JEE Main 2023 Mathematics question
Let the tangent at any point P on a curve passing through the points (1, 1) and , intersect positive x-axis and y-axis at the points A and B respectively. If PA : PB = 1 : k and y = y(x) is the solution of the differential equation , y(0) = k, then 4y(1) – 5 is equal to .
Answer: 0
Solution
TOPIC NOT IN SYLLABUS: the source's own working derives k=2, then 2y(1) = 3(ln3 – 1) + 5, giving 4y(1) in terms of ln3 that does not
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