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Binomial Theorem question

2025 · 4 Apr · Shift 1 · Q39
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  5. /2025 · 4 Apr · Shift 1 · Q39

Binomial Theorem question

2025 · 4 Apr · Shift 1 · Q39

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
For an integer n≥2n \geq 2n≥2, if the arithmetic mean of all coefficients in the binomial expansion of (x+y)2n−3(x+y)^{2 n-3}(x+y)2n−3 is 16 , then the distance of the point P(2n−1,n2−4n)\mathrm{P}\left(2 n-1, n^2-4 n\right)P(2n−1,n2−4n) from the line x+y=8x+y=8x+y=8 is
  1. A
    2\sqrt{2}2​
  2. B
    222 \sqrt{2}22​
  3. C
    525 \sqrt{2}52​
  4. D
    323 \sqrt{2}32​
View written solutionFree

Correct answer: D

  1. Find the arithmetic mean of the coefficients

For the binomial expansion of

(x+y)2n−3,(x+y)^{2n-3},(x+y)2n−3,

the number of coefficients is

(2n−3)+1=2n−2.(2n-3)+1=2n-2.(2n−3)+1=2n−2.

The sum of all coefficients in a binomial expansion is obtained by putting x=1,y=1x=1, y=1x=1,y=1:

(1+1)2n−3=22n−3.(1+1)^{2n-3}=2^{2n-3}.(1+1)2n−3=22n−3.

So the arithmetic mean of all coefficients is

22n−32n−2=16.\frac{2^{2n-3}}{2n-2}=16.2n−222n−3​=16.

That is,

22n−32(n−1)=16\frac{2^{2n-3}}{2(n-1)}=162(n−1)22n−3​=16 22n−4=16(n−1).2^{2n-4}=16(n-1).22n−4=16(n−1).

Since 16=2416=2^416=24,

22n−4=24(n−1).2^{2n-4}=2^4(n-1).22n−4=24(n−1).

Rewriting,

22n−8=n−1.2^{2n-8}=n-1.22n−8=n−1.

Now check integer solutions for n≥2n\ge 2n≥2:

  • If n=5n=5n=5, then 22=4=n−12^{2}=4=n-122=4=n−1, true.

So,

n=5.n=5.n=5.
  1. Coordinates of point PPP

Given

P(2n−1, n2−4n).P(2n-1,\, n^2-4n).P(2n−1,n2−4n).

Substitute n=5n=5n=5:

P=(2⋅5−1, 52−4⋅5)=(9,5).P=(2\cdot 5-1,\, 5^2-4\cdot 5)=(9,5).P=(2⋅5−1,52−4⋅5)=(9,5).
  1. Distance from point to line x+y=8x+y=8x+y=8

Write the line in standard form:

x+y−8=0.x+y-8=0.x+y−8=0.

Distance of point (x1,y1)(x_1,y_1)(x1​,y1​) from line Ax+By+C=0Ax+By+C=0Ax+By+C=0 is

∣Ax1+By1+C∣A2+B2.\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.A2+B2​∣Ax1​+By1​+C∣​.

Here, A=1,B=1,C=−8A=1, B=1, C=-8A=1,B=1,C=−8, and point is (9,5)(9,5)(9,5). Thus,

distance=∣9+5−8∣12+12=62=32.\text{distance} = \frac{|9+5-8|}{\sqrt{1^2+1^2}} =\frac{6}{\sqrt{2}} =3\sqrt{2}.distance=12+12​∣9+5−8∣​=2​6​=32​.
  1. Option check

The distance is

32.3\sqrt{2}.32​.

So the correct option is:

D: 323\sqrt{2}32​

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