JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Continuity and Differentiability

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}

(f \circ g^{- 1}) (x) = (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}

x = 2

is ____.

Answer: 0.25

Solution

Let

h (x) = f (g^{- 1} (x))

. By the chain rule:

h^{'} (x) = f^{'} (g^{- 1} (x)) \cdot (g^{- 1})^{'} (x)

Note that

g (0) = \frac{4}{1 + e ^{0}} = \frac{4}{2} = 2

, so

g^{- 1} (2) = 0

.

Thus:

h^{'} (2) = f^{'} (0) \cdot (g^{- 1})^{'} (2)

Compute

f^{'} (x)

f^{'} (x) = \frac{2 x + 2}{x ^{2} + 2 x + 4} \Rightarrow f^{'} (0) = \frac{2}{4} = \frac{1}{2}

For

(g^{- 1})^{'} (2)

, use the inverse function theorem:

(g^{- 1})^{'} (2) = \frac{1}{g ^{'} ( g ^{- 1} ( 2 ))} = \frac{1}{g ^{'} ( 0 )}

Compute

g^{'} (x)

g (x) = \frac{4}{1 + e ^{- 2 x}} \Rightarrow g^{'} (x) = \frac{4 \cdot 2 e ^{- 2 x}}{( 1 + e ^{- 2 x} ) ^{2}} = \frac{8 e ^{- 2 x}}{( 1 + e ^{- 2 x} ) ^{2}}

g^{'} (0) = \frac{8 \cdot 1}{( 1 + 1 ) ^{2}} = \frac{8}{4} = 2

Thus:

(g^{- 1})^{'} (2) = \frac{1}{2}

Therefore:

h^{'} (2) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} = 0.25

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer