Question

Solve  $$\frac{1}{\left|x\right|-3}\le \frac{1}{2}$$

Answer 1

$$\text{ As },\frac{1}{\left|x\right|-3}\le \frac{1}{2}$$
$$\Rightarrow \frac{1}{\left|x\right|-3}-\frac{1}{2}\le 0$$
$$\Rightarrow \frac{2-(\left|x\right|-3)}{2(\left|x\right|-3)}\le 0$$
$$\Rightarrow \frac{2-\left|x\right|+3}{2(\left|x\right|-3)}\le 0$$
$$\Rightarrow \frac{5-\left|x\right|}{\left|x\right|-3}\le 0$$
$$\text{ Case I: When }x\ge 0,\left|x\right|=x,$$
$$\frac{5-x}{x-3}\le 0$$
$$\Rightarrow (5-x\le 0\text{ and }x-3>0)\text{ or }(5-x\ge 0\text{ and }x-3<0)$$
$$\Rightarrow (x\ge 5\text{ and }x>3)\text{ or }(x\le 5\text{ and }x<3)$$
$$\Rightarrow x\ge 5orx<3$$
$$\Rightarrow x\in (0,3)\cup [5,\mathrm{\infty })$$
$$\text{ Case II: When }x<0,\left|x\right|=-x,$$
$$\frac{5+x}{-x-3}\le 0$$
$$\Rightarrow \frac{x+5}{x+3}\ge 0$$
$$\Rightarrow (x+5>0\text{ and }x+3>0)or(x+5<0\text{ and }x+3<0)$$
$$\Rightarrow (x>-5\text{ and }x>-3)\text{ or }(x<-5\text{ and }x<-3)$$
$$\Rightarrow x>-3\text{ or }x<-5$$
$$\Rightarrow x\in (-\mathrm{\infty },-5)\cup (-3,\mathrm{\infty })$$
$$\text{ So, from both the cases, we get }$$
$$x\in (-\mathrm{\infty },-5)\cup (-3,\mathrm{\infty })\cup (0,3)\cup [5,\mathrm{\infty })$$
$$\therefore x\in (-\mathrm{\infty },-5]\cup (-3,3)\cup [5,\mathrm{\infty })$$