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# Let A=R-{3},B=R-{1}.Let $f \;:\;A \rightarrow B$ be defined by $f(x)=\Large{\frac{x-2}{x-3}}\normalsize \forall x \in A$.Then show that f is bijective.

Toolbox:
• A function$f :A \to B$ is bijective if it is one-one ie $f(x)=f(y)=>x=y \qquad x \in A$ and onto ie for every $y \in B$ then exists $x \in A$ such that $f(x)=y$
Given :Let $A=R-\{3\}\qquad B=R -\{1\}$

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

Step1: Injective or One-One function:

Let $f(x)=f(y) \qquad x,y \neq 3$

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

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

$xy-2y-3x+6=xy-2x-3y+6$

$-2x+3y=-2x+3x$

$x=y$

Hence f is one -one

Step 2: Surjective or On-to function:

Let $y \in B$ such that $y \neq 1$

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

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

$yx-3y=x-2$

$yx-x=3y-2$

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

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

$x \in A$ since $y \neq 1$

Hence there exists $x \in A$ for every $y \in B$ such that $f(x)=y$

f is onto

Solution:Hence f is bijective

edited Mar 27, 2013 by meena.p