Question

The solution set of inequality |x -1| + |x + 1| < 4 is

• A. $(-\infty ,-2)$
• B. $(-2,2)$
• C. $(2,\infty )$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is B $(-2,2)$

Best way to solve such problem is by checking at different values of x.
At x = -3, |x -1| + |x + 1| = 6 is not less than 4.
At x = 3, |x -1| + |x + 1| = 6 is not less than 4.
Hence, option (a), (c), and (d) can be eliminated.
$\therefore$ Option (b) is correct answer.

Alternatively:
The solution will consist of three intervals
$x<-1,-1\le x<1andx\ge 1Ifx<-1,then|x-1|=-(x-1)and|x+1|=-(x+1)Weget-(x-1)-(x+1)<4\Rightarrow x>-2\therefore Wehavex<-1andx>-2\Rightarrow -2<x<-1\Rightarrow xϵ(-2,-1)If-1\le x<1,then|x-1|=-(x-1)and|x+1|=x+1.$
We get, $-(x-1)+(x+1)<4\Rightarrow 2<4,$ which is valid
$\therefore$ Inequality holds true for all x satisfying
$-1\le x<1\Rightarrow xϵ[-1,1)Ifx\ge 1,then|x-1|=(x-1)and|x+1|=x+1wegetx-1+x+1<4\Rightarrow x<2.\therefore Wehavex\ge 1andx<2\Rightarrow 1\le x<2\Rightarrow xϵ[1,2)$
Combining the solutions of the above three cases, we get
$xϵ(-2,-1)\cup [-1,1)\cup [1,2)\Rightarrow (-2,2)$