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

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Aicharya · Accepted Answer

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

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