JEE Main 2023 — Mathematics Calculus

Given,

$${({\mathrm{log}}_{e}(\mathrm{cos}y))}^{2}\mathrm{cos}ydx-(1+3x{\mathrm{log}}_{e}(\mathrm{cos}y))\mathrm{sin}ydy=0$$(loge​(cosy))2cosydx−(1+3xloge​(cosy))sinydy=0

$$\Rightarrow \frac{dx}{dy}+(\frac{-3\mathrm{tan}y}{\mathrm{lncos}y})x=\frac{\mathrm{sin}y}{{(In\mathrm{cos}y)}^{2}\cdot \mathrm{cos}y}$$⇒dydx​+(lncosy−3tany​)x=(Incosy)2⋅cosysiny​

Which is a linear differential equation,

So, Integrating factor will be, $\displaystyle IF={e}^{3\int \frac{-\mathrm{tan}y}{\mathrm{lncos}y}dy}={(\mathrm{lncos}y)}^{3}$IF=e3∫lncosy−tany​dy=(lncosy)3

So, solution of differential equation is given by,

$$x{(\mathrm{lncos}y)}^{3}=\int \frac{\mathrm{sin}y}{\mathrm{cos}y}(\mathrm{lncos}y)dy$$x(lncosy)3=∫cosysiny​(lncosy)dy

$$\Rightarrow x{(\mathrm{lncos}y)}^{3}=\frac{-{(\mathrm{lncos}y)}^{2}}{2}+c$$⇒x(lncosy)3=2−(lncosy)2​+c

Now using the given value of $x(\frac{\pi }{3})=\frac{1}{2\mathrm{ln}2}$x(3π​)=2ln21​

We get, $c=0$c=0

Hence, $x=\frac{-1}{2\mathrm{ln}\mathrm{cos}y}$x=2lncosy−1​

Now finding, $x(\frac{\pi }{6})=-\frac{1}{2\mathrm{ln}(\frac{\sqrt{3}}{2})}=\frac{1}{\mathrm{ln}4-\mathrm{ln}3}$x(6π​)=−2ln(23​​)1​=ln4−ln31​

Now on comparing with $\frac{1}{{\mathrm{log}}_{e}m-{\mathrm{log}}_{e}n}$loge​m−loge​n1​ we get, $m=4,n=3$m=4,n=3

Hence, $mn=12$mn=12