Let `A=R-{3}` and `B=R-{1}`. Consider the function `f: A to B` defined by `f(x)=((x-2)/(x-3))`.
Is `f` is one-one and onto? Justify your answer

Explanation

`A=R-{3}`
`B=R-{1}`
`f : A to B`
`f(x)=frac(x-2)(x-3)`
`A={-1,0,1,2},B={-4,-2,0,2}`
`(x_1-2)/(x_1-3)=(x_2-2)/(x_2-3)`
`x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6`
`-3x_1-2x_2=-3x_2-2x_1`
`-x_1=-x_2`
`x_1=x_2`
So,`f(x)` is one-one
`f(x)=(x-2)/(x-3)`
`y=(x-2)/(x-3)`
`y(x-3)=x-2`
`x=(3y-2)/(y-1) `
`f(x)=(x-2)/(x-3)`
`=\frac{\frac(3y-2)(y-1)-2}{\frac(3y-2)(y-1)-3}`
`={3y-2-2(y-1)}/{3y-2-3(y-1)}`
`={3y-2-2y+2}/{3y-2-3y+3}`
`implies y`
`f(x)=y`
`f(x)` is onto.
So, f(x) is bijective and invertible.