TN samacheer kalvi 10th Maths:
Chapter 1 Relations and Functions – Exercise 1.1 solutions
1. Find A × B, A × A and B × A
i) A = {2, -2, 3} and
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} ; B = Φ
A × B = { }
A × A =
{
(m, m), (m, n),
(n, m), (n, n)
}
B × A = { }
2. Let A= {1,2,3} and B = {× | x is a prime number less than 10}.
Find A × B and B × A.
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)
}
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}
4. If A= {5, 6}, B = {4, 5
,6}, C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C)
A ={5,6}
B = {4,5,6}
C = {5, 6,7}
L.H.S:
A × A =
{
(5, 5), (5, 6)
(6, 5), (6, 6)
} -----------------------------------(1)
R.H.S:
B × B =
{
(4, 4), (4, 5), (4, 6),
(5, 4), (5, 5), (5, 6),
(6, 4), (6, 5), (6, 6)
}
C × C =
{
(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) we get
L.H.S
= R.H.S
A × A = (B × B) ∩ (C × C), Hence verified
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?
A ={1, 2, 3}
B = {2, 3, 5}
C = {3, 4}
D = {1, 3, 5}
L.H.S: {(A∩C) × (B∩D)
A ∩C = {3}
B ∩D = {3, 5}
(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} ------------(1)
R.H.S: = (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)
From (1) and (2) we get
(A ∩ C) x (B ∩ D) = (A × B) ∩ (C × D)
L.H.S = R.H.S, Hence verified
6. Let A = {x ∑ W | x < 2}, B = {x ∑ N | 1 < 1 < × < 4}
and C = {3,5}. Verify that
A = {0, 1}
B = {2,3,4}
C = {3,5}
(i) A × (B ∪ C) = (A × B) ∪ (A × c)
L.H.S:
B ∪ C = {2, 3,4} ∪ {3,5} = {2, 3, 4, 5}
A × (B ∪ C) = {0, 1} × {2, 3, 4, 5}=
{
(0, 2), (0, 3), (0, 4), (0, 5),
(1, 2), (1, 3), (1, 4), (1, 5)
} --------------------------------------(1)
R.H.S:
A × B = {0, 1} × {2,3,4}=
{
(0,2), (0,3), (0,4),
(1,2), (1,3), (1,4)
}
A × C = {0, 1} × {3, 5}=
{
(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) we get
A × (B ∪ C) = (A × B) ∪ (A × C)
L.H.S = R.H.S, Hence verified
(ii) A × (B n C) = (A × B) n (A × C)
L.H.S:
B ∩ C = {2,3,4} ∩ {3,5} = {3}
A × (B ∩ C) = {0, 1} × {3}
= {(0,3) (1,3)} -----------------------(1)
R.H.S:
A × B = {0,1} × {2,3,4}=
{
(0, 2), (0, 3), (0, 4),
(1,2), (1,3), (1,4)
}
A × C = {0,1} × {3,5}=
{
(0, 3), (0, 5),
(1,3),
(1,5)
}
(A × B) n (A × C) = {(0, 3) (1, 3)} --------------------(2)
From (1) and (2) we get
A × ( B n C) = (A × B) n (A × C)
L.H.S = R.H.S, Hence verified
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
L.H.S:
A ∪ B = {0, 1} ∪ {2,3,4}
= {0,1, 2, 3, 4}
(A ∪ B) × C = {0, 1,2, 3,4} × {3,5}=
{
(0, 3), (0, 5)
(1, 3), (1, 5),
(2, 3), (2, 5),
(3, 3),(3, 5),
(4, 3), (4, 5)
} ----------------------------------------------------(1)
R.H.S:
A × C = {0, 1} × {3,5}=
{
(0,3), (0,5),
(1,3), (1,5)
}
B × C = {2,3,4} × {3,5}=
{
(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) we get
(A ∪ B) × C = (A × C) ∪ (B × C)
L.H.S = R.H.S, Hence verified
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
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
(i)(A ∩ B) × C = (A × c) ∩ (B × C)
L.H.S: (A ∩ B) × C
A ∩ B = {2, 3, 5, 7}
(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} -----------
(1)
R.H.S: (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)
From (1) and (2) we get
(A ∩ B) × C = (A × c) ∩ (B × C)
L.H.S = R.H.S, Hence verified
(ii) A × (B – C) = (A × B) – (A × C)
L.H.S: 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)
R.H.S: (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)
From (1) and (2) we get
A ×
(B – C) = (A × B) – (A × C)
L.H.S = R.H.S, Hence verified