Question

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

• A. 5
• B. 10
• C. 0
• D. 15

Answer 1

Answer: (A)

5

$F\left(x\right)={\left\{f\left(\frac{x}{2}\right)\right\}}^2+{\left\{g\left(\frac{x}{2}\right)\right\}}^2$
Hence,
$F^{\prime }\left(x\right)=f\left(\frac{x}{2}\right)f^{\prime }\left(\frac{x}{2}\right)+g\left(\frac{x}{2}\right)g^{\prime }\left(\frac{x}{2}\right)$
$\Rightarrow F^{\prime }\left(x\right)=f\left(\frac{x}{2}\right)f^{\prime }\left(\frac{x}{2}\right)+f^{\prime }\left(\frac{x}{2}\right)f^{\prime \prime }\left(\frac{x}{2}\right)$
$\Rightarrow F^{\prime }\left(x\right)=f\left(\frac{x}{2}\right)f^{\prime }\left(\frac{x}{2}\right)-f\left(\frac{x}{2}\right)f^{\prime }\left(\frac{x}{2}\right)$
$\Rightarrow F^{\prime }\left(x\right)=0$
Hence, $F\left(x\right)=k$
$F\left(5\right)=5$
Hence, $F\left(x\right)=5$, for all $x$.
Hence, $F\left(10\right)=5$