Question

Let \[ A = \{x : |x^2 - 10| \le 6\} \quad \text{and} \quad B = \{x : |x - 2| > 1\}. \] Then 

• A. $A - B = [2, 3]$
• B. $A \cap B = [-4, -2] \cup [3, 4]$
• C. $B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$
• D. $A \cup B = (-\infty, 1] \cup (2, \infty)$

Hint:

Solve inequalities involving absolute values by splitting them into two cases: $|f(x)| \le a \iff -a \le f(x) \le a$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to determine the sets $A$ and $B$ by solving the given inequalities and then verify the operations.

Step 2: Detailed Explanation:
For Set $A$:
\[ -6 \le x^2 - 10 \le 6 \implies 4 \le x^2 \le 16 \]
\[ 2 \le |x| \le 4 \implies x \in [-4, -2] \cup [2, 4]. \]
For Set $B$:
\[ x - 2>1 \text{ or } x - 2<-1 \implies x>3 \text{ or } x<1 \]
\[ B = (-\infty, 1) \cup (3, \infty). \]
Now evaluate $B - A$:
$B - A$ contains elements in $B$ that are NOT in $A$.
The complement of $A$ is $A^c = (-\infty, -4) \cup (-2, 2) \cup (4, \infty)$.
$B \cap A^c = ((-\infty, 1) \cup (3, \infty)) \cap ((-\infty, -4) \cup (-2, 2) \cup (4, \infty))$.
This gives: $(-\infty, -4) \cup (-2, 1) \cup (4, \infty)$.
Note: Option (A) $A - B = A \cap [1, 3] = [2, 3]$ is also technically correct, but (C) is often the preferred notation for such problems.

Step 3: Final Answer:
Option (C) is the correct representation.