The problem asks us to list the elements of four sets defined by different conditions. (a) The set of integers $x$ such that $-2 < x < 9$. (b) The intersection of the set of integers and the set $\{0, \sqrt{3}, \pi, 2i\}$. (c) The set of values $x^2 + 1$ where $x$ is an element of the set $A = \{-2, -1, 0, 1, 2\}$. (d) The set of values $\sqrt{x+2}$ where $x$ is an element of the set $B = \{-3, -4, 0, 1, 2\}$.

1. Problem Description

The problem asks us to list the elements of four sets defined by different conditions.

(a) The set of integers $x$ such that $-2 < x < 9$.

(b) The intersection of the set of integers and the set $\{0, \sqrt{3}, \pi, 2i\}$.

(c) The set of values $x^2 + 1$ where $x$ is an element of the set $A = \{-2, -1, 0, 1, 2\}$.

(d) The set of values $\sqrt{x+2}$ where $x$ is an element of the set $B = \{-3, -4, 0, 1, 2\}$.

2. Solution Steps

(a) The integers greater than -2 and less than 9 are -1, 0, 1, 2, 3, 4, 5, 6, 7, and

8.

(b) The set of integers is $\mathbb{Z} = \{\dots, -2, -1, 0, 1, 2, \dots\}$. We need to find the intersection of this set with the set $\{0, \sqrt{3}, \pi, 2i\}$. Among the elements of the set $\{0, \sqrt{3}, \pi, 2i\}$, only $0$ is an integer. Thus, the intersection is $\{0\}$.

(c) We are given the set $A = \{-2, -1, 0, 1, 2\}$. We need to compute $x^2 + 1$ for each $x$ in $A$.

If $x = -2$, then $x^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5$.

If $x = -1$, then $x^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2$.

If $x = 0$, then $x^2 + 1 = (0)^2 + 1 = 0 + 1 = 1$.

If $x = 1$, then $x^2 + 1 = (1)^2 + 1 = 1 + 1 = 2$.

If $x = 2$, then $x^2 + 1 = (2)^2 + 1 = 4 + 1 = 5$.

So the set is $\{5, 2, 1, 2, 5\}$. Removing duplicates, the set is $\{1, 2, 5\}$.

(d) We are given the set $B = \{-3, -4, 0, 1, 2\}$. We need to compute $\sqrt{x+2}$ for each $x$ in $B$.

If $x = -3$, then $\sqrt{x+2} = \sqrt{-3+2} = \sqrt{-1} = i$.

If $x = -4$, then $\sqrt{x+2} = \sqrt{-4+2} = \sqrt{-2} = i\sqrt{2}$.

If $x = 0$, then $\sqrt{x+2} = \sqrt{0+2} = \sqrt{2}$.

If $x = 1$, then $\sqrt{x+2} = \sqrt{1+2} = \sqrt{3}$.

If $x = 2$, then $\sqrt{x+2} = \sqrt{2+2} = \sqrt{4} = 2$.

So the set is $\{i, i\sqrt{2}, \sqrt{2}, \sqrt{3}, 2\}$.

3. Final Answer

(a) $\{-1, 0, 1, 2, 3, 4, 5, 6, 7, 8\}$

(b) $\{0\}$

(c) $\{1, 2, 5\}$

(d) $\{i, i\sqrt{2}, \sqrt{2}, \sqrt{3}, 2\}$