Question

If $a^2+b=2$, then the maximum value of the term independent of $x$ in the expansion of ${(a{x}^{\frac{1}{6}}+b{x}^{\frac{-1}{3}})}^9$ is, where $(a>0,b>0)$.

• A. $48$
• B. $84$
• C. $42$
• D. $168$

Answer 1

The correct option is B $84$

$\begin{array}{c}{\left(a{x}^{\frac{1}{6}}+b{x}^{\frac{-1}{3}}\right)}^9\\ {\left(r+1\right)}^{th}term{=}^q{C}_r{\left(a{x}^{\frac{1}{6}}\right)}^r{\left(b{x}^{\frac{-1}{3}}\right)}^{q-r}\\ {\Rightarrow }^q{C}_r{a}^r{b}^{q-r}{x}^{\left(\frac{ax}{6}-3+\frac{r}{3}\right)}\\ {\Rightarrow }^q{C}_r{a}^r{b}^{q-r}{x}^{\frac{r}{2}-3}\\ \frac{r}{2}-3=0\\ \Rightarrow r=6\\ \therefore Independentterm{=}^9{C}_6{a}^6{b}^3\\ =\frac{9\times 8\times 7}{3\times 2}=84a^6{b}^3\\ Now,\\ a^2+b=2\\ \Rightarrow a=1\&b=1\\ \therefore Independentterm=84\end{array}$.