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Let \(\mathrm{A}=\{x: x \in \mathrm{R},|x| < 1\}\);
\(\mathrm{B}=\{x: x \in \mathrm{R},|x-1| \geq 1\} \text { and } \mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D} \text {, }\)
then the set \(\mathrm{D}\) is
Options:
\(\mathrm{B}=\{x: x \in \mathrm{R},|x-1| \geq 1\} \text { and } \mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D} \text {, }\)
then the set \(\mathrm{D}\) is
Solution:
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Verified Answer
The correct answer is:
\(\{x: 1 \leq x < 2\}\)
\(\begin{aligned}
& \mathrm{A}=[x: x \in \mathrm{R},-1 < x < 1] \\
& \mathrm{B}=[x: x \in \mathrm{R}: x-1 \leq-1 \text { or } x-1 \geq 1] \\
&=[x: x \in R: x \leq 0 \text { or } x \geq 2] \\
& \therefore A \cup B=R-D,
\end{aligned}\)
where \(D=[x: x \in R, 1 \leq x < 2]\)
& \mathrm{A}=[x: x \in \mathrm{R},-1 < x < 1] \\
& \mathrm{B}=[x: x \in \mathrm{R}: x-1 \leq-1 \text { or } x-1 \geq 1] \\
&=[x: x \in R: x \leq 0 \text { or } x \geq 2] \\
& \therefore A \cup B=R-D,
\end{aligned}\)
where \(D=[x: x \in R, 1 \leq x < 2]\)
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