1
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
$$\frac{1}{e}$$
2
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
$$\frac{1}{2}$$
3
JEE Main 2024 (Online) 6th April Morning Shift
MCQ (Single Correct Answer)
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0$$ and $$y\left(e^{-1}\right)=0$$. Then, $$y(e)$$ is equal to

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

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

A
$$\frac{1}{4}\left(e^{\pi / 2}-1\right)$$
B
$$\frac{1}{2}\left(1-e^{\pi / 2}\right)$$
C
$$\frac{1}{4}\left(1-e^{\pi / 2}\right)$$
D
$$\frac{1}{2}\left(e^{\pi / 2}-1\right)$$
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