JEE Advanced 2025 Paper 2Question 1
Let $\mathbb{R}$ denote the set of all real numbers. Let $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to(0,4)$ be functions defined by $$f(x)=\log_e(x^2+2x+4),\quad\text{and}\quad g(x)=\frac{4}{1+e^{-2x}}.$$ Define the composite function $f\circ g^{-1}$ by $(f\circ g^{-1})(x)=f(g^{-1}(x))$, where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $f\circ g^{-1}$ at $x=2$ is ______.
Functions
Answer$0.2 to 0.3$
Solution
$g(x)=2\Rightarrow e^{-2x}=1\Rightarrow g^{-1}(2)=0$. $(f\circ g^{-1})'(2)=\frac{f'(0)}{g'(0)}=\frac{2/4}{8e^{0}/4}=\frac{1/2}{2}=\frac14$.