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  • <img alt="\! \!\!\!\!\!\!\!Let f(x)=\left\{\begin{array}{cc}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{array}\right.\\ \\Then\; at\; x=0" src="https://entrancecorner.on

\! \!\!\!\!\!\!\!Let f(x)=\left\{\begin{array}{cc}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{array}\right.\\ \\Then\; at\; x=0

Option: 1

f is continuous but not differentiable


Option: 2

f and\: f' both are continuous


Option: 3

f'is continuous but not differentiable


Option: 4

f is continuous but f'is not continuous


Answers (1)

best_answer

\begin{aligned} & f(x)=\left\{\begin{array}{cc} x^2 \sin \frac{1}{x} & x \neq 0 \\ 0 & x=0 \end{array}\right. \\ & \rightarrow \text { cont. of } \mathrm{f}(\mathrm{x}) \text { at } \mathrm{x}=0 \end{aligned}



\rightarrow \text { Diff. of } \mathrm{f}(\mathrm{x}) \text { at } \mathrm{x}=0
\begin{aligned} & \mathrm{RHD}=\underset{\mathrm{ht} \rightarrow 0}{ } \frac{\mathrm{h}^2 \sin \left(\frac{1}{\mathrm{~h}}\right)-0}{\mathrm{~h}}=\underset{\mathrm{ht} \rightarrow 0}{\mathrm{dt}} \sin \frac{1}{\mathrm{~h}}=0 \\ & \mathrm{LHD}=\mathrm{dt}_{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}^2 \sin \left(\frac{-1}{\mathrm{~h}}\right)-0}{-\mathrm{h}}=\underset{h \rightarrow 0}{\mathrm{dt}} \mathrm{h} \sin \frac{1}{\mathrm{~h}}=0 \end{aligned} \\ \text Hence\; \mathrm{f}(\mathrm{x}) \;is\; di\!f\!f\!.\; at\; \mathrm{x}=0\\ \text Now \;di\!f\!f\!.\; \mathrm{f}(\mathrm{s})\; at\; x=0
\begin{aligned} & f^{\prime}(x)=\left[\begin{array}{cc} 2 x \sin \left(\frac{1}{x}\right)+x^2 \cos \left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right) & x \neq 0 \\ 0 & x=0 \end{array}\right. \\ & f^{\prime}(x)=\left[\begin{array}{cc} 2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x=0 \end{array}\right. \end{aligned} \\ hence \;f^{\prime}(x)\; limit\; ossicilate\; at\; x=0 \\hence\; f^{\prime}(x) \;is\; D.C. \;at \;x=0

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Deependra Verma

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