The equation `∣∣x−1∣+a∣=4` can have real solutions for `x` if a belongs to the interval.

Answer

To solve the equation \( ||x - 1| + a| = 4 \) for real solutions of \( x \), we need to determine the values of \( a \) for which this equation has real solutions. Here’s a step-by-step solution:

### Step 1: Understand the Equation
The equation we are dealing with is \( ||x - 1| + a| = 4 \). The outer modulus indicates that the expression inside can take both positive and negative values.

### Step 2: Set Up the Cases
From the modulus, we can set up two cases:
1. \( |x - 1| + a = 4 \)
2. \( |x - 1| + a = -4 \)

### Step 3: Analyze the First Case
For the first case \( |x - 1| + a = 4 \):
- Rearranging gives us \( |x - 1| = 4 - a \).
- For this to have real solutions, \( 4 - a \) must be non-negative:
\[ 4 - a \geq 0 \implies a \leq 4 \]

### Step 4: Analyze the Second Case
For the second case \( |x - 1| + a = -4 \):
- Rearranging gives us \( |x - 1| = -4 - a \).
- Since the modulus cannot be negative, we require \( -4 - a \geq 0 \):
\[ -4 - a \geq 0 \implies a \leq -4 \]

### Step 5: Combine the Results
From the two cases, we have:
1. \( a \leq 4 \)
2. \( a \leq -4 \)

Thus, \( a \) must satisfy both conditions. The second condition is more restrictive.

### Step 6: Write the Interval
The values of \( a \) that satisfy \( a \leq -4 \) can be expressed in interval notation as:
\[ (-\infty, -4] \]

### Step 7: Conclusion
The equation \( ||x - 1| + a| = 4 \) can have real solutions for \( x \) if \( a \) belongs to the interval:
\[ (-\infty, -4] \cup [4, \infty) \]