The function f(x) given by
`f(x)=|{:(x-1", "x+1", "2x+1),(x+1", "x+3", "2x+3),(2x+1", "2x-1", "4x+1):}|` has

Options

A. one point of maximum and one point of minimum
B. one point of maximum only
C. one point of maximum only
D. none of these

Answer

To solve the problem, we need to evaluate the determinant given by the function \( f(x) \) and determine whether it has any maxima or minima. Let's go through the steps systematically.

### Step 1: Write the determinant
The function \( f(x) \) is given by the determinant of the following matrix:

\[ f(x) = \begin{vmatrix} x - 1 & x + 1 & 2x + 1 \\ x + 1 & x + 3 & 2x + 3 \\ 2x + 1 & 2x - 1 & 4x + 1 \end{vmatrix} \]

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

1. **Row Operation**: Replace \( R_3 \) with \( R_3 - R_1 + R_2 \):
\[ R_3 = (2x + 1) - (x - 1) + (x + 3) = 2x + 1 - x + 1 + x + 3 = 5 \]
The new matrix becomes:
\[ \begin{vmatrix} x - 1 & x + 1 & 2x + 1 \\ x + 1 & x + 3 & 2x + 3 \\ 0 & 0 & 5 \end{vmatrix} \]

### Step 3: Calculate the determinant
Now we can compute the determinant. Since the last row has two zeros, we can expand along that row:
\[ f(x) = 5 \cdot \begin{vmatrix} x - 1 & x + 1 \\ x + 1 & x + 3 \end{vmatrix} \]

Calculating the 2x2 determinant:
\[ \begin{vmatrix} x - 1 & x + 1 \\ x + 1 & x + 3 \end{vmatrix} = (x - 1)(x + 3) - (x + 1)(x + 1) \]
\[ = (x^2 + 3x - x - 3) - (x^2 + 2x + 1) = 2x - 4 \]

Thus, we have:
\[ f(x) = 5(2x - 4) = 10x - 20 \]

### Step 4: Find the derivative
To find maxima or minima, we need to take the derivative of \( f(x) \):
\[ f'(x) = 10 \]

### Step 5: Analyze the derivative
Since \( f'(x) = 10 \) is a constant and never equal to zero, it indicates that the function is always increasing.

### Conclusion
Since the derivative does not equal zero and is positive, \( f(x) \) has no maximum or minimum points.

### Final Answer
The function \( f(x) \) has no maxima or minima.

---