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  • Let    <img alt="\mathrm{ f(x)=\left\{\begin{array}{c} x^2\left|\cos \frac{\pi}{x}\right|:|x| \neq 0 \\ 0 \quad ; \quad x=0 \end{array}, x \in R\right.}" src="https://entrancecorner.

Let    \mathrm{ f(x)=\left\{\begin{array}{c} x^2\left|\cos \frac{\pi}{x}\right|:|x| \neq 0 \\ 0 \quad ; \quad x=0 \end{array}, x \in R\right.}  Then \mathrm{f}  is :

Option: 1

differentiable both at  \mathrm{x=0}  and at  \mathrm{x=2} 


Option: 2

differentiable at  \mathrm{x=0}  but not differentiable at \mathrm{x=2}


Option: 3

not differentiable at \mathrm{x=0}  but differentiable at \mathrm{x=2}


Option: 4

differentiable neither at \mathrm{x=0} nor at \mathrm{x=2}


Answers (1)

best_answer

\mathrm{f(x)=\left\{\begin{array}{c} x^2\left|\cos \frac{\pi}{x}\right| ; x \neq 0 \\ 0 \quad ; x=0 \end{array}\right.}

\mathrm{ \begin{gathered} \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^2\left|\cos \frac{\pi}{x}\right|=0 \\ f(0)=0 \end{gathered} }
Hence \mathrm{f(x)} is continuous at \mathrm{x=0}

\mathrm{f(0) =\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{x^2\left|\cos \frac{\pi}{x}\right|-0}{x} \\ }

\mathrm{ =\lim _{x \rightarrow 0} x\left|\cos \frac{\pi}{x}\right|=0 }

Hence, differentiable at  \mathrm{x=0}
Now at \mathrm{x=2 f(x)}  is continuous.


\mathrm{ f(2)=4\left(\left|\cos \frac{\pi}{x}\right|\right)=0 }
\mathrm{f^{\prime}(2)=\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2} \frac{x^2 \mid \cos \frac{\pi}{x \mid}}{x-2}}
\mathrm{f^{\prime}\left(2^*\right)=\lim _{x \rightarrow 2^{+}} \frac{x^2\left|\cos \frac{\pi}{x}\right|}{(x-2)}=\lim _{x \rightarrow 2^*} x^2 \frac{\cos \frac{\pi}{x}}{(x-2)}}

\mathrm{=\lim _{x \rightarrow 2^{+}} \frac{x^2 \sin \left(\frac{\pi}{2}-\frac{\pi}{x}\right)}{(x-2)}=\lim _{x \rightarrow 2^{+}} \frac{x^2 \sin \left(\frac{\pi(x-2)}{2 x}\right)}{\frac{\pi}{2 x}(x-2)} \frac{\pi}{2 x}}

=\pi \\ \mathrm{f^{\prime}\left(2^{-}\right) =\lim _{x \rightarrow 2^{-}} \frac{x^2\left|\cos \frac{\pi}{x}\right|}{(x-2)}=\lim _{x \rightarrow 2^{-}}-\frac{x^2 \cos \frac{\pi}{x}}{(x-2)} \\ }

\mathrm{=\lim _{x \rightarrow 2^{-}}-\frac{x^2 \sin \left(\frac{\pi}{2}-\frac{\pi}{x}\right)}{(x-2)}=\lim _{x \rightarrow 2^{-}}-\frac{x^2 \sin \left(\frac{\pi(x-2)}{2 x}\right)}{\frac{\pi}{2 x}(x-2)} \frac{\pi}{2 x} \\ }

=-\pi

Hence not derivable at  \mathrm{x=0 }.

Posted by

Shailly goel

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