Given $$f:\left[ { - 2a,\,2a} \right] \to R$$    is an odd function such that the left hand derivative at $$x = a$$  is zero and $$f\left( x \right) = f\left( {2a - x} \right)\forall \,x\, \in \left( {a,\,2a} \right),$$       then its left had derivative at $$x = - a$$   is :

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