JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $y = y(x)$ be a solution curve of the differential equation $(1 - x^2y^2)dx = ydx + xdy$ . 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 the value of $\alpha$ is:

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Explanation

Equation ko aise likhte hain: $dx = \frac{ydx + xdy}{1 - (xy)^2}$ . Hum jante hain ki $ydx + xdy = d(xy)$ .

$\int dx = \int \frac{d(xy)}{1 - (xy)^2}$

$x + C = \frac{1}{2} \ln \left| \frac{1 + xy}{1 - xy} \right|$
Point $(1, 2)$ par $C$ nikalte hain:

$1 + C = \frac{1}{2} \ln \left| \frac{1 + 2}{1 - 2} \right| = \frac{1}{2} \ln(3) \implies C = \frac{1}{2} \ln(3) - 1$
Point $(2, \alpha)$ par value rakhte hain:

$2 + \frac{1}{2} \ln(3) - 1 = \frac{1}{2} \ln \left| \frac{1 + 2\alpha}{1 - 2\alpha} \right|$

$1 = \frac{1}{2} \ln \left| \frac{1 + 2\alpha}{3(1 - 2\alpha)} \right| \implies e^2 = \frac{1 + 2\alpha}{3 - 6\alpha}$

$3e^2 - 6\alpha e^2 = 1 + 2\alpha \implies \alpha(2 + 6e^2) = 3e^2 - 1$

$\alpha = \frac{3e^2 - 1}{2(3e^2 + 1)}$

Choose the correct answer:

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