Explanation:
B Given,
\(A=\{x: x \in R,|x|\lt 1\}\)
\(B=\{x: x \in R, \mid x-1\} \geq 1\}\)
And \(\quad \mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D}\)
Then, \(A\) is also written as -
\(A=\{x: x \in R,-1\lt x\lt 1\}\)
And, \(\mathrm{B}\) is also written as -
\( B =\{x: x \in R, x-1 \geq 1 \text { or } x-1 \leq-1\}\)
\(\text { i.e., } B =\{x: x \in R, x \geq 2 \text { or } x \leq 0\}\)
\(\therefore A =\text { Range set }=(-1,1)\)
\( B =\text { Range set }=x \geq 2 \text { or } x \leq 0\)
\(=R (0,2)=(-\infty, 0] \cup[2, \infty)\)
\( A \cup B=(-1,1) \cup(-\infty, 0] \cup[2, \infty)\)
\(\text { So, } =(-\infty, 1) \cup[2, \infty)\)
So,
\(\mathrm{A} \cup \mathrm{B} =(-1,1) \cup(-\infty, 0] \cup[2, \infty)\)
\(=(-\infty, 1) \cup[2, \infty)\)
Then, \(\mathrm{A} \cup \mathrm{B}=\mathrm{R}-\{(\mathrm{x}: 1 \leq \mathrm{x}\lt 2)\}\)
Since, \(\mathrm{R}=(-\infty, \infty)\)
Hence, \(\mathrm{A} \cup \mathrm{B}=\mathrm{R}-\mathrm{D}\)
By comparing -
\(\mathrm{A} \cup \mathrm{B}=\mathrm{R}-\{\mathrm{x}: 1 \leq \mathrm{x}\lt 2\}\)Hence, \(\mathrm{D}=\{\mathrm{x}: 1 \leq \mathrm{x}\lt 2\}\)