17. (This question is worth 12 points)

A minibus has 5 seats numbered $1,2,3,4,5$. Passengers $P _ { 1 } , P _ { 2 } , P _ { 3 } , P _ { 4 } , P _ { 5 }$ have assigned seat numbers $1,2,3,4,5$ respectively. They board in order of increasing seat numbers. Passenger $P _ { 1 }$ did not sit in seat 1 due to health reasons. The driver then requires the remaining passengers to be seated according to the following rule: if their own seat is empty, they must sit in their own seat; if their own seat is occupied, they can choose any of the remaining empty seats. (1) If passenger $P _ { 1 }$ sits in seat 3 and other passengers are seated according to the rule, there are 4 possible seating arrangements. The table below shows two of them. Please fill in the remaining two arrangements (enter the seat numbers where passengers sit in the blank cells);
(2) If passenger

Passenger	$P _ { 1 }$	$P _ { 2 }$	$P _ { 3 }$	$P _ { 4 }$	$P _ { 5 }$
\multirow{3}{*}{Seat Number}	3	2	1	4	5
	3	2	4	5	1

$P _ { 1 }$ sits in seat 2, and other passengers are seated according to the rule, find the probability that passenger $P _ { 5 }$ sits in seat 5.

Let the set $A = \{1,2,3\}$, $B = \{2,3,4\}$, then $A \cup B =$
A. $\{1,2,3,4\}$
B. $\{1,2,3\}$
C. $\{2,3,4\}$
D. $\{1,1,4\}$

Given sets $A = \{ 0,2 \}$ and $B = \{ - 2 , - 1,0,1,2 \}$, then $A \cap B =$
A. $\{ 0,2 \}$
B. $\{ 1,2 \}$
C. $\{ 0 \}$
D. $\{ - 2 , - 1,0,1,2 \}$

Given sets $A = \{ x \mid x - 1 \geqslant 0 \} , B = \{ 0,1,2 \}$, then $A \cap B =$
A. $\{ 0 \}$
B. $\{ 1 \}$
C. $\{ 1,2 \}$
D. $\{ 0,1,2 \}$

Given sets $A = \{ 1,3,5,7 \} , B = \{ 2,3,4,5 \}$, then $A \cap B =$
A. $\{ 3 \}$
B. $\{ 5 \}$
C. $\{ 3,5 \}$
D. $\{ 1,2,3,4,5,7 \}$

Given set $A = \left\{ ( x , y ) \left| x ^ { 2 } + y ^ { 2 } \leqslant 3 , x \in \mathbf { Z } , y \in \mathbf { Z } \right. \right\}$, the number of elements in $A$ is
A. 9
B. 8
C. 5
D. 4

Chinese mathematician Chen Jingrun achieved world-leading results in research on Goldbach's conjecture. Goldbach's conjecture states that ``every even number greater than 2 can be expressed as the sum of two prime numbers'', such as $30 = 7 + 23$. Among prime numbers not exceeding 30, if two different numbers are randomly selected, the probability that their sum equals 30 is
A. $\frac { 1 } { 12 }$
B. $\frac { 1 } { 14 }$
C. $\frac { 1 } { 15 }$
D. $\frac { 1 } { 18 }$

3. After the examination ends, submit both this test paper and the answer sheet. Section I: Multiple Choice Questions: This section has 12 questions, each worth 5 points, for a total of 60 points. For each question, only one of the four options is correct.
1. Given sets $M = \{ x \mid - 4 < x < 2 \} , N = \left\{ x \mid x ^ { 2 } - x - 6 < 0 \right\}$ , then $M \cap N =$
A. $\{ x \mid - 4 < x < 3 \}$
B. $\{ x \mid - 4 < x < - 2 \}$
C. $\{ x \mid - 2 < x < 2 \}$
D. $\{ x \mid 2 < x < 3 \}$
2. Let complex number $z$ satisfy $| z - \mathrm { i } | = 1$ , and the point corresponding to $z$ in the complex plane is $( x , y )$ , then
A. $( x + 1 ) ^ { 2 } + y ^ { 2 } = 1$
B. $( x - 1 ) ^ { 2 } + y ^ { 2 } = 1$
C. $x ^ { 2 } + ( y - 1 ) ^ { 2 } = 1$
D. $x ^ { 2 } + ( y + 1 ) ^ { 2 } = 1$
3. Given $a = \log _ { 2 } 0.2 , b = 2 ^ { 0.2 } , c = 0.2 ^ { 0.3 }$ , then
A. $a < b < c$
B. $a < c < b$
C. $c < a < b$
D. $b < c < a$

4. According to historical records, the ``Hundred Family Names'' was written in the early Northern Song Dynasty. Table 1 records the top 24 surnames from the beginning of the ``Hundred Family Names'':
\begin{table}[h]
Table 1

赵	钱	孙	李	周	吴	郑	王	冯	陈	褚	卫
蒋	沈	韩	杨	朱	秦	尤	许	何	吕	施	张

\end{table}
Table 2 records the top 25 most populous surnames in China in 2018:
\begin{table}[h]
Table 2

1: Li	2: Wang	3: Zhang	4: Liu	5: Chen
6: Yang	7: Zhao	8: Huang	9: Zhou	10: Wu
11: Xu	12: Sun	13: Hu	14: Zhu	15: Gao
16: Lin	17: He	18: Guo	19: Ma	20: Luo

\end{table}

21: Liang

22: Song

23: Zheng

24: Xie

25: Han

If one surname is randomly selected from the top 24 surnames in the ``Hundred Family Names'', the probability that this surname is among the top 24 most populous surnames in China in 2018 is
A. $\frac { 5 } { 12 }$ B. $\frac { 11 } { 24 }$ C. $\frac { 13 } { 24 }$ D. $\frac { 1 } { 2 }$

5. After the examination ends, submit both this test paper and the answer sheet together. I. Multiple Choice Questions: This section has 12 questions, each worth 5 points, for a total of 60 points. For each question, only one of the four options is correct.
1. Given sets $M = \{ x \mid - 4 < x < 2 \} , N = \left\{ x \mid x ^ { 2 } - x - 6 < 0 \right\}$, then $M \cap N =$
A. $\{ x \mid - 4 < x < 3 \}$
B. $\{ x \mid - 4 < x < - 2 \}$
C. $\{ x \mid - 2 < x < 2 \}$
D. $\{ x \mid 2 < x < 3 \}$
2. Let complex number $z$ satisfy $| z - \mathrm { i } | = 1$, and the point corresponding to $z$ in the complex plane is $( x , y )$, then
A. $( x + 1 ) ^ { 2 } + y ^ { 2 } = 1$
B. $( x - 1 ) ^ { 2 } + y ^ { 2 } = 1$
C. $x ^ { 2 } + ( y - 1 ) ^ { 2 } = 1$
D. $x ^ { 2 } + ( y + 1 ) ^ { 2 } = 1$
3. Given $a = \log _ { 2 } 0.2 , b = 2 ^ { 0.2 } , c = 0.2 ^ { 0.3 }$, then
A. $a < b < c$
B. $a < c < b$
C. $c < a < b$
D. $b < c < a$
4. In ancient Greece, people believed that the most beautiful human body has the ratio of the length from the top of the head to the navel to the length from the navel to the sole of the foot equal to $\frac { \sqrt { 5 } - 1 } { 2 } \left( \frac { \sqrt { 5 } - 1 } { 2 } \approx 0.618 \right.$, called the golden ratio), and the famous ``Venus de Milo'' exemplifies this. Furthermore, the ratio of the length from the top of the head to the throat to the length from the throat to the navel is also $\frac { \sqrt { 5 } - 1 } { 2 }$. If a person satisfies both golden ratio proportions, with a shoulder width of 105 cm and the length from the top of the head to the chin of 26 cm, then their height could be [Figure]
A. 165 cm
B. 175 cm
C. 185 cm
D. 190 cm
5. The graph of the function $f ( x ) = \frac { \sin x + x } { \cos x + x ^ { 2 } }$ on $[ - \pi , \pi ]$ is approximately
A. [Figure]
B. [Figure]
C. [Figure]
D. [Figure]

1. Let $M = \{ 1,3,5,7,9 \} , N = \{ x \mid 2 x > 7 \}$, then $M \cap N = ( )$
A. $\{ 7,9 \}$
B. $\{ 5,7,9 \}$
C. $\{ 3,5,7,9 \}$
D. $\{ 1,3,5,7,9 \}$

1. Let sets $M = \{ x \mid 0 < x < 4 \}, N = \left\{ x \left\lvert \, \frac { 1 } { 3 } \leq x \leq 5 \right. \right\}$, then $M \cap N =$
A. $\left\{ x \left\lvert \, 0 < x \leq \frac { 1 } { 3 } \right. \right\}$
B. $\left\{ x \left\lvert \, \frac { 1 } { 3 } \leq x < 4 \right. \right\}$
C. $\{ x \mid 4 \leq x < 5 \}$
D. $\{ x \mid 0 < x \leq 5 \}$

2. Let $U = \{ 1,2,3,4,5,6 \} , A = \{ 1,3,6 \} , B = \{ 2,3,4 \}$. Then $A \cap \left( \complement_U B \right) =$
A. $\{ 3 \}$
B. $\{ 1,6 \}$
C. $\{ 5,6 \}$
D. $\{ 1,3 \}$
【Answer】B 【Solution】 【Analysis】Use the definitions of intersection and complement to find $A \cap \left( \complement_U B \right)$. 【Detailed Solution】From the given conditions, $\complement_U B = \{ 1,5,6 \}$, so $A \cap \left( \complement_U B \right) = \{ 1,6 \}$, Therefore, the answer is: B.

Let set $A = \{ - 2 , - 1,0,1,2 \} , B = \left\{ x \left\lvert \, 0 \leq x < \frac { 5 } { 2 } \right. \right\}$ , then $A \cap B =$（ ）
A. $\{ 0,1,2 \}$
B. $\{ - 2 , - 1,0 \}$
C. $\{ 0,1 \}$
D. $\{ 1,2 \}$

Set $M = \{ 2,4,6,8,10 \} , N = \{ x \mid - 1 < x < 6 \}$ , then $M \cap N =$
A. $\{ 2,4 \}$
B. $\{ 2,4,6 \}$
C. $\{ 2,4,6,8 \}$
D. $\{ 2,4,6,8,10 \}$

Let the universal set $U = \{ 1,2,3,4,5 \}$, and set $M$ satisfies $C_U M = \{ 1,3 \}$. Then
A. $2 \in M$
B. $3 \in M$
C. $4 \notin M$
D. $5 \in M$

Let $U = \{ 1,2,3,4,5,6 \} , A = \{ 1,3,6 \} , B = \{ 2,3,4 \}$. Then $A \cap \left( \complement_U B \right) =$
A. $\{ 3 \}$
B. $\{ 1,6 \}$
C. $\{ 5,6 \}$
D. $\{ 1,3 \}$

5. From 7 integers from 2 to 8, two different numbers are randomly selected. The probability that these two numbers are coprime is
A. $\frac { 1 } { 6 }$
B. $\frac { 1 } { 3 }$
C. $\frac { 1 } { 2 }$
D. $\frac { 2 } { 3 }$

Let $A = \{ x \mid x = 3k + 1 , k \in Z \} , B = \{ x \mid x = 3k + 2 , k \in Z \} , U$ be the set of integers, then $C_{U}(A \bigcap B) =$
A. $\{ x \mid x = 3k , k \in Z \}$
B. $\{ x \mid x = 3k - 1 , k \in Z \}$
C. $\{ x \mid x = 3k - 1 , k \in \mathrm{Z} \}$
D. $\varnothing$

Let the universal set $U = \mathbb{R}$, set $M = \{ x \mid x < 1 \}$, $N = \{ x \mid - 1 < x < 2 \}$, then $\{ x \mid x \geqslant 2 \} =$
A. $C _ { U } ( M \cup N )$
B. $N \cup C _ { U } M$
C. $C _ { U } ( M \cap N )$
D. $M \cup C _ { U } N$

Let $A=\{0,-a\}$, $B=\{1,a-2,2a-2\}$. If $A\subseteq B$, then $a=$
A. 2
B. 1
C. $\frac{2}{3}$

Given sets $M = \{ x \mid - 4 < x \leq 1 \} , N = \{ x \mid - 1 < x < 3 \}$, then $M \cup N =$ \_\_\_\_

Given sets $A = \left\{ x \mid - 5 < x ^ { 3 } < 5 \right\} , B = \{ - 3 , - 1,0,2,3 \}$ , then $A \cap B =$
A. $\{ - 1,0 \}$
B. $\{ 2,3 \}$
C. $\{ - 3 , - 1,0 \}$
D. $\{ - 1,0,2 \}$

Let the universal set $U = \{x \mid x \text{ is a positive integer less than } 9\}$, and set $A = \{1,3,5\}$. Then the number of elements in $\complement_U A$ is
A. $2$
B. $3$
C. $5$
D. $8$

Let the universal set $U = \{1,2,3,4,5,6,7,8\}$, and set $A = \{1,3,5\}$. The number of elements in $C_U A$ is
A. $0$
B. $3$
C. $5$
D. $8$