If `f(x)="sin ln" ((sqrt(4-x^(2)))/(1-x))` then

Options

A. domain of `f(x)` is `(-2,1)`
B. domain of `f(x)` is `[-1,1]`
C. range of `f(x)` is `[-1,1]`
D. range of `f(x)` is `[-1,1)`

Explanation

To solve the given problem, we need to determine the domain and range of the function \( f(x) = \sin \left( \ln \left( \frac{\sqrt{4 - x^2}}{1 - x} \right) \right) \).

### Step 1: Determine the domain of the function

The function \( f(x) \) is defined when the argument of the logarithm is positive. Therefore, we need to ensure that:

\[ \frac{\sqrt{4 - x^2}}{1 - x} > 0 \]

This inequality implies two conditions:

1. The numerator \( \sqrt{4 - x^2} \) must be greater than 0.
2. The denominator \( 1 - x \) must be greater than 0.

#### Condition 1: \( \sqrt{4 - x^2} > 0 \)

For the square root to be positive, we need:

\[ 4 - x^2 > 0 \implies x^2 < 4 \implies -2 < x < 2 \]

#### Condition 2: \( 1 - x > 0 \)

This condition simplifies to:

\[ x < 1 \]

### Step 2: Combine the conditions

Now we combine the two conditions:

1. From \( -2 < x < 2 \)
2. From \( x < 1 \)

The intersection of these intervals is:

\[ -2 < x < 1 \]

Thus, the domain of \( f(x) \) is:

\[ \text{Domain: } (-2, 1) \]

### Step 3: Determine the range of the function

Next, we need to find the range of \( f(x) \). The function \( f(x) \) can be expressed as:

\[ f(x) = \sin \left( \ln \left( \frac{\sqrt{4 - x^2}}{1 - x} \right) \right) \]

The sine function has a range of \( [-1, 1] \). Therefore, we need to find the range of the argument of the sine function, which is:

\[ \ln \left( \frac{\sqrt{4 - x^2}}{1 - x} \right) \]

#### Step 3.1: Analyze the argument

The expression \( \frac{\sqrt{4 - x^2}}{1 - x} \) must be positive for the logarithm to be defined. Since we have already established that \( -2 < x < 1 \), we can analyze the behavior of this expression within this interval.

1. As \( x \) approaches -2, \( \sqrt{4 - x^2} \) approaches \( \sqrt{4 - 4} = 0 \), making the logarithm approach \( -\infty \).
2. As \( x \) approaches 1, \( 1 - x \) approaches 0 from the positive side, making the logarithm approach \( +\infty \).

Thus, the argument \( \ln \left( \frac{\sqrt{4 - x^2}}{1 - x} \right) \) will take all real values from \( -\infty \) to \( +\infty \).

### Step 4: Determine the range of \( f(x) \)

Since the argument of the sine function can take any real value, the sine function will oscillate between -1 and 1. Therefore, the range of \( f(x) \) is:

\[ \text{Range: } [-1, 1] \]

### Final Results

- **Domain**: \( (-2, 1) \)
- **Range**: \( [-1, 1] \)