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

To solve this problem, we need to evaluate the sets \( A \) and \( B \) based on their given conditions and understand what each option states. Let's begin by analyzing each set:

Step 1: Understanding Set A

Set \( A \) is defined as:

\(A = \{x : |x^2 - 10| \le 6\}\) 

This inequality breaks down into:

\(-6 \le x^2 - 10 \le 6\)

Simplifying each part:

• \(x^2 - 10 \ge -6 \Rightarrow x^2 \ge 4 \Rightarrow x \le -2 \, \text{or} \, x \ge 2\)
• \(x^2 - 10 \le 6 \Rightarrow x^2 \le 16 \Rightarrow -4 \le x \le 4\)

Combining these conditions, we find:

\(A = [-4, -2] \cup [2, 4]\)

Step 2: Understanding Set B

Set \( B \) is defined as:

\(B = \{x : |x - 2| > 1\}\)

This inequality breaks down into:

• \(x - 2 > 1 \Rightarrow x > 3\)
• \(x - 2 < -1 \Rightarrow x < 1\)

Thus, we have:

\(B = (-\infty, 1) \cup (3, \infty)\)

Step 3: Evaluating the Options

Let's analyze each given option:

1. \( A - B = [2, 3] \)
    

Elements in \( A \) but not in \( B \):

• For \([2, 4]\) in \( A \), exclude overlap with \( B \): \([3, 4]\) is in \( B \), leaving \([2, 3]\) for \( A - B\).

1. \( A \cap B = [-4, -2] \cup [3, 4] \)
    

Elements both in \( A \) and \( B \):

• For \([3, 4]\) (intersection of \([2, 4]\) and \((3, \infty)\)).
• \([-4, -2] \cap (-\infty, 1) = [-4, -2]\)

1. \( B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty) \)
    

Elements in \( B \) but not in \( A \):

• \(B - A\) overlaps:\)

1. \( A \cup B = (-\infty, 1] \cup (2, \infty) \)
    

Union of \( A \) and \( B \) should consider every open and closed segment, which disagrees here: false.

Hence, the correct answer, as identified from our results, is:

\( B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty) \)