For $\frac{\left(\right|x-1\left|\right)}{(x+2)}-1<0$, $x$ lies in the interval

The correct option is A

$(-\infty ,–2)\cup (-\frac{1}{2},\infty )$

Step1: When $\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}$then

Given, $\frac{\left(\right|x-1\left|\right)}{(x+2)}-1<0$

$\Rightarrow$ $\frac{(1-x)}{(x+2)}-1<0$

$\Rightarrow$$\frac{(1-x)-(x+2)}{(x+2)}<0$

$\Rightarrow$ $\frac{-2x-1}{x+2}<0$

$\Rightarrow$ $\frac{2x+1}{x+2}>0$

Now,

$x+2>0$

$\Rightarrow$ $x<-2$

$\Rightarrow$$2x+1>0$

$\Rightarrow$ $x>\frac{-1}{2}$

Examining the expression $\frac{2x+1}{x+2}>0$

When $x\in (–\infty ,–2);x\in (-1/2,\infty )$here $x<1$

$\therefore x\in (-\infty ,–2)\cup (–(1/2),1)\mathbf{}\mathbf{}\mathbf{}\mathbf{—}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Step 2: When $\mathit{x}\mathbf{}\mathbf{\ge }\mathbf{}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{|}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{|}\mathbf{}\mathbf{=}\mathbf{}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}$

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

$\Rightarrow$$\frac{(x-1)-(x+2)}{(x+2)}<0$

$\Rightarrow$ $\frac{-3}{(x+2)}<0$

$\Rightarrow$ $x+2<0$

$\Rightarrow$ $x>-2$

But $x\ge 1$

$\therefore x\in [1,\infty )\mathbf{-}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Step 3. Combining equation (1) and (2) , we get

$\mathit{x}\mathbf{}\mathbf{\in }\mathbf{}\mathbf{(}\mathbf{-}\mathbf{}\mathbf{\infty }\mathbf{,}\mathbf{}\mathbf{–}\mathbf{}\mathbf{2}\mathbf{)}\mathbf{}\mathbf{\cup }\mathbf{}\mathbf{(}\mathbf{}\mathbf{–}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{/}\mathbf{}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{\infty }\mathbf{)}$

Hence, option “A” is correct .