 # NCERT Solutions for Class 11 Maths Chapter 2 - Relations and Functions Exercise 2.3

The solutions of NCERT are provided here with the aim of making the correct answers to the problems given in the textbook accessible to all the students. Chapter 2 Relations and Functions of Class 11 Maths is categorised under the CBSE Syllabus for 2022-23. Moving on, we come across the third exercise of the second chapter, which contains questions covering a variety of topics. Exercise 2.3 of NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions is based on the following topics:

1. Functions
1. Some functions and their graphs
• Identity function
• Constant function
• Polynomial function
• Rational functions
• The Modulus function
• Signum function
• Greatest integer function
2. Algebra of real functions
• Addition of two real functions
• Subtraction of a real function from another
• Multiplication by a scalar
• Multiplication of two real functions
• The quotient of two real functions

Scoring high in Maths is slightly tough but not impossible. Solving and practising the questions given in the NCERT textbook of Class 11 with the help of the NCERT Solutions of Class 11 Maths is the best method of scoring high in the Maths board exam.

### Download the PDF of NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions Exercise 2.3   ### Access other exercise solutions of Class 11 Maths Chapter 2 – Relations and Functions

Exercise 2.1 Solutions 10 Questions

Exercise 2.2 Solutions 9 Questions

Miscellaneous Exercise on Chapter 2 Solutions 12 Questions

### Access Solutions for Class 11 Maths Chapter 2.3 Exercise

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 – x2

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 – 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

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) = 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 .

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 – 3xx ∈ R, x > 0

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

(iii) f(x) = xx is a real number.

Solution:

(i) Given,

f(x) = 2 – 3xx ∈ 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) = x2 + 2, x is a real number We know that

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) 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

#### 1 Comment

1. Shikha Kumari

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