QJEEM24Mab69f: JEE Main 2024 Mathematics - Class 11 JEEM24Mab69f Solution: If the constant term in the expansion of , is , then is equal…

$\begin{aligned}& \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \\& T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r\end{aligned}$

For constant term $-12+r+r=0$

$\begin{aligned}& \Rightarrow \quad r=6 \\& \therefore \quad \text { Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6\end{aligned}$

$\begin{aligned}& ={ }^{12} C_6 \times \frac{2^6}{25} \times 3.3^{\frac{1}{5}} \\& =\frac{231}{25} \times 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \\& =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \\& \therefore \quad \alpha=\frac{693}{25} \\& 25 \alpha=693\end{aligned}$