Question

If \( x = -1 \) and \( x = 2 \) are extreme points of \( f(x) = \alpha \log|x| + \beta x^2 + x \), \( x \neq 0 \), then:

• A. \( \alpha = -6, \beta = \frac{1}{2} \)
• B. \( \alpha = -6, \beta = -\frac{1}{2} \)
• C. \( \alpha = 2, \beta = -\frac{1}{2} \)
• D. \( \alpha = 2, \beta = \frac{1}{2} \)

Hint:

Setting the first derivative to zero at the given extreme points yields a system of equations that can be solved for the unknown parameters.

Answer 1

The Correct Option is C

Step 1: Find the first derivative of \( f(x) \).
The function is given by \( f(x) = \alpha \log|x| + \beta x^2 + x \). The derivative \( f'(x) \) is: \[ f'(x) = \frac{d}{dx} (\alpha \log|x| + \beta x^2 + x) = \frac{\alpha}{x} + 2\beta x + 1 \]
Step 2: Use the fact that extreme points occur where the first derivative is zero.
Given that \( x = -1 \) and \( x = 2 \) are extreme points, we have \( f'(-1) = 0 \) and \( f'(2) = 0 \). For \( x = -1 \): \[ f'(-1) = \frac{\alpha}{-1} + 2\beta (-1) + 1 = -\alpha - 2\beta + 1 = 0 \] For \( x = 2 \): \[ f'(2) = \frac{\alpha}{2} + 2\beta (2) + 1 = \frac{\alpha}{2} + 4\beta + 1 = 0 \]
Step 3: Solve the system of linear equations for \( \alpha \) and \( \beta \).
We have the system:
1) \( -\alpha - 2\beta = -1 \)
2) \( \alpha + 8\beta = -2 \) (multiplying the second equation by 2)
Adding the two equations: \[ (-\alpha - 2\beta) + (\alpha + 8\beta) = -1 + (-2) \] \[ 6\beta = -3 \] \[ \beta = -\frac{1}{2} \] Substitute \( \beta = -\frac{1}{2} \) into the first equation: \[ -\alpha - 2\left(-\frac{1}{2}\right) = -1 \] \[ -\alpha + 1 = -1 \] \[ -\alpha = -2 \] \[ \alpha = 2 \] Thus, \( \alpha = 2 \) and \( \beta = -\frac{1}{2} \).