## Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Ex 2.1

Question 1.

If A = {1, 2, 3}, B = {4, 5, 6}, then which of the following is relation from A to B? Justify your answer also.

(i) {(1, 4), (3, 5), (3, 6)

(ii) {(1, 6), (2, 6), (3, 6)}

(iii) {(1, 5), (3, 4), (5, 1), (3, 6)}

(iv) {(2, 4), (2, 6), (3, 6), (4, 2)}

(v) A × B

Solution:

Here, A = {1, 2, 3}, B = {4, 5, 6}

Then A × B = {(1, 4), (1, 5), (1,6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(i) Let R_{1} = {(1, 4), (3, 5), (3, 6)}

(1, 4) ∈ A × B

(3, 5) ∈ A × B

(3, 6) ∈ A × B

R_{1} ⊆ A × B

Hence, R_{1} is a relation from A to B.

(ii) Let R_{2} = {(1, 6), (2, 6), (3, 6)}

(1, 6) ∈ A × B

(2, 6) ∈ A × B

(3, 6) ∈ A × B

R_{2} ⊆ A × B

R_{2} = {(1, 6), (2, 6), (3, 6)}

Hene, R_{2} is a relation from A to B.

(iii) R_{3} = {(1, 5), (3, 4), (5, 1), (3, 6)}

(1, 5) ∈ A × B

(3, 4) ∈ A × B

(5, 1) ∉ A × B

R_{3} ⊄ A × B

Hence, R_{3} is not a relation from A to B.

(iv) Let R_{4} = {(2, 4), (2, 6), (3, 6), (4, 2)}

(2, 4) ∈ A × B but (4, 2) ∉ A × B

R_{4} ⊄ A × B

Hence, R_{4} is not a relation from A to B.

(v) A × B ⊆ A × B

Hence, it is a relation.

Question 2.

Express the following relations in the rules form defined in N:

(i) {(1, 3), (2, 5), (3, 7), (4, 9), …}

(ii) {(2, 3), (4, 2), (6, 1)}

(iii) {(2, 1), (3, 2), (4, 3), (5, 4), …}

Solution:

(i) N = (1, 2, 3, …}

The relation from N to N is given by: {(1, 3), (2, 5), (3, 7), (4, 9), …}

when, x = 1 then y = 3

x = 2 then y = 5

x = 3 then y = 7

x = 4 then y = 9

3, 5, 7, 9, … is an A.P.

Hence, its nth term = a + (n – 1 ).d, where a is first term and d, is a common difference.

T_{n} = 3 + (n – 1) × 2 = 3 + 2n – 2 = 2n + 1

Here, we get the requied rule by putting n = x and T_{n} = y

{(x, y) | x, y ∈ N and y = 2x + 1}.

(ii) Relation in N is expressed as :

{(2, 3), (4, 2), (6, 1)} = {(6, 1), (4, 2), (2, 3)}

Here, 6, 4, 2 are in an A.P.

Its general term T_{n} = 6 + (n – 1) × (-2)

T_{n} = 6 – 2n + 2

T_{n} = 8 – 2n

Here, we get the required rule by putting x = y and T_{n} = x

{(x, y) | x, y ∈ N, x = 8 – 2y or x + 2y = 8} and y < 4

(iii) Relation in N is expressed as:

{(2, 1), (3, 2), (4, 3), (5, 4), …}

Here, 2, 3, 4, 5, … are in an A.P.

So, nth term T_{n} = 2 + (n – 1) × 1 = 2 + n – 1 = n + 1

Here, by putting n = x and T_{n} = y

Required rule = {(x, y) | x, y ∈ N, x = y + 1 or y = x – 1}

Question 3.

A relation R from set A = {2, 3, 4, 5} to set B = {3, 6, 7, 10} is defined in such a way that xRy ⇔ x is a prime number related to y. Write relation R in the set form or order pairs and also find domain and range of if.

Solution:

Given, A = {2, 3, 4, 5}, B = (3, 6, 7, 10}

Relation R from A to B is defined as:

xRy ⇔ x, y is a prime number related by ∀ x, y ∈ R

(Numbers a and b are called co-prime if there is no common factor except 1 for example 3 is co-prime related to 10)

x = 2 ∈ A then 2 is a prime number related to 3 and 7.

Then (2, 3) ∈ R and (2, 7) ∈ R

when x = 3 ∈ A

then 3 is a prime number related to 7 and 10.

Then (3, 7) ∈ R and (3, 10) ∈ R

when x = 4 ∈ A

then 4 is a prime number related to 3 to 7.

then (4, 3) ∈ R and (4, 7) ∈ R

when x = 5 ∈ A then 5 is a prime number related to 3, 6 and 7.

then (5, 3) ∈ R, (5, 6) ∈ R and (5, 7) ∈ R

Hence R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

Domain of R = {2, 3, 4, 5}

and range of R = {3, 6, 7, 10}.

Question 4.

If in a set of integers Z, a relation R is defined in such a way that xRy ⇔ x^{2} + y^{2} = 25, then write R and R^{-1} in the form of a set of order pairs and also write their domain.

Solution:

Given set = 2

Z = {Set of integers} = {0, ± 1, ± 2, ± 3, …}

Relation in Z i.e., a relation R from Z to Z is defined as

xRy ⇔ x^{2} + y^{2} = 25 ∀ x, y ∈ Z

when x = 0 then y = ± 5 because from 0^{2} + y^{2} = 25

⇒ (0, 5) ∈ R and (0, -5) ∈ R

when x = ±3

(±3)^{2} + y^{2} = 25

⇒ y^{2} = 25 – 9 = 16

⇒ y = ±4

then (3, 4) ∈ R and (-3, 4) ∈ R

(-3, 4) ∈ R (-3, -4) ∈ R

when x = ± 4 then y = ± 3

from (±4)^{2} + y^{2} = 25

⇒ y^{2} = 25 – 16 = 9

⇒ y = ± 3

⇒ (4, 3) ∈ R, (4, -3) ∈ R,

(-4, 3) ∈ R, (-4, -3) ∈ R

when x = ± 5 then y = 0

(±5)^{2} + y^{2} = 25

⇒ y^{2} = o

⇒ y = 0

⇒ (5, 0) ∈ R and (-5, 0) ∈ R

Hence,

R={(0, 5), (0, -5), (3, 4), (-3, 4), (3, -4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5, 0), (-5, 0)}

and R^{-1} = {(5, 0), (-5, 0), (4, 3), (4, -3), (-4, 3), (-4, -3), (3, 4), (-3, 4), (3, -4), (-3, -4), (0, 5), (0, -5)}

Domain of R = {0, 3, -3, 4, -4, 5, -5}

and domain of R^{-1} = {5, -5, 4, -4, 3, -3, 0}.

Question 5.

If a relation Φ from set C of complex numbers to a set R of real numbers is so defined that

x Φ y ⇔ |x|= y.

(i) (1 + i)Φ3

(ii)3Φ(-3)

(iii) (2 + 3i)Φ13

(iv) (1 + i)Φ1

Solution:

Given set

C = {a + ib : a ∈ R, b ∈ R, i = √-1} = Set of complex number

R = Set of real number

Relation Φ from C to R is defined as:

x Φ y ⇔ |x|= y ∀ x ∈ c, y ∈ R

(i) (1 + i)Φ3

|1 + i|= √(1^{2} + 1^{2}) = √2 ≠ 3

Hence, (1 + i)Φ3 is false.

(ii) 3Φ(-3)

|3|= 3 ≠ -3

Relation Φ is false.

(iii) (2 + 3i)Φ13

|2 + 3i|= √(2^{2} + 3^{2}) = √(4 + 9) = √13 ≠ 13

Relation is false.

(iv) (1 + i)Φ1

|1 + i| = √(1^{2} + 1^{2}) = √2 ≠ 1

Hence, relation is false.

Question 6.

If a relation R from set A = {1, 2, 3, 4, 5} to set R = {1, 4, 5} is defined such that x < y, then write R in the form of set of order pairs. Also find R^{-1}

Solution:

Given that, a relation R from A to B is defined as:

xRy ⇔ x < y ∀ x ∈ A, y ∈ B

1 ∈ A

1 < 4, 1 < 5

So, (1, 4) ∈ R and (1, 5) ∈ R

Again 2 ∈ A and 2 < 4, 2 < 5

So, (2, 4) ∈ R, (2, 5) ∈ R

Again 3 ∈ A and 3 < 4, 3 < 5

So, (3, 4) ∈ R, (3, 5) ∈ R

Again 4 ∈ A and 4 < 5

So, (4, 5) ∈ R

R = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)}

and R^{-1} = {(4, 1), (5, 1), (4, 2), (5, 2), (4, 3), (5, 3), (5, 4)}

Question 7.

Express the following relations in the form of sets or orders pairs:

(i) R_{1} is relation from set A = {1, 2, 3, 4, 5, 6} to set B = (1, 2, 3} such that “x = 2y”.

(ii) R_{2} is a relation set A = {8, 9, 10, 11} to set B = {5, 6, 7, 8} such that “y = x – 2”.

(iii) R_{3} is a relation in set A = {0, 1, 2,…, 10} defined by 2x + 3y = 12.

(iv) R_{4} is a relation from set A = {5, 6, 7, 8} to set B = {10, 12, 15, 16, 18} is defined such that “x is divisor of y”

Solution:

(i) Given relation R1 is defined as “x = 2y” and A = {1, 2, 3, 4, 5, 6} and in B = {1, 2, 3}

x = 2y

y = \(\frac { x }{ 2 }\)

when x = 2 then y = 1. So, (2, 1) ∈ R_{1}

when x = 4 then y = 2. So, (4, 2) ∈ R_{1}

and x = 6 then y = 3. So, (6, 3) ∈ R_{1}

R_{1} = {(2, 1), (4, 2), (6, 3)}

(ii) Given relation R_{2} is defined as “y = x – 2” and A = {8, 9, 10, 11} and B= {5,6, 7,8}

So, from y = x – 2

when x = 8 then y = 8 – 2 = 6 So, (8, 6) ∈ R_{2}

when x = 9 then y = 9 – 2 = 7 So, (9, 7) ∈ R_{2}

when x = 10 then y = 10 – 2 = 8 So, (10, 8) ∈ R_{2}

and when x = 11 then y = 11 – 2 = 9 So, (11, 9) ∈ R_{2}

R_{2} = {(8, 6), (9, 7), (10, 8)}

(iv) Given relation R4 is defined as “x, y”

A = {5, 6, 7, 8}

and B = {10, 12, 15, 16, 18}

So, xR_{4}y ⇔ x is divisor of y

x ∈ A, y ∈ B

when x = 5 then 5 is a divisor of 10 and 15

So, (5, 10), (5, 15) ∈ R_{4}

when x = 6 then 6 is a divisor of 12 ahd 18

So, (6, 12), (6, 18) ∈ R_{4}

when x = 8 then 8 is a divisor of 16

So, (8, 16) ∈ R_{4}

R_{4} = (5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}

Question 8.

Find the inverse of each of the following relation:

(i) R = {(2, 3), (2, 4), (3, 3), (3, 2), (4, 2)}

(ii) R = {(x, y) | x, y ∈ N; x < y}

(iii) R is defined by 2x + 3y = 12 in set A = (0, 1, 2, …10}.

Solution:

(i) Given that

R = {(2, 3) (2, 4), (3, 3), (3, 2), (4, 2)}

R^{-1} = {(3, 2), (4, 2), (3, 3), (2, 3), (2, 4)}

(ii) Given that

R = {(x, y) | x, y ∈ N; x < y}

R^{-1} = {(y, x) | x, y ∈ N; x > y}

(iii) Relation in set A is defined by

2x + 3y = 12

R = {(0, 4), (3, 2), (6, 0)}

R^{-1} = {(4, 0), (2, 3), (0, 6)}.