Given,
The 1011th term from the end in the binomial expansion of (54x−2x5)2022 is 1024 times 1011th term from the beginning,
So, according to question we have, T1011T1013=1024
⇒C10102022(54x)1012(2x−5)1010C10122022(54x)1010⋅(2x−5)1012=1024
⇒(54x)2(2x−5)2=1024
⇒26x454=210
⇒(x5)4=216
⇒∣x∣=165
⇒32∣x∣=10
Note this question was bonus in Jee Main 2023 April session as answer key was not matching, so we have modified the question.