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

Options

A. f(50)=50
B. f(X)=x
C. f(100)=0
D. f(0)=0

Answer

To solve the problem, we need to evaluate the determinant given in the function \( f(x) \) and analyze its behavior for specific values of \( x \).

### Step-by-Step Solution:

1. **Write the Determinant**:
The function is defined as:
\[ f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{vmatrix} \]

2. **Expand the Determinant**:
We will expand this determinant using the properties of determinants. We can perform row operations to simplify it.

3. **Factor Out Common Terms**:
Notice that in the second and third rows, there are common factors of \( x \) and \( (x-1) \). We can factor these out:
\[ f(x) = x(x-1) \begin{vmatrix} 1 & 1 & 1 \\ 2 & (x-1) & (x+1) \\ 3(x-2) & (x-1)(x-2) & (x+1) \end{vmatrix} \]

4. **Perform Row Operations**:
Now we can perform row operations to simplify the determinant further. We will subtract the first row from the second and third rows:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & (x-2) & (x) \\ 2(x-2) & (x-1)(x-2) & (x) \end{vmatrix} \]

5. **Calculate the Determinant**:
Now we can calculate the determinant using the formula for 3x3 matrices:
\[ f(x) = x(x-1) \left( 1 \cdot \left( (x-2)(x) - (x-1)(x-2) \right) - 1 \cdot \left( 1 \cdot (x) - 1 \cdot 2(x-2) \right) + 1 \cdot \left( 1 \cdot 2(x-2) - 1 \cdot (x-1) \right) \right) \]

6. **Simplify the Result**:
After performing the calculations, we find that the determinant simplifies to 0:
\[ f(x) = 0 \]

7. **Evaluate Specific Values**:
Now, we can evaluate \( f(x) \) for specific values:
- \( f(50) = 0 \)
- \( f(100) = 0 \)
- \( f(0) = 0 \)
- \( f(x) = x \) is not true since \( f(x) = 0 \).

### Conclusion:
The correct options are:
- \( f(100) = 0 \)
- \( f(0) = 0 \)