If $$f\left( x \right)$$  is continuous and differentiable function and $$f\left( {\frac{1}{n}} \right) = 0\,\forall \,n \geqslant 1$$    and $$n \in I,$$   then-

Solution :
Given that $$f\left( x \right)$$  is a continuous and differentiable function and $$f\left( {\frac{1}{x}} \right) = 0,\,x = n,\,n \in \,I$$
$$\therefore f\left( {{0^ + }} \right) = f\left( {\frac{1}{\infty }} \right) = 0$$
Since R.H.L. $$=0$$
$$\therefore f\left( 0 \right) = 0$$   for $$f\left( x \right)$$  to be continuous.
$$\eqalign{ & {\text{Also }}f'\left( 0 \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - f\left( 0 \right)}}{{h - 0}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right)}}{h} = 0 \cr & = 0\left[ {{\text{Using }}f\left( 0 \right) = 0{\text{ and }}f\left( {{0^ + }} \right) = 0} \right] \cr & {\text{Hence }}f\left( 0 \right) = 0,\,f'\left( 0 \right) = 0 \cr} $$