## 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 ∧ y^{2} = x}

(ix) {(x, y) | x, y ∈ R ∧ x^{2} = y}

(x) {(x, y) | x, y ∈ R ∧ x = y^{3}}

(xi) {(x, y) | x, y ∈ R ∧ y = x^{3}}

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, y^{2} = x}

Here y^{2} = 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, x^{2} = y}

It is a function because for y = x^{2}, each real value of x there is a unique image in R for each element of R

(x) {(x, y) : x, y ∈ R, x = y^{3}}

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

(xi) {(x, y) : x, y ∈ R, y = x^{3}}

It is also a function because for y = x^{3} ∀ x ∈ R unique image is in set 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

Given: 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

(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) = 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, Range of f = set of f(x) = {-4, -3, 0, 5}

(ii) Let pre-image of 6 is x

f(x) = 6

⇒ x^{2} + 2x – 3 = 6

⇒ x^{2} + 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

⇒ x^{2} + 2x – 3 = -3

⇒ x^{2} + 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

⇒ x^{2} + 2x – 3 = 5

⇒ x^{2} + 2x – 8 = 0

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