Let `f(x) = |x -1| + |x-2| + |x-3|` , match the column I for the value of k column II.
`{:("Column - I" , " Column -II"),("(A) f(x) = k has no solution " ,"(p) 1 " ),("(B) f(x)=k has only one solution" , "(q) 2"),(" (C) f(x) = k has two solution of same sign " , "(r) 4"),("(D) f(x)= k has two solution of opposite sign" , "(s) 5"),(,"(t) 8"):}`

Answer

To solve the problem, we need to analyze the function \( f(x) = |x - 1| + |x - 2| + |x - 3| \) and determine how many solutions the equation \( f(x) = k \) has for various values of \( k \).

### Step-by-Step Solution:

1. **Identify the critical points**: The critical points of the function occur where the absolute values change, which are at \( x = 1, 2, 3 \). These points will help us divide the number line into intervals.

2. **Analyze the intervals**: The intervals created by these points are:
- \( (-\infty, 1) \)
- \( [1, 2) \)
- \( [2, 3) \)
- \( [3, \infty) \)

3. **Determine the function behavior in each interval**:
- For \( x < 1 \):
\[ f(x) = -(x - 1) - (x - 2) - (x - 3) = -3x + 6 \]
- For \( 1 \leq x < 2 \):
\[ f(x) = (x - 1) - (x - 2) - (x - 3) = -x + 4 \]
- For \( 2 \leq x < 3 \):
\[ f(x) = (x - 1) + (x - 2) - (x - 3) = x + 2 \]
- For \( x \geq 3 \):
\[ f(x) = (x - 1) + (x - 2) + (x - 3) = 3x - 6 \]

4. **Evaluate the function at critical points**:
- \( f(1) = |1 - 1| + |1 - 2| + |1 - 3| = 0 + 1 + 2 = 3 \)
- \( f(2) = |2 - 1| + |2 - 2| + |2 - 3| = 1 + 0 + 1 = 2 \)
- \( f(3) = |3 - 1| + |3 - 2| + |3 - 3| = 2 + 1 + 0 = 3 \)

5. **Determine the minimum value of \( f(x) \)**:
- The minimum value occurs at \( x = 2 \), where \( f(2) = 2 \).

6. **Determine the behavior of \( f(x) \)**:
- As \( x \) approaches \( -\infty \), \( f(x) \) approaches \( +\infty \).
- As \( x \) approaches \( +\infty \), \( f(x) \) also approaches \( +\infty \).
- The function is symmetric around \( x = 2 \).

7. **Match the values of \( k \) with the number of solutions**:
- \( k < 2 \): \( f(x) = k \) has no solution (A matches with p).
- \( k = 2 \): \( f(x) = k \) has only one solution (B matches with q).
- \( 2 < k < 3 \): \( f(x) = k \) has two solutions of the same sign (C matches with r).
- \( k = 3 \): \( f(x) = k \) has two solutions of opposite sign (D matches with s).
- \( k > 3 \): \( f(x) = k \) has two solutions of opposite sign (D matches with t).

### Final Matching:
- (A) \( f(x) = k \) has no solution → (p) 1
- (B) \( f(x) = k \) has only one solution → (q) 2
- (C) \( f(x) = k \) has two solutions of the same sign → (r) 4
- (D) \( f(x) = k \) has two solutions of opposite sign → (s) 5
- \( k = 8 \) → (t) 8