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 {x}^{7} in the expansion of ${\left(2{x}^{2} + \dfrac{1}{2x}\right)}^{11} is equal to
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
The given expansion is .
The general term in the binomial expansion of {T}_{r + 1} = Cr n {x}^{n - r}{a}^{r}⇒ {T}_{r + 1} = Cr 11 {\left(2{x}^{2}\right)}^{11 - r}{\left(\dfrac{1}{2x}\right)}^{r}= Cr 11 \left({2}^{11 - r - r}\right){x}^{22 - 2r - r}{x}^{7}$
Coefficient of C5 11 \left({2}^{11 - 5 \times 2}\right){x}^{10} ⇒ 22 - 3r = 10⇒ r = 4{x}^{10} is
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|= \left|924 - 2640\right| = 1716= {12}^{3} - 12= {12}^{3} - 12$.