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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

20

NTA final key.

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

Same idea in other shifts

Asked 2× in all

28 Jun 2022, Shift 2 · Q11Let the slope of the tangent to a curve $y=f(x)$ at $(x, y)$ be given by $2\tan x(\cos x-y)$ . If the curve passes through the point…HardSingle 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 108 of the 174 JEE Main shifts held from 2017 to 2026.