If `f(x)=|(0, x-a, x-b), (x+a, 0, x-c),(x+b, x+c, 0)|`, then

Options

A. `f(a)=0`
B. `f(b)=0`
C. `f(0)=0`
D. `f(1)=0`

Answer

To solve the problem, we need to evaluate the determinant \( f(x) = \left| \begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array} \right| \).

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ f(x) = \left| \begin{array}{ccc} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{array} \right| \]

2. **Expand the Determinant**:
We can expand the determinant using the first row:
\[ f(x) = 0 \cdot \left| \begin{array}{cc} 0 & x-c \\ x+c & 0 \end{array} \right| - (x-a) \cdot \left| \begin{array}{cc} x+a & x-c \\ x+b & 0 \end{array} \right| + (x-b) \cdot \left| \begin{array}{cc} x+a & 0 \\ x+b & x+c \end{array} \right| \]

The first term is zero, so we focus on the other two terms.

3. **Calculate the 2x2 Determinants**:
- For the second term:
\[ \left| \begin{array}{cc} x+a & x-c \\ x+b & 0 \end{array} \right| = (x+a)(0) - (x-c)(x+b) = -(x-c)(x+b) = - (x^2 + bx - cx - bc) \]
- For the third term:
\[ \left| \begin{array}{cc} x+a & 0 \\ x+b & x+c \end{array} \right| = (x+a)(x+c) - 0 = (x+a)(x+c) = x^2 + (a+c)x + ac \]

4. **Combine the Results**:
Now substituting back into \( f(x) \):
\[ f(x) = -(x-a)(-(x^2 + (b-c)x + bc)) + (x-b)(x^2 + (a+c)x + ac) \]
Simplifying this will yield a polynomial in \( x \).

5. **Evaluate \( f(a) \), \( f(b) \), and \( f(0) \)**:
- For \( f(a) \):
\[ f(a) = 0 \quad \text{(since one term will contain \( a-a \))} \]
- For \( f(b) \):
\[ f(b) = 0 \quad \text{(since one term will contain \( b-b \))} \]
- For \( f(0) \):
\[ f(0) = abc \quad \text{(after substituting \( x = 0 \))} \]

6. **Conclusion**:
The function \( f(x) \) evaluates to zero for both \( f(a) \) and \( f(b) \), and a non-zero value for \( f(0) \). Thus, the correct option is that \( f(x) \) has roots at \( x = a \) and \( x = b \).