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Term Independent of x

If,a2+b=2, th...

Question

If , $a^{2}$ +b=2, then maximum value of the term independent of x in the expansion of ${(a x^{1 / 6} + b x^{- 1 / 3})}^{9}$ is (a>0,b>0)

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Solution

The correct option is B
84

Let $(K + 1)^{t h}$ term be independent of x.

$∴ t_{k + 1} =^{9} C_{k} {(a x^{1 / 6})}^{9 - k} {(b x^{- 1 / 3})}^{k} =^{9} C_{K} a^{9 - k} b^{k} x^{\frac{9 - k}{6}} - \frac{k}{6} A s t_{k + 1} is independent of x; \frac{9 - k}{6} - \frac{k}{3} = 0 \Rightarrow k = 3; t_{4} =^{9} C_{3} a^{6} b^{3} =^{9} C_{3} {(\sqrt{a^{2} b})}^{6} = 84 {(\sqrt{a^{2} b})}^{6} \leq 84 {(\frac{a^{2} + b}{2})}^{6}; (G . M \leq A . M) = 84 (∵ a^{2} + b = 2)$
$∴ m a x i m u m v a l u e = 84$

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