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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 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) A = {2, -2, 3} and B = {1, -4}
A × B = {2,-2, 3} × {1,-4}
= {(2, 1), (2, -4)(-2, 1) (-2, -4) (3, 1) (3,-4)}
A × A = {2,-2, 3} × {2,-2, 3}
= {(2, 2)(2, -2)(2, 3)(-2, 2)
(-2, -2)(-2, 3X3,2) (3,-2) (3,3)}
B × A = {1,-4} × {2,-2, 3}
= {(1, 2)(1, -2)( 1, 3)(-4, 2) (-4,-2)(-4, 3)}

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

(iii) A = {m, n} × B = Φ
Note: B = Φ or {}
A × B = {m, n) × { }
= { )
A × A = {m, n) × (m, n)}
= {(m, m)(w, w)(n, m)(n, n)}
B × A = { } × {w, n}
= { }

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.
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) × (B ∩ D) = (A × B) ∩ (C × D) is true?
A = {1,2, 3}, B = {2, 3, 5}, C = {3,4} D = {1,3,5}
A ∩ c = {1,2,3} ∩ {3,4}
= (3}
B ∩ D = {2,3, 5} ∩ {1,3,5}
= {3,5}
(A ∩ C) × (B ∩ D) = {3} × {3,5}
= {(3, 3)(3, 5)} ….(1)
A × B = {1,2,3} × {2,3,5}
= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5)}
C × D = {3,4} × {1,3,5}
= {(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) we get
(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D)
This 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)
(iv) 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}

(i) (A ∩ B) × C = (A × C) ∩ (B × C)
A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7}
(A ∩ B) × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)} ….(1)
A × C = {1,2, 3, 4, 5, 6, 7} × {2}
= {(1,2) (2, 2) (3, 2) (4, 2)
(5.2) (6, 2) (7, 2)}
B × C = {2, 3, 5, 7} × {2}
= {(2, 2) (3, 2) (5, 2) (7, 2)}
(A × C) ∩ (B × C) = {(2, 2) (3, 2) (5, 2) (7, 2)} ….(2)
From (1) and (2) we get
(A ∩ B) × C = (A × C) ∩ (B × C)

(ii) A × (B – C) = (A × B) – (A × C)
B – C = {2, 3, 5, 7} – {2}
= {3,5,7}
A × (B – C) = {1,2, 3, 4, 5, 6,7} × {3,5,7}
= {(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)
A × B = {1,2, 3, 4, 5, 6, 7} × {2, 3, 5,7}
= {(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, 3,4, 5, 6, 7} × {2}
= {(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)
From (1) and (2) we get
A × (B – C) = (A × B) – (A × C)

Relations
Let A and B be any two non-empty sets. A “relation” R from A to B is a subset of A × B satisfying some specified conditions.

Note:

1. The domain of the relations R = {x ∈ A/xRy, for some y ∈ B}
2. The co-domain of the relation R is B
3. The range of the ralation

R = (y ∈ B/xRy for some x ∈ A}