Question

Solve $\frac{1}{\left|x\right|-3}\le \frac{1}{2}$ [NCERT EXEMPLAR]

Answer 1

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