Find the range of root[(1 - mod x)/(2 - mod x)]. Please reply fast its very urgent.

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$$\begin{gathered} \mathbf{Always}\mathbf{ }\mathbf{keep}\mathbf{ }\mathbf{one}\mathbf{ }\mathbf{thing}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{ }\mathbf{mind}\mathbf{.} \\ \mathbf{Whenever}\mathbf{ }\mathbf{question}\mathbf{ }\mathbf{asks}\mathbf{ }\mathbf{about}\mathbf{ }\mathbf{domain}\mathbf{,}\mathbf{ }\mathbf{we}\mathbf{ }\mathbf{need}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{find}\mathbf{ }\mathbf{condition}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{x} \\ \mathbf{and}\mathbf{ }\mathbf{whenever}\mathbf{ }\mathbf{question}\mathbf{ }\mathbf{asks}\mathbf{ }\mathbf{about}\mathbf{ }\mathbf{range}\mathbf{ }\mathbf{we}\mathbf{ }\mathbf{need}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{find}\mathbf{ }\mathbf{condition}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{y}\mathbf{.} \\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{y}=\sqrt{\frac{1-\left|\mathrm{x}\right|}{2-\left|\mathrm{x}\right|}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \left|\mathrm{x}\right|=\left\{\begin{array}{ll}\mathrm{x}& \mathrm{x}\ge 0\\ -\mathrm{x}& \mathrm{x}<0\end{array}\right. \\ \mathrm{Hence} \mathrm{for} \mathrm{x}\ge 0 \\ \mathrm{y}=\sqrt{\frac{1-\mathrm{x}}{2-\mathrm{x}}} \\ \mathrm{As} \mathrm{sqaure} \mathrm{root} \mathrm{cannot} \mathrm{be} \mathrm{negative}. \\ \Rightarrow \mathrm{y}\ge 0 ;\mathrm{condition}\left(\mathrm{i}\right) \\ \mathrm{Now} \\ {\mathrm{y}}^2=\frac{1-\mathrm{x}}{2-\mathrm{x}} \\ \Rightarrow {\mathrm{y}}^2\left(2-\mathrm{x}\right)=1-\mathrm{x} \\ \Rightarrow 2{\mathrm{y}}^2-{\mathrm{xy}}^2=1-\mathrm{x} \\ \Rightarrow 2{\mathrm{y}}^2-1=\mathrm{x}\left({\mathrm{y}}^2-1\right) \\ \Rightarrow \mathrm{x}=\frac{2{\mathrm{y}}^2-1}{{\mathrm{y}}^2-1} \\ \mathrm{As} \mathrm{x}\ge 0 \\ \Rightarrow \frac{2{\mathrm{y}}^2-1}{{\mathrm{y}}^2-1}\ge 0 \\ \Rightarrow \frac{{\mathrm{y}}^2-{\displaystyle \frac{1}{2}}}{{\mathrm{y}}^2-1^2}\ge 0 \\ \Rightarrow \frac{{\mathrm{y}}^2-{\left(\frac{1}{\sqrt{2}}\right)}^2}{{\mathrm{y}}^2-1^2}\ge 0 \\ \Rightarrow \frac{\left(\mathrm{y}-{\displaystyle \frac{1}{\sqrt{2}}}\right)\left(\mathrm{y}+{\displaystyle \frac{1}{\sqrt{2}}}\right)}{\left(\mathrm{y}-1\right)\left(\mathrm{y}+1\right)}\ge 0 \\ \mathrm{By} \mathrm{using} \mathrm{method} \mathrm{of} \mathrm{interval} \left(\mathrm{wavy} \mathrm{curve}\right) \mathrm{we} \mathrm{get}, \\ \mathrm{y}\in \left(-\infty ,-1\right)\cup \left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]\cup \left(1,\infty \right) ; \mathrm{condition}\left(\mathrm{ii}\right) \\ \mathrm{Taking} \mathrm{intersection}\left(\mathrm{common}\right) \mathrm{of} \mathrm{condition}\left(\mathrm{i}\right) \mathrm{and} \mathrm{condition}\left(\mathrm{ii}\right) \mathrm{we} \mathrm{get}, \\ \mathrm{y}\in \left[0,\frac{1}{\sqrt{2}}\right]\cup \left(1,\infty \right) \\ \mathrm{Now} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\frac{1-\left|\mathrm{x}\right|}{2-\left|\mathrm{x}\right|}} \\ \Rightarrow \mathrm{f}\left(-\mathrm{x}\right)=\sqrt{\frac{1-\left|-\mathrm{x}\right|}{2-\left|-\mathrm{x}\right|}} \\ \Rightarrow \mathrm{f}\left(-\mathrm{x}\right)=\sqrt{\frac{1-\left|\mathrm{x}\right|}{2-\left|\mathrm{x}\right|}} \\ \Rightarrow \mathrm{f}\left(-\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{an} \mathrm{even} \mathrm{function} \mathrm{which} \mathrm{is} \mathrm{symmetric} \mathrm{about} \mathrm{y}-\mathrm{axis}. \\ \Rightarrow \mathrm{We} \mathrm{will} \mathrm{have} \mathrm{same} \mathrm{range} \left(\mathrm{y}\in \left[0,\frac{1}{\sqrt{2}}\right]\cup \left(1,\infty \right)\right) \mathrm{for} \mathrm{x}<0. \\ \mathbf{Hence}\mathbf{ }\mathbf{range}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\sqrt{\frac{\mathbf{1}\mathbf{-}\left|\mathbf{x}\right|}{\mathbf{2}\mathbf{-}\left|\mathbf{x}\right|}}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{y}\mathbf{\in }\left[\mathbf{0}\mathbf{,}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\right]\mathbf{\cup }\left(\mathbf{1}\mathbf{,}\mathbf{\infty }\right)\mathbf{.} \end{gathered}$$

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