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QJEEM23Mabdcd: JEE Main 2023 Mathematics - Class 11 JEEM23Mabdcd Solution: If the coefficient of in and the coefficient of in are equal,…

If the coefficient of x7{x^7} in (ax−1bx2)13{\left( {ax - {1 \over {b{x^2}}}} \right)^{13}} and the coefficient of x−5{x^{ - 5}} in (ax+1bx2)13{\left( {ax + {1 \over {b{x^2}}}} \right)^{13}} are equal, then a4b4{a^4}{b^4} is equal to :

Multiple Choice Options

  • A 22
  • B 33
  • C 44
  • D 11

Step-by-Step Verified Solution

The given expression is (ax−1bx2)13\left(a x-\frac{1}{b x^2}\right)^{13}

So,

Tr+1amp;=13Cr(ax)13−r(−1bx2)ramp;=13Cr(a)13−rx13−r−2r(−1/b)ramp;=13Cr(a)13−r(−1b)rx13−3r\begin{aligned}T_{r+1} & ={ }^{13} C_r(a x)^{13-r}\left(-\frac{1}{b x^2}\right)^r \\\\& ={ }^{13} C_r(a)^{13-r} x^{13-r-2 r}(-1 / b)^r \\\\& ={ }^{13} C_r(a)^{13-r}\left(-\frac{1}{b}\right)^r x^{13-3 r}\end{aligned}

For coefficient of x7x^7 in (ax−1bx2)13\left(a x-\frac{1}{b x^2}\right)^{13}

amp;13−3r=7amp;⇒3r=6⇒r=2amp;∴ Coefficient of x7=13C2⋅(a)11⋅1b2\begin{aligned}& \quad 13-3 r=7 \\\\& \Rightarrow 3 r=6 \Rightarrow r=2 \\\\& \therefore \text { Coefficient of } x^7={ }^{13} C_2 \cdot(a)^{11} \cdot \frac{1}{b^2}\end{aligned}

Again, the another expression is (ax+1bx2)13\left(a x+\frac{1}{b x^2}\right)^{13}

So, Tn+1=13Cr(ax)13−r(1bx2)r=13Cr(a)13−r(1b)rx13−3rT_{n+1}={ }^{13} C_r(a x)^{13-r}\left(\frac{1}{b x^2}\right)^r={ }^{13} C_r(a)^{13-r}\left(\frac{1}{b}\right)^r x^{13-3 r}

For coefficient x−5x^{-5} in (ax+1bx2)13\left(a x+\frac{1}{b x^2}\right)^{13}

amp;13−3r=−5amp;⇒r=6\begin{aligned}&13-3 r =-5 \\\\&\Rightarrow r =6\end{aligned}

So, coefficient of x−5=13C6(a)71b6x^{-5}={ }^{13} C_6(a)^7 \frac{1}{b^6}

Now, according to the question,

13C2(a)111b2=13C6(a)71b6{ }^{13} C_2(a)^{11} \frac{1}{b^2}={ }^{13} C_6(a)^7 \frac{1}{b^6}

amp;⇒a4b4=13C613C2amp;∴a4b4=22\begin{aligned}& \Rightarrow a^4 b^4=\frac{{ }^{13} C_6}{{ }^{13} C_2} \\\\& \therefore a^4 b^4=22\end{aligned}

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