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Question 4:
Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) =
, x ∈R is one-one and onto function.
It is given that f: R → {x ∈ R: −1 < x < 1} is defined as f(x) =, x ∈R.
Suppose f(x) = f(y), where x, y ∈ R.
It can be observed that if x is positive and y is negative, then we have:
Since x is positive and y is negative:
x > y ⇒ x − y > 0
But, 2xy is negative.
Then, 
.
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:
When x and y are both negative, we have:
∴ f is one-one.
Now, let y ∈ R such that −1 < y < 1.
If x is negative, then there exists
such that
If x is positive, then there exists
such that
∴ f is onto.
Hence, f is one-one and onto.
