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Step-by-Step Solution
1. Understand the Function
We are given the function
$f(x) = |x - \pi| \, \bigl(e^{|x|} - 1\bigr) \, \sin\bigl(|x|\bigr).$
We want to determine the set
$S = \{\,t \in \mathbb{R}\colon f \text{ is not differentiable at } t\}.$
The problem statement claims that $S$ is the empty set, i.e., $\phi.$
We will verify the differentiability of $f$ at all points, focusing especially on $x = 0$ and $x = \pi,$ where there could be potential non-differentiability due to absolute values.
2. Check Differentiability at x = 0
We first compute the left-hand derivative (LHD) and the right-hand derivative (RHD) at $x = 0.$
2.1 Left-Hand Derivative (LHD) at x = 0
By definition,
$$
\text{LHD} = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-\,h}.
$$
Substitute $f(x) = |x - \pi|\,(e^{|x|} - 1)\,\sin(|x|)$ into the expression:
$$
f(-h) = |(-h) - \pi| \, \bigl(e^{|-h|} - 1\bigr) \, \sin\bigl(|-h|\bigr).
$$
At $x = 0,$ $f(0) = |0 - \pi|\,(e^{|0|} - 1)\,\sin(|0|) = 0,$
because $\sin(0) = 0.$ Thus,
$$
\text{LHD} = \lim_{h \to 0} \frac{|{-\,h - \pi}| \, (e^{|{-\,h}|} - 1) \,\sin\bigl(|-h|\bigr) - 0}{-\,h}.
$$
Notice that as $h \to 0,$ $\sin\bigl(|-h|\bigr) \approx 0,$ and $(e^{|\,{-h}\,|} - 1)$ also approaches 0 linearly in $|h|.$ Multiplying by the factor $|{-h-\pi}|$ and dividing by $-\,h$ yields 0 in the limit. Therefore,
$$
\text{LHD} = 0.
$$
2.2 Right-Hand Derivative (RHD) at x = 0
By definition,
$$
\text{RHD} = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h}.
$$
Substitute $f(h) = |h - \pi|\,(e^{|h|} - 1)\,\sin\bigl(|h|\bigr).$
Again $f(0)=0,$ so
$$
\text{RHD} = \lim_{h \to 0} \frac{|h - \pi| (e^{|h|} - 1)\,\sin\bigl(|h|\bigr)}{h}.
$$
The same reasoning applies: as $h \to 0,$ $|h - \pi|$ is approximately $|\pi|,$ but more crucially, $\sin\bigl(|h|\bigr)$ and $(e^{|h|}-1)$ both go to 0 at $h=0.$ The expression goes to 0. Hence,
$$
\text{RHD} = 0.
$$
2.3 Conclusion at x = 0
Since LHD = RHD = 0 at $x = 0,$ the function $f$ is differentiable at $x = 0.$
3. Check Differentiability at x = $\pi$
We next check the derivatives at $x = \pi.$
3.1 Left-Hand Derivative (LHD) at x = $\pi$
By definition,
$$
\text{LHD} = \lim_{h \to 0} \frac{f(\pi - h) - f(\pi)}{-\,h}.
$$
Substitute $f(\pi - h) = |\pi - h - \pi|\,(e^{|\pi - h|} - 1)\,\sin\bigl(|\pi - h|\bigr).$
At $x = \pi,$ $f(\pi) = 0$ (since $\sin|\pi| = \sin(\pi) = 0$). Thus,
$$
\text{LHD} = \lim_{h \to 0} \frac{|{-\,h}|\,(e^{|\pi - h|} - 1)\,\sin\bigl(|\pi - h|\bigr)}{-\,h}.
$$
As $h \to 0,$ $\sin\bigl(|\pi - h|\bigr) = \sin(\pi - h) \approx 0,$ and $(e^{|\pi - h|} - 1)$ becomes $(e^{\pi} - 1)$ up to a small deviation. However, multiplied by $|-h|$ in the numerator and divided by $-\,h,$ the limit goes to 0. Hence,
$$
\text{LHD} = 0.
$$
3.2 Right-Hand Derivative (RHD) at x = $\pi$
By definition,
$$
\text{RHD} = \lim_{h \to 0} \frac{f(\pi + h) - f(\pi)}{h}.
$$
Here, $f(\pi + h) = |\pi + h - \pi|\,(e^{|\pi + h|} - 1)\,\sin\bigl(|\pi + h|\bigr).$
Again $f(\pi)=0,$ so
$$
\text{RHD} = \lim_{h \to 0} \frac{|h|\,(e^{|\pi + h|} - 1)\,\sin\bigl(|\pi + h|\bigr)}{h}.
$$
The reasoning is similar: $\sin\bigl(|\pi + h|\bigr)$ near $\pi$ is small, and multiplying by $|h|$ in the numerator leads to a limit of 0. Therefore,
$$
\text{RHD} = 0.
$$
3.3 Conclusion at x = $\pi$
Since LHD = RHD = 0 at $x = \pi,$ the function $f$ is differentiable at $x = \pi.$
4. Differentiability at Other Points
Away from $x = 0$ and $x = \pi,$ the components $|x - \pi|,$ $(e^{|x|} - 1),$ and $\sin(|x|)$ do not introduce any additional discontinuities or non-differentiable points (absolute value terms affect potentially at 0 and $\pi,$ already checked). Therefore, $f(x)$ is differentiable everywhere on $\mathbb{R}.$
5. Final Conclusion
Since there are no points at which $f(x)$ is non-differentiable, the set
$S = \{\,t \in \mathbb{R}\colon f(t) \text{ is not differentiable}\}$
must be the empty set, denoted by $\phi.$
Answer: $\phi$ (the empty set).
