Question

paragraph for question
an even periodic function (f : r \
ightarrow r) with period 4 is such that
(f(x) = \

$$\begin{cases} \\text{max. } (|x|, x^2) & ; 0 \\le x < 1 \\\\ x & ; 1 \\le x \\le 2 \\end{cases}$$

)

1. the value of (\\{f(x)\\}) at (x = 5.12) (where (\\{\\}) represents fractional part), is

(a) (\\{f(7.88)\\})
(b) (\\{f(3.26)\\})
(c) (\\{f(2.12)\\})
(d) (\\{f(5.88)\\})

Explanation:

🆕 New Concept Discovered: Periodic and Even Functions
Using symmetry and repetition to simplify inputs

Step 1: Simplify the input using periodicity

Since \( f(x) \) is a periodic function with period \( 4 \), we have \( f(x) = f(x - 4k) \) for any integer \( k \).

For \( x = 5.12 \):

$$ f(5.12) = f(5.12 - 4) = f(1.12) $$

Therefore, the target value is:

$$ \{f(5.12)\} = \{f(1.12)\} $$

Step 2: Simplify the options using periodicity and evenness

An even function satisfies \( f(-x) = f(x) \). Let's evaluate each option to see which one matches \( \{f(1.12)\} \).

Option (a): \( \{f(7.88)\} \)

Subtract the period \( 8 \) (since \( 2 \times 4 = 8 \)):

$$ f(7.88) = f(7.88 - 8) = f(-0.12) $$

Since \( f(x) \) is even:

$$ f(-0.12) = f(0.12) $$

Thus, \( \{f(7.88)\} = \{f(0.12)\} \).

Option (b): \( \{f(3.26)\} \)

Subtract the period \( 4 \):

$$ f(3.26) = f(3.26 - 4) = f(-0.74) $$

Since \( f(x) \) is even:

$$ f(-0.74) = f(0.74) $$

Thus, \( \{f(3.26)\} = \{f(0.74)\} \).

Option (c): \( \{f(2.12)\} \)

Subtract the period \( 4 \):

$$ f(2.12) = f(2.12 - 4) = f(-1.88) $$

Since \( f(x) \) is even:

$$ f(-1.88) = f(1.88) $$

Thus, \( \{f(2.12)\} = \{f(1.88)\} \).

Option (d): \( \{f(5.88)\} \)

Subtract the period \( 4 \):

$$ f(5.88) = f(5.88 - 4) = f(1.88) $$

Subtract another period \( 4 \):

$$ f(1.88) = f(1.88 - 4) = f(-2.12) $$

Since \( f(x) \) is even:

$$ f(-2.12) = f(2.12) $$

Thus, \( \{f(5.88)\} = \{f(2.12)\} \).

None of the simplified options directly match \( \{f(1.12)\} \) yet. We need to calculate the actual values using the definition of \( f(x) \).

Step 3: Calculate the target value \( \{f(1.12)\} \)

For \( 1 \le x \le 2 \), the function is defined as:

$$ f(x) = x $$

Since \( 1.12 \) lies in the interval \( [1, 2] \):

$$ f(1.12) = 1.12 $$

The fractional part of a number \( y \) is defined as \( \{y\} = y - \lfloor y
floor \):

$$ \{f(1.12)\} = \{1.12\} = 0.12 $$

Step 4: Calculate the values of the options

Now we find which option yields a fractional part of \( 0.12 \).

• From Option (a): We need \( \{f(0.12)\} \).

For \( 0 \le x < 1 \), the function is:

$$ f(x) = \max(|x|, x^2) $$

For \( x = 0.12 \):

$$ |0.12| = 0.12 \quad \text{and} \quad (0.12)^2 = 0.0144 $$

$$ f(0.12) = \max(0.12, 0.0144) = 0.12 $$

Taking the fractional part:

$$ \{f(0.12)\} = \{0.12\} = 0.12 $$

This matches our target value of \( 0.12 \).

Answer:

(a) {f (7.88)}