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

(A) Given $A = R - \{3\}$,$B = R - \{1\}$ and $f: A \rightarrow B$ defined by $f(x) = \frac{x-2}{x-3}$.
For one-one:
Let $x, y \in A$ such that $f(x) = f(y)$.
$\Rightarrow \frac{x-2}{x-3} = \frac{y-2}{y-3}$
$\Rightarrow (x-2)(y-3) = (y-2)(x-3)$
$\Rightarrow xy - 3x - 2y + 6 = xy - 3y - 2x + 6$
$\Rightarrow -3x - 2y = -3y - 2x$
$\Rightarrow x = y$.
Since $f(x) = f(y) \Rightarrow x = y$,$f$ is one-one.
For onto:
Let $y \in B = R - \{1\}$. Then $y \neq 1$.
We need to find $x \in A$ such that $f(x) = y$.
$\frac{x-2}{x-3} = y$
$\Rightarrow x - 2 = y(x - 3)$
$\Rightarrow x - 2 = xy - 3y$
$\Rightarrow x - xy = 2 - 3y$
$\Rightarrow x(1 - y) = 2 - 3y$
$\Rightarrow x = \frac{2 - 3y}{1 - y}$.
Since $y \neq 1$,$x$ is well-defined. Also,$x \neq 3$ because if $\frac{2 - 3y}{1 - y} = 3$,then $2 - 3y = 3 - 3y$,which implies $2 = 3$,a contradiction. Thus $x \in A$.
Since for every $y \in B$,there exists $x \in A$ such that $f(x) = y$,$f$ is onto.
Therefore,$f$ is both one-one and onto.