Question

Prove that the function \(f:R\to\{x\in R:-1<x<1\}\), where \(f(x)=\dfrac{2x}{1+|x|}\), \(x\in R\), is one-one.

Hint:

Split into cases based on the sign of x and show f(x1)=f(x2) forces x1=x2 in every case.

Answer 1

Step 1: Understanding the Concept:
A function is one-one (injective) if different inputs always give different outputs.
This means we must show that whenever \(f(x_1)=f(x_2)\), it forces \(x_1=x_2\).
Since the formula involves \(|x|\), we split the proof into cases based on the sign of x.

Step 2: Case When Both Inputs Are Non-negative:
Let \(x_1,x_2\ge0\). Then \(|x_1|=x_1\) and \(|x_2|=x_2\), so \(f(x)=\dfrac{2x}{1+x}\).
Assume \(f(x_1)=f(x_2)\).
\[ \dfrac{2x_1}{1+x_1}=\dfrac{2x_2}{1+x_2} \]
Cross multiplying gives \(x_1(1+x_2)=x_2(1+x_1)\), which simplifies to \(x_1+x_1x_2=x_2+x_1x_2\).
This reduces to \(x_1=x_2\).

Step 3: Case When Both Inputs Are Negative:
Let \(x_1,x_2<0\). Then \(|x_1|=-x_1\) and \(|x_2|=-x_2\), so \(f(x)=\dfrac{2x}{1-x}\).
Assume \(f(x_1)=f(x_2)\).
\[ \dfrac{2x_1}{1-x_1}=\dfrac{2x_2}{1-x_2} \]
Cross multiplying gives \(x_1(1-x_2)=x_2(1-x_1)\), which simplifies to \(x_1-x_1x_2=x_2-x_1x_2\).
This again reduces to \(x_1=x_2\).

Step 4: Case When Inputs Have Opposite Signs:
Let \(x_1>0\) and \(x_2<0\) (the other order is similar). Then \(f(x_1)=\dfrac{2x_1}{1+x_1}>0\) while \(f(x_2)=\dfrac{2x_2}{1-x_2}<0\).
Since a positive number can never equal a negative number, \(f(x_1)\ne f(x_2)\) in this case.
So equal outputs are impossible unless \(x_1\) and \(x_2\) lie in the same case, which already forces \(x_1=x_2\).

Final Answer:
In every case \(f(x_1)=f(x_2)\) implies \(x_1=x_2\), so f is one-one, hence proved. \[ \boxed{f \text{ is one-one}} \]