Let \(f:(0,1) \rightarrow R\) be defined by \(f(x)=\frac{b-x}{1-b x}\), where \(b\) is a constant such that \(0 < b < 1\). Then,

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, \(0 < x < 1\)
\[
\begin{gathered}
\therefore \quad 0 < \frac{b-y}{1-b y} < 1 \\
\frac{b-y}{1-b y}>0 \text { and } \frac{b-y}{1-b y} < 1 \\
\quad+\quad-\quad+ \\
\Rightarrow \quad b \\
y < b \text { or } y>\frac{1}{b} \\
\frac{(b-1)(y+1)}{1-b y} < -1 < y < \frac{1}{b}
\end{gathered}
\]
From Eqs. (i) and (ii), we get \(y \in\left(-1, \frac{1}{b}\right) \subset\) Codomain
Thus, \(f(x)\) is not invertible.