Question

Let $f\left(x\right)={\int }_2^x\frac{dt}{\sqrt{1+t^4}}$ and $g$ be the inverse of $f$. Then the value of $g^{\prime }\left(0\right)$ is

• A. $1$
• B. $17$
• C. $\sqrt{17}$
• D. none of these

Answer 1

The correct option is C $\sqrt{17}$

$f\left(x\right)={\int }_2^x\frac{dt}{\sqrt{1+t^4}}$ $[\because f\left(2\right)=0]$
$g\left(f\left(x\right)\right)=x$
$g^{\prime }\left(f\left(x\right)\right)f^{\prime }\left(x\right)=1$
$g^{\prime }\left(0\right)f^{\prime }\left(2\right)=2$
$g^{\prime }\left(0\right)=\frac{1}{f^{\prime }\left(2\right)}=\sqrt{17}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\sqrt{1+x^4}}\Rightarrow f^{\prime }\left(2\right)=\frac{1}{\sqrt{17}}$