Question

Show that the function f : R → {x ∈ R : –1 < x < 1} defined by f(x) = `x/(1 + |x|)`, x ∈ R is one-one and onto function.

Answer 1

It is given that f : R → {x ∈ R : –1 < x < 1} is defined as f(x) = `x/(1+ |x|)`, x ∈ R.

Suppose f(x) = f(y), where x, y ∈ R.

⇒ `x/(1 +| x|) = y/(1 - |y|)`

⇒ 2xy =  x – y

⇒ `x/(1 + x) = y/(1 - y) `

⇒ x + xy = y + xy

⇒ x = y

Since x is positive and y is negative:

x > y 

⇒ x − y > 0

But, 2xy is negative.

Then, 2xy ≠ x – y

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out

∴ x and y have to be either positive or negative.

When x and y are both positive, we have:

⇒ f(x) = f(y)

⇒ `x/(1 + |x|) = y/(1 - |y|) `

⇒ `x/(1 + x) = y/(1 + y)`

⇒ x + xy = y + xy

⇒ x = y

When x and y are both negative, we have:

f(x) = f(y)

⇒ `x/(1 - x) = y/(1 - y) `

⇒ x – xy = y – yx 

⇒ x = y

∴ f is one-one.

Now, let y ∈ R such that –1 < y < 1.

If x is negative, then there exists x = `y/(1 + y) ∈ R` such that

`f(x) = f(y/(1 + y)) `

= `((y/(1 + y)))/(1+ |y/(1 + y)|)`

= `(y/(1 + y))/(1 + (-y)/(1 + y))`

= `y/(1 + y - y)`

= y

If x is positive, then there exists x = `y/(1 - y) ∈ R` such that

`f(x) = f(y/(1 - y)) = (y/(1 - y))/(1 + |(y/(1 - y))|)`

= `(y/(1 - y))/(1 + y/(1 - y))`

= `y/(1 - y + y)`

= y

∴ f is onto.

Hence, f is one-one and onto.