Let f be a real-valued function defined on the interval (-1,1) su… - Vidyadip

MCQ

Let $\mathrm{f}$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t$, for all $\mathrm{x}$ $\in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then $\left(f^1\right)^{\prime}(2)$ is equal to

A
$1$
✓
$1 / 3$
C
$1 / 2$
D
$1 / e$

✓

Answer

Correct option: B.

$1 / 3$

b
$ \mathrm{e}^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t \ldots \text { (i) } $

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

$ \Rightarrow f^{\prime}\left(f^{-1}(x)\right)\left(f^{-1}(x)\right)^{\prime}=1 \Rightarrow\left(f^{-1}(2)\right)^{\prime}=\frac{1}{f^{\prime}\left(f^{-1}(2)\right)} \Rightarrow f(0)=2 \Rightarrow f^{-1}(2)=0 $

$ \left(f^{-1}(2)\right)^{\prime}=\frac{1}{f^{\prime}(0)} $

$ e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} $

$ P u t x=0 \Rightarrow f^{\prime}(0)-2=1 \Rightarrow f^{\prime}(0)=3 $

$ \left(f^{-1}(2)\right)^{\prime}=1 / 3$

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