Q44 / 90 · JEE Main 2024 (05 Apr Shift 2) · Mathematics

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

  • A
    724
  • B
    742
  • C
    639
  • Correct answer
    D
    693

Solution

$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^{\mathrm{r}} \\ & \mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}} \\ & \mathrm{r}=6 \\ & \mathrm{~T}_7=\frac{{ }^{12} \mathrm{C}_6(3)^{6 / 5}(2)^6}{5^2}=\left(\frac{9 \times 11 \times 7}{25}\right) 2^8 \cdot 3^{1 / 5} \\ & 25 \alpha=693\end{aligned}$

Hence, the correct answer is option: 'D'.

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