Verified Solution • 2025–26 Session moderate

QJEEM23Ma012f: JEE Main 2023 Mathematics - Class 11 JEEM23Ma012f Solution: If the coefficient of in the expansion of is equal to the…

If the coefficient of x15x^{15} in the expansion of (ax3+1 bx1/3)15\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15} is equal to the coefficient of x−15x^{-15} in the expansion of (ax1/3−1bx3)15\left(a x^{1 / 3}-\frac{1}{b x^{3}}\right)^{15} , where aa and bb are positive real numbers, then for each such ordered pair (a,b)(\mathrm{a}, \mathrm{b}) :

Multiple Choice Options

  • A a = 3b
  • B ab = 1
  • C ab = 3
  • D a = b

Step-by-Step Verified Solution

For (ax3+1bx13)\left( {a{x^3} + {1 \over {b{x^{{1 \over 3}}}}}} \right)

Tr+1=15Cr(ax3)15−r(1bx13)1{T_{r + 1}} = {}^{15}{C_r}{(a{x^3})^{15 - r}}{\left( {{1 \over {b{x^{{1 \over 3}}}}}} \right)^1}

∴\therefore x15→3(15−r)−r3=15{x^{15}} \to 3(15 - r) - {r \over 3} = 15

⇒30=10r3⇒r=9\Rightarrow 30 = {{10r} \over 3} \Rightarrow r = 9

Similarly, for (ax13−1bx3)15{\left( {a{x^{{1 \over 3}}} - {1 \over {b{x^3}}}} \right)^{15}}

Tr+1=15Cr(ax13)15−r(−1bx3)2{T_{r + 1}} = {}^{15}{C_r}{\left( {a{x^{{1 \over 3}}}} \right)^{15 - r}}{\left( { - {1 \over {b{x^3}}}} \right)^2}

∴\therefore For x−15→15−r3−3r=−15⇒r=6{x^{ - 15}} \to {{15 - r} \over 3} - 3r = - 15 \Rightarrow r = 6

∴\therefore 15C9a6b9=15C6a9b6⇒ab=1{}^{15}{C_9}{{{a^6}} \over {{b^9}}} = {}^{15}{C_6}{{{a^9}} \over {{b^6}}} \Rightarrow ab = 1

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