question_answer

\[f(x)\sin \left| x \right|.f(x)\]is not differentiable at

A)  \[x=0\]only                

B)  all x

C)  multiples of \[\pi \]                  

D)  multiples of\[\frac{\pi }{2}\]

Correct Answer: C

Solution :

 The function breaks at x = 0 and multiples of x. Hence the function is differentiable at all other points as the function is continuous at all these pts. At x = 0, for f (x) to be continous \[\underset{x\to 0}{\mathop{\lim }}\,f({{0}^{-}})=f(x=0)=\underset{x\to 0}{\mathop{\lim }}\,f({{0}^{+}})\] \[f(x)=0\]at x = 0 \[RHL=\underset{x\to 0}{\mathop{\lim }}\,\sin (x+h)=\sin \,h>0\] \[L.H.L.=\underset{x\to 0}{\mathop{\lim }}\,\sin (x-h)=\sin (-h)<0\] Hence, not differentiable at x = 0 Similarly, f(x) is not differentiable at all multiples of \[\pi \], i.e, \[n\pi \]where n=0,1,2..............