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.

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

Show that the relation R in the set R of real numbers, defined as R= { (a,b):ab2}is neithe rreflexive norsymmetric nor transitive.

Ans.

We have R = {(a, b): a b 2 } where a, bR We can see that 1 3 ( 1 3 ) 2 is not valid . So,{ 1 3 , 1 3 }R Then, R is not reflexive. Since, ( 2,3 )Rand​ 2<3 2 But 3 2 is not less than 2.

So, ( 3,2 )R

Thus, R is not symmetric.Again, let (5,6),(6,2)RAs5<(6)2 and 6<22But 5 is not less than 4.So, (5,2)RThus, R is not transitive.Therefore, R is neither reflexive nor symmetric nor transitive.

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}

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} We can find (1,1)R, where 1A. R is not reflexive. It can be observed that (1, 2)R, but (2, 1)R. R is not symmetric. Now, (1, 2), (2, 3)R but ( 1,3 )R. R is not transitive. Hence, R is neither reflexive nor symmetric nor transitive. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1F5B@

 

Q.4

Show that the relation R in Rdefined as R= {(a,b):ab}is reflexive, symmetric or transitive.

Ans.

We haveR = {(a, b); ab}We can see that (12,12)R as 12=12R is reflexive.Since (1,2)R as 2 is greater than 1.but1​ is not greater than2, so    (2,1)RR is not symmetric.Again, let (a, b), (b, c)R.Then according to condition,ab and bc  ac (a, c)R


R is transitive.

Hence,R is reflexive and transitive but not symmetric.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaqGibGaaeyzaiaab6gacaqGJbGaaeyzaiaabYcacaqGsbGaaeiiaiaabMgacaqGZbGaaeiiaiaabkhacaqGLbGaaeOzaiaabYgacaqGLbGaaeiEaiaabMgacaqG2bGaaeyzaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaaeiDaiaabkhacaqGHbGaaeOBaiaabohacaqGPbGaaeiDaiaabMgacaqG2bGaaeyzaiaabccacaqGIbGaaeyDaiaabshacaqGGaGaaeOBaiaab+gacaqG0bGaaeiiaiaabohacaqG5bGaaeyBaiaab2gacaqGLbGaaeiDaiaabkhacaqGPbGaae4yaiaab6caaaa@6878@

Q.5

Check whether the relation R in R defined as R={(a,b):ab3}is reflexive, symmetric or transitive.

Ans.

We haveR = {(a, b); ab3}We can see that 12(12)3

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.

Since, A={ 1,2,3 } A relation R on A is defined as R = {(1, 2), (2, 1)}. Since (1, 1), (2, 2), (3, 3)R. R is not reflexive. Now, as (1, 2)R and (2, 1)R, R is symmetric. Now, (1, 2) and (2, 1)R But( 1,1 )R R is not transitive. Hence, R is symmetric but neither reflexive nor transitive. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@297D@

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.

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

R is reflexive.

If (x, y)R where x and y have the same number of pages. y and x have the same number of pages. (y, x)R R is symmetric. Again, let (x, y)R and (y, z)R. x and y and have the same number of pages and y and z have the same number of pages. x and z have the same number of pages. (x, z)R Therefore, R is transitive. Since, R is reflexive, symmetric and transitive, so R is an equivalence relation. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@9507@

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.

Since,A = {1, 2, 3, 4, 5} R={ ( a,b ):| ab |iseven } For any element aA, we have | aa |=0,which is even number. R is reflexive. Let (a, b)R.


| ab |is even.

| ( ba ) |iseven. | ba |iseven. ( b,a )R So,R is symmetric. Again, let ( a,b )Rand( b,c )R. | ba |is even number and | cb |is also even number. ( ab )iseven and ( bc )is even. ( ac )=( ab )+( bc )is even. | ac |is even. ( a,c )is even. ( a,c )R R is transitive. Hence, R is an equivalence relation. Now, all elements of the set {1, 3, 5} are related to each other as the modulus of difference between any two elements of this set is even. Similarly, all elements of the set {2, 4} are related to each other as the difference between the two elements is even. Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even.The modulus of difference between odd and even is again 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Q.9

Show that each of the relation R in the set A = xZ,0x12 , given by i R= a,b : ab 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.

Here,A={ xZ,0x12 }={ 0,1,2,3,4,5,6,7,8,9,10,11,12 } ( i )R={ ( a,b ):| ab | is a multiple of 4 } For any element aA, we have ( a,a )Ras| aa |=0whichis multiple of 4. R is reflexive. Let ( a,b )Ras​ | ab |=multipleof4 and| ba |=| ( ab ) | =| ab |=multiple of 4 So, ( b,a )R Therefore,R is symmetric. Let​​( a,b ),( b,c )R | ab |=multipleof4and| cb |=multipleof4 ( a,b ) is multiple of 4and( b,a ) is multiple of 4

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

| ( ac ) | is also multiple of 4 ( a,c )R R is transitive. Hence, R is equivalence relation. Let x be an element of A such that (x,1)R Then |x1| is a multiple of 4 |x1|=0,4,8,12 x=1,5,9 [ 13 is not a part of setA ] Hence the set of all elements of A which are related to 1 is {1,5,9}. ( ii )R = {(a, b): a = b} For any element aA, we have (a, a)R, since a = a. R is reflexive. Now, let (a, b)R. a=b b=a ( b,a )R R is symmetric. Let (a, b)R and (b, c)R a = b and b = c a = c (a, c)R R is transitive. Hence, R is an equivalence relation. The elements in R that are related to 1 will be elements from set A which are equal to 1. Hence, the set of elements related to 1 is {1}. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGLbGaaeiBaiaabwgacaqGTbGaaeyzaiaab6gacaqG0bGaae4CaiaabccacaqGPbGaaeOBaiaabccacaqGsbGaaeiiaiaabshacaqGObGaaeyyaiaabshacaqGGaGaaeyyaiaabkhacaqGLbGaaeiiaiaabkhacaqGLbGaaeiBaiaabggacaqG0bGaaeyzaiaabsgacaqGGaGaaeiDaiaab+gacaqGGaGaaeymaiaabccacaqG3bGaaeyAaiaabYgacaqGSbGaaeiiaiaabkgacaqGLbGaaeiiaiaabwgacaqGSbGaaeyzaiaab2gacaqGLbGaaeOBaiaabshacaqGZbGaaeiiaaqaaiaabAgacaqGYbGaae4Baiaab2gacaqGGaGaae4CaiaabwgacaqG0bGaaeiiaiaabgeacaqGGaGaae4DaiaabIgacaqGPbGaae4yaiaabIgacaqGGaGaaeyyaiaabkhacaqGLbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqGSbGaaeiiaiaabshacaqGVbGaaeiiaiaabgdacaqGUaaabaGaaeisaiaabwgacaqGUbGaae4yaiaabwgacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGZbGaaeyzaiaabshacaqGGaGaae4BaiaabAgacaqGGaGaaeyzaiaabYgacaqGLbGaaeyBaiaabwgacaqGUbGaaeiDaiaabohacaqGGaGaaeOCaiaabwgacaqGSbGaaeyyaiaabshacaqGLbGaaeizaiaabccacaqG0bGaae4BaiaabccacaqGXaGaaeiiaiaabMgacaqGZbGaaeiiaiaabUhacaqGXaGaaeyFaiaab6caaaaa@AD29@

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.

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

Thus (P,P)R for all PA

So R is reflexive Now, Let (P, Q)R. OP=OQ OQ=OP ( Q,P )R R is symmetric. Again, let ( P,Q ),( Q,S )R OP=OQ and OQ=OS OP=OS ( P,S )R R is transitive. Thus, R is an equivalence relation. Again,Let P be a fixed point in set A and Q be a point in set A such that (P,Q)R. Then OP=OQ Q moves in the plane in such a way that its distance from the origin (0,0) is alwaysequal and is equal to OP Locus of Q is a circle with centre at the origin and radius OP. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@27A5@

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.

Since, every triangle is similar to itself. R = {(T 1 , T 2 ): T 1 is similar to T 2 } (T,T)R for all TA R is Reflexive. Let (T 1 , T 2 )R, T 1 is similar to T 2 . T 2 is similar to T 1 . (T 2 , T 1 )R R is symmetric. Now, (T 1 ,T 2 ,T 3 ) A such that (T 1 ,T 2 ) R and (T 2 ,T 3 )R T 1 is similar to T 2 and T 2 is similar to T 3 . T 1 is similar to T 3 . (T 1 , T 3 )R R is transitive. Thus, R is an equivalence relation. In Triangles T 1 and T 3 we observe that the corresponding angles are equal and thecorresponding sides are proportional i.e. 3 6 = 4 8 = 5 10 = 1 2 Hence, T 1 is related to T 3 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@20F4@

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.

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

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.

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

Since, R = {(L 1 , L 2 ): L 1 is parallel to L 2 } (L 1 ,L 1 )R as every line is parallel to itself R is reflexive. Now, Let (L 1 , L 2 )R. L 1 is parallel to L 2 . L 2 is parallel to L 1 . (L 2 , L 1 )R R is symmetric. Let (L 1 , L 2 ), (L 2 , L 3 )R. L 1 is parallel to L 2 . And L 2 is parallel to L 3 . L 1 is parallel to L 3 . R is transitive. Hence, R is an equivalence relation. The set of all lines related to the line y = 2x + 4 must be parallel line only and we know that in parallel lines’ equations only the constant value changes the co-efficient of x and y remains same set of parallel lines is y=2x+c, where c can be any constant. 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Q.15

Let R be the relation in the set {1, 2, 3, 4} given byR={(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}.

Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.(B) R is reflexive and transitive but not symmetric.(C) R is symmetric and transitive but not reflexive.(D) R is an equivalence relation.

Ans.

(B) is the correct answer    Since {(1,1),(2,2),(3,3),(4,4)}R R is reflexive    Now (1,2)R but (2,1)R R is not symmetricAnd      (1,2) and (2,2)R(1,2)R R is transitiveHence the given Relation R is reflexive and transitive but not symmetric

Q.16

Let R be the relationin theset  N  given byR={(a,b):a=b2,  b>6}.Choose the correct answer.(A)(2, 4)R  (B)(3, 8)R  (C)(6, 8)R  (D)(8, 7)R

Ans.

R = {(a, b): a = b2, b > 6}Now, since b > 6, (2, 4)ROption(A) cannot be the correct answerAlso by putting 8 we will not get 3 as difference Option (B) is also not the correct optionAnd, as 8 72 (8, 7)RNow, consider (6, 8).We have 8 > 6 and also, 6 = 82.(6, 8)RThe correct answer is option  (C).

Q.17

Show that the function f : R *R* definedby f(x)=1x  isoneoneand onto,where R*is these t of all nonzerorealnumbers. Is there sulttrue, if the domain R * is  replaced by Nwithcodomain being same as R * ?

Ans.

Let x,yR*such f(x)=f(y)oneone:f(x)=f(y)          1x=1y          x=ySo,  f:R*R*is oneone.Onto:

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

        1x=y  or  x=1yNow, f(1y)=1(1y)=ySince Range = Codomainf is onto.Thus, the given function (f) is oneone and onto.Consider function f: NR*defined byf(x)=1xFor any x,yN we see thatf(x)=f(y)    1x=1y      x=ySo f:NR* is oneone functionSince fractional numbers like 23,25etc.in codomain R*haveR*have no pre image indomain N.So,  f:NR* is not onto.

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

(ii) f: Z Z given by f(x)= x 2 (iii) f: R R given by f(x)= x 2 (iv) f: N N given by f(x)= x 3 (v) f: ZZ given by f(x)= x 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaaykW7ieqacaWFOaGaa8xAaiaa=LgacaWFPaGaa8hiaiaa=zgacaWF6aGaa8hiaiaa=PfacaWFGaGaeyOKH4Qaa8Nwaiaa=bcacaaMc8UaaGPaVlaaykW7caWFNbGaa8xAaiaa=zhacaWFLbGaa8NBaiaa=bcacaWFIbGaa8xEaiaa=bcacaWFMbGaa8hkaiaa=HhacaWFPaGaa8xpaiaa=bcacaWF4bWaaWbaaSqabeaacaWFYaaaaaGcbaGaa8hkaiaa=LgacaWFPbGaa8xAaiaa=LcacaWFGaGaa8Nzaiaa=PdacaWFGaGaa8Nuaiaa=bcacqGHsgIRcaWFsbGaa8hiaiaaykW7caaMc8Uaa83zaiaa=LgacaWF2bGaa8xzaiaa=5gacaWFGaGaa8Nyaiaa=LhacaWFGaGaa8Nzaiaa=HcacaWF4bGaa8xkaiaa=1dacaWFGaGaa8hEamaaCaaaleqabaGaa8NmaaaaaOqaaiaa=HcacaWFPbGaa8NDaiaa=LcacaWFGaGaa8Nzaiaa=PdacaWFGaGaa8Ntaiaa=bcacqGHsgIRcaWFobGaa8hiaiaaykW7caaMc8Uaa83zaiaa=LgacaWF2bGaa8xzaiaa=5gacaWFGaGaa8Nyaiaa=LhacaWFGaGaa8Nzaiaa=HcacaWF4bGaa8xkaiaa=1dacaWFGaGaa8hEamaaCaaaleqabaGaa83maaaaaOqaaiaaykW7caWFOaGaa8NDaiaa=LcacaWFGaGaa8Nzaiaa=PdacaWFGaGaa8NwaiabgkziUkaa=PfacaWFGaGaa8hiaiaa=bcacaWFNbGaa8xAaiaa=zhacaWFLbGaa8NBaiaa=bcacaWFIbGaa8xEaiaa=bcacaWFMbGaa8hkaiaa=HhacaWFPaGaa8xpaiaa=bcacaWF4bWaaWbaaSqabeaacaWFZaaaaaaaaa@ABD5@

Ans.

(i) f:NN is given by,f( x )= x 2 To check for injectivity let f( x )=f( y ) x 2 = y 2 x=y ( There are no negative natural numbers. ) f( x ) is injective. To check for onto: Let f( x )=y x 2 =y x= y f( y )=y Since range is not equal to co-domain. f(x) is not onto or surjective (ii) f: ZZ is given by, f(x)= x 2 For injectivity let f(x)=f(y) x 2 = y 2 x=±y f is not injective. For onto: Let (x)=y x 2 =y x= y f( y )=y Hence, function f is neither injective nor surjective. ( iii )f:RR is given by, f( x )= x 2 For injectivity: f( x )=f( y ) x 2 = y 2 x=±y ( x can not take more than one value for injectivity. ) fis not injective. For onto: Let f( x )=y x 2 =y x= y f( y )=y Since range is not equal to co-domain. fis not surjective. ( iv )f:NN is given by, f( x )= x 3 For injectivity: f( x )=f( y ) x 3 = y 3 x=y fis injective. For onto: Let f( x )=y x 3 =y x= y 3 f( y 3 )=y ( yNbut y 3 N ) Since range is not equal to co-domain. fis not surjective. Thus, given function is injective but not surjective. ( v )f:ZZ is given by, f( x )= x 3 For injectivity: f( x )=f( y ) x 3 = y 3 x=y fis injective. For onto: Let f( x )=y x 3 =y x= y 3 f( y 3 )=y ( yZbut y 3 Z ) Since range is not equal to co-domain. fis not surjective. Thus, given function is injective but not surjective. 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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.

Since, f:RR is given by, f( x )=[ x ] We​ see that f( 0.1 )=0,f( 0.3 )=0. f( 0.1 )=f( 0.3 ),but 0.10.3 So, f is not injective i.e., one-one. Now, let 0.6R. It is given that f( x )=[ x ]is always an integer. Thus, there does not exist any element xR such that f( x )=0.6. Then, f is not onto. Therefore, the greatest integer function is neither one-one nor onto. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6345@

Q.20

Show that the Modul us Function f :RR  given by f(x)=|x|,is neither oneonen or onto, where|x|is  x,ifx  is positiveor 0 and |x| is x, if x  is negative.

Ans.

f: R R is given by,Modulus function is defined byf(x)={x, if x0x, if x<0Here, f(1)=|1|=1,f(1)=|1|=1, f(1)=f(1),  but 11So, f is not oneone.Let 1R.Since, f(x)=|x| is always positive. So, there does not existany element in x in domain R such that f(x)=1.f(x)​ is not onto.Therefore, the modulus function is neither oneone nor onto.

Q.21

Show that the Signum Function f : R R, given byf(x)={  1,if x>0  0,if x=01,if x<0is neither oneone noronto.

Ans.

f: R R is given by,Modulus function is defined byf(x)={1, if x>0  0, if x=01, if x<0Here, f(1)=1,f(3)=1, but 13So, f is not oneone.Since, range of f is (1,0,1) for element 2 in codomain R,there does not exist any value in domain R such thatf(x)=2.f is not onto.Hence, the signum function is neither oneone nor onto.

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.

It is given that A = { 1, 2, 3 }, B={ 4, 5, 6, 7 } f:AB is defined asf = {(1, 4), (2, 5), (3, 6)}. f (1) = 4, f (2) = 5, f (3) = 6 Since every element has a unique value. The given function f is one-one. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@DD6B@

Q.23

In each of the following cases, state whether the function is oneone, onto or bijective. Justify your answer. (i) f: RR defined by f(x) = 34x (ii) f: RR defined by f(x) = 1 + x 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=LeacaWFUbGaa8hiaiaa=vgacaWFHbGaa83yaiaa=HgacaWFGaGaa83Baiaa=zgacaWFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=zgacaWFVbGaa8hBaiaa=XgacaWFVbGaa83Daiaa=LgacaWFUbGaa83zaiaa=bcacaWFJbGaa8xyaiaa=nhacaWFLbGaa83Caiaa=XcacaWFGaGaa83Caiaa=rhacaWFHbGaa8hDaiaa=vgacaWFGaGaa83Daiaa=HgacaWFLbGaa8hDaiaa=HgacaWFLbGaa8NCaiaa=bcacaWF0bGaa8hAaiaa=vgacaWFGaGaa8Nzaiaa=vhacaWFUbGaa83yaiaa=rhacaWFPbGaa83Baiaa=5gacaWFGaGaa8xAaiaa=nhacaWFGaaabaGaa83Baiaa=5gacaWFLbGaa8xlaiaa=9gacaWFUbGaa8xzaiaa=XcacaWFGaGaa83Baiaa=5gacaWF0bGaa83Baiaa=bcacaWFVbGaa8NCaiaa=bcacaWFIbGaa8xAaiaa=PgacaWFLbGaa83yaiaa=rhacaWFPbGaa8NDaiaa=vgacaWFUaGaa8hiaiaa=PeacaWF1bGaa83Caiaa=rhacaWFPbGaa8Nzaiaa=LhacaWFGaGaa8xEaiaa=9gacaWF1bGaa8NCaiaa=bcacaWFHbGaa8NBaiaa=nhacaWF3bGaa8xzaiaa=jhacaWFUaaabaGaa8hiaiaa=HcacaWFPbGaa8xkaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFsbGaeyOKH4Qaa8Nuaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5bGaa8hiaiaa=zgacaWFOaGaa8hEaiaa=LcacaWFGaGaa8xpaiaa=bcacaWFZaGaa8xlaiaa=rdacaWF4baabaGaa8hkaiaa=LgacaWFPbGaa8xkaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFsbGaeyOKH4Qaa8Nuaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5bGaa8hiaiaa=zgacaWFOaGaa8hEaiaa=LcacaWFGaGaa8xpaiaa=bcacaWFXaGaa8hiaiaa=TcacaWFGaGaa8hEamaaCaaaleqabaGaa8Nmaaaaaaaa@D15C@

Ans.

(i) f: R R is defined as f(x) = 34x. For one-one Let f(x)=f(y) 34x=34y x=y f is one-one. for onto Let f(x)=y 34x=y x= 3y 4 f( 3y 4 )=34( 3y 4 )=y Since Range = Co-domain f is onto. Hence, f is bijective. (ii) f: R R is defined as f(x)=1+x 2 For one-one Let f(x)=f(y ) 1+x 2 = 1+y 2 x 2 = y 2 x=±y f is not one-one. For onto: Let f( x )=y 1+x 2 =y x=± y1 Since Range is not equal to Co-domain. Then given function f(x) is not onto. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaabIcacaqGPbGaaeykaiaabccacaqGMbGaaeOoaiaabccacaqGsbGaaeiiaiabgkziUkaabccacaqGsbGaaeiiaiaabMgacaqGZbGaaeiiaiaabsgacaqGLbGaaeOzaiaabMgacaqGUbGaaeyzaiaabsgacaqGGaGaaeyyaiaabohacaqGGaGaaeOzaiaabIcacaqG4bGaaeykaiaabccacaqG9aGaaeiiaiaabodacqGHsislcaqG0aGaaeiEaiaab6caaeaacaqGGaGaaeOraiaab+gacaqGYbGaaeiiaiaab+gacaqGUbGaaeyzaiaab2cacaqGVbGaaeOBaiaabwgaaeaacaqGmbGaaeyzaiaabshacaqGGaGaaGPaVlaaykW7caqGMbGaaeikaiaabIhacaqGPaGaeyypa0JaaeOzaiaabIcacaqG5bGaaeykaaqaaiabgkDiElaabodacqGHsislcaqG0aGaaeiEaiabg2da9iaabodacqGHsislcaqG0aGaaeyEaaqaaaqaaiabgkDiElaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabIhacqGH9aqpcaqG5baabaGaeyinIWLaaeiiaiaabAgacaqGGaGaaeyAaiaabohacaqGGaGaae4Baiaab6gacaqGLbGaaeylaiaab+gacaqGUbGaaeyzaiaab6caaeaacaqGMbGaae4BaiaabkhacaqGGaGaae4Baiaab6gacaqG0bGaae4BaaqaaiaabYeacaqGLbGaaeiDaiaabccacaqGMbGaaeikaiaabIhacaqGPaGaaeypaiaabMhaaeaacqGHshI3caqGZaGaeyOeI0IaaeinaiaabIhacaqG9aGaaeyEaaqaaiabgkDiElaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaeiEaiaab2dadaWcaaqaaiaabodacqGHsislcaWG5baabaGaaGinaaaaaeaacqGHshI3caqGMbWaaeWaaeaadaWcaaqaaiaabodacqGHsislcaWG5baabaGaaGinaaaaaiaawIcacaGLPaaacaqG9aGaae4maiabgkHiTiaabsdadaqadaqaamaalaaabaGaae4maiabgkHiTiaadMhaaeaacaaI0aaaaaGaayjkaiaawMcaaiaab2dacaqG5baabaGaae4uaiaabMgacaqGUbGaae4yaiaabwgacaqGGaGaaeOuaiaabggacaqGUbGaae4zaiaabwgacaqGGaGaaeypaiaabccacaqGdbGaae4Baiaab2cacaqGKbGaae4Baiaab2gacaqGHbGaaeyAaiaab6gaaeaacqGH0icxcaqGMbGaaeiiaiaabMgacaqGZbGaaeiiaiaab+gacaqGUbGaaeiDaiaab+gacaqGUaaabaGaaeisaiaabwgacaqGUbGaae4yaiaabwgacaqGSaGaaeiiaiaabAgacaqGGaGaaeyAaiaabohacaqGGaGaaeOyaiaabMgacaqGQbGaaeyzaiaabogacaqG0bGaaeyAaiaabAhacaqGLbGaaeOlaaqaaaqaaiaabIcacaqGPbGaaeyAaiaabMcacaqGGaGaaeOzaiaabQdacaqGGaGaaeOuaiabgkziUkaabkfacaqGGaGaaeyAaiaabohacaqGGaGaaeizaiaabwgacaqGMbGaaeyAaiaab6gacaqGLbGaaeizaiaabccacaqGHbGaae4CaiaabccacaqGMbGaaeikaiaabIhacaqGPaGaaeypaiaabgdacaqGRaGaaeiEamaaCaaaleqabaGaaeOmaaaaaOqaaiaabAeacaqGVbGaaeOCaiaabccacaqGVbGaaeOBaiaabwgacaqGTaGaae4Baiaab6gacaqGLbaabaGaaeitaiaabwgacaqG0bGaaeiiaiaabAgacaqGOaGaaeiEaiaabMcacaqG9aGaaeOzaiaabIcacaqG5bGaaeiiaiaabMcaaeaacqGHshI3caqGXaGaae4kaiaabIhadaahaaWcbeqaaiaabkdaaaGccqGH9aqpcaqGXaGaae4kaiaabMhadaahaaWcbeqaaiaabkdaaaaakeaacqGHshI3caWLjaGaaGPaVlaaykW7caqG4bWaaWbaaSqabeaacaqGYaaaaOGaeyypa0JaaeyEamaaCaaaleqabaGaaeOmaaaaaOqaaiabgkDiElaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaeiEaiabg2da9iabgglaXkaabMhaaeaacqGH0icxcaqGGaGaaeOzaiaabccacaqGPbGaae4CaiaabccacaqGUbGaae4BaiaabshacaqGGaGaae4Baiaab6gacaqGLbGaaeylaiaab+gacaqGUbGaaeyzaiaab6caaeaacaqGgbGaae4BaiaabkhacaqGGaGaae4Baiaab6gacaqG0bGaae4BaiaabQdaaeaacaqGmbGaaeyzaiaabshacaqGGaGaaGPaVlaabAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaWG5baabaGaeyO0H4TaaGPaVlaaykW7caaMc8UaaeymaiaabUcacaqG4bWaaWbaaSqabeaacaqGYaaaaOGaeyypa0JaaeyEaaqaaiabgkDiElaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caqG4bGaeyypa0JaeyySae7aaOaaaeaacaWG5bGaeyOeI0IaaGymaaWcbeaaaOqaaiaabofacaqGPbGaaeOBaiaabogacaqGLbGaaeiiaiaabkfacaqGHbGaaeOBaiaabEgacaqGLbGaaeiiaiaabMgacaqGZbGaaeiiaiaab6gacaqGVbGaaeiDaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiBaiaabccacaqG0bGaae4BaiaabccacaqGdbGaae4Baiaab2cacaqGKbGaae4Baiaab2gacaqGHbGaaeyAaiaab6gacaqGUaaabaGaaeivaiaabIgacaqGLbGaaeOBaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGaaeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGMbGaaeikaiaabIhacaqGPaGaaeiiaiaabMgacaqGZbGaaeiiaiaab6gacaqGVbGaaeiDaiaabccacaqGVbGaaeOBaiaabshacaqGVbGaaeOlaaaaaa@EF5D@

 

Q.24

Let A and B be sets. Show that f : A×BB×A such thatf(a,b)=f(b,a) is bijective function.

Ans.

f: A×BB× A is defined as f(a, b) = (b, a). Let (a1,  b1),(a2,  b2)A×B such that f(a1,  b1)=f(a2,  b2)(b1,  a1)=(b2,  a2)b1=b2anda1=a2(a1,b1)=(a2,b2)f is oneone.Now, let (b, a)B×A be any element.Then, there exists (a, b)A×B such that f(a, b) = (b, a). [By definition of f]f is onto.Hence, f is bijective.

Q.25

Let f: NN be defined by f( n )={ n+1 2 , if n is odd n 2 , if n is even for all nN. State whether the function f is bijective. Justify your answer. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=XeacaWFLbGaa8hDaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFobGaeyOKH4Qaa8Ntaiaa=bcacaWFIbGaa8xzaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5baabaGaa8NzamaabmaabaGaa8NBaaGaayjkaiaawMcaaiaa=1dadaGabaabaeqabaWaaSaaaeaacaWFUbGaa83kaiaa=fdaaeaacaWFYaaaaiaa=XcacaWLjaGaa8xAaiaa=zgacaWFGaGaa8NBaiaa=bcacaWFPbGaa83Caiaa=bcacaWFVbGaa8hzaiaa=rgaaeaadaWcaaqaaiaa=5gaaeaacaWFYaaaaiaa=XcacaWLjaGaaCzcaiaa=LgacaWFMbGaa8hiaiaa=5gacaWFGaGaa8xAaiaa=nhacaWFGaGaa8xzaiaa=zhacaWFLbGaa8NBaaaacaGL7baacaaMc8UaaGPaVlaaykW7caaMc8Uaa8Nzaiaa=9gacaWFYbGaa8hiaiaa=fgacaWFSbGaa8hBaiaa=bcacaWFUbGaeyicI4Saa8Ntaiaa=5caaeaacaWFtbGaa8hDaiaa=fgacaWF0bGaa8xzaiaa=bcacaWF3bGaa8hAaiaa=vgacaWF0bGaa8hAaiaa=vgacaWFYbGaa8hiaiaa=rhacaWFObGaa8xzaiaa=bcacaWFMbGaa8xDaiaa=5gacaWFJbGaa8hDaiaa=LgacaWFVbGaa8NBaiaa=bcacaWFMbGaa8hiaiaa=LgacaWFZbGaa8hiaiaa=jgacaWFPbGaa8NAaiaa=vgacaWFJbGaa8hDaiaa=LgacaWF2bGaa8xzaiaa=5cacaWFGaaabaGaa8Nsaiaa=vhacaWFZbGaa8hDaiaa=LgacaWFMbGaa8xEaiaa=bcacaWF5bGaa83Baiaa=vhacaWFYbGaa8hiaiaa=fgacaWFUbGaa83Caiaa=DhacaWFLbGaa8NCaiaa=5caaaaa@B686@

Ans.

Let f: NN is defined as f( n )={ n+1 2 , if n is odd n 2 , if n is even for all nN. It can be observed that: f( 1 )= 1+1 2 =1,f( 2 )= 2 2 =1 [ By definition of f ] f( 1 )=f( 2 ),where12. f is not one-one. Consider a natural number (n) in co-domain N. Case I: n is odd n = 2m + 1 for some mN. Then,there exists 4m + 1N such that f( 4m+1 )= 4m+1+1 2 =2m+1 CaseII: n is even n=2mfor some mN. Then,there exists 4mN such that f( 4m )= 4m 2 =2m f is onto. Hence, f is not a bijective function. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqaaiaa=XeacaWFLbGaa8hDaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFobGaeyOKH4Qaa8Ntaiaa=bcacaWFPbGaa83Caiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=fgacaWFZbaabaaabaGaa8NzamaabmaabaGaa8NBaaGaayjkaiaawMcaaiaa=1dadaGabaabaeqabaWaaSaaaeaacaWFUbGaa83kaiaa=fdaaeaacaWFYaaaaiaa=XcacaWLjaGaa8xAaiaa=zgacaWFGaGaa8NBaiaa=bcacaWFPbGaa83Caiaa=bcacaWFVbGaa8hzaiaa=rgaaeaadaWcaaqaaiaa=5gaaeaacaWFYaaaaiaa=XcacaWLjaGaaCzcaiaa=LgacaWFMbGaa8hiaiaa=5gacaWFGaGaa8xAaiaa=nhacaWFGaGaa8xzaiaa=zhacaWFLbGaa8NBaaaacaGL7baacaaMc8UaaGPaVlaaykW7caaMc8Uaa8Nzaiaa=9gacaWFYbGaa8hiaiaa=fgacaWFSbGaa8hBaiaa=bcacaWFUbGaeyicI4Saa8Ntaiaa=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@D481@

 

Q.26

Let A = R{3} and B = R{1}. Consider the function f: AB defined byf( x )= x2 x3 . Is f oneone and onto? Justify your answer. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=XeacaWFLbGaa8hDaiaa=bcacaWFbbGaa8hiaiaa=1dacaWFGaGaa8NuaiabgkHiTiaa=ThacaWFZaGaa8xFaiaa=bcacaWFHbGaa8NBaiaa=rgacaWFGaGaa8Nqaiaa=bcacaWF9aGaa8hiaiaa=jfacqGHsislcaWF7bGaa8xmaiaa=1hacaWFUaGaa8hiaiaa=neacaWFVbGaa8NBaiaa=nhacaWFPbGaa8hzaiaa=vgacaWFYbGaa8hiaiaa=rhacaWFObGaa8xzaiaa=bcacaWFMbGaa8xDaiaa=5gacaWFJbGaa8hDaiaa=LgacaWFVbGaa8NBaiaa=bcaaeaacaWFMbGaa8Noaiaa=bcacaWFbbGaeyOKH4Qaa8Nqaiaa=bcacaWFKbGaa8xzaiaa=zgacaWFPbGaa8NBaiaa=vgacaWFKbGaa8hiaiaa=jgacaWF5bGaaGPaVlaa=zgadaqadaqaaiaa=HhaaiaawIcacaGLPaaacaWF9aWaaSaaaeaacaWF4bGaa8xlaiaa=jdaaeaacaWF4bGaa8xlaiaa=ndaaaGaa8Nlaiaa=bcacaWFjbGaa83Caiaa=bcacaWFMbGaa8hiaiaa=9gacaWFUbGaa8xzaiabgkHiTiaa=9gacaWFUbGaa8xzaiaa=bcacaWFHbGaa8NBaiaa=rgacaWFGaGaa83Baiaa=5gacaWF0bGaa83Baiaa=9dacaWFGaaabaGaa8Nsaiaa=vhacaWFZbGaa8hDaiaa=LgacaWFMbGaa8xEaiaa=bcacaWF5bGaa83Baiaa=vhacaWFYbGaa8hiaiaa=fgacaWFUbGaa83Caiaa=DhacaWFLbGaa8NCaiaa=5caaaaa@A35D@

Ans.

A = R{3} and B = R{1} f: AB is defined as f( x )= x2 x3 Let x,yA such that f( x )=f( y ). x2 x3 = y2 y3 ( x2 )( y3 )=( x3 )( y2 ) xy 3x2y+ 6 = xy 2x3y+ 6 3x2y=2x3y x=y f if one-one. yB=R-{ 1 }.Then,y1 The function f is onto if there exists xA such that f(x) = y x2 x3 =y x2=y( x3 ) x2=yx3y xyx=23y x( 1y )=23y x= 23y ( 1y ) A [ y1 ] Thus, for any yB, there exists 23y ( 1y ) Asuchthat f( 23y 1y )= 23y ( 1y ) 2 23y ( 1y ) 3 = 23y2( 1y ) 1y 23y3( 1y ) 1y = 23y2+2y 23y3+3y = y 1 f( 23y 1y )=y f is onto. Hence, function f is one-one and onto. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqaaiaa=feacaWFGaGaa8xpaiaa=bcacaWFsbGaeyOeI0Iaa83Eaiaa=ndacaWF9bGaa8hiaiaa=fgacaWFUbGaa8hzaiaa=bcacaWFcbGaa8hiaiaa=1dacaWFGaGaa8NuaiabgkHiTiaa=ThacaWFXaGaa8xFaaqaaiaabAgacaqG6aGaaeiiaiaabgeacqGHsgIRcaqGcbGaaeiiaiaabMgacaqGZbGaaeiiaiaabsgacaqGLbGaaeOzaiaabMgacaqGUbGaaeyzaiaabsgacaqGGaGaaeyyaiaabohacaqGGaGaaeOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiEaiabgkHiTiaaikdaaeaacaWG4bGaeyOeI0IaaG4maaaaaeaacaWGmbGaamyzaiaadshacaqGGaGaaeiEaiaabYcacaqG5bGaeyicI4SaaeyqaiaabccacaqGZbGaaeyDaiaabogacaqGObGaaeiiaiaabshacaqGObGaaeyyaiaabshacaqGGaaabaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaWGMbWaaeWaaeaacaWG5baacaGLOaGaayzkaaGaaiOlaaqaaiabgkDiElaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWG4bGaeyOeI0IaaGOmaaqaaiaadIhacqGHsislcaaIZaaaaiabg2da9maalaaabaGaamyEaiabgkHiTiaaikdaaeaacaWG5bGaeyOeI0IaaG4maaaaaeaacqGHshI3caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaeWaaeaacaWG4bGaeyOeI0IaaGOmaaGaayjkaiaawMcaamaabmaabaGaamyEaiabgkHiTiaaiodaaiaawIcacaGLPaaacqGH9aqpdaqadaqaaiaadIhacqGHsislcaaIZaaacaGLOaGaayzkaaWaaeWaaeaacaWG5bGaeyOeI0IaaGOmaaGaayjkaiaawMcaaaqaaiabgkDiEpaaKqaabaGaamiEaiaadMhaaaGaeyOeI0IaaG4maiaadIhacqGHsislcaaIYaGaamyEaiabgUcaRmaaKqaabaGaaGOnaaaacqGH9aqpdaajeaqaaiaadIhacaWG5baaaiabgkHiTiaaikdacaWG4bGaeyOeI0IaaG4maiaadMhacqGHRaWkdaajeaqaaiaaiAdaaaaabaGaeyO0H4TaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabgkHiTiaaiodacaWG4bGaeyOeI0IaaGOmaiaadMhacqGH9aqpcqGHsislcaaIYaGaamiEaiabgkHiTiaaiodacaWG5baabaGaeyO0H4TaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG4bGaeyypa0JaamyEaaqaaiabgsJiCjaaykW7caWGMbGaaeiiaiaabMgacaqGMbGaaeiiaiaab+gacaqGUbGaaeyzaiaab2cacaqGVbGaaeOBaiaabwgacaqGUaaabaGaaGPaVlaaykW7caqG5bGaeyicI4SaaeOqaiabg2da9iaabkfacaqGTaWaaiWaaeaacaaIXaaacaGL7bGaayzFaaGaaiOlaiaaykW7caWGubGaamiAaiaadwgacaWGUbGaaiilaiaaykW7caWG5bGaeyiyIKRaaGymaaqaaiaabsfacaqGObGaaeyzaiaabccacaqGMbGaaeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGMbGaaeiiaiaabMgacaqGZbGaaeiiaiaab+gacaqGUbGaaeiDaiaab+gacaqGGaGaaeyAaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeOCaiaabwgacaqGGaGaaeyzaiaabIhacaqGPbGaae4CaiaabshacaqGZbGaaeiiaiaabIhacqGHiiIZcaqGbbGaaeiiaiaabohacaqG1bGaae4yaiaabIgacaqGGaGaaeiDaiaabIgacaqGHbGaaeiDaiaabccaaeaacaWLjaGaaCzcaiaaykW7caaMc8UaaeOzaiaabIcacaqG4bGaaeykaiaabccacaqG9aGaaeiiaiaabMhaaeaacqGHshI3caWLjaGaaCzcamaalaaabaGaamiEaiabgkHiTiaaikdaaeaacaWG4bGaeyOeI0IaaG4maaaacqGH9aqpcaWG5baabaGaeyO0H4TaaCzcaiaaxMaacaWG4bGaeyOeI0IaaGOmaiabg2da9iaadMhadaqadaqaaiaadIhacqGHsislcaaIZaaacaGLOaGaayzkaaaabaGaeyO0H4TaaCzcaiaaxMaacaWG4bGaeyOeI0IaaGOmaiabg2da9iaadMhacaWG4bGaeyOeI0IaaG4maiaadMhaaeaacqGHshI3caWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGHsislcaWG5bGaamiEaiabg2da9iaaikdacqGHsislcaaIZaGaamyEaaqaaiabgkDiElaaxMaacaaMc8UaaGPaVlaaykW7caWG4bWaaeWaaeaacaaIXaGaeyOeI0IaamyEaaGaayjkaiaawMcaaiabg2da9iaaikdacqGHsislcaaIZaGaamyEaaqaaiabgkDiElaaxMaacaaMc8UaaGPaVlaaykW7caWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamiEaiabg2da9maalaaabaGaaGOmaiabgkHiTiaaiodacaWG5baabaWaaeWaaeaacaaIXaGaeyOeI0IaamyEaaGaayjkaiaawMcaaaaacqGHiiIZcaWGbbGaaCzcaiaaxMaadaWadaqaaiaadMhacqGHGjsUcaaIXaaacaGLBbGaayzxaaaabaGaaeivaiaabIgacaqG1bGaae4CaiaabYcacaqGGaGaaeOzaiaab+gacaqGYbGaaeiiaiaabggacaqGUbGaaeyEaiaabccacaqG5bGaeyicI4SaaeOqaiaabYcacaqGGaGaaeiDaiaabIgacaqGLbGaaeOCaiaabwgacaqGGaGaaeyzaiaabIhacaqGPbGaae4CaiaabshacaqGZbGaaGPaVpaalaaabaGaaGOmaiabgkHiTiaaiodacaWG5baabaWaaeWaaeaacaaIXaGaeyOeI0IaamyEaaGaayjkaiaawMcaaaaacqGHiiIZcaWGbbGaaGPaVlaadohacaWG1bGaam4yaiaadIgacaaMc8UaaGPaVlaadshacaWGObGaamyyaiaadshaaeaaaeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamOzamaabmaabaWaaSaaaeaacaaIYaGaeyOeI0IaaG4maiaadMhaaeaacaaIXaGaeyOeI0IaamyEaaaaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaamaalaaabaGaaGOmaiabgkHiTiaaiodacaWG5baabaWaaeWaaeaacaaIXaGaeyOeI0IaamyEaaGaayjkaiaawMcaaaaacqGHsislcaaIYaaabaWaaSaaaeaacaaIYaGaeyOeI0IaaG4maiaadMhaaeaadaqadaqaaiaaigdacqGHsislcaWG5baacaGLOaGaayzkaaaaaiabgkHiTiaaiodaaaaabaGaaCzcaiaaxMaacaWLjaGaeyypa0ZaaSaaaeaadaWcaaqaaiaaikdacqGHsislcaaIZaGaamyEaiabgkHiTiaaikdadaqadaqaaiaaigdacqGHsislcaWG5baacaGLOaGaayzkaaaabaGaaGymaiabgkHiTiaadMhaaaaabaWaaSaaaeaacaaIYaGaeyOeI0IaaG4maiaadMhacqGHsislcaaIZaWaaeWaaeaacaaIXaGaeyOeI0IaamyEaaGaayjkaiaawMcaaaqaaiaaigdacqGHsislcaWG5baaaaaaaeaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiaaikdacqGHsislcaaIZaGaamyEaiabgkHiTiaaikdacqGHRaWkcaaIYaGaamyEaaqaaiaaikdacqGHsislcaaIZaGaamyEaiabgkHiTiaaiodacqGHRaWkcaaIZaGaamyEaaaaaeaacaWLjaGaaCzcaiaaxMaacqGH9aqpdaWcaaqaaiabgkHiTiaadMhaaeaacqGHsislcaaIXaaaaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamOzamaabmaabaWaaSaaaeaacaaIYaGaeyOeI0IaaG4maiaadMhaaeaacaaIXaGaeyOeI0IaamyEaaaaaiaawIcacaGLPaaacqGH9aqpcaWG5baabaGaeyinIWLaaGPaVlaadAgacaqGGaGaaeyAaiaabohacaqGGaGaae4Baiaab6gacaqG0bGaae4Baiaab6caaeaacaqGibGaaeyzaiaab6gacaqGJbGaaeyzaiaabYcacaqGGaGaaeOzaiaabwhacaqGUbGaae4yaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaaeOzaiaabccacaqGPbGaae4CaiaabccacaqGVbGaaeOBaiaabwgacaqGTaGaae4Baiaab6gacaqGLbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGVbGaaeOBaiaabshacaqGVbGaaeOlaaaaaa@CC9F@

Q.27

Let:RR be defined as f(x)=x4. Choose the correctanswer.(A) f is oneone onto            (B)fismanyone onto.(C )f is oneone but not onto  (D)f is n either oneone noronto.

Ans.

f: RR is defined as f(x)=x4.Let x,yR such that f(x)=f(y).Then,    f(x)=f(y)        x4=y4        x=±y          f(x)=f(y)butxySo, f is not oneone.Consider an element 3 in codomain R. It is clear that there does not exist any x in domain R such that f(x) = 3.f is not onto.Hence, function f is neither oneone nor onto.The correct answer is D.

Q.28

Let f:RR be defined asf(x)=3x.Choose the correctanswer.(A)f  isoneone onto       (B)f  is manyone on to(C)f  isoneone but not onto (D)fis neither oneone nor onto.

Ans.

f:RRbedefinedasf(x)=3x.Let x, yR such that f(x) = f(y).3x=3y        x=yf is oneone.Also, for any real number (y) in codomain R, there exists y3  in R such that  f(y3)=3(y3)=yf is onto.Therefore, function f is oneone and onto.The correct answer is A.

Q.29

Let f: {1,3,4}{1, 2, 5} and g:{1, 2, 5}{1, 3}be givenby f={(1, 2),(3, 5),(4, 1)} and g={(1, 3), (2, 3), (5,1)}.Write down go f.

Ans.

Let f: {1,3,4}{1, 2, 5} and g:{1, 2, 5}{1, 3}be givenby f={(1, 2),(3, 5),(4, 1)} and g={(1, 3), (2, 3), (5,1)}.Write down go f.

Q.30

Let f: {1,3,4}{1, 2, 5} and g:{1, 2, 5}{1, 3}be givenby f={(1, 2),(3, 5),(4, 1)} and g={(1, 3), (2, 3), (5,1)}.Write down go f.

Ans.

Let ( ( f+g )oh )( x )=( foh )( x )+( goh )( x ) = f{ h( x ) }+g{ h( x ) } = ( foh )( x )+( goh )( x ) = { ( f+g )oh }( x ) = { ( foh )+( goh ) }( x ) xR Hence, ( f+g )oh=( foh )+( goh ). Now, { ( f.g )oh }( x )=( f.g )( h( x ) ) =f( h( x ) ).g( h( x ) ) =( foh )( x ).( goh )( x ) ={ ( foh ).( goh ) }( x ) { ( f.g )oh }( x )={ ( foh ).( goh ) }( x ) xR Hence, ( f.g )oh=( foh ).( goh ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6FA4@

Q.31

Find gof and fog, if (i)f( x )=| x |andg( x )=| 5x2 | ( ii )f( x )=8 x 3 andg( x )= x 1 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=zeacaWFPbGaa8NBaiaa=rgacaWFGaGaa83zaiaa=9gacaWFMbGaa8hiaiaa=fgacaWFUbGaa8hzaiaa=bcacaWFMbGaa83Baiaa=DgacaWFSaGaa8hiaiaa=LgacaWFMbaabaGaa8hkaiaa=LgacaWFPaGaaGPaVlaaykW7caWFMbWaaeWaaeaacaWF4baacaGLOaGaayzkaaGaa8xpamaaemaabaGaa8hEaaGaay5bSlaawIa7aiaaykW7caWFHbGaa8NBaiaa=rgacaaMc8Uaa83zamaabmaabaGaa8hEaaGaayjkaiaawMcaaiaa=1dadaabdaqaaiaa=vdacaWF4bGaa8xlaiaa=jdaaiaawEa7caGLiWoaaeaadaqadaqaaiaa=LgacaWFPbaacaGLOaGaayzkaaGaaGPaVlaa=zgadaqadaqaaiaa=HhaaiaawIcacaGLPaaacaWF9aGaa8hoaiaa=HhadaahaaWcbeqaaiaa=ndaaaGccaaMc8UaaGPaVlaa=fgacaWFUbGaa8hzaiaaykW7caaMc8Uaa83zamaabmaabaGaa8hEaaGaayjkaiaawMcaaiaa=1dacaWF4bWaaWbaaSqabeaadaWcaaqaaiaa=fdaaeaacaWFZaaaaaaaaaaa@82A6@

Ans.

( i )f( x )=| x |andg( x )=| 5x2 | ( gof )( x )=g( f( x ) ) =g( | x | ) =| 5| x |2 | ( fog )( x )=f( g( x ) ) =f( | 5x2 | ) =| | 5x2 | | =| 5x2 | ( ii )f( x )=8 x 3 andg( x )= x 1 3 ( gof )( x )=g( f( x ) ) =g( 8 x 3 ) = ( 8 x 3 ) 1 3 =2x ( fog )( x )=f( g( x ) ) =f( x 1 3 ) =8 ( x 1 3 ) 3 =8x MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaamaabmaabaGaamyAaaGaayjkaiaawMcaaiaaykW7caaMc8UaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9maaemaabaGaamiEaaGaay5bSlaawIa7aiaaykW7caWGHbGaamOBaiaadsgacaaMc8UaaGPaVlaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaabdaqaaiaaiwdacaWG4bGaeyOeI0IaaGOmaaGaay5bSlaawIa7aaqaaiabgsJiCnaabmaabaGaam4zaiaad+gacaWGMbaacaGLOaGaayzkaaWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0Jaam4zamaabmaabaGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaaGaayjkaiaawMcaaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpcaWGNbWaaeWaaeaadaabdaqaaiaadIhaaiaawEa7caGLiWoaaiaawIcacaGLPaaaaeaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaaqWaaeaacaaI1aWaaqWaaeaacaWG4baacaGLhWUaayjcSdGaeyOeI0IaaGOmaaGaay5bSlaawIa7aaqaaiaaykW7caaMc8UaaGPaVlaaykW7daqadaqaaiaadAgacaWGVbGaam4zaaGaayjkaiaawMcaamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaaaiaawIcacaGLPaaaaeaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0JaamOzamaabmaabaWaaqWaaeaacaaI1aGaamiEaiabgkHiTiaaikdaaiaawEa7caGLiWoaaiaawIcacaGLPaaaaeaaaeaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaaqWaaeaadaabdaqaaiaaiwdacaWG4bGaeyOeI0IaaGOmaaGaay5bSlaawIa7aaGaay5bSlaawIa7aaqaaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaabdaqaaiaaiwdacaWG4bGaeyOeI0IaaGOmaaGaay5bSlaawIa7aaqaamaabmaabaGaamyAaiaadMgaaiaawIcacaGLPaaacaaMc8UaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaiIdacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaaGPaVlaaykW7caWGHbGaamOBaiaadsgacaaMc8UaaGPaVlaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaWG4bWaaWbaaSqabeaadaWcaaqaaiaaigdaaeaacaaIZaaaaaaaaOqaaiabgsJiCnaabmaabaGaam4zaiaad+gacaWGMbaacaGLOaGaayzkaaWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0Jaam4zamaabmaabaGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaaGaayjkaiaawMcaaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpcaWGNbWaaeWaaeaacaaI4aGaamiEamaaCaaaleqabaGaaG4maaaaaOGaayjkaiaawMcaaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaqadaqaaiaaiIdacaWG4bWaaWbaaSqabeaacaaIZaaaaaGccaGLOaGaayzkaaWaaWbaaSqabeaadaWcaaqaaiaaigdaaeaacaaIZaaaaaaaaOqaaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpcaaIYaGaamiEaaqaaiaaykW7caaMc8UaaGPaVlaaykW7daqadaqaaiaadAgacaWGVbGaam4zaaGaayjkaiaawMcaamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaaaiaawIcacaGLPaaaaeaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0JaamOzamaabmaabaGaamiEamaaCaaaleqabaWaaSaaaeaacaaIXaaabaGaaG4maaaaaaaakiaawIcacaGLPaaaaeaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0JaaGioamaabmaabaGaamiEamaaCaaaleqabaWaaSaaaeaacaaIXaaabaGaaG4maaaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaiodaaaaakeaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0JaaGioaiaadIhaaaaa@39DE@

Q.32

If f( x )= 4x+3 6x4 ,x 2 3 , show that fof( x )=x,for all x 2 3 . What is the inverse of f? MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=LeacaWFMbGaa8hiaiaa=zgadaqadaqaaiaa=HhaaiaawIcacaGLPaaacaWF9aWaaSaaaeaacaWF0aGaa8hEaiaa=TcacaWFZaaabaGaa8Nnaiaa=HhacaWFTaGaa8hnaaaacaWFSaGaaGPaVlaa=HhacqGHGjsUdaWcaaqaaiaa=jdaaeaacaWFZaaaaiaa=XcacaWFGaGaa83Caiaa=HgacaWFVbGaa83Daiaa=bcacaWF0bGaa8hAaiaa=fgacaWF0bGaa8hiaiaa=zgacaWFVbGaa8NzamaabmaabaGaa8hEaaGaayjkaiaawMcaaiaa=1dacaWF4bGaa8hlaiaaykW7caWFMbGaa83Baiaa=jhacaWFGaGaa8xyaiaa=XgacaWFSbGaa8hiaiaa=HhacqGHGjsUdaWcaaqaaiaa=jdaaeaacaWFZaaaaiaa=5caaeaacaWFxbGaa8hAaiaa=fgacaWF0bGaa8hiaiaa=LgacaWFZbGaa8hiaiaa=rhacaWFObGaa8xzaiaa=bcacaWFPbGaa8NBaiaa=zhacaWFLbGaa8NCaiaa=nhacaWFLbGaa8hiaiaa=9gacaWFMbGaa8hiaiaa=zgacaWF=aaaaaa@803D@

Ans.

It is given thatf( x )= 4x+3 6x4 ,then fof( x )=f( f( x ) ) =f( 4x+3 6x4 ) = 4( 4x+3 6x4 )+3 6( 4x+3 6x4 )4 = ( 16x+12+18x12 6x4 ) ( 24x+1824x+16 6x4 ) = 34x 34 fof( x )=x, for all x 2 3 . fof=I Hence, the given function f is invertible and the inverse of f is f itself. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaadMeacaWG0bGaaeiiaiaabMgacaqGZbGaaeiiaiaabEgacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabshacaqGObGaaeyyaiaabshacaaMc8UaaGPaVlaaykW7caaMc8ocbaGaa8NzamaabmaabaGaa8hEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaa8hnaiaa=HhacaWFRaGaa83maaqaaiaa=zdacaWF4bGaa8xlaiaa=rdaaaGaaiilaiaaykW7caWG0bGaamiAaiaadwgacaWGUbaabaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaa8Nzaiaa=9gacaWFMbWaaeWaaeaacaWF4baacaGLOaGaayzkaaGaeyypa0JaamOzamaabmaabaGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaaGaayjkaiaawMcaaaqaaiaaxMaacaWLjaGaaCzcaiabg2da9iaadAgadaqadaqaamaalaaabaGaa8hnaiaa=HhacaWFRaGaa83maaqaaiaa=zdacaWF4bGaa8xlaiaa=rdaaaaacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacaWLjaGaeyypa0ZaaSaaaeaacaaI0aWaaeWaaeaadaWcaaqaaiaa=rdacaWF4bGaa83kaiaa=ndaaeaacaWF2aGaa8hEaiaa=1cacaWF0aaaaaGaayjkaiaawMcaaiabgUcaRiaaiodaaeaacaaI2aWaaeWaaeaadaWcaaqaaiaa=rdacaWF4bGaa83kaiaa=ndaaeaacaWF2aGaa8hEaiaa=1cacaWF0aaaaaGaayjkaiaawMcaaiabgkHiTiaaisdaaaaabaGaaCzcaiaaxMaacaWLjaGaeyypa0ZaaSaaaeaadaqadaqaamaalaaabaGaaGymaiaaiAdacaWG4bGaey4kaSIaaGymaiaaikdacqGHRaWkcaaIXaGaaGioaiaadIhacqGHsislcaaIXaGaaGOmaaqaaiaaiAdacaWG4bGaeyOeI0IaaGinaaaaaiaawIcacaGLPaaaaeaadaqadaqaamaalaaabaGaaGOmaiaaisdacaWG4bGaey4kaSIaaGymaiaaiIdacqGHsislcaaIYaGaaGinaiaadIhacqGHRaWkcaaIXaGaaGOnaaqaaiaa=zdacaWF4bGaa8xlaiaa=rdaaaaacaGLOaGaayzkaaaaaaqaaiaaxMaacaWLjaGaaCzcaiabg2da9maalaaabaGaaG4maiaaisdacaWG4baabaGaaG4maiaaisdaaaaabaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaa8Nzaiaa=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@446B@

Q.33

State with reason whether following functionshave inverse(i) f:{1,2,3,4}{10} with    f={(1,10),(2,10),(3,10),(4,10)}(ii)  g:{5,6,7,8}{1,2,3,4} with      g={(5,4),(6,3),(7,4),(8,2)}(iii)h:{2,3,4,5}{7,9,11,13} with      h={(2,7),(3,9),(4,11),(5,13)}

Ans.

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

Q.34

Show that f: [1, 1]R, given by f( x )= x x+2 isoneone. Find the inverse of the function f: [1, 1]Range f.MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=nfacaWFObGaa83Baiaa=DhacaWFGaGaa8hDaiaa=HgacaWFHbGaa8hDaiaa=bcacaWFMbGaa8Noaiaa=bcacaWFBbGaeyOeI0Iaa8xmaiaa=XcacaWFGaGaa8xmaiaa=1facqGHsgIRcaWFsbGaa8hlaiaa=bcacaWFNbGaa8xAaiaa=zhacaWFLbGaa8NBaiaa=bcacaWFIbGaa8xEaiaa=bcacaWFMbWaaeWaaeaacaWF4baacaGLOaGaayzkaaGaa8xpamaalaaabaGaa8hEaaqaaiaa=HhacaWFRaGaa8NmaaaacaWFGaGaa8xAaiaa=nhacaaMc8Uaa83Baiaa=5gacaWFLbGaa8xlaiaa=9gacaWFUbGaa8xzaiaa=5cacaWFGaaabaGaa8Nraiaa=LgacaWFUbGaa8hzaiaa=bcacaWFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=LgacaWFUbGaa8NDaiaa=vgacaWFYbGaa83Caiaa=vgacaWFGaGaa83Baiaa=zgacaWFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=zgacaWF1bGaa8NBaiaa=ngacaWF0bGaa8xAaiaa=9gacaWFUbGaa8hiaiaa=zgacaWF6aGaa8hiaiaa=TfacqGHsislcaWFXaGaa8hlaiaa=bcacaWFXaGaa8xxaiabgkziUkaa=jfacaWFHbGaa8NBaiaa=DgacaWFLbGaa8hiaiaa=zgacaWFUaaaaaa@9568@

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=xxy =x 1y x= 2y 1y ,y1 Now, let us define g: Range f[1, 1] as g y = 2y 1y ,y1. Now, gof x =g f x =g x x+2 = 2 x x+2 1 x x+2 = 2x x+2x = 2x 2 gof x =x fog y =f g y =f 2y 1y = 2y 1y 2y 1y +2 = 2y 2y+22y = 2y 2 fog y =y go f 1 = I 1,1 andfo g 1 = I Rangef f 1 =g f 1 y = 2y y1 ,y1. f 1 x = 2x x1 ,x1. 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Q.35

Consider f : R Rgiven by f(x)=4x+3. Show that f isinvertible. Find the inverse off.

Ans.

f: RR is given by, f(x) = 4x + 3Oneone:Let f(x) = f(y).4x + 3=4y + 3        4x=4y            x=yf is a oneone function.Onto:For yR, let y = 4x + 3.          x=y34RTherefore, for any yR, there exists,f1 exists.Let us define g: RR by gy=y34Now,gofx=gfx        =g4x+3        =4x+334gofx=xfogy=fgy        =fy34        =4y34+3fogy=ygof=fog=IRHence, f is invertible and the inverse of f is given by      f1y=gy        =y34.        f1x=x34

Q.36

Consider f: R + [4,) given by f(x) = x 2 + 4. Show that f is invertible with the inverse f 1 of given f by, where R + is the set of all nonnegative realnumbers. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=neacaWFVbGaa8NBaiaa=nhacaWFPbGaa8hzaiaa=vgacaWFYbGaa8hiaiaa=zgacaWF6aGaa8hiaiaa=jfadaWgaaWcbaGaa83kaaqabaGccqGHsgIRcaWFBbGaa8hnaiaa=XcacqGHEisPcaWFPaGaa8hiaiaa=DgacaWFPbGaa8NDaiaa=vgacaWFUbGaa8hiaiaa=jgacaWF5bGaa8hiaiaa=zgacaWFOaGaa8hEaiaa=LcacaWFGaGaa8xpaiaa=bcacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaa8hiaiaa=TcacaWFGaGaa8hnaiaa=5cacaWFGaGaa83uaiaa=HgacaWFVbGaa83Daiaa=bcacaWF0bGaa8hAaiaa=fgacaWF0bGaa8hiaiaa=zgacaWFGaGaa8xAaiaa=nhacaWFGaaabaGaa8xAaiaa=5gacaWF2bGaa8xzaiaa=jhacaWF0bGaa8xAaiaa=jgacaWFSbGaa8xzaiaa=bcacaWF3bGaa8xAaiaa=rhacaWFObGaa8hiaiaa=rhacaWFObGaa8xzaiaa=bcacaWFPbGaa8NBaiaa=zhacaWFLbGaa8NCaiaa=nhacaWFLbGaa8hiaiaa=zgadaahaaWcbeqaaiaa=1cacaWFXaaaaOGaa8hiaiaa=9gacaWFMbGaa8hiaiaa=DgacaWFPbGaa8NDaiaa=vgacaWFUbGaa8hiaiaa=zgacaWFGaGaa8Nyaiaa=LhacaWFSaGaa8hiaiaa=DhacaWFObGaa8xzaiaa=jhacaWFLbGaa8hiaiaa=jfadaahaaWcbeqaaiaa=TcaaaGccaWFGaGaa8xAaiaa=nhacaWFGaGaa8hDaiaa=HgacaWFLbGaa8hiaiaa=nhacaWFLbGaa8hDaiaa=bcaaeaacaWFVbGaa8Nzaiaa=bcacaWFHbGaa8hBaiaa=XgacaWFGaGaa8NBaiaa=9gacaWFUbGaa8xlaiaa=5gacaWFLbGaa83zaiaa=fgacaWF0bGaa8xAaiaa=zhacaWFLbGaa8hiaiaa=jhacaWFLbGaa8xyaiaa=XgacaaMc8UaaGPaVlaa=5gacaWF1bGaa8xBaiaa=jgacaWFLbGaa8NCaiaa=nhacaWFUaaaaaa@C381@

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 asx=y R + fisoneone function. Onto: For y [4,), let y= x 2 + 4 x= y4 0 Therefore, for any yR, there existsx= y4 0, such that f x =f y4 = y4 2 +4 =y4+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 = y4 Now, gof x =g f x =g x 2 +4 = x 2 +4 4 gof x =x fog y =f g y =f y4 = y4 2 +4 =y4+4 fog y =y gof=fog= I R + . Hence, f is invertible and the inverse of f is given by f 1 y =g y = y4 f 1 x = x4 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabAgacaqG6aGaaeiiaiaabkfadaWgaaWcbaGaae4kaaqa baGccqGHsgIRcaqGGaGaae4waiaabsdacaqGSaGaaeiiaiabg6HiLk aabMcacaqGGaGaaeyAaiaabohacaqGGaGaae4zaiaabMgacaqG2bGa aeyzaiaab6gacaqGGaGaaeyyaiaabohacaqGGaGaaeOzaiaabIcaca qG4bGaaeykaiaabccacaqG9aGaaeiiaiacycyG4bWaiGjGCaaaleqc ycyaiGjGcGaMagOmaaaakiacycyGGaGaiGjGbUcacGaMagiiaiacyc yG0aGaaeOlaaqaaiaab+eacaqGUbGaaeyzaiaab2cacaqGVbGaaeOB aiaabwgacaqG6aaabaGaaGPaVlaaykW7caaMc8UaaGPaVlaabYeaca qGLbGaaeiDaiaabccacaqGMbGaaeikaiaabIhacaqGPaGaaeiiaiaa b2dacaqGGaGaaeOzaiaabIcacaqG5bGaaeykaaqaaiabgkDiElaayk W7caaMc8UaaGPaVlaabIhadaahaaWcbeqaaiaabkdaaaGccaqGGaGa 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Q.37

Consider f: R + [5, ) given by f(x) = 9 x 2 + 6x5. Show that f is invertible with f 1 ( y )=( ( y+6 )1 3 ).MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaGqabiaa=neacaWFVbGaa8NBaiaa=nhacaWFPbGaa8hzaiaa=vgacaWFYbGaa8hiaiaa=zgacaWF6aGaa8hiaiaa=jfadaWgaaWcbaGaa83kaaqabaGccqGHsgIRcaWFGaGaa83waiaa=1cacaWF1aGaa8hlaiaa=bcacqGHEisPcaWFPaGaa8hiaiaa=DgacaWFPbGaa8NDaiaa=vgacaWFUbGaa8hiaiaa=jgacaWF5bGaa8hiaiaa=zgacaWFOaGaa8hEaiaa=LcacaWFGaGaeyypa0Jaa8hiaiaa=LdacaWF4bWaaWbaaSqabeaacaWFYaaaaOGaa8hiaiaa=TcacaWFGaGaa8Nnaiaa=HhacaWFTaGaa8xnaiaa=5cacaWFGaaabaGaa83uaiaa=HgacaWFVbGaa83Daiaa=bcacaWF0bGaa8hAaiaa=fgacaWF0bGaa8hiaiaa=zgacaWFGaGaa8xAaiaa=nhacaWFGaGaa8xAaiaa=5gacaWF2bGaa8xzaiaa=jhacaWF0bGaa8xAaiaa=jgacaWFSbGaa8xzaiaa=bcacaWF3bGaa8xAaiaa=rhacaWFObGaa8hiaiaa=zgadaahaaWcbeqaaiabgkHiTiaa=fdaaaGcdaqadaqaaiaa=LhaaiaawIcacaGLPaaacqGH9aqpdaqadaqaamaalaaabaWaaeWaaeaadaGcaaqaaiaa=LhacaWFRaGaa8NnaaWcbeaaaOGaayjkaiaawMcaaiabgkHiTiaa=fdaaeaacaWFZaaaaaGaayjkaiaawMcaaiaa=5caaaaa@8D83@

Ans.

f: R + [5, ) is given as f(x) = 9x 2 + 6x5. Let y be an arbitrary element of [5,). Let y = 9x 2 + 6x5 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 ∵y6y+60 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 + 6x5 = 9x 2 + 6x5 +6 1 3 = 9x 2 + 6x+1 1 3 = 3x+1 2 1 3 = 3x+11 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+62 y+6 +1 9 +2 y+6 25 =y+62 y+6 +1+2 y+6 25 fog y =y gof= I R andfog= 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

Let f :X→Y be an invertible function.Show that f has unique inverse.

Ans.

Let f: XY be an invertible function.Also, suppose f has two inverses (say g1 and g2)Then, for all yY, we have:fog1y=IYy=fog2y    fg1y=fg2y    g1y=g2yf is invertiblef is oneone          g1=g2g is oneoneHence, f has a unique inverse.

Q.39

Consider f : {1,2,3} {a, b, c} given by f (1)=a, f(2)=band f (3)=c. Find f1and show that (f1)1=f.

Ans.

Function f: {1, 2, 3}{a, b, c} is given by,f(1) = a, f(2) = b, and f(3) = cIf we define g: {a, b, c}{1, 2, 3} as g(a) = 1, g(b) = 2, g(c) = 3, then we have:(fog)(a)=f(g(a))=f(1)=a(fog)(b)=f(g(b))=f(2)=b(fog)(c)=f(g(c))=f(3)=cand(gof)(1)=g(f(1))=g(a)=1(gof)(2)=g(f(2))=g(b)=2(gof)(3)=g(f(3))=g(c)=3gof=IX and fog=IY, where X={1,2,3} and Y={a,b,c}.Thus, the inverse of f exists and f1 = g.f1:{a,b,c}{1,2,3} is given by,f1(a)=1,f1(b)=2,f1(c)=3 The inverse of f1 The inverse of gWe define, h:{1,2,3}{a,b,c}  ash(1) = a, h(2) = b, h(3) = c, then we have:(goh)(1)=g(h(1))=g(a)=1(goh)(2)=g(h(2))=g(b)=2(goh)(3)=g(h(3))=g(c)=3and(hog)(a)=h(g(a))=h(1)=a(hog)(b)=h(g(b))=h(2)=b(hog)(c)=h(g(c))=h(3)=cgoh=IXand  hog=IY,where X={1,2,3} and={a,b,c}.Thus, the inverse of g exists andg1 = h(f1)1= h.It can be noted that h = f.Hence, (f1)1 = f.

Q.40

Let f:XY be aninvertible function. Show that the inverseof f1 is f,i.e., (f1)1=f.

Ans.

Let f: XY be an invertible function.Then, there exists a function g: YX such that gof = IXand fog = IY.Here, f1 = g.Now, gof = IXand fog = IYf1of = IXand fof1= IYHence, f1: YX is invertible and f is the inverse of f1i.e., (f1)1 = f.

Q.41

If f: RR be given by, f(x)=(3x3)13then f of (x) is(A)x13  (B)x3(C) x(D) (3x3).

Ans.

f: RR is given as  f(x)=(3x3)13fof(x)=f(f(x))      =f{(3x3)13}      =[3{(3x3)13}3]13      ={3(3x3)}13      =(33+x3)13  fof(x)=xThe correct answer is (C).

Q.42

Let  f:R{43}R be a function defined as f(x) =4x3x+4.The inverse of f is map g: R ange fR{43} given by(A)g(y)=3y34y(B)g(y)=4y43y(C)g(y)=4y34y(D)g(y)=3y43y

Ans.

It is given thatf:R{43}Ris defined as f(x)=4x3x+4.Lety be an arbitrary element of Range f.Then, there exists xR{43}suchthaty=f(x).    y=4x3x+43xy+4y=4x3xy+4x=4yx(43y)=4y        x=4y43yLet us define g: Range  fR{43}  asg(y)=4y43yNow,(gof)(x)=g(f(x))        =g(4x3x+4)        =4(4x3x+4)43(4x3x+4)        =16x12x+1612x        =xAnd,(fog)(y)=f(g(y))      =f(4y43y)      =4(4y43y)3(4y43y)+4      =16y12y+1612y      =16y16      =ygof=IR{43}  and  fog=IRangefThus, g is the inverse of f i.e., f1 = g.Hence, the inverse of f is the map g: Range,  RR{43},which is given by  g(y)=4y43y.The correct answer is B.

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.

( i ) On Z + , * is defined by a * b = ab. Here, ( 1,2 ) Z + so, 1*2=12=1 Z + Thus,* is not a binary operation. ( ii ) On Z + , * is defined by a*b = ab. Here, a,b Z + and ab Z + This means that * carries each pair (a, b) to a unique element a * b = ab in Z + . Therefore, *is a binary operation. ( iii ) On R, * is defined by a * b = ab 2 . It is seen that for each a, bR, there is a unique element ab 2 in R. This means that * carries each pair (a, b) to a unique element a * b = ab 2 in R. Therefore, * is a binary operation. ( iv ) On Z + , * is defined by a*b = |ab|. It is seen that for each a, b Z + , there is a unique element |a b| in Z + . This means that * carries each pair (a, b) to a unique element a*b =|a b| in Z + . Therefore, * is a binary operation. ( v ) On Z + , * is defined by a*b = a. * carries each pair (a, b) to a unique element a * b = a in Z + . Therefore, * is a binary operation. 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Q.44

For each binary operation*defined below, determinewhether *is commutative or associative.  (i) On Z, definea*b=ab(ii) On Q, definea*b=ab+1(iii) OnQ, definea*b=ab2 (iv) OnZ+, definea*b=2ab(v) OnZ+, definea*b=ab(vi)OnR{1}, definea*b=ab+1

Ans.

(i) On Z, * is defined by a * b = ab.      It can be observed that 1 * 2 = 12 = 1 and 2 * 1 = 21 = 1.    1*22 * 1; where 1, 2ZHence, the operation * is not commutative.Also we have:(1 * 2) * 3 = (12)*3 =1 * 3 =13 =41 * (2 * 3) = 1 * (23) = 1 * 1 = 1 (1) = 2(1 * 2) * 31 * (2 * 3) ; where 1, 2, 3ZHence, the operation * is not associative.(ii)On Q, * is defined by a * b = ab + 1. It is known that:              ab = ba a, bQab + 1 = ba + 1                  a, bQ    a * b = a * ba, bQTherefore, the operation * is commutative.It can be observed that:  (1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10  1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8(1 * 2) * 31 * (2 * 3) ; where 1, 2, 3QTherefore, the operation * is not associative.(iii)On Q, * is defined by a*b=ab2      It is known that:ab = baa, bQ    ab2=ba2a, bQa*b=b*aa, bQTherefore, the operation * is commutative.For all a, b, cQ, we have:(a*b)*c=(ab2)*c=(ab2)c2=abc4a*(b*c)=a*(bc2)=a(bc2)2=abc4So,  (a*b)*c=a*(b*c)a, b, cQTherefore, the operation * is associative.(iv)On Z+, * is defined by a*b = ab.It can be observed that:    1*2=12=1 and 2*1=22=4  1*22*1;    where  1,2Z+Therefore, the operation * is not commutative.For Associative Law:(3*4)*5=(32)*5=(32)2=813*(4*5)=3*(42)=32=9(3*4)*53*(4*5)  where 2, 3, 4Z+Therefore, the operation * is not associative.(vi)On R, *{1} is defined by  a*b=ab+1It can be observed that  2*3=23+1=24=12and          3*2=32+1=33=1    2*33*2 ; where 2, 3R{1}Therefore, the operation * is not commutative.For​ Associative Law:(2*3)*4=(23+1)*4=(24)*4=(12)*4=(12)4+1=1102*(3*4)=2*(34+1)=2*(35)=2(35)+1=285=54(2*3)*42*(3*4);  where 2,3,4R{1}Therefore, the operation * is not associative.

Q.45

Consider the binary operation on theset {1, 2, 3, 4, 5}definedby  ab=min{a,b}.Write the operation tableof the operation .

Ans.

The binary operationon the set {1, 2, 3, 4, 5} is defined as ab= min{a,b} a,b{1, 2, 3, 4, 5}.Thus, the operation table for the given operationcan be given as:

  ^ 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

Conside ra binary operation*on the set {1,2,3,4,5} given      by the following multiplicationtable  (Table1.2)(i) Compute(2*3)*4and2*(3*4)(ii) Is*commutative?(iii) Compute(2*3)*(4*5)

* 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.

(i)(2*3)*4=1*4=1    2*(3*4)=2*1=1(ii)For every a, b{1, 2, 3, 4, 5}, we have a * b = b * a.     Therefore, the operation * is commutative.(iii)(2 * 3) = 1 and (4 * 5) = 1(2 * 3) * (4 * 5) = 1 * 1 = 1

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.

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

*’ 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

Let * be the binary operation on Ngiven by a*b=L.C.M. of aand b. Find(i) 5*7,20*16(ii) Is*commutative?(iii) Is*associative?(iv) Find the identity of*in N(v) Which element so f Narein vertible f or the operation*?

Ans.

The binary operation * on N is defined as a * b = L.C.M. of a and b.(i) 5 * 7 = L.C.M. of 5 and 7     = 3520 * 16 = L.C.M of 20 and 16 = 80(ii) It is known that:L.C.M of a and b = L.C.M of b and a       a, bN.                      a * b = b * aThus, the operation * is commutative.(iii) For a, b, cN, we have:      (a * b) * c = (L.C.M of a and b) * c           = LCM of a, b, and c      a * (b * c)= a * (LCM of b and c)           = L.C.M of a, b, and c  (a * b) * c = a * (b * c)Thus, the operation * is associative.(iv) It is known that:L.C.M. of a and 1 = a = L.C.M. 1 and a a N                    a * 1 = a = 1 * a a NThus, 1 is the identity of * in N.(v) An element a in N is invertible with respect to the operation * if there exists an element b in  N, such that a * b = e = b * a.Here,       e = 1This means that:L.C.M of a and b = 1 = L.C.M of b and aThis case is possible only when a and b are equal to 1.Thus, 1 is the only invertible element of N with respect to the operation *.

Q.49

Is*defined on the set {1, 2, 3, 4, 5} by a*b= L.C.M. of aand b a binary operation? Justify your answer.

Ans.

The operation * on the set A = {1, 2, 3, 4, 5} is defined asa * b = L.C.M. of a and b.Then, the operation table for the given operation * can be givenas:

* 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.

The binary operation * on N is defined as: a * b = H.C.F. of a and b Since, H.C.F. of a and b= H.C.F. of b and a a, bN. a*b=b*a Thus, the operation * is commutative. For a, b, cN, we have: (a * b)* c=(H.C.F. of a and b) * c =H.C.F. of a, b, and c a *(b * c)=a *(H.C.F. of b and c) =H.C.F. of a, b, and c (a*b)*c=a*(b*c) Thus, the operation * is associative. Letan element eN will be the identity for the operation * if a*e=a=e* a aN. But this relation is not true for all aN. Thus, the operation * does not have any identity in N. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@2014@

Q.51

Let * be a binary operation on the set  Q of rationalnumbers as follows:(i) a*b=ab (ii) a*b=a2+b2(iii )a*b=a+ab (iv)a*b=ab2(v)  a*b=ab4  via*b=ab2Find which of the binary operation s are commutativeand which are as sociative.

Ans.

(i) Since,  a*b=ab a, bQ=(ba) a, bQ=b*a a, bQSo, the operation * is not cummutative.Now, let a, b, cQ, then  (a*b)*c=(ab)*c=abcand        a*(b*c)=a*(bc)=a(bc)=ab+cSo,  (a*b)*ca*(b*c)Thus, the operation * is not associative.(ii)Since,  a*b=a2+b2 a,bQ=b2+a2 a,bQ=b*a a,bQSo, the operation * is cummutative.Now, let a,b,cQ, then  (a*b)*c=(a2+b2)*c=(a2+b2)2+c2and        a*(b*c)=a*(b2+c2)=a2+(b2+c2)2So,  (a*b)*ca*(b*c)Thus, the operation * is not associative.(iii)Since,  a*b=a+ab a,bQand  b*a=b+ba a,bQa*bb*a a,bQSo, the operation * is not cummutative.Now, let a,b,cQ, then  (a*b)*c=(a+ab)*c=(a+ab)+(a+ab)c=a+ab+ac+abcand        a*(b*c)=a*(b+bc)=a+a(b+bc)=a+ab+abcSo,  (a*b)*ca*(b*c)Thus, the operation * is not associative.(iv)Since,  a*b=(ab)2 a,bQ  ={(ba)}2                      =(ba)2 a,bQ  =b*aa*bb*a         a,bQSo, the operation * is not cummutative.Now, let a,b,cQ, then  (a*b)*c={(ab)2}*c={(ab)2c}2=(a22ab+b2c)2and        a*(b*c)=a*(bc)2={a(bc)2}2=(ab2+2bcc2)2So,  (a*b)*ca*(b*c)Thus, the operation * is not associative.(v)Since,  a*b=ab4a,bQ  =ba4a,bQ  =b*aa*b=b*a          a,bQSo, the operation * is cummutative.Now, let a,b,cQ, then  (a*b)*c=(ab4)*c=(ab4)c4=abc16and        a*(b*c)=a*(bc4)=a(bc4)4=abc16So,  (a*b)*c=a*(b*c)Thus, the operation * is associative.(vi)Since,  a*b=ab2a,bQ  b*a=ba2a,bQ  a*bb*aa,bQSo, the operation * is not cummutative.Now, let a,b,cQ, then  (a*b)*c=(ab2)*c=(ab2)c2=ab2c2and        a*(b*c)=a*(bc2)=a(bc2)2=ab2c4So,  (a*b)*ca*(b*c)Thus, the operation * is not associative.Hence, the operations defined in (ii), (iv), (v) are commutative and the operation defined in (v) is associative.

Q.52

Let * be a binary operation on the set  Q  of rationalnumbersas follows:(i) a*b=ab (ii)a*b=a2+b2(iii) a*b=a+ab (iv)a*b=ab2(v)  a*b=ab4via*b=ab2

Find which of the operations given above has identity.

Ans.

An element eQ will be the identity element for the operation * if  a * e = a = e * a, aQ.However, there is no such element eQ with respect to each of the six operations satisfying theabove condition.Thus, none of the six operations in above question has identity.

Q.53

Let A=N×Nand*be the binary operation on A defined by(a, b)*(c, d)=(a+c, b+d)Show that *is commutative and associative. Find the identity element for*onA, if any.

Ans.

A = N × N* is a binary operation on A and is defined by:(a, b) * (c, d) = (a + c, b + d)Let (a, b), (c, d)AThen,   a, b, c, dNSince,(a, b) * (c, d) = (a + c, b + d)and (c, d) * (a, b) = (c + a, d + b)   = (a + c, b + d)[Addition is commutative in the set of natural numbers](a, b)*(c, d) = (c, d)*(a, b)Therefore, the operation * is commutative.For associativity:Now, let (a, b), (c, d), (e, f)AThen, a, b, c, d, e, fNSo,    {(a, b) * (c, d)}*(e,f)={(a + c, b + d)}*(e,f)           =(a + c+e, b + d+f )    (a, b) * {(c, d)*(e,f)}=(a, b) * {(c + e, d+ f)}           =(a + c + e, b + d + f){(a, b) * (c, d)}*(e,f)=(a, b) * {(c, d)*(e,f)}Therefore, the operation * is associative.An element e=(e1+e2)Awill be an identity element for the operation * if a*e=a=e*aa=(a1,a2)Ai.e.,  (a1+e1,a2+e2)=(a1,a2)=(e1+a1,e2+a2)which is not true for any element in A.Therefore, the operation * does not have any identity element.

Q.54

State whether the following statements are true or false.Justify.(i)For an arbitrary binary operation*on a set N,a*a=aa*N.(ii) If*is acommutative binary operationon N, thena*(b*c)=(c*b)*a

Ans.

(i) Define an operation * on N as:      a * b = a + b a, bNThen, in particular, for b = a = 5, we have:      5 * 5 = 5 + 5=105Therefore, statement (i) is false.(ii) R.H.S. = (c * b) * a    = (b * c) * a [* is commutative]    = a * (b * c) [Again, as * is commutative]    = L.H.S. a * (b * c) = (c * b) * aTherefore, statement (ii) is true.

Q.55

Consider a binary operation*on N  defined as a*b=a3+b3.Choose the correct answer.(A )Is*both a ssociative and commutative?(B) Is*commutative but not associative?(C)Is*associative but not commutative?(D)Is*neither commutative nor associative?

Ans.

On N, the operation * is defined as a*b = a 3 + b 3 . For, a, bN, we have: a*b= a 3 + b 3 = b 3 + a 3 =b*a [Addition is commutative in N] Therefore, the operation * is commutative. For associative: Since, 2,3,4N So,( 2*3 )*4=( 2 3 + 3 3 )*4 =35*4 = 35 3 + 4 3 =42939 and 2*( 3*4 )=2*( 3 3 + 4 3 ) =2*( 91 ) = 2 3 + 91 3 =8+753571 =753579 ( 2*3 )*42*( 3*4 ); where 2,3,4N Therefore, the operation * is not associative. Hence, the operation * is commutative, but not associative. Thus, the correct answer is B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@00E8@

Q.56

Let f:RR be defined as f(x)=10x+7.Find the functiong:RR such that go f=fog=IR.

Ans.

It is given that f: RR is defined as f(x) = 10x + 7.Oneone:Let f(x) = f(y), where x, yR. 10x + 7 = 10y + 7 x = y f is a oneone function.Onto:For yR, let y=10x + 7.        x=y710RThen,  f(x)=f(y710)[yR]  =10(y710)+7  =yf is onto.Therefore, f is oneone and onto.Thus, f is an invertible function.For inverse:Let g:RR as g(y)=y710So,gof(x)=g{f(x)}=g(10x + 7)=10x + 7710=xandfog(y)=f{g(y)}=f(y710)=10(y710)+7=ygof=IR and fog=IRHence, the required function g: RR is defined asg(y)=y710

Q.57

Letf:WW bede fine d as f(n)=n1,if n is odd andf(n)=n+1,if n is even. Show tha tf is invertible. Findthe inverse off.Here, W is the set of all whole numbers.

Ans.

Given:f:WW is defined as f(n)={n1,    if n is oddn+1, if n is evenOneone:Let  f(n)=f(m)n1=m+1[If n is odd and m is even.]    nm=2,which is impossible.Let  f(n)=f(m)n+1=m1[If n is even and m is odd.]    nm=2,which is also impossible.Now, if both n and m are odd, then we have:f(n)=f(m)n1=m1      n=mAgain, if both n and m are even, then we have:f(n) = f(m) n + 1= m + 1     n= mf is oneone.Onto:For any odd number, 2r+1Nf(2r+1)=2r+11        =2rNFor any even number, 2rN      f(2r)=2r+1Nf is onto function.Therefore, f is invertible function.Let us define g: WW as:g(m)={m+1,  if m is evenm1,  if m is odd Now, when n is oddgοf(n)=g(f(n))    =g(n1)    =n1+1    =nAnd, when n is even:gοf(n)=g(f(n))    =g(n+1)    =n+11    =nWhen m is odd:gοf(m)=g(f(m))    =g(m1)    =m1+1    =mAnd, when n is even:gof(m)=g(f(m))    =g(m+1)    =m+11    =mgof=IWand  gof=IWThus, f is invertible and the inverse of f is given by f1= g, which is the same as f. Hence, the inverse of f is f itself.

Q.58

If f : R → R is defined by f(x)=x23x+2, find f (f(x)).

Ans.

Given:f :R R is defined as f(x)=x23x+2.f (f(x))=f(x23x+2)      =(x23x+2)23(x23x+2)+2      =x4+9x2+46x312x+4x23x2+9x6+2      =x46x3+10x23x

Q.59

Show that function f :R{xR:1<x<1} defined by f(x)=x1+|x|,xR isoneone and onto function.

Ans.

Given:f:R{xR:1<x<1}isdefinedasf(x)=x1+|x|, xR.Ler   f(x)= f(y), where x, yR.  x1+|x|=y1+|y|If x is positive and y is negative, then we have:        x1+x=y1yx(1y)=y(1+x)    xxy=y+xyxy=2xySince x is positive and y is negative:x>yxy>0But, 2xy is negative.Then, 2xyxyThus, the case of x being positive and y being negative can not be considered.Similarly, the case of x being negative and y being positive also can not be considered.x and y have to be either positive or negative.When x and y are both positive, we have:f(x)=f(y)    x1+x=y1+yx+xy=y+xy          x=yWhen x and y are both negative, we have:f(x)=f(y)    x1x=y1yxxy=yxy          x=yf is oneone.Now, let yR such that 1 < y < 1.If y is negative, then there exists x=y1+yR such thatf(x)=f(y1+y)=y1+y1+|y1+y|=y1+y1+y1+y=yIf y is positive, then there exists x=y1+yRsuchthatf(x)=f(y1y)=y1y1+|y1y|=y1y1+y1y=yf is onto.Hence, f is oneone and onto.

Q.60

Show that the function f : R→ R given by f(x)=x3is injective.

Ans.

f: RR is given as f(x) = x3.Suppose f(x) = f(y), where x, yR.    x3 = y3 ... (1)Now, we need to show that x = y.Suppose xy, their cubes will also not be equal.    x3y3However, it will be a contradiction to equation(1).      x=yHence, f is injective.

Q.61

Give examples of two functions f: NZ and g:ZZ suchthat gοf is injective but g is not injective.(Hint: Consider f(x)=x and g(x)=|x|)

Ans.

Define f: NZ as f(x) = x and g: ZZ as g(x) =|x|Let us first show that g is not injective.It can be observed that:    g(1) =|1|=1        g(1) =|1|=1g(1) =g(1), but -11g is not injective.Now, gof: NZ is defined asgοf(x)=g(f(x))=g(x)=|x|Let x, yN such that gof(x) = gof(y).      x=ySince x and yN, both are positive|x|=|y|x=yHence, gοf is injective.

Q.62

Given example s of two functions f : N N and g : N N suchthat go f is onto but f is not onto.(Hint:Consider f(x)=x+1andg(x)={x1,    ifx>1x+1,    ifx<1

Ans.

Define f: NN by,  f(x) = x + 1And, g: N N by,  g(x)={x1,  if x>11,  if x=1We first show that f is not onto:For this, consider element 1 in codomain N. It is clear that this element is not an image of any of the elements in domain N.f is not onto.Now, gof: NN is defined by,gοf(x)=g(f(x))    =g(x+1)    =x+11[xN(x+1)>1]    =xThen, it is clear that for yN, there exists x = yN such that gof(x) = y.gof is onto.

Q.63

Given an on empty set X, consider P (X) which is the set ofall sub sets of X. Define therelation Rin P (X) as follows:For sub sets A, Bin P(X), ARB if and only if AB. Is Ranequivalence relation on P (X) ? Justify you answer.

Ans.

Since every set is a subset of itself, ARA for all AP(X).R is reflexive.Let ARBAB.For instance, if A = {3, 4} and B = {3, 4, 5}, then it cannot be implied that B is related to A.R is not symmetric.Further, if ARB and BRC, then AB and BC.ACARCR is transitive.Hence, R is not an equivalence relation since it is not symmetric.

Q.64

Given a nonempty set X, consider the binary operation*:P(X)×P(X)P(X)given by A*B=AB  A, Bin PX,where PX is the power set of X.Show that X is the identityelement for this operation and X is the only invertible elementin P(X) with respect to the operation*.

Ans.

Given that:P(XP(X)P(X) is defined as A * B = AB  A,BP(X)Since, AX=A=XA    A,BP(X)  A*X=A=X*A          A,BP(X)Thus, X is the identity element for the given binary operation *.Now, an element  AP(X)  is invertible if there exists BP(X)such that A*B=X=B*A[As X is the identity element.]i.e.,  AB=X=BAThis case is possible only when A = X = B.Thus, X is the only invertible element in P(X) with respect to the given operation*.Thus, the given result is proved.

Q.65

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

Ans.

Onto functions from the set {1, 2, 3, ,n} to itself is simply a permutation on n symbols 1, 2, , n. Thus, the total number of onto maps from {1, 2, , n} to itself is the same asthe total number of permutations on n symbols 1, 2, , n, which is n. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@150B@

Q.66

Let S={a,b,c} and T={1,2,3}. Find F1 of the followingfunctions F from S to T, if it exists.(i) F={(a, 3), (b, 2), (c,1)} (ii) F={(a, 2),(b, 1),(c, 1)}

Ans.

Given:S = {a, b, c}, T = {1, 2, 3}(i) F: ST is defined as:F = {(a, 3), (b, 2), (c, 1)} F (a) = 3, F (b) = 2, F(c) = 1Therefore, F1: T S is given byF1= {(3, a), (2, b), (1, c)}.(ii) F: S T is defined as:F = {(a, 2), (b, 1), (c, 1)}Since F (b) = F (c) = 1, F is not oneone.Therefore, F is not invertible i.e., F1 does not exist.

Q.67

Consider the binary operations*:R × Rando:R×RRdefined asa*b=|ab|and aοb=a,a,bR.Show that* iscommutative but notassociative,ο is associative but notcommutative. Further, show that   a, b, c R,a*(bοc)=(a*b)ο(a*c).[If it is so,we say that theoperation*distributes over the operationο].Does οdistributeover*? Justif y your answer.

Ans.

It is given that *: R×Rand o: R×RR isdefined asa*b=|ab|andaοb=a    a,bR.For a, bR, we have:a*b=|ab|b*a=|ba|=|(ab)|=|ab|a*b=b*aThe operation * is commutative.For associative, we see that(2*3)*4=(|23|)*4=1*4=|14|=32*(3*4)=2*|34|  =2*1  =|21|=1(2*3)*42*(3*4)[where  2,3,4R]Thus, the operation * is not associative.Now, consider the operation ο:It can be noticed that 1ο2 = 1 and 2ο1 = 2.1ο22ο1 (where 1, 2R)The operation ο is not commutative.Let a, b, cR. Then, we have:      (aοb)οc=aοc=a      aο(bοc)=aοb=a(aοb)οc=aο(bοc)The operation o is associative.Now, let a, b, cR, then we have:    a*(bοc)=a*b=|ab|(a*b)ο(a * c)=(|ab|)ο(|ac|)=|ab|  a*(bοc)=(a*b)ο(a * c)Now,         2ο(3*4)=2ο(|34|)=2ο1=2and    (2*3)ο(2*4)=(|23|)ο(|24|)=1                    2ο(3*4)(2*3)ο(2*4)      (where 1, 2, 3R)The operation ο does not distribute over *.

Q.68

Given a nonempty set X, let *:P (X)×P (X)P (X) be defined asA*B=(AB)(BA),A,BP(X).Show that the emptyse t is the identity for the operation*and all thee lements A ofP (X) are invertible with A1=A.(Hint:(A)(A)=Aand (AA)(AA)=A*A=).

Ans.

It is given that *: P(X)×P(X)P(X) is defined asA * B = (AB)(BA)    A, BP(X).Let AP(X). Then, we have:      A * =(A)( A)= A=A       * A=(A)(A)=A=A  A * = A = * A.    AP(X)Thus, is the identity element for the given operation*.Now, an element AP(X) will be invertible if there exists BP(X) such thatA * B==B * A. (As is the identity element)Now, we observed thatA*A=(AA)(AA)==    AP(X).Hence, all the elements A of P(X) are invertible with A1=A.

Q.69

Define a binary operation*on the set {0,1, 2, 3, 4, 5}  asa*b={a+b,  ifa+b<6a+b6,  ifa+b6Show that zero is the identity for this operation and eachelement a0 of these t is invertible with 6a being theinverse of a.

Ans.

Let X = {0, 1, 2, 3, 4, 5}.The operation * on X is defined as:a*b={a+b,  if a+b<6a+b6,  if a+b6An element eX is the identity element for the operation *,ifa*e=a=e*aaX.For aX, we see that:a*0=a+0=a[aXa+0<6]0*a=0+a=a[aX0+a6]a*0=0*aaXThus, 0 is the identity element for the given operation *.An element aX is invertible if there exists bX such that a * b = 0 = b * a.i.e., {a+b=0=b+a,ifa+b<6a+b6=0=b+a6,ifa+b6a=b or b=6aBut, X = {0, 1, 2, 3, 4, 5} and a,bX. Then, ab.b = 6a is the inverse of a aX.Hence, the inverse of an element aX, a0 is 6a i.e., a1 = 6a.

Q.70

Let A={1, 0, 1, 2}, B={4,2,0,2} and f, g :AB befunctions defined by f(x)=x2x, xA  and  g(x)=2|x12|1, xA. Are f and g equal?Justify your answer.(Hint: One may note that two function f : A B and g : ABsuch that f(a)=g(a)    aA,are called equal functions).

Ans.

It is given that A = {1,  0,  1,  2}, B={4,2, 0, 2}.Also, it is given that f, g: AB  are defined by f(x) = x2x, xA and g(x)=2|x12|1,xA.It is observed that:f(1)=(1)2(1)=1+1=2g(1)=2|(1)12|1=2(32)1  =31=2f(1)=g(1)f(0)=(0)2(0)=0g(0)=2|012|1=11=0f(0)=g(0)f(1)=(1)2(1)=0g(1)=2|112|1=11=0f(1)=g(1)f(2)=(2)2(2)=2g(2)=2|212|1=31=2f(2)=g(2)f(a)=g(a)aAHence, the functions f and g are equal.

Q.71

Let A={1,2,3}. Then number of relations containing (1,2)and (1,3) which are reflexive and symmetric but nottransitive is(A)1(B)2(C)3(D)4

Ans.

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

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.

It is given that A = {1, 2, 3}. The smallest equivalence relation containing (1, 2) is given by, R 1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1). If we odd any one pair [say (2, 3)] to R 1 , then for symmetry we must add (3, 2). Also, for transitivity we are required to add (1, 3) and (3, 1). Hence, the only equivalence relation (bigger than R 1 ) is the universal relation.This shows that the total number of equivalence relations containing (1, 2) is two. The correct answer is B. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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jeacaqGUaaaaaa@0113@

Q.73

Letf:RR be the Signum Function defined asf(x)={1,  ifx>00,if  x=01,    ifx<0Also, g:RR be the Greatest Integer Function given byg(x)=[x], where [x] is great est integerless than o requaltox.Then does fog and go f co incide in(0,1]?

Ans.

Given:f:RRf(x)={1,  if x>00,  if​ x=01,  if x<1Also, g: RR is defined as g(x) = [x],where [x] is the greatest integer less than or equal to x.Now, let x(0, 1].Then, we have:[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1.fog(x)=f(g(x))=f([x])={f(1),  if  x=1f(0),  ifx(0,1)