Mathematics - Functions Question with Solution | TestHub

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