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 $f"\left(x\right)=-f\left(x\right)$ and $g\left(x\right)=f^{\prime }\left(x\right)$
and given that F(5)=5,then F(10) is equal to.

• A. 5
• B. 10
• C. 0
• D. 15

Answer 1

The correct option is A 5

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

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

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

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

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

using equation (2) in equation (1)

$f^{\prime }\left(x\right)=f\left(\frac{x}{2}\right)\cdot f^{\prime }\left(\frac{x}{2}\right)-f^{\prime }\left(\frac{x}{2}\right)f\left(\frac{x}{2}\right)=0$