Let `f(x)={:{(4, :, x lt-1),(-4x,:,-1 lex le 0):}`. If f(x) is an even function in R, then the defination of `f(x) " in " (0, +oo)` is :

Options

A. `{:f(x)={(4x, :, 0 ltx le 1),(4, :, x gt1):}`
B. `{:f(x)={(4x, :, 0lt x le 1),(4, :, x gt1):}`
C. `{:f(x)={(4, :, 0lt x le 1),(4, :, x gt1):}`
D. None of these

Answer

To solve the problem, we need to determine the definition of the function \( f(x) \) in the interval \( (0, +\infty) \) given that \( f(x) \) is an even function.

### Step-by-Step Solution:

1. **Understanding Even Functions**:
An even function satisfies the property \( f(-x) = f(x) \) for all \( x \) in its domain. This means that the function is symmetric about the y-axis.

2. **Analyzing the Given Function**:
The function is defined as:
\[ f(x) = \begin{cases} 4 & \text{if } x < -1 \\ -4x & \text{if } -1 \leq x \leq 0 \end{cases} \]

3. **Finding \( f(-x) \)**:
We need to find \( f(-x) \) for \( x \) in the interval \( (0, +\infty) \):
- For \( x > 0 \), \( -x < 0 \). Thus, we use the second case of the function:
\[ f(-x) = -4(-x) = 4x \quad \text{for } -1 \leq -x \leq 0 \text{ (which is true for } 0 < x \leq 1\text{)} \]

4. **Finding \( f(x) \) for \( x \) in \( (0, +\infty) \)**:
- For \( x \in (0, 1) \), since \( f(-x) = 4x \), we need \( f(x) \) to be equal to \( f(-x) \) to maintain the even function property. Therefore, we define:
\[ f(x) = 4x \quad \text{for } 0 < x < 1 \]
- For \( x \geq 1 \), we observe that \( f(-x) \) should also equal \( f(x) \). Since \( f(-x) = 4x \) and we need \( f(x) \) to be constant (as per the first case when \( x < -1 \)):
\[ f(x) = 4 \quad \text{for } x \geq 1 \]

5. **Final Definition of \( f(x) \)**:
Combining all parts, we can define \( f(x) \) for \( x \in (0, +\infty) \) as:
\[ f(x) = \begin{cases} 4x & \text{if } 0 < x < 1 \\ 4 & \text{if } x \geq 1 \end{cases} \]

### Summary:
The definition of \( f(x) \) in the interval \( (0, +\infty) \) is:
\[ f(x) = \begin{cases} 4x & \text{if } 0 < x < 1 \\ 4 & \text{if } x \geq 1 \end{cases} \]