Let `f (x)= {{:(x ^(2),0lt x lt2),(2x-3, 2 le x lt3),(x+2, x ge3):}` then the tuue equations:

Options

A. `f (f(f ((3)/(2)))=f((3)/(2))`
B. `1+f (f(f((5)/(2))))=f((5)/(2))`
C. `f(f(f(2)))=f(1)`
D. `ubrace(f (f(f(......f(4))...)))=2012`

Answer

To solve the problem, we need to analyze the piecewise function defined as follows:

\[ f(x) = \begin{cases} x^2 & \text{if } 0 < x < 2 \\ 2x - 3 & \text{if } 2 \leq x < 3 \\ x + 2 & \text{if } x \geq 3 \end{cases} \]

We need to determine which of the given equations involving \( f \) are true. Let's denote the equations as follows for clarity:

1. \( f(f(3/2)) = 3/2 \)
2. \( 1 + f(f(f(5/2))) = f(5/2) \)
3. \( f(f(f(2))) = f(1) \)
4. \( f(f(f(3))) = 6 \)

We will check each equation step by step.

### Step 1: Check Equation 1: \( f(f(3/2)) = 3/2 \)

1. **Calculate \( f(3/2) \)**:
- Since \( 3/2 \) lies in the interval \( (0, 2) \), we use \( f(x) = x^2 \).
- Thus, \( f(3/2) = (3/2)^2 = 9/4 = 2.25 \).

2. **Calculate \( f(f(3/2)) = f(2.25) \)**:
- Since \( 2.25 \) lies in the interval \( [2, 3) \), we use \( f(x) = 2x - 3 \).
- Thus, \( f(2.25) = 2(2.25) - 3 = 4.5 - 3 = 1.5 = 3/2 \).

3. **Conclusion**:
- LHS = \( f(f(3/2)) = 3/2 \) matches RHS.
- Therefore, **Equation 1 is true**.

### Step 2: Check Equation 2: \( 1 + f(f(f(5/2))) = f(5/2) \)

1. **Calculate \( f(5/2) \)**:
- Since \( 5/2 = 2.5 \) lies in the interval \( [2, 3) \), we use \( f(x) = 2x - 3 \).
- Thus, \( f(5/2) = 2(5/2) - 3 = 5 - 3 = 2 \).

2. **Calculate \( f(f(5/2)) = f(2) \)**:
- Since \( 2 \) lies in the interval \( [2, 3) \), we again use \( f(x) = 2x - 3 \).
- Thus, \( f(2) = 2(2) - 3 = 4 - 3 = 1 \).

3. **Calculate \( f(f(f(5/2))) = f(1) \)**:
- Since \( 1 \) lies in the interval \( (0, 2) \), we use \( f(x) = x^2 \).
- Thus, \( f(1) = 1^2 = 1 \).

4. **Combine results**:
- LHS = \( 1 + f(f(f(5/2))) = 1 + 1 = 2 \).
- RHS = \( f(5/2) = 2 \).
- Therefore, **Equation 2 is true**.

### Step 3: Check Equation 3: \( f(f(f(2))) = f(1) \)

1. **Calculate \( f(2) \)**:
- Since \( 2 \) lies in the interval \( [2, 3) \), we use \( f(x) = 2x - 3 \).
- Thus, \( f(2) = 2(2) - 3 = 1 \).

2. **Calculate \( f(f(2)) = f(1) \)**:
- Since \( 1 \) lies in the interval \( (0, 2) \), we use \( f(x) = x^2 \).
- Thus, \( f(1) = 1^2 = 1 \).

3. **Calculate \( f(f(f(2))) = f(f(1)) = f(1) \)**:
- We already calculated \( f(1) = 1 \).

4. **Combine results**:
- LHS = \( f(f(f(2))) = 1 \).
- RHS = \( f(1) = 1 \).
- Therefore, **Equation 3 is true**.

### Step 4: Check Equation 4: \( f(f(f(3))) = 6 \)

1. **Calculate \( f(3) \)**:
- Since \( 3 \) lies in the interval \( [3, \infty) \), we use \( f(x) = x + 2 \).
- Thus, \( f(3) = 3 + 2 = 5 \).

2. **Calculate \( f(f(3)) = f(5) \)**:
- Since \( 5 \) lies in the interval \( [3, \infty) \), we use \( f(x) = x + 2 \).
- Thus, \( f(5) = 5 + 2 = 7 \).

3. **Calculate \( f(f(f(3))) = f(7) \)**:
- Since \( 7 \) lies in the interval \( [3, \infty) \), we use \( f(x) = x + 2 \).
- Thus, \( f(7) = 7 + 2 = 9 \).

4. **Combine results**:
- LHS = \( f(f(f(3))) = 9 \).
- RHS = \( 6 \).
- Therefore, **Equation 4 is false**.

### Final Conclusion:
The true equations are:
- Equation 1: True
- Equation 2: True
- Equation 3: True
- Equation 4: False