Relations and Functions — Class 11 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Relations and Functions" — 8 important questions with detailed answers for CBSE board exam preparation.

TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Relations and Functions" — 8 important questions with detailed answers for CBSE bo…

By Syllab.in · Updated Jun 14, 2026

Key Questions Covered:

Let A = {1, 2, 3} and B = {4, 5}. Define a relation R from A to B as R = {(x,…
Determine whether the relation R = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 3)} o…
Check whether f(x) = 2x + 3 is a function from ℝ to ℝ. If yes, find f(-2), f(…
For the function f(x) = x² - 2x + 3, find (i) f(-1), (ii) f(a + 1), (iii) the…
Determine whether the function f(x) = x³ from ℝ to ℝ is one-one (injective) a…
Given f(x) = 2x - 1 and g(x) = x + 3, find (i) (f ∘ g)(x), (ii) (g ∘ f)(x), (…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Let A = {1, 2, 3} and B = {4, 5}. Define a relation R fro…	✓ Solved
Determine whether the relation R = {(1, 1), (1, 2), (2, 2…	✓ Solved
Check whether f(x) = 2x + 3 is a function from ℝ to ℝ. If…	✓ Solved
For the function f(x) = x² - 2x + 3, find (i) f(-1), (ii)…	✓ Solved
Determine whether the function f(x) = x³ from ℝ to ℝ is o…	✓ Solved
Given f(x) = 2x - 1 and g(x) = x + 3, find (i) (f ∘ g)(x)…	✓ Solved

Showing 6 of 8 questions

Q1: Let A = {1, 2, 3} and B = {4, 5}. Define a relation R from A to B as R = {(x, y) : x < y}. Write R and determine its domain and range.

Step 1: Find all ordered pairs (x, y) where x ∈ A, y ∈ B, and x < y. Check (1, 4): 1 < 4? Yes ✓ Check (1, 5): 1 < 5? Yes ✓ Check (2, 4): 2 < 4? Yes ✓ Check (2, 5): 2 < 5? Yes ✓ Check (3, 4): 3 < 4? Yes ✓ Check (3, 5): 3 < 5? Yes ✓ Step 2: Write the relation. R = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} Step 3: Find the domain (all first elements). Domain(R) = {1, 2, 3} Step 4: Find the range (all second elements). Range(R) = {4, 5} Final Answer: R = {(1,4), (1,5)...

Q2: Determine whether the relation R = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 3)} on set A = {1, 2, 3} is reflexive, symmetric, and transitive.

Step 1: Check reflexivity. A relation is reflexive if (a, a) ∈ R for all a ∈ A. Check (1, 1): Yes ✓ Check (2, 2): Yes ✓ Check (3, 3): Yes ✓ R is reflexive. Step 2: Check symmetry. A relation is symmetric if (a, b) ∈ R ⟹ (b, a) ∈ R. Check (1, 2): (1, 2) ∈ R, but (2, 1) ∉ R R is not symmetric. Step 3: Check transitivity. A relation is transitive if (a, b) ∈ R and (b, c) ∈ R ⟹ (a, c) ∈ R. Check (1, 1) and (1, 2): (1, 2) ∈ R ✓ Check (1, 2) and (2, 2): (1, 2) ∈ R ✓ Check (1, 2) and (2, 3): (1, 3) ∉...

Q3: Check whether f(x) = 2x + 3 is a function from ℝ to ℝ. If yes, find f(-2), f(0), and f(1).

Step 1: Verify f is a function. For f to be a function, each x ∈ ℝ must have exactly one image in ℝ. For any x ∈ ℝ, 2x + 3 is uniquely defined and is a real number. Therefore, f is a function from ℝ to ℝ. Step 2: Calculate f(-2). f(-2) = 2(-2) + 3 = -4 + 3 = -1 Step 3: Calculate f(0). f(0) = 2(0) + 3 = 0 + 3 = 3 Step 4: Calculate f(1). f(1) = 2(1) + 3 = 2 + 3 = 5 Final Answer: Yes, f is a function. f(-2) = -1, f(0) = 3, f(1) = 5

Q4: For the function f(x) = x² - 2x + 3, find (i) f(-1), (ii) f(a + 1), (iii) the domain and range.

Step 1: Calculate f(-1). f(-1) = (-1)² - 2(-1) + 3 = 1 + 2 + 3 = 6 Step 2: Calculate f(a + 1). f(a + 1) = (a + 1)² - 2(a + 1) + 3 = a² + 2a + 1 - 2a - 2 + 3 = a² + 2 Step 3: Find the domain. f(x) = x² - 2x + 3 is a polynomial, defined for all real numbers. Domain = ℝ Step 4: Find the range. Rewrite f(x) by completing the square: f(x) = x² - 2x + 3 = (x² - 2x + 1) + 2 = (x - 1)² + 2 Since (x - 1)² ≥ 0 for all x ∈ ℝ, (x - 1)² + 2 ≥ 2 The minimum value is 2 (when x = 1). Range = [2, ∞) Final An...

Q5: Determine whether the function f(x) = x³ from ℝ to ℝ is one-one (injective) and onto (surjective).

Step 1: Check if f is one-one (injective). f is one-one if f(a) = f(b) ⟹ a = b. Assume f(a) = f(b): a³ = b³ Taking cube root: a = b Therefore, f is one-one. Step 2: Check if f is onto (surjective). f is onto if for every y ∈ ℝ, there exists x ∈ ℝ such that f(x) = y. Let y ∈ ℝ be arbitrary. We need x³ = y This gives x = ∛y (cube root of y) For any real y, ∛y is a real number. Therefore, every y ∈ ℝ has a pre-image x = ∛y ∈ ℝ. So f is onto. Step 3: Conclusion. Since f is both one-one and onto, f...

Q6: Given f(x) = 2x - 1 and g(x) = x + 3, find (i) (f ∘ g)(x), (ii) (g ∘ f)(x), (iii) (f ∘ g)(2).

Step 1: Find (f ∘ g)(x) = f(g(x)). (f ∘ g)(x) = f(x + 3) = 2(x + 3) - 1 = 2x + 6 - 1 = 2x + 5 Step 2: Find (g ∘ f)(x) = g(f(x)). (g ∘ f)(x) = g(2x - 1) = (2x - 1) + 3 = 2x + 2 Step 3: Find (f ∘ g)(2). (f ∘ g)(2) = 2(2) + 5 = 4 + 5 = 9 Alternatively: g(2) = 2 + 3 = 5, then f(5) = 2(5) - 1 = 9 Step 4: Note that (f ∘ g)(x) ≠ (g ∘ f)(x). Composition of functions is not commutative. Final Answer: (i) (f ∘ g)(x) = 2x + 5, (ii) (g ∘ f)(x) = 2x + 2, (iii) (f ∘ g)(2) = 9

Showing 6 of 8 questions. Visit the full page for complete solutions.