Let `f(x)=|x-1|`. Then,

Options

A. `f(x^(2))=[f(x)]^(2)`
B. `f(|x|)=|f(x)|`
C. `f(x+y)=f(x)+f(y)`
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) = |x - 1| \) and evaluate the given options one by one.

### Step 1: Evaluate Option A
**Option A:** \( f(x^2) = f(x)^2 \)

1. Calculate \( f(x^2) \):
\[ f(x^2) = |x^2 - 1| \]

2. Calculate \( f(x)^2 \):
\[ f(x) = |x - 1| \implies f(x)^2 = (|x - 1|)^2 = (x - 1)^2 \]

3. Now we need to check if \( |x^2 - 1| = (x - 1)^2 \) for all \( x \).

- For \( x = 2 \):
\[ f(2^2) = |4 - 1| = 3 \]
\[ f(2)^2 = (|2 - 1|)^2 = 1^2 = 1 \]
They are not equal.

- Thus, \( f(x^2) \neq f(x)^2 \).

**Conclusion for Option A:** False

### Step 2: Evaluate Option B
**Option B:** \( f(|x|) = |f(x)| \)

1. Calculate \( f(|x|) \):
\[ f(|x|) = ||x| - 1| \]

2. Calculate \( |f(x)| \):
\[ f(x) = |x - 1| \implies |f(x)| = ||x - 1| \]

3. Check if \( ||x| - 1| = ||x - 1| \).

- For \( x = 1 \):
\[ f(|1|) = |1 - 1| = 0 \]
\[ |f(1)| = |1 - 1| = 0 \]
They are equal.

- For \( x = -1 \):
\[ f(|-1|) = |1 - 1| = 0 \]
\[ |f(-1)| = |-1 - 1| = 2 \]
They are not equal.

**Conclusion for Option B:** False

### Step 3: Evaluate Option C
**Option C:** \( f(x + y) = f(x) + f(y) \)

1. Calculate \( f(x + y) \):
\[ f(x + y) = |(x + y) - 1| = |x + y - 1| \]

2. Calculate \( f(x) + f(y) \):
\[ f(x) + f(y) = |x - 1| + |y - 1| \]

3. Check if \( |x + y - 1| = |x - 1| + |y - 1| \).

- For \( x = 1 \) and \( y = 1 \):
\[ f(1 + 1) = |2 - 1| = 1 \]
\[ f(1) + f(1) = 0 + 0 = 0 \]
They are not equal.

**Conclusion for Option C:** False

### Final Conclusion
Since all options A, B, and C are false, the correct answer is:

**Option D:** None of the above.