NCERT Solutions Class 11 Maths
The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions Exercise 2.3 has been provided here to help the students in solving the questions from this exercise.
Chapter 2 (Relations and Functions)
| Exercise – 2.3 |
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)}
Here, each element in domain is having unique/distinct image. So, the given relation is a function.
Domain = {2, 5, 8, 11, 14, 17}
Range of the function = {1}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
Here, each element in domain is having unique/distinct image. So, the given relation is a function.
Domain = {2, 5, 8, 11, 14, 17}
Range of the function = {1}
(iii) {(1, 3), (1, 5), (2, 5)}
This relation is not a function since an element 1 corresponds to two elements/images i.e, 3 and 5.
Hence, this relation is not a function.
2. Find the domain and range of the following real function:
(i) f(x) = –|x|
Solution –
(ii) f(x) = √(9 – x2)
Solution –
(i) Given, f(x) = –|x|, x ∈ R
We know that, |x| =
| x | if x >= 0 |
| -x | if x < 0 |
Here f(x) = -x =
| -x | x >= 0 |
| x | 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)
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.
|x| <=3
So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or Domain of f = [–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 we can say Range of f = [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 – Given, Function, 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 – Here in ques , it is given that :
t(C) = 9C/5 + 32
So,
(i) t(0) = 9(0) / 5 + 32
= 0 + 32
= 32
(ii) t(28) = 9(28) / 5 + 32
Taking LCM and solving,
= ( 252 +160 ) / 5
= 412 / 5
(iii) t(-10) = 9(-10) / 5 + 32
= -18 + 32
= 14
(iv) Here , in this ques we have to find the value of C.
Given that , t(C) = 212,
9C / 5 + 32 = 212
9C / 5 = 180
9C = 180 × 5
C = (180 × 5)/9
C = 100
The value of C 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) Given 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. that is range of
f = (- ∞, 2)
(ii) Given, 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. that is range of
f = [2, ∞]
(iii) Given, f(x) = x, x is a real number
It is clear that the range of f is the set of all real numbers. Therefore, the Range of f = R.
