We have $$f\left( x \right) = \left[ {{{\tan }^2}x} \right]$$
$$\tan \,x$$   is an increasing function for $$ - \frac{\pi }{4} < x < \frac{\pi }{4}$$
$$\eqalign{ & \therefore \,\tan \left( { - \frac{\pi }{4}} \right) < \tan \,x < \tan \left( {\frac{\pi }{4}} \right) \cr & \Rightarrow - 1 < \tan \,x < 1 \cr & \Rightarrow 0 < {\tan ^2}x < 1 \cr & \Rightarrow \left[ {{{\tan }^2}x} \right] = 0 \cr & {\text{Hence, }}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \left[ {{{\tan }^2}x} \right] = 0 \cr & {\text{Also }}f\left( 0 \right) = 0 \cr & \therefore f\left( x \right)\,\,{\text{is continuous at}}\,x = 0 \cr} $$

$$\eqalign{ & {\text{Given, }}f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a - h} \right) - f\left( a \right)}}{{ - h}} = 0.....\left( 1 \right) \cr & {\text{Now, }}f'\left( { - {a^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( { - a - h} \right) - f\left( { - a} \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - f\left( {a + h} \right) + f\left( a \right)}}{{ - h}}\,\,\,\,\,\left[ {\because \,f\left( x \right)\,{\text{is odd function}}} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - f\left( {a - h} \right) + f\left( a \right)}}{{ - h}}\,\,\,\,\left[ {\because \,f\left( {2a - x} \right) = f\left( x \right) \Rightarrow f\left( {a + x} \right) = f\left( {a - x} \right)} \right] \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a - h} \right) - f\left( a \right)}}{h} \cr & = 0\,\,\,\,\,\,\left[ {{\text{From }}\left( 1 \right)} \right] \cr} $$

Given function is :
\[f\left( x \right) = \sin \left| x \right| = \left\{ \begin{array}{l} \sin \left( x \right),\,\,\,\,\,\,\,x \ge 0\\ \sin \left( { - x} \right),\,\,\,x < 0 \end{array} \right. = \left\{ \begin{array}{l} \,\,\,\,\sin \,x,\,\,\,\,\,x \ge 0\\ - \sin \,x,\,\,\,\,x < 0 \end{array} \right.\]
$$\eqalign{ & {\text{L}}{\text{.H}}{\text{.D}}{\text{. at}}\,x = 0 = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{0 - h - 0}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin \left( { - h} \right) - 0}}{{ - h}} \cr & = - 1 \cr & {\text{R}}{\text{.H}}{\text{.D}}{\text{. at}}\,x = 0 = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( x \right)}}{{0 + h - 0}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {h - 0} \right)}}{h} \cr & = 1 \cr & {\text{L}}{\text{.H}}{\text{.D}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} \cr} $$
$$f\left( x \right)$$  is not differentiable at $$x = 0.$$

$$\eqalign{ & {\text{RH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{\log \left| {\frac{1}{2} + h - 1} \right| - \log \left| {\frac{1}{2} - 1} \right|}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\log \left| {h - \frac{1}{2}} \right| - \log \frac{1}{2}}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\log \left( {\frac{1}{2} - h} \right) + \log \,2}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{\frac{1}{2} - h}}\left( { - 1} \right) + 0}}{1} \cr & = - 2 \cr & {\text{LH derivative}} = \mathop {\lim }\limits_{h \to 0} \frac{{\log \left| {\frac{1}{2} - h - 1} \right| - \log \left| {\frac{1}{2} - 1} \right|}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\log \left( {\frac{1}{2} + h} \right) - \log \frac{1}{2}}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{\frac{1}{2} + h}} - 0}}{{ - 1}} \cr & = - 2 \cr & {\bf{Alternate\,method :}} \cr & {\text{Around}}\,\,x = \frac{1}{2},\,f\left( x \right) = \log \left( {1 - x} \right) \cr & {\text{So, }}f'\left( x \right) = \frac{{ - 1}}{{1 - x}} \cr & \therefore f'\left( {\frac{1}{2}} \right) = \frac{{ - 1}}{{1 - \frac{1}{2}}} = - 2 \cr} $$

$$\eqalign{ & U = {x^3} \cr & \frac{{dU}}{{dx}} = 3{x^2}......(1) \cr & V = {x^2} \cr & \frac{{dV}}{{dx}} = 2x......(2) \cr & {\text{From }}(1){\text{ and }}(2) \cr & \frac{{dU}}{{dV}} = \frac{{3{x^2}}}{{2x}} = \frac{3}{2}x \cr} $$

Suppose, there are two points $${x_1}$$ and $${x_2}$$ in $$\left( {a,\,b} \right)$$  such that $$f'\left( {{x_1}} \right) = f'\left( {{x_2}} \right) = 0.$$     By Rolle's theorem applied to $$f'$$ on $$\left[ {{x_1},\,{x_2}} \right],$$  there must be a $$c\, \in \,\left( {{x_1},\,{x_2}} \right)$$   such that $$f''\left( c \right) = 0.$$   This contradicts the given condition $$f''\left( x \right) < 0,\forall \,x\, \in \left( {a,\,b} \right).$$
Hence, our assumption is wrong. Therefore, there can be at most one point in $$\left( {a,\,b} \right)$$  at which $$f'\left( x \right)$$  is zero.