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…

Question

The slope of the tangent to a curve $C: y=y(x)$ at any point $(x, y)$ on it is $$\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}}$$.

If $C$ passes through the points $$\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$$ and $$\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right)$$, then $$\mathrm{e}^{\alpha}$$ is equal to :

Accepted Answer

$\frac{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}=e^{2 x}-\frac{6 e^{-x}}{2+9 e^{-2 x}}$

$$
\begin{aligned}
&\int d y=\int e^{2 x} d x-3 \int \underbrace{1+\left(\frac{3 e^{-x}}{\sqrt{2}}\right)^{2}}_{\text {put } e^{-x}=t} d x \\
&=\frac{e^{2 x}}{2}+3 \int \frac{d t}{1+\left(\frac{3 t}{\sqrt{2}}\right)^{2}} \\
&=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1} \frac{3 t}{\sqrt{2}}+C
\end{aligned}
$$

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

It is given that the curve passes through

$$
\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)
$$

$$
\begin{aligned}
& \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}=\frac{1}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)+C
\end{aligned}
$$

$\Rightarrow \quad C=\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)$

Now if $\left(\alpha, \frac{1}{2} e^{2 \alpha}\right)$ satisfies the curve, then

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

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

$$
\frac{\frac{3}{\sqrt{2}}-\frac{3 e^{-\alpha}}{\sqrt{2}}}{1+\frac{9}{2} e^{-\alpha}}=1
$$

$$
\frac{3}{\sqrt{2}} e^{\alpha}-\frac{3}{\sqrt{2}}=e^{\alpha}+\frac{9}{2}
$$

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

Solutions · 1