1
JEE Main 2024 (Online) 30th January Evening Shift
Numerical
+4
-1

Let $$Y=Y(X)$$ be a curve lying in the first quadrant such that the area enclosed by the line $$Y-y=Y^{\prime}(x)(X-x)$$ and the co-ordinate axes, where $$(x, y)$$ is any point on the curve, is always $$\frac{-y^2}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) \neq 0$$. If $$Y(1)=1$$, then $$12 Y(2)$$ equals __________.

2
JEE Main 2024 (Online) 30th January Morning Shift
Numerical
+4
-1

Let $$y=y(x)$$ be the solution of the differential equation $$\left(1-x^2\right) \mathrm{d} y=\left[x y+\left(x^3+2\right) \sqrt{3\left(1-x^2\right)}\right] \mathrm{d} x, -1< x<1, y(0)=0$$. If $$y\left(\frac{1}{2}\right)=\frac{\mathrm{m}}{\mathrm{n}}, \mathrm{m}$$ and $$\mathrm{n}$$ are co-prime numbers, then $$\mathrm{m}+\mathrm{n}$$ is equal to __________.

3
JEE Main 2024 (Online) 29th January Morning Shift
Numerical
+4
-1

If the solution curve $$y=y(x)$$ of the differential equation $$\left(1+y^2\right)\left(1+\log _{\mathrm{e}} x\right) d x+x d y=0, x > 0$$ passes through the point $$(1,1)$$ and $$y(e)=\frac{\alpha-\tan \left(\frac{3}{2}\right)}{\beta+\tan \left(\frac{3}{2}\right)}$$, then $$\alpha+2 \beta$$ is _________.

4
JEE Main 2024 (Online) 27th January Evening Shift
Numerical
+4
-1

If the solution curve, of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y-2}{x-y}$$ passing through the point $$(2,1)$$ is $$\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _{\mathrm{e}}\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log _{\mathrm{e}}|x-1|$$, then $$5 \beta+\alpha$$ is equal to __________.