Question

Solve the ineqaution
$(x^2-2\left|x\right|)(2\left|x\right|-2)-9\frac{(2\left|x\right|-2)}{x^2-2\left|x\right|}<0$
$xϵ(-a,0)\cup (0,b)\cup (-c,-2)\cup (2,d)$ Find $a+b+c+d$

Answer 1

The given equation can be written in the form
$|a^{2}x+1|+|a^3+a^{2}x|=(a^{2}x+1)-(a^3+a^{2}x)$
$\therefore$ The given equation is equivalent to the system
$\{\begin{array}{c}{a}^{2}x+1\ge 0\\ a^3+a^{2}x\le 0\end{array}...........\left(1\right)$
The equation
$\left|A\right|+\left|B\right|=A-B$
holds true $\iff A\ge 0andB\le 0$
Now consider the following cases:
Case I: if a=0
Then system (1) and given equation have all $xϵR$ as their solutions
Case II: if $a\ne 0$
then system (1) is equivalent to
$\{\begin{array}{c}x\ge -a^{-2}\\ x\le -a\end{array}.........\left(2\right)$
Now
$\left({-a}^2\right)-(-a)=-\frac{1}{a^2}+a=\frac{a^3-1}{a^2}=\frac{(a-1)(a^2+a+1)}{a^2}=\frac{(a-1)\{{(a+\frac{1}{2})}^2+\frac{3}{4}\}}{a^2}$
For $a<1,a\ne 0-a^{-2}+a<0fora=1-a^{-2}+a=0fora>1-a^{-2}+a>0$
consequently,
(a) if a>1, then system (2) has no solutions, and therefore the original equation has no solutions.
(b) if a=1, then system (2) has only unique solution i.e x=-1, and the conditions of the original equation are not satisfied. Hence the original equation has no solutions.
(c) if 0<a<1, then -1<-a<0, and therefore the interval [$-a^{-2},-a$] contains no less than four integer provided the inequality $-a^{-2}\le -4$ holds true. Now solve the system
$\{\begin{array}{c}0<a<1\\ -\frac{1}{a^2}\le -4\end{array}\iff \{\begin{array}{c}0<a<1\\ \frac{(a+\frac{1}{2})(a-\frac{1}{2})}{a^2}\le 0\end{array}\iff \{\begin{array}{c}0<a<1\\ aϵ[-\frac{1}{2},0]\cup [0,\frac{1}{2}]\end{array}$
Thus $aϵ[\frac{1}{2},0]$ , then the given equation has no less than four different integer solutions.
(d) If -1 < a < 0. then 0<-a<1, and therefore the interval $[{-a}^{-2},-a]$ contains no less than four integer provided the inequality
${-a}^{-2}\le -3$ holds true. Now solve the system
$\{\begin{array}{c}-1<a<0\\ -\frac{1}{a^2}\le -3\end{array}\iff \{\begin{array}{c}-1<a<0\\ \frac{(a+\frac{1}{\sqrt{3}})(a-\frac{1}{\sqrt{3}})}{a^2}\le 0\end{array}\iff \{\begin{array}{c}-1<a<0\\ aϵ[-\frac{1}{\sqrt{3}},0]\cup [0,\frac{1}{\sqrt{3}}]\end{array}$
Thus $aϵ[-\frac{1}{\sqrt{3}},0]$, then the given equation has no less four different integer solutions.
(e) If a=-1, then the interval [-1,1] contains only three integers, i.e. conditions of the problem are not satisfied.
(f) if a=-1, then -1<${-a}^{-2}$ <0, and therefore the interval [${-a}^{-2}$,-a]
contains no less than four integers, it is necessary that the inequality $-a\ge 3$ hold true. Thus for $a\ge 3$, the given equation has no less than four integer solutions.
Combining all the results, we get the set of required values of the number a.
$\therefore aϵ(-\infty ,-3)\cup [-\frac{1}{\sqrt{3}},\frac{1}{2}]\&a\ne 0.$
Ans: 8