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