JEE Main 2022 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 25 June 2022, Shift 1, Maths Q15

Question 15 of 90 in this shift, Maths question 15 of 30, Section A.

Let

y=y(x)

be the solution of the differential equation

(x+1)y'-y=e^{3x}(x+1)^2

, with

y(0)=\frac{1}{3}

. Then, the point

x=-\frac{4}{3}

for the curve

y=y(x)

is :

(1)not a critical point
(2)a point of local minimaOfficial answer
(3)a point of local maxima
(4)a point of inflection

Official answer

Option 2

NTA final key.

Chapter: Differential Equations
Topic: Linear Differential Equations
Idea tested: Finding Extrema directly from LDE

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 108 of the 174 JEE Main shifts held from 2017 to 2026.