Verified Solution • 2025–26 Session moderate

QJEEM20Maf01c: JEE Main 2020 Mathematics - Class 11 JEEM20Maf01c Solution: Let > 0, > 0 be such that 3 + 2 = 4. If the maximum value of…

Let α\alpha > 0, β\beta > 0 be such that
α\alpha 3 + β\beta 2 = 4. If the maximum value of the term independent of x in
the binomial expansion of (αx19+βx−16)10{\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}} is 10k,
then k is equal to :

Multiple Choice Options

  • A 176
  • B 336
  • C 352
  • D 84

Step-by-Step Verified Solution

General term

Tr + 1 = 10Cr α10−r.(x)10−r9.βr(x)−r6{\alpha ^{10 - r}}.{\left( x \right)^{{{10 - r} \over 9}}}.{\beta ^r}{\left( x \right)^{ - {r \over 6}}}

= 10Cr α10−rβr.(x)10−r9−r6{\alpha ^{10 - r}}{\beta ^r}.{\left( x \right)^{{{10 - r} \over 9} - {r \over 6}}}

If Tr + 1 is independent of x

∴\therefore 10−r9−r6{{{10 - r} \over 9} - {r \over 6}} = 0

⇒\Rightarrow r = 4

∴\therefore T5 = 10C4 α6β4{\alpha ^6}{\beta ^4}

Also given, α\alpha 3 + β\beta 2 = 4

By AM-GM inequality

α3+β22≥(α3β2)12{{{\alpha ^3} + {\beta ^2}} \over 2} \ge {\left( {{\alpha ^3}{\beta ^2}} \right)^{{1 \over 2}}}

⇒\Rightarrow (2)2 ≥\ge α3β2{{\alpha ^3}{\beta ^2}}

⇒\Rightarrow α6β4{\alpha ^6}{\beta ^4} ≤\le 16

∴\therefore 10k = 10C4 (16)

⇒\Rightarrow k = 336

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