1. Given function

$f(x)=2x^3-9ax^2+12a^2x+1,\, a>0$

We are told that the function attains its maximum at $p$ and minimum at $q$, with

$p^2=q.$

We need to find $a$.

2. Find critical points

Differentiate:

$f'(x)=6x^2-18ax+12a^2$

Factor:

$f'(x)=6(x^2-3ax+2a^2)=6(x-a)(x-2a).$

So the critical points are:

$x=a,\quad x=2a.$

3. Identify maximum and minimum points

Use second derivative:

$f''(x)=12x-18a.$

Now check at the critical points:

• At $x=a$: $f''(a)=12a-18a=-6a<0$ since $a>0$. Hence, $x=a$ is a point of local maximum.

• At $x=2a$: $f''(2a)=24a-18a=6a>0$ Hence, $x=2a$ is a point of local minimum.

Therefore,

$p=a,\qquad q=2a.$

4. Use the condition $p^2=q$

Given:

$p^2=q$

Substitute $p=a$ and $q=2a$:

$a^2=2a$

$a(a-2)=0$

Since $a>0$, we must have

$a=2.$

5. Check options

• A: $\frac12$ ❌
• B: $3$ ❌
• C: $1$ ❌
• D: $2$ ✅

So the correct answer is:

$\boxed{2}$