JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 19)
Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively
in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}$ is :
in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}$ is :
$\frac{1}{9}$
$\frac{1}{4}$
$\frac{4}{9}$
$\frac{9}{4}$
Explanation
$\begin{aligned} & \mathrm{t}_7={ }^{18} \mathrm{C}_6\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^6={ }^{18} \mathrm{C}_6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^6} \\\\ & \mathrm{t}_{13}={ }^{18} \mathrm{C}_{12}\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^6\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} \mathrm{C}_{12} \frac{1}{(3)^6} \cdot \frac{1}{2^{12}} \cdot \mathrm{x}^{-6}\end{aligned}$
$$ \therefore $$ $m={ }^{18} C_6\left(\frac{1}{3}\right)^{12}\left(\frac{1}{2}\right)^6$
$n={ }^{18} C_{12}\left(\frac{1}{3}\right)^6\left(\frac{1}{2}\right)^{12}$
$\begin{aligned}\left(\frac{m}{n}\right)^{\frac{1}{3}} & =\left(\frac{{ }^{18} C_6\left(\frac{1}{3}\right)^{12}\left(\frac{1}{2}\right)^6}{{ }^{18} C_{12}\left(\frac{1}{3}\right)^6\left(\frac{1}{2}\right)^{12}}\right)^{\frac{1}{3}} \\\\ & =\left(\frac{\left(\frac{1}{3}\right)^6}{\left(\frac{1}{2}\right)^6}\right)^{\frac{1}{3}}=\left(\left(\frac{2}{3}\right)^6\right)^{\frac{1}{3}}=\frac{4}{9}\end{aligned}$
$\therefore\left(\frac{n}{m}\right)^{\frac{1}{3}}=\frac{9}{4}$
$$ \therefore $$ $m={ }^{18} C_6\left(\frac{1}{3}\right)^{12}\left(\frac{1}{2}\right)^6$
$n={ }^{18} C_{12}\left(\frac{1}{3}\right)^6\left(\frac{1}{2}\right)^{12}$
$\begin{aligned}\left(\frac{m}{n}\right)^{\frac{1}{3}} & =\left(\frac{{ }^{18} C_6\left(\frac{1}{3}\right)^{12}\left(\frac{1}{2}\right)^6}{{ }^{18} C_{12}\left(\frac{1}{3}\right)^6\left(\frac{1}{2}\right)^{12}}\right)^{\frac{1}{3}} \\\\ & =\left(\frac{\left(\frac{1}{3}\right)^6}{\left(\frac{1}{2}\right)^6}\right)^{\frac{1}{3}}=\left(\left(\frac{2}{3}\right)^6\right)^{\frac{1}{3}}=\frac{4}{9}\end{aligned}$
$\therefore\left(\frac{n}{m}\right)^{\frac{1}{3}}=\frac{9}{4}$
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