Which of the following is true for `f(x) = sgn(x) xx sin x`

Options

A. (a) Discontinuous no where
B. (b) An even function
C. (c) Discontinuous everywhere
D. (d) Continuous everywhere

Answer

To determine the properties of the function \( f(x) = \text{sgn}(x) \cdot \sin(x) \), we will analyze its continuity and whether it is an even function. The signum function \( \text{sgn}(x) \) is defined as follows:

\[ \text{sgn}(x) = \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases} \]

### Step 1: Define the function based on the signum function
Using the definition of the signum function, we can express \( f(x) \) in piecewise form:

- For \( x < 0 \):
\[ f(x) = \text{sgn}(x) \cdot \sin(x) = -1 \cdot \sin(x) = -\sin(x) \]

- For \( x = 0 \):
\[ f(0) = \text{sgn}(0) \cdot \sin(0) = 0 \cdot 0 = 0 \]

- For \( x > 0 \):
\[ f(x) = \text{sgn}(x) \cdot \sin(x) = 1 \cdot \sin(x) = \sin(x) \]

Thus, we can summarize the function as:
\[ f(x) = \begin{cases} -\sin(x) & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ \sin(x) & \text{if } x > 0 \end{cases} \]

### Step 2: Analyze continuity
To check the continuity of \( f(x) \) at \( x = 0 \), we need to evaluate the left-hand limit, right-hand limit, and the function value at that point.

- **Left-hand limit** as \( x \to 0^- \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -\sin(x) = -\sin(0) = 0 \]

- **Right-hand limit** as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \sin(x) = \sin(0) = 0 \]

- **Function value at \( x = 0 \)**:
\[ f(0) = 0 \]

Since both limits are equal to the function value at \( x = 0 \), \( f(x) \) is continuous at \( x = 0 \).

For \( x < 0 \) and \( x > 0 \), \( f(x) \) is continuous because it is composed of continuous functions (sine function). Therefore, \( f(x) \) is continuous everywhere.

### Step 3: Check if the function is even
A function is even if \( f(-x) = f(x) \) for all \( x \).

- For \( x < 0 \):
\[ f(-x) = f(x) = \sin(x) \quad \text{(since } -x > 0\text{)} \]

- For \( x = 0 \):
\[ f(-0) = f(0) = 0 \]

- For \( x > 0 \):
\[ f(-x) = f(x) = -\sin(x) \quad \text{(since } -x < 0\text{)} \]

Since \( f(-x) = f(x) \) holds for all cases, \( f(x) \) is an even function.

### Conclusion
The function \( f(x) = \text{sgn}(x) \cdot \sin(x) \) is continuous everywhere and is an even function.

### Final Answer
The correct options are:
- Continuous everywhere
- Even function