Range of `f(x) =1/(1-2cosx)` is uu[1/3,oo)` (4) `[-1/3,1]`

Options

A. `[1/3,1]`
B. `[-1,1/3]`
C. `(-oo,-1]uu[1/3,oo)`
D. `[-1/3,1]`

Answer

To find the range of the function \( f(x) = \frac{1}{1 - 2 \cos x} \), we will follow these steps:

### Step 1: Identify where the function is undefined
The function \( f(x) \) is undefined when the denominator is zero:
\[ 1 - 2 \cos x = 0 \]
Solving for \( \cos x \):
\[ 2 \cos x = 1 \implies \cos x = \frac{1}{2} \]
The values of \( x \) for which \( \cos x = \frac{1}{2} \) are:
\[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
Thus, the function is undefined at these points.

### Step 2: Determine the range of \( \cos x \)
The range of \( \cos x \) is:
\[ [-1, 1] \]
Therefore, the range of \( 2 \cos x \) will be:
\[ [-2, 2] \]

### Step 3: Analyze the expression \( 1 - 2 \cos x \)
The range of \( 1 - 2 \cos x \) can be determined by evaluating the endpoints:
- When \( \cos x = -1 \):
\[ 1 - 2(-1) = 1 + 2 = 3 \]
- When \( \cos x = 1 \):
\[ 1 - 2(1) = 1 - 2 = -1 \]
Thus, the range of \( 1 - 2 \cos x \) is:
\[ [-1, 3] \]

### Step 4: Exclude points where the function is undefined
From Step 1, we found that \( 1 - 2 \cos x = 0 \) when \( \cos x = \frac{1}{2} \). This corresponds to:
\[ 1 - 2 \cos x = 0 \implies 1 - 2 \cdot \frac{1}{2} = 0 \]
Thus, we need to exclude the value \( 0 \) from the range of \( 1 - 2 \cos x \).

### Step 5: Determine the range of \( f(x) \)
Now, we can find the range of \( f(x) = \frac{1}{1 - 2 \cos x} \):
- As \( 1 - 2 \cos x \) approaches \( 0 \) from the left, \( f(x) \) approaches \( -\infty \).
- As \( 1 - 2 \cos x \) approaches \( 0 \) from the right, \( f(x) \) approaches \( +\infty \).
- The minimum value occurs when \( 1 - 2 \cos x = 3 \):
\[ f(x) = \frac{1}{3} \]
- The maximum value occurs when \( 1 - 2 \cos x = -1 \):
\[ f(x) = -1 \]

### Final Range
Thus, the range of \( f(x) \) is:
\[ (-\infty, -1) \cup \left[\frac{1}{3}, \infty\right) \]

### Conclusion
The correct answer is:
\[ (-\infty, -1) \cup \left[\frac{1}{3}, \infty\right) \]