JEE 2023 — Mathematics PYQ

In the given figure, let $AB=CD=x$, then $DE=x\tan {\theta }_2$ and $AC=x\tan {\theta }_1$.  
Also given that ${\theta }_1+{\theta }_2=\frac{\pi }{2}$ and $\sqrt{3}BE=4AB$.

Area of $△CAB=2\sqrt{3}-3$,

$\frac{1}{2}\times AB\times AC=2\sqrt{3}-3$

$\Rightarrow \frac{1}{2}x\cdot x\tan {\theta }_1=2\sqrt{3}-3...(i)$

Now, $\sqrt{3}BE=4AB$

$\Rightarrow \sqrt{3}(x\tan {\theta }_2+BD)=4AB$

$\Rightarrow \sqrt{3}(x\tan {\theta }_2+x\tan {\theta }_1)=4x$

$\Rightarrow \sqrt{3}x(\tan {\theta }_1+\tan {\theta }_2)=4x$

$\Rightarrow \tan {\theta }_1+\tan \left(\frac{\pi }{2}-{\theta }_1\right)=\frac{4}{\sqrt{3}}$

$\Rightarrow \tan {\theta }_1+\cot {\theta }_1=\frac{4}{\sqrt{3}}=\sqrt{3}+\frac{1}{\sqrt{3}}$

$\Rightarrow \tan {\theta }_1=\sqrt{3}$  
or ${\theta }_1=\frac{\pi }{3}$ and ${\theta }_2=\frac{\pi }{6}$  

or ${\theta }_1=\frac{\pi }{6}$ and ${\theta }_2=\frac{\pi }{3}$

$\therefore \frac{{\theta }_2}{{\theta }_1}$ is largest when  
${\theta }_1=\frac{\pi }{6}$ and ${\theta }_2=\frac{\pi }{3}$

So from eqn $(i)$,

$\frac{1}{2}{x}^2\times \frac{1}{\sqrt{3}}=2\sqrt{3}-3$

$\Rightarrow x^2=12-6\sqrt{3}=(3-\sqrt{3}{)}^2$

$\Rightarrow x=3-\sqrt{3}$

Hence, perimeter of $△CED=CE+ED+CD$

$=x\sec {\theta }_2+x\tan {\theta }_2+x$

$=x\left(\sec \frac{\pi }{3}+\tan \frac{\pi }{3}+1\right)$

$=(3-\sqrt{3})(2+\sqrt{3}+1)$

$=(3-\sqrt{3})(3+\sqrt{3})$

$=9-3=6\text{units}$