JEE Main 2025 Jan 22 Shift 1, Mathematics Q2: Area as Definite Integral

JEE Main2025Jan 22, Shift 1Mathematics

Let $f : R \to R$ be a twice differentiable function such that $f (x + y) = f (x) f (y)$ for all $x, y \in R$ . If $f^{'} (0) = 4 a$ and $f$ satisfies $f^{''} (x) - 3 a f^{'} (x) - f (x) = 0$ , $a > 0$ , then the area of the region $R = {(x, y) ∣ 0 \leq y \leq f (a x), 0 \leq x \leq 2}$ is :

A
$e^{2} - 1$
B
$e^{4} + 1$
C
$e^{4} - 1$
D
$e^{2} + 1$

Solution

The Cauchy multiplicative relation $f (x + y) = f (x) f (y)$ together with differentiability forces $f (x) = e^{λ x}$ for some constant $λ$ . From $f^{'} (0) = λ = 4 a$ , hence $f (x) = e^{4 a x}$ .

Substituting into $f^{''} (x) - 3 a f^{'} (x) - f (x) = 0$ gives the auxiliary equation $λ^{2} - 3 aλ - 1 = 0$ with $λ = 4 a$ :

$(4 a)^{2} - 3 a (4 a) - 1 = 0 \Rightarrow 16 a^{2} - 12 a^{2} = 1 \Rightarrow a^{2} = \frac{1}{4}$ . Since $a > 0$ , $a = \frac{1}{2}$ .

So $f (a x) = e^{4 a \cdot a x} = e^{4 a^{2} x} = e^{x}$ .

Area $= \int_{0}^{2} e^{x} d x = e^{2} - 1$ .

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q2: Area as Definite Integral

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