Mathematics - Binomial Theorem Question with Solution | TestHub

$$\begin{array}{rl}& T_{r+1}\text{ in the expansion }{\left[a{x}^2+\frac{1}{bx}\right]}^{11}={}^{11}{C}_r{\left(a{x}^2\right)}^{11-r}{\left(\frac{1}{bx}\right)}^r\\ & ={}^{11}{C}_r(a{)}^{11-r}(b{)}^{-r}(x{)}^{22-2r-r}\\ & \Rightarrow 22-3r=7\Rightarrow r=5\\ & \therefore \text{ coefficient of }{x}^7={}^{11}{C}_5(a{)}^6(b{)}^{-5}....(i)\end{array}$$ Again $T_{r+1}$ in the expansion ${\left[ax-\frac{1}{b{x}^2}\right]}^{11}={}^{11}{C}_r(ax{)}^{11-r}{\left(-\frac{1}{b{x}^2}\right)}^r$ $={}^{11}{\mathrm{C}}_{\mathrm{r}}{\mathrm{a}}^{11-\mathrm{r}}(-1{)}^{\mathrm{r}}\times (\mathrm{b}{)}^{-\mathrm{r}}(\mathrm{x}{)}^{-2\mathrm{r}}(\mathrm{x}{)}^{11-\mathrm{r}}$ Now $11-3r=-7\Rightarrow 3r=18\Rightarrow r=6$ $$\begin{array}{rl}& \therefore \text{ coefficient of }{x}^{-7}={}^{11}{C}_6{a}^5\times 1\times (b)-6\\ & \Rightarrow {}^{11}{C}_5(a{)}^6(b{)}^{-5}={}^{11}{C}_6{a}^5\times (b{)}^{-6}\\ & \Rightarrow ab=1\end{array}$$