## Relations and Functions – Class XI – Exercise 2.1

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

Solution:

It is given that (x/3 +1, y – 2/3) = (5/3, 1/3)

since the ordered pairs are equal the corresponding elements will also be equal.

Therefore (x/3 +1 = 5/3) and (y – 2/3 = 1/3)

x/3 +1 = 5/3  and y – 2/3 = 1/3

x/3 = 5/3 – 1 and y = 2/3 + 1/3

x/3 = 2/3 and y = 1

⇒ x = 2 and y = 1

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

Solution:

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3

Number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)

= 3 × 3

= 9

Thus, the number of elements in (A × B) is 9.

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

Solution:

G = {7, 8} and H = {5, 4, 2}

We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as

P × Q = {(p, q): p∈ P, q ∈ Q}

∴ 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)}

4: State whether each of the following statement 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 ∩ Φ) = Φ.

Solution:

(i) False If P = {m, n} and Q = {n, m}, then P × Q = {(m, m), (m, n), (n, m), (n, n)}

(ii) True

(iii) True

1. If A = {–1, 1}, find A × A × A.

Solution:

It is known that for any non-empty set A, A × A × A is defined as A × A × A = {(a, b, c): a, b, c ∈ A}

It is given that A = {–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.

Solution:

It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets P and Q is defined as

P × Q = {(p, q): p ∈ P, q ∈ Q}

∴ A is the set of all first elements and B is the set of all second elements. Thus, A = {a, b} and 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)

(ii) A × C is a subset of B × D

solution:

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

∴ L.H.S. = A × (B ∩ C) = A × Φ = Φ

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

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

∴ R.H.S. = (A × B) ∩ (A × C) = Φ

∴ L.H.S. = R.H.S

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

(ii) To verify: A × C is a subset of B × D

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

A × 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 observe that all the elements of set A × C are the elements of set B × D.

Therefore, A × C is a subset of B × D.

1. 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:

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

We know that,

A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and 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.

Solution:

We know that if n(A) = p and n(B) = q, then n(A × B) = pq.

∴ n(A × A) = n(A) × n(A)

It is given that n(A × A) = 9

∴ n(A) × n(A) = 9

⇒ n(A) = 3

The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A×A.

We know that A × A = {(a, a): a ∈ A}.

Therefore, –1, 0, and 1 are elements of A.

Since n(A) = 3, it is clear that A = {–1, 0, 1}.

The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1).