JEE Main 2024 Mathematics question
Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e⁻ᵃ + 4a² + a − 1. Then the differential equation, whose general solution is y = c₁f(x) + c₂, where c₁ and c₂ are arbitrary constants, is:
Answer: (C)
Solution
Differentiating the area gives f(x) = −e⁻ˣ + 8x + 1, so f′ = e⁻ˣ + 8 and f″ = −e⁻ˣ. From
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