Question

If $|x-1|+|x+1|=2$, then $x$ belongs to

• A. $(0,\infty )$
• B. $[-1,1]$
• C. $\{-1,1\}$
• D. $(-\infty ,0]$

Answer 1

The correct option is B $[-1,1]$

$|x-1|+|x+1|=2$

Case $-1:$ When $x\ge 1$
$x-1+x+1=2$
$\Rightarrow x=1\in [1,\infty )$

Case $-2:$ When $-1<x<1$
$-(x-1)+x+1=2$
$\Rightarrow -x+1+x+1=2$
$\Rightarrow 2=2$
$\therefore$ $x$ can have any value as it will satisfy the equation, but here the interval is $(-1,1)$
i.e. $-1<x<1$ this whole internal would satisfy case $2.$

Case $-3:$ When $x\le -1$
$-(x-1)-(x+1)=2$
$\Rightarrow -x+1-x-1=2$
$\Rightarrow -2x=2$
$\Rightarrow x=-1\in (-\infty ,-1]$
From $1\cup 2\cup 3$
$x\in [-1,1]$