If x = -1 and x = 2 are extreme points of f(x) = `alpha log|x| + beta x^2 + x`, then

Options

A. `alpha =-6,beta=1/3`
B. `alpha =-6,beta=-1/2`
C. `alpha =2,beta=-1/2`
D. `alpha = 2,beta=1/2`

Explanation

To solve the problem, we need to find the values of \(\alpha\) and \(\beta\) given that \(x = -1\) and \(x = 2\) are extreme points of the function \(f(x) = \alpha \log|x| + \beta x^2 + x\). This means that the derivative \(f'(x)\) must equal zero at these points.

### Step-by-Step Solution:

1. **Find the derivative of \(f(x)\)**:
\[ f(x) = \alpha \log|x| + \beta x^2 + x \]
The derivative \(f'(x)\) is given by:
\[ f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \]

2. **Set up the equations for the extreme points**:
Since \(x = -1\) and \(x = 2\) are extreme points, we have:
\[ f'(-1) = 0 \quad \text{and} \quad f'(2) = 0 \]

3. **Calculate \(f'(-1)\)**:
\[ f'(-1) = \frac{\alpha}{-1} + 2\beta(-1) + 1 = -\alpha - 2\beta + 1 = 0 \]
Rearranging gives:
\[ \alpha + 2\beta = 1 \quad \text{(Equation 1)} \]

4. **Calculate \(f'(2)\)**:
\[ f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = \frac{\alpha}{2} + 4\beta + 1 = 0 \]
Rearranging gives:
\[ \alpha + 8\beta + 2 = 0 \quad \text{(Equation 2)} \]

5. **Solve the system of equations**:
From Equation 1:
\[ \alpha = 1 - 2\beta \]
Substitute \(\alpha\) into Equation 2:
\[ (1 - 2\beta) + 8\beta + 2 = 0 \]
Simplifying gives:
\[ 1 - 2\beta + 8\beta + 2 = 0 \implies 6\beta + 3 = 0 \]
Solving for \(\beta\):
\[ 6\beta = -3 \implies \beta = -\frac{1}{2} \]

6. **Substitute \(\beta\) back to find \(\alpha\)**:
Substitute \(\beta = -\frac{1}{2}\) into Equation 1:
\[ \alpha + 2\left(-\frac{1}{2}\right) = 1 \implies \alpha - 1 = 1 \implies \alpha = 2 \]

### Final Values:
Thus, the values are:
\[ \alpha = 2, \quad \beta = -\frac{1}{2} \]