Let 0,0 be such that ^3+^2=4.

Question

Let $\alpha>0,\beta>0$ be such that $\alpha^3+\beta^2=4$. If the maximum value of the term independent of $x$ in the binomial expansion of $\left(\alpha x^{1/9}+\beta x^{-1/6}\right)^{10}$ is $10k$, then $k$ is equal to:

Accepted Answer

General term

Tr + 1 = 10Cr$${\alpha ^{10 - r}}.{\left( x \right)^{{{10 - r} \over 9}}}.{\beta ^r}{\left( x \right)^{ - {r \over 6}}}$$

= 10Cr$${\alpha ^{10 - r}}{\beta ^r}.{\left( x \right)^{{{10 - r} \over 9} - {r \over 6}}}$$

If Tr + 1 is independent of x

$$ \therefore $$ $${{{10 - r} \over 9} - {r \over 6}}$$ = 0

$$ \Rightarrow $$ r = 4

$$ \therefore $$ T5 = 10C4 $${\alpha ^6}{\beta ^4}$$

Also given, $$\alpha $$3 + $$\beta $$2 = 4

By AM-GM inequality

$${{{\alpha ^3} + {\beta ^2}} \over 2} \ge {\left( {{\alpha ^3}{\beta ^2}} \right)^{{1 \over 2}}}$$

$$ \Rightarrow $$ (2)2 $$ \ge $$ $${{\alpha ^3}{\beta ^2}}$$

$$ \Rightarrow $$ $${\alpha ^6}{\beta ^4}$$ $$ \le $$ 16

$$ \therefore $$ 10k = 10C4 (16)

$$ \Rightarrow $$ k = 336

Solutions · 1