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.
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…
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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…
- + 5 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 |
Showing 3 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 symm...
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...
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...
Showing 3 of 8 questions. Visit the full page for complete solutions.