JEE Advanced 2025Paper 2
- Exam
- JEE Advanced
- Year
- 2025
- Paper
- Paper 2
- Subject
- Mathematics
- Chapter
- Differentiation
JEE Advanced 2025 · Paper 2 · Mathematics · Differentiation
Let ℝ denote the set of all real numbers. Let f: ℝ arrow ℝ and g: ℝ...
Let denote the set of all real numbers. Let and be functions defined by , and
Define the composite function by \left(f \circ g^{-1}\right)(x)=\left(g^{-1}(x)\right)g^{-1} is the inverse of the function .
Then the value of the derivative of the composite function at is ________ .
Correct answer
0
Step-by-step solution
\begin{aligned} & \text{ Let } \mathrm{h}(\mathrm{x})=\mathrm{f}\left(\mathrm{g}^{-1}(\mathrm{x})\right) \text{ and } \mathrm{g}(0)=2 \\ & \begin{aligned} \mathrm{h}^{\prime}(\mathrm{x}) & =\mathrm{f}^{\prime}\left(\mathrm{g}^{-1}(\mathrm{x})\right) \cdot\left(\mathrm{g}^{-1}(\mathrm{x})\right)^{\prime} \\ \mathrm{h}^{\prime}(2) & =\mathrm{f}^{\prime}\left(\mathrm{g}^{-1}(2) \cdot\left(\mathrm{g}^{-1}\right)^{\prime}(2)\right. \\ & =\mathrm{f}^{\prime}(0) \cdot\left(\mathrm{g}^{-1}\right)^{\prime}(2)\end{aligned}\end{aligned}$\begin{array}{l|l}\left(\left(\mathrm{g}^{-1}\right)^{\prime}(\mathrm{g}(\mathrm{x}))\right) \mathrm{g}^{\prime}(\mathrm{x})=1 & \mathrm{ g}(\mathrm{x})=\frac{4}{1+\mathrm{e}^{-2 \mathrm{x}}}\left(\mathrm{g}^{-1}\right)^{\ppprime}(2)=\ffrac{1}{\mathrm{ g}^{\prime}(0)}& \mathrm{g}^{\prime}( \mathrm{x})=\frac{8 \mathrm{e}^{-2 \mathrm{x}}}{\left(1+\mathrm{e}^{-2 \mathrm{x}}\right)^2}\\=\frac{1}{2}& \mathrm{ g}^{\pprime}(0)=\frac{8}{4}=2 \So \mmathrm{h}(2)=\frac{1}{4}=0.25 &\end{array}$