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JEE Main 2024 Mathematics question

MathematicsDifferential EquationJEE Main 2024

Let Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y − y = Y′(x)(X − x) and the co-ordinate axes, where (x, y) is any point on the curve, is always , Y′(x) ≠ 0. If Y(1) = 1, then 12 Y(2) equals .

Answer: 20

Solution

The tangent at (x, y) meets the axes at (x − y/y′, 0) and (0, y − xy′). The area of the triangle is ½(y − xy′)(x − y/y′) = −y²/(2y′) + 1. Simplifying gives −x²y′ +

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Question ID VK-017014