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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 8 April · Evening Shift · Mathematics · Binomial Theorem

The absolute difference of the coefficients of x ^ 10 and x ^ 7 in...

The absolute difference of the coefficients of x10and{x}^{10} and{x}^{7} in the expansion of ${\left(2{x}^{2} + \dfrac{1}{2x}\right)}^{11} is equal to

Options

  1. A133−13{13}^{3} - 13
  2. B113−11{11}^{3} - 11
  3. C103−10{10}^{3} - 10
  4. D123−12{12}^{3} - 12

Correct answer

D. 123−12{12}^{3} - 12

Step-by-step solution

The given expansion is (2x2+12x)11{\left(2{x}^{2} + \dfrac{1}{2x}\right)}^{11}.

The general term in the binomial expansion of (x+a)nisgivenby{\left(x + a\right)}^{n} is given by{T}_{r + 1} = Cr n {x}^{n - r}{a}^{r}.</p><p>.</p><p>⇒ {T}_{r + 1} = Cr 11 {\left(2{x}^{2}\right)}^{11 - r}{\left(\dfrac{1}{2x}\right)}^{r}</p><p></p><p>= Cr 11 \left({2}^{11 - r - r}\right){x}^{22 - 2r - r}</p><p>Nowforcoefficientof</p><p>Now for coefficient of{x}^{7}$⇒22−3r=7⇒ 22 - 3r = 7

⇒r=5⇒ r = 5

Coefficient of x7is{x}^{7} isC5 11 \left({2}^{11 - 5 \times 2}\right)</p><p>Nowforcoefficientof</p><p>Now for coefficient of{x}^{10} ⇒ 22 - 3r = 10</p><p></p><p>⇒ r = 4</p><p>Coefficientof</p><p>Coefficient of{x}^{10} is C411(211−2×4)C4 11 \left({2}^{11 - 2 \times 4}\right)

We need the absolute difference, we get it by \left|C5 11 \left({2}^{11 - 5 \times 2}\right) - C4 11 \left({2}^{11 - 4 \times 2}\right)\right|.</p><p>.</p><p>= \left|924 - 2640\right| = 1716</p><p></p><p>= {12}^{3} - 12</p><p>Hencetheabsolutedifferenceis</p><p>Hence the absolute difference is= {12}^{3} - 12$.

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