If `f(x)`=`|(a, -1, 0), (ax, a,-1 ) ,(ax^(2), ax, a)|`, then `f(2x)-f(x)` is divisible by
1) `a`
2) `b`
3) `c ,d ,e`
4). none of these

Options

A. x
B. a
C. `2a+3x`
D. `x^(2)`

Explanation

To solve the problem, we need to evaluate the determinant function \( f(x) \) and then find \( f(2x) - f(x) \). We will also check the divisibility of the resulting expression.

### Step 1: Define the determinant function
We start with the function defined as:
\[ f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix} \]

### Step 2: Perform row operations
We will perform row operations to simplify the determinant.

1. **Row Operation 1**: Replace \( R_3 \) with \( R_3 - x \cdot R_2 \):
\[ R_3 \rightarrow R_3 - xR_2 \]
This gives us:
\[ R_3 = (ax^2 - x \cdot a, ax - x \cdot a, a - (-1)x) = (ax^2 - ax, ax - ax, a + x) = (a(x - 1)x, 0, a + x) \]

2. **Row Operation 2**: Replace \( R_2 \) with \( R_2 - x \cdot R_1 \):
\[ R_2 \rightarrow R_2 - xR_1 \]
This gives us:
\[ R_2 = (ax - ax, a - (-1)x, -1 - 0) = (0, a + x, -1) \]

Now the determinant looks like this:
\[ f(x) = \begin{vmatrix} a & -1 & 0 \\ 0 & a + x & -1 \\ a(x - 1)x & 0 & a + x \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula, we can calculate \( f(x) \):
\[ f(x) = a \begin{vmatrix} a + x & -1 \\ 0 & a + x \end{vmatrix} - (-1) \begin{vmatrix} 0 & -1 \\ a(x - 1)x & a + x \end{vmatrix} \]
Calculating the first determinant:
\[ = a((a + x)(a + x) - 0) = a(a + x)^2 \]
Calculating the second determinant:
\[ = -1(0 - (-1)(a(x - 1)x)) = a(x - 1)x \]
Thus,
\[ f(x) = a(a + x)^2 + a(x - 1)x \]

### Step 4: Calculate \( f(2x) \)
Now we need to calculate \( f(2x) \):
\[ f(2x) = a(a + 2x)^2 + a(2x - 1)(2x) \]

### Step 5: Find \( f(2x) - f(x) \)
Now we compute \( f(2x) - f(x) \):
\[ f(2x) - f(x) = a(a + 2x)^2 + a(2x - 1)(2x) - (a(a + x)^2 + a(x - 1)x) \]

### Step 6: Factor the expression
After simplification, we factor out common terms:
\[ = a \cdot x \cdot (2a + 3x) \]

### Step 7: Check divisibility
Now we check the divisibility of \( f(2x) - f(x) \):
- The expression \( a \cdot x \cdot (2a + 3x) \) is divisible by \( a \) and \( x \).

### Conclusion
Thus, the correct options for divisibility are:
1) \( a \)
2) \( x \)
3) \( 2a + 3x \)