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

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

Komal MiglaniUpdated on 05 May 2025, 04:19 PM IST

Think of a vending machine! When you press one button, you will get exactly one snack. You wouldn’t expect one button to give two different snacks, right? This is how a function works! It gives only one output for each input. A function is a special type of relation where every element in the domain is linked to exactly one element in the codomain. A series of questions related to functions are covered in NCERT, where you will learn to identify functions from ordered pairs, arrow diagrams, and understand how they behave.

This Story also Contains

  1. Class 11 Maths Chapter 2 Exercise 2.2 Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 2: Exercise 2.2
  3. Topics covered in Chapter 2 Relations and Functions Exercise 2.2
  4. Class 11 Subject-Wise Solutions
NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.2 - Relations and Functions
2.2

The NCERT Solutions Chapter 2 Exercise 2.2 are designed in a way that it will help students understand the concepts easily through simple calculations. These NCERT solutions will strengthen your command over the topic and will also prepare you to tackle similar questions in board exams and competitive tests.

Class 11 Maths Chapter 2 Exercise 2.2 Solutions - Download PDF

Download PDF


NCERT Solutions Class 11 Maths Chapter 2: Exercise 2.2

Question 1: Let A = {1, 2, 3,...,14}. Define a relation R from A to A by $R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \epsilon A \right \}$ .Write down its domain, codomain and range.

Answer:

It is given that
$A = \left \{ 1, 2, 3, ..., 14 \right \} \ and \ R = \left \{ (x, y) : 3x - y = 0, \ where \ x, y \ \epsilon \ A \right \}$
Now, the relation R from A to A is given as
$R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \epsilon A \right \}$
Therefore,
The relation in roaster form is , $,R = \left \{ (1, 3), (2, 6), (3, 9), (4, 12) \right \}$
Now,
We know that the Domain of R = set of all first elements of the order pairs in the relation
Therefore,
Domain of $R = \left \{ 1, 2, 3, 4 \right \}$
And
Codomain of R = the whole set A
i.e. Codomain of $R = \left \{ 1, 2, 3, ..., 14 \right \}$
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of $R = \left \{ 3, 6, 9, 12 \right \}$

Question 2: Define a relation R on the set N of natural numbers by $R =\{ ( x,y ) : y = x +5 , x\}$ is a natural number less than $4 ; x , y \epsilon N \left. \right \}$. Depict this relationship using roster form. Write down the domain and the range.

Answer:

As x is a natural number which is less than 4.
Therefore,
the relation in roaster form is, $R = \left \{ (1,6), (2,7), (3,8) \right \}$
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ 1, 2, 3 \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R = \left \{ 6, 7, 8 \right \}$

Therefore, domain and the range are $\left \{ 1,2,3 \right \} \ \ and \ \ \left \{ 6, 7, 8 \right \}$ respectively

Question 3: A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by $R = \left \{ ( x,y ) : the \: \: diffrence \: \: between \: \: x \: \: and \: \: y \: \: is \: \: odd ; x \epsilon A , y \epsilon B \right \}$. Write R in roster form.

Answer:

It is given that
A = {1, 2, 3, 5} and B = {4, 6, 9}
And
$R = \left \{ ( x,y ) : the \: \: diffrence \: \: between \: \: x \: \: and \: \: y \: \: is \: \: odd ; x \epsilon A , y \epsilon B \right \}$
Now, it is given that the difference should be odd. Let us take all possible differences.
(1 - 4) = - 3, (1 - 6) = - 5, (1 - 9) = - 8(2 - 4) = - 2, (2 - 6) = - 4, (2 - 9) = - 7(3 - 4) = - 1, (3 - 6) = - 3, (3 - 9) = - 6(5 - 4) = 1, (5 - 6) = - 1, (5 - 9) = - 4
Taking the difference which are odd we get,

Therefore,
the relation in roaster form, $R = \left \{ (1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6) \right \}$

Question 4: (i) The Fig2.7 shows a relationship between the sets P and Q. Write this relation in set-builder form

Answer:

It is given in the figure that

P = {5,6,7}, Q = {3,4,5}

Therefore,
the relation in set builder form is ,
$R = \left \{ {(x, y): y = x-2; x \ \epsilon \ P} \right \}$
OR
$R = \left \{ {(x, y): y = x-2; \ for \ x = 5, 6, 7} \right \}$

Question 4: (ii) The Fig2.7 shows a relationship between the sets P and Q. Write this relation roster form. What is it domain and range?

Answer:

From the given figure. we observe that

P = {5,6,7}, Q = {3,4,5}

And the relation in roaster form is , $R = \left \{ {(5,3), (6,4), (7,5)} \right \}$

As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ {5, 6, 7} \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R = \left \{ {3, 4, 5} \right \}$

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

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)} \right \}$

Question 5: (ii) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$ Find the domain of R

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ {1, 2, 3, 4, 6} \right \}$

Question 5: (iii) Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by $\left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$ Find the range of R.

Answer:

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As the range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R = \left \{ {1, 2, 3, 4, 6} \right \}$

Question 6: Determine the domain and range of the relation R defined by $R = \left \{ ( x , x +5 ): x \epsilon \left \{ 0,1,2,3,,4,5 \right \} \right \}$

Answer:

It is given that
$R = \left \{ ( x , x +5 ): x \epsilon \left \{ 0,1,2,3,,4,5 \right \} \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)} \right \}$

Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R =\left \{ {0, 1, 2, 3, 4, 5} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Range of $R =\left \{ {5, 6, 7, 8, 9, 10} \right \}$

Therefore, the domain and range of the relation R is $\left \{ 0,1,2,3,4,5 \right \} \ \ and \ \ \left \{ {5, 6, 7, 8, 9, 10} \right \}$ respectively

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

Answer:

It is given that
$R = \left \{ \right.(x, x^3) : x\: \: is\: \: a\: \: prime\: \: number \: \: less\: \: than\: \: 10\: \: \left. \right \}$
Now,
As we know the prime number less than 10 are 2, 3, 5 and 7.
Therefore,
the relation in roaster form is , $R =\left \{ {(2,8), (3,27), (5,125), (7,343)} \right \}$

Question 8: Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer:

It is given that
A = {x, y, z} and B = {1, 2}
Now,
$A \times B = \left \{ {(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)} \right \}$
Therefore,
$n(A \times B) = 6$
Then, the number of subsets of the set $(A \times B) = 2^n = 2^6$

Therefore, the number of relations from A to B is $2^6$

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

Answer:

It is given that
$R = \left \{ ( a,b) : a , b \epsilon Z , a-b\: \: is \: \: an \: \: integer \right \}$
Now, as we know that the difference between any two integers is always an integer.
And
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
The domain of R = Z

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of R = Z

Therefore, the domain and range of R is Z and Z respectively

Also Read

Topics covered in Chapter 2 Relations and Functions Exercise 2.2

1) Relation

A relation is just a way to show how things from one set are connected to things in another set.

Example: If set $A=\{1,2\}$ and set $B=\{a, b\}$, then a relation could be $\{(1, a),(2, b)\}$.


2) Domain and Range

This will help you Identify the domain and range of a given relation.
Domain- The set of first elements in the relation (the "inputs").
Range-The set of second elements in the relation (the "outputs").
Example: For $\{(1, a),(2, b)\}$, domain $=\{1,2\}$, range $=\{a, b\}$

3) Types of Relations

Reflexive Relation

It is a type of relation where every element is related to itself.
Example: $\{(1,1),(2,2),(3,3)\} \rightarrow$ Reflexive, because each number appears with itself.

Symmetric Relation

If "a is related to b ", then " b is related to a " too.
Example: $\{(1,2),(2,1)\} \rightarrow$ Symmetric, because both $(1,2)$ and $(2,1)$ are there.

Transitive Relation
If "a is related to $b$ " and "b is related to $c$ ", then "a is related to $c$ ".
Example: $\{(1,2),(2,3),(1,3)\} \rightarrow$ Transitive, because it connects all the way through.

Also Read

Class 11 Subject-Wise Solutions

Do follow the links below to get NCERT solutions for all the subjects. Also, check out the exemplar solutions for effective learning.

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions


Frequently Asked Questions (FAQs)

Q: What is relation ?
A:

Relation R defined from a nonempty set A to nonempty set B is a subset of cartesian product A x B.

Q: A relation R is defined from set A to set B than the what is the domain of the relation R ?
A:

The relation R is a set of ordered pairs from cartesian product A x B and the set of all first elements of the ordered pairs in a relation R is called the domain of the relation R.

Q: A relation R is defined from set A to set B than the what is the co-domain of the relation R ?
A:

The relation R is a set of ordered pairs from cartesian product A x B and the whole set B is called the codomain of the relation R.

Q: A relation R is defined from set A to set B than the what is the range of the relation R ?
A:

The relation R is a set of ordered pairs from cartesian product A x B and the set of all second elements of the ordered pairs in a relation R is called the range of the relation R.

Q: Does range is a subset of co-domain ?
A:

Yes, the range is always a subset of co-domain.

Q: Find the total number of relations can be defined from set A to set B where n(A) = 2 and n(B) = 3 ?
A:

n(A) = 2 and n(B) = 3 

n(A x B) = 2 x 3= 6

 The total number of relations = 2^(6) = 64

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