OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Ex 2(b)

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S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(b)

Question 1.
Write down the relation shown by the arrow diagram, by listing the ordered pairs. State the domain, co-domain, and the range of the relation.

Solution:
Relation R is given by {{p, 3), (r, 5), (q, 5), (s, 7)}
∴ Domain of R = {p, q, r, s}
Range of R = {3, 5, 7}
Codomain of R = {1, 3, 5, 7, 9}

Question 2.
Which of the following are relations from B to A, where A = {a, b, c, d} and B = {x, y, z) ?
(i) {(z, x), (z, y), (x, a)}
(ii) {(z, a), (z, b), (z, c), (z, d)}
(iii) {(x, b), (y, a)}
(iv) {(b, y), (c, z), (a, x)}
(v) {(x, d), (y, c), (z, a)}
Solution:
Given A = {a, b, c, d) and B = {x, y, z}
We know that a relation R from set A to set B is any subset of A x B i.e. RcAxB where A and B are non-empty sets.
(i) (z, x) ∈ R but z ∈B while x ∉ A. ∴ given set is not a relation from B to A

(ii) Yes, since in each ordered pair, first component belongs to B while second component belongs to A.

(iii) Yes, In ordered pair (x, b), x∈B and be A and in ordered pair (y, a), y ∈ B and a ∈ A

(iv) No, In ordered pair (b, y), b ∈ A and y ∈ B
while for relation from B to A, first component of each ordered pair ∈ B and second component ∈ A.

(v) Yes, since first component of each ordered pair belongs to B and second component of each ordered pair belongs to A.

Question 3.
In each of the following, state which of the ordered pairs belong to the given relations ?
(i) {(x, y) : x > y + 5} ; (1, 0), (8, 2), (0,1), (2, 8), (9, 3), (10, 7), (123, 4)
(ii) {(x, y) : xy = 12} ; (3, 4), (4, 3), (12, 0), (0,12), (12, 1), (6, 2), (7, 5)
(iii) y) : y = \(\frac{x+3}{x-1}, x \neq 1\)}; (o, 1), (2, 5), (5, 2), (3, 3), (7, 5), (7, \(\frac { 5 }{ 3 }\))
Solution:
(i) R = {(x, y), x > y + 5}

Question 4.
Let N be the set of natural numbers. Describe the following relations in words, giving their domain and the range.
(a) {(2, 1), (4, 2), (10, 5), (18, 9), (20, 10)}
(b) (3, 1) (6, 2), (15, 5)}
(c) {(1, 4), (5, 16), (7, 22), (12, 37)}
Solution:
(a) Given R = {(2, 1), (4, 2), (10, 5), (18, 9), (20, 10)}
In given relation, in each ordered pair, first component is twice the second component
∴ R = {(x, y) : x = 2y, y ∈ {1,2, 5, 9, 10}}
Domain (R) = {2, 4, 10, 18, 20} and Range (R) = {1, 2, 5, 9, 10}

(b) In given relation R, first component is 3 times the second component in each ordered pair.
∴ R= {(x, y) : x = 3y, y ∈ {1, 2, 5})
Domain (R) = {3, 6, 15} and Range (R) = {1, 2, 5}

(c) Clearly in given relation R, second component of each ordered pair one more than thrice the first component.
∴ R= {(x, y) : y = 3y + 1, x ∈ {1, 5, 7, 12}}
Domain of R = {1, 5, 7, 12} and Range (R) = {4, 16, 22, 37}

Question 5.
Z is the set of integers. Describe the following relation in set builder form, given its domain and range.
{(0, – 7), (2, – 5), (4, – 3), (- 13, – 20), …}
Solution:
In given relation, we observe that, difference between first and second component of each ordered pairs is 7.
∴ In set builder form, R = {(x, y) ; x – y = 7, x ∈ Z}
Thus domain of R = {0, 2, 4, – 13, ….} and Range of R = {- 7, – 5, – 3, – 20, ….}

Question 6.
Write down the domain and range of the relation (x, y) : x – 3y and x and y are natural numbers less than 10.
Solution:
Given relation R = {(x, y) ; x = 3y, x, y ∈ N, x, y < 10} given x = 3y, x, y ∈ N, x, y < 10
∴ x, y ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}
When y = 1 ⇒ x = 3 ∴ (3, 1) ∈ R
When y = 2 ⇒ x = 6 ∴ (6, 2) ∈ R
When y = 3 ⇒ x = 9 ∴ (9, 3) ∈ R
∴ R = {(3, 1), (6, 2), (9, 3)}
Thus, domain of R = {3, 6, 9} and Range of R = {1, 2, 3}

Question 7.
Determine the domain and range of the relation R.
(a) R = {(x + 1, x + 5) | x ∈ {0, 1, 2, 3, 4, 5}}. Draw the graph of R.
(b) R = {(x, x³} | x is a prime number less than 10}.
Solution:
(a) Given R = {(x + 1, x + 5 | x ∈ {0, 1,2, 3, 4, 5}} given x ∈ {0, 1,2, 3, 4, 5}
When x =0 ; (0 + 1, 0 + 5) i.e. (1, 5) ∈ {R}
When x = 1 ; (1 + 1, 1 + 5) i.e. (2, 6) ∈ {R}
When x = 2 ; (2 + 1, 2 + 5) i.e. (3, 7) ∈ {R}
When x = 3 ; (3 + 1, 3 + 5) i.e. (4, 8) ∈ {R}
When x = 4 ; (4 + 1, 4 + 5) i.e. (5, 9) ∈ {R}
When x = 5 ; (5 + 1, 5 + 5) i.e. (6, 10) ∈ {R}
∴ R = {(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)}
Domain (R) = {1, 2, 3, 4, 5, 6} and Range (R) = {5, 6, 7, 8, 9, 10}

(b) Given R = {(x, x³) | x is a prime number < 10}
Since x is a prime number < 10 ∴ x = 2, 3, 5, 7
When x = 2, (2, 2³) i.e. (2, 8) ∈ R
When x = 3, (3, 3³) i.e. (3, 27) ∈ R
When x = 4, (4, 4³) i.e. (4, 64) ∈ R
When x = 5, (5, 5³) i.e. (5, 125) ∈ R
∴ R = {(2, 8), (3, 27), (5, 64), (5, 125)}
domain (R) = (2, 3, 4, 5} and Range (R) = {8, 27, 64, 125}

Question 8.
Given A = {- 2, – 1, 0, 1, 2}, list the ordered pairs determined by each of the following relations applied on A :
(i) R₁ = “is less than”
(ii) R₂ = “is the square of”
(iii) R₃ = “is the additive inverse of”
(iv) R₄ = “is equal to”
Solution:
Given A = {-2,- 1,0, 1,2}
(i) R₁ = {(a, b) | a < b, a, b ∈ A}
since
– 2 < – 1 (-2,-1) ∈ R₁
– 2 < 0 ∴ (- 2, 0) ∈ R₁
– 2 < 1 ∴ (-2, 1) ∈ R₁
– 2 < 2 ∴ (- 2, 2) ∈ R₁
– 1 < o ∴ (- 1, 0) ∈ R₁
– 1 < 1 ∴ (- 1, 1) ∈ R₁
– 1 < 2 ∴ (- 1, 2) ∈ R₁
0 < 1 ∴ (0, 1) ∈ R₁
0 < 2 ∴ (0, 2) ∈ R₁
1 < 2 ∴ (1, 2) ∈ R₁
Thus R₁ = {(- 2, – 1), (- 2, 0), (-2, 1), (- 2, 2), (-1, 0), (-1, 1), (-1, 2), (0, 1), (0, 2), (1, 2)}

(ii) R₂ = {(a, b) | a = b², a, b ∈ A}
since
1 = (- 1)² ∴ (1, – 1) ∈ R₂
1 = 1² ∴ (1, 1) ∈ R₂
0 = 0² ∴ (0, 0) ∈ R₂
Thus, R₂ = {(1, – 1), (1, 1), (0, 0)}

(iii) R₃ = {(a, b) | a is the additive inverse of 6}b
We know that – a be the additive inverse of a i.e. R₃ = {(a, b) | a + b = 0, a, b ∈ A}
since
2 + (- 2) = 0 ⇒ (2, – 2) ∈ R₃
– 2 + (2) = 0 ⇒ (- 2, 2) ∈ R₃
1 + (- 1) = 0 ⇒ (1, – 1) ∈ R₃
-1 + 1=0 ⇒ (-1, 1) ∈ R₃
0 + 0 = 0 ⇒ (0,0) ∈ R₃
Thus, R₃ = {(2, – 2), (- 2, 2), (- 1, 1), (1, – 1), (0, 0)}

(iv) R₄ = {(a, b) | a = b, a, b ∈ A}
since
1 = 1 ∴ (1, 1) ∈ R₄
2 = 2 ∴ (2, 2) ∈ R₄
0 = 0 ∴ (0,0) ∈ R₄
– 1 = – 1 ∴ (- 1, – 1) ∈ R₄
– 2 = – 2 ∴ (- 2, – 2) ∈ R₄
Thus R₄ = {(2, 2), (1, 1), (0, 0), (- 1, – 1), (- 2, – 2)}

Question 9.
Given A = {2, 3, 4, 5, 6}. List the elements of each of the following relations :
(i) {(x, y) ∈ A x A : x = y}
(ii) {(x, y) ∈ A x A : x > y, \(\frac {x}{y}\) ∉ W}
(iii) {(x, y) ∈ A x A : x is a divisor of y and x ≠ y}.
Solution:
Given A = {2, 3, 4, 5, 6}
(i) Given R = {(x, y) ∈ A x A : x = y}
since (x, y) ∈ A x A, where x = y
∴ (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∈ R
Thus R = {(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

(ii) Given R = {(x, y) ∈ A x A : x > y, \(\frac {x}{y}\) ∉ W}

Therefore R = {(3, 2), (5, 2), (4, 3), (5, 4), (5, 3), (6, 4), (6, 5)}

(iii) Given R = {(x, y) ∈ A x A : x is a divisor of y and x ≠ y}
Since 2 is a divisor of 4 and 6
∴ (2, 4), (2, 6) ∈ R
Since 3 is a divisor of 6 (3, 6) ∈ R
Hence R = {(2, 4), (2, 6), (3, 6)}

Question 10.
(i) If A is the set of even natural numbers less than 8 and B is the set of prime numbers les than 7, then the number of relations from A to B is
(a) 2⁹
(b) 9²
(c) 3²
(d) 2⁹ – 1
(ii) Let A be a finite set. The number of relations on A where A has 3 elements are :
(a) 9
(b) 0.81
(c) 24.3
(d) 512
Solution:
(i) A = set of even natural numbers less than 8 = {2, 4, 6}
B = set of prime numbers less than 7 = {2, 3, 5}
∴ O (A) = 3 ; O (B) = 3
Thus, the no. of relations from A to B = \(2^{\mathrm{O}(\mathrm{A} \times \mathrm{B})}=2^{\mathrm{O}(\mathrm{A}) \times \mathrm{O}(\mathrm{B})}=2^{3 \times 3}=2^9\) = 512

(ii) Since A has 3 elements ∴ O (A) = 3
No. of relations on A = \(2^{\mathrm{O}(\mathrm{A} \times \mathrm{A})}=2^{\mathrm{O}(\mathrm{A}) \times \mathrm{O}(\mathrm{A})}=2^{3 \times 3}=2^9\) = 512

Question 11.
Let n (A) = p. Then the number of all relations on A is
(i) 2P
(b) 2p¹
(c) 2p²
(d) None of these
Solution:
Given n(A) = p
∴ No. of relations on A = \(2^{n(\mathrm{~A} \times \mathrm{A})}=2^{n(\mathrm{~A}) \times n(\mathrm{~A})}=2^{p \times p}=2^{p^2}\)