JEE 2023 — Mathematics PYQ

Given ${\left(a{x}^2+\frac{1}{2bx}\right)}^{11}$

$T_{r+1}{=}^{11}{C}_r\cdot {\left(a{x}^2\right)}^{11-r}+{\left(\frac{1}{2bx}\right)}^r$

and ${\left(ax-\frac{1}{3bx}\right)}^{11}$

$T_{r+1}{=}^{11}{C}_r\cdot {\left(ax\right)}^{11-r}{\left(\frac{-1}{3bx}\right)}^r$

According to question

${}^{11}{C}_5{\left(a\right)}^6\left(\frac{1}{2^5{b}^5}\right){=}^{11}{C}_6\frac{a^5}{5^6{b}^6}$

$\Rightarrow ab=\frac{2^5}{3^6}$

$\Rightarrow 729ab=32$