JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q10

← JEE Main 2022 · 25 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 10 of 90

The slope of the tangent to a curve $C:y=y(x)$ at any point $(x,y)$ on it is $\dfrac{2e^{2x}-6e^{-x}+9}{2+9e^{-2x}}$ . If C passes through the points $\left(0,\dfrac12+\dfrac\pi{2\sqrt2}\right)$ and $\left(\alpha,\dfrac12e^{2\alpha}\right)$ , then $e^\alpha$ is equal to :

\dfrac{3+\sqrt2}{3-\sqrt2}

\dfrac3{\sqrt2}\left(\dfrac{3+\sqrt2}{3-\sqrt2}\right)

\dfrac1{\sqrt2}\left(\dfrac{\sqrt2+1}{\sqrt2-1}\right)

\dfrac{\sqrt2+1}{\sqrt2-1}

Answer and worked solution

Official answer: B

Simplify. $y'=e^{2x}-\dfrac{6e^x}{2e^{2x}+9}$ , so $y=\dfrac{e^{2x}}2-\sqrt2\tan^{-1}\dfrac{\sqrt2e^x}3+C$ ; the first point gives $C=\frac\pi{2\sqrt2}+\sqrt2\tan^{-1}\frac{\sqrt2}3$ .

At α. $\tan^{-1}\frac{\sqrt2e^\alpha}3=\frac\pi4+\tan^{-1}\frac{\sqrt2}3$ , so $\dfrac{\sqrt2e^\alpha}3=\dfrac{3+\sqrt2}{3-\sqrt2}$ .

(B)

Topics:Variable separable Substitution

← Previous question Next question →

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 10