Question

Given the sets $ A = \{ x \mid |x - 2|<3 \} $ and $ B = \{ x \mid |x + 1| \leq 4 \} $, find $ A \cap B $.

• A. \((-2, 4)\)
• B. \((-2, 3)\)
• C. \((-1, 4)\)
• D. \((-1, 3\)]

Hint:

Remember to carefully check open and closed intervals when finding intersections, especially when inequality signs differ (< or \(\leq\)).

Answer 1

The Correct Option is D

To determine the intersection of sets \( A \) and \( B \), the intervals for each set must first be established.

Set A: \( A = \{ x \mid |x - 2|<3 \} \)

Solving the absolute value inequality:

\(|x - 2| < 3\)

This results in:

\(-3 < x - 2 < 3\)

Adding 2 to all parts of the inequality:

\(-3 + 2 < x < 3 + 2\)

\(-1 < x < 5\)

Therefore, \( A = (-1, 5) \).

Set B: \( B = \{ x \mid |x + 1| \leq 4 \} \)

Solving the absolute value inequality:

\(|x + 1| \leq 4\)

This results in:

\(-4 \leq x + 1 \leq 4\)

Subtracting 1 from all parts of the inequality:

\(-4 - 1 \leq x \leq 4 - 1\)

\(-5 \leq x \leq 3\)

Therefore, \( B = [-5, 3] \).

Intersection \( A \cap B \):

The intersection \( A \cap B \) is found by identifying the overlapping region of the intervals \( (-1, 5) \) and \( [-5, 3] \).

The overlap spans from \( x = -1 \) to \( x = 3 \). The lower bound \( x = -1 \) is excluded, while the upper bound \( x = 3 \) is included due to its presence in set \( B \).

Consequently, \( A \cap B = (-1, 3] \).