If `f(x) = |[x+a,x+2,x+1],[x+b,x+3,x+2],[x+c,x+4,x+3]|` and `a - 2b + c = 1` then

Options

A. `f(-50) = 501`
B. `f(-50) = -1`
C. `f(50) = 1`
D. `f(50) = -501`

Answer

To solve the problem, we need to find the determinant of the given matrix and analyze it based on the condition \( a - 2b + c = 1 \).

### Step 1: Write down the determinant
The function is defined as:
\[ f(x) = \left| \begin{array}{ccc} x + a & x + 2 & x + 1 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{array} \right| \]

### Step 2: Apply row operations
We will simplify the determinant using row operations. We can perform the operation \( R_1 \rightarrow R_1 + R_3 - 2R_2 \):
\[ R_1 = (x + a) + (x + c) - 2(x + b) \]
This simplifies to:
\[ R_1 = (x + a + x + c - 2x - 2b) = a + c - 2b \]
Thus, the first row becomes:
\[ a + c - 2b, \quad x + 4 - 6, \quad x + 1 - 4 \]
This results in:
\[ a + c - 2b, \quad 0, \quad -3 \]

### Step 3: Rewrite the determinant
Now the determinant can be rewritten as:
\[ f(x) = \left| \begin{array}{ccc} a + c - 2b & 0 & -3 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{array} \right| \]

### Step 4: Calculate the determinant
Using the properties of determinants, we can expand this determinant:
\[ f(x) = (a + c - 2b) \left| \begin{array}{cc} x + 3 & x + 2 \\ x + 4 & x + 3 \end{array} \right| \]
Calculating the 2x2 determinant:
\[ = (x + 3)(x + 3) - (x + 2)(x + 4) = (x + 3)^2 - (x^2 + 6x + 8) \]
Expanding and simplifying:
\[ = x^2 + 6x + 9 - x^2 - 6x - 8 = 1 \]
Thus, we have:
\[ f(x) = (a + c - 2b) \cdot 1 = a + c - 2b \]

### Step 5: Use the condition
Given the condition \( a - 2b + c = 1 \), we can substitute:
\[ f(x) = 1 \]

### Conclusion
Since \( f(x) = 1 \) for all \( x \), it is a constant function. Therefore, for any value of \( x \), including \( x = 50 \):
\[ f(50) = 1 \]

### Final Answer
The correct option is:
\[ f(50) = 1 \]