Question

The solution set of the inequality $||x|-1<1-x,2\forall xϵR$ is equal to

• A. $(0,\infty )$
• B. $(-1,\infty )$
• C. $(-1,1)$
• D. $(-\infty ,0)$

Answer 1

The correct option is D $(-\infty ,0)$

$(-\infty ,0)$
Here because of |x| which is equalto either x or -x we shall consider the cases x>0 or x<0
Then we will have |x - 1| = |x + 1| when x < 0
Case 1: x > 0, then |x-1|<(1-x)
or |1-x| < (1-x) i.e. |t|<t
Above is not true for any value of t.
Case 2: $x<0,|-x-1|<1-x$
or $|x+1|<(1-x)\dots \dots \left(1\right)$
If $x+1\ge 0i.ex\ge -1$ and $x<0$
i.e. $-1\le x<0$ i.e. $xϵ[-1,0)$ then (1) reduces to
$x+1=1-x$ or $2x<0$ or $x<0$
If $x+1<0$ i.e. $x<-1$ then (1) reduce to
$-(x+1)<(1-x)$ or $-2<0$ is true for all $x<-1\dots \dots \left(2\right)$
Hence from (1) and (2) $xϵ(-\infty ,0)$