The slope of the tangent to a curve $C: y=y(x)$ at any point $[x, y)$ on it is $\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}$. If $C$ passes through the points $\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$ and $\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$ then $\mathrm{e}^{\alpha}$ is equal to :

(B) $\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

(A) $\frac{3+\sqrt{2}}{3-\sqrt{2}}$

(C) $\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$

(D) $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Differential Equations

The slope of the tangent to a curve

C : y = y (x)

at any point

[x, y)

on it is

\frac{2 e ^{2 x} - 6 e ^{- x} + 9}{2 + 9 e ^{- 2 x}}

. If

C

passes through the points

(0, \frac{1}{2} + \frac{π}{2 2})

and

(α, \frac{1}{2} e^{2 α})

then

e^{α}

is equal to :

Answer: (B)

Given

\frac{d y}{d x} = \frac{2 e ^{2 x} - 6 e ^{- x} + 9}{2 + 9 e ^{- 2 x}}

.

Rewrite:

\frac{d y}{d x} = e^{2 x} - \frac{6 e ^{x}}{2 e ^{2 x} + 9}

Integrating:

y = \frac{e ^{2 x}}{2} - tan^{- 1} (\frac{2 e ^{x}}{3}) + c

Using the point

(0, \frac{1}{2} + \frac{π}{2 2})

\frac{1}{2} + \frac{π}{2 2} = \frac{1}{2} - tan^{- 1} (\frac{2}{3}) + c

c = \frac{π}{2 2} + tan^{- 1} (\frac{2}{3})

Using the point

(α, \frac{1}{2} e^{2 α})

\frac{1}{2} e^{2 α} = \frac{e ^{2 α}}{2} - tan^{- 1} (\frac{2 e ^{α}}{3}) + c

This gives

tan^{- 1} (\frac{2 e ^{α}}{3}) = \frac{π}{2 2} + tan^{- 1} (\frac{2}{3})

Solving yields

e^{α} = \frac{3}{2} (\frac{3 + 2}{3 - 2})

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option