Question

If the coefficient of $x^7$ in the expansion of ${(a{x}^2+\frac{1}{bx})}^{11}$ and the coefficient of $x^{-7}$ in the expansion of ${(ax-\frac{1}{b{x}^2})}^{11}$ are equal, then the value of $(ab{)}^2$ is

• A. $\frac{1}{3}$
• B. $5$
• C. $1$
• D. $4$

Answer 1

The correct option is C $1$

For ${(a{x}^2+\frac{1}{bx})}^{11},$ $T_{r+1}={}^{11}{C}_r(a{x}^2{)}^{11-r}{\left(\frac{1}{bx}\right)}^r$
$={}^{11}{C}_r\cdot \frac{a^{11-r}}{b^r}\cdot x^{22-3r}$
For coefficient of $x^7,$
$22-3r=7$
$\Rightarrow r=5$
$\therefore T_6={}^{11}{C}_5\cdot \frac{a^6}{b^5}\cdot x^7$
$\Rightarrow$ Coefficient of $x^7$ is ${}^{11}{C}_5\frac{a^6}{b^5}.$

Similarly, coefficient of $x^{-7}$ in ${(ax-\frac{1}{b{x}^2})}^{11}$ is ${}^{11}{C}_6\frac{a^5}{b^6}.$

According to question,
${}^{11}{C}_5\frac{a^6}{b^5}={}^{11}{C}_6\frac{a^5}{b^6}\Rightarrow ab=1$
$\therefore (ab{)}^2=1$