Question

If $f\text{'}\text{'}\left(x\right)=-f\left(x\right)$ and$g\left(x\right)=f\text{'}\left(x\right)$ and$F\left(x\right)={\left[f\left(\frac{x}{2}\right)\right]}^2+{\left[g{\left(\frac{x}{2}\right)}^{}\right]}^{2^{}}$ and$F\left(5\right)=5$ then $F\left(10\right)=?$

• A. $5$
• B. $10$
• C. $0$
• D. $15$

Answer 1

The correct option is A

$5$

Explanation for the correct option:

Find the value of $F\left(10\right)$:

We have,

$$\begin{gathered} g\left(x\right)=f\text{'}\left(x\right)...\left(i\right) \\ g\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)=–f\left(x\right)...\left(ii\right) \\ F\left(x\right)={\left[f\left(\frac{x}{2}\right)\right]}^2+{\left[g\left(\frac{x}{2}\right)\right]}^2 \end{gathered}$$

Differentiate the above function with respect to $x$.

$\begin{array}{rcl}F\text{'}\left(x\right)& =& 2f\left(\frac{x}{2}\right)f\text{'}\left(\frac{x}{2}\right)\left(\frac{1}{2}\right)+2g\left(\frac{x}{2}\right)g\text{'}\left(\frac{x}{2}\right)\left(\frac{1}{2}\right)\mathbf{[}\mathbf{\because }\frac{\mathbf{d}\mathbf{\left(}\mathbf{u}\mathbf{v}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{v}\mathbf{u}\mathbf{\text{'}}\mathbf{+}\mathbf{u}\mathbf{v}\mathbf{\text{'}}\mathbf{]}\\ & =& f\left(\frac{x}{2}\right)f\text{'}\left(\frac{x}{2}\right)+g\left(\frac{x}{2}\right)g\text{'}\left(\frac{x}{2}\right)\\ & =& 0\end{array}$

Since $F\text{'}\left(x\right)=0$ it means $F\left(x\right)=cons\tan t$

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

Hence, the correct option is A.