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:
Correct Option: , 3
Coefficient of $x^{7}$ in $\left(x^{2}+\frac{1}{b x}\right)^{11}$
${ }^{11} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{11-\mathrm{r}} \cdot\left(\frac{1}{\mathrm{bx}}\right)^{\mathrm{r}}$
${ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{22-3 \mathrm{r}} \cdot \frac{1}{\mathrm{~b}^{\mathrm{r}}}$
$22-3 r=7$
$r=5$
$\therefore{ }^{11} \mathrm{C}_{5} \cdot \frac{1}{\mathrm{~b}^{5}} \cdot \mathrm{x}^{7}$
Coefficient of $x^{-7}$ in $\left(x-\frac{b}{b x^{2}}\right)^{11}$
${ }^{11} C_{r}(x)^{11-r} \cdot\left(-\frac{1}{b x^{2}}\right)^{r}$
${ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{11-3 \mathrm{r}} \cdot \frac{(-1)^{\mathrm{r}}}{\mathrm{b}^{\mathrm{r}}}$
$11-3 r=-7 \quad \therefore r=6$
${ }^{11} C_{6} \cdot \frac{1}{b^{6}} x^{-7}$
${ }^{11} C_{5} \cdot \frac{1}{b^{5}}={ }^{11} C_{6} \cdot \frac{1}{b^{6}}$
Since $b \neq 0 \therefore b=1$