Mathematics Test
(Examination time 100 minutes, full mark 150 minutes)
Note to candidates:
1. This paper contains three major questions, including ***25 questions;
2. When answering questions, candidates must answer in the position specified in the answer sheet according to the answer requirements, and the answer on the draft paper and this test paper is invalid;
3. Except for the first and second questions, unless otherwise specified, the main steps of proof or calculation must be written in the corresponding position on the answer sheet.
1. Multiple choice questions: (There are 6 questions in this big question, 4 points for each question, out of 24 points).
[Of the four options in the following question, one and only one is correct. Select the code of the correct item and fill it in the corresponding position on the answer sheet]
1. In the following quadratic roots, and are similar quadratic roots.
(1); (b) and: (c) and: (4).
2. In the image of the following function, there is no common point with the axis.
(1); (b) and: (c) and: (4).
3. Given a given point P (- 1, 3), the coordinates of the point symmetric with respect to the origin O are
(A)(- 1，-3)； (B)( 1，-3)； (C)( 1，3)； (D)(3，- 1)。
4. As shown in the figure, given the directional quantity,,, then the following conclusion is correct.
(1); (b) and: (c) and: (4).
5. What is wrong in the following proposition is
(a) The two diagonal lines of a rectangle are equal;
(b) The two diagonals of the isosceles trapezoid are perpendicular to each other;
(c) The two diagonals of the parallelogram are bisected;
(d) The two diagonal lines of a square are perpendicular and equal to each other.
6. Xiaojie investigated the weight of his classmates and drew the histogram of frequency distribution. So the following conclusion is incorrect.
The total number of students in each class is 45;
The number of people weighing between 50 kg and 55 kg is the largest;
(c) The mode of students' weight is14;
(d) The number of people weighing 60 kg to 65 kg accounts for the whole class.
Of the total number of people
Fill in the blanks: (4 points for each question, out of 48 points)
7. Calculation: _ _ _ _ _ _.
8. Factorize the factors in the real number range: _ _ _ _ _ _ _ _ _ _ _ _ _ _.
9. The domain of the function is _ _ _ _ _ _ _ _.
10. The solution of the equation is _ _ _ _ _ _ _ _ _.
1 1. It is known that the image of the proportional function (k ≠ 0) passes through the point (-4,2), so the function value y will be _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
12. Four cards with the same size and texture are marked with the numbers 1, 2, 3 and 4 respectively. Now put the numbered side on the table and choose two cards at random. Then the probability that the product of the numbers on two cards is even is _ _ _ _ _ _ _ _.
13. A school randomly selected 50 students to ask "How much do you know about the World Expo?" Through the investigation, it is found that 45 of them have a comprehensive understanding of the World Expo, from which it can be estimated that among the 0/500 students in the whole school, the comprehensive understanding of the World Expo is about _ _ _ _ _ _ _ _.
14. As shown in the figure, in the rectangle ABCD-EFGH, the sides are perpendicular to the plane ADHE.
* * * There are _ _ _ _ _ _.
15. Simplification: _ _ _ _ _ _.
16. In trapezoidal ABCD, AD // BC, E and F are the midpoint of AB side and CD side respectively. If AD = 5,
EF = 1 1, then BC = _ _ _ _ _ _ _ _ _
17. in Rt△ABC, ∠ C = 90, ∠ A = 30 and AB = 8. If the circle centered on point C is tangent to the AB side, then the radius of ⊙C is equal to _ _ _ _ _ _ _ _ _ _ _ _.
18. In △ABC, ∠ A = 45 and BC = 5, then AC = _ _ _ _ _ _ _ _
Third, answer: (This big question is ***7 questions, out of 78 points)
19. (The full mark of this question is 10)
Solve the inequality group: and express the solution set on the number axis.
20. (The full mark of this question is 10)
Solve the equation:
2 1. (This title is ***2, item (1) is 4 points, item (2) is 6 points, and the full score is 10).
According to previous sales experience, the daily selling price and sales volume of a commodity are as follows:
Price per kilogram (RMB) 38 37 36 35...20
Daily sales (kg) 50 52 54 56 86
Suppose that when the unit price is reduced from 38 yuan/kg to X yuan, the sales volume is Y kg. It is known that the functional relationship between y and x is a linear function.
(1) Find the resolution function of y and x;
(2) If the cost price of a commodity is 20 yuan/Jin, what is the selling price of that day to make the profit of a certain day reach 780 yuan? (Profit = Total Sales-Cost)
22. (This question ***2 is a minor issue, the first question (1) is 5 points, the second question (2) is 5 points, and the full score is 10).
As shown in the figure, point P is a point in ∠AOB ∠, passing through point P is PC // OB, PD // OA, OA and OB meet at point C and D respectively, PE⊥OA,PF⊥OB, and the footsteps are point E and point F respectively.
(1) Verification:;
(2) When the point P is located at ∠AOB, is the quadrangle CODP a diamond? And prove your conclusion.
23 (This topic is ***2, the first item (1) is 5 points, and the second item (7 points, full score 12 points).
As shown in the figure, it is known that in △ABC, AB = AC = 8, where D is the midpoint of BC side, points E and F are on AB side and AC side, and ∠EDF =∠B connects EF.
(1) if BE = 4, find the length of CF;
(2) If EF // BC, find the length of EF.
24. (This question ***2 is a minor issue, with the score of (1) being 5, the score of (2) being 8 and the full score being 12).
It is known that the image of quadratic function passes through point m (1, 0).
(1) Find the analytic expression of this quadratic function and the vertex coordinates of the function image;
(2) It is known that the image of the linear function intersects with the X axis and the Y axis at points A and B, respectively. The symmetry axis of the image of the quadratic function obtained in (1) intersects with the image of the linear function at point C, and the symmetry axis intersects with the X axis at point D. If so, find the value of b. 。
25. (This title is ***3, item (1) is 4 points, item (2) is 5 points, item (3) is 5 points, and the full score is 14).
As shown in the figure, it is known that in the square ABCD, AB = 2, P is any point on the BC side, and E is a point on the extension line of the BC side, connecting AP. The crossing point p is PF⊥AP, intersects with the bisector CF of ∠DCE at point f, connects AF, intersects with edge CD at point g, and connects PG.
(1) verification: AP = FP；;
(2) The radii ⊙P and ⊙G are PB and GD, respectively. Try to judge the positional relationship between ⊙ P and ⊙G, and explain the reasons;
(3) When BP takes any value, PG // CF
Minhang district 2008 school year second semester ninth grade quality survey exam
Reference answers and grading standards of mathematics test papers
1. Multiple choice questions: (There are 6 questions in this big question, 4 points for each question, out of 24 points).
1.c； 2.a； 3.b； 4.d； 5.b； 6.C。
Fill in the blanks: (4 points for each question, out of 48 points)
7.； 8.； 9.； 10 . x = 2； 1 1. decrease; 12.；
13. 1350; 14.4; 15.； 16. 17; 17.； 18. 1 or 7.
Third, answer: (This big question is ***7 questions, out of 78 points)
19. (The full mark of this question is 10)
Solution: From ① ....................................................... (2 points)
From ② ............................................. (2 points)
Solution ............................................ (2 points)
Therefore, the solution set of the original inequality group is ................................................. (2 points).
If the solution set of the inequality group is expressed on the number axis, you will get 2 points correctly, without removing the end point deduction 1 point.
20. (The full mark of this question is 10)
Solution: Multiply both sides by the simplest common denominator, and you get
................................. (2 points)
Upon completion, you will get a .................................................. (3 points).
Solution, ....................................... (2 points)
It is proved to be the root of the original equation, so it is discarded; Is the root of the original equation ........... (2 points).
Therefore, the root of the original equation is x = 4 .................................... (1min).
2 1. (This title is ***2, item (1) is 4 points, item (2) is 6 points, and the full score is 10).
Solution: (1) Let the resolution function between y and x be (k ≠ 0).
According to the meaning of the question, get ............................... (2 points)
Solution ............................... (1)
Therefore, the resolution function sought is ..................................... (1 min).
(2) Let the sales price of this day be X yuan .......................................................... (1 min).
According to the meaning of the question, get ................................. (2 points)
Upon completion, you will receive .......................................... (1 minute).
Solution, .................................. (1 min)
A: The sales price of this day should be 33 yuan or ............................................................., 50 yuan (1 min).
22. (This question ***2 is a minor issue, the first question (1) is 5 points, the second question (2) is 5 points, and the full score is 10).
Proof: (1)∵PC // OB, PD // OA,
∴ Quadrilateral OCPD is a parallelogram, and ∠ECP =∠O, ∠ FDP = ∠ O...( 1 min)
∴PC = OD, PD = OC, ∠ ECP = ∠ FDP ...................... (1min)
∴∠pec ∵pe⊥oa,pf⊥ob =∠pfd = 90。
∴△ PCE ∽△ PDF ....................................... (1min)
That is ∴ ................................ (1min).
∴ ..................................... (1min)
(2) When the point p is on the bisector of ∠AOB∞, the quadrilateral CODP is a rhombic ................................................................................................................ (1 min).
When point P is on the bisector of ∠AOB, PE = PF. It is obtained from PE⊥OA,PF⊥OB.
Then, from △PCE∽△PDF, PC = PD ............................................................. (2 points).
∵ Quadrilateral CODP is a parallelogram, and∴ Quadrilateral CODP is a rhombic .............................. (1 min).
When point P is not on the bisector of ∠AOB, PE ≠ PF. You can get it, that is, PC ≠ PD. Available.
∴ When the point P is not on the bisector of ∞∠AOB, the quadrilateral CODP is not a diamond ... (1 min)
23 (This topic is ***2, the first item (1) is 5 points, and the second item (7 points, full score 12 points).
Solution: (1) Add ad.
Ab = ac = 8, D is the midpoint of the side length, BC ∴ AD BC ⊥ ................................... (1min).
In Rt△ABD, ∴ BD = CD = 5 ................. (1).
∠∠EDC =∠b+∠BED =∠EDF+∠CDF，∠EDF =∠B，
∴∠ Bed =∠ CDF .............................................................. (1min)
∫ab = ac,∴∠b =∠c
∴△ BDE ∽△ CFD ..................................... (1min)
BE = 4，...........................................................( 1)
(2)∫△bde∽△CFD, ∴ ........................... (1min)
Bd = cd, ∴ .............................................. (1min)
∠EDF =∠B，∴△ BDE ∽△ DFE。 ∴∠ bed =∠ def ............................................ (1min)
∫ef//BC, ∴∠ BDE = ∠ def ..................................................................... (1min)
∴∠ BDE = ∠ bed. ∴ Be = BD = 5 ............................................................................ (1min)
Therefore, from AB = 8, AE = 3. ..
∵EF // BC，∴............................................( 1)
∫ BC = 10, ∴ ............................... (1min)
24. (This question ***2 is a minor issue, with the score of (1) being 5, the score of (2) being 7 and the full score being 12).
Solution: (1)∫ The image of quadratic function passes through point m (1, 0),
∴ ...................................... (1min)
∴ m =-3 .......................................... (1min)
The analytical formula of the function is ................................................................................................................................... (1 min).
In addition, the coordinate of vertex ∴ is (2,1) ....................... (2 points).
(2) The symmetry axis of the quadratic function image obtained from (1) is the straight line x = 2, ∴ d (2 2,0) ................... (1min).
Judging from the meaning of the question, A (0), B(0, B), c (2 2,4+B) ................................ (2 points).
∵ Symmetry axis straight line x = 2 is parallel to the y axis,
∴△ AOB ∽△ ADC ........................................... (1min)
That is ∴ ................................. (1min).
Solution, ......................................... (2 points)
It is verified that,, are all the values of m that meet the conditions.
25. (This title is ***3, item (1) is 4 points, item (2) is 5 points, item (3) is 5 points, and the full score is 14).
(1) Proof: Cut the line segment AH on the AB side, make AH = PC, and connect pH. 。
From the square of ABCD, we get ∠ B = ∠B =∠BCD =∠D = 90, AB = BC = AD...( 1).
∠∠APF = 90° ,∴∠apf =∠b
∠∠APC =∠b+∠BAP =∠APF+∠FPC，
∴∠ PAH =∠ FPC ........................................... (1min)
∠∠BCD =∠DCE = 90, CF ∠DCE, ∴∠ FCE = 45.
∴∠PCF = 135。
And ∵AB = BC, AH = PC, ∴BH = BP, that is ∠ BPH = ∠ BHP = 45.
∴∠∠ AHP = 135, that is ∠ AHP = ∠ PCF .................................................................... (1min).
In △AHP and △PCF, ∠PAH =∠FPC, AH = PC, ∠AHP =∠PCF,
∴△ Analytic Hierarchy Process△ ?△PCF. ∴ AP = PF .............................. (1min)
(2) solution: circumscribe the positional relationship between ⊙P and ⊙ g.
Extend CB to point m, make BM = DG, and connect AM.
By AB = AD, ∠ abm = ∠ d = 90, BM = DG,
Get △ adg △ abm, namely AG = AM, ∠ mab = ∠ gad ........................... (1min).
ap = fp，∠APF = 90° ,∴∠paf = 45°。
∠∠bad = 90, ∴∠BAP +∠DAG = 45, that is ∠ Map = ∠ PAG = 45. (1 min).
Then, AM = AG, ∠MAP =∠PAG, AP = AP,
Get △ APM△ APG. ∴ PM = PG。
That is Pb+DG = PG ................................................. (2 points).
∴⊙P and ⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙ ...............................................................................
(3) solution: from PG // CF, ........................................... (∠ GPC = ∠ FCE = 45) was obtained.
Therefore, from < BCD = 90, < GPC = < PGC = 45.
∴ PC = GC。 DG = BP ................................................. (1min)
Let BP = x, DG = X. From AB = 2, PC = GC = 2–X.
∫p b+ DG = pg,∴pg = 2 x
At Rt△PGC, ∠ PCG = 90, ............. (1) is obtained.
I see. Get its .............................. (1 min)
∴ (if applicable), pg//cf ...............................................................................................................................................................
It's not on the map