Let alpha 0, beta 0 be such that alpha^{3 | vedclass

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(B) The general term $T_{r+1}$ in the expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (\alph

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$336$

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(B) The general term $T_{r+1}$ in the expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (\alpha x^{\frac{1}{9}})^{10-r} (\beta x^{-\frac{1}{6}})^{r} = {}^{10}C_{r} \alpha^{10-r} \beta^{r} x^{\frac{10-r}{9} - \frac{r}{6}}$.For the term to be independent of $x$,the exponent of $x$ must be zero:$\frac{10-r}{9} - \frac{r}{6} = 0$ $\Rightarrow 2(10-r) - 3r = 0$ $\Rightarrow 20 - 5r = 0$ $\Rightarrow r = 4$.The independent term is $T_{5} = {}^{10}C_{4} \alpha^{6} \beta^{4} = 210 \alpha^{6} \beta^{4}$.We are given $\alpha^{3} + \beta^{2} = 4$. By $AM \geq GM$ inequality for positive numbers:$\frac{\frac{\alpha^{3}}{2} + \frac{\alpha^{3}}{2} + \frac{\beta^{2}}{2} + \frac{\beta^{2}}{2}}{4} \geq \sqrt[4]{\frac{\alpha^{3}}{2} \cdot \frac{\alpha^{3}}{2} \cdot \frac{\beta^{2}}{2} \cdot \frac{\beta^{2}}{2}}$$\frac{4}{4} \geq \sqrt[4]{\frac{\alpha^{6} \beta^{4}}{16}}$ $\Rightarrow 1 \geq \frac{\alpha^{6} \beta^{4}}{16}$ $\Rightarrow \alpha^{6} \beta^{4} \leq 16$.The maximum value of $T_{5}$ is $210 	imes 16 = 3360$.Given $10k = 3360$,we get $k = 336$.

Let $\alpha > 0, \beta > 0$ be such that $\alpha^{3} + \beta^{2} = 4$. If the maximum value of the term independent of $x$ in the binomial expansion of $(\alpha x^{\frac{1}{9}} + \beta x^{-\frac{1}{6}})^{10}$ is $10k$,then $k$ is equal to

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