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

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## NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1 - Free PDF Download

The NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions includes solutions to all exercise 1.1 problems. These NCERT Solutions for Class 12 Maths are prepared by experts to be simple as per NCERT (CBSE) books guidelines, allowing students to grasp the topics fully. Students can use the NCERT Solutions for Class 12 Mathematics to improve their math skills.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.1 - Free PDF Download
2. Glance on NCERT Solutions for Exercise 1.1 Class 12 Maths Chapter 1
3. Topics Covered in Class 12 Maths Chapter 1 Exercise 1.1
4. Access NCERT Class 12 Maths Chapter 1 Exercise 1.1 Solutions – Relations and Functions
4.1Exercise 1.1
5. NCERT Solutions for Class 12 Maths Chapter 1 Exercises
6. CBSE Class 12 Maths Chapter 1 Other Study Materials
FAQs

## Glance on NCERT Solutions for Exercise 1.1 Class 12 Maths Chapter 1

• Chapter 1 of Class 12 Maths, and specifically Exercise 1.1, covers the basic terminology and applications of Relations and Functions.

• Ex 1.1 Class 12 maths NCERT Solutions has over all 16 Questions with 14 short answers and 2 multiple choice questions.

• Relations are a collection of ordered pairs from a set A to a set B. In other words, a relation R on a set A is a subset of the Cartesian product A x A.

• Class 12 Ex 1.1 likely focuses on identifying different types of relations, including:

• Reflexive: Every element in the set is related to itself.

• Symmetric: If (a, b) is in the relation, then (b, a) must also be in the relation.

• Transitive: If (a, b) and (b, c) are in the relation, then (a, c) must also be in the relation.

• Equivalence: A relation that is reflexive, symmetric, and transitive.

• The exercise provides various questions where you will be asked to analyze relations based on the above concepts. Such as:

• Identify the Sets: Recognize the sets involved in the relation.

• Examine the Relationship: Understand the rule that defines how elements are related in the given relation.

• Verify Properties: Check if the relation satisfies the properties of reflexive, symmetric, or transitive based on the rule and the sets.

## Topics Covered in Class 12 Maths Chapter 1 Exercise 1.1

• Relation: Understanding what a relation is, its representation, and the difference between relations and functions.

• Empty and Universal Relations: Identifying and working with empty relations (where no elements are related) and universal relations (where all elements are related).

• Properties of Relations: These include reflexive, symmetric, and transitive.

These concepts are used in many areas of math, including algebra, calculus, and discrete mathematics.

Competitive Exams after 12th Science

## Access NCERT Class 12 Maths Chapter 1 Exercise 1.1 Solutions – Relations and Functions

### Exercise 1.1

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

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

Ans: The given relation is: $\text{R = }\left\{ \left( \text{1, 3} \right)\text{, }\left( \text{2, 6} \right)\text{, }\left( \text{3, 9} \right)\text{, }\left( \text{4, 12} \right) \right\}$

Since $\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{ }...$ and $\left( \text{14, 14} \right)\notin R$.

We conclude that $\text{R}$ is not reflexive.

Since $\left( \text{1, 3} \right)\in \text{R}$, but $\left( \text{3, 1} \right)\notin \text{R}$. since $\text{3}\left( \text{3} \right)\text{-1}\ne \text{0}$

We conclude that $\text{R}$ is not symmetric.

Since $\left( \text{1, 3} \right)$ and $\left( \text{3, 9} \right)\in \text{R}$, but$\left( \text{1, 9} \right)\notin \text{R}\text{. }\left[ \text{3}\left( \text{1} \right)\text{-9}\ne \text{0} \right]$.

We conclude that $\text{R}$ is not transitive.

Therefore, the relation $\text{R}$ is not reflexive, symmetric or transitive.

1. Relation $\text{R}$ in the set $\text{N}$ of natural numbers defined as $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: y = x + 5}$ and $\text{x<4 }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\left\{ \left( \text{1, 6} \right)\text{, }\left( \text{2, 7} \right)\text{, }\left( \text{3, 8} \right) \right\}$.

Since $\left( \text{1, 1} \right)\notin \text{R}$.

We conclude that $\text{R}$ is not reflexive.

Since $\left( \text{1, 6} \right)\in \text{R}$ but $\left( \text{6, 1} \right)\notin \text{R}$.

We conclude that $\text{R}$ is not symmetric.

In the given relation $\text{R}$ there is not any ordered pair such that $\left( \text{x, y} \right)$ and $\left( \text{y, z} \right)$ both $\in \text{R}$, therefore we can say that $\left( \text{x, z} \right)$ cannot belong to $\text{R}$.

Therefore $\text{R}$ is not transitive.

Hence, the given relation $\text{R}$ is not reflexive, symmetric or transitive.

1. Relation $\text{R}$in the set $\text{A = }\!\!\{\!\!\text{ 1, 2, 3, 4, 5, 6 }\!\!\}\!\!\text{ }$ as $\text{R= }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: y}$is divisible by $\text{x }\!\!\}\!\!\text{ }$

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{ : y}$ is divisible by $\text{x }\!\!\}\!\!\text{ }$

As we know that any number except $\text{0}$ is divisible by itself, therefore $\left( \text{x, x} \right)\in \text{R}$.

We conclude that $\text{R}$ is reflexive.

Since $\left( \text{2, 4} \right)\in \text{R}$ (because 4 is divisible by 2, but $\left( \text{4, 2} \right)\notin \text{R}$ (since 2 is not divisible by 4.

We conclude that $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)$ and $\left( \text{y, z} \right)\in \text{R}$, $\text{y}$ is divisible by $\text{x}$ and $\text{z}$ is divisible by $\text{y}$. Hence $\text{z}$ is divisible by $\text{x}\Rightarrow \left( \text{x, z} \right)\in \text{R}$.

We conclude that $\text{R}$ is transitive.

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

1. Relation $\text{R}$ in the set $\text{Z}$ of all integers defined as $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x - y }\!\!\}\!\!\text{ }$ is as integer

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x - y}$ is an integer$\text{ }\!\!\}\!\!\text{ }$

If $\text{x}\in \text{Z, }\left( \text{x, x} \right)\in \text{R}$ because $\text{x-x = 0}$ is an integer.

Hence, we conclude that $\text{R}$ is reflexive.

For $\text{x, y}\in \text{Z}$, if $\left( \text{x, y} \right)\in \text{R}$, then $\text{x - y}$ is an integer and therefore $\left( \text{y-x} \right)$ is also an integer.

Therefore, we conclude that $\left( \text{y, x} \right)\in \text{R}$and hence $\text{R}$ is symmetric.

Assuming that $\left( \text{x, y} \right)$ and $\left( \text{y, z} \right)\in \text{R}$, where $\text{x, y, z}\in \text{Z}$.

We can say that $\left( \text{x-y} \right)$ and $\left( \text{y-z} \right)$ are integers.

so, $\left( \text{x, z} \right)\in \text{R}$

Hence, we conclude that $\text{R}$ is transitive.

Therefore the given relation $\text{R}$ is reflexive, symmetric, and transitive.

1. Relation $\text{R}$ in the set $\text{A}$ of human beings in a town at a particular time given by

1. The relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}$ work at the same place$\}$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ and $\text{y}$ work at the same place$\text{ }\!\!\}\!\!\text{ }$

This implies that $\left( \text{x, x} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is reflexive.

Now, $\left( \text{x, y} \right)\in \text{R}$, then $\text{x}$ and $\text{y}$ work at the same place, which means $\text{y}$ and $\text{x}$ also work at the same place. Therefore, $\left( \text{y, x} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is symmetric.

Let us assume that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$.

Then, we can say that $\text{x}$ and $\text{y}$ work at the same place and $\text{y}$ and $\text{z}$ work at the same place. Which means that $\text{x}$ and $\text{z}$ also work at the same place.

Therefore, $\left( \text{x, z} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is reflexive, symmetric and transitive.

1. The relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}$ live in the same locality$\text{ }\!\!\}\!\!\text{ }$

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x and y}$ live in the same locality$\text{ }\!\!\}\!\!\text{ }$

Since, $\left( \text{x, x} \right)\in \text{R}$.

Therefore, we conclude that $\text{R}$ is reflexive.

Since $\left( \text{x, y} \right)\in \text{R}$, $\text{x}$ and $\text{y}$ live in the same locality. Therefore, $\text{y}$ and $\text{x}$ also live in the same locality, so, $\left( \text{y, x} \right)\in \text{R}$.

Hence, $\text{R}$ is symmetric.

Let $\left( \text{x, y} \right)\in \text{R}$ and $\left( \text{y, z} \right)\in \text{R}$. Hence $\text{x}$ and $\text{y}$ live in the same locality and $\text{y}$ and $\text{z}$ also live in the same locality. Which means that $\text{x}$ and $\text{z}$ also live in the same locality.

Therefore, $\left( \text{x, z} \right)\in \text{R}$.

Hence, we conclude that $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is reflexive, symmetric and transitive.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is exactly $\text{7}$cm taller than $\text{y }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is exactly $\text{7}$ cm taller than $\text{y }\!\!\}\!\!\text{ }$

Since, $\left( \text{x, x} \right)\notin \text{R}$.

Therefore, we conclude that $\text{R}$ is not reflexive.

Let $\left( \text{x, y} \right)\in \text{R}$ , Since $\text{x}$ is exactly $\text{7}$ cm taller than $\text{y}$, therefore $\text{y}$ is obviously not taller than $\text{x}$, so, $\left( \text{y, x} \right)\notin \text{R}$.

Hence, $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$, we can say that $\text{x}$ is exactly $\text{7}$ cm taller than $\text{y}$and $\text{y}$ is exactly $\text{7}$ cm taller than $\text{z}$. Which means that $\text{x}$ is exactly $\text{14}$ cm taller than $\text{z}$. So, $\left( \text{x, z} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is not reflexive, symmetric or transitive.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is wife of $\text{y }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is the wife of $\text{y }\!\!\}\!\!\text{ }$.

Since, $\left( \text{x, x} \right)\notin \text{R}$

Therefore, we conclude that $\text{R}$ is not reflexive.

Let $\left( \text{x, y} \right)\in \text{R}$ , Since $\text{x}$ is the wife of $\text{y}$, therefore $\text{y}$ is obviously not the wife of $\text{x}$, so, $\left( \text{y, x} \right)\notin \text{R}$.

Hence, $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$, we can say that $\text{x}$ is the wife of $\text{y}$and $\text{y}$ is the wife of $\text{z}$, which is not possible. So, $\left( \text{x, z} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore the given relation $\text{R}$is not reflexive, symmetric or transitive.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is father of $\text{y }\!\!\}\!\!\text{ }$

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x, y} \right)\text{: x}$ is the father of $\text{y }\!\!\}\!\!\text{ }$

Since, $\left( \text{x, x} \right)\notin \text{R}$

Therefore, we conclude that $\text{R}$ is not reflexive.

Let $\left( \text{x, y} \right)\in \text{R}$ , Since $\text{x}$ is the father of $\text{y}$, therefore $\text{y}$ is obviously not the father of $\text{x}$, so, $\left( \text{y, x} \right)\notin \text{R}$.

Hence, $\text{R}$ is not symmetric.

Assuming that $\left( \text{x, y} \right)\text{, }\left( \text{y, z} \right)\in \text{R}$, we can say that $\text{x}$ is the father of $\text{y}$and $\text{y}$ is the father of $\text{z}$, then $\text{x}$ is not the father of  $\text{z}$. So, $\left( \text{x, z} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore the given relation $\text{R}$is not reflexive, symmetric or transitive.

2. Show that the relation $\text{R}$ in the set $\text{R}$ of real numbers, defined $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{2}}} \right\}$ is neither reflexive nor symmetric nor transitive.

Ans: The given relation is: $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{2}}} \right\}$

Since $\left( \dfrac{\text{1}}{\text{2}}\text{, }\dfrac{\text{1}}{\text{2}} \right)\notin \text{R}$. (Since $\frac{1}{2}$ is not less than $\frac{1}{4}$)

Therefore, $\text{R}$ is not reflexive.

Since $\left( \text{1, 4} \right)\in \text{R}$ as $\text{1}{{\text{4}}^{\text{2}}}$, but $\left( \text{4, 1} \right)\notin \text{R}$ as ${{\text{4}}^{2}}$ is not less than ${{\text{1}}^{2}}$.

Therefore $\text{R}$ is not symmetric.

Assuming that $\left( \text{3, 2} \right)\text{, }\left( \text{2, 1}\text{.5} \right)\in \text{R}$, so, $\text{3}{{\text{2}}^{\text{2}}}\text{=4}$ and $\text{2}{{\left( \text{1}\text{.5} \right)}^{\text{2}}}\text{=2}\text{.25}$ but $\text{3}{{\left( \text{1}\text{.5} \right)}^{\text{2}}}\text{=2}\text{.25}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is neither reflexive, nor symmetric, nor transitive.

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

Ans: The given relation is $\text{R = }\left\{ \left( \text{a, b} \right)\text{: b=a+1} \right\}$ defined in the set $\text{A=}\left\{ \text{1, 2, 3, 4, 5, 6} \right\}$.

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

Since, $\left( \text{a, a} \right)\notin \text{R,a}\in \text{A}$.

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

Therefore, $\text{R}$ is not reflexive

Since, $\left( \text{1, 2} \right)\in \text{R}$, but $\left( \text{2, 1} \right)\notin \text{R}$.

Therefore $\text{R}$ is not symmetric.

Since $\left( \text{1, 2} \right)\text{, }\left( \text{2, 3} \right)\in \text{R}$, but $\left( \text{1, 3} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is neither reflexive, nor symmetric, nor transitive

4. Show that the relation $\text{R}$ in $\text{R}$ defined as $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le \text{b} \right\}$ is reflexive and transitive but not symmetric.

Ans: The given relation is $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le \text{b} \right\}$.

Since, $\left( \text{a, a} \right)\in R$.

Therefore, $\text{R}$ is reflexive.

Since, $\left( \text{2, 4} \right)\in R$ (as $\text{2,4}$), but $\left( \text{4, 2} \right)\notin R$(as $\text{4,2}$).

Therefore $\text{R}$ is not symmetric.

Assuming that $\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$, $\text{a}\le \text{b}$ and $b\le c$, therefore, $a\le c$.

Hence, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is reflexive and transitive but not symmetric.

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

Ans: The given relation is: $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a}\le {{\text{b}}^{\text{3}}} \right\}$

Since $\left( \dfrac{\text{1}}{\text{2}}\text{, }\dfrac{\text{1}}{8} \right)\notin \text{R}$.  (Since $\frac{1}{2}$ is not less than $\frac{1}{8}$)

Therefore, $\text{R}$ is not reflexive.

Since $\left( \text{1, 4} \right)\in \text{R}$ as $\text{1}{{\text{4}}^{\text{3}}}$, but $\left( \text{4, 1} \right)\notin \text{R}$ as $\text{4}$ is not less than ${{\text{1}}^{\text{3}}}$.

Therefore $\text{R}$ is not symmetric.

Assuming that $\left( \text{3, }\dfrac{\text{3}}{\text{2}} \right)\text{, }\left( \dfrac{\text{3}}{\text{2}}\text{, }\dfrac{\text{6}}{\text{5}} \right)\in \text{R}$, so, $\text{3}{{\left( \dfrac{\text{3}}{\text{2}} \right)}^{\text{3}}}$ and $\dfrac{\text{3}}{\text{2}}\text{}{{\left( \dfrac{\text{6}}{\text{5}} \right)}^{\text{3}}}$ but $\text{3}{{\left( \dfrac{\text{6}}{\text{5}} \right)}^{\text{3}}}\notin R$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is neither reflexive, nor symmetric, nor transitive.

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

Ans: The given relation is $\text{R = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 1} \right) \right\}$ on the set $\text{A=}\left\{ \text{1, 2, 3} \right\}$.

Since $\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\notin \text{R}$

Therefore, $\text{R}$ is not reflexive.

Since, $\left( \text{1, 2} \right)\in R$ and $\left( \text{2, 1} \right)\in R$.

Therefore $\text{R}$ is symmetric.

Since, $\left( \text{1, 2} \right)\in R$ and $\left( \text{2, 1} \right)\in R$, but $\left( \text{1, 1} \right)\notin \text{R}$.

Hence, $\text{R}$ is not transitive.

Therefore, the given relation $\text{R}$ is symmetric but neither reflexive nor transitive.

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

Ans: The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{x,}\,\,\text{y} \right)\text{: x}$ and $\text{y}$ have the same number of pages$\text{ }\!\!\}\!\!\text{ }$

Since  $\left( \text{x, x} \right)\in \text{R}$ as $\text{x}$ and $\text{x}$ have the same number of pages.

Therefore, $\text{R}$ is reflexive.

Let $\left( \text{x, y} \right)\in R$, so $\text{x}$ and $\text{y}$ have the same number of pages, $\left( \text{y, x} \right)\in \text{R}$ therefore $\text{y}$ and $\text{x}$ will also have the same number of pages.

Therefore $\text{R}$ is symmetric.

Assuming $\left( \text{x, y} \right)\in R$ and $\left( \text{y, z} \right)\in R$. $\text{x}$ and $\text{y}$ have the same number of pages and $\text{y}$ and $\text{z}$ also have the same number of pages. Therefore, $\text{x}$ and $\text{z}$ will also have the same number of pages. So, $\left( \text{x, z} \right)\in R$.

Hence, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

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

Ans: Let $\text{a}\in \text{A}$,

So, $\left| \text{a-a} \right|\text{ = 0}$  (which is an even number).

Therefore, $\text{R}$ is reflexive.

Let $\left( \text{a, b} \right)\in \text{R}$,

Now, $\left| \text{a-b} \right|$ is even,

Hence $\left| \text{a-b} \right|$ and $\left| \text{b-a} \right|$ are both even

Therefore, $\left( \text{b, a} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric

Let $\left( \text{a, b} \right)\in \text{R}$and $\left( \text{b, c} \right)\in \text{R}$,

$\Rightarrow \left| \text{a-b} \right|$ is even and $\left| \text{b-c} \right|$ is even

$\Rightarrow \left| \text{a-c} \right|$ is even.

$\Rightarrow \left( \text{a, c} \right)\in \text{R}$

Therefore, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

All the elements of the set $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ are all odd. Hence, the modulus of the difference of any two elements will be an even number. So, all the elements of this set are related to each other.

All elements of $\left\{ \text{2, 4} \right\}$ are even while all the elements of $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ are odd so no element of $\left\{ \text{1, }\!\!~\!\!\text{ 3, 5} \right\}$ can be related to any element of$\left\{ 2,\text{ }4 \right\}$.

Therefore, the absolute value of the difference between the two elements (from each of these two subsets) will not be an even value.

9. Show that each of the relation $\text{R}$ in the set $\text{A = }\!\!\{\!\!\text{ x}\in \text{Z: 0}\le \text{x}\le \text{12 }\!\!\}\!\!\text{ }$ , is an equivalence relation. Find the set of all elements related to 1 in each case.

1. $\text{R = }\!\!\{\!\!\text{ }\left( \text{a, b} \right)\text{ : }\left| \text{a -- b} \right|$ is a multiple of $\text{4 }\!\!\}\!\!\text{ }$

Ans: The given set $\text{A = }\!\!\{\!\!\text{ x}\in \text{Z: 0}\le \text{x}\le \text{12 }\!\!\}\!\!\text{ = }\left\{ \text{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} \right\}$

The given relation is: $\text{R = }\!\!\{\!\!\text{ }\left( \text{a, b} \right)\text{ : }\left| \text{a -- b} \right|$ is a multiple of $\text{4 }\!\!\}\!\!\text{ }$.

Let $a\in A$,

$\left( \text{a, a} \right)\in R$ as $\left| \text{a-a} \right|\text{=0}$ is a multiple of $\text{4}$.

Therefore, $\text{R}$ is reflexive.

Let, $\left( \text{a, b} \right)\in \text{R}\Rightarrow \left| \text{a-b} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left| \text{-}\left( \text{a-b} \right) \right|\text{=}\left| \text{b-a} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{b, a} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

$\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$.

$\Rightarrow \left| \text{a-b} \right|$ is a multiple of $\text{4}$ and $\left| \text{b-c} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{a-b} \right)$ is a multiple of $\text{4}$ and $\left( \text{b-c} \right)$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{a-c} \right)\text{=}\left( \text{a-b} \right)\text{+}\left( \text{b-c} \right)$ is a multiple of $\text{4}$.

$\Rightarrow \left| \text{a-c} \right|$ is a multiple of $\text{4}$.

$\Rightarrow \left( \text{a, c} \right)\in \text{R}$

Therefore, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

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

1. $\text{ }\!\!~\!\!\text{ R = }\left\{ \left( \text{a, b} \right)\text{ : a = b} \right\}$

Ans: The given relation is: $\text{R = }\left\{ \left( \text{a, b} \right)\text{ : a = b} \right\}$.

$a\in A,\left( \text{a, a} \right)\in R$, since $\text{a = a}$.

Therefore, $\text{R}$ is reflexive.

Let $\left( \text{a, b} \right)\in \text{R}\Rightarrow \text{a=b}$.

$\Rightarrow b\text{=a}\Rightarrow \left( \text{b, a} \right)\in R$

Therefore $\text{R}$ is symmetric.

$\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$

$\Rightarrow a\text{=b}$ and $b\text{=c}$

$\Rightarrow a\text{=c}$

$\Rightarrow \left( \text{a, c} \right)\in R$

Therefore, $\text{R}$ is transitive.

Therefore, the given relation $\text{R}$ is an equivalence relation.

The set of elements related to $\text{1}$ is $\left\{ \text{1} \right\}$.

10. Give an example of a relation. Which is

1. Symmetric but neither reflexive nor transitive.

Ans: Let us assume the relation $\text{R= }\left\{ \left( \text{5, 6} \right)\text{, }\left( \text{6, 5} \right) \right\}$ in set $\text{A= }\left\{ \text{5, 6, 7} \right\}$.

So, the relation $\text{R}$ is not reflexive as $\left( \text{5, 5} \right)\text{, }\left( \text{6, 6} \right)\text{, }\left( \text{7, 7} \right)\notin \text{R}$.

The relation $\text{R}$ is symmetric as $\left( \text{5, 6} \right)\in R$ and $\left( \text{6, 5} \right)\in R$.

The relation $\text{R}$ is not transitive as $\left( \text{5, 6} \right)\text{, }\left( \text{6, 5} \right)\in R$ , but $\left( \text{5, 5} \right)\notin \text{R}$.

Therefore, the given relation $\text{R}$ is symmetric but not reflexive or transitive.

1. Transitive but neither reflexive nor symmetric.

Ans: Let us assume the relation $\text{R = }\left\{ \left( \text{a, b} \right)\text{ : a b} \right\}$

So, the relation $\text{R}$ is not reflexive because for $a\in R$, $\left( \text{a, a} \right)\notin \text{R}$ since a cannot be strictly less than itself.

Let $\left( {1,2} \right) \in R\left( {as1 < 2} \right)$

Since $\text{2}$ is not less than $\text{1}$, $\left( \text{2, 1} \right)\notin \text{R}$.

Therefore $\text{R}$ is not symmetric.

Let $\left( \text{a, b} \right)\text{, }\left( \text{b, c} \right)\in R$.

$\Rightarrow \left( \text{a, c} \right)\in R$

Therefore, $\text{R}$ is transitive.

So, the relation $\text{R}$ is transitive but not reflexive and symmetric.

1. Reflexive and symmetric but not transitive.

Ans: Let us assume the relation $R=\left\{ \left( \text{4, 4} \right)\text{, }\left( \text{6, 6} \right)\text{, }\left( \text{8, 8} \right),\text{ }\left( \text{4, 6} \right),\text{ }\left( \text{6, 4} \right),\text{ }\left( \text{6, 8} \right),\text{ }\left( \text{8, 6} \right) \right\}$ in set $\text{A= }\left\{ \text{4, 6, 8} \right\}$.

The relation $\text{R}$ is reflexive since for $a\in R$, $\left( \text{a, a} \right)\in R$.

The relation $\text{R}$ is symmetric since $\left( \text{a, b} \right)\in R\Rightarrow \left( \text{b, a} \right)\in R$ for $a,b\in R$.

The relation $\text{R}$ is not transitive since $\left( \text{4, 6} \right)\text{, }\left( \text{6, 8} \right)\in R$, but $\left( \text{4, 8} \right)\notin \text{R}$.

Therefore the relation $\text{R}$ is reflexive and symmetric but not transitive.

1. Reflexive and transitive but not symmetric.

Ans: Let us take the relation $\text{ }\!\!~\!\!\text{ R = }\left\{ \left( \text{a, b} \right)\text{ : }{{\text{a}}^{\text{3}}}\ge {{\text{b}}^{\text{3}}} \right\}$.

Since $\left( \text{a, b} \right)\in \text{R}$.

Therefore $\text{R}$ is reflexive.

Since $\left( \text{2, 1} \right)\in \text{R}$, but $\left( \text{1, 2} \right)\notin \text{R}$,

Therefore $\text{R}$ is not symmetric.

Let $\left( {a,b} \right),\left( {b,c} \right) \in R$

$\Rightarrow {{\text{a}}^{\text{3}}}\ge {{\text{b}}^{\text{3}}}$ and ${{\text{b}}^{\text{3}}}\ge {{\text{c}}^{\text{3}}}$

$\Rightarrow {{\text{a}}^{\text{3}}}\ge {{\text{c}}^{\text{3}}}$

$\Rightarrow \left( \text{a, c} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is reflexive and transitive but not symmetric.

1. Symmetric and transitive but not reflexive.

Ans: Let us take a relation $\text{R=}\left\{ \left( \text{-5, -6} \right)\text{, }\left( \text{-6, -5} \right)\text{, }\left( \text{-5, -5} \right) \right\}$ in set $\text{A=}\left\{ \text{-5, -6} \right\}$.

The relation $\text{R}$ is not reflexive as $\left( \text{-6, -6} \right)\notin \text{R}$.

Since $\left( \text{-5, -6} \right)\in \text{R}$ and $\left( \text{-6, -5} \right)\in \text{R}$.

Therefore $\text{R}$ is symmetric.

Since $\left( \text{-5, -6} \right)\text{, }\left( \text{-6, -5} \right)\in \text{R}$ and $\left( \text{-5, -5} \right)\in \text{R}$.

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is symmetric and transitive but not reflexive.

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

Ans: The given relation is $\text{ }\!\!~\!\!\text{ R = }\!\!\{\!\!\text{ }\left( \text{P, Q} \right)\text{ :}$ Distance of $\text{P}$ from the origin is the same as the distance of $\text{Q}$ from the origin}

Since, $\left( \text{P, P} \right)\in \text{R}$.

The relation $\text{R}$ is reflexive.

Let $\left( \text{P, Q} \right)\in \text{R}$, distance of $\text{P}$ from the origin is the same as the distance of $\text{Q}$ from the origin similarly distance of $\text{Q}$ from the origin will be the same as the distance of $\text{P}$ from the origin. So, $\left( \text{Q, P} \right)\in \text{R}$.

Therefore $\text{R}$ is symmetric.

Let $\left( \text{P, Q} \right),\left( \text{Q, S} \right)\in \text{R}$.

Distance of $\text{P}$ from the origin is the same as the distance of $\text{Q}$ from the origin and distance of $\text{Q}$ from the origin is the same as the distance of $\text{S}$ from the origin. So, the distance of $\text{S}$ from the origin will be the same as the distance of $\text{P}$ from the origin. So, $\left( \text{P, S} \right)\in \text{R}$.

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

The set of points related to $\text{P}\ne \left( \text{0, 0} \right)$ will be those points whose distance from origin is same as distance of $\text{P}$ from the origin and will form a circle with the centre as origin and this circle passes through $\text{P}$.

12. Show that the relation $\text{R}$ is defined in the set $\text{A}$ of all triangles as $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}} \right)\text{ : }{{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{2}}}\text{ }\!\!\}\!\!\text{ }$, is equivalence relation. Consider three right angle triangles ${{\text{T}}_{\text{1}}}$ with sides $\text{3, 4, 5}$ and ${{\text{T}}_{\text{2}}}$ with sides $\text{5, 12, 13}$  and ${{\text{T}}_{\text{3}}}$ with sides $\text{6, 8, 10}$. Which triangles among ${{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}}$ and ${{\text{T}}_{\text{3}}}$ are related?

Ans: The given relation is $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}} \right)\text{ : }{{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{2}}}\text{ }\!\!\}\!\!\text{ }$ since, every triangle is similar to itself.

The relation $\text{R}$ is reflexive.

If $\left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}} \right)\in \text{R}$, then ${{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{2}}}$.

$\Rightarrow {{\text{T}}_{\text{2}}}$ is similar to ${{\text{T}}_{\text{1}}}$.

$\Rightarrow \left( {{\text{T}}_{\text{2}}}\text{, }{{\text{T}}_{\text{1}}} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

Let $\left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{2}}} \right)\text{,}\left( {{\text{T}}_{\text{2}}}\text{, }{{\text{T}}_{\text{3}}} \right)\in \text{R}$.

$\Rightarrow {{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{2}}}$ and ${{\text{T}}_{\text{2}}}$ is similar to ${{\text{T}}_{\text{3}}}$.

$\Rightarrow {{\text{T}}_{\text{1}}}$ is similar to ${{\text{T}}_{\text{3}}}$.

$\Rightarrow \left( {{\text{T}}_{\text{1}}}\text{, }{{\text{T}}_{\text{3}}} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

$\dfrac{\text{3}}{\text{6}}\text{=}\dfrac{\text{4}}{\text{8}}\text{=}\dfrac{\text{5}}{\text{10}}\left( \text{=}\dfrac{\text{1}}{\text{2}} \right)$

Since, the corresponding sides of triangles ${{\text{T}}_{\text{1}}}$ and ${{\text{T}}_{\text{3}}}$ are in the same ratio, therefore triangle ${{\text{T}}_{\text{1}}}$ is similar to triangle ${{\text{T}}_{\text{3}}}$.

Hence, ${{\text{T}}_{\text{1}}}$ is related to ${{\text{T}}_{\text{3}}}$.

13. Show that the relation $\text{R}$ defined in the set $\text{A}$ of all polygons as $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\text{: }{{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides$\text{ }\!\!\}\!\!\text{ }$, is an equivalence relation. What is the set of all elements in $\text{A}$ related to the right angle triangle $\text{T}$ with sides $\text{3, 4}$  and $\text{5}$?

Ans: $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\text{: }{{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides$\text{ }\!\!\}\!\!\text{ }$.

Since $\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{1}}} \right)\in \text{R}$ , as same polygon has same number of sides.

The relation $\text{R}$ is reflexive.

Let $\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\in \text{R}$ .

$\Rightarrow {{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides.

$\Rightarrow {{\text{P}}_{\text{2}}}$ and ${{\text{P}}_{\text{1}}}$ have same number of sides.

$\Rightarrow \left( {{\text{P}}_{\text{2}}}\text{, }{{\text{P}}_{\text{1}}} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

Let $\left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{2}}} \right)\text{,}\left( {{\text{P}}_{\text{2}}}\text{, }{{\text{P}}_{\text{3}}} \right)\in \text{R}$.

$\Rightarrow {{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{2}}}$ have same number of sides.

$\Rightarrow {{\text{P}}_{\text{2}}}$ and ${{\text{P}}_{\text{3}}}$ have same number of sides.

$\Rightarrow {{\text{P}}_{\text{1}}}$ and ${{\text{P}}_{\text{3}}}$ have same number of sides.

$\Rightarrow \left( {{\text{P}}_{\text{1}}}\text{, }{{\text{P}}_{\text{3}}} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

The elements in $A$ related to right-angled triangle $\left( \text{T} \right)$ with sides $\text{3, 4}$ and $\text{5}$ are the polygons having $\text{3}$ sides.

14. Let $\text{L}$ be the set of all lines in $\text{XY}$ plane and $\text{R}$ be the relation in $\text{L}$ defined as $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{2}}} \right)\text{: }{{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{2}}}$$\text{ }\!\!\}\!\!\text{ }$. Show that $\text{R}$ is an equivalence relation. Find the set of all lines related to the line $\text{y=2x+4}$.

Ans: $\text{R = }\!\!\{\!\!\text{ }\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{2}}} \right)\text{: }{{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{2}}}\text{ }\!\!\}\!\!\text{ }$.

The relation $\text{R}$ is reflexive as any line ${{\text{L}}_{\text{1}}}$ is parallel to itself, so, $\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{1}}} \right)\in \text{R}$.

Let $\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{2}}} \right)\in \text{R}$.

$\Rightarrow {{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{2}}}$, therefore ${{\text{L}}_{\text{2}}}$ is parallel to ${{\text{L}}_{\text{1}}}$.

$\Rightarrow \left( {{\text{L}}_{\text{2}}}\text{, }{{\text{L}}_{\text{1}}} \right)\in \text{R}$

Therefore $\text{R}$ is symmetric.

Let $\left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{2}}} \right)\text{,}\left( {{\text{L}}_{\text{2}}}\text{, }{{\text{L}}_{\text{3}}} \right)\in \text{R}$.

$\Rightarrow {{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{2}}}$

$\Rightarrow {{\text{L}}_{\text{2}}}$ is parallel to ${{\text{L}}_{\text{3}}}$

$\Rightarrow {{\text{L}}_{\text{1}}}$ is parallel to ${{\text{L}}_{\text{3}}}$

$\Rightarrow \left( {{\text{L}}_{\text{1}}}\text{, }{{\text{L}}_{\text{3}}} \right)\in \text{R}$

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is an equivalence relation.

Set of all lines related to the line $\text{y=2x+4}$ is the set of all lines that are parallel to the line $\text{y=2x+4}$.

Slope of line $\text{y=2x+4}$ is $\text{m = 2}$. Therefore, lines parallel to the given line are of the form $\text{y=2x+c}$, where $\text{c}\in \text{R}$.

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

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

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

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

4. $\text{R}$ is an equivalence relation

Ans: $\text{R = }\left\{ \left( \text{1, 2} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{1, 1} \right)\text{, }\left( \text{4, 4} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{3, 2} \right) \right\}$.

Since $\left( \text{a, a} \right)\in \text{R}$, for every $\text{a}\in \left\{ \text{1, 2, 3, 4} \right\}$

The relation $\text{R}$ is reflexive.

Since $\left( \text{1, 2} \right)\in \text{R}$ , but $\left( \text{2, 1} \right)\notin \text{R}$ .

Therefore $\text{R}$ is not symmetric.

$\left( \text{a, b} \right)\text{,}\left( \text{b, c} \right)\in \text{R}\Rightarrow \left( \text{a, c} \right)\in \text{R}$ for all $\text{a, b, c}\in \left\{ \text{1, 2, 3, 4} \right\}$.

Therefore $\text{R}$ is transitive.

Therefore the relation $\text{R}$ is reflexive and transitive but not symmetric.

The correct answer is ($B$) $\text{R}$ is reflexive and transitive but not symmetric.

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

1. $\left( \text{2, 4} \right)\in \text{R}$

2. $\left( \text{3, 8} \right)\in \text{R}$

3. $\left( \text{6, 8} \right)\in \text{R}$

4. $\left( \text{8, 7} \right)\in \text{R}$

Ans: The given relation is $\text{R = }\left\{ \left( \text{a, b} \right)\text{: a = b - 2, b > 6} \right\}$

Now,

Considering $\left( \text{2, 4} \right)\in \text{R}$.

Since, $\text{b 6}$, so, $\left( \text{2, 4} \right)\notin \text{R}$.

Considering $\left( \text{3, 8} \right)\in \text{R}$.

Since $\text{3 }\ne \text{ 8 - 2}$, so $\left( \text{3, 8} \right)\notin \text{R}$.

Considering $\left( \text{6, 8} \right)\in \text{R}$.

Since $\text{86}$ and $\text{6=8-2}$, so $\left( \text{6, 8} \right)\in \text{R}$.

Therefore, the correct answer is ($C$)$\left( \text{6, 8} \right)\in \text{R}$.

## Conclusion

In Ex 1.1 Class 12 Maths Chapter 1 - Relations and Functions, you're diving into the foundational concepts of relations and functions. This section is crucial as it lays the groundwork for understanding more complex mathematical concepts later on. Pay close attention to defining relations, understanding how to represent them using different methods like arrow diagrams and ordered pairs, and grasping the notion of functions and their types. Ensure you are clear on the difference between a relation and a function, as well as the importance of domains and ranges. Mastering these basics will pave the way for smoother learning ahead.

## NCERT Solutions for Class 12 Maths Chapter 1 Exercises

 Chapter 1 Functions and Relations All Exercises in PDF Format Exercise 1.2 12 Questions and Solutions Miscellaneous Exercise 7 Questions and Solutions

## CBSE Class 12 Maths Chapter 1 Other Study Materials

 S.No Important Links for Chapter 1 Relations and Functions 1 Relations and Functions Revision Notes 2 Relations and Functions Important Questions 3 Relations and Functions Formula 4 Relations and Functions NCERT Exemplar Solution

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 1 Relations and Function - Exercise 1.1

1. Which site provides NCERT Solutions for Maths Chapter 1 Relations and Functions Ex 1.1 Class 12?

Vedantu provides the free PDF of NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Exercise 1.1. Exercise-wise NCERT Solutions for Chapter 1 Relations and Functions are designed by expert Maths tutors at Vedantu. The material is designed to help students revise the complete syllabus and score well in exams. NCERT Solutions for Class 12 Maths Chapter 1 are prepared as per the latest NCERT guidelines and exam pattern. It provides students with a thorough knowledge of the chapter and simple explanations to the exercise problems designed to help students in doubt clearance.

2. What are the salient features of Vedantu’s NCERT exercise 1.1 Class 12 maths solutions and other exercises?

Vedantu’s NCERT Solutions for Class 12 Maths for Chapter 1 Relations and Functions are available on its site. Following are some of the key features of NCERT Solutions for Class 12 Maths Chapter 1 offered by Vedantu:

1. Step-wise solutions.

2. Designed by subject matter experts.

3. Available in the free PDF format.

4. Provides coverage of the NCERT questions.

5. Designed as per the latest syllabus and the exam pattern.

3. Why must I download exercise-wise NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions?

CBSE Class 12 Maths Chapter 1 Relations and Functions is an important chapter of Class 12 Maths syllabus. The chapter explains the concepts related to Relations and Functions and how to solve problems based on the chapter. It is important to solve each question of the exercises to score well in exams. NCERT Solutions for Class 12 Maths Chapter 1 are the most reliable online resource for students facing any doubt in the chapter. These solutions are provided by experts and help clear all the doubts. The experts have years of experience in the field of teaching. NCERT Solutions for Class 12 Chapter 1 Relations and Functions for Exercise 1.1 and other exercises is a great way to clear be a top scorer in the Maths exam.

4. What are the important learning outcomes of Class 12 Maths Exercise 1.1 solutions of Relations and Functions?

In Exercise 1.1 of Class 12 Maths Chapter 1 Relations and Functions, students will learn about the various types of Relations. Exercise 1.1 includes problems based on the types of Relations. Students can avail the solutions by experts for a better understanding of the concepts. After referring to NCERT Solutions for Class 12 Maths Chapter 1, students will be able to solve the Exercise 1.1 questions confidently without any mistakes. Exercise 1.1 of Class 12 Maths Chapter 1 is important for understanding the basics of Relations and Functions. The solutions are available in the free PDF format on Vedantu’s site.

5. Apart from the NCERT questions, where can I find Class 12 Maths questions to practice?

Apart from the NCERT Solutions for Class 12 Maths Chapter 1, this chapter contains other exercises with numerous questions. Vedantu’s in-house subject specialists solve all of these questions. As a result, these are guaranteed to be high quality, and anyone can use them to study for exams. It is critical to grasp all of the topics in the textbooks and solve the questions from the exercises given next to them to achieve the highest possible grade in the class.

6. Where can I download the PDF of the NCERT book of Maths Class 12?

Students can download NCERT Books for Class 12 Maths PDF from Vedantu’s website (vedantu.com). They will find chapter-wise PDFs of the Class 12 Maths textbook. Hence, they can download the chapter of their choice for using it offline. These solutions are available free of cost on Vedantu (vedantu.com). You can download using the Vedantu app as well.

7. What are the most important formulae that I need to remember in Chapter 1 Maths Class 12?

Chapter 1 of Maths Class 12 doesn’t have a lot of formulae that need to be memorised. The definitions of various types of relations, types of sets, functions and operations are the most important, and they are followed by their respective formula and how that is obtained. This process and the basic numerical representation of the formula is to be memorised.

8. What are the most important definitions in Chapter 1 Maths Class 12?

The most important definitions of Chapter 1 of Class 12 Mathematics are the ones that are given in the blueprint, where it starts by ‘Definition’ and is followed by the statements. Some of the most important definitions are those of different types of relations and the types of sets. The definitions of various functions and operations are also quite important and have a high probability of coming in exams.

9. Are NCERT questions enough for Class 12th board exams?

NCERT questions form a great base, and you must solve them numerous times to ensure maximum marks in Class 12th board examinations. However, they are not sufficient as CBSE asks for a variety of questions, and you must be familiar with all of them. You can find plenty of questions and exercises to practice on Vedantu (vedantu.com), where the course wise exercises are given, and their expert-written solutions are also provided.