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# Let f be the twice differentiable function such that $f''\left( x \right)=-f\left( x \right)\text{ and }f'\left( x \right)=g\left( x \right).$If $h'\left( x \right)=\left[ f{{\left( x \right)}^{2}}+g{{\left( x \right)}^{2}} \right],h\left( 1 \right)=8,h\left( 0 \right)=2$, then h(2) equals toA. 1B. 2C. 3D. None of these  Hint: Differentiate the function $h'\left( x \right)=f{{\left( x \right)}^{2}}+g{{\left( x \right)}^{2}}$. Then substitute the values and given conditions to the differentiate function.

Given that $h'\left( x \right)=f{{\left( x \right)}^{2}}+g{{\left( x \right)}^{2}}..............\left( 1 \right)$
Differentiate equation (1) with respect to x.
We get;
$h''.\left( x \right)=2f\left( x \right)f'\left( x \right)+2g\left( x \right)g'\left( x \right)................\left( 2 \right)$
It’s given that f ’(x) = g(x)
$\therefore h''\left( x \right)=2f\left( x \right)g\left( x \right)+2g\left( x \right)g'\left( x \right)$
Given f ”(x) = -f(x)
f ’(x) = g’(x), take the differentiation
then f ” (x) = g’(x)
Substitute, the values for f ’(x) and f ”(x)
\begin{align} & \therefore h''\left( x \right)=2f\left( x \right)g\left( x \right)+2g\left( x \right)f''\left( x \right) \\ & =2f\left( x \right)g\left( x \right)-2g\left( x \right)f\left( x \right)=0 \\ \end{align}
Thus h’(x) = k, a constant for all $x\in R$
Let h(x) = kx + m, forms the equation of line
i.e. h(0) = 2, from question
$\Rightarrow$ We get a possible equation as $2=kx+m$
$2=k\times 0+m\Rightarrow m=2,\text{ where }x=0$
Then, h(1) = 8
$\Rightarrow$8 = k x 1 + m
8 = k x 1 +2
$\Rightarrow$k = 8 – 2 = 6
Hence, we got the values of m and k
m = 2, k = 6
h(2) , where x = 2
Substitute these to get value of h(2);
$\therefore$ h(2) = 6 x 2 + 2 = 14
Hence, value of h(2) = 14.

Note: f ’(x) = g(x), is differentiated again to form f ”(x) = g’(x). As we got h”(x) = 0, h’(x) = k is constant for all $x\in R$.
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