1
JEE Main 2022 (Online) 25th July Morning Shift
+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
$$\frac{\sqrt{2}+1}{\sqrt{2}-1}$$
2
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
$$\left|\left(y^{2}+2 x\right)^{3}\right|=C\left(2 y^{2}+x\right)^{4}$$
3
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1
Out of Syllabus

Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$$, represents a circle with center ($$\alpha$$, $$\beta$$). Then, $$\alpha$$ + 2$$\beta$$ is equal to :

A
$$-$$1
B
0
C
1
D
2
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
EXAM MAP
Medical
NEET