If $A = {x \in R : |x| < 2}$ and $B = {x \in R : |x – 2| \geq 3}$; then :
(A)
$A – B = [–1, 2)$
(B)
$A \cup B = R – (2, 5)$
(C)
$A \cap B = (–2, –1)$
(D)
$B – A = R – (–2, 5)$
MEDIUM
Correct Answer:D
Explanation
Given $A = {x \in R : |x| < 2}$ and $B = {x \in R : |x – 2| \geq 3}$.
We can rewrite the sets as follows:
$A = {x \in R : -2 < x < 2} = (-2, 2)$
$B = {x \in R : x – 2 \geq 3 \text{ or } x – 2 \leq -3} = {x \in R : x \geq 5 \text{ or } x \leq -1} = (-\infty, -1] \cup [5, \infty)$
Now, we want to find $B – A$, which is the set of elements in $B$ but not in $A$.
$B – A = B \cap A^c$, where $A^c$ is the complement of $A$.
$A^c = (-\infty, -2] \cup [2, \infty)$
Then, $B – A = ((-\infty, -1] \cup [5, \infty)) \cap ((-\infty, -2] \cup [2, \infty))$
$B - A = (-\infty, -1] \cup [5, \infty) - (-2,2) = (-\infty, -1] \cup [5, \infty)$.
In other words, $B-A$ is the set of all real numbers except those in the interval $(-2, 2)$.
So, $B-A = R - (-2, 5)$.