The values of `alpha and beta` for which `|e^(|x-beta|)-alpha|=2` has four distinct solutions are

Options

A. `a in (-2, oo), b=0`
B. `a in (2, oo), b =0`
C. `a in (3,oo) , b in R`
D. `a in (2, oo) b =0`

Answer

To solve the equation \( |e^{|x - \beta|} - \alpha| = 2 \) for the values of \( \alpha \) and \( \beta \) such that there are four distinct solutions, we can break down the problem step by step.

### Step 1: Rewrite the equation
The given equation can be rewritten as:
\[ e^{|x - \beta|} - \alpha = 2 \quad \text{or} \quad e^{|x - \beta|} - \alpha = -2 \]
This leads to two cases:
1. \( e^{|x - \beta|} = \alpha + 2 \)
2. \( e^{|x - \beta|} = \alpha - 2 \)

### Step 2: Analyze the first case
For the first case \( e^{|x - \beta|} = \alpha + 2 \):
- Since the exponential function is always positive, we require:
\[ \alpha + 2 > 0 \quad \Rightarrow \quad \alpha > -2 \]

### Step 3: Analyze the second case
For the second case \( e^{|x - \beta|} = \alpha - 2 \):
- Again, since the exponential function is always positive, we require:
\[ \alpha - 2 > 0 \quad \Rightarrow \quad \alpha > 2 \]

### Step 4: Combine the conditions on \( \alpha \)
From the analysis of both cases, we find that:
\[ \alpha > 2 \]
This means \( \alpha \) must be in the interval \( (2, \infty) \).

### Step 5: Determine the values of \( \beta \)
The term \( |x - \beta| \) can take any non-negative value, which means:
- The expression \( e^{|x - \beta|} \) can cover all positive values.
- Therefore, \( \beta \) can be any real number.

### Conclusion
Thus, the values of \( \alpha \) and \( \beta \) for which the equation \( |e^{|x - \beta|} - \alpha| = 2 \) has four distinct solutions are:
\[ \alpha \in (2, \infty) \quad \text{and} \quad \beta \in \mathbb{R} \]