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 _________.
Answer & explanation
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