gaokao

Papers (71)
2025
national-I 18 national-II 18 national-I_eol 18
2024
beijing 19 national-II 18 national-I_github 18
2023
national-A-science 20 national-B-science 15 national-II 7
2022
national-A-arts 13 national-A-science 11 national-B-arts 19 national-B-science 17 national-I 18 national-II 7
2021
national-A-arts 19 national-A-science 16 national-I 12 national-II 10
2020
national-I-arts 21 national-I-science 9 national-II-science 14 national-III-arts 22 national-III-science 16 shanghai 14
2019
beijing-science 17 national-I-arts 17 national-I-science 13 national-I-science_gkztc 18 national-II-arts 13 national-II-science 8 national-II-science_gkztc 14 national-III-arts 11 national-III-science 16 national-III-science_gkztc 15
2018
national-I-arts 14 national-I-science 13 national-II-arts 20 national-II-science 19 national-III-science 17
2017
national-I-arts 12 national-I-science 13 national-II-arts 14 national-II-science 9
2016
national-I 7
2015
anhui-arts 14 beijing-arts 18 beijing-science 16 chongqing-arts 16 chongqing-science 13 fujian-science 18 hubei-arts 18 hubei-science 15 hunan-arts 20 hunan-science 11 jiangsu 21 national-II-arts 16 national-II-science 16 shaanxi-science 18 sichuan-arts 16 sichuan-science 21 tianjin-arts 18 tianjin-science 17 zhejiang-arts 14 zhejiang-science 15
2011
shanghai-arts 21
2010
shanghai-arts 20 shanghai-science 12
2004
shanghai-science 17
Unknown
sample_questions 3 sample_questions_testmocks 6
2019 national-I-arts

17 maths questions

Q3 Complex Numbers Arithmetic Powers of i or Complex Number Integer Powers View
3. The main content of this test paper covers all content of the college entrance examination.
Section I
I. Multiple Choice Questions: This section contains 12 questions, each worth 5 points, totaling 60 points. For each question, only one of the four options is correct.
1. The conjugate of the complex number $z = \mathrm { i } ^ { 9 } ( - 1 - 2 \mathrm { i } )$ is
A. $2 + \mathrm { i }$
B. $2 - \mathrm { i }$
C. $- 2 + \mathrm { i }$
D. $- 2 - \mathrm { i }$
2. Let sets $A = \{ a , a + 1 \} , ~ B = \{ 1,2,3 \}$. If $A \cup B$ has 4 elements, then the set of possible values of $a$ is
A. $\{ 0 \}$
B. $\{ 0,3 \}$
C. $\{ 0,1,3 \}$
D. $\{ 1,2,3 \}$
3. For the hyperbola $C : \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > 0 , b > 0 )$, the length of the real axis and the focal distance are 2 and 4 respectively. The asymptote equations of hyperbola $C$ are
A. $y = \pm \frac { \sqrt { 3 } } { 3 } x$
B. $y = \pm \frac { 1 } { 3 } x$ C. $y = \pm \sqrt { 3 } x$
D. $y = \pm 3 x$
Q5 Curve Sketching Exponential Equation Solving View
5. The sum of the zeros of the function $f ( x ) = \left\{ \begin{array} { l } 6 ^ { x } - 2 , x > 0 , \\ x + \log _ { 6 } 12 , x \leq 0 \end{array} \right.$ is
A. $-1$ B. $1$ C. $-2$ D. $2$
Q6 Trig Graphs & Exact Values View
6. The monotonically increasing interval of the function $f ( x ) = \cos \left( 3 x + \frac { \pi } { 2 } \right)$ is
A. $\left[ \frac { \pi } { 6 } + \frac { 2 k \pi } { 3 } , \frac { \pi } { 2 } + \frac { 2 k \pi } { 3 } \right] ( k \in \mathbb{Z} )$
[Figure]
Front View
[Figure]
Top View
B. $\left[ \frac { \pi } { 6 } + \frac { k \pi } { 3 } , \frac { \pi } { 2 } + \frac { k \pi } { 3 } \right] ( k \in \mathbb{Z} )$
C. $\left[ \frac { \pi } { 6 } + \frac { k \pi } { 3 } , \frac { \pi } { 6 } + \frac { k \pi } { 3 } \right] ( k \in \mathbb{Z} )$
D. $\left[ - \frac { \pi } { 6 } + \frac { 2 k \pi } { 3 } , \frac { \pi } { 6 } + \frac { 2 k \pi } { 3 } \right] ( k \in \mathbb{Z} )$
Q8 Vectors Introduction & 2D Volume of a Region Defined by Inequalities in 3D View
8. Two unit vectors $e _ { 1 } , e _ { 2 }$ have an angle of $60 ^ { \circ }$ between them. Vector $m = t e _ { 1 } + 2 e _ { 2 } ( t < 0 )$. Then
A. The maximum value of $\frac { | m | } { t }$ is $\frac { \sqrt { 3 } } { 2 }$
B. The minimum value of $\frac { | m | } { t }$ is $- 2$
C. The minimum value of $\frac { | m | } { t }$ is $\frac { \sqrt { 3 } } { 2 }$
D. The maximum value of $\frac { | m | } { t }$ is $- 2$
Q9 Differentiating Transcendental Functions Evaluate derivative at a point or find tangent slope View
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. $\frac { 1 } { 2 }$
Q10 Inequalities Geometric properties of tangent lines (intersections, lengths, areas) View
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. $1 + \sqrt { 3 }$
Q11 Exponential Functions Quadratic Inequality Holding for All x (or a Restricted Domain) View
11. If the range of the function $f ( x ) = a \cdot \left( \frac { 1 } { 3 } \right) ^ { x } \left( \frac { 1 } { 2 } \leq x \leq 1 \right)$ is a subset of the range of the function $g ( x ) = \frac { x ^ { 2 } - 1 } { x ^ { 2 } + x + 1 } ( x \in \mathbb{R} )$, then the range of positive number $a$ is
A. $(0,2]$
B. $(0,1]$
C. $( 0,2 \sqrt { 3 } ]$
D. $( 0 , \sqrt { 3 } ]$
Q12 Sine and Cosine Rules Inscribed/Circumscribed Circle Computations View
12. In $\triangle A B C$, the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given $10 \sin A - 5 \sin C = 2 \sqrt { 6 }$ and $\cos B = \frac { 1 } { 5 }$, then $\frac { c } { a } =$
$$\text { A. } \frac { 6 } { 7 } \quad \text{B.} \frac { 7 } { 6 } \quad \text{C.} \frac { 5 } { 6 } \quad \text{D.} \frac { 6 } { 5 }$$
Section II
II. Fill-in-the-Blank Questions: This section contains 4 questions, each worth 5 points, totaling 20 points. Write your answers on the answer sheet.
Q13 Measures of Location and Spread View
13. A school will select one person from three candidates (A, B, C) to participate in the city-wide middle school boys' 1500-meter race. The mean and variance of their 10 recent training times (in seconds) are shown in the following table:
ABC
Mean280280290
Variance201616

Based on the data in the table, the school should select \_\_\_\_ to participate in the race.
Q14 Addition & Double Angle Formulae Addition/Subtraction Formula Evaluation View
14. Given $\tan \left( \alpha + \frac { \pi } { 4 } \right) = 6$, then $\tan \alpha = $ \_\_\_\_.
Q15 Circles Sphere and 3D Circle Problems View
15. A quadrangular pyramid $P - A B C D$ has all vertices on the surface of sphere $O$. $PA$ is perpendicular to the plane containing rectangle $A B C D$. $AB = 3$, $AD = \sqrt { 3 }$. The surface area of sphere $O$ is $13 \pi$. The length of segment $PA$ is \_\_\_\_.
Q16 Straight Lines & Coordinate Geometry Intersection and Distance between Curves View
16. A line $l$ with slope $k ( k < 0 )$ passes through point $F ( 0,1 )$ and intersects the curve $y = \frac { 1 } { 4 } x ^ { 2 } ( x \geq 0 )$ and the line $y = - 1$ at points $A$ and $B$ respectively. If $| F B | = 6 | F A |$, then $k = $ \_\_\_\_.
III. Solution Questions: This section contains 6 questions, totaling 70 points. Show your working, 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 the requirements. (I) Required Questions: Total 60 points.
Q17 12 marks Sequences and series, recurrence and convergence Sequence Defined by Recurrence with AP Connection View
17. (12 points)
A sequence $\left\{ a _ { n } \right\}$ satisfies $\frac { 1 } { a _ { n + 1 } } - \frac { 2 } { a _ { n } } = 0$, and $a _ { 1 } = \frac { 1 } { 2 }$.
(1) Find the general term formula of the sequence $\left\{ a _ { n } \right\}$;
(2) Find the sum $S _ { n }$ of the first $n$ terms of the sequence $\left\{ \frac { 1 } { a _ { n } } + 2 n \right\}$.
Q20 12 marks Conic sections Vector and Dot Product Conditions on Conics View
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
Q21 1 marks Stationary points and optimisation Determine parameters from given extremum conditions View
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
Q22 10 marks Parametric curves and Cartesian conversion View
22. Solution: (1) From the problem conditions, $|a| = 1$. Thus the parametric equation of $l$ is $\left\{\begin{array}{l} x = 4t + 1 \\ y = 3t - 1 \end{array}\right.$ ($t$ is the parameter). The parametric equation of circle $C$ is $\left\{\begin{array}{l} x = 1 + \cos\theta \\ y = -2 + \sin\theta \end{array}\right.$ ($\theta$ is the parameter). & \hline \end{tabular}
Eliminating parameter $t$, the ordinary equation of $l$ is $3x - 4y - 7 = 0$. ..... 3 marks
Eliminating parameter $\theta$, the ordinary equation of $C$ is $(x-1)^2 + (y+2)^2 = 1$. ..... 5 marks
(2) The equation of $l'$ is $y = \frac{3}{4}(x + m) - \frac{7}{4}$, i.e., $3x - 4y + 3m - 7 = 0$. ..... 6 marks
Since circle $C$ has only one point at distance $\mathbf{1}$ from $l'$, and the radius of circle $C$ is $\mathbf{1}$, the distance from $C(1, -2)$ to $l'$ is 2. ..... 8 marks
That is, $\frac{|3 + 8 + 3m - 7|}{5} = 2$. Solving, we get $m = 2$ ($m = -\frac{14}{3} < 0$ is rejected). ..... 10 marks
Q23 10 marks Inequalities Absolute Value Inequality View
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]