Question

Let f be a twice differentiable function such that $f^{\prime \prime }\left(x\right)=-f\left(x\right)$ and $f^{\prime }\left(x\right)=g\left(x\right)$. If $h^{\prime }\left(x\right)={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2$, $h\left(1\right)=8andh\left(0\right)=2,$ then $h\left(2\right)$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. none of these

Answer 1

Answer: (D)

none of these

$h^{\prime }\left(x\right)={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2$

Differentiate with respect to $x$
$h^{\prime \prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)+2g\left(x\right)g^{\prime }\left(x\right)$
$=2f\left(x\right)g\left(x\right)+2g\left(x\right)f^{\prime \prime }\left(x\right)$ $[\because f^{\prime }\left(x\right)=g\left(x\right)]$
$=2f\left(x\right)g\left(x\right)-2g\left(x\right)f\left(x\right)=0$ $[\because f^{\prime \prime }\left(x\right)=-f\left(x\right)]$
Thus $h^{\prime }\left(x\right)=k$, a constant, for all $xϵR$.

Hence $h\left(x\right)=kx+m$, so that from $h\left(0\right)=2$, we get $m=2$ and from $h\left(1\right)=8$, we get $k=6$.
Therefore, $h\left(2\right)=14$.