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 $f^2\left(4\right)+g^2\left(4\right)$ is equal to

• A. $8$
• B. $16$
• C. $32$
• D. $64$

Answer 1

The correct option is A

$8$

Explanation for the correct option:

Find the value of $f^2\left(4\right)+g^2\left(4\right)$:

Given that,

$f\text{'}\left(x\right)=g\left(x\right)$

$g\text{'}\left(x\right)=–f\left(x\right)$

Let us consider $r\left(x\right)=f^2\left(x\right)+g^2\left(x\right)$

Differentiating $r\left(x\right)$ with respect to $x$ we get,

$$\begin{gathered} r\text{'}\left(x\right)=2f\left(x\right)f\text{'}\left(x\right)+2g\left(x\right)g\text{'}\left(x\right) \\ =2f\left(x\right)g\left(x\right)+2g\left(x\right)(-f(x\left)\right) \\ =0 \end{gathered}$$

Since, $r\text{'}\left(x\right)=0$

$\begin{array}{rcl}r\left(x\right)& =& cons\tan t\Rightarrow r\left(4\right)=r\left(2\right)\\ r\left(2\right)& =& f^2\left(2\right)+f{\text{'}}^2\left(2\right)\\ & =& 2^2+2^2\\ & =& 8\end{array}$

$r\left(2\right)=r\left(4\right)=8$

Hence, the correct option is A.