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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 R\mathbb{R} denote the set of all real numbers. Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} and g:R→(0,4)g: \mathbb{R} \rightarrow(0,4) be functions defined by f(x)=log⁡e(x2+2x+4)f(x)=\log _e\left(x^2+2 x+4\right), and g(x)=41+e−2xg(x)=\frac{4}{1+e^{-2 x}} Define the composite function f∘g−1f \circ g^{-1} by \left(f \circ g^{-1}\right)(x)=\left(g^{-1}(x)\right),where, whereg^{-1} is the inverse of the function gg. Then the value of the derivative of the composite function f∘g−1f \circ g^{-1} at x=2x=2 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}NowNowf(x)=log⁡0(x2+2x+4)f′(x)=2x+2x2+2x+4f′(0)=12g(x)=41+e−2x,g(0)=2g−1(g(x))=x\begin{aligned} & f(x)=\log _0\left(x^2+2 x+4\right) \\ & f^{\prime}(x)=\frac{2 x+2}{x^2+2 x+4} \\ & f^{\prime}(0)=\frac{1}{2} \\ & g(x)=\frac{4}{1+e^{-2 x}}, g(0)=2 \\ & g^{-1}(g(x))=x \end{aligned}$\begin{array}{l|l}\left(\left(\mathrm{g}^{-1}\right)^{\prime}(\mathrm{g}(\mathrm{x}))\right) \mathrm{g}^{\prime}(\mathrm{x})=1 & \mathrm{ g}(\mathrm{x})=\frac{4}{1+\mathrm{e}^{-2 \mathrm{x}}}\left(\mathrm{g}^{-1}\right)^{\ppprime}(2)=\ffrac{1}{\mathrm{ g}^{\prime}(0)}& \mathrm{g}^{\prime}( \mathrm{x})=\frac{8 \mathrm{e}^{-2 \mathrm{x}}}{\left(1+\mathrm{e}^{-2 \mathrm{x}}\right)^2}\\=\frac{1}{2}& \mathrm{ g}^{\pprime}(0)=\frac{8}{4}=2 \So \mmathrm{h}(2)=\frac{1}{4}=0.25 &\end{array}$
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