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

Answer

To solve the problem step by step, we will evaluate the determinant and apply the given condition \( a - 2b + c = 1 \).

### Step 1: Write the determinant
We start with the determinant:
\[ f(x) = \begin{vmatrix} x + a & x + 2 & x + 1 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{vmatrix} \]

### Step 2: Apply row operations
We will perform the row operation \( R_1 \rightarrow R_1 + R_3 - 2R_2 \):
\[ R_1 = (x + a) + (x + c) - 2(x + b) = (x + a + x + c - 2x - 2b) = a - 2b + c \]
For the second column:
\[ R_1 = (x + 2) + (x + 4) - 2(x + 3) = (x + 2 + x + 4 - 2x - 6) = 0 \]
For the third column:
\[ R_1 = (x + 1) + (x + 3) - 2(x + 2) = (x + 1 + x + 3 - 2x - 4) = 0 \]
Now, the determinant looks like:
\[ f(x) = \begin{vmatrix} a - 2b + c & 0 & 0 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{vmatrix} \]

### Step 3: Substitute the condition
Given \( a - 2b + c = 1 \), we can substitute this into the determinant:
\[ f(x) = \begin{vmatrix} 1 & 0 & 0 \\ x + b & x + 3 & x + 2 \\ x + c & x + 4 & x + 3 \end{vmatrix} \]

### Step 4: Expand the determinant
Since the first row has two zeros, we can expand the determinant along the first row:
\[ f(x) = 1 \cdot \begin{vmatrix} x + 3 & x + 2 \\ x + 4 & x + 3 \end{vmatrix} \]
Calculating this 2x2 determinant:
\[ = (x + 3)(x + 3) - (x + 2)(x + 4) = (x + 3)^2 - (x^2 + 6x + 8) \]
\[ = (x^2 + 6x + 9) - (x^2 + 6x + 8) = 1 \]
Thus, we have:
\[ f(x) = 1 \]

### Step 5: Conclusion
Since \( f(x) = 1 \) for all \( x \), it is a constant function. Therefore, regardless of the value of \( x \), \( f(x) \) will always equal 1.

### Final Answer
The correct option is \( f(50) = 1 \).