Let `f(x) = x(2-x), 0 le x le 2`. If the definition of `f(x)` is extended over the set `R-[0,2]` by `f (x+1)= f(x)`, then f is a

Options

A. periodic function with period 1
B. non-periodic function
C. periodic function with period 2
D. periodic function with period `1//2`

Explanation

To solve the problem, we need to analyze the function \( f(x) = x(2-x) \) defined for \( 0 \leq x \leq 2 \) and how it behaves when extended to the entire set of real numbers using the rule \( f(x+1) = f(x) \).

### Step 1: Evaluate the function \( f(x) \) on the interval \( [0, 2] \)

The function is defined as:
\[ f(x) = x(2 - x) \]
We can calculate the values of \( f(x) \) at the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = 0(2 - 0) = 0 \]
- At \( x = 2 \):
\[ f(2) = 2(2 - 2) = 0 \]
- At \( x = 1 \):
\[ f(1) = 1(2 - 1) = 1 \]

### Step 2: Identify the maximum value of \( f(x) \)

The function is a quadratic function that opens downwards (since the coefficient of \( x^2 \) is negative). The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \) where \( f(x) = ax^2 + bx + c \). Here, \( a = -1 \) and \( b = 2 \):
\[ x = -\frac{2}{2(-1)} = 1 \]
At \( x = 1 \), we already calculated \( f(1) = 1 \). Thus, the maximum value of \( f(x) \) on the interval \( [0, 2] \) is 1.

### Step 3: Extend the function to \( \mathbb{R} \)

The extension of \( f(x) \) is defined by the rule \( f(x+1) = f(x) \). This means that the function repeats its values every 2 units. Therefore, we can express:
\[ f(x) = f(x - 2) \text{ for any } x \in \mathbb{R} \]

### Step 4: Determine the periodicity of \( f(x) \)

Since \( f(x) \) repeats its values every 2 units, we conclude that:
\[ f(x) \text{ is periodic with period } 2. \]

### Conclusion

Thus, the function \( f(x) \) is periodic with a period of 2.

---