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

Answer: (A)

$x^2 - 6x + 8 = 0$

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: $x^2 - 6x + 8 = 0.$