An even periodic functin `f : R to R` with period 4 is such that
`f (x)= [{:(max. (|x|"," x ^(2)),, 0 le x lt 1),( x,, 1 le x le 2):}`
The value of `{f (5.12)}` (where {.} denotes fractional part function), is :

Options

A. `{f(3.26)}`
B. `{f (7.88)}`
C. `{f(2.12)}`
D. `{f (5.88)}`

Answer

To solve the problem, we need to find the value of \( f(5.12) \) for the given even periodic function \( f \) with period 4. The function is defined piecewise as follows:

1. \( f(x) = \max(|x|, x^2) \) for \( 0 \leq x < 1 \)
2. \( f(x) = x \) for \( 1 \leq x \leq 2 \)

Since \( f \) is an even periodic function with period 4, we can use the property of periodicity to simplify our calculations.

### Step 1: Reduce \( 5.12 \) using periodicity
Since the function has a period of 4, we can reduce \( 5.12 \) by subtracting 4 until we get a value within the range of one period (0 to 4).

\[ 5.12 - 4 = 1.12 \]

### Step 2: Determine which piece of the function to use
Now we need to evaluate \( f(1.12) \). Since \( 1.12 \) falls within the interval \( [1, 2] \), we use the second piece of the function:

\[ f(x) = x \quad \text{for } 1 \leq x \leq 2 \]

### Step 3: Calculate \( f(1.12) \)
Now we can directly substitute \( 1.12 \) into the function:

\[ f(1.12) = 1.12 \]

### Step 4: Find the fractional part
The fractional part function \( \{x\} \) is defined as \( x - \lfloor x \rfloor \). To find the fractional part of \( f(5.12) \):

\[ \{f(5.12)\} = \{1.12\} = 1.12 - \lfloor 1.12 \rfloor \]

Since \( \lfloor 1.12 \rfloor = 1 \):

\[ \{1.12\} = 1.12 - 1 = 0.12 \]

### Final Answer
Thus, the value of \( f(5.12) \) is \( 0.12 \).

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