If f(x) = `|{:(1, x, x +1),(2x , x (x - 1), (x + 1)x),(3x(x - 1), x (x - 1)(x-2),(x + 1)x (x - 1)):}|`
Then f (100) is equal to

Options

A. 0
B. 1
C. 100
D. 10

Answer

To solve the problem, we need to evaluate the determinant \( f(x) \) defined as:

\[ f(x) = \begin{vmatrix} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x - 1) & x(x - 1)(x - 2) & (x + 1)x(x - 1) \end{vmatrix} \]

### Step 1: Factor out common terms from the rows

Let's analyze the rows of the determinant. We can factor out common terms from each row.

1. From the first row, there are no common factors.
2. From the second row, we can factor out \( x \):
\[ 2x, x(x - 1), (x + 1)x \Rightarrow x(2, x - 1, x + 1) \]
3. From the third row, we can factor out \( x(x - 1) \):
\[ 3x(x - 1), x(x - 1)(x - 2), (x + 1)x(x - 1) \Rightarrow x(x - 1)(3, x - 2, x + 1) \]

Thus, we can rewrite the determinant as:
\[ f(x) = x^2(x - 1) \begin{vmatrix} 1 & x & x + 1 \\ 2 & x - 1 & x + 1 \\ 3 & x - 2 & x + 1 \end{vmatrix} \]

### Step 2: Simplify the determinant

Now, we will simplify the determinant:
\[ \begin{vmatrix} 1 & x & x + 1 \\ 2 & x - 1 & x + 1 \\ 3 & x - 2 & x + 1 \end{vmatrix} \]

We can perform row operations to simplify this determinant. Let's subtract the first row from the second and third rows:
- \( R_2 \leftarrow R_2 - 2R_1 \)
- \( R_3 \leftarrow R_3 - 3R_1 \)

This gives us:
\[ \begin{vmatrix} 1 & x & x + 1 \\ 0 & -2 & -1 \\ 0 & -3 & -2 \end{vmatrix} \]

### Step 3: Calculate the determinant

Now we can calculate the determinant:
\[ = 1 \cdot \begin{vmatrix} -2 & -1 \\ -3 & -2 \end{vmatrix} \]

Calculating the 2x2 determinant:
\[ = 1 \cdot (-2 \cdot -2 - (-1) \cdot -3) = 1 \cdot (4 - 3) = 1 \]

### Step 4: Combine results

Now substituting back into our expression for \( f(x) \):
\[ f(x) = x^2(x - 1) \cdot 1 = x^2(x - 1) \]

### Step 5: Evaluate \( f(100) \)

Now we can evaluate \( f(100) \):
\[ f(100) = 100^2(100 - 1) = 10000 \cdot 99 = 990000 \]

Thus, the final answer is:
\[ \boxed{990000} \]