Is `f:R rarr R`, given by `f(x)=|x-1|` one-one? Give reason

Answer

To determine whether the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = |x - 1| \) is one-one, we will follow these steps:

### Step 1: Understand the Definition of a One-One Function
A function \( f \) is said to be one-one (or injective) if for every pair of distinct elements \( x_1 \) and \( x_2 \) in the domain, \( f(x_1) \neq f(x_2) \). In other words, if \( f(x_1) = f(x_2) \), then it must follow that \( x_1 = x_2 \).

### Step 2: Analyze the Function
The function is given as \( f(x) = |x - 1| \). This function takes the absolute value of \( x - 1 \), which means it will output the same value for both \( x \) and \( 2 - x \).

### Step 3: Find Specific Values
Let’s consider two specific values:
- Let \( x_1 = 3 \)
- Let \( x_2 = -1 \)

Calculating the function values:
- \( f(3) = |3 - 1| = |2| = 2 \)
- \( f(-1) = |-1 - 1| = |-2| = 2 \)

### Step 4: Compare the Function Values
We see that:
- \( f(3) = 2 \)
- \( f(-1) = 2 \)

Here, \( f(3) = f(-1) \) but \( 3 \neq -1 \). This shows that distinct inputs \( x_1 \) and \( x_2 \) yield the same output.

### Step 5: Conclusion
Since we found distinct elements \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \), we conclude that the function \( f(x) = |x - 1| \) is **not one-one**.

### Summary
The function \( f(x) = |x - 1| \) is not a one-one function because there exist distinct values \( x_1 \) and \( x_2 \) (e.g., \( 3 \) and \( -1 \)) such that \( f(x_1) = f(x_2) \).

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