Mathematics 25 questions in Shanghai senior high school entrance examination

In 2009, Shanghai junior high school graduation unified academic examination.

Mathematics volume

(Full score 150, test time 100 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. Answering questions on draft paper and this 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).

One and only one of the four options in the following question is correct. Select the code of the correct item and fill it in the corresponding position on the answer sheet.

1. The result of calculation is (b

)

2. The solution set of inequality group is (c)

3. When solving the fractional equation with method of substitution, if it is assumed that the original equation is reduced to an integral equation, the integral equation is (a).

4. The vertex coordinates of parabola (constant) are (

5. In the regular polygon below, the central angle is equal to the inner angle (c).

)

A. Regular hexagon, regular pentagon, regular quadrangle and regular triangle

6. As shown in figure 1, the following conclusion is correct (a).

)

Two. Fill in the blanks: (This topic is entitled *** 12, with 4 points for each question, out of 48 points)

Please fill in the results in a straight line in the corresponding position on the answer sheet.

7. The denominator is reasonable.

8. The root of the equation is x=2.

9. If the equation (constant) about has two equal real roots, then.

10. So, the known function

— 1/2

1 1. The two branches of the inverse proportional function image are in the first i.

Quadrant three.

12. After translating the parabola up by one unit, a new parabola can be obtained. Then the expression of the new parabola is

13. If 1 student is selected from 6 students including Xiao Ming as volunteers for the World Expo, then the probability that Xiao Ming is selected is 1/6.

14. Original price of a commodity 100 yuan. If the price is reduced twice and the percentage of each price reduction is 0, then the current price of the goods is 100 * (1-m) 2.

Meta (the result is represented by the contained algebraic expression).

15. as shown in figure 2, in, it is the center line on the side, let the vector,

If a vector is used to represent a vector, then =+(/2).

16. In a circle, the length of a chord is 6, and its corresponding chord center distance is 4, so the radius.

five

17. In a quadrilateral, the diagonal and the intersection are equal to. Without any auxiliary lines, it is necessary to add a condition to make the quadrilateral rectangular, which can be AC=BD or the internal angle is equal to 90 degrees.

18. In, the points on the side are connected (as shown in Figure 3). If the point falls at the midpoint of an edge after being folded into a straight line, the distance from the point to it is

2 .

Third, answer: (This big question is ***7 questions, out of 78 points)

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

Calculation:.

= — 1

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

Solve the equation:

(X=2 y=3)

(x=- 1

y=0)

2 1. (The full score of this question is 10, and the full score of each item is 5 points. )

As shown in fig. 4, in the trapezoid, the connection.

( 1);

(2) If they are midpoints, connect and find out the lengths of the line segments.

The radical sign of (1) two-thirds 3

(2)8

22. (The full mark of this question is 10, (1) item 2, (2) item 3, (3) item 2, (4) item 3).

In order to understand the physical quality of junior high school boys in a school, some students from grade six to grade nine were selected for "pull-ups" test. The number of "pull-ups" of all subjects is shown in table 1. The percentage of the test population in each grade to all the test population is shown in Figure 5 (the relevant data of grade six are not marked).

Multiply by 0 1 23455 6789 10.

Number of people1122342201

Table 1

According to the above information, answer the following questions (write the results directly):

(1) Students in Grade 6 account for 20% of all students.

;

(2) In all subjects, the number of students in Grade 9 is

six

;

(3) In all subjects, the proportion of people who have at least six "pull-ups" is 35%.

;

(4) In all the times of "pull-ups", the mode is 5.

23. (The full score of this question is 12, and the full score of each small question is 6.)

It is known that the line segment intersects with the midpoint of the sum of points, connections and (as shown in Figure 6).

(1) addition conditions,

Verification:

Proof: from the known condition: 2oe = 2ochob = oc.

Angle AOB= angle DOC, so all triangles ABO are equal to triangle DOC.

therefore

(2) Write ①, ② and ③ respectively, add conditions ① and ③, and make a proposition 1 with ② as the conclusion, and make a proposition 2 with ① as the conclusion. Proposition 1 Yes

real

Proposition, proposition 2 is

wrong

Proposition (choose "right" or "wrong" to fill in the blanks)

24. (The full score of this question is 12, and the full score of each small question is 4 points.)

On the rectangular coordinate plane, it is the origin, the coordinate of the point is, and the coordinate of the point is, the axis of the straight line (as shown in Figure 7). Points are symmetrical about the origin, and a straight line (constant) passes through the point and intersects and connects with the straight line at the point.

(1) and the coordinates of the point;

(2) Set a point on the positive semi-axis of the shaft, and if it is an isosceles triangle, find the coordinates of the point;

(3) Under the condition of (2), if a circle with radius is circumscribed with a circle, find the radius of the circle.

Solution: (1) point B (- 1, 0), and substitute it to get the straight line BD: Y = X+ 1.

Substitute Y=4 into point D (3, 1) where x=3.

(2) 1, PO=OD=5, then P (5 5,0)

2. If PD=OD=5, then PO=2*3=6, then point P (6 6,0).

3、PD=PO

Let P(x, 0)

D(3，4)

It is solved by Pythagorean theorem.

X=25/6, and then point P (25/6,0)

(3) From the coordinates of p and d, we can calculate:

1、PD=2

r=5—2

2、PD=5

r= 1

3、PD=25/6

r=0

25. (The full mark of this question is 14, (1) item 4, (2) item 5, (3) item 5).

It is called the moving point on the line segment, which is on the ray and satisfies (as shown in Figure 8).

(1) Find the length of the line segment when the points coincide (as shown in Figure 9);

(2) In Figure 8, add. When, and points are on a straight line segment, set the distance between points as,, where the area represented by, the area represented by, find the resolution function about, and write the function definition domain;

(3) When, and the point is on the extension line of the line segment (as shown in Figure 10), find the size.

Solution: (1)AD=2, and point Q coincides with point B. According to the meaning of the question, ∠PBC=∠PDA because ∠A=90. PQ/PC=AD/AB= 1, so: △PQC is an isosceles right triangle, BC=3, so: PC=3 /2,

(2) Add auxiliary lines as shown in the figure. According to the meaning of the question, the area of two triangles can be expressed as S 1, S2, and the height is h, h,

Then: s1= (2-x) h/2 = (2 * 3/2)/2-(x * h/2)-(3/2) * (2-h)/2.

S2=3*h/2 because of two S 1/S2=y, eliminate h, h, and get:

Y=-( 1/4)*x+( 1/2)，

Domain: When point P moves to coincide with point D, the value of X is the largest. When PC is perpendicular to BD, then X=0. Connect DC to make QD a vertical DC. According to the known conditions, the circle with four points B, Q, D and C can be found. It can be inferred from the theorem of circle angle that the triangle QDC is similar to the triangle ABD.

QD/DC=AD/AB=3/4, let QD=3t and DC=4t, then: QC=5t, which is obtained by Pythagorean theorem:

In the right triangle AQD: (3/2) 2+(2-x) 2 = (3t) 2.

QBC: 3 2+x 2 = (5t) 2 in a right triangle.

Finishing: 64x 2-400x+30 1 = 0.

(8x-7)(8x-43)=0

X 1=7/8。

X2=(43/8)>2 (rounding) So the function:

The domain of Y=-( 1/4)*x+ 1/2 is [0,7/8].

(3) Because PQ/PC=AD/AB, assuming PQ is not perpendicular to PC, we can draw a straight line PQ' perpendicular to PC and intersecting with AB at Q'.

Then, four-point * * circles of B, Q', P and C are obtained from the theorem of the angle of circle and the properties of similar triangles;

PQ′/PC = AD/AB，

Since PQ/PC=AD/AB and point Q' coincide with point Q, the angle ∠ QPC = 90.