JEE Main 2024 (8 Apr, Shift 1) Maths Q12: Area Under Curves

Question

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

Accepted Answer

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

Answer

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

Answer

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

Answer

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

Answer

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

JEE Main 8 April 2024, Shift 1, Maths Q12

Same idea in other shifts