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

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

Page No 44:

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)}

Answer:

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

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

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

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

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

Question 2:

Find the domain and range of the following real function:

(i) f(x) = –|x| (ii) 

Answer:

(i) f(x) = –|x|, x ∈ R

We know that |x| = 

Since f(x) is defined for x ∈ R, the domain of f is R.

It can be observed that the range of f(x) = –|x| is all real numbers except positive real numbers.

∴The range of f is (–, 0].

(ii) 

Sinceis defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].

For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

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)

Answer:

The given function is f(x) = 2x – 5.

Therefore,

(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

Question 4:

The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by.

Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212

Answer:

The given function is.

Therefore,

(i) 

(ii) 

(iii) 

(iv) It is given that t(C) = 212

Thus, the value of t, when t(C) = 212, is 100.

Question 5:

Find the range of each of the following functions.

(i) f(x) = 2 – 3x, x ∈ R, x > 0.

(ii) f(x) = x² + 2, x, is a real number.

(iii) f(x) = x, x is a real number

Answer:

(i) f(x) = 2 – 3x, x ∈ R, x > 0

The values of f(x) for various values of real numbers x > 0 can be written in the tabular form as

x	0.01	0.1	0.9	1	2	2.5	4	5	…
f(x)	1.97	1.7	–0.7	–1	–4	–5.5	–10	–13	…

Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2.

i.e., range of f = (–, 2)

Alter:

Let x > 0

⇒ 3x > 0

⇒ 2 –3x < 2

⇒ f(x) < 2

∴Range of f = (–, 2)

(ii) f(x) = x² + 2, x, is a real number

The values of f(x) for various values of real numbers x can be written in the tabular form as

x	0	±0.3	±0.8	±1	±2	±3		…
f(x)	2	2.09	2.64	3	6	11		…..

Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2.

i.e., range of f = [2,)

Alter:

Let x be any real number.

Accordingly,

x² ≥ 0

⇒ x² + 2 ≥ 0 + 2

⇒ x² + 2 ≥ 2

⇒ f(x) ≥ 2

∴ Range of f = [2,)

(iii) f(x) = x, x is a real number

It is clear that the range of f is the set of all real numbers.

∴ Range of f = R