Question

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 $\left(\alpha, \frac{1}{2} e ^{2 \alpha}\right)$ 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{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}} $
$ \frac{d y}{d x}=e^{2 x}-\frac{6 e^x}{2 e^{2 x}+9} $
$ y=\frac{e^{2 x}}{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 $e ^\alpha=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$