If the function `f(x)=2x^3-9ax^2+12a^2x+1` attains its maximum and minimum at p and q respectively such that `p^2=q` , then a equals

Options

A. 0
B. 1
C. 2
D. none of these

Answer

To solve the problem, we need to find the value of \( a \) such that the function \( f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 \) attains its maximum and minimum at points \( p \) and \( q \) respectively, with the condition that \( p^2 = q \).

### Step-by-Step Solution:

1. **Find the derivative of the function**:
We start by differentiating the function \( f(x) \).
\[ f'(x) = \frac{d}{dx}(2x^3 - 9ax^2 + 12a^2x + 1) = 6x^2 - 18ax + 12a^2 \]

2. **Set the derivative equal to zero**:
To find the critical points where the function attains maximum and minimum values, we set the derivative to zero:
\[ 6x^2 - 18ax + 12a^2 = 0 \]
Dividing the entire equation by 6 simplifies it to:
\[ x^2 - 3ax + 2a^2 = 0 \]

3. **Use the quadratic formula**:
We will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots \( p \) and \( q \):
\[ x = \frac{3a \pm \sqrt{(-3a)^2 - 4 \cdot 1 \cdot 2a^2}}{2 \cdot 1} = \frac{3a \pm \sqrt{9a^2 - 8a^2}}{2} = \frac{3a \pm \sqrt{a^2}}{2} \]
This simplifies to:
\[ x = \frac{3a \pm a}{2} \]
Thus, we have two solutions:
\[ p = \frac{4a}{2} = 2a \quad \text{and} \quad q = \frac{2a}{2} = a \]

4. **Apply the condition \( p^2 = q \)**:
We know from the problem statement that \( p^2 = q \). Substituting the values we found:
\[ (2a)^2 = a \]
This simplifies to:
\[ 4a^2 = a \]

5. **Rearranging the equation**:
Rearranging gives us:
\[ 4a^2 - a = 0 \]
Factoring out \( a \):
\[ a(4a - 1) = 0 \]

6. **Finding the values of \( a \)**:
This gives us two possible solutions:
\[ a = 0 \quad \text{or} \quad 4a - 1 = 0 \implies a = \frac{1}{4} \]

7. **Determine valid solutions**:
Since \( a = 0 \) does not yield a valid maximum and minimum for the function, we discard it. Thus, we have:
\[ a = \frac{1}{4} \]

### Final Answer:
The value of \( a \) is \( \frac{1}{4} \).