Question

If f(x) is a function such that $f^{\prime \prime }\left(x\right)+f\left(x\right)=0$ and $g\left(x\right)=\left[f\right(x){]}^2+[f^{\prime }\left(x\right){]}^2$ and g(3)=8, then $g\left(8\right)=$_____

• A. $0$
• B. $3$
• C. $5$
• D. $8$

Answer 1

The correct option is D $8$

We have, $g\left(x\right)=\left[f\right(x){]}^2+[f^{\prime }\left(x\right){]}^2$

Differentiate the function g(x)

$\Rightarrow g^{\prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)+2f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)$, use chain rule

$=2f^{\prime }\left(x\right)\left[f\right(x)+f^{\prime \prime }(x\left)\right]=2f^{\prime }\left(x\right)\left(0\right)=0$, use the given condition

Hence $g\left(x\right)$ is a constant function

$\Rightarrow g\left(x\right)=c$, constant

But $g\left(3\right)=8$, so $g\left(x\right)=8$, for all real x

Hence $g\left(8\right)=8$