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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Binomial Theorem

If the constant term in the expansion of ( [5] 3 x + 2 x [3] 5 )^...

If the constant term in the expansion of (35x+2x53)12,x≠0\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0, is α×28×35\alpha \times 2^8 \times \sqrt[5]{3}, then 25α25 \alpha is equal to :

Options

  1. A724
  2. B742
  3. C639
  4. D693

Correct answer

D. 693

Step-by-step solution

Tr+1=12Cr(31/5x)12−r(2x51/3)rTr+1=12Cr(3)12−r5(2)r(x)2r−12(5)r/3r=6 T7=12C6(3)6/5(2)652=(9×11×725)28⋅31/525α=693\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^{\mathrm{r}} \\ & \mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}} \\ & \mathrm{r}=6 \\ & \mathrm{~T}_7=\frac{{ }^{12} \mathrm{C}_6(3)^{6 / 5}(2)^6}{5^2}=\left(\frac{9 \times 11 \times 7}{25}\right) 2^8 \cdot 3^{1 / 5} \\ & 25 \alpha=693\end{aligned}
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