1
JEE Main 2023 (Online) 12th April Morning Shift
+4
-1 Let $$y=y(x), y > 0$$, be a solution curve of the differential equation $$\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x$$. If $$y(0)=1$$ and $$y(2 \sqrt{2})=\beta$$, then

A
$$e^{\beta^{-1}}=e^{-2}(3+2 \sqrt{2})$$
B
$$e^{3 \beta^{-1}}=e(5+\sqrt{2})$$
C
$$e^{3 \beta^{-1}}=e(3+2 \sqrt{2})$$
D
$$e^{\beta^{-1}}=e^{-2}(5+\sqrt{2})$$
2
JEE Main 2023 (Online) 11th April Evening Shift
+4
-1 Let $$y=y(x)$$ be the solution of the differential equation $$\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x > 0$$. If $$y(1)=2$$, then $$y(2)$$ is equal to :

A
$$\frac{693}{128}$$
B
$$\frac{697}{128}$$
C
$$\frac{637}{128}$$
D
$$\frac{679}{128}$$
3
JEE Main 2023 (Online) 11th April Morning Shift
+4
-1 Let $$y=y(x)$$ be a solution curve of the differential equation.

$$\left(1-x^{2} y^{2}\right) d x=y d x+x d y$$.

If the line $$x=1$$ intersects the curve $$y=y(x)$$ at $$y=2$$ and the line $$x=2$$ intersects the curve $$y=y(x)$$ at $$y=\alpha$$, then a value of $$\alpha$$ is

A
$$\frac{1+3 e^{2}}{2\left(3 e^{2}-1\right)}$$
B
$$\frac{3 e^{2}}{2\left(3 e^{2}-1\right)}$$
C
$$\frac{1-3 e^{2}}{2\left(3 e^{2}+1\right)}$$
D
$$\frac{3 e^{2}}{2\left(3 e^{2}+1\right)}$$
4
JEE Main 2023 (Online) 10th April Morning Shift
+4
-1 Let $$f$$ be a differentiable function such that $${x^2}f(x) - x = 4\int_0^x {tf(t)dt}$$, $$f(1) = {2 \over 3}$$. Then $$18f(3)$$ is equal to :

A
160
B
210
C
150
D
180
JEE Main Subjects
Physics
Mechanics
Electricity
Optics
Modern Physics
Chemistry
Physical Chemistry
Inorganic Chemistry
Organic Chemistry
Mathematics
Algebra
Trigonometry
Coordinate Geometry
Calculus
EXAM MAP
Joint Entrance Examination