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]


Similar Questions

If A = {x,y}, then the power set of A is

If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to

Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets A\(\times\)B having 3 or more elements is

The set A = [x : x \(\in\) R, \(x^2\) = 16 and 2x = 6] equal

If N is a set of first 10 natural numbers and a relation R is defined as a + 2b = 10 where a, b \(\in\) N. find inverse of R.

Leave a Comment

Your email address will not be published. Required fields are marked *

Ezoicreport this ad