# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

In this chapter, there are 4 exercises with 55 questions and one miscellaneous exercise with 19 questions. In this article, you will find the detailed solutions of NCERT for class maths 12 chapter 1 Relations and functions. These CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions are explained in a step-by-step method, so it will be very easy to understand the concepts. Still if you are in a doubt anywhere, you can contact our subject matter experts who are available to help you out and make learning easier for you.

## What is the Relation?

The meaning of the term ‘relation’ in mathematics is the same as the meaning of 'relation' in the English language. Relation means two quantities or objects are related if there is a link between them. In other words, we can say that it is a connection between or among things.

Let's understand with an example - let A is the set of students of class XII of a school and B is the set of students of class XI of the same school. Then some of the examples of relations from A to B are-

(i) {(a, b) ∈A × B: a is a brother of b},

(ii) {(a, b) ∈A × B: a is a sister of b},

(iii) {(a, b) ∈A × B: age of a is less than the age of b}.

If (a, b) ∈ R, we can say that ‘a’ is related to ‘b’ under the relation ‘R’ and we write as  ‘a R b’. To understand the topic in-depth, after every concept, some topic wise questions are given in the textbook of CBSE class 12. In this article, you will find solutions of NCERT for class 12 maths chapter 1 Relations and Functions for such type of questions also.

## Topics of NCERT Grade 12 Maths Chapter-1  Relations and Functions

1.1 Introduction

1.2 Types of Relations

1.3 Types of Functions

1.4 Composition of Functions and Invertible Function

1.5 Binary Operations

## CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.1

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

$A = \{1,2,3 ...,13 ,14\}$

$R = \{(x,y): 3x - y = 0\}$ $= \left \{ \left ( 1,3 \right ),\left ( 2,6 \right ),\left ( 3,9 \right ),\left ( 4,12 \right ) \right \}$

Since,  $\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 14,14 \right ) \notin R$ so $R$ is not reflexive.

Since, $\left ( 1,3 \right ) \in R$ but  $\left ( 3,1 \right ) \notin R$ so $R$ is not symmetric.

Since, $\left ( 1,3 \right ),\left ( 3,9 \right ) \in R$ but $\left ( 1,9 \right ) \notin R$ so $R$ is not transitive.

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

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

$R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\}$ $= \left \{ \left ( 1,6 \right ),\left ( 2,7 \right ),\left ( 3,8 \right ) \right \}$

Since, $\left ( 1,1 \right ) \notin R$

so $R$ is not  reflexive.

Since, $\left ( 1,6 \right )\in R$ but $\left ( 6,1 \right )\notin R$

so $R$ is not symmetric.

Since there is no pair in  $R$ such that $\left ( x,y \right ),\left ( y,x \right )\in R$ so this is not transitive.

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

### Question1(iii) Determine whether each of the following relations are reflexive, symmetric and transitive:

(iii) Relation R in the set $A = \{1,2,3,4,5,6\}$ as $R = \{(x,y) : y \; \textup{is} \; divisible \; by\; x\}$

$A = \{1,2,3,4,5,6\}$

$R = \left \{ \left ( 2,4 \right ),\left ( 3,6 \right ),\left ( 2,6 \right ),\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 6,6 \right )\right \}$

Any number is divisible by itself  and  $\left ( x,x \right ) \in R$.So it is  reflexive.

$\left ( 2,4 \right ) \in R$ but $\left ( 4,2 \right ) \notin R$ .Hence,it is not symmetric.

$\left ( 2,4 \right ),\left ( 4,4 \right ) \in R$ and  4 is divisible by 2 and 4 is divisible by 4.

Hence, it is transitive.

Hence, it is reflexive and transitive but not symmetric.

### Question.1(iv) Determine whether each of the following relations are reflexive, symmetric and transitive:

(iv). Relation R in the set Z of all integers defined as $R = \{(x,y): x - y \;is\;an\;integer\}$

$R = \{(x,y): x - y \;is\;an\;integer\}$

For $x \in Z$$\left ( x,x \right ) \in R$ as $x-x = 0$ which is an integer.

So,it is reflexive.

For $x,y \in Z$ ,$\left ( x,y \right ) \in R$ and $\left ( y,x \right ) \in R$ because $x-y \, \, and \, \, y-x$  are both integers.

So, it is symmetric.

For $x,y,z \in Z$ ,$\left ( x,y \right ),\left ( y,z \right ) \in R$ as $x-y \, \, and \, \, y-z$ are both integers.

Now, $x-z = \left ( x-y \right )+\left ( y-z \right )$ is also an integer.

So,$\left ( x,z \right ) \in R$ and hence it is transitive.

Hence, it is reflexive, symmetric and transitive.

### Question:1(v) Determine whether each of the following relations are reflexive, symmetric and transitive:

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

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

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

$\left ( x,x \right )\in R$,so it is reflexive

$\left ( x,y \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ .

$y \;and\; x\;work\;at\;the\;same\;place$ i.e. $\left ( y,x \right )\in R$ so it is symmetric.

$\left ( x,y \right ),\left ( y,z \right )\in R$ means $x \;and\; y\;work\;at\;the\;same\;place$ also $y \;and\; z\;work\;at\;the\;same\;place$.It states that $x \;and\; z\;work\;at\;the\;same\;place$ i.e. $\left ( x,z \right )\in R$.So, it is transitive.

Hence, it is reflexive, symmetric and transitive.

### Question:1(v) Determine whether each of the following relations are reflexive, symmetric and transitive:

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

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

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

$\left ( x,x \right )\in R$ as $x$ and $x$ is same human being.So, it is reflexive.

$\left ( x,y \right )\in R$ means  $x\;and\;y\;live\;in\;the\;same\;locality$.

It is same as $y\;and\;x\;live\;in\;the\;same\;locality$  i.e. $\left ( y,x \right )\in R$.

So,it is symmetric.

$\left ( x,y \right ),\left ( y,z \right )\in R$ means  $x\;and\;y\;live\;in\;the\;same\;locality$ and $y\;and\;z\;live\;in\;the\;same\;locality$.

It implies that $x\;and\;z\;live\;in\;the\;same\;locality$ i.e. $\left ( x,z \right )\in R$.

Hence, it is reflexive, symmetric and
transitive.

### Question:1(v) Determine whether each of the following relations are reflexive, symmetric and transitive:

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

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

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

$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ but $x\;is\;not\;\;taller\;than\;x$ i.e. $\left ( x,x \right )\notin R$.So, it is not reflexive.

$\left ( x,y\right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$  but $y\;is\;not\;\;taller\;than\;x$ i.e  $\left ( y,x \right )\notin R$.So, it is not symmetric.

$\left ( x,y\right ),\left ( y,z \right )\in R$ means $x\;is\;exactly\;7\;cm\;taller\;than\;y$ and $y\;is\;exactly\;7\;cm\;taller\;than\;z$.

$x\;is\;exactly\;14\;cm\;taller\;than\;z$  i.e.  $\left ( x,z \right )\notin R$.

Hence, it is not reflexive,not symmetric and
not transitive.

### Question:1(v) Determine whether each of the following relations are reflexive, symmetric and transitive:

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

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

$R = \{(x, y) : x\;is\;wife\;of\;y\}$

$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but  $x\;is\;not\, wife\;of\;x$ i.e.$\left ( x,x \right ) \notin R$.

So, it is not reflexive.

$\left ( x,y \right ) \in R$ means $x\;is\;wife\;of\;y$ but  $y\;is\;not\, wife\;of\;x$ i.e.$\left ( y,x \right ) \notin R$.

So, it is not symmetric.

Let, $\left ( x,y \right ),\left ( y,z \right ) \in R$ means $x\;is\;wife\;of\;y$ and $y\;is\;wife\;of\;z$.

This case is not possible so it is not transitive.

Hence, it is not  reflexive, symmetric and
transitive.

### Question:1(v) Determine whether each of the following relations are reflexive, symmetric and transitive:

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

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

$R = \{(x, y) : x \;is \;father \;of \;y \}$

$(x, y) \in R$ means $x \;is \;father \;of \;y$ than $x \;cannot \, be \;father \;of \;x$  i.e. $(x, x) \notin R$.So, it is not reflexive..

$(x, y) \in R$ means $x \;is \;father \;of \;y$ than  $y \;cannot \, be \;father \;of \;x$ i.e. $(y, x) \notin R$.So, it is not symmetric.

Let, $(x, y),\left ( y,z \right )\in R$ means $x \;is \;father \;of \;y$ and $y \;is \;father \;of \;z$ than $x \;cannot \, be \;father \;of \;z$ i.e. $(x, z) \notin R$.

So, it is not transitive.

Hence, it is neither reflexive nor symmetric and nor transitive.

## Question:2 Show that the relation R in the set R of real numbers defined as $R = \{(a, b) : a \leq b^2 \}$ is neither reflexive nor symmetric nor transitive.

$R = \{(a, b) : a \leq b^2 \}$

Taking

$\left ( \frac{1}{2},\frac{1}{2} \right )\notin R$

and

$\left ( \frac{1}{2} \right )> \left ( \frac{1}{2} \right )^{2}$

So, R is not reflexive.

Now,

$\left ( 1,2 \right )\in R$ because    $1< 4$.

But, $4\nless 1$  i.e. 4 is not less than 1

So,$\left ( 2,1 \right )\notin R$

Hence, it is not symmetric.

$\left ( 3,2 \right )\in R\, \, and \, \, \left ( 2,1.5 \right )\in R$   as $3< 4\, \, and \, \, 2< 2.25$

Since $\left ( 3,1.5 \right )\notin R$  because $3\nless 2.25$

Hence, it is not transitive.

Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive.

### Question: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.

R defined in the set $\{1, 2, 3, 4, 5, 6\}$

$R = \{(a, b) : b = a + 1\}$

$R=\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}$

Since, $\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 6,6 \right ) \right \}\notin R$ so it is not reflexive.

$\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}\in R$   but $\left \{ \left ( 2,1 \right ),\left ( 3,2 \right ),\left ( 4,3 \right ),\left ( 5,4 \right ),\left ( 6,5 \right ) \right \}\notin R$

So, it is not symmetric

$\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}\in R$   but $\left \{ \left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 4,6 \right )\right \}\notin R$

So, it is not transitive.

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

transitive but not symmetric.

$R = \{(a, b) : a \leq b\}$

As $\left ( a,a \right )\in R$ so it is reflexive.

Now we take an  example

$\left ( 2,3 \right )\in R$    as $2< 3$

But $\left ( 3,2 \right )\notin R$  because $2 \nless 3$.

So,it is not symmetric.

Now if we take,$\left ( 2,3 \right )\in R\, \, and\, \, \left ( 3,4 \right )\in R$

Than, $\left ( 2,4 \right )\in R$ because $2< 4$

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

### Question:5 Check whether the relation R in R defined by $R = \{(a, b) : a \leq b^3 \}$ is reflexive, symmetric or transitive.

$R = \{(a, b) : a \leq b^3 \}$

$\left ( \frac{1}{2},\frac{1}{2} \right )\notin R$     because     $\frac{1}{2}\nleqslant (\frac{1}{2}) ^{3}$

So, it is not symmetric

Now, $\left ( 1,2 \right ) \in R$    because $1< 2^{3}$

but $\left ( 2,1 \right )\notin R$  because $2\nleqslant 1^{3}$

It is not symmetric

$\left ( 3,1.5 \right ) \in R\, \, and \, \, \left ( 1.5,1.2 \right ) \in R$   as   $3< 1.5^{3} \, \, and \, \, 1.5< 1.2^{3}$.

But, $\left ( 3,1.2 \right )\notin R$   because $3 \nleqslant 1.2^{3}$

So it is not transitive

Thus, it is neither reflexive, nor symmetric, nor transitive.

### Question: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.

Let A= $\{1, 2, 3\}$

$R = \{(1, 2), (2, 1)\}$

We can see $\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right )\notin R$  so it is not reflexive.

As $\left ( 1,2 \right )\in R \, and \, \left ( 2,1 \right )\in R$ so it is symmetric.

$(1, 2) \in R \, and\, (2, 1)\in R$

But  $(1, 1)\notin R$ so it is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

### Question: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.?

A = all the books in a library of a college

$R = \{(x, y) : x \;and\;y\;have\;same\;number\;of\;pages\}$

$(x,x) \in R$  because x and x have the same number of pages so it is reflexive.

Let  $(x,y) \in R$  means x and y have same number of pages.

Since y and x have the same number of pages so  $(y,x) \in R$  .

Hence, it is symmetric.

Let  $(x,y) \in R$  means x and y have the same number of pages.

and  $(y,z) \in R$  means y and z have the same number of pages.

This states,x and z also have the same number of pages i.e.$(x,z) \in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence
relation.?

## Question: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 $\{2, 4\}$.

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

$R = \{(a, b) : |a - b| \;is\;even\}$

$R=\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 3,1 \right ),\left ( 5,1 \right ),\left ( 4,2 \right ),\left ( 5,3 \right )\right \}$

Let there be $a\in A$ then $(a,a)\in R$ as $\left | a-a \right |=0$  which is even number. Hence, it is reflexive

Let  $(a,b)\in R$ where $a,b\in A$ then $(b,a)\in R$ as $\left | a-b \right |=\left | b-a \right |$

Hence, it is symmetric

Now, let $(a,b)\in R \, and\, (b,c)\in R$

$\left | a-b \right | \, and \, \left | b-c \right |$   are even number i.e. $(a-b)\, and\,(b-c)$ are even

then, $(a-c)=(a-b)+(b-c)$ is even                  (sum of even integer is even)

So, $(a,c)\in R$. Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The elements of $\{1, 3, 5\}$ are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.

The elements of $\{2, 4\}$are related to each other because the difference of even number is even number and in this set, all numbers are even.

The element of $\{1, 3, 5\}$ is not related to  $\{2, 4\}$ because a difference of odd and even number is not even.

## Question:9(i) Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$, given by

(i) $R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

$A = \{x \in Z : 0 \leq x \leq 12\}$

$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$

$R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}$

For $a\in A$ , $(a,a)\in R$  as $\left | a-a \right |=0$ which is multiple of 4.

Henec, it is reflexive.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4.

then  $\left | b-a \right |$ is also multiple of 4 because $\left | a-b \right |$  =  $\left | b-a \right |$  i.e.$(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $\left | a-b \right |$ is multiple of 4   and    $(b,c)\in R$  i.e. $\left | b-c \right |$ is multiple of 4 .

$( a-b )$ is multiple of 4  and   $(b-c)$ is multiple of 4

$(a-c)=(a-b)+(b-c)$  is multiple of 4

$\left | a-c \right |$ is multiple of 4 i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is  $\left \{1,5,9 \right \}$

$\left | 1-1 \right |=0$ is multiple of 4.

$\left | 5-1 \right |=4$ is multiple of 4.

$\left | 9-1 \right |=8$ is multiple of 4.

## Question:9(ii) Show that each of the relation R in the set $A = \{x \in Z : 0 \leq x \leq 12\}$, given by

(ii)  $R = \{(a, b) : a = b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

$A = \{x \in Z : 0 \leq x \leq 12\}$

$A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}$

$R = \{(a, b) : a = b\}$

For $a\in A$ , $(a,a)\in R$  as  $a=a$

Henec, it is reflexive.

Let, $(a,b)\in R$ i.e. $a=b$

$a=b$ $\Rightarrow$  $b=a$  i.e.$(b,a)\in R$

Hence, it is symmetric.

Let, $(a,b)\in R$ i.e. $a=b$   and    $(b,c)\in R$  i.e. $b=c$

$\therefore$  $a=b=c$

$a=c$ i.e. $(a,c)\in R$

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is {1}

## Question:10(i) Give an example of a relation.

(i) Which is Symmetric but neither reflexive nor transitive.

Let

$A = \left \{ 1,2,3 \right \}$

$R = \left \{ \left ( 1,2 \right ),\left ( 2,1 \right )\right \}$

$\left ( 1,1 \right ),\left ( 2,2 \right ),(3,3) \notin R$ so it is not reflexive.

$(1,2)\in R$   and  $(2,1)\in R$  so it is symmetric.

$(1,2)\in R \, and\, (2,1)\in R$  but  $(1,1)\notin R$ so it is not transitive.

Hence, symmetric but neither reflexive nor transitive.

## Question:10(ii) Give an example of a relation.

(ii) Which is transitive but neither reflexive nor symmetric.

Let

$R = \left \{ \left ( x,y \right ): x> y \right \}$

Now for $x\in R$ ,$(x,x)\notin R$ so it is not reflexive.

Let $(x,y) \in R$  i.e. $x> y$

Then $y> x$ is not possible i.e. $(y,x) \notin R$ . So it is not symmetric.

Let $(x,y) \in R$  i.e. $x> y$    and  $(y,z) \in R$ i.e.$y> z$

we can write this as $x> y> z$

Hence,$x> z$  i.e. $(x,z)\in R$. So it is transitive.

Hence, it is transitive but neither reflexive nor symmetric.

### Question:10(iii) Give an example of a relation.

(iii) Which is Reflexive and symmetric but not transitive.

Let

$A = \left \{ 1,2,3 \right \}$

Define a relation R on A as

$R= \left \{ (1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2) \right \}$

If   $x\in A$ ,$(x,x)\in R$ i.e.$\left \{ (1,1),(2,2),(3,3)\right \} \in R$. So it is reflexive.

If  $x,y\in A$  ,  $(x,y)\in R$   and  $(y,x)\in R$ i.e.$\left \{(1,2),(2,1),(2,3),(3,2) \right \}\in R$. So it is symmetric.

$(x,y)\in R$  and $(y,z)\in R$  i.e. $(1,2)\in R$.  and $(2,3)\in R$

But $(1,3)\notin R$ So it is not transitive.

Hence, it  is Reflexive and symmetric but not transitive.

### Question:10(iv) Give an example of a relation.

(iv) Which is Reflexive and transitive but not symmetric.

Let there be a relation R in R

$R=\left \{ (a,b):a\leq b \right \}$

$(a,a)\in R$  because $a=a$

Let $(a,b)\in R$  i.e.$a\leq b$

But $(b,a)\notin R$ i.e.$b\nleqslant a$

So it is not symmetric.

Let $(a,b)\in R$  i.e.$a\leq b$   and $(b,c)\in R$  i.e. $b\leq c$

This can be written as $a\leq b\leq c$  i.e. $a\leq c$  implies $(a,c)\in R$

Hence, it is transitive.

Thus, it  is Reflexive and transitive but not symmetric.

### Question:10(v) Give an example of a relation.

(v) Which is Symmetric and transitive but not reflexive.

Let there be a relation A in R

$A= \left \{ 1,2 \right \}$

$R=\left \{ (1,2),(2,1),(2,2)\right \}$

$(1,1)\notin R$  So R is not reflexive.

We can see  $(1,2)\in R$  and  $(2,1)\in R$

So it is symmetric.

Let $(1,2)\in R$     and  $(2,1)\in R$

Also  $(2,2)\in R$

Hence, it is transitive.

Thus, it  Symmetric and transitive but not reflexive.

### Question:11 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 points related to a point $P \neq (0, 0)$ is the circle passing through P with origin as centre.

$R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}$

The distance of point P from the origin is always the same as the distance of same point P from origin i.e.$(P,P)\in R$

$\therefore$ R is reflexive.

Let $(P,Q)\in R$ i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

this is the same as distance of the point Q from the origin is the same as the distance of the point P from the origin i.e. $(Q,P)\in R$

$\therefore$R is symmetric.

Let    $(P,Q)\in R$     and    $(Q,S)\in R$

i.e. the distance of point P from the origin is the same as the distance of point Q from the origin, and also the distance of point Q from the origin is the same as the distance of the point S from the origin.

We can say that the distance of point P, Q, S  from the origin is the same. Means distance of point P from the origin is the same as the distance of point S from origin i.e.  $(P,S)\in R$

$\therefore$  R is transitive.

Hence, R  is an equivalence relation.

The set of all points related to a point $P \neq (0, 0)$ are points whose distance from the origin is the same as the distance of point P from the origin.

In other words, we can say there be a point O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.

Hence, these sets of points form a circle with the centre as the origin and this circle passes through the point.

### Question:12 Show that the relation R defined in the set A of all triangles as $R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$, 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?

$R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$

All triangles are similar to itself, so it is reflexive.

Let,

$(T_1,T_2) \in R$  i.e.T1 is similar to T2

T1 is similar to T2 is the same asT2 is similar to T1 i.e. $(T_2,T_1) \in R$

Hence, it is symmetric.

Let,

$(T_1,T_2) \in R$  and  $(T_2,T_3) \in R$  i.e. T1 is similar to T2  and T2 is similar toT3 .

$\Rightarrow$T1 is similar toT3   i.e. $(T_1,T_3) \in R$

Hence, it is transitive,

Thus,  $R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}$, is equivalence relation.

Now, we see the ratio of sides of triangle T1 andT3 are as shown

$\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$

i.e. ratios of sides of T1 and T3 are equal.Hence, T1 and T3 are related.

## Question:13 Show that the relation R defined in the set A of all polygons as $R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$, 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?

$R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$

The same polygon has the same number of sides with itself,i.e. $(P_1,P_2) \in R$, so it is reflexive.

Let,

$(P_1,P_2) \in R$  i.e.P1 have same number of sides as  P2

P1 have the same number of sides as Pis the same as P2 have same number of sides as P1 i.e. $(P_2,P_1) \in R$

Hence,it is symmetric.

Let,

$(P_1,P_2) \in R$  and  $(P_2,P_3) \in R$  i.e. P1 have the same number of sides as P2  and Phave same number of sides as P3

$\Rightarrow$ Phave same number of sides as P3   i.e. $(P_1,P_3) \in R$

Hence, it is transitive,

Thus,  $R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}$, is an equivalence relation.

The elements in A related to the right angle triangle T with sides 3, 4 and 5 are those polygons which have 3 sides.

Hence, the set of all elements in A related to the right angle triangle T is set of all triangles.

### Question:14 Let L be the set of all lines in XY plane and R be the relation in L defined as $R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$. Show that R is an equivalence relation. Find the set of all lines related to the line $y = 2x + 4.$

$R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$

All lines are parallel to itself, so it is reflexive.

Let,

$(L_1,L_2) \in R$  i.e.L1 is parallel to L2.

L1 is parallel to L2 is same as L2 is parallel to L1 i.e. $(L_2,L_1) \in R$

Hence, it is symmetric.

Let,

$(L_1,L_2) \in R$  and  $(L_2,L_3) \in R$  i.e. L1 is parallel to L2  and L2 is parallel  to L3 .

$\Rightarrow$L1 is parallel to L3   i.e. $(L_1,L_3) \in R$

Hence, it is transitive,

Thus,  $R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}$ , is equivalence relation.

The set of all lines related to the line $y = 2x + 4.$ are lines parallel to $y = 2x + 4.$

Here,  Slope = m = 2  and constant = c = 4

It is known that the slope of parallel lines are equal.

Lines parallel to this ( $y = 2x + 4.$ )  line  are $y = 2x + c$  , $c \in R$

Hence, set of all parallel lines to $y = 2x + 4.$ are $y = 2x + c$.

### Question:15 Let R be the relation in the set $\{}1, 2, 3, 4\}$ given by $R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}$. Choose the correct answer.

(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.

A = $\{}1, 2, 3, 4\}$

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

For every  $a \in A$  there is  $(a,a) \in R$

$\therefore$ R is reflexive.

Given, $(1,2) \in R$  but  $(2,1) \notin R$

$\therefore$ R is not symmetric.

For  $a,b,c \in A$ there are $(a,b) \in R \, and \, (b,c) \in R$  $\Rightarrow$ $(a,c) \in R$

$\therefore$ R is transitive.

Hence, R  is reflexive and transitive but not symmetric.

The correct answer is option B.

### Question:16 Let R be the relation in the set N given by $R = \{(a, b) : a = b - 2, b > 6\}$. Choose the correct answer.

(A) $(2, 4) \in R$
(B) $(3,8) \in R$
(C) $(6,8) \in R$
(D) $(8,7) \in R$

$R = \{(a, b) : a = b - 2, b > 6\}$

(A) Since, $b< 6$ so $(2, 4) \notin R$

(B) Since, $3\neq 8-2$  so  $(3,8) \notin R$

(C) Since, $8> 6$  and  $6=8-2$  so  $(6,8) \in R$

(d) Since,$8\neq 7-2$ so  $(8,7) \notin R$

The correct answer is option C.

## NCERT solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.2

Given, $f: R_* \longrightarrow R_{*}$ is defined by  $f(x) = \frac{1}{x}$.

One - One :

$f(x)=f(y)$

$\frac{1}{x}=\frac{1}{y}$

$x=y$

$\therefore$  f is one-one.

Onto:

We have  $y \in R_*$, then there exists $x=\frac{1}{y} \in R_*$     ( Here $y\neq 0$) such that

$f(x)= \frac{1}{(\frac{1}{y})} = y$

$\therefore f is \, \, onto$.

Hence, the function is one-one and onto.

If the domain R is replaced by N with co-domain being same as R∗   i.e.  $g: N \longrightarrow R_{*}$  defined by

$g(x)=\frac{1}{x}$

$g(x_1)=g(x_2)$

$\frac{1}{x_1}=\frac{1}{x_2}$

$x_1=x_2$

$\therefore$  g is one-one.

For   $1.5 \in R_*$ ,

$g(x) = \frac{1}{1.5}$   but there does not exists any x in N.

Hence, function g is one-one but not onto.

(i) $f : N\rightarrow N$ given by $f(x) = x^2$

$f : N\rightarrow N$

$f(x) = x^2$

One- one:

$x,y \in N$  then $f(x)=f(y)$

$x^{2}=y^{2}$

$x=y$

$\therefore$  f is one- one i.e. injective.

For $3 \in N$ there is no x in N such that $f(x)=x^{2}=3$

$\therefore$  f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

(ii) $f : Z \rightarrow Z$ given by $f(x) = x^2$

$f : Z \rightarrow Z$

$f(x) = x^2$

One- one:

For   $-1,1 \in Z$  then $f(x) = x^2$

$f(-1)= (-1)^{2}$

$f(-1)= 1$   but  $-1 \neq 1$

$\therefore$  f is not one- one i.e. not injective.

For  $-3 \in Z$  there is no x in Z such that $f(x)=x^{2}= -3$

$\therefore$  f is not onto i.e. not surjective.

Hence, f is neither injective nor surjective.

(iii) $f: R \rightarrow R$ given by $f(x) = x^2$

$f: R \rightarrow R$

$f(x) = x^2$

One- one:

For   $-1,1 \in R$  then $f(x) = x^2$

$f(-1)= (-1)^{2}$

$f(-1)= 1$   but  $-1 \neq 1$

$\therefore$  f is not one- one i.e. not injective.

For  $-3 \in R$  there is no x in R such that $f(x)=x^{2}= -3$

$\therefore$  f is not onto i.e. not surjective.

Hence, f is not injective and not surjective.

(iv) $f: N \rightarrow N$ given by $f(x) = x^3$

$f : N\rightarrow N$

$f(x) = x^3$

One- one:

$x,y \in N$  then $f(x)=f(y)$

$x^{3}=y^{3}$

$x=y$

$\therefore$  f is one- one i.e. injective.

For $3 \in N$ there is no x in N such that $f(x)=x^{3}=3$

$\therefore$  f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

(v)  $f : Z \rightarrow Z$ given by $f(x) = x^3$

$f : Z \rightarrow Z$

$f(x) = x^3$

One- one:

For   $(x,y) \in Z$  then $f(x) = f(y)$

$x^{3}=y^{3}$

$x=y$

$\therefore$  f is one- one i.e. injective.

For  $3 \in Z$  there is no x in Z such that $f(x)=x^{3}= 3$

$\therefore$  f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

$f : R\longrightarrow R$

$f (x) = [x]$

One- one:

For   $1.5,1.7 \in R$   then $f(1.5)=\left [ 1.5 \right ] = 1$     and    $f(1.7)=\left [ 1.7 \right ] = 1$

but   $1.5\neq 1.7$

$\therefore$  f is not one- one i.e. not injective.

For  $0.6 \in R$  there is no x in R such that $f(x)=\left [ 0.6 \right ]$

$\therefore$  f is not onto i.e. not surjective.

Hence, f is not injective but not surjective.

$f : R \rightarrow R$

$f (x) = | x |$

$f (x) = | x | = x \, if\, x\geq 0 \,\, and \, \, -x\, if\, x< 0$

One- one:

For   $-1,1 \in R$   then $f (-1) = | -1 |= 1$

$f (1) = | 1 |= 1$

$-1\neq 1$

$\therefore$  f is not one- one i.e. not injective.

For  $-2 \in R$,

We know $f (x) = | x |$   is always positive there is no x in R such that $f (x) = | x |=-2$

$\therefore$  f is not onto i.e. not surjective.

Hence,  $f (x) = | x |$, is neither one-one nor onto.

$f : R \rightarrow R$  is given by

$f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right.$

As we can see    $f(1)=f(2)=1$   ,  but $1\neq 2$

So it is not one-one.

Now, f(x) takes only 3 values (1,0,-1) for the element -3 in codomain $R$ ,there does not exists x in domain $R$ such that $f(x)= -3$.

So it is not onto.

Hence, signum function is neither one-one nor onto.

$A = \{1, 2, 3\}$

$B = \{4, 5, 6, 7\}$

$f = \{(1, 4), (2, 5), (3, 6)\}$

$f : A \rightarrow B$

$\therefore$  $f(1)=4,f(2)=5,f(3)=6$

Every element of A has a distant value in f.

Hence, it is one-one.

(i) $f: R\rightarrow R$ defined by $f(x) = 3 -4x$

$f: R\rightarrow R$

$f(x) = 3 -4x$

Let  there  be  $(a,b) \in R$  such that $f(a)=f(b)$

$3-4a = 3 -4b$

$-4a = -4b$

$a = b$

$\therefore$ f is one-one.

Let there be $y \in R$,    $y = 3 -4x$

$x = \frac{(3-y)}{4}$

$f(x) = 3 -4x$

Puting value of x,    $f(\frac{3-y}{4}) = 3 - 4(\frac{3-y}{4})$

$f(\frac{3-y}{4}) = y$

$\therefore$  f is onto.

f is both one-one and onto hence, f is bijective.

(ii) $f : R\rightarrow R$ defined by $f(x) = 1 + x^2$

$f : R\rightarrow R$

$f(x) = 1 + x^2$

Let  there  be  $(a,b) \in R$  such that $f(a)=f(b)$

$1+a^{2} = 1 +b^{2}$

$a^{2}=b^{2}$

$a = \pm b$

For   $f(1)=f(-1)=2$   and  $1\neq -1$

$\therefore$ f is not one-one.

Let there be $-2 \in R$   (-2 in codomain of R)

$f(x) = 1 + x^2 = -2$

There does not exists any x in domain R  such that  $f(x) = -2$

$\therefore$  f is not onto.

Hence, f is neither one-one nor onto.

$f : A \times B \rightarrow B \times A$

$f (a, b) = (b, a)$

Let $(a_1,b_1),(a_2,b_2) \in A\times B$

such that  $f (a_1, b_1) = f(a_2, b_2)$

$(b_1,a_1)=(b_2,a_2)$

$\Rightarrow$     $b_1= b_2$   and   $a_1= a_2$

$\Rightarrow$          $(a_1,b_1) = (a_2,b_2)$

$\therefore$    f is one- one

Let,  $(b,a) \in B\times A$

then there exists $(a,b) \in A\times B$  such that  $f (a, b) = (b, a)$

$\therefore$ f is onto.

Hence, it is bijective.

$f : N \rightarrow N$   ,  $n\in N$

$f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;evem \end{matrix}\right.$

Here we can observe,

$f(2)=\frac{2}{2}=1$           and       $f(1)=\frac{1+1}{2}=1$

As we can see $f(1)=f(2)=1$  but $1\neq 2$

$\therefore$     f is not one-one.

Let,$n\in N$    (N=co-domain)

case1   n be even

For $r \in N$,      $n=2r$

then there is $4r \in N$ such that $f(4r)=\frac{4r}{2}=2r$

case2   n be odd

For  $r \in N$,   $n=2r+1$

then there is $4r+1 \in N$ such that $f(4r+1)=\frac{4r+1+1}{2}=2r +1$

$\therefore$  f is onto.

f is not one-one but onto

hence, the function f is not bijective.

$A = R - \{3\}$

$B = R - \{1\}$

$f : A\rightarrow B$

$f(x) = \left (\frac{x-2}{x-3} \right )$

Let $a,b \in A$ such that  $f(a)=f(b)$

$\left (\frac{a-2}{a-3} \right ) = \left ( \frac{b-2}{b-3} \right )$

$(a-2)(b-3)=(b-2)(a-3)$

$ab-3a-2b+6=ab-2a-3b+6$

$-3a-2b=-2a-3b$

$3a+2b= 2a+3b$

$3a-2a= 3b-2b$

$a=b$

$\therefore$  f is one-one.

Let,  $b \in B = R - \{1\}$      then   $b\neq 1$

$a \in A$ such that   $f(a)=b$

$\left (\frac{a-2}{a-3} \right ) =b$

$(a-2)=(a-3)b$

$a-2 = ab -3b$

$a-ab = 2 -3b$

$a(1-b) = 2 -3b$

$a= \frac{2-3b}{1-b}\, \, \, \, \in A$

For any $b \in B$ there exists  $a= \frac{2-3b}{1-b}\, \, \, \, \in A$   such that

$f(\frac{2-3b}{1-b}) = \frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}$

$=\frac{2-3b-2+2b}{2-3b-3+3b}$

$=\frac{-3b+2b}{2-3}$

$= b$

$\therefore$  f is onto

Hence, the function is one-one and onto.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

$f : R \rightarrow R$

$f(x) = x^4$

One- one:

For   $a,b \in R$  then $f(a) = f(b)$

$a^{4}=b^{4}$

$a=\pm b$

$\therefore f(a)=f(b)$ does not imply that $a=b$

example: and $2\neq -2$

$\therefore$  f is not  one- one

For  $2\in R$  there is no x in R such that $f(x)=x^{4}= 2$

$\therefore$  f is not onto.

Hence, f is neither one-one nor onto.

Option D is correct.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

$f : R\rightarrow R$

$f(x) = 3x$

One - One :

Let $\left ( x,y \right ) \in R$

$f(x)=f(y)$

$3x=3y$

$x=y$

$\therefore$  f is one-one.

Onto:

We have  $y \in R$, then there exists $x=\frac{y}{3} \in R$    such that

$f(\frac{y}{3})= 3\times \frac{y}{3} = y$

$\therefore f is \, \, onto$.

Hence, the function is one-one and onto.

The correct answer is A .

## CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions: Exercise: 1.3

Given :       $f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}$             and       $g : \{1, 2, 5\} \rightarrow \{1, 3\}$

$f = \{(1, 2), (3, 5), (4, 1)\}$         and      $g = \{(1, 3), (2, 3), (5, 1)\}$

$gof(1) = g(f(1))=g(2) = 3$                   $\left [ f(1)=2 \, and\, g(2)=3 \right ]$

$gof(3) = g(f(3))=g(5) = 1$                   $\left [ f(3)=5 \, and\, g(5)=1 \right ]$

$gof(4) = g(f(4))=g(1) = 3$                 $\left [ f(4)=1 \, and\, g(1)=3 \right ]$

Hence,   $gof$ = $\left \{ (1,3),(3,1),(4,3) \right \}$

To prove : $\\(f + g) o h = foh + goh$

$((f + g) o h)(x)$

$=(f + g) ( h(x) )$

$=f ( h(x) ) +g(h(x))$

$=(f o h)(x) +(goh)(x)$

$=\left \{ (f o h) +(goh) \right \}(x)$                      $x\forall R$

Hence, $\\(f + g) o h = foh + goh$

To prove:$(f \cdot g) o h = (foh) \cdot (goh)$

$((f . g) o h)(x)$

$=(f . g) ( h(x) )$

$=f ( h(x) ) . g(h(x))$

$=(f o h)(x) . (goh)(x)$

$=\left \{ (f o h) .(goh) \right \}(x)$                      $x\forall R$

$\therefore$   $(f \cdot g) o h = (foh) \cdot (goh)$

Hence, $(f \cdot g) o h = (foh) \cdot (goh)$

(i) $f (x) = | x |$  and  $g(x) = \left | 5x-2 \right |$

$f (x) = | x |$ and $g(x) = \left | 5x-2 \right |$

$gof$ $= g(f(x))$

$= g( | x |)$

$= |5 | x |-2|$

$fog$ $= f(g(x))$

$=f( \left | 5x-2 \right |)$

$=\left \| 5x-2 \right \|$

$=\left | 5x-2 \right |$

Question:3(ii) Find gof and fog, if

(ii) $f (x) = 8x^{3}$and $g(x) = x^{\frac{1}{3}}$

The solution is as follows

(ii)  $f (x) = 8x^{3}$and $g(x) = x^{\frac{1}{3}}$

$gof$  $= g(f(x))$

$= g( 8x^{3})$

$= ( 8x^{3})^{\frac{1}{3}}$

$=2x$

$fog$ $= f(g(x))$

$=f(x^{\frac{1}{3}} )$

$=8((x^{\frac{1}{3}} )^{3})$

$=8x$

$f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}$

$fof (x) = x$

$(fof) (x) = f(f(x))$

$=f( \frac{4x + 3}{6x - 4})$

$=\frac{4( \frac{4x + 3}{6x - 4}) +3}{6( \frac{4x + 3}{6x - 4}) -4}$

$= \frac{16x+12+18x-12}{24x+1824x+16}$

$= \frac{34x}{34}$

$\therefore fof(x) = x$                ,  for all  $x \neq \frac{2}{3}$

$\Rightarrow fof=Ix$

Hence,the given function $f$is invertible and the inverse of $f$ is $f$ itself.

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$

with $f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$ with
$f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

From the given definition,we have:

$f\left ( 1 \right )=f\left ( 2 \right )=f\left ( 3 \right )=f(4)=10$

$\therefore$ f is not one-one.

Hence, f do not have an inverse function.

(ii)  $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

(ii)  $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

From the definition, we can conclude :

$g(5)=g(7)=4$

$\therefore$ g is not one-one.

Hence, function g does not have inverse function.

Question:5(iii) State with reason whether following functions have inverse

(iii)   $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

(iii)   $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

From the definition, we can see the set $\left \{ 2,3,4,5 \right \}$ have distant values under h.

$\therefore$ h is one-one.

For every element y of set $\left \{ 7,9,11,13 \right \}$,there exists an element x  in $\left \{ 2,3,4,5 \right \}$ such that  $h(x)=y$

$\therefore$ h is onto

Thus, h is one-one and onto so h has an inverse function.

$f : [-1, 1] \rightarrow R$

$f(x) = \frac{x}{(x + 2)}$

One -one:

$f(x)=f(y)$

$\frac{x}{x+2}=\frac{y}{y+2}$

$x(y+2)=y(x+2)$

$xy+2x=xy+2y$

$2x=2y$

$x=y$

$\therefore$   f is one-one.

It is clear that  $f : [-1, 1] \rightarrow Range f$ is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in  $Range f\rightarrow [-1, 1]$

$g: Range f\rightarrow [-1, 1]$

let y be an arbitrary element of range f

Since, $f : [-1, 1] \rightarrow R$ is onto, so

$y=f(x)$  for $x \in \left [ -1,1 \right ]$

$y=\frac{x}{x+2}$

$xy+2y=x$

$2y=x-xy$

$2y=x(1-y)$

$x = \frac{2y}{1-y}$     ,$y\neq 1$

$g(y) = \frac{2y}{1-y}$

$f^{-1}=\frac{2y}{1-y},y\neq 1$

$f : R \rightarrow R$  is given by  $f (x) = 4x + 3$

One-one :

Let  $f(x)=f(y)$

$4x + 3 = 4y+3$

$4x=4y$

$x=y$

$\therefore$ f is one-one function.

Onto:

$y=4x+3\, \, \, , y \in R$

$\Rightarrow x=\frac{y-3}{4} \in R$

So, for $y \in R$ there is  $x=\frac{y-3}{4} \in R$   ,such that

$f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3$

$= y-3+3$

$= y$

$\therefore$ f  is onto.

Thus, f is one-one and onto so $f^{-1}$ exists.

Let, $g:R\rightarrow R$ by $g(x)=\frac{y-3}{4}$

Now,

$(gof)(x)= g(f(x))= g(4x+3)$

$=\frac{(4x+3)-3}{4}$

$=\frac{4x}{4}$

$=x$

$(fog)(x)= f(g(x))= f(\frac{y-3}{4})$

$= 4\times \frac{y-3}{4}+3$

$= y-3+3$

$= y$

$(gof)(x)= x$             and     $(fog)(x)= y$

Hence, function f is invertible and inverse of f is $g(y)=\frac{y-3}{4}$.

It is given that
$f : R^+ \rightarrow [4,\infty)$  ,  $f(x) = x^2+4$  and

Now, Let f(x) = f(y)

⇒ x2 + 4 = y2 + 4

⇒ x2 = y2

⇒ x = y

⇒ f is one-one function.

Now, for y $\epsilon$[4, ∞), let y = x2 + 4.

⇒ x2 = y -4 ≥ 0

⇒ for any y $\epsilon$ R, there exists x =  $\epsilon$ R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

Now, gof(x) = g(f(x)) = g(x2 + 4) =

And, fog(y) = f(g(y)) =  =

Therefore, gof = gof = IR.

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) =

$f : R_+ \rightarrow [- 5, \infty)$

$f (x) = 9x^2 + 6x - 5$

One- one:

Let       $f(x)=f(y)$    $for \, \, x,y\in R$

$9x^{2}+6x-5=9y^{2}+6y-5$

$9x^{2}+6x=9y^{2}+6y$

$\Rightarrow$              $9(x^{2}-y^{2})=6(y-x)$

$9(x+y)(x-y)+6(x-y)= 0$

$(x-y)(9(x+y)+6)=0$

Since, x and y are positive.

$(9(x+y)+6)> 0$

$\therefore x=y$

$\therefore$   f is one-one.

Onto:

Let for  $y \in [-5,\infty)$  , $y=9x^{2}+6x-5$

$\Rightarrow$         $y=(3x+1)^{2}-1-5$

$\Rightarrow$             $y=(3x+1)^{2}-6$

$\Rightarrow$                  $y+6=(3x+1)^{2}$

$(3x+1)=\sqrt{y+6}$

$x = \frac{\sqrt{y+6}-1}{3}$

$\therefore$ f is onto  and range is $y \in [-5,\infty)$.

Since f is one-one and onto so it is invertible.

Let  $g : [-5,\infty)\rightarrow R_+$    by  $g(y) = \frac{\sqrt{y+6}-1}{3}$

$(gof)(x)=g(f(x))=g(9x^{2}+6x)-5=g((3x+1)^{2}-6)\\=\sqrt{{(3x+1)^{2}}-6+6} -1$

$(gof)(x)=\frac{3x+1-1}{3}=\frac{3x}{3}= x$

$(fog)(x)=f(g(x))=f(\frac{\sqrt{y+6}-1}{3})$

$=[3(\frac{\sqrt{y+6}-1}{3})+1]^{2}-6$

$=(\sqrt{y+6})^{2}-6$

$=y+6-6$

$=y$

$\therefore gof=fog=I_R$

Hence,  $f$ is invertible with the inverse $f^{-1}$ of $f$ given by  $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

Let $f : X \rightarrow Y$be an invertible function

Also, suppose f has two inverse $g_1 and g_2$

For $y \in Y$, we have

$fog_1(y) = I_y(y)=fog_2(y)$

$\Rightarrow$     $f(g_1(y))=f(g_2(y))$

$\Rightarrow$             $g_1(y)=g_2(y)$                     [f is invertible implies f is one - one]

$\Rightarrow$                     $g_1=g_2$                        [g is one-one]

Thus,f has a unique inverse.

It is given that
$f : \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$

$f(1) = a, f(2) = b \ and \ f(3) = c$

Now,, lets define a function g :
$\left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$  such that

$g(a) = 1, g(b) = 2 \ and \ g(c) = 3$
Now,

$(fog)(a) = f(g(a)) = f(1) = a$
Similarly,

$(fog)(b) = f(g(b)) = f(2) = b$

$(fog)(c) = f(g(c)) = f(3) = c$

And

$(gof)(1) = g(f(1)) = g(a) = 1$

$(gof)(2) = g(f(2)) = g(b) = 2$

$(gof)(3) = g(f(3)) = g(c) = 3$

Hence, $gof = I_X$ and$fog = I_Y$, where $X = \left \{ 1,2,3 \right \}$ and$Y = \left \{ a,b,c \right \}$

Therefore, the inverse of f exists and $f^{-1} = g$

Now,
$f^{-1} : \left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}$  is given by

$f^{-1}(a) = 1, f^{-1}(b) = 2 \ and \ f^{-1}(c) = 3$

Now, we need to  find the inverse of  $f^{-1}$,

Therefore, lets define$h: \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}$  such that

$h(1) = a, h(2) = b \ and \ h(3) = c$

Now,

$(goh)(1) = g(h(1)) = g(a) = 1$

$(goh)(2) = g(h(2)) = g(b) = 2$

$(goh)(3) = g(h(3)) = g(c) = 3$

Similarly,

$(hog)(a) = h(g(a)) = h(1) = a$

$(hog)(b) = h(g(b)) = h(2) = b$

$(hog)(c) = h(g(c)) = h(3) = c$

Hence,  $goh = I_X$ and $hog = I_Y$, where$X = \left \{ 1,2,3 \right \}$ and$Y = \left \{ a,b,c \right \}$

Therefore,  inverse of  $g^{-1} = (f^{-1})^{-1}$  exists and  $g^{-1} = (f^{-1})^{-1} = h$

$\Rightarrow h = f$

Therefore,  $(f^{-1})^{-1} = f$

Hence proved

$f : X \rightarrow Y$

To prove: $(f^{-1})^{-1} = f$

Let  $f:X\rightarrow Y$  be a invertible function.

Then there is    $g:Y\rightarrow X$  such that   $gof =I_x$  and $fog=I_y$

Also,       $f^{-1}= g$

$gof =I_x$  and $fog=I_y$

$\Rightarrow$        $f^{-1}of = I_x$            and         $fof^{-1} = I_y$

Hence, $f^{-1}:Y\rightarrow X$    is invertible function and f is inverse of $f^{-1}$.

i.e. $(f^{-1})^{-1} = f$

(A) $x^{\frac{1}{3}}$

(B) $x^3$

(C) $x$

(D) $(3 - x^3)$

$f(x) = (3 - x^3)^{\frac{1}{3}}$

$fof(x)$$=f(f(x))=f((3-x^{3})^{\frac{1}{3}})$

$=[3- ((3-x^{3})^{\frac{1}{3}})^{3}]^{\frac{1}{3}}$

$=([3- ((3-x^{3})]^{\frac{1}{3}})$

$= (x^{3})^{\frac{1}{3}}$

$=x$

Thus,$fof(x)$ is x.

Hence, option c is correct answer.

(A) $g(y) = \frac{3y}{3 -4y}$

(B) $g(y) = \frac{4y}{4 -3y}$

(C) $g(y) = \frac{4y}{3 -4y}$

(D) $g(y) = \frac{3y}{3 -4y}$

$f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$

$f(x) = \frac{4x}{3x + 4}$

Let f inverse  $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$

Let y be the element of range f.

Then there is $x \in R - \left\{-\frac{4}{3}\right\}$    such that

$y=f(x)$

$y=\frac{4x}{3x+4}$

$y(3x+4)=4x$

$3xy+4y=4x$

$3xy-4x+4y=0$

$x(3y-4)+4y=0$

$x= \frac{-4y}{3y-4}$

$x= \frac{4y}{4-3y}$

Now , define$g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$     as $g(y)= \frac{4y}{4-3y}$

$gof(x)= g(f(x))= g(\frac{4x}{3x+4})$

$= \frac{4(\frac{4x}{3x+4})}{4-3(\frac{4x}{3x+4})}$

$=\frac{16x}{12x+16-12x}$

$=\frac{16x}{16}$

$=x$

$fog(y)=f(g(y))=f(\frac{4y}{4-3y})$

$= \frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y}) + 4}$

$=\frac{16y}{12y+16-12y}=\frac{16y}{16}$

$=y$

Hence, g is inverse of f and $f^{-1}=g$

The inverse of f is given by   $g(y)= \frac{4y}{4-3y}$.

The correct option is B.

## Solutions of NCERT for class 12 maths chapter 1 Relations and Functions: Exercise 1.4

(i) On $Z^+$ , define ∗ by $a * b = a - b$

(i) On $Z^+$ , define ∗ by $a * b = a - b$

It is not a binary operation as the image of $(1,2)$ under * is $1\ast 2=1-2$ $=-1 \notin Z^{+}$.

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

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

We can observe that for $a,b \in Z^+$,there is a unique element ab in $Z^+$.

This means * carries each pair $(a,b)$  to a unique element $a * b = ab$ in $Z^+$.

Therefore,* is a binary operation.

(iii) On $R$, define ∗ by $a * b = ab^2$

(iii) On $R$, define ∗ by $a * b = ab^2$

We can observe that for $a,b \in R$,there is a unique element $ab^{2}$  in $R$.

This means * carries each pair $(a,b)$  to a unique element $a * b = ab^{2}$ in $R$.

Therefore,* is a binary operation.

(iv) On $Z^+$, define ∗ by $a * b = | a - b |$

(iv) On $Z^+$, define ∗ by $a * b = | a - b |$

We can observe that for $a,b \in Z^+$,there is a unique element $| a - b |$  in $Z^+$.

This means * carries each pair $(a,b)$  to a unique element $a * b = | a - b |$ in $Z^+$.

Therefore,* is a binary operation.

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

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

* carries each pair $(a,b)$  to a unique element $a * b = a$ in  $Z^+$.

Therefore,* is a binary operation.

(i)On $Z$, define $a * b = a-b$

a*b=a-b

b*a=b-a

$a*b\neq b*a$

so * is not commutative

(a*b)*c=(a-b)-c

a*(b*c)=a-(b-c)=a-b+c

(a*b)*c not equal to a*(b*c), so * is not associative