JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 5)

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 :
176
336
352
84

Explanation

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

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