Question

The set of all values of ${}^{\prime }x$ which satisfies the inequation$|1-\frac{|x|}{1+|x|}|\ge \frac{1}{2}$ belongs to

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

Answer 1

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

$|1-\frac{|x|}{1+|x|}|\le \frac{1}{2}$

$+ve$

$1-\frac{|x|}{1+|x|}\ge \frac{1}{2}$,

$\frac{|x|}{1+|x|}\le \frac{1}{2}$

$\frac{|x|}{1|x|}-\frac{1}{2}\le 0$

$\frac{2|x|-1-|x|\le 0}{2(1+|x|)}$

$\frac{|x|-1}{(1+|x|)}\le 0$

$|x|\in (-1,2]$

$-ve$

$1-\frac{|x|}{1+|x|}\le \frac{-1}{2}$

$\frac{|x|}{1+|x|}\le \frac{3}{2}$

$\frac{|x|}{1+|x|}-\frac{3}{2}\le 0$

$\frac{2|x|-3-|x|}{(1+|x|)}\le 0$

$\frac{3+|x|}{1+|x|}\ge 0$

$|n|\in (-\infty ,-3]\cup (-1,\infty )$

$|n|\in [-1,1]$