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

We are given that the slope of the tangent to the curve $$C: y = y(x)$$ is $$\dfrac{dy}{dx} = \dfrac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$$.

Multiply numerator and denominator by $$e^{2x}$$:

$$\dfrac{dy}{dx} = \dfrac{2e^{4x} - 6e^{x} + 9e^{2x}}{2e^{2x} + 9}$$

Let $$u = e^x$$. Then the expression becomes:

$$\dfrac{dy}{dx} = \dfrac{2u^4 + 9u^2 - 6u}{2u^2 + 9}$$

Performing polynomial division: $$2u^4 + 9u^2 = (2u^2 + 9) \cdot u^2$$, so:

$$\dfrac{2u^4 + 9u^2 - 6u}{2u^2 + 9} = u^2 - \dfrac{6u}{2u^2 + 9} = e^{2x} - \dfrac{6e^x}{2e^{2x} + 9}$$

$$y = \int e^{2x}\, dx - \int \dfrac{6e^x}{2e^{2x} + 9}\, dx$$

First integral: $$\displaystyle\int e^{2x}\, dx = \dfrac{e^{2x}}{2}$$.

Second integral: Substitute $$t = e^x$$, $$dt = e^x\, dx$$:

$$\int \dfrac{6e^x}{2e^{2x} + 9}\, dx = 6\int \dfrac{dt}{2t^2 + 9} = \dfrac{6}{2}\int \dfrac{dt}{t^2 + \frac{9}{2}}$$

$$= 3 \cdot \dfrac{1}{\frac{3}{\sqrt{2}}} \tan^{-1}\left(\dfrac{t}{\frac{3}{\sqrt{2}}}\right) = \sqrt{2}\, \tan^{-1}\left(\dfrac{\sqrt{2}\, e^x}{3}\right)$$

Therefore: $$y = \dfrac{e^{2x}}{2} - \sqrt{2}\, \tan^{-1}\left(\dfrac{\sqrt{2}\, e^x}{3}\right) + C$$.

$$\dfrac{1}{2} + \dfrac{\pi}{2\sqrt{2}} = \dfrac{1}{2} - \sqrt{2}\, \tan^{-1}\left(\dfrac{\sqrt{2}}{3}\right) + C$$

$$C = \dfrac{\pi}{2\sqrt{2}} + \sqrt{2}\, \tan^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$$

$$\dfrac{e^{2\alpha}}{2} = \dfrac{e^{2\alpha}}{2} - \sqrt{2}\, \tan^{-1}\left(\dfrac{\sqrt{2}\, e^\alpha}{3}\right) + C$$

$$\sqrt{2}\, \tan^{-1}\left(\dfrac{\sqrt{2}\, e^\alpha}{3}\right) = C = \dfrac{\pi}{2\sqrt{2}} + \sqrt{2}\, \tan^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$$

Dividing by $$\sqrt{2}$$:

$$\tan^{-1}\left(\dfrac{\sqrt{2}\, e^\alpha}{3}\right) = \dfrac{\pi}{4} + \tan^{-1}\left(\dfrac{\sqrt{2}}{3}\right)$$

Taking tangent of both sides and using $$\tan\left(\dfrac{\pi}{4} + \theta\right) = \dfrac{1 + \tan\theta}{1 - \tan\theta}$$:

$$\dfrac{\sqrt{2}\, e^\alpha}{3} = \dfrac{1 + \dfrac{\sqrt{2}}{3}}{1 - \dfrac{\sqrt{2}}{3}} = \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}$$

Solving for $$e^\alpha$$:

$$e^\alpha = \dfrac{3}{\sqrt{2}} \cdot \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}$$

The answer is Option B: $$\dfrac{3}{\sqrt{2}} \cdot \dfrac{3 + \sqrt{2}}{3 - \sqrt{2}}$$.