Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x} f(x)=2+\int\limits_0^x \sqrt{t^4+1} d t$, for all $x \in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then, $\left(f^{-1}\right)'(2)$ is equal to

$1 / 3$

We have,

$e^{-x} f(x)=2+\int\limits_0^x \sqrt{t^4+1} d t$              ......(i)

Differentiating both sides w.r.to $x$, we get

$\Rightarrow -e^{-x} f(x)+e^{-x} f'(x)=\sqrt{x^4+1}$           ......(ii)

Now,

$fof^{-1}(x)=x$

$\Rightarrow \frac{d}{d x}\left(fof^{-1}(x)\right)=1$

$\Rightarrow \frac{d}{d x}\left\{f\left(f^{-1}(x)\right)\right\}=1$

$\Rightarrow f'\left(f^{-1}(x)\right) \times \frac{d}{d x}\left\{f^{-1}(x)\right\}=1$

$\Rightarrow f'\left(f^{-1}(2)\right)\left(f^{-1}\right)'(2)=1$          [Putting x = 2]

$\Rightarrow f'(0)\left(f^{-1}\right)(2)=1$           [∵ f(0) = 2 ⇒ f^-1(2) = 0]

$\Rightarrow \left(f^{-N_1}\right)'(2)=\frac{1}{f'(0)}$            [Putting x = 0 in (ii)     -f(0) + f'(0) = 1 ⇒ f^-1(0) = 3]

$\Rightarrow\left(f^{-1}\right)'(2)=\frac{1}{3}$