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 = \(\frac{2}{3}\) ⇒ x = 2
and y – \(\frac{2}{3}\) = \(\frac{1}{3}\) ⇒ y = \(\frac{2}{3}\) + \(\frac{1}{3}\) = 1.
∴ x = 2 and y = 1.

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

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).
Solution:
Set A has 3 elements and set B also has 3 elements.
∴ Number of elements in A × B = 3 × 3
= 9.

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

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

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

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 × Q = {(m, n), (n, m)}
  2. If A and B are non empty sets, then A × B is a non empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.
  3. If A = {1, 2}, B = {3, 4}, then A × (B ∩ ϕ) = ϕ.

Solution:

  1. False, P × Q = {(m, n), (m, m)){(n, n), (n, m)}
  2. False, A × B is a non-empty set of ordered pairs such that x ∈ A and y ∈ B.
  3. True, B ∩ ϕ = ϕ ⇒ A × (B ∩ ϕ) = A’ × ϕ = ϕ.

Question 5.
If A = {- 1, 1), find A × A × A.
Solution:
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)

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

Question 6.
If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B?
Solution:
A = [a, b}, B = {x, y} since A × 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 × (B ∩ C) = (A × B) ∩ (A × C)
  2. A × C is a subset of B × D.

Solution:
1. B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = ϕ
L.H.S. = A × (B ∩ C) = (1, 2) × 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 × C = {1, 2} × {5, 6} = {(1, 5), (1, 6), (2, 5), (2, 6)}.
∴ R.H.S = (A × B) ∩ (A × C) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)} ∩{(1, 5), (1, 6), (2, 5), (2, 6)} = ϕ

2. A × C = {1, 2} × {5, 6} = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {1, 2, 3, 4} × {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 × C belong to set B × D.
∴ (A × C) ⊂ (B × D).

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

Question 8.
Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets write A × B have? List them?
Solution:
A × B = (1, 2} × {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}
No. of subsets of A × B = 24 = 16.
Subsets of A × B 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, 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 × B, find A and B, where x, y and z are distinct elements?
Solution:
Elements x, y, z ∈ A and so A = {x, y, z}.
Elements 1, 2 ∈ B and so B = {1, 2}.

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

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:
(- 1, 0) ∈ A × A. Also, (0, 1) ∈ A × A
⇒ – 1, 0 ∈ A.
Further, (0, 1) ∈ A × A
⇒ 0, 1 ∈ A.
So, 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) which includes (- 1, 0) and (0, 1) both.

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