Let `y=sgn (x)`, then

Options

A. `|x| = x Sgn (x)`
B. `sgn (sgn(x)) = sgn(x)`
C. `x = |x| Sgn (x)`
D. All of these

Answer

To solve the problem, we need to analyze the function \( y = \text{sgn}(x) \) and verify the given options based on the properties of the signum function. The signum function is defined as follows:

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

Now, let's check each option one by one.

### Option 1: \( |x| = x \cdot \text{sgn}(x) \)

1. **Case 1: \( x > 0 \)**
- Here, \( |x| = x \) and \( \text{sgn}(x) = 1 \).
- Therefore, \( |x| = x \cdot 1 = x \). This is true.

2. **Case 2: \( x = 0 \)**
- Here, \( |0| = 0 \) and \( \text{sgn}(0) = 0 \).
- Therefore, \( |0| = 0 \cdot 0 = 0 \). This is true.

3. **Case 3: \( x < 0 \)**
- Here, \( |x| = -x \) (since \( x \) is negative) and \( \text{sgn}(x) = -1 \).
- Therefore, \( |x| = -x \cdot (-1) = x \). This is true.

Thus, Option 1 is correct.

### Option 2: \( \text{sgn}(\text{sgn}(x)) = \text{sgn}(x) \)

1. **Case 1: \( x > 0 \)**
- Here, \( \text{sgn}(x) = 1 \).
- Therefore, \( \text{sgn}(\text{sgn}(x)) = \text{sgn}(1) = 1 \). This is true.

2. **Case 2: \( x = 0 \)**
- Here, \( \text{sgn}(0) = 0 \).
- Therefore, \( \text{sgn}(\text{sgn}(0)) = \text{sgn}(0) = 0 \). This is true.

3. **Case 3: \( x < 0 \)**
- Here, \( \text{sgn}(x) = -1 \).
- Therefore, \( \text{sgn}(\text{sgn}(x)) = \text{sgn}(-1) = -1 \). This is true.

Thus, Option 2 is correct.

### Option 3: \( x = |x| \cdot \text{sgn}(x) \)

1. **Case 1: \( x > 0 \)**
- Here, \( |x| = x \) and \( \text{sgn}(x) = 1 \).
- Therefore, \( x = x \cdot 1 = x \). This is true.

2. **Case 2: \( x = 0 \)**
- Here, \( |0| = 0 \) and \( \text{sgn}(0) = 0 \).
- Therefore, \( 0 = 0 \cdot 0 = 0 \). This is true.

3. **Case 3: \( x < 0 \)**
- Here, \( |x| = -x \) and \( \text{sgn}(x) = -1 \).
- Therefore, \( x = -x \cdot (-1) = x \). This is true.

Thus, Option 3 is correct.

### Conclusion
All options are correct.