Explanation

(d)
$h^{'}(x) = \lbrack f(x)^{2} + g(x)^{2}\rbrack$
$\Rightarrow \ \ h^{''}(x) = 2f(x)f^{'}(x) + 2g(x)g^{'}(x)$
$\ \begin{bmatrix}
\because g(x) = f^{'}(x) \\
\Rightarrow g^{'}(x) = f^{''}(x)
\end{bmatrix}$
$\therefore\ h^{''}(x) = 2f(x)g(x) + 2g(x)\left( - f(x) \right)$
$\lbrack\because f^{''}(x) = - f(x)\rbrack$
$\Rightarrow h^{''}(x) = 0$
$\Rightarrow h^{'}(x) = C$, a constant for all $x \in R$
$\Rightarrow \ h(x) = Cx + C_{1}$
$\Rightarrow h(0) = C_{1}\ and\ h(1) = C + C_{1}$
$\Rightarrow 2 = C_{1}\ and\ 8 = C + C_{1}$
$\Rightarrow C_{1} = 2\ and\ C = 6$
$\therefore h(x) = 6x + 2$
$\Rightarrow h(2) = 6 \times 2 + 2 = 14$