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 number of solution of `f (x)= | 3 sin x|` for ` x in (-6, 6)` are :

Options

A. 5
B. 3
C. 7
D. 9

Answer

To solve the problem, we need to analyze the function \( f(x) \) and the equation \( f(x) = |3 \sin x| \) over the interval \( x \in (-6, 6) \).

### Step 1: Define the function \( f(x) \)

The function \( f(x) \) is defined piecewise as follows:

1. For \( 0 \leq x < 1 \):
\[ f(x) = \max(|x|, x^2) \]
Since \( |x| = x \) in this interval, we can simplify this to:
\[ f(x) = \max(x, x^2) \]
In this range, \( x^2 \) is less than \( x \) for \( 0 < x < 1 \), and they are equal at \( x = 0 \) and \( x = 1 \). Therefore:
\[ f(x) = x \quad \text{for } 0 < x < 1 \]
At \( x = 0 \), \( f(0) = 0 \) and at \( x = 1 \), \( f(1) = 1 \).

2. For \( 1 \leq x \leq 2 \):
\[ f(x) = x \]

3. Since \( f(x) \) is an even periodic function with a period of 4, we can extend this definition to other intervals:
- For \( 2 < x < 3 \), \( f(x) = 4 - x \) (due to periodicity and symmetry).
- For \( 3 < x < 4 \), \( f(x) = 0 \) (as it mirrors the behavior of \( f(x) \) in the first interval).

### Step 2: Graph the function \( f(x) \)

The graph of \( f(x) \) over one period (from \( 0 \) to \( 4 \)) looks like this:
- From \( 0 \) to \( 1 \): linear increase from \( (0, 0) \) to \( (1, 1) \).
- From \( 1 \) to \( 2 \): linear increase from \( (1, 1) \) to \( (2, 2) \).
- From \( 2 \) to \( 3 \): linear decrease from \( (2, 2) \) to \( (3, 1) \).
- From \( 3 \) to \( 4 \): linear decrease from \( (3, 1) \) to \( (4, 0) \).

Due to periodicity, this pattern repeats for \( x \in (-6, 6) \).

### Step 3: Analyze \( |3 \sin x| \)

The function \( |3 \sin x| \) oscillates between \( 0 \) and \( 3 \). The sine function has a period of \( 2\pi \), so \( |3 \sin x| \) will also repeat every \( 2\pi \).

### Step 4: Find intersections

We need to find the number of intersections between \( f(x) \) and \( |3 \sin x| \) in the interval \( (-6, 6) \).

1. **Determine the number of periods**:
- The interval \( (-6, 6) \) covers \( 12 \) units.
- The period of \( f(x) \) is \( 4 \), so there are \( 3 \) full periods of \( f(x) \) in this interval.

2. **Intersections in one period**:
- In each period, \( f(x) \) reaches a maximum of \( 2 \) at \( x = 2 \) and a minimum of \( 0 \) at \( x = 0 \) and \( x = 4 \).
- The function \( |3 \sin x| \) oscillates between \( 0 \) and \( 3 \).
- The intersections will occur where \( f(x) \) is less than or equal to \( 3 \).

### Step 5: Count the solutions

- In each period from \( 0 \) to \( 4 \), \( f(x) \) intersects \( |3 \sin x| \) at:
- \( 0 \) (once)
- Between \( 0 \) and \( 1 \) (once)
- Between \( 1 \) and \( 2 \) (once)
- Between \( 2 \) and \( 3 \) (once)
- At \( 3 \) (once)
- Between \( 3 \) and \( 4 \) (once)

Thus, in one period, there are \( 5 \) intersections. Since there are \( 3 \) periods in \( (-6, 6) \):
\[ \text{Total intersections} = 5 \times 3 = 15 \]

### Final Answer

The number of solutions of \( f(x) = |3 \sin x| \) for \( x \in (-6, 6) \) is \( 15 \).