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
Answer & explanation
Correct answer: option 2
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]
$\Rightarrow\left(f^{-1}\right)'(2)=\frac{1}{3}$