JEE Advanced · Mathematics · 24. Differentiation
If the function \(f(x)=x^3+e^{\frac{x}{2}}\) and \(g(x)=f^{-1}(x)\), then the value of \(g^{\prime}(1)\) is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
Given, \(g\{f(x)\}=x\)
\[
\begin{array}{lc}
\Rightarrow & g^{\prime}\{f(x)\} f^{\prime}(x)=1 \\
\text { If } & f(x)=1 \Rightarrow x=0, f(0)=1
\end{array}
\]
Substitute \(x=0\) in Eq. (i), we get
\[
\begin{aligned}
& g^{\prime}(1)=\frac{1}{f^{\prime}(0)} \\
& \Rightarrow \quad g^{\prime}(1)=2 \\
& {\left[\because f^{\prime}(x)=3 x^2+\frac{1}{2} e^{x / 2} \Rightarrow f^{\prime}(0)=\frac{1}{2}\right]} \\
&
\end{aligned}
\]
\[
\begin{array}{lc}
\Rightarrow & g^{\prime}\{f(x)\} f^{\prime}(x)=1 \\
\text { If } & f(x)=1 \Rightarrow x=0, f(0)=1
\end{array}
\]
Substitute \(x=0\) in Eq. (i), we get
\[
\begin{aligned}
& g^{\prime}(1)=\frac{1}{f^{\prime}(0)} \\
& \Rightarrow \quad g^{\prime}(1)=2 \\
& {\left[\because f^{\prime}(x)=3 x^2+\frac{1}{2} e^{x / 2} \Rightarrow f^{\prime}(0)=\frac{1}{2}\right]} \\
&
\end{aligned}
\]
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