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If the coefficient of $x^7$ in the expansion of $(a^2x+b^{-1}x^{-1})^{11}$ is equal to the coefficient of $x^{-7}$ in $(ax-b^{-1}x^{-2})^{11}$ then $ab=$

$\begin{array}{1 1}(A)\;1\\(B)\;2\\(C)\;3\\(D)\;4\end{array} $

1 Answer

  • $T_{r+1}=nC_r a^{n-r} b^r$
$T_{r+1}$ in $(ax^1+b^{-1}x^{-1})^{11}$
$\Rightarrow 11C_r(ax^2)^{11-r}(\large\frac{1}{bx})^r$
$\Rightarrow 11C_r a^{11-r} b^{-r} x^{22-3r}$
Let $T_{r+1}$ contains $x^7$
$\therefore 22-3r=7$ or $5$
$\therefore T_{r+1}=T_{5+1}=11C_5a^{11-5}b^{-5} x^7$
$\therefore$ Coefficient of $x^7=11C_5a^6b^{-5}$
$T_{r+1}$ in $(ax-b^{-1}x^{-2})^{11}$
$\Rightarrow 11C_r (ax) ^{11-r} (-b^{-1}x^{-2})^r$
$\Rightarrow 11C_r(-1)^ra^{11-r}b^{-r} x^{11-3r}$
Let $T_{r+1}$ contains $x^{-7}$
$\therefore 11-3r=-7$ or 6
$\therefore T_{r+1}=T_{6+1}=11C_6 (-1)^6a^{11-6} b^{-6}x^{-7}$
$\therefore$ Coefficient of $x^{-7}=11C_6a^5b^{-5}$
$\therefore 11C_5a^6b^{-5}=11C_6a^5b^{-6}$
$\Rightarrow a=b^{-1}\Rightarrow ab=1$
Hence (A) is the correct answer.
answered Jun 25, 2014 by sreemathi.v

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