1
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1

The area enclosed by the closed curve $$\mathrm{C}$$ given by the differential equation

$$\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$$ is $$4 \pi$$.

Let $$P$$ and $$Q$$ be the points of intersection of the curve $$\mathrm{C}$$ and the $$y$$-axis. If normals at $$P$$ and $$Q$$ on the curve $$\mathrm{C}$$ intersect $$x$$-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$R S$$ is :

A
$$\frac{4 \sqrt{3}}{3}$$
B
$$2 \sqrt{3}$$
C
2
D
$$\frac{2 \sqrt{3}}{3}$$
2
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1

If $$y=y(x)$$ is the solution curve of the differential equation

$$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to

A
$$\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)$$
B
$$\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2 \sqrt{3}}{e}\right)$$
C
$$\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)$$
D
$$\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)$$
3
JEE Main 2023 (Online) 31st January Evening Shift
+4
-1
Let $y=y(x)$ be the solution of the differential equation

$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$

such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to :
A
64
B
$16 \sqrt{2}$
C
32
D
$32 \sqrt{2}$
4
JEE Main 2023 (Online) 31st January Morning Shift
+4
-1
Out of Syllabus

Let a differentiable function $$f$$ satisfy $$f(x)+\int_\limits{3}^{x} \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3$$. Then $$12 f(8)$$ is equal to :

A
19
B
34
C
17
D
1
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