Question

Let f be a twice differentiable such that $f^{\prime \prime }\left(x\right)=-f\left(x\right)$ and $f^{\prime }\left(x\right)=g\left(x\right)$. If $h\left(x\right)={\left\{f\right(x\left)\right\}}^2+{\left\{g\right(x\left)\right\}}^2$, where $h\left(5\right)=11$. Find $h\left(10\right)$

• A. $1$
• B. $10$
• C. $11$
• D. $100$

Answer 1

Answer: (C)

$11$

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

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

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

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

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

$=0$

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

$\Rightarrow \frac{d}{dx}h\left(x\right)=0$

$\Rightarrow \int dh\left(x\right)=\int 0\times dx$

$\Rightarrow h\left(x\right)=0+c$

$\Rightarrow h\left(x\right)=c$

now, $h\left(5\right)=11$

$\therefore h\left(10\right)=h\left(5\right)=11$