Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • $T_{r+1}=nC_r a^{n-r} b^r$
$T_{r+1}$ 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App