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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 01:14 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Maths Chapter 1 Exercise 1.1

NCERT Solutions for Exercise 1.1 Class 12 Maths Chapter 1 Relations and Functions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 12 Maths ex 1.1 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.

12th class Maths exercise 1.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

##### VMC VIQ Scholarship Test

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

##### Pearson | PTE

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer! Trusted by 3,500+ universities globally

## Assess NCERT Solutions for Class 12 Maths Chapter 1 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.

(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.

(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.

(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.

(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.

(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.

(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.

(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.

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

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.

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.

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

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.

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

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

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.

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.?

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

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

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

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.

(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.

(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.

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.

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.

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.

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.

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

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.

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

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.

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

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.

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

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 $y = 2x + c$ .

(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.

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.

(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.

## 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. Students can get access of NCERT Notes For Class 12 Mathematics Chapter 1 which can be used for quick revision of important concepts of this chapter.

## 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.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

## Key Features Of NCERT Solutions for Exercise 1.1 Class 12 Maths Chapter 1

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 1.1 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 1.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 1.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 1.1 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for ex 1.1 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for class 12 maths ex 1.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

## Subject wise NCERT Exemplar solutions

Happy learning!!!

1. Which concepts are covered in Exercise 1.1 Class 12 Maths?

Concepts related to symmetric, reflexive and transitive, equivalence relations etc, are discussed in the Exercise 1.1 Class 12 Maths

2. What are the important topics in chapter relations and functions ?

Topics like

• Introduction to relations and functions,
• types of relations, types of functions,
• composition of functions etc. are mentioned in this chapter.
3. What is the weightage of the chapter relations and functions for CBSE board exam ?

Two chapters 'relation and function' and 'inverse trigonometry' combined has 10 % weightage in the CBSE final board exam.

4. How are the NCERT solutions helpful in the board exam ?

From the analysis of previous year questions of Board exams, it is clear that direct questions are asked from the NCERT questions. Also Some of the questions are repeated year after year. Hence it is said that NCERT solutions are low hanging fruits. Every serious student must practice NCERT questions to score well in the exam.

5. What are relations in Class 12 Maths?

In maths, relation defines the relationship between sets of values of ordered pairs

6. What are some types of relations discussed in Exercise 1.1 Class 12 Maths

symmetric, reflexive and transitive, equivalence relations etc, are discussed in the Exercise 1.1 Class 12 Maths

7. How many questions are covered in Exercise 1.1 Class 12 Maths ?

There are 15 questions in Exercise 1.1 Class 12 Maths

8. Mention the total number of exercises in NCERT class 12 maths chapter 1 relations and functions?

In  NCERT class 12 maths chapter 1 relations and function, there are a total of 5 exercises which includes a miscellaneous exercise also.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:05 August,2024 - 20 September,2024

#### Meghalaya Board of Secondary School Leaving Certificate Examination

Application Date:05 September,2024 - 23 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

#### National Institute of Open Schooling 10th examination

Admit Card Date:13 September,2024 - 07 October,2024

#### National Institute of Open Schooling 12th Examination

Admit Card Date:13 September,2024 - 07 October,2024

Edx
1113 courses
Coursera
804 courses
Udemy
394 courses
Futurelearn
222 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD

### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

• Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
• Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
• Practice Regularly: Consistent practice is key to mastering chemistry.
2. Consider Professional Help:

• Tutoring: A tutor can provide personalized guidance and support.
• Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
3. Explore Alternative Options:

• Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
• Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
4. Focus on NEET 2025 Preparation:

• Stay Dedicated: Continue your NEET preparation with renewed determination.
• Utilize Resources: Make use of study materials, online courses, and mock tests.
5. Seek Support:

• Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
• Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9