Question

If $g^{\prime }\left(x^3\right)=e^{x^3}+x^6-x^{-3/2}\forall x>0,g\left(0\right)=0,$ then $g\left(x\right)$ is

• A. $e^x+\frac{x^3}{3}-2\sqrt{x}$
• B. $e^{2x}+\frac{x^2}{3}-2x+1$
• C. $e^x+\frac{x^2}{2}-x$
• D. $e^x+\frac{x^3}{3}-2\sqrt{x}-1$

Answer 1

The correct option is D $e^x+\frac{x^3}{3}-2\sqrt{x}-1$

Given $g^{\prime }\left(x^3\right)=e^{x^3}+x^6-x^{-3/2}\forall x>0$
Let $x^3=t$
$\Rightarrow g^{\prime }\left(t\right)=e^t+t^2-t^{-1/2}$
$\Rightarrow \int g^{\prime }\left(t\right)dt=\int (e^t+t^2-t^{-1/2})dt$
$\therefore g\left(t\right)=e^t+\frac{t^3}{3}-2\sqrt{t}+C$
$\therefore g\left(0\right)=0\Rightarrow C=-1$
$\therefore g\left(x\right)=e^x+\frac{x^3}{3}-2\sqrt{x}-1$