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

\(\mathrm{F}(x)=\left(\mathrm{f}\left(\frac{x}{2}\right)\right)^2+\left(\mathrm{g}\left(\frac{x}{2}\right)\right)^2 \)
\( \therefore \mathrm{~F}^{\prime}(x)= 2 \mathrm{f}\left(\frac{x}{2}\right) \cdot \mathrm{f}^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}+2 \mathrm{~g}\left(\frac{x}{2}\right) \cdot \mathrm{g}^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2} \)
\( = \mathrm{f}\left(\frac{x}{2}\right) \cdot \mathrm{f}^{\prime}\left(\frac{x}{2}\right)+\mathrm{g}\left(\frac{x}{2}\right) \cdot \mathrm{g}^{\prime}\left(\frac{x}{2}\right) \)
\( = \mathrm{f}\left(\frac{x}{2}\right) \cdot \mathrm{g}\left(\frac{x}{2}\right)+\mathrm{g}\left(\frac{x}{2}\right) \cdot \mathrm{f}^{\prime \prime}\left(\frac{x}{2}\right) \)
\( \ldots\left[\because \mathrm{g}(x)=\mathrm{f}^{\prime}(x) \Rightarrow \mathrm{g}^{\prime}(x)=\mathrm{f}^{\prime \prime}(x)\right] \)
\( = \mathrm{f}\left(\frac{x}{2}\right) \cdot \mathrm{g}\left(\frac{x}{2}\right)+\mathrm{g}\left(\frac{x}{2}\right) \cdot\left(-\mathrm{f}\left(\frac{x}{2}\right)\right) \)
\( = 0\)
\(\Rightarrow \mathrm{F}(x)\) is a constant for all \(x\)
\(\begin{aligned}
& F(5)=5 \\
& \Rightarrow F(x)=5 \text { for all } x \\
& \Rightarrow F(10)=5
\end{aligned}\)