Let `a_(2),a_(3)epsilonR` be such that `|a_(2)-a_(3)|=6`, Let
`f(x)=|(1,a_(3),a_(2)),(1,a_(3),2a_(2)-x),(1,2a_(3)-x,a_(2))|, x epsilon R` The maximum value of f(x) is

Options

A. 6
B. 9
C. 12
D. 36

Answer

To find the maximum value of the function \( f(x) = |(1, a_3, a_2), (1, a_3, 2a_2 - x), (1, 2a_3 - x, a_2)| \), we will follow these steps:

### Step 1: Write the determinant
The determinant can be expressed as:
\[ f(x) = \begin{vmatrix} 1 & a_3 & a_2 \\ 1 & a_3 & 2a_2 - x \\ 1 & 2a_3 - x & a_2 \end{vmatrix} \]

### Step 2: Apply row operations
We can simplify the determinant by performing row operations. Subtract the first row from the second and third rows:
\[ f(x) = \begin{vmatrix} 1 & a_3 & a_2 \\ 0 & 0 & 2a_2 - x - a_2 \\ 0 & 2a_3 - x - a_3 & a_2 - a_2 \end{vmatrix} \]
This simplifies to:
\[ f(x) = \begin{vmatrix} 1 & a_3 & a_2 \\ 0 & 0 & a_2 - x \\ 0 & a_3 - x & 0 \end{vmatrix} \]

### Step 3: Calculate the determinant
Since the first column has a leading 1, we can expand the determinant:
\[ f(x) = (a_2 - x) \begin{vmatrix} 1 & a_3 \\ 0 & a_3 - x \end{vmatrix} \]
Calculating the determinant:
\[ = (a_2 - x)(a_3 - x) \]
Thus, we have:
\[ f(x) = (a_2 - x)(a_3 - x) \]

### Step 4: Use the given condition
Given that \( |a_2 - a_3| = 6 \), we can express \( a_2 - a_3 \) as either \( 6 \) or \( -6 \).

### Step 5: Find the maximum value
To find the maximum value of \( f(x) \), we can rewrite it as:
\[ f(x) = -x^2 + (a_2 + a_3)x - a_2 a_3 \]
This is a quadratic function in standard form \( ax^2 + bx + c \) with \( a = -1 \), which opens downwards. The maximum occurs at:
\[ x = -\frac{b}{2a} = \frac{a_2 + a_3}{2} \]

### Step 6: Substitute \( x \) back into \( f(x) \)
Substituting \( x = \frac{a_2 + a_3}{2} \) into \( f(x) \):
\[ f\left(\frac{a_2 + a_3}{2}\right) = \left(a_2 - \frac{a_2 + a_3}{2}\right)\left(a_3 - \frac{a_2 + a_3}{2}\right) \]
This simplifies to:
\[ = \left(\frac{2a_2 - a_2 - a_3}{2}\right)\left(\frac{2a_3 - a_2 - a_3}{2}\right) = \frac{(a_2 - a_3)(a_3 - a_2)}{4} \]
Since \( |a_2 - a_3| = 6 \):
\[ f\left(\frac{a_2 + a_3}{2}\right) = \frac{6^2}{4} = \frac{36}{4} = 9 \]

### Conclusion
Thus, the maximum value of \( f(x) \) is \( \boxed{9} \).