Let `f: RvecR` be a function defined by `f(x+1)=(f(x)-5)/(f(x)-3)AAx in Rdot` Then which of the following statement(s) is/are ture? `f(2008)=f(2004)` `f(2006)=f(2010)` `f(2006)=f(2002)` `f(2006)=f(2018`

Options

A. `f(2008)=f(2004)`
B. `f(2006)=f(2010)`
C. `f(2006)=f(2002)`
D. `f(2006)=f(2018)`

Explanation

To solve the problem, we start with the given functional equation:

\[ f(x + 1) = \frac{f(x) - 5}{f(x) - 3} \]

### Step 1: Analyze the functional equation

We will first rewrite the functional equation for clarity:

1. **Equation**:
\[ f(x + 1) = \frac{f(x) - 5}{f(x) - 3} \]

### Step 2: Substitute \( x \) with \( x - 1 \)

Next, we will substitute \( x \) with \( x - 1 \) in the original equation:

2. **Substitution**:
\[ f(x) = \frac{f(x - 1) - 5}{f(x - 1) - 3} \]

### Step 3: Establish a relationship between \( f(x) \) and \( f(x + 2) \)

Now, we can derive a relationship between \( f(x) \) and \( f(x + 2) \):

3. **Using the previous equations**:
\[ f(x + 2) = \frac{f(x + 1) - 5}{f(x + 1) - 3} \]
Substitute \( f(x + 1) \) from Step 1:
\[ f(x + 2) = \frac{\frac{f(x) - 5}{f(x) - 3} - 5}{\frac{f(x) - 5}{f(x) - 3} - 3} \]

### Step 4: Simplify the expression

We can simplify the expression derived in Step 3:

4. **Simplification**:
\[ f(x + 2) = \frac{(f(x) - 5) - 5(f(x) - 3)}{(f(x) - 5) - 3(f(x) - 3)} \]
This leads to:
\[ f(x + 2) = \frac{-4f(x) + 10}{-2f(x) + 4} \]

### Step 5: Establish periodicity

Continuing this process, we can find that:

5. **Periodic Function**:
By continuing to substitute and simplify, we can show that:
\[ f(x + 4) = f(x) \]
This indicates that \( f(x) \) is periodic with a period of 4.

### Step 6: Evaluate the given statements

Now we can evaluate the statements given in the problem:

1. **Statement 1**: \( f(2008) = f(2004) \)
- True, because \( 2008 - 2004 = 4 \).

2. **Statement 2**: \( f(2006) = f(2010) \)
- True, because \( 2010 - 2006 = 4 \).

3. **Statement 3**: \( f(2006) = f(2002) \)
- True, because \( 2006 - 2002 = 4 \).

4. **Statement 4**: \( f(2006) = f(2018) \)
- True, because \( 2018 - 2006 = 12 \), which is a multiple of 4.

### Conclusion

All statements are true:
- \( f(2008) = f(2004) \)
- \( f(2006) = f(2010) \)
- \( f(2006) = f(2002) \)
- \( f(2006) = f(2018) \)