Question

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

• A. 5
• B. 10
• C. 0
• D. 15

Answer 1

The correct option is A

5

$F\left(x\right)=\left(f{\left(\frac{x}{2}\right)}^2\right)+{\left(g\left(\frac{x}{2}\right)\right)}^2$
$\Rightarrow F^{\prime }\left(x\right)=2f\left(\frac{x}{2}\right).f^{\prime }\left(\frac{x}{2}\right).\frac{1}{2}+2g\left(\frac{x}{2}\right).g^{\prime }\left(\frac{x}{2}\right).\frac{1}{2}$
$=f\left(\frac{x}{2}\right).f^{\prime }\left(\frac{x}{2}\right)+f^{\prime }\left(\frac{x}{2}\right).f^{\prime }\left(\frac{x}{2}\right)$
$[\because g(x)=f^{\prime }(x)\Rightarrow g(x)=f"(x\left)\right]$
$=f\left(\frac{x}{2}\right).f^{\prime }\left(\frac{x}{2}\right)-f^{\prime }\left(\frac{x}{2}\right)f\left(\frac{x}{2}\right)$
$=0[\because f^{\prime }(x)=-f(x\left)\right]$
$\Rightarrow F\left(x\right)$ is a constant fucntion.
$\therefore F\left(x\right)=F\left(5\right)=5\forall x\in R\Rightarrow F\left(10\right)=5$