# $Let \; \alpha >0, \beta>0\; be \;such \; that \; \alpha^{3}+\beta^{2}=4.$ If the maximum value of the term independent of x in the binomial expansion of $\left ( \alpha x^{\frac{1}{9}}+\beta x^{-\frac{1}{6}} \right )^{10}\; is \; 10k,$then k is equal to : Option: 1 336 Option: 2 352 Option: 3 84 Option: 4 176

$\\\text{Let } t_{r+1} \text{ denotes} \\\mathrm{r}+1^{\text {th }} \text { term of }\left(\alpha \mathrm{x}^{1 / 9}+\beta \mathrm{x}^{-1 / 6}\right)^{10}$

\begin{aligned} t_{r+1} &={ }^{10} C_{r} \alpha^{10-r}(x)^{\frac{10-r}{9}} \cdot \beta^{r} x^{-\frac{r}{6}} \\ &={ }^{10} C_{r} \alpha^{10-r} \beta^{r}(x)^{\frac{10-r}{9}-\frac{r}{6}} \end{aligned}

\begin{aligned} &\text { If } t_{r+1} \text { is independent of } x\\ &\frac{10-r}{9}-\frac{r}{6}=0 \Rightarrow r=4 \end{aligned}

$\\\text{maximum value of }t_{5}\; is \;10 \mathrm{~K}\; (given) \\\Rightarrow{ }^{10} \mathrm{C}_{4} \alpha^{6} \beta^{4}\text{ is maximum}\\ \text{By } \mathrm{AM} \geq \mathrm{GM}\; \text{(for positive numbers)}$

$\\\frac{\frac{\alpha^{3}}{2}+\frac{\alpha^{3}}{2}+\frac{\beta^{2}}{2}+\frac{\beta^{2}}{2}}{4} \geq\left(\frac{\alpha^{6} \beta^{4}}{16}\right)^{\frac{1}{4}}\\\Rightarrow \alpha^6\beta^4\leq16\\ \text { So, } 10 \mathrm{~K}={ }^{10} \mathrm{C}_{4} 16 \\ \Rightarrow \mathrm{K}=336$

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