Question

We have `f:(0,oo)->RR;f(x)=(x+a)/(x+b);a,binRR`.How you verify if `f` is a bijective function?

Answer 1

f(x1) = f(x2)

Explanation:

`f(x_1)`=`f(x_2)`
`(x_1+a)/(x_1+b)` = `(x_2+a)/(x_2+b)`

Now, just do cross multiplication and simplify to obtain

`x_1=x_2`

For surjectivity,

let, y = `(x+a)/(x+b)`

Then, `yx+yb=x+a`
`\implies yb=x(y-1)+a`
`\implies(yb-a)/(y-1)+a`

Therefore, y can assume any real value except 1, which proves that it is not surjective

Answer 2

See the Explanation.

Explanation:

It is injective, as shown by Abjo D.

But it is not surjective. Consider, `y=1 in RR.`

We claim that, corresponding to this

`y=1 in RR," there is no "x in (0,oo)," such that, "f(x)=y=1.`

Suppose, to the contrary, that `f(x)=1," for some "x in (0,oo).`

`f(x)=1 rArr (x+a)/(x+b)=1 rArr x+a=x+b, or, a=b,` a contradiction.

Enjoy Maths.!