Let `f(x) = x^(2) - 3x + 2` be a function defined from `R rarr R` and `|x| = {{:(x,, x ge 0),(-x ,, x le 0):}`, then answer the following
Range of value of x for which `|f|x|| - f|x| = `0 has no solution

Options

A. [1,2]
B. `[-2, -1] cup [1, 2]`
C. `(-2, -1) cup (1, 2)`
D. `(-oo, 0) cup (1, 2)`

Answer

To solve the problem, we need to analyze the function \( f(x) = x^2 - 3x + 2 \) and find the range of values of \( x \) for which the equation \( |f(|x|)| - f(|x|) = 0 \) has no solutions.

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

The function is given by:
\[ f(x) = x^2 - 3x + 2 \]
We can factor this quadratic function:
\[ f(x) = (x - 1)(x - 2) \]
The roots of the function are \( x = 1 \) and \( x = 2 \).

### Step 2: Analyze the expression \( |f(|x|)| - f(|x|) = 0 \)

This equation simplifies to:
\[ |f(|x|)| = f(|x|) \]
This means that \( f(|x|) \) must be non-negative for the absolute value to equal the function itself.

### Step 3: Determine when \( f(x) \) is non-negative

The function \( f(x) \) is a quadratic that opens upwards (since the coefficient of \( x^2 \) is positive). The roots are \( x = 1 \) and \( x = 2 \).

We can analyze the sign of \( f(x) \):
- For \( x < 1 \), \( f(x) > 0 \)
- For \( 1 < x < 2 \), \( f(x) < 0 \)
- For \( x > 2 \), \( f(x) > 0 \)

Thus, \( f(x) \) is non-negative when:
\[ x \leq 1 \quad \text{or} \quad x \geq 2 \]

### Step 4: Analyze \( |x| \)

Since we are dealing with \( |x| \), we need to consider two cases:
1. When \( x \geq 0 \) (then \( |x| = x \))
2. When \( x < 0 \) (then \( |x| = -x \))

### Step 5: Solve for \( |x| \)

- For \( x \geq 0 \):
- We need \( |x| \leq 1 \) or \( |x| \geq 2 \), which simplifies to \( 0 \leq x \leq 1 \) or \( x \geq 2 \).

- For \( x < 0 \):
- Here, \( |x| = -x \), so we need \( -x \leq 1 \) or \( -x \geq 2 \), which simplifies to \( x \geq -1 \) or \( x \leq -2 \).

### Step 6: Combine the intervals

Combining the intervals:
- From \( x \geq 0 \): \( [0, 1] \cup [2, \infty) \)
- From \( x < 0 \): \( (-\infty, -2] \cup [-1, 0) \)

### Step 7: Identify the range where there are no solutions

The function \( |f(|x|)| - f(|x|) = 0 \) has no solutions in the intervals where \( f(|x|) < 0 \):
- From the analysis, \( f(|x|) < 0 \) when \( |x| \) is in the interval \( (1, 2) \).

Thus, the ranges for \( x \) where this occurs are:
- For \( |x| \) in \( (1, 2) \):
- This corresponds to \( x \in (-2, -1) \cup (1, 2) \).

### Final Answer

The range of values of \( x \) for which \( |f(|x|)| - f(|x|) = 0 \) has no solutions is:
\[ x \in (-2, -1) \cup (1, 2) \]