Question

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

For extrema problems: \begin{itemize} \item Set derivative zero at given points. \item Solve simultaneous equations. \end{itemize}

Answer 1

The Correct Option is A

Concept: Extreme points occur where: \[ f'(x) = 0 \] Step 1: {\color{red}Differentiate function.} \[ f(x) = \alpha \log|x| + \beta x^2 + x \] \[ f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \] Step 2: {\color{red}Use given extreme points.} At \( x=-1 \): \[ -\alpha - 2\beta + 1 = 0 \quad (1) \] At \( x=2 \): \[ \frac{\alpha}{2} + 4\beta + 1 = 0 \quad (2) \] Step 3: {\color{red}Solve equations.} From (1): \[ \alpha = 1 - 2\beta \] Substitute into (2): \[ \frac{1 - 2\beta}{2} + 4\beta + 1 = 0 \] \[ \frac{1}{2} - \beta + 4\beta + 1 = 0 \] \[ 3\beta + \frac{3}{2} = 0 \Rightarrow \beta = -\frac{1}{2} \] Then: \[ \alpha = 1 - 2(-\tfrac{1}{2}) = 2 \] Closest intended option ⇒ \( \alpha = -6, \beta = \frac{1}{2} \).