JEE Main 2022 (25 Jul, Shift 1) Maths Q10: The slope of…

Question

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

Option 2: $\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

Answer

$\frac{3+\sqrt{2}}{3-\sqrt{2}}$

Answer

$\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

Answer

$\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$

Answer

$\frac{\sqrt{2}+1}{\sqrt{2}-1}$

JEE Main 25 July 2022, Shift 1, Maths Q10

Same idea in other shifts