Let `f(x) = abs(x – 1)`. Then :

Options

A. `f(x^2) = (f(x))^2`
B. `f(x+y)= f(x) + f(y)`
C. `f(absx)= abs(f(x))`
D. None of these

Answer

To solve the problem, we need to analyze the function \( f(x) = |x - 1| \) and evaluate the given options based on this function. Let's break it down step by step.

### Step 1: Define the function
The function is defined as:
\[ f(x) = |x - 1| \]
This means that \( f(x) \) gives the distance of \( x \) from 1 on the number line.

### Step 2: Evaluate \( f(x^2) \)
Next, we need to find \( f(x^2) \):
\[ f(x^2) = |x^2 - 1| \]

### Step 3: Evaluate \( (f(x))^2 \)
Now, we calculate \( (f(x))^2 \):
\[ (f(x))^2 = (|x - 1|)^2 = (x - 1)^2 \]

### Step 4: Compare \( f(x^2) \) and \( (f(x))^2 \)
We need to check if \( f(x^2) \) is equal to \( (f(x))^2 \):
\[ |x^2 - 1| \quad \text{and} \quad (x - 1)^2 \]
To see if these are equal, we can analyze their graphs or values.

### Step 5: Analyze the graphs
1. The graph of \( |x^2 - 1| \) will have a V-shape with points of intersection at \( x = -1 \) and \( x = 1 \).
2. The graph of \( (x - 1)^2 \) is a parabola opening upwards with its vertex at \( x = 1 \).

Since these two graphs have different shapes and intersections, we can conclude that:
\[ f(x^2) \neq (f(x))^2 \]

### Step 6: Evaluate \( f(x+y) \) and \( f(x) + f(y) \)
Now we check if \( f(x+y) = f(x) + f(y) \):
\[ f(x+y) = |(x+y) - 1| = |x + y - 1| \]
\[ f(x) + f(y) = |x - 1| + |y - 1| \]
These two expressions are not necessarily equal for all \( x \) and \( y \) since the absolute value function can change based on the values of \( x \) and \( y \).

### Step 7: Evaluate \( f(|x|) \) and \( |f(x)| \)
Next, we compare \( f(|x|) \) and \( |f(x)| \):
\[ f(|x|) = ||x| - 1| \]
\[ |f(x)| = ||x - 1| \]
For \( x \geq 0 \):
\[ f(|x|) = |x - 1| \quad \text{and} \quad |f(x)| = |x - 1| \]
For \( x < 0 \):
\[ f(|x|) = |-x - 1| = |-(x + 1)| = |x + 1| \quad \text{and} \quad |f(x)| = |-(x - 1)| = |1 - x| \]
These two expressions are not equal for \( x < 0 \).

### Conclusion
Since all three options are incorrect, the answer is:
\[ \text{None of these} \]