# Let $A=R-\{3\}$ and $B=R-\{1\}$. Consider the function $$f:A \to B$$ defined by $f(x)=\large\frac {x-2}{ x-3}$. Is $f$ one-one and onto?

$\begin{array}{1 1} \text{one-one and onto} \\ \text{neither one -one nor onto} \\ \text{only one-one} \\ \text{only onto}\end{array}$

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $A=R- \{3\}$ $A=R- \{3\}$ $f(x)=(\frac{x-2}{x-3})$
Let $x,y \in A$ such that $f(x)=f(y)$
Step1: Injective or One-One function:
$\frac{x-2}{x-3}=\frac{y-2}{y-3}$
$(x-2)(y-3)=(x-3)(y-2)$
$xy-3x-2y+6 =xy -3y-2x+6$
$3x-2x=3y-2y$
$=> x=y$
Therefore $A=R- \{3\}$ $A=R- \{3\}$ $f(x)=(\frac{x-2}{x-3})$ is one-one.
Step 2: Surjective or On-to function:
Let $y \in B=R-[-1] =>y \neq 1$
$f(x)=y$
$(\frac{x-2}{x-3})=y$
$(x-2)=y(x-3)$
$x(1-y)=-3y+2$
$x=\frac{2-3y}{1-y} \in A\qquad since \;y \neq 1$
Therefore there exist an element x in A such that f(x)=y, $y \in B$
$f\bigg[\large\frac{2-3y}{1-y}\bigg]=\frac{(\frac{2-3y}{1-y})-2}{(\frac{2-3y}{1-y})-3}=\frac{-y}{-1}=y$
Therefore $A=R- \{3\}$ $A=R- \{3\}$ $f(x)=(\frac{x-2}{x-3})$ is onto
Solution:Therefore $A=R- \{3\}$ $A=R- \{3\}$ $f(x)=(\frac{x-2}{x-3})$ is one one and onto

edited Mar 20, 2013 by meena.p