Question
Let \[f\left( x \right) = \left\{ \begin{array}{l}
\left( {x - 1} \right)\sin \frac{1}{{x - 1}}\,{\rm{if}}\,\,x \ne 1\\
0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x = 1
\end{array} \right.\]
Then which one of the following is true?
A.
$$f$$ is neither differentiable at $$x =0$$ nor at $$x=1$$
B.
$$f$$ is differentiable at $$x=0$$ and at $$x=1$$
C.
$$f$$ is differentiable at $$x =0$$ but not at $$x=1$$
D.
$$f$$ is differentiable at $$x = 1$$ but not at $$x=0$$
Answer :
$$f$$ is differentiable at $$x =0$$ but not at $$x=1$$
Solution :
We have \[f\left( x \right) = \left\{ \begin{array}{l}
\left( {x - 1} \right)\sin \frac{1}{{x - 1}}\,{\rm{if}}\,\,x \ne 1\\
0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{if}}\,\,x = 1
\end{array} \right.\]
$$\eqalign{
& Rf\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{h\,\sin \frac{1}{h} - 0}}{h} \cr
& = \mathop {\lim }\limits_{h \to 0} \,\sin \frac{1}{h} = {\text{a finite number}} \cr} $$
Let this finite number be $$l$$
$$\eqalign{
& Lf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{ - h\,\sin \left( {\frac{1}{{ - h}}} \right)}}{{ - h}} \cr
& = \mathop {\lim }\limits_{h \to 0} \,\sin \left( {\frac{1}{{ - h}}} \right) \cr
& = - \mathop {\lim }\limits_{h \to 0} \,\sin \left( {\frac{1}{h}} \right) = - \left( {{\text{a finite number}}} \right) = - l \cr} $$
Thus $$Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$$
$$\therefore f$$ is not differentiable at $$x = 1$$
Also,
$$\eqalign{
& {\left. {f'\left( 0 \right) = \sin \frac{1}{{\left( {x - 1} \right)}} - \frac{{x - 1}}{{{{\left( {x - 1} \right)}^2}}}\cos \left( {\frac{1}{{x - 1}}} \right)} \right]_{x = 0}} \cr
& = - \sin \,1 + \cos \,1 \cr} $$
$$\therefore f$$ is differentiable at $$x =0$$