20 10 Guangdong junior high school graduates' academic examination
mathematics
1. Multiple-choice questions (5 small questions in this big question, 3 points for each small question, *** 15 points) Only one of the four options listed in each small question is correct. Please black out the options of the corresponding topic on the answer sheet.
1.(20 10 Dongguan, Guangdong 1, 3 points) -3 is the reciprocal of ().
3rd century BC to 3rd century BC.
Analyze the definition of inverse number: only two numbers with different signs are called reciprocal numbers, so the inverse of -3 is 3.
Answer a
It involves defining the opposite number of knowledge points.
The comment on this topic belongs to the basic topic, which mainly examines the grasp of the concept of reciprocal.
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2.(20 10, Dongguan, Guangdong, 2,3 points) The following operation is correct ()
A.B.
C.D.
Analysis is not that similar items cannot be combined. When applying the law of multiplication and distribution, the factors outside brackets should be multiplied by the factors inside brackets respectively, and the multiplication cannot be omitted.
Answer c
Similar items involving knowledge points, operation of algebraic expressions and multiplication formulas.
Comment on this topic belongs to the basic topic, mainly examining the relevant knowledge in the operation of algebraic expressions. There must be three similarities in the same category: they contain the same letters, and the indexes of the same letters are the same; The theoretical basis of the rule of removing brackets is the law of multiplication distribution, and the difference between square difference formula and complete square formula should be distinguished in the application of multiplication formula. The basic operation knowledge points of algebraic expressions are examined comprehensively and have high reliability.
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3.(20 10, Dongguan, Guangdong, 3,3 points) As shown in the figure, it is known that ∠ 1 = 70 If CD ∠Be, then ∠ B's degree is ().
a . 70 b . 100 c . 1 10d . 120
According to the analysis of "two straight lines are parallel and the same angle is equal", the adjacent complementary angles are equal to ∠B,
So ∠ b =180-70 =110.
Answer c
Relates to the properties of parallel lines and adjacent complementary angles of knowledge points.
Comment on this question to examine the property theorem of parallel lines, the knowledge point is single, it is a simple question with high credibility.
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4.(20 10, Dongguan, Guangdong, 4, 3 points) Seven students from a study group donated money to Yushu earthquake-stricken area, and the amount of donations was 5 yuan, 6 yuan, 6 yuan, 7 yuan, 8 yuan and 9 yuan 10 yuan, so the median and mode of this set of data were () respectively.
A.6，6 B.7，6 C.7，8 D.6，8
The sequence of this set of data from small to large is: 5, 6, 6, 7, 8, 9, 10. The number of data is 7, so the number of bits is the fourth, that is, 7; Among them, data 6 appears the most times, so the mode is 6.
Answer b
Involving the median and mode of knowledge points.
Comment on the median and mode of data in this question, which is a basic conceptual problem and relatively simple. As long as you master the concept, you can score.
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5.(20 10, Dongguan, Guangdong, 5,3 points) The left figure is the geometry of the main direction, and its top view is ().
According to the placement of geometry, the top view should be the fourth.
Answer d
Three views on geometry involving knowledge points
There is only one knowledge point in this question, which requires candidates to have a certain spatial imagination and belongs to the basic question.
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Fill-in-the-blank questions (5 small questions in this big question, 4 points for each small question, ***20 points) Please fill in the correct answers to the following questions in the corresponding positions on the answer sheet.
6.(20 10, Dongguan, Guangdong, 6,4 points) According to Xinhua Online Hai June 1 Japan, the passenger flow in China International Import Expo(CIIE) has been stable since its opening one month ago, and the cumulative number of visitors on that night 19 exceeded 8 million. The scientific symbol means 8,000,000 =.
Analysis 8000000 = 8× 100000, 100000 = 106, so 800000000 = 8×106.
Answer 8× 106
It involves the scientific notation of knowledge points.
comment
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7. The solution of (7.(20 10/0, Dongguan, Guangdong, 7,4 points) fractional equation =.
The simplest common denominator of analysis is, so both sides are multiplied by () at the same time to get:, solve and test:, so it is the solution of the equation.
answer
Fractional equation involving knowledge points
The key to solving the fractional equation is to remove the denominator by using the properties of the equation and transform the fractional equation into a linear equation, which embodies the mathematical idea of transformation. Another point to be noted in solving fractional equations is that it is necessary to check to prevent the root from increasing.
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8.(20 10, Dongguan, Guangdong, 8,4 points) As shown in the figure, it is known that in Rt△ABC, if the height on the hypotenuse BC is AD = 4 and cosB= =, then AC =.
By analyzing ∠ B =∠ CAD, we can get ∠ cosCAD= =, because AD = 4, so AC = 5.
Answer 5
Right triangle involving knowledge solution
As one of the required knowledge points in the annual senior high school entrance examination, the right triangle problem is generally not difficult to solve, and it is mainly based on basic concepts, but if the concepts are confused, it is difficult to distinguish.
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9.(20 10, Dongguan, Guangdong, 9,4 points) The average price per square meter of commercial housing in a city in 2007 and 2009 was 4,000 yuan and 5,760 yuan respectively. Suppose that the annual growth rate of the average price of commercial housing per square meter is X in the two years after 2007, try to list the equation about X:
According to the analysis, the average price of commercial housing per square meter in 2008 was, and that of commercial housing per square meter in 2009 was.
answer
Using a quadratic equation involving knowledge points to solve practical problems.
This topic mainly focuses on solving practical problems with a list of quadratic equations in one variable, which is a conventional topic and not difficult.
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10.(20 10, Dongguan, Guangdong, 10, 4 minutes) As shown in the figure 1, it is known that the area of the small square ABCD is 1, and the sides are doubled to obtain a new square A1b/kloc-. Square a1b1c1d1is doubled according to the original method to obtain square A2B2C2D2 (as shown in Figure 2); In this way, the area of the square A4B4C4D4 is.
Aa 1 = 1, AB 1 = 2, so a1b1=; A 1A2= =, A 1B2= =, so A2B2 = 5 = According to the law, we can find that the side length of a square AnBnCnDn is, so its area is.
Answer 625
Pythagorean theorem involves knowledge points, square area
The comment on this topic skillfully combines the square area with Pythagorean theorem and adopts the form of regular inquiry, which requires higher thinking ability of candidates and is slightly more difficult.
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Iii. Answering Questions (1) (This big question has five small questions, each with 6 points and ***30 points)
1 1.(20 10 Dongguan, Guangdong, 1 1, 6 points) Calculation:.
The original answer = 2+2-2×+1= 4-1+1= 4.
It involves real number operation of knowledge points, trigonometric function value of special angle, and zero exponential power.
Comment on real number operation has always been an important content of senior high school entrance examination. It is often combined with negative integer exponential power, zero exponential power and absolute value, and trigonometric function value of special angle to solve the problem. The topic is not difficult, mainly to examine the candidates' mastery of basic concepts and basic skills of operation.
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12.(20 10 Dongguan, Guangdong, 12, 6 points) Simplify first, then evaluate:, in which.
The original answer is =; When, the original formula =
It involves factorization of knowledge points, multiplication and division of fractions and simplification of quadratic roots.
The operation of annotation score is always inseparable from factorization. This topic is relatively simple, but we should pay attention to the premise of simplification before evaluation, and we can't directly evaluate it and substitute it into the formula. The final result should also be reduced to the simplest quadratic root.
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13.(20 10 Dongguan, Guangdong, 13, 6 minutes) As shown in the figure, each small square in the grid paper is a square with a side length of 1 unit, and the vertices of Rt △ABC are all on the grid points. After the plane rectangular coordinate system is established, the coordinate of point A is (-6,65438).
(1) Translate Rt△ABC by 5 units along the positive direction of the X axis to obtain Rt△A 1B 1C 1. Try to draw the graph of Rt△A 1B 1C 1 and write the coordinates of point A 1.
⑵ Rotate the original Rt△ABC 90 clockwise around point B to get Rt△A2B2C2, and try to draw the graph of Rt△A2B2C2 on the graph.
answer
A 1(- 1， 1)
Knowledge points related to translation, rotation and plane rectangular coordinate system.
Comment on this topic to realize the translation and rotation of graphics in plane rectangular coordinate system. The topic is relatively simple and belongs to the subtitle.
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14.(20 10, Dongguan, Guangdong 14, 6 minutes) As shown in the figure, PA and ⊙O are tangent to point A, and the chord AB⊥OP has a vertical foot of C, and OP and ⊙O intersect at point D. It is known that OA = 2.
(1) Find the degree of ∠POA;
⑵ Calculate the chord length AB.
Analytic (1) PA tangency ∠ pao = 90; ∠ apo = 30 from OA = 2 and OP = 4,
So ∠ POA = 60
2 △ AOC is a right triangle according to AB⊥OP, and OC = 1 is obtained from ∠ POA = 60 and AO = 2, so AC =；; According to the vertical diameter theorem, there is CB = AC =, so AB =
The answers (1) ∵ PA and ⊙O are tangent to point a.
∴∠PAO=90
OA=2，OP=4
∴∠APO=30
∴∠POA=60
⑵∵AB⊥OP
∴△AOC is a right triangle, AC = BC.
∫∠POA = 60
∴∠AOC=30
AO = 2
∴OC= 1
∴ in Rt△AOC,
∴AB=AC+BC=
It involves the vertical diameter theorem of knowledge points, the nature of tangent, the right angle side of 30 is equal to half of the hypotenuse, and the pythagorean theorem.
This topic belongs to the basic application of vertical diameter theorem and tangent property, and integrates the relevant knowledge of right triangle. It's not difficult and easy to use. As long as you master the basic concepts and make careful calculations, you can get points.
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15.(20 10, Dongguan, Guangdong 15, 6 minutes) As shown in the figure, the image of the linear function y = KX- 1 and the image of the inverse proportional function intersect at point A and point B, where the coordinate of point A is (2 1).
(1) Try to determine the values of k and m;
⑵ Find the coordinates of point B 。
Analysis (1) Substitute the coordinates of point A into two function expressions, and the solution can be obtained; ⑵ Two analytical expressions are combined into equations, and two coordinates can be obtained by solving the equations. Because point B is in the third quadrant, the coordinates of point B can be determined.
Answer (1) Substitute the point (2, 1) into the: of the resolution function to get the solution.
(2) The solution can be obtained according to the meaning of the question, so the coordinate of point B is (-1, -2).
Analytic functions, functions and equations (groups) are solved by undetermined coefficient method involving knowledge points.
Comment on using undetermined coefficient method to find analytic function and finding the intersection coordinates of function images is a common knowledge point in the senior high school entrance examination over the years. This question focuses on the application of basic concepts and methods, which is relatively simple and can get full marks with a little attention.
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Iv. Answer (2) (There are 4 small questions in this big question, with 7 points for each small question and 28 points for * * *)
16.(20 10 Dongguan, Guangdong, 16, 7 minutes) Divide the circular dials A and B with pointers into four equal sectors and three equal sectors respectively, and mark numbers on each sector (as shown in the figure). Huanhuan and Lele play roulette. The rules of the game are: rotate two turntables at the same time, when the turntables stop. If the digital product of the two areas pointed by the pointer is even, Lele wins; If the pointer falls on the dividing line, it is invalid and the turntable needs to be rotated again.
(1) Try to list or draw a tree diagram to find the probability of winning the prize;
(2) Is this game rule fair to Huanhuan and Lele? Try to explain why.
Answer (1) list:
1 2 3 5
1 1 2 3 5
2 2 4 6 10
3 3 6 9 15
So p (odd number) = 1
⑵ P (even number) = table, so p (odd number) = P (even number), so the rules of the game are fair to both sides.
Probability involving knowledge points
It is a common question in the senior high school entrance examination to find probability by list method or tree diagram. As long as you master the basic method of finding probability, you will generally not lose points. This question is relatively simple.
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17.(20 10, Dongguan, Guangdong, 17, 7 minutes) The image of the known quadratic function is shown in the figure, and the coordinates of its intersection with the axis are (-1 0) and (0,3).
(1) Find the values of b and c, and write the analytical formula of quadratic function at this time;
⑵ According to the image, write the range of independent variable X with function value y as positive.
Answer (1) According to the meaning of the question, we can get: and get the solution, so the analytical formula of parabola is
(2) sequence and solution; According to the image, when the function value y is positive, the range of the independent variable x is-1 < < 3.
It involves the idea of undetermined coefficient method of knowledge points, quadratic function, quadratic equation of one variable and combination of numbers and shapes.
This topic not only examines the method of undetermined coefficients and the solution of equations (groups), but also involves the important idea of combining numbers and shapes in mathematics. The second sub-question is more difficult, and quite a few candidates may list an unsolvable quadratic inequality, but it is more intuitive and convenient to solve it with images.
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18.(20 10 Dongguan, Guangdong 18, 7 minutes) As shown in the figure, the right-angled side AC and the hypotenuse AB of Rt△ABC are made into equilateral △ACD and equilateral △ Abe respectively. It is known that ∠ BAC = 30, EF⊥AB, and the foot is vertical.
(1) Try to explain that AC = ef
⑵ Verification: Quadrilateral ADFE is a parallelogram.
Analysis (1) from equilateral △ABE, ∠ Abe = 60, AB = BE, from EF ∠ AB, ∠ BFE = 90, it can be proved that △ ABC △ EF ⊥ AB, AC = EF.
⑵ AD = AC from equilateral △ACD, ∠ CAD = 60, so ∠ Bad = 90, then AD‖EF, AD=EF from AC = EF, so the quadrilateral AD = EF is a parallelogram.
Answer (1) ∵ equilateral △ Abe
∴∠ABE=60，AB=BE
∵EF⊥AB ∴∠BFE=∠AFE=90
∠∠BAC = 30，∠ACB=90
∴∠ABC=60
∴∠ABC=∠ABE,∠ACB=∠BFE=90
∴△ABC≌△EFB，
∴AC=EF
⑵ equilateral triangle△ △ACD
∴AD=AC,∠CAD=60
∴∠BAD=90 ,∴AD‖EF
AC = EF
∴AD=EF
Quadrilateral ADFE is a parallelogram.
It involves the judgment of equilateral triangle, right triangle and parallelogram.
Comments on special triangles and parallelograms have always been a must in the senior high school entrance examination. This question combines them skillfully, which is not difficult, and it is a good question.
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19.(20 10 Dongguan, Guangdong, 19, 7 points) A school organized 340 teachers and students to carry out long-distance investigation activities, with 70 pieces of luggage/kloc-0, and planned to rent two types of cars *** 10. It should be understood that each vehicle may
(1) Please help the school design all feasible car rental schemes;
(2) If the rental of a car is 2000 yuan per car and the rental of b car is 1800 yuan, which feasible scheme can save the rental cost?
Analysis (1) can be analyzed in tables:
Quantity can carry the number of people and luggage.
Jiache
40
16
Car B 10-
30( 10- )
20( 10- )
The inequality relation is implied in the problem: the loading capacity is not less than the loading demand, that is:
Number of people that can be carried by car A+number of people that can be carried by car B ≥ 340;
The number of baggage that can be loaded by car A+the number of baggage that can be loaded by car B ≥ 170.
According to the relationship between two inequalities, the inequality group is listed, the solution set of this inequality group is obtained, and the scheme can be obtained by taking a positive integer solution;
⑵ The total rental fee can be expressed as w = 2000+1800 (10-) = 200+18000, which is a linear function. According to the increase or decrease of linear function, we can get the scheme to minimize the car rental cost.
Answer (1) If you rent a model car, rent a model car (10-). According to the meaning of the question, you get:
Solution: 4 ≤≤≤. Because it is a positive integer, * * * has four schemes, namely: Scheme 1: Rent 4 Class A cars and 6 Class B cars; Option 1: Rent 5 A-type cars and 5 B-type cars; Option 1: Rent 6 Type A cars and 4 Type B cars; Scheme 1: lease 7 A-type cars and 3 B-type cars.
(2) If the total rental fee is W, then W = 2000+1800 (10-) = 200+18000, > 0, and W increases with the increase, so you can immediately choose the first scheme to save rent.
Inequalities involving knowledge points and linear functions
The practical application of commenting on inequality groups has always been one of the compulsory test points in the senior high school entrance examination. The key to solve the problem is to find out the inequality relationship in the problem correctly, so as to get the inequality group and then determine its positive integer solution. The problem of choosing the best scheme between them is usually solved by increasing or decreasing a linear function.
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Verb (abbreviation of verb) solution (3) (This big question has three small questions, each with 9 points and * * * 27 points)
20.(20 10, Dongguan, Guangdong, 20, 9 minutes) It is known that two congruent right-angled triangular pieces of paper ABC and DEF are placed as shown in the figure (1). Point B and point D coincide, point F is on BC, and AB and EF intersect at point G ∠ C = ∠ EFB = 90, ∠ E =.
(1) Verification: △EGB is an isosceles triangle;
⑵ If the paper DEF does not move, it is required that △ABC rotate counterclockwise for a minimum degree around point F, and the quadrilateral ACDE becomes a trapezoid with ED as the bottom (as shown in Figure 2). Find the height of this trapezoid.
Analytically (1) prove isosceles triangle, only need to prove ∠ EBA = ∠ E = 30; ⑵ FC = known by rotation. When the quadrangle ACDE becomes a trapezoid with ED as the base, ED‖AC, then ED⊥CB. At this time, the rotation angle ∠ DFB = 30, and the distance from DF = 2 to F point is 0, from which the height of the trapezoid can be obtained.
Answer (1)∫∠EFB = 90, ∠ ABC = 30.
∴∠EBG=30
∠∠E = 30
∴∠E=∠EBG
∴EG=BG
∴△EGB is an isosceles triangle
(2) In Rt△ABC, ∠ C = 90, ∠ ABC = 30, AB = 4.
∴bc=；
In Rt△DEF, ∠ EFD = 90, ∠ E = 30, DE = 4.
∴DF=2
∴CF=。
∫ Quadrilateral ACDE becomes a trapezoid with ED as the base.
∴ED‖AC
∠∠ACB = 90°
∴ED⊥CB
∠∠EFB = 90°，∠E = 30°
∴∠EBF=60
∵DE=4∴DF=2
∴ The distance from F to ED is
∴ The height of the trapezoid is
Knowledge point solutions involving right triangle, rotation, isosceles triangle judgment and trapezoid
The essence of comment rotation is that rotation does not change the shape and size of graphics. Grasping this point, we can easily find the length of CF, which is also the key to find the height of trapezoid in this question. This question is not difficult, but it is compatible with many knowledge points and requires candidates' comprehensive application ability of knowledge.
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2 1.(20 10 Guangdong dongguan 2 1, 9 points) read the following materials:
1×2= ( 1×2×3-0× 1×2),
2×3= (2×3×4- 1×2×3),
3×4= (3×4×5-2×3×4),
Adding up the above three equations, we can get
1×2+2×3+3×4= ×3×4×5=20.
After reading the above materials, please calculate the question:
(1)/kloc-0 /× 2+2× 3+3× 4+…+10×1(writing process);
⑵ 1×2+2×3+3×4+…+n×(n+ 1)=；
⑶ 1×2×3+2×3×4+3×4×5+…+7×8×9= .
analyse
The answer (1)/kloc-0 /× 2+2× 3+3× 4+…+10×1.
= ×( 1×2×3-0× 1×2+2×3×4- 1×2×3…+ 10× 1 1× 12-9× 10× 1 1)
= × 10× 1 1× 12
=440
⑵ 1×2+2×3+3×4+…+n×(n+ 1)
= ×[ 1×2×3-0× 1×2+2×3×4- 1×2×3+…
+ ]
=
⑶ 1×2×3+2×3×4+3×4×5+…+7×8×9
= ×[ 1×2×3×4-0× 1×2×3×4+2×3×4×5- 1×2×3×4+…+7×8×9× 10-6×7×8×9]
= ×7×8×9× 10
= 1260
Operations involving real number knowledge points
The key to evaluating routine operation test questions is to find out the internal rules. The first two questions are moderately difficult, and the third question is somewhat difficult. Only by careful analysis and truly finding out the law can we determine whether the previous score is right or wrong.
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22.(20 10, Dongguan, Guangdong, 22, 9 minutes) As shown in Figures (1) and (2), the side length of rectangular ABCD is AB = 6, BC = 4, point F is on DC, and DF = 2. The moving points m and n start from points d and b at the same time, respectively, and stop moving simultaneously along rays DA and b, m and n. Connect FM, MN and FN. When f, n and m are not in a straight line, the midpoint of △FMN and △FMN can be obtained as △ pqw. The speed of moving points M and N is 1 unit/second, and the moving time of M and N is x seconds. Try to answer the following questions:
(1) Description of △ FMN ∽△ QWP;
⑵ Set 0 ≤ 4 (that is, the time period when M moves from D to A). Why is △PQW a right triangle? In what range is △PQW not a right triangle?
(3) When asked what the value is, the line segment MN is the shortest? Find the value of MN at this time.
Analysis (1) PQ‖FN, PW‖MN and WQ‖MF can be obtained from the midline theorem, and it can be proved that ∠ PQW = ∠ MFN and ∠ PWQ = ∠ FMN are similar to two triangles according to the properties of parallel lines. ⑵ No matter how the point moves, when the point M is on the line segment DA, MD = BN =, then AM =, an =. We can first express the square of the line segment MN, MF and NF with the inclusion formula, and then discuss that when M, N and F are right-angled vertices, they correspond to W, P and Q. According to Pythagorean theorem, we can list the equations and find the corresponding values. (3) Since point N is on the AB line and point M is on the ABDA ray, according to "the line between a point outside the straight line and all points on the straight line is the shortest", it can be known that when point M moves to coincide with point A, MN is the shortest. At this time, DM = BN = 4 and Mn = 2.
The answer (1) ∵ p, q and w are the midpoint of the triangle △FMN respectively.
∴PQ‖FN,PW‖MN
∴∠MNF=∠PQM=∠QPW
Similarly: ∠ nfm = ∠ pqw
∴△FMN ∽ △QWP
⑵
△FMN ∽ △QWP is obtained by (1), so when △FMN is a right triangle, △QWP is also a right triangle. As shown in the figure, if n crosses the point, it is the NECD in E. According to the meaning of the question, DM = BM =, ∴ AM = 4-, An = DE = 6-
∵DF=2,∴EF=4-
∴mf2=22+x2=x2+4,mn2=(4-x)2+(6-x)2=2x2-20x+52,nf2=(4-x)2+42=x2-8x+32，
(1) If ∠ MNF = 90, there is 2x2-20x+52+x2-8x+32 = x2+4, and the solution is x2= 10 = 4, x2= 10 (missing);
② If ∠ NMF = 90, there is 2x2-20x+52+x2+4 = x2-8x+32, which is simplified as: x2-6x+ 12 = 0, △ =- 12 < 0, and the equation has no real root;
③ If ∠ MFN = 90, there is 2x2-20x+52 = x2+4+x2-8x+32, and x =.
∴ When it is 4 or △PQW is a right triangle; When 0 ≤