Let $Z$ be the set of integers. If $A = {x \in Z : 2(x + 2) (x^2 - 5x + 6) = 1}$ and $B = {x \in Z : -3 < 2x - 1 < 9}$, then the number of subsets of the set $A \times B$, is
(A)
$2^{12}$
(B)
$2^{18}$
(C)
$2^{10}$
(D)
$2^{15}$
MEDIUM
Correct Answer:D
Explanation
Given $A = {x \in Z : 2(x+2)(x^2 - 5x + 6) = 1}$.
Since $2(x+2)(x^2 - 5x + 6) = 1$, we can rewrite it as $2(x+2)(x^2 - 5x + 6) = 2^0$.
This implies that $x = -2, 2, 3$, so $A = {-2, 2, 3}$.
Also, $B = {x \in Z : -3 < 2x - 1 < 9}$.
Adding 1 to all sides, we get $-2 < 2x < 10$.
Dividing by 2, we get $-1 < x < 5$.
Thus, $B = {0, 1, 2, 3, 4}$.
Now, $A \times B$ has $3 \times 5 = 15$ elements.
The number of subsets of $A \times B$ is $2^{15}$.