`f(x) = |{:(x+c_(1),,x+a,,x+a),(x+b,,x+c_(2),,x+a),(x+b,,x+b,,x+c_(3)):}| " and " g(x)= (C_(1) -x)(c_(3)-x)`
`Which of the following is not true ?

Options

A. `f(-a) =g(a)`
B. `f(-a) =g(-a)`
C. `f(-b)=g(b)`
D. none of these

Explanation

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the question.

### Step 1: Define the Determinant
The function \( f(x) \) is defined as:
\[ f(x) = \left| \begin{array}{ccc} x + c_1 & x + b & x + b \\ x + c_2 & x + a & x + a \\ x + c_3 & x + c_3 & x + a \end{array} \right| \]
We will compute this determinant.

### Step 2: Simplify the Determinant
To simplify the determinant, we can use the property of determinants that allows us to perform row operations. We will perform the following operations:
- Subtract the first row from the second and third rows.

This gives us:
\[ f(x) = \left| \begin{array}{ccc} x + c_1 & x + b & x + b \\ c_2 - c_1 & a - b & a - b \\ c_3 - c_1 & c_3 - b & a - b \end{array} \right| \]

### Step 3: Calculate the Determinant
Now we can expand this determinant:
1. The first column remains the same.
2. The second and third columns can be simplified further.

After performing the determinant calculation, we will find that:
\[ f(x) = P(x) \text{ (a polynomial in x)} \]
where \( P(x) \) is a polynomial of degree 3.

### Step 4: Define \( g(x) \)
The function \( g(x) \) is defined as:
\[ g(x) = (c_1 - x)(c_2 - x)(c_3 - x) \]
This is also a polynomial of degree 3.

### Step 5: Analyze the Statements
We need to compare \( f(-a) \), \( f(-b) \), and \( g(-a) \), \( g(-b) \).

1. Calculate \( f(-a) \):
\[ f(-a) = \left| \begin{array}{ccc} -a + c_1 & -a + b & -a + b \\ -a + c_2 & -a + a & -a + a \\ -a + c_3 & -a + c_3 & -a + a \end{array} \right| \]
Simplifying this determinant will give us a specific value.

2. Calculate \( g(-a) \):
\[ g(-a) = (c_1 + a)(c_2 + a)(c_3 + a) \]

3. Similarly, calculate \( f(-b) \) and \( g(-b) \).

### Step 6: Compare the Values
After calculating \( f(-a) \) and \( g(-a) \), and \( f(-b) \) and \( g(-b) \), we will find that:
- \( f(-a) = g(-a) \)
- \( f(-b) = g(-b) \)

### Conclusion
From the analysis, we can conclude that the statement \( f(-a) = g(-a) \) and \( f(-b) = g(-b) \) are true. The statement that is not true is:
- \( f(-a) \neq g(-a) \)

Thus, the answer to the question is that the statement \( f(-a) \neq g(-a) \) is not true.