Mathematics (a)

Precautions:

1. Full volume *** 150, examination time 120 minutes.

2. Candidates must fill in (scribble) personal information such as school, name, admission ticket number, examination room and seat number on the answer sheet.

In the corresponding position.

Candidates must fill in (paint) the answers directly in the corresponding positions on the answer sheet.

1. Multiple choice questions (this question 15, 4 points for each question, ***60 points. Give four options for each question.

Only one of them meets the requirements of the topic. )

1. Observe the following bank signs. As can be seen from the patterns, some bank signs are both axisymmetric and centrally symmetric.

1。

2. The value range of the independent variable X in the function y =+ is

A.x ≤ 2b.x = 3c.x < 2 and x≠3d.x ≤ 2 and x ≠ 3.

3. Given three views of the geometry as shown on the right, this

Geometry is

A. cylinder

B. cone

C. Scope

D. cube

4. There are four propositions as follows: ① Diameter is chord; 2 After three points, you can definitely make a circle; ③ The distance from the outer center of the triangle to each vertex of the triangle is equal; ④ Two semicircles with equal radii are equal arcs. Among them, the correct ones are

A.4 B.3 C.2 D. 1

5. The vertex coordinates of the quadratic function image are

A.(- 1，8)b .( 1，8)c .( 1，2)d .( 1，-4)

6. It is known that the radii r and r of two circles are the two roots of the equation, and the distance between the centers of the two circles is 1. The positional relationship between two circles is

A. Externalization B. Insectization C. Intersection D. Externalization

7. Place the protractor on the triangular cardboard as shown in the figure, so that point C is on the semicircle. When the readings at point A and point B are 86 and 30 respectively, the size of ∠ACB is

A. 15

Figure 7, Figure 8

8. There are 20 people in a shooting group, and the coach draws a statistical chart as shown in the figure according to their shooting data. And then this group

The mode and median of the data are [source: Zxxk.Com] respectively.

A.7, 7b.8, 7.5c.7, 7.5d.8, 6 [Source: Xue&; Departments and networks]

9. There is a fan-shaped piece of paper with a central angle of and a radius of, which is just used to enclose the side of a cone (ignoring the seam). The radius of the circle at the bottom of the cone is

A.B. C. D。

10. As shown in the figure, the radius of the inscribed circle of a regular triangle is 1, so the side length of this regular triangle is

A.B. C. D。

[Source: Science, Science, Internet]

/DrawingNo. KLOC-0/0/DrawingNo. KLOC-0/1

1 1. As shown in the figure, the circumference of diamond-shaped ABCD is 20, DE⊥AB, the vertical foot is E, and A=, so the following conclusion is correct.

① ② ③ The area of the diamond is ④.

1。

12. The original price of a souvenir of the Shanghai World Expo was 168 yuan, and the price was 128 yuan after two consecutive price reductions. The following equation is correct.

A.B.

C.D.

13. Parabolic images are translated by 2 units to the right, and then translated by 3 units downward, and the obtained images are analyzed.

Type is, then the values of b and c are.

Answer. b=2，c=2 B. b=2，c=0 C . b= -2，c=- 1 D. b= -3，c=2

14. The known points (-1), (2) and (3) are on the image of the inverse proportional function. The following conclusion is correct.

A.B. C. D。

15. Parabolic image as shown, linear function and inverse proportional function.

Images in the same coordinate system are roughly

15 topic map [Source: Subject Network]

Fill in the blanks (5 small questions in this question, 4 points for each small question, ***20 points)

16. It is known that the univariate quadratic equation about x has real roots, so the range of m is.

17. As shown in the figure, in the right-angled trapezoidal ABCD, ad∑BC, AB⊥BC, AD = 2, rotate the waist cd 90 counterclockwise around D to DE, and the area connecting AE, CE and △ADE is 3, then the length of BC is.

18. As shown in the figure, when fan-shaped OAB, ∠AOB=90, ⊙P and OA, OB are tangent to points F and E respectively, and arc AB is tangent to point C, the area ratio of fan-shaped OAB to ⊙ P is.

17 map 18 map

19. As shown in the figure, on the physical education class, when two students, A and B, stand in the positions of C and D respectively, the shadow of B happens to be in the shadow of A.

In it, it is known that students A and B are 1 m apart. A is height1.8m, and b is height1.5m, so the shadow length of A is meters.

20. As shown in the picture, Xiaoming's father tied a rope between two trees 2 meters apart and made a simple swing for Xiaoming. The place where the rope is tied is 2.5 meters from the ground, and the rope naturally hangs in a parabolic shape. Xiao Ming, whose height is 1 m, just touches the rope when he is 0.5 m away from the nearby tree, so the lowest point of the rope is meters away from the ground.

[Source: Zxxk.Com]

Drawingno. 19 drawing no.20

Third, solve the problem (this problem is 8 small questions, ***70 points. Write the necessary text description, proof process or calculation steps when answering. )

2 1. (The full mark of this question is 10)

(1) (Full score for this small question)-+[Source: Zxxk.Com]

(2) (The full score of this small question is 6) It is known that y = y 1+y2, y 1 is directly proportional to x2, and y2 is inversely proportional to x, and

When x = 1, y = 3；; When x =- 1, y = 1. When x =-, find the value of y.

22. There is a rectangular clearing in front of Xiaoming's house. There are three trees, A, B and C, in the open space. Xiao Ming

I want to build a round flower bed so that all three trees are beside the flower bed.

(1) (4 points for this small question) Please help Xiao Ming draw the position of the flower bed (ruler drawing, no writing, package)

Leave it as a map trace).

(2) If AB = 8m, AC = 6m, and ∠BAC= in △ABC, try to find the area of Xiaoming's round flower bed.

Question 22

23. Xiaoli's parents bought a ticket to Shanghai to see the World Expo in July this year. She and her brother are two years old.

Everyone wants to see it, but there is only one ticket. My ninth-grade brother thought of a way and took eight playing cards.

Give Xiao Li four cards numbered 1, 2, 3 and 5, and keep four cards numbered 4, 6, 7 and 8 for yourself, and

According to the following rules of the game: Xiaoli and her brother randomly draw one card from their respective four cards, and then draw two cards.

Add up the numbers of playing cards. If the sum is even, Xiaoli will leave. If the total is odd, my brother will go.

(1) Please find out the probability that Xiaoli will go to Shanghai to see the World Expo by figures or lists.

(2) Are the rules of the game designed by my brother fair? If it is fair, please explain the reasons; If it is unfair, please design a public

Simple rules of the game.

24. (Full mark for this question is 8) The picture shows the plan of transporting goods at a freight terminal. In order to improve the safety of the conveying process,

The master wants to reduce the angle between the conveyor belt and the ground from 45 to 30. It is known that the original conveyor belt AB is 4 meters long.

(1) Find the length of the new conveyor belt AC;

(2) If it is necessary to leave a 2-meter passage on the left side of landing point C, try to judge whether goods MNQP 4 meters away from point B need to be removed, and explain the reasons. (Note: (1) (2) The calculation result is accurate to 0. 1.4 1, (1.73).

Question 24

25. As shown in the figure, P 1 is a point of the inverse proportional function on the first quadrant image, and the coordinate of point A 1 is (2,0).

(1) When the abscissa of point P 1 increases gradually, the area of △P 1O A 1.

How will it change?

(2) if △P 1O A 1 and △P2 A 1 A2 are equilateral triangles, find

The analytical expression of this inverse proportional function and the coordinates of point A2.

Question 25

26. (The full mark of this question is 10) As shown in the figure, AB is known to have a diameter of ⊙O, point C is on ⊙O, and the straight line passing through point C intersects with the extension line of AB at point P, AC=PC, ∠COB=2∠PCB.

(1) Verification: PC is the tangent of ⊙O;

(2) Verification: BC = AB；;

(3) point m is the midpoint of arc AB, and CM intersects AB at point n, if

AB=4, find MN? The value of MC

Question 26

27. (The full mark of this question is 10) In the known parallelogram ABCD, diagonal AC and BD intersect at point O, and AC= 10.

BD=8。

(1) If AC⊥BD, try to find the area of quadrilateral ABCD;

(2) If the included angle between AC and BD is ∠AOD=, find the area of quadrilateral ABCD;

(3) Discussion: If "parallelogram ABCD" in the title is changed to "quadrilateral ABCD" and ∠AOD=

AC=, BD=, try to find the area of quadrilateral ABCD (expressed by algebraic expression containing,,,).

Question 27

28. (The full mark of this question is 1 1) As shown in figure 1, it is known that vertex A of rectangular ABCD coincides with point O, and AD and AB are on the X axis and Y axis respectively, with AD=2 and AB = 3；; The throwing line passes through the coordinate origin O and another point E (4,0) on the X axis.

(1) What is the maximum value of this parabola when x is taken?

(2) Move the rectangular A BCD from the position shown in Figure 1 at the speed of 1 unit length per second along the positive direction of the X axis, and at the same time move the moving point P from point A to point B at the same speed. Let their movement time be t seconds (0≤t≤3), and the intersection of straight line AB and parabola is n (as shown in Figure 2).

(1) If yes, judge whether the point P is on the straight line ME and explain the reason;

② Whether the area of a polygon with P, N, C and D as vertices is possible to be 5, and if possible, find out the coordinates of N points at this time; If not, please explain why.

Figure 1 Figure 2

Question 28

20 10 Lanzhou junior high school graduates' academic papers

Mathematics (1) Reference Answers and Grading Criteria [Source: Science # Subject # Net]

Reviewer: Zhang Hao Proofread: Chen Liang.

First, multiple-choice questions (this question 15, 4 points for each question, ***60 points)

Title 1234567 [Source: zxxk.com] 891012131415.

Answer bb bb bb bb bb bb bb

Fill in the blanks (5 questions in this question, 4 points for each question, ***20 points)

16. And m ≠ 1 17.5 18.

19.6 20.

Third, the answer (this question is 8 small questions, ***70 points. Write the necessary text description, proof process or calculation steps when solving problems. )

2 1. (The full mark of this question is 10)

(1) (Full score for this small question)

Solution: The original formula = ... 2 points.

= ... 3 points.

= 5 points and 4 points.

(2) (The full score of this small question is 6 points)

Solution: Solution: y 1 is directly proportional to x2, and y2 is inversely proportional to X.

Let y 1 = k 1x2, y2 =, y = k 1x2+ ..............................................................................................................................

Substitute x = 1, y = 3, x =- 1, y = 1 into the above formula to obtain

∴ ……………………………………………………………………………………………………………………………………………………………………………………………………………………………… 5 points.

When x =-, y = 2× (-) 2+=-2 =-... 6 points.

22. (The full mark of this question is 6 points)

(1) (Full score for this small question)

Draw 2 points of perpendicular bisector and ............................ on both sides with a ruler.

Form a circle, ........................., 3 points.

⊙O is the location of the garden to be built. (Sketch) 4 o'clock.

(2) (Full score for this small question)

Solution: ∫∠BAC =, AB = 8m, AC = 6m, BC = 10m.

The radius of abc circumscribed circle is 5 meters and 5 points.

Xiaoming's round flower bed covers an area of 2 square meters, 6 points in ........................................

23. (The full mark of this question is 6 points)

(1) All possible results are as follows:

A * * * has 16 results, and each result appears.

The possibility is the same.

..................................., two points.

The probability that the sum is even is

Therefore, the probability that Xiao Li will go to Shanghai to see the World Expo is

(2) From the result of (1) list, we can know that the probability of Xiaoli going is 0, and the probability of her brother going is 0, so we play a game.

That's not fair. It's good for my brother. Four points for .............................................................................................................................

The rules of the game are changed to: if the sum is even, Xiaoli gets 5 points; if the sum is odd, my brother gets 3 points, and the game is over.

Fair ........................................, 6 points.

There are various changes in the rules of the game, and the marking teacher will give points according to the situation. )

24. (The full mark of this question is 8)

(1) As shown in the figure, let AD⊥BC be at point D .............................. 1.

In Rt△A BD,

Ad = ABS in 45 = 4...2 points.

In Rt△ACD, ∫∠ACD = 30.

∴ AC = 2ad = material .................................... 3 points.

That is, the length of the new conveyor belt AC is about m.

(2) Conclusion: goods MNQP should cancel ..................................... 5 points.

Solution: in Rt△ABD, BD = ABCOS 45 = 4.

In Rt△ACD, CD = AC cos 30 =

∴CB=CD—BD= ≈2. 1

∫PC = p B- CB≈4-2. 1 = 1.9 < 2 ........................................................................................................................

∴ Goods should be deleted. ........................................................... 8 points.

25. (The full mark of this question is 9 points)

(1) solution: (1)△P 1OA 1, the area will gradually decrease. ....................................................................................................................

(2) Let P 1C⊥OA 1 and the vertical foot be C, because △P 1O A 1 is an equilateral triangle.

Therefore, OC= 1, P 1C=, so P 1 .....................................................................................................................................

The result of substitution is k=, so the analytical formula of inverse proportional function is .............................................................................................................................................................

Let P2D⊥A 1 A2, the footing is d, and A 1D=a, then OD=2+a and p2d = a.

Therefore, P2 ..................................................................................................... scored 6 points.

Substitution, acquisition and simplification

Solution: A =- 1 7 points.

∵ a > 0 ∴ ..............................................................................................................................................................

So the coordinates of point A2 are |, 0 | ... 9 o'clock.

26. (The full mark of this question is 10)

Solution: (1)∵OA=OC,∴∠A=∠ACO.

∠∠COB = 2∠A，∠COB=2∠PCB

∴∠∠

∵AB is the diameter of⊙ O.

∴∠∠ACO+∣ACO

∴∠ PCB+∠ OCB = 90, that is, OC ⊥ CP ................................................... scored 3 points.

∵OC is the radius of⊙ O [Source: Subject Network]

∴PC is the tangent of ⊙∵∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴87

(2)∫PC = communication ∴∠A=∠P

[Source: Xue. Ke. Net Z.X.X.K]

∠∠COB =∠A+∠ACO，∠CBO=∠P+∠PCB

∴∠ CBO = ∠ COB ........................................................................ 5 points.

∴BC=OC

Bc = ab.

(3) Connect mA and mA

Point m is the midpoint of arc AB.

∴ ARCAM = ARCBM ∴∠ ACM = ∠ BCM ............................................................ 7 points.

∠∠ACM =∠abm ∴∠bcm=∠abm

≈BMC =≈BMN

∴△MBN∽△MCB

∴

∴BM2=MC？ [Source: Theme Network]

∵AB is the diameter⊙ O, and the arc AM= the arc BM.

∴∠AMB=90，AM=BM

∵ AB = 4 ∴ BM = ... 9 points.

∴MC？ Mn = bm2 = 8 ......................................10.

27. (The full mark of this question is 10)

Solution: (1)∵AC⊥BD

∴ area of quadrilateral ABCD

...................................., two points.

(2) Point A is AE⊥BD, and foot is E. ........................................................................ 3 points [Source: Xue+Ke+Net]

∵ Quadrilateral AB CD is a parallelogram.

In Rt⊿AOE,

........................... 4 points.

Five points.

The ∴ area of quadrilateral ABCD is 6 points.

(3) As shown in the figure, point A and point C are AE⊥BD and CF⊥BD respectively, and the vertical feet are E and F. .........................................................................................................................

In Rt⊿AOE,

∴

In the same way.

…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

∴ area of quadrilateral ABCD

2 8. (The full mark of this question is 1 1)

Solution: (1) Because the parabola passes through the coordinate origin O (0,0) and point E (4,0).

Therefore, we can get c = 0 and b = 4.

So the analytical formula of parabola is ................................ 1.

pass by

When x=2, the maximum value of this parabola is 4. ..............................................................................................................................................................

(2)① point p is not on the straight line ME.

It is known that the coordinates of point M are (2,4) and the coordinates of point E are (4,0).

Let the relationship of straight line ME be y = kx+b.

So go ahead, go ahead.

Therefore, the relationship of the straight line ME is y =-2x+8. ..........................................................................................................................................................

It can be easily obtained from known conditions. When it is correct, OA=AP=, ... 4 o'clock.

The coordinate of point ∵ P does not satisfy the relation of straight line ME, y=-2x+8.

∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴∴875

② The area of a polygon with vertices P, N, C and D may be 5.

∵ point a is on the non-negative semi-axis of the x axis, and n is on the parabola.

∴ OA=AP=t

The coordinates of points p and n are (t, t), (t, -t 2+4t). .....................................................................................................................................

∴ AN=-t 2+4t (0≤t≤3)，

∴an-AP =(-T2+4t)-t =-T2+3t = t(3-t)≥0，∴ PN=-t 2+3 t

Seven points

(1) When PN=0, that is, t=0 or t=3, a polygon with points P, N, C and D as its vertices is a triangle, and the height of this triangle is AD, ∴ S= DC? AD= ×3×2=3。

(ii) When PN≠0, a polygon with points P, N, C and D as vertices is a quadrilateral.

* pn∥cd,ad⊥cd，

∴ S= (CD+PN)？ ad =[3+(-T ^ 2+3t)]×2 =-T ^ 2+3t+3……

When-t 2+3t+3 = 5, the solution is t= 1, 2 ... 9 points.

And 1 and 2 are both in the range of 0≤t≤3, so the polygon area with P, N, C and D as vertices is 5.

To sum up, when t = 1 2, the area of a polygon with points p, n, c and d as vertices is 5.

When t= 1, the coordinate of point n is (1, 3). .......................................................................................................................................

When t=2, the coordinate of point N is (2,4) .....................................11.

Note: When t=0 and t=3, the relationship in (II) also applies. (So there is no (i) when marking papers, only (II) can be used, and no points will be deducted. )

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