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 ${\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}}$ is 10k,
then k is equal to :

General term <br><br>T_{r + 1} = ¹⁰C_r$${\alpha ^{10 - r}}.{\left( x \right)^{{{10 - r} \over 9}}}.{\beta ^r}{\left( x \right)^{ - {r \over 6}}}$$ <br><br>= ¹⁰C_r$${\alpha ^{10 - r}}{\beta ^r}.{\left( x \right)^{{{10 - r} \over 9} - {r \over 6}}}$$ <br><br>If T_{r + 1} is independent of x <br><br>$\therefore$ ${{{10 - r} \over 9} - {r \over 6}}$ = 0 <br><br>$\Rightarrow$ r = 4 <br><br>$\therefore$ T₅ = ¹⁰C₄ ${\alpha ^6}{\beta ^4}$ <br><br>Also given, $\alpha$³ + $\beta$² = 4 <br><br>By AM-GM inequality <br><br>$${{{\alpha ^3} + {\beta ^2}} \over 2} \ge {\left( {{\alpha ^3}{\beta ^2}} \right)^{{1 \over 2}}}$$ <br><br>$\Rightarrow$ (2)² $\ge$ ${{\alpha ^3}{\beta ^2}}$ <br><br>$\Rightarrow$ ${\alpha ^6}{\beta ^4}$ $\le$ 16 <br><br>$\therefore$ 10k = ¹⁰C₄ (16) <br><br>$\Rightarrow$ k = 336