NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.1 - Relations and Functions

NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.1 - Relations and Functions

Vishal kumarUpdated on 05 May 2025, 04:18 PM IST

Have you ever tried to make a chart linking your classmates to their favorite subjects or your friends to their birthdays? That’s the core idea behind relations in mathematics! The relation is defined as the link between a pair of objects from two sets and the function is a special type of relation where each input is related to exactly one output. These ideas serve as the cornerstone for understanding the relationships between variables in both mathematical and practical contexts. In this exercise of NCERT, you will get questions related to basic definitions of relations and Cartesian products of sets.

This Story also Contains

  1. Class 11 Maths Chapter 2 Exercise 2.1 Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 2: Exercise 2.1
  3. Topics covered in Chapter 2 Relations and Functions Exercise 2.1
  4. Class 11 Subject-Wise Solutions
  5. NCERT Solutions of Class 11 Subject Wise
NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.1 - Relations and Functions
2.1

The NCERT Solutions for Chapter 2 Exercise 2.1 will help you grasp these concepts by offering detailed explanations and step-by-step calculations. The terms like domain, codomain and range are introduced in this exercise in order to identify different types of relations. Students can also download the PDF available for the NCERT Solutions for feasible learning.

Class 11 Maths Chapter 2 Exercise 2.1 Solutions - Download PDF

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NCERT Solutions Class 11 Maths Chapter 2: Exercise 2.1

Question1: If $\left ( \frac{x}{3}+1 , y - \frac{2}{3} \right ) = \left ( \frac{5}{3},\frac{1}{3} \right )$ , find the values of x and y.

Answer:

It is given that
$\left ( \frac{x}{3}+1 , y - \frac{2}{3} \right ) = \left ( \frac{5}{3},\frac{1}{3} \right )$
Since the ordered pairs are equal, the corresponding elements will also be equal
Therefore,
$\frac{x}{3}+1= \frac{5}{3} \ \ \ and \ \ \ y - \frac{2}{3}= \frac{1}{3}$
$\frac{x}{3}= \frac{5}{3}-1 \ \ \ and \ \ \ y = \frac{1}{3}+ \frac{2}{3}$
$\frac{x}{3}= \frac{5-3}{3} \ \ \ and \ \ \ y = \frac{1+2}{3}$
$\frac{x}{3}= \frac{2}{3} \ \ \ and \ \ \ y = \frac{3}{3}$
$x= 2 \ \ \ and \ \ \ y = 1$

Therefore, values of x and y are 2 and 1 respectively

Question 2: If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in $( A \times B )$.

Answer:

It is given that set A has 3 elements and the elements in set B are 3 , 4 , and 5
Therefore, the number of elements in set B is 3
Now,
Number of elements in $(A \times B)$
= ( Number of elements in set A ) $\times$ ( Number of elements in set B)
= 3 $\times$ 3
= 9
Therefore, number of elements in $(A \times B)$ is 9

Question 3: If G = {7, 8} and H = {5, 4, 2}, find $G \times H \: \: and \: \: H \times G$

Answer:

It is given that
G = {7, 8} and H = {5, 4, 2}
We know that the cartesian product of two non-empty sets P and Q is defined as
P $\times$ Q = {(p,q) , where p $\epsilon$ P , q $\epsilon$ Q }
Therefore,
G $\times$ H = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
And
H $\times$ G = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

Question 4:(ii) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x \epsilon A$ and $y \epsilon B$

Answer:

It is a TRUE statement

$\because$ If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x \epsilon A$ and $y \epsilon B$

Question 4: (iii) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. If A = {1, 2}, B = {3, 4}, then$A \times (B \cap \phi ) = \phi$

Answer:

This statement is TRUE

$\because$ If A = {1, 2}, B = {3, 4}, then

$B\cap\phi=\phi$

There for

$A \times (B \cap \phi ) = \phi$

Question 5: If A = {–1, 1}, find $A \times A \times A$

Answer:

It is given that
A = {–1, 1}
A is an non-empty set
Therefore,
Lets first find $A \times A$
$A \times A = \left \{ -1,1 \right \} \times \left \{ -1,1 \right \} = \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$
Now,

$A \times A \times A=A \times(A \times A)=\{-1,1\} \times\{(-1,-1),(-1,1)$,$(1,-1),(1,1)\}$

$=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1)$,$(1,-1,1),(1,1,-1),(1,1,1)\}$

Question 6: If $A \times B$ = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Answer:

It is given that
$A \times B$ = {(a, x),(a , y), (b, x), (b, y)}
We know that the cartesian product of two non-empty set P and Q is defined as

$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we know that A is the set of all first elements and B is the set of all second elements
Therefore,
$A= \left \{ a,b \right \} \ \ \ and \ \ \ B = \left \{ x , y \right \}$

Question 7: (i) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$.

Answer:

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
$B \cap C = \left \{ 1,2,3,4 \right \} \cap \left \{ 5,6 \right \} = \Phi$
Now,
$A \times ( B \cap C ) = A \times \phi = \phi \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

$A \times B = \left \{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4) \right \}$
And
$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
Now,
$(A \times B)\cap (A \times C) =\phi \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
From equation (i) and (ii) it is clear that
$L.H.S. = R.H.S.$
Hence,
$A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$

Question 7: (ii) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A \times C$ is a subset of $B \times D$

Answer:

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,

$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
And
$B \times D$ = {(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),
(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)}
We can clearly observe that all the elements of the set $A \times C$ are the elements of the set $B \times D$
Therefore, $A \times C$ is a subset of $B \times D$

Question 8: Let A = {1, 2} and B = {3, 4}. Write $A \times B$. How many subsets will $A \times B$ have?
List them.

Answer:

It is given that
A = {1, 2} and B = {3, 4}
Then,
$A \times B = \left \{ (1,3),(1,4),(2,3),(2,4) \right \}$
$\Rightarrow n\left ( A \times B \right ) = 4$
Now, we know that if C is a set with $n(C) = m$
Then,
$n[P(C)]= 2^m$
Therefore,
The set $A \times B$ has $2^4=16$ subsets.

Question 9: Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in $A \times B$, find A and B, where x, y and z are distinct elements.

Answer:

It is given that
n(A) = 3 and n(B) = 2 and If (x, 1), (y, 2), (z, 1) are in A × B.

By definition of Cartesian product of two non-empty Set P and Q:
$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we can see that
P = set of all first elements.

And
Q = set of all second elements.
Now,
$\Rightarrow$ (x, y, z) are elements of A and (1,2) are elements of B
As n(A) = 3 and n(B) = 2
Therefore,
A = {x, y, z} and B = {1, 2}

Question 10 :The Cartesian product $A \times A$ has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of $A \times A$

Answer:

It is given that Cartesian product A × A having 9 elements among which are found (–1, 0) and (0,1).
Now,
Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)
$n(A \times A) = n(A) \times n(A)$
It is given that $n(A \times A) = 9$
Therefore,
$n(A) \times n(A) = 9$
$\Rightarrow n(A) = 3$
Now,
By definition A × A = {(a, a): a ? A}
Therefore,
-1, 0 and 1 are the elements of set A
Now, because, n(A) = 3 therefore, A = {-1, 0, 1}
Therefore,
the remaining elements of set (A × A) are
(-1,-1), (-1,1), (0,0), (0, -1), (1,1), (1, -1) and (1, 0)

Also read,

Topics covered in Chapter 2 Relations and Functions Exercise 2.1

1) Cartesian product of sets

If $A$ and $B$ are two non-empty sets, the Cartesian product is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.
$$
A \times B=\{(a, b) \mid a \in A, b \in B\}
$$
If set $A$ has $m$ elements and set $B$ has $n$ elements, then $A \times B$ will have $m \times n$ elements.

2) Definition of a relation
A relation from set $A$ to set $B$ is a subset of the Cartesian product $A \times B$.
Each element in the relation is an ordered pair $(a, b)$ where $a \in A, b \in B$, and $a$ is related to $b$.
So, relation $R \subseteq A \times B$.

Domain, codomain, and range of a relation
Domain- Set of all first elements (inputs) in the relation.
Codomain- The entire set $B$ (i.e., all possible outputs).
Range- Actual set of second elements (outputs) that are related to elements of $A$.
$$
\text { Range } \subseteq \text { Codomain }
$$

3) Types of relations (empty, universal, identity, etc.)
1. Empty (Void) Relation
No element of set $A$ is related to any element of $A$.

$$
R=\emptyset
$$

2. Universal Relation
Every element of $A$ is related to every element of $A$.

$$
R=A \times A
$$

3. Identity Relation
Each element is related only to itself.

$$
R=\{(a, a) \mid a \in A\}
$$

4. Reflexive Relation
Every element is related to itself, i.e., $(a, a) \in R$ for all $a \in A$.

5. Symmetric Relation
If $(a, b) \in R$, then $(b, a) \in R$ must also be true.

6. Transitive Relation
If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ should also be in $R$.

Also Read

Class 11 Subject-Wise Solutions

Students can also check the NCERT textbook and exemplar solutions from the links below and can excel in their exam preparations.

Frequently Asked Questions (FAQs)

Q: Do the ordered pairs (2,1) and (1,2) are equal ?
A:

No, two ordered pairs are equal, if and only if the corresponding elements of both the pairs are equal.

Q: If (x + 1, 1) = (3,1), find the values of x ?
A:

x +1 = 3

x = 3 -1 =2

Q: If the P and Q are two different sets, does cartesian product P x Q and Q x P are same ?
A:

No, cartesian product  P x Q and Q x P are not same. The number of elements in each set are the same.

Q: If P = {1, 2}, than find the set P x P ?
A:

P = { 1,2} 

P x P = { (1,1), (1,2), (2,1), (2,2) }

Q: If P = {1, 3}, than find the set P x P x P ?
A:

P = { 1,3} 

P x P x P=  {(1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3)}.

Q: If the number of elements in the set A is 2 and in set B is 3 than find the number of elements in the A x B ?
A:

n(A) = 2 and n(B) = 3

n(A × B) = 2x3 = 6

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