Interactive OP Malhotra Class 11 Solutions Chapter 2 Relations and Functions Ex 2(a) engage students in active learning and exploration.
S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(a)
Question 1.
x ∈ {2, 4,6,9} and y ∈ {4, 6, 18, 27, 54}. From all ordered pairs (x, y) such that x is a factor of y and x < y.
Solution:
x ∈ {2, 4, 6, 9} and y ∈ {4, 6, 18, 27, 54}
we want to find all ordered pairs (x, y) such that x is a factor of y and x < y
Now 2 be a factor of 4, 6, 18, 54 and 2 < 4, 2 < 6, 2 < 18 and 2 < 54
∴ possible ordered pairs are ; (2, 4), (2, 6), (2, 18), (2, 54)
Now 4 is a factor of 4 but 4 ≮ 4
Now 6 is a factor of 18 and 54 and 6 < 18, 6 < 54
Then possible ordered pairs in this case are (6, 18) and (6, 54).
9 is a factor of 18, 27, 54, s.t 9 < 18 ; 9 < 27 and 9 < 54.
Then possible ordered pairs in this case are :
(9, 18), (9, 27) and (9, 54).
Hence required possible ordered pairs are :
{(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)}
Question 2.
Find the numbers x and y if (x + 3, y – 5) = (5, 0).
Solution:
Given (x + 3, y – 5) = (5, 0)
∴ x + 3 = 5
⇒ x = 2 and y – 5 = 0
⇒ y = 5
Question 3.
If A = {1, 3, 5, 7} and B = {2, 4, 6}, find
(i) A x A
(ii) A x B
(iii) B x A
(iv) B x B
(v) n (A x A)
(vi) n (A x B)
(vii) n (B x A)
(viii) n (B x B)
Solution:
Given A = {1, 3, 5, 7} and B = {2, 4, 6}
(i) ∴ A x A = {1,3,5, 7} x {1,3, 5,7}
= {(1, 1), (1, 3), (1, 5), (1, 7), (3, 1), (3, 3), (3, 5), (3, 7), (5, 1), (5, 3), (5, 5), (5, 7), (7, 1), (7, 3), (7, 5), (7, 7)} [∵ A x B = {(x, y); x ∈ A and y ∈ B}]
(ii) A x B = {1, 3, 5, 7} x {2, 4, 6}
= {(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6), (7, 2), (7, 4), (7, 6)}
(iii) B x A = {2, 4, 6} x {1,3, 5, 7}
= {(2, 1), (2, 3), (2, 5), (2, 7), (4, 1), (4, 3), (4, 5), (4, 7), (6, 1), (6, 3), (6, 5), (6, 7)}
(iv) B x B = {2, 4, 6} x {2, 4, 6}
= {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
(v) n (A x A) = no. of distinct elements in A x A = 16
(vi) n (A x B) = n (A) x n (B) = 4 x 3 = 12
(vii) n (B x A) = n (B) x n (A) = 3 x 4 = 12
(viii) n (B x B) = n (B) x n (B) = 3 x 3 = 9
Question 4.
Answer true or false :
(i) If P = {m, n} and Q = {m, n}, then P x Q = {(m, n), (n, m)}
(ii) {(a, x), (a, y), (b, x), (b, y)} is a product set.
(iii) If n (A) = x and n (B) = y and A ∩ B = Φ, then n (A x B) = xy.
(iv) If A and B are non-empty sets, then A x B is a non-empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.
Solution:
(i) Given P = {m, n) and Q = {n, m}
∴ P x Q = {{m, n), (m, m), (n, n), (n, m)}
Thus, given statement is false.
(ii) Let A = {a, b} and B = {x, y}
∴ A x B = {{a, x), (a, y), (b, x), (b, y)} which is the cartesian product of set A and set B.
Thus given statement is true.
(iii) n (A x B) = n (A) x n (B) = xy
∴ given statement is true.
(iv) False, ∵ A x B = {(x, y); x ∈ A and y ∈ B}
Question 5.
Given A = {1, 2}, B = {3}, C = {4, 5}, test whether the following are true :
(i) A x (B ∪ C) = (A ∪ B) x (A ∪ C)
(ii) A x (B ∩ C) = (A x B) ∩ (A x C).
Solution:
Given A = {1, 2} ; B = {3} and C = {4, 5}
(i) L.H.S = A x (B ∪ C) = {1, 2} x {3, 4,5}
= {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}
Now A ∪ B = {1, 2, 3}
and A ∪ C = {1, 2, 4, 5}
∴ R.H.S = (A ∪ B) x (A ∪ C)
= {1, 2, 3} x {1, 2, 4, 5}
= {(1, 1), (1, 2), (1, 4), (1, 5), (2, 1), (2, 2), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5)}
∴ L.H.S ≠ R.H.S
Thus given statement is not true.
(ii) ∴ B ∩ C = Φ
L.H.S = A x (B ∩ C) = {1, 2} x Φ = Φ
∴ A x B = {1, 2} x {3} = {(1, 3), (2, 3)}
A x C = {1, 2} x {4, 5} = {(1, 4), (1, 5), (2, 4), (2, 5)}
R.H.S. = (A x B) ∩ (A x C) = Φ
∴ L.H.S. = R.H.S
Thus given statement is true.
Question 6.
If A = {1, 2, 3, 4}, B = {5, 7, 9}, C = {2, 4, 6}, find
(i) A x B
(ii) (B x C)
(iii) C X A and draw their graphs.
Solution:
Given A = {1, 2, 3, 4} ; B = (5, 7, 9} and C = {2, 4, 6}
(i) A x B = {(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)}
(ii) B x C = {5, 7, 9} x {2, 4, 6}
= {(5, 2), (5, 4), (5, 6), (7, 2), (7, 4), (7, 6), (9, 2), (9, 4), (9, 6)}
(iii) C x A = {2, 4, 6} x {1,2, 3, 4}
= {(2, 1), (2, 2), (2, 3), (2, 4), (4, 1), (4, 2), (4, 3), (4, 4), (6, 1), (6, 2), (6, 3), (6,4)}
Question 7.
Some elements of A x B are (a, x), (c, y), (d, z). If A = {a, b, c, d], find the remaining elements of A x B such that n (A x B) is least.
Solution:
Given A = {a, b, c, d}
Since some elements of A x B are (a, x), (c, y), (d, z)
Thus B must be taken as {x, y, z} s.t n (A x B) is least.
∴ A x B = {a, b, c, d) x {x, y, z} = {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z), (d, x), (d, y), (d, z)}
Thus remaining elements of A x B are {(a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, z), (d, x), (d y)}
Question 8.
The ordered pairs (1, 1), (2, 2), (3, 3) are among the elements in the set A x B. If A and B have 3 elements each, how many elements in all does the set A x B have ? Also find the remaining elements.
Solution:
Given n (A) = 3 ; n (B) = 3
∴ n (A x B) = n (A) x n (B) = 3 x 3 = 9
Now (1,1), (2, 2), (3, 3) ∈ A x B
∴ first components of A x B i.e. 1, 2, 3 ∈ A
∴ A = {1, 2, 3} since A has exactly 3 elements and B = {1, 2, 3}
∴ A x B = {(x, y) : x ∈ A and y ∈ B}
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
Thus remaining elements of A x B are {(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)}
Question 9.
If A = {1, 4}, B = {2, 3, 6} and C = {2, 3, 7}, then verify that
(i) A x (B ∪ C) = (A x B) ∪ (A x C)
(ii) A x (B n C) = (A x B) ∩ (A x C)
Solution:
Given A= {1, 4} ; B = {2, 3, 6} and C = (2, 3, 7}
(i) B ∪ C = {2, 3, 6} ∪ {2, 3, 7} = {2, 3, 6, 7}
L.H.S = A x (B ∪ C) = {1, 4} x {2, 3, 6, 7}
= {(1, 2), (1, 3), (1, 6), (1, 7), (4, 2), (4, 3), (4, 6), (4, 7)}
A x B = {1, 4} x {2, 3, 6} = {(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)}
A x C = {1, 4} x (2, 3, 7} = {(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)}
∴ R.H.S. = (A x B) ∪ (A x C)
= {(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)} u {(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)}
= {(1, 2), (1, 3), (1, 6), (1, 7), (4, 2), (4, 3), (4, 6), (4, 7)}
Thus L.H.S. = R.H.S.
∴ A x (B ∪ C) = (A x B) ∪ (A x C)
(ii) ∴ B ∩ C = {2, 3, 6} ∩ {2, 3, 7} = {2, 3}
L.H.S = A x (B ∩ C) = {1, 4} x {2, 3} = {(1, 2), (1, 3), (4, 2), (4, 3)}
R.H.S = (A x B) ∩ (A x C) = {(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)} ∩ {(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)}
= {(1, 2), (1, 3), (4, 2), (4, 3)}
∴ L.H.S = R.H.S
Thus A x (B ∩ C) = (A x B) ∩ (A x C)
Question 10.
If A = {2, 3}, B = {1, 2, 3}, C = {2, 3, 4} show that A x A = (B x B) ∩ (C x C).
Solution:
Given A = {2, 3} ; B = {1, 2, 3} and C = {2, 3, 4}
L.H.S = A x A = {2, 3} x {2, 3} = {(2, 3). (2, 3), (3, 2), (3, 3)}
Now B x B = {1, 2, 3} x {1,2,3}
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
and C x C = {2, 3, 4} x {2, 3, 4}
= {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}
∴ R.H.S = (B x B) ∩ (C x C)
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} ∩ {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)} = {(2, 2), (2, 3), (3, 2), (3, 3)}
∴ L.H.S = R.H.S
Question 11.
If A and B be two sets such that n (A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y, z are distinct elements.
Solution:
Given n (A) = 3 and n (B) = 2
since, (x, 1), (y, 2), (z, 1) ∈ A x B
Since first components of A x B are in set A and second components are in set B.
Since x, y, z are given to be distinct elements.
∴ A = {x, y, z} similarly B = {1, 2}
Since A x B = {(x, y) : x ∈ A and y ∈ B}
Question 12.
The Cartesian product A x A has 9 elements among which are found (- 1, 0) and (0, 1). Find the set A are the remaining elements of A x A.
Solution:
Given n (A x A) = 9 ⇒ n (A) x n (A) = 9
⇒ n (A) = 3
Thus no. of distinct elements in set A be 3.
Now (- 1, 0), (0, 1) ∈ A x A
So first and second components of each ordered pair of A x A are belonging to A
∴ – 1, 0, 1 ∈ A
Thus A ={-1,0,1} and A x A = {- 1, 0, 1} x {- 1, 0, 1}
∴ A x A = {(- 1, – 1), (- 1, 0), (- 1, 1), (0, – 1), (0, 0), (0, 1), (1, – 1), (1, 0), (1, 1)}
Hence remaining elements of A x A are (- 1, – 1), (- 1, 1), (0, – 1), (0,0), (1, – 1), (1,0), (1, 1).