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Answer: $[0]$

Given: $f(x)=|\sin x|$

Solution: $f(x)=|\sin x|$

$f(x)= \begin{cases}\sin x & x \geq 0 \\ -(\sin x) & x<0\end{cases}$

\begin{aligned} &R f^{\prime}(0)=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h} \\\\ &=\lim _{h \rightarrow 0^{+}} \frac{|\sinh |-0}{h} \\\\ &=\lim _{h \rightarrow 0^{+}} \frac{\sinh }{h} \\\\ &=1 \end{aligned}

\begin{aligned} &\text { As } h \rightarrow 0^{+},|\sinh |=\sinh \\\\ &\begin{aligned} L f^{\prime}(0) &=\lim _{h \rightarrow 0^{-}} \frac{f(0-h)-f(0)}{h} \\\\ &=\lim _{h \rightarrow 0^{-}} \frac{|\sinh |-0}{h} \end{aligned} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0^{-}} \frac{(-\sinh )}{h} \\ &=-1 \end{aligned}

\begin{aligned} &\text { As } h \rightarrow 0^{-},|\sinh |=-\sinh \\\\ &\text { So, } R f^{\prime}(0) \neq L f^{\prime}(0) \end{aligned}

Hence, $f(x)=|\sin x|$ is not differentiable at $x=0$.

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