If `f(x)||x|-1|`, then fof(x) equals

Options

A. `f(x)={{:(|x|-2,|x| le 2),(2-|x|,1 lt |x| lt 2),(|x|,|x| le 1):}`
B. `f(x)={{:(|x|+2,|x| le 2),(|x|-2,1 le |x| le 2),(|x|,|x| le 1):}`
C. `f(x)={{:(|x|+2, |x| ge2),(2+|x|,1 le |x| le 2), (|x|,|x| le1):}`
D. none of these

Answer

To solve the problem, we need to find \( f(f(x)) \) where \( f(x) = ||x| - 1| \).

### Step 1: Understand the function \( f(x) \)
The function \( f(x) = ||x| - 1| \) involves taking the absolute value twice.

### Step 2: Analyze the inner absolute value
First, we need to analyze \( |x| - 1 \):
- If \( |x| \geq 1 \), then \( |x| - 1 \geq 0 \) and \( f(x) = |x| - 1 \).
- If \( |x| < 1 \), then \( |x| - 1 < 0 \) and \( f(x) = -( |x| - 1 ) = 1 - |x| \).

### Step 3: Define the cases for \( f(x) \)
Thus, we can write \( f(x) \) as:
1. If \( |x| \geq 1 \):
\[ f(x) = |x| - 1 \]
2. If \( |x| < 1 \):
\[ f(x) = 1 - |x| \]

### Step 4: Calculate \( f(f(x)) \)
Now we need to find \( f(f(x)) \).

#### Case 1: \( |x| \geq 1 \)
In this case, \( f(x) = |x| - 1 \). We need to evaluate \( f(f(x)) \):
- Since \( |x| \geq 1 \), we have \( |x| - 1 \geq 0 \). Therefore, we use the first case for \( f \):
\[ f(f(x)) = f(|x| - 1) = (|x| - 1) - 1 = |x| - 2 \]

#### Case 2: \( |x| < 1 \)
In this case, \( f(x) = 1 - |x| \). We need to evaluate \( f(f(x)) \):
- Since \( |x| < 1 \), we have \( 1 - |x| > 0 \). Therefore, we again use the first case for \( f \):
\[ f(f(x)) = f(1 - |x|) = (1 - |x|) - 1 = -|x| \]

### Step 5: Combine the results
We can summarize our findings:
- If \( |x| \geq 1 \), then \( f(f(x)) = |x| - 2 \).
- If \( |x| < 1 \), then \( f(f(x)) = -|x| \).

### Final Answer
Thus, the final answer for \( f(f(x)) \) is:
\[ f(f(x)) = \begin{cases} |x| - 2 & \text{if } |x| \geq 1 \\ -|x| & \text{if } |x| < 1 \end{cases} \]