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Papers (71)
2025
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2024
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2023
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2022
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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
2010 shanghai-science

12 maths questions

Q1 Inequalities Solve Polynomial/Rational Inequality for Solution Set View
1. The solution set of the inequality $\frac { 2 - x } { x + 4 } > 0$ is $\_\_\_\_$ $( - 4,2 )$.
Analysis: This examines the method for solving fractional inequalities. $\frac { 2 - x } { x + 4 } > 0$ is equivalent to $( x - 2 ) ( x + 4 ) < 0$, so $- 4 < x < 2$
Q2 Complex Numbers Arithmetic Complex Division/Multiplication Simplification View
2. If the complex number $z = 1 - 2 i$ ($i$ is the imaginary unit), then $z \cdot \bar { z } + z =$ $\_\_\_\_$ $6 - 2i$.
Analysis: This examines basic operations with complex numbers. $z \cdot \bar { z } + z = ( 1 - 2 i ) ( 1 + 2 i ) + 1 - 2 i = 6 - 2 i$
Q3 Circles Circle-Related Locus Problems View
3. A moving point $P$ has equal distance to the point $F ( 2,0 )$ and to the line $x + 2 = 0$. Then the locus equation of $P$ is $y ^ { 2 } = 8 x$.
Analysis: This examines the definition and standard equation of a parabola. By the definition, the locus of $P$ is a parabola with focus $F ( 2,0 )$. Since $p = 2$, its equation is $y ^ { 2 } = 8 x$.
Q4 Reciprocal Trig & Identities View
4. The value of the determinant $\left| \begin{array} { c c } \cos \frac { \pi } { 3 } & \sin \frac { \pi } { 6 } \\ \sin \frac { \pi } { 3 } & \cos \frac { \pi } { 6 } \end{array} \right|$ is $\_\_\_\_$ $0$.
Analysis: This examines the rules for computing determinants. $\left| \begin{array} { c c } \cos \frac { \pi } { 3 } & \sin \frac { \pi } { 6 } \\ \sin \frac { \pi } { 3 } & \cos \frac { \pi } { 6 } \end{array} \right| = \cos \frac { \pi } { 3 } \cos \frac { \pi } { 6 } - \sin \frac { \pi } { 3 } \sin \frac { \pi } { 6 } = \cos \frac { \pi } { 2 } = 0$
Q5 Circles Distance from Center to Line View
5. The distance from the center of circle $C : x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 4 = 0$ to the line $l : 3 x + 4 y + 4 = 0$ is $d =$ $\_\_\_\_$ $3$. Analysis: This examines the point-to-line distance formula. The distance from the center $(1,2)$ to the line $3 x + 4 y + 4 = 0$ is $\frac { | 3 \times 1 + 4 \times 2 + 4 | } { 5 } = 3$
Q6 Discrete Probability Distributions Probability Distribution Table Completion and Expectation Calculation View
6. The probability distribution of the random variable $\xi$ is given in the table below:
$\mathbf { x }$7$\mathbf { 8 }$$\mathbf { 9 }$$\mathbf { 1 0 }$
$\mathbf { P } ( \xi = x )$0.30.350.20.15

Then the expected value of the random variable $\xi$ is $\_\_\_\_$ $8.2$
Analysis: This examines the definition of expected value. $\mathrm { E } \xi = 7 \times 0.3 + 8 \times 0.35 + 9 \times 0.2 + 10 \times 0.15 = 8.2$
Q8 Composite & Inverse Functions Find or Apply an Inverse Function Formula View
8. For any positive number $a$ not equal to 1, the graph of the inverse function of $f ( x ) = \log _ { a } ( x + 3 )$ always passes through point $P$. Then the coordinates of point $P$ are $\_\_\_\_$ $(0, -2)$
Analysis: The graph of $f ( x ) = \log _ { a } ( x + 3 )$ passes through the fixed point $( - 2,0 )$, so the graph of its inverse function passes through the fixed point $( 0 , - 2 )$
Q9 Probability Definitions Probability Using Set/Event Algebra View
9. A card is randomly drawn from a shuffled deck of playing cards (52 cards). Event $A$ is ``drawing the King of Hearts'', and event $B$ is ``drawing a Spade''. Then the probability $P ( A \cup B ) = $ $\_\_\_\_$ $\frac { 7 } { 26 }$ (express the result as a fraction in lowest terms).
Analysis: This examines the probability formula for mutually exclusive events. $P ( A \cup B ) = \frac { 1 } { 52 } + \frac { 13 } { 52 } = \frac { 7 } { 26 }$
Q11 Straight Lines & Coordinate Geometry Area Computation in Coordinate Geometry View
11. Let the area of the closed region bounded by the lines $l _ { 2 } : n x + y - n = 0$, $l _ { 3 } : x + n y - n = 0$ ($n \in \mathbb{N} ^ { * } , n \geq 2$), the $x$-axis, and the $y$-axis be denoted by $S _ { n }$. Then $\lim _ { n \rightarrow \infty } S _ { n } =$ $\_\_\_\_$ $1$.
Analysis: $B \left( \frac { n } { n + 1 } , \frac { n } { n + 1 } \right)$, so $OB \perp AC$, $S _ { n } = \frac { 1 } { 2 } \times \sqrt { 2 } \times \frac { n } { n + 1 } \sqrt { 2 } = \frac { n } { n + 1 }$, so $\lim _ { n \rightarrow \infty } S _ { n } = 1$ [Figure]
Q13 Conic sections Vector Algebra and Triple Product Computation View
13. As shown in the figure, the line $x = 2$ intersects the asymptotes of the hyperbola $\Gamma : \frac { x ^ { 2 } } { 4 } - y ^ { 2 } = 1$ at points $E _ { 1 }$ and $E _ { 2 }$. Let $\overrightarrow { O E _ { 1 } } = \overrightarrow { e _ { 1 } }$ and $\overrightarrow { O E _ { 2 } } = \overrightarrow { e _ { 2 } }$. For any point $P$ on the hyperbola $\Gamma$, if $\overrightarrow { O P } = a \overrightarrow { e _ { 1 } } + b \overrightarrow { e _ { 2 } }$ ($a , b \in \mathbb{R}$), then $a$ and $b$ satisfy the equation $\_\_\_\_$ $4 a b = 1$.
Analysis: $E _ { 1 } ( 2,1 )$, $E _ { 2 } ( 2 , - 1 )$
$$\overrightarrow { O P } = a \overrightarrow { e _ { 1 } } + b \overrightarrow { e _ { 2 } } = ( 2 a + 2 b , a - b ), \text{ point } P \text{ is on the hyperbola}$$
$\therefore \frac { ( 2 a + 2 b ) ^ { 2 } } { 4 } - ( a - b ) ^ { 2 } = 1$, which simplifies to $4 a b = 1$ [Figure]
Q14 Combinations & Selection Subset Counting with Set-Theoretic Conditions View
14. From the subsets of the set $U = \{ a , b , c , d \}$, select 2 different subsets that must satisfy both of the following conditions:
(1) Both $a$ and $b$ must be selected;
(2) For any two selected subsets $A$ and $B$, we must have $A \subseteq B$ or $B \subseteq A$. Then there are $\_\_\_\_$ $36$ different ways.
Analysis: By enumeration, there are 36 ways in total.
II. Multiple Choice Questions (Total Score: 20 points) This section contains 4 questions. Each question has exactly one correct answer. Candidates must shade the box corresponding to the correct answer on the answer sheet. Each correct answer is worth 5 points; otherwise, zero points are awarded.
15. ``$x = 2 k \pi + \frac { \pi } { 4 } ( k \in \mathbb{Z} )$'' is a \_\_\_\_ condition for ``$\tan x = 1$''. [Answer] (A)
(A) Sufficient but not necessary condition.
(B) Necessary but not sufficient condition.
(C) Sufficient condition.
(D) Neither sufficient nor necessary condition.
Analysis: $\tan \left( 2 k \pi + \frac { \pi } { 4 } \right) = \tan \frac { \pi } { 4 } = 1$, so it is sufficient; However, the converse does not hold. For example, $\tan \frac { 5 \pi } { 4 } = 1$, so it is not necessary.
Q16 Vectors 3D & Lines Parametric Representation of a Line View
16. The parametric equation of line $l$ is $\left\{ \begin{array} { l } x = 1 + 2 t \\ y = 2 - t \end{array} ( t \in \mathbb{R} ) \right.$. Then a direction vector $\overrightarrow { Therefore $\frac { ( 1 - \cos \theta ) ^ { 2 } } { 4 } + \frac { ( 1 + \sin \theta ) ^ { 2 } } { 4 } < 1$ , simplifying gives $\sin \theta - \cos \theta < \frac { 1 } { 2 } , \sin \left( \theta - \frac { \pi } { 4 } \right) < \frac { \sqrt { 2 } } { 4 }$ , Also $0 < \theta < \pi$ , that is $- \frac { \pi } { 4 } < \theta - \frac { \pi } { 4 } < \frac { 3 \pi } { 4 }$ , so $- \frac { \pi } { 4 } < \theta - \frac { \pi } { 4 } < \arcsin \frac { \sqrt { 2 } } { 4 }$ , Thus the range of values for $\theta$ is $\left( 0 , \frac { \pi } { 4 } + \arcsin \frac { \sqrt { 2 } } { 4 } \right)$ .