Question

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

• A. $5$
• B. $10$
• C. $0$
• D. $3$

Answer 1

Answer: (A)

$5$

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

$F^{\prime }\left(x\right)=2\times f\left(\frac{x}{2}\right)\times f^{\prime }\left(\frac{x}{2}\right)+\frac{1}{2}\times g\left(\frac{x}{2}\right)\times g^{\prime }\left(\frac{x}{2}\right)\times \frac{1}{2}$

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

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

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

$\therefore F^{\prime }\left(x\right)=(f\left(\frac{x}{2}\right)\times f^{\prime }\left(\frac{x}{2}\right))+(f^{\prime }\left(\frac{x}{2}\right)\times (-f\left(\frac{x}{2}\right)))=0$

$\therefore F\left(x\right)$ is fixed constant function

$\therefore F\left(10\right)=F\left(5\right)=5$