If $$F\left( x \right) = {\left( {f\left( {\frac{x}{2}} \right)} \right)^2} + {\left( {g\left( {\frac{x}{2}} \right)} \right)^2}$$     where $$f''\left( x \right) = - f\left( x \right)$$   and $$g\left( x \right) = f'\left( x \right)$$   and given that $$F\left( 5 \right) = 5,$$   then $$F\left( {10} \right)$$  is equal to

Solution :
$$\eqalign{ & F\left( x \right) = {\left( {f\left( {\frac{x}{2}} \right)} \right)^2} + {\left( {g\left( {\frac{x}{2}} \right)} \right)^2} \cr & \Rightarrow F'\left( x \right) = 2f\left( {\frac{x}{2}} \right).f'\left( {\frac{x}{2}} \right).\frac{1}{2} + 2g\left( {\frac{x}{2}} \right).g'\left( {\frac{x}{2}} \right).\frac{1}{2} \cr & = f\left( {\frac{x}{2}} \right).f'\left( {\frac{x}{2}} \right) + f'\left( {\frac{x}{2}} \right).f''\left( {\frac{x}{2}} \right)\,\left[ {\because g\left( x \right) = f'\left( x \right) \Rightarrow g'\left( x \right) = f''\left( x \right)} \right] \cr & = f\left( {\frac{x}{2}} \right).f'\left( {\frac{x}{2}} \right) - f'\left( {\frac{x}{2}} \right)f\left( {\frac{x}{2}} \right) \cr & = 0\,\,\left[ {\because f''\left( x \right) = - f\left( x \right)} \right] \cr & \Rightarrow F\left( x \right)\,{\text{is}}\,{\text{a}}\,{\text{constant function}}{\text{.}} \cr & \therefore F\left( x \right) = F\left( 5 \right) = 5\forall x \in R \Rightarrow F\left( {10} \right) = 5 \cr} $$