1
JEE Main 2023 (Online) 10th April Morning Shift
+4
-1

Let $$f$$ be a differentiable function such that $${x^2}f(x) - x = 4\int\limits_0^x {tf(t)dt}$$, $$f(1) = {2 \over 3}$$. Then $$18f(3)$$ is equal to :

A
160
B
210
C
150
D
180
2
JEE Main 2023 (Online) 6th April Evening Shift
+4
-1

If the solution curve $$f(x, y)=0$$ of the differential equation

$$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$$,

passes through the points $$(1,0)$$ and $$(\alpha, 2)$$, then $$\alpha^{\alpha}$$ is equal to :

A
$$e^{\sqrt{2} e^{2}}$$
B
$$e^{2 e^{\sqrt{2}}}$$
C
$$e^{e^{2}}$$
D
$$e^{2 e^{2}}$$
3
JEE Main 2023 (Online) 1st February Evening Shift
+4
-1

Let $$\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$$ be the solution of the differential equation $$2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0, x > 0,y(2)=\sqrt{\log _{e} 2}$$. Then $$\alpha+\beta-\gamma$$ equals :

A
1
B
0
C
3
D
$$-1$$
4
JEE Main 2023 (Online) 1st February Morning Shift
+4
-1

The area enclosed by the closed curve $$\mathrm{C}$$ given by the differential equation

$$\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$$ is $$4 \pi$$.

Let $$P$$ and $$Q$$ be the points of intersection of the curve $$\mathrm{C}$$ and the $$y$$-axis. If normals at $$P$$ and $$Q$$ on the curve $$\mathrm{C}$$ intersect $$x$$-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$R S$$ is :

A
$$\frac{4 \sqrt{3}}{3}$$
B
$$2 \sqrt{3}$$
C
2
D
$$\frac{2 \sqrt{3}}{3}$$
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