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If $f^{\prime \prime}(x)=-f(x)$, where $f(x)$ is a continuous double differentiable function and $g(x)=f^{\prime}(x)$. If $F(x)=\left(f\left(\frac{x}{2}\right)\right)^2+\left(g\left(\frac{x}{2}\right)\right)^2$ and $F(5)=5$, then $F(10)$ is
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5
5
As, $f(x)=-f(x) \Rightarrow \frac{d}{d x}\left(f^{\prime}(x)\right)=-f(x)$ $\Rightarrow \quad g^{\prime}(x)=-f(x)$ and $f^{\prime}(x)=g(x)$ where, $\therefore \quad F^{\prime}(x)=2\left(f\left(\frac{x}{2}\right)\right) \cdot f^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}+2\left(g\left(\frac{x}{2}\right)\right) \cdot g^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}=0$ [using Eq. (1)] $\therefore F(x)$ is constant $\Rightarrow F(10)=F(5)=5$
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