NCERT Solutions for Class 11 Maths Chapter 2: Relations and Functions

NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions are presented in a detailed and easy-to-understand format in the PDF linked below. These solutions cover every exercise question comprehensively and are designed to help students prepare effectively for their exams. Developed by highly experienced teachers, the solutions strictly follow the latest CBSE Syllabus 2025–26 and focus on simplifying complex concepts for better understanding. 

These solutions enable Class 11 students to build a strong foundation in the topic of Relations and Functions. Before attempting the exercise problems, students can develop a clear grasp of the fundamental ideas. With consistent practice, they’ll confidently distinguish between relations and functions and apply the correct techniques to solve related problems.

The NCERT textbook includes several examples ahead of each exercise to demonstrate the appropriate problem-solving methods. By referring to the NCERT Solutions PDF for Chapter 2, students can review key concepts, reinforce their learning, and approach their final exams with confidence.

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions

Here is the PDF of NCERT Solutions for class 11 maths chapter 2:

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions - Question with answers

Q1. If set A = {1, 2, 3} and B = {4, 5}, then find the number of elements in A × B.

Solution:
Number of elements in A = 3
Number of elements in B = 2
So, number of elements in A × B = 3 × 2 = 6
A × B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

Q2. Let A = {a, b}, B = {1, 2}, define all possible relations from A to B.

Solution:
A × B = {(a,1), (a,2), (b,1), (b,2)} → Total = 4 elements
Number of relations = 2⁴ = 16
So, there are 16 possible relations from A to B (each subset of A × B is a relation).

Q3. Let R be a relation on set A = {1, 2, 3} defined by R = {(a, b) ∈ A × A | a < b}. Write R.

Solution:
We include only pairs where the first element is less than the second:
R = {(1, 2), (1, 3), (2, 3)}

Q4. Determine the domain and range of the relation R = {(1, 2), (2, 3), (3, 4)}.

Solution:
Domain = Set of first elements = {1, 2, 3}
Range = Set of second elements = {2, 3, 4}

Q5. State whether the following relation is a function: R = {(1, 2), (2, 3), (1, 4)}

Solution:
The first element 1 maps to both 2 and 4, which violates the definition of a function (one input → one output).
Hence, R is not a function.

Q6. Determine if the function f: R → R defined by f(x) = x² is one-one and onto.

Solution:

• One-one: No, because f(2) = 4 and f(-2) = 4 → not injective
• Onto: No, negative numbers are not in the range of x²
  So, f(x) = x² is neither one-one nor onto when domain and codomain are R.

Q7. Let f: A → B be defined by f(x) = 3x + 2, where A = {1, 2, 3}, B = {5, 8, 11}. Show f is one-one and onto.

Solution:

• f(1) = 5, f(2) = 8, f(3) = 11 → All unique → One-one
• Range = {5, 8, 11} = Codomain → Onto
  So, f is both one-one and onto.

Q8. Find the number of functions from A = {1, 2} to B = {a, b, c}.

Solution:
Each element in A can map to any of 3 elements in B.
So, number of functions = 3 × 3 = 9

Q9. Let A = {1, 2, 3} and B = {4, 5, 6}. Is f = {(1, 4), (2, 5), (3, 6)} a function?

Solution:
Yes, because every element in A appears once as the first element in each pair.
So, f is a function.

Q10. Define an identity function with an example.

Solution:
An identity function is defined as f(x) = x for all x in the domain.

Q11. Determine whether the function f: R → R defined by f(x) = 2x + 3 is one-one and onto.

Solution:

• One-one: f(x₁) = f(x₂) → 2x₁ + 3 = 2x₂ + 3 → x₁ = x₂ → One-one

Q12. Find the domain and range of the function f(x) = √(x - 2).

Solution:

• √(x - 2) is defined only when x - 2 ≥ 0 → x ≥ 2

Q13. If f(x) = x² - 1 and g(x) = x + 2, find (f ∘ g)(x).

Solution:
(f ∘ g)(x) = f(g(x)) = f(x + 2) = (x + 2)² - 1 = x² + 4x + 4 - 1 = x² + 4x + 3

Q14. If A = {1, 2, 3}, define a relation R on A such that R is reflexive.

Solution:
For reflexive relation, (a, a) ∈ R for all a ∈ A
So, R = {(1,1), (2,2), (3,3)}

Q15. Let f: R → R be defined by f(x) = 1 / (x - 1). Find its domain.

Solution:
f(x) is undefined when denominator = 0 → x ≠ 1
So, Domain = R – {1}

Q16. Let f(x) = x² and g(x) = √x. Find (g ∘ f)(x) and its domain.

Solution:
(g ∘ f)(x) = g(f(x)) = g(x²) = √(x²) = |x|
Domain = R, since x² is always ≥ 0 and √(x²) is defined for all real numbers.

Q17. Prove that the function f: R → R defined by f(x) = x³ is one-one and onto.

Solution:

• One-one: f(x₁) = f(x₂) → x₁³ = x₂³ → x₁ = x₂

Q18. Define a constant function and give an example.

Solution:
A constant function is a function that assigns the same value to every element in the domain.
Example: f(x) = 5 for all x ∈ R
So, f = {(x, 5) | x ∈ R}

Q19. What is the range of f(x) = 1 / x when x ∈ R, x ≠ 0?

Solution:

• f(x) is undefined at x = 0

Q20. If f(x) = 3x - 4 and f⁻¹(x) is the inverse function, find f⁻¹(x).

Solution:
To find the inverse:
Let y = f(x) = 3x - 4
⇒ x = (y + 4) / 3
So, f⁻¹(x) = (x + 4)/3