JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

General term of $(2 x^{2} + \frac{1}{2 x})^{11}$ is:

$T_{r + 1} =^{11} C_{r} (2 x^{2})^{11 - r} (\frac{1}{2 x})^{r}$

$=^{11} C_{r} 2^{11 - r} x^{22 - 2 r} 2^{- r} x^{- r}$

$=^{11} C_{r} 2^{11 - r} x^{22 - 3 r}$

Now, $22 - 2 r = 10$ and $22 - 3 r = 7$

$\Rightarrow 3 r = 12$

$\Rightarrow 3 r = 15$

$\Rightarrow r = 4$

$\Rightarrow r = 5$

$∴$ Coeff. of $x^{10} =^{11} C_{4}$ , $2^{11 - 8} =^{11} C_{4} \times 8$

Coeff. of $x^{7} =^{11} C_{5}$ , $2^{11 - 10} =^{11} C_{4} \times 2$

Now, required difference

$=^{11} C_{4} \times 8 -^{11} C_{5} \times 2$

$= \frac{11 \times 10 \times 9 \times 8 \times 7 !}{4 ! \times 7 !} \times 8 - \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 ! \times 2}{5 ! \times 6 !}$

\begin{aligned}
& =\frac{11\times10\times9\times8\times8}{24}-\frac{11\times10\times9\times8\times7\times2}{120} \\
& =11\times10\times8\times3-11\times3\times4\times7 \\
& =11\times3\times4[20-7] \\
& =11\times12\times13=(12-1)\times12\times(12+1) \\
& =12(12^{2}-1)=12^{3}-12
\end{aligned}