If the 1011^ th term from the end in the binominal expansion of (\frac4 x5-\frac52 x)^2022 is 1024 times 1011^ th R term from the beginning, then |x| is equal to

Binomial Theorem PYQ — JEE Main 2023 · 11 Apr · Shift 2 · Q29 with Solution

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Correct answer: A

General term from the beginning

For $\left(\frac{4x}{5}-\frac{5}{2x}\right)^{2022},$ the general term from the beginning is $T_{r+1}=\binom{2022}{r}\left(\frac{4x}{5}\right)^{2022-r}\left(-\frac{5}{2x}\right)^r.$

So the $1011^{\text{th}}$ term from the beginning corresponds to $r=1010.$ Hence, $T_{1011}=\binom{2022}{1010}\left(\frac{4x}{5}\right)^{1012}\left(-\frac{5}{2x}\right)^{1010}.$

$1011^{\text{th}}$ term from the end

In an expansion of $(a+b)^n$ , the $k^{\text{th}}$ term from the end is the $(n-k+2)^{\text{th}}$ term from the beginning.

Here $n=2022$ and $k=1011$ , so $\text{term from beginning} = (2022-1011+2)^{\text{th}} = 1013^{\text{th}}.$ Thus the $1011^{\text{th}}$ term from the end is $T_{1013}$ from the beginning, i.e. $r=1012$ : $T_{1013}=\binom{2022}{1012}\left(\frac{4x}{5}\right)^{1010}\left(-\frac{5}{2x}\right)^{1012}.$

Given relation

It is given that $T_{1013}=1024\,T_{1011}.$ So, $\frac{T_{1013}}{T_{1011}}=1024.$

Now,

\frac{\binom{2022}{1012}}{\binom{2022}{1010}} \cdot \left(\frac{4x}{5}\right)^{-2} \cdot \left(-\frac{5}{2x}\right)^2.$$ Since $$\binom{2022}{1012}=\binom{2022}{2022-1012}=\binom{2022}{1010},$$ the binomial coefficient ratio is $1$. Also, $$\left(\frac{4x}{5}\right)^{-2}=\left(\frac{5}{4x}\right)^2,$$ and $$\left(-\frac{5}{2x}\right)^2=\left(\frac{5}{2x}\right)^2.$$ Therefore, $$\frac{T_{1013}}{T_{1011}}=\left(\frac{5}{4x}\right)^2\left(\frac{5}{2x}\right)^2 =\frac{625}{64x^4}.$$ Given this equals $1024$: $$\frac{625}{64x^4}=1024.$$ 4. **Solve for $x$** $$625=1024\cdot 64\,x^4.$$ Since $$1024\cdot 64=2^{10}\cdot 2^6=2^{16}=65536,$$ we get $$x^4=\frac{625}{65536}=\frac{5^4}{2^{16}}.$$ Thus, $$|x|=\frac{5}{2^4}=\frac{5}{16}.$$ 5. **Check options** $$|x|=\frac{5}{16}$$ which matches **Option A**.

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View written solutionFree

Correct answer: A

General term from the beginning

For $\left(\frac{4x}{5}-\frac{5}{2x}\right)^{2022},$ the general term from the beginning is $T_{r+1}=\binom{2022}{r}\left(\frac{4x}{5}\right)^{2022-r}\left(-\frac{5}{2x}\right)^r.$

So the $1011^{\text{th}}$ term from the beginning corresponds to $r=1010.$ Hence, $T_{1011}=\binom{2022}{1010}\left(\frac{4x}{5}\right)^{1012}\left(-\frac{5}{2x}\right)^{1010}.$

$1011^{\text{th}}$ term from the end

In an expansion of $(a+b)^n$ , the $k^{\text{th}}$ term from the end is the $(n-k+2)^{\text{th}}$ term from the beginning.

Given relation

It is given that $T_{1013}=1024\,T_{1011}.$ So, $\frac{T_{1013}}{T_{1011}}=1024.$

Now,

\frac{\binom{2022}{1012}}{\binom{2022}{1010}} \cdot \left(\frac{4x}{5}\right)^{-2} \cdot \left(-\frac{5}{2x}\right)^2.$$ Since $$\binom{2022}{1012}=\binom{2022}{2022-1012}=\binom{2022}{1010},$$ the binomial coefficient ratio is $1$. Also, $$\left(\frac{4x}{5}\right)^{-2}=\left(\frac{5}{4x}\right)^2,$$ and $$\left(-\frac{5}{2x}\right)^2=\left(\frac{5}{2x}\right)^2.$$ Therefore, $$\frac{T_{1013}}{T_{1011}}=\left(\frac{5}{4x}\right)^2\left(\frac{5}{2x}\right)^2 =\frac{625}{64x^4}.$$ Given this equals $1024$: $$\frac{625}{64x^4}=1024.$$ 4. **Solve for $x$** $$625=1024\cdot 64\,x^4.$$ Since $$1024\cdot 64=2^{10}\cdot 2^6=2^{16}=65536,$$ we get $$x^4=\frac{625}{65536}=\frac{5^4}{2^{16}}.$$ Thus, $$|x|=\frac{5}{2^4}=\frac{5}{16}.$$ 5. **Check options** $$|x|=\frac{5}{16}$$ which matches **Option A**.

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