Question

Let $\alpha \left(a\right)$ and $\beta \left(a\right)$ be the roots of the equation
$(\sqrt[3]{31+a}-1)x^2+(1-\sqrt{1+a})x+\sqrt[3]{31+a}=\sqrt[6]{61+a},$ where $a>0,\alpha \left(a\right)>\beta \left(a\right).$ If minimum value of expression $f\left(a\right)=\alpha \left(a\right)+\left(12\beta \right(a){)}^2+\frac{4}{\alpha \left(a\right)\left(3\beta \right(a){)}^2},a>0$ is $m$, then value of $\left[m\right]$ is
[ Note: $\left[m\right]$ represents the greatest integer less than or equal to $m$ ]

Answer 1

Let $(1+a{)}^{1/6}=t$
$(t^2-1)x^2+(1-t^3)x+t^2=t$
$\because a>0\Rightarrow t-1\ne 0$
$\Rightarrow x^2(t+1)+(-t^2-t-1)x+t=0$
$\Rightarrow x(t+1)(x-t)-1(x-t)=0$
$\Rightarrow [x-t]\left[x\right(t+1)-1]=0$
we get $\alpha \left(a\right)=t=(1+a{)}^{1/6}$
$\beta \left(a\right)=\frac{1}{t+1}=\frac{1}{1+(1+a{)}^{1/6}}$
$\therefore f\left(a\right)=(1+a{)}^{1/6}+{\left(\frac{12}{1+(1+a{)}^{1/6}}\right)}^2+\frac{4[1+(1+a{)}^{1/6}{]}^2}{9(1+a{)}^{1/6}}$
By using $A.M.\ge G.M.$
$\frac{f\left(a\right)}{3}\ge \sqrt[3]{3(1+a{)}^{1/6}\times \frac{144}{[1+(1+a{)}^{1/6}{]}^2}\times \frac{4[1+(1+a{)}^{1/6}{]}^2}{9(1+a{)}^{1/6}}}$
$\Rightarrow \frac{f\left(a\right)}{3}\ge \sqrt[3]{364}$
$\Rightarrow f\left(a\right)\ge 12$