Question

Let $A=R-\left\{3\right\},B=R-\left\{1\right\}$. Let $f:A\to B$ defined by $f\left(x\right)=\frac{x-2}{x-3}.$ Show that $f$ is bijective.

Answer 1

To show $f$ is bijective , we need to prove

(i) $f$ is injective i.e. one-one.

(ii) $f$ is surjective i.e. onto.

(i) Let $x_1,x_2\in A$

Let $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$

$\Rightarrow (x_1-2)(x_2-3)=(x_1-3)(x_2-2)$
$\Rightarrow x_1{x}_2-2x_2-3x_1+6=x_1{x}_2-2x_1-3x_2+6$
$\Rightarrow x_1=x_2$

Hence, $f$ is one-one.

(ii) Let $y$ be any arbitrary element of $B$. Suppose there exists an $x$ such that $f\left(x\right)=y,$

i.e. $\frac{(x-2)}{(x-3)}=y$

$x-2=xy-3y.$

$\Rightarrow x=\frac{(3y-2)}{(y-1)}$
Since $y\ne 1,$ $x$ is real.
Further $x\ne 3,$

For if $x=3$,

Then, $3=\frac{(3y-2)}{(y-2)}$

$3y-3=3y-2,$ which is false.

It follows that $x=\frac{(3y-2)}{(y-1)}\in A$ such that $f\left(x\right)=y$ and so $f$ is surjective.

Thus $f$ is bijective.