Question

If the function \(f(x) = 2x^3 - 9ax^2 + 12a^2x + 1, \, a>0\) has a local maximum at \(x = \alpha\) and a local minimum at \(x = \alpha^2\), then \(\alpha\) and \(\alpha^2\) are the roots of the equation:

• A. \(x^2 - 6x + 8 = 0\)
• B. \(8x^2 + 6x - 8 = 0\)
• C. \(8x^2 - 6x + 1 = 0\)
• D. \(x^2 + 6x + 8 = 0\)

Answer 1

The Correct Option is A

\( f(x) = 6x^2 - 18ax + 12a^2 = 0 < a^2 \) 

\( \alpha + \alpha^2 = 3a \, \text{and} \, \alpha \times \alpha^2 = 2a^2 \)

\( \Rightarrow (\alpha + \alpha^2)^3 = 27a^3 \)

\( \Rightarrow 2a^2 + 4a^4 + 3(3a)(2a^2) = 27a^3 \)

\( \Rightarrow 2 + 4a^2 + 18a = 27a \)

\( \Rightarrow 4a^2 - 9a + 2 = 0 \)

\( \Rightarrow 4a^2 - 8a - a + 2 = 0 \)

\( \Rightarrow (4a - 1)(a - 2) = 0 \Rightarrow a = 2 \)

\( \text{So, } 6x^2 - 36x + 48 = 0 \)

\( \Rightarrow x^2 - 6x + 8 = 0 \, \text{(1)} \)

If we take \( a = \dfrac{1}{4} \), then \( \alpha = \dfrac{1}{2} \), which is not possible.

Answer 2

Given the function: \[ f(x) = 2x^3 - 9ax^2 + 12a^2 x + 1. \]

To find the critical points, we differentiate: \[ f'(x) = 6x^2 - 18ax + 12a^2. \] 

Given that \( f'(x) = 0 \) at \( x = \alpha \) (local maximum) and \( x = \alpha^2 \) (local minimum), 

we have: \[ 6\alpha^2 - 18a\alpha + 12a^2 = 0 \quad \text{and} \quad 6(\alpha^2)^2 - 18a\alpha^2 + 12a^2 = 0. \] 

Factoring out 6 from both equations: \[ \alpha^2 - 3a\alpha + 2a^2 = 0 \quad \text{and} \quad \alpha^4 - 3a\alpha^2 + 2a^2 = 0. \] 

From these equations, we observe that \(\alpha\) and \(\alpha^2\) are the roots of the quadratic equation: \[ x^2 - 3ax + 2a^2 = 0. \] 

Given the relationships: \[ \alpha + \alpha^2 = 3a \quad \text{and} \quad \alpha \times \alpha^2 = 2a^2. \] 

Substituting these into the expression \((\alpha + \alpha^2)^3\): \[ (\alpha + \alpha^2)^3 = 27a^3. \] 

Expanding: \[ 2a^2 + 4a^4 + 3(3a)(2a^2) = 27a^3. \] 

Simplifying: \[ 2 + 4a^2 + 18a = 27a. \] 

Rearranging terms: \[ 4a^2 - 9a + 2 = 0. \] 

Factoring: \[ (4a - 1)(a - 2) = 0. \] Since \( a > 0 \), we have: \[ a = 2. \] Substituting \( a = 2 \) back into the equation for \(\alpha\) and \(\alpha^2\): \[ 6x^2 - 36x + 48 = 0. \] 

Dividing by 6: \[ x^2 - 6x + 8 = 0. \] 

Therefore: \[ x^2 - 6x + 8 = 0. \]