Binomial Theorem Question - None

Question

Let a and b are two positive real numbers such that a² + b = 2, then the maximum value of term independent of x in the expansion of $\left(a x^{\frac{1}{6}}+b x^{\frac{-1}{3}}\right)^{9}$ is:

Accepted Answer

$\left(\operatorname{ax}^{\frac{1}{6}}+b x^{\frac{-1}{3}} ight)^{9}$
(r + 1)^th term = $C_{r}\left(\operatorname{ax}^{\frac{1}{6}} ight)^{r}\left(b x^{\frac{-1}{3}} ight)^{q-r}$
$\Rightarrow$ $\mathrm{q}_{\mathrm{C}_{\mathrm{r}}} \mathrm{a}^{\mathrm{r}} \mathrm{b}^{\mathrm{q}-\mathrm{r}} \mathrm{x}^{\left(\frac{\mathrm{ax}}{6}-3+\frac{\mathrm{r}}{3} ight)}$
$\Rightarrow$ ${ }^{\mathrm{q}} {C}_{\mathrm{r}} {a}^{\mathrm{r}} {b}^{\mathrm{q}-\mathrm{r}} {x}^{\frac{{r}}{2}-3}$
$\frac{{r}}{2}-3$ = 0
$\Rightarrow$ r = 6
$ herefore$ Independent term = ⁹C₆a⁶b³
= $\frac{9 imes 8 imes 7}{3 imes 2}$= 84a⁶b³
Now, a² + b = 2
$\Rightarrow$ a = 1 & b = 1
$ herefore$ Independent term = 84

Question:

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