Question

Let $A=R-\left\{3\right\}$ and $B=R-\left\{1\right\}$ consider the function $f:A\to B$ defined by $f\left(x\right)=\left(\frac{x-2}{x-3}\right)$.Is $f$ one -one and onto?Justify your answer.

Answer 1

$\because A=R-\left\{3\right\};B=R-\left\{1\right\}f:A⟶Bisdefinedasf\left(x\right)=\left(\frac{x-2}{x-3}\right)Leta,b\in Asuchthatf\left(a\right)=f\left(b\right)\Rightarrow \left(\frac{a-2}{a-3}\right)=\left(\frac{b-2}{b-3}\right)\Rightarrow (a-2)(b-3)=(a-3)(b-2)\Rightarrow a=b\therefore fisone-oneLety\in B=R-\left\{1\right\},then,y\ne 1Thefunctionisontoifthereexistsx\in Asuchthatf\left(x\right)=yNow,f\left(x\right)=y\Rightarrow \frac{x-2}{x-3}=y\Rightarrow (x-2)=y(x-3)\Rightarrow x=\frac{2-3y}{1-y}\in A[y\ne 1]\therefore fisontohencefunctionisone-oneandonto.$