Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>For \( x \in R \), \( f(x) = |\log 2 - \sin x| \) and \( g(x) = f(f(x)) \), then:</p>
<p>\(g\) is differentiable at \(x = 0\) and \(g'(0) = -\sin(\log 2)\).</p>
<p>\(g\) is not differentiable at \(x = 0\).</p>
<p>\(g'(0) = \cos(\log 2)\).</p>
<p>\(g'(0) = -\cos(\log 2)\).</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = |log 2 - sin x| is always positive (since log 2 ≈ 0.693 and sin x ∈ [-1,1]), so f(f(x)) applies f to a positive argument. The continuity and differentiability of g(x) depend on where the inner absolute value expression equals zero.
<p><strong>Step 1:</strong> Analyze f(x) = |log 2 - sin x|. Since log 2 ≈ 0.693 and sin x ∈ [-1, 1], we need to check if log 2 - sin x ever equals zero.</p><p><strong>Step 2:</strong> For log 2 - sin x = 0, we need sin x = log 2 ≈ 0.693. Since -1 ≤ sin x ≤ 1 and log 2 ≈ 0.693 ∈ [-1, 1], there exist real values where sin x = log 2.</p><p><strong>Step 3:</strong> At points where sin x = log 2, the function f(x) = |log 2 - sin x| is non-differentiable (it has a sharp corner).</p><p><strong>Step 4:</strong> For g(x) = f(f(x)), since f(x) > 0 always (as |log 2 - sin x| ≥ 0 and equals 0 at isolated points), we apply f again. The differentiability of g depends on whether f∘f has any corner points.</p><p><strong>Step 5:</strong> At points where f'(x) doesn't exist, the chain rule fails for g'(x) = f'(f(x))·f'(x). This occurs at finitely many points where sin x = log 2.</p><p><strong>Step 6:</strong> Therefore, g(x) is continuous everywhere but not differentiable at the points where sin x = log 2 (and corresponding image points under f).</p><p>∴ Answer: C</p>
Correct Answer: C