Question

If $f^{\prime }\left(x\right)=g\left(x\right)$ and $g^{\prime }\left(x\right)=-f\left(x\right)$ and $f\left(2\right)=4=f^{\prime }\left(2\right)$ then $f^2\left(16\right)+g^2\left(16\right)$ is

• A. 16
• B. 32
• C. 64
• D. None of these

Answer 1

Answer: (B)

32

$\frac{d}{dx}(f^2\left(x\right)+g^2\left(x\right))=2[f\left(x\right)f^{\prime }\left(x\right)+g\left(x\right)g^{\prime }\left(x\right)]=2[f\left(x\right)-g\left(x\right)f\left(x\right)]=0$
Thus $f^2\left(x\right)+g^2\left(x\right)$ is constant.
Therefore $f^2\left(16\right)+g^2\left(16\right)=f^2\left(2\right)+g^2\left(2\right)=f^2\left(2\right)+{\left(f^{\prime }\left(2\right)\right)}^2=16+16=32$