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(A) The general term in the expansion of $(x^{2/3} + 2x^{-3})^{30}$ is given by:$T_{r+1} = {}^{30}C_{r} (x^{2/3})^{30-r} (2x^{-3})^{r}$$T_{r+1} = {}^{

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$2$

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(A) The general term in the expansion of $(x^{2/3} + 2x^{-3})^{30}$ is given by:$T_{r+1} = {}^{30}C_{r} (x^{2/3})^{30-r} (2x^{-3})^{r}$$T_{r+1} = {}^{30}C_{r} \cdot 2^{r} \cdot x^{(60-2r)/3} \cdot x^{-3r}$$T_{r+1} = {}^{30}C_{r} \cdot 2^{r} \cdot x^{(60-11r)/3}$We are given that the term is $\beta x^{-\alpha}$,where $\alpha > 0$ and $\beta \in \mathbb{N}$.Thus,$-\alpha = \frac{60-11r}{3}$,which implies $\alpha = \frac{11r-60}{3}$.Since $\alpha > 0$,we have $11r - 60 > 0$,so $r > \frac{60}{11} \approx 5.45$.Since $r$ must be an integer and $0 \le r \le 30$,the smallest integer $r$ is $6$.For $r = 6$,$\alpha = \frac{11(6)-60}{3} = \frac{66-60}{3} = \frac{6}{3} = 2$.For $r = 6$,$\beta = {}^{30}C_{6} \cdot 2^{6}$,which is a natural number.Therefore,the smallest value of $\alpha$ is $2$.

Let $\alpha > 0$ be the smallest number such that the expansion of $(x^{2/3} + 2x^{-3})^{30}$ has a term $\beta x^{-\alpha}$,where $\beta \in \mathbb{N}$. Then $\alpha$ is equal to $.............$.

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