Verified Solution • 2025–26 Session moderate

QJEEM23Ma8666: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8666 Solution: If the term from the end in the binominal expansion of is 1024…

If the 1011th 1011^{\text {th }} term from the end in the binominal expansion of (4x5−52x)2022\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{2022} is 1024 times 1011th 1011^{\text {th }} R term from the beginning, then ∣x∣|x| is equal to

Multiple Choice Options

  • A 516\frac{5}{16}
  • B 8
  • C 12
  • D 15

Step-by-Step Verified Solution

T1011\mathrm{T}_{1011} from beginning =T1010+1=\mathrm{T}_{1010+1}

=2022C1010(4x5)1012(−52x)1010={ }^{2022} \mathrm{C}_{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}

T1011\mathrm{T}_{1011} from end

=2022C1010(−52x)1012(4x5)1010={ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010}

amp; Given : 2022C1010(−52x)1012(4x5)1010amp;=210⋅2022C1010(−52x)1010(4x5)1012\begin{aligned}& \text { Given : }{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} \\\\& =2^{10} \cdot{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}\end{aligned}

amp;⇒(−52x)2=210(4x5)2amp;⇒x4=54216amp;⇒∣x∣=516\begin{aligned}&\Rightarrow \left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2 \\\\&\Rightarrow x^4=\frac{5^4}{2^{16}} \\\\& \Rightarrow |x|=\frac{5}{16}\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring