Question

If $f\text{'}\left(x\right)=g\left(x\right)$ and $g\text{'}\left(x\right)=-f\left(x\right)$ for all $x$ and$f\left(2\right)=4=f\text{'}\left(2\right)$, then $f^2\left(16\right)+g^2\left(16\right)$ is

• A. $16$
• B. $32$
• C. $64$
• D. $noneofthese$

Answer 1

The correct option is B

$32$

Explanation for the correct option:

Step 1: Finding the first derivative of $f^2\left(x\right)+g^2\left(x\right)$

$\begin{array}{rcl}D\left[f^2\right(x)+g^2(x\left)\right]& =& 2f\left(x\right)f\text{'}\left(x\right)+2g\left(x\right)g\text{'}\left(x\right)\\ & =& 2(-g\text{'}(x\left)\right)g\left(x\right)+2g\left(x\right)g\text{'}\left(x\right)\\ & =& 0\end{array}$

Step 2: Calculate the value of $f^2\left(16\right)+g^2\left(16\right)$

$$\begin{gathered} D\left[f^2\right(x)+g^2(x\left)\right]=0 \\ \Rightarrow f^2\left(x\right)+g^2\left(x\right)=cons\tan t \\ \Rightarrow f^2\left(16\right)+g^2\left(16\right)=f^2\left(2\right)+g^2\left(2\right)=4^2+4^2 \\ \Rightarrow f^2\left(16\right)+g^2\left(16\right)=32. \end{gathered}$$

Hence, the correct option is B.