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Question

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

Vedclass · Accepted Answer

$336$

Vedclass · Answer

Let $t_{\mathrm{r}}+1$ denotes

$\mathrm{r}+1 \mathrm{th}$ term of $\left(\alpha \mathrm{x}^{\frac{1}{9}}+\beta \mathrm{x}^{-\frac{1}{6}}\right)^{10}$

$t_{\mathrm{r}+1}={ }^{10} \mathrm{C}_{\mathrm{r}} \alpha^{10-\mathrm{r}}(\mathrm{x})^{\frac{10-\mathrm{r}}{9}} \cdot \beta^{\mathrm{r}} \mathrm{x}^{-\frac{\mathrm{r}}{6}}$

$={ }^{10} \mathrm{C}_{\mathrm{r}} \alpha^{10-\mathrm{r}} \beta^{\mathrm{r}}(\mathrm{x})^{\frac{10-\mathrm{r}}{9}-\frac{\mathrm{r}}{6}}$

If $t_{\mathrm{r}}+1$ is independent of $\mathrm{x}$

$\frac{10-r}{9}-\frac{r}{6}=0 \Rightarrow r=4$

maximum value of $\mathrm{t}_{5}$ is $10 \mathrm{K}$ (given)

$\Rightarrow{ }^{10} \mathrm{C}_{4} \alpha^{6} \beta^{4}$ is maximum

By $\mathrm{AM} \geq \mathrm{GM}$ (for positive numbers)

$\frac{\frac{\alpha^{3}}{2}+\frac{\alpha^{3}}{2}+\frac{\beta^{2}}{2}+\frac{\beta^{2}}{2}}{4} \geq\left(\frac{\alpha^{6} \beta^{4}}{16}\right)^{\frac{1}{4}}$

$\Rightarrow \alpha^{6} \beta^{4} \leq 16$

So, $10 \mathrm{K}={ }^{10} \mathrm{C}_{4} 16$

$\Rightarrow \mathrm{K}=336$

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