Which of the following functions is not differentiable at $$x = 1\,?$$

Solution :
$$\eqalign{ & f\left( x \right) = \left( {{x^2} - 1} \right)\left| {\left( {x - 1} \right)\left( {x - 2} \right)} \right| \cr & f'\left( {{1^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {1 + h} \right)}^2} - 1} \right)\left| {h \cdot \left( {1 + h - 2} \right)} \right| - 0}}{h} = 0, \cr & f'\left( {{1^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {1 - h} \right)}^2} - 1} \right)\left| { - h \cdot \left( {1 - h - 2} \right)} \right| - 0}}{{ - h}} = 0 \cr & {\text{Hence, it is differentiable at }}x = 0 \cr & {\text{For, }}f\left( x \right) = \sin \left( {\left| {x - 1} \right|} \right) - \left| {x - 1} \right| \cr & f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h - h - 0}}{h} = 0, \cr & f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,\left| { - h} \right| - \left| { - h} \right|}}{{ - h}} = 0 = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \,h - h}}{{ - h}} = 0 \cr & {\text{Hence, }}f\left( x \right){\text{ is differentiable at }}x = 0 \cr & {\text{For }}f\left( x \right) = \tan \left( {\left| {x - 1} \right|} \right) + \left| {x - 1} \right| \cr & f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h + h - 0}}{h} = 2, \cr & f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,\left| { - h} \right| + \left| { - h} \right|}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{\tan \,h + h}}{{ - h}} = - 2 \cr} $$