RBSE Solutions for Class 11 Maths Chapter 2 Relations and Functions

RBSE Solutions for Class 11 Maths Chapter 2 Relations and Functions

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 R₁ = {(1, 4), (3, 5), (3, 6)}
(1, 4) ∈ A × B
(3, 5) ∈ A × B
(3, 6) ∈ A × B
R₁ ⊆ A × B
Hence, R₁ is a relation from A to B.

(ii) Let R₂ = {(1, 6), (2, 6), (3, 6)}
(1, 6) ∈ A × B
(2, 6) ∈ A × B
(3, 6) ∈ A × B
R₂ ⊆ A × B
R₂ = {(1, 6), (2, 6), (3, 6)}
Hene, R₂ is a relation from A to B.

(iii) R₃ = {(1, 5), (3, 4), (5, 1), (3, 6)}
(1, 5) ∈ A × B
(3, 4) ∈ A × B
(5, 1) ∉ A × B
R₃ ⊄ A × B
Hence, R₃ is not a relation from A to B.

(iv) Let R₄ = {(2, 4), (2, 6), (3, 6), (4, 2)}
(2, 4) ∈ A × B but (4, 2) ∉ A × B
R₄ ⊄ A × B
Hence, R₄ 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 ⇔ x² + y² = 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 ⇔ x² + y² = 25 ∀ x, y ∈ Z
when x = 0 then y = ± 5 because from 0² + y² = 25
⇒ (0, 5) ∈ R and (0, -5) ∈ R
when x = ±3
(±3)² + y² = 25
⇒ y² = 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 + y² = 25
⇒ y² = 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 + y² = 25
⇒ y² = 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|= √(1² + 1²) = √2 ≠ 3
Hence, (1 + i)Φ3 is false.

(ii) 3Φ(-3)
|3|= 3 ≠ -3
Relation Φ is false.

(iii) (2 + 3i)Φ13
|2 + 3i|= √(2² + 3²) = √(4 + 9) = √13 ≠ 13
Relation is false.

(iv) (1 + i)Φ1
|1 + i| = √(1² + 1²) = √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) R₁ is relation from set A = {1, 2, 3, 4, 5, 6} to set B = (1, 2, 3} such that “x = 2y”.
(ii) R₂ is a relation set A = {8, 9, 10, 11} to set B = {5, 6, 7, 8} such that “y = x – 2”.
(iii) R₃ is a relation in set A = {0, 1, 2,…, 10} defined by 2x + 3y = 12.
(iv) R₄ 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 = x2
when x = 2 then y = 1. So, (2, 1) ∈ R₁
when x = 4 then y = 2. So, (4, 2) ∈ R₁
and x = 6 then y = 3. So, (6, 3) ∈ R₁
R₁ = {(2, 1), (4, 2), (6, 3)}

(ii) Given relation R₂ 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) ∈ R₂
when x = 9 then y = 9 – 2 = 7 So, (9, 7) ∈ R₂
when x = 10 then y = 10 – 2 = 8 So, (10, 8) ∈ R₂
and when x = 11 then y = 11 – 2 = 9 So, (11, 9) ∈ R₂
R₂ = {(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, xR₄y ⇔ 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) ∈ R₄
when x = 6 then 6 is a divisor of 12 ahd 18
So, (6, 12), (6, 18) ∈ R₄
when x = 8 then 8 is a divisor of 16
So, (8, 16) ∈ R₄
R₄ = (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)}.

Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Ex 2.2

Question 1.
Examine the reflexivity, symmetricity, and transitivity of the following relations:
(i) mR₁n ⇔ m and n both are odd, ∀ m, n ∈ N
(ii) In the power set P(A) of set AR₂B ⇔ A ⊆ B, ∀ A, B ∈ P(A)
(iii) In set B of straight lines situated in three dimensions space, L₁R₃L₂ ⇔ L₁ and L₂ are coplanar ∀ L₁, L₂ ∈ S
(iv) aR₄b ⇒ b is divisible by a, ∀ a, b ∈ N.
Solution:
(i) Given, set N = {1, 2, 3, 4, …}
A relation R₁ in N is defined as
mR₁n ⇔ m and n both are odd ∀ m, n ∈ N
Reflexivity: Let m ∈ N
m ∈ N ⇒ m : even or odd number
⇒ (m, m) ∈ R₁ If m is odd
But (m, m) ∉ R₁ If m is even
So, (m, m) ∉ R₁
R₁ is not reflexivity.
Symmetricity: Let m, n ∈ N, mR₁ is true,
So, mR₁n ⇒ m and n are both odd.
⇒ n and m are both odd
⇒ nR₁m
R₁ is a symmetric relation.
Transitivity: Let m, n, r ∈ N
mR₁n and nR₁r is true, so,
mR₁n and nR₁r ⇒ m and n are both odd and n, r are both odd.
⇒ m, n and r are odd.
⇒ mR₁r
R₁ is transitive relation.

(ii) Given set is
AR₂B ⇔ A ⊆ B ∀ A, B ∈ P(A)
Reflexivity: Let A ∈ P(A)
A ∈ P(A) ⇒ A ⊆ A (∵ every set is a subset of itself)
⇒ AR₂A
Hence, R₂ reflexive relation.
Symmetricity: Let A, B ∈ P(A), AR₂B are true, so,
AR₂B ⇒ A ⊆ B
⇒ B ⊄ A (until A = B)
⇒ B ⊄ A
R₂ is not a symmetric relation.
Transitivity: A, B, C ∈ P(A),
AR₂B and BR₂C are true. So,
AR₂B and BR₂C
⇒ A ⊆ B and B ⊆ C
⇒ A ⊆ C(A ⊆ B ⊆ C)
⇒ AR₂C
R₂ is transitive relation.

(iii) Given set is
S = {Set of straight lines situated in three dimensional space}
A relation R₃ in S is defined as
L₁R₃L₂ ⇒ L₁ and L₂ are co-planar lines ∀ L₁, L₂ ∈ S
Reflexivity: Let L ∈ S
L ∈ S ⇒ L and L are coplanar
⇒ (L, L) ∈ R₃ ∀ L ∈ S
R₃ is a reflexive relation.
Symmetricity: Let L₁, L₂ ∈ S
In this way (L₁, L₂) ∈ R₃
(L₁, L₂) ∈ R₃
⇒ L₁ and L₂ are coplanar
⇒ L₂ and L₁ are coplanar
⇒ (L₂, L₁) ∈ R₃
So, (L₁, L₂) ∈ R₃
⇒ (L₂, L₁) ∈ R₃ ∀ L₁, L₂ ∈ S
R₃ is a symmetric relation.
Transitivity: Let L₁, L₂, L₂ ∈ S is given as
(L₁, L₂) ∈ R₃ and (L₂, L₃) ∈ R₃
(L₁, L₂) ∈ R₃ ⇒ L₁ and L₂ are coplanar ……(1)
(L₂, L₃) ∈ R₃ ⇒ L₂ and L₃ are coplanar …(2)
So, from (1) and (2) we can not say that L₁ and L₃ are also coplanar.
From figure L₁ and L₂ are situated in a plane (1),
L₂ and L₃ are situated in a plane (2)
but L₁ and L₃ are not coplanar.

So, (L₁, L₃) ∉ R₃
R₃ is not a transitive relation.

(iv) Given set is N = {1,2, 3, 4, …}
Relation R₄ in N is defined as
aR₄b ⇔ b is divided by a ∀ a, b ∈ N
⇔ 1/2 = k ∈ I
where I is the set of integers ∀ a, b ∈ N
⇔ ∀ a is divided by b
∀ a, b ∈ N
Reflexivity: Let a ∈ N
a ∈ N ⇒ a is divided by a
⇒ (a, a) ∈ R₄ ∀ a ∈ N
R₄ is a reflexive relation.
Symmetricity: Let a, b ∈ N is in this way (a, b) ∈ R₄
(a, b) ∈ R₄ ⇔ b is divided by a
⇔ a cannot be divided by b until a = b
⇔ (b, a) ∉ R₄
(a, b) ∈ R₄ ⇔ (b, a) ∉ R₄
So, R₄ is not a symmetric relation.
Transitivity: Let a, b, c ∈ N
In this way
(a, b) ∈ R₄ and (b, c) ∈ R₄
(a, b) ∈ R₄ ⇒ b is divided by a
⇒ b/a = k (Let) where k ∈ I …(1)
(b, c) ∈ R₄ ⇒ c is divided by b
⇒ c/b = m (Let) where m ∈ I …(2)
Putting the value of b from equation (1) in equation (2) we have
m ∈ I, k ∈ I
mk ∈ I
c/ak = m ⇒ c/a = mk ∈ I
⇒ (a, c) ∈ R₄
(a, b) ∈ R₄, (b, c) ∈ R₄
⇒ (a, c) ∈ R₄ ∀ a, b, c ∈ N
Hence, R₄ is a transitive relation.

Question 2.
Any relation P is defined in set R0 of non zero real numbers by following ways:
(i) xPy ⇔ x² + y² = 1
(ii) xPy ⇔ xy = 1
(iii) xPy ⇔ (x + y) is a rational number
(iv) xPy ⇔ x/y is a rational number
Test the reflexivity, symmetricity and transitivity of these relations.
Solution:
(i) Given set
R0 = Set of non-zero real number
Relation P in R0 is defined as
xPy = x² + y² = 1 ∀ x, y ∈ R₀
Reflexivity: P is not reflexivie because when 2 ∈ R₀
Then (2)² + (2)² ≠ 1 so, (2, 2) ∉ R₀
Similarly a² + a² ≠ 1 so, (a, a) ∉ R₀
Hence, P is not reflexive.
Symmetricity: Let a, b ∈ R₀
In this way (a, b) ∈ P
(a, b) ∈ P
⇒ a² + b² = 1
⇒ b² + a² = 1
⇒ (b, a) ∈ P
So, (a, b) ∈ P
⇒ (b, a) ∈ P ∀ a, b ∈ R₀
P is symmetric relation.
Transitivity: P is not transitive relation because

P is not transitive relation.
(ii) Given set R₀ = {Set of real numbers}
Relation P in R₀ is defined as
xPy ⇔ xy = 1 ∀ x, y ∈ P
Reflexivity: Let 2 ∈ R₀
But 2 × 2 = 4 ≠ 1
⇒ (2, 2) ∉ P
Hence, P is not reflexive
Symmetricity: In this way (a, b) ∈ R₀
In this way (a, b) ∈ P
(a, b) ∈ P
⇒ ab = 1
⇒ ba = 1 [∵ Multiplication of real number is commutative]
⇒ b . a ∈ P
⇒ (a, b) ∈ P
⇒ (b, a) ∈ P ∀ a, b ∈ R₀
P is a symmetric relation.
Transitivity: Let a, b, c ∈ R₀ is in this way
a, b ∈ P and b, c ∈ P
(a, b) ∈ P ⇒ a.b = 1 ……(1)
(b, c) ∈ P ⇒ b.c = 1 …..(2)
From equation (1) and (2),
ab/bc = 1/1
⇒ a/c = 1
⇒ (a, c) ≠P ……..(3)
So, (a, b) ∈ P, (b, c) ∈ P but (a, c) ∉ P
Hence, P is not transitive relation.
(iii) Given set R₀ = Set of real number
Relaton P in R₀ is defined as
xPy ⇔ x + y is a rational number ∀ x, y ∈ R₀
Reflexivity : Let x ∈ R₀
x ∈ R₀ ⇒ x + x is need not to be rational
For example √3 ∈ R₀ ⇒ √3 + √3 = 2√3
is an irrational number
So, (x, x) ∉ P
P is not reflexive relation.
Symmetricity: Let a, b ∈ R₀ then (a, b) ∈ P
(a, b) ∈ P
⇒ a + b is a rational number
⇒ (b + a) is also a rational number
⇒ (b, a) ∈ P
So, (a, b) ∈P
⇒ (b, a) ∈ P ∀ a, b ∈ R₀
P is a symmetric relation.
Transitivity: Let a, b, c ∈ R₀ are in this way
(a, b) ∈ P ⇒ a + b is a rational number
(b, c) ∈ P ⇒ b + c is a rational number
⇒ a + c is not necessary a rational number
For example: 2 + √3, -√3 + 6, √3 + √7 ∈ Ro and (2 + √3, -√3 + 6) ∈ P because
2 + √3 – √3 + 6 = 8 is a rational number
(-√3 + 6, √3 + 7) ∈ P because
-√3 + 6 + √3 + 7 = 13 is a rational number
But (2 + √3, √3 + 7) ∈ P because
2 + √3 + √3 + 7 = 2√3 + 9 is an irrational number
Hence, P is not a transitive relation.
(iv) Given: Set : R₀ = Set of real numbers
Relation P in R₀ is defined as a rational number
xPy = x/y is a rational number ∀ x, y ∈ R₀
Reflexivity: Let a ∈ R₀
a ∈ R₀ ⇒ a/a = 1 is a rational number
⇒ (a, a) ∈ P
P is a reflexive number.
Symmetricity: Let a, b ∈ R₀ is in this way (a, b) ∈ P
(a, b) ∈ P
⇒ a/b is a rational number
⇒ b/a is a rational number
(From infination of set of rational numbers Q)
⇒ (b, a) ∈ P
(a, b) ∈ P
⇒ (b, a) ∈ P ∀ a, b ∈ R₀
P is a symmetric relation
Transitivity: Let a, b, c ∈ R₀ is in this way
(a, b) ∈ P and (b, c) ∈ P
(a, b) ∈ P ⇒ a/b is a rational number
(b, c) ∈ P ⇒ b/c is a rational number
⇒ (a/b)(b/c)is also a rational number.
[∵ Multiplication of rational number is also a rational number]
⇒ a/c is a rational number
⇒ (a, c) ∈ P
(a, b) ∈ P, (b, c) ∈ P
⇒ (a, c) ∈ P ∀ a, b, c ∈ R₀
P is a transitive relation.

Question 3.
A relation R1 is defined in set R if of real numbers in the following way:
(a, b) ∈ R₁ ⇔ 1 + ab > 0, ∀ a, b ∈ R
Prove that R₁ is reflexive and symmetric but not transitive.
Solution:
Given: Set R = set of real numbers.
Relation R₁ in R is defined as
(a, b) ∈ R₁ ⇒ 1 + ab > 0 ∀ a, b ∈ R
(i) Reflexivity: Let a ∈ R
a ∈ R ⇒ 1 + a.a > 0
⇒ a.a ∈ R₁ ∀ a ∈ R
R₁ is a reflexive relation.
(ii) Symmetricity: Let a, b ∈ R in this way
(a, b) ∈ R₁
(a, b) ∈ R₁
⇒ 1 + ab > 0
⇒ 1 + b.a > 0 [ab = ba]
⇒ (b, a) ∈ R₁
So, (a, b) ∈ R₁
⇒ (b, a) ∈ R₁ ∀ a, b ∈ R
R₁ is a symmetric relation.
(iii) Transitivity: Let a, b, c ∈ R is in this way
(a, b) ∈ R₁ and (b, c) ∈ R₁
(a, b) ∈ R₁ ⇒ 1 + ab > 0
(b, c) ∈ R₁ ⇒ 1 + bc > 0
But 1 + ac > 0 is not necessary.
For example: 1, 1/2, -1 ∈ R
and (1, 1/2) ∈ R₁ because

R₁ is not a transitive relation.
Hence Proved.

Question 4.
N is a set of natural numbers. If a relation R is defined in set N × N such that (a, b) R(c, d) ⇔ ad = bc ∀ (a, b), (c, d) ∈ N × N, then prove that R is an equivalence relation.
Solution:
Given:
Set N = {1, 2, 3, 4, …} = Set of natural numbers
A relation R in N × N is defined as
(a, b) R(c, d) ⇔ ad = bc where a, b,c, d ∈ N ∀ (a, b), (c, d) ∈ N × N
Here, to prove R is a equivalence relation, we have to show that R is reflexive, symmetric and transitive.
(i) Reflexivity: Let
(a, b) ∈ N × N
{a, b) ∈ N × N
⇒ a.b = ba (Commutative law of multiplication)
⇒ (a, b) R(a, b) ∀ (a, b) ∈ N × N
R is reflexive relation.
(ii) Symmetricity: Let (a, b) (c, d) ∈ N × N is in this way (a, b) R(c, d)
(a, b) R(c, d)
⇒ ad = bc
⇒ bc = ad
⇒ c.b = d.a
⇒ (c, d) R(a, b)
So, (a, b) R(c, d) ⇒ (c, d) R(a, b) ∀ (a, b)(c, d) ∈ N × N
R is a symmetric relation.
(iii) Transitivity: Let
(a, b), (c, d), (e, f) ∈ N × N
is in this way
(a, b) R(c, d) and (c, d) R(e, f)
(a, b) R(c, d) ⇒ ad = bc
(c, d) R(e, f) ⇒ cf = de
(a, b)R (c, d) and (c, d)R (e, f)
⇒ (ad) (cf) = (bc)(de) (on multiplication)
⇒ af = be
⇒ (a.b) R(e, f)
So, (a, b) and R(c, d) and (c, f) R(e, f)
⇒ (a, b) R(e, f) ∀ (a, b), (c, d), (e, f) ∈ N × N
R is a transitive relation.
Hence, according to (i), (ii) and (iii), the given relation is an equivalence relation.
Hence Proved.

Question 5.
A relation R is defined in a set Q₀ set of non zero rational numbers such that aRb ⇔ a = 1/b, ∀ a, b ∈ Q₀. Is R is equivalence relations.
Solution:
Given: Set Q₀ = Set of non zero rational numbers
A relation in Q₀ is defined as
aRb ⇔ a = 1/b ∀ a, b ∈ Q₀
If R is reflexive, symmetric and transitive than R is an equivalence relation.
(i) Reflexivity: Let a ∈ Q₀
a ∈ Q₀ ⇒ a ≠ 1/a (a ≠ 1)
⇒ (a, a) ∉ R ∀ a ∉ Q₀
So, R is not a reflexive relation
R is also not equivalenced relation.

Question 6.
Let {(a, b) | a, b ∈ R} where I is set of integers. Relations R₁ on x is defined in the following way
(a, b) R₁(c, d) ⇒ b – d = a – c
Prove that R₁ is an equivalence relation.
Solution:
Given : Set X = {(a, b) : a, b ∈ I}
where I is the set of integers.
A relation R in X is defined as:
(a, b) R(c, d) ⇔ b – d = a – c ∀ (a, b) (c, d) ∈ X
To prove that R is equivalence relation, we have to prove that R is reflexive, symmetric and transitive.
(i) Reflexivity: Let (a, b) ∈ X
(a, b) ∈ X ⇒ (a, b) ∈ I
⇒ b – b = a – a = 0
⇒ (a, b) R(a, b) ∀ (a, b) ∈ X
R is a reflexive relation.
(ii) Symmetricity: Let (a, b), (c, d) ∈ X is in this way
(a, b) R(c, d)
(a, b) R(c, d)
⇒ b – d = a – c
⇒ -(d – b) = -(c – a)
⇒ d – b = c – a
⇒ (c.d) R(a.b)
(a, b) R(c, d) ⇒ (cd) R(ab) ∀ (a, b), (c, d) ∈ X
R is a symmetric relations.
(iii) Transitivity: Let (a, b), (c, d), (e, f) ∈ X is in this way (a, b) R(c, d) and (c, d) R(e, f)
(a, b) R(c, d) ⇒ b – d = a – c …(1)
(c, d) R(e, f) ⇒ d – f = c – e …(2)
Adding equation (i) and (2), we have
b – d + d – f = a – c + c – e
⇒ b – f = a – e
⇒ (a, b) R(e, f)
So, (a, b) R(c, d) and (c, d) R(e, f)
⇒ (a, b) R(e, f) ∀ (a, b), (c, d), (e, f) ∈ X
R is a transitive relation.
Hence, according to (i), (ii) and (iii), the given relation is equivalence relation.
Hence Proved.

Question 7.
A relation R is defined in a set T of triangles situated in a plane such that xRy ⇔ x is similar to y. Prove that R is an equivalence relation.
Solution:
Given : Set T = {Set of similar triangles}
A relation R in T is defined as:
xRy ⇔ x is similarly to y ∀ x, y ∈ T
To prove R is an equivalence relation, we have to prove that R is reflexive, symmetric and transitive.
(i) Symmetricity: Let x ∈ T
x ∈ f ⇒ x is similar to x.
⇒ (x, x) ∈ R ∀ x ∈ T
R is a reflexive relation.
(ii) Symmetricity: Let x, y ∈ T is in this way
(x, y) ∈ R
(x, y) ∈ R ⇒ x is similar to y
⇒ y is similar to x
⇒ (y, x) ∈ R
So, (x, y) ∈ R ⇒ (y, x) ∈ R ∀ x, y ∈ T
R is a symmetric relation.
(iii) Transitivity: Let x, y, z ∈ T is in this way
(x, y) ∈ R, (y, z) ∈ R
(x, y) ∈ R ⇒ x is similar to y.
(y, z) ∈ R ⇒ y is similar to z.
set x is similar to z.
So, (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R ∀ x, y, z ∈ T
R is a transitive relation.
Hence, according to (i), (ii) and (iii), the given relation is an equivalence relation.
Hence Proved.

Question 8.
Let a relation R is defined in a set A = {1, 2, 3} as : R = {(1, 1), (1, 2), (2, 1) (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)} Examine the reflexivity,
symmetricity and transitivity of R.
Solution:
Given set A = {1, 2, 3}
Reiation R In A is defined as:
R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)}
(i) Reflexivity: Here (1, 1), (2, 2), (3, 3) ∈ R
So, ∀ a ∈ A ⇒ (a, a) ∈ R
R is reflexive.
(ii) Symmetricity: Here, R is symmetric, because
(1, 2) ∈ R ⇔ (2, 1) ∈ R
(1, 3) ∈ R ⇔ (3, 1) ∈ R
(3, 2) ∈ R ⇔ (2, 3) ∈ R
So, (a, b) ∈ R ⇔ (b, a) ∈ R ∀ a, b ∈ A
So, R is a transitive relation.
(iii) Transitivity: (1, 2) ∈ R, (2, 1) ∈ R
⇒ (1, 1) ∈ R
(2, 3) ∈ R, (3, 2) ∈ R
⇒ (2, 2) ∈ R
Hence, by definition of Transitivity.
(a, b) ∈ R, (b, c) ∈ R
⇒ (a, c) ∈ R ∀ a, b, c ∈ A
So, R is a transitive relation.

Question 9.
A relation R in a set C₀ of non zero complex numbers is defined as:

Prove that R is an equivalence relation.
Solution:
Given: Set C₀ = set of the non-zero complex number and a relation R in C₀ is defined as:



Hence from (i), (ii) and (iii), the given relation is an equivalence relation.
Hence Proved.

Question 10.
If R is a relation in set X of subsets defined as “A is disjoint to B” then examine the reflexivity, symmetricity and transitivity of R.
Solution:
Set X = set of subsets A relation R in X is defined as
ARB ⇔ A is disjoint to B ∀ A, B ∈ X
⇔ A ∩ B = Φ ∀ A, B ∈ X (where Φ is a null set)
(i) Reflexivity: Let A ∈ X
A ∈ X ⇒ A ∩ A = A
⇒ A ∩ A ≠ Φ (untill A becomes Φ)
⇒ A is not disjoint to A
⇒ (A, A) ∉ R ∀ A ∈ X
R is not reflexive relation.
(ii) Symmetricity: Let A, B ∈ X in this way
(A, B) ∈ R
(A, B) ∈ R
⇒ A is disjoint to B thus A ∩ B = Φ
⇒ B ∩ A thus, B is disjoint to A (By commutative)
⇒ (B, A) ∈ R
So, (A, B) ∈ R ⇒ (B, A) ∈ R ∀ A, B ∈ X
R is a symmetric relation.
(iii) Transitivity: Let A, B, C ∈ X is in this way
(A, B) ∈ R and (B, C) ∈ R
A, B ∈ R ⇒ A ∩ B = Φ
B, C ∈ R ⇒ B ∩ C = Φ
Then A ∩ C = Φ is not necessary.
For example:
A = {1, 2, 3}, B = {4, 5, 6}, C = {1, 2, 7}
Here A, B ∈ R because A ∩ B = Φ
B, C ∈R because B ∩ C ⇒ Φ
But (A, C) ∉ R because A ∩ C = {1, 2} ≠ Φ
R is not transitive relation.
From (i), (ii) and (iii) given relation R is symmetric but relative R is not reflexive and transitive.
Hence Proved.

Question 11.
A relation R is defined in a set N of natural numbers such that aRb if a is a divisor of b, Prove that R is a partially ordered relation but not a total ordered relation.
Solution:
Given set N = {1, 2, 3, 4,…} = Set of natural numbers
Where a relation is define as
aRb ⇔ a is divisor of b ∀ a, b ∈ N
There R is a partial ordered relation if it is reflexive, anti symmetric and transitive.
(i) Reflexivity: Let a ∈ N
a ∈ N ⇒ a is divisor of a
⇒ (a, a) ∈ R ∀ a ∈ N
R is reflexive relation.
(ii) Anti Symmetricity: Let a, b ∈ N is in this way (a, b) ∈ R
(a, b) ∈ R ⇔ a is divisor of b.
⇔ b is disivor of a.
if a = b ∀ a, b ∈ N ⇔ (b, a) ∈ R; a = b, ∀ a, b ∈ N
(a, b) ∈ R ⇔ (b, a) ∈ R ⇒ a = b ∀ a, b ∈ N
So, R is anti-symmetric relation.
(iii) Transitivity: Let a, b, c ∈ N is in this way
(a, b) ∈ R and (b, c) ∈ R

So, (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R ∀ a, b, c ∈ N
So, R is a transitive relation.
Hence, from (i), (ii) and (iii) the given relation is a partially ordered relation.
Hence Proved.

Question 12.
Whether following subsets of N are total ordered set for relaton “x divides y or not” :
(i) {2, 4, 6, 8,……}
(ii) {0, 2, 4, 6,……..}
(iii) (3, 9, 5, 15,…}
(iv) (5, 15, 30}
(v) {1, 2, 3, 4}
(vi) {a, b, ab} ∀ a, b ∈ R.
Solution:
(i) Let subset A = {2, 4, 6, 8, …} ∈ N
A relation given in N



Hence, the second rule is not followed then the given relation is not a total ordered relation. For this it is clear that to prove a relation is a total ordered relative must be reflexive, symmetric and transitive and also every a, b ∈ A, (a, b) ∈ R or (b, a) ∈ R or a = b is true.
Hence, the given relation and their results are:

Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Ex 2.3

Question 1.
Examine which of the following is/are functions:
(i) {(1, 2), (2, 3), (3, 4), (2, 1)}
(ii) {(a, 0), (b, 0), (c, 1), (d, 1)}
(iii) {(1, a), (2, b), (1, b), (2, a)}
(iv) {(a, a), (b, b), (c, c)}
(v) {(a, b)}
(vi) {(4, 1), (4, 2), (4, 3), (4, 4)}
(vii) {(1, 4), (2, 4), (3, 4), (4, 4)}
(viii) {(x, y) | x, y ∈ R ∧ y² = x}
(ix) {(x, y) | x, y ∈ R ∧ x² = y}
(x) {(x, y) | x, y ∈ R ∧ x = y³}
(xi) {(x, y) | x, y ∈ R ∧ y = x³}
Solution:
(i) {(1, 2), (2, 3), (3, 4), (2, 1)}
It is not a function because element 2 corresponds to two elements 3 and 1.

(ii) {(a, 0), (b, 0), (c, 1), (d, 1)}
It is a function because under this each element corresponds to one and only one element.

(iii) {(1, a), (2, b), (1, b), (2, a)}
It is not a function because element 1 corresponds to two elements a and b.

(iv) {(a, a), (b, b), (c, c)}
It is a function because first element of ordered pair set is not same.

(v) {a, b}
It is a function because a corresponds to b

(vi) {(4, 1), (4, 2), (4, 3), (4, 4)}
It is not a function because first element of ordered pair set is same.

(vii) {(1, 4), (2, 4), (3, 4), (4, 4)}
It is a function because first element of ordered pair set is unequal.

(viii) {(x, y) : x, y ∈ R, y² = x}
Here y² = x ⇒ y = ±√x and if x = 4 then y = ±2
Hence, element ofy is related with 2 and -2 so, it is not a function.

(ix) {(x, y) : x, y ∈ R, x² = y}
It is a function because for y = x², each real value of x there is a unique image in R for each element of R

(x) {(x, y) : x, y ∈ R, x = y³}
It is a function because y = x^1/3, ∀ x ∈ R unique image is the set B.

(xi) {(x, y) : x, y ∈ R, y = x³}
It is also a function because for y = x³ ∀ x ∈ R unique image is in set B.

Question 2.
If f : R → R, f(x) = x², then find
(i) Range of f,
(ii) {x | f(x) = 4},
(iii) {y | f(y) = -1}
Solution:
(i) Given, f : R → R
and f(x) = x²
then if x < 0 ⇒ x² > 0
x = 0 ⇒ x² = 0
x > 0 ⇒ x² > 0
So, f(x) = x² ≥ 0 ∀ x ∈ R
Hence, range of R = R⁺ ∪ {0} or {x ∈ R | 0 ≤ x < ∞}
(ii) f(x) = 4
⇒ x² = 4
⇒ x = ±2
Hence, {x : y (x) = 4} = {-2, 2}.
(iii) f(y) = -1
⇒ y² = -1
⇒ y = ±√1
⇒ (y : f(y) = -1} = Φ null set.

Question 3.
Let A = {-2, -1, 0, 1, 2} and function f is defined in A to R by f(x) = x² + 1. Find the range of f.
Solution:
Given, A = {-2, -1, 0, 1, 2}
and R = set of real numbers
Given: f(x) = x² + 1
then f(-2) = (-2)² + 1 = 5
f(-1) = (-1)² + 1 = 2
f(0) = (0)² + 1 = 1
f(1) = (1)² + 1 = 2
f(2) = (2)² + 1 = 5
Hence, range of f = (1, 2, 5}.

Question 4.
Let A = {-2, -1, 0, 1, 2} and f : A → Z where f(x) = x² + 2x – 3, then find
(i) Range of f
(ii) Pre image of 6, -3 and 5
Solution
(i) Given,
A = {-2, -1, 0, 1, 2} and Z = {0, ± 1, ± 2, …}
from f(x) = x² + 2x – 3
f(-2) = (-2)² + 2(-2) – 3 = 4 – 4 – 3 = -3
f(-1) = (-1)² + 2(-1) – 3 = 1 – 2 – 3 = – 4
f(0) = 0² + 2(0) – 3 = -3
f(1) = 1² + 2(1) – 3 = 1 + 2 – 3 = 0
f(2) = (2)² + 2(2) – 3 = 4 + 4 – 3 = 5
Hence, Range of f = set of f(x) = {-4, -3, 0, 5}
(ii) Let pre-image of 6 is x
f(x) = 6
⇒ x² + 2x – 3 = 6
⇒ x² + 2x – 9 = 0

As there is no pre-image of 6 in A.
Hence, pre-image of A is Φ
Again, let the pre-image of A be -3 in x, then
f(x) = -3
⇒ x² + 2x – 3 = -3
⇒ x² + 2x = 0
⇒ x(x + 2) = 0
⇒ x = 0, x = – 2
Hence, pre-image of A in -3 is -2 or 0 {-2, 0}
Similarly to find the pre-image of 5
f(x) = 5
⇒ x² + 2x – 3 = 5
⇒ x² + 2x – 8 = 0


The domain of the signum function is the set of real numbers and the range is {-1, 0, 1}.

Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Ex 2.4

Question 1.
Classify the following functions is the form of one-one, many one, into and onto, also give reason to support your answer.
(i) f : Q → Q, f(x) = 3x + 7
(ii) f : C → R, f(x + iy) = x
(iii) f : if R → [-1, 1], f(x) = sin x
(iv) f : N → Z, f(x) = |x|
Solution:
(i) Given f : Q → Q and f(x) = 3x + 7
where Q is set of rational numbers.
Let x₁, x₂ ∈ Q are in this way f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ 3x₁ + 7 = 3x₂ + 7
⇒ 3x₁ = 3x₂
⇒ x₁ = x₂
⇒ f(x₁) = f(x₂)
⇒ x₁ = x₂
x₁, x₂ ∈ Q
Hence, f is one-one function.
Now, Let y ∈ Q co-domain
If possible, then let pre-image of y is x in domain Q then
f(x) = y
⇒ 3x + 7 = y
⇒ x = y−7/3 ∈ Q
So, every element in the co-domain of Q has pre-image in Q
Hence, f is onto function.
Thus, f is a one-one, onto function.

(ii) f : C → R : f(x + iy) = x
Given : function
f : C → R and f(x + iy) = x
Here C = set of complex numbers
R = set of real numbers
Let x + iy and x – iy (y ≠ 0) are different elements in domain C.
f(x + iy) = x and f (x – iy) = x
⇒ f(x + iy) = f(x – iy)
So, two different elements of domain R have the same image.
So, f is a many-one function.
Here Range of f = (x : x + iy ∈ C} – R Co-domain
[Range of x + iy in x ∈ R, y ∈ R and i = √-1 ]
f is a onto function.
Thus, f is a many-one, onto function.

(iii) f : R → [-1, 1], f(x) = sin x
Given: f : R → [-1, 1] and f(x) = sinx
where R is set of real numbers.
Let x₁, x₂ ∈ R
If f(x₁) = f(x₂) ⇒ sin x₁ = sin x₂
⇒ x₁ = nπ + (-1)ⁿx₂, n ∈ I (General value of sin θ)
⇒ x₁ ≠ x₂
f is not one-one function so, it is many-one function
Again, Let y ∈ Y
If possible then let pre-image of y under f is x then
f(x) = y
⇒ sin x = y
⇒ x = sin^-1y
So, -1 ≤ y ≤ 1
Many value of sin^-1y present in R, then
x ∈ R ∀ y ∈ [-1, 1]
f is onto function.
Thus, f is many-one, onto function.

(iv) f : N → Z, f(x) = |x|
Given f : N → Z and f(x) = |x|
where N = Set of natural number = {1, 2, 3, 4,…}
and Z = set of natural numbers = {0, ± 1, ± 2, ± 3, …}
Let x₁, x₂ ∈ N
If f(x₁) = f(x₂) ⇒ |x₁| = |x₂| [∵ x₁ > 0, x₂ > 0, x₁, x₂ ∈ N]
⇒ x₁ = x₂
f is one-one function.
Again ∵ Range of f = {|x| : x ∈ N], Z ≠ Z (co-domain)
So, f is not onto function.
Hence, f is into function.
Thus, f is one-one into function.

Question 2.
If A = {x | -1 ≤ x ≤ 1} = B, then find out which function is one-one, into or one-one onto defined from A to B
(i) f(x) = x/2
(ii) g(x) = |x|
(iii) h(x) = x²
(iv) k(x) = sinπx
Solution:
Given equation
A = {x : -1 ≤ x ≤ 1)
and B ={x : -1 ≤ x ≤ 1}
(i) Function is defined from A to B.

f is not onto function.
Hence, it is proved that f is one-one, into function,

(ii) g : A → B, g (x) = |x|
Let x₁, x₂ ∈ A
If f(x₁) = f(x₂)
⇒ g(x₁) = g(x₂)
⇒ |x₁| = |x₂|
⇒ x₁ = ±x₂
g is not one-one function so, g is many one function.
Again, range of g = { x : -1 ≤ x ≤ 1} ≠ B (co-domain)
i.e., pre-image of negative number is not exist in codomain B.
Thus, it is proved that f is many-one, into function.
g is not one-one function.

(iii) h : A → B, h(x) = x²
Let x₁, x₂ ∈ A.
Thus, h(x₁) = h(x₂)
h (x₁) = h (x₂)
⇒ x₁² = x₂²
⇒ x₁ = ± x₂
h is not one-one function.
For example 1 ≠ -1 but h(1) = h(-1) = 1
So, h is a many-one function.
Let y ∈ B if possible pre-image of y is x exist in B, then
h(x) = y
⇒ x² = y
⇒ x = ±√y
when y is positive then, x does not exist
i.e., pre-image of a negative number does not exist.
So, Range of h ={x : 0 ≤ x ≤ 1} ≠ B (co-domain)
So, h is a into function
Hence, it is proved that h is many-one, into function,

(iv) k : A → B, k(x) = sin πx
-1, 1 ∈ A are such numbes
-1 ≠ 1
But k(-1) = sin (-π) = 0
⇒ k(1) = sin π = 0
So, -1 ≠ 1
⇒ k(-1) = k(1)
i.e., different elements have same image.
k is many-one function.
Again, range of k = {sin πx : x ∈ A} = {x : -1 ≤ x ≤ 1} = B (co-domain)
k is onto function
Hence, it is proved that k is many-one, onto function.

Question 3.
If f : C → C, f(x + iy) = (x – iy), then prove that f is an one-one onto function.
Solution:
Given : f : C → C and f(x + iy) = x – iy
where C is set of complex numbers.
Let x₁ + iy₁ and x₂ + iy₂ ∈ C thus.
f(x₁ + iy₁) = f (x₂ + iy₂)
⇒ x₁ – iy₁ = x₂ – iy₂


So, f is one-one function.
Range of f = {x – iy : x + iy ∈ C} = C (co-domain)
f is onto function.
Hence, f is one-one, onto function.
Hence Proved.

Question 4.
Give one example of each of the following function:
(i) One-one into
(ii) Many-one onto
(iii) Onto but not one-one
(iv) One-one but not onto
(v) Neither one-one nor onto
(vi) One-one onto
Solution:
(i) f : N → N, f(x) = 2x
(ii) f : R₀ → R+, f(x) = x²
(iii) f : z₀ → N, f(x) = |x|
(iv) f : Z → Z, f(x) = 2x
(v) f : R → R₁, f(x) = x²
(vi) f : Z → f(z) = -x

Question 5.
Prove that f : R → R, f(x) = cos x is a many- one into function. Change the domain and co-domain of f such that I become:
(i) One-one into
(ii) Many-one onto
(iii) One-one onto
Solution:
Given, f : R → R, f(x) = cos x
Let x₁, x₂ ∈ R
Thus, f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ cos x₁ = cos x₂
⇒ x₂ = 2nπ ± x₂, x₁ ∈ I
f is not one-one function.
Again, let, y ∈ R (co-domain), if possible let pre-image of y is x is domain R, then
f(x) = y
⇒ cos x = y
⇒ x = cos^-1y
x only exist when,
-1 ≤ y ≤ 1
when y ∈ R – [-1, 1]
Pre-image of y is not present in domain R.
So, f is not onto function
Hence, f is many-one, into function.
(i) One-one, into function
f : [0, π] → R, f(x) = cos x
(ii) Many-one, onto function
f : R → [-1, 1], f(x) = cos x
(iii) One-one, onto function.
f : [0, π] → [-1, 1], f(x) = cos x

Question 6.
If N= {1, 2, 3, 4, …), O = (1, 3, 5, 7, …}, E = (2, 4, 6, 8,…..) and f₁, f₂ are function defined as
f1 : N → O, f₁(x) = 2x – 1; f₂ : N → E, f₂(x) = 2x
Then prove that f₁ and f₂ are one-one onto.
Solution:
(i) Given N = {1, 2, 3, …}
O = {1, 3, 5, …}
Function f(x) = 2x – 1
Let x₁, x₂ ∈ N
Thus f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ 2x₁ – 1 = 2x₂ – 1
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ N
So, f₁ is one-one function.
Again, let y ∈ O (co-domain), if possible let pre-image of y is x exist in domain N, then f(x) = y.
f(x) = y
⇒ 2x – 1 = y
⇒ x = y+1/2 ∈ N ∀ y ∈ O
So, pre-image of every element of O (co-domain) exist in domain N
So, f₁ is onto function.
Hence, f is one-one onto function

(ii) Given set are
N = {1, 2, 3, …}
E = {2, 4, 6, …}
Function f₂ : N → E, f₂(x) = 2x ∀ x ∈ N
Let x₁, x₂ ∈ N
Thus f₂(x₁) = f₂(x₂)
f₂(x₁) = f₂(x₂)
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂ ∀ a, b ∈ N
f₂ is one-one function.
Again, let y ∈ E (co-domain), if possible, let pre-image of y is x exist in domain N then f₂(x) = y
f₂(x) = y
⇒ 2x = y
⇒ x = y/2 ∈ N ∀ y ∈ E
So, pre-image of every element of E (co-domain) exist in domain N.
Hence, f₂ is one-one, onto function.

Question 7.
If function f is defined from set of real numbers R to R in the following way then classify them in the form of one-one, many-one, into or onto.
(i) f(x) = x²
(ii) f(x) = x³
(iii) f(x) = x³ + 3
(iv) f(x) = x³ – x
Solution:
(i) Given, f : R → R, f(x) = x²
where, R is set of real number
Let x₁, x₂ ∈ R
Thus f (x₁) = f (x₂)
f(x₁) = f(x₂)
⇒ x₁² = x₂²
⇒ x₁ = ±x₂
f is not one-one function. So, f is many-one function.
Let y ∈ R (co-domain), if possible, let pre-image of y is x in domain R, then f(x) = y
f (x) = y
⇒ x² = y
⇒ x = ±√y
If is clear, x is not exist for negative values of y.
So, f is into function.
Hence, f is many-one into function.

(ii) Given: f : R → R and f(x) = x³
Let x₁, x₂ ∈ R
Thus f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ x₁³ = x₂³
⇒ x₁ = x₂ ∀ x₁, x₂ ∈ R
f is one-one function
Again, let y ∈ R (co-domain), if possible, let pre-image ofy is x in domain R, then
f (x) = y
f(x) = y
⇒ x³ = y
⇒ x = y^1/3 ∈ R
So, f is onto function.
Hence, f is one-one onto function.

(iii) Given: f : R → R and f(x) = x³ + 3
Where R is set of real number
Let x₁, x₂ ∈ R
Thus f (x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ x₁³ + 3 = x₂² + 3
⇒ x₁³ = x₂³ ∀ x₁, x₂ ∈ R
So, f is one-one function.
Again let y ∈ R (co-domain)
If possible, let pre-image of y is x in f then
f (x) = y
⇒ x³ + 3 = y
⇒ x = (y – 3)^1/3 ∈ R ∀ y ∈ R
So, pre-image of every element R (co-domain) is exist in domain R.
Hence, f is one-one, onto function.

(iv) Given: f : R → R, f(x) = x³ – x
Where R is set of real numbers.
Let 0, 1 ∈ R are such numbers
0 ≠ 1
0 ∈ R ⇒ f(0) = 0 – 0 = 0
1 ∈ R ⇒ f(1) = 1³ – 1 = 0
0 ≠ 1 and f(0) = f(1) = 1
So, f is many-one function.
Again, range of f = f(R) = {x³ – x : x ∈ R) = R (domain)
So, f is onto function
Hence, f is many-one, onto function.

Rajasthan Board RBSE Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise

Question 1.
If A = {a, b, c, d} and B = {p, q, r, s} then relation from A to B is
(A) {(a, p), (b, s, (c, r)}
(B) {(a, p), (b, q), (c, r), (s, d)}
(C) {{b, a), (q, b), (c, r)}
(D) {(c, s), (d, s), (r, a), (q, b)}
Solution:
A × B = {(a, p), (a, q), (a, r), (a, s), {b, p), (b, q), {b, r), (b, s), (c, p), (c, q), (c, r) (c, s) (d, p), (d, q), (d, r), (d, s)}
{(a, p), (b, r), (c, r)}
Because {(a, p), (b, r), (c, r)} ⊆ A × B.
Hence, option (A) is correct.

Question 2.
A relation R in N is defined such that xRy ⇔ x + 4y = 16, then the range of R is
(A) {1, 2, 4}
(B) {1, 3, 4}
(C) {1, 2, 3}
(D) {2, 3, 4}
Solution:
Given,set N = set of natural number = {1, 2, 3,…}
A relation R in Nto N is defined as
xRy ⇔ x + 4y = 16 ∀ x, y ∈ N
or xRy ⇔ y = 16−x/2 ∀ x, y ∈ N
when x = 4 then

So, R = {(4, 3), (8, 2), (12, 1)}
So, range of R= {3, 2, 1} and {1, 2, 3}
Hence, option (C) is correct.

Question 3.
Rule form of relation {(1, 2), (2, 5), (3, 10), (4, 17), …} in N
(A) {(x, y) : x, y ∈ N, y = 2x + 1}
(B) {(x, y) : x, y ∈ N, y = x² + 1}
(C) {(x, y) : x, y ∈ N, y = 3x – 1}
(D) {(x, y) : x, y ∈ N, y = x + 3}
Solution:
{(x, y) : x, y ∈ N, y = x² + 1}
when x = 1, then y = (1)² + 1 = 2 ⇒ (1, 2) ∈ N
when x = 2, then y = (2)² + 1 = 4 + 1 = 5 ⇒ (2, 5) ∈ N
when x = 3, then y = (3)² + 1 = 9 + 1 = 10 ⇒ (3, 10) ∈ N
when x = 4, then y = (4)² + 1 = 16 + 1 = 17 ⇒ (4, 17) ∈ N
So, R = {(1, 2), (2, 5), (3, 10), (4, 17), …}
= {(x, y) : x, y ∈ N, y = x² + 1}
Hence, option (B) is correct.

Question 4.
If A = {2, 3, 4} and B = {3, 4, 5, 6, 7, 8} : A relation if from A to B is defined such that “x divides y” then R^-1 is
(A) {(4, 2), (6, 2), (8, 2), (3, 3), (6,3), (4, 4), (8, 4)}
(B) {(2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (4, 4), (4, 8)}
(C) {(3, 3), (4, 4), (8, 4)}
(D) {(4, 2), (6, 3), (8, 4)}
Solution:
Given sets are
A = {2, 3, 4} and B = {3, 4, 5, 6, 7, 8}
Relation R from A to B is defined as
“x divides y” ∀ x ∈ A, y ∈ B
So, when x = 2 ∈ A
Then 2 divides the element 4, 6, 8 of B.
So, (2, 4), (2, 6), (2, 8) ∈ R
when x = 3 ∈ A, then 3 divides elements 3, 6 of B
So, (3, 3), (3, 6) ∈ R
when x = 4 ∈ A then 4 divides the elements 4, 8 of B.
So, (4, 4), (4, 8) ∈ R
R = {(2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (4, 4), (4, 8)}
R^-1 = {(4, 2), (6, 2), (8, 2), (3, 3), (6, 3), (4, 4), (8, 4)}
Hence, option (A) is correct.

Question 5.
In the set of real numbers, Relation “x is smaller than y” will be
(A) Reflexive and Transitive
(B) Symmetric and Transitive
(C) Anti-symmetric and Transitive
(D) Reflexive and Anti-symmetric
Solution:
Because this relationship is only transitive
If “x is less than or equal to y” then this relation is reflexive and anti-symmetric.
Hence, option (C) is correct.

Question 6.
A relation R inset of non zero integers is defined as xRy ⇔ xy = yx then R is
(A) reflexive and symmetric but not transitive
(B) reflexive and anti-symmetric but not transitive
(C) reflexive, anti-symmetric and transitive
(D) reflexive, symmetric and transitive
Solution:
Given, Set I₀ = set of non-zero integers = {±1, ±2, …}
A relation in I₀xRy ⇔ xy = yx ∀ x, y ∈ I₀
⇔ y log x = x log y ∀ x, y ∈ I₀
To prove R is equivalence relation we have to prove that R is reflexive, symmetric and transitive relation.
(i) Reflexivity: Let a ∈ I₀
a ∈ I₀
⇒ a log a = a log a ∀ a ∈ I₀
R is reflexive.

(ii) Symmetricity: Let a, b ∈ I₀
Now, (a, b) ∈ R
(a, b) ∈ R
⇒ b log a = a log b
⇒ (b, a) ∈ R
So, (a, b) ∈ R
⇒ (b, a) ∈ R ∀ a, b ∈ I₀
R is a symmetric relation.

(iii) Transitivity: Let a, b, c ∈ I₀
Now, (a, b) ∈ R, (b, c) ∈ R
(a, b) ∈ R ⇒ b log a = a log b ……(1)
(b, c) ∈ R ⇒ c log b = b log c
⇒ b log c = c log b …(2)
By dividing equation (1) from equation (2).

R is a transitive relation.
Thus, it is proved that R is reflexive, symmetric and transitive relation.
Hence, the option (D) is correct.

Question 7.
If relation R is defined as “x is divisor of y” then from the following subet of N. Which is a total ordered set ?
(A) {36, 3, 9}
(B) {7, 77, 11}
(C) {3, 6, 9, 12, 24}
(D) {1, 2, 3, 4, …}
Solution:
Given : N = set of natural numbers = (1, 2, 3, 4, …}
A relation R is N is defined as
xRy ⇒ “x is divisor of y” ∀ x, y ∈ N
⇒ y/x = k ∈ N
where N is a set of natural numbers ∀ x, y ∈ N
From option ‘A’, 36/9, 36/3, 9/3 all are natural numbers
From option ‘B’, 11/7 ∉ N, from option ‘C’, 9/6 ∉ N
From option ‘D’, 3/2 ∉ N, Hence, option (A) is correct.

Question 8.
From the following relations defined on set Z of integers, which of the relation is not equivalence relation
(A) aR₁b ⇔ (a + b) is an even integer
(B) aR₂b ⇔ (a – b) is an even integer
(C) aR₃b ⇔ a < b
(D) aR₄b ⇔ a = b
Solution:
Correct option (C)
aR₃b ⇔ a < b, R₃ is not an equivalence relation because it is not reflexive and symmetric.
Hence, option (C) is correct.

Question 9.
A relation R is defined on set A = {1, 2, 3}, where R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is
(A) reflexive but not transitive
(B) reflexive but not symmetric
(C) symmetric and transitive
(D) neither symmetric nor reflexive
Sotution:
Given : Set A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
(i) Reflexivity: (1, 1) ∈ R, (2, 2) ∈ R, (3, 3) ∈ R
(a, a) ∈ R
R is reflexive.

(ii) Symmetricity: (1, 2) = (2, 1) ∉ R
(1, 3) = (3, 1 ) ∉ R
(2, 3) = (3, 2) ∉ R
(a, b) ∈ R ⇒ (b, a) ∉ R So, R is not symmetric
Hence, option (B) is correct.

Question 10.
If A = {a, b, c}, then number of possible non-zero relations in A is
(A) 511
(B) 512
(C) 8
(D) 7
Solution:
A = {a, b, c} Number of elements in A = n(A) = 3
Then, numbers of elements in A × A = n(A × A) = 3² = 9
So, the number of relations in A are = 2ⁿ – 1 = 2⁹ – 1 = 512 – 1 = 511.
Hence, option (A) is correct.

Question 11.
If A = {1, 2, 3, 4} then which of the following is a function in A
(A) f₁ = {(x, y) : y = x + 1}
(B) f₂ = {(x, y) : x + y > 4}
(C) f₃ = {(x, y) : y < x}
(D) f₄ = {(x, y) : x + y = 5}
Solution:
Here, f₁ = {(x, y) : y = x + 1} = {(1, 2), (2, 3), (3, 4)}
f₂ = {(x, y) : x + y > 4} = (1, 4), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
f₃ = {(x, y) : y < x} = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}
f₄ = {(x, y) : x + y = 5} = {(x, y) : x = 5 – y} = {(1, 4), (2, 3), (3, 3), (4, 1)}
It is clear that only f₄ is a function because eveiy element of A corresponts to unique element in B.
Hence, the option (D) is correct.

Question 12.
Function f : N → N, f(x) = 2x + 3 is
(A) One-one onto
(B) One-one into
(C) Many one-onto
(D) Many-one into
Solution:
Given: f : N → N and f(x) = 2x + 3
where N = set of natural numbers
Let x₁, x₂ ∈ N is such that f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ 2x₁ + 3 = 2x₂ + 3
⇒ 2x₁ = 2x₂
⇒ x₁ = x₂
f(x₁) = f(x₂) ⇒ x₁ = x₂ ∀ x₁, x₂ ∈ N
f is one-one function.
Again, let y ∈ N (co-domain) if possible than let pre image of x is in domain N then f(x) = y
f(x) = y
⇒ 2x + 3 = y
⇒ x = y−3/2 ∈ N
At y = 1, then x = 1−3/2 = -1 ∉ N
If this way, y has many values for which x is not exist in domain A. So, f is into function.
Hence, option (B) is correct.

Question 13.
Which one of the following is onto function define from R to R.
(A) f(x) = |x|
(B) f(x) = e^-x
(C) f(x) = x³
(D) f(x) = sin x
Solution:
f(x) = x³
Given, f : R → R and f(x) = x³
Let y ∈ R (co-domain) if possible, then let pre-image of y is x in domain R, then
f(x) = y
⇒ x³ = y
⇒ x = y^1/3 ∈ R ∀ y ∈ R
So, pre-image of each value ofy is exist in domain R.
So, R is onto function.
Hence, option (C) is correct.

Question 14.
Which of the following is one-one function defined from R to R
(A) f(x) = |x|
(B) f(x) = cos x
(C) f(x) = e^x
(D) f(x) = x²
Solution:
(C) f(x) = e^x
Let x, y ∈ R are such that
f(x) = f(y)
⇒ e^x = e^y
⇒ x log_e e = y log_e e
⇒ x = y
f(x) = f(y)
⇒ x = y [∵ log_e e = 1]
f is one-one function ∀ x, y ∈ R

Question 15.
f : R → R, f(x) = x² + x is:
(A) One-one one
(B) One-one into
(C) Many-one onto
(D) Many one onto
Solution:
Given: f : R → R and f(x) = x² + x
where R is a set of real numbers.
Let x₁, x₂ ∈ R are such that f(x₁) = f(x₂)
f(x₁) = f(x₂)
⇒ x₁² + x₁ = x₂² + x₂
⇒ x₁² – x₂² + x₁ – x₂ = 0
⇒ (x₁ – x₂)(x₁ + x₂) + 1 (x₁ – x₂) = 0
⇒ (x₁ – x₂) (x₁ + x₂ + 1} = 0
⇒ x₁ = x₂, x₁ = -(x₂ + 1) ∀ x₁, x₂ ∈ R
Here, element of set A relates to two elements of set B.
So, it is a many-one function.
Again, let y ∈ R (co-domain)
If possible then let pre-image of y is x in domain R.
then f(x) = y
⇒ x² + x = y
⇒ x(x + 1) = y
⇒ x = y, x = y – 1
If y < 1, then there is no real value of x.
So, pre-image of many elements of R does not exist in domain R, so, f is an into function.
Thus, f is many-one, into function.
Hence, the option (D) is correct.

Question 16.
Which of the following is onto function—
(A) f : Z → Z, f(x) = |x|
(B) f : N → Z, f(x) = |x|
(C) f : R₀ → R⁺, f(x) = |x|
(D) f : C → R, f(x) = |x|
Solution:
f : R₀ → R⁺, f(x) = |x|
Pre-image of every positive real number is exists in domain R₀. So, function is onto.
Hence, option (C) is correct.

Question 17.
Domain of function
f(x)=1/√|x|−x
(A) R⁺
(B) R^–
(C) R₀
(D) R
Solution:

It is clear that function is defined is x < 0 because [x ≥ 0 then f(x) = ∞]
Required domain of f(y) = R^–
Hence, option (B) is correct.

Question 18.
If x is real number than the range of

Solution:

Question 19.
Range of function f(x) = cos  x/3 is
(A) (0, ∞)
(B) (−1/3, 1/3)
(C) [-1, 1]
(D) [0, 1]
Solution:
Let y = cos  x/3
-1 ≤ cos  x/3 ≤ 1
-1 ≤ y ≤ 1
x is defined if -1 ≤ y ≤ 1
Hence, required range = [-1, 1]
Hence, option (C) is correct.

Question 20.
From ]−π/2,−π/2[ which of the following is one-one onto function defined in R
(A) f(x) = tan x
(B) f(x) = sin x
(C) f(x) = cos x
(D) f(x) = e^x + e^-x
Solution:
f(x) = tan x

Hence, option (A) is correct.

Question 21.
Find the domain and range of the relaton R
R = {(x + 1, x + 5)} : x ∈ {0, 1, 2, 3, 4, 5}
Solution:
Given relation
R = {(x + 1, x + 5) : x ∈ (0, 1, 2, 3, 4, 5}
Then, domain of relation R {(x + 1) : x ∈ {0, 1, 2, 3, 4, 5}
Domain of R = {1, 2, 3, 4, 5, 6}
and Range of R = {(x + 5) : x ∈ (0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Question 22.
If A = {1, 2}, then write all non-zero relations defined in A.
Solution:
All non-zero relation are:
{(1, 1)}, {(2, 2)}, {(1, 2)}, {(2, 1)}
{(1, 1), (1, 2)}, {(1, 1), (2, 1)}, {(2, 2), (1,2)}, {(2, 2), (2, 1)}, {(1, 2), (2, 1)}
{(1, 1), (2, 2), (1, 2)}, {(1, 1), (2, 2), (2, 1)}, {(1, 1), (1, 2), (2, 1)}, {(2, 2), (1, 2), (2, 1)}
{(1, 1), (1, 2), (2, 1), (2, 2)}

Question 23.
Find the domain and range of the following relations
(i) R₁ = {(x, y) : x, y ∈ N, x + y = 10}
(ii) R₂ = {(x, y) : y = |x – 1|, x ∈ z and |x| ≤ 3}
Solution:
(i) Given set N = {1, 2, 3,…}
A relation R₁ in N is defined as
R₁ = {(x, y) : x, y ∈ N, x + y = 10} ∀ x, y ∈ N
So, xR₁y ⇔ x + y = 10 ⇔ y = 10 – x
when x = 1, then y = 10 – 1 = 9 ∈ N then (1, 9) ∈ R₁
when x = 2, then y = 10 – 2 = 8 ∈ N then (2, 8) ∈ N
when x = 3, then y = 10 – 3 = 7 ∈ N then (3, 7) ∈ R₁
when x = 4, then y = 10 – 4 = 6 ∈ N then (4, 6) ∈ R₁
Similarly, (5, 5) ∈ R₁, (6, 4) ∈ R₁, (7, 3) ∈ R₁, (8, 2) ∈ R₁, (9, 1) ∈ R₁
R₁ = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}
Domain of R₁ ={1, 2, 3, 4, 5, 6, 7, 8, 9}
Range of R₁ = {9, 8, 7, 6, 5, 4, 3, 2, 1}

(ii) Given set Z = set of integers = {0, ±1, ±2, ±3,…}
A relation R in Z is defined as
R₂ = {(x, y), y = |x – 1|, x ∈ z and x ≤ 3}
or xRy ⇔ y = |x – 1|, |x| ≤ 3 ∀ y ∈ Z
Here |x| ≤ 3 ⇒ -3 ≤ x ≤ 3, x ∈ Z

Domain of R₂ = {-3, -2, -1, 0, 1, 2, 3}
Range of R₂ = {4, 3, 2, 1, 0}

Question 24.
In the set of real numbers R, two relations R₁ and R₂ can be defined as
(i) aR₁b ⇔ a – b > 0
(ii) aR₂b ⇔ |a| ≤ b
Also test the reflexivity, symmetricity and transitivity of R₁ and R₂.
Solution:
(i) Given set R = Set of real numbers
Reflexivity R aR₁b ⇔ a – b > 0 ∀ a, b ∈ R
Reflexivity : Let a ∈ R
a ∈ R
⇒ a – a = 0 > 0
⇒ a – a = 0 > 0
⇒ (a, a) ∉ R₁ a ∈ R
R is not reflexive relation.
Symmetricity : Let a, b ∈ R are such that (a, b) ∈ R₁
(a, b) ∈ R₁
⇒ a – b > 0
⇒ b – a < 0
⇒ (b, a) ∉ R₁
So, (a, b) ∈ R₁
⇒ (b, a) ∉ R₁
R₁ is not symmetric.
Transitivity : Let a, b, c ∈ R are such that
(a, b) ∈ R₁, (b, c) ∈ R₁
(a, b) ∈ R₁
⇒ a – b > 0 …(1)
(b, c) ∈ R₁ ⇒ b – c > 0 …(2)
and a – c = (a – b) + (b – c) > 0 [∴ From (1) and (2)]
⇒ a – c > 0
⇒ (a, c) ∈ R₁
So, (a, b) ∈ R₁, (b, c) ∈ R₁
⇒ (a, c) ∈ R₁ ∀ a, b, c ∈ R
R₁ is transitive.
From above it is clear that R₁ is not reflexive and symmetric relation.
It is only a transitive relation.

(ii) Given: Relation R = set of real numbers
A relation R₂ in R is defined as
aR₂b ⇔ |a| ≤ b ∀ a, b ∈ R
Reflexivity: Let a ∈ R
a ∈ R ⇒ |a| ≤ a is not necessary
For example a = -2 ∈ R
and |-2| ≤ -2 ⇒ (-2, -2) ∈ R₂
So, R₂ is not a reflexive relation.
Symmetricity: Let (a, b) ∈ R are such that (a, b) ∈ R₂
(a, b) ∈ R₂ ⇒ |a| ≤ b
Then |b| ≤ a is not necessary.
For example : a = -2, b = 3
and (-2, 3) ∉ R₂ as |-2| ≤ 3
But (3, -2) ∉ R₂ because |3| ≤ -2
R₂ is not symmetric.
Transitivity : Let a, b, c ∈ R are such that
(a, b) ∈ R₂ and (b, c) ∈ R₂
(a, b) ∈ R₂ ⇒ |a| ≤ b …(1)
(b, c) ∈ R₂ ⇒ |b| ≤ c …(2)
From equation (1) and (2),
|a| ≤ b ≤ |b| ≤ c
⇒ |a| ≤ c
⇒ (a, c) ∈ R₂
So, (a, b) ∈ R₂, (b, c) ∈ R₂
⇒ (a, c) ∈ R₂ ∀ a, b, c ∈ R
R₂ is transitive.
Hence, from above it is clear that R₂ is not reflexive and symmetric relation.
It is only a transitive relation.

Question 25.
In a set of natural numbers, a relation R is defined as
aRb ⇔ a² – 4ab + 3b² = 0, (a, b ∈ N)
Prove that R is reflexive but not symmetric and transitive.
Solution:
Given set N = {1, 2, 3, 4,….}
A relation R in N is defined as
aRb ⇔ a² – 4ab + 3b² = 0, ∀ a, b ∈ N

Question 26.
In the set of real numbers R, two relations R₁ and R₂ are defined as
(i) aR₁b ⇔ |a| = |b|
(ii) aR₂b ⇔ |a| ≤ |b|
Prove that R₁ is an equivalence relation but R₂ is not.
Solution:
(i) Given set R = set of real numbers.
A relation R₁ in R is defined as
aR₁b = |a| = |b| ∀ a, b ∈ R
For proving is equivalence relation we have to prove that R₁ is reflexive, symmetric and transitive relation.

Question 27.
A relation R in set A = {1, 2, 3} is defined as:
R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)}
Test the reflexivity, symmetricity and transitivity of R.
Solution:
Given : Set A = {1, 2, 3}
Relation R in A is defined as
R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1) (2, 3), (3, 2)}
(i) Reflexivity : Here
(1, 1) ∈ R
(2, 2) ∈ R
(3, 3) ∈ R
So, ∀ a ∈ A ⇒ (a, a) ∈ R
R is not reflexive.

(ii) Symmetricity :
Here
(1, 2) ∈ R ⇔ (2, 1) ∈ R
(2, 3) ∈ R ⇔ (3, 2) ∈ R
(1, 3) ∈ R ⇔ (3, 1) ∈ R
So, (a, b) ∈ R ⇒ (b, a) ∈ R
R is symmetric relation.

(iii) Transitivity:
(1, 2) ∈ R, (2, 1) ∈ R ⇔ (1, 1) ∈ R etc.,
So, by definition of transitive relation.
R is transitive if
(a, b) ∈ R, (b, c) ∈ R
⇒ (a, c) ∈ R ∀ a, b, c ∈ A

Question 28.
Find the domain of function
1/√(x+1)(x+2)
Solution:
Let f(x) = 1/√(x+1)(x+2)
Function f(x) is defiined if (x + 1) (x + 2) > 0
x > -1, x > -2
⇒ x ∈ (-∞, -2) ∪ (-1, ∞)
Hence, required domain (-∞, -2) ∪ (-1, ∞).