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 (0,$\frac{1}{2}$+$\frac{\pi}{2\sqrt2}$) and (α,$\frac{1}{2e^{2\alpha}}$), then e α is equal to

• A. $\frac{3+√2}{3-√2}$
• B. $\frac{3}{\sqrt2}$($\frac{3+√2}{3-√2}$)
• C. $\frac{1}{\sqrt2}$($\frac{\sqrt2+1}{\sqrt2-1}$)
• D. ($\frac{\sqrt2+1}{\sqrt2-1}$)

Answer 1

Answer: (B)

$\frac{3}{\sqrt2}$($\frac{3+√2}{3-√2}$)

Answer 2

$\frac{dy}{dx}$=$\frac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}$=$\frac{e^{2x}-6e^{-x}}{2+9e^{-2x}}$
It is given that the curve passes through
(0,$\frac{1}{2}$+$\frac{\pi}{2\sqrt2}$)
$\frac{1}{2}$+$\frac{\pi}{2\sqrt2}$=$\frac{1}{2}$+$\sqrt2$tan−1⁡($\frac{3}{\sqrt2}$)+C
⇒ C=$\frac{\pi}{2\sqrt2}$−2tan−1⁡($\frac{3}{\sqrt2}$)
Now if
(α,$\frac{1}{2}$+$\frac{\pi}{2\sqrt2}$)
satisfies the curve, then
tan−1⁡($\frac{3}{\sqrt2}$)−tan−1⁡($\frac{3e6^{−α}}{\sqrt2}$)=$\frac{\pi}{2\sqrt2}$×$\frac{1}{\sqrt2}$=$\frac{\pi}{4}$
$e^\alpha=\frac{\frac{9}{2}+\frac{3}{\sqrt2}}{\frac{3}{\sqrt2}-1}$=$\frac{3}{\sqrt2}(\frac{3+\sqrt2}{3-\sqrt2})$