1
JEE Main 2022 (Online) 26th July Evening Shift
+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
+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 Morning Shift
+4
-1

The general solution of the differential equation $$\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$$ is :

A
$$\left(y^{2}+x\right)^{4}=\mathrm{C}\left|\left(y^{2}+2 x\right)^{3}\right|$$
B
$$\left(y^{2}+2 x\right)^{4}=C\left|\left(y^{2}+x\right)^{3}\right|$$
C
$$\left|\left(y^{2}+x\right)^{3}\right|=\mathrm{C}\left(2 y^{2}+x\right)^{4}$$
D
$$\left|\left(y^{2}+2 x\right)^{3}\right|=C\left(2 y^{2}+x\right)^{4}$$
4
JEE Main 2022 (Online) 29th June Evening Shift
+4
-1

If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to

A
2
B
$$-$$2
C
$$-$$4
D
$$-$$1
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