Let `A=R-{2}` and `B=R-{1}` . If `f: A->B` is a mapping defined by `f(x)=(x-1)/(x-2)` , show that `f` is bijective.

Answer

N/a

Explanation

Given that , `A=R-{3}, B=Rj-{1}.`
`f:A to B` is defined by `f(x)= (x-2)/(x-3), AA x in A`
For injectivity
Let `f(x_(1))=f(x_(2))implies(x_(1)-2)/(x_(1)-3)=(x_(2)-2)/(x_(2)-3)`
`implies(x_(1)-2)(x_(2)-3)=(x_(2)-2)(x_(1)-3)`
`implies x_(1)x_(2)-3x_(1)-2x_(2)+6= x_(1)x_(2)-3x_(2)-2x_(1)+6`
`implies -3x_(1)-2x_(2)= -3x_(2)-2x_(1)`
`implies -x_(1)= -x_(2)implies x_(1)= x_(2)`
So, f(x) is an injective function.
For surjectivity
Let `y=(x-2)/(x-3)impliesx-2=xy-3y`
`impliesx(1-y)=2-3yimplies x=(2-3y)/(1-y)`
`implies x=(3y-2)/(y-1) in A, AA y in B " " ` [codomain]
So, f(x) is surjective function.
Hence, f(x) is a bijective function.