10TH Maths Relations and Functions EXERCISE 1.1

10TH Maths Relations and Functions EXERCISE 1.1

10TH Maths Relations and Functions EXERCISE 1.1

1. Find A × B, A × A and B × A
(i) A = {2, – 2, 3} and B = {1, – 4}
(ii) A = B = {p, q} (iii) A = {m, n} ; B = φ

(i) Given A = {2, – 2, 3}, B = {1, – 4}.
A × B = {(2, 1), (2, – 4), (– 2, 1), (–2, – 4),
(3, 1), (3, – 4)}A × A = {(2, 2), (2, – 2), (2, 3), (– 2, 2),
(– 2, – 2), (– 2, 3), (3, 2), (3, – 2), (3, 3)}
B × A = {(1, 2), (1, – 2), (1, 3), (– 4, 2),
(– 4, – 2), (– 4, 3)}

(ii) Given A = B = {p, q}
A × B = {(p, p), (p, q), (q, p), (q, q)}
A × A = {(p, p), (p, q), (q, p), (q, q)}
B × A = {(p, p), (p, q), (q, p), (q, q)}

(iii) A = {m, n}, B = φ
If A = φ (or) B = φ, then A × B = φ.
and B × A = φ
A × B = φ and B × A = φ
A × A = {(m, m), (m, n), (n, m), (n, n)}

2. Let A = {1, 2, 3} and B = {x | x is a prime
number less than 10}. Find A × B and

Solution :

Given A = {1, 2, 3}, B = {x | x is a prime
number less than 10}.
B = {2, 3, 5, 7}
A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2),
(2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2),
(3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}

3. If B × A = {(– 2, 3), (– 2, 4), (0, 3), (0, 4),
(3, 3), (3, 4)} find A and B.

Solution :

Given B × A = {(–2, 3), (– 2, 4), (0, 3),
(0, 4), (3, 3), (3, 4)}
∴ B = {– 2, 0, 3}, A = {3, 4}

4. If A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}.
Show that A × A = (B × B) ∩ (C × C).

Solution :

Given A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}
LHS : A × A = {5, 6} × {5, 6}
= {(5, 5), (5, 6), (6, 5), (6, 6)} ...(1)
RHS : B ×B = {4, 5, 6} ×{4, 5, 6}
= {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5),
(5, 6), (6, 4), (6, 5), (6, 6)}
C × C = {5, 6, 7} × {5, 6, 7}
= {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),
(6, 7), (7, 5), (7, 6), (7, 7)}
∴ (B × B) ∩ (C × C) = {(5, 5), (5, 6), (6, 5),
(6, 6)} ...(2)
∴ From (1) and (2).
LHS = RHS

5. Given A = {1, 2, 3}, B = {2, 3, 5},
C = {3, 4} and D = {1, 3, 5}, check if
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is
true ?

Solution :

Given A = {1, 2, 3}, B = {2, 3, 5},
C = {3, 4}, D = {1, 3, 5}
A ∩ C = {3}, B ∩ D = {3, 5}
∴ (A ∩ C) × (B ∩ D) = {(3, 3), (3, 5)} ... (1)
A × B = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3),
(2, 5), (3, 2), (3, 3), (3, 5)}
C × D = {(3, 1), (3, 3), (3, 5), (4, 1),
(4, 3), (4, 5)}
∴ (A × B) ∩ (C × D) = {(3, 3), (3, 5)} ...(2)
∴ From (1) and (2)
LHS = RHS.

6. Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x
≤ 4} and C = {3, 5}. Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

Solution :
Given A = {x ∈ W | x < 2} ⇒ A = {0, 1}
B = {x ∈ N | 1 < x ≤ 4}
⇒ B = {2, 3, 4}
C = {3, 5}

(i) To verify :
A × (B ∪ C) = (A × B) ∪ (A × C)
B ∪ C = {2, 3, 4, 5}
∴ A × (B ∪ C) = {(0, 2), (0, 3), (0, 4), (0, 5),

(1, 2), (1, 3), (1, 4), (1, 5)} ...(1)

A × B = {(0, 2), (0, 3), (0, 4), (1, 2),
(1, 3), (1, 4)}
A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
∴ (A × B) ∪ (A × C) = {(0, 2), (0, 3), (0, 4),
(0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} ...(2)
∴ From (1) and (2) LHS = RHS.

(ii) To verify : A × (B ∩ C) = (A × B) ∩ (A × C)
B ∩ C = {3}
∴ A × (B ∩ C) = {(0, 3), (1, 3)} ...(1)
A × B = {(0, 2), (0, 3), (0, 4), (1, 2),
(1, 3), (1, 4)}
A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
∴ (A × B) ∩ (A × C) = {(0, 3), (1, 3)} ...(2)
∴ From (1) and (2), LHS = RHS.

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
A ∪ B = {0, 1, 2, 3, 4}
∴ (A ∪ B) × C = {(0, 3), (0, 5), (1, 3), (1, 5),
(2, 3), (2, 5), (3, 3), (3, 5),
(4, 3), (4, 5)} ...(1)
A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
B × C = {(2, 3), (2, 5), (3, 3), (3, 5),
(4, 3), (4, 5)}
∴ (A × C) ∪ (B × C) = {(0, 3), (0, 5), (1, 3),
(1, 5), (2, 3), (2, 5)
(3, 3), (3, 5), (4, 3),
(4, 5)} ...(2)
∴ From (1) and (2) LHS = RHS.

7. Let A = The set of all-natural numbers
less than 8, B = The set of all prime num-
be less than 8, C = The set of even prime

number. Verify that
(i) (A ∩ B) × C = (A × C) ∩ (B × C)
(ii) A × (B – C) = (A × B) – (A × C)

Solution :

Given A = {1, 2, 3, 4, 5, 6, 7}
B = {1, 3, 5, 7}
C = {2}
(i) To verify : (A ∩ B) × C = (A × C) ∩ (B × C)
A ∩ B = {1, 3, 5, 7}
∴ (A ∩ B) × C = {(1, 2), (3, 2), (5, 2), (7, 2)}
... (1)
A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2),
(6, 2), (7, 2)}
B × C = {(1, 2), (3, 2), (5, 2), (7, 2)} ... (2)
∴ From (1) and (2), LHS = RHS.

(ii) To verify : A × (B – C) = (A × B) – (A × C)
B – C = {1, 3, 5, 7}
∴ A × (B – C) = {(1, 1), (1, 3), (1, 5), (1, 7),
(2, 1), (2, 3), (2, 5), (2, 7),
(3, 1), (3, 3), (3, 5), (3, 7),
(4, 1), (4, 3), (4, 5), (4, 7),
(5, 1), (5, 3), (5, 5), (5, 7),
(6, 1), (6, 3), (6, 5), (6, 7),
(7, 1), (7, 3), (7, 5), (7, 7),}
A × B = {(1, 1), (1, 3), (1, 5), (1, 7),
(2, 1), (2, 3), (2, 5), (2, 7),
(3, 1), (3, 3), (3, 5), (3, 7),
(4, 1), (4, 3), (4, 5), (4, 7),

(5, 1), (5, 3), (5, 5), (5, 7),
(6, 1), (6, 3), (6, 5), (6, 7),
(7, 1), (7, 3), (7, 5), (7, 7),}
A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2),
(6, 2), (7, 2)}
∴ (A × B) – (A × C)
= {(1, 1), (1, 3), (1, 5), (1, 7),
(2, 1), (2, 3), (2, 5), (2, 7),
(3, 1), (3, 3), (3, 5), (3, 7),
(4, 1), (4, 3), (4, 5), (4, 7),
(5, 1), (5, 3), (5, 5), (5, 7),
(6, 1), (6, 3), (6, 5), (6, 7),
(7, 1), (7, 3), (7, 5), (7, 7),}

...(2)

∴ From (1) and (2), LHS = RHS.




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