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7.Binomial Theorem
normal
If $a^3 + b^6 = 2$, then the maximum value of the term independent of $x$ in the expansion of $(ax^{\frac{1}{3}}+bx^{\frac{-1}{6}})^9$ is, where $(a > 0, b > 0)$
A
$42$
B
$68$
C
$84$
D
$148$
Solution
$r=\frac{3}{\frac{1}{3}+\frac{1}{6}}=6$
$Tr + 1 = T6 + 1 = {\,^9}{c_6}{a^3}{b^6} = 84{a^3}{b^6}$
$A \geq G$
$\frac{a^{3}+b^{6}}{2} \geq \sqrt{a^{3} b^{6}}$
$1 \geq a^{3} b^{6} \Rightarrow a^{3} b^{6} \leq 1$
So $84 a^{3} b^{6} \leq 84$
Standard 11
Mathematics
