Let $$S = \{t \in R : f(x) = |x - \pi| \cdot (e^{|x|} - 1)\sin|x|$$ is not differentiable at $$t\}$$. Then the set S is equal to:

We have the real-valued function

$$f(x)=|x-\pi|\;\bigl(e^{|x|}-1\bigr)\;\sin|x|.$$

Non-differentiability can come only from the two absolute-value expressions $$|x-\pi| \text{ and } |x|.$$ Hence the only candidate points are $$x=0 \quad\text{and}\quad x=\pi.$$ At every other point each factor is either smooth or the product of smooth factors, so the function must be differentiable there.

Checking differentiability at $$x=0$$

For $$x>0$$ we have $$|x|=x,\;|x-\pi|=\pi-x$$ and $$\sin|x|=\sin x.$$ Therefore

$$f(x)=\bigl(\pi-x\bigr)\,(e^{x}-1)\,\sin x.$$ Using the standard Maclaurin expansions $$e^{x}-1 = x+\dfrac{x^{2}}{2}+O(x^{3}),\qquad \sin x = x-\dfrac{x^{3}}{6}+O(x^{5}),$$ we obtain

$$f(x)=\bigl(\pi-x\bigr)\;\bigl(x+\dfrac{x^{2}}{2}+O(x^{3})\bigr)\; \bigl(x+O(x^{3})\bigr) =\bigl(\pi-x\bigr)\;(x^{2}+O(x^{3}))\\ =\pi x^{2}+O(x^{3}).$$

Hence for $$x>0$$

$$\dfrac{f(x)-f(0)}{x-0} =\dfrac{\pi x^{2}+O(x^{3})}{x} =\pi x+O(x^{2})\xrightarrow[x\to0^{+}]{}0.$$

For $$x<0$$ we have $$|x|=-x,\;|x-\pi|=\pi-x$$ and $$\sin|x|=\sin(-x)=-\sin x.$$ Thus

$$f(x)=\bigl(\pi-x\bigr)\,\bigl(e^{-x}-1\bigr)\,\bigl(-\sin x\bigr).$$ Again expanding, $$e^{-x}-1 = -x+\dfrac{x^{2}}{2}+O(x^{3}),\qquad -\sin x = -\bigl(x-\dfrac{x^{3}}{6}+O(x^{5})\bigr)= -x+O(x^{3}),$$ we get

$$f(x)=\bigl(\pi-x\bigr)\;(x^{2}+O(x^{3})) =\pi x^{2}+O(x^{3}).$$

Therefore for $$x<0$$

$$\dfrac{f(x)-f(0)}{x} =\dfrac{\pi x^{2}+O(x^{3})}{x} =\pi x+O(x^{2})\xrightarrow[x\to0^{-}]{}0.$$

Both the right-hand and left-hand derivatives at $$x=0$$ exist and are equal to $$0,$$ so $$f$$ is differentiable at $$x=0.$

Checking differentiability at $$x=\pi$$

Put $$x=\pi+h$$ with $$h\to0.$$ Then $$|x-\pi|=|h|,\;|x|=|\,\pi+h\,|=\pi+h$$ (because $$\pi+h>0$$ for small $$h$$), and $$\sin|x|=\sin(\pi+h)=-\sin h.$$ Thus

$$f(\pi+h)=|h|\;\bigl(e^{\pi+h}-1\bigr)\;(-\sin h) =-|h|\;\sin h\;\bigl(e^{\pi+h}-1\bigr).$$

Write $$K=e^{\pi}-1\;(\text{a positive constant}).$$ For small $$h,$$

$$e^{\pi+h}-1 = K + h\,e^{\pi}+O(h^{2}),\qquad \sin h = h + O(h^{3}).$$ Therefore

$$f(\pi+h)= -|h|\;h\;\bigl(K+O(h)\bigr) = -K\,h\,|h| + O(h^{3}).$$

Now examine the derivative. For $$h>0$$ (i.e. $$x\to\pi^{+}$$)

$$\dfrac{f(\pi+h)-f(\pi)}{h} =\dfrac{-K\,h^{2}+O(h^{3})}{h} =-K\,h+O(h^{2})\xrightarrow[h\to0^{+}]{}0.$$

For $$h<0$$ (i.e. $$x\to\pi^{-}$$)

$$\dfrac{f(\pi+h)-f(\pi)}{h} =\dfrac{-K\,h\,|h|+O(h^{3})}{h} =\dfrac{-K\,h\,(-h)+O(h^{3})}{h} =K\,h+O(h^{2})\xrightarrow[h\to0^{-}]{}0.$$

The two one-sided limits are equal; hence $$f$$ is differentiable at $$x=\pi$$ as well.

Since $$f$$ is differentiable at both possible trouble points and is obviously differentiable everywhere else, there is no point at which differentiability fails. Therefore

$$S=\varnothing.$$

Hence, the correct answer is Option B.