Question

If the coefficients of $x^{7}$ in $\left(x^{2}+\frac{1}{b x}\right)^{11}$ and $x^{-7}$ in $\left(x-\frac{1}{b x^{2}}\right)^{11}, b \neq 0$, 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 $\left(x^{2}+\frac{1}{b x}\right)^{11}$
${ }^{11} C_{r}\left(x^{2}\right)^{11-r} \cdot\left(\frac{1}{b x}\right)^{r}$
${ }^{11} C_{r} x^{22-3 r} \cdot \frac{1}{b^{r}}$
$22-3 r=7$
$r=5$
$\therefore { }^{11} C_{5} \cdot \frac{1}{b^{5}} \cdot x^{7}$
Coefficient of $x^{-7} \text { in }\left(x-\frac{b}{b x^{2}}\right)^{11}$
${ }^{11} C_{r}(x)^{11-r} \cdot\left(-\frac{1}{b x^{2}}\right)^{r}$
${ }^{11} C_{r} x^{11-3 r} \cdot \frac{(-1)^{r}}{b^{r}}$
$11-3 r=-7$
$ \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 $b \neq 0$
$\therefore b=1$