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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Binomial Theorem

For an integer n ≥ 2 , if the arithmetic mean of all coefficients...

For an integer n≥2\mathrm{n} \geq 2, if the arithmetic mean of all coefficients in the binomial expansion of (x+y)2n−3is16,thenthedistanceofthepoint(x+y)^{2 n-3} is 16 , then the distance of the pointP(2n−1,n2−4n)P\left(2 n-1, n^2-4 n\right) from the line x+y=8x+y=8 is:For an integer n ≥ 2 , if the arithmetic mean of all coefficients in the binomial expansion of (x+y)^ 2 n-3 is 16 ,

Options

  1. A2\sqrt{2}
  2. B222 \sqrt{2}
  3. C525 \sqrt{2}
  4. D323 \sqrt{2}

Correct answer

D. 323 \sqrt{2}

Step-by-step solution

No. of terms in (x+y)^{(2 n-3)} \Rightarrow\left[(2n−3+1)(2n−2)\begin{array}{c}(2 n-3+1) \\ (2 n-2)\end{array}\right]. \therefore$ sum of all coefficients $=2^{2 n-3}$ (Put $\mathrm{x}=\mathrm{y}$=1$ ) $\therefore$ Arithmetic mean of all coefficients\begin{aligned} & =\left(\frac{2^{2 n-3}}{2 n-2}\right)=16 \\ & \Rightarrow 2^{2 n-3}=2^5(n-1) \Rightarrow n=5 \\ & \therefore P \left(2 n-1, n^2-4 n\right)=(9,5) \end{aligned}<imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/zQOPegNaGroihIXObDobjfcSVyZC8NsiEeh7ci6YInA.original.fullsize.png"/><br/><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/zQOPegNaGroihIXObDobjfcSVyZC8NsiEeh7ci6YInA.original.fullsize.png"/><br/>\begin{aligned} & x+y=8 \\ & \therefore P M=\left|\frac{9+5-8}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=\frac{3 \times 2}{\sqrt{2}}=3 \sqrt{2}\end{aligned}$$
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