1
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
$$\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{e}\left(\frac{2}{e \sqrt{3}}\right)$$
2
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
$32 \sqrt{2}$
3
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
4
JEE Main 2023 (Online) 30th January Evening Shift
+4
-1
The solution of the differential equation

$\frac{d y}{d x}=-\left(\frac{x^2+3 y^2}{3 x^2+y^2}\right), y(1)=0$ is :
A
$\log _e|x+y|+\frac{x y}{(x+y)^2}=0$
B
$\log _e|x+y|-\frac{x y}{(x+y)^2}=0$
C
$\log _e|x+y|+\frac{2 x y}{(x+y)^2}=0$
D
$\log _e|x+y|-\frac{2 x y}{(x+y)^2}=0$
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