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← JEE Advanced 2025 · Paper 2
MathematicsNumerical+4Question 14 of 48

JEE Advanced 2025 · Paper 2, Mathematics question 14

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

f(x)=log⁡e(x2+2x+4),andg(x)=41+e−2x.f(x) = \log_e\left(x^2 + 2x + 4\right),\quad\text{and}\quad g(x) = \frac{4}{1 + e^{-2x}}.

Define the composite function f∘g−1f\circ g^{-1} by (f∘g−1)(x)=f(g−1(x))(f\circ g^{-1})(x) = f\left(g^{-1}(x)\right), where g−1g^{-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 _____.

Answer and worked solution

Official answer: 0.2

Where to evaluate. g(x)=2g(x)=2 gives e−2x=1e^{-2x}=1, so g−1(2)=0g^{-1}(2)=0.

Derivative of g−1g^{-1}.
g′(x)=8e−2x(1+e−2x)2,g′(0)=2,(g−1)′(2)=12.g'(x)=\frac{8e^{-2x}}{(1+e^{-2x})^2},\qquad g'(0)=2,\qquad (g^{-1})'(2)=\frac12.

Derivative of f. f′(x)=2x+2x2+2x+4f'(x)=\frac{2x+2}{x^2+2x+4}, so f′(0)=12f'(0)=\frac12.

Chain rule.
(f∘g−1)′(2)=f′(0)⋅(g−1)′(2)=14=0.25.(f\circ g^{-1})'(2)=f'(0)\cdot(g^{-1})'(2)=\frac14=0.25.