Let A = {x : x in R, |x|

Question

(B) Given $A = \{x : x \in R, |x| < 1\}$. This implies $-1 < x < 1$,so $A = (-1, 1)$.Given $B = \{x : x \in R, |x - 1| \ge 1\}$. This implies $x - 1 \

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$\left\{x : 1 \le x < 2\right\}$

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(B) Given $A = \{x : x \in R, |x| Given $B = \{x : x \in R, |x - 1| \ge 1\}$. This implies $x - 1 \le -1$ or $x - 1 \ge 1$,which simplifies to $x \le 0$ or $x \ge 2$. So $B = (-\infty, 0] \cup [2, \infty)$.Now,$A \cup B = (-1, 1) \cup (-\infty, 0] \cup [2, \infty)$.Combining these intervals,we get $A \cup B = (-\infty, 1) \cup [2, \infty)$.We are given $A \cup B = R - D$. This means $D$ is the complement of $A \cup B$ in $R$.The complement of $(-\infty, 1) \cup [2, \infty)$ is the set $[1, 2)$.Therefore,$D = \{x : x \in R, 1 \le x < 2\}$.

Let $A = \{x : x \in R, |x| < 1\}$; $B = \{x : x \in R, |x - 1| \ge 1\}$ and $A \cup B = R - D$,then the set $D$ is

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