NCERT Solutions for Class 11 Maths Chapter 2: Relations and Functions - Exercise 2.2

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 down its domain, codomain and range.

Ans: The relation \[{\rm{R}}\] from \[{\rm{A}}\]to \[{\rm{A}}\]is given as \[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x - y = 0}}} \right\}\] where \[{\rm{x,y}} \in {\rm{A}}\]

i.e., \[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x = y}}} \right\}\] where \[{\rm{x,y}} \in {\rm{A}}\]

\[\therefore {\rm{R = }}\left\{ {\left( {{\rm{1,3}}} \right){\rm{,}}\left( {{\rm{2,6}}} \right){\rm{,}}\left( {{\rm{3,9}}} \right){\rm{,}}\left( {{\rm{4,12}}} \right)} \right\}\]

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

Therefore, domain of \[{\rm{R = }}\left\{ {{\rm{1,2,3,4}}} \right\}\].

The whole set \[{\rm{A}}\]is the codomain of the relation \[{\rm{R}}\].

Therefore, codomain of \[{\rm{R = }}\left\{ {{\rm{1,2,3}}...{\rm{14}}} \right\}\]

The range 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 \[{\rm{R}}\]on the set \[{\rm{N}}\] of natural numbers by\[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:y = x + 5, \text{x  is  a  natural  number  less  than  4}; x,y}} \in {\rm{N}}} \right\}\]. Depict this relationship using roster form. Write down the domain and the range.

Ans: We are given that,

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

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

Therefore, \[{\rm{R = }}\left\{ {\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,7}}} \right){\rm{,}}\left( {{\rm{3,8}}} \right)} \right\}\]

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

Therefore, domain of \[{\rm{R = }}\left\{ {{\rm{1,2,3}}} \right\}\]

The range 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.\[{\rm{A = }}\left\{ {{\rm{1,2,3,5}}} \right\}\]and \[{\rm{B = }}\left\{ {{\rm{4,6,9}}} \right\}\]. Define a relation \[{\rm{R}}\]from \[{\rm{A}}\] to \[{\rm{B}}\] by \[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{\text{: the difference between x and y is odd; x}}} \in {\rm{A, y}} \in {\rm{B}}} \right\}\]. Write \[{\rm{R}}\] in roster from. 

Ans: We are given that, 

\[{\rm{A = }}\left\{ {{\rm{1,2,3,5}}} \right\}\] and \[{\rm{B = }}\left\{ {{\rm{4,6,9}}} \right\}\]

The relation is given by,

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

Therefore, \[{\rm{R = }}\left\{ {\left( {{\rm{1,4}}} \right){\rm{,}}\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,9}}} \right){\rm{,}}\left( {{\rm{3,4}}} \right){\rm{,}}\left( {{\rm{3,6}}} \right){\rm{,}}\left( {{\rm{5,4}}} \right){\rm{,}}\left( {{\rm{5,6}}} \right)} \right\}\]

4. The given figure shows a relationship between the sets \[{\rm{P}}\] and \[{\rm{Q}}\]. Write this relation 

(i) In set-builder form

(ii) In roster form

What is its domain and range?

(Image will be Uploaded Soon) 

Ans: 

(i) According to the given diagram, \[{\rm{P = }}\left\{ {{\rm{5,6,7}}} \right\}\]

And \[{\rm{Q = }}\left\{ {{\rm{3,4,5}}} \right\}\]

Therefore, the set builder form of relation is 

\[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{: y = x - 2; x}} \in {\rm{P}}} \right\}\] or

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

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

And \[{\rm{Q = }}\left\{ {{\rm{3,4,5}}} \right\}\]

Therefore, the roster form of relation is 

\[{\rm{R = }}\left\{ {\left( {{\rm{5,3}}} \right){\rm{,}}\left( {{\rm{6,4}}} \right){\rm{,}}\left( {{\rm{7,5}}} \right)} \right\}\]

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 roster form 

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

(iii) Find the range of \[{\rm{R}}\]

Ans: 

(i) We are given that \[{\rm{A = }}\left\{ {{\rm{1,2,3,4,6}}} \right\}\]

and \[{\rm{R}} = \left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\rm{A, \text{b is exactly divisible by a}}}} \right\}\]

Therefore, the roster form of relation \[{\rm{R}}\] is

\[{\rm{R = }}\left\{ {\left( {{\rm{1,1}}} \right){\rm{,}}\left( {{\rm{1,2}}} \right){\rm{,}}\left( {{\rm{1,3}}} \right){\rm{,}}\left( {{\rm{1,4}}} \right){\rm{,}}\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,2}}} \right){\rm{,}}\left( {{\rm{2,4}}} \right){\rm{,}}\left( {{\rm{2,6}}} \right){\rm{,}}\left( {{\rm{3,3}}} \right){\rm{,}}\left( {{\rm{3,6}}} \right){\rm{,}}\left( {{\rm{4,4}}} \right){\rm{,}}\left( {{\rm{6,6}}} \right)} \right\}\]

(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 range of the relation \[{\rm{R}}\] defined by \[{\rm{R = }}\left\{ {\left( {{\rm{x,x + 5}}} \right){\rm{:x}} \in \left\{ {{\rm{0,1,2,3,4,5}}} \right\}} \right\}\]

Ans: We are given that the relation \[{\rm{R}}\] is given by

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

Therefore, \[{\rm{R = }}\left\{ {\left( {{\rm{0,5}}} \right){\rm{,}}\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,7}}} \right){\rm{,}}\left( {{\rm{3,8}}} \right){\rm{,}}\left( {{\rm{4,9}}} \right){\rm{,}}\left( {{\rm{5,10}}} \right)} \right\}\]

Domain of \[{\rm{R = }}\left\{ {{\rm{0,1,2,3,4,5}}} \right\}\]

Range of \[{\rm{R = }}\left\{ {{\rm{5,6,7,8,9,10}}} \right\}\]

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 roster form.

Ans: We are given that \[{\rm{R = }}\left\{ {\left( {{\rm{x,}}{{\rm{x}}^{\rm{3}}}} \right){\rm{\text{:x is a prime number less than 10}}}} \right\}\]

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

Therefore, \[R = \left\{ {\left( {{\rm{2,8}}} \right){\rm{,}}\left( {{\rm{3,27}}} \right){\rm{,}}\left( {{\rm{5,125}}} \right){\rm{,}}\left( {{\rm{7,343}}} \right)} \right\}\] is the roster form.

8. Let \[{\rm{A = }}\left\{ {{\rm{x,y,z}}} \right\}\] and \[{\rm{B = }}\left\{ {{\rm{1,2}}} \right\}\]. Find the number of relations from \[{\rm{A}}\]to \[{\rm{B}}\].

Ans: It is given that \[{\rm{A = }}\left\{ {{\rm{x,y,z}}} \right\}\] and \[{\rm{B = }}\left\{ {{\rm{1,2}}} \right\}\].

Therefore, \[{\rm{A \times B = }}\left\{ {\left( {{\rm{x,1}}} \right){\rm{,}}\left( {{\rm{x,2}}} \right){\rm{,}}\left( {{\rm{y,1}}} \right){\rm{,}}\left( {{\rm{y,2}}} \right){\rm{,}}\left( {{\rm{z,1}}} \right){\rm{,}}\left( {{\rm{z,2}}} \right)} \right\}\]

Since, \[{\rm{n(A \times B) = 6}}\], the 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 range of \[{\rm{R}}\].

Ans: We are given that \[{\rm{R = }}\left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\rm{Z, a - b \text{ is an integer}}}} \right\}\]

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

Therefore, domain of \[{\rm{R = Z}}\]

And the range of \[{\rm{R = Z}}\]

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

Opting for the NCERT solutions for Ex 2.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 2.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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