# NCERT Solutions for Exercise 1.1 Class 12 Maths Chapter 1 - Relations and Functions

NCERT Solutions for Class 12 maths chapter 1 deals with questions related to various concepts of Relations and Functions which includes types of relations, functions, binary operations etc. Exercise 1.1 Class 12 Maths will help students to grasp the basic concepts of sets and relations. It is highly recommended to students to practise the NCERT Solutions for Class 12 Maths chapter 1 exercise 1.1 to score well in CBSE class 12 board exam. In competitive exams also like JEE main ,some questions can be asked from Class 12 Maths chapter 1 exercise 1.1. Concepts related to functions discussed in Class 12th Maths chapter 1 exercise 1.1 are important for Board examination also. The NCERT chapter Relations and Functions also has the following exercise for practice.

• Relations and Functions Exercise 1.2

• Relations and Functions Exercise 1.3

• Relations and Functions Exercise 1.4

• Relations and Functions Miscellaneous Exercise

## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions: Exercise 1.1

Question1(i) . 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 $R = \{(x,y): 3x - y = 0\}$

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

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

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.

Question 1(ii) . Determine whether each of the following relations are reflexive, symmetric and
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.

Question:4 Show that the relation R in R defined as $R = \{(a, b) : a \leq b\}$ , is reflexive and

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 T 1 with sides 3, 4, 5, T 2 with sides 5, 12, 13 and T 3 with sides 6, 8, 10. Which triangles among T 1 , T 2 and T 3 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.T 1 is similar to T2

T 1 is similar to T2 is the same asT2 is similar to T 1 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. T 1 is similar to T2 and T2 is similar toT 3 .

$\Rightarrow$ T 1 is similar toT 3 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 T 1 andT 3 are as shown

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

i.e. ratios of sides of T 1 and T 3 are equal.Hence, T 1 and T 3 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.P 1 have same number of sides as P 2

P 1 have the same number of sides as P 2 is the same as P 2 have same number of sides as P 1 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. P 1 have the same number of sides as P 2 and P 2 have same number of sides as P 3

$\Rightarrow$ P 1 have same number of sides as P 3 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.L 1 is parallel to L 2 .

L1 is parallel to L 2 is same as L 2 is parallel to L 1 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 L 2 and L 2 is parallel to L 3 .

$\Rightarrow$ L 1 is parallel to L 3 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 $y = 2x + c$ .

### 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.

Also Read| NCERT Notes For Class 12 Mathematics Chapter 1

More About NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1

The NCERT Class 12 maths chapter Relations and Functions has a total of 5 exercises including miscellaneous. Exercise 1.1 Class 12 Maths covers solutions to 16 main questions and their sub-questions. The initial 10 questions are based on concepts like symmetric, reflexive and transitive relation and subsequent questions upto 15 are based in equivalence relation etc. NCERT Solutions for Class 12 Maths chapter 1 exercise 1.1 is good source to learn concepts related to symmetric relations, equivalence of a relation etc.

Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1

• The Class 12th maths chapter 1 exercise provided here is in detail which is solved by subject matter experts .

• Students are recommended to practice Exercise 1.1 Class 12 Maths to prepare for exams, direct questions are asked in Board exams.

• These NCERT text book Class 12 Maths chapter 1 exercise 1.1 solutions can be referred by students to revise just before the exam.

• NCERT Syllabus Class 12 Maths chapter 1 exercise 1.1 provided here are one stop solutions for students aspiring to score well in examinations.

### Also see-

• NCERT exemplar solutions class 12 maths chapter 1

• NCERT solutions for class 12 maths chapter 1

NCERT Solutions Subject Wise

• NCERT solutions class 12 chemistry

• NCERT solutions for class 12 physics

• NCERT solutions for class 12 biology

• NCERT solutions for class 12 mathematics

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• NCERT Exemplar Class 12th Maths
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