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

## 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 24, 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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