1
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1

If the solution curve of the differential equation $$\frac{d y}{d x}=\frac{x+y-2}{x-y}$$ passes through the points $$(2,1)$$ and $$(\mathrm{k}+1,2), \mathrm{k}>0$$, then

A
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
B
$$\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
C
$$2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)$$
D
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right)$$
2
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1

Let $$y=y(x)$$ be the solution curve of the differential equation $$\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1$$, which passes through the point $$(0,1)$$. Then $$y(1)$$ is equal to :

A
$$\frac{1}{2}$$
B
$$\frac{3}{2}$$
C
$$\frac{5}{2}$$
D
$$\frac{7}{2}$$
3
JEE Main 2022 (Online) 29th July Morning Shift
+4
-1

Let the solution curve $$y=y(x)$$ of the differential equation $$\left(1+\mathrm{e}^{2 x}\right)\left(\frac{\mathrm{d} y}{\mathrm{~d} x}+y\right)=1$$ pass through the point $$\left(0, \frac{\pi}{2}\right)$$. Then, $$\lim\limits_{x \rightarrow \infty} \mathrm{e}^{x} y(x)$$ is equal to :

A
$$\frac{\pi}{4}$$
B
$$\frac{3\pi}{4}$$
C
$$\frac{\pi}{2}$$
D
$$\frac{3\pi}{2}$$
4
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
$$19-6 \log _{\mathrm{e}} 3$$
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