Mathematics - Binomial Theorem Question with Solution | TestHub

Coefficient of $x^7$ in ${\left(x^2+\frac{1}{bx}\right)}^{11}$

$={}^{11}C_r{\left(x^2\right)}^{11-r}·{\left(\frac{1}{bx}\right)}^{\mathrm{r}}$

$={}^{11}C_r{x}^{22-3r}·\frac{1}{{ b}^r}$

So, $22-3r=7$

$\Rightarrow r=5$

So, the term is ${}^{11}C_5·\frac{1}{{ b}^5}·x^7$

Coefficient of $x^{-7}$ in ${\left(x-\frac{1}{b{x}^2}\right)}^{11}$

$={}^{11}C_r{\left(x\right)}^{11-r}·{\left(-\frac{1}{ b{x}^2}\right)}^r$

$={}^{11}C_r{x}^{11-3r}·\frac{{\left(-1\right)}^r}{b^r}$

So, $11-3r=-7 \therefore r=6$

So, the term is ${}^{11}C_6·\frac{1}{{ b}^6}{x}^{-7}$

Hence, ${}^{11}C_5·\frac{1}{{ b}^5}={}^{11}C_6·\frac{1}{{ b}^6}$

Since $b\ne 0 \therefore b=1$