# Let ( A=R-{3} ) and ( B=R-{1} ) Let ( f: A rightarrow B: f(x)=frac{x-2}{x-3} ) for all values of ( x in A ) Show that fis one-one and onto.

( quad f ) is one-one, since

[

begin{aligned}

fleft(x_{1}right)=fleft(x_{2}right) & Rightarrow frac{x_{1}-2}{x_{1}-3}=frac{x_{2}-2}{x_{2}-3}

& Rightarrowleft(x_{1}-2right)left(x_{2}-3right)=left(x_{1}-3right)left(x_{2}-2right)

& Rightarrow x_{1} x_{2}-3 x_{1}-2 x_{2}+6=x_{1} x_{2}-2 x_{1}-3 x_{2}+6

& Rightarrow x_{1}=x_{2}

end{aligned}

]

Let ( y in B ) such that ( y=frac{x-2}{x-3} )

Then, ( (x-3) y=(x-2) Rightarrow x=frac{(3 y-2)}{(y-1)} )

Clearly, ( x ) is defined when ( y eq 1 ) Also, ( x=3 ) will give us ( 1=0, ) which is false.

( therefore quad x eq 3 )

[

text { And, } f(x)=frac{left(frac{3 y-2}{y-1}-2right)}{left(frac{3 y-2}{y-1}-3right)}=y

]

Thus, for each ( y in B ), there exists ( x in A ) such that ( f(x)=y ).

( therefore f ) is onto.

Hence, ( f ) is one-one onto.