Question

Write the solution of set of $\left|x+\frac{1}{x}\right|>2$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \left|x+\frac{1}{x}\right|>2 \\ \Rightarrow \left|x+\frac{1}{x}\right|-2>0 \\ \mathrm{CASE}1:\mathrm{When}x+\frac{1}{x}>0,\mathrm{then}\left|x+\frac{1}{x}\right|=x+\frac{1}{x} \\ \mathrm{Now},\left|x+\frac{1}{x}\right|-2>0 \\ \Rightarrow x+\frac{1}{x}-2>0 \\ \Rightarrow \frac{x^2+1-2x}{x}>0 \\ \Rightarrow \frac{(x-1{)}^2}{x}>0 \\ \Rightarrow x>0\mathrm{and}x\ne 1 \\ \Rightarrow x\in (0,1)\mathrm{U}(1,\infty )...\left(\mathrm{i}\right) \\ \mathrm{CASE}2:\mathrm{When}x+\frac{1}{x}<0,\mathrm{then}\left|x+\frac{1}{x}\right|=-(x+\frac{1}{x}) \\ \mathrm{Now},\left|x+\frac{1}{x}\right|-2>0 \\ \Rightarrow -x-\frac{1}{x}-2>0 \\ \Rightarrow \frac{-x^2-1-2x}{x}>0 \\ \Rightarrow \frac{x^2+1+2x}{x}<0\left[\mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}-1\right] \\ \Rightarrow \frac{(x+1{)}^2}{x}<0 \\ \Rightarrow x<0\mathrm{and}x\ne -1 \\ \Rightarrow x\in (-\infty ,-1)\mathrm{U}(-1,0)...\left(\mathrm{ii}\right) \\ \mathrm{Thus},\mathrm{the}\mathrm{solution}\mathrm{set}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{inequation}\mathrm{is}\mathrm{the}\mathrm{un}\mathrm{ion}\mathrm{of}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ (0,1)\mathrm{U}(1,\infty )\cup (-\infty ,-1)\mathrm{U}(-1,0)=R-\left\{-1,0,1\right\} \\ \therefore x\in R-\left\{-1,0,\right\} \end{gathered}$$