Question

The range of the function \( f(x) = \frac{1}{2 - \cos 3x} \) is

• A. \( (-2,\infty) \)
• B. \( [-2,3] \)
• C. \( \left(\frac{1}{3},2\right) \)
• D. \( \left(\frac{1}{2},1\right) \)
• E. \( \left[\frac{1}{3},1\right] \)

Hint:

When taking reciprocals, always reverse inequality signs if all terms are positive.

Answer 1

Answer: (E)

Understanding the Concept: The cosine function has range: \[ -1 \leq \cos 3x \leq 1 \] We transform this to find the range of the function.

Step 1: Apply cosine bounds
\[ -1 \leq \cos 3x \leq 1 \]

Step 2: Transform denominator
\[ 2 - \cos 3x \] When \( \cos 3x = 1 \): \[ 2 - 1 = 1 \] When \( \cos 3x = -1 \): \[ 2 - (-1) = 3 \] So, \[ 1 \leq 2 - \cos 3x \leq 3 \]

Step 3: Take reciprocal
\[ f(x) = \frac{1}{2 - \cos 3x} \] Taking reciprocal reverses inequality: \[ \frac{1}{3} \leq f(x) \leq 1 \]

Step 4: Check endpoints
Both values are attainable since cosine reaches both -1 and 1.

Step 5: Final Answer
\[ \boxed{\left[\frac{1}{3},1\right]} \]