4. For optional questions: First blacken the question number of your chosen question at the designated location on the answer sheet using a 2B pencil. Write your answer in the corresponding answer area on the answer sheet. Answers written on the test paper, scratch paper, and non-answer areas on the answer sheet are invalid.

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 bottom of the feet equal to $\frac { \sqrt { 5 } - 1 } { 2 } \left( \frac { \sqrt { 5 } - 1 } { 2 } \approx 0.618 \right.$ , called the golden ratio), as exemplified by the famous ``Armless Venus''. Furthermore, the ratio of the length from the top of the head to the base of the neck to the length from the base of the neck to the navel is also $\frac { \sqrt { 5 } - 1 } { 2 }$ . If a person satisfies both golden ratio proportions above, with leg length 105 cm and head to base of neck length 26 cm, then their height is likely to be [Figure]
A. 165 cm
B. 175 cm
C. 185 cm
D. 190 cm

4. Drawings may first be made with a pencil, and after confirmation must be traced with a black signature pen.

4. For drawings, you may first use pencil, and after confirmation, you must trace over with a black pen.

5. Keep the card surface clean, do not fold, do not tear or wrinkle, and do not use correction fluid, correction tape, or scraping knives. I. Multiple Choice Questions: This section has 12 questions, 5 points each, 60 points total. For each question, only one of the four options is correct.
1. Given sets $A = \{ x \mid x > - 1 \} , B = \{ x \mid x < 2 \}$, then $A \cap B =$
A. $( - 1 , + \infty )$
B. $( - \infty , 2 )$
C. $( - 1,2 )$
D. $\varnothing$
2. Let $z = \mathrm { i } ( 2 + \mathrm { i } )$, then $\bar { z } =$
A. $1 + 2 \mathrm { i }$
B. $- 1 + 2 \mathrm { i }$
C. $1 - 2 \mathrm { i }$
D. $- 1 - 2 \mathrm { i }$
3. Given vectors $\boldsymbol { a } = ( 2,3 ) , \boldsymbol { b } = ( 3,2 )$, then $| \boldsymbol { a } - \boldsymbol { b } | =$
A. $\sqrt { 2 }$
B. 2
C. $5 \sqrt { 2 }$
D. 50
4. A biology laboratory has 5 rabbits, of which only 3 have been measured for a certain indicator. If 3 rabbits are randomly selected from these 5 rabbits, the probability that exactly 2 have been measured for this indicator is
A. $\frac { 2 } { 3 }$
B. $\frac { 3 } { 5 }$
C. $\frac { 2 } { 5 }$
D. $\frac { 1 } { 5 }$
5. After a knowledge test on the ``Belt and Road'', three people A, B, and C made predictions about their scores. A: My score is higher than B's.
B: C's score is higher than both mine and A's. C: My score is higher than B's. After the scores were announced, the three people's scores are all different and only one person's prediction is correct. Then the three people in order from highest to lowest score are
A. A, B, C
B. B, A, C
C. C, B, A
D. A, C, B

7. The three-view drawing of a certain solid is shown in the figure. The volume of this solid is
A. $24 \pi - 6$
B. $8 \pi - 6$
C. $24 \pi + 6$
D. $8 \pi + 6$

7. Let $\alpha , \beta$ be two planes. Then $\alpha / / \beta$ is a necessary and sufficient condition for
A. There are infinitely many lines in $\alpha$ parallel to $\beta$
B. There are two intersecting lines in $\alpha$ parallel to $\beta$
C. $\alpha , \beta$ are both parallel to the same line
D. $\alpha , \beta$ are both perpendicular to the same plane

7. Let $\alpha , \beta$ be two planes. Then $\alpha / / \beta$ is a necessary and sufficient condition for
A. $\alpha$ contains infinitely many lines parallel to $\beta$
B. $\alpha$ contains two intersecting lines parallel to $\beta$
C. $\alpha , \beta$ are both parallel to the same line
D. $\alpha , \beta$ are both perpendicular to the same plane

8. The figure shows a flowchart for computing $\frac { 1 } { 2 + \frac { 1 } { 2 + \frac { 1 } { 2 } } }$ . The blank box should be filled with [Figure]
A. $A = \frac { 1 } { 2 + A }$
B. $A = 2 + \frac { 1 } { A }$
C. $A = \frac { 1 } { 1 + 2 A }$
D. $A = 1 + \frac { 1 } { 2 A }$

9. If the line $y = k x - 2$ is tangent to the curve $y = 1 + 3 \ln x$, then $k =$
A. $2$ B. $\frac { 1 } { 3 }$
C. $3$ D. $\frac { 1 } { 2 }$

A certain geometric solid is obtained by removing a quadrangular prism from a cube. Its three-view drawing is shown in the figure. If the side length of each small square on the grid paper is 1, then the volume of this geometric solid is $\_\_\_\_$.

Let $l , m$ be two different lines outside plane $\alpha$. Three propositions are given: (1) $l \perp m$; (2) $m \| \alpha$; (3) $l \perp \alpha$.
Using two of these propositions as conditions and the remaining one as the conclusion, write out a correct proposition: $\_\_\_\_$.

12. Given that the four vertices of tetrahedron $P - A B C$ lie on the surface of sphere $O$, with $P A = P B = P C$, $\triangle A B C$ is an equilateral triangle with side length 2, $E , F$ are the midpoints of $P A , A B$ respectively, and $\angle C E F = 90 ^ { \circ }$, then the volume of sphere $O$ is
A. $8 \sqrt { 6 } \pi$
B. $4 \sqrt { 6 } \pi$
C. $2 \sqrt { 6 } \pi$
D. $\sqrt { 6 } \pi$
II. Fill-in-the-Blank Questions: This section has 4 questions, each worth 5 points, for a total of 20 points.

12. Given that the four vertices of tetrahedron $P - A B C$ lie on the surface of sphere $O$ , with $P A = P B = P C$ , $\triangle A B C$ is an equilateral triangle with side length 2, $E , F$ are the midpoints of $P A , A B$ respectively, and $\angle C E F = 90 ^ { \circ }$ , then the volume of sphere $O$ is
A. $8 \sqrt { 6 } \pi$
B. $4 \sqrt { 6 } \pi$
C. $2 \sqrt { 6 } \pi$
D. $\sqrt { 6 } \pi$
Section II: Fill-in-the-Blank Questions: This section has 4 questions, each worth 5 points, for a total of 20 points.

Li Ming started his own business and operates a fruit shop online, selling strawberries, Chinese pears, watermelons, and peaches at prices of 60 yuan/box, 65 yuan/box, 80 yuan/box, and 90 yuan/box respectively. To increase sales, Li Ming offers a promotion: when the total price of a purchase reaches 120 yuan, the customer pays $x$ yuan less. After a customer successfully pays online, Li Ming receives 80\% of the payment. (1) When $x = 10$, if a customer purchases 1 box each of strawberries and watermelons, the amount to be paid is $\_\_\_\_$ yuan; (2) In the promotion activity, to ensure that Li Ming receives at least 70\% of the pre-promotion total price for each order, the maximum value of $x$ is $\_\_\_\_$.

Students engage in labor practice at a factory using 3D printing technology to create models. As shown in the figure, the model is a rectangular prism $ABCD - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ with a square pyramid $O - EFGH$ removed, where $O$ is the center of the rectangular prism, and $E , F , G , H$ are the midpoints of the respective edges. $AB = BC = 6 \mathrm{~cm} , AA _ { 1 } = 4 \mathrm{~cm}$. The density of the 3D printing material is $0.9 \mathrm{~g} / \mathrm{cm} ^ { 3 }$. Disregarding printing losses, the mass of material needed to create this model is \_\_\_\_\_\_.

16. China has a long history of stone and seal culture, and seals are representatives of this culture. Seals are usually shaped as rectangular prisms, cubes, or cylinders, but the seal of the official Dugu Xin from the Northern and Southern Dynasties is shaped as a ``semi-regular polyhedron'' (Figure 1). A semi-regular polyhedron is a polyhedron formed by two or more types of regular polygons. Semi-regular polyhedra embody the symmetry beauty of mathematics. Figure 2 shows a semi-regular polyhedron with 48 edges, with all vertices on the surface of a cube with edge length 1. This semi-regular polyhedron has $\_\_\_\_$ faces, and its edge length is $\_\_\_\_$. (The first blank is worth 2 points, the second blank is worth 3 points.)
[Figure]
Figure 1
[Figure]
Figure 2
III. Solution Questions: 70 points total. Solutions should include written explanations, proofs, or calculation steps. Questions 17--21 are required questions that all candidates must answer. Questions 22 and 23 are optional questions; candidates should answer according to requirements. (I) Required Questions: 60 points total.

16. China has a long history of stone and seal culture, and seals are a representative of this culture. Seals are usually shaped as rectangular prisms, cubes, or cylinders, but the seal of the official Dugu Xin from the Northern and Southern Dynasties is shaped as a ``semi-regular polyhedron'' (Figure 1). A semi-regular polyhedron is a polyhedron formed by two or more types of regular polygons. Semi-regular polyhedra embody the symmetry of mathematics. Figure 2 shows a semi-regular polyhedron with 48 edges. All its vertices lie on the surface of a cube with edge length 1. This semi-regular polyhedron has $\_\_\_\_$ faces and edge length $\_\_\_\_$ . (The first blank is worth 2 points, the second blank is worth 3 points.)
[Figure]
Figure 1
[Figure]
Figure 2

III. Solution Questions: 70 points total. Show your work, proofs, or calculation steps. Questions 17-21 are required for all students. Questions 22 and 23 are optional; students answer according to requirements.

(A) Required Questions: 60 points total.

16. Students participate in labor practice at a factory using 3D printing technology to make models. As shown in the figure, the model is a rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ with a square pyramid $O - E F G H$ removed, where $O$ is the center of the rectangular prism, and $E , F , G , H$ are the midpoints of their respective edges. $A B = B C = 6 \mathrm {~cm} , ~ A A _ { 1 } = 4 \mathrm {~cm}$ . The density of the 3D printing material is $0.9 \mathrm {~g} / \mathrm { cm } ^ { 3 }$ . Disregarding printing waste, the mass of material needed to make this model is $\_\_\_\_$ g. [Figure]
III. Solution Questions: Total 70 points. Solutions should include written explanations, proofs, or calculation steps. Questions 17--21 are required questions that all candidates must answer. Questions 22 and 23 are optional questions; candidates answer according to requirements.
(A) Required Questions: Total 60 points.

16. Students participate in labor practice at a factory using 3D printing technology to make models. As shown in the figure, the model is the solid obtained by removing the square pyramid $O - E F G H$ from the rectangular prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ , where $O$ is the center of the rectangular prism, and $E , F , G , H$ are the midpoints of their respective edges. Given $A B = B C = 6 \mathrm {~cm} , ~ A A _ { 1 } = 4 \mathrm {~cm}$ , and the density of the 3D printing material is $0.9 \mathrm {~g} / \mathrm { cm } ^ { 3 }$ . Disregarding printing waste, the mass of material needed to make this model is $\_\_\_\_$ g. [Figure]

III. Solution Questions: Total 70 points. Solutions should include written explanations, proofs, or calculation steps. Questions 17-21 are required for all students. Questions 22 and 23 are optional; students answer according to requirements.

(A) Required Questions: Total 60 points.

\section*{18. (12 Let $CF = a$. In $\triangle AFM$, $AM = \sqrt{3}$, $FM = \sqrt{a^2 + 1}$, $AF = \sqrt{a^2 + 4}$.
Then $\frac{1}{2} \times \sqrt{3} \times \sqrt{a^2 + 1} = \frac{1}{2} \times \sqrt{a^2 + 4} \times \frac{\sqrt{30}}{5}$. Solving, we get $a = 1$.
Thus $V_{C-AFM} = V_{F-ACM} = \frac{1}{3} \times 1 \times \frac{1}{2} \times \frac{\sqrt{3}}{4} \times 2^2 = \frac{\sqrt{3}}{6}$.

18. (12 points) As shown in the figure, the right prism $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ has a rhombus base, with $A A _ { 1 } = 4 , A B = 2 , \angle B A D = 60 ^ { \circ }$ . Let $E , M , N$ be the midpoints of $B C$ , $B B _ { 1 Therefore $f(x)$ has a unique zero point on $\left[\frac{\pi}{2}, \pi\right]$.
(iv) When $x \in (\pi, +\infty)$, $\ln(x+1) > 1$, so $f(x) < 0$, thus $f(x)$ has no zero points on $(\pi, +\infty)$.
In conclusion, $f(x)$ has exactly 2 zero points.

Given a sequence $\left\{ a _ { n } \right\}$, if we select the $i _ { 1 }$-th term, the $i _ { 2 }$-th term, $\cdots$, the $i _ { m }$-th term $\left( i _ { 1 } < i _ { 2 } < \cdots < i _ { m } \right)$, and if $a _ { i _ { 1 } } < a _ { i _ { 2 } } < \cdots < a _ { i _ { m } }$, then the new sequence $a _ { i _ { 1 } } , a _ { i _ { 2 } } , \cdots, a _ { i _ { m } }$ is called an increasing subsequence of length $m$ of $\left\{ a _ { n } \right\}$. By convention, any single term of the sequence $\left\{ a _ { n } \right\}$ is an increasing subsequence of length 1 of $\left\{ a _ { n } \right\}$. (I) Write out an increasing subsequence of length 4

20. (1) Solution: $\because e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{m}{m+2}} = \sqrt{\frac{2}{m+2}}$
Since $m > 1$, $\therefore 0 < e < \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$.
\begin{tabular}{|l|l|} \hline $\therefore e \in \left(0, \frac{\sqrt{6}}{3}\right)$ & (2) Proof: Since the major axis length of the ellipse is $2\sqrt{m+2} = 4$, $\therefore m = 2$. \hline The equation of line $BM$ is $y = -\frac{y_0}{4}(x-2)$, i.e., $y = -\frac{y_0}{4}x + \frac{1}{2}y_0$. & Substituting into the ellipse equation $x^2 + 2y^2 = 4$, \hline By Vieta's formulas, $2x_1 = \frac{4(y_0^2 - 8)}{y_0^2 + 8}$, & 9 marks \hline 10 marks & \hline $\therefore x_1 = \frac{2(y_0^2 - 8)}{y_0^2 + 8}$, $\therefore y_1 = \frac{8y_0}{y_0^2 + 8}$, $\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$ $\therefore \overrightarrow{OP} \cdot \overrightarrow{OM} = -2x_1 + y_0y_1 = -\frac{4(y_0^2 - 8)}{y_0^2 + 8} + \frac{8y_0^2}{y_0^2 + 8} = \frac{4y_0^2 + 32}{y_0^2 + 8} = 4 = \mathbf{2m}$. & 12 marks \hline

21. Solution: (1) When $a = 0$, it clearly does not satisfy the problem conditions, so $a \neq 0$. & 1 mark \hline $f'(x) = 3ax^2 - 2x$. Setting $f'(x) = 0$, we get $x = 0$ or $x = \frac{2}{3a}$. & $f'(x) = 3ax^2 - 2x$. Setting $f'(x) = 0$, we get $x = 0$ or $x = \frac{2}{3a}$. \hline From the problem conditions, $1 < \frac{2}{3a} < 3$. Solving, we get $\frac{2}{9} < a < \frac{2}{3}$, i.e., the range of $a$ is $\left(\frac{2}{9}, \frac{2}{3}\right)$. & 4 marks \hline (2) When $a = 0$, $f(x) = -x^2$ has minimum value $f(2) = -4$ on $[-1, 2]$. & 5 marks \hline Since $x \in [-1, 2]$, $\therefore 3x - \frac{2}{a} \leq 0$. & When $0 < a \leq \frac{1}{3}$, $\frac{2}{a} \geq 6$. $f'(x) = ax\left(3x - \frac{2}{a}\right)$. \hline 6 marks & \hline $\because f(2) - f(-1) = (8a - 4) - (-a - 1) = 9a - 3 \leq 0$, $\therefore f(x)_{\min} = f(2) = 8a - 4$. & 7 marks \hline When $a > \frac{1}{3}$, $f'(x) = ax\left(3x - \frac{2}{a}\right)$, $0 < \frac{2}{3a} < 2$. When $x \in [-1, 0) \cup \left(\frac{2}{3a}, 2\right]$, $f'(x) > 0$; & \hline When $x \in \left(0, \frac{2}{3a}\right)$, $f'(x) < 0$. & 8 marks \hline $\therefore f(x)_{\min} = \min\left\{f(-1), f\left(\frac{2}{3a}\right)\right\}$. & \hline 9 marks & \hline Since $a > \frac{1}{3}$, $\therefore 27a^3 + 27a^2 - 4 > 0$, $\frac{27a^3 + 27a^2 - 4}{27a^2} > 0$. & 10 marks \hline $\therefore f(x)_{\min} = f(-1) = -a - 1$. & 11 marks \hline In summary, when $0 \leq a \leq \frac{1}{3}$, $f(x)_{\min} = 8a - 4$; when $a > \frac{1}{3}$, $f(x)_{\min} = -a - 1$. & 12 marks \hline 1 mark & \hline