Key Concepts and Formulas

• First Derivative Test: Critical points of a function $f(x)$ occur where $f'(x) = 0$ or is undefined. These are potential locations for local extrema.
• Second Derivative Test: If $x=c$ is a critical point, then:
  • If $f''(c) > 0$, then $f(x)$ has a local minimum at $x=c$.
  • If $f''(c) < 0$, then $f(x)$ has a local maximum at $x=c$.
• Quadratic Equation Formation: If $r_1$ and $r_2$ are the roots of a quadratic equation, then the equation is given by $x^2 - (r_1 + r_2)x + r_1r_2 = 0$.

Step-by-Step Solution

Step 1: Find the first derivative of $f(x)$ and find critical points.

• Why: The first derivative helps us identify points where the function's slope is zero, which are potential locations for local maxima or minima.
• Working: Given $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$, we differentiate with respect to $x$: $f'(x) = \frac{d}{dx}(2x^3 - 9ax^2 + 12a^2x + 1) = 6x^2 - 18ax + 12a^2$
• To find the critical points, we set $f'(x) = 0$: $6x^2 - 18ax + 12a^2 = 0$
• Divide by 6 to simplify: $x^2 - 3ax + 2a^2 = 0$
• Factor the quadratic: $(x - a)(x - 2a) = 0$
• Therefore, the critical points are $x = a$ and $x = 2a$.

Step 2: Classify the critical points using the Second Derivative Test.

• Why: The Second Derivative Test helps us determine whether each critical point is a local maximum or a local minimum by examining the concavity of the function at those points.
• Working: Find the second derivative of $f(x)$: $f''(x) = \frac{d}{dx}(6x^2 - 18ax + 12a^2) = 12x - 18a$
• Evaluate $f''(x)$ at each critical point:
  • At $x = a$: $f''(a) = 12a - 18a = -6a$
  • At $x = 2a$: $f''(2a) = 12(2a) - 18a = 24a - 18a = 6a$
• Since $a > 0$:
  • $f''(a) = -6a < 0$, so $x = a$ is a local maximum.
  • $f''(2a) = 6a > 0$, so $x = 2a$ is a local minimum.

Step 3: Relate the critical points to $\alpha$ and $\alpha^2$.

• Why: The problem states that the local maximum occurs at $x = \alpha$ and the local minimum at $x = \alpha^2$. We use this information to relate $\alpha$ to $a$.
• Working:
  • The local maximum occurs at $x = a$, and we are given it occurs at $x = \alpha$. Therefore, $\alpha = a$.
  • The local minimum occurs at $x = 2a$, and we are given it occurs at $x = \alpha^2$. Therefore, $\alpha^2 = 2a$.

Step 4: Solve for $\alpha$ using the given conditions.

• Why: We have two equations with two unknowns, $\alpha$ and $a$. We can solve this system to find the value of $\alpha$.
• Working: We have the equations:
  1. $\alpha = a$
  2. $\alpha^2 = 2a$
• Substitute $a = \alpha$ from equation (1) into equation (2): $\alpha^2 = 2\alpha$
• Rearrange and solve for $\alpha$: $\alpha^2 - 2\alpha = 0$ $\alpha(\alpha - 2) = 0$
• This gives two possible solutions: $\alpha = 0$ or $\alpha = 2$.
• Since $a > 0$ and $\alpha = a$, we must have $\alpha > 0$. Therefore, $\alpha \neq 0$, and we conclude that $\alpha = 2$.

Step 5: Determine the roots of the required quadratic equation.

• Why: We need to find the quadratic equation whose roots are $\alpha$ and $\alpha^2$. We have already found the value of $\alpha$.
• Working:
  • We found $\alpha = 2$.
  • Then, $\alpha^2 = 2^2 = 4$.
• Thus, the roots of the required quadratic equation are 2 and 4.

Step 6: Construct the quadratic equation.

• Why: Knowing the roots, we can construct the quadratic equation using the relationship between the roots and the coefficients of the quadratic.
• Working:
  • Sum of the roots: $S = 2 + 4 = 6$
  • Product of the roots: $P = 2 \times 4 = 8$
• The quadratic equation is: $x^2 - Sx + P = 0$ $x^2 - 6x + 8 = 0$

This matches option (A).

Common Mistakes & Tips

• Remember the condition $a > 0$: This condition is crucial for eliminating the solution $\alpha = 0$.
• Verify max/min: Do not assume that the critical points are maxima or minima. Use the Second Derivative Test (or First Derivative Test) to confirm their nature.
• Factor correctly: Ensure correct algebraic manipulation when finding the roots of the quadratic equation.

Summary

This problem requires finding the local extrema of a cubic function using derivatives, relating them to given variables, and then constructing a quadratic equation from its roots. The key steps involve finding the first and second derivatives, classifying critical points, solving for the unknown variable $\alpha$, and constructing the quadratic equation. The constraint $a>0$ is essential for selecting the correct value of $\alpha$.

The final answer is \boxed{x^2-6x+8=0}, which corresponds to option (A).