Answer :(-infinity, 0)

Answer :(-infinity, 0) Find the solution of llxl - II < II -xl.

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$$\begin{gathered} To Solve : \left|\left|x\right|-1\right|<\left|1-x\right| \\ Solution : Here we will have 4 Cases \\ Case 1 : x\le -1 \\ Checking modulus to be positive or negative by taking a value less than -1 \\ take -2 \\ \left|\left|-2\right|-1\right|<\left|1-(-2)\right| \\ =\left|-(-2)-1\right|<\left|1+2\right| \\ = \left|2-1\right|<\left|1+2\right| \\ = 1<3\left(true\right) \\ So, \\ \left(-\left(x\right)-1\right)<\left(1-x\right) \\ \Rightarrow \cancel{-x}-1<1\cancel{-x} \\ \Rightarrow -1<1 which is always true \\ \Rightarrow x can take every value in this set \\ \Rightarrow x\in (-\infty , -1] \\ . \\ Case 2 : -1<x<0 \\ Checking modulus to be positive or negative \\ by taking a value less than 0 and greater than -1 \\ take \frac{-1}{2} \\ \left|\left|\frac{-1}{2}\right|-1\right|<\left|1-\left(\frac{-1}{2}\right)\right| \\ =\left|-\left(\frac{-1}{2}\right)-1\right|<\left|1+\frac{1}{2}\right| \\ = \left|\frac{1}{2}-1\right|<\left|\frac{2+1}{2}\right| \\ = \left|\frac{1-2}{2}\right|<\left|\frac{3}{2}\right| \\ = \left|\frac{-1}{2}\right|<\frac{3}{2} \\ = -\left(\frac{-1}{2}\right)<\frac{3}{2} \\ = \frac{1}{2}<\frac{3}{2}\left(true\right) \\ So, \\ -\left(-\left(x\right)-1\right)<\left(1-x\right) \\ \Rightarrow -\left(-x-1\right)<1-x \\ \Rightarrow x+1<1-x \\ \Rightarrow x+x<1-1 \\ \Rightarrow 2x<0 \\ \Rightarrow x<0 \\ and the set we are examining is (-1, 0) \\ So, Solution will be x\in (-1, 0) \\ . \\ Case 3 : 0\le x<1 \\ Checking modulus to be positive or negative \\ by taking a value less than 1 and greater than or equal to 0 \\ take \frac{1}{2} \\ \left|\left|\frac{1}{2}\right|-1\right|<\left|1-\left(\frac{1}{2}\right)\right| \\ =\left|\frac{1}{2}-1\right|<\left|1-\frac{1}{2}\right| \\ = \left|\frac{1-2}{2}\right|<\left|\frac{2-1}{2}\right| \\ = \left|\frac{-1}{2}\right|<\left|\frac{1}{2}\right| \\ = -\left(\frac{-1}{2}\right)<\frac{1}{2} \\ = \frac{1}{2}<\frac{1}{2}(not true) \\ However, \\ -\left(x-1\right)<\left(1-x\right) \\ \Rightarrow \cancel{-x}+1<1\cancel{-x} \\ \Rightarrow 1<1 \\ \Rightarrow which is not true so, in this set there exist no solution \\ \Rightarrow x \in \{\varnothing \} \\ . \\ Case 4 : x\ge 1 \\ Checking modulus to be positive or negative \\ by taking a value greater than or equal to 1 \\ take 2 \\ \left|\left|2\right|-1\right|<\left|1-2\right| \\ =\left|2-1\right|<\left|-1\right| \\ = \left|1\right|<-(-1) \\ = 1<1(not true) \\ However, \\ \left(\right(x)+1)<-(1-x) \\ \Rightarrow \cancel{x+}1<-1\cancel{+x} \\ \Rightarrow 1<-1 \\ \Rightarrow which is a false Statement. so, in this set there exist no solution \\ \Rightarrow x \in \{\varnothing \} \\ So, Complete solution is \\ x\in (-\infty , -1] \cup (-1, 0) \\ =x\in (-\infty , 0] \end{gathered}$$
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