1. Given function

We need the points where

$$f(x)=|x-\pi|\,(e^{|x|}-1)\sin |x|$$

is not differentiable.

Let us simplify first.

2. Use the identity

Since $\sin$ is odd,

$$\sin|x|= \begin{cases} \sin x, & x\ge 0 \\ \sin(-x)=-\sin x, & x<0 \end{cases}$$

But this is exactly equal to $|\sin x|$ for all real $x$. Hence

$$\sin|x|=|\sin x|.$$

So

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

Thus possible non-differentiability can occur only where one of the absolute value factors may create a corner/cusp, namely around:

• $x=0$ from $|x|$,
• $x=\pi$ from $|x-\pi|$,
• zeros of $\sin x$ because of $|\sin x|$, i.e. $x=n\pi$.

We now check these carefully.

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3. Check differentiability at $x=0$

Near $x=0$, we have:

• $|x-\pi|=\pi-x$ for $x$ near $0$,
• $e^{|x|}-1 \sim |x|$,
• $\sin|x| \sim |x|$.

Therefore,

$$f(x)\sim (\pi-x)|x|\,|x|=(\pi-x)x^2.$$

So near $0$, $f(x)$ behaves like a smooth multiple of $x^2$, hence should be differentiable.

More explicitly:

For $x>0$ near $0$,

$$f(x)=(\pi-x)(e^x-1)\sin x.$$

For $x<0$ near $0$,

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

Now

$$f(0)=|- \pi|(e^0-1)\sin 0=0.$$

So

$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{f(h)}{h}.$$

Using $(e^{|h|}-1)\sim |h|$ and $\sin|h|\sim |h|$,

$$f(h)\sim |h-\pi|\,|h|\,|h|\sim \pi h^2.$$

Hence

$$\frac{f(h)}{h}\sim \pi h\to 0.$$

So $f$ is differentiable at $0$.

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4. Check differentiability at $x=\pi$

At $x=\pi$,

$$f(\pi)=|\pi-\pi|(e^{|\pi|}-1)\sin|\pi|=0.$$

To test differentiability,

$$f'(\pi)=\lim_{h\to 0}\frac{f(\pi+h)-f(\pi)}{h}= \lim_{h\to 0}\frac{f(\pi+h)}{h}.$$

Now near $x=\pi$:

• $|x-\pi|=|h|$,
• $e^{|x|}-1=e^{\pi+h}-1$ (since $\pi+h>0$ near $\pi$), which tends to $e^\pi-1$,
• $\sin|x|=\sin(\pi+h)\sim -h$ in magnitude, so actually $|\sin(\pi+h)|\sim |h|$.

Thus

$$f(\pi+h)\sim |h|\,(e^\pi-1)\,|h|=(e^\pi-1)h^2.$$

Hence

$$\frac{f(\pi+h)}{h}\sim (e^\pi-1)h\to 0.$$

So $f$ is differentiable at $\pi$.

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5. Check other multiples of $\pi$

If $t=n\pi$ with $n\ne 0,1$, then only $|\sin x|$ may be suspicious, since $|x-\pi|$ and $e^{|x|}-1$ are differentiable at such points and are finite constants there.

Near $x=t=n\pi$, we have

$$|\sin x|\sim |x-t|.$$

Therefore

$$f(x)\sim |t-\pi|\,(e^{|t|}-1)\,|x-t|.$$

Since the coefficient

$$|t-\pi|\,(e^{|t|}-1)$$

is nonzero for $t=n\pi$ with $n\ne 0,1$, this behaves like a nonzero multiple of $|x-t|$, which is not differentiable at $x=t$.

So all such multiples of $\pi$ are points of non-differentiability.

Examples:

• at $x=-\pi$,
• at $x=2\pi$,
• at $x=-2\pi$, etc.

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6. Hence the set $S$

Therefore,

$$S=\{n\pi: n\in \mathbb Z,\ n\ne 0,1\}.$$

In particular, $S$ is not any of the listed finite sets, and certainly not empty.

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7. Evaluate options

• A: $\{0,\pi\}$ — false, since function is differentiable at both $0$ and $\pi$.
• B: $\phi$ — false, since it is not differentiable at $x=-\pi,2\pi,\dots$.
• C: $\{0\}$ — false.
• D: $\{\pi\}$ — false.

So none of the given options is correct.

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8. Comparison with stored correct answer

Stored correct answer is B (empty set), but that is incorrect because, for example, at $x=2\pi$,

$$f(x)=|x-\pi|(e^{|x|}-1)|\sin x|,$$

and near $2\pi$ this is a nonzero constant times $|x-2\pi|$, which is not differentiable at $2\pi$.

Thus the correct set is

$$S=\{n\pi: n\in \mathbb Z,\ n\ne 0,1\}.$$