If `f(x)=(x^(2)+2alphax+alpha^(2)-1)^(1//4)` has its domain and range such that their union is set of real numbers, then `alpha` satisfies

Options

A. `-1 lt alpha lt 1`
B. `alpha le -1`
C. `alpha ge 1`
D. `alpha le 1`

Answer

To solve the problem, we need to analyze the function \( f(x) = (x^2 + 2\alpha x + \alpha^2 - 1)^{1/4} \) and determine the conditions on \( \alpha \) such that the union of the domain and range of \( f(x) \) is the set of real numbers.

### Step-by-Step Solution:

1. **Rewrite the Function**:
We start with the function:
\[ f(x) = (x^2 + 2\alpha x + \alpha^2 - 1)^{1/4} \]
Notice that the expression inside the parentheses can be factored as:
\[ f(x) = ((x + \alpha)^2 - 1)^{1/4} \]

2. **Set the Condition for the Domain**:
For \( f(x) \) to be defined (i.e., the expression inside the fourth root must be non-negative), we require:
\[ (x + \alpha)^2 - 1 \geq 0 \]
This simplifies to:
\[ (x + \alpha - 1)(x + \alpha + 1) \geq 0 \]

3. **Find Critical Points**:
The critical points of the inequality are:
\[ x + \alpha - 1 = 0 \quad \Rightarrow \quad x = 1 - \alpha \]
\[ x + \alpha + 1 = 0 \quad \Rightarrow \quad x = -1 - \alpha \]

4. **Analyze the Sign of the Expression**:
We analyze the sign of the product \( (x + \alpha - 1)(x + \alpha + 1) \):
- The product is non-negative when \( x \) is in the intervals:
- \( (-\infty, -1 - \alpha] \) or \( [1 - \alpha, \infty) \)

5. **Determine the Union of Domain and Range**:
The domain of \( f(x) \) is \( (-\infty, -1 - \alpha] \cup [1 - \alpha, \infty) \).
The range of \( f(x) \) is \( [0, \infty) \) because the fourth root of a non-negative number is non-negative.

6. **Condition for Union to be Real Numbers**:
For the union of the domain and range to cover all real numbers, we need:
\[ -1 - \alpha \leq 0 \quad \text{and} \quad 1 - \alpha \geq 0 \]
This leads to two inequalities:
- From \( -1 - \alpha \leq 0 \):
\[ \alpha \geq -1 \]
- From \( 1 - \alpha \geq 0 \):
\[ \alpha \leq 1 \]

7. **Combine Conditions**:
Combining these conditions gives:
\[ -1 \leq \alpha \leq 1 \]

### Conclusion:
Thus, the values of \( \alpha \) that satisfy the condition such that the union of the domain and range of \( f(x) \) is the set of real numbers are:
\[ \alpha \in [-1, 1] \]