A) f is not differentiable only at 0
B) f is differentiable at 9 only
C) f is differentiable everywhere
D) f is non-differentiable at many points
Correct Answer: A
Solution :
[a] Given function is : \[f(x)=\sin \left| x \right|\] \[=\left\{ \begin{matrix} \sin (x), & x\ge 0 \\ \sin (-x), & x<0 \\ \end{matrix} \right.\] \[=\left\{ \begin{matrix} \sin \,x, & x\ge 0 \\ -\sin \,x, & x<0 \\ \end{matrix} \right.\] LHD at \[x=0=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{0-h-0}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{-\sin (-h)-0}{-h}=-1\] RHD at \[x=0=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(x)}{0+h-0}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin (h-0)}{h}=1\] \[LHD\ne RHD\] \[f(x)\] is not differentiable at \[x=0\].
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