Let a function `f:(2,oo)rarr[0,oo)"defined as " f(x) = (|x-3|)/(|x-2|),` then f is

Options

A. injective & surjective
B. not injective but surjective
C. injective but not surjective
D. neither injective nor surjective

Answer

To determine the nature of the function \( f(x) = \frac{|x-3|}{|x-2|} \) defined on the interval \( (2, \infty) \) and mapping to \( [0, \infty) \), we need to analyze its properties, specifically whether it is injective (one-to-one) or surjective (onto).

### Step-by-Step Solution:

1. **Identify the Domain and Codomain:**
- The domain of \( f \) is \( (2, \infty) \).
- The codomain is \( [0, \infty) \).

2. **Break Down the Function Based on Cases:**
- We will analyze the function in two cases based on the value of \( x \):
- **Case 1:** \( 2 < x \leq 3 \)
- **Case 2:** \( x > 3 \)

3. **Case 1: \( 2 < x \leq 3 \)**
- In this range, \( |x-3| = 3-x \) and \( |x-2| = x-2 \).
- Thus, the function becomes:
\[ f(x) = \frac{3-x}{x-2} \]
- As \( x \) approaches \( 2 \), \( f(x) \) approaches \( +\infty \) (since the denominator approaches \( 0 \) from the positive side).
- At \( x = 3 \), \( f(3) = 0 \).
- Therefore, in this case, \( f(x) \) decreases from \( +\infty \) to \( 0 \).

4. **Case 2: \( x > 3 \)**
- Here, \( |x-3| = x-3 \) and \( |x-2| = x-2 \).
- The function simplifies to:
\[ f(x) = \frac{x-3}{x-2} \]
- As \( x \) approaches \( 3 \), \( f(x) \) approaches \( 0 \).
- As \( x \) approaches \( +\infty \), \( f(x) \) approaches \( 1 \) (since the leading coefficients of the numerator and denominator are both \( 1 \)).
- Therefore, in this case, \( f(x) \) increases from \( 0 \) to \( 1 \).

5. **Combine the Results:**
- From Case 1, \( f(x) \) covers the interval \( (0, +\infty) \) as \( x \) moves from \( 2 \) to \( 3 \).
- From Case 2, \( f(x) \) covers the interval \( (0, 1) \) as \( x \) moves from \( 3 \) to \( +\infty \).
- The overall range of \( f(x) \) is \( [0, +\infty) \).

6. **Determine Injectivity and Surjectivity:**
- Since \( f(x) \) decreases to \( 0 \) in the first case and increases to \( 1 \) in the second case, there are multiple \( x \) values that can yield the same \( f(x) \) value in the range \( (0, 1) \).
- Therefore, the function is **not injective**.
- However, since the function covers the entire codomain \( [0, +\infty) \), it is **surjective**.

### Conclusion:
The function \( f(x) = \frac{|x-3|}{|x-2|} \) is surjective but not injective.