Question

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

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

Answer 1

Answer: (B)

$\frac{1}{3}$

$e^{-x}f\left(x\right)=2+{\int }_0^x\sqrt{t^4+1}dt$ ...(i)
Now, $f\left(f^{-1}\left(x\right)\right)=x$
$\Rightarrow f^{{}^{\prime }}\left(f^{-1}\left(x\right)\right){\left(f^{-1}\left(x\right)\right)}^{{}^{\prime }}=1\Rightarrow {\left(f^{-1}\left(2\right)\right)}^{{}^{\prime }}$
$=\frac{1}{f^{{}^{\prime }}\left(f^{-1}\left(2\right)\right)}\Rightarrow f\left(0\right)=2\Rightarrow f^{-1}\left(2\right)=0$
$\Rightarrow {\left(f^{-1}\left(2\right)\right)}^{{}^{\prime }}=\frac{1}{f\left(0\right)}$
$e^{-x}\left(f^{\prime }\right(x)-f\left(x\right))=\sqrt{x^4+1}$
Put $x=0\Rightarrow f^{\prime }\left(0\right)-2=1\Rightarrow f^{\prime }\left(0\right)=3{\left(f^{-1}\left(2\right)\right)}^{{}^{\prime }}=\frac{1}{3}$