Class 11 Maths NCERT Solutions for Chapter 2 Relations and Functions Exercise 2.3
Relations and Functions Exercise 2.3 Solutions
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) {(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.
2. Find the domain and range of the following real function:
(i) f(x) = –|x|
(ii) f(x) = √(9 – x2)
Solution
(i) f(x) = - |x| , x ∈ R 
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) f(x) = √(9 – x2)
Since √(9 – x2) is 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].
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
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
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
The given function is t(C) = 9C/5 + 32 .
Therefore, 
(iv) It is given that t(C) = 212 
Thus, the value of t, when t(C) = 212, is 100.
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
Solution
(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 |
… |
i.e., range of f = (–∞, 2)
Alter:
Let x > 0
⇒ 3x > 0
⇒ 2 –3x < 2
⇒ f(x) < 2
∴Range of f = (–∞, 2)
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 |
... |
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,)
It is clear that the range of f is the set of all real numbers.
∴ Range of f = R