Question

Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1)x^2+(1-\sqrt{1+a})x+\sqrt[3]{1+a}=\sqrt[6]{1+a}$, where $a>0$, $\alpha(a) > \beta(a)$. If minimum value of expression $f(a)=\alpha(a)+12\beta(a)+\frac{4}{\alpha(a)(3\beta(a))^2},a>0$ is m, then value of [m] is

[Note: [m] denotes the largest integer less than or equal to m]

Answer 1

8

Answer 2

Let $y = \sqrt[6]{1+a}$. Since $a>0$, $y>1$. The given equation can be written as: $(y^2-1)x^2 + (1-y^3)x + y^2 = y$ $(y^2-1)x^2 - (y^3-1)x + (y^2-y) = 0$ Dividing by $y-1$ (since $y>1$, $y-1 \neq 0$): $(y+1)x^2 - (y^2+y+1)x + y = 0$ We observe that $x=y$ is a root. The product of roots is $\alpha \beta = \frac{y}{y+1}$. If $\alpha = y$, then $\beta = \frac{1}{y+1}$. Since $y>1$, we have $y > \frac{1}{y+1}$, so $\alpha(a) = y = \sqrt[6]{1+a}$ and $\beta(a) = \frac{1}{y+1} = \frac{1}{\sqrt[6]{1+a}+1}$.

The expression to minimize is $f(a)=\alpha(a)+12\beta(a)+\frac{4}{\alpha(a)(3\beta(a))^2}$. Substituting $\alpha=y$ and $\beta=\frac{1}{y+1}$: $f(a) = y + 12\left(\frac{1}{y+1}\right) + \frac{4}{y\left(3\cdot\frac{1}{y+1}\right)^2}$ $f(a) = y + \frac{12}{y+1} + \frac{4(y+1)^2}{9y}$

Let $g(y) = y + \frac{12}{y+1} + \frac{4(y+1)^2}{9y}$ for $y>1$. To find the minimum value, we compute the derivative with respect to $y$: $g'(y) = 1 - \frac{12}{(y+1)^2} + \frac{4}{9} \cdot \frac{2(y+1)y - (y+1)^2}{y^2}$ $g'(y) = 1 - \frac{12}{(y+1)^2} + \frac{4}{9} \cdot \frac{(y+1)(2y - (y+1))}{y^2}$ $g'(y) = 1 - \frac{12}{(y+1)^2} + \frac{4}{9} \cdot \frac{(y+1)(y-1)}{y^2}$ This derivative calculation seems complicated. Let's re-evaluate $g(y)$ first. $g(y) = y + \frac{12}{y+1} + \frac{4(y^2+2y+1)}{9y} = y + \frac{12}{y+1} + \frac{4y}{9} + \frac{8}{9} + \frac{4}{9y}$ $g(y) = \frac{13y}{9} + \frac{12}{y+1} + \frac{4}{9y} + \frac{8}{9}$. Now, $g'(y) = \frac{13}{9} - \frac{12}{(y+1)^2} - \frac{4}{9y^2}$. Setting $g'(y)=0$: $\frac{13}{9} = \frac{12}{(y+1)^2} + \frac{4}{9y^2}$. By inspection, $y=2$ satisfies this equation: $\frac{13}{9} = \frac{12}{(2+1)^2} + \frac{4}{9(2^2)} = \frac{12}{9} + \frac{4}{36} = \frac{12}{9} + \frac{1}{9} = \frac{13}{9}$. The second derivative $g''(y) = \frac{24}{(y+1)^3} + \frac{8}{9y^3}$. For $y>1$, $g''(y)>0$, indicating a minimum. The minimum value occurs at $y=2$. $m = g(2) = 2 + \frac{12}{2+1} + \frac{4(2+1)^2}{9(2)} = 2 + \frac{12}{3} + \frac{4(9)}{18} = 2 + 4 + 2 = 8$. The value of $[m] = [8] = 8$.