The range of the function ` f(x) = (1)/(2-sin 3x)` is

Options

A. `(1//3, 1 )`
B. `[1//3, 1 )`
C. `[1//3, 1]`
D. none of these

Explanation

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

### Step 1: Understand the Range of the Sine Function
The sine function, \( \sin(3x) \), oscillates between -1 and 1 for all real values of \( x \). Therefore, we can write:
\[ -1 \leq \sin(3x) \leq 1 \]

### Step 2: Substitute the Range of Sine into the Function
Now, we substitute the range of \( \sin(3x) \) into the function \( f(x) \):
\[ 2 - \sin(3x) \]
This gives us:
\[ 2 - 1 \leq 2 - \sin(3x) \leq 2 - (-1) \]
So,
\[ 1 \leq 2 - \sin(3x) \leq 3 \]

### Step 3: Find the Range of \( f(x) \)
Next, we need to find the range of \( f(x) = \frac{1}{2 - \sin(3x)} \). Since \( 2 - \sin(3x) \) varies between 1 and 3, we can take the reciprocal:
- The maximum value of \( 2 - \sin(3x) \) is 3, which gives the minimum value of \( f(x) \):
\[ f(x) \text{ is minimum when } 2 - \sin(3x) = 3 \Rightarrow f(x) = \frac{1}{3} \]
- The minimum value of \( 2 - \sin(3x) \) is 1, which gives the maximum value of \( f(x) \):
\[ f(x) \text{ is maximum when } 2 - \sin(3x) = 1 \Rightarrow f(x) = 1 \]

### Step 4: Combine the Results
Thus, the range of \( f(x) \) is:
\[ \frac{1}{3} \leq f(x) < 1 \]
In interval notation, the range is:
\[ \left[\frac{1}{3}, 1\right) \]

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

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