Question

Let A = R $-$ {3} and B = R $-$ {1}. Consider the function f : A $\to$ B defined by f(x) = $\frac{x-2}{x-3}$. Show that f is one-one and onto and
hence find f^$-$1. [CBSE 2012, 2014]

Answer 1

We have,

A = R $-$ {3} and B = R $-$ {1}
The function f : A $\to$ B defined by f(x) = $\frac{x-2}{x-3}$

$$\begin{gathered} \mathrm{Let}x,y\in A\mathrm{such}\mathrm{that}f\left(x\right)=f\left(y\right).\mathrm{Then}, \\ \frac{x-2}{x-3}=\frac{y-2}{y-3} \\ \Rightarrow xy-3x-2y+6=xy-2x-3y+6 \\ \Rightarrow -x=-y \\ \Rightarrow x=y \\ \therefore f\mathrm{is}\mathrm{one}-\mathrm{one}. \\ \mathrm{Let}y\in B.\mathrm{Then},y\ne 1. \\ \mathrm{The}\mathrm{function}f\mathrm{is}\mathrm{onto}\mathrm{if}\mathrm{there}\mathrm{exists}x\in A\mathrm{such}\mathrm{that}f\left(x\right)=y. \\ \mathrm{Now}, \\ f\left(x\right)=y \\ \Rightarrow \frac{x-2}{x-3}=y \\ \Rightarrow x-2=xy-3y \\ \Rightarrow x-xy=2-3y \\ \Rightarrow x\left(1-y\right)=2-3y \\ \Rightarrow x=\frac{2-3y}{1-y}\in A\left[y\ne 1\right] \\ \mathrm{Thus},\mathrm{for}\mathrm{any}y\in B,\mathrm{there}\mathrm{exists}\frac{2-3y}{1-y}\in A\mathrm{such}\mathrm{that} \\ f\left(\frac{2-3y}{1-y}\right)=\frac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}=\frac{2-3y-2+2y}{2-3y-3+3y}=\frac{-y}{-1}=y \\ \therefore f\mathrm{is}\mathrm{onto}. \\ \mathrm{So},f\mathrm{is}\mathrm{one}-\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{fucntion}. \\ \mathrm{Now}, \\ \mathrm{As},x=\left(\frac{2-3y}{1-y}\right) \\ \mathrm{So},f^{-1}\left(x\right)=\left(\frac{2-3x}{1-x}\right)=\frac{3x-2}{x-1} \end{gathered}$$