Question

If the coefficients of $x^7$ in ${\left(x^2+\frac{1}{bx}\right)}^{11}$ and $x^{-7}$ in ${\left(x-\frac{1}{b{x}^2}\right)}^{11},b\ne 0$, are equal, then the value of $b$ is equal to:

• A. 2
• B. -1
• C. 1
• D. -2

Answer 1

Answer: (C)

Coefficient of $x^7$ in ${\left(x^2+\frac{1}{bx}\right)}^{11}$
${}^{11}{C}_r{\left(x^2\right)}^{11-r}\cdot {\left(\frac{1}{bx}\right)}^r$
${}^{11}{C}_r{x}^{22-3r}\cdot \frac{1}{b^r}$
$22-3r=7$
$r=5$
$\therefore {}^{11}{C}_5\cdot \frac{1}{b^5}\cdot x^7$
Coefficient of $x^{-7}\text{ in }{\left(x-\frac{b}{b{x}^2}\right)}^{11}$
${}^{11}{C}_r(x{)}^{11-r}\cdot {\left(-\frac{1}{b{x}^2}\right)}^r$
${}^{11}{C}_r{x}^{11-3r}\cdot \frac{(-1{)}^r}{b^r}$
$11-3r=-7$
$\therefore r=6$
${}^{11}{C}_6\cdot \frac{1}{b^6}{x}^{-7}$
${}^{11}{C}_5\cdot \frac{1}{b^5}={}^{11}{C}_6\cdot \frac{1}{b^6}$
Since $b\ne 0$
$\therefore b=1$