The value (s) of k for which
`|x-1|+|x-2|+|x+1|+|x+2|=4k`
has integer solutions, is (are)

Options

A. 1,2,3,4,5,…..
B. 2,3,4,5,6,……
C. 1,3,5,7,…..
D. 0,1,2,3,4,5,….

Explanation

To solve the equation \( |x-1| + |x-2| + |x+1| + |x+2| = 4k \) for integer values of \( k \), we will analyze the left-hand side function \( f(x) = |x-1| + |x-2| + |x+1| + |x+2| \) by breaking it down into different intervals based on the critical points where the absolute values change.

### Step 1: Identify the critical points
The critical points are \( x = -2, -1, 1, 2 \). We will analyze the function \( f(x) \) in the intervals defined by these points:
1. \( x < -2 \)
2. \( -2 \leq x < -1 \)
3. \( -1 \leq x < 1 \)
4. \( 1 \leq x < 2 \)
5. \( x \geq 2 \)

### Step 2: Evaluate \( f(x) \) in each interval

1. **For \( x < -2 \)**:
\[ f(x) = -(x-1) - (x-2) - (x+1) - (x+2) = -4x \]

2. **For \( -2 \leq x < -1 \)**:
\[ f(x) = -(x-1) - (x-2) + (x+1) + (x+2) = -2x + 4 \]

3. **For \( -1 \leq x < 1 \)**:
\[ f(x) = -(x-1) - (x-2) + (x+1) + (x+2) = 6 \]

4. **For \( 1 \leq x < 2 \)**:
\[ f(x) = (x-1) + (x-2) + (x+1) + (x+2) = 2x + 4 \]

5. **For \( x \geq 2 \)**:
\[ f(x) = (x-1) + (x-2) + (x+1) + (x+2) = 4x \]

### Step 3: Determine the values of \( f(x) \)
Now we summarize the values of \( f(x) \) in each interval:
- For \( x < -2 \): \( f(x) = -4x \)
- For \( -2 \leq x < -1 \): \( f(x) = -2x + 4 \)
- For \( -1 \leq x < 1 \): \( f(x) = 6 \)
- For \( 1 \leq x < 2 \): \( f(x) = 2x + 4 \)
- For \( x \geq 2 \): \( f(x) = 4x \)

### Step 4: Find the ranges of \( f(x) \)
- As \( x \to -\infty \), \( f(x) \to \infty \).
- At \( x = -2 \), \( f(-2) = 8 \).
- At \( x = -1 \), \( f(-1) = 6 \).
- At \( x = 1 \), \( f(1) = 6 \).
- At \( x = 2 \), \( f(2) = 8 \).
- As \( x \to \infty \), \( f(x) \to \infty \).

### Step 5: Determine integer solutions for \( 4k \)
From the evaluation, we see that:
- \( f(x) = 6 \) for \( -1 \leq x < 1 \).
- \( f(x) \) takes values from \( 6 \) to \( 8 \) at the endpoints.

Thus, \( 4k \) must be equal to \( 6 \) or \( 8 \):
- For \( 4k = 6 \) → \( k = \frac{6}{4} = 1.5 \) (not an integer)
- For \( 4k = 8 \) → \( k = \frac{8}{4} = 2 \) (integer)

### Conclusion
The only integer value of \( k \) for which the equation has integer solutions is \( k = 2 \).