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

Chapter 2, Relations and Functions of Class 11 Maths, is categorised under the CBSE Syllabus for 2023-24. The miscellaneous exercise contains extra questions covering the entire topic in the chapter. The last Exercise of NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions based on all the topics in the chapter, which are as follows:

1. Cartesian Product of Sets
2. Relations
3. Functions
   1. Some functions and their graphs
   2. Algebra of real functions

Every student aims to score high marks in their exam. Learn with the help of NCERT Solutions for Class 11 to understand the method of answering a question, which will help in scoring high marks in the board exam.

NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions Miscellaneous Exercise

Download PDF

carouselExampleControls112

Previous





Next

Access other exercise solutions of Class 11 Maths Chapter 2 – Relations and Functions

The exercise-wise problems of NCERT Class 11 Maths Solutions can be accessed below.

Exercise 2.1 Solutions 10 Questions

Exercise 2.2 Solutions 9 Questions

Exercise 2.3 Solutions 5 Questions

Access Solutions for Class 11 Maths Chapter 2 Miscellaneous Exercise

1. The relation f is defined by 

The relation g is defined by 

Show that f is a function and g is not a function.

Solution:

The given relation f is defined as

It is seen that for 0 ≤ x < 3,

f(x) = x² and for 3 < x ≤ 10,

f(x) = 3x

Also, at x = 3

f(x) = 3² = 9 or f(x) = 3 × 3 = 9

i.e., at x = 3, f(x) = 9 [Single image]

Hence, for 0 ≤ x ≤ 10, the images of f(x) are unique.

Therefore, the given relation is a function.

Now,

In the given relation g is defined as

It is seen that, for x = 2

g(x) = 2² = 4 and g(x) = 3 × 2 = 6

Thus, element 2 of the domain of the relation g corresponds to two different images, i.e., 4 and 6.

Therefore, this relation is not a function.

2. If f(x) = x², find

Solution:

Given,

f(x) = x²

Hence,

3. Find the domain of the function 

Solution:

Given function,

.

It is clearly seen that the function f is defined for all real numbers except at x = 6 and x = 2, as the denominator becomes zero otherwise.

Therefore, the domain of f is R – {2, 6}.

4. Find the domain and the range of the real function f defined by f(x) = √(x – 1).

Solution:

Given real function,

f(x) = √(x – 1)

Clearly, √(x – 1) is defined for (x – 1) ≥ 0.

So, the function f(x) = √(x – 1) is defined for x ≥ 1.

Thus, the domain of f is the set of all real numbers greater than or equal to 1.

Domain of f = [1, ∞).

Now,

As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0

Thus, the range of f is the set of all real numbers greater than or equal to 0.

Range of f = [0, ∞).

5. Find the domain and the range of the real function f defined by f (x) = |x – 1|.

Solution:

Given, a real function,

f (x) = |x – 1|

Clearly, the function |x – 1| is defined for all real numbers.

Hence,

Domain of f = R

Also, for x ∈ R, |x – 1| assumes all real numbers.

Therefore, the range of f is the set of all non-negative real numbers.

6. Let be a function from R into R. Determine the range of f.

Solution:

Given function,

Substituting values and determining the images, we have

The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.

Or,

We know that, for x ∈ R,

x² ≥ 0

Then,

x² + 1 ≥ x²

1 ≥ x² / (x² + 1)

Therefore, the range of f = [0, 1)

7. Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.

Solution:

Given, the functions f, g: R → R is defined as

f(x) = x + 1, g(x) = 2x – 3

Now,

(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

Thus, (f + g) (x) = 3x – 2

(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

Thus, (f – g) (x) = –x + 4

f/g(x) = f(x)/g(x), g(x) ≠ 0, x ∈ R

f/g(x) = x + 1/ 2x – 3, 2x – 3 ≠ 0

Thus, f/g(x) = x + 1/ 2x – 3, x ≠ 3/2

8. Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Solution:

Given, f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

And the function defined as, f(x) = ax + b

For (1, 1) ∈ f

We have, f(1) = 1

So, a × 1 + b = 1

a + b = 1 …. (i)

And for (0, –1) ∈ f

We have f(0) = –1

a × 0 + b = –1

b = –1

On substituting b = –1 in (i), we get

a + (–1) = 1 ⇒ a = 1 + 1 = 2

Therefore, the values of a and b are 2 and –1, respectively.

9. Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b²}. Are the following true?

(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

Justify your answer in each case.

Solution:

Given relation R = {(a, b): a, b ∈ N and a = b²}

(i) It can be seen that 2 ∈ N; however, 2 ≠ 2² = 4

Thus, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3²

Now, 3 ≠ 9² = 81; therefore, (3, 9) ∉ N

Thus, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) It’s clearly seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4² and 4 = 2²

Now, 16 ≠ 2² = 4; therefore, (16, 2) ∉ N

Thus, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

10. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?

(i) f is a relation from A to B (ii) f is a function from A to B.

Justify your answer in each case.

Solution:

Given,

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

So,

A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

Also, given that

f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It’s clearly seen that f is a subset of A × B.

Therefore, f is a relation from A to B.

(ii) As the same first element, i.e., 2 corresponds to two different images (9 and 11), relation f is not a function.

11. Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.

Solution:

Given, the relation f is defined as

f = {(ab, a + b): a, b ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.

As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f

i.e., (12, 8), (12, –8) ∈ f

It’s clearly seen that the same first element, 12, corresponds to two different images (8 and –8).

Therefore, the relation f is not a function.

12. Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Solution:

Given,

A = {9, 10, 11, 12, 13}

Now, f: A → N is defined as

f(n) = The highest prime factor of n

So,

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

Thus, it can be expressed as

f(9) = The highest prime factor of 9 = 3

f(10) = The highest prime factor of 10 = 5

f(11) = The highest prime factor of 11 = 11

f(12) = The highest prime factor of 12 = 3

f(13) = The highest prime factor of 13 = 13

The range of f is the set of all f(n), where n ∈ A.

Therefore,

Range of f = {3, 5, 11, 13}