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

Explanation

Here `A = R - {3} , B = R - {1}`
Now, in `f: A to B, f(x) = (x-2)/(x-3)`
Let ` x, y in A and f(x) = f(y)`
`rArr " " (x- 2)/(x-3) = (y-2)/(y-3)`
`rArr (x-2) (y-3) = (x-3)(y-2)`
`rArr xy - 3x - 2y + 6 = xy - 2x - 3y + 6 `
`rArr x=y`
` therefore f` is one -one.
`rArr ` Again let `f(x) = y` where `y = in B ( y ne 1)`
`rArr " " (x-2)/(x-3) = y`
`rArr " "x -2 = xy - 3y`
`rArr " " x(1-y) = 2-3y`
`rArr " " x = (2-3y)/(1-y) in A `
Now `f(x) = f((2-3y)/(1-y)) = ((2-3y)/(1-y)-2)/((2-3y)/(1-y)-3) `
`" " = (2-3y- 2 + 2y )/( 2-3y - 3 + 3y ) = y`
` therefore f ` is onto
Therefore, `f` is one-one onto functions.