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  • At , the function <img alt="\mathrm{f(x)= \left|\sin \frac{2 \pi x}{L}\right|,(-\infty<x<\

At \mathrm{x=0}, the function \mathrm{f(x)= \left|\sin \frac{2 \pi x}{L}\right|,(-\infty<x<\infty, L>0)} is

Option: 1

Continuous and differentiable
 


Option: 2

Not continous and not differentiable
 


Option: 3

Not continuous but differentiable
 


Option: 4

Continuous but not differentiable


Answers (1)

We have                     \mathrm{f(x)=\left|\sin \left(\frac{2 \pi x}{L}\right)\right|}

                                           \mathrm{= \begin{cases}-\sin \left(\frac{2 \pi x}{L}\right), & x<0 \\ +\sin \left(\frac{2 \pi x}{L}\right), & x \geq 0\end{cases}}

\because  At  \mathrm{x=0, L H L=R H L=f(0)=0} so \mathrm{f(x)} is continuous at \mathrm{x=0}

Again                      \mathrm{f^{\prime}(x)= \begin{cases}\frac{-2 \pi}{L} \cos \left(\frac{2 \pi x}{L}\right), & x<0 \\ +\frac{2 \pi}{L} \cos \left(\frac{2 \pi x}{L}\right), & x>0\end{cases}}

                               \mathrm{L H D=f^{\prime}\left(0^{-}\right)=-\frac{2 \pi}{L}}

\mathrm{\text { \& }}                           \mathrm{R H D=f^{\prime}\left(0^{+}\right)=+\frac{2 \pi}{L}}

\because LHD \neq RHD so \mathrm{ f(x)} is NOT differentiable at \mathrm{x=0}

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Kshitij

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