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

Options

A. `(0,2)`
B. `[0,2]`
C. `[1,2]`
D. `[1,3]`

Answer

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

### Step 1: Analyze the numerator
The numerator is \( \sqrt{4 - x^2} \). For the square root to be defined, the expression inside 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:
\[ -2 \leq x \leq 2 \]
So, the values of \( x \) must be in the interval \([-2, 2]\).

### 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:
\[ -1 \leq 2 - x \leq 1 \]
Breaking this 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 two inequalities, we find:
\[ 1 \leq x \leq 3 \]

### Step 3: Find the intersection of the intervals
Now, we have two intervals:
1. From the numerator: \([-2, 2]\)
2. From the denominator: \([1, 3]\)

To find the domain of \( f(x) \), we take the intersection of these two intervals:
\[ [-2, 2] \cap [1, 3] = [1, 2] \]

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