JEE Advanced · Mathematics · 24. Differentiation
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
- A
1
- B
\(\frac{1}{3}\)
- C
\(\frac{1}{2}\)
- D
\(\frac{1}{e}\)
Answer & Solution
Correct Answer
(B)
\(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
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
\[
\begin{array}{ll}
& e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} \\
\Rightarrow \quad & f^{\prime}(x)=f(x)+\sqrt{x^4+1} e^x \\
\because & f^{-1} \text { is the inverse of } f \\
\therefore & f^{-1}(f(x))=x \\
\Rightarrow & f^{-1^{\prime}}(f(x)) f^{\prime}(x)=1 \\
\Rightarrow & f^{-1^{\prime}}(f(x))=\frac{1}{f^{\prime}(x)} \\
\Rightarrow \quad & f^{-1^{\prime}}(f(x))=\frac{1}{f(x)+\sqrt{x^4+1} e^x} \\
\text { At } \quad & x=0, f(x)=2 \\
& f^{-1^{\prime}}(2)=\frac{1}{2+1}=\frac{1}{3}
\end{array}
\]
On differentiating w.r.t. \(x\), we get
\[
\begin{array}{ll}
& e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} \\
\Rightarrow \quad & f^{\prime}(x)=f(x)+\sqrt{x^4+1} e^x \\
\because & f^{-1} \text { is the inverse of } f \\
\therefore & f^{-1}(f(x))=x \\
\Rightarrow & f^{-1^{\prime}}(f(x)) f^{\prime}(x)=1 \\
\Rightarrow & f^{-1^{\prime}}(f(x))=\frac{1}{f^{\prime}(x)} \\
\Rightarrow \quad & f^{-1^{\prime}}(f(x))=\frac{1}{f(x)+\sqrt{x^4+1} e^x} \\
\text { At } \quad & x=0, f(x)=2 \\
& f^{-1^{\prime}}(2)=\frac{1}{2+1}=\frac{1}{3}
\end{array}
\]
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