- Home
- Standard 11
- Mathematics
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 $10 k$ then $\mathrm{k}$ is equal to
$176$
$336$
$352$
$84$
Solution
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, $10 \mathrm{K}={ }^{10} \mathrm{C}_{4} 16$
$\Rightarrow \mathrm{K}=336$
