Let \[f\left( x \right) = \left\{ \begin{array}{l} {x^2}\left| {\cos \frac{\pi }{x}} \right|,\,\,\,\,x \ne 0\\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \end{array} \right.,\,x \in R\]        then $$f$$ is-

Solution :
$$\eqalign{ & {\text{We have }}f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\left| {\cos \frac{\pi }{h}} \right|}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} h\left| {\cos \frac{\pi }{h}} \right| \cr & = 0 \times {\text{ some finite value}} = 0 \cr & {\text{Also }}f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\left| {\cos \frac{\pi }{{ - h}}} \right|}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} - h\left| {\cos \frac{\pi }{h}} \right| \cr & = 0 \times {\text{ some finite value}} = 0 \cr & \because f'\left( {{0^ + }} \right) = f'\left( {{0^ - }} \right) \cr & \Rightarrow f\,{\text{is differentiable at }}x = 0 \cr & {\text{Now }}f'\left( {{2^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {2 + h} \right) - f\left( 2 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}\left| {\cos \frac{\pi }{{2 + h}}} \right| - 4\left| {\cos \frac{\pi }{2}} \right|}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}\left( {\cos \frac{\pi }{{2 + h}}} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}}}{h}\sin \left( {\frac{\pi }{2} - \frac{\pi }{{2 + h}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}}}{h}\sin \left( {\frac{{\pi h}}{{2\left( {2 + h} \right)}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}}}{h} \times \frac{{\sin \left( {\frac{{\pi h}}{{2\left( {2 + h} \right)}}} \right)}}{{\left( {\frac{{\pi h}}{{2\left( {2 + h} \right)}}} \right)}} \times \frac{{\pi h}}{{2\left( {2 + h} \right)}} = \pi \cr & {\text{Also }}f'\left( {{2^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {2 - h} \right) - f\left( 2 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 - h} \right)}^2}\left| {\cos \frac{\pi }{{2 - h}}} \right| - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - {{\left( {2 - h} \right)}^2}\cos \left( {\frac{\pi }{{2 - h}}} \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 - h} \right)}^2}\sin \left( {\frac{\pi }{2} - \frac{\pi }{{2 - h}}} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 - h} \right)}^2}}}{h} \times \frac{{\sin \left( {\frac{{ - \pi h}}{{2\left( {2 - h} \right)}}} \right)}}{{\left( {\frac{{ - \pi h}}{{2\left( {2 - h} \right)}}} \right)}} \times \frac{{ - \pi h}}{{2\left( {2 - h} \right)}} = - \pi \cr} $$
As $$f'\left( {{2^ + }} \right) \ne f'\left( {{2^ - }} \right)\,\, \Rightarrow f$$     is not differentiable at $$x=2.$$