If `f(x)=sgn(cos 2x - 2 sin x + 3)`, where sgn () is the signum function, then f(x)

Options

A. is continuous over its domain
B. has a missing point discontinuity
C. has isolated point discontinuity
D. has irremovable discontinuity

Answer

To solve the problem, we need to analyze the function \( f(x) = \text{sgn}(\cos(2x) - 2\sin(x) + 3) \) and determine its continuity.

### Step-by-step Solution:

1. **Understand the Signum Function**:
The signum function, \( \text{sgn}(x) \), is defined as:
\[ \text{sgn}(x) = \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases} \]
Therefore, \( f(x) \) will be discontinuous at points where \( \cos(2x) - 2\sin(x) + 3 = 0 \).

2. **Set the Function to Zero**:
We need to find the values of \( x \) for which:
\[ \cos(2x) - 2\sin(x) + 3 = 0 \]

3. **Use Trigonometric Identities**:
We can rewrite \( \cos(2x) \) using the identity \( \cos(2x) = 1 - 2\sin^2(x) \):
\[ 1 - 2\sin^2(x) - 2\sin(x) + 3 = 0 \]
Simplifying this gives:
\[ -2\sin^2(x) - 2\sin(x) + 4 = 0 \]
Rearranging:
\[ 2\sin^2(x) + 2\sin(x) - 4 = 0 \]
Dividing the entire equation by 2:
\[ \sin^2(x) + \sin(x) - 2 = 0 \]

4. **Factor the Quadratic**:
We can factor this quadratic equation:
\[ (\sin(x) + 2)(\sin(x) - 1) = 0 \]
This gives us two cases:
- \( \sin(x) + 2 = 0 \) (not possible since sine values range from -1 to 1)
- \( \sin(x) - 1 = 0 \) which leads to \( \sin(x) = 1 \)

5. **Find Values of \( x \)**:
The solution to \( \sin(x) = 1 \) occurs at:
\[ x = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
This means that \( f(x) \) will be discontinuous at these points.

6. **Determine the Type of Discontinuity**:
Since the function is discontinuous at isolated points (i.e., \( x = \frac{\pi}{2} + 2n\pi \)), we conclude that \( f(x) \) has isolated point discontinuities.

### Conclusion:
The correct option is that \( f(x) \) has isolated point discontinuities.

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