If `|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|` then the set of values of x is

Options

A. `(-oo , 0] cup [ 3,00)`
B. `[0,1/2] cup [2,3]`
C. `(-oo,0]cup [1/2,2] cup [3,oo)`
D. `[0,2] cup [3,oo)`

Explanation

To solve the equation \( |x^2 - 2x + 2| - |2x^2 - 5x + 2| = |x^2 - 3x| \), we will analyze the expressions inside the absolute values and determine the conditions under which they are positive or negative.

### Step 1: Analyze the expressions

1. **Identify the expressions:**
- Let \( A = x^2 - 2x + 2 \)
- Let \( B = 2x^2 - 5x + 2 \)
- Let \( C = x^2 - 3x \)

### Step 2: Find the roots of the expressions

2. **Find the roots of \( A \):**
\[ A = x^2 - 2x + 2 = 0 \]
The discriminant \( D = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \) (no real roots, always positive).

3. **Find the roots of \( B \):**
\[ B = 2x^2 - 5x + 2 = 0 \]
The discriminant \( D = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \) (real roots).
\[ x = \frac{5 \pm 3}{4} \Rightarrow x = 2 \text{ or } x = \frac{1}{2} \]

4. **Find the roots of \( C \):**
\[ C = x^2 - 3x = 0 \Rightarrow x(x - 3) = 0 \]
Roots are \( x = 0 \) and \( x = 3 \).

### Step 3: Determine intervals for analysis

5. **Intervals based on roots:**
The critical points from the roots of \( B \) and \( C \) are \( x = \frac{1}{2}, 0, 2, 3 \). We will analyze the sign of \( B \) and \( C \) in the intervals:
- \( (-\infty, \frac{1}{2}) \)
- \( (\frac{1}{2}, 0) \)
- \( (0, 2) \)
- \( (2, 3) \)
- \( (3, \infty) \)

### Step 4: Test the intervals

6. **Test each interval:**
- For \( x < \frac{1}{2} \): Choose \( x = 0 \)
- \( A > 0, B > 0, C < 0 \) → \( |A| - |B| = |C| \) does not hold.
- For \( \frac{1}{2} < x < 0 \): Choose \( x = \frac{1}{4} \)
- \( A > 0, B < 0, C < 0 \) → \( |A| + |B| = |C| \) does not hold.
- For \( 0 < x < 2 \): Choose \( x = 1 \)
- \( A > 0, B > 0, C < 0 \) → \( |A| - |B| = |C| \) does not hold.
- For \( 2 < x < 3 \): Choose \( x = 2.5 \)
- \( A > 0, B < 0, C < 0 \) → \( |A| + |B| = |C| \) holds.
- For \( x > 3 \): Choose \( x = 4 \)
- \( A > 0, B > 0, C > 0 \) → \( |A| - |B| = |C| \) does not hold.

### Step 5: Conclusion

7. **Final solution:**
The solution set is \( x \in \left[\frac{1}{2}, 2\right] \).