Let `n_(k)` be the number of real solution of the equation `|x+1| + |x-3| = K`, then

Options

A. `n_(k) = 0`, if `K lt 4`
B. `n_(k) = 2` if `K gt 4`
C. `n_(k)` is infinitely many if K = 4
D. Minimum value of `f(x) = |x+1| + |x-3|` is 2

Answer

To solve the equation \( |x + 1| + |x - 3| = K \) and find the number of real solutions \( n_k \), we will analyze the behavior of the function \( f(x) = |x + 1| + |x - 3| \).

### Step 1: Identify the critical points
The absolute value function changes at the points where the expressions inside the absolute values are zero. Therefore, we need to find the points:
- \( x + 1 = 0 \) → \( x = -1 \)
- \( x - 3 = 0 \) → \( x = 3 \)

These points divide the real number line into three intervals:
1. \( (-\infty, -1) \)
2. \( [-1, 3] \)
3. \( (3, \infty) \)

### Step 2: Define the function \( f(x) \) in each interval
Now, we will express \( f(x) \) in each of these intervals.

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

2. **For \( -1 \leq x < 3 \)**:
\[ f(x) = (x + 1) - (x - 3) = x + 1 - x + 3 = 4 \]

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

### Step 3: Analyze the function in each interval
Now we analyze the function \( f(x) \) in each interval:

1. **For \( x < -1 \)**:
- The function \( f(x) = -2x + 2 \) is a linear function with a negative slope. As \( x \) decreases, \( f(x) \) increases.

2. **For \( -1 \leq x < 3 \)**:
- The function \( f(x) = 4 \) is constant.

3. **For \( x \geq 3 \)**:
- The function \( f(x) = 2x - 2 \) is a linear function with a positive slope. As \( x \) increases, \( f(x) \) increases.

### Step 4: Determine the minimum value of \( f(x) \)
- At \( x = -1 \), \( f(-1) = 4 \).
- At \( x = 3 \), \( f(3) = 4 \).
- The minimum value of \( f(x) \) occurs at both critical points and is \( 4 \).

### Step 5: Analyze the number of solutions based on \( K \)
1. **If \( K < 4 \)**:
- The line \( y = K \) will not intersect \( f(x) \) since the minimum value of \( f(x) \) is \( 4 \). Thus, \( n_k = 0 \).

2. **If \( K = 4 \)**:
- The line \( y = 4 \) will touch the graph of \( f(x) \) at \( x = -1 \) and \( x = 3 \). Thus, \( n_k = \infty \) (infinitely many solutions).

3. **If \( K > 4 \)**:
- The line \( y = K \) will intersect the graph of \( f(x) \) at two points (one in the interval \( (-\infty, -1) \) and one in the interval \( (3, \infty) \)). Thus, \( n_k = 2 \).

### Conclusion
The number of real solutions \( n_k \) can be summarized as follows:
- \( n_k = 0 \) if \( K < 4 \)
- \( n_k = \infty \) if \( K = 4 \)
- \( n_k = 2 \) if \( K > 4 \)