Question

A function f: `R - {3/5} -> R - {3/5}` is defined as `f(x) = (3x + 2)/(5x - 3)`. Show that f is one-one and onto.

Answer 1

`f(x) = (3x + 2)/(5x - 3)`

One-one:

Let f(x₁) = f(x₂),

`(3x_1 + 2)/(5x_2 - 3) = (3x_2 + 2)/(5x_2 - 3)`

Cross-multiply:

(3x₁​ + 2)(5x₂ ​− 3) = (3x₂ + 2)(5x₁ − 3)

x₁​ = x₂

So, f is one-one.

Onto: Let,

`y = (3x + 2)/(5x - 3)`

y(5x − 3) = 3x + 2

5xy − 3y = 3x + 2

5xy − 3x = 3y + 2

x(5y − 3) = 3y + 2

`x = (3y + 2)/(5y - 3)`

Since `y ≠ 3/5`, x exists in `R - {3/5}`

Therefore, f is onto,

Hence, f is one-one and onto.