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Q.
Let $f$ be a twice differentiable function such that $f''(x)=-f(x)$ and $f'(x)=g(x)$, $h(x)=(f(x))^{2}+(g(x))^{2}$. If $h(17)=23$, then find $h(51)$.
Continuity and Differentiability
Solution:
$h(x)=(f(x))^{2}+(g(x))^{2}$
Differentiate with respect to $x$, we get
$h'(x)=2 f(x) f'(x)+2 g(x) g'(x)$
$=2 f(x) g(x)+2 g(x) g'(x)$
$=2 f(x) g(x)+2 g(x)\left(f''(x)\right)$
$=2 f(x) g(x)-2 f(x) g(x)$
$=0$
$\Rightarrow h$ is a constant function
$h(17)=23$
$\Rightarrow h(x)=23$, for all $x$
$\Rightarrow h(51)=23$