The slope of the tangent to a curve C :   y = y x at any point [ x , y ) on it is 2 e 2 x - 6 e - x + 9…

Given,

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

On rearranging we get,

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

Integrating both side we get,

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

If the curve passes through the point $\left(0,\frac{1}{2}+\frac{\pi }{2\sqrt{2}}\right)$

Then $c=\sqrt{2}\left(\frac{\pi }{4}+{\tan }^{-1}\frac{\sqrt{2}}{3}\right)$

So, curve will be $y=\frac{e^{2x}}{2}-\sqrt{2}\left({\tan }^{-1}\left(\frac{\sqrt{2}{e}^x}{3}\right)-\frac{\pi }{4}-{\tan }^{-1}\frac{\sqrt{2}}{3}\right)$

Again curve passes through the point

$\left(\alpha ,\frac{1}{2}{\mathrm{e}}^{2\alpha }\right)$ 

Putting the value in curve equation we get,

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

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

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

$\Rightarrow \frac{\sqrt{2}{e}^{\alpha }}{3}=\left(\frac{\tan \frac{\pi }{4}+{\mathrm{tantan}}^{-1}\frac{\sqrt{2}}{3}}{1-\tan \frac{\pi }{4}\times {\mathrm{tantan}}^{-1}\frac{\sqrt{2}}{3}}\right)$

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

$\Rightarrow {\mathrm{e}}^{\alpha }=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$