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1.Set Theory
hard
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
A
$\{x:1 < x \le 2\}$
B
$\{x:1 \le x < 2\}$
C
$\{x:1 \le x \le 2\}$
D
None of these
Solution
(b) $A = [x:x \in R,\, – 1 < x < 1]$
$B = [x:x \in R:x – 1 \le – 1$ or $x – 1 \ge 1]$
= $[x:x \in R:x \le 0{\rm{ or }}x \ge 2]$
$\therefore A \cup B = R – D$, where $D = [x:x \in R,\,1 \le x < 2]$.
Standard 11
Mathematics
