JEE Main 2024 — 9 April, Shift 2 — Question 12

$\quad T_{r+1}={ }^{9} C_{r}\left(x^{2 / 3}\right)^{9-r}\left(\frac{x^{-2 / 5}}{2}\right)^{r}$

$={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{2}\right)^{\mathrm{r}}(\mathrm{r})^{\left(6-\frac{2 \pi}{3-2 r}\right)}$

for coefficient of $\mathrm{x}^{2 / 3}$,

put $6-\frac{2 r}{3}-\frac{2 r}{5}=\frac{2}{3}$ $\Rightarrow \mathrm{r}=5$

$\therefore$ Coefficient of $\mathrm{x}^{2 / 3}$ is $={ }^{9} \mathrm{C}_{5}\left(\frac{1}{5}\right)^{5}$

For coefficient of $\mathrm{x}^{-2 / 5}$,

put $6-\frac{2 \mathrm{r}}{3}-\frac{2 \mathrm{r}}{5}=-\frac{2}{5}$

$\Rightarrow \mathrm{r}=6$

Coefficient of $\mathrm{x}^{-2 / 5}$ is ${ }^{9} \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6}$

Sum $={ }^{9} \mathrm{C}_{5}\left(\frac{1}{2}\right)^{5}+{ }^{9} \mathrm{C}_{6}\left(\frac{1}{2}\right)^{6}=\frac{21}{4}$