JEE Main 2023

Question

If the $1011^{	ext{th}}$ term from the end in the binomial expansion of $\left(\dfrac{4x}{5}-\dfrac{5}{2x}ight)^{2022}$ is 1024 times the $1011^{	ext{th}}$ term from the beginning, then $|x|$ is equal to A). $12$ B). $8$ C). $\dfrac{5}{16}$ D). $15$

Edvaya · Accepted Answer

The correct answer is $\dfrac{5}{16}$. **Solution:** The $r^{	ext{th}}$ term from the end in the expansion of $(a+b)^n$ is equal to the $r^{	ext{th}}$ term from the beginning in the expansion of $(b+a)^n$. Therefore, the $1011^{	ext{th}}$ term from the end in $\left(\dfrac{4x}{5}-\dfrac{5}{2x}ight)^{2022}$ is the $1011^{	ext{th}}$ term from the beginning in $\left(-\dfrac{5}{2x}+\dfrac{4x}{5}ight)^{2022}$. $T_{1011}$ from the end $= {}^{2022}C_{1010}\left(-\dfrac{5}{2x}ight)^{1012}\left(\dfrac{4x}{5}ight)^{1010}$ $T_{1011}$ from the beginning $= {}^{2022}C_{1010}\left(\dfrac{4x}{5}ight)^{1012}\left(-\dfrac{5}{2x}ight)^{1010}$ Given that the term from the end is 1024 times the term from the beginning: ${}^{2022}C_{1010}\left(-\dfrac{5}{2x}ight)^{1012}\left(\dfrac{4x}{5}ight)^{1010} = 1024 	imes {}^{2022}C_{1010}\left(\dfrac{4x}{5}ight)^{1012}\left(-\dfrac{5}{2x}ight)^{1010}$ Dividing both sides by ${}^{2022}C_{1010}\left(\dfrac{4x}{5}ight)^{1010}\left(-\dfrac{5}{2x}ight)^{1010}$ yields: $\left(-\dfrac{5}{2x}ight)^2 = 1024 	imes \left(\dfrac{4x}{5}ight)^2$ $\dfrac{25}{4x^2} = 1024 	imes \dfrac{16x^2}{25}$ $x^4 = \dfrac{25 	imes 25}{4 	imes 1024 	imes 16} = \dfrac{5^4}{2^2 	imes 2^{10} 	imes 2^4} = \dfrac{5^4}{2^{16}}$ $|x| = \dfrac{5}{2^4} = \dfrac{5}{16}$

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