Verified Solution • 2025–26 Session moderate

QJEEM23Macabe: JEE Main 2023 Mathematics - Class 11 JEEM23Macabe Solution: If the coefficient of in and in are equal, then :

If the coefficient of x7{x^7} in (ax2+12bx)11{\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}} and x−7{x^{ - 7}} in (ax−13bx2)11{\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}} are equal, then :

Multiple Choice Options

  • A 243ab=64243ab = 64
  • B 32ab=72932ab = 729
  • C 64ab=24364ab = 243
  • D 729ab=32729ab = 32

Step-by-Step Verified Solution

General term of (ax2+12bx)11\left(a x^2+\frac{1}{2 b x}\right)^{11} is

Tr+1=11Cr(ax2)11−r(12bx)r=11Cr(a)11−r(12b)rx22−3rT_{r+1}={ }^{11} C_r\left(a x^2\right)^{11-r}\left(\frac{1}{2 b x}\right)^r={ }^{11} C_r(a)^{11-r}\left(\frac{1}{2 b}\right)^r x^{22-3 r}

amp; Now, 22−3r=7amp;⇒15=3ramp;⇒r=5\begin{array}{rlrl}&\text { Now, } 22-3 r =7 \\\\&\Rightarrow 15 =3 r \\\\&\Rightarrow r =5\end{array}

and general term of (ax−13bx2)11\left(a x-\frac{1}{3 b x^2}\right)^{11} is

Tr+1amp;=11Cr(ax)11−r(−13bx2)ramp;=11Cra11−r(−13b)rx11−3r\begin{aligned}T_{r+1} & ={ }^{11} C_r(a x)^{11-r}\left(-\frac{1}{3 b x^2}\right)^r \\\\& ={ }^{11} C_r a^{11-r}\left(-\frac{1}{3 b}\right)^r x^{11-3 r}\end{aligned}

Now, 11−3r=−711-3 r=-7

⇒18=3r⇒r=6\Rightarrow 18=3 r \Rightarrow r=6

amp; Since, coefficient of x7 in (ax2+12bx)11amp;= Coefficient of x−7 in (ax−13bx2)11amp;⇒11C5(a)6(12b)5=11C6(a)5(−13b)6amp;⇒a32b5=1729b6⇒729ab=32\begin{aligned} & \text { Since, coefficient of } x^7 \text { in }\left(a x^2+\frac{1}{2 b x}\right)^{11} \\\\ & =\text { Coefficient of } x^{-7} \text { in }\left(a x-\frac{1}{3 b x^2}\right)^{11} \\\\ & \Rightarrow{ }^{11} C_5(a)^6\left(\frac{1}{2 b}\right)^5={ }^{11} C_6(a)^5\left(-\frac{1}{3 b}\right)^6 \\\\ & \Rightarrow \frac{a}{32 b^5}=\frac{1}{729 b^6} \Rightarrow 729 a b=32\end{aligned}

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