If A = R - {3} and B = R - {1}. Let f: A → B such that f(x) = (x-2)/(x-3) for all x ∈ A. Then show that f is bijective.

Given: 

$A=\mathbb{R}-\{3\}$

$B=\mathbb{R}-\{1\}$

$f:A\to B$ such that $f(x)$ $=$ $\frac{x-2}{x-3}$

To Prove:

$$ (i) $f(x)$ is one-one.

$$ (ii) $f(x)$ is onto.

Proof:

# Condition for One-One:

$$If $f(x_1)=f(x_2)$ then 

$x_1=x_2\mathrm{\forall }{x}_1,x_2\in A$

$$ OR $$If $x_1\ne x_2$ then 

$f(x_1)\ne f(x_2)\mathrm{\forall }{x}_1,x_2\in A$

(i) Let $x_1,x_2\in A$ be any element such that:

$f(x_1)=f(x_2)$

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

$(x_1-2)(x_2-3)=(x_1-3)(x_2-2)$

$x_1{x}_2-3x_1-2x_2+6$

$=x_1{x}_2-2x_1-3x_2+6$

$-3x_1-2x_2=-2x_1-3x_2$

$-3x_1+2x_1=+2x_2-3x_2$

$-x_1=-x_2$

$x_1=x_2\mathrm{\forall }{x}_1,x_2\in A$

$\boxed{{\displaystyle \text{So, }f(x)\text{ is one-one.}}}$

# Condition for Onto:

$$For every $y\in B$ there exist $x\in A$ such that

$y=f(x)$

$$OR $$ Range $=$ Codomain.

(ii) Let $y\in B$ be any element

We need to find an element $x\in A$ such that:

$y=f(x)$

Finding suitable value of $x$:

$y$ $=$ $\frac{x-2}{x-3}$

$(x-3)y=(x-2)$

$xy-3y=x-2$

$xy-x=3y-2$

$x(y-1)=3y-2$

$\boxed{{\displaystyle x=\frac{3y-2}{y-1}}}(1)$

If $x=3$ then,

$3$ $=$ $\frac{3y-2}{y-1}$

$3(y-1)=(3y-2)$

$3y-3=3y-2$

$-3=-2$

which cannot be possible.

So, $x\in \mathbb{R}-\{3\}=A$

Using $(1)$, we get

 $f(x)$ $=$ $\frac{x-2}{x-3}$

$=$ $\frac{\frac{3y-2}{y-1}-2}{\frac{3y-2}{y-1}-3}$

$=$ $\frac{(3y-2)-2(y-1)}{(3y-2)-3(y-1)}$

$=$ $\frac{3y-2-2y+2}{3y-2-3y+3}$

$=$ $\frac{y}{1}$

$f(x)=y\mathrm{\forall }y\in B$

$\boxed{{\displaystyle \text{So, }f(x)\text{ is onto.}}}$