The set of all values of the parameters $a$ for which the points of minimum of the function $y=1+a^{2}x$ 

$-x^3$ satisfy the inequality $\frac{x^2+x+2}{x^2+5x+6}\le 0$ is

 

 

 

 

$\frac{\mathrm{d}y}{ \mathrm{d}x}=a^2-3x^2=0\iff x=\pm a/\sqrt{3}\text{. Since }\frac{{\mathrm{d}}^{2}y}{ \mathrm{d}{x}^2}$

$=-6x\text{ so }y$ is minimum for $x=-a/\sqrt{3}$

Since $x^2+x+2>0\text{ for all }x\text{ so for }\frac{x^2+x+2}{x^2+5x+6}\le 0$we must

have $x^2+5x+6<0.\text{ If }x=-a/\sqrt{3}$ we have $a^2/3-5a/\sqrt{3}$

$+6<0\text{ i.e. }{a}^2-5\sqrt{3}a+18<0\iff (a-2\sqrt{3})(a-3\sqrt{3})$

$<0\text{ i.e. }a\in (2\sqrt{3},3\sqrt{3})$