Key Concepts and Formulas

• First Derivative Test: A critical point $c$ of a function $f(x)$ occurs when $f'(c) = 0$ or $f'(c)$ is undefined. These points are potential locations for local maxima or minima.
• Second Derivative Test: If $f'(c) = 0$, then:
  • If $f''(c) > 0$, $f(x)$ has a local minimum at $x = c$.
  • If $f''(c) < 0$, $f(x)$ has a local maximum at $x = c$.
  • If $f''(c) = 0$, the test is inconclusive.

Step-by-Step Solution

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

We need to find the critical points of $f(x)$, so we start by finding its first derivative $f'(x)$. $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ $f'(x) = 6x^2 - 18ax + 12a^2$

Step 2: Find the critical points by setting $f'(x) = 0$

To find the critical points, we set the first derivative equal to zero and solve for $x$. $6x^2 - 18ax + 12a^2 = 0$ Divide by 6: $x^2 - 3ax + 2a^2 = 0$ Factor the quadratic: $(x - a)(x - 2a) = 0$ So, the critical points are $x = a$ and $x = 2a$.

Step 3: Find the second derivative of $f(x)$

To determine whether the critical points are local maxima or minima, we find the second derivative $f''(x)$. $f'(x) = 6x^2 - 18ax + 12a^2$ $f''(x) = 12x - 18a$

Step 4: Apply the second derivative test to determine the nature of the critical points

Evaluate $f''(x)$ at the critical points $x = a$ and $x = 2a$.

• At $x = a$: $f''(a) = 12a - 18a = -6a$ Since $a > 0$, $f''(a) = -6a < 0$. Therefore, $f(x)$ has a local maximum at $x = a$. So, $p = a$.

• At $x = 2a$: $f''(2a) = 12(2a) - 18a = 24a - 18a = 6a$ Since $a > 0$, $f''(2a) = 6a > 0$. Therefore, $f(x)$ has a local minimum at $x = 2a$. So, $q = 2a$.

Step 5: Use the given relationship $p^2 = q$ to solve for $a$

We are given that $p^2 = q$. Substituting $p = a$ and $q = 2a$, we get: $a^2 = 2a$ $a^2 - 2a = 0$ $a(a - 2) = 0$ So, $a = 0$ or $a = 2$. Since we are given that $a > 0$, we must have $a = 2$.

Step 6: Verify the solution

If $a=2$, then $p=a=2$ and $q=2a=4$. Then $p^2 = 2^2 = 4 = q$, so the condition $p^2=q$ is satisfied.

Common Mistakes & Tips

• Remember to consider the condition $a > 0$ when solving for $a$. Discard any non-positive solutions.
• Carefully calculate the first and second derivatives. Double-check your work to avoid errors.
• Don't forget to factor the quadratic equation correctly to find the critical points.

Summary

We found the critical points of the given function by taking the first derivative and setting it to zero. We then used the second derivative test to classify these critical points as local maxima or minima. Finally, we used the given relationship between the locations of the maximum and minimum ($p^2 = q$) to solve for the value of $a$.

The final answer is \boxed{2}, which corresponds to option (D).