JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If the $1011^{th}$ term from the end in the binomial expansion of $\left(\frac{4x}{5} - \frac{5}{2x}\right)^{2022}$ is $1024$ times $1011^{th}$ term from the beginning, then $|x|$ is equal to:

Choose the correct answer:

Explanation

Solution

Binomial expansion $(a+b)^n$ mein, $(k+1)^{th}$ term from beginning $T_{k+1} = \binom{n}{k} a^{n-k} b^k$ hota hai.

$(k+1)^{th}$ term from end, $(n-k+1)^{th}$ term from beginning ke barabar hota hai.

Step 1: Terms ko likhna

Yahan $n = 2022$ . $1011^{th}$ term ke liye $k = 1010$ .

Term from beginning ( $T_{1011}$ ):

$T_{1011} = \binom{2022}{1010} \left(\frac{4x}{5}\right)^{1012} \left(-\frac{5}{2x}\right)^{1010}$
Term from end ( $T'_{1011}$ ): Yeh shuruat se $(2022 - 1011 + 2) = 1013^{th}$ term hogi.

$T'_{1011} = \binom{2022}{1012} \left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1012}$

Step 2: Condition apply karna

Humein diya gaya hai $T'_{1011} = 1024 \cdot T_{1011}$ .

Kyonki $\binom{2022}{1012} = \binom{2022}{1010}$ , ye cancel ho jayenge:

\left(\frac{4x}{5}\right)^{1010} \left(-\frac{5}{2x}\right)^{1012} = 1024 \cdot \left(\frac{4x}{5}\right)^{1012} \left(-\frac{5}{2x}\right)^{1010}

\frac{\left(-\frac{5}{2x}\right)^2}{\left(\frac{4x}{5}\right)^2} = 1024 \implies \left(\frac{25}{4x^2} \cdot \frac{25}{16x^2}\right) = 1024

\frac{625}{64x^4} = 1024 \implies x^4 = \frac{625}{64 \times 1024} = \frac{5^4}{2^6 \times 2^{10}} = \frac{5^4}{2^{16}}

x^2 = \frac{5^2}{2^8} \implies |x| = \frac{5}{2^4} = \frac{5}{16}

Sahi Option: (3) $\frac{5}{16}$