Question

If$f\text{'}\text{'}\left(x\right)=-f\left(x\right)$ and $f\text{'}\left(x\right)=g\left(x\right),m\left(x\right)={\left[f\right(x\left)\right]}^2+{\left[g\right(x\left)\right]}^2$ then find $m\left(20\right)$, where $m\left(10\right)=22$

• A. $22$
• B. $11$
• C. $0$
• D. $5$

Answer 1

The correct option is A

$22$

Explanation for the correct option:-

Finding the value of $m\left(20\right)$:

Given,

$$\begin{gathered} f’’\left(x\right)=-f\left(x\right) \\ f’\left(x\right)=g\left(x\right) \\ m\left(x\right)={[f(x\left)\right]}^2+{[g(x\left)\right]}^2 \end{gathered}$$

Differentiate the given function with respect to $x$.

$$\begin{gathered} m\left(x\right)={[f(x\left)\right]}^2+{[g(x\left)\right]}^2 \\ \Rightarrow m\text{'}\left(x\right)=2f\left(x\right)f\text{'}\left(x\right)+2g\left(x\right)g\text{'}\left(x\right) \\ =2g\left(x\right)\left[f\right(x)+g\text{'}(x\left)\right] \\ =2g\left[f\right(x)+f\text{'}\text{'}(x\left)\right]\left[\because f\text{'}\text{'}\left(x\right)=g\text{'}\left(x\right)\right] \\ =g\left[f\right(x)-f(x\left)\right] \\ =0 \end{gathered}$$

Here, $m\text{'}\left(x\right)=0$ so

$m\left(x\right)=\mathrm{constant}\mathrm{for}\mathrm{all}\mathrm{x}$

$m\left(10\right)=m\left(20\right)=22$

Hence, the correct option is A.