Let `A=R-{3}`and `B=R-[1]dot`Consider the function `f: AvecB`defined by `f(x)=((x-2)/(x-3))dot`Show that `f`is one-one and onto and hence find`f^(-1)`

Let `A=R-{3}`and `B=R-[1]dot`Consider the function `f: AvecB`defined b

1.	Let `A=R-{3}`and `B=R-[1]dot`Consider the function `f: AvecB`defined by `f(x)=((x-2)/(x-3))dot`Show that `f`is one-one and onto and hence find`f^(-1)`
Answer» f is one-one since `f(x_(1)) =f(x_(2)) rArr (x_(1) -2)/(x_(1) -3) =(x_(2)-2)/(x_(2)-3)` `rArr (x_(1)-2) (x_(2) -3) =(x_(1) -3) (x_(2)-2)` `rArr x_(1)x_(2) -3x_(1) -2x_(2)+6 =x_(1)x_(2)-2x_(1)-3x_(2) +6` `rArr x_(1)=x_(2)` Let `y inB` such that `y= (x-2)/(x-3)` Then `(x-3) y= (x-2) rArr x=((3y-2))/((y-1))` Clearly x is defined when `y ne ` Also x=3 will give us 1 = 0 which is false `:. x ne 3` And `f(x) = (((3y-2)/(y-1)-2))/(((3y-2)/(y-1)-3))=y` Thus for each `y in B` there exists `x in A` such that `f(x) =y` `:. ` f is into. Hence if is one-one onto.