Let's analyze the given sets \(A\) and \(B\) and determine the correct option.
Set A: The set is defined as \(A = \{x : |x^2 - 10| \le 6\}\).
Set B: The set is defined as \(B = \{x : |x - 2| > 1\}\).
We need to determine \(A \cup B\).
From \(A = [-4, -2] \cup [2, 4]\) and \(B = (-\infty, 1) \cup (3, \infty)\), we can find the union:
Thus, the correct answer is \(A \cup B = (-\infty, 1] \cup (2, \infty)\).
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
If $ \left(\frac{1}{^{15}C_0} + \frac{1}{^{15}C_1}\right)$ $\left(\frac{1}{^{15}C_1} + \frac{1}{^{15}C_2}\right)$ $\dots$ $ \left(\frac{1}{^{15}C_{12}} + \frac{1}{^{15}C_{13}}\right)$ = $\frac{\alpha^{13}}{^{14}C_0 ^{14}C_1 \dots ^{14}C_{12}} $, then $30\alpha$ is equal to __________.
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.