Question

Let A = R – {2} and B = R – {1}. If f : A → B is a function defined by $f\left(x\right)=\frac{x-1}{x-2},$ show that f is one-one and onto. Find f^–1.

Answer 1

Given: $f\left(x\right)=\frac{x-1}{x-2}$

To show f is one-one:

$$\begin{gathered} \mathrm{Let}f\left(x_1\right)=f\left(x_2\right) \\ \Rightarrow \frac{x_1-1}{x_1-2}=\frac{x_2-1}{x_2-2} \\ \Rightarrow \left(x_1-1\right)\left(x_2-2\right)=\left(x_2-1\right)\left(x_1-2\right) \\ \Rightarrow x_1{x}_2-2x_1-x_2+2=x_1{x}_2-2x_2-x_1+2 \\ \Rightarrow -2x_1-x_2=-2x_2-x_1 \\ \Rightarrow -2x_1+x_1=-2x_2+x_2 \\ \Rightarrow -x_1=-x_2 \\ \Rightarrow x_1=x_2 \\ \mathrm{Hence},f\mathrm{is}\mathrm{one}-\mathrm{one}. \\ \mathrm{To}\mathrm{show}\mathit{}f\mathrm{is}\mathrm{onto}: \\ \mathrm{Let}y\in B \\ \therefore y=f\left(x\right) \\ \Rightarrow y=\frac{x-1}{x-2} \\ \Rightarrow y\left(x-2\right)=x-1 \\ \Rightarrow xy-2y=x-1 \\ \Rightarrow xy-x=2y-1 \\ \Rightarrow x\left(y-1\right)=2y-1 \\ \Rightarrow x=\frac{2y-1}{y-1} \\ \mathrm{Thus},\mathrm{for}\mathrm{every}\mathrm{value}\mathrm{of}y\mathrm{in}R-\left\{1\right\},\mathrm{there}\mathrm{exists}\mathrm{a}\mathrm{pre}-\mathrm{image}x=\frac{2y-1}{y-1}\mathrm{in}R-\left\{2\right\}. \\ \mathrm{Hence},f\mathrm{is}\mathrm{onto}. \end{gathered}$$

Since, f is one-one and onto
Therefore, f is invertible with $f^{-1}\left(y\right)=\frac{2y-1}{y-1}$.

$\mathrm{Hence},f^{-1}\left(x\right)=\frac{2x-1}{x-1}.$