If `a << b << c << d` and `x in R` then the least value of the function, `f(x)=|x-a|+|x-b|+|x-c|+|x-d|` is

Options

A. c-d+b-a
B. c+d-b-a
C. c+d-b+a
D. c+d-d+A

Answer

To find the least value of the function \( f(x) = |x-a| + |x-b| + |x-c| + |x-d| \), we can analyze the behavior of the function based on the values of \( x \) in relation to \( a \), \( b \), \( c \), and \( d \).

### Step-by-step Solution:

1. **Identify the critical points**: The critical points are \( a \), \( b \), \( c \), and \( d \). These points divide the real number line into intervals where the expression inside the absolute values can be simplified.

2. **Consider the intervals**:
- **Case 1**: \( x < a \)
\[ f(x) = (a-x) + (b-x) + (c-x) + (d-x) = (a+b+c+d) - 4x \]
- **Case 2**: \( a \leq x < b \)
\[ f(x) = (x-a) + (b-x) + (c-x) + (d-x) = (b+c+d-a) - 2x \]
- **Case 3**: \( b \leq x < c \)
\[ f(x) = (x-a) + (x-b) + (c-x) + (d-x) = (c+d-a-b) + 0 = c+d-a-b \]
- **Case 4**: \( c \leq x < d \)
\[ f(x) = (x-a) + (x-b) + (x-c) + (d-x) = (d-a-b-c) + 2x \]
- **Case 5**: \( x \geq d \)
\[ f(x) = (x-a) + (x-b) + (x-c) + (x-d) = 4x - (a+b+c+d) \]

3. **Analyze the function in each interval**:
- In **Case 1**, \( f(x) \) is decreasing as \( x \) increases.
- In **Case 2**, \( f(x) \) is also decreasing.
- In **Case 3**, \( f(x) \) is constant.
- In **Case 4**, \( f(x) \) is increasing.
- In **Case 5**, \( f(x) \) is increasing.

4. **Determine the minimum value**:
The minimum value of \( f(x) \) occurs in **Case 3** where \( f(x) = c + d - a - b \). Since this is a constant value in that interval, it is the least value of the function.

5. **Conclusion**:
The least value of the function \( f(x) \) is:
\[ \text{Least value} = c + d - a - b \]