JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 18)
Let $$\alpha>0$$, be the smallest number such that the expansion of $$\left(x^{\frac{2}{3}}+\frac{2}{x^{3}}\right)^{30}$$ has a term $$\beta x^{-\alpha}, \beta \in \mathbb{N}$$. Then $$\alpha$$ is equal to ___________.
Answer
2
Explanation
$\mathrm{T}_{\mathrm{r}+1}={ }^{30} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2 / 3}\right)^{30-\mathrm{r}}\left(\frac{2}{\mathrm{x}^{3}}\right)^{\mathrm{r}}$
$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
$\frac{60-11 \mathrm{r}}{3}<0 $
$\Rightarrow 11 \mathrm{r}>60 $
$\Rightarrow \mathrm{r}>\frac{60}{11} $
$\Rightarrow \mathrm{r}=6$
$\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}$
We have also observed $\beta={ }^{30} \mathrm{C}_{6}(2)^{6}$ is a natural number.
$\therefore \alpha=2$
$={ }^{30} \mathrm{C}_{\mathrm{r}} \cdot 2^{\mathrm{r}} \cdot \mathrm{x}^{\frac{60-11 \mathrm{r}}{3}}$
$\frac{60-11 \mathrm{r}}{3}<0 $
$\Rightarrow 11 \mathrm{r}>60 $
$\Rightarrow \mathrm{r}>\frac{60}{11} $
$\Rightarrow \mathrm{r}=6$
$\mathrm{T}_{7}={ }^{30} \mathrm{C}_{6} \cdot 2^{6} \mathrm{x}^{-2}$
We have also observed $\beta={ }^{30} \mathrm{C}_{6}(2)^{6}$ is a natural number.
$\therefore \alpha=2$
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