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

Solution :

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 or x \(\ge\) 2]

\(\therefore\) A \(\cup\) B = R – D

where D = [x : x \(\in\) R, 1 \(\le\) x < 2]

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