Q21 / 57 · JEE Advanced 2010 (Paper 2) · Mathematics
Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t, \quad$ for all $x \in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then $\left(f^{-1}\right)^{\prime}(2)$ is equal to
- A1
- Correct answerB$\frac{1}{3}$
- C$\frac{1}{2}$
- D$\frac{1}{e}$
Solution
We have, $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t x \in(-1,1)$
On differentiating w.r.t. $x$, we get
$$
\begin{array}{ll}
& e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} \\
\Rightarrow \quad & f^{\prime}(x)=f(x)+\sqrt{x^4+1} e^x \\
\because & f^{-1} \text { is the inverse of } f \\
\therefore & f^{-1}(f(x))=x \\
\Rightarrow & f^{-1^{\prime}}(f(x)) f^{\prime}(x)=1 \\
\Rightarrow & f^{-1^{\prime}}(f(x))=\frac{1}{f^{\prime}(x)} \\
\Rightarrow \quad & f^{-1^{\prime}}(f(x))=\frac{1}{f(x)+\sqrt{x^4+1} e^x} \\
\text { At } \quad & x=0, f(x)=2 \\
& f^{-1^{\prime}}(2)=\frac{1}{2+1}=\frac{1}{3}
\end{array}
$$
Hence, the correct answer is option: 'B'.