The equation `||x-1|+a | =4,a in R,` has :

Options

A. 3 distinet real roots for unique value of a.
B. 4 distinct real roots for `a in (-oo, -4)`
C. 2 distinct real roots for `|a| lt 4`
D. no real roots for `a gt 4`

Answer

To solve the equation \( ||x-1| + a| = 4 \) for \( a \in \mathbb{R} \), we will analyze the expression step by step.

### Step 1: Break down the absolute value
The expression \( ||x-1| + a| \) can be simplified by considering the inner absolute value \( |x-1| \). We have two cases based on the value of \( x \):

1. **Case 1:** \( x \geq 1 \)
- Here, \( |x-1| = x-1 \).
- Thus, the equation becomes \( |(x-1) + a| = 4 \).

2. **Case 2:** \( x < 1 \)
- Here, \( |x-1| = 1-x \).
- Thus, the equation becomes \( |(1-x) + a| = 4 \).

### Step 2: Solve Case 1
For **Case 1** where \( x \geq 1 \):
- The equation is \( |(x-1) + a| = 4 \).
- This gives us two sub-cases:
1. \( (x-1) + a = 4 \)
- Rearranging gives \( x + a = 5 \) or \( x = 5 - a \).
2. \( (x-1) + a = -4 \)
- Rearranging gives \( x + a = -3 \) or \( x = -3 - a \).

Now, we need to ensure that \( x \geq 1 \):
- For \( x = 5 - a \):
\[ 5 - a \geq 1 \implies a \leq 4 \]
- For \( x = -3 - a \):
\[ -3 - a \geq 1 \implies -a \geq 4 \implies a \leq -4 \]

### Step 3: Solve Case 2
For **Case 2** where \( x < 1 \):
- The equation is \( |(1-x) + a| = 4 \).
- This gives us two sub-cases:
1. \( (1-x) + a = 4 \)
- Rearranging gives \( -x + a = 3 \) or \( x = a - 3 \).
2. \( (1-x) + a = -4 \)
- Rearranging gives \( -x + a = -5 \) or \( x = a + 5 \).

Now, we need to ensure that \( x < 1 \):
- For \( x = a - 3 \):
\[ a - 3 < 1 \implies a < 4 \]
- For \( x = a + 5 \):
\[ a + 5 < 1 \implies a < -4 \]

### Step 4: Summary of Conditions
From both cases, we have the following conditions:
1. From Case 1:
- \( a \leq 4 \) and \( a \leq -4 \)
2. From Case 2:
- \( a < 4 \) and \( a < -4 \)

### Step 5: Conclusion
Now, we can summarize the results:
- For \( a > 4 \): No solutions.
- For \( a \leq -4 \): Four distinct real roots.
- For \( -4 < a < 4 \): Two distinct real roots.
- For \( a = 4 \): One distinct real root.

### Final Answer
The equation \( ||x-1| + a| = 4 \) has:
- No real roots for \( a > 4 \).
- Four distinct real roots for \( a < -4 \).
- Two distinct real roots for \( -4 < a < 4 \).
- One distinct real root for \( a = 4 \).