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Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $\mathrm{f}(\mathrm{x})=\sin (|\mathrm{x}|)$
Which one of the following is correct?
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Which one of the following is correct?
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The correct answer is:
fis not differentiable only at 0
Given function is $: \mathrm{f}(\mathrm{x})=\sin |\mathrm{x}|$
$=\left\{\begin{array}{ll}\sin (\mathrm{x}), & \mathrm{x} \geq 0 \\ \sin (-\mathrm{x}), & \mathrm{x} < 0\end{array}\right.$
$=\left\{\begin{array}{cl}\sin \mathrm{x}, & \mathrm{x} \geq 0 \\ -\sin \mathrm{x}, & \mathrm{x} < 0\end{array}\right.$
$\mathrm{LHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{0-\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-\sin (-h)-0}{-h}=-1$
$\mathrm{RHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{0+\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\sin (\mathrm{h}-0)}{\mathrm{h}}=1$
$\mathrm{LHD} \neq \mathrm{RHD}$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.
$=\left\{\begin{array}{ll}\sin (\mathrm{x}), & \mathrm{x} \geq 0 \\ \sin (-\mathrm{x}), & \mathrm{x} < 0\end{array}\right.$
$=\left\{\begin{array}{cl}\sin \mathrm{x}, & \mathrm{x} \geq 0 \\ -\sin \mathrm{x}, & \mathrm{x} < 0\end{array}\right.$
$\mathrm{LHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{0-\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{-\sin (-h)-0}{-h}=-1$
$\mathrm{RHD}$ at $\mathrm{x}=0=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{0+\mathrm{h}-0}$
$=\lim _{h \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\sin (\mathrm{h}-0)}{\mathrm{h}}=1$
$\mathrm{LHD} \neq \mathrm{RHD}$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.
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