Let `f:R in R` be a continuous function such that f(x) is not identically equal to zero. If `int_(0)^(x) |x-2|dx,x ge 0`. Then, f'(x) is

Options

A. an even function
B. an odd function
C. a periodic function
D. none of these

Answer

To solve the problem, we need to find the derivative \( f'(x) \) of the function defined by the integral

\[ f(x) = \int_{0}^{x} |x - 2| \, dx, \quad x \geq 0. \]

### Step 1: Analyze the integrand

The integrand \( |x - 2| \) is a piecewise function. We need to determine how it behaves over the interval \( [0, x] \):

- For \( x < 2 \), \( |x - 2| = 2 - x \).
- For \( x \geq 2 \), \( |x - 2| = x - 2 \).

### Step 2: Define the function piecewise

Thus, we can define \( f(x) \) piecewise:

1. **For \( 0 \leq x < 2 \)**:
\[ f(x) = \int_{0}^{x} (2 - t) \, dt \]

2. **For \( x \geq 2 \)**:
\[ f(x) = \int_{0}^{2} (2 - t) \, dt + \int_{2}^{x} (t - 2) \, dt \]

### Step 3: Calculate the integrals

**For \( 0 \leq x < 2 \)**:
\[ f(x) = \int_{0}^{x} (2 - t) \, dt = [2t - \frac{t^2}{2}]_{0}^{x} = 2x - \frac{x^2}{2} \]

**For \( x \geq 2 \)**:
First, calculate the integral from 0 to 2:
\[ \int_{0}^{2} (2 - t) \, dt = [2t - \frac{t^2}{2}]_{0}^{2} = 4 - 2 = 2 \]

Now calculate the integral from 2 to \( x \):
\[ \int_{2}^{x} (t - 2) \, dt = \left[\frac{t^2}{2} - 2t\right]_{2}^{x} = \left(\frac{x^2}{2} - 2x\right) - \left(\frac{2^2}{2} - 2 \cdot 2\right) = \frac{x^2}{2} - 2x + 2 \]

Combining these results for \( x \geq 2 \):
\[ f(x) = 2 + \left(\frac{x^2}{2} - 2x + 2\right) = \frac{x^2}{2} - 2x + 4 \]

### Step 4: Find the derivative \( f'(x) \)

Now we can find \( f'(x) \) for both cases:

1. **For \( 0 \leq x < 2 \)**:
\[ f'(x) = \frac{d}{dx}\left(2x - \frac{x^2}{2}\right) = 2 - x \]

2. **For \( x \geq 2 \)**:
\[ f'(x) = \frac{d}{dx}\left(\frac{x^2}{2} - 2x + 4\right) = x - 2 \]

### Step 5: Combine results

Thus, we have:
\[ f'(x) = \begin{cases} 2 - x & \text{if } 0 \leq x < 2 \\ x - 2 & \text{if } x \geq 2 \end{cases} \]

### Final Answer
The derivative \( f'(x) \) is given by:
\[ f'(x) = \begin{cases} 2 - x & \text{if } 0 \leq x < 2 \\ x - 2 & \text{if } x \geq 2 \end{cases} \]