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 […]