Tamilnadu Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.1

Question 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)
Solution:
(i) 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) 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} × Φ
A × B = { }
A × A = {(m,m), (m,n), (n, m), (n, n)}
B × A = { }

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

Question 3.
If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4),(3, 3), (3, 4)} find A and B.
Solution:
B × A ={(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)}
A = {3, 4), B = { -2, 0, 3}

Question 4.
If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C).
Solution:
A = {5,6}, B = {4,5,6},C = {5,6, 7}
A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} …(1)
B × B = {(4, 4), (4, 5), (4, 6), (5, 4),
(5,5), (5,6), (6,4), (6,5), (6,6)} …(2)
C × C = {(5,5), (5,6), (5,7), (6,5), (6,6),
(6, 7), (7, 5), (7, 6), (7, 7)} …(3)
(B × B) ∩ (C × C) = {(5, 5), (5,6), (6, 5), (6,6)} …(4)
(1) = (4)
A × A = (B × B) ∩ (C × C)
It is proved.

Question 5.
Given A ={1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A ∩ C) x (B ∩ D) = (A × B) ∩ (C × D) is true?
Solution:
LHS = {(A∩C) × (B∩D)
A ∩C = {3}
B ∩D = {3,5}
(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} …(1)
RHS = (A × B) ∩ (C × D)
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)
∴ (1) = (2) ∴ It is true.

Question 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)
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
Solution:
A = {x ∈ W|x < 2} = {0,1}
B = {x ∈ N |1 < x < 4} = {2,3,4}
C = {3,5}
LHS =A × (B ∪ C)
B ∪ C = {2,3,4} ∪ {3,5}
= {2, 3, 4, 5}
A × (B ∪ C) = {(0, 2), (0, 3), (0,4), (0, 5), (1.2) , (1,3), (1,4),(1,5)} …(1)
RHS = (A × B) ∪ (A × C)
(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), (1, 2), (1.3), (1,4), (0, 5), (1,5)} ….(2)
(1) = (2), LHS = RHS
Hence it is proved.

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
LHS = A × (B ∩ C)
(B ∩ C) = {3}
A × (B ∩ C) = {(0, 3), (1, 3)} …(1)
RHS = (A × B) ∩ (A × C)
(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)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
LHS = (A ∪ 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)
RHS = (A × C) ∪ (B × C)
(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)
(1) = (2)
∴ LHS = RHS. Hence it is verified.

Question 7.
Let A = The set of all natural numbers less than 8, B = The set of all prime numbers 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)
A = {1,2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
Solution:
(i)(A ∩ B) × C = (A × c) ∩ (B × C)
LHS = (A ∩ B) × C
A ∩ B = {2, 3, 5, 7}
(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} …(1)
RHS = (A × C) ∩ (B × C)
(A × C) = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}
(B × C) = {2,2), (3,2), (5,2), (7,2)}
(A × C) ∩ (B × C) = {(2,2), (3,2), (5,2), (7,2)} …(2)
(1) = (2)
∴ LHS = RHS. Hence it is verified.

(ii) A × (B – C) = (A × B) – (A × C)
LHS = A × (B – C)
(B – C) = {3,5,7}
A × (B – C) = {(1,3), (1, 5), (1,7), (2,3), (2,5), (2.7) , (3,3), (3,5), (3,7), (4,3), (4,5), (4,7), (5,3), (5,5), (5,7), (6,3) , (6,5), (6,7), (7,3), (7,5), (7.7)} …(1)
RHS = (A × B) – (A × C)
(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),
(4, 2), (4, 3), (4, 5), (4, 7),
(5, 2), (5, 3), (5, 5), (5, 7),
(6, 2), (6, 3), (6, 5), (6, 7),
(7, 2), (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, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3), (6, 5), (6, 7), (7, 3), (7, 5), (7,7) } …(2)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.