1
JEE Main 2022 (Online) 26th July Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let the solution curve $$y=f(x)$$ of the differential equation $$\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}$$, $$x\in(-1,1)$$ pass through the origin. Then $$\int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$$ is equal to

A
$$\frac{\pi}{3}-\frac{1}{4}$$
B
$$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$$
C
$$\frac{\pi}{6}-\frac{\sqrt{3}}{4}$$
D
$$\frac{\pi}{6}-\frac{\sqrt{3}}{2}$$
2
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1

If $${{dy} \over {dx}} + 2y\tan x = \sin x,\,0 < x < {\pi \over 2}$$ and $$y\left( {{\pi \over 3}} \right) = 0$$, then the maximum value of $$y(x)$$ is :

A
$${1 \over 8}$$
B
$${3 \over 4}$$
C
$${1 \over 4}$$
D
$${3 \over 8}$$
3
JEE Main 2022 (Online) 25th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus

Let a smooth curve $$y=f(x)$$ be such that the slope of the tangent at any point $$(x, y)$$ on it is directly proportional to $$\left(\frac{-y}{x}\right)$$. If the curve passes through the points $$(1,2)$$ and $$(8,1)$$, then $$\left|y\left(\frac{1}{8}\right)\right|$$ is equal to

A
$$2 \log _{e} 2$$
B
4
C
1
D
$$4 \log _{e} 2$$
4
JEE Main 2022 (Online) 25th July Morning Shift
MCQ (Single Correct Answer)
+4
-1

The slope of the tangent to a curve $$C: y=y(x)$$ at any point $$(x, y)$$ on it is $$\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}$$. If $$C$$ passes through the points $$\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$$ and $$\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$$, then $$\mathrm{e}^{\alpha}$$ is equal to :

A
$$\frac{3+\sqrt{2}}{3-\sqrt{2}}$$
B
$$\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$$
C
$$\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$$
D
$$\frac{\sqrt{2}+1}{\sqrt{2}-1}$$
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