Question

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

• A. 1
• B. 1/3
• C. 1/2
• D. 1/e

Answer 1

The correct option is B 1/3

$e^{-x}f\left(x\right)=2+{\int }_0^x\sqrt{t^4+1}dt\cdots$ (i)
$f\left(f^{-1}\right(x\left)\right)=x$...(ii)
$\Rightarrow f^{\prime }\left(f^{-1}\right(x\left)\right)\left(f^{-1}\right(x){)}^{\prime }=1\Rightarrow (f^{-1}\left(2\right){)}^{\prime }=\frac{1}{f^{\prime }\left(f^{-1}\right(2\left)\right)}$
Putting $x=0$ in equation (i), we get $f\left(0\right)=2$....(iii)
From (ii) and (iii), $f^{-1}\left(2\right)=0$
$\Rightarrow (f^{-1}\left(2\right){)}^{\prime }=\frac{1}{f^{\prime }\left(0\right)}$
Now $e^{-x}\left(f^{\prime }\right(x)-f(x\left)\right)=\sqrt{x^4+1}$ (differentiating equation (i))
Put $x=0\Rightarrow f^{\prime }\left(0\right)-2=1\Rightarrow f^{\prime }\left(0\right)=3$
$\Rightarrow \left(f^{-1}\right(2){)}^{\prime }=1/3$