JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 19)
Explanation
$$ \begin{aligned} & T_{r+1}={ }^9 C_r\left(\frac{x^{-2 / 5}}{2}\right)^r\left(x^{2 / 3}\right)^{9-r} \\ & ={ }^9 C_r \frac{1}{2^r} x^{\frac{2}{3}(9-r)+\left(\frac{-2 r}{5}\right)} \\ & ={ }^9 C_r \cdot \frac{1}{2^r} \cdot x^{6-\frac{16 r}{15}} \end{aligned} $$
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 $$x^{2 / 3}$$ & $$x^{-2 / 5}$$
$$\begin{aligned} & ={ }^9 C_5 \cdot \frac{1}{2^5}+{ }^9 C_6 \cdot \frac{1}{2^6} \\ & =\frac{9!}{5!4!}\left(\frac{1}{2^5}\right)+\frac{9!}{6!3!}\left(\frac{1}{2^6}\right)=\frac{21}{4} \end{aligned}$$
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