 27), (5, 125), (7, 343)} \ Range = {8, 27, 125, 343} Reason is false as Domain takes only prime values.

10 Oswaal CUET (UG) Chapterwise Question Bank 5.  Option (1) is correct. Explanation: Given, n(A) = 4 Total number of reflexive relations = 2n(n–1) = 24×3 = 212 6.  Option (4) is correct. Explanation: We have, A = {1, 5, 8, 9}, B = {4, 6} and f = {(1, 4), (5, 6), (8, 4), (9, 6)} So, all elements of B has a domain element on A or we can say elements 1 and 8 & 5 and 9 have same range 4 & 6, respectively. Therefore, f : A → B is a surjective function not one to one function. Also, for a bijective function, f must be both one to one and onto. 7.  Option (1) is correct. Explanation: Here, point (1, 5) is repeated twice and (7, −8) is written thrice. We can rewrite it by writing a single copy of the repeated ordered pairs. So, ‘A’ is a function. 8.  Option (4) is correct. Explanation: Assertion: We have, f ( x ) = 2 − 3 x, x ∈ R , x > 0 Let f ( x ) = y, then y = 2 − 3 x ⇒ 3 x = 2 − y 2− y ⇒x= 3 ⇒ x > 0 ⇒



[C] COMPETENCY BASED QUESTIONS 1.  Option (4) is correct. Explanation: (X, Y) ∈ R.   X exercised his voting right while, Y did not cast her vote in general election-2019 And R = {(V1, V2) : V1 V2 ∈ I and both use their voting right in general election-2019} 2.  Option (1) is correct. 3.  Option (1) is correct. 4.  Option (3) is correct. xplanation: R is reflexive, since every person is friend or E itself. i.e., (F1, F2) ∈ R Further, (F1, F2) ∈ R ⇒ F1 is friend of F2 ⇒ F2 is friend of F1 ⇒ (F2, F1) ∈ R ⇒ R is symmetric ⇒ F1 is friend of F2 and F2 is friend of F3. ⇒ F1 is a friend of F3.

2− y >0⇒2− y >0⇒2> y 3

⇒ (F1, F3) ∈ R Therefore, R is an equivalence relation. 5.  Option (1) is correct. Explanation: Equivalence class [1] is the set of elements related to 1 = {1, 5, 9} 6.  Option (1) is correct. Explanation:

Let y = f ( x ) , then

y = x2 + 2 ⇒ x =

y−2 x assumes real values, if y − 2 ≥ 0

Here, B has 3 elements and G has 2 elements. So,

⇒ y ≥ 2 ⇒ y ∈ 2, ∞ ) ∴ Range of f = 2,∞ )

Number of relations from B to G

ence, Assertion is false, Reason is true. H 9.  Option (2) is correct. Explanation: Assertion: Given,

{

hen, T y = x −1 ∵ x ≥ 1 \  Range of f = [0,∞ ). Hence, Assertion is false, Reason is true

Moreover, (F1, F2) (F2, F3) ∈ R

  y 1 1 xy

(xv) cosec–1x + sec–1x = , |x| 1 2 –1 –1 2 x (xvii) 2 tan x = tan , –1 −1  1 + xy  

Property VI

1 1 (i) 2sin −1x = sin −1  2 x 1 − x 2  , − ≤x≤   2 2

(

)

(ii) 2cos −1x = cos −1 2 x 2 − 1 , 0 ≤ x ≤ 1 (iii) 2 tan 1x  tan 1  2 x  , 1  x  1    1  x2 

16 Oswaal CUET (UG) Chapterwise Question Bank 

MATHEMATICS/APP. MATH.

Property VII





sin 1 x 1  y 2  y 1  x2 , if  1  x, y  1and x2  y 2  1or xy  0 and x2  y 2  1   (i) sin −1 x + sin −1 y =   sin 1 x 1  y 2  y 1  x2 , if 0 < x, y  1if  1  x, y  0 and x2  y 2  1     sin 1 x 1  y 2  y 1  x2 , if  1  x, y  0 and x2  y 2  1 











 1 2 2 sin x 1  y  y 1  x    sin 1 x 1  y 2  y 1  x 2  (ii) sin 1 x  sin 1 y    1 2 2    sin x 1  y  y 1  x  









, if  1  x, y  1 and x 2  y 2  1 or xy  0 and x 2  y 2  1 , if 0  x  1, 1  y  0 and x 2  y 2  1



, if 1  x  0, 0  y  1and x 2  y 2  1

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Match List I with List II.

LIST I

[CUET 2023]

LIST II π 2 π 6 -

A. sin −1x + cos −1x, x ∈ [−1,1] I. B.

tan -1 3 - cos -1 (- 3 )

II.

13  C. cos  cos  6  

π III. 2

−1  1  D. sin  −   2

IV.

1 

(3) x + y + z + xyz = 0

π 6

(1)

(2) A-IV, B-I, C-II, D-III (4) A-I, B-II, C-III, D-IV



  3    is [CUET 2023] 2. The value of tan 2sin 2cos −1   2         π π 2π  (1) (2) (3)  (4) 3 6 3 3 −1 

9 2 2 9 1 1 3. The value of sin 1  sin is: 4 3 4 3  (1)

9 4

(2)

9π 4

(3)

π 8

[CUET 2023] (4)

9π 8

4. Match List I with List II. LIST I

LIST II

(1)

3π 5

(3) -

π 10

3π 5

(2)

-7π 5

(3)

π 10

(1)

3+ 5 2

(2)

3- 5 2

(3)

−3 + 5 2

(4)

-3 - 5 2

(4)

-3π 5

(4)

-π 10

1 9. The value of expression 2sec −1 2 + sin −1   is 2 (1)

The range of sin x is

I.

B.

The range of tan–1 x is

II.

    ,    2 2



C.

The range of cosec–1 x is

III.

 [0, ]    2

(1)

IV.

     2 , 2  

The range of sec–1 x is

(2)

1  5  8. The value of tan  cos −1   is  3    2  

A.

D.

π 10

  33   7. The value of sin 1 cos    is   5 

     2 , 2   {0}  

–1

(4) xy + yz + zx + 1 = 0

3π   6. The value of sin −1  cos  is 5  

Choose the correct answer from the options given below: (1) A-III, B-I, C-IV, D-II (3) A-II, B-III, C-IV, D-I

Choose the correct answer from the options given below: (1) A-IV, B-III, C-II, D-I (2) A-IV, B-I, C-III, D-II (3) A-I, B-IV, C-II, D-III (4) A-IV, B-II, C-I, D-III  [CUET 2022]  1 1 1 5. If tan x  tan y  tan z  , then  [CUET 2021] 2 (1) x + y + z − xyz = 0 (2) xy + yz + zx − 1 = 0

π 6

(2)

5π 6

(3)

7π 6

(4) 1

10. The principal value of  1  1 1 cos −1   + 2sin −1   + 4 tan −1   is 2 2      3

π 3

(2)

π 6

(3)

4π 3

(4)

3π 4

(4)

π 3

11. The principal value of cot -1 (- 3 ) is (1)

5π 6

(2)

π 2

(3)

π 4

17

INVERSE TRIGONOMETRIC FUNCTIONS

12. Which of the following is the principal value branch of cos–1 x?  π π (1)  − ,  (2) [-π , π ]  2 2   (3) [0, π] (4) (0,  )     2 13. Which of the following is the principal value branch of cosec–1 x? π π  (1)  ,  2 2

π  (2) [0, π ] −   2

 π π (3)  − ,   2 2

 π π (4)  − ,  − {0}  2 2

−1

−1

14. If 3tan x + cot x = π , then x equals 1 2  −1 2 −1  1 5. If cos  sin + cos x  = 0 , then x is equal to 5   (1) 0

(2) 1

(3) –1

(4)

1 2 (2) (3) 0 (4) 1 5 5 –1 16. The value of [2tan (0.75)] is equal to (1) 0.75 (2) 1.5 (3) 0.96 (4) sin 1.5 (1)

5 +2 2

5 -2

1 (2) 1, . 2

(2) -

π 3

π , then x is equal to 2 1 (3) 0. (4) . 2

(3)

π 3

(4)

2π 3

x x− y 21. tan −1   − tan −1 is equal to y x +y   (1)

π 2

(2)

π 3

(3)

π 4

(2) -

π 2

23. If cos −1x > sin −1x , then

(3) 0

1 2

(2)

1 3

(3)

1 4

(4) 1

 2x  25. If | x | ≤ 1, then 2 tan −1x + sin −1   is equal to  1 + x2  (1) 4 tan -1x (2) 0

(3)

π 2

(4) π

5π 2

(4)

3π   26. The value of cos −1  cos  is 2   (1)

π 2

(2)

(

3π 2

(3)

7π 2

)

27. sin tan −1x ,| x |< 1 is equal to (1)

x 1- x

2

1 1+ x

2



(2)



(4)

π 5

(2)

1 1 - x2 x 1 + x2



4π , then cot −1x + cot −1 y equals to 5

2π 5

(3)

3π 5

(4) π

π 3

(4)

7π   29. cos −1  cos  is equal to 6   (1)

7π 6

(2)

5π 6

(3)

π  6

(4)

-3π 4

(3) (-∞, ∞) (4) None of these

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is not the correct explanation of A (3) A is true but R is false (4) A is false and R is true 2  2  1. Assertion (A): sin 1  sin   

3 

2. Assertion (A): The domain of the function sec–1 2x is

1 1     ,     ,   . 2 2   (4) 2 3

3

 −   Reason (R): sin (sinθ ) = θ , if  ∈  ,   2 2 −1

22. tan -1 3 - cot -1 (- 3 ) is equal to (1) π

(1)

(1) [-1, p+1] (2) [-1, 1]

20. The value of tan -1 3 - sec -1 (-2) is equal to (1) π

(4) x > 0

30. The domain of the function given as f(x) = sin-1x + cos x is

(4) 5 + 2 

19. Solve sin −1 (1 − x) − 2sin −1x = 1 (1) 0, . 2

1 2

1 2

π  1  24. sin  − sin −1  −   is equal to 3  2  

(1)

1 2  1 8. The value of the expression tan  cos −1  is 2 5 

(3)

(2) 0 ≤ x
sin x where x ∈ [−1,1] Now, we have

...(i) cos 1x  sin 1x  1 1   sin x  sin x 2    2sin 1x 2

23

INVERSE TRIGONOMETRIC FUNCTIONS

cos 1x  sin 1x    sin 1x  sin 1x 2    2sin 1x 2  ...(ii)  sin 1 x  4   But   sin 1 x  2 2 π −1 From (i) and (ii) sin x < 4 π  π ⇒ sin  −  ≤ x < sin  2 4 −1≤ x
0, n > 0, a > 0 and a ≠ 1 n (iii) log a mn = n log a m

(ii) It is differentiable on (a, b). Then, there exists a real number c ∈ (a, b) such that f c 

f b   f  a 

. ba  Geometrical Interpretation of Lagrange’s Mean Value Theorem The geometrical interpretation of Lagrange’s mean value theorem is that the chord passing through the points of the graph corresponding to the ends of the segments a and b has a slope equal to: f b   f  a  tan   ba Then there is a point c in (a, b) such that the tangent to the graph at point x is equal to c is parallel to the chord. y

(iv) log a a  1; a  0 and a ≠ 1 log m a ; a  0, b  0, b  1, m  0 and m ≠ 1 (v) logb a  log m b (vi) aloga m  m; a  0, m  0, a  1

C

rd

f(b)

B

f(a)

A

d  dy  f  x  ,   is called the second order derivative of y w.r.t x dx  dx  d2y

0

or y″ or y2.

dx2 L-Hospital’s rule states that for functions f and g which are differentiable on an open interval I except possibly at a point c contained in I, it limx→c f(x) = limx→c g(x) = 0 or ≠ ∞; g′ (x) ≠ 0 for all x in I with x ≠ c and f  x lim exists, then x c g   x  f c f  x lim  lim x c g  x  x c g   c  The differentiation of numerator and denominator often simplifies the quotient or converts it to a limit that can be evaluated directly.  Rolle’s Theorem: Let f be real valued function defined on the closed interval [a, b] such that, (i) It is continuous on the closed interval [a, b]. (ii) It is differentiable on the open interval (a, b). (iii) f (a) = f (b). Then, there exists a real number c ∈ (a, b) such that f ′ (c) = 0.  Geometrical Interpretation of Rolle’s Theorem The geometrical interpretation of Rolle’s Theorem is that if f(x) is a continuous function in [a, b] and a differentiable function in (a, b) then there is a point c ∈ (a, b) where the tangent to curve f (x) is horizontal or we can say it is parallel to the X-axis. y M

a

c

b

 Lagrange’s Mean Value Theorem Let f (x) be a function defined on [a, b], such that (i) It is continuous on [a, b].

a

c

b

Example 1: Find the value of c in Lagrange’s mean value theorem for the function f (x) = x2 + x + 1, x ∈ [0, 4]. (1) +1

(2) –1

(3) +2

(4) –2

Sol. Option (3) is correct. Explanation: Given, f (x) = x2 + x + 1  f  x  2x  1 f  4  f 0  f c  40 21  1 2c  1  4  2c  4  c2 Example 2: It is given that Rolle’s theorem holds good for the 4 function, f(x) = x3 + ax2 + bx, x ∈ [1, 2] at the point x = . 3 Find the values of a and b. (1) a = 5, b = 8 (3) a = 5, b = 8 Sol. Option (4) is correct. Explanation:

(2) a = –5, b = –8 (4) a = –5, b = 8

Given, f  x   x3  ax2  bx, x  1, 2 , f is continuous for x  1, 2 f   x   3 x2  2ax  b exists in x  1, 2 f (1)  a  b  1 f (2)  8  4a  f(2) = 28b+ 4a + 2b

Now,   a + b + 1 = 8 + 4a + 2b ⇒ 3a + b + 7 = 0 ⇒  3a + b = –7 

tangent

y = f(x)

0

α

cho

 Second Order Derivative dy  f   x  is called the first derivative of y or Let y = f (x), then dx

and it is denoted by

t

gen

tan

...(i)

16 4 4 f     3   2a   b  0 3 9 3  

x

⇒

8a + 3b = -16  Solving (i) and (ii), we get, ∴ a = –5, b = 8

...(ii)

54 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Which of the following statements is/are correct?  [CUET 2023] (A) |x| is continuous every where in its domain. (B) |x| is differentiable every where in its domain. (C) [x] is continuous at every integral point. (D) [x] is discontinuous at every integral point. Choose the correct answer from the options given below. (1) (A), (C) Only (2) (B), (D) Only (3) (B), (C) Only (4) (A), (D) Only 1  sinx  cosx 

  with ,  x  4 4 

2. The derivative of cos  2  respect to x is 1 (1) . (2) 1. 2 -1 (3) -1. (4) . 2

[CUET 2023]

[CUET 2023]

dy    Derivative  dx  

(A) y = cos x

(I)



 



sec tan x tan tan x sec2 x 2 x

    



  

(D) y  cos sin x3



x x2  1

, x  1, f 1  1, is [CUET 2023]

discontinuous at Exactly one point. Exactly two points. Exactly three points. No point.

(1)

1 - x2 1- y

2

1  x2 1 y

2

.

(2)

.

(4)

dy = [CUET 2023] dx

1 - y2

.

1 - x2 1  x2

.

1  y2

  , x   1,1, is sin  sin 1 x  cos1 x  e sin tan 1 x  cot 1 x

10. Value of

e

[CUET 2023]

π . 2

(1) 0.

(2)

(4) 

(IV)

-1 sin x 2 x

11. The derivative of sin tan 1 e2 x with respect to x is



 (1)



2e2 x sin tan 1 e2 x 1 e 2x

(3)

4x



2e sin tan 1 e2 x 1 e

x2

 .

(4) f (x) = x4, x ∈[0, 2]

3 3

6. If x = at and y = a t , then 

2

d y dx2

will be

3 (2) . 4at

(4)

3a 2t . 2

[CUET 2023]

2 . t2 1 (3) - . t (1)

 . 2



[CUET 2023] 2e2 x cos tan 1 e2 x (2) . 1  e4 x



 .

12. If x = log t and y =

(3) f (x) = x3, x ∈[-1, 2]

3a . 4t



x 1

(3) 1.

Then number of points of discontinuity for this function is (1) 0. (2) 1. (3) 2. (4) 3. 5. Which of the following functions satisfies Rolle’s theorem?  [CUET 2023] (1) f (x) = x, x ∈[1, 2] (2) f (x) = x2, x ∈[-1, 1]

(3)

8. The function f  x  

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

3 . 8

(4)

cos x 2 sin x

 3 f  x    x  3, if x  0 if x  0 4,

3 at (1) . 2

2 . 27

[CUET 2023]

(III)

Choose the correct answer from the options given below. (1) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) (2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (3) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (4) (A)-(II), (B)-(I), (C)-(III), (D)-(IV) 4. The function f is given by [CUET 2023]

2

(3)

(3)

(II) 3 x2cos x3 sin sin x3

(B) y = sin x



2 . 9

List - II

(Function-y)

(C) y  sec tan x

(1)

9. If 1  x2  1  y 2  a  x  y , then

3. Match List - I with List - II. List - I

1 d2y , then at x = 2 is  x 1 dx2 3 (2) . 2

7. If y 

(4) 1 t2

, then





2e2 x cos tan 1 e2 x 1 e 2

d y

dx2 4 (2) 2 . t 4 (4) - 2 . t

is

 x3  d2y 13. If y  loge  3  , then is equal to e  dx2   3 2 (1) 2 . (2) - 2 . x x 3 2 (3) - 2 . (4) - . x x

2x

.

[CUET 2023]

[CUET 2023]

55

CONTINUITY AND DIFFERENTIABILITY





14. If y  sin 1x , then 1  x2 y2 is equal to (1) xy1 .

(2) xy .

(4) x2 .

(3) xy2 .

15. If y = enx, then nth derivative of y is  (1) e .

(2) n e .

nx

(1) [CUET 2022]

(3) ny.

2 nx

(4) n y. n

 1  cos 2 x , x0  16. If f  x    , then the value of k will x 2  , k x  0  make function f continuous at x = 0 is  [CUET 2022] (1) 1. (2) –1. (3) 0. (4) No such value exists dy x2 17. If y  log sec e , then =  [CUET 2022] dx





2

2

(2) e x tan e x .

2

2

2

(4) xe x tan e x .

(1) x2e x tan e x .

2

(3) 2 xe x tan e x . 18. If y  elog sin

1

x

2

 elog cos

1

x

2

, 0  x  1, then

 (1)

[CUET 2022] dy = 0. dx

(2)

dy   . dx 2

dy  dy   . (4) does not exist.  dx 3 dx 3 19. Let [xr] denotes the greatest integer of xr and |x| denotes the modulus of x.  [CUET 2021] (3)



Then lim x 0

 r 100 r 1  x  (2) 0 is –1. (4) is 100.

(3)

dx

2

(4) None of these

d y

2

dy  0. dx

2

dy  2 y  0. dx

2

(3)

dx

2

(3)

1 t2

(2)

d2y dx2

d y

2

dy  2 y  0. dx

2

(4)

(4)

.

27. If y  sin x  y , then

(1) 2.

22. If y  e x  A cos x  B sin x , then y is a solution of d2y

1 - x4 -4 x3

.

. (4) . 4 - x4 1 - x4 24. Let f (x) = |sin x|, then  (1) f is everywhere differentiable. (2) f is everywhere continuous but not differentiable at x = nπ, n ∈Z. (3) f is everywhere continuous but not differentiable at  x   2n  1 , n  Z . 2 (4) None of these d4y 25. If y = x3 log x, then is dx4 6 (1) 6x. (2) . x x (3) . (4) log 6. 6 d2y 26. Given that x = at 2 and y = 2at then is dx2 -1 -1 (1) . (2) . 2at 3 2at 2 -2a . t

dy is equal to  dx cos x (2) . 1- 2y sin x (4) . 2y -1



. (4) - 2 . x2 x 21. The set of points where the function f given by f(x) = |2x – 1|sin x is differentiable is  1  (1) R. (2) R    . 2

(1)

-4 x



28. The derivative of cos1 2 x2  1 w.r.t. cos-1x is

2

(3) (0, ∞).

(2)

.

cos x . 2y -1 sin x (3) . 1- 2y

 x2  d2y 20. If y  loge  2  , then equals e  dx2   1 1 (1) - . (2) - 2 . x x 2

1- x 1

4

(1)

1 x

(1) does not exist. (3) is 1.

(3)

4 x3

dx2

 1  x2  dy 23. If y  log  , then is equal to  1  x2  dx  

 2 y  0.

(2)

-1 2 1 - x2

.

2 . (4) 1 - x2 . x 29. The function f (x) = cot x is discontinuous on the set (3) (1)

 x  n; n  Z  .

(2)

 x  2n; n  Z  .

 n     (3)  x   2n  1 ; n  Z  . (4)  x  ;n  Z  . 2 2     30. The derivative of m = 13x4 + 5x3 -12y3 + 24x2 + y2 -2x 156 w.r.t. x is dm  52 x3  15 x2  48 x  2 . (1) dx (2)

dm  52 x3  15 x2  36 y 2  48 x  2 x  2. dx

(3)

dm  52 x3  15 x2  36 y 2  48 x  2 y  3 . dx

(4)

dm  36 x3  2 y . dx

56 Oswaal CUET (UG) Chapterwise Question Bank [B] ASSERTION REASON QUESTIONS

MATHEMATICS/APP. MATH.

[C] COMPETENCY BASED QUESTIONS

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. (1) Both A and R are true, and R is the correct explanation of A.

I. Read the following passage and answer the questions from 1 to 5. Ms. Renuka 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 of x and f(x) = g(h(x)), then

(2) Both A and R are true, but R is not the correct explanation of A. (3) A is true but R is false. (4) A is false and R is true. 1. Assertion (A): Let y = t + 1 and x = t + 1, then d2y = 20t 8. dx2 10



8

1.

d2y

d  dy  dt Reason (R): 2    . dt  dx  dx dx 3

3

2. Assertion (A): The derivative of e x w.r.t. log x is 3 x3e x . 

3

3

Reason (R): The derivative of e x w.r.t. log x is 3 x3e x + e2

3. Assertion (A): Function f  x   ( x  1)

cot x

is continuous

at x = 0 and value of f (0) is e.



x2  3x  4

is not continuous at -4 and 1. x2  3 x  4 Reason (R): The function f(x) is rational function. Rational function can be continuous unless there is a discontinuity in the from of division by zero.

5. Assertion (A): If f  x   xn , n  0 is differentiable for all x, then x can be any element of the interval 1, .

Reason (R): The derivative of f(x) = xn with the respect to

x is given by f´(x) = n xn-1.

6. Assertion (A): The function f  x   x is continuous at 3



all x. Reason (R): The function is continuous at a point. When

L.H.L. = f(x) = R.H.L. 7. Assertion (A): The function f (x) = |cos x| is continuous function. Reason (R): The absolute value function, denoted as |x|, is a continuous function for all real value of x. 8. Assertion (A): The function defined by f (x) = cos (x2) is a continuous function. Reason (R): The sine function is continuous in its domain, i.e., x ∈R. 9. Assertion (A): Every differentiable function is continuous but converse is not true. Reason (R): Function f (x) = |x| is continuous. 10. f (x) = [x - 1] + |x - 2|, where [⋅] denotes the greatest integer function. Assertion (A): f (x) is discontinuous at x = 2. Reason (R): f (x) is non-derivable at x = 2.

by derivative of the inside

d cos x5 = dx





(1) x4 sin x5

(2) -5 x4 sin x5

(3) 5 x sin x d 2. sin  cos x  = dx (1) cos(cos x) (3) –sin x⋅cos(cos x)

(4) 4 x5 sin x4

4

3.

Reason (R): Apply L-Hospital rule to find value of f (0). 4. Assertion (A): The values of x for which the function f  x 

- keep the inside - take derivative of outside

5

(2) sin x cos(cos x) (4) cos x sin(cos x)

d sin3 x  dx





(1) cos3 x

(2) 3sin x cos x

(3) 3sin x cos x

(4) −cos3 x

2

4.

d   sin 2 x  at x  is dx 2

(3) 2. (4) -2. dx 5. If sin y  x cos  a  y , then is dy cos a  cos a (1) . (2) . 2 cos  a  y  cos2  a  y  (1) 0.

(3)

(2) 1.

cos a sin y 2

.

(4)

- cos a sin 2 y

.

II. Read the following passage and answer the questions from 6 to 10. 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.

6. (1) (3) 7. (1) (3)

When x > 4 what will be the height in terms of x? x - 2 (2) x - 3 2x - 5 (4) 5 - 2x Will the slope vary with x value? Yes (2) No Can’t say (4) Incomplete data

57

CONTINUITY AND DIFFERENTIABILITY

8. What is

dy at = 3? dx

(1) 2 (2) -2 (3) Function is not differentiable (4) 1 9. If the potter is trying to make a pot using the function f (x) = [x], will he get a pot or not? Why? (1) Yes, because it is a continuous function. (2) Yes, because it is not continuous.

(3) No, because it is a continuous function. (4) No, because it is not continuous. 10. The value of ‘k’ for which the function 1  cos 4 x , if x  0  f  x    8 x2 is continuous at x = 0 is  k, if x  0   (1) 0.

(2) -1.

(3) 1.

(4) 2.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (4)

2. (3)

3. (2)

4. (2)

5. (2)

6. (3)

7. (3)

8. (3)

9. (2)

10. (3)

11. (2)

12. (2)

13. (3)

14. (1)

15. (4)

16. (1)

17. (3)

18. (1)

19. (1)

20. (4)

21. (2)

22. (3)

23. (2)

24. (2)

25. (2)

26. (1)

27. (1)

28. (1)

29. (1)

30. (1)

8. (2)

9. (3)

10. (2)

8. (3)

9. (4)

10. (3)

[B] ASSERTION REASON QUESTIONS 1. (4)

2. (3)

3. (1)

4. (1)

5. (1)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (3)

3. (3)

4. (4)

5. (1)

6. (3)

7. (1)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (4) is correct. Explanation: |x| is continuous every where in its domain. |x| is not differentiable everywhere in its domain.

(B) Let y = sin x     y 2  sin x Differentiating w.r.t x

dy = cos x dx dy cos x cos x    dx 2y 2 sin x 2y

[x] is not continuous at every integral point. [x] is discontinuous at every integral point. 2. Option (3) is correct.

 sin x  cos x  y  cos1   2   1  1  1 sin x  cos x   cos  2  2     1   cos  sin sin x  cos cos x  4 4      1   cos  cos cos x  sin sin x  4 4     1   cos  cos   x   4    y x 4 dy  1 dx  3. Option (2) is correct. Explanation: (A) Let y  cos x dy d   sin x dx dx sin x  2 x



(C) Let y  sec tan x

Explanation:

 x



Differentiate with respect to x dy d  sec tan x dx dx

  

 sec tan

 x   tan  tan x   (sec

x )2 

                

1 2 x

3 (D) yy  cos cos sin sin xx3

dd dy dy 3 3  sin   sin sin sin sin xx3 sin xx3 dx dx dx dx 3 3 dd  sin xx33 cos xx3 sin sin sin xx3 cos dx dx 3 3  33xx22sin sin sin sin xx3 cos cos xx3

4. Option (2) is correct. The given function is discontinuous at x = 0. 5. Option (2) is correct. Let f be real valued function defined on the closed interval [a, b] such that, (i) It is continuous on the closed interval [a, b]. (ii) It is differentiable on the open interval (a, b). (iii) f(a) = f(b). ⇒ f (x) = x2, x ∈[-1,1] verifies Rolle’s theorem.

58 Oswaal CUET (UG) Chapterwise Question Bank 6. Option (3) is correct. x = at2 and y = a3t3 On differentiating both sides w.r.t. t, we get, dx dy = 2= at and 3a3t 2 dt dt dy 3a3t 2 3a 2t Now, = = dx 2at 2   2 2 d y 3a dt   2 dx dx2 d 2 y 3a 2 1   2 2at dx2 d 2 y 3a  dx2 4t 7. Option (3) is correct. Explanation: 1 x 1 On differentiating w.r.t. x, we get dy 1  dx ( x  1)2 Again, differentiating w.r.t. x, we get 1 d2y  (1)  (2)  ( x  1)3 dx2 d 2 y  1  2 2  dx x  2 (2  1)3 2  27 8. Option (3) is correct. x 1 Explanation: We have, f  x   . Given function f(x) x x2  1

On differentiating w.r.t. x, we get 1 1 dy   0 2 2 dx 1 x 1 y 1 1 y

2



dy 1  dx 1  x2

dy 1  y2  dx 1  x2 10. Option (3) is correct.

  sin  sin 1 x  cos1 x  e sin tan 1 x  cot 1 x

Explanation:

e

[Since, tan–1x + cot–1x =

π π and sin–1x + cos–1 x = ] 2 2  sin



y





is discontinuous when f(x) is not defined i.e., when denominator is equal to zero.

∴  

x(x2 - 1) = 0  x(x - 1)(x + 1) = 0 x = 0, x = 1 and x = -1 Therefore, given function is not defined at three values of x. Hence, f (x) is discontinuous at exactly three points. 9. Option (2) is correct. Explanation: 1  x2  1  y 2  a  x  y  Put x = sin A and y = sin B 1  sin 2 A  1  sin 2 B  a sin A  sin B  cos A  cos B  a sin A  sin B   A B  A B  A B  A B 2 cos   cos    a 2 cos   sin   2 2      2   2   A B  A B cos    a sin   2   2    A B 1 tan    2  a A B 1  tan 1   2 a 1 A  B  2 tan 1   a 1 sin 1 x  sin 1 y  2 tan 1   a

MATHEMATICS/APP. MATH.

 11. Option (2) is correct. Explanation:









y  sin tan 1 e2 x

e

2

 sin e 2 1

e

e1 1

On differentiating w.r.t. x, we get dy 1  cos tan 1 e2 x   e2 x  2 dx 1  e4 x 2e2 x cos tan 1 e2 x  1  e4 x





12. Option (2) is correct. Explanation: Given, x = log t

dx 1 = …(i) dt t 1 and y = 2 t On differentiating w.r.t. t, we get dy 2 …(ii)  dt t 3 On dividing eq (i) by eq (ii), we get dy 2 t              3  dx t 3 1 dy 2  dx t 22             On differentiating w.r.t. x, we get d2y 2 dt  2  3  dx dx2 t 4 t  3 [from eq (i)] t 1 4  2 t 13. Option (3) is correct. Explanation:  x3  y  loge  3  e    On differentiating w.r.t. x, we get dy 1 1    3x2 dx x3 e3 On differentiating w.r.t. x, we get x

e3 dy 3  dx x

59

CONTINUITY AND DIFFERENTIABILITY

Again, differentiating w.r.t. x, we get d2y 14. Option (1) is correct.

dx2



x2

Explanation: y  sin 1x dy 1 dy   1  x2  1 2 dx dx 1 x Again, differentiating both sides with respect to x, we get d 2 y dy  2 x  1  x2 2   0 dx  2 1  x2  dx Simplifying, we get 1  x2 y2  xy1





15. Option (4) is correct. Explanation: Given, y  enx y  nenx  ny y  n2enx  n2 y y  n3enx  n3 y

⇒

1  cos 2 x x 2



1  1  2sin 2 x

 lim



x 2 2sinx  lim x 0 x 2 sinx  lim x 0 x 1 17. Option (3) is correct. x 0

  tan e    e   2 x

Explanation: Given, y = log sec e x sec e   

1 dy  dx sec e x2 2

x2

x2

2

 2 xe x  tan e x 18. Option (1) is correct. Explanation: We have, 1

2

x2

[Using chain rule of differentiation]

 r 1 x 100

r

 1 x  x    x2    x3   .   x100  L.H.L.  lim 1 x x  0  x    x2    x3      x100   lim 1 x x  0 lim

x 0

 

d2y

On differentiating both sides w.r.t., x we get dy  e  x  A cos x  B sin x   e  x   A sin x  B cos x  dx dy   y  e  x   A sin x  B cos x  dx Again, differentiating both sides w.r.t. x, we get d2y dx 

1

y  elog sin x  elog cos x , 0  x  1 y  sin 1 x  cos1 x, 0  x  1   1 1 y ∵ sin x  cos x  2  dy  0 dx 19. Option (1) is correct. Explanation: We have,

dy 2 = dx x

2  dx2 x2 21. Option (2) is correct. Explanation: Given that, f (x) = |2x - 1| sin x The function sin x is differentiable. The function |2x – 1| is differentiable, except 2x  1  0 ⇒ 1 x 2 1  Thus, the given function is differentiable R   . 2 22. Option (3) is correct. Explanation: Given that, y = e-x (A cos x + B sin x)

16. Option (1) is correct. Explanation: Since, f (x) is continuous at x = 0. x 0

   

⇒

nth derivative of y  yn   nnenx  nn y

 k  lim

1  0  1  0  1..0  50 1 0  x    x2    x3   .   x100  R.H.L.  lim 1 x x  0 0  0  0  0  0  0 1 0 L.H.L.  R.H.L. So, limit does not exist. 20. Option (4) is correct. Explanation:  x2  Given, Given , y  loge  2  e    22 ⇒  y loge exx-log log  y =22log ee ee  y =22log ⇒  y loge exx-2 2 

3

  2 



2

d2y dx

2

d2y dx

2

d2y



 dy  e  x   A cos x  B sin x   e  x   A sin x  B cos x  dx



dy dy y y dx dx

 2 2

dy  2y dx

dy  2y  0 dx

dx 23. Option (2) is correct. 2

Explanation: Given,  1  x2  y  log    1  x2   









 y  log 1  x 2  log 1  x 2 . Differentiating with respect to x, we get









dy d  d   log 1  x 2   log 1  x 2       dx dx dx  2 x 2x   1  x2 1  x2   2  

60 Oswaal CUET (UG) Chapterwise Question Bank dy d  d  2  2 













24. Option (2) is correct. Explanation: Given, f  x   sin x

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.

27. Option (1) is correct. Explanation: Given that, y  sin x  y  y 2  sin x  y. On differentiating with respect to x, we have dy dy  cos x  dx dx dy  cos x   2 y  1 dx dy cos x   dx 2 y  1 

2y

28. Option (1) is correct. Explanation: Let 1 2 uu   cos  11 cos 11 22 xx 22  du du      dx dx

⇒

Thus, f is everywhere continuous but not differentiable at x  n , n  Z . Explanation: x3 log x If yy   x3 log x dy d 3 dy  d x 3 log x dx  dx x log x dx dx dy  dy  x 22 1  3 log x   dx  x 1  3 log x  dx



d2y dx 2 d y 2

dx 2 d3y dx3 d3y dx3 d4y dx 4

 x2

⇒



And,

d d 1  3 log x   1  3log x  x 2 dx dx

d d  5  6 log x    5  6log x  x dx dx

 11  6 log x 

     





44 xx

 

2 11   22 xx 22   11

44 xx

22 2

4 2 11   44 xx 44   44 xx 22   11

1  4x  4x  1 44 xx 44 xx 444   44 xx 222  22

du   du     dx dx  xx 22 11  v  cos 1x dv 1  dx 1  x2

⇒

 x  5  6 log x  x

du du    dx dx du du    dx dx

⇒

25. Option (2) is correct.

MATHEMATICS/APP. MATH.



 log 1  x log 1  x  dx   dx dx  2 x 2x   1  x2 1  x2   2    2 x  2 2   1 x 1 x    4x  1  x4 

du =2 dv 29. Option (1) is correct. Explanation: Given that, Thus,

cos x sin x f (x) is discontinuous at sin x = 0  x  n , n  Z f  x   cot x 

Thus, the given function is discontinuous at  x  n : n  Z . 30. Option (1) is correct.

6 x

26. Option (1) is correct.

Explanation: Given

Explanation:

m  13 x 4  5 x3  12 y 3  24 x 2  y 2  2 x  156 dm   52 x3  15 x 2  48 x  2 dx

dy dy / dt  dx dx / dt 2a  2at 1  t d2y dx 2

 

1 t2



[B] ASSERTION REASON QUESTIONS 1. Option (4) is correct. Explanation: Here, dt dx



1 2at 3

 Now,,

y  t10  1and x  t 8  1 dy dx  10t 9 and  8t 7 dt dt dy 10t 9 5 2   t dx 8t 7 4 d2y dx

2

d2y

d  dy  d  5 2  t  dx  dx  dx  4  dt 5   2t  dx 4



5

1

y  t10  1and x  t 8  1 dy dx  10t 9 and  8t 7 dt dt



61

CONTINUITY AND 9 DIFFERENTIABILITY

dy 10t 5   t2 dx 8t 7 4



d2y

Now,,

dx 2

6. Option (1) is correct.

d  dy  d  5 2  t  dx  dx  dx  4  dt 5   2t  dx 4

1/ 3 Explanation: Assertion: Given, f  x   3 x or f  x   ( x)



Now, we check the continuity of the function at x = 0 . L.H.L.  f  0  0   lim f  0  h  h 0

d2y

1 5  t  7 2 2 dx 8t

 lim (0  h)1/ 3

d2y

 (0  0)1/ 3  0

h 0

5



R.H.L.  f  0  0   lim f  0  h 

dx 16t 2. Option (3) is correct. 2

6

h 0 1/ 3

 lim (0  h)

Explanation: Let,

h 0 1/ 3

and f  0   (0)

u  e x and v  log x 3 du dv 1   3 x 2e x and dx dx x 3



3

cos x, x  0 Explanation: Assertion: We have, f  x   cos x   x0  1, Continuity at x = 0,

cot x Explanation: Given, f  x   ( x  1)

L.H.L.  lim f  0  h   lim cos  0  h   cos 0  1

Taking log on both sides, we get



= lim

log ( x + 1)

and f  0   1

[Using L-Hospital Rule] = lim

x→0

1

( x + 1) sec2 x

cos 2 x x→0 x + 1 =1

So, f (x) is continuous at x = 0.

Hence, f  x  is continuous everywhere. Reason: The absolute value function is continuous function. Hence, f  x  is continuous everywhere.

 

f (0) = e1 = e

4. Option (1) is correct. Explanation: Given, f  x  

 L.H.L.  R.H.L.  f  0 

Hence, both Assertion and Reason are true, Reason is the correct explanation of Assertion. 8. Option (2) is correct. 2 Explanation: Assertion: We have, f  x   cos x

= lim

⇒

h 0

h 0

0  m  form  0

tan x

x→0

h 0

 L.H.L.  R.H.L.

lim log f ( x ) = lim cot x log ( x + 1) x→0

h 0

 lim cos h  cos 0  1

For checking continuity at x = 0

)

h 0

R.H.L.  lim f  0  h   lim cos  0  h 

log f  x   cot x log  x  1

(

L.H.L.  R.H.L.  f  0 

∵

3 du 3 x 2e x   3 x3e x dv 1/ x 3. Option (1) is correct.

x→0

0

So, function is continuous at x = 0. 7. Option (1) is correct.

Thus,



 (0  0)1/ 3  0

Here, x  3x  4 2

x  3x  4 2

The function f  x  is not defined at x  3x  4  0 2

L.H.L.  lim cos (c  h) 2  cos c 2 h 0

R.H.L.  lim cos (c  h) 2  cos c 2 h 0

and f  c   cos c 2

  x  4   x  1  0

 L.H.L.  R.H.L.  f  c 

 x  4 and x  1

So, f (x) is continuous at x = c. Hence, f (x) is continuous for every value of x. All of the trigonometric function sin x, cos x, tan x, sec x, csc x, cot x are continuous on their domains. Hence, both Assertion and Reason are true and Reason is not the correct explanation of Assertion. 9. Option (3) is correct.

Thus, f  x  is not continuous at –4 and 1. 5. Option (1) is correct. Explanation: Given, f  x   xn , n  0

f   x   nx n 1 For f (x) to be differentiable for all values of x, n 1  0  n  1 Required interval is [1, ∞) .

Explanation: Assertion: Differentiability implies continuity but continuity does not imply differentiability. Hence, A is true. Reason: We have, f  x   x At x = 0,

62 Oswaal CUET (UG) Chapterwise Question Bank L.H.L.  lim h 0



 lim

f 0  h   f 0 h 0h 0

 lim

h

h  0  h



h  0

f 0  h   f 0 h 0h 0

h  lim  1  h h 0 h 0 h Here, L.H.L. ≠ R.H.L., hence f (x) is not continuous at x = 0.  lim



10. Option (2) is correct. Explanation: Assertion: L.H.L. = lim f ( x ) = lim f ( 2 − h ) x → 2−



 3 sin 2 x cos x

 1

and R.H.L.  lim

3. Option (3) is correct. Explanation: d d sin 3 x  3 sin 2 x  sin x  dx dx

h

h  0

h→0

= lim [ 2 − h − 1] + 2 − h − 2 h→0

= lim [1 − h ] + − h = 0 h→0

Similarly, lim f ( x) = 0 x→2

+

and f ( 2) = [ 2 − 1] + 2 − 2 = [1] + 0 = 1 ∴ L.H.L. = R.H.L. ≠ f (2) Hence, f(x) is discontinuous at x = 2. Reason: f  2  h   f  2 f   2   lim h 0 h 2  h  1  2  h  2  2  1  2  2  lim h 0 h 0  h 1 0    lim 1  h   0  ∵ hlim h 0 0 h   1  lim 1    h 0 h (not defined)  f  x  is not differentiable at x = 2. Hence, both Assertion and Reason are true and Reason is not a correct explanation of Assertion.

4. Option (4) is correct. Explanation: d d  sin 2 x   cos 2 x  2 x  dx dx  2 cos 2 x  d At x  ,  sin 2 x   2 cos   2 2 dx 5. Option (1) is correct.

Explanation: Given, sin y  x cos  a  y  sin y 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 cos 2  a  y 





dx cos  a  y  cos y  sin y   sin  a  y    dy cos 2  a  y 



dx cos  a  y  cos y  sin y sin  a  y   dy cos 2  a  y 



dx cos  a  y   y   dy cos 2  a  y 



dx cos a  2 dy cos  a  y 

Explanation: The given function can be written as 5  2 x, if x  2  f  x   1, if 2  x  3 2 x  5, if x  3  When x  4, f  x   2 x  5

1. Option (2) is correct.

7. Option (1) is correct.

Explanation:



    sin x  5 x 

d 5 d x cos x5   sin x5 dx dx 5

4



6. Option (3) is correct.

[C] COMPETENCY BASED QUESTIONS



MATHEMATICS/APP. MATH.

Explanation: 2, if x  2  f   x   0, if 2  x  3 2, if x  3 

 5 x 4 sin x5 2. Option (3) is correct.

8. Option (3) is correct.

Explanation:

9. Option (4) is correct.

d d sin  cos x   cos  cos x   cos x  dx dx  cos  cos x    sin x 

Explanation: [x] is not continuous at integral values of x.

  sin x  cos  cos x 

Explanation: f (x) is not differentiable at x = 2 and x = 3.

10. Option (3) is correct. Explanation: The function f is continuous at x = 0 if lim x 0 f  x   f  0 

CONTINUITY AND DIFFERENTIABILITY

 lim

x 0

We have f  0   k and lim x 0 f  x   lim x 0  lim

x 0

2 sin 2 2 x 8x2

 sin 2 x   lim   x 0 2 x 



 lim

x 0

2

sin 2 2 x 4 x2

1  cos 4 x 8x2

[Applying L-Hospital Rule]

2 sin 2 2 x 8x

2

 sin 2 x   lim   x 0 2 x   Hence, k = 1.

 lim

x 0

2

sin 2 2 x 4x

2

63

Study Time Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

6

Revision Notes  

 Rate of Change of Quantities � If y = f (x) is a function, where y is dependent variable and x is independent dy variable. Then, [or f ′(x)] represents the dx rate of change of y w.r.t. x and  dy  or f   x0   represents the rate of    dx  x  x0 change of y w.r.t. x at x = x0. y (i) Average rate of change of y w.r.t. x  . x (ii) Instantaneous rate of change of y w.r.t. x =

APPLICATIONS OF DERIVATIVES Scan to know more about this topic

Rate of Change of Quantities

dy . dx

A normal is a straight line perpendicular to a tangent to the curve y = f (x) intersecting at the point of contact.  Slope of Tangent and Normal If a tangent line to the curve y = f(x) makes an angle q with dy X-axis in the positive direction, then = Slope of tangent dx 1 1  tan  and slopeof normal   slope of tangent dy /dx  Equation of Tangent and Normal � Let y = f (x) be a curve and P(x1, y1) be a point on it. Then, (a)  equation of tangent at P(x1, y1) is, dy   y  y1      x  x1   dx   x , y  1

dy dy Note: Here, is positive, if y increases as x increases and dx dx is negative, if y decreases as x increases.  Increasing and Decreasing Functions

(b)  equation of normal at P(x1, y1) is,

(i) Increasing functions: Let I be an open interval contained in the domain of a real valued function f. Then, f is said to be (a) Increasing on I, if x1 < x2

 Maxima and Minima � Let f be a real valued function and c be an interior point in the domain of f. Then,

⇒ ⇒

(b) Strictly increasing on I, if x1 < x2

 f  x1   f  x2  , x1, x2  I

⇒

 y  y1  

1  x  x1   dy   dx     x1 , y1 

(i) Point c is called a local maxima, if

 f  x1   f  x2  , x1, x2  I

(ii) Decreasing functions: Let I be an open interval contained in the domain of a real valued function f. Then, f is said to be (a)   Decreasing on I, if x1 < x2 ⇒

Scan to know more about this topic



Increasing and Decreasing Functions

   f  x1   f  x2  , x1, x2  I

(b) Strictly decreasing on I, if x1 < x2   f  x1   f  x2  , x1, x2  I

 Theorem- Let f  be continuous on [a, b] and differentiable on (a, b). Then, (a)   If f ′(x) > 0 for each x ∈ (a, b), then f (x) is said to be increasing in [a, b] and strictly increasing in (a, b). (b)   If f ′(x) < 0 for each x ∈ (a, b), then f (x) is said to be decreasing in [a, b] and strictly decreasing in (a, b). (c)   If f ′(x) = 0 for each x ∈ (a, b), then f is said to be a constant function in [a, b].  Tangents and Normals A tangent is a straight line, which touches the curve y = f (x) at a dy point. represents the gradient or slope of a curve y = f (x). dx

1

Scan to know more about this topic

there is a h > 0 such that f  c   f  x  x in  c  h, c  h  .

Here, value f  c  is called the local Maxima and Minima maximum value of f. (ii) Point c is called a point of local minima, if there is a h > 0 such that f  c   f  x  x in  c  h, c  h  .

Here, value f (c) is called the local minimum value of f.  Critical Point A point c in the domain of a function f at which either f   c   0 or f is not differentiable, is called a critical point of f.  First Derivative Test Let f be a function defined on an open interval I and let f be continuous at a critical point c in I. Then, (i) If f   x  change sign from positive to negative as x increases through c, then c is a point of local maxima. (ii) If f   x  change sign from negative to positive as x increases through point c, then c is a point of local minima. (iii) If f   x  does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection.

APPLICATIONS OF DERIVATIVES

65

66 Oswaal CUET (UG) Chapterwise Question Bank  Second Derivative Test Let f be a function defined on an interval I and c ∈ I. Let f be twice differentiable at c, then (i) x = c is a point of local maxima, if f   c   0 and f   c   0. The value f (c) is local maximum value of f. (ii) x = c is a point of local minima, if f   c   0 and f   c   0 . Then, f (c) is local minimum value of f. (iii) If f   c   0 and f   c   0, then the test fails.  Absolute Maxima and Absolute Minima � Let f  be a differentiable function on [a, b] and c be a point in [a, b] such that f   c   0 . Then, find f  a  , f  b  and f (c). The maximum of these values gives a maxima or absolute maxima and minimum of these values gives a minima or absolute minima. Example 1: For the function f (x) = 2x3 - 3x2 - 12x + 4, find the point of local maxima and minima. Sol. We have, f (x) = 2x3 - 3x2 - 12x + 4 f′(x) = 6x2 - 6x - 12  Now, f′(x) = 0 ⇒ 6(x2 - x - 2) = 0 ⇒ 6(x + 1)(x - 2) = 0 ⇒  x = -1 and x = +2 On number line for f′(x), we get + –

+

–1 2 Hence, x = -1 is point of local maximum and x = 2 is point of local minima. Example 2: If x is real, then find the minimum value of x2 - 8x + 17. Sol. Let f (x) = x2 - 8x + 17 On differentiating with respect to x, we get    f ′(x) = 2x - 8

MATHEMATICS/APP. MATH.

So,   f (x) = 0 ⇒ 2x - 8 = 0 ⇒ 2x = 8 \  x = 4 Now, again on differentiating with respect to x, we get f ′′(x) = 2 > 0, ∀x 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 Example 3: Find the intervals in which the function  f(x) =

x4  x3  5 x 2  24 x  12 4

(a) strictly increasing, (b) strictly decreasing. Sol. f ′(x) = x3 - 3x2 - 10x + 24 = (x - 2) (x - 4) (x + 3) f (x) = 0 ⇒ x = -3, 2, 4. sign of f′(x): – + – + –∞ –3 2 4 ∞ \ f (x) is strictly increasing on (-3, 2) ∪ (4, ∞) and f (x) is strictly decreasing on (-∞, -3) ∪ (2, 4) Example 4: 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...(i) π  cos x < 0 for each x ∈  , π  2  \ f′(x) < 0 π  Hence, function is strictly decreasing in  , π  . 2 

[from (i)]

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. The edge of a cube is increasing at a rate of 7 cm/s. Find the rate of change of area of the cube when a = 3 cm.  (1) 252 cm2/s. (3) 498 cm2/s.

(2) 504 cm2/s. (4) 287 cm2/s.

2. The maximum value of the f  x    (1)

4 . 3

(2)

3 . 4

(3)

-4 . 3

1

is

4 x2  2 x  1 [CUET 2023] (4) -

3 . 4

3. The rate of change of area of square is 40 cm2/s. What will be the rate of change of side if the side is 5 cm.  (1) 2 cm/s.

(2) 4 cm/s. (3) 6 cm/s. (4) 8 cm/s.

4. The minimum value of f (x) = 4x3 - 48x + 105 in the interval [1, 3] is [CUET 2023] (1) 61. (2) 41. (3) 32. (4) 69. 5. Match List I with List II  [CUET 2023] LIST I Maximum value of A f (x) = -x + 1 +3

LIST II I

6

II

5

C Maximum value of f (x) = 6 - x2

III

no maximum value

D Maximum value of f (x) = x3 + 1

IV

3

B

Minimum value of f (x) = (2x - 1)2 + 5

Choose the correct answer from the options given below: (1) A-IV, B-II, C-I, D-III (2) A-III, B-IV, C-I, D-II (3) A-I, B-II, C-III, D-IV (4) A-II, B-III, C-IV, D-I

67

APPLICATIONS OF DERIVATIVES

6. If f  x  

1 , then for x > 1, f (x) is  1 x

[CUET 2023]

(1) decreasing. (3) increasing. 7. (1) (3) 8. (1) (3) 9. (1)

(2) constant. (4) neither decreasing nor increasing. The approximate volume of a cube of side a metres on increasing the side by 4% is  [CUET 2023] 1.04a3 m3 (2) 1.004a3 m3 1.12a3 m3 (4) 1.12a2 m3 Two positive numbers x and y whose sum is 25 and the product x3 y2 is maximum are  [CUET 2023] x = 10, y = 15 (2) x = 15, y = 10 x = 12, y = 13 (4) x = 16, y = 9 The function f (x) = x2 - 2x is strictly decreasing in the interval  [CUET 2022] (-∞, -1). (2) (-1, ∞). (3) (-∞, 1). (4) (-1, ∞).

10. The total revenue (in `) received by selling ‘x’ units of a certain products is given by: R  x   4 x 2  10 x + 3. What

is the marginal revenue on selling 20 such units?  [CUET 2022] (1) ` 130 (2) ` 170 (3) ` 173 (4) ` 360 2 11. If x is a real, then minimum value of x  8 x  17 is

[CUET 2022]



(1) -1 (2) 0 (3) 1 (4) 2 12. If f (x) = ax2 + 6x + 5 attains its maximum value at x = 1, then the value of a is [CUET 2021] (1) 0 (2) 5 (3) 3 (4) -3 13. The equation of the tangent to the curve given by x = a  sin3t, y = b cos3 t at a point where t  is [CUET 2021] 2 (1) y = 1.

(2) y = 0.

(3) x = 0.

(4) x = 1.

14. The function f  x   2 x  3 x  12 x  4, has 3

2

(1) two points of local maximum. (2) two points of local minimum. (3) one maxima and one minima. (4) no maxima or minima. 15. The total revenue in rupees received from the sale of x units of a product is given by R  x   3 x 2  26 x  15. The marginal revenue, when x = 15 is (1) ` 100. (2) ` 116. (3) ` 123. (4) None 16. The maximum profit that a company can make, if the profit 2 function is given by p  x   41  72 x  18 x is

(1) 111. (2) 112. (3) 113. (4) 114. 17. What is the slope of the tangent to the curve 2x y 2 at (1, 1)? ( x  1) (1) 3 (2) 2 (3) 1 (4) 0 sin x ⋅ cos x 18. The maximum value of is (1) 1 . 4

(2) 1 . 2

(3)

2 .

(4) 2 2 .

x

1 19. The maximum value of   is x

1/ e

1 (4)   e 20. The minimum value of f (x) = x3 - 3x in [0, 2] is (1) e.

(3) e1/e.

(2) ee.

.

(1) 2 (2) -2 (3) 0 (4) None of these 21. The rate of change of area of circle with respect to its radius r at r = 3 cm is (1) 6p (2) 8p (3) 12p (4) 3p 22. The least value of a so that the function f (x) = x2 + ax + 1 is strictly increasing on [1, 2] is (1) 2 (2) -2 (3) 1 (4) -1 23. The demand function of a toy is, x = 75 - 3p and its total cost function is TC  100  3 x . For maximum profit the value of x is (1) 33 (2) 31 (3) 29 (4) 24 24. Suppose that you have the following utility function, u  x  u  x   x . Find  . u  x  1 1 (2) (3) 2x (4) -2x 2x 2x 25. Find the second derivative of the following function: (1)

f  x   5 x 2  x  47 

(1) f   x   30 x  470

(2) f   x   30 x  470

2 (3) f   x   15 x  235

2 (4) f   x   15 x  470 x

26. The absolute maximum value of y = x3 – 3x + 2 in 0 ≤ x ≤ 2 is (1) 0 (2) 2 (3) 4 (4) 6 5 , f  x   2 sin 3 x  3 cos 3 x is 27. At x  6 (1) maximum. (2) minimum. (3) zero. (4) neither maximum nor minimum. 3 2 28. The second derivative of x  5 x  x  0 will be (1) 10 x - 5 . (2) 6 x - 10 . (3) 3 x + 10 . (4) 3 x 2 - 5 x . 2 3 29. If f  x   5 x (4 x  9) then f   x    3 2 2 (1) 10 x(4 x  9)  60 x (4 x  9) 3 2 2 (2) 10 x(4 x - 9) - 60 x (4 x - 9) 3 2 2 (3) 10 x(4 x  9)  60 x (4 x  9) 3 2 2 (4) 10 x(4 x + 9) + 60 x (4 x + 9) 3 30. If y = x log x, then

(1) 3 . x

(2)

4 . x

d4y dx 4

is (3) 6 . x

(4) x . 6

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. (1) Both A and R are true, and R is the correct explanation of A.

68 Oswaal CUET (UG) Chapterwise Question Bank (2) Both A and R are true, but R is not the correct explanation of A. (3) A is true but R is false. (4) A is false and R is true. 1. Assertion (A): The average rate of change of the function

y  15  x 2 between x = 2 and x = 3 is -5. Reason (R): Average rate of change  y  yat x  3  yat x = 2.

2. Assertion (A): A particle moving in a straight line covers a distance of x cm in t second, where x = t3 + 3t2 - 6t + 18. The velocity of particle at the end of 3 seconds is 39 cm/s. Reason (R): Velocity of the particle at the end of 3 dx seconds is at t = 3. dt 3. Assertion (A): If 3 ≤ x ≤ 10 and 5 ≤ y ≤ 15 , then



x maximum value of   is 2.  y Reason (R): If 3 ≤ x ≤ 10 and 5 ≤ y ≤ 15, then minimum

x 1 value of   is .  y 5 4. Assertion (A): Minimum value of (x - 5) (x - 7) is -1. 4ac - b 2 . 4a 5. Assertion (A): If the circumference of the circle is changing at the rate of 10 cm/s, then the area of the circle changes at the rate 30 cm2/s, if radius is 3 cm. Reason (R): If A and r are the area and radius of the circle, respectively, then rate of change of area of the dA dr  2 r circle is given by . dt dt 6. Let f :  →  be a function such that

Reason (R): Minimum value of ax2 + bx + c is

f  x   x  x  3 x  sin x. Assertion (A): f is one-one. Reason (R): f (x) is neither increasing nor decreasing function. 3 2 7. Assertion (A): f  x   2 x  9 x  12 x  3 is increasing outside the interval (1, 2). Reason (R): f   x   0 for x  1, 2 .

3

2

8. Assertion (A): The curves x = y2 and xy = k cut at right angle, if 8k2 = 1. Reason (R): Two curves intersect at right angle, if the tangents to the curves at the point of intersection are perpendicular to each other i.e., product of their slope is -1. 9. Assertion (A): If the radius of a sphere is measure as 9 m with an error of 0.03 m, then the approximate error in calculating its surface area is 2.16p m2.  ds  Reason (R): We have, S    r where, DS =  dr  Approximate error in calculating the surface area, Dr = Error in measuring radius r. 10. Assertion (A): If the length of three sides of a trapezium other than base are equal to 10 cm, then the area of

trapezium when it is maximum, is 75 3 cm 2. Reason (R): Area of trapezium is maximum at x = 5.

MATHEMATICS/APP. MATH.

[C] COMPETENCY BASED QUESTIONS I. Read the below passage and solve the questions from 1-5 A cable network provider in a small town has 500 subscriber and he used to collect ` 300 per month from each subscriber. He proposes to increase the monthly charges and it is believed from the past experiences that for every increase of ` 1, one subscriber will discontinue the service.  [CUET 2022] 1. If ` x is the monthly increase in subscription amount, then the number of subscribers is (1) x (2) 500 - x (3) x - 500 (4) 500 2. Total revenue ‘R’ is given by (in `) (1) R  300 x  300  500  x  . (2) R   300  x   500  x . (3) R   300  x   500  x . (4) R  300 x  500  x  1. 3. The number of subscribers which gives the maximum revenue is (1) 100

(2) 200

(3) 300

(4) 400

4. What is increase in changes per subscriber that yields maximum revenue? (1) 100

(2) 200

(3) 300

(4) 400

5. The maximum revenue generated is (1) ₹ 200000 (2) ₹ 180000 (3) ₹ 160000 (4) ₹ 150000 II. Read the below passage and solve the questions from 6-10 In a school, a auditorium was used for its cultural activities. The shape of the floor of the auditorium is rectangular with dimensions x and y (x > y) has fixed perimeter p. [CUET 2022] 6. If x and y represent the length and breadth of the rectangular region, then (1) p  x  y .

(2) p 2  x 2  y 2 .

(3) p  2  x  y  .

(4) p  x  2 y .

7. The area (A) of the floor, as a function of x can be expressed as (1) A  x   px 

x . 2

(2) A  x  

px  x 2 . 2

px  2 x 2 x2 . (4) A  x    px 2 . 2 2 8. The value of x, for which area of floor of auditorium is maximum is (3) A  x  

(1)

p . 4

(2)

p . 2

(3) p.

(4)

p . 3

9. The value of y, for which area of floor of auditorium is maximum is (1)

p . 2

(2)

p . 3

(3)

p . 4

(4)

p . 16

(3)

p2 . 28

(4)

p2 . 64

10. Maximum area of floor is (1)

p2 . 4

(2)

p2 . 16

69

APPLICATIONS OF DERIVATIVES

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (1)

3. (2)

4. (2)

5. (1)

6. (1)

7. (3)

8. (2)

9. (3)

10. (2)

11. (3)

12. (4)

13. (2)

14. (3)

15. (2)

16. (3)

17. (4)

18. (2)

19. (3)

20. (2)

21. (1)

22. (2)

23. (1)

24. (1)

25. (2)

26. (3)

27. (4)

28. (2)

29. (1)

30. (3)

8. (1)

9. (1)

10. (1)

8. (1)

9. (3)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (1)

3. (2)

4. (1)

5. (1)

6. (3)

7. (2)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (3)

3. (4)

4. (1)

5. (3)

6. (3)

7. (3)

ANSWERS WITH EXPLANATION 1. Option (1) is correct. Explanation: Let the edge of the cube be a. The rate of change of edge of the cube is given by The area of the cube is A = 6a2

dA . dt

Differentiating w.r.t. ‘ t ‘, we get dA da  12a dt dt dA  12a  7  84a dt

∴

dA = 40 cm 2 /s dt dxdx 40 \  402 2x x dtdt dxdx 2020 ⇒   dtdt x x dx 20 = = 4cm/s Thus, dt x=5 5  

[A] MULTIPLE CHOICE QUESTIONS

4. Option (2) is correct. [Since, given

da = 7 cm / s] dt

cm 2 dA  84  3  252 Thus, s dt a 3

 Let f(x) = 4x3 – 48x + 105   f’(x) = 12x2 – 48   f’(x) = 12(x2 − 4)   f’(x) = 12(x − 2)(x + 2) Now, f’(x) = 0 gives x = 2 or x = −2.

2. Option (1) is correct. Explanation: If we can find the minimum of the expression

We consider only x = 2 ϵ [1, 3] f(1) = 4(1)3 – 48(1) + 105 = 4 – 48 + 105 = 61

4 x 2 + 2 x + 1, that will be the maximum value of the given expression. We can write the expression as,

f(2) = 4(2)3 – 48(2) + 105 = 32 – 96 + 105 = 41

 x 1 3 1 3  4  x2      4  x     2 16 16  4  16     This expression is always positive and has minimum value at

5. Option (1) is correct. Explanation: A. Maximum value of f  x    x  1  3 is 3 when x  1 such that A belongs to IV. 1 2 B. Minimum value of f  x   (2 x  1)  5 is 5 when x = . 2 Such that option B from list I belongs to option II of list II.

2

x

1 4

Thus, at x   3 3 4    16  4

1 we get the value of the expression as 4

4 So, the maximum value of expression is . 3 3. Option (2) is correct. Explanation: Let the side of the square be x. Area of square, A = x2 Differentiating w.r.t. ‘t’, we get dA dx = 2x   dt dt Since, given

f(3) = 4(3)3 – 48(3) + 105 = 108 – 144 + 105 = 69 So the minimum value of f on [1, 3] is 41 occurring at x = 2

2 C. Maximum value of f  x   6  x is 6 when x = 0. Such that option C from list I belongs to option I of list II. 3 D. There is no maximum value of f  x   x  1 Such that option D from list I belongs to option III of list II. 6. Option (1) is correct. Explanation:

1 1 x For x  1,  x  1 1 x  11 Here, f ( x) is decreasing as1  x  0 f ( x) 

70 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH.

7. Option (3) is correct. Explanation: Volume (V) of a cube with side a is given by, V = a3 dV   3a 2  V  3a 2 a, da

i.e.,

2x  2  0

or,

2  x  1  0

or,

x a ) is given by

x=b

If the curve under consideration lies below X-axis, then f(x) < 0 from x=a to x=b, the area bounded by the curve y=f(x) and the ordinates x=a, x=b and X-axis is negative



94 Oswaal CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH.

∴ Area ∆ ACE = APPLICATION OF INTEGRALS

=

⇒ or

1 2

∫ ( x − 1) dx

∫

1 = 2

3

ydx

95

3

1

 1  x2  − x 2  2 1

2

 y  0  1 0 x 1 3 1 y 1   x 1 2 1 y   x  1 2

∴ Area ∆ ACE =

1

3

3 2   4 3  2      2 2   3  2 Equation of line between A (1,0) and C(3,1) is 2

3

∫

=

  12   1  32  − 3 −  − 1  2  2   2    1  9   1  =  − 3 −  − 1     2  2 2

=

1 3 1 1 = × 2 =1 + 2  2 2  2

=

Hence, area of triangle ABC = Area ∆ ABD + Area

ydx

1

3 square units 2

Area of ABC 

3

∫ ( x − 1) dx

3   BDCE − Area D ACE  1   1 2  

1

3

 1  x2 =  − x 2  2 1

OBJECTIVE TYPE QUESTIONS

2    1QUESTIONS [A] MULTIPLE 1 CHOICE 32

 − 3 −  − 1  2  2   2 x2 y 2  1. The area enclosed by the ellipse  2  1 is 2 6 1  9   1 9 =  − 3 −  − 1   [CUET 2023]   2  2  2 (1) 15 π (2) (4) 36 π 1 54 3 π 1  (3) 1 18 π =  +  = × 2 =1 2. The area of the by the lines x = 2y + 3, 2 region 2 2 bounded 2 x = 0, y = 1 and y = − 1 is [CUET 2023] (1) 4 sq. units (2) 6 sq. units (3) 8 sq. units (4) 3 sq. units 2 =

3. The area of the region bounded by the parabola y2 = 4ax and its latus rectum is [CUET 2023] (1)

4a 2 sq. units 3 2a 3

(2)

2

8a 2 1 sq. units 3 3

9a sq. units 5 4. The area of the region bounded by the line y = 4 and the curve y = x2 is  [CUET 2021] (3)

sq. units

32 square units 3 (3) 1 square unit (1)

(4)

p p p (3) (4) 2 3 4 8. The bounded by the curve 3x - 5y = 15, x = 4, x = 7 and x-axis is (1) p

19 sq. units. 10 10 (3) sq. units. 3

(4) 32 square units

5. The area of the region bounded by the circular pizza, x2 + y2

(1) 2(π – 2) (2) π – 2

(3)

(4)

1 sq. units 7

7. Area lying in the first quadrant and bounded by circle x2 + y2 = 4 and the lines x = 0 and x = 2 is









4 8  3 3

(4)









4 8  3 3

12. The area lying between the curve y2 = 4x and y = 2x is 2 4 7 1 (2) (3) (4) 2 3 3 12 3 13. The area of circle of radius ‘a’ can be calculated by the formula: (1)

49 sq. units 3

(4) 2(π + 2)

11. Area bounded by the curve y = x3, the x-axis and the ordinates x = −2 and x = 1 is 17 15 15 (1) −9 (2) - (3) (4) 4 4 4

6. The area bounded by the curve y2 = x, line y = 7 and y-axis is (2)

(3) 2π – 1

10. The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is 4 4 (1) 4  3 (2) 4  3 3 3

(1)

343 sq. units 3 1 (3) sq. units 3

(2)

9. Smaller area enclosed by the circle x2 + y2 = 4 and the line x+y=2

= 4 and the edge of the knife, x = 3 y in the first quadrant is (in sq. units) p 2p 5p p (1) (2) (3) (4) 3 2 3 3

(1)

9 sq. units. 10 3 (4) sq. units. 10

(1)

2

(2) 0 square unit

(2)

 a b

2

a

(3) 4 

 a a

0



 x 2 dx 2

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



 x 2 dx

2

2

2

0

a

2

2

0

14. The area bounded by the curve y = |x − 1| and y = 1, using integration is (in sq. units) 1 (1) 5. (2) . (3) 2. (4) 1. 2

96 Oswaal CUET (UG) Chapterwise Question Bank 15. The area of the region bounded by curve 4x2 = y and the line y = 8x + 12, is (in sq. units) 128 32 1 (2) 4 (3) (4) 3 3 32 16. The area of the figure bounded by the curve y = logex, the x-axis and the straight line x = e is (1)

(1) 5 - e (2) 3 + e (3) 1 (4) None of these 17. The area of the region enclosed by the curves 1 = y 2 x= ,x , y = 0 and x = 1, is (in sq. units) 4 1 1 (1) . (2) . 8 4 7 2 (3) (4) . . 12 3 18. Shown below is a parabola.

MATHEMATICS/APP. MATH.

The area of the shaded region is (Note: You need not evaluate the square roots.)  10   3  sq. units. (1)    3 

 5 10   3  sq. units. (2)    3 

 5 10   3  sq. units. (4)  2 10  11 sq. units. (3)     31    3  20. The area of the region bounded by the curves x2 + y2 = 4 and X-axis in the first quadrant is (in sq. units) (1) 3p . (2) 4. (3) 4. (4) 2p . 4 3 21. The area of the region bounded by the y-axis, y = cos x and y  sin x,0  x   / 2 is  [NCERT Exemplar Pg 177 Q 24] (1) (3)

2 ysq.  units 16  x 2





2  1 sq. units

 2  1 sq. units (4)  2 2  1 sq. units (2)

22. The area of the region bounded by the curve x2 = 4y and the straight-line x = 4y − 2 is  [NCERT Exemplar Pg 177 Q 25] 3 5 (1) sq. units (2) sq. units 8 8 (3)

7 sq. units 8

(4)

9 sq. units 8

23. The area of the region bounded by the curve y  16  x 2 and X-axis is [NCERT Exemplar Pg 177 Q 26]

The area of the shaded region is (Note: Take (1)

2 as 1.4 and

4 2 sq. units. 3

(3) 4.27 sq. units.

(1) 8π sq. units

(2) 20π sq. units

(3) 16π sq. units

(4) 256π sq. units

24. Area of the region in the first quadrant enclosed by the

3 as 1.7.) (2) 2 sq. units. (4) 5.27 sq. units. 2

19. Shown below is the graph of f (x) = 2x in the first quadrant.

X-axis, the line y = x and the circle x2 + y2 = 32 is  [NCERT Exemplar Pg 178 Q 27] (1) 16π sq. units (2) 4π sq. units (3) 32π sq. units

(4) 24π sq. units

25. Area of the region bounded by the curve y = cos x between x = 0 and x = π is [NCERT Exemplar Pg 178 Q 28] (1) 2 sq. units (2) 4 sq. units (3) 3 sq. units

(4) 1 sq. unit

26. The area of the region bounded by parabola y2 = x and the straight line 2y = x is [NCERT Exemplar Pg 178 Q 29] 4 sq. units 3 2 (3) sq. units 3

(2) 1 sq.units

(3) 3 sq. units

(4) 1 sq. units

(1)

(4) 1 sq. units 3 27. The area of the region bounded by the curve y = sin x  between the ordinates x  0, x  and the axis is 2  [NCERT Exemplar Pg 178 Q 30] (1) 2 sq. units (2) 4 sq. units

97

APPLICATION OF INTEGRALS

2 2 28. The area of the region bounded by the ellipse x  y  1 25 16 is [NCERT Exemplar Pg 178 Q 31]

(1) 20p sq. units

(2) 20p 2 sq. a units

(3) 16p 2 sq. units

xdy (4) 25 sq.4 units



0 29. The area of the region bounded by the circle x 2  y 2  1 is a

 (1) 2p sq. units (3) p sq. units



Reason (R): Area of the curve y = f(x) is given by



x2

f  x  dx x1 Where, x1 and x2 are the end-points between which the area is required. 3. Assertion (A): The area bounded by the circle x2 + y2 = 16 is 16p sq. units. Reason (R): We have x2 + y2 = 16, which is a circle having centre at (0, 0) and radius 4 units. A

[NCERT Exemplar Pg 178 Q 32]  4 a 2  y 2 dy (2) 3p sq. units 0 (4) 4p sq. units  a  y 2  16  x 2  y  16  x 2 2  y y =2x + 12andathe 1 y 30. The area of the region bounded by thecurve 4 a y  sin  2 a   2 lines x = 2 and x = 3 is [NCERT Exemplar Pg 178 Q 34] 0 9 7 2 (1) sq. units (2) sq. units    2  4  a  0  a sin 11  0  2    2  2   (3) 11 sq. units (4) 13 sq. units 2 2 a2  4 X' 2 2 [B] ASSERTION REASON QUESTIONS O 2   a A statement of Directions: In the following questions, Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as (1) Both A and R are true and R is the correct explanation of A.



(2) B  oth A and R are true but R is not the correct explanation of A. (3) A is true but R is false. (4) A is false but R is true. 1. Assertion (A): The area enclosed by the circle x 2  y 2  a 2 is p a 2.

X

Y' From figure, area of shaded region 4

A4



42  x 2 dx

0

4. Assertion (A): The area of region bounded by the curve 45 y = x + 5 and the lines x = 1 and x = 4 is sq. units. 2

B (0,a)



Reason (R): Required area, A 



5

4

. 4 ydyA

1

A(a,0)



42  x 2 dx

0

5. Assertion (A): The region bounded by the curve 8 y 2 = 16 x, Y -axis and the line y = 2 is . 3

Reason (R): Required area 

2

 xdy. 0

6. Assertion (A): The area of the smaller region bounded by

x2 y 2   1 and the straight line 3 x  4 y  12 16 9 is  3  6  sq. units. the ellipse

Reason (R): The area enclosed by the circle

a

 4  xdy 0



a

 4  a 2  y 2 dy

Reason (R):

0

a

y 2 a2 y  4 a  y 2  sin 1  2 a 0 2 2  a   a  4   0  sin 1 1   0  2  2   2 a  4 2 2   a2 2. Assertion (A): The area of curve y = 4x3 between the end points x = [-1, 2] is 17 sq. units.

\ Required area 



43 04

16  x 2 dx 

1 4

4

 12  3x  dx 0

98 Oswaal CUET (UG) Chapterwise Question Bank 7. Assertion (A): The area of the region bounded by the lines AB : 2 x  y  4, BC : 3 x  2 y  6 and CA : x  3 y  5  0 7 is sq units. 2 A(1,

d

d





A  xdy  g ( y )dy

C(4, 3)

2)

MATHEMATICS/APP. MATH.

c

c

Reason (R): If the curve under consideration lies below x -axis, then f (x) < 0 from x = a to x = b, the area bounded by the curve y = f (x) and the ordinates x = a, x = b and X-axis is negative. But if the numerical value of the area is to be taken into consideration, then b

Area 

Reason (R): Required area 

41

3 1

2

1

[C] COMPETENCY BASED QUESTIONS

 x  5 dx    4  2 x  dx

4





 f  x  dy



1  3x  6  dx 2

a

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

2

8



Assertion (A): The area of the region in the first quadrant

enclosed by the line x  y  2, the parabola y 2 = x and 7 X-axis is square units. 6

Reason (R): Solving x  y  2 and y 2 = x simultaneously, we get the points of intersection as (1, 1) and (4, 2).

The bridge connects two hills 100 feet apart. The arch on the bridge is in a parabolic form. The highest point on the bridge is 10 feet above the road at the middle of the bridge as shown in the figure. 1. The equation of the parabola designed on the bridge is

The required area = The shaded area 1







2

xdx 

0

 2  x  dx

x2  2  3/ 2 1  x  2 x   0  3 2   1 2 1 7    square units 3 2 6 

9. Assertion (A): The area of the region { x, y  : 0  y  11 . x 2 , 0  y  x, 0  x  2 is 6





Reason (R): The points of intersection of the parabola y = x 2 and the line y = x are (1, 0) and (1, 4) and Required area 

2

4

 x dx   xdx 0

2

(2) y 2 = 250 x.

(3) x 2  250 y.

(4) y 2 = 250 y.

1

2



(1) x 2 = 250 y.



(1) 1000 . 3

(3) 1200.

(2) 250 . 3

3. The integrand of the integral (1) Even



50 50

(4) 0.

x 2 dx is ............. function.

(2) Odd

(3) Neither odd nor even (4) None of these 4. The area formed by the curve x 2 = 250 y, X -axis, y = 0 and y = 10 is (1)

1000 2 4 . (2) . 3 3

(3)

2

10. Assertion (A): The area A of the region bounded by the curve x = g(y), Y-axis and the lines y = c and y = d is given by

x2 dx is 50 250

50

2. The value of the integral

5. The value of the integral (1)

25 . 3

(2)

25 . 4

1000 . 3

(4) 0.

x2 dx is 10 250

10



(3)

1 . 250

8 (4) . 3

99

APPLICATION OF INTEGRALS

II.  Read the following text and answer the following questions on the basis of the same: In the figure O  0, 0  is the centre of the circle. The line y = x meets the circle in the first quadrant at the point B.

7. The co-ordinates of B are (1) (1, 1). (2) (2, 2).





(3) 4 2 , 4 2 .

(4) (4, 4).

8. Area of OBM 1, 1 is (in sq. units) (1) 8.

(2) 16.

(3) 32.

(4) 32p.

9. The area of Ar  BAMB  4, 4  is (1) 32p.

(2) 4p.

(3) 8.

(4) 4p -8. 2

10. The area bounded by the circles x + y = r2, r = 1, 2 and the rays 2x2 – 3xy – 2y2 = 0 (y > 0) is

6. The equation of the circle is (1) x 2  y 2  4 2 . (3) x 2  y 2  32.

(1)

p sq. units 4

(2)

p sq. units 2

(3)

3p sq. units 4

(4)

3p sq. units 2

x2 dx 10 250 2 (4) ( x  4 2 )  0. 10



2 2 (2) x  y  16.

2

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (2)

3. (2)

4. (1)

5. (4)

6. (1)

7. (1)

8. (2)

9. (2)

10. (3)

11. (3)

12. (1)

13. (3)

14. (4)

15. (1)

16. (3)

17. (3)

18. (4)

19. (2)

20. (4)

21. (3)

22. (4)

23. (1)

24. (2)

25. (1)

26. (1)

27. (4)

28. (1)

29. (3)

30. (1)

8. (1)

9. (3)

10. (2)

8. (1)

9. (4)

10. (3)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (1)

3. (1)

4. (3)

5. (4)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (3)

2. (1)

3. (1)

4. (3)

5. (4)

6. (3)

7. (4)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

Y

1. Option (2) is correct.

y= 1

x2

0

1



y2

X'

 1 with above

92 62 ellipse equation, we get a = 9 and b = 6, then Area enclosed by the ellipse = πab  96  54 2. Option (2) is correct. Explanation: Required area, A    2 y  3 dy 1 1  2 y2    3y  2 

X

3

y

b2

y = –1

+

On comparing given ellipse equation



2y

x

a2 We know that, Area of ellipse = πab

2

x=

Explanation: Equation of ellipse

2

Y'

3. Option (2) is correct. Explanation: The equation of parabola is y2 = 4ax Y

L

1

–6, 0

O

S –4, 0

X

1

1

  y 2  3 y  1  1  3  1  3  6 sq units.

L‘

Let O be the vertex, S be the focus and L' — be the latus rectum of parabola.

100 Oswaal CUET (UG) Chapterwise Question Bank The equation of latus rectum is x = a. Also, we know that parabola is symmetric about X-axis. ∴ Required area = 2( area of OSLO) ∴ Required area a

a





2 ydx  2 2 a xdx 0

0

a

   x3/ 2  1/ 2  2  2 a x dx  4 a    3  0  2  0 a





7



MATHEMATICS/APP. MATH.

y 2 dy

0

7

 y3  1 343 sq.units     [ 7 3  0]  3 3 3   0 7. Option (1) is correct. Explanation: The area bounded by the circle and the lines in the first quadrant is represented as:

2  3/ 2  a 8 a  3/ 2 x  a  0 0  3 3  8 a 3/ 2 8 2 a  a sq. units  3 3 4. Option (1) is correct. 4 a

Explanation: Required area  2

4

 xdy

2

0

A=  ydx 0 2

Y y=x

2

  4  x2 dx 0

y=4 X′

2

4 x x   4  x2  sin 1  2 2 2  0   sq. units

X

8. Option (2) is correct.

Y′ 4

Explanation: Given curve, 3x - 5y = 15 3 x  15 ⇒ y 5

 2  ydy 0 4

 3 2  2 ( y ) 2   3  0 3

 2(4) 2 

2 3

  

Required area 

4  8 3 32 units2  3 5. Option (4) is correct. Explanation:

   

7

y dx  

y dx

4 7  3 x  15  4

4 7  3 x  15  4

7

5

 dx 



3 5

 



7

4

 dx 

5

 x  5 dx 7

3 3 9 3   x  5   2 2   7  5    4  5     4  1  sq. u x   4     10 10 10 5 2 Required area = ∫ dx + ∫  4 7 3 0 3   x  5 2  9 3 3 3 2 2   3  2    7  5   4  5    4  1  sq. units x 1  x2   2 5 4 2−1  x   10  10 10   =   +  4 − x + sin   4  2 3  2 0  2  3 2 9. Option (2) is correct.     1 3 3 3 Explanation: The smaller area enclosed by the circle x2 + y2 and  + 2sin −1 = − 0  +  2sin −1 (1) −     2   3 2   the line, x + y = 2 is represented by the shaded area ACBA as:  2 3

2

3  2 5 4 − x dx

3 2 3 2    2 2 2 3   sq. units 3 6. Option (1) is correct. Explanation: Given curves, y2 = x, y = 7 and y-axis 

Required area 



7

0

x dy



7

 x  5 dx

2



101

APPLICATION OF INTEGRALS

It can be observed that Area of ACBA = Area of OACBO - Area of ∆AOB Thus, Area of ACBA 2

2

0

0

  4  x2 dx    2  x  dx x   2 x  2 2  2  1 

2

2

22 1 x   x2  sin   2 x   2 2   2  0 0 2 2 2  0 2 1 x  4  x  2 sin  2 2   20   2  0 2  2 2 1 2   0 2 1 0  4  2  2 sin 4  0  2 sin  4200 2   2 2  0  2 sinn 1 1  0 4  2  0   2 22  x 2 

 2 sin 1 1  2   2  2 2

4 Required area  16  (4  3 ) 3 4  16  (4  3 ) 3 32 4 3   3 3 4  [8  3 ] sq. units 3 11. Option (3) is correct. Explanation: Required area, 1

A=

∫ ydx

−2 1

=

∫ x dx 3

−2

Y

y = x³

10. Option (3) is correct. Explanation:

X’ B

B(1, 1)

O

C

A

X

D (–2, –8) x=–2

Y’ x=1

4 1

x    4  2

Area bounded by the circle and parabola = 2[area(OAMO) + area( AMBA)] 2

∫

= 2

4

6 xdx + 2

0

16 − x 2 dx

2

2

= 2 6

∫

1     4 4   15  4 15 15  Area    sq.units 4 4

∫ 0

4

∫

xdx + 2

16 − x 2 dx

2

4

= 2 6× =

x 2  3/2  2 16  x x + 2  16 − x 2 + sin −1    0  4  3 2 2  2

{

12. Option (1) is correct. Explanation: The area lying between the curve y2 = 4x and y = 2x is represented by the shaded area OBAO as

}

  4 6  1   (2 2 − 0) + 2  0 + 8sin −1 (1) − 2 3 + 8sin −1      2    3   16 3 π π  + 2 8 × − 2 3 − 8 ×  3 2 6 

= =

4π  16 3  + 2  4π − 2 3 −  3 3 

16 3 8π + 8π − 4 3 − 3 3 16 3 + 24π − 4 3 − 8π = 3 4 = [4π + 3]sq.units 3

=

Area of circle = π(r )2 = π(4)2 = 16π sq. units

The points of intersection of the curves are O(0,0) and A(1,2) We draw AC perpendicular to X-axis such that coordinate of C is (1, 0). Area of OBAO = Area of OCABO −Area of ∆OCA 1

1





A  2 xdx  2 xdx 0

0

1

  1  x2   x3/ 2   2  2     2  0  3   2  0

∫

∫

0

1

1

2

  x2   x2 =  x −  +  − x 2   2 1  1 1 0 102 Oswaal CUET (UG) Chapterwise MATH. 0   4Bank MATHEMATICS/APP.  1   Question 1  A  2 xdx  2 xdx =  1 −  −  0 −  +  − 2  −  − 1 2 2 2 2         0 0 1 1 1 = +   1 2 2  x2   x3/ 2  = 1 sqq. unit  2  2    2  0  3  15. Option (1) is correct.  2  0 Explanation: Given curve is 4x2 = y and line is y = 8x + 12. On 4     1 solving both equations, we get 3  1  sq. unit 3 13. Option (3) is correct. y = 8x+12 Y Explanation: Here, equation of circle is x2 + y2 = a2 (3, 36)





B

4x2 = y

(–1, 4) X'

3 2

(0, 0)

Y'

4 x  8 x  12 2

Area of first quadrant  

 a a

2

x

0

2

 dx

Area of circle  4 



a

0

 a a



y dx

2

0

x2  2 x  3

 x2  2 x  3  0  x  3,  1



 x 2 dx

3

Required area 

8x  12  4 x  dx 2

1 3

14. Option (4) is correct. Explanation:

4

  2 x  3  x  dx 2

1

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  32 128  4  sq. units 2 3 3

We have,

y   x  1 y  x  1, if x  1  0 y   x  1, if x  1  0 Required Area = Area of shaded region 2 2

A= = ∫ ydx ydx A

16. Option (3) is correct. Explanation: Given curve, y = logex At, x = 1, y = loge (1) = 0 At, x = e, y = loge (e) = 1

0 0 1 1

e

2 2

+ ∫( xx − − 11) dx = ∫(11 − − xx ) dx dx + dx = 0 0

1 1

11 x22 

 x22

22

  x x = =  xx − −  + +  − − xx  2 2  1 2 0  2 0 1 0   44  11     11  0 = 1 − 0 − − + − 22  − = 1 −  −  0 −  +  − − − − 11  22     22  22   22 1 11 =1+ + = 22 22 qq.. unit = 11 sq = sq unit

Required area, A    loge x dx 1

Using integration by parts, A = [x loge x – x]1e = [e – e – 0 + 1] =1

X

103

APPLICATION OF INTEGRALS

17. Option (3) is correct. 2

Explanation: The area of the region bounded by the curve, y = x, 1 the lines x = and x = 4 and y = 0 (i.e., X-axis) is the ABEF 4

4 4 2  2 3   1.4  2  1.7 3 3 =1.87 + 3.4 = 5.27 sq. units (Approx.) 19. Option (2) is correct. Explanation: 

dx

Thus, required area = area of ABCD 1

1

  ydx   xdx 1 4

1 4

1

 3 3  3  x2  2  2  1 2      (1)    3 4   3     1  2 

0

4

2  1 1 3  8  2 7    3 8 7  sq.units 12 18. Option (4) is correct. Explanation: 

Area bounded region = ar ( OACD ) − ar rect ( ABDF ) 5

  xdy  length  breath 0



5

y

0

2

dy  1  3

[Since, given y = 2x2 or x = 5

 3 1  y2     3 2 3   2 0 

2 3 2

y ] 2

3

(5) 2  3

 5 10  25 5  3    3  sq. units. 3 3 2   20. Option (4) is correct. 

Area bounded region = 2ar (OBC) + ar (CBDE)  2

2 0

xdx   2

3 2

xdx 3

 3  3  x2   x2   2      3   3   2 0  2  2 3 3 3 2 2  2  (2) 2   (3) 2  (2) 2  3  3    2 4   2 2  3 3  2 2  3 3 4 2 2  2 2    3 3 3 3 3 2 2 2 2 3 3

Explanation: Solving y = 3 x and x2 + y2 = 4 We get x2 + 3x2 = 4 ⇒ x2 = 1 ⇒x=1

2 104 Oswaal CUET (UG) Chapterwise Bank  x  2 x2Question 

1

2

0

1

Required Area  3  xdx  

A    dx 4   4 1 2 1  x2 x3     2x   4  2 3  1 1 1 8 1  2  4      2   3 4 3  2 1 10 7      4  3 6 9  sq. units 8 23. Option (1) is correct.

22  x2 dx 2



3  2 1  x 2  x  x  2  x2  2sin 1    2  0  2  2  1



3   3   2    2  2  2 2 6 

2 sq. units 3 21. Option (3) is correct. 

 Explanation: We have y = cos x and y = sin x, where 0  x  2 We get cos x = sin x   x  4 From the figure, area of the shaded region,

MATHEMATICS/APP. MATH.

Explanation: We have y  16  x2 y 2  16  x2 , y  0 y 2  x2  16, y  0 Graph of above function is semi-circle lying above the X-axis as shown in the adjacent figure. From the figure, area of the shaded region,

0

 4

A    cos x  sin x  dx 0 

 [sin x  cos x]04      sin  cos  sin 0  cos 0  4 4   1 1    1  2  1 sq. units 2 2 22. Option (4) is Correct. Explanation:



A=∫ =∫



−4 4 −4

( 16 − x ) dx ( 4 − x ) dx 2

2

2

4

x 42 x =  42 − x2 + sin −1  2 2 4   −4 4 2 −1 4  2 =  4 − 4 + 8sin 4  2 −1 −1 = 0 + 8sin 1 − 0 − 8sin ( −1)

x2  x  2 x x20  x  2   x  1  0 x  1, 2 2

 4  4  8 sq. units

1 and for x = 2, y = 1 4 1  Points of intersection are  1,  and (2, 1). 4  Graphs of parabola x2 = 4y and x = 4y − 2 are shown in figure: For x  1, y 

x2

0 2  x  2 x2  A    dx 4   4 1 2 2 1x x3     2x   4  2 3 

1



4

1 8 1 1 24  2  4  3  2 3

24. Option (2) is Correct. Explanation: We have y = 0, y = x and the circle x2 + y2 = 32 in the first quadrant. Solving y = x with the circle x2  x2  32 x2  16 x4 When x = 4, y = 4 for the point of intersection of the circle with the x X-axis. Put y = 0 x2  0  32 x  4 2 So, the circle intersects the X-axis at 4 2 , 0 .





105

APPLICATION OF INTEGRALS

– 0 0

From the figure, area of the shaded region, 4

x  A   x   dx 2 

 0

From the above figure, area of the shaded region, 4

A   xdx   0

4 2 4

4

(4 2 )  x dx 2

2

2 4

4 2

x x (4 2 )2 1 x       ( 4 2 )2  x 2  sin  2 4 2  4  2 0  2 4 1  16       0  16sin 11  (4 2 )2  16 sin 1  2 2 2      8  8  8  4  4 sq. units 25. Option (1) is correct. Explanation: We have y  cos x, x  0, x  

2 1 x2    x3/ 2    2 2   3 0 2 16 0   (4)3/ 2  3 4 16  4 3 4  sq.units 3 27. Option (4) is correct.

 Explanation: We have y  sin x , 0  x  . 2 2 sin 1

0 –1

From the figure, area of the shaded region, From the figure, area of the shaded region,  2 cos x dx  0

A



 cos x dx 2

 2



 sin x 0   sin x   2

    sin  sin 0     2  1  0   0  1 11  2 sq. units

  sin   sin 2   

A  

/2 0

sin xdx

 [ cos x]0 / 2

      cos  cos 0  2    0 1  1sq.unit 28. Option (1) is correct. Explanation: We have

26. Option (1) is correct.

axes as coordinate axes.

x2 5

2



y2 42

 1, which is ellipse with its

Y

Explanation: When y2 = x and 2y = x Solving we get y2 = 2y  y  0, 2 and when= y 2= ,x 4 So, points of intersection are (0, 0) and (4, 2).

X

X

2

Graphs of parabola y = x and 2y = x are as shown in the following figure: Y

106 Oswaal CUET (UG) Chapterwise Question Bank y2 42

1

From the figure, area of the shaded region,

x2

3

52

 x2  y  16 1    25   4 2  5  x2 5 From the figure, area of the shaded region,

A

3

 x2     x  2  2 4 9     3   2 2 2   7  sq.units 2

5

5

4

52  x 2 dx

0



16  x 2 52 x 5  x 2  sin 1   2 5  5  2



 16  5 1 0  sinn 1  0  0  5  2 

5

0

 x  1 dx 2

2

A4

MATHEMATICS/APP. MATH.

[B] ASSERTION REASON QUESTIONS 1. Option (1) is correct. Explanation: Assertion (A) and Reason (R) both are correct., Reason (R) is the correct. explanation of Assertion (A).

2

2. Option (1) is correct. Explanation: Given, f (x) = 4x3

16 25    5 2 2  20  sq.units 

\ Required area, A 

29. Option (3) is correct. Explanation: We have, x 2  y 2  1, which is a circle having



centre at (0, 0) and radius ‘1’ unit.  y 2  1  x2



2



0

1

1

4 x3 dx 4 x3 dx 



2

4 x3 dx

0

0

2

  x 4   x 4   4    4  4   4    1  0

y  1  x2



 





 1)  (16  17 sq. unnits 3. Option (1) is correct. Explanation:

1

–1

4

A  4  4 2  x 2 dx

–1

0

From the figure, area of the shaded region,

4

16 x x  4  16  x 2  sin 1  2 4 2  0    4 0  8   2   16  sq . units

1

A  4  12  x 2 dx 0

1

x 12 x  4  12  x 2  sin 1  2 1 0 2 2   1   4 0    0  0  2 2    sq.units

4. Option (3) is correct. Explanation: From figure, Area of shaded region,

30. Option (1) is correct. Explanation:

0



 4 4  4 4    0    1 |  |  2    0    

1

A



4

1

 

4

y dx 

 x  5 dx 1 4

 x2     5x  2 1

  16  1    20     5    2  2

4

107

 x2  OF INTEGRALS APPLICATION    5x  2 1

8. Option (1) is correct. Explanation: Assertion (A) and Reason (R) both are correct., Reason (R) is the correct. explanation of Assertion (A).

 16  1     20     5   2  2  56 11   2 2 45  sq.uniits 2 5. Option (4) is correct.

9. Option (3) is correct. Explanation:

Y

Explanation:

X 2

Required area 

 2



2

The points of intersection of the parabola y = x 2 and the line y = x are  0, 0  and 1,1 .

xdy

0



y2 dy 16

Required Area 

0

 

1  y3     16  3 



2



04

2

yline dx

1

2

 x dx   xdx 2

1

2

10 Option (2) is correct. Explanation: Assertion is true. Y

\ Required area 1 4

1



 x3   x 2  1 3 11        3 2   0  1 3 2 6

0

Explanation:

16  x 2 dx 

0

1

1 8 1   sq. units 16 3 6 6. Option (1) is correct.

43

parabola dx 

0





1

y

4

 12  3x  dx

x = g(y)

0

3 x x 1 3x 2      16  x 2  8sin 1   12 x   4  4  2    4  2 0

d

 3    8   6 2 4 

c

4

A

X O

  3  6  sq. units

Reason is also true, but not the explanation of Assertion.

7. Option (1) is correct. Explanation: Required area

1. Option (3) is correct. Explanation:

 

41

2

41

1

1

22

 3  x  5 dx    4  2 x  dx   4

 

[C] COMPETENCY BASED QUESTIONS

2

 3x  6  dx 4

 (2  x)2  1  ( x  5) 2  3  ( x  2) 2     2     3  2   2 1 2  2  2 1

X 10

3  81 36         0  1   4 6 6 4    

15 7  1  3  sq. units. 2 2

Y

(50,–10)

100

x2 = –4ay Standard form of this parabola is x2 = -4ay At point (50, -10), we get

∫ 250 dx = 250  3 

−10

−10

1 1 × (10)3 − ( −10)3   250 3  108 Oswaal CUET (UG) 1Chapterwise Question Bank MATHEMATICS/APP. MATH. = [1000 + 1000] 750 2 (50) = -4a (-10) 2000 = 125 750 a 2 8 = Therefore required equation is 3 125 6. Option (3) is correct. 2 x  4 y 2 Explanation: =

x 2  250 y 2. Option (1) is correct. Explanation: 50

50

1  x3  x2 dx    250 250  3  50 50



1 1  (50)3  (50)3   250 3  1  125000  125000 750  1000  3 3. Option (1) is correct. 

Explanation: f  x   x2 f   x   x2 \ f (x) is even function. 4. Option (3) is correct. Explanation: x = 250 y 1 2 y= x 250 at y = 0, x = 0 at y = 10, x = 50, −50 \ Area formed by curve 50



50

1 2 x dx 250 50

1 1   x3  250 3   0 1  2, 50, 000 750  1000  sq. units 3 5. Option (4) is correct. 

Explanation: 10

x

2

x 2  y 2  (4 2 )2  x 2  y 2  32 7. Option (4) is correct. Explanation: x 2  y 2  32 ...(i) yx ...(ii) Solving (i) and (ii),  x 2  y 2  32  x 2  16  x4  yx4  B  (4, 4) 8. Option (1) is correct. Explanation:



ar  OBM   xdx 0

4

 x2     2  0  8 sq. units 9. Option (4) is correct. Explanation: ar  BAMB  

4 2



32  x 2 dx

4

4 2

x  x   32  x 2  16sin 1  2 4 2 4  4 1    . 0  16 sin 1 1  32  16 2  16 sin 1  2 2    4   8  sq . units. 10. Option (3) is correct.

1 x

10 3

∫ 250 dx = 250  3 

−10

r4 2 Equation of circle is

4

2



Centre   0, 0 

−10

1 1 = × (10)3 − ( −10)3   250 3  1 = [1000 + 1000] 750 2000 = 750

Explanation: We have, 2(x)2 – 3xy – 2(y)2 = 0 2(x)2 – 4xy + xy – 2(y)2 = 0 2x(x – 2y) + y(x – 2y) = 0  

(2x + y) (x – 2y) = 0 (2x + y) = 0 or (x – 2y) = 0

109

APPLICATION OF INTEGRALS

y = - 2x or y = x/2

We know that, the circle with radius 2 is bigger than the circle x with radius 1. Now, we have two lines, y = - 2x and y = . 2 The slope of the line y = - 2x is - 2 and passes through the origin. x 1 Similarly, the slope of the line y = is . 2 2 Thus, product of slopes of lines  2 

1  1 2

So, the lines are perpendicular. Area lying between the circles = π (22 ) -π (12 ) = 3π sq. units So the required area (between the circles and both lines for y>0) Area lying between thecircles = 4 3  sq. units 4

Course of Action

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

9

DIFFERENTIAL EQUATIONS

  Revision Notes

 Fundamentals  An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation, e.g., (i)

x

dx + y = 0, dy

(ii) 2

d2y

+ y3 = 0 dx 2  The order of the highest order derivative of dependent variable with respect to the independent variable involved in the differential equation is called the order of the differential equation, e.g., dx d 2 y dx  y  c, 2  yk dy dy dx involve derivatives whose highest orders are 1 and 2 respectively.  When a differential equation is a polynomial equation in derivatives, the highest power (positive integral index), of the highest order derivative is known as the degree of the differential equation, e.g., (i)

2

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Overview of differential equations

dy  dy  + y = c , the highest order derivative is   + dx  dx 

dy and its positive integral power is 2. dx ∴ Its degree is two. (ii)

d3y dx

3

derivative is

+

d2y dx 2

d3y

+

dx + 4 = 0 . Here, the highest order dy

. Its positive integral power is 1. dx3 ∴ Its degree is one.  Formation of Differential Equations  For any given differential equation, the solution is of the form f (x, y, c1, c2, …….,cn) = 0 where x and y are the variables and c1, c2 ……. cn are the arbitrary constants. To obtain the differential equation from this equation we follow the following steps: (i) Differentiate the given function with respect to the independent variable present in the equation. (ii) Keep differentiating times in such a way that (n+1) equations are obtained. (iii) Using the (n+1) equations obtained, eliminate the constants (c1, c2 … …. cn).  Solution of a Differential Equation  A solution of a differential equation is a relation between the variables (independent and dependent), which is free of

derivatives of any order, and which satisfies the differential equation identically. A solution which contains as many arbitrary constants as the order of the differential equation is called a general solution. A particular solution of a differential equation is a solution obtained from the general solution by assigning specific values to the arbitrary constants.  Methods of solving first order, first degree differential equations Scan to know (i)  Variables Separable Method: “When more about this topic

the equation may be expressed as dy = h ( y ) g ( x ), then we can write it as dx dy = g ( x ) dx. h( y)

Variable Separable Method

Integrating, we get the solution as dy � = � g ( x ) dx + C h( y)

∫

∫

Example 1: Solve the following differential equation: dy = x3 cosec y, given that y(0) = 0. dx x4 x3 (1) cos y  1  (2) sin y  1  4 3 x3 x4 (4) cos y  1  3 4 Sol. Option (1) is correct. Explanation: dy  x3cosec y ; y  0   0 dx dy  x3dx cosec y (3) sin y  1 

   sin ydy  x dx 3

 cos y 

x4  c ∵ y  0, when x  0  4

1  c 4

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Homogeneous

x4 Differential Equations 4  Homogeneous Differential Equations  Homogenous functions are defined as functions in which the total power of all the terms of the function is constant. For example, f (x, y) = (x2 + y2 – xy) is a homogeneous function of degree 2. The general form of the homogeneous differential equation is f (x, y).dy + g(x, y).dx = 0, where f (x, y) and h(x, y) is a homogenous function. cos y  1 









dy dv vx dx dx

3.dx

  2 x. e 

3.dx

dx  c  ye3 x  2  xe3 x dx  c.

First Level

Second Level



Trace the Mind Map



( (

Formation of Differential Equations

Definition

ial Equa ent r t e

Linear Differential Equations

Homogeneous Differential Equations

Variable Separation Method

Third Level

To form a differential equation from a given function, we differentiate the function successively as many times as the no. of arbitrary constants in the given function, and then eliminate the arbitrary constants. eg: Let the function be y=ax+b, then we have to differentiate it two times, since there are 2 arbitrary constants a and b. ∴ y ' = a ⇒ y " = 0. Thus y " = 0 is the required differential equation.

It is used to solve such an equation in which variables can be separated completely. dx dy eg : ydx = xdy can be solved as  ; x y Integrating both sides x log x  log y  log c   c  x  cy, y is the solution.

solution ye 

The differential equation of the form

dy  Py  Q, dx where P, Q are constants or functions of 'x' only is called a first order linear differential equation. Its solution is given as ye  P.dx  Q. e  P.dx dx  c . eg : dy  3 y  2 x has  dx

To solve this, we substitute y  vx, and

eg : x 2  xy dy  x 2  y 2 dx



Solution of a Differential Equation

Degree of a Differential Equation

Order of a Differential Equation

It is the order of the highest order derivative occurring in the differential equation dy  ex eg: The order of dx is one and order of d2y  x  0 is two. dx 2

y = mx, m = constant, then, y '  m dy dy y  y ' x  y  x  x – y  0. dx dx

The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves. eg: Let the family of curves be

x x x x Since y '  e and y " e  y "– y '  e – e  0.

eg: y  e x  1 is a solution of y” –y' = 0 .

A function which satisfies the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution.

( (

only) of the highest order derivative. 3 d2y dy   0 is three eg: The degree of dx dx 2 Order and degree (if defined) of a differential equation are always positive integers.

defined as the highest power (positive integer

polynomial equation in its derivatives, and is

It is defined if the differential equation is a

An equation involving derivatives of the dependent variable with respect to independent variable (variables) is called a differential equation. If there is only one independent variable, then we call it as an ordinary differential d2y dy equation. eg: 2 2   0. dx dx

Dif f

A differential equation which can be expressed dx dy in the form  f x, y  or  g x, y , dy dx where, f (x, y) and g (x , y) are homogeneous functions of degree zero is called a homogenous differential equation

DIFFERENTIAL EQUATIONS

111

s ion

112 Oswaal CUET (UG) Chapterwise Question Bank We can solve a homogeneous differential equation of the form dx/dy = f (x, y) where, f (x, y) is a homogeneous function, by simply replacing x/y to v or putting y = vx. Then after solving the differential equation, we put back the value of v to get the final solution. Example 2: Find the general solution of the differential

(

dx (2)  When the equation is of the form + Px = Q , where P dy and Q are the functions of y. ∴ Solution is I.F.  e 

)

equation: x3 + y 3 dy = x 2 ydx (1) (3)

x3 x3 y

3

(2)

 log x  c

3 y3

(4)

 log y  c

xe  x2 3 y3 x3 3 y3

 log y  c  log y  c

Sol. Option (4) is correct. Explanation: Given differential equation is

x

3



...(i)

v+ y

dv v3 + 1 = 2 dy v

      x  (3) cosec x  2  x  tan       2 tan   8    4 2  2  

Explanation: The differential equation of a linear differential equation. I.F. = e∫ cot xdx = elog sin x = sin x The general solution is given by

dy



Linear Differential Equations

= log y + c

(iii)  Linear Differential Equation: dy  Py  Q, (1)  The linear differential equation is of the form, dx where P and Q are the functions of x. This is a first order, first degree differential equation. To solve the equation, we find the integrating factor I.F. = e∫ Pdx Then, the solution is

ye 

Pdx

    1 dx  y sin x  2 1      1  cos  2  x          1 dx  y sin x  2 1  x   2 2 cos   4 2     

Scan to know more about this topic

v3  log y  c 3 x Putting v = , we get y 3 y3

sin x

 1  sin x dx sin x  1  1  y sin x  2  1  sin x dx 1    y sin x  2 1    1  sin x dx y sin x  2

v dv   y

x3

dy  C.

  π   π x π (1) cos x  2  x + tan  −  −  + 2 tan   8    4 2  2  

dv v3 + 1 = 2 −v dy v dv 1 y = dy v 2 dy v 2 dv = y (Using variable separation method) Integrating both sides, we get

Pdy

Example 3: The particular solution of the differential equation 2 dy   y cot x  , given that y = 0 when x  , is 4 1  sin x dx

y

2



Sol. Option (4) is correct.

dv (vy )3 + y 3 v+ y = dy (vy )2 y dv v3 y 3 + y 3 = dy v2 y3

 Q  e

      x  (4) cosec x  2  x  tan       2 tan   4 2 2 8       

From eq. (i), we have

v+ y

Pdy

Pdy

      x  (2) cosec x  2  x  tan       2 tan   4 2 2 8       

 y 3 dy  x 2 ydx

dx x3  y 3  dy x2 y Put x  vy dx dv v y ⇒ dy dy

MATHEMATICS/APP. MATH.



 Pdx dx  C. Qe





 1   x   y sin x  2 1  sec 2    dx 2  4 2      x   y sin x   2  x   tan     C  c  4 2  





 

Given that y = 0, when x =

π 4

    0  2   tan   C  C     2 tan  8 8 4 2

Hence, the particular solution is  

   x 

y  cosec x  2  x  tan       2 tan   4 2 2 8       

113

DIFFERENTIAL EQUATIONS

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 2

d y dy   1. The degree of the differential equation 1 +  =  2  dx    dx  is [CUET 2023] (1) 1. (2) 3. (3) 2. (4) 1.

(1) x2 – y2 = C.

(2)

(3) xy = C.  3. Particular solution



2

dy x + = 0 , represents the dx y

2. The differential equation family of curves is

x = C. y 2

2

(4) x + y = C. [CUET 2023, 2021] of the differential equation

(1) e x  e  y  2. (2) e  x  e y  2.

(

)(

)

(1) tan 1 y  x 

x3  c. 3

[CUET 2023]

(2) tan 1 y  x 

x3  c. 3

x x (3) tan y  x  (4) tan 1 y  x 2   c.  c. 3 3 5. If a and b are order and degree of differential equation 3

1

3

2

y′′ + ( y′ ) + 2 y = 0, then value of 2a + 6b, is  [CUET 2022] (1) 3. (2) 4. (3) 6. (4) 10. 2

6. The solution of the differential equation xdy – ydx = 0 represent family of [CUET 2022] (1) Circles passing through origin. (2) Straight line parsing through (–1, 6). (3) Straight line passing through the origin. (4) Circle whose centre is at the origin. x y

x y

  7. For differential equation ye dx =  xe + y 2  dy , y(0) = 1,     the value of x(e) is equal to [CUET 2022] (1) 0.

(2) 1.

(3) 2.

(4) e.

8. The order of the differential equation whose general solution is y  e x  a cos x  b sin x , where a and b are [CUET 2022]

arbitrary constants is (1) 1.

(2) 3.

(3) 2.

(4) 6. dx  Px  Q, 9. The solution of the differential equation dy where P and Q are constants or functions of y, is given by 

[CUET 2021]

(1) xe 

Pdx

 Qe 



Pdx

dx  c. (2) ye 

Pdy

 Qe 



Pdx



dx  c. (4) xe 

 Qe 

Pdy

Pdy

dy  c.

Pdy



(1) y = vx

(2) v = xy

(3) x = vy

dy  c.

(4) x = v

11. Which of the following is a second-order differential equation?

( y′)2 + x = y 2 (2)

(3) y   y   y  0

(4) y′ = y 2

2

12. The

yy  y  sin x

integrating factor of dy 1 − x2 − xy = 1 is dx (1) –x. (2)

(

differential

)

equation

x . 1 + x2 1 (3) 1 − x 2 . (4) log 1  x 2 . 2 13. y = ae mx + be − mx satisfies which of the following differential equation? dy dy + my = 0 (2) (1) − my = 0 dx dx



(3) e x  e y  2. (4) e  x  e  y  2. dy = 1 + x 2 1 + y 2 is dx

 Qe 

10. A homogeneous differential equation of the form x dx  h   can be solved by making the substitution. dy  y

(1)

 dy  log   = x + y given that when= x 0= , y 0 is  dx  [CUET 2023]

4. Solution of

(3) ye 

Pdx

4

(3)

d2y

(1)

x e

(4)

− m2 y = 0

dx 14. The integrating dy 1+ y +y= is dx x 2

(2)

. x

d2y

+ m2 y = 0 dx 2 of differential equation

factor

ex . x



(3) xe x .

(4) e x .

15. The numbers of arbitrary constants in the particular solution of a differential equation of third order are (1) 3. (2) 2. (3) 1. (4) 0. dy = ex+ y 16. The general solution of the differential equation dx is (1) ex + e–y = C. (3) e–x + ey = C.

(2) ex + ey = C. (4) e–x + e–y = C. dy

2 xy

1

17. The solution of differential equation + = dx 1 + x 2 is 1 + x2





(1) y 1  x 2  C  tan 1 x. (2)





y 1  x2



(

)

2

 C  tan 1 x.



(3) y log 1  x 2  C  tan 1 x. (4) y 1  x 2  C  sin 1 x. 18. Which of the following is a homogeneous differential equation? (1) ( 4 x + 6 y + 5 ) dy − ( 3 y + 2 x + 4 ) dx = 0 (2) (3)

( xy ) dx − ( x3 + y3 ) dy = 0

(x

3

)

+ 2 y 2 dx + 2 xydy = 0

(

)

(4) y 2 dx + x 2 − xy − y 2 dy = 0

114 Oswaal CUET (UG) Chapterwise Question Bank 19. The solution of differential equation (1) y  tan 1x.

dy 1  y 2  is dx 1  x 2

(2) x  tan 1 y.

(3) y  x  k 1  xy  .

(4) tan  xy   k .

(1) family of hyperbolas.

dy + x = C represents dx (2) family of parabolas.

(3) family of ellipses.

(4) family of circles.

20. The differential equation y

MATHEMATICS/APP. MATH.

29. Family y = Ax + A3 of curves is represented by the differential equation of degree (1) 1. (2) 2. (3) 3. (4) 4. 30. The integrating factor of (1) x.

(2) 1 . x

xdy  y  x 4  3 x is dx (3) log x.

(4) –x.

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, A statement of 2 Assertion (A) is followed by a statement of Reason (R). Mark  d 2 y   dy 2  dy  21. The degree of the differential equation  sin  x   the correct choice as:      dx 2   dx   dx    2 2 (1) Both A and R are true and R is the correct explanation of A. 2  d y   dy   dy   2      x sin   is (2) B  oth A and R are true but R is not the correct explanation  dx   dx   dx    of A. (3) A is true but R is false. (1) 1. (2) 2. (3) 3. (4) not defined. (4) A is false but R is true. 22. The degree of the differential equation 1. Assertion (A): The order of differential equation 3/ 2   dy 2  d2y 3/ 4 1 +    = is   dy 2  dy d3y   2   dx   dx 1     cos     is 3.   dx  dx3  dx   3 (1) 4. (2) . Reason (R): The degree of differential equation 2 (3) not defined.

(4) 2.

3/ 4

23. The order and degree of the differential equation 1/ 4 d 2 y  dy  + + x1/ 5 = 0 respectively, are   dx 2  dx  (1) 2 and 4. (2) 2 and 2.

(3) 2 and 3.

d2y dx 2

dy  0. dx

(2)

d2y dx 2

2

dy  2 y  0. dx

dy d2y  2  0 . (4)  2 y  0. y dx dx 2 dx 2 25. The differential equation for y = A cos α x + B sin α x, where A and B are arbitrary constants, is (3)

(1) (3)

d2y

2

d2y dx 2

d2y dx 2

2

  2 y  0. (2)

  y  0. (4)

d2y dx 2 d2y dx 2

  2 y  0.   y  0.

dy 3e 2 x  3e 4 x is......  x dx e  e x (2) y = e -3x + C

26. The solution of differential equation

3

d3y

(4) 3 and 3.

24. If y  e  x  A cos x  B sin x , then y is a solution of (1)

  dy 2 d3y  dy   1     cos     is not defined.   dx  dx3  dx   2. Assertion (A): The order of differential equation  dy  + 4 ×   + y 2 = 0 is 3. dx  dx  Reason (R): The sum of order and degree of differential 3



equation

d3y

3

 dy  + 4 ×   + y 2 is 5. dx  dx  3

3. Assertion (A): The solution of differential equation dy y = with initial condition x = 1 and y = 1 is x = y . dx x Reason (R): Separation of variable method can be used to solve the differential equation. 4. Assertion (A): The solution of the equation 3 yy′ + 4 x = 0 represents family of ellipses.

Reason (R): Equation of ellipse is

x2 a2

+

y2 b2

= 1.

5. Assertion (A): The solution of the differential equation dQ + Q = 1 with Q = 0 at t = 0 is Q ( t ) = 1 − e −t . -x x dt (3) y = e + C (4) y = e + C Reason (R): The given differential equation is a linear 27. The integrating factor of differential equation differential equation. dy cos x  y sin x  1 is dy  y dx 6. Assertion (A): = f   is homogeneous differential dx x (1) cos x. (2) tan x. (3) sec x. (4) sin x. equation 28. Solution of the differential equation tan y sec 2 xdx  tan x sec 2 ydy Reason 0 (R): To solve the homogeneous differential (1) y = e3x + C

an y sec 2 xdx  tan x sec 2 ydy  0 is:

(1) tan x  tan y  k . (3)

tan x = k . tan y



(2) tan x  tan y  k . (4) tan x  tan y  k .

dy  y = f   , we put y = νx. dx x 7. Assertion (A): The particular solution of the differential equation equation of the form

115

DIFFERENTIAL EQUATIONS



y xe x



dy  y  y − ysin   + x sin   = 0, for = x 1,= y 0 is x dx x  

  y  y  sin    cos    e x  x    

y x

From eq. (i), we get vx

 log x 2  1

dv v2 v2  v2  v  v  dx v  1 v 1 v 1 dx v dv  or dv  x dx v  1 v x



Reason (R): Given differential equation is homogeneous. dy dv ∴ Put y = νx ⇒   =v+x dx dx y  y y sin    xe x dy x  dx  y x sin   x

x





dv v sin v  ev  dx sin v

or

vx



or

dv ev vx v dx sin v



or





or





I1   sin ve

or

I1  

v



 cos ve

v

Putting (i),  sin v  cos v  e

y  x e

  y y or sin    cos    x  x    



...(i)

Hence, solution is



  y  y  sin    cos    e x  x    

y x

 log x  2C1

 log x 2  C2

Reason (R): Given differential equation is



dy y2  = dx xy − x 2



Let F  x, y  



Now, on replacing x by λx and y by λy, we get



F ( λ x, λ y ) =



Thus, the given differential equation is a homogeneous differential equation. Now, to solve it, put y = νx dy dv ⇒ � =v+x dx dx



...(i) y2 xy  x 2

λ

2

( xy − x ) 2

v  log v  log x  C

y2 xy − x

2



  m ∵log    log m  log n  n     y   log y  C x

9. Assertion (A): The solution of

dy − y = 1, y ( 0 ) = 1 is dx

[C] COMPETENCY BASED QUESTIONS

 log x 2  1

= λ0

or

dy Reason (R): The solution of − y = 1, y ( 0 ) = 1 is dx y = 2ex – 1. dy y + 1 = 10. Assertion (A): The number of solutions of , dx x − 1 when y (1) = 2 is none. dy y + 1 = Reason (R): The number of solutions of , dx x − 1 when y (1) = 2 is one.

dy y2 y is  log y  C. = dx xy − x 2 x

λ 2 y2





2

8. Assertion (A): The solution of the differential equation



dv

given by xy  e x .

For = x 1,= y 0 or C2 = 1

1

 1  v  dv   x

v

 sin ve dv

1  sin v  cos v  ev 2 v





 y y y  log  log x  C  put v   x x x  y or − log y + log x = log x + C x

 x

or



dv v v −1 dx  or dv = dx v 1 v x On integrating both sides, we get or x

or

dv ev  dx sin v dx sin ve v dv   or I1   log x  C1  x I1   sin v  e v  cos ve v dv  



dv v2 x2 v2  2  dx vx  x 2 v  1

= λ 0 F ( x, y )

I. Read the following text and answer the following questions on the basis of the same. A veterinary doctor is examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11:30 pm which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation. It was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton law of cooling which is governed dT by the differential equation: ∝ (T − 70 ) , where 70° F is the dt room temperature and T is the temperature of the object at time t. Substituting the two different observations of T and t made, in dT the solution of the differential equation = k (T − 70 ) where dt k is a constant of proportion, time of death is calculated. 1. What will be the degree of the above given differential equation? (1) 2 (2) 0 (3) 1 (4) 3

116 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH.

2. If the temperature was measured 2 hours after 11:30 pm, what will be the change in time of death? (1) No change (2) Death time increased (3) Death time decreased (4) Death time always constant dT 3. The solution of the differential equation = k (T − 70 ) dt is given by

dy = k ( 50 − y ) where x denotes the number of weeks and y dx the number of children who have been given the drops. 6. State the order of the above given differential equation. (1) 2 (2) 0 (3) 1 (4) Can’t define

(1) log T  70  kt  C.

(1) log 50  y  kx  C.

(2) log T  70  log kt  C.

(3) log 50  y  log kx  C. (4) 50  y  kx  C.

(3) T  70  kt  C. (4) T  70  ktC.

8. The value of C in the particular solution given that y(0) = 0 and k = 0.049 is 1 (1) log 50. (2) log . (3) 50. (4) –50. 50 9. Which of the following solutions may be used to find the number of children who have been given the polio drops?

4. If t = 0 when T is 72 , then the value of C is (1) –2.

(2) 0.

(3) 2.

(4) log 2.

5. Write the order of the above given differential equation. (1) 1

(2) 0

(3) 2

(4) 3

II. R  ead the following text and answer the following questions on the basis of the same. Polio drops are delivered to 50 K children in a district. The rate at which polio drops are given is directly proportional to the number of children who have not been administered the drops. By the end of 2nd week half the children have been given the polio drops. Given that the drops by the end of 3rd week can be estimated using the solution to the differential equation

dy 7. The solution of the differential equation = k ( 50 − y ) is dx given by (2) log 50  y  kx  C.

(1) y = 50 − e kx (2) y = 50 − e − kx

(

)

(

(3) y = 50 1 − e − kx

)

(4) y = 50 e − kx − 1

10. What will be the degree of the above given differential equation? (1) 2 (2) 0 (3) 1 (4) 3

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (4)

3. (1)

4. (1)

5. (4)

6. (3)

7. (4)

8. (3)

9. (4)

10. (2)

11. (2)

12. (3)

13. (3)

14. (2)

15. (4)

16. (1)

17. (1)

18. (4)

19. (3)

20. (4)

21. (4)

22. (4)

23. (1)

24. (3)

25. (2)

26. (3)

27. (3)

28. (4)

29. (1)

30. (3)

8. (1)

9. (4)

10. (4)

8. (2)

9. (3)

10. (3)

[B] ASSERTION REASON QUESTIONS 1. (2)

2. (2)

3. (1)

4. (1)

5. (1)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (3)

2. (1)

3. (1)

4. (4)

5. (1)

6. (3)

7. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

1. Option (3) is correct. Explanation: We know that degree of differential equation is the power of the highest derivative in a differential equation. Therefore, degree of given differential equation is 2. 2. Option (4) is correct. dy x Explanation: + =0 dx y dy x =− dx y ydy = − xdx − ydy = xdx Integrating both sides





 ydy  xdx 2

y C x2   2 2 2

C x2 y 2   2 2 2

x2  y 2  C 3. Option (1) is correct. Explanation:

\

 dy  log    x  y  dx   dy  x y  dx   e   dy  e x dx ey



e y



x

x

dx xe y  y 2  x dy ye y

 dy  log    x  y  dx  DIFFERENTIAL  dy  EQUATIONS x y  e dx   dy  e x dx ey



x



or,

e y  e x  c1 1

Put

e x  e  y  c  c  c1 

⇒

e x  e y  c

 when 0 x 0= 0 and y = 0 e e c 0 e0  1 e 1   cc 1  c1   c2 c2 Put in eq.  i  Now, from Put in eq .  ieq.  (i), the required particular solution is e x  e y  2 e x  e y  2 4. Option (1) is correct. Explanation: dy = 1 + x2 1 + y 2 dx

...(i)

v

ev  c  y x

y  ey c y ( 0)  1

\

2

1  e0  c  c  0 y  ex/ y x log y  log e  y log y x y or, log e For value at x(e), put y = e e log e e ( x)(e)  log e

2

2

Integrating both sides,

 1  y   1  x  dx dy



2

2

x3 C 3   1 dy  tan 1 y  ∵ 2 1 y   5. Option (4) is correct. tan 1 y  x 



 y  e x  a cos x  b sin x  

2

Order of differential equation is 2. Thus, a = 2. Degree of differential equation is 1 . Thus, b = 1 Therefore, 2a  6b  2  2  6  1  4  6  10. 6. Option (3) is correct. Explanation: xdy  ydx  0

log y  log  xc  y  cx which represent the straight line passing through origin. 7. Option (4) is correct. Explanation: We have, x  x  ye y dx   xe y  y 2  dy   x y

dx xe  y 2  x dy ye y x

dx xe y y  x  x dy ye y e y y dy x   dx y e xy x  vy

[Taking log both sides]

8. Option (3) is correct. Explanation: General solution is

Explanation: Given differential equation is: y′′ + ( y′ ) + 2 y = 0

xdy  ydx dy dx  y x log y  log x  log c

117

e dv  dy

( )( ) dy = (1 + x ) (1 + y ) dx dy = (1 + x ) dx 1+ y 2

\

dx xe y y   dy ye xy e xy dy x y   x dx y e y x  vy dx dv v y dy dy dv y v v v y dy e dv y vv v y dy e y dv  y dy ev

...(i)

y  e x  a cos x  b sin x   e x   a sin x  b cos x  or, y  y  ae x sin x  be x cos x or, y  y  e x   a sin x  b cos x  

...(ii)

Differentiating again w.r.t. x d2y

dy  e x ( a sin x  b cos x)  e x ( a cos x  b sin x) dx dy   e x ( a sinn x  b cos x)  y  ...(iii) dx Subtracting eq. (ii) from eq. (iii), we get dx 2



d2y dx 2

⇒

d2y dx 2

−2

−

dy dy = − 2y dx dx

dy + 2y = 0 dx

Thus, order of differential equation is 2 . 9. Option (4) is correct. Explanation: Solutions of

dx  Px  Q is given by dy xe 

Pdy

 Qe 



Pdy

dy  c

dy  1  x  1  y  dx  x  x (1  x) 1 ,Q  P x x

118 Oswaal CUET (UG) Chapterwise Question Bank x 1 MATHEMATICS/APP. MATH. 1 x Pdx  dx  e  x dx e x IF  e  10. Option (2) is correct. Explanation:

e

For solving the homogeneous equation of the form:

 e x  log x



x dx  h   , we need to make the substitution as x = vy. dy  y 11. Option (2) is correct. Explanation : The second-order differential equation is yy  y  sin x. 12. Option (3) is correct. Explanation : dy Given that, 1 − x2 − xy = 1 dx

(

)

1 dy x − y= dx 1 − x 2 1 − x2 which is a linear differential equation. ⇒

⇒ 

IF

x − dx − 1 x2 =e

∫

Put 1 − x 2 = t ⇒ −2 xdx = dt ⇒ xdx = − 1 dt



1

1

log t

Now, IF  e 2 t  e 2  e2 13. Option (3) is correct. Explanation :



log 1 x 2

dt 2



1  x2

On differentiating both sides w.r.t. x, we get dy = mae mx − bme − mx dx Again, differentiating both sides w.r.t.x, we get 2 dd22yy mx 22 mx mx  mx 22 mx mae ae bm ee mx  m bm  m bm 2e 22ae 2 dx dx dx 22 d 2 y dd 2yy  mx mx mx mx m22ae ae bemx m ⇒  m 22 ae be  mx be 2 dx dx dx 22 d 2 y dd 2yy  m22 y ⇒  m 22y  m y dx dx 2 dx 22 2 d y dd 2 yy  m22 y  0 m ⇒  m y 0 y0 dx22 dx 2 dx 14. Option (2) is correct. Explanation : dy 1+ y Given that, +y= dx x dy 1  y  y ⇒ dx x dy 1  y  xy  ⇒ dx x dy 1 y (1  x) ⇒   dx x x dy  1  x  1 ⇒  y  dx  x  x (1  x) 1 Here, P  ,Q  x x

d2y

IF  e 

Pdx



e

e

1

1 x  dx

 e x log x

x  1 dx x



1 log  

 ex  e  x  1  ex  x 15. Option (4) is correct. Explanation: In the particular solution of a differential equation, there are no arbitrary constants. 16. Option (1) is correct. Explanation: dy = ex+ y   dx = ex ⋅ e y   dy x+ y ⇒⇒ � y = e dx edx ⇒⇒ � e − y dy==e xe x⋅ dx ey Integrating dy both sides, we get: ⇒ � y = e x dx −ey � e dy = � e x dx ⇒ � e − y dy = e x dx e− y = e x + k ⇒ ⇒�

∫

∫

⇒ � e x + e− y = −k ⇒

Given that, y = ae mx + be − mx



1

1 x  dx





 e



x 1 dx x



⇒ � e x + e− y = C ⇒

(where, C = − k ) 17. Option (1) is correct. Explanation : dy 2 xy Given that, + = dx 1 + x 2 Here,  P =

2x 1 + x2

1

(1 + x ) 2

2

1

and Q =

(1 + x ) 2

2

which is a linear differential equation. 2x

∴ IF = e

∫ 1+ x2 dx

Put 1 + x 2 = t ⇒ 2 xdx = dt



dt

∴ IF  e t  elog t  e The general solution is

(

  1  x2

1 ) ∫(11++ xx ) 1 +1x (1 + x )

2 yy⋅⋅ 11++ xx2 ==

22

( ) ∫11++11xx dxdx++CC tan xx ++ C C yy (11++ xx ) == tan

2 ⇒ yy 11++ xx2 ==

⇒



log 1 x 2

22

+C

+C 22 22

22

−−11

18. Option (4) is correct. Explanation : Function F(x, y) is said to be the homogenous function of degree n, if F(l x, l y) = l” F(x, y) for any non-zero constant (l). Consider the equation given in alternative 4: y2dx + (x2-xy-y2) dy = 0

119

DIFFERENTIAL EQUATIONS



dy y2 y2  2  2 2 dx x  xy  y y  xy  x 2

y2 Let F( x , y )  2 y  xy  x 2  F ( x , y ) 

2

( y ) ( y )  ( x )( y )  ( x )2 2

2 y2  2 2  ( y  xy  x 2 )   y2  0  2 2   y  xy  x    0 .F( x , y ) 19. Option (3) is correct. Explanation : dy 1 + y 2 = Given that, dx 1 + x 2 ⇒ ⇒  

dy 1 + y2

=

dx 1 + x2

On integrating both sides, we get ⇒   tan −1 y = tan −1x + C ⇒  tan 1 y  tan 1 x  C  yx  ⇒  tan 1  C  1  xy  yx  tan C ⇒   1  xy ⇒ y  x  tan C 1  xy  ⇒ y  x  k 1  xy  where,

 

 k = tan C

20. Option (4) is correct. Explanation: dy Given that, y xC dx ⇒ y

dy Cx dx

 ydy   C  x  dx ⇒   On integrating both sides, we get y2 x2  Cx  K 2 2

⇒

x2 y 2   Cx  K 2 2

x2 y 2   Cx  K 2 2 which represent family of circles. 21. Option (4) is correct. ⇒

Explanation : The degree of above differential equation is not  dy  defined because when we expand sin   we get an infinite  dx  dy series in the increasing powers of . Therefore, its degree is dx not defined.

22. Option (4) is correct. Explanation : Given that,   dy 2  1 +      dx   On squaring both sides, we get

3/ 2

=

d2y dx 2

3

  dy 2   d 2 y 2 1 +    =     dx    dx 2  So, the degree of differential equation is 2 . 23. Option (1) is correct. Explanation : Given that, ⇒ ⇒

d2y

1/ 4

d2y

1/ 4

 dy  +  2 dx  dx 

 dy  +  dx 2  dx 

= − x1/ 5 = − x1/ 5

 d2y  = −  x1/ 5 + 2   dx   On squaring both sides, we get 1/ 4

 dy  ⇒      dx 

 d2y  =  x1/ 5 + 2   dx   Again, on squaring both sides, we have 1/ 2

 dy   dx   

dy  1/ 5 d 2 y  =x + 2  dx  dx 

2

4

Order = 2, degree = 4 24. Option (3) is correct. Explanation : x Given that, y  e  A cos x  B sin x  On differentiating both sides w.r.t., x we get dy  e  x ( A cos x  B sin x)  e  x ( A sin x  B cos x) dx dy   y  e  x ( A sin x  B cos x) dx Again, differentiating both sides w.r.t. x, we get d2y dx 2



 dy  e  x ( A cos x  B sin x)  e  x ( A sin x  B cos x) dx

 ⇒

d2y dy  dy    y    y dx dx dx 2  

 ⇒

d2y dy dy  y y 2 dx dx dx

d 2 y dy  2  2  2 y ⇒ dx dx 2 dy d y  2 2  2y  0 ⇒ dx dx 25. Option (2) is correct. Explanation : Given, y  A cos  x  B sin  x ⇒ 

dy   A sin  x   B cos  x dx

120 Oswaal CUET (UG) Chapterwise Question Bank 29. Option (1) is correct. Explanation :

Again, differentiating both sides w.r.t. x, we get d y 2

  A 2 cos  x   2 B sin  x

dx 2

Given that, y = Ax + A3

  d2y 2 B xsin  x     xAcos   A2 2 cos  2Bxsin 2 dx dx 2 d 2 y d y2   2 y ⇒    2  A cos  x  B sin  x  2 dx dx 2 2 d y d y2 ⇒ 2y  0    2 y 2 dx dx d2y

[We can differential above equation only once because it has only one arbitrary constant.] ∴ Degree =1 30. Option (2) is correct. Explanation : dy Given that, x − y = x 4 − 3x dx dy y ⇒   ⇒ − = x3 − 3 dx x 1 Here,   P = − , Q = x3 − 3 x

d2y

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



e2 x  e4 x

e

dy  3

e



3 ye

x

Pdx



e x  e x

x

e e

3x

∴ IF = e ∫

 e x

3x



Here,

Pdx

sec 2 x −sec 2 y dx = dy tanx tany

On integrating both sides, we have

Put tan x = t in LHS integral,  sec 2 xdx  dt and put tan y = u in RHS integral, ⇒ sec 2 ydy = du On substituting these values in eq. (i), we get du

∫u

log t   log u  log k

⇒ 

log  t  u   log k

 log  tan x tan y   log k ⇒  ⇒

tan x tan y  k

d3y

dx3

...(i)

3

 dy   4    y 2 dx3  dx 

⇒



d3y

 dy   4    y2  0 dx3  dx 

⇒  tan y sec xdx   tan x sec ydy

∫

=

3

d3y

2

sec 2 x sec 2 y dx = − dy tanx tany

3/ 4

3

tan xdx

2

=−

x

2    3 4 ⇒ 1 +  dy  + cos  dy   =  d y        dx   dx    dx3  Thus, order is 3, but degree is not defined as equation is not a polynomial. 2. Option (2) is correct. Explanation: We have,

IF  e   e  eln sec x ∴   IF = sec x 28. Option (4) is correct. Explanation: Given that, tan y sec 2 xdx  tan x sec 2 ydy  0

dt

1

∫ x dx = e−logx = 1

  dy 2  dy   1 +   + cos      dx   dx  

P = tan x and Q = sec x

∫t

−

1. Option (2) is correct. Explanation: Given,

27. Option (3) is correct. Explanation : dy dy Given that, cos x  y sin x  1   y tan x  sec x dx dx

∫

=e

[B] ASSERTION REASON QUESTIONS

x

C

⇒  

dy =A dx

⇒ ⇒

2y  0 dx 2 26. Option (3) is correct. Explanation : ⇒

MATHEMATICS/APP. MATH.

2  dy 3   d3y   3   16    y 2   dx   dx     Thus, order = 3 and degree = 2 Hence, sum of order and degree is 5. 3. Option (1) is correct. Explanation: We have, dy y  dx x dy dx ⇒  y x dy dx ⇒  y x ⇒ log y  log x  log C At x  1, y  1, weget C  1 ∴ yx





121

DIFFERENTIAL EQUATIONS

4. Option (1) is correct. Explanation: We have, 3 yy′ + 4 x = 0 dy ⇒ = −4 x 3y dx 3 ydy = −4 xdx ⇒

∫

⇒

∫

3 ydy = −4 xdx  x2   y2  3  = −4   + C  2   2     

Thus,

3 2 y + 2 x2 = C 2

x2 y2 + = 1, 1 2 C C 2 3     which represents a equation of family of ellipses. 5. Option (1) is correct. Explanation: We have, dQ + Q =1 dt Comparing above differential equation by or

When x = 0 and y = 1, then    log 2  0  C ⇒  C  log 2 The required solution is log 1  y   x  log 2 1+ y  ⇒ log  =x  2  1+ y ⇒ = ex 2 ⇒ 1 + y = 2e x y = 2e x − 1

⇒

10. Option (4) is correct. Explanation: Given that, ⇒ ⇒

dy y + 1 = dx x − 1

dy dx = y +1 x −1

On integrating both sides, we get log  y  1  log  x  1  log C



C ( y + 1) = ( x − 1)



x −1 y +1

dy + Py = M dx Here,= P 1,= M 1

⇒ C =

∴ I.F.  e   et Required solution is

So, the required solution is x − 1 = 0 Hence, only one solution can exist.

1dt



t t Q Q  tt   eet   eet dt dt  C C

⇒ At ∴

t t Q Q  tt   eet  C  eet  C Q tt   00,, Q  00 0  ee00  C C C C  11 00  ee0 

Thus solution is Q ( t ) = 1 − e −t. 6. Option (1) is correct. Explanation: Both A and R are true and R is the correct explanation of A 7. Option (1) is correct. Explanation: Both A and R are true and R is the correct explanation of A 8. Option (1) is correct.

When x = 1 and y = 2 , then C = 0

[C] COMPETENCY BASED QUESTIONS 1. Option (3) is correct. Explanation: The highest order derivative present in the given dy differential equation is . Thus, degree of given differential dx equation is 1. 2. Option (1) is correct. Explanation: No, the time of death would not changed. 3. Option (1) is correct. Explanation: dT  k T  70  dt  1  dT k     T 70  dt 

 log 1  y   x  C

1

log T  70  kt  C

9. Option (4) is correct Explanation: Given that, dy dy − y =−1y = 1 dx dx dy dy ⇒ = 1 +=y1 + y dx dx dy dy = dx= dx ⇒ 1 + y1 + y On integrating both sides, we get



  T  70  dT  k dt

Explanation: Both A and R are true and R is the correct explanation of A

4. Option (4) is correct. Explanation: Given, t = 0, when T = 72 Now, log T  70  kt  C or, or,

log 72  70  k  0   C

C  log 2 5. Option (1) is correct. Explanation: The highest order derivative present in the given dy differential equation is , hence its order is 1. dx

122 Oswaal CUET (UG) Chapterwise Question Bank 6. Option (3) is correct. Explanation: The highest order derivative present in the given dy differential equation is . Thus, order of given differential dx equation is 1. 7. Option (2) is correct. Explanation: dy = k ( 50 − y ) dx dy = kdx 50 − y

∫

∫

− log 50 − y = kx + C 8. Option (2) is correct. y  0   0 and k  0.049 Explanation:  log 50  y  ykx 0 C Given,  0 and k  0.049 log 50  y 50  kx We have,  log y  Ckx  C  log 50log  0 50  0 y C  kx  C

 log 50  0  0[∵  Cx  0, K  0.049, y  0   0] log 50  C [∵ x  0, K  0.049, y  0   0] 1 50  C C log  log 50 1 C  log 50



MATHEMATICS/APP. MATH.

9. Option (3) is correct. Explanation: We have −log 50 − y = kx + C



1 −log 50 − y = kx + log 50 50 log − y = −kx 50 50 − y − kx = e − kx 50  

50 − y = 50e − kx y = 50 − 50e − kx

(

y = 50 1 − e − kx

)

10. Option (3) is correct. Explanation: The highest order derivative present in the given dy differential equation is . Thus, degree of given differential dx equation is 1.

Course of Action Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

10

VECTORS

Revision Notes  

 Fundamentals { A quantity that has magnitude as well as direction is called a vector.   a { The unit vector in the direction of a is given by  and is a represented by a . { Position vector of a point P  x, y, z  is given as   OP  xi  y j  zk and its magnitude as OP  x 2  y 2  z 2 , where, O is the origin. { The scalar components of a vector are its direction ratios, and represent its projections along the respective axes. { The magnitude r, direction ratios (a, b, c) and direction cosines (l, m, n) of any vector are related as: a b c Scan to know l  , m  , n  and l 2  m2  n2  1 more about r r r this topic  Example: A vector r is inclined at equal angles to the three axes. If the magnitude of   r is 2 3 units, find r.  The Vector Sol. We have, r = 2 3 Cross Product  Since, r is equally inclined to the three  axes, r so direction cosines of the unit vector r will be same i.e., l = m = n. Scan to know We know that, more about this topic

l 2  m2  n2  1  l2  l2  l2  1 1  l2  3  So, 

 1  l     3 1  1  1  k r   i j 3 3 3    r  r  r r ∵ r    r    1  1  1    i j k2 3 3 3 3    2i  2 j  2k



 2 i  j  k

Parallel Vectors

Scan to know more about this topic

Collinear Vectors

{ The sum of the vectors representing the three sides of a triangle taken in order is 0 .

 Addition of Vectors   If a  a1i  a2 j  a3 k and b  b1i  b2 j  b3 k are two vectors   then a  b   a1  b1  i   a2  b2  j   a3  b3  k .   Example: Find the sum of vectors a  4i  j  k and b  2 j  k .    Sol. Let c denote the sum of a and b .    We have c  a  b

 4i  j  k  2 j  k  4i  j  2k

 Scalar Multiplication   If a is a given vector and λ is a scalar, then λ a is a vector    whose magnitude is  a   a . The direction of λ a is same   as that of a if λ is positive and, opposite to that of a if λ is negative.   If a  a1i  a2 j  a3 k and b  b1i  b2 j  b3 k are two vectors, then   λ a   λ a1  i   λ a2  j   λ a3  k and λb   λb1  i   λb2  j   λb3  k.  Vector Joining Two Points

If P1  x1, y1, z1  and P2  x2 , y2 , z2  are any two points, then  P1P2   x2  x1  i   y2  y1  j   z2  z1  k  2 2 2 Magnitude  P1P2   x2  x1    y2  y1    z2  z1   Section Formula The position vector of a point R dividing the line segment  joining the points P and Q whose position vectors are a and b   mb + na (i) in the ratio m : n internally, is given by m+n   mb − na (ii) in the ratio m : n externally, is given by m−n Triangle Law of VectorAddition  Let the vectors a and b be so positioned Scan to know that initial point of one coincides with     a AB terminal point of the other, = = , b BC .   Then, the vector a + b is represented by   side      the third  AC  of ∆ABC  i.e., AB  AB BC AC       AC  BC  AB BC AC    AB       BC AC    BC    AC AB      Or,     AC  AB  BC AC AB BC     AC AB BC  BC  AB       BC Or,  AC  AC   AB AC BC AB  AC    BC    AB         Or,   AB  AB BC CA   BC   CA 0  AB  BC  CA  000 AB as BCtriangle  CA  This is known law of addition.

more about this topic

Vectors Addition

→

→

→ →

Second Level

Third Level

Cross product of two vectors

Trace the Mind Map

First Level

a × b = a b sin θ n , n is a unit vector perpendicular to line joining a, b.



→



→

them, then their scalar product a.b = a b cos θ →→ a.b ⇒ cos θ = → → a b

→ →

Scalar product of two vectors

→→

If a, b are the vectors and 'θ', angle between

→ →

Vector

Positions Vectors

Direction ratios and direction cosines

A quantity that has both magnitude and direction is called a vector. The distance between the initial and terminal points of a vector is→called → its magnitude. Magnitude of vector AB is |AB|.

mb − na na + mb , (ii) externally is (i) internally is m–n m+n

→

The position vector of a point R dividing a line segment → joining P, Q whose position vectors are→a, b respectively, in the ratio m : n. →

l=

a b c ,m = ,n = Where r = a 2 + b 2 + c 2 r r r  e.g. : If AB = i + 2 j + 3k , then r = 1 + 4 + 9 = 14 Direction ratios are (1, 2, 3)  1 2 3  , , and direction cosines are  14 14 14 

The scalar components of a vector are its direction ratios, and represent its projections along the respective axes. The magnitude (r) direction ratios (a, b, c) and direction cosines (l,m,n) of vector ai + b j + ck are related as:

→

5i = i ,which is a unit vector. 5

→

direction of a. e.g.‚ If a=5i, then a =

→

→

a gives the unit vector in the a

→

For a given vector a, the vector a =



Position vector

Vectors

Unit vector

Properties of Vector

Types of Vectors

(

→

) (

)

j a2 b2

k a3 b3

C

D

sides of a triangle taken in If ABC is given triangle, then    → AB + BC + CA = 0. B

order is 0 . i.e.,

A

  If AB , AC are the given vectors,    then AB + AC = AD

A

B

C

The vector sum of two co-initials vectors is given by the diagonal of the parallelogram whose adjacent sides are given vectors.

→

i (iv) a × b = a1 b1 →

The vector sum of the three →

) (

(iii) → a.b = a1 b1 + a2 b2 + a3 b3 and

→

(i) a→± b = a ± b i + a ± b j + a ± b k 1 1 2 2 3 3 (ii) λ→ a = (λa1 )i + (λa2 ) j + (λa3 )k

If we have two vectors → a = a1 i + a2 j + a3 k , b = b1 i + b2 j + b3 k and λ is any scalar, then-

→

(i) Zero vector (initial and terminal points coincide) (ii) Unit vector (magnitude is unity) (iii) Coinitial vectors (same initial points) (iv) Collinear vectors (parallel to the same line) (v) Equal vectors (same magnitude and direction) (vi) Negative of a vector (same magnitude, opp. direction)

and its magnitude is 22 +32 +52 = 38.

Position vector of a point P (x, y, z) is xi + y j + zk and its magnitude is OP (r ) = x 2 + y 2 + z 2 . eg: Position vector of P (2, 3, 5) is 2i + 3 j + 5kˆ

124 Oswaal CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH.

125

VECTORS

 Parallelogram Law ofVector Addition  If the two vectors a and b are represented by the two adjacent sides OA and OB of a parallelogram OACB , then their sum    a + b is represented by the diagonal OC of parallelogram both in magnitude and direction through their common point O.    i.e.,OA  OB  OC

O  Projection

    a ⋅b Projection of vector a along vector b is  b  Scalar or Dot Product   The scalar or dot product of two given vectors a and b having an angle θ between them is defined as:     a  b  a b cosθ   If a  a1i  a2 j  a3 k and b  b1i  b2 j  b3 k are two vectors,   then a  b  a1b1  a2b2  a3b3.  Properties of Scalar Product :   (1) a ⋅ b is a scalar quantity.     (2) When θ  0, a  b  a b       (3) When   , a  b  a b cos  0 2 2     ⇒ When a  b , a  b  0       (4) When either a = 0 or b  0, a  b  0  Vector or Cross Product   The cross product of two vectors a and b having angle θ between them is given as     a  b  a b sinθ n , where n is a unit vector perpendicular to the plane containing     a and b . If a  a1i  b1 j  c1 k and b  a2 i  b2 j  c2 k are two vectors, then i   a  b  a1 a2

j b1 b2

k c1   b1c2  b2c1  i   a2c1  a1c2  j   a1b2  a2b1  k c2

 Properties of Vector Product:     (1) a  b  a b sinθ .n      (i) If a = 0 or b  0, a  b  0      (ii) a  b , a  b  0 (or θ = 0 )

      (2) a × b is a vector i.e., a  b  b  a     ∴ a  b  b  a   ⇒  a  b is not commutative.      (3) When   , a  b  a  b  n 2     or a  b  a b  (4)   i  i  j  j  k  k  0 and i  j  k , j  k  i, k  i  j j  i   k , k  j  i, i  k   j

⇒

  a b (5) sin     a b   (6) If a and b represent the adjacent sides of a parallelogram,   then its area  a  b .   (7) If a and b represents the adjacent sides of a triangle, then 1   its area  a  b . 2 Example: Using vectors, find the value of k such that the points  k,  10, 3 , 1,  1, 3 and  3, 5, 3 are collinear. Sol. Let the points are A  k ,  10, 3 , B 1,  1, 3 and C  3, 5, 3 .    So, AB  OB  OA



 

 i  j  3k  k i  10 j  3k



 1  k  i   1  10  j   3  3 k

 

 1  k  i  9 j  0k

  A B  (1  k ) 2  (9) 2  0  (1  k ) 2  81 Similarly,    BC  OC  OB  3i  5 j  3k  i  j  3k  2i  6 j  0k   BC  22  62  0



 



 2 10    and AC  OC  OA  3i  5 j  3k  ki  10 j  3k   3  k  i  15 j  0k  2 AC   3  k   225



 



If A, B and C are collinear, then sum of modulus of any two vectors will be equal to the modulus of third vectors.    For AB  BC  AC , (1  k ) 2  81  2 10  (3  k ) 2  225 (3  k ) 2  225  (1  k ) 2  81  2 10 9  k 2  6k  225  1  k 2  2k  81  2 10

k 2  6k  234  40  2 k 2  6k  234 . 2 10  k 2  2k  82  6k  234  40  k MATH. 2k  82  4 126 Oswaal CUET (UG) Chapterwise Question Bankk MATHEMATICS/APP. 2

k )2 2 81 k ) 2  225 (1  k ) 2(181 102 10 (3 k ) 2(3225

2

10 k 2 

4k  192  4 10 k 2  234  6k

⇒ (3  k ) 2(3225 k ) 2  225 k )2 2 81 (1 k ) 2(181 10 2 10

⇒

2 ⇒ 9  k 2 9 6kk2 22 5 k2 1   65k  22  2 k 1 2kk 81 2 81 10 2 10

On squaring both sides, we get

 k  48  10 k 2  234  6k





 k 2  4k  16  16  260 [Divided by 9 both sides] 2 ⇒ 2   2 2  k  2 k  82 k 2 6k  234 k2 106k 234 k22 2k222k10 82 6k  234  2 10  k  2k  82 k  6k  234  2 10 k k 6k2k 234  82 2 10  k 2  2k  82  k 2  4k  4 2 2 2 2 2 2       . k 6 k 234 40 2 k 6 k 234 ⇒ k 2 6k  234  40  2 k 22 6k 234 6k . 234  40 22 k  6k  234 . k  6k  234  40  2 2kk  6kk  234 234 . 40  2 k  6k  234 . k 2  4k  4  0 10  k  2k2 10 82 2 10  k 2 2 2k2 82  k 2  2k  82 2 2 10  k  2k  82 2 10   k  2k  82 (k  2) 2  0 2  6 2 82  42 10 k 2  234 2404k 210 2 234 2kk k k 2 2 6k  234 k406kk2 2234 k 22k k82  234  6  6  40   2  82  4 10  234  6k k k k k ⇒ k  6k  234  40  k k 2k 6k82  4 10 k  4 10 k 2  234  6k k  2  234 40 k 2k 2 234 2k 682 2 4k k2192  410  234 6k k 2  234  6k 4k  192  4 10  234 64kk k192  4 10 ⇒  4k  192  4 10 k 2 234 6k  4 10 k 2  234  6k 4k 192 2 8   10 6k k 2  234  6k  k  48  10 kk2 24234 6k  kk 48234  10  k  48  10 k  234 k 6k48  10 k 2  234  6k

OBJECTIVE TYPE QUESTIONS

[A] MULTIPLE CHOICE QUESTIONS 1. (1) 2.  (1) 3.

 

a  b  c

(1) 4.

(1) 5. (1) 6. (1) 7. (1) 8. (1) 9.

(1)

 If cos ,cos ,cos  are the direction cosines of vector a, then value of cos 2  cos 2  cos 2  is equal to [CUET 2022] 3. (2) 0. (3) 2. (4) -1.    If x i  j  k is a unit vector then value of x is [CUET 2022] 1 1 ± 3 . (2) ± . (3) � ± 3. (4) ± . 3 3





  10. Let a  3i  2 j  2k and b  i  2 j  2k be two vectors.   If a vector perpendicular to both the vectors a + b and   a − b has the magnitude 12 then one such vector is

  (3) 4  2i  2 j  k  . (1) 4 2i  2 j  k .

  (4) 4  2i  2 j  k  . (2) 4 2i  2 j  k .

      11. Let      2  a  b and    4  2  a  3b be two   3  given vectors where a and b are non collinear vectors. Let a , b and c be three unit vectors such that a  b  c  b  c   2 The value of λ for which vectors a and b are collinear, 3      b  c . If b is not parallel to c, then the angle is 2 (1) −4. (2) −3. (3) 4. (4) 3.   between a and c is [CUET 2021]       12. Let a , b and c be three unit vectors such that a  b  c  0. If 5π π 0 (2) 2π (3) (4)              6 6   a  b  b  c  c  a and d  a  b  b  c  c  a , then the      ordered pair,  , d is equal to Let   i  2i  k ,   2i  j  3k ,   2i  j  6k . If a and           b are both perpendicular to a vector δ and     10, then  3 3 (1)  , 3 (c  b )  . (2)  , 3 (a  c )  .  2 2     the magnitude of δ is



 







 

   3   3 (3)  , 3 (a  b )  . (4)  , 3 (b  c ) . 2   2   13. |λ| times the magnitude of vector a is denoted as The area of parallelogram determined by vectors 2i and     3 j is [CUET 2023] (1) |λ a |. (2) λ| a |. (3) |λ | a . (4) λ a .  6. (2) 5. (3) 1. (4) 8.   14. Let a  2i  1 j  3k , b  4i   3   2  j  6k and c  3i  6 j   3  1 k       The value of i  j  k  j  k  i is [CUET 2023]     c  3i  6 j   3  1 k be three vectors such that b = 2a and a is 6. (2) 2. (3) 1. (4) 8.  perpendicular to c. Then a possible value of  1,  2 , 3  is The value of a for which the vectors 2i  3 j  4k and  1  (1) 1, 3, 1 . (2)  , 4, 0  . ai  6 j  8k are collinear? 2   3. (2) 2. (3) 4. (4) 12. 1  (3)  , 4, 2  . (4) 1, 5, 1. The value of λ for which the vectors 2i + j + k and 4  2i  4 j   k are perpendicular is [CUET 2023]    15. Let a  i  j , b  i  j  k and c be a vector such that 1. (2) 2. (3) 4. (4) 0.    2       a  c  b  0 and a  c  4, then | c | is equal to  If a  b  c  12i  16 j , where a  14i  14 j and   19 17 (1) . (2) 9. (3) 8. (4) . b  8i  10 j , then vector c is 2 2 3i  j .  (2)  2 3i  j .  (3)  2 3i  j .  (4)   3i  j . 3 . 2

(2) 2 3 .

(3)

 











3 .

(4)

1 . 3











127

VECTORS

  16. The projection vector of a on b is      a b  a ⋅b  b . (1) (2)  .  b  b       a  b   a ⋅b (3)  . (4)   2  b . |a|  a          17. If a , b and c are three vectors such that a  b  c  0    a 2= and = , b 3 and c = 5, then the value of       a  b  b  c  c  a is (1) 0. (2) 1. (3) −19. (4) 38. 18. Find λ and μ if (i  3 j  9k )  (3i   j   k )  0. (1)    9,   27

(4)   3,   9   19. The angle between two vectors a and b with magnitudes   3 and 4 , respectively, and a  b  2 3 is π . 6

5π . 2 20. The value of λ for which the vectors 3i  6 j  k and 2i  4 j   k are parallel is

π . 3



(2)

3 . 2

(3)

π . 2

(3)



5 . 2

(4)

2 . 5 21. The vector in the direction of the vector i  2 j  2k that has magnitude 9 is  [NCERT Exemplar Pg 216 Q 19] i  2 j  2k (1) i  2 j  2k . (2) . 3 (3) 3 i  2 j  2k . (4) 9 i  2 j  2k . (1)

2 . 3

(2)



(4)



22. The position vector of the point which divides the join of     points 2a − 3b and a + b in the ratio 3 : 1 is     3a − 2b 7 a − 8b (1) . (2) . 2 4   3a 5a (3) . (4) . 4 4 23. The vector having initial and terminal points as (2, 5, 0) and (−3, 7, 4), respectively is (1) i  12 j  4k . (2) 5i  2 j  4k . (3) 5i  2 j  4k .

(4) i + j + k .

[NCERT Exemplar Pg 217 Q 20] (2) 1.

(3)

3 . 2

(4)

−5 . 2

 c

        (1) a and c (2) a and b (3) c and d (4) a and d 27. The vectors from origin to the points A and B are   a  2i  3 j  2k and b  2i  3 j  k respectively, then the area of triangle OAB is [NCERT Exemplar Pg 217 Q 25] 1 229 . 2     28. For any vector a , the value of (a  i ) 2  (a  j ) 2  (a  k ) 2 (2)

(3)

25 .

229 . (4)

is equal to [NCERT Exemplar Pg 218 Q 26] 2 2   (1) a . (2) 3a . (3) 4a 2 . (4) 2a 2 .       a 10 29. If = = , b 2 and a  b  12 , then the value of a × b is  (1) 5

[NCERT Exemplar Pg 218 Q 27] (3) 14 (4) 16   30. If a = 4 and 3    2 , then the range of λa is [NCERT Exemplar Pg 218 Q 32] (1) [0, 8]. (2) [−12, 8]. (3) [0, 12]. (4) [8, 12]. (2) 10

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. (1) Both A and R are true, and R is the correct explanation of A. (2) Both A and R are true, but R is not the correct explanation of A. (3) A is true but R is false. (4) A is false and R is true. 1. Assertion (A): The acute angle between the line π  r  i  j  2k   i  j and the X-axis is . 4 Reason (R): The acute angle θ between the lines  r  x1i  y1 j  z1 k   a1i  b1 j  c1 k and  r  x2 i  y2 j  z2 k   a2 i  b2 j  c2 k is given by

 



cos  

24. For two vectors to be collinear, which of the following is true? (1) The vectors should be parallel to the same line. (2) The vectors should have the same initial point. (3) The vectors should have the same magnitude. (4) The vectors should have the magnitude 1 and 0 respectively. 25. The value of λ such that the vectors a  i   j  k and  b  i  2 j  3k are orthogonal is (1) 0.

 a

(1) 340.

(2)   3,   9

(3)   9,   3

(1)

26. Which of the following vectors are coinitial in the figure given below?  b



a1a2  b1b2  c1c2 a12



 b12  c12 a22  b22  c22



.

2. Assertion (A): If two vectors are inclined at an angle, so 3 . 2 Reason (R): If two vectors are inclined at an angle, so that 1 their resultant is also a unit vector, then sinθ is . 2    3. Assertion (A): If vectors a  3i  4 j  2k and  b  2i  3 j  p k are mutually perpendicular then p is −9.   Reason (R): For perpendicular vectors a  b  0. that their resultant is also a unit vector, then sin θ is

128 Oswaal CUET (UG) Chapterwise Question Bank  4. Assertion (A): If force F  3i  4 j  2k acts on a particle having position vector 2i + j + 2k then, the torque about

the origin is 10i  10 j  5k .    Reason (R): Torque,   r  F .  5. Assertion (A): For a vector a with initial point P  4, 0, 2   and terminal point Q  6, 1, 2  , the value of a is 5 .   Reason (R): a  PQ  2i  j .    6. Assertion (A): In ABC , AB  BC  CA  0.        Reason (R): If= OA a= , OB b , then AB  a  b (triangle law of addition).   7. Assertion (A): a  i  p j  2k and b  2i  3 j  qk are 3 = , q 4. 2   Reason (R): If a  a1i  a2 j  a3 k and b  b1i  b2 j  b3 k

parallel vectors if= p

a1 a2 a3 are parallel = = . b1 b2 b3 8. Assertion (A): The projection of the vector a  2i  3 j  2k  5 on the vector b  i  2 j  k is 6.  3  Reason (R): The projection of vector a on vector b is  1   a b . a 9. Consider the shown figure.

 

  Assertion (A): If a and b represent the adjacent sides of 1   a triangle as shown, then its area is a × b . 2 1   Reason (R): Area of triangle ABC  a b sin  where, 2   θ is the angle between the adjacent sides a and b (as shown in figure). 10. Assertion (A):   a  b   b c  b c  i   a c  a c  j   a b  a b  k

1 2



Reason (R): j i   a  b  a1 b1 a2 b2

2 1

2 1

1 2

1 2

2 1

k c1 c2

[C] COMPETENCY BASED QUESTIONS

I.  Read the following text and answer the following questions on the basis of the same: Solar Panels have to be installed carefully so that the tilt of the roof, and the direction to the Sun, produce the largest possible electrical power in the solar panels.

MATHEMATICS/APP. MATH.

A surveyor uses his instrument to determine the coordinates of the four corners of a roof where solar panels are to be mounted. In the picture, suppose the points are labelled counter clockwise from the roof corner nearest tothecamerainunitsofmeters P1  6, 8, 4  , P2  21, 8, 4  , P3  21,16,10  and P4  6,16,10  . 1. What are the components to the two edge vectors defined by   A = PV of P2 − PV of P1 and B = PV of P4 − PV of P1 ? (where PV stands for position vector) (1) 0, 0,15 : 0, 8, 6

(2) 15, 0, 0 : 0, 8, 6

(3) 0, 8, 6 : 0, 0,15 2. (1) 3. (1) (3) 4. (1) 5. (1)

(4) 15, 0, 0 : 6, 8, 8   What are the magnitudes of the vectors A and B ? 9, 10 (2) 115, 50 (3) 15, 10 (4) 100, 200  What are the components to the vector N , perpendicular   to A and B and the surface of the roof ? 0, –90, 90 (2) 0, 120, 18 0, –90, 100 (4)  0, –90, 120 What is the magnitude of N ? 100 (2) 150 (3) 50 (4) 90   Find the projection of vector A on vector B. 15 (2) 10 (3) 30 (4) 0

II. Read the following text and answer the following questions on the basis of the same: A class XII student appearing for a competitive examination was asked to attempt the following questions.    Let a , b and c be three non zero vectors.       6. If a and b are such that a  b  a  b then       (1) a ⊥ b . (2) a  b . (3) a = b . (4) None of these        7. Let a , b and c be unit vectors such that a  b  a  c  0 and    π angle between b and c is then a = 6     (1) 2 b  c . (2) 2 b  c .     (3) 2 b  c . (4) 2 b  c .   8. If a  i  2 j , b  2i  j  3k then       2a  b   2a  b  a  2b    





  

 



 









(1) 0 (2) 2 (3) 3 (4) 4   9. The area of the parallelogram formed by a and b as diagonals is (1) 70 sq. units (3)

70 sq. units 2

(2) 35 sq. units (4)

70 sq. units

129

VECTORS

  10. If a  4i  6 j  12k , then find the unit vector in direction of a .         2i 3j 6k (1)         7  7   7       

 5i   3 j   2k  (2)         7  7   7       

 2i   3 j   6k  (3)         5  5   5       

 2i   3 j   6k  (4)         7  7   7       

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (4) 11. (1)

2. (4) 12. (3)

3. (4) 13. (1)

4. (2) 14. (2)

5. (1) 15. (1)

6. (2) 16. (2)

7. (3) 17. (3)

8. (4) 18. (1)

9. (2) 19. (2)

10. (1) 20. (1)

21. (3)

22. (4)

23. (3)

24. (1)

25. (4)

26. (1)

27. (4)

28. (4)

29. (4)

30. (3)

8. (3)

9. (1)

10. (1)

8. (1)

9. (3)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (3)

3. (1)

4. (1)

5. (1)

6. (3)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (3)

3. (4)

4. (2)

5. (4)

6. (1)

7. (3)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (4) is correct. Explanation: Since, cos α, cos b and cos g are direction cosines of vector a, then Now, cos 2  cos 2  cos 2  = (2cos2a - 1) + (2cos2β - 1) + (2cos2g - 1) = 2(cos2a + cos2β + cos2g) - 3 =2×1-3 =2-3 = -1 2. Option (4) is correct. Explanation: Given, x ijjkk 11 i x ji     x k  1         x ixx i ji  jkjkk 111  j k  1 xxji i  kj k1  1   kx i  1 ⇒ x ixx i ji  j jkk 11 2 2 2 x2x 11122111221121 11 2122x21212  221 ⇒ x x1x2  11 11 x11 3111 ⇒ x 3  1 x 3x  31  1 xx 3311 11 ⇒ 1 11x  xx1 3 x xx  3 3 3 33 11 ⇒ 1 x11 xx1 3 x x x 3 3 3 33 3. Option (4) is correct.    3   Explanation: a  b  c  b c 2       3 3 a  c b  a  b c  b  c 2 2 on compariing we get,

       







 

 

  a  c     a c cos   cos    



3 2 3 2 3 2  6



4. Option (2) is correct.  Explanation: Let   ai  b j  ck     Given,    and    a  2b  c  0 2 a  b  3c  0 a b c   6  1 2  3 1  4 c a b   5 5 5 a b c   1 1 1    k i  j  k       10 k i  j  k  2i  j  6 k  10











k  2  1  6   10 10  2 k 5    2i  2 j  2 k    22  22  22  2 3 units 5. Option (1) is correct.

Explanation: Area of parallelogram   = a × b = 2i × 3j = 6 k = 6 sq.units 6. Option (2) is correct. Explanation:

(i ×j )⋅ k + (j ×k )⋅i = k ⋅ k +i ⋅i = 1+1 =2

130 Oswaal CUET (UG) Chapterwise Question Bank 7. Option (3) is correct. Explanation: Vectors are collinear when 2 3 4   a 6 8 ∴ a4 8. Option (4) is correct. Explanation: Vectors are perpendicular when

 2  2   1  4   1    0



44 0 0 9. Option (2) is correct Explanation:      Here, c  12i  16 j  b  a  12i  16 j  8i  10 j  14i  4 j

 

      6i  2 j  2  3i  j 



 





 16i  16 j  8k   c  12





 16i  16 j  8k  12  162  162  82  12 24  12 1  2  1  ∴ c = 16i − 16j − 8 k 2 = 4 2i − 2j −  k



( (

)

)

11. Option (1) is correct.       4  2  a  3b ,        2 a  b        Because  &  are collinear        2  a  b    4  2  a  3b  On comparing, we get   2    4  21 and 1 = 3µ    2    4  2  3 1 1 ⇒    2    344  2  3 and    2  44  2   1 ⇒    22  44  2   3 ⇒     44    2   4  2     4

 .



i j k  c 4 4 0 2 0 4

Given,

12. Option (3) is correct.    Explanation: a  b  c  0    Squaring (a  b  c ) 2  0       a 2  b 2  c 2  2(a  b  b  c  c  a )  0        1  1  1  2(a  b  b  c  c  a )  0       3 a b  b c  c a   2 3  2    [(| a |  1,| b | 1,| c | 1, (givenn)]        d  ab  b c  c a    c  a  b          d  a  b  b  (a  b )  (a  b )  a            a b b a b b  a a b a        ab  ab  0  0  ab    3(a  b )   a a  0     b  b  0       a  b  b  a    3    ( , d )   , 3 a  b   2 





10. Option (1) is correct. Explanation:          A vector ⊥ to both a  b & a  b is c   a  b  a  b   a  b  4i  4 j   a − b = 2i + 4 k

MATHEMATICS/APP. MATH.





13. Option (1) is correct.  Explanation:  We know that the magnitude of vector a is   denoted by | a |. If we multiply the magnitude of vector a with  magnitude of λ, then we get |λ a |.   Thus, |λ| times the magnitude of vector a is denoted as |λ a | = |λ|  |a | 14. Option (2) is correct. Explanation:   b  2a



4i   3   2  j  6k  2 2i  1 j  3k



3   2  21 

21   2  3   ∴ a⊥c

...(i)

  a c  0

 2i  1 j  3k  3i  6 j   3  1 k   0

6  61  3  3  1  0 21  3  1 

...(ii)

From equations (i) and (ii) by trial method, it satisfied the option (2). 15 Option (1) is correct. Explanation:     c  b  0 a  caa bc 0b  0   ci  c j  c k c  ccc1i Let Let cc12i j c2c3j k c3 k Let  1  2  3  acc   aa  b  cbb ⇒   i ij kj k i j k    i  jikj ck  cc1 c2 c3 i  j  k1  12 c23 c3 011 00 1 111    k  c  c  jii kjj ckk i  c i   i  ck1 c112  c22  j cc33kjj  3 cc333i  c c c    1 &     c  c31&1c&c1  c12(ii11) (i (iii)) 3

ac  4

3

1

12

2

Let c  c1i c21j  c32 k 3    b  c  a  b  cb  a a c  a  b  c  a c b     j kii jj kk j i k j k iii j  k  c1 c2 c3    jccc1k2 c ccc2311 cc322 c33 i i j jikkj ci kc1 1 2 3 1 1 0 VECTORS 1 1 0 1 11 01 01 0  j  k  c i  c j  k  c  c  j   ii c j c33kicc33 jc k c11  c22  ij cck3i   i  k     k3 j c k31cc12  c21 2  ⇒  i  j i k j c3ki3 c j  c33  313 & c11  1c22 21 (ii)  c  1 &c13c& (ii)2   1c &c1c2  (ii)1(ii) ⇒ ...(i) 1(i  c3 3a  c13c& i)1c  4c1 1 c2c121 a  c a 4c a4 c  4 a  c  4    c2 j  c3 k  4    i cjj j c11cii  2 3 i i j jicci1ji ic 3jkccjk4 c34k  4  2ji  c c  1 c 1 2 1 2 3 k 234   c11  c22  4 ... ii  ⇒   ... ii   ii ... ii    c  c c4c1...(ii)  c1 1c2c12 4 2 4c2  4... ii ... eqns.(i) & (ii), Solving eqns. (i) & (ii),eqSolving Solving ns.(i) &eq(ii), Solving ns.(i) & (ii), Solving eq& ns.(i) & (ii), Solving eqns.(i) (ii), 3 5 3 3c11  3 5& c225  5 c1 3c & cc1&c52 & 2 c1   2 2 c2   2 1 c2 2 2& 2 22 2 2 3 5 c  53i  5 j  k   c 3 3 i 535  j 2kij  2k j  k c  2ic 2cji  k 2 22 2 9 225 c25   9  25  1  c  9 9 25 14  4  1 c9125 c  c 4 4 4 14 4 44 438 19 38 38 193819 19 19 38    4  2  4 4 2 4 2 2 2 19 2 4  2 19 | c19|22  19 2 |c |  2 | c2 | |19 | c |  c |2  2 2 2 16. Option (2) is correct.

              

17. Option (3) is correct. Explanation:        Here, a  b  c  0 and a 2  4, b 2  9, c 2  25         a b c  a b c 0                  a2  a  b  a  c  b  a  b 2  b  c  c  a  c  b  c 2  0           a2  b 2  c 2  2 a  b  b  c  c  a  0     [∵ a  b  b  a ]        4  9  25  2 a  b  b  c  c  a  0















      38 a b  b c  c a   2  19 18. Option (1) is correct. Explanation: (i  3 j  9k )  (3i   j   k )  0 i j k 1 3 9 0 3  

or

or i (3  9)  j (  27)  k (  9)  0 or 3  9  0 or   27  0 or 9  0 From eqn. (ii)) and (iii),   27 and   9

...(i) ...(ii) ...(iii)

19. Option (2) is correct.     Explanation: Here, = |a| = 3 ,| b | 4 and a  b  2 3 [Given] We know that,     a  b | a || b | cos   2 3  3  4  cos  2 3 4 3 1  2 

 cos  

    a  b | a || b | cos 

131

 2 3  3  4  cos  2 3 4 3 1  2    3 20. Option (1) is correct.     Explanation:  Let a  3i  6 j  k and b  2i  4 j   k Since,      a  b 3 6 1  ⇒  2 4  2  ⇒ 3  cos  

21. Option (3) is correct.  Explanation: Let a  i  2 j  2k

  a Any vector in the direction of a vector a is given by  . a i  2 j  2k i  2 j  2k   2 3 12  2  22

 

 \ Vector in the direction of a with magnitude 9 . i  2 j  2k 9 3   3 i  2 j  2k





22. Option (4) is correct. Explanation: Let R be the position vector of the point which     divides the join of points 2a − 3b and a + b .     3 a  b  1 2a  3b ∴ Position vector, R  3 1 [Since, the position vector of a point R dividing the line segments   joining the points P and Q, whose position vectors are p and q in the   mq + np ration m : n internally, is given by .] m+n  5a R  4 23. Option (3) is correct. Explanation: Required vector   3  2  i   7  5  j   4  0  k   5i  2 j  4k



 



24. Option (1) is correct. Explanation: Two or more vectors are called collinear vectors if they are parallel to the same line. It is not necessary that they should have the same magnitude and direction. 25. Option (4) is correct.   Explanation: Since, two non-zero vectors a and b are   orthogonal, i.e., a  b  0







 2i   j  k  i  2 j  3k  0  

2  2  3  0 5  2

132 Oswaal CUET (UG) Chapterwise Question Bank 26. Option (1) is correct. Explanation: The vectors which start from the same initial  point are called coinitial vectors. In the given figure, vectors a  and c start from the same point. 27. Option (4) is correct. Explanation: 1   Area of OAB  OA  OB 2 1    2i  3 j  2k  2i  3 j  k 2 i j k 1  2 3 2 2 2 3 1



 



1  i  3  6   j  2  4   k  6  6    2  1  9i  2 j  12k 2 1  81  4  144  2 1 = 229 sq. units 2 28. Option (4) is correct.  Explanation: Let a  xi  y j  zk  

 a 2  x2  y 2  z 2 i j k   a i  x y z 1 0 0





  (a  i ) 2  z j  y k z j  y k y z 2





2



12 3  20 5

9  sinθ  1  cos θ  1  25 4 sinθ   5

     a  b  a b sin

 10  2 

4 = 16 5



  2 x 2  y2  z2  2 a 2

12  10  2cosθ

2

1. Option (1) is correct. Explanation: The equation of the X-axis may be written as  r = ti. Hence, the acute angle q between the given line and the x-axis is given by cos  

1  1   1  0  0  0 1  (1) 2  02  12  02  02 2

 1  4 2 2. Option (3) is correct.     Explanation: Let a  b  a  b       a  b |2  a |2  | b |2 2 a  b   1  1  1  2 a  b cos  



29. Option (4) is correct Explanation:     a 10 Here, = = , b 2 and a  b  12 [Given]     \  a  b  a b cosθ

 cosθ 

[B] ASSERTION REASON QUESTIONS

3 2

3. Option (1) is correct.    Here, a  b  0Here, a  b  0 Explanation:     [Since, a and b[Since, are perpendicular] a and b are perpendicular] i  3 j  pk 0   3i  4 j  2 k 3i2 4 j  2k  2i  3 j  pk  0

 

    

 and a  2 4  8, at   2  So, the range of λa is 0,12.

Now, sin120 =

2

2  Similarly a  j  x 2  z 2  ( a  k )2  x 2  y 2 and 2 2     a  i  a  j  a  k

30. Option (3) is correct.  Explanation: We have, a = 4 and 3    2    a   a   4   a   3 4  12, at   3  a  0 4  0, at   0

1  2  2  1  1cos  1 2  120 Thus, cos      2 3

 i 0  j  z   k  y   z j  y k

MATHEMATICS/APP. MATH.

  

 

 

 



6  12  2 p  06  12  2 p  0

 4. Option (1) is correct.

2 p  18 p  9

2 p  18 p  9

Explanation: We know that,    Torque,   r  F i j k   rF  2 1 2 3 4 2  i  2  8   j  4  6   k 8  3  10i  10 j  5kk 5. Option (1) is correct.   Explanation: Here, a  PQ   6  4  i   1  0  j   2  2  k  or a  2i  j  Thus, a  22  (1) 2  4  1  5

133

VECTORS

6. Option (3) is correct. Explanation: Assertion is true as it is a triangle law of vector addition.    We know that, AB  OB   OA  b a Therefore, reason is incorrect. 7. Option (1) is correct.   Explanation: Given, a and b are parallel then a1 a2 a3 = = b1 b2 b3 1 p 2 = = 2 3 q 3 and q = 4 . 2 8. Option (3) is correct.  Explanation: We know that projection of vector a on vector    a ⋅b b is  . b On solving, p =

 2 1   3 2    2 1  12  22  12

10 6 5  6 3 9. Option (1) is correct.   a b     Explanation: sin     a  b  a b sin a b   If a and b represents the adjacent sides of a triangle, then its area 1    ab 2 10. Option (1) is correct. 

i Explanation: a  b  a 1 a2

j b1 b2

k c1 c2

[C] COMPETENCY BASED QUESTIONS





 15i  0 j  0k  B  PV of P4  PV of P1  6i  16 j  10k  6i  8 j  4k



 0i  8 j  6k





 64  36  100  10 3. Option (4) is correct. Explanation:    N  A B i j k  N  15 0 0 0 8 6  90 j  120k 4. Option (2) is correct.  Explanation: N 90jj 120 120kk N  90   90))22 ((120 120))22 NN  ((90 810014400 14400  8100 22500  22500 150 150

5. Option (4) is correct.

 Explanation: We know that projection of vector A on vector    A⋅ B B is  B 

On solving determinants we get,   a  b   b1c2  b2c1  i   a2c1  a1c2  j   a1b2  a2b1  k

1. Option (2) is correct. Explanation:  A  PV of P2  PV of P1  21i  8 j  4k  6i  8 j  4k

2. Option (3) is correct. Explanation:  | A | (15) 2  02  02  15  | B | 02  82  62



\ Components are A 15, 0, 0  and B  0, 8, 6  .



15i  0 j  0k    0i  8 j  6k  0 2  82  6 2 15 0    0 8   0  6  100

0  10  = 0 6. Option (1) is correct.     Explanation: Here, a  b |2  a  b |2   Or, 4  a  b  0   Or, a ⊥ b 7. Option (3) is correct.     Explanation: Given, a  b  a  c  0    Thus, a is perpendicular to b and c.   A unit vector perpendicular to b and c

134 Oswaal CUET (UG) Chapterwise Question Bank   b c    b c   b c     b  c sin 6     b c   2 b  c 1 2 8. Option (1) is correct.         Explanation: Here, 2aHere,  b 2a4ib3 j 4i3 k 3 j  3k     and a  and 2b a32i b 4j3i6k 4 j  6k





   



j i   30 k j  25k  30i  15 2515       Now, 2a  b   2a  b  a  2b   





 

 



 4i  3 j  3k  30i  15 j  25k





 120  45  75  0 9. Option (3) is correct. Explanation: We know that, The area of a parallelogram 1 = × cross product of its diagonals  2

  \ The area of the parallelogram formed by a and b as diagonals 1        a  b 2







   2i  j  3k 

1  i  2 j 2 i j 1  1 2 2 2 1 



j i k j i k        2a  b 2a a b 2b a  42b 3 4 3 3 3 3 4 36 4 6

MATHEMATICS/APP. MATH.

k 0 3

1   6i  3 j  5k 2 70 1 36  9  25  sq. units  2 2 

10. Option (1) is correct.  Given, a  4i  6 j  12k  a  42  62  122  14  Therefore, the unit vector in direction of a is given by  a 4i  6 j  12k a    a 14 4  6  12  i j k 14 14 14 2 3  6   i j k 7 7 7 

Course of Action Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

11

THREE-DIMENSIONAL GEOMETRY  Angle

Revision Notes  Fundamentals  Direction cosines of a line are the cosines of the angles made by the line with positive directions of the coordinate axes.  If

l , m, n are the direction cosines of a line, then

l 2  m2  n2  1  Direction cosines of a line joining two points P  x1, y1, z1  and PQ 

Q  x2 , y2 , z2 

are

x2 - x1 y2 - y1 z2 - z1 , , PQ PQ PQ

where,

 x2  x1 2   y2  y1 2   z2  z1 2

 If

l , m, n are the direction cosines and a, b, c are the direction ratios of a line, then a b c l ,m ,n 2 2 2 2 2 2 2 a b c a b c a  b2  c2 Example: Find the direction ratio and direction cosines of a line parallel to the line whose equations are 6 x  2  3 y  1  2 z  2.  Sol. Given line L : 6x  2  3y  1  2z  2 1 y 1   3  z 1   3  1 1 1 6 3 2 1 1 x y 3  3  z 1 1 2 3 x

1 1 1 , , or, 1, 2, 3 6 3 2 ∴ Direction cosines of line L are: 1 2 3 , , 2 2 2 2 2 2 2 1 +2 +3 1 +2 +3 1 + 22 + 32 1 2 3 i.e., , , 14 14 14 Now, since the parallel lines have the proportional d.r.’s and direction cosines so, d.r.’s of required line passing through 1 2 3  1 1  , , .  , , 1 are 1, 2, 3 and direction cosines are 14 14 14 3 3  So, d.r.’s of line L are: 1,2,3

 Skew lines are the lines in the space which are neither parallel

nor intersecting. They lie in the different planes.  Angle between skew lines is the angle between two intersecting lines drawn from any point (preferably through the origin) parallel to each of the skew lines.

between two lines cos   l1l2  m1m2  n1n2

or cos  

a1a2  b1b2  c1c2 a12

 a22  a32  b12  b22  b32

Note:  For two perpendicular lines: a1a2 + b1b2 + m1m2 + n1n2 = 0. a1 b1 c1 l1  For two parallel lines: = = ;= a2 b2 c2 l2  Line

+ c1c2 = 0, l1l2 m1 n1 = .. m2 n2

 Vector

equation of a line that passes through a given point   whose position vector is a and parallel to a given vector b is    r  a  λ b.  Equation

of a line through a point  x1, y1, z1  and having

direction cosines l, m, n (or, direction ratios a, b and c) is x  x1 y  y1 z  z1  x  x1 y  y1 z  z1  .     or  c  l m n b  a  The

vector equation of a line that passes through two points       whose position vectors are a and b is r  a   b  a .  Cartesian equation of a line that passes through two points xx yy zz  x1, y1, z1  and  x2 , y2 z2  is x  x1  y  y1  z  z1 . 2 1 2 1 2 1     If θ is the acute angle between the lines r  a1   b1 and    r  a2  b2 , then θ is given by     b1  b2 b1  b2 cos     or   cos 1   b1 b2 b1 b2 x y z Example: Find the acute angle between the lines = = 3 4 5 x  1 y  1 z  10 .  and  3 4 5  Sol. Vector in the direction of first line b  3i  4 j  5k  Vector in the direction of second line d  4i  3 j  5k



Angle θ between two lines is given by   bd    cos    bd cos  

(3i  4 j  5k )  (4i  3 j  5k ) | (3i  4 j  5k ) || (4i  3 j  5k ) |

12  12  25 9  16  25 16  9  25 25  cos   50 50 1  cos   2

 cos  











(

)

)

=

b1 b2 2

c1 c2

2

2

(b1 c2 – b2 c1 ) + (c1 a2 – c2 a1 ) + (a1 b2 – a2 b1 )

a1 a2

x – x1 y – y1 z – z1 x – x 2 y – y2 z – z2 is and = = = = a1 b1 c1 a2 b2 c2 x 2 – x 1 y 2 – y 1 z 2 – z1

)(

(

)

)(

)

A +B +C

Ax + By + Cz + D plane Ax + By + Cz + D = 0 is 1 2 1 2 1 2

 The distance of a point with position vector a from the plane ^  r . n = d is d – a . nˆ point ( x1 , y1 , z1 ) . The distance from a to the

(

      Two lines r = a1 + λ b1 , r = a2 + µb2 are coplanar if     a2 – a1 . b1 × b2 = 0. Equation of a plane that cuts x y z co-ordinate axes at (a,0,0), (0,b,0), (0,0,c) is a + + c = 1. b

(

( ) ( )( )

cos θ = l1 . l2 + m1 . m2 + n1 . n2

if

x – x 1 y – y 1 z – z1 = = and l1 m1 n1 x – x 2 y – y2 z – z2 = = l2 m2 n2 are the equations of two lines, then acute angle between them is

b .b then, cos θ = 1 2 b1 . b2

Equation of line Vector in 3D

between them, then cos θ = l1 l2 + m1 m2 + n1 n2 =

a12 + b12 + c12

a22 + b22 + c 22

a1 a2 + b1 b2 + c1 c 2

are the D.Rs of the two lines and 'θ' is the acute angle

if l1, m1, n1 , l2, m2, n2 are the D.Cs and a1, b1, c1, a2, b2, c2

First Level

Second Level

Trace the Mind Map Third Level

Equation of a line through point (x1, y1, z1) and x – x1 y – y1 z – z1 Also, = = l m n equation of a line that passes through two points is x – x1 y – y1 z – z1 = = x 2 – x 1 y 2 – y 1 z 2 – z1 having D.Cs l, m, n is

( )

Vector equation of a line which passes through   two points whose position vectors are a and b is     r = a +λ b– a

Vector equation of a line passing through the given  point whose position vector is a and parallel to a given     vector b is r = a + λ b

Angle between the two lines

Skew lines

These are the lines in space which are neither parallel nor intersecting. They lie in different planes. Angle between skew lines is the angle between two intersecting lines drawn from any point (origin) parallel to each of the skew lines.

line if l, m, n are the D.Cs and a, b, c are D.Rs of a line, then a b c ,m= ,n= a2 + b2 + c 2 a2 + b2 + c 2 a2 + b2 + c 2

l=

then l2 + m2 + n2 = 1. D. Cs of a line joining P (x1, y1, z1) and Q (x2, y2, z2) are x 2 – x1 , y 2 – y1 , z2 – z1 , where PQ = (x – x )2 + (y – y )2 + (z – z )2 2 1 2 1 2 1 PQ PQ PQ D.Rs of a line are the no.s which are proportional to the D.Cs of the

positive direction of the co-ordinate axes. If l, m, n are the D. Cs of a line,

D. Cs of a line are the cosines of the angles made by the line with the

If 'θ' is the acute angle       between r = a1+λ b1 , r = a2 + λb2

Angle between two lines

on ensi al G m

Parallel lines

Direction ratios and direction cosines of a line

Characteristics of planes

Vector equation of a plane

Equation of a plane

(i) which contains three non-collinear points having position          vectors a , b , c is r – a .  b – a × c – a  = 0.     (ii) That passes through the intersection of planes r . n1 = d1 &      r . n2 = d2 is r . n1 + λ n2 = d1 + λd2 , λ − non-zero constant.

(i) which is at distance 'd' from origin and D.C.s of the normal to the plane as l,m,n is lx+my+nz=d. (ii) ⊥r to a given line with D.Rs. A,B,C and passing through (x1,y1,z1) is A (x–x1) + B (y–y1) +C (z–z1) = 0 (iii) Passing through three non-collinear points x – x 1 y – y 1 z – z1 (x1,y1,z1) (x2,y2,z2), (x3,y3,z3) is x 2 – x1 y 2 – y1 z2 – z1 = 0. x 3 – x 1 y 3 – y 1 z 3 – z1

)

Di Three

   b × a2 − a1       is (iv) Distance between parallel lines r = a1 + λ b and r = a2 + µb  b

(iii)

(

(i) two skew lines is the line segment perpendicular to both the lines           (ii) r = a1 + λ b1 and r = a 2 + µb 2 is b1 × b 2 . a 2 – a 1   b1 × b 2



etry 

m eo



136 Oswaal CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH

bd cos  

(3i  4 j  5k )  (4i  3 j  5k ) | (3i  4 j  5k ) || (4i  3 j  5k ) |

137

THREE-DIMENSIONAL 12  12GEOMETRY  25

 cos  

9  16  25 16  9  25 25  cos   50 50 1  cos   2    3

{ If

 PLANE { Plane:

x  x1 y  y1 z  z1 x  x2 y  y2 z  z2     and l1 m1 n1 l2 m2 n2

are equations of two lines, then the acute angle θ between the two lines is given by cos  l1l2  m1m2  n1n2 . { The shortest distance between two skew lines is the length of

the line segment perpendicular to both the lines.    { The shortest distance between the lines r  a1   b1 and    r  a2  b2 is     b1  b2   a2  a1    . b1  b2 x  x1 y  y1 z  z1   { Shortest distance between the lines: a1 b1 c1 x  x2 y  y2 z  z2   and is a2 b2 c2



d



x2  x1 a1 a2

y2  y1 b1 b2

z2  z1 c1 c2

 b1c2  b2c1 2   c1a2  c2a1 2   a1b2  a2b1 2

{ Distance between parallel lines

   b   a2  a1   is d  . b

A plane is a surface such that a line segment joining any two points of it lies wholly on it. A straight line which is perpendicular to every line lying on a plane is called a normal to the plane. { Equations of a Plane in Normal form Vector form: The equation of plane in normal form is given    by r  n  d , where n is a vector which is normal to the plane. Cartesian form: The equation of the plane is Scan to know given by ax + by + cz = d, where a, b and c are more about this topic the direction ratios of plane and d is the distance of the plane from origin. Another equation of the plane is lx + my + nz = p, where l, m and n are direction cosines of the Definition of perpendicular from origin and p is a distance of Plane a plane from origin. Scan to know Note: If d is the distance from the origin and l, more about this topic m and n are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is (ld, md, nd). { Equation of a plane perpendicular to a given vector and passing through a given point

Equation of Plane

Vector form: If a plane passes through a point A with position      vector a and perpendicular to the vector n , then  r  a   n  0 Cartesian form: Equation of plane passing through point  x1, y1, z1  is given by a  x  x1   b  y  y1   c  z  z1   0

.

      r  a1  b and r  a2  b

{ Co-planarity

of Two Lines: Assume that the given lines are L1 : r  a1  b1 and L2 : r  a2  b2 such that L1 passes through A(x1, y1, z1) with     position vector a1 and is parallel to b1 with d.r.’s a1, b1, c1. Also   L2 passes through B(x2, y2, z2) with position vector a2 and is   parallel to b2 with the d.r.’s a2, b2, c2.

where, a, b and c are the direction ratios of normal to the plane. { Equation of plane passing through three non-collinear points    Vector form: If a , b and c are the position vectors of three given points, then equation of a plane passing through three      non-collinear points is  r  a   b  a   c  a   0 .



Cartesian form: If



 x1, y1, z1  ,  x2 , y2 , z2 



and

 x3 , y3 , z3 

are three non-collinear points, then equation of the plane is x  x1 y  y1 z  z1 x2  x1 y2  y1 z2  z1  0 x3  x1 y3  y1 z3  z1 (a) Vector form of co-planarity of lines: { Equation of Plane in Intercept Form: If a, b and c are      We know that AB  a2  a1 . Now the lines L1 and L2 are coplanar x-intercept, y-intercept and z-intercept, respectively made by the           plane on the coordinate axes, then equation of plane is if AB is perpendicular to b1 × b2. That implies, AB· b1  b2  0 x y z    1. or a b c         a2  a1 · b1  b2  0 Note; Equation of XY-plane: z = 0, (b) Cartesian form of co-planarity of lines:   Equation of YZ-plane: x = 0, We know that AB   x 2  x1  i   y 2  y1  j   z2  z1  k , b1  a1 i  b1 j  c1 k Equation of ZY-plane: y = 0.       z2  z1  k , b1  a1 i  b1 j  c1 k and b2  a2 i  b2 j  c 2 k . So by using { Angle between Two Planes in Vector Form:         For planes r  n1  d1 and r  n 2  d2 acute angle θ is, a2  a1 · b1  b2  0 , we get















x 2  x1 a1 a2

y 2  y1 b1 b2



z 2  z1 c1  0 c2

   Note that only coplanar lines can intersect each other in the plane they exist.

cos  { In

n1  n 2 n1  n 2

Cartesian Form: For planes A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0 acute angle θ is,

138 Oswaal CUET (UG) Chapterwise Question Bank A1 A2  B1B2  C1C2

cos  

A1  B1  C1 2

2

2

A2  B2  C2 2

2

2

{ Angle

between Line and Plane in Vector Form: For line r  a  b and plane r , n = d, acute angle θ is, sin  

b n | b || n |

{ In

Cartesian Form: x  x3 y  y1 z  Z1 For line   and plane A2 x + B2 y + C2 z A1 B1 C1 + D2 = 0 acute angle θ is,

sin θ 

MATHEMATICS/APP. MATH

A1 A2  B1B2  C1C2 A  B12  C12  A22  B22  C22 2 1

{ Distance

of a Point from a Plane in Vector Form: Distance of point with position vector a from paine i  n  d is d  a  n In Cartesian Form: Distance of point(x1, y1, z1) from palne Ax + By + Cz + D = 0 is Ax1 + By1 + z1 + D A 2 + B2 + C 2

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

x2 y 3 z 5   1. The angle between the line and the 3 2 6 plane 2 x  10 y  11z  5 is 1  8  1  8  (1) cos   . (2) sin   .  21   21  1 

21  (3) cos  .  82 

1 

21  (4) sin   .  82 

2. If a line makes angles 90°, 60° and θ with X, Y and Z axis respectively, where θ is acute, then value of θ is π π π π (1) . (2) . (3) . (4) . 6 4 3 2 x 1 y  2 z - 3   3. The point of intersection the lines 2 3 4 x  4 y 1   2 , is and 5 2 (1) 1, 1, 1  (2)  1,  1,  1  (3)   1, 1,  1  (4)   1,  1,  1

4. The distance between the point (3, 4, 5) and the point where x3 y 4 z 5   the line meets the plane x + y + z = 17, 1 2 2 is (1) 3. (2) 2. (3) 1. (4) 0. 5. The equation of plane passing through the point (0, 7, −7) x 1 y  3 z  2   and containing the line is 3 2 1 (1) x  y  z  0.

(2) 4 x  y  z  0.

(3) x  y  z  0. (4) 3 x  2 y  2 z  0 . x  2 y  3 z 1   6. A line L: is perpendicular to a plane 1 2 1 (P), which passing through the point (4,3, 9). If the mirror image of point ‘S’ on the line (L) in the given plane (P) is (2, 3, 1), then co-ordinates of point S, is (1) (1, 0, 3)   (2)   0, 1, 3   (3)   2, 3, 1   (4)   9, 7,  1 7. Equation of the line passing through the point (1, 2, 3) and parallel to the vector 3i  2 j  2k is x 1 y  2 z  3 x3 y 2 z 2     (1) . (2) . 3 2 2 1 2 3 x 1 y  2 z  3 x3 y2 z2     (3) . (4) . 3 2 2 1 2 3 8. Equation of line passing through (1, 2, 3) and parallel to the x3 4 y z 8   line given by is 3 5 6

(1)

x 1 y  2 z  3   . 3 5 6

(2)

x 1 y  2 z  3   . 5 3 6

x 1 y  2 z  3 x3 y 2 z 2     . (4) . 3 2 5 1 2 3 9. Equation of line passing through the origin and (5, -2, 3) is (3)

(1)

x y z   . 6 2 3

(2)

x y z   . 9 2 3

(3)

x y z   . 5 2 3

x y z (4) = = . 5 2 3

10. Equation of plane passing through the point (1, 2, 3) and perpendicular to the line with direction ratio’s 2, 3, −1 is (1) 2  x  1  3  y  2   1 z  3  0. (2) 2  x  1  3  y  2   1 z  3  0. (3) 1 x  2   2  y  3  3  z  1  0. (4) 2  x  1  3  y  2   1 z  3  0. 11. Equation of plane with intercepts 2, 3 and 4 on X, Y and Z-axes respectively is (1) 2 x  3 y  4 z  1. (2) 6 x  4 y  3 z  12. (3) x  y  z  1.

(4) 2 x  3 y  4 z  0.

12. The equation ax  by  c  0 represents a straight line (1) for all real numbers a, b and c. (2) only when b ≠ 0. (3) only when a ≠ 0. (4) only when at least one of a and b is non-zero.   13. Let the vectors a  i  3 j  2k , b  2i  j  k and  c  3i  5 j  2 k be coplanar. Then λ is equal to (1) −1. (2) 1. (3) −2. (4) 2. 14. The equation of a plane containing the line of intersection of the planes 2x − y − 4 = 0 and y + 2z − 4 = 0 and passing through the point (1, 1, 0) is (1) x  3 y  z  4.

(2) 2 x  z  2.

(3) x  3 y  2 z  2.

(4) x  y  z  0.

15. The plane through the intersection of the planes x  y  z  1 and 2 x  3 y  z  4  0 and parallel to y-axis also passes through the point (1)  3, 0,  1 . (2)  3, 1, 1 . (3)  3, 3,  1.

(4)  3, 2, 1.

139

THREE-DIMENSIONAL GEOMETRY

16. The vector equation of the plane through the line of intersection of the planes x  y  z  1 and 2 x  3 y  4 z  5 which is perpendicular to the plane x  y  z  0 is   (1) r  i  k  2  0. (2) r  i  k  2  0.   (3) r  i  k  2  0. (4) r  i  k  2  0.

   

   

17. The plane passing through the point  4,  1, 2  and parallel to the lines

x  2 y  2 z 1 x2 y 3 z 4     and 1 3 2 1 2 3

also passes through the point (1) 1, 1,  1. (2) 1, 1, 1. (3)  1,  1,  1 .

18. The plane which bisects the line segment joining the points  3,  3, 4  and  3, 7, 6  at right angles, passes through which one of the following points? (1) (−2, 3, 5) (2) (4, −1, 7) (3) (2, 1, 3) (4) (4, 1, −2) 1 and 3

2 units from the planes 4 x  2 y  4 z    0 and 3 2 x  y  2 z    0 , respectively, then the maximum value of    is equal to (1) 22. (2) 20. (3) 18. (4) 24. 20. The direction ratios of normal to the plane through the π points  0,  1, 0  and  0, 0, 1 and making an angle with 4 the plane y  z  5  0 (3)

(4) 2 3 , 1, − 1.

2 , 1, − 1. (2) β .

(3)    . (4)

2

2

  .

 2 3 6  23. The distance of the plane r   i  j  k   1 from the 7 7 7   origin is 1 (1) 1. (2) 7. (3) . (4) None of these. 7  [NCERT Exemplar Pg 238 Q31] 24. The sine of the angle between the straight line x2 y 3 z 4   and the plane 2 x  2 y  z  5 is 3 4 5 10 4 (1) . (2) . 6 5 5 2 (3) 

2 3

.

(4)  ,  ,  .



[NCERT Exemplar Pg 238 Q33]

26. The area of the quadrilateral ABCD, where A(0, 4, 1), B  2, 3,  1 , C  4, 5, 0  and D  2, 6, 2 , is equal to  (1) 9 sq. units. (3) 27 sq. units. (1) (2) (3) (4)

[NCERT Exemplar Pg 238 Q34] (2) 18 sq. units. (4) 81 sq. units.

A pair of perpendicular lines. A pair of parallel lines. A pair of parallel planes. A pair of perpendicular planes.

(1)

3 . 2

(2)

2 (4) . 10 [NCERT Exemplar Pg 238 Q32]

2 . 3

2 (3) . 7

3 (4) . 7





 29. The angle between the planes r  2i  3 j  k  1 and  [NCERT Exemplar Pg 238 Q35] r  i  j  4 is

 

-1 (1) cos

5

.

-1 (2) sin

5 7 . 2

-1 (4) sin

2 7

5 2 7

.

5 7 . 2  30. The angle between the lines r  3i  2 j  6k  -1 (3) cos



 [NCERT Exemplar Pg 238 Q29] 22. If the directions cosines of a line are k, k, k then (1) k > 0. (2) 0 < k < 1. 1 1 (3) k = 1. (4) k = or . 3 3  [NCERT Exemplar Pg 238 Q30]

5

(3)   ,   ,  .

 2i  j  2k

21. Distance of the point  ,  ,   from y-axis is (1) β .

(2)  0, 0,  .

1 28. The plane 2 x  3 y  6 z  11  0 makes an angle sin   with X-axis. The value of α is equal to

19. If the plane 2 x  y  2 z  6  0 has the distances

(2) 2, 2 , − 2 .

(1)  ,  ,0 .

27. The locus represented by xy  yz  0 is

(4)  1,  1, 1.

(1) 2, − 1, 1.

25. The reflection of the point  ,  ,   in the xy-plane is





 



 and r  2 j  5k   6i  3 j  2k is.

-1 (1) cos

21 . 19

-1 (2) sin

19 . 21

-1 (3) cos

19 . 21

-1 (4) sin

21 . 19

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices. (1) Both A and R are true, and R is the correct explanation of A. (2) Both A and R are true, but R is not the correct explanation of A. (3) A is true but R is false. (4) A is false and R is true. 1. Assertion (A): The vector equation of a line passing through the points (−1, 2, 7) and (4, 6, 11) is

 i  2 j  7k    5i  4 j  4k .

Reason (R): The vector equation of a line passing through       the points A and B is r  a   b  a , where a and b .





be the position vectors of points A and B, respectively. 2. Assertion (A): For every point P  x, y, z  on XY plane, Z-component is zero.

140 Oswaal CUET (UG) Chapterwise Question Bank Reason (R): The XY plane is a two-dimensional plane and the cartesian co-ordinates of all the point on this plane, z is always zero. 3. Assertion (A): Let P be a point on the line joining the points and B  4,  2, 1. If X-coordinate of P is 3, then its



1 . 4 Reason (R): The equation of line passing through two Y-coordinate is -

points A  x1, y1, z1  and B  x2 , y2 , z2  is given by x  x1 y  y1 z  z1 .   x2  x1 y2  y1 z2  z1

4. Assertion (A): Two lines

x  x1 y  y1 z  z1   and a1 b1 c1

x  x2 y  y2 z  z2 a1 b1 c1   are parallel, = = . a2 b2 c2 a2 b2 c2 Reason (R): Two lines are always be parallel. When the ratio of their direction ratios are equal. 5. Assertion (A): If two lines are in the same plane, i.e., they are coplanar, they will intersect each other if they are nonparallel. Hence the shortest distance between them is zero. If the lines are parallel then the shortest distance between them will be the perpendicular distance between the lines i.e., the length of the perpendicular drawn from a point on one line onto the other line. Reason (R): The angle between the lines with direction ratio  a1, b1, c1  and  a2 , b2 , c2  is given by: cos  

MATHEMATICS/APP. MATH

13 13 , ,13 units. 5 2 10. Assertion (A): Distance of a point with position vector a         from a plane r  N  d is given by a  N  d . ∴ Length of intercepts are

Reason: The length of perpendicular from origin O to the  d     plane r  N  d is   . N

[C] COMPETENCY BASED QUESTIONS I. R  ead the below passage and solve the questions from passage: A ball is thrown upwards from the plane surface of the ground. Suppose the plane surface from which the ball is thrown also consists of the points A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1) on it. The highest point of the ball takes, is D(2, 3, 1) as shown in the figure. Using this information answer the questions. D

(2, 3, 1)

B (3, –1, 1)

a1a2  b1b2  c1c2 a12

A (1, 0, 2)

 b12  c12 a22  b22  c22

6. Assertion (A): Direction cosines of a line are the sines of the angles made by the line with the negative directions of the coordinate axes. Reason (R): The acute angle between the lines x  2  0 

and 3 x  y  2  0 is 30 . 7. Assertion (A): P is a point on the line segment joining the points (3, 2, –1) and (6, –4, –2). If x coordinate of P is 5, then its y coordinate is –2. Reason (R): The two lines x  ay  b, z  cy  d and x  ay  b, z  cy  d  will be perpendicular, iff a′ bb  cc  0 . 8. Assertion (A): The angle between the straight lines x 1 y  2 z  3 x 1 y  2 z  3      and is 90 . 3 2 5 4 1 2 Reason (R): Skew lines are lines in different planes which are parallel and intersecting. 9. Assertion (A): The length of the intercepts on the co-ordinate 13 13 axes made by the plane 5 x  2 y  z  13  0 are , ,13 5 2 units. Reason (R): Given, equation of plane 5 x  2 y  z  13  0  5 x  2 y  z  13 5x  2 y  z 1  13 x y z    1 13 13 13 5 2 1

C (1, 2, 1)

1. The equation of the plane passing through the points A, B and C is (1) 3 x  2 y  4 z  11.

(2) 3 x  2 y  4 z  11.

(3) 3 x  2 y  4 z  11.

(4) 3 x  2 y  4 z  11.

2. The maximum heights of the ball from the ground is 5 7 (1) units. (2) units. 29 29 6 8 units. (4) units. 29 29 3. The equation of the perpendicular line drawn from the maximum height of the ball to the ground, is x 1 y  3 z  5 x  2 y  3 z 1 (1) . (2) .     2 1 2 3 2 4 (3)

x  2 y  3 z 1 x 1 y  3 z  5 . (4) .     2 1 3 2 4 2 4. The co-ordinates of the foot of the perpendicular drawn from the maximum height of the ball to the ground are  9 1 10   43 77 9  ,  . (1)  , (2)  , , . 7 7 7   29 29 29  (3)

 43 77 9  (3)  , ,  .  29 29 29  5. The area of ∆ABC

7 19   13 (4)   ,  ,   .  29 29 29 

(1)

29 sq. units.

(2)

(3)

1 29 sq. units. 16

1 29 sq. units. 4 1 29 sq. units. (4) 2

141

THREE-DIMENSIONAL GEOMETRY

II. R  ead the following text and answer the following questions: The Indian coast guard, while patrolling, saw a suspicious boat with people. They were nowhere looking like fishermen. The coast guards were closely observing the movement of the boat for an opportunity to seize the boat. They observed that the boat is moving along a planar surface. At an instant of time, the coordinates of the position of the coast guard helicopter and the boat is (1, 3, 5) and (2, 5, 3), respectively.

8. If the coast guard decides to shoot the boat at that given instant of time, when the speed of bullet is 36 m/s, then what is the time taken for the bullet to travel and hit the boat? 1 1 (1) seconds (2) seconds 8 14 (3)

1 seconds 10

(3) x  2 y  2 z  6.

(4) x  2 y  2 z  6.

7. If the coast guard decides to shoot the boat at that given instant of time, then what is the distance (in meters) that the bullet has to travel? (1) 5m (2) 3m (3) 6m (4) 4m

1 seconds 12

9. At that given instant of time, the equation of line passing through the positions of the helicopter and boat is x 1 y  3 z  5 x 1 y  3 z  5 (1) . (2) .     1 2 2 2 1 2 (3)

6. If the line joining the positions of the helicopter and the boat is perpendicular to the plane in which the boat moves, then the equation of the plane is (1)  x  2 y  2 z  6. (2) x  2 y  2 z  6.

(4)

x 1 y  3 z  5 .   2 1 2

(4)

x 1 y  3 z  5 .   1 2 2

10. At a different instant of time, the boat moves to a different position along the planar surface. What should be the coordinates of the location of the boat if the coast guard shoots the bullet along the line whose equation is x y 1 z  2   for the bullet to hit the boat? 1 2 1  8 19 14  (1)  , ,   3 3 3 

 8 19 14  , (2)  ,  3  3 3

 8 19 14  ,  (3)  , 3 3 3 

(4) None of these

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (1)

3. (4)

4. (1)

5. (3)

6. (2)

7. (1)

8. (2)

9. (3)

10. (4)

11. (2)

12. (4)

13. (4)

14. (4)

15. (4)

16. (3)

17. (2)

18. (4)

19. (1)

20. (2)

21. (4)

22. (4)

23. (1)

24. (4)

25. (4)

26. (1)

27. (4)

28. (3)

29. (1)

30. (3)

8. (3)

9. (1)

10. (4)

8. (4)

9. (1)

10. (4)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (1)

3. (1)

4. (1)

5. (2)

6. (4)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (1)

3. (3)

4. (3)

5. (4)

6. (3)

7. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (2) is correct. Explanation: We know that the angle q between the line x  x1 y  y1 z  z1 and plane a2x + b2y + c2z + d = 0 is,   a1 b1 c1 a1a2  b1b2  c1c2 2 a a2  b2 b 2 c c 2 2 sin   a1  b11 2 c11 2a2 1 b22  c2 2 2 2 2 2 b13can  cz1be  c22 x  2 aof1 yline  awritten 25 b2 as Equation sin  

 

 

 y 2 3 z  6 5    6 3 2 and equation of plane is 2x + 10y - 11z - 5 = 0 Here, a1 = 3, b1 = 2, c1 = 6 and a2 = 2, b2 = 10, c2 = -11, Now, angle between the line and plane x  3 2 

sin  

 3 2    2 10    6   11 (3)  (2) 2  (6) 2 (2) 2  (10) 2  (11) 2 2

6  20  66 40  49 225 7  15 8 sin   21  8    sin 1    21  2. Option (1) is correct. Explanation: If α, β and g are the angles made by the line with the X, Y and Z axes respectively, then sin  

cos 2  cos 2   cos 2  1 Therefore, cos 2 90  cos 2 60  cos 2 q  1

142 Oswaal CUET (UG) Chapterwise Question Bank 0

   

1  cos 2   1 4 3 cos 2   4 3 cos   2  π   30  6

3. Option (4) is correct. Explanation: Let x -1 y - 2 z - 3 = = =r 2 3 4 ⇒ x = 2r + 1, y = 3r + 2, z = 4r + 3 ...(i) x - 4 y -1 z - 0 Line 2: = = 2 5 1 Substitute value of x, y and z from eq. (i) to line 2, we get 2r  1  4 3r  2  1 = = 4r + 3 2 5 Equation first and last term, 2r  1  4 = 4r + 3 5 ⇒ 2r – 3 = 20r + 15 ⇒   18r = –18 ⇒     r = –1 Required point of intersection, from eq. (i) x = –1, y = –1 and z = –1 4. Option (1) is correct. Explanation: Let the point of intersection of line x3 y 4 x5   1 2 2 x  y  z 17 and the plane be  x0 , y0 , z0  As  x0 , y0 , z0  is the Line 1:

point of intersection of the line and plane, so it must satisfy both of the equations of line and plane. x  3 y0  4 z0  3    k (let) i.e., 0 1 2 2 or, x0  k  3, y0  2k  4, z0  2k  5 Put these values in equation of plane, we get

 k  3   2k  4    2k  5  17 5k  12  17 5k  5 k 1 Thus, required point of intersection is= x0 4= , y0 6, z0 = 7 Now, distance between (3, 4, 5) and (4, 6, 7) is d  (3  4) 2  (4  6) 2  (5  7) 2   1  4  4 = 3 units 5. Option (3) is correct. Explanation: Any plane passing through (0, 7, –7) is a(x – 0) + b(y – 7) + c(z + 7) = 0 ...(i) If plane (i) contains thee given line, then it must passes through the point (–1, 3, –2) and must be parallel to the given line. If (i) passes through (–1, 3, –2), then a(–1 – 0) + b(3 – 7) + c(–2 + 7) = 0

MATHEMATICS/APP. MATH

a + 4b – 5c = 0  If (i) is parallel to the given line, then (–3)a + 2.b + 1.c = 0 –3a + 2b + c = 0  Solving eqs. (ii) and (iii), we get a b c   4  10 15  1 2  12 a b c   14 14 14 a b c   k 1 1 1 a  k , b  k and c  k substituting these values in eq. (i), we get

...(ii)

…(iii)

k  x  0  k  y  7   k  z  7   0

kx + ky + kz – 7k + 7k = 0    kx + ky + kz = 0      k(x + y + z) = 0 or   (x + y + z) = 0 6. Option (2) is correct.



Explanation: The direction ratios of given line are i, 2 j ,  k ATQ, given line is perpendicular to the plane. ⇒ The direction ratios of plane are also i, 2 j ,  k







\ The equation of plane passing through (4, 3, 9) is a  x  4   b  y  3  c  z  9   0 1  x  4   2  y  3  1 z  9   0 x  2y  z  4  6  9  0 x  2 y  z  1 ...(i)



x  2y  z 1  0



x  2 y  3 z 1 k   1 2 1 x  k  2, y  2k  3, z   k  1 ∴ Put the values of x, y and z in eq. (i), we get Let

 

 k  2   2  2k  3   k  1  1

6k  7  1 k  1 Now,   x  k  2  1  2  1 y  2k  3  2  1  3  1 z   k  1    1  1  2

 x, y, z   1, 1, 2 

Let N  x2 , y2 , z2  be the foot of perpendicular from a given point  x1, y1, z1  to the given plane x  2 y  z  1  0 Now, foot of the perpendicular N  x2 , y2 , z2   1, 1, 2  S (x1, y1, z1)

N (x2, y2, z2)

ax+by+cz+d=0 S' (2, 3, 1)

Since, Point N  x2 , y2 , z2  is midpoint of point S  x1, y1, z1  and point S   x3 , y3 , z3 , x2 

x1  x3 2

143

THREE-DIMENSIONAL GEOMETRY

x1  2 2 x = 0   1 y  y3   y2  1 2 y1  3  1  2 y1  1 z z z2  1 3 2 z1  1   2 2

x y z   1 a b c x y z   1 2 3 4 6 x  4 y  3z 1 12 6 x  4 y  3 z  12 12. Option (4) is correct. Explanation: Any equation of the form ax + by + c = 0 will represent a straight line on the xy-plane at least one of a and b is non-zero. 13. Option (4) is correct.    Explanation: a  i  3 j  2k , b  2i  j  k and c  3i  5 j  2 k

  z1 = 3

are coplanar.

 1 

Coordinates of point S are  0,  1, 3 7. Option (1) is correct. Explanation: Equation of line passing through a point (1, 2, 3) and parallel to the vector 3i  2 j  2k is x 1 y  2 z  3   . 3 2 2 8. Option (2) is correct. Explanation: Equation of line passing through (1, 2, 3) and parallel to the line given by x   3 3



y 4 z 8  is 5 6

x  x1 y  y1 z  z1     a1 b1 c1 = y1 2, z1 = 3   x1 1,=

Here,

a1  3, b1  5, c1  6 \ Equation of required line  is

x 1 y  2 z  3   5 3 6

9. Option (3) is correct. Explanation: Equation of a line, passing through the origin  x1, y1, z1    0, 0, 0  and  x2 , y2 , z2    5,  2, 3 is x  x1 y  y1 z  z1   x2  x1 y2  y1 z2  z1 x0 y0 z0   5  0 2  0 3  0 x y z     5 2 3 10. Option (4) is correct. Explanation: Equation of plane passing through the point  x1, y1, z1   1, 2, 3 and perpendicular to the line with direction ratios  a, b, c    2, 3, 1 is a  x  x1   b  y  y1   c  z  z1   0 2  x  1  3  y  2    1  z  3  0 2  x  1  3  y  2    z  3  0 11. Option (2) is correct. Explanation: Equation of plane with intercepts  a, b, c    2, 3, 4  on x, y, z axis is

 [a b c] = 0 1 3 2 1  0  2 1 3 5 2

1(2  5)  3(4  3)  2(10  3)  0 2  5  12  9  14  0   2  5  12  23  0 14  28  0 14  28   2 14. Option (4) is correct. Explanation: Required equation is:

 2x  y  4    y  2 z  4  0

Since, it passes through the point (1, 1, 0). ∴ (2 − 1 −4) + λ (1 + 0 −4) = 0    1 Plane  2 x  y  4    y  2 z  4   0 x yz 0 15. Option (4) is correct. Explanation: Required equation of plane is

 x  y  z  1  λ  2 x  3 y  z  4   0 x 1  2λ   y 1  3λ   z 1  λ   1  4λ  0

But the above plane is parallel to y-axis 1  3λthen  0 its normal plane is perpendicular to y-axis. 1 λ ⇒ (1  x 2y)iz(1 13λ) j 2x(13y )k z( 04ij030 k )  0 3λ1x31z)101λ x 1[(1 2λ2) y 01]  [( ⇒ ] z[( 11447λ) 000]  0  3 13  33λ  0 x  4z  7  0 1 λ 3 1 4 7 x 0 z   0 3 3 3 x  4z  7  0 Here point (3, 2, 1) satisfies this equation. 16. Option (3) is correct. Explanation: Plane passing through intersection of x  y  z  1 and 2 x  3 y  4 z  5



is



( x  y  z  1)   (2 x  3 y  4 z  5)  0 x(1  2 )  y (1  3 )  z (1  4 )  1  5 = 0

144 Oswaal CUET (UG) Chapterwise Question Bank Given that above plane is perpendicular to plane x  y  z = 0. (1  2 )  1  (1  3 )(1)  (1  4 )  1 = 0 1 ⇒   λ 3 So equation of plane is 5  4  2 x 1    y (1  1)  z 1    1   0 3  3  3 2 x z    0y    0 3 3 3  r  (i  k )  2  0

Distance,

∴

17. Option (2) is correct. Explanation: Plane passing through  4,  1, 2  a  x  4   b  y  1  c  z  2   0

Parallel to the lines x  2 y  2 z  1 ...(i)     3 1 2 x2 y 3 z 4 and   1 2 3 ∴ 3a  b  2c  0 a  2b  3c  0 a b c   3  4 2  9 6  1 a b c    1 1 1 Substituting in (i), 

1 x  4   1 y  1  1 z  2   0 x  y  z 1  0 By option it also passes (1, 1, 1). 18. Option (4) is correct. Explanation: Mid-point (0, 2, 5) D’r of line (6, 10, 2) So equation of plane is

6  x  0   10  y  2   2  z  5   0 6 x  10 y  2 z  30  0 By option, (4, 1, -2) satisfy. 19. Option (1) is correct. Explanation:  yy   22 zz   66   00   22 xx  ...(i)  22 yy   44 zz    0   44 xx  ...(ii)  0 As the planes (i) and (ii) are parallel and distance between them 1 is equal to . 3 d1  d 2d1  d 2 Distance = Distance = 2 a  b 2a 2 c 2b 2  c 2   6 61 2 1 2   3 4  1  44 1  34      6  16  1 2 2   14, 10    14,10 2 x  y 2x2z y 6 2z0 6  0 ...(i) Similarly, 2 x  y 2x2z y  2z 0   0 ...(iii) As the planes (i) and (ii) are parallel and distance between them 2 is equal to . 3

MATHEMATICS/APP. MATH

  ⇒

 

 6 2  3 4 1 4   6  2   8, 4     6  2

⇒ Max. value of     14  8 = 22 20. Option (3) is correct. Explanation: Equation of plane passing (0, -1, 0) is a  x  0   b  y  1  c  z  0   0 ax  by  cz  b  0 Plane also passes through (0, 0, 1) ⇒     0+0+c+b=0 ⇒   c = -b  π Making angle with y  z  5  0 4  a  0  b  1  c  1 cos   4 2 a 2  b2  c2

...(i)

 a 2  b 2  c 2  (b  c) 2 From2 (i), 2 a  b  b 2  (2b) 2    a2 + b2 + b2 = (2b)2  a 22 =  22b b22 ⇒  a  ⇒

a   2b

Equation of plane when a = 2b is 2b  x  0   b  y  1  b  z  0   0 ⇒

2x  y 1 z  0

 

2x  y  z 1  0 d rs 2 ,1, 1, option  3 , When a   2b So, direction ratios are: ⇒

 2b  x  0   b  y  1  b  z  0   0  2x  y  z 1  0 2x  y  z 1  0

So, direction ratios are:

2 , − 1, 1

21. Option (4) is correct. Explanation: The foot of perpendicular from point P  ,  ,   on y-axis is Q(0, b, 0). \ Required distance, PQ  (  0) 2  (    ) 2  (  0) 2   2   2 22. Option (4) is correct. Explanation: Since, direction cosines of a line are k, k and k.  l  k , m  k and n = k We know that, l 22  m 22  n 22  1  k  k  k 1  k k22+kk22 +kk22  ⇒ = 11 1  k 22  1 ⇒    k 3 3 1 \ k  1   k  3 3

145

THREE-DIMENSIONAL GEOMETRY

23. Option (1) is correct. Explanation:

 2 3 6  The distance of the plane r   i  j  k   1 form the 7 7  7 origin is 1.  [Since r  n  d is the form of above equation, where d represents the distance of plane from the origin, i.e., d = 1] 24. Option (4) is correct. Explanation: We have, the equation of line as x2 y 3 z 4   3 4 5  This line is parallel to the vector, b  3i  4 j  5k . Equation of plane is 2 x  2 y  z  5. Normal to the plane is n  2i  2 j  k . Let angle between line and plane is ‘q’. Then,   3i  4 j  5k  2i  2 j  k b n sin      b n 32  42  52 4  4  1





685 50 9







3 1  15 2 5 2

2 10 25. Option (4) is correct. 26. Option (1) is correct Explanation:

Clearly yz = 0 represents the plane passing through the origin and perpendicular to X-axis. Hence xy + yz = 0 represents the locus a pair of perpendicular planes. 28. Option (3) is correct. Explanation: We have equation of plane as 2 x  3 y  6 z  11  0.  Normal to the plane is n  2i  3 j  6k .  Also X-axis is along the vector a  i  0i  0k . According to the question,   an sin    a n 



i  2i  3 j  6k 1 4  9  36

 2 7

29. Option (1) is correct. Explanation: Normal to the plane   r  2i  3 j  k  1 is n1  2i  3 j  k





 

  Normal to the plane r  i  j  4 is n2  i  j

sin  

\ Angle between the planes is given by   n n cos   1 2 n1 n2

We have, A  0, 4, 1 , B  2, 3,  1 , C  4, 5, 0  and D  2, 6, 2 

 cos  

 AB   2  0  i   3  4  j   1  1 k

 2i  j  2k  BC   4  2  i   5  3 j   0  1 k  2i  2 j  k  CD   2  4  i   6  5  j   2  0  k  2i  j  2k  DA   0  2  i   4  6  i  1  2  k  2i  2 j  k Thus quadrilateral formed is parallelogram. \ Area of quadrilateral ABCD      AB  BC i j k    2 1 2 2 2 1

4  9 1 11 23 5  cos    14 2 2 7  5  \    cos 1   2 7  30. Option (3) is correct. Explanation:  We have line, r  3i  2 j  6k   2i  i  2i , which is parallel to   the vector b1  2i  j  2k and r  2 j  5k   6i  3 j  2k ,  which is parallel to the vector b2  6i  3 j  2k . If q is an angle between the lines, then   b1  b2 cos    b1 b2



and perpendicular to Z-axis.





 

2i  j  2k    6i  3 j  2k    2i  j  2k 6i  3 j  2k

   3i  6 j  6k    9  36  36  = 9 sq. units 27. Option (4) is correct. Explanation: Given equation is xy  yz  0 Clearly xy = 0 represents the plane passing through the origin

 2i  3j  k   i  j 



\

12  3  4

9 49 19   cos 1 21



19 21

[B] ASSERTION REASON QUESTIONS 1. Option (1) is correct. Explanation: Here,   a  i  2 j  7 k and b  4i  6 j  11k



 tan  

1  m1  m2

m2 1 m1  tan   146 Oswaal CUET (UG) Chapterwise 1 Question Bank MATHEMATICS/APP. MATH  m2 m1 \ Vector equation is: 1  tan   3 i  2 j  7 k    4  1 i   6  2  j  11  7  k       30  i  2 j  7 k   5i  4 j  4k 7. Option (3) is correct. 2. Option (1) is correct. Explanation: Assertion (A) is correct. Explanation: Since P   5, y, z  Y Equation of line joining  3, 2,  1 and  6,  4,  2  is P=(x, y, 0)







 



x3 y2 z 1 x  3 y  2 z 1      3 6 1 6  3 4  2 2  1

X

x3 y2 z 1 x  3 y  2 z 1      6 1 6  3 4  2 2  1 3

Z 3. Option (1) is correct. Explanation: Equation of line passing through the points A  0, 5,  4  and B  4,  2, 1 is

x0 y 5 z4   r 4  0 2  5 1  4 x y 5 z 4   r 7 4 5 Given, X-coordinate of P is 3. 3  4r  3  r  \ 4 or

Therefore, Y-coordinate  7 r  5 3 5 4 21  20 1   4 4  7 

So if point P lies on the line then it must satisfy the above equation 5  3 y  2 z 1   6 1 3 53 y 2  6 3 Hence y co-ordinate of P is –2 . Reason (R) is false. Since, the two lines x  ay  b, z  cy  d and  ay  b, z  cy  d   ay  b, z  cy  d  will be perpendicular, if aa  cc  1  0. 8. Option (3) is correct. Explanation: Assertion (A) is correct. x 1 y  2 z  3   Given: 4 2 5 x 1 y  2 z  3 and   3 1 2 a1 2= , b1 5, c1 = 4 and a2  1, b2  2, c2  3 Direction ratios of lines are=

4. Option (1) is correct. a2  1, b2  2, c2  3 Explanation: If two lines have direction ratios a1, b1, c1 and As we know, the angle between the lines is given by a2 , b2 , c2, respectively. a1a2  b1b2  c1c2 cos   a1 b1 c1  a 2  b2  c2    a 2  b2  c2  Then, condition for parallel lines: = . =  1 1 1   2 2 2  a2 b2 c2     5. Option (2) is correct. 2  1  5  2  4  3  cos   Explanation: Assertion (A) and Reason (R) both are  22  52  42    12  22  (3) 2      individually correct.     6. Option (4) is correct.   =0 Explanation: Assertion (A) is wrong. \      90 Since, direction cosines of a line are the cosines of the angles Reason (R) is wrong. made by the line with the positive directions of the coordinate In the space, the lines that are neither intersecting nor parallel, axes. are called non-coplanar or skew lines. Reason (R) is correct. 9. Option (1) is correct. Since, the slope of the line x - 2 = 0 is ∞. Explanation: Assertion (A) and Reason (R) both are correct The slope of line 3 x  y  2  0 is 3 . and Reason (R) is the correct explanation of Assertion (A). 10. Option (4) is correct. Let m1  , m2  3 and the angle between the given lines is q. Explanation:    m2  m1 For a plane whose equation is given by r .N = D, the distance of  tan    1  m1  m2 a point A whose position vector is given by a to the plane is m2 given by 1 m1     tan   aN D 1     m2 N m1  tan   

1 3

  30

147

THREE-DIMENSIONAL GEOMETRY

[C] COMPETENCY BASED QUESTIONS 1. Option (2) is correct. Explanation: The equation of plane passing through three noncollinear points is given by x  x1 x2  x1 x3  x1

y  y1 y2  y1 y3  y1

z  z1 z2  z1  0 z3  z1

⇒

x 1 y  0 z  2  3  1 1  0 1  2  0 11 2  0 1 2

⇒

x 1 y z  2   2 1 1  0 0 2 1

5. Option (4) is correct. Explanation: Clearly area of DABC 

1   AB  AC 2



1    2i  j  k  0i  2 j  k 2





1 i 1  2   j  2  0   k  4  0  2



1  3i  2 j  4k 2

 3  x  1  2 y  4  z  2   0



1 2 9  22  42 2

3 x  2 y  4 z  11  0 3 x  2 y  4 z  11

=

1 29 sq. units 2

 

2. Option (1) is correct. Explanation: Height of the ball = Perpendicular distance from point (2,3,1) to the plane 3x + 2y + 4z =11 So,

6  6  4  11



5 5 units  29 29

3 2 4 3. Option (3) is correct. Explanation: D.R.s of perpendicular are  3, 2, 4  [Since, perpendicular is parallel to the normal to the plane] Since, perpendicular is passing through the point (2, 3, 1), therefore its equation is x  2 y  3 z 1   3 2 4 4. Option (3) is correct. Explanation: Let the coordinate of foot of perpendicular are  3λ  2, 2λ  3, 4λ  1 2

2

2

Since, this point lie on the plane 3 x  2 y  4 z  11, therefore we get 3  3  2   2  2  3  4  4  1  11   9  6  4  6  16  4  11 29  16  11 29  11  16 5  29 Thus, coordinates of foot of perpendicular are   5     5     5    3     2  ,  2     3  ,  4     1     29     29     29     15   20     10  1   3 ,    2 , 29 29     29     43 77 9   , ,   29 29 29 



i j k 1  2 1 1 2 0 2 1

  x  1 1  2   y  2  0    z  2   4  0   0 ⇒

 

6. Option (3) is correct.

Explanation: Let P 1, 3, 5  and Q  2, 5, 3 be the positions of helicopter and boat respectively. Now, direction ratios of PQ are proportional to 2–1, 5–3, 3–5, i.e., 1, 2, –2. So, equation of plane passing through Q  2, 5, 3 and perpendicular to PQ is 1 x  2   2  y  5    2   z  3  0 ⇒

 x  2 y  2z  6

7. Option (2) is correct. Explanation:     n  i  2 j  2k

Distance that the bullet has to travel  1 4  4  3m 8. Option (4) is correct. Explanation: Time taken for the bullet to travel and hit the boat 3m 36 m/s 1 = seconds 12 =

9. Option (1) is correct. Explanation: Here, direction cosines are 1, 2, –2. Equation of line passing through the positions of the helicopter and boat is x  x1 y  y1 z  z1   l m n x 1 y  3 z  5   2 1 2 

x 1 y  3 z  5   2 1 2

148 Oswaal CUET (UG) Chapterwise Question Bank 10. Option (4) is correct.

3  8

x y 1 z  2  Explanation: Any point on the line  given by 1 2 1



 λ , 2λ  1, λ  2 . Now, on substituting this point in the equation of plane x  2 y  2 z  6, we get   2  2  1  2    2   6

MATHEMATICS/APP. MATH

8 3

8  8  8  Thus, the required point is  , 2    1,    2  , i.e., 3 3 3      8 19 14   , , . 3 3 3 

Course of Action Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

12   Revision Notes

LINEAR PROGRAMMING

 A. Introduction and Related Terminology:  Linear Programming (LP) is a mathematical optimization technique used to find the best outcome in a mathematical model with linear relationships. Before diving into LP problems, it’s important to familiarize yourself with some key terminology: 1. Decision Variables: These are the unknown quantities you want to find in an LP problem. They represent the choices or decisions you need to make. 2. Objective Function: A linear function in the form Z = ax + by.  a and b: Constants in the linear function.  Minimized or Maximized: The goal is to either minimize or maximize this linear function.  Decision Variables: x and y are the variables that need to be determined to optimize the objective function. The example Z = 175x + 150y demonstrates a linear objective function where x and y are the decision variables. This function would be used in linear programming to make decisions that maximize or minimize some quantity (Z) while considering the constraints placed on x and y. 3. Constraints: Constraints are linear inequalities or equations that restrict the possible values for the decision variables. They represent limitations or restrictions in the problem. For example, 5x + y ≤ 100; x + y ≤ 60 are constraints. 4. Feasible Region: The feasible region is the set of all possible combinations of decision variables that satisfy all the constraints. It is usually depicted graphically as a polygon or a polyhedron. 5. Optimal Solution: The optimal solution is the point within the feasible region that maximizes or minimizes the objective function.  B. Mathematical Formulation of Linear Programming Problem: To formulate an LP problem mathematically, you need to define the following components: 1. Decision Variables: Define the decision variables and represent them with symbols (e.g., x1, x2). 2. Objective Function: Write down the objective function that needs to be maximized or minimized. 3. Constraints: Write down all the constraints as linear inequalities or equations. 4. Objective (Maximization/Minimization): Specify whether you want to maximize or minimize the objective function.  C. Different Types of Linear Programming Problems: There are several types of LP problems, including:

1. Linear Maximization Problem: The objective is to maximize the objective function while satisfying constraints. 2. Linear Minimization Problem: The objective is to minimize the objective function while satisfying constraints. 3. Transportation Problem: Involves optimizing the transportation of goods from multiple sources to multiple destinations while minimizing transportation costs. 4. Assignment Problem: Focuses on assigning a set of tasks or jobs to a set of workers or machines in a way that minimizes the total cost or time.  D. Graphical Method of Solution for Problems in Two Variables: For LP problems involving two variables (x and y), the graphical method is a visual approach to finding the optimal solution. Here are the steps: 1. Plot the constraints on a 2D graph to form a feasible region. 2. Identify the feasible region’s vertices (corner points). 3. Calculate the objective function’s value at each vertex. 4. Determine the vertex that maximizes or minimizes the objective function. This vertex represents the optimal solution.  E. Fundamental Theorems for Solving Linear Programming Theorem 1: Let R be the feasible region for a linear programming problem and let z = ax + by be the objective function. When z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities. This optimal value must occur at a corner point (vertex) of the feasible region. Theorem 2: Let R be the feasible region for a linear programming problem and let z = ax + by be the objective function. If R is bounded, then z has both a maximum and a minimum value on R and each of these recurs at a corner point of R. Example 1. Given linear programming problem is Minimize Z = 200 x + 500 y , subject to the constraints: x  2 y  10 3 x  4 y  24 x  0, y  0 Minimum value of Z is (1) 2300 (2) 2200 (3) 1200 Sol. Option (1) is correct. Explanation: Given objective function is: Minimize Z  200 x  500 y

(4) 2500

First Level

Second Level



Trace the Mind Map Third Level

Find the feasible region of the linear programming problem and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point. 1. Evaluate the objective function Z = ax+by at each corner point. Let M and m, respectively denote the largest and smallest values of these points. 2. (i) When the feasible region is bounded, M and m are the maximum and minimum values of Z. (ii) In case, the feasible region is unbounded, we have: (a) M is the maximum value of Z, if the open half plane determined by ax+by>M has no point in common with the feasible region. Otherwise, Z has no maximum value. (b) Similarly, m is the minimum value of Z, if the open half plane determined by ax+by 0. Then, the condition on p and (1) 50. (2) 30. (3) 25. (4) 60. q so that the maximum of Z occurs at both the points (15, 15) 5. The minimum value of the objective function Z  30 x  10 y and (0, 20), is subject to the constraints x + 2 y ≤ 30, 3 x  y  30, 4 x  3 y  60, x(1) (2) p = 2q (3) q = 2p (4) q = 3p , y  p0 = q 14. The feasible region for an LPP is shown in the given figure.  y  30, 4 x  3 y  60, x, y  0 is Bounded in first quadrant but has no solution. Unbounded in first quadrant but has a solution. Unbounded in first quadrant and has no solution. Bounded and has a solution= x 0= , y 0, Z = 0. The maximum value of Z = 3 x + y subject to the constraints x + y ≤ 30, 2 x + y ≤ 40, x, y ≥ 0 is

(1) 100

(2) 450

(3) 300

Y

(4) 1200

6. Match List I with List II. LIST I A.

B.

LIST II

The set of decision variables whose values do not satisfy all the constrains and nonnegativity condition of a LP is called

I.

In a LPP, the objective function is always

II.

B(0, 6)

Linear. C(0, 3)

C.

In a LPP, the linear inequalities on variable are called

III.

Infeasible solution.

D.

The feasible region for a LPP is always a

IV.

Constraints.

Choose the correct answer from the options given below: (1) A-I, B-III, C-IV, D-II (2) A-IV, B-III, C-I, D-II

E(9,0)

X-’

Convex polygon.

( )

9,3 A 2,2 D (6,0)

O

_6 x + y< Y-’

Then, the minimum value of Z = 11x + 7y is (1) 21. (2) 47. (3) 20. (4) 31. 15. The feasible solution for a LPP is shown in following figure. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at Y

(3) A-III, B-I, C-IV, D-II (4) A-III, B-I, C-II, D-IV 7. The corner points of the feasible region for a LPP are ( 0, 4 ) , ( 2, 3) , ( 4, 5) , ( 7, 0 ). If objective function is Z = px + qy; p, q > 0

(4,10) (6,8)

(0,8)

Z = px + qy; p, q > 0 then the condition on p and q so that the

(6,5)

minimum of Z occurs at (2, 3) and (7, 0) is (1) 7 p = 4q (2) 5 p = 3q (3) 4 p = q

(1) (2) (3) (4) 11.

There are two feasible regions. There are infinite feasible regions. There is no feasible region. None of the above Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q>0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is q (2) p = (3) p = 3q (4) p = q 2

(1) p = 2q

(0,0)

(4) 3 p = 3q

8. The maximum value of z = 4 x + 2 y subject to constraints 2 x + 3 y ≤ 28, x + y ≤ 10, x, y ≥ 0 is 100 . (1) 36. (2) 40. (3) (4) 32. 3 9. Corner points of the feasible region for an LPP, are (0, 2), (3, 0), (6, 0) and (6, 8). If Z = 2x + 3y is the objective function of LPP then max. (Z) - min. (Z) is equal to (1) 30. (2) 24. (3) 21. (4) 9. 10. For the constraint of a linear optimizing function z  x1  x2, given by x1 + x2 ≤ 1, 3 x1 + x2 ≥ 3 and x1, x2 ≥ 0.

X _9 x + 3y >

X

X' (5,0) Y'

(1) (0, 0). (2) (0, 8). (3) (5, 0). (4) (4, 10). 16. The feasible region (shaded) for a LPP is shown in following figure. Y

B

x+ D

2y =

76 E

X'

X O

A

C

2x + y = 104 Y'

Then, the maximum value of Z = 3x + 4y is (1) 156. (2) 196. (3) 152. (4) 180.

153

LINEAR PROGRAMMING

17. The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y. Compare the quantity in column A and column B. Column A

Column B

325 Maximum of Z The quantity in column A is greater. The quantity in column B is greater. The two quantities are equal. The relationship cannot be determined on the basis of the information supplied. 18. The feasible region (shaded) for a LPP is shown in following figure (1) (2) (3) (4)

Z = ax + 2by where a, b > 0. The condition on a and b such that the maximum Z occurs at Q and S is (1) a − 5b = 0 (2) a − 3b = 0 (3) a − 2b = 0 (4) a − 8b = 0 23. The corner points of the shaded unbounded feasible region of an LPP are (0, 4), (0.6, 16) and (3, 0) as shown in the figure. The minimum value of the objective function Z = 4 x + 6 y occurs at

A

B Q P

C R

O

S

Then, the maximum value of Z = x + 2y is (1) 1. (2) 5. (3) 9. (4) 8. 19. The linear programming problem minimize Z = 3x + 2y subject to constraints x + y ≥ 8, 3x + 5y ≤ 15, x ≥ 0 and y ≥ 0 has (1) one solution. (2) no feasible solution. (3) two solution. (4) infinitely many solution. 20. If constraints in a linear programming problem are changed (1) solution is not defined. (2) the objective function on has to be modified. (3) the problem is to be re-evaluated. (4) None of these 21. Determine the maximum value of Z = 4x + 3y if the feasible region for an LPP is shown in Fig. Y

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

(0.6, 1.6) only. (3, 0) only. (0.6, 1.6) and (3, 0) only. at every point of the line-segment joining the points (0.6, 1.6) and (3, 0). 24. The solution set of the inequality 3 x + 5 y < 4 is (1) an open half-plane not containing the origin. (2) an open half-plane containing the origin. (3) the whole XY-plane not containing the line 3x + 5y = 4. (4) a closed half plane containing the origin. 25. A Linear Programming Problem is as follows: Minimise Z  2 x  y Subject to the constraints x  3, x  9, y  0 x  y  0, x  y  14 The feasible region has (1) 5 corner points including  0, 0  and  9, 5  . (2) 5 corner points including  7, 7  and  3, 3 . (3) 5 corner points including 14, 0  and  9, 0  (4) 5 corner points including  3, 6  and  9, 5  26. The feasible region for an LPP is shown in the given figure. Let F = 3x – 4y be that objective function. Maximum value of F is Y

(0, 40) (0, 24) C

(12, 6)

B (16, 16) (0, 4)

X (48, 0) A (25, 0) (1) 100 (2) 111 (3) 121 (4) 112 22. The corner points of the feasible region determined by a set of constraints (linear inequalities) are P(0, 5), Q(3, 5), R(5, 0) and S(4, 1) and the objective function is O

X (6,0)

(1) 0. (2) 8. (3) 12. (4) -18. 27. A Linear Programming Problem is as follows: Maximise/Minimise objective function Z = 2x – y + 5

154 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH.

Subject to the constraints 3 x  4 y  60 x  3 y  30 x  0, y  0 In the corner points of the feasible region are A(0, 10), B(12, 6), C(20, 0) and O(0, 0), then which of the following is true? (1) Maximum value of Z is 40 (2) Minimum value of Z is -5 (3) Difference of maximum and minimum values of Z is 35 (4) At two corner points, value of Z are equal 28. Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. The minimum value of F occurs at (1) (0, 2) only (2) (3, 0) only (3) the mid-point of the line segment joining the points (0, 2) and (3, 0) only (4) any point on the line segment joining the points (0, 2) and (3, 0) 29. The corner points of the feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4), and (0, 5). If the maximum value of Z = ax + by, where a, b > 0 occurs at both (2, 4) and (4, 0), then (1) a = 2b. (2) 2a = b. (3) a = b (4) 3a = b



30. Z  7 x  y, subject to 5 x  y  5, x  y  3, x  0, y  0. The maximum value of Z occurs at 1 5 (1) (3, 0). (2)  ,  . 2 2



Assertion (A): The minimum value of Z is 15.



Reason (R): The maximum value of Z is 72.

(3) (7, 0).

(4) (0, 5).

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as (1) Both A and Rare true and R is the correct explanation of A. (2) Both A and R are true but R is not the correct explanation of A. (3) A is true but R is false. (4) A is false but R is true. 1. Assertion (A): The constraints of LPP given by x1  2 x2  2000

x1  x2  1500 x2  600 and

x1, x2 ≥ 0 The points (1000, 0), (0, 500), (2, 0) lie in the positive bounded region. Reason (R): The point (2000, 0) does not lie in the positive bounded region. 2. Assertion (A): The feasible solution of a LPP belongs to any quadrant. Reason (R): The feasible solution of a LPP belongs to only first quadrant. 3. Assertion (A): The graph of x ≤ 2 and y ≥ 2 will be situated in the first and second quadrants.

Reason (R): The graph of x ≤ 2 and y ≥ 2 will be situated in first and third quadrants. 4. Assertion (A): The vertices of a feasible region by the linear constraints 3 x  4 y  18 2x  3y  3 and x, y ≥ 0 are (0, 4.5), (6, 0) and (0, 0). Reason (R): Y (0, 4.5) 3x + 4y = 18 2x + 3y = 3 (0,1) O

( 3,0 2 )

(6, 0)

X

5. Assertion (A): Every LP problem has a unique solution.

Reason (R): If an LP problem has two optimal solutions, then it has infinitely many solutions.

6. For the LP problem Minimize Z = 2x + 3y the coordinates of the corner points of the bounded feasible region are A(3, 3), B(20, 3), C(20, 10), D(18, 12) and E(12, 12).

7. Corner points of the bounded feasible region for an LP problem are (0, 4), (6, 0), (12, 0), (12, 16) and (0, 10). Let Z = 8x + 12y be the objective function.

Assertion (A): Minimum value of Z occurs at (0, 4) and (6, 0).



Reason (R): Maximum value of Z occurs at (12, 0) and (12, 16).

8. Assertion (A): For the inequality 2x - 3y > -5, out of the points (1, 1), ( -1, 1), (1, -1), (-1, -1), (-2, 1), (2, -1), (-1, 2) and (-2, -1); only 5 points satisfy the inequality.

Reason (R): For the inequality 2x - 3y > -5, out of the points (1, 1), (-1, 1), (1, -1), (-1, -1), (-2, 1), (2, -1), (- 1, 2) and (-2, -1); only 4 points satisfy the inequality.

9. Assertion (A): Feasible region is the set of points which satisfy all of the given constraints and objective function too.

Reason (R): The optimal value of the objective function is attained at the points on X-axis only..

10. Corner point of feasible region are (0, 0), (3, 0) and (0, 3) and objective function Z = 4x + 7y. Assertion (A): Minimum value of Z is 12. Reason (R): Maximum value of Z is 21. Y B (0, 3) A O

(3, 0)

X

155

LINEAR PROGRAMMING

[C] COMPETENCY BASED QUESTIONS I.  Read the following text and answer the following questions on the basis of the same. An aeroplane can carry a maximum of 200 passengers. A profit of ` 1000 is made on each executive class ticket and a profit of ` 600 is made on each economy class ticket. The airline reserves at least 20 seats for the executive class. However, at least 4 times as many passengers prefer to travel by economy class, than by executive class. It is given that the number of executive class tickets is x and that of economy class tickets is y.

II. R  ead the following text and answer the following questions on the basis of the same. A manufacturer produces two Models of bikes Model X and Model Y. Model X takes a 6 man hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and Marketing costs are ` 2,000 and ` 1,000 per unit for Models X and Y respectively. The total funds available for these purposes are ` 80,000 per week. Profits per unit for Models X and Y are ` 1,000 and ` 500, respectively.

The feasible region of L.P.P. is shown in the following graph. Y D (0,80) B (0,45)

(25,30) (75,0)

O

C (40,0)

X A

6. The equation of line AB is 1. The maximum value of x + y is (1) 100.

(2) 200.

(3) 20.

(4) 80.

2. The relation between x and y is (1) x < y.

(2) y > 80. (3) x ≥ 4 y. (4) y ≥ 4 x.

3. Which among these is not a constraint for this LPP? (1) x ≥ 0 (2) x + y ≤ 200 (3) x ≥ 80 (4) 4 x − y ≤ 0 4. The profit when x = 20 and y = 80 is (1) ` 60,000. (3) ` 64,000. (1) 1,36,000. (3) 68,000.

(2) 2 x  y  80.

(3) 5 x  3 y  225.

(4) x  2 y  80.

(1) 3 x  5 y  225.

(2) 2 x  y  80.

(3) 5 x  3 y  225.

(4) x  2 y  80.

7. The equation of line CD is

8. The coordinates of point E are

(1) (25, 30). (2) (30, 25). (3) (25, 25). (4) (30, 30).

9. The constraints for L.P.P. are

(1) 3 x  5 y  225, 2 x  y  80, x  0, y  0.

(2) ` 68,000. (4) ` 1,36,000.

5. The maximum profit is ₹

(1) 3 x  5 y  225.

(2) 3 x  5 y  225, 2 x  y  80, x  0, y  0. (3) 3 x  5 y  225, 2 x  y  80, x  0, y  0. (4) 3 x  5 y  225, 2 x  y  80, x  0, y  0.

(2) 1,28,000. (4) 1,20,000.

10. How many bikes of model X and model Y should the manufacturer produce so as to yield a maximum profit? (1) (40, 0)

(2) (25, 30) (3) (25, 45) (4) (40, 25)

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (3)

3. (3)

4. (4)

5. (3)

6. (3)

7. (2)

8. (2)

9. (1)

10. (3)

11. (2)

12. (1)

13. (4)

14. (1)

15. (2)

16. (2)

17. (2)

18. (2)

19. (2)

20. (3)

21. (4)

22. (4)

23. (4)

24. (2)

25. (2)

26. (3)

27. (2)

28. (4)

29. (1)

30. (3)

8. (3)

9. (3)

10. (4)

8. (1)

9. (1)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (4)

3. (3)

4. (4)

5. (4)

6. (2)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (4)

3. (3)

4. (2)

5. (1)

6. (1)

7. (2)

156 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH.

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (2) is correct.

Y

Explanation: At ( x, y ) = (10,15 ) Z = 2x + 3y  

= 2 (10 ) + 3 (15 ) = 20 + 45 = 65

40

Here, objective function is Max. at x = 10 and y = 15 . 2. Option (3) is correct.

(0, 30) 30 (10, 20)

20

Explanation: Shaded region is bounded between four corner

10

points  0, 0  ,  40, 0  ,  30, 20  and  0, 50  . Hence,

(0, 40)

(20, 0) 0 (0,0)

x + y ≤ 50

2 x  y  80 x, y  0 are required equations.

X 10

20

30

2x + y = 40



(x, y)

3. Option (3) is correct. Explanation: WE have, Max Z = 3x + 2y s.t.   x ≥ 0, y ≥ 0 x − 2y ≤ 3 x − 2y = 3 or, x

3

9

13

y

0

3

5

40

50

x + y = 30

Z = 3x + y

(0, 0)

0

(20, 0)

60

(10, 20)

50

(0, 30)

30

Here, at x = 20 and y = 0. Maximum value of Z = 3x + y is 60. 5. Option (3) is correct. Explanation: For x + 2y ≤ 30, 3x + y ≥ 30, 4x + 3y ≥ 60 and x, y ≥ 0 For, x + 2 y = 30

Y

For, 3 x + y = 30

10 9 8 7 6 5 4 3 2 1

(30, 0)

y= x–2

3

  4 x + 3 y = 60

x

0

30

6

y x

15

0

12

0

10

6

y x

30

0

12

0

15

6

y

20

0

12

Y

(13, 0) (9, 3)

30

(3, 0) X

0

Unbounded in first quadrant and has no solution. 4. Option (4) is correct.

(0, 15)15

Explanation: For x + y ≤ 30, 2 x + y ≤ 40 and x, y ≥ 0

2 x + y = 40

25 (0, 20)20

1 2 3 4 5 6 7 8 9 10 11 12 13

x + y = 30

(0, 30)

(6, 12)

10

x

0

30

10

5

y

30

0

20

0

x

0

20

10

y

40

0

20

(30, 0) X 5 10 15 20 25 30 35 (10, 0) (15, 0) x + 2y = 30 3x + y = 30 4x + 3y = 60

157

LINEAR PROGRAMMING

Z(x, y) = 30x + 10y Z(6,12) = 300, Z(15,0) = 450, Z(30, 0) = 900 Here, at x = 6 and y = 12 minimum value of the objective function Z = 30x + 10y is 300. 6. Option (3) is correct. Explanation: A → III B→I C → IV D → II 7. Option (2) is correct. Explanation: Z ( x, y )  px  qy 4q Z (2, 3)  2 p  3q Z (4, 5)  4 p  5q Z (7, 0)  7 p Given, minimum of Z occurs at (2, 3) and (7, 0) Therefore, 2 p + 3q = 7 p 5 p = 3q 8. Option (2) is correct. Explanation: Given LPP is Max Z  4 x  2 y Subject to constraints: 2 x  3 y  28 x  y  10 x, y  0 Y

( )

B 0, 28 3

(0, 10) C(2, 8)

(14, 0) X’ Y’

Corner points

X A(10, 0) x + y = 10 2x + 3y = 28

Value of Z  4 x  2 y

A(10, 0)

Z  40  Maximum

 28  B  0,   3 

Z=

C(2, 8)

Z = 24

56 3

9. Option (1) is correct. Explanation:

 6, 0 

Z = 12

 6, 8

Z = 36 → Maximum

 max  Z   min  Z   36  6  30 10. Option (3) is correct. Explanation: Clearly from graph there is no feasible region.

B

A C O

11. Option (2) is correct. Explanation: Corner points

Corresponding Value Z = px + qy; p, q > 0

(0, 3)

3q

(1, 1)

p+q

(3, 0)

3p

of

So, condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) is 3p  p  q ⇒ 2p  q q p ∴ 2 12. Option (1) is correct. Explanation: The variables used in a Linear Programming Problem are called decision variables. 13. Option (4) is correct. Explanation: The maximum value of Z occurs at (15, 15)  p  15  q  15  15( p  q ) and the maximum value of Z occur at (0, 20)  p  0  q  20  20q For finding the relation is p and q, 15( p + q ) = 20q ⇒ 15 p + 15q = 20q ⇒ 20q − 15q = 15 p 5q = 15 p ⇒ q = 3p ⇒

Corner points

Value of Z  2 x  3 y

 0, 2 

Z = 6 → Minimum

14. Option (1) is correct. Explanation: Given, x + 3 y = 9

 3, 0 

Z = 6 → Minimum

and    x + y = 5

158 Oswaal CUET (UG) Chapterwise Question Bank Solving eq. (i) and (ii), we get = x 3= and y 2 Now both lines intersect at point (3, 2) For line

x + y = 5,

Required points are (0, 5) and (5, 0). For line

x + 3 y = 9,

Required points are (0, 3) and (9, 0). ∴ Shaded area belongs to the points (3, 2), (0, 3) and (0, 5). Corner points

Value of Z = 11x + 7y

(3, 2)

47

(0, 3)

21 (Min)

(0, 5)

35

Hence, the maximum value of Z is 21. 15. Option (2) is correct. Explanation: Corner points

Value of Z = 3x – 4y

(0, 0)

0

(5, 0)

15

(6, 5)

-2

(6, 8)

-14

(4, 10)

-28

(0, 8)

-32 (Minimum)

Hence, the minimum of Z occurs at (0, 8). 16. Option (2) is correct. Explanation: Given x + 2y = 76 ...(i)  2x + y = 104...(ii) and Solving eq. (i) and (ii), we get and x = 44 y = 16 So, required point is E(44, 16). From eq. (i) If x = 0, then y = 38. So, required point is D(0, 38). From eq. (ii) If y = 0, then x = 52. So, required point is A(52, 0). Corner points Value of Z = 3x + 4y 0(0, 0) 0 196 (Max.) E(44, 16) 152 D(0, 38) 156 A(52, 0) Hence, the maximum value of Z is 196. 17. Option (2) is correct. Explanation: Corner points

Value of Z = 4x + 3y

(0, 0) (0, 40) (20, 40) (60, 20) (60, 0)

0 120 200 300 (Max) 240

Here, 325 > 300 Hence, the quantity is column B is greater

MATHEMATICS/APP. MATH.

18. Option (2) is correct. Explanation: Corner points

Value of Z = x + 2y

 18 2   ,   7 7

22 7

7 3  ,  2 4

5

 3 15   ,  2 4 

9(Max)

 3 24   ,   13 13 

51 13

Hence, the maximum value of Z is 9. 19. Option (2) is correct. Explanation: Given, subject to constraints: x + y ≥ 8 3x + 5y ≤ 15 x ≥ 0 and y ≥ 0 Convert into linear equation form, x+ y =8 3 x + 5 y = 15 x = 0 and y = 0 From eq. (i), If x = 0, then y = 8 and if y = 0, then x = 8 So, required points are (0, 8) and (8, 0). From eq.(ii), If x = 0, then y = 3 and if y = 0, then x = 5 So, required points are (0, 3) and (5, 0). Now, we construct a graph:

...(i) ...(ii)

We can see that there is no common points which satisfy all the constraints. 20. Option (3) is correct. 21. Option (4) is correct. Explanation: The feasible region is bounded. Therefore, maximum of Z must occur at the corner point of the feasible region (Fig. 12.1). Corner points

Value of Z

O, (0, 0)

4(0) + 3(0) = 0

A(25, 0)

4(25) + 3(0) = 100

B(16, 16)

4(16) + 3(16) = 112

C(0, 24)

4(0) + 3(0) = 72

Hence, the maximum value of Z is 112.

159

LINEAR PROGRAMMING

22. Option (4) is correct. Explanation: Given, Max. Z = ax + 2by Max. value of Z on Q(3, 5) = Max. value of Z on S(4, 1) ⇒ 3a + 10b = 4a + 2b ⇒   a - 8b = 0 23. Option (4) is correct. Explanation: The minimum value of the objective function occurs at two adjacent corner points (0.6, 1.6) and (3, 0) and there is no point in the half plane 4x + 6y < 12 in common with the feasible region. So, the minimum value occurs at every point of the line-segment joining the two points 24. Option (2) is correct. Explanation: The strict inequality represents an open half plane and it contains the origin as (0, 0) satisfies it. 25. Option (2) is correct. Explanation: Y (0, 14) (7, 7) . D (3, 3) X'

E (9, 5)

C

O (0, –1)

B (3, 0) (1, 0)

A (9, 0)

(14, 0)

X x+y=14

x–y=0 Y’ x = 3

26. Option (3) is correct. Explanation: The feasible region as shown in the figure, has objective function F = 3x – 4y. Corresponding value of F = 3x - 4y

(0, 0)

0

(12, 6)

12 ← Maximum

(0, 4)

-16 ← Minimum

Hence, the maximum value of F is 12. 27. Option (2) is correct. Explanation: Corner Points

Corner points Corresponding value of F = 4x + 6y (0, 0) 12 ← Minimum (3, 0) 12 ← Minimum (6, 0) 24 (6, 8) 72 ← Maximum (0, 5) 30 Hence, minimum value of F occurs at any points on the line segment joining the points (0, 2) and (3,0). 29. Option (1) is correct. Explanation: Value of Z at (2, 4) = Value of Z at (4, 0)  2a  4b  4a  2a  4b a  2b 30. Option (3) is correct. Explanation: Corner Points

Z = 7x + y

(3, 0)

21

1 5  ,  2 2

6

(7, 0) (0, 5)

49 (Maximum) 5

[B] ASSERTION REASON QUESTIONS

x=9

On plotting the constraints x = 3, x = 9, x = y and x + y =14, we get the following graph. From the graph given below it, clear that feasible region is ABCDEA, including corner points A(9, 0), B(3, 0), C(3, 3), D(7, 7) and E(9, 5). Thus feasible region has 5 corner points including (7, 7) and (3, 3).

Corner points

28. Option (4) is correct. Explanation:

Value of Z = 2 x − y + 5

A(0, 10)

Z = 2 ( 0 ) − 10 + 5 = −5((Minimum)

B(12, 6)

Z = 2 (12 ) − 6 + 5 = 23

C(20, 0)

Z = 2 ( 20 ) − 0 + 5 = 45 (Maximum)

O(0, 0)

Z = 0 (0) − 0 + 5 = 5

So, the minimum value of Z is -5.

1. Option (1) is correct Explanation: X2 x1 + x2 = 1500 (0, 1500) (0,1000)

(900, 600)

x2 = 600

(1000, 500)

(2000,0) X1 x1 + 2x2 = 2000 From the graph, it is clear that the point (2000, 0) is outside, where as point (1000, 0), (0, 500), (2, 0) lie in the positive bounded region. 2. Option (4) is correct. Explanation: The feasible solution of a LPP belongs to only first quadrant because of non-negativity condition of variables associated with LPP 3. Option (3) is correct. Explanation: O

(1500,0)

Y

y=2

X O x=2

It is clear from the graph, Assertion is true.

160 Oswaal CUET (UG) Chapterwise Question Bank 4. Option (4) is correct Explanation: It is clear from the graph given in Reason (R), that 3  the vertices of feasible region are (0, 0), (1, 0) and  , 0  . So, 2  Assertion (A) is wrong. 5. Option (4) is correct. Explanation: Every LPP has an optimal solution. The optimal solution of an LPP either exists uniquely, does not exist or exists infinitely. If an LPP has two optimal solutions, then it has infinitely many optimal solutions. 6. Option (2) is correct. Explanation: Corner Points

Value of Z = 2x + 3y

A(3,3)

15 (minimum)

B(20, 3)

49

C(20, 10)

70

D(18, 12)

72(maximum)

E(12, 12) 60 7. Option (3) is correct. Explanation: Corner Points

Value of Z = 8x + 12y

(0,4)

48 (minimum)

(6, 0)

48 (minimum)

(12, 0)

96

(12, 16)

288 (maximum)

(0, 10) 120 8. Option (2) is correct. Explanation: Points

Inequality 2x – 3y > – 5 satisfy

(1, 1)

2 – 3 = -1 > -5

Yes

(-1, 1)

-2 -3 = -5 > - 5

No

(1, - 1)

2+3 =5>5

Yes

(-1, -1)

-2 + 3 = 1 > - 5

Yes

(-2, 1)

-4 -3 = -7 > - 5

No

(2, -1)

4+3=7>-5

Yes

(-1, 2)

-2 - 6 = - 8 > - 5

No

MATHEMATICS/APP. MATH.

At least 20 seats are reserved for the executive class. It is given that at least 4 times as many passengers prefer to travel by economy class than by executive class. So, let’s set up an inequality: x (executive class)  y (economy class)  200 x (executive class) +4x (economy class) ≥ 200 (since at least 4 times as many passengers prefer economy class) 5 x ≤ 200 x ≤ 200 / 5 x ≤ 40 y 4= x 160 . Now, if x = 40 , then = So, the maximum value of x + y is 40 + 160 = 200. 2. Option (4) is correct. Explanation: The relation between x and y: We know that at least 20 seats are reserved for the executive class. It is given that at least 4 times as many passengers prefer to travel by economy class than by executive class. So, the relationship between x and y is: y ≥ 4x 3. Option (3) is correct. 4. Option (2) is correct. Explanation: The profit when x = 20 and y = 80: Profit from executive class = 20 tickets × 1000/ ticket = ` 20,000 Profit from economy class = 80 tickets × ` 600/ ticket = ` 48,000 Total profit = ` 20,000 + ` 48,000 = ` 68,000 5. Option (1) is correct. Explanation: Objective function: Maximize Z = 1000x + 600y Constraints: x  y  200 x  20, y  4x x  0, y  0

(-2, -1) -4 + 3 = - 1 > - 5 Yes 9. Option (3) is correct. Explanation: The optimal value of the objective function is attained at the corner points of feasible region. 10. Option (4) is correct Explanation: Thus, corner points are: O(0, 0), A(3, 0) and B(0, 3). Max. value of Z at corner points

Z o  4  0   7  0   0 Min.

Z A  4  3  7  0   12 Z B  4  0   7  3  21  Max.

[C] COMPETENCY BASED QUESTIONS 1. Option (2) is correct. Explanation : The maxımum value of x + y: We know that an airplane can carry a maximum of 200 passengers.

The corner points are A(20, 180), B(40, 160), C(20, 80) Evaluating the objective function Z  1, 000 x  600 y at A, B and C At A  20,180  ,

Z  1, 000  20  600  180  20, 000  1, 08, 000  `1, 28, 000 AT B  40,160 , Z  1, 000  40  600  160  40, 000  96, 000

Z  1, 000 x  600 y at A, B and C At A  20,180  ,

Z  1, 000  20  600  180

161

LINEAR PROGRAMMING  20, 000  1, 08, 000

 `1, 28, 000 AT B  40,160 , Z  1, 000  40  600  160  40, 000  96, 000  `1, 36, 000 (max) AT C  20, 80 , Z  1000  20  600  80  20, 000  48, 000  ` 68, 000 So, Z is maximum, when = x 40 = , y 160. So, 40 tickets of executive class and 160 tickets of economy class should be sold to get the maximum profit of ` 1,36,000. 6. Option (1) is correct. Explanation: From the given graph OA= 75 and OB = 45 . x y  1 The equation of line AB is 75 45 i.e., 3 x  5 y  225 7. Option (2) is correct. Explanation: From the given graph OC = 40 and OD = 80. x y The equation of line CD is   1 i.e., 40 80 2 x  y  80 8. Option (1) is correct. Explanation: On solving the equations of lines AB and CD, we get the coordinates of point E i.e., (25, 30). 9. Option (1) is correct. Explanation: Let the manufacturer produces x number of model X and y number of model Y bikes. Model X takes 6 manhours to make per unit and model Y takes 10 man-hours to make per unit. There is total of 450 man-hours available per week.  6 x  10 y  450  3 x  5 y  225 ...(i)

For models X and Y, handling and marketing costs are ` 2,000 and ` 1,000, respectively, total funds available for these purposes are ` 80,000 per week.  2, 000 x  1, 000 y  80, 000 ...(ii)  2 x  y  80 Also, x  0, y  0 Alternate Solution: As (0, 0) lies in the region 3 x  5 y  225 and also (0, 0) lies in the region 2 x  y  80, therefore the constraints for the LPP is 3 x + 5 y ≤ 225, 2 x + y ≤ 80, x ≥ 0, y ≥ 0 10. Option (1) is correct. Explanation: Let 10. Option (2) is correct. Explanation: Y D (0,80) B (0,45)

(25,30) (75,0)

O

C (40,0)

X A

Corner points of feasible region are: (0, 0), (40, 0), (0, 45) and (25, 30). Corner Points

Value of Z = 1000x + 500y

(0, 0)

0

(40, 0)

40,000 (Maximum)

(0, 45)

22,500

(25, 30)

40,000 (Maximum)

So, the manufacturer should produce 25 bikes of model X and 30 bikes of model Y to get a maximum profit of ` 40,000. Since, in question it is asked that each model bikes should be produced.

Course of Action Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

13

 Revision Notes

PROBABILITY AND PROBABILITY DISTRIBUTIONS

 Conditional Probability: If E and F are two events associated with the same sample space of a random experiment, then the conditional probability of the event E under the condition that the event F has occurred, written as P (E | F), is P � (� E ∩ F ) given by P(E | F) = , P(F) ≠ 0 P(F )  Properties of Conditional Probability: Let E and F be events associated with the sample space S of an experiment. Then: (i) P(S|F) = P(F|F) = 1 (ii) P [(A ∪ B) | F] = P (A | F) + P (B | F) – P [(A ∩ B | F)], where A and B are any two events associated with S. (iii) P (E ' | F ) = 1 – P (E | F)  Multiplication Theorem on Probability { Let E and F be two events associated with a sample space of an experiment. Then P (E ∩ F) = P (E) P(F|E), P (E) ≠ 0 = P (F) P (E | F), P (F) ≠ 0 If E, F and G are three events associated with a sample space, then P (E ∩ F ∩ G) = P (E) P (F | E) P (G | E ∩ F)  Independent Events { Let E and F be two events associated with a sample space S. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we say that the two events are independent. Thus, two events E and F will be independent, if (a)  P (F | E) = P (F), provided P (E) ≠ 0 (b)  P (E | F) = P (E), provided P (F) ≠ 0. Using the multiplication theorem on probability, we have (c)  P (E ∩ F) = P (E) P (F) { Three events A, B and C are said to be mutually independent if all the following conditions hold:

P(A ∩ B) = P (A) P (B)



P (A ∩ C) = P (A) P (C) P (B ∩ C) = P (B) P (C) and

P (A ∩ B ∩ C) = P (A) P (B) P (C)  Partition of a Sample Space { A set of events E1, E2,...., En is said to represent a partition of a sample space S, if (a) Ei ∩ Ej = φ, i ≠ 0; i, j =1, 2, 3, ...... , n (b) E1∪ E2∪ . . . ∪ En = S, and (c) Each Ei ≠ φ, i.e., P (Ei ) > 0 for all i = 1, 2, ..., n

Example 1: If A and B are two events such that P ( A) ≠ 0 and P ( B | A) = 1, then (1) A ⊂ B (2) B ⊂ A (3) B   (4) A   Sol. Option (1) is correct. Explanation: P( A)  0 and P( B | A)  1 P ( B  A) P( B | A)  P ( A) P ( B  A) 1 P ( A) P ( A)  P ( B  A)  A B  Theorem of Total Probability { Let {E1, E2, ..., En} be a partition of the sample space S. Let A be any event associated with S, then P(A) =

 j 1 P  E j  P( A | E j ) n

 Bayes’ Theorem { If E1, E2,..., En are mutually exclusive and exhaustive events associated with a sample space, and A is any event of non-zero probability, then



P  Ei | A  

P  Ei  P  A | Ei 

n

 P  Ei  P  A | Ei  i 1

 Random Variable: A random variable (r.v.) is a real valued function defined on a sample space S and taking value in (– ∞, ∞) or whose possible values are numerical outcomes of a random experiment. Types of Random Variable: Random variables are classified into two types namely discrete and continuous random variables. • Discrete random variable: A variable which can assume finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable. Examples of discrete random variable: { Marks obtained in a test. { Number of red marbles in a jar. { Number of telephone calls at a particular time. { Number of cars sold by a car dealer in one month, etc. • Continuous random variable: A random variable X which can take on any value (integral as well as fraction) in the interval is called continuous random variable.

PROBABILITY AND PROBABILITY DISTRIBUTIONS

163

164 Oswaal CUET (UG) Chapterwise Question Bank Examples of continuous random variable { The amount of water in a 10 ounce bottle. { The speed of a car. { Electricity consumption in kilowatt hours. { Height of population. { Weight of students in a class. { The amount of time taken by a truck driver to go from Chennai to Madurai, etc. Probability Mass Function of Discrete Random Variable: If X is a discrete random variable with distinct values x1, x2 , …, xn ,…, then the function, denoted by PX ( x) and defined by  P  X  xi   pi  p  xi  if x  xi , i  1, 2,, n, PX ( x)  p ( x)   if x  xi 0 This is defined to be the probability mass function or discrete probability function of X. The probability mass function p(x) must satisfy the following conditions: (i) p  xi   0 i, 

Scan to know more about this topic

(ii)  p  xi   1 i 1

 Expected value: The expected value Expectation and Variance is a weighted average of the values of a random variable may assume. The weights are the probabilities. Let X be a discrete random variable with probability mass function (p.m.f.) p(x). Then, its expected value is defined by E ( X )  xp ( x) x

 Variance: The variance is a weighted average of the squared deviations of a random variable from its mean. The weights are the probabilities. The variance of X is defined by Var ( X )  [ x  E ( X )]2 p ( x)

Expected value of [X – E(X)]2 is called the variance of the random variable. i.e., Var(X) = E[X – E(X)]2 = E(X2) – [E(X)]2  Bernoulli Trials { Trials of a random experiment are called Bernoulli trials, if they satisfy the following four conditions: (a) The trials should be finite in numbers. (b) The trials should be independent of each other. (c) Each of the trial yields exactly two outcomes i.e., success or failure. (d) The probability of success or the failure remains the same in each of the trial.  Binomial Distribution Scan to know { Let E be an event. Let p = probability more about of success in one trial (i.e., occurrence this topic of event E in one trial) and, q = 1 – p = probability of failure in one trial (i.e., non-occurrence of event E in one trial). Let X = number of successes (i.e., number Binomial of times event E occurs in n trials) Distribution Then, probability of X successes in n trials is given by the relation:

P ( X  r )  P (r ) n Cr p rq n  r , where r = 0, 1, 2, 3,..., n;p = probability of success in one trial and q = 1 - p = probability of failure in one trial.

MATHEMATICS/APP. MATH.

 Special Cases • P(X = r) or P(r) is also called the probability of occurrence of event E exactly r times in n trials. n! • Here, n Cr = r !( n − r )! n

• Note that

Cr p r q n - r is the (r + 1)th term in the binomial

expansion of (q + p ) n . • Mean =

n

r  P(r )  np

r 0

n

• Variance  r 2  P (r )  (Mean )2  npq r 0

• A Binomial distribution with n Bernoulli trials and

probability of success in each trial as p is denoted by B(n, p). Here, n and p are known as the parameters of binomial distribution. • The expression P(x = r) or P(r) is called the probability function of the binomial distribution. Example 2 : Eight coins are tossed together. The probability of getting exactly 3 heads is 7 9 1 5 (1) . (2) . (3) . (4) . 32 32 32 32 Sol. Option (1) is correct. Explanation: We know that, binomial probability distribution P(X = r) = nCr (p)r qn-r 1 1 Here, n = 8, r = 3, p = and q = 2 2 1

3

\ Required probability = 8C3   2

1   2

8





8!  1    3!5!  2 



=

8.7.6 1 7 = 3.2 28 32

5

Scan to know more about this topic

 Poisson Distribution Poisson { It is a discrete probability distribution. Distribution The Poisson distribution is a particular limiting form of Binomial distribution when p (or q) is very small and n is large enough so that np (or nq) is a finite constant say m. A probability distribution of a random variable x is called Poisson distribution if x can assume non-negative integral values only and the distribution is given by  e m m r , r  0,1, 2  P(r )  P( X  r )   r !  0, r  0,1, 2 

Here, the value of e = 2.7183, it is the base of the natural system of logarithms. Examples of Poisson Variates (i) The number of cars passing through a certain street in time t. (ii) The number of defective screws per box of 100 screws. (iii) The number of deaths in a district in one year by a rare disease. (iv) The number of printing mistakes at each page of the book. Constants of the Poisson Distribution (i) Mean, E(X) = m = np (ii) Variance, V(X) = m = np

165

PROBABILITY AND PROBABILITY DISTRIBUTIONS

Example 3 : If P(X = 2) = 9P(X = 4) + 90 P(X = 6) in Poisson distribution, then E(X) is (1) 6. (2) 3. (3) 1. (4) 0. Sol. Option (3) is correct. Explanation: emmr r  0,1, 2, r! P ( X  2)  9 P ( X  4)  90 P ( X  6)

P( X  r ) 

Let Given ,

e m m2 e m m4 e  m m6 m 2 9m 4 90m6 9  90    2! 4! 6! 2 24 720

 



m 4  3m 2  4  0





 m4  4 m2  1  0  Thu us,

m 1 E( X )  m  1

(∵other values are not admissible)

 Difference between Binomial and Poisson Distributions • The binomial distribution is affected by the sample size n and the probability p, whereas the poisson distribution is affected only by the mean m. • In a binomial distribution the possible values of the random variable x are 0, 1, ... n but a poisson distribution has possible x values of 0, 1, ...., with no upper limit.  Normal distribution Scan to know It is a continuous probability more about this topic distribution in which the relative frequencies of a continuous variable are distributed according to normal probability. In simple words, it is a symmetrical distribution in which the Normal frequencies are distributed evenly about Distribution the mean of distribution. e.g.: (i) Height and intelligence are approximately normally distributed. (ii) Measurement of errors also often have a normal distribution. The normal (or Gaussian) distribution is defined by the probability density function 1  x     

2



  1 f ( x)  e 2  2



where,  and   0 are parameters of the distribution.



, for    x  

Clearly, f ( x)is non-negative and



The notation N , 2









f ( x)dx  1.

means normally distributed with



Normal distribution is limiting case of binomial distribution under the following conditions: (i) n, the number of trials is infinitely large, i.e., n → ∞ (ii) neither p(nor q) is very small, The normal distribution of a variable when represented graphically, takes the shape of a symmetrical curve, known as the Normal Curve. The curve is asymptotic to X-axis on its either side. Chief characteristics or properties of normal probability distribution and normal probability curve. The normal probability curve with mean m and standard deviation s has the following properties: (i)  The curve is bell-shaped and symmetrical about the line x=m (ii)  Mean, median and mode of the distribution coincide. (iii)  X-axis is an asymptote to the curve, (tails of the curve never touches the horizontal (X) axis). (iv)  No portion of the curve lies below the X-axis as f(x) being the probability function can never be negative.

(v) The points of inflection of the curve are x     (vi) T  he curve of a normal distribution has a single peak i.e., it is a unimodal. (vii) As x increases numerically, f(x) decreases rapidly, the maximum probability occurring at the point x   and 1 is given by [ p ( x)]max   2 (viii) The total area under the normal curve is equal to unity and the percentage distribution of area under the normal curve is given below. (a) About 68.27% of the area falls between m - s and m + s. P (    X    )  0.6826 (b) About 95.5% of the area falls between   2 and   2. P(  2  X    2)  0.9544 (c) About 99.7% of the area falls between   3 and   3. P(  3  X    3)  0.9973



mean m and variance σ 2 . If we say x  N , 2 we mean that





x is distributed N , 2 .

 Normal distribution is diagrammatically represented as follows:

 Standard Normal Distribution { A random variable Z  ( X   ) /  follows the standard normal distribution. Z is called the standard normal variate with mean 0 and standard deviation 1 i.e., Z ∼ N (0,1). Its probability density function is given by:

166 Oswaal CUET (UG) Chapterwise Question Bank 1  e 2

( z ) 

z2 2

MATHEMATICS/APP. MATH.

 Standard Normal Distribution Table This table provides the area between the mean and some Z score.

,   z  

1. The area under the standard normal curve in equal to 1. 2. 68.26% of the area under the standard normal curve lies between Z = -1 and Z = 1

For example, when Z score = 1.45, the area = 0.4265.

3. 95.44% of the area lies between Z = -2 and Z = 2 4. 99.74% of the area lies between Z = -3 and Z = 3

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Probabilities to solve a specific problem by A, B and C are 1 1 1 respectively. Probability that at least one will , and 2 3 4 solve the problem is  [CUET 2023] (1)

1 . 24

(2)

1 . 4

(3)

23 . 24

(4)

3 . 4

2. In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed and X = 1 if he is in favour. Then, E (X) is [CUET 2023] (1)

7 . 10

(2)

1 . 2

(3)

1 . 3

(4)

3. Match List - I with List - II.

7 . 11

[CUET 2023]

List - I The probability distribu- (I)  tion is applied for discrete random variable

List - II Normal distribution

(B)

 normal distribution is (II) A symmetric about

Standard deviation

(C)

 his probability distribu- (III) Mean T tion is applied for continuous random variable

(A)

7. A coin is tossed 7 times. The probability of getting at least 4 heads is [CUET 2023] 1 5 3 1 (1) . (2) . (3) . (4) . 2 8 4 4 8. Between 3 p.m. and 5 p.m. the average number of phone calls per minute coming into the help line desk of a bank is 5. The probability that during one particular minute there will be only one phone call is  [CUET 2023] (1) 0.5e−5. (2) 5e−5. (3) e−5. (4) 25e−5. 9. If the sum and product of the mean and variance of a Binomial distribution are 18 and 72 respectively, then the probability of obtaining at most one success is [CUET 2023] 24

(D)

(4) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

[CUET 2023]

4. In a box containing 100 bulbs. 10 are defective. Then the probability, that out of a sample of 5 bulbs none is defective, is[CUET 2023] 5

(1) 10-1.

1 (2)   . 2

5

 9   9  (3)   . (4)   .  10   10 

5. The mean number of heads in two tosses of a coin is  [CUET 2023] 1 3 (1) 2. (2) . (3) 1. (4) . 2 2 6. Two dice are thrown simultaneously. If X denotes the number of sixes, then the variance of X is [CUET 2023] 5 (1) . 18

7 (2) . 18

1 (3) . 3

2 (4) . 3

24

23

10. Match List I with List II

[CUET 2023]

B.

LIST I The variance of a poisson distribution with mean λ is The standard deviation of a Poisson distribution with mean λ is

C.

In a poisson distribution, if mean is 4, then the standard deviation is

III.

λ

D.

In a poisson distribution, if mean is 4, then the variance is

IV.

2

A.

 he shape of normal curve (IV) Poisson distribution T depends on Choose the correct answer from the options given below: (1) (A)-(II), (B)-(I), (C)-(III), (D)-(IV) (2) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) (3) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

24

1 1 1 1 (1) 25   . (2)   . (3) 24   . (4) 24   . 2 2 2 2 LIST II I.

√λ

II.

4

Choose the correct answer from the options given below: (1) A-III, B-I, C-II, D-IV (2) A-III, B-I, C-IV, D-II (3) A-I, B-III, C-II, D-IV (4) A-I, B-III, C-IV, D-II 11. Let X be a discrete random variable whose probability distribution is defined as [CUET 2023]



, if x  0  0.5   k ( x  1) , if x  1 or 2 P( X  x)   k (6  x) , if x  3 or 4  0 , otherwisse



The value of k is 1 1 1 1 (1) . (2) . (3) . (4) . 10 20 2 4 12. The probability distribution of a discrete random variable X is given as: [CUET 2023] X

0

1

2

3

P(X)

2k2

k2

3k2

k

167

PROBABILITY AND PROBABILITY DISTRIBUTIONS

The mean of the distribution is 4 5 7 16 (1) . (2) . (3) . (4) . 3 3 6 9 13. The probability distribution of X is

[CUET 2022]

x

0

1

2

3

4

P(X = x)

0.1

2k

k

k

2k

Then var (X) = 3 141 159 9 (1) (2) . (3) (4) . . . 20 20 80 4 14. If m is mean of random variable X, with probability distribution. [CUET 2022] x

0

1

2

P (X = x)

4 9

4 9

1 9

1

P(X = x)

k

2

3

4

k 2

2k

8k2

5 1 – 5k

2 15

(2)

13 15

6 k 2

where k > 0. Then consider the following statements: (A) P ( X = 3) (B) P( X ≤ 2) (C) P ( X ≥ 5) (D) P ( X = 4)

(3)

(d) A’ and B’ are also independent Choose the correct answer from the options given below: (1) (a) and (b) only (2) (b) and (c) only (3) (c) and (d) only

Then value of 9k + 4 is (1) 4. (2) 9. (3) 10. (4) 17. 15. In a game, a child will win ₹ 5 if he gets all heads or all tails when three coins are tossed simultaneously and he will lose ₹ 3 for all other cases. The expected amount to lose in the game is [CUET 2022] (1) ₹ 0. (2) ₹ 0.8. (3) ₹ 1. (4) ₹ 2. 16. A biased dice is thrown once. If X denotes the number appearing on it and have probability distribution. [CUET 2022] x

2 1 (4) 13 5 7 5 3 20. For two events A, B, P  A  B   , P  A   , P  B   . 12 12 12 Then P(A ∩ B) =  [CUET 2022] 1 1 1 1 (1) (2) (3) (4) 6 12 3 2 21. If events A and B are independent, then identify the correct statements (a) A and B must be mutually exclusive (b) The sum of their probabilities must be equal to 1 (c) P  A   P  B   P  A  B  (1)

(4) (a) and (d) only

22. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is 1 1 1 (1) 0. (2) . (3) (4) . . 3 12 36 23. Which of the following is false about the Central Limit Theorem ? (1) It states that a distribution of sample approaches a normal distribution. (2) It shows bell shaped curve. (3) It does not depend on the shape of population distribution. (4) It can be represented by using a pie-chart 24. If the mean of a binomial distribution is 24 and its standard deviation is 4, then the probability of getting success is 2 1 1 1 . (2) . (3) . (4) . 3 5 6 3 25. One of the following is true of relation between sample mean (x) and population mean (m). (1)

(E) P ( X = 1) + P ( X = 5)

(1) x   increases when increases the size of sample

Choose the correct answer from the options given below: (1) C>D>B>A>E (2) E>C>D>A>B (3) E>C>A>B>D (4) C>E>A>B>D 3 17. If A and B are two independent events with P ( A ) = and 5 4 P ( B ) = , then P ( A′ ∩ B′ ) is equal to [CUET 2022] 9

(2) x = m(ma), for all sample sizes

1 2 4 8 (2) (3) (4) 15 3 9 45 18. A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is either ace or a king?  [CUET 2021] 1 2 4 (1) (2) (3) (4) None of these 13 13 13 19. Let E1 and E2 be two independent events. Let P(E) denotes the probability of the occurrence of the event E. Further, let E1´and E2´ denote the complements of E1 and E2, respectively. (1)





If P E1'  E2  





2 1 and P E1  E2'  , then P(E1) is 15 6 [CUET 2021]

(3) x   do not change with size of sample (4) x   decreases when increases the size of sample 26. The probability mass functions of random variable X is:  P(X = x) = (0.6)x(0.4)1–x; x = 0, 1 The variance of X is (1) 0.60. (2) 0.124. (3) 0.244. (4) 0.240. 27. If X is a Poisson variable such that P(X = 1) = 2P(X = 2), then P(X = 0) is (1) e. (2) 1/e. (3) 1. (4) e2. 28. If X is a Poisson variate such that 3P(X = 2) = 2P(X = 1) then the mean of the distribution is equal to 4 3 4 3 (1) . (2) − . (4) − . . (3) 3 4 3 4 29. If random variable X represents the number of heads when a coin is tossed twice, then mathematical expectation of X is 1 1 (1) 0. (2) . (3) . (4) 1. 4 2

168 Oswaal CUET (UG) Chapterwise Question Bank 30. In a Poisson distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is (1) m = np. (2) m = (np)2. (3) m = np (1− p). (4) m = p.

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: (1)  Both A and R are true and R is the correct explanation of A (2)  Both A and R are true but R is not the correct explanation of A (3)  A is true but R is false (4)  A is false and R is true 1. Assertion (A) : The probability of an impossible event is 1. Reason (R) : If A is a perfect subset of B and P(A) < P(B), then P(B – A) is equal to P(B) – P(A). 2. Assertion (A): If A and B are two independent events with 1 1 1 P ( A ) = and P  B   then P  A / B  is . 5 3 5 P ( A′ ∩ B ) Reason (R): P ( A′ / B ) = . P ( B) 3. A can solve 80% of the problems given in a book and B can solve 60%. Assertion (A): The probability that atleast one of them will solve a problem is 0.92. Reason (R): The probability that none of them will solve a problem is 0.08. 4. Assertion (A): If A and B are two events such that P ( B ) > 0 and P  A   1 , then P ( B′ / A′ ) is

Reason (R): P ( B′ / A′ ) =

1 − P( A ∪ B) 1 − P ( A)

.

P ( A′ ∩ B′ ) . P ( A′ )

5. Assertion (A): If P  A   0.6 and P ( B ) = 0.4 and P  A | B   0.4 , then P ( A ∪ B ) = 0.72 .



Reason (R): P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) .

6 Assertion (A): Poisson distribution is applied for discrete random variable. Reason (R): In a Poisson distribution, the mean and variance are equal. 7. Assertion (A): The range of normal distribution is – ∞ to ∞.  Reason (R): In normal distribution, Mean = Mode = Median 8. Assertion (A): The area under a standard normal curve is 1. Reason (R): The area under a standard normal curve is 2. 9. Assertion (A): The standard normal curve is symmetric about the value of mean = 0. Reason (R): For a standard normal variate, the value of standard deviation is 1. 10. Assertion (A): In a poisson distribution, if mean, m = e, e( x - m ) x! Reason (R): In a poisson distribution, if mean, m = e, then e− m P(X) is given by P(X) = x! then P(X) is given by P(X) =

MATHEMATICS/APP. MATH.

[C] COMPETENCY BASED QUESTIONS I.  Read the following text and answer the following questions on the basis of the same: Mr. Rama Roy a owner of a factory lives in Karnataka. In his factory, he manufactures razor blades. There is a small chance 1 that for any blade to be defective. The blades are in the 500 packets of 10.

1. The probability that packet contain no defective blade is (1) e–0.02. (2) 0.02. (3) 0.02 e–0.02. (4) 0.002 e–0.02. 2. The probability that packet contain one defective blade is (1) e–0.02. –0.002

(2) 0.02e–0.02.

(3) 0.02e . (4) 0.002e–0.02. 3. Find the probability that packet contain two detective blade, when it is given that e–0.02 = 0.9802. (1) 0.0002 (2) 0.002 (3) 0.2 (4) None of these 4. Find the approximate number of packets containing no defective blade, when there are 10000 packets in a consignment. (Use e–0.02 = 0.9802) (1) 9802 (2) 9800 (3) 2 (4) 196 5. Find the approximate number of packets containing one defective blade, when there are 20,000 packets in the consignment. (Use e–0.02 = 0.9802) (1) 196 (2) 392 (3) 932 (4) 619 II. Read the following text and answer the following questions on the basis of the same: Today in mathematics class, Mr. Lal teaches the topic of Binomial distribution. Four friends Rohan, Rohit, Roshan and Raman have few doubts in this topic, so they decided for group study at Rohan’s place. They are trying to solve a question in which a pair of dice is thrown 7 times. If getting a total 7 is considered a success, then answer the following questions:

169

PROBABILITY AND PROBABILITY DISTRIBUTIONS

6. If X denote the number of success in 7 throws of a pair of dice, then X Binomial variate with parameters: n 7= ,p (1) =

1 1 = n 2= ,p (2) 6 6

n 7= ,p (3) =

5 5 = n 2= ,p (4) 6 6

7. What is the probability of no success? 7

7

7

6 5 1 5 (1)   (2)   (3)   (4) 6 5 6 5        

8. What is the probability of 6 success? 7

7

5 1 1 1 (1)   (2) 35   (3) 35   (4) 35   6 6 6 6

7

9. What is the probability of at least 6 success ? 5

7

6

1 1 1 5 (1)   (2)   (3)   (4)   6 6 6 6

5

10. What is the probability of at most 6 success? 5

7

6

1 1 1 5 (1) 1 −   (2) 1 −   (3) 1 −   (4) 1 −   6 6 6       6

5

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (4)

2. (1)

3. (3)

4. (3)

5. (3)

6. (1)

7. (4)

8. (2)

9. (1)

10. (2)

11. (2)

12. (4)

13. (4)

14. (3)

15. (3)

16. (3)

17. (4)

18. (3)

19. (4)

20. (2)

21. (3)

22. (4)

23. (4)

24. (4)

25. (4)

26. (4)

27. (1)

28. (1)

29. (4)

30. (1)

8. (3)

9. (2)

10. (3)

8. (3)

9. (1)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (4)

2. (4)

3. (2)

4. (1)

5. (1)

6. (2)

7. (2)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (2)

3. (1)

4. (1)

5. (2)

6. (1)

7. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (4) is correct. Explanation: P (Probability of solving problem by A), 1 P  A  2 P (problem not solve by A),

 

P A  1  P  A 1 

1 2

1 4

 1  P C 3 4 P (at least one will solve the problem) = 1 - P (Problem not solve by any one among A, B and C).  1  P  A P  B  P  C  

1 2 3   2 3 4 1 1 4 3  4 1

1 2

P(Probability of solving problem by B), 1 P B  3 P(Probability not solve by B),

 

P B  1  P B 1

 

 P C 1

1 3

2 3 P(Probability of solving problem by C), 1 P (C ) = 4 1 P (Probability not solve by C)  P C  1  4  1  P C 

 



3 4

2. Option (1) is correct. Explanation: Case 1: If selected member opposed 30 P(X = 0) = 100 Case 2: If selected member favour 70 P(X = 1) = 100 Required probability distribution is 0 1 X 30 70 P(X) 100 100

170 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH.

6. Option (1) is correct. Explanation: Here, X represents the number of sixes obtained when two dice are thrown simultaneously. Therefore, X can take the value of 0, 1 or 2. \ P(x = 0) = P (not getting six on any of the dice)

Then, 30 70 E(X) = 0× + 1× 100 100 70 = 100 7 = 10 3. Option (3) is correct. (A) The Poisson distribution is a discrete distribution that measures the probability of a given number of events happening in a specified time period.

5 5 × 6 6 25 = 36 P(X = 1) = P (six on first dice and number after than six on second dice) + P (number other than six on first dice and six on second dice)

(B) A normal distribution bell curve is always symmetrical about the mean.



=



=

=

Explanation:

(C) A continuous random variable is one that cannot be counted but rather is estimated to be in a range or interval. The normal distribution is one of the many probability distributions that a continuous random variable can possess. (D) The shape of a normal (bell) curve depends on two parameters, µ and σ, which correspond, respectively, to the mean and standard deviation of the population for the associated random variable. 4. Option (3) is correct.

1 5 5 1    6 6 6 6 5 5 + = 36 36 10 = 36 P(X = 2) = P (six on both the dice)



Explanation: The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denote the number of defective bulbs out of a sample of 5 bulbs.

X P(X)

Probability of getting a defective bulb, = p

10 1 = 100 10



Clearly, X has a binomial distribution with n = 5 and p = n

 P ( X  r )  Cr q

\

 9  P ( X  0)  5C0    10  5

1

2

25 36

10 36

1 36

5 r

5 0

 1     10 

 1     10 

0

5

5. Option (3) is correct. Explanation: Total possible outcomes = {TT, TH, HT, HH}    No Head, P(0) = 1 (TT)   One Head, P(1) = 2 (HT, TH)   Two Head, P(2) = 1 (HH) 1

2

1/4

2/4

1/4

1 2 1 Mean = ∑ Xi Pi = 0 × + 1 × + 2 × 4 4 4 1 1 =0+ + =1 2 2

25 1 10 +1 × +2× 36 36 36 10 2 12 1 = + = = 36 36 36 3 =0×

Variance (s2) = ∑P(Xi) (Xi)2 − (Mean)2

p

 9   9   (1)   (1)     10   10 

0



1 10

nr r

 9  P ( X  r )  5Cr    10  P (none of the bulb is defective) = P(X = 0)

Pi

0

Mean (expected value) m = Xi P(Xi)

1 9 \ q 1  10 10

So, xi

1 1 × 6 6 1 = 36

r



= ∑P(Xi) (Xi)2 − ∑(Xi P(Xi))2

2

1  1 25 10   12   4    = ( 0 )2  36 36 36   3   10 1 1 = + − 36 9 9 5 10 = = 18 36 7. Option (4) is correct. Explanation: Use the Binomial distribution directly. Let us assume that the number of heads is represented by x (where a result of heads is regarded at success and in this case x ≥ 4.

Assuming that the coin is unbiased you have a probability of 1 failure ‘q’ is (where q is considered as failure). The number 2 of trials is represented by the letter n and for this question n = 7, you have a probability of success ‘p’ (where p is 1 considered as success) is . 2 Now just use the probability function for a binomial distribution P(X = x) = nCx Px qn−x

171

PROBABILITY AND PROBABILITY DISTRIBUTIONS

Using the information in the problem we get,

Therefore, the Binomial distribution is 24

P(x ≥ 4) = P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) 7 = C4

1   2

1   2 7

3

4

1 1 7   + C5   2 2

5

2

1 7 1   + C6   2 2

6

1 1  (p + q)n =    2 2 Now, Probability of obtaining almost one success P (x ≤ 1) = P(x = 0) + P(x = 1)

1

1 7   + C7 2

0

1   2

1

=   (7C4+ 7C5+ 7C6+ 7C7) 2



=  1  (35+21+7+1) 2

1 1 = × 64 = 2 128 8. Option (2) is correct. Explanation: Let X be denoting the number of phone calls per minutes. Given that, mean l= 5 and we need to find P(X=1) Now, e  (  )1  e  (  )r  Since, P( X  r )   P(X=1) = 1!  r !  e −5 ( 5)1 = 5e−5 1! 9. Option (1) is correct. Explanation: Let the Binomial distribution be (p+q)n we have, Mean = np Variance = npq Given, np + npq = 18  np (1+q) = 18 On squaring both sides n2p2 (1+q)2 = 324 ... (i) and (np) (npq) = 72 =



2 2

n p q



324 72

2



1  q  2q 9  q 2 2

2q − 5q + 2 = 0 2q2 − 4q − q + 2 = 0 2q(q − 2) − 1(q − 2) = 0 (q − 2)(2q − 1) = 0 1 q = 2 (not possible), q = 2 We know p+q=1 1  p + =1 2 1  p = 2 So by eq (ii), we get 11 n2     = 72 42 n2 = 72 × 8 n = 24

23

1 = 25   2 10. Option (2) is correct.

 

n 2 p 2 (1  q ) 2

1

1 1 + C1     2 2 24



7

 n2p2q = 72 On dividing eq (i) by eq (ii), we get

24

24

7



0

1 1 = C0     2 2 24

... (ii)

Explanation: A. We know, Variance of a Poisson distribution with mean l=l Here, A Match with III. B.  Standard deviation of a Poisson distribution with mean l= l Here, B Match with I. C. S.D. when l = 4 is √4 = 2 Here, C Match with IV. D. Variance when l = 4 = 4 Here, D Match with II. 11. Option (2) is correct. Explanation: P(0) + P(1) + P(2) + P(3) + P(4) = 1 0.5 + 2k + 3k + 3k + 2k = 1 1 +10k = 1 2 1 10k = 2 1 k= 20 12. Option (4) is correct. Explanation: 2k2 + k2 + 3k2 + k = 1 6k2 + k − 1 = 0 6k2 + 3k − 2k − 1 = 0



3k(2k + 1) −1(2k + 1) = 0

(2k + 1)(3k − 1) = 0

−1 1 (not possible), k = 2 3 Now, k=

X

0

1

2

3

P(X)

2 9

1 9

1 3

1 3

2 1 1 1 +1× +2× +3× 9 9 3 3 2 1 =0+ + +1 3 9

Mean = 0 ×

1+ 6 + 9 16 = 9 9 13. Option (4) is correct.

=

Explanation: Since,

SP(X) = 1

172 Oswaal CUET (UG) Chapterwise Question Bank 16. Option (3) is correct.

\ 0.1 + 2k + k + k + 2k = 1

Explanation: ∑[P(X = x)] = 1

 6k = 1 – 0.1



0.9 k= = 0.15 6

So, distribution is: 0

1

2

3

4

P(X = x)

0.1

0.30

0.15

0.15

0.30

xiP(X)

0

0.30

0.30

0.45

1.20

2.25

0

0.30

0.60

1.35

4.80

7.05

Total

Var(X) = Sxi2P(X) – (SxiP(X))2





= 7.05 – (2.25)2



= 7.05 – 5.0625



= 1.9875

19875 159 = = 1000 80 14. Option (3) is correct. P(X = x)

⇒ 16k2 – 2k = 0 ⇒ 2k(8k – 1) = 0 1 ⇒ k = 0 or k = 8 1 ⇒  k = (Rejecting k = 0) 8 Probability distribution is: x

1

2

3

4

5

6

P(X = x)

1 8

1 16

1 4

1 8

3 8

1 16

1 4 (B) P(X ≤ 2) = P(X = 1) + P(X = 2) (A) P(X = 3) =

Explanation: x

k k  2 k  8 k 2  (1  5 k )  = 1 2 2 ⇒ 2k + k + 4k + 16k2 + 2 – 10k + k = 2 ⇒ k

x

xi2P(X)

x.P(X = x)

=

0

4 9

1

4 9

4 9

=

2

1 9

2 9

(D) P(X = 4) =

0

Total Since,

\ Now,

Sx. P(X = x) =

6 9

6 9

6 9m + 4 = 9.  4  6  4  10 9

15. Option (3) is correct. Explanation: Let X be the amount received by the person. Then X can take values 5 and –3 such that P(X = 5) : Probability of getting all heads or all tails when three coins are tossed 2 1 = 8 4 P(X = –3) : Probability of getting one or two heads

 P(X = 5) =

6 3 = 8 4 Therefore, expected amount to win, on the average, per game  X = ∑ XP (X)

\

P(X = –3) =

1 3  (3)  4 4 5 9 = − 4 4 = –1 Thus, the person will, on average lose ` 1 per toss of the coin.

= 5

3 1 1 2 +1 + = = 16 16 8 16

(C) P(X ≥ 5) = P(X = 5) + P(X = 6)

Mean (m) = Sx.P(X = x)

m=

MATHEMATICS/APP. MATH.

3 1 6 1 7    8 16 16 16 1 8

(E) P(X = 1) + P(X = 5) 

1 3 4 1    8 8 8 2

Thus, E > C > A > B > D 17. Option (4) is correct. Explanation:

P ( A′ ∩ B′ ) = P ( A ∪ B )′ = 1 − P( A ∪ B) = 1 − [ P ( A) + P ( B ) − P ( A ∩ B )] = 1 − [ P ( A) + P ( B ) − P ( A) ⋅ P ( B )] [ A and B are independent, therefore, P ( A  B )  P ( A)  P ( B )] 3 4 3 4 =1−  + − ×  5 9 5 9  27 + 20 12  =1−  −  45   45  47 12  =1−  −   45 45  35 =1− 45 10 = 45 2 = 9 18. Option (3) is correct. Explanation: No. of kings in a pack = 4 No. of ace in a pack = 4

173

PROBABILITY AND PROBABILITY DISTRIBUTIONS

Favourable events n(E) = 4 + 4 = 8 Total events n(S) = 52 Favourableevents Probability of (Ace or King) = Totalevents 8 = 52 2 = 13 19. Option (4) is correct. Explanation: 2 P  E '1  E2    Given  15 2 P  E '1  .P  E2    Independent events  15 2 1  P  E1   .P  E2   15 2 ...(i) P  E2   P  E1  .P  E2   15 1 P  E1  E '2   6 1 P  E1  .P  E '2   6 1 P  E1  1  P  E2   6 1 P  E1   P  E1  P  E2   6 1 2 54 1   P  E1   P  E2    From eq. (i) 6 15 30 30  1 P  E2   P  E1    ii  30 Now, from equations (i) and (ii), 1 1  2  P  E1    P  E1   P  E1     30 30  15 





Let P  E1   x 1 1  2  x  x x    30 30  15  x 1 2 x  x2   30 30 15 31x 5  0  x2  30 30 2 30 x  31x  5  0

or 30 x2  25x  6 x  5  0  6 x  5  5 x  1  0 5 If 6 x  5  0 , then x  6 1 If 5 x  1  0 , then x  5 20. Option (2) is correct. Explanation: We have, 7 5 3 P  A  B   , P  A  , P  B   12 12 12 ∵ P  A  B   P  A  P  B   P  A  B  5 3 7 P A  B    12 12 12 8 7   12 12 1  12

21. Option (3) is correct. Explanation: (a) If two events are independent, they cannot be mutually exclusive. (b) To test if probability of independent events is 1 or not. Let A be the event of getting head. 1  P  A  2 Let B be the event of getting 5 on a die. 1  PB  6 Here, A and B are independent events. 1 1 4 1 Therefore, P  A   P  B      1 2 6 12 3 Thus, statement (b) is also incorrect. 22. Option (4) is correct. Explanation: When two dices are rolled, the number of outcomes is 36. The only even prime number is 2. Let E be the event of getting an even prime number on each die. ∴ E = {(2, 2)} 1 P(E) = 36 23. Option (4) is correct. Explanation: In probability theory, the central limit theorem (CLT) states that the distribution of a sample variable approximates a normal distribution (i.e., a ‘bell curve’) as the sample size becomes larger, assuming that all samples are identical in size and regardless of the population’s actual distribution shape. 24. Option (4) is correct. Explanation: We know that, for Binomial distribution Mean, m = np  Standard deviation, s = npq Therefore, np = 24 and

npq = 4

Now, npq = 16

⇒

 24 q = 16

⇒

=  q

16 2 = 24 3

Thus, probability of success, p = 1 – q = 1  25. Option (4) is correct.

2 1  3 3

Explanation: As a sample size increases the sample mean become better estimator of the population mean and sampling variability decreases. 26. Option (4) is correct. Explanation: Given, P(X = x) = (0.6)x (0.4)1–x, x = 0, 1 Here,

p = 0.6, q = 0.4, n = 1  

Variance = npq

= 1 × 0.6 × 0.4 = 0.24 27. Option (1) is correct. Explanation: Given,   P(X = 1) = 2P(X = 2)

P  A / B   

P  A  B  P  B

P  A   P  B 

 B  Bank 174 Oswaal CUET (UG) Chapterwise PQuestion  P  A  m 1 m 2  1  P  A e m e m 

2

1!

2!

1

m 1



Now, P ( X  0) 

e  m m0 e11  e 0! 1



Explanation: 3P(X = 2) = 2P(X = 1) m 2e − m me − m =2 2! 1! 4 ⇒ m= 3 ⇒3

   

 1   P A  P B      1  1  0.80   1  0.60  

Explanation: Sample Event

P  xi   pi

TT

1 4

1

HT,TH

1 2

1 2

2

HH

1 4

1 2

xi pi

0

Mathematical expectation E ( X )   pi xi  1. 30. Option (1) is correct. Explanation: The mean value or the expected value for a discrete probability function, is given by Mean (μ) =



n xp x 0

 x

For Poisson distribution P(x) = Substitute in above equation and solve to get m = m = np.

4 5

Therefore, P (atleast one of them will solve a problem)

29. Option (4) is correct.

0

1 5

3. Option (2) is correct. Explanation: Here, 80 60 P  A   0.80 and P  B    0.60 100 100

28. Option (1) is correct.

xi

MATHEMATICS/APP. MATH.

 1   0.20  0.40 

 1  0.0800  0.92 Also, P (none of them solve a problem ) = 1 - P (atleast one of them will solve a problem) = 1 - 0.92 = 0.08 4. Option (1) is correct. Explanation: We have, P(B) > 0 and P(A) ≠ 1 P  B/A   

P  A  B  P  A  1  P A  B 1  P  A

5. Option (1) is correct. Explanation:

e- m m x x!

P A | B  0.4 

P A  B P B

P A  B

0.4 P  A  B   0.16

[B] ASSERTION REASON QUESTIONS



1. Option (4) is correct.

Now , P  A  B   P  A   P  B   P  A  B 

Explanation: Assertion (A) is wrong. If the probability of an event is 0, then it is called as an impossible event. But Reason (R) is correct. From Basic Theorem of Probability, P(B - A) = P(B) - P(A), this is true only if the condition given in the question is true. 2. Option (4) is correct. Explanation: P  A / B   

P  A  B  P  B

P  A   P  B  P B

 P  A   1  P  A 1

1 5

 0.6  0.4  0.16  1  0.28  0.72 6. Option (2) is correct. Explanation: Poisson distribution is a discrete function, means that the variable can only take specific values i.e., 0, 1, 2, ... etc. with no fraction or decimals. Also, for Poisson distribution mean = variance = np Thus, both assertion and reason are correct. 7. Option (2) is correct. Explanation: The area under the standard (i.e, mean = 0, standard deviation = 1) normal distribution from negative infinity to positive infinity is equal to 1. The normal distribution is symmetrical, bell-shaped distribution in which the mean, median and mode are all equal.

175

PROBABILITY AND PROBABILITY DISTRIBUTIONS

8. Option (3) is correct. Explanation: For any probability distribution, the sum of all probabilities is 1. Area under normal curve refers to sum of all probabilities. 9. Option (2) is correct. Explanation: If the mean and standard deviation of a normal variate are 0 and 1, respectively it is called as standard normal variate. Thus, both the given statements are individually true. 10. Option (3) is correct. Explanation: Since, for poisson distribution e− m m x x! Put m = e, we get



P( X ) =

P( X ) =



= 20,000 × 0.0196

= 392 6. Option (1) is correct. Explanation: Let p denote the probability of getting a total of 7 in a single throw of a pair of dice. Then 6 1 = 36 6 [ The sum can be 7 in any one of the ways: (1, 6), (6, 1), = p

(2, 5), (5, 2), (3, 4) and (4, 3)] So, the parameters are: 1 6 7. Option (2) is correct. n 7= and p =

e− me x x!

Explanation: The X is a binomial variate with parameters n  7 and p 

e( x − m ) = x!

r

1 5 P ( X  r )  7C r     6 6 r  0, 1, 2

[C] COMPETENCY BASED QUESTIONS 1. Option (1) is correct.

1 Explanation: Here, given N = 1000, n = 10,= p = 0.002 500 ⇒ m = np = 10 × 0.002 ⇒

m = 0.02

0

1 5 7 C0     6 6

 = e-0.02 (0.002)0 = e-0.02

2. Option (2) is correct. Explanation: P(one defective blade) = P= ( X 1) =

e −0.02 (0.02)1 1!

5   6 8. Option (3) is correct.

Explanation: P(two defective blade) = P= ( X 2) 0.02 02)) 22 ee 0.02 ((00..02   22!!  00..0004 0..9802 9802  0004 0   22   00..000196 000196 ∼ ∼ 00..0002 0002

4. Option (1) is correct. Explanation: P (no. defective blade) = e–0.02 = 0.9802 So, the approximate number of packets containing no defective blade

= 10,000 × 0.9802 = 9802

5. Option (2) is correct. Explanation:  P(one defective blade) = 0.02e–0.02

= 0.02 × 0.9802

= 0.0196 So, the approximate number of packets containing 1 defective blade

7 0

7

Explanation: From eq. (i), Probability of 6 success = P(X = 6)

= 0.02e −0.02 3. Option (1) is correct.

7r

...(i)

1 5  ∵ q  1  p  1  6  6    Probability of no success  P ( X  0))

P(No. defective blade) = P(X = 0)

1 6

6

1 5 =7 C6     6 6



1 = 35   6 9. Option (1) is correct.

7 −6

7

Explanation: Probability of at least 6 success = P(X ≥ 6) = P(X = 6) + P(X = 7) 6

1 5  7C6     6 6

7 6

7

1 5  7 C7     6 6

6

7

1 5 1 5  7C6      7 C7     6 6 6 6 6

1 5 1  7       6 6 6 6

7

 1   35 1   1        6  6 6 6

5

0

7 7

176 Oswaal CUET (UG) Chapterwise Question Bank 10. Option (2) is correct.

= 1 – P(X = 7)

Explanation:

1 5 = 1 −7 C7     6 6

From eq. (i), Probability of at most 6 success = P(X ≤ 6) = 1 – P(X > 6)



7

1 =1−   6

7

7 −7

MATHEMATICS/APP. MATH.

Course of Action Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

14

Revision Notes  

NUMBERS, QUANTIFICATION AND NUMERICAL APPLICATION

 Modulo Arithmetic Modulo Arithmetic is a special type of arithmetic that involves only integers and the operations used are addition, subtraction, multiplication and division. The only difference between modulo arithmetic and simple arithmetic that we have learned in our primary classes is that is modulo arithmetic all operations are performed regarding a positive integer i.e., the modulus. Let’s revise the division theorem that tells us that for any two integers a and b where b ≠ 0, there always exists unique integer a and r such that a = qb + r and 0 ≤ r < |b|. e.g., a = 39, b = 5, we can find q = 7 and r = 4. So, that 39 = 7 × 5 + 4. Here, a is called dividend b is called divisor q is called quotient r is called remainder If r = 0, the we say b divides a or a is divisible by b. This established a natural congruence relation on the integers.  Congruence Modulo Let m be a positive integer and I be the set of all integers. The relation “congruence modulo m” is defined on all a, b ∈ I by a ≡ b (mod m) if and only if m divides (a – b). The symbol “ ≡ (mod m)” is read as “congruence modulo m”. “m divides (a – b)” or “m is a factor of (a – b)” is usually denoted by “m|(a – b)”. Clearly, a ≡ b (mod m) if (a – b) is a multiple of m, i.e., a = b + km for some integer k. Thus, • 69 ≡ 25 (mod 4), since 4|(69 - 25) or 4 divides (69 - 25) or (69 - 25) = 44 is a multiple of 4. • 8 = - 8 (mod 3), since 8 - (-8) = 16, which is not a multiple of 3.  Properties of Congruence Modulo Property I: If a ≡ b (mod m), then b ≡ a (mod m)

Property VII: If a ≡ b (mod m), and c ≡ d (mod m), then, ac ≡ bd (mod m) Property VIII: If x, is a solution of ax ≡ Scan to know b (mod m), then any other integer x2 ≡ x1, more about this topic (mod m) is also a solution. Property IX: The congruence modulo ax ≡ b (mod m) has a solution iff the greatest common divisor (GCD) of a and m divides Alligation and b. If the greatest common divisor d, of a Mixture and m, divides b, then congruence modulo has exactly d incongruent solutions.  Alligation and Mixture: It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price. There are two types of questions asked in alligation. Type 1: If two ingredients are mixed, then  Quantity of cheaper   C.P. of dearer    Mean Price     Quantity of dearer  Mean Price   C.P. of cheaper 

C.P. of a unit quantity of cheaper  C.P. of a unit quantity of dearer c d

m (d–m)

(m–c)

∴ (Cheaper quantity): (Dearer quantity) = (d - m): (m - c) Here, C.P. = Cost Price Mean Price (m): The cost price of a unit quantity of the mixture Example 1: In what ratio must rice at `9.30 per kg be mixed with rice at `10.80 per kg so that the mixture be worth `10 per kg? Sol. By the rule of alligation, we have

C.P. of 1 kg rice of 1st kind (in paise)  C.P. of 1 kg rice of 2nd kind (in paise)

d

c

Property II: If a ≡ b (mod m), then a (mod m) = b (mod m). Property III: If a ≡ b (mod m), then (a ± c) = (b ± c) (mod m). Property IV: If a ≡ b (mod m), then ac ≡ bc (mod m). Property V: If a ≡ b (mod m) and c ≡ d (mod m), then, (a + c) ≡ (b + d) (mod m). Property VI: If a ≡ b (mod m), and c ≡ d (mod m), then, (a - c) ≡ (b - d) (mod m)

(d – m)

∴ Required ratio = 80 : 70 = 8 : 7

(m– c)

Second Level

Trace the Mind Map

First Level





Third Level

m

e r i c al Ap

Congruence modulo

ca pli



u an d N

• If a  b (mod m), then b  a (mod m) • If a  b (mod m), then (a + c)  (b + c) (mod m) • If a  b (mod m), then ac  bc (mod m) • If a  b (mod m) and c  d (mod m), then (a + c)  (b + d) (mod m) • If a  b (mod m) and c  d (mod m), then ac  bd (mod m) • If x1 is a solution of ax  b (mod m), then any other integer x2  x1 is also a solution.

Properties of congruence modulo

Let a, b  Z, m  Z+ Then, a is congruent to b modulo m written as 'a  b (mod)m' iff 1. a mod m = b mod m 2. m |(a – b) i.e., (a – b) mod m = 0

178 Oswaal CUET (UG) Chapterwise Question Bank

tion

MATHEMATICS/APP. MATH.

179

NUMBERS, QUANTIFICATION AND NUMERICAL APPLICATION

Type 2: Suppose a container contains x units of liquids from which y units are taken out and replaced by water. n   y  After n operations, the quantity of pure liquid =  x 1      x  units    Boats and Streams Stream is the moving water in the river Upstream is the direction against the stream. Downstream is the direction along the stream. Still water is the state where water is considered to be stationary and the speed of the water in this case is zero. • If the speed of a boat in still water is u km/h and the speed of the stream is v km/h, then: Speed of boat in downstream = (u + v) km/h Speed of boat in upstream = (u - v) km/h • If the speed of boat in downstream is a km/h and the speed of boat in upstream is b km/h, then: 1 Speed of boat in still water = (a + b) km/h 2 1 Speed of stream = (a - b) km/h 2 Example 2: In a stream running at 2 km/h, a motorboat goes 6 km upstream and back again to the starting point in 33 minutes. Find the speed of the motorboat in still water. Sol. Let the speed of the motorboat in still water be x km/h. Then, Speed downstream = (x + 2) km/h   Speed upstream = (x - 2) km/h 6 6 33 11 ∴    x  2 x  2 60 20 2 ⇒  11x - 240x - 44 = 0 ⇒  11x2 - 242x + 2x - 44 = 0 ⇒ (x - 22) (11x + 2) = 0 ⇒  x = 22 Hence, speed of motorboat in still water is 22 km/h.  Pipes and Cistern Scan to know Inlet: A pipe connected with a tank or a more about this topic cistern or a reservoir, that fills it, is known as inlet. Outlet: A pipe connected with a tank or cistern or a reservoir, emptying it, is called an outlet. Pipes and • If a pipe can fill a tank in hours, then Cistern 1 part of tank filled in 1 hour = . x • If a pipe can empty a full tank in y hours, then part of tank 1 emptied in 1 hour = y • If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both 1 1 pipes, the net part filled in 1 hour     x y • If a pipe can fill a tank in hours and another pipe can empty the full tank in y hours (where x > y), then on opening both 1 1 pipes, the net part emptied in 1 hour      y x Example 3: A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously then after how much time will the cistern get filled ?

Sol. Part filled by first tap in 1 hour = Part emptied by second tap 1 hour =

1 4

1 9

1 1 5 So, net part filled in 1 hour       4 9  36 36 ∴ The cistern will be filled in hours i.e., 7.2 hours or 7 hours 5 12 minutes  Races and Games Races: A contest of speed in running, Scan to know riding, driving, sailing or rowing is called more about this topic a race. Race Course: The ground or path on which contests are made is called race course. Starting Point: The point from which a race begins is known as a starting point. Races and Games Winning Point or Goal: The point set to bound a race is called a winning point or a goal. Winner: The person who first reaches the winning point is called winner. Dead Heat Race: If all the persons contesting a race reach the goal exactly at the same time, then the race is said to be a dead heat race. Start: Suppose A and B are two contestants in a race. If before the start of the race, A is at the starting point and B is ahead of A by 15 meters, then we say that ‘A gives B, a start of 15 metres. So, in case to cover a race of 100 metres, A will have to cover 100 metres while B will have to cover only (100 - 15) metres = 85 metres. In other words, we can say that to cover a race of 100 metres ‘A can give B, 15 meters’ or ‘A can give B a start of 15 meters’ or ‘A beats B, by 15 metres. Games: A game of 100, means that the person among the contestants who scores 100 points first is the winner. If A scores 100 points while B scores only 80 points, then we say that ‘A can give B 20 points’. Example 4: Aman can run 1 km in 3 min 10 s and Binay can cover the same distance in 3 min 20 s. By what distance can A beat B ? Sol. Clearly, Aman beats Binay by 10 s. Distance covered by Binay in 10 s  1000    10  m  50 m  200  [ 1 km = 1000 m and 3 min 20 s = 200 s] Thus, A beats B by 50 metres.  Inequalities: Two real numbers or two algebraic expression related by the symbol ‘’,‘≤’, or ‘≥’, form an inequality  Types of inequalities Numerical inequality: An inequality which does involve any variable is called a numerical inequality. e.g.: 4 >2, 8 < 21 Literal inequality: An inequality which have variable is called a literal inequality. e.g.: x < 7, y ≥11, x ≤ 4 Strict inequality: An inequality which have only < or> is called strict inequality. e.g.: 3x + y < 0, x 7

180 Oswaal CUET (UG) Chapterwise Question Bank Slack inequality: An inequality which have only ≥ or ≤ is called slack inequality. e.g.: 3x + 2y ≤ 0, y ≤ 4 Linear inequality: An inequality is said to be linear, if each variable occurs in first degree only and there is no term involving the product of the variables. e.g.: ax + b ≤ 0, ax + by + c > 0, ax ≤ 4 An inequality in one variable in which degree of variable is 2, is called quadratic inequality in one variable. e.g.: ax² + bx + c ≥ 0, 3x² + 2x + 4 ≤ 0  Linear inequality in one variable A linear inequality which has only one variable, is called linear inequality in one variable. e.g.: ax + b < 0, where a ≠ 0 4c + 7 ≥ 0 Replacement Set: The set from which values of the variables (involved in the inequality) are chosen is called the ‘replacement set’. Solution Set: A solution to an inequality is a number (chosen from replacement set) which, when substituted for the variable, make the inequality true. The set of all solutions of an inequality is called the ‘solution set’ of the inequality. e.g.: Consider the inequality x < 5 Replacement Set Solution Set (i) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} {1, 2, 3, 4} (ii) {-1, 0, 1, 2, 3, 5, 8} {-1, 0, 1, 2, 3} (iii) {-5, 1, 9, 10} {-5, 1} (iv) (6, 7, 8, 9, 10) ϕ Rules for solving inequalities in one variable: The rules for solving inequalities are similar to those for solving equations except for multiplying or dividing by a negative number. (i) If a ≥ b, then a ± k ≥ b ± k, where k is any real number. (ii) If a ≥ b, then ka is not always ≥ kb

MATHEMATICS/APP. MATH.

(iii) If k > 0 (i.e., positive), then a ≥ b ⇒ ka ≥ kb (iv) If k < 0 (i.e., negative), then a ≥ b ⇒ ka ≤ kb Thus, always reverse the sign of inequality while multiplying or dividing both sides of an inequality by a negative number. Procedure to solve a linear inequality in one variable: (i) Simplify both sides by collecting like terms. (ii) Remove fraction (or decimals) by multiplying both sides by appropriate factor (L.C.M. of denominator or a power of 10 in case of decimals) (iii) Isolate the variables on one side and all constant on the other side. (iv) Make the coefficient of the variable equal to 1. (v) Choose the solution set from the replacement set. 3x  4 x  1   1 Show the graph of the Example 5: Solve 2 4 solutions on number line. 3x  4 x  1  1 Sol. We have, 2 4 3x  4 x  3  or, 2 4 or, 2(3x-4) ≥ (x -3) or, 6x – 8 ≥ x - 3 or, 5x ≥ 5 or, x≥1 The graphical representation of solution is given as: –4 –3 –2 –1 0 1 2 3 4 5  Linear inequality in two variables The inequality of form ax + by + c > 0, ax + by + c = 0, or ax + by + c < 0 etc. where a ≠ 0, b ≠ 0 is called a linear equality in two variables x and y.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS

1. The cost of type A cement is ` 100/kg and that of Type B cement is ` 120/kg. If both are mixed in the ratio of 2 : 3, the price of the cement mixture per kg will be:  [CUET 2023] (1) ` 118 (2) ` 115 (3) ` 112 (4) ` 110 2. Two inlet pipes can fill a tank in 20 minutes and 24 minutes, respectively. An outlet pine can empty 30 litres of water per minute. If all 3 pipes working together can fill the tank in 15 minutes. The capacity of the tank is:  [CUET 2023] (1) 120 litres (2) 1200 litres (3) 240 litres (4) 2400 litres 15 3. Evaluate : 3 mod (7) [CUET 2023] (1) 3 (2) 2 (3) 6 (4) 25 4. Jeep A and Jeep B are competing in a motor race. After starting together, Jeep B covers the target of 30 km in 30 min 04 sec. Jeep A covers the target in 30 min 04 sec. By what distance will Jeep A beat Jeep B? [CUET 2023] (1) 30 m

(2) 40 m

(3) 50 m

(4) 60 m

5. A motor boat can row at the speed of 12 km/h in still water. If the river is flowing at 4 km/h and it takes 12 hours for a round trip, then the distance between the two places is:  [CUET 2023] (l) 48 km (2) 32 km (3) 64 km (4) 36 km 6. Pure honey costs ` 300 per litre. A shopkeeper adds water to 10 litres of pure honey and sells the resulting syrup at ` 250 per litre. The quantity of water added by the shopkeeper is: [CUET 2023] (1) 2 litres (2) 5 litres (3) 3 litres (4) 1.5 litres 7. Three persons A, B and C enter into a partnership to run a 4 5 6 business. They invested their capitals in the ratio : : . 3 2 5 After 5 months B increases his share by 40%. If the total profit at the end of a year is ₹ 50,550, then A’s share in the profit is: [CUET 2023] (1) ₹8,000 (2) ₹10,000 (3) ₹20,000 (4) ₹12,000 8. If 57 = x (mod 5). Then the least positive value of x is:  [CUET 2023] (1) 57 (2) 5 (3) 4 (4) 2

181

NUMBERS, QUANTIFICATION AND NUMERICAL APPLICATION

9. The speed of a motor boat in still water is 14.4 times the speed of the current of water. If the motor boat covers a certain distance upstream in 6 hours 25 minutes, then the time taken by the motor boat to come back is:  [CUET 2023] (1) 5 hours 35 minutes (2) 5 hours 25 minutes (3) 5 hours 10 minutes (4) 5 hours 55 minutes 10. The longest side of a triangle is four times the shortest side. The third side of the triangle is 3 cm shorter than the longest side. If the perimeter of the triangle is at least 69 cm, then its:  [CUET 2023] (1) Shortest-side < 8 cm (2) Shortest-side > 8 cm (3) Shortest-side ≤ 8 cm (4) Shortest-side ≥ 8 cm 11. Match List I with List II  [CUET 2023] List I A. The solution set of the inequality 3 x + 7 >12

I.

B. The solution set of the inequality 3x  5  1, x  R 2

II.

List II [-1, ∞) 17   8 ,   

C. The solution set of the inequality III. 5   2x + 5 5 has solution:  [CUET 2022] (1) (2, ∞) (2) (2, 8) (3) (8, ∞) (4) (-∞, 8) 20. The value of x which satisfied 3 x  6  3 x  [CUET 2022] (A) [0, 1] (B) [l, 4] (C) (4, ∞) (D) (-1, 0) (E) (-∞, 0) Choose the correct answer form the options given below: (1) A and B only (2) C and E only (3) B and C only (4) D and E only 21. What is the least value of ‘x’ that satisfies x = 27 (mod 4), when 27 < x ≤ 36? (1) 27 (2) 30 (3) 31 (4) 35 22. Let p > 0 and q < 0 and p, q ∈ Z, then choose the correct inequality from the given below options to complete the statement: p + q ...... p - q (1) > (2) ≥ (3) ≤ (4) < 23. A person can row a boat along the stream of the river at 10 km/h and against the stream in 6 km/h. What is the speed of the stream flow? (1) 1 km/h (2) 2 km/h (3) 4 km/h (4) 5 km/h 24. Two water supplying trucks - A and B supply water to remote areas. Truck A is carrying 100 litres of water to a village 1.5 km away and truck B is delivering 80 litres of water to another village, 1 km away. Due to bad road conditions, each truck loses 20 ml water while travelling each metre distance. Which truck is able to deliver more water and by how much more? (1) Truck A, 20 litres (2) Truck B, 20 litres (3) Truck A, 10 litres (4) Truck B, 10 litres 25. In what ratio shall I add water to the liquid detergent costing ` 480 per litre to get resulting mixture worth ` 300 per litre? (1) 5 : 3 (2) 3 : 8 (3) 3 : 5 (4) 5 : 8 26. If 100 = x (mod 7), then the least positive value of x is:  (1) 2 (2) 3 (3) 6 (4) 4

182 Oswaal CUET (UG) Chapterwise Question Bank 27. The remainder when 561 is divided by 7 is: (1) 1 (2) 2 (3) 4 (4) 5 28. If a man rows 32 km downstream and 14 km upstream in 6 hours each, then the speed of the stream is: (1) 2 km/h

(2) 1.5 km/h

MATHEMATICS/APP. MATH.

8. Assertion (A): In what ratio water is mixed with milk to 2 gain 16 % on selling the mixture at cost price. 3 Reason (R): To solve the above statement we use the rule of alligation i.e.,

(3) 2.5 km/h (4) 2.25 km/h 29. In a 2 km race, P can give Q a start of 200 m and R a start of 560 m. Then in the same race, Q can give R a start of (1) 360 m

(2) 380 m (3) 400 m (4) 430 m x3  0, x  5, x  ℝ is: 30. The solution of x5 (1) x > 3 (2) x < -5 (3) x < -5 or x > 3

(4) no solution

[B] ASSERTION REASON QUESTIONS Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: (1)  Both A and R are true and R is the correct explanation of A (2)  Both A and R are true but R is not the correct explanation of A (3)  A is true but R is false (4)  A is false and R is true 1. Assertion (A): (186 × 93) mod 7 = 1

Reason (R): Assertion can be solved using the formula a . b (mod n) = a (mod n) . b (mod n)

2. Assertion (A): When 14 is divisible by 5, then remainder is 4. So, we write 14 (mod 5) = 4.

Reason (R): 37 ≡ 12(mod 5)



Here, 37 - 12 ≡ 25 is an integral multiple of 5.

3. Assertion (A): The last two digits of the product 2345 × 6789 are 05.

Reason (R): To find the last two digits of the product 2345 × 6789, we find 2345 × 6789(mod 100).

4. Assertion (A): When 783 × 657 × 594 × 432 × 346 × 251 is divided by 5, then remainder is 3.

Reason (R): To find the remainder, we find (783 × 657 × 594 × 432 × 346 × 251) ≡ 3 (mod 5)

5. Assertion (A): The last three digits of the product 1234 × 5678 are 6, 5 and 2, respectively.

Reason (R): To find the last two digits of the product 1234 × 5678, we find 1234 × 5678(mod 1000).

6. Assertion (A): Let x and b are real numbers. If b > 0 and |x| > b, then x ∈ [- ∞, - b) ∪ (b, ∞).

Reason (R): Let x and b are real numbers. If b > 0 and |x| > b, then x ∈ [- ∞, - b) ∪ (b, ∞).

7. Assertion (A): If in a 1000 m race, A can beat B by 100 m. In a race of 400 m, B can beat C by 40 m then in a 500 m race A beats C by 95 m.

Reason (R): A beat C by ‘x’ meters means C is behind A by x meters, when A reached of finishing point.



e.g. To cover a race of 100 meters in this case, A has covered 100 meters while C has covered only (100 - x) meters.

(Cheaper Quantity): (Dearer Quantity) = (d – m) : (m – c) 9. Assertion (A): If the boat takes 12 hours to row 48 km upstream and 8 hours to row the same distance downstream, then the boat’s speed in still water is 5 km/h and speed of river is 1 km/h. Reason (R): Let speed of boat and speed of river X km/h and Y km/h, respectively Downstream speed = X + Y Upstream speed = X - Y Distance covered Also, Downstream speed = TimeTaken Distance covered Upstream speed = TimeTaken 10. Tap P alone fills a cistern in 2 hours; while tap Q alone fills the same cistern in 3 hours. A new tap R is attached to the bottom of the cistern which can empty the completely filled cistern in 6 hours. Assertion (A): Sunny started all three taps together at 9:00 a.m. then the tank will be full at 10:30 a.m. Reason (R): Sunny started all three taps together at 9:00 a.m. then the tank will be full at 09:30 a.m.

[C] COMPETENCY BASED QUESTIONS I. Read the following text and answer the following questions on the basis of the same: The 12 hour clock is most commonly represented on an analogue clock with number 1-12, when shown on a digital clock, it usually accompanied by a.m. or p.m. The 24 hour clock is more often shown on digital clocks and is written in a 4 digit form, with the first two digits representing the hours and the last two digits representing the minutes. There is no need for a.m. or p.m. as each time represents each hour in a 24 hour day.

183

NUMBERS, QUANTIFICATION AND NUMERICAL APPLICATION

1. It is currently 7 : 00 a.m. in 12 hour clock. The time in next 493 hours will (1) 8 : 00 a.m. (2) 8 : 00 p.m. (3) 20 : 00 a.m. (4) 20 : 00 p.m. 2. If the time after 640 hours from now will be 9 a.m., then the current time is (1) 5 : 00 a.m. (2) 5 : 00 p.m. (3) 4 : 00 a.m. (4) 4 : 00 p.m. 3. Which of two times are same in 12 hours and 24 hours clocks respectively? (1) 1 : 00 p.m., 13 : 00 (2) 3 : 00 p.m., 03 : 00 (3) 02 : 00 a.m., 14 : 00 (4) 04 : 00 a.m., 16 : 00 4. 4. a.m. stands for: (1) ante meridiem (2) anti meridiem (3) after mid-day (4) None of these 5. What time in 24 hour clock is equivalent to 7 a.m. in 12 hour clock? (1) 00 : 07 (2) 07 : 00 (3) 19 : 00 (4) 14 : 00 II. Read the following text and answer the following questions on the basis of the same: Sohil is good in Mathematics. His friend gave him a question from the topic Pipes and Cisterns i.e., “Pipe R can empty a full tank in 30 hours. But two pipes P and Q can fill a tank in 15

hours and 10 hours respectively. Ram unknowingly opened all three taps. After 2 hours Shyam realized it and closed Pipe R.”

6. In 1 hour tank filled by pipe P is 1 1 1 (1) (2) (3) 15 30 10

(4)

5 30

7. In 1 hour tank emptied by pipe R is 1 1 1 (1) (2) (3) 15 30 10

(4)

5 30

8. When P, Q and R all are open, then tank filled in 1 hour is 1 1 4 5 (1) (2) (3) (4) 15 30 30 30 9. If R is closed, then entire tank get filled in how much time? (1) 6 hours (2) 4 hours (3) 8 hours (4) 2 hours 10. When P, Q and R all are open, then tank filled in 2 hours is (1)

1 15

(2)

8 30

(3)

4 30

(4)

5 30

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (2)

3. (3)

4. (3)

5. (3)

6. (1)

7. (4)

8. (4)

9. (1)

10. (4)

11. (2)

12. (1)

13. (2)

14. (2)

15. (1)

16. (2)

17. (1)

18. (2)

19. (3)

20. (3)

21. (3)

22. (4)

23. (2)

24. (3)

25. (3)

26. (1)

27. (4)

28. (2)

29. (3)

30. (3)

8. (1)

9. (1)

10. (3)

8. (3)

9. (1)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (2)

3. (1)

4. (1)

5. (1)

6. (3)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (2)

3. (1)

4. (1)

5. (2)

6. (1)

7. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (3) is correct. Explanation: Type A 100

Type B 120

m

(120 – m)

(m – 100)

120  m 2  m  100 3 ⇒ 360 - 3m = 2m - 200 ⇒ 5m = 560 ⇒ m = ` 112 2. Option (2) is correct. Explanation: Let the outlet pipe empty the tank in x minutes. 1 1 Portion of tank filled by both inlet pipes in 1 hour   20 24 1 Portion of tank emptied by outlet pipe in 1 hour = x According to question, we have Now,

184 Oswaal CUET (UG) Chapterwise Question Bank 1 1 1 1    20 24 x 15 9 1 1   80 x 15 1 1 1 1 60  50  80 30 1       x 20 24 15 1200 1200 40 ⇒ x = 40 minutes Outlet pipe can empty 30 litres per minute. In 40 minutes, it can empty = 30 × 40 = 1200 litres Thus, capacity of the tank = 1200 litres 3. Option (3) is correct. Explanation: We know that, 5 ≡ 5 (mod 7) 5 3 (3 ) ≡ (5)3 (mod 7) 315 ≡ 125 (mod 7) 315 ≡ 6 (mod 7) 4. Option (3) is correct. Explanation: 30 minutes 04 seconds = 1804 seconds 30 minutes 01 seconds = 1801 seconds In 181 seconds Jeep A covers = 30 km = 30,000 m Time is taken 3 second more than Jeep A. In 3 seconds, Jeep B covers = (30000/1804) × 3 = 49.89 m ~ 50 m 5. Option (3) is correct. Explanation: Speed in still water =12 km/h Speed of river (stream) = 4 km/h Speed of upstream = (12 - 4 ) km/h = 8 km/h Speed of downstream = (12 + 4) km/h = 16 km/h Total time taken for round trip = 12 h Let the distance be x km. x x    12 8 16 ⇒ 3x = 12 × 16 ⇒ x = 64 km 6. Option (1) is correct Explanation: Given, Quantity of pure honey = 10 litre. Per litre Price of pure honey = ` 300 Total selling price of 10 litre honey = 10 × 300 = ` 3000 After adding x litre of water into 10 litre of honey, Now new quantity of syrup (mixture of water and honey) = (10 + x) litre Given, Shopkeeper selling syrup at the rate of ` 250 per litre, \ Total sealing price = 250 (10 + x) According to the question, 3000 = 250 (10 + x) 3000 = 2500 + 250 x 500 = 250 x x = 2 litre 7. Option (4) is correct Explanation: One month ratio of investment of three persons 4 5 6 A, B and C = : : 3 2 5 = 40 : 75 : 36 A’s Investment for a year = 40x × 12  = 480x

MATHEMATICS/APP. MATH.

140  75 x  7 100   = 375x + 735x  = 1110x C’s Investment for a year = 36x × 12  = 432x Yearly ratio of capital investment of three persons A, B and C = 480x : 1110x : 432x = 80 : 185 : 72 Given, Total profit at the end of a year = ` 50,550 80  50550 A’s profit share  80  185  72 80   50550 337 = ` 12000 8. Option (4) is correct Explanation: We know that, when 57 is divided by 5, we get 2 as reminder, Hence x = 2 9. Option (1) is correct Explanation: Let the speed of current of water = x km/h speed of motor boat in still water = 14.4x km/h Speed in upstream = (14.4x - x) km/h Speed of motor boat in down stream = (14.4x × x) km/h  Time taken by boat in upstream = 6 h 25 min 25 6 60 77 = h 12 B’s Investment for a year = 75x × 5 

Distance time Distance 14.4 x  x  77 12 12 D 13.4 x  77 77 D  13.4  x 12 Now, time taken by motor in Downstream Distance = Speed of motor boat in Downstream D  14.4 x  x 13.4 x 77   15.4 x 12 134 1   2 12  5 hours35minutes 10. Option (4) is correct. Explanation: Let the shortest side = x cm Longest side = 4x cm Third side = (4x-3) cm Given, Perimeter ≥ 69  x + 4x + 4x -3 ≥ 69 9x ≥ 72 x≥8 Shortest side ≥ 8   Speed of motor boat in upstream =

185

NUMBERS, QUANTIFICATION AND NUMERICAL APPLICATION

11. Option (2) is correct. Explanation: A. 3x + 7 > 12 3x > 5 5 x  3 5  x  ,  3  Here, A Match with III 3x  5 B. 1 2 3x + 5 ≥ 2 ⇒ 3x ≥ -3 x ≥ -1 ⇒ x ∈ [-1, ∞) Here, B Match with I C. 2x + 5 < 7x + 9 5 - 9 < 7x - 2x ⇒ -4 < 5x 4  4    x x   ,  5  5  Here, C Match with IV D. 6x -5 ≥ -2x + 12 6x + 2x ≥ 12 + 5 ⇒ 8x ≥ 17 17 17  x  x   ,  8 8  Here D Match with II 12. Option (1) is correct Explanation:

Clearly, the gap between P and R is 300 m. 13. Option (4) is correct. Explanation: 17123 = 17123 (The result will remain the same) 7123 = 74×30+3 7123 = (74) 30 × 73 7123 = 130 × 73 7123 = 1 × 343 7123 = 343 14. Option (2) is correct. Explanation: A. 33 ≡ b(mod 9) Here, 33 ≡ 0(mod 9) B. 25 ≡ b(mod 15) Here, 25 ≡ 2(mod 15) C. 43 ≡ b (mod 10) Here, 43 ≡ 4(mod 10) D. 53 ≡ b(mod 12) Here, 53 ≡ 5(mod 12) There, A-II, B-III, C-I, D-IV 15. Option (1) is correct. Explanation: Ratio of milk and water in original mixture i.e., M1 Milk : Water 8:x After adding 3 litres of water in 33 litres of mixture the ratio of mixture is M2 Milk : Water 2:1 or 8:4

Since, the ratio of milk does not change as we are adding water only. Thus, the ratio of water increases by 3(= 4 - 1). 16. Option (2) is correct. Explanation: Speed of motor boat in still water = 15 km/h  Speed of current = 3 km/h Speed of upstream = speed of motor boat - speed of current = 15 - 3 km/h = 12 km/h Distance = speed × time \



Time =

36 = 3 hours 12

17. Option (1) is correct. Explanation: Hari beats Ram by (45 - 36) seconds = 9 seconds 100   Speed of Ram = 45 Distance covered by Ram in 9 seconds 100  9 45  20 m So, Hari is 20 m ahead from Ram. 18. Option (2) is correct. Explanation: 5 Emptied part in 20 minutes = 6 5 Emptied part in 1 minutes  6  20 59 3  Emptied part in 9 minutes  6  20 8 19. Option (3) is correct. Explanation: 2x - 1 ≥ 3 and x - 3 > 5 2x ≥ 4 and x > 8 x ≥ 2 and x > 8 x ∈ [2, ∞ ) and x ∈ (8, ∞) Therefore, x ∈ (8 ∞) 20. Option (3) is correct. Explanation: | 3 x||3x| 6| | 63 x|3 x | (| 3 x(||)32x |) 2(|6 (| 63 x|)32x |) 2 9x2 ≥ (6 - 3x)2

9x2 ≥ 36 - 36x + 9x2 0 ≥ 36 - 36x  0 ≥ 36(1 - x) 0≥1-x x≥1 x ∈ [1, ∞) Only options B and C lies in range of x. 21. Option (3) is correct. Explanation: x ≡ 27(mod 4) ⇒ x - 27 = 4k, for some integer k ⇒ x = 31 as 27 < x ≤ 36

186 Oswaal CUET (UG) Chapterwise Question Bank 22. Option (4) is correct. 23. Option (2) is correct. Explanation: Speed of boat downstream = u = 10 km/h And, speed of boat upstream = v = 6 km/h 1 ⇒ Speed of stream = (u - v) = 2 km/h 2 24. Option (3) is correct. Explanation: 20  1500 Truck A carries water  100   70l 1000 20  1000 Truck B carries water  80   60l 1000 25. Option (3) is correct. Explanation:

⇒ 180 : 300 = 3 : 5 26. Option (1) is correct. Explanation: We know that, a = b(mod c) Then b is the remainder, when a is divided by c. 100 = x(mod 7) 7) 100 (14 7 30 28 2 → Remainder Thus, x = 2 27. Option (4) is correct. Explanation: 561 = (53)20 × 51 = (126 - 1)20 × 51 = [12620 - 20C112619 + 20C212618 + ......... + 20C19 1261 + 20C201260120] × 5 = 126[12619 - 20C112618 + 20C212617+ ............. + 20C19] × 5 +1×5 = 126 × 5 [12619-20C112618 + 20C212617 + .............. + 20 C19] + 5 Here, 126 × 5 = 630, which is divisible by 7, thus remainder is 5. 28. Option (2) is correct. 14 Explanation: Upstream speed = km/h 6 32 Downstream speed = km/h 6 32 14 18  6  6   Speed ot the stream  6 2 2 18 3     6 2 2 = 1.5 km/h

MATHEMATICS/APP. MATH.

29. Option (3) is correct. Explanation: Let P is complete the race 2000 m and Q is at 1800 m and R is at 1440 m ⇒ In a race of 1800 m Q R P 1440 1800 2000 Q gives R a start at = 1800 - 1440 = 360 360  ⇒ 1m race = 1800  360 2000 m race =  2000 ⇒ 1800 = 400 m 30. Option (3) is correct. x3  0, x  5 Explanation: Given, x5 2 Multiply by (x + 5) (x + 5) (x - 3) > 0 + + – –∞ –5 –3 ∞ \ Solution (- ∞, -5) ∪ (3, ∞) or x < -5 x>3

[B] ASSERTION REASON QUESTIONS 1. Option (1) is correct. Explanation: We know that, a . b(mod n) ≡ a(mod n) . b(mod n) So, (186 × 93)mod 7 ≡ 186(mod 7) . 93(mod 7) ≡ 4(mod 7) . 2(mod 7) ≡ 4.2 (mod 7) ≡ 8(mod 7) ≡ 1(mod 7) 2. Option (2) is correct. Explanation: Both statements are individually correct. Both the statements are of Modulo arithmetic which is system of arithmetic for integers, which considers the remainder. 3. Option (1) is correct. Explanation: Since 2345 = 45(mod 100) and 6789 ≡ 89(mod 100)  So, 2345 × 6789 ≡ 45 × 89(mod 100) ≡ 4005(mod 100) ≡ 05(mod 100) Hence, the last two digits of the product 2345 × 6789 are 05. 4. Option (1) is correct. Explanation: Since, 783 ≡ 3(mod 5)  657 ≡ 2(mod 5)   594 ≡ 4(mod 5)   432 ≡ 2(mod 5)   346 ≡ 1(mod 5) and 251 ≡ 1(mod 5) So, (783 × 657 × 594 × 432 × 346 × 251)mod 5   ≡ (3 × 2 × 4 × 2 × 1)mod 5   ≡ 48(mod 5)   ≡ 3(mod 5) Hence, the remainder when (783 × 657 × 594 × 432 × 346 × 251) is divided by 5 is 3.

187

NUMBERS, QUANTIFICATION AND NUMERICAL APPLICATION

5. Option (1) is correct. Explanation: Since, 1234 ≡ 234(mod 1000) and 5678 ≡ 678(mod 1000) So, 1234 × 5678 ≡ 234 × 678(mod 1000) ≡ 158652(mod 1000) ≡ 652(mod 1000) Hence, the last three digits of the product 1234 × 5678 are 6, 5 and 2, respectively. 6. Option (3) is correct. Explanation: Given, |x| > b, b > 0 ⇒ x < - b or x > b ⇒ x ∈ (- ∞, - b) ∪ (b, ∞) 7. Option (1) is correct. Explanation: A : B = 1000 : 900,  B : C = 400 : 360 = 100 : 90 = 900 : 810 ⇒ A : B : C = 1000 : 900 : 810 ⇒  A : C = 1000 : 810 ⇒  A : C = 500 : 405 ⇒ In a 500 m race, A beats C by (500 - 405) m = 95 m 8. Option (1) is correct. Explanation: Let C.P. of 1 litre milk be ` 1. 50 S.P. of 1 litre of mixture = ` 1, Gain = % 3 \

3    1 C.P. of 1 litre of mixture  100  350   6  7

By the rule of alligation, we have 1 6 \ Ratio of water and milk = = : 1: 6 7 7 9. Option (1) is correct. Explanation: Let speed of a boat is X km/h and speed of river is Y km/h Distance covered Down stream speed = Time taken  =

48 = 6 km/h 8

Distance covered Time taken 48 = = 4 km/h 12    X + Y = 6 km/h    and X - Y = 4 km/h On adding them we get,  X + Y + X - Y = 10 km/h \   X = 5 km/h    Upstream speed =

= Speed of Boat Y = 6 - 5 = 1 km/h = Speed of river 10. Option (3) is correct. Explanation: 1 Cistern filled or work done by Tap P in 1 hour = 2 1 Cistern filled or work done by Tap Q in 1 hour = 3 1 Tank emptied or work done by Tap R in 1 hour = 6 Tank filed or work done by all three pipes in 1 hour 1 1 1 2     2 3 6 3 3 So Tank gets completely filled in hours 2  = 1.5 hours = 1 hr 30 min So time will be 9:00 a.m. + 1 hour 30 min = 10:30 a.m.

[C] COMPETENCY BASED QUESTIONS 1. Option (2) is correct. Explanation: We know that time repeat after every 24 hours. So, we find 493 (mod 24) So, 493 (mod 24) = 13 \ 493 hours are equivalent to 13 hours. Now, 7 : 00 a.m. + 13 hours = 8 p.m. 2. Option (2) is correct. Explanation: Here, we find 640 (mod 24) 640 (mod 24) ≡ 16 \ 640 hours is equivalent to 16 hours Now, 9 : 00 a.m. - 16 hours = 5 p.m. 3. Option (1) is correct. Explanation: Here, 3 : 00 p.m. in 12 hour clock = 15 : 00 in 24 hour clock 2 : 00 a.m. in 12 hour clock = 02 : 00 in 24 hour clock 4 : 00 a.m. in 12 hour clock = 04 : 00 in 24 hour clock So, only option (1) is correct i.e., 1 : 00 p.m. in 12 hour clock is same as 13 : 00 in 24 hour clock. 4. Option (1) is correct. Explanation: a.m. stands for ante meridien and p.m. stands for post meridien. 5. Option (2) is correct. Explanation: 7 : 00 hour in 24 hour clock is equivalent to 07 : 00 a.m. in 12 hour clock. 6. Option (1) is correct. Explanation: Pipe P can fill the tank in 15 hours. 1 So, in 1 hour tank filled by pipe P = 15 7. Option (2) is correct. Explanation: Pipe R can empty the tank in 30 hours. 1 So, in 1 hour tank emptied by pipe R = 30 8. Option (3) is correct. Explanation: When P, Q and R all are open, then tank filled in 1 1 1 4 1 hour     15 10 30 30

188 Oswaal CUET (UG) Chapterwise Question Bank 9. Option (1) is correct. Explanation: If pipe R is closed, then in one hour tank filled 1 1 5   15 10 30 30 So entire tank gets filled in = 6 hours 5 

MATHEMATICS/APP. MATH.

10. Option (2) is correct. Explanation: When P, Q and R all are open, then tank filled in 1 hour 1 1 1 4     15 10 30 30 4 8 So, in 2 hours  2   30 30

Course of Action Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

15

INDEX NUMBERS AND TIME-BASED DATA

 Revision Notes:  Index Numbers Index numbers are a special type of average used in statistics and economics to represent the relative changes in a set of data points over time, across different categories, or with respect to some base period. They are particularly useful for tracking trends and making comparisons. Scan to know more about Index numbers are essentially a weighted this topic average, where each data point is assigned a weight based on its relative importance or relevance. The formula for a simple index number is:

Where,

 P1 I   P0 

   100  

Broadly, the calculation of price index can be divided into two subgroups, namely (a) Simple (unweighted) Aggregate method. (b) Weighted Aggregates method. 1. Relative Index Number A relative index number is a type of index number used to express the relative change in a variable, such as prices, quantities, or values, in comparison to a base period or reference point. It is a straightforward method for measuring and comparing changes in data over time, across different categories, or in different locations. The formula for calculating a relative index number is as follows:

Index number

is the index number. is the sum of values in the current period.  ∑P0 is the sum of values in the base period.  I

 ∑P1

Use of Index Numbers 1. Index numbers help to measure changes in the standards of living as well as prices fluctuations. 2. Government policies are framed on the basis of index numbers of prices. 3. Index Numbers not only help in the study of past and present behavior, they are also used for forecasting economic and business activities. 4. Index numbers facilitate comparative study with respect to time and place, especially where units(weights) are different.

 Types of Index Numbers

Consider some price indexes that are important measure of business and economy. 1. Value Index is the measure of the average value for a particular period with that of the average period of the base period. It is used to keep stock of inventory, sales and trading etc. 2. Quantity Index is the measure of change in the quantity of goods (produced/ consumed/ sold) within a stipulated period of time. An example of quantity index is the Index of Industrial Production, known as IIP. 3. Price Index is the measure relative price change over a period of time. An example of price index is the Consumer Price Index, known as CPI.

Where:

 P R =   × 100  P0 

 is the relative index number.  is the current value or observation.  P0  is the value in the base period or reference point. In the same way, we can compute quantity index using the formula: Relative Quantity Index number in time period  R  P

R = In =

Q ×100 Qo

NOTE: An index series is a list of index numbers for two or more periods of time, on the basis of same base period. NOTE: For the sake of comparison, we shall consider the price level for the base period as 100 and the price level of a particular year in consideration is expressed relative to the base period. For example, if the relative index number for a specific item is 120% , it indicates that the price of that item has increased by 20% compared to the base period. Conversely, if the index number is 80% , it means the price has decreased by 20% . Example 1: A departmental store paid annually for newspaper and television advertisements in 2012 and 2022 as shown below: Expenditure

2012

2022

Newspaper (in ten thousand ₹)

1.3

2.9

Advertisement (in ₹)

1.8

3

Using 2012 as the base year, the price index for newspaper and television advertisement prices for year 2022 is (1) 223,186 (2) 223,167 (3) 222,168 (4) 222,167

First Level

Second Level

Third Level

How to understand Mind Map?

It helps to make forecast when all statistical information we take into account the last few data points, e.g: Data for last three, four, five and six periods. Three period moving average is defined as a value by adding numerical data for three periods and then dividing by 3. Four period moving average is defined as a value by adding numerical data for four periods and then dividing by 4

Additive model: Y =T + S + C + I Multiplicative models:Y =T S C I where, Y = original value T = Trend value S = Seasonal variations C = Cyclic variations I = Irregular variations

an ers

Measurement of Trends

-

 p1q0  100  p0 q 0

 p1q0   p1q1  100  p0 q0   p0 q1 





P01P

P01L

where p0 = base year price p1 = current year price q0 = base year quantity q1 = current year quantity

P01F

• Fisher's price Index Number

• Passche's price Index Number  p1q1 P01P   100  p0 q1

P01L 

where, p0 = price of base year, p1= price of current year, N = number of items, w = weight related to item

4. Weighted average of price relative method p    1  100   100  p2  P01  w

1. Simple aggregative method p1 P01 100 p0 2. Simple average of price relative method p    1  100   p0  P01  N 3. Weighted aggregative method  p1w  100 P01   p0 w

1. Unweighted (simple) Index Number • Simple aggregative method • Simple average of price relative method 2. Weighted Index Number • Weighted average of price relative method • Weighted aggregative method

Index numbers are used as an indicator to indicate the changes in economic activity. They also provide frame work for decision making and to predict future events. There are three types of index numbers (i) Price index, (ii) quantity index and (iii) Value index

• Laspeyre's price Index Number

Some other important formulae

Formulae

Methods of Constructing Index Numbers

Index Number

d Time

Time Reversal Test

where P01F is Fisher's price index number

P01F  P10F  1

Index number satisfies the relationship

The straight line trend in represented by Y=aX+b where, Y = Trend value to be computed X = Unit of time (Independent variable) a = Constant to be calculated b = Constant to be calculated Y  XY Also, a=  2 and b  2 X ∑ X2 X n = number of items

Mathematical models for time series

Time Series

Its componants:A time series is a set of observations taken at specified times, usually at equal time intervals

ex Numb Ind

It has 4 components:(i) Secular (simple) Trend (ii) Seasonal variations (iii) Cyclic variations (iv) Irregular or random variations

ta



s ed D a



Ba



190 Oswaal CUET (UG) Chapterwise Question Bank MATHEMATICS/APP. MATH

191

INDEX NUMBERS AND TIME-BASED DATA

Solution: Option (2) is correct. Explanation: I 2022 (Newspaper) I 2022 (Television)

Therefore, the index number of year 2022 on the base year 2014 = I 2022 =

2.9 × 100 = 223 1.3 3 × 100 = 167 1.8



a. Select a Base Period: Choose a reference period (the base period) for which the index number will be set to 100. b. Calculate the Index: Calculate the index for each period (other than the base period) by dividing the sum of the values in that period by the sum of the values in the base period and then multiplying by 100.

 P 

n Price index number in time period n = I n     100 P  0   Where,  In

is the index for period n .

∑Pn  ∑P0 

200.4 × 100 187 = 107.165 ≈ 107.2 =

Clearly, newspaper advertising cost increased at a greater rate as compared to television advertisement cost. 2. Simple (Unweighted) Aggregative Method This method consists of expressing aggregate of prices in any year as a percentage of their aggregate in base year. Here’s a basic outline of how the Simple Aggregative Method works:

i.e.,

∑ pn × 100 ∑ p0

From the above example, we can conclude that the price index number of year 2022 has only increased by 7.2% over the period of 2014 to 2022. 3. Weighted Aggregative Method Scan to know more about In this method appropriate weights are this topic assigned to different commodities to make them comparable and thus compatible for summation. The advantage of this method of computing index number is that the allotment of weights enables the Index Numbers and Timecommodities of greater importance to have Based Data more impact on index number. A weighted aggregative index is constructed as follows: ∑ PQ i i ×100 Index number in time period n = I n = ∑ PQ i i where, I n represents weighted aggregative index Qi is the quantity of usage for commodity i

is the sum of values in period n .

P0 represents unit price for the base period 0

is the sum of values in the base period.

c. Repeat for Other Periods: Calculate the index numbers for all the other periods, keeping the base period as the reference point (i.e., the index for the base period is always 100). d. Interpretation: The resulting index numbers indicate how much the values in each period have changed relative to the base period. If the index is greater than 100, it suggests an increase, and if it’s less than 100, it suggests a decrease. Example 2: A manufacturer purchases four distinct raw materials, that differ in unit price as given below: Commodity

Unit Price (₹) Year 2014

Unit Price (₹) Year 2022

A

3.20

3.8

B

1.70

2.1

C

148.10

149.50

D

34

45

An unweighted aggregate price index for year 2022 using year 2014 as the base period is (1) 7.2% (2) 7% (3) 7.4% (4) 7.6% Solution: Option (1) is correct. Explanation: Commodity

Unit Price (₹) Year 2014

Unit Price (₹) Year 2022

A

3.20

3.8

B

1.70

2.1

C

148.10

149.50

D

34

45

Total

∑p0 = 187

∑pn = 200.4

And, Pi represents unit price of commodity i for the current period n NOTE: Here, quantity is fixed and assumed not to vary over time as prices change. Example 3: What will happen to the index number for year 2014 if the commodities are used in different weights (quantities)? Commodity

Quantity (weights)

Unit Price (₹) Year 2014

Unit Price (₹) Year 2022

A

100

3.20

3.8

B

20

1.70

2.1

C

15

148.10

149.50

D

50

34

45

(1) 14% increased (2) 15% decreased (3) 15% increased (4) 14% decreased Solution: Option (3) is correct Explanation: 3.8  100  21  20  19.5  15  4  50  100 3.2  100  1.7  00  148.1  15  34  50 4914.3   100 4273.5  115

I 2022 

Manufacturer has increased by 15% over the period from year 2014 to year 2022.

192 Oswaal CUET (UG) Chapterwise Question Bank 4. Laspeyre’s Index In a special case of the fixed-quantity weights considered from base period usage, the weighted aggregative index is known by a new name, Laspeyres Index. The Laspeyres Price Index is defined by the following formula: Index number in time period



Pi  Q0  100   P0  Q0 

n = I n= L   Where, 

L is the Laspeyre’s Price Index.

∑Pi is the total expenditure on the basket of goods and services in the current period.  P0 is the price of the goods and services in the base period.  Q0 is the quantity of goods and services in the base period. 5. Paasche’s Index Paasche suggested that for determining quantity(weights) is to revise the quantity over time. When the fixed-quantity weights are considered from current period usage, the weighted aggregative index is known by another name, Paasche Index The Paasche Price Index is defined by the following formula: 

 Pi  Qi P  P0  Qi

   100 

Where,

Scan to know more about this topic

P is the Paasche Price Index.  Pi is the total expenditure on the basket of goods and services in the current period.  P0 is the price of the goods and services in the base period. Calculation of Weighted  Qi is the quantity of goods and services Average Cost in the current period. of Capital Example 4: Following table shows the data on energy consumption and expenditure at Badarpur Thermal Power Station, in Delhi region. 

Sector

Quantity (weights) Year 1994

Unit Price (v/kWh)

Year 2022

Year 1994

Year 2022

Commercial

5416

6015

1.97

10.92

Residential

15293

20262

2.32

6.16

Industrial

21287

17832

0.79

5.13

Agriculture

9473

8804

2.25

8.10

What is the different of Laspeyre’s index and Paasche’s index? (1) 29 (2) 20 (3) 23 (4) 22 Solution: Option (4) is correct. Explanation: (Laspeyre’s Index) I 2015 =



10.92 × 5416 + 6.16 × 15293 + 5.13 × 21287 + 8.10 × 9473 × 100 1.97 × 5416 + 2.32 × 15293 + 0.79 × 21287 + 2.25 × 9473

339281.21 × 100 84280.26 = 403 =

MATHEMATICS/APP. MATH

(Paasche’s Index) I 2015 =

10.92 × 6015 + 6.16 × 20262 + 5.13 × 17832 + 8.10 × 8804 × 100 1.97 × 6015 + 2.32 × 20262 + 0.79 × 17832 + 2.25 × 8804

353288.28 × 100 92753.67 = 381 Different between Laspeyres’ index and Paasche index = 403 − 381 = 22 NOTE: Paasche value being less than the Laspeyres indicates usage has increased faster in the lower priced sectors. 6. Fisher’s Ideal method This index calculation gives the geometric mean* of Laspeyre’s and Paasche’s methods. Index number in time period

=

n = In =

∑ P1Q0 ∑ P1Q1 × × 100 ∑ P0 Q0 ∑ P1Q1

7. Marshall-Edgeworth’s Method The statistician duo, Marshall and Edgeworth proposed that index number is to be calculated by taking the average of the base year and the current year. ∑ P  Q  Qi  Index number in time period n  I n ME  i 0  100 ∑ P0  Q0  Qi  8. Weighted Average of Relatives This method makes use of price relatives. When the base and current prices of a number of commodities with varying weights or quantities are given, then this method to construct index number is recommended. P ∑ i ( P0 Q0) P0 Index number in time period n = I n = × 100 ∑ P0Q0

 Test of Adequacy of Index Numbers

The following tests are available for checking the adequacy of index number1. Unit test 2. Time reversal test 3. Factor reversal test 4. Circular test These tests maintain consistency by verifying their adequacy. Here, we consider only time reversal test. The time-reversal test is used to test whether the method of constructing index number will work with any consideration of time period. This test says that the method used should give the same ratio between one point or another for comparison; no matter which time period is taken as base period. Basically, if the time subscripts ( Po and Pn ) of a price or quantity index number are interchangeable then the resulting price/quantity relative should be reciprocal of the original price/quantity relative - i.e. if P0 represents price of wheat in year 2017 and P1 represent price in year 2022; then P01  P10  1 Here, P01 is the index for current year ‘ 1 ‘ on the basis of base year ‘ 0 ‘ And, P10 is the index for year ‘ 0 ‘ based on year ‘ 1 ‘

193

INDEX NUMBERS AND TIME-BASED DATA

Clearly this test of adequacy cannot be tested on Laspeyers’ method and Paasche’s method because Clearly this test cannot be tested on Laspeyres’ method of index number because

 PQ   P Q  P Q  PQ 1 0

0 1

0 0

1 1

1

Also the Paasche method of index number cannot be tested for adequacy using this test as

 PQ   P Q  P Q  PQ 1 1

0 0

0 1

1 0

1

Whereas, the Fisher’s Ideal index number satisfies the timereversal test

 Time Series

A time series is a sequence of data points collected at regular time intervals, used to study trends and patterns over time, enabling predictions and analysis of time-dependent phenomena. Such data analysis is considered in three types: Time series data: when data of the variable is collected at distinct time intervals, for a specified period of time.  Cross-sectional data: when data for one or more variables is collected at the same point in time.  Pooled data: when data in a combination of time series data and cross-sectional data is collected. Forecasting methods can be classified as quantitative and qualitative. Quantitative method of forecasting can be used: 

When the past information about the variable is available. When information and data of the variable can be quantified.  On the assumption that the pattern of the past will continue in the future.  The variable has a cause-and-effect relationship with one or more other variables When forecasting is done based on historic data of past values, it is called a time series method. Qualitative method is generally based on Scan to know more about expert judgement and analytical opinion to this topic develop forecasts. One of the benefits of using these methods is that they can be applied when information on the data of the variable cannot be quantified or historic information Time-Based is neither available nor applicable.  

 Time Series Analysis

Data (Time Series)

A time series in which data of only one variable is varying over time is called a univariate time series/data set. For example, data collected from a temperature sensor measuring the temperature of a place every second, the data will show us only one-dimensional value - temperature. 1. Secular trend component: also known as trend series, is the smooth, regular and long-term variations of the series, observed over a long period of time shows an upward trend for annual electricity consumption per household in a certain residential locality from years 2016 - 2022. In general, trend variations can be either linear or non-linear.

2016

2017

2018

2019

2020

2021

2022

2. Seasonal component - when a time series captures the periodic variability in the data, capturing the regular pattern of variability; within one-year periods. The main causes of such fluctuations are usually climate changes, seasons, customs and habits which people follow at different times. e.g., seasonal electricity consumption and variations of peak demand in Nepal and India 3. Cyclical component - when a time series shows an oscillatory movement where period of oscillation is more than a year where one complete period is called a cycle. 4. Irregular component - these kinds of fluctuations are unaccountable, unpredictable or sometimes caused by unforeseen circumstances like - floods, natural calamities, labor strike etc. Such random.

 Trend Analysis by fitting Linear Trend Line

Among the four components of the time series as discussed above, the secular trend analysis (also known as trend analysis) depicts the long-term direction of the series. One of the most widely used in practice mathematical techniques of finding the trend values is the method of least squares. It plays an important role in finding the trend forecasts for the future economic and business time series data Trend can be measured using the following methods: 1. Moving averages method 2. Method of least squares We shall be studying two methods to compute trend line from the above list. 1. Trend Analysis by Moving Average Method This method is used to draw smooth curve for a time series data. It is mostly used for eliminating the seasonal variations for a given variable. The moving average method helps to establish a trend line by eliminating the cyclical, seasonal and random variations present in the time series. The period of the moving average depends upon the length of the time series data. As shown in the figure the smooth curve is the trendcycle, which is noticeably smoother than the original data and captures the main movement of the time series ignoring the minor fluctuations. The order of the moving average determines the smoothness of the trend-cycle estimate. Annual electricity sales: Asia While using this method: 1. Choice of the length of moving average is very important 2. Appropriate length plays an important role in smoothing the variations

194 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH

Let us take an example for n = 3 years moving averages to understand the procedure.

s

Procedure for calculating Moving average for odd number of years (n = odd)

1. Add up the values of the first 3 years and place the yearly sum against the median (middle) year. (This sum is called 3-year moving total) 2. Continue this process by leaving the first-year value, add up the next three year values and place it against its median year. 3. This process must be continued till all the values of the data are taken for calculation. 4. Calculate the n-year average by dividing each n-yearly moving total by n to get the n-year moving averages, which is our required trend values. 5. There will be no trend value for the beginning period and the ending period in this method.

Example 5: Compute centered averages at 2018. Assuming that 4-year cycle present in the following series is : Year

2013

2014

2015

2016

2017

2018

2019

2020

2021

2022

Index number

400

470

450

410

432

475

461

500

480

430

(1) 455.75 (2) 442.38 (3) 437.75 (4) 455 Solution. Option (1) is correct. Explanation: The 4-year moving averages are shown in the last column as centered average. Year

Index Number

2013

400

2014

4-year Moving Average

–

–

470 →

2015

4-year Moving total

1730

2016

2017

2018

2019

2020

500

–

875.5

875.5 = 437.75 2

884.75

884.75 = 442.38 2

886.25

443.13

911.5

455.75

946

473

946.75

473.38

1868 = 467 4

461 1916

–

1778 = 444.5 4

475 1868

–

1767 = 441.75 4

432 1778

–

1772 = 443 4

410 1767

Centered moving average

1730 = 432.5 4

450 1762

Centered total

1916 = 479 4

195

INDEX NUMBERS AND TIME-BASED DATA

1871 2021

1871 = 467.75 4

480 –

2022

–

–

–

–

–

–

430

2. Computation of Straight-Line Trend by using Method of Least Squares Method of least squares is a technique for finding the equation which best fits a given set of observations. In this technique, the sum of squares of deviations of the actual and computed values is least and eliminates personal bias. Suppose we are given n number of observations and it is required to fit a straight line to these data. Note that n, the number of observations can be odd or even. Recall that the general linear equation to represent a straight line is: y = a + bx

1. The sum of the deviations of the actual values of y and yˆ (estimated value of  y) is zero ⇒ ∑ ( y − yˆi ) =0 2. The sum of squares of the deviations of the actual values of  2

y  and  yˆ  (estimated value of  y) is least ⇒ ∑ ( y − yˆi ) is least For the purpose of plotting the best fitted line for trend analysis, the real values of constants ‘ a ’ and ‘ b ’ are estimated by solving the following two equations: ∑ Y = na + b ∑ X ∑ XY = a ∑ X + b ∑ X 2

Where ‘ n ‘ = number of years given in the data. 3. Remember that the time unit is usually of successive uniform duration. Therefore, when the middle time period is taken as the point of origin, it reduces the sum of the time variable x to zero Which means that by taking the mid-point of the time as the origin,

–– Predicted price

Price

Actual price

we get ∑ X = 0

4. When ∑ X = 0 , the equations (ii) and (iii) reduce to: ∑ Y = na + b ( 0 ) Time

where y is the actual value, x is time; a and b are real numbers. In order to fit the best fitted trend line with the help of general equation y = a + bx for the given time series, we will try to find the estimated values of yi say yˆi close to the observed values yi for i = 1, 2,…, n . According to the principle of least squares, the best fitting equation is obtained by minimizing the sum of squares of differences which leads us to two conditions:

⇒

a=

⇒

b=

Scan to know more about this topic

∑Y

n And, ∑ XY = a ( 0 ) + b ∑ X 2

∑XY 2 ∑X

Least Square Method

5. By substituting the obtained values of ‘ a ‘ and ‘ b ‘ in equation (i), we get the trend line of best fit.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Consider the following data: Commodity

Price year 2010

Price year 2016

Quantity Quantity year 2010 year 2016

A

1

2

10

13

B

5

10

12

16

C

6

10

15

18

The Laspeyre’s price index number for year 2016 with year 2010 as base year is: [CUET 2023] (1) 181. (2) 200. (3) 150. (4) 170. 2. The price relatives and weights of a set of commodities are given as:

Commodity

A

B

C

Price Relative

150

130

180

Weight

x

3x

y

If the sum of weights is 30 and the index for the set is 144, then the values of x and y are [CUET 2023] (1) x = 6, y = 8. (2) x = 8, y = 4. (3) x = 6, y = 6. (4) x = 5, y = 10. 3. Given the data for the sales of a product in a state is as follows: Year

2005

2006

2007

2008

2009

Sales(In lakh `)

150

130

160

170

200

The equation of the straight-line trend by method of least squares is: [CUET 2022] (1) 14 + 162x. (2) 126 + 15x. (3) 128 + 14x. (4) 162 + 14x.

196 Oswaal CUET (UG) Chapterwise Question Bank 4. Match List I with List II. LIST I

LIST II

A.

Quantity index

I.

Measures relative price change over a period of time.

B.

Time series

II.

Measures change in quantity of consumption of goods over a specific period of time.

C.

Price index

III. Measures average value of goods for specific time period.

D.

Value index

IV.

Statistical observation taken at different points of time for specific period of time. Choose the correct answer from the options given below:  [CUET 2022] (1) A-III, B-I, C-II, D-IV (2) A-II, B-III, C-I, D-IV (3) A-III, B-IV, C-I, D-II (4) A-II, B-IV, C-I, D-III 5. Given that, ∑ p0q0 = 700, ∑ p0q1 = 1450, ∑ p1q0 = 855 and ∑ p1q1 = 1300. Where subscripts 0 and 1 are used for base year and current year respectively. The Laspeyer’s price index number is  [CUET 2022] (1) 118.46 (2) 119.35 (3) 120.23 (4) 122.14 6. If y = a + b(x − 2005) fits the time series date x (year) :

2018

2019

2020

2021

2022

y (yield in tons) :

6

13

17

20

14

Then the value of a + b is: (1) 16. (2) 20.3. (3) 43. (4) 80.3. 7. A price index which is based on the prices of the items in the composite, weighted by their relative index is called (1) price relatives. (2) Consumer price index. (3) Weighted aggregative price index. (4) Simple aggregative index. 8. Which of the following is the correct option? If the CPI was 120 this year and 100 last year, then (1) the cost of the CPI basket of goods and services increased by 20 percent this year. (2) the price level increased by 120 percent this year. (3) the inflation rate for this year was 20 percent higher than the inflation rate for last year. (4) all of the above are correct. 9. Which of the following are limitations of using index numbers? (1) The use of each index number is restricted to a specific object (2) It ignores the quality of commodities (3) It is useful only for short term comparison (4) All of the above 10. A price index which is based on the prices of the items in the composite, weighted by their relative index is called: (1) price relatives. (2) Consumer price index. (3) Weighted aggregative price index. (4) Simple aggregative index. 11. Which of the following is an example of time series problem?

MATHEMATICS/APP. MATH

(i) Estimating numbers of hotel rooms booking in next 6 months. (ii) Estimating the total sales in next 3 years of an insurance company. (iii) Estimating the number of calls for the next one week. (1) Only (iii) (2) (i) and (ii) (3) (i), (ii) and (iii) (4) (ii) and (iii) 12. In Paasche’s price index number weight is considered as (1) Quantity in base year. (2) Quantity in current year. (3) Prices in base year. (4) Prices in current year. 13. Moving average method is used for measurement of trend when: (1) Trend is linear. (2) Trend is non-linear. (3) Trend is curvilinear. (4) None of these. 14. A time series has ______ components. (1) 1 (2) 2 (3) 3 (4) 4 15. The prices of group of commodities is given in the following table: Commodities

A

B

C

D

p0 [Price (`) in 2019]

40

28

120

112

p1 [Price (`) in 2019]

50

35

135

120

The price index for 2020 taking 2019 as base year using simple aggregative method is: (1) 88.23%. (2) 113.33%. (3) 120.5%. (4) 36%. 16. For data regarding some commodities, the price indexes using Laspeyre’s and Paasche’s method are 118.4 and 117.5 respectively. The Fishers price index for the data is (1) 115.95. (2) 117.95. (3) 120.84. (4) 121.45. 17. The price and quantities of certain commodities are shown in the following table: A

B

p0

1

1

q0

10

5

p1

2

x

q1

5

2

If ratio of Laspeyre’s (L) and Paasche’s (P) index number i.e., L : P = 28 : 27, then the value of x is (1) 2. (2) 3. (3) 4. (4) 5. 18. To find the Index number by weighted average of price relatives, we use the formula  p  ∑  p1  ( p0q0 )  0 (1) × 100 ∑ ( p0q0 ) (2) (3)

∑ p1 ( p0q0 ) × 100 ∑ ( p0q0 ) ∑ p0 ( p0q0 ) × 100 ∑ ( p0q0 )  p 

(4)

∑  p1  ( p1q1 ) 

0



∑ ( p1q1 )

× 100

197

INDEX NUMBERS AND TIME-BASED DATA

19. Time reversal test is satisfied by (1) Laspeyres index only. (2) Paasches index only. (3) Both Laspeyres and Paasches index numbers. (4) Fishers ideal index. 20. An observed set of the population that has been selected for analysis is called (1) a sample (2) a process (3) a forecast (4) a parameter. 21. A weighted aggregate price index in which the weight for each variable is considered its currentperiod quantity is: (1) Aggregative index. (2) Consumer Price index. (3) Laspeyres Index. (4) Paasche’s index. 22. An index constructed to measure changes in quantities over a period of time is (1) Quantity index. (2) Time series index. (3) Quality index. (4) Value index. 23. For calculating the weighted index number, which of the following uses quantities consumed in the base period as weights: (1) Fisher’s method. (2) Paasche’s method. (3) Laspeyres method. (4) Aggregative method. 24. What is the index number of the base period? (1) 200 (2) 300 (3) 10 (4) 100 25. Which of the following is not an example of a time series model? (1) Naive approach (2) Exponential smoothing (3) Moving average (4) None of these 26. Which of the following can’t be a component for a time series plot? (1) Seasonality (2) Trend (3) Noise (4) None of these 27. Increase in the number of patients in the hospital due to heat stroke is (1) Secular trend. (2) Irregular variation. (3) Seasonal variation. (4) Cyclical variations. 28. The graph of time series is called: (1) Histogram. (2) Straight line. (3) Historigram. (4) Ogive. 29. A statement made about a population parameter for testing purpose is called (1) statistic (2) parameter (3) hypothesis (4) level of significance 30. Which of the following is the cyclic behavior of time series? (1) Level (2) Trend (3) Seasonality (4) Noise

[B] ASSERTION REASON QUESTIONS

1. Assertion(A): The impact of change in the price of a commodity with little weight in the index will be small. Reason(R): The impact of change in the price of a commodity with little weight in the index will be large.  2. Assertion(A): A index number is called as a simple index when it is computed from single variable. Reason(R): Index number work as Economic Barometers. 3. Assertion(A): When the prices of rice are to be compared, we compute, price index. Reason(R): Index number is always expressed in Proportion. 4. Assertion(A): Laspeyre’s index number is called as ideal index number. Reason(R): Fisher’s index number is called as ideal index number. 5. Assertion(A): Index numbers may be categorized in terms of variables. Reason(R): Index number may be categorized in terms of constant. 6. Assertion(A): “Naive approach is not an example of a time series model.” Reason(R): “Trend can’t be a component for a time series plot.” So, A is false and R is true. 7. Assertion(A): The data of the consumer price index (CPI) is released every month. Reason(R): The data of the consumer price index (CPI) is issued by Ministry of Commerce and Industry. 8. Assertion(A): In the theory of time series, shortage of certain consumer goods before the annual budget is due to seasonal variation. Reason(R): In the measurement of the secular trend, the moving averages Smooth out the time series. 9. Assertion(A): The most commonly use mathematical method for measuring the trend is Method of least squares. Reason(R): A time series is a set of observations taken at specified times, usually at equal intervals. 10. Assertion(A): Purchasing power of money can be accessed through Consumer price index. Reason(R): Cost of living at two different cities can be compared with the help of Consumer price index. [C] COMPETENCY BASED QUESTIONS I.  Read the following text and answer the following questions on the basis of the same: To fit a straight line by the method of least squares, Rohan constructed the following table: Year (t)

Profit (y)

x = ti − 2018

x2

xy

2015

114

−3

9

−342

2016

130

−2

4

−260

(1)  Both A and R are true and R is the correct explanation of A

2017

126

−1

1

−126

(2)  Both A and R are true but R is not the correct explanation of A

2018

144

0

0

0

2019

138

1

1

138

(3)  A is true but R is false

2020

156

2

4

312

(4)  A is false and R is true

2021

164

3

9

492

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as:

198 Oswaal CUET (UG) Chapterwise Question Bank The trend equation can be considered as yt = a + bx. 1. The value of ‘ a ’ in the trend equation is: (1) 138.86. (2) 7.64. (3) 183.86. (4) 7.46.

Year (t)

Sales (y)

2. The value of ‘ b ’ in the trend equation is: (1) 138.86. (2) 7.64. (3) 183.86. (4) 7.46.

2015

6.7

2016

5.3

2017 2018

3. The trend equation is: (1) yt = 138.86 + 7.64 x.

(2) yt = 7.64 + 138.86 x.

(3) yt = 138.86 − 7.64 x.

(4) None of these

4. The trend value for year 2015 is: (1) 161.78. (2) 154.14. (3) 115.94. (4) 138.86. 5. The trend value for year 2018 is: (1) 161.78. (2) 154.14. (3) 115.94. (4) 138.86. II.  Read the following text and answer the following questions on the basis of the same: Today in class Mr. Agarwal is teaching the Method of Least Squares to measure the trend in time series. After explaining the method, he had taken an example which is as follows: Given below are the data relating to the sales of a product in a district. Fit a straight line trend by the method of least squares and tabulate the trend values. Year 2015 2016 2017 2018 2019 2020 2021 2022 Sales

6.7

5.3

4.3

6.1

5.6

7.9

5.8

6.1

To solve the given example, he constructed the following table: Computation of trend values by the method of least squares.

MATHEMATICS/APP. MATH

(t − 2018.5) 0.5

xy

x2

−7

−46.9

49

−5

−26.5

25

4.3

−3

−12.9

9

6.1

−1

−6.1

1

2019

5.6

1

5.6

1

2020

7.9

3

23.7

9

2021

5.8

5

29.0

25

2022

6.1

7

42.7

49

x=

Here, n = 8 (even) So, origin is mean of two middle years i.e.,

2018 + 2019 = 2018.5 2

6. If the straight line is yt = a + bx , then value of ‘ a ’ is: (1) 5.615.

(2) 5.759.

(3) 5.829.

(4) 5.975.

7. If the straight line is yt = a + bx , then value of ‘ b ’ is: (1) 0.051. (3) 0.411. 8.

(2) 0.011. (4) None of these

The trend equation is:

(1) yt = 5.975 + 0.051x.

(2) yt = 0.051 + 5.975 x.

(3) yt = 5.975 − 0.051x. (4) None of the above 9. The trend value for year 2015 is: (1) 5.691. (2) 5.618. (3) 5.987. (4) 5.719. 10. The trend value for year 2016 is: (1) 5.451. (2) 5.616. (3) 5.966. (4) 5.720.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (3)

3. (4)

4. (4)

5. (4)

6. (2)

7. (3)

8. (1)

9. (4)

10. (3)

11. (3)

12. (2)

13. (1)

14. (4)

15. (2)

16. (2)

17. (3)

18. (1)

19. (4)

20. (3)

21. (4)

22. (1)

23. (1)

24. (4)

25. (1)

26. (4)

27. (3)

28. (3)

29. (1)

30. (3)

8. (2)

9. (2)

10. (1)

8. (1)

9. (2)

10. (4)

[B] ASSERTION REASON QUESTIONS 1. (3)

2. (2)

3. (3)

4. (4)

5. (3)

6. (4)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (2)

3. (1)

4. (3)

5. (4)

6. (4)

7. (1)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

Commodity

1. Option (1) is correct. Explanation: P0

A

1

P1 Q0

Q1

P0Q0

B

5 10 12

16

C

6 10 15

18

P1Q0

P01 = 2 10

13

10

20

 =

∑ PQ 1 0 ∑ P0Q0

60

120

90

150

∑P0Q0=160

∑P1Q0=290

× 100

290 × 100 =181.25 or 181 (approx.) 160

199

INDEX NUMBERS AND TIME-BASED DATA

2. Option (3) is correct. Explanation: Commodity

Weight (w)

Price relative (p)

wp

A

x

150

150x

B

3x

130

390x

C

y

180

180y

Sum

4x + y

540x + 180y

Sum of weights = 30 4x + y = 30 ... (i) Index Number = 144 540 x + 180 y = 144 30 9x + 3y = 72 3x + y = 24 ... (ii) On solving eq. (i) and eq. (ii) to get x and y, we get x = 6 and y = 6. 3. Option (4) is correct. Explanation: Calculating for the straight line trend by least square method Year (X)

Sales (y)

x = X − 2007

X 2

XY

2005

150

−2

4

−300

2006

130

−1

1

−130

2007

160

0

0

0

2008

170

1

1

170

2009

200

2

4

400

∑x = 10

∑xy = 140

n=5

∑y = 810

2

∑x = 0

a = 

∑ y = 810  = 162



∑ xy = 140 ∑ x 2 10

b = 

n

855 × 100 700

=



= 122.14

y(yield in tons)

X = xi – 2005

X2

Xy

2018 2019 2020 2021 2022

6 13 17 20 24

–2 –1 0 1 2

4 1 0 1 4

–12 –13 0 20 48

∑y = 80

∑X = 0

∑X 2 = 10

∑Xy = 23

Now,

a =



b =

= 14

y ∑= n

80 = 18 (n = 5) 5  

∑ Xy= ∑X2

23 = 2.3 10

a + b = 18 + 2.3 = 20.3

⸫

7. Option (4) is correct. 8. Option (1) is correct. Explanation : If the CPI was 120 this year and 100 last year, then the cost of the CPI basket of goods and services increased by 20 percent this year. 9. Option (4) is correct. 10. Option (3) is correct. 11. Option (3) is correct. Explanation: All the component are associated with time. 12. Option (2) is correct. Explanation: Using current year quantities as weight is known as Paasche’s formula. 13. Option (1) is correct. Explanation: When trend is linear, then only we use moving average method for measurement. 14. Option (4) is correct. Explanation: A time series has 4 components : Level, Trend, Seasonality and Noise. 15. Option (2) is correct. Explanation: Price Index = 340 ∑ p1 × 100 = × 100 ∑ p0 300

5

Y = a + bX = 162 + 14x 4. Option (4) is correct. 5. Option (4) is correct. Explanation: Laspeyre’s price index ∑ p1q0 × 100 = ∑ p0q0

x(year)

= 113.33% 16. Option (2) is correct. Explanation: F P01 =

(P

L P 01 × P01

)

= 118.4 × 117.5 = 117.95 17. Option (3) is correct. Explanation: Items A B

p0 1 1

q0 10 5

p1 2 x

6. Option (2) is correct. Explanation: Since,

L:P = 28 : 27

q1 5 2

p0q0 10 5 Ʃp0q0 = 15

p1q0 20 5x Ʃp1q0 = 20 + 5x

p0q1 5 2 Ʃp0q1 =7

p1q1 10 2x Ʃp1q1 = 10 + 2x

200 Oswaal CUET (UG) Chapterwise Question Bank ⸫

[B] ASSERTION REASON QUESTIONS

∑ p1q0 × ∑ p0q1 = 28 ∑ p0q0 ∑ p1q1 27

⇒ 9x + 36 = 40 + 8x ⇒ x=4 18. Option (1) is correct. Explanation: To calculate Index number by weighted average of price relatives, we apply formula  p 

  p1   p0q0  

0



  p0q0 

MATHEMATICS/APP. MATH

 100

19. Option (4) is correct. Explanation: Time reversal Test is satisfied by Fishers ideal index because it is a weighted geometric mean of the Laspeyre’s and Paasche’s indices. 20. Option (1) is correct. Explanation: A selection of a group of individuals from a population in such a way that it represents the population is called as sample and the number of individuals in the sample is called the sample size. 21. Option (4) is correct. Explanation: The Paasche’s Price Index is commonly referred to as the “current weighted index.” 22. Option (1) is correct. Explanation: A quantity index number is used to measure changes in the volume or quantity of goods that are produced, consumed, and sold within a stipulated period. 23. Option (1) is correct. Explanation: Laspeyre’s method uses base year’s quantities as weights, Paasche’s method uses current year’s quantities as weights and Fisher’s method uses both base year as well as current year’s quantities as the base. 24. Option (4) is correct. Explanation: Conventionally, index numbers are expressed in terms of percentage of two periods, of two periods, the period with which the comparison is to be made, is known as the base period. The value in the base period is given the index number 100. 25. Option (1) is correct. Explanation: A Naive approach is one in which the forecast for a particular period is just the same as the preceding period’s value e.g., you sold 150 computers last month and estimate that you would sell 150 computers again this month. 26. Option (4) is correct. 27. Option (3) is correct. Explanation: In seasonal variation, tendency movements are due to the nature which repeat themselves periodically in every season. 28. Option (3) is correct. Explanation: A historigram is a graphical representation of a time series that reveals the changes that occurred at different time periods. 29. Option (3) is correct. Explanation: Statistical hypothesis is some assumption or statement, which may or may not be true, about a population. 30. Option (3) is correct. Explanation: Seasonality is the cyclic behavior of time series.

1. Option (3) is correct. Explanation: An equal rise in the price of an item with little weight will have lower implications for the overall change in the price index than that of an item with more weight. 2. Option (2) is correct. Explanation: Simple index number is the index number which measures the relative change in the single variable with respect to its base. Also, index numbers are economic barometers since they help in understanding the changes in economic conditions of society. 3. Option (3) is correct. Explanation: Price indices are commonly used to compare the relative changes in the prices of goods and services, including commodity (rice). So, Assertion (A) is true. Index numbers are expressed as percentages, indicating relative changes in the values they represent. So, Reason (R) is false. 4. Option (4) is correct Explanation: The Laspeyre’s index number is not called the “ideal index number.” It is just one of many index number formulas used for measuring changes in prices or quantities. The Fisher’s index is considered an ideal index number because it overcomes some of the weaknesses of the Laspeyre’s and Paasche’s indexes. The assertion is false but R is true. 5. Option (3) is correct Explanation: An Index Number is a tool to evaluate the differences within a variable or a set of variables for things like time, geographical location or quantity. 6. Option (4) is correct. Explanation: The assertion “Naive approach is not an example of a time series model” is false, and the reason “Trend can’t be a component for a time series plot” is true. The Naive approach is indeed a simple time series model, and time series data can include trends. So, A is false and R is true. 7. Option (3) is correct. Explanation: CPI data is published monthly, with the index value representing an estimate of the price level for the month as a whole, rather than a specific date. The data of the consumer price index (CPI) is issued by Ministry of Statistics and Program Implementation. 8. Option (2) is correct. Explanation: Seasonal variation is variation in a time series within one year that is repeated more or less regularly. So, shortage of consumer goods before the annual budget is due to seasonal variation. Moving averages is a series of arithmetic means of variate values of a sequence. This is another way of drawing a smooth curve for a time series data. 9. Option (2) is correct. Explanation: Assertion: “The most commonly used mathematical method for measuring the trend is method of least squares” is true and Reason “A time series is a set of observations taken at specified times, usually at equal intervals” is also true but not the explanation of A. 10. Option (1) is correct Explanation: Both A and R are true, and R is the correct explanation of A. Assertion is true because CPI tracks changes in the purchasing power of money.

201

INDEX NUMBERS AND TIME-BASED DATA

Reason is true, and it explains how CPI is used to compare the cost of living in different cities. [C] COMPETENCY BASED QUESTIONS 1. Option (1) is correct. Explanation: ∑y Here, a= n n  7 and ∑ y  972

So,

972 7  138.86

a

2. Option (2) is correct. Explanation: ∑ xy b= 2 Here, ∑x and

n  7,

7. Option (1) is correct. Explanation: ∑ xy  46.9  26.5  12.9  6.1 + 5.6 + 23.7  29.0 + 42.7  8.6 ∑xy b  2 ∑x Here, n = 7 and ∑xy = 8.6 and ∑x2= 168 So, b  8.6 168 = 0.051

 x y  214  x 2  28

214 28  7.64

b 

3. Option (1) is correct. Explanation: Since, a = 138.86 and b = 7.64 Thus, trend equation is given by yt = 138.86 + 7.64 x 4. Option (3) is correct. Explanation: For year 2015, x = −3  yt  138.86  7.64  3  115.94 5. Option (4) is correct. Explanation: For year 2018, x = 0 \ yt = 138.86 + 7.64(0) = 138.86 6. Option (4) is correct.

Σ y = 6.7 + 5.3 + 4.3 + 6.1 + 5.6 + 7.9 + 5.8 + 6.1 Σy a= n

Here, n = 8 and ∑ y = 47.8 47.8 So, a= 8 = 5.975

8. Option (1) is correct. Explanation: Since,

a = 5.975 b = 0.051

and Thus, trend equation is given by

yt = 5.975 + 0.051x 9. Option (2) is correct. Explanation: For year 2015, x = −7 ∴

yt = 5.975 + 0.051( −7 ) = 5.618

10. Option (4) is correct. Explanation: For year 2016, x  5 yt  5.975  0.051 5    5.975  0.255  5.720

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

16

INFERENTIAL STATISTICS

  Revision notes

Inferential Statistics is mainly used to derive estimates about a large group (or population) and draw conclusions on the data, based on hypothesis testing methods.  Population: The group of individuals considered under study is called as population. The word population here refers not only to people but to all items that have been chosen for the study. Thus in statistics, population can be number of bikes manufactured in a day or week or month, number of cars manufactured in a day or week or month, number of fans, TVs, chalk pieces, people, students, girls, boys, any manufacturing products, etc. Finite and Infinite Population: When the number of observations/individuals/products is countable in a group, then it is a finite population. Example: Weights of students of class XII in a school. When the number of observations/individuals/products is uncountable in a group, then it is a finite population. Example: Number of grains in a sack, number of germs in the body of a sick patient.  Sample and Sample Size: A selection of a group of individuals from a population in such a way that it represents the population is called as sample and the number of individuals included in a sample is called the sample size.  Sampling: Sampling is the procedure or Scan to know process of selecting a sample from a popumore about this topic lation. Sampling is quite often used in our day-to-day practical life.  Parameter: The statistical constants of the population like mean (μ), variance (s2) are referred as population parameters. Sampling Statistic: Any statistical measure computed from sample is known as statistics. n

x

i

i 1

 ( mean ) 

N

n

 2 ( variance ) 



 xi   2 N

i 1

n

 ( standard deviation, SD ) 

 i 1

 xi   2 N

where N is the population size.  Errors in a Sample: A sample is a part of the whole population. A sample drawn from the population depends upon chance

and as such all the characteristics of the population may not be present in the sample drawn from the same population. The errors involved in the collection, processing and analysis of the data may be broadly classified into two categories namely, (i) Sampling Errors (ii) Non-Sampling Errors (i) Sampling Errors: Errors, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors. Sampling errors are inherent in the method of sampling. They may arise accidentally without any bias or prejudice. (ii) Non-Sampling Errors: The errors that arise due to human factors which always vary from one investigator to another in selecting, estimating or using measuring instruments (tape, scale) are called Non-Sampling errors. Types of Sampling: There are various techniques of sampling, but they can be broadly grouped into two: (i) Random or probability sampling. (ii) Non-Random or Non-Probability sampling. We will consider only random (Probability) sampling. Random sampling or Probability sampling: Random sampling refers to selection of samples from the population in a random manner. A random sample is one where each and every item in the population has an equal chance of being selected. The following are different types of probability sampling: (i) Simple Random Sampling: Simple random sampling is the randomised selection of a small segment of individuals or members from a whole population. It provides each individual or member of a population with an equal and fair chance of being chosen. This is one of the most common method of sampling. e.g., There are 1000 students in a school, we want to select a simple random sample of 100 students. We can assign a number to every student in the school from 1 to 1000 and select randomly 100 numbers. (ii) Systematic Sampling: Systematic sampling is the selection of specific individuals or members from entire population. The selection often follows a predetermined interval (k). The systematic sampling method is comparable to the simple random sampling method, however it is less complicated to conduct. e.g., Out of 1000 students of school, we want to select a sample of 100 students. All students of the school are arranged in alphabetical order and assigned a number 1 to 1000. Now, we randomly select a number (say 4) from first ten numbers and then every tenth student in the list is selected i.e., 4, 14, 24, ...

Two Sample t-test

Null Hypothesis

Sampling Errors

A healthcare consultant wants to compare the patient satisfaction ratings of two hospitals. The consultant collects ratings from 20 patients for each of the hospitals. The consultant performs a 2-sample t-test to determine whether there is a difference in the patient ratings between the hospitals.

n1n2 x−y The t-test formula is given as : t = s n1 + n2

Σ( x − x )2 + Σ( y − y )2 Thus, standard error can be given by s = n1 + n2 − 2

Let the two independent samples be x1, x2, x3, ..... xn1 and Σy Σx y1, y2, y3, ....., yn with means x = and y = from two 2 n2 n1 normal population with means µ1 and µ2 and common variance σ 2 (unknown). 1 1 s12 = Σ( x − x )2 and s22 = Σ( y − y )2 Let n1 − 1 n2 − 1

The null hypothesis for a 1-sample t-test is : H0 : µ = µ0 where : • µ = the population mean • µ0 = the hypothesized mean You can choose any one of three alternative hypotheses : H1: µ > µ0 One-tailed test H1: µ < µ0 One-tailed test H1: µ ≠ µ0 Two-tailed test

a

Population

Inferential statistics

Standard Error

Parameter & Statistics

A sample is a smaller group of members of a population selected to represent the population. A random sample is one in which every member of a population has an equal chance of being selected.

σ=

Σ( xi − µ)2 N

σ2 2 / n

σ 2 / 2n

PQ / n

σ / n

Standard Error

First Level

Second Level

Trace the Mind Map Third Level

N = the size of the population xi = each value from the population µ = population mean P = Population Proportion Q = 1-P

σ = population standard deviation

where,

4. Sample variance (s2 )

3. Sample standard deviation (s)

2. Observed sample proportion (p)

1. Sample mean ( x )

Sl Statistic

Parameter: The statistical constants of the population like mean (µ), variance (σ2) are referred as population parameters. Statistic : Any statistical measure computed from sample is known as statistic.

Sample

A population is a group of phenomena that have something in common. The term often refers to a group of people, as in the following examples: • All registered voters in India. • All members of the International Cricket Team.

The One-Sample t-Test procedure tests whether the mean of a single variable differs from a specified constant. Example : A researcher might want to test whether the average IQ score for a group of students differs from 100. x −µ Formula for One Sample t-test t = / n s where, t = the t statistic x = the mean of the sample µ = the comparison mean s = the sample standard deviation n = the sample size

One Sample t-test

Inferential Statistics

ntial St

Sampling

e

Infe r

It use measurements from the sample of subjects in the experiment to compare the treatment groups and make generalizations about the larger population of subjects.



Sampling is the process of selecting observations (a sample) to provide an adequate description and robust inferences of the population. Sampling can be broadly grouped into two : • Random (Probability) Sampling • Non Random Sampling.



ics it st



• Sampling Errors Errors caused by the act of taking a sample. Make sample results inaccurate. • Random Sampling Error Errors caused by the chance in selecting a random sample. • Nonsampling Error Errors not related to the act of selecting a sample from the population. They can even be present in a census.

INFERENTIAL STATISTICS

203

Example

204 Oswaal CUET (UG) Chapterwise Question Bank and end up with sample of 100 members. (iii) Stratified Sampling: Stratified sampling includes the partitioning of a population into subclasses with notable distinctions and variances. This method is useful when population is dispersed. e.g., Suppose a school has its branches in 15 cities. We want to select the sample of 100 students. It is difficult to select students from each branch. So first select any 5 branches and then select students from each branch.  Sampling Distribution: Sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population. For instance, if we draw a sample of size n from a given finite population of size N, then the total number of possible samples N!  k(say). For each of these k samples we are N Cn  n!( N  n)! can compute some statistic t = t(x1, x2, x3, … xn), in particular the mean x, the variance S2, etc., is given below: Sample Number

Statistic t

x

S2

1

t1

x1

S12

2

t2

x2

S22

3

t3

x3

S32







k

tk

xk

.. Sk2

The set of the values of the statistic so obtained, one for each sample, constitutes the sampling distribution of the statistic.  Standard Error: The standard deviation of the sampling distribution of statistic known as its Standard Error abbreviated as S.E. The Standard Errors (S.E.) of some of the well-known statistics, for large samples, are given below, where n is the sample size, s2 is the population variance. S. No

Statistic

Standard Error

1.

Sample mean ( x )

s n

2.

Observed sample proportion (p)

PQ n

3.

Sample standard deviation (s)

σ2 2n

4.

Sample variance (s2)

s

MATHEMATICS/APP. MATH

sample and test that value for the population. This is done by the two important classifications in statistical inference, (i) Estimation (ii) Testing of Hypothesis (i) Estimation: The method of obtaining the most likely value of the population parameter using statistic is called estimation. (ii) Estimator: Any sample statistic which is used to estimate an unknown population parameter is called an estimator i.e., an estimator is a sample statistic used to estimate a population parameter.  Estimate: To estimate an unknown parameter of the population, concept of theory of estimation is used. There are two types of estimation namely.  Point Estimation: When a single value is used as an estimate, it is called as point estimation.  Interval Estimation: An interval within which the parameter would be expected to lie is called interval estimation.  Statistical Hypothesis: Statistical hypothesis is some assumption or statement, which may or may not be true, about a population. There are two types of statistical hypothesis: Scan to know more about (i) Null hypothesis this topic (ii) Alternative hypothesis (i) Null Hypothesis: According to Prof. R.A. Fisher; Null hypothesis is the hypothesis which is tested for possible Hypothesis rejection under the assumption that it is Testing true”, and it is denoted by H0. For example: If we want to find the population mean having a specified value m0, then the null hypothesis H0 is set as follows H0 : µ = µ0 (ii) Alternative Hypothesis: Any hypothesis which is complementary to the null hypothesis is called as the alternative hypothesis and is usually denoted by H1. For example: If we want to test the null hypothesis that the population has specified mean µ i.e., H0: µ = µ0, then the alternative hypothesis could be any one among the following: (a) H1 : m ≠ m0 (b) H1 : m > m0 (c) H1 : m < m0 The alternative hypothesis in H1: m ≠ m0 is known as two tailed alternative test. Two tailed test is one where the hypothesis about the population parameter is rejected for the value of sample statistic falling into either tails of the sampling distribution. When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as one-tailed test. Here H1: m > m0 and H1: m < m0 are known as one-tailed alternative.

2 n

 Statistical Inferences: One of the main objectives of any statistical investigation is to draw inferences about a population from the analysis of samples drawn from that population. Statistical Inference provides us how to estimate a value from the

Two tailed diagram Right tailed test: H1 : m > m0 is said to be right tailed test where the rejection region or critical region lies entirely on the right

205

INFERENTIAL STATISTICS

tail of the normal curve.

Right tailed diagram Left tailed test: H1 : m < m0 is said to be left tailed test where the critical region lies entirely on the left tail of the normal curve.

Left tailed diagram  Types of Errors in Hypothesis: There is every chance that a decision regarding a null hypothesis may be correct or may not be correct. There are two types of errors. They are Type I error: The error of rejecting H0 when it is true. Type II error: The error of accepting H0 when it is false.  Critical Region or Rejection Region: A region corresponding to a test statistic in the sample space which tends to rejection of H0 is called critical region or region of rejection. The region complementary to the critical region is called the region of acceptance.  Level of Significance: The probability of type I error is known as level of significance and it is denoted by a. The levels of significance which are usually employed in testing of hypothesis are 5% and 1%. The level of significance is always fixed in advance before collecting the sample information.  Critical Values or Significant Values: The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value. It depends upon: (i) The level of significance.

(ii) The alternative hypothesis whether it is two-tailed or single tailed.  t-Test: The t-test is a test in statistics that is used for testing hypothesis regarding the mean of a small sample of the population when the standard deviation of the population is not known. t-Test was first invented by William Sealy Gosset, in 1908. Since he used the pseudo name as ‘Student’ when publishing his method in the paper titled Biometrika’, the test came to be known as Student’s T Test. There are many types of t-test. Two of them are:  The one-sample t-test, which is used to compare the mean of a population with a theoretical value.  The two-sample t-test, which is used to compare the mean of two independent given samples. One Sample t-Test Procedure: Step 1: Define the Null Hypothesis (H0) and Alternate Hypothesis (H1) Example: H0: Sample mean (X) = Hypothesized Population mean (m)

H1: Sample mean (X) ≠ Hypothesized Population mean (m) The alternate hypothesis can also state that the sample mean is greater than or less than the comparison mean. Step 2: Compute the test statistic (t) t

Z X    S .E n

Scan to know more about

where S.E. is the standard error this topic Step 3: Find the t-critical from the t-Table Use the degree of freedom and the alpha level (0.05) to find the t-critical. Step 4: Determine if the computed test staStudent’s t-test tistic falls in the rejection region. Alternately, simply compute the P-value. If it is less than the significance level (0.05 or 0.01), reject the null hypothesis.

t Table Cum.prob

t.50

t.75

t.80

t.85

t.90

t.95

t.975

t.99

t.995

t.999

t.9995

One-tail

0.50

0.25

0.20

0.15

0.10

0.05

0.025

0.01

0.005

0.001

0.0005

1.00

0.50

0.40

0.30

0.20

0.10

0.05

0.02

0.01

0.002

0.001

1

0.000

1.000

1.376

1.963

3.078

6.314

12.71

31.82

63.66

318.31

636.62

2

0.000

0.816

1.061

1.386

1.885

2.920

4.303

6.965

9.925

22.327

31.599

3

0.000

0.765

0.978

1.250

1.638

2.353

3.182

4.541

5.841

10.215

12.924

4

0.000

0.741

0.941

1.190

1.533

2.132

2.776

3.747

4.604

7.173

8.610

5

0.000

0.727

0.920

1.156

1.476

2.015

2.571

3.365

4.032

5.893

6.869

6

0.000

0.718

0.906

1.134

1.440

1.943

2.447

3.143

3.707

5.208

5.959

7

0.000

0.711

0.896

1.119

1.415

1.895

2.365

2.998

3.499

4.785

5.408

8

0.000

0.706

0.889

1.108

1.397

1.860

2.306

2.896

3.335

4.501

5.041

9

0.000

0.703

0.883

1.100

1.389

1.833

2.262

2.821

3.250

4.297

4.781

Two-tails df

206 Oswaal CUET (UG) Chapterwise Question Bank

MATHEMATICS/APP. MATH

10

0.000

0.700

0.879

1.093

1.372

1.812

2.228

2.764

3.169

4.144

4.587

11

0.000

0.697

0.876

1.088

1.363

1.796

2.201

2.718

3.106

4.025

4.437

12

0.000

0.695

0.873

1.083

1.356

1.782

2.179

2.681

3.055

3.930

4.318

13

0.000

0.694

0.870

1.079

1.350

1.771

2.160

2.650

3.012

3.852

4.221

14

0.000

0.692

0.870

1.079

1.350

1.771

2.160

2.650

3.012

3.852

4.221

15

0.000

0.691

0.866

1.074

1.341

1.753

2.131

2.602

2.947

3.733

4.073

16

0.000

0.690

0.865

1.071

1.337

1.746

2.120

2.583

2.921

3.686

4.015

17

0.000

0.689

0.863

1.069

1.333

1.740

2.110

2567

2.898

3.646

3.965

18

0.000

0.688

0.862

1.067

1.330

1.734

2.101

2.552

2.878

3.610

3.922

19

0.000

0.688

0.861

1.066

1.328

1.729

2.093

2.539

2.861

3.579

3.883

20

0.000

0.687

0.860

1.064

1.325

1.725

2.086

2.528

2.845

3.552

3.850

21

0.000

0.686

0.859

1.063

1.323

1.721

2.080

2.518

2.831

3.527

3.819

22

0.000

0.685

0.858

1.061

1.321

1.717

2.074

2.508

2.819

3.505

3.792

23

0.000

0.685

0.858

1.060

1.319

1.714

2.069

2.500

2.807

3.485

3.768

24

0.000

0.685

0.857

1.059

1.318

1.711

2.064

2.492

2.797

3.467

3.745

25

0.000

0.684

0.856

1.058

1.316

1.708

2.060

2.485

2.787

3.450

3.725

26

0.000

0.684

0.56

1.058

1.315

1.706

2.056

2.479

2.779

3.435

3.707

27

0.000

0.684

0.855

1.057

1.314

1.703

2.052

2.473

2.771

3.421

3.690

28

0.000

0.683

0.855

1.056

1.313

1.701

2.048

2.467

2.763

3.408

3.674

29

0.000

0.683

0.854

1.055

1.311

1.699

2.045

2.462

2.756

3.396

3.659

30

0.000

0.686

0.854

1.055

1.310

1.697

2.042

2.457

2.750

3.385

3.646

40

0.000

0.681

0.851

1.050

1.303

1.684

2.021

2.423

2.704

3.307

3.551

60

0.000

0.679

0.848

1.045

1.296

1.671

2.000

2.390

2.660

3.232

3.460

80

0.000

0.678

0.846

1.043

1.292

1.664

1.990

2.374

2.639

3.195

3.416

100

0.000

0.677

0.845

1.042

1.290

1660

1.984

2.364

2.626

3.174

3.390

1000

0.000

0.675

0.842

1.037

1.282

1.646

1.962

2.330

2.581

3.098

3.300

Z

0.000

0.674

0.842

1.036

1.282

1.645

1.960

2.326

2.576

3.090

3.297

0%

50%

60%

70%

80%

90%

95%

98%

99%

99.8%

99.9%

Confidence Level Two Sample t-Test The two sample t-test is a test that is used to compare the mean of two groups of samples. It is meant for evaluating whether the means of the two sets of data are statistically and significantly different from each other. Let the two independent samples be x1, x2, x3, ....., xn and y1, 1

Σx Σy y2, y3, ....., yn with means x = and y = from two 2 n1 n2 normal population with means m1 and m2 and common variance s2(unknown). 1 1 Let s12 =  ( x  x )2 and s22 = n  1  ( y  y )2 n1  1 2 Thus, standard error can be given by s=

 ( x  x )2   ( y  y )2 n1  n2  2

The t-test formula is given as: t =

n1n2 xy n1  n2 s

The significance of t is tested in the same way as in one sample t-test with degree of freedom n1 + n2 – 2 to test the hypothesis m1 = m2.  Central Limit Theorem (CLT) The central limit theorem states that whenever a random sample of size n is taken from any distribution with mean and variance, then the sample mean will be approximately a normal distribution with mean and variance. The larger the value of the sample size, the better the approximation of the normal. The word “Central” comes from “fluctuations around centre (=average)”.

207

INFERENTIAL STATISTICS

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Which of the following is not a statistic? (1) The mean marks of 100 students of a school of strength 2000. (2) The median height of 100 students in a school of strength 2000. (3) The mean salary of all workers of an organisation. (4)  The mean salary of randomly selected half of all employees of an organisation. [CUET 2023] 2.  T-Test : A t-test is a test of difference for parameter data T=

x1 − x 2 sp 1 + 1 x1 x 2

Then read the following statements and choose the correct statements (A) The null hypothesis and the alternative hypothesis have the same view points. (B) In t-test; testing the significance of mean value is done, when sample size is small. (C) t-test for two independent groups when variance is equal (D) Testing is a process used by statisticians, to accept or reject the hypothesis. (E) If the value of test statistic is greater than the table values, we do not reject the null hypothesis Choose the correct answer from the options given below:  [CUET 2023] (1) (B), (C) and (D) Only (2) (A), (B) and (D) Only (3) (A), (B) and (E) Only (4) (D), and (E) Only 3.  Out of 1000 employees, 100 have to be selected for a survey. After being arranged in the alphabetical order, each one is assigned a number from 1 to 1000. A number 4 is randomly selected and then every tenth person is selected (i.e., 4,14,24 .. ) Which form of sampling is this an example of?  [CUET 2023] (1) Cluster sampling (2) Snowball sampling (3) Voluntary response sampling (4) Systematic sampling 4. Consider the following hypothesis test : H0 : m ≤ 24 Ha : m > 24 A sample of 64 provided a sample mean of 24.3. The population standard deviation is 2 . The value of the test statistic is: [CUET 2023] (1) 1.2 (2) 2.1 (3) 1.3 (4) –1.2 5. A simple random sample consists of five observations 2, 4, 6, 7, 6. The point estimate of population standard deviation is:  [CUET 2023] (1) 4 (2) 2.5 (3) 5 (4) 2 6. Consider the following hypothesis test: H0 : m ≥ 20 H1 : m < 20 A sample of 64 provided a sample mean of 19.5 . The population standard deviation is 2 . The value of the test statistic is: [CUET 2023]

(1) –2.5 (2) –2 7. Match List I with List II

(3) 2

(4) –1.5

List I

List II

A.

A special characteristic of a population is known as a

I.

Statistic

B.

A special characteristic of sample is known as a

II.

Confidence interval

C.

The uncertainty of a sampling process is expressed by

III.

Estimation

D.

The process by which one makes the inferences about a population based on the information obtained from a sample is known as

IV.

Parameter

 (1) (3) 8.

Choose the correct answer from the options given below: [CUET 2023] A-II, B-III, C-IV, D-I (2) A-I, B-IV, C-II, D-III A-IV, B-I, C-II, D-III (4) A-IV, B-I, C-III, D-II Which of the following is a statistic? (1) m (2) x (3) s2 (4) None 9. In one sample test, the estimation for population mean is x −m x −m x −m (1) (2) (3) s 2 (4) None s s n n n 10.  If the calculated value of |t| < tv (a), then the null hypothesis is: (1) rejected (2) accepted (3) cannot be determined (4) neither accepted nor rejected 11.  For testing the significance of difference between the means of two independent samples, the degree of freedom (v) is taken as: (1) n1 − n2 + 2 (2) n1 − n2 − 2 (3) n1+ n2 − 2 (4) n1 + n2 − 1 12.  One of the following is true of relation between sample mean ( x) and population mean (m). [CUET 2022] (1) | x − m | increases when increases the size of sample (2) x  , , for all sample sizes (3) | x − m | do not change with size of sample (4) | x − m | decreases when increases the size of sample 13. Below are the stages for Drawing statistical inferences. (A) Sample (B) Population (C) Making Inference (D) Data tabulation (E) Data Analysis Choose the correct answer from the options given below:  [CUET 2022] (1) (B), (D), (A), (E), (C) (2) (A), (B), (D), (C), (E) (3) (B), (A), (D), (E), (C) (4) (D), (B), (A), (C), (E) 14. A machine makes car wheels and in a random sample of 26 wheels, the test statistic is found to be 3.07. As per

208 Oswaal CUET (UG) Chapterwise Question Bank the t-distribution test (of 5% level of significance), what can you say about the quality of wheels produced by the machine? (Use t25(0.05) = 2.06) (1) Superior quality (2) Inferior quality (3) Same quality (4) Cannot say 15. For the purpose of t-test of significance, a random sample of size (n) 34 is drawn from a normal population, then the degree of freedom (v) is: 1 (1) (2) 33 (3) 34 (4) 35 34 16. Standard deviation of a sample from a population is called a: (1) Standard error (2) Parameter (3) Statistic (4) Central limit 17.  A grain whole-seller visits the granary market. While going around to make a good purchase, he takes a handful of rice from random sacks of rice, in order to inspect the quality of farmers produce. The handful of rice taken from a sack of rice for quality inspection is a: (1) statistic (2) population (3) parameter (4) sample 18. In a school, a random sample of 145 students is taken to check whether a student’s average calory intake is 1500 or not. The collected data of average calories intake of sample students is presented in a frequency distribution, which is called a: (1) Statistics (2) Sampling distribution (3) Parameter (4) Population sampling 19.  Which of the following values is used as a summary measure for a sample, such as a sample mean ? (1) Population Parameter (2) Sample Parameter (3) Sample Statistic (4) Population mean 20. Which of the following is a branch of statistics ? (1) Descriptive Statistics (2) Inferential Statistics (3) Industrial Statistics (4) Both (1) and (2) 21.  A statement made about a population parameter for testing purpose is called ............... . (1) statistic (2) parameter (3) hypothesis (4) level of significance 22. If we reject the null hypothesis, we might be making: (1) Type-I error (2) Type-II error (3) A correct decision (4) A wrong decision. 23.  A sample of 50 pens is taken at random. Out of 50 pens we found 15 pens are of Cello, 17 are of Parker of population proportion of Parker. (1) 0.3 (2) 0.34 (3) 0.36 (4) 0.4 24.  A simple random sample consist of four observation 1, 3, 5, 7. What is the point estimate of population standard deviation ? (1) 2.3 (2) 2.52 (3) 0.36 (4) 0.4 25. Which of the following statements are true? I: The mean of a population is denoted by x . II: The population mean is a statistic. (1) I only (2) II only (3) Both I and II (4) None 26. A machine produces washers of thickness 0.50 mm. To determine whether the machine is in proper working

MATHEMATICS/APP. MATH

order, a sample of 10 washers is chosen for which the mean thickness is 0.53 mm and the standard deviation is 0.03 mm. As per the t-distribution test (of 5% level of significance), what you say about the working of machine ? [Given t0.025 = 2.262 at 9 degree of freedom] (1) Proper working (2) Improper working (3) Cannot say (4) none of these 27.  The mean weekly sales of a four-wheeler were 50 units per agency in 20 agencies. After an advertising campaign, the mean weekly sales increased to 55 units per agency with standard deviation of 10 units. As per the t-distribution test (of 5% level of significance), what you say about the advertising campaign ? (Use t0.005 = 1.729 for 19 d.f.) (1) Unsuccessful (2) Successful (3) Cannot say (4) none of these 28. A random sample of 60 observations was drawn from a large population and its standard deviation was found to be 2.5. The suitable standard error that this sample is taken from a population with standard deviation 3 is: (1) 0.2 (2) 0.27 (3) 0.31 (4) 0.35 29. A sample of 1000 students whose mean weight is 119 lbs (pounds) from a school in Uttar-Pradesh State was taken and their average weight was found to be 120 lbs with a standard deviation of 30 lbs. The standard error of the mean is : (1) 0.95 (2) 0.83 (3) 0.75 (4) 0.73 30. In a sample of 400 population from a village 230 are found to be eaters of vegetarian items and the rest non-vegetarian items. The standard error assuming that both vegetarian and non-vegetarian foods are equally popular in that village is : (1) 0.015 (2) 0.252 (3) 0.025 (4) 0.125

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: (1)  Both A and R are true and R is the correct explanation of A (2)  Both A and R are true but R is not the correct explanation of A (3)  A is true but R is false (4)  A is false and R is True 1. Assertion (A): A group of objects, events experiments, observations etc is referred to as population. Reason (R): A statistic is a numerical characteristic of a sample and a parameter is a numerical characteristic of population. 2. Assertion (A): t-distribution is also known as student’s t-distribution. Reason (R): t-distribution is also known as teacher’s t-distribution. 3. Assertion (A): The assumption are opposite of what is made in null hypothesis is called alternative hypothesis. Reason (R): For computing t-statistics we use the following formula x t  n 

209

INFERENTIAL STATISTICS

4. Assertion (A): The mean of the t-distribution is zero. Reason (R): The mean of t-distribution is 1.  5.  Assertion (A): A server channel monitored for an hour was found have an estimated mean of 20 transactions transmitted per minute. If the variance is known to be 4, then standard error is 0.258. s Reason (R) : Standard Error  n 6. Assertion (A): If the standard deviation of a sample of size 50 is 6.3, then the standard error whose population standard deviation is 6 is 0.5.

Reason (R) : Standard Error =

2  2n

7.  Assertion (A): Standard error is always non-negative. Reason (R) : Sampling error increases as we increase the sampling size. 8. Assertion (A): A subset of population is called sample. Reason (R) : The sampling error is defined as difference between population and sample. 9.  Assertion (A): Any population which we want to study is referred as target population. Reason (R) : Suppose we want to make a voters list for the general elections 2024 then we require census.  10. Assertion (A): If the Critical region is evenly distributed then the test is referred as Two tailed. Reason (R) : The point where the Null Hypothesis gets rejected is called as Critical Value.

4. The standard error of this is: (1) 4.65 (2) 4.55 (3) 46.5 (4) 4.10 5.  Using the fact the 5% value of t for 20 degrees of freedom is 2.09, whether diets A and B differ significantly as regards the effect on increases in weight. (1) Significant (2) not significant (3) can’t say (4) None of the above II.  Read the following text and answer the following questions on the basis of the same: Eight pots growing three barley plants each were exposed to high tension discharge, while nine similar pots were enclosed in an earthed wire cage. The numbers of tillers in each pot were as follows: Caged

17

26

18

25

27

28

26

23 17

Electrified

16

16

22

16

21

18

15

20

Given that the value of t for 15 degree of freedom at 5% level of significance is 2.131.

[C] COMPETENCY BASED QUESTIONS I.  Read the following text and answer the following questions on the basis of the same: For a random sample of 10 pigs fed on diet A, the increases in weight in pounds in a certain period were 10, 6, 16, 17, 13, 12, 8, 14, 15, 9 lbs For another random sample of 12 pigs fed on diet B, the increases in the same period were 7, 13, 22, 15, 12, 14, 18, 8, 21, 23, 10, 17 lbs

6. (1) 7. (1) 8.

Mean tillers of caged sample is: 23 (2) 18 (3) 9 Mean tillers of electrified sample is: 23 (2) 18 (3) 9 The formula for standard error is:

(1) s 

 (x  x )   (y  y )

(2) s 

 (x  x )2   (y  y )2

(3) s 

 (x  x )   (y  y )

(4) s 

 (x  x )   (y  y )

2

2

n1  n2  2

2

n1  n2  2 2

Mean of increases in weight on diet A is 12 pounds (2) 15 pounds 20 pounds (4) 27 pounds Sum of square of deviation for diet B is: 120 (2) 314 (3) 150 (4) 214 The number of degrees of freedom is: 10 (2) 12 (3) 15 (4) 20

(4) 8

n1  n2  2

2

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

(4) 8

2

n1  n2

9. The formula for two sample t-test is: (1) t 

x  y n1n2 s n1  n2

(2) t 

xy s

n1  n2 n1  n2

(3) t 

xy s

(4) t 

xy s

n1n2 n1  n2

n1n2 n1  n2

210 Oswaal CUET (UG) Chapterwise Question Bank 10. If the calculated value for t-test is 2.751, then which of the following statement is true: (1) The electrification does exert some effect on the tillering.

MATHEMATICS/APP. MATH

(2)  The electrification does not exert some effect on the tillering. (3) Can’t say. (4) None of these.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (1)

3. (4)

4. (1)

5. (4)

6. (2)

7. (4)

8. (2)

9. (1)

10. (2)

11. (3)

12. (2)

13. (3)

14. (2)

15. (2)

16. (2)

17. (4)

18. (2)

19. (3)

20. (4)

21. (3)

22. (1)

23. (2)

24. (2)

25. (3)

26. (2)

27. (2)

28. (2)

29. (1)

30. (3)

8. (2)

9. (2)

10. (2)

8. (1)

9. (4)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (2)

2. (3)

3. (3)

4. (3)

5. (1)

6. (4)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (2)

3. (4)

4. (1)

5. (2)

6. (1)

7. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (2) is correct. Explanation: The average (or mean) of sample values is a statistic. The term statistic is used both for the function and for the value of the function on a given sample. 2. Option (1) is correct. Explanation: Option (A) is incorrect as the null hypothesis is often stated as the assumption that there is no change, no difference between two groups, or no relationship between two variables on the other hand, alternative hypothesis, stated that there is a change, difference, or relationship. Option (E) is incorrect because if the test statistic is greater than the critical (table) value, then the null hypothesis is rejected in favour of the alternative hypothesis. 3. Option (4) is correct. Explanation: This is example of systematic sampling. In systematic sampling, every member of population is listed with a number, and individuals are chosen at regular intervals. 4. Option (1) is correct. Explanation:



X   n



24.3  24  2 64 0.3  2 8  0.3  4  1.2

t

5. Option (4) is correct Explanation: 24676 5 25  5 x 5

Mean 



Standard deviation





 x  x 





(2  5) 2  (4  5) 2  (6  5) 2  (7  5) 2  (6  5) 2 5



9 11 4 1 5



16  1.78 ∼ 2 5



i

2

f

6. Option (2) is correct Explanation: Test statistic

X − 0 /  19.5 − 20 −0.5 = = = −2 2/8 2 / 64

t=

7. Option (4) is correct Explanation: A → IV B→I C → III D → II

211

INFERENTIAL STATISTICS

8. Option (2) is correct. Explanation: x is a statistic. 9. Option (1) is correct. Explanation: The one sample t test, also referred to as a single sample t test, is a statistical hypothesis test used to determine whether the mean calculated from sample data collected from a single group is different from a designated value specified by the researcher. x It is calculated using t  s n





10. Option (2) is correct. Explanation: Given, | t |< tv (a ) Accept Reject t v () 11. Option (3) is correct. Explanation: The degree of freedom is the number of independent estimates of variance on which mean squared error is based. This is equal to (n1 − 1) + (n2 − 1), where n1 is the sample size of the first group and n2 is the sample size of the second group. 12. Option (2) is correct. Explanation: The sample mean is a sample statistic and the population mean is a population parameter. The sample mean is used to estimate the population mean when the sample size is large. According to the central limit theorem, the sample mean tends to the population mean as the sample size increases. x → μ as n → ∞ 13. Option (3) is correct. Explanation: In statistical inferences, first we need population, and then we take a sample from it. After that data is tabulated and analysed. Finally, we make an inference (conclusion). 14. Option (2) is correct. Explanation: n = 26 |t | = 3.07 > t25 (0.05) = 3.07 > 2.06 ⇒ 15. Option (2) is correct. Explanation: n = 34 ⇒ v = 34 – 1 = 33 16. Option (2) is correct. Explanation: Standard error is used to refer to the standard deviation of various sample statistics such as mean or median. 17. Option (4) is correct. Explanation: A sample refers to a smaller, manageable version of a larger group. It is a subset containing the characteristics of a larger population. 18. Option (2) is correct. Explanation: A sampling distribution refers to a probability distribution of a statistic that comes from choosing random samples of a given population. 19. Option (3) is correct. Explanation: A numerical value used as a summary measure for a sample, such as a sample mean, is known as a sample statistic.

20. Option (4) is correct. Explanation: Study of statistics can be categorised in two types: (i) Descriptive Statistics (ii) Inferential Statistics. 21. Option (3) is correct. Explanation: Statistical hypothesis is an assumption or statement, which may or may not be true, about a population. 22. Option (1) is correct. Explanation: Type I error: The error of rejecting hypothesis (H0) when it is true. 23. Option (2) is correct. 17 = 0.34 Explanation: Population proportion of Parker = 50 24. Option (2) is correct. Explanation: The point estimation of population standard deviation is sample deviation.



S

=

 xi  x 2 n 1

(1 − 4 )2 + ( 3 − 4 )2 + ( 5 − 4 )2 + ( 7 − 4 )2 4 −1

[ ∵ x =

=

9 +1+1+ 9 3

=

x 16  4] n 4

20 = 2.52 3

25. Option (3) is correct. Explanation: The symbol used for the mean of a sample is x , while the symbol used for a population mean is μ. A statistic is a number that summarizes data from a sample. x is a statistic since it is the mean of a sample. 26. Option (2) is correct. Explanation: Define Null hypothesis H0, Alternate hypothesis H1 as follows : H0 : m = 0.50 mm H1 : m ≠ 0.50 mm Thus, a two-tailed test is applied under hypothesis H0, we have x m 0.53  0.50 n 1  t 3  3 s 0.03 Since, the calculated value of t i.e., tcal(= 3) > ttab(= 2.262), the null hypothesis H0 can be rejected. Hence, we conclude that machine is not working properly. 27. Option (2) is correct. Explanation: Sample size, n = 20 agencies Sample mean x = 55 units Sample SD, s = 10 units Population mean, m = 50 units Null Hypothesis: The advertising campaign is not successful i.e., H0 : m = 50 ( There is no significant difference between the mean weekly Sales of units in agencies before and after advertising campaign) Alternate Hypothesis, H1 : m > 50. The advertising campaign was successful. Test statistic, x m 55  50 5  4.358 t n 1   19   2.179 s 10 10 

212 Oswaal CUET (UG) Chapterwise Question Bank Since, the calculated value of t i.e., tcal(= 2.173) > ttab(=1.729), the null hypothesis H0 can be rejected. Hence, we conclude that the advertising campaign was successful. 28. Option (2) is correct. Explanation: Given sample size n = 60 Sample standard deviation = 2.5 Population standard deviation σ = 3 The S.E. is given by

s2 = 2n

9 = 120

0.075 = 0.2739 29. Option (1) is correct. Explanation: Given n = 1000, mean = 119, σ = 30  30 30 = = 0.9487 Standard Error = = 31 .623 √ √ n 1000 30. Option (3) is correct. Explanation: Given sample size 400 and 230 are vegetarian eaters. 230 So sample proportional = = 0.575 400 Population proportion P = Prob. (vegetarian eaters from the 1 village) = 2 (Since vegetarian and non-vegetarian foods are equally popular), 1 1 Q=1–P=1   2 2 The standard error S.E. = PQ  N

1 1  2 2  0.25  0.000625  0.025 400 400

[B] ASSERTION REASON QUESTIONS 1. Option (2) is correct. Explanation: A population is a complete set of people with a specialized set of characteristics and a sample is a subset of the population. A statistic is a numerical description of a sample characteristic. 2. Option (3) is correct. Explanation: t-distribution is also known as student’s t-distribution gets its name from Willam Sealy Gosset who first published it in English in 1908 in scientific journal ̒Biometrika̕ using his pseudonym ‘Student’. 3. Option (3) is correct. Explanation: Correct formula for calculating t-statistic is t

x  n

4. Option (3) is correct. Explanation: The t-distribution is symmetric like normal distribution. So, its mean = 0.

MATHEMATICS/APP. MATH

5. Option (1) is correct. Explanation: Given s2 = 4 which implies s = 2, n = 1 hour = 60 minutes, X = 20/minutes σ 2 Standard Error    0.2582 √ n √60 6. Option (4) is correct. Explanation: Sample size n = 50 Sample S.D. (s) = 6.3 Population S.D. (s) = 6 The standard error for sample S.D. is given by

Standard Error =

=

σ2

= 2n

62 = 2 × 50

2 2n 6 = 0.6 100

Thus standard error for sample S.D = 0.6. 7. Option (3) is correct. Explanation: Assertion is true because when we square the mean for standard deviations any negative value becomes positive. The addition of all the positive values results in a positive value. Then the square root of the positive value is also positive. Hence all standard deviations are non-negative. Reason is false because sampling error is inversely proportional to the sampling size. As the sampling size increases the sampling error decreases. 8. Option (2) is correct. Explanation: Assertion is true. In sampling distribution we take a subset of population which is called as a sample. The main advantage of this sample is to reduce the variability present in the statistics. Reason is true. In sampling distribution the sampling error is defined as the difference between population and the sample. Sampling error can be reduced by increasing the sample size. 9. Option (2) is correct. Explanation: Assertion is true. In sampling distribution we take a part of a population under study which is called as target population. Target population is also called as a sample. Reason is true. Study of population is called a Census. Hence for making a voter list for the general elections 2024 we require Census. 10. Option (2) is correct. Explanation: Assertion is true. In two tailed test the critical region is evenly distributed. One region contains the area where Null Hypothesis is accepted and another contains the area where it is rejected. Reason is true. The point where the Null Hypothesis gets rejected is called as Critical Value. It is also called as dividing point for separation of the regions where hypothesis is accepted and rejected.

[C] COMPETENCY BASED QUESTIONS 1. Option (1) is correct. Explanation: Mean of increases in weight on diet A

x

x n1

213

INFERENTIAL STATISTICS

10 + 6 + 16 + 17 + 13 + 12 + 8 + 14 + 15 + 9 120 = = = 12 pounds 10 10 2. Option (2) is correct. Explanation: Sum of squares of deviations of diet B



 y  y 

Here, y =

2

∑

y

n2

 y  y

2

7 + 13 + 22 + 15 + 12 + 14 +18 + 21 + 23 + 10 + 17 180 = = = 15 pounds 12 12

 25 + 4 = 314 3. Option (4) is correct. Explanation: The number of degrees of freedom (n) = n1 + n2 – 2 = 10 + 12 – 2 = 20 4. Option (1) is correct. Explanation: ∵ Standard error,

2

∑( x − x ) + ∑( y − y ) n1 + n2 − 2

2

=

120 + 314 20

= 21.7 = 4.65

[∵   x  x   120 and   y  y   314 ] 2

x~y S .E .



n1n2 12 ~ 15 12  10   1.5 n1  n2 4.65 12  10

For 20 degrees of freedom we are given that the value of t at 5% level of significance is 2.09. The calculated value t is less than this value. Hence the difference between the sample means is not significant.

x n1

17  26  18  25  27  28  26  23  17 207   23 9 9

7. Option (2) is correct. Explanation: Number of terms in second set i.e., n2 = 8 Mean tillers of electrified sample is y y n2 16 + 16 + 22 + 16 + 21 + 18 + 15 + 20 144 = = = 18 8 8 8. Option (1) is correct. Explanation: The formula for standard error is

 (x  x )   (y  y ) 2

s

2

n1  n2  2

9. Option (4) is correct. Explanation: The formula for two sample t-test is

2

5. Option (2) is correct. Explanation: Here, t

x



 64 + 4 + 49 + 0 + 9 + 1 + 9 + 49 + 36 + 64 +

S .E . =

6. Option (1) is correct. Explanation: Number of terms in first set i.e., n1 = 9 Mean tillers of caged sample is

t

xy s

n1n2 n1  n2

10. Option (1) is correct. Explanation: It is given that the value of t for 15 degree of freedom at 5% level of significance is 2.131 which is less than the calculated value of t i.e., 2.751. Hence the difference between the mean tillers of the two samples is significant i.e., the electrification does exert some effect on the tillering.

Study Time

CHAPTER

17

  Revision notes

Max. Time: 1:50 Hours Max. Questions: 50

FINANCIAL MATHEMATICS

 Perpetuity: Perpetuity in the financial system is a situation where a stream of cash flow payments continues indefinitely or is an annuity that has no end. In valuation analysis, perpetuities are used to find the present value of a company’s future projected cash flow stream and the company’s terminal value. Essentially, a perpetuity is a series of cash flows that keep paying out forever. Scan to know  Perpetuity Formula: There are two difmore about ferent annual perpetual valuations, perpetuthis topic ity with flat or constant annuity and perpetuity with a growing annuity. Perpetuity gives a business the value of its cash flow, an essential slice of data as it aids in determining Perpetuity the firm’s total cash flow in a single year. (1) Flat Perpetuity: Perpetuity is also known as regular flat perpetuity. Here, cash flow is constant for each year. This perpetuity formula is the simplest, and it is straight forward as it doesn’t include terminal value. It is the basic formula for the price of perpetuity. For calculating the present value of flat perpetuity we only need to divide the cash flows/payments by the discount rate.

Cash flow Interest rate or yield (2) Growing Perpetuity: A growing perpetuity is the same as regular or flat perpetuity, but the difference is that the cash flow is growing each year. The present value of growing perpetuity formula factors in long term growth. This version is used to calculate the terminal value in a stream of cash flows for valuation purposes which is always more complicated. Payment Present value of Perpetuity = Interest rate − Growth rate Present value of Perpetuity =

 Sinking Fund: A sinking fund is a type Scan to know of fund that is created and set up purposely more about this topic for repaying debt. The owner of the account sets aside a certain amount of money regularly and uses it only for a specific purpose. Often, it is used by corporations for bonds and deposit money to buy back issued Sinking Fund bonds or parts of bonds before the maturity date arrives. It is also one way of enticing investors because the fund helps convince them that the issuer will not default on their payments. Example 1: Let us consider a franchisee of 7-Eleven who issues ₹ 50,000 worth bonds with a sinking fund provision and establishes a sinking fund where in the franchisee regularly deposits ₹ 500, with the intent of using it to buy back bonds slowly before they mature.

The provision will then allow him to buy back the bonds at a lower price if the market price is lower or at face value if the market price goes higher. Eventually, the principal amount owed will be lower, depending on how much was bought back. However, it is important to remember that there is a certain limit to how many bonds can be bought back before the maturity date. The formula to calculate the sinking fund is given below:    r   m n   1       1   m    P Sinking Fund, A   r   m where, P = Periodic contribution to the sinking fund   r = Annualised rate of interest n = No. of year m = No. of payments per year  Difference between Sinking Fund and Saving Account Sinking funds and savings accounts are two financial tools that are used to create financial stillness and make personal finance as simple as possible. The key difference between the sinking fund and savings account is that a sinking fund is recognised for a specific purpose and timeframe, while a savings account can be used for any purpose. Savings accounts are set up for long-term savings. A savings account is intended to sit and receive deposits for long enough to accumulate a substantial amount of money whereas sinking funds are building up for a variety of reasons and doesn’t have a time limit. Scan to know  Bond: A written agreement between more about a borrower and a lender (bond holder) this topic is known as bond. Upon maturity of this agreement, the borrower promises to pay a specified sum at a specific future date or to pay interest payments at a specific rate Bond Valuation at equal intervals of time. Bonds are also known as debentures. A bond is characterised by following terms:  Face Value or Par Value: The face value of a bond is the price at which the bond is sold to buyers, at the time of issue. Face value is also considered as the price at which the bond is redeemed at maturity.  Redemption Price: It is the amount the paid by the issuer to the bond holder at the time of maturity. In case the bond is redeemed at par, the redemption price is usually equal to the face value.  Discount: We say bond is selling at a discount when the market price of bond is less than its face value.

FINANCIAL MATHEMATICS

215

216 Oswaal CUET (UG) Chapterwise Question Bank  Premium:

We say bond is selling at a premium when the market price of bond is greater than its face value.  Bond valuation: Bond valuation is a technique for determining the theoretical fair value of a particular bond. It involves calculating the present value of a bond’s expected future coupon payments, or cash flow, and the bond’s value upon maturity, or face value.  Nominal rate of interest: Nominal rate of interest is also known as coupon rate. It is the rate at which a bond yields interest.  Coupon Rate: The interest rate that is fixed and is surely received by the bondholder as payment for interest is known as Coupon rate. The value of the coupon rate remains the same until the maturation of the bond. It is simply the coupon payment C as a percentage of the face value F. C Coupon Rate = F  Current Yield: The current yield is the coupon payment C as

a percentage of the (current) bond price P0. C Coupon yield = P0  Yield

to Maturity (YTM): In the case of a bond, YTM is defined as the total rate of return that a bond holder expects to earn if a bond is held till maturity. The YTM formula for a single bond is:  Yield to Maturity    FV  Price    Annual Inetrest      Maturity     FV  Price     2   The yield to maturity (YTM) is the discount rate which returns the market price of a bond. To achieve a return equal to YTM, the bond owner must: (1) Buy the bond at a price P0 (2) Hold the bond until maturity (3) Redeem the bond at par  Relationship between Bond, YTM and Coupon yield  When a bond sells at a discount, YTM > current yield > coupon yield.  When a bond sells at a premium, coupon yield > current yield > YTM  When a bond sells at par, YTM = current yield = coupon yield  Formula to Calculate Bond Value  1  1  i  n    C 1  i  n V  R i     Where, C = Redemption price or Maturity value i = Yield rate or interest rate per period R= Coupon payment or periodic interest (dividend) payment n = No. of periods Example 1: Find the purchase price of a ₹ 1200, 6% bond, dividends payable semi-annually redeemable at par in 6 years, if the yield rate is to be 6% compounded semi-annually. (1) 660 (2) 600 (3) 700 (4) 760

MATHEMATICS/APP. MATH

Sol. Option (1) is correct Explanation: Face value of the bond C = ₹ 600 Nominal rate of interest i = 6% or 0.06 As dividends are paid semi-annually 0.06 = 0.03 Therefore, Rate of interest per period id = 2 Therefore, periodic dividend payment R = C × id = 600 × 0.04 = 24 So, semi-annual dividend R is ₹ 24. Yield rate is 6% = 0.06, compounded semi annually 0.06 Therefore, i = = 0.03 2 Number of years = 6 Therefore, number of dividend periods (n) = 6 × 2 = 12 Purchase price (V) of the bond is given by 1  1  0.0312  12   600 1  0.03  V  24  0.03   1  1.0312  12   600 1.03  24  . 0 03   1  0.7014   24  Scan to know   600  0.7014  0.03  more about this topic  24  9.954  420.84  238.89  420.84 = 659.74 Therefore, purchase price of bond is ₹ 660 (approx). Equated  EMI (Equated Monthly Installment): Monthly An equated monthly installment (EMI) is Installment a fixed payment amount made bya borrower to a lender at a specified date each calendar month. Equated monthly installments are used to pay off both interest and principal each month so that over a specified number of years, the loan is paid off in full. With most common types of loans such as real estate mortgages, auto loans, and student loans the borrower makes fixed periodic payments to the lender over the course of several years with the goal of retiring the loan. The EMI system of loan repayment has following features: (1) Each installment contains both components of principal repayment and interest charges. (2) Interest is calculated on reducing balance method. (3) Interest component is higher in the beginning and progressively lower towards the end. That means, the principal component of an EMI is lower during initial periods and higher during later periods. (4) The amount of EMI depends on: (i) The period of compounding i.e., whether the compounding is yearly, half yearly, quarterly or monthly. If the compounding is more frequent, then the amount of EMI would be higher and vice-versa. (ii) The rate of interest. (iii) Period of repayment if the repayment period is more, then EMI would be lower and vice versa.  Flat Rate EMI: In the flat-rate method, each interest charge is calculated based on the original loan amount, even though the loan balance outstanding is gradually being paid down.

217

FINANCIAL MATHEMATICS

The formula for Flat rate EMI is: p+I EMI = n Where, P = Principal of the loan I = Total interest on the principal n = number of months in loan period respectively. Reduced Balance Method EMI: It is also known as Amortisation. When one is amortising a loan, at the beginning of any period, the principal outstanding is the present value of the remaining payments. The formula for Reduced Balance Method EMI is: Installment Amount 

1  i n  P  i   1  i n  1

monthly interest rate annual 12  100 n = number of instalment P = principal amount of the loan  Calculation of Returns: All financial decisions involve some risk. One may expect to get a return of 15% per annum in his investment but the risk of not able to achieve 15% return will always be there. Return is simply a reward for investing as all investing involves some risk. A debt investment is a loan and the return is just the loan’s interest rate. This is simply the ratio of the interest paid to the loan principal. int erest paid Re turn , k = loan amount where, i 

This formulation leads to the convenient idea that a return is what the investor receives divided by what he or she invests.  Rate of Return (ROR) or Nominal Rate of Return (NROR): A Rate of Return (ROR) is the gain or loss of an investment over a certain period of time. In other words, the rate of return is the gain (or loss) compared to the cost of an initial investment, typically expressed in the form of a percentage. When the ROR is positive, it is considered a gain and when the ROR is negative, it reflects a loss on the investment.

The standard formula for calculating ROR is as follows:

Ending valueof investment − Initial valueof investment Rate of Re turn = ×100 Beginning valueof investment

Scan to know

more about Depreciation allows a this topic portion of the cost of a fixed asset to the revenue generated by the fixed asset. This is mandatory under the matching principle as revenues are recorded with their associated expenses in the accounting period when Depreciation by the asset is in use. This helps in getting a Linear Method complete picture of the revenue generation transaction. For Example: If a delivery truck is purchased by a company with a cost of ₹ 1,00,000 and the expected usage of the truck is 5 years, the business might depreciate the asset under depreciation expense as ₹ 20,000 every year for a period of 5 years. There are three methods commonly used to calculate depreciation. They are: (1) Linear or Straight line method (2) Unit of production method (3) Double-declining balance method Three main inputs are required to calculate depreciation:  Useful life: This is the time period over which the organisation considers the fixed asset to be productive. Beyond its useful life, the fixed asset is no longer cost-effective to continue the operation of the asset.  Salvage value: Post the useful life of the fixed asset, the company may consider selling it at a reduced amount. This is known as the salvage value of the asset.  The cost of the asset: This includes taxes, shipping, and preparation/setup expenses. Here, we have discussed only Linear or Straight line method for calculating depreciation.  Linear or Straight-line Depreciation Method: This is the simplest method of all. It involves simple allocation of an even rate of depreciation every year over the useful life of the asset. The formula for straight line depreciation is: Asset cos t − Re sidual value Annual Depreciation Expense = Useful life of the asset  Depreciation:

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Which of the following in NOT true? [CUET 2023] (1)  Sinking fund is established to set aside revenue for a future expense (2) Sinking funds are the same as savings account (3) Sinking fund is set up for a particular purpose (4) Sinking funds are to be used at a particular time 2.  Mr Jain takes a personal loan of ₹ 10,00,000 at 12% rate of interest per annum for 3 years. His EMI by flat rate method is : [CUET 2023] (1) ₹ 28,120.77 (2) ₹ 37,777.77 (3) ₹ 28,777.77 (4) ₹ 28,888.77 3.  A sum of ₹ 60,000 invested at r % compounded quarterly will provide payments at ₹ 600 each at the end of every three months. Then the value of r is: [CUET 2023]

(1) 8% (2) 4% (3) 2% (4) 5% 4.  The present value of a perpetuity of ₹ 1200 payable at the beginning of each year, if money is worth 5% per annum is: [CUET 2023] (1) ₹ 25,500 (2) ₹ 24,000 (3) ₹ 24,200 (4) ₹ 25,200 5.  An asset costing ₹ 2,00,000 is expected to have a useful life of 10 years and a final scrap value of ₹ 40,000. The book value of the machine at the end of sixth year is:  [CUET 2023] (1) ₹ 1,36,000 (2) ₹ 1,04,000 (3) ₹ 1,20,000 (4) ₹ 88,000 6.  A person takes a car loan of ₹ 9,00,000 at the rate of 12% per annum for 5 years from a bank. The EMI under flat rate system is: [CUET 2023]

218 Oswaal CUET (UG) Chapterwise Question Bank (1) ₹ 24,000 (2) ₹ 20,000 (3) ₹ 16,000 (4) ₹ 28,000 7.  If the cash equivalent of a perpetuity of ₹ 300 payable at the end of each quarter is ₹ 24000 then rate of interest converted quarterly is (1) 5% (2) 4% (3) 3% (4) 2% 8.  A machine costing ₹ 30,000 is expected to have a useful life of 4 years and a final scrap value of ₹ 4000. The annual depreciation is (1) ₹ 55,00 (2) ₹ 65,00 (3) ₹ 75,00 (4) ₹ 85,00 9.  The present value of a perpetuity of ₹ R payable at the end of each payment period, when the money is worth i per period, is given by:  (1) Ri

(2) R +

R i

(3)

R i

(4) R – Ri

10.  A vehicle has a scrap value of ₹ 7,50,000 after 6 years of its purchase. If the annual depreciation charge is ₹ 55,000, then the original cost of the vehicle is: [CUET 2022] (1) ₹ 12,00,000 (2) ₹ 11,60,000 (3) ₹ 10,80,000 (4) ₹ 9,30,000 11.  Which of the following statements are correct? (A) If discount rate is greater than coupon rate, then present value of a bond is greater than face value. (B) An annuity in which the periodic payment begins on a fixed date and continues forever is called perpetuity. (C) The issuer of bond pays interest at fixed interval at fixed rate of interest to investor is called coupon payment. (D) A sinking fund is a fixed payment made by a borrower to a lender at a specific date every month to clear off the loan. (E) The issues of bond repays the principle i.e., face value of the bond to the investor at a later date termed as maturity date. Choose the correct answer from the options given below:  [CUET 2022] (1) (A), (C), (E) only (2) (A), (B), (D) only (3) (B), (C), (E) only (4) (A), (B), (C) only 12. Which of the following statements is true? (A) EMI in flat rate method, Principle  Interest EMI  Number of payment (B) EMI in reducing balance method, i EMI  P  n 1  1  i  where, P = Principle, i = interest rate, n = no. of payments (C) In sinking fund, a fixed amount at regular intervals is deposited. (D) Approximate Yield to Maturity Face Value  Present Value Coupen Payment  Number of Payment  Face Value  Present Value 2 Choose the correct answer from the options given below:  [CUET 2022] (1) (A) and (B) only (2) (B) and (C) only

MATHEMATICS/APP. MATH

(3) (A) and (C) only (4) (C) and (D) only 13.  Mr. Dev wishes to purchase an AC for ₹ 45,000 with a down payment of ₹ 5000 and balance in EMI for 5 years. If Bank charges 6% per annum compounded monthly then monthly EMI is: [CUET 2022]   0.005  0.0194   use 60 1  (1.005)   (1) ₹ 776 (2) ₹ 700 (3) ₹ 737 (4) ₹ 673 14.  The cost of a machine is ₹ 20,000 and its estimated useful life is 10 years. The scrap value of the machine, when its value depreciates at 10% p.a, is: use (0.9)10 = 0.35  [CUET 2022] (1) ₹ 9,672 (2) ₹ 7,000 (3) ₹ 6,982 (4) ₹ 35,00 15. What is the face value of a sinking fund that yields a dividend of ₹ 18,00 at 10% semi-annually? (1) ₹ 36,00 (2) ₹ 18,000 (3) ₹ 24,000 (4) ₹ 36,000 16.  A newspaper printing machine costs ₹ 4,80,000 and estimated scrap value of ₹ 25,000 at the end of its useful life of 10 years. What is its annual depreciation as per linear method? (1) ₹ 4,550 (2) ₹ 45,500 (3) ₹ 50,500 (4) ₹ 61,500 17. At what rate of interest will the present value of perpetuity of ₹ 1500 payable at the end of every 6 months be ₹ 20,000 ? (1) 10% (2) 12% (3) 15% (4) 16.5% 18. If Mr Surya borrows a sum of ₹ 5,00,000 with total interest paid ₹ 2,00,000 (flat) and he is paying an EMI of ₹ 12,500, then the loan tenure is  (1) 36 months (2) 48 months (3) 56 months (4) 60 months 19. Mitul invested ₹ 3,50,000 in a fund. At the end of the year the value of the fund is ₹ 4,37,500. The nominal rate of interest, if the market price is same at the end of the year is (1) 15% (2) 20% (3) 23% (4) 25% 20. The value of a bond and debenture is (1) Present value of interest payments it gets (2) Present value of contractual payments it gets till maturity (3) Present value of redemption amount (4) None of the above 21.  The money is needed to endure a series of lectures costing ₹ 15,00 at the beginning of each year indefinitely, if money is worth 4% compounded annually is : (1) ₹ 39,000 (2) ₹ 40,000 (3) ₹ 42,000 (4) ₹ 45,000 22.  If Raj borrows ₹ 6,00,000 with 9% annual interest rate for 4 years, then EMI under Reducing Balance method is:  [Use(1.0075)–48 = 0.6986] (1) ₹ 14,500 (2) ₹ 14,930 (3) ₹ 15,436 (4) ₹ 16,580 23.  Under straight line methods of charging depreciation, value of depreciation charge on (1) Original cost (2) Written down value (3) Scrap value (4) None of the above 24.  Under straight line methods of charging depreciation, value of depreciation (1) Increase every year (2) Decrease every year (3) Constant every year (4) None of the above 25.  Depreciation is a process of (1) Allocation (2) Valuation

219

FINANCIAL MATHEMATICS

(3) Both (4) None of these 26.  If C is the original cost of an asset, S is the salvage value of an asset and n is the number of year estimated for useful life of an asset, an annual depreciation of an asset is given by (1) D 

CS n

(2) D 

S C n

(3) � D 

S C n

(4) � D 

S n C

27.  Depreciation is deducted from the concerned ………...… (1) Asset (2) Liability (3) Expense (4) Income 28.  An asset is purchased for ₹ 70,000 on which depreciation is to be provided annually according to the straight line method. The useful life to the asset is 12 years and residual value is ₹ 10000 the rate of depreciation is: (1) 18% (2) 15% (3) 9% (4) 7% 29.  ₹ 5000 is invested in a Term Deposit Scheme that fetches interest 6% per annum compounded quarterly. The interest after one year will be Given that (1.015)4 = 1.0613 (1) ₹ 307 (2) ₹ 315 (3) ₹ 401 (4) ₹ 417 30. Assume that Shyam holds a perpetual bond that generates an annual payment of ₹ 500 each year. He believes that the borrower is creditworthy and that an 8% interest rate will be suitable for this bond. The present value of this perpetuity is:  (1) ₹ 6520 (2) ₹ 6250 (3) ₹ 5620 (4) ₹ 2650

[B] ASSERTION REASON QUESTIONS

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as: (1) Both A and R are true and R is the correct explanation of A (2) Both A and R are true but R is not the correct explanation of A (3) A is true but R is false (4) A is false and R is true 1. Assertion (A): If the nominal rate of interest is 12.5% and the inflation is 2%, then the effective rate of interest is 10.5% Reason (R): If the interest is calculated only at the end of a year, then the effective rate of interest is same as the nominal rate of interest. 2. Assertion (A): If the rate of return on an investment is negative, then it indicates profit. Reason (R): If the rate of return on an investment is negative, then it indicates loss. 3. Assertion (A): An annuity in which the periodic payment begin on a fixed date and continue forever is called perpetuity. Reason (R): The amount or future value of perpetuity is always defined. 4. Assertion (A): A loan is said to be amortised if each instalment is used to pay interest and part of principal. Reason (R): If the market value of a share is less than its nominal value, then share is called at premium. 5. Assertion (A): If ₹2,50,000 cash is equivalent to a perpetuity of ₹ 7,500 payable at the end of each quarter,

then the rate of interest convertible quarterly is 12%. Reason (R): To calculate the rate of interest we apply R formula, P =  i 6. Assertion (A):10 years ago, Mr. Bhalla set up a sinking fund to save for his daughter’s higher studies. At the end of 10 years, he has received an amount of ₹ 10,21,760. Thus, the amount he put in the sinking fund at the end of every 6 months for the tenure is ₹ 40,000, which paid him 5% p.a. compounded semi-annually. [Use (1.025)20 = 1.6386] Reason (R): To calculate the amount that Mr. Bhalla set aside at the end of every six months,we apply formula  1  i n  1   S  R i     7. Assertion (A): If Reena makes an investment of ₹ 3,000 for two years and she gets a rate of interest of 12%, then the future value of the investment is ₹ 3763.20.[Use (1.12)2 = 1.2544]  Reason (R): To calculate the future value, we use the formula F.V. = C.F. (1+i)n. 8. Assertion (A): Radhika wants to retire from her job and get hold of ₹ 5,000 per month. She wants the money to go to the future generation after she dies. If she will earn an interest rate of 8% compounded annually, then the total amount she will need to achieve the perpetuity goal is ₹ 746269.  Reason (R): To calculate the perpetuity goal, we apply the formula P = 

i . R

9. Assertion (A): Anita plans a trip to Jaisalmer and estimated that he requires ₹ 55,000 for the same, She takes a loan from her credit card with annual interest rate of 12% for 4 years. The EMI under flat rate system is 1600. P +1 Reason (R): Flat rate EMI is given by EMI =  n 10. Assertion (A): EMI is the monthly payment done towards a loan that has been lent by the borrower. Reason (R): EMI is same as the sinking fund.

[C] COMPETENCY BASED QUESTIONS I. Read the following text and answer the following questions on the basis of the same:  Rohan has completed his MBA and now he wants to start a new business. So, he approaches to many banks. One bank is agreed to give loan to Rohan. So, Rohan has borrowed ₹ 5 lakhs from a bank on the interest rate of 12 per cent for ten years. 1. EMI stands for: (1) Equated Monthly Instalments (2) Emerging Monthly Instalments (3) Easy Monthly Instalments (4) None of these 2. To calculate monthly instalment, we use the following formula: (1  i ) n (1) Instalment Amount   ( P  i) (1  i ) n (2) Instalment Amount 

(1  i ) n

(1  i ) n  1

 ( P  i)

220 Oswaal CUET (UG) Chapterwise Question Bank (3) Instalment Amount 

(1  i ) n

(1  i ) n 1

MATHEMATICS/APP. MATH

 ( P  i)

(4) None of the above 3. Calculate monthly instalment using (1.01)120 = 3.300 (1) ₹ 7,100 (2) ₹ 7,174 (3) ₹ 7,147 (4) ₹ 7,200 4.  Find the amount of total payment made by Rohan. (1) ₹ 86,088 (2) ₹ 8,80,880 (3) ₹ 8,60,000 (4) ₹ 8,60,880 5. Find the amount of interest paid by Rohan. (1) ₹ 36,088 (2) ₹ 3,60,880 (3) ₹ 36,000 (4) ₹3,60,088 II.  Read the following text and answer the following questions on the basis of the same: Ajay is a service man. He lives in a joint family. There are 6 members in his family. He is planning to purchase a car so he is searching for a bank for a loan. He take a loan of ₹ 2,50,000 at the interest rate of 6% p.a. compounded monthly is to be amortise by equal payment at the end of each month for five years. Given that (1.005)60 = 1.3489, (1.005)21 = 1.1104.

6.  The size of each monthly payment is: (1) ₹ 4,833 (2) ₹ 8,432 (3) ₹ 4,383 (4) ₹ 4,338 7.  The principal outstanding at beginning of 40th month is: (1) ₹ 60,069 (2) ₹ 96,097 (3) ₹ 96,000 (4) ₹ 98,670 8. The interest paid in 40th payment is: (1) ₹ 480 (2) ₹ 408 (3) ₹ 840 (4) ₹ 482 9. Principal amount paid in 40th payment is: (1) ₹ 4,325 (2) ₹ 4,359 (3) ₹ 4,352 (4) ₹ 4,532 10. Total interest paid is: (1) ₹ 39,961 (2) ₹ 3,6991 (3) ₹ 31,699 (4) ₹ 3,9691

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2) 11. (3) 21. (1)

2. (2) 12. (3) 22. (2)

3. (2) 13. (1) 23. (1)

4. (4) 14. (2) 24. (3)

5. (2) 15. (4) 25. (1)

6. (1) 16. (2) 26. (1)

7. (1) 17. (3) 27. (1)

8. (2) 18. (3) 28. (4)

9. (3) 19. (4) 29. (1)

10. (3) 20. (2) 30. (2)

8. (3)

9. (4)

10. (3)

8. (1)

9. (3)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (2)

2. (4)

3. (3)

4. (3)

5. (1)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (2)

3. (2)

4. (4)

5. (2)

6. (1)

7. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

Alternate Solution:

 1 1. Option (2) is correct. Flat rate EMI  P  i    n Explanation: Sinking funds and savings accounts go hand in hand, but  they are not the same thing. A savings account is Here, P = ₹ 10,00,000 where we save our money whereas a sinking fund is how we 12 = i rate = of interest per rupee per month = 0.01 save our money. 1200 2. Option (2) is correct. n  Number of instalments  3  12  36 Explanation: We know that, Flat rate EMI is given by: 1   P+I EMI  10, 00, 000  0.01   EMI = 36   n Where,  10, 00, 000  0.01  0.02777777  P = Principal of the loan  10, 00, 000  0.03777777  `37777.77 I = Total interest on the principal 3. Option (2) is correct. n = number of months in loan period respectively. Explanation: Here, P = ₹ 10,00,000, I = 10, 00, 000  EMI =

12  3  3, 60, 000 and n = 12 × 3 = 36 100

10, 00, 000 + 3, 60, 000 = ₹ 37777.77 36

R   A  P 1    100  Here, P = ₹ 60,000; R = r,

n

221

FINANCIAL MATHEMATICS

A = ₹ 60,000 + 2,400 (in 1 year) = ₹ 62,400 [Since, in three months payment received ₹ 600] n = 1 year 1  r   62, 400  60, 000 1    100   62, 400 r 624 r ⇒  1 ⇒    1 60, 000 400 600 400 24 r 24 ⇒   = ⇒   = r ⇒ r = 4% 600 100 6



4. Option (4) is correct Explanation: PV = ₹ 25,200 1, 200 R P  R   1, 200  5 I = 1,200 + 24,000 = ₹ 25,200 5. Option (2) is correct Explanation: Cost Price − Scrap Value Annual Depreciation = Estimated life of machine

  =

2, 00, 000 − 40, 000 10

= ₹ 16,000 Book value = C  ost Price – no. of years after which book value is to be computed × Annual Depreciation = 2,00,000 – 6 × 16,000 = 2,00,000 – 96,000 = ₹ 1,04,000 6. Option (1) is correct Explanation: P = 9,00,000, r = 12%, A = 5 years Total Interest in five years,

I= 

P× r ×t 100

9, 00, 000  12  5 = ₹ 5,40,000 100

Total Amount Payable,  A = P + I = 9,00,000 + 5,40,000 = ₹ 14,40,000 14, 40, 000 EMI under flat rate system = 60 = ₹ 24,000 7. Option (1) is correct. r Explanation:  i = 400 R  P = i 120   24000  300  400 ⇒=  r = 5% 24 r 8. Option (2) is correct. cs Explanation: D  n 30, 000  4, 000 ⇒   D  4 ⇒

⇒

= D

26, 000 = 6, 500 4

Hence, the depreciation is ₹ 6,500 9. Option (3) is correct. Explanation: Present value at the end of each payment period R is . i 10. Option (3) is correct. Explanation: Amount of annual depreciation 

 ⇒

55,000  =

Cost of machine  Scrap Value of machine Life in years

Cost of machine − 7, 50, 000 6

⇒ Cost of machine = 55,000 × 6 + 7,50,000 = ₹ 10,80,000 11. Option (3) is correct. Explanation: Option (1) is incorrect because if the discount rate (or required yield) is greater than the coupon rate, the present value of a bond is less than its face value, not more. This is because the fixed payments from the bond (the coupons and the face value at maturity) are being discounted at a higher rate, which reduces their present value. Option (4) is incorrect because a sinking fund is not a fixed payment made by a borrower to a lender every month to clear off the loan. A sinking fund is a means of repaying funds borrowed through a bond issue through periodic payments to a trustee who retires part of the issue by purchasing the bonds in the open market. 12. Option (3) is correct. Pi Explanation: Option B, EMI  1  (1  i )  n where, P = Principle, i = interest rate n = number of payments c  {( F  P.V .) / N } Option D, Approx. YTM = ( F  P.V .) 2 Where, F is face value c is coupon rate P.V. is present value N is number of payments 13. Option (1) is correct. Explanation: Given, Cost of AC = ₹ 45,000  Down payment = ₹ 5,000 ⸫ Balance = ₹ 40,000 So, P = ₹ 40,000 6   i   0.005, 12  100  n = 5 × 12 = 60 Pi EMI = 1  (1  i )  n 40, 000  0.005



     



     40, 000 



1  (1  0.005) 60 0.005

1  (1.005) 60   = 40,000 × 0.0194

  0.005  0.0194  Given, 60 1  (1.005)    = ₹ 776

222 Oswaal CUET (UG) Chapterwise Question Bank 14. Option (2) is correct. Explanation: Amount on depreciation r  = Original amount 1    100 

n

10

10   = 20, 000  1    100  10 = 20,000 × (0.9) = 20,000 × 0.35 [Given (0.9)10 = 0.35] = ₹ 70,000 15. Option (4) is correct. Explanation: Let the face value of the sinking fund be x 10 Then, x = 1, 800 200 ⇒ x = ₹ 36,000 16. Option (2) is correct. Explanation:  D  C  S n 4, 80, 000  25, 000    = ₹ 45,500 10

MATHEMATICS/APP. MATH

21. Option (1) is correct. Explanation: We have, R = 1,500, i = 0.04 Money needed to endure a series of lectures costing ₹ 1500 at the beginning of each year means the present value of a perpetuity of ₹ 1500 payable at the beginning of each year R P=R+ i 1, 500  1, 500  37, 500 ` 39, 000 = 1,500 + 0.04 22. Option (2) is correct. Explanation: By Reducing Balance method, Pi EMI = 1  (1  i )  n P = ₹ 6,00,000, i = EMI 

9  0.0075, n = 4 × 12 = 48 100  12

6, 00, 000  0.0075 1  1  0.0075 

48



4, 500 1  1.0075 

48



4, 500 1  0.6986

19. Option (4) is correct. Explanation: Vf = 4,37,500, Vi = 3,50,000

4, 500  ` 14, 930 0.3014 23. Option (1) is correct. Explanation: Under straight line method, the amount of depreciation is deducted from the original cost of an asset 24. Option (3) is correct. Explanation: The amount of depreciation will remain constant every year when straight line method is followed. 25. Option (1) is correct. Explanation: Depreciation is the process of allocating  an asset’s cost over the course of its useful life in order to align its expenses with revenue generation. 26. Option (1) is correct. Explanation: The formula to calculate annual depreciation using the straight-line method is (cost – salvage value) / useful life. 27. Option (1) is correct. Explanation: Depreciation represents the decrease in the value of an asset due to its continuous deterioration through its useful life. Companies calculate depreciation to estimate how much their assets have decreased in value over time. Depreciation is carried out for tangible assets which are the physical assets. 28. Option (4) is correct. Explanation: Depreciation = (Cost price - scrap value)/ Useful life of an asset Rate of depreciation = Depreciation value/ Cost price of asset Given, Cost price = ₹ 70,000, Scrap value = ₹ 10,000, Useful life of asset = 12 years

Nominal rate =

Depreciation =

17. Option (3) is correct. Explanation: Let rate of interest be r% per annum, then i =

r 200

Given R = ₹ 1,500 and P = ₹ 20,000 R P= i R 1, 500 1, 500 r = ⇒ i= ⇒ = P 20, 000 200 20, 000 r =15%

⇒

18. Option (3) is correct. Explanation: Here, P = 5,00,000, I = 2,00,000 EMI = 12,500 P+I   EMI = n  12, 500  5, 00, 000  2, 00, 000 n n=

⇒



=

7, 00, 000  = 56 months. 12, 500

V f −Vi ×100 Vi

4, 37, 500  3, 50, 000  100  25% 3, 50, 000

20. Option (2) is correct. Explanation : The present value of a bond (debenture) is the total value of the bond’s (debenture’s) future interest payments and its face value (the value at maturity), discounted back to the present using a rate of return (or discount rate) that represents the investor’s required rate of return.

 ( 70, 000 − 10, 000 ) 12 years

Depreciation = ₹ 5,000

 5, 000  Rate of depreciation =    × 100  70, 000  Rate of depreciation = 7.14% ~ 7% (approx.) 29. Option (1) is correct. Explanation: We know that, Compound interest, I = P [(1 + i)n – 1]

223

FINANCIAL MATHEMATICS

Here, P = ₹ 5,000, i = 6% p.a. = 0.06 p.a. = 0.06 × (1/4) per quarter = 0.015 per quarter \ I = 5,000[(1 + 0.015)4 – 1] = 5,000{(1.015)4 – 1} = 5,000(1.0613 – 1) = 5,000 × 0.0613 = 306.82 30. Option (2) is correct. Explanation: PV of perpetuity =

=

Annual Payment Interest rate 500 500 = = ` 6, 250 8 0.08 100

[B] ASSERTION REASON QUESTIONS 1. Option (2) is correct. Explanation: Effective rate of interest = Nominal rate – inflation rate = 12.5 – 2 = 10.5% 2. Option (4) is correct. Explanation: Positive rate of return represents profit where as negative rate of return represents loss. 3. Option (3) is correct. Explanation: The amount of future value of perpetuity is undefined. 4. Option (3) is correct. Explanation: If the market value of a share is more than its nominal value, then share is called at premium. 5. Option (1) is correct. r Explanation: P = 2,50,000, R = 7,500, i = 400 75, 00 × 400 ⇒ 2,50,000 = r ⇒ r = 12 % p.a. 6. Option (1) is correct. Explanation: n = 10 × 2 = 20, S = 10,21,760, i = 52,00 = 0.025, R = ?  (1  i ) n  1  We know that, S = R   i   ⇒

 (1  0.025) 20  1  10,21,760 = R   0.025  

⇒

 (1.025) 20  1  10,21,760 = R    0.025 

⇒

1.6386  1  10,21,760 = R   0.025 

⇒

 0.6386  10,21,760 = R   0.025 

10, 21, 760 × 0.025 ⇒  R = 0.6386 ⇒  R = ₹ 40,000 7. Option (1) is correct. Explanation: We know that, F.V. = C.F. (1+i)n Where, F.V. = Future Value C.F. = Cash Flow

i = Rate of interest = 0.12 n = Time Period = 2 Hence, F.V. = 3,000 (1 + 0.12)2 = ₹ 3,000 × 1.2544 = ₹ 3,763.20 8. Option (3) is correct. Explanation: Given R = ₹ 5,000, i = 008/12 = 0.00667 Using the values in the formula, we get P=

5, 000 R = ₹ 7,46,269. = i 0.00667

9. Option (4) is correct. Explanation: We know that, Flat rate EMI is given by : p+I EMI = n Where, P = Principal of the loan I = Total interest on the principal n = number of months in loan period respectively. Here, P = ₹ 55,000, I = 55, 000 

12  4  26, 400 and 100

n = 12 × 4 = 48 55, 000 + 26, 400 EMI = = ₹ 1,696 48 10. Option (3) is correct. Explanation: EMI is the monthly payment at a fixed rate done towards a loan that has been borrowed by borrower from the lender. A sinking fund is a sum of money created by saving small amounts over a period of time and is completely separate from the saving account or emergency fund. Thus, EMI is not same as sinking fund.

[C] COMPETENCY BASED QUESTIONS 1. Option (1) is correct. Explanation: The full form of EMI is Equated Monthly Instalment. EMI is a fixed amount of money we need to pay to a lender or merchant in case we have taken a loan or any other credit. The EMI component includes both the principal amount and the interest charged. 2. Option (2) is correct. Explanation: Formula to calculate monthly instalment is: Instalment Amount 

(1  i ) n  ( P  i) (1  i ) n  1

3. Option (2) is correct. Explanation:   annual rate  12 i   100  

Given,





    

  12     12     = 1 = 0.01   100  100  n = 10 × 12 = 120  P = ₹ 5,00,000

224 Oswaal CUET (UG) Chapterwise Question Bank Instalment Amount 

(1  i ) n  ( P  i) (1  i ) n  1

Instalment Amount 

(1  0.01)120 × (5,00,000 × 0.01) (1  0.01)120  1

3.300   5, 000 3.300  1 16, 500 = 2.300



= ₹ 7,173.91 ~ ₹ 7,174 So, EMI that Rohan has to pay is ₹ 7,174 4. Option (4) is correct. Explanation: Total payment made by Rohan to the bank in 10 years = (EMI × Total tenure in months) = ₹ (7,174 × 120) = ₹ 8,60,880 5. Option (2) is correct. Explanation: The total interest amount payable will be (Total payment – loan amount) = ₹ 8,60,880 – ₹ 5,00,880 = ₹ 3,60,880 6. Option (1) is correct. Explanation: Given, P = ₹ 25,000 6 i  12 100 = 0.005 and n = 5 × 12 = 60 EMI 

p  i  (1  i ) n (1  i ) n  1

25, 000  0.005  (1.005)60 (1.005)60  1 25, 000  0.005  1.3489  0.3489 

     = ₹ 4,832.69 = ₹ 4,833 7. Option (2) is correct. Explanation: Principal outstanding at beginning of 40th month

=

MATHEMATICS/APP. MATH

EMI (1 + i ) n−40+1 − 1 i (1 + i ) n−40+1

=

4, 832.69 (1 + 0.005)60−40+1 − 1

=

4, 832.69 (1.005) 21 − 1

0.005(1 + 0.005)60−40+1

0.005 × (1.005) 21

=

4, 832.69[1.1104 − 1] 0.005 × 1.1104

=

4, 832.69 × 0.1104 0.005 × 1.1104

= ₹ 96,096.72 ≈ ₹ 96,097 8. Option (1) is correct. Explanation: Interest paid in 40th payment 

EMI (1  i ) n401  1 (1  i ) n401



4, 832.69 (1  i )60401 



4, 832.69  (1.005) 21  1

(1  i )60401

(1.005) 21 4, 832.69  0.1104  1.1104

= ₹ 480.48 = ₹ 480 9. Option (3) is correct. Explanation: Principal paid in 40th payment = EMI – Interest paid in 40th payment = 4,832.69 – 480.48 = ₹ 4,352.21 = ₹ 4,352 10. Option (1) is correct. Explanation: Total interest paid = 60 × 4832.69 – 2,50,000 = 2,89,961.40 – 2,50,000 = ₹ 39,961.40 = ₹ 39,961