Question

Let $f\left(x\right)={\int }_3^x\frac{dt}{\sqrt{t^4+3t^2+13}}$. If $g\left(x\right)$ is the inverse of $f\left(x\right)$ then $g^{\prime }\left(0\right)$ has the value equal to

• A. $\frac{1}{11}$
• B. $11$
• C. $\sqrt{13}$
• D. $\frac{1}{\sqrt{13}}$

Answer 1

The correct option is B $11$

$\frac{dy}{dx}=\frac{1}{\sqrt{x^4+3x^2+13}}$ when $y=f\left(x\right)$
$\therefore g^{\prime }\left(y\right)=\frac{1}{dy/dx}=\sqrt{x^4+3x^2+13}$
When $y=0$, $x=3$
Hence $g^{\prime }\left(0\right)=\sqrt{3^4+27+13}=\sqrt{121}=11$