If the 1011 th term from the end in the binomial expansion of 4 x 5 - 5 2 x 2022 is 1024 times 1011 th term…

Given,

The ${1011}^{\mathrm{th}}$ term from the end in the binomial expansion of ${\left(\frac{4x}{5}-\frac{5}{2x}\right)}^{2022}$ is $1024$ times ${1011}^{\mathrm{th}}$ term from the beginning,

So, according to question we have, $\frac{T_{1013}}{T_{1011}}=1024$
$\Rightarrow \frac{{}^{2022}C_{1012}{\left(\frac{4x}{5}\right)}^{1010}·{\left(\frac{-5}{2x}\right)}^{1012}}{{}^{2022}C_{1010}{\left(\frac{4x}{5}\right)}^{1012}{\left(\frac{-5}{2x}\right)}^{1010}}=1024$

$\Rightarrow \frac{{\left({\displaystyle \frac{-5}{2x}}\right)}^2}{{\left({\displaystyle \frac{4x}{5}}\right)}^2}=1024$

$\Rightarrow \frac{5^4}{2^6{x}^4}=2^{10}$

$\Rightarrow {\left(\frac{5}{x}\right)}^4=2^{16}$

$\Rightarrow \left|x\right|=\frac{5}{16}$

$\Rightarrow 32\left|x\right|=10$

Note this question was bonus in Jee Main 2023 April session as answer key was not matching, so we have modified the question.