NCERT Solutions for Exercise 2.3, Class 11, Maths

Exercise 2.3 Relations and Functions
Question and Answers
Class 11 – Maths

Class	Class 11
Subject	Mathematics
Chapter Name	Relations and Functions
Chapter No.	Chapter 2
Exercise	Exercise 2.3
Category	Class 11 Maths NCERT Solutions

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)}

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

As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called 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)}

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

As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called 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)}.

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

It’s seen that the same first element, i.e., 1, corresponds to two different images, i.e., 3 and 5; this relation cannot be called a function.

Question 2 Find the domain and range of the following real functions:

(i) f(x) = – x

(i) We have the given function as, f (x) = − |x|

It is also know that, | x | =  x , if x ≥ 0

                                        −x ,if x < 0

Thus, f(x) = − | x | = − x ,if x ≥ 0

                                    x ,if x < 0  

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

It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.

Therefore, the range of f is given by (–∞, 0].

(ii) f(x) =  √9 − x² 

(ii) As √(9 – x²) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x² ≥ 0.

So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3]

For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.

Therefore, 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)

(i) Function, f(x) = 2x – 5

f (0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f (7)

(ii) Function, f(x) = 2x – 5

f(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) f (–3)

(iii) Function, f(x) = 2x – 5

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 t(C) = 9C/ 5 + 32. Find

(i) t(0)

(i) We have our given function as, t(C)= (9C/5) +32

To find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

(ii) We have our given function as, t(C)= (9C/5) +32

To find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

t (28) =[  (9 × 28 ) /5 ] + 32

t (28 ) =( 252 + 160) /5

t (28 ) =(412 /5 ] 

t (28 ) = 82.4

(iii) t(–10)

(iii) We have our given function as, t(C)= (9C/5) +32

To find our needed values of the function we just have to put the values in the given function and simplify it.

So, we get now,

t ( -10 ) = [ (9 × (-10) ) /5 ] + 32

t (-10) =−18+32

t (-10) = 14

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

(iv) We have our given function as, t(C)= (9C/5) +32

we are given that, t(C)=212

212 = (9C/5) +32

9C/5 = 212−32

9C/5 = 180

C = 900 / 9 = 100

Thus , it can be said that, for t(C)=212 the value of t is 100

Question 5 Find the range of each of the following functions.

(i) f (x) = 2 – 3x, x ∈ R, x > 0.

(i) f(x) = 2 – 3x, x ∈ R, x > 0

We can now see, it can be seen that the elements of the range is less than 2.

So, the range will be, f = (−∞ ,2 )

x > 0,  So ,  3x > 0

-3x < 0 [Multiplying by -1 on both sides, the inequality sign changes]

2 – 3x < 2

Therefore, the value of 2 – 3x is less than 2.

Hence, Range = (–∞, 2)

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

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

So, we see that the range of the function f is the set of all numbers which are greater than or equal to 2.

Thus we can conclude that the range of the function is, [2,∞]

x² ≥ 0

So,

x² + 2 ≥ 2 [Adding 2 on both sides]

Therefore, the value of x² + 2 is always greater or equal to 2, for x is a real number.

Hence, Range = [2, ∞)

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

(iii) f(x) = x, x is a real number

Clearly, the range of f is the set of all real numbers.

Thus,

Range of f = R