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(D) 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

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(D) 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 substitute these into the derivative: $h^{\prime}(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$. Since $h^{\prime}(x) = 0$,$h(x)$ is a constant function. Therefore,$h(5) = h(10) = 1$.

Let $f$ be a twice differentiable function such that $f^{\prime \prime}(x) = -f(x)$,$f^{\prime}(x) = g(x)$ and $h(x) = [f(x)]^2 + [g(x)]^2$. If $h(5) = 1$,then $h(10)$ is $\qquad$

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