**NCERT
Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1**

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1
are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT
Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1.

**Ex 2.1
Class 11 Maths Question 1.**

If (x/3 + 1, y − 2/3) = (5/3, 1/3), find the values of x and
y.

**Solution.**

If the ordered pairs are equal, the corresponding elements are also equal.

∴ x/3 + 1 = 5/3 and y − 2/3 = 1/3

⇒ x/3 = 5/3 − 1 and y = 1/3 + 2/3

⇒ x = 2 and
y = 1.

**Ex 2.1 Class 11
Maths Question 2.**

If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of
elements in (A × B).

**Solution.**

According to the question, n(A) = 3 and n(B) = 3.

∴ n(A × B) = n(A) × n(B) = 3 × 3 = 9

∴ There are a total of 9 elements in (A ×
B).

**Ex 2.1 Class 11
Maths Question 3.**

If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

**Solution.**

We have G = {7, 8} and H = {5, 4, 2}. Then, by the definition of the cartesian
product, we have

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.

**Ex 2.1 Class 11
Maths Question 4.**

State whether each of the following statements are true or false. If the
statement is false, rewrite the given statement correctly.**(i)**If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.

**(ii)**If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

**(iii)**If A = {1, 2}, B = {3, 4}, then A × (B ∩ Ï†) = Ï†

**Solution.**

**(i)** False,
if P = {m, n} and Q = {n, m}

Then P × Q = {(m, n), (m, m), (n, n), (n, m)}.

**(ii)** True,
by the definition of cartesian product.

**(iii)** True,
we have A = {1, 2} and B = {3, 4}

Now, B ∩ Ï† = Ï†

∴ A × (B ∩ Ï†) = A × Ï† = Ï†

**Ex 2.1 Class 11
Maths Question 5.**

If A = {-1, 1}, find A × A × A.

**Solution.**

A = {-1, 1}

Then, A × A = {-1, 1} × {-1, 1} = {(-1, -1), (-1, 1), (1, -1), (1, 1)}

A × A × A = ((-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, 1, 1)}

**Ex 2.1 Class 11 Maths Question
6.**

If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.

**Solution.**

Given, A × B = {(a, x), (a, y), (b, x), (b, y)}

If (p, q) ∈ A × B, then p ∈ A and q ∈ B

∴ A = {a, b} and B = {x, y}.

**Ex 2.1 Class 11
Maths Question 7.**

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D.

**Solution.**

Given, A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}, D = {5, 6, 7, 8}

**Ex 2.1 Class 11
Maths Question 8.**

Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have?
List them.

**Solution.**

Given, A = {1, 2} and B = {3, 4}

Then, A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

i.e., A × B has 4 elements. So, it has 2^{4}, i.e., 16 subsets.

The subsets of A × B are as follows:

Ï†, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1,
3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1,
4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (2,
3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

**Ex 2.1 Class 11
Maths 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 × B, find A and B, where x, y and z are distinct elements.

**Solution.**

Given, n(A) = 3 and n(B) = 2.

Now (x, 1) ∈ A × B ⇒ x ∈
A and 1 ∈ B,

(y, 2) ∈ A × B ⇒ y ∈
A and 2 ∈ B

(z, 1) ∈ A × B ⇒ z ∈
A and 1 ∈ B

∴ x, y, z ∈ A and 1, 2 ∈ B

Hence, A = {x, y, z} and B = {1, 2}.

**Ex 2.1 Class 11
Maths Question 10.**

The Cartesian product A × A has 9 elements among which are found (-1, 0) and
(0, 1). Find the set A and the remaining elements of A × A.

**Solution.**

Since, we have n(A × A) = 9

⇒ n(A) × n(A) = 9 [ ∵ n(A × B) = n(A) × n(B)]

⇒ (n(A))^{2} = 9 ⇒ n(A) = 3

Also, given (-1, 0) ∈ A × A ⇒ -1, 0 ∈ A ,

and (0, 1) ∈ A × A ⇒ 0, 1 ∈ A

∴ -1, 0, 1 ∈ A

Hence, A = {-1, 0, 1} [∵ n(A) = 3]

The remaining elements of A × A are (-1, -1), (-1, 1), (0, -1), (0, 0), (1,
-1), (1, 0), (1, 1).

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