RBSE Maths Chapter 2 – Relation and function Class 11 Important questions and solutions are available here. The important questions and solutions of Chapter 2, available at BYJU’S, contain step by step answers which help the students in understanding the concepts better. All these important questions are based on the new pattern designed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 11 solutions.

RBSE Maths Chapter 2 solutions of the RBSE Class 11 Maths will help the students to solve problems related to open sentence, ordered pair, a cartesian product of two sets, relation, domain and range of a relation, inverse relation, identity relation, types of relations, functions, function as a set of ordered pairs, domain, codomain and range of a function, types of functions, algebra of real functions, kinds of functions.

### RBSE Maths Chapter 2: Exercise 2.1 Textbook Important Questions and Solutions

**Question 1: If A = {1, 2, 3}, B = {4, 5, 6}, then which of the following is the 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)}**

**Solution: **

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

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

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

(ii) 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

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

**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 arithmetic progression.

Hence, its n^{th} 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

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

{(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 arithmetic progression.

T_{n} = 6 + (n – 1) × (-2)

T_{n} = 6 – 2n + 2

T_{n} = 8 – 2n

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

{(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 arithmetic progression.

T_{n} = 2 + (n – 1) × 1

= 2 + n – 1

= n + 1

By putting n = x and T_{n} = y, the required rule obtained is

= {(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 the domain and range of R. **

**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 ∀ x, y ∈ R

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 ordered pairs and also define their domain. **

**Solution:**

Given set = Z

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

Relation in Z is a relation R from Z to Z which is defined as xRy ⇔ x^{2} + y^{2} = 25 ∀ x, y ∈ Z

When x = 0

0^{2} + y^{2} = 25

y = ± 5

⇒ (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} = 0

⇒ 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 defined such that x Φ y ⇔ |x| = y. **

**State whether the following relations are true or false. Justify your answer.**

**(i) (1 + i) Φ 3 **

**(ii) 3 Φ (-3)**

**Solution:**

The given set C = {a + ib : a ∈ R, b ∈ R, i = √-1} = Set of complex numbers

R = Set of real numbers

Relation Φ from C to R is defined as x Φ y ⇔ |x| = y ∀ x ∈ c, y ∈ R

(i) (1 + i) Φ 3

|1 + i| = √(1^{2} + 1^{2} ) = √2 ≠ 3

Hence, (1 + i) Φ 3 is false.

(ii) 3 Φ (-3)

|3| = 3 ≠ -3

So, the 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 a set of ordered pairs. Also find R ^{-1}. **

**Solution:**

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 ordered 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. **

**Solution:**

(i) The given relation R_{1} is defined as x = 2y and A = {1, 2, 3, 4, 5, 6} to B = {1, 2, 3}.

x = 2y

y = 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

When x = 6 then y = 3.

So, (6, 3) ∈ R_{1}

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

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

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), (11, 9)}

**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)}.

### RBSE Maths Chapter 2: Exercise 2.2 Textbook Important Questions and Solutions

**Question 1. Examine the reflexivity, symmetricity, and transitivity of the following relations: **

**(i) mR _{1}n ⇔ m and n both are odd, ∀ m, n ∈ N **

**(ii) In the power set P(A) of set AR _{2}B ⇔ A ⊆ B, ∀ A, B ∈ P(A)**

**Solution:**

(i) Given, set N = {1, 2, 3, 4, …}

A relation R_{1} in N is defined as mR_{1}n ⇔ m and n both are odd ∀ m, n ∈ N

Reflexivity:

Let m ∈ N

m ∈ N ⇒ m: even or odd number

⇒ If m is odd, (m, m) ∈ R_{1}

If m is even, (m, m) ∉ R_{1}

So, (m, m) ∉ R_{1}

R_{1} is not reflexive.

Symmetricity:

Let m, n ∈ N, mR_{1} is true.

So, mR_{1}n ⇒ m and n are both odd.

⇒ n and m are both odd

⇒ nR_{1}m

R_{1} is a symmetric relation.

Transitivity:

Let m, n, r ∈ N

mR_{1}n and nR_{1}r are true,

So, mR_{1}n and nR_{1}r ⇒ m and n are both odd and n, r are both odd.

⇒ m, n and r are odd.

⇒ mR_{1}r

R_{1} is a transitive relation.

(ii) Given set is AR_{2}B ⇔ A ⊆ B, ∀ A, B ∈ P(A)

Reflexivity:

Let A ∈ P(A)

A ∈ P(A) ⇒ A ⊆ A (∵ every set is a subset of itself)

⇒ AR_{2}A

R_{2} is a reflexive relation.

Symmetricity:

Let A, B ∈ P(A)

AR_{2}B are true,

So, AR_{2}B ⇒ A ⊆ B

⇒ B ⊄ A (until A = B)

⇒ B ⊄ A

R_{2} is not a symmetric relation.

Transitivity:

A, B, C ∈ P(A)

AR_{2}B and BR_{2}C are true.

So, AR_{2}B and BR_{2}C ⇒ A ⊆ B and B ⊆ C

⇒ A ⊆ C(A ⊆ B ⊆ C)

⇒ AR_{2}C

So, R_{2} is a transitive relation.

**Question 2: Any relation P is defined in set R _{0} of non zero real numbers by the following ways:**

**(i) xPy ⇔ (x + y) is a rational number**

**(ii) xPy ⇔ x / y is a rational number**

**Test the reflexivity, symmetricity and transitivity of these relations.**

**Solution:**

(i) The given set R_{0} = Set of real numbers.

Relation P in R_{0} is defined as xPy ⇔ x + y is a rational number ∀ x, y ∈ R_{0}

Reflexivity :

Let x ∈ R_{0}

x ∈ R_{0} ⇒ x + x need not be rational

For example √3 ∈ R_{0} ⇒ √3 + √3 = 2√3 is an irrational number.

So, (x, x) ∉ P

P is not a reflexive relation.

Symmetricity:

Let a, b ∈ R_{0} 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 ∈ R_{0}

P is a symmetric relation.

Transitivity:

Let a, b, c ∈ R_{0}

(a, b) ∈ P ⇒ a + b is a rational number

(b, c) ∈ P ⇒ b + c is a rational number

⇒ a + c is not necessarily a rational number

For example: 2 + √3, -√3 + 6, √3 + √7 ∈ R_{0} 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.

(ii) The given Set: R_{0} = Set of real numbers

Relation P in R_{0} is defined as a rational number

xPy = x / y is a rational number ∀ x, y ∈ R_{0}

Reflexivity:

Let a ∈ R_{0}

a ∈ R_{0}

⇒ [a / a] = 1 is a rational number

⇒ (a, a) ∈ P

P is a reflexive number.

Symmetricity:

Let a, b ∈ R_{0} is in this way (a, b) ∈ P

(a, b) ∈ P

⇒ [a / b] is a rational number

⇒ [b / a] is a rational number

⇒ (b, a) ∈ P

(a, b) ∈ P

⇒ (b, a) ∈ P ∀ a, b ∈ R_{0}

P is a symmetric relation.

Transitivity:

Let a, b, c ∈ R_{0} 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 two rational numbers 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 ∈ R_{0}

P is a transitive relation.

**Question 3: A relation R _{1} is defined on the set R of real numbers in the following way: **

**(a, b) ∈ R _{1} ⇔ 1 + ab > 0, ∀ a, b ∈ R **

**Prove that R _{1} is reflexive and symmetric but not transitive. **

**Solution:**

The given set is

Set R = set of real numbers.

Relation R_{1} in R is defined as

(a, b) ∈ R_{1} ⇒ 1 + ab > 0 ∀ a, b ∈ R

Reflexivity:

Let a ∈ R

a ∈ R ⇒ 1 + a.a > 0

⇒ a . a ∈ R_{1} ∀ a ∈ R

R_{1} is a reflexive relation.

Symmetricity:

Let a, b ∈ R in this way

(a, b) ∈ R_{1}

(a, b) ∈ R_{1}

⇒ 1 + ab > 0

⇒ 1 + b . a > 0 [ab = ba]

⇒ (b, a) ∈ R_{1}

So, (a, b) ∈ R_{1}

⇒ (b, a) ∈ R_{1} ∀ a, b ∈ R

R_{1} is a symmetric relation.

Transitivity:

Let a, b, c ∈ R is in this way

(a, b) ∈ R_{1} and (b, c) ∈ R_{1}

(a, b) ∈ R_{1} ⇒ 1 + ab > 0

(b, c) ∈ R_{1} ⇒ 1 + bc > 0

But 1 + ac > 0 is not necessary.

For example: 1, ½ , -1 ∈ R and (1, ½) ∈ R_{1} because

1 + 1 * ½ = 1 + ½ = [3 / 2] > 0

(½ , -1) ∈ R_{1} because

1 + ½ * (-1) = 1 – ½ = ½ > 0

But (1, -1) ~~∈~~ R_{1} because

1 + 1 * (-1) = 1 – 1 = 0 [not greater than 0]

R_{1} is not a transitive relation.

**Question 4: N is a set of natural numbers. If a relation R is defined on 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:**

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 an equivalence relation, it should be proved that R is reflexive, symmetric and transitive.

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 a reflexive relation.

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.

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.

Thus, from the above three results, the given relation is an equivalence relation.

**Question 5: A relation R is defined in a set Q _{0} set of non zero rational numbers such that aRb ⇔ a = 1 / b, ∀ a, b ∈ Q_{0}. Is R an equivalence relation? **

**Solution:**

Set Q_{0} = Set of non zero rational numbers

A relation in Q_{0} is defined as aRb ⇔ a = 1 / b ∀ a, b ∈ Q_{0}

If R is reflexive, symmetric and transitive, then R is an equivalence relation.

Reflexivity:

Let a ∈ Q_{0}

a ∈ Q_{0} ⇒ a ≠ (1 / a) (a ≠ 1)

⇒ (a, a) ∉ R ∀ a ∉ Q_{0}

Hence, R is not a reflexive relation.

So, R is not an equivalence relation.

**Question 6. Let {(a, b) | a, b ∈ R} where I is a set of integers. Relation R _{1 }on x is defined in the following way: (a, b) R_{1}(c, d) ⇒ b – d = a – c. **

**Prove that R _{1} is an equivalence relation. **

**Solution:**

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, it should be proved that R is reflexive, symmetric and transitive.

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.

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 relation.

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 (1) 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.

Thus, from the above three results, the given relation is an equivalence relation.

**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:**

Set T = {Set of similar triangles}

A relation R in T is defined as xRy ⇔ x is similar to y ∀ x, y ∈ T

To prove R is an equivalence relation, it should be proved that R is reflexive, symmetric and transitive.

Reflexivity:

Let x ∈ T

x ∈ f

⇒ x is similar to x.

⇒ (x, x) ∈ R ∀ x ∈ T

R is a reflexive relation.

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.

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.

Thus, from the above three results, the given relation is an equivalence relation.

**Question 8. Let a relation R be 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:**

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)}

Reflexivity:

Here (1, 1), (2, 2), (3, 3) ∈ R

So, ∀ a ∈ A ⇒ (a, a) ∈ R

R is reflexive.

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.

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.

### RBSE Maths Chapter 2: Exercise 2.3 Textbook Important Questions and Solutions

**Question 1: Examine whether the following 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)}**

**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.

**Question 2: If f : R → R, f (x) = x ^{2}, 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) = x^{2} then

If x < 0 ⇒ x^{2} > 0

x = 0 ⇒ x^{2} = 0

x > 0 ⇒ x^{2} > 0

So, f (x) = x^{2} ≥ 0 ∀ x ∈ R

Hence, range of R = R^{+} ∪ {0} or {x ∈ R | 0 ≤ x < ∞}

(ii) f (x) = 4

⇒ x^{2} = 4

⇒ x = ± 2

Hence, {x : y (x) = 4} = {-2, 2}.

(iii) f (y) = -1

⇒ y^{2} = -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) = x ^{2} + 1. Find the range of f. **

**Solution:**

Given, A = {-2, -1, 0, 1, 2} and R = set of real numbers

f (x) = x^{2} + 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) = x ^{2} + 2x – 3, then find the range of f. **

**Solution:**

A = {-2, -1, 0, 1, 2} and Z = {0, ± 1, ± 2, …}

f (x) = x^{2} + 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) = 0^{2} + 2 (0) – 3 = -3

f (1) = 1^{2} + 2 (1) – 3 = 1 + 2 – 3 = 0

f (2) = (2)^{2} + 2 (2) – 3 = 4 + 4 – 3 = 5

Hence, the range of f = set of f (x) = {-4, -3, 0, 5}.

### RBSE Maths Chapter 2: Exercise 2.4 Textbook Important Questions and Solutions

**Question 1: Classify the following functions into one-one, many one, into and onto. Justify your answer. **

**(i) f : Q → Q, f (x) = 3x + 7 **

**(ii) f : C → R, f (x + iy) = x**

**Solution:**

(i) f : Q → Q and f (x) = 3x + 7 where Q is a set of rational numbers.

Let x_{1}, x_{2} ∈ Q are in this way

f (x_{1}) = f (x_{2})

⇒ 3x_{1} + 7 = 3x_{2} + 7

⇒ 3x_{1 }= 3x_{2}

⇒ x_{1} = x_{2}

⇒ f (x_{1}) = f (x_{2})

⇒ x_{1} = x_{2}

x_{1}, x_{2} ∈ Q

Hence, f is one-one function.

Let y ∈ Q be the co-domain and the pre-image of y be x in domain Q then

f (x) = y

⇒ 3x + 7 = y

⇒ x = [y – 7] / [3] ∈ Q

Hence, 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 ; C = set of complex numbers and 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, the two different elements of domain R have the same image.

Hence, f is a many-one function.

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.

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) g (x) = |x|

(ii) h (x) = x^{2}

Solution:

(i) g : A → B, g (x) = |x|

Let x_{1}, x_{2} ∈ A

If f (x_{1}) = f (x_{2}),

⇒ g (x_{1}) = g (x_{2})

⇒ |x_{1}| = |x_{2}| ⇒ x_{1} = ± x_{2}

g is not one-one function

g is many one function

Range of g = {x: -1 ≤ x ≤ 1} ≠ B (co-domain)

The pre-image of a negative number does not exist in codomain B.

Thus, it is proved that f is many-one, into function.

g is not a one-one function.

(ii) h : A → B, h (x) = x^{2}

Let x_{1}, x_{2} ∈ A

Thus, h (x_{1}) = h (x_{2})

⇒ x_{1}^{2} = x^{2}_{2}

⇒ x_{1} = ± x_{2}

h is not a one-one function.

h is a many-one function.

Let y ∈ B if possible let the pre-image of y be x which exists in B, then

h (x) = y

⇒ x^{2} = y

⇒ x = ±√y

When y is positive, x does not exist.

The pre-image of a negative number does not exist.

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.

**Question 3: If f : C → C, f (x + iy) = (x – iy), then prove that f is a one-one onto function. **

**Solution:**

f : C → C and f (x + iy) = x – iy where C is a set of complex numbers.

Let x_{1} + iy_{1} and x_{2} + iy_{2} ∈ C

f (x_{1} + iy_{1}) = f (x_{2} + iy_{2})

⇒ x_{1} – iy_{1} = x_{2} – iy_{2}

([x_{1} – iy_{1}] * [x_{1} + iy_{1}]) / [x_{1} + iy_{1}] = ([x_{2} – iy_{2}] * [x_{2} + iy_{2}]) / [x_{2} + iy_{2}]

x_{1} + iy_{1} = x_{2} + iy_{2}

f (x_{1} + iy_{1}) = f (x_{2} + iy_{2}) = x_{1} + iy_{1} = x_{1} + iy_{1} + x_{2} + iy_{2}

So, f is a 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.

**Question 4: Give one example for each of the following functions: **

**(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 : R_{0} → R^{+}, f (x) = x^{2}

(iii) f : z_{0} → N, f (x) = |x|

(iv) f : Z → Z, f (x) = 2x

(v) f : R → R_{1}, f (x) = x^{2}

(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 it becomes: **

**(i) One-one into **

**(ii) Many-one onto **

**(iii) One-one onto **

**Solution:**

f : R → R, f (x) = cos x

Let x_{1}, x_{2} ∈ R

Thus, f (x_{1}) = f (x_{2})

⇒ cos x_{1} = cos x_{2}

⇒ x_{2} = 2nπ ± x_{2}

f is not one-one function.

Let y ∈ R (co-domain) and let the pre-image of y be x in domain R, then

f (x) = y

⇒ cos x = y

⇒ x = cos^{-1} y

x exist only 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 f _{1}, f_{2} are functions defined as f_{1} : N → O, f_{1}(x) = 2x – 1; f_{2} : N → E, f_{2}(x) = 2x Then prove that f_{1} and f_{2} are one-one onto. **

**Solution:**

(i) N = {1, 2, 3, …} O = {1, 3, 5, …}

f (x) = 2x – 1

Let x_{1}, x_{2} ∈ N

Thus f (x_{1}) = f (x_{2})

⇒ 2x_{1} – 1 = 2x_{2} – 1

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2} ∀ x_{1}, x_{2} ∈ N

So, f_{1} is a one-one function.

Again, let y ∈ O (co-domain) and let the pre-image of y be x which exists in domain N, then

f (x) = y

⇒ 2x – 1 = y

⇒ x = [y + 1] / [2] ∈ N ∀ y ∈ O

So, the pre-image of every element of O (co-domain) exist in domain N

f1 is onto function.

So, f is one-one onto function.

(ii) N = {1, 2, 3, …} E = {2, 4, 6, …}

f_{2} : N → E, f_{2}(x) = 2x ∀ x ∈ N

Let x_{1}, x_{2} ∈ N

Thus f_{2} (x_{1}) = f_{2} (x_{2})

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2} ∀ a, b ∈ N

f_{2} is one-one function.

Again, let y ∈ E (co-domain) and let the pre-image of y be x which exists in domain N then

f_{2 }(x) = y

⇒ 2x = y

⇒ x = [y / 2] ∈ N ∀ y ∈ E

So, pre-image of every element of E (co-domain) exist in domain N.

f_{2} is one-one, onto function.