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 2023-24. 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 are 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.

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

For more problems regarding the NCERT Class 11 Solutions Maths Chapter 2, refer to the links here.

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)}

Itâ€™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)Â f(x) = â€“|x| (ii) f(x) = âˆš(9 â€“ x2)Â

Solution:

(i)Â Given,

f(x) = â€“|x|,Â xÂ âˆˆÂ R

We know that

AsÂ f(x) is defined forÂ xÂ âˆˆÂ R, the domain ofÂ fÂ isÂ R.

It is also seen that the range ofÂ f(x) = â€“|x| is all real numbers except positive real numbers.

Therefore, the range ofÂ f is given by (â€“âˆž, 0].

(ii)Â f(x) = âˆš(9 â€“ x2)

As âˆš(9 â€“ x2)Â is defined for all real numbers that are greater than or equal to â€“3 and less than or equal to 3, for 9 â€“ x2 â‰¥ 0

So, the domain ofÂ f(x) is {x: â€“3 â‰¤Â x â‰¤ 3} or [â€“3, 3]

Now,

For any value ofÂ xÂ in the range [â€“3, 3], the value ofÂ f(x) will lie between 0 and 3.

Therefore, the range ofÂ f(x) is {x: 0 â‰¤Â x â‰¤ 3} or [0, 3]

3. A functionÂ fÂ is defined byÂ f(x) = 2xÂ â€“ 5. Write down the values of

(i)Â f(0), (ii)Â f(7), (iii)Â f(â€“3)

Solution:

Given,

Function, f(x) = 2x â€“ 5

Therefore,

(i)Â f(0) = 2 Ã— 0 â€“ 5 = 0 â€“ 5 = â€“5

(ii)Â f(7) = 2 Ã— 7 â€“ 5 = 14 â€“ 5 = 9

(iii)Â f(â€“3) = 2 Ã— (â€“3) â€“ 5 = â€“ 6 â€“ 5 = â€“11

4. The function â€˜tâ€™, which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by.

Find (i)Â tÂ (0) (ii)Â tÂ (28) (iii)Â tÂ (â€“10) (iv) The value of C, whenÂ t(C) = 212

Solution:

5. Find theÂ range of each of the following functions.

(i)Â f(x) = 2 â€“ 3x,Â xÂ âˆˆÂ R,Â x > 0

(ii)Â f(x) =Â x2Â + 2,Â x is a real number.

(iii)Â f(x) =Â x,Â xÂ is a real number.

Solution:

(i) Given,

f(x) = 2 â€“ 3x,Â xÂ âˆˆÂ R,Â x > 0

We have,

x > 0

So,

3x > 0

-3x < 0 [Multiplying by -1 on both sides, the inequality sign changes.]

2 â€“ 3x < 2

Therefore, the value of 2 â€“ 3x is less than 2.

Hence, Range = (â€“âˆž, 2)

(ii) Given,

f(x) =Â x2Â + 2,Â x is a real number

We know that

x2Â â‰¥ 0

So,

x2 + 2 â‰¥ 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, âˆž)

(iii) Given,

f(x) =Â x, xÂ is a real number

Clearly, theÂ range ofÂ fÂ is the set of all real numbers.

Thus,

Range ofÂ fÂ =Â R