If satisfies `|x-1| + |x-2|+|x-3|gt6,` then :

Options

A. `x in (-oo, 1)`
B. `x in (-oo, 0)`
C. `x in (4,oo)`
D. `(2,oo)`

Answer

To solve the inequality \( |x-1| + |x-2| + |x-3| > 6 \), we will analyze the expression by breaking it down into different intervals based on the critical points where the absolute values change, which are \( x = 1 \), \( x = 2 \), and \( x = 3 \).

### Step 1: Identify the intervals
The critical points divide the number line into four intervals:
1. \( (-\infty, 1) \)
2. \( [1, 2) \)
3. \( [2, 3) \)
4. \( [3, \infty) \)

### Step 2: Analyze each interval

#### Interval 1: \( (-\infty, 1) \)
In this interval, all expressions inside the absolute values are negative:
\[ |x-1| = -(x-1) = -x + 1, \quad |x-2| = -(x-2) = -x + 2, \quad |x-3| = -(x-3) = -x + 3 \]
Thus, the inequality becomes:
\[ (-x + 1) + (-x + 2) + (-x + 3) > 6 \]
Simplifying this:
\[ -3x + 6 > 6 \]
Subtracting 6 from both sides:
\[ -3x > 0 \implies x < 0 \]
Since \( x < 1 \) is satisfied, the solution in this interval is:
\[ x < 0 \]

#### Interval 2: \( [1, 2) \)
In this interval, \( x-1 \) is non-negative, while \( x-2 \) and \( x-3 \) are negative:
\[ |x-1| = x - 1, \quad |x-2| = -x + 2, \quad |x-3| = -x + 3 \]
The inequality becomes:
\[ (x - 1) + (-x + 2) + (-x + 3) > 6 \]
Simplifying:
\[ -x + 4 > 6 \]
Subtracting 4 from both sides:
\[ -x > 2 \implies x < -2 \]
This solution does not satisfy \( x \geq 1 \), so there are no solutions in this interval.

#### Interval 3: \( [2, 3) \)
In this interval, \( x-1 \) and \( x-2 \) are non-negative, while \( x-3 \) is negative:
\[ |x-1| = x - 1, \quad |x-2| = x - 2, \quad |x-3| = -x + 3 \]
The inequality becomes:
\[ (x - 1) + (x - 2) + (-x + 3) > 6 \]
Simplifying:
\[ x > 6 \]
This solution does not satisfy \( x < 3 \), so there are no solutions in this interval.

#### Interval 4: \( [3, \infty) \)
In this interval, all expressions inside the absolute values are non-negative:
\[ |x-1| = x - 1, \quad |x-2| = x - 2, \quad |x-3| = x - 3 \]
The inequality becomes:
\[ (x - 1) + (x - 2) + (x - 3) > 6 \]
Simplifying:
\[ 3x - 6 > 6 \]
Adding 6 to both sides:
\[ 3x > 12 \implies x > 4 \]
This solution satisfies \( x \geq 3 \), so the solution in this interval is:
\[ x > 4 \]

### Step 3: Combine the solutions
The solutions from the intervals are:
1. From interval 1: \( x < 0 \)
2. From interval 4: \( x > 4 \)

Thus, the final solution set is:
\[ x < 0 \quad \text{or} \quad x > 4 \]

### Final Answer
The solution can be expressed in interval notation as:
\[ (-\infty, 0) \cup (4, \infty) \]