A) \[x=0\] only
B) all\[x\]
C) multiples of\[\pi \]
D) multiples of\[\frac{\pi }{2}\]
Correct Answer: A
Solution :
Key Idea: Any function \[f(x)\] is derivable at\[x=a\], if \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f(a+h)-f(a)}{h}=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(a-h)-f(a)}{-h}\] Given, \[f(x)=\sin |x|\] \[\Rightarrow \] \[f(x)=\left\{ \begin{matrix} \sin x & x>0 \\ 0 & x=0 \\ -\sin x & x<0 \\ \end{matrix} \right.\] \[RHD=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin |(0+h)|-\sin (0)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{h}=1\] \[LHD=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin |(0-h)|-\sin (0)}{-h}\] \[=\frac{-\sin h}{h}=-1\] \[\therefore \]\[LHD\ne RHD\]at\[x=0\] \[\therefore \]\[f(x)\]is not derivable at\[x=0\].
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