Question

If the range of values of parameter $a$ for which the point of minimum of the function $f\left(x\right)=1-a^{2}x+2x^3$ satisfy the inequality $\frac{x^2+2x+4}{x^2-\sqrt{6}x-36}<0$ is given by $(-b,b)-\left\{0\right\}$, then $b$ is

Answer 1

$\frac{x^2+2x+4}{x^2-\sqrt{6}x-36}<0$
$\Rightarrow x^2-\sqrt{6}x-36<0$
$\Rightarrow (x+2\sqrt{6})(x-3\sqrt{6})<0$
$\Rightarrow -2\sqrt{6}<x<3\sqrt{6}$

Now, $f\left(x\right)=1-a^{2}x+2x^3$
$f^{\prime }\left(x\right)=-a^2+6x^2$
$f^{\prime \prime }\left(x\right)=12x$
$f^{\prime }\left(x\right)=0$ gives $x=\pm \frac{a}{\sqrt{6}}$

When $a>0$, $f^{\prime \prime }\left(a/\sqrt{6}\right)>0$
$\Rightarrow x=a/\sqrt{6}$ is a point of minima.
So, $0<a/\sqrt{6}<3\sqrt{6}$
$\Rightarrow 0<a<18$

When $a<0$, $f^{\prime \prime }(-a/\sqrt{6})>0$
$\Rightarrow x=-a/\sqrt{6}$ is a point of minima.
So, $0<-a/\sqrt{6}<3\sqrt{6}$
$\Rightarrow -18<a<0$

Hence, $a\in (-18,18)-\left\{0\right\}$
So, the value of $b=18$