Question

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

• A. 8
• B. 3
• C. 0
• D. 5

Answer 1

Answer: (A)

We have, $g(x)=[f(x){]}^2+{\left[f^{\prime }(x)\right]}^2$
Differentiate the function $g(x)$
$\Rightarrow g^{\prime }(x)=2f(x)f^{\prime }(x)+2f^{\prime }(x)f^{\prime \prime }(x)$, use chain rule
$=2f^{\prime }(x)\left[f(x)+f^{\prime \prime }(x)\right]=2f^{\prime }(x)(0)=0$, use the given condition
Hence $g(x)$ is a constant function
$\Rightarrow g(x)=c$, constant
But $g(3)=8,\mathrm{so}g(x)=8$, for all real $x$
Hence $g(8)=8$