# NCERT Solutions for Mathematics Education 11 Chapter 2 Relations and Functions (Ex. 2.2) Exercise 2.2 (2023)

### Exercise 2.2

1. Let ${\rm{A = }}\left\{ {{\rm{1,2,3}}...{\rm{14}}} \right\}$. Define a relation ${\rm{R}}$ from ${\rm{A}}$ to ${\rm{A}}$ by ${\rm{R = }}\ left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x - y = 0}}} \right\}$, where ${\rm{x, y}} \in {\rm{A}}$. Write your domain, reach and image.

Answer:The ratio ${\rm{R}}$ of ${\rm{A}}$ for ${\rm{A}}$ is given as ${\rm{R = }} \left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x - y = 0}}} \right\}$ at ${\rm{x, y}} \in {\rm{A}}$

oder seja, ${\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x = y}}} \right\} \ ] wobei \[{\rm{x,y}} \in {\rm{A}}$

$\so {\rm{R = }}\left\{ {\left( {{\rm{1,3}}} \right){\rm{,}}\left( {{\rm{2 ,6}}} \right){\rm{,}}\left( {{\rm{3,9}}} \right){\rm{,}}\left( {{\rm{4,12 }}} \gut gut\}$

The domain of ${\rm{R}}$ is the set of all first elements of ordered pairs in the relation.

Also Definitionsbereich von ${\rm{R = }}\left\{{{\rm{1,2,3,4}}} \right\}$.

The complete set ${\rm{A}}$ is the domain of the relation ${\rm{R}}$.

Hence condom of ${\rm{R = }}\left\{ {{\rm{1,2,3}}...{\rm{14}}} \right\}$

The domain of ${\rm{R}}$ is the set of all second elements of the ordered pairs in the relation.

Therefore range of ${\rm{R = }}\left\{{{\rm{3,6,9,12}}} \right\}$

2. Define a relation ${\rm{R}}$over the set ${\rm{N}}$ ​​​​​​​​​​​​​of natural numbers by ${\ rm{ R = }}\left \{ { \ left( {{\rm{x,y}}} \right){\rm{:y = x + 5, \text{x is an integer less than 4 }; x,y}} \in {\rm{N}}} \right\}$. Describe this relationship using the list form. Write the domain and image.

(Video) Chapter 2 Exercise 2.2 (Q1,Q2,Q3) Relations and Functions Class 11 Maths NCERT

${\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:y = x + 5,\text{x is a natural one number less than 4}; x,y}} \in {\rm{N}}} \direita\}$

The natural numbers less than 4 are 1, 2, and 3.

Daher ist ${\rm{R = }}\left\{ {\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm {2 ,7}}} \right){\rm{,}}\left( {{\rm{3,8}}} \right)} \right\}$

The domain of ${\rm{R}}$ is the set of all first elements of ordered pairs in the relation.

Also Definitionsbereich von ${\rm{R = }}\left\{{{\rm{1,2,3}}} \right\}$

The domain of ${\rm{R}}$ is the set of all second elements of the ordered pairs in the relation.

Therefore range of ${\rm{R = }}\left\{{{\rm{6,7,8}}} \right\}$

3.${\rm{A = }}\left\{ {{\rm{1,2,3,5}}} \right\}$y ${\rm{B = }}\left \{ {{\rm{4,6,9}}} \right\}$. Define a relation ${\rm{R}}$from ${\rm{A}}$ to ${\rm{B}}$ times ${\rm{R = }}\ left\{ {\left( {{\rm{x,y}}} \right){\rm{\text{: the difference between x and y is odd; x}}} \in {\rm{A, y}} \in {\rm{B}}} \right\}$. Enter ${\rm{R}}$ in the list.

${\rm{A = }}\left\{ {{\rm{1,2,3,5}}} \right\}$ y ${\rm{B = }}\left\{ {{\rm{4,6,9}}} \derecho\}$

The relationship is given by

${\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{\text{: the difference between x and y is odd; x}}} \in {\rm{A, y}} \in {\rm{B}}} \direita\}$

Daher ist ${\rm{R = }}\left\{ {\left( {{\rm{1,4}}} \right){\rm{,}}\left( {{\rm {1 ,6}}} \right){\rm{,}}\left( {{\rm{2,9}}} \right){\rm{,}}\left( {{\rm{3 ,4 }}} \right){\rm{,}}\left( {{\rm{3,6}}} \right){\rm{,}}\left( {{\rm{5,4 }} } \right){\rm{,}}\left( {{\rm{5,6}}} \right)} \right\}$

4. The given figure shows a relation between the sets ${\rm{P}}$ and ${\rm{Q}}$. Write that relationship down

(i) In the form of a set constructor

(Video) Class 11 math chapter 2 Relations and functions exercise 2.2 NCERT solutions || exercise 2.2

(ii) In Listenform

What is your domain and image?

(UE)According to the given diagram ${\rm{P = }}\left\{ {{\rm{5,6,7}}} \right\}$

mi ${\rm{Q = }}\left\{ {{\rm{3,4,5}}} \right\}$

Hence the relational form of the set constructor

${\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{: y = x - 2; x}} \in {\rm{P}}} \right\}$ oder

${\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{: y = x - 2 \text{for} x = 5 ,6,7}}} \direita\}$

(ii)According to the given diagram ${\rm{P = }}\left\{ {{\rm{5,6,7}}} \right\}$

mi ${\rm{Q = }}\left\{ {{\rm{3,4,5}}} \right\}$

Hence the list form of relationships

${\rm{R = }}\left\{ {\left( {{\rm{5,3}}} \right){\rm{,}}\left( {{\rm{6,4 }}} \right){\rm{,}}\left( {{\rm{7.5}}} \right)} \right\}$

5. Let ${\rm{A = }}\left\{ {{\rm{1,2,3,4,6}}} \right\}$. Let ${\rm{R}}$ be the relation on ${\rm{A}}$ defined by $\left\{ {\left( {{\rm{a,b}}} \ right){\rm{: a,b}} \in {\rm{A, \text{ b is exactly divisible by a}}}} \right\}$.

(i) Write ${\rm{R}}$ in list form

(Video) Relations and Functions|Exercise 2.2 Solutions|Class 11|Maths|Chapter2|Plus One|Malayalam|CBSE|NCERT

(ii) Find the domain of ${\rm{R}}$

(iii) Find the image of ${\rm{R}}$

(UE)We know that ${\rm{A = }}\left\{{{\rm{1,2,3,4,6}}} \right\}$

e ${\rm{R}} = \left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\rm{A , \text{b é exatamente dividido por a}}}} \right\}$

Therefore the list form of the relation is ${\rm{R}}$

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

(ii)The domain of ${\rm{R}}$ is $\left\{ {{\rm{1,2,3,4,6}}} \right\}$

(iii)The range of ${\rm{R}}$ is $\left\{ {{\rm{1,2,3,4,6}}} \right\}$

6. Determine the domain and domain of the relation ${\rm{R}}$ defined by ${\rm{R = }}\left\{ {\left( {{\rm{x,x + 5 }}} \right){\rm{:x}} \in \left\{ {{\rm{0,1,2,3,4,5}}} \right\}} \right\}\ ] Answer:We know that the relation \[{\rm{R}}$ is given by

${\rm{R = }}\left\{\left( {{\rm{x,x + 5}}} \right){\rm{:x}} \in \left\{ {{ \ rm{0,1,2,3,4,5}}} \derecho\}} \derecho\}$

Daher ist ${\rm{R = }}\left\{ {\left( {{\rm{0.5}}} \right){\rm{,}}\left( {{\rm{1 ,6 }}} \right){\rm{,}}\left( {{\rm{2,7}}} \right){\rm{,}}\left( {{\rm{3,8 }} } \right){\rm{,}}\left( {{\rm{4,9}}} \right){\rm{,}}\left( {{\rm{5,10}} } \ gut gut\}$

Dominio de ${\rm{R = }}\left\{{{\rm{0,1,2,3,4,5}}} \right\}$

Rango de ${\rm{R = }}\left\{ {{\rm{5,6,7,8,9,10}}} \right\}$

(Video) Relations and Functions|Exercise 2.2 Solutions|Class 11|Maths|Chapter2|Plus One|Malayalam|CBSE|NCERT

7. Write the relation ${\rm{R = }}\left\{ {\left( {{\rm{x,}}{{\rm{x}}}^{\rm{3}} }} \right){\rm{\text{:x is a prime number less than 10}}}} \right\}$ in list form.

Answer:if you give us this ${\rm{R = }}\left\{\left( {{\rm{x,}}{{\rm{x}}^{\rm{3}}}} \right){\ rm {\text{:x is a prime less than 10}}}} \right\}$

The prime numbers less than 10 are 2, 3, 5, and 7.

Por lo tanto, $R = \left\{ {\left( {{\rm{2.8}}} \right){\rm{,}}\left( {{\rm{3.27}}} \ right) {\ rm{,}}\left( {{\rm{5,125}}} \right){\rm{,}}\left( {{\rm{7,343}}} \right)} \right\ }\ ] es el formelio de lista. 8. Let \[{\rm{A = }}\left\{ {{\rm{x,y,z}}} \right\}$ and ${\rm{B = }}\left\ { {{\rm{1,2}}} \right\}$. Find the number of ratios from ${\rm{A}}$ to ${\rm{B}}$.

Answer:Angenommen, ${\rm{A = }}\left\{ {{\rm{x,y,z}}} \right\}$ und ${\rm{B = }}\left \{ {{\rm{1,2}}} \right\}$.

Daher ist ${\rm{A \times B = }}\left\{ {\left( {{\rm{x,1}}} \right){\rm{,}}\left( {{\ rm{x,2}}} \right){\rm{,}}\left( {{\rm{y,1}}} \right){\rm{,}}\left( {{\rm{ y,2}}} \right){\rm{,}}\left( {{\rm{z,1}}} \right){\rm{,}}\left( {{\rm{z, 2}}} \direita)} \direita\}$

Since ${\rm{n(A \times B) = 6}}$, or number of subsets of ${\rm{A \times B}}$ is ${{\rm{2 } }^{\rm{6}}}$.

9. Let ${\rm{R}}$ be the relation on ${\rm{Z}}$ defined by ${\rm{R = }}\left\{ {\left( { { \rm{a,b}}} \right){\rm{: a,b}} \in {\rm{Z, a - b \text{is an integer}}}} \right\}$ . Find the domain and image of ${\rm{R}}$.

Answer:Tenemos ${\rm{R = }}\left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\ rm{ Z , a - b \text{es un número entero}}}} \right\}$

It is well known that the difference between two integers is always an integer.

Hence the domain of ${\rm{R = Z}}$

And the range of ${\rm{R = Z}}$

### NCERT Solutions for Class Math 11 Chapter 2 Relations and Functions Exercise 2.2

Opting for NCERT solutions for Ex 2.2 Grade 11 Math is considered to be the best option for CBSE students when it comes to exam preparation. This chapter consists of many exercises. We provide solutions of NCERT Mathematics Exercise 2.2 Class 11 in PDF format on this page. You can download this solution at your convenience or study it directly from our online website/application.

Practice issues were resolved by Vedantu's in-house experts with the utmost care and in compliance with all CBSE guidelines. 11th grade students finished with all the concepts of the math book and well versed in all the problems of the exercises in it so that every student can easily achieve the highest possible grade in the exam completion. With the help of these solutions of Class 11 Mathematics Chapter 2 Exercise 2.2, students can easily understand the pattern of questions that can be asked in the exam of this chapter and also learn the weight of the chapter marks. So that they can optimally prepare for the final exam.

(Video) Class - 11 Ex - 2.2, Q1 to Q10 (Relation and Functions) Maths Chapter 2 || CBSE NCERT || Green Board

Besides these NCERT solutions for Class 11 Mathematics Chapter 2 Exercise 2.2, there are many exercises in this chapter which also contain numerous questions. All these questions will be solved/answered by our in-house experts as mentioned above. Therefore, they are all of the highest quality and can be consulted by anyone during the exam preparation period. In order to get the best possible grades in class, it is very important to understand all the concepts in the textbook and solve the problems in the following exercises.

Don't hesitate any longer. Download NCERT Solutions for Class 11 Mathematics Chapter 2 Exercise 2.2 from Vedantu website now to better prepare for the exam. If you have the Vedantu app on your phone, you can also download it through the app. The best thing about these solutions is that they are accessible both online and offline.

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