Question

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

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

Answer 1

Answer: (B)

Let ${\left(k+1\right)}^{th}$ term be independent of $x$
$T_{k+1}{=}^9{C}_k{\left(a{x}^{\frac{1}{6}}\right)}^{9-k}{\left(b{x}^{-\frac{1}{3}}\right)}^k$
${=}^9{C}_k{a}^{9-k}{b}^k{x}^{\left(\frac{9-k}{6}-\frac{k}{3}\right)}$
For this to be independent of $x,$
$\frac{9-k}{6}-\frac{k}{3}=0\Rightarrow \frac{9}{6}=\frac{k}{6}+\frac{k}{3}=\frac{k}{2}\Rightarrow k=3.$
$\Rightarrow$ $T_{k+1}{=}^9{C}_3{a}^6{b}^3=84{\left(\sqrt{a^{2}b}\right)}^6$
Using $A.M.\ge G.M.$ we get $\frac{a^2+b}{2}\ge \sqrt{a^{2}b}$
$\Rightarrow$ $\sqrt{a^{2}b}\le 1$
$\Rightarrow$ $T_{k+1}\le 84$