# NCERT Solutions Class 12 Mathematics Chapter 1- Relation and Function

NCERT Solutions for Class 12 Mathematics Chapter 1 are available on the Extramarks website for all students preparing for the term one exam. The chapter is about Relation and Function, and our solutions offer theoretical knowledge and answer to all questions from the NCERT textbook. We provide a step-by-step guide and solutions for students to understand each concept thoroughly. In addition, our solutions are derived from the NCERT books CBSE syllabus for the latest year.

Chapter 1 Class 12 Mathematics dives into the basic concepts of Relations and Function. In this chapter, students can review their previous Class 11 Mathematics learning. The chapter covers the introduction, types of relations, functions, and binary operations. Students will learn about the nature of the concepts and undergo preparation to cover Chapter 1 for Class 12 Mathematics. Besides, Extramarks solutions cover vital concepts, theories, and solved exercises to comprehend the topic. If you are looking for perfect study material for Relations and Function, you may refer to Extramarks NCERT Solutions Class 12 Mathematics Chapter 1.

The primary purpose of delivering the solutions is to help students score perfectly. With the help of Extramarks NCERT Solutions for Class 12 Mathematics Chapter 1, students can eventually substitute a set of numbers with a binary process. This chapter will also help students understand the concepts better by explaining the formula of the pair that relates to the elements.

As a student, you can visit the Extramarks website for the latest information and syllabus updates. You can also view articles on notes for NCERT Solutions Class 1, NCERT Solutions Class 2, and NCERT Solutions Class 3.

### Key Topics Covered In NCERT Solutions for Class 12 Mathematics Chapter 1

In Extramarks NCERT Solutions Class 12 Mathematics Chapter 1, students can expect all exercises and concepts to be explained in detail. The NCERT Solution begins with introducing Relation and Functions, and students will get to hold on to the basic concepts and properties of functions. Experts have updated the latest syllabus, and the main topics covered in NCERT Solution for Class 12 Mathematics Chapter 1 are:

 Exercise Topic 1.1 Introduction 1.2 Recall 1.3 Types of Function 1.4 Composition of Function and Invertible Function 1.5 Binary Operations

1.1 Introduction

In this introductory part, students will get a complete idea regarding concepts of Relations and Functions. It will also recap their learning from Class 11 Mathematics Chapter 1 and through all chapters. However, in the introduction, students will get a clear idea of what is present in the curriculum. Students can also get a proper set of instructions to understand the relationship between two objects belonging to the sets.

Students will get a proper set of instructions which will enable them to understand the relationship between two objects belonging to the sets. So, students get clarity and essential elements of the syllabus. It includes the concepts of relation and function, properties, formulas, and definitions. Chapter 1 Mathematics Class 12 is necessary to prepare for the integers and binary numbers part of their syllabus.

Some of the main features of this chapter are as follows:

• Empty relation
• Symmetric relation
• Equivalence relation
• Transitive relation

1.2 Recall

In NCERT Solutions for Class 12 Mathematics Chapter 1, students can learn and revise previous concepts and topics of Class 11 Mathematics. Overall, they can expect one quick revision and a deeper understanding of real numbers in this section. In addition, students will get more clarity on vertible and invertible functions, usage of addition, multiplication, division, and subtraction. This section is essential as it covers all the basics and paves the way towards complex concepts. Students get a good experience and the overall idea of the main topics, including polynomial function and modulus function.

1.3 Types of Functions

The students will grab all the essential elements of Relation and Functions, including identity, constant, modulus, signum, and rational functions. In addition, they will get proper knowledge and in-depth concepts of the injective and the subjective part. In NCERT Solutions Class 12 Mathematics Chapter 1, students can acquire accurate information of the elements of three different numbers and understand finite and infinite sets.

Some of the main functions explained in the chapter are as follows:

• A function f: X → Y is one-one. For example if f (x1) = f(x2) ⇒ x1 = x2 ∀ x1, x2 ∈ X.
• A function f: X → Y is onto. For example, if given any y ∈ Y, ∃ x ∈ X such that f(x) = y.
• A function f: X → Y is one-one and onto. For example, if f is both one-one and onto.

1.4 Composition of Functions and Invertible Function

The topic is one of the most critical sections, as it covers the composition of the function. Therefore, students can benefit from a complete understanding of the sets and the codes. In addition, examples are present in this section with short and long questions to practice.

1.5 Binary Operations

Studying binary operation is essential for students because it covers the integration and derived concepts. Students appearing for JEE Mains shall learn this section more precisely. Binary operations consist of integers, commutativity, associativity, rational numbers, and positive integers.

NCERT Solution for Class 12 Mathematics Chapter 1- covers all the important elements of the binary operation. Students can get a proper explanation of the arbitrary number set with concerns about the binary process. Further to this, the subject matter expert also elaborated the binary functions in correlation to two integers into one. Extramarks has covered all the essential formulae in NCERT Solutions Class 12 Mathematics Chapter 1 in the notes provided for reference.

List of NCERT Solutions Class 12 Mathematics Chapter 1 Exercise & Answer Solutions

NCERT Solutions for Class 12 Mathematics Chapter 2 Relation and Functions is available on the Extramarks website for free. It has step-by-step solutions for the examples present in the NCERT textbooks. The solution covers all the essential concepts and theories and is based on the latest CBSE 2022-2023 Syllabus guidelines. Students can also view the NCERT Solutions of other chapters from the Extramarks website.

Click on the below links to view NCERT Solutions Class 12 Mathematics Chapter 1:

• Chapter 1: Exercise 1.1 Solutions: 16 Questions (14 Short Answers, 2 MCQ)
• Chapter 1: Exercise 1.2 Solutions: 12 Questions (10 Short Answers, 2 MCQ)
• Chapter 1: Exercise 1.3 Solutions: 14 Questions (12 Short Answers, 2 MCQ)
• Chapter 1: Exercise 1.4 Solutions: 13 Questions (12 Short Answer, 1 MCQ)
• Chapter 1: Exercise 1.5 Solutions: 16 Questions (12 Short Answer, 1 MCQ)

Students can also view and explore other NCERT Solutions on our Extramarks website:

• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10

### NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematics is available on the Extramarks website for all the CBSE students. As the Term One exam approaches, students must consider referring to the NCERT Exemplar to gather complete mathematics concepts. Each exercise consists of solutions and problems with proper explanation.

Exemplar books play a vital role in preparing for competitive exams and help score more in CBSE exams. If students want to score more in the examination, referring to NCERT Exemplar Class 12 Mathematics is ideal. Students can start by preparing from Chapter 1 - Relation and Function. It becomes a great companion in the learning journey for preparing for competitive exams such as NEET and JEE Mains. At Extramarks, students get worksheets and best practising materials for the preparation.

### Key Features of NCERT Solutions Class 12 Mathematics Chapter 1

NCERT Solutions Class 12 Mathematics Chapter 1 provides in-depth solutions to various problems mentioned in the syllabus. To compete and score dynamically in exams like NEET and JEE, students must have a strong command of Mathematics. NCERT books cover all the challenging topics that help boost the brain with fast calculations.

Students get more exposure to all kinds of questions, pushing them to attempt the most challenging question in the competitive questions. For this purpose, students can refer to NCERT Solutions Class 12 Mathematics Chapter 1. Some of the other reasons include:

• Extramarks NCERT Solutions Class 12 Mathematics Chapter 1 is prepared by Subject Matter Experts.
• The NCERT Solutions are explained such that it helps the students enjoy the learning process.
• With the help of Relation and Functions chapter solutions, students can easily attempt complex problems in the exam.
• Students will score more in the exams as the answers are short, self-explanatory, and well structured.

Q.1 Determine whether each of the following relations are reflexive, symmetric and transitive:

(i ) Relation R in the set A = {1, 2, 3…13, 14}
defined as R = {(x, y): 3x y = 0}

(ii) Relation R in the set N of natural numbers
defined as R = {(x, y): y = x + 5 and x < 4}

(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as

R = {(x, y): y is divisible by x}

(iv) Relation R in the set Z of all integers defined as

R = {(x, y): x y is as integer}

(v) Relation R in the set A of human beings in a
town at a particular time given by

(a) R = {(x, y): x and y work at the same place}

(b) R = {(x, y): x and y live in the same locality}

(c) R = {(x, y): x is exactly 7 cm taller than y}

(d) R = {(x, y): x is wife of y}

(e) R = {(x, y): x is father of y}

Ans.

i) A = {1, 2, 3 … 13, 14}

R = {(x, y): 3x y = 0 or y=3x}

∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Given relation R is not reflexive because
(1, 1), (2, 2), (3,3)… (14, 14) ∉ R.

Also, R is not a symmetric relation as
(2, 6) ∈R, but (6,2) ∉ R.

Also, R is not transitive as (1, 3), (3, 9) ∈R,
but (1, 9) ∉ R.

Hence, R is neither reflexive, nor symmetric, nor transitive.

$\begin{array}{l}\text{(ii) R}=\text{}\left\{\left(x,y\right):y=x+\text{5 and}x<\text{4}\right\}\\ \text{R}=\text{}\left\{\left(\text{1},\text{6}\right),\text{}\left(\text{2},\text{7}\right),\text{}\left(\text{3},\text{8}\right)\right\}\\ \text{Since}\left(\text{1},\text{1}\right)\notin \text{R}.\\ \therefore \text{R is not reflexive}.\\ \left(\text{1},\text{6}\right)\in \text{R But}\left(\text{6},\text{1}\right)\notin \text{R}.\\ \therefore \text{R is not symmetric}.\\ \text{Now},\text{since there is no pair in R such that}\left(a,\text{}b\right)\text{and}\\ \left(b,c\right)\in \text{R},\text{then}\left(a,c\right)\text{cannot belong to R}.\\ \therefore \text{R is not transitive}.\\ \text{Hence},\text{R is neither reflexive},\text{nor symmetric},\text{nor transitive}.\\ \text{(iii) A = {1, 2, 3, 4, 5, 6}}\\ \text{R = {(x, y): y is divisible by x}}\\ \text{Since every number is divisible by itself}.\\ \therefore \text{(x, x)}\in \text{R}\\ \text{Hence R is reflexive}\text{.}\\ \text{Now,}\\ \text{(2, 4)}\in \text{R [as 4 is divisible by 2]}\\ \text{But,}\\ \text{(4, 2)}\notin \text{R}\text{. [as 2 is not divisible by 4]}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{. Then, y is divisible by x and z is divisible}\\ \text{by y}\text{.}\\ \therefore \text{z is divisible by x}\text{.}\\ \text{Hence (x, z)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive and transitive but not symmetric}\text{.}\\ \text{(iv) R = {(x, y): x}-\text{y is an integer}}\\ \text{For every x}\in \text{Z, (x, x)}\in \text{R as x}-\text{x = 0 which is an integer}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now, for every x, y}\in \text{Z if (x, y)}\in \text{R, then x}-\text{y is an integer}\text{.}\\ \text{For every (y,x)}\in \text{R by definition}\\ \left(y-x\right)=-\left(x-y\right)\text{\hspace{0.17em}}\text{is also an integer}\text{.}\\ ⇒\left(y-x\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is also an integer}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now,}\\ \text{Let (x, y) and (y, z)}\in \text{R, where x, y, z}\in \text{Z}\text{.}\\ ⇒\text{(x}-\text{y) = integer}\\ ⇒\text{(y}-\text{z) = integer}\\ \text{On adding both, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x-z\right)=\text{integer}\text{.}\\ \therefore \left(x,z\right)\in R\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric, and transitive}\text{.}\\ \\ \text{(v) (a) R = {(x, y): x and y work at the same place}}\\ \text{Every individual worker belongs to itself}\\ ⇒\text{(x, x)}\in \text{R}\\ \text{\hspace{0.17em}}\therefore \text{R is reflexive}\text{.}\\ \text{\hspace{0.17em}}\text{If (x, y)}\in \text{R, then x and y work at the same place}\text{.}\\ \text{Similarly y and x work at the same place}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\text{.}\\ \text{Hence R is symmetric}\text{.}\\ \text{Now, let (x, y), (y, z)}\in \text{R}\\ ⇒\text{x and y work at the same place and y and z work at}\\ \text{the same place}\text{.}\\ \therefore \text{x and z also work at the same place}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric and transitive}\text{.}\\ \text{(b) R = {(x, y): x and y live in the same locality}}\\ \text{Every individual belongs to itself}\\ \text{Thus (x, x)}\in \text{R}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Since (x, y)}\in \text{R it means x and y live in the same locality}\text{.}\\ \text{Which implies y and x also live in the same locality}\text{.}\\ \therefore \text{(y, x)}\in \text{R}\\ \text{Hence R is symmetric}\text{.}\\ \text{Now, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ ⇒\text{x and y live in the same locality and y and z live in the same}\\ \text{locality}\text{.}\\ \text{Which implies that x and z also live in the same locality}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \text{Hence R is transitive}\text{.}\\ \text{Hence, R is reflexive, symmetric and transitive}\text{.}\\ \text{(c) R = {(x, y): x is exactly 7 cm taller than y}}\\ âˆµ\text{An individual height can be equal to itself but cannot}\\ \text{be more by 7 cm}\\ \text{Hence (x, x)}\notin \text{R}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{Now, let (x, y)}\in \text{R}\text{.}\\ ⇒\text{x is exactly 7 cm taller than y}\\ \text{Then y should be shorter than x}\\ \therefore \text{(y, x)}\notin \text{R}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Again,}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{.}\\ ⇒\text{x is exactly 7 cm taller than y and y is exactly 7 cm}\\ \text{taller than z}\text{.}\\ \\ \text{Which implies x is exactly 14 cm taller than z}\text{.}\\ \therefore \text{(x, z)}\notin \text{R}\\ so,\text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\\ \text{(d) R = {(x, y): x is the wife of y}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Since x cannot be the wife of herself}\text{.}\\ \therefore \text{(x, x)}\notin \text{R}\\ \text{Then, R is not reflexive}\text{.}\\ \text{Let}\text{\hspace{0.17em}}\left(x,y\right)\in R\\ which\text{implies that x is wife of y}\text{.}\\ i.e.,\text{y is husband of x}\text{.}\\ \text{so, y can not be wife of x}\text{.}\\ ⇒\left(y,x\right)\notin R\\ Then,\text{\hspace{0.17em}}\text{R is not symmetric}\text{.}\\ \text{Let}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(x, y), (y, z)}\in \text{R}\\ ⇒\text{x is the wife of y and y is the wife of z}\text{.}\\ \text{A husband can never become wife of anyone}.\\ so,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x,z\right)\notin R\\ Thus,\text{â€‹}\text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\\ \text{(e) R = {(x, y): x is the father of y}}\\ \text{Since, x cannot be the father of himself}\text{.}\\ \text{So,}\left(x,x\right)\notin R\\ Then,\text{\hspace{0.17em}}\text{R is not reflexive}\text{.}\\ \text{Now, let (x, y)}\in \text{R}\text{.}\\ \\ \text{If x is the father of y}\text{.}\\ \text{Then y cannot be the father of x}\text{.}\\ \text{so,}\left(y,x\right)\notin R\\ Then,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{R is not symmetric}\text{.}\\ \text{Again, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ ⇒\text{x is father of y and y is father of z}\text{.}\\ ⇒\text{x can not be father of z}\text{.}\\ ⇒\text{x will be grandfather of z}\\ \therefore \left(x,z\right)\notin R\\ \therefore \text{\hspace{0.17em}}R\text{is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric not transitive}\text{.}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Show}\mathrm{}\mathrm{that}\mathrm{}\mathrm{the}\mathrm{}\mathrm{relation}\mathrm{}\mathrm{R}\mathrm{}\mathrm{in}\mathrm{}\mathrm{the}\mathrm{}\mathrm{set}\mathrm{}\mathrm{R}\mathrm{}\mathrm{of}\mathrm{}\mathrm{real}\mathrm{}\mathrm{numbers},\mathrm{}\\ \mathrm{defined}\mathrm{}\mathrm{as}\mathrm{}\mathrm{R}=\mathrm{}\left\{\mathrm{}\left(\mathrm{a},\mathrm{b}\right):\mathrm{ }\mathrm{a}\le {\mathrm{b}}^{2}\right\}\mathrm{ }\mathrm{is}\mathrm{neithe}\mathrm{rreflexive}\mathrm{nor}\\ \mathrm{symmetric}\mathrm{nor}\mathrm{transitive}.\end{array}$

Ans.

$\begin{array}{l}\text{We have R = {(a, b): a}\le {b}^{\text{2}}\text{} where a, b}\in \text{R}\\ \text{We can see that}\frac{\text{1}}{3}\le {\left(\frac{\text{1}}{3}\right)}^{\text{2}}\text{is not valid}\text{. So,}\left\{\frac{\text{1}}{3}\text{,}\frac{\text{1}}{3}\right\}\notin R\\ Then,\text{\hspace{0.17em}}\text{R is not reflexive}\text{.}\\ \text{Since,}\left(-2,3\right)\in R\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{â€‹}-{\text{2<3}}^{\text{2}}\\ But{\text{3}}^{\text{2}}\text{is not less than}-\text{2}\text{.}\\ \end{array}$

So, ( 3,2 )R

$\begin{array}{l}\mathrm{Thus},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Again},\mathrm{let}\left(5,-6\right),\left(-6,2\right)\in \mathrm{R}\\ \mathrm{As}\mathrm{ }5<{\left(-6\right)}^{2}\mathrm{and}-6{<2}^{\mathrm{2}}\\ \mathrm{But}5\mathrm{is}\mathrm{not}\mathrm{less}\mathrm{than}4\mathrm{.}\\ \mathrm{So},\left(5,2\right)\notin \mathrm{R}\\ \mathrm{Thus},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{.}\\ \mathrm{Therefore},\mathrm{R}\mathrm{is}\mathrm{neither}\mathrm{reflexive}\mathrm{nor}\mathrm{symmetric}\mathrm{nor}\mathrm{transitive}.\end{array}$

Q.3 Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.

Ans.

Given A = {1, 2, 3, 4, 5, 6}. A relation R is defined
on A as: R = {(a, b): b = a + 1} $\begin{array}{l}\therefore \text{R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}}\\ \text{We can find (1,1)}\notin \text{R, where 1}\in \text{A}\text{.}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{It can be observed that (1, 2)}\in \text{R, but (2, 1)}\notin \text{R}\text{.}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Now, (1, 2), (2, 3)}\in \text{R but}\left(1,3\right)\notin R.\\ \therefore \text{R is not transitive}\text{.}\\ \text{Hence, R is neither reflexive nor symmetric nor transitive}\text{.}\end{array}$

Q.4

$\begin{array}{l}\mathbf{Show}\mathbf{}\mathrm{}\mathbf{that}\mathbf{}\mathrm{}\mathbf{the}\mathbf{}\mathrm{}\mathbf{relation}\mathbf{}\mathrm{}\mathbf{R}\mathbf{}\mathrm{}\mathbf{in}\mathrm{}\mathbf{R}\mathrm{ }\mathbf{defined}\mathrm{}\mathbf{as}\mathrm{}\mathbf{R}=\mathrm{}\left\{\left(\mathbf{a},\mathbf{b}\right):\mathrm{ }\mathrm{a}\le \mathrm{b}\right\}\mathrm{ }\\ \mathrm{is}\mathrm{reflexive},\mathrm{symmetric}\mathrm{or}\mathrm{transitive}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{R}= \left\{\left(\mathrm{a},\mathrm{b}\right);\mathrm{a}\le \mathrm{b}\right\}\\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\left(\frac{\mathrm{1}}{2}\mathrm{,}\frac{\mathrm{1}}{2}\mathrm{\right)}\in \mathrm{R}\mathrm{as}\frac{\mathrm{1}}{2}=\frac{\mathrm{1}}{2}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{.}\\ \mathrm{Since}\left(1,2\right)\in \mathrm{R}\mathrm{as}2\mathrm{is}\mathrm{greater}\mathrm{than}1\mathrm{.}\\ \mathrm{but}1\mathrm{â€‹}\mathrm{is}\mathrm{not}\mathrm{greater}\mathrm{than}2,\mathrm{so}\\ \mathrm{ }\left(2,1\right)\notin \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Again},\mathrm{let}\left(\mathrm{a},\mathrm{b}\right), \left(\mathrm{b},\mathrm{c}\right)\in \mathrm{R}\mathrm{.}\\ \mathrm{Then}\mathrm{according}\mathrm{to}\mathrm{condition},\\ \mathrm{a}\le \mathrm{b}\mathrm{and}\mathrm{b}\le \mathrm{c} ⇒\mathrm{a}\le \mathrm{c}\\ ⇒\left(\mathrm{a},\mathrm{c}\right)\in \mathrm{R}\end{array}$

R is transitive.

$\text{Hence,R is reflexive and transitive but not symmetric}\text{.}$

Q.5

$\begin{array}{l}\mathrm{Check}\mathrm{whether}\mathrm{the}\mathrm{relation}\mathrm{R}\mathrm{in}\mathrm{R}\mathrm{defined}\mathrm{as}\mathrm{R}=\left\{\left(\mathrm{a},\mathrm{b}\right):\mathrm{a}\le {\mathrm{b}}^{3}\right\}\\ \mathrm{is}\mathrm{reflexive},\mathrm{symmetric}\mathrm{or}\mathrm{transitive}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have}\\ \mathrm{R}= \left\{\left(\mathrm{a},\mathrm{b}\right);\mathrm{a}\le {\mathrm{b}}^{\mathrm{3}}\mathrm{\right\}}\\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\frac{\mathrm{1}}{2}\le {\left(\frac{\mathrm{1}}{2}\right)}^{3}\mathrm{ }\end{array}$

so,( 1 2 , 1 2 )R R is not reflexive. Since ( 1,2 ) R as 1<2 3 But( 2,1 ) Ras2is not smaller than 1 3 . R is not symmetric. Again, let (5, 5 2 ), ( 5 2 , 5 4 )R. Then according to condition, 5< ( 5 2 ) 3 and 5 2 < ( 5 4 ) 3 (5, 5 4 )Ras5> ( 5 4 ) 3 R is not transitive. Hence,R is neither reflexive nor symmetric nor transitive.

Q.6 Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Ans.

$\begin{array}{l}Since,\\ A=\left\{1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3\right\}\\ \text{A relation R on A is defined as R = {(1, 2), (2, 1)}}\text{.}\\ \text{Since (1, 1), (2, 2), (3, 3)}\notin \text{R}\text{.}\\ \therefore \text{R is not reflexive}\text{.}\\ \text{Now, as (1, 2)}\in \text{R and (2, 1)}\in \text{R,}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now, (1, 2) and (2, 1)}\in \text{R}\\ But\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1,1\right)\notin R\\ \therefore R\text{is not transitive}\text{.}\\ \text{Hence, R is symmetric but neither reflexive nor transitive}\text{.}\end{array}$

Q.7 Show that the relation R in the set A of all the books in a library of a college, given by
R = {(x, y): x and y have same number of pages} is an equivalence relation.

Ans.

$\begin{array}{l}\text{Set A is the set of all books in the library of a college}\text{.}\\ \text{R = {x, y): x and y have the same number of pages}}\\ \text{since (x, x)}\in \text{R as x and x has the same number of pages}\text{.}\end{array}$

R is reflexive.

$\begin{array}{l}\text{If (x, y)}\in \text{R where x and y have the same number of pages}\text{.}\\ ⇒\text{y and x have the same number of pages}\text{.}\\ ⇒\text{(y, x)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Again, let (x, y)}\in \text{R and (y, z)}\in \text{R}\text{.}\\ ⇒\text{x and y and have the same number of pages and y and z}\\ \text{have the same number of pages}\text{.}\\ ⇒\text{x and z have the same number of pages}\text{.}\\ \therefore \text{(x, z)}\in \text{R}\\ \text{Therefore, R is transitive}\text{.}\\ \text{Since, R is reflexive, symmetric and transitive, so R is an}\\ \text{equivalence relation}\text{.}\end{array}$

Q.8 Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Ans.

$\begin{array}{l}Since,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A = {1, 2, 3, 4, 5}}\\ R=\left\{\left(a,b\right):|a-b|\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even\right\}\\ For\text{any element a}\in \text{A, we have}|a-a|=0,\text{\hspace{0.17em}}which\text{is even number}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Let (a, b)}\in \text{R}\text{.}\end{array}$

| ab |is even.

$\begin{array}{l}⇒\text{\hspace{0.17em}}|-\left(b-a\right)|\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even.\\ ⇒|b-a|\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even.\\ ⇒\left(b,a\right)\in R\\ So,\text{\hspace{0.17em}}\text{R is symmetric}\text{.}\\ \text{Again, let}\left(a,b\right)\in R\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\left(b,c\right)\in R.\\ ⇒|b-a|\text{\hspace{0.17em}}is\text{even number and}|c-b|\text{\hspace{0.17em}}is\text{also even number}\text{.}\\ ⇒\left(a-b\right)\text{\hspace{0.17em}}is\text{\hspace{0.17em}}even\text{and}\left(b-c\right)\text{is even}\text{.}\\ ⇒\left(a-c\right)=\left(a-b\right)+\left(b-c\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}is\text{even}\text{.}\\ ⇒|a-c|\text{\hspace{0.17em}}is\text{even}\text{.}\\ ⇒\left(a,c\right)\text{\hspace{0.17em}}is\text{even}\text{.}\\ ⇒\left(a,c\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\\ \text{Now, all elements of the set {1, 3, 5} are related to each}\\ \text{other as the modulus of difference between any two elements}\\ \text{of this set is even}\text{.}\\ \text{Similarly, all elements of the set {2, 4} are related to each}\\ \text{other as the difference between the two elements is even}\text{.}\\ \text{Also, no element of the subset {1, 3, 5} can be related to any}\\ \text{element of {2, 4} as all elements of {1, 3, 5} are odd and all}\\ \text{elements of {2, 4} are even}\text{.}\text{\hspace{0.17em}}\text{The modulus of difference between}\\ \text{odd and even is again odd}.\end{array}$

Q.9

$Show that each of the relation R in the set A = x∈Z,0≤x≤12 , given by i R= a,b : a–b is a multiple of 4 ii R= a,b :a=b is an equivalence relation. Find the set of all elements related to 1 in each case. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaahofacaWHObGaaC4BaiaahEhacaqGGaGaaCiDaiaahIga caWHHbGaaCiDaiaabccacaWHLbGaaCyyaiaahogacaWHObGaaeiiai aah+gacaWHMbGaaeiiaiaahshacaWHObGaaCyzaiaabccacaWHYbGa aCyzaiaahYgacaWHHbGaaCiDaiaahMgacaWHVbGaaCOBaiaabccaca WHsbGaaeiiaiaahMgacaWHUbGaaeiiaiaahshacaWHObGaaCyzaiaa bccacaWHZbGaaCyzaiaahshacaaMc8oabaGaaeyqaiaabccacaqG9a GaaeiiamaacmaabaGaamiEaiabgIGiolaadQfacaGGSaGaaGimaiab gsMiJkaadIhacqGHKjYOcaaIXaGaaGOmaaGaay5Eaiaaw2haaiaacY cacaGGGcGaaC4zaiaahMgacaWH2bGaaCyzaiaah6gacaqGGaGaaCOy aiaahMhacaaMc8UaaGPaVdqaaiaaxMaacaWLjaGaaGPaVlaaykW7ca aMc8+aaeWaaeaaieqacaWFPbaacaGLOaGaayzkaaGaaGPaVlaa=jfa caWF9aWaaiWaaeaadaqadaqaaiaa=fgacaWFSaGaa8NyaaGaayjkai aawMcaaiaa=Pdadaabdaqaaiaa=fgacaWFTaGaa8NyaaGaay5bSlaa wIa7aiaaykW7caWFPbGaa83Caiaa=bcacaWFHbGaa8hiaiaa=1gaca WF1bGaa8hBaiaa=rhacaWFPbGaa8hCaiaa=XgacaWFLbGaa8hiaiaa =9gacaWFMbGaa8hiaiaa=rdaaiaawUhacaGL9baaaeaacaWLjaGaaC zcaiaaykW7daqadaqaaiaa=LgacaWFPbaacaGLOaGaayzkaaGaaGPa Vlaa=jfacaWF9aWaaiWaaeaadaqadaqaaiaa=fgacaWFSaGaa8Nyaa GaayjkaiaawMcaaiaa=PdacaWFHbGaa8xpaiaa=jgaaiaawUhacaGL 9baaaeaacaWFPbGaa83Caiaa=bcacaWFHbGaa8NBaiaa=bcacaWFLb Gaa8xCaiaa=vhacaWFPbGaa8NDaiaa=fgacaWFSbGaa8xzaiaa=5ga caWFJbGaa8xzaiaa=bcacaWFYbGaa8xzaiaa=XgacaWFHbGaa8hDai aa=LgacaWFVbGaa8NBaiaa=5cacaWFGaGaa8Nraiaa=LgacaWFUbGa a8hzaiaa=bcacaWF0bGaa8hAaiaa=vgacaWFGaGaa83Caiaa=vgaca WF0bGaa8hiaiaa=9gacaWFMbGaa8hiaiaa=fgacaWFSbGaa8hBaiaa =bcacaWFLbGaa8hBaiaa=vgacaWFTbGaa8xzaiaa=5gacaWF0bGaa8 3Caaqaaiaa=bcacaWFYbGaa8xzaiaa=XgacaWFHbGaa8hDaiaa=vga caWFKbGaa8hiaiaa=rhacaWFVbGaa8hiaiaa=fdacaWFGaGaa8xAai aa=5gacaWFGaGaa8xzaiaa=fgacaWFJbGaa8hAaiaa=bcacaWFJbGa a8xyaiaa=nhacaWFLbGaa8Nlaaaaaa@FED7@$

Ans.

$\begin{array}{l}Here,\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left\{x\in Z,0\le x\le 12\right\}=\left\{0,1,2,3,4,5,6,7,8,9,10,11,12\right\}\\ \left(i\right)\text{\hspace{0.17em}}R=\left\{\left(a,b\right):|a-b|\text{is a multiple of 4}\right\}\\ For\text{any element a}\in \text{A,}\\ \text{we have}\left(a,a\right)\in R\text{\hspace{0.17em}}\text{\hspace{0.17em}}as\text{\hspace{0.17em}}|a-a|=0\text{\hspace{0.17em}}which\text{\hspace{0.17em}}\text{\hspace{0.17em}}is\text{multiple of 4}\text{.}\\ \therefore \text{\hspace{0.17em}}\text{R is reflexive}\text{.}\\ \text{Let}\left(a,b\right)\in R\text{\hspace{0.17em}}as\text{â€‹}\text{}|a-b|=multiple\text{\hspace{0.17em}}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ and\text{\hspace{0.17em}}|b-a|=|-\left(a-b\right)|\\ \text{}\text{}\text{\hspace{0.17em}}=|a-b|=\text{multiple of 4}\\ \text{So,}\text{\hspace{0.17em}}\left(b,a\right)\in R\\ Therefore,\text{\hspace{0.17em}}R\text{is symmetric}\text{.}\\ Let\text{â€‹}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{â€‹}\left(a,b\right),\text{\hspace{0.17em}}\left(b,c\right)\in R\\ ⇒|a-b|=multiple\text{\hspace{0.17em}}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}|c-b|=multiple\text{\hspace{0.17em}}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ ⇒\left(a,b\right)\text{is multiple of 4}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(b,a\right)\text{is multiple of 4}\end{array}$

( ac )=( ab )+( bc )is multiple of 4

$\begin{array}{l}⇒|\left(a-c\right)|\text{is also multiple of 4}\\ ⇒\left(a,c\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Hence, R is equivalence relation}\text{.}\\ \text{Let x be an element of A such that (x,1)}\in \text{R}\\ \text{Then}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{|x}-\text{1| is a multiple of 4}\\ ⇒\text{\hspace{0.17em}}\text{|x}-\text{1|}=0,4,8,12\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=1,5,9\text{}\text{}\left[13\text{is not a part of set}\text{\hspace{0.17em}}\text{A}\right]\\ \text{Hence the set of all elements of A which are related to 1}\\ \text{is {1,5,9}}.\\ \left(ii\right)\text{R = {(a, b): a = b}}\\ \text{For any element a}\in \text{A, we have (a, a)}\in \text{R, since a = a}\text{.}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now, let (a, b)}\in \text{R}\text{.}\\ ⇒\text{a}=\text{b}\\ ⇒\text{b}=\text{a}\\ \left(b,a\right)\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Let (a, b)}\in \text{R and (b, c)}\in \text{R}\\ ⇒\text{a = b and b = c}\\ ⇒\text{a = c}\\ ⇒\text{(a, c)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\end{array}$ $\begin{array}{l}\text{The elements in R that are related to 1 will be elements}\\ \text{from set A which are equal to 1}\text{.}\\ \text{Hence, the set of elements related to 1 is {1}}\text{.}\end{array}$

Q.10 Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Ans.

$\begin{array}{l}\text{Let O denote the origin in the given plane}\text{. Then}\\ \text{R = {(P, Q): OP=OQ}}\\ \text{We can observe that for any point P in set A we have}\\ \text{OP=OP}\\ ⇒\text{(P,P)}\in \text{R}\end{array}$

Thus (P,P)R for all PA

$\begin{array}{l}\text{So R is reflexive}\\ \text{Now,}\\ \text{Let (P, Q)}\in \text{R}\text{.}\\ ⇒\text{OP=OQ}\\ ⇒\text{OQ=OP}\\ ⇒\left(Q,\text{\hspace{0.17em}}\text{\hspace{0.17em}}P\right)\in \text{R}\\ \therefore R\text{is symmetric}\text{.}\\ \text{Again, let}\left(P,Q\right),\left(Q,S\right)\in \text{R}\\ ⇒\text{OP=OQ and OQ=OS}\\ ⇒\text{OP=OS}\\ ⇒\left(P,S\right)\in R\\ \therefore R\text{is transitive}\text{.}\\ \text{Thus, R is an equivalence relation}\text{.}\end{array}$ $\begin{array}{l}Again,\text{Let P be a fixed point in set A and Q be a point in set A}\\ \text{such that (P,Q)}\in \text{R}\text{. Then}\\ ⇒\text{OP=OQ}\\ ⇒\text{Q moves in the plane in such a way that its distance from}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{the origin (0,0) is always}\text{\hspace{0.17em}}\text{equal and is equal to OP}\\ ⇒\text{Locus of Q is a circle with centre at the origin and radius OP}\text{.}\end{array}$

Q.11 Show that the relation R defined in the set A of all triangles as R = {(T1, T2): T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1, T2 and T3 are related?

Ans.

$\begin{array}{l}Since,\text{every triangle is similar to itself}\text{.}\\ {\text{R = {(T}}_{\text{1}}{\text{, T}}_{\text{2}}{\text{): T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{}}\\ \text{(T,T)}\in \text{R for all T}\in \text{A}\\ \therefore \text{R is Reflexive}.\\ {\text{Let (T}}_{\text{1}}{\text{, T}}_{\text{2}}\text{)}\in \text{R,}\\ ⇒{\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{.}\\ ⇒{\text{T}}_{\text{2}}{\text{is similar to T}}_{\text{1}}\text{.}\\ ⇒{\text{(T}}_{\text{2}}{\text{, T}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Now,}\\ {\text{(T}}_{\text{1}}{\text{,T}}_{\text{2}}{\text{,T}}_{\text{3}}\text{)}\in {\text{A such that (T}}_{\text{1}}{\text{,T}}_{\text{2}}\text{)}\in {\text{R and (T}}_{\text{2}}{\text{,T}}_{\text{3}}\text{)}\in \text{R}\\ ⇒{\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{2}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{T}}_{\text{2}}{\text{is similar to T}}_{\text{3}}\text{.}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{T}}_{\text{1}}{\text{is similar to T}}_{\text{3}}\text{.}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{(T}}_{\text{1}}{\text{, T}}_{\text{3}}\text{)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Thus, R is an equivalence relation}\text{.}\\ {\text{In Triangles T}}_{\text{1}}\text{\hspace{0.17em}}{\text{and T}}_{\text{3}}\text{we observe that the corresponding angles}\\ \text{are equal and the}\text{\hspace{0.17em}}\text{corresponding sides are proportional}\\ \text{i}\text{.e}\text{.}\text{\hspace{0.17em}}\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}\\ Hence,\text{\hspace{0.17em}}{\text{T}}_{\text{1}}\text{is related to}\text{\hspace{0.17em}}{\text{T}}_{\text{3}}\text{.}\end{array}$

Q.12 Give an example of a relation. Which is

(i) Symmetric but neither reflexive nor transitive.

(ii) Transitive but neither reflexive nor symmetric.

(iii) Reflexive and symmetric but not transitive.

(iv) Reflexive and transitive but not symmetric.

(v) Symmetric and transitive but not reflexive.

Ans.

$\begin{array}{l}\text{(i) Let A = {1,2,3}}\text{.}\\ \text{Define a relation R on A as R = {(1, 2), (2, 1)}}\text{.}\\ \text{Relation R is not reflexive as (1,1), (2, 2), (3, 3)}\notin \text{R}\text{.}\\ \text{Now, as (1, 2)}\in \text{R and also (2, 1)}\in \text{R, R is symmetric}\text{.}\\ ⇒\text{(1, 2), (2, 1)}\in \text{R, but (1, 1)}\notin \text{R}\\ \therefore \text{R is not transitive}\text{.}\\ \text{Hence, relation R is symmetric but neither reflexive or transitive}\text{.}\\ \text{(ii) Consider a relation R in R defined as:}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{R = {(x, y): x < y}}\\ \text{Since x cannot be less than x}.\\ \therefore \text{R is not reflexive}\text{.}\\ Now,\text{\hspace{0.17em}}\text{}\text{(1, 2)}\in \text{R}\text{}\text{}\text{(as1 < 2)}\\ \text{But, 2 is not less than 1}\text{.}\\ \therefore \text{(2, 1)}\notin \text{R}\\ \therefore \text{R is not symmetric}\text{.}\\ \\ \text{Again, let (a, b), (b, c)}\in \text{R}\text{.}\\ ⇒\text{a < b and b < c}\\ ⇒\text{a < c}\\ ⇒\left(a,c\right)\in R\\ \therefore \text{\hspace{0.17em}}\text{}R\text{is transitive}\text{.}\\ \text{Hence, relation R is transitive but neither reflexive nor symmetric}\text{.}\\ \left(iii\right)\text{\hspace{0.17em}}\text{Let A = {2, 4, 6}}\text{.}\\ \text{Define a relation R on A as:}\\ \text{A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)}}\\ \text{Relation R is reflexive as (2,2), (4,4), (6,6)}}\in \text{R}\text{.}\\ \text{Relation R is symmetric as (2,4), (4,2), (4,6), (6,4)}\in \text{R}\text{.}\\ \text{Relation R is not transitive since (2,4), (4,6)}\in \text{R, but (2,6)}\notin \text{R}\text{.}\\ \text{Hence, relation R is reflexive and symmetric but not transitive}\text{.}\\ \left(iv\right)\text{A = {1, 2, 3, 4, 5, 6}}\\ \text{R = {(x, y): y is divisible by x}}\\ \text{Since every number is divisible by itself}\\ \therefore \text{(x, x)}\in \text{R}\\ \text{Hence R is reflexive}\text{.}\\ \text{Now,}\\ \text{(2, 4)}\in \text{R}\text{}\text{[as 4 is divisible by 2]}\\ \text{But,}\\ \text{(4, 2)}\notin \text{R}\text{.}\text{}\text{[as 2 is not divisible by 4]}\\ \therefore \text{R is not symmetric}\text{.}\\ \text{Let (x, y), (y, z)}\in \text{R}\text{.}\\ \\ \text{Then, y is divisible by x and z is divisible by y}\text{.}\\ \therefore \text{z is divisible by x}\text{.}\\ \text{Hence (x, z)}\in \text{R}\text{.}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is reflexive and transitive but not symmetric}\text{.}\\ \text{(v) Let A = {1, 2}}\text{.}\\ \text{Define a relation R on A as:}\\ \text{R = {(1,2), (2,1), (1,1)}}\\ \text{Relation R is not reflexive as (2,2)}\notin \text{R}\text{.}\\ \text{Relation R is symmetric as (1,2)}\in \text{R and (2,1)}\in \text{R}\text{.}\\ \text{It is seen that (1,2), (2,1)}\in \text{R}\text{. Also, (1,1)}\in \text{R}\text{.}\\ \therefore \text{The relation R is transitive}\text{.}\\ \text{Hence, relation R is symmetric and transitive but not reflexive}\text{.}\end{array}$

Q.13 Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Ans.

$\begin{array}{l}Since,\\ {\text{R = {(P}}_{\text{1}}{\text{, P}}_{\text{2}}{\text{): P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have same the number of sides}}\\ {\text{Since (P}}_{\text{1}}{\text{, P}}_{\text{1}}\text{)}\in \text{R polygon with same number of sides belongs}\\ \text{to itself}\text{.}\\ \therefore \text{R is Reflexive}\text{.}\\ {\text{Let (P}}_{\text{1}}{\text{, P}}_{\text{2}}\text{)}\in \text{R}\text{.}\\ ⇒{\text{P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have the same number of sides}\text{.}\\ ⇒{\text{P}}_{\text{2}}{\text{and P}}_{\text{1}}\text{have the same number of sides}\text{.}\\ ⇒{\text{(P}}_{\text{2}}{\text{, P}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ \text{Again, let}\left({P}_{1},{P}_{2}\right),\left({P}_{2},{P}_{3}\right)\in \text{R}\\ ⇒{\text{P}}_{\text{1}}{\text{and P}}_{\text{2}}\text{have the same number of sides}{\text{. Also, P}}_{\text{2}}{\text{and P}}_{\text{3}}\text{have}\\ \text{the same number of sides}\text{.}\\ ⇒{\text{P}}_{\text{1}}{\text{and P}}_{\text{3}}\text{have the same number of sides}\text{.}\\ ⇒{\text{(P}}_{\text{1}}{\text{, P}}_{\text{3}}\text{)}\in \text{R}\\ \therefore \text{R is transitive}\text{.}\\ \text{Therefore, R is an equivalence relation}\text{.}\\ \text{Hence, the set of all elements in A related to triangle T is the}\\ \text{set of all triangles}\text{.}\end{array}$

Q.14 Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is
an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Ans.

$\begin{array}{l}Since,\\ {\text{R = {(L}}_{\text{1}}{\text{, L}}_{\text{2}}{\text{): L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}\text{}}\\ {\text{(L}}_{\text{1}}{\text{,L}}_{\text{1}}\text{)}\in \text{R as every line is parallel to itself}\\ \therefore \text{R is reflexive}\text{.}\\ \text{Now,}\\ {\text{Let (L}}_{\text{1}}{\text{, L}}_{\text{2}}\text{)}\in \text{R}\text{.}\\ ⇒{\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}\text{.}\\ ⇒{\text{L}}_{\text{2}}{\text{is parallel to L}}_{\text{1}}\text{.}\\ ⇒{\text{(L}}_{\text{2}}{\text{, L}}_{\text{1}}\text{)}\in \text{R}\\ \therefore \text{R is symmetric}\text{.}\\ {\text{Let (L}}_{\text{1}}{\text{, L}}_{\text{2}}{\text{), (L}}_{\text{2}}{\text{, L}}_{\text{3}}\text{)}\in \text{R}\text{.}\\ ⇒{\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{2}}{\text{. And L}}_{\text{2}}{\text{is parallel to L}}_{\text{3}}\text{.}\\ ⇒{\text{L}}_{\text{1}}{\text{is parallel to L}}_{\text{3}}\text{.}\\ \therefore \text{R is transitive}\text{.}\\ \text{Hence, R is an equivalence relation}\text{.}\\ \text{The set of all lines related to the line y = 2x + 4 must be parallel}\\ \text{line only and we know that in parallel lines’ equations only the}\\ \text{constant value changes the co-efficient of x and y remains same}\\ \therefore \text{set of parallel lines is y=2x+}\text{\hspace{0.17em}}\text{c, where c can be any constant}\text{.}\end{array}$

Q.15

$\begin{array}{l}\mathrm{Let}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{relation}\mathrm{in}\mathrm{the}\mathrm{set}\left\{1,2,3,4\right\}\mathrm{given}\mathrm{by}\\ \mathrm{R}=\left\{\left(1,2\right),\left(2,2\right),\left(1,1\right),\left(4,4\right),\left(1,3\right),\left(3,3\right),\left(3,2\right)\right\}.\mathrm{}\end{array}$

Choose the correct answer.

$\begin{array}{l}\left(\mathrm{A}\right)\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{symmetric}\mathrm{but}\mathrm{not}\mathrm{transitive}.\\ \left(\mathrm{B}\right)\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{transitive}\mathrm{but}\mathrm{not}\mathrm{symmetric}.\\ \left(\mathrm{C}\right)\mathrm{R}\mathrm{is}\mathrm{symmetric}\mathrm{and}\mathrm{transitive}\mathrm{but}\mathrm{not}\mathrm{reflexive}.\\ \left(\mathrm{D}\right)\mathrm{R}\mathrm{is}\mathrm{an}\mathrm{equivalence}\mathrm{relation}.\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{B}\right)\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \mathrm{ }\mathrm{Since}\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\right\}\in \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}\\ \mathrm{ }\mathrm{Now}\left(1,2\right)\in \mathrm{R}\mathrm{but}\left(2,1\right)\notin \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\\ \mathrm{And}\\ \left(1,2\right)\mathrm{and}\left(2,2\right)\in \mathrm{R}⇒\left(1,2\right)\in \mathrm{R}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{transitive}\\ \mathrm{Hence}\mathrm{the}\mathrm{given}\mathrm{Relation}\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{and}\mathrm{transitive}\\ \mathrm{but}\mathrm{not}\mathrm{symmetric}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Let}\mathrm{R}\mathrm{be}\mathrm{the}\mathrm{relationin}\mathrm{theset} \mathrm{N} \mathrm{given}\mathrm{by}\\ \mathrm{R}=\left\{\left(\mathrm{a},\mathrm{b}\right):\mathrm{a}=\mathrm{b}-2, \mathrm{b}>6\right\}.\\ \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}.\\ \left(\mathrm{A}\right)\left(2,4\right)\in \mathrm{R} \left(\mathrm{B}\right)\left(3,8\right)\in \mathrm{R} \left(\mathrm{C}\right)\left(6,8\right)\in \mathrm{R} \mathrm{ }\left(\mathrm{D}\right)\left(8,7\right)\in \mathrm{R}\end{array}$

Ans.

$\begin{array}{l}\mathrm{R}= \left\{\left(\mathrm{a},\mathrm{b}\right):\mathrm{a}=\mathrm{b}-2,\mathrm{b}> 6\right\}\\ \mathrm{Now},\mathrm{since}\mathrm{b}> 6, \left(2, 4\right)\in \mathrm{R}\\ \therefore \mathrm{Option}\left(\mathrm{A}\right)\mathrm{cannot}\mathrm{be}\mathrm{the}\mathrm{correct}\mathrm{answer}\\ \mathrm{Also}\mathrm{by}\mathrm{putting}8\mathrm{we}\mathrm{will}\mathrm{not}\mathrm{get}3\mathrm{as}\mathrm{difference}\end{array}$ $\begin{array}{l}\therefore \mathrm{Option}\left(\mathrm{B}\right)\mathrm{is}\mathrm{also}\mathrm{not}\mathrm{the}\mathrm{correct}\mathrm{option}\\ \mathrm{And},\mathrm{as}8\ne \mathrm{7}-2\\ \therefore \left(8, 7\right)\notin \mathrm{R}\\ \mathrm{Now},\mathrm{consider}\left(6, 8\right)\mathrm{.}\\ \mathrm{We}\mathrm{have}8 > 6\mathrm{and}\mathrm{also}, 6 = 8-2.\\ \therefore \left(6, 8\right)\in \mathrm{R}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option} \left(\mathrm{C}\right)\mathrm{.}\end{array}$

Q.17

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{f}:\mathrm{R}{}_{*}\to {\mathrm{R}}_{*}\mathrm{definedby}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}} \mathrm{is}\\ \mathrm{one}–\mathrm{one}\mathrm{ }\mathrm{and}\mathrm{onto},\mathrm{ }\mathrm{where}{\mathrm{R}}_{*}\mathrm{ }\mathrm{is}\mathrm{these}\mathrm{t}\mathrm{of}\mathrm{all}\mathrm{non}–\mathrm{zero}\\ \mathrm{realnumbers}.\mathrm{Is}\mathrm{there}\mathrm{sulttrue},\mathrm{if}\mathrm{the}\mathrm{domain}\mathrm{R}{}_{*}\mathrm{is} \\ \mathrm{replaced}\mathrm{by}\mathrm{Nwithco}–\mathrm{domain}\mathrm{being}\mathrm{same}\mathrm{as}{\mathrm{R}}_{*}?\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{x},\mathrm{y}\in {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{such}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \mathrm{one}-\mathrm{one}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \mathrm{So}, \mathrm{f}:{\mathrm{R}}_{\mathrm{*}}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Onto}:\end{array}$

Letf( x )=ywhere y be any element of R * ( Codomain )

$\begin{array}{l}⇒ \frac{1}{\mathrm{x}}=\mathrm{y} \mathrm{or} \mathrm{x}=\frac{1}{\mathrm{y}}\\ \mathrm{Now},\mathrm{ }\mathrm{f}\left(\frac{1}{\mathrm{y}}\right)=\frac{1}{\left(\frac{1}{\mathrm{y}}\right)}=\mathrm{y}\\ \mathrm{Since}\mathrm{Range}=\mathrm{Co}–\mathrm{domain}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{Consider}\mathrm{function}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\\ \mathrm{For}\mathrm{any}\mathrm{x},\mathrm{y}\in \mathrm{N}\mathrm{we}\mathrm{see}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \mathrm{ }\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \mathrm{So}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{function}\\ \mathrm{Since}\mathrm{fractional}\mathrm{numbers}\mathrm{like}\frac{2}{3},\frac{2}{5}\mathrm{ }\mathrm{etc}.\mathrm{ }\mathrm{in}\mathrm{co}–\mathrm{domain}{\mathrm{R}}_{*}\mathrm{ }\mathrm{have}\\ {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{have}\mathrm{no}\mathrm{pre}\mathrm{image}\mathrm{in}\mathrm{ }\mathrm{domain}\mathrm{N}.\\ \mathrm{So}, \mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{*}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\end{array}$

Q.18(i) f: N N given by f(x)= x2

$\begin{array}{l}\text{\hspace{0.17em}}\left(ii\right)f:Z\to Z\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}givenbyf\left(x\right)={x}^{2}\\ \left(iii\right)f:R\to R\text{\hspace{0.17em}}\text{\hspace{0.17em}}givenbyf\left(x\right)={x}^{2}\\ \left(iv\right)f:N\to N\text{\hspace{0.17em}}\text{\hspace{0.17em}}givenbyf\left(x\right)={x}^{3}\\ \text{\hspace{0.17em}}\left(v\right)f:Z\to Zgivenbyf\left(x\right)={x}^{3}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f:N}\to \text{N is given by,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f}\left(x\right)={x}^{2}\\ \text{To check for injectivity let f}\left(x\right)=\text{f}\left(y\right)\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}={y}^{2}\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\text{}\left(\begin{array}{l}There\text{are no negative}\\ \text{natural numbers}\text{.}\end{array}\right)\\ \therefore f\left(x\right)\text{is injective}\text{.}\\ \text{To check for onto}:\\ \text{Let}\text{\hspace{0.17em}}\text{f}\left(x\right)\text{=y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}\text{=y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x=}\sqrt{y}\\ \therefore \text{f}\left(\sqrt{y}\right)\text{=y}\\ \text{Since range is not equal to co-domain}\text{.}\\ \text{f(x) is not onto or surjective}\\ \text{(ii) f: Z}\to \text{Z is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f(x)}={\text{x}}^{\text{2}}\\ \text{For injectivity let}\text{\hspace{0.17em}}\text{f(x)}=\text{f(y)}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}={y}^{2}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±y\\ \\ \therefore \text{f is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{(x)}=\text{y}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{x}}^{\text{2}}=\text{y}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=\sqrt{\text{y}}\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\\ \text{Hence, function f is neither injective nor surjective}\text{.}\\ \\ \left(iii\right)\text{\hspace{0.17em}}f:R\to R\text{\hspace{0.17em}}\text{is given by, f}\left(x\right)={x}^{2}\\ For\text{injectivity:}\\ \text{}\text{}\text{}\text{}\text{f}\left(x\right)=f\left(y\right)\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}={y}^{2}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±y\text{}\text{}\left(\begin{array}{l}x\text{can not take more than one}\\ \text{value for injectivity}\text{.}\end{array}\right)\\ \therefore f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{f}\left(x\right)=y⇒{x}^{2}=y\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{y}\\ \therefore \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{\hspace{0.17em}}\text{is not surjective}\text{.}\\ \left(iv\right)\text{\hspace{0.17em}}f:N\to N\text{\hspace{0.17em}}\text{is given by, f}\left(x\right)={x}^{3}\\ For\text{injectivity:}\\ \text{}\text{}\text{}\text{}\text{f}\left(x\right)=f\left(y\right)\\ \\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{3}={y}^{3}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\text{}\text{}\\ \therefore f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{f}\left(x\right)=y⇒{x}^{3}=y\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{y}\\ \therefore \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\text{}\text{}\text{}\left(y\in N\text{\hspace{0.17em}}\text{\hspace{0.17em}}but\text{\hspace{0.17em}}\sqrt{y}\notin N\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{\hspace{0.17em}}\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\\ \\ \left(v\right)\text{\hspace{0.17em}}f:Z\to Z\text{\hspace{0.17em}}\text{is given by, f}\left(x\right)={x}^{3}\\ For\text{injectivity:}\\ \text{}\text{}\text{}\text{}\text{f}\left(x\right)=f\left(y\right)\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{3}={y}^{3}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\\ \therefore f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{f}\left(x\right)=y⇒{x}^{3}=y\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{y}\\ \therefore \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\text{}\text{}\text{}\left(y\in Z\text{\hspace{0.17em}}\text{\hspace{0.17em}}but\text{\hspace{0.17em}}\sqrt{y}\notin Z\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{\hspace{0.17em}}\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\end{array}$

Q.19 Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Ans.

$\begin{array}{l}Since,\\ f:R\to R\text{is given by,}\\ \text{f}\left(x\right)=\left[x\right]\\ We\text{â€‹}\text{see that f}\left(0.1\right)=0,\text{f}\left(0.3\right)=0.\\ \therefore \text{f}\left(0.1\right)=\text{f}\left(0.3\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}but\text{0}\text{.1}\ne \text{0}\text{.3}\\ \text{So, f is not injective i}\text{.e}\text{., one-one}\text{.}\\ \text{Now, let 0}\text{.6}\in \text{R}\text{.}\\ \text{It is given that f}\left(\text{x}\right)\text{=}\left[x\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is always an integer}\text{. Thus, there does}\\ \text{not exist any element x}\in \text{R such that f}\left(\text{x}\right)\text{=0}\text{.6}\text{.}\\ \text{Then, f is not onto}\text{.}\\ \text{Therefore, the greatest integer function is neither one-one}\\ \text{nor onto}\text{.}\end{array}$

Q.20

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{Modul}\mathrm{us}\mathrm{Function}f:\mathrm{R}\to \mathrm{R} \mathrm{given}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=|\mathrm{x}|,\\ \mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{onen}\mathrm{or}\mathrm{onto},\mathrm{where}|\mathrm{x}|\mathrm{is} \mathrm{x},\mathrm{if}\mathrm{ }\mathrm{x} \mathrm{is}\mathrm{positive}\\ \mathrm{or}0\mathrm{and}|\mathrm{x}|\mathrm{is}–\mathrm{x},\mathrm{if}\mathrm{x} \mathrm{is}\mathrm{negative}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}\ge \mathrm{0}\\ -\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}\left(-1\right)=|-1|=1,\mathrm{ }\mathrm{f}\left(1\right)=|1|=1,\mathrm{}\\ \therefore \mathrm{f}\left(-1\right)=\mathrm{f}\left(1\right), \mathrm{ }\mathrm{but}-\mathrm{1}\ne \mathrm{1}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Let}\mathrm{}-\mathrm{1}\in \mathrm{R}\mathrm{.}\\ \mathrm{Since},\mathrm{f}\left(\mathrm{x}\right)\mathrm{=}|\mathrm{x}|\mathrm{is}\mathrm{always}\mathrm{positive}.\mathrm{So},\mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\\ \mathrm{any}\mathrm{element}\mathrm{in}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=-1.\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{â€‹}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{modulus}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.21

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{Signum}\mathrm{Function}\mathrm{f}:\mathrm{R}\to \mathrm{R},\mathrm{given}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l} \mathrm{ }1,\mathrm{ }\mathrm{if}\mathrm{x}>0\\ 0,\mathrm{ }\mathrm{if}\mathrm{x}=0\\ –1,\mathrm{ }\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{noronto}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}1,\mathrm{}\mathrm{if}\mathrm{x}>0\\ 0,\mathrm{}\mathrm{if}\mathrm{x}=0\\ -1,\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}\left(1\right)=1,\mathrm{ }\mathrm{f}\left(3\right)=1,\mathrm{but}1\ne \mathrm{3}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \\ \mathrm{Since},\mathrm{range}\mathrm{of}\mathrm{f}\mathrm{is}\left(1,0,-1\right)\mathrm{for}\mathrm{element}-2\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\\ \mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{value}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=-2.\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{signum}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.22 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Ans.

$\begin{array}{l}\text{It is given that}\\ \text{A}=\text{}\left\{\text{1},\text{2},\text{3}\right\},\text{B}=\left\{\text{4},\text{5},\text{6},\text{7}\right\}\text{}\text{\hspace{0.17em}}\\ \text{f}:\text{A}\to \text{B is defined as}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f = {(1, 4), (2, 5), (3, 6)}}\text{.}\\ \therefore \text{f (1) = 4, f (2) = 5, f (3) = 6}\\ \text{Since every element has a unique value}\text{.}\\ \text{The given function f is one-one}\text{.}\end{array}$

Q.23

$\begin{array}{l}Ineachofthefollowingcases,statewhetherthefunctionis\\ one–one,ontoorbijective.Justifyyouranswer.\\ \left(i\right)f:R\to Rdefinedbyf\left(x\right)=3–4x\\ \left(ii\right)f:R\to Rdefinedbyf\left(x\right)=1+{x}^{2}\end{array}$

Ans.

$\begin{array}{l}\text{(i) f: R}\to \text{R is defined as f(x) = 3}-\text{4x}\text{.}\\ \text{For one-one}\\ \text{Let}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f(x)}=\text{f(y)}\\ ⇒\text{3}-\text{4x}=\text{3}-\text{4y}\\ \\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=\text{y}\\ \therefore \text{f is one-one}\text{.}\\ \text{for onto}\\ \text{Let f(x)=y}\\ ⇒\text{3}-\text{4x=y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x=}\frac{\text{3}-y}{4}\\ ⇒\text{f}\left(\frac{\text{3}-y}{4}\right)\text{=3}-\text{4}\left(\frac{\text{3}-y}{4}\right)\text{=y}\\ \text{Since Range = Co-domain}\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is bijective}\text{.}\\ \\ \text{(ii) f: R}\to {\text{R is defined as f(x)=1+x}}^{\text{2}}\\ \text{For one-one}\\ \text{Let f(x)=f(y )}\\ ⇒{\text{1+x}}^{\text{2}}={\text{1+y}}^{\text{2}}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{x}}^{\text{2}}={\text{y}}^{\text{2}}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=±\text{y}\\ \therefore \text{f is not one-one}\text{.}\\ \text{For onto:}\\ \text{Let}\text{\hspace{0.17em}}\text{f}\left(x\right)=y\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{1+x}}^{\text{2}}=\text{y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=±\sqrt{y-1}\\ \text{Since Range is not equal to Co-domain}\text{.}\\ \text{Then given function f(x) is not onto}\text{.}\end{array}$

Q.24

$\begin{array}{l}\mathrm{Let}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{be}\mathrm{sets}.\mathrm{Show}\mathrm{that}\mathrm{f}:\mathrm{A}×\mathrm{B}\to \mathrm{B}×\mathrm{A} \mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\mathrm{f}\left(\mathrm{b},\mathrm{a}\right) \mathrm{is}\mathrm{bijective}\mathrm{function}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{A}×\mathrm{B}\to \mathrm{B}×\mathrm{A}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{a},\mathrm{b}\right) = \left(\mathrm{b},\mathrm{a}\right). \\ \mathrm{Let}\mathrm{}\left({\mathrm{a}}_{1}, {\mathrm{b}}_{1}\right),\left({\mathrm{a}}_{2}, {\mathrm{b}}_{2}\right)\in \mathrm{A}×\mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}\left({\mathrm{a}}_{1}, {\mathrm{b}}_{1}\right)=\mathrm{f}\left({\mathrm{a}}_{2}, {\mathrm{b}}_{2}\right)\\ ⇒\left({\mathrm{b}}_{1}, {\mathrm{a}}_{1}\right)=\left({\mathrm{b}}_{2}, {\mathrm{a}}_{2}\right)\\ ⇒{\mathrm{b}}_{1}={\mathrm{b}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{a}}_{1}={\mathrm{a}}_{2}\\ ⇒\left({\mathrm{a}}_{1},{\mathrm{b}}_{1}\right)=\left({\mathrm{a}}_{2},{\mathrm{b}}_{2}\right)\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\left(\mathrm{b},\mathrm{a}\right)\in \mathrm{B}×\mathrm{A}\mathrm{be}\mathrm{any}\mathrm{element}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\left(\mathrm{a},\mathrm{b}\right)\in \mathrm{A}×\mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{a},\mathrm{b}\right) = \left(\mathrm{b},\mathrm{a}\right).\\ \left[\mathrm{By}\mathrm{definition}\mathrm{of}\mathrm{f}\right]\\ \therefore \mathrm{ }\mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{bijective}\mathrm{.}\end{array}$

Q.25

$\begin{array}{l}Letf:N\to Nbedefinedby\\ f\left(n\right)=\left\{\begin{array}{l}\frac{n+1}{2},\text{}ifnisodd\\ \frac{n}{2},\text{}\text{}ifniseven\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}foralln\in N.\\ Statewhetherthefunctionfisbijective.\\ Justifyyouranswer.\end{array}$

Ans.

$\begin{array}{l}Letf:N\to Nisdefinedas\\ \\ f\left(n\right)=\left\{\begin{array}{l}\frac{n+1}{2},\text{}ifnisodd\\ \frac{n}{2},\text{}\text{}ifniseven\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}foralln\in N.\\ \text{It can be observed that:}\\ f\left(1\right)=\frac{1+1}{2}=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(2\right)=\frac{2}{2}=1\text{}\text{}\left[By\text{definition of f}\right]\\ \therefore f\left(1\right)=f\left(2\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}where\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\ne 2.\\ \therefore \text{f is not one-one}\text{.}\\ \text{Consider a natural number (n) in co-domain N}\text{.}\\ \text{Case I: n is odd}\\ \therefore \text{n = 2m + 1 for some m}\in \text{N}\text{. Then,there exists 4m + 1}\in \text{N}\\ \text{such that}\\ \text{f}\left(4m+1\right)=\frac{4m+1+1}{2}=2m+1\\ Case\text{\hspace{0.17em}}II:\text{n is even}\\ \therefore \text{\hspace{0.17em}}\text{n=2}m\text{\hspace{0.17em}}\text{for some}m\in \text{N}\text{. Then,there exists 4}m\in \text{N such that}\\ \text{f}\left(4m\right)=\frac{4m}{2}=2m\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is not a bijective function}\text{.}\end{array}$

Q.26

$\begin{array}{l}LetA=R-\left\{3\right\}andB=R-\left\{1\right\}.Considerthefunction\\ f:A\to Bdefinedby\text{\hspace{0.17em}}f\left(x\right)=\frac{x–2}{x–3}.Isfone-oneandonto?\\ Justifyyouranswer.\end{array}$

Ans.

$\begin{array}{l}A=R-\left\{3\right\}andB=R-\left\{1\right\}\\ \text{f: A}\to \text{B is defined as f}\left(x\right)=\frac{x-2}{x-3}\\ Let\text{x,y}\in \text{A such that}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f}\left(x\right)=f\left(y\right).\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{x-2}{x-3}=\frac{y-2}{y-3}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x-2\right)\left(y-3\right)=\left(x-3\right)\left(y-2\right)\\ ⇒\overline{)xy}-3x-2y+\overline{)6}=\overline{)xy}-2x-3y+\overline{)6}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3x-2y=-2x-3y\\ ⇒\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\\ \therefore \text{\hspace{0.17em}}f\text{if one-one}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}\in \text{B}=\text{R-}\left\{1\right\}.\text{\hspace{0.17em}}Then,\text{\hspace{0.17em}}y\ne 1\\ \text{The function f is onto if there exists x}\in \text{A such that}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f(x) = y}\\ ⇒\text{}\text{}\frac{x-2}{x-3}=y\\ ⇒\text{}\text{}x-2=y\left(x-3\right)\\ ⇒\text{}\text{}x-2=yx-3y\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-yx=2-3y\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\left(1-y\right)=2-3y\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{2-3y}{\left(1-y\right)}\in A\text{}\text{}\left[y\ne 1\right]\\ \text{Thus, for any y}\in \text{B, there exists}\text{\hspace{0.17em}}\frac{2-3y}{\left(1-y\right)}\in A\text{\hspace{0.17em}}such\text{\hspace{0.17em}}\text{\hspace{0.17em}}that\\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\frac{2-3y}{1-y}\right)=\frac{\frac{2-3y}{\left(1-y\right)}-2}{\frac{2-3y}{\left(1-y\right)}-3}\\ \text{}\text{}\text{}=\frac{\frac{2-3y-2\left(1-y\right)}{1-y}}{\frac{2-3y-3\left(1-y\right)}{1-y}}\\ \text{}\text{}\text{}=\frac{2-3y-2+2y}{2-3y-3+3y}\\ \text{}\text{}\text{}=\frac{-y}{-1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\frac{2-3y}{1-y}\right)=y\\ \therefore \text{\hspace{0.17em}}f\text{is onto}\text{.}\\ \text{Hence, function f is one-one and onto}\text{.}\end{array}$

Q.27

$\begin{array}{l}\mathrm{Let}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{4}.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{onto} \mathrm{ }\left(\mathrm{B}\right)\mathrm{fismany}–\mathrm{one}\mathrm{onto}.\\ \left(\mathrm{C}\right)\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto} \left(\mathrm{D}\right)\mathrm{f}\mathrm{is}\mathrm{n}\mathrm{either}\mathrm{one}–\mathrm{one}\\ \mathrm{noronto}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{4}.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right).\\ \mathrm{Then}, \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \mathrm{ }{\mathrm{x}}^{4}={\mathrm{y}}^{4}\\ ⇒ \mathrm{ }\mathrm{x}=±\mathrm{y}\\ \therefore \mathrm{ }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{ }\mathrm{but}\mathrm{ }\mathrm{x}\ne \mathrm{y}\\ \mathrm{So},\mathrm{ }\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Consider}\mathrm{an}\mathrm{element}3\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R}.\mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{there}\\ \mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right) = 3\mathrm{.}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.28

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{asf}\left(\mathrm{x}\right)=3\mathrm{x}.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f} \mathrm{isone}–\mathrm{one}\mathrm{onto} \mathrm{ }\left(\mathrm{B}\right)\mathrm{f} \mathrm{is}\mathrm{many}–\mathrm{one}\mathrm{on}\mathrm{to}\\ \left(\mathrm{C}\right)\mathrm{f} \mathrm{isone}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto}\left(\mathrm{D}\right)\mathrm{f}\mathrm{ }\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{Rbedefinedasf}\left(\mathrm{x}\right)=3\mathrm{x}.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right) =\mathrm{f}\left(\mathrm{y}\right)\mathrm{.}\\ ⇒3\mathrm{x}=3\mathrm{y}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{for}\mathrm{any}\mathrm{real}\mathrm{number}\left(\mathrm{y}\right)\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\mathrm{there}\mathrm{exists}\\ \frac{\mathrm{y}}{3} \mathrm{in}\mathrm{R}\mathrm{such}\mathrm{that} \mathrm{f}\left(\frac{\mathrm{y}}{3}\right)=3\left(\frac{\mathrm{y}}{3}\right)=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \\ \mathrm{Therefore},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.29

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\left\{1,3,4\right\}\to \left\{1,2,5\right\}\mathrm{and}\mathrm{g}:\left\{1,2,5\right\}\to \left\{1,3\right\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\left(1,2\right),\left(3,5\right),\left(4,1\right)\right\}\mathrm{and}\mathrm{g}=\left\{\left(1,3\right),\left(2,3\right),\left(5,1\right)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\left\{1,3,4\right\}\to \left\{1,2,5\right\}\mathrm{and}\mathrm{g}:\left\{1,2,5\right\}\to \left\{1,3\right\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\left(1,2\right),\left(3,5\right),\left(4,1\right)\right\}\mathrm{and}\mathrm{g}=\left\{\left(1,3\right),\left(2,3\right),\left(5,1\right)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Q.30

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\left\{1,3,4\right\}\to \left\{1,2,5\right\}\mathrm{and}\mathrm{g}:\left\{1,2,5\right\}\to \left\{1,3\right\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\left(1,2\right),\left(3,5\right),\left(4,1\right)\right\}\mathrm{and}\mathrm{g}=\left\{\left(1,3\right),\left(2,3\right),\left(5,1\right)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Ans.

$\begin{array}{l}\text{Let}\left(\left(\text{f}+\text{g}\right)\text{oh}\right)\left(\text{x}\right)=\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{f}\left\{\text{h}\left(\text{x}\right)\right\}+\text{g}\left\{\text{h}\left(\text{x}\right)\right\}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ \\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left\{\left(\text{f}+\text{g}\right)\text{oh}\right\}\left(\text{x}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left\{\left(\text{foh}\right)+\left(\text{goh}\right)\right\}\left(\text{x}\right)\text{}\forall x\in R\\ Hence,\text{}\left(f+g\right)oh=\left(\text{foh}\right)+\left(\text{goh}\right).\\ Now,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\left(f.g\right)oh\right\}\left(x\right)=\left(f.g\right)\left(h\left(x\right)\right)\\ \text{}\text{}\text{}\text{}=f\left(h\left(x\right)\right).g\left(h\left(x\right)\right)\\ \text{}\text{}\text{}\text{}=\left(foh\right)\left(x\right).\left(goh\right)\left(x\right)\\ \text{}\text{}\text{}\text{}=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\left(f.g\right)oh\right\}\left(x\right)=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\text{}\text{}\forall x\in R\\ Hence,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(f.\text{\hspace{0.17em}}g\right)oh=\left(foh\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left(goh\right)\end{array}$

Q.31

$\begin{array}{l}Findgofandfog,if\\ \left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=|x|\text{\hspace{0.17em}}and\text{\hspace{0.17em}}g\left(x\right)=|5x–2|\\ \left(ii\right)\text{\hspace{0.17em}}f\left(x\right)=8{x}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)={x}^{\frac{1}{3}}\end{array}$

Ans.

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=|x|\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)=|5x-2|\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=g\left(|x|\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|5|x|-2|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=f\left(|5x-2|\right)\\ \\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=||5x-2||\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|5x-2|\\ \left(ii\right)\text{\hspace{0.17em}}f\left(x\right)=8{x}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)={x}^{\frac{1}{3}}\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=g\left(8{x}^{3}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(8{x}^{3}\right)}^{\frac{1}{3}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=f\left({x}^{\frac{1}{3}}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8{\left({x}^{\frac{1}{3}}\right)}^{3}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8x\end{array}$

Q.32

$\begin{array}{l}Iff\left(x\right)=\frac{4x+3}{6x–4},\text{\hspace{0.17em}}x\ne \frac{2}{3},showthatfof\left(x\right)=x,\text{\hspace{0.17em}}forallx\ne \frac{2}{3}.\\ Whatistheinverseoff?\end{array}$

Ans.

$\begin{array}{l}It\text{is given that}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=\frac{4x+3}{6x–4},\text{\hspace{0.17em}}then\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof\left(x\right)=f\left(f\left(x\right)\right)\\ \text{}\text{}\text{}=f\left(\frac{4x+3}{6x–4}\right)\\ \text{}\text{}\text{}=\frac{4\left(\frac{4x+3}{6x–4}\right)+3}{6\left(\frac{4x+3}{6x–4}\right)-4}\\ \text{}\text{}\text{}=\frac{\left(\frac{16x+12+18x-12}{6x-4}\right)}{\left(\frac{24x+18-24x+16}{6x–4}\right)}\\ \text{}\text{}\text{}=\frac{34x}{34}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof\left(x\right)=x,\text{for all x}\ne \frac{2}{3}.\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof=I\\ \text{Hence, the given function f is invertible and the inverse of f}\\ \text{is f itself}\text{.}\end{array}$

Q.33

$\begin{array}{l}\mathrm{State}\mathrm{with}\mathrm{reason}\mathrm{whether}\mathrm{following}\mathrm{functions} \mathrm{have}\mathrm{inverse}\\ \left(\mathrm{i}\right)\mathrm{f}:\left\{1,2,3,4\right\}\to \left\{10\right\}\mathrm{with}\\ \mathrm{ }\mathrm{f}=\left\{\left(1,10\right),\left(2,10\right),\left(3,10\right),\left(4,10\right)\right\}\\ \left(\mathrm{ii}\right) \mathrm{g}:\left\{5,6,7,8\right\}\to \left\{1,2,3,4\right\}\mathrm{with}\\ \mathrm{g}=\left\{\left(5,4\right),\left(6,3\right),\left(7,4\right),\left(8,2\right)\right\}\\ \left(\mathrm{iii}\right)\mathrm{h}:\left\{2,3,4,5\right\}\to \left\{7,9,11,13\right\}\mathrm{with}\\ \mathrm{ }\mathrm{h}=\left\{\left(2,7\right),\left(3,9\right),\left(4,11\right),\left(5,13\right)\right\}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{f}: \left\{1, 2, 3, 4\right\}\to \left\{10\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{ }\mathrm{f}= \left\{\left(1, 10\right), \left(2, 10\right), \left(3, 10\right), \left(4, 10\right)\right\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{f},\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{f}\mathrm{is}\mathrm{a}\mathrm{many}\\ \mathrm{one}\mathrm{function}\mathrm{as}:\mathrm{f}\left(1\right) =\mathrm{f}\left(2\right)=\mathrm{f}\left(3\right) =\mathrm{f}\left(4\right) = 10\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{g}: \left\{5, 6, 7, 8\right\}\to \left\{1, 2, 3, 4\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{g}= \left\{\left(5, 4\right), \left(6, 3\right), \left(7, 4\right), \left(8, 2\right)\right\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{g},\mathrm{it}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{g}\mathrm{is}\mathrm{a}\mathrm{many}\mathrm{one}\\ \mathrm{function}\mathrm{as}:\mathrm{g}\left(5\right) =\mathrm{g}\left(7\right) = 4\mathrm{.}\\ \therefore \mathrm{g}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{g}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{h}: \left\{2, 3, 4, 5\right\}\to \left\{7, 9, 11, 13\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{h}= \left\{\left(2, 7\right), \left(3, 9\right), \left(4, 11\right), \left(5, 13\right)\\ \mathrm{It}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{all}\mathrm{distinct}\mathrm{elements}\mathrm{of}\mathrm{the}\mathrm{set}\left\{2, 3, 4, 5\right\}\\ \mathrm{have}\mathrm{distinct}\mathrm{images}\mathrm{under}\mathrm{h}\mathrm{.}\\ \therefore \mathrm{Function}\mathrm{h}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{h}\mathrm{is}\mathrm{onto}\mathrm{since}\mathrm{for}\mathrm{every}\mathrm{element}\mathrm{y}\mathrm{of}\mathrm{the}\mathrm{set}\\ \left\{7, 9, 11, 13\right\},\mathrm{there}\mathrm{exists}\mathrm{an} \mathrm{element}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{set}\\ \left\{2, 3, 4, 5\right\} \mathrm{such}\mathrm{that}\mathrm{h}\left(\mathrm{x}\right) =\mathrm{y}\mathrm{.}\\ \mathrm{Thus},\mathrm{h}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{function}.\mathrm{Hence},\mathrm{h}\mathrm{has}\mathrm{an}\mathrm{inverse}\mathrm{.}\end{array}$

Q.34

$\begin{array}{l}Showthatf:\left[-1,1\right]\to R,givenbyf\left(x\right)=\frac{x}{x+2}is\text{\hspace{0.17em}}one–one.\\ Findtheinverseofthefunctionf:\left[-1,1\right]\to Rangef.\end{array}$

Ans.

$f: [−1, 1]→R is given as f x = x x+2 Let f x =f y ⇒ x x+2 = y y+2 ⇒ x y+2 =y x+2 ⇒ xy+2x=xy+2y ⇒ x=y ∴f is one-one function. It is clear that f: [−1, 1]→Range f is onto. ∴f: [−1, 1]→Range f is one-one and onto and therefore, the inverse of the function: f: [−1, 1]→ Range f exists. Let g: Range f→[−1, 1] be the inverse of f. Let y be an arbitrary element of range f. Since f: [−1, 1]→Range f is onto, we have: f x =y for some x∈ −1,1 ⇒ y= x x+2 ⇒xy+2y=x ⇒ 2y=x−xy =x 1−y ⇒ x= 2y 1−y , y≠1 Now, let us define g: Range f→[−1, 1] as g y = 2y 1−y , y≠1. Now, gof x =g f x =g x x+2 = 2 x x+2 1− x x+2 = 2x x+2−x = 2x 2 gof x =x fog y =f g y =f 2y 1−y = 2y 1−y 2y 1−y +2 = 2y 2y+2−2y = 2y 2 fog y =y ∴go f −1 = I −1,1 and fo g −1 = I Range f ∴ f −1 =g ⇒ f −1 y = 2y y−1 , y≠1. ⇒ f −1 x = 2x x−1 , x≠1. 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Q.35

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\mathrm{R}\to \mathrm{Rgiven}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=4\mathrm{x}+3.\mathrm{Show}\mathrm{that}\mathrm{f}\mathrm{is}\\ \mathrm{invertible}.\mathrm{Find}\mathrm{the}\mathrm{inverse}\mathrm{off}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}\left(\mathrm{x}\right) = 4\mathrm{x}+ 3\\ \mathrm{One}–\mathrm{one}:\\ \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{.}\\ ⇒4\mathrm{x}+ 3=4\mathrm{y}+ 3\\ ⇒ \mathrm{ }4\mathrm{x}=4\mathrm{y}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{function}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{y}\in \mathrm{R},\\ \mathrm{let}\mathrm{y}= 4\mathrm{x}+ 3\mathrm{.}\\ ⇒ \mathrm{ }\mathrm{x}=\frac{\mathrm{y}-3}{4}\in \mathrm{R}\\ \\ \mathrm{Therefore},\mathrm{for}\mathrm{any}\mathrm{y}\in \mathrm{R},\mathrm{there}\mathrm{exists},{\mathrm{f}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{by}\mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-3}{4}\\ \mathrm{Now},\mathrm{ }\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \mathrm{ }=\mathrm{g}\left(4\mathrm{x}+3\right)\\ \mathrm{ }=\frac{4\mathrm{x}+3-3}{4}\\ \left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{ }\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \mathrm{ }=\mathrm{f}\left(\frac{\mathrm{y}-3}{4}\right)\\ \mathrm{ }=4\left(\frac{\mathrm{y}-3}{4}\right)+3\\ \mathrm{ }\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{y}\\ \therefore \mathrm{gof}=\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{given}\mathrm{by}\\ {\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\mathrm{g}\left(\mathrm{y}\right)\\ \mathrm{ }=\frac{\mathrm{y}-3}{4}.\\ ⇒ {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{x}-3}{4}\end{array}$

Q.36

$\begin{array}{l}Considerf:{R}_{+}\to \left[4,\infty \right)givenbyf\left(x\right)={x}^{2}+4.Showthatfis\\ invertiblewiththeinverse{f}^{–1}ofgivenfby,where{R}^{+}istheset\\ ofallnon–negativereal\text{\hspace{0.17em}}\text{\hspace{0.17em}}numbers.\end{array}$

Ans.

$f: R + → [4, ∞ ) is given as f(x) = x 2 + 4. One-one: Let f(x) = f(y) ⇒ x 2 + 4= y 2 + 4 ⇒ x 2 = y 2 ⇒ x=y as x=y∈ R + ∴ f is one−one function. Onto: For y∈ [4,∞), let y= x 2 + 4 x= y−4 ≥0 Therefore, for any y∈R, there exists x= y−4 ≥0, such that f x =f y−4 = y−4 2 +4 =y−4+4 f x =y ∴ f is onto. Thus, f is one-one and onto and therefore, f -1 exists. Let us define g: [4,∞)→ R + by, g y = y−4 Now, gof x =g f x =g x 2 +4 = x 2 +4 −4 gof x =x fog y =f g y =f y−4 = y−4 2 +4 =y−4+4 fog y =y ∴gof= fog= I R + . 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Q.37

$\begin{array}{l}Considerf:{R}_{+}\to \left[–5,\infty \right)givenbyf\left(x\right)=9{x}^{2}+6x–5.\\ Showthatfisinvertiblewith{f}^{-1}\left(y\right)=\left(\frac{\left(\sqrt{y+6}\right)-1}{3}\right).\end{array}$

Ans.

$f: R + → [−5, ∞ ) is given as f(x) = 9x 2 + 6x−5. Let y be an arbitrary element of [−5,∞). Let y = 9x 2 + 6x−5 ⇒ 9x 2 + 6x− 5+y =0 ⇒ x= −6± 6 2 −4×9×− 5+y 2×9 = −6± 36+36 5+y 18 = −6±6 1+5+y 18 x= −1± 6+y 3 = 6+y −1 3 âˆµy≥−6⇒y+6≥0 ∴f is onto, thereby range f = [−5, ∞). Let us define g: [−5, ∞)→ R + as g y = y+6 −1 3 Now, gof x =g f x =g 9x 2 + 6x−5 = 9x 2 + 6x−5 +6 −1 3 = 9x 2 + 6x+1 −1 3 = 3x+1 2 −1 3 = 3x+1−1 3 =x And, fog y =f g y =f y+6 −1 3 =9 y+6 −1 3 2 + 6 y+6 −1 3 −5 =9 y+6−2 y+6 +1 9 +2 y+6 −2−5 =y+6−2 y+6 +1+2 y+6 −2−5 fog y =y ∴ gof= I R and fog= I −5,∞ Hence, f is invertible and the inverse of f is given by f −1 y =g y = y+6 −1 3 . 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Q.38

$\text{Let f :X→Y be an invertible function.Show that f has unique inverse.}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Also},\mathrm{suppose}\mathrm{f}\mathrm{has}\mathrm{two}\mathrm{inverses}\left(\mathrm{say}{\mathrm{g}}_{\mathrm{1}}\mathrm{and}{\mathrm{g}}_{\mathrm{2}}\mathrm{\right)}\\ \mathrm{Then},\mathrm{for}\mathrm{all}\mathrm{y}\in \mathrm{Y},\mathrm{we}\mathrm{have}:\\ {\mathrm{fog}}_{\mathrm{1}}\left(\mathrm{y}\right)={\mathrm{I}}_{\mathrm{Y}}\left(\mathrm{y}\right)={\mathrm{fog}}_{2}\left(\mathrm{y}\right)\\ ⇒ \mathrm{f}\left({\mathrm{g}}_{1}\left(\mathrm{y}\right)\right)=\mathrm{f}\left({\mathrm{g}}_{2}\left(\mathrm{y}\right)\right)\\ ⇒ {\mathrm{g}}_{1}\left(\mathrm{y}\right)={\mathrm{g}}_{2}\left(\mathrm{y}\right)\left[\mathrm{f}\mathrm{is}\mathrm{invertible}⇒\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ ⇒ {\mathrm{g}}_{1}={\mathrm{g}}_{2}\left[\mathrm{g}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ \mathrm{Hence},\mathrm{f}\mathrm{has}\mathrm{a}\mathrm{unique}\mathrm{inverse}\mathrm{.}\end{array}$

Q.39

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\left\{1,2,3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\mathrm{given}\mathrm{by}\mathrm{f}\left(1\right)=\mathrm{a},\mathrm{f}\left(2\right)=\mathrm{b}\\ \mathrm{and}\mathrm{f}\left(3\right)=\mathrm{c}.\mathrm{Find}{\mathrm{f}}^{–1}\mathrm{and}\mathrm{show}\mathrm{that}{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Function}\mathrm{f}: \left\{1, 2, 3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}\left(1\right) =\mathrm{a},\mathrm{f}\left(2\right) =\mathrm{b},\mathrm{and}\mathrm{f}\left(3\right) =\mathrm{c}\\ \mathrm{If}\mathrm{we}\mathrm{define}\mathrm{g}: \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\to \left\{1, 2, 3\right\}\mathrm{as}\mathrm{g}\left(\mathrm{a}\right) = 1,\mathrm{g}\left(\mathrm{b}\right) = 2,\\ \mathrm{g}\left(\mathrm{c}\right) = 3,\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{fog}\right)\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{f}\left(1\right)=\mathrm{a}\\ \left(\mathrm{fog}\right)\left(\mathrm{b}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{f}\left(2\right)=\mathrm{b}\\ \left(\mathrm{fog}\right)\left(\mathrm{c}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{f}\left(3\right)=\mathrm{c}\\ \mathrm{and}\\ \left(\mathrm{gof}\right)\left(1\right)=\mathrm{g}\left(\mathrm{f}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{gof}\right)\left(2\right)=\mathrm{g}\left(\mathrm{f}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{gof}\right)\left(3\right)=\mathrm{g}\left(\mathrm{f}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{where}\mathrm{X}=\left\{1,2,3\right\}\mathrm{and}\mathrm{Y}=\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{exists}\mathrm{and}{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \therefore {\mathrm{f}}^{-1}:\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\to \left\{1,2,3\right\}\mathrm{is}\mathrm{given}\mathrm{by},\\ {\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{a}\right)=1,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{b}\right)=2,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{c}\right)=3\\ \mathrm{The}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{–1}\mathrm{}⇒\mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{g}\\ \mathrm{We}\mathrm{define},\mathrm{h}:\left\{1,2,3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\} \mathrm{as}\\ \mathrm{h}\left(1\right) =\mathrm{a},\mathrm{h}\left(2\right) =\mathrm{b},\mathrm{h}\left(3\right) =\mathrm{c},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{goh}\right)\left(1\right)=\mathrm{g}\left(\mathrm{h}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{goh}\right)\left(2\right)=\mathrm{g}\left(\mathrm{h}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{goh}\right)\left(3\right)=\mathrm{g}\left(\mathrm{h}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \mathrm{and}\\ \left(\mathrm{hog}\right)\left(\mathrm{a}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{h}\left(1\right)=\mathrm{a}\\ \left(\mathrm{hog}\right)\left(\mathrm{b}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{h}\left(2\right)=\mathrm{b}\\ \left(\mathrm{hog}\right)\left(\mathrm{c}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{h}\left(3\right)=\mathrm{c}\\ \therefore \mathrm{goh}={\mathrm{I}}_{\mathrm{X}}\mathrm{ }\mathrm{and} \mathrm{hog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{ }\mathrm{where}\mathrm{ }\mathrm{X}=\left\{1,2,3\right\}\mathrm{and}=\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{g}\mathrm{exists}\mathrm{and}\mathrm{ }{\mathrm{g}}^{-1}=\mathrm{h}⇒{\left({\mathrm{f}}^{-\mathrm{1}}\right)}^{-\mathrm{1}}=\mathrm{h}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{noted}\mathrm{that}\mathrm{h}=\mathrm{f}\mathrm{.}\\ \mathrm{Hence},{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}\mathrm{.}\end{array}$

Q.40

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{aninvertible}\mathrm{function}.\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}{\mathrm{f}}^{–1}\mathrm{is}\mathrm{f},\mathrm{i}.\mathrm{e}.,{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{a}\mathrm{function}\mathrm{g}:\mathrm{Y}\to \mathrm{X}\mathrm{such}\mathrm{that}\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\\ \mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\mathrm{.}\\ \mathrm{Here},{\mathrm{f}}^{-1}=\mathrm{g}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\\ ⇒{\mathrm{f}}^{–1}\mathrm{of}={\mathrm{I}}_{\mathrm{X}}\mathrm{ }\mathrm{and}{\mathrm{fof}}^{–1}={\mathrm{I}}_{\mathrm{Y}}\\ \mathrm{Hence},{\mathrm{f}}^{-\mathrm{1}}:\mathrm{Y}\to \mathrm{X}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{-\mathrm{1}}\\ \mathrm{i}.\mathrm{e}., \left({\mathrm{f}}^{-\mathrm{1}}{\mathrm{\right)}}^{-1}=\mathrm{f}\mathrm{.}\end{array}$

Q.41

$\begin{array}{l}\mathrm{If}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{given}\mathrm{by},\mathrm{f}\left(\mathrm{x}\right)={\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\mathrm{ }\mathrm{then}\mathrm{f}\mathrm{of}\left(\mathrm{x}\right)\mathrm{is}\\ \left(\mathrm{A}\right){\mathrm{x}}^{\frac{1}{3}} \left(\mathrm{B}\right){\mathrm{x}}^{3}\left(\mathrm{C}\right)\mathrm{x}\left(\mathrm{D}\right)\left(3-{\mathrm{x}}^{3}\right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{as} \mathrm{f}\left(\mathrm{x}\right)={\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\\ \therefore \mathrm{fof}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \mathrm{ }=\mathrm{f}\left\{{\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\right\}\\ \mathrm{ }={\left[3-{\left\{{\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\right\}}^{3}\right]}^{\frac{1}{3}}\\ \mathrm{ }={\left\{3-\left(3-{\mathrm{x}}^{3}\right)\right\}}^{\frac{1}{3}}\\ \mathrm{ }={\left(3-3+{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\\ \mathrm{ }\mathrm{fof}\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\left(\mathrm{C}\right)\mathrm{.}\\ \end{array}$

Q.42

$\begin{array}{l}\mathrm{Let} \mathrm{f}:\mathrm{R}-\left\{-\frac{4}{3}\right\}\to \mathrm{R}\mathrm{be}\mathrm{a}\mathrm{function}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{map}\mathrm{g}:\mathrm{R}\mathrm{ange}\mathrm{f}\to \mathrm{R}-\left\{-\frac{4}{3}\right\}\mathrm{given}\mathrm{by}\\ \left(\mathrm{A}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{B}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \left(\mathrm{C}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{D}\right)\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{4-3\mathrm{y}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{ }\mathrm{f}:\mathrm{R}-\left\{-\frac{4}{3}\right\}\to \mathrm{R}\mathrm{ }\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{Let}\mathrm{ }\mathrm{y}\mathrm{be}\mathrm{an}\mathrm{arbitrary}\mathrm{element}\mathrm{of}\mathrm{Range}\mathrm{f}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{x}\in \mathrm{R}-\left\{-\frac{4}{3}\right\}\mathrm{ }\mathrm{such}\mathrm{ }\mathrm{that}\mathrm{ }\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right).\\ ⇒ \mathrm{ }\mathrm{y}=\frac{4\mathrm{x}}{3\mathrm{x}+4}\\ ⇒3\mathrm{xy}+4\mathrm{y}=4\mathrm{x}\\ ⇒-3\mathrm{xy}+4\mathrm{x}=4\mathrm{y}\\ ⇒\mathrm{ }\mathrm{x}\left(4-3\mathrm{y}\right)=4\mathrm{y}\\ ⇒ \mathrm{ }\mathrm{x}=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{Range} \mathrm{f}\to \mathrm{R}-\left\{-\frac{4}{3}\right\} \mathrm{as}\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\mathrm{ }\\ \mathrm{Now},\mathrm{ }\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \\ \mathrm{ }=\mathrm{g}\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)\\ \mathrm{ }=\frac{4\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}{4-3\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}\\ \mathrm{ }=\frac{16\mathrm{x}}{12\mathrm{x}+16-12\mathrm{x}}\\ \mathrm{ }=\mathrm{x}\\ \mathrm{And},\mathrm{ }\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \mathrm{ }=\mathrm{f}\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)\\ \mathrm{ }=\frac{4\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)}{3\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)+4}\\ \mathrm{ }=\frac{16\mathrm{y}}{12\mathrm{y}+16-12\mathrm{y}}\\ \mathrm{ }=\frac{16\mathrm{y}}{16}\\ \mathrm{ }=\mathrm{y}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{R}-\left\{-\frac{4}{3}\right\}} \mathrm{and} \mathrm{fog}={\mathrm{I}}_{\mathrm{Range}\mathrm{ }\mathrm{f}}\\ \mathrm{Thus},\mathrm{g}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{i}.\mathrm{e}.,{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{map}\mathrm{g}:\mathrm{Range}, \mathrm{R}\to \mathrm{R}-\left\{-\frac{4}{3}\right\},\\ \mathrm{which}\mathrm{is}\mathrm{given}\mathrm{by} \mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}.\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{B}\mathrm{.}\end{array}$

Q.43

Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On Z+, define * by a * b = a b

(ii) On Z+, define * by a * b = ab

(iii) On R, define * by a * b = ab2

(iv) On Z+, define * by a * b = |a b|

(v) On Z+, define * by a * b = a

Ans.

$\begin{array}{l}\left(i\right){\text{On Z}}^{\text{+}}\text{, * is defined by a * b = a}-\text{b}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Here,}\left(\text{1,2}\right)\in {Z}^{+}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}so,\text{1*2}=\text{\hspace{0.17em}}1-2=-1\notin {Z}^{+}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Thus,\text{\hspace{0.17em}}\text{\hspace{0.17em}}*\text{is not a binary operation}\text{.}\\ \left(ii\right)\text{\hspace{0.17em}}{\text{On Z}}^{\text{+}}\text{, * is defined by a*b = ab}\text{.}\\ \text{Here, a,b}\in {\text{Z}}^{\text{+}}\text{and ab}\in {\text{Z}}^{\text{+}}\\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{This means that * carries each pair (a, b) to a unique element}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{a * b = ab in Z}}^{\text{+}}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Therefore, *}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is a binary operation}\text{.}\\ \left(iii\right)\text{\hspace{0.17em}}{\text{On R, * is defined by a * b = ab}}^{\text{2}}\text{.}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is seen that for each a, b}\in \text{R, there is a unique element}\\ {\text{ab}}^{\text{2}}\text{in R}\text{.}\\ \text{}\text{\hspace{0.17em}}\text{This means that * carries each pair (a, b) to a unique}\\ {\text{element a * b = ab}}^{\text{2}}\text{in R}\text{.}\\ \text{}\text{\hspace{0.17em}}\text{Therefore, * is a binary operation}\text{.}\\ \left(iv\right)\text{\hspace{0.17em}}{\text{On Z}}^{\text{+}}\text{, * is defined by a*b = |a}-\text{b|}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{It is seen that for each a, b}\in {\text{Z}}^{\text{+}}\text{, there is a unique element}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{|a}-{\text{b| in Z}}^{\text{+}}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{This means that * carries each pair (a, b) to a unique element}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{a*b =}\text{\hspace{0.17em}}\text{|a}-{\text{b| in Z}}^{\text{+}}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Therefore, * is a binary operation}\text{.}\\ \left(v\right)\text{\hspace{0.17em}}{\text{On Z}}^{\text{+}}\text{, * is defined by a*b = a}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{* carries each pair (a, b) to a unique element a * b = a in Z}}^{\text{+}}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Therefore, * is a binary operation}\text{.}\end{array}$

Q.44

$\begin{array}{l}\mathrm{For}\mathrm{each}\mathrm{binary}\mathrm{operation}*\mathrm{defined}\mathrm{below},\mathrm{determine}\\ \mathrm{whether}*\mathrm{is}\mathrm{commutative}\mathrm{or}\mathrm{associative}.\\ \mathrm{ }\left(\mathrm{i}\right)\mathrm{On}\mathrm{Z},\mathrm{definea}*\mathrm{b}=\mathrm{a}-\mathrm{b}\\ \mathrm{ }\left(\mathrm{ii}\right)\mathrm{}\mathrm{On}\mathrm{Q},\mathrm{definea}*\mathrm{b}=\mathrm{ab}+1\\ \left(\mathrm{iii}\right)\mathrm{OnQ},\mathrm{definea}*\mathrm{b}=\frac{\mathrm{ab}}{2}\\ \mathrm{}\left(\mathrm{iv}\right){\mathrm{OnZ}}^{+},\mathrm{definea}*\mathrm{b}={2}^{\mathrm{ab}}\\ \left(\mathrm{v}\right){\mathrm{OnZ}}^{+},\mathrm{definea}*\mathrm{b}={\mathrm{a}}^{\mathrm{b}}\\ \left(\mathrm{vi}\right)\mathrm{ }\mathrm{OnR}-\left\{-1\right\},\mathrm{definea}*\mathrm{b}=\frac{\mathrm{a}}{\mathrm{b}+1}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{On}\mathrm{Z}, *\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{a}-\mathrm{b}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}\\ \mathrm{ }1* 2 = 1-2= 1\mathrm{and}2 * 1 = 2-1= 1\mathrm{.}\\ \therefore 1*2\ne 2* 1;\mathrm{where}1, 2\in \mathrm{Z}\\ \mathrm{Hence},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{Also}\mathrm{we}\mathrm{have}:\\ \left(1* 2\right) * 3 = \left(1-2\right)*3 =-1* 3 =-\mathrm{1}-3=-\mathrm{4}\\ 1* \left(2 * 3\right) = 1 * \left(2-3\right) = 1 *-1= 1-\left(-1\right) = 2\\ \therefore \left(1* 2\right) * 3\ne 1* \left(2 * 3\right) ;\mathrm{where}1, 2, 3\in \mathrm{Z}\\ \mathrm{Hence},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{On}\mathrm{Q}, *\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{ab}+ 1\mathrm{.}\\ \mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \mathrm{ }\mathrm{ab}=\mathrm{ba}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ ⇒\mathrm{ab}+ 1 =\mathrm{ba}+ 1 \mathrm{ }\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ ⇒ \mathrm{ }\mathrm{a}*\mathrm{b}=\mathrm{a}*\mathrm{b}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}:\\ \mathrm{ }\left(1* 2\right) * 3 = \left(1 × 2 + 1\right) * 3 = 3 * 3 = 3 × 3 + 1 = 10\\ \mathrm{ }1* \left(2 * 3\right) = 1 * \left(2 × 3 + 1\right) = 1 * 7 = 1 × 7 + 1 = 8\\ \therefore \left(1* 2\right) * 3\ne 1* \left(2 * 3\right) ;\mathrm{where}1, 2, 3\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{On}\mathrm{Q}, *\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{2}\\ \mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \mathrm{ab}=\mathrm{ba}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ ⇒ \mathrm{ }\frac{\mathrm{ab}}{2}=\frac{\mathrm{ba}}{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ ⇒\mathrm{ }\mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{all}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{we}\mathrm{have}:\\ \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\left(\frac{\mathrm{ab}}{2}\right)*\mathrm{c}=\frac{\left(\frac{\mathrm{ab}}{2}\right)\mathrm{c}}{2}=\frac{\mathrm{abc}}{4}\\ \mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}*\left(\frac{\mathrm{bc}}{2}\right)=\frac{\mathrm{a}\left(\frac{\mathrm{bc}}{2}\right)}{2}=\frac{\mathrm{abc}}{4}\\ \mathrm{So}, \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)\forall \mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iv}\right)\mathrm{ }\mathrm{On}{\mathrm{Z}}^{\mathrm{+}},*\mathrm{is}\mathrm{defined}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{ab}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}:\\ \mathrm{ }1*2={\mathrm{1}}^{\mathrm{2}}=1\mathrm{and}2*1={\mathrm{2}}^{\mathrm{2}}=4\\ \therefore 1*2\ne 2*1; \mathrm{where} 1,2\in {\mathrm{Z}}^{\mathrm{+}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{Associative}\mathrm{Law}:\\ \left(3*4\right)*5=\left({3}^{2}\right)*5\\ ={\left({3}^{2}\right)}^{2}\\ =81\\ 3*\left(4*5\right)=3*\left({4}^{2}\right)\\ ={3}^{2}\\ =9\\ \therefore \left(3*4\right)*5\ne 3*\left(4*5\right) \mathrm{where}2, 3, 4\in {\mathrm{Z}}^{\mathrm{+}}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{vi}\right)\mathrm{ }\mathrm{On}\mathrm{R}, *-\mathrm{\left\{}-1\right\}\mathrm{is}\mathrm{defined}\mathrm{by} \mathrm{a}*\mathrm{b}=\frac{\mathrm{a}}{\mathrm{b}+1}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that} \\ 2*3=\frac{2}{3+1}\\ =\frac{2}{4}\\ =\frac{1}{2}\\ \mathrm{and} \mathrm{ }3*2=\frac{3}{2+1}\\ =\frac{3}{3}\\ =1\\ \therefore 2*3\ne 3*2 ;\mathrm{where}2, 3\in \mathrm{R}-\mathrm{\left\{}-1\right\}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{â€‹}\mathrm{Associative}\mathrm{Law}:\\ \left(2*3\right)*4=\left(\frac{2}{3+1}\right)*4=\left(\frac{2}{4}\right)*4=\left(\frac{1}{2}\right)*4=\frac{\left(\frac{1}{2}\right)}{4+1}=\frac{1}{10}\\ 2*\left(3*4\right)=2*\left(\frac{3}{4+1}\right)=2*\left(\frac{3}{5}\right)=\frac{2}{\left(\frac{3}{5}\right)+1}=\frac{2}{\frac{8}{5}}=\frac{5}{4}\\ \therefore \left(2*3\right)*4\ne 2*\left(3*4\right); \mathrm{ }\mathrm{where}2,3,4\in \mathrm{R}-\left\{-1\right\}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\end{array}$

Q.45

$\begin{array}{l}\mathrm{Consider}\mathrm{the}\mathrm{binary}\mathrm{operation}\wedge \mathrm{on}\mathrm{theset}\left\{1,2,3,4,5\right\}\\ \mathrm{definedby} \mathrm{a}\wedge \mathrm{b}=\mathrm{min}\left\{\mathrm{a},\mathrm{b}\right\}.\mathrm{Write}\mathrm{the}\mathrm{operation}\mathrm{table}\\ \mathrm{of}\mathrm{the}\mathrm{operation}\wedge .\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{binary}\mathrm{operation}\wedge \mathrm{on}\mathrm{the}\mathrm{set}\left\{\mathrm{1},2,3,4,5\right\}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{a}\wedge \mathrm{b}=\mathrm{min}\left\{\mathrm{a},\mathrm{b}\right\}\forall \mathrm{a},\mathrm{b}\in \left\{1, 2, 3, 4, 5\right\}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}\mathrm{table}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{operation}\wedge \mathrm{can}\mathrm{be}\\ \mathrm{given}\mathrm{as}:\\ \begin{array}{c}\end{array}\end{array}$

 ^ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5

Q.46

$\begin{array}{l}\mathrm{Conside}\mathrm{ra}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{the}\mathrm{set}\left\{1,2,3,4,5\right\}\mathrm{given}\\ \mathrm{by}\mathrm{the}\mathrm{following}\mathrm{multiplication}\mathrm{ }\mathrm{table} \left(\mathrm{Table}\mathrm{ }1.2\right)\\ \left(\mathrm{i}\right)\mathrm{Compute}\left(2*3\right)*4\mathrm{and}2*\left(3*4\right)\\ \left(\mathrm{ii}\right)\mathrm{Is}*\mathrm{commutative}?\\ \left(\mathrm{iii}\right)\mathrm{Compute}\left(2*3\right)*\left(4*5\right)\\ \begin{array}{}\end{array}\end{array}$

 * 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{ }\left(2*3\right)*4=1*4=1\\ \mathrm{ }2*\left(3*4\right)=2*1=1\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{For}\mathrm{every}\mathrm{a},\mathrm{b}\in \left\{1, 2, 3, 4, 5\right\},\mathrm{we}\mathrm{have}\mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}.\\ \mathrm{ }\mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \left(\mathrm{iii}\right)\left(2* 3\right) = 1\mathrm{and}\left(4 * 5\right) = 1\\ \therefore \left(2* 3\right) * \left(4 * 5\right) = 1 * 1 = 1\end{array}$

Q.47

Let *′ be the binary operation on the set {1, 2, 3, 4, 5} defined by a *′ b = H.C.F. of a and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

Ans.

$\begin{array}{l}\text{The binary operation *’R on the set {1, 2, 3 4, 5} is defined as}\\ \text{a *’ b = H}\text{.C}\text{.F of a and b}\text{.}\\ \text{The operation table for the operation *’ can be given as:}\\ \end{array}$

 *’ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 1 1 1 4 1 2 1 4 1 5 1 1 1 1 1

We see that the operation tables for the operations * and *’ are the same. Thus, the operation *’ is same as the operation*.

Q.48

$\begin{array}{l}\mathrm{Let}*\mathrm{be}\mathrm{the}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{Ngiven}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\\ \mathrm{and}\mathrm{b}.\mathrm{Find}\\ \left(\mathrm{i}\right)5*7,20*16\\ \left(\mathrm{ii}\right)\mathrm{Is}*\mathrm{commutative}?\\ \left(\mathrm{iii}\right)\mathrm{Is}*\mathrm{associative}?\\ \left(\mathrm{iv}\right)\mathrm{Find}\mathrm{the}\mathrm{identity}\mathrm{of}*\mathrm{in}\mathrm{N}\\ \left(\mathrm{v}\right)\mathrm{Which}\mathrm{element}\mathrm{so}\mathrm{f}\mathrm{Narein}\mathrm{vertible}\mathrm{f}\mathrm{or}\mathrm{the}\mathrm{operation}*?\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{N}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\\ \mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{.}\\ \left(\mathrm{i}\right) 5 * 7 =\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}5\mathrm{and}7 =35\\ \mathrm{ }20* 16 =\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}20\mathrm{and}16 = 80\\ \left(\mathrm{ii}\right)\mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{b}\mathrm{and}\mathrm{a} \forall \mathrm{a},\mathrm{b}\in \mathrm{N}\mathrm{.}\\ \therefore \mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \\ \left(\mathrm{iii}\right)\mathrm{For}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{N},\mathrm{we}\mathrm{have}:\\ \left(\mathrm{a}*\mathrm{b}\right) *\mathrm{c}= \left(\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\right) *\mathrm{c}\\ =\mathrm{LCM}\mathrm{of}\mathrm{a},\mathrm{b},\mathrm{and}\mathrm{c}\\ \mathrm{ }\mathrm{a}* \left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}* \left(\mathrm{LCM}\mathrm{of}\mathrm{b}\mathrm{and}\mathrm{c}\right)\\ =\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a},\mathrm{b},\mathrm{and}\mathrm{c}\\ \therefore \mathrm{ }\left(\mathrm{a}*\mathrm{b}\right) *\mathrm{c}=\mathrm{a}* \left(\mathrm{b}*\mathrm{c}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iv}\right)\mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}:\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\mathrm{and}1 =\mathrm{a}=\mathrm{L}.\mathrm{C}.\mathrm{M}.1\mathrm{and}\mathrm{a}\mathrm{}\forall \mathrm{a}\in \mathrm{N}\\ ⇒ \mathrm{ }\mathrm{a}* 1 =\mathrm{a}= 1 *\mathrm{a}\mathrm{}\forall \mathrm{a}\in \mathrm{N}\\ \mathrm{Thus}, 1\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{of}*\mathrm{in}\mathrm{N}\mathrm{.}\\ \left(\mathrm{v}\right)\mathrm{An}\mathrm{element}\mathrm{a}\mathrm{in}\mathrm{N}\mathrm{is}\mathrm{invertible}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\\ \mathrm{operation}*\mathrm{ }\mathrm{if}\mathrm{there}\mathrm{exists}\mathrm{an}\mathrm{element}\mathrm{b}\mathrm{in} \mathrm{N},\mathrm{such}\\ \mathrm{that}\mathrm{a}*\mathrm{b}=\mathrm{e}=\mathrm{b}*\mathrm{a}\mathrm{.}\\ \mathrm{Here}, \mathrm{e}= 1\\ \mathrm{This}\mathrm{means}\mathrm{that}:\\ \mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}= 1 =\mathrm{L}.\mathrm{C}.\mathrm{M}\mathrm{of}\mathrm{b}\mathrm{and}\mathrm{a}\\ \mathrm{This}\mathrm{case}\mathrm{is}\mathrm{possible}\mathrm{only}\mathrm{when}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{equal}\mathrm{to}1\mathrm{.}\\ \mathrm{Thus}, 1\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{invertible}\mathrm{element}\mathrm{of}\mathrm{N}\mathrm{with}\mathrm{respect}\mathrm{to}\\ \mathrm{the}\mathrm{operation}*\mathrm{.}\end{array}$

Q.49

$\begin{array}{l}\mathrm{Is}*\mathrm{defined}\mathrm{on}\mathrm{the}\mathrm{set}\left\{1,2,3,4,5\right\}\mathrm{by}\mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\\ \mathrm{and}\mathrm{b}\mathrm{a}\mathrm{binary}\mathrm{operation}?\mathrm{Justify}\mathrm{your}\mathrm{ }\mathrm{answer}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{operation}*\mathrm{on}\mathrm{the}\mathrm{set}\mathrm{A}= \left\{1, 2, 3, 4, 5\right\}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{a}*\mathrm{b}=\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}.\mathrm{Then},\mathrm{the}\mathrm{operation}\mathrm{table}\mathrm{for}\mathrm{the}\mathrm{given}\\ \mathrm{operation}*\mathrm{can}\mathrm{be}\mathrm{givenas}:\\ \end{array}$

 * 1 2 3 4 5 1 1 2 3 4 5 2 2 2 6 4 10 3 3 6 3 12 15 4 4 4 12 4 20 5 5 10 15 20 5

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6A, 5 * 2 = 2 * 5 = 10A,

3 * 4 = 4 * 3 = 12A

3 * 5 = 5 * 3 = 15A, 4 * 5 = 5 * 4 = 20A

Hence, the given operation * is not a binary operation.

Q.50

Let * be the binary operation on N defined by
a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?

Ans.

$\begin{array}{l}\text{The binary operation * on N is defined as:}\\ \text{a * b = H}\text{.C}\text{.F}\text{. of a and b}\\ Since,\\ \text{H}\text{.C}\text{.F}\text{. of a and b}=\text{H}\text{.C}\text{.F}\text{. of b and a}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\forall \text{a, b}\in \text{N}\text{.}\\ \therefore \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{a*b}=\text{b*a}\\ Thus,\text{the operation * is commutative}\text{.}\\ \text{For a, b, c}\in \text{N, we have:}\\ \text{(a * b)* c}=\text{(H}\text{.C}\text{.F}\text{. of a and b) * c}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{H}\text{.C}\text{.F}\text{. of a, b, and c}\\ \\ \text{a *(b * c)}=\text{a *(H}\text{.C}\text{.F}\text{. of b and c)}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{H}\text{.C}\text{.F}\text{. of a, b, and c}\\ \therefore \text{(a*b)*c}=\text{a*(b*c)}\\ \text{Thus, the operation * is associative}\text{.}\\ Let\text{\hspace{0.17em}}\text{an element e}\in \text{N will be the identity for the operation}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{* if a*e}=\text{a}=\text{e* a}\forall \text{a}\in \text{N}\text{.}\\ \text{But this relation is not true for all a}\in \text{N}\text{.}\\ \text{Thus, the operation * does not have any identity in N}\text{.}\end{array}$

Q.51

$\begin{array}{l}\mathrm{Let}*\mathrm{be}\mathrm{a}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{the}\mathrm{set} \mathrm{Q}\mathrm{ } \mathrm{of}\mathrm{ } \mathrm{rational}\\ \mathrm{numbers}\mathrm{as}\mathrm{follows}:\\ \left(\mathrm{i}\right)\mathrm{a}*\mathrm{b}=\mathrm{a}–\mathrm{b}\left(\mathrm{ii}\right)\mathrm{a}*\mathrm{b}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}\\ \left(\mathrm{iii}\right)\mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{ab}\left(\mathrm{iv}\right)\mathrm{a}*\mathrm{b}={\left(\mathrm{a}–\mathrm{b}\right)}^{2}\\ \left(\mathrm{v}\right) \mathrm{ }\mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{4} \mathrm{ }\left(\mathrm{vi}\right)\mathrm{ }\mathrm{a}*\mathrm{b}={\mathrm{ab}}^{2}\\ \mathrm{Find}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{binary}\mathrm{operation}\mathrm{s}\mathrm{are}\mathrm{commutative}\\ \mathrm{and}\mathrm{which}\mathrm{are}\mathrm{as}\mathrm{sociative}.\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{ }\mathrm{Since}, \mathrm{a}*\mathrm{b}=\mathrm{a}-\mathrm{b}\mathrm{}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ =-\left(\mathrm{b}-\mathrm{a}\right)\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ =-\mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\left(\mathrm{a}-\mathrm{b}\right)*\mathrm{c}\\ =\mathrm{a}-\mathrm{b}-\mathrm{c}\\ \\ \mathrm{and}\mathrm{ }\\ \mathrm{ }\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}*\left(\mathrm{b}-\mathrm{c}\right)\\ =\mathrm{a}-\left(\mathrm{b}-\mathrm{c}\right)\\ =\mathrm{a}-\mathrm{b}+\mathrm{c}\\ \mathrm{So}, \mathrm{ }\left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}\ne \mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{Since}, \mathrm{a}*\mathrm{b}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ ={\mathrm{b}}^{2}+{\mathrm{a}}^{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ =\mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)*\mathrm{c}\\ ={\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)}^{2}+{\mathrm{c}}^{2}\\ \mathrm{and}\mathrm{ }\\ \mathrm{ }\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}*\left({\mathrm{b}}^{2}+{\mathrm{c}}^{2}\right)\\ ={\mathrm{a}}^{2}+{\left({\mathrm{b}}^{2}+{\mathrm{c}}^{2}\right)}^{2}\\ \mathrm{So}, \mathrm{ }\left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}\ne \mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{Since}, \mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{ab}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{and} \mathrm{ }\mathrm{b}*\mathrm{a}=\mathrm{b}+\mathrm{ba}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \therefore \mathrm{ }\mathrm{a}*\mathrm{b}\ne \mathrm{b}*\mathrm{a}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \\ \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\left(\mathrm{a}+\mathrm{ab}\right)*\mathrm{c}\\ =\left(\mathrm{a}+\mathrm{ab}\right)+\left(\mathrm{a}+\mathrm{ab}\right)\mathrm{c}\\ =\mathrm{a}+\mathrm{ab}+\mathrm{ac}+\mathrm{abc}\\ \mathrm{and}\mathrm{ }\\ \mathrm{ }\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}*\left(\mathrm{b}+\mathrm{bc}\right)\\ =\mathrm{a}+\mathrm{a}\left(\mathrm{b}+\mathrm{bc}\right)\\ =\mathrm{a}+\mathrm{ab}+\mathrm{abc}\\ \mathrm{So}, \mathrm{ }\left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}\ne \mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{iv}\right)\mathrm{Since}, \mathrm{a}*\mathrm{b}={\left(\mathrm{a}-\mathrm{b}\right)}^{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{ }={\left\{-\left(\mathrm{b}-\mathrm{a}\right)\right\}}^{2}\\ ={\left(\mathrm{b}-\mathrm{a}\right)}^{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{ }=\mathrm{b}*\mathrm{a}\\ \therefore \mathrm{ }\mathrm{a}*\mathrm{b}\ne \mathrm{b}*\mathrm{a} \mathrm{ } \forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\left\{{\left(\mathrm{a}-\mathrm{b}\right)}^{2}\right\}*\mathrm{c}\\ ={\left\{{\left(\mathrm{a}-\mathrm{b}\right)}^{2}-\mathrm{c}\right\}}^{2}\\ ={\left({\mathrm{a}}^{2}-2\mathrm{ab}+{\mathrm{b}}^{2}-\mathrm{c}\right)}^{2}\\ \mathrm{and}\\ \mathrm{ }\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}*{\left(\mathrm{b}-\mathrm{c}\right)}^{2}\\ ={\left\{\mathrm{a}-{\left(\mathrm{b}-\mathrm{c}\right)}^{2}\right\}}^{2}\\ \\ ={\left(\mathrm{a}-{\mathrm{b}}^{2}+2\mathrm{bc}-{\mathrm{c}}^{2}\right)}^{2}\\ \mathrm{So}, \mathrm{ }\left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}\ne \mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \left(\mathrm{v}\right)\mathrm{Since}, \mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{4}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{ }=\frac{\mathrm{ba}}{4}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{ }=\mathrm{b}*\mathrm{a}\\ \therefore \mathrm{ }\mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a} \forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\left(\frac{\mathrm{ab}}{4}\right)*\mathrm{c}\\ =\frac{\left(\frac{\mathrm{ab}}{4}\right)\mathrm{c}}{4}\\ =\frac{\mathrm{abc}}{16}\\ \mathrm{and}\mathrm{ }\\ \mathrm{ }\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}*\left(\frac{\mathrm{bc}}{4}\right)\\ =\frac{\mathrm{a}\left(\frac{\mathrm{bc}}{4}\right)}{4}\\ =\frac{\mathrm{abc}}{16}\\ \\ \mathrm{So}, \mathrm{ }\left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \\ \left(\mathrm{vi}\right)\mathrm{Since}, \mathrm{a}*\mathrm{b}={\mathrm{ab}}^{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{ }\mathrm{b}*\mathrm{a}\mathrm{ }={\mathrm{ba}}^{2}\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \therefore \mathrm{ }\mathrm{a}*\mathrm{b}\ne \mathrm{b}*\mathrm{a}\mathrm{ }\forall \mathrm{a},\mathrm{b}\in \mathrm{Q}\\ \mathrm{So},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{cummutative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{Q},\mathrm{then}\\ \left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}=\left({\mathrm{ab}}^{2}\right)*\mathrm{c}\\ =\left({\mathrm{ab}}^{2}\right){\mathrm{c}}^{2}\\ ={\mathrm{ab}}^{2}{\mathrm{c}}^{2}\\ \mathrm{and}\mathrm{ }\\ \mathrm{ }\mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\mathrm{a}*\left({\mathrm{bc}}^{2}\right)\\ =\mathrm{a}{\left({\mathrm{bc}}^{2}\right)}^{2}\\ ={\mathrm{ab}}^{2}{\mathrm{c}}^{4}\\ \mathrm{So}, \mathrm{ }\left(\mathrm{a}*\mathrm{b}\right)*\mathrm{c}\ne \mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{operations}\mathrm{defined}\mathrm{in}\left(\mathrm{ii}\right), \left(\mathrm{iv}\right), \left(\mathrm{v}\right)\mathrm{are}\\ \mathrm{commutative}\mathrm{and}\mathrm{the}\mathrm{operation}\mathrm{defined}\mathrm{in}\left(\mathrm{v}\right)\mathrm{is}\mathrm{associative}\mathrm{.}\end{array}$

Q.52

$\begin{array}{l}\mathrm{Let}*\mathrm{be}\mathrm{a}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{the}\mathrm{set} \mathrm{Q} \mathrm{of}\mathrm{ } \mathrm{rational}\\ \mathrm{numbersas}\mathrm{follows}:\\ \left(\mathrm{i}\right)\mathrm{a}*\mathrm{b}=\mathrm{a}–\mathrm{b}\left(\mathrm{ii}\right)\mathrm{a}*\mathrm{b}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}\\ \left(\mathrm{iii}\right)\mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{ab}\left(\mathrm{iv}\right)\mathrm{a}*\mathrm{b}={\left(\mathrm{a}–\mathrm{b}\right)}^{2}\\ \left(\mathrm{v}\right) \mathrm{ }\mathrm{a}*\mathrm{b}=\frac{\mathrm{ab}}{4} \mathrm{ }\left(\mathrm{vi}\right)\mathrm{ }\mathrm{a}*\mathrm{b}={\mathrm{ab}}^{2}\end{array}$

Find which of the operations given above has identity.

Ans.

$\begin{array}{l}\mathrm{An}\mathrm{element}\mathrm{e}\in \mathrm{Q}\mathrm{will}\mathrm{be}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\\ \mathrm{operation}*\mathrm{if} \mathrm{ }\mathrm{a}*\mathrm{e}=\mathrm{a}=\mathrm{e}*\mathrm{a},\forall \mathrm{a}\in \mathrm{Q}\mathrm{.}\\ \mathrm{However},\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{such}\mathrm{element}\mathrm{e}\in \mathrm{Q}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{each}\\ \mathrm{of}\mathrm{the}\mathrm{six}\mathrm{operations}\mathrm{satisfying}\mathrm{the}\mathrm{ }\mathrm{above}\mathrm{condition}\mathrm{.}\\ \mathrm{Thus},\mathrm{none}\mathrm{of}\mathrm{the}\mathrm{six}\mathrm{operations}\mathrm{in}\mathrm{above}\mathrm{question}\mathrm{has}\mathrm{identity}\mathrm{.}\end{array}$

Q.53

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\mathrm{N}×\mathrm{Nand}*\mathrm{be}\mathrm{the}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{A}\mathrm{defined}\mathrm{by}\\ \left(\mathrm{a},\mathrm{b}\right)*\left(\mathrm{c},\mathrm{d}\right)=\left(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d}\right)\\ \mathrm{Show}\mathrm{that}*\mathrm{is}\mathrm{commutative}\mathrm{and}\mathrm{associative}.\mathrm{Find}\mathrm{the}\mathrm{}\\ \mathrm{identity}\mathrm{element}\mathrm{for}*\mathrm{onA},\mathrm{if}\mathrm{any}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{A}=\mathrm{N}×\mathrm{N}\\ *\mathrm{is}\mathrm{a}\mathrm{binary}\mathrm{operation}\mathrm{on}\mathrm{A}\mathrm{and}\mathrm{is}\mathrm{defined}\mathrm{by}:\\ \left(\mathrm{a},\mathrm{b}\right) * \left(\mathrm{c},\mathrm{d}\right) = \left(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d}\right)\\ \mathrm{Let}\left(\mathrm{a},\mathrm{b}\right), \left(\mathrm{c},\mathrm{d}\right)\in \mathrm{A}\\ \mathrm{Then}, \mathrm{ }\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathrm{N}\\ \mathrm{Since},\\ \left(\mathrm{a},\mathrm{b}\right) * \left(\mathrm{c},\mathrm{d}\right) = \left(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d}\right)\\ \mathrm{and}\\ \mathrm{ }\left(\mathrm{c},\mathrm{d}\right) * \left(\mathrm{a},\mathrm{b}\right) = \left(\mathrm{c}+\mathrm{a},\mathrm{d}+\mathrm{b}\right)\\ \mathrm{} =\left(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d}\right)\\ \left[\mathrm{Addition}\mathrm{is}\mathrm{commutative}\mathrm{in}\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{natural}\mathrm{numbers}\right]\\ \therefore \left(\mathrm{a},\mathrm{b}\right)*\left(\mathrm{c},\mathrm{d}\right) = \left(\mathrm{c},\mathrm{d}\right)*\left(\mathrm{a},\mathrm{b}\right)\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{associativity}:\\ \mathrm{Now},\mathrm{let}\left(\mathrm{a},\mathrm{b}\right), \left(\mathrm{c},\mathrm{d}\right), \left(\mathrm{e},\mathrm{f}\right)\in \mathrm{A}\\ \mathrm{Then},\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d},\mathrm{e},\mathrm{f}\in \mathrm{N}\\ \\ \mathrm{So},\\ \left\{\left(\mathrm{a},\mathrm{b}\right) * \left(\mathrm{c},\mathrm{d}\right)\right\}\mathrm{*}\left(\mathrm{e},\mathrm{f}\right)=\left\{\left(\mathrm{a}+\mathrm{c},\mathrm{b}+\mathrm{d}\right)\right\}*\left(\mathrm{e},\mathrm{f}\right)\\ \mathrm{ } =\left(\mathrm{a}+\mathrm{c}+\mathrm{ }\mathrm{e},\mathrm{b}+\mathrm{d}+\mathrm{f} \right)\\ \left(\mathrm{a},\mathrm{b}\right) *\left\{\left(\mathrm{c},\mathrm{d}\right)*\left(\mathrm{e},\mathrm{f}\right)\right\}=\left(\mathrm{a},\mathrm{b}\right) *\left\{\left(\mathrm{c}+\mathrm{e},\mathrm{d}+\mathrm{f}\right)\right\}\\ \mathrm{ } =\left(\mathrm{a}+\mathrm{c}+\mathrm{e},\mathrm{b}+\mathrm{d}+\mathrm{f}\right)\\ \therefore \left\{\left(\mathrm{a},\mathrm{b}\right) * \left(\mathrm{c},\mathrm{d}\right)\right\}\mathrm{*}\left(\mathrm{e},\mathrm{f}\right)=\left(\mathrm{a},\mathrm{b}\right) *\left\{\left(\mathrm{c},\mathrm{d}\right)*\left(\mathrm{e},\mathrm{f}\right)\right\}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{associative}\mathrm{.}\\ \mathrm{An}\mathrm{element}\mathrm{e}=\left({\mathrm{e}}_{1}+{\mathrm{e}}_{2}\right)\in \mathrm{A}\mathrm{ }\mathrm{will}\mathrm{be}\mathrm{an}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\\ \mathrm{operation}*\mathrm{if}\\ \mathrm{a}*\mathrm{e}=\mathrm{a}=\mathrm{e}*\mathrm{a}\forall \mathrm{a}=\left({\mathrm{a}}_{1},{\mathrm{a}}_{2}\right)\in \mathrm{A}\\ \mathrm{i}.\mathrm{e}., \mathrm{ }\left({\mathrm{a}}_{1}+{\mathrm{e}}_{1},{\mathrm{a}}_{2}+{\mathrm{e}}_{2}\right)=\left({\mathrm{a}}_{1},{\mathrm{a}}_{2}\right)=\left({\mathrm{e}}_{1}+{\mathrm{a}}_{1},{\mathrm{e}}_{2}+{\mathrm{a}}_{2}\right)\\ \mathrm{which}\mathrm{is}\mathrm{not}\mathrm{true}\mathrm{for}\mathrm{any}\mathrm{element}\mathrm{in}\mathrm{A}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{operation}*\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{any}\mathrm{identity}\mathrm{element}\mathrm{.}\end{array}$

Q.54

$\begin{array}{l}\mathrm{State}\mathrm{whether}\mathrm{the}\mathrm{following}\mathrm{statements}\mathrm{are}\mathrm{true}\mathrm{or}\mathrm{false}.\\ \mathrm{Justify}.\\ \left(\mathrm{i}\right)\mathrm{For}\mathrm{an}\mathrm{arbitrary}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{a}\mathrm{set}\mathrm{N},\\ \mathrm{a}*\mathrm{a}=\mathrm{a}“\mathrm{a}*\mathrm{N}.\\ \left(\mathrm{ii}\right)\mathrm{If}*\mathrm{is}\mathrm{acommutative}\mathrm{binary}\mathrm{operationon}\mathrm{N},\mathrm{then}\\ \mathrm{a}*\left(\mathrm{b}*\mathrm{c}\right)=\left(\mathrm{c}*\mathrm{b}\right)*\mathrm{a}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Define}\mathrm{an}\mathrm{operation}*\mathrm{on}\mathrm{N}\mathrm{as}:\\ \mathrm{a}*\mathrm{b}=\mathrm{a}+\mathrm{b}\forall \mathrm{a},\mathrm{b}\in \mathrm{N}\\ \mathrm{Then},\mathrm{in}\mathrm{particular},\mathrm{for}\mathrm{b}=\mathrm{a}= 5,\mathrm{we}\mathrm{have}:\\ 5* 5 = 5 + 5=10\ne \mathrm{5}\\ \mathrm{Therefore},\mathrm{statement}\left(\mathrm{i}\right)\mathrm{is}\mathrm{false}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{R}.\mathrm{H}.\mathrm{S}.= \left(\mathrm{c}*\mathrm{b}\right) *\mathrm{a}\\ =\left(\mathrm{b}*\mathrm{c}\right) *\mathrm{a}\left[*\mathrm{is}\mathrm{commutative}\right]\\ =\mathrm{a}* \left(\mathrm{b}*\mathrm{c}\right)\left[\mathrm{Again},\mathrm{as}*\mathrm{is}\mathrm{commutative}\right]\\ =\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \therefore \mathrm{a}* \left(\mathrm{b}*\mathrm{c}\right) = \left(\mathrm{c}*\mathrm{b}\right) *\mathrm{a}\\ \mathrm{Therefore},\mathrm{statement}\left(\mathrm{ii}\right)\mathrm{is}\mathrm{true}\mathrm{.}\end{array}$

Q.55

$\begin{array}{l}\mathrm{Consider}\mathrm{a}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{N} \mathrm{defined}\mathrm{as}\mathrm{a}*\mathrm{b}={\mathrm{a}}^{3}+{\mathrm{b}}^{3}.\\ \mathrm{Choose}\mathrm{the}\mathrm{correct}\mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{Is}*\mathrm{both}\mathrm{a}\mathrm{ssociative}\mathrm{and}\mathrm{commutative}?\\ \left(\mathrm{B}\right)\mathrm{Is}*\mathrm{commutative}\mathrm{but}\mathrm{not}\mathrm{associative}?\\ \left(\mathrm{C}\right)\mathrm{Is}*\mathrm{associative}\mathrm{but}\mathrm{not}\mathrm{commutative}?\\ \left(\mathrm{D}\right)\mathrm{Is}*\mathrm{neither}\mathrm{commutative}\mathrm{nor}\mathrm{associative}?\end{array}$

Ans.

$\begin{array}{l}{\text{On N, the operation * is defined as a*b = a}}^{\text{3}}{\text{+ b}}^{\text{3}}\text{.}\\ \text{For, a, b}\in \text{N, we have:}\\ \text{a*b}={\text{a}}^{\text{3}}{\text{+ b}}^{\text{3}}={\text{b}}^{\text{3}}{\text{+ a}}^{\text{3}}=b*a\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{[Addition is commutative in N]}\\ \text{Therefore, the operation * is commutative}\text{.}\\ \text{For associative:}\\ Since,\text{}2,3,4\in N\\ So,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(2*3\right)*4=\left({2}^{3}+{3}^{3}\right)*4\\ \text{}\text{}\text{}\text{}=35*4\\ \text{}\text{}\text{}\text{}={35}^{3}+{4}^{3}\\ \text{}\text{}\text{}\text{}=42939\\ and\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2*\left(3*4\right)=2*\left({3}^{3}+{4}^{3}\right)\\ \text{}\text{}\text{}\text{}=2*\left(91\right)\\ \text{}\text{}\text{}\text{}={2}^{3}+{91}^{3}\\ \text{}\text{}\text{}\text{}=8+753571\\ \text{}\text{}\text{}\text{}=753579\\ \therefore \text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(2*3\right)*4\ne 2*\left(3*4\right);\text{where 2,3,4}\in \text{N}\\ \text{Therefore, the operation * is not associative}\text{.}\\ \text{Hence, the operation * is commutative, but not associative}\text{.}\\ \text{Thus, the correct answer is B}\text{.}\end{array}$

Q.56

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=10\mathrm{x}+7.\mathrm{Find}\mathrm{the}\mathrm{function}\\ \mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{go}\mathrm{f}=\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right) = 10\mathrm{x}+ 7\mathrm{.}\\ \mathrm{One}–\mathrm{one}:\\ \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right) =\mathrm{f}\left(\mathrm{y}\right),\mathrm{where}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{.}\\ ⇒10\mathrm{x}+ 7 = 10\mathrm{y}+ 7\\ ⇒\mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{function}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{y}\in \mathrm{R},\mathrm{let}\mathrm{y}=10\mathrm{x}+ 7\mathrm{.}\\ \mathrm{ }\mathrm{x}=\frac{\mathrm{y}-7}{10}\in \mathrm{R}\\ \mathrm{Then}, \mathrm{ }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{y}-7}{10}\right)\left[\mathrm{y}\in \mathrm{R}\right]\\ \\ \mathrm{ }=10\left(\frac{\mathrm{y}-7}{10}\right)+7\\ \mathrm{ }=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Therefore},\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{For}\mathrm{inverse}:\\ \mathrm{Let}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{as}\mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-7}{10}\\ \mathrm{So},\\ \mathrm{gof}\left(\mathrm{x}\right)=\mathrm{g}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=\mathrm{g}\left(10\mathrm{x}+ 7\right)=\frac{10\mathrm{x}+ 7-7}{10}=\mathrm{x}\\ \mathrm{and}\\ \mathrm{fog}\left(\mathrm{y}\right)=\mathrm{f}\left\{\mathrm{g}\left(\mathrm{y}\right)\right\}=\mathrm{f}\left(\frac{\mathrm{y}-7}{10}\right)=10\left(\frac{\mathrm{y}-7}{10}\right)+7=\mathrm{y}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{R}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{function}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-7}{10}\end{array}$

Q.57

$\begin{array}{l}\mathrm{Letf}:\mathrm{W}\to \mathrm{W}\mathrm{bede}\mathrm{fine}\mathrm{d}\mathrm{as}\mathrm{f}\left(\mathrm{n}\right)=\mathrm{n}-1,\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\mathrm{and}\\ \mathrm{f}\left(\mathrm{n}\right)=\mathrm{n}+1,\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}.\mathrm{Show}\mathrm{tha}\mathrm{tf}\mathrm{is}\mathrm{invertible}.\mathrm{Find}\\ \mathrm{the}\mathrm{inverse}\mathrm{off}.\mathrm{Here},\mathrm{W}\mathrm{is}\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{all}\mathrm{whole}\mathrm{numbers}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}:\mathrm{W}\to \mathrm{W}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{n}\right)\mathrm{=}\left\{\begin{array}{l}\mathrm{n}-1, \mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\\ \mathrm{n}+1,\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}\end{array}\\ \mathrm{One}-\mathrm{one}:\\ \mathrm{Let} \mathrm{f}\left(\mathrm{n}\right)=\mathrm{f}\left(\mathrm{m}\right)\\ ⇒\mathrm{n}-1=\mathrm{m}+1\left[\mathrm{If}\mathrm{n}\mathrm{is}\mathrm{odd}\mathrm{and}\mathrm{m}\mathrm{is}\mathrm{even}\mathrm{.}\right]\\ ⇒ \mathrm{ }\mathrm{n}-\mathrm{m}=2,\\ \mathrm{which}\mathrm{is}\mathrm{impossible}\mathrm{.}\\ \mathrm{Let} \mathrm{f}\left(\mathrm{n}\right)=\mathrm{f}\left(\mathrm{m}\right)\\ ⇒\mathrm{n}+1=\mathrm{m}-1\left[\mathrm{If}\mathrm{n}\mathrm{is}\mathrm{even}\mathrm{and}\mathrm{m}\mathrm{is}\mathrm{odd}\mathrm{.}\right]\\ ⇒ \mathrm{ }\mathrm{n}-\mathrm{m}=-2,\\ \mathrm{which}\mathrm{is}\mathrm{also}\mathrm{impossible}\mathrm{.}\\ \mathrm{Now},\mathrm{if}\mathrm{both}\mathrm{n}\mathrm{and}\mathrm{m}\mathrm{are}\mathrm{odd},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \mathrm{ }\mathrm{f}\left(\mathrm{n}\right)=\mathrm{f}\left(\mathrm{m}\right)\\ ⇒\mathrm{n}-1=\mathrm{m}-1\\ ⇒ \mathrm{ }\mathrm{n}=\mathrm{m}\\ \mathrm{Again},\mathrm{if}\mathrm{both}\mathrm{n}\mathrm{and}\mathrm{m}\mathrm{are}\mathrm{even},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \mathrm{f}\left(\mathrm{n}\right) =\mathrm{f}\left(\mathrm{m}\right)\\ ⇒\mathrm{n}+ 1=\mathrm{m}+ 1\\ ⇒ \mathrm{ }\mathrm{n}=\mathrm{m}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{any}\mathrm{odd}\mathrm{number}, 2\mathrm{r}+1\in \mathrm{N}\\ \mathrm{f}\left(2\mathrm{r}+1\right)=2\mathrm{r}+1-1\\ =2\mathrm{r}\in \mathrm{N}\\ \mathrm{For}\mathrm{any}\mathrm{even}\mathrm{number}, 2\mathrm{r}\in \mathrm{N}\\ \mathrm{ }\mathrm{f}\left(2\mathrm{r}\right)=2\mathrm{r}+1\in \mathrm{N}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{function}\mathrm{.}\\ \mathrm{Therefore},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{W}\to \mathrm{W}\mathrm{as}:\\ \mathrm{g}\left(\mathrm{m}\right)=\left\{\begin{array}{l}\mathrm{m}+1, \mathrm{if}\mathrm{m}\mathrm{is}\mathrm{even}\\ \mathrm{m}-1, \mathrm{if}\mathrm{m}\mathrm{is}\mathrm{odd}\end{array}\mathrm{}\\ \mathrm{Now},\mathrm{when}\mathrm{n}\mathrm{is}\mathrm{odd}\\ \mathrm{gοf}\left(\mathrm{n}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{n}\right)\right)\\ \mathrm{ }=\mathrm{g}\left(\mathrm{n}-1\right)\\ \mathrm{ }=\mathrm{n}-1+1\\ \mathrm{ }=\mathrm{n}\\ \mathrm{And},\mathrm{when}\mathrm{n}\mathrm{is}\mathrm{even}:\\ \mathrm{gοf}\left(\mathrm{n}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{n}\right)\right)\\ \mathrm{ }=\mathrm{g}\left(\mathrm{n}+1\right)\\ \mathrm{ }=\mathrm{n}+1-1\\ \mathrm{ }=\mathrm{n}\\ \mathrm{When}\mathrm{m}\mathrm{is}\mathrm{odd}:\\ \mathrm{gοf}\left(\mathrm{m}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{m}\right)\right)\\ \mathrm{ }=\mathrm{g}\left(\mathrm{m}-1\right)\\ \mathrm{ }=\mathrm{m}-1+1\\ \mathrm{ }=\mathrm{m}\\ \mathrm{And},\mathrm{when}\mathrm{n}\mathrm{is}\mathrm{even}:\\ \mathrm{gof}\left(\mathrm{m}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{m}\right)\right)\\ \mathrm{ }=\mathrm{g}\left(\mathrm{m}+1\right)\\ \mathrm{ }=\mathrm{m}+1-1\\ \mathrm{ }=\mathrm{m}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{W}}\mathrm{ }\mathrm{and} \mathrm{gof}={\mathrm{I}}_{\mathrm{W}}\\ \mathrm{Thus},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{given}\mathrm{by}{\mathrm{f}}^{–1}=\mathrm{g},\\ \mathrm{which}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{f}.\mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{f}\mathrm{itself}\mathrm{.}\end{array}$

Q.58

${\text{If f : R → R is defined by f(x)=x}}^{2}–3\mathrm{x}+2,\mathrm{find}\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right).$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{2}-3\mathrm{x}+2.\\ \mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{f}\left({\mathrm{x}}^{2}-3\mathrm{x}+2\right)\\ ={\left({\mathrm{x}}^{2}-3\mathrm{x}+2\right)}^{2}-3\left({\mathrm{x}}^{2}-3\mathrm{x}+2\right)+2\\ ={\mathrm{x}}^{4}+9{\mathrm{x}}^{2}+4-6{\mathrm{x}}^{3}-12\mathrm{x}+4{\mathrm{x}}^{2}-3{\mathrm{x}}^{2}+9\mathrm{x}-6+2\\ ={\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+10{\mathrm{x}}^{2}-3\mathrm{x}\end{array}$

Q.59

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{function}\mathrm{f}:\mathrm{R}\to \mathrm{\left\{}\mathrm{x}\in \mathrm{R}:–1<\mathrm{x}<1\right\}\mathrm{defined}\mathrm{by}\mathrm{}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1+|\mathrm{x}|},\mathrm{x}\in \mathrm{R}\mathrm{}\mathrm{isone}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{function}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}:\mathrm{R}\to \left\{\mathrm{x}\in \mathrm{R}:–1<\mathrm{x}<1\right\}\mathrm{isdefinedasf}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1+|\mathrm{x}|},\mathrm{x}\in \mathrm{R}.\\ \mathrm{Ler} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right),\mathrm{where}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{.}\\ ⇒ \mathrm{ }\frac{\mathrm{x}}{1+|\mathrm{x}|}=\frac{\mathrm{y}}{1+|\mathrm{y}|}\\ \mathrm{If}\mathrm{x}\mathrm{is}\mathrm{positive}\mathrm{and}\mathrm{y}\mathrm{is}\mathrm{negative},\mathrm{then}\mathrm{we}\mathrm{have}:\\ ⇒ \frac{\mathrm{x}}{1+\mathrm{x}}=\frac{\mathrm{y}}{1-\mathrm{y}}\\ ⇒\mathrm{x}\left(1-\mathrm{y}\right)=\mathrm{y}\left(1+\mathrm{x}\right)\\ ⇒ \mathrm{ }\mathrm{x}-\mathrm{xy}=\mathrm{y}+\mathrm{xy}\\ ⇒\mathrm{x}-\mathrm{y}=2\mathrm{xy}\\ \mathrm{Since}\mathrm{x}\mathrm{is}\mathrm{positive}\mathrm{and}\mathrm{y}\mathrm{is}\mathrm{negative}:\\ \mathrm{x}>\mathrm{y}⇒\mathrm{x}-\mathrm{y}>0\\ \mathrm{But}, 2\mathrm{xy}\mathrm{is}\mathrm{negative}\mathrm{.}\\ \mathrm{Then},2\mathrm{xy}\ne \mathrm{x}-\mathrm{y}\\ \mathrm{Thus},\mathrm{the}\mathrm{case}\mathrm{of}\mathrm{x}\mathrm{being}\mathrm{positive}\mathrm{and}\mathrm{y}\mathrm{being}\mathrm{negative}\mathrm{can}\mathrm{not}\\ \mathrm{be}\mathrm{considered}\mathrm{.}\\ \mathrm{Similarly},\mathrm{the}\mathrm{case}\mathrm{of}\mathrm{x}\mathrm{being}\mathrm{negative}\mathrm{and}\mathrm{y}\mathrm{being}\mathrm{positive}\\ \mathrm{also}\mathrm{can}\mathrm{not}\mathrm{be}\mathrm{considered}\mathrm{.}\\ \therefore \mathrm{x}\mathrm{and}\mathrm{y}\mathrm{have}\mathrm{to}\mathrm{be}\mathrm{either}\mathrm{positive}\mathrm{or}\mathrm{negative}\mathrm{.}\\ \mathrm{When}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{are}\mathrm{both}\mathrm{positive},\mathrm{we}\mathrm{have}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \frac{\mathrm{x}}{1+\mathrm{x}}=\frac{\mathrm{y}}{1+\mathrm{y}}\\ ⇒\mathrm{x}+\mathrm{xy}=\mathrm{y}+\mathrm{xy}\\ ⇒ \mathrm{ }\mathrm{x}=\mathrm{y}\\ \mathrm{When}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{are}\mathrm{both}\mathrm{negative},\mathrm{we}\mathrm{have}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \frac{\mathrm{x}}{1-\mathrm{x}}=\frac{\mathrm{y}}{1-\mathrm{y}}\\ ⇒\mathrm{x}-\mathrm{xy}=\mathrm{y}-\mathrm{xy}\\ ⇒ \mathrm{ }\mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}-1<\mathrm{y}< 1\mathrm{.}\\ \mathrm{If}\mathrm{y}\mathrm{is}\mathrm{negative},\mathrm{then}\mathrm{there}\mathrm{exists}\mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}\in \mathrm{R}\mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{y}}{1+\mathrm{y}}\right)=\frac{\frac{\mathrm{y}}{1+\mathrm{y}}}{1+|\frac{\mathrm{y}}{1+\mathrm{y}}|}=\frac{\frac{\mathrm{y}}{1+\mathrm{y}}}{1+\frac{-\mathrm{y}}{1+\mathrm{y}}}=\mathrm{y}\\ \mathrm{If}\mathrm{y}\mathrm{is}\mathrm{positive},\mathrm{then}\mathrm{there}\mathrm{exists}\mathrm{x}=\frac{\mathrm{y}}{1+\mathrm{y}}\in \mathrm{R}\mathrm{ }\mathrm{such}\mathrm{ }\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{y}}{1-\mathrm{y}}\right)=\frac{\frac{\mathrm{y}}{1-\mathrm{y}}}{1+|\frac{\mathrm{y}}{1-\mathrm{y}}|}=\frac{\frac{\mathrm{y}}{1-\mathrm{y}}}{1+\frac{\mathrm{y}}{1-\mathrm{y}}}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\end{array}$

Q.60

${\text{Show that the function f : R→ R given by f(x)=x}}^{3}\mathrm{is}\mathrm{injective}.$

Ans.

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right) ={\mathrm{x}}^{\mathrm{3}}\mathrm{.}\\ \mathrm{Suppose}\mathrm{f}\left(\mathrm{x}\right) =\mathrm{f}\left(\mathrm{y}\right),\mathrm{where}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{.}\\ ⇒ {\mathrm{x}}^{\mathrm{3}}={\mathrm{y}}^{\mathrm{3}}\mathrm{}...\left(1\right)\\ \mathrm{Now},\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{show}\mathrm{that}\mathrm{x}=\mathrm{y}\mathrm{.}\\ \mathrm{Suppose}\mathrm{x}\ne \mathrm{y},\mathrm{their}\mathrm{cubes}\mathrm{will}\mathrm{also}\mathrm{not}\mathrm{be}\mathrm{equal}\mathrm{.}\\ ⇒ {\mathrm{x}}^{\mathrm{3}}\ne {\mathrm{y}}^{\mathrm{3}}\\ \mathrm{However},\mathrm{it}\mathrm{will}\mathrm{be}\mathrm{a}\mathrm{contradiction}\mathrm{to}\mathrm{equation}\left(1\right)\mathrm{.}\\ \therefore \mathrm{x}=\mathrm{y}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{injective}\mathrm{.}\end{array}$

Q.61

$\begin{array}{l}\mathrm{Give}\mathrm{examples}\mathrm{of}\mathrm{two}\mathrm{functions}\mathrm{f}:\mathrm{N}\to \mathrm{Z}\mathrm{and}\mathrm{g}:\mathrm{Z}\to \mathrm{Z}\mathrm{such}\\ \mathrm{that}\mathrm{g}\mathrm{ }\mathrm{ο}\mathrm{ }\mathrm{f}\mathrm{is}\mathrm{injective}\mathrm{but}\mathrm{g}\mathrm{is}\mathrm{not}\mathrm{injective}.\\ \left(\mathrm{Hint}:\mathrm{Consider}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)=|\mathrm{x}|\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{Define}\mathrm{f}:\mathrm{N}\to \mathrm{Z}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right) =\mathrm{x}\mathrm{and}\mathrm{g}:\mathrm{Z}\to \mathrm{Z}\mathrm{as}\mathrm{g}\left(\mathrm{x}\right) =|\mathrm{x}|\\ \mathrm{Let}\mathrm{us}\mathrm{first}\mathrm{show}\mathrm{that}\mathrm{g}\mathrm{is}\mathrm{not}\mathrm{injective}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{observed}\mathrm{that}:\\ \mathrm{g}\left(-1\right) =|-1|=1\\ \mathrm{g}\left(1\right) =|1|=1\\ \therefore \mathrm{g}\left(-1\right) =\mathrm{g}\left(1\right),\mathrm{but}-1\ne \mathrm{1}\\ \therefore \mathrm{g}\mathrm{is}\mathrm{not}\mathrm{injective}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}:\mathrm{N}\to \mathrm{Z}\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{gοf}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{g}\left(\mathrm{x}\right)=|\mathrm{x}|\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{N}\mathrm{such}\mathrm{that}\mathrm{gof}\left(\mathrm{x}\right) =\mathrm{gof}\left(\mathrm{y}\right)\mathrm{.}\\ ⇒ \mathrm{ }\mathrm{x}=\mathrm{y}\\ \mathrm{Since}\mathrm{x}\mathrm{and}\mathrm{y}\in \mathrm{N},\mathrm{both}\mathrm{are}\mathrm{positive}\\ \therefore |\mathrm{x}|=|\mathrm{y}|⇒\mathrm{x}=\mathrm{y}\\ \mathrm{Hence},\mathrm{}\mathrm{gοf}\mathrm{is}\mathrm{injective}\mathrm{.}\end{array}$

Q.62

$\begin{array}{l}\mathrm{Given}\mathrm{example}\mathrm{s}\mathrm{of}\mathrm{two}\mathrm{functions}\mathrm{f}:\mathrm{N}\to \mathrm{N}\mathrm{and}\mathrm{g}:\mathrm{N}\to \mathrm{N}\mathrm{such}\\ \mathrm{that}\mathrm{go}\mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{but}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}.\\ \left(\mathrm{Hint}:\mathrm{Consider}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+1\mathrm{and}\mathrm{ }\mathrm{g}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{x}–1, \mathrm{ }\mathrm{ifx}>1\\ \mathrm{x}+1, \mathrm{ }\mathrm{ifx}<1\end{array}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Define}\mathrm{f}:\mathrm{N}\to \mathrm{N}\mathrm{by}, \mathrm{f}\left(\mathrm{x}\right) =\mathrm{x}+ 1\\ \mathrm{And},\mathrm{g}:\mathrm{N}\to \mathrm{N}\mathrm{by}, \mathrm{g}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{x}-1, \mathrm{if}\mathrm{x}>1\\ 1, \mathrm{ }\mathrm{if}\mathrm{x}=1\end{array}\\ \mathrm{We}\mathrm{first}\mathrm{show}\mathrm{that}\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}:\\ \mathrm{For}\mathrm{this},\mathrm{consider}\mathrm{element}1\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{N}.\mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{this}\\ \mathrm{element}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{image}\mathrm{of}\mathrm{any}\mathrm{of}\mathrm{the}\mathrm{elements}\mathrm{in}\mathrm{domain}\mathrm{N}\mathrm{.}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}:\mathrm{N}\to \mathrm{N}\mathrm{is}\mathrm{defined}\mathrm{by},\\ \mathrm{gοf}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \mathrm{ }=\mathrm{g}\left(\mathrm{x}+1\right)\\ \mathrm{ }=\mathrm{x}+1-1\left[\mathrm{x}\in \mathrm{N}⇒\left(\mathrm{x}+1\right)>1\right]\\ \mathrm{ }=\mathrm{x}\\ \mathrm{Then},\mathrm{it}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{for}\mathrm{y}\in \mathrm{N},\mathrm{there}\mathrm{exists}\mathrm{x}=\mathrm{y}\in \mathrm{N}\mathrm{such}\mathrm{that}\\ \mathrm{gof}\left(\mathrm{x}\right) =\mathrm{y}\mathrm{.}\\ \therefore \mathrm{ }\mathrm{gof}\mathrm{is}\mathrm{onto}\mathrm{.}\end{array}$

Q.63

$\begin{array}{l}\mathrm{Given}\mathrm{an}\mathrm{on}\mathrm{empty}\mathrm{set}\mathrm{X},\mathrm{consider}\mathrm{P}\left(\mathrm{X}\right)\mathrm{which}\mathrm{is}\mathrm{the}\mathrm{set}\mathrm{ofall}\mathrm{sub}\mathrm{sets}\mathrm{of}\mathrm{X}.\mathrm{ } \mathrm{Define}\mathrm{the}\\ \mathrm{relation}\mathrm{Rin}\mathrm{P}\left(\mathrm{X}\right)\mathrm{as}\mathrm{follows}:\mathrm{For}\mathrm{sub}\mathrm{sets}\mathrm{A},\mathrm{Bin}\mathrm{P}\left(\mathrm{X}\right),\mathrm{ARB}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}\mathrm{A}\subset \mathrm{B}.\mathrm{Is}\mathrm{Ran}\\ \mathrm{equivalence}\mathrm{relation}\mathrm{on}\mathrm{P}\left(\mathrm{X}\right)?\mathrm{Justify}\mathrm{you} \mathrm{answer}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since}\mathrm{every}\mathrm{set}\mathrm{is}\mathrm{a}\mathrm{subset}\mathrm{of}\mathrm{itself},\mathrm{ARA}\mathrm{for}\mathrm{all}\mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right)\mathrm{.}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{reflexive}.\\ \mathrm{Let}\mathrm{ARB}⇒\mathrm{A}\subset \mathrm{B}\mathrm{.}\\ \mathrm{For}\mathrm{instance},\mathrm{if}\mathrm{A}= \left\{3, 4\right\}\mathrm{and}\mathrm{B}= \left\{3, 4, 5\right\},\mathrm{then}\mathrm{it}\mathrm{cannot}\mathrm{be}\\ \mathrm{implied}\mathrm{that}\mathrm{B}\mathrm{is}\mathrm{related}\mathrm{to}\mathrm{A}\mathrm{.}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\\ \mathrm{Further},\mathrm{if}\mathrm{ARB}\mathrm{and}\mathrm{BRC},\mathrm{then}\mathrm{A}\subset \mathrm{B}\mathrm{and}\mathrm{B}\subset \mathrm{C}\mathrm{.}\\ ⇒\mathrm{A}\subset \mathrm{C}\\ ⇒\mathrm{ARC}\\ \therefore \mathrm{R}\mathrm{is}\mathrm{transitive}\mathrm{.}\\ \mathrm{Hence},\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{an}\mathrm{equivalence}\mathrm{relation}\mathrm{since}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{symmetric}\mathrm{.}\end{array}$

Q.64

$\begin{array}{l}\mathrm{Given}\mathrm{a}\mathrm{non}–\mathrm{empty}\mathrm{set}\mathrm{X},\mathrm{consider}\mathrm{the}\mathrm{binary}\mathrm{operation}*:\\ \mathrm{P}\left(\mathrm{X}\right)×\mathrm{P}\left(\mathrm{X}\right)\to \mathrm{P}\left(\mathrm{X}\right)\mathrm{given}\mathrm{by}\mathrm{A}*\mathrm{B}=\mathrm{A}\cap \mathrm{B} \forall \mathrm{A},\mathrm{Bin}\mathrm{P}\left(\mathrm{X}\right),\\ \mathrm{where}\mathrm{P}\left(\mathrm{X}\right)\mathrm{is}\mathrm{the}\mathrm{power}\mathrm{set}\mathrm{of}\mathrm{X}.\mathrm{Show}\mathrm{that}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\\ \mathrm{element}\mathrm{for}\mathrm{this}\mathrm{operation}\mathrm{and}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{invertible}\mathrm{element}\\ \mathrm{in}\mathrm{P}\left(\mathrm{X}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\mathrm{operation}*.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}\mathrm{that}:\\ \mathrm{P}\left(\mathrm{X}\right)×\mathrm{P}\left(\mathrm{X}\right)\to \mathrm{P}\left(\mathrm{X}\right)\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{A}*\mathrm{B}=\mathrm{A}\cap \mathrm{B} \mathrm{ }\forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{Since},\mathrm{A}\cap \mathrm{X}=\mathrm{A}=\mathrm{X}\cap \mathrm{A} \forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ ⇒ \mathrm{A}*\mathrm{X}=\mathrm{A}=\mathrm{X}*\mathrm{A} \forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{Thus},\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{binary}\mathrm{operation}*\mathrm{.}\\ \mathrm{Now},\mathrm{an}\mathrm{element} \mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right) \mathrm{is}\mathrm{invertible}\mathrm{if}\mathrm{there}\mathrm{exists}\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{such}\mathrm{that}\\ \mathrm{A}*\mathrm{B}=\mathrm{X}=\mathrm{B}*\mathrm{A}\left[\mathrm{As}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{.}\right]\\ \mathrm{i}.\mathrm{e}., \mathrm{A}\cap \mathrm{B}=\mathrm{X}=\mathrm{B}\cap \mathrm{A}\\ \mathrm{This}\mathrm{case}\mathrm{is}\mathrm{possible}\mathrm{only}\mathrm{when}\mathrm{A}=\mathrm{X}=\mathrm{B}\mathrm{.}\\ \mathrm{Thus},\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{only}\mathrm{invertible}\mathrm{element}\mathrm{in}\mathrm{P}\left(\mathrm{X}\right)\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{the}\\ \mathrm{given}\mathrm{operation}*\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.65

Find the number of all onto functions from the set {1, 2, 3, …, n) to itself.

Ans.

$\begin{array}{l}\text{Onto functions from the set {1, 2, 3,}\dots \text{,n} to itself is simply a}\\ \text{permutation on n symbols 1, 2,}\dots \text{, n}\text{.}\\ \text{Thus, the total number of onto maps from {1, 2,}\dots \text{, n} to itself}\\ \text{is the same as}\text{\hspace{0.17em}}\text{the total number of permutations on n symbols}\\ \text{1, 2,}\dots \text{\hspace{0.17em}}\text{, n, which is n}\text{.}\end{array}$

Q.66

$\begin{array}{l}\mathrm{Let}\mathrm{S}=\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\mathrm{and}\mathrm{T}=\left\{1,2,3\right\}.\mathrm{Find}{\mathrm{F}}^{–1}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{functions}\mathrm{F}\mathrm{from}\mathrm{S}\mathrm{to}\mathrm{T},\mathrm{if}\mathrm{it}\mathrm{exists}.\\ \left(\mathrm{i}\right)\mathrm{F}=\left\{\left(\mathrm{a},3\right),\left(\mathrm{b},2\right),\left(\mathrm{c},1\right)\right\}\mathrm{ }\left(\mathrm{ii}\right)\mathrm{F}=\left\{\left(\mathrm{a},2\right),\left(\mathrm{b},1\right),\left(\mathrm{c},1\right)\right\}\end{array}$

Ans.

$\begin{array}{l}Given:\\ S= \left\{a,b,c\right\},T= \left\{1, 2, 3\right\}\\ \left(i\right)F:S\to Tisdefinedas:\\ F= \left\{\left(a, 3\right), \left(b, 2\right), \left(c, 1\right)\right\}\\ ⇒F\left(a\right) = 3,F\left(b\right) = 2,F\left(c\right) = 1\\ Therefore,{F}^{-1}:T\to Sisgivenby\\ {F}^{-1} =\left\{\left(3,a\right), \left(2,b\right), \left(1,c\right)\right\}.\\ \left(ii\right)F:S\to Tisdefinedas:\\ F= \left\{\left(a, 2\right), \left(b, 1\right), \left(c, 1\right)\right\}\\ SinceF\left(b\right) =F\left(c\right) = 1,Fisnotone–one.\\ Therefore,Fisnotinvertiblei.e.,{F}^{–1}doesnotexist.\end{array}$

Q.67

$\begin{array}{l}\mathrm{Consider}\mathrm{the}\mathrm{binary}\mathrm{operations}*:\mathrm{R}×\mathrm{R}\to \mathrm{ando}:\mathrm{R}×\mathrm{R}\to \mathrm{R}\\ \mathrm{defined}\mathrm{as}\mathrm{ }\mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\mathrm{ }\mathrm{and}\mathrm{aοb}=\mathrm{a},\forall \mathrm{a},\mathrm{b}\in \mathrm{R}.\mathrm{Show}\mathrm{that}*\mathrm{is}\\ \mathrm{commutative}\mathrm{but}\mathrm{notassociative},\mathrm{ο}\mathrm{is}\mathrm{associative}\mathrm{but}\mathrm{not}\\ \mathrm{commutative}.\mathrm{Further},\mathrm{show}\mathrm{that} \forall \mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R},\\ \mathrm{a}*\left(\mathrm{bοc}\right)=\left(\mathrm{a}*\mathrm{b}\right)\mathrm{ο}\left(\mathrm{a}*\mathrm{c}\right).\left[\mathrm{If}\mathrm{it}\mathrm{is}\mathrm{so},\mathrm{we}\mathrm{say}\mathrm{that}\mathrm{the}\\ \mathrm{operation}*\mathrm{distributes}\mathrm{over}\mathrm{the}\mathrm{operationο}\right].\\ \mathrm{Does}\mathrm{οdistributeover}*?\mathrm{Justif}\mathrm{y}\mathrm{your}\mathrm{answer}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}*:\mathrm{R}×\mathrm{R}\to \mathrm{and}\mathrm{o}:\mathrm{R}×\mathrm{R}\to \mathrm{R}\mathrm{isdefined}\mathrm{as}\\ \mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{aοb}=\mathrm{a} \forall \mathrm{a},\mathrm{b}\in \mathrm{R}\mathrm{.}\\ \mathrm{For}\mathrm{a},\mathrm{b}\in \mathrm{R},\mathrm{we}\mathrm{have}:\\ \mathrm{ }\mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\\ \mathrm{b}*\mathrm{a}=|\mathrm{b}-\mathrm{a}|=|-\left(\mathrm{a}-\mathrm{b}\right)|=|\mathrm{a}-\mathrm{b}|\\ \therefore \mathrm{a}*\mathrm{b}=\mathrm{b}*\mathrm{a}\\ \therefore \mathrm{The}\mathrm{operation}*\mathrm{is}\mathrm{commutative}\mathrm{.}\\ \mathrm{For}\mathrm{associative},\mathrm{we}\mathrm{see}\mathrm{that}\\ \left(2*3\right)*4=\left(|2-3|\right)*4\\ =1*4\\ =|1-4|=3\\ \mathrm{ }2*\left(3*4\right)=2*|3-4|\\ =2*1\\ =|2-1|=1\\ \left(2*3\right)*4\ne \mathrm{ }2*\left(3*4\right)\left[\mathrm{where} 2,3,4\in \mathrm{R}\right]\\ \mathrm{Thus},\mathrm{the}\mathrm{operation}*\mathrm{is}\mathrm{not}\mathrm{associative}\mathrm{.}\\ \mathrm{Now},\mathrm{consider}\mathrm{the}\mathrm{operation}\mathrm{ο}\mathrm{:}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{noticed}\mathrm{that}1\mathrm{ο}2= 1\mathrm{and}2\mathrm{ο}1= 2\mathrm{.}\\ \therefore \mathrm{1}\mathrm{ο}\mathrm{2}\ne \mathrm{2}\mathrm{ο}1\left(\mathrm{where}1, 2\in \mathrm{R}\right)\\ \therefore \mathrm{The}\mathrm{operation}\mathrm{ο}\mathrm{is}\mathrm{not}\mathrm{commutative}\mathrm{.}\\ \mathrm{Let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R}.\mathrm{Then},\mathrm{we}\mathrm{have}:\\ \left(\mathrm{aοb}\right)\mathrm{οc}=\mathrm{aοc}=\mathrm{a}\\ \mathrm{aο}\left(\mathrm{bοc}\right)=\mathrm{aοb}=\mathrm{a}\\ ⇒\left(\mathrm{aοb}\right)\mathrm{οc}=\mathrm{aο}\left(\mathrm{bοc}\right)\\ \therefore \mathrm{The}\mathrm{operation}\mathrm{o}\mathrm{is}\mathrm{associative}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\mathrm{a},\mathrm{b},\mathrm{c}\in \mathrm{R},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \mathrm{ }\mathrm{a}*\left(\mathrm{bοc}\right)=\mathrm{a}*\mathrm{b}=|\mathrm{a}-\mathrm{b}|\\ \left(\mathrm{a}*\mathrm{b}\right)\mathrm{ο}\left(\mathrm{a}*\mathrm{c}\right)=\left(|\mathrm{a}-\mathrm{b}|\right)\mathrm{ο}\left(|\mathrm{a}-\mathrm{c}|\right)=|\mathrm{a}-\mathrm{b}|\\ \therefore \mathrm{ }\mathrm{a}*\left(\mathrm{bοc}\right)=\left(\mathrm{a}*\mathrm{b}\right)\mathrm{ο}\left(\mathrm{a}*\mathrm{c}\right)\\ \mathrm{Now}, 2\mathrm{ο}\left(3*4\right)=2\mathrm{ο}\left(|3-4|\right)=2\mathrm{ο}1=2\\ \mathrm{and} \mathrm{ }\left(2*3\right)\mathrm{ο}\left(2*4\right)=\left(|2-3|\right)\mathrm{ο}\left(|2-4|\right)=1\\ \therefore \mathrm{2}\mathrm{ο}\left(3*4\right)\ne \left(2*3\right)\mathrm{ο}\left(2*4\right) \left(\mathrm{where}1, 2, 3\in \mathrm{R}\right)\\ \therefore \mathrm{The}\mathrm{operation}\mathrm{ο}\mathrm{does}\mathrm{not}\mathrm{distribute}\mathrm{over}*\mathrm{.}\end{array}$

Q.68

$\begin{array}{l}\mathrm{Given}\mathrm{a}\mathrm{non}–\mathrm{empty}\mathrm{set}\mathrm{X},\mathrm{let}*:\mathrm{P}\left(\mathrm{X}\right)×\mathrm{P}\left(\mathrm{X}\right)\to \mathrm{P}\left(\mathrm{X}\right)\mathrm{be}\mathrm{defined}\mathrm{as}\\ \mathrm{A}*\mathrm{B}=\left(\mathrm{A}-\mathrm{B}\right)\cup \left(\mathrm{B}-\mathrm{A}\right),\forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right).\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{empty}\\ \mathrm{se}\mathrm{t}\varnothing \mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{for}\mathrm{the}\mathrm{operation}\mathrm{ }*\mathrm{and}\mathrm{all}\mathrm{thee}\mathrm{lements}\mathrm{A}\mathrm{of}\\ \mathrm{P}\left(\mathrm{X}\right)\mathrm{are}\mathrm{invertible}\mathrm{with}{\mathrm{A}}^{–1}=\mathrm{A}.\left(\mathrm{Hint}:\left(\mathrm{A}-\varnothing \right)\cup \left(\varnothing –\mathrm{A}\right)=\mathrm{A}\\ \mathrm{and}\left(\mathrm{A}–\mathrm{A}\right)\cup \left(\mathrm{A}–\mathrm{A}\right)=\mathrm{A}*\mathrm{A}=\varnothing \right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}*:\mathrm{P}\left(\mathrm{X}\right)×\mathrm{P}\left(\mathrm{X}\right)\to \mathrm{P}\left(\mathrm{X}\right)\mathrm{is}\mathrm{defined}\mathrm{as}\\ \mathrm{A}*\mathrm{B}= \left(\mathrm{A}-\mathrm{B}\right)\cup \left(\mathrm{B}-\mathrm{A}\right) \forall \mathrm{A},\mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\mathrm{.}\\ \mathrm{Let}\mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right).\mathrm{Then},\mathrm{we}\mathrm{have}:\\ \mathrm{A}*\varnothing =\left(\mathrm{A}-\varnothing \mathrm{\right)}\cup \mathrm{\left(}\varnothing -\mathrm{A}\right)=\mathrm{A}\cup \varnothing =\mathrm{A}\\ \varnothing *\mathrm{A}=\mathrm{\left(}\varnothing -\mathrm{A}\right)\cup \left(\mathrm{A}-\varnothing \mathrm{\right)}=\varnothing \cup \mathrm{A}=\mathrm{A}\\ \therefore \mathrm{ }\mathrm{A}*\varnothing =\mathrm{A}=\varnothing *\mathrm{A}. \forall \mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right)\\ \mathrm{Thus},\varnothing \mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{operation}*\mathrm{.}\\ \mathrm{Now},\mathrm{an}\mathrm{element}\mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right)\mathrm{will}\mathrm{be}\mathrm{invertible}\mathrm{if}\mathrm{there}\mathrm{exists}\\ \mathrm{B}\in \mathrm{P}\left(\mathrm{X}\right)\mathrm{such}\mathrm{that}\\ \mathrm{A}*\mathrm{B}=\varnothing =\mathrm{B}*\mathrm{A}.\left(\mathrm{As}\varnothing \mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\right)\\ \mathrm{Now},\mathrm{we}\mathrm{observed}\mathrm{that}\\ \mathrm{ }\mathrm{A}*\mathrm{A}=\left(\mathrm{A}-\mathrm{A}\right)\cup \left(\mathrm{A}-\mathrm{A}\right)=\varnothing \cup \varnothing =\varnothing \mathrm{ }\forall \mathrm{A}\in \mathrm{P}\left(\mathrm{X}\right).\\ \mathrm{Hence},\mathrm{all}\mathrm{the}\mathrm{elements}\mathrm{A}\mathrm{of}\mathrm{P}\left(\mathrm{X}\right)\mathrm{are}\mathrm{invertible}\mathrm{with}{\mathrm{A}}^{–1}=\mathrm{A}\mathrm{.}\\ \end{array}$

Q.69

$\begin{array}{l}\mathrm{Define}\mathrm{a}\mathrm{binary}\mathrm{operation}*\mathrm{on}\mathrm{the}\mathrm{set}\left\{0,1,2,3,4,5\right\} \mathrm{as}\\ \mathrm{a}*\mathrm{b}=\left\{\begin{array}{l}\mathrm{a}+\mathrm{b}, \mathrm{ifa}+\mathrm{b}<\mathrm{ }6\\ \mathrm{a}+\mathrm{b}–6, \mathrm{ifa}+\mathrm{b}\ge 6\end{array}\\ \mathrm{Show}\mathrm{that}\mathrm{zero}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{for}\mathrm{this}\mathrm{operation}\mathrm{and}\mathrm{each}\\ \mathrm{element}\mathrm{a}\ne 0\mathrm{of}\mathrm{these}\mathrm{t}\mathrm{is}\mathrm{invertible}\mathrm{with}6–\mathrm{a}\mathrm{being}\mathrm{the}\\ \mathrm{inverse}\mathrm{of}\mathrm{a}.\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{X}= \left\{0, 1, 2, 3, 4, 5\right\}\mathrm{.}\\ \mathrm{The}\mathrm{operation}*\mathrm{on}\mathrm{X}\mathrm{is}\mathrm{defined}\mathrm{as}:\\ \mathrm{a}*\mathrm{b}=\left\{\begin{array}{l}\mathrm{a}+\mathrm{b}, \mathrm{if}\mathrm{a}+\mathrm{b}<6\\ \mathrm{a}+\mathrm{b}-6, \mathrm{if}\mathrm{a}+\mathrm{b}\ge \mathrm{6}\end{array}\\ \mathrm{An}\mathrm{element}\mathrm{e}\in \mathrm{X}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{operation}*,\\ \mathrm{if}\mathrm{ }\mathrm{a}*\mathrm{e}=\mathrm{a}=\mathrm{e}*\mathrm{a}\forall \mathrm{a}\in \mathrm{X}.\\ \mathrm{For}\mathrm{a}\in \mathrm{X},\mathrm{we}\mathrm{see}\mathrm{that}:\\ \mathrm{a}*0=\mathrm{a}+0=\mathrm{a}\left[\mathrm{a}\in \mathrm{X}⇒\mathrm{a}+0<6\right]\\ 0*\mathrm{a}=0+\mathrm{a}=\mathrm{a}\left[\mathrm{a}\in \mathrm{X}⇒0+\mathrm{a}\ge 6\right]\\ \therefore \mathrm{a}*0=0*\mathrm{a}\forall \mathrm{a}\in \mathrm{X}\\ \mathrm{Thus}, 0\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{element}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{operation}*\mathrm{.}\\ \mathrm{An}\mathrm{element}\mathrm{a}\in \mathrm{X}\mathrm{is}\mathrm{invertible}\mathrm{if}\mathrm{there}\mathrm{exists}\mathrm{b}\in \mathrm{X}\mathrm{such}\mathrm{that}\\ \mathrm{a}*\mathrm{b}= 0 =\mathrm{b}*\mathrm{a}\mathrm{.}\\ \mathrm{i}.\mathrm{e}.,\left\{\begin{array}{l}\mathrm{a}+\mathrm{b}=0=\mathrm{b}+\mathrm{a},\mathrm{if}\mathrm{ }\mathrm{a}+\mathrm{b}<6\\ \mathrm{a}+\mathrm{b}-6=0=\mathrm{b}+\mathrm{a}-6,\mathrm{if}\mathrm{ }\mathrm{a}+\mathrm{b}\le 6\end{array}\\ ⇒\mathrm{a}=-\mathrm{b}\mathrm{or}\mathrm{b}=6-\mathrm{a}\\ \mathrm{But},\mathrm{X}= \left\{0, 1, 2, 3, 4, 5\right\}\mathrm{and}\mathrm{a},\mathrm{b}\in \mathrm{X}.\mathrm{Then},\mathrm{a}\ne -\mathrm{ }\mathrm{b}\mathrm{.}\\ \therefore \mathrm{ }\mathrm{b}= 6-\mathrm{a}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{a}\forall \mathrm{a}\in \mathrm{X}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{an}\mathrm{element}\mathrm{a}\in \mathrm{X},\mathrm{a}\ne 0\mathrm{is}6-\mathrm{a}\\ \mathrm{i}.\mathrm{e}.,{\mathrm{a}}^{-\mathrm{1}}= 6-\mathrm{a}\mathrm{.}\end{array}$

Q.70

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left\{–1,0,1,2\right\},\mathrm{B}=\left\{-4,-2,0,2\right\}\mathrm{and}\mathrm{f},\mathrm{g}:\mathrm{A}\to \mathrm{B}\mathrm{be}\\ \mathrm{functions}\mathrm{defined}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{2}-\mathrm{x},\mathrm{x}\in \mathrm{A} \mathrm{and} \\ \mathrm{g}\left(\mathrm{x}\right)=2|\mathrm{x}-\frac{1}{2}|-1,\mathrm{ }\mathrm{x}\in \mathrm{A}.\mathrm{Are}\mathrm{f}\mathrm{and}\mathrm{g}\mathrm{equal}?\\ \mathrm{Justify}\mathrm{your}\mathrm{answer}.\\ \left(\mathrm{Hint}:\mathrm{One}\mathrm{may}\mathrm{note}\mathrm{that}\mathrm{two}\mathrm{function}\mathrm{f}:\mathrm{A}\to \mathrm{B}\mathrm{and}\mathrm{g}:\mathrm{A}\to \mathrm{B}\\ \mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{a}\right)=\mathrm{g}\left(\mathrm{a}\right) \forall \mathrm{a}\in \mathrm{A},\mathrm{are}\mathrm{called}\mathrm{equal}\mathrm{functions}\right).\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{A}=\left\{-1, 0, 1, 2\right\},\mathrm{}\mathrm{B}=\left\{-\mathrm{ }4,-2,\mathrm{}0,\mathrm{}2\right\}\mathrm{.}\\ \mathrm{Also},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{f},\mathrm{g}:\mathrm{A}\to \mathrm{B} \mathrm{are}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right) ={\mathrm{x}}^{\mathrm{2}}-\mathrm{x},\mathrm{x}\in \mathrm{A}\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)=2|\mathrm{x}-\frac{1}{2}|-1,\mathrm{ }\mathrm{x}\in \mathrm{A}.\\ \mathrm{It}\mathrm{is}\mathrm{observed}\mathrm{that}:\\ \mathrm{f}\left(-1\right)={\left(-1\right)}^{\mathrm{2}}-\left(-1\right)=1+1=2\\ \mathrm{g}\left(-1\right)=2|\left(-1\right)-\frac{1}{2}|-1=2\left(\frac{3}{2}\right)-1\\ \mathrm{ }=3-1=2⇒\mathrm{f}\left(-1\right)=\mathrm{g}\left(-1\right)\\ \mathrm{f}\left(0\right)={\left(0\right)}^{2}-\left(0\right)=0\\ \mathrm{g}\left(0\right)=2|0-\frac{1}{2}|-1\\ =1-1=0\\ ⇒\mathrm{f}\left(0\right)=\mathrm{g}\left(0\right)\\ \mathrm{f}\left(1\right)={\left(1\right)}^{2}-\left(1\right)=0\\ \mathrm{g}\left(1\right)=2|1-\frac{1}{2}|-1\\ =1-1=0⇒\mathrm{f}\left(1\right)=\mathrm{g}\left(1\right)\\ \mathrm{f}\left(2\right)={\left(2\right)}^{2}-\left(2\right)=2\\ \mathrm{g}\left(2\right)=2|2-\frac{1}{2}|-1\\ =3-1=2\\ ⇒\mathrm{f}\left(2\right)=\mathrm{g}\left(2\right)\\ \therefore \mathrm{f}\left(\mathrm{a}\right)=\mathrm{g}\left(\mathrm{a}\right)\forall \mathrm{a}\in \mathrm{A}\\ \mathrm{Hence},\mathrm{the}\mathrm{functions}\mathrm{f}\mathrm{and}\mathrm{g}\mathrm{are}\mathrm{equal}\mathrm{.}\end{array}$

Q.71

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left\{1,2,3\right\}.\mathrm{Then}\mathrm{number}\mathrm{of}\mathrm{relations}\mathrm{containing}\left(1,2\right)\\ \mathrm{and}\left(1,3\right)\mathrm{which}\mathrm{are}\mathrm{reflexive}\mathrm{and}\mathrm{ }\mathrm{symmetric}\mathrm{but}\mathrm{not}\\ \mathrm{transitive}\mathrm{is}\\ \left(\mathrm{A}\right)1\left(\mathrm{B}\right)2\left(\mathrm{C}\right)3\left(\mathrm{D}\right)4\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{set}\mathrm{is}\mathrm{A}= \left\{1, 2, 3\right\}\mathrm{.}\\ \mathrm{The}\mathrm{smallest}\mathrm{relation}\mathrm{containing}\left(1, 2\right)\mathrm{and}\left(1, 3\right)\mathrm{which}\mathrm{ }\mathrm{is}\\ \mathrm{reflexive}\mathrm{and}\mathrm{symmetric},\mathrm{but}\mathrm{not}\mathrm{transitive}\mathrm{is}\mathrm{given}\mathrm{by}:\\ \mathrm{R}= \left\{\left(1, 1\right), \left(2, 2\right), \left(3, 3\right), \left(1, 2\right), \left(1, 3\right), \left(2, 1\right), \left(3, 1\right)\right\}\\ \mathrm{This}\mathrm{is}\mathrm{because}\mathrm{relation}\mathrm{R}\mathrm{is}\mathrm{reflexive}\mathrm{as}\left(1, 1\right), \left(2, 2\right), \left(3, 3\right)\in \mathrm{R}\mathrm{.}\\ \mathrm{Relation}\mathrm{R}\mathrm{is}\mathrm{symmetric}\mathrm{since}\left(1, 2\right), \left(2, 1\right)\in \mathrm{R}\mathrm{and}\\ \left(1, 3\right), \left(3, 1\right)\in \mathrm{R}.\mathrm{But}\mathrm{relation}\mathrm{R}\mathrm{is}\mathrm{not}\mathrm{transitive}\mathrm{as}\\ \left(3, 1\right), \left(1, 2\right)\in \mathrm{R},\mathrm{but}\left(3, 2\right)\notin \mathrm{R}\mathrm{.}\\ \mathrm{Now},\mathrm{if}\mathrm{we}\mathrm{add}\mathrm{any}\mathrm{two}\mathrm{pairs}\left(3, 2\right)\mathrm{and}\left(2, 3\right) \left(\mathrm{or}\mathrm{both}\right)\mathrm{to}\\ \mathrm{relation}\mathrm{R},\mathrm{then}\mathrm{relation}\mathrm{R}\mathrm{will}\mathrm{become} \mathrm{transitive}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{desired}\mathrm{relations}\mathrm{is}\mathrm{one}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.72 Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1 (B) 2 (C) 3 (D) 4

Ans.

$\begin{array}{l}\text{It is given that A = {1, 2, 3}}\text{.}\\ \text{The smallest equivalence relation containing (1, 2) is given by,}\\ {\text{R}}_{\text{1}}\text{= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}}\\ \text{Now, we are left with only four pairs i}\text{.e}\text{., (2, 3), (3, 2), (1, 3),}\\ \text{and (3, 1)}\text{.}\\ {\text{If we odd any one pair [say (2, 3)] to R}}_{\text{1}}\text{, then for symmetry}\\ \text{we must add (3, 2)}\text{. Also, for transitivity we are required to}\\ \text{add (1, 3) and (3, 1)}\text{.}\\ {\text{Hence, the only equivalence relation (bigger than R}}_{\text{1}}\text{) is the}\\ \text{universal relation}\text{.This shows that the total number of}\\ \text{equivalence relations containing (1, 2) is two}\text{.}\\ \text{The correct answer is}B\text{.}\end{array}$

Q.73

$\begin{array}{l}\mathrm{Letf}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{the}\mathrm{Signum}\mathrm{Function}\mathrm{defined}\mathrm{as}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}1, \mathrm{ifx}>0\\ 0,\mathrm{if} \mathrm{x}=0\\ –1, \mathrm{ifx}<0\end{array}\\ \mathrm{Also},\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{the}\mathrm{Greatest}\mathrm{Integer}\mathrm{Function}\mathrm{given}\mathrm{by}\\ \mathrm{g}\left(\mathrm{x}\right)=\left[\mathrm{x}\right],\mathrm{where}\left[\mathrm{x}\right]\mathrm{is}\mathrm{great}\mathrm{est}\mathrm{integer}\mathrm{ }\mathrm{less}\mathrm{than}\mathrm{o}\mathrm{requal}\\ \mathrm{tox}.\mathrm{Then}\mathrm{does}\mathrm{fog}\mathrm{and}\mathrm{go}\mathrm{f}\mathrm{co}\mathrm{incide}\mathrm{in}\left(0,1\right]?\end{array}$

Ans.

$\begin{array}{l}\mathrm{Given}:\mathrm{ }\mathrm{f}:\mathrm{R}\to \mathrm{R}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}1, \mathrm{if}\mathrm{x}>0\\ 0, \mathrm{if}â€‹ \mathrm{x}=0\\ -1, \mathrm{if}\mathrm{x}<1\end{array}\\ \mathrm{Also},\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{g}\left(\mathrm{x}\right) = \left[\mathrm{x}\right],\mathrm{where}\left[\mathrm{x}\right]\mathrm{is}\mathrm{the}\mathrm{greatest}\\ \mathrm{integer}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}\mathrm{x}.\mathrm{Now},\mathrm{let}\mathrm{x}\in \left(0, 1\right]\mathrm{.}\\ \mathrm{Then},\mathrm{we}\mathrm{have}:\\ \left[\mathrm{x}\right] = 1\mathrm{if}\mathrm{x}= 1\mathrm{and}\left[\mathrm{x}\right] = 0\mathrm{if}0 <\mathrm{x}< 1\mathrm{.}\\ \therefore \mathrm{fog}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{f}\left(\left[\mathrm{x}\right]\right)=\left\{\begin{array}{l}\mathrm{f}\left(1\right), \mathrm{if} \mathrm{x}=1\\ \mathrm{f}\left(0\right), \mathrm{if}\mathrm{ }\mathrm{x}\in \left(0,1\right)\end{array}\end{array}$