JEE Main 2024 — Mathematics Calculus

Given,

$${\mathrm{sec}}^{2}xdx+({e}^{2y}{\mathrm{tan}}^{2}x+\mathrm{tan}x)dy=0,$$sec2xdx+(e2ytan2x+tanx)dy=0,

$$\Rightarrow {\mathrm{sec}}^{2}\frac{dx}{dy}+{e}^{2y}{\mathrm{tan}}^{2}x+\mathrm{tan}x=0$$⇒sec2dydx​+e2ytan2x+tanx=0

Now, let $\mathrm{tan}x=t\Rightarrow {\mathrm{sec}}^{2}x\frac{dx}{dy}=\frac{dt}{dy}$tanx=t⇒sec2xdydx​=dydt​

$$\Rightarrow \frac{dt}{dy}+{e}^{2y}\times {t}^{2}+t=0$$⇒dydt​+e2y×t2+t=0

$$\Rightarrow \frac{dt}{dy}+t=-{t}^{2}.{e}^{2y}$$⇒dydt​+t=−t2.e2y

$$\Rightarrow \frac{1}{{t}^{2}}\frac{dt}{dy}+\frac{1}{t}=-{e}^{2y}$$⇒t21​dydt​+t1​=−e2y

Now, taking $\frac{1}{t}=u\Rightarrow \frac{-1}{{t}^{2}}\frac{dt}{dy}=\frac{du}{dy}$t1​=u⇒t2−1​dydt​=dydu​

$$\Rightarrow \frac{-du}{dy}+u=-{e}^{2y}$$⇒dy−du​+u=−e2y

$$\Rightarrow \frac{du}{dy}-u={e}^{2y}$$⇒dydu​−u=e2y

Now, finding an integrating factor $\displaystyle \mathrm{IF}={e}^{-\int dy}={e}^{-y}$IF=e−∫dy=e−y

So, the solution is given by,

$$u{e}^{-y}=\int {e}^{-y}\times {e}^{2y}dy$$ue−y=∫e−y×e2ydy

$$\Rightarrow \frac{{e}^{-y}}{t}={e}^{y}+c$$⇒te−y​=ey+c

$$\Rightarrow \frac{1}{\mathrm{tan}x}\times {e}^{-y}={e}^{y}+c$$⇒tanx1​×e−y=ey+c

Now, using the given condition at $x=\frac{\pi }{4},y=0$x=4π​,y=0 we get,

$$\Rightarrow c=0$$⇒c=0

$$\Rightarrow \frac{1}{\mathrm{tan}x}={e}^{2y}$$⇒tanx1​=e2y

Now, putting the value $x=\frac{\pi }{6},y=\alpha$x=6π​,y=α we get,

$${e}^{2\alpha }=\sqrt{3}$$e2α=3​

$$\Rightarrow {e}^{8\alpha }=9$$⇒e8α=9