Question

The solution set of the inequality $|x+1|+|x|>3$ is

• A. $(-\infty ,-2)$
• B. $(1,\infty )$
• C. $(-\infty ,-2)\cup (1,\infty )$
• D. $(-\infty ,-2]\cup [1,\infty )$

Answer 1

Answer: (C)

On LHS of the given inequality, we have two terms both containing modulus. By equating the expression with in the modulus to zero, we get $x=-1,0$ as critical points. These critical points divide the real line in three parts as $(-\infty ,-1),[-1,0),[0,\infty )$
Case I When $-\infty <x<-1$,
$|x+1|+|x|>3\Rightarrow -x-1-x>3\Rightarrow x<-2$
Case II When $-1\le x<0$,
$|x+1|+|x|>3\Rightarrow x+1-x>3\Rightarrow 1>3\text{ (not possible) }$
Case III When $0\le x<\infty$,
$|x+1|+|x|>3\Rightarrow x+1+x>3\Rightarrow x>1\text{. }$
Combining the results of cases (I), (II) and (III), we get
$x\in (-\infty ,-2)\cup (1,\infty )$