(1) $176$

(2) $336$

(3) $352$

(4) $84$

**Ans. (2)** 336

**Sol.** $T_{r+1}=\ ^{10}\!C_r(\alpha x^{1/9})^{10–r}(\beta x^{-1/6})^r$

$T_{r+1}=\ ^{10}\!C_r\alpha^{10–r}\beta^r(x)^{{10-r\over9}-{r\over6}}$

Term independent of $x$

${10-r\over9}-{r\over6}=0\implies r=4$

$T_5=\ ^{10}\!C_4\alpha^6\beta^4$

Now Let $\alpha^3,\ \beta^2$ are 2 numbers.

$A\geq G$

$\implies{\alpha^3+\beta^2\over2}\geq(\alpha^3\beta^2)^{1/2}$

$\implies\alpha^3\beta^2\leq4$

$\implies\alpha^6\beta^4\leq16$

$\implies{T_5\over\ ^{10}C_4}\leq16$

$\implies T_5\leq16.\ ^{10}C_4$

$\implies T_{5\text{ max}}=16×\ ^{10}C_4=10 K$

$\implies K=336$