Question

Solve $\left|x+1\right|+\left|x\right|>3$ [NCERT EXEMPLAR]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\left|x+1\right|+\left|x\right|>3 \\ \mathrm{As},\left|x+1\right|=\left\{\begin{array}{l}\left(x+1\right),x\ge -1\\ -\left(x+1\right),x<-1\end{array}\right. \\ \mathrm{and}\left|x\right|=\left\{\begin{array}{l}x,x\ge 0\\ -x,x<0\end{array}\right. \\ \mathrm{Case}\mathrm{I}:\mathrm{When}x<-1, \\ \left|x+1\right|+\left|x\right|>3 \\ \Rightarrow -\left(x+1\right)-x>3 \\ \Rightarrow -2x-1>3 \\ \Rightarrow -2x>4 \\ \Rightarrow x<\frac{4}{-2} \\ \Rightarrow x<-2 \\ \mathrm{So},x\in \left(-\infty ,-2\right) \\ Case\mathrm{II}:\mathrm{When}-1\le x<0, \\ \left|x+1\right|+\left|x\right|>3 \\ \Rightarrow \left(x+1\right)-x>3 \\ \Rightarrow 1>3,\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{possible} \\ \mathrm{So},x\in \varphi \\ Case\mathrm{III}:\mathrm{When}x\ge 0, \\ \left|x+1\right|+\left|x\right|>3 \\ \Rightarrow \left(x+1\right)+x>3 \\ \Rightarrow 2x+1>3 \\ \Rightarrow 2x>2 \\ \Rightarrow x>\frac{2}{2} \\ \Rightarrow x>1 \\ So,x\in \left(1,\infty \right) \\ \mathrm{From}\mathrm{all}\mathrm{the}\mathrm{cases},\mathrm{we}\mathrm{get} \\ x\in \left(-\infty ,-2\right)\cup \left(1,\infty \right) \end{gathered}$$