Let f(x)`=int_(0)^(x) |xx-2|dx, ge 0`. Then, f'(x) is

Options

A. continuous and non differentiable at x=2
B. discontinuous at x=4
C. neither continuous nor differentiable at x=2
D. non-differentiable at x=4

Answer

To solve the problem, we need to find \( f'(x) \) for the function defined as:

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

### Step 1: Analyze the function inside the integral

The expression \( |x - 2| \) can be rewritten based on the value of \( x \):

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

### Step 2: Define \( f(x) \) piecewise

We can express \( f(x) \) as a piecewise function:

\[ f(x) = \begin{cases} \int_{0}^{x} (2 - x) \, dx & \text{if } x < 2 \\ \int_{0}^{x} (x - 2) \, dx & \text{if } x \geq 2 \end{cases} \]

### Step 3: Calculate \( f(x) \) for each case

**Case 1: \( x < 2 \)**

\[ f(x) = \int_{0}^{x} (2 - t) \, dt = \left[ 2t - \frac{t^2}{2} \right]_{0}^{x} = 2x - \frac{x^2}{2} \]

**Case 2: \( x \geq 2 \)**

\[ f(x) = \int_{0}^{x} (t - 2) \, dt = \left[ \frac{t^2}{2} - 2t \right]_{0}^{x} = \frac{x^2}{2} - 2x \]

### Step 4: Differentiate \( f(x) \)

Now we differentiate \( f(x) \) for both cases:

**For \( x < 2 \):**

\[ f'(x) = \frac{d}{dx} \left( 2x - \frac{x^2}{2} \right) = 2 - x \]

**For \( x \geq 2 \):**

\[ f'(x) = \frac{d}{dx} \left( \frac{x^2}{2} - 2x \right) = x - 2 \]

### Step 5: Evaluate \( f'(x) \) at \( x = 2 \)

To check the differentiability at \( x = 2 \), we need to find the left-hand and right-hand derivatives:

- **Left-hand derivative** as \( x \to 2^- \):

\[ f'(2^-) = 2 - 2 = 0 \]

- **Right-hand derivative** as \( x \to 2^+ \):

\[ f'(2^+) = 2 - 2 = 0 \]

### Step 6: Conclusion

Since both left-hand and right-hand derivatives at \( x = 2 \) are equal, \( f'(2) = 0 \). Therefore, the function \( f'(x) \) is defined as:

\[ f'(x) = \begin{cases} 2 - x & \text{if } x < 2 \\ x - 2 & \text{if } x \geq 2 \end{cases} \]

### Final Answer

Thus, \( f'(x) \) is given by:

\[ f'(x) = \begin{cases} 2 - x & \text{if } x < 2 \\ x - 2 & \text{if } x \geq 2 \end{cases} \]