In 2009, Yunnan Province took the senior high school entrance examination mathematics examination paper.

Yunnan Province in 2009 high school (technical secondary school) unified entrance examination.

Mathematics Test

(The whole volume contains three big questions, ***23 small questions, ***8 pages; Full mark 120, test time 120 minutes)

note:

1. This volume is a test paper. Candidates must answer questions on the answer sheet (answer sheet), and the answers should be written in the corresponding position on the answer sheet (answer sheet). The answers on the test paper and draft paper are invalid.

After the exam, please return the test paper and the answer sheet (answer sheet) together.

First, multiple-choice questions (there are 7 small questions in this big question, and each small question has only one correct option, with 3 points for each small question, out of 2 1 point).

1. The following calculation is correct ()

A.B.(-2)3 = 8

C.D.

2. In the function, the value range of the independent variable is ()

A.x ≠ 3 B. x>3

C.x0), and the obtained circle is called the circle of motion P. If the diameter and length of the circle of motion P are AC, then the two tangents of the circle of motion P passing through point D are points E and F. Please find out whether there is the minimum area S of the quadrilateral DEPF, and if so, find the value of S; If it does not exist, please explain why.

Note: Question (3) should be answered with another picture.

In 2009, Yunnan Province took the senior high school entrance examination mathematics examination questions.

Reference answer

First, multiple-choice questions (3 points for each small question, out of 2 1 point)

1.C 2。 D 3。 C 4 explosive A 5。 B 6。 B 7。 A

Fill in the blanks (3 points for each small question, out of 24 points)

8.7 9. 10.2 1 1.6.96× 107

12.13.914. △ MBD or △MDE or △ EAD 15. 2,2)

Third, answer questions.

16. Solution:

0.6 points

After testing, it is the solution of the original equation. 7 points

17. solution: passing through point a is the extension line of AE‖BD AC DC at point e,

Then AEC = BDC = 90

∵ , ,

Three points

∵ ,

Six points

(meter)

Answer: The height of the tree is about 0.8 meters.

18. Proof: (1) As shown in the figure, in △ABC and △DCB,

∫AB = DC，AC=DB，BC=CB，

∴△ ABC△ dcb.4 points

(2) BN = CN is known, which is proved as follows:

∫CN‖BD，BN‖AC，

∴ Quadrilateral BMCN is a parallelogram. 6 points

From (1), ∴bm=cm∠MBC =∠MCB,

∴ Quadrilateral BMCN is a diamond. ∴ BN = CN.9 points.

19. Solution: (1) Let the price of type A washing machine be RMB and the price of type B washing machine be RMB.

According to the meaning of the question, the equations can be listed as four points.

solve

∴ The price of a washing machine is 1 100 yuan, and the price of B washing machine is 1600 yuan. Six cents.

(2) The actual payment made by Xiao Li is: (yuan);

The actual amount paid by Xiao Wang is: (Yuan).

∴ Xiao Li and Xiao Wang actually paid 957 yuan and 1392 yuan respectively. Nine points

20. Solution: (1) According to the meaning of the question, it is 30%×450= 135 (Zhang).

Li Hong's votes: 36%×450= 162 (Zhang)

Number of votes: 34%×450= 153 (Zhang) 3 points.

(2) Average score of Wang Rui: (points)

Average score of Li Hong: (points)

Average score of Sharla Cheung: (points)

∴ Sharla Cheung was elected to participate in the provincial finals. Nine points

2 1. solution:

Red, red, yellow and blue

Red (red, red) (red, red) (red, yellow) (red, blue)

Yellow (yellow, red) (yellow, red) (yellow, yellow) (yellow, blue)

Blue (blue, red) (blue, red) (blue, yellow) (blue, blue)

5 points

According to the tree diagram or table above, there are 16 possible results.

P (Xiaoming won) =, P (Xiao Liang won) =.

∴: This game is unfair to both sides. Liang Xiao has a good chance of winning. Eight points.

(Note: When answering questions, you only need to use tree diagram or list method for analysis. )

22. Solution: (1) As shown in the figure, draw △△ ao1b1; +0;

B 1(4，2)，O 1(4，4)； 4 points

(2) Let the function relation corresponding to parabola be y=a(x-m)2+n,

From the AO 1‖x axis, m = 2.

∴y=a(x-2)2+n.

∫ The parabola passes through points A and B,

Get a solution

∴ The functional relationship corresponding to parabola is,

That's 0.9 points.

The drawn parabolic image is shown in the figure. 1 1.

23. Solution: (1) If the points are connected and intersect, when the points move to the points, the straight line bisects the area of the rectangle. The reason for this is the following：

∵ A rectangle is a central symmetric figure, and a point is the symmetric center of the rectangle.

According to any straight line passing through the symmetry center of the central symmetric figure, the area of the central symmetric figure is divided equally. Because the straight line passes through the symmetrical center point of the rectangle, the straight line bisects the area of the rectangle. ...........................................................................................................................

The coordinates of this point are known as.

Let the resolution function of a straight line be.

Then there is a solution.

So the resolution function of the straight line is: .5 points.

(2) Existence makes and resembles.

As shown in the figure, it is suggested to set a straight line and the positive and half axes of the shaft at the point.

Because, if △DOM is similar to △ABC, there is an or.

When, that is, solution. So the point satisfies the condition.

From the symmetry point of view, this point also meets the conditions.

To sum up, there are three similarities, namely 0 and 0.9.

(3) As shown in the figure, DP⊥AC passing through D is point P, with P as the center, the radius length as a circle, the tangents DE and DF passing through D as tangents respectively, and points E and F as tangents. Except point P, take any point P 1 on the straight line AC, the radius length is a circle, and the tangents DE 1 and DF 65448 are the tangents passing through D, respectively.

In △DEP and △DFP, ∠ PED = ∠ PFD, PF = PE, PD = PD, ∴△DPE≌△DPF. ..

∴S quadrilateral DEPF = 2s△ DPE = 2×.

When DE takes the minimum value, the value of S quadrilateral DEPF is the minimum.

∵ , ,

∴ .

∵ ,∴ .

From the arbitrariness of points, we know that virtue is

The minimum length of the straight line connecting this point and the tangent point ... 12 points.

In △ADP and △AOC, ∠ DPA = ∠ AOC,

Democratic Action Party = Cao,

That's it.

∴ .