1
JEE Main 2024 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let $$y=y(x)$$ be the solution curve of the differential equation $$\sec y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0$$. Then $$y(\sqrt{3})$$ is equal to:

A
$$\frac{\pi}{6}$$
B
$$\frac{\pi}{12}$$
C
$$\frac{\pi}{3}$$
D
$$\frac{\pi}{4}$$
2
JEE Main 2024 (Online) 8th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

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 $$e^{-a}+4 a^2+a-1$$. Then the differential equation, whose general solution is $$y=c_1 f(x)+c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is

A
$$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$$
B
$$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$$
C
$$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$$
D
$$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$$
3
JEE Main 2024 (Online) 8th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$(1+y^2) e^{\tan x} d x+\cos ^2 x(1+e^{2 \tan x}) d y=0, y(0)=1$$. Then $$y\left(\frac{\pi}{4}\right)$$ is equal to

A
$$\frac{1}{e^2}$$
B
$$\frac{2}{e^2}$$
C
$$\frac{2}{e}$$
D
$$\frac{1}{e}$$
4
JEE Main 2024 (Online) 6th April Evening Shift
MCQ (Single Correct Answer)
+4
-1

Suppose the solution of the differential equation $$\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}$$ represents a circle passing through origin. Then the radius of this circle is :

A
$$\sqrt{17}$$
B
2
C
$$\frac{\sqrt{17}}{2}$$
D
$$\frac{1}{2}$$
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