Question

Let A = R - {3}, B = R - {1}. Let $f:A\to B$ be defined by $f\left(x\right)=(x-2)/(x-3)$. Is $f$ bijective? Give reasons.

Answer 1

$A=R-\left\{3\right\}B=R-\left\{1\right\}f:A⟶Bisdefinedasf\left(x\right)=\left(\frac{x-2}{x-3}\right)Letx,yϵAsuchthatf\left(x\right)=f\left(y\right)⟹\frac{x-2}{x-3}=\frac{y-2}{y-3}⟹(x-2)(y-3)=(x-3)(y-2)⟹xy-3x-2y+6=xy-3y-2x+6⟹-3x-2y=-3y-2x⟹3x-2x=3y-2y⟹x=y$

$\therefore f$ is one-one.

$LetyϵB=R-\left\{1\right\}.Theny\ne 1$.

The function $f$ is onto if there exists $x$ $ϵ$ $A$ such that $f\left(x\right)=y$.

$Now,f\left(x\right)=y⟹\frac{x-2}{x-3}=y⟹x-2=xy-3y⟹x(1-y)=2-3y⟹x=\frac{2-3y}{1-y}ϵA[y\ne 1]$

Thus, for any $y$ $ϵ$ $B$, there exists $\frac{2-3y}{1-y}$ $ϵ$ $A$ such that

$f\left(\frac{2-3y}{1-y}\right)=\frac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}=\frac{2-3y-2+2y}{2-3y-3+3y}=\frac{-y}{-1}=y$

$\therefore f$ is onto.

Hence, the function $f$ is one-one and onto.