# Relations and Functions Exercise 2.1 Solutions – NCERT Class 11 Mathematics Chapter 2

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

$$\left(\frac{x}{3} + 1, y – \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)$$

Comparing both sides of this equation we get

$$\frac{x}{3} + 1 = \frac{5}{3} \\ y – \frac{2}{3} = \frac{1}{3} \\$$

Simplifying x/3 + 1 = 5/3

$$\frac{x}{3} + 1 = \frac{5}{3} \\ \frac{x}{3} = \frac{5}{3} – 1 \\ \frac{x}{3} = \frac{5 – 3}{3} \\ \frac{x}{3} = \frac{2}{3} \\ x = 2$$

Simplifying y – 2/3 = 1/3

$$y – \frac{2}{3} = \frac{1}{3} \\ y = \frac{1}{3} + \frac{2}{3} \\ y = \frac{1 + 2}{3} = \frac{3}{3} \\ y = 1$$

Thus if (x/3 + 1, y – 2/3) = (5/3, 1/3) then value x = 2 and y = 1.

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

We know that n(A × B) = n(A) × n(B)
As per question its given that n(A) = 3 and n(B) = 3
Hence n(A × B) = n(A) × n(B) = 3 × 3
n(A × B) = 3 × 3
n(A × B) = 9

Thus number of elements in A × B is 9 if set A have 3 elements and set B also have 3 elements.

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

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

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)}

False
P × Q = {{m, n} × {n, m}} = {(m, n), (m, m), (n, n), (n, m)}
Hence if P = {m, n} and Q = {n, m} then P × Q = {(m, n), (m, m), (n, 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

True

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

True
Here’s why it’s True
B ∩ φ = {3, 4} φ = φ
A × (B ∩ φ) = A × φ = φ

5. If A = {–1, 1} then Find A × A × A
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}
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)}

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

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)

B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = ϕ
A × (B ∩ C) = {1, 2} × ϕ = ϕ
A × (B ∩ C) = ϕ

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

A × C = {1, 2} × {5, 6} = {(1, 5), (1, 6), (2, 5), (2, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

(A × B) ∩ (A × C) = ϕ

Hence A × (B ∩ C) = (A × B) ∩ (A × C) = ϕ

(ii) A × C is a subset of B × D
A × C = {1, 2} × {5, 6}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {1, 2, 3, 4} × {5, 6, 7, 8}
B × 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)}

As elements in set A × C are also in set B × D hence A × C is a subset of B × D

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

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

Basic formula to find out how many subsets does a set have is 2n where n is number of elements in set.
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
=> Number of elements in A × B = 4
Hence number of subsets of A × B = 24 = 16

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.

Given that n(A) = 3 and n(B) = 2
And (x, 1), (y, 2), (z, 1) ∈ A × B
This means that
x, y, z ∈ A
1, 2 ∈ B

As A × B = {(a, b): a ∈ A, b ∈ B}
This implies
A = {x, y, z}
B = {1, 2}

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

Number of elements in set A × A = 9
Elements in cartesian product (-1, 0) and (0, 1)
(-1, 0), (0, 1) ∈ A × A
A = {-1, 0, 1}

A × A = {{-1, 0, 1} × {-1, 0, 1}}
{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}

Hence remaining elements in cartesian production A × A are (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)