QJEEM24Ma7a88: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7a88 Solution: The sum of the coefficient of and in the binomial expansion of is

$\begin{aligned}& T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\& ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\& ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}}\end{aligned}$

For coefficient of $x^{2 / 3}$

$\begin{aligned}& \Rightarrow 6-\frac{16 r}{15}=\frac{2}{3} \\& \Rightarrow 90-16 r=10 \\& \Rightarrow r=5\end{aligned}$

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

$\begin{aligned}& \Rightarrow 6-\frac{16 r}{15}=\frac{-2}{5}\\& \Rightarrow 90-16 r=-6 \\& \Rightarrow r=6\end{aligned}$

Sum of coefficient of $x^{2 / 3}$ & $x^{-2 / 5}$

$\begin{aligned}& ={ }^9 C_5 \cdot \frac{1}{2^5}+{ }^9 C_6 \cdot \frac{1}{2^6} \\& =\frac{9!}{5!4!}\left(\frac{1}{2^5}\right)+\frac{9!}{6!3!}\left(\frac{1}{2^6}\right)=\frac{21}{4}\end{aligned}$