# GSEB Solutions Class 11 Maths Chapter 2 Relations and Functions Ex 2.1

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 2 Relations and Functions Ex 2.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 2 Relations and Functions Ex 2.1

Question 1.
If ($$\frac{x}{3}$$ + 1, y – $$\frac{2}{3}$$) = ($$\frac{5}{3}$$, $$\frac{1}{3}$$), find the values of x and y?
Solution:
$$\frac{x}{3}$$ + 1 = $$\frac{5}{3}$$ or $$\frac{x}{3}$$ = $$\frac{5}{3}$$ – 1 = $$\frac{2}{3}$$ â‡’ x = 2
and y – $$\frac{2}{3}$$ = $$\frac{1}{3}$$ â‡’ y = $$\frac{2}{3}$$ + $$\frac{1}{3}$$ = 1.
âˆ´ x = 2 and 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 Ã— B).
Solution:
Set A has 3 elements and set B also has 3 elements.
âˆ´ Number of elements in A Ã— B = 3 Ã— 3
= 9.

Question 3.
If G = {7, 8} and H = {5, 4, 2}, find G Ã— H and H Ã— G?
Solution:
(i) G Ã— H = {7, 8} Ã— {5, 4, 2}
= {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

(ii) H Ã— G = {5, 4, 2} Ã— (7, 8}
= {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

Question 4.
State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly?

1. If P = {m, n} and Q = {n, m), then P Ã— Q = {(m, n), (n, m)}
2. If A and B are non empty sets, then A Ã— B is a non empty set of ordered pairs (x, y) such that x âˆˆ B and y âˆˆ A.
3. If A = {1, 2}, B = {3, 4}, then A Ã— (B âˆ© Ï•) = Ï•.

Solution:

1. False, P Ã— Q = {(m, n), (m, m)){(n, n), (n, m)}
2. False, A Ã— B is a non-empty set of ordered pairs such that x âˆˆ A and y âˆˆ B.
3. True, B âˆ© Ï• = Ï• â‡’ A Ã— (B âˆ© Ï•) = A’ Ã— Ï• = Ï•.

Question 5.
If A = {- 1, 1), find A Ã— A Ã— A.
Solution:
A Ã— A = {-1, 1} Ã— {-1, 1} = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
A Ã— A Ã— A = {- 1, 1} Ã— {(- 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)

Question 6.
If A Ã— B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B?
Solution:
A = [a, b}, B = {x, y} since A Ã— B = {(a, x), (a, y), (b, x), (b, y)}

Question 7.
Let A = (1, 2}, B = (1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}?
Verify that:

1. A Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C)
2. A Ã— C is a subset of B Ã— D.

Solution:
1. B âˆ© C = {1, 2, 3, 4} âˆ© {5, 6} = Ï•
L.H.S. = A Ã— (B âˆ© C) = (1, 2) Ã— Ï•
(A Ã— B) = {1, 2} Ã— {1, 2, 3, 4} = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3) (2, 4)}.
A Ã— C = {1, 2} Ã— {5, 6} = {(1, 5), (1, 6), (2, 5), (2, 6)}.
âˆ´ R.H.S = (A Ã— B) âˆ© (A Ã— C) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)} âˆ©{(1, 5), (1, 6), (2, 5), (2, 6)} = Ï•

2. A Ã— C = {1, 2} Ã— {5, 6} = {(1, 5), (1, 6), (2, 5), (2, 6)}
B Ã— D = {1, 2, 3, 4} Ã— {5, 6, 7, 8} = {(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)}
All the elements of set A Ã— C belong to set B Ã— D.
âˆ´ (A Ã— C) âŠ‚ (B Ã— D).

Question 8.
Let A = {1, 2} and B = {3, 4}. Write A Ã— B. How many subsets write A Ã— B have? List them?
Solution:
A Ã— B = (1, 2} Ã— {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}
No. of subsets of A Ã— B = 24 = 16.
Subsets of A Ã— B are 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, 3), (2, 3), (2, 4)}, {(1, 4), (2, 1), (2, 4)}, {(1, 3) (1, 4), (2, 3), (2, 4)}.

Question 9.
Let A and B be two sets such that n(A) = 3, n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A Ã— B, find A and B, where x, y and z are distinct elements?
Solution:
Elements x, y, z âˆˆ A and so A = {x, y, z}.
Elements 1, 2 âˆˆ B and so B = {1, 2}.

Question 10.
The cartesian product A Ã— A has 9 elements among which are found (- 1, 0) and (0, 1). Find the set A and the remaining elements of A Ã— A?
Solution:
(- 1, 0) âˆˆ A Ã— A. Also, (0, 1) âˆˆ A Ã— A
â‡’ – 1, 0 âˆˆ A.
Further, (0, 1) âˆˆ A Ã— A
â‡’ 0, 1 âˆˆ A.
So, A = {- 1, 0, 1).
âˆ´ A Ã— A = {- 1, 0, 1} Ã— {- 1, 0, 1}
= {(- 1, – 1), (- 1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, – 1), (1, 0), (1, 1) which includes (- 1, 0) and (0, 1) both.