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 find the intersection of the sets \( A \) and \( B \), we must first determine the intervals represented by each set. 

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

We solve the absolute value inequality:

\(|x - 2| < 3\)

This implies:

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

Adding 2 to all sides:

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

\(-1 < x < 5\)

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

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

We solve the absolute value inequality:

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

This implies:

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

Subtracting 1 from all sides:

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

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

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

Intersection \( A \cap B \):

To find \( A \cap B \), take the overlap of the intervals \( (-1, 5) \) and \( [-5, 3] \).

The overlap is from \( x = -1 \) to \( x = 3 \), where \( x = -1 \) is not included but \( x = 3 \) is included in set \( B \).

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

Answer 2

Step 1: Solving Set A
The inequality for set A is: \[ |x - 2|<3 \]

• The absolute value inequality $|x - a|<b$ is equivalent to $-b<x - a<b$
• Applying this property: \[ -3<x - 2<3 \]
• Adding 2 to all parts: \[ -3 + 2<x<3 + 2 \] \[ -1<x<5 \]

Thus, set A is the open interval: \[ A = (-1, 5) \] Step 2: Solving Set B
The inequality for set B is: \[ |x + 1| \leq 4 \]

• The absolute value inequality $|x + a| \leq b$ is equivalent to $-b \leq x + a \leq b$
• Applying this property: \[ -4 \leq x + 1 \leq 4 \]
• Subtracting 1 from all parts: \[ -4 - 1 \leq x \leq 4 - 1 \] \[ -5 \leq x \leq 3 \]

Thus, set B is the closed interval: \[ B = [-5, 3] \] Step 3: Finding the Intersection
We need to find all $x$ values that satisfy both:
 

• From set A: $x>-1$ AND $x<5$
• From set B: $x \geq -5$ AND $x \leq 3$

Combining these conditions:
 

• The lower bound is the maximum of $-1$ (from A) and $-5$ (from B), which is $-1$
• The upper bound is the minimum of $5$ (from A) and $3$ (from B), which is $3$
• The inequality becomes $-1<x \leq 3$ because:
  • $x>-1$ from set A (strict inequality)
  • $x \leq 3$ from set B (includes endpoint)

Therefore, the intersection is: \[ A \cap B = (-1, 3] \] Step 4: Verifying with Options
Comparing with the given options:

• (A) $(-1, 3]$ - Exactly matches our solution
• (B) $(-2, 3)$ - Incorrect bounds and missing endpoint
• (C) $(-1, 4)$ - Incorrect upper bound
• (D) $(-2, 4)$ - Incorrect bounds

Conclusion:
The correct answer is $(-1, 3]$.