JEE Main 2024 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 4 April 2024, Shift 1, Maths Q14

Question 14 of 90 in this shift, Maths question 14 of 30, Section A.

If the solution

y=y(x)

of the differential equation

(x^4+2x^3+3x^2+2x+2)dy-(2x^2+2x+3)dx=0

satisfies

y(-1)=-\frac{\pi}{4}

, then

y(0)

is equal to :

(1) $\frac{\pi}{4}$ Official answer
(2) $\frac{\pi}{2}$
(3) $0$
(4) $-\frac{\pi}{12}$

Official answer

Option 1

NTA final key.

Chapter: Differential Equations
Topic: Variable Separable Form
Idea tested: Partial Fractions in Separated Integrals

Same topic in other shifts

All Variable Separable Form questions

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8 Apr 2024, Shift 1 · Q13Let $y=y(x)$ be the solution of the differential equation $(1+y^2)e^{\tan x}dx+\cos^2x\,(1+e^{2\tan x})dy=0$ , $y(0)=1$ . Then…MediumSingle correct
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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.