The domain of the function f defined by `f(x)= sqrt(a-x)+(1)/( sqrt(x^(2)-a^(2))` is

Options

A. `(-oo,a]`
B. `(-oo,a]`
C. `(-oo,-a)`
D. `(a,oo)`

Answer

To find the domain of the function \( f(x) = \sqrt{a - x} + \frac{1}{\sqrt{x^2 - a^2}} \), we need to ensure that the function is defined for all values of \( x \). This involves checking the conditions for both the square root and the denominator.

### Step-by-Step Solution:

1. **Condition for the square root \( \sqrt{a - x} \)**:
- The expression inside the square root must be non-negative:
\[ a - x \geq 0 \]
- Rearranging gives:
\[ x \leq a \]
- This means \( x \) can take values from \( -\infty \) up to \( a \).

2. **Condition for the denominator \( \sqrt{x^2 - a^2} \)**:
- The expression inside the square root must be positive (since it is in the denominator):
\[ x^2 - a^2 > 0 \]
- This can be factored as:
\[ (x - a)(x + a) > 0 \]
- To solve this inequality, we find the critical points, which are \( x = -a \) and \( x = a \). We will test the intervals determined by these points:
- **Interval 1**: \( x < -a \) (Choose \( x = -a - 1 \)):
\[ (-a - 1 - a)(-a - 1 + a) > 0 \implies (-2a - 1)(-1) > 0 \text{ (True)} \]
- **Interval 2**: \( -a < x < a \) (Choose \( x = 0 \)):
\[ (0 - a)(0 + a) > 0 \implies (-a)(a) < 0 \text{ (False)} \]
- **Interval 3**: \( x > a \) (Choose \( x = a + 1 \)):
\[ (a + 1 - a)(a + 1 + a) > 0 \implies (1)(2a + 1) > 0 \text{ (True)} \]

- From this analysis, the solution to \( (x - a)(x + a) > 0 \) is:
\[ x < -a \quad \text{or} \quad x > a \]

3. **Combining the conditions**:
- From step 1, we have \( x \leq a \).
- From step 2, we have \( x < -a \) or \( x > a \).
- The only overlapping condition that satisfies both is:
\[ x < -a \quad \text{and} \quad x \leq a \]

4. **Final Domain**:
- Therefore, the domain of the function \( f(x) \) is:
\[ (-\infty, -a) \cup (-a, a] \]