

NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.3: NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.3 focus on the concept of functions and their various types. This exercise introduces key functions such as identity functions, constant functions, polynomial functions, rational functions, modulus functions, signum functions, and greatest integer functions.
These solutions help students understand how to perform these operations effectively, providing a deeper understanding of the behavior of different types of functions and their graphs. By solving this exercise, students gain the necessary skills to handle more complex mathematical problems related to functions and are better prepared for exams.Class 11 Maths Chapter 2 Exercise 2.3 Questions and Answers PDF
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)} 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)} 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)} 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.2. Find the domain and range of the following real function.
(i) f ( x ) = –| x | (ii) f(x) = √(9 – x 2 )
Solution:
(i) Given, f ( x ) = –| x |, x ∈ R We know that
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 – x
2
)
As √(9 – x
2
) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x
2
≥ 0
So, the domain of
f
(
x
) is {
x
: –3 ≤
x
≤ 3} or [–3, 3]
Now,
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]
3. A function f is defined by f ( x ) = 2 x – 5. Write down the values of
(i) f (0), (ii) f (7), (iii) f (–3)
Solution:
Given, Function, f ( x ) = 2 x – 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
.
Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t (C) = 212
Solution:
5. Find the range of each of the following functions.
(i) f ( x ) = 2 – 3 x , x ∈ R, x > 0
(ii) f ( x ) = x 2 + 2, x is a real number.
(iii) f ( x ) = x , x is a real number.
Solution:
(i) Given, f(x) = 2 – 3 x , x ∈ R, x > 0
We have,
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) Given,
f
(
x
) =
x
2
+ 2,
x
is a real number
We know that
x
2
≥ 0
So,
x
2
+ 2 ≥ 2 [Adding 2 on both sides]
Therefore, the value of
x
2
+ 2 is always greater or equal to 2, for x is a real number.
Hence, Range = [2, ∞)
(iii) Given,
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
