Let S = {t in R : f(x) = |x-pi|(e^{|x|}-1)sin | vedclass

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(D) The function is given by $f(x) = |x-\pi|(e^{|x|}-1)\sin|x|$.We check the differentiability of $f(x)$ at the points where the absolute value functi

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$\emptyset$

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(D) The function is given by $f(x) = |x-\pi|(e^{|x|}-1)\sin|x|$.We check the differentiability of $f(x)$ at the points where the absolute value functions might not be differentiable,namely $x=0$ and $x=\pi$.At $x=\pi$:$f(\pi) = 0$.$RHD = \lim_{h 	o 0^+} \frac{|\pi+h-\pi|(e^{|\pi+h|}-1)\sin|\pi+h| - 0}{h} = \lim_{h 	o 0^+} \frac{h(e^{\pi+h}-1)\sin(\pi+h)}{h} = (e^{\pi}-1)\sin(\pi) = 0$.$LHD = \lim_{h 	o 0^+} \frac{|\pi-h-\pi|(e^{|\pi-h|}-1)\sin|\pi-h| - 0}{-h} = \lim_{h 	o 0^+} \frac{h(e^{\pi-h}-1)\sin(\pi-h)}{-h} = -(e^{\pi}-1)\sin(\pi) = 0$.Since $RHD = LHD = 0$,$f(x)$ is differentiable at $x=\pi$.At $x=0$:$f(0) = 0$.$RHD = \lim_{h 	o 0^+} \frac{|h-\pi|(e^{|h|}-1)\sin|h| - 0}{h} = \lim_{h 	o 0^+} \frac{|h-\pi|(e^h-1)\sin(h)}{h} = |-\pi| \cdot (1) \cdot (0) = 0$.$LHD = \lim_{h 	o 0^+} \frac{|-h-\pi|(e^{|-h|}-1)\sin|-h| - 0}{-h} = \lim_{h 	o 0^+} \frac{|-h-\pi|(e^h-1)\sin(h)}{-h} = |-\pi| \cdot (1) \cdot (0) = 0$.Since $RHD = LHD = 0$,$f(x)$ is differentiable at $x=0$.Since $f(x)$ is differentiable at all points,the set $S$ of points where $f(x)$ is not differentiable is empty,i.e.,$S = \emptyset$.

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

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