Let sgn (x) denote the signum function of x. Let `A = {x|x ne 1/2npi, n in Z}` Define `f : A to R` by `f(x) = sgn(cos x) + sgn (sinx) + sgn (tan x) + sgn (cot x)` Then range of f is

Options

A. {-2,0,4}
B. {0,4}
C. {-4,-2,0,4}
D. {-2,4}

Answer

To find the range of the function \( f(x) = \text{sgn}(\cos x) + \text{sgn}(\sin x) + \text{sgn}(\tan x) + \text{sgn}(\cot x) \) defined on the set \( A = \{ x | x \neq \frac{1}{2} n \pi, n \in \mathbb{Z} \} \), we will analyze the contributions of each term in the function.

### Step 1: Analyze the signum function values
The signum function, \( \text{sgn}(x) \), can take the following values:
- \( 1 \) if \( x > 0 \)
- \( 0 \) if \( x = 0 \)
- \( -1 \) if \( x < 0 \)

### Step 2: Identify the intervals for \( \cos x \) and \( \sin x \)
The function \( f(x) \) is dependent on the signs of \( \cos x \) and \( \sin x \):
- \( \cos x \) is positive in the intervals \( (2k\pi - \frac{\pi}{2}, 2k\pi + \frac{\pi}{2}) \) for \( k \in \mathbb{Z} \).
- \( \sin x \) is positive in the intervals \( (2k\pi, 2k\pi + \pi) \) for \( k \in \mathbb{Z} \).

### Step 3: Determine the values of \( f(x) \)
1. **Case 1: \( x \) in \( (0, \frac{\pi}{2}) \)**
- \( \cos x > 0 \) → \( \text{sgn}(\cos x) = 1 \)
- \( \sin x > 0 \) → \( \text{sgn}(\sin x) = 1 \)
- \( \tan x > 0 \) → \( \text{sgn}(\tan x) = 1 \)
- \( \cot x > 0 \) → \( \text{sgn}(\cot x) = 1 \)
- Thus, \( f(x) = 1 + 1 + 1 + 1 = 4 \).

2. **Case 2: \( x \) in \( (\frac{\pi}{2}, \pi) \)**
- \( \cos x < 0 \) → \( \text{sgn}(\cos x) = -1 \)
- \( \sin x > 0 \) → \( \text{sgn}(\sin x) = 1 \)
- \( \tan x < 0 \) → \( \text{sgn}(\tan x) = -1 \)
- \( \cot x < 0 \) → \( \text{sgn}(\cot x) = -1 \)
- Thus, \( f(x) = -1 + 1 - 1 - 1 = -2 \).

3. **Case 3: \( x \) in \( (\pi, \frac{3\pi}{2}) \)**
- \( \cos x < 0 \) → \( \text{sgn}(\cos x) = -1 \)
- \( \sin x < 0 \) → \( \text{sgn}(\sin x) = -1 \)
- \( \tan x > 0 \) → \( \text{sgn}(\tan x) = 1 \)
- \( \cot x < 0 \) → \( \text{sgn}(\cot x) = -1 \)
- Thus, \( f(x) = -1 - 1 + 1 - 1 = -2 \).

4. **Case 4: \( x \) in \( (\frac{3\pi}{2}, 2\pi) \)**
- \( \cos x > 0 \) → \( \text{sgn}(\cos x) = 1 \)
- \( \sin x < 0 \) → \( \text{sgn}(\sin x) = -1 \)
- \( \tan x < 0 \) → \( \text{sgn}(\tan x) = -1 \)
- \( \cot x > 0 \) → \( \text{sgn}(\cot x) = 1 \)
- Thus, \( f(x) = 1 - 1 - 1 + 1 = 0 \).

### Step 4: Compile the results
From the analysis, we find that:
- The value \( f(x) = 4 \) occurs in \( (0, \frac{\pi}{2}) \).
- The value \( f(x) = -2 \) occurs in \( (\frac{\pi}{2}, \pi) \) and \( (\pi, \frac{3\pi}{2}) \).
- The value \( f(x) = 0 \) occurs in \( (\frac{3\pi}{2}, 2\pi) \).

### Conclusion: Range of \( f \)
Thus, the range of \( f \) is \( \{-2, 0, 4\} \).