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 $h^{\prime }\left(x\right)={\left[f\left(x\right)\right]}^2+{\left[g\left(x\right)\right]}^2,h\left(1\right)=6$ and $h\left(0\right)=4$ then $h\left(4\right)$ is equal to?

• A. $16$
• B. $12$
• C. $13$
• D. None of these

Answer 1

Answer: (B)

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

Differentiating both side w.r.t $x$, we get

$h^{\prime \prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)+2g\left(x\right)g^{\prime }\left(x\right)=2f\left(x\right)g\left(x\right)+2g\left(x\right)g^{\prime \prime }\left(x\right)[\because f^{\prime }\left(x\right)=g\left(x\right)]=2f\left(x\right)g\left(x\right)-2g\left(x\right)f\left(x\right)=0[\because f^{\prime \prime }\left(x\right)=-f\left(x\right)]$

Thus $h^{\prime }\left(x\right)=k,$ a constant for all $x\in R$.

Hence $h\left(x\right)=ax+b,$ so that form $h\left(0\right)=4,$ we get $b=4$

and from $h\left(1\right)=6$ we get $a=2$

Therefore $h\left(4\right)=12$