Let $f(x)$ be a positive function such that the area bounded by $y=f(x)$, $y=0$ from $x=0$ to $x=a>0$ is $e^{-a}+4a^2+a-1$. Then the differential equation, whose general solution is $y=c_1f(x)+c_2$, where $c_1$ and $c_2$ are arbitrary constants, is

Option 3: $(8e^x+1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$

$(8e^x-1)\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$

$(8e^x-1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$

$(8e^x+1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$

$(8e^x+1)\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$

JEE Main 2024 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 8 April 2024, Shift 1, Maths Q12

Question 12 of 90 in this shift, Maths question 12 of 30, Section A.

Let

f(x)

be a positive function such that the area bounded by

y=f(x)

y=0

from

x=0

x=a>0

e^{-a}+4a^2+a-1

. Then the differential equation, whose general solution is

y=c_1f(x)+c_2

, where

c_1

and

c_2

are arbitrary constants, is

(1) $(8e^x-1)\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$
(2) $(8e^x-1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$
(3) $(8e^x+1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$ Official answer
(4) $(8e^x+1)\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$

Official answer

Option 3

NTA final key.

Chapter: Differential Equations
Topic: Area Under Curves
Idea tested: Curves from Area Properties

Same topic in other shifts

All Area Under Curves questions

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.