First, multiple-choice questions (this big question * *10 small questions, 3 points for each small question, out of 30 points)

1. If 60m means "Go 60m north", then "Go 40m south" can be expressed as

A.-20 meters wide -40 meters wide

Answer B.

The number of test sites is opposite.

This paper analyzes two concepts that north and south are in opposite directions, north is+south is negative. So according to the definition of reciprocal, we can get the result directly.

2. In the figure below, there are both axisymmetric and central symmetric figures.

Answer C.

Axisymmetric figure of test center, central symmetric figure.

Analysis According to the definitions of axisymmetric figure and centrosymmetric figure, we can know that A is centrosymmetric figure rather than axisymmetric figure. B is also a centrally symmetric figure rather than an axisymmetric figure; C is not only an axisymmetric figure, but also a centrally symmetric figure. It has four axes of symmetry, namely, a horizontal line and a vertical line connecting three small circular segments, and two bisectors between the horizontal line and the vertical line. D is neither an axisymmetric figure nor a centrally symmetric figure.

3. The result of calculation is

A.3 B.3 C. 3 D.3

Answer D.

Cubic root of test center.

According to the definition of cube root, because 33=27, so.

4. The length of the following three line segments cannot form a triangle.

A.3、8、4

C. 15，20，8

Answer a.

Composition conditions of test center triangle.

According to the condition that the sum of any two sides of a triangle is greater than the third side, 3+4 < 8, so the three segments of A cannot form a triangle.

5. As shown in the figure, AB∨CD, ∠ DCE = 80, then ∠ BEF =

a . 120 b . 1 10 c . 100d . 80

Answer C.

The nature of parallel lines in test sites.

According to the nature of parallel lines with complementary internal angles on the same side, since AB∨CD, ∠DCE and ∠BEF are internal angles on the same side, ∠ BEF =

6. In the following horizontally placed geometry, the top view is rectangular.

Answer B.

Three views of test center geometry.

Analysis According to the law of geometric top plan, the top plans of A and D are circular, B is rectangular, and C is.

The top view is triangular.

7. If 3 is one root of the equation x2-5x+c =, then the other root of this equation is

A.-2b 2c-5d 5

Answer B.

The relationship between roots and coefficients of a quadratic equation in one variable.

The analysis is based on the relationship between the roots and coefficients of a quadratic equation: the sum of the two roots is equal to the reciprocal of the quotient of the coefficients of the first term and the coefficients of the second term, so there is.

8. As shown in figure ⊙O = 8, the chord AB, m is the midpoint of AB, and OM = 3, then the radius of ⊙O is equal to

A.8 B.4 C. 10 D.5

Answer 5.

Test the diameter of the central circle to bisect the chord vertically, Pythagorean theorem.

Analyze the theorem of bisecting chords vertically according to the diameter of a circle. OAM is a right triangle, in Rt? Pythagorean theorems used in OAM include.

9. Party A and Party B advance at a constant speed from place A to place B along the same route, with a distance of 20 kilometers ... The distance they advance is s(km), and the time after Party A leaves is t(h). The function image of distance and time between Party A and Party B is shown in the figure. According to the image information, the following statement is correct.

The speed of A.A is 4 km/h and the speed of b is10 km/h.

C.B leaves a later than a 1 hour and arrives at b 3 hours later than b.

Answer a.

Test center linear function.

The analysis is based on the given linear function image: the speed of A. A is; B.b's speed is; C. B leaves later than a; D. A arrives at B later than B.

10. let m > n > 0, m2+N2 = 4mn, then =

A.2 B. C. D.3

Answer a.

Test center algebraic transformation, complete square formula, square difference formula, radical calculation.

From the analysis of m2+N2 = 4mn, because m > n > 0, then.

2. Fill in the blanks (this topic is entitled ***8 small questions, with 3 points for each small question, out of 24 points)

1 1. Given = 20, the complementary angle of is equal to.

Answer 700.

Corner of the test center.

According to the definition of complementary angle, the direct result is: 900-200=700.

12. Calculation:-=.

The answer.

Seek the roots at the test site.

Analysis uses radical calculation rules to directly export results:.

13. In the function y =, the range of the independent variable x is.

The answer.

Score definition of test center.

According to the definition of a fraction, the denominator cannot be 0, so the conclusion is drawn.

14. The weight (unit: kg) of the seven girls are 36, 42, 38, 42, 35, 45 and 40 respectively, so the figure of the seven girls is

The median weight is kilograms.

Answer 40.

Median testing center.

According to the definition of median, median refers to arranging data in order of size to form a sequence.

Data in the middle of a series. Therefore, the rights of seven girls should be rearranged first: 35, 36, 38, 40, 42, 42,

45, so the median is 40.

15. As shown in the figure, in rectangular paper ABCD, AB = 2 cm, and point E is on BC, AE.

= =CE .. If the paper is folded along AE and point B coincides with point B 1 on AC, AC

= cm。

Answer 4.

Rectangular nature of test center, folding, isosceles triangle nature, right triangle nature, 300-degree right triangle nature.

The analysis shows that ∠B=900 from the rectangle attribute and ∠BAC=∠EAC from the folding. Equilateral equivalence of isosceles triangle

The nature of the angle is ∠EAC=∠ECA from AE = Ce. According to the complementary nature of two acute angles of a right triangle, we can get

∠ECA=300. So according to the property that the right side of a 300-angle right triangle is half of the hypotenuse, Rt? American Broadcasting Company Inc (ABC)

AC=2AB=4。

16. Decomposition factor: 3m (2x ― y) 2 ― 3mn2 =.

The answer.

The common factor method of extracting test sites and the factor decomposition of applying formula method.

Analysis.

17. As shown in the figure, in order to measure the river width AB (assuming that the two banks of the river are parallel), the measured ACB is ∠ 30.

∠ ADB = 60, CD = 60m, then the river width AB is m (reserved root sign).

Answer a.

Test point solution right triangle, special angle trigonometric function, radical calculation.

Analysis in Rt? ABD and Rt? At ABC

As shown in the figure, the three semicircles are circumscribed in turn, the centers of which are all on the X axis and tangent to the straight line Y = X. Let half of the three semicircles be.

The diameters are r 1, r2 and r3 in turn, so when r 1 = 1, R3 =.

Answer 9.

Linear function of test center, properties of right triangle, similar triangles. Let the straight line y = x and three semicircles be tangent to a respectively,

B, c, if the AEX axis is in E, then in Rt? In AEO 1, it is easy to get ∠AOE=∠EAO 1=300, while EO= from r 1 = 1

AE=, OE=, OO 1=2. Then. In the same way.

Third, answer the question (this big question * * 10 small question, out of 96 points)

19.( 10) (1) Calculation: 22+(-1) 4+(-2) 0-|-3 |;

(2) Simplify first and then evaluate: (4ab3-8a2b2) ÷ 4ab+(2a+b) (2a-b), where a = 2 and b = 1.

Solution: (1) The original formula = 4+1+1-3 =1.

(2) The original formula = 4ab (b2-2ab) ÷ 4ab+4a2-b2 = b2-2ab+4a2-b2 = 4a2-2ab.

When a = 2 and b = 1, the original formula = 4× 22-2× 2×1=16-4 =12.

Even power, zeroth power, absolute value, algebraic simplification, square difference formula of negative number of test sites.

Analysis (1) uses the definitions of even power, zero power and negative absolute value to get the results directly.

(2) By extracting the common factor to simplify the score, the polynomial is multiplied by the square difference formula, and then similar terms are merged and substituted. [Source: Subject Network]

20.(8 points) Find the solution set of the inequality group and write its integer solution.

Answer: from ①, x 1 is obtained, and from ②, X < 4 is obtained.

So the solution set of the inequality group is. Its integer solution is 1, 2, 3.

Test center-unary linear inequality group.

Analyze the solution method of one-dimensional linear inequality group, get the result directly, and then write its integer solution.

2 1.(9 points) In order to understand the students' love for ball games in a middle school, several students were randomly selected to conduct a questionnaire survey (each student was required to fill in only one kind of ball games he liked), and the survey results were drawn into the following two incomplete statistical charts.

Please answer the following questions according to the information provided in the picture:

(1) Some students participated in the survey. In the pie chart, the central angle of the fan representing "other ball games" is degrees;

(2) Supplementary bar chart;

(3) If there are 2000 students in this school, it is estimated that there are * * * students who like basketball.

Answer: (1) 300,36.

(2) There are 300- 120-60-30 = 90 people who like football, so the histogram is supplemented accordingly (as shown on the right).

(3) Among the students who participated in the survey, 120 people like basketball, accounting for.

120300 = 40%, so among the 2000 students in this school, it is estimated that 2000×40%=800 students like basketball.

Fan chart, bar chart, frequency, test center frequency.

Analysis (1) As can be seen from the figure, 60 people like table tennis, accounting for 20%, so 60.20% = 300 students (people) participated in the survey.

There are 30 people who like other ball games, accounting for 30300 = 10%, so the central angle of the fan representing "other ball games" is 3600× 10%=360.

(2) Subtract other items from the total number of students participating in the survey (1) to get the number of people who like football and complete the histogram.

(3) We can estimate the number of students who like basketball in the whole school by the percentage of students who like basketball in the survey.

22.(8 points) As shown in the figure, point A AM cuts ⊙O, point D BD⊥AM, and BD crosses ⊙ O.

Divide ∠ AOB at point C and point OC. Find the degree of ∠ B.

The answer is: ∫oc divides equally ∠AOB, ∴∠ AOC = ∠ COB.

At point a, namely OA⊥AM and BD⊥AM,

∴OA∥BD,∴∠AOC=∠OCB

oc = ob，∴∠ OCB = ∠ B，∴∠ B = ∠ OCB = ∠ COB = 600。

Test the tangent, angle bisector, parallel line, internal angle and sum of triangles of the central circle.

The analysis requirement is ∠B, because OC = OB, so we can know ∠ OC=OB = ∠ b according to the equilateral corner. Because OA and BD are perpendicular to the same straight line AM, OA∨BD has the same internal angle ∠ AOC = ∠ OCB according to the parallelism of the two straight lines. but

OC bisects ∠AOB, and ∠ B = ∠ OCB = ∠ COB can be obtained through equivalent substitution, so ∠ B = = 600 can be obtained from the sum of the interior angles of the triangle.

23.(8 points) In community fitness activities, the father and son participated in a skipping competition. At the same time, the father jumped 180 and the son jumped 2 10. As we all know, the son jumps 20 times more than his father every minute. How many times do father and son jump every minute?

Answer: Let the father jump X times a minute and the son jump X+20 times a minute.

According to the meaning of the question. X = 120 is obtained by solving.

X = 120 is the root of the equation.

When x = 120, x+20 = 140.

A: The father jumps 120 times a minute and the son jumps 140 times a minute.

Test center series equation solves application problems and fractional equation.

The key to solving application problems by analyzing sequence equations is to find out the equivalence relation: father jumps 180 and son jumps 2 10 at the same time. That is, the time for the father to jump 180 times = the time for the son to jump 2 10 times, and the time = the amount of exercise and the speed of exercise.

24.(8 points) Comparing regular pentagons with regular hexagons, we can find their similarities and differences. For example:

They have one thing in common: the sides of a regular Pentagon are equal, and so are the sides of a regular hexagon.

One difference between them is that a regular Pentagon is not a centrally symmetric figure, while a regular hexagon is a centrally symmetric figure.

Please write down the similarities and differences between them:

Similarities:

① ;

② .

Difference:

① ;

② .

Answer: Similarities: ① Regular pentagons and regular hexagons are axisymmetric figures.

② The internal angles of regular pentagons and regular hexagons are equal.

Differences: ① The diagonals of regular pentagons are all equal; Diagonal lines of regular hexagons are not equal.

② Diagonal lines of regular pentagons do not intersect at the same point; The three diagonals of a regular hexagon intersect at the same point.

The test sites are regular pentagons and regular hexagons.

Similarity analysis: ① A regular Pentagon has five symmetrical axes, which are straight lines connecting the vertices and the midpoint of their opposite sides; The six symmetry axes of a regular hexagon are straight lines connected by diagonal vertices and straight lines connected by opposite midpoints.

② Each internal angle of a regular Pentagon is1080; Every inner corner of a regular hexagon is 1200.

Difference: ① A triangle formed by the diagonal of a regular pentagon and two adjacent sides.

Is consistent; The three straight lines passing through the center on the diagonal of a regular hexagon are equal in length (red in the figure).

Lines), but the six in the middle are of equal length (blue lines in the picture).

(2) As can be seen from the figure.

25.(9 points) Guangming Middle School attaches great importance to the eye hygiene of middle school students and conducts regular eye examinations. There are two test sites, A and B, and three students, A, B and C, randomly choose one of them to test their eyes.

(1) Find the probability that three students, A, B and C, will have their eyesight tested in the same place;

(2) Find out the probability that at least two of the three students A, B and C will have an eyesight test in B. 。

Answer: (1) List all situations in which students A, B and C each randomly choose one of them to test their eyesight:

If three people don't choose a place, then three people choose b place, which is 1 example.

Among the three people, one chooses a place and the other two choose b place, which is divided into three situations; A choose location a, b and c choose location b; B choose location a, a and c choose location b; C chooses a place, and A and B choose B place.

Among the three people, two choose a place and one chooses b place, which is divided into three situations; Party A and Party B choose location A, and Party C chooses location B; Party A and Party C choose location A, and Party B chooses location B; B, C choose location A, A choose location B. ..

If all three people choose A, then all three people don't choose B, which is 1 example.

There are eight possible situations. There are two situations in which students A, B and C have their eyesight tested in the same place: they all choose A or they all choose B. Therefore, the probability that students A, B and C have their eyesight tested in the same place is

(2) In four cases, at least two of the three students A, B and C have their eyesight tested at B: two of them choose B, and all three choose B. So the probability that at least two of the three students A, B and C will have their eyesight tested at B is.

Test site probability.

Analyze and list all the situations, analyze the conditions and find out the probability.

26.( 10) As shown in figure 1, o is the center of the square ABCD.

Extend OA and OD to points f and e, respectively, so that of = 2oa,

OE = 2od, connect EF. Turn △EOF counterclockwise around o point.

The rotation angle is △E 1OF 1 (Figure 2).

(1) Explore and prove the quantitative relationship between AE 1 and BF 1;

(2) When = 30, it is proved that △AOE 1 is a right triangle.

Answer: (1) AE 1 = BF 1, which is proved as follows:

∵O is the center of the square ABCD, ∴ OA = OB = OD, ∴ OE = of.

∫△e 1 of 1 is obtained by the counterclockwise rotation angle of△△ eof around point O, ∴ OE 1 = of 1.

∠∠AOB =∠eof = 900，∴∠e 1oa = 900-∠f 1oa =∠f 1ob

OE 1=OF 1

In △E 1OA and △F 1OB, ∠ E 1OA = ∠ F 1OB, ∴△ e1OA ≌△ f/kloc.

OA=OB

∴ AE 1=BF 1 .

(2) Take the midpoint g of OE 1 and connect AG.

∠∠aod = 900，=30，∴ ∠E 1OA=900-=60 .

∵oe 1=2oa,∴oa=og,∴∠e 1oa =∠ago =∠OAG = 60 .

∴ag=ge 1,∴∠gae 1=∠ge 1a=30 .∴ ∠E 1AO=90 .

∴△AOE 1 is a right triangle.

The nature and determination of central square, rotation, determination and property of congruent triangles, determination of right triangle.

Analysis (1) To prove AE 1 = BF 1, we should first consider that they are the corresponding edges of congruent triangles. Examining △E 1OA and △F 1OB, it is found that the property bisected by the diagonal of the square is OA = OB, and then look at OE 1 and OF 1. They are obtained by rotating OE and OF, which are known to be equal. Finally, the included angles ∠E 1OA and ∠GE 1A are complementary to ∠F 1OA. So as to obtain a certificate.

(2) To prove that △AOE 1 is a right triangle, we should consider proving ∠ E 1ao = 90. Considering OE 1 = 2oa, as the auxiliary line AG, we get ∠ Ago = ∠ OAG, and because of ∠E 1OA and redundancy, we get ∠ E 1oa = 60, so all three angles of △AOG are equal. And from ag = ge 1, ∠ gae 1 = ∠ ge 1a = 30. So ∠ E 1AO = 90 proves this point.

27.( 12) It is known that A (1 0), B(0,-1), C (- 1 2), D(2,-1), E (4.

(1) proves that c and e cannot be on the parabola Y = A (X- 1) 2+K (A > 0) at the same time;

(2) Is point A on the parabola y = a (x- 1) 2+k (a > 0)? Why?

(3) Find the values of a and k 。

Solution to the answer: (1) Proof: Through the reduction to absurdity. Let C (- 1, 2) and E (4 4,2) both lie in the parabola y = a (x-1) 2+k.

(a > 0), simultaneous equation,

The solution is a = 0 and k = 2. This does not match the required a > 0.

∴C and e cannot be on the parabola y = a (x- 1) 2+k (a > 0) at the same time.

(2) point a is not on the parabola, y = a (x- 1) 2+k (a > 0). This is because if point A is on a parabola, then k = 0. B(0,-1) is on the parabola, and A =- 1, and D(2,-1) is on the parabola, and A =- 1, which is inconsistent with the known A > 0. According to (1), C and E cannot be on a parabola at the same time.

So point a is not on the parabola y = a (x- 1) 2+k (a > 0).

(3) Synthesis (1)(2) Discuss in two situations:

① parabola y = a (x- 1) 2+k (a > 0) passes through three points: B(0,-1), c (- 1, 2) and D(2,-1).

a(0- 1)2+k=- 1

Simultaneous equation a (- 1- 1) 2+k = 2,

a(2- 1)2+k=- 1

The solution is a = 1 and k =-2.

② parabola y = a (x- 1) 2+k (a > 0) passes through three points: B(0,-1), D(2,-1) and e (4 4,2).

a(0- 1)2+k=- 1

Simultaneous equation a (2- 1) 2+k =- 1,

a(4- 1)2+k=2

The solution is a =, k =.

Therefore, when the parabola passes through B, C and D, A = 1 and K =-2. When the parabola passes through B, D and E,

a=，k= .

Quadratic function of test center, binary linear equations.

In the analysis of proof by reduction to absurdity (1), as long as the conclusion is assumed to be established first, you can get a conclusion that is contradictory to what is known.

(2) Prove that point A is not on a parabola, as long as point A and any other two points are not on the same parabola.

(3) List all the cases of any three points on the parabola, except point A in (2), there are four points B, C, D and E. The possible cases are 1B, C, D, 2B, C, E, 3B, D, E and 4C, D, E .. However, through (/) This leaves only ①B, c and d.

And ③B, d and e, the equations can be solved simultaneously.

28. As shown in the figure, it is known that the straight line L passes through point A (1 0) and the hyperbola Y =

(x > 0) passes through point b (2, 1). The intersection P(p, P- 1) (P > 1) is the plane of the x axis.

The straight line intersects the hyperbola y = (x > 0) and y =-(x < 0) at m and n points respectively.

(1) Find the value of m and the analytical formula of straight line L;

(2) If the point P is on the straight line Y = 2, it is verified that: △ PMB ∽△ PNA; ? From: China.

(3) Is there a real number p that makes s △ AMN = 4s △ amp? If it exists, request all values of p that meet the conditions; if

Does not exist, please explain why.

Solution: (1) Starting from point B (2, 1) on y =, there is 2 =, that is, m = 2.

Let the analytical formula of the straight line L be, from point A (1, 0) and point B (2, 1), we get

Solve it and get it.

The analytical formula of ∴ straight line L is.

(2) point P(p, p- 1) is on the straight line y = 2, and ∴P is on the straight line l, which is the intersection of the straight line y = 2 and l, as shown in figure (1).

According to the conditions, the coordinates of each point are n (- 1, 2), m (1, 2) and p (3, 2).

∴np=3-(- 1)=4,mp=3- 1=2,ap=，

Blood pressure =

∴, ∠ MPB = ∠ NPA, in △PMB and △PNA.

∴△PMB∽△PNA。

(3)S△AMN=. The following points are discussed:

When 1 < P < 3, extend MP across the x axis to q, as shown in Figure (2). Let the straight line MP be

solve

Then the straight MP is

When y = 0, x =, that is, the coordinate of point Q is (,0).

Then,

If there is 2 = 4, solve, p = 3 (disagree, give up), p =.

When p = 3, see figure (1) s △ amp = = s △ AMN. This is beside the point.

When p> is at 3 o'clock, extend the intersection of PM and X axis to Q, as shown in Figure (3).

At this time, S△AMP is greater than the triangle area S△AMN when p = 3. So there is no real number p, so s △ AMN = 4s △ amp.

To sum up, when p =, s △ AMN = 4s △ amp.

Inverse proportional function of test center, linear function, undetermined coefficient method, binary linear equations, Pythagorean theorem, similar triangles unary quadratic equation.

Analytic (1) Substitute the coordinates of point B (2, 1) into y = to get the value of m, and use the undetermined coefficient method to solve the binary linear equations to get the analytical expression of straight line L.

(2) The point P(p, p- 1) is on the straight line Y = 2, which actually means that the point is the intersection of the straight line Y = 2 and L, so it is necessary to prove that △PMB∽△PNA is only proportional to the corresponding line segment.

(3) The position of point P should be considered first. In fact, when P = 3, it is easy to find out S△AMP = S△AMN. When p>3, note that when p = 3, S △ amp is larger than the triangle area, so it is larger than S △ AMN. So as long as we mainly study the situation when 1 < P < 3. After making the necessary auxiliary lines, first find the equation of the straight line MP, then find the coordinates of each point (expressed by P), then find the expression of the area, and then find the value of P after substituting S △ AMN = 4S △ AMP.