If `f : R -> R` is defined by `f(x) = 1 /(2-cos3x)` for each `x in R` then the range of `f` is

Options

A. `[-1//3,0]`
B. R
C. [1/3,1]
D. none of these

Answer

To find the range of the function \( f(x) = \frac{1}{2 - \cos(3x)} \), we will follow these steps:

### Step 1: Determine the range of \( \cos(3x) \)

The cosine function oscillates between -1 and 1 for all real numbers. Therefore, we can write:
\[ -1 \leq \cos(3x) \leq 1 \]

### Step 2: Transform the inequality for \( 2 - \cos(3x) \)

Now, we will manipulate the inequality to find the range of \( 2 - \cos(3x) \):
\[ 2 - 1 \leq 2 - \cos(3x) \leq 2 - (-1) \]
This simplifies to:
\[ 1 \leq 2 - \cos(3x) \leq 3 \]

### Step 3: Take the reciprocal of the inequality

Since \( 2 - \cos(3x) \) is always positive (it ranges from 1 to 3), we can take the reciprocal of the entire inequality. Remember that taking the reciprocal reverses the inequality:
\[ \frac{1}{3} \leq \frac{1}{2 - \cos(3x)} \leq 1 \]

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

From the previous step, we see that:
\[ \frac{1}{3} \leq f(x) \leq 1 \]
Thus, the range of \( f(x) \) is:
\[ \left[\frac{1}{3}, 1\right] \]

### Conclusion

The range of the function \( f(x) = \frac{1}{2 - \cos(3x)} \) is \( \left[\frac{1}{3}, 1\right] \).