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Exercise - 2.1

Relations and Functions

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

In the Ordered Pairs, the corresponding values are equal. So, x/3 + 1 = 5/3 x/3 = 5/3 - 1 x/3 = (5 - 3)/3 3x = 2/3 x = 2/3 x 3 x = 2 y - 2/3 = 1/3 y = 1/3 + 2/3 y = (1 + 2)/3 y = 3/3 y = 1

**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).

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

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

Given that G = {7, 8} and H = {5, 4, 2}. G x H = {7, 8} x {5, 4, 2} = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 3)} H x G = {5, 4, 2} x {7, 8} = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

**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 ∩ φ) = φ.

(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

**Question-5 :-** If A = {–1, 1}, find A × A × A.

Given that A = {–1, 1}. A x A = {–1, 1} x {–1, 1} = {(-1, -1), (-1, 1), (1, -1), (1, 1)} A x A x A = {(-1, -1), (-1, 1), (1, -1), (1, 1)} x {-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 × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

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

**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.

Given that A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. (i) A × (B ∩ C) = (A × B) ∩ (A × C) L.H.S = B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ = A × (B ∩ C) = A × Φ = Φ R.H.S 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)} = (A × B) ∩ (A × C) = Φ L.H.S. = R.H.S Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

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

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

Given that A = {1, 2} and B = {3, 4}. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)} n(A × B) = 4 We know that if C is a set with n(C) = m, then n[P(C)] = 2ᵐ. Therefore, the set A × B has 2⁴ = 16 subsets. These are {Φ, {(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, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)} }

**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.

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

**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.

We know that if n(A) = p and n(B) = q, then n(A × B) = pq. ∴ n(A × A) = n(A) × n(A) 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}. A × A = {(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), (1, 1)}.

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