A mapping `f: X to Y `is one-one, if

Options

A. `f(x_(1)) ne f(x_(2))` for all `x_(1), x_(2) in X`
B. `f(x_(1)) = f(x_(2))Rightarrow x_(1)=x_(2)` for all `x_(1), x_(2) in X`
C. `x_(1)=x_(2) Rightarrow f(x_(1))=f(x_(2))` for all `x_(1), x_(2) in X`
D. none of these

Explanation

To determine whether a mapping \( f: X \to Y \) is one-one (or injective), we need to follow these steps:

### Step-by-Step Solution:

1. **Definition of One-One Function**: A function \( f \) is said to be one-one (injective) if different elements in the domain map to different elements in the codomain. This means that if \( f(x_1) = f(x_2) \), then it must follow that \( x_1 = x_2 \).

2. **Assume \( f(x_1) = f(x_2) \)**: To check if the function is one-one, we start by assuming that \( f(x_1) = f(x_2) \) for some \( x_1, x_2 \in X \).

3. **Conclude \( x_1 = x_2 \)**: If our assumption leads us to conclude that \( x_1 = x_2 \), then the function satisfies the condition for being one-one.

4. **Final Statement**: Therefore, the mapping \( f: X \to Y \) is one-one if and only if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

### Conclusion:
The correct answer is that a function is one-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

---