The domain of the function f(x) given by `f(x)=(sqrt(4-x^(2)))/(sin^(-1)(2-x))` is

Options

A. [0,2]
B. [0,2)
C. [1,2)
D. [1,2]

Answer

To find the domain of the function \( f(x) = \frac{\sqrt{4 - x^2}}{\sin^{-1}(2 - x)} \), we need to consider the conditions under which both the numerator and denominator are defined.

### Step 1: Analyze the Numerator
The numerator is \( \sqrt{4 - x^2} \). For this square root to be defined and real, the expression inside the square root must be non-negative:
\[ 4 - x^2 \geq 0 \]
This can be rearranged to:
\[ x^2 \leq 4 \]
Taking the square root of both sides gives:
\[ -2 \leq x \leq 2 \]
Thus, the domain from the numerator is:
\[ x \in [-2, 2] \]

### Step 2: Analyze the Denominator
The denominator is \( \sin^{-1}(2 - x) \). The inverse sine function is defined for inputs in the range \([-1, 1]\). Therefore, we need:
\[ -1 \leq 2 - x \leq 1 \]
This can be split into two inequalities:
1. \( 2 - x \geq -1 \) which simplifies to \( x \leq 3 \)
2. \( 2 - x \leq 1 \) which simplifies to \( x \geq 1 \)

Combining these inequalities gives:
\[ 1 \leq x \leq 3 \]

### Step 3: Combine the Results
Now we need to find the intersection of the two intervals obtained from the numerator and denominator:
- From the numerator: \( x \in [-2, 2] \)
- From the denominator: \( x \in [1, 3] \)

The intersection of these two intervals is:
\[ x \in [1, 2] \]

### Step 4: Exclude Points Where Denominator is Zero
Next, we need to ensure that the denominator is not equal to zero. The denominator \( \sin^{-1}(2 - x) \) is zero when \( 2 - x = 0 \), or \( x = 2 \). Since \( x = 2 \) is included in our interval, we must exclude it.

Thus, the final domain of the function \( f(x) \) is:
\[ x \in [1, 2) \]

### Conclusion
The domain of the function \( f(x) \) is:
\[ \boxed{[1, 2)} \]