JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 7)
If the $$1011^{\text {th }}$$ term from the end in the binominal expansion of $$\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{2022}$$ is 1024 times $$1011^{\text {th }}$$R term from the beginning, then $$|x|$$ is equal to
$$
\frac{5}{16}
$$
8
12
15
Explanation
$\mathrm{T}_{1011}$ from beginning $=\mathrm{T}_{1010+1}$
$$ ={ }^{2022} \mathrm{C}_{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010} $$
$\mathrm{T}_{1011}$ from end
$$ ={ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} $$
$$ \begin{aligned} & \text { Given : }{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} \\\\ & =2^{10} \cdot{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012} \end{aligned} $$
$$ \begin{aligned} &\Rightarrow \left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2 \\\\ &\Rightarrow x^4=\frac{5^4}{2^{16}} \\\\ & \Rightarrow |x|=\frac{5}{16} \end{aligned} $$
$$ ={ }^{2022} \mathrm{C}_{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010} $$
$\mathrm{T}_{1011}$ from end
$$ ={ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} $$
$$ \begin{aligned} & \text { Given : }{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} \\\\ & =2^{10} \cdot{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012} \end{aligned} $$
$$ \begin{aligned} &\Rightarrow \left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2 \\\\ &\Rightarrow x^4=\frac{5^4}{2^{16}} \\\\ & \Rightarrow |x|=\frac{5}{16} \end{aligned} $$
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