Question

Let $f$ be 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 $h^{\prime }\left(x\right)={\left[f\right(x\left)\right]}^2+{\left[g\right(x\left)\right]}^2$,$h\left(1\right)=8$ and $h\left(0\right)=2$, then$h\left(2\right)=$

• A. $10$
• B. $12$
• C. $14$
• D. none of these

Answer 1

The correct option is C $14$

Given $f^{\prime \prime }\left(x\right)=-f\left(x\right),f^{\prime }\left(x\right)=g\left(x\right)$

$\Rightarrow g^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)=-f\left(x\right)$

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

Differentiate on both sides w.r.t x

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

$=2f\left(x\right)g\left(x\right)+2g\left(x\right)(-f(x\left)\right)$

$(\because f^{\prime }(x)=g(x),g^{\prime }(x)=f^{\prime \prime }(k)=-f(x\left)\right)$

$h^{\prime \prime }\left(x\right)=0$

$\therefore h^{\prime }\left(x\right)=a$

$\therefore h\left(x\right)=ax+k$

$h\left(0\right)=2$

$\therefore k=2$

$h\left(1\right)=8$

$a+2=8\Rightarrow a=6$

$\therefore h\left(x\right)=6x+2$

$h\left(2\right)=14$