Let f:(0,1) arrow R be defined by f(x)=(b-x)/(1-b x) , where b is a...

Here, $f(x)=\frac{b-x}{1-b x}$ where, $0 < b < 1,0 < x < 1$ For function to be invertible it should be one-one onto. $\therefore$ Check range : Let $\quad f(x)=y \Rightarrow y=\frac{b-x}{1-b x}$$\begin{aligned}& \Rightarrow y-b x y=b-x \Rightarrow x(1-b y)=b-y \\ & \Rightarrow \quad x=\frac{b-y}{1-b y} \end{aligned}$$$where,$0 < x < 1 \begin{gathered} \therefore \quad0 < \frac{b-y}{1-b y} < 1 \\ \frac{b-y}{1-b y}>0 \text{\text{ and } } \frac{b-y}{1-b y}< 1 \\ \quad+\quad-\quad+ \\ \Rightarrow \quad b \\ y < b \text{\text{ or } } y>\frac{1}{b}\\ \frac{(b-1)(y+1)}{1-b y}< -1 < y < \frac{1}{b} \end{gathered}$From Eqs. (i) and (ii), we get$y \in\left(-1, \frac{1}{b}\right) \subset$Codomain Thus,$f(x)$ is not invertible.