A) \[f\] is differentiable at \[x=0\] but not at \[x=1\]
B) \[f\] is differentiable at \[x=1\] but not at \[x=0\]
C) \[f\] is neither differentiable at \[x=0\] nor at \[x=1\]
D) \[f\] is differentiable at \[x=0\] and at \[x=1\]
Correct Answer: A
Solution :
\[f\left( x \right)=\left\{ \begin{matrix} \left( x-1 \right)\sin \frac{1}{x-1} & if\,x\ne 1 \\ 0 & if\,x=1 \\ \end{matrix}; \right.\] \[Rf'\left( 1 \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\sin \frac{1}{h}-0}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{h}\], which does not exist. \[{{\left. f'\left( 0 \right)=\sin \frac{1}{x-1}-\frac{\left( x-1 \right)}{{{\left( x-1 \right)}^{2}}}\cos \frac{1}{x-1} \right|}_{x=0}}\] \[=-\sin 1+\cos 1\]. So \[f\left( x \right)\] is differentiable at \[x=0\] but not at \[x=1\].
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