JEE Advanced 2025 Paper 2, Mathematics Section 3 Q6: Some Basic Applications Of Derivatives

JEE Advanced2025Paper 2Mathematics Section 3

Let $R$ denote the set of all real numbers. Let $f : R \to R$ and $g : R \to (0, 4)$ be functions defined by

$f (x) = lo g_{e} (x^{2} + 2 x + 4), and g (x) = \frac{4}{1 + e ^{- 2 x}} .$

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 ________.

Your answer

Solution

Step 1: Find $g^{- 1} (2)$ .

Solve $g (x_{0}) = 2$ : $\frac{4}{1 + e ^{- 2 x_{0}}} = 2 \Rightarrow 1 + e^{- 2 x_{0}} = 2 \Rightarrow x_{0} = 0$ .

Step 2: Chain rule.

$(f \circ g^{- 1})^{'} (2) = f^{'} (g^{- 1} (2)) \cdot (g^{- 1})^{'} (2) = f^{'} (0) \cdot \frac{1}{g ^{'} ( 0 )}$ .

Step 3: Compute $f^{'} (0)$ .

$f^{'} (x) = \frac{2 x + 2}{x ^{2} + 2 x + 4}$ , so $f^{'} (0) = \frac{2}{4} = \frac{1}{2}$ .

Step 4: Compute $g^{'} (0)$ .

$g^{'} (x) = \frac{8 e ^{- 2 x}}{( 1 + e ^{- 2 x} ) ^{2}}$ , so $g^{'} (0) = \frac{8}{4} = 2$ .

Step 5: Combine.

$(f \circ g^{- 1})^{'} (2) = \frac{1/2}{2} = \frac{1}{4} = 0.25$ .

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