Question

Show that function f : R→{ x ∈ R:−1< x <1 } defined by f(x)= \(\frac{x}{1+\mid x\mid}\), x∈R is one-one and onto function.

Answer 1

It is given that f: R \(\to\) {x ∈ R: −1 < x < 1} is defined as f(x) = \(\frac{x}{1+\mid x\mid }\) , x ∈R. 
Suppose f(x) = f(y), where x, y ∈ R.
\(\Rightarrow \frac {x}{1+\mid x \mid}=\frac{y}{1+\mid y \mid}\)
It can be observed that if x is positive and y is negative, then we have:
\(\frac {x}{1+x}=\frac{y}{1-y}=\frac{2xy}{x-y}.\)
Since x is positive and y is negative:
x > y \(\Rightarrow\) 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) \(\Rightarrow \frac{x}{1+x}=\frac{y}{1+y}\) 
\(\Rightarrow\) x + xy = y + xy \(\Rightarrow\) x = y.
When x and y are both negative, we have:
f(x)=f(y) \(\Rightarrow \frac {x}{1-x}=\frac{y}{1-y}\) 
\(\Rightarrow\) x-xy = y-xy \(\Rightarrow\) x = y.
∴ f is one-one.
Now, let y ∈ R such that −1 < y < 1.
If y is negative, then there exists x= \(\frac{y}{1+y}\) ∈R such that 
\(f(x)=f(\frac{y}{1+y})=\frac{(\frac{y}{1+y})}{1+\mid \frac{y}{1+y} \mid}=\frac{(\frac{y}{1+y})}{1+(-\frac{y}{1+y})}=\frac{y}{1+y-y}=y\)
If y is positive, then there exists \(x=\frac{y}{1-y}\) ∈R such that 
\(f(x)=f(\frac{y}{1-y})=\frac{(\frac{y}{1-y})}{1+\mid \frac{y}{1-y}\mid}=\frac{(\frac{y}{1-y})}{1+(-\frac{y}{1-y})}=\frac{y}{1-y+y}=y.\)
∴ f is onto.

Hence, f is one-one and onto.