Question

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

• A. $\frac{3+\sqrt{2}}{3-\sqrt{2}}$
• B. $\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$
• C. $\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$
• D. $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

Answer 1

Answer: (B)

$\frac{dy}{dx}=\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}$
$\frac{dy}{dx}=e^{2x}-\frac{6e^x}{2e^{2x}+9}$
$y=\frac{e^{2x}}{2}-{\tan }^{-1}\left(\frac{\sqrt{2}{e}^x}{3}\right)+c$
If $C$ passes through the point $\left(0,\frac{1}{2}+\frac{\pi }{2\sqrt{2}}\right)$
$c=-\frac{\pi }{4}-{\tan }^{-1}\frac{\sqrt{2}}{3}$
Again C passes through the point $\left(\alpha ,\frac{1}{2}{e}^{2\alpha }\right)$
then $e^{\alpha }=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$