Let f be a twice differentiable function such that f''(x) = -f(x) and f'(x) = g(x). If $h'(x)=[f(x)]^2+[g(x)]^2$, h(1) = 8 and h(0) = 2, then h(2) =

none of these

We have,

$h'(x)=[f(x)]^2+[g(x)]^2$

$\Rightarrow h''(x)=2 f(x) f'(x)+2 g(x) g'(x)$

$\Rightarrow h''(x)=2 f(x) g(x)+2 g(x) f''(x)$               $\left[\begin{array}{l}∵ g(x)=f'(x) \\ ∴ g'(x)=f''(x)\end{array}\right]$

$\Rightarrow h''(x)=2 f(x) g(x)+2 g(x)(-f(x))$            $\left[∵ f''(x)=-f(x)\right]$

$\Rightarrow h''(x)=0$

$\Rightarrow h'(x)=c$, a constant for all $x \in R$.

$\Rightarrow h(x)=c x+c_1$

$\Rightarrow h(0)=c_1$ and $h(1)=c+c_1$

$\Rightarrow 2=c_1$ and $8=c+c_1$

$\Rightarrow c_1=2$ and $c=6 $

∴  $h(x)=6 x+2 \Rightarrow h(2)=6 \times 2+2=14$