Question

Let $a$ and $b$ are two positive real numbers such that $a^2+b=2$, 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

• A. 168
• B. 98
• C. 84
• D. 42

Answer 1

Answer: (C)

$T_{r+1}={}^9{C}_r{a}^{a-r}{b}^r{x}^{\frac{3-r}{2}}\Rightarrow r=3$ (for term independent of $x$ )
$\therefore$ Term independal at $x$ is equal to ${}^9{C}_3{a}^6{b}^3\dots .$. (i)
to get the maximum value of $a^6{b}^3$ use $AM\ge GM$
$\text{ i.e., }\frac{3\left(\frac{a^2}{3}\right)+3\pi \left(\frac{b}{3}\right)}{6}\ge {\left[\frac{a^6{b}^3}{3^6}\right]}^{\frac{1}{3}}$
$\therefore {\left(a^6{b}^3\right)}_{\max }=1$
$\therefore \text{ Answer }=9C_3$