Calculus

As known: $f''(x)+f(x)=0 \Rightarrow f(x)=\alpha \sin(x) + \beta \cos(x) = a\cos(x+b) \Rightarrow g(x)=-a\sin(x+b)$ So, replacing: $F(x)=a(\cos^2(\frac{x+b}{2})+\sin^2(\frac{x+b}{2}))=a$ Given the condition: $F(5)=5\Rightarrow F(x)=5 \Rightarrow F(10)=5$

Given $f''(x) = - f(x)$ $\implies f(x) = A\sin x + B \cos x$ , where $A$ and $B$ are arbitrary constants, as shown below.

$\begin{aligned} f(x) & = A\sin x + B \cos x \\ f'(x) & = A \cos x - B \sin x = g(x) \\ f''(x) & = - A\sin x - B \cos x = - f(x) \end{aligned}$

Then, we have:

$\begin{aligned} F(x) & = \left(f \left(\frac x2\right)\right)^2+\left(g \left(\frac x2\right)\right)^2 \\ & = \left(A\sin \frac x2 + B \cos \frac x2 \right)^2+\left(A\cos \frac x2 - B \sin \frac x2 \right)^2 \\ & = A^2\sin^2 \frac x2 + 2AB \sin \frac x2 \cos \frac x2 + B^2 \cos^2 \frac x2 + A^2\cos^2 \frac x2 - 2AB \sin \frac x2 \cos \frac x2 + B^2 \sin^2 \frac x2 \\ & = A^2 \left(\sin^2 \frac x2 + \cos^2 \frac x2\right) + B^2 \left(\sin^2 \frac x2 + \cos^2 \frac x2\right) \\ & = A^2 + B^2 \quad \quad \small \color{#3D99F6} \text{Note that }F(x) \text{ is independent of }x. \\ \implies F(5) & = A^2 + B^2 = 5 \\ \implies F(10) & = \boxed{5} \end{aligned}$

Solving for the functions $f(x)$ & $g(x)$ yields:

$f''(x) + f(x) = 0 \Rightarrow f(x) = A cos(x) + B sin(x)$ and $g(x) = f'(x) = -A sin(x) + B cos(x)$ (where $A,B \in \mathbb{R}).$

Solving for $F(x)$ produces:

$F(x) = [f(x/2)]^{2} + [g(x/2)]^{2} = (Acos(x/2) + Bsin(x/2))^{2} + (-Asin(x/2) + B cos(x/2))^{2}$ ;

or $A^{2}cos^{2}(x/2) + 2ABcos(x/2)sin(x/2) + B^{2}sin^{2}(x/2) + A^{2}sin^{2}(x/2) - 2ABcos(x/2)sin(x/2) + B^{2}cos^{2}(x/2);$

or $(A^2 + B^2)(cos^{2}(x/2) + sin^{2}(x/2));$

or $\boxed{F(x) = A^2 + B^2 = constant}$ for all $x \in \mathbb{R}.$ Hence, $F(10) = F(5) = \boxed{5}.$