If ${\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)\text{ and }{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)=-\mathrm{f}\left(\mathrm{x}\right)\text{ for all }\mathrm{x}$ and $\mathrm{f}\left(2\right)=4={\mathrm{f}}^{\mathrm{\prime }}\left(2\right),\mathrm{then} \left(\mathrm{f}\right(24){)}^2+(\mathrm{g}\left(24\right){)}^2 \mathrm{is}$

Solution:

We have,$\frac{\mathrm{d}}{\mathrm{dx}}\left\{\left(\mathrm{f}\right(\mathrm{x}){)}^2+(\mathrm{g}\left(\mathrm{x}\right){)}^2\right\}$

$\begin{array}{l}=2\mathrm{f}\left(\mathrm{x}\right)\cdot {\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)+2\mathrm{g}\left(\mathrm{x}\right)\cdot {\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\\ =2\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+2\mathrm{g}\left(\mathrm{x}\right)(-\mathrm{f}(\mathrm{x}\left)\right)\\ =2\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)-2\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\\ =0\end{array}$

$\therefore \left(\mathrm{f}\right(\mathrm{x}){)}^2+(\mathrm{g}\left(\mathrm{x}\right){)}^2$ is constant.

Again, $\begin{array}{l}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\\ \mathrm{g}\left(2\right)={\mathrm{f}}^{\mathrm{\prime }}\left(2\right)=4\\ \left(\mathrm{f}\right(24){)}^2+(\mathrm{g}\left(24\right){)}^2\end{array}$

$\begin{array}{l}=\left(\mathrm{f}\right(2){)}^2+(\mathrm{g}\left(2\right){)}^2\\ =(4{)}^2+(4{)}^2\\ =16+16=32\end{array}$