gaokao

Papers (66)
2025
national-I 19 national-II 18 national-I_eol 19
2024
beijing 21 national-II 19 national-I_github 19
2023
national-A-science 18 national-B-science 16 national-II 8
2022
national-A-arts 15 national-A-science 14 national-B-arts 20 national-B-science 17 national-I 19 national-II 7
2021
national-A-arts 19 national-A-science 18 national-I 14 national-II 12
2020
national-I-arts 20 national-I-science 9 national-II-science 17 national-III-arts 22 national-III-science 16 shanghai 17
2019
beijing-science 15 national-I-arts 13 national-I-science 12 national-I-science_gkztc 14 national-II-arts 13 national-II-science 9 national-II-science_gkztc 13 national-III-arts 14 national-III-science 16 national-III-science_gkztc 14
2018
national-I-arts 14 national-I-science 13 national-II-arts 20 national-II-science 18 national-III-science 17
2017
national-I-arts 13 national-I-science 14 national-II-arts 15 national-II-science 8
2016
national-I 7
2015
anhui-arts 14 beijing-arts 15 beijing-science 13 chongqing-arts 20 chongqing-science 15 fujian-science 18 hubei-arts 16 hubei-science 15 hunan-arts 21 hunan-science 11 jiangsu 19 national-II-arts 15 national-II-science 16 shaanxi-science 17 sichuan-arts 15 sichuan-science 21 tianjin-arts 17 tianjin-science 17 zhejiang-arts 15 zhejiang-science 15
Unknown
sample_questions 7
2015 hubei-science

15 maths questions

Q5 Complex Numbers Arithmetic Powers of i or Complex Number Integer Powers View
5. After the examination ends, please submit both this examination paper and the answer sheet.
I. Multiple Choice Questions: This section has 10 questions, each worth 5 points, for a total of 50 points. For each question, only one of the four options is correct.
1. Let $i$ be the imaginary unit. The conjugate of $\mathrm{i}^{607}$ is
A. $i$
B. $-i$
C. $1$
D. $-1$
2. In the ancient Chinese mathematical classic ``Mathematical Treatise in Nine Sections,'' there is a problem on ``grain and millet separation.'' A grain warehouse receives 1534 stones of rice. Upon inspection, the rice contains mixed millet. A sample of rice is taken, and among 254 grains, 28 are millet. Approximately how much millet is in this batch of rice?
A. 134 stones
B. 169 stones
C. 338 stones
D. 1365 stones
3. In the expansion of $(1+x)^n$, the binomial coefficients of the 4th term and the 8th term are equal. The sum of the binomial coefficients of odd-numbered terms is
A. $2^{12}$
B. $2^{11}$
C. $2^{10}$
D. $2^9$
4. Let $X \sim N(\mu_1, \sigma_1^2)$ and $Y \sim N(\mu_2, \sigma_2^2)$. The density curves of these two normal distributions are shown in the figure. Which of the following conclusions is correct?
A. $P(Y \geq \mu_2) \geq P(Y \geq \mu_1)$
B. $P(X \leq \sigma_2) \leq P(X \leq \sigma_1)$
C. For any positive number $t$, $P(X \leq t) \geq P(Y \leq t)$
D. For any positive number $t$, $P(X \geq t) \geq P(Y \geq t)$
[Figure]
Figure for Question 4
5. Let $a_1, a_2, \ldots, a_n \in \mathbf{R}$, $n \geq 3$. If $p$: $a_1, a_2, \ldots, a_n$ form a geometric sequence; $q$: $(a_1^2 + a_2^2 + \cdots + a_{n-1}^2)(a_2^2 + a_3^2 + \cdots + a_n^2) = (a_1a_2 + a_2a_3 + \cdots + a_{n-1}a_n)^2$, then
A. $p$ is a sufficient but not necessary condition for $q$
B. $p$ is a necessary but not sufficient condition for $q$
C. $p$ is a sufficient and necessary condition for $q$
D. $p$ is neither a sufficient nor a necessary condition for $q$
Q6 Modulus function Algebraic identities and properties of modulus View
6. The sign function is defined as $\operatorname{sgn} x = \begin{cases} 1, & x > 0, \\ 0, & x = 0, \\ -1, & x < 0. \end{cases}$ Let $f(x)$ be an increasing function on $\mathbf{R}$, and $g(x) = f(x) - f(ax)$ where $a > 1$. Then
A. $\operatorname{sgn}[g(x)] = \operatorname{sgn} x$
B. $\operatorname{sgn}[g(x)] = -\operatorname{sgn} x$
C. $\operatorname{sgn}[g(x)] = \operatorname{sgn}[f(x)]$
D. $\operatorname{sgn}[g(x)] = -\operatorname{sgn}[f(x)]$
Q7 Geometric Probability View
7. Two numbers $x$ and $y$ are randomly chosen from the interval $[0,1]$. Let $p_1$ be the probability of the event ``$x + y \geq \frac{1}{2}$'', $p_2$ be the probability of the event ``$|x - y| \leq \frac{1}{2}$'', and $p_3$ be the probability of the event ``$xy \leq \frac{1}{2}$''. Then
A. $p_1 < p_2 < p_3$
B. $p_2 < p_3 < p_1$
C. $p_3 < p_1 < p_2$
D. $p_3 < p_2 < p_1$
Q8 Circles Circle Identification and Classification View
8. The eccentricity of hyperbola $C_1$ is $e_1$. Both the semi-major axis $a$ and semi-minor axis $b$ (where $a \neq b$) are increased by $m$ units (where $m > 0$) to obtain hyperbola $C_2$ with eccentricity $e_2$. Then
A. For any $a, b$, we have $e_1 > e_2$
B. When $a > b$, $e_1 > e_2$; when $a < b$, $e_1 < e_2$
C. For any $a, b$, we have $e_1 < e_2$
D. When $a > b$, $e_1 < e_2$; when $a < b$, $e_1 > e_2$
Q9 Parametric curves and Cartesian conversion View
9. Let $A = \{(x,y) \mid x^2 + y^2 \leq 1, x, y \in \mathbf{Z}\}$, $B = \{(x,y) \mid |x| \leq 2, |y| \leq 2, x, y \in \mathbf{Z}\}$. Define $A \oplus B = \{(x_1 + x_2, y_1 + y_2) \mid (x_1, y_1) \in A, (x_2, y_2) \in B\}$. The number of elements in $A \oplus B$ is
A. 77
B. 49
C. 45
D. 30
Q10 Trig Graphs & Exact Values View
10. Let $x \in \mathbf{R}$ and $[x]$ denote the greatest integer not exceeding $x$. If there exists a real number $t$ such that $[t] = 1$, $[t^2] = 2$, $\ldots$, $[t^n] = n$ all hold simultaneously, then the maximum value of the positive integer $n$ is
A. 3
B. 4
C. 5
D. 6
II. Fill-in-the-Blank Questions: This section has 6 questions. Candidates must answer 5 of them, each worth 5 points, for a total of 25 points. Write your answers in the corresponding positions on the answer sheet. Answers in wrong positions, illegible writing, or ambiguous answers will receive no credit.
(A) Compulsory Questions (Questions 11-14)
Q11 Vectors Introduction & 2D Dot Product Computation View
11. Given vectors $\overrightarrow{OA} \perp \overrightarrow{AB}$ and $|\overrightarrow{OA}| = 3$, then $\overrightarrow{OA} \cdot \overrightarrow{OB} = $ $\_\_\_\_$ .
Q12 Curve Sketching Number of Solutions / Roots via Curve Analysis View
12. The number of zeros of the function $f(x) = 4\cos^2\frac{x}{2}\cos\left(\frac{\pi}{2} - x\right) - 2\sin x - |\ln(x+1)|$ is $\_\_\_\_$ .
Q13 Sine and Cosine Rules Heights and distances / angle of elevation problem View
13. As shown in the figure, a car is traveling due west on a horizontal road. At point $A$, the mountain peak $D$ on the north side of the road is measured to be in the direction $30°$ west of north. After traveling 600 m to reach point $B$, the peak is measured to be in the direction $75°$ west of north with an elevation angle of $30°$. The height of the mountain $CD = $ $\_\_\_\_$ m.
[Figure]
Figure for Question 13
[Figure]
Figure for Question 14
Q14 Circles Chord Length and Chord Properties View
14. As shown in the figure, circle $C$ is tangent to the $x$-axis at point $T(1,0)$ and intersects the positive $y$-axis at two points $A$ and $B$ (with $B$ above $A$), and $|AB| = 2$. (I) The standard equation of circle $C$ is $\_\_\_\_$ ; (II) A line is drawn through point $A$ intersecting circle $O: x^2 + y^2 = 1$ at points $M$ and $N$. Consider the following three conclusions:
(1) $\frac{|NA|}{|NB|} = \frac{|MA|}{|MB|}$ ;
(2) $\frac{|NB|}{|NA|} - \frac{|MA|}{|MB|} = 2$ ;
(3) $\frac{|NB|}{|NA|} + \frac{|MA|}{|MB|} = 2\sqrt{2}$ .
The correct conclusion(s) is/are $\_\_\_\_$ . (Write the numbers of all correct conclusions)
(B) Optional Questions (Choose one of questions 15 and 16 to answer. First fill in the box after the question number you choose on the answer sheet with a 2B pencil. If you choose both, only question 15 will be graded.)
Q15 Circles Tangent Lines and Tangent Lengths View
15. (Elective 4-1: Geometric Proof) As shown in the figure, $PA$ is tangent to the circle at point $A$, and $PBC$ is a secant line with $BC = 3PB$. Then $\frac{AB}{AC} = $ $\_\_\_\_$ .
Q16 Polar coordinates View
16. (Elective 4-4: Coordinate Systems and Parametric Equations)
[Figure]
Figure for Question 15
In the rectangular coordinate system $xOy$, establish a polar coordinate system with $O$ as the pole and the positive $x$-axis as the polar axis. The polar equation of line $l$ is $\rho(\sin\theta - 3\cos\theta) = 0$. The parametric equation of curve $C$ is $\begin{cases} x = t - \frac{1}{t}, \\ y = t + \frac{1}{t} \end{cases}$ (where $t$ is the parameter). If $l$ and $C$ intersect at points $A$ and $B$, then $|AB| = $ $\_\_\_\_$ .
III. Solution Questions: This section has 6 questions, for a total of 75 points. Show your work, proofs, or calculation steps.
Q17 11 marks Completing the square and sketching Determining coefficients from given conditions on function values or geometry View
17. (11 points)
A student uses the ``five-point method'' to sketch the graph of $f(x) = A\sin(\omega x + \varphi)$ (where $\omega > 0$, $|\varphi| < \frac{\pi}{2}$) over one period and creates a table with partial data filled in as follows:
\begin{tabular}{ | c | c | c | c | c | c | } \hline $\omega x + \varphi$ & 0 & $\frac{\pi}{2}$ & $\pi$ & $\frac{3\pi}{2}$ & $2\pi$ \hline $x$ & & $\frac Thus $\overrightarrow { P B } \cdot \overrightarrow { D E } = 0$, that is, $P B \perp D E$. Since $E F \perp P B$ and $D E \cap E F = E$, we have $P B \perp$ plane $D E F$. Because $\overrightarrow { P C } = ( 0,1 , - 1 ) , \overrightarrow { D E } \cdot \overrightarrow { P C } = 0$, then $D E \perp P C$, so $D E \perp$ plane $P B C$. From $D E \perp$ plane $P B C$ and $P B \perp$ plane $D E F$, we know that all four faces of tetrahedron $B D E F$ are right triangles, that is, tetrahedron $B D E F$ is an orthocentric tetrahedron, with right angles on its four faces being $\angle D E B , \angle D E F , \angle E F B , \angle D F B$ respectively.
[Figure]
Solution diagram 1 for Question 19
[Figure]
Solution diagram 2 for Question 19
(II) Since $P D \perp$ plane $A B C D$, we have $\overrightarrow { D P } = ( 0,0,1 )$ is a normal vector to plane $A B C D$; From (I), $P B \perp$ plane $D E F$, so $\overrightarrow { B P } = ( - \lambda , - 1,1 )$ is a normal vector to plane $D E F$. If the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, then $\cos \frac { \pi } { 3 } = \left| \frac { \overrightarrow { B P } \cdot \overrightarrow { D P } } { | \overrightarrow { B P } | \cdot | \overrightarrow { D P } | } \right| = \left| \frac { 1 } { \sqrt { \lambda ^ { 2 } + 2 } } \right| = \frac { 1 } { 2 }$, solving gives $\lambda = \sqrt { 2 }$. Therefore $\frac { D C } { B C } = \frac { 1 } { \lambda } = \frac { \sqrt { 2 } } { 2 }$. Thus when the dihedral angle between plane $D E F$ and plane $A B C D$ is $\frac { \pi } { 3 }$, we have $\frac { D C } { B C } = \frac { \sqrt { 2 } } { 2 }$.
Q20 12 marks Completing the square and sketching Optimization on a constrained domain via completing the square View
20. (12 marks) (I) Let the daily production quantities of products $A$ and $B$ be $x$ and $y$ respectively, with corresponding profit $z$. Then we have
$$\left\{ \begin{array} { l } 2 x + 1.5 y \leq W \\ x + 1.5 y \leq 12 \\ 2 x - y \geq 0 \\ x \geq 0 , \quad y \geq 0 \end{array} \right.$$
The objective function is $z = 1000x + 1200y$.
[Figure]
Solution diagram 1 for Question 20
[Figure]
Solution diagram 2 for Question 20
[Figure]
Solution diagram 3 for Question 20
When $W = 12$, the planar region represented by (1) is shown in Figure 1, with three vertices $A ( 0,0 ) , B ( 2.4,4.8 ) , C ( 6,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 2.4 , y = 4.8$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 2.4 \times 1000 + 4.8 \times 1200 = 8160$. When $W = 15$, the planar region represented by (1) is shown in Figure 2, with three vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 7.5,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 3 , y = 6$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 3 \times 1000 + 6 \times 1200 = 10200$. When $W = 18$, the planar region represented by (1) is shown in Figure 3, with four vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 6,4 ) , D ( 9,0 )$ respectively. Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$, when $x = 6 , y = 4$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept, maximum profit $Z = z _ { \max } = 6 \times 1000 + 4 \times 1200 = 10800$. Thus the distribution of maximum profit $Z$ is
$Z$81601020010800
$P$0.30.50.2

Therefore, $E ( Z ) = 8160 \times 0.3 + 10200 \times 0.5 + 10800 \times 0.2 = 9708$. (II) From (I), the probability that daily maximum profit exceeds 10000 yuan is $p _ { 1 } = P ( Z > 10000 ) = 0.5 + 0.2 = 0.7$. By the binomial distribution, the probability that at least one day out of 3 days has maximum profit exceeding 10000 yuan is $p = 1 - \left( 1 - p _ { 1 } \right) ^ { 3 } = 1 - 0.3 ^ { 3 } = 0.973$.
Q21 14 marks Circles Circle-Related Locus Problems View
21. (14 marks) (I) Let point $D ( t , 0 ) ( | t | \leq 2 ) , N \left( x _ { 0 } , y _ { 0 } \right) , M ( x , y )$. According to the problem,
$$\overrightarrow { M D } = 2 \overrightarrow { D N } \text {, and } | \overrightarrow { D N } | = | \overrightarrow { O N } | = 1,$$
[Figure]
so $( t - x , - y ) = 2 \left( x _ { 0 } - t , y _ { 0 } \right)$, and $\left\{ \begin{array} { l } \left( x _ { 0 } - t \right) ^ { 2 } + y _ { 0 } ^ { 2 } = 1 , \\ x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } = 1 . \end{array} \right.$ That is, $\left\{ \begin{array} { l } t - x = 2 x _ { 0 } - 2 t , \\ y = - 2 y _ { 0 } . \end{array} \right.$ and $t \left( t - 2 x _ { 0 } \right) = 0$. Since when point $D$ is fixed, point $N$ is also fixed, $t$ is not identically zero, thus $t = 2 x _ { 0 }$, so $x _ { 0 } = \frac { x } { 4 } , y _ { 0 } = - \frac { y } { 2 }$. Substituting into $x _ { 0 } ^ { 2 } + y _ { 0 } ^ { 2 } = 1$, we obtain $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$, that is, the equation of the required curve $C$ is $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$. (II) (1) When the slope of line $l$ does not exist, line $l$ is $x = 4$ or $x = - 4$, and we have $S _ { \triangle O P Q } = \frac { 1 } { 2 } \times 4 \times 4 = 8$.
(2) When the slope of line $l$ exists, let line $l : y = k x + m \left( k \neq \pm \frac { 1 } { 2 } \right)$. From $\left\{ \begin{array} { l } y = k x + m , \\ x ^ { 2 } + 4 y ^ { 2 } = 16 , \end{array} \right.$ eliminating $y$, we obtain $\left( 1 + 4 k ^ { 2 } \right) x ^ { 2 } + 8 k m x + 4 m ^ { 2 } - 16 = 0$. Since line $l$ always has exactly one common point with ellipse $C$, we have $\Delta = 64 k ^ { 2 } m ^ { 2 } - 4 \left( 1 + 4 k ^ { 2 } \right) \left( 4 m ^ { 2 } - 16 \right) = 0$, that is, $m ^ { 2 } = 16 k ^ { 2 } + 4$.
From $\left\{ \begin{array} { l } y = k x + m , \\ x - 2 y = 0 , \end{array} \right.$ we obtain $P \left( \frac { 2 m } { 1 - 2 k } , \frac { m } { 1 - 2 k } \right)$; similarly, we obtain $Q \left( \frac { - 2 m } { 1 + 2 k } , \frac { m } { 1 + 2 k } \right)$. From the distance from origin $O$ to line $P Q$ being $d = \frac { | m | } { \sqrt { 1 + k ^ { 2 } } }$ and $| P Q | = \sqrt { 1 + k ^ { 2 } } \left| x _ { P } - x _ { Q } \right|$, we obtain $S _ { \triangle O P Q } = \frac { 1 } { 2 } | P Q | \cdot d = \frac { 1 } { 2 } | m | \left| x _ { P } - x _ { Q } \right| = \frac { 1 } { 2 } \cdot | m | \left| \frac { 2 m } { 1 - 2 k } + \frac { 2 m } { 1 + 2 k } \right| = \left| \frac { 2 m ^ { 2 } } { 1 - 4 k ^ { 2 } } \right|$.
Substituting (1) into (2), we obtain $S _ { \triangle O P Q } = \left| \frac { 2 m ^ { 2 } } { 1 - 4 k ^ { 2 } } \right| = 8 \frac { \left| 4 k ^ { 2 } + 1 \right| } { \left| 4 k ^ { 2 } - 1 \right| }$. When $k ^ { 2 } > \frac { 1 } { 4 }$, $S _ { \triangle O P Q } = 8 \left( \frac { 4 k ^ { 2 } + 1 } { 4 k ^ { 2 } - 1 } \right) = 8 \left( 1 + \frac { 2 } { 4 k ^ { 2 } - 1 } \right) > 8$; When $0 \leq k ^ { 2 } < \frac { 1 } { 4 }$, $S _ { \triangle O P Q } = 8 \left( \frac { 4 k ^ { 2 } + 1 } { 1 - 4 k ^ { 2 } } \right) = 8 \left( - 1 + \frac