Let `f (x) =2-|x-3| , 1 le x le 5 and ` for rest of the values f (x) can be obtained by unsing the relation `f (5x)=alpha f (x) AA x in R.`
The vlaue of f (2007), taking `alpha =5,` is :

Options

A. 1118
B. 2007
C. 1250
D. 132

Answer

To solve the problem step by step, we will follow the given function and the relationship defined in the question.

### Step 1: Define the function for the given range
The function is defined as:
\[ f(x) = 2 - |x - 3| \text{ for } 1 \leq x \leq 5 \]

### Step 2: Calculate the function values for the range [1, 5]
1. **Calculate \( f(1) \)**:
\[ f(1) = 2 - |1 - 3| = 2 - 2 = 0 \]

2. **Calculate \( f(2) \)**:
\[ f(2) = 2 - |2 - 3| = 2 - 1 = 1 \]

3. **Calculate \( f(3) \)**:
\[ f(3) = 2 - |3 - 3| = 2 - 0 = 2 \]

4. **Calculate \( f(4) \)**:
\[ f(4) = 2 - |4 - 3| = 2 - 1 = 1 \]

5. **Calculate \( f(5) \)**:
\[ f(5) = 2 - |5 - 3| = 2 - 2 = 0 \]

### Step 3: Use the relation \( f(5x) = \alpha f(x) \)
Given \( \alpha = 5 \), we can express \( f(2007) \) using the relation:
\[ f(2007) = f\left(\frac{2007}{5}\right) \cdot 5 \]

### Step 4: Calculate \( \frac{2007}{5} \)
\[ \frac{2007}{5} = 401.4 \]

### Step 5: Use the relation iteratively
We need to keep applying the relation until we reach a value within the range [1, 5].

1. **Calculate \( f(401.4) \)**:
\[ f(401.4) = f\left(\frac{401.4}{5}\right) \cdot 5 \]
\[ \frac{401.4}{5} = 80.28 \]

2. **Calculate \( f(80.28) \)**:
\[ f(80.28) = f\left(\frac{80.28}{5}\right) \cdot 5 \]
\[ \frac{80.28}{5} = 16.056 \]

3. **Calculate \( f(16.056) \)**:
\[ f(16.056) = f\left(\frac{16.056}{5}\right) \cdot 5 \]
\[ \frac{16.056}{5} = 3.2112 \]

Now, since \( 3.2112 \) is within the range [1, 5], we can find \( f(3.2112) \):
\[ f(3.2112) = 2 - |3.2112 - 3| = 2 - 0.2112 = 1.7888 \]

### Step 6: Backtrack to find \( f(2007) \)
Now we can substitute back:
\[ f(16.056) = 5 \cdot f(3.2112) = 5 \cdot 1.7888 = 8.944 \]
\[ f(80.28) = 5 \cdot f(16.056) = 5 \cdot 8.944 = 44.72 \]
\[ f(401.4) = 5 \cdot f(80.28) = 5 \cdot 44.72 = 223.6 \]
\[ f(2007) = 5 \cdot f(401.4) = 5 \cdot 223.6 = 1118 \]

### Final Answer
Thus, the value of \( f(2007) \) is:
\[ \boxed{1118} \]