Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>Let <br>\(f(x) = |x - \pi| \cdot (e^{|x|} - 1)\sin|x|\)<br>Then which of the following is true?</p><p>(The set <em>S</em> of points where <em>f</em> is not differentiable)</p>
<p>\(S = \{0, \pi\}\)</p>
<p>\(S = \{0\}\)</p>
<p>\(S = \{\pi\}\)</p>
<p>\(S = \phi\) (empty set)</p>
Step-by-Step Solution
Key Concept: Differentiability requires checking all points where absolute value expressions change sign (x = 0 and x = π) and verifying left/right derivatives exist and are equal at each point.
<p><strong>Step 1:</strong> Identify critical points where differentiability may fail: x = 0 and x = π (where expressions inside absolute values change sign).</p><p><strong>Step 2:</strong> Analyze f(x) = |x - π| · (e^|x| - 1)sin|x|</p><p><strong>Step 3:</strong> Check x = 0: Factor (e^|x| - 1) → 0 as x → 0 and sin|x| → 0 as x → 0. The product (e^|x| - 1)sin|x| is differentiable at x = 0 (both factors have finite derivatives). So |x - π|·g(x) is differentiable at x = 0 since g(0) is smooth.</p><p><strong>Step 4:</strong> Check x = π: We have |x - π| with a corner at x = π. The factor (e^|π| - 1)sin|π| = (e^π - 1)·0 = 0. When a non-differentiable term is multiplied by a factor that equals 0 at that point, we must check derivatives carefully. Left derivative: lim(h→0⁻) [|π+h-π|·g(π+h) - 0]/h = lim(h→0⁻) h·g(π+h)/h = g(π) = 0. Right derivative: lim(h→0⁺) [|π+h-π|·g(π+h) - 0]/h = g(π) = 0. Both equal 0.</p><p><strong>Step 5:</strong> Since both critical points have matching left and right derivatives (both equal 0), f is differentiable everywhere.</p><p>∴ Answer: D (S is empty or f is differentiable everywhere)</p>
Correct Answer: D