Sets and Relations Part 1

Concept:Solving modulus inequalities gives the range of $x$ for sets $A$ and $B$.Set operations like union, intersection, and difference are then computed.Explanation:For set $A$: $|x^2 - 10| \le 6$ means $-6 \le x^2 - 10 \le 6$.Add $10$ to all parts: $4 \le x^2 \le 16$.Take square root: $2 \le |x| \le 4$.This gives $A = [-4, -2] \cup [2, 4]$.For set $B$: $|x-2| > 1$ gives $x-2 < -1$ or $x-2 > 1$.Solving: $x < 1$ or $x > 3$.So $B = (-\infty, 1) \cup (3, \infty)$.Now check $B - A$: elements in $B$ not in $A$.From $(-\infty, 1)$, remove $[-4, -2]$ (belongs to $A$).Remaining: $(-\infty, -4) \cup (-2, 1)$.From $(3, \infty)$, remove $[2, 4]$ (overlap up to $4$).Remaining: $(4, \infty)$.Combine: $B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$.This matches Option B.Answer:Option B is correct.