Maxima and MinimaEasy
Question
If a3 + b6 = 2, then the maximum value of the term independent of x in the expansion of (ax1/3 + bx-1/6)9 is, where (a > 0, b > 0).
Options
A.42
B.68
C.84
D.148
Solution
Tr+1 = T6+1 = 9c6 a3b6 = 84a3b6
A > G
1 ≥ a3b6 ⇒ a3b6 ≤ 1
So 84a3b6 ≤ 84
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