Relations and Functions — Class 12 Mathematics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Relations and Functions" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 12 Mathematics chapter "Relations and Functions" — 8 important questions with detailed answers for CBSE bo…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Check whether the relation R = {(a, b) : a ≤ b²} on the set ℕ is reflexive, s…
Let f: ℝ → ℝ be defined by f(x) = x + 1 and g: ℝ → ℝ be defined by g(x) = x².…
Determine whether the function f(x) = 2x/(x² + 1) defined on ℝ is one-one (in…
If f(x) = √(x - 1) and g(x) = x², find the domain of (f ∘ g).
Show that f(x) = 3x + 5 is a bijection from ℝ to ℝ and find its inverse.
Verify whether the function f: {1, 2, 3} → {4, 5, 6} defined by f = {(1, 4), …
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Check whether the relation R = {(a, b) : a ≤ b²} on the s…	✓ Solved
Let f: ℝ → ℝ be defined by f(x) = x + 1 and g: ℝ → ℝ be d…	✓ Solved
Determine whether the function f(x) = 2x/(x² + 1) defined…	✓ Solved
If f(x) = √(x - 1) and g(x) = x², find the domain of (f ∘…	✓ Solved
Show that f(x) = 3x + 5 is a bijection from ℝ to ℝ and fi…	✓ Solved
Verify whether the function f: {1, 2, 3} → {4, 5, 6} defi…	✓ Solved

Showing 6 of 8 questions

Q1: Check whether the relation R = {(a, b) : a ≤ b²} on the set ℕ is reflexive, symmetric, and transitive.

We need to check three properties for R = {(a, b) : a ≤ b²} on ℕ. REFLEXIVE: For R to be reflexive, (a, a) ∈ R for all a ∈ ℕ. This means a ≤ a². For a = 1: 1 ≤ 1² = 1 ✓ For a = 2: 2 ≤ 2² = 4 ✓ For all a ≥ 1: a ≤ a² is true (since a² - a = a(a-1) ≥ 0). So R is REFLEXIVE. SYMMETRIC: For R to be symmetric, if (a, b) ∈ R then (b, a) ∈ R. If a ≤ b², does this imply b ≤ a²? Counter-example: (2, 3) ∈ R since 2 ≤ 3² = 9 ✓ But (3, 2) ∉ R since 3 ≤ 2² = 4 is false. So R is NOT SYMMETRIC. TRANSITIVE: Fo...

Q2: Let f: ℝ → ℝ be defined by f(x) = x + 1 and g: ℝ → ℝ be defined by g(x) = x². Find (f ∘ g) and (g ∘ f) and check if they are equal.

Given: f(x) = x + 1 and g(x) = x² FINDING (f ∘ g): (f ∘ g)(x) = f(g(x)) = f(x²) = x² + 1 FINDING (g ∘ f): (g ∘ f)(x) = g(f(x)) = g(x + 1) = (x + 1)² Expanding: (x + 1)² = x² + 2x + 1 COMPARISON: (f ∘ g)(x) = x² + 1 (g ∘ f)(x) = x² + 2x + 1 Clearly (f ∘ g)(x) ≠ (g ∘ f)(x) for all x ∈ ℝ (except at x = 0 where both equal 1). CONCLUSION: (f ∘ g) and (g ∘ f) are NOT equal. Function composition is not commutative.

Q3: Determine whether the function f(x) = 2x/(x² + 1) defined on ℝ is one-one (injective).

To check if f(x) = 2x/(x² + 1) is one-one, we need to verify if f(a) = f(b) implies a = b. Assume f(a) = f(b): 2a/(a² + 1) = 2b/(b² + 1) Cross-multiplying: 2a(b² + 1) = 2b(a² + 1) a(b² + 1) = b(a² + 1) ab² + a = ba² + b ab² - ba² = b - a ab(b - a) = b - a (ab - 1)(b - a) = 0 So either ab = 1 or a = b. If ab = 1, then b = 1/a. Let's check with a = 2, b = 1/2: f(2) = 2(2)/(4 + 1) = 4/5 f(1/2) = 2(1/2)/(1/4 + 1) = 1/(5/4) = 4/5 So f(2) = f(1/2) but 2 ≠ 1/2. CONCLUSION: f is NOT one-one (not i...

Q4: If f(x) = √(x - 1) and g(x) = x², find the domain of (f ∘ g).

We need to find the domain of (f ∘ g)(x) = f(g(x)) = f(x²) = √(x² - 1). For f(g(x)) to be defined: 1. g(x) must be defined (this is true for all x ∈ ℝ since g(x) = x² is defined everywhere) 2. g(x) must be in the domain of f Domain of f is {x : x - 1 ≥ 0}, i.e., x ≥ 1 So we need g(x) ≥ 1 x² ≥ 1 x ≥ 1 or x ≤ -1 Also, for (f ∘ g)(x) = √(x² - 1) to be defined: x² - 1 ≥ 0 x² ≥ 1 x ≥ 1 or x ≤ -1 CONCLUSION: Domain of (f ∘ g) = (-∞, -1] ∪ [1, ∞)

Q5: Show that f(x) = 3x + 5 is a bijection from ℝ to ℝ and find its inverse.

To show f: ℝ → ℝ defined by f(x) = 3x + 5 is a bijection, we need to prove it is both injective and surjective. INJECTIVITY: Let f(a) = f(b) 3a + 5 = 3b + 5 3a = 3b a = b So f is injective (one-one). SURJECTIVITY: For any y ∈ ℝ, we need to find x ∈ ℝ such that f(x) = y. 3x + 5 = y 3x = y - 5 x = (y - 5)/3 Since y is arbitrary and (y - 5)/3 is always a real number, f is surjective (onto). So f is a bijection. FINDING THE INVERSE: Let y = f(x) = 3x + 5 Solving for x: y = 3x + 5 3x = y - 5 x = ...

Q6: Verify whether the function f: {1, 2, 3} → {4, 5, 6} defined by f = {(1, 4), (2, 5), (3, 6)} is a bijection.

Given f: {1, 2, 3} → {4, 5, 6} with f = {(1, 4), (2, 5), (3, 6)} CHECKING INJECTIVE: Domain elements: 1, 2, 3 have images 4, 5, 6 respectively. All images are different, so distinct elements map to distinct elements. f is INJECTIVE. CHECKING SURJECTIVE: Codomain: {4, 5, 6} Range (set of actual outputs): {4, 5, 6} Every element in the codomain is the image of some domain element. f is SURJECTIVE. CONCLUSION: Since f is both injective and surjective, f is a BIJECTION. The inverse function is: ...

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