Consider the equation is real number x and a real parameter `lamda, |x-1| -|x-2| + |x-4| =lamda ` Then for `lamda ge 1,` the number of solutions, the equation can have is/are :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( |x-1| - |x-2| + |x-4| = \lambda \) for \( \lambda \geq 1 \), we will analyze the expression by considering different intervals based on the critical points where the absolute values change, which are \( x = 1, 2, 4 \).

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

### Step 2: Solve in each interval

#### Interval 1: \( (-\infty, 1) \)
In this interval:
- \( |x-1| = -(x-1) = -x + 1 \)
- \( |x-2| = -(x-2) = -x + 2 \)
- \( |x-4| = -(x-4) = -x + 4 \)

Substituting these into the equation:
\[ -x + 1 - (-x + 2) + (-x + 4) = \lambda \]
Simplifying:
\[ -x + 1 + x - 2 - x + 4 = \lambda \]
\[ -x + 3 = \lambda \]
Thus, the equation becomes:
\[ -x + 3 = \lambda \quad \Rightarrow \quad x = 3 - \lambda \]
For \( x \) to be in the interval \( (-\infty, 1) \):
\[ 3 - \lambda < 1 \quad \Rightarrow \quad \lambda > 2 \]

#### Interval 2: \( [1, 2) \)
In this interval:
- \( |x-1| = x - 1 \)
- \( |x-2| = -(x-2) = -x + 2 \)
- \( |x-4| = -(x-4) = -x + 4 \)

Substituting:
\[ (x - 1) - (-x + 2) + (-x + 4) = \lambda \]
Simplifying:
\[ x - 1 + x - 2 - x + 4 = \lambda \]
\[ x + 1 = \lambda \quad \Rightarrow \quad x = \lambda - 1 \]
For \( x \) to be in the interval \( [1, 2) \):
\[ 1 \leq \lambda - 1 < 2 \quad \Rightarrow \quad 2 \leq \lambda < 3 \]

#### Interval 3: \( [2, 4) \)
In this interval:
- \( |x-1| = x - 1 \)
- \( |x-2| = x - 2 \)
- \( |x-4| = -(x-4) = -x + 4 \)

Substituting:
\[ (x - 1) - (x - 2) + (-x + 4) = \lambda \]
Simplifying:
\[ x - 1 - x + 2 - x + 4 = \lambda \]
\[ - x + 5 = \lambda \quad \Rightarrow \quad x = 5 - \lambda \]
For \( x \) to be in the interval \( [2, 4) \):
\[ 2 \leq 5 - \lambda < 4 \quad \Rightarrow \quad 1 < \lambda \leq 3 \]

#### Interval 4: \( [4, \infty) \)
In this interval:
- \( |x-1| = x - 1 \)
- \( |x-2| = x - 2 \)
- \( |x-4| = x - 4 \)

Substituting:
\[ (x - 1) - (x - 2) + (x - 4) = \lambda \]
Simplifying:
\[ x - 1 - x + 2 + x - 4 = \lambda \]
\[ x - 3 = \lambda \quad \Rightarrow \quad x = \lambda + 3 \]
For \( x \) to be in the interval \( [4, \infty) \):
\[ \lambda + 3 \geq 4 \quad \Rightarrow \quad \lambda \geq 1 \]

### Step 3: Summary of solutions
- For \( \lambda < 2 \): No solutions.
- For \( 2 \leq \lambda < 3 \): 2 solutions (one from interval 2 and one from interval 3).
- For \( \lambda = 3 \): 3 solutions (one from each of intervals 2, 3, and 4).
- For \( \lambda > 3 \): 4 solutions (one from each interval).

### Conclusion
For \( \lambda \geq 1 \), the number of solutions can be 0, 1, 2, 3, or 4 depending on the value of \( \lambda \).