Key Concepts and Formulas

• Fermat's Theorem: If a function $f(x)$ has a local extremum (maximum or minimum) at $x=c$, and $f(x)$ is differentiable at $x=c$, then $f'(c) = 0$.
• Derivative of Logarithmic Function: $\frac{d}{dx}(\log|x|) = \frac{1}{x}$ for $x \neq 0$.
• Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.

Step-by-Step Solution

Step 1: Find the derivative of $f(x)$

We are given the function: $f(x) = \alpha \log|x| + \beta x^2 + x$ We need to find its derivative, $f'(x)$. We will differentiate each term with respect to $x$.

• Derivative of $\alpha \log|x|$: $\frac{d}{dx} (\alpha \log|x|) = \alpha \cdot \frac{1}{x} = \frac{\alpha}{x}$
• Derivative of $\beta x^2$: $\frac{d}{dx} (\beta x^2) = \beta \cdot 2x = 2\beta x$
• Derivative of $x$: $\frac{d}{dx} (x) = 1$

Therefore, the first derivative of $f(x)$ is: $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$

Step 2: Apply the condition for extreme points

We are given that $x=-1$ and $x=2$ are extreme points. According to Fermat's Theorem, this implies $f'(-1) = 0$ and $f'(2) = 0$.

• Substitute $x = -1$ into $f'(x)$: $f'(-1) = \frac{\alpha}{-1} + 2\beta(-1) + 1 = 0$ $-\alpha - 2\beta + 1 = 0$ Rearranging the equation: $\alpha + 2\beta = 1 \quad \ldots(i)$

• Substitute $x = 2$ into $f'(x)$: $f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = 0$ $\frac{\alpha}{2} + 4\beta + 1 = 0$ Multiplying the entire equation by 2 to eliminate the fraction: $\alpha + 8\beta + 2 = 0$ Rearranging the equation: $\alpha + 8\beta = -2 \quad \ldots(ii)$

Step 3: Solve the system of linear equations

We have the following system of equations: $\alpha + 2\beta = 1 \quad \ldots(i)$ $\alpha + 8\beta = -2 \quad \ldots(ii)$

We can solve this system using the elimination method. Subtract equation $(i)$ from equation $(ii)$: $(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - 1$ $6\beta = -3$ $\beta = -\frac{3}{6} = -\frac{1}{2}$

Now substitute the value of $\beta$ into equation $(i)$ to find $\alpha$: $\alpha + 2\left(-\frac{1}{2}\right) = 1$ $\alpha - 1 = 1$ $\alpha = 2$

Common Mistakes & Tips:

• Sign Errors: Pay close attention to signs during substitution and algebraic manipulation.
• Derivative of $\log|x|$: Remember that the derivative of $\log|x|$ is $\frac{1}{x}$ for all $x \neq 0$.
• Checking the solution: After finding the values of $\alpha$ and $\beta$, substitute them back into the original equations to verify the solution.

Summary

By finding the first derivative of the given function and applying the condition that the derivative is zero at extreme points, we derived a system of two linear equations. Solving this system allowed us to determine the values of $\alpha$ and $\beta$. The values are $\alpha = 2$ and $\beta = -\frac{1}{2}$.

The final answer is $\boxed{\alpha = 2,\beta = - {1 \over 2}}$, which corresponds to option (A).