**NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.3**

**Ex 2.3 Class 11 Maths Question 1.**

Which of the following relations are functions? Give reasons. If it is a
function, determine its domain and range.**(i)**{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

**(ii)**{{2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

**(iii)**{(1, 3), (1, 5), (2, 5)}

**Solution.**

**(i) **Given relation R = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1),
(17, 1)}

Since 2, 5, 8, 11, 14, 17 are the elements of domain of R having their
unique images.

∴ The given relation is a function.

Hence, domain = {2, 5, 8, 11, 14, 17) and Range = {1}.

**(ii)** Given relation R = {(2, 1), (4, 2),
(6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Since 2, 4, 6, 8, 10, 12, 14 are the elements of domain of R having their
unique images.

∴ The given relation is a function.

Hence, domain = {2, 4, 6, 8, 10, 12, 14} and Range = {1, 2, 3, 4, 5, 6, 7}.

**(iii)** Given relation R = {(1, 3), (1, 5),
(2, 5)}

Since the distinct ordered pairs (1, 3) and (1, 5) have the same first element,
i.e., 1 does not have a unique image under R.

∴ It is not a function.

**Ex 2.3 Class 11 Maths Question 2.**

Find the domain and range of the following real functions:**(i)**f(x) = −|x|

**(ii)**f(x) = √9 − x

^{2}

**Solution.**

**Ex 2.3 Class 11
Maths Question 3.**

A function f is defined by f(x) = 2x – 5. Write down the values of**(i)**f(0)

**(ii)**f(7)

**(iii)**f(-3)

**Solution.**

Given f(x) = 2x – 5

**(i)** f(0)
= 2(0) – 5 = 0 – 5 = –5

**(ii)** f(7)
= 2(7) – 5 = 14 – 5 = 9

**(iii)** f(-3)
= 2(–3) – 5 = –6 – 5 = –11

**Ex 2.3 Class 11 Maths Question 4.**

The function ‘t’ which maps temperature in degree Celsius into temperature in
degree Fahrenheit is defined by t(C) = 9C/5 + 32.Find:

**(i)**t(0)

**(ii)**t(28)

**(iii)**t(-10)

**(iv)**The value of C, when t(C) = 212.

**Solution.**

**Ex
2.3 Class 11 Maths Question 5.**

Find the range of each of the following functions.**(i)**f(x) = 2 – 3x, x ∈ R, x > 0.

**(ii)**f(x) = x

^{2 }+ 2, x is a real number.

**(iii)**f(x) = x, x is a real number.

**Solution.**

**(i)** Given
f(x) = 2 – 3x, x ∈ R, x > 0

∵ x > 0 ⇒ -3x < 0 ⇒ 2 – 3x < 2 + 0 ⇒ f(x) < 2

∴ The range of f(x) is (-2).

**(ii)** Given f(x) = x^{2} +
2, x is a real number

We know x^{2 }≥ 0 ⇒
x^{2} + 2 ≥ 0 + 2

⇒ x^{2} + 2 > 2 ∴ f(x) ≥ 2

∴ The range of f(x) is [2, ∞).

**(iii)** Given f(x) = x, x is a real number.

Let y = f(x) = x ⇒ y = x

∴ Range of f(x) = Domain of f(x)

∴ Range of f(x) is R.

**Related Links:**

**NCERT Solutions for Maths Class 9**