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(C) Given $h(x)=(f(x))^2+(g(x))^2$. Taking the derivative with respect to $x$: $h^{\prime}(x)=2f(x)f^{\prime}(x)+2g(x)g^{\prime}(x)$. Since $f^{\prime

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$2$

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(C) Given $h(x)=(f(x))^2+(g(x))^2$. Taking the derivative with respect to $x$: $h^{\prime}(x)=2f(x)f^{\prime}(x)+2g(x)g^{\prime}(x)$. Since $f^{\prime}(x)=g(x)$, we have $g^{\prime}(x)=f^{\prime \prime}(x)$. Given $f^{\prime \prime}(x)=-f(x)$, we have $g^{\prime}(x)=-f(x)$. Substituting these into the derivative of $h(x)$: $h^{\prime}(x)=2f(x)g(x)+2g(x)(-f(x)) = 2f(x)g(x)-2f(x)g(x) = 0$. Since $h^{\prime}(x)=0$, $h(x)$ is a constant function. Given $h(1)=2$, it follows that $h(x)=2$ for all $x$. Therefore, $h(2)=2$.

$f(x)$ is a twice differentiable function such that $f^{\prime \prime}(x)=-f(x)$ and $f^{\prime}(x)=g(x)$. If $h(x)=(f(x))^2+(g(x))^2$ and $h(1)=2$, then $h(2)=$

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