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

We have,

$f''(x)=-f(x)$  and  $g(x)=f'(x) \Rightarrow g'(x)=-f(x)$  and  $g(x)=f'(x)$

Now,

$F(x)=\left\{f\left(\frac{x}{2}\right)\right\}^2+\left\{g\left(\frac{x}{2}\right)\right\}^2$

$\Rightarrow F'(x)=f\left(\frac{x}{2}\right) f'\left(\frac{x}{2}\right)+g\left(\frac{x}{2}\right) g'\left(\frac{x}{2}\right)$

$\Rightarrow F'(x)=f\left(\frac{x}{2}\right) g\left(\frac{x}{2}\right)-g\left(\frac{x}{2}\right) f\left(\frac{x}{2}\right)=0$

⇒ F(x) = Constant ⇒ F(10) = F(5) = 5