Explanation

(b)
We have,
$e^{- x}f(x) = 2 + \int_{0}^{x}\sqrt{t^{4} + 1}\ dt,x\epsilon( - 1,1)$
On differentiating w.r.t $x$, we get
$e^{- x}\left( f^{'}(x) - f(x) \right) = \sqrt{x^{4} + 1}$
$\Rightarrow f^{'}(x) = f(x) + \sqrt{x^{4} + 1}\ e^{x}$
$\because f^{- 1}$ is the inverse of $f$
$\therefore f^{- 1}\left( f(x) \right) = x$
$\Rightarrow f^{- 1^{'}}\left( f(x) \right)f'(x) = 1$
$\Rightarrow \ {f^{- 1}}^{'}\left( f(x) \right) = \frac{1}{f^{'}(x)}$
$\Rightarrow \ {f^{- 1}}^{'}\left( f(x) \right) = \frac{1}{f(x) + \sqrt{x^{4} + 1}\ e^{x}}$
As $x = 0,\ \ f(x) = 2$
and $f^{- 1}(2) = \frac{1}{2 + 1} = \frac{1}{3}$