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 $\left(k + 1\right)^{t h}$ 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 $A.M.\geq G.M.$ we get $\frac{a^{2} + b}{2}\geq \sqrt{a^{2} b}$
$\Rightarrow $ $\sqrt{a^{2} b}\leq 1$
$\Rightarrow $ $T_{k + 1}\leq 84$