The range of the function `f(x)=sin{log_(10)((sqrt(4-x^(2)))/(1-x))}` ,is

Options

A. [0,1]
B. (-1,0)
C. [-1,1]
D. (-1,1)

Answer

To find the range of the function \( f(x) = \sin\left(\log_{10}\left(\frac{\sqrt{4 - x^2}}{1 - x}\right)\right) \), we will follow these steps:

### Step 1: Determine the Domain of the Function
The function involves a square root and a logarithm, so we need to ensure that the arguments of both are valid.

1. **Square Root Condition**: For \( \sqrt{4 - x^2} \) to be defined, we require:
\[ 4 - x^2 \geq 0 \]
This simplifies to:
\[ x^2 \leq 4 \quad \Rightarrow \quad -2 \leq x \leq 2 \]

2. **Logarithm Condition**: For \( \log_{10}\left(\frac{\sqrt{4 - x^2}}{1 - x}\right) \) to be defined, the argument must be positive:
\[ \frac{\sqrt{4 - x^2}}{1 - x} > 0 \]
This implies:
- \( \sqrt{4 - x^2} > 0 \) (which is true for \( -2 < x < 2 \))
- \( 1 - x > 0 \quad \Rightarrow \quad x < 1 \)

Combining these conditions, the domain of \( f(x) \) is:
\[ -2 < x < 1 \]

### Step 2: Analyze the Argument of the Logarithm
Next, we need to analyze the expression inside the logarithm:
\[ y = \frac{\sqrt{4 - x^2}}{1 - x} \]

As \( x \) approaches -2, \( y \) approaches:
\[ y \to \frac{\sqrt{4 - (-2)^2}}{1 - (-2)} = \frac{0}{3} = 0 \]

As \( x \) approaches 1, \( y \) approaches:
\[ y \to \frac{\sqrt{4 - 1^2}}{1 - 1} = \frac{\sqrt{3}}{0} \to +\infty \]

### Step 3: Determine the Range of the Logarithm
Since \( y \) varies from 0 to \( +\infty \) as \( x \) varies from -2 to 1, we can find the range of the logarithm:
\[ \log_{10}(y) \text{ varies from } \log_{10}(0) \text{ (undefined) to } +\infty \]
Thus, the range of \( \log_{10}(y) \) is:
\[ (-\infty, +\infty) \]

### Step 4: Determine the Range of the Sine Function
The sine function, \( \sin(z) \), where \( z = \log_{10}(y) \), has a range of:
\[ [-1, 1] \]
for any real number \( z \).

### Final Answer
Therefore, the range of the function \( f(x) \) is:
\[ \boxed{[-1, 1]} \]