<p>Let \( S = \{t \in R : f(x) = |x - \pi| \cdot (e^{|x|} - 1)\sin|x| \) is not differentiable at \(t\}\). Then the set <em>S</em> is equal to:</p>
Step-by-Step Solution
Key Concept: A product of functions is non-differentiable at points where at least one factor is non-differentiable, or where the product's derivative has a corner/cusp. Here, |x - π| is non-differentiable at x = π, and |x| is non-differentiable at x = 0, but their combined effect with the smooth factor sin|x| and (e^|x| - 1) requires careful analysis at these points.
<p><strong>Step 1:</strong> Identify candidate points where f might fail to be differentiable: x = 0 (where |x| is non-differentiable) and x = π (where |x - π| is non-differentiable).</p><p><strong>Step 2:</strong> Analyze x = 0: f(x) = |x - π|(e^|x| - 1)sin|x|. As x → 0, the factor (e^|x| - 1) ~ |x| and sin|x| ~ |x|, so (e^|x| - 1)sin|x| ~ |x|². This makes the second part behave like |x|², which is differentiable at 0. Thus f is differentiable at x = 0.</p><p><strong>Step 3:</strong> Analyze x = π: f(x) = |x - π| · (e^|x| - 1)sin|x|. Near x = π, |x - π| is non-differentiable but (e^π - 1)sin(π) = 0. However, the derivative from left and right of |x - π| creates different slopes multiplied by a non-zero limiting factor, making the overall product non-differentiable.</p><p><strong>Step 4:</strong> More precisely at x = π: the left derivative involves -(x - π) and the right involves (x - π), creating a corner in f(x) since |x - π| has a cusp and (e^|x| - 1)sin|x| is smooth and non-zero near π.</p><p>∴ Answer: S = {π}</p>
Correct Answer: D