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) = x^{2} + 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) = x^{2} + 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,
x^{2} ≥ 0
⇒ x^{2} + 2 ≥ 0 + 2
⇒ x^{2} + 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