If f(x) is a function such that f^{prime prim | vedclass

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(D) Given,$f^{\prime \prime}(x)+f(x)=0$ ... $(i)$ And $g(x)=[f(x)]^{2}+[f^{\prime}(x)]^{2}$. Differentiating $g(x)$ with respect to $x$: $g^{\prime}(x

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$8$

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(D) Given,$f^{\prime \prime}(x)+f(x)=0$ ... $(i)$ And $g(x)=[f(x)]^{2}+[f^{\prime}(x)]^{2}$. Differentiating $g(x)$ with respect to $x$: $g^{\prime}(x) = 2f(x)f^{\prime}(x) + 2f^{\prime}(x)f^{\prime \prime}(x)$ Substitute $f^{\prime \prime}(x) = -f(x)$ from equation $(i)$: $g^{\prime}(x) = 2f(x)f^{\prime}(x) + 2f^{\prime}(x)(-f(x))$ $g^{\prime}(x) = 2f(x)f^{\prime}(x) - 2f(x)f^{\prime}(x) = 0$. Since the derivative of $g(x)$ is $0$,$g(x)$ is a constant function. Given $g(3) = 8$,therefore $g(x) = 8$ for all $x$. Thus,$g(8) = 8$.

If $f(x)$ is a function such that $f^{\prime \prime}(x)+f(x)=0$ and $g(x)=[f(x)]^{2}+[f^{\prime}(x)]^{2}$ and $g(3)=8$,then $g(8)$ is equal to

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