Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>\(f(x) = 1 + |\sin x|\)</p>
<p>(a) is continuous nowhere</p>
<p>(b) is continuous everywhere</p>
<p>(c) is differentiable nowhere</p>
<p>(d) \(f'(0)\) does not exist</p>
Step-by-Step Solution
Key Concept: The absolute value function creates a sharp corner at x = 0 where sin(x) = 0, making the left and right derivatives unequal at that point. Even though f is continuous everywhere, differentiability requires the derivative to exist from both sides.
<p><strong>Step 1:</strong> Analyze continuity of f(x) = 1 + |sin x|</p><p>Since |sin x| is continuous everywhere, f(x) is continuous for all x ∈ ℝ.</p><p><strong>Step 2:</strong> Check differentiability at x = nπ (where sin x = 0)</p><p>Left derivative: lim(h→0⁻) [|sin(nπ + h)| - 0]/h = lim(h→0⁻) |sin h|/h</p><p>For h < 0: |sin h|/h → -1 (since sin h < 0 when h < 0)</p><p><strong>Step 3:</strong> Compute right derivative</p><p>Right derivative: lim(h→0⁺) |sin h|/h → +1 (since sin h > 0 when h > 0)</p><p><strong>Step 4:</strong> Conclusion</p><p>Left derivative ≠ Right derivative at x = nπ, so f is not differentiable at x = 0, ±π, ±2π, ...</p><p>f is continuous everywhere but NOT differentiable at x = nπ where n ∈ ℤ.</p><p>∴ Answer: D</p>
Correct Answer: D