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Step-by-Step Solution
Step 1: Write down the function
The given function is
$f(x) = 2x^3 - 9a\,x^2 + 12a^2 x + 1$,
where $a>0$.
Step 2: Find the first derivative
To locate potential maxima or minima, we compute the first derivative and set it to zero:
$f'(x) = \frac{d}{dx} \bigl(2x^3 - 9a\,x^2 + 12a^2 x + 1\bigr) = 6x^2 - 18a\,x + 12a^2.$
We then solve $f'(x) = 0$ for $x$.
Step 3: Solve for critical points
Set $6x^2 - 18a\,x + 12a^2 = 0$. Dividing by 6 gives
$x^2 - 3a\,x + 2a^2 = 0.$
This quadratic factors as $(x - a)(x - 2a) = 0,$ so
$x = a \quad \text{or} \quad x = 2a.$
Step 4: Find the second derivative and determine max/min
The second derivative is
$f''(x) = \frac{d}{dx} \bigl(6x^2 - 18a\,x + 12a^2\bigr) = 12x - 18a.$
At $x = a$, $f''(a) = 12a - 18a = -6a$. Since this is negative, $f$ has a local maximum at $x=a$.
At $x = 2a$, $f''(2a) = 12(2a) - 18a = 24a - 18a = 6a$. Since this is positive, $f$ has a local minimum at $x=2a$.
Hence, $p = a$ (maximum) and $q = 2a$ (minimum).
Step 5: Apply the given condition
The problem states: $p^2 = q.$ Substituting $p=a$ and $q=2a$ gives
$a^2 = 2a.$
Rewriting this, we get
$a(a - 2) = 0.$
Therefore,
$a = 0 \quad \text{or} \quad a = 2.$
Since $a > 0,$ we choose $a = 2.$
Step 6: Final result
Thus, the value of $a$ that satisfies the given conditions is
$a = 2.$
