JEE Main 2025 Mathematics question
Let ƒ : R→R be a twice differentiable function such that ƒ(x + y) = ƒ(x) ƒ(y) for all x, y ∈ R. If ƒ'(0) = 4a and ƒ satisfies ƒ''(x) – 3a ƒ'(x) – ƒ(x) = 0, a > 0, then the area of the region R = {(x,y) | 0 ≤ y ≤ ƒ(ax), 0 ≤ x ≤ 2} is :
Answer: (1)
Solution
f(x)=e^(4ax), substituting into the ODE gives a=1/2, so f(ax)= on [0,2]. Area = ∫₀² eˣdx = e²-1.