As shown in the figure, the set of points consisting of $\triangle ABC$ and its interior is denoted as $D$. For any point $P(x, y)$ in $D$, the maximum value of $z = 2x + 3y$ is

Let $a , b > 0 , a + b = 5$. The maximum value of $\sqrt { a + 1 } + \sqrt { b + 3 }$ is $\_\_\_\_$ .

14. If $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { c } x + y - 5 \leq 0 \\ 2 x - y - 1 \geq 0 \\ x - 2 y + 1 \leq 0 \end{array} \right.$, then the maximum value of $z = 2 x + y$ is $\_\_\_\_$ .

If $\mathrm { x } , \mathrm { y }$ satisfy the constraint conditions $\left\{ \begin{array} { l } x - y + 1 \geqslant 0 , \\ x - 2 y \leqslant 0 , \\ x + 2 y - 2 \leqslant 0 , \end{array} \right.$ then the maximum value of $z = x + y$ is $\_\_\_\_$ .

14. Given that real numbers $x , y$ satisfy $x ^ { 2 } + y ^ { 2 } \leq 1$ , then the maximum value of $| 2 x + y - 4 | + | 6 - x - 3 y |$ is $\_\_\_\_$ .

Let $A = \left\{ x \mid x ^ { 2 } - 4 x + 3 < 0 \right\} , B = \{ x \mid 2 x - 3 > 0 \}$, then $A \cap B =$
(A) $\left( - 3 , - \frac { 3 } { 2 } \right)$
(B) $\left( - 3 , \frac { 3 } { 2 } \right)$
(C) $\left( 1 , \frac { 3 } { 2 } \right)$
(D) $\left( \frac { 3 } { 2 } , 3 \right)$

Given sets $A = \{ x \mid x < 2 \}$, $B = \{ x \mid 3 - 2x > 0 \}$, then
A. $A \cap B = \{ x \mid x \leq \frac{3}{2} \}$
B. $A \cap B = \varnothing$
C. $A \cup B = \left\{ x \left\lvert \, x < \frac{3}{2} \right. \right\}$
D. $A \cup B = \mathbf{R}$

Let $x, y$ satisfy the linear constraints $\left\{\begin{array}{l} 2x - 3y + 3 \geq 0, \\ y + 3 \geq 0, \\ 3x - 3 \leq 0 \end{array}\right.$ and let $z = 2x + y$. The minimum value of $z$ is
A. $-15$
B. $-9$
C. $1$
D. $9$

Let $x$ and $y$ satisfy the constraints $\left\{ \begin{array} { l } x + 2 y \leqslant 1, \\ 2 x + y \geqslant - 1, \\ x - y \leqslant 0, \end{array} \right.$ then the minimum value of $z = 3 x - 2 y$ is \_\_\_\_

[Optional 4-5: Inequalities] (10 points)
Given functions $f(x) = -x^2 + ax + 4$ and $g(x) = |x + 1| + |x - 1|$.
(2) If the solution set of the inequality $f(x) \geq g(x)$ contains $[-1, 1]$, find the range of values for $a$.

Given the set $A = \left\{ x \mid x ^ { 2 } - x - 2 > 0 \right\}$, then $\mathrm { C } _ { \mathrm { R } } A =$
A. $\{ x \mid - 1 < x < 2 \}$
B. $\{ x \mid - 1 \leqslant x \leqslant 2 \}$
C. $\{ x \mid x < - 1 \} \cup \{ x \mid x > 2 \}$
D. $\{ x \mid x \leqslant - 1 \} \cup \{ x \mid x \geqslant 2 \}$

If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x - 2 y - 2 \leqslant 0 , \\ x - y + 1 \geqslant 0 , \\ y \leqslant 0 , \end{array} \right.$ then the maximum value of $z = 3 x + 2 y$ is $\_\_\_\_$

If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x - 2 y - 2 \leq 0 , \\ x - y + 1 \geq 0 , \\ y \leq 0 , \end{array} \right.$ then the maximum value of $z = 3 x + 2 y$ is \_\_\_\_

If $x , y$ satisfy the constraints $\left\{ \begin{array} { l } x + 2 y - 5 \geqslant 0 , \\ x - 2 y + 3 \geqslant 0 , \\ x - 5 \leqslant 0 , \end{array} \right.$ then the maximum value of $z = x + y$ is $\_\_\_\_$.

Let $A = \left\{ x \mid x ^ { 2 } - 5 x + 6 > 0 \right\} , B = \{ x \mid x - 1 < 0 \}$, then $A \cap B =$
A． $( - \infty , 1 )$
B． $( - 2,1 )$
C． $( - 3 , - 1 )$
D． $( 3 , + \infty )$

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. $b < c < a$

If $x , y$ satisfy $| x | \leqslant 1 - y$ and $y \geqslant - 1$, then the maximum value of $3 x + y$ is (A) $- 7$ (B) 1 (C) 5 (D) 7

5. Keep the answer sheet clean, do not fold it, do not tear or wrinkle it, and do not use correction fluid, correction tape, or scrapers.

I. Multiple Choice Questions: 12 questions in total, 5 points each, 60 points total. For each question, only one of the four options is correct.

1. Let $A = \left\{ x \mid x ^ { 2 } - 5 x + 6 > 0 \right\} , B = \{ x \mid x - 1 < 0 \}$, then $A \cap B =$
A. $( - \infty , 1 )$
B. $( - 2,1 )$
C. $( - 3 , - 1 )$
D. $( 3 , + \infty )$
2. Let $z = - 3 + 2 \mathrm { i }$. In the complex plane, the point corresponding to $\bar { z }$ is located in
A. the first quadrant
B. the second quadrant
C. the third quadrant
D. the fourth quadrant
3. Given $\overrightarrow { A B } = ( 2,3 ) , \overrightarrow { A C } = ( 3 , t ) , | \overrightarrow { B C } | = 1$, then $\overrightarrow { A B } \cdot \overrightarrow { B C } =$
A. $- 3$
B. $- 2$
C. $2$
D. $3$
4. On January 3, 2019, the Chang'e-4 probe successfully achieved humanity's first soft landing on the far side of the moon, marking another major achievement in China's space program. A key technical challenge in achieving soft landing on the far side of the moon is maintaining communication between the ground and the probe. To solve this problem, the Queqiao relay satellite was launched, which orbits around the Earth-Moon Lagrange point $L _ { 2 }$. The $L _ { 2 }$ point is an equilibrium point located on the extension of the Earth-Moon line. Let the Earth's mass be $M _ { 1 }$, the Moon's mass be $M _ { 2 }$, the Earth-Moon distance be $R$, and the distance from the $L _ { 2 }$ point to the Moon be $r$. According to Newton's laws of motion and the law of universal gravitation, $r$ satisfies the equation: $\frac { M _ { 1 } } { ( R + r ) ^ { 2 } } + \frac { M _ { 2 } } { r ^ { 2 } } = ( R + r ) \frac { M _ { 1 } } { R ^ { 3 } }$. Let $\alpha = \frac { r } { R }$. Since $\alpha$ is very small, in approximate calculations $\frac { 3 \alpha ^ { 3 } + 3 \alpha ^ { 4 } + \alpha ^ { 5 } } { ( 1 + \alpha ) ^ { 2 } } \approx 3 \alpha ^ { 3 }$. Then the approximate value of $r$ is
A. $\sqrt { \frac { M _ { 2 } } { M _ { 1 } } } R$
B. $\sqrt { \frac { M _ { 2 } } { 2 M _ { 1 } } } R$
C. $\sqrt [ 3 ] { \frac { 3 M _ { 2 } } { M _ { 1 } } } R$
D. $\sqrt [ 3 ] { \frac { M _ { 2 } } { 3 M _ { 1 } } } R$
5. In a speech competition, 9 judges each give an original score to a contestant. When determining the contestant's final score, 1 highest score and 1 lowest score are removed from the 9 original scores, leaving 7 valid scores. Compared with the 9 original scores, the numerical characteristic that remains unchanged for the 7 valid scores is
A. median
B. mean
C. variance
D. range

6. If $a > b$, then
A. $\ln ( a - b ) > 0$
B. $3 ^ { a } < 3 ^ { b }$
C. $a ^ { 3 } - b ^ { 3 } > 0$
D. $| a | > | b |$

10. The system of inequalities $\left\{ \begin{array} { l } x - 1 \geq 0 , \\ k x - y \leq 0 , \\ x + \sqrt { 3 } y - 3 \sqrt { 3 } \leq 0 \end{array} \right.$ represents a planar region that is an equilateral triangle. The minimum value of $z = x + 3 y$ is
A. $2 + 3 \sqrt { 3 }$ B. $1 + 3 \sqrt { 3 }$ C. $2 + \sqrt { 3 }$ D. $1 + \sqrt { 3 }$

11. Let the plane region represented by the system of inequalities $\left\{ \begin{array} { l } x + y \geq 6 , \\ 2 x - y \geq 0 \end{array} \right.$ be $D$ . Proposition $p : \exists ( x , y ) \in D , 2 x + y \geq 9$ ; Proposition $q : \forall ( x , y ) \in D , 2 x + y \leq 12$ . Four propositions are given below:
(1) $p \vee q$
(2) $\neg p \vee q$
(3) $p \wedge \neg q$
(4) $\neg p \wedge \neg q$
The numbers of all true propositions among these four are
A. (1)(3)
B. (1)(2)
C. (2)(3)
D. (3)(4)

13. If variables $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { l } 2 x + 3 y - 6 \geq 0 , \\ x + y - 3 \leq 0 , \\ y - 2 \leq 0 , \end{array} \right.$ then the maximum value of $z = 3 x - y$ is $\_\_\_\_$ .

23. Solution: (1) When $a = 1$, $f(x) = \left\{\begin{array}{l} 5 - 2x, & x \leq 1 \\ 3, & 1 < x < 4 \\ 2x - 5, & x \geq 4 \end{array}\right.$ ..... 3 marks
Thus the solution set of the inequality $f(x) < x$ is $(3, 5)$. ..... 5 marks
(2) $f(x) = |x - a| + |x - 4| \geq |(x - a) - (x - 4)| = |a - 4|$. ..... 6 marks
$\therefore |a - 4| \geq \frac{4}{a} - 1 = \frac{4 - a}{a}$. ..... 7 marks
When $a < 0$ or $a \geq 4$, the inequality clearly holds. ..... 8 marks
When $0 < a < 4$, $\frac{1}{a} \leq 1$, then $1 \leq a < 4$. ..... 9 marks
Thus the range of $a$ is $(-\infty, 0) \cup [1, +\infty)$. ..... 10 marks
Zizzsw Online was founded in 2014 and is a platform dedicated to independent recruitment, subject competitions, and national college entrance examinations. It operates a matrix of websites and WeChat media with over one million followers. The user base covers teachers, parents, and students from over 90\% of key secondary schools nationwide, attracting attention from many key universities.
To obtain first-hand information and preparation guides, please follow the official WeChat account of Zizzsw Online: zizzsw.
Official WeChat public account: zizzsw
Official website: www.zizzs.com
Consultation hotline: 010-5601 9830
WeChat customer service: zizzs2018
[Figure]

23. Solution: (1) When $a = 1$, $f ( x ) = | x - 1 | x + | x - 2 | ( x - 1 )$ .
When $x < 1$, $f ( x ) = - 2 ( x - 1 ) ^ { 2 } < 0$ ; when $x \geq 1$, $f ( x ) \geq 0$ .
Therefore, the solution set of the inequality $f ( x ) < 0$ is $( - \infty , 1 )$ .
(2) Since $f ( a ) = 0$, we have $a \geq 1$ .
When $a \geq 1 , x \in ( - \infty , 1 )$, $f ( x ) = ( a - x ) x + ( 2 - x ) ( x - a ) = 2 ( a - x ) ( x - 1 ) < 0$ .
Therefore, the range of values for $a$ is $[ 1 , + \infty )$ .

Let sets $A = \left\{ x \mid x ^ { 2 } - 4 \leqslant 0 \right\}$, $B = \{ x \mid 2 x + a \leqslant 0 \}$, and $A \cap B = \{ x \mid - 2 \leqslant x \leqslant 1 \}$. Then $a =$
A. $- 4$
B. $- 2$
C. 2
D. 4