Question

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 ${(a{x}^{\frac{1}{9}}+\beta x^{\frac{1}{6}})}^{10}$ is $10k$, then $k$ is equal to:

• A. $176$
• B. $336$
• C. $352$
• D. $84$

Answer 1

The correct option is B $336$

For term independent of $x$

$T_{r+1}={}^{10}{C}_r{\alpha }^{10-r}{\beta }^r\cdot x^{\frac{10-r}{9}}\cdot x^{\frac{r}{6}}$

$\because \frac{10-r}{9}-\frac{r}{6}=0\Rightarrow r=4$

$T_5={}^{10}{C}_r{\alpha }^6.{\beta }^4$

$\because AM\ge GM$

Now

$\frac{(\frac{{\alpha }^3}{2}+\frac{{\alpha }^3}{2}+\frac{{\beta }^2}{2}+\frac{{\beta }^2}{2})}{4}\ge \sqrt[4]{4\frac{{\alpha }^6.{\beta }^4}{2^4}},\because ({\alpha }^3+{\beta }^2=4)$

$\Rightarrow {\left(\frac{4}{4}\right)}^4\ge \frac{{\alpha }^6{\beta }^4}{2^4}$

$\Rightarrow {\alpha }^6.{\beta }^4\le 2^4$
$\Rightarrow {}^{10}{C}_4.{\alpha }^6.{\beta }^4\ge {}^{10}{C}_4{2}^4$

$\Rightarrow T_5\le {10}_{C_4}{2}^4$

$\therefore T_5\le \frac{{\mathrm{10.9.8.7.2}}^4}{\mathrm{4.3.2.1}}$

maximum value of
$T_5=\mathrm{10.3.7.16}=10k$
$\therefore k=336$