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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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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.

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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

 

 

answered Mar 4, 2013 by meena.p
edited Mar 27, 2013 by meena.p
 

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