If `|x^(2) - 3 x + 2| gt x^(2) - 3 x + 2`. Then which one of the following is correct ?

Options

A. `x le or x ge 2`
B. `1 le x le 2`
C. `1 lt x lt 2`
D. x is any real value except 3 and 4

Answer

To solve the inequality \( |x^2 - 3x + 2| > x^2 - 3x + 2 \), we will analyze the expression in three different cases based on the critical points where the expression inside the absolute value changes sign.

### Step 1: Identify the critical points
The expression \( x^2 - 3x + 2 \) can be factored as:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]
The critical points are \( x = 1 \) and \( x = 2 \). These points will help us divide the number line into intervals for testing.

### Step 2: Analyze the intervals
We will consider three cases based on the intervals created by the critical points:

1. **Case 1**: \( x \leq 1 \)
2. **Case 2**: \( 1 < x < 2 \)
3. **Case 3**: \( x \geq 2 \)

### Step 3: Case 1: \( x \leq 1 \)
In this interval, \( x^2 - 3x + 2 \) is non-positive (since it is zero at \( x = 1 \) and negative for \( x < 1 \)). Therefore, we have:
\[ |x^2 - 3x + 2| = -(x^2 - 3x + 2) = -x^2 + 3x - 2 \]
The inequality becomes:
\[ -x^2 + 3x - 2 > x^2 - 3x + 2 \]
Simplifying this gives:
\[ -2x^2 + 6x - 4 > 0 \quad \Rightarrow \quad 2x^2 - 6x + 4 < 0 \]
Factoring or using the quadratic formula, we find the roots:
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} = \frac{6 \pm \sqrt{36 - 32}}{4} = \frac{6 \pm 2}{4} \]
This gives us the roots \( x = 2 \) and \( x = 1 \). The quadratic opens upwards, so it is negative between the roots:
\[ 1 < x < 2 \]
However, since we are in the case \( x \leq 1 \), there are no valid solutions here.

### Step 4: Case 2: \( 1 < x < 2 \)
In this interval, \( x^2 - 3x + 2 \) is negative. Thus:
\[ |x^2 - 3x + 2| = -(x^2 - 3x + 2) = -x^2 + 3x - 2 \]
The inequality becomes:
\[ -x^2 + 3x - 2 > x^2 - 3x + 2 \]
This simplifies to:
\[ -2x^2 + 6x - 4 > 0 \quad \Rightarrow \quad 2x^2 - 6x + 4 < 0 \]
We already found the roots are \( x = 1 \) and \( x = 2 \). The quadratic is negative between these roots:
\[ 1 < x < 2 \]
This case is valid.

### Step 5: Case 3: \( x \geq 2 \)
In this interval, \( x^2 - 3x + 2 \) is non-negative. Therefore:
\[ |x^2 - 3x + 2| = x^2 - 3x + 2 \]
The inequality becomes:
\[ x^2 - 3x + 2 > x^2 - 3x + 2 \]
This simplifies to:
\[ 0 > 0 \]
This is false, so there are no solutions in this case.

### Conclusion
The only valid solutions are in Case 2, where \( 1 < x < 2 \).

### Final Answer
The correct option is that \( x \) lies in the interval \( (1, 2) \).