Find the number of solution of equation (where sgn represent signum function)
`(i) sgn (x)=|x|` `(ii) sgn(x^(2)-1)=(x+1)^(2)`

Answer

To solve the given equations involving the signum function, we will analyze each equation step by step.

### (i) Solve the equation: \( \text{sgn}(x) = |x| \)

1. **Understanding the Signum Function**:
The signum function, \( \text{sgn}(x) \), is defined as:
- \( \text{sgn}(x) = 1 \) if \( x > 0 \)
- \( \text{sgn}(x) = 0 \) if \( x = 0 \)
- \( \text{sgn}(x) = -1 \) if \( x < 0 \)

2. **Understanding the Absolute Value Function**:
The absolute value function, \( |x| \), is defined as:
- \( |x| = x \) if \( x \geq 0 \)
- \( |x| = -x \) if \( x < 0 \)

3. **Setting Up Cases**:
We will analyze the equation \( \text{sgn}(x) = |x| \) for three cases based on the value of \( x \).

- **Case 1**: \( x > 0 \)
- Here, \( \text{sgn}(x) = 1 \) and \( |x| = x \).
- The equation becomes \( 1 = x \).
- Thus, \( x = 1 \) is a solution.

- **Case 2**: \( x = 0 \)
- Here, \( \text{sgn}(x) = 0 \) and \( |x| = 0 \).
- The equation becomes \( 0 = 0 \).
- Thus, \( x = 0 \) is also a solution.

- **Case 3**: \( x < 0 \)
- Here, \( \text{sgn}(x) = -1 \) and \( |x| = -x \).
- The equation becomes \( -1 = -x \) or \( x = 1 \).
- Since \( x < 0 \), this case does not yield any valid solutions.

4. **Conclusion**:
The solutions we found are \( x = 1 \) and \( x = 0 \). Therefore, the total number of solutions is **2**.

### (ii) Solve the equation: \( \text{sgn}(x^2 - 1) = (x + 1)^2 \)

1. **Understanding the Signum Function for \( x^2 - 1 \)**:
- The expression \( x^2 - 1 \) is zero at \( x = -1 \) and \( x = 1 \).
- For \( x < -1 \), \( x^2 - 1 > 0 \) (so \( \text{sgn}(x^2 - 1) = 1 \)).
- For \( -1 < x < 1 \), \( x^2 - 1 < 0 \) (so \( \text{sgn}(x^2 - 1) = -1 \)).
- For \( x > 1 \), \( x^2 - 1 > 0 \) (so \( \text{sgn}(x^2 - 1) = 1 \)).

2. **Setting Up Cases**:
We will analyze the equation \( \text{sgn}(x^2 - 1) = (x + 1)^2 \) for different intervals.

- **Case 1**: \( x < -1 \)
- Here, \( \text{sgn}(x^2 - 1) = 1 \) and \( (x + 1)^2 \geq 0 \).
- The equation becomes \( 1 = (x + 1)^2 \).
- This gives \( (x + 1)^2 - 1 = 0 \) or \( (x + 1 - 1)(x + 1 + 1) = 0 \).
- Thus, \( x + 1 = 0 \) or \( x + 1 = 2 \) leading to \( x = -1 \) or \( x = 1 \).
- However, \( x = -1 \) is not valid in this case since \( x < -1 \).

- **Case 2**: \( -1 < x < 1 \)
- Here, \( \text{sgn}(x^2 - 1) = -1 \) and \( (x + 1)^2 \geq 0 \).
- The equation becomes \( -1 = (x + 1)^2 \).
- This is impossible since \( (x + 1)^2 \) is always non-negative.

- **Case 3**: \( x > 1 \)
- Here, \( \text{sgn}(x^2 - 1) = 1 \) and \( (x + 1)^2 \geq 0 \).
- The equation becomes \( 1 = (x + 1)^2 \).
- This gives \( (x + 1)^2 - 1 = 0 \) leading to \( x + 1 = 0 \) or \( x + 1 = 2 \).
- Thus, \( x + 1 = 2 \) gives \( x = 1 \), but \( x = 1 \) is not valid in this case since \( x > 1 \).

3. **Conclusion**:
In all cases, we find that there are no valid solutions. Therefore, the total number of solutions is **0**.

### Summary of Solutions:
- For \( \text{sgn}(x) = |x| \): **2 solutions** (at \( x = 0 \) and \( x = 1 \)).
- For \( \text{sgn}(x^2 - 1) = (x + 1)^2 \): **0 solutions**.