JEE Main 2024 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 30 January 2024, Shift 2, Maths Q27

Question 27 of 90 in this shift, Maths question 27 of 30, Section B.

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

\frac{-y^2}{2Y'(x)} + 1

Y'(x) \neq 0

. If

Y(1) = 1

, then

12Y(2)

equals ________.

Official answer

NTA final key.

Chapter: Differential Equations
Topic: Linear Differential Equations
Idea tested: Forming LDEs from Geometric Conditions

Same topic in other shifts

All Linear Differential Equations questions

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8 Apr 2026, Shift 2 · Q17Let $y = y(x)$ be the solution of the differential equation $x\sqrt{1-x^2}\,dy + \left(y\sqrt{1-x^2} - x\cos^{-1}x\right)dx = 0$ ,…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.