# RBSE Maths Class 11 Chapter 2: Relation and function Important Questions and Solutions

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 R1 = {(1, 4), (3, 5), (3, 6)}

(1, 4) ∈ A × B

(3, 5) ∈ A × B

(3, 6) ∈ A × B

R1 ⊆ A × B

So, R1 is a relation from A to B.

(ii) 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

So, R3 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 nth term = a + (n – 1 ) * d, where a is first term and d is a common difference.

Tn = 3 + (n – 1) × 2

= 3 + 2n – 2

= 2n + 1

The required rule by putting n = x and Tn = 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.

Tn = 6 + (n – 1) × (-2)

Tn = 6 – 2n + 2

Tn = 8 – 2n

The required rule by putting x = y and Tn = 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.

Tn = 2 + (n – 1) × 1

= 2 + n – 1

= n + 1

By putting n = x and Tn = 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 ⇔ x2 + y2 = 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 ⇔ x2 + y2 = 25 ∀ x, y ∈ Z

When x = 0

02 + y2 = 25

y = ± 5

⇒ (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 = 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| = √(12 + 12 ) = √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) 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.

Solution:

(i) The given relation R1 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) ∈ R1

When x = 4, then y = 2.

So, (4, 2) ∈ R1 and

When x = 6 then y = 3.

So, (6, 3) ∈ R1

R1 = {(2, 1), (4, 2), (6, 3)}

(ii) The given relation R2 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) ∈ 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), (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) 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)

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

⇒ If m is odd, (m, m) ∈ R1

If m is even, (m, m) ∉ R1

So, (m, m) ∉ R1

R1 is not reflexive.

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 are 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 a 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

R2 is a 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

So, R2 is a transitive relation.

Question 2: Any relation P is defined in set R0 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 R0 = Set of real numbers.

Relation P in R0 is defined as xPy ⇔ x + y is a rational number ∀ x, y ∈ R0

Reflexivity :

Let x ∈ R0

x ∈ R0 ⇒ x + x need not be rational

For example √3 ∈ R0 ⇒ √3 + √3 = 2√3 is an irrational number.

So, (x, x) ∉ P

P is not a 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

(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 ∈ R0 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: 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

⇒ (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 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 ∈ R0

P is a transitive relation.

Question 3: A relation R1 is defined on the set R 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:

The given set is

Set R = set of real numbers.

Relation R1 in R is defined as

(a, b) ∈ R1 ⇒ 1 + ab > 0 ∀ a, b ∈ R

Reflexivity:

Let a ∈ R

a ∈ R ⇒ 1 + a.a > 0

⇒ a . a ∈ R1 ∀ a ∈ R

R1 is a reflexive relation.

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.

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, ½ , -1 ∈ R and (1, ½) ∈ R1 because

1 + 1 * ½ = 1 + ½ = [3 / 2] > 0

(½ , -1) ∈ R1 because

1 + ½ * (-1) = 1 – ½ = ½ > 0

But (1, -1) R1 because

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

R1 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)

⇒ 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 Q0 set of non zero rational numbers such that aRb ⇔ a = 1 / b, ∀ a, b ∈ Q0. Is R an equivalence relation?

Solution:

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, then R is an equivalence relation.

Reflexivity:

Let a ∈ Q0

a ∈ Q0 ⇒ a ≠ (1 / a) (a ≠ 1)

⇒ (a, a) ∉ R ∀ a ∉ Q0

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 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:

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) = 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

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 the range of f.

Solution:

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

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, 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 x1, x2 ∈ Q are in this way

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.

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) = x2

Solution:

(i) 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

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) = x2

Let x1, x2 ∈ A

Thus, h (x1) = h (x2)

⇒ x12 = x22

⇒ x1 = ± x2

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

⇒ x2 = 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 x1 + iy1 and x2 + iy2 ∈ C

f (x1 + iy1) = f (x2 + iy2)

⇒ x1 – iy1 = x2 – iy2

([x1 – iy1] * [x1 + iy1]) / [x1 + iy1] = ([x2 – iy2] * [x2 + iy2]) / [x2 + iy2]

x1 + iy1 = x2 + iy2

f (x1 + iy1) = f (x2 + iy2) = x1 + iy1 = x1 + iy1 + x2 + iy2

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 : 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 it becomes:

(i) One-one into

(ii) Many-one onto

(iii) One-one onto

Solution:

f : R → R, f (x) = cos x

Let x1, x2 ∈ R

Thus, f (x1) = f (x2)

⇒ cos x1 = cos x2

⇒ x2 = 2nπ ± x2

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 f1, f2 are functions 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) N = {1, 2, 3, …} O = {1, 3, 5, …}

f (x) = 2x – 1

Let x1, x2 ∈ N

Thus f (x1) = f (x2)

⇒ 2x1 – 1 = 2x2 – 1

⇒ 2x1 = 2x2

⇒ x1 = x2 ∀ x1, x2 ∈ N

So, f1 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, …}

f2 : N → E, f2(x) = 2x ∀ x ∈ N

Let x1, x2 ∈ N

Thus f2 (x1) = f2 (x2)

⇒ 2x1 = 2x2

⇒ x1 = x2 ∀ a, b ∈ N

f2 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

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.

f2 is one-one, onto function.