Question

If $f^{\prime \prime }\left(x\right)=-f\left(x\right)$, $g\left(x\right)=f^{\prime }\left(x\right)$, $F\left(x\right)={\left(f\left(\frac{x}{2}\right)\right)}^2+{\left(g\left(\frac{x}{2}\right)\right)}^2$ and given that $F\left(5\right)=5$, then $F\left(10\right)$ is

• A. $5$
• B. $10$
• C. $0$
• D. $15$

Answer 1

The correct option is A $5$

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

Here, $g\left(x\right)=f^{\prime }\left(x\right)$ and $g^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)=-f\left(x\right)$
So, $F^{\prime }\left(x\right)=f\left(\frac{x}{2}\right)g\left(\frac{x}{2}\right)-f\left(\frac{x}{2}\right)g\left(\frac{x}{2}\right)=0$
$\Rightarrow F\left(x\right)$ is a constant function.
$\Rightarrow F\left(10\right)=5(\because F(5)=5)$