Question

If $C = \{x \in \mathbb{R} : |x+1| \leq 3\}$ and $D = \{x \in \mathbb{R} : |x-1| > 2\}$, then which of the following is true?

• A. $C - D = [-4,-1)$
• B. $D - C = \mathbb{R} - [-4,3]$
• C. $C \cup D = \mathbb{R} - (-1,3)$
• D. $C \cap D = (3, 4]$

Answer 1

Answer: (B)

(b)

Answer 2

First, determine the interval representation of set $C$. The inequality $|x+1| \leq 3$ is equivalent to $-3 \leq x+1 \leq 3$. Subtracting 1 from all parts, we get $-3 - 1 \leq x \leq 3 - 1$, which simplifies to $-4 \leq x \leq 2$. So, $C = [-4, 2]$.

Next, determine the interval representation of set $D$. The inequality $|x-1| > 2$ is equivalent to $x-1 > 2$ or $x-1 < -2$.

Case 1: $x-1 > 2 \implies x > 2+1 \implies x > 3$. Case 2: $x-1 < -2 \implies x < -2+1 \implies x < -1$. So, $D = (-\infty, -1) \cup (3, \infty)$.

Now, let's evaluate each option.

(a) $C - D = [-4,-1)$ $C - D$ is the set of elements in $C$ but not in $D$. $C - D = [-4, 2] \setminus ((-\infty, -1) \cup (3, \infty))$. This is equivalent to the intersection of $C$ with the complement of $D$. The complement of $D$, $D^c$, is $\mathbb{R} \setminus ((-\infty, -1) \cup (3, \infty)) = [-1, 3]$. So, $C - D = C \cap D^c = [-4, 2] \cap [-1, 3]$. The intersection of $[-4, 2]$ and $[-1, 3]$ is $[-1, 2]$. Thus, $C - D = [-1, 2]$. Option (a) states $C - D = [-4,-1)$, which is incorrect.

(b) $D - C = \mathbb{R} - [-4,3]$ $D - C$ is the set of elements in $D$ but not in $C$. $D - C = ((-\infty, -1) \cup (3, \infty)) \setminus [-4, 2]$. This is equivalent to the intersection of $D$ with the complement of $C$. The complement of $C$, $C^c$, is $\mathbb{R} \setminus [-4, 2] = (-\infty, -4) \cup (2, \infty)$. So, $D - C = D \cap C^c = ((-\infty, -1) \cup (3, \infty)) \cap ((-\infty, -4) \cup (2, \infty))$. We can find the intersection by distributing: $D - C = ((-\infty, -1) \cap (-\infty, -4)) \cup ((-\infty, -1) \cap (2, \infty)) \cup ((3, \infty) \cap (-\infty, -4)) \cup ((3, \infty) \cap (2, \infty))$.

$(-\infty, -1) \cap (-\infty, -4) = (-\infty, -4)$. $(-\infty, -1) \cap (2, \infty) = \emptyset$. $(3, \infty) \cap (-\infty, -4) = \emptyset$. $(3, \infty) \cap (2, \infty) = (3, \infty)$.

So, $D - C = (-\infty, -4) \cup \emptyset \cup \emptyset \cup (3, \infty) = (-\infty, -4) \cup (3, \infty)$.

Now let's evaluate the right side of option (b): $\mathbb{R} - [-4,3]$. $\mathbb{R} - [-4,3]$ is the set of all real numbers excluding the closed interval $[-4,3]$. This means $x < -4$ or $x > 3$. So, $\mathbb{R} - [-4,3] = (-\infty, -4) \cup (3, \infty)$. Comparing $D - C = (-\infty, -4) \cup (3, \infty)$ with $\mathbb{R} - [-4,3] = (-\infty, -4) \cup (3, \infty)$, we see that they are equal. Thus, option (b) is true.

(c) $C \cup D = \mathbb{R} - (-1,3)$ $C \cup D = [-4, 2] \cup ((-\infty, -1) \cup (3, \infty))$. Combining the intervals: $(-\infty, -1) \cup [-4, 2] = (-\infty, 2]$. So, $C \cup D = (-\infty, 2] \cup (3, \infty)$. The right side of option (c) is $\mathbb{R} - (-1,3)$, which is the set of all real numbers excluding the open interval $(-1,3)$. This means $x \leq -1$ or $x \geq 3$. So, $\mathbb{R} - (-1,3) = (-\infty, -1] \cup [3, \infty)$. Comparing $C \cup D = (-\infty, 2] \cup (3, \infty)$ with $\mathbb{R} - (-1,3) = (-\infty, -1] \cup [3, \infty)$, we see that they are not equal. For example, $x=0$ is in $C \cup D$ but not in $\mathbb{R} - (-1,3)$. Option (c) is incorrect.

(d) $C \cap D = (3, 4]$ $C \cap D = [-4, 2] \cap ((-\infty, -1) \cup (3, \infty))$. $C \cap D = ([-4, 2] \cap (-\infty, -1)) \cup ([-4, 2] \cap (3, \infty))$. $[-4, 2] \cap (-\infty, -1) = [-4, -1)$. $[-4, 2] \cap (3, \infty) = \emptyset$. So, $C \cap D = [-4, -1) \cup \emptyset = [-4, -1)$. Option (d) states $C \cap D = (3, 4]$, which is incorrect.

Based on the evaluation of all options, only option (b) is true.