Question

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

Answer 1

Solve for one-one.
Given: $A=R-\left\{3\right\}$ and $B=R-\left\{1\right\}$
$f:A\to B$ defined by $f\left(x\right)\frac{x-2}{x-3}$
Let $(x,y)\in A,$ then
$f\left(x\right)=\frac{x-2}{x-3}$ and $f\left(y\right)=\frac{y-2}{y-3}$
For one-one, $f\left(x\right)=f\left(y\right)$
$\frac{x-2}{x-3}=\frac{y-2}{y-3}$
$(x-2)(y-3)=(y-2)(x-3)$
$xy-3x-2y+6=xy-3y+2x+6$
$-3x-2y=3y-2x$
$-3x+2x=-3y+2y$
$-x=-y$
$x=y$
If $f\left(x\right)=f\left(y\right),$ then $x=y$
$\therefore f$ is one-one.

Solve for onto
$\because f\left(x\right)=\frac{x-2}{x-3}$
Let $y=f\left(x\right)$
$y=f\left(x\right)=\frac{x-2}{x-3}$
$y(x-3)=x-2$
$xy-3y=x-2$
$x(y-1)=3y-2$
$x=\frac{3y-2}{y-1}$
Now,
$f\left(\frac{3y-2}{y-1}\right)=\frac{\frac{3y-2}{y-1}-2}{\frac{3y-2}{y-1}-3}=y$
$f\left(x\right)=y$
Therefore, $f$ is onto function.
Hence, $f$ is one-one and onto.