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 ${ }^9 C _3 a ^6 b ^3 \ldots .$. (i)
to get the maximum value of $a^6 b^3$ use $A M \geq G M$
$\text { i.e., } \frac{3\left(\frac{ a ^2}{3}\right)+3 \pi\left(\frac{ b }{3}\right)}{6} \geq\left[\frac{ a ^6 b ^3}{3^6}\right]^{\frac{1}{3}} $
$\therefore \left( a ^6 b ^3\right)_{\max }=1$
$\therefore \text { Answer }=9 C _3 $