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ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.1
Question 1.
Find a and b if
(i) (a + 1, b – 2) = (3, 1)
(ii) (\(\frac{a}{3}\) + 1, b – \(\frac{2}{3}\)) = (\(\frac{5}{3}, \frac{1}{3}\))
(iii) (2a, a + b) = (6, 2)
(iv) (a + b, 3b – 2) = (7, – 5)
Solution:
(i) Given (a + 1, b – 2) = (3, 1)
∴ a + 1 = 3
⇒ a = 2
and b – 2 = 1
⇒ b = 3
(ii) Given (\(\frac{a}{3}\) + 1, b – \(\frac{2}{3}\)) = (\(\frac{5}{3}, \frac{1}{3}\))
i.e. \(\frac{a}{3}\) + 1 = \(\frac{5}{3}\)
⇒ \(\frac{a}{3}=\frac{2}{3}\)
⇒ a = 2
and b – \(\frac{2}{3}\) = \(\frac{1}{3}\)
⇒ b = \(\frac{2}{3}\) + \(\frac{1}{3}\) = 1
(iii) Given (2a, a + b) = (6, 2)
∴ 2a = 6
⇒ a = 3
and a + b = 2
⇒ b = 2 – 3 = – 1
(iv) Given (a + b, 3b – 2) = (7, – 5)
∴ a + b = 7
and 3b – 2 = – 5
∴ 3b = – 3
∴ b = – 1
∴ a = 7 – b
= 7 + 1 = 8.
Question 2.
Find x and y if
(ij) (4x + 3, y) = (3x + 5, – 2)
(ii) (x – y, x + y) = (6, 10)
Solution:
(i) Given (4x + 3, y) = (3x + 5, – 2)
∴ 4x + 3 = 3x + 5
⇒ x = 2 and y = – 2
(ii) (x – y, x + y) = (6, 10)
∴ x – y = 6 ………………(1)
and x + y = 10 …………………(2)
On adding (1) and (2), we have
2x = 16
⇒ x = 8
∴ from (1) ;
y = 8 – 6 = 2
(ii) Given (2x + y, x – y) = (8, 3)
∴ 2x + y = 8 ……………..(1)
x – y = 3 ………………(2)
On adding ( ) and (2) ; we have
3x = 11
⇒ x = \(\frac{11}{3}\)
∴ from (2) ; we have
y = x – 3
= \(\frac{11}{3}\) – 3
= \(\frac{11-9}{3}\)
= \(\frac{2}{3}\)
Question 3.
If the ordered pars (a, – 1) and (5, b) belong to {(x, y) : y = 2.x – 3}, find the values of a and b.
Solution:
Since (a, – 1) ∈ {(x, y) : y = 2x – 3}
∴ – 1 = 2a – 3
⇒ 2a = 2
⇒ a = 1
(5, b) ∈ {(x, y) : y = 2x – 3}
∴ b = 2 × 5 – 3 = 7
Question 4.
If A = {- 1, 0, 1} and B = {3, 5), write the following:
(i) A × B
(ii) B × A
(iii) B × B.
Solution:
Given A = {- 1, 0, 1}
and B = {3, 5}
(i) A × B = {(- 1, 3), ( 1, 5), (0, 3), (0, 5), (1, 3), (1, 5)}
(ii) B × A = {(3, – 1), (3, 0), (3, 1), (5, – 1), (5, 0), (5, 1)}
(iii) B × B = {(3, 3), (3, 5), (5, 3), (5, 5)}
Question 5.
If A is a set such that n (A) = 3 and B = {3, 4, 5) then what is the number of elements in A × B ?
Solution:
Given n (A) = 3
and B = {3, 4, 5}
∴ n(B) = 3
Thus, n (A × B) = n (A) × n (B)
= 3 × 3 = 9
Question 6.
If A = {- 3, – 1, 0, 4} and B = {- 1, 0, 1, 2, 3}, then write the number of elements in each of the following cartesian products:
(i) A × B
(ii) B × A
(iii) A × A
(iv) B × B.
Solution:
Given A = {- 3, – 1, 0, 4}
∴ n(A) = 4
and B = {- 1, 0, 1, 2, 3}
n(B) = 5
(i) ∴ no. of elements in A × B = n (A × B)
= n (A) × n (B)
= 4 × 5 = 20
(ii) ∴ no. of elements in B × A = n (B × A)
= n (B) × n (A)
= 5 × 4 = 20
(iii) ∴ no. of elements in A × A = n (A × A)
= n (A) × n (A)
= 4 × 4 = 16
(iv) ∴ no. of elements in B × B = n (B × B)
= n (B) × n (B)
= 5 × 5 = 25
Question 7.
If A = {1, 2} and B = {3, 4}, then how many ubets will A × B have?
Solution:
Given A = {1, 2}
and B = {3, 4}
∴ n (A) = 2 ;
n (B) = 2
Thus number of subsets of A × B = 2n (A × B)
= 2n (A) × n (B)
= 22 × 2
= 24 = 16
Question 8.
If x ∈ {2, 3, 5} and y ∈ {2, 4, 6}, form the set of all ordered pairs (x, y) such that x < y.
Solution:
Given x ∈ {2, 3, 5} and y ∈ {2, 4, 6}
We want to find all ordered pair (x, y) for which x < y.
since 2 < 4 ;
2 < 6 ;
3 < 4 ;
3 < 6 ;
5 < 6 ;
Hence, required set of all such ordered pairs (x, y) s.t. x < y be {(2, 4) (2, 6), (3,4) (3, 6) (5, 6)}.
Question 9.
If x ∈ {- 1, 2, 3, 4, 5} and y ∈ {0, 3, 6}, form the set of all ordered pairs (x, y) such that x + y = 5.
Solution:
Given x ∈ {- 1, 2, 3, 4, 5}
and y ∈ {0, 3, 6}
Hence the set of all ordered pair (x, y) for which x + y = 5 be given by {(- 1, 6), (2, 3), (5, 0)}.
Question 10.
If x ∈ (2, 3, 4) and y ∈ {4, 6, 9, 10}, form the set of all ordered pairs (x, y) such that x is a factor of y.
Solution:
Given x ∈ {2, 3, 4}
and y ∈ {4, 6, 9, 10}
since 2/4, 2/6, 2/10, 3/6, 3/9, 4/4
Hence the set of all ordered pair (x, y) for which x be a factor of y be given by
{(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}
Question 11.
State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.
(iii) If A = (1, 2), B = {3, 4}, then A × (B + Φ) = Φ.
Solution:
(i) false, true statement is
P × Q = {(m, n), (m, m), (n, n), (n, m)}
(ii) false, If A and B are non empty sets then A × B is a non-empty set of ordered pairs (x, y) s.t x ∈ A and y ∈ B.
(iii) True, A × (B ∩ Φ) = A × Φ = Φ
Question 12.
If A = {- 1, 0, 1, 2, 3), write the subset S of A × A such that the second component of the elements of S is 0.
Solution:
Given A = {- 1, 0, 1, 2, 3}
∴ A × A = {(- 1,- 1), (- 1, 0), (- 1, 1), (-1, 2), (- 1, 3), (0, – 1), (0, 0), (0, 1), (0, 2), (0, 3), (1, – 1), (1, 0) (1, 1), (1, 2), (1, 3), (2, – 1), (2, 0), (2, 1), (2, 2), (2, 3), (3, – 1), (3, 0), (3, 1), (3, 2), (3, 3)}
Thus, A × A has 25 different ordered pairs
∴ S = {(- 1, 0), (0, 0), (1, 0), (2, 0). (3,0)} ⊆ A × A.
Question 13.
If A × B = {(p, q), (p, r), (m, q), (m, r)}, find A and B.
Solution:
Given A × B = {(p, q), (p, r), (m, q), (m, r)}
since A × B = {(x, y), x ∈ A, y ∈ B}
∴ A = {p, m}
and B = {q, r}
Question 14.
If A = {x, y, z} and some elements of A × B are (x, 1), (y, 2), (z, 1) then write the set B such that n (A × B) = 6.
Solution:
Given A = {x, y, z}
since A × B = {(x, y) : x ∈ A and y ∈ B} ……………(1)
Some elements of A × B are (x, 1), (y, 2) and (z, 1)
also, n (A × B) = 6
∴ A × B = {x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Thus B = {1, 2}.
Question 15.
If A × B = {(x, 1), (y, 2), (x, 3), (y, 3), (y, 1), (x, 2)}, then find B × A.
Solution:
Given A × B = {(x, 1), (y, 2), (x, 3), (y, 3), (y, 1), (x, 2)}
∴ A = {x, y}
and B = {1, 2, 3}
Since A × B = {(x, y) : x ∈ A, y ∈ B}
Thus, B × A = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)}.
Long answer questions (16 to 22) :
Question 16.
If A = {1, 2, 3, 4} and B = {5, 7, 9}, find
(i) A × B
(ii) B × A
(iii) Is A × B = B × A ?
(iv) Is n (A × B) = n (B × A) ?
Solution:
Given A = {1, 2, 3, 4}
and B = {5, 7, 9}
(i) A × 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 × A = {(5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)}
(iii) Clearly A × B ≠ B × A
(iv) n (A × B) = number of distinct elements in A × B
= n (A) × n (B)
= 4 × 3 = 12
n (B × A) = n(B) × n(A)
= 3 × 4 = 12
∴ n (A × B) = n (B × A
Question 17.
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × C is a subset of B × D.
Solution:
Given A = {1, 2},
B = {1, 2, 3, 4}
C = {5, 6}
and D = {5, 6, 7, 8}
(i) (B ∩ C) = {1, 2, 3, 4} ∩ {5, 6} = Φ
A × (B ∩ C) = Φ
Now, A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5). (2, 6)}
∴ (A × B) ∩ (A × C) = Φ …………………….(2)
∴ from (1) and (2) ; we have
A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) Now A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
and B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Clearly all members of set A × C are the members of B × D.
∴ A × C be a subset of B × D.
Question 18.
If A = {x : x ∈ W, x < 2), B = {x : x ∈ N, 1 < x < 5} and C = {3, 5}, find
(i) A × (B ∩ C)
(ii) A × (B ∪ C)
Solution:
(i) Given A = {x : x ∈ W, x < 2) = {0, 1}
B = {x : x ∈ N, 1 < x < 5}
= {2, 3, 4}
and C = {3, 5}
B ∩ C = {2, 3, 4) ∩ {3, 5) = {3}
∴ A × (B ∩ C) = {0, 1} × {3}
= {(0, 3), (1, 3)}
(ii) B ∪ C = {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)}
Question 19.
If A = {x : x ∈ N and x ≤ 3}, B = {x : x ∈ I, – 1 ≤ x ≤ 1} and C = {1, 2}, 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 : x ∈ N and x ≤ 3} = {1, 2, 3}
B = {x : x ∈ I and – 1 ≤ x ≤ 1} = {- 1, 0, 1}
and C = {1, 2}
(i) B ∪ C = {- 10, 1, 2}
L.H.S. A × (B ∪ C) = {1, 2, 3} ∪ {- 1, 0, 1, 2}
= {(1, – 1), (1, 0), (1, 1), (1, 2), (2, – 1), (2, 0), (2, 1), (2, 2), (3, – 1), (3, 0), (3, 1), (3, 2)}
Now A × B = {(1, – 1), (1, 0), (1, 1), (2, – 1), (2, 0), (2, 1), (3, – 1), (3, 0), (3, 1)}
A × C = {1, 2, 3) × {1, 2}
= {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
∴ R.H.S. = (A × B) ∪ (A × C)
= ((1, – 1), (1, 0), (1, 1), (1, 2), (2, – 1), (2, 0), (2, 1), (2, 2), (3, 1), (3, 2)}
∴ R.H.S. = (A × B) ∩ (A × C)
= {(1, 1), (2, 1), (3, 1)}
Thus L.H.S = R.H.S.
(iii) A – B = {2, 3}
L.H.S. = (A – B) × C
= {(2, 1), (2, 2), (3, 1), (3, 2)}
A × C = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
and B × C = {(- 1, 1), (- 1, 2), (0, 1), (0, 2), (1, 1), (1, 2)}
∴ R.H.S. = A × C – B × C
= {(2, 1), (2, 2), (3, 1), (3, 2)}
Thus L.H.S. = R.H.S.
Question 20.
If P = (1, 2), form the set P × P × P.
Solution:
Given P = {1, 2}
∴ P × P = {(1, 1), (1, 2), (2, 1), (2, 2)}
∴ P × P × P = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
[using definition]
Question 21.
If A = [a, b, c] and some elements of A × B are (a, p), (b, q), (c, p), write the set B and find the remaining ordered pairs of A × B such that n (A × B) = 6.
Solution:
Given A = {a, b, e} ;
n(A) = 3
Also, n(A × B) = 6
n(A). n(B) = 6
⇒ n(B) = 2 so B must contain., exactly two distinct elements.
Also, (a, p), (b, q), (c, p) ∈ A × B
a, b, c ∈ A and p, q ∈ B
Thus B = {p, q)
A × B = {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)}
Question 22.
Given B = (2, 3, 5) and some elements of A × B are (a, 2), (b, 3), (c, 5). Find the set A and the remaining ordered pairs of A × B such that A × B is least.
Solution:
Given B = {2, 3, 5)
and Some elements of A × B are (a, 2), (b, 3), (c, 5)
and a, b, c are distinct elements
∴ a, b, c ∈ A.
Thus, A = {a, b, e)
∴ A × B = {(a, 2), (a, 3), (a, 5), (b, 2), (b, 3), (b, 5), (c, 2), (e, 3), (c, 5)}
so that A × B is least.
Thus, remaining elements of A × B are (a, 3), (a, 5), (b, 2), (b, 5), (c, 2), (c, 3).