Question

Let $f$ be a real-valued function defined on the intrval $(-1,1)$ such that $e^{-x}f\left(x\right)=2+{\int }_0^x\sqrt{t^4+1}dt$, for all $xϵ(-1,1)$, and let $f^{-1}$ be the inverse function of $f$. Then $\left(f^{-1}\right)‘\left(2\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $1$
• D. $\frac{1}{e}$

Answer 1

The correct option is B $\frac{1}{3}$

We have,
$e^{-x}f\left(x\right)=2{\int }_0^x\sqrt{t^4+1}txϵ(-1,1)$
Differentiating w.r.t.$x,$ we get
$e^{-x}(f‘(x)-f9x))=\sqrt{x^4+1}$
$f‘\left(x\right)=f\left(x\right)+\sqrt{x^2+1}{e}^x$
$\because f^{-1}$ is the inverse of $f$
$\therefore f^{-1‘}\left(f\right(x\left)\right)f‘\left(x\right)=\Rightarrow f^{-1‘}\left(f\right(x\left)\right)=\frac{1}{f‘\left(x\right)}$
$\Rightarrow f^{-1‘}\left(f\right(x\left)\right)=\frac{1}{f\left(x\right)+\sqrt{x^4+1}{e}^x}$
at $x=0,f\left(x\right)=2$
$f^{-1‘}\left(2\right)=\frac{1}{2+1}=\frac{1}{3}$