1
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1 Let $$y=y(x)$$ be the solution curve of the differential equation $$\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}$$, $$x >1$$ passing through the point $$\left(2, \sqrt{\frac{1}{3}}\right)$$. Then $$\sqrt{7}\, y(8)$$ is equal to :

A
$$11+6 \log _{e} 3$$
B
19
C
$$12-2 \log _{\mathrm{e}} 3$$
D
$$19-6 \log _{\mathrm{e}} 3$$
2
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1 The differential equation of the family of circles passing through the points $$(0,2)$$ and $$(0,-2)$$ is :

A
$$2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0$$
B
$$2 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0$$
C
$$2 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0$$
D
$$2 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0$$
3
JEE Main 2022 (Online) 28th July Morning Shift
+4
-1 Let the solution curve of the differential equation $$x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :

A
$$\frac{1}{2}$$
B
$$\frac{3}{2}$$
C
$$-$$$$\frac{3}{2}$$
D
$$\frac{5}{2}$$
4
JEE Main 2022 (Online) 28th July Morning Shift
+4
-1 If $$y=y(x), x \in(0, \pi / 2)$$ be the solution curve of the differential equation

$$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$,

with $$y(\pi / 4)=\mathrm{e}^{-\pi}$$, then $$y(\pi / 6)$$ is equal to :

A
$$\frac{2}{\sqrt{3}} e^{-2 \pi / 3}$$
B
$$\frac{2}{\sqrt{3}} \mathrm{e}^{2 \pi / 3}$$
C
$$\frac{1}{\sqrt{3}} e^{-2 \pi / 3}$$
D
$$\frac{1}{\sqrt{3}} e^{2 \pi / 3}$$
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