1.Set Theory
hard

જો  $A = \{x:x \in R,\,|x|\, < 1\}\,;$ $B = \{x:x \in R,\,|x - 1| \ge 1\}$ અને  $A \cup B = R - D,$ તો ગણ $D$ એ . . .

A

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

B

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

C

$\{x:1 \le x \le 2\}$

D

એકપણ નહિ.

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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.